Nathan Jacobson
Lectures in Abstract
Algebra
III. Theory of Fields and Galois Theory
Springer-Verlag New York Heidelberg Berlin

Nathan Jacobson
Yale University
Department of Mathematics
New Haven, Connecticut 06520
Managing Editor
P. R. Halmos
Indiana University
Department of Mathematics
Swain Hall East
Bloomington, Indiana 47401
Editors
F. W. Gehring C. C. Moore
University of Michigan University of California at Berkeley
Department of Mathematics Department of Mathematics
Ann Arbor, Michigan 48104 Berkeley, California 94720
AMS Subject Classification
12-01
Library of Congress Cataloging in Publication Data
Jacobson, Nathan, 1910—
Lectures in abstract algebra.
(Graduate texts in mathematics; v. 32)
Reprint of the 1951-1964 ed. published by Van Nostrand, New York in The
University series in higher mathematics.
Bibliography: v. 3, p.
Includes indexes.
CONTENTS: 2. Linear algebra. 3. Theory of fields and Galois theory. 1. Algebra,
Abstract. I. Title. II. Series.
QA162.J3 1975 512'.02 75-15564
Second corrected printing
All rights reserved
No part of this book may be translated or reproduced in any form without written
permission from Springer-Verlag.
© 1964 by Nathan Jacobson
Originally published in the University Series in Higher Mathematics (D. Van
Nostrand Company): edited by M. H. Stone, L. Nirenberg and S. S. Chern.
Printed in the United States of America.
ISBN 0-387-90168-X Springer-Verlag New York Heidelberg Berlin
ISBN 3-540-90168-X Springer-Verlag Berlin Heidelberg New York

TO
POLLY

PREFACE
The present volume completes the series of texts on algebra
which the author began more than ten years ago. The account
of field theory and Galois theory which we give here is based on
the notions and results of general algebra which appear in our first
volume and on the more elementary parts of the second volume,
dealing with linear algebra. The level of the present work is
roughly the same as that of Volume II.
In preparing this book we have had a number of objectives in
mind. First and foremost has been that of presenting the basic
field theory which is essential for an understanding of modern
algebraic number theory, ring theory, and algebraic geometry.
The parts of the book concerned with this aspect of the subject
are Chapters I, IV, and V dealing respectively with finite dimen-
dimensional field extensions and Galois theory, general structure theory
of fields, and valuation theory. Also the results of Chapter III on
abelian extensions, although of a somewhat specialized nature,
are of interest in number theory. A second objective of our ac-
account has been to indicate the links between the present theory of
fields and the classical problems which led to its development.
This purpose has been carried out in Chapter II, which gives
Galois' theory of solvability of equations by radicals, and in
Chapter VI, which gives Artin's application of the theory of real
closed fields to the solution of Hilbert's problem on positive defi-
definite rational functions. Finally, we have wanted to present the
parts of field theory which are of importance to analysis. Partic-
Particularly noteworthy here is the Tarski-Seidenberg decision method
for polynomial equations and inequalities in real closed fields
which we treat in Chapter VI.
As in the case of our other two volumes, the exercises form an
important part of the text. Also we are willing to admit that
quite a few of these are intentionally quite difficult.

VH1 PREFACE
Again, it is a pleasure for me to acknowledge my great indebted-
indebtedness to my friends, Professors Paul Cohn and George Seligman,
for their care in reading a preliminary version of this material.
Many of their suggestions have been incorporated in the present
volume. I am indebted also to Professors Cohn and James Reid
and to my wife for help with the proof reading. Finally, I wish to
acknowledge my appreciation to the U. S. Air Force Office of
Scientific Development whose support during a summer and half
of an academic year permitted the completion of this work at an
earlier date than would have been possible otherwise.
N.J.
New Haven, Conn.
January 20,1964

CONTENTS
INTRODUCTION
SECTION PAGE
1. Extension of homomorphisms 2
2. Algebras 7
3. Tensor products of vector spaces 10
4. Tensor product of algebras 15
CHAPTER I: FINITE DIMENSIONAL EXTENSION FIELDS
1. Some vector spaces associated with mappings of fields ... 19
2. The Jacobson-Bourbaki correspondence 22
3. Dedekind independence theorem for isomorphisms of a field . 25
4. Finite groups of automorphisms 27
5. Splitting field of a polynomial 31
6. Multiple roots. Separable polynomials 37
7. The "fundamental theorem" of Galois theory 40
8. Normal extensions. Normal closures 42
9. Structure of algebraic extensions. Separability 44
10. Degrees of separability and inseparability. Structure of
normal extensions 49
11. Primitive elements 54
12. Normal bases 55
13. Finite fields 58
14. Regular representation, trace and norm 62
15. Galois cohomology 75
16. Composites of fields 83
CHAPTER II: GALOIS THEORY OF EQUATIONS
1. The Galois group of an equation 89
2. Pure equations 95
3. Galois' criterion for solvability by radicals 98
ix

X CONTENTS
SECTION PAGE
4. The general equation of w-th degree 102
5. Equations with rational coefficients and symmetric group as
Galois group 105
CHAPTER III: ABELIAN EXTENSIONS
1. Cyclotomic fields over the rationals 110
2. Characters of finite commutative groups 116
3. Kummer extensions 119
4. Witt vectors 124
5. Abelian ^-extensions 132
CHAPTER IV. STRUCTURE THEORY OF FIELDS
1. Algebraically closed fields 142
2. Infinite Galois theory 147
3. Transcendency basis 151
4. Liiroth's theorem 157
5. Linear disjointness and separating transcendency bases . . . 160
6. Derivations 167
7. Derivations, separability and ^-independence 174
8. Galois theory for purely inseparable extensions of exponent one 185
9. Higher derivations 191
10. Tensor products of fields 197
11. Free composites of fields 203
CHAPTER V: VALUATION THEORY
1. Real valuations 211
2. Real valuations of the field of rational numbers 214
3. Real valuations of $(x) which are trivial in $ 216
4. Completion of a field 216
5. Some properties of the field of p-adic numbers 222
6. Hensel's lemma 230
7. Construction of complete fields with given residue fields . . 232
8. Ordered groups and valuations 236
9. Valuations, valuation rings, and places 239
10. Characterization of real non-archimedean valuations .... 243
11. Extension of homomorphisms and valuations 246
12. Application of the extension theorem: Hilbert Nullstellensatz 251
13. Application of the extension theorem: integral closure . . . 255

CONTENTS XI
SECTION PAGE
14. Finite dimensional extensions of complete fields 256
15. Extension of real valuations to finite dimensional extension
fields 262
16. Ramification index and residue degree 265
CHAPTER VI: ARTIN-SCHREIER THEORY
1. Ordered fields and formally real fields 270
2. Real closed fields 273
3. Sturm's theorem 278
4. Real closure of an ordered field 284
5. Real algebraic numbers 287
6. Positive definite rational functions 289
7. Formalization of Sturm's theorem. Resultants 295
8. Decision method for an algebraic curve 300
9. Equations with parameters 307
10. Generalized Sturm's theorem. Applications 312
11. Artin-Schreier characterization of real closed fields 316
Suggestions for further reading 319
Index 321

Introduction
In this book we shall assume that the reader is familiar with
the general notions of algebra and the results on fields which
appear in Vol. I, and with the more elementary parts of Vol. II.
In particular, we presuppose a knowledge of the characteristic of
a field, prime field, construction of the field of fractions of a com-
commutative integral domain, construction of simple algebraic and
transcendental extensions of a field. These ideas appear in
Chaps. II and III of Vol. I. We shall need also the elementary
factorization theory of Chap. IV. From Vol. II we require the
basic notions of vector space over a field, dimensionality, linear
transformation, linear function, compositions of linear trans-
transformations, bilinear form. On the other hand, the deeper results
on canonical forms of linear transformations and bilinear forms
will not be needed.
In this Introduction we shall re-do some things we have done
before. Our motivation for this is twofold. In the first place,
it will be useful for the applications that we shall make to sharpen
some of the earlier results. In the second place, it will be con-
convenient to list for easy reference some of the results that will be
used frequently in the sequel. The topics that we shall treat
here are: extension of homomorphisms (cf. Vol. I, Chap. Ill),
algebras (Vol. II, Chap. VII), and tensor products * of vector
spaces and algebras (Vol. II, Chap. VII). The notion of extension
of homomorphism is one of the main tools in the theory of fields.
The concept of an algebra arises naturally when one studies a
field relative to a selected subfield as base field. The concept of
tensor product is of lesser importance in field theory and it per-
* In Vol. II this notion was called the Kronecker product. Current usage favors the
term tensor product, so we shall adopt this in the present volume. Also we shall use the
currently standard notation ® for the X of Vol. II.
1

INTRODUCTION
haps could be avoided altogether. However, this notion has
attained enormous importance throughout algebra and algebraic
topology in recent years. For this broader reason it is a good
idea for the student to become adept in handling tensor products,
and we shall use these freely when it seems appropriate.
1. Extension of homomorphisms. Throughout this book we
shall adopt the convention that the rings we consider all have
identity elements 1 /0. The term subring will therefore mean
subring in the old sense (as in Vol. I) containing 1, and by a
homomorphism of a ring 21 into a ring SB we shall understand a
homomorphism in the old sense sending the 1 of 21 into the 1 of SB.
Now let o be a subring of a field P and let i> be the subfield of P
generated by o. We recall that the elements of $ can be ex-
expressed as simple fractions a/? of elements a, /8 e o (/3 j^ 0).
Hence "£> is the subring of P generated by o and the inverses of
the elements of the set o* of non-zero elements of o. The set o*
contains 1 and is closed under the multiplication of o. It is some-
sometimes useful to generalize this situation in the following way: We
are given a subring o of P and a subset M of o* containing 1 and
closed under multiplication. We shall refer to such a subset as a
sub-semigroup of the multiplicative group of the field. We are
interested in the subring om generated by o and the inverses of
the elements of M. For example, we could take P to be the field
Ro of rational numbers and M = {2k\k = 0, 1, 2, • ■ •}. Then
oaf is the subring of rational numbers whose denominators are
powers of 2. In the general case,
Ojif = {a.$~x\a CO, $
for, if we denote the set on the right-hand side of this equation by
o', then clearly o' C om and o' contains o = {a = al~1}. Also
o' contains every /3 = ljS for /3 e M. One checks directly
that o' is a subring of P. Then it follows that o' = om-
Now suppose P' is a second field and we have a homomorphism
s of o into P' such that /3* ?* 0 for every j8 e M. Our first homo-
homomorphism extension theorem concerns this situation. This is the
following result.
I. Let 0 be a subring {with 1) of afield P, M a subset of non-zero
elements of 0 containing 1 and closed under multiplication, om the

INTRODUCTION
subring of P generated by o and the inverses of the elements of M.
Let s be a homomorphism of o into a field P' such that ft' p* 0 for
every /? e M. Then s has a unique extension to a homomorphism
S of om into P'. Moreover, S is an isomorphism if and only if s
is an isomorphism.
Proof. Let auSi = a2p2~1, «; e o, j3,- e M. Then «ij32 =
a2pi and consequently ai8/328 = a^Pi'. This relation in P' gives
ai'GSi8) = a2*(/32*)~1. Hence the mapping
S:aP~l -* a'O?8), aeo, K e M
which is defined on the whole of oat = {a/S} is single-valued.
One checks that S is a homomorphism (Vol. I, p. 92). If a e o,
then as = (al ~l)s = a'Y = a", so S is the same as s on o. Hence
S is a homomorphism of om which extends the given homomor-
homomorphism of o. Now let <S" be any such extension. Then the relation
ftS-1 = 1 for ^eM gives /3s'OS)8' = 1, so (fl-1)*' = (/3s')-
If aeo, then we have (a/3)^ = o8'^') = aW =
(a^I5. Hence 6" = S and 61 is unique. Clearly, if S is an iso-
isomorphism, then its restriction s to o is an isomorphism. Now
assume s is an isomorphism and let a/3 be in the kernel of the
homomorphism S: 0 = (a^M = a8^8). Then a8 = 0, a = 0,
and a^~1 = 0. This shows that the kernel of S is 0; hence S is an
isomorphism.
We consider next an arbitrary commutative ring 21 and the
polynomial ring 21[a?], x an element which is transcendental rela-
relative to 21 (Vol. I, p. 93). The elements of 2tM have the form
aQ + axx + a2x2 + • • ■ + anxn where the at e 21 and a0 + axx +
• —\- anxn = 0 only if all the «,• = 0. We now have the follow-
following homomorphism theorem.
II. Let 21 be a commutative ring, 2l[,v] the polynomial ring over 21
in a transcendental element x and let s be a homomorphism of 21 into
a commutative ring SB. If u is any element of S3 there exists a unique
homomorphism S of 2l[x] into SB such that: as = a', a e 21, Xs = u.
The reader is referred to Vol. I, p. 97, for the proof. This result
has an immediate extension to a polynomial ring 2l[#i, x2, • • •, xr]
where the #,• are algebraically independent elements. We recall
that the algebraic independence of the xt means the following:

INTRODUCTION
If (mi, m2, • • •, mr) is an r-tuple of non-negative integers ;»,-, then
a relation 52 ami.. .mrXimi ■ ■ ■ xrmr = 0, ami...mr e 21, can hold
ml
only if every ami...mr = 0. From now on we shall refer to ele-
elements #,• which belong to a commutative ring and are algebraically
independent relative to a subring 21 as indeterminates (relative to
21). Then we have
III. Let 2l[#i, • • •, xr] be a commutative polynomial ring in #,-
which are indeterminates (relative to 21) and let s be a homomorphism
of 21 into a commutative ring SB. If u\, u2, • • •, ur are arbitrary
elements of SB, then there exists a unique homomorphism S of 2l[#,-]
into SB such that 1) as = a',ae 21; 2) xf = «,-, i = 1, 2, • • •, r.
We now suppose we have a commutative ring &, 21 a subring,
j a homomorphism of 21 into another commutative ring SB. Let
h, h, • •' y tr be elements of S and let 2l[/i, t2, ■ • •, tr] be the sub-
subring of S generated by 21 and the /,-. Under what conditions can
s be extended to a homomorphism S of 2l[/t] = 2l|/i, t2, • • •, tr]
into SB so that f,-5 = «,-, 1 < i < r, where the «,■ are prescribed
elements of SB ? The answer to this basic question is
IV. Let SB and S fo commutative rings, 21 a subring of ($,, s a
homomorphism of 21 *'«/o S3. L# /1} ■ • • ,tr be elements of(£.,Ui, • • •,
ur elements of SB. Then there exists a homomorphism S of 2l[A, • • •,
tr] into SB such that as - a',az%. and tts = «,-, i = 1, 2, • • •, r, if
and only if for every polynomial f(xx, • • •, xr) e 2l[^J, #,• indeter-
indeterminates, such that f(ti, ■ ■ -, tr) = 0 w<? A<zt^ f(ui, • ■ -,ur) =0.
/ffr<? /*(^i, • • •, xr) is obtained by applying s to the coefficients of
f(xu ■ • -, xr). IfS exists, it is unique.
Proof. The set $ of polynomials f(xi, • • •, xr) such that
f(h, ■ ■ ■, tr) = 0 is the kernel of the homomorphism h(xi, • • •, xr)
—> h(t\, • • •, tr) of %[xi\ into 2l|/,-]. Hence we have the isomor-
isomorphism T:h(tu ■ ■ -,tr) —» h(xu ■ ■ ■, xr) + $t of 2l[/,] onto the dif-
difference ring 2l[*t-]/$. Next we consider the homomorphism h(x1}
■ ■ •, xr) -* h'(uu ■ ■ -,ur) of 2l[^t] into SB (cf. III). Assume that
/"(«!, • • -,ur) =0 for every fe$. Then every/e^ is mapped
into 0 by the homomorphism h(xu • ■ -,xr) —» h' («1} ■ • -,ur) so
$ is contained in the kernel of this homomorphism. It follows
(Vol. I, p. 70) that we have the homomorphism h(xiy • • •, xr) +

INTRODUCTION
St -» ha(uly ■■ -,ur) of %[xi]/St into SB. Combining this with the
isomorphism t we obtain the homomorphism
A) S:h(tu--,tr) -> h'{uu ...,«r)
of 2l[A] into SB. This is the required extension of s. If S' is any
extension of s to a homomorphism of 2l[/»] into £9 such that a8' —
as and tf = ui} then h(tu ■ ■ ■, tr)s> = h°(uu ■■, «r); hence S' =
S and 6" is unique. Also, it is trivial that, if/(/i, •••,tr) = 0,
then 0 = f(tu • • •, Os = /*(«i> • • ■, «r) if «y is a homomorphism
of 2l[/i, •••,/,•] satisfying our conditions. Hence it is clear that
the condition stated in the theorem is necessary for the existence
of the extension S.
We have noted in the proof that the set $ of polynomials
/(*i> • • • j *r) such that /(/i, • • •, tr) = 0 is the kernel of a homo-
homomorphism. Hence this is an ideal in the polynomial ring 2l[.*i, x2,
■ • • y xr]. Now let X = \g] be a set of generators of ®: X c $
and every element/ e$ has the form S«,(xi, • • •, xr)gi(xi, ■ ■ •, xr)
where the a,- (xi, ■ ■ ■, xr) e 2l[#i, #2j • • • > #r] and the gi(x\, • • •, xr)
e X. It is clear that, if gs(#i, ••-,«,) =0 holds for every gz X,
then also /'(«i, ■ • ■ ,ur) = 0 for every / e $. Hence we can obtain
from IV the following result which is often easier to apply than
IV itself:
IV. Let SB and £ be commutative rings, 21 a subring of E, and s
a homomorphism of 21 into SB. Let X be a set of generators of the
ideals of polynomials f in 2l[^i, x2, • • •, xr], Xi indeterminates, such
that fit i, t2, • • ■, tr) = 0. Then there exists a homomorphism S of
2l[/x, t2, • • •, tr] into SB such that as = a\ az 21, and t{s = «,-,
1 < i < r, if and only if g*{uu ■ ■ -,ur) = Ofor every g e X. If S
exists, then it is unique.
We now consider the important special case of IV in which
21 = $ a field and r = 1. Then we know that <£>[#] is a principal
ideal domain (Vol. I, p. 100). Hence the ideals = (/(#)), where
(/(*)) denotes the ideal of polynomial multiples of the poly-
polynomial/(#) e$. It is clear that $ ?* A) = $[x) since, otherwise,
0 = *[*]/fi ^ <t>[t] 3 $ which contradicts 1^0. Since (a)
= A) if a is a non-zero element of <1>, it is clear that the possibili-
possibilities for $ areS = @) or$ = (/(*)) where f(x) is a non-zero poly-

INTRODUCTION
nomial in <&[x] of positive degree. In the first case we have #[#]
S *W and / is transcendental. Then II (or IV) is applicable and
shows that s can be extended to a homomorphism S sending /
into any u e 93. Now suppose that fix) ?* 0. In this case we
call the element /eg algebraic over # since we have a non-zero
polynomial/(#) such that/(/) = 0. The ideal $ is, by definition,
the set of polynomials g(x) such that g(t) = 0. The polynomial
f(x) is a polynomial of least degree in $ and every other poly-
polynomial contained in $ = (/(#)) has the form g(x)f(x). We can
normalize fix) by multiplying it by the inverse of its leading
coefficient to obtain a polynomial with leading coefficient 1. If
we let/(#) be this polynomial, then clearly/ can be characterized
by the properties that it is the polynomial of least degree belong-
belonging to &[x] with leading coefficient 1 satisfying/(/) = 0. We shall
call f(x) the minimum polynomial (over $) of the algebraic element
/ e E. We can now state the following result which is a special
case of IV.
V. Let 93 and & be commutative rings, * a subfield of <S., t an ele-
element q/"S which is algebraic over®, and s an isomorphism of$ into 93:
* \
■V' •
93
Then s can be extended to a homomorphism S of $[t] into 93 so that
Is = u, if and only if /*(«) = Ofor the minimum polynomial f(x)
of t over$. When the extension exists it is unique.
Remarks. The condition one has to put on u to insure the
existence of S can be stated also in the following way: u is alge-
algebraic over the image $* of $ and its minimum polynomial over
$• is a factor of fix). The equation A) giving the form of S
now becomes
B) S:g(i) -» *•(«).
It is immediate from this that S is an isomorphism if and only if
fix) is the minimum polynomial of «.

INTRODUCTION
2. Algebras. We recall the definition of an algebra 21 over a
field * (Vol. II, p. 36 and p. 225): 21 is a vector space over * in
which a product xy e SI is defined for x, y in 21 such that
(#i + x2)y = xxy + x2y, x(j! + y2) - xyx + xy2
C)
a(xy) = (ax)y = x(ay), ae*.
We shall be interested only in algebras which have identities 1 and
which are associative; hence in this volume "algebra" will always
mean just this.
We shall usually encounter algebras in the following way: We
are given a ring 21 and a subfield * of the center of 21. Then
we can consider 21 as a vector space over * by taking ax, a e *,
x e 21, to be the ring product of a and x in 21. Clearly this makes
21 a vector space over *. Also C) is clear since a is in the center.
Hence we have an algebra 21/$ B1 over *).* This procedure for
defining an algebra will be used in studying a field P relative to a
subfield *. Then we obtain the algebra P/*.
Another algebra which is basic is the algebra 8$( 9W) of linear
transformations of a vector space Wl over a field *. Here A + B,
AB and aA for A, B e 8*( m) and a e * are defined by x(A + B)
= xA + xB, x(AB) = (xA)B, x(aA) = a(xA) = (ax)A. The
dimensionality [8*( 2)?) :$] of 8*( 93?) over * is finite if and only
if [ m:$>] is finite. If [ 3ft:<i>] = m, then [8*( m) :<*>] = m2 (Vol. II,
p. 41).
Evidently an algebra is a ring relative to the + of the vector
space and the multiplication ab. A subalgebra S3 of an algebra 21
over $ is a subspace of 21 which is also a subring. An ideal of
21/* is a subspace which is an ideal of 21 as a ring. A homomorphism
s of the algebra 2l/4> into the algebra 83/* is a mapping of 21 into
93 which is ^-linear and a ring homomorphism. Isomorphisms
and automorphisms are defined in a similar fashion. If $ is an
ideal in 21/$, then the factor space 21/$ is an algebra over * rela-
relative to its vector space compositions and the multiplication
(a + ®)(l> + ®) = ab + $. We have the algebra homomorphism
a —> a + $ of 2l/# onto 21/$ over <£>. If s is a homomorphism of
21/* into 33/*, then the image 21* is a subalgebra of S3 and the
* We shall use the notation 21/33 also for the difference ring of 21 relative to the ideal S3-
Which of these meanings is intended will always be clear from the context.

8 INTRODUCTION
kernels of s is an ideal in 21. We have the isomorphism a + ® —>
a* of 21/$ onto 21s. The basic results on ring homomorphisms ex-
extend to algebras and we shall use these without comment.
We shall now record some elementary results on finite dimen-
dimensional algebras which will be used frequently in the sequel. The
first concerns a dimensionality relation for 21/<J> and 21/E, where
E is a subfield of $. Evidently if E is a subfield of $, then we can
restrict the multiplication ax, a e $>, x e 31 to a in E. This turns
SI into an algebra 21 over E. Also since E is a subfield of $ we
can define the algebra $/E. We now have
VI. Let 21 be an algebra overQ?, E a subfield of$. Suppose [21:$] <
oo and [$:E] < °o. Then
D) [21: E] = [«:*][*: E].
Proof. Let («»•), 1 < i < n, be a basis for %/$, (yy), 1 < j <m,
a basis for $/E. Then D) will follow if we can show that (t/K,-)
n
is a basis for 21/E. First let a e 21. Then a = ^ «,«,-, a,- e$,
1
and a; = ^ ey-yy where e,-y e E. Then « =Se,y7y«j is a linear
combination of the elements 7y«,- with coefficients ety in E. Now
suppose litifijUi = 0 where the ety e E. Then we have Sat«t- = 0
for at- = ^2 tiffj in <3?. Since the «,- are •I'-independent, this gives
3
at = 0, 1 < i < n. Then the formulas a,- = 2«</yy and the E-
independence of the y,- give eti — 0 for all i, j. This proves that
the elements yjUt are E-independent and so these form a basis
for 21/E.
VII. Let 21 be a finite dimensional algebra over a fields. Then 21
is a division ring if and only if 21 is an integral domain.
Proof. We know that division rings are integral domains (Vol.
I, p. 54). Now suppose 21 is an integral domain and let a be any
non-zero element of 21. Consider the right multiplication aR:
x —» xa determined by a. This is a linear transformation in 21/*
and, since ba = 0 in 21 implies b — 0, the null space of aR is 0.
It follows that aR is surjective (that is, maps 2t onto 21). Hence
there exists an element a' such that a'a = a'aR = 1. Thus a

INTRODUCTION
has a left inverse. A similar argument using the left multiplica-
multiplication «£, shows that a has a right inverse. Hence every non-zero
element of 21 is a unit and 21 is a division ring.
We consider next algebras 21 = $[/] which have a single genera-
generator / (cf. § 1). We have the homomorphism g{x) —> g(t) of
•I'M, x an indeterminate, onto 21. If S is the kernel, then 21 =
*M/t. Also we have seen in § 1 that$ = (/(#)) where /(*) = 0
or is a non-zero polynomial with leading coefficient 1. In the
first case, t is transcendental and the homomorphism we indicated
is an isomorphism. In the second case, t is algebraic and/(,v) is
its minimum polynomial. Then we have
VIII. Let 21 = $[/] be an algebra over <3? generated by a single
algebraic element t whose minimum polynomial is f{x). Then
E) [«:*] = deg/W,
the degree off(x).
Proof. Let n = deg/(#). Then we assert that A, /, • • •, tn~l)
is a basis for 2t/«t>. Thus let a be any element of 21 = $[/]. This
has the form g(t), g(x) in $[*]. By the division process in f>[x]
we can write g(x) = f(x)q(x) + r(x) where degr(x) < deg/(#).
Then if we apply the homomorphism of <£[#]/<£ onto $[/]/$ send-
sending x into /, we obtain a = g(t) = Oq(t) + r(f). Since deg r{x) <
n, this shows that a = r(t) is a ^-linear combination of 1, t, • • •,
tn~1. Next we note that 1, /, • • •, tn~l are linearly independent
over $ since otherwise we would have a polynomial g(x) ?* 0 of
degree < n such that g(t) = 0. This contradicts the hypothesis
that/(x) is the minimum polynomial. Hence A, t, • • •, tn~l) is a
basis and E) holds.
We recall that <£[/] ^ ^[x]/{f{x)),f{x) a. polynomial of positive
degree, is a field if and only if/(x) is irreducible (Vol. I, p. 101).
Otherwise, *[/] is not an integral domain. It is useful to have a
more complete analysis of the structure of <£[/] in terms of the
minimum polynomial/(#). We shall indicate the results in the
following exercises.
EXERCISES
1. An algebra 21 is a direct sum of ideals 21,- if 21 is a vector space direct sum of
the subspaces 21,-. Let 21 = $[/], t algebraic with minimum polynomial /(#).

10 INTRODUCTION
Suppose/(*) =/i(*!/,(«) • ••/,(*) where (/,•(*),/,(*)) - 1 if » * j. Set ?<(*) -
/(#)//•(#)• Show that there exist polynomials «,•(*) such that
r
i
Set <?,- = ai{t)qi(f) and show that
'i + *2 H h <fr = 1, e? - a, <?,<?,- = 0, » 7^ _/.
Show that 21 = 2fei © %e2 ©• • •© 2l<rr and that the ideal 2k = {ae(\a e2T}
considered as an algebra with identity if< has the form *[/i?J and is isomorphic to
2. Let 21 = $[/], / algebraic with minimum polynomial /(#)• Let /(*) =
^iM^PjM*2 ' • • PrW*11, .PiM irreducible, ^><(#) ^ />,(#), / ?* j. Show that if z =
Pi(t)pz(l) " • • .PrW, then the ideal 9t = 2lz in 21 is nilpotent in the sense that there
exists an integer k such that every product of k elements of SH is 0. Show that
21 = 2l/9t = #[/], 1 = t + 1R, and t is algebraic with minimum polynomial
fM = Pi(*)pt(x) ■ ■ ■ Prix). Show that I = Ii © I2 © • • ■ © Ir where %
is an ideal which as an algebra is isomorphic to the field *[*]/(?,•(*)).
3. Let 21/$ be an algebraic algebra in the sense that every element of 21 is
algebraic. Prove that, if 21 is an integral domain, then 21 is a division ring.
3. Tensor products of vector spaces. Let 2ft, 9<l and $ be
vector spaces over the same field <£. Then a bilinear mapping of
9K, 91 into $ is a mapping of the product set SO? X 9t into ty such
that, if x X y denotes the image of the pair (x,y), x e SW, y e 31,
then
(xi + x2) X y = xi X y + x2 X y,
F) x X Oi + y2) = x X yx + x X y2
a(x X y) = ax X y = x X ay, a e *.
It is clear that the product xy in any algebra 21 is bilinear from
21, 21 to 21. We shall say that a vector space "p and a bilinear
mapping <g> of 3ft, 91 into ty is a tewjor product of Eft and 9i and
we write <$ = 2ft ® 91 if the pair (<g>,$) is "universal" for bi-
bilinear mappings in the sense that the following condition is ful-
fulfilled:
If <P' is any vector space and X' is a bilinear mapping of 2ft, 91
into ^J', then there exists a unique linear mapping w of $ into $'
such that (x <g> y)ir = x X'y.
This notion is a special case of the general concept of the tensor
product of a right module 2ft over a ring 21 and a left module
91 over 21. The special case we have defined for vector spaces is
treated under slightly different but equivalent hypotheses in Vol.

INTRODUCTION 11
II, Chap. VII. In particular, a proof of the existence of a tensor
product of vector spaces and nearly all the basic properties we
shall require were given in Vol. II. At this point we shall give
another derivation of some of these basic results which is more in
keeping with the spirit of the now standard treatment of the
module case.
We first give a construction of a tensor product. To do this one
begins with a vector space g having as basis the product set
2ft X 9t of pairs (x, y), x e 2ft, y e 9fc. Thus the elements of ft are
the expressions £i(*i,.yi) + Z2(x2,y2) H 1- %m(xm,ym) where
£t- e*, x{ e 2ft, yt e % and the pairs (^,-,jy.) are distinct. If two
elements are given we can introduce terms with 0 coefficients and
m m
thus suppose that the elements are 2 $»(#»> 7») and ]£ '7»(#»>.)'«)«
1 1
Then equality holds if and only if £,• = ?;,•, / = 1, 2, • • •, m. Addi-
tion is defined by £ £»•(*»> yd + Hvi(xi,yi) = Z) (fc + Vi)(xi,yi)
111
and multiplication by a in $ by «2£,•(*,-,yt) = S(a?,)(x,-,_?,-)• It
is immediate that g is a vector space over $. Since 2ft X 51 is
usually infinite, 5 is usually an infinite dimensional space. Now
let 9? be the subspace of g spanned by all the vectors of the follow-
following forms:
Oi + *2,.y) - (xuy) - {x2>y)
(*>yi +yz) (x,yi) - (x,y2)
(ax,y) - (x,ay)
x e 2ft, y eSfl, a e <£>. Let $ be the factor space %/dt and set
x ® y =5 (x, y) + % the coset of (x,y) in g/9t Then we have:
(#1 + x2) ® y - X! ® y — x2 ®y
= Oi + x2, y) - (xu y) - (x2, y) + 9t = 9t
■v ® (jt + y2) — x ® yi — x ® y2
= (^^1 + J2) - O,.yi) - {x,y2) + 9? = 9*
a(x ® y) — ax ® y = a(xt y) — (ax, y) + 9? = 9J
ax®y — x®ay = (ax, y) — (x, ay) + 9? = 9*.

12 INTRODUCTION
Hence x ® y is bilinear. Since the vectors (x, y) generate g, the
cosets x ® y generate 5JJ = g/9J.
Now let X' be a bilinear mapping of 2ft, 5ft into the vector space
5P'. Since the vectors (x, y) form a basis for g, there exists a
linear mapping it' of g into 5C' such that (x, y)v' = x X' y. Let
$ be the kernel of ir'. Then ((*i + #2, jy) — (*i,y) — (x2}y))ir' =
(#i + #2) X'jy - *i X'jy - #2 X'jy = 0; so (#1 + x2,y)
- (x,y) - (x2,y) e®. Similarly, {x,yx + y2) - {x,yx) - (x,y2)
e.f, (ax,y) — a(x,y) e$, and (ax,y) — (x, ay) e®. This implies
that 9? C $ and, consequently, we have the linear mapping w
of 5C = g/SR into 5p' such that (* ®jyOr = ((^,jy) + m)ir = * X'jy.
Since the space 5|3 = ft/dt is generated by the elements x ® y,
it is clear that tt is uniquely determined by the linearity property
and (x ® y)w = x X' y. We have therefore shown that EC, ®) is
a tensor product of 2ft and 5ft and accordingly we shall write 5J3 =
2ft ® 5ft (or 2ft ®* 5ft, if it is necessary to indicate the base field
<$). It is immediate from the definition that if ECi, ®i) and EC2,
<g>2) are two tensor products, then we have a linear mapping of
5|3i into 5p2 such that x ®iy —> x ®2 y and we have a linear map-
mapping of 5K2 into 5Ci such that x ®2y —> x ®iy. Since the x ®iy
generate 5C,-, the products in both orders of the two linear mappings
are identity mappings. It follows that both mappings are sur-
jective (onto) linear isomorphisms. In this sense the tensor
product is uniquely determined and so we may speak of the tensor
product of 2ft and 5ft.
Let {ea} and {fp} be sets of generators for 2ft and 5ft respec-
m
tively. Then any x e 2ft has the form x = 53 £%ex where {ei\
1
is a finite subset of \ea) and any^ e 5ft has the form^ = ]>
Hence, by the bilinearity of ® we have x ® y =
Since the elements x ® y generate 2ft ® 5ft, we
see that the products ea ® f$ generate 2ft ® 5ft. Now suppose
that the {<?„} and {/$} are independent as well as generators, that
is, these form bases for their respective spaces. We assert that
the set of products {ea ® f$} is a basis for 2ft ® 5ft. Since these
are generators we just need to show that they are linearly in-
independent. For this purpose we form a vector space 5$' with

INTRODUCTION 13
basis gap in 1-1 correspondence with the product set (a, /3) of the
index sets of a and of j3. If x = 2£;<?; and y = Zyjfj, then we
define x X' y = ^^iViia- It is easy to check that the product X'
is bilinear, so we have the linear mapping ir of 3ft <8> 5ft into 5p'
sending x <g> y —> x X'y. In particular, ea ® fp —> ea X' fp =
gap. Since the gap are linearly independent, the same holds for
the ea ® fp and we have proved
IX. Let {ea} and \fp\ be generators for 3ft over § and 5ft over $
respectively. Then the set \ea ® fp) generates 3ft ® 5ft. More-
Moreover, if the {ea) and {fp} are bases, then the same holds for {ea ®fp}.
The second property actually characterizes the tensor product
among the bilinear mappings of 3ft and 5ft. More precisely, let
X' be a bilinear mapping from 3ft and 5ft to a space 5E' and suppose
there exists a basis (ea) for 3ft over <t> and a basis (fp) for 5ft over
* such that (<?« X fp) is a basis for 5p'. Then EE', X') is a ten-
tensor product. Thus we have the linear mapping of 3ft ® 5ft into
5C' sending ea ® fp into ea X' fp. Since the ea X' fp generate
5p', the mapping is surjective and, since the ea X' fp are linearly
independent, the mapping is 1-1. Thus we have a linear iso-
isomorphism of 3ft ® 5ft onto 5p', mapping x ® y into x X' y. This
implies that $', X') is a tensor product.
In the case of finite dimensional spaces we have the following
simple criterion.
X. Let X' be a bilinear mapping of the finite dimensional spaces
3ft and 5ft into 5C' and suppose that 5p' is generated by the products
x X'y. Then the dimensionality [<$':Q < [3ft:$][5ft:<J?] and equality
holds if and only if ($', X') is a tensor product of 3ft and 5ft.
Proof. Let (<?*), (/,•) be bases for 3ft and 5ft respectively. Then
every x X' y is a linear combination of the elements <?* X' // and
so every element of 5p' is a linear combination of these elements.
This implies $':$] < [3fl :<*>][5ft :$]. Ep', X') is the tensor product
if and only if the set (*,- X' /,■) is a basis. This is the case if and
only if the equality holds in the dimensionality relation.
We recall that, if A is a linear mapping of 3ft into 5tfti and B
is a linear mapping of 5ft into 5fti, then there exists a uniquely de-
determined linear mapping A <8> B of 3ft ® 5ft into 3fti ® 5fti such

14 INTRODUCTION
that (x ®y)(A ® B) = xA ® yB (Vol. II, p. 211). We recall
also that, if P is an extension field of the field * so that Pisa vector
space over <i> and 9Ji is any vector space over $, then P ®* 3ft
can be considered as a vector space over P by means of the
product pBpi ® Xi) = Sppt- ® xi} p, pt- e P, xt e 3ft (Vol. II, p.
221). We denote this vector space as 3ftp and we refer to it as
the space obtained from 3ft by extending the base field to P. If
A is a linear transformation in 9ft over $, then \ ® A (defined by
(Zpi ® Xi) A ® A) = Sp,- ® XiA) is a linear transformation in
2ttp over P which may be considered as the extension of A to Wlp.
We shall use the same letter A to denote this extension. If (ea)
is a basis for 2ft over $, then A ® ea) is a basis for 2ftP over P, so 9Ji
over <J? and 9Ttp over P have the same dimensionality. If $ft is finite
dimensional with basis (*,-), 1 < / < n, and yf is the linear trans-
transformation with matrix (a,y) relative to this basis, then dA =
'Zaijej and A ® e^A = Sa,y(l ® e,). Hence the extension A has
the same matrix relative to the basis A ® <?,).
We recall also that the tensor product is commutative in the
sense that there exists a 1-1 linear transformation such that x ®
y —»jy<g>,vof3ft<g>5ft onto 5ft ® 3ft. Moreover, associativity
holds in the sense that there is a linear isomorphism of Cft ® 5ft)
® <S onto 3ft ® Eft ® @) mapping (x ® y) ® z into x ® (y ® z).
These results have been established in Vol. II, pp. 209-210. We
shall indicate alternative proofs in some of the following exercises.
EXERCISES
1. Show that, if {//jj is a set of generators for 5ft, then every element of 3ft ® 5ft
has the form 2#,- ®fi, {/<} a finite subset of {f$} and Xi e 5DJ. Show that, if the
{/(j) are linearly independent, then 2*,- ® fi = 0 if and only if every Xi = 0.
2. Show that, if 3fti is a subspace of 3ft, then the subspace Sfti ® 5ft generated
by all vectors #i ® y, xi e 2rti, y e 5ft is the tensor product of 3fti and 5ft relative
to the ® defined in 3ft ® 5ft.
3. Let$ be a subspace of 3ft, 8 a subspace of 5ft. Show that Bft/$) ® Eft/S)
and Elft ® 5ft)/(^ ® 5ft + 3ft ® ?) are isomorphic under a linear mapping such
that (* +1) ® (y + 8) -» x ® y + (f <g» 5ft + 3ft ® 2).
4. Let Sfti, 3ft2, • • •, 3ftr and 5[J be vector spaces over $. Define an r-linear
mapping (#i, ■• -,xr) —> xi X xt X ■ • • X xr e *$, xt e 3ft<, by the properties:
*i X • ■ • X (*/ + xi") X • ■ • X *v = *i X • • • X */ X • • ■ X xT
+ XI X • • • X Xi" X • • • X *r
<*(*! X • • • X #r) = *1 X • • • X OCXi X • • • X Xr.

INTRODUCTION 15
Show that there exists a "*|3 and an r-linear mapping of 2fti, • • •, 2ftr into 5P such
that: if (#1, • • •, xr) —* xi X' *2 X'■ • ■ X' xr is an r-linear mapping of 2fti, • • •,
2ftr into 5JJ', then there exists a unique linear mapping x of 5(J into IP' such that
(#1 ® ■ ■ • ® xr)r = xi X'• ■ • X' xr. Denote this $ together with its product as
the tensor product 2fti ® 2ft2 ® • • • ® 2ftr.
5. Show that 2ft ® 5ft ® 50 is isomorphic to 2ft ® Eft ® 50) and Bft ® 1ft) ® 5(J
by means of linear mappings such that x ® y ® z —* x ® (y ® z) and
(x ® y) ® z respectively. Generalize to r factors.
6. Show that 2ft ® 5ft is isomorphic to 5ft ® 50? under a linear mapping send-
sending x ® y -» y ® x. (Hint: Given 5ft ® 2ft, define x X'y = y ® x, x 8 2ft,
y e 5ft. Show that this gives a bilinear mapping of 2ft, 5ft into 5ft ® 2ft and
apply the defining property of 2ft ® 5ft. Then reverse the roles of 2ft and 5ft.)
4. Tensor product of algebras. We recall that, if 2lx and 2l2
are algebras over f>, then the vector space 21 = 2li ® 2l2 is an
algebra relative to its vector space compositions and the multi-
multiplication
(8) ( X au ® a2i
au, bij e 2li, a2i, hj e 2l2 (Vol. II, p. 225). The associativity of
2li and 2l2 implies associativity of 2li <8i 2l2 and li ® 12 is the
identity 1 of 21 = 2li ® 2l2 if 1,- is the identity of 21,-. Also 21 is
commutative if the 21, are commutative. The basic property of
the tensor product of algebras is the following homomorphism
theorem.
XI. Let 21,, / = 1, 2, be algebras over $, .?,• a homomorphism of
2lf into an algebra S3 such that a^az'* = «2*2«i*', «i e 2li, a2 e 2l2.
Then there exists a homomorphism s of 21 = 2li ® 2l2 into 33 such
that
(9) (Hau ® a2i)' = ^au^a2i'\
Proof. The algebra product a\ X' a2 = fliSlfl2S2 e S3 defines a
bilinear mapping of 2li, 2l2 into 33. This is clear from the linearity
of the Si and the properties of the multiplication composition in 33.
Hence the definition implies that we have a linear mapping s of
2li ® 2I2 into 33 such that («i ® a2)' = ax'a2\ Then s has the
form (9). We have {{ax ® a2){bx ® b2))s = {axbx ® a2b2)' =
(aihyK^h)'* = a^b^a2'*b2'* = a^a2HxH2'* = ((«x ® a2)'
{b\ ® b2)"). This implies that s is an algebra homomorphism.

16 INTRODUCTION
Suppose now that the following condition holds in 93:
(i) If (ea) is a basis for 2li over $> and (fp) is a basis for 2t2 over
<i>, then the set \ea*lfpS2} is linearly independent.
An equivalent condition for this which we shall sometimes find
more convenient is
(i') If (fp) is a basis for 2l2 over <f>, then a relation a^fi'2 +
a28lf231 -\ h amHfm82 = 0 for a{ e 211 and/, e (/^) implies that
every «,• = 0 (cf. ex. 1 of § 3).
Now we have seen that, if (i) or (i') holds, then the mapping
s given by (9) is an isomorphism of 21 = 2li ® 2t2 as vector space
into 93. Since this is an algebra homomorphism, clearly it is an
algebra isomorphism. We remark that (i) cannot hold unless sx
and s2 are isomorphisms.
The result we have obtained actually gives an internal charac-
characterization of 2li ® 2t2- For this we note that ax —> «iSl = «i <8> I2
and a2 —> a22 = li ® a2 are homomorphisms of 2li and 2t2 re-
respectively into 2li ® 2l2, since the linearity of the mappings we
have indicated follows from the bilinearity of ax ® a2, and the
homomorphism for multiplication is clear from (9). The com-
mutativity condition: a^a^2 = a^a^1 is clear, since ax*la2l =
(«i ® I2)(li ® a2) = «i ® a2 = (li ® «2)(«i ® 12) = a^'ai'1.
Finally, if (?„) and (fp) are bases for 2li and 2l2 respectively, then
the set {eaSlfpa2} = [ea ® fp} is linearly independent. It follows
that (eaSl) is a basis for %x = \ax ® 1} and (fp82) is a basis for
2I282. Also sx and s2 are isomorphisms and we can identify Hi*1
with 2Ii, 21/2 with 2I2- Our results evidently lead to the following
internal characterization of the tensor product of algebras:
XII. Let 21 be an algebra, 211 and 2l2 subalgebras such that
(i) «i«2 = a2au ai e 21,-.
(ii) If (ea) is a basis for 2li and (fp) is a basis for 2l2, then {eafp}
is a linearly independent set.
(iii) 21 is generated by 2li and 2l2.
Then 'Lau ® a2i —> 2«1,«2f is an isomorphism of 2li <8> 2l2 o»/o 21.
Because of this result and the situation we noted in 2li ® 2l2
itself, we shall say that 21 is the tensor product of its subalgebras 2li
and 2l2 if the above conditions (i)-(iii) are fulfilled. As we have

INTRODUCTION 17
seen, the condition (ii) can be replaced by the equivalent condi-
condition:
(ii') If (fp) is a basis for 2I2, then axfx + a2f2 H h amfm = 0
for at e 2li, /,- e (/^) implies every a{ = 0.
Of course, the roles of Stx and 2l2 can be interchanged in this.
We remark also that (ii) and (iii) can be combined in a single con-
condition: If (ea) is a basis for 211 and (f?) is a basis for 2l2, then (eafp)
is a basis for 21. For finite dimensional algebras this is equivalent
to the dimensionality condition: [21 :$] = [§li :<*][2C2:*] (cf. X).
EXERCISES
1. Let 21 be an algebra over the field $ and let 2t[#] be the algebra of polynom-
polynomials in an indeterminate x over 21. Show that 2l[*] is the tensor product of its
subalgebra 21 (constants of 2I[#]) and its subalgebra $[x] of polynomials in x with
coefficients in $. Use this to prove that <!>[#, y], x, y indeterminates, is the tensor
product of its subalgebras $[x] and $[>].
2. Let $(x,y) be the field of rational expressions in the indeterminates x,y,
that is, the field of fractions of $[#,.}']. Let 21 be the subset of fractions with
denominators of the form f{x)g{y),f{x) e $[#], g(y) e $[y]. Show that 21 is a sub-
subalgebra of 3?(x,y) which contains the subalgebras 4K#), $(y) where these are the
fields of fractions of <£[#] and $[y] respectively. Show that 21 is the tensor prod-
product of these subalgebras and that 21 is not a field.

Chapter I
FINITE DIMENSIONAL EXTENSION FIELDS
If i> is a subfield of a field P, then we have seen that we can
consider P as an algebra over $>. In this chapter we shall be con-
concerned primarily with the situation in which P is finite dimen-
dimensional over the subfield <£. We shall be concerned particularly
with the general results of Galois theory that are of importance
throughout algebra and especially in the theory of algebraic num-
numbers. We shall consider the notions of normality, separability,
and pure inseparability for extension fields, Galois cohomology,
regular representations, traces, and norms. Also the basic results
on finite fields will be derived and the notion of composites of two
extension fields will be considered.
In most of our considerations, and indeed throughout this book,
we shall usually be given a field <£ and we shall be concerned with
extension fields P/$. The ways of obtaining such extensions have
already been indicated in Vol. I, pp. 100-104. At the beginning
of this chapter we adopt a different point of view. Here we are
given the top field P and we look down at its various subfields;
moreover, we do not insist that these contain any particular sub-
subfield (except, of course, the prime field). The treatment here will
be abstract in the sense that no knowledge of the structure of an
extension is required. In spite of this we can give a survey of the
subfields which are of finite co-dimension in the given field P and
those which are Galois in P. These surveys are given in two general
"Galois correspondences." After these rather abstract considera-
considerations we shall go down to <i> and we shall apply the general results
to the extension P/<i> in terms of polynomial equations with co-
coefficients in P.
18

FINITE DIMENSIONAL EXTENSION FIELDS 19
1. Some vector spaces associated with mappings of fields.
Let E and P be two fields, and let ?(E, P) denote the set of homo-
morphisms of the additive group (E, +) of E into (P, +). The
set 8(E, P) is a group relative to the composition A + B defined
by e(A + B) = eA + eB for e in E. One checks that A + B e
2(E,P) and that the group conditions hold. The 0 of 8(E, P) is
the mapping 0 such that eO = 0, the 0 of P, for all e in E, and —A
is given by e(-A) = -eA (cf. Vol. I, § 2.13 and Vol. II, § 2.2).
If A is a third field and ^e8(E,P) and B e S(P, A), then the
resultant AB defined by e(AB) = (eA)B is an element of ?(E, A).
Both distributive laws hold for this composition. In combined
form they say that, if A1} A2 e ?(E, P) and Blt B2 e 8(P, A), then
(Ax + A2){BX + B2) = AXBX + AXB2 + A2BX + A2B2. Fi-
Finally, we note that the associative law of multiplication holds:
If T is another field and ^eg(E,P), B e S(P, A), Ceg(A, T),
then (AB)C = A(BC) e 8(E, T). All of these assertions are
readily verified and they are very similar to facts about composi-
composition of linear mappings which we have considered in Vol. II,
§ 2.2. We leave it to the reader to carry out the verifications.
The results we have indicated imply that 2(E, E) is a ring
under the compositions of addition and multiplication. This is
just the ring of endomorphisms of the additive group (E, +)
which has been considered in the general case in Vol. I, § 2.13.
If p eP, then the mappingpr'.I* —> £p(= p£) inP belongs to ?(P,P).
Since AB e g(E, P) for A in ?(E, P) and B in 2(P, P), we see that
Apn e8(E,P). This observation permits us to convert ?(E, P)
into a right vector space over the field P. For this purpose we
define Ap = ApR for A e 8(E, P) and p e P. Then we have
(A + B)P = {A + B)PB = APR + BPB = Ap + Bp
A(P + a-) = A(p + <t)r = A(PB + <rR)
= Apr + Ao-r = Ap + A<r
A(pa) = A{p<r)R = A{pr<tb) = (ApR)<TR = (Ap)a
A\ = A\R = A,
which shows that ?(E, P) is a right vector space over P.
We note next that if tR denotes the mapping rj —* ?je in E, then
e« e 8(E, E). Hence, if A e 8(E, P), then tRA e 8(E, P). We can

20 FINITE DIMENSIONAL EXTENSION FIELDS
now consider 8(E, P) also as a left vector space over E by defining
tA = €rA. It should be remarked that, if we do this, then there
is an ambiguity in writing eA which can mean either the image of
e under A or the endomorphism eRA. For this reason we shall
avoid considering 8(E, P) as a left vector space over E and use
instead the product 6rA when this will be needed.
All that we have just said applies also to fields over a given
field <£. Consider the fields E/<£ and P/$. In this connection it is
natural to consider the subset ?*(E, P) of 8(E, P) of linear trans-
transformations of E as vector space over $ into P over <i>. If a e <J?
and £, peP, then (ag)pB = (a£)p = a(fr) = «(&>«)> which implies
that pR e 8*(P, P). If A e 8»(E, P), then Ap = ApR e 8*(E, P); so
it is clear that 8*(E, P) is a subspace of the right vector space
8(E, P) over P. If 21 is any right vector space over P, we denote its
dimensionality over P as [2l:P]/j. Then we have the following im-
important result on [8*(E, P) :P]fi.
Theorem 1. Let E/$, P/$ be fields over $ and let 8*(E, P) be the
right vector space over P of linear mappings of E/<£ into F/<&. Then
[E:<j>] is finite if and only if [8*(E, P):P]fl is finite and when both
are finite then
A) [E:*] = [8.(E,P):P]«.
Proof. Let i)U -q2, • ■ -,Vn be elements of E which are linearly
independent over*. Then we may imbed this set in a basis {r/a}
for E over $ (Vol. II, p. 239). If we choose a correspondent ra
eP for each r)a, then there exists a unique element A e 8*(E, P)
such that t\aA = ra for every r)a. This implies that for each i =
1,2, •••,», there exists a linear mapping Ei (not necessarily
unique) such that ?;,■£,■ = 1, ijy£< = 0 if_; 5^ /. Then if pt- eP,
/ n \ n
( Z) Etpi) = 23 (viEi)pi = pj.
\ 1 / t-i
Hence X) -^iPi = 0 implies every pt- = 0, which shows that, if
1
[E:<i>] is infinite, then for every n there exist n right P-in-
dependent elements of 8*(E, P). Then [8*(E, P) :7]R > n for every
», so this dimensionality is infinite. Next suppose [E:«i>] = n <

FINITE DIMENSIONAL EXTENSION FIELDS 21
00 and that the ?j's constitute a basis. Let ^e8*(E, P) and set
t\iA = p^ Then t\j ( J2 EiP%) = Pi — Vi-d- Thus -^ and 2£,p,-
have the same effect on the basis G71, tj2, ••-,17™) for E/$. It
n
follows that // = 22 -Et'Pi andj since the £,• are right independent
1
over P, these form a basis for 8*(E, P) over P. Hence [8*(E, P) :P]B
= n = [E:$]. This completes the proof.
We now drop <t> and consider again E and P arbitrary fields
and 8(E, P) the group of homomorphisms of (E, +) into (P, +).
We consider this as a right vector space over P as before. Let
21 be a subspace of this space. Let e be a fixed element of E.
Then e determines a mapping /, of 21 into P by the rule that
ff(A) = eA eP. We have ft(A + B) = e(A + B) = eA + tB
= ft(A) + ft(B) and, if p eP, then /,(^p) = «(yfp) = (eA)P =
ft{A)p. Thus we see that/e is a P-linear mapping of the right
vector space 21 over P into the one dimensional space P over P,
that is, ft e 21*, the conjugate space of 21. Of course, 21* is a left
vector space over P. The process we have just indicated produces
a collection {ft | e e E} of linear functions. This collection is
"total" in the sense that, if ft(A) = 0 for all e, then A = 0.
This is clear since the requirement is that tA = 0 for all e and this
is just the definition of A = 0. We can now prove the following
useful
Lemma. Let %bea subspace of 8(E, P) overT such that [2l:P]ij =
n < 00. Then there exist elements ei, e2, ■■■,«,eE and a right
basis Ei, E2, ••-,£„ for 21 over P such that tJLj = StJ- Ei;- = 0 if
1 *j,t« = 1).
Proof. We are given that [2l:P]ie = n < <x>. This implies that
the conjugate space 21* is «-dimensional. Let S3* be the subspace
of 21* spanned by the linear functions /«, e e E. Since f(A) = 0
for all/ e S3* implies that A = 0, it follows that S3* = 21* (Vol.
2, § 2.10). Hence we can find n linear functions/«„ ffl, ••-,/•„
which form a basis for 21*. Since 21 can be considered as the con-
conjugate space of 21*, we can find a basis Eu E2, • • -,En for 21 over
P such that/e<(£3) = 5»j. Recalling the meaning of/€ we see that
we have uEj = S,y as required.

22 FINITE DIMENSIONAL EXTENSION FIELDS
2. The Jacobson-BourbaM correspondence. Let P be a field
and let ?(P, P) be the ring of endomorphisms of the additive
group (P, +). As before, we consider S(P, P) as a right vector
space over P. If $ is a subfield, then 8*(P, P) the ring of linear
transformations of P/$ is a subring of 8(P, P) and a subspace of
8(P, P) over P. Moreover, we have seen (Th. 1) that, if $ is of
finite co-dimension in P in the sense that [P:$] = n < °o, then
[8*(P, P): P]« = ». These properties of 8*(P, P) in no way refer
to the subfield $. We shall now show that they are charac-
characteristic of the sets ?*(P, P). This is a consequence of the follow-
following
Theorem 2 (Jacobson-Bourbaki). Let P be a field and 21 a set
of endomorphisms of (P, +) such that:
(i) 21 is a subring of'8(P, P) the ring ofendomorphisms of (P, +)
(containing the identity mapping, by our convention, Introd.
p.2).
(ii) SI is a subspace of 8(P, P) as right vector space over P.
(Hi) [%:P]R = n < oo.
Let <£ be the subset of P of elements a such that oirA = Aolr for all
Az%. Then <i> is a subfield of P, [P:*] = n and 21 = S*(P, P) the
complete set of linear transformations of
Proof (Hochschild). The verification that * is a subfield is
immediate and will be omitted. Next we apply the lemma of § 1
to obtain elements pi, p2, • • •, pn in P and a right basis (E\, E2}
• • •, En) for 21 over P such that piEj = 8,y. Since pror = <trpr
for any p, a in P, it is clear that $ is the set of a e P satisfying
i = 1, 2, •••,». Also it follows from p,-£y = 8,-,-
n
that, if we express the element A of 21 as A = 22 &&*> then
i
PjA = J3 (PjEi)ffi = <tj. Hence the representation of any A in
i n n
terms of the basis reads: A = 23 Ei(piA) or A = 23
i i
We shall now use this formula to show that every Ei maps P into
$. For this purpose let <r be any element of P and consider the
mappingEj<rREk,j, k = 1,2, •••,», which belongs to 21, since 2t is

FINITE DIMENSIONAL EXTENSION FIELDS 23
a subring of the ring of endomorphisms. The formula we obtained
can be applied for A = Ej<TREk to give
= Ej{\<rREk)R
= Ej{trEk)R.
This means that for arbitrary p in P we have
pEj<rREk- = pEj{aEk)R.
In other words, {{pE,)a)Ek = (pE,){aEk). Then
{a(pEj))Ek = {ffEk)\pEj).
If we think of a as the argument, this gives the operator identity
(pEj)REk = Ek(pEj)Ry which implies that pEj e<i>, and this holds
for all p e P. We can now show that the pt- we started with form
a basis for P/<i>. Let a e P and consider the element a' = a —
52 {<rEj)pj in P. Since oEj e <I> and a.REk — EkaR for a in *, we
j ( \ ( \
have d'Ek = <*Ek — I 2I1 (ff-^y)py) -^* = ff-^fc — ( aL Pii^^j/R ) Ek
\ j / \ j /
= o-Ejt — 52 {pjEk)(aEj)R = <rEk — <rEk = 0. Since 1 e 21, 1 =
2Ek\k for suitable Xfc e P. Then a'Ek = 0 implies <r'l = 0 so
<r' = 0. We therefore see that cr = S(<r£y)py is a$-linear combina-
combination of the pj. If Haipi = 0, at e f", then ay = (Satpt)£y = 0.
Hence (pi, p2, ---jpn) is a basis for P over 4> and [P:<3?] = n.
Since a^// = AaR for every a e $ and A e 21, every yf e 21 is a
linear transformation of P over <J>. Hence 21 C 8*(P, P). Since
[8*(P, P):P]ie = »by Theorem 1, and [2l:P]jj = «, we see that
21 = 8»(P, P).
Theorem 2 permits us to establish our first and most general
"Galois correspondence" for a field P. This concerns two collec-
collections of objects: the collection fF of subfields $ which are of
finite co-dimension in P and the collection S% of sets of endomor-
phisms of (P, +) having the properties (i), (ii), (iii) of the theo-
theorem. To each $e^" we associate i?($) = 8*(P, P). This is a
subring of 2(P, P), a subspace of 8(P, P) over P and satisfies
[8*(P, P): P]K < 00. Hence R($) = 8*(P, P) e ^. On the other

24 FINITE DIMENSIONAL EXTENSION FIELDS
hand, if 21 e ^ then we can associate the subfield .FBl) = $ =
{a|a e P, aRA = AaR, A e 21}. This is of finite co-dimension in
P and so it belongs to #" By Theorem 2, we have R(F(fC)) = %•
If $eST and 21 = #(<*>) = 8»(P, P), then [21: P]* = [P:$] by
Theorem 1 and [21: P]B = [T:F(X)] by Theorem 2. If a e*, we
certainly have a^yf = ^/ajj for yf e 21. Hence $ c .FBl) by the
definition of F. Since [P:#] = [P:.F(a)][F(a):*] (VI, Introd.)
and[P:$] = [P:FBl)], we have [jF(a):#] = 1 and so* =
F(R($)). The two relations
= 21, 21 e 3t
imply that the mappings 2? and F are inverses and are 1-1 of &~
onto ^ and ^ onto «^~ respectively. It should be noted that the
definitions of R and F show that these mappings are order revers-
reversing for the inclusion relation: $! c <J>2 for subfields implies R($i)
3 /?(#a) and 2li c 2l2 for 21,- e ^ implies FBli) 2 F(a2).
In § 4 we shall establish a Galois correspondence between finite
groups of automorphisms of a field P and certain subfields of
finite co-dimension in P. Later (§ 8, Chap. IV) we shall establish
a similar correspondence between certain Lie algebras of deriva-
derivations in P and certain subfields of P. Both of these correspond-
correspondences will be derived from the general "Jacobson-Bourbaki cor-
correspondence" which we have just given. In addition to this we
shall need some information on special generators for some of the
rings 21 e ^. For the automorphism theory the generators are
automorphisms of P. The results we require for these will be
derived in the next section.
EXERCISES
1. Let 21 be a set of endomorphisms of (P, +) satisfying conditions (i) and (ii)
of Theorem 2. Show that 21 is an irreducible ring of endomorphisms (Vol. II, p.
259). Apply the density theorem for such rings (Vol. II, p. 274) to show that, if
Pi» Pit • • • j Pm are ^-independent elements ($ as in Th. 2) and cri, ffa, • ■ •, <rm are
arbitrary in P, then there exists an A e 21 such that piA = <n, i = 1,2, ■ • -,m.
Use this result to give another proof of Theorem 2.
2. Let P be an arbitrary extension field of the field $. Show that, if a e P satis-
satisfies <xrA = AaR for all A e ?*(P, P), then a 8 <i>.
3. Let (pi,P2, • • •,Pn) be a basis of P/$, (A\, At, ••-, An) a right basis for
8»(P, P) over P. Show that the n X » matrix (piAj) has an inverse in Pn.

FINITE DIMENSIONAL EXTENSION FIELDS 25
3. Dedekind independence theorem for isomorphisms of a field.
Let s be an isomorphism of a field E into a field P. Then s is an
isomorphism of the additive group (E, +) of E into (P, +)
satisfying the multiplicative condition (etj)' = eV- We can
write this in operator form as:
B) vrs = j(ijf)«
where rm is the multiplication by r\ in E and (vs)r is the multiplica-
multiplication by 77s in P. If both E and P are fields over <!?, then an iso-
isomorphism of E/$ into P/f> is an algebra isomorphism of the first
algebra into the second. Hence, in addition to the conditions:
(e + vy = e> + 77% (er,)8 = eV, Is = 1, s is 1-1, we have (««)* =
ae8 for a e <&. The first and last of these are just the conditions
that s e ?#(E, P). Hence if s is an isomorphism of E/$ into P/S",
then a8 = (al)' = al8 = a holds for every ae«i>. Conversely,
this condition implies that (ae)8 = ae8, e e E. Thus an isomor-
isomorphism of E/<£> into P/3> is just an isomorphism of E into P which
is the identity mapping on <i>.
We shall now derive two basic results on linear relations con-
connecting isomorphisms of E into P (no $).
Theorem 3 (Dedekind). Let E and P be fields and let su s2, • • •,
sn be distinct isomorphisms q/E into P. Then the s{ are right linearly
independent over P: Sj,pt- = 0, pt- e P, implies every pt- = 0. Here
•fp = spR.
Proof. If the assertion is false, then we have a shortest relation,
which by suitable ordering reads:
C) J1P1 + J2P2 H h srpr = 0,
where every p,- t± 0. Suppose r > 1. Since Ji ^ s2 there exists
V e E such that j? 5^ ,7"*. Now multiply C) on the left by tjr.
If we take into account B), this gives: Ji»7Slpi + s2r)'2P2 +• —h
srV'rPr = 0. Next we multiply C) on the right by rj81 and obtain
■fiPiV1 + S2P2V'1 + • • • + JrPiV1 = 0. Subtraction of the two new
relations gives
S2P2(V'* ~ V'1) + S3P3(r," ~ V'1) + • • • = 0.
Since P2O7*2 — v'1) ^ 0> this is a non-trivial relation which is
shorter than C). Hence we are forced to conclude that r = 1,

26 FINITE DIMENSIONAL EXTENSION FIELDS
that is, J1P1 = 0. Since pir~1 exists, this gives jx = 0 contrary
to the assumption that jt is an isomorphism.
We can combine Theorem 1 and Dedekind's theorem to obtain
the following
Corollary. Let E and P be fields over § such that [E :$] = »<<».
Then there exist at most n distinct isomorphisms of E/$ into P/#.
Proof. Let su s2, • ■ ■, sr be distinct isomorphisms of E/$ into
P/f>. Then these are elements of 8*(E, P) which are right P-
independent. Since [8»(E, P): P]r = n, we must have r < n.
In the next section we shall be concerned with right P-vector
spaces spanned by a finite number of automorphisms of a field.
More generally, let Ji, s2, ■ ■ ■, sn be distinct isomorphisms of E
into P and let 21 be the set of endomorphisms of the form
D) JiPi + s2p2 H h snpn, pi e P.
Evidently, 21 is a subspace of the right P-vector space 8(E, P).
Moreover, if e e E, then eRSi = J,-(«*')«> by B), so
n \ n n
l / l l
This shows that 21 is closed under left multiplication by arbitrary
€r, e e E. We shall require the following
Theorem 4. Let E and P be fields, sly s2, • • •, J» isomorphisms of
E into P, and let 21 be the right ~P-subspace of 8(E, P) of endomor-
endomorphisms 'SiSiPi, pi e P. Let S3 be a V-subspace of 21 which is invariant
under left multiplication by elements €r, e e E. Then $5 — j^P +
shY + h j,vP ( = j E SijpJj where [stiy si2, ■■-, sir} = » fl
f r I
Proof. It is clear that I Yl *iiPij\pij e P| c 58. To prove the
n
opposite inclusion it suffices to show that, if X stP< e ®j then the
i
Ji for which pt- ^ 0 are contained in S3. Suppose this is not the
case. Then we have an element sklpkl + J*jP*2 "I 1" J*.P*. m

FINITE DIMENSIONAL EXTENSION FIELDS 27
23 in which every p*, 7^ 0 and Sk, i S3. We can then argue as in
the proof of Dedekind's theorem. We assume s minimal. If
s > 1, we apply the process we used before to obtain a shorter
element of the same type contained in S3. Then sj^Pki e S3 which
implies that s^ e S3 contrary to assumption.
EXERCISE
1. Let E = $@) where 6 is algebraic over $ and (/(*■)) is the kernel of the
homomorphism g(x) -* g(ff) (Vol. I, p. 103). Then [E:$] = deg/. Use the ex-
extension theorem V of Introduction to show that the number of isomorphisms of
E/$ into P/3> does not exceed deg/. Extend this result to obtain an alternative
proof of the Corollary to Theorem 3.
4. Finite groups of automorphisms. Let G be a group of auto-
automorphisms of a field P and let <£ be the subset of P of elements a
such that a" = a for every s eG. We shall call <3? the set of G-
invariants of P. Since the invariants (or fixed elements) of an
automorphism form a subfield, $ is a subfield of P. We denote
<J? = I(G) (or /p(G) if it is necessary to indicate P) and we call a
subfield which has this form, that is, which is the subfield of in-
invariants of a group of automorphisms, Galois in P. We shall
also say that P is Galois over f> or P/<£ is Galois.
The process we have just indicated associates with groups of
automorphisms G, subfields I(G), and we have the mapping
G —> I(G) of these groups into subfields of P. We now define
a mapping in the opposite direction. If $ is any subfield of P,
then we associate with $ the set A($>) (or yfp($)) consisting of the
automorphisms of P/*, that is, the automorphisms s of P such
that as = a for all a e $. Evidently, A(<£>) is a subgroup of the
group A of all the automorphisms of P. We call A(p) the Galois
group of P/«i>. We have the subfield-group mapping $ —> A(<£>).
The following properties of the mappings G —»■ I(G), $ —>
are clear from the definitions:
(a) G1^G2 =» /(GO C I(G2) (=* denotes "implies").
03) *i 2 $2 =
G) /(^(#)) 2 #.
(S) A(I(G)) 2 G.

28 FINITE DIMENSIONAL EXTENSION FIELDS
These relations have the following consequences:
(«) I{A{I{G))) = /(G).
The proofs of these two are identical so we consider («) only.
Here we use G) for * = I(G) and obtain I(A(I(G))) 2 /(G).
On the other hand, if we apply / to A(I(G)) 3 G, we obtain I(G) 3
I(A(I(G))). Hence (e) holds. A consequence of («) is that $ is
Galois in P if and only if <i> is the set of invariants of the Galois
group of P/$, that is, $ = I(A(<&)). Clearly this condition is
sufficient. On the other hand, if $ = /(G) for some group of auto-
automorphisms G, then $ = /(G) = I(A(I(G))) = I{A{3>)).
We shall now study the Galois correspondences $ —> A($),
G —> /(G) starting with finite groups of automorphisms. We
denote the order of a group G by (G:l) and, more generally, the
index of a subgroup // in G by (G:H). We shall deduce all the
results on the subfield-group correspondence from the Jacobson-
Bourbaki theorem (Th. 2) via the following
Lemma. Let G be a finite group of automorphisms in the field P
f n I
and let 31 = \^2 SiPi\Si eG,p,-e P|. Then 21 satisfies the hypoth-
l 1 J
eses (i), (ii), (iii) of Theorem 2, [2t:P]« = (G:l), and the subfield
$ given in Theorem 2 is the subfield of G-invariants. If S3 is a
subring of 21 and a subspace of 21 over P, then
where H = {/,} /j « subgroup of G.
Proof. If p e P and j is an automorphism, then B) shows that
Prs = s(p')R. Hence {siPi)(siP)) = s^piRS^pjR = SiSj(pi'')BpjR -
SiSjpi"'pj e 21 since jtjy e G. This implies that 21 is a subring of the
ring of endomorphisms 8(P, P). Since 1 e G and G C 21, 1 e 21.
It is clear that 21 is a subspace of 8(P, P) as right vector space
over P. Since the Si are independent over P by Dedekind's
theorem, [21: P]« = (G:l) < 00. The subfield * of Theorem 2 is
the set of a e P such that aRA = AaB for all A e 21. Since ocrPr =

FINITE DIMENSIONAL EXTENSION FIELDS29
p e P, anyhow, the condition is equivalent to aRSt =
Si e G. Since aRSi = Ji(a*0«> this is equivalent to Si(a")B = SiaB,
Si e G. Since j, exists, this becomes (a*0« = aR or a>< = «
which shows that <xrA = yfafl, A e SI is equivalent to: a is G-
invariant. Now let S3 be a subring of SI which is a P-subspace.
Then 93 3 IP = {pr\p e P} and consequently 93 is invariant
under left multiplication by the pr. Hence, by Theorem 4,
93 = >iP + h? H h A-P where # = {/,•} = G D 93. Evidently
// = G D 93 is closed under multiplication so this is a finite sub-
semigroup of G. Hence H is a subgroup of G.
The main result on finite groups of automorphisms of a field is
Theorem 5. Let F be a field and let s& be the collection of finite
groups of automorphisms in P, S the collection of subfields of P
which are Galois and of finite co-dimension in P. If $ e^ let
A{$) be the Galois group of P/$ and, ifGej< let I(G) be the sub-
field of P of G-invariants. Then: (i) If $ e J^ yf($) e ^ W */
G e J< 7(G) e ./ Moreover, I(A@>)) = * W A{I{G)) = G.
(ii) // G e J< rtt» (G: 1) = [P:/(G)]. (iii) If $ e S and E is a
subfield of P containing $, Afow Ee^ (iv) 7w /A*\r situation H =
yf(E), wA*VA « a subgroup of G = ^f($), /j invariant in G if and
only if E *j Galois over®. Then the Galois group Ae($) of E/$ is
isomorphic to G/H.
Proof, (i)-(ii). If G e */ and 31 = {2j,-p,|j< e G, Pi e P}, then
[P:/(G)] = [«:P]B = (G:l), by the lemma and Theorem 2. If
we set * = I(G) and G' = ^($) the Galois group of P/$, then
the corollary to Dedekind's theorem shows that (G': 1) < [P:$] =
(G:l). Since G C G' is evident, G' = G. Thus ^(/(G)) = G.
Next let $ be Galois and of finite co-dimension in P. Then $ =
7(G) where G is the Galois group of P/$. This is finite by the
corollary to Dedekind's theorem. Hence A($) e s& and I{A($))
= <£. This completes the proof of (i) and (ii). (iii) Let * e ^
and let SI be the ring of endomorphisms defined by the Galois
group G of P/<S>. By Theorem 2, SI = g*(P, P). Now let E be a
subfield of P containing $. Then 93 = 8E(P, P) is a subring of SI
of the sort considered in the lemma. Hence 93 = /iP + • • • + trV
where H = {/,•} is a subgroup of G. Since E = \e\eRB = BeR,
B e 93}, it follows that E is the subfield of //-invariants. This

30 FINITE DIMENSIONAL EXTENSION FIELDS
proves (iii). (iv) If s e G, E* the image of E under s is another
subfield of P containing <£ and it follows directly from the defini-
definition that yf(E") = s~1Hs. Hence H is invariant in G if and only
if E* = E for every ;eG. We proceed to show that this holds
if and only if E is Galois over 4> and then ^e($) = G/H. Assume
first that E* = E and let G' be the group of restrictions s' to E
of the s e G. Then G' is a finite group of automorphisms in E
and J(G') = *. Hence $ is Galois in E and G' = JE($) by (i)
applied to E. The mapping s —> s' is a homomorphism of G onto
G'. The kernel is the set of s e G such that s' = 1 on E. This
is H. Hence G' ^ G/H. Next let E be Galois over $>. Then we
have [E:4>] distinct automorphisms of E over $ and these can be
considered as isomorphisms of E/<£ into P/3>. On the other hand,
by the corollary to Dedekind's theorem there are at most [E:<J?]
isomorphisms of E/<i> into P/<l> so these must coincide with the
automorphisms of E/$. If s e G, the restriction of s to E is an
isomorphism of E/$ into P/<£; hence this is an automorphism.
This implies that Es = E for all s e G.
Theorem 5 establishes, in particular, a bijection A-1, onto
mapping) between the collection of subfields E of P which con-
contain a fixed subfield $, which is Galois and of finite co-dimension
in P, and the collection of subgroups H of the Galois group G of
P/$. This correspondence satisfies the properties in (iii) and
(iv). We remark also that \H] is finite, which implies that the
collection of fields beween P and $ is finite. At this point there
is one serious gap in our theory: We have given no conditions
that P be finite dimensional Galois over <£. The next three sec-
sections will be devoted to filling this gap and to forging the link
between the present "abstract" Galois theory and the theory of
equations.
EXERCISES
1. Let C be the field of complex numbers and let P = C(£), a simple transcend-
transcendental extension of C (Vol. I, p. 101). Let s be the automorphism of P/C such
that $* = e£ where « is a primitive M-th root of 1 and let t be the automorphism of
P/C such that £' = £-1. Show that sn = 1, t2 = 1, st = ts~x and that the group
G of automorphisms generated by s, t is of order In. Show that the subfield of
G-invariants is C{i\), i\ = ijn + £~".
2. Determine the Galois group of $(p) over $ where $ is the field of rational
numbers and p4 = 2.

FINITE DIMENSIONAL EXTENSION FIELDS 31
3. Let P be finite dimensional Galois over $ and suppose the Galois group G =
Gi X G2, Gi subgroups of G. Show that, if P< is the subfield corresponding to
d, then Pt/$ is Galois and P = Pi ® P2 (over <£).
4. Let $ be a field of characteristic 5^ 2 and let P be an extension such that
[P:<i>] = 2. Show that P = <£>@) where 9! = ae$. Use this to prove that P is
Galois over f>.
5. Show that, if $1 and $2 are Galoisin P, then $1 D $2 is Galois in P. Let <5
be a field of characteristic 0, P = $(£), £ transcendental. Let $1 = $(i;2),
$2 = $(£(£ + 1)). Show that [P:$i] = 2 = [P:$2] but [P:$! fl $2] is infinite.
6. Show that, if /?o is the field of rational numbers, then R0(\/2) is not Galois
over Rq.
7. Let G be an arbitrary group of automorphisms in a field P and let 21
= {S Sipi I Si e G, p<eP). Show that 21 satisfies the hypotheses (i) and (ii) of
Theorem 2. Show that $ = {a.\ctRA = Aolr, A e2l} is the subfield of G-in-
variants. Use these results and ex. 1, § 2, to prove that, if E is a subfield of P
containing <£> such that [E:<t>] < °o, then any isomorphism of E/<i> into P/$ can
be extended to an automorphism of P/$.
8. (Kaplansky). P, <£, E, and G as in ex. 7. Prove that E is Galois in P. (In
other words, if $ is Galois in P and E 2 $ satisfies [E:<J>] < °o, then E is Galois
in P.) (Hint: Set H = G f\ A(E). Let A be a finite dimensional subspace of
P/$ containing E. Use ex. 1, § 2, to show that §*(A, P) has a right P-basis of
the form (/1, J2, • • •, /„), /< the restriction to A of jt- e G. Use Theorem 4 to show
that 8E(A, P) has a P-basis (h, h, ■ • ■, 7r), U e H. Use this and ex. 2, § 2, to
prove that E = /(#).)
5. Splitting field of a polynomial. Let $ be a given field and
f(x) a non-zero polynomial contained in the polynomial ring
$M, x an indeterminate. We recall that an element p of $ is called
a root of f(x) or of the equation/(#) = 0 if/(p) = 0. We know
that this is the case if and only if/(#) = (x — p)g(x) in $[x] (Vol.
I, p. 99); and if deg f{x) = «, then/(^) has at most n roots in $
(Vol. I, p. 104). If p1} p2, • • ■, pr are distinct roots, then
f{x) = ix - pi)(# - p2) •••(# — Pr)£(#).
In Vol. I, pp. 101-102, we have given a construction for an ex-
extension P/$ in which a given irreducible polynomial/^) e $[x] has
a root. If we apply this to an irreducible factor of any non-zero
f(x) e $[x], we obtain an extension P/<£> containing a root of fix).
We shall now establish the existence of a minimal field extension P
in which a given polynomial/^), deg/(#) > 0, decomposes as a
product of linear factors. Unless otherwise indicated we shall
assume our polynomials have leading coefficients 1. Then we re-
require an extension P/«i> such that
E) /(*) = (* - Pl)(* - P2) • • • (* - p.)

32 FINITE DIMENSIONAL EXTENSION FIELDS
in P[#]. If<i>(pi, p2, • • •, pn) denotes the subfield of P/$ generated
by the p,-, then evidently the factorization E) is valid also in
$(pi> P2, •••>Pn)M« Hence, if P/$ is to be minimal, then we
must have P = $(pi, P2, • • -,pn)- We recall also that the fac-
factorization E) is unique in F[x] apart from factors in <i> (Vol. I,
p. 100, p. 123). From this it follows that the set {p,} is the com-
complete set of roots of fix) in P and that, if 2/<£> is a subfield of $(p1}
■ • •» Pn)/® such that/(x) is a product of linear factors in 2[#],
then 2 = #(pi, P2, • • •, Pn). This leads us to give the following
Definition 1. Let <i> be a field and fix) a polynomial of positive
degree with coefficients in <i> [leading coefficient 1). Then an extension
field P/f> is called a splitting field of fix) if a factorization E) holds
in P[x] and P = $(pi, p2, • • •, pn).
We shall now state two immediate results which will be used
frequently.
Lemma 1. A) If P/<I> is a splitting field of fix) e*[*] and?/<S>
is a subfield of P/<i>, then P/2 is a splitting field of fix). B) If P/2
/j a splitting field for fix) e$[x] and 2 = $(ffi, • • •, <rr) where fi<jj)
= 0, then P/$ m a splitting field of fix).
Proof. A) This is an immediate consequence of the definition.
B) By assumption we have P = 2(pi, ■ • -,pn) where E) holds
in P[x]. Also 2 = *((rx, • ■ •, <rr) and/(<ry) = 0. It follows that
every <jj is one of the p,-; hence P = $(pi, P2, • • •, Pn)-
We can now prove the following existence theorem.
Theorem 6. Any polynomial fix) e $[x] of positive degree has a
splitting field
Proof. Let/(*) = fi(x)f2(x) •••/*(*) be the factorization of
/(*) into irreducible factors (with leading coefficients 1). Evi-
Evidently k < n = deg/(#). We use induction on n — k. If n — k
= 0, the fiix) are all of degree 1 and this means that $ itself is a
splitting field. Now assume n — k > 0 so that some/,-, say f\{x),
is of degree > 1. Then there exists an extension field E/f> such
that E = <Kp) and /j(p) = 0. Then /x(#) = (* - p)fx*{x) in
E[*] and sof(x) is a product of / > k irreducible factors in E[#].

FINITE DIMENSIONAL EXTENSION FIELDS 33
Then n — I < n — k and the induction hypothesis permits us to
conclude that there exists a splitting field P/E for/(x). Since E
= $(p), the lemma shows that P/$ is a splitting field for/(#).
We consider some examples of splitting fields.
A) /(*) = *2 + ax + /3- If/ is reducible in $[*], then * is a splitting field.
Otherwise, we let P = $[#]/(/(#)), which is a field since/ is irreducible. If we set
pi = x + (/(#)) where, as usual, (/(#)) denotes the principal ideal generated by
this polynomial, then/(pi) = 0 in P so/(#) = (x — pi)(x — P2) in P[x]. Thus
P = $[pi] = $(pi) is a splitting field. Since /(#) is the minimum polynomial
of pi, [P:$] = 2, by VIII of the Introduction.
B) Letp be a prime and let/(*) = *" - 1 = (* - l)^ + x"'* -\ h 1),
$ = Ro the field of rational numbers. Then P is a splitting field of/(*) if and
only if it is a splitting field of g(x) = xp~l + tf" -\ hi. It is known that
g(x) is irreducible (Vol. I, ex. 2, p. 127), so P = Ro[x]/(g(x)) is a field over Ro and
P = R0[p] = Roip), p = x + (g(x)). We have p" = 1 and p 7^ 1, which implies
that p is of order p in the multiplicative group P* of P. Hence, l,p, p2, • • ■,pp~1
are distinct and all of these are roots of xv — 1 =0. It follows that xv — 1 =
p-1
II (# — P*) and so P = Ro(p) is a splitting field.
*=C) /(*) = (*2 - 2)U2 - 3), $ = /?0. We first form E = R0(p) where p2 = 2.
We know that x2 - 2 is irreducible in R0[x] (Euclid). In E we have x2 - 2 =
(x — p)(x + p). However, x2 — 3 is irreducible in E[*]. Otherwise, there exists
V e E such that t?2 = 3. But y = a + /3p, a, /S rational and 172 = (a2 + 2/S2) +
2a|8p; so, if this = 3, then afi = 0 and a2 + 2/32 = 3. If j8 = 0, we must have
a2 = 3 and, if a = 0, we have /32 = 3/2. Both of these are impossible for ra-
rational numbers. We now form P = Efo) where j/2 = 3. We have P = /?o(p, v)
and in P[*], (*2 - 2)(*2 - 3) = (* - p)(* + p)(x - y)(x + y), so P/Ro is a
splitting field. Using VI and VIII of the Introduction, one sees that [P: $] = 4.
Before continuing with our discussion of splitting fields it will
be well to fix some notations on field extensions and algebra exten-
extensions of a field $ which to some extent have already been used.
If S is a subset of a field P/$, then we let $[S] and $(<S") respec-
respectively denote the subalgebra and the subfield over $ generated by
S. By definition, the first of these is the intersection of all sub-
algebras of P/f> containing S and the second is the intersection
of all subfields of P/$ containing S. It is clear that $[S] is the
subspace of P over $ spanned by 1 and all monomials <ri<r2 • ■ • am,
<Ti e S, and that *(<?) is the set of elements aP~l, a, /3 in $[S],
P t* 0. It follows directly from the definition that, if S\ and S2
are subsets of P, then (*F11))F12) = *(Si U <S) where the first
of these is, of course, the subfield of P/*F"i) generated by .SV
If p is an algebraic element of P/<i>, then we know that [<i>[p] :#]

34 FINITE DIMENSIONAL EXTENSION FIELDS
= deg f{x) where f(x) is the minimum polynomial of p over $
(Introd. VIII). In fact, we have seen that, if deg f{x) = », then
A, p, p2, • • •, pn~1) is a basis for P/<i>. Since the dimensionality
[<£>[p]:<l>] is finite, we know that <3>[p] is a field (Introd. VII).
Hence it is clear that $(p) = f>[p]. We shall now generalize these
results in proving the following key lemma on successive algebraic
extensions.
Lemma 2. Let P = *(pi, P2, • • •, pm) and assume that p»- is alge-
algebraic over <£(pi, p%, • • •, Pt-i)> /= 1, 2, •••,*». Then [P:$] < <»
and P = #[pi, p2, •• -,pm].
(We shall see later (p. 254) that, conversely, if P = $[pi, P2,
• • •, pn] is a field then the p»- are algebraic over $.)
Proof. We have seen that this holds for m = 1. Suppose m
> 1 and assume the result holds for r < m. Then $(pi, • • •, pr)
= *[pi, • • •, pr] and this is finite dimensional over f>. Since pr+i
is algebraic over *(pl5 ••■,pr), we have $(p!, • • •, pr)(pr+i) =
• ■ *j Pr)[pr+i] and the dimensionality of this extension over
• • •> Pr) is finite. It follows that
F) [*(p!, • • •, pr+i) :#] = [*(pi, • • •, Pr) (p
is finite. Also ^(pi, • • •, pr+i) = $(pi, • • -,Pr)[Pr+i] and
• • "j Pr) = *[pij ■ ■ ■, Pr] imply that every element of #(p1} • • •,
Pr+i) is a polynomial in the p,-, for 1 < / < r + 1. Hence
*(pi> • • •jPr+i) = *[pi, • • -,Pr+i]. The lemma now follows by
induction.
This result is applicable in particular if the p< are all algebraic
over *. Since the roots p,- off(x) are algebraic over #, it is ap-
apparent that if P/# is a splitting field off(x) e$[x], then [P:<i>] < oo.
We shall now show that any two splitting fields of a polynomial
are isomorphic over $. In order to carry out an inductive argu-
argument (and for other reasons, too) it is useful to generalize the
result as follows. Let $ and $ be fields which are isomorphic and
let a —* a be an isomorphism of <i> onto $. We know that this
can be extended to a unique isomorphism f(x) —> J{x) of &[x]

FINITE DIMENSIONAL EXTENSION FIELDS 35
onto $[x] so that x —> x (Introd. II). We wish to consider a
splitting field over $ of a polynomial f(x) and a splitting field over
$ of the corresponding polynomial fix) in $[x]. The following
theorem will imply uniqueness of the splitting field and gives an
important result on the number of isomorphisms of a splitting
field.
Theorem 7. Let a —> a be an isomorphism of a field <£ onto
the field <5 and letf(x) be a polynomial of positive degree with leading
coefficient l,f(x) in $>[#], and let f(x) be the corresponding polynomial
in $[#]. Let P and P be splitting fields over $ and $ off{x) and fix)
respectively. Then there exists an isomorphism of P onto P which
coincides with the given isomorphism on <£. Moreover, if fix) is a
product of distinct linear factors in P[x], then the number of extensions
of the given isomorphism on <£ to an isomorphism of P into P is
[P:*].
Proof. Both assertions will be proved by induction on [P:3>].
If [P:f>] = 1, P = $ and /(*) = n(# - pi) in $[*]. Applying
the isomorphism hix) —> Ti(x) off>[*] we obtain fix) = II(,v — pt)
and the p.- e*. It follows that these are the roots of fix) = 0 in
P so P = $ and both results hold in this case. Now assume
[P:<£] > 1. Then fix) is not a product of linear factors in $[#]
and so it has an irreducible factor gix) of degree r > 1. Then gix)
r
is a factor of fix). Also we may assume that^(^) = II (# — P*)>
i
I(*) = II (* - *<) where/(tf) = II (* - Pi) and/(*) = II (* - 9<).
i 11
Since gix) is irreducible, this is the minimum polynomial over
$ of pi, and E = $(pi) is r-dimensional over $ (r = degg > 1).
The extension theorem V of the Introduction implies that the
isomorphism a —> a can be extended to a unique isomorphism
of E = *[px] = *(pj) into P so that px —> ai} i = 1, • • -,r.
We observe next that the indicated isomorphisms of E into P
are the only extensions of a —> a. Again by the extension
theorem V, in any isomorphism of E extending the given iso-
isomorphism, pi is mapped into an element a such that giff) = 0.
r
Since gix) = II (# — *»)> it follows that a = <t,- for some i,

36 FINITE DIMENSIONAL EXTENSION FIELDS
1 < i < r. Then the isomorphism coincides with one of those
we indicated. Thus we see that a —> a can be extended to an
isomorphism of E = #(pi) into P and the number of such exten-
extensions is the number of distinct elements in {du • • -,dr}. In
particular, if /(#) is a product of distinct linear factors, then this
is true also forg(x) and the number is then r = degf(x) = [E:<i>].
We now replace the base field # by E and let E be its image under
one of the chosen extensions of the isomorphism on <£ to an iso-
isomorphism of E into P. We denote this extension by e —> «. Then
P/E is a splitting field of /(*) and P/E is a splitting field of
/(#). Moreover, [P:E] < [P:$] since [E:S>] = r > 1. Hence the
induction hypothesis shows that « —» l can be extended to an
isomorphism of P onto P and the number of such extensions is
[P: E] \ij{x) is a product of distinct linear factors in P[#]. If we
take into account the first result on the extension of a —> a to
e —>«, we see that there exists an extension of the isomorphism
a —> a to an isomorphism of P onto P and, if /(#) splits into
distinct linear factors, then we obtain [P:E][E:$] = [P:$] dis-
distinct isomorphisms since we have [E:$] extensions to E and each
of these has [P:E] extensions to P. Thus we obtain [P:#] =
[P: E][E:$] distinct extensions. It is clear that we have accounted
for every extension in our enumeration (cf. also Cor. to Th. 3)
and so the proof is complete.
We now specialize the result we have just proved by taking
$ = # and the identity mapping a —> a in*. Then the conclusion
is that, if P/$ and P/# are two splitting fields of the same poly-
polynomial/^), then P/<i> and P/<£ are isomorphic. Moreover, the
second part of the result is that, iff{x) has distinct roots, then
the number of automorphisms of P/# is [P:$]. In other words,
(G:l) = [P:$] for G the Galois group of P/$.
EXERCISES
1. Construct a splitting field over the rationals for x6 — 2. Find the dimen-
dimensionality.
2. Let P/$ be a splitting field of /(#) ^ 0 in *[*] and let E be a subfield of
P/$. Show that any isomorphism of E/$ into P/3> can be extended to an
automorphism of P.
3. Show that the dimensionality of a splitting field P/$ of a polynomial/(*)
of degree n cannot exceed n!.

FINITE DIMENSIONAL EXTENSION FIELDS 37
6. Multiple roots. Separable polynomials. Let/(#) be a poly-
polynomial of positive degree in $[x] and let P/# be a splitting field.
We now write
G) /(*) = (X - Pl)H* ~ P2)*1 •••(*- Pr)\
Pi e P, pi 9* pj if *' t1* j, and we say that p< is a root of multiplicity
ki off(x) = 0. If kt =1, then p,- is called a simple root; otherwise
pi is a multiple root. If we have a second splitting field P over $,
r
then /(#) = II (x — p~i)hi in P where pj —> p,- in an isomorphism
1
of P/<£ onto P/$. It is clear that the existence of multiple roots
for / is independent of the particular choice of a splitting field.
We shall now carry over a classical criterion for multiple roots
which can be tested in <£>[#] itself. For this we need the standard
formal derivative (or derivation) in $[x]. Thus we define a linear
mapping / —» /' in $[x] by specifying that (#*)' = ixi~1, i = 0,
1, 2, • • •, *° = 1. Since A, x, x2, • ■ •) is a basis for $[x] over #,
this defines a unique linear mapping / —> /' in #[#] over <l>. We
call /' the (formal) derivative of/and we note the basic rule:
(8) (fgY-fg + tf.
Because of the linearity of the derivative, it suffices to check
this for/ = xi, g = Xs in the basis (#*) for $[x]. Then jg = xi+i
so that (Jg)f = (* +j)xi-"~l, fg = ixi+'-\ fg1 =jxi+i~\ so
(8) is valid. We can now prove
Theorem 8. Iff{x) e $[x] and deg / > 0, then all the roots of /
(in its splitting field) are simple if and only if (/, /') = 1 (that is,
1 is the highest common factor of f and /').
Proof. Let d(x) be the highest common factor (/, /') of / and
/' in *[*] (cf. Vol. I, p. 100, p. 122). Suppose f(x) has a multiple
root in P[#], so f(x) = (x — p)kg(x), k > 1. If we take deriva-
derivatives in P[#], this gives f' = (x — p)kg' + k(x — p)k~1g which
is divisible by x — p since k — 1 > 1. Thus (x — p) \f (i.e. x — p
is a factor of /) and (x — p)\f, so (x — p)\d. Hence d(x) ?± 1.
Next, suppose all the roots of / are simple. Then we have f(x) =
n
H (x — pi), pi t£ pj, i j£ j. The usual extension of (8) to several
i

38 FINITE DIMENSIONAL EXTENSION FIELDS
factors gives
n
/'(*) = 22 (* - Pl) • • ■ (* - Pi-l)(.X - Pj+l) ••• (X - Pn).
It is clear from this that (x — p,) )( f'(x) and this implies that
(/,/') = 1-
Iff is irreducible in *[*•], then (/,/') ^ 1 implies that f\f.
By degree considerations this can happen only if /' = 0. If the
characteristic is 0, this evidently implies that/ is an element of
*. If the characteristic is p ^ 0 and f(x) = aoxn + aixn~1 +
a2xn~2 A h «„, then /'(*) = Ka^" + (n - l)aixn~2 +
(» — 2)a2x"~3 H , so /'(#) = 0 implies that (« — /)«»• = 0.
This implies that a,- = 0 if the integer n — i is not divisible by ^>.
Hence we see that /(*) = 0oxmp + PiX<m~1)p +---+0m = g(xp)
where g(x) = $oxm + pxxm~l -\ \- /3m. This condition is also
clearly sufficient that /' = 0 since (#**)' = kpx1'1'-1 = 0. In the
characteristic p ^ 0 case we shall see that the conditions: / ir-
irreducible of positive degree, /' = 0, can be fulfilled. This is a
basic difference between fields of characteristic 0 and those of
characteristic p tA 0 and this is the root of a host of complications
in the latter case.
Let us now look more closely at fields of characteristic p t* 0.
We recall that, if* is of this type, then we have
(9) (« + PY = a? + Pp, {a®* = apPp
in * (Vol. I, ex. 3, p. 120). The second of these is clear and the
first is a consequence of the binomial theorem and the fact that
the binomial coefficient I . J = p\/i\{p — i)\ is divisible by p for
1 < i < P — 1 since this is an integer and p occurs in the nu-
numerator of the fraction but not in the denominator. We note
also that, if a? = CP, then (a - 0)p = ap - (8P = 0 so a = /?.
Thus we see from this and (9) that the mapping a —* <xp is an
isomorphism of* into itself. The image *" = {a^lae*} is a
subfield, the subfield of p-th. powers. We can iterate the mapping
a —> ap and obtain the isomorphism a —» ap'} e — 1, 2, • • • of
* onto the subfield *pe of p'-th powers.

FINITE DIMENSIONAL EXTENSION FIELDS 39
We prove next the following general result which will be useful
later on.
Lemma. If<i> is a field of characteristic p ^ 0, then xp — a is
irreducible in $[x] unless a = fip, fi e $, in which case xp — a —
(x - /3)p.
Proof. Let P be a splitting field for xp — a. If jC is a root of
xv - a = 0, then a = /3P. Hence xp - a = xp - /3P = (x - 0)p
in P[x]. Now suppose xp — a = g(x)h(x) in $[#] where deg^ =
k and 1 < k < p — 1. Then in P[#] we must have g = (x — /?)*
= xk — k^xk~1 +• • •. This implies that ^e$; hence Ce$.
Then xp - a = (x - /3)p holds in $[*].
We now consider the following example. We let Ip = I/{p)
the field of residues modulo p, and we let $ = /p(£), £ transcen-
transcendental. Then we claim that ^$p. Now, if y e«i>, we can write
7 = a(f)/3(^) ~x where a(£) and /3(£) are polynomials. Then 7P =
"(F^CP), since a(f) = a0 + «iH implies a(£)p = aop +
aipP H = «o + «iP H (by Fermat's theorem). Hence
7P = f implies that a(£p) = /3(£p)£ and this is impossible since 1,
£, • • • are /p-independent. Thus we see that £ ^ <i>p and hence, by
the lemma, xp — f is irreducible in $[x]. On the other hand, we
have seen that xp — £ has /> equal roots in its splitting field. We
note also that (xp — £)' = 0.
We shall now call a polynomial / (of positive degree) separable
if it is a product of irreducible polynomials in $[x] all of which
have only simple roots in a splitting field. Our discussion shows
that, if $ is of characteristic 0, then every f(x) e $[x] is separable,
whereas for characteristic p ^ 0 there exist inseparable poly-
polynomials.
EXERCISES
1. Prove the following extension of the lemma: xp' — a is irreducible unless
2. Let f>o be a finite field of q elements and let P = $o(£), £ transcendental over
$<>• Let G be the finite group of the automorphisms of P over <£>o such that
HHa,<«e*o- Show that <i> = I{G) = $0(£9 - £).
3. Let * be a field of characteristic p ^ 0 and let f i, £2, • • •, £n be indeter-
minates over <i>, P = $(£i, £2, ■ • •, £„) the field of fractions of €>[£i, £2, • • •, £n]-
Show that [P:$(£ip, £2", • • •, £„")] = pn. Show also that the Galois group of P
over $(£1", £2", • • •, £n") is the identity.

40 FINITE DIMENSIONAL EXTENSION FIELDS
4. Let P = <£(£i, £2, • • •, £n) of characteristic p j* 0 and suppose that £<p'* e $
for < = 1,2, • • •, n, ti a positive integer. Show that the Galois group of P over $
is the identity.
5. Let $ be a field of characteristic p 9* 0. A polynomial with coefficients in $
is called a p-polynomial if it has the form xpm + aixJ>m~1 + a2X1*n~* -\ 1- <W.
Show that a polynomial (with leading coefficient 1) is a ^-polynomial if and only
if its roots form a subgroup of the additive group of the splitting field and all the
roots have the same multiplicity p'. Show that the roots of the displayed p-
polynomial are all simple if and only if am ^ 0.
6. Let/(#) be irreducible in <t>[x], $ of characteristic p y* 0. Show that/(#) can
be written in the form g(xp') where g(x) is irreducible and has distinct roots. Use
this to show that every root off{x) has the same multiplicity p' (in a splitting
field).
7. Let $ be a field of characteristic 0, /(*) a polynomial of positive degree con-
contained in $[*]. Show that if d{x) is the highest common factor off(x) and/'(*),
then g(x) = f(x)d(x)~1 has simple roots which are the distinct roots of/(#).
7. The "fundamental theorem" of Galois theory. We now take
up again the abstract Galois theory of § 4 and we shall answer
first the question which we raised at the end of § 4: that of charac-
characterizing finite dimensional Galois extensions. The result is the
following
Theorem 9. Afield P/4> is finite dimensional Galois over $ if and
only P is1 a splitting field over $ of a separable polynomial f(x) e
Proof. Let /(*) be separable and let f(x) = /i(tf)*1- • -/i(*)"
where the /,(#) are irreducible in $[#] and /,- ?■* fj if i 9* j. Then
fi(x) has only simple roots. Moreover, since /,• and fj for / ?* _/'
are distinct irreducible polynomials, their highest common factor
is 1. Hence 1 = a(x)fi(x) + b(x)fj{x) for a(x), b{x) in $[#], and
this implies that /,• and /,• have no common roots in any extension
field. It follows that g{x) = fi(x)f2(x) • • -fi(x) has no multiple
roots, and it is clear that, if P/«i> is a splitting field of /, then it
is a splitting field also for g. Now we know that any splitting
field is finite dimensional and Theorem 6 implies that, if G is the
Galois group of P/*, then (G:l) = [P:*]. Let $' = I(G) the
set of G-invariants. Then, by Theorem 5 (ii), (G:l) = [P:«*>'].
Since <i>' 2 $ we have $ = <$', which shows that f> is Galois in P
or P is Galois over <$. This proves the sufficiency of the condition.
Next assume P is finite dimensional Galois over $ and let G be
the Galois group. We know that G is finite and we indicate it as

FINITE DIMENSIONAL EXTENSION FIELDS 41
G = {su s2, ■ • •, sn}. If p e P, we shall call the images p'< under
Si e G the conjugates of p in P/<t>. We may assume that p'1, • • •,
p'T are distinct and that this set includes all the conjugates. Then
r
we assert that h(x) = H (x — p'f) e #[*]. To see this let s e G
and let s be its extension to P[x] such that x' = x. Then we have
r
h'(x) = II (# — p'1')- Since the elements p'1', • • -, p*>* are dis-
i
tinct conjugates, this set is the complete set of conjugates and so
h"(x) = h(x), seG. Hence h(x) e$[x]. This shows that the
minimum polynomial over $ of any p e P (this is actually h(x)) is
separable and splits as a product of linear factors in P[x]. Now
let (pi, P2, • • •, pn) be a basis for P/<£> and let /,•(#) be the minimum
polynomial of pt- over $. Then f(x) = Hfi(x) is separable and
clearly P is a splitting field over $ of /.
The main Galois correspondence (Th. 5) can now be applied to
state the following result that is known classically as the
Fundamental Theorem of Galois Theory. Let P be a splitting
field over $ of a separable polynomial and let G be the Galois group of
P/$>. With each subgroup H of G we associate the subfield E of P
over * of H-invariants and with each subfield E over $ we associate
the subgroup H of G of elements t such that e' = e for all « in E.
Then these two correspondences are inverses and are bijections of the
set of subgroups of G and the set of subfields of P over #. The corre-
correspondences are order inverting relative to inclusion and
A0) (#:1) = [P:E], (G:H) = [E:*].
Moreover, H is invariant in G if and only if the corresponding field
E is Galois over * and in this case the Galois group of E/$ is iso-
morphic to the factor group G/H.
All of this can be read off directly from Theorem 5 and the
remarks which follow it. The only part which has not been made
explicit before is A0). Now it is clear from the definition that H
is the Galois group of P/E. Hence G7:1) = [P: E]. Also (G:l)
= [P:*],so(G:#) = (G:1)/(H:1) = [P:*]/[P:E] = [E:#], We
note also that (G:H) — [E:<£] is the number of distinct isomor-

42 FINITE DIMENSIONAL EXTENSION FIELDS
phisms of E/<£> into P/*. To see this we consider the restrictions
s to E of the elements s e G. If s = 1 for s, t eG, then j/-1 = 1
which means that st~* e H. Then the cosets Hs and Ht are
identical. The converse follows by retracing the steps. Hence
we see the collection {j|jeG} contains (G:H) distinct isomor-
isomorphisms of E/<S> into P/<3?. We know also that there are no more
than [E:«i>] = (G:H) isomorphisms of E/<E> into P/$ (Cor. to
Th. 3). Hence we have caught them all. Incidentally, we have
shown also that every isomorphism of E/<i> into P/<3? is a restric-
restriction of an automorphism of P/<£>. In other words, any such iso-
isomorphism can be extended to an automorphism (cf. ex. 7, §4).
8. Normal extensions. Normal closures. At the beginning of
the last section we gave an abstract characterization of splitting
fields of separable polynomials: these are just the finite dimen-
dimensional Galois extensions. We shall now give two abstract charac-
characterizations of arbitrary splitting fields.
Theorem 10. The following three conditions on a finite dimen-
dimensional extension P/$ are equivalent:
A) P/<i> is a splitting field of a polynomial f(x) e$[x].
B) Any isomorphism s of P/<I> into an extension field A/* is an
automorphism.
C) Every irreducible polynomial g{x) e $[x] which has a root in
P is a product of linear factors in P[#].
Proof. A) => B) ("=»" means "implies"): Let P = *(pi,p2,
• • -,Pn) where f{x) = Il(x — pf) in P[.v] and f(x) e$[x]. Sup-
Suppose A 2 P 3 * and let s be an isomorphism of P/$ into A/*.
Since /(p<) = 0, we have /(p,-*) = 0 and, since {p,-} is the com-
complete set of roots of f(x) in A, p/ is one of the p,-. Hence s maps
every generator pt- of P = <£>(pi, p2, • • •, p«) into P. Hence P" c
P. Since s is 1-1 ^-linear and [P:<i>] < », we have P3 = P and
s is an automorphism. B) => C): Assume every isomorphism
of P/$ into any extension field A/$ is an automorphism. Let
g(x) be irreducible in $[#] and have a root <r in P. Write P =
P2, • ■ • j Pm) and let /,•(*) be the minimum polynomial of
m
Pi over <t>. Set f{x) = g{x) TJ /i(«) and let A/P be a splitting

FINITE DIMENSIONAL EXTENSION FIELDS 43
field over P off(x). Since p,- e P and f(x) e <£>[#], A is also a split-
splitting field over $ of f(x) and it contains a splitting field over $ of
g{x). Hence it will follow that g(x) is a product of linear factors
in P[#] if we can show that every root a' of g(x) contained in A
is contained in P. To prove this we note that, since g(x) is ir-
irreducible in$M a.ndg(a) = 0 = g(<r'), there exists an isomorphism
s of <J?(<r)/$ into <£>(</)/* such that a' = a'. We now observe that
we can consider A as a splitting field over <£(o-) and over $(a') of
/(*) = /"(•*■)• Hence the main isomorphism theorem for split-
splitting fields (Th. 6) shows that s can be extended to an automor-
automorphism s of A. Since a' = a, a e <£, s is a ^-automorphism of A.
Its restriction to P is an isomorphism of P/f> into A/$. Hence,
by hypothesis, this restriction is an automorphism of P/$.
Since <r e P, it follows that <r' = a' e P. This proves C). C) =>
A). Write P = *(pi, P2> • ■ -,Ptr) and let/,(x) be the minimum
polynomial of p,- over $. If we assume C), then/,•(*) is a product
of linear factors in P[x]. Hence P/$ is a splitting field of/(x) =
n/,-(,v). This completes the proof.
A finite dimensional extension P/$ satisfying any one (hence
all) of the conditions of Theorem 10 is called a normal extension. It
is clear from the condition A) that, if P is normal over $ and E is
a subfield of P/<J>, then P is normal over E. On the other hand,
if P D E 2 f, then it may well happen that P/ E and E/$ are
normal and P/<3? is not (ex. 1 below). Let P = *(<ri, • • •, <rm) be
an arbitrary finite dimensional extension of # and let f{x) =
n/»(.v) where /,•(*) is the minimum polynomial of a, over $. Let
A/P be a splitting field of fix) z$\x\. Then A/* is a splitting
field off(x); hence A/* is normal. Now let A'/$ be any normal
extension of P/#. Since A' contains at and /,•(#) is irreducible in
<£>, condition B) of Theorem 8 shows that A'/* contains a splitting
field of /(*•). Hence we have an isomorphism of A/* into A'/*.
This implies that no proper subfield of A containing P is normal
over #. We now define a normal closure of P/<£ as a normal
extension of $ containing P and having the property that no
proper subfield containing P is normal over <£. Then we can say
that A/* is a normal closure of P/<$ and the remark about A and
A' shows that such an extension is determined up to <i>-isomor-
phism by

44 FINITE DIMENSIONAL EXTENSION FIELDS
EXERCISE
1. Let P = RoiVT), Ro the rationals, and let E = R0(VT) C P. Show that
P/E and E//?o are normal but P/2?o is not normal.
9. Structure of algebraic extensions. Separability. The struc-
structure theory of fields will be taken up in detail in Chapter IV.
However, at this point it is convenient to derive the basic theo-
theorems on algebraic extensions and more generally on the set of
algebraic elements of any field P/<i>. We have shown in Vol. I
(p. 183) that, if P is a field over <£>, then the subset A of elements
of P which are algebraic over $ form a subfield over $ and every
element of P which is algebraic over A is contained in A. The
subfield A/* is called the algebraic closure of * in P and $ is called
algebraically closed in P if <i> = A. The field P/<£> is algebraic if
P = A, that is, every element of p is algebraic over <l>. Thus the
second part of the result we have quoted above is that, if A is the
algebraic closure of $ in P, then A is algebraically closed in P. We
shall now indicate another proof of these results which is based
on the Lemma 2 of § 5: If pt- is algebraic over $(pi, • • •, Pi—i) then
$(pi> P2j • • -,Pn)/* is finite dimensional. Now let A be the set
of dements of P which are algebraic over <i> and let p, a e A. Then
<J>(p, a) is finite dimensional. Since $(t)/$ is infinite dimensional
for transcendental t, it follows that every element of $(p, a) is
algebraic over $. In particular, p ± a, pa, and p~x are algebraic
if p 5* 0. Since p and a are arbitrary in P, this implies that A is
a subfield of P. Also it is clear that A3*. Now let p be an
element of P which is algebraic over A and let f{x) = xn +
a1xn~l + • • • + an be its minimum polynomial over A. Then the
a* e A and so are algebraic over $. Moreover, it is clear that p
is algebraic over $(«i, a2, • ■ •, <xn). It now follows thaft^ai, • • •,
an, p) is finite dimensional over $. Hence p is algebraic over *
and consequently A is algebraically closed in the field P. The
result we proved on the algebraic closure of A in P implies the
following transitivity property: if B/A is algebraic and A/<i> is
algebraic, then B/«i> is algebraic. To see this let r/f> be the sub-
subfield of B/$ of elements which are algebraic over $. Clearly T 3
A and we have seen that, if $ e B is algebraic over T, then it belongs
to P. On the other hand, if /3 is any element of B, then /3 is alge-

FINITE DIMENSIONAL EXTENSION FIELDS 45
braic over A, hence over F, so /3 e T. This shows that B = T is
algebraic over $.
There are several other useful remarks on algebraic elements
which are worth recording here for future reference. The first of
these, which is implicit in what we have proved before, is that,
if 2l/$> is a subalgebra of a field P/31, then every element of 21 is
algebraic over <t> if and only if every finite subset X of 21 is con-
contained in a finite dimensional subalgebra £/<$. This implies that
every £ e 21 is algebraic over $ since it implies that <£[£] is
finite dimensional. On the other hand, if every element of 21 is
algebraic and X — {£i, £2j • • •,£.•}, then the lemma we quoted
shows that #(£i, • • •, £r)/* is finite dimensional. We recall also
that $[£i, • • •, £r] = *(£i, • • •, i-r), which implies that, if every
element of 21 is algebraic, then 21 is a subfield of P/<i>. We note
also that, if E/f> is an algebraic subfield of P/$ and A/* is an
arbitrary subfield, then the subalgebra EA/$ generated by E and
A is a subfield which is algebraic over A. To see this we observe
that EA is the set of elements of the form 2«i5,-, e,- e E, St- e A.
Hence, if X is a finite subset of EA, then there exists a finite sub-
subset {e,} such that every element of X is a A-linear combination of
the «,-. Since E/$ is algebraic, we may imbed the set {e,} in a
finite dimensional subalgebra. If we express the «,- in terms of a
basis {77/} for this subalgebra, then we see that every element of X
has the form 25^-, 5y e A. Since r\p\k = ~2yjkim> "Ym e <£>, it is clear
that the set 2 Arij of A-linear combinations of the t]j is a subalgebra
of P/A. We have therefore proved that every finite subset of
EA is contained in a finite dimensional subalgebra over A. Hence
every element of EA is algebraic over A and EA is a subfield.
An algebraic element p e P/$ is called separable {algebraic) over
$ if its minimum polynomial over <E> is separable. It is clear that
p is separable over $ if and only if there exists a polynomial f{x)
e$[x] with distinct roots such that /(p) = 0. Also p is separable
if and only if there exists a polynomial f{x) e$[x] with (/, /') =
1 such that /(p) = 0. If <£' is an extension field of $, we shall
again have (/,/) = 1 in *'[*] (since (/,/) = af+bf, ex. 3, p.
122 of Vol. I). It follows that, if $'/* is a subfield of P/$ and
P e P is separable over <£, then p is separable over $'. We have
seen (§6) that every polynomial with coefficients in a field of

46 FINITE DIMENSIONAL EXTENSION FIELDS
characteristic 0 is separable. Consequently, the results we shall
now consider become trivial in the characteristic 0 case.
An extension P/<3? will be called separable {algebraic) if every
element p e P is separable over <3>. Let A/$ be algebraic and let
2 be the subset of A of elements which are separable over <!>. We
wish to show that 2 is a subfield containing $ and that every ele-
element of A which is separable over 2 is contained in 2. For this
we shall need the following
Lemma 1. Let PDED$ where E and $ are subfields of P and
E/$ is finite dimensional Galois. Then any element 6 e P which
is separable algebraic over E is separable algebraic over $.
Proof. Let g(x) be the minimum polynomial of 6 over E. If
s e G the Galois group of E/$, then s has a unique extension to
EM satisfying xs = x. Let gH(x), g"*(x), ■ ■ -,g'r(x) be the
r
distinct images of g(x) under seG and let f(x) = H £**(#).
i
Then fB(x) = f{x) for all s e G, which implies that f{x) e$[x].
Since g{x) is irreducible in E[#] and (g, g') = 1, the same is true
for every gH. Hence every gH(x) has distinct roots. We note also
that, if i ?£ j, then gH and g*> are relatively prime, since other-
otherwise (gs<, gs>) = gSi(x) = g*'(x) because these are irreducible in
EM. This contradicts gH 5* ga> for * 7^ j. Thus 1 = (g'^g'1)
and consequently these have no common roots in a splitting field
for/(.v). It is now clear that/(^) has distinct roots. Since /@) =
0 and / e $[#], we see that 9 is separable over $.
Clearly if p e E, then p is separable algebraic over E (with
minimum polynomial x — p). Hence Lemma 1 shows that p is
separable over 4>. In other words, we have the
Corollary. Any finite dimensional Galois extension is separable.
We can now prove the main result on separability.
Theorem 11. If A/* is algebraic, then the set 2 of elements of A
which are separable over $ is a subfield containing $. Moreover, 2
contains every element of A. which is separable algebraic over 2.
Proof. Let p, a e 2 and let g{x) and h(x) be the minimum poly-
polynomials over <i> of p and a respectively. Then f(x) = g(x)h(x) is
separable. If A is a splitting field over <£(p, <r) of f(x), then A is

FINITE DIMENSIONAL EXTENSION FIELDS47
also a splitting field over <i> of/(*) (the normal closure of <i>(p,
Hence A/<3? is Galois, so by the corollary above, every element of
A is separable over*. In particular, p ± a, p<r, p (if p 9± 0) and
every element of $ are separable over <i>. This proves that 2 is a
subfield containing $>. Now let 6 be an element of A which is
separable algebraic over 2 and let xn + p\xn~l -\— • + pn, p,- e 2,
be its minimum polynomial over 2. The subfield 4>(pi, P2, • • •,
pn; 6) is finite dimensional over $. Let A/$ be its normal closure.
Let /»•(#) be the minimum polynomial of p»- over <i>. Then A con-
n
tains a splitting field E/$ of /(#) = IJ /»(#) and this is Galois
i
over $ since f(x) is separable. Evidently E 3 $(pi, P23 • • •, P«)-
Also 0 is separable over $(pi, p2, • • -,pn) since xn + pixn-1 +
• • • + Pn is its minimum polynomial. Hence 6 is separable alge-
algebraic over E. Then 6 is separable algebraic over <3? by Lemma 1.
This proves the second statement.
If the only elements of an algebraic extension A/# which are
separable are the elements of $, then we say that A/$ is purely
inseparable. Similarly, an algebraic element p is purely insepa-
inseparable over $ if $(p)/$ is purely inseparable. It is clear from the
definitions that, if p is at the same time separable and purely in-
inseparable over "i>, then p e<l>. Also, it should be remarked that
an element can be inseparable (= not separable) without being
purely inseparable (cf. ex. 3 below). If A/$ is algebraic and 2/<l>
is the maximal separable subfield of A/# (that is, the subfield of
all the separable elements), then the second half of Theorem 11
states that A/2 is purely inseparable. This shows that every
algebraic extension A/* can be built up in two "pure" stages:
first, a separable extension 2/$> and next a purely inseparable
extension A/2. The second part of Theorem 11 and the argu-
argument we used before for algebraic extensions (p. 44) implies the
transitivity: If A/# is separable algebraic and B/A is separable
algebraic, then B/$ is separable algebraic. We are going to prove
a similar transitivity for purely inseparable extensions. Since
everything is trivial for characteristic 0, we shall assume in the
rest of this section that the characteristic is p t^ 0. We shall
need the following important criterion for separable and purely
inseparable elements.

48 FINITE DIMENSIONAL EXTENSION FIELDS
Lemma 2. Let $ be of characteristic p 5* 0. (i) Then an alge-
algebraic element p of an extension field is separable over <£ if and only if
<i>(p) = $(pp) = $(pp2) = • • ■. (ii) If p is purely inseparable, then
its minimum polynomial has the form x"' — a, a e<i>. On the other
hand, if p satisfies an equation of the form xv" = a e <!>, e > 0,
then p is purely inseparable over $>.
Proof. Let g(x) be the minimum polynomial of p over $. (i)
Suppose first that p is not separable. Then g(x) = h(xp) and pp
is a root of h(x). Hence [3>(pp) :<£] ^ deg h(x) < deg g(x) =
[$(p):f>]. Consequently, <&(pp) <z$(p). Next suppose p is sepa-
separable so that g(x) has distinct roots. Let h(x) be the minimum
polynomial of p over 3>(pp). Then h(x) \g(x), so h{x) has distinct
roots. Also p is a root of the polynomial xp — pp e$(pp)[x], so
h(x)\xp — pp = (x — p)p. Since h(x) has distinct roots, this im-
implies that h(x) = x — p. Hence p e $(pp) = *[pp] and p is a poly-
polynomial in pp with coefficients in <£>. Taking p-th powers shows
that pp is a polynomial in pp2 with coefficients in $. Hence p e
>). A repetition of the argument shows that <l>(p) = $(pp) =
) = • • •• This proves (i). (ii) Let p be purely inseparable
over $ and write g(x) = h{xp') where e is maximal for this. Then
h'(x) 7^ 0 since, otherwise, h(x) = k{xp) and j'(^) = k(xp°+1) con-
contrary to the choice of e. We have h(ppe) = 0, so pp* is a root of a
separable polynomial. Since p was assumed purely inseparable,
this implies that pp* = a e3> and p is a root of xp' — a. Since g(x)
— h(xp') is the minimum polynomial of p over <£, it is clear that
g(x) = xp' — a. Next assume that pp' = a e$ for some non-
negative integer e. Let a e $(p) = $[p] so that a = a0 + «iP +
• • • + amp"\ at e #. Then <rp> = aQp' + afff +■■■+ amp'(pp')m
e$. If <r is separable, then $@-) = $(<rp')> by (i). Hence $(<r) =
$ and uef. Thus p is purely inseparable.
The second part of this lemma shows that A/$ is purely insepa-
inseparable if and only if every element of A satisfies an equation of the
form xp' = a e ■*>. Since (xp')pf = xp'+/, this implies that if B/A
is purely inseparable and A/$ is purely inseparable, then B/<i>
is purely inseparable. Also it is clear from the second part of the
lemma that if A is purely inseparable over $, then it is purely
inseparable over any subfield E of P/<1>.

FINITE DIMENSIONAL EXTENSION FIELDS 49
EXERCISES
1. Let A/$ be algebraic. Verify that the set of elements y e A which are
purely inseparable over $ form a subfield containing $.
2. Let Ip be the field I/(p) and let P = /,,(£, rj) where £, t\ are indeterminates
(cf. ex. 3, § 6). Let * = /„(£*, vp - V - I). Show that [P:$] = pi and deter-
determine the maximal separable subfield of P/$.
3. (J. D. Reid). Let P be as in ex. 2 and let E = P(£) where J"a = ttp + £?.
Show that [E :P] = p2 and E/P is inseparable. Show that E/P contains no purely
inseparable element over P not contained in P.
4. Let P/$ be algebraic and let P(£i, £2, • • •, £r) be the field of fractions of
P[£i) fa, • • • > £r], & indeterminates. Let A be the set of elements in P(£<) having
the form Fg~\ where F e Pfo, ■ • •, gr], £ e $[£i, • ■ •, |r]. Prove that A is a sub-
subfield of P(£i, ••-,£,-)/$(£i, ---.fr) which is algebraic. Prove that A =
P(fi> ■' '> ?r). Hence prove that every non-zero polynomial with coefficients in
P has a non-zero multiple with coefficients in $.
10. Degrees of separability and inseparability. Structure of
normal extensions. We assume throughout this section that the
characteristic is p 5* 0 and we consider finite dimensional exten-
extensions. For such an extension P/<t> with maximal separable sub-
subfield S/# we consider the dimensionalities [2:f>] and [P:S], which
we call the separability degree and inseparability degree respectively
ofP/$. We write [S:$] = [P:$],, [P:2] = [P :#],-. Thenwehave
A1) [P:#] = [P:*].[P:*]<.
We shall now show that [P:$]» = pf> which amounts to saying
that the dimensionality of a purely inseparable extension is a
power of the characteristic. If P = $, this is clear since [P: $] =
1 = p°. Otherwise, let p e P, i $. Then Lemma 2 of § 9 shows
that the minimum polynomial of p over $ has degree pe, e > 0.
Then [#(p):*] = p" and [P:*(p)] < [P:$]. Since P is purely in-
inseparable over $(p), we may assume (using induction on the
dimensionality) that [P:$(p)] = pl. Then [P:<i>] = p"pl = pe+g.
We now consider successive finite dimensional extensions: A/P
is finite dimensional and P/<£ is finite dimensional; hence A/* is
finite dimensional. We have seen that, if P/* and A/P are sepa-
separable (purely inseparable), then A/$ is separable (purely insepa-
inseparable). If P/$ is separable and A/P is purely inseparable, then
one sees easily that P/€> is the maximal separable subfield of A/<i>.
Then [A:*], = [P:<£>] and [A:$]t- = [A:P]. We now consider the
interesting combination: P/<l> purely inseparable and A/P sepa-

50 FINITE DIMENSIONAL EXTENSION FIELDS
rable. We shall show that the maximal separable subfield of the
result A/<i> will have the same dimensionality as A/P. A con-
considerably sharper statement is the following
Lemma. Let A/P be separable, P/3> purely inseparable. Then
A/$ = P ®<t 2 where 2/<l> is the maximal separable subfield of
A/$. Moreover, [A:P] = [2:$].
Proof. It follows easily from XII, Introduction that the state-
statement A = P ®* 2 is equivalent to: there exists a 4>-basis for
2 which is at the same time a P-basis for A. This implies that
[2:$] = [A:P]. We proceed to determine the required type of
basis. First, let (8x, 82, • • •, Sn) be a basis for A/P and write
StSj = ^ PijkSk, Pijk e P. If 5 is any element of A and g(x) =
k
h(xp') is its minimum polynomial over <I> such that h(x) is separ-
separable, then 8p' is separable over <D. Hence also 8pe+/ = (Sp')pf is
separable. It follows that we can choose e so that every 8/' and
every pakp' is separable over <i>. Since P/$ is purely inseparable,
this implies that aijk = Pijkp' e$. We have if if = 2piy/V;
hence, if we put 8,p" = <n, then we have <r,- e 2 and <Xi<x}- = 2a,yj;o-j;,
aijk e>i>. We claim that (<ru <r2, • ■ •> <rn) is a basis for A/P and
for 2/$. We note first that the multiplication table for the o-t-
shows that J2 *ff» ^s a ^"-subalgebra of 2/$ and X P°"» is a P-
subalgebra of A/P. Also the number of o-,- is «; so to show that
(o"ij ^25 • • • > O is a P-basis for A it is enough to prove that every
5 e A is a P-linear combination of the «r,-. To prove that (au a2,
■ • ■, <rn) is a *-basis for 2 it will be enough to show that every
a e 2 is a ^-linear combination of the o-,- since, if these are P-
linearly independent, then they are certainly ^-linearly independ-
independent. Now let 5 e A. Then 5 is separable over P so S e P[5pe]. We
have 8 = 2pi5i, Pi e P, since (Su ■ ■ ■, Sn) is a P-basis. Then 8pe =
2pip88tJ"i = 2pipVi e-^ Pa,-. Since 8 e P[8pe], this implies that
8 e ^ P^t- Next let <r e 2. Then, as we have just shown, a =
i
2pi<Ti, Pi e P. If / is large enough, then p?1 e $ and apf =
2p,J)/<7ip/ e 22 *f »• Since <r is separable over $, <r e $[<rp/] C ^2
This completes the proof.

FINITE DIMENSIONAL EXTENSION FIELDS 51
We can now prove
Theorem 12. If A/P and P/* are finite dimensional, then
A2) [A:*]. = [A:P].[P:*]., [A:#]f = [A:P]<[P:*],-.
Proof. It is enough to prove the first of these equations since
the second will follow from it and A1). Assume first that A/P is
purely inseparable. Then any element of A/$ which is separable
over * is separable over P and so belongs to P. Hence the maxi-
maximal separable subfield 2/* of A/$ is contained in P and so this is
the maximal separable subfield of P/<i>. Hence [A:$]s = [P:$]s.
On the other hand, since A/P is purely inseparable, [A:P], = 1.
Hence A2) holds in this case. Next assume A/P is separable.
Considering the maximal separable subfield 2/<i> of P/$> as base
field, we apply the lemma to the separable extension A/P and
the purely inseparable extension P/2. This gives [A: P] = [2':2]
where 2'/2 is the maximal separable subfield of A/2. Since
separability is transitive, it is clear that 2'/<£> is the maximal sep-
separable subfield of A/*. Hence, by definition, [A:$]s = [2': <£>]
and
[A:*], = [2':*] = [2':2][2:#]
Clearly, [2:f>] = [P:^>]8 and, since A/P is separable, [A:P] =
[A: P]3. Substituting in the above equation gives A2) in this case.
Finally, let A/P be arbitrary. Let E/P be the maximal separable
subfield of A/ P, so A/ E is purely inseparable. Then, on consider-
considering E 2 P 2* where E/P is separable, we see that [E:«i>], =
[E: PJ^P:*],. Since A/E is purely inseparable, the first case ap-
applied toADED* gives [A:*], = [A: E],[E:$],. Similarly, con-
considering A 3 E 3 P, we obtain [A:P]4 = [A: E]S[E: P]s. Com-
Combining, we obtain
[A:*]. = [A:E],[E:*].
which is A2) in the general case.

52 FINITE DIMENSIONAL EXTENSION FIELDS
We have seen that, if P/i> is Galois, then P/<i> is separable (Cor.
to Lemma 1, § 9). Also, since P/$ is a splitting field, this exten-
extension is normal. Conversely, if P/«I> is normal and separable, then
P/$ is a splitting field of a polynomial which is separable. Hence
P/<I> is Galois. Thus the condition: P/«i> Galois is equivalent to
P/<i> is separable and normal. We claim also that any purely in-
inseparable extension P/«i> is normal. To see this let g(x) e$[x]
be irreducible and suppose g(a) = 0 for a e P. Then g(x) is the
minimum polynomial of <r over*, so g(x) = xp' — a. Since ap' =
a, we have the factorization g(x) = xp' — a = xp' — <rp' =
(x — o)p', so g(x) is a product of linear factors in P[x]. Hence
P/# is normal. The following theorem gives a rather precise
description of the structure of normal extensions.
Theorem 13. 7/"P/<£> is finite dimensional normal, then P = 2 ®*
F where 2/<i> is Galois and T/$ is purely inseparable. Conversely,
if F/$ is a finite dimensional purely inseparable extension and 2/<l>
is finite dimensional separable extension, then the algebra P/<£ =
F ® * 2 is afield and this is normal if 2/$ is Galois.
Proof. Assume P/# finite dimensional normal. Let T be the
set of purely inseparable elements over $>, so F is a subfield over
$ (ex. 1, § 9). Let p e P and letg(x) be the minimum polynomial
of p over <i> and write g(x) = h(xp') where h(x) is separable. Since
g(x) is irreducible in $[*•], it is clear that h(x) is irreducible in
<£>[#]. Since h(pp') — 0, the normality of P/<i> implies that h{x) =
fl (* - ft) in P[*]. Also g(x) = h(xp') = II (xp' - ^) is a
i i
product of linear factors in P[x], Hence xp' — &■ has a root p,-
in P and consequently xp' — ft = xp' — pf = (x — p,)p*. Then
^*) = h(xp') = ft (^' - ft) = fl (x - Pl)p\ Now set k(x) =
i i
II (x — Pi). Since the ft = p,p' are distinct, the pt- are dis-
i
tinct. We have k(x)p' = H(xp' - Pip') = IL(xp' - ft) = g(x). If
k{x) = xm + o-i* H \- <rm, then the cry e P and k{x)p° =
xmP> _|_ ^^(m-Df _| 1_ ^p' = ^ which shows that o-/' e
<3>; hence the 07 e T. Hence k(x) e T[x]. Since p is a root of k(x)
and k{x) has distinct roots, p is separable over T. Since p was any

FINITE DIMENSIONAL EXTENSION FIELDS 53
element of P, this proves that P/r is separable. The factoriza-
factorization P = F ®* 2 where 2 is the maximal separable subfield of
P/«i> therefore follows from the lemma at the beginning of this
section. If a e 2 and f{x) is its minimum polynomial over $, then
f(x) = n(# — ffk) in P[#]. Evidently the o-* are separable over $
so they are contained in 2. Hence the factorization f(x) =
H(x — ah) holds in 2[#]. This proves that 2/<l? is normal as well as
separable. Hence it is Galois over <£. This proves the first state-
statement. Now let F/<I> be finite dimensional purely inseparable,
finite dimensional separable. We shall show first that, if
ff2j • • •> O is a basis for 2/$, then the same is true for (aip',
02p'j • • ■> ^m"')) * > 1. Clearly it suffices to show that every ele-
element a e 2 is a ^-linear combination of the <rf. Now for any
j > 0, a' is a linear combination of the <r,-. Taking p"-th powers
shows that (o**)' is a 4>-linear combination of the <r,p" for j =
0, 1, 2, • • •. Since a is separable, we know that a e $[<rp*\ (Lemma
2 of § 9). Hence <r e £ $<r,p' and (en11*, o-2p', • • •, amp') is a basis
i
for 2/$. Now consider P = F <g>* 2. This is a commutative al-
algebra and any element of this algebra can be written in the form
27,- ® <Ti, fi e F, and (o-j, • • •, <rm), the basis for 2/<i>. Now if
P = 27,- ® o-f t^ 0, then one of the 7,-, say, 71 5^ 0. Since r/i> is
purely inseparable, we can choose e > 0 so that yf — a{ e $,
1 < i < m. Then pp" = 2af ® <?*' = 1 ® Sa,^'. Since
("■ip'> • • • > ffmp') is a basis for 2/$ and ax t± 0, 2a,GiJ'' is a non-
nonzero element of 2. Since this has an inverse in 2, p has an in-
inverse in P. Thus P is a field. Now assume 2/$ is Galois. Then
2/$ is a splitting field of a polynomial g(x) e $[x] and F/* is a
splitting field of h(x) e *M. Since P is generated by subfields
isomorphic to F and 2, it follows that P/$ is a splitting field of
g(x)h(x). Hence P/$ is normal.
EXERCISES
1. Let E/$ be finite dimensional, P/$ an arbitrary field extension. Show that
the number of distinct isomorphisms of E/$ into P/<£> does not exceed [E :<£],.
Show that this number is attained if P/$ is the normal closure of E/$.
2. Let P/3> be finite dimensional normal, 2/4? the maximal separable subfield,
G the Galois group of P/*. Show that G maps S into itself and that the mapping
s —* J the restriction of s 8 G to 2 is an isomorphism of G onto the Galois group
of 2/$. Show that /(G) = F/$, the maximal purely inseparable subfield of P/$.

54 FINITE DIMENSIONAL EXTENSION FIELDS
11. Primitive elements. In this section and the next we shall
obtain some special generations of finite dimensional extensions
P/$. The results are valid for arbitrary <£. However, the proofs
in these two sections will require $ to be infinite; the validity of
the results for finite <3> will be established in § 13.
If P = $F), that is, P is generated over $ by 8, then we have
called P a simple extension of $ (Vol. I, p. 101). We shall now
say also that 6 is a primitive element of P/$. We shall prove two
results on existence of primitive elements.
Theorem 14. Let <£> be an infinite field and let P = $(£, v) be a
field generated over $ by a separable algebraic element £ and an alge-
algebraic element 77. Then P/$ has a primitive element.
Proof. Let f(x) and g(x) be the minimum polynomial over $
of £ and 7) respectively and let A/P be a splitting field of f(x)g(x).
Then A/<i> is a splitting field of f(x)g(x) containing P. Let £i =
f> £2, • • •> £m be the distinct roots of f(x), r)lt rj2, ■ ■ •, r?r those of
g{x). Then the £,- are all the roots of f{x) and we may assume m >
1 since, otherwise, £e<i> and P = $G7). Consider one of the
linear equations *£i + r/i = #£,- + t\j, i = 2, • • •, m,j = 1, • • •, r.
This has at most one solution in <E>. Hence, since $ is infinite, we
can avoid the finite set of solutions of these equations and choose
x = 7 e# so that y£x + Vi ^ 7£i + Vi, i = 2, ■ ■ ■, m;j = 1, • • •,
r. We assert that 8 = y^ + 771 is a primitive element of P. Thus
consider the polynomial gF — yx) which evidently belongs to
<S>@)[#]. We haveg(8 - y^) = g{yn) = 0 and, since 8 - y^ 9* i\j
fori = 2,---,m and./ = 1, 2, • • •, r, g(8 — ?£,) ^ 0. Hence the
m
highest common factor of g{8 — 7*) and /(#) = II (x — £,■) is
1
x — £ = x — £1. Hence there exist A{x), B{x) e$(8)[x] so that
(*-£)= A{x)f(x) + B(x)g(8 - yx), which implies that£ e#@).
Then 77 = 8 — 7£ e $@) and 0 is a primitive element.
The result just proved has an immediate extension, by induc-
induction on k to show that, if P = $(£1, •••,&, 17) where the £,- are
separable algebraic and 77 is algebraic, then P has a primitive ele-
element. In particular, we see that any finite dimensional separable
extension has a primitive element. We note that the number of
intermediate fields of such an extension is finite. This is clear

FINITE DIMENSIONAL EXTENSION FIELDS 55
since P can be imbedded in an extension A/$ which is finite di-
dimensional Galois, and the set of intermediate fields between A
and $ are in 1-1 correspondence with the set of subgroups of a
finite group—the Galois group of A/$. The theorem on primitive
elements for finite dimensional separable extensions is therefore
also a consequence of the following
Theorem 15 (Artin). Let <£ be an infinite field and P a finite di-
dimensional extension field of <£. Then P/& is a simple extension if
and only if there are only a finite number of intermediate fields be-
between P and$.
Proof. Suppose first that P = $@) and let E be an inter-
intermediate field. Let g(x) be the minimum polynomial of 6 over E
and let E'/$ be the field generated by the coefficients of g{x).
Then E' C E, but g(x) is also the minimum polynomial over E'
of 0. Hence [P:Ef] = degg(x) = [P:E\. Hence E = E' is
generated by the coefficients of g(x). Now g(x) is a factor of the
minimum polynomial f{x) of 0 over $ and both g(x), f(x) e P[*].
Since f(x) has only a finite number of distinct factors in ~P[x]
with leading coefficients 1, the number of E is finite. Next assume
that there are only a finite number of intermediate fields between
P and $>. It suffices to show that, if £, 77 e P, then $(£, 77) is simple.
Now let ae$ and consider the subfield Pa = $(£ + <xr\). We
have an infinite number of a e <$ and a finite number of Pa. Hence
there exist a, C in $, a. 9^ C, such that Pa = Pp. Then 77 = (a —
P)-1^ + at) - £ - fa) e Pa and hence £ = £ + aV - oa\ e P«.
Thus Pa = <£>(£, 77) and this is generated by £ + m\.
EXERCISES
1. Let $0 be of characteristic p t£ 0 and let P = $o(?, v) the field of fractions
of $[£, 77], £, 77 indeterminates. Let $ = $o(Pp) the subfield over $0 generated by-
all the^>-th powers. Show that [P:<£] = p1 and that P does not have a primitive
element over <£.
2. Let P be the splitting field over the rationals of (#2 - 3)(*2 - 2). Find a
primitive element for P.
3. Do the same for #B — 2.
12. Normal bases. If P is finite dimensional Galois over $
with Galois group G = {j1} s2, • ■ ■, sn}, then p e P has a minimum
polynomial of degree n over $ if and only if the elements pSi, i =

56 FINITE DIMENSIONAL EXTENSION FIELDS
1, •••,«, are distinct. This is clear since, if p'1, • • -, p'r are the
r
distinct conjugates, then f(x) = II (# — P*') is the minimum
7 = 1
polynomial of p over $. It is clear also that p is a primitive element
of P/<i> if and only if the degree of its minimum polynomial is
n = (G:l). Hence p is a primitive element if and only if the
pH, i = 1, •••,», are distinct. A stronger condition than this is
evidently that these elements are linearly independent. Then we
have the basis (pSl, p*2, • • •, p*n) of P over $>. Such a basis, con-
consisting of the conjugates of a single element, is called a normal
basis for the Galois extension. We shall now show that such
bases always exist if $ is infinite. Our proof of this fact will be
based on the notion of algebraic independence of isomorphisms
which is of considerable interest on its own. We define this as
follows:
Definition 2. Let E be afield over <£ and Q an extension field of
E. Let si, ■ • •, smbe isomorphisms of E/$ into Q/$. Then we shall
call the Si algebraically independent over Q,- if the following is true:
The only polynomial f(xu • • -, xm) e Q,[xi, • • •, xm], *,- indetermi-
nates, such that f(t]'1, V2, • • ■, 77s-) = Ofor allrjeE is / = 0.
We require the following
Lemma. Let Q be an extension field of an infinite field <b and let
ftxu ••-,#») e Q[#i, ••-,#*>] satisfy /(&, ••-,£„) = 0 for all &
e #. Then f = 0.
Proof. Let (wa) be a basis for S2 over <£. Then we can write
r
/(*i, • • -,xm) = J2 />(#i> • •', Xm)o>j where {w,} is a finite sub-
1
set of (ua) and the/,- e &[xu • • •, #m]. Then 0 = /(&, •••,£„,) =
2/X£i) • • • j £m)wy for all & e $. Since the wy are ^-independent and
the /y(|i, • • •, {«) e $, this implies that every /,(^, • • •, £m) =0.
Hence, by a result proved in Vol. I, p. 112,/y(*i, • • •, #m) = 0 for
ally and so/(xi, ■ ■ -,xm) =0.
We can now prove the following theorem on algebraic independ-
independence of isomorphisms.
Theorem 16. Let P be finite dimensional Galois over an infinite
field $, E a subfield of P/$, and 0 an arbitrary extension field of

FINITE DIMENSIONAL EXTENSION FIELDS 57
P. Let jj, s2, ■ ■ ■, sm be the different isomorphisms of E over $ into
P over $. Then the Si are algebraically independent over Q.
Proof. We recall that the number m of isomorphisms is [E:$]
(§ 7). We note next that, if («i, e2, ■ • ■, em) is a basis for E/*,
then the determinant det («,••') of the matrix whose (i,j) entry
is e,-8' is not 0. Otherwise, the rows of this matrix are P-dependent,
so there exist py not all 0 in <£ such that Spye,-*' = 0, i = 1, 2, • • ■,
wz. If e is any element in E, we can write e = S/3,€i, /3* e$, and
we obtain ^ &iPj*i'> = 0. Since &■"> = fo, we have X) P^8' = 0-
This states that the operator Sjypyj? = 0 contrary to Dedekind's
independence theorem for isomorphisms. We have therefore
established that det (e;8)) y* 0. Now suppose /e Q[xi, #2, • • •> *m]
and /(e81, e82, • • •, e8™) = 0 for all « e E. Then /(ZfteA S^e,-8*,
• • •, S/3tei8m) = 0 for all & in the infinite field *. Now let ^(^i,
•■-,*„) = /B^te,-8', S*^-8*, • • •, -Lxit^) e Of*!, ■ • •, xm]. This
vanishes for all Xi = |3,- e«J>, so by the lemma, g(xly • • ■, xm) = 0.
Now det («,•*') 5^ 0, so the matrix (e,*0 has an inverse (py). Then
f{xi, ■ ■ ■, xm) = gBxinn, 2xini2, • • •, ZxiMn) = 0. This proves
that Ji, • • •, sm are algebraically independent over 12.
We can use the result just proved to establish
Theorem 17. Let P be finite dimensional Galois over an infinite
<i>. Then P/4> has a normal basis.
Proof. Let G = {su • • •, sn} be the Galois group of P/«i>. We
have just seen that, if (pi, • • •, pn) is a basis of P over $, then
det (p/') 5^ 0. Conversely, this condition is sufficient for a basis;
for, if S/3iPi = 0 where the /?,■ 8$, then 2/3t-pi8'' = 0,j = 1, • • •, m.
This implies that the columns of (p,*') are ^-dependent unless
every /3,- = 0. Our criterion shows that, for a particular p, (p81,
P82, ■ • •, p8n) is a normal basis if and only if det (p8*'8') 9^ 0. We
now write j,J/ = sif and we know that (ly, 2y, • • •, »y) is a permuta-
permutation of A,2, • • •, n). Consider the matrix whose (i,j) entry is the
indeterminate #t-y (/y = 1,2, •••,«) in P[^1} • • •, #„]. We assert
that the polynomial d(xly • ■ •, xn) = det (*,-,) ^ 0. To see this
we specialize X\ = 1, x2 = ■ ■ ■ = xn = 0. Since each row and
column of (*,-,) contains exactly one xu it follows that det (**,■) =
± 1 if the Xi are specialized as indicated. Therefore, d(xi} • • •, xn)

58 FINITE DIMENSIONAL EXTENSION FIELDS
9* 0, and so by the algebraic independence of the x{ we can
find a p e P such that det (p"8') = det (psv) ^ 0. Then p de-
determines a normal basis.
There is another, more sophisticated, formulation of the normal
basis theorem which we shall now indicate. For this we introduce
the group algebra $(G) of the group G: $(G) has the basis G =
{si, ■••,sn} and multiplication is defined by (Sa,j,)B/8;-j,-) =
H,a.i$jSiSj = XciiPjSij (cf. Vol. I, ex. 2, p. 95). We consider two
right modules for <i>(G): The first of these is <I>(G) itself considered
in the usual way: xa, x e$(G), a e<i>(G) is the algebra product.
Next we consider P as «f>(G)-module by defining pa = Safp** for
a = Satjt- in <£(G). It is immediate that the module axioms hold
for this multiplication. The normal basis theorem is just the
statement that these two modules are isomorphic. Thus, let
(p8<) be a normal basis and consider the linear mapping of 4>(G)
over <£ into P over <i> sending j,- into pH. This is a ^-linear iso-
isomorphism and, if x = S^,j», then xsj = 2£,\f,(. —* ^ £iP*<j =
i
2 £tP*<S) = B£,Ps0-^- Hence, if we denote the image of x by
»
x', then x'sj = {xs})r. This implies that x'a = {xa)' for all a e
<£(G) so we have a <J?(G)-isomorphism. It is easy to check that,
conversely, if a? —> x' is a $(G)-isomorphism of<£(G) onto P, then
the image of (jj = 1, s2y • • •, sn) is a normal basis for P/<i>.
EXERCISE
1. Prove the following generalization of Theorem 16: Let/(*i(l), • ••,*mA),
*iB), • ■ •, #mB), • ■ •, #i(r), • ■ -, #m(r)) be a non-zero polynomial in indeterminates
Xi'-'K Then there exist i/i, • • •, i\r e E such that /(jji*1, • • •, 171* m, • • •, »7r*S • • •,
ij,*») ^ 0.
13. Finite fields. The main results on finite fields are readily
obtained as applications of Galois theory. We proceed to derive
these. At the same time we shall establish the validity of the
theorem on primitive elements and normal basis for finite base
fields.
We remark first that any finite field P is of characteristic p p* 0,
since otherwise P contains a subfield isomorphic to the field of
rational numbers. Hence the prime field *0 °f P is isomorphic to

FINITE DIMENSIONAL EXTENSION FIELDS 59
Iv = I/(p). If $ is any subfield, of course, [P:$] = n < °o. If
(pij ■ ■ • > Pn) is a basis for P over <£, every element p e P can be
n
written in one and only one way as ^ a<Pi> ai e *• If the cardinal
i
number 1*1= q, then it is clear from this that |P| = qn. In
particular, if [P:*o] = N, <J>0 the prime field, then \V\ = pN.
This shows that the number of elements in any finite field is a
power of its characteristic.
We show next that for any prime power pN there exists one
and, in the sense of isomorphism, only one field with pN elements.
We consider the uniqueness first. Let P be a field with | P| = pN.
Then it is clear that the prime field $0 of P is isomorphic to Ip.
If p is a non-zero element of P, then ppN~1 = 1 since the order of
the multiplicative group P* of non-zero elements of P is pN — 1.
We have also that ppN = p, an equation which is valid for every
p e P. Thus every element of P is a root of xp — x = 0 and
xpN — x e <£oM> $o the prime field. Then
A3) xpN - x = n c* - p.-)
1
where the pt- are the elements of P. This shows that P/#o is a
splitting field of the polynomial xpN — x. Now suppose P' is a
second field such that | P'| = pN; then P' has characteristic p so
its prime field *</ = 3>0- Also P'/$o is a splitting field of xpN — x.
Hence P' = P by the uniqueness theorem on splitting fields.
The method just used also gives the existence of a field P with
| P | = pNy p a prime. For this we begin with $0 = Ip which has
p elements and we let P be a splitting field over $0 of xp — x.
Let S be the set of roots of xpN — x = 0 contained in P. Since
the derivative (xpN — x)' = — 1, xpN — x = 0 has distinct roots
so \X\ = pN. We note next that S is a subfield, since, if £, t\ e
2, then e" = & r,pN = r, so (£ - r,)pN = ?* - r,pN = $ - ,,
(«,)"* = e%pN = &, and or1);" = c^*)-1 = v'1 if v * o. it
now follows that 2d$0 and, since P is a splitting field of xpN —
x, P = *0(S) = 2. Thus we have | P| = |2| = pN.
We prove next the theorem on primitive elements: If f> is a
finite field and P is a finite dimensional extension of*, then P =
Clearly, under the given conditions P is finite. We now

60 FINITE DIMENSIONAL EXTENSION FIELDS
show that the multiplicative group P* is cyclic. This is a con-
consequence of the following useful general result.
Lemma 1. Any finite subgroup A of the multiplicative group of
afield is cyclic.
Proof. Let m be the order of A and let m' be the highest order
for the elements of A. It is known that, if a and b are two ele-
elements of a finite commutative group, then there exists a c in the
group whose order is the least common multiple of the orders of a
and b (Vol. II, ex. 1, p. 69). It follows that, if m' is the highest
order, then bm' = 1 for every b. On the other hand, we know
that the equation xm> — 1=0 has at most m' roots in a field.
Since m' \ m, we have m' = m. Moreover, if a is an element of
order m} then the order of the cyclic group [a] generated by a is m.
Hence A = [a].
Now if P is a finite field and $ is a subfield, then surely P =
<I>@) if d is chosen to be a generator of the cyclic group P*.
We consider next the automorphisms of a finite field P. If the
characteristic is p, then we know that the mapping ir:£ —> £p is
an isomorphism of P into P. Since P is finite, this is an auto-
automorphism. If | P| = pN, thenppN = p for every p e P. Evidently
7T* is the automorphism £ —» £**, so we have irN = 1. On the
other hand, if 0 is a geneiator of the group P*, then 6pm ?£ d if
m < N. This implies that irm ^ 1. Hence the cyclic group G =
[ir] has the order N. Let $ be the set of G-invariants of P. Then
we know that $ is a subfield and [P:<J>] = N. On the other hand,
we know that [P:<$o] = PN if $0 is the prime field. Hence $ =
"i>o- We now see that the field P is Galois over its prime field <£0
and the Galois group is G = [ir]. The Galois correspondence now
gives a correspondence between the collection of subfields of P
and the collection of subgroups of G. Since G is cyclic of order N,
for each divisor n of N there exists one and only one subgroup H
of index n. We have H — [r] where t = ir". The correspond-
corresponding field * of //-invariants (or of r-invariants) has dimension-
dimensionality n over $o- Hence |<i>| = pn and we have shown that the
subfields $ of P have order pn where n | N and for each such order
there is precisely one subfield of P of this order. The Galois
group of P over <i> is the cyclic group H = [r] as before. In general,

FINITE DIMENSIONAL EXTENSION FIELDS 61
we shall call an extension field P/f> cyclic, abelian, or solvable if P/<3>
is finite dimensional Galois and its Galois group is respectively
cyclic, commutative, or solvable. Hence we can say that any
finite field is a cyclic extension of any of its subfields. We shall
therefore have the normal basis theorem also for finite base
fields by proving
Lemma 2. Any cyclic extension P/$ has a normal basis over #.
Proof. Let s be a generator of the Galois G group of P/f>.
We consider s as a linear transformation in P over $ and let n(x) e
$[x] be its minimum polynomial. Now Dedekind's independence
theorem implies that the automorphisms 1,j, •••,j"~1 are P-
independent if (G:l) = n. It follows that these are also *-in-
dependent and consequently deg n(x) > n. On the other hand,
[P:<l>] = n so the degree of n(x) cannot exceed n (Vol. II, p. 69).
Hence deg/i(#) = n. Since sn = 1, we see that n(x) = xn — 1.
Now we know that there exists a p e P whose order polynomial
relative to the linear transformation s is the minimum polynomial
(Vol. II, p. 67). Then p, p', • • •, p'n ' are ^-independent and so
these elements form a normal basis for
EXERCISES
Note: A set of exercises on finite fields is given in Vol. I, pp. 112-113.
1. Let $ be a finite field of order of g(= p1*). Show that an irreducible poly-
polynomial/^) e $[*] is a factor of #9" — * if and only if deg/(#) \n. (Hint: consider
the field $[.*]/(/(*)).) Show that #9" — x = Ufi(x) where/<(#) runs over the ir-
irreducible polynomials with leading coefficients 1 of degrees divisors of n. Let
N(g, r) denote the number of these polynomials of degree r. Derive the formula
where /x is the Mobius function (cf. Vol. I, ex. 5, p. 120).
2. Let 3JJ be an n dimensional vector space over a finite field $ of odd order q
(characteristic ?* 2) and let ,j(.v,.y) be a non-degenerate symmetric bilinear form
on 9)t over <i>. Show that if n > 2, then there exists a vector u in 9Ji such that
g{u, u) = 1. Apply this and the reduction theory of Vol. II, pp. 152-154, to
prove that SO? has an orthogonal basis («i, «2, • • •, «n) such that g(uu «i) =
S 7* 0, g(ui, Ui) = 1 if; > 1. Use this to prove that any two non-singular sym-
symmetric n X n matrices with entries in $ are cogredient if and only if their
determinants differ by a multiplicative factor which is a square E = 5'p2) in $.
Hence show that there are just two cogredience classes of non-singular sym-
symmetric matrices.

62
FINITE DIMENSIONAL EXTENSION FIELDS
3. Let <i>, 2JJ, g be as in ex. 2. If (ei, <?2, •••,*».) is a basis, then 5 =
det (g(et, e,)) is called a discriminant of £. For £ e $ let JVfo b) be the number
of vectors « e 9JJ satisfying #(«, «) = b. Show that
|V ~X - ?' + tf*, if » = 2k and
(—1)'5 is not a square
»-i + q* _ ^-^ if „ = 2v and
(—1)'8 is a square
L?2". if n = 2k + 1
?2v-i + ^-^ if ^ 5«£ 0, w = 2k, and
(—1) is not a square
?2.-i _ f-it ■l(b^Qn = 2v> ancj
( —1) is a square
?2' - f,\fb^0,n = 2v+l, and
(—1)'5£ is not a square
?2" + if, if ^ ^ 0, « = 2p + 1, and
(—l)"8b is a square
4. Let O(»,j-) denote the orthogonal group determined by g-O(n,g) is the
group of linear transformations A of M such that g{xA, yA) = g{x,y) for all
x,y eM. If « is a non-isotropic vector, let Ou be the subgroup of 0(n,g) leaving u
fixed. Show that Ou is isomorphic to O(» — 1,1'') where g7 is the restriction of £
to ($«)-■-. Use Witt's theorem to show that the number of cosets OUA of Ou in
O(n, g) is the number of vectors v satisfying g(v, v) = g(u, u). Use this result and
ex. 3 to establish the following formulas for the order (O(w, g): 1):
N(g,b) =
nl)/2
g JJ (?2i _ 1)) if „ ;s odd
.-1
(n-2)/2
2.?n(n-2)/4(?n/2 _ e) JJ (?« - 1), if „ js even
l
Here e = 1 if ( — 1)' 5 is a square; otherwise « = — 1.
14. Regular representation, trace and norm. In this section
we consider a finite dimensional extension field P/$ and we shall
define certain mappings of P into f> called the trace and the norm.
These functions can be defined just as easily for arbitrary finite
dimensional algebras and are of importance for these also. We
shall therefore begin by considering a finite dimensional algebra
Sl/4> with basis («i, u2, • • •, un) over <i>. We define a {finite di-
dimensional) representation of §l/4> to be a homomorphism of 2l/4>
into the algebra ?*(9JJ) of linear transformations of a finite di-
dimensional vector space 9K/<E>. If S is such a representation: a —>

FINITE DIMENSIONAL EXTENSION FIELDS 63
as> then the defining conditions are
{a + b)s = as + bs, (aa)s = aas,
A4)
(ab)s = asbs, I5 = 1,
a, b e 21, a e <!?. If (#i, #2, • • ■, #;v) is a basis for 9ft over $, then
we can determine the matrix as relative to this basis in the usual
manner: We write
A5) Xias = £ aii{a)xh i = l,2,---,N.
3=1
This gives the matrix a{a) = (a,y(«)) and the mapping a —* a(a)
of 21/$ into the algebra $n/$ of N X N matrices with entries in
<£. Since the mapping A —> (a) of the linear transformation A
into its matrix (a) relative to (x\, x2, • • ■, xn) is an isomorphism,
the mapping a —» a(«) is a homomorphism of 21/4? into $jv/4?.
Such a homomorphism is called a matrix representation. We recall
that, if we change the basis (xly • • ■, Xn) to another basis (ji, ■ • •,
^jv) where jt- = HpijXj, then the matrix representation defined by
£ and this basis is a —* (/*)«(«)(m) where (ju) = (na) (cf. Vol.
II, p. 42).
The most important representation of 21/* is the so-called
regular representation R. Here aR = aR the right multiplication
x -* xa defined by a. One checks directly that an is a linear
transformation in 21 over $ and that a —■> #r is an algebra homo-
homomorphism (cf. Vol. I, p. 82). Since 21 has an identity, a —> aR
is 1-1 and so is an isomorphism of 2f/4> into 8#Bl). Since xaR =
xa, we obtain the matrix representation associated with the basis
(«i, #2j • • •>««) of 21/3? by writing the products «ta as ^-linear
combinations of the «;:
A6) «i« = Sp,y(a)«y, _/ = 1, 2, •••,».
We write p(a) = (pij(a)) and we have the matrix representation
a —> p(a) which is 1-1. Also since lR = 1, p(l) = 1, the identity
matrix. As in the general case, a change to the basis (vi, v2, ■ • ■,
vn), where v, = ~ZmjUj, gives the new matrix representation a —*
a(a), where
A7) a(a) =

64
FINITE DIMENSIONAL EXTENSION FIELDS
As an example of this we consider an algebra 21 = $[a] with a
single generator. Since [21:*] < °°, $[a] =$[x]/(f(x)) where/(#) is
a non-zero polynomial with leading coefficient 1. We have /(a) =
0 and f{x) is the non-zero polynomial of least degree (leading
coefficient 1) having a as a root. Thus the polynomial/^) is the
minimum polynomial of a (Introduction, p. 6). Also $[a] has
the basis A, a, ■ • -, an~l) where n = [2l:$] = deg/(#). Suppose
A8) /(*) = xn - axxn-x H 1- (-l)nan, on e*.
Then we have the relations
A9)
an~la =
\a = a, aa = a2, ■ • •,
an~2a = a"-1
- a2an
These show that, if p(a) denotes the matrix of ur relative to
A, «, • • •, an~1)} then we have
B0)
P(a) =
0 1
0 0
(-1)"'
0
1
which is called the companion matrix of the polynomial f(x). In
theory, once we know this matrix, we know p{b) for any element b
in <£[«] since b is a polynomial in a.
We now consider the general case again and we define the charac-
characteristic polynomial of the element a e 21 to be the characteristic
polynomial
B1) /»(*) = det (*1 - p{a))
of the linear transformation aR in 21 or of the corresponding matrix
p(a). By A7) we have
which shows that
det (*1 - <r(a))
= det (M)( p
= det (M) det (*1 - P{a)) det
= det (xl — p(a)).

FINITE DIMENSIONAL EXTENSION FIELDS 65
Thus we see that /<,(#) is independent of the choice of the basis
for SI/*. We can write the characteristic polynomial as
B2) /„(*) = xn - T{a)xn~l +■■■+ (-l)nN(a).
n
We have T{a) — trace p(a) = £ Pu(a), N(a) — det (p(a)) and we
1
call these respectively the trace and norm of a in 21 over $. We
shall find it necessary at times to specify the base field of the alge-
algebra and also the algebra itself. In these cases we shall write
T%\<t,(a) for T(a), Nz\*(a) for N(a). Since the trace is a linear
function of matrices and since a —> p{a) is linear, it is clear that
a -» Tai*(«) is a linear mapping of SI into $. Also since p(l) =
1, we have 7Vi«(l) = »1 and T%\*(aa) — aT%\<t{a). Thus we
have the following relations:
Tn\*(a + b) = Tn\*(a) + T«i.(J)
B3) T*\*(aa) = a7V|.(«), ae*
r(D = «i-
Since a —> p(«) is multiplicative and ^/ —> det A is a multiplica-
multiplicative mapping of the set of matrices, we have N%\*(a&) =
iVai*(a)iVai»(^). Also, it is clear that iVai*(afl) = aniVai»(«)
and iVai*(l) = 1. Thus we have:
B4)
7VA) = 1.
We recall that according to the Hamilton-Cayley theorem p(a)
is a root of /„(*) = 0. If we apply the isomorphism p(b) —» ^, we
see that /o(«) = 0. Thus we have
B5) an - TWa*-1 H h (-l)nN(a)l = 0.
Let ma(x) be the minimum polynomial of p{a) (or of «#)• Since
^ —> p{b) is an isomorphism, it is clear that ma(x) is the minimum
polynomial of a. We recall that the minimum polynomial of a
matrix is a factor of its characteristic polynomial and these two
have the same irreducible factors in $[*], differing only in the
multiplicities of these factors (Vol. II, p. 99, or p. 102).

66 FINITE DIMENSIONAL EXTENSION FIELDS
The trace function can be used to define an important bilinear
form on the algebra 2l/<£>. This is the regular trace form
B6) {a, b) = T*Uab).
Evidently, we have the following rules governing this function
whose values are in <£:
(«, bx + b2) = (a, bx) + (a, b2)
(«x + a2, b) = (au b) + (a2, b)
B7)
a(a, b) = (aa, b) = (a, ab)
(ab,c) = (a, be) ( = Tn\z(abc)).
Also we recall that, if M and N are matrices, then the tr MN =
tr NM (Vol. II, p. 104). This implies that
B8) (a,b) = (b,a),
so (a, b) is a symmetric bilinear form. We shall define also the
discriminant of 21 over * relative to the basis («1} u2, • ■ ■, un) to
be
B9) 5(«.) = det ((«,, «y)) = det (Tai*(«<«,•)).
It is immediate that, if we replace («i, ••-,«„) by the basis
(tfi, • • •, Vn), Vi = ZfiijUj, then the matrix ((z<t-, uj)) is replaced by
(fa, vj)) = M((«,-, «y))M', M = (M,v) (Vol. II, P. 149). Hence the
discriminant relative to fa) is 8(ui)n2, M = det M.
We now suppose that E is a subfield of <i> of finite co-dimension
inS. Then 21 3 $ 3 E and [21: E] = [«:*][$: E] is finite so, if we
consider 21 as a vector space over E, this is finite dimensional.
Hence 21 is a finite dimensional algebra over E. We can therefore
carry out all of the above considerations for the algebra 2I/E. We
can also consider $ as an algebra over E and we can define T%\-&,
7*Ie, iVgiB, A^ie as well as Tai*, Mil*. We shall now proceed
to develop the following fundamental transitivity relations con-
connecting these functions: If $ 2 E,
C0)
C1) NnUa) = iV#|B(iV«i#(«)).

FINITE DIMENSIONAL EXTENSION FIELDS 67
As before, we let («i, • • •, un) be a basis for 2t/<i> and we suppose
(yi> 72} • • •, 7a) is a basis for */E. Then
C2) Gi«i, 72«i, • • •> 7a«i; 7i«2, ■ • •
is a basis for 21/E. If p e $, we write
C3) 79P = SX9<(PO«, ?,/ = 1, •••,*
so we have the isomorphism p —» (X(p)) where X(p) is the matrix
(X9<(p)) with entries in E. Then T*|E(p) = 2X99(p) and iV*|E(p)
= det (X(p)). We now combine the relations C3) and A6) to
write
C4) (yqUi)a = 2yqpi}{a)Uj = 2\t(Pij(a))ytuh
i, j = 1, • • •, », y, f = 1, • • •, A. This shows that if the basis
(y9h») is ordered as in C2), then the matrix of aR in 9I/E is
X(pu) X(pi2) •
,_f. .,. X(p2i) X(p22) ••• X(p2n)
C5) A(«) =
■X(pnl) X(pn2) • • • X(pnn)J
where X(p,y) is an h X h matrix with entries in E and we have ab-
abbreviated pij = pijia). It is clear from the form of C5) that
= tr A(«) (tr = trace)
= tr X(pn) + tr X(p22) H h tr X(pnn)
= tr X(pu + P22 -\ h Pnn)
= tr X(T«i.(a))
This establishes C0).
For the proof of C1) we require a general transitivity property
of determinants (Vol. II, ex. 2, p. 135) which we proceed to derive.
We suppose we have an nh X nh matrix with entries in a field E
and we assume that, if we partition this as an n X « matrix A =
(Xy) where each X# is an h X h matrix, then the X,y all commute.
This is equivalent to assuming that the Xy all belong to a com-
commutative subalgebra 93 of the matrix algebra *a. This is precisely

68
FINITE DIMENSIONAL EXTENSION FIELDS
the situation for the matrix A(a) and the blocks X(p,y) of C5).
Since the X,y e 33 and 33 is commutative, the usual definition and
properties of determinants hold and we can consider
C6) detn (A) = X} «pXltlX2,-2 • • • X»ln,
p
where the summation is over the permutations (iii2 • • • in) of
A2 • • • n) and ep = 1,-1 according as P is even or odd. Now
detn (A) as defined above is an element of $*. Hence we can take
the usual determinant of this. We shall now establish the follow-
following formula:
C7) det (detn (A)) = det A
where det A is the usual determinant of the nh X nh matrix.
To prove this result we extend the base field E to a splitting
field over E of the product of the characteristic polynomials of all
the matrices X,y. It suffices to prove the result in this field. Hence
without loss of generality we may assume that E contains the
characteristic roots of all the X,y. The theory of sets of commuting
linear transformations (Vol. II, pp. 133-134) shows that there
exists a matrix fieEj such that every m~1X,-,m is triangular:
C8) m:
Hence if we set
C9)
then
D0) Jl
0
Pijh
M =
111 9l2
121 922
9nl 9n2
9ln
Vnn-i

FINITE DIMENSIONAL EXTENSION FIELDS
69
We have det A = det M 'AM and, since 77,7 =
detn M
Hence det (detn M~lKM) = det (detn A); so it suffices to verify
that
D1)
det (detnM-xAM) = det M~lKM.
Now it follows directly from the definition of detn and from the
way triangular matrices are multiplied and added that
D2) detn M~XAM =
det
det
P2
0
where
D3)
Pk =
Piifc
P2U
-Pnlife
Pl2*
P22*
Pn2*
Pink
P2nk
Pnnk-
det pa-
Hence
D4) det (detn Af-'AM) = det pi det p2 • ■ • det pA.
We need to calculate next det M~lkM. For this we make the
following permutations of rows and columns:
column (i — Y)h + j —* column (J — 1)« + /
row (i — \)h + j —> row (y — 1)» + /'
for / = 1, 2, • • •, n and y = 1, 2, • • •, h. This gives the matrix
D5)
Pi
P2
PA-1

70 FINITE DIMENSIONAL EXTENSION FIELDS
where p& is as in D3). Hence, by Laplace's expansion, det M~1KM
= det px det p2 • • ■ det ph = det (detn M-1AM). This proves C7),
as required.
We now apply this to norms. Here we have iVal*(«) =
det (pij(a)) e * and
A^El*A^ai*(«) = det(X(detpw)).
Since p —* X(p) is an isomorphism, we have
(X(det Pi])) = det, (X(py))
so
detX(detp,;) = det detn (X(piy)) = detA(a),
by C7). Since det A.(a) = A^ie(«) we have the norm formula C1).
We shall now specialize all of this to the case: SI = P a field.*
We know that the minimum polynomial ma{x) of any a e P is ir-
irreducible. Hence the characteristic polynomial fa{x) = ma(x)r.
We have [P:<f>] = n = deg fa(x) and [*(«):*] = deg ma(x); there-
therefore r = deg/«(*)/degw«(*) = [P:*]/[*(a):$] = [P:*(a)]. Hence
we have
D6) /„(*) = ma{x)V-*^\
We shall now obtain some important formulas for the norm
and trace of a field and we look first at the separable case. Thus
let P/$ be finite dimensional separable, O/$ the normal closure of
P/<E>. Then fi/«i> is Galois and [fl:$] = (G:l) for the Galois group
G of fl/$. Let H be the subgroup of G corresponding to P/<1> (the
Galois group of £2/P). Since [P:<i>] = n, the index (G:H) = n
and we have n distinct cosets Hs\, Hs2'} • • •, Hsn'. If jt- denotes
the restriction of s/ to P, then j1} j2) • • •> -fn are distinct isomor-
isomorphisms of P/<J> into 0/$ and these are all the isomorphisms of P/<3?
into fi/<i> (§ 7). Next let p e P and let K be the subgroup of G
corresponding to $(p). Then G 2 K 3 //. Let //, • • •, tm' be a
complete set of representatives of the cosets Kt' in G and let
«i'j •••>«/ be a complete set of representatives of the cosets
Ha' in #. Then we have G = UX//, K = UHuk' so G =
* A simplified version of the proof of the transitivity formula for norms in this case will
be indicated in ex. 2 below.

FINITE DIMENSIONAL EXTENSION FIELDS 71
k't/ and the mr elements «*'// form a complete set of rep-
representatives of the cosets of H in G. We may assume that these
are the s/ which we indicated before. The restrictions of the //
to f>(p) give all the isomorphisms of $(p)/$ into £2/<l> and these
are distinct. Since p generates ^(p), it follows that the elements
Ph> > P*1'> '"' j Ptm> are distinct and these include all the conjugates
p"', s' e G. Hence the minimum polynomial of p over$ is mp(x) =
m
II (x - pV). Also we have p"*'«/ = p'»' for all k and./. Hence
i-i
n r m
II (* - P8<) = II II (^ - P"*''O = »,(*)'. On the other hand,
«=i k=i y=i
r = [P:$(p)], so by D6) (for a = p), we see that the charac-
characteristic polynomial
D7) /,(*) = f[ (x - ff*)
1
where ji, s2, ■ • •, sn are the different isomorphisms of P/<£ into
its normal closure Q/$. Comparison of this formula with B2)
gives the following formulas for the trace and norm in the sepa-
separable case:
D8) 7Vi#(p) = i>s WpI.(p) = fl P'*-
Next let P/<£ be purely inseparable of characteristic p^O.
Then [P:€>] = pf. If p e P, the minimum polynomial mp(x) has
the form xp' — a = (x — p)p\ Since P/#(p) is purely insepa-
inseparable, [P:$(p)] =^g and pf = [P:$] = [P:$(p)][$(p):f>] = pe^.
Hence f = g -\- e. By D6), the characteristic polynomial is
D9) /,(*) = (**>• - ay = {x- Py.
This shows that
E0) TP|*(p) = [P:*>]p, iVp|*(P) - Pv-*K
Now let P/$ be arbitrary, S/* the maximal separable sub-
field, fl/€> the normal closure of P/<i>. Then fl/$ contains the
normal closure A/$ of S/4>. Again we assume the characteristic
is p 5^ 0. Then [P:2] = p', / > 0, and this is the degree of in-

72 FINITE DIMENSIONAL EXTENSION FIELDS
separability [P:*],- (§ 10). If p e P we have, by E0) and D7),
ivPi*(p) = iWiVpizGO) = nMp[P:*u)
where Ji, j2> • • •> Jn are the different isomorphisms of S/# into
A/*. Now it is easily seen that every j,- is the restriction of an
isomorphism of P/f> into Q/$ and distinct isomorphisms of P/$
in fl/$ have distinct restrictions to S/$ and map this field into
A/$ (ex. 1, § 10). It follows that the foregoing formula can be re-
rewritten as
E1) iVP|*(p) = (pV-p'f:}li
where su s2, • • •, sn are now considered as the different isomor-
isomorphisms of P/f> into fl/<£. In exactly the same way we obtain
E2) 7pi*(p) = [P:$],-(p81 + p« + ■ ■ ■ + p-).
If P is not separable over *, then / > 0 and [P:$]i = pf is
divisible by p. Hence we see that 7p|*(p) = 0 for inseparable
We obtain next some formulas for the discriminant of
relative to a basis (pi, p2, • • •, pn). This is
E3) 5 = det Gp|*(pfp,)).
If P/* is inseparable, 7p|* = 0 so 5 = 0. Now assume P/f> sepa-
separable and, as before, let Ji, s2, • • •, sn be the isomorphisms of P/f>
into 12/$. Consider the matrix
E4) ^ = (p,'O, /,y = 1,2, ■••,«.
We have shown in the proof of Theorem 16 that det A j£ 0. We
consider now the matrix AA', A' the transpose of A, whose (z, ^)-
entry is
E5) p*P* + pi'W* +■■■+ Pi'"Pi'n = Zp|#G>tf»y)-
Hence 5 = det ^/^/' and we have
E6) 5 = (det AJ, A = (p/0-
Since det A ?* 0, this shows that 5^0. We recall that this im-
implies that the trace bilinear form (p, a) = Tv\<t(fio) is non-de-
non-degenerate (Vol. II, p. 140). We therefore have the following

FINITE DIMENSIONAL EXTENSION FIELDS
73
Theorem 18. If P/# is finite dimensional separable, then the
trace form (p, ai) = Tp|*(p<r) is non-degenerate and the discriminants
S of P/<i> are j* 0.
Let 0 be a primitive element of the finite dimensional separable
extension. Then it is clear from D6) that the characteristic poly-
polynomial f(x) of 0 is the same as the minimum polynomial. In Q[x]
we have f(x) = (x — di)(x — 02) ■ ■ • (x — 0n), 0X = 0, and the
Bi are distinct. If fix) is the derivative of fix), then
E7) fie) = @ - e2)(e - e3) ■ • • (e - en)
and this element is contained in P = $(d) since fix) e<f>[*]. The
element f{6) is called the different of 6. We shall show that the
discriminant 5 determined by the basis A, 0, 02, • • ■, 0n-1) is
E8) 5 = (-ir
We have 8 = det TP|*@<~10y~1). Now it is clear that we have an
isomorphism of *@)/* into Q/<£> sending $ into 0,-, 1 < i < n.
Hence the 6t are the conjugates 6" of 0 and @*)** = 0,-*. The
matrix /f of E4) for the basis A, 0, 02, • • •, 0n-1) now becomes
fl
A =
1
1
0
n~1
It is well known that det A, a so-called Vandermonde determinant,
has the value H @,- — 0y). Consequently E6) gives the formula
E9) 5= II (*.-- *;J, iyj" 1,2, ••-,«,
«<>
for the discriminant. On the other hand, /'@) = TLi^i ~ ^»)-
tvi
Applying jy which sends 0X into 0,- we obtain fF)'i — JJ @y — 0,).
It follows that
F0)
0,J.
Comparison of E9) and F0) proves E8).

74 FINITE DIMENSIONAL EXTENSION FIELDS
EXERCISES
1. Let 21 be the algebra $[*]/(*" - 1), so that 21 has the basis A,0, • • •, 0"-1)
where 6 is the coset x + (#n — 1). Show that, if a = a<> + a\d -\ h
an-i6n~1, a< e <l>, then the matrix of or relative to the basis A,6, •■-,dn~1) is
the circulant matrix
«o ai • • • «n-i"
»_i ao «i • • an-2
Lai a« «i • • ao
Show that, if $ contains n distinct »-th roots ft of 1, then
N(a) ~ dtt A-iL(Z <*t A
,-_i \ o /
2. Let P 2 * 2 E be finite dimensional extension fields of the field E. Let
a eP and let xn — on*"-1 -\ h ( —l)nan be the minimum polynomial of a
over $ so that B0) defines a matrix representation of $(a)/$. Show that one ob-
obtains a matrix representation of $(a)/E by replacing the entries 0, 1 which
appear by the h X h zero and identity matrices respectively and the a< by the
representing matrices X(a») for a matrix representation of $/E. Use Laplace's
expansion to verify that the determinant of the resulting matrix is
/eW = detX(an).
Since an = iV* (<,)/*(<?), this gives
Next show that N-p\s(i) = A^oj/eW, r = [P:$(«)]. Use these results to prove
C1) for 31 = P.
3. Let 21/* be an algebra with the basis («i, «2, • • •, «n) and let 3£ =
^"(li. £2, • • •, ?n) the field of rational expressions in indeterminates &. Consider
the algebra B1 <8>* 36) over 3£ which has the basis («i, «2, • • •, «n) over 36. Show
that, if X = 53 £•"•> then the characteristic polynomial/xW of AT is a homogene-
1
ous polynomial of degree n in $[x, £1, • • •, f«], #, £i, indeterminates. Use this
and the arithmetic theory of polynomial rings of Vol. I, pp. 124—127, to show that
the minimum polynomial nx{x) of X has the form xm — /(£i, • ■ ■, ifn)* + • • •
+ (— l)m»(£i, •••,£„) where the coefficient of jrm~* is a homogeneous poly-
polynomial of degree * in the £'s. If a = Sa,«j e 21, set MaM = *m — '("i, • • •,
an)xn~l H h (- l)m»(ai, •■-,«„). Prove that Mo(a) = 0.
4. Let the notations be as in 3 and assume 21 = P is a field. Show that nx(x)
is irreducible (Hint: Use ex. 4, § 9); hence show that/x(*) is a power of nx(x).
Show that »(£i, • ■ ■, £„) is irreducible and that the norm form
N(X)

FINITE DIMENSIONAL EXTENSION FIELDS 75
15. Galois cohomology. One is often interested in studying
mappings of the Galois group G of a finite dimensional Galois
extension P/# into P or into the multiplicative group P* of non-
nonzero elements of P. More generally, one encounters mappings of
the product sets G X G, G X G X G, • • • (functions of several
variables in G) into P or P*. A particularly important type of
mapping of G into P*, s —> m« e P*, is one which satisfies Emmy
Noether's equations
F1) M«« = M.V/-
If the n, e#, then m«* = m« and this reads: n,t = m»m« which is just
a character or multiplicative mapping of G into <£>. If G is cyclic
with generatorg: G = {l,gy ■ • •,gn-1},gn = 1, then any element
M e P such that N(n) = mm* • • • M*" * = 1 defines a mapping j —>
/*, satisfying F1) if we define
F2) Ml = 1, Mg = M, M«* = MM*? • • •) M«—i = MM* • • ' M*""*-
Then /*g<+i = mm* • • • Mg> = (mm* • •" /**'"')* M = M«<V« holds for
« = 1, •••,« — 2. Also F1) is clear for / = 1 since mi = 1 and
1 = m«» = l^g'-^^g — Mg • • • M*" V = N{n). Hence F1) holds
for all s = gl and t = g. It is easy to check by induction that it is
valid for all s and all / = g>.
In the general case, if y is any element of P*, we can set m» =
7G*)-1 and we have
Thus m« = 7G*)-1 satisfies Noether's equations for any non-zero
7 in P. We proceed to show that this "trivial" solution of Noe-
Noether's equations is the only possible one, for we have
Theorem 19. Let s —> m« be a mapping of G into P* such that
M«« = Pi'fit, s,teG. Then there exists a non-zero element y in P
such that m« = 7G*) -1.
Proof. Since the m« are ?* 0 and the automorphisms are right
linearly independent over P, we see that the operator 2jm« ( =

76 FINITE DIMENSIONAL EXTENSION FIELDS
is t^ 0. Thus we can find aCeP such that y = /3(Sjju,)
» 9* 0. We now calculate
since j/ ranges over G if s ranges over G. Hence we have 7' =
ynt~l and m« = tCt')" as required.
We have seen that, if G is cyclic with generator g and n is an
element of norm one in P (A^.|*(/x) = 1), then Noether's equations
hold for Hgi = nng • ■ • m**, I < i < n — 1, jui = 1. The theorem
now states that there exists a 7 e P* such that /* = /t« = tCt*)-
This gives the following corollary which is referred to in the
literature as "Hilbert's Satz 90":
Corollary. Let P/$ be a finite dimensional cyclic extension field
and let g be a generator of the Galois group of P over <£. Then any
element /ieP such that iVp|*(ju) = 1 has the form n = 7G*)-1 for
a suitable 7 e P.
The two results which we have just obtained have analogues
for the additive group of the Galois extension P/$. We consider
a mapping s —* 5, of G into P. The additive analogue of Noe-
Noether's equation is:
F3) 5,, = 8,' + 5t, s,teG.
If 7 e P and we set S, = 7 — 7% then S,t = 7 — 7" and 8,' +
St = 7' — 7" + 7 — 7' = 8H; so F3) holds. The direct analogue
of Theorem 19 is valid:
Theorem 20. Let Sa, s e G, be elements of P satisfying F3).
Then there exists a 7 e P such that 8, = 7 — 7".

FINITE DIMENSIONAL EXTENSION FIELDS 77
Proof. We choose an element p e P such that 7p|*(p) = 2p* ?*
0. This can be done since ^ s ^ 0 by the Dedekind inde-
independence theorem. Set7 = ^ T(p)~1S,pa. Then
8tG
y-7' =
which is what we want.
In the cyclic case, G generated by g, the analogue of the condi-
condition iVp|*Gu) = 1 is Tpi*(m) =0. If n is such an element, then
we set Si = 0, 5g.- = m + Mg H h M*'1 and it is easy to check
that F3) holds. We therefore have the following additive analogue
ofHilbert'sSatz90:
Corollary. Let P/# be finite dimensional cyclic, g a generating
automorphism of the Galois group of P/<£>. Then any element y. e P
such that 7p|*(m) = 0 has the form y — ygfor a suitable y e P.
We recall that, if P/$ is finite dimensional Galois with G as the
Galois group, then the set S*(P) of linear transformations of P
as vector space over $ coincides with the set 21 of operators of the
form E sp8 = 2 SP»R (Lemma of § 4). We know also that by
seG s
the Dedekind independence theorem the group elements (s),
s eG, form a basis for 8*(P) as right vector space over P. We shall
now show that Noether's equations arise in considering the follow-
following question: What are the automorphisms of the ring ?*(P) which
leave fixed every element of the subring P« (= IP)? Let A be
such an automorphism and set sA = u,,se G. Then if we apply A

78FINITE DIMENSIONAL EXTENSION FIELDS
to the basic relation prs = s(p')r (eq. B)), we obtain
F4) pRu, = us(p')R.
Hence,
Thus we see that s~1us is an endomorphism of the additive group
of P which commutes with every right multiplication pR. This
implies that s~lus is itself a right multiplication (Vol. I, p. 83).
Hence s~1ua = naR and ua = spa, s eG. We now use the fact that
s —» sA = ua is a homomorphism of G. This implies that u,t =
u,ut, s,teG and so we have st/j,at = (j/x,)(tnt) — J//i«W Hence
the ns e P* (since ua ^ 0) satisfy Noether's equations. Con-
Conversely, it is easy to see by reversing the steps that, if the na ?* 0
satisfy Noether's equations, then the mapping
A: Yi, sp» -*■ H uapa, u, = Sfi,
is an automorphism of 8$(P) which is the identity on Pr. We now
recall that any automorphism of £*(P) which is the identity on <£#
(acting in P) is an inner automorphism (Vol. II, ex. 5, p. 237).
Hence there exists an element C e 8*(P) such that XA = C^XC
holds for all Ze?»(P). In particular, we have pr = pra =
C~1prC for all p e P, that is, C commutes with every pR. This
implies that C = yR, y a non-zero element of P. Then
sn, = a. = sA = C^sC = yn
= -f(G-x)s7)ie,
which implies that n, = 7G*)-1. This gives another proof of
Theorem 19. Of course, this is considerably less elementary than
the first proof. However, the method is useful in related con-
contexts in which the first method is not applicable.
The representation of 8$(P) as 2t = {2jp,} suggests a construc-
construction of a more general kind of ring, called a crossed product of
the field P and its Galois group G. For this purpose we consider
a right vector space SB over P with basis («„) in 1-1 correspondence
s —* u, with the group G. Thus the elements of SB can be written
in one and only one way in the form Y, uapa, pa e P, so that [S3: P]«

FINITE DIMENSIONAL EXTENSION FIELDS 79
= (G:l). We now suppose we have a mapping of G X G into
P* so that for each ordered pair (s, t) of group elements we have a
corresponding ju,t< e P*. We use these to define a multiplication
in 58 according to the formula
««P« )( X) UtGt ) = £ «»ePs
F5)
It is easy to verify that the multiplication is both ways distribu-
distributive relative to addition. Hence 33 will be a ring if and only if the
associative law of multiplication holds. Also because of the dis-
distributive laws it suffices to have (ab)c = a(bc) for a = u,p, b =
ut<r, c = uvt, s,ttve G. Now
(ab)c =
=
a{bc) =
Hence associativity holds if and only if
F6) M«,«V»«,t; = m«,«»m«,«, s,t,veG.
A set of non-zero ne,t, s,teG, satisfying these conditions is
called a (G, P*) factor set. Our argument shows that such a set
defines a ring S8 by means of F5), the associativity conditions cor-
corresponding precisely to the conditions F6). The ring 58 is called
the crossed product of G and P with respect to the factor set ns,t.
We shall write 58 = (G, P, n) to indicate the ingredients G, P and
the factor set n = (/*»,«)•
If we consider again the representation of £*(P) as 21 = {Sjps},
we see that 21 is isomorphic to the crossed product (G, P, 1) where
1 is the factor set fxs,t = 1, s,t eG, 1 the identity of P. This is
clear if we compare F5) with the multiplication of elements of 21.
We now replace the right basis (s) of 21 over P by (us) where u, =
J7»j 7« a non-zero element of P. Then we have
u,ut = (sys)(tyt) = styfyt

80 FINITE DIMENSIONAL EXTENSION FIELDS
Thus we see that 21 is also isomorphic with the crossed product
(G, P, m) where
F7) fi,,t = 7«<~17.<7«-
It is easy to check that these satisfy the factor set conditions but
this is unnecessary since these are equivalent to the associative
law. A factor set m which is obtained from a function s —» y, e P*
by means of F7) is said to be equivalent to I (n ~ 1). The result
we have established is that, if m ~ 1, then (G, P, /*) is isomorphic
to 8*(P). One might be tempted to guess that the analogue to
Theorem 18 is valid for factor sets. However, this is not the case
and we shall indicate this by considering the special case of a
cyclic group.
Let G be cyclic with g as generator and let (G: 1) = n. We set
for 0 < i, j < n — 1,
F8) ixgi<gi = . . .
la^O in <±> it i +j > n.
We have to check the factor set conditions F6). Since 1, a e$
these "simplify to
F9) /V,«'/V+>\«* = /V,«'+*M«',«*-
There are three cases: i + j + k<n, n < i + j + k < In, and
i: + j + k > In. In the first case, both sides reduce to 1. In
the second, both are a; and in the third, both are a2. A crossed
product (G, P, n) where G is cyclic and ju is of the type just de-
defined is called a cyclic algebra or cyclic crossed product. The condi-
condition that n ~ 1 is that there exist non-zero elements yg< such that
JV,« = 7«<+1~17«'g7«. This gives for y = yg> yg* = 77*, ••■,
7g»-i = 77* • • • 7«"~2, a = 7i~17«»-lg7 = 7i~177* • • • 7*" =
7i~WP|*G). Also 1 = /ui.i = 7i~17i17i gives 71 = 1 so we must
have a = JVp|*G). It is easily seen also that this condition implies
that n ~ 1. Thus we see that we can get a factor set m 9^ 1
simply by choosing an a e $ which is not the norm of any element
7 of P. For example, let $ be the field of real numbers and P
the field of complex numbers, P = $>@> i2 = — 1. Then if 7 =
7i + *72, 7i, 72 in $, yl - 7 = 7i - *72 and iVfr) = 712 +
722 ^ 0. Hence if a < 0, then a is not a norm. We remark that

FINITE DIMENSIONAL EXTENSION FIELDS 81
it is easy to see that the cyclic crossed product constructed with
such an a is isomorphic to Hamilton's quaternion algebra over $.
The notions with which we have been dealing are all special
cases of notions in the cohomology theory of groups. We shall
now indicate briefly the general situation. We begin with an
arbitrary group G and the group ring I(G) of G over the integers.
The elements of I(G) are the elements J2 m>s where the m, are
integers and m, t± 0 for a finite subset of G (cf. Vol. I, ex. 2, p. 95).
We consider 2m,s = 2n,s if and only if m, = », for all s. Addi-
Addition in I(G) is by components: 2mas + 2«,j = 2(w, + n,)s.
Multiplication is defined by ( 52 w,j ) I ^2 ntf ) = ]C m»*ttst.
\ kG / \teO / s, teQ
Since G is associative, I(G) is an associative ring. Let 9ft be a
right 7(G)-module so that 9ft is a commutative group under addi-
addition and a product xa, x e 9ft, a e I(G), xa e 9ft is defined so that
(x + y)a — xa + ya, x(a -\- b) = xa + #^, a;(«^) = (xa)b, x\ =
1.
Let Cr(G, 9ft) denote the set of mappings of the r-fold product
GXG X -XG into 9ft. The elements of Cr(G, 9ft) will be
called r-cochains of G relative to the module 9ft. These are the
mappings (su s2, ■■■, sr) -> f(su s2, ■■■, sT) e 9ft, s{ e G. We
make a commutative group out of C(G, 9ft) by defining / + g
in the usual way by (/ + g)(su s2, ■•-, sT) = f(su • • •, sr) +
g(su • • -,Sr). We shall now define a homomorphism d, the co-
boundary operator of C(G, 3ft) into Cr+1(G, 9ft). We do this by
defining 4/for /in Cr = Cr(G, 9ft) by
2, • • ', Jr + l) = /(JS, • • •, Jr + l)
r
G0) + X) (-l)V(Ji> ■••, s<-u SiSi+i, •••,sr+1)
»•—1
It is clear that d{J + g) = df + dg, so d is a homomorphism of
Cr into C+1. Strictly speaking we should denote the d we de-
defined by G0) by dr\ however, it is convenient to use the same nota-
notation for all of these homomorphisms which are defined on Cr, r =
1, 2, • • •. It is convenient to include also the group C° of 0-

82 FINITE DIMENSIONAL EXTENSION FIELDS
cochains which we take to be the module HJJ itself. Then if
x e C° = Wl, dx is the element of C1 such that (dx)(s) = x — xs.
The kernel of d (acting on C) is denoted as Zr and its elements
are called r-cocycles of G relative to the module 9tt. The image
in CT of Cr~l under d is denoted as Br and its elements are called
r-coboundaries. Both Zr and Br are subgroups of Cr and it can
be shown that Zr 2 -Sr. This is equivalent to showing that d2 —
0 for the coboundary operator d. We shall leave the verification
as an exercise (ex. 1 below). The factor group Hr(G, 9JI) =
Zr/Br is called the r-th cohomology group of G relative to the module
5DJ. Here we take r = 0, 1, 2, • • •, and we adopt the convention
that -8° = 0, so H° = Z°, the group of 0-cocycles. The elements
of this group are just the elements x of 9JJ such that xs — x = 0
for all s e G. Evidently these are just the set of invariants of 9K
relative to G.
We shall now show that the notions we have been considering
in this section fit into this general picture. We take G to be the
Galois group of the field P/f> where P/"J> is finite dimensional
Galois. For the module W we take either the multiplicative
group P* of P or the additive group (P, +) of P. In the first
case we make P* a module for I(G) by defining pa, p e P*, a =
m8s to be the element XI (ps)m*- Since G is a finite group
t?
there is no difficulty in defining this product. It is trivial to
check the module axioms and we leave this to the reader. A 1-
cochain s —> jus = n(s) e P* is a cocycle if and only if (dn)(s, t) =
M«M«(~Vs' = 1 (the 0 of P*). This is equivalent to nst = ntH*'
which are Emmy Noether's equations F1). If y e P*, then y is
a 0-cochain and its coboundary is the 1-cochain f(s) = 7G*) ~1.
Theorem 19 can now be re-interpreted as the statement that
every 1-cocycle of G relative to P* is a coboundary. In other
words, Z1/B1 = 1, or the first cohomology group of G relative #0
P* is the identity.
If (j, i) —> n^t is a 2-cochain, then the coboundary definition
gives
(dn)(s,t,u) = M«,«Ms«,«~V»,(«(Ms,«u)~1-
It follows that juSt( is a 2-cocycle if and only if /*„««/*«,« =
M8«,uM«,e"> s, t, u eG and these are just the conditions F6) defin-

FINITE DIMENSIONAL EXTENSION FIELDS 83
ing a factor set. Thus the 2-cocycles are just the factor sets. If
s —> 7, is a 1-cochain, its coboundary dy is given by (dy)(s, t) =
yt(y>t)~1y*t' Hence 2-coboundaries are just the factor sets
equivalent to 1. The general considerations imply that the set of
factor sets form a group under multiplication: (jw)«,t = ix,,tv,,t.
The factor sets equivalent to 1 form a subgroup and the second co-
homology group H2(G, P*) is the factor group of the first of these
groups relative to the second. As we have seen, in general the co-
homology group H°(G, P*) is the set of G-invariants of P*. Thus
this is the multiplicative group $* of the subfield $>.
An entirely analogous discussion can be made for the additive
group (P, +) considered as an /(G)-module by means of the
definition pa = ^ m»p'- It is easily seen that Theorem 20 states
seQ
that the first cohomology group of G relative to (P, +) is 0. It can be
shown that if the characteristic of P is either 0 or not a divisor of
the order n of G, then all the cohomology groups Hr(G, P) = 0,
r > 1. This is an immediate consequence of ex. 2 below.
EXERCISES
1. Prove tP = 0.
2. Let G be a finite group of order n and let 9JJ be a module for I(G) which is
uniquely n-divisible in the sense that for any y e 9J? there exists a unique x
(written as — y) such that nx = y. Prove that the groups HT(G, 9J}) = 0 for
n /
r> 1.
3. Let P/<£ be a cyclic field extension, [P:$] = n, r a divisor of n, y a non-zero
element of $ such that yT = 7Vp|*G>), p e P. Prove that y = Ne\*(v) where
E/$ is the (unique) subfield of P/$ such that [P:E] = r andijeE. (Hint: Set
n = mr and consider the element /8 = pp' • • • />**" \ Show that A^pieOS) = 7r
and apply Hilbert's Satz 90 to P~*y)
16. Composites of fields. In this section we consider a prob-
problem which can be formulated roughly in the following manner.
Given two extension fields E and P over #, to determine the ways
these can be put together to form another extension field of $.
More precisely we seek to determine the composite fields of E and
P over $ in the following precise sense.
Definition 3. Let E and P be two fields over i>. Then a composite
field of E/$ and P/<£ is a triple (T, s, t) where T is afield over * and
s and t are isomorphisms of E/<£ and P/# respectively into r/$ such

84 FINITE DIMENSIONAL EXTENSION FIELDS
that F is generated as a field by the images Es and P'. The com-
composites (F, s, i) and (F', s', t') of E/«l> and P/<i> are equivalent if there
exists an isomorphism u of r/<i> onto T'/$ such that su = j' and tu =
t'.
The problem is to determine the equivalence classes of com-
composites. We shall consider this question now under the assump-
assumption that one of the fields, say P, is finite dimensional over $. In
Chapter IV (§ 10) we shall investigate the problem for infinite
dimensional extensions.
Suppose (F, s, t) is a field composite of E/<i> and P/<£> where
[P:<J>] = n < oo. We consider the subset
Clearly this is the subalgebra of F/$ generated by the two sub-
algebras Ey<£> and P'/$. Also it is immediate that, if (pi, p2, • • ■,
Pn) is a basis for P/$, then ESP( = EV + EV H h E V,
the set of Es-linear combinations of the p/, 1 < i < n. Since F
and hence E"P' is commutative, ESP( is an algebra over Es and
[E"P':ES] < « < oo. Since E'P' is contained in a field, it has no
zero divisors; hence by VII of the Introduction, ESP( is a field.
Since F is the subfield of F generated by Es and Pf, we see that
F = ESP(. This important relation leads us to look at the tensor
product algebra E ®* P whose elements we indicate in the
original notation: 2e< <g> p,-. The basic property of the tensor
product is that the mapping 2e* ® pt- —* Se,-*p»* is a homomor-
phism of E ®* P onto F = E'P(. If 3 is the kernel of this homo-
morphism, then F ^ (E <8>4, P)/3. Since F is a field, this implies
that 3 is a maximal ideal: 3 is a proper subset of E ® P and there
exists no ideal 3*' such that E ® P D 3' 3 3. Conversely, if 3
is a maximal ideal in E <g> P, then F = (E ® P)/3 ^ 0 and this
has no ideals ^ 0, E ® P. Hence F is a field (Vol. I, p. 77). We
can now state the following result.
Theorem 21. Let E/<i> and P/<i> be fields such that [P:$] < oo
and let 3 be a maximal ideal in E <g># P. Let s be the mapping e —*
e ® 1 + 3 of E into F = (E ® P)/3 and let t be the mapping
p—> 1 ® p + 3 c/ P into F. Then (F, s, t) is a field composite

FINITE DIMENSIONAL EXTENSION FIELDS 85
of E/$ and P/<£. Distinct maximal ideals S, 3f' /» E ® P give
rise in this way to inequivalent composites. Moreover, every field
composite of E/<£ and P/<i> is equivalent to one of the (P, s, t) given by
a maximal ideal 3f *'» E ® P.
Proof. If 3 is a maximal ideal in E ® P, then e —> e ® 1 is a
homomorphism into E ® P so j:e—> e ® 1 + 3 is a homo-
morphism into r = (E ® P)/3f. Since 1 -> 1 + 3 and E is a
field, s is an isomorphism. Similarly f.p —■» 1 <g> p + $ is an iso-
isomorphism of P/$ into F. Any element of P has the form 2et- ®
Pl- + 3 and ct- ® Pi + 3 = («,■ ® 1 + 30A ® Pi + 3) = e,-V;
hence P is generated by Es and P(. Also P is a field since 3 is
maximal. Hence (P, s, t) is a composite. Next let 3 and 3' be
two maximal ideals, (P, s, t), (P, s', t') the associated composites
and assume that there exists an isomorphism u of P/$ onto P'/$
such that s' = su, t' = tu. Let 2et- ® p,- e 3. Then the defini-
definitions of s, t give the relation 2ei*p.-' = 0 in P. Applying u we ob-
obtain Se/'pt*' = 0 which means that Se,- ® pt e 3'- Thus we see
that 3 C 3'. Since 3f is maximal we have 3 = 3'. We have
therefore proved that, if the composites (P, s, i), (P, s', /') are
equivalent, then 3 = &'• Finally, let (P', s', t') be a composite of
E/4? and P/$ constructed in any way. We have seen that the
mapping Se, ® pt- —* Se.-'p/ is a homomorphism of E ® P onto
P' whose kernel 3isa maximal ideal in E ® P. We have the in-
induced isomorphism u: 2et- ® Pi + 3 -» 2«,spi' of P = (E ® P)/3
onto P'. One checks that this is an equivalence of the composite
(P, s, t) defined by 3 with (P', s', t'). This completes the proof.
We have now established a bijection of the collection of equiva-
equivalence classes of composites with the collection {3} of maximal
ideals in the tensor product E ®* P. Since E ®* P can be con-
considered as a finite dimensional algebra over E (Introduction), the
following result implies that there are only a finite number of
equivalence classes of composites of E/$ and P/$.
Theorem 22. A finite dimensional algebra with an identity ele-
element has only a finite number of distinct maximal ideals. If these
h
are <h, 32, • • •, 3h and 91 = fl 3y, then I = «/8t S Pi ® P2 0
■■■QTh where T, =

86 FINITE DIMENSIONAL EXTENSION FIELDS
Proof. The direct sum I\ © r2 ©• • •© Th is just the set of h-
tuples (yi, 72> • • ■> 7a)> 7» e F,-, where equality is defined by
equality of components and addition and multiplication are also
by components. Evidently the dimensionality of the direct sum
is the sum of the dimensionalities of the T}. Now let 3i, • • •, 3a
be any distinct maximal ideals and let 33 = Fx ©F2©- • •© Th,
Tj = 21/3,. We define a homomorphism of 21 into 99 by mapping
a -» (a + 3i, a + 52, •••,« + 3a)- The fact that this map-
mapping is a homomorphism is immediate. The kernel 9i of this
homomorphism is the set of elements a such that a + 3;; = 3/
for every _/'. Hence 5R = fl 3/- We shall now show that the
i
homomorphism is surjective. We show first that 3i + S2S3 ■ • • 3a
= 21. Since the 3,- are distinct maximal ideals, 21 = 3i + 32
and 21 = 3i + 33- Multiplication gives 21 = 2l2 = 3i2 +3i3s +
323i + 323a = 3i + 323s- Now suppose we already have 21 =
31 + 32 • • • 3*- Since 21 = 3X + 3t+i a similar multiplication
gives 21 = 3i + 323s • • • 3fc+i. Hence we have 21 = 3i +
32 • • • 3a and this implies that 21 = 3i + C2 0 ■ • • fl 3a) since
32 ■ ■ • 3a Q 32 n ■ ■ • fi 3A. If a is any element of 21, then our
relation shows that a = b + c where b e 3i, c e 32 fl • • • fl 3a-
Hence the image of c in our homomorphism is (c + 3i> c + 32»
• • •, c + 3a) = (a + 3i, 32, • ■ ■, 3a), which shows that, if 71 is
any element of r1} then the element G1,0, • • •, 0) is in the image
of the homomorphism. ki a similar fashion, if 7, is any element
of rt, then @, • • • 0, 7,-,0, • • • 0) is in the image. Addition shows
that any element G1, 72, • ■ •, 7a) is in the image so the homo-
homomorphism is surjective. It is now clear that, if 3i» 32» • • •> 3a
are distinct maximal ideals, then the dimensionality [21:*] >
A
£ [iy*] where Ty = a/3/. Since every [iy*] > 0, this of
1
course puts a bound on the number of 3/- Also we have seen
that, if 3i> 32> • • ■, 3a are distinct maximal ideals and dt = f) 3a,
then 21/9? ^r!©---©rA.
We shall now obtain more precise information on composites
under the assumption that P = $(8) is a simple algebraic exten-
extension of <i>. Let f(x) be the minimal polynomial of 0 over €>. Then
A, 6, e2, ■ ■ •, 0"-1) is a basis for P/$ and Bn = a0 + ax0 H h

FINITE DIMENSIONAL EXTENSION FIELDS 87
an-i0n~l, if f(x) = xn — an-iXri~1 — • • ■ — a0. Now consider
E ®* P. The elements of this algebra can be written in one
n-l
and only one way as 2 e« ® 0*> «» e E. We consider E ®* P
o
as an algebra over E by defining ^(Se,- ® 9*) = S^e,- <g> 0*, t; e E (cf.
Introd.). Then it is clear that 1 ® 6 is a generator of E ®# P
over E and the minimum polynomial over E of this element is f(x).
Thus we see that E ®* P ^ E[x]/(/(x)). The ideals of this alge-
algebra have the form (/>(*) )/(/(*)) where p(x) is a divisor of fix)
and such an ideal is maximal if and only if p{x) is irreducible.
Then the difference algebra (E[*]/(/(*))/((p(x)/(/(*))) is iso-
morphic to the field E[x]/(p(x)).
Suppose finally that P is a finite dimensional separable exten-
extension of <£>. Then we know that P = $@) where the minimum poly-
polynomial f(x) of 6 is irreducible and separable. The derivative
criterion shows that in E[x] we have the factorization f(x) =
pi(x)p2(x) • • • ph(x) where pj{x) is irreducible of positive degree
and pi{x) 7* pj(x) if i ^ j. Thus we see that we have h inequiva-
lent field composites of P and E over <i>. These have the form
(Fh sh t,) where T, ^ E[#]/(py(*)). Also by ex. 1, § 2 of Introd.,
h
E <g>* P s* E[x]/(f(x)) « Tx © T2 ©• • ■© Th and £ [r,:E] =
i
[P:#]. We state this result as
Theorem 23. Let P/<£> ^«? finite dimensional separable and let
i? «« arbitrary extension field. If 6 is a primitive element of P
(*) its minimum polynomial over <!>, then the field composites
(F, s, i) are in 1-1 correspondence with the irreducible factors p{x)
of fix) in E[x]. If iTj,Sj,tj), j = 1, • ■ •, h are the inequivalent
h
composites of P/$ and E/*, M<?w [P:4>] = X) [r^E].
7=1
EXERCISES
1. Show that, if P/$ is finite dimensional Galois, then there are n = [P:$] in-
equivalent composites of P with itself and, if (F, s, t) is one of these, then T =
P« = P(. Use this to prove that P®*P£ P<1> © P<2> ©• • •© P(n) where
p(«) ^ p_
2. Let P be finite dimensional Galois over $ and let E be a subfield of P over
$. Show that P <g>* E ^ P'1' © P<2> © • • • © P<m) where Pw ^ P and » =
[E:*J.

FINITE DIMENSIONAL EXTENSION FIELDS
3. Let P/$ be finite dimensional separable. Show that P/$ and E/<i> have
only one (in the sense of equivalence) composite if either 1) E/f> is purely in-
inseparable, or 2) E = $(£1, £2, • • •, £„) the field of rational expressions in indeter-
minates £».
4. Define a composite (F, j 1, s%, • • •, •*>•) of r extension fields Pi/$, 1 < i < r, as
a field F/$ and isomorphisms j,- of Pj into F such that F is generated by the sub-
fields PA Call two such composites (F, su •••,Jr), (F', j/, •••,*/) of P<
equivalent if there exists an isomorphism a of F/$ into F'/$ such that V = Silt,
1 < » < r. Assume every P,/$ is finite dimensional and prove that Theorem 21
generalizes to composites (F, s\, ■ ■ ■, sr) and the tensor product Pi <S> P2 ® • ■ ■ ®
P,.
5. Let P/$ be finite dimensional Galois and let (si, J2, • • •, sT) be an ordered
r-tuple of automorphisms of P/<l>. Then (P, Ji, S2, ■ • •, sT) is an r-fold composite
of P/$. Show that (si, ■ ■ •, sr), (si, ■ ■ -, j/) determine equivalent composites if
and only if s,' = S{U is an automorphism of P/<J>. Let !$(si, ■ ■ ■, sr) be the ideal
in P (g> P ® ■ ■ ■ <8> P (r factors) associated with (P, si, ■ ■ ■, jr). Use the fact
that there are [P:*]' distinct ideals Q(su ■ ■ ■, sr) and that
(P®---® P)/3(xi,--.,j,)^P
to prove that the 3i(ji, • • •, sr) are the only maximal ideals in P(r) = P(S> • • ■ ®P
and that every r-fold composite of P/$ is equivalent to one of the composites
(P, si, S2, • • •, Jr)- Note that $(su '' '> sr) is the kernel of the homomorphism
of P(r)/<i> into P/$ such that
Pi <8> P2 <8> ■ • • ® Pr —» Pl*^*2 • • • Pr*r-

Chapter II
GALOIS THEORY OF EQUATIONS
In this chapter we shall consider the classical application of
Galois theory: Galois' criterion for solvability by radicals of a
polynomial equation f{x) = 0. To say that an equation is solv-
solvable by radicals means roughly that its roots can be obtained
from the coefficients by rational operations and root extractions.
A criterion for this was given by Galois after Abel and Ruffini had
proved that the general equation of the fifth degree is not solvable
by radicals. Galois was led to the development of his theory in
order to give the criterion for solvability by radicals. Besides the
fundamental group-field correspondence which we gave in the last
chapter and whose scope goes far beyond the theory of equations,
some results of a more special nature are needed. These concern
cyclotomic fields, that is, fields of the roots of 1, and "pure" ex-
extensions P = $(9), 6n = a in <i>. The study of these fields is
interesting also beyond the theory of equations and we shall under-
undertake a detailed study of such fields in the next chapter. In the
present one we confine ourselves to the minimum which is needed
for the theory of equations.
1. The Galois group of an equation. Let <£ be a field, f{x) a poly-
polynomial of positive degree in <i>[x] having leading coefficient 1. Let
P/<1> be a splitting field off(x), so P = $(/>i, • • •, pm) and/(#) =
(x — Pi)Cl(x — p2N2 • • • (x — pm)em in PM where the pi are dis-
distinct and the <?,• are positive integers. Since the p; are generators
of P/$>, any automorphism s of P/4> is completely determined by
its action on the finite set of roots R = {pi, p2, • • ■, pm\- Also
each pi is again a root off(x), so p/ e R and the restriction Sf of s
to R is a permutation of this finite set. Thus we see that every s
89

90 GALOIS THEORY OF EQUATIONS
of the Galois group G of P/$ defines a permutation sf of R. The
mapping s —> sf is a homomorphism of G into the symmetric
group S(R) of 1-1 mappings of R. Moreover, if s e G has the
property that p,* = p,-, 1 < i < m, then j = 1 in P = $(pi, • • •,
pm). Consequently, s —* Sf is an isomorphism of G with a sub-
subgroup Gf = {j/} of the symmetric group <S"(i?). In view of this
isomorphism we are led, in studying the equation f(x) = 0, to
shift our attention from the group G to the permutation group
Gf. Accordingly, we give the following
Definition 1. If $ is afield andf(x) is a non-zero polynomial in
<£>[#], then the Galois group of the equation/(*•) = 0 over $ is the
group Gf induced by the Galois group G ofa splitting field P/$ in the
set of roots of fix) = 0 in P.
Since any two splitting fields are isomorphic, Gf is essentially
uniquely determined by $ and/(#).
It is convenient to identify the permutation p,- —> p/ of R with
the permutation / —» V of {1, 2, • • •, m\. In this way we can
consider Gf as a subgroup of the symmetric group Sm of per-
permutations of {1, 2, • ■ •, m}. We shall do this from now on.
Moreover, we assume in the sequel that/(#) has simple roots, that
is, the <?,- = 1. This implies that P/# is Galois. Hence we have
the fundamental correspondence between the collection of sub-
subgroups of the Galois group G and the collection of subfields E/$ of
P/<i». Combining this with the isomorphism of G onto Gf we ob-
obtain a 1-1 correspondence between the collection of subgroups of
Gf with the collection of subfields E/$. We shall refer to the sub-
field E/<$ corresponding to a subgroup Hf of Gf as the "field of in-
invariants ofHf." In reality, of course, E/# is the field of invariants
of the subgroup H of G corresponding to Hf. In the other direc-
direction, Hf is the Galois group oif{x) = 0 over the subfield E.
We recall that the symmetric group Sm contains the alternating
group Am as an invariant subgroup of index 2. Am is the set of
even permutations, that is, the set of permutations which can be
written as products of an even number of transpositions (if) (Vol.
I, pp. 35-36). If Gf is the Galois group of the equation/(*) = 0
over <£, then Gf fl Am is a subgroup of index 1 or 2 in Gf. We shall
now give an identification of the corresponding subfield of P/*,

GALOIS THEORY OF EQUATIONS 91
assuming the characteristic is not two (see ex. 1 below for the
characteristic 2 case). The result is the following
Theorem 1. Let $ be afield of characteristic ?* 2 andf{x) a non-
nonzero polynomial e $[x] without multiple roots. Let P/$ be a splitting
field o//(x), pi, p2, • • •, pm its roots, Gf the Galois group of the equa-
equation fix) = 0 considered as a permutation group of {1, 2, • • -,m}.
Then the subfield of invariants of Gf 0 Am is $(A), where
(i) a = n (p.- - pd.
Proof. We recall a standard characterization of the alternating
group. For this one considers the ring $[#i, x2y • • •, xm], #,• in-
determinates. If / —> /' is a permutation of 1, 2, • • ,m, then we
have the automorphism A(o) of $[xu • • •, xm] over <£> such that
XM«) = Xf (Vol. I, p. 107, and Introd.). Let X = ]J (x{ - #,).
Then XAM = x(v)X where x(c) = 1 or — 1 according as a is even
or odd (cf. Vol. I, ex. 2, p. 110). Now let ir be the homomorphism
of $[#i, x2, • • •, xm] over * into P/$ such that x* = pi} 1 < / <
m. Let s be in the Galois group of P/f>, sf the corresponding per-
permutation of the pi. Then if we apply x to the relation XM*f) =
x(s/)X we obtain A* = x(J/)A where A is given by A). Since
A ^ 0 we see that A* = A if and only if sf e Gf 0 An. Thus the
Galois group of the equation f(x) = 0 over $(A) is Gf C\ Am.
Consequently, by the Galois correspondence, *(A) is the set of
invariants of Gf D Am.
We have seen that for any s e G, A" = ±A; hence if 5 = A2,
then 8" = 5 for all s and so S e <£. We have seen also that the
statement of the theorem is equivalent to the assertion that the
Galois group of the equation f(x) = 0 over #(A) is Am D G.
This implies the
Corollary. The Galois group off(x) = 0 over $ is a subgroup of
the alternating group if and only if 5 is the square of an element of 4>.
Proof. Clearly the condition Gf C Am or Gf = Gf D Am is
that #(A) = $. If this holds, then Ae$ and 5 = A2 is a square
of an element of f>. Conversely, if 5 = a2, a e #, then 8 = A2
gives A = iac$. Hence #(A) = <£>.

92
GALOIS THEORY OF EQUATIONS
We recall that, if 6 is a separable algebraic element over $ and
0x = 0, 02, • • •, 8n are the distinct images of 0 under the isomor-
isomorphisms of $@) into its normal closure, then 5 = JJ @t- — 0yJ
is a discriminant of the field #(»)/* (cf. § 1.14).* If/(*) =
(x — pi)(x — p2) • ■ ■ (x — pm), as before, then we shall call 5 = A2
= JJ (p{ — pjJ the discriminant of the polynomial f(x) or of the
equation f(x) — 0. It follows from the main theorem on symmetric
polynomials (Vol. I, p. 109) that 5 can be expressed as a poly-
polynomial with coefficients in the prime field in the coefficients of
B)
a2xm~2
We shall now indicate how this can be done. We begin with the
Vandermonde formula:
C)
1
Pi
1
P2
1
Pn
m—1
= n (p< - Pi).
Squaring we get
D)
8 =
m
<T2
O'm+l
<Jm—\
02m— 2
where <n = pi1 + p2l + • ■ ■ + Pm- Since the power-sums can be
expressed as polynomials in the at with coefficients in the prime
field, D) will give the same kind of expression for 5.*
We shall now carry this out for the cases m = 2, 3.
m = 1. We have/(,v) = x2 — a±x + a2 = (x — pi)(x — p2) so
o"i = Pi + P2 = «i and pip2 = «2- Then <r2 = pi2 + P22
(pi + P2J — 2pjp2 = «i2 — 2a2. The formula D) gives
22 =
E)
5 =
* That this can be done follows from the fundamental theorem on symmetric poly-
polynomials (Vol. I, p. 109). Explicit recursion formulas, the so-called Newton's identities
can be used to express the <r< in terms of a,- (Vol. I, ex. 4, p. 110).

GALOIS THEORY OF EQUATIONS 93
m = 3. Here f{x) = x3 — ctix2 + a2x — a3 = (x — p\){x —
P2)(X — P3) SO ffi = Pi + p2 + p3 = «1> PlP2 + P2P3 + PlP3 = OC2,
P1P2P3 = «3- Then <r2 = Pi2 + p22 + P32 = (pi + P2 + P3J —
2(piP2 + P1P3 + P2P3) = ai2 — 2a2. To calculate <r3 and <r4 we
use the relations pk3 = a\pk2 — a2pk + oc3, pk* = axpk3 — a2pk2 +
«3Pfc. Then
<r3 = Pi3 + P23 + Ps3
= «i(pi2 + P22 + P32) — «2(pi + P2 + P3) + 3a3
— 2a2) —
0-4 = «iO-3 — a2<r2
= ai(«i3 — 3«ia2 + 3a3) — a2(«i2 — 2a2)
= «i4 — 4«i2a2 + 4«ia3 + 2a22.
Using D) and these formulas we obtain
F) 8 = 3G2<T4 + 2<ri<T2<T3 — <T23 — 3<T32 — (Ti20
— 4a23 — 27a32.
W.e obtain next a criterion on Gf as permutation group of
{1, 2, • ■ •, m} that f(x) be irreducible in *[#]. This is the following
Theorem 2. Letf{x) e *[x] A«y^ wo multiple roots in its splitting
field P. Then f(x) is irreducible in <i>[.v] if and only if the Galois
group Gf of f{x) = 0 over $ is a transitive permutation group.
Proof. We recall that a transformation group of a set M is
called transitive if given any pair (x, y), x,y e M there exists a a
in the group such that x" = y. Suppose first that/(,v) is irreduci-
irreducible in $[x] and let pi, p2 be two of its roots in P. Since f(x) is ir-
irreducible and/(pi) = 0 =/(p2)> there exists an isomorphism of
3>(pi)/$ onto $(p2)/$ mapping px on p2. This isomorphism can
be extended to an automorphism s of P/$. Then s e G and
Pi' = P2- This implies that Gf is transitive. Conversely, suppose
Gf is transitive. Let/i(#) be an irreducible factor of f(x) of
positive degree and let pi be one of its roots. Let p2 be any root of
f(x). Then there exists an s e G such that pi* = p2. Then

94 GALOIS THEORY OF EQUATIONS
fiiPz) =/i(pi*) =/i(pi)* = 0. This shows that every root of
fix) is a root of/i(x). Hence/(x) = /i(x) is irreducible.
The two results which we have derived make it trivial to calcu-
calculate the Galois groups of quadratic and cubic equations. Similar
ideas can be applied to quarries. We shall look at the first two
cases now and will indicate how quarries can be handled in the ex-
exercises which follow. We assume that the characteristic of * is
not 2 and that/(x) has distinct roots. If f(x) is a quadratic:
f{x) = x2 — aix + a2, then the group is the symmetric group S2
or A2 = 1 according as 5 = «i2 — 4a2 is not or is a square in $.
Next let f(x) = x3 — aix + a2x — a3. If f{x) = (x — p)g(x) in
$[x], then the Galois group of/(x) = 0 is the same as that of the
quadratic g(x). Hence we may assume f{x) irreducible in #[#].
Since the only transitive subgroups of S3 are Ss and A3i the Galois
group Gf is one of these. The corollary to Theorem 1 shows that
Gf = Az if 5 = —4«i3a3 + ai2a22 + 18aia2a3 — 4a23 — 27a32 is
a square in <£>. Otherwise, Gf = S3.
EXERCISES
1. Assume $ has characteristic 2 and/Cx) e <£[*] has distinct roots pi, pi, • ••,Pm
in its splitting field P. Let A' = X) pj»B~1P2'n~2 • • • Pfr-ty- Show that the
subfield of invariants of Gf C\ Am is
2. Let $ be a finite field, /(#) an irreducible polynomial of »-th degree with co-
coefficients in <£. Show that Gf consists of the powers of an »-cycle which may be
taken to be A23 •••«).
In the remainder of the exercises we assume the characteristic of the base field
$ is 5^ 2, /(#) = x* — aix* + atx2 — a^x + cm has distinct roots pi, pi, ps, p\ in
the splitting field P/$, G the Galois group of P/<£.
3. Show that the subgroup V (Klein's Vierergruppe) = {1, A2)C4), A3)B4),
A4)B3)) is invariant in S*.
4. Show that the subfield of invariants relative to G/D V is $(ti, n, tj) where
Tl = PlP2 + P8P4, T2 = plP3 + P2P4, T* = pjfii + pip3.
5. Let g(x) = (x — Ti)(x — Tt)(x — 73) (the resolvent cubic of/(#)). Verify
that
G) g(x) =xi- ft*2 + fae - ^
where
(81 = as
(8) & = «ia8 - 4at
0s = ai2a4 + as2 —
and tha.tg(x) and/(*) have the same discriminant.

GALOIS THEORY OF EQUATIONS 95
6. Prove that the transitive subgroups of <?4 are (i).?«, (ii) //4, (iii) V, (iv) C =
{1, A234), A3)B4), A432)) and its conjugates, (v) D = V U {A2), C4), A423),
A324)) a Sylow 2-group (subgroup of order 8) and its conjugates.
7. Show that the Galois group Gg of g(x) = 0 is isomorphic to G//(Gf D V).
Assume /(#) is irreducible and verify that, if-(i) G/ = Siy then Ge is of order 6, (ii)
Gf = Aiy Gg is of order 3, (iii) Gf = V, Gg = 1, (iv) G/ = C or one of the con-
conjugates (that is, any cyclic subgroup of order 4 of St), then G4 is of order 2, (v)
G/ = D or one of its conjugates (any Sylow subgroup of order 8 in S4), then G£ is
of order 2. Note that these results identify Gf if we know G, unless G/ is either as
in (iv) or (v).
8. Prove that, if G, is of order 2, then G/S D or G/^ C according as/(#) is or
is not irreducible in $(V?) where 8 is the discriminant of/(*).
9. Determine the Galois group of x* + 3x* — 3* — 2 = 0 over the field of
rational numbers.
2. Pure equations. In this section we shall derive the special
results which are needed for Galois' criterion. We shall formulate
these in the invariant fashion in terms of splitting fields rather
than, as in the last section, of the groups of equations. The re-
results we need concern equations of the form xn — a = 0 (or xn =
a) which are called pure (or binomial) equations. Occasionally,
we use the notation p = Wo. or p = a1/n to indicate that p is a
root of xn = a. We consider first the case a = 1. The roots of
xn = 1 are called the »-th roots of 1 and a splitting field P of this
equation is called a cyclotomic field of order n over *. The
derivative (xn — 1)' = nxn~x is not relatively prime to xn — 1 if
and only if the characteristic is p j± 0 and p \ n. Then we can write
« = p'nr, (n',p) = 1 and we have xn - 1 = (xn> - l)p'. Hence
the cyclotomic field of order n coincides with that of order »'.
We shall therefore assume from now on that p X n m tne char-
characteristic p j* 0 case.
Let P/$ be a cyclotomic field of order n over $. Because of
our assumption on the characteristic, the set Z{n) = {ft-} of »-th
roots of 1 contains n elements. If m \ n, then i\m = 1 implies
r}n = 1; hence the cyclotomic field of order n contains that of
order m for every divisor m of n. We observe next that Z(») is a
subgroup of the multiplicative semigroup of P. This is clear since
Tin = 1 = r2n imply (^2)" = 1, (ri)" = 1- Since Z(«) is
finite, this is a cyclic group (Lemma 1, § 1.13). Hence there
exists a feZ(») such that Z(») = {f'|i = 0, 1, •••,» — 1}.
Such a f is called a primitive n-th root of 1. Since P/<i> is generated

96 GALOIS THEORY OF EQUATIONS
by the f,-, we have P = $(f) so f is a primitive element of the field
P/<£. We shall now prove
Theorem 3. If the characteristic of $ is not a divisor of n @ in-
included), then the Galois group G of the cyclotomic field P/<l> of order
n is isomorphic to a subgroup of the multiplicative group U(n) of
units in I/(n), I the ring of integers.
Proof. As in § 1 let G/ denote the group of permutations of the
set Z(«) of roots induced by G. Since the elements of G/ are
restrictions of automorphisms, it is clear that they are automor-
automorphisms of the multiplicative group of Z(n). Hence G/ (^ G) is
isomorphic to a subgroup of the group of automorphisms of Z(«).
Now Z(«) is a cyclic group of order « and it is well known that the
group of automorphisms of such a group is isomorphic to U{n)
(Vol. I, ex. 3, p. 47, and ex. 1, p. 82). Hence the Galois group G
is isomorphic to a subgroup of U(n).
It is important to note that G is commutative since U(n) is
commutative. Moreover, we observe that, if / is a prime, then
U(l) is just the multiplicative group (of order / — 1) of the field
//(/) and this is a cyclic group. Hence G which is isomorphic to a
subgroup of £/(/) is cyclic. We therefore have the
Corollary. If the notation is as in Theorem 3, then G is a com-
commutative group and G is cyclic if n is a prime.
Next we consider the Galois group of any pure equation xn = a
under the assumption that the base field <3? contains n distinct
w-th roots of 1. We have seen that this implies that the charac-
characteristic is not a divisor of n. Then xn — a is prime to its deriva-
derivative nxn~1 if a 5^ 0 (the case a = 0 is trivial) and so xn — a has n
distinct roots. We have the following
Theorem 4. If <3? contains n distinct n-th roots of 1 then the
Galois group of the equation xn = a over $ is cyclic of order a divisor
of n.
Proof. Let P/<i> be a splitting field over * of xn — a, G its
Galois group. We have to show that G is cyclic. If a = 0, we
have P = <i>, G = 1. Hence we assume a ^ 0. Let p be one of
the roots of xn — a in P. If Z(») = {^ = 1, f2, • • •, fn} is the
set of »-th roots of 1 contained in *, then we know that Z(«) is a

GALOIS THEORY OF EQUATIONS 97
cyclic group under multiplication and, since p ?^ 0, it is clear that
{pfi> Pf2> • • j Pfn} is the set of roots of xn — a. Evidently P =
$(p); so an automorphism s e G is completely determined by its
effect on p. We have ps = UwP where f t(s) is one of the f's e Z{n)
and is uniquely determined by s. If / e G and pl = UwP> then
This shows that the mapping j —> f,-(,) is a homomorphism of G
into Z(n). If f,-(«) = 15 we have p* = p so s = 1. Hence j—>
ft-(j) is an isomorphism. Thus G is isomorphic to a subgroup of
the cyclic group Z(n) and the result is clear.
We shall need one more special result for the proof of Galois'
criterion. This is the following converse of Theorem 4.
Theorem 5. Assume $ has n distinct n-th roots of 1 and let P/#
be a cyclic n dimensional extension field. Then P = <£(£) where
£" = a e $.
Proof. The hypothesis on P is that P/f> is Galois with Galois
group G which is cyclic of order n. Since P is separable over $ it
has a primitive element so P = $F). Let j be a generator of G
and let J be the "Lagrange resolvent":
(9) £ = $ + 6*rl + 0s'r~2 H r- fl'"-1^"-1)
where f is a primitive »-th root of 1. Then
F = es + e*^-1 -\ h flf-<»-«
Then J** = f*£ so J has » distinct conjugates and hence its mini-
minimum polynomial is of degree n. Consequently P = $>(£). Set
f1 = a. Then (£")s = (££)" = T implies that a' = asoae$ and
the proof is complete.
EXERCISES
1. Let p be a prime not equal to the characteristic of the field $. Show that, if
«e$, then xp — a is irreducible in $[*] or it has a root in this field.
2. Let $ be the field of rational numbers, p a prime, and let xp — a be ir-
irreducible in $[#]. Show that the Galois group of the equation xp = a over $ is
isomorphic to the group of transformations in I/(j>) which have the form y —*

98 GALOIS THEORY OF EQUATIONS
3. Let $ be a field of characteristic p ?* 0. Show that xp — x — a is ir-
irreducible in "I'M unless a = /8P — /3, /3 in $. Show also that, if xp — x — a is
irreducible, then the group of the equation xp = x + a is cyclic of order p.
(Hint: Show that, if p is a root of xp — x — a = 0, then p, p + 1, p + 2, ■ • •,
P + (p — l)are roots. Hence show that the Galois group of the equation is iso-
morphic to a subgroup of the additive group of //(/>).)
4. Let $ be of characteristic p j± 0, P/$ cyclic and/> dimensional. Show that
P can be generated over $ by an element £ such that f — £ = a e $.
3. Galois' criterion for solvability by radicals. It is essential
first to have a precise formulation of the statement that an equa-
equation/(#) = 0 is solvable by radicals over a field <I>. We give this in
the following
Definition 2. Let <i> be afield and letf{x) e <£[*] be of positive de-
degree. Then the equation f(x) = 0 is said to be solvable by radicals
over $ // the splitting field P/$ can be imbedded in afield S which
possesses a tower of sub fields:
A0) * = #i C *2 c #3 C • • • C $r+1 = S
where each f>l+i = <£,•(£,) ««^ £»•"' = a,- e *,-. ^ chain of fields such
as B) is called a root tower for 2/$.
For the sake of simplicity we restrict our attention to fields of
characteristic 0. This will avoid the complications of insepara-
inseparability and some difficulties with roots of 1 in the characteristic
p t* 0 case. Our objective is to establish the following criterion
of Galois:
An equation fix) = 0 is solvable by radicals over a field 4> of
characteristic 0 if and only if its Galois group is solvable.
We recall that a group G is defined to be solvable if it has a
chain of subgroups G = Gx 2 G2 2 G3 2 • • • 3 Gr+1 = 1 such
that each Gi+1 is invariant in Gt- and Gi/Gi+1 is commutative.
Every subgroup and homomorphic image of a solvable group is
solvable. Moreover, if G contains an invariant subgroup H
such that H and G/H are solvable, then G is solvable. A finite
group G is solvable if and only if it has a composition series G =
Gi 3 G2 3 • • • 3 G,+i = 1 whose composition factors G,-/G,-+i
are cyclic of prime order. We recall also that the alternating
group An, n > 5, is simple and this implies that the symmetric
group <S"n on n letters is not solvable if n > 5. A proof of the
statement about An is given in Vol. I, p. 139. All of the other

GALOIS THEORY OF EQUATIONS 99
results which we have stated are easy consequences of the theory
of normal series and most of these have been given as exercises in
Vol. I, pp. 139, 143. At any rate we shall assume all of these
results.
In order to prove the necessity of Galois' criterion we shall need
the following
Lemma. If 2 has a root tower over $ of characteristic 0, then
2 has an extension field fi which is finite dimensional Galois over <£>
and also has a root tower over $.
Proof. We are given $ = <$! c $2 Q • • ■ Q *r+i = 2 where
$,-+i = #»•(£,), &"' = a,- e #,-. We shall show that there exists a
field A,- 2 2 which also contains a subfield S2» such that 1) S2,- 2 $,•,
2) Q,- is Galois over <i>, 3) Q,- has a root tower over $:
Now for (' = 1 we take Ax = 2, Qi = $i and we suppose we are
given A,- and Q,- for a certain /. Let G< be the Galois group of £2,-
over $ and let a,*1, • • •, a,-**< be the conjugates of the element a,-
*•■
under the automorphisms Sj e G,-. Set ^<(*) = H (xn* — oci'i).
i-i
Then g{(x) e $[*]. Let Al+1 be a splitting field over A,- of gt(x) and
let £,-, f/, ^,", • • • be the roots of gi(x) in A,+i. Note that one of
these is the & such that <l>,+i = #,(£,) since £,(£»•) = 0 and
A,+i 2 A,- 3 S. Set fl,+1 = Q,^, {/, fe", • • •)• Since Q,/* is a
splitting field of a polynomial /»•(#) e #[x], Qt+i/$ is a splitting
field offi(x)gi(x) and (since the characteristic is 0) ft,+i is Galois
over <i>. Since Q,+i 2 Qt- and ^< e Q,-, Qt+i 2 #<+i = *i(^i). Let
{,(fc) be any one of the elements £,-, £,', f,", • • • ; then gi(£iW) = 0
and gi(x) = n(#n< — a/') show that (£,-w)n'' is one of the a,*».

100 GALOIS THEORY OF EQUATIONS
Hence fil+i = fl,(£,-, (■/, £/', • • •) has a root tower over $. This
shows that Ai+i and fl,+i satisfies the conditions 1), 2), 3). We
now take fl = fir+i and this satisfies the conditions stated in the
lemma.
Remark. Note that the integers nt for the root tower for £2/$
are the same as those for the given tower for 2/<i>.
We can now prove the necessity of Galois' condition. Thus let
f{x) = 0 be solvable by radicals over $ (of characteristic 0) so the
splitting field P/$ off(x) can be imbedded in a field 2 which has a
root tower over f>. By the lemma we may assume that 2/<l> is
Galois. Let n be the least common multiple of the exponents «,•
occurring in a root tower for 2 and let A be a splitting field over 2
of xn - 1 so A = 2(f) where f is a primitive w-th root of 1 and A
is Galois over $ and has a root tower over <3>. Moreover, it is
clear that we can obtain a root tower for A which has the form:
A1) $ = $! C $2 = *i(f) c $3
= *a(*i) C • • • C *r+1(£r) = A
where £,"•' e $,-+1. If// is the Galois group of A over $, then the
chain of subfields A1) gives rise to a decreasing chain of sub-
subgroups
A2) H = Hi 3 H2 2 ■ • • 2 #r+2 = 1
where Ht is the Galois group of A over <£,-. By Theorem 3, *2 is
Galois over $! with commutative Galois group and since <£2 con-
contains the necessary roots of 1, $i+1 is cyclic over $,■ if i > 1. This
implies that Hj+1 is an invariant subgroup of Hj forj > 1. The
factor group Hi/H2 is isomorphic to the Galois group of f>2 over
$i and so is commutative while the factor group Hi/Hf+iy
i > 2, is isomorphic to the Galois group of $,-+1 over <£»• and so is
cyclic. Thus the sequence of groups A2) shows that H is solvable.
Now we have A z> P 2 $ where P/* is the splitting field of/(#).
Hence if K is the subgroup corresponding to P, then K is invariant
in H and ///if ^ G, the Galois group of P/$. Since //is solvable,
this shows that G is solvable; hence the Galois group G/ of the
equation/(#) = 0 is a solvable group.
In order to prove the sufficiency of Galois' condition we require
the following result which is of independent interest.

GALOIS THEORY OF EQUATIONS 101
Theorem 6. Let P/$ be finite dimensional Galois over $ and let
P' be an extension field of P such that P' is generated by P and a
second subfield <£' 2 #. Then P'/$' zj finite dimensional Galois
and its Galois group G' is isomorphic to a subgroup of the Galois
group G of P/f>.
P'
Proof. We know that P = $(£i, •••,£„) where the £,- are the
roots of a separable polynomial/(x) e $[#]. Since P' is generated
by $' 2 * and P, we have P' = $'(&, • • -,£„). Hence P' is a
splitting field over $>' off(x). Since separability is invariant under
extension of the base field, f(x) is separable over f>' and conse-
consequently P' is Galois over $'. Let s' belong to the Galois group G'
of P'/$'. Then s' is the identity mapping in <t> C <£>' and j' maps
the set /? = {?i, &> • • •> ?n} into itself. Hence s' maps P = <£>(#)
into itself and so the restriction of s' to P is an element s of the
Galois group of P over f>. The mapping s' —> s is a homomor-
phismof G' into G. Since j = 1 implies that £,* = £,-, 1 < i < n,
and this implies that s' = 1, we see that s' —> j is an isomorphism,
so G' is isomorphic to a subgroup of G.
We can now give the proof of the sufficiency of Galois' condi-
condition. We assume that/(a1) = 0 has a solvable Galois group G/;
hence the Galois group G of the splitting field P/$ of f(x) is
solvable. We are assuming also that "i" is of characteristic 0. Let
n = (G: 1) and let P' = P(f) where f is a primitive »-th root of 1.
Then P' is generated by P and the subfield <£' = $(£■)• Hence, by
Theorem 6, P' is Galois over $>' and its Galois group G' over <£>' is
isomorphic to a subgroup of G. Hence G' is solvable and has a
composition series G' = G/ 3 G2' zd ■ ■ • z> Gs+1' = 1 whose com-
composition factors Gi'/Gi+i' are cyclic of prime order. Evidently
these orders are divisors of n = (G: 1). Let <£' = $i' c $2' c: • • •
d f>s+i' = P' be the chain of subfields corresponding to the com-
composition series for G'. Since Gl+1' is invariant in G/ and Gi'/Gi+i'

102 GALOIS THEORY OF EQUATIONS
is cyclic, $»+i' is Galois over $,•' with cyclic Galois group whose
order «,- is a divisor of n. Now <£>/ contains a primitive wt-th root
of 1 since */ 2 $' = $(f). Hence, by Theorem 5, $t+i' =
*«(&) where £<"' = on e *,. Thus "V c $2' <= • • • <= $,+i' = P'
is a root tower for P' over $'. Since $' = $(f), f" = 1, $ £ *i' c
<£2' c • • • c $,+/ = P' is a root tower for P' over <l>. Since
P' 2 Pj this shows that f(x) = 0 is solvable by radicals over $.
EXERCISES
1. Let P/$ be a splitting field over $ of characteristic 0 for xv — 1, p a prime.
Prove that P/$ can be imbedded in a field S/$ which has a root tower A0) for
which the w< are primes and [<I>i+i :<!>,■] = «<. Call such a root tower normalized.
(Hint: Use induction on p and ex. 1 of § 2.)
2. Obtain normalized root tower fields over the cyclotomic fields of Sth and 7th
roots of 1 over the field i?o of rational numbers.
3. Prove that, iff(x) — 0 has a solvable Galois group over a field of charac-
characteristic 0, then its splitting field can be imbedded in an extension which has a
normalized root tower.
4. Let $ be of characteristic p ^ 0. Call an equation f(x) = 0, f(x) e <£[*],
solvable by equations xp — x — a if its splitting field P/$ can be imbedded in a
field S which has a tower of fields $i = $ C $2 C ... C $r+1 = S where
*<+i = $<(£<), Si" — & = a,-e<l>,-. Show that, if /(x) has distinct roots, then
/(*) = 0 is solvable by equations xp — x = a if and only if its Galois group is of
order />'. (Hint: Use ex. 3, 4 of § 2 and the fact that a finite group of prime
power order is solvable.)
4. The general equation of n-th degree. The formula x =
(a ± Va2 — 4£)/2 for the solutions of the quadratic equation
x2 — ax + b = 0 (characteristic j£ 2) is valid if a, £ are con-
considered as indeterminates. When this is done one has a "general
quadratic equation." Particular quadratic equations are obtained
by specializing the coefficients. The corresponding specialization
for the solutions gives the solutions of the particular equations.
Similar solutions for general cubic and quartic equations by radi-
radicals are known (ex. 3, 4 below). We shall now consider the ques-
question of solvability by radicals of the general equation of »-th de-
degree for any w.
Let $ be a field and let S = *(/x, t2, ■ ■ ■, tn) be the field of
rational expressions in indeterminates tt over $. Then the equa-
equation
A3) /(*) = #n - fi*"-1 + t2xn~2 h (-!)"'» = 0

GALOIS THEORY OF EQUATIONS 103
is called the general equation of the n-th degree over <t>. We wish to
determine the Galois group G/ over 2 of this equation. Let P =
2(^i, x2, •■•j.Vn) be a splitting field over 2 of f(x) such that
f(x) = (x — xi)(x — x2) • • • (x — xn) in P[x]. Then
A4) ti = Xxi} t2 = X) *f#« ■ ■ ->tn = #1*2 • • ■ xn;
hence
A5) P = 2(*i, x2, ••-,#») = *(/1} t2, ■■■,tn;xu ■■ -,xn)
= *(#i,#2, ••■,xn).
In order to determine Gf we consider first a simpler problem.
We introduce new indeterminates £i, £2, • • •>£n over # and the
field P = $(fi, £2, • • •, £n) of rational expressions in the £,-. Con-
Consider the polynomial
A6) /(*) = (X- fc)(* -&)•••(*- ?n)
in PM. We have
A7) f(x) = xn - r^" + r2xn~2 + (-l)nrn
where
A8) n = 2^,-, r2 = £ ^-, • • -, rn =
We now consider the subfield 2 = $(ri, t2, • • •, rn) of P/f> and
we note that the relation P = 2(£i, f2j •■•)^n) and A6) show
that P is a splitting field over 2 of f(x). We assert that the Galois
group Gj of the equation f(x) = 0 over 2 is the symmetric group.
Thus we have to show that, if £t- —> £,/ is any permutation of the
£,-, then there exists an s-f e G-f such that £/' = (-,■». Now we know
that we have an automorphism J of the polynomial algebra
*[£i> £2, • • •, £«] over $ such that £/ = £,•», 1 < i < n. We know
also that J has an extension to an automorphism s of the field
P = $(£1, i-2> • • • > £n) over $. Finally J can be extended to an
automorphism s of P[x] so that x' = x. Then we have f'(x) =
C* - &')(# - £2') • • ■ (* - f«') = /(*) which, by A7), implies
Ti* = Ti} 1 < i < n. (This can be seen also by using the expres-
expression A8) for the t,-.) Now t/ = t,- implies that the elements of
2 = *(ti, r2, • • •, rn) are fixed under s. Hence s is in the Galois
group of P/2 and the induced mapping Sj satisfies |/' = £»•-,
1 < i < n, as required.

104 GALOIS THEORY OF EQUATIONS
We shall now carry over the result we have just obtained on the
pair of fields P, 2 to the pair P, 2 by establishing an isomorphism
of P onto P which maps 2 onto 2. We consider first the algebra
homomorphism 77 over $ of
*['i> h, ■-, tn] -+ $[ti, t2, • • •> rn]
such that t? = Ti, 1 < i < n. The existence of 77 is clear since
the ti are indeterminates. We assert that 77 is an isomorphism. To
see this we note that we have the homomorphism f over $ of
so that £/ = #,-. Again this is clear since the £t- are indeterminates.
Note also that $[£1, £2> • • • > £n] 2 *[ti, t2, • • •, rn] so r;f is de-
defined. Now the formulas A8) and A4) show that r/ = /,-.
Hence t^ = r/ = /,■ and consequently g1* = g for every g in
*['ij h, ■ ■, tn]- This implies that our first mapping 77 is an iso-
isomorphism since gv = 0 gives g = g1* = 0 for g in $[/1} • • •, /„].
We are now in a position to extend 17 to an isomorphism 77 of
2 = *(*!, /2, ■ ■ •, /„) onto 2 = $(ti, t2, • • •, tb) and this extends
to an isomorphism 77 of 2[#] onto 2[x] so that a:11 = x. Then
/(*)* = (^n - hx"-1 H )« = xn - 7Xxn-x -\ = /(*)•
On the other hand, P is a splitting field over 2 off(x) and P is a
splitting field over 2 of f{x). Hence the general uniqueness
theorem for splitting fields (Th. 1.7) provides an isomorphism 77 of
P onto P which coincides with the given 77 on 2. It is immediate
from the existence of such an isomorphism that the Galois group
G of P/2 is isomorphic to the Galois group G of P/2. In fact, it is
clear that the mapping s —> 77 "^77 is an isomorphism of G onto G.
The fact that G-t = Sn now implies that the Galois group G/ of
/(#) = 0 over 2 is Sn. It is clear also that the roots of/(x) are
distinct and Theorem 2 shows that/(,v) is irreducible in 2[^]. The
results we have obtained can be stated as
Theorem 7. The general equation of the n-th degree A3) is ir-
irreducible in 2 = <S>(/i, t2, ■ ■ ■, tn) and has distinct roots. The
Galois group off(x) = 0 is the symmetric group Sn.
Since Sn is not solvable if n > 4 this implies the
Theorem of Abel-Rumni. The general equation of the n-th degree
is not solvable by radicals if n > 4 (characteristic 0).

GALOIS THEORY OF EQUATIONS 105
EXERCISES
1. Use the fact that every finite group is isomorphic to a subgroup of Sn to con-
construct a field P whose Galois group over a suitable field $ is isomorphic to a given
finite group G. (The construction of P for a given $ and G is an open problem.
In fact, for $ the field of rational numbers this is a classical problem which is still
unsolved.)
2. Use the Galois theory to prove that, if r(xi, #2, • • •, #„) e $(#1, #2, • • •, xn)
the field of rational expressions in indeterminates #,• over the field $ and r is
symmetric in the sense that r{xy, xy, • ■ •, xn') = r(xi, #2, ■ • •, xn) for every per-
permutation Xi —> Xf of the x's, then r is a rational expression with coefficients in $
in the elementary symmetric polynomials h = 2,v,-, fa = ^T, XiX,-, •••,(„ =
i<j
X1X2 • • • xn. (Compare the fundamental theorem on symmetric polynomials,
Vol. I, p. 109.)
3. Assume the characteristic of $ is not two or three and consider the general
cubic x3 — hx2 + hx — /» = (# — xi)(x — xi)(x — x3). Here the /,• are in-
indeterminates and the Xi are in a splitting field P over S = $(/i, h, 1$). Nothing is
lost in replacing Xi by yt = #,• — -$(xi + #2 + #3) = #» — %h. Then the
given equation is replaced by y% + py + g = 0 whose roots are y 1,^2, yz where
71+^2+^3 = 0. Then the formula F) for the discriminant gives 5 = — 4p3 —
27g2. The group of P over S(\/5) is the alternating group A$ which is cyclic of
order 3. Let f be a primitive cube root of 1 (e.g., f = —§ + i\/~3) and set
zi = yx + t~ly* + f"bi = yi + Zbz + £>, z2 = yx + f~2y2 + i 4jv3 = y\ +
J>i + f% 23 =yi+yi + yz = 0. Verify that Zl3 = -^ - fv^^ V5
if f = -i + IV-3 and z23 = -2£-q + f \A=3 v^, ztZ2 = -3p. Hence
where the determination of V—35 is the same in both formulas and that of
*Z/~ is such that Z1Z2 = —3p. Solve the equations zi = _yi + ^y^ + fj»3, Z2 =
^i + &2 + ^yz, z3 = j-i + jy2 + ^3 for y\, yi, y3 to obtain Cardan's solution of
the equation y3 -\- py -{- g = 0.
4. Assume the characteristic is not two or three and consider the general
quartic xi — hx3 + hx2 — hx + U = (x — xi)(x — X2)(x — ^3)(* — #4). Re-
Replacing xi by yi = Xi — -j^ gives an equation /(jy) = jy4 + py2 + gy + r = 0
whose roots are y\, yi, y%, yt. Show that the resolvent cubic of f{y) = 0 is
^(z) = z3 - 2pz2 + (p2 - 4r)z + q2 = 0 (cf. the exercises in § 1). Show that the
Galois group of P = $(xu xi, xs, xt) = $(yi, ys, ys, yd over $(zi, Z2, z3), Zi the
roots of g(z) = 0 is the Vierergruppe. Obtain formulas for yi, yz, yz, yt in terms
of zi, z2, Z3 and square roots of elements of $(zi, z2) Z3).
5. Consider a splitting field P over S = $(/i, • ■ •, /„), ti indeterminates, of the
general equation A3) and let #1, #2, • • -,xn be the roots. Assume $ contains n
distinct elements a, cz, • • •, c„. Prove that 0 = ciX\ + c^xi -\ (- CnXn is a
primitive element of P/2.
5. Equations with rational coefficients and symmetric group as
Galois group. The theorem of Abel-Ruffini shows that equations

106 GALOIS THEORY OF EQUATIONS
of degree >5 with indeterminate coefficients are not solvable by
radicals. On the other hand, it is clear that for certain fields <£>,
e.g., the field of real numbers or the field of complex numbers,
every equation with coefficients in <£ is solvable by radicals. We
shall now show that there exist equations with rational coeffi-
coefficients which are not solvable by radicals. We shall do this by
showing that there exist rational equations of any prime degree p
with Galois group the symmetric group Sv. We prove first the
following result on permutation groups.
Lemma. If G is a permutation group on p elements where p is a
prime and G contains an element of order p and a transposition, then
Proof. We recall that any permutation can be written as a
product of disjoint cycles (Vol. I, p. 35). Moreover, the order of a
cycle is the number of letters it contains. This implies that, if
<r e G has order p, then a is a cycle containing all the letters 1,
2, • • ■ yp. By re-ordering the elements 1, 2, • • • suitably, we may
assume that G contains the transposition A2). Since a suitable
power of the^>-cycle <r has the form A2 • ■ • ) further re-ordering
of the elements 12, • • •, p, if necessary, permits us to assume that
G contains A2) and a = A23 • • • p). We recall that, if r is any
element of Sp (or Sn), then r~1(ij)r = (iTjT) where iT, jT are the
images of i,j respectively under r. This shows that <t~1(\2)(t =
B3), (r-2A2>2 = C4), • • •, O - 1, p) and (pi) are contained in
G. Since
A3) = A2)B3)A2)
A4) = A3)C4)A3)
(\p) = (lp
all of these elements are contained in G. Since (if) = (li)(lf)(U)
if 1, i,j are different, this shows that every transposition is con-
contained in G. Since every element of Sp is a product of transposi-
transpositions, we have G = Sp.
We shall now prove the following

GALOIS THEORY OF EQUATIONS
107
Theorem 8. Let f{x) be a polynomial of prime degree with ra-
rational coefficients which is irreducible in the rational field. Suppose
f{x) = 0 has exactly two non-real roots in the field C of complex
numbers. Then the group G/ of f(x) = 0 over the rationals is the
symmetric group.
Proof. The fundamental theorem of algebra asserts that
f(x) = (x — pi)(x — p2) • • • (x — pp) in C[x]. Then the subfield
P = i?o(pi> P2» • • • j Pp)> Ro the rationals, of C is a splitting field of
/(*) over Ro. Since P 2 Ro(pi) and [/2o(pi)^o] = deg/(#) = p,
[P:i?o] is divisible by p. Hence p is a divisor of (G:l), G the
Galois group of P over Ro. It follows from Sylow's theorem that G
contains an element of order p. Now consider the automorphism
a — a + PV^I —> a — /SV—1 = a, a, /? real, of C over the
field of real numbers. This maps f(x) into itself since the co-
coefficients off(x) are real. Hence it maps the set {pu p2, • • •, pp) of
the roots off(x) belonging to C into itself. Let pu p2 be the non-
real roots off(x). Then a —> a interchanges pi and p2 and leaves
fixed all the p,-, / > 2. Thus the restriction of the automorphism
a —» a of C to the set of roots is an element of G/ which is a
transposition. Hence G/ contains an element of order p and a
transposition and Gf = Sp by the lemma.
We shall now indicate how one can construct polynomials
satisfying the conditions of the theorem.* Let m be a positive
integer, »j < »2 < • • • < »r-2 be r — 2 even integers where r
is odd and >3. Consider the polynomial
B0) g{x) = (x2 + m){x - n1)(x - n2) ■ ■ ■ (x - »,_,).
The real roots of g(x) are »i, »2, • ■ •, wr-2 and the graph of y =
g(x) has the form:
* The construction we give is due to R. Brauer.

108 GALOIS THEORY OF EQUATIONS
This has (r — 3)/2 relative maxima and, since \g(k) | > 2 for any
odd integer k, it is clear that the values of these relative maxima
are >2. This implies that f(x) = g(x) — 2 has (r — 3)/2 posi-
positive relative maxima between wt and wr-2- It follows that f{x)
has r — 3 real roots in the interval («l3 »r_2). Since /(«r-2) =
—2 and/(«=) = °°, there is also a real root >nr_2- This gives
r — 2 real roots forf(x). Let «i, a2, • • •, ar be the complex roots
of f{x). Then /(*) = n(* - en) = (*2 + m){x - »0 • • • (* -
wr_2) — 2, and equating coefficients of xr~1 and •v'", we obtain
r r-2
B1) 23 «»• = Z) «*> Z) «f«y = X) «*»* + »».
1 1 i<i h<l
Hence
B2) S«t-2 = (SaO2 - 2 £ aW
i
If we choose m sufficiently large, B2) shows that 2a,-2 < 0 and
this implies that not every at is real. If ax is a non-real root, then
«i ?* ai is another such root so we have at least two non-real roots.
Since in any case we have r — 1 real roots, we see that/(a1) has
exactly r — 1 real roots. We now write f(x) = xr + aixr~1
+ •••+«,. Clearly the «,• are even integers. Moreover, since
the constant term of g(x) is divisible by 4, that of/(x) = g{x) — 2
is not divisible by 4. It follows by Eisenstein's criterion applied
to the prime q = 1 that/(.v) is irreducible in the rational field.
We therefore see that we can satisfy the conditions of the theorem
for every prime p = r > 5. It is easy to see that this holds also
for p = 2, 3. Hence the conditions hold for every prime, so we see
that there exist rational equations of every prime degree p with
Galois group the symmetric group iS"p.
EXERCISES
1. Let/(#) 8 €>[#] have distinct roots pi, p2, • • •, pn in a splitting field P/$ and
let Gf C Sn be the Galois group of the equation /. Let yi, yz, • • •, yn be in-
determinates and set
F(.*) = II (* - (Piyi + P»yi H r- Pn>y*))
teSn
= II (* — (pvi' + pty* H 1- p».y»0)-

GALOIS THEORY OF EQUATIONS 109
Show that F(x)e$[yi,yi, ■ ■ -,yn, x]. Let F(x) = Fi(#)F2(*) • • • Fr(x) be the
factorization of F(x) into irreducible factors with leading coefficient 1 in $(yi, yz,
• • -,yn)[x]. Show that, if* — 2p>tV« ls a factor of Fi(x), then
Hence show that deg Fi(x) = (<?/:l).
2. Same notations as 1. Assume, moreover, that $ = Ro the field of rational
numbers and that/(.v) has integer coefficients and leading coefficient 1. Assume
p is a prime such that the polynomial f(x) obtained by replacing the_coefficients
o(/(x) by their residues modulo p has distinct roots in a splitting field P//j>. Show
that F(x) = II (* — Ptyt') m P[*>yi, ' • •> JVn] where pi, p2, • • •, J5» are the roots
off(x) in P. Use this and ex. 1 to prove that, if the pi are suitably ordered, then
Gf is a subgroup of Gf.
3. Show that any transitive subgroup of £„ which contains an (» — l)-cycle
and a transposition coincides with Sn.
4. Show that the equation
*6 + 22x8 - 9xi + 12*3 - 37*2 - 29* - 15
over Ro has the group S%. (Hint: Apply ex. 2 using the primes p = 2, 3, 5.)

Chapter III
ABELIAN EXTENSIONS
In this chapter we shall investigate several types of abelian ex-
extension fields. First, we shall consider cyclotomic fields over the
field of rational numbers and we shall determine their dimen-
dimensionalities and Galois groups. Next we shall consider Kummer
extensions, which are obtained by adjoining the roots of a finite
number of pure equations xm = a to a field containing m distinct
wz-th roots of 1. Finally, we shall study the so-called abelian p-
extensions, which are defined to be abelian extensions of pf
dimensions of a field of characteristic p ^ 0. The theory of
characters of finite commutative groups is a basic tool for the in-
investigation of Kummer extensions and abelian ^-extensions. Be-
Besides this, our study of abelian ^-extensions will be based on a cer-
certain type of ring, a ring of Witt vectors which can be constructed
from any commutative algebra 31 over a field of characteristic
p 9^ 0. For any such 21 and integer m = 1, 2, • • •, we have a ring
of Witt vectors SBmBl) of characteristic pm. In the theory of
valuations it is useful to pass to the limit as m —* °o and to con-
consider also rings SB(Sl) of infinite Witt vectors. This will be con-
considered in Chapter V. A number of the results of this chapter will
be needed for an application to the theory of formally real fields
which we shall take up in Chapter VI.
1. Cyclotomic fields over the rationals. We have defined the
cyclotomic field of order m over a field f> to be the splitting field
over <£> of the polynomial xm — 1 (§ 2.2). We have shown that, if
the characteristic of <i> is not a divisor of m, then the Galois group
of the cyclotomic field is isomorphic to a subgroup of the group
U(m) of units in the ring I/irri) (Th. 2.3). We now assume that
the base field <$ = Ro, the field of rational numbers, and we let
110

ABELIAN EXTENSIONS 111
P(m) denote the cyclotomic field of the m-th roots of 1 over Ro.
Let Z[m) be the multiplicative group of the wz-th roots of 1. We
recall that Z(rri) is cyclic and its generators are called primitive
w-th roots of 1. Also P(m) = R§{£), where f is any primitive w-th
root of 1; hence the dimensionality [T(m):R0] is the degree of the
minimum polynomial of f over Ro. If f is a primitive m-th root of
1, then any other primitive m-th root of 1 has the form f* where
(k, m) = 1. Hence the number of primitive m-th roots of 1 is
<p(m) the number of positive integers not exceeding m which are
relatively prime to m. This is also the order of the group U(m).
Now let
A) \*(x) = n c* - r).
f primitive
This is a polynomial of degree <p{m) with coefficients in P(m). If s
is in the Galois group G of P(m) over Ro, then clearly s maps the
set of primitive m-th roots of 1 into itself. Hence we have the
relation \m'(x) = \m(x) for every s e G. Since P("° is Galois over
RQ, we see that \m(x) e Ro[x], that is, \m(x) has rational coeffi-
coefficients. We can see this also in a more elementary way which, at
the same time, gives an inductive procedure for calculating
\m(x). Since the order of any m-th root of 1 is a divisor of m and
since every d-th root of 1 for d\ m is an w-th root of 1, we clearly
have the formula
B) *" -1 = n xd(*).
d\m
l<d<m
Evidently we have \i(x) = x — 1 and, assuming that \d(x) e
Ro[x] for all d such that 1 < d < m, then the formula B) gives
C) \m{x) = (xm - 1)/ II A,(*)
d\m
which shows that \m(x) e R0[x], This gives a practical way of
calculating \m(x). For example, we have \i(x) — x — 1,
X2(*) = (x2 -
\3(x) = (^3 -
X4W = (x* - l)/Mx)\2(x) =x2
X8W = (*6 -

112 ABELIAN EXTENSIONS
and
X12W = (*12 - l)Ai(*)X2(*)Xs(*)X4(*)A6(*) = xi - x2 + 1.
If p is a prime, we have
D) \p(x) = (xp - l)/(* - 1) = x'-1 + xp~2 + • ■ ■ + 1
and it is easy to see, using Eisenstein's criterion, that \p(x) is ir-
irreducible in R0[x] (Vol. I, ex. 2, p. 127). We shall now prove the
following general result
Theorem 1. \m(x) is irreducible in the rational field.
Proof. We observe first that Xm(A;) has integer coefficients.
For, assuming this holds for every X«*(,v), d < m, and setting
p(x) — II Xd(#), we obtain by the usual division algorithm
d\m
l<d<m
that xm — 1 = p{x)q{x) + r(x) where q(x) and r(x) e I[x] and
deg r(x) < degp(x). On the other hand, we have xm — 1 =
p(x)\m(x), so by the uniqueness of the quotient and remainder,
Xm(#) = q(x) has integer coefficients. Now suppose that \m(x) =
h{x)k{x) where k(x) is irreducible in RQ[x] and degh(x) > 1. By
Gauss' lemma (Vol. I, p. 125) we may assume that h{x) and k(x)
have integer coefficients and leading coefficients 1. Let p be a
prime integer such that p )( m and let f be a root of h{x). We
shall show that fp is a root of h(x). Since (p, m) = 1, fp is a
primitive w-th root of 1 and, if fp is not a root of A(x), fp is a root
of k(x); consequently f is a root of k(xp). Since h{x) is irre-
irreducible in Rq[x] and has f as a root, h{x)\k{xp). It follows (as
above) that k{xp) = h{x)l{x), where l{x) has integer coefficients
and leading coefficient 1. Also we have xm — 1 = \m{x)p{x) =
h(x)k{x)p{x) and all of these polynomials have integer coefficients
and leading coefficients 1. We now pass to congruences modulo p
or, what is the same thing, to relations in the polynomial ring
Ip[x], Then we obtain
E) xm - I = l{x)l(x)p(x)
where in general, iff{x) = aoxn + a1xn~1 + ■ • ■ + an e I[x], then
J(x) = aoxn + axxn~x_A h dni at = at + (p) in Ip. Similarly,
we have £(#*) = H(x)l(x). On the other hand, using a? — a for

ABELIAN EXTENSIONS 113
every integer a, we see that
J(x)p = (doxn H 1- any = dopxpn H \- dnp
= doxpn + •••+*„ = ]{xp)
for any polynomial/(#). Hence %(x)p = l{xp) = Ji(x)l{x) which
implies that (h{x), £(#)) 9^ 1. Then E) shows that xm — I has
multiple roots in its splitting field over Ip. Since p Jf m this is
impossible and so we have proved that fp is a root of h(x) for every
primes satisfying/* X w. A repetition of this process shows that
f is a root of h(x) for every integer r prime to m. Since any primi-
primitive #z-th root of 1 has the form f, (r, m) = 1 we see that every
primitive «z-th root of 1 is a root of h(x). Hence h{x) = \m(x) and
XmC*1) is irreducible in Rq[x].
We now see that \m{x) is the minimum polynomial over Ro of
any primitive w-th root of 1. Since P(m) = Z?o(r)> f primitive
we have established the formula
F) [P<~>:*0] = *(»»).
This implies that (G: 1) = <p(m) for the Galois group G of P(m)/^o.
Since (£/(#*) :1) = v'(w) and G is isomorphic to a subgroup of
U(m)} this proves
Theorem 2. Z>/ P(m) ^ the cyclotomic field of order m over the
rationals Ro. Then the Galois group of P(m)/^o is isomorphic to
U(m), the multiplicative group of units in the ring I/(rn).
We shall now proceed to determine the structure of the Galois
group G or, what is the same thing, that of U{m). It is easy to see
that, if m = pieip2H • • • p/' where the pi are distinct primes, then
U(m) is isomorphic to the direct product of the U(piei). For this
reason we shall confine Our attention to the case m = pe a prime
power. Then U(pe) is a commutative group of order <p(p') =
p> _ p'-i — pe~1(p — 1). We prove first
Theorem 3. If p is an odd prime, then the multiplicative group
U(pe) of units in I/(pe) is cyclic.
Proof. Since the order of this group is pe~l(j> — 1), U(pe) is a
direct product of its subgroup H of order pe~1 consisting of the
elements which satisfy #p'~l = 1 and the subgroup K of order
p — 1 of the elements satisfying xp~1 = 1. It suffices to show

114 ABELIAN EXTENSIONS
that both H and K are cyclic since the direct product of cyclic
groups having relatively prime orders is cyclic. If e = 1, then
U(p) = K is the multiplicative group of the field //(/>) and this is
cyclic. Hence we can choose an integer a such that a + (p), a2 +
(p), ■ ■ ■}ap~1 + (p) are distinct in I/(p). Set b = ap"~\ Since
(a,p) = 1, (b,pe) = 1 and b + (pe) and a + (pe) e £/(/>*). Also
#•-1 = (^p*-1)?-1 = av(p') = 1 (mod pe) so b + (pe) e K. Since
b = a1" = a (mod p), £ + (p), b2 + (p), • • •, ^~J + [p) are
distinct. Hence also b + (pe), b2 + (pe)y • ■ -,bv-x + (pe) are
distinct. This implies that the order of b + (pe) is precisely
p — 1. Since (i£:l) = p — 1, it follows that X" is cyclic with
generator b + (pe). It remains to prove that H is cyclic, and we
may assume that e > 2, since, otherwise, H = A) and the result
is clear. Assuming e > 2, we can conclude that H is a direct
product of & > 1 cyclic groups of order pei, et > 1. Then the
number of solutions of the equation xv = 1, x e H is ph. Hence it
will be enough to show that the number of integers n, 0 < n < pe,
satisfying np = 1 (mod />e) does not exceed p. Now if n satisfies
these conditions, then, since nv = n (mod p), we have n = 1
(mod p). Then if « j± 1, we may write » = 1 -\- ypf + zpf+l
where 1 </<<? — 1, 0<_y<^>, and z is a non-negative
integer. Then
0- + *p)pppf
If np = 1 (mod p") and / < e — 1, this gives ypf+l = 0 (mod
pf+2) so y = 0 (mod />) contrary to 0 < _y < p. Hence we see
that, if 1 < n < pe satisfies np = 1 (mod pe), then » = 1 +
yp"~l, 0 < y < p. This gives altogether at most p solutions in-
including 1 and completes the proof of the theorem.
We consider next the case of the prime 2 in the following
Theorem 4. GB) and UD) are cyclic and, if e > 3, then UBe)
is a direct product of a cyclic group of order 2 and one of order 2e~2.
Proof. The order of UBe) is <pBe) = 2—1. lie = 1, A7B): 1)
= 1 and if e = 2, UBe) = £7D) has only two elements and so is

ABELIAN EXTENSIONS 115
cyclic. Suppose e > 3. We show first that there are four distinct
elements x e U(le) satisfying x2 = 1. This will imply that U{2")
is a direct product of at least two distinct cyclic groups ?± 1. Set
ax = 1, a2 = -1, a3 = 1 + 2—*, «4 = -1 + 2"-1, xt = «,• +
B*). Then the xt are distinct and satisfy x2 = 1, which proves
our assertion. Also since UBe) is a direct product of at least two
cyclic groups 5^ 1 and the order of UBe) is 2"~1, we see that, if
x e UBe), then x2'~* = 1 or, what is the same thing, if a is an odd
integer, then a2* = 1 (mod 2e). The proof will be completed by
displaying an x such that x2"~* ?± 1. Then we shall have a cyclic
subgroup of order 2e~2 and this can happen only if UBe) is a
direct product of a cyclic group of this order and one of order 2.
We proceed to show that we may take x = 5 + Be). Note first
that, if e = 3, then 52*~' = 5*1 (mod 2e) but 52<~' s 1 (mod
2e-1). Now let/ > 3 and let k(f) be the largest integer k such
that 52/"' a 1 (mod 2*). Then we have *C) = 2. Also for any
/ > 3 we have 52/" = 1 + jy2*(/) where _y is odd. This gives
which shows first that *(/+ 1) •> k(f), so k(f) > 2 if / > 3.
Then the relation shows that 52</+1)~* = 1 + 22*<')+1 where
z = y + 2k^~1y2 is odd. Hence *(/+ 1) = k(f) + 1. This and
*C) = 2 imply that k(J) =/ - 1 for all/ > 3. Thus 52^ * 1
(mod 2') if <? > 3 which is what we needed. This completes the
proof.
Theorems 2, 3, and 4 give a description of the Galois group of
the field of the ^>e-th roots of 1 over the rationals. The result is
the following
Theorem 5. Let m = p',pa prime, and let P(m) be the field of the
m-th roots of 1 over the field Ro of rational numbers. Then the Galois
group G of P(m)/RQ is cyclic unless p = 2 and e > 3, in which case
G is a direct product of a cyclic group of order 2 and one of order
EXERCISES
1. Use the Mobius inversion formula (Vol. I, ex. 5, p. 120) to prove that

116 ABELIAN EXTENSIONS
2. Letp be a prime and let P(p) be the cyclotomic field of p-th roots of 1 over
the field i?o of rational numbers. Let g + (p) be a generator of the cyclic group
U(p) and let s be the automorphism of P(p)/i?0 such that f = fg, f a fixed
primitive p-th root of 1. Show that (f, f*, f*2, • • •, f8") form a basis (normal
basis) for P(p)/i?o. Suppose/) — 1 = ef, <?,/positive integers and let E/2?o be the
subfield of ^dimensions of PM/R0. Show that, if / = se and rj = f + f' + f'2
-I h f(/ ', then (»?,r\\ ■ ■ ■ ,-if' L) is a basis for E//?0- Show that the multipli-
multiplication table for this basis has integer coefficients.
3. Let P be the field of the 17-th roots of 1 over i?o. Determine the subfields
Rh i = 1, 2, 3 such that Ro C Rx C R2 C R3 C Rt = P and [Rf :Ri-i] = 2.
Find an element w< in /?< so that Ri = /?,_i(a),), a><2 e /?<—1> 1 < * < 4.
4. (O. Todd) Let P be the field of p-th roots of 1 over i?0 where p is a prime of
the form 4» + 3. Show that P is a tensor product of a quadratic subfield and a
subfield of odd dimensionality. Show that the quadratic subfield is not real (if P
is considered as a subfield of the field of complex numbers).
5. Let E(m) be the cyclotomic field of degree m over $0 = Ip- Write m =
m'p°, (rri, p) = 1. Show that [E(m) :$o] is the order of the element p + (mr) in
the group U{mr).
2. Characters of finite commutative groups. In the remainder
of this chapter we shall study two classes of abelian extension
fields: Kummer extensions and abelian extensions of pe dimen-
dimensions over a field of characteristic p. For both of these the theory
of characters of finite commutative groups is basic, so we shall
develop this first.
Let A and B be two commutative groups (written multipli-
catively) and let x and \p be homomorphisms of A into B. We de-
define the product xP by ax^ = axa*. One checks that this is again a
homomorphism and that the set Horn {A, B) of all the homo-
homomorphisms of A into B is a commutative group under the prod-
product x'A (cf. Vol. I, p. 78). We shall be interested particularly in
the case A finite and B = Z a finite cyclic group whose order is
divisible by the orders of all the elements of A. We shall call the
maximum order of the elements of A the exponent of A. We recall
that the order of every element is a divisor of the exponent (Vol.
II, ex. 1, p. 69), so the condition we have imposed on Z is equiva-
equivalent to: the order of Z is divisible by the exponent of A.
We wish to determine Horn (A, Z) from a particular decom-
decomposition of A as A = Ax X A2 X • • • X Ar where the At are cyclic
subgroups. Thus we are assuming that A = A\ • • • AT and
At D Ai • • • Ai_xAi+i • ■ • Ar = 1. Let n{ = (Ai'A) and le{
Cf be the subgroup of Z of elements z satisfying zn< = 1. Since «,•

ABELIAN EXTENSIONS 117
is a divisor of the order of Z, Ct is the subgroup of order «* of Z.
Let C be the group of r-tuples (cu c2, ■ ■ •, cr) where ct e C1,- and
multiplication is defined componentwise. Hence C is the (ex-
(external) direct product of the groups Cu C2, ■ • •, Cr and C ^ A
(cf. Vol. I, p. 144). We shall now obtain an isomorphism of
Horn (Ay Z) onto C. For this purpose we choose a generator a{
of A{, i = 1, 2, • • •, r. If x e Horn (/f, Z), then «,•* = d satisfies
CtUi = 1, since atni = 1. Thus ct e C\. We now map x into the
element (cu c2, ■ ■ -,cr) = (a1x,a2x, ■ ■ -,arx)eC. If x^eHom (A,Z),
then
= (ai*, ■■■,a*)(a1*, •••,«/),
so x —* (^ix5 • • •, <^rx) is a homomorphism of Horn (A, Z) into C
If a{x = 1, / = 1, • • ■, r, then «x = 1 for every a e A, since the «,•
are generators of A. Thus a? = 1 for all i implies x = 1> which
shows that x ~* («i*> • ■ ■, #rx) is an isomorphism into C. It re-
remains to show that this mapping is surjective. Let ct be any ele-
element of Ct. Then dni = 1 and it is clear that we have a homo-
homomorphism Xi of Ai onto d such that atXi = Ci. Since A = Ai X
A2 X • • • X AT the mapping X1.V2 ■ ■ ■ xr —> ^iXia;2X2 ■ • • *vXr, #i e
yfi, is a homomorphism x of A into Z. Clearly x —* («ix5 ■ • •> «rx)
== (^i> ^2> • • ■ j ^r)- This shows that the mapping of Horn (//, Z) in-
into C is surjective. Thus we have shown that A ^C ^ Horn (//, Z).
Theorem 6. Let A be a finite commutative group and let Z be a
finite cyclic group whose order is divisible by the exponent of A.
Then the group Horn (A, Z) is isomorphic to A.
If Z satisfies the condition of the theorem, then we shall call the
group Horn (A, Z) a character group of the group A and we shall
call the elements of this group characters of A.
We are now in a position to derive in quick succession the results
on characters which we need. We note first the following
Corollary 1. If a ^ 1 in A, then there exists a character x b
Horn (A, Z) such that ax j±\.
Proof. Let B be the subgroup of A of elements b such that b* =
1 for all x e Horn (A, Z). Then we see immediately that our

118 ABELIAN EXTENSIONS
assertion will follow if we can show that 5 = 1. Now let x e
Horn {A, Z). Since bx = 1, b e B, B is in the kernel of x and
so we have an induced homomorphism x of A/B into Z defined by
{aB)* = ax. If x, ^ e Horn (A, Z) and x = t, then the definition
shows that x = ^- Hence the mapping x —* x of Horn (A, Z)
into Horn (A/B, Z) is 1-1. Since (A:l) = (Horn (A, Z):l) and
(A/B:\) = (Horn (A/B, Z):l), by Theorem 6, we must have
equality of all of these numbers. This implies that 5 = 1, which
is what we needed.
If a is a fixed element of A, then we can define a mapping ?/o of
Horn (A, Z) into Z by x"° = «*• If X, ^ e Horn (^, Z) we have
(xpy* = «** = «x<^ = x^'V", which shows that r/o is a homo-
homomorphism of Horn (A, Z) into Z. Thus ?7O is a character of the
group Horn (A, Z). Then we have the basic
Corollary 2. For az A define a mapping -qa of Horn (A, Z) into
Z by xv° = «*• Then i)a e Horn (Horn (A, Z), Z) and the mapping
a —> T)a is an isomorphism of A onto Horn (Horn (A, Z), Z).
Proof. Observe first that a —» rja is a homomorphism since
xi.i = (^)x = ax£x = x'-x1" = Xv°'"' (the last equation by the
definition of the product in a character group). Next suppose
»7O = 1. Then ax = 1 for all x so, by Cor. 1, a = 1. This
shows that the kernel of the homomorphism a —> r)a is the iden-
identity. Hence the mapping is an isomorphism. Since (A'A) =
(Horn (A,Z):\) = (Horn (Horn (A, Z), Z):l), by Theorem 6,
a —* rja is surjective and the proof is complete.
Corollary 2 permits us to identify A with the character group
(relative to Z) of Horn (A, Z). By virtue of this result we have a
perfect duality between A and Horn (A, Z). We use this in the
proof of
Corollary 3. A set {xi, X2> • • •> Xr} of characters generate the
character group Horn (A, Z) if and only if the only a e A satisfying
a*< = 1, i = 1,2, ■■•Jris a = 1.
Proof. This is equivalent to the dual statement {ai, a2, ■ • ■, aT\
generate A if and only if atx = 1, for * = 1, 2, • • •, r holds only
for the character 1. This is easy; for, if a1} • • •, ar generate A and
atx = 1 holds for the character x» then ax = 1 holds since x is a

ABELIAN EXTENSIONS 119
homomorphism. This implies that x = 1- On the other hand, if
the subgroup B generated by a1} • • •, ar is a proper subgroup,
then there exists a character x^ 1 for A/B. Ifa e A the mapping
defined by a —> aB —* {aBY is an element f* 1 of Horn {A, Z)
satisfying atx = 1, / = 1, • • •, r.
3. Kummer extensions. It is generally a difficult problem to
obtain a survey of the abelian extensions of a given field <£. For
example, if $ is the field of rational numbers, this requires deep
arithmetic considerations. However, there are two types of
abelian extensions which can be quite exhaustively studied by
comparatively elementary algebraic means. One of these, which
we shall call abelian ^-extensions, are the abelian extensions of pe
dimensions over a field of characteristic p 9^ 0. We shall con-
consider these in § 5. In the present section we shall develop the
theory of Kummer extensions, which are defined as follows.
Definition 1. Let P be an abelian extension of afield $. Then
P/* is called a Kummer w-extension if the Galois group ofP/$ is of
exponent m and f> contains m distinct m-th roots of 1.
We shall now suppose that f> is a given field which contains m
distinct w-th roots of 1. The field $ and the integer m will be fixed
throughout our discussion. We are interested in obtaining a sur-
survey of the Kummer w'-extensions P/* where m'\m. We recall
that the condition that <£ contain m distinct m-th roots of 1 im-
implies that the characteristic is not a divisor of m (§ 2.2). If P/$
is a Kummer #z'-extension where m'\m, then [P:<£] = (G:l) and,
since the exponent and order of a finite commutative group are
divisible by the same primes, we see that the characteristic is not
a divisor of [P:$].
Let $ and m be as indicated and let P/* be a Kummer w'-ex-
tension, m'\m. Let P* and $>* be the multiplicative groups of
non-zero elements of P and * respectively. For p e P*, the
mapping p —> pm is an endomorphism of P* which maps 4>* into
itself. The kernel of p —» pm is Z(m) the group of order m of m-th.
roots of 1 and Z{m) c <i>*. Let
G) M(P) = {peP*|pme$*}
(8) JV(P) = {p-|peM(P)}.

120 ABELIAN EXTENSIONS
Thus M(P) consists of the m-th roots in P of the elements of <£*
and iV(P) is the set of elements of $* which are m-th powers of
elements of P. It is clear that M(P) is a subgroup of P* con-
containing 3>* and N(P) is a subgroup of <£>* containing $*m =
{am\ae $*}.
Let p e M(P) and set XpCO = PSP~1, s eG. Since pm = a e $,
(p')m = a so p'p e Z(m). Moreover, since Z(m) C $>,
Thus we see that Xp e Horn (G, Z), Z = Z(m), which is a character
group of the finite commutative group G since the exponent of G
is a divisor of m. Conversely, let x be any element of Horn (G, Z).
Then we have x(J^) = xOOxW = xWxD so Noether's equa-
equations are satisfied. Consequently, by Noether's theorem (Th.
1.19), there exists a non-zero element p e P such that xCO =
pap~x. Since pV e Z we have (ps)m = pm or (pm)s = pm for every
s eG. This implies that pm e $ and so p e M(P). We have there-
therefore shown that every element of the character group Horn (G, Z)
is of the form xOO = PSP~1, P in M(P). If pi, p2 e M(P) and
Xpu Xpj are the corresponding characters of G, then Xpi^W =
(piP2)"(piP2)~1 = Pi*Pi~1P28P2~1 = Xpi(-f)Xp,(-f). Hence the map-
mapping p —> XpCO is a homomorphism of M(P) onto Horn (G, Z).
The kernel of this homomorphism is the set of elements p e M(P)
such that pV = ly s eG. This is just the set of elements satis-
satisfying ps = p, s e G, p ?± 0 and so it is $*.
It is convenient to state the result which we have just obtained
on the homomorphism of M(P) onto Horn (G, Z) as a result on
exact sequences of group homomorphisms. If Gly G2, • • ■ ,Gk are
groups and rn is a homomorphism of Gt into Gi+i, then we say
that the sequence
is f^<2f/ if for each /' = 1, 2, • • •, k — 2 the image of Gt under rn
coincides with the kernel of rn+1. If 1 denotes the group consisting
of 1 alone then the only homomorphism of 1 into any group G is
1 —> 1. It follows from this and the definition of exactness that
1 —> Gi —> G2 is exact if and only if i) is 1-1 and G\ —» G2 —* 1
is exact if and only if 77 is surjective.

ABELIAN EXTENSIONS 121
Using this terminology we can state the following theorem.
Theorem 7. Let $> be afield containing m distinct m-th roots of 1
and let P/$ be a Kummer m'-extension where m'\m. Let M(P) be
defined by G) where P* is the multiplicative group of P and <£* is the
multiplicative group of <£. Then we have the exact sequence of
multiplicative groups
1 -+ $* -> M(P) -> Horn (G, Z) -> 1
wAm? /A* homomorphism of <£* /j £fo inclusion mapping and that of
M(P) is p —» Xp> Xp(j) = P*P~1- The factor group M(P)/P* /j
finite and isomorphic to G. We have P = $(M(P)) ««^/ P =
i P2> • ■ • j Pr), P% in M(P), if and only if the cosets p,-$* generate
Proof. The first statement on the exactness of the displayed
sequence means that «£* is the kernel of the mapping p —* xP and
this mapping is surjective on Horn (G,Z). Both of these facts
were established above. Consequently, we have Horn (G, Z) =
M(P)/<t>*. Since Horn (G, Z) = G, by Theorem 6, we have
M(P)/<i>* = G. This proves the second statement. Now let
Pij • • y Pr be elements of M(P) such that the cosets Pt$* generate the
finite group M(P)/4?*. Clearly the homomorphism p —> xP of M(P)
gives the isomorphism p$* —> xP of M(P)/$* onto Horn (G, Z).
Hence we see that the characters xPi generate Horn (G, Z). Now let
P' = $(pi, P2, • ■ •, Pr) and let H be the subgroup of G corre-
corresponding to P' (the Galois group of P/P'). If t e H, we have
p,' = pi} 1 < z < r, so XpiW = 1- This implies that xW = 1 for
every x e Horn (G, Z). It follows from Corollary 1 to Theorem 6
that / = 1. Thus H = \ which implies that P' = $(pi, • • •, pr)
= P and P = $(M(P)). Conversely, let pu ■ ■ •, pr e M(P)
satisfy *(pi, • • •, pr) = P and let s e G. Then p,-* = p»-, 1 < i < r,
will imply that p* = p, p e P. Hence we see that XpXj) = 1j
1 < i < r, implies s = 1. Then Corollary 3 to Theorem 6 im-
implies that the xPi generate Horn (G, Z). In particular, if p e M(P),
then Xp = Xp1*1Xp2*2 ' ' " Xprk'- Thus for every j e G we have
Hence (ppi^pa"*2 • • • Pr~*r)s = PPi~klPz~kl • • ■ Pr~kr, s e G. It

122ABELIAN EXTENSIONS
follows that p = /Spi*1 • • • prk', /3 e <!>*; Since p was any element of
M(P), this shows that the cosets pi$* generate M(P)/$*. This
completes the proof.
Next we consider the mapping p —> pm$*m of Af(P) onto the
factor group N(P)/$*m. This is a homomorphism whose kernel
is the set of elements of M(P) such that pm = am where o e <£*.
Then p = $a where fm = 1. Since Z C $*, these are just the ele-
elements of <£*. Hence we have the isomorphism p$* —> pm$*m of
M(P)/<i>* onto iV(P)/<i>*m. Since M(P)/$* is isomorphic to the
Galois group of P/3>, it is clear that N(P)/$*m is a finite subgroup
of <£>*/<|>*m and we have
(9) N(P)/$*m ^ M(P)/$* ^ G.
We shall now shift our attention to the subgroup N(P) of $*.
This satisfies the two conditions: N(P) 3 f>*m and iV(P)/$*m is
finite. We shall see that these subgroups, which are defined by $
and m, can be used to give a survey of the Kummer extensions
P/<l>. We observe first that, if «i, <x2, • • ■ ,ar are elements of
iV(P) such that the cosets a,$*m generate N(P)/$*m, then P/$
is the splitting field of
A0) /(*) = (*• - ai)(xm - a2) ■ ■ ■ (xm - ar).
For, we have p,m = at where p, e M(P) and the isomorphism
p$* —> pm$*m ofM(P)/$* with N(P)/$*m implies that the cosets
Pi«l>* generate M(P)/$*. Hence, by Theorem 7, P = *(pi, • • •,
pr). If Z = {£,-}, then the roots off(x) are p,-fy, so we see that
P = *(p,-fy) is a splitting field over $ off(x).
We proceed to show next that any subgroup iV of $* satisfying
the stated conditions arises from a Kummer extension. The pre-
precise result is the following
Theorem 8. Let # be afield containing m distinct m-th roots of 1
and let N be a subgroup of ** containing $*m such that N/$*m is
finite. Then there exists a Kummer m'-extension P/<i> with m'\m
such that JV(P) = N where N(T) and M(P) are defined by (8) and
G).
Proof. The foregoing analysis of Kummer extensions gives the
clue to the definition of P/$. In view of this, we are led to choose

ABELIAN EXTENSIONS 123
«ij «2, • • • j «r in the given group N so that the cosets a,#*m
generate N/$*m. Let P/<f> be a splitting field of the polynomial
f(x) given in A0). Since xm — a{ has m distinct roots, f{x) is
separable and P/$ is finite dimensional Galois. Let G be the
Galois group. If p,- is a root of #"" — a,-, then all the roots of this
polynomial are the elements ptf,-, £,- in the group Z of »z-th
roots of 1. Hence if s e G, then p,-8 = £i(s)pi, f,-(j) e Z. If j, / e G,
we have Pt'* = (f.COp,)' = Us)Pi* = f<(j)r<Wp<- Hence p<" =
PiU, i = 1, 2, • • •, r, and since it is clear that P = 3>(pi, p2, • • •,
pr), st = is for all s, t e G. This shows that G is a commutative
group. Also we have p,-** = f,-(j)*pt> £ = 1> 2, • • •, and conse-
consequently p*m = pi which implies that sm = 1, s e G. Thus the ex-
exponent of G is a divisor of m and P/$ is a Kummer w'-extension
since Z c <£. It remains to show that, if iV(P) is defined by (8),
then iV(P) = N. Since Pim = a* e $, Pi e M(P) defined by G).
Since P = *(pi, • • ^Pr), Theorem 7 shows that the cosets pt-$*
generate M^P)/**. Applying the isomorphism of M(P)/$* with
iV(P)/$*m we see that the cosets a^*m generate N(P)/$*m. On
the other hand, we know that the cosets a^*m generate N/$*m.
This implies that iV(P) = N
We now consider two Kummer w,-extensions P,-/$, / = 1, 2,
where »»,-| tw. It is clear from the definitions of M(P,), iV(P») that,
if Pi/$ ^ P2/*, then the subgroups iV(Pi) and iV(P2) of $* coin-
coincide. Conversely, suppose we have iV(Pi) = iV(P2). We have
seen that, if au a2, ■ ■ ■, ar are elements of A^(Pt) such that the
cosets a,$*m generate N(Pi)/&*m, then P,- is a splitting field over i>
of/(*) = (xm — ai){xm — ot2) • • • (xm — ar). The uniqueness of
splitting fields implies that Pi/* ^ P2/$ if N(Pi) = iV(P2).
Next we look at the Kummer m'-extensions, P/3*, m' \ m, which
are contained in one extension field 0/$ (e.g., the algebraic closure
of $ in the sense of § 4.1). We have seen that, if P»/$ is one of our
extensions, then P,- = $(M(Pt)). Hence it is clear that Pi 2 P2
if and only if M(PX) 2 M(P2). Also it is clear that M(Pi) 2
M(P2) if and only if iV(P0 3 A^(P2). Hence Px 2 P2 if and
only if the A^(PX) 3 A^(P2). It is apparent that our results give a
completely satisfactory internal description of the Kummer ex-
extensions P/$ by means of the subgroups iVof $* satisfying the two
conditions: N 2 #*m, N/$*m is finite.

124 ABELIAN EXTENSIONS
EXERCISES
1. Show that there exist an infinite number of non-isomorphic quadratic ex-
extensions of the field of rational numbers.
2. Assume <£> contains m distinct *w-th roots of 1 and let P/$ be cyclic of m
dimensions over $,ja generator of the Galois group of P/$. Show that P =
$(p) where p" = as$ andp* = fp where f is a primitive w-th root of 1. Show
that, if a e P satisfies <rm e $, then a — /3p* where |3 e $ and 1 < k < m.
3. (Albert). Let P be cyclic of n = I" dimensions over $ where /is a prime and
$ contains / distinct /-th roots of 1. Let j be a generator of the Galois group G of
P/<E>, H the subgroup of order / of G generated by / = sm, m — le~x, E the sub-
field of //-invariants, so E/$ is the unique subfield of m dimensions in P/$. By
2, P = E(p) where pl = a e E and p' = fp, f a primitive /-th root of 1. Show
that p* = /3p* where /3 e E and 1 < k < 1. Show that p' = Yp*" = fp where
7 8 E and hence that km = 1 (mod /) and k = 1. Show that ./VeI#03) = f and
4. (Albert). Assume $ has / distinct /-th roots of 1, / a prime, and that E/$ is
cyclic of m = le~l dimensions over 3>, e > 1. Suppose E contains an element
/3 such that TVe 1*08) = fa primitive /-th root of 1. Show that there exists an
a e E such that a8a-1 = /3' where j is a generator of the Galois group of E/$.
Show that a is not an /-th power in E so that, if P = E(p) with p! = a, then
[P:E] = /. Show that P is cyclic of/" dimensions over <£>.
5. Note that ex. 3 and 4 imply the following: If $ contains /distinct /-th roots
of 1, / a prime, and E/$ is cyclic of P > 1 dimensions, then E/$ can be im-
imbedded in an extension P/$ which is cyclic of P+l dimensions if and only if
the primitive /-th root of 1, f is a norm of an element of E. Use this to prove
that, if $ is of characteristic ?* 2, the quadratic extension E = $(«), e2 =7 8$,
can be imbedded in a quartic cyclic extension of $ if and only if 7 is a sum of two
squares of elements of $. In particular, show that, if i?o is the field of rational
numbers, then an imaginary quadratic extension i?o(€)» e2 = 7 < 0 in i?o cannot
be imbedded in a cyclic quartic extension.
6. (O. Todd). Let P be the field of p-th roots of 1 over Z?o where p is a prime
of the form 4» + 1. Show that P contains a real quadratic subfield.
7. Assume $ contains four distinct fourth roots of 1. Show that any quadratic
extension E/$ can be imbedded in a cyclic quartic extension
4. Witt vectors. We have defined abelian ^-extensions of a field
4> of characteristic p 7^ 0 to be abelian extensions of pe dimen-
dimensions of $. Cyclic ^-extensions of dimensionality p and p2 were
encountered first by Artin and Schreier in connection with a prob-
problem on real fields (see § 6.9). Their construction was generalized by
Albert to give an inductive construction of cyclic ^-extensions of
p" dimensions. Slightly later Witt gave a direct construction
and survey of abelian ^-extensions along the lines of the theory of
Kummer extensions which we have just considered. Witt's
method is based on an ingenious definition of a ring of vectors

ABELIAN EXTENSIONS 125
defined by a given field of characteristic p. This construction has
important application in other connections (e.g., valuation theory)
and we shall consider it now in its general form.
We shall begin first with the polynomial ring 36 = Ro[xi, jyy, z*J
in indeterminates Xi,yj, Zk, i,j, k = 0, 1, • • •, m — 1, over the
field Ro of rational numbers. Let X(m) be set of w-tuples (a0, aiy
• • ■, am_i), cii e 36, with the usual definition of equality and with
addition and multiplication by components. If a = (a0, ••',
#m-i)> b = (b0, • • •, bm-i), then we denote the sum and product
by a © b, a Q b, so that a © b = (a0 + t>o, • • •, am-i + bm_^),
a Q b = (aobOy ■ • •, «m_1^m_1). Let p be a fixed prime number.
We use this to define a mapping <p in 36(m) by the rule that, if a =
(a0, au ■ ■ -, am_y), then of = («@), a™, ■ ■ ■, «(m~1>) where
A1) aM = a/ + paf-1 + ■■■ + p"ap, v = 0, 1, • • ■, m - 1.
Thus am = a0, aw = aop + p#i, • • • . We introduce also the
mapping P\a —> ap = {aop,aipi •••,«m_ip). Then the defini-
definition A1) gives
A2) aw = «0> «(r) = {aPYv~l) +P"av, v > 1.
Next let ^ = («@), «A), •••,«(m-1)) be arbitrary and define a
mapping ^ by A* = («0, «i, • • •, ««-0 where
«o = «@),
A3)
(w T *''^) > 1
One checks directly that «** = a, ^p = ^/, which shows that <p is
1-1 surjective with ^ as its inverse.
We shall now use the mapping <p and \j/ = (p~l to define a new
addition and multiplication composition in y}m). These are re-
respectively
ab = (cP © b*y~\
These provide another ring structure in 3E(m) (Vol. I, ex. 6, p. 71).
We denote the new ring as Hm so 3£m and £(m) coincide as sets and
a —> a*1 is an isomorphism of 3Em onto 3£(m). Hence 3Em, like 3£(m), is
commutative.

126 ABELIAN EXTENSIONS
We now examine the formulas for x + y, xy, and x — y for the
"generic" vectors x = (x0, xu ■ • -, xm-x),y = {yo,yi, ■ ■ -,ym-i),
Xi,y}- the given indeterminates. For example, we have
— Z ( . )
p <-i W
(x + yH = x0 + y0, (
(xyH = xoyo, (xy)i =
In general, if ° denotes any one of the compositions +, •, —, then
it is clear from the definitions that the p-th component (x o y)p
of x o y is a polynomial with rational coefficients in xo,yo, x1)y1,
■ ■ -, x,,yP. Also one sees easily that
A5) (* + y)v = xv + y, +f,(xo,yo, ••-, *,_i,>_i)
where/, is a polynomial in the indicated indeterminates. The
basic result which we shall now establish is that (x o y)v is a poly-
polynomial in #Oj yo, " • • > xv, yv with integer coefficients.
Throughout our discussion we write a? = (a@), «A), •••,
a(-m~V)) if a = {a0, aXy • • •, «m_i) etc. Let I[xiy y}] be the ring of
polynomials in Xo, y0} • • •, xm_i, ym-i with coefficients in the ring
of integers /. If n is a non-negative integer we denote the ideal
p"I[xi} y,) by (/>") and we write c = d (pM) for c — d e (jf). Then
we have
Lemma 1. Let ju > 1, 0 < k < m — 1, a = {a,), b
0 < v < m - 1, «„, bvzl[xhy,]. Write cf = («w), #" =
TA<?« /A^ system of congruences
A6) *„ = Ki^), 0 <v <k
is equivalent to
A7) «w s ^)(^+'), 0 <»/</&.
Proof. We have «@) = «0> ^@) = ^o, so the result is clear for
k = 0. To prove the result by induction on k we may assume
that both sets A6) and A7) hold for 0 < v < k — 1 and prove
that under these conditions ah = bk(jf) if and only if aw = b(k)
(p"+*). It is clear that ak = bk(j/) if and only if pkak s pkbk
{p"+k). Hence, using A2), it is enough to show that (ap)(k~1) =
(bp)<-k~1)(jf+k) holds under the induction hypothesis. We have

ABELIAN EXTENSIONS 127
av =
), Q<v<k-\. Using \\\ = 0 (p), 1 < i < p - 1,
this gives arp = brp(p»+1), 0 < v < k — 1. Hence the induction
on k applies to ap and bp to give (a*)**"" s (^)(*-1)O"+1+*-1)
which is what is required.
We can now prove the basic
Theorem 9. If x o y denotes x+y,xyorx—y, then (x ° y)v is a
polynomial in xo,yo, xlyyu ■ • • ,xv,yv with integer coefficients.
Proof. Since (x o y)v is a polynomial in xo,yo, • • •, xv,yv with
rational coefficients, it suffices to prove that (x o y)v el[xi,yj\.
This is clear for (x o y)Q and we assume it for (x ° y)k, 0 < k <
v — 1. We have
A8) f{x o y), = (* o y)M - ((* o y)^'^,
by A2) and (x oy)M = xM ±yM el[xi}y,]. The induction
hypothesis implies that ((# ° y)p)(p_1) el[xi,yjl. Hence, by A8),
it suffices to show that (x o _y)W = ((x o yYY''^ (p"). We have
*W = (**")<'-" (/>') and ^("> = (yp)(r~1) O"), by A2). Hence
A9) (* o jf)« = *W ± jfW s
We are assuming that (x o y)k e I[xi,y}], 0 < k < v — 1. For any
polynomial with integer coefficients one has f(x0, y0, • • • )p =
f(xop,yop, •••)(?)• It follows that (* o j,)/ = (^ o jO* (?),
0 < k < v — 1. Hence, by Lemma 1, we have
B0) ((* o yW~» s (xp o ^)('-
By A9) and B0), (x oy)™ = ((«./)''-" (p"), which is what
was needed.
It is convenient to write the result we have proved as follows:
(x +y)v = s,(xo,yo, ■■, xv,yP) el[xi,y}]
B1) (xy)p = mr(x0,y0, ■ • -,xv,yp) eI[xi,y3]
(x — y)v = dr(x0,y0, • • -,xv,yv) eI[xi,y,].
We note also that, since @, • • •, 0) and A, •■•,!) are the zero and
identity elements of 3E(m) and @, • • •, 0)*" = @, • • •, 0), A, 0, • • •,

128 ABELIAN EXTENSIONS
0)" = A, • • •, 1), then @, • • •, 0) and A, 0, • • •, 0) are the zero
and identity of £m. Let y be an algebra homomorphism of £ over
Ro into itself and assume that xj1 = av, yj> = bvy 0 < v < m — 1.
Then we have Ow)" = aw, (jyw)" = ^w, ((# + _y)w)" = «w +
*« and ((* +jO,)' = (« + *)r. Hence, by B1), (a + b), =
sv(a0, b0, ■ • •, av, bv) and similar formulas hold for (ab)v and
(a — £)„. Since there exists a homomorphism 77 of X over i?0 such
that xyv a.ndy,v are arbitrary elements of X, these formulas hold for
all a, b e Hm. Evidently they imply that, if 33 is any subring of
Ro[xi,yj, Zfc], then the set 33m of vectors (b0, biy • • •, bm_i) with the
bv e S3 is a subring of Hm. In particular, this holds for 33 = g) =
I[xt,yh 2*] and for 33 = W = I[xi,y}]._
We are now ready to define the ring 9Bm(SI) of Witt vectors.
Here 21 is any commutative algebra over the field Ip ofp elements,
where p is the prime used above. The elements of9BmBl) are the
"vectors" (a0, au ■ ■ ■, am_i), av e 21, with equality defined as
usual. If a = (a0, ■ ■ ■, am_i), b = (b0, • ■ ■, bm^), then we define
addition and multiplication in 2BmBl) by
(a + b)v = sy(a0, b0, ■ • •, ay, bv)
(ab)y = mv(a0, b0, ■ ■ -,av, bv).
Here we understand that a + b = ((a + bH, •••,(« + ^)m_i),
ab = ((abH, ■ • •, (tf£)m_i) and, if /(^o,Jo, ■ ■ •) is a polynomial
with integer coefficients, then f(a0, au • • •) is the element of 21
obtained by replacing the integer coefficients off(x0,yo, • • •) by
their cosets in Ip, xv by ay, yv by bv, 0 < v < m — \. These re-
replacements amount to applying the homomorphism of I[xi,y}]
into 2t such that »—>« = « + (p), n el, xr —> av,yv —> <£„.
Now suppose « = («„), ^ = (^), c = (cv) are any three elements
ofSBmBl). We have a homomorphism of/[*,-,_yy, z*] into 21 such
that « —» n,n e I,Xy —> «v, ^v —> ^, z,, —> £•„. Consider the sub-
ring 3m of Hm of vectors (o>0, wi, • • •, wm_1) where wF e /[#;,jy;,
z*]. We have seen that, if / = (/„, • • -, tm-i) e 3m, then (w + /)„
= Sy(W0, tQ, •••, Wy, ty), (wt)y = OT,,(tt>0> ^0, ' ' ' J W,, /V) . It folloWS
that the mapping (w0, •••, Wm-i) -> (»o'j •••, »m-i') is a
homomorphism of 3m into the system (9BmBI), +, •) where +
and • are defined by B2). Note that our homomorphism maps x
into a, y into b, z into c. We remark also that any element w =

ABELIAN EXTENSIONS 129
(«;„) such that the wv e (p) is in the kernel of the homomorphism
of Qm into 21.
We can now prove
Theorem 10. 2BmCl) is a commutative ring.
Proof. Let a = («„), b = (bv), c = (cv) be any three elements
of 3BmBl). Then we have just seen that we have a homomorphism
of 3m into 2BmBt) such that * = (#„) -> a, j' = (y,) -> £, z =
(z,) —» f. Then the associative, commutative, and distributive
laws of addition and multiplication in $m give the same rules for
the elements a, b> c (e.g., {ab)c = a{bc)). The image of 0 = @,
• • •, 0) and 1 = A,0, • • •, 0) under our homomorphism are 0 =
@, • • •, 0) and 1 = (I, 0, • • •, 0) and the relations x + 0 = x, xl
= x give a + 0 = ay a\ = a inS33mBl). Also if we write a' for the
image of — x under the homomorphism, then we have a + a' = 0.
Since a, b, c are arbitrary in 2J3mBl), these remarks show that
SBmB0 is a commutative ring with 0 = @, ■ • •, 0), 1 = (I, 0, • • •,
0) as 0 and identity elements.
We shall call SBmBl) the ring of Witt vectors of length m over 21.
We remark thatSBiBI) can be identified with 21 itself since we have
the isomorphism a —» («) of 21 onto S3B1Bl).
Now let 8 be a subalgebra of 21 over Ip and form the ring
SZBmOB) of Witt vectors over S3. Then it is clear that b = {bv) —> b
is an isomorphism of 233mC3) into 9J3mBl). In this way we can
identify SBmOB) with the subring of 2BmBl) of the Witt vectors b
with i, e 58. In particular, if we take 58 = Ip we obtain the sub-
ring SEBOT(/p) of vectors with components in Ip. This subring
evidently consists of pm elements.
We define the mapping P of SBmBl) into itself by ap = (aop,
aip> '' "> «m-iJ>) for a = (a0, ax, ■ • ■, «m_i). We have noted that,
if f(xo,yo, ■■■) el[xi,yj\, then /(«„, K •■•)*_ = 7(«op, V, • • •)•
This and the definitions of addition and multiplication in SBmBl)
imply that
B3) (a + b)p = ap + bp; (ab)p = apbp.
We shall call P the Frobenius endomorphism in 9BmBl). We intro-
introduce the restriction mappingi? of2BmBl) intoSBm_iBl) by (a0, • • •,
am-i)R = («OJ • • -,am_2) and the shift mapping V of SBm_iBt)

130 ABELIAN EXTENSIONS
into 2BmBl) by (a0, • • -,am_2)r = @, a0, ■■-, am_2). It is im-
immediate that R is a ring homomorphism and we shall see that V
is a homomorphism of the additive group of 28»»_iBl) into that of
SB»(8(). We have
(«o> • • •> «m-i)riJ = @, «o, • • -, am-2) = («o, • • ■, am_x)RV.
Also it is clear that PV = FP, £P = Pi?, and (FR)m = 0 hold
We prove next the important
Lemma 2. The following relations hold in Witt rings:
, p ,
B4) pi = 1 + 1 + 1 • • • + 1 = 1VR
B5) (a + b)v = av + iv
B6) ayb={abr*)y, «effim(a), £e2B
B7) pa = aPVR.
Proof. Consider the subrings Sm_i, 3m, 3m+i of ^«-i> x»»
Im+i of elements with components in /[*», jyy, z*] and define the
mappings R and V for these in the same way as for the Witt rings.
Also we have the mapping P defined before. Consider the ele-
element 1 = A, 0, • • •, 0) of /. Set p = pi. We have V = A, 1,
••■,1) and
' P '
Hence pw =p, 0<v<m-l. On the other hand, lVR =
@, 1, 0, • • •, 0), so the definition of v gives A™)«» = 0, A™)M
= P,l < v < m - 1. Then we have A™)« = pw (p'+1), 0 < v
< m — 1.. By Lemma 1, this implies that (lrR), = py (p). We
have seen that there is a homomorphism of Sm into SBTOBI) such
that every w = (wv), wv e (p), is in the kernel. If we apply this
to 1VR and to p and use the foregoing relations on components,
we obtain B4) in 9BmBt). Next we note that xv = @, x0, ••-,
#m-i), yv = @,jo, • ■ -,ym-i) for x = (*0, • • •,#m_i), y = (jo,
• • -,ym-i) in 3fm. Then
(xr)M = px,?"-1 + p2x/~* + ••■ + p"xr_u 1 < v < m,

ABELIAN EXTENSIONS 131
by A1); hence
B8) (xr)W = px<>-l\ 1 < v < m.
Since (* + y)w = xw + /">, this and (xr)m = (yryo) = ((x +
y)y)W = o give ((# + y)v)w = (xv)M + (yr)M, 0 < v < m.
Hence (x + y)y = xv + yr holds in 3Wi- ^ we apply the
homomorphism of I[xi,yj, zk] into SI such that n —»• « = n + (/>),
*c —> «», jy^ —* K, zv —> o, to the components of (x + y)v and
^r + yv> we obtain B5) for a, b e«Bm(8). To prove B6) we shall
show that
B9) (xvy)v = {{xyPR)y)v (/>), 0 < , < m
if x = (x0, xi, ■ ■ ■, xm_x) e 3m and jy = iyo,yi, • • -,jym) e ^m+i,
^,-, yy indeterminates. Set x^jy = (t^0, tp1} ■••Jte;m), (xyPB)y =
(*o, h, • • •> '«)• Then we have to show that wv = tv (p), 0 < v <
m. By Lemma 1, this is equivalent to a>w = t(r) (j>"+1). This
holds for v = 0 since w@) = 0 = /@). For ^ > 1, we have, by B8),
that ww = pxl'-»yw and^tM = px^-^fj^Y'-^. Since jw =
g}ves tne congruences
Hence B9) holds. Applying a suitable homomorphism into SI,
we obtain B6). If we apply R to both sides of B6), we obtain
aVRbR = («^)™. Setting a = 1 and bR = c e«B.(a), we ob-
obtain \VRc = cPVR. Since 1™ = />1, by B4), this gives/* = cPVR.
Since c = IP can be taken to be any element of SBm(Sl), this is
equivalent to B7).
We can now derive the basic properties of 2Bm(SI) which we shall
need. We prove first
Theorem 11. SBm(Sl) is a ring of characteristic pm.
Proof. It suffices to show that the order of 1 in the additive
group of $EBmBl) is pn. We have seen that p\ = lVR = @, 1, 0,
• • •, 0) and by iterating B7) we obtain pH = @, 0, 1, 0, • • •)
etc. This shows that/)! = @, • • •, 0,1) ^ 0 bntpm\ = 0, as
required.
We have seen that, if S3 is a subalgebra of 31, then we can con-

132 ABELIAN EXTENSIONS
siderSBmE8) as a subring of3BmBl). In particular this holds if we
take 23 = Ip. Then Z = 2Bm(/p) is the set of vectors with com-
components in the field Ip and so the number of elements in Z is pm.
On the other hand, Theorem 11 shows that there are pm distinct
elements of the form kl, k an integer, in2BmBl) and these belong to
Z. Hence it is clear that Z is just the set of integral multiples of
the identity of 9BmBl). Evidently Z is isomorphic to the ring
I/(pm) of residues modulo pm. The following result gives an in-
insight into the structure of 3BmBl).
Theorem 12. The mapping a = (a0, aly ■ ■ ■, «m_i) —> a0 is a
homomorphism q/2BmBl) onto 21 whose kernel 91 is a nilpotent ideal.
Proof. We have seen that R is a homomorphism of 3BmBI) onto
SBm_iBt). Iteration of this shows that Rm~1 is a homomorphism
of9BmBl) onto9BiBt) = 21. Evidently tf7" is the mapping we
have indicated. The kernel of our homomorphism is the ideal 9?
of elements of the form @, a0, a1} •••,am_2). Hence 9? =
2BmBl)™. If we apply R to B6) we obtain aVRbR = {abPR)VR.
Since bR can be taken to be any element c in 2BmBl), this gives the
relation aVRc = {acp)VR in SBmBI). Then aVRcVR = (acPVR)VR =
(apcpyVR)* eSBmBI)(™>\ Thus 5TC2 = (9S«Bl)y-BJ C SR™. Now
assume that for some * > 2, ftk c gER<™>*"* c $R<™>M. Then if
</ = «rie e Sft and ^ e 9?*, we have * = c{VR)\ c c3BmBI), since * e
BI)(^)i. Hence db = a^f^*)* e 9i9f(^)*-1 and so
c 9l3l(Fa)M. Moreover, if a, c e3BmBI), then aVRc(rR)k =
e (g^ano*-)™ g (^(ra)*-1)^ = sr(^*. Hence
SR^*)* and so 9i*+1 c 9l(ra)*. This shows that
 holds for all k > 2. Since 9f = 9BmBtOi? and
= 0 this gives 9T = 0.
Corollary. An element a = (a0, ax, ■•■,am_1) is a unit in
2SmBl) if and only if a0 is a unit in 21.
Proof. This follows from Theorem 12 and the remark that, if 9?
is a nilpotent ideal in a ring SB, then an element a e 9B is a unit in SB
if and only if the coset a + 9i is a unit in 95/91. We leave the
proof as an exercise.
5. Abelian ^-extensions. It will be instructive to consider first
briefly the abelian extensions of a field $ of characteristic p ^ 0

ABELIAN EXTENSIONS 133
whose Galois groups G have exponent p (cf. ex. 3, 4, p. 98). In
this case we let Z be the additive cyclic group generated by the
element 1 of $ and we consider the character group Horn (G, Z)
where G is the Galois group of an extension P of the type specified.
The elements x e Horn (G, Z) are the mappings of G into Z
satisfying x(st) = x(J) + xW- Since xCO e Z C $, this can be
written also in the form x(-tf) = xW + xW so that we have an
instance of the additive analogue of Noether's equations. Hence
by Theorem 1.20, there exists ape? such that x(-0 — p' — P- Since
xO) e Z, x0)p = xW so (p5 - p)p = p* - p. This gives the
equation (pp — p)' = pp — p, s e G; hence pp — p = a e <£. Con-
Conversely, let p be any element of P such that p" — p = a e <£> and de-
define xW = Ps - P. Then xW* - xM = (pp - p)' - (pp - p)
= a' — a = 0. Hence x(s)p — x(-f) and this implies that xCO is
in the prime field, so xCO £ 2. Also we have x(-tf) = P3t — P —
(P8 - p)' + (p' - p) = (ps - P) + (p' - p) = xW + xW; hence
X e Horn (G,Z). Following the pattern of the Kummer theory
this leads us to consider the subset £(P) of P of elements p such that
pp — p e $. This is a subgroup of the additive group (P, +) con-
containing ($, +) and we have the mapping p —> xP) where xP(-0 =
p" — p, of S(P) onto Horn (G, Z). Since Z is an additive group the
composition in Horn (G, Z) is (x + ^)(J) = x(-f) + ^W- More-
Moreover, if p, aeS(P), then Xp+<rW = (p + <0s - (p + <0 = Xp(-f) +
Xff(-f); hence p —» Xp is a homomorphism of S(P) onto Horn (G, Z).
It is clear that the kernel of this homomorphism is <£. Hence
J(P)/# ^ Horn (G, Z) s G.
The next step in the discussion is to consider the subset j?(P) of
$ of elements of the form pp — p, p e <S"(P). This is a subgroup of
the additive group (<£, +) containing the subgroup of elements of
the form ap — a, a e $. One sees easily that the factor group of
Q(¥) relative to the last subgroup is isomorphic to «y(P)/$, hence
to Horn (G, Z) and to G. It can be shown that any subgroup of
($, +) containing the subgroup of elements a? — a, a e $, and
having a finite factor group relative to this subgroup is a group
Q(P) for an abelian /(-extension with Galois group of exponent
< p. The groups Q give a survey of these extensions in the same
manner that the group ./V(P) gave a survey of the Kummer ex-
extensions. We shall not work out the details here but instead we

134 ABELIAN EXTENSIONS
shall proceed to the general case of arbitrary p-extensions. The
idea here is to work in the ring 28m(P) of Witt vectors over the
given extension P where m > ey pe the exponent of G. Then the
subgroup Z of the additive group of 2Bm(P) generated by 1 is
cyclic of order pm; hence, Horn (G, Z) is a character group of G.
We shall need first of all the generalization of Theorem 1.20 to the
ring of Witt vectors and we proceed to derive this result.
Suppose first that P is a finite dimensional Galois extension field
of the field <£ of characteristic p 9^ 0 with Galois group G. Let
2Bm(P) be the ring of Witt vectors of length m > 1 over P. We
have seen that we can identify 28m(<£>) with the subset of£Bm(P) of
vectors 0 = (p0, Pi, ■ • ■, Pm-i) with the ft e $. If p = (p0, • • •,
pm_i) e28m(P) and s e G, we define p" = (p0*, • • •, pm_i')- It is
clear that p —» p" is an automorphism of 5!Bm(P) and that the set
of these automorphisms is a group isomorphic to G. We denote
this group again as G. Evidently p" = p if and only if p,'=py,
0 < v < m — 1. Hence 2Bm(f>) can be characterized as the sub-
ring of G-invariants of the ring 2Bm(P).
If p e SBm(P) we define its trace T{p) = £ p*. Evidently
seG
Tip)' = T(p), j e G, so T(p) eSBm(#). If p = (Po, Pl, • ■ •, p^O,
then the first component of T(p) is T(p0) (trace in P over <£),
since first components are added in forming a sum in 28m(P). We
recall that the automorphisms s e G in P are P-independent and
this implies that there exists a p0 e P such that T(p0) ^ 0. If p0
is chosen in this way and p = (p0, •■•)> then T(p) = (T(p0),
• • •) has non-zero first component. It follows from the corollary
to Theorem 12 that T(p) is a unit in 2Bm($). Hence we have
proved the following
Lemma 1. There exist pe9Bm(P) such that T(p)-1 exists in
We use this to prove the following key cohomology result.
Theorem 13. Let s —» m« be a mapping of G into SBm(P) such
that ft,t = fit* + nh s,tsG. Then there exists an element <r e 3Bm(P)
such that n, = a* — <r. Conversely, if a e 9Bm(P), then n, = a' — a
satisfies the given equations.

ABELIAN EXTENSIONS 135
Proof. The proof is identical with that of the special case of
Galois extension fields treated in Theorem 1.20. We choose p in
2Bm(P) so that T(p) exists in 2Bm($) and we let t =
. Then
T-T* =
Hence if we take a = — t, then we have /i, = ir' - ir as required.
Conversely, if we take n, — <r' — a where a is any element of P,
then we have n,* + y.t = <T8i — a* + <rl — a = <rst — <r = /*««•
We recall that the Frobenius mapping p —* pp = (pop, pip,
■ • ■, pmp) is an endomorphism of the ring 2Bm(P). We shall now
introduce the mapping ^ in SBm(P) defined by
C0) W) = PP -P.
It is clear that ty is an endomorphism of the additive group of
2Bm(P) (but not of the ring 3Bm(P)). The kernel of $ is the set of
vectors (p0, p\, ■ • -,pm_i) such that pvp = pv, 0 < p < m — 1.
Evidently this is just the set of vectors with components p» in the
prime field #0 (= IP)- Hence the kernel of <P is the set of Witt
vectors (p0, Pi, • • •, pm_i) with the pt- e #0- We have seen (after
Th. 11) that this is just the set Z of integral multiples of the
identity 1, and Z is a cyclic group of order pm under addition.
We now assume that the Galois group G is an abelian group of
order pf and that m > e where pe is the exponent of G. Let
C1) «y(SB»(P)) = \p e5Bm(P)|<P(P)
Then SQBm(F)) is a subgroup of the additive group BBm(P), +)
containing SBOTD>). If p e «S\2Bm(P)), then we define the mapping
X, of G by xp« = P4 - P. Then x,«p = p'p - PP = PP* - PP
= (p« + a) - (p + a) if <P(p) = a. Hence x,WP = P8 - P =

136 ABELIAN EXTENSIONS
XP(j). We have seen that this implies that xP(-0 e Z. Also we
havexp(^) = Pst ~ P = PH ~ P* + Pf ~ P = (ps ~ P) + (p1 - p)
= XpO) + XpO). Hence xP e Horn (G,Z). Next let />,«
J(SBm(P)). Then p + <r e £(SBm(P)) and Xp+aW = (p + <0a -
(p + <r) = (p« - p) + (<rs - a) = xp(-f) + XaO). This shows that
the mapping p —> xP is a homomorphism of 61CBm(P)) into
Horn (G, Z). If Xp(-f) = 0 for all s e G, then we have p* = p,
j e G, and this implies that p e SBm($). Hence the kernel of p —> Xp
isSBmCf1). Finally, we note that our homomorphism is surjective.
For, let xeHom {G,Z). Then x(st) = xO) + xW and, since
the x(-0 e -Z5 we have also x(-tf) = x(-f)f + x(/)- Hence, by Th. 13,
there exists a p e3Sm(P) such that x(-f) = Ps — P- Since xCO e Z,
x(-f)P = xO) and this gives (pF - p)' = pp - p. Hence $(p) =
Pp - p e2Bm($) and so p e ^(aBm(P)). We now see that the map-
mapping p -^ Xp of <S"(9!Bm(P)) into Horn (G, Z) is surjective and since
the kernel is 2Bm(<i>) we have ^BSm(P))/SBm(*) 2* Horn (G, Z) s
G. We have therefore proved the first two statements of the
following theorem which is a perfect analogue of Theorem 7:
Theorem 14. Let § be a field of characteristic p ^ 0, P/* ««
abelian p-extension whose Galois group G is of exponent pe and let
SBm(P) be the ring of Witt vectors of length m over P where m > e.
Let <SX393m(P)) be defined by C1). Then we have the exact sequence of
additive groups
0 -► 2Bm($) -> S(SB»(P)) ~» Horn (G, Z) -» 0
the homomorphism of 2Bm(<£) /j the inclusion mapping and
that ofSCSm(F)) into Horn (G, Z) is p -> xP, Xp(j) = p" - p. The
factor group 6'(aBm(P))/SBm($) is finite and is isomorphic to G. TA<?
^?f/^ P/$ w generated by the components of the vectors p e S(9&m(P))
and
p = #(po<», • • -.p^Wjpo1", ■ • •,pm-iB); • • -;pow, • • -,pm-i(r))
if and only if the cosets p(<> + ©„(*), p(i) = (p0(i), • • •, pm-i(i)),
The proof of the last statement is exactly like that of the corre-
corresponding statement of Theorem 7. We leave it to the reader to
check the details.

ABELIAN EXTENSIONS 137
Following the pattern of our treatment of the Kummer theory
we introduce next the set
C2) 0(SBM(P)) = {%0|p e6-(SBm(P))} = SBm($) D
This is a subgroup of the additive group (SBmC*), +) containing
$(SBm($)) the subgroup of vectors ty(a), a e2Bm($). Consider the
homomorphism
) +
of £BBm(P)) onto 0BBm(P))/$BBm(*)). An element p is in the
kernel of this homomorphism if and only if 'ip(p) = $(a), a e
%&„($). This is equivalent to $(p — a) = 0 which means that p —
a e Z. Hence it is clear that the kernel of the homomorphism is
2Bm($) and we have the isomorphism
C3) 0(SB»(P))/?(«B»(*)) S J(«Bw(P))/©»(*).
This implies that j?CB.»(P))/$BSm($)) is a finite group isomorphic
to Horn (G, Z) and to G. We wish to show next that, if Q is any
subgroup ofS33m(^>) containing l>pCB,n(ti>)) as a subgroup of finite
index, then Q = <2(9Bm(P)) for an abelian p-extension P over <£>.
For this we need
Lemma 2. Let j8 = (/30,/31} • • •, |8m_i)
wt j/j « finite dimensional separable extension field ¥ of 3> such that
P = #(p) = #(p0, pi, • • •, pm_i) a«^ /A<? element p = (p0, pls • • •,
pm-i) o/
Proof. If w = 1 we just have to construct a separable exten-
extension P = $(p) generated by a root p of an equation xp — x = /3, 0
a given element in <1>. Since the derivative (xp — x — /3)' = —1
the given equation has distinct roots so any field generated by a
root of this equation will satisfy the condition. Now suppose
we have already constructed a separable extension E = *(p0,
• • •, pm_2) so that the vector <r = (p0, ■ ■ •, pm_2) of 5ZBm_2 (E)
satisfies ^(tr) = (|30, ••-,/3m_2). Consider the polynomial ring
EM and the Witt ring SBm(E|>]). We take the vector y =
(po> • • •> Pm-2, *) in this ring and we form
(POP, • • •, Pm-2P, *P) ~ (p0, • • •> Pm-2, *).

138 ABELIAN EXTENSIONS
Then <P0) = (ft, 0i, • • •, 0»-2,/(*)), /(*) e E[*]. Hence @O,
01, • • •, 0m_2,/W) + (Po, • • •, Pm-2, *) = (p0P, • • •, Pm-2P, *P).
Using the formula A5) we see that
C4) x»=f(x) + x + y
where y e E. Hence f(x) = xp — x — y. The derivative argu-
argument shows that/(.v) = 0m_i has distinct roots. If P = E(pm_i)
where/(pm_i) = 0m-i, then P is separable over E so P = $(p0)
■••,pm_i) is separable over <£. Moreover, it is clear from the
formulas given above that p = (p0, • • •, pm-i) is an element of
mm(P) such that $(p) = 0.
We can now prove
Theorem 15. Let Q be a subgroup of BBm($), +) containing
^(SBmW) and having the property that 0/$(9Bm(*)) is finite. Then
there exists an abelian p-extension P of $ such that the exponent of
the Galois group is pe, e S m, and j?BB«(P)) = Q.
Proof. Let /3A), 0B), • • •, /3(r) be elements of Q such that the
cosets j8(i) + ^S(9Sm(*)) generate j?/Sp(S3Bm(*)). By Lemma 2, we
can construct a field P which is finite dimensional separable over $
and is generated by elements p,(i), 1 < i < r, 0 < v < m — 1,
such that ^_(p0(i), • • •, pm_i(i)) = 03o(i), • • •, 0m-iw) in SB»(P).
Let 0 be a finite dimensional Galois extension field of <£> containing
P. We form 2Bm(O) and let the Galois group G of Q/$ act in 3Bm(S2)
as before. If s e G and p(i) = (Po(i\ ■ ■ -,pm-i(i)), then <p(p(i)) =
0<o gives <P(p(i)s) = 0(i). Hence ^(p(i)s - p(i)) = 0 so p(i)<1 -
p(i) eZcS33m($). This implies that P'CP,;eG. It follows
that P is Galois over $ and so we may take fl = P. If s, t are in
the Galois group G of P over *, then p(i)s = p(i) + y{i) and p(>;)f =
p(i) _|_ j@ where r60, 5(i) e2Bm(«i>). Hence p(i)8( = p(i) + 7(i) +
g(») _ p(i)(« which implies that G is commutative. Also p(i)s* =
p(i) + ky{i), so p(i)8!'"' = p(i) since 2Bm(P) has characteristic pm.
This shows that jp™ = 1 and so G is of order pf and of exponent pe,
e < m. Let Xi be the character of G determined by p(i): Xi(s) =
p@» _ p(i)# Then it is clear that x*W = 1, 1 < « < r, implies
that s = 1. It follows that the x» generate the character group
Horn (G, Z). Hence if p is any element of 9Bm(P) such that
e 9Bm($), then we have xP = IIx,-"". This implies that p =

ABELIAN EXTENSIONS 139
Sw,-p(i) + (8, j3eSBm(<l>), m{ integers. Then <JJ(p) =
$(/3) e £?. Since p is any element of 6'(aBm(P)) this shows that
(?BBm(P)) C B. The converse is clear so the proof is complete.
The results which we have now obtained correspond to the
main results on Kummer extensions. They have the consequence
that two abelian ^-extensions Pi/*, P2/* with Galois groups of
exponent pe, e < m, are isomorphic if and only if B(SBm(Pi)) =
(?B8m(P2)) (ex. 2 below). We have also the order preserving cor-
correspondence between the subfields P/$ of a particular Q/3? and
the subgroups (?B8m(P)) of the additive group (SBm(<i>), +) (ex. 1
below). We shall now consider the special case of cyclic ^-exten-
^-extensions. We note first that it is an immediate consequence of our
results that the cyclic extensions of p dimensions of <t> have the
form #(p) where pp — p = @ e$ and /3 i y($), that is, 0 ^ a? —
a, a e <£. We shall now show that, if such an extension exists
over $, which is equivalent to the condition <£ j± $($), then there
exist cyclic extensions of pm dimensions over $ for any m = 1, 2,
• • •. This will follow from
Lemma 3. If /30, • • •, /3m_x e $, then /30 e $(<£) if and only if
P = 00, 0i, • • •, 0»_i) satisfies pm~ll3 e ?(©»(*)).
Proof. By B7), p*"-^ = @, • • •, 0, /So5). We have @, ■ ■ •,
0, /30) - @, • • •, 0, /So"") = @, • • •, 0, j80) - @, ■ • •, 0, ft>p) +
@, • • -, 0, W) - @, • • -, 0, /V') + • • • + @, • ■ •, 0, ^op"") ~ @,
• •., 0, /V") e ?(«B»(#)). Hence pm~^ = @, • • •, 0, iSo11"")
e $BBm($)) if and only if @, • • •, 0, /30) e ^(S3Sm($)). Suppose
this holds, say, @, • • •, 0, /30) = ap — a where a = (a0, «i, • • •,
am_x). Then a1"* - «« = @, • • •, 0, ^oM = 0 so aiep = a*
and, if 7 = (a0, «i, • • •, am_2, 0), then yp = 7 so 5 = a — 7
satisfies Sp - 8 = @, • • •, 0, 0O). Moreover, 5 = @, • • -,0,
Sm_0. This implies that Sm_ip - 8m_! = ^0 so ft, e ?(*)•
Conversely, if this condition holds so that /30 = am-ip — cem-i,
then o^ - a = @, • • •, 0, 0O) for a = @, • • •, 0, am_x).
We can now prove
Theorem 16. Let $ be afield of characteristic p j± 0. TA^« M«r
(»*/j/ cyclic extensions of pm dimensions, m = 1, 2, 3, • • • odw $
7/ «»</ o«/y i/ there exist such extensions of p dimensions. The con-
condition for this is $ j

140 ABELIAN EXTENSIONS
Proof. We have seen that there exists a cyclic extension of p
dimensions over <£ if and only if $ 9^ $($). Suppose this condi-
condition holds and choose /30 e <S>, i $($). Let 0 = (/30, ft, • • •,&„_!)
where the /3,-, z > 0, are any elements of "f>. We have shown that
pm~lP i $($) and this implies that the subgroup Q of 2
generated by /? and $CBm($)) has the property that (?/$BB
is cyclic of order pm. By Theorem IS, Q = (?(P) for an abelian p-
extension P. Moreover, we have seen that the Galois group G of
P/$ is isomorphic to Q/ty(%£m($)) and so this is cyclic of pm di-
dimensions over <i>.
EXERCISES
1. Let Pi and P2 be two abelian ^-extensions of $ contained in the same field
0. Show that Pi 3 P2 if and only if (?(9Bn(Pi)) 2 !?(SBm(P2)) where m > et,
pei, the exponent of the Galois group of P»/$.
2. Let Pi, m be as in 1, but do not assume that the P< are contained in the
samefi. Show that Pi and P2 are isomorphic over $ if and only if C(SBm(Pi)) =
m
3. Prove that if 0 is an element of 2Bm($) such that pm~^ e ^}(SBm($)), then
there exists a 7 in SBmOl1) such that py = |8. Use this to prove that any cyclic
extension of pm~1 dimensions over $ of characteristic p can be imbedded in a
cyclic extension of pm dimensions over $.

Chapter IV
STRUCTURE THEORY OF FIELDS
In this chapter we shall analyze arbitrary extension fields of a
field $. A study of finite dimensional extension fields and a partial
study of algebraic extensions has been made in Chapter I. In this
chapter our primary concern will be with infinite dimensional ex-
extensions and we shall begin again with the algebraic ones. We
define algebraically closed fields and prove the existence of an
algebraic closure of any field. We shall extend the classical Galois
theory to apply to infinite dimensional normal and separable ex-
extensions. After this we shall consider arbitrary extension fields
and we shall show that these can be built up in two stages: first a
purely transcendental one and then on top of this an algebraic ex-
extension. The invariant of this mode of generating a field is the
transcendency degree which is the cardinal number of a transcend-
transcendency basis. We shall obtain conditions for the existence of a
transcendency basis such that the extension is separable algebraic
over the purely transcendental extension determined by the
basis. We shall also give a definition of separability of an ex-
extension field that generalizes the notion of algebraic separability.
The notion of a derivation plays an important role in these con-
considerations. Moreover, this notion can be used to develop a
Galois theory for finite dimensional purely inseparable extensions
of exponent one. We shall consider also briefly the notion of a
higher derivation that is useful for purely inseparable extensions
of exponent greater than one. At the end of the chapter we con-
consider the tensor product of extension fields, neither of which is
algebraic, and we apply this to the study of free composites of
fields.
141

142 STRUCTURE THEORY OF FIELDS
1. Algebraically closed fields. The "fundamental theorem of
algebra" states that every algebraic equation/(#) = 0 with co-
coefficients in the field of complex numbers has a root in this field.
Any field that has this property is called algebraically closed. If $
is an algebraically closed field, then every polynomial/(#) e $[x]
of positive degree has a linear factor x — p in $[x] and, conse-
consequently, every/(.v) can be written as a product of linear factors in
$[#]. Clearly, the converse holds also: If every polynomial of
positive degree in $[x] is a product of linear factors in $[x], then $
is algebraically closed. We recall that a field $ is called alge-
algebraically closed in an extension field P if the only elements of P
which are algebraic over <E> are the elements belonging to $ (§ 1.9).
We now note that a field * is algebraically closed if and only if it is
algebraically closed in every extension field. Thus let $ be alge-
algebraically closed and let P be an extension field. Let p e P be
algebraic over <i> and suppose f{x) is its minimum polynomial.
Since/(#) is irreducible and $ is algebraically closed,/^) is of first
degree. Hence pe$. Conversely, suppose 4> is algebraically
closed in every extension field and let/(tf) be an irreducible poly-
polynomial of positive degree belonging to $[x]. We can form the ex-
extension field P = $[*]/(/(#)) whose dimensionality is the degree
of/(x). Since P is algebraic over $ and $ is algebraically closed
in P, P = <f>. Hence degf(x) = 1, which shows that the only ir-
irreducible polynomials of positive degree in $[x] are the linear ones.
This means that $ is algebraically closed.
Let $ be an arbitrary field and let P be an algebraically closed
extension field of <£. Let A/$ be the subfield of elements of P/$> of
algebraic elements. If f{x) z A[x], we have/(,v) = II (# — p,) in
T[x] and the pt are evidently algebraic over A. Since A is alge-
algebraically closed in P (§ 1.9), the p* e A. Hence we see that every
polynomial of positive degree in A[x] is a product of linear factors
in A[^]. This implies that A is algebraically closed. It is there-
therefore clear that, if there exists an algebraically closed field con-
containing a given, field *, then there exists such a field which is,
moreover, algebraic over $. This leads to the definition: An ex-
extension field A/<£ is called an algebraic closure of $ if: 1) A is alge-
algebraic over $ and 2) A is algebraically closed. We proceed to
prove the existence and uniqueness in the sense of isomorphism of
an algebraic closure for any field <£.

STRUCTURE THEORY OF FIELDS 143
If $ is countable there is a rather straightforward way of con-
constructing an algebraic closure of $. Thus, in this case, it is easy to
enumerate the polynomials of positive degree with leading co-
coefficients 1. Let/i(^),/2(*),/3(^), • • • be such an enumeration.
Then we begin with <£0 = $ and we construct $,• inductively as a
splitting field over <£>,-_i of/,•(*). There is a simple way of making
precise the notion of the union A = U $,- of all the <$,-. Once this
has been done, one can prove that A is an algebraic closure of $>
in the following way. First, it is clear that A/$ is algebraic. Let
P be an algebraic extension of A and let p e P. Since p is algebraic
over A and A is algebraic over <£>, p is algebraic over #. Hence the
minimum polynomial/(•*•) over $ is one of the polynomials fi(x),
say f(x) = fn(x). Since #„ contains all the roots off(x), p e $„
C A. This shows that A is algebraically closed.
The procedure just sketched can be used also in the general case
by invoking transfinite induction. However, we prefer to give
another construction which will be based on Zorn's lemma.* We
shall need also the following
Lemma. If A. is an algebraic extension of an infinite field $, then
the cardinal number | A | = | <£> |.
Proof. Let 2 be the subset of $[x] of polynomials of positive
degree with leading coefficients 1 and let 2(n) be the subset of 2 of
polynomials of degree n + 1 = 1, 2,3, • • •. The elements of
2(n) have the form xn+1 + axxn + a2xn~l -\ \- an, a,- e *, so
2(n) has the same cardinal number as the «-fold product set
$ X $ X-X $. Since $ is infinite, |2(n)| = |$ X---X $| =
|*|. Also |2| = |U2(n)| = |*|.f We now map each/(*) e2
into the finite set R/ (possibly vacuous) of its roots in A. Since
every element of A is algebraic, U R/ = A. Since each R/ is finite
the cardinal number of the collection {R/} of these subsets of A is
the same as |A|. Hence |A| = \{R/}\ < |S| = |*|. This
implies that | A | = | $ |.
* This has been used in several places in Vol. II. An adequate account of this lemma or
"maximum principle" can be found in Kelley's General Topology, D. Van Nostrand Co.,
Inc., Princeton, N. J., 1955, p. 33.
t For properties of cardinal numbers, see Sierpinski's Lecons sur les Nombret Transfinis,
Paris, 1928.

144 STRUCTURE THEORY OF FIELDS
We can now prove
Theorem 1. Any field has an algebraic closure.
Proof. If <£ is a given field, then we can imbed <E> in a set Q which
is very large compared to <E> in the following sense: if <£ is finite,
then 0 is not countable and, if <E> is infinite, then \Q\ > | $|. We
now make extension fields out of subsets E of 0 containing <£.
More precisely, we consider the collection T of all triples (E, +, •)
where E is a subset of fi containing f>, and + and • are binary
compositions in E such that E with these compositions as addition
and multiplication is an algebraic extension field of $. We par-
partially order T by defining (Ei, +i, -0 < (E2, +2, -2) if E2 is an
extension field of Ex. Any linearly ordered subcollection (Ea,
+a, -a) of T has an upper bound whose underlying set is the union
of the Ea and whose addition and multiplication are defined in the
obvious way. Thus Zorn's lemma is applicable and it gives a
maximal element (A, +, •) in the collection r. We assert that A
is algebraically closed. Otherwise, A has a proper algebraic ex-
extension B. By the lemma, |B| = |A| = |$| if $ is infinite. For
finite $, |A| and hence |B| is countable. Hence, in both cases,
|B| < |Q|. This implies that there exists a 1-1 mapping of B
into fi which is the identity on A. We can use this mapping to
convert the image B' in fl into a field over A isomorphic to E over
A. Then (B', +, •), where the + and • are the addition and
multiplication obtained by carrying over the + and • of B, is in the
collection P. Moreover B'dA and this contradicts the maxi-
mality of (A, +, •). Hence A is algebraically closed. Since A is
algebraic over $, A is an algebraic closure of <i>.
We shall generalize next the notion of a splitting field and we
shall prove a result for these which will give the uniqueness of
algebraic closures as a special case. For this we consider a collec-
collection fl of polynomials of positive degree with coefficients in $.
We shall say that an extension field P/$ is a splitting field of ft if
A) every polynomial in fl is a product of linear factors in T?[x] and
B) no proper subfield of P/<f> satisfies A). Let A be an algebraic
closure of $ and let P be the subfield of A/* generated by all the
roots of the polynomials/eQ. Evidently P is a splitting field
over $ of the set fl. It is clear also that A itself is a splitting field

STRUCTURE THEORY OF FIELDS 145
over "£> of the complete set of polynomials of positive degree be-
belonging to $[#]. The isomorphism over $ of any two algebraic
closures of $ is clearly a consequence of the following extension of
the theorem on splitting fields of a polynomial (Th. 1.8).
Theorem 2. Let a —> a be an isomorphism of a field $ onto a
field<b and let Q,be a set of polynomials of positive degree contained in
<£[*•], 0 the set of images of the f e Q, under the isomorphism g(x) —»
g(x) of <£[#] onto $[#]. Let P/f> be a splitting field of 0 and P/f a
splitting field of J2. Then the isomorphism of $ onto $ can be ex-
extended to an isomorphism of P onto P.
Proof. We consider the collection A of isomorphisms s of sub-
fields of P/$> onto subfields of P/$ which coincide with the given
isomorphism a —> a of $ onto$. We can partially order A = {s}
by defining st < s2 if s2 is an extension of su Then it is clear that
A is inductive, that is, every linearly ordered subset of A has an
upper bound. We may therefore invoke Zorn's lemma to obtain a
maximal element / e A. We assert that / is an isomorphism of P
onto P extending a —» a. Otherwise, the domain of definition oft
is a proper subfield E of P/$. Since P/$ is a splitting field of Q
and E c P, there exists a polynomial/(#) e fl that does not have
all of its roots in E. Hence if Pi, P2, • • ■, pn are these roots in P,
then E(pi, p2) • • •> Pn) 3 E and evidently E(pi, ■•-,pn) is a
splitting field over E of/(*). On the other hand, E = E' can be
imbedded in a subfield of P which is a splitting field over E of
f(x) e 0. The theorem on a single polynomial can now be applied
to give an extension of t to an isomorphism of E(pi, • • •, pn) onto
the splitting field over E of f(x). This contradicts the maximality
of t, so we see that E = P. Evidently the image P' is a splitting
field over $ of fi. Hence P' = P and the theorem is proved.
If we take $ = # and as«in this result we see that any two
splitting fields over $ of a set of polynomials are isomorphic over
<£. In particular, we have the
Corollary. Any two algebraic closures of afield $ are isomorphic
over #.
Let A be an algebraic closure of a field <£. There are two sub-
fields of A/# which are of particular interest. The first of these is

146 STRUCTURE THEORY OF FIELDS
the subfield 2 of separable elements over $. This can be defined
also as a splitting field over $ of the set of separable polynomials
belonging to $[x]. We shall call 2 a separable algebraic closure
of <£. Next let <£ be of characteristic p 9^ 0 and let <£>p~" be the
subfield of elements of A which are purely inseparable over <£>.
By Lemma 2 of § 1.9, these are the elements of A which are roots
of equations of the form xp° — a = 0, a in <£. If $ is of charac-
characteristic p j^ 0, we call <l?p~" a perfect closure of $>, and if $ is of
characteristic 0, then the perfect closure of $ is taken to be $ it-
itself. It is immediate that for p j^ 0 every element of $p~" is a p-
th power so the mapping a —» dP is an automorphism of $p"".
Moreover, $*"" is the smallest subfield of A over <i> which has the
property that all of its elements are ^>-th powers in this sub-
subfield.
A field <£ is called perfect if every algebraic extension of <i> is
separable. The perfect closure of any field which we have just de-
defined is a perfect field; for we have the following
Theorem 3. Any field of characteristic 0 is perfect and a field $
of characteristic p 5^ 0 is perfect if and only if 3? = $p, that is, every
element of Q is a p-th power in <i>.
Proof. The first statement is clear since inseparable poly-
polynomials exist only for characteristic p 5^ 0. Now let <3? be of
characteristic p 9^ 0 and suppose $p c <i>. Let a be an element
of $ which is not a p-xkv power in $>. Then we know that xp — a
is irreducible and inseparable in $[x] (Lemma in § 1.6). Then
P = &[x]/(xp — a) is an inseparable extension of f> different from
$ so * is not perfect. Conversely, assume that $p = $ and let
f(x) be a polynomial in $[x] such that/'(#) = 0. Then we can
write /(*) = g(xp) where g(x) = xm + iSi* H \- /3m. Let
7ip = fr, * = 1, ■ • •, m, and set h(x) = xm + y^™'1 H \- ym.
Then we have/(#) = g(xp) = h(x)p. Thus every polynomial in
*M having zero derivative is a p-th power and so there exist no
irreducible inseparable polynomials of positive degree in $[x].
Hence $ has no proper inseparable algebraic extension field.
If $ is a finite field of characteristic p> then the isomorphism a
—> ap of $ into <£>p is necessarily an automorphism. It follows
from Theorem 3 that every finite field is a perfect field.

STRUCTURE THEORY OF FIELDS 147
EXERCISES
1. Let E be an algebraic extension of a field $ and let A be an algebraic closure
of $. Show that E/<£> is isomorphic to a subfield of A/$. (Hint: Consider the
algebraic closure of E and note that this is an algebraic closure of <£.)
2. Show that, if $ is of characteristic p ^ 0 and | is transcendental over $,
then $(£) is not perfect.
3. Prove that any algebraic extension of a perfect field is perfect.
4. Let $ be a field, <f>* its perfect closure. Prove that either <t>* = $ or
[#*:<!>] ;s infinite.
5. Prove that any algebraically closed field is infinite.
A field is called absolutely algebraic if it is algebraic over its prime field.
Examples are finite fields. We recall that for every prime p and every integer n
there exists one and in the sense of isomorphism only one field of cardinality pn
(§ 1.13). This result was generalized by Steinitz to arbitrary absolutely alge-
algebraic fields of characteristic p 5^ 0. We indicate this in the following exercise.
6. A Steinitz number is a formal product A*" = lip,-** over all primes pi where
ki = 0, 1, 2, • ■ •, or w. If M = Hpi'i is a second Steinitz number, we say that
M is a divisor of N (M\N) if /< < ki for all i. This leads in an obvious way to a
definition of the least common multiple (L.C.M.) of any collection of Steinitz
numbers. Let $ be absolutely algebraic of characteristic p. Define deg $ to
be the Steinitz number L.C.M. of the degrees of the minimum polynomials over
the prime field (= Ip) of the elements of $. Note that if $ is finite, then | $ | =
/>dei*. Prove that for any given prime p and Steinitz number N there exists an
absolutely algebraic field $p,jv of characteristic p and deg $Pijv = N. (Hint: Let
rn be the highest common factor of N and »!, so that rn\rn+i, n — 1, 2, • • •.
Let $„ be a field of cardinality pr» and suppose "^n £ $n+1 C • • •. Then
*j>,at =.U$B.) Show that any two absolutely algebraic fields having the same
prime characteristic and Steinitz degree are isomorphic. Prove that $p,m is iso-
isomorphic to a subfield of $p,n if and only if M \ N.
2. Infinite Galois theory. In this section we shall give a
generalization of the fundamental theorem of Galois theory to
certain infinite dimensional algebraic extensions P/#. We
assume that P is a splitting field over * of a set fi of separable
polynomials and we prove first the following
Lemma 1. Any finite subset of P is contained in a subfield E/$
which is finite dimensional Galois.
Proof. Let/ be a polynomial which is a product of a finite
number of polynomials contained in the set 0. Then it is clear
that P contains a splitting field P//$ off. Moreover, we know
that P/ is finite dimensional Galois over <i> (Th. 1.10). It is clear
also that, if/ and g are both products of polynomials belonging to
Q, then Pfg is the subfield of P generated by P/ and Pg. It

148 STRUCTURE THEORY OF FIELDS
follows that UP/, the union of all of these subfields, is a subfield
of P over $. Since UP/ contains a splitting field of every g e ft,
it is clear that UP/ = P. This implies that any p,- e P is contained
in a subfield P/{ and consequently any finite subset {pi, p2, • • •, pm\
is contained in the subfield Pn/; which is finite dimensional
Galois over $.
Let G be the Galois group of P/4>. The following result gives
essentially the first half of the Galois correspondence.
Lemma 2. $ = I(G), that is, the only elements of P which are
G-invariant are the elements
Proof. We have to show that, if p e P, i$, then there exists
an automorphism s of P over $ such that p' j£ p. By Lemma 1,
p is contained in a subfield E/<& which is finite dimensional Galois
over $>. Since p $ $ there exists an element s of the Galois group
of E/$ such that p* ^ p. On the other hand, it is clear that P is a
splitting field over E of the set of polynomials 0 and consequently,
by Theorem 2, the automorphism s of E can be extended to an
automorphism s of P/<£>. Evidently s eG and p' = p' 9^ p.
The full intermediate subfield-subgroup correspondence which
holds in the finite dimensional case fails if P is of infinite dimen-
dimensionality. As an example of this we consider the algebraic closure
P of the field $ = Iv of p elements. Since <£ is perfect, all poly-
polynomials of $[x] are separable and so P is a splitting field over $
of a set Q of separable polynomials contained in <£[#]. Let G be
the Galois group of P/$ and let H be the subgroup generated by
the automorphism ir'.p —> pp. (It is clear that this is an auto-
automorphism.) The subfield I{H) of //-invariants is <£> since the
only elements p such that pp = p are the p elements of $. We
shall now show that H is a proper subgroup of G; then we shall
have two subgroups of G, namely, G and H which have the same
subfield of invariants. To do this we note that, if p" is any power
of p, e > 1, then P contains a subfield $e of order p'. We recall
also that f>e C $/ if and only if e \ f (§ 1.13). Now let / be a prime
and let $z« denote the union of the fields in the sequence <£>* c <i>j»
c$;i c---. It is immediate that <!>*" is a proper subfield of P
and P is a splitting field over <!>*« of the set Q. Hence Lemma 2
shows that there exists an automorphism s of P over <£j» such that

STRUCTURE THEORY OF FIELDS 149
s 9^ 1. Now s is not a power of the automorphism ir; for if s =
ir*, then the subfield of j-invariants is the finite set of elements
satisfying ppk = p. This set must include$i» and this is impossible
since <£*« is an infinite subfield of P.
This type of difficulty in the infinite Galois theory was first ob-
observed (in the field of algebraic numbers of the rationals) by
Dedekind. The way out of the difficulty was found by Krull who
saw that it was necessary to restrict the correspondence to sub-
subgroups of the Galois group which are closed in a certain topology
which we shall now define.
The topology one needs is essentially the same as the finite
topology which we introduced in Vol. II, p. 248, for the set of
linear transformations of one vector space into a second one.
We consider the set Pp of (single-valued) mappings of the field
P into itself. If (£1} £2, • • ■, £«) and (rn, tj2, • ■ •, Vm) are finite
sequences of elements of P, then we let O(f;, t/,) be the subset of
Pp of all s such that &' = rn, i = 1, • • •, m. The sets O(&, jj,-)
can be used as a basis for a set of open sets which make Pp a
topological space (cf. Vol. II, p. 248). This topology of Pp is
called the finite topology.
If G is any subset of Pp, that is, any set of mappings of P into
itself, then we topologize G as a subspace of Pp. In particular, we
shall do this for the Galois group G of P/$. We now prove that
the fact that P is algebraic over $> implies that G is a closed sub-
subset of Pp. Thus let s belong to the closure of G and let £, r\ e P,
a e <i>. Then there exists an j e G such that a" = a', !■' = £%
n' = rf, (f + v)° = tt + V)S, (&)' = (£?)'"• Since a* = <*,($ +
i?)* = Is + Vsy (&)' = £V we have the same relations for J and
these show that s is an isomorphism of P/$ into itself. To see
that J is surjective we let £ be any element of P and we let E be
the subfield of P/$ generated by all the roots £' in P of the mini-
minimum polynomial /(*) of £ over <i>. Clearly [E:f>] < oo. Since J
is an isomorphism of P/<J? into itself, E1 C E. Hence the restric-
restriction of J is a linear isomorphism of E/$ into itself and so this map-
mapping is surjective. Thus there exists an rj e E such that rf = £.
Hence s is an automorphism of P/*, so I e G and G is closed.
We can prove the following fundamental theorem of the in-
infinite Galois theory.

150 STRUCTURE THEORY OF FIELDS
Theorem 4. Let P/<£> be a splitting field of a set SI of separable
polynomials with coefficients in $ and let G be the Galois group of
P/<i>. With each closed subgroup H of G we associate the subfield
E = I{H) of H-invariants and with each subfield Eo/P over $ we
associate the Galois group ^/(E) of P over E. Then these two cor-
correspondences are inverses of each other. Moreover, a closed subgroup
H is invariant in G if and only if the corresponding field E = I{H) is
Galois over <£> and, in this case, the Galois group of E/<£ is isomorphic
to G/H.
Proof. If E is a subfield of P/<t>, then P is a splitting field over
E of fi. Hence if H = J(E) the Galois group of P/E, then H is
closed and Lemma 2 shows that I(J(E)) = E. Next let H be a
closed subgroup of G and let E = I{H). We have to show that,
if s is an automorphism of P/E, then s e H. Since H is closed it is
enough to show that s is in the closure of H, that is, if pi, ■ • •, pn e
P, then there exists a /eH such that pi = p', 1 < i < n. Let
A/E be a subfield of P/E which is finite dimensional Galois and
contains {p»} (Lemma 1). Then j and the / e H map A into itself
and so their restrictions are elements of the Galois group of A/E.
If the restriction s' of j to A coincides with no restriction / of
teHtok, then the group H' of the restrictions of / e H is a proper
subgroup of the Galois group of A over E. Consequently, there
exists an element £ e A, ^ E such that £' = £ for every t eH. This
contradicts the definition of E as I{H). This proves the first state-
statement. If H is a closed subgroup and s e G, then s~lHs is closed,
and if E = I(H), then E3 = /(s^Hs). It follows that H is
invariant in G if and only if Es = E for every s e G. If this con-
condition holds, then the set of restrictions to E of the s e G is a
group of automorphisms G in E whose set of invariants is<J>. Hence
E is Galois over <$. Conversely, assume E is Galois over $ and let
G be the Galois group of E/<£. If e e E and s eG, then e and e*
have the same minimum polynomial over $>. Hence e has only
a finite number of conjugates e% s eG. If these are i\ = e,
«2) • • -j «r, then the polynomial f(x) = n(# — «,■) has coefficients
in $ and e is a root of /(#) =0. If s eG, then es is also a root of
/(#) = 0 so e" = e,- e E. Since e is arbitrary, this shows that
E* Cj E for s eG. This implies that /f is invariant in G. Since

STRUCTURE THEORY OF FIELDS 151
P is a splitting field over E of a set of polynomials belonging to
<£[#], any automorphism of E/<i> can be extended to an automor-
automorphism of P/<i>. It follows easily from this that, if E is Galois over
<f>, then the mapping s —> J the restriction of s to E is a homo-
morphism of G onto G. The kernel is H = ^(E) so G ^ G/H.
EXERCISES
1. Show that the hypothesis of Theorem 4 can be replaced by: P/4> is a
separable and normal algebraic extension, where we define normality by the
condition that, if/(.v) is irreducible in $[x] and has a root in P, then/(*) is a
product of linear factors in P[«], (cf. § 1.8).
1. A topological space is called discrete if every subset is open. Let P be a
splitting field over $ of a set of separable polynomials and let G be the Galois
group of P over <i>. Show that G is discrete if and only if [P:$] < ».
3. Let P, <£, and G be as in ex. 2. Use the fact that every p e P has only a
finite number of conjugates and the Tychonoff theorem to prove that G is a com-
compact group.
4. Let P be the algebraic closure of the field $ = Ip and let G be the Galois
group of P over f>. Show that G is a commutative group. Let $;" be the sub-
field of P defined for the prime / as in the text. Let v be the automorphism
p -^> p? restricted to $ f. Show that wl —* 1 in the sense that, if S is any finite
subset of $j", then there exists a positive integer N such that £*'* = £ for all
£ e S provided k > N. Let mi, m%, • • • be a sequence of integers such that for
any positive integer k there exists an N such that mT = ms (mod /*) if r, s > iV.
Show that the sequence of automorphisms irmi, ir1, • • • converges to an auto-
automorphism er of 4?*" over $ in the sense that irmicr~l —> 1.
5. Let G be a group of automorphisms in a field P and let $ = I(G). Assume
G is a compact subset of Pp. Show that this implies that for every £ £ P the set
{ij* | j e G} is finite. Hence prove that P is the splitting field over $ of a set of
separable polynomials and that G is the Galois group of P/<1>.
6. Let G be the Galois group of P/<i> where P is algebraic over $ and let \Ga}
be the collection of invariant subgroups of finite index in G. Show that C\Ga
= 1.
7. Let $ be a finite field, A its algebraic closure, and G the Galois group of
A/$. Show that G has no elements of finite order ^ 1.
8. Let Ap be the algebraic closure of the field Ip of p elements, Gp the Galois
group of Ap/Ip. Show that Gp = GQ for any two primes p, q.
9. Let P = <J?(£i, £2, • ■ ■) the field of rational expressions in an infinite number
of indeterminates. Show that the Galois group of P/<i> is not closed in the finite
topology.
3. Transcendency basis. We have defined the property of
algebraic independence over $ for a finite subset {£1, £2> ■••,!;«}
of a field P over $ in Vol. I and this definition has been repeated
in Introduction, p. 4. We now extend this notion to arbitrary

152 STRUCTURE THEORY OF FIELDS
subsets by stating that such a set S is algebraically independent if
every finite subset of S is algebraically independent. A set which
is not algebraically independent will be called algebraically
dependent; hence a set is algebraically dependent if and only if
it contains a non-vacuous algebraically dependent finite subset.
We shall now introduce another notion, which we shall see in our
first theorem is intimately related to those just given.
Definition 1. Let S be a subset of P over <£ and let p be an element
of P; then p is said to be algebraically dependent over $ on S if p
is algebraic over ${S).
We note first that, if p is algebraically dependent over $ on S
and f(x) e$(S)[x] is the minimum polynomial of p over
then the coefficients of f(x) are contained in a subfield
where F is a finite subset of S. Hence it is clear that p is algebrai-
algebraically dependent over $ona set S if and only if p has this property
for a finite subset F of S.
Theorem 5. A non-vacuous subset S of a field P/$ is algebrai-
algebraically dependent over <i> if and only if there exists an element £ e S
which is algebraically dependent over $ on the complementary set
Proof. The remarks we have made show that it is sufficient to
assume 6* is finite, say, S = {£1} £2j • • •, £»}• Assume the condi-
condition stated holds. Then we may suppose that £„ is algebraic over
*($ij • • -i £n_i). Let /(#) e<l>($i, • • •, £n_i)M be the minimum
polynomial of £„ over $(&, ■ ■ •, £n_i) and let ft, 182, ••-,&„ be its
coefficients. Now every element of $(|i, • • •, £n-i) has the form
£(£1, • • •, $n-i)h(£1} • ■ •, fn-i)-1 where g, h e$[#i, • • ■,xn_1], x{
indeterminates, and h(£u • • -, £n-i) ^ 0. In particular, 0,- =
gjtiu ■■■> *.-i)A/(«i, • • •, £«-i)~\ Ay(fi, • • •, €.-1) ^ 0. Set
• • •,#»_!) and
H h gm(Xi, • • ■ , Xn-.i)hm(X\, ■ • •, ^n-i)}.
Then F is a non-zero element in $[#1, • • •, xn], Xi indeterminates,
and we have -F(£i, • • •, £n) =0. This means that the £,• are alge-

STRUCTURE THEORY OF FIELDS 153
braically dependent. Conversely, assume there exists a non-zero
polynomial F(x1} •••,#„) e$[xu ■ ■ •, xn] such that F(tu ••-,£„)
= 0, which amounts to saying that the £,- are algebraically depend-
dependent over €>. We may assume that n is minimal and we may write
F(X1} •••,#„) = /oOl, ■ • •, Xn-^Xn™ + /i(^i, ■ • •, tfn-O*,, +
• • •+fm(xi, ■ • -,.vB_i) where /0^0 and m > 1. Since n is
minimal, /0(£i, • • ■, $»_i) ^ 0. Then
is a non-zero element of $(fi, • • •, £n-i)M such that /(£„) = 0.
Hence £n is algebraically dependent on £l5 • • •, £n_i.
The relation of algebraic dependence in a field P/<i> is a special
kind of relation between elements of P and subsets of P. Another
relation of a similar type is that of linear dependence of a vector
in a vector space on a subset of the space, and we shall encounter
still others. It is therefore worthwhile to treat such relations
axiomatically and we shall do this by considering an arbitrary set
P. A relation < between elements of P and subsets S of P (£ < S)
is called a dependence relation if the following conditions hold.
I. If £ e S, then £ < S.
II. If £ < S, then £ < F for some finite subset F of S.
III. If £ < S and every 77 in S satisfies 77 < T, then £ < T.
IV. If £ < £ and £ < S - {17} where 77 eS, then 77 < (S -
M) U {£} (Exchange axiom).
Now let P be a field over $ and let £ < £ for £ in P, S a subset
of P, mean that f is algebraically dependent on S over $. Then
we have
Theorem 6. Algebraic dependence in P/<i> is a dependence rela-
relation in the sense of I—IV.
Proof. I. This is evident. II. This was proved before. III.
Let £ be algebraic over $(£) and suppose every 77 e S is algebraic
over $(T). Consider the subset A of P of elements which are
algebraic over *(T). Then we know that A is a subfield of P/^T)
and A is algebraically closed in P. Now S C A and £ is alge-
algebraic over $(<?)> so over A. Hence £ e A which means that £ < T.

154 STRUCTURE THEORY OF FIELDS
IV. Suppose £ < S and £ < T = S - {77} where tj e S. Let E =
*(T). Then £ is transcendental over E and algebraic over E(i7).
Hence there exists a polynomial f(x, y) e E[x, y], x, y indetermi-
nates over E, such that f(x,y) 9^ 0 and /(£,??) = 0. We write
f(x,y) = ao(x)ym + a1{x)ym~1 -\ f- am(x) where the at(x) e
E[x] and ao(x) ^ 0. Then ao(!-) 5* 0 and m > 0, since £ is trans-
transcendental over E. The polynomial f(!-,y) is a non-zero poly-
polynomial belonging to E(£)[y] and rj is a root of f(!-,y) = 0. Hence
t\ is algebraic over E(£), which implies that j? is algebraic over
#(r u {$}). Thus 7? < r u {£}.
We now return to the general theory of dependence relations.
As before, P is an arbitrary set. We define a subset S of P to be
independent (relative to <) if no £ e^ is dependent on S — {£}.
Then we have the following
Lemma. If B is independent and £ is not dependent on B, then
B U {£} is independent.
Proof. Otherwise, we have an i\ e B such that -q < (B U {£}) —
{ri}. Since i\ < B — {??} the exchange axiom implies that £ < B
= {B — {rj}) U {rj} contrary to hypothesis.
A subset B of P will be called a basis for P (relative to <) if A)
B is independent and B) every £ in P is dependent on S. The
main result on dependence relations is the following
Basis theorem. The set P has a basis. Moreover, any two bases
have the same cardinal number.
Proof. To prove the existence of a basis we consider the collec-
collection / of subsets of P which are independent. (It may happen
that the vacuous set is the only member of /.) We order / by the
inclusion relation. If \S\ is a linearly ordered subset of/, then
US is contained in /. Otherwise, there is a £ e US which is
dependent on US — {£}. Then £ < F where F is a finite subset
of US — {£} and F U {£} is a finite subset which is not independ-
independent. Since {S} is linearly ordered, F U {£} c Tfor some T e {S}
and this contradicts the assumption that T is an independent
set. We now see that / is inductive and so, by Zorn's lemma, there
exists a maximal element B in /. Now let £ be any element of P.
Then £ is dependent on B since otherwise B U {£} is independent,
by the lemma. This would contradict the maximality of B.

STRUCTURE THEORY OF FIELDS 155
Hence every £ in P is dependent on B. This means that B is a
basis.
Now let B and C be two bases for P. We have to show that the
cardinal numbers \B\ = \C\. Assume first that B is finite, say,
B = {01, 02, • • ■, Pn}- We assert that there is a y = 71 e C such
that 7 is not dependent on {02, • • •, 0«}. Otherwise, by III, every
element of P is dependent on {/32, ••-,0n}. In particular, 0X
has this property, contrary to the independence of B. Now if yx
is not dependent on {02, • • •> 0™}, then {71,/32, ■■■,&*,) is in-
independent. Moreover, the exchange axiom shows that 0i <
{71, 02, •••>&»} so every 0,- < {71, 02, • • •> 0»}- Thus {71, 02,
• • -,0n} is a basis. We can repeat this process and obtain 72
in C so that {71, 72, 03, • • •, 0n} is a basis. Continuing in this
way we obtain a basis {71, • • •, yn} which is a subset of C and has
the same cardinal number as B. Since C is independent, this is
all of C and we have \C\ = \B\. Next assume \C\ and \B\ are
infinite. In this case we use a counting argument which is due to
Lowig (cf. Vol. II, p. 241). Let 7 e C. Then 7 is dependent on a
finite subset By of B. Consequently | {By} \ < \C\ and
y —
Next we note that U By = B. Otherwise, we have a 0 e B,
i \JBy. Since 0 < C and every 7 e C satisfies 7 < U5r, we have
0 < I) By which does not contain 0. This contradicts the in-
independence of B. Thus UBy = B and the above relation on
cardinals gives |5|<|C|. By symmetry \C\ < \B\; hence \B\
= |C|.
This result is applicable in particular to algebraic dependence
in P/f>. In this case a basis B is a set of algebraically independent
elements of P/€> such that every £ in P is algebraically dependent
on B. Such a set B is called a transcendency basis for P/f> and its
cardinal number, which is the same for all bases, is called the trans-
transcendency degree (tr. d.) of P/<i>. An extension P/«i> is algebraic if and
only if P has a vacuous transcendency basis over <£, hence if and
only if the transcendency degree is 0. If a field P has a trans-
transcendency basis B over * such that P = *E), then P is called a purely
transcendental extension of <i>. The theorem on the existence of a

156 STRUCTURE THEORY OF FIELDS
transcendency basis can be interpreted in the following manner:
Every field can be obtained as an algebraic extension of a purely
transcendental extension $(B) of the base fields. If X\, x2, ■ ■ ■, xr
are indeterminates, then the field of fractions of the algebra
$[#i, • • •, xr] is a purely transcendental extension $(# i, • • •, xr) of
$ with the transcendency basis {#,}. Moreover, it is clear that
any purely transcendental extension of degree r < <x> is essentially
identical with $(^i, • • •, xr).
Of particular interest in algebraic geometry are the fields P =
*(£i> £2, • • •, £n) which are generated over the base field $ by a
finite set of elements £,-. If B is a maximal algebraically independ-
independent subset of the set {£1, £2) ■ • •, £*}, then B is a transcendency
basis. We may assume B = {£1} £2, • • •, £r}. A field of the form
P — $(£15 £2, • • •} £n) is called a field of algebraic functions over
<J> and the transcendency degree r (< n) is called the number of
variables of P. If {£1} • • •, £r} is a transcendency basis, then P is a
finite dimensional extension of $(£1} £2, • • ■, £»•)• If this is sepa-
separable over $(£1, ■ • •, £r)j then one of the theorems on primitive
elements shows that P = $(£1, ■ • •, £r, n) for a suitable 7/ in P.
This is always the case for characteristic 0 and we shall see in § 5
that simple conditions can be given to insure the existence of a
basis {£»■} for a field of algebraic functions such that P is separable
algebraic over $(£1, £2, ■■•, £r).
EXERCISES
1. Show that, if C is a subset of P/$ such that every element of P is alge-
algebraically dependent on C, then C contains a transcendency basis. Show also
that, if D is an algebraically independent subset of P/$, then D can be imbedded
in a transcendency basis.
2. Let E/$ be a subfield of P/$. Show that the transcendency degree tr. d.
P/E < tr. d. P/$ and that tr. d. E/$ < tr. d. P/<£.
3. Let E/$ be a subfield of P/$ and let 5 and C be transcendency bases for
E/$ and P/E respectively. Show that B U C is a transcendency basis for
Hence prove the formula
A) tr. d. P/# = tr. d. P/E + tr. d. E/$.
Note that ex. 2 is a consequence of this. (Hint: Since E is algebraic over <£>(.B)
the subalgebra generated by E/<J> and $(C) is a field which is algebraic over
<E>(B, C) (p. 45). Hence E(C) is algebraic over f(B, C).)
4. Prove that, if $ is a field of characteristic ?* 3 and P = $(£, V) where £
is transcendental and 173 + £3 = 1, then P is not purely transcendental over $.

STRUCTURE THEORY OF FIELDS 157
5. Let P be the field of complex numbers, $ the subfield of rationals. Show
that tr. d. P/$ = c = | P|. Show that, if B is a transcendency basis of P/$,
then any 1-1 surjective mapping of B can be extended to an automorphism of
P/<£. Hence show that P has as many automorphisms as 1-1 surjective map-
mappings.
6. Prove that, if P is finitely generated over <£, then this holds for any subfield
E/*.
4. Luroth's theorem. The purely transcendental extensions
P = $(£1, £2, • • •, £r) appear to be the simplest types of extension
fields. Nevertheless, it is easy to ask difficult questions about
such extensions, particularly about subfields of P/4> if r > 1. If
r = 1 the situation is comparatively simple and we shall look at
this in this section.
Let P = $>(£), £ transcendental, and let 77 be an element of P
which is not contained in f>. We can write 77 = /(£)jr(£)-1 where
/(£) and g(£) are polynomials in £ which we may assume have no
common factor of positive degree in £. We may write /(£) =
«o + «i« + • • • + «„£", g(Q = 0o + 0i* + • ■ • + j8n€" where either
an j* 0 or (8n -^ 0, so n is the larger of the degrees of / and g. The
relation 77 = /(£),?(£)-1 gives /(£) - vg(£) = 0 and
0 = («„ - lift,)*" + («n-l - TJlSn-Op-1 + • ■ • + @0 - 77/8o).
Moreover, an — t]fin ?± 0 since an or f}% ^ 0 and 77 ^ *. Thus we
n
see that £ is a root of the equation of degree n: ^2 (at- — yP^x* = 0
with coefficients in $(tj). We proceed to show that X^ (at- — rift^x1
0
is irreducible in *(?j)M. First, it is clear that 77 is transcendental
over <i>, since £ is algebraic over $G7); hence 77 algebraic over *
implies £ algebraic over $, contrary to assumption. The ring
*[*?> *] = *WM is tne polynomial ring in two indeterminates 77, x
and we know that this ring is Gaussian, that is, the theorem on
unique factorization into irreducible elements holds in $[77, x]
(Vol. I, p. 126). We recall also that a polynomial in $[77, x] of posi-
positive degree in x is irreducible in $G7)[x] if it is irreducible in $[77, x]
Now 7G7, x) = S(a,- — 77/8j)Ar* = f{x) — 77^(^) is of degree 1 in 77.
Hence if /(?7, x) is reducible in 4>(t7)M, then it has a factor h{x)
of positive degree in x. This implies that f{x) and £(*•) are divisible
by h(x) contrary to assumption. We have therefore shown that

158 STRUCTURE THEORY OF FIELDS
f(t), x) is irreducible in $(»;)[*]. Thus £ is algebraic of degree n
over $G7). This proves
Theorem 7. Let P = $(£), £ transcendental over $ and let 77 be an
element of P not in <£. Write 77 = f(Og(£)-1 w^m1 /(£) and g(^)
are polynomials in £ with no common factor of positive degree in £.
Let n = max (deg /, deg g). Then £ is algebraic over $G7) and
[$(£) :$(rj)] = «. Moreover, f(x, 77) = /(#) — rjg(x) is irreducible
in $A7) M.
This result enables us to determine the automorphisms of #(£)
over <£>. Such an automorphism is completely specified by the
image ij of the generator £. For, if £ —> 77, then «(£)»(£)-1 —>
u(v)v(v) ~1 f°r w> w polynomials in £. It is clear also that, if 77 is the
image of £ under an automorphism, then $G7) = $(£). If 77 =
/(£)£(£)-1 as above, then [$(£):*(r/)] = » = max (deg/, deg^).
This shows that $G7) = $(£) if and only if max (deg /, deg,g-) = 1.
Then we have
B) 77 = >
w t£ + «
where a ^ 0 or 7^0 and a£ + (8, t£ + 5 have no common
factor of positive degree. It is easy to see that these conditions
are equivalent to the single condition:
C) ad - 0y ?* 0.
If this condition holds, then $G7) = $>(£) and the mapping
«(£)»(£)-1 —> #G7HG7)-1 is an automorphism of P/<£>.
The condition C) is equivalent to the requirement that the
matrix
[; a
is non-singular. With each such matrix we associate the auto-
automorphism of <£(£) over* such that £ —* 77 given by B). One veri-
verifies directly that the mapping of the non-singular matrix into the
corresponding automorphism is a group homomorphism. The
such that (a£ + 0) G£ + 5)-1
7 SJ
= £ or a£ + |8 = £(?£ + 8). This implies 7 = 0, /3 = 0, a = 5.

STRUCTURE THEORY OF FIELDS 159
Hence the kernel is the set of scalar matrices ^ 0. It
lO a\
is now clear that the group of automorphisms of $(£) is isomorphic
to the factor group of the group Z,(<£, 2) of 2 X 2 non-singular
matrices relative to the subgroup of scalar matrices. This factor
group is called the projective group PL(fb, 2).
We now consider an arbitrary subfield E of <1>(£)/<I>. We may
assume E^$. Then E contains an element r) not in $ so P =
<£(£) is algebraic over $G7) and hence is algebraic over E 3
Let the minimum polynomial of £ over E be f(x) — xn +
+ • • • + 7«. The 7i have the form M;(£)J'i(£)~1 where m, Vi are
polynomials in the transcendental element £. Multiplication of
f(x) by a suitable polynomial in £ will give a polynomial
E) /(£, x) = *„(«)*" + 'itt)*—1 + • • • + *.(«)
in <£[£, #], £, * indeterminates, which is a primitive polynomial in *
in the sense that the highest common factor of the r,-(£) is 1. Also
we have 7,- = ci(£)c0(£) ~'eE and not all of these are in <3? since
I is transcendental over <£. Thus one of the 7's has the form 7 =
g(^)h(^)~1 where ,§"(£), A(£) have no common factor of positive
degree in £ and max (deg g, deg h) = m > 0. We have seen before
that g(x) — yh(x) is irreducible in <D(y)[#] and [P:O(y)] = w.
Since E 2O(y) and [P:E] = «, clearly m ^ «. We shall show
that m= n and this will prove that E = <t>(y).
Since £ is a root of g{x) — yh(x) = 0 and the coefficients of this
polynomial are contained in E, we have g(x) — yh(x) = f(x)q(x)
in E[#]. We have 7 = g{^)h{^)~1 and we can replace the coef-
coefficients of / and q by their rational expressions in £ and then
multiply by a suitable polynomial in £ to obtain a relation in
$[£, x] of the form
F) k{Q\g{x)h® - g{k)h{x)] = M x)q& x),
where /(£, x) is the primitive polynomial given in E). It now
follows that k(£) is a factor of q{£, x) and so cancelling this we may
assume the relation is
G) g(x)h(& - g(&h(x) = /«, x)q{S, x).
Now the degree in f of the left-hand side is at most m. Since

160 STRUCTURE THEORY OF FIELDS
Y = KfKO with (K0» K*)) = 1 and max(deg£> deg h) = m,
the f-degree of/(£, #) is at least *». It follows that it is exactly m
and y(|, *) = y(*) e <D|V|. Then the right hand side of G) is
primitive as a polynomial in x. This holds also for the left hand
side. By symmetry, the left hand side is primitive as a poly-
polynomial in g also, and this implies that q(x) = q is a non-zero
element of®. Then G) implies that the ^-degree and f-degree
of/(f, #) are the same. Thus m— n and E = $(y). As we saw
before, EDO implies that y is transcendental. We have
proved the following
Theorem 8 (Liiroth). If P = #(£), £ transcendental over $, /A<?»
<z«y subfield E D $ » «/jo <? simple transcendental extension: E =
#G), 7 transcendental.
The theorem of Liiroth is not valid for purely transcendental
extensions P/$ of transcendency degree r > 1. The best positive
result in this direction is a theorem of Castelnuovo-Zariski which
states that, if <£> is algebraically closed and r = 2, then a subfield
E/$ of tr. d. 2 such that P/E is separable is a purely transcendental
extension.*
EXERCISES
1. Show that, if P = $(£, ij) where £ is transcendental and ij2 + £2 = 1, then P
is purely transcendental.
2. Let $ be a finite field, |$| = q = />m. Determine the order of the Galois
group of <$(£)/<£, I transcendental.
3. Give an example of a subalgebra of <!>[£], f transcendental, which does not
have a single generator.
5. Linear disjointness and separating transcendency bases.
Let * be of characteristic p ^ 0 and let P = <£(£, 17) where £ is
transcendental and t\v = £. Then {£} is a transcendency basis
for P/$ and P is inseparable over <!>(£). On the other hand, P =
$(»7) is separable over P. This simple example shows that certain
transcendency bases B for an extension may be preferable to
others in that P/<J?(-5) is separable algebraic. We remark also
that such bases may not always exist, as is shown by the example
* See O. Zariski, On Castelnuovo's criterion of rationality pa = Pt = 0, Illinois Jour, of
Math., Vol. 2 A958), pp. 303-315.

STRUCTURE THEORY OF FIELDS 161
of any P/f> which is algebraic and not separable over #. A trans-
transcendency basis B for P/<i> such that P is separable algebraic over
$(B) is called a separating transcendency basis. If P/$ has such
a basis, then we shall say that P/$ is separably generated. In
this section we shall derive a criterion that P/$ is separably
generated, based on the following important notion.
Definition 2. Let Ex and E2 be subalgebras of an arbitrary field
P/i>. Then Ei and E2 are said to be linearly disjoint over $ if the sub-
algebra Ei E2 generated by Ei and E2 is the tensor product E\ ®* E2.
More precisely, what is meant here is that the canonical homo-
morphism of Ex ®# E2 into EiE2 sending ei ® e2 /»/o e1e2 « aw
isomorphism (see Introd., § 3).
It is well to recall the conditions that EXE2 = Ex 0* E2 which
we obtained in the Introduction. We recall first that a sufficient
condition is that there exist bases (#<*)> (vp) of Ex and E2 over <£
respectively such that (uavp) is a basis for EXE2 over f>. Since
every element of the subalgebra EiE2 is anyhow a linear combina-
combination of the elements uavp, we see that a sufficient condition for
EiE2 = Ei <g>* E2 is that there exist bases («„), (vp) for Ex/$ and
E2/<i> such that {uavp} is ^-independent. If Ex is a subfield of
P/$, then EiE2 = Ei <g»* E2 if there exists a basis (vp) for E2
over $ such that the set (vp) is Ei-independent.
Conversely, assume Ex and E2 are linearly disjoint subalgebras.
Then (uavp) is a basis for EiE2/f> for any basis («a) of Ei/# and
any basis (up) of E2/<i>. Also if Ei is a subfield, then (vp) is a basis
for E1E2/Ei. We note also that linear disjointness of the sub-
subalgebras Ex and E2 implies that, if {ua} is any linearly independent
subset of Ei/$ and {vp} is any linearly independent subset of
E2/<£, then {uavp} is ^-independent.
We note next that Ei and E2 are linearly disjoint subalgebras
over $ if and only if the subfields (?i and Q2 generated by Ei and
E2 respectively are linearly disjoint over $. For, if Q\ and Q2
have the property, then so do Ex and E2, since subalgebras of
linearly disjoint algebras are clearly linearly disjoint by our
criteria. Conversely, suppose Ex and E2 are linearly disjoint.
Let (•!, £2, ■ ■ -, i;m be elements of (?i which are ^-independent and
*h> V2> • ■ ■ y Vn elements of Q2 which are ^-independent. We can

162 STRUCTURE THEORY OF FIELDS
write & = S/f, y = v/v'1 where £/, £ e Els r?/, i\ e E2. Then
{^i'j^'j ■"•>£m'} is a "^-independent subset of Ex and {771', ?;2',
'' '>Vn'} is a ^-independent subset of E2. Hence the w« elements
(j/77/ are ^-independent and so also the elements £,•%■ are *-in-
dependent. This implies that Qi and Q2 are linearly disjoint over*.
We are interested primarily in subfields of P and we shall
require the following lemma which will enable us to prove linear
disjointness by steps.
Lemma. Let Ex and E2 be subfields of P/*, Ax a subfield of
Ei/*. Then Ex and E2 are linearly disjoint over * if and only if the
following two conditions hold: A) A! and E2 are linearly disjoint
over$ and B) the field Ai(E2) and Ei are linearly disjoint over Ax.
Ei(E2)
Proof. Assume A) and B). Let (ua) be a basis for E2/*. By
A), the ua are linearly independent over Ai. Since these elements
are contained in Ai(E2), and Ai(E2) and Ei/A! are linearly dis-
disjoint over Ai, by B), the ua are linearly independent over Ex.
Hence Ex and E2 are linearly disjoint over *. Conversely,
assume this holds. Then it is clear that Ax and E2 are linearly
disjoint over *, that is, A) holds. Also the hypothesis implies
that, if (ua) is a basis for Ei over Ai, (vp) a basis for At over *,
(wy) a basis for E2 over *, then (uavpwy) is a basis for EiE2 over
*. This implies that, if we have a relation Sc,«a< = 0, Ci e E2Ai,
then every d = 0. Now suppose we have S5t«a,. = 0, 5,- e Ai(E2).
Then we can write 5,- = Cid~1, a, d e E2Ax and we obtain St,«a,. =
0, Ci = 0 and di = 0. Thus we have shown that the basis (ua) of
Ei/Ai is Ai(E2)-independent. This implies that Ei and A!(E2)
are linearly disjoint over A!.
We now embark on the study of linear disjointness and sepa-
separability. We assume that P is an extension field of a field * of

STRUCTURE THEORY OF FIELDS 163
characteristic/) 5^ 0 and we shall operate in an algebraic closure
A of P (which contains an algebraic closure of <£>). We consider
the subset <£>p 1 of A of elements y such that yp e <i>. This is a
subfield over $. We shall be interested in linear disjointness of P
and <I?P l over <£. In studying this question it is useful to note the
following simple criterion: A set {pi, p2, ■ ■ -yPn}, Pi e P is <t>p 1-
independent if and only if {pip,p2p, • • ■, Pnp} is ^-independent;
for, suppose we have XPiPi = 0, 0,- e $p~\ Then 2a;pip = 0 for
«» = Pip in <i>. On the other hand, if 2aiPip = 0, a,- in $, then,
since A contains an algebraic closure of $, a,- = /3,p, /3,- in $""'.
Then SfrW = 0. Hence (S/3tPl)p = 0 and 2ftp,- = 0. Also it is
clear in both situations that a,- = 0 if and only if /3< = 0. We shall
now establish the following criterion.
Theorem 9. If P « «« algebraic extension of $ {possibly infinite
dimensional), then P /j separable over <£ // «w^/ o«/y // P is linearly
disjoint to $p-1 over $.
Proof. We recall that an algebraic element p of P over * is
separable if and only if p ef>(pp) (Lemma 2 of § 1.9). Suppose
first that P and *p~1 are linearly disjoint over * and let p e P.
Let [<i>(p) :<£>] = n. Then A, p, p2, • • •, p") is a basis for <£>(p) =
$[p] and hence these elements are 3>p~'-independent. This im-
implies that the elements l,pp, p2p, ■ • ■)P^n~1)p are ^-independent.
Since there are n of these and they are contained in <t>(p), they
form a basis for <i>(p). Evidently this implies that p ef>[pp] so p
is separable over f>. Conversely, assume P separable over * and
let {pi, • • •, pn} be a finite "^-independent subset of P. We may
imbed this set in a subfield E/«i> which is finite dimensional, and
we can choose a basis (pi, • • -,pn, Pn+i, ■ • ',Pq) for E/$. Any
element e of E is a ^-linear combination of the py, 1 < j < q.
Then tv is a ^-combination of the pf. The same holds for e2p =
(«2)p, e3p = (e3)p, • • •. On the other hand, since e is separable,
e e$(«p) = #[€*]. Consequently, e itself is a$-linear combination
of the pyp. Since [E:$] = q this implies that (pip, p2p, • ■ ■, p,p)
is a basis for E/<i>. Hence {p/, • • -,pnp} is ^-independent and
[piy • • • > Pn} is 4>p'-independent. Since {pi, •••,pn} was an
arbitrary finite ^-independent subset of P, this proves that P
and *pl are linearly disjoint over *.

164 STRUCTURE THEORY OF FIELDS
We prove next the following
Theorem 10. If P is purely transcendental over $, then P is
linearly disjoint to <£p ' over 3>.
Proof. Our assumption is that P = $(B) where B is an alge-
algebraically independent set. We have seen also that P is linearly
disjoint to <i>p ' over $ if and only if the subalgebra $[B] of poly-
polynomials in the elements of B is linearly disjoint to $p '. To prove
that the latter holds, it suffices to give a basis for $[5]/* which is
$p '-independent. For this we take the basis M consisting of the
monomials in £a e B. Now it is clear that, if Wj and m2 are dis-
distinct monomials, then m^ and m2p are distinct monomials. Hence
it is clear that the set Mv of p-ti\ powers of the elements of M is a
^-independent set. We have seen that this implies that M is
•^"'-independent. Hence $[B] is linearly disjoint to $p~l and
the proof is complete.
We can now prove our main result which is
MacLane's Criterion. If P/<£ is separably generated {of
characteristic p), then P and <l?p ' are linearly disjoint over $. On
the other hand, if P is finitely generated over $ and P and $pl are
linearly disjoint over $>, then P is separably generated over <£.
Proof. Suppose first that P is separably generated over $,
which means that P has a transcendency basis B over $ such that
P is separable algebraic over 2 = $(B). Then, by Theorem 10,
2 and <£P~I are linearly disjoint over $. Also, by Theorem 9, P
and 2P * are linearly disjoint over 2. Hence P and 2(<i>p~1)
which is a subfield of 2P ' over 2 are linearly disjoint over 2.
The lemma now shows that P and$p * are linearly disjoint over <£.
Next we assume P = $(£ 1} £ 2) • • - > £m) and P and 4>pl are
linearly disjoint over <i>. We may assume also that {£i, £2? • ■ ■> £r}
is a transcendency basis. Suppose we know already that £r+i>
•••,£, are separable algebraic over $(£1, ■ • •, £r). If s = m, then
{li> •••>£>•} is a separating transcendency basis. Hence we
suppose that %3+1 is inseparable algebraic over 2 = $(£1, • • •, £r).
Let f(x) be the minimum polynomial of £s+1 over 2. If we
multiply / by a suitable polynomial in £x, • • -,£r we obtain a
polynomial Ffa, ■ • ■, £r, x) e $[£1, • ■ •, £r, x] which is irreducible

STRUCTURE THEORY OF FIELDS 165
in $[£u • • •, £r, x] and satisfies F(ZU •••,£,-, £,+i) = 0. Since / is
inseparable it is a polynomial in xp; hence jF(£i, • • •, £r> #) is a
polynomial in xp. We assert that there exists a £,-, 1 < i < r,
such that .F is not a polynomial in £,•*. Otherwise, the monomials
i° £u ?25 • • • j £r> * which actually occur in F are all />-th powers
and this implies that F(l-U • • •, £r, #) = //(£ 1} • • •, £r, #)p in
^"'[Si, • • •, €r, #]• Then //(^, ••-,£,., £,+i) = 0, so the mono-
monomials in £i, ■••,£r, £«+i occurring in H are linearly dependent
over $"""*. The assumption of linear disjointness of P and $p *
therefore implies that the same monomials are ^-dependent. This
implies that £,+i is a root of a polynomial h(x) e 2[#], A(#) ?■* 0
of lower degree than /, contradicting the fact that / is the mini-
minimum polynomial of £8+1 over 2. This shows that we may suppose
that F(l-i> ■ ■ ■, !~r, x) is not a polynomial in £ip. The relation
F(j;i, ■ ■ •, £r> £»+i) = 0 shows that £i is algebraic over $(!-2, • • •,
kr, £«+i)- Since ^2j • ■ •, %r are also algebraic over this subfield, it
is clear that {£2j • • •> fr} £«+i} is a transcendency basis.
We shall show that £x is separable over 2' = <£(£2, • • •, £r, £8+i)-
We recall that F(£u ■ • -,kr,y) is irreducible in $[^1} • ■ •, £r, j].
Hence F(x, x2, • • •, xr,y) is irreducible in $[x, x2, • • •, xr,y],
x,Xi,y, indeterminates, and consequently this polynomial is ir-
irreducible in 4>(x2, • • •, xr,y)[x] where $(x2> ■ ■ -,xr,y) is the field
of fractions of $[x2, • • -, xr,y\. Since £2, • • •, £r, £«+i are alge-
algebraically independent over $, #(£2, • • •, £r, £«+i) =*(^2, • • •, xr,
y) under a ^-isomorphism such thatx,- —> £,-, 2 < i < r,y —* £«+i.
It follows that /^(x, f2, • • •, fr, £s+i) is irreducible in <i>(£2, • • -,
£r, £«+i)M and so this is a multiple of the minimum polynomial of
£x over 2'. Since this polynomial is not a polynomial in xp, £i
is separable algebraic over 2'. Also £,-, 1 < i < s, are separable
algebraic over *(fi, • • •, £r, £s+i) and, since £x is separable alge-
algebraic over 2', £; is separable algebraic over 2'. If we re-number
the £'s we now have a transcendency basis £i, • • •, £r such that
every £y, 1 < j < s + 1, is separable algebraic over $(£i, • • •, £r)«
This establishes the inductive step to show that we can choose
£i> • • • j £r among the generators £i, ••-,£„» so that every £,• is
separable algebraic over 2 = €>(£!, • • •, £r)• Then {£1} • • •, f r} is a
separating transcendency basis for P/<£>. This completes the
proof of the second assertion of the theorem.

166 STRUCTURE THEORY OF FIELDS
It is important to note for future use that, if P = $(£1, •••,£»,)
is linearly disjoint to <&p ' over <$, then we have shown that a
separating transcendency basis can be extracted from the set
{£i> £2; ••■»£«}• We note also the following
Corollary (F. K. Schmidt). If $ is perfect, then any field of
algebraic functions $(£1, • • ■, £m) has a separating transcendency
basis over $.
This is an immediate consequence of MacLane's criterion since
P is certainly linearly disjoint to $p l = $.
The results which we have proved, particularly Theorem 9,
make it natural to extend the notion of separability to arbitrary
(not necessarily algebraic) field extensions in the following way.
Definition 3. Afield P is separable over <1> if it is either of charac-
characteristic 0 or if it is of characteristic p ^ 0 and P is linearly disjoint
to f^ over <£>.
Theorem 9 shows that this is equivalent to the usual notion of
separability if P is algebraic over <£. Also MacLane's criterion
shows that, if P is finitely generated over <$, then it is separable
over <i> if and only if P is separably generated over <£>. The follow-
following theorem gives two other properties of separability which are
familiar in the algebraic case.
Theorem 11. A) If P is separable over * and E is a subfield of
P over f>, then E is separable over <£>. B) If P is separable over E
and E is separable over <£, then P is separable over <£.
Proof. We may assume the characteristic is p t± 0. A) This
is clear since the linear disjointness of P and <£p ' implies the
linear disjointness of E and $p \ B) We are assuming that $J>~1
is linearly disjoint to E over <J> and W~x is linearly disjoint to P
over E. Then E^) which is a subfield of W~x is linearly dis-
disjoint over E to P. The lemma now applies to show that #pl
and P are linearly disjoint over $. Hence P is separable over $.
We close the present discussion with two negative results.
First, we recall that, if P is separable algebraic over <$, then P
is separable algebraic over any intermediate field. This fails in
the general case since, for £ transcendental, $(£) is separable over
# of characteristic p 5^ 0, but <!>(£) is not separable over <£(£").

STRUCTURE THEORY OF FIELDS 167
Next we note that a field may be separable over $ and not sepa-
separably generated. An example of this is given in ex. 1 below.
EXERCISES
1. Let <£ be of characteristic p 7* 0 and let P = <£(£, I", £p~*, • • •) where £
is transcendental over $. Show that P is separable over $ but not separably
generated over $.
2. Let ^p~m be the subfield^of the algebraic closure of P 3 $ of elements £
such that f"e$ and Jet $p~" = U^"". Show that P is separable over $ if
and only if P and $p are linearly disjoint over <5.
3. Let E/$ and A/$ be subfields of P/$ such that E/$ is purely transcen-
transcendental and A/<£ is algebraic. Show that E and A are linearly disjoint over $.
4. Let P = <!>(£, 17, f, t) where $ is of characteristic p ?* 0, £, 17, f are alge-
algebraically independent and tp = ££" + 77. Show that P is not separably gener-
generated over E = $(£,17).
5. (MacLane). Let$ be a perfect field of characteristic p 5^ 0, P an imperfect
extension field of $ such that tr. d. P/O = 1. Show that P is separably gener-
generated over $.
6. Derivations. We have found it useful to introduce the usual
formal derivative of a polynomial in considering multiple roots
(§ 1.6). The mapping of the polynomial algebra $[#] into itself
defined by: f(x) —> /'(#) the formal derivative of /(#), is an
example of a derivation in the algebra $[x]. More generally it is
convenient to consider derivations from a subalgebra into an
algebra. This general notion, which is of great importance in
algebra, is given in the following
Definition 4. If % is a subalgebra of an algebra SB, a derivation
D of 21 into S3 is a linear mapping of 21 into SB such that
(8) (ab)D = (aD)b + a(bD), a,bz%.
If 21 = SB, then we speak of a derivation in 21.
We shall be interested mainly in derivations in fields of alge-
algebraic functions. In this section we consider some general results
on extension of derivations and on the algebraic system consisting
of all the derivations of an algebra into itself. We begin our con-
considerations by noting first that the study of derivations is equiva-
equivalent to the study of a certain type of algebra isomorphisms. This
will enable us to derive the main facts-about derivations as con-
consequences of corresponding results on homomorphisms. For this

168 STRUCTURE THEORY OF FIELDS
purpose we introduce the algebra £ with basis A,/) over the
base field $ and multiplication rule t2 = 0. Thus % = <£>[#]/(#2),
x an indeterminate, and / is the coset x + (x2). If 33 is an arbi-
arbitrary algebra, then we form the algebra 33 ® S. If we identify 33
in the usual way with the subalgebra of elements b ® \ and X
with the subalgebra of elements 1 ® #, u e 2, then we see that
the elements of 33 ® % can be written in one and only one way
in the form b0 + bxt, bt e 93, the generator t of Z. We have bt =
tb and in general the multiplication rule in 33 ® X is
(9) (b0 + b!t)(c0 + cxt) = Vo + (Vi + hco)t,
b{, d e 33. The algebra 33 ® £ is called the algebra of dual numbers
over 33.
Now let D be a derivation of 21 into 33. Then we can use this
to define a mapping s = s(D) of 21 into 33 <8> X by
A0) a -> a' = a + (aD)t.
Evidently s is linear. Furthermore, if a, b e 21, then
a'b' = (a + (aD)t)(b + (bD)t)
= ab + (a(bD) + (aD)b)t
= ab + {{ab)D)t
= (ab)>.
Hence s is a homomorphism of the algebra 21 into the algebra of
dual numbers 33 ® Z. The homomorphism s has a simple charac-
characterization. For this we introduce the mapping ir: a + bt —> a,
a, b e 33, of 33 ® X into 33. It is clear that this is a homomorphism
of 33 ® X into 33 which is the identity mapping on the subalgebra
33. Now we see that, if a e 21 and s is defined by the derivation
D of 21 into 33 as before, then a'* = (a + (aD)t)r = a. Evidently,
this requirement guarantees that s is an isomorphism.
Conversely, let s be any homomorphism of 21 into 33 ® X such
that a'T = a, a e 21. Then a' = a + bt, a, b e 33 and b is uniquely
determined by a. Hence we have the mapping D: a —» by and
we may write a' = a + (aD)t. It is clear that the linearity of s
implies the linearity of D. Also since (ab)' = a'b' for any a, b in 21,
(a + (aD)t)(b + (bD)t) = ab + (a(bD) + (aD)b)t
= ab + ((ab)D)t.

STRUCTURE THEORY OF FIELDS 169
Hence we have (ab)D = (aD)b + aibU) so D is a derivation. We
can therefore state the following
Theorem 12. If 31 is a subalgebra of S3 and D is a derivation of
2t into S3, then s'.a —> a + (aD)t is an isomorphism of 31 into the
algebra of dual numbers S3 ® !£ over 33 such that a" — a. Con-
Conversely, any homomorphism of 21 into S3 ® £ satisfying this condition
has the form a —» a + («D)/ a?Ai?r<? D /j « derivation of 21 *'»/o S3.
We shall now obtain some simple consequences of this connec-
connection between derivations and isomorphisms. First, let % be a
set of generators of the subalgebra 21 of the algebra S3 and let D\
and D2 be derivations of 21 into S3. Suppose xDx = xD2 for every
x eTH. Then x'1 = xH for the associated isomorphisms Si = s(D1),
s2 = s(D2) of 21 into S3 <g> 1. It follows that a81 = a for every
a e 21 so sx = s2 and D\ = D2. This shows that, if two deriva-
derivations coincide on a set of generators of 21, then they are identical
on 21. We remark next that, if s is a homomorphism of 21 into
S3 <8> £ such that x8r = x for x e 36 a set of generators, then a'T =
a for all a e 31. Hence s defines a derivation in the manner in-
indicated.
An element c of 31 such that cD = 0 is called a D-constant.
Evidently c is a D-constant if and only if c' = c for the isomor-
isomorphism s = s(D). It follows from this—or directly—that the set
of D-constants is a subalgebra of 31. In particular, 1 is a D-
constant for every derivation D. If 31 is commutative and $ is
of characteristic p, then every p-th. power in 21 is a D-constant.
For, in any commutative algebra the basic property (8) for D
implies that (ak)D = ka^^aD). Hence if k = p, then (ap)D = 0.
We note also that, if 31 = P is a field, then the set of D-constants
of P forms a subfield F of P. This is clear from the consideration
of s = s(D) or it follows directly, by noting the rule for the deriva-
derivative of y~1:y~1D — —(yD)y~2, which follows by taking the
derivative of the relation ty = 1. If p e P and y eF, then
(yp)D = y(pD) for the derivation D. This shows that D e
3)r(P, S3) the set of derivations of P/F into S3/F. In considering
a particular derivation D of a field, it is often convenient to shift
from the original base field to the field of constants F of D or to
some subfield E/*

170 STRUCTURE THEORY OF FIELDS
We shall now carry over to derivations the two basic results
I and IV on extensions of homomorphisms of commutative rings
which we derived in the Introduction. We remark that these
results are valid for algebras over a field $ and we shall use them
in this form. Our first result on extension of derivations is
Theorem 13. Let P be a field over $, SI a subalgebra of
{containing 1), Ma multiplicatively closed subset of non-zero elements
of 21 containing 1, and let 21 jf be the subalgebra of P of elements of
the form ab~l, a e 21, b e M. Let D be a derivation of 21 into P.
Then D can be extended in one and only one way to a derivation of
Hi* into P.
•
Proof. Let s be the isomorphism a —> a + (aD)t of SI into
P <8> £. If a ?± 0, then a* = a + (aD)t has the inverse a~l —
a~2(aD)t since
(a + (aD)t)(a-1 - a~*(aD)t) = 1 + (aD)a~1t - a~\aD)t = 1.
By I of the Introduction, s can be extended to an isomorphism
of %m into P. The extension is unique and maps ab -1 —> a'(b')~l.
We have
*•(*•)-! = (a + (aD)t)(b~l - b~\bD)t)
= ab-1 + ((aD^-1 - ab~2(bD))t.
This formula shows that, if s denotes the extension of s to ^.u,
then tab-1)™ = (a^b8)-1)* = ab~\ It follows that s defines the
derivation:
A1) ab~l -> (aD^-1 - ab~2(bD)
of Stjw into P. The argument shows also that this extension of
D is unique.
Next we consider a subalgebra 21 of P/f> and an extension of this
which has the form 2l[£i, £2> • • 5 £m]> £i elements of the field P.
We suppose we are given a derivation D of 21 into P and elements
Vi) V2, • • • > Vm of P. We seek conditions on D and the rn which
insure that D can be extended to a derivation D of St[£i, £2, • • •,
£m] such that %{D = t)i, i = 1,2, • • •, m. Since 21 and the &
generate Sl[£i, • • •, £m] it is clear that, if the extension exists, then

STRUCTURE THEORY OF FIELDS 171
it is unique. As before, we consider the isomorphism s\a —> a +
{aU)t of 21 in P ® X satisfying a8w = a. Then D can be extended
to a derivation of 2l[£i, • • •, £m] into P so that £,• —> t\i if and only
if s can be extended to an isomorphism of 2l[£i, • • •, £m] into P <g> 2:
so that £» —* £»■ + Tjj/. Clearly the condition is necessary and, if
it holds, we have a" = a and £/* = £»• for the extension s. Hence s
will give rise to a derivation of 2l[£i, £2> • • • > £m]9 as before. The con-
conditions for the extension of s have been given in IV of the Introduc-
Introduction. We recall that the set S of polynomials f(xly x2, • • ■, xm) e
%[xu x2, ■ • •, xm], Xi indeterminates, such that /(£i, f2, ••-,?«) =
0 is an ideal in Slf^j, x2, • • •, xm]. The condition IV for an exten-
extension s of s such that £,•• = f,-, 1 < / < m, is that £*(£i, J-2, • • •, tm)
= 0 for every g e X, a set of generators of $. Hence we see that
D can be extended to a derivation of &[£i, • • •, £m] into P such
that £t- —> ?;,•, /' = 1, 2, • • •, m, if and only if
g*(£i + Vit, & + ifcf, •••,£» + ilW) = 0
for every g in a set 36 of generators of the ideal S of polynomials
/(#i, • • • > xm) e 8l[xi, • • •, *m] such that /(^, • • •, £m) = 0.
We proceed to work out these conditions in detail. Let a e 21
and consider the monomial M(xi, • • •, ,vm) = axiklx2ki • • • xmkm.
Then
A**(€i + »?!/, £2 + 112/, •",€» + W)
If we define the formal partial derivative of / = 2at1...kmxikl
xmkm relative to Xi as
S*tfX*' • • • AT,*'-1 • • • Xmk",

172 STRUCTURE THEORY OF FIELDS |
— ) , then the
above calculation shows that * x'~'
Af*tti H- ifi/, •••,€» +W)
)
where, in general fD(x\, • • ■, xm) is the polynomial obtained from
/ by replacing the coefficients by their images under D. Hence
if / e 2l[*i, • • •, xm], then we have
A2) f(h + vit, b + vat, ■ ■ •, fc, +
It is now clear that /s(£i + ^i/, ••-,?« + W) = 0 if and only if
•••,«») = 0and
A3)
The criterion which we gave can now be stated in the following
manner.
Theorem 14. Let 21 be a subalgebra over $ of the field P/$ and
let £i, £2j • • • j Sm, *?i> 92J • • ■, Vm be elements of P, D a derivation of
21 into P . L#$ fo the ideal of polynomials /(*i, • • •, ^fm) e 2[[#1} • • •,
A;m] j«^A that f(i-1, • • •, $m) = 0 «»*/ /i?/ 36 be any set of generators
for ®. Then D can be extended to a derivation D of 2I[£i, • • •, £m]
into P such that £,D = in, i = 1, 2, • • •, w, //"««^/ o«/y //
A4)
for every g e X. i/" /A^ extension exists, then it is unique.
A special case of the result is the following: If the £,• are alge-
algebraically independent over 21, then there exists a derivation D
extending D on 21 and mapping £,• —> rn where ?7i, • • -,Vm are
arbitrary in P. This is clear since in this case the ideal $ = 0,
so the condition for extension is trivially satisfied.

STRUCTURE THEORY OF FIELDS 173
We consider next an arbitrary algebra S3, a subalgebra 21, and
the set £>*Bl, S3) of derivations of 21/$ into 33/*. If Du D2 e
jD*B1, S3) and a e *, then aDi and Di + D2 are linear mappings
of 21 into S3. Moreover, if a, b e 21,
= (a(aD1))b + a(b(aD1))
b + a{bBx) + (aD2)b + a(bD2)
= (a(D1 + D2))b + a{b{D, + D2)).
This shows that aDi and Di + D2 are derivations. Hence
2)#Bl, S3) is a subspace of the space 8$Bl, S3) of linear mappings
of 21/$ into 33/$. Next let c be an element of the center of 33
and, as usual, let cr denote the mapping x —> xc = ex in 33. We
assert that, if D is a derivation of 21 into S3, then Dcr is also a
derivation of 21 into 33. For, it is clear that Dcr is linear and we
have
(ab)DcB = ((aD)b + a{bD))cR
= (aDcB)b + a((bD)cR).
Hence D<rfl e 5D*B1, S3).
Next let S3 = 21 and let £)*Bl) = £>*Bt, 21) the set of derivations
in 21. Let Dly D2 e £)*Bl). Then DtD2 is a linear transformation
of the space 21. However,
(aDx)b)D2
+ (aDaXWi) + (aDO^Da) +
Since {aD2){bDl) + («Di)(^D2) may be ?* 0, it is clear that
DXD2 need not be a derivation. The "obstruction" (aD2)(bDi) +
(aDi)(bD2) is symmetric in Dx and D2 so we obtain the same
obstruction for D2Dx. These cancel off if we form [DiD2] =
DiDa - D2I>i. Hence it is clear that [D^] e £>*Bl). The expres-
expression [DiD2] is called the Lie or additive commutator of Dx and D2.
Our result is that £)*Bl) is a subspace of the space of linear trans-
transformations of 21 closed under Lie commutators, that is, if Dly D2 e
£>*Bl), then [DiAs] e £)*BI). A subspace of ?*Bt) having this

174 STRUCTURE THEORY OF FIELDS
property is called a Lie algebra of linear transformations. The Lie
product [DiD2] is bilinear but it is not associative. The basic
properties which it has are
A5) [DD] = 0, [[ADJAd + [[DaAJZM + [[DsDiM = 0.
The first of these is clear and the second follows from a straight-
straightforward calculation which we leave to the reader. We note next
the following Leibniz formula for the &-th power of a derivation:
A6) (ab)Dk = £ (*) («D W-<), * = 1, 2, • ■ •.
,-=o W
This is readily proved by induction on k. Now suppose the base
field is of characteristic p t± 0. Then f .1^ = 0 for / = 1, 2, ■ ■ •,
p — 1 and any a e 31, so in this case, A6) for k = p reduces to
A7) {ab)D* = (aDp)b + a(bDp),
which shows that 3)*Bl) is also closed under ^>-th powers, that is,
if D e $)*(»), then Dp e £)*Bl). A Lie algebra of linear trans-
transformations in a vector space over a field <£ of characteristic p 5* 0
having this extra closure property is called a restricted Lie algebra
of characteristic p.
EXERCISES
1. Let 31 be an algebra over $ and let d e.%. Verify that the mapping a —*
[ad] = ad — da is a derivation in 21. Such a derivation is called an inner deriva-
derivation of SI. Prove that, if Id denotes the inner derivation determined by d, then
laidi+aidz = ctxldi + ot-ildz, o>-i in $ and I\didi\ = [Id\Id^- Show also that, if $ is of
characteristic p 9* 0, then I& = (Id)p.
1. Let 21 be a subalgebra of an algebra 33. Verify that a mapping D of 21 into
SB is a derivation if and only if the mapping
aD-\
\
if 21 into the matrix algebra 332 of 2 X 2 matrices over SB is an isomorphism.
7. Derivations, separability and /(-independence. We shall
now take up the study of derivations in a field P/$. We note
first that, if 21 is a subalgebra of P/3>, then any derivation D of
2l/<£> into P/«l> has a unique extension to a derivation D of the sub-
field E of P generated by 21. This is a special case of Theorem 13

STRUCTURE THEORY OF FIELDS 175
since E = $Lm for M the set of non-zero elements of 21. Suppose
next that E is a subfield of P/<$ and D is a derivation of E/$ into
P/$. Let £ e P. Then if £ is transcendental over P, D can be
extended to E[£] so that £Z) = rj is any element of P. This is a
consequence of Theorem 14. Moreover, D can be extended to
the field E(£) so that £Z) = rj. Next assume £ is algebraic over E,
so E[£] = E(£) and let f(x) be the minimum polynomial of £ over
E. Then the ideal $ in E|#] of polynomials h(x) such that A(£)
= 0 is the principal ideal (/(#)). Hence Theorem 14 shows that
D can be extended to a derivation of E(£) such that £ —»■ 17 if and
only if
A8) />(£) + /'(£)»? = 0,
f'{x) the usual derivative of f{x) (cf. V of Introd.). If £ is sepa-
separable, then /'(£) ^ 0 and A8) gives n = -/>(£)/'(£) -1. Hence
there is only one choice possible for 77 to give an extension of D.
Thus we see that, if E(£) is separable algebraic over E, then a
derivation of E/<E> into P/<i> can be extended in one and only one
way to a derivation of E(£) over f>. In particular, if D = 0 on
E, then the only extension of D to a derivation in E(£) is D = 0
on E(£). If £ is inseparable, then /'(£) = 0. Hence D can be
extended to a derivation in E(£) if and only if/°(£) = 0 and, when
this condition is fulfilled, then 77 is arbitrary so D can be extended
to E(£) in such a way that £Z) = t\ is any chosen element of P. If
/(*) = *" + ax*" H , then fD{x) = (axD)*" + (a2D)xn~2
+ • • • and since f(x) is the- minimum polynomial, the condition
/°(£) = 0 holds if and only if every a,D = 0. Thus, a necessary
and sufficient condition for the extendability of D to E(£), £ in-
inseparable algebraic over E is that the coefficients of the minimum
polynomial of £ over E are D-constants. We shall need this
criterion particularly in the case f{x) = xp — a. Then the con-
condition is simply that aD = 0.
Now let P = $(£1, £2, • • •, £m) a finitely generated extension
field of* (that is, a field of algebraic functions). Let M be the
ideal in ^[xi, x2, • • •, xm] of polynomials f(xu x2, • • ■, xm) such
that /(£i, £2, • • •, £m) = 0 and let £ be a basis for $. If D is a
derivation of the algebra $[£x, £2, • • •, £m]/f> into P/$, D has a
unique extension to P/$. Theorem 14, applied to the derivation

176 STRUCTURE THEORY OF FIELDS
D = 0 on $>, shows that there exists a derivation D of $[l-u £2> •
£m]/$ into P/$—hence of P/<f> into itself—such that £tD =
/ = 1, 2, • • •, m, if and only if
A9)
for every £ e 36.
We have considered the system 35#(P) of derivations of P/<i>
in the last section and we have seen that this is a Lie algebra of
linear transformations which is restricted in the characteristic
p j± 0 case, and 35*(P) is closed under right multiplication by
elements pr, p e P. Hence we see that 35$(P) is a subspace of the
right vector space 8#(P) over P (see § 1.1). We shall now in-
investigate £)*(P) as right vector space over P for P = $(£1, £2, • • •>
For this purpose we introduce the right vector space P(m) of
w-tuples (pi, p2, ■ • •, pm), Pi e P, with the usual addition and
multiplication by elements of P. If D e 35 = 35*(P), then we
map D into the element (£iD, £2D, • • •, %mD) e P(m). This map-
mapping is P-linear and so its image 5)' is a subspace of P(m)/P. If
£tZ) = 0, 1 < / < m, then D = 0 since the £,- are generators of
P = 3*(£i) ?2> • • • > £?»)• This shows that the kernel of the mapping
D —> (£iD) of 2) onto 2)' is 0 and so the mapping is a P-linear
isomorphism of 35 onto 35'.
Next we shall give a description of the subspace 55' of P(m) in
terms of the ideal $ defined before. We note first that, if / e
*[#i> #25 • • -j xm], then the mapping
(rif
dXi
is a linear function on P(m), that is, an element of the conjugate
space P(m) * of P(m). Let d% denote the subspace of P(m) * spanned
by the elements dg, fel,a set of generators for M. The condition
A9) on G71,7/2, • • •, Vm) is that (rn)dt = 0 for all g e 36. Hence we
see that there exists an element D e 35*(P) such that £,-D = ij»,
1 < i < m, if and only if (i/,-)</4 = 0 for all g eX. This clearly im-
implies that 35'' is the subspace of P(m) of vectors incident with the
subspace dX of p(m>* (Vol. II, p. 55). We recall that the sum of

STRUCTURE THEORY OF FIELDS
177
the dimensionalities of £)' and dUL is m. If we replace £ by the
complete ideal $, then we have dH C dSt; but, since both spaces
have the same dimensionality m — [2)': P]r, it is clear that dH =
d®. This shows that dUl is the same for any two sets of generators,
a fact which is easy to see directly also. The result that we have
obtained is the following
Theorem 15. Let P = <i>(£i, £2, • • •, £m) a field of algebraic func-
functions over®. Let Hbe a set of generators for the ideal ft of polynomials
/(#!, x2, •■•,xm) such that /(&, £2) • • •, £m) = 0 and let 5D*(P) be
the right P-vector space of derivations in P/#. Then
B1) [£>*(P):P] = m- [dX:F]R
where dH is the set of linear functions de, g e I, defined by B0).
If HI = {gi, g2, ■ • •, gr}, then it is clear from the definition of dt
and from the relation between dimensionality and determinantal
rank (Vol. II, p. 22) that \dH\ Pjs is the rank of the matrix
B2)
to/ x]=i} ■>
Hence the rank of this "Jacobian" matrix and B1) give the di-
dimensionality of 2)*(P) over P.
We shall now look at these questions in a different way from
the point of view of the structure of P/3> and we prove first the
following
Lemma. Let P = $(£i, $2, • • •, £™)- Then 0 is the only deriva-
derivation of P/$ into itself if and only if P is separable algebraic over #.
Proof. If p is a separable algebraic element of P and D is a
derivation of P/<$, then we have seen that pD = 0. Hence it is
clear that, if P/$ is separable algebraic, then D = 0 is the only
derivation in P/<$. Next suppose P is not separable algebraic over

178 STRUCTURE THEORY OF FIELDS
<i>. We may suppose {£1, £2> ■ • > £r} is a transcendency basis (r —
0 if P is algebraic). If P is not separable over $(£i, £2> • • > £r),
then the characteristic is p 9^ 0 and, if 2 is the subfield of ele-
elements of P which are separable over *(£i, £2, • • •, £r), then P Z) 2
and P is purely inseparable over 2. We assert that there exists a
subfield E 3 2 such that P = E(p), where the minimum poly-
polynomial of p over E is xp — /3, 0 e E. We have [P:2] < 00 and
we can take E to be a maximal proper subfield of P containing 2.
If a e P, i E, P = E(<r) by the maximality of E. Since P is purely
inseparable over 2, hence over E, the minimum polynomial of a
over E has the form xpk - /3, k > 0. Then p = av"~l $ E so E(p)
3 E. By the maximality of E we have E(p) = P. Moreover,
pP = ffpk = f}} so xp — /3 is the minimum polynomial of p over E.
Now we have seen that there exists a derivation D of P/E such
that pD is any chosen element of P. If we take pD 5^ 0, D is a
non-zero derivation of P/$. Next assume P is separable algebraic
over <!>(£!, £2, • • ■, £»•). Then since P is not separable algebraic over
f>, r > 0, and there exists a non-zero derivation of $[£1, • • •, £P]
over $ into P over $. This can be extended to P; hence in this
case also we obtain a non-zero derivation in P/4>.
We can now prove the following result on the dimensionality
of £)*(P) over P.
Theorem 16. If P = #(£1, £2, • • -,&»), M<?» [SD*(P): P]« w
smallest integer s such that there exists a subset {£,-„ £,„ •••,£(,}
{£i> £2} ■■■>!»«} Juch that P /j separable algebraic over
€4).
Proof. As before, we consider the mapping D —> (£iD,
• • ■, £mD) of SD = £)*(P) into P(m). We know that this is a P
isomorphism into P(m)/P- Let (Dl5 D2, • • •, Ds) be a right basis
of 2) over P. Then s < m and the image of 3) in P(m) has the
basis (iiDj, faDj, • • ■, i-mD]), 1 < 7 < -f • The rank of the s X m ma-
matrix A-iDj) is j, so we can choose the order of the £'s so that det (ZiDj)
?* 0,1 < /,./ < j. Set E = $($!, ?2, • • •, ?«) and let D be a deriva-
S
tion of P/E into itself. Then D e 35 so D = 2 A-My> Mi e P.
* >-1
Also £,-D = 2^ (£,-Dy)My = 0 for / = 1, 2, • • •, j. Since det (&£>,-)

STRUCTURE THEORY OF FIELDS 179
^ 0, this implies that every juy = 0 so D = 0. We therefore see
that the only derivation of P/E is D = 0. Hence, by the lemma,
P is separable algebraic over E = $(£i, £2, • • > £»)• Next suppose
{£»'o £»» ' ' "> £*'<} is a subset of the £'s such that P is separable
algebraic over $(^tl, ••-,£;,). If we re-order the £'s we may assume
the given set is {£i, £2> • •-,£«}. We now use these £'s to map 2)
into P(() by means of the mapping D —» (&D), 1 <j <t.
Again this is P-linear. If (£yZ)) = 0, then D maps E = #(£i, £2,
•••,£*) into 0 and so D is a derivation of P/E into itself. Since
P is separable algebraic over E, the lemma shows that D = 0.
Hence we see that the mapping D —» (£,Z)) is an isomorphism
and consequently s = [35:P]r < t. This completes the proof.
Corollary. // P = *(fc, fc, ■ ■ •, U), then [5D.(P):Pfe > r =
tr. d. P/3> aw*/ equality holds if and only if P is separably generated
over 3>.
Proof. The theorem shows that, if s = [T>:P]r, then we may
assume that P is separable algebraic over E = $(£i, £2, • • -j ?«)•
Since P is algebraic over E, it follows that {£1, £2> •••»£»} con-
contains a transcendency basis; hence s > r. If s = r, then since P
is separable over E, the set {£1, f25 • • •> £r} is a separating trans-
transcendency basis. Conversely, suppose P is separably generated.
Then we know that we may select a separating transcendency
basis from the set of £'s. We may assume this is {£1, £2> * • •> £>■}•
Then P is separable algebraic over $(£1, £2> • • • > £r) and the theorem
shows that r > [S):P]ft. Since we have shown that [2):P]ij > r
always, we see that [3D: P]R = r.
In the remainder of this section we shall assume the charac-
characteristic of the field P is p 5* 0. We shall see that the theory of
derivations in this case is closely connected with the study of
purely inseparable extensions of a simple type. We assume P
is purely inseparable (algebraic) over *. If p e P, then the mini-
minimum polynomial of p over * has the form xp' — /3 (Lemma 2,
§ 1.9). We call e the exponent of the purely inseparable element p.
Evidently the exponent is 0 if and only if p e <i>. If there exists a
maximum k for the exponents of the elements of P, then we say
that P is of exponent k over <£>; otherwise, the exponent of P/$ is
infinite.

180 STRUCTURE THEORY OF FIELDS
We shall be interested particularly in purely inseparable exten-
extensions of exponent <1. P has this property relative to <£ if and
only if Pp c <l> where Pp is the subfield of p-th powers of elements
of P. Hence it is clear that, if P is any extension of* of charac-
characteristic p, then P is purely inseparable of exponent < 1 over <!>' =
f>(Pp). We shall now say that an element p e P is p-dependent in
P over $ on the subset S of P if p e$'(S) where $' = *(PP). We
indicate this relation by p <PS (assuming P and $ are fixed in
our discussion). We proceed to show that this is a dependence
relation in the sense of § 3. First, it is clear that, if p e^, then
p e$'(S); hence p <p S. If p <PS we have p e$'(S) and, since
$'(S) is the union of its subfields &(F) where F is a finite subset of
S, then p <p F for some finite subset F of S. If p z&{S) and
every a e S is contained in &(T), then p e$'(T). Hence if p <p <S"
and every a eS satisfies it <p T, then p <p T. It remains to
check the exchange axiom. This states that, if pe&(S) and
P**'(£ - {a-}) for some er in S, then «re#'((J - {a}) U {p}).
Set T = S - \<i} and consider the subfields &{T, p, or), $'(T, p),
^'(T'j <0j *'(^0 ^or which we have the diagram:
We have *'(r, p) => *'(T) and $'(r, «r) =) *'(T). Also pp e $'(D
and <7p e *'(T). It follows that [$'(^, <r) :*'(T)] = 1> = [*'(r, p):
*'G0]. Since p e*'(T, <r), *'(T, p, <r) = $'(T, <r) so [*'(T, p, <r):
&(T)] = p. It follows that *'(T, p, <r) = $'(T, p) = $'(T, <r) so
<r <p T U {p} = (.S1 - {<r}) U {p}. This completes the verifica-
verification of the axioms for a dependence relation.
We can now apply the general theory of dependence relations.
Accordingly, we call a subset S of P p-independent if a <CPS — {a}
for every o- e S. The general basis theorem implies that there
exists a ^-independent subset B of P such that every element is
^-dependent on B. The latter condition is equivalent to P =
#'E). The set B is called a p-basis for P over $. Any two p-
bases have the same cardinal number.

STRUCTURE THEORY OF FIELDS 181
If F = {pi, p2, • • •, Pm\ is a ^-independent set, then p,p = /3* e
*' and Pi: tf $'(pi, p2, • ■ •, Pi-i)- Hence [$'(pi, • • •, p*):$'(pi, • • •,
Pt-i)] = ^ and [*'(pi, • ■ ■, pm)'-&] = pm- It follows that the pm
elements
B3) Pl*V>2*2 • • • pj", 0 < kt < p,
form a basis for <3?'(pi, P2S • • 'jPm) over <£'. Conversely, if this
condition holds, then it is immediate that F is a ^-independent set.
We shall find it useful to apply this criterion in the following
equivalent form: F is ^-independent if and only if the only relation
of the form
B4) 2akl...kmPlk> • • • pj" = 0, 0 <ki<p
with the as in $' is the trivial one in which every akl- • • *m = 0.
We note also that any ^-independent subset A can be imbedded
in a maximal /(-independent set B and such a set is necessarily a
basis.
We return to the consideration of derivations in any field P/f>
of characteristic p. If E is a subfield of P/$ and D is a derivation
of E/# into P/<i>, then t?D = p<?-l(eD) = 0 for any e in E. The
set of ZXconstants is a subfield F of E over $ and the remark just
made shows that F 2 $(EP). If y e T and e e E, then (ye)D =
y(eD). This shows that D is a derivation of E/F into P/F. Since
#(EP) C F, every derivation of E/<£> into P/<l> is a derivation of
E/<£>(EP) into P/$(EP) and the converse is clear. Hence, in con-
considering the derivations of E over $ into P over *, we may as well
replace <i> by #(EP) and so we may assume Ep C <i>. In other
words, we may assume E is purely inseparable of exponent < 1 over
f>. It is now an easy matter to determine the derivations of E
over $ into P over <£. This is given in the following
Theorem 17. Let P be an arbitrary field of characteristic j* 0,
* a subfield and E an intermediate field. Let B be a p-basis of E
over <£. Let S be an arbitrary mapping of B into P. Then there
exists one and only one derivation D of E over $ into P over <i> such
that eD — 5(e) for every e e B.
Proof. As we indicated, there is no loss in generality in assum-
assuming E is purely inseparable of exponent < 1 over #. Also, we may

182 STRUCTURE THEORY OF FIELDS
suppose E 3 $ which means that B is non-vacuous and the ex-
exponent of E/<£> is exactly one. Let e e B and set B( = B — {e}.
Then e $. &(Be) so the minimum polynomial of e over $(Bf) is of
the form xp — /3. Hence there exists a derivation De of E =
$(Be, e) over $(Bf) into P/<£> sending e into the element 5(e). If
■f = {«i> «2, • • •> «r} is a finite subset F of 5, then DF = Dtl +
Df2 + • • • + DCr is a derivation of E/<i> into P/$ such that
eiDp = 5(ei), * = 1, 2, • • -,r. If G is a finite subset of 5 con-
containing F, then the restriction of Dg to $(F) coincides with the
restriction of DF to $(F). Now if £ is an arbitrary element of E,
we can choose a finite subset F such that £ e ^CF) and we can
map £ —> £DF. Then it is clear that %DF is the same for any
finite subset such that £e$(F). Hence the mapping D:£ —>
£DP is single-valued. It is immediate that D is a derivation of
into P/$ such that eD = 5(e) for every ee5. Since E =
), Z) is unique.
Let £>*(E, P) denote the set of derivations of E/$ into P/$.
We consider jD$(E, P) as right vector space over P as we did
before for £)#(P) (cf. § 1.1 and p. 176). Then we have
Corollary 1. [£)*(E, P): P]^ < °o */ and only if E/$ has a
finite p-basis. Then [£>*(E, P): P]B = | B \.
Proof. Let B be a ^-basis for E over $. Let A(B, P) be the set
of mappings of B into P which we consider as a right vector space
over P in the obvious way: EX + 52)(/3) = SiO) + 82(/3), 8,- e A,
0e£ and Ep)(^) = 8@)p, 5 e A, (8 e B, p e P. We now map
©*(E, P) into A(B, P) by sending D e £>*(E, P) into its restric-
restriction S to B. This mapping is linear and, since E = $(B), D —> 5
is an isomorphism. Moreover, the theorem shows that the map-
mapping is surjective. Now it is clear that [A(jB, P): P]r is infinite
(even uncountable) if B is infinite. Moreover, if B is finite, say,
B = {(8i, /32, ■ ■ •, pr}> then the r-mappings St- such that 5t(&) =
8ij (the Kronecker 5»y) form a basis for [AE, P): P]«. Hence r =
[ACB,P):Pfe=.[S).(E,P):Pk.
In the special case P = E = <J>(£;i, £2> • • •, £m) this corollary
gives, in addition to Theorem 15 and 16, still a third way of
evaluating [35*(P):P]« in the characteristic p 5^ 0 case, namely,
this dimensionality is the number of elements in a p-basis for

STRUCTURE THEORY OF FIELDS 1&3
P/<i>. A second consequence of Theorem 17 is the following
Corollary 2. Every derivation of E/$ into P/$ can be extended to
a derivation of P/$ if and only if the elements of any p-basis B of
E/$ are p-independent in P/$.
Proof. If the condition holds, then B can be imbedded in a p-
basis C of P over $. If D is a derivation of E/<£ into P/$, then
the restriction 5B of D to B can be extended to a mapping Sc of
C to P. The corresponding derivation D' of P/$ into itself is an
extension of D. On the other hand, suppose B is not/"-independent
in P over $ and let /3 be an element of B which is ^-dependent in
P on Bp = B — {/3}. If D' is any derivation in P such that
P'D' = 0 for all j8' e Bp, then since j8 e$(Pp, £„), 0D' = 0. The
theorem shows that there exist derivations D of E over <£ into P
over <£> such that P'D = 0, 0' e 2?$ but 0Z> ^ 0. Clearly, such a
derivation cannot be extended to P.
Our next two corollaries will deal with the special case E = P.
The proofs are quite similar to those we have just given so we
leave these as exercises.
Corollary 3. Let P be any field of characteristic p j£ 0 over $.
Then an element p e P is in <£' = <£(PP) if and only if pD = 0 for
every derivation D of V over <£.
Corollary 4. A subset S of P is p-independent if and only if for
every p e S there exists a derivation D of P over <i> such that pD 9^ 0
and oD = Ofor every a 7^ p in S.
We shall now specialize our results by taking $ to be the prime
field $0 (= IP). A derivation of E/<J>0 into P/*o will simply be
called a derivation of E into P. We remark that, if D is a mapping
of E into P such that («i + e2)D = etD + e2D and (eie2)D =
(eiD)e2 + e1(e2S), then D is a derivation of E into P in the
present sense, since (ae)D = a(eD) for ae$oisa consequence of
the first property. We note also that $o(Ep) = W. Hence Corol-
Corollary 3 gives a criterion for an element to be a p-th. power. We
shall now investigate the criterion given in Corollary 2 that the
derivations of E into P be extendable to derivations in P. We
proceed to show that the condition given is equivalent to sepa-
separability, in the general sense, of P over E. First assume the con-

184 STRUCTURE THEORY OF FIELDS
dition: every p-basis of E (over <J>0) is ^-independent in P. Let
Pi, P2, • • •, Pn eP and suppose we have 2etp,p = 0 for «,• 5^ 0 in E.
If B is ap-basis for E, then we can write e, = S7,-*l*,...trj8i*lj82*1 • • •
(8r*r where the C,e5, 0 < kj < p, yikl...kreEp. We have
2 «*,... ^i*1 ••■ i3rk' = 0 where
Since the /3's are p-independent in P we have 5^...^ = 0. We
can write yikl...kr = Vik1---krp, Viki---hr in E. Then 0 = 8*,...*, =
23 Viki- ■ krpPip gives J2 7}ikl.. -krPi = 0 for all kj. Since the «,• ^ 0
i t
one of these relations is non-trivial, so we have shown that any
non-trivial relation of the form Sttp,-5 = 0, et- e E, p,- e P implies
one of the form 27?,pi = 0, t\i e E. This is equivalent to sepa-
separability of P/E. Conversely, assume P separable over E and let
Pi, P2, • • •, fir be elements of E for which we have a relation
Zlkf-kAV* ■ ■ ■ &rK = 0, ykl...kr = Vkl...krp e Pp, 0 < kt < p.
n
Let {pa} be a basis for P/E and write r}hi---kr = 13
X's in E. Then we have *~
0 =
where m = ^,\kl.. -m^i*1/^*2 • • • Prkf- Since P is separable over
k
E, S/ifp/1 = 0 implies every m = 0 so
2X*1...W»!81*V32*1 • • • Prkr = 0, i = 1, 2, • • •, n.
If some yicv--kr ^ 0, one of the X*!-..^.- 9^ 0 and so we obtain a
non-trivial relation with coefficients in W involving the powers
/3i*' • • • Prkr. This implies that, if {jSx, /32, • • •, Pr} is p-depend-
ent in P, then it is ^-dependent in E. It is clear that this implies
the condition of Corollary 2. This corollary therefore gives the
following
Theorem 18. The following two conditions on afield P/E of
characteristic p are equivalent: A) P/E is separable, B) every deriva-
derivation of E into P can be extended to a derivation in P.

STRUCTURE THEORY OF FIELDS 185
EXERCISES
1. Let P = <i>(p) where p^ e $ but pp $ <£. Show that {pip} is a p-independent
subset of E = ^(fip)/^ but {pp} is not a ^-independent subset of P/$.
2. Let 5 be a p-basis for P over $ of characteristic p^O. Show that for
every positive integer k, P = ^P11*, 5).
In ex. 3, 4, P is purely inseparable of exponent one over $ and [P:O] = pm <
«.
3. (Baer) Show that there exists a derivation D of P/$ such that the only D-
constants are the elements of $. (Hint: Let E be a proper subfield of P and sup-
suppose we already have a derivation D in E/$ satisfying the condition. Let
p e P, $ E. We can choose a|3eE which is not of the form eD, « in E and ex-
extend D to E(p) by specifying that pD = /3. Then the only Z)-constants of E(p)
are the elements of 4>.)
4. Show that the D in ex. 3 can be chosen so that D is nilpotent.
5. (Faith) Let P be a field of algebraic functions over $ (any characteristic)
and let E over $ be a subfield. Show that [£>*(P): P]s > [£)*(E) :E]B.
6. Let P = <£(£i, f2) • • •, £m), $ of characteristic p ^ 0. Show that tr. d. P/$
does not exceed the number of elements in a p-basis for P/€>.
7. Let P and $ be as in ex. 6. Show that, if (Di, D2, • • •, Dr) is a right P-basis
for £>*(P), then the elements pi, P2, ■ • ■, pr form a ^>-basis for P over $ if and only
if the matrix (piDj) is non-singular. Show also that, if pi, p2, •• -,pr is a p-basis,
then the elements Di, Di, • ■ ■, Dr form a right P-basis of J)*(P) if and only if
(fiiDj) is non-singular.
8. Let P = $(£1, £2, • • •, £m). Show that P/$ is separable algebraic if and
only if there exist m polynomials g\(x\, #2, • • •, xm), • • -, gm(xi, #2, • • •, xm) in
1, #*, ■ • •, #m] such that gi(£i, £2, • • •, £m) =0 and the Jacobian
9. Let D be a derivation in P/$, F the subfield of D-constants. Prove that
the elements pi, P2, • ■ •, pm are F-dependent if and only if the Wronskian deter-
determinant
Pi Pi • •• Pm
piD pj) ••• pmD
P1D-1 P2D
0.
8. Galois theory for purely inseparable extensions of exponent
one. In this section we shall develop a Galois theory for purely
inseparable extensions of exponent one in which the role of the
Galois group of the classical theory is taken by the Lie algebra of
derivations.
First, let P be purely inseparable of exponent < 1 over <£ and
suppose P has a finite p-basis B = {pi, p2, • • •, pm} over*. Then

186 STRUCTURE THEORY OF FIELDS
[P:*] = pm and the elements Piklp2ki • • • pmkm, 0 < k{ < p, form
a basis for P/<J>. We have pf = ft e #. As before, we let 35*( P)
denote the set of derivations of P/<£ and we recall that 35#(P) is a
restricted Lie algebra of linear transformations in P over <l>. This
means that 35*(P) is a subspace of the space S*(P) of linear trans-
transformations of the vector space P over $ such that, if Di, D2 e 35*(P),
then [£>iD2] = DJJ - D2Dt e 35*(P) and Ap e 35*(P). We
have seen also that 35#(P) is# right vector space over P relative
to Dp = DpR for p in P. Also we know that [35*(P):P]jj = m
(Cor. 1 to Th. 17) and, if p is an element of P such that pD = 0
for every D e 35*(P), then p e$ (Cor. 3 to Th. 17). This last
result gives one half of the Galois correspondence which we shall
establish.
To obtain the second half of this correspondence we now sup-
suppose that P is any field of characteristic p ?* 0 and we do not
specify any subfield as base field. As at the end of the last sec-
section, we consider derivations in P, which can be defined either as
derivations of P over its prime field or as endomorphisms D of
(P, +) such that {pa)D = {pD)<r + p(<tD), p, a e P. We suppose
now that we are given a set 35 of derivations in P with the follow-
following closure properties: A) 35 is closed under addition. B) 35 is
closed under Lie commutation [DiD2]- C) 35 is closed under p-th.
powers. D) 3) is closed under right multiplication by elements
pr, p e P. The conditions A) and D) amount to saying that 35 is a
subspace of the right vector space of endomorphisms of the addi-
additive group (P, +) considered as a space over P relative to Ap =
Apr. Any set of endomorphisms of (P, +) which satisfy A) to
D) will be called a restricted F-Lie algebra of endomorphisms of
(P, +).* We can now state the following theorem.
Theorem 19 (Jacobson). Let P be afield of characteristic p j* 0
and let 35 be a restricted F-Lie algebra of derivations in F such that
[£>:F]R = m < oo. Then: A) if * is the subfield of ^-constants,
then F is purely inseparable of exponent <1 over $ and [P:<i>] =
pm; B) ifD is any derivation of F over®, then D e 35; C) // (Du D2,
• ■ •, Dm) is any right basis for 35 over P, then the set of monomials
* It should not be inferred from this terminology that 35 is an algebra over P as base
field. One of the conditions for an algebra is that [D\Ddp = [Dip, D2] = [Diylhp] and
this does not hold for every p (see equation B6) given subsequently).

STRUCTURE THEORY OF FIELDS 187
B5) AW • • • DJ-, 0<k{< p, (D/> = 1)
is a right basis for the ring 8*(P) of linear transformations of P over
<& considered as a right vector space over P.
Proof. The idea of the proof we shall give is basically the same
as that we used for the Galois theory of automorphisms: we shall
use the given set 3D to define a set of endomorphisms 21 satisfying
the hypotheses of the Jacobson-Bourbaki theorem (Th. 1.2). In
the present case we let 21 be the set of right P-linear combinations
of the endomorphisms given in B5). Then it is clear that 21 is a
right vector space over P and that [2l:P]« < °°. It remains to
show that 21 is a subring of the ring of endomorphisms of (P, +)
and for this it is enough to show that 1 e 21 and that 2[ is closed
under multiplication. The first of these is clear since 21 contains
Di°D2° • • • Dm° = 1. To show closure under multiplication it is
enough to prove that every product (DihlD2k* ■ • • Dmkmp)Dj e 21
for p e P; for, if this holds, then one sees easily that every product
(ZV1 • • • DmS>)(ZVl • • ■ Dj-a), a e P, is contained in 21. If D
is a derivation in P, the condition (£p)D = (£D)p + £(pD) can
be written in operator form as:
B6) PRD = DPB + (PD)R.
This implies that (ZV1 • ■ • Dmk-P)D3- = ZV' • • • DJ^Djp +
Dikl ■ ■ ■ Dmkm(pDj). Hence to show that 21 is closed under multipli-
multiplication it is enough to show that D\kl • • • DmkmDj e 21 for every./ =
1, • • • }m and 0 < ki < p — 1. We shall now assign an (apparent)
degree N = ki + k2 + • • ■ + km to the monomial DiklD2k2 • • •
Dmkm and we shall show that Dxkl • • • DmkmDj is a right P-linear
combination of monomials B5) of degree <iV + 1. This is clear
if N = 0 so we assume it holds for every Dih ■ ■ ■ Dmlm of degree
2/,- < N. Suppose first that j = m. Then if km < p — 1, Dxkl ■ • •
DmkmDm is one of the monomials B5) of degree iV + 1 so the
result holds in this case. If km = p — 1, then (D^1 ■ ■ ■ Dmkm)Dm
= ZV1 • • • Dm_1k'-iDmp and Dmp = 2D,-/j™ since D is closed
under p-th powers. Hence
ZV' • • • Dmk"Dm = 2ZV' • • • Dm-.1k*>-iDinim
so the induction hypothesis applies to show that Df1 ■ ■ • DmkmDm

188 STRUCTURE THEORY OF FIELDS
is a right P-linear combination of monomials B5) of degree
<N + 1. Thus we have the result if j = m and so we can now
make an additional induction hypothesis, namely, that the result
asserted holds for DJ1 ■ ■ ■ Dmk™Di for all / > j. Since N = 2£,- >
0, some ki ^ 0 so we may assume kr ^ 0 and k, = 0 if s > r.
Then we have (D^ ■ ■ ■ Dj^Dj = (D^ ■ ■ ■ D/OA'. U j > r,
the product is a monomial B5) so the result holds in this case. If
j = r, the argument given before for^' = m is applicable to prove
the assertion. Hence it remains to consider the case: j < r.
Since 35 is closed under commutation, DrDj — DjDr +
vhtj e P. Then
and every Di*1 • • • Drkr~1Dh is a P-linear combination of monomi-
monomials B5) of total degree <N. Also this holds for Dxkl ■ ■ ■ Df'^Dj
and, since r > j, multiplication on the right by Dr gives a
P-linear combination of terms B5) of total degree <iV + 1.
This completes the proof of our assertion and shows that 31 is a
subring of the ring of endomorphisms of (P, +). It is clear from
the definition of 21 that [21:P]r < pm and equality holds only if
the monomials B5) are right P-independent and thus form a
basis. We can now apply the Jacobson-Bourbaki theorem (Th.
1.2) to SI and we obtain the following conclusions: If $ is the sub-
field of elements a of P such that ajtA = Aolr for all A e 21, then
[P:$] = [21: P]R and 21 = ?*(P). Now it is clear that aRA = AaB
holds for all A e 21 if and only if cxrD = Dcxr for all D in 2), and
since agD = Dolr + (<xD)r the condition for this is aD = 0 for
all D e SD. Hence we see that 4> is the subfield of ©-constants.
If p is any element of P, then pp is a ©-constant. Hence P is
purely inseparable of exponent <1 over * so we have [P:€>] =
pm> where m' is the number of elements in a ^-basis of P/f>, and
m' <m since [P:$] = [2l:P]B < pm. Also we know that, if
£)*(P) is the set of derivations of P/$, then [35*(P): P]r = m'. If
a e $ and DeS, then {ap)D = a(pD) + (aD)p = a(pD) so D e
2)»(P). Thus £) c ©4,(P) and, since [£>:P]ie = m, we must
have 35 = 35#(P) and m = m'. Then £) contains every deriva-
derivation of P/# and [P:<i>] = pm. This completes the proof.

STRUCTURE THEORY OF FIELDS 189
We can now establish a Galois type correspondence between the
following two collections determined by an arbitrary field P of
characteristic p: Let <% be the collection of subfields $ of P such
that P is purely inseparable of exponent < 1 over * and [P :<£>] < oo
and let & denote the collection of derivation algebras S in P
which are restricted P-Lie algebras of finite dimensionality over
P. If © e ^, let CED) be the subfield of 2)-constants, and, if
$ e «?, let 5D*(P) be the set of derivations of P/<i>. Then C(SD»(P))
= <£ and 35c(D) (£(£>)) = ©• In particular, we obtain a 1-1 cor-
correspondence between the collection of intermediate fields of P/$,
P purely inseparable of exponent <1 over $, [P:4>] < oo and the
restricted P-Lie subalgebras of the Lie algebra £)*(P) of deriva-
derivations in P over $.
EXERCISES
1. Let $[#,.)'] be the polynomial ring in indeterminates x,y over a field of
characteristic/), 31 any algebra over<£. Use the identities (x — y)v — xp — yp and
(x — y)p~1 = 2 x<y' >n *[*>>! to prove the following identities in 31:
<i
B7) [■ ■ ■ [ia]a] ■ ■ ■ a] = [tap]
r-p-l-,
B8) [••• \ba\a\ ■ ■ ■ a] = £ JbaK
i+i-p-i
(Hint: Note that [ba] — b(oR — a{) where or and ai, are the right and left multi-
multiplications determined by a in 31. Specialize the indicated identities by taking
* = &R, y = at in the commutative algebra of linear transformations generated
by ag and «l.)
2. Let 31 be as in ex. 1, SIM the polynomial algebra over 31 in an indeterminate
x. Let a, b e 31 and write
B9) {a + bxY = a" + ^ Si(a,b)x' + fit'.
i
Use the fact that 2a,-** —* S/a.MC* is a derivation in 3l[*] and B9) to obtain
C0) £ {a + bx)%a + bx)i = *£ ist(a, i)**.
+l <l
Use this relation and B8) to prove the following identity
C1) (a + b)p = a? + b* + 5 Si(.a, b)
i
where tSi(a, b) is the coefficient of a;* in
[■■■[[b,a

190 STRUCTURE THEORY OF FIELDS
3. LetP = $0>i, • • -,pm),$ of characteristics j* 0,pi" = /?,• e $, [P:$] = pm.
Let D be a derivation in P/$ such that $ is the subfield of D-constants (see ex. 3
in § 7). Show that the minimum polynomial of D as linear transformation in P
over $ is a p-polynomial of the form
C2) *p"+/8i*pra~1+ft*pm~'1+" • + &»*, ft e$.
Show that there exists an element p e P such that (p,pD, ■ ■ •,pDpm~v) is a basis
for P over $. (This is an analogue of the normal basis theorem for separable
normal extensions.) Show that every element in the algebra £*(P) of linear
transformations in P over $ can be written in one and only one way in the form
C3) l<r0 + Do-i + DV2 H h D^-^-i, o-.eP.
4. Let P, $ be as in ex. 3, 2)*(P) the set of derivations of P/$. Let % be a
subspace of the right vector space 35*(P) over P which is closed under p-th.
powers. Prove that ^ is also closed under commutation so g satisfies all the
conditions of Theorem 19.
5. Show that if D is a derivation in P and rj E P, then rmD*
* = 52 ( .) DJ
j-0 V/
6. Let D be a derivation in an algebra 21 and let %[t, D] be the set of formal
m
polynomials 23 t%ai-> ai e 8t- Equality, addition, and multiplication by elements
o
of $ are defined as for ordinary polynomials. Multiplication is defined by
C4)
Verify the associative law and hence show that 211/, D] is an algebra.
7. Let D be a derivation in a field P of characteristic p^0,$ the subfield of
D-constants and assume [P:$] = />m < °o. Then ex. 3 implies that there exists
a ^-polynomial C2) such that Z>>m + /SiD"™ -| 1- /SmD = 0, & e *. Let
P[/, D] be the algebra of differential polynomials defined as in 6. Verify that if y
is any element of $, then x(y) = tpm + /p™~ /3i + ■ ■ • + //3m — 7 is in the center of
P[/, D]. Let (xG)) denote the ideal generated by ir(y). Show that, if 2lT =
P[/, Z)]/Gr(y)), then [2IT:$] = ^2m. Show that 2to ^ ?*(P).
8. Same notations as 7. Let p be any element of P. Show that there exists an
automorphism of P[/, D] such that / —> / + p and i\ —» 77 for every 77 8 P. Note
that [p, /] = pt — /p = pD by C4) and deduce from this and C1) that (/ + p)p =
tp + (p" + pD"). More generally prove that
C5)
where
C6) pi*1'] = p* + (piy-y-1 + (piy-y-* +■■■+
9. Continuation of ex. 7 and 8. Show that the automorphism of P[/, D] such
thnt t —> / + p, r; —> ??, 17 e P, sends the ideal generated by ir(y) into itself if and
only if p satisfies
C7) P^1"! + /S^t*"! + fate""*! + ■ ■ ■ + pmp = 0.

STRUCTURE THEORY OF FIELDS 191
10. Continuation of ex. 7 through 9. Prove that there exists an automor-
automorphism of 8*(P) sending every 17 e P into itself and sending D —> D + lp, p e P,
if and only if p satisfies C7). Use this to prove the following analogue of Hil-
bert's Satz 90: An element p satisfies C7) if and only if it is a "logarithmic deriva-
derivative" {pD)<r~l of some a in P.
11. Prove the following analogue of the result that the first cohomology
group IP(G, P*) = 0 in the Galois case (cf. § 1.15). Let P be a purely insepa-
inseparable extension of exponent one over $, [P:$] = pm < » and let 35 be the re-
restricted P-Lie algebra of derivations of P over <£. Let D —> fi(D) be a P-linear
mapping of 3D into P (thus an element of the conjugate space 35* of 35) such that
C8) M(D») = v{DY + n{B)iy>-\
Then there exists a a in P such that n(D) = (<rD)o—x for all D.
12. Show that, if Sly is as in ex. 7, then 2lT ^ Sis if
C9) 5 - y = p[*ml + foto"-1! + fol*°>-*i +... + fimp
(as in C7)) for some p e P. Hence use ex. 7 to show that SI = <£,,»• if there exists a
p 8 P such that y = p1*1™1 + /3ipIp">~11 H h PmP- (The conditions given here
are also necessary.)
13. Apply ex. 1 to prove the following result on polynomials with integer co-
coefficients: Let g(x) be any such polynomial and define gk(x) = gk—i(,x)g'(x),
gi(x) = g(x) where ' is the standard derivative. Show that for any prime p,
gp-i'(x) = r](xp) (mod p) where r;(#) is a polynomial with integer coefficients.
14. Let y and S be elements of $ which are not p-ih. powers in $ of charac-
characteristic p 7^ 0. Use ex. 12 (both necessity and sufficiency) to prove that
D0) (*o» + *p_i) + xfy + xfy2 +■■■+ *P_iIO!>-1 = 5
has a solution for Xi e $ if and only if
D1) (y<? + yp_f> + yi'8 + y?8* +■■■ + ^-i'S*-1 = 7
has a solution for yt e "$.
15. Let D be a non-zero derivation in a field P of characteristic p. Show that
the operators 1, D, • • ■, Dp~1 are right linearly independent over P and that, if
pi e P, then p<> + Dpi H \- Dp~1pp-i is a derivation only if every p< = 0,
t-i
(VI. Show that, if p e P, then (Dp)* = DV + UpE)*-1 + £ D<p.- where
2
p,- e P and E = Dp (= Dps). Use these results to prove the following formula
which is due to Hochschild:
E* = {DpY = DV + DOoE"-1).
16. Investigate the possibility of a Krull type Galois theory for purely in-
inseparable extensions of exponent one of infinite dimensionality.
9. Higher derivations. The notion of a derivation can be
generalized in the following way.
Definition 5. Let %be a subalgebra of an algebra S3 overQ. Then
a sequence of mappings D(m) = {Do = 1, Du • • •, Dm} of SI into

192 STRUCTURE THEORY OF FIELDS
58 is called a higher derivation of rank m of 21 into 23 */ every Di is
^-linear and
D2) {ab)Di = £ (aDiWDtJ), j = 0,l,---,m
»=o
holds for every a, b e 21. A higher derivation of infinite rank is an
infinite sequence {Do = 1, Du • • •) of linear mappings of 21 into
23 such that D2) holds for all j = 0, 1, 2, ■ ■ ■.
Clearly, if {Do, D\, D2, • • •} is a higher derivation of infinite
rank, then the section {Do, Di, ■ ■ ■, Dm} is a higher derivation of
rank m and any section {Do, Du ■ ■ ■, Dq}, q < m, of the higher
derivation {Do, •••yDm} is a higher derivation. The mapping
Di is a derivation of 21 into 23.
Let 21 = 23 = $[x] where x is transcendental and let Di be the
linear mapping in 21 whose effect on the basis (I,*,*2, • • •) is
given by
D3) xmDi
where we agree that I 1 = 0 if i > m. Then
)xm+n-'
xm+nDj = [
A V
and
.
Since 2_j I . 11 . . I = I . I, we have )> (xmDi) (xnDj^i
i=o\t /\j - t/ \ j / ,-_0
y. This shows that A, Du D2, • ■ •) is a higher deriva-
derivation of infinite rank in $[x].
If f> is of characteristic 0, then D3) shows that i\Di = Z)i*
where Di is the usual standard derivation in $[x]. Thus D,- =
— Di*. More generally, if Di is a derivation in any algebra of
characteristic 0 and we define Di = — ZV, then {1, Di, D2, • • •}
is a higher derivation of infinite rank in 21. This follows im-

STRUCTURE THEORY OF FIELDS 193
mediately from Leibniz' formula: (ab)LP = Tl(J.)(aDi)(bD>-i)
which gives (abXEP/jl) = 2(aDi/i\)£&Di-i/(j - *H- This is
D2) for Di = - Di\
2!
The device we used for reducing the study of derivations to
homomorphisms can be generalized so as to apply to higher deriva-
derivations. Let £(m) be the algebra over $ with basis A,/, • • -,tm)
such that tm+l = 0. Hence £(m) e**[*]/(*m+1). Let 93(m) =
23 ®*£(m). If D(m) = {l,Di, ■■■,Dm) is a higher derivation
of rank m of 21 into 33, then we introduce the mapping j(D(m)) of
21 into 58(m) as
D4) a -> a + («DX)/ + («D2)/2 + • • • + (aDm)tm.
Evidently s — s(Dim)) is linear. Also
a'i' = L
0
= {ab)'.
This shows that j is a homomorphism of 21 into S8(m). We have
the homomorphism ir:a0 + «i/ + <32^2 + • • • + amtm —> a0, at e S3
and a8' = « for every a e 21. As in the special case of derivations,
this property is characteristic of the homomorphisms s obtained
from higher derivations of rank m.
Similar considerations apply to higher derivations of infinite
rank. The place of the algebra S8(m) is now taken by the algebra
©[[/]] of power series
D5) 0O + ait + a2t2 +■•■
where the a{ e S3 (cf. Vol. I, p. 95). As before, if {1, Du ■ ■ ■} is
a higher derivation of infinite rank, then the mapping s'.a —» a +
(aDi)t + (aD2)t2 +■ ■ • is a homomorphism of 21 into 93[|/]] such

194 STRUCTURE THEORY OF FIELDS
that aST = a for all a e 21 where -k is the homomorphism S«
a0- Conversely, if a —> a3 is a homomorphism of 2f into
such that a" = a, a e 21, then we write as = a + (aDJt +
(aD2)t2 +■ • ■ and {Z)o = 1> -Dij As, • • •} is a higher derivation
of 21 into SB.
If {Dt} is a higher derivation of rank m (infinite rank) of 21 into
58, an element a e 21 is a constant relative to the higher derivation
if aDi = 0 for all i > 0. This simply means that a* = a for the
homomorphism associated with the higher derivation. Hence it is
clear that the set of constants is a subalgebra of the algebra 21.
Our purpose in this section is to give just an introduction to
higher derivations and to examine briefly higher derivations of
purely inseparable fields. We suppose now that P/<£> is a field of
characteristic p ^ 0. Let E be a subfield of P/$ and let D(m) =
{1, Di, ■ • •, Dm} be a higher derivation of rank m of E/$ into
P/$. In general, if Z)x = ZJ = • ■ • = £>e_i = 0 but Dq ^ 0,
then we shall say that the higher derivation is of order q and Z>(m)
is called proper if Dx 5* 0. If the order is q, the associated homo-
homomorphism s = s(D(m)) of E into P(m) has the form
D6) e -> e + (eDq)t« + («Z)g+1)/«+1 + • ■ • + (eDm)tm,
where eDg ^ 0 for some e in E. We shall use this to prove the
following
Theorem 20. Let P/$ be afield of characteristic p 9^ 0, E a sub-
field of P/$, Z)(m) a higher derivation of rank m and order q of E/$
into P/<J>. Let Y be the subfield of T>(m)-constants of E and let p" be
m
the smallest power of p > — . Then E is purely inseparable of
q
exponent e over V.
Proof. We have to show that ep* e T for every e e E and that
there exists an e e E such that ep' 1 i T. The first is clear from
D6) since
= 6"' + (eDqy'tp'« +■■■= epe.
Hence ev' e T. Now choose e so that eDq 9^ 0. Then (eO3 =
ept~l + (tDg)p'~1tp'~lq H . Since .p*? < m it is clear that
(eP'-'y ^ €pe~\ Hence e**1 F.

STRUCTURE THEORY OF FIELDS 195
We consider next a purely inseparable simple extension field
P = <£(£) where xp' — a is the minimum polynomial of £ over <1>.
Let {D{} be the higher derivation in the polynomial algebra $[x]
defined by D3) and let £><*•-» = {l,Du ■■•,DP._1} be the
higher derivation of rank pe — 1 which is a section of this higher
derivation. We have (xp' — a)Dj = 0 for 1 < j < pe — 1 which
together with the defining relations D2) imply that the principal
ideal 3 = C*p' — «) is mapped into itself by every Dj. Hence
every Dj induces a linear mapping, which we denote again by Dj,
in P = <!>(£) = *M/3. The conditions in <£[*•] for Dj go over to
the same conditions D2) for the Dj in <£(£). Hence we obtain a
higher derivation D(p'~1) in <*>(£) such that
D7) rDi = (™) r~\ m = 0, 1, • • •, f - 1.
We shall now show that the subfield F of {D,}-constants for
jy,v'-\) js $ Thus suppose * d F. Then the minimum poly-
polynomial of £ over F is xpf — /3 with / < e and /3 in F. Then £p/ e F.
On the other hand, the definition D7) gives ^plDpt = 1. This
proves our assertion.
We assume next that P is a purely inseparable extension of*
which is a tensor product of simple extensions. Pi, P2, • • •, Pr,
P,- = $(£;). This means that P = <£(£i, £2, • • ■, £r) and the mono-
monomials fi*1^*2 • • • £r*r, 0 < ki < pei, form a basis for P over <£>.
If we set $; = *(£i, • • •, £,-_!, f,-+1, • • •, |r), then P = $,-(f,-) and
*i H *2 n ■ ■ ■ n $r = <!>. There exists a higher derivation in P
whose constants are the elements of <£,. Hence it is clear that * is
the subset of P of elements which are constants relative to all the
higher derivations of finite rank in P over <i>.
EXERCISES
1. Let {Di} be the higher derivation in <&[#], x transcendental, defined by D3).
Show that
/(* + a) =/(a) + (/Di)(a)* + (/A)(a)*2 +• • •.
2. Let #[#1, #2, • • •, xm] be the algebra of polynomials in indeterminates *,■
over a field $. If (£1, ki, • • ■, &m) is a sequence of non-negative integers ki we de-
define a linear operator ©*,*,...*„ in $[*i, xi, • • •,*„] by its action in the basis

196 STRUCTURE THEORY OF FIELDS
(*iB»#2"« • • • *„"«) as follows:
0 if any *,• > »,•
• • • *„"»-*»
if ki ^ »i
Show that, if f(xi, x2, •••, xm) E $[#1, #2, • • •, xm] and an, a-2j • • •, am e<J>, then
f(x\ + ai, #2+0% • • -,xm + aw) = 23 (JDkfo-■ •*„)*,• -a,-*!*1^*2 • • • *m*m.
3. Let /(#i, ■ • •, #m) be a homogeneous polynomial of degree n < m in
$[#i, #2, ■ • •, xm]- Suppose there exists an a = (ai, a2, • • ■, aw), a,- e <£, such
that <JDk1ki...km)x,-.ai = 0 if Ski < n - 2 and
;o.
Show that the equation f(xi, x%, ■ • •, *m) = /8 has a solution in $ for any /3 e $•
Use this to prove that
x* + y% + z* - 3*yz = /3
is solvable in any field of characteristic ^ 3.
4. A higher derivation in 31 of infinite rank is called iterative if DtDj =
( . j Di+1- and a higher derivation £>(m) = {A'} is called iterative if DtDj =
(' . J) Di+i for i +j < m and DiD,- = 0 if i +j>m. Verify that the higher
derivations defined by D3) and D7) are iterative.
5. Let P = $(£) where $ is of characteristic p ^ 0 and the minimum poly-
polynomial of £ over $ is xp' — a. Show that the subfields of P/<£ are the fields
<$(£p/) where 0 < / < e, and that the indicated e + 1 subfields are distinct.
6. (Weisfeld). Let $o be a field of characteristic p ?* 0, $ = $o(«,0,7)
where ap, jSp, 7P £$o and these elements are ^-independent over $o([3>:<J>o] = P*)-
Let P = $(£, v) where £"* = a, yj» = 0^ + 7. Show that [P:$] = p3. Show
that [#(£):$] = p2, [#A7):$] = p2 and $(f) fl $(j?) = $. Show that P 5*
$(£> f) and P 5^ *('?, f) where f is any element such that fp e $. Hence show
that P/$ is not a tensor product of simple extensions.
7. Show that {A! is a higher derivation if and only if orDj = 23 A(<*A'-»)«j
i-0
j = 0, 1, • • ■. Show that, if D\ 5^ 0 in the higher derivation {1, Di, Ai, • • •,
Dm], then the endomorphisms A, D\, •••, Dm) in (P, +) are right P-independent.
8. Let D(v'~v> be an iterative higher derivation of rank p' — 1 in a field P of
characteristic p. Assume T^~v> is proper and that $ is the subfield of con-
p'-i
stants. Show that every 1 inear transformation in P over $ has the form 23 DiPi =
0
SDiPig, pi e P, and that P = <£(£) where the minimum polynomial of £ over $ is
xp' - a.
9. Continuation of 8. Show that a sequence of linear transformations {do,
d\, ••', dp'— 1} in P/$ satisfies pndj = 23 ^i{pA'-i)>J = 0> 1> '"iP* ~ 1> '^ and

STRUCTURE THEORY OF FIELDS 197
only if there exists a vector (<ro, <ri, • • -jOy-i), 00 = 1, ffi E P such that d< =
jEWo + Di-iffi + • • • + \<Ji. Use this to obtain necessary and sufficient condi-
conditions that a vector (<ro, a\, ■ ■ ■, <v_i) be a "logarithmic derivative" in the sense
that there exists a p e P such that Oi = p~1(pDi)> i = 0, 1, • • -,p' — 1.
10. Tensor products of fields. In Chapter I we considered
tensor products of two fields, one of which was finite dimensional
over the base field. We saw that it was necessary to know the
maximal ideals of P ®* E in order to survey the composites of
the field P/4> and E/$ where [P:$] < «. In this section and the
next we shall obtain the extension of these results to arbitrary
fields. We shall first collect a number of results for the case in
which one of the fields is algebraic. In our statements, sepa-
separability will mean separability in the general sense defined on p.
166; pure inseparability will be used for an extension which is
purely inseparable algebraic. Also we shall say that a subfield
$ is algebraically closed (separably algebraically closed) in P if
every algebraic (separable algebraic) element of P/f> is contained
in <i>. We can now state the following
Theorem 21. Let P/* and E/$ be extension fields 0/$.
A) If P/$ is separable and E/$ is purely inseparable, then P
<g>* E is a field. On the other hand, if P/<£> is not separable, then
there exists a purely inseparable extension E/$ of exponent 1 such
that P ®# E contains a non-zero nilpotent element.
B) If P/<l> is separable algebraic, then P <g># E has no non-zero
nilpotent elements for arbitrary E/<i>, and P <g>* E is afield if & is
separably algebraically closed in E.
C) The elements of P <g>* E are either units or nilpotents if
either P/$ is purely inseparable and E/$ is arbitrary, or P/<£ is
algebraic and <i> is separably algebraically closed in E.
Proof. In A) and the first part of C) we may assume the
characteristic is p 9^ 0. In all cases we write P ® E for P ®* E
and we identify P and E with subalgebras of P ® E = PE.
These are linearly disjoint and consequently they satisfy the
various linear independence properties which we have noted for
this relation.
A) Assume P/$ is separable and E/<£> is purely inseparable. The
separability implies that, if px, p2, • • ■, pm are ^-independent ele-

198 STRUCTURE THEORY OF FIELDS
ments of P, then the elements pipt, p2p', ■ • •, pmp' are *-independ-
m
ent for every e = 0, 1, 2, • • •. Now let z = 23 P&i e P ® E
i
where the pt- e P and a, e E. We may assume the p* are <£-inde-
pendent and, if z ^ 0, then we may assume also that every <r,- j£ 0.
Since E/$ is purely inseparable there exists a positive integer e
such that <r*' = a,- e$ for 1 < i < m. Then zp' = 'Leap*' e P
and, if z 5^ 0, then the a,- 9^ 0 and zp< is a non-zero element of P.
Hence zp' and consequently z has an inverse. Thus P ® E is a
field. Next assume P/* is not separable. Then there exist ele-
elements pi, p2, • • ■, pm in P which are ^-independent but for which
there exist ?,- ^ 0 in $ such that 2yipf = 0. Not all the 7,- are
/>-th powers in <£ so an extension field of the form E = $(<ri, (r2, • • •,
^m), fft" = 7tj is of exponent 1 over <£. The element z = 2pi<r,- of
P <8> E is not zero since the pt- are ^-independent and the <rt- e E.
On the other hand, zp = 2p*W = '2yipt3> = 0.
B) Assume P/$ is separable algebraic, E/$ is arbitrary. We
have to show that P ® E has no non-zero nilpotents and that
P ® E is a field if <£> is separably algebraically closed in E. If
m
z e P <8> E, z = ^2 pitFi where the p* e P and a,- e E. Since P/€>
1
is algebraic, the pt- generate a finite dimensional extension and we
may clearly replace P by this extension in proving our result.
Hence it suffices to assume that [P:<£>] < °°. Then the sepa-
separability of P implies that P = $F) ^ *[*]/(/(#)) where f(x) is
separable and irreducible in $[x]. As we saw in Chapter I (p. 87)
P ® E^E[x]/(/(x)). Hence our result will follow if we can
show that E[x]/(/(x)) has no non-zero nilpotents and this is a
field if # is separably algebraically closed in E. Now we have
seen in Chapter I that E[x]/(/(^)) is a direct sum of fields, and it
is easy to verify that an algebra having this structure contains no
non-zero nilpotent elements. This proves the first statement.
Next assume E[#]/(/(#)) is not a field. Then f(x) = g{x)h{x) in
~E[x] where deg g > 0 and deg h > 0. Let 0 be a splitting field
over # of /(#) and let f{x) = !!(# — wt) in Q,[x). Since the wt-
are roots of f{x), they are separable algebraic over *. It follows
that the coefficients ofg-(#) and h(x) are separable algebraic over f>.
These are elements of E and they are not all contained in $

STRUCTURE THEORY OF FIELDS 199
since f(x) is irreducible in $[x]. Thus $ is not separably alge-
algebraically closed in E.
C) Assume first that P/* is purely inseparable and E/* is
arbitrary. Let z = ^ pto-,- e P ® E where p,- e P, <rt- e E. Choose
1
e > 0 so that p,-*' = a,- e$. Then zp< = 2aiaip' e E. Either zp'
= 0 or zp* has an inverse in E. In the latter case z is a unit in
P ® E. Next assume P/$ is algebraic and $ is separably alge-
algebraically closed in E. Let 2/$ be the maximum separable sub-
field of P*/#. The subalgebra 2E of PE = P ® E over $ is the
tensor product of 2/$ and E/$. Since $ is separably algebraically
closed in E, we have, by B), that 2E = 2<g>Eisa field. Let
{pa} be a basis for P/2, {o-p} a basis for 2/$. Then {pa<rp} is a
basis for P/3> and these elements are E-independent in P ® E. It
follows that the elements pa are 2E-independent. This implies
that, if P and 2E are regarded as algebras over 2, then PBE) =
P ®j2E. On the other hand, PBE) is the same algebra over
<£ as PE = P ®* E; hence it suffices to show that every element of
P <g>z 2E is either nilpotent or a unit. Since P/2 is purely in-
inseparable, this follows from the first part of the present proof.*
Our next task is to obtain some information on tensor products
of two fields, one of which is purely transcendental. The result
we shall prove for these in the following
Theorem 22. Let P be purely transcendental over <£>, say, P =
$(B) where B is a transcendency basis and let E/# be arbitrary.
Then P ®* E has no zero-divisors, and if Q is its field of fractions,
then $2 = EB?) is purely transcendental over E with B as transcend-
transcendency basis. Moreover, if # is algebraically closed {separably alge-
algebraically closed) in E, then P = $(B) is algebraically closed (sepa-
(separably algebraically closed) in 0, = EE).
Proof. As usual, we consider P and E as subalgebras of P ®* E.
Since B is an algebraically independent set, the set M of
distinct monomials jSi*^*1 • • • /3r*r, h > 0 in the /3 e B forms
a basis for the subalgebra $[B] generated by B. Since $[B] and
* The identification of P<8>* E with P®* B®* E) which was used in the proof can be
established also by general formulas on tensor products. One has the associativity:
P®2B®*E)^(P®s2)®*E (cf. ex. 5, p. 15). Moreover, P®S2^P. Hence
P®s B®* E) S* P®« E.

200 STRUCTURE THEORY OF FIELDS
E are linearly disjoint, the set M is E-independent. Hence B is
algebraically independent in E[5]. Then we know that, if F is a
finite subset of B, E[F] has no zero divisors (Vol. I, p. 106). Hence
E[5] is an integral domain and so it has a quotient field 12 whose
elements have the form PQ~X where P,Q e E[2?]. Thus we see
that 12 = E(i?) and, since B is an algebraically independent set
over E, clearly 12 is purely transcendental over E with B as a
transcendency basis. We observe next that 12 contains the sub-
algebra fli of elements of the form Pq~l where P e E[Z?] and q e
$[B]. We proceed to show that this subalgebra can be identified
with P <g>* E. First, we have the identity isomorphism of E[5] C
12 into E[.5] C P ®$ E and this can be extended, by I of the
Introduction, to a unique isomorphism of 12i = \Pq~1\P e E[5],
q t* 0 in $[B]} into P <g># E, since the element q-1 exists in P =
$(B). Let z be any element of P ®* E and write z = 2p*e,-,
Pi e P = #(.S), «,- e E. We can write p,- = p*? where piy q e
$[5]. Then z = (Sj&.-e,-)?-1 = Pq~x where PeE[5], It follows
that z is in the image of the isomorphism of I2i, so 12X is isomorphic
to P <g># E. Hence if we identify P <g>* E with Qx and observe
that 0 is also the field of fractions of i2x since Ox 3 $[5], we ob-
obtain the first statement. To prove the second we shall show that,
if £2 = E(B) contains an element which is algebraic (separable
algebraic) over $(B) which is not contained in 4>E), then E con-
contains an element which is algebraic (separable algebraic) over $
not contained in <£. Clearly, if an element of the type indicated
exists in 0 = EE), then it exists in ~E(F) for a finite subset F of
B. Hence we may take B finite and an induction argument shows
that it is enough to prove the following result: Let E/$ be arbi-
arbitrary and let £ be transcendental over E. If E(£) contains an ele-
element which is algebraic (separable algebraic) over $(£) and not
contained in $(£), then E contains an element which is algebraic
(separable algebraic) over * and not contained in $. Thus let t\ be
an element of E(|) which is algebraic over <£(£) and let xn +
fi\Xn~1 + • • • + j8n be its minimum polynomial over <$(£). We
can write & = piq~% wherep,-, q e $[£] (e.g., q can be taken to be the
product of the denominators of the /3<). Then H = qt) is alge-
algebraic over <£(£) with minimum polynomial xn + p\Xn~x + p%xn~2
H \-pH. If H = PQ-1 where P and Q e E[fl and are rela-

STRUCTURE THEORY OF FIELDS 201
tively prime polynomials, then the equation for H gives
P" = -PxP^Q - p2Pn-2Q2 pnQ*.
If Q is of positive degree, then Q has an irreducible factor and the
displayed relation shows that this is a factor of P", hence of P,
contrary to the assumption on P and Q. It follows that Q is a
unit and so H e E[£]. We now write H = «o + «i? + «2?2 + • • ■ +
«m£m where the et- e E and we shall show that the relation 0 =
mr + pii&mi:)*-1 +■■■ + *»(& h = m&, pt = p,® e
$[£], implies that the coefficients e;- are algebraic over $. Thus let
a e * and consider the homomorphism of E[£] over E into E send-
sending £ —> a. Such a homomorphism exists since £ is transcendental.
As usual we denote the image of Q{£) by Q{a). Then we have the
relation H(a)n + p1(a)H(a)n-1 -\ 1- />„(«) = 0- Since the
pi{a) e $, this shows that the element /3 = //(a) is algebraic over
$>. Suppose first that $ contains m -\- \ distinct elements ai,
m
«2> • • ■> «m+i- Then H(ak) = 23 €ia*; = i®* is algebraic over
3=0
$ for ^ = 1, 2, • • •, m + 1. Since the Vandermonde determinant
det («ftJ) 5^ 0, these equations for the «,- have a unique solution
which is given by the usual determinant formulas. These show
that the e's are algebraic over $. If # does not have m + 1 ele-
elements, we have to modify this argument slightly in the following
manner. If p is the characteristic, we choose r so that pT > m
and we let E be a splitting field over E of xv' — 1. We let 5 be
the subfield of E of elements which are algebraic over $. Evi-
Evidently this contains m -j- 1 distinct a^. We now make the argu-
argument with these elements using E in place of E, $ in place of <£.
Then we can conclude as before that the tj are algebraic over $,
hence, over <£. Now it is clear that, if the t\ we started with
t $(£), then H #$(£) and consequently not every «,- in H — Ze,?*
is in <£. Thus there exists an e in E algebraic over $ which is not
contained in E. Next assume t\ $ <£>(£) and rj is separable over
<£(£). Then H$$(£) and is separable algebraic over $(£). Then
the a arc algebraic and the field $(eu e2, • • •, O contains a
separable algebraic element not in <i>. Otherwise, the charac-
characteristic is p and we have e/' e $ for some e = 1, 2, • • •. Then we

202 STRUCTURE THEORY OF FIELDS
have Hp' e$(!;) contrary to the separability of H over $(£). This
completes the proof.
We are now ready to handle the "mixed" cases in which the
fields need not be either algebraic or purely transcendental. We
prove first the following extension of a part of Theorem 21:
Theorem 23. If P/$ is separable and E/$ is arbitrary, then
P ®* E has no non-zero nilpotent elements.
Proof. It is clear that it suffices to prove this result under the
additional assumption that P is finitely generated. Then P is
separably generated, so that P has a transcendency basis B such
that P is separable algebraic over $(B). We now consider the
subalgebra $E)E = $E) <g>* E generated by $(B) and E and
we regard this as well as P as an algebra over the field $(B). One
sees easily that, if {pa} is a basis for P over $(B), then the only
relations of the form 2f,Pt = 0, ct e3>(Z?)E are the trivial ones for
which every ct = 0. This implies that P ®* E = P ®$<.b) <3?(.B)E.*
We now apply Theorem 22 to the factor $E)E = $E) ®* E.
According to this result <i>(.B)E can be imbedded in a field 0 =
EE). Then P ®*(b) *E)E is a subalgebra of P ®#(s) 0 where
fl is a field over $(B) and it suffices to prove that P ®*(b> fl has
no non-zero nilpotent elements. Since P is separable algebraic
over $(B), this follows from Theorem 21 B).
We assume next that P is arbitrary and that $ is separably
algebraically closed in E. Let B be a transcendency basis for P
over *. As in the foregoing proof we have P ®* E = P ®*(b)
and this is a subalgebra of P ®*(B) fl where fl is a field
By Theorem 22 we know that $(B) is separably algebrai-
algebraically closed in 12. Since P is algebraic over $(B), Theorem 21C)
shows that every element of P ®*<b) ^ is either nilpotent or a
unit. Now let z be any element of P ®* E C P ®*(S) Q. Either
z is nilpotent or it is a unit in P ®*(s) fl- In the latter case z is
not a zero divisor in P ®* E. We can therefore state the following
Theorem 24. If P is an arbitrary extension field of afield * and
$ is separably algebraically closed in E then every zero divisor of
P ®* E is nilpotent.
* A more sophisticated argument can be used to establish this. See the footnote on
p. 199.

STRUCTURE THEORY OF FIELDS 203
Clearly the last two theorems have the following immediate
consequence.
Corollary 1. Let P and E be extension fields of $ such that A)
either P/$ or E/<J> is separable, B) $ is separably algebraically
closed in either P or E. Then P <g># E is an integral domain.
In particular, we see that, if P/<£ is separable and $ is alge-
algebraically closed in P, then P <g># E is an integral domain for any
E/#. An extension P/$ satisfying these two conditions is
called regular. If $ is algebraically closed, then it is perfect so
any extension P/$ is separable. Moreover, it is clear that $ is
algebraically closed in P. Hence every extension of an alge-
algebraically closed field is regular and consequently we have
Corollary 2. If <£> is algebraically closed, then P ®* E is an
integral domain for arbitrary extension fields P and Eo/$.
11. Free composites of fields. We recall that a composite of
two fields E and P over <£ is a triple (F, s, t) where T is a field over
$ and s and t are isomorphisms of E over # and P over <£ respec-
respectively into F such that T is generated by the images E* and P*
(§ 1.16). The composites (T,s,t) and (T',s',tr) of E and P are
equivalent if there exists an isomorphism u of T onto I" such that
s' = us, t' = ut. In § 1.16 we studied composites of a finite di-
dimensional extension P and another extension. In algebraic
geometry one is interested in composites of fields which need not
be algebraic but one restricts the notion in the following way.
Definition 6. A field composite (F, s, t) of E/<£> and P/<i> is
called free if for any algebraically independent subsets C and D of
E and P respectively, the sets C", Dl are non-overlapping and C U Dl
is algebraically independent in
Since any algebraically independent set can be imbedded in a
transcendency basis, it is clear that the condition that (T, s, t) is
free is equivalent to the following: for every pair of transcendency
bases B and B' of E/$ and P/*, respectively, B" and Bn are non-
overlapping and Bs U B" is algebraically independent. We now
observe that the word "every" can be replaced by "some" in

204 STRUCTURE THEORY OF FIELDS
this criterion. Thus suppose there exists a transcendency basis
B for E/* and a transcendency basis B' for P/<i> such that Bs and
Bn are non-overlapping and B* U Bn is algebraically independent.
We assert that this implies that the composite F is free. It
clearly suffices to establish the condition of the definition for
finite sets, C, D. Now we can find a finite subset F of B such that
C is algebraically dependent over <£> on F. Since F is a subset of
B, F° is algebraically independent in F over P*. Hence F* is
algebraically independent in $(.F% D() over $(Dl). This implies
that the transcendency degree of $(F", C, Dl) over $ is / + d
where /is the cardinal number \F\ and ^ = \D\ (cf. ex. 3, § 3).
Since the transcendency degree of $(Fa, Cs) over $ is / and C8 is
algebraically independent, the transcendency degree of $(.F8, C)
over f>(Cs) is / — c where c = \C\. It follows that the tran-
transcendency degree of $(FS, C, Dl) over $(C3, D() does not exceed
f — c. This and the formula for the transcendency degree of
S, C, Dl) over <1> imply that the transcendency degree of
", Dl) over $ is at least (/ + d) - (f - c) = d + c. It follows
that C, Dl are not overlapping and C U Dl is algebraically in-
independent. We state this result as the following
Lemma 1. Let (T, s, t) be a field composite of the fields E over
$ and P over$. Suppose there exists a transcendency basis B for E
over $ and a transcendency basis B' for P over $ such that B", Bn
are non-overlapping and B' U Bn is algebraically independent.
Then (T, s, t) is a free composite of E/<i> and P/«i>.
We remark also that if the condition of the lemma holds for B
and B', then B' U Bn is a transcendency basis for F. For, it is
clear that the elements of E* and of P' are algebraic over $E8 U
Bn). Since F is generated by E8 and Pf, it follows that F is
algebraic over ${B" U Bn). Hence B* U Bn is a transcendency
basis.
We can use the criterion of the lemma to prove the existence of
a free composite for any two fields E and P over $>. Let B and B'
be transcendency bases for E and P over $ respectively. If B
and B' are finite, say B = {£l5 ■ • •, £„,}, B' = {r?i, • ■ -,»;„}, then
we construct the polynomial algebra $[xu x2, • • •, xm+n] in m + n
indeterminates x\, x2, • • •, xm+n, and we form the field of frac-

STRUCTURE THEORY OF FIELDS205
tions $(xi, x2, • • -,xm+ri). We have an isomorphism s of
into $(*i, x2, ■ ■ •, xm+n) such that £t- —» #,•> < = 1, 2, •••,/» and
an isomorphism / of $(B') into $(^i, x2, •••ixm+n) such that
>7y —* #m+ij i = 1> 2, •••,«. Now let 0 be an algebraic closure
of #(#i, x2, • ■ -yXm+n). Then we know that the isomorphisms s
and t can be extended to isomorphisms s and / of the algebraic ex-
extensions E and P of <t>(i?) and *(£') into fi (cf. ex. 1, p. 147). From
the lemma, then, if F is the subfield of 12 generated by E* and P',
(F, s, t) is a free composite of E and P. If either B or B' is infinite
a similar procedure can be employed, or we can modify it slightly
by defining 1-1 mappings of B and of B' into the one of these, say
By which has the larger cardinal number, in such a way that the
images are disjoint. These mappings can be extended to iso-
isomorphisms s and /of $[JS] and$[Z?'] intof>E). Then they can be
extended to isomorphisms s and / of $E) and ${B') into $(B),
which can then be extended to isomorphisms s and t of E and P
into an algebraic closure $2 of$(B). Then (F, s, t), where T is the
subfield generated by Es and P(, is a free composite of P and E.
We shall now extend the considerations of § 1.16 to obtain a
survey of all the composites and all free composites (in the sense
of equivalence) of two given fields E and P over $. As before,
we form the tensor product E ®* P and we identify E and P with
their images in E ®# P. Let $ be a prime ideal in E ®# P (Vol.
I, p. 173); hence (E <g>* P)/$ is an integral domain as well as an
algebra over $. We can imbed this in its field of fractions F. Let
s denote the canonical homomorphism e —> e + ^J of E (C E <g> P)
into (E <g>* P)/93. Since E is a field and Is = 1, this is an iso-
isomorphism. Also since (E ®* T?)/V £ F, we can consider s as
an isomorphism of E/# into F/$. Similarly, we have the iso-
isomorphism f.p —> p + $ of P into F. Now E and P generate
E <g> P. Consequently E" and P' generate the algebra (E <g> P)/$.
Since F is the field of fractions of (E <g> P)/1^ we see that the
field F is generated by its subfields E" and P*. Hence (F, s, i) is
a composite of E/# and P/#.
Next let $' be a second prime ideal in E ®* P and let (F', j', t')
be the corresponding composite constructed in the manner just
given. Suppose (F', s', t') is equivalent to (F, s, t). Then we
have an isomorphism a of F onto F' so that s' = su, t1 — tu.

206 STRUCTURE THEORY OF FIELDS
Thus u maps e" = e + $ -» e8' = e + $', p' = p + '"P -» p + $'.
Consequently the restriction of « to the subalgebra E8P'/$
sends 2«,p; + $ —»• 2e<p,- + $' for e,- e E, pt- e P. It follows as in
§ 1.16 that Ze,p; e <$ implies 2«8p; e $'. Hence $ C $', and if we
repeat the argument with u~1 we see that ty' C l>p. Thus we see
that distinct prime ideals in E <g>* P give rise to inequivalent com-
composites of E/$ ««</ P/<i>.
Now let (F', j', /') be any composite of E/$ and P/$. Then we
can combine the isomorphisms s', t' of E/$ and P/$ into I" to
obtain the homomorphism 2e,-p,- —> 2e/'p/' of E <g>$P into I".
The image under this homomorphism is the subalgebra E8'Pf/
generated by E8'/* and P''/*> This is an integral domain.
Hence if $ is the kernel of the homomorphism, then (E ® P)/$ =
E3'P(' and (E ® P)/$ is an integral domain. Hence $ is a prime
ideal in E ® P so this can be used to construct the composite
(F, j, /) as before. Now the homomorphism of E ® P onto
E*'P'' gives rise to the isomorphism u of (E <g> P)/$ onto E*'P('
such that ~LtiPi + ^P —> S«,*'p«''. This has a unique extension to
an isomorphism « of the field of fractions r of (E <g> P)/^J onto
r'. We have es" = (« + ^)" = e8', e e E and p(u = (p + $)" =
p*', p e P. Hence u is an equivalence of (T, s, t) and (F', j', /'). Our
considerations therefore establish a 1-1 surjective mapping from
the set of prime ideals $ in E ®* P to the set of equivalence
classes of composites in E/«i> and P/<£.
In § 1.16 we established a 1-1 surjective correspondence be-
between the set of maximal ideals in E ®# P for [P:<t>] < <» and
the equivalence classes of composites of E/<S> and P/$. We can
now see that this is a special case of the present more general
considerations. We recall that an integral domain which is a
finite dimensional algebra is a field (Introd., p. 8). This im-
implies that any prime ideal in a finite dimensional algebra is maxi-
maximal. If P/$ is finite dimensional, then E ®* P can be considered
as a finite dimensional algebra over E. Hence the prime ideals
in this algebra are maximal and the present correspondence re-
reduces to the earlier one for [P :$] < °°.
It remains to sort out the prime ideals ^3 in E <g> P for which the
corresponding composites (F, s, t) are free. Let B and B' be
transcendency bases for E and P respectively. We know that

STRUCTURE THEORY OF FIELDS 207
the set M of monomials in the /3 e B are ^-independent. A similar
statement holds for the set M' of monomials in the /3' e B'. More-
Moreover, if M = {nti} and M' = {»,}, then the set of products
{w,»y} is ^-independent. This implies that the sets B and B' are
not overlapping and B U B' is an algebraically independent set.
The same statement can be made about the images Bs = {/3 + $}
and Bn = {/3' + $} if and only if no non-zero element of the
subalgebra $[B U B'] is mapped into 0 in the canonical homo-
morphism of E ® P into (E <g> P)/$. This is equivalent to the
condition that $[B U B'] D $ = 0. Hence we obtain our first
condition: The composite (r, s, t) determined by the prime ideal
% in E ® P is free if and only if <f>[B U B'] D $ = 0. It is con-
convenient to change this slightly by replacing $[B U B'] by the
subalgebra $(B)$(B') generated by the subfields *E) and $(B')
of E and P respectively. It is easily seen that the elements of
this subalgebra of E ® P have the form Pq~xr~x where P e
$[B U B'], getyB], re$[B']. It is clear that $[B U B'] is an
integral domain and this and the form of the elements of$(B)$(B')
imply that $(B)$(B') is an integral domain. If Pq~lr~x 5^ 0 is
in <p 0 $(B)$(B'), then P ^ 0 and P e $(B)$(B') fi <JJ. Hence
<P 0 <f?(B)$(B') ^ 0 implies $ 0 §[5 U 5'] ^ 0. Since the con-
converse is clear the foregoing condition gives the following
Lemma 2. The composite field (T, s, t) defined by a prime ideal
$ in E <g) P is free if and only if ^ H $(B)$(B') = 0 where B and
B' are transcendency bases for E/€> and P/<£ respectively.
We recall that if o is a commutative ring and ® is a subring, then
an element a e o is called integral over ® if there exists a polynomial
g(x) e ©[*] such that g(x) has leading coefficient 1 and g(a) = 0
(Vol. I, p. 181). We have proved in Vol. I, p. 182, that, if ® is
Noetherian, then the set of ©-integral elements of o form a sub-
ring containing ®. We shall see later (§ 5.13) that this result is
valid also for any commutative integral domain o. However, the
Noetherian case is adequate to prove the following result which
we require.
Lemma 3. Let B and B' be transcendency bases for E/4> and
P/<i> respectively. Then every element of E <g>* P is integral over

208 STRUCTURE THEORY OF FIELDS
Proof. Since E and P are algebraic over $(B) and $(B') respec-
respectively, it is clear that the elements of E and of P are integral over
$(B)$(B'). Since E ® P is generated by E and P, the result will
follow if we can show that the set of $(B)$(B')-'mtegral elements
is a subring. Hence we have to show that, if a, /3 are $(B)$(B')-
integral, then so are a — /3 and a/3. Since any pair a, /3 are both
integral over a subalgebra $(F)$(F') where F and F' are finite
subsets of B and B', it suffices to prove this for B and B' finite.
In this case we can apply Hilbert's basis theorem for polynomial
rings (Vol. I, p. 172) to conclude that$(.5)[Z?'] is Noetherian. We
shall show next that $(B)$(B') is Noetherian. Thus let 3 be
an ideal in *E)$E'). Then 3' = 3H *(B)[B/] is an ideal in
$(B)[B']y so it has a finite set of generators Pu P2> ■■■)Pm.
Any element of <t>(B)$(B') has the form Pq~x where P e$(B)[B']
and qe$[B']. If this element is in 3, then P = {Pq~l)q e 3' so
P = XAfPi where 4ie*(B)[B']. Hence Pq'1 = ^{Aiq-l)Pi.
This shows that Pi, P2, • • •, Pm is a set of generators for 3.
Hence $(B)$(B') is Noetherian. It follows that a — /S and afi
are <f>(.5)<£>(.5')-integral and this completes the proof.
We can now prove the following
Theorem 25. The composite (T, s, t) of E and P over $ de-
determined by the prime ideal % in E <8># P is free if and only if all the
elements of % are zero divisors in E <g>* P.
Proof. In view of Lemma 2 one has to show that
0 = 0
for B and B' transcendency bases for E/O and P/$ if and only
if every element of $ is a zero divisor. Suppose first that $ con-
contains only zero divisors and let P e ty D $(B)$(B'). Then P is
an element of $(B)$(B') which is a zero divisor in E <g> P. We
shall show that P is a zero divisor in $(B)$(B'). For this pur-
purpose we choose d basis \ua} for E over $B?) and a basis \vp} for
P over $E'). Then it is easily seen that every element of E <g> P
can be written as a sum 2Qapuavp, Qap e$E)$E') and 2QapuaVp
= 0 only if every (?a/s = 0. (We leave this as an exercise.) Since
P is a zero divisor in E ® P we have an element 2Qapuavp ^ 0
such that PBQaf,uaVf,) = 0. Then ^PQafiuave = 0 and since

STRUCTURE THEORY OF FIELDS 209
PQa0 e *E)*E') we have PQa& = 0 and some Qa? * 0. Thus P
is a zero divisor in $(B)$(B'). Since $(B)$(B') is an integral
domain, this implies P = 0. Hence <$ 0 $(B)$(B') = 0. Con-
Conversely, suppose ty D $(B)$(B') = 0. Let P be any element of
$. Then Lemma 3 implies that there exists a relation of the form
pn + flp»-i _|_ C2p«-2 ^ 1_ Cn = o where the d e$(B)$(B').
We may assume « minimal. This relation shows that cn = — P"
- dP71-1 rn_!P e $ D 3>(B)$>{B'). Hence <:„ = 0. Then
we have P(Pn~l + ClPn~2 -\ h cn_x) = 0 and since « was
minimal P" + ^P" H \- cn^ ^ 0. Hence P is a zero
divisor and we have shown that any P e ty is a zero divisor. This
completes the proof.
The set of nilpotent elements of a commutative ring o forms an
ideal called the (nil)radical 3} of o (Vol. I, p. 173). If *$ is a prime
ideal in o and z e SR, then zm e $ for some integer wj. This implies
that ze^J. Hence 9? is contained in every prime ideal ty of o.*
We have shown in the last section that, if E is any field over $
and $ is separably algebraically closed in P, then the zero divisors
of E <g>* P are nilpotent. This and the result just noted implies
that the radical $R of E ® P is the only prime ideal in E ®# P all
of whose elements are zero divisors. Hence we can conclude from
Th. IS and the fact that every composite of E and P over $ is
equivalent to one determined by a prime ideal in E <8> P the follow-
following
Theorem 26. If E is an arbitrary extension field of $ and <£ is
separably algebraically closed in P, then in the sense of equivalence
there is only one free composite of E/<£> and P/#.
* We shall see in Chapter V that 9J is the intersection of all the prime ideals of o.

Chapter V
VALUATION THEORY
The notion of a valuation of a field arises when one attempts to
assign magnitudes to the elements of a field. The classical case is
that of the absolute value | a | in the field of real numbers or in the
field of rational numbers. Of basic importance for the study of
arithmetic properties of the rational and more generally of number
fields (finite algebraic extensions of the rationals) are the p-adic
valuations of the field of rational numbers. For a given prime p
the valuation <pp(a) of the rational number a indicates the power
of p which divides the rational number a. Valuations play a
fundamental role also in the study of algebraic function fields. For
these it is necessary to generalize the notion somewhat so that it
becomes equivalent to the notion of a place, which was first intro-
introduced by Dedekind and Weber in giving a purely algebraic defini-
definition of Riemann surfaces for algebraic functions. Valuation
theory forms a solid link between algebra and analysis. On the
one hand, it permits a precise study of algebraic functions and, on
the other hand, it leads to the introduction of analytic notions
(convergence, integration) in the study of arithmetic questions.
We shall begin our discussion with real valued valuations. One
can distinguish two types of these: archimedean and non-archime-
dean. The latter lead to the extension in which the values are
taken from an ordered commutative group rather than the field of
real numbers. We shall determine the valuations of the simplest
types of fields and consider in some detail the problem of extension
of valuations. Applications to the Hilbert Nullstellensatz and to
the study of the integral closure of a commutative integral domain
will be given.
210

VALUATION THEORY 211
1. Real valuations. We shall consider first valuations which are
real valued and we shall call these real valuations. It is possible
to give a development of the theory which gives at the same time a
development of the real number system from the point of view of
convergence. This adds a small complication, so we shall avoid it
and assume familiarity on the part of the reader with the basic no-
notions on real numbers which will be needed.
Definition 1. A real valuation <p of afield $ is a mapping a —>
<p(a) of <£ into the field of real numbers such that
(i) <p(a) > 0, <p(a) = 0 if and only if a = 0
(ii) v(a0) = V{a)<p(fi)
(iii) <p(a + ]8) < <p{a)
Examples.
A) $ the field of complex numbers, <p(a) the usual absolute value
of the complex number a = a + by/ — \, a, b real. This gives a valuation on
any subfield, in particular, on the field of real numbers and on the rational field.
B) $ the field of rational numbers. Let^> be a prime integer. If a y* 0 in $,
we write a = a'pk where k = 0, ±1, ±2, ■ • • and a! is a rational number prime
to p (notation: (a',p) = 1) in the sense that its numerator and denominator in
some representation are prime to p. The integer k is uniquely determined by a
and we write vP{a) = k, <pp(a) = p-'r<<*\ Also, we set vp@) = °o, <pp@) = 0.
Then (i) is evident and (ii) and (iii) are valid. This is obvious if either « = 0 or
/3 = 0. Suppose a 9* 0, j8 ^ 0, and let a = a'p*, 0 = /8>! where (a', p) = 1 =
(fi',p). Then aP = a'P'pk+l and (a'0',p) = 1. Hence vp{aff) = k + I =
vp(a) + vp@), so <pp(ap) = <pp{a)<pp(£). If k < I, then a + p = £*(<*' + |Sy-*)
and vp(a + ft) > min (^(a), vp(ffj). Hence ^.(a + 0) < max {<pp{a), <f>p(ff>)
which is a stronger relation than (iii). Hence <pp(a) is a valuation. This is called
the p-adic valuation of the rational field.
C) P = $(*■) the extension field of $ by a transcendental element x. Let ir(x)
be an irreducible polynomial in <£>[#]. If a is a non-zero rational expression, we
write a = ir(x)ka' where k is an integer and a! is a rational expression which is
prime to 7r {{a', x) = 1) in the sense that it has a representation with numerator
and denominator prime to x. We set vr(a) = k and <pr = ck where c is a real
number, 0 < c < 1. Also we set vx @) = <x>,<pT@) = 0. One checks as in exam-
example 2 that <pT is a valuation. A classical case of this type of valuation is that
in which $ is the field of complex numbers and $(#) is identified with the field
of rational functions on $. Here ir(*) has the form x — r and c» (a(x)) describes
the behavior of the rational function a(x) in the neighborhood of the point x = r.
One sees that, if vT(a) = k > 0, then a has a zero of order ka.tr and, if »v(a)
= — k, k > 0, then «(*■) has a pole of order k at # = r. If v»(a) = 0, then a has
neither a zero nor pole at x = r. It is of interest also to consider the behavior
oiaix) at infinity. This can be done by introducing another valuation in $(#).
If a(x) * 0, we write a{x) =@o+Pi*-\ h PmXm)(yo + 7i* -\ h 7n*n)-1

212 VALUATION THEORY
where pm * 0, 7n * 0. Then a(x) = (-)"~"G8o (-)" + • • • + AJfro (-)"
(_j V.*/ \x/ \x/
- ) H h 7n)-1 and a(,v) has a zero of order « — m at infinity if
n — m > 0, a pole of order »j — wifw — »>0, and has neither a zero nor pole
at infinity if » = m. We define va{a{x)) = n — m, <pa(a(x)) = fn-m, 0 < c < 1,
PooCO) = °°, *>oo@) = 0. This gives a valuation. This procedure is applicable to
any $(#), x transcendental.
D) Any field $ with <p(a) = 1 if a ^ 0 and p@) = 0. Such a valuation is
called trivial. We remark that the valuations <pT and £>„ of example 3 are all
trivial on <£.
We now list some immediate consequences of the definition of a
real valuation. We note first that (ii) implies that <p(l) = 1,
<p(-l) = 1, and<p(-a) = <p{a). Also ^(a) = ^(a) if a ^ 0,
and <p(£) = 1 if f is a root of unity. This implies that the only
valuation in a finite field is the trivial one. Also we note that
A) \<p(a) -<p(
where | | is the ordinary absolute value. All these assertions are
readily established and we leave their verification to the reader.
Definition 2. The real valuations <pi and <p2 are called equivalent
if <pi(a) > ^i(/3) holds for a, C e $ if and only if <p2(oi) > <p2(fi)-
It is natural from the point of view of convergence which we
shall consider in § 4 to identify valuations that are related as in the
foregoing definition. This relation leads to the following some-
somewhat surprising consequence.
Theorem 1. If <pi is equivalent to <p2, then there exists a positive
real number s such that <p2(a) = <f>i(a)' for all a e <$.
Proof. We may assume that one of the valuations is non-trivial
and, since the conclusion is symmetric in <pi and <p2(<pi = <p2 '),
we may suppose that <pi is non-trivial. Then there exists an a0 in
* such that 0 < <Pi(ao) < 1 = ^i(l). Then also 0 < <p2(a0) < 1,
so <p2 is non-trivial. Moreover, we can write <p2(a0) = <pi(a0)'
where s > 0. In fact, this relation is equivalent to s =
log ^2(«o)/log^i(«o) which is positive since logvi(«o) < 0 and
log P2(«o) < 0- We wish to show that
B) log <p2(a)
log <p2(a0)

VALUATION THEORY 213
if a is any element in * such that 0 < <pi(a) < 1 and so 0 < <p2(a)
< 1. The two ratios in B) are positive. Let m and n be positive
integers such that m/n > log <p1(a)/log <pi(a0). Then m log (pi(a0)
< n log ^(a), log^i(«om) < log<pi(a") and <pi(aom) < <pi(an).
Hence <p2(a0m) < <p2(a") so, if we re-trace the steps, then we see
that m/n > log <p2(a)/log <p2(a0). By symmetry (<p2 is non-trivial),
if m/n > log <p2(a)/log (p2(a0), then m/n > log ^(oO/log <pi(a0).
Since these relations hold for all positive rationals r = m/n, we
have the equality B). Hence
log <p2{a) _ log <p2(ao) _
log<pi(a) log<pi(a0)
and <P2(od — <Pi(od" holds for all a with <?i(a) < 1. By taking a
we see that this holds also if <pi(a) > 1. Moreover, it is clear
that, if <pi(a) = 1 = <pi(l), then <f>2(a) = 1. Hence <p2(a) =
<f>i(a)s for all a.
Definition 3. A real valuation is called archimedean if <p(n) > 1
for some integer «(= n\ = 1 + 1 ••• +1, » times) in the prime
field. Otherwise the valuation is non-archimedean.
If $ has characteristic p 9^ 0, then any n ^ 0 in the prime field
is a root of unity; hence <p{ri) = 1. Consequently, every valuation
of a field of characteristic p is non-archimedean. We note also
that any valuation which satisfies <p(a + 0) < max (<p(a), <p(P)) is
non-archimedean. For, this can be extended by induction to
give <p(a! + a2 +■ ■ ■+<*„) < max (<p(«i)> • ' > *>(«»)) and this
implies that <p(n) < <p(l) = 1. The converse of this result is valid
also since we have
Theorem 2. If <p is a non-archimedean real valuation, then
<p{a + ff) < max (<p(a), ip{E)) for every a, E in <£.
Proof. We have
<p(a + C)n = p(a»
max

214 VALUATION THEORY
Hence we have <p(a + /3) < in + lI/n max (<p{a), v>(/S)). Since
lim (» + lI/n = 1, this implies that
C) <p(a + /3) < max (?(<*), <p(fi)).
EXERCISES
In these exercises "valuation" will mean "real valuation."
1. Show that, if <p is a valuation and s is a real number such that 0 < s < 1,
then a —* <p(a)' is a valuation. Show also that, if <p is non-archimedean, then
a —* <p(a)' is a valuation for any j > 0.
2. Establish the following properties of non-archimedean valuations:
D) *>(« + 0) = <o(a) if p(a) > ?>(/S).
E) If cti + <*2 H \- an = 0, then ^>(a<) = <p(a,) for some / 5^ j.
3. Let ¥> be a valuation in P such that (p is trivial on a subfield $ of P such that
P is algebraic over $. Show that <p is trivial on P.
4. Let p be a non-trivial valuation of $ and let /S be a non-zero element of $
such that^>(/3) < 1. Show thatp(a) < 1 if and only if <p(fia.n) < 1, n = 1, 2, • • •.
Use this to prove that, if ^ is a valuation such that ^G) < 1 implies ^G) < 1,
then also <p(y) > 1 implies ^G) > 1 and <p(y) = 1 implies ^G) = 1. Hence
show that <p and ^ are equivalent.
5. Show that, if <pi,<pt, • • •,<pn are inequivalent non-trivial valuations of a
field $, then there exists an a in $ such that <pi(a) > 1 and <pi(a) < 1 for i =
2, 3, ■■•,». (Hint: The case n = 2 is an easy consequence of ex. 4. Using this
and induction one obtains /8 such that <pi(fi) > 1, <p,ifi) < i,j = 2, • • •, n — 1,
and 7 such that <pi(y) > 1, <pn(y) < 1- If Vn(|8) < 1, one can take a = (8*7 for a
sufficiently large integer k. 1{ <pn(fi) > 1, one can take a = 7/3*(l + (8*) for k
sufficiently large.)
2. Real valuations of the field of rational numbers. We begin
by determining the archimedean valuations of the rationals. The
result is the following
Theorem 3. Any archimedean real valuation of the rationals is
equivalent to the absolute value valuation.
Proof (Artin). Let n and »' be integers > 1 and write «' = *0
+ a-ji -\ 1- «*»*, 0 < at < n, ak y* 0. Then,
*>(»') < <pM + <p(ai)<p(n) -\ h <p(ak)<p(n)h.
Since 0 < (p(ai) < a,- < n, this gives
*>(«') < »A + <p(n) +■■■+ <p(n)k) < n(k + 1) max A, <p(n)k).
We have n' > »* so k < log »'/log n and

VALUATION THEORY 215
F) <p(n') < n (•; h 1 ) max ( 1, <p(n)log " ) •
\ log» / \ /
If we replace »' by (n')r, r a positive integer, we obtain from F):
max
Taking r-th roots we obtain
T /rlosn M1/r / lo8"'
G) |(l jj ^
Since lim (ra + b)llT = 1 if a ^ 0, G) implies
r—*co
(logn'
l,v(«)logn
Since p is archimedean, »' can be chosen so that <p(n') > 1; hence
by (8),
log n'
(9) 1 < <p{n') < ^(wI^".
Hence <p(n) > 1, so we can interchange the roles of n and »' to ob-
obtain
A0) *>(»)logn = *>(
for any two positive integers n, n'. Then log ^>(«)/log n is a posi-
positive real number s independent of n and <p{n) = n'. It follows
that (f>(a) = | a |" for every rational number a. Evidently cp(a) is
equivalent to the absolute value valuation.
Theorem 4. Any non-trivial non-archimedean real valuation of
the rationals is equivalent to a p-adic valuation for some prime p.
Proof. We have <p(n) < 1 for every integer n. If <p(n) = 1 for
every integer, then <p is trivial. Hence there exist non-zero integers
b such that <p(l>) < 1. Let $ be the collection of integers b satisfy-
satisfying this condition. This set is an ideal in the ring of integers /
since <p(bi — b2) < max (v(^i), <p(b2)) < 1 if bt e fy, and <p(nb) =
<p{n)ip(b) < 1 if n e /, b e ip. Also ty is prime since <p(n) = 1 =
<p(n') implies <f>{nn') = 1. Hence fy = (p) where p is a prime.

216 VALUATION THEORY
We can write v(p) = p * where s > 0 since 0 < <p(p) < 1. Let
n be any integer and write n = n'pk where k > 0, in',p) = 1.
Then n' ffy so <p{n') = 1; hence <p(n) = p~h\ It follows that <p
is the J-th power of the ^>-adic valuation determined by p.
3. Real valuations of *(x) which are trivial on #. We suppose x
is transcendental in P = $(x) and we shall determine the real
valuations <p which are trivial on <£>. Since the prime field is con-
contained in $, <p(n) = 1 for every integer =^ 0 in the prime field.
Hence <p is non-archimedean. We distinguish two cases:
I. <p(x) < 1. In this case <p(f(x)) < 1 for every f(x) = a0 +
ayX + • • • + dnxn e <£>[#]. This is clear from the non-archimedean
property of <p. From now on we assume <p is non-trivial and this
implies that there exists a polynomial/(#) such that <p(f) < 1.
Let ty be the subset of <£>[#] of polynomials/ such that <p(f) < 1.
As in the proof of Theorem 4 one sees that ty is a prime ideal, ty =
(>(*)), in $[*]. We have <p(tt(x)) = c, 0 < c < 1. If/(*) =
ir(x)kg(x) where {t(x), g(x)) = 1, then <p(f) = ch. Hence <p is the
valuation <pT discussed in example 3 of § 1.
II. <p(x) > 1. Let/(#) = a0 + 0LXx + ■ • • + amxm where am ^
0. Then <p(amxm) = <p(x)m > ^(a^) for / < m. Hence <f>(f) =
<p{x)m (cf. ex. 2 in § 1). If we set <p{x) = c~\ 0 < c < 1, then
^(/) = c~m. It is easy to check that <p is a valuation (p^ as de-
defined in example 3 of § 1.
4. Completion of a field. One of the most important aspects of
a real valuation is that it leads to the introduction of metric space
notions for a field. The most convenient form for these is based on
sequences and convergence. The basic definitions are patterned
after those of ordinary analysis.
Definition 4. Let $ be afield with a real valuation <p. A sequence
{a/e}, k = 1, 2, • • • is said to converge in f> {relative to <p) if there
exists an a in $ such that for any real e > 0 there exists an integer
N = N(e) such that
A1) <p(ct -«„)<€
for all n > N. Then a is unique and is called the limit of {ak}. If
a = 0, {ak} is a null sequence. A sequence {ak} is called a Cauchy
sequence if for any e > 0 there exists an integer N = N(e) such
that

VALUATION THEORY 217
A2) <p(am — an) < e
for all m,n > N(e).
00
Convergence of series X ak is denned as usual by the conver-
gence of the sequence {sk} of partial sums sk = ^2 a,. For ex-
ample, in the rational field with the />-adic valuation, the series
00
Y!jPk~l converges to 1/A — p) since
fpCfZ Sn) = Vp^/v^1 ~ P) = P~n < *
if n is sufficiently large.
It is easy to see, as in the real case, that any convergent sequence
is a Cauchy sequence, but the converse need not hold. This leads
to the following
Definition 5. Afield <£> is said to be complete with respect to a real
valuation <p if every Cauchy sequence of elements of <£> is convergent
in $.
We shall now carry out for any field $ with a real valuation <p a
construction of a completion $ of $. This is a field $ with the
following properties:
1. 5 is an extension field of $ and has a real valuation <p which
is an extension of the valuation <p of $.
2. $ is ^-complete.
3. The subfield $ is dense in $ in the sense that every element
of $ is a limit of a convergent sequence of elements of <!>.
We consider first the set C of Cauchy sequences {ak} of ele-
elements a*, e <£. We shall show that C is a ring relative to the com-
compositions {ak} + {pk} = {ak + /?*}, {ak}{pk\ = {<xkpk}. For this
and a later application we require the following
Lemma 1. If \ak), \fa) eC, then \ak + fo] and |«A) e C.
If {ak} e C and is not a null sequence, then there exists i\ > 0 and an
integer N such that tp(an) > vfor all n > N.
Proof. Given e > 0, determine A^i so that <p(am — «„) < e/2
if m, n > Nu and A^2 such that <p(PP — /34) < e/2 if p, q > N2.

218 VALUATION THEORY
LetiV = max (Nu N2). Then<p(am + 0m - an - 0n) < <p(am -
«„) + v(Pm - 0«) < e/2 + e/2 = e if m, n > N. Hence {ak +
I3k} e C. We note next that there exist positive real numbers s, t
such that <p{ak) < j for all k and <p@k) < t for all k; for, we have
(f>(am — aN) < 1 if m > N and N is sufficiently large. Hence
<p(am) — (p(aN) < <p(am — aN) < 1 so (p(am) < <p(aN) + 1 for all
m >N. Then if s = max (p(a,-) + 1), / = 1, 2, • • •, iV, then
via*) < s for all k. Similarly we can find / > 0 such that <p((}k) <
t for all k. Then
A3) <f>(ampm - «„/?„)
= <p((XmPm — am&n + Otml3n — <XnPn)
< <p(am)<p(pm - 0n) + <p(l3n)<p(am - an)
< s<p@m - ft,) + t<p{am - an).
If we take iVi so that <p@m — 0„) < e/2s for m, n > Ni, and N2
so that <p(am — an) < e/2t for m, n > N2, then A3) shows that
<p(<xmPm — an(ln) < e if m, n > N = max (N1} N2). Hence
{akl3k} e C. Now suppose {ak} e C and this is not a null sequence.
Then there exists e > 0 such that <p(ak) > « for an infinite number
of k. Also there is an N such that (p(am — an) < e/2 for all m,
n > N. There is a p > N such that (p(ap) > e. Then if n > p,
<p(an) = <p(ap — (ap — «„)) > <p(ap) — <p(ap — an) > e/2 = rj.
This completes the proof.
To see that C is a ring under the indicated compositions we re-
recall that the set of unrestricted sequences of elements of $ is a
ring under component addition and multiplication. This is just
the complete direct sum of a countable number of copies of <$.
The 0 element of the ring is {0} and the identity is {1} where here
we write {a} for the sequence {<xk} with «& = a for k = 1, 2, • • •.
We call this the constant sequence {a}. It is clear that the set of
constant sequences is a subring of the ring of sequences and this
subring is isomorphic to $ under the mapping a —* {a}. Lemma
1 implies that the set C of Cauchy sequences is a subring of the
ring of sequences and evidently C contains the ring of constant
sequences. Thus we see that C is a commutative ring with an ele-
element 1 = {1} and C contains the subring of constant sequences
which is isomorphic to $.

VALUATION THEORY 219
We consider next the subset Z of C consisting of the null se-
sequences. We have the following
Lemma 2. Z is a maximal ideal in C.
Proof. It is easy to see that the difference {«&} — {/3*} =
{a* — /3fc} of two null sequences is a null sequence. Now let {a*}
be a null sequence and let {74} be a Cauchy sequence. The proof
of Lemma 1 shows that there exists a positive real s such that
<p(yk) < s for all k. If e > 0 we choose N so that <?(«„) < e/s for
all n > N. Then (p(anyn) < e for all n > N, so {a*?*} is a null
sequence. Hence Z is an ideal in C. To show that Z is maximal
we have to show two things: Z 7* C and, if B is any ideal in C con-
containing Z and an element {aj} ^ Z, then B = C. The first of these
is clear since no constant sequence 7^ {0} is contained in Z. Next
let B be an ideal in C containing Z and containing the element
{«&} ^ Z. Lemma 1 shows that there exists a positive t\ and an
integer p such that <p(an) > V for all n > p. Let /?& = 1 if k < p
and /8fc = a* if k > p. Then {«&} — {0*} e Z. Consider the
sequence {fa-1}. We have C^1 ^1)
0n) < -5 ^(am — «n) if m, n > p. This implies that {(S*} eC.
Since {«*} - \fa\ e Z C 5 and {«*} e 5, {0*} e B. Hence 1 =
{|8*-1}{^}e5andsoJS = C.
Lemma 2 implies that the difference ring $ = C/Z is a field.
We proceed to show that $ is a field with a valuation which has the
properties of a completion of <£.
Theorem 5. Let # be afield with a real valuation <p and let $ =
C/Z the difference ring of the ring of Cauchy sequences with respect to
the ideal Z of null sequences. Then $ is afield which contains a sub-
field isomorphic to <i> such that, if # is identified with this subfield,
then $ is a completion of <£.
Proof. We have seen that the mapping a —> [a] is an iso-
isomorphism of* with a subring of C. The canonical homomor-
phism {a} —* {a} + Z is an isomorphism since the only con-
constant sequence contained in Z is {0}. Hence we have the iso-
isomorphism a —* {a} + Z of $ into $ = C/Z. From now on we

220 VALUATION THEORY
shall identify a with \a} + Z, $ with its image in $. We show
next that $ has a valuation <p which is an extension of the valua-
valuation <p in <£. Now let {ak} e C. Then {(p(ak)} is a Cauchy se-
sequence of real numbers since | <p(am) — <p(an) \ < (p(am — an) and
{ak} is a Cauchy sequence. Hence, by the completeness of the
field of real numbers with respect to the absolute value valuation,
lim <p(ak) exists. Next let {ak'} be another Cauchy sequence such
that {ak} + Z = {ak'\ + Z. This means that <p(an — an') < e
for a given e > 0 provided n > N(e). Then \<p(an) — <p(an')\ <
<p{an — an') < e if n > N(e). Hence lim <p(ak) = lim <p(ak), so
this real number is independent of the choice of the element {ak}
in the coset A = {ak} + Z. We now set ip{A) = lim <p(ak) and
we proceed to show that ^ is a valuation in $. First, it is clear
that <p{A) > 0. If A = {ak\ + Z and <p{A) = 0, then lim <p(ak) =
0; hence {ak} is a null sequence, so {ak} eZ and A = 0. If B =
{^} + Z, then AB = j«A| + Z and <p(AB) = lim <p(akpk) =
lim p(afc)*G3*) = <p{A)?{B). Also ^ + 5 = {ak + $k) + Z and
<p(A + B) = lim ^(afc + /3*) < lim {<p{ak) + <f>(pk)) = <p(A) +
<p{B). Hence <p is a valuation. If yf = a e $, so A = {a} + Z,
then ^(^/) — lim ^>(a) = <p{a); hence ^ is an extension of the
valuation <p on $. We shall show next that $ is dense in $. Let
yf = {ak} + Z be an element of 3>. Let ak be the constant se-
sequence all of whose terms are ak. Then we have identified Ak =
ak + Z with ak. We assert that lim Ak = A. For, if e > 0 is
given, we can find iV" such that (p(am — an) < e'\im, n > N. Then
lim <p(am — an) exists and this is < e. On the other hand, <p(A —
n—k»
yfm) = lim ^(aOT — an), so ^(yf — Am) < e if m > N. Hence
71—»W
lim Ak = A and $ is dense in $. It remains to show that $ is
complete. Let \Ak\ be a Cauchy sequence of elements of $. For
each k we can choose ate^cl such that ^{A^ — ak) < —.-•
Then <p(am - an) = <p(am — Am + Am - An + An — an) <
<p{am - An) + ip(An - An) + <p(An - an) < ?(Am - An) +
— H—-• Since \Ak\ is a Cauchy sequence, this shows that {ak}
is a Cauchy sequence of elements of $. If we now go back to the

VALUATION THEORY 221
original $ and take the element A = \ak\ -\- Z, akin the original
$>, then we see easily that lim Ak = A.
From now on we shall write <p for the valuation <p in $.
We now take up the question of uniqueness of the field 3>.
More generally, let$*, i = 1, 2, be a complete field with a valua-
valuation <pi and let <$, be a dense subfield. Suppose s is an isomorphism
of $i onto f>2 which is isometric in the sense that <p2(«s) = <f>i{oi),
a e $i. Let yf e?i and let {aft} be a sequence of elements of $x
such that lim <xk = A. Then {a*;"} is a Cauchy sequence in <£2j so
it has a limit B. If {a*/} is a second sequence such that lim ak =
A, then lim (ak — <xk) = 0, lim cpi(ak — ak) = 0; hence lim
^2(«fcs ~ «fc'3) = 0 and lim {ak — ak's) = 0. This implies that
lim ak's = B. Hence the mapping s\A -> 5 of $i into $2 is
single-valued. It is easy to check that this is a homomorphism.
Clearly s = s on $j. Similarly, we can extend J to a homo-
homomorphism s~1 of $2 into$i which is defined in the same way as s.
Then one sees that A'33'1 = // for all A e$! and 5s'" = 5 for
all 5 e$2- This implies that J is surjective and an isomorphism.
We remark finally that, if Jx and s2 are isometric isomorphisms of
$i onto $2 which coincide on $1} then Ji = s2. The proof is clear.
We have therefore established the following
Theorem 6. Let $,-, i = 1, 2, fo a complete field with a valuation
(pi and $,■ a dense subfield of $j. Let s be an isometric isomorphism of
$i onto $2- Then s has a unique extension to an isometric isomor-
isomorphism o/$i onto $2-
This result implies, in particular, that, if $i and $2 are com-
completions of the same field <$, then there exists an isometric isomor-
isomorphism of #!/$ onto $2/^- We just have to apply the theorem to
the identity mapping in <i>. In this sense the completion is unique
and we have the right to use the term: the completion of the field
$ relative to the real valuation <p.
EXERCISES
1. Let $ be a field with a real valuation <p. Show that the sum, product, and
difference are continuous functions on $ in the usual sense. Show also that the
mapping a —> a" is continuous on $*, the set of non-zero elements of f>.
2. Let 5 be the completion_of $. Show that the identity mapping is the only
continuous automorphism of $ over $.

222 VALUATION THEORY
5. Some properties of the field of />-adic numbers. We look
first at some properties of any field relative to a non-trivial non-
archimedean real valuation <p. If $ is such a field, then the subset
o of elements a e $ such that <p(a) < 1 is a subring of <£; for, if a,
f3 e o, then <p(af$) = <p(d)<p{($) < 1 and <p(a — 0) < max (<p(a),
<p(P)) < 1- The ring o is called the valuation ring of<p. The subset
p of o of elements /3 such that <p({3) < 1 is an ideal in o, since
<p{^) < 1, <p(p2) < 1, *>(«) < 1 imply pGSj - /32) < 1 and
<p(af3i) < 1. The elements a in o which are not in p satisfy <p(ci) =
1; hence <p(a~i) = 1 and a eo. Conversely, if a is a unit in o,
then <p(a) < 1, ^(a-1) < 1 and #>(«)¥>(« "^ = 1 imply that <p(a)
= lysoa$p. Thus we see that p is the set of non-units of o. This
implies that any ideal of o properly containing p contains a unit
and so coincides with o. Hence p is a maximal ideal in o. If q is
any ideal properly contained in o, then q contains no units of o;
consequently qCp, Hence p is the only maximal ideal of o. The
difference ring o/p is a field which is called the residue field of $
relative to <p.
The set T = {<p(oi), a ^ 0 in <£>} is clearly a subgroup of the
multiplicative group of positive real numbers. T is called the
value group of <p. The valuation is called discrete if T is a cyclic
group. It is easy to see that a subgroup T ^ 1 of the positive reals
is cyclic if and only if the subset I" of elements < 1 has a maximal
element. This element is a generator of T. Let y> be discrete and
let ir be an element of $ such that <p(-n) is the largest element < 1 in
F. Then tt e p the maximal ideal of the valuation ring o and, if /3
is any element of p, then <p(fi) < ^(ir), ^OStt) < 1, so jStt =
a eo and /3 = ocir. Then p is the principal ideal (tt). Conversely,
if p is a principal ideal: p = (ir), then any /3 e p has the form air,
a e o, so <p{&) = (p(a)<p(ir) < <p(ir). Hence <p(ir) is the largest ele-
element < 1 in T and <p is discrete. Since <p(w) is a generator of r,
we have for any non-zero a in <£, <p(a) = <p(ir)k for some integer k.
Then if « = air~k, (f>(e) = <f>(a)<p(ir) ~~h = 1, so e is a unit in o.
Consequently, any non-zero element of * has the form eirk, k —
0, ±1, ±2, • ■ • where e is a unit in o.
Let <£> be any field with a non-archimedean real valuation <p and
let $ be the completion of $ relative to <p. We shall now show that
the value group of # and $ are the same and in a certain sense the

VALUATION THEORY 223
same statement can be made for the residue fields. Let a. e $.
Then the density of * in $ implies that there exists an a in * such
that tp(a — a) < <p(a). Since the valuation is non-archimedean
we have <p(a) = <p(a + (a — a)) = max (,<p(a), <p(a — a)) =
<p{a). Thus we have an a e 3> such that <p(a) = <p(a) and clearly
this means that $ and $ have the same value group F. Next let 5
be the valuation ring of $, p its maximal ideal of non-units. If o
and p are the corresponding subsets of <1>, then o = o C\ <£, p =
p fl f. If a e 5 we choose a e $ so that <p(a — a) < 1. Then a —
aep and so a e o. Hence a = a (mod p") which shows that o +
5 = 5. We have the standard isomorphism:
o/p" = (o + P)/P S o/(o fl p") = o/p.
By means of this isomorphism we can identify the residue field of $
with that of <i>.
The theory of convergence of series in a complete field with a
non-archimedean valuation is strikingly simple. The complete-
00
ness implies that ^ «& converges if and only if for any e > 0 there
i
exists an integer N such that <p(am+i + • • • + <xm+k) < e if m > N
and k = 1,2, ■•-. Since the valuation is non-archimedean,
(p(am+i + ■ • • + am+k) < max <p(am+i). Hence the condition is
equivalent to (p(am+i) < e for m > N, i = 1,2, • • •. This is
equivalent to lim an = 0. This shows that a series converges if
00 m
and only if its «-th term converges to 0. Since ^ ak = 22 ««
1 i
«; we have <p( £ «* ) < max(^( X) «*) ' ^( ^ a0)and'
M V 1 / V \ 1 / Vm+1 //
<p ( ^Z aj) can be made arbitrarily small by taking m large
1 /" \ /m \
enough, we have <pI ^ «& ) = <pl ^2 ai)[{m is large enough. If,
11 (m \
in addition, we have <p(ai) > <p(a2) > <p{a3) > • ■ ■, then <p ( 23 «,- 1
= tp{ai). Hence <p ( 2Z «* ) = v(«i) in this case.
We now consider the special case of the field R(p) which is the
completion of the rational field Ro with respect to thep-adic valua-
m+l
since

224 VALUATION THEORY
tion <pp(a) = p~k for a = pka', (a', p) = 1. The field R<-p) is
called the field of p-adic numbers. Evidently the value group of
Ro relative to tpv is the cyclic group generated by p ~x; hence the
same result holds for R(p) and the valuations of Ro and Rip) are
discrete. Let 5 be the valuation ring of R(p). The elements of 5
are the p-adic numbers a such that <pp(a) < 1 and these are called
p-adic integers. It is clear from the definition of <pp that the ra-
rational numbers a which are p-adic integers are those which can be
written in the form m/n where (n, p) = 1. If o is the valuation
ring of Ro, p its maximal ideal, then o = 5 D Ro, p = p fl Ro. We
have seen that 5 = o -f- p. Hence, if d is any p-adic integer, then
there exists a rational number m/n with (n, p) = 1 such that d —
m/n e p". There exist integers a, b such that na + pb = 1. Then
m/n = ma + pibm/n) so m/n = ma (mod p). Hence a = ma
(mod p"), which shows that we have 5 = I + p", where / is the
ring of integers. It is clear that p A / = (p). Hence the residue
field o/p ^ //(p) is just the field of p elements.
Let a be a p-adic integer. Then our argument shows that there
exists an element a of I (that is, an ordinary integer) such that a —
a e p\ If a = b (mod p), then « — b e p and so d — £ ep. This
shows that for every p-adic integer a we can choose a0 in {0, 1, 2,
• • •, p — 1} such that a — a0 ep. We assert that di = — (d — «0)
is a p-adic integer. We note first that p is an element of p such
that <p(p) is maximal. Since the value groups of Ro and 25(p) are
identical, it follows that p is an element of p with maximal <p(p).
Consequently, the ideal p is principal with p as generator, so if J3
satisfies ^>(JS) < 1, then /3 = yp where y is a p-adic integer. In
particular, d — a0 = pax where dx e 5. Hence ax = - (d — «0)
e o. We can repeat the argument with &i. Thus we can find ax =
0, 1, 2, • ■ •, p — 1 such that dx — at e p" and d2 = - (di — «x) e 5.
P
Then d = a0 + pa.x = a0 + axp + oc2p2. Continuing this process
we obtain
d = a0 + axp + a2p2 -\ h akpk + ak+1p
k+1

VALUATION THEORY 225
where 0 < at < p — 1 and ak+i e o. Then ak+1pk+1 —*■ 0 and
so we have
A4) a = ao + aip + a2p2 H , 0 < a{ < p - 1.
Conversely, consider any series of this form. We may suppose
this is ampm + am+ipm+l +■ ■ ■ where m > 0, am ^ 0. Then
this series converges. If d is its limit, then yp(d) = <pp(pm) =
p~m. Hence d e 5. We see also that d is a unit in 5 if and only if
m = 0. Hence the units of 5 are the elements A4) with a0 9^ 0.
We have seen that the ideal jj is the principal ideal generated by p.
It follows that every element of i?(p) has the form pkl where e is a
unit and k = 1, ±1, ±2, • ■ •. Hence every element has the
form pk(a0 + axp + • • ■) where 0 < a{ < p — 1.
Let U be the multiplicative group of units in o. We wish to
analyze the structure of U, and first we shall show that U con-
contains a subgroup isomorphic to the multiplicative group of non-
nonzero elements of I/(p), that is, a cyclic group of order p — 1. Let
a be one of the numbers 1, 2, • • -,p — 1. We know that ap = a
+ xp where x e /. It follows by induction that apk = apk l
av" _ ap*-i
(mod pk) so e / C o. It follows that
P
(is) r._.
is a well-defined element of R(p). This is a limit of the sequence
{Uk)} where
Now in any field with a valuation one can prove as for the
reals that lim ak = d, lim bk = J8 imply lim (ak ± bk) = d ± J8
and lim a^*; = d^ (cf. ex. 1, § 4). Hence lim fow = fo implies
lim (fa(J;))p = T/. Since fa(A;) = «p* we have lim of = fo and
lim («p*)p = lim apk+1 = fap. But evidently lim apk+1 = fo. Hence
we have fop = fo- Also fo = a (mod p) is clear from A5), and
since a & 0 (mod jj), ta ^ 0. Hence fop-1 = 1. The same argu-
argument shows that, if a =^ b in the set {1, 2, • ■ -,p — 1}, then
To 5^ ft. Hence we have constructed p — 1 distinct (p — l)-st

226 VALUATION THEORY
roots of 1. This is all we can have in a field. We know also that
{fa} is a cyclic group (Lemma 1, § 1.13).
Let i e Uy so that « = a + &\p + a2p2 + ■ • • where 0 < a < p
and 0 < «,• < p. Then f o = g (mod £) and «i = fa* = 1
(mod {)). We shall call ap-adic integer a \-unit (Einseinheit) if it
is congruent to 1 modulo p. We have shown that every unit in
5 has the form « = «ifo where «! is a 1-unit. Let U\ be the set of 1-
units. Then Ui is a subgroup of £7. To see this, let ifi, 1J e C/i so
9< = 1 + Pi, pt e p. Then we have 9^s = 1 + 0i + fo + frfo =
1 (mod jj), since J3i + /32 + /S^ e ?• Also, it is easy to see that
1 - Pi + Pi2 = A + iSi)-1. Clearly, 1 - 3i + Pi2
= 1 (mod p). Hence Th = A + ^i) e Ut.
In order to study the subgroup Ui of U more closely we find it
convenient to introduce the exponential function in the field of p-
adic numbers. We define this by means of the series:
A6) exp*=l+^ + ^+---
which we shall show converges for all x e }5 if p 7^ 2. As in § 1, we
write <f>p(x) = p~"*^ and we know that vp(x) is an integer. The
condition j>p(x) = / > 0 is equivalent to: x e $l. For a rational
number, vp(x) is the power of p which divides x in the sense that x
= p'pWy where (y, p) = 1. In order to prove convergence of A6)
we evidently need a formula for i>p(k\). For this we note that —
of the numbers 1, 2, • •, k are divisible by p where, as usual [z]
denotes the integral part of the real number z. (These are p, 2p,
3p> " 'i\ — \P') Similarly, — of the numbers 1, 2, • • •, k are
divisible by p2, — are divisible by p3, etc. This implies that
Hence vp(k\) < k( 1 - -\ ) = k( )• We now
\p p* / \p -

VALUATION THEORY 227
assume p ^ 2. Then if vp(x) = / > 1 and k > 0, vp(xk/k\) >
k[l ) > 0. We can include x = 0 in this by taking
\ p — 1/
vp@) = oo. The inequalities show that xk/k\efp if & > 0 and
lim xk/k\ = 0. Hence A6) is convergent and exp x is defined for
all x e jj. Moreover, since **/&! e p for & > 0, exp x is an element
of 5 and exp x = 1 (mod &). Hence exp x e Ui. If x ^ 0 and
",(*) = / > 1, then vp(x*/2!) = 2/ > / and, if * > 2, then
f,(**/*0 > * (' ) > /• It follows that, if x e p!, / > 1,
\ p 1/
p-\
then
A7) exp*- 1 - xepl+1.
We shall now show that, if x,y e ji, then
A8) exp (x+y) = (exp *)(exp^).
n xk n vk n (x _i_ v\*
t-,. v-V— V- V— 7 — T^v J;
Let Xn~t^' y""r^' n"r *i
It T. j. I
K Y VK l
Since — (x + y)K = 2^ ^7
=o /! (* -
J>nor
The inequalities noted before imply lim (Z2n — XnYn) = 0.
Since lim Xn = exp *, lim Yn = expjy, lim Z2n = exp (x + y),
this gives A8). This equation and the fact that exp#el7i
establish a homomorphism of the additive group E, +) into Ui.
We shall show that the mapping x —> exp * is in fact an iso-
isomorphism of E, +) onto C7"i. To see that the mapping is an iso-
isomorphism, it suffices to show that x 9* 0 implies exp x 9* 1. Thus
suppose vp(x) = / ^ 00, so x e ji1, ^ j)I+1. Then it is clear from A7)
that exp* p£ 1.
Next consider any element of Uu This has the form 1 + y,
y ef>. Set Xi = y and consider A + y) exp (—#1). By A7),

228 VALUATION THEORY
exp (—xi) = 1 — Xi + zx where zx e p2; hence
A + y) exp (-*!) = A + #x)(l - *i + Zl)
= 1 + (Zj - A?!2 + #lZl)
= 1 +x2
where #2 = Zj — #i2 + #iZj e p2. Suppose we have already deter-
determined elements xu x2, • • •, xk such that *,- e p* and
A + y) exp (-*x - x2 xk) = 1 +
where #*+i e jj*+1. Then
= A + y) exp (-#! — #2 ,v;b) exp (~xk+1)
= A + Xk+i) exp (—
where zfc+i e p4 and A + #*+i)(l - xk+1 + zk+1) = 1 + xk+2
where xk+2 = zk+i — xk+12 + Xk+iZk+i e ph+2. This shows that
for any integer n > 1 we have #i, ^2> • • •> xn, *i e p4, such that
71 00
A + y) exp (-J3*,-) = 1 (mod p"+1). Then x = ^2 xk is an
1 !
element of p and we assert that exp x = 1 + y. Let Xn = 23 *»•
i
Then exp (—x) expXn = exp (Xn — x) = 1 (mod p"+1) since
Xn — x e pB+1. Now one verifies as for p that, if Zi = 1 (mod
pn+1) and z2 = 1 (mod p"+1), then Z!Z2 = 1 (mod pn+1). Hence
we can conclude from A +jy) exp (—Xn) = 1 (mod p™+1) and
exp (—x) exp Z»sl (mod pn+1) that
A +jy) exp (-*) s 1 (mod pn+1).
Since n is arbitrary, this gives A + y) exp ( — x) = 1 and 1 + y =
exp # as required. This shows that x —» exp «■, x e p is surjective
on U\. Hence we have proved the following
Theorem 7. Let p be the maximal ideal in the ring 5 oj p-adic
integers, p ?± 2, and let Ui be the group of elements = 1 (mod p).
Then the exponential mapping x —* exp x is an isomorphism of the
additive group (p, +) onto the multiplicative group Ui ofl-units oft.

VALUATION THEORY 229
Remark. It is natural to establish the fact that x —> exp x is
surjective by giving the inverse function log A + y) = y —
y2 y3
1 • • •, which is defined for all y e p (ex. 4 below). Then
one has to show that exp (log A + y)) = 1 + y. The details of
this are somewhat lengthy. For this reason we have preferred the
above proof that x —> exp x is surjective since it does not require
the explicit definition of the inverse. The reader may refer to
Hasse's Zahlentheorie, Berlin 1949, pp. 188-199, for a complete
treatment of these questions.
It is clear from Theorem 7 that the group U\ has no elements of
finite order. Hence if Z denotes the group of (j> — l)-st roots
of 1 which we constructed before, then U\ D Z = 1. We have
seen that every element of the group of units t/of 6 is a product of
an element of Z and an element of U\. Hence we have U = Ui X
Z (direct product).
As an application of these results we consider the question of
solvability of equations of the form x2 = m in p-adic fields where
m is an ordinary integer prime to p and p ^ 2. Then m e U and
we can write m = rj%a where rj e U\ and m = a (mod p), 0 < a <
p. It is clear that, if a2 = m for a e R^p\ then <pp(a) = 1, so if a
solution of x2 = m exists in R^p\ then this solution must belong to
U. Hence it has the form \£b where f b is one of the (p — l)-st
roots of 1 and X e Uu It follows from U = Ux X Z that X2 = rj,
fb2 = fo- We now note that the equation x2 = rj has a solution
for any rj e U\. Using the isomorphism of Ux with (p, +), it
suffices to see that the mapping x —» 2x is an automorphism of the
latter group. This is clear since 1~x e U and x —> 2~xx maps p
into itself and is the inverse of the mapping x —> 2x. Hence we
see that the equation x2 = m is solvable in R(p) if and only if
tb2 = fo is solvable. It is easy to see that the condition for this is
that x2 = m or x2 = a (mod p) is solvable, that is, m is a quadratic
residue modulo p. Hence x2 = m is solvable in R{p\ p ^ 2
(w, />) = 1 if and only if x2 = m (mod p) is solvable in integers,
that is, if and only iff — J = 1 where f — 1 is the Legendre symbol.
For example, if p = 5 and m = — 1, then 22 = — 1 (mod 5) so

230 VALUATION THEORY
f —— j = 1. Hence V—1 exists in the 5-adic field. On the other
hand, (-J = —1, so V3 does not exist in this field.
EXERCISES
1. Obtain the S-adic expansion of the form A4) for the 5-adic integer f.
2. Show that the field of p-ad\c numbers is uncountable for any p = 1, 3,
5, • • ■. Use this to prove the existence of ^>-adic numbers which are transcen-
transcendental over the rational subfield.
3. Use the binomial expansion of A — 2*)M to obtain a convergent series for
= i(l ~ 10) * in the 5-adic field.
4. Define log A + y) = y —r- + ^r . Show that this series converges
for all y e Jj if p ^ 2. Show that log A + j>i)(l + yt) = log A
log(l + yi),yiep'.
5. Show that the equation #8 = 4 is solvable in the field of 5-adic numbers.
6. Show that in the field of 2-adic numbers the exponential mapping is an
isomorphism of {j2 onto the group of elements of 0 which are = 1 (mod p2).
6. Hensel's lemma. There is another, more powerful, method
for handling equations in ^>-adic fields and more generally in com-
complete fields with a discrete non-archimedean real valuation. This
is based on a fundamental reducibility criterion for polynomials
which is known as
Hensel's lemma. Let $ be a complete field relative to a non-
archimedean discrete real valuation <p. Let o be the valuation ring of
$, p its maximal prime ideal, A = o/p the residue field and let a —*
a* = a + P be the canonical homomorphism of o onto A. Suppose
f(x) eo[x] has the property that its image f*(x) = y{x)i\{x) in A[x]
where (y(x),r)(x)) = 1 and the leading coefficient of y(x) is 1. Then
/(*) = g(x)h(x) in o[x] where g*{x) = y(x), h*(x) = r,(x), deg ,§•(#)
= deg y(x) and g(x) has leading coefficient 1.
Proof. Let deg/(«) = n, deg y{x) = r < n. We can choose
gi(x), hi(x) eo[x] so that£i*(.v) = y(x), hx*(x) = r){x), deg^i(jf)
= r, deg h\(x) < n — r, leading coefficient of £i(#) is 1. Then we
have/(#) = gi(x)hi(x) (mod p) in the sense that the coefficients
are congruent (mod p). We proceed to determine two sequences
of polynomials {gk(x)}} {hk(x)}, k = 1,2, • • •, in o[x] such that:
(i) gh(x) = gk+i(x) (mod p"), hk(x) = hk+1(x) (mod pk), (ii)/(*) =

VALUATION THEORY 231
gh{x)hk{x) (mod p*), (iii) deg gk(x) = r, deg hh(x) < n - r,
leading coefficient of gk(x) = 1. We can begin these sequences
with the ^i(*) and hx{x) which we have chosen. Hence we may
suppose that the sequences have already been constructed for
k < s. We set ga+i(x) = gB(x) + u(x)ir°, h.+i(x) = h,(x) +
v(x)ir", where p = (ir) (as in § 5). Then (i) will hold for any choice
of u(x) and v(x) in o[x]. We seek to satisfy (ii). This requires that
= ga(x)h,(x) + \g,(x)v(x) + h.(xMx)]r' (mod ps+1)
or
A9) Ax) - g.(x)h.(x) = \g.(x)v(x) + h.(x)u(x)]*' (mod p8+1).
Since/^) = gs(x)hs(x) (mod ps) we can write/(,*) — gs(x)h8(x) =
ir'w(x) where u(x) eo[x]. Since degAx) = n an(i deg g,(x)hK(x)
< n we may suppose deg u(x) < n. It is clear that A9) will hold
if
B0) g,(x)v(x) + h,(x)u(x) = u(x) (mod p).
Now it is clear from (i) that £,*(.*) = gi*(x) = y(x) and h,*{x) =
v(x)> so we consider the equation
B1) y(x)v*(x) + ,(*)«•(*) = «•(*)
in A[x]. Since G(x), ri(x)) = 1, there exist polynomials «(*), (S(^)
in Afx] such that a(x)y(x) + fi(x)ri(x) = 1. Multiplication by
w*(x) gives polynomials k(x), \(x) such that ic(x)y(x) + \(x)ri(x)
= u)*(x). We can write X(^) = y(x)n(x) + p(x) where deg p(x)
< r and then we obtain
+ p(x)v(x).
Then deg p(x)i](x) < n while deg w*(x) < n. Since deg y(x) = r,
the foregoing relation shows that the degree of (k(x) + n(x)t}(x))
does not exceed n — r. If we call this polynomial <t(x), we have
<r(x)y(x) + p(x)t)(x) = »*(#),
where deg p(x) < r and deg a{x) < n — r. Then we can choose
u(x) and v(x) e o[x] so that u*(x) = p(a:), y*(^) = <r(x), deg «(x)

232 VALUATION THEORY
= deg p{x) < r, deg v(x) < n — r. Then B1) holds and
gs+i(x) = gs(x) + u(x)ws, hs+x{x) + hs(x) + v(x)ir" satisfy also
(iii). This completes the proof of the existence of the sequence
j^Wj, {hk{x)}, k = 1,2, ■■• satisfying (i), (ii), and (iii). The
conditions (i) and (iii) and the completeness of $ imply that the
sequences {gk(x)}> \hk(x)} converge to polynomials g(x), h(x) in
the sense that the sequences of coefficients of like powers of x con-
converge to those of g(x), h{x). Moreover, we have deg g(x) = r,
deg h{x) < n — r, leading coefficient of g(x) = 1. It follows also
from (ii) tha.tf(x) = g{x)h{x) and this completes the proof.
EXERCISES
1. Use Hensel's lemma to prove that in the field of ^>-adic numbers there
exists a fo such that f ap~1 — 1, f a = a (mod J3) where a is any integer prime to p.
Also use this to obtain another proof of the existence of y/—1 and \/A in the
5-adic field.
2. Hypotheses on $ as in Hensel's lemma. Let/(#) = aoxn + a\xn~1 + • • • +
an e o[x] satisfy: «o, an e {) but there exists ar, 1 < r < n — 1, such that ar & p.
Then f(x) is reducible in o[*]. Use this to show that, if g(x) =*" + aix"~1
+ ■ • • + an is an irreducible polynomial in $[#] and an e 0, then all the a< E 0.
7. Construction of complete fields with given residue fields.
Let A be a given field. We consider the problem of constructing
complete fields with non-archimedean real valuations such that
the residue field is the given field A. We shall give two construc-
constructions: the first, in which the complete field contains A and so has
the same characteristic as A; the second, in which A is perfect of
characteristic p ^ 0 and the complete field is of characteristic 0.
A special case of the latter is A = Ip and the complete field is
the field of/>-adic numbers.
We consider first the field $ = A(£) where £ is transcendental
over A. We introduce the order function v by v(a(£)) = k if a(£)
= ?A:/3(£O(£)~1 where /?(£) and y(£) are polynomials not divisible
by £. We define a valuation <p by <p(a(£)) = c"(a({)), c a fixed real
number 0 < c < 1 (cf. example 3, § 1). Let $ be the completion
of $ relative to (p. Since <p is trivial on A, it is clear that <p is non-
archimedean. Hence its extension to <5, which we shall denote by
<p also, is non-archimedean. The value group V of <& and of $
consists of the powers of c, so the valuation is discrete. Let 5 be
the valuation ring of*, p its maximal ideal, and let o = 5 D <£,

VALUATION THEORY 233
p = j fl §. It is clear that £ is an element of p for which <p(£) = c
is maximal. Hence, as in the p-adic case, every element a of $ has
the form £*e where e e 5, $. p and k is an integer. We can therefore
define v{£ke) = k and it is clear that this coincides on <l> with the
order v which we denned originally in $.
We have seen in § 5 that 5 = o + p and this permits us to
identify the residue fields o/p and o/p. The ring o is the set of
rational expressions in £ with coefficients in A which are "finite at
0" in the sense that a(£) = /3(£)'y(£)~1 where j8 and y are poly-
polynomials and 7@) ^ 0. The argument in the ^>-adic case showing
that 5 = / + p (p. 224) can be used in the present situation to
prove that 5 = A[£] + p. Since £ e p", this gives o = A + p" and,
since p D A = 0, we have the isomorphism 5 —> 5 + p in o/p" of A
with the residue field 5/p. In this sense we can say that A is the
residue field of $.
Now let a be any element of 5. Then o = A + P shows that we
can find So e A such that a — 80 e p. Then ai = (a — 8o)£-1 e 5
and we can repeat the argument with this obtaining 8X e A such
that dx — Si e p and a2 = («i — 8i)£-1 e 5. We have a = 60 +
Si? + a2%2, a2 e 5. As in the ^>-adic case, we can continue this
process and obtain
B2) a = So + 5i{ + hi2 +■■■+ «*{* + a*+i€*+1
where the 5; e A and ajt+i e 5. Since v(ak+il-k+1) > k + 1, it is
clear that the sequence {a^} is a null sequence. Hence we have
B3) & = S0 + 5i£ + S2£2+---, MA,
for any a e o. If 3 is any element of $ we can write J9 = d£~*
where k is a non-negative integer and a e 5. Then we have
B4) J3 = $-*(«o + «i€ + S2£2 +•••)•
This shows that $ is the set of power series of the form B4) in £
with coefficients in the field A. It is easy to see that the expression
B4) for J3 is unique, that is, k and the 5,- e A are uniquely deter-
determined by J3. Moreover, the addition and multiplication of ele-
elements of <& are the usual ones for formal power series based on the
compositions in A. For example, we have E0 + 5X£ +•••) +
(«o + *ik + •••) = («o + «o) + (Si + ei)£ + • • • for 8,, e,- in A and

234 VALUATION THEORY
E *<?)( E *i?) = E ij*{* where Vk = £ S.e^-. It is clear
0 / \ 0 ' 0 t=0
that we have a good hold on the field $ as the field of formal power
series B4).
We consider next the case in which A is a perfect field of charac-
characteristic p 5* 0 and we shall use this to construct a field $ which is a
generalization of the field of ^>-adic numbers. The construction
we shall give is based on Witt vectors which we considered in
§ 3.4. We begin with the definition of the ring 93B1) of Witt
vectors {of infinite length) based on a commutative algebra 21 over
Ip. The elements of 23B1) are the infinite sequences
B5) (ao,aua2, •••), «» e 21,
where equality is defined component-wise. We can define addi-
addition and multiplication by the formulas B2) of Chap. Ill which
were used to define these compositions in 93mBl) the ring of Witt
vectors of length m defined by 21. Then one can verify that SB B1)
is a ring. It is more convenient, however, to adopt an equivalent
but slightly different approach which is a special case of the defini-
definition of an inverse limit of rings. In the present case we are deal-
dealing with such a limit for the rings 21 = 93iBI), S3J2Bt), • • • with
the restriction homomorphism R of 9BmBl) into 23m_iBl). We
associate with the element A = (a0, aly ■ • •) of 23B1) its projec-
projection A*™ = (a0, au ■ ■ •, «m_0 in 23mBt). Then A*™R = (a0, ■■■,
am-z) = A"m-\ On the other hand, let \Am\m - 0, 1, 2, • • ■} be
any sequence of elements Am where Am z 38mBl) and AmR = Am_u
m = 1, 2, • • •. Then it is clear that {Am} = ATm for a unique
A e9BB1). Hence we can identify the elements of 933B1) with the
sequences {Am}, ^me3BmBl) such that AmR = Am^x. If A —
{Am } and B = {Bm\ are two such sequences, we define A -\- B =
\Am + Bm}, AB = {AmBm}. Since R is a ring homomorphism,
(Am + Bm)R = AmR + BmR = Am_x + Bm_, and (AmBm)R =
AmRBmR = Am^Bm-i. Hence A + B and AB e 33B1). It is
trivial to check that 93B1) is a commutative ring relative to these
compositions and that 0 = @, 0, • • •), 1 = A, 0, 0, • • •)■ For a
fixed m, the mapping irm: A —» ATm is a homomorphism of 93B1)
onto 9BmBl). Since Vm has order pm, it is clear that the identity 1
of 93B1) has infinite order in the additive group (93B1), +).

VALUATION THEORY 235
Let ftm denote the kernel of irm. Then ftm is the set of elements
of 3BBl) of the form @, • ■ •, 0, am, *m+i, • • • )• Hence
B6)
m=l
We can use the set {ftm} to define convergence in 23B1). If
{Ak\k = 1, 2, • • ■} is a sequence of elements of 935B1), then we
say that [Ak] converges to the element A of 933B1) (Ak —> A) if for
any positive integer m there exists a positive integer N{m) such
that A — Ak e ftjv(m) for all k > N{m). It is easy to see that the
limit A is unique and that Ak —> A, Bk —> B imply Ak ± Bk —>
A ± B and AkBk —> ^5. Suppose {Cjt|& = 0, 1, 2, • • •} is a
sequence such that Ck e ft*, & = 1, 2, • • •. Set yffc = Co +
Ci + ■ ■ ■ + Ck. Then Am*«*R = Am*<" = Am_f»; hence the
sequence of elements {AmTm+1, in = 0,1, •••} where AmTm+l e
2Bm+i(Sl) can be identified with an element ^e 233B1). One
checks that Ak —> A. Since Ak = Co + Ci H h Cfc, we
shall indicate the convergence Ak —> A by writing ^ C* = ^f.
We recall that, if ft is the ideal of elements @, aiy ■ • •, am_i) in
2BmBl), then ft is nilpotent (Th. 3.12). In fact, the proof of this
result shows that ft* is contained in the set of vectors of the form
@, • • •, 0, ak+u • • •, tf»_i). This implies that fti* C ft*, in
» /m \
Hence, if Z e ft1} then X) z* is defined. Since ^Z4 A - Z)
*=o V o /
= i _ zm+1, it follows that 1 - Z is a unit in 28C1) with X) Zk as
o
inverse. Since (a0, • • •)(ao~1, ■ • •) = A, • • •) this implies that,
if aQ is a unit in SI, then (a0, au ■ • •) is a unit in 933B1).
Now let 21 = A a perfect field of characteristic p. The formula
P(ao, aly ■ ■ •, «w_i) = @, aop, axv, ■ ■ ■, «m_2p) in 933mBl) which
we established in § 3.4 (Equation B7)) implies that p(a0} «1} • • •)
= @, aop, aip, • • •) holds in 9BBl). Iteration of this formula gives
B7) p\a0, «i, • • ■) = @, • • •, 0, ao*\ a?\ ■ ■ ■).
Since 21 = A is perfect, the elements a?k can be taken to be arbi-

236 VALUATION THEORY
trary elements of A. Hence we have pk%&( A) = 3tk where 9?*, is the
ideal we defined before. Also B7) shows that, if A 9^ 0, then
pkA 7^ 0 for k = 1, 2, • • •. Now let A and 5 be any non-zero
elements of 9B(A). Then we can write A = pkC, B = plD where
C, D $ Sfti. Then C = (c0, ■ ■ •) and D = (d0, ■ ■ •) where c0 ^ 0,
do ^ 0. Hence CD = (ro^o, • • •) ^ 0 and AB = pk+lCD ^ 0.
This shows that 2B(A) is an integral domain. Let i> be the field of
fractions of 3B(A) and consider the subset <£' of $> of elements of
the form pkC where Ce2B(A) and * = 0, ±1, ±2, • • -. Since
any CeSB(A), # p9£(A) is a unit in 28(A), it is clear that the non-
nonzero elements of <!?' form a group under multiplication. Since <&'
is a subring of $ which contains 3B(A), it follows that <£' = 4>.
If A = p*C, C e 3B(A), ^ ^j, then we define the order v{A) = k
and we define <p(^) = p~k, <p@) = 0. Then <p is a real non-
archimedean valuation of $. The subring 28(A) is the set of ele-
elements satisfying <p(A) < 1 and 9^1 is the ideal of elements B of
SBB(A) such that <p(B) < 1. The residue ring is 2B(A)/9ti which is
isomorphic to A. The result we noted before on convergence of
sequences in SBBl) implies that $ is complete relative to the valua-
valuation tp. We leave it to the reader to check this. Since 1 is of in-
infinite order, $ is of characteristic 0. Thus i> has all the properties
we required: completeness relative to a non-archimedean real
valuation, characteristic 0, residue field the given perfect field A
of characteristic p. If we start with A = Ip, then the field $ we
obtain in this way is the field of ^>-adic numbers.
8. Ordered groups and valuations. A non-archimedean real val-
valuation satisfies <p(a + j8) < max (<p(a), <p(/3)), <p{aff) = <p(a) <p@).
Hence it is clear that in considering such a valuation the addi-
addition of the reals plays no. role. Only the multiplication and
order of the non-negative reals are involved in the defining proper-
properties. As we shall see, this leads to a generalization of the concept
of a non-archimedean real valuation to (non-archimedean) valua-
valuations with values in any ordered commutative group. Besides the
increased generality which results from this extension, the
generalization is essentially simpler and more natural than the
original concept. We consider first the notion of an ordered
commutative group.

VALUATION THEORY 237
Definition 6. An ordered (commutative) group G is a commuta-
commutative group G together with a subset H satisfying the three conditions:
l)\^H,2)ifaeG either a e H, a = 1 or a~x e H, 3) H is closed
under the multiplication in G.
If (G, H) is an ordered group, then we let H~x = \b~x \b e H}.
Then condition 2 states that G = H U {1} U H~x. Moreover,
these sets are non-overlapping. This is assumed for H and {1} in
condition 1 and it follows for H~x and {1} on observing that, if
1 sH~1, then 1 eH contrary to condition 1. Finally, if a eH f\
H~x, then a~x e H and 1 = aa~x e H by condition 3. This again
contradicts condition 1.
The positive reals form an ordered group if we take H to be the
set of elements < 1. We can take H equally well to be the set of
elements > 1. In fact, if G is any ordered group, then H~x is
closed under multiplication and satisfies conditions 1 and 2 of
Definition 6, so we can obtain another ordered group on replacing
H by H~x. In any ordered group G we define a < b to mean that
ab~x e H. This defines a linear ordering in G, that is, we have the
following properties: 1. a < b, b < c implies a < c. 2. For any
pair {a, b),a, b e G, one and only one of the following holds: a < by
a = b, b < a (as usual we write b > a for a < b). The order in G
is invariant under multiplication, that is, we have: 3. If a < b,
then ac < be. Conversely, if a relation a < b is defined in a group
G so that properties 1, 2, and 3 hold, then G is ordered by the
subset H = \a\a < 1}. Clearly condition 1 of Definition 6 holds
for H. To prove conditions 2 and 3 we note first that, if a < b
and c < d, then ac < be < bd so ac < bd\ hence, a < b if and only
if a > b~x. In particular, a < 1 if and only if a~x > 1. Since
any a satisfies one of the conditions: a < 1, a = 1, a > 1, it is
clear that condition 2 of Definition 6 holds. Finally, a < 1,
£ < 1 imply <z£ < 1, so H is closed under the multiplication in G.
We remark also that the ordering defined by H in the manner
indicated: a < b if ab~x e H is the same as the original ordering
since ab~l e H means ab~x < 1 and this holds if and only if a < b.
If Gi is a subgroup of an ordered group G ordered by the set
H = {a\a eG, a > \), then Gx has an induced ordering defined
by Hi = Gi 0 H. This can be verified directly, or it can be seen

238 VALUATION THEORY
by noting that the relation > defined in G gives a relation in Gi
which satisfies the conditions stated before. If G is ordered by H
and G' is a second ordered group, ordered by H', then an isomor-
isomorphism r) of G into G' is called an order-isomorphism if Hr\ C H'.
Also G and G' are order-isomorphic if there exists an order iso-
isomorphism 77 of G onto G'. In this case one necessarily has Hi\ =
H'. For example, the group of positive reals under multiplication
with H defined as before is order isomorphic to the additive group
of all the real numbers ordered by the set H' of negative reals.
The mapping a —> log a (natural logarithm) is an order isomor-
isomorphism of the first group onto the second one.
If G is an ordered group, G contains no elements 7^ 1 of finite
order; for, if a < 1 {a > 1), then an < 1 (an > 1), so an ^ 1 for
every positive integer n. A consequence of this property of G is
that for any fixed integer n the mapping x —> xn of G is an iso-
isomorphism of G onto a subgroup of G, which is order preserving if
n > 1.
To define general valuations we shall need to consider ordered
groups V with 0. We define such a system to be an ordered group
G to which a 0 element has been adjoined: F = GU {0}. The
ordering in G is extended to V by defining 0 < a for every a e G
and we define aO = 0 for all a. We can now give the following
Definition 7. Let <i> be afield and let V be an ordered {commuta-
{commutative') group with 0. A mapping <p: a —* <p(a) of $ into V is called a
valuation //
(i) <p(a) = 0 if and only if a = 0.
(ii) <p(ap) = <p(a)<p@).
(iii) v(a + 0) < max (?(«), v@)).
The exact sweep of this definition will become apparent soon.
At this point it is clear that real non-archimedean valuations are a
special case in which V is the set of non-negative real numbers.
On the other hand, it should be noted that the real archimedean
valuations are not valuations in the present sense. This incon-
inconsistency in terminology will cause no real difficulty. We shall now
give an example of a valuation for which V is not the non-negative
reals.

VALUATION THEORY 239
Example. In this example we shall find it convenient to use the additive
notation in the group G. The modifications in Definition 7 which are necessi-
necessitated by this change are obvious, so we shall not write these down. The group G
we shall consider is the additive group of integer pairs (k, /). We introduce the
lexicographic order in G, that is, we define (k, f) < (k1,11) if either k < k' or
k = k' and / < /'. One checks that this is a linear ordering preserved under
addition; hence G is an ordered (additive) group. We let V = G U {00} where
the ordering is extended to V by setting 00 > (k, I) for every (£, /) e G. Also we
define (k, I) + 00 = 00. Now let P = <!>(£, rj), a purely transcendental extension
of a field 4> where {£, r;} is a transcendency basis for P over $. If a E P and
a ^ 0, we can write a = £"Vp(|, '!)?($.'?)-1 where />(£,»?) and ?(£,i?) are poly-
polynomials in £,»; with non-zero constant terms, and w and n are integers. Then
we define <p(a) = {m, n). Also we set ^j@) = «. Then (i) holds. It is easy to
check that <p{ab) = <p{a) + <pU>) and ^>(a + £) > min (tp{a), <p(i)). The first of
these is (ii) in the additive notation and the second can be changed to (iii) by re-
reversing the ordering (writing > for <). Hence our function is essentially a
valuation.
EXERCISES
1. Let G be the additive ordered group of integer pairs (k, I) given in the fore-
foregoing example. Let c and <? be real numbers such that 0 < c < 1 and e is posi-
positive and irrational. Show that the mapping (k, I) —* ck+el is an isomorphism of
G into the ordered multiplicative group of positive real numbers P. Show that G
is not order isomorphic to a subgroup of P.
2. Let P = <!>(£, rj) and a = ^tfpii, f)?(£, v)~1 where p and q are polynomials
in £, 77 with non-zero constant terms, as in the example above. Define ^(a) =
cm+en wnere f and c are real numbers, 0 < c < 1, e positive irrational. Show
that i/' is a non-archimedean real valuation which is not discrete.
3. Define a valuation <p of an integral domain 0 by replacing the field $ in
Definition 7 by the integral domain 0. Show that any valuation ^ of 0 into V
has a unique extension to a valuation of the field of fractions $ of 0.
4. Let G be an arbitrary (commutative) ordered group and let 0 = $o(G) be
the group ring over a field $0 of G (Vol. I, ex. 2, p. 95). Show that 0 is an integral
r
domain. If a = ^ <*<£«> a« 5^ 0 in $0, gi E G, define <p(a) = min gi (in the
1
ordering < defined in G). Define ^>@) = 0. Show that <p is a valuation of 0.
Use exs. 3 and 4 to show that if V is any ordered group with 0, then there exists a
field $ with a valuation <f> of $ into F such that
9. Valuations, valuation rings, and places. In this section we
shall establish an equivalence between the concepts of a valuation
in the sense of Definition 7 and two other concepts: valuation ring
and place. The first of these, valuation ring, is an intrinsic notion
in the sense that its definition does not require any system external
to the given field $. Moreover, the valuation rings give the link
between valuations and places. We have already encountered
these for real non-archimedean valuations.

240 VALUATION THEORY
Now let <£ be any field and let ^ be a valuation with values in
the ordered group V with 0. We note first that <p(lJ = <p(l2) =
<p(l) and, since G contains no elements of finite order ^ 1, <p(l)
= 1. Also <p(-\J = <p{\) = 1, so tp(-l) = 1 and <p(-a) =
<p( — l)<p(<x) = <p(a). From aa~x = 1 we obtain (p(a~1) — <p(a)~1
and <p(a[i~r) = <p(a)(p(ff)~1. Now let o be the subset of $ of ele-
elements a such that <p(a) < 1. Then, if a, 0 e o, <p(a — /8) < max
(<p(d), <p(f3)) < 1 and <p(a/3) = ip(<x)<p(C) < 1. Hence o is a sub-
ring. Now suppose a^o, then tp(a) > 1 and ^(a) = <p(a)-1
< 1. Hence a eo. We therefore see that o is a valuation ring
(in <J>) in the sense of the following
Definition 8. If $ is afield, a valuation ring o in $ is a subring
of $ (containing 1) such that every element of $ is either in o or is the
inverse of an element of o.
If o is the subring of elements a satisfying <p(a) < 1 for the
valuation <p, then o is called the valuation ring of <p. This is a direct
generalization of the definition we gave before for non-archime-
dean real valuations. We shall now show that any valuation ring
gives rise to a valuation <p' for which the given ring is the valua-
valuation ring. Suppose o is a valuation ring in <£. Let U be the set of
units of o, p the set of non-units, p* the set of non-units ?* 0, $*
the multiplicative group of non-zero elements of <£. Then U is a
subgroup of the commutative group <£* and we shall take G' =
#*/ U for our group. We introduce an ordering in G' by letting H'
be the set of cosets fiU, /8 e p*. It is clear that the product of a
non-unit of o with any element of o is a non-unit. Hence if /8i,
Pa e p*, then ftjSa e p*; so if faU, 02U,eH', then (pxU)(&2U)
= fafaUeH'. If PU is any element of G' = **/C7, then
0 ?* 0, and if /3 ^ p*, then either /3 e U or ,5 ?! C7 and C ^ p*. In the
first case I3U = U, and in the second /3 ^ o, so /3 e o and, since
iS e 1/ implies 0 e £7, we have fi~l e p*. Hence ($U)~X =
p-tUeH'. Thus we see that G' = H' \J {1} (J (fl7) holds.
Also 1 = [/ ^ /f. Hence i/' makes G' an ordered group as in
Definition 6. Next we adjoin a 0 to G', obtaining V = G' U {0},
and we define a mapping <p' of $ into V by
B8) ^'@) = 0, v'(a) = olUzG' if «^0.

VALUATION THEORY 241
The conditions (i) and (ii) for a valuation are clearly satisfied.
Also (iii) is clear if either a = 0 or 0 = 0. If a ^ 0, j8 j± 0,
either afi~l e 0 or /3a e 0 and we may as well assume the former.
Then we have a = Py where 7 e 0 and <p'{a) = <p'@)<p'(y) < <p'(P)
since <p'(y) = yU < 1 = U. Also aP~x + leo, so v'iafi-1 +
1) < 1 and v'(a + 0) = ^(a^-1 + 1 V(/3) < v'(j8) = max (*'(«),
p'(/3)). Hence (iii) holds. It is clear from B8) and the defini-
definition of G' and H' that <p'(a) < 1 is equivalent to a e 0. Hence 0
is the valuation ring of the valuation <p'. We shall call the valua-
valuation <p' the canonical valuation of the valuation ring 0.
Now consider again an arbitrary valuation <p of $ into V =
(G, 0) where G is a commutative group ordered by H. Let 0 be
the valuation ring of <p and <p' the canonical valuation of $ into
V = (Gf, 0) where G' = $*/U is ordered by #' = {pU\p e p*}.
The definition B8) gives <p'@) = 0, <p'(a) = aC/ if a ^ 0. We
have the homomorphism a —> p(a) of the multiplicative group <!>*
into G whose kernel is the subgroup U. Hence we have the in-
induced isomorphism ii:<p'(a) = aU —> <p(a) of G' = $*/U into G.
This is an order isomorphism since, if fiU zH', then /3 e p*, so
£>(/3) < 1. We now see that the given valuation can be factored
as <p = ip'j) where 77 is an order isomorphism of G' into G (more pre-
precisely, V into V).
These considerations make it natural to lump together the
valuations of $ which have the same valuation ring 0. Accord-
Accordingly, we shall say that such valuations are equivalent.
There is a third concept, that of a place which is also equivalent
to the concepts of valuation and valuation ring. We define this as
follows:
Definition 9. If $ is a field, a place & is a homomorphism of a
subring o of & into a field A such that, if a $ a then a e o and
^(a-1) = 0. {We recall that 1 e o and &A) = 1 by our conven-
conventions on subrings and homomorphisms.)
It is clear from the definition that, if & is a place, then the sub-
ring o given by 0* is a valuation ring. On the other hand, suppose
o is any valuation ring and let p be the set of non-units of o. Then
it is clear that, if /3 e p and aeo, a@ e p. In particular, —£ =
( —I)j8ep. If 0! and /32 e p, we may assume that j8i/32~1eo.

242 VALUATION THEORY
Then Pxp2~l + 1 e o, so 01 + 02 = Oifts + 1H2 e p. Hence p
is an ideal in o. Since p is the set of non-units of o, it is clear that p
is maximal and A' = o/p is a field. Let 0*' be the canonical homo-
morphism of o onto A' = o/p. Then it is clear that 0*' and o
satisfy the defining conditions for a place. We shall call this place
the canonical place of the valuation ring o. The image of o under
&' is A' = o/p where p is the ideal of non-units of o. As in the
special case of real non-archimedean valuation, we shall call A'
the residue field of the valuation ring o.
Now consider again an arbitrary place 0* of <l> into the field A
and let o be the valuation ring on which 0> is defined. Let p be the
ideal of non-units of o. If a e p and a^O, then a" $ o, so the
hypothesis on 0* gives 0>{ot) = 0. This holds" also if a = 0.
Hence we see that p is contained in the kernel of 0t Since p is a
maximal ideal, this shows that p is the kernel of 0t The homo-
morphism a —» 0>{a), a e o, therefore gives an isomorphism 0"(a)
= a + p —» 0>(a), and so the place 0* is the resultant of the
canonical place 0>l and an isomorphism of A' into A. As for
valuations, it is natural to consider as equivalent places that have
the same valuation ring.
We have now established the procedures for passing from one of
the concepts: valuation, valuation ring, place, to any other.
Clearly, a result on one of these can be translated to the other two.
In the sequel we shall apply this idea to obtain extensions of
valuations via extensions of places. The latter amounts to ex-
extensions of homomorphisms, for which we have available the basic
extension theorems of the Introduction.
EXERCISES
1. Let 0> be a place on $ with values in A. Adjoin a new element » to A and
define 00 + 5 = 00 = 8 + 00, 5 e A, woo = «, »S = « = 5°o if 5 ?* 0 in A.
Extend 0* to the whole of $ by defining 0>(pi) = w if a $ 0. Verify that
, 0>(a+f})=0>(a)+0>(fi)
whenever the right-hand sides are defined. Conversely, assume that 0* is a
function defined on $ with values in (A, «),Aa field where A fl {«} = 0 and »
obeys the rules indicated. Assume B9) hold whenever the right-hand sides are
defined. Let 0 be the inverse image 0'~1 (A). Show that the restriction of 0> to
0 is a place. This gives an alternative definition of a place.

VALUATION THEORY 243
2. Let 0* be a place with valuation ring 0. Assume 0* is an isomorphism.
Show that 0 = $ and that the canonical valuation of 0 is trivial in the sense that
<p'@) =0,<p'(a) = 1 if a 5*0.
10. Characterization of real non-archimedean valuations. In
order to apply the general theory of valuations to the case of non-
archimedean real valuations it is necessary to characterize these
among all possible valuations of a field. In view of the foregoing
discussion this is equivalent to the problem of characterizing the
ordered groups which are order isomorphic to subgroups of the
multiplicative group of positive reals or, equivalently, to the
additive group of all the real numbers with the usual order in this
group. Hence we seek a characterization of the ordered groups
which are order isomorphic to subgroups of the additive group of
real numbers. It will be convenient to use the additive notation
in all the groups which we shall consider in this section.
Let G be an ordered group: If a e G, we define \a\ = a if a > 0
and \a\ = — a if a < 0. We define an is olated subgroup K of G as
a subgroup such that, if a e K and \b\ < \a\y then b e K. Let
K\ and K2 be isolated subgroups. Then we assert that either
^i Q K2 or K2 Q Kx. For, if neither of these inclusions holds,
then there exists a bx e Ku i K2, and a b2 e K2, ft Ku and we may
suppose that bt > 0. If b2 > bu then bx e K2 contrary to assump-
assumption. Hence b2 > bt and similarly bx > b2 which contradicts the
fact that G is an ordered group. Thus we have either K\ 2 K2 or
K2 3 K\ so the set of isolated subgroups is linearly ordered by the
inclusion relation. The order type of the set of isolated subgroups
is called the rank of G.* The simplest situation is that of a group
of rank one in which G^0 and G has no isolated subgroup 5^
0, G. These groups can be characterized by the archimedean
property which is familiar for real numbers:
Lemma. An ordered group G G* 0) is of rank one if and only if
given any ay b e G with a > 0 there exists a positive integer n such
that na > b.
Proof. Suppose first that G contains two elements a, b such
that a > 0 and na < b for all positive integers ». Let K+ denote
the subset of G of elements u such that 0 < u < ma for some
positive integer m. K+ is not vacuous since a < 2a and clearly
• Cf., for example, F. Hausdorff, Mengenlehre, 3rd Ed., Chap. 3, de Gruyter & Co., 1937.

244 VALUATION THEORY
K + is closed under addition. Moreover, K+ contains every v such
that 0 < v < u for some u in K+. Hence if u,\ and «2 e K+ and
«! < u2, then 0 < u2 — «i < #2 so u2 — Ui e K+. It follows
that the union of 7C+, 0, and — K+, the set of negatives of the ele-
elements of K+, is a subgroup K of G. Now K is isolated, since, if
u e K and « > 0, then every v such that 0 < v < u is in K. Also
K 9^ G since b $K. Hence G is not of rank one. Conversely,
assume G not of rank one and let K be an isolated subgroup ^ 0,
G. Since K ^ G there exists a positive element £ such that b > a
for every a e K. Choose a > 0 in K, then na < b for all « =
1, 2, 3, • • •. Hence the archimedean property fails in G.
It is clear from this criterion that if G is of rank one, then any
non-zero subgroup of G is of rank one. In particular, any non-zero
subgroup of the additive group of real numbers is of rank one.
Moreover, these are essentially all the ordered groups of rank one,
since we have the following
Theorem 8. Any ordered group G of rank one is order isomorphic
to a subgroup of the additive group of real numbers.
Proof. We shall define an order isomorphism 77 of G into the
additive group R of real numbers. For this purpose we choose a
u > 0 in G. If v > 0, then there exist pairs (m, n) of positive
integers m, n such that nv > mu. Thus we may take m = 1 and,
by the archimedean property, determine n so that nv > u = \u.
If q tP, the collection of positive integers, then qnv > qmu if and
only if nv > mu. Hence if r = m/n = m''/»', m, n, m', n' e P,
then nv > mu if and only if n'v > m'u. The rational numbers r =
m/n satisfying this condition form a set which we denote as Rv.
If r = ffz/« and j = m'In' < r, w', «' e P, then w'w < w«'. If
r e /?„, then nv > mu and ww'y > mn'u > m'nu. Hence n'v >
m'w so that s e 2?,,. We note next that the set of positive rationals
Rv is bounded above. Otherwise, the result just proved implies
that Rv is the complete set of positive rationals. Hence every
positive integer k is in Rv which means that v > ku, k e P. This
contradicts the archimedean property of G. We now define vv to
be the positive real number sup Rv. Since Rv contains every s < r
for every r e Rv, it is clear that Rv and its complementary set Rv'
in the set of positive rationals defines a Dedekind cut. Hence

VALUATION THEORY 245
sup Rv = inf /?„'. Now let vu v2 be positive elements of G and let
7»i/«i e RVl, m2/n2 e RVt where mi} »t- e P. Then nxv\ > mxu,
n2v2 > ^2« and «i»2£'i > m1n2u, n^n2v2 > nxm2u. Hence
»i»2(t>i + ^2) > (^iw2 + m2ny)u and so mx/ni + m2/n2 e
/?,,I+,,2. This implies that (yx + i^)* > 0i' + 02'- On the other
hand, a repetition of the argument just given shows that, if
»»i/«i e Rn' (that is, Wi^ < mxu) and m2/n2 e /?„/, then m^Ki +
m2/n2 e /?»,+„/. Since p11 = inf Rv', this implies that (&! + £/2)' <
fi" + v2v. Hence
C0) (»! + v2)" = Ol' + o2"
holds for Vi, v2 positive in G. We extend the mapping r/ to all of G
by defining 0' = 0 and ( — »)'= — o*1 if" y is positive. It is imme-
immediate that C0) holds if either vt > 0, v2 > 0, or vx < 0, v2 < 0.
Suppose Vi > 0 and v2 < 0. If v 1 + v2 > 0, we write vx — (vi +
^2) + ( — ^2) and obtain Vi1 = (vx + p2)' + ( — 02)' = (^i + ^ —
v2v. Then (y 1 + v2)'1 = t;^ + V2V. If »i + p2 < 0, then we write
— v2=—(v1 + v2) + Vx and obtain ( — v2)v = ( —(»i + ^2))' +
»i'. Thus —v2v= — (vi + v2)v + vi" and again C0) holds.
Similarly, C0) holds if v\ < 0 and v2 > 0. Thus v is a group
homomorphism of G into R. If v > 0, then 0' > 0; hence no
positive element is in the kernel of rj. It follows that the kernel is 0
and 17 is an isomorphism. Since positive elements are mapped into
positive elements by rj, 77 is an order isomorphism of G into R.
There are several observations which should be made on the
foregoing proof. In the first place it is clear from the definition of
the isomorphism t\ that uv = 1. We note next that rj is deter-
determined by this property, that is, if f is any order isomorphism of G
into R such that «f = 1, then f = rj. Thus let b > 0 and let
m/n, m, n positive integers, satisfy m/n > v*. Then ml > nvv
and muv > nvv, (mu)v > (nv)v. Hence mu > nv and re-tracing the
steps we obtain m/n > or. Similarly, m/n > t/ implies m/n >
vv. Since this holds for arbitrary rationals it follows that if = ^;
hence r\ — f. If ^ is any positive real number, then the mapping
x —» fix is an order preserving automorphism of R mapping
1 —> /3. It follows from this that there exists an order isomor-
isomorphism of G mapping the given positive element u into any positive
i8 in R. Moreover, such an isomorphism is unique.

246 VALUATION THEORY
A group of rank one is called discrete if it is order isomorphic
to the ordered group of integers (positivity as usual). We have
noted before (§ 5) that a subgroup of the multiplicative group of
positive reals is discrete if and only if it contains a largest element
< 1. This and Theorem 8 imply that an ordered group of rank
one is discrete if and only if it contains a least positive element.
EXERCISES
1. Let 2?<n) denote the additive group of »-tuples x = (fi, • • -,{"„) of real
numbers f». Define the set of positive elements of R^ by the condition that
x > 0 if the first non-zero fi is > 0. Show that this gives an ordered group.
Determine the isolated subgroups.
2. Call an ordered group G of rank n, n a positive integer, if n is the cardinal
number of the set of non-zero isolated subgroups. Show that any ordered group
of rank n is order isomorphic to a subgroup of the group i?(n) of ex. 1.
3. Call an ordered group G of rank n discrete if the factor groups of successive
isolated subgroups are all infinite cyclic groups. Show that any such group is
isomorphic to the subgroup of 2?(n) of «-tuples a = (ai, at, • ■ ■, an) such that
the at are integers.
11. Extension of homomorphisms and valuations. In this
section we shall prove a fundamental theorem on extension of a
homomorphism denned on a subring of a field. This result leads
to a general theorem on extension of valuations from a subfield to a
field. We prove first the following key lemma.
Lemma 1. Let o be a subring of a field * and let m be a proper
ideal in o. If a is a non-zero element of <£ and o[a] is the subring of <$
generated by o and a, then either mo[a], the ideal generated by m in
o[a], is proper in o[a] or tno[a~x] is proper in o[a-1].
Proof. Suppose the contrary: mo[a] = o[a], mo[a~x] = o[a-1].
Then 1 e mo[a] and 1 e mofa], so we have relations of the form:
C1) 1 = Mo«m + Mi«m-1 H h/«», m e m,
C2) 1 = voa-n + v1a-^-1) H h vn, Vj e m.
Since tn ^ o, we have m > 0 and n > 0 and we may assume m, n
are minimal for the relations C1) and C2). Also we may assume
m > n. Then C2) implies that am = voam~n + via.m-n+l -\
+ vnam; hence
C3) am(l - vn)

VALUATION THEORY 247
Multiplication of C1) by 1 — vn gives
C4) 1 - Vn = M0(l - VnHLm + ^A - Vn)*™-1
+ ■■■+ Mm(l - vn).
Hence, by C3),
1 - vn = Mo("o«m-" + • • • + i-.-ia"-1)
+ Mi(l - v^a™-1 +■■■+ nm(l - vn).
Since the m, vj e nt, this gives another relation like C1) with m re-
replaced by m — 1 contrary to the minimality of m. Hence the
proof is complete.
If 0* is a place which is a homomorphism of a subring o of the
field $ into the field A, then we shall say that 0* is ^-valued. Our
main result is an extension theorem for homomorphisms to places,
as follows.
Theorem 9. Let oo be a subring of afield $ and let 0q be a homo-
homomorphism of oo into an algebraically closed field Q. Then 0*^ can be
extended to an Qr-valued place & on $>.
Proof. We consider the collection of extensions 0*' of the homo-
homomorphism ^o where &' is a homomorphism into 0 of a subring o'
of $ containing oo. These can be partially ordered in the usual
manner: &' < 0i" if 0>" is an extension of &'. Then, as usual, we
can apply Zorn's lemma to obtain a maximal extension 0* which is
defined on a subring o of <£. The proof will be completed by show-
showing that o is a valuation ring. Then & will be an 0-valued place
for <£. Let m be the kernel of 0. Since 1 —> 1, m 9^ o. Since 0
has no zero-divisors 9^ 0, m is a prime ideal on o. Consequently,
the complementary set M of m in o is multiplicatively closed and
0 i M. Let o' be the subset of $ of elements of the form a^~l
where a, 0 e o and /3 e M. Then o' is a subring of $ containing o
and & can be extended to a homomorphism &' of o' into fi by de-
defining 0>'(al3-1) = ^(a)^(j8)-1(loflntrod.). Since ^ is maximal
we have o' = o. This implies that the image of o under 0* is a
subfield E of 0; for, if 0 ^ y = ^@), (8 e o, then CeM,so /3 e
o' = o and 7-1 = 0>(fi~1) is in the image of o. Now let a be any
element 5^ 0 of f>. We shall show that either a or a~l e 0, which
is what is needed to prove that 0 is a valuation ring and 0* is a place.

248 VALUATION THEORY
Now Lemma 1 shows that mo[a] C o[a] or mo[a *] c o[a *] and
we may as well assume the former. Then we shall show that & can
be extended to a homomorphism of o[a] into fi. This and the
maximality of 0* will imply that a e o. We consider the poly-
polynomial rings o[x] and EM, x an indeterminate, and we extend 0* to a
homomorphism of t>M onto EM sending x —> x. Let 21 be the
ideal of polynomials g(x) e o[x] such that g(a) = 0 and let 21' be its
image in EM under the extension of 0>. Since the homomorphism
ofoM is surjective, 21'is an ideal in EM- Also 21' c E[*]. Other-
wise, there exists a polynomial ^ #«#* e o[#] such that 2/3,-a1 = 0
0
and 2 ^(jS,)** = 1. Then ^(C0) = 1 and 0>(fii) = 0 if / > 0, so
o
1 — /30 e m and /9,- e tn for / > 0. Then the relation S/3,a* = 0
gives 1 = 1- Sfta1' = A - j80) + Z (-P<W. Since 1 - 0O,
i>0
j8,- e m, this implies that 1 e nto[a] contrary to hypothesis. Hence
we see that 21' is a proper ideal in EM and, since EM is a principal
ideal domain, 21' = (f(x)) where f{x) is either 0 or a polynomial of
positive degree. In the first case, we choose any element y in £2
and in the second case we choose y e 12 so that f(y) = 0. This
can be done since ft is algebraically closed. Now our choice of y
amounts to this: If g(x) is any polynomial in oM such that^(a) =
0, then g0* (y) = 0 for the image g& [x] in EM- Hence the exten-
extension theorem IV of the Introduction shows that & can be ex-
extended to a homomorphism of o[a] into A sending a into 7. This
completes the proof.
Suppose now that <po is a valuation of a subfield $0 of the field <£.
Let oo be the valuation ring of <p0, p0 the ideal of non-units, Uo the
multiplicative group of units of oo. We have seen that <p0 is
equivalent to the canonical valuation p0' into the group $0*/U0
where the positive elements of this group are the cosets 0OUO,
/30 9^ 0 in p0. We also have the canonical place ^V of $0 deter-
determined by oo. This is the homomorphism a0 —> a0 + Po of oo
into the residue field oo/po- We can imbed oo/po in an algebrai-
algebraically closed field Q. Then ^V can be considered as an fl-valued
place ^0 on $0- Since ft is algebraically closed, the extension
theorem states that ^0 can be extended to an fi-valued place 0* on

VALUATION THEORY 249
<&. Let o be the valuation ring in <i> on which & is defined and let p
be the ideal of non-units of o. Since 0* is an extension of &$, o 2 Oo
and since p and p0 are respectively the kernels of & and &Q, p 2 p0-
Hence we have o D $0 2 oo and p D $0 2 Po^ If /8 e o D $0 and
|8 ^ oo, then jS e p0 £ P5 but this implies that /3 ^ o. Hence o fl
<E>o = oo. Since p and p0 are the ideals of non-units of o and oo
respectively, the relation o fl <£0 = t>o implies p fl $oc p0.
Hence p D <J>0 = po and £/ fl $0 = t/0 where C/ is the set of units
of o. These relations imply that /30C/0 —*■ PoU, /30 e $0*, is an
order isomorphism of the ordered group $0*/U0 into $*/U
ordered by the set of elements /3C7, P e p. If we apply this iso-
isomorphism to the canonical valuation p0', w^ obtain an equivalent
valuation <p0" of $0 into the group $*/U. We also have the
canonical valuation <p' of $ into $*/U and the definitions show
that <p' is an extension of the valuation <p0". In this sense we have
obtained an "extension" of the given valuation of $0 to a valua-
valuation on $.
We shall be interested particularly in the case in which $ is
finite dimensional over <J>0 and the given valuation <p0 is of rank 1.
In the general case, if <p is a valuation of a field f> into V = (G, 0),
then the subgroup of G of values <p(a), a 9^ 0 in $, is called the
group of p. We shall need the following
Lemma 2. Z,^/ <p be a valuation of a field f>, f^ a subfield of
finite co-dimension in $. Then the value group of f> is order iso-
morphic to a subgroup of the value group of $0 {relative to the re-
restriction of <f>).
Proof. Let f e * and let a^ni + a2£n> -\ h <**£"* = 0 where
the at 5^ 0 in $0 and «i > w2 >•••>»*. As in the case of
non-archimedean real valuations, if ^(/3x) > ^(/8;), y 5^ 1, then
<pBj3i) = viPx). Hence our relation implies that there exist * < j
such that ?(«<$"*) = #>(«>rO- Then *>(€"'""') = ^(aya*). If
[<i>:$o] = », then we may assume that »,■ — «y < w; hence ^>(?)"!
is in the value group of <E>0. This shows that for any a in the value
group G of <i>, anl is in the value group Go of $0- On the other
hand, we have seen that a —> «"' is an order preserving iso-
isomorphism of G onto a subgroup. Hence G is order isomorphic to a
subgroup of Go.

250 VALUATION THEORY
This result and Theorem 8 imply that the value group of $ is of
rank 1 (discrete of rank 1) if and only if the same is true for the
value group of $0-
We shall now see how all of this applies to real valuations. Let
<p0 be a non-trivial non-archimedean valuation of a field *0 into the
non-negative reals and let $ be a finite dimensional extension of
<t>0. Then we know that <p0 = <Po'v where <p0' is the canonical
valuation of <£0 associated with the valuation ring oo of <p0 and r\ is
an order isomorphism of the value group Go' of <p0' into the posi-
positive reals P. Also we have just seen that we have a valuation ring
o of <£ and an order isomorphism f of Go' into the value group G' of
the canonical valuation <p' determined by o such that <p'(«o) =
(<p'J")(a0) for all a0 e *0- Since Go' is of rank 1 the same is true of
G' and consequently we have an order isomorphism X of G' into
P. Thus we have the following diagram of mappings:
Go'
<Po
<p'
where i is the inclusion mapping and the first rectangle is commu-
commutative: tip' = <po'f. Assume Go' ^ 1 and let 5 be some element ?*
1 in Go'. Then we can choose X so that 5rx = 571 and then we shall
have 7o'rx = To'" for all y0' e Go' (§ 10). This means that the
second rectangle in our diagram is also commutative. Then
<p = <p'\ is a real non-archimedean valuation which extends the
given valuation <p0 on $0; for if aQ e $0) then <po(«o) = (^o'n)(«o)
= (WfX)(a0) = (?'X)Oo) = «p(«o)- If Go' = 1, then Lemma 2
shows that necessarily G' = 1. Then rj and X are unique and
commutativity holds. This case is, of course, trivial at the out-
outset, since it is the one in which <p0 is a trivial valuation. We have
therefore proved the following
Theorem 10. Let <p0 be a non-archimedean real valuation on a
field 4>o and let $ be a finite dimensional extension field of <I>. Then
there exists a real valuation on $ which is an extension of <p.

VALUATION THEORY 251
12. Application of the extension theorem: Hilbert Nullstel-
lensatz. Before continuing our study of valuations we digress
slightly to take up some important applications of the homomor-
phism extension theorem (Th. 9). The first of these, Hilbert's
Nullstellensatz, plays an important role in algebraic geometry.
We shall give it in its original ideal-theoretic form.
We consider a polynomial algebra $[xly #2, • • •, xn] in indeter-
minates Xi over a field $. Let fl be the algebraic closure of #. If
f{x\, • • •, xn) e $[x1} • • •, xn] and the £t- are elements of 0 such that
/(£i> • • • > in) =0, then we shall call (j-u ■ ■ •, £„) an {algebraic) zero
of/(xi, • • •, xn). If S is a set of polynomials contained in $[xi,
• ■ ■ j *n]} then we define a zero of S to be an »-tuple (£1} • ■ •, £„),
£,- e fl, which is a zero for every f eS. Our main result concerns
the zeros of a proper prime ideal $ in $[#1, • • •, xn]. This is the
following
Theorem 11. Let ^ be a prime ideal in $[xx, • • ■, xn], 4? afield,
and suppose ty ^ A) (= *|>i, • • •, xn]). Let g{xu ■ ■ -,xn) be a
polynomial not contained in %. Then there exist £t- in the algebraic
closure Q, of & such that (£1, • • •, £„) is a zero for <$ and is not a zero
forg(xu ■■•,xn).
Proof. Since $ 5^ A), *[#i, • • -,^™]/^ is an algebra over <!>
which is 5^ 0 and this is generated over f> by the cosets %• = #,•
+ % i = 1,2, ••■,». Also #[71, •••,7«] = *[*i, •••,*«]/?
is an integral domain so this can be imbedded in its field of frac-
fractions P = $G1, 72, • •',■¥")• Suppose first that all the y{ are
algebraic. Then P is an algebraic extension of $ so we have an
isomorphism of P/<£> into the algebraic closure Q/f>. Suppose
ft —> li in this isomorphism. Then if/(#i, ■ • •, xn) e ^P,/Gi, • • •,
7n) = 0 and so /(&, ••-,£„) =0. Hence (fi, £2, • • •, fn) is a
zero of $. On the other hand, g(xu ■ ■ ■, xn) i ty so #G1, • • •, yn)
9^ 0; hence g(^i, •••,?„) 9^ 0. This proves the theorem in this
case. Next assume that not all the 7,- are algebraic. We may
suppose that the'7's are ordered so that {71} y2, ■ ■ •, yr} (r > 1) is
a transcendency basis for P/$. Since g(x 1, • • •, xn) $ ^, ^G1, • • •,
7n) 7^ 0 in P so £G1, • • •, 7n)-1 exists in P. This element and the
elements 7r+i, • • •, 7n are algebraic over $G1, • • •, yr) and so

252 VALUATION THEORY
they satisfy algebraic equations of the form
C5) «oGi, • • -,7r)*m + «iGi, • • •, Yr)*"*-1
H \-am(yu •••,7r) = 0
where the «,• are polynomials in the y,-, j — 1, • • • ,r, and #oGi>
■■■,yr)^0. For each 7r+i,---,7n and g(yu ■ • •, yn)~l we
choose such an equation and we let a(yu • ■ •, yT) be the product
of the leading coefficients of these equations. Since a(xx, • • •, xr)
7^ 0, we may choose £i, • • •, £r in the infinite field 0 so that
«(£i> • • •, ?r) 5^ 0 (Vol. I, p. 112). Since the yh 1 < j < r, are
algebraically independent, we have an algebra homomorphism of
$[7i> • ■ • j 7r] into n/$ such that y, —> £/. By the extension
theorem (Theorem 9) this homomorphism can be extended to an
Q-valued place 0* on P. Since 0* is an extension of an algebra
homomorphism, 0* is the identity on $ and so 0* is an algebra
homomorphism into fl/$. We note next that the y^, r + \ <
k < n, are in the valuation ring o of 0*. Otherwise, 0>{yh-1) = 0.
On the other hand, we have an equation of the form
•, 7r) + «iGi, •■•, 7rO
~1
fc
H h am(yu ■■■, 7rOfc~™ = 0,
and applying^3, we obtain «0(£i, • • •, fr) = •^2>(«oGi, ■ • •, 7r)) = 0.
This contradicts the facts that a(%u ■■■,&) ^0 and a(yu • • -,
yr) has ao(yi, • ■ •, yr) as a factor. A similar argument shows that
0>(g(yi, ■ ■ -, 7-)) ^ 0. Now let £* = ^Gfc), r + 1 < * < n.
Then we assert that (£x, £2, • • •, ^n) satisfies the conditions of the
theorem. In the first place, if/(*i, • • ■, xn) e $, then/G!, • • •,
yn) = 0 and applying 0* we have/(£i, • • •, £n) =0. Next we see
that j(fc, ••-,{.) = ^(fGi, •' •, 7-)) ^ 0.
The Hilbert Nullstellensatz is the extension of Theorem 11 from
prime ideals to arbitrary ideals in $[xi, • • •, xn]. To obtain this
we need a characterization of the (nil)radical of an ideal of a
commutative ring (Vol. I, p. 173). The result we require is that,
if 21 is an ideal in a commutative ring o, then the radical 31B1) is the
intersection D <$ of the prime ideals ty containing 21. If o is
Noetherian, this result is an easy consequence of the decomposi-
decomposition theorem for ideals into primary ideals (Vol. I, p. 176, ex. 2, p.

VALUATION THEORY 253
181). Although this is all we need here it is of interest to estab-
establish this result in the general case. We prove first the following
Lemma 1. Let o be a commutative ring, 21 an ideal in o and S a
non-vacuous multiplicatively closed subset of o such that 21 D S = 0.
Then there exists a prime ideal ^ in o such that ty 3 21 and ty D S
= 0.
Proof. Let U be the collection of ideals 58 in o such that: 1.
33 2 21, 2. S3 0 S = 0. Then U is non-vacuous since 21 e U. We
order the elements of U by inclusion. Let V be a linearly ordered
subset of U and let @ = U S3. Then S f] S = 0 and g 2 31.
SSeV
Moreover, it is easy to check that £ is an ideal. Hence £ e U and
S is an upper bound for the set V. Thus U is an inductive set and
so we can apply Zorn's lemma to conclude that U contains a
maximal element ty. Let a,-, i = 1, 2, be elements of o not con-
contained in $. Then the ideal 21, generated by «,• and $ properly
contains ty and contains 21. Since $ is maximal in U, it follows
that 21,- ^ t/ which means that 2T,- fl <? ^ 0. Let s{ e 2tt- fl S. If
we take into account the form of the elements of 2lt- we see that
Si = Xiai + pi where #,• e o and pi e 'ip. Then
C6) s — SiS2 = XiX2aia2 + p
where p e $. Since «S" is multiplicatively closed, j e <S\ If «i«2 e $,
then C6) implies that s e $ contrary to $ fl 61 = 0. Hence we see
that «i«2 tf ^ so we have shown that ax ^ $, «2 tf $ implies «i«2 ^ $•
Hence ^5 is a prime ideal satisfying the required conditions.
We can now prove
Theorem 12. Let 21 be an ideal in the commutative ring o. Then
the radical 9?Bl) = f\ ty the intersection of the prime ideals ty con-
containing 21.
Proof. Let a e 9?Bl) and let $ be a prime ideal containing 21. A
suitable power an e 21 so an e ^8. Since ty is prime, this implies
that a e $. Hence 91(81) C <p and 91(81) C D $ for the prime ideals
<P containing 21. Next let a i 9JBl) and let .? = {an,n = 1,2,
• • • }. Then S D 21 = 0 and £ is multiplicatively closed. Hence
the lemma implies that there exists a prime ideal ^J containing 21

254 VALUATION THEORY
such that a $ ty. Hence a is not contained in the intersection of the
prime ideals containing 21. Thus we have proved that D ty c
5R(Sl) and so by the earlier inclusion, 9?Bl) = D $.
Theorems 11 and 12 imply the
Hilbert Nullstellensatz. Let 21 be an ideal in the polynomial
algebra Q\x\, x2, ■ • •, xn], $ a field, Xi indeterminates, and let 0 be
the algebraic closure of <£. Then a polynomial g{x\, • • •, xn) e 91B0
if and only ifgfa, •••,£„) = Ofor every zero fa, • • -,£„), £,• e 0, of
the ideal 21.
Proof. Let V denote the set of zeros fa, ■■-, £„), £* e B, of 21.
Suppose g(x1} ■ • •, xn) e 9?Bt). Then gn e 21 for some positive
integer n. Hence gfa, • • •, £n)" = 0 for every fa) e V and
gfa, ■ ■ ■, ^n) = 0 for every (£,-) e F. Conversely, let^(xi, • • •, xn)
be a polynomial such that ^(^i, • • •, £„) = 0 for every fa) e V.
Let $ be a prime ideal containing 21 and let W be the set of zeros of
% Since $ 3 21, W C F and consequently £(£l5 ••-,{„) = 0 for
every (£,-) e /iF. It follows from Theorem 11 that ^(^i, • • •, xn) e
^3. Thus g is contained in every prime ideal containing 21 and so,
by Theorem 12, g e 9JBt). This completes the proof.
We shall give next an application of the existence of an alge-
algebraic zero of a prime ideal to a theorem on finite generation of a
field. We recall that we saw long ago (Lemma 2, § 1.5) that, if
7i> f2, ■ • •, 7n are algebraic over <£, then the field P = *Gi, y2,
■ ■ • > 7n) coincides with the algebra $[71, y2, • • •, 7n] generated
by the 7,-. We can now prove the following converse of this result.
Theorem 13. If the algebra P = $[71, 72, • • •, yn] over $ gener-
generated by the 7,- is afield, then the 7,- are algebraic over f>.
Proof. Let $[#1, x2, • ■ •, xn] be the polynomial algebra over $ in
indeterminates x,- and consider the homomorphism of this algebra
onto P/$ mapping *,- —> yiy 1 < i < n. Let ^3 be the kernel of
the homomorphism. Since P is a field, ^ is a maximal ideal. If 12
is the algebraic closure of <£, then we have seen that we can find
(!i, &,•••, £») in 0 such that ffa, ••■,£„)= 0 for every /e*.
By IV of the Introduction we have a homomorphism of P =
$[7i, 72, • • •, yn] over * onto $[£l5 £2, • • •, £„] such that 7,- -> £,-,
!</<». Since P is a field, this homomorphism is an isomor-

VALUATION THEORY 255
phism. Since £,■ is algebraic, it follows that y,- is algebraic, 1 <
* < n.
EXERCISE
1. Let P = $[71,72, • • •, 7n] be a finitely generated commutative algebra
over <$ and let 9i be the ideal of nilpotent elements. Show that 9? is the inter-
intersection of the maximal ideals of fy.
13. Application of the extension theorem: integral closure. We
shall apply the extension theorem next to obtain an important
characterization of the integral closure of a subring of a field. Let
g be a subring of the field <£. We recall that an element a e <£> is
called integral over g or g-integral if there exists a polynomial
f{x) eq[x] with leading coefficient 1 such that /(a) = 0. The set
® of elements of # which are g-integral is called the integral
closure of g in <f>. We shall characterize this set. In the proof we
shall need the following
Lemma 1. If o is a commutative ring {with an identity 1), any
proper ideal % of o can be imbedded in a maximal ideal.
Proof. The proof is obtained as a special case of the argument
in the proof of Lemma 1 of § 12. We let S = {1}, so S is multipli-
catively closed and S f\ 2t = 0. Let U be the set of ideals 93 such
that 93 2 21 and 93 is proper (so that 93 D S = 0). Then U con-
contains a maximal element $. It is immediate that ty is a maximal
ideal containing 21.
Theorem 14 (Rrull). Let $be a subring containing 1 in a field <£.
Then the integral closure ®of$in$ is f) o, the intersection of all the
valuation rings of <J> which contain g.
Proof. Let a e ® so that we have a relation an + Tia"
+ • • • + Tn = 0, « > 1, 7i e g. Let <p be a valuation whose valua-
valuation ring o contains g. If a i o, then ^(a"-1) < 1. But 1 =
— 7i«-1 — • • • — yna~n and <p(yi) < 1. Hence every <p(yia~l) <
1 and this is impossible since the relation gives 1 = <p(l) <
max (<p(yia~1)) < 1. Hence a e o so we have proved that ® is con-
contained in H o for the valuation rings containing g. Next suppose
a tf ®. Then a is not a unit in the ring gfa], since otherwise
its inverse a = 7O1 + 7i«-1 + • • • + yn-i<x~<'n~1\ 7, eg, and hence
an = To"" + 7i«"~2 + • —I- 7»-i so a e ®. Since a" is not

256 VALUATION THEORY
a unit in g[a *], the principal ideal a \[a x] is properly contained
in gfa]. By Lemma 1 there exists a maximal ideal nt in gfa]
containing a^**]. Then g[a-1]/m is a field which can be im-
imbedded in an algebraically closed field Q. The canonical homo-
morphism of g[a-1] onto g[a-1]/m can be considered as a homo-
morphism of gla] into Q. The extension theorem gives an 12-
valued place 0* whose valuation ring o contains gfa]. The
ideal p of non-units of o contains m, hence, a. It follows that
a $ o. Thus a $ ® implies aflflo for the valuation rings o con-
containing g. We therefore have ® = fl o and the proof is complete.
The subring g is called integrally closed in <& if ® = g. Then
we have the following
Corollary. If g is a subring of <$, then the set ® of ^-integral ele-
elements is a subring of <£ containing g and ® is integrally closed in <i>.
Proof. The first statement is clear since ® is an intersection of
subrings of <£ and since ® certainly contains g, Also the set of ®-
integral elements is the intersection D o for the valuation rings
containing ® and hence containing g. On the other hand, if o is a
valuation ring containing g, then o3®. Hence the intersection
of the valuation rings containing ® is the same as that of the
valuation rings containing g, so this is ®. Hence ® is integrally
closed.
EXERCISES
1. (Artin). Let g be a subring of a field and let ai, as, ■ • •, acr be elements of $.
Suppose that for each ; there exists a positive integer »< such that otini =
Pi(ai, at, • ■ ■, ar) where Pi is a polynomial of total degree < «j. Show that every
ai is g-integral.
2. (Artin). Let g be as in ex. 1 and let 3JI be a subring of $ which is a finitely
generated g-module. Show that every element of 9ft is g-integral (cf. Vol. I, p.
182).
3. A commutative integral domain g is called integrally closed if it is integrally
closed in its field of fractions. Show that if g is Gaussian (that is, unique facto-
factorization holds), then g is integrally closed.
4. (Cohn). Show that a subalgebra 21 of $[*■], $ a field, x an indeterminate,
has a single generator if and only if 31 is integrally closed. (Hint: Use Luroth's
theorem and Th. 14.)
14. Finite dimensional extensions of complete fields. In the
remainder of this chapter we return to the consideration of real
valuations (archimedean as well as non-archimedean). We shall

VALUATION THEORY 257
begin by considering the problem of extending a valuation on a
complete field $ to a finite dimensional extension field. Our first
objective is to prove uniqueness of the extension. For this we re-
require
Lemma 1. Let f> be complete with respect to a non-trivial real
valuation tp and let P be an extension field of $ with a valuation <p
which is an extension of that of <$. Suppose uly u2, ■ • • ,ur are ele-
elements of P which are ^-independent. Then a sequence {an}, an =
r
^2 aniUi, ani e *, is a Cauchy sequence in P if and only if the r
sequences {«„,-}, i = 1,2, ■ • -,r, are Cauchy sequences in <£>.
Proof. It is immediate that, if the \ani\ are Cauchy sequences,
then so is \an). Conversely, suppose {«„} is Cauchy. If r = 1,
then it is clear that \an\\ is Cauchy. We shall now prove our
assertion for arbitrary r by induction. If the sequence \anT\ is a
Cauchy sequence, then the sequence {bn},bn = an — anrur is a
r-1
Cauchy sequence. Since bn = 2 UnjUj the required result follows
i
by induction. The proof will now be completed by showing that
the assumption that \anr} is not Cauchy leads to a contradiction.
We make this assumption. Then there exists a real e > 0 such
that for any positive N there exist p, q > N such that <p(apr —
aqr) > e. Hence there exist pairs of positive integers {pk, q^), pi
< P2 < ■ ■ ■, <Ji < <J2 < • • • such that <p(aPkr — aqkr) > «. Then
(cw — "str) exists and we can form the sequence {b^} where
C7) h = (aPkr - a^r)^ - agk).
We have <p{aPkr — "g^) < - and [aPh — aQk} is a null sequence.
e r-i
Hence {bk} is a null sequence. On the other hand, bk = ]C Pkjiij
j=\
+ ur and this implies that, if ck = 2^,«y, then {ck} is a Cauchy
sequence. Then the r — 1 sequences {/3&y}, j = 1, 2, • • •, r — 1,
are Cauchy sequences. Since $ is complete, lim Ck)- = jfy exists.
r—1 r—1
Since lim bk = 0 we get from £* = 23 £*y«/ + «r> 0 = 53
i i

258 VALUATION THEORY
+ «,. This contradicts the linear independence of the «'s and
completes the proof.
We note two important consequences of this lemma: A) If {«„}
is a null sequence, then all the sequences {ani} are null sequences.
B) If [P:f>] < oo3 then P is complete. The first of these is clear
since the {«„,} are Cauchy sequences. Hence lim ant- = a,- exists
and "liociUi = 0. Hence every a, = 0 by the linear indepen-
independence of the «,-. To prove the second statement we suppose
that («!, u2, • • •, «r) is a basis. Then if [an) is Cauchy, every
{«„,} is Cauchy and so lim ani = a,- exists and lim an = 2aj#t-.
We can now prove
Theorem 15. Let P be a finite dimensional extension field of a
field which is complete with respect to a non-trivial real valuation <p.
Then if <p can be extended to a real valuation of P, this valuation is
unique and is given by the formula
C8) v(p) = ^(TVpi^p)I'", n = [P:*].
Proof. Assume the extension <p exists and suppose there exists a
p e P such that C8) does not hold. Then <p(pn) ^ <p(N(p)), so
p^O and either <p{pn) < <p(N(p)) or <p(pn) > <p(N(p)). By re-
replacing p by p~1, if necessary, we may suppose <p(pn) < <p(N(p)).
Set <r = pnN(p)-\ Then v(o) < 1 and N(<r) = iV(pnOV(p)~"
= 1. Since <p(a) < 1, we have lim ak = 0. If («i, «2, •••,«») is a
n
basis and <rk = 22 «ti«i> then lim ak = 0 implies that lim au = 0
for every i. Since the norm of an element e = XjiUi, yt e 4>, is a
homogeneous polynomial of the «-th degree in the 7* with fixed
coefficients, it is clear that lim aki = 0 for every i implies that
limAV) = 0. This contradicts AV) = N(<r)k = 1.
We have seen before that any non-archimedean real valuation
on a subfield can be extended. Hence in the non-archimedean
case the formula C8) provides a valuation for the finite dimen-
dimensional extension P. It remains to consider the archimedean case.
The extension theorem in this case will be obtained by a complete
determination of the fields which are complete with respect to an
archimedean real valuation. We shall show that the only such
fields are the field of real numbers and the field of complex numbers.

VALUATION THEORY 259
Lemma 2. Let P be a quadratic extension of a field <£> which is
complete with respect to a real archimedean valuation <p. Then <p can
be extended to a valuation of P.
Proof. We recall that the existence of an archimedean valua-
valuation implies that the characteristic is 0. Hence P is Galois over <l>.
Let a —* a be the automorphism of P/$ which is not the identity.
Then the trace and norm of a e P are T{a) = a + a, N(a) = act
and we have a2 — T{a)a + N{a) = 0 for any a e P. We shall
show that <p(a) = <p{N{a)YA defines a valuation of P. If a e $,
N(a) = a2. This implies that the mapping <p defined on P is an
extension of the <p which is given on <$. We evidently have <p(a) =
0 only if a = 0 and the multiplicative property of the norm implies
that <p(a/3) = <p(a)<p(fi). Hence all one needs to show is: <p(a + /3)
< <p(a) + <p{$)- This will follow if we can show that <p(a + 1) <
<p(a) + 1; for, <p(a + 0) < <p(a) + <p@) is clear if 0 = 0, and if
|8 ?* 0, then
Hence, if «?(a0 + 1) < *5(a0-1) + 1, then
Now fpia. + 1) < <p(a) + 1 holds if a e # so we suppose that
a ^ <3>. Then P = $(a) and #2 — T(a)^ + N(a) is the min-
minimum polynomial of a and iV(a +1) = (a+l)(a+l) = aa.
+ a + a + 1 = N(a) + T(a) + 1. Hence *>(a + 1) < <p(a)
+ 1 is equivalent to <p(a + IJ < ^(aJ + 2<p(a) + 1 and to
C9) *A + r(a) + N(a)) < 1 + 2v{N{a))* + <p(N(a)).
If we use the addition property of <p in $ it is clear that C9) will
hold if <p(T(a)) < 2<p(N(a))*. Hence we suppose that <p(T(a))
> 2<p{N(a))\ or <p(T(a)J > 4<p(N{a)). We write a = T(a),
b = N(a), so we are assuming p(«J > 4^(^). We shall show
that this implies that ae$ which will contradict our assumption.
Hence the proof will be completed by proving
Lemma 3. Let $ be a field which is complete relative to a real
valuation <p and let x2 — ax + b = 0 be an equation with coefficients
«, b in <£> such that <p(aJ > 4<p(b). Then the equation has roots in $.

260 VALUATION THEORY
Proof. A non-zero root a of this equation will be a root of
a = a — ba~1. We shall obtain such a root as a limit of a
sequence {an} where an is denned recursively by ax = \a, an+i =
a — ban~l. We show first that no an = 0 so the definition works
for all n. We have <p(^i) = h<f(a) > 0 ar>d we mav suppose that
<p(an) > hvia). Then
<p(an+1) = <p(a - ban'1) > <p(a) - <p(b)<p(an)
~l
Hence <p(an) > %<p(a) > 0 holds for all n = 1, 2, 3, ■ • • and every
an -^ 0. Now we have an+2 - an+1 = ban+i ~lan~1{an+-i. — an);
and <p(an+1)~1<p(an)~1 < A<p{a)~2\ hence
D0) <p{an+2 - an+1) <
If we set r — 4<p(b)/<p(aJ we have 0 < r < 1 and we may iterate
D0) to obtain <p(an+2 — an+i) < rnc where c = <p(a2 — a{). This
inequality implies easily that {«„} is a Cauchy sequence. Hence
a = lim an exists and since <p(an) > %<p(a) > 0, a 5* 0. Hence
the recursion formula an+i = a — ban~x gives a. = a — ba~x so
a2 - aa + b = 0.
We are now ready to prove
Theorem 16 (Ostrowski). The only fields which are complete
relative to a real archimedean valuation are the field of real numbers
and the field of complex numbers.
Proof. Let $ be complete relative to the archimedean valua-
valuation <p. Then <t> is of characteristic 0 and so it contains the ra-
rationals. Since any real archimedean valuation of the rationals is
equivalent to the absolute value valuation and f> is complete, it is
clear that $ contains the field of real numbers. If $ contains an
element i such that i2 = — 1, then * contains the field C of com-
complex numbers. Otherwise, we adjoin *" to $ and obtain $(/) which
contains C. By Lemma 2, <p can be extended to a real valuation
of $(/')• Also we have seen that $(«) is complete. The theorem
will therefore follow if we can show that, if <i> is complete with

VALUATION THEORY 261
respect to an archimedean valuation and $2C, then <£ = C.
Since the restriction of <p to the real subfield (the completion of
the rationals) is equivalent to the absolute value valuation,
Theorem 15 shows that <p is equivalent to the absolute value
valuation on C.
Now suppose $dC and let a e <£>, ^ C. Let r — inf <p(a — c)
for c e C. Then we claim that there exists a c0 e C such that
<p(a — c0) = r. First, it is clear that r = inf <p(a — c) for all c
such that (p(a — c) < r + 1 and, if cx and c2 are two complex
numbers satisfying <p(a — Ci) < r + 1, <p(a — <r2) < r + 1, then
¥>(fi — r2) < 2r + 2. Hence the c satisfying <p(a — c) < r + 1
form a closed and bounded set in C. Since <p(a — c) is a con-
continuous function of c it is clear that there exists a c0 such that
<p(a — c0) = r. Since o^Cwe have r > 0. If we replace a by
a — c0 we may assume that c0 = 0. Then we have <f>(a) = r > 0
and <p(a — c) > r for every feC. We shall now show that we
have p(a — c) = r for every complex c with ip{c) < r. To see
this we let n be any positive integer and we consider an — cn =
(a — c)(a — ec) ■ • ■ (a — en~lc) where e is a primitive w-th root
of 1 contained in C. Then
<p{a — c)<p(a — ec) • ■ ■ v(a — en~1c)
= <p(an - cn) < <p(a)n + <p(c)\
Since <p(a — ekc) > r, we obtain
Hence
so if <p(c) < r, then lim ( 1 + ( ) ) = 1 gives the asserted re-
relation <p(a — c) = r. We can now replace a by a. — c for any c
such that <p(c) < r and we obtain <p{a. — 2c) = r. If we repeat
this process we obtain <p(<x — nc) = r for all n = 1,2, • ■ • and all c
such that ip{c) < r. This amounts to saying that <p(a. — c) = r if
(p(c) < nr and, since n is arbitrary, we have <p(a — c) = r for all

262 VALUATION THEORY
c e C. Then if cx, c2 e C, <p(ci — t2) < v»(« — ^1) + *»(« — ^2) =
2r which is absurd since <p is equivalent to the absolute value
valuation on C. Thus we must have C = <£> and the theorem is
proved.
The extension theorem for valuations for complete fields relative
to an archimedean valuation becomes trivial in view of Ostrowski's
theorem. If $ is complete relative to an archimedean valuation,
then €> is either the reals or the complexes. In the first case, the
only finite dimensional extensions are $ and the field of complex
numbers. In the second, the only possibility is #. In all cases the
extension theorem is clear. If we combine this with the earlier
results we obtain the following
Theorem 17. If $ is complete relative to a real valuation <p and P
is a finite dimensional extension of <£>, then the valuation can be ex-
extended in one and only one way to P. The extension is given by the
formula C8). Moreover, P is complete relative to its valuation.
15. Extension of real valuations to finite dimensional extension
fields. We now take up the problem of determining all the ex-
extensions of a real valuation defined in a field $ to a finite dimen-
dimensional extension field P/$. The case in which 4> is complete has
been treated in the last section. We shall use the result obtained
there to treat the general case. Let $ be the completion of <£ rela-
relative to <p and denote the valuation in $ which extends that in <p
by <p. Now suppose (E, s, t) is a field composite of P/# and
E is a field over <l>, s and t are isomorphisms of P/$ and
respectively into E/#, and E is generated by P* and $K Since
[P :<$] = «< 00 we have [E:$'] < » < °o. The valuation <p in $
can be transferred to 3>' by defining ?«(«') = ?(a), a. e $. Clearly
q>t coincides with <p on $. Since $ is complete relative to £, it is
clear that $' is complete relative to pt. Since E is a finite dimen-
dimensional extension of$l, the real valuation $t has a unique extension
to a real valuation ^ on E. Let ^, be the restriction of $ to the
subfield P* and transfer \f/, to P by ^(p) = ^,(p*). Then it is clear
that ^ is a real valuation on P which extends <p.
Thus we have a process for associating with every composite
(E, s, t) of P and $ a real valuation ^ on P which extends <p. We
shall show that this correspondence between the composites and

VALUATION THEORY 263
the extensions of the valuations is 1-1 and surjective, if we identify
equivalent composites. First, suppose the two composites
(Ei, ji, ^i) and (E2, s2, t2) of P/<t> and $/<!> are equivalent. Then
we have an isomorphism u of Ei/$ onto E2/<£ such that a'1" =
a'2, 5 e $ and p" = p82, p e P. For the valuations on <$(l and $'2
we have ^2(a<2) = *(a) = ^(a'O- Hence <pB(a<>") = ^(a*1).
Let ^i and i?2 be the valuations of Et and E2 respectively, which
extend <ptl and £<„. Now ^'(yi") — !?iGi)> Yi e E1} defines a real
valuation on E2 such that for ahu(e $h) we have ^2'(a'lU) =
#i(a'O = ^(i(«(l) = !Ph{oitlU). Thus ^2' is an extension of the
valuation <p<2 on <5'2. Since <$'2 is complete, this extension is unique
and so it coincides with $2. Hence we have 1/1G1) = $2G1") for
every 71 e Ex. This implies that the restrictions ^/H and ^»2 to
P" and P82 satisfy ^ai(p81) = ^S2(pSi") = iWp*1)- Hence the cor-
corresponding valuations ipi and ^2 on P satisfy ^i(p) = ^8l(p!>1) =
yi/»i{pH) ~ 1^2(p)- Thus equivalent composites give the same val-
valuation.
Conversely, assume ^i(p) = ^2(p) for the valuations fi, ^2 of P
determined by the composites (Ex, j1} t{) and (E2, s2, t2). Then
we have \p8l(pSi) = ^«j(p")j P e P. Next we observe that Et-, / = 1,2,
is the closure of P8i in the topology defined by the valuation in E,-.
Clearly, this closure contains $'< and P% hence E,-, since this field
is generated by $'* and P8*'. It is now clear that E,- is a completion
of P8' relative to the valuation ^,,. in the sense of Definition 5.
Consequently, by Theorem 6, the isomorphism p'1 —> p82 of P*1
onto P has a unique extension to an isometric isomorphism u of
Ei onto E2. We have ^1G1) = ^2Gi") for the valuations f,- of Et-
and p81" = p. Since &* is the closure of $ in Et- and since u is the
identity on <£, it is clear that u maps <5'1 onto $'2. Hence the re-
restriction of u to $*' is an isometric isomorphism which is the
identity on <J>. On the other hand, the mapping a'1 —» a'2 has
these same properties since ^(a*1) = p(a) = ^i2(a'2). Hence by
Theorem 6, a'1 —* a'2 coincides with the mapping u. Hence we
have a'1" = a'2 and so (Ex, s1} /x) and (E2, s2, t2) are equivalent.
It remains to show that every valuation \p on P which is an ex-
extension of <p can be obtained from a composite in the manner
indicated. To see this we let E be the completion of P relative to
\j/ and let s denote the canonical imbedding (isomorphism) of P

264 VALUATION THEORY
into E. Now we have an isomorphism t of the completion $ into
the closure of <£ in E. The subfield of E generated by <bl and Ps
is a finite dimensional extension of $', so it is complete relative to
the valuation obtained from E. It follows that this coincides
with E. Hence we have a composite (E, s, t) and one checks that
the valuation of P obtained from this composite is the given
valuation \p. We can now state the following
Theorem 18. Let P be a finite dimensional extension field of a
field <£ with a real valuation <p and let <5 be the completion of $. Then
the extensions of ip to valuations ty in P are in 1-1 correspondence
with the equivalence classes of composites (E, s, t) of P/$ and <5.
In § 1.16 we have established a 1-1 correspondence between the
equivalence classes of composites (E, s, t) and the maximal ideals
of the algebra # <g>$ P. We have seen that, if 3s is a maximal ideal
in <5 ® P, then this determines a composite whose field is E =
($ <g> P)/3f. Distinct 3 give inequivalent composites and every
composite is equivalent to one obtained from a maximal ideal $.
We have seen also that the number of maximal ideals is finite and,
if 3i, $2, ■ • •, Sh are the distinct maximal ideals in $ <g> P and
9? = D Si, then (<S <g> P)/9J = Ej © E2 ©• • -0 Eh where Ey 2*
($ <g> P)/3y. The field Ey is the completion of P relative to a
valuation #,-. We shall call [E>: <5] = »,- the local dimensionality of
P determined by #,-. Then we have
D1) = [P:$] - [«:*]
= « — [JR:f] < n.
Moreover, S»y = » if and only if 9i = 0. Since $ <g> P can be
considered a finite dimensional algebra over $, VII of the Intro-
Introduction implies that C> ® P)/-3 is a field if and only if it is an
integral domain. Hence 3 is maximal in ($ <8> P) if and only if 31
is prime. Hence, by Theorem 12, D 3/ = $R is the radical of the
algebra $ ® P, that is, 9? is the set of nilpotent elements of $ ® P
and 9? = 0 if and only if $ <g> P has no non-zero nilpotent ele-
elements. If P is separable over $ we have $ ®* P = ~Ei © E2
©■ • •© EA where the Ey/$ are fields which can be determined ex-

VALUATION THEORY 265
plicitly from the minimum polynomial/(#) of a primitive element
6 of P over <£ (§ 1.16). Since a direct sum of fields contains no
non-zero nilpotents, it is clear that <S <g> P has zero radical dt if P is
separable over 4>. Consequently, the formula D1) becomes
D2) n = 2«,-
in this case.
EXERCISE
1. Determine the number of extensions of the p-adic valuation of the rationals
to the cyclotomic field of 5-th roots of 1 for p — 3, 5, 11.
16. Ramification index and residue degree. Let * be a field
with a non-trivial non-archimedean real valuation <p and let y
be the value group, o/p the residue field of <£> relative to <p (§ 5).
Suppose P is a finite dimensional extension field, ^ an extension of
the valuation <p to P, F the corresponding value group, D/ty the
residue field of P. Since £) and ty are the sets of elements p satis-
satisfying i[/(p) < 1, yp{p) < 1 respectively it is clear that o C £) and
p = o fl $. Hence we can identify the residue field o/p with the
subfield (o + ty/ty of the residue field O/ty. In this way we can
consider the dimensionality [O/'iP'.o/p] = / which we shall call the
residue degree of the valuation ^ of the extension P/f>. It is clear
also that the value group 7 is a subgroup of F and we shall call the
index e of 7 in F the ramification index of \p. If p e P then we can
multiply p by a suitable non-zero element of p to obtain an
element of 1JJ. Hence we can choose elements of $ as representa-
representatives of the cosets of 7 in F. Both the residue degree and the
ramification index are finite and, in fact, we have
Lemma 1. ef < n = [P :<£>].
Proof. Let pu p2, • • ■, p/t be elements of £> which are linearly
independent over (o + ^3)/^3. Thus if at- are elements of o and
~ZaiPi e $, then every on e p. Let tu x2, • ■ •, irei be elements of ty
such that the cosets ^(irxO, • ■ •, \f/(irei)y are distinct in r/7. We
assert that the e^fi elements p,Xj- are ^-independent. Thus sup-
suppose 2aijPiiTj = 0 where the an e $. We shall show first that, if
the «i e * and 2a,p» 5^ 0, then ^-(Saj-pi) e 7. If ~2,aiPi 5^ 0, then
some ai 7* 0 and we may assume that 0 7^ *K«i) ^ ^(«<)- Then

266 VALUATION THEORY
if ft = «,«! 1, ^(ft) < 1, so ft is in the valuation ring o of <p. We
have 2a,p,- = a^Sftp,). Also ^(Sftp,) < 1 and since ft = 1 and
the ft e o, it is clear that ^Bftpt) < 1 would contradict the
linear independence of the pt- over (o + $)/$. Hence we see that
^Bftp.) = I and so fBaiPi) = ^(aOWSftp,)) = ^(«x) e y. We
now return to our relation Sa^-piTTy = 0, a,y e «$. Assume there
exists a j so that ^(Sa,yp,) ?^ 0. Then we have distinct y, say
j = 1, 2, so that^CSaixp.xj) = $Bai2PiTr2) ^ 0 (ex. 2, § 1). Then
ipi)^Gri) = ^(Sa^Pi)^^) ^ 0 and the cosets 7^G1"!) =
) by the result we have proved. This contradicts the choice
of the tt's. Hence we see that we must have ^Ba,-7p,) = 0 or
2<XijPi = 0 for every j. The argument used before based on the
linear independence over @ + fy)/fy of the p,- now implies that
every a,y = 0. This proves our assertion that the e\f\ elements
PiTj are ^-independent. Hence ^i/i < ». Evidently the defini-
definitions of ex and/i now imply that ef < n.
Lemma 2. ef = n if <p is discrete and # is complete relative to <p.
Proof. Since <p is discrete the valuation ^ in P is discrete.
Moreover, P is complete. The groups 7 and T are cyclic and T/y
is cyclic of order e. Let tt and P be elements of $ and p respec-
respectively such that yp{ir) and ^-((8) = <p@) are maximal. Any non-
nonzero element of P has the form eirk where ^(e) = 1 and k = 0,
±1, ±2, • • •. Hence i//(w) is a generator of F. If/8 = tjtt6' where
^G7) = 1 and e' > 0 since /3 e p C <p, then ^(ttN' e 7 so / is divis-
divisible by the order e of the coset ^{tf)y. On the other hand, ^(tt)" =
H^e) = W) for some /3' e p and ^' = f/3* where ^(f) = 1.
Hence f(ire) = ^(^*) = t({rnre')k) = iK*-6'*). Hence ^ = e'k. It
follows that £ = 1, e' = <?, and so we have the relation /3 = rjire,
tin) = 1> e the order of Vjy. Let pu p2, • • •, P/ be elements of O
such that the cosets p,- + ^JJ form a basis for the field £>/$ over the
subfield @ + fy)/y = o/p. We shall show that the elements
Pitc\ 1 < i <f, 0 < j < e — 1 form a basis for P over #. Since
\KxO is of order <?, ^A), ^(V), • • •, ^Gre~x) are in distinct cosets
relative to 7; hence the proof of Lemma 1 shows that the elements
Piirj are ^-independent. It remains to show that every element of
P is a ^-linear combination of these elements and we shall show
first that every element of O is a linear combination with co-

VALUATION THEORY 267
efficients in o of the elements p^. Let v eO. Then \p(v) =
\l/(wk) for some k > 0. We can write k = mxe +ji where mx >
0, 0 <jx < e - 1. Then }{v) = ^((8mVl) so M = (jS"*1^) "^
satisfies ^(m) = 1. The definition of the p.- shows that there exist
elements an e o such that m ~ S «i«P» e $. Then
= 1 and, if vx = Pmi^(n - Saup,), then tK"i) < iK").
We have
D3) y = (S^VV = /3mVl(SaltPl) + vl
We may repeat this argument with vx and obtain a sequence j>i,
^2, ••• such that
D4) »*_! = ?»****&<***<) + vh
where the aki e o, w^ > 0, 0 < _/'fc < ^ — 1, ^Bafc,p,) = 1 and
<K"fc) < Hn-i)- Then D4) implies that H"k-i) = ^mk^jk). It
follows that vk -► 0, j8m* -> 0, and BakiPi)l3m* -» 0. The last
implies that every infinite series whose terms form a subse-
subsequence of the sequence (LakiPi)^mt, k — 1, 2, ■ • ■, converges.
By D3) and D4) we have
D5) v = /3mV'BaliPt) + j8-V»Baa<p0 + • • •
Since j'i —> 0 and the coefficients of the various powers ir\ 0 <
j < e — 1, in D5) converge, we obtain from D5) that v = S/3,ypiirJ,
0 <_;<£ — 1, where j8t;- e o. Now let v be any element of P.
Then we can find a power of 0 so that j^"* e O. Then we obtain
v = /3*(S/3lypi-7Ty) where /3,-y e o so every element of P is a ^-linear
combination of the p,-7rJ.
We can now prove
Theorem 19. Let $ be afield with a non-archimedean real valua-
valuation. Let P be a finite dimensional extension field of <$, ^i, ^2, • • •,
xf/h the different valuations of P which extend <p and let e^ fi be the
ramification index and residue degree of P/€> relative to ^j. Then
D6) 2></«<» = [P:*]

268 VALUATION THEORY
and
D7) Z etfi = n
i
holds if P is separable over <£ and <p is discrete.
Proof. Let E,- be the completion of P relative to ipi- Then for $
the completion of <£> and »,- = [E,:$] we have S«» < n and 2»,- =
n for P separable over $. Also we have seen in § 5 that E,- and P
have the same value group relative to \pi and $ and $ have the
same value group relative to <p. Hence the ramification index ei of
P over <i> relative to \f/i is the same as that of E,- over $. Similarly,
§ 5 and the definitions show that the residue degree /,• of P/$
relative to \f/i is the same as that of E/<f>. By Lemmas 1 and 2 we
have eifi < »» and *,-/,• = »i if the valuation is discrete. Hence
2*?i/i < 2«i < n in every case and 2<?t/j = 2»; = w if P/f> is
separable and <p is discrete.
EXERCISE
1. Determine the residue degrees and ramification indices in the cases given in
ex. 1 of § IS.

Chapter VI
ARTIN-SCHREIER THEORY
In this chapter we shall consider the theory of formally real
fields which is due to Artin and Schreier. A basic algebraic
property of the field of real numbers is that the only relations of
the form 2a,2 = 0 which can hold in this field are the trivial ones:
02 + 02 + • • • + 02 = 0. This observation led Artin and Schreier
to call any field having this property formally real. Any such
field can be ordered and, on the other hand, any ordered field is
formally real. Of central interest in the theory are the real closed
fields, which are the formally real fields maximal under algebraic
extension. A real closed field has a unique ordering which can
be specified by the requirement that a > 0 in such a field if and
only if a = /32 j£ 0. Also, if P is real closed, then P(Vr—T) is
algebraically closed. Any formally real field can be imbedded in
a real closed field which is algebraic over the given field. More-
Moreover, if the original field is ordered, then the imbedding can be
made so that the (unique) ordering in the real closed algebraic
extension is an extension of that of the given field. Such a real
closed extension of an ordered field is essentially unique and is
called the real closure of the ordered field.
The classical application of the Artin-Schreier theory is to the
problem of determining which elements of a field are representable
as sums of squares of elements of the field. For finite algebraic
extensions of the rationals this has a simple answer which is due
to Hilbert and to Landau (Th. 11). The theory of formally real
fields led Artin to the solution of Hilbert's problem on the resolu-
resolution of positive definite rational functions as sums of squares.
We shall give a proof of Artin's theorem (Th. 12).
269

270 ARTIN-SCHREIER THEORY
The most important development of the theory of formally real
fields subsequent to the original work of Artin and Schreier is the
metamathematical principle due to Tarski which asserts that any
elementary statement of algebra which is valid for one real closed
field is valid for every real closed field. This is based on an algo-
algorithm for deciding the solvability in a real closed field of a finite
system of polynomial equations and inequalities with rational
coefficients. Such a decision method was given originally by
Tarski. We shall give an alternative one due to Seidenberg.
In the last section we shall establish the Artin-Schreier charac-
characterization of real closed fields as the fields which are not alge-
algebraically closed but are of finite co-dimension in algebraically
closed fields.
1. Ordered fields and formally real fields. We have defined
ordered groups in the last chapter (§ 5.7). In a similar manner
one has the following
Definition 1. An ordered field <t> is a fields together with a subset
P (the set of positive elements) of$ such that: A) 0 $ P, B) If a e <£>,
then either a e P, a = 0, or —a e P, C) P is closed under addition
and multiplication.
Since any field contains more than one element, it is clear that
the subset P is not vacuous. If N denotes the set { — a\a e P],
then B) states that * = P U {0} UiV. Moreover, it is clear
from A) that P fl {0} = 0 and N D {0} = 0. Also P D TV =
0 since, if a e P D N, then —a e P C\ N and so 0 = a + ( — a) e
P contrary to A). Hence the decomposition <i> = P U {0} UN
is one into non-overlapping sets. It is clear that N is closed
under addition since ( — a) + (—/8) = — (a + #) e N if a, & e P.
On the other hand, (-a)(-/8) = a@ e P if -a, -fieN.
We can introduce a partial ordering in the ordered field <£> (or
more precisely <J>, P) by defining a>/3 if a — /? e P. Then if
a, /3 are any two elements of*, we have the trichotomy: one and
only one of the relations a>/3, a = P, /3>a holds. Thus * is
linearly ordered by the relation a > /?. If a > /?, then a + y >
j3 + 7 and ad > /35 if 5 > 0. Conversely, we can define an ordered
field by means of a linear ordering > such that a > /3 implies

ARTIN-SCHREIER THEORY 271
a + y > 0 + 7 and ad > 05 if 5 > 0. Let P denote the set of
elements a > 0. Then it is immediate that $, P is an ordered
field in the original sense and that the relation > defined by <I>, P
is the given ordering relation.
As usual, it is convenient to write a < 0 for 0 > a. The
elementary properties of the ordering in the field of real numbers
are readily established. We list some of these: a > 0 implies
a > 0 and « > K > 0 implies 0 > a > 0. If a > 0, then
— a < —0 and, if a > 0 and y > 6, then a + y > 0 + 8. As
usual, one defines |a| = a if a > 0 and \a\ = —a if a < 0, and
one proves that |a + /3|<|a| + |/3| and j «0 \ = | a | | /31.
If <£>' is a subfield of an ordered field <£, P, then <i>' is ordered
relative to P' = <£' D P. We shall call this the induced ordering
in <*>'. Evidently a' > 0' in $', P' if and only if a' > 0' in <S>, P.
If <i>, P and 3?', P' are any two ordered fields, then an isomorphism
s of $ into $' is called an order isomorphism (or an isomorphism
of the ordered fields) if Ps c P'. This implies that N8 c JV7, the
set of negatives of the elements of P' and, if j is surjective, then
Ps = P' and N' = Nr.
In any ordered field $,a^0 implies a2 > 0. Hence if aly a2,
• • -,ar are ?^0, then 2«,-2 > 0. This shows that any ordered
field is formally real in the sense of the following
Definition 2. A field$ is called formally real if the only relations
r
of the form ^ a,-2 = 0 in $ are those for which every a,- = 0.
»=i
It is immediate that * is formally real if and only if —1 is not
a sum of squares of elements of $. If the characteristic of $ is
p 5* 0, then 0 = I2 + I2 + • • • + I2 (p terms); hence it is clear
that formally real fields are necessarily of characteristic 0.
In any field $> let S(f>) denote the subset of elements which are
sums of squares. Evidently S(*) contains 0 and is closed under
addition and multiplication. Moreover, we have seen that "£ is
formally real if and only if — 1 £ 2($). If 0 =^ 0 is in 2($), then
0-1 e2(*); for, we have 0 = 20/ and so 0 = 0C0J =
2@i0-1J. We note also that, if* is not formally real and not of
characteristic two, then 2(€>) = <£>; for, —1 e2(<i>) and, if a is any

272 ARTIN-SCHREIER THEORY
element of <I>, then
e 2D?) since 2(<J>) is closed under addition and multiplication. It
will be useful to state these results on 2D?) in the following
Lemma. Let <i> be afield and let 2 ($) be the subset of$of elements
which are sums of squares. Then 2(<i>) is closed under addition and
multiplication and contains ff~1for every P ?* 0 in 2D>). If$ is not
formally real and not of characteristic two, then 2D?) = 4>.
EXERCISES
1. Show that the field of rational numbers can be ordered in one and only one
way.
2. Show that the field i?o(V2) where Ro is the field of rational numbers has
exactly two distinct orderings.
3. Let $ be an ordered field, f(x) = x" + a\xn~1 + ■ • • + an a polynomial
with coefficients in $. Let M = max A, |aij + \ott\ +■—h |an|). Show that
every root of/(#) in <i> is contained in the interval — M < x < M.
4. Show that any purely transcendental extension of a formally real field is
formally real.
5. Let Ro be the rationals and let $ = Ro(Q where £ is transcendental. Show
that $ has a non-countable number of distinct orderings.
6. Let $ be a formally real field and let ^($n) denote the set of n X n sym-
symmetric matrices with entries in $. Show that §($n) is formally real in the sense
that ~2Ai2 = 0, Ai e ^($n) implies that every Ai = 0.
7. Let (x, y) be a symmetric bilinear form on an n dimensional vector space 5D?
over $ where $ is an ordered field. Let {jSi, fa, ■ • ■, fin} be a diagonal matrix for
(x,y). Prove the following extension of Sylvester's theorem (Vol. II, p. 156):
The number of positive /S; is an invariant of (x, y).
8. An ordered field is called archimedean if, given any a > 0, /3 > 0, there
exists an integer n such that na > /3 (equivalently, given a > 0, there exists an
integer n such that n > a). Let P be an ordered field, $ a subfield with the in-
induced ordering. Show that P is archimedean if: 1) $ is archimedean and 2)
[P:$] < oo. (Hint: Use ex. 3.)
9. Prove that any archimedean ordered field is order isomorphic to a sub-
field of the field R of real numbers (cf. Th. 5.8).
10. (Cohn). Let $ be ordered with P as the set of positive elements. Show
that $(£), £ transcendental over * can be ordered by choosing as set Pj of posi-
positive elements those elements which have the form fi^r/g~1 where fi £ P and/ and g
are polynomials in £ with constant term 1. Show that $(£) is not archimedean
ordered.
11. (Cohn). Let $ be ordered and let £, r\ be algebraically independent over $
in $(£, r;). Order $(£) as in ex. 10 and then repeat the process for $(£, tj) con-

ARTIN-SCHREIER THEORY 273
sidering this as the purely transcendental extension $(£)(»?) of $(£). Show that
every element of $(£, rf) is majorized by an element of $(tj) but that there exists
no element of $G7) between f and £2.
12. Let P be an ordered field, <£> a subfield. Let p be the set of elements /3 of P
such that |/}| < |a| for every a y* 0 in <£ and let 0 = {7 e P|7p C p}. Show
that 0 is a valuation ring in P containing $ and that p is the ideal of non-units of
0. Show that the residue field o/p can be ordered by defining 7 + p > 0 if
7 ^ p and 7 > 0 in P. Show that o/p is an extension of $ (identified with
($ + p)/p) which is an archimedean extension of $ in the sense that every interval
{a, b), a, it o/p, contains an element of <I>.
2. Real closed fields. The deeper properties of formally real
fields concern real closed fields which are defined as follows.
Definition 3. A field <£ is called real closed if Q is formally real
and no proper algebraic extension of$ is formally real.
We shall show first that any real closed field can be ordered in
one and only one way. This is an easy consequence of the following
Theorem 1. If ^ is real closed, then any element of$ is either a
square or the negative of a square.
Proof. Let a be an element of <£> which is not a square. Then
we can construct the proper algebraic extension Q, = $(-s/a).
This field is not formally real, so there exist ft, 7,- not all 0 in $
such that 2 (ft + 7,-VaJ = 0. This gives 2 (ft2 + 7;2a) +
2Bft7t)Va = 0. Since Va ?<<i> we have 22ft7* = 0 and 2ft2 +
«27i2 = 0. Since $ is formally real, 27/ ^ 0. Then — a =
Bft2)B7,-2). Using the properties of the set 2($) of sums of
squares stated in the lemma of § 1, it follows that -ae2($).
Since —1 t2(<&) by the formal reality of $ this implies that
a $ 2($). Thus we have shown that, if an element of $ is not a
square, then it is not a sum of squares. In other words, if a e 2($),
then a is a square. Moreover, we have seen that, if a is not a
square, then — a e 2(€>) and this now implies that —a is a square.
This is what we wished to show.
We can now prove
Theorem 2. Any real closed field can be ordered in one and only
one way. Any automorphism of such afield is an order isomorphism.
Proof. Let P be the subset of non-zero squares in the real
closed field <i>. Then 0 i P and, if a ^ 0 and a # P, then —aeP

274 ARTIN-SCHREIER THEORY
by Theorem 1. If a = /32 and y = S2 e P, then a + y e P. Other-
Otherwise, a + y = — e2 where ee$. This gives /32 + 52 + e2 = 0
contrary to the formal reality of <£>. Hence we see that the sub-
subset P satisfies the conditions 1, 2, 3 for an ordered field and #, P
is such a field. Let P' be any subset of $ which gives an ordering.
If a e P, a = /32 ?± 0. Then a > 0 in the ordering given by P'.
Hence P' 2 P. This implies that P' = P so the ordering in #
is uniquely determined. If s is an automorphism of <£, then it is
clear that s maps the set P of non-zero squares into itself. Hence
s is an order isomorphism of <£.
The question of the existence of real closed fields is easily
settled. In fact, we have the following
Theorem 3. Let $ be a formally real field and let 12 be an algebraic
closure of$. Then ft contains a real closed field A containing $.
Proof. We consider the collection of formally real subfields of
Q containing $. This collection is not vacuous since it contains $.
Moreover, it is clear that the collection is inductive, so, by Zorn's
lemma, it contains a maximal element A. If A is not real closed,
then it has a proper algebraic extension A' which is formally real.
Since 0 is algebraically closed, we may suppose that A' C S2
(ex. 1, p. 147). This contradicts the maximality of A in 12. Hence
A is real closed.
Evidently Theorems 2 and 3 and the existence of an algebraic
closure for any field imply the following corollaries.
Corollary 1. Any formally real field can be imbedded in a real
closed field which is algebraic over the given field.
Corollary 2. Any formally real field can be ordered.
If $ is real closed, then — 1 is not a square in $ so $(\/ — 1)
3 #. We shall show that <$(■%/—1) is algebraically closed and we
shall see that this property is characteristic of real closed fields.
For this purpose we prove first the following result.
Theorem 4. If $ is real closed, then every polynomial of odd
degree with coefficients in <i> has a root belonging to $.
Proof. The result is clear for polynomials of degree 1 and we
use induction on the degree n of fix). If fix) is reducible, one of

ARTIN-SCHREIER THEORY 275
its factors is of odd degree so it has a root in $. Hence we may
assume f(x) is irreducible. Let A = <J>@) where /@) = 0. Then
A D §, so A is not formally real. Hence we have a relation
2<pi(dJ = — 1 where <Pi(x) is a polynomial in x of degree <n — 1.
The relation indicated implies that 2<p,(xJ = — 1 + f(x)g(x).
The leading coefficient of S^t(xJ is positive in the ordering in <i>
and the degree of this polynomial is even and <2(« — 1). It
follows that degg(x) is odd and <2(w — 1)— n = n — 2.
Hence there exists a /3 e <i> such that g(fi) = 0. Substituting this
0 in the relation 2<f>{(xJ = -1 + f(x)g(x) gives 2p;(|3J = -1
contrary to the formal reality of <£>.
We shall prove next the following generalization to real closed
fields of the so-called fundamental theorem of algebra. The proof
is patterned rather closely after one of Gauss' proofs of the classical
result.
Theorem 5. Let <£ be an ordered field such that: A) positive ele-
elements in $ have square roots in <J>, B) any polynomial of odd degree
with coefficients in <1> has a root in <i>. Then y/ — l $ <£ and$(y/ — 1)
is algebraically closed.
Proof. Since $ is real, it is clear that y/ — 1 i <£. Consider
<$(a/ — 1) 3 $. Let p —> p be the automorphism of $(\/ — 1)
over $ such that *' = — * for * = \/ —1. If /(a:) e <i>(\/— l)[x],
then f(x)J(x) e$[x], and if this has a root in $(V — 1), then /(x)
has a root in $('\/ —1). Hence the algebraic closure of 4>(\/ — 1)
will follow if we can show that every non-constant polynomial with
coefficients in * has a root in $(V — 1). This holds by B) if the
degree of the polynomial is odd. We show next that every ele-
element of <i>(\/—-I) has a square root in this field. First, if a e *
and a > 0, then, by A), a = /32, 0 e <f>. Next if a e $ and a < 0,
then — a = /32 and a = ("s/^TJ/32. Now let p = a + 0/, / =
y/ — 1, a, j8 in <£>, C ?^ 0. Consider the element £ + ij», £, 17 in <$.
We have (£ + viJ = S2 - *?2 + 2$iji so (f + r,iJ = a + /3* is
equivalent to
A) £2-,2 = «, 2^ = 0.
Since ^0 we may (by multiplying by a suitable element of f>)
assume that 0 = 2, so the second equation becomes £ij = 1. This

276 ARTIN-SCHRETER THEORY
holds if rj = £-1. Then the first equation becomes £2 — t-~2 = a
or X — X" = a for X = £2. Then we have X2 — aX — 1 =0
which has the solution (a + Va2 + 4)/2 in 4> since a2 + 4 > 0.
Also a + V<x2 + 4 > 0 since a + Va2 + 4 < 0 leads to 4 < 0.
Hence there exists a £ ?^ 0 in <i> such that £2 = ^(a + Va2 + 4).
Then £4 - a£2 = 1 and £2 - £~2 = «• Hence £ and rj = S-1
satisfy A) with /3 = 2. We have therefore proved that every
element of <£("n/ — 1) has a square root in this field. Consequently
there exists no extension field A of $(V' — 1) such that
[A-MV=i)] = 2.
We proceed to use this fact to prove that every polynomial of
positive degree with coefficients in <£ has a root in <&(\/ — 1). Let
f(x) be such a polynomial and let E be a splitting field over $ of
(x2 + 1 )/(#). We may assume that E 2 $(V —1). Since the
the characteristic is 0, E is Galois over <£. Let G be its Galois group
and let (G:l) = 1em where m is odd. By Sylow's theorem G has
a subgroup H of order 2". Let A be the subfield over $ of //-
invariants. Then [E:A] = 2e and [A:«i>] = m. Since $ has no
proper odd dimensional extension field we must have A = $ and
m = 1. Hence G = H has order 2e. Such a group is solvable. If
£ > 1, it follows easily from Galois theory that E contains a sub-
field r over ^(v^T) such that |T:$(v^T)] = 2. This con-
contradicts what we proved before. Hence e = 1, so [E:$] = 2 and
E = $(V —1). This shows that f>(V —1) is a splitting field of
(x2 + 1)/O) and that /(*) has a root of$(\/^T). Hence *(V—I)
is algebraically closed.
If$> is a real closed field, then we have seen that€> can be ordered
in exactly one way. The proof of Theorem 2 shows that this
ordering is obtained by specifying that a > 0 if a = /32, /? j£ 0.
Hence we see that every real closed field is ordered and satisfies
condition A) of Theorem 5. Theorem 4 shows that every real
closed field satisfies condition B) of Theorem 5. Hence we have
the following
Corollary. If $ is a real closed field, then y/ — 1 $
is algebraically closed.
We shall prove next the converse of this, namely,

ARTIN-SCHREIER THEORY 277
Theorem 6. If $ is a field such that V —1 i $ and #( V— 1) is
algebraically closed, then <i> is real closed.
Proof. Suppose <3? satisfies the conditions. We note first that
the irreducible polynomials of positive degrees in $[x] have degree
1 or 2. Let f{x) be such a polynomial and let 8 be a root of f(x)
contained in £2 = <I?(V — 1). Then [<i>@) :$] = deg /(*•) and
[$@):<t>] < [fl:$] = 2. Hence deg /(*) = 1 or 2 as asserted. Now
let a, j8 5^ 0 e <J? and consider the polynomial
B) £(*) = (*2 - aJ + /32 = (x2 - a - /3/)(*2 - a + #)
= (*- (a+ j80 «)(*+(«+ j80><)-
(* - (a - pi)*)(x + (a - jSO^),
where « = \/—1. This polynomial belongs to $[#] and has no
linear factors in $[x] since ±a ± /3/ $ <£. Hence ^(*) is a product
of two irreducible quadratic polynomials. The one divisible by
x — (a + &iYA cannot be
(*-(« + #)*)(* + (« + /W)H) = ^2 - (a + Pi);
for, this would imply that a + 0i e $. Hence the polynomial in
question is either
(*-(« + #)*)(* - (a - ^)H)
or
(x-(a + fH)iA){x + (a - ^/)M).
Either possibility implies that (a2 + /32)^ e$. Since a and /? were
arbitrary non-zero elements of <£, we have proved that the sum
of two squares of elements in $ is a square. Induction shows
that every sum of squares is a square in <£>. Since — 1 is not a
square, this implies that — 1 is not a sum of squares in <J> and so *
is formally real. If P is a proper algebraic extension of $, then
P is isomorphic to U = 4?("\/ — 1). Then P is not formally real
and so $ is real closed. This completes the proof of Theorem 6.
The corollary to Theorem 5 and Theorem 6 give the charac-
terization of real closed fields by the properties that V —1 t *
and <£(\/ — 1) is algebraically closed. We remark also that there
is another characterization involved in our discussion, namely,
an ordered field is real closed if and only if it satisfies conditions

278 ARTIN-SCHREIER THEORY
A) and B) of Theorem 5, that is, positive elements of €> have
square roots in <1> and polynomials of odd degree with coefficients in
$ have roots in $. This is easily deduced from our results. We
derive next the following useful consequence of one of our charac-
characterizations of real closed fields.
Corollary. If V is a real closed extension field of a field <£, then
the subfield A of elements of P which are algebraic over $ is real
closed.
Proof. Let 0 = P(v^-T). Then Q is algebraically closed.
Hence the subfield F of elements of 0 which are algebraic over 4>
is algebraically closed. If a + /3\/ — 1, a, |3 e P, is in F, then so is
a - pV^l. Hence a = ^(a + (SV^T + a - /3\/^T) e F.
Then 0 e F. Since a, /3 e P we see that a, /? e A. It follows that
F = A(a/ —1). Since V —1 i A, we see that A fulfills the condi-
conditions of Theorem 6. Hence A is a real closed field.
EXERCISE
1. Let Q/$ be algebraically closed, $ formally real. Show that Q/f> contains a
real closed subfield P/$ such that 0 = P( y/ — 1 ). In particular, show that
every algebraically closed field of characteristic 0 contains a real closed subfield
P such that fi = P(V=T).
3. Sturm's theorem. In this section we shall derive a classical
result, Sturm's theorem, which permits us to determine the exact
number of roots in a real closed field of a polynomial equation
fix) = 0. This result is fundamental in the sequel. In deriving
it we shall follow rather closely Weber's exposition in Lehrbuch der
Algebra A898), Vol. I, pp. 301-313. We shall need first the follow-
following basic result.
Lemma. Let $ be a real closed field and f(x) a polynomial with
coefficients in <i>. Suppose a and /3 are elements of f> such that f(a)
< 0 while /(/8) > 0. Then there exists a y between a and /8 such
that /G) = 0.
Proof. We recall that the only irreducible polynomials in $[x]
are the linear ones and the quadratic ones. Let g(x) = x2 +
nx -\- v e $[x] be irreducible. We assert that necessarily y? — \v <
0. This is clear from the formulas for the roots of a quadratic

ARTIN-SCHREIER THEORY 279
equation. We can now set Av — p.2 = 452 where S is a non-zero
element of $ and we have
C) g(x) = x2 + nx + v = (x + 0 + S2.
Evidently this formula shows that g(rj) > 0 for every t\ in <t>.
Now let f(x), a, /3 be as in the statement of the theorem. In
$[x] we have the factorization
D) /(#) = p(x - pi)(x - p2) •••(*- Pk)gi(x) ■ ■ ■ gi(x)
where gt(x) is an irreducible quadratic with leading coefficient 1.
Suppose none of the p{ is between a and 0. Then for each i,
a — pi and /3 — p,- have the same sign (both positive or both
negative). Since gj(a) > 0 and gj{0) > 0, 1 <j<l, this implies
that /(a) and /(/3) have the same sign, contrary to hypothesis.
Hence there is a p,- between a and /3. This completes the proof.
Let $ be a real closed field and let /(,*) be a polynomial of posi-
positive degree with coefficients in <i>. Following Weber, we shall say
that a sequence of polynomials
E) /„(*) = /(*), /iW,--,/.W
is a Sturm sequence of polynomials for f(x) for the interval [a, /3]
(that is, a < x < 0) if the /,-(*) e $[x] and satisfy the following
conditions:
(i) /,(#) has no roots in [a, 0].
(ii) /0(a) * 0, /0(j8) ^ 0.
(iii) If y e [a, /3] is a root of /,•(#), 0 < ,; < s, then
/y-.W/y+iG) < 0
(iv) If A"f) = 05 7 e [a, 0], then there exist intervals ji < x <
y and 7 < x < y2 such that Mx)fi(x) < ^ ^or * m tne
first of these and /o(#)/i(#) > 0 for ^ in the second.
(This amounts to saying that Mx)/i(x) ls an increasing
function of x at x = 7.)
We shall establish the existence of such sequences for any
polynomial with distinct roots, but first we shall see how such a
sequence can be used to determine the number of roots of Ax) m

280 ARTIN-SCHREIER THEORY
the open interval (a, /3) (that is, a < x < j3). We consider the
number of variations in sign of the sequence
/o(«), /!(«),-••,/.(«)
/o@),/i(iS), ■••
of elements of $. If 7 = {71, 72, • • -, 7m} is a finite sequence of
non-zero elements of $, then we define the number of variations in
sign of 7 to be the number of i, 1 < i < m — 1, such that 7iY,-+i
< 0. If 7 = {71,72, • • -,7m} is an arbitrary sequence of ele-
elements of $, then we define the number of variations in sign of 7
to be the number of variations in sign of the abbreviated sequence
7' obtained by dropping the 0's in 7. For example,
{1,0,0,2,-1,0,3,4,-2}
has three variations in sign.
We can now state
Theorem 7. Let f(x) be a polynomial of positive degree with co-
coefficients in a real closed field <i> and let fo(x) = f(x), fi(x), ■ ■ ■,
fs(x) be a Sturm sequence for f{x) for the interval [a, 0\. Then the
number of distinct roots of f(x) in (a, /3) is Va — V$ where, in
general Vy denotes the number of variations in sign of the sequence
{My),My), ■■',/.&)}•
Proof. The interval [a, /3] is decomposed into subintervals by
the roots of the polynomials fj(x) of the given Sturm sequence.
Thus we have a sequence a = a0 < «i < • • • < <*m = P such
that none of the f,(x) has a root in (a,-, aI+i). Choose a/ e
(a,-_i, a,-), 1 < i < m (e.g., a/ = -|(a;_i + a,)) and let Vai, be
the number of variations in sign of the sequence {/>(«/), j =
0,1, ■ ■ •, s}. Evidently,
m-l
Vtt-Vz=Va- Va, + £ (Vat. - VaiW) + Fam, - Fp,
at.
1
so we shall try to compute Va — Vai>> Vai, — Vai+l>, Vam, — V$.
We have /„(«) * 0, fo@) * 0, /,(«,) ^ 0, /,(«/) ^ 0. Suppose
first that no /,(«) = 0, 0 < j < s. Then M<*)M"i) > ° for
k = 0, • • ■, s, since, otherwise, by the lemma, one of the /&(#) has
a root in (a, a/) contrary to the property of the intervals (a,-,
a,+i). Hence we have Va = Vax> in the case under consideration.

ARTIN-SCHREIER THEORY 281
Next let/;-(a) = 0 for some j, 0 < j < s. Then fj_i{a)fi+1(a)
< 0, by (iii). Since fj_i(x) and /,•(#) have no roots in (a,ai),
we have /,--i(a)/i-i(«i') > 0 and /,+1(a)/,+1(ai') > 0. Hence
/,_!(«/)/,•+!(«/) < 0- ^ follows that /y_x(«), 0, /,+1(«) and
/j_i(ai'), /;(ai')> /y+i(«iO contribute the same number of varia-
variations of sign to ^a and Fai> respectively. Taking into account all
they we see that Fa — Fai< = 0. A similar argument shows that
Vaw! — Fp = 0. Also the same argument shows that, if a,-, 1 <
/ < m — 1, is not a root of f{x) = fo(x) then again ^av — Vat+X>
= 0. It remains to consider what happens if /(«») = 0 for 1 <
i < m — 1. Then, by (iv) and the choice of the a/, we have
/o(a/)/i(*/) < 0 and /0(ai+1')/i(«t+i') > 0. Then the se-
sequence /0(a/)j /i(«/) has one variation in sign while the sequence
/o(ai+i')j /i(«»+iO has none. The argument used before shows
that /;_i(a/), /,<«»'), /y+i(a/) and /y_i(af+1'), /i(«i+i'),
/;+i(«f+i') have the same number of variations of sign if j > 1.
Hence we see that Vai, — Fai+l' = 1 if /(a,) = 0. We have
therefore shown that Fai> — Fa.+l, = 0 or 1 according as /(a,-)
?* 0 or /(a*) = 0. Hence
Va-Vt=Va- Vax. + m£ {Vai, - Fai+1.) + Fam, - F0
i
is the number of a,- such that /(a,) = 0.
Now let f(x) be an arbitrary polynomial. We define the
standard sequence for f(x) by
/o(*) = /(*)> /iW = /'(*) (formal derivative of /(*)),
/o(«) = ?i(*)/iM - /2W, deg/2 < deg/i
G) /<_i(*) = ?.■(*)/<(*)-/<+i(*), deg /,-+1 < deg /,•
Thus the /,•(*) are obtained by modifying the Euclid algorithm
for finding the highest common factor of f(x) and f'{x) in such a

282 ARTIN-SCHREIER THEORY
way that the last polynomial obtained at each stage is the nega-
negative of the remainder in the division process. Clearly, /,(#) is
the highest common factor of f(x) and /'(#) and this is a divisor
of all the fi(x). Now set g%(x) = /i(x)/a(x)~1 and consider the
sequence
(8) go(x), gi(x), ■■■,gs(x).
We proceed to show that this is a Sturm sequence for go(x) for
any interval [a, /3] such that go(<x) ^ 0, go(&) ^ 0. Clearly (ii)
in the definition of Sturm sequences is satisfied. Also (i) holds
since ga(x) = 1. Dividing the polynomials in G) by fa(x) gives
the relation g,-i{x) = qj(x)gj{x) - gj+x{x), 0 < j < s. Suppose
gi(y) = 0- Then ,§7-1G) 5^ 0 and gj+i(y) ?* 0, since otherwise
the relations indicated would imply that all the ^G) = 0 from
a certain point on contrary to ge(x) = 1. Thus gj-i(,y)gj+i(y) ^
0 and, since £,—1G) = gj{y)qj{y) -^+1G) = -^+1G), we have
gi—i(y)gj+i(y) < 0 and (iii) holds. Now suppose that £0G) = 0 for
7 in [a, |8]. Then we have f{x) = (x - y)eh(x), e > 0, h(y) ^ 0
and /'(*) = (x - y)eh'(x) + e{x - yY~lh{x). Also /.(*) = (* -
yY^Hx) where k(y) ?* 0. Hence h(x) = k{x)l(x) where /G) ^
0 and h'{x) = k(x)m(x). These relations give
*o(*) = (* - 7)/W, l(y) * 0
gi(x) = (x - y)m(x) + e!{x)
so ^1G) = ^G) 9^ 0. Now choose an interval [71, 72] containing
7 in its interior such that l{x) 9^ 0 a.n<\gx{x) ^ 0 in [7^ 72]. Then
the lemma implies that £i(.tf) and l(x) are either both positive or
both negative in [yu y2] so gi(x)l(x) > 0 in [71} 72]. Hence
go(x)gi(x) = (x - y)gx{x)l{x) has the same sign as x - 7 in [7!, 72]
so go{x)gx(x) < 0 in 71 < x < y andfoWf iC*) > 0 in 7 < a: < 72.
This shows that (iv) holds and so (8) is a Sturm sequence for go(x).
If f(x) has no multiple roots, then f(x) and /'(*) have 1 as
highest common factor. Then the sequence {/o(x), fi(x), ■ ■ •,
fa(x)} differs from {go(x),gi(x), • ■ -,gs(x)} by a non-zero multi-
multiplier in $. Hence the sequence of /,•(#) is a Sturm sequence for
/M = fo(x). If f{x) has multiple roots, then the standard se-
sequence G) will not be a Sturm sequence for an interval contain-
containing a multiple root. Nevertheless, we can still use the standard

ARTIN-SCHREIER THEORY 283
sequence to determine the number of distinct roots of f(x) in
(a, 0). This is the content of
Sturm's theorem. Let f{x) be any polynomial of positive degree
with coefficients in a real closed field <i> and let {fo(x) = f(x), f\ (x) =
/'(#)> • • *j /»(#)} be the standard sequence G) for f(x). Assume
[a, 0] is an interval such that /(a) ^ 0, /(/3) ^ 0. Then the number
of distinct roots of f{x) in (a, 0) is Va — V$ where Fy denotes
the number of variations in sign of {fo(y), fi(y), ••-, f.(y)}.
Proof. Let gi(x) = fi(x)f,(x)~1 as above. Then apart from
multiplicities, the polynomials/^) zndgo(x) have the same roots in
in (a, 0) (ex. 7, p. 40). Since the sequence {gi(x)} is a Sturm se-
sequence for go(x), the number of these roots is Fa(g) — Fp{g) where
Vyig) is the number of variations in sign in \gi(y)}. Since
My) = g<(y)My) and /,(«) * o, M0) * o
it is clear that Fa(g) = Va and F?(g) = Fp. Hence Va - Vfi
gives the number of distinct roots of f(x) in (a, 0).
We have seen that the roots of f(x) = xn + aixn~1 +•••+«„
in * are in the interval [ — M,M] where M = max A, |«i| +
|«2| + --- + kn|)(ex.3,§l). IfwesetM = 1 + |«i| + • • • + \an\,
then the roots of f{x) in $ are in (—p., n). If fo(x) = f(x), fi(x),
• • "j /•(*) is the standard sequence G) for f(x), then the number
of roots of f(x) in $ is F_ll — ¥„, where Fy is the number of varia-
variations in sign in {fo(y), /i(t), • • -,f,(y)}- This gives a construc-
constructive way of determining the number of roots of f(x) in <£>. Some-
Sometimes it is preferable to use instead of /x a bound 17 which is a
polynomial in the «,-. For this purpose we note that 1 + a? >
|«i|, so we can take r, = 1 + 2A + a{2) = (» + 1) + 2«,-2.
Then the roots in * lie in (—ij, »j).
EXERCISES
In all of these exercises $ is a real closed field.
1. Prove Rolle's theorem: If/(.v) E$[x] has roots a, |3 in $, a < /3, then there
exists a7in$, a<7</3 such that/'G) = 0.
2. Prove the mean value theorem for polynomials: If a < j3, then there exists a
7, a < y < 0 such that/(|8) -/(a) = (/3 - a)/'G).
3. Prove that/(#) has a maximum on any closed finite interval, [a, /3].
4. (Budan's theorem). Let/(«) have degree n and assume a < C in $ are not
roots of/(#). Let /F,, denote the number of variations in sign in the sequences:

284 ARTIN-SCHREIER THEORY
/(t),/'(t), • • •,/in)(y)- Prove that Wa — Wq exceeds the number of roots of
f(x) in $ in (a,j3) counting the multiplicities of these roots by a non-negative
even integer.
5. Deduce from ex. 4 Descartes' rule: Let /(#) = aoxn + a\xn~1 -\ h
ctixn~l, ao r* 0,ai ?* 0, a* e $. Let P denote the number of variations in sign in
the sequence (ao, ai, • • • , «;)• Show that P exceeds the number of positive roots
o{/(x), counting multiplicities, by a non-negative even integer.
4. Real closure of an ordered field. We have seen that every
formally real field can be imbedded in a real closed field. In
particular, this applies to ordered fields. We shall now show that,
if $ is an ordered field, then there exists a real closed algebraic
extension field A of $ whose (unique) ordering is an extension
of that of $. Moreover, we shall see that A is essentially unique.
To prove the existence of A we need the following
Lemma. Let <!> be an ordered field, Q, an algebraic closure of $ and
let E be the subfield of Q/"J> obtained by adjoining to $ the square
roots of the positive elements of<&. Then E is formally real.
Proof. Suppose we have a relation 2£;2 = 0 in E. Then the
£t- are contained in a finite dimensional extension field of the form
^"(v^, V^, • • •, '*/&■) where the & are positive elements of <i>.
Hence it suffices to show that every subfield <£>('s//3^, a//32, • • •,
V^), Pi > 0, of E is formally real. We prove this by induction
on the dimensionality of the subfield and for this it is convenient
to prove the apparently stronger statement that, if Zy&i2 = 0
for Yi > 0 in * and £,- in $(v^, V/32, • • ■, V^), then every
£i = 0. This is clear for $> since this is an ordered field.
Suppose it holds for subfields of the indicated form of lower
dimensionality than that of T = $(V|8^, • • •, V%-)- We may
assume that F 3 H = $(\/]3i, • • •, "s//3r_i), so the result holds
for H. Now assume 2y£i2 = 0, £t- e T, yt > 0 in $. Write £* =
Vi + UVfi'r, m, U e H. Then Vy^2 + Z^y^i2 + 2(SyMdVWr
= 0. Since VWt i H, Zy^mU = 0, so Hynn2 + S/3r7ifi2 = 0.
Since yiy /3r7i e $, 7; > 0, ^ryt > 0, and rn, f,- e H, every rn and
f i = 0. Then every £* = 0 and the result is valid for T.
Definition 4. Let $ be an ordered field. Then an extension field
A of $ is called a real closure of $ if A) A is real closed, B) A is
algebraic over <i>, C) the ordering of A is an extension of that of 3>.
We can now prove the following basic result.

ARTIN-SCHREIER THEORY 285
Theorem 8. Every ordered field <i> has a real closure. If $1 and
$2 are ordered fields with the real closures At and A2, respectively,
then any order isomorphism of $1 onto $2 has a unique extension to
an isomorphism of A1 onto A2. The extension is an order isomor-
isomorphism.
Proof. Let <£ be an ordered field, Q, an algebraic closure of <£.
Let E be the subfield of Q obtained by adjoining to $ the square
roots of all the positive elements of $. Then E is formally real
and ft is an algebraic closure of E. We have seen that there exists
a real closed subfield A of 12/E (Th. 3). Suppose 0 e $ and /3 > 0.
Then /8 = p2, p e A. Hence /3 > 0 in A so the ordering in A is an
extension of that of <i>. Hence A is a real closure of <i>.
Next let <t\, / = 1, 2, be ordered fields, A,- a real closure of <£,•
and let a —> a be an order isomorphism of $1 onto "i^. We wish
to extend the given isomorphism to an isomorphism of Ax onto
A2. We note first that, if f(x) e^ifx1], then f(x) and its image
f(x) under a —> a have the same number of roots in At and A2
respectively. We have seen that there exists a p. > 0 in $x such
that every root of f{x) in Ax is contained in ( — p., p.). Moreover,
by Sturm's Theorem, the number of roots of f{x) in At in the in-
interval ( — p., p.), hence the total number of roots off(x) in Ax, is
given by V_v. — V^ where Vy is the number of variations in sign of
the standard sequence G) for / at 7. Since the standard sequence
of / is contained in $i[x], all of this carries over to f(x) and A2.
Hence the number of roots of ]{x) in A2 is the same as the number
of roots of f(x) in Aj. We note next that, if F = {pl5 p2, ■ ■ •, pn)
is a finite subset of A1} then there exists a subfield Yx of Ax/<£> con-
containing F and an isomorphism r of Ti into A2 which extends a —>
a and is such that, if p1 < p2 < ■ ■ ■ < pn, then pxT < p2r < • ■ • <
pnT. For this purpose let f(x) be a polynomial in &i[x] which has
the elements p,, 1 < i < n, a, ==■ "s/p,+i — py, 1 < j < n — \,
among its roots. We note that the <rj e Ax since Ai is real closed
and Pj+i — pj > 0 (proof of Th. 2). Let Tx be the finite dimen-
dimensional extension of $! generated by the roots of f(x) in Ax. Then
Ti = *i(^i) and, if g(x) is the minimum polynomial of 0i over $1}
^(x) has a root 02 in A2. We have an isomorphism t of rx onto
f>2(^2) such that aT = a, a e$u and BXT = 62. Then pj+1r — p/ =
(pj+1 — pj)T = (o/J > 0. Hence pxT < p2T < • • • < pnT in A2 as

286 ARTIN-SCHREIER THEORY
required. We shall now define a mapping 77 of Ax into A2 in the
following way: Let p be an element of Ax and let h{x) be its mini-
minimum polynomial over #1. Let the roots of h{x) in Ax be px <
P2 < • • • < Pm and suppose pk = p. Then h~(x) has exactly w*
roots pi' < p2' < • ■ • < pm' in A2 and we now set pv = p*/. Evi-
Evidently a.r' = a,a e<i>i, and it is easy to see that 77 is 1-1 and surjec-
tive. We assert that, if p, <r e A1} then (p + <t)v = pv + an,
(p<r)v = pV so that 77 is an isomorphism of Ax into A2 extending
a —* a. Let F be a finite subset of Ax which includes the roots
in Aj of the minimum polynomials over <!>! of p, <r, p + a and pa-.
Then we have seen that there exist a subfield rx of Ax over *!
containing i*1 and an isomorphism r of I\ in A2 extending a —* a
such that t preserves the order of the elements of F. As before,
let h(x) be the minimum polynomial of p over $1 and let px <
p2 < ■ • • < pm be the roots of h(x) contained in Ax. Then p,- e F
and PiT < p2T < • • ■ < Pm- We have ^(p,T) = 0 and it follows
from the definition of 77 that pv = pT. Similarly, we see that crv =
<rT, (p + a)v = (p + <r)T, (ptrI' = (pa)T. Since r is an isomorphism,
this implies that (p + a)v = pv + <rn, (p<r)v = pV. Hence 77 is an
isomorphism of Ax onto A2 extending the given isomorphism of $x
onto <i>2. Now let 77' be any isomorphism of Ax onto A2. Since 77'
maps squares into squares it is clear that 77' is an order isomor-
isomorphism. Suppose also that 77' extends the mapping a —> a. Let
p e Ax and let pi < p2 < • • • < pm be the roots in Ax of the mini-
minimum polynomial h(x) of p over <I?i. Then p^' < P2'' < • • • < p*/
are the roots in A2 of Jt{x). It follows that pv' = pv. Hence the
extension 77 is unique. This completes the proof of the theorem.
If Ax and A2 are two real closures of a given ordered field <£,
then the identity mapping on €> can be extended to an order iso-
isomorphism of Ax onto A2. In this sense real closures are equivalent
and we may therefore speak of the real closure of $.
EXERCISES
1. Let $ be an ordered field, A an extension field such that the only relations
of the form ~Ly&? = 0 with yt positive in $ and £,• in A are those in which every
£,■ = 0. Show that A can be ordered in such a way that its ordering is an ex-
extension of that of $.
2. Let $ be an ordered field, A a real closed extension field whose order is an ex-
extension of that of <£. Show that A contains a real closure of $.

ARTIN-SCHREIER THEORY 287
5. Real algebraic numbers. We have seen that the field Ro of
rational numbers has a unique ordering (ex. 1, § 1). This ordered
field has a real closure Ao which is determined up to isomorphism.
We shall call any real closure Ao of RQ the field of real algebraic
numbers. Clearly fi0 = A0(V —1) is an algebraic closure of Ro,
and we shall call this field the field of algebraic numbers.
Now let F = Rq{0) be a finite dimensional extension field of the
rationals. Then if n = [F:/?o], we have n distinct isomorphisms
of r/i?0 into SIq/Rq. These are determined by mapping 0 —> 0,-,
1 < i < n, where {0i, 02, ■ • •, 6n] is the set of roots of the mini-
minimum polynomial g(x) of 0 in fl0- Let 0x, 62, • ■ ■, 0r be those 0,-
which belong to Ao. We shall call these the real conjugates of 0.
We agree to set r = 0 if 0 has no real conjugates. Let r,-, 1 < / <
r, be the isomorphism of T/Ro into AQ/R0 such that (P* = 0,-. Then
the ordering of Ro@i) Q A0 imposed by the unique ordering in
Ao provides an ordering of T: We define p > 0 for p e F if and
only if pT* > 0. We shall refer to this ordering of F as the order-
ordering determined by t,-. Now suppose we have any ordering of F
and let A be a real closure of F relative to this ordering. Since F
is algebraic over Ro, it is clear that A is a real closure of Ro. Con-
Consequently, we have an order isomorphism t of A/Ro onto Ao/Ro.
The restriction of t to F coincides with one of the r,- and it is clear
that the given ordering of F is the same as the ordering determined
by t,-. Finally, suppose t,- and 77 provide the same ordering of F.
Then we have an order preserving isomorphism of Ro(8i) onto
Ro@j) such that 0,- —> 0y. Since A0 is a real closure of Ro@i) and
of i?o@y), by Theorem 8, we have an automorphism of AO (over
Ro) sending 0t- into 0y. On the other hand, since Ao is a real closure
of Ro, Theorem 8 shows also that the identity is the only auto-
automorphism of Ao over Ro. Hence we must have 0}- = 0,-. These
results establish the following
Theorem 9. Let T be a finite dimensional extension of the field
of rational numbers. Then the number of distinct orderings of F is
the same as the number of isomorphisms ofT/Ro into the field Ao/Ro
of real algebraic numbers.
In particular, this number cannot exceed [F:/20] and there are
no orderings of F = i?o@) if and only if the minimum polynomial
of 0 over Ro has no real roots, that is, no roots in Ao.

288 ARTIN-SCHREIER THEORY
We shall now apply this result to obtain a theorem of Hilbert
and Landau which gives a necessary and sufficient condition that
an element of F = Ro@), d algebraic, is a sum of squares in this
field. First, let $> be any field of characteristic ^2 and, as in
§ 1, let 2(<i>) be the subset of* of elements of the form 2at-2,
a.i e<i>. Next, we introduce the following definition.
Definition 5. An element p of a field is called totally positive if
p > 0 in every ordering of the field.
In particular, it will be understood that in a field which has no
ordering, then every element is totally positive. Thus every
element of a field which is not formally real is totally positive.
We have the following general criterion
Theorem 10. Let <i> be a field of characteristic 9^1. Then an
element p 9^ 0 in <t> is totally positive in ? if and only if p is a sum of
squares of elements o/*.
Proof. If 0 7^ p = 2a,2, then clearly p > 0 in every ordering of
$. Conversely, assume p 7^ 0 is not a sum of squares in <$. Let
£2 be an algebraic closure of $ and consider the collection of sub-
fields E of Sl/$ in which p is not a sum of squares. This collection
contains €> and is inductive; hence it contains a maximal element
P. Now P is formally real; otherwise, the lemma of § 1 shows
that every element of P is a sum of squares, but we know that p
is not a sum of squares in P. Then P can be ordered. We note
next that — p is a square in P. Otherwise, we have the field
P(a/—p) in 0 and this properly contains P. Hence in this field
we must have p = 2(£,- + 'm's/ — pJ, £i, m in P- This gives p =
2k2 - p277i2 + 2B£i77i\/^). It follows that Sfcij, = 0 so
p(l + 2rji2) = 2k2. Then 1 + 2rjt-2 ^ 0 by the formal reality
of P; hence p = B^-2)(l + 2rji2)~1 is a sum of squares in P by
the lemma of § 1. This contradicts the choice of P. Hence we
see that — p = /32, j3 e P. This implies that — p > 0 and p < 0 in
every ordering of P. Since P can be ordered, the induced order-
ordering in $ gives an ordering of 3? in which p < 0. Thus p is not
totally positive.
This criterion and the result we obtained before on the form of
the orderings of a finite dimensional extension of the rationals
evidently imply the following

ARTIN-SCHREIER THEORY 289
Theorem 11 (Hilbert-Landau). Let T be a finite dimensional
extension field of rationals and let t\, t2, • • •, rr (r > 0) be the dif-
different isomorphisms ofT/R0 into the field of real algebraic numbers.
Then an element p ?± 0 of T is a sum of squares in T if and only if
pTi > Ofori = 1,2, -^r.
EXERCISES
1. Let <J? be an ordered field and A a formally real field over $. Let p be an
element of A which cannot be written in the form
A0) Wi2, ft>0in*,
£,• in the larger field. Show that there exists an algebraic extension P of A such
that p is not of the form A0) in P (£» in P) but p has this form in every proper
algebraic extension P' of P. Show that every positive element of $ is a square in
P and hence that p is a sum of squares in any P' 3 P, P' algebraic over P.
Prove that P is real closed and that the ordering in P is an extension of that of $.
Prove that p < 0 in the ordering of P. Hence prove the following theorem: A
necessary and sufficient condition that an element p of A have the form A0) in A
is that p > 0 in every ordering of A which extends the ordering of $.
2. Let $ be an ordered field, A the real closure of $ and T a finite dimensional
extension of $. Prove the following generalization of Theorem 9: If r is the
number of isomorphisms of F/$ into A/$, then r is the number of ways of ex-
extending the ordering of $ to an ordering of T.
6. Positive definite rational functions. One of the problems
proposed by Hilbert in his address to the 1900 Paris Congress of
Mathematicians was the following: Let Q be a rational function
of n variables with rational coefficients such that Q(l-i, ■ ■ ■, £„) >
0 for all real (£i, •■-,£„) for which Q is defined. Then is Q neces-
necessarily a sum of squares of rational functions with rational coef-
coefficients? * By a rational function with rational coefficients we
mean a mapping (£1} • ■ •, £„) -> Q{$u •••,£„) where Q(xu • • •,
xn) is a rational expression in indeterminates X{ with rational coef-
coefficients. The domain of definition of Q is the set of real «-tuples
(£i5 • • -j £«) for which the denominator of Q(xu ■ ■ ■, xn) is not 0.
In 1927 Artin gave an affirmative answer to Hilbert's question
and proved the following stronger result.
Theorem 12 (Artin). Let * be afield of real numbers {that is, a
subfield of the field R of ordinary real numbers) which has a unique
* This is known as Hilbert's 17th problem. See D. Hilbert, Mathematische ProHeme,
Gottinger Nachrichten, 1900, p. 284 or Gosammelte Abhandlungen, Vol. 3, p. 317.

290 ARTIN-SCHREIER THEORY
ordering and let Q be a rational function with coefficients in <f> which
is rationally definite in the sense that (?(£i, •■■>£») > 0 for all
rational (£,) for which Q is defined. Then Q is a sum of squares of
rational functions with coefficients in <£.
Instances of fields $ which satisfy the condition are: the rational
field, any real closed subfield of real numbers, the field of all real
numbers. If we take $ to be the first of these, then Artin's
theorem gives a stronger result than that suggested by Hilbert.
Let $ be as in the theorem and consider the field $(xi) = $(xi, x2,
• • ■, xn) in indeterminates #,• with coefficients in $. This field is
formally real (ex. 4, § 1). According to Theorem 10, Q(xu ■ ■ ■ ,xn)
j£ 0 in $(Xi) is a sum of squares in this field if and only if Q > 0
in every ordering of $(#*). Hence Theorem 12 will follow if we
can show that, if Q ^ 0 is rationally definite, then Q > 0 in every
ordering of *(#,•). This will follow from the following
Theorem 13. Let $ be afield of real numbers, <$(*,•) = $(#1, • • •,
xn) the field of rational expressions in n indeterminates xt with
coefficients in $ and suppose an ordering has been given to <£(#,)
which extends the ordering of $ as a subfield of the field of real num-
numbers. Suppose fi(xu ■ ■ •,*„), • • -,fk(xu ■ ■ ■, xn) is a finite set of
elements of$(xi). Then there exists a rational n-tuple (aly • • •, an)
such that for every j, 1 < j < k, /y(*i, ■ • •, xn) is defined at («,-)
and fj and fj(au ■ • •, an) have the same sign in the sense that
f}(au • • •» an) < 0 according as /,• <■ 0 in the given ordering qf$(xi).
Suppose that this result holds and let <$ be as in Theorem 12.
Let Q 9^ 0 be an element of <$(#;) which is not a sum of squares in
$(x{). Then we know that there exists an ordering of $(^t) for
which Q < 0. Since $ has only one ordering, the ordering of
$(#,) is an extension of that of f>. Hence Theorem 13 gives a set
(a,), a( rational, such that Q(at) < 0. Then Q is not rationally
definite. Hence we see that, if Q is rationally definite, then it is a
sum of squares of elements of f>(xt) and this is Artin's theorem.
We shall prove Theorem 13—after some necessary preliminaries
—by induction on the number n of xt. The result is clear if n = 0
since in this case $(^i) = <!>, so the functions are just constant
functions. It remains to prove the inductive step, so we assume

ARTIN-SCHREIER THEORY 291
the result for $(#i, • • •, xn) and we shall prove it for $(#1, • • •,
*n,y), where y is an extra indeterminate. We shall see also that
it is, in essence, sufficient to consider polynomials in y with coef-
coefficients in <£(#,). Let Fx{xi,y), • • -,Fk(Xi,y) e$(*,•)[y]. Then we
shall call a property P of this set of polynomials in y rationally
specializable if there exists a set of elements fa(xi), ■ • ■, fa.(xi) in
$(xi) such that, if (aiy • ■ •, an) is any rational «-tuple for which
fa, • • •> th is defined and fa(ai), 1 < / < h, has the same sign as
fa (in a given ordering of $(*,-))> then the coefficients of all the
F}(xi,y) are defined at (au •••,«„) and the polynomials F^a^y),
• • •, Fk(ai,y) have the property P. We shall require two results
on specializable properties which we state as lemmas.
Lemma 1. The property that F(xt,y) = ym + ¥>i(#,)y~1 +
• • ■ + <pm(xi) has precisely r roots in the real closure of $>(#,•) is
rationally specializable.
Proof. We are assuming that $ is a subfield of the field of real
numbers and <£ is ordered by the ordering of the field of real num-
numbers. The subfield of the latter of elements which are algebraic
over <i> is a real closure A of $ (Cor. to Th. 6 and ex. 2, § 4). The
assertion of the lemma is that there exist fa, fa, • • •, fa in #(*,•)
such that if {au • • ■, an) is any rational «-tuple such that every
fa is defined at (au ••-,«„) and fa(au ■ • -,an) has the same sign
as fa, then F(ai,y) is defined and the number of its real roots (or
roots in A) is r. Let Fo = F(xi,y), F\, • • •, F, be the standard
sequence for F(xi, y) (as in G)). If (aly • • •, an) is a rational «-
tuple for which the non-zero coefficients of the Fj and of the quo-
quotients Qj as in G) are defined and have non-zero values in <£>,
then F0(ai,y), ■ • -,F,(ai,y) is the standard sequence for F{aiyy)
m
= F0(ai,y). Let tjC*,) = ]£ <Pj{xiJ + (m + 1). Then we have
seen (p. 283) that the r roots of F(xi,y) in the real closure of
$(#t) lie in the interval (—*7,17). By Sturm's theorem the differ-
difference in the variations in sign between the two sequences
Fo(xiy —n), Fx{Xi, -17), • • •, F,(xi} -ri) and F0(xi, if), F^x^ri),
•••,F,(xi,v) is r. Now let {^i(.v,-), •••,fa(xi)} be the set of
elements of <$(#,•) consisting of the coefficients of the standard
sequence for F and of the quotients Qj in this sequence, and the

292 ARTIN-SCHREIER THEORY
elements Fj(xi, -7?(#i, • ■ •, xn)), Fj(xi} r)(xu • • •,#„)), 0 <j < s.
Then it is clear from Sturm's theorem that, if (au • • •, an) is a
rational w-tuple such that the ^'s are defined at (aly ■ • •, an) and
every ^i{at, •■•,an) has the same sign as \[/i, then F(aiyy) is
defined and has exactly r roots in A in (—17(^1, • • •, «„), v(ai, • • •>
an)). If we refer to the result on bounds for the roots again we see
that there are no real roots of F(a{, y) outside of the indicated
interval. Hence r is the number of real roots of F{at, y).
Lemma 2. Let {Fi(xi,y), • • •, Ft{xiyy)} be a sequence of poly-
polynomials {not necessarily distinct) belonging to $(xi)[y]. Assume
the leading coefficients are 1. The property that F,{xi,y) has a
root pj in the real closure P of $(#,•) and pi < p2 < • • • < pt is
rationally specializable.
Proof. The elements pk and (py+1 — pjYA, 1 <j<t — \, are
contained in P and these generate a finite dimensional extension
field A of $(xi) which has a primitive element 6. Let g(xi,y) be
the minimum polynomial of 8 over $(#,-). We have pk = <Ph(xi, 0),
(py+i - pjYa = <Tj(xi, 6) where <pk(xi,y), <Tj{Xi,y) e*(*,-)|>]. Since
Fk(xi,pk) = 0, Fk(xi,<pk{xi',y)) has 6 as root and since g(xiyy)
is the minimum polynomial of d, we have
A1) Fk(xi,<pk(xi.,y)) = Gk(xi,y)g(xi,y), 1 < k < t.
Similarly the relation pj+i — py = aj{xiy BJ or <pj+i(xi} 0) —
<Pi(xi, 9) = ffj(xi, ffJ gives a relation
A2)
= H]{xi,y)g{xi,y), 1 <j < t - 1,
in ^{Xi)[y\. Since aj(xiy 6) 5^ 0 it has an inverse ry(^,-, 9) in A and
so we have relations of the form
A3) <rj(Xi,y)rj(Xi,y) ~ 1 = ki(xi,y)g(xi,y)
in $(xi)[y]. Let {^i(xu ■ • -,xn)} be a finite set of elements of
$(#,-) which includes the coefficients of the Fk(xi,y), all the coef-
coefficients of the polynomials in y appearing in A1), A2), and A3)
and a set of elements given in Lemma 1 to insure that g(ai,y)
has a real root 7. Moreover, if the «,• are chosen so that every
^j(«t) is defined, then substitution of y for y in every polynomial

ARTIN-SCHREIER THEORY 293
which occurs in A1), A2), and A3) is permissible. Substituting
y = y in A1) we see that Fk(aiyy) has the root /3fc = <Pk(#i, */)•
Substituting^ = y in A2) we see that /9,-+I — /9,- = p/+1(«,-, 7) —
¥>>(«., 7) = °v(«<> lJ > 0. By A3), we have a}{ai} i)Tj{aiy 7) =
1. Hence <ry(«,-, 7) 5^ 0 and /3y+1 > /3y. Thus we see that -FyOz,-, j>)
has the real root £,• and /3i < |82 <•••< |8< as required.
We can now give the
Proof of Theorem 13. As we have seen before, it suffices to
prove that, if the theorem holds for $(*i, • • •, xn), then it holds
for $(#i, • ■ ■, xn>y), y an extra indeterminate. Let P' be a real
closure of the ordered field $(xiyy), P the real closure of $(x,-) con-
contained in P'. We are given a finite set of elements F(xiyy) of
*(*»■> y) and we have to show that rational «,• and £ can be chosen
so that F(ai} b) is defined and has the same sign as F(xiyy) in the
ordering in P' and this holds for every F in the given set. We can
write F = <p(xu ■ • ■, xn)Pi(xiyy) ■ ■ ■ Ph{xi,yYh where <p(xu ■■■,
xn) e$(xi)y ej is an integer, Pj(xiyy) is irreducible in $(x,)[^] and
has leading coefficient 1, and the Pj(xi,y) are distinct. If «», b
have the property that <p(ai> • • ">«n), Pj{aiyV) are defined and
have the same sign as <p and Pj, 1 < j < h, then F(au ■ ■ •, an, b)
is defined and has the same sign as F. This remark shows that
we may as well suppose that the given set consists of elements
<p e $(xi) and F e $(xpi)[jy] such that every F is irreducible in $(#,-)[y]
and has leading coefficient 1. Let pi < p2 < • • • < Pt be the roots
in P of the given set {F} of polynomials in y. We can form a
sequence Fly F2, ■ ■ -,Ft whose terms are in {F} so that pj is a
root of Fj. Since the F are irreducible and the field is of charac-
characteristic 0, the roots of F are distinct. Also distinct F's are relatively
prime. Hence if G(x{, y) is the product of the distinct .F's, then
G has distinct roots. By Lemma l,we can find elements ^i, ■ • -,^a
in $(xi) such that, if au ■ ■ ■, an are rational and every if/i is
defined at (au • • •, an) and tyi{au ■ ■ ■, an) has the same sign
as ^j, 1 < / < h, then G(ai} y) is defined and has / real roots.
By Lemma 2, we have elements th+i, • • •, ^k so that, if the a's
are rational and i>m{a\, • • ■, an) is defined and has the same sign
as \f/m, h + 1 < m < k, then Fj(at,y) is defined and has a real
root ^j so that 0i < /32 < • • • < fr. We now add to the i^'s al-
already given all the elements <p of the set given initially and the

294 ARTIN-SCHREIER THEORY
discriminant 8 of G{Xiy y), which is different from 0, since G
has distinct roots. By the induction hypothesis, we can choose
rational a{ so that all the conditions given by the ^'s, the <p's, and
S are satisfied. Now in P[y] we have the factorization
A4) Fj(xity) = (y - Ph)(y - Ph) ■ ■ ■ (y - pjtt)Qi(y) ■ ■ ■ Q.,(y)
where the Q's are irreducible quadratics with leading coefficients
1 and {ji, j2, • ■ ■, jtj} are distinct and are a subset of {1, 2, • • •,
t}. Our choice of the a's insures that in the field of real numbers
we have
A5) Fjia^y) = (y - ph)(y - 0h) • • ■ (y - &„) S1{y) ■ ■ ■ S.t(y)
where the »S"s are irreducible quadratics with leading coefficients
1. Since y is transcendental over $(#,•) and the pj are algebraic
over this field, it is clear that y is contained in one of the follow-
following open intervals in P': (-», Pl), (pu p2), • • •, (jpt-u Pt), (pt, °°)-
Also we have seen that an irreducible quadratic with coefficients
in a real closed field and leading coefficient 1 has the form (y — yJ
+ S2, S ^ 0 (see C)). This implies that every Q(y) > 0 in P',
and for any real number b, S{b) > 0 for the «S"s in A5). It now
follows from A4) and A5) that, if y is in the k-th. interval of the
sequence ( —<», Pi), (pi, P2), ■ ■ ■ > (pi, °°) and b is any real num-
number contained in the &-th interval ( —oo, ^j), @U /32), • ■ •, (f}h «),
then Fj{a{, b) and Fj(x{, y) have the same sign and this will hold
for every j. Now it follows from the archimedean property of the
real field that every open real interval contains a rational num-
number. Hence we can choose a rational b so that Fj{cii, b) has the
same sign as F,-(xi,y) for all the^'. This completes the proof.
Remark. It is natural to ask if a result like Artin's holds for
polynomials with coefficients in a fields such as in Artin's theorem.
In view of this theorem one can formulate the question in the
following way: Let P(xi, • ■ ■, xn) e$[xi, • • -,xn] such that P =
XRi(xi, • • •, xnJ where the i?»- are rational expressions in the .v's
with coefficients in $. Does this imply that P = SPy(xi, • • •, xnJ,
Pj e^[xi, • • •, xn]. Artin has shown that this is correct if n = 1
and # is any field of real numbers. On the other hand, some ex-
examples due to Hilbert show that the result is false for n > 1 even
for * the field of real numbers.

ARTIN-SCHREIER THEORY 295
EXERCISE
1. Let $ be an ordered field and $(#i, ■ • •, *n) the field of rational expressions
in indeterminates xy, ■ ■ -,xn over $. Suppose Q E $(**) satisfies Q(l-1, • • •, £m)
> 0 for all £< in a real closure A of $ for which Q{%\, • ■ ■, £„) is defined. Prove
that Q = 2/3,\F,(.*i, • • •, #nJ where the 0; are non-negative elements of $ and the
. (Hint: See ex. 1 of § 5 and prove a suitable analogue of Th. 13.)
7. Formalization of Sturm's theorem. Resultants. In the
next few sections we shall develop an algorithm, due to Tarski
and to Seidenberg, for testing the solvability in a real closed
field of a finite system of polynomial equations and inequalities (in
several variables). The ultimate test (Th. 16) will consist in the
verification of a finite system of polynomial equations and in-
inequalities in the coefficients of the given system. In this section
we shall consider first a reformulation of Sturm's theorem in this
manner. We shall develop also an elimination method based on
resultants which will be essential in the sequel.
To obtain the formalized version of Sturm's theorem it is con-
convenient to begin with the ring 21M where 21 = R0[ti, • ■ ■, tr], x
and the ti are indeterminates and Ro is the field of rational num-
numbers. Let F{tu ■ ■ ■, tr; x) e 2l[#], so F(ti; x) = anxn + an_xxn-x
+ • ■ • + a0 where the at e 21 and an 5* 0. If the tt are specialized
to ti = Ti e $, then we obtain the polynomial f(x) = F{Tt\ x) e
$[x] where deg f(x) < n. We shall now obtain a number of se-
sequences E: \F0(ti\ x), Fiid; x), ■ ■ •, Fa{ti\ x)} such that for any
Ti, 1 < i < r, in $ there exists one of these sequences E such that
the specialized sequence {F0(Ti\x), _Fi(t,-; #),•••, F,(t,-; #)} is
essentially the standard Sturm sequence G) for f{x).
Our choice for F0(ti; x) is any one of the polynomials amxm +
am-iXm~1 + • • • + a0, am 5^ 0, m < n obtained by dropping lead-
leading terms anxn + ■ • • + am+ixm+1 of F(ti;x). Next we take
Fi(ti; x) = F0'(tiw, x) the formal derivative of Fo considered as a
polynomial in x. Suppose we have already defined Fo, Fx, • • •,
Fk. If Fk = 0, we break off" the sequence with Fo, Fu ■ ■ ■, Fs =
Fk-i. Otherwise, let Fk_r = bpxv -\ h i0, Fk = cgxq -\ \-
c0 where bv 9^ 0, cq ^ 0, and p > q. The usual division process
shows that we can find polynomials Q{x), R{x) in %[x] such that
+fc_i = QFk - R where degl R(d; x) < deg, Fk(/n x).

296 ARTIN-SCHREIER THEORY
For our purposes it is preferable to replace p — q + 1 by the
smallest even integer e > p — q + 1. Thus, changing notation,
we write cqeFk-i = QFk — R and we call Q and R the quotient
and remainder on dividing Fk_x by Fk. It is clear that these are
unique. We now take Fk+i to be R or one of the polynomials ob-
obtained from R by dropping leading terms in the manner that
Fo(ti', x) was obtained from F(/*; x). The sequences {Fo, Fi, ■ • •,
Fs} obtained in this way will be called generic standard sequences
for F(t{; x). Clearly the number of these is finite.
Let E: {Fo, Fi, ■■■,FS} be one of the generic standard se-
sequences for F. Then we associate with E two finite subsets 8(E)
and X(£) of 21 = Ro[ti]- The set S(E) is the set of coefficients of
the terms dropped in forming the sequence. Thus, S(E) consists
of the coefficients (of the powers of x) in F{tt\ x) — F0(ti; x) and
those of R — .Ffc+i, k > 1, as above. We let \{E) be the set con-
consisting of the leading coefficients of the Fi in the sequence.
Now let T,e<i>, 1 < i < r, and consider the polynomial f(x) =
F{Ti;x). We assume f{x) ?± 0. Suppose deg f(x) = m < n.
Then we shall take F0{ti\ x) = amxm + • • • + a0 and we have
F0(Ti; x) = f{x), am(T{) ^ 0. The condition -F0(t;; x) = f{x)
gives an(ri) = • • • = am+1(n) = 0. One sees easily by an induc-
inductive argument that there exists a generic standard sequence
{Fo, Fiy ■ ■ ■ ,FS] for F such that, if 4 is the leading coefficient of
Fk, then
4(r;) ^0, 0 < k < s
A6) F0(Ti;x) = f(x)
lk(Tiy"Fk^i(ri;x) = Fk{ri-x)Qk{ri-x) - F*+1(t,-; *),
where 0 < k < s, Fa+i = 0 and ek is an even integer, and Qk
is the quotient on dividing Fk_i by Fk. Since -Fo(t,-; x) — f(x),
Fi(Tu x) = f'(x), lk{Ti)ek > 0 and the degrees of the Fk(n; x) are
decreasing, it is clear that the terms of {F0{ri\ x), -Fi(t,; x), ■ ■ ■,
F,(ri, x)} differ from those of the standard sequence G) for f(x)
by positive multipliers in f>. Hence the sequence {F0(Ti4, x), ■ ■ ■}
can be substituted for the standard sequence in applying Sturm's
theorem.
We shall now formalize the conditions given in this theorem by
considering the finite collection of systems of equations and in-

ARTIN-SCHREIER THEORY 297
equalities made up of relations of the following types:
Fk{h;y)Fl{ti;y) < 0,
Fk+1(ti;y) = Fi^ti-y) = 0, k < I
Fp{t{- z)Fg(ti; z) > 0,
Fp+1(ti; z) = F,_i(/<; z) = 0, p <q,
where y and z are indeterminates and we require that the pairs
(k, /), • ■ ■ ,{p, q), ■ • • used in the two sets of relations are such
that, if ek = 0, 1, — 1, ijp = 0, 1, — 1 satisfy the same conditions,
namely, ekei < 0, ek+i = • • • = «j_i = 0, ■ • -,i}priq > 0, t/p+1 =
• • • = ijg_! = 0, • • •, then the number of variations of sign of
{e0, 6i, ••■,€«,} exceeds that of {770,171, • • ■, 77,}. Now let (/3, 7)
be an interval in $, 0 < 7, such that /(/3) ^ 0, /(?) ^ 0. Then
it is clear from Sturm's theorem that f(x) has a root in (/3, 7) if
and only if /,• = t,-, _y = /3, z = 7 satisfies one of the systems of
relations A7).
If we take into account all the generic sequences E and observe
that A6) is equivalent to the conditions /(tj) 9^ 0, d{ji) = 0 for
all / eX(£) and dz 8(E), we see that A6) and A7) give a finite
collection of conditions {G1} G2, ■ • •, Gh}, where each G3- is a finite
set of polynomial equations and inequalities with rational coef-
coefficients, such that f(x) has a root in (/?, 7) if and only if /t- = r,-,
y = j8, z = 7 satisfies all the conditions of one of the systems G,.
Now let /(*) = amxm + a™-!* H h «o where am 5^ 0.
Then we know that all the roots in <E> of f(x) are contained in the
ire—1
interval ( — 77,77) where 77 = m + 1 + 2 «;2«m~2. We have
0
F0(ti; x) = amxm + am_xxm~x -\ \- a0 where am = am(tu • • •,
m-l
/r) r^ 0. If we substitute^ = — (m + 1) — J2 a?am~2 in Fh{U\
m-i °
y) and z = m + 1 + 23 a?am~2 in F^fe z) and clear of fractions
0
by multiplying by a suitable even power of am, we obtain poly-
polynomial relations like those in A7), for the tt alone. In this way
one obtains a finite collection {Gx, G2, • • •, Gh} where each Gj is a
finite system of polynomial equations and inequalities with

298
ARTIN-SCHREIER THEORY
rational coefficients in the ti alone such that the statement that
f(x) has a root in $ is equivalent to the validity of one of the
systems Gj for /,- = r,-.
We shall consider next a classical determinant criterion for the
existence of a common factor of positive degree for two poly-
polynomials. We consider the polynomials f(x) = anxn + an^.1xn~1
+ • ■ ■ + «o, g(x) = Pmxm + Pm-xx™-1 + • • • + 0«> in <%] where
<l> is an arbitrary field. We assume m > 0, n > 0, but we shall
allow an = 0 or 0m = 0. The result we require is the following
Theorem 14. Let /(*) = anxn + an_xxn~l H \- a0) g(x) =
$mxm + 0m_i#m"~1 H h 0o wAi?r<? m,n > 0 and put
A8)
m rows
» rows
R(f> g) = 0 if and onh if either an = 0 = 0m or f(x
g(x) have a common factor of positive degree in x.
Proof. If «„ = 0 = 0m, then the first column of the determi-
determinant is 0. Hence R(f, g) = 0. Next assume f(x) and g(x) have
a common factor h(x) of positive degree and that either an j* 0
or 0m 7± 0. Then /(#) = fi(x)h(x), g(x) = ^i(^)A(x) and either
fi(x) 9^ 0 or ^i(^) ^ 0, according as an ^ 0 or 0m ^ 0. By sym-
symmetry, we may assume an ?± 0, f\{x) ?± 0. We have f{x)g\{x) =
g(*)/i(x), /W = /i(*)A(*) ^ 0. If degA(*) = r, then deg/j =
n — r. If g(x) = 0, we have gi(x) = 0; otherwise the relation
f(x)gi(x) — g(x)fi(x) gives deg^i(^) < m — r. Hence we may
write fx{x) = -7n_i^n~1 - yn^2xn~2 7o, gi(x) =
bm-\Xm~1 + Sm_2^m~2 + • • • + 80, so that we have
A9) (anxn H h ao)(dm_1xm-1 -\ \- 80)
+ ■ —h 0o)Gn-i^n~1 + • • ■ + 7o) =0.

ARTIN-SCHREIER THEORY 299
If we equate to 0 the coefficients of ^m+"-1,^m+n~25 • • •, 1 in A9),
we obtain the following equations:
«nSm_l + /3m7n-l = 0
Cin&m-2 + an_i5m_! + $m1n-2 + Pm-lfn-l = 0
B0)
ao8o
We consider this as a system of homogeneous equations in the
7's and 5's taken in the order 5m_i, 5m_2, • • •, 50, Tn-i, • • •, To-
Since not all the y's are 0, the determinant of the coefficients of
the y's and 5's is 0. If we take the transpose of this determinant
obtained by ordering the y's and 5's as indicated, we obtain A8).
Hence R(f, g) = 0. Conversely, assume R{f, g) = 0. Then we
can re-trace the steps through B0) and A9) and conclude that
there exist fi(x), gi(x) such that f(x)gi(x) = g{x)fi{x) where
deg /i < » — 1, deg gi < m — \, and either ft 5* 0 or £1 1A 0.
Assume /1 7^ 0. If gx = 0, then g = 0, /3m = 0, and either f{x)
is a non-zero common factor of / and g or an = 0. If gi 7^ 0 and
g = 0 the foregoing argument applies. Now suppose gi ?* 0 and
g * 0. Then the relations j{x)gM = g(x)Mx), fx ?* 0, ^ ^ 0,
^^0 imply / 5^ 0. Either an = 0 = /3m or we may assume «„ 5^
0 which implies that deg /(#) = w. Since deg/iW < » — 1,
the relation f(x)gi(x) = g(x)fi(x) and the factorization of the
non-zero polynomials /, fi, g, g\ into irreducible factors implies
that f(x) and g(x) have a common factor of positive degree.
We shall call R(f,g) the resultant of / and g (relative to x).
If either highest coefficient of /or of g is not 0, then the vanishing
of R(f, g) is a polynomial relation on the a,-, /37- with integer coef-
coefficients which is equivalent to the statement that / and g have a
common factor of positive degree.
EXERCISE
1. Prove that, if/(*) = xn + a^x"-1 -\ \-a0, then R(J, /) is the dis-
discriminant of/(x) (cf. § 3.1).

300 ARTIN-SCHREIER THEORY
8. Decision method for an algebraic curve. In this section we
shall give Seidenberg's method for deciding the solvability of an
equation f(x,y) = 0 in a real closed field. This will be based on
the result which we shall now establish, that if f(x, y) = 0 has a
solution in $, then the equations f(x,y)=0, (y — 8)
df dX
(x — y) — = 0, have a common solution in <£ for any y, 5 in $. The
geometric idea underlying this result will be clear from the two
lemmas which are used to prove it.
Lemma 1. Let f(x,y) e$[x,y], x,y indeterminates, $ a real
closed field. Then if f(x, y) = 0 has a solution in <£, it has a solu-
solution (a, /3) nearest the origin.
Proof. We consider the intersection in the space i>B) of pairs
(£> v)> £> V e*j of the curve C: f(x,y) = 0 with the circle x2 + y2
= 72j 7 > 0. Our hypothesis implies that there exist y for which
this intersection is not vacuous, and we have to show that the
set S of y > 0 such that C meets x2 + y2 = y2 in $B) has a
minimum. We now consider the polynomials f(x,y) and x2 +
y2 — c2 as polynomials in y with coefficients in <£(<:, x), where c
and x are regarded as indeterminates, and we form the resultant
g(c, x) of these two polynomials. The formula A8) shows that
g(c, x) is a polynomial in c and x with coefficients in *. If (a, |8)
is a point of intersection of the circle x2 + y2 = y2 and the curve
C, then f(a,y) and y2 + a2 — y2 have a common factor
y — [1. Hence ^G, a) = 0 and g(y, x) has the root a e $. More-
Moreover, —7 < a < y. Conversely, assume that for 7 > 0, g(y, x)
has a root a in $, — 7 < a < y. Since the leading coefficient of
y my2 + a2 — y2 is 1, it follows from Theorem 14 thatjy2 + a2 —
y2 and f(a,y) have a common factor in 4>[jy]. Since the
factors of y2 + a2 — y2 are y ± @ where /3 = G2 — a2)H, it
follows that (a, /3) or (a, — ^) is a point of intersection of the two
curves. Hence we see that the set S of 7 > 0 such that C meets
x2 + y2 = y2 in <i>B) is the same as the set of 7 > 0 such that
g(y, x) = 0 has a root * = aef, — 7 < a < 7. Let <S" be the
subset of S of the 7 such that g(y, ±7) 7^ 0. For these the condi-
condition is that #G, x) has a root in (—7, 7). It is clear that we can

ARTIN-SCHREIER THEORY 301
obtain g(y, x) by a suitable specialization of a polynomial with
rational coefBcients such that one of the parameters is specialized
to 7. Hence we can apply the result obtained in the last section
to conclude that S' is the union of a finite number of sets defined
by polynomial equations and inequalities of the form p(c) = 0,
q(c) = 0, r{c) ^ 0 where p, q, r are polynomials with coefficients
in <i>. One sees easily that such a set is a union of a finite number
of intervals which may be open, closed, half open, single points or
extend to infinity. Since the set of 7 such that ^G, ±7) = 0 is
either finite or all 7 > 0, it is clear that S has the same structure
as S'. The result will now follow by showing that the complement
of S in the set of non-negative elements is the union of open inter-
intervals; for, this implies that S is the union of a finite number of
closed intervals and hence has a minimal element. Thus, let
5 > 0, 5 i S. Then g(8, x) = 0, — 8 < x < 8 has no solution x in
<*. Write g{c, x) = go(x) + gl(x)(c - 8) + ■ ■ ■ + gm{x){c - h)m
where the gi(x) are polynomials in x. Then go(x) 5* 0 in —5 <
x < 8. It follows that there exists a 8' > 8 such that ^o(^) ** 0
in -8' < x < 8'. Then there exist b > 0, B > 0 such that
|jTo(*) I > b, \gi(x) I < B for all x in [-5', 5'] (ex. 3, § 3). Then,
if \c — 8\< |and \c - 8\ < b/\B and x e[-5', 5],
\g(c, x) I > \go(x) I - \gi(x)(c - 5) + • ■ • + gm{x){c - 8)™\
>b-2B\c-8\>b-b- = b--
This implies that every 8" satisfying S" < 5', 8" < 8 + £, 5" <
8 + b/4B is contained in the complement of S. Hence this com-
complement contains an open interval containing 8 and the proof is
complete.
As in the classical case of the field of real numbers, a point
(a, /3) on f{x, y) = 0 is called a simple point if
£) Wo,.
Then the normal vector at (a, K) is (( — ) , ( — ) ) and the
\\dx/(a,l)) \dy/(a,i3)/

302 ARTIN-SCHREIER THEORY
tangent line to the curve at (a, 0) is defined by the equation
Now let (a, j8) be a point on C: f(x,y) = 0 in <l>B) nearest the
origin. We wish to show that P[ — ) — a( — ) =0.
\dx/(a,fi) \dy/(a,f))
This is clear if (a, /3) = @, 0) or (a, /3) is not a simple point.
Otherwise, the equation states that the vector joining @, 0) to
(a, /3) and the normal vector are linearly dependent. Hence C
and the circle with center at the origin and radius (a2 + /32)^
have the same tangent at (a, j8). If this is not the case, then the
tangent to C at (a, /3) contains interior points of the circle whereas
C itself does not. The result will therefore follow from
Lemma 2. Let p be a point of intersection {coordinates in $) of a
circle and a curve C:f{x,y) = 0, f(x,y) e$[x,y]. Assume p is a
simple point and the tangent at p to C has points interior to the
circle. Then C itself has points interior to the circle.
Proof. We take p = @, 0) and the tangent to C at p to be the
/df\
#-axis. Then /@, 0) = 0 and ( — ) = 0, and we may suppose
that (—) @, 0) = 1. The center of the circle is not on the x-
\dy/
axis, so we may denote it as (a, b) with «^0. We have f(x,y) =
dx/o \dy/o 2!
—2 ) y%\ "l"'''> so taking into account the conditions on / we
dy /o J
see that we can write f(x,y) = y(l + h(x,y)) + g(x) where
h@, 0) = 0 and g{x) is a polynomial in x divisible by x2. Since
h@, 0) = 0 we may choose a 5 > 0 such that \h(x,y)\ < % if
|*| < 5 and |.y| < S. Then \ < 1 + h(x,y) < f and 8A +
h(x, 8)) is between -|5 and fS while —5A + h(x, —5)) is between
—%8 and —18 for all x satisfying |*| < 5. Since ^@) = 0 there
exists a 5', 0 < 5' < 5 such that /(*, 5) = 5A + h(x, 5)) +
g(x) > 0 and f(x, -5) < 0 if |*| < 5'. Then for every x0,

ARTIN-SCHREIER THEORY 303
|*o I < 3' there exists a.y0 e [ — 8, 8] such that f(xo,yo) = 0. Then
(a - x0J + (b - y0J
gOo) \2
1 + h(xo,yo)/
(f(*o)J
1 + A(*o, Jo) A +
Since ^(*o) is divisible by x02, it is clear that, if we take x0 suf-
sufficiently small so that ax0 > 0, then (a — x0J + (i> — y0J <
a2 + b2. Hence (x0, y0) is a point on C interior to the given circle.
Our results now show that, if C:/(x,y) = 0 has a solution in
df df
$, then there exists a solution in <i> which is also iny x — = 0.
dx dy
If we replace the origin by G, 5) where 7, 5 e f>, then we see in
the same way that the intersection of C and the curve D:
(y — 8) (x — y) — = 0 contains a point in <i>B).
dx dy
We shall now apply this to obtain Seidenberg's procedure for
deciding the solvability in $ of f(x, y) = 0. First, we can obtain
the highest common factor of the coefficients of the powers
of y in f(x,y) and put /(*, y) = d(x)fi(x, y) where f^x, y) is
not divisible by a polynomial of positive degree in x alone.
Evidently f{x,y) = 0 has a solution in $ if and only if either d{x)
= 0 or fi(x,y) = 0 has such a solution. This reduces the discus-
discussion to polynomials which are not divisible by polynomials of
positive degree in x alone. Next we can compute a highest
common factor in <!>(.*) fy] of f(x,y) and — /(*,jy) by using the
Euclidean algorithm. We may assume this belongs to $[x, y] and
is not divisible by polynomials of positive degree in x alone. Then
we can divide out by this highest common factor and obtain a
polynomial g(x,y) which is a factor of f{x,y), has the same
irreducible factors as/(*,jy) and has no multiple factors in $[x, y).
Clearly, f(x, y) = 0 is solvable in $> if and only if g(x, y) = 0 is
solvable in <f>.

304 ARTIN-SCHREIER THEORY
If we replace / by g and change the notation back to / we may
suppose that f{x,y) has no multiple factors of positive degree
and no factors of positive degree in x alone. The first of these
d
conditions implies that f(x,y) and —f(x,y) have no common
dy
factor contained in*(#)[y] of positive degree in^y.
d f d f
We now consider the polynomial g{x, y) = y (x — y) —
dx dy
where 7 is any element of <t>. We know that, if the curve C:f(x,
y) = 0 contains a point in <£>B), then the intersection of C and
D:g(x,y) = 0 contains such a point. Before we can make use
of this it is necessary to arrange matters, by choosing a suitable
7 so that the intersection of C and D is a finite set. To do this we
introduce another indeterminate c and we consider the poly-
polyps 1 a f
nomial g(c: x,y) = y (x — c) — . Let R(c; x) be the result-
resulted dy
ant relative to y (that is, considering the polynomials as poly-
polynomials in^) of f(x,y) and g(c; x,y). We claim that R(c; x) j* 0.
Otherwise, R(y; x) = 0 for all 7. Now, Theorem 14 shows that
if 7 has this property, then g(y; x,y) and f(x,y) have a common
factor in$(x)[y] and, consequently, in $[#,_)'] of positive degree in_y
(see Vol. I, p. 125). Hence if R(y; x) = 0 for all 7, then there exist
distinct 7, say, yt and 72 such that f(x,y), g(yy, x,y) and ^G2;
x, y) all have a common factor of positive degree in y. This follows
from the fact that to within associates f(x,y) has only a finite
number of different factors in $[#,.)']. We can then conclude
that f{x,y) and G1 - 72) — f(x,J>) = g(ji;x,y) - g(y2; x,y)
have a common factor of positive degree in y and this
contradicts the fact that f{x,y) and —f{x,y) have no such
dy
common factor. This proves that R(c; x) 9^ 0.
We can now choose a 7 e$ such that R(x) = R(y; x) 9* 0. Set
g(x,y) = £G;#,.y). Then f{x,y) and g(x,y) have no common
factor of positive degree in y and no common factor of positive
degree in x alone; hence they have no common factors except units
in $[x,y]. It follows that the resultant Q{y) of f(x,y) and g(x,y)

ARTIN-SCHREIER THEORY 305
relative to x is not 0. Let V be the intersection in fiB), where
0 = <1>(\/ — 1) is the algebraic closure of <i>, of C:f(x,y) = 0 and
D:g(x,y) = 0. If (p, <r) e P, /(P, <r) = 0 = j(p, a) imply that R(p)
= 0 and Q(<r) = 0. Since #(*) ^ 0, 000 ^ 0, this gives only a
finite number of possibilities. Hence V is a finite set. We know
that, if C contains a point in f>B), then V has such a point and,
consequently, R(x) has a root in <l>. Conversely, suppose R(x)
has a root a in <£. If a is not a root of the polynomial in x which
is the coefficient of the highest power of y in f{x,y), then R(a) =
0 implies the existence of a a e fl such that (a, <r) e V. If a =
/3 e <!?, then we have the desired result that V, and hence C, has a
point in <i>B). Otherwise, (a, <r) e V where a t^ <t is the conjugate
of o- under the automorphism ^ 1 of Q/<J>. Then we have two
points in V: (a, <r) and (a, <r) with the same abscissa.
We can easily overcome^—by a suitable choice of axes—the two
difficulties which we have noted which may prevent concluding
that F, and hence C, has a point in <i>B) from the fact that R(x)
has a root in $. We shall change to an ^',jy'-system where x =
n(x' + y'), y = y' and ft ^0 will be chosen suitably in <£. The
equation of C in the x',j'-system is f(fi(x' +y'),y') = 0.
Let fn(x,y) be the homogeneous part of highest degree w (>0)
in x and_y in the polynomial f(x,y). Then the coefficient of (y')n
in f(fj.(x' + y'),y') is/n(M> !)• Since/„(*■, 1) ?* 0, we can choose
H e <£> so that/n(/u, 1) 5^ 0. Since the total degree of f(x, y) is
n, it will follow that the constant/B(/x, 1) j* 0 is the polynomial
in x' which is the coefficient of the highest power of y' in
f(n(x' + y'),y'). This will take care of one of the difficulties.
To take care of the other we compute, by using the Euclidean
algorithm, applied to R(x) and R'(x), a polynomial r(x) which
has simple roots that are the same as those of R(x). Similarly, we
compute a polynomial q{y) having simple roots the same as those
°f Q(y)- We note next that we can compute a polynomial s(x)
whose roots are (p,- — pv){(Tj — <tj')~1 where p1} ■ • •, ps are the
roots of r{x) and <ru ■ • •, at are those of q(y), i 7^ i',j 9* j'. For
this we introduce indeterminates £,-, 1 < i < s, r/j, 1 < j < t, and
we consider the polynomial
n Km - vj')x - («v - mi

306 ARTIN-SCHREIER THEORY
This is invariant under all permutations of the £'s and r?'s so the
coefficients of the powers of x are polynomials with integer coef-
coefficients in the elementary symmetric polynomials of the £'s and
the 17's (Vol. I, p. 109). If we replace these elementary symmetric
polynomials by the corresponding coefficients of r{x) and q{y)
normalized to have leading coefficients 1, we obtain a polynomial
s(x) whose roots are (p; — Pi')(oj — <fj')~1> ' ^ l>■> J ^ j' • Assume
now that y. is not a root of s(x) (as well as not a root of fn(x, 1))
and consider the set of points (p,, 07). This contains V and no
two distinct points in this set have the same abscissa in the x', y'-
system since (p, a) is the point (m-1p — o", o) in the x', y-system.
Hence n~lpt - a, ^ n~lPi- — <tj> if (*, j) ^ (*', /)•
We now choose ju as indicated and we replace f(x,y), g(x,y) by
K*>y) = /(m(* +y),y) and k(x,y) = g(n(x +y),y). Let /(#)
be the resultant relative to y of h{x,y) and k{x,y). Then the
argument shows that f(x,y) = 0 is solvable in <1> if and only if
f(x) has a root in 3>. The latter problem can be decided by Sturm's
theorem.
In order to carry this over to more than two variables it is
necessary to consider polynomials involving parameters and to
apply an inductive procedure. This necessitates an extension of
the decision method we have just given to take care of an equa-
equation f(x,y) = 0 restricted by an inequality g(x) 5^ 0. To handle
this we first obtain a highest common factor d{x) of g(x) and the
coefficients of the powers of y in f(x,y) by the Euclidean algo-
algorithm. Write f(x,y) = d{x)fi{x,y), g(x) = d(x)g1(x). Then
the pair of conditions /(a, /?) = 0, g(a) 7^ 0 is equivalent to the
pair: fi(a, /3) = 0, g(a) ^ 0. This remark permits us to reduce
the consideration to the case in which g{x) and f(x, y) have no
common factor of positive degree. To avoid considering trivial
cases we assume also that deg g(x) > 0 and deg* f(x, y) > 0.
Let T(y) be the resultant relative to x of f{x,y) and g(x). Then
T(y) 7^ 0 since, otherwise, f(x, y) and g(x) have a common factor
of positive degree in x in "t^jj] contrary to our arrangement.
We now choose t in <i> so that T(r) 5^ 0 and we replace f(x, y) by
h(x,y) = f(x,y + r). Then the resultant relative to x of h(x,y)
and g(x) is T(y + r) which is not 0 for y = 0. This implies that
g(x) and h(x, 0) are relatively prime. Clearly we can replace the

ARTIN-SCHREIER THEORY 307
pair f{x,y), g(x) by the pair h(x, y), g(x) for the problem of test-
testing the existence of a solution in <t> of f{x, y) = 0, g(x) 9^ 0. Now
let k{x,y) = h(x,g(x)y). Then if (a, 0) satisfies h(a, 0) = 0,
g(a) 5^ 0, we have k(a, 7) = 0 for 7 = %(a)-1. On the other
hand, if k(a, 7) = 0, h(a, g(a)y) = 0, so g(a) ^ 0 since h(x, 0)
and #(#) are relatively prime. Hence a and /3 = £(aO satisfy
A(a, (8) = 0,£(a) ^ 0. This shows that f{x,y) = 0,g(x) 9* 0 has
a solution (a, /?) in $B) if and only if k(x, y) = 0 has a solution in
$B) and this is the situation we handled before.
9. Equations with parameters. If one attempts to extend the
method which we have given in the last section to more than two
variables, one is led to treat all but two of the variables as parame-
parameters and to seek a reduction of the number of variables by means
of the method. This leads to the consideration of polynomials in-
involving parameters. Since the parameters will be allowed to take
on any values in the real closed field, there is no loss in generality
in assuming that the coefficients of the polynomials are rational
numbers. Moreover, the result one obtains in this way will be
applicable impartially to all real closed fields, and this can be used
to establish an important principle due to Tarski which states
that any elementary statement of algebra (this has to be made
precise) which is valid for one real closed field is valid for all real
closed fields. The main result in Seidenberg's method for treat-
treating these questions is the following
Theorem 15. Let F(ti; x,y) eR0[tx, ■ ■-,tr;x,y], G(/*; x) e
Ro[h> • • •> tr\ x], /,-, x,y indeterminate s, Ro the field oj rational num-
numbers. Then one can determine in a finite number oj steps a finite set
of pairs of polynomials {F^tf, x), Gy(/,-)), Fj e R0[ti; x], Gj e R0[ti],
j = 1, 2, • • •, h, such that, if $ is any real closed field, then r,- e#,
1 < i < r, has the property that
B1) F(rtix,y) =0, G(ri;*)^0-
is solvable for x,y ef> if and only if one of the conditions:
B2) G,(t<) 5^ 0 and F^n; x) = 0
is solvable for x in $, is satisfied.

308 ARTIN-SCHREIER THEORY
The proof of this theorem is essentially a formalization of the
decision method of the last section. We consider first some neces-
necessary preliminary notions.
We shall call the set $>(r) of r-tuples (ti, t2, • • •, rr), t,- ef>, the
parameter space. A finite set of pairs of finite subsets Ey, Xy) of
21 = Ro[ti\, j = 1, 2, • • ■, A, will be called a rational cover if for any
$ of characteristic 0, <t»(r) is the union of the sets <Sy where S}- is
defined by Ey, Ay) in the sense that it is the set of (t.) e$(r) satis-
satisfying d(n) =0,de Sh l(n) * 0, / e Xy. If («/, A/), i = 1, •••,*,
Eft", X&"), & = 1, ■ • •, q are rational covers, then so is (8/ U 5k",
X,' U Xj/O, j = 1, • ■ ■, h, k = 1, • • •, q. The corresponding sets
are the intersections of those of the two given rational covers.
We shall call this a refinement of the two rational covers.
We have noted that, if F = anxn -\ \- a0, G = bmxm +■ ■ ■
+ b0, a,-, bj e R0[ti], an ^ 0, bm t± 0, n > m, then we have a
uniquely determined division algorithm which yields an even
integer e > n — m + 1 and a quotient Q and remainder R in
Roid; x] = SIM, 21 = Ro[ti\, such that bm'F = QG - R where
degx /? < degx G. This can be extended to the case n < mor F — 0
by taking e = 0, Q = 0, R = —F. We now associate with the
pair (F, G) a number of generic Euclidean sequences Fo, Fi, • • •, F,
determined by the following rules Fo and Fi are F and G or are
obtained from these respectively by dropping leading terms.
Thus, Fo = apxp + \- a0 where 0 < p < n, Fj = bqxq -\
+ b0, 0 < q < m. If Fi = 0 we take s = 0 and let the sequence
consist of Fo alone. Otherwise, we divide Fo by Fi and we let F2
be the remainder or a polynomial obtained from the remainder
by dropping leading terms. If F2 = 0, we stop with F0,Fi;
otherwise, we repeat the process. Clearly, this process breaks off
in a finite number of steps and, since we have only a finite number
of choices for every Fk, we obtain a finite number of generic
Euclidean sequences E for (Fy G). Set £)(/,-; x) = F,{ti\ x) the
last term in the sequence E. Then D ?* 0 unless Fo = Fi = 0
and, except in this case, we can divide F and G by D obtaining
m(ti)eF = FWD - Rw, m(tiYG = GA)D - S™ where «(/<) is
the leading coefficient of D, e and / are the even integers, and
Fw, GA) the quotients, Rw, Sw the remainders obtained in the
division. With each E we associate also the pair of subsets

ARTIN-SCHREIER THEORY 309
X(is)) of i?oW where S(E) is the set of coefficients of the dropped
terms in the process of forming E (e.g., the coefficients of F — Fo
and G — Go) and \(E) is the set of leading coefficients of the F^.
Now let $ be any field of characteristic 0, let (rf) e <i>(r) and set
/(*) = F(ri; x),g(x) = G(ri; x). It is easily seen that there exists
a generic Euclidean sequence E for (F, G) such that <^(t,) = 0
for all d e S(£), /(t,-) ^ 0 for all / e \(E). Hence the set of pairs
(§(£), X(£)) for all generic £ is a rational cover. If E is chosen
as indicated for (t,), then d{x) = D(tj-; at) is a highest common
factor in <£>[#] of f(x) and #(#) and, if £)(/,; x) ^ 0, we have the
polynomials i?A)(/i; *), GA)(/*; *) such that m{r,)ef{x) = </(*)/i(*),
»(t*)'*(*) = </(*)fi(*) where AW = /^(t,;*),^*) = G'1'^;*)
and m{ti) is the leading coefficient of £)(/»■; x). We have w(t,) 5^ 0
since »»(/,■) e X(£).
The procedure we have just indicated can be extended in an
obvious way to any finite set of polynomials. We shall need the
process also for polynomials in two indeterminates x,y (besides
the ti). Here we begin with F(ti; x,y) and G{tt; x,y) in 2l[#,jy] =
Ro[h\ x,y] and we treat x like one of the t{. The division algo-
algorithm with respect tojy gives /(/;; x)'F = QG — R where degy R <
degj, G. If we observe that a relation d(ji\ x) = 0 for </(/»•; x) e
Ro[U\ x] is equivalent to /*(t,-) = 0 for all the coefficients <f*(/,-)
of i/(/i; x) and /(t,-; x) ?^ 0, /(/,-, x) e i?0[^; ■"■]> holds if and only if
h(ji) j± 0 for one of the coefficients 4, we see that we can de-
determine a rational cover E;-, X3), j = 1,2, ■ ■ •, h, and polynomials
A('<; *»^) and ^iA)('<; *,J')» G,A)(/i; *,jr) if A' ^ 0, such that if
(t,) is, in the subset 61,- defined by (8,-, Xy), then d{x,y) = Dy(T,-;
•v,_y) is a highest common factor in $(#)[)>] of f{x,y) = F{Tt;
x,y) and g(x,y) — G{ri\x,y). Moreover, if D(/t-; x,y) ^ 0 and
m(ti\ x) is its leading coefficient regarding D as a polynomial in y,
then m(x) = »»(/,•;*) ^ 0 and m(x)e/(x,y) = d(xfy)f1(x,y),
m(x)fg(x,y) = d(x, y)gi(x, y) where /i(x,j) = iV1^; *,jO,
^(^,J) = G/1»(ri;^7)-
There is one more device we shall need which will take the place
of the step in the decision method of choosing an element y in
$ such that for a given polynomial f{x) ^ 0 one has f(y) ?± 0.
Let F(ti; x) = Fq(d)xq -\ h F0(ti) where Fq(tt) ^ 0. Assume
first that (t,) in $(r) satisfies ^(t,) ^ 0. If we recall the bound

310 ARTIN-SCHREIER THEORY
for the roots in <i> of a polynomial given in § 3 we see that 77 =
9-1
(.9 + 1) + Z ftW^W is not a root of F{Ti; x). Hence
0
if we set Q{ti) = (g + l)Fq(tt)* + *£ Fk(ti), P(U) = Fq(tl)\
0
then P(Vt) ^ 0, Q(tx) ^ 0 for all (V<) satisfying ^9(t,-) ^ 0 and
7; = B(t,-)P(t,-) -1 is not a root of -F(t,-; *•). Next assume ^(tj) =
0 and Fp(Ti) ^ 0 for the first non-zero coefficient Fp(ti) after
Fg(ti). Then we can repeat the argument with p replacing q.
Continuing in this way we obtain a rational cover Ey, \}), j =
1, 2, ■ • •, h, such that .F(t,-; #) = 0 for (t,) e <?/, and for j < h we
have Pyfa), 0y(A-) such that P^n) * 0, 0y(Ti) ^ 0 and F(n;
j?y(r0Py(T0-1)^0for(T0e^.
We are now ready to give the
Proof of Theorem 15. We note first that it is sufficient to give a
rational cover E*, \k), k = 1, • ■ •, m, such that for each k one
defines a finite set of pairs of polynomials Gkj(ti) e Ro[ti\, Fkj{ti, x)
e Ro[ti", x] having the property that, if (ri) e Sk, the subset of $(r)
defined by Ek, Xj), then .F(t,-; x, y) = 0, G(V;; x) 5* 0 is solvable
in f> if and only if one of the conditions: G*j(rt-) ?^ 0 and i^t;(r,-; #)
= 0 is solvable in <i>, is satisfied. If we have this situation, we
put Fkj*{ti;x) = Fki(ti-xJ + £</(/,J, Gw*fe) = Gkj(t{) R/((<).
deSk ieX/t
Then the finite set of pairs {Fkj*(ti\ x), Gk]*(ti)) satisfies the con-
condition for the set of pairs (Fjfc; x), Gy(/,-)) in the statement of the
theorem.
We consider next the reduction of the theorem from the pair
of conditions F(t^x,y) = 0, G(/j; x) t^ 0 to a single condition
F(tiix>y) = 0. (This corresponds to the second half of the argu-
argument given in the last section.) We shall use an induction on
degz F and we note that the result is trivial if F does not involve
x. Then we can take F(tt; x) to be the polynomial obtained by re-
replacing y by the missing x and take G(ti) to be the sum of the
squares of the coefficients of G(/t-; x). We now assume dega;
F(k\x,y) > 0 and we apply the considerations on highest com-
common factors to G(/,-; x) and the coefficients of the powers of y in
F(ti\x,y). Accordingly, we obtain a rational cover such that
for each member (S, X) of the cover we can determine polynomials

ARTIN-SCHREIER THEORY311
m(ti), D(ti\x), FA)((i;x,y), GA)(/,;x) with rational coefficients
such that D(ri; x) is a highest common factor of G(r,-; x) and the
coefficients of the y terms in F(ri;x,y) and w(t.) ?■* 0,
ix) = D(Ti; x)G™(t<; x)
for all (t.) in the set S defined by E, X). We can replace the pair
F(ti\ x,y), G(({; x) by the pair -FA)(/t-; x,y), GA)(>;; x) in the set S so
if degj FA) < degx F the induction can be used. Hence we may
assume equality of the degrees indicated, which means that we
have degx D = 0. Then £)(/,•; x) = >»(/,•), and G(t,-; #) and the
coefficients of F(ri;x,y) are relatively prime. Now let T{ti\y)
be the resultant relative to x of F(ti\x,y) and O* + G(/,-; #).
Then T(r{;y) 5^ 0 for all (tj) e <£ and by passing to a refinement of
the rational cover we may assume also that we can find -P(/,•),
Q(ti) e R0[ti] such that P(rt)^0, 0(tO?*O, and T(r,-; Q(Ti)P(u) ~')
* 0 for (tO in S. We replace F(/i5 x,y) by ^(/,-; *,jr) = PWF^
*>y + Q{h)P(ti)~l) where / = degv F(t{; x,y). The resultant of
H(?n x,y) and G(/t-; *) relative to x has the form P{tt)gT{ti\y +
Q{ti)P{ti)-1) and this is not 0 for (rt) e S, y = 0. It follows that
H{Tt; x,y) = 0, G(n; x) ?± 0 is solvable in $ if and only if ^(r,-;
*jjO = 0 is solvable in * for K(ti\ x,y) = Hfc; x, G(t{;y)y).
We now consider a single equation F{ti\x,y) = 0. By con-
considering the highest common factor of the coefficients of the powers
of y of F we reduce the consideration to subsets S defined by a
rational cover and polynomials F(ti;x,y) such that F(ri;x,y) is
not divisible by a polynomial of positive degree in x for (t.) e S.
dF
Next we consider the highest common factor of F and — and
dy
after a refinement we may assume that we have determined poly-
polynomials m{ti\x), D(ti;x,y), Fi(ti;x,y) with rational coefficients
such that D(ri;x,y) is a highest common factor in &(x)[y] of
Firv, x,y) and — F(Ti; x,y), »j(t,-; x) 9^ 0 and m(n; x)eF(ri; x,y)
dy
= D(th x,y)Fi(ri; x,y). Then -Fi(t,-; x,y) has no multiple fac-
factors of positive degree in y and F and Fx have the same irreducible
factors of positive degree in y in $[x, y]. Again we can determine

312ARTIN-SCHREIER THEORY
*(/,•), Lfax), F2(ti;x,y) such that *(>,•)^(t,-; x,y) = L(n;
x)F2{Ti\ x,y) where F2{ri\ x,y) is not divisible by a polynomial of
positive degree in x. Then it is clear that we may replace F by F2
and so we may assume that for (t,-) eS, F(ji\ x,y) has no mul-
multiple factors of positive degree in y and no factor of positive
d
degree in y alone. Then F(n;x,y) and —F(ri;x,y) have no
dy dF
common factors of positive degree. Set G(ti} c:x,y) = y
dF dX
(x — c) — where c is another indeterminate and let R(ti, c\ x)
dy
be the resultant relative toy of G(/,-, c; x,y) and F{ti\ x,y). Then
one can argue as in the decision method itself that 2?(t,-, c; x) ^ 0.
By going to a refinement of the rational cover we can obtain
PQi), Q(U) e Raid] such that P(n) * 0, Q{rt) * 0, R{n, Q{r%)
P(ri)~1; x) 5^0. If we replace G(/,-, c\ x,y) by G(/,;x,j) =
P(ti)G(ti, Q{t%)P{tl)~1;x,y), we see that the resultant R{tvyx) of
F{ti\x,y) and G(ti;x, y) relative to y satisfies R(tj;x) 5^0,
(t,-) e 6". As before, we can argue that also the resultant j2(/t-; jy) of
F and G relative to x satisfies (?(t,;^) t^ 0. The remainder of
the proof can be made along the lines of the decision method itself.
We leave it to the reader to carry this out.
10. Generalized Sturm's theorem. Applications. We can now
prove the following generalization of Sturm's theorem which is due
to Tarski.
Theorem 16. Let <p be a finite set of polynomial equations and
inequalities of the form F(tu ■ ■ •, tr; xu • • ■, xn) = 0, G(/1} • • •,
tr; #i, • • ■, xn) ^ 0 or H{tu ■ ■ ■, tr; x1} ■ ■ ■, xn) > 0 where F, G,
H z Ro[tit ■ ■ •, tr; X\, • • •, xn], Then one can determine in a finite
number of steps a finite collection of finite sets \ftj of polynomial equa-
equations and inequalities of the same type in the parameters tt alone
such that, if$ is any real closed field, then the set <p has a solution for
the x's in $for ti = t,-, 1 < i < r, if and only if the rt satisfy all the
conditions of one of the sets yj/j.
Proof. We show first that we can reduce the system <p to a
single equation of the form F{tt; Xj) = 0 where the number of #'s
may have to be increased. First it is clear that an inequality

ARTIN-SCHREIER THEORY 313
G 7^ 0 is equivalent to G2 > 0. Next we can replace an inequality
H > 0 by the equivalent equation z2H —1=0 where z is an
extra indeterminate. Finally, a number of equations Ft = 0 can
be replaced by the single equation ~ZiF2 = 0. These observations
prove the assertion, so we take <p to be a single equation F(ti\ x,) =
0. We show first by induction on the number n of x's that we
can determine a finite number of sets of equations of the form
Fk(ti; x) = 0, Gk(ti) 9^ 0 such that a set tx, ■ ■■, rr, t* in $, has
the property that F(ri; x}) = 0 is solvable for x's in $ if and only
if for some k one has Gk(ri) 7^ 0 and Fk(ji\ x) = 0 is solvable in $>.
This is trivial for n = 1 and it is a consequence of Theorem 15
if n = 2. Assume it holds for n — 1 > 2. Then treating xn as
one of the parameters we conclude that we can determine a
finite number of pairs of polynomials (Fk(ti, xn; x), Gk{ti, xn))
with rational coefficients such that, if the t,- and £„ e $, then
F(tj; x1} • • •, xn-i, £«) = 0 is solvable for xi} ■ ■ ■, xn_\ in $ if and
only if, for some k, Gk(ri, £„) ^ 0 and Fkiu, £n;jy) = 0 is solvable
in $. By Theorem 15, for each k one can find a finite set of pairs of
polynomials (Fki(tim, x)> Gkj(ti)) with rational coefficients such that
Fk(ri, x;y) = 0, Gk(n; x) y^ 0 is solvable in $ if and only if for
some j we have Gkj{ri) ?± 0 and Fk](Ti; x) = 0 is solvable in $. It
follows that the set of pairs (Fkj(ti; x), Gki{ti)) satisfies the re-
required condition for F(/,-; xu ■ ■ •, xn). We now denote these pairs
as (Fj(U\ x), Gj(ti)). For each iv(/,-; x) the version of Sturm's
theorem we considered in § 7 shows that a finite set of polynomial
equations and inequalities with rational coefficients in the /,- can
be found such that these are satisfied by /,- = t,- e $ if and only
if Fj{ri\ x) is solvable in <t>. If we add to each set the inequality
Gj(ji) ^Owe obtain the sets \f/ satisfying the requirement of the
theorem.
Suppose now that we have a system of equations and inequali-
inequalities with rational coefficients which have a solution in one real
closed field <J>i. It is clear that we can introduce parameters and
change our assertion to one that a certain system with parameters
and rational coefficients has a solution in <!>! for certain rational
values of the parameters. Then Theorem 16 implies that these
rational numbers satisfy one of a certain set of rational equations
and inequalities. Then if* is any other real closed field we can

314 ARTIN-SCHREIER THEORY
apply Theorem 16 again in the reverse direction and conclude that
the original system has a solution in $.
Again, suppose we have a system of equations and inequalities
with rational coefficients involving parameters and suppose that
for one real closed field <£>i it is true that the system has a solution
in <i>! for all choices of the parameters in $1# Then one concludes
from Theorem 16 that this is equivalent to the statement that
every set of values for the parameters in f>x satisfies one of a cer-
certain finite collection of finite sets of equations and inequalities.
It is easy to see that this in turn is equivalent to the statement
that there are no solutions in $t of any one of another finite col-
collection of finite sets of rational equations and inequalities for the
parameters. The foregoing result shows that this carries over to
every real closed field <i>. Hence we see that the original system
has a solution in f> for all choices of the parameters in $ where «i>
is any real closed field.
We shall now consider an application of these results to an
important theorem on division algebras.
A long time ago, before real closed fields were invented, Fro-
benius proved the following theorem: The only finite dimensional
division algebras over the field R of real numbers are: A) R it-
itself, B) R(\/ — 1), C) Hamilton's quaternion algebra over R.
The known proofs of this theorem are algebraic and give the same
result for any real closed field. The reader may refer to Dickson's
Algebras and Their Arithmetics, p. 62, for an elementary proof of
this type. We now drop the assumption of associativity which we
have made throughout this book and consider non-associative
algebras. These are defined to be vector spaces over a base field
$ in which a multiplication xy is defined satisfying the distribu-
distributive laws and the rule a(xy) = (ax)y = x(ay), ae$. Such an
algebra which is finite dimensional is called a division algebra if
it has no zero divisors: xy = 0 implies either x = 0 or y = 0 in
the algebra. Besides the examples noted above there is one other
important example of a non-associative division algebra, namely,
an algebra of eight dimensions of octonions which was discovered
by Cayley and by Graves. The known examples of finite dimen-
dimensional non-associative division algebras over the field of real
numbers have dimensions 1, 2, 4, and 8. It was conjectured for

ARTIN-SCHREIER THEORY 315
a long time that these are the only possible dimensions and this
was finally established by deep topological considerations by Bott
and Milnor. It would be a hardy task to attempt to carry over
the proof to the case of real closed fields. Moreover, this is un-
unnecessary since it is quite easy to conclude the result for arbitrary
real closed fields from its validity for the field of real numbers.
Assuming the Bott-Milnor result for the field of real numbers,
we shall prove that, if n j± 1, 2, 4, 8, <» and * is a real closed field,
then there exists no n dimensional non-associative division algebra
over <£>. To prove this let 21 be a non-associative algebra
with the basis (uly u2, • • ■, un) over <l> and suppose «,-«y = DjijkUk
where the 7,^ e#. If x = 2|t-«t-, £,• e<J>, then the mapping y —> xy
in 21 is a linear one whose matrix relative to the basis («l5 ••-,«„)
is (pyfc) where pjk = J^ £»7»7&- The existence of a y 9^ 0 such that
i
xy = 0 is equivalent to the statement that y —> xy is a singular
linear transformation and this is the case if and only if F(yijk; £,)
= det (pjjc) — 0. To show that 21 is not a division algebra we
have to show that there exists an* ^ 0 such that F(yijk; £,•) = 0.
We now see that our assertion is equivalent to the following: Let
Xi) = det ( ^ Xj/,7fc) which can be considered as a poly-
polynomial in indeterminates /,•,■*, #,• with rational coefficients. Then
for all choices %* = 7,^ e $ there exists a solution #,• = £, in $ of
the system F(yijk; #,) = 0, S*,-2 9^ 0. Now by the Bott-Milnor
theorem this holds for <i> = R the field of real numbers. Hence our
results show that it holds for every real closed field.
Another example of the same type is a theorem of Hopf 's which
states that the only possible finite dimensionalities for real non-
associative commutative division algebras are n = 1, 2. Com-
mutativity, of 21 is equivalent to the condition yijk = yjik for all
/, j. Hence in the foregoing argument we consider indeterminates
tijk for / < j and define tjik = tijk for j > i. Then detf ]£ #</»,,■& )
is a polynomial with rational coefficients in the indeterminates
tijk, i < j- The rest of the argument carries over and shows that
Hopf's theorem is valid for all real closed fields.

316 ARTIN-SCHREIER THEORY
There is a general class of statements on real closed fields which
can be treated in the foregoing manner. These are the so-called
elementary sentences of algebra. We shall not attempt to give
the precise definition for these but refer the reader to the literature
(see the bibliographic notes on this chapter). The results we have
considered are special cases of the general principle of Tarski that
any elementary sentence of algebra is either true for all real
closed fields or is false for all real closed fields.
EXERCISES
1. Assuming the result for the field of real numbers prove that, if $ is any real
closed field and F\{x\, •••,«„)= 0, ■••, Fk{x\t • ■ ■, xn) = 0 where the F's e
$[xi, • • • > xn] has a solution Xi = £; 8 $, then it has a solution nearest the origin.
2. Prove the analogue of Theorem 16 for algebraically closed fields $ of charac-
characteristic 0 and finite sets of equations F(h, • ■ ■ ,tr;x\, • • •, xn) = 0 and inequalities
G(h, ■ ■ ■, tr; xi, ■ ■ ■, xn) ?* 0 where the F,G e Ralk; ■*»]■ (Hint: A simple proof of
this can be based on the generic Euclidean sequences and the following simple
observation due to Tarski: if/(#), g(x) e $[x] and deg/ > 0, deg g > 0, thenTW
= 0, g(x) t* 0 has a solution in $ if and only if/M is not a divisor of ^(^)deg/(a;)).
3. Prove the result of ex. 2 also for $ of characteristic^) 5^ 0 by developing the
corresponding results on generic Euclidean sequences of /?[/,■; x], IP = I/(p).
11. Artin-Schreier characterization of real closed fields. We
shall complete our discussion of real closed fields by proving a
beautiful characterization of real closed fields which is due to
Artin and Schreier. We recall that, if $ is a field not containing
•%/ —1 and $(y/ — 1) is algebraically closed, then $ is real closed
(Th. 6). We shall now prove
Theorem 17. Let Q be an algebraically closed field and $ a
proper subfield which is of finite co-dimension in fl. Then <i> is real
closed and £2 = <I>(V —1).
Proof. Let <i>' = $(V —1) Q ft. The theorem will follow from
the result quoted if we can show that $' = 12. Hence we suppose
that 12 3 $'. Let E be an algebraic extension of $'. Then E is
isomorphic to a subfield of £2 over <£' and so [E:4>'] < [0:$'].
Hence the dimensionalities of algebraic extensions of <£' are
bounded. This implies that $>' is perfect. Otherwise, the charac-
characteristic is p ?* 0 and there exists a /? e $' which is not a p-th power.
Then for every e > 0, xp' — /3 is irreducible in <£'[#] (ex. 1, § 1.6)
and this provides an algebraic extension of p' dimensions over $'.

ARTIN-SCHREIER THEORY 317
Since e is arbitrary, this contradicts what we proved. Thus #'
is perfect and so £2 is separable over $'. Since 0 is algebraically
closed it is Galois over #' and its Galois group G over #' is 5^1 be-
because 0 3 $>'. Hence G contains a cyclic subgroup of prime order
q and consequently there exists a subfield E3*' such that 0 is
cyclic of q dimensions over E. Since fl is an algebraic closure of E
and [0: E] = q, it is clear that fl and E are the only algebraic exten-
extensions of E. It now follows that the characteristic of * is not q.
Otherwise, Q is a cyclic y-extension of E, and the existence of such
an extension of E implies the existence of cyclic ^-extensions of
E for every m (Th. 3.16). This has been ruled out and so the
characteristic is not q. This implies that fl, which is algebraically
closed, contains q distinct roots of 1. Since these are roots of
(x — \)(xq~1 + x9~2 + ■ ■ • + 1) and since the irreducible poly-
polynomials in E[#] have degrees 1 or q, all the q-th. roots of 1 are con-
contained in E. Since 0 is cyclic y-dimensional over E, Q = E(Va)
where a e E and is not a ^-th power in E (Th. 2.5). Consider the
2
polynomial g(x) = JJ (x — f'p) where f is a primitive
i
of 1 and p is an element of fi such that p9* = a. Since the inclu-
inclusion flp e E implies that E contains an element (f*p)s = # such
that /89 = a, we see that no f*p e E. Since g(x) = xQ' — a e E[#],
it follows that all of its irreducible factors in E[x] are of degree q.
If/8 is the constant term of one of these, then /3 = p9rj where 77 is a
power of f. Since (p«)« = a and Q = E(v^), p« t E and Q =
E(p9) = E@p~q) = E(ij). Since E contains all the q-th. roots of
1, we see that r\ is a primitive ^2-root of 1. Let $0 be the prime
field of S2 and now consider the subfield <£o07) of & If $0 is the
field Ro of rational numbers we know that the dimensionality of
the field of ^r-th roots of 1 is <p{qr) (Th. 3.2) and this goes to in-
infinity with r. If $0 has characteristic p 7^ q, then the field of the
qr-ih roots of 1 over f>0 contains at least qr elements, so again the
dimensionality of this field over $0 approaches infinity. In any
case it now follows that there exists a positive integer r such that
$0G7) contains a primitive /-th root of 1 but no primitive /+1-st
root of 1. Since t\ is a primitive ^2-root of 1, r > 2. The field 0
contains a primitive ?r+1-st root of 1, say f. Let h(x) be the

318 ARTIN-SCHREIER THEORY
minimum polynomial of £ over E. Since rj $ E, £ $ E, so deg h(x)
qr+l
= q. Also h(x) is a factor of x«r+1 - 1 = JJ (# — f )> so the
1
coefficients of A(#) are contained in $o(£); hence they are con-
contained in the field r = $0(£) H E. It follows that [*<,(£): r] = q.
Next we consider the subfield r' = $0G), 7 = £9, of $0(i;). Evi-
Evidently 7 is a primitive ^r-th root of 1, so r' contains q distinct <?-th
roots of 1. On the other hand, 3>o(£) = !"(£) where £9 = 7 e r',
so either $>o(£) = r' or *o(£) is cyclic of y dimensions over r'. If
$o(£) = r' = *oGK we have *„(£) <= *0(»?) since f>0(»?) contains
all the ^r-th roots of 1. Then $0(ri) contains £, a primitive ^r+1-st
root of 1, contrary to hypothesis. Thus we have [<J?0(£): r'] = q.
Now r' 9^ r. Otherwise, T contains a primitive qr-th root of 1,
so r and E contain rj contrary to Q = E(jj) 3 E. We have there-
therefore proved that the field <I>o(£) of the #r+1-st roots of 1 over the
prime field contains two distinct subfields r and r' over which it is
^-dimensional. It follows that the Galois group of $0(£) over $0
is not cyclic. By Lemma 1 of § 1.13 and Theorem 3.5, this is the
case only if the characteristic is 0 and q = 2. Then the element
t) considered before is a primitive 4-th (q2 with q = 2) root of 1.
On the other hand, E contains <£' which contains V' —1 and this is
a primitive 4-th root of 1. Hence we have fl = E(jj) = E con-
contrary to Q 3 E. This contradiction shows that <£' = $(V —1) =
fl and * is real closed.

SUGGESTIONS FOR FURTHER READING
Chapter I. The classical Galois correspondence between groups of
automorphisms and subfields has been extended in a number of different
directions. First, one has Krull's Galois theory of infinite dimensional
extensions which is considered in Chapter VI. Next one has the Galois
theory of division rings which is due (independently) to H. Cartan and
the present author. An account of this can be found in the author's
Structure of Rings, A.M.S. Colloquium Vol. 37 A956), Chapter VII.
(Our development of the Galois theory in Chapter I is based on the
methods which were developed originally to handle the non-commutative
theory.) A Galois theory of finite dimensional separable extensions
based on the notion of a self-representation of a field is due to Kaloujnine.
This is contained in a more general theory given by the present author
in two papers in Am. J. Math., Vol. 66 A944), pp. 1-29 and pp. 636-644.
See also two papers by Hochschild and by Dieudonne in the same
journal, Vol. 71 A949), pp. 443-460 and Vol. 73 A951), pp. 14-24.
Quite recently a Galois theory of automorphisms of commutative
rings has been developed jointly by S. U. Chase, D. K. Harrison, and
A. Rosenberg. This paper will appear in Transactions A.M.S.
A general cohomology theory of fields has been given by Amitsur in
Trans. A.M.S., Vol. 90 A959), pp. 73-112. See also the paper by
Rosenberg and Zelinsky on this subject in Trans. A.M.S., Vol. 97 (I960),
pp. 327-356, and Amitsur's paper in /. Math. Soc. Japan, Vol. 14 A962),
pp 1-25.
Chapter II. We have indicated in the text the unsolved problem of
the existence for a given field $ and a given finite group G of a Galois
extension P/$ whose Galois group is isomorphic to G. A closely related
question is that of the existence of an equation with coefficients in <£
having a given subgroup of Sn as group. These problems have been
studied extensively for $ the field of rational numbers and more generally
for algebraic number fields (finite dimensional extensions of the ra-
tionals). Two methods have been developed for this problem: one based
on arithmetic properties of number fields, and a second more elementary
method based on an irreducibility criterion due to Hilbert. The deepest
results thus far obtained in the arithmetic theory are due to Safarevic.
A summary of his results is given in Math. Reviews, Vol. 16 A955), pp.
The Hilbert method (which was used by Hilbert to prove the existence
of rational equations with Sn as Galois group) has two stages. Given
a field $ one requires first a purely transcendental extension field
319

320 SUGGESTIONS FOR FURTHER READING
®{h, ■ ■ ■ ,tr) and a Galois extension P of <£(/;) with Galois group iso-
morphic to the given group G. This problem is still open except for
special cases (Sn, alternating group and some others). Next one needs
to know that * is a Hilbertian field in the sense that Hilbert's irreduci-
bility theorem holds for $. (For example, the rational field is Hil-
Hilbertian; the field of p-adic numbers and finite fields are not.) A discus-
discussion of this theorem and its relation to Galois theory is given in S. Lang's
book Diophantine Geometry, New York, 1962, Chapter VIII.
An interesting aspect of the classical Galois theory of equations is
Klein's theory of form problems. A development of this from the point
of view of algebras, particularly crossed products, is due to R. Brauer in
Math. Annalen, Vol. 110 A934), pp. 437-500. Reference to the classical
works on the subject is given in this paper.
A general reference book for Galois theory of equations is Tschebota-
row's Gurndziige der Galois'schen Theorie, Groningen, 1950 (translated
from Russian by Schwerdtfeger).
Chapter III. D. K. Harrison has given a general theory of abelian
extension fields in Trans. A.M.S., Vol. 106 A963), pp. 230-235.
Chapter IV. Some of the deeper results of this chapter have been
developed to meet the needs of algebraic geometry. The reader may con-
consult S. Lang's Introduction to Algebraic Geometry, 1958, or A. Weil's
Foundations of Algebraic Geometry, A.M.S. Colloquium Vol. 29, Provi-
Providence, 1st. Ed., 1946, 2nd. Ed., 1962, for these connections.
Chapter V. There are several directions that one may take in pursuing
the subject matter of this chapter. First, one can study the general
theory of valuations as given in Zariski-Samuel's Commutative Algebra
Vol. II, D. Van Nostrand Co., Inc., Princeton, I960. Chapter VI.
Secondly, this chapter leads to the arithmetic theory of number fields
and fields of algebraic functions of one variable. For this the reader
may consult Chevalley's book Algebraic Functions of One Variable,
Princeton, 1951, Artin's book Theory of Algebraic Numbers, Gottingen,
1959, and E. Weiss' book Algebraic Number Theory, New York, 1963.
A third direction which one can take after studying Chapter V is local
class field theory. For this the reader may consult Serre's book Corps
Locaux, Paris, 1962.
Chapter VI. The original Artin-Schreier theory is given in papers by
Artin and Schreier and by Artin in the Hamburg Abhandl., Vol. 5 A927).
Our exposition follows these papers rather closely. Seidenberg's work
is in Annals of Math., Vol. 60 A954), pp. 365-374. This contains also a
statement of Tarski's principle and, of course, a reference to Tarski's
earlier paper. Much of the present chapter can be developed also as a
part of mathematical logic, more exactly, as an aspect of the theory of
models. The reader may consult A. Robinson's book, Model Theory,
Amsterdam, 1963, particularly Chapter VIII. Also references to the
literature are given in this book.

INDEX
Abelian extension field, 61, Chapter III
Abelian ^-extensions, 132-140
Algebraically closed field, 142-147
Algebraic closure, 142
separable, 146
uniqueness of, 145
Algebraic element, 6
Algebraic field extension, 44
absolutely, 147
Algebraic functions, 156
Algebraic independence, 4, 151-157
of isomorphisms, 56
Algebras, 7-9
algebraic, 10
homomorphism of, 7
ideals of, 7
of dual numbers, 168
tensor products of, 15-17
Artin's theorem on positive definite
rational functions, 289
Bilinear mapping, 10
Bott-Milnor theorem, 315
Character, 75
Character group, 117
of finite commutative group, 116—
119
Characteristic polynomial, 64
Cohomology groups, 82
Complete field (relative to a real valua-
valuation), 217
finite dimensional extensions of,
256-262
Completion of a field (relative to a real
valuation), 216-221
Composites of fields, 83-89, 262-264
free, 203-209
Constant, 169, 194
321
Crossed product, 79
Cyclic algebra, 80
Cyclic extension field, 61
Cyclic ^-extensions, 139-140
Cyclotomic field, 95, 110-116
Decision method, 300-307
Dedekind independence theorem, 25
Degree of separability and insepara-
inseparability, 49
Dependence relations, 153-155
algebraic, 151-157
Derivations, 167-174, 183
constant relative to, 169
Galois theory of, 185-191
higher, 191-197
iterative higher, 196
Different, 73
Direct sum, 9, 85
Discriminant:
of an algebra, 66
of a polynomial, 92
Equations with symmetric group as
Galois group, 105-109
Exponential function in ^>-adic num-
numbers, 226-228
Extension of derivations, 170-172,
174-185
Extension of homomorphisms, 2-6,
246-248
Extension of valuations, 246-250,
256-265
Factor set, 79
Finite fields, 58-62
Finite topology, 149
Formally real field, 271
Frobenius' theorem, 314

322
INDEX
Fundamental theorem of Galois
theory, 41
for infinite dimensional extensions
(Krull's theorem), ISO
for purely inseparable extensions of
exponent one, 186
Galois cohomology, 75-83
Galois correspondence, 23
for subgroups and subfields, 29
Galois' criterion for solvability by
radicals, 98-102
Galois extension field, 27
Galois group:
of an equation, 89-97
of an extension field, 27
of cyclotomic extensions, 96, 113,
115
of general equation, 104
of quartic equations, 94-95
of simple transcendental extensions,
158-159
Galois theory for purely inseparable
extensions of exponent one, 185—
191
General equation of w-th degree, 102-
105
Groups of automorphisms of fields,
27-31
Hensel's lemma, 230-232
Hilbert Nullstellensatz, 254
Hilbert's "Satz 90," 76
Hilbert's 17th problem, 289
Homomorphism:
of an algebra, 7
of additive group of a field, 19
Hopf's theorem, 315
Ideal, 7
imbedding in maximal ideal, 255
imbedding in prime ideal, 253
radical of, 209, 253
Indeterminates, 4
Infinite Galois theory, 147-151
Integral closure, 255-256
Isometric mapping, 221
Jacobson-Bourbaki theorem, 22
Kronecker product, see tensor product
Kummer extensions, 119-124
Lie algebra of linear transformations,
174
restricted, 174
Lie commutator, 173
Linear disjointness, 160-167
Local dimensionality, 265
Luroth's theorem, 157-160
MacLane's criterion, 164
Minimum polynomial, 6
Multiple roots, 37
Noether's equations, 75
Norm, 65
transitivity of, 66
Normal basis, 56, 61
Normal closure, 43
Normal extension, 43, 52-53
Number of solutions of quadratic equa-
equations in finite fields, 62
Order isomorphism, 238, 271
Ordered field, 270
archimedean, 272
Ordered group, 237
rank of, 243
of rank one, 244-246
Ostrowski's theorem, 260
^-adic numbers, 222-230, 234-236
^>-basis, 180
p-independence, 180
Perfect closure, 146
Perfect field, 146
Place, 241
Positive definite rational functions,
289-295
Power series, 233-234
Primitive elements, 54-55, 59
Pure equation, 95
Pure transcendental extension, 155
Purely inseparable extension, 47
exponent of, 179
Galois theory of, 185-191

INDEX
323
Ramification index, 265
Rationally specializable property, 291
Real algebraic numbers, 287-289
Real closed field, 273
characterization of, 276-278, 316—
318
Real closure, 284-286
Representation:
matrix, 63
regular, 63
Residue degree, 265
Residue field, 222
Resultant, 298-299
Root tower, 98
Roots of unit, 95, 110-116
Seidenberg's decision method, 300-
307
Separable:
algebraic closure, 146
element, 45
extension, 46, 166
polynomial, 39
Separating transcendency bases, 161,
164-167
and derivations, 178-179, 184
Solvable extension field, 61
Splitting field of a polynomial, 31
isomorphism theorem for, 35
Standard sequence, 281
Sturm sequence, 279
Sturm's theorem, 283, 295
generalized (Tarski's theorem), 312
Subalgebra, 7
Tarski's theorem, 312
Tensor products, 10-17
of algebras, 15-17
of fields, 52, 84-87, 197-203
of subalgebras, 16
of vector spaces, 10-15
Theorem of Abel-Ruffini, 104
Theorem of Hilbert-Landau, 289
Trace, 65
transitivity of, 66
Trace form, 66
Transcendency basis, 151-157
separating, 161, 164-167
Transcendency degree, 155
Transitivity theorem for determinants,
68
Unit group in ^>-adic numbers, 225-
230
Valuations:
archimedean, 213
discrete, 222
equivalence of, 212
general, 238
of field of rational numbers, 214-216
of simple transcendental extensions,
216
p-adic, 211
real, 211
Valuation ring, 222, 240
Witt vectors, 124-132, 234-236
Wronskian, 185