ELEMENTS OF THE
THEORY OF ALGEBRAIC CURVES
A. SEIDENBERG
University of California, Berkeley, California
.
T
ADDISON- WESLEY PUBLISHING COMPANY
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This book is in the
ADDISON-WESLEY SERIES IN Mi\THEMATICS
Consulting Editor: Lynn H. Loomis
Copyright cg 1968 by Addison-Wesley Publishing Company, Inc. All rights reserved.
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Catalog Card No. 68-19345.

PREFACE
A large part of the present book has been offered as a graduate course
several times, and, moreover, with a preliminary presentation of the ring and
field theory necessary for complete proofs. Several reasons, however, have
led us to abandon, as part of the course, this preliminary material, to enter
directly into the geometry, and to bring the subject within the scope of an
undergraduate audience.
First, there are separate courses for the algebra, and, although we have
little confidence in "prerequisites," the curriculum can ill bear the duplica-
tion. Moreover, we find that a real understanding of the ring and field
theory is not necessary for the geometry, but that a clear statement of the
basic facts is sufficient. The student accepts that the complex number field
is algebraically closed; in the same way, he may accept that every field has
an algebraic closure. Except, then, for some of the subtleties of the algebra,
the geometry becomes quite accessible.
Second, there is a tendency to reserve the basic graduate course in
Algebraic Geometry for the general theory of algebraic varieties. This,
together with the other basic algebra, commutative and otherwise, consti-
tutes a heavy load for the first years of graduate study; and at the same
time there is a deficiency in Algebra in the undergraduate program (relative
to Analysis, say). We think that Algebraic Curves is no harder than the
much more widely offered undergraduate course in Differential Geometry,
and can compete with it in usefulness and interest.
The typical student we have in mind, then, at least for the first half of
the book, is a senior mathematics major who has taken a course in algebra.
We would hope that he has taken a course in Projective Geometry and
knows what homogeneous coordinates are, but ,ve have supplied a chapter
on this. Probably he kno,vs what a ring is, but we have tried to put down
the definitions of all terms, except those he has encountered in the basic
calculus sequence. Sometimes this is done in the notes, but often in the text
with the euphemism "Recall that." (The notes are signaled by a superior
iii

IV PREFACE
dagger.) If a term is not known, the index should be first consulted. We
refer the student to G. Birkhoff and S. MacLane, A Survey of Modern
Algebra, for definitions we may have inadvertently omitted.
As for the organization of the book, in the first eighteen chapters, except
for preliminary remarks, we work over the field C of complex numbers. The
arguments hold word for word for arbitrary algebraically closed fields K of
characteristic zero; but we do not have to make constant reference to the
base field, or bother the student with the essentially spurious generality.
The restriction to characteristic zero is made initially to a void secondary
issues; but the student will also want the results for arbitrary characteristic.
The modifications needed for characteristic not equal to 0 are given in
Chapter 19. In this way, the student will not have to unlearn, or reJearn,
the subject.
Two equivalent definitions, i( r, ; P) and j( r, ; P), are given for the
intersection multiplicity of two curves r,  at a point P. The definition
of i is given by means of resultants. It is an awkward definition for several
reasons, but has the advantage that it can be explained easily and gives
some of the basic projective properties of intersection multiplicity without
excessive difficulty. It is, however, definitely unsuited for a birational study.
One goes over to a definition, j(r, ; P), by means of branches. To avoid
making the equivalence i = j central in our exposition, we show in Chapters
12 and 14 that i and j share certain basic properties. The proof of the
equivalence is postponed to Chapter 21, but logically is not really needed
even then.
The branches of a curve at a point can be obtained quite rapidly, in the
case of characteristic zero, by using Hensel's Lemma, as van der Waerden
does in his Einfiihrung in die Algebraische Geometrie. Our method in Chapter
12 is not so short; but one gets the branches as a side result of an analysis
of a point by means of locally quadratic transformations-an analysis that
in no way involves power-series (as Hensel's Lemma does). It also, inci-
dentally, in no way uses the characteristic. This method is not really new,
but it does not occur in the textbooks.
Chapters 9 and 10 deal with a special topic, namely, cubics. They introduce
several general ideas and provide illustrative material, but are not strictly
necessary for the later developments.
Originally we learned the subject from lectures by Professor Oscar Zariski.
Since then we have studied S. Le£chetz's Algebraic Geometry, B. L. van der
Waerden's Einfiihrung, F. Severi's Vorlesungen iiber Algebraische Geometrie
(not to mention works on algebraic functions of one variable, a closely related
subject); and, more recently, have consulted R. J. Walker's Algebraic
Curves, and J. G. Semple and G. T. Kneebone's Introduction to Algebraic
Curves. There is no reason why the student cannot learn the subject,

PREFACE V
and very well indeed, from these books. We can only hope that here
and there there may be an especially attractive element in our presenta-
tion.
An invitation by the University of Mexico in 1962 to deliver some lectures
on a not too advanced level was the occasion for my deciding to write the
present book. Professor E. Lluis wrote up my lectures and my wife trans-
lated them from Spanish into English for me. I thank them both for helping
me to get started.
The visit to the University of Mexico was sponsored by the State Depart-
ment upon recommendation by the Conference Board of Associated Research
Councils, Committee on International Exchange of Persons.
Professor Jack Ohm read the manuscript, made helpful comments, and
pointed out slips and unfelicitous passages. I wish to express my gratitude
to him.
Berkeley, California
May 1968
A. S.

CONTENTS
1
2
List of Symbols
Introductory Remarks .
Homogeneous Coordinates
Projective planes
The line at infinity
Projective planes over an extension field of K
Algebraic Curves
The equations of a curve
Intersections
Affine algebraic curves
Resultants .
1
6
7
8
9
10
11
15
17
21
27
2U
30
31
:38
39
43
48
55
60
65
viii
3
4
5
Linear Transformations
Coordinate systems
Transformations in the affine plane
Simple Points and Singular Points .
The polar .
Some computational rules .
Bezout's Theorem
6
7
8
9
The Basic Inequality
Cubics
Rational functions on an irreducible curve
Abelian varieties
VI

10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
CONTENTS VB
Cubics (Continued)
Points of inflection
Cubics through eight points
The modulus of a nonsingular cubic
Expansion at a Simple Point
Formal power series .
Branches
Imprimitive branch representations
Generic Points
Rational mappings
Places
Zeros and Poles.
Differen tials
The genus
Infinitely near points .
Noetherian Conditions .
Linear Series
Linear Systems .
The Theorem of Riemann-Roch
Extension and reformulation of the theorem of Riemann-Roch
The Question of the Characteristic
Linear Series and Rational Mappings; Space Curves
Analytic Branches
Noetherian Conditions and Intersection Multiplicity
Specialization; Space Curves .
Specializations and residue maps ("places") .
Valuations and morphisms ("places") .
Infinitely Near Points .
Notes
Index
67
67
73
75
82
83
89
101
103
105
III
113
115
118
127
129
135
143
147
149
152
160
173
186
193
196
199
203
209
213

LIST OF SYMBOLS
Page Symbol
1 k[ X, Y]: ring of polynomials in two letters
2 0: complex number field
3 ft: is not an element of
6 AN K: affine number plane over K
7 K - 0: set of elements of K with the exception of 0
7 (x o : Xl : x 2 ): the class {(tx o , tx l , tx 2 )}
8 PNK: projective number plane over K
13 U: set-theoretic union
14 q:: is not contained in
14 == 0: is divisible by
16 G.C.D.: greatest common divisor
16 O(X o , Xl): field of quotients fig with f, g E O[X o , Xl]
22 R(J, g): resultant of J and g
27 IAI: determinant of the matrix A
27 LA: linear transformation of matrix A
43 i( r, ; P): intersection multiplicity of r and  at P
55 min {a, b}: minimum of the numbers a, b
57 S X T: Cartesian product of Sand T
83 K[[X, Y]]: ring of formal power series in two letters
84 K((X, Y)): fieldofquotientsflgwithf,gEK[[X, Y]]
99 j(, r; P): j-multiplicity of  and r at P
104 K ((:J..1' . . . , (:J..s): field adjunction of (:J..1'. . ., (:J..s to K
110 (r; (, 'YJ)): model of curve r with generic point (, 'YJ)
113  - 0: set of elements in  not in 0
113 [: O()]: degree of  over O()
116 dO: differential of 0
118 . d'l +  · d'2 + · · .: set of linear combinations of dl' d2' .
with coefficients in 
129 r () Ll: intersection of rand Ll
129 (F, G): ideal generated by F, G
133 . r: divisor cut out on r by Ll
135 ()o, ()oo: zeros, poles of 
135 A == B: equivalence of divisors A, B
140 g: a linear series of effective divisors
141 101: linear series of effective divisors equivalent to 0
viii

Chapter 1
INTRODUCTORY REMARKS
Algebraic Geometry is a study in which algebraic and geometric consider-
ations come together. It has two aspects: on the one hand one studies
geometrical problems by means of algebra, and on the other, algebraic prob-
lems by means of geometry.
Thus let us consider the assertion tb.t in the plane a circle is cut
by a straight line in at most two points. This is a purely geometrical
assertion, but it can be established algebraically: The circle is given by an
equation (X - a)2 + (Y - b)2 === r 2 , and the line, by an equation
uX + v Y + w === 0, with u -=1= 0 or v -=1= 0; supposing that v -=1= 0, we may
rewrite the equation as Y === mX + c. Now the problem becomes to solve
simultaneously
(X - a)2 + (Y - b)2 === r 2 ,
Y === mX + c.
Eliminating Y, one obtains an equation in X. This equation may have no
(real) roots and can have at most two. Each root will yield exactly one
intersection of the circle and line. Thus the two curves intersect in at most
two points.
Here we have studied a geometrical problem algebraically. On the other
hand, let us consider the circle X2 + Y2 - 1 === o. Associated with the
circle is the polynomial X 2 + Y 2 - 1 in the ring k[ X, Y] of polynomials in
two inde"terminates, or letters, over (i.e., with coefficients in) the field k of
real numbers. t Here we may ask a purely algebraic question: Is the poly-
nomial X2 + Y2 - 1 irreducible in k[X, Y] 1 To answer this question, let
us suppose that X2 + Y2 - 1 === F(X, Y) . G(X, Y), F, G in k[X, Y].
Since the degree of FG is the sum of the degrees of F and of G, we obtain
deg F + deg G === 2. If deg F === 0, then F is in k, F has a reciprocal in
k[X, Y], and we do not consider that F · G is really a factorization in
k[X, Y] (since such factorizations can be produced at will). Similarly, we
exclude the possibility that deg G === o. This leads to deg F === 1, deg G === 1.
I

2 INTRODUCTORY REMARKS
But a polynomial of degree one has a line as locus. The conclusion is that if
X2 + Y2 - 1 factors in k[X, Y], then the circle X2 + Y2 - 1 == 0 is
made up of two lines. Since every line has at least three points on it, and
since a line and a circle have at lnost two points in common, this cannot
happen, and hence X2 + Y2 - 1 is irreducible.
These two examples are typical. In the first we answered a geometrical
question algebraically; in the second, an algebraic question geometrically.
By an algebraic curve (in the plane) we mean the locus determined by an
equation F(X, Y) == O. Our initial point of view is that F has coefficients
in k, the field of real numbers, and by a point (x, y) we mean a pair of numbers
in k. However, here we at once meet a difficulty. Let us suppose that we
wish to study X2 + Y2 + 1 in the same way that previously we studied
X2 + Y2 - 1. Presumably we wish to do this by considering the locus of
X 2 + Y2 + 1 == O. However, this locus has no points. Naturally, we can
obtain little information about X2 + Y2 + 1 if we start from the empty set.
Hence we will change our point of view: we will allow the coefficients
of F to be complex numbers, and by a point we will mean an ordered pair
of numbers (x, y), x, y in the field G of complex numbers. Bya line (or complex
line) we will mean the locus determined by an equation uX + v Y + W == 0;
here u, v, W may be any complex numbers with u *- 0 or v *- o.
This change in point of view has many repercussions. On one hand, we
cannot be guided so easily by our geometrical intuition, since this has been
formed relative to k, !!ather than to G. However, with some precaution, our
intuition retains its usefulness. Another change is that the questions we ask
are different. Thus, previously we asked whether X2 + Y2 - 1 was
irreducible in k[X, Y]; now we will ask whether it is irreducible in G[X, Y].
To answer this question, we can make use of our previous result, but first
we need a lemma.
Lemma 1.1. The real points of a complex line are on a real line.
Proof. Let uX + v Y + W == 0 be the given line; let u == u l + iu 2 ,
v == VI + iv 2 , and W == w l + iw 2 , with u l , V v W v u 2 , V 2 , and w 2 in k and
i == vi -1. We ask for the real numbers (x, y) such that ux + vy + W == o.
We obtain the conditions
uix + vly + WI == 0,
u 2 x + v 2 y + W 2 == O.
As u -=1= 0 or v -=1= 0, at least one of u l , u 2 , V v V 2 is -=1= 0; and (x, y) is on
ulX + vlY + WI == 0 and on u 2 X + v 2 Y + W 2 == 0, of which equations
at least one is that of a real line.
Now we have the following theorem.
Theorem 1.2. X 2 + Y 2 - 1 is irreducible in ()[ X, Y].

INTRODUCTORY REMARKS 3
Proof. If X2 + Y2 - 1 were reducible, then reasoning as before, one would
have a decomposition into two factors
X2 + Y2 - 1 == (uX + vY + w)(u'X + v'Y + w').
Hence the (complex) locus of X2 + Y2 - 1 === 0 would be made up of t,\tO
(complex) lines. But then the real points of X 2 + Y2 - 1 == 0 would lie
on two real lines, and this is not so.
Theorem 1.3. X 2 + Y 2 + 1 is reducible in O[ X, Y].
Proof. Let us consider the transformation
t : (x, y)  (ix, i y ) .
This transformation is univalent (that is, distinct points are transformed
into distinct points) and is onto the plane (that is, every point of the plane is
an image of some point under the transformation t); moreover it transforms
lines into lines and X2 + Y2 + 1 == 0 into X2 + Y2 - 1 == 0 [the
equations of t are X' == iX, Y' == iY, from which the transformed locus
is obtained as (X' Ji)2 + (Y' Ji)2 + 1 == 0 and we omit the primes]. Thus if
X2 + Y2 + 1 factors, then also X2 + Y2 - 1 factors, and we know that
this is not so.
We have changed our point of view, but someone else may say that he
is still interested in the field k of real numbers: he wants answers to his
original questions, for example, is X2 + Y2 + 1 irreducible in k[X, Y] 1
This is easy to answer: X2 + Y2 + 1 is irreducible in C[X, Y] and hence,
a fortiori, is irreducible in k[X, Y].
Let us observe, however, that factorization in O[X, Y] is not always
the same as in k[X, Y]. For example, X2 + Y2 is irreducible in k[X, Y]
but is reducible in O[X, Y]; in fact, X2 + Y2 == (X + iY)(X - iY).
This example shows that we ought to be careful in using our geometric
intuition, that is, technically speaking, in passing from k[X, Y] to O[X, Y].
Thus X2 + Y2 and X2 - Y2 are essentially different in k[X, Y] (one
factors but the other does not), but are essentially the same in O[X, Y]
(both factor into distinct linear factors; moreover, the transformation
X' == X, Y' == i Y takes one into the other).
Initially we passed from k to 0 so that the locus of X 2 + Y2 + 1 = 0
would not be empty. This leads us to our first general theorem.
Theoreln 1.4. On every (plane, algebraic) curve there are infinitely many
poi nts .
[Let us repeat, but with more precision: by a plane algebraic curve
we mean the set of points (x, y) with x, y in 0 that satisfy an equation
F(X, Y) == 0, where F is in C[X, Y] but not in 0 (in symbols: F ft 0).
We insist that F t! O. If F == CEO, the locus of F == 0 contains no point

4 INTRODUCTORY REMARKS
if c =1= 0 and contains every point if c = 0; we do not wish to call either of
these loci a curve.]
Proof. Let the curve be given by F(X, Y) = o. As F ft 0, either X or Y
(or both) occur in F, let us say Y occurs. We write F as a polynomial in Y
with coefficients in O[ X] :
F = ao(x)ym + a1(X)ym-l + ... + am(X),
where
ai(X) E O[X],
i = 0, . . . , m,
and
ao(X) =1= o.
Let deg ao(X) = n. Then ao(X) has at most n roots. Let a E 0 be such that
ao(a) =1= 0; there are infinitely many ways to choose a. Now consider the
equation
ao(a) ym + a1(a) ym-l + · · · + am(a) = o.
Here m > o. Now we use the basic fact about 0, namely, that every poly-
nomial in one letter with coefficients in 0 of positive degree has at least one
zero in O. Hence there exists a number b E 0 such that
ao(a)b m + a1(a)b m - 1 + · · · + am(a) = o.
Hence the point (a, b) is on F(X, Y) = O. Since there are infinitely many
possibilities for a, one obtains infinitely many points (a, b) on F(X, Y) = o.
The same argument shows tha.t there are infinitely many points not on
F(X, Y) = O. Let us take a E 0 as above. Now we use the fact that
ao(a) ym + · · · + am(a) = 0 has at most m roots to obtain a b in 0 such
that. ao(a)b m + · · · + am(a) =1= o.
Our next question is: Is it true that every pair of curves has a point
in common 1 Clearly no, 'as parallel lines (for example, X = 1 and X = 2)
do not meet. We know how to remedy this difficulty: we pass to the pro-
jective plane. In it we define an algebraic curve as the locus determined by
an equation F(X, Y, Z) = 0 with F a homogeneous polynomial in
O[X, Y, Z], F ft O. The question now is: Is it true that every curve in the
projective plane over the field 0 of complex numbers has with every other
such curve at least one point in common 1
Here we have still another change in point of view: we pass from the
affine plane to the projective plane. I t is not necessary to do this; one can
study curves in the affine plane and one ,vill find a criterion (more or less
complicated) for two curves to meet. However, it is very convenient to pass
to the projective plane, where we will find simpler formulations for many of
our t.hel)rems.
Remark. We said above that we could learn little about X2 + Y2 + 1
from the locus of real points of X 2 + Y 2 + 1 = 0, since this locus is empty.

INTRODUCTORY REMARKS 5
Actually, we can prove that X2 + Y2 + 1 is irreducible in lc[X, Y] already
from this fact: if X2 + Y2 + 1 factored, it would factor into two linear
factors, the locus X2 + Y2 + 1 = 0 would consist of two straight lines,
and such a locus cannot be empty. A better example is X4 + Y4 + 1 = O.
A priori, this could factor in k[X, Y] into a linear factor and a factor of
degree three, or into two factors of degree two. The first possibility is
excluded by the argument just given, but the second cannot be similarly
excluded. Later, however, we shall define the notion of a singular point of
a curve, and we shall see that X4 + Y4 + 1 is irreducible in C[X, Y] be-
cause the curve X4 + Y4 + 1 = 0 has no singular points (in the projective
plane over C), while any reducible curve does have singularities. A fortiori
X 4 + Y4 + 1 does not factor in k[X, Y].
Exercises.
1. Show that any line meets the curve Y2 - X2 - X3 = 0 in at most three
points. (Consider separately lines of the form Y - mX - b = 0 and lines
of the form X - a = 0.)
2. Does Y2 - X2 - X3 factor (properly) in C[X, Y] 1 If not, explain.
3. Consider the curve Y2 - X2 - X3 = 0 as a real locus. Sketch the curve,
especially near the origin.
4. Do Exercises 1, 2, and 3 for y2 - X3 = o.
5. Let
PICX, Y) = X2 + Y2 + a1X + b1Y + C!!,
P 2 (X, Y) = X2 + y2 + a 2 X + b 2 Y + C2'
with PI =1= P 2 . The locus of points (x, y) in the complex affine plane such that
PI (x, y) = P 2 (x, y)
is a straight line through the intersections of the curves PI (X, Y) = 0 and
P 2 (X, Y) = O. [If P1(X, Y) = 0 is a circle in the Euclidean plane and
(x, y) is a real point, then P 1 (x, y) is called the power of (x, y) with respect
to the circle. It has a simple geometrical interpretation.]
· Declarative sentences in the exercises (as, for example, the second sentence
in Exercise 5) are assertions to be established.

Chapter 2
HOMOGENEOUS COORDINA TES
In this chapter we recall some of the basic facts about affine and projective
planes. We start with the Euclidean plane and take in it a coordinate system.
Each point, then, is associated with an ordered pair of numbers (x, y), its
coordinates; every ordered pair (x, y) gives the coordinates of some point.
Every line has a linear equation aX + bY + c = 0 with a =F 0 or b =F 0,
i.e., a point (x, y) is on the line if and only if ax + by + c = 0; the equation
is essentially unique, i.e., if a' X + b' Y + c' = 0 is another equation for
the sanie line, then there is a number A such that a' = Aa, b' = Ab, c' = AC.
Conversely, every linear equation aX + bY + c = 0 with a =F 0 or b i= 0
is the equation of some line.
In the Euclidean plane, the coordinates x, yare real numbers, and one
has the notion of the distance between two points. If one wishes to disregard
this notion, one speaks of the affine plane (over the real-number field k).
But then, more generally, one can start from an arbitrary field K, define a
point as a pair (x, y) of elements from K and a line as the set of points satisfy-
ing a linear equation aX + bY + c = 0 with a =F 0 or b =F 0, and, on the
whole, retain the geometry of points and lines of the real affine plane.
For example, we shall still have the following theorems:
I) On any two (distinct) points there is one and only one line.
II) (The Parallel Postulate) For any line l and any point P not on l, there
is one and only one line on P that does not meet l.
The proofs of these theorems are algebraic arguments which are familiar
for k, but which work equally well for any field K.
The object we have just described is called the affine (number) plane
over K and is denoted AN K.
Let (x, y) be a point in AN K. By homogeneous coordinates for (x, y)
we mean the triple (1, x, y) or any multiple (t, tx, ty) thereof provided that
6

PROJECTIVE PLANES 7
t i= 0 (t E K). Any triple (X o , Xl' X 2 ) with Xo i= 0 is a set of homogeneous
coordinates for just one point, namely (x1/X O ' X2/XO). The relations between
the original, or nonhomogeneous coordinates, and the homogeneous coordinates
are
X == XI/XO'
y == X2/XO.
Hence if bX + c Y + a == 0 is a line, the condition on the homogeneous
coordinates (x o , Xv x 2 ) of a point P to lie on the line is
b(xl/x O ) + c(xl/X O ) + a == 0,
or since Xo i= 0, bX I + cX 2 + ax o == 0, so that the homogeneous equation
for the line is bX I + cX 2 + aX o == o.
A linear equation aoXo + alX I + a 2 X 2 == 0 can be regarded either
from a homogeneous point of view or from a nonhomogeneous point of view.
Regarding it in the first way, we do not count (0, 0, 0) as a solution, or
(for emphasis) we call it the trivial solution, where the word trivial is being
used in a technical sense. If the triple (X o , Xl' X 2 ) satisfies the equation, then
so does (tx o , tx l , tx 2 ) for any t i= 0 (in K); if xO' Xv x 2 are not all zero, one
calls the whole class (tx o , tx l , tx 2 ), where t runs over K - 0, a (single) solution
of the (homogeneous) equation. [Notation. (x o : Xl : x 2 ) stands for this class.]
Assuming, then, that a l *- 0 or a 2 *- 0, one sees that there is a one-to-one
correspondence between the solutions of a o + alX + a 2 Y == 0 and those of
aoXo + alX I + a 2 X 2 == 0 in which Xo *- O. Considering now only triples
(xo, Xv x 2 ) in which Xo == 0, one observes that (0, -a 2 , a l ) satisfies the equa-
tion, hence so does (0, -ta 2 , tal); and one proves that no other triple satisfies
the equation. Thus there is one and only one solution to
aoXo + alX I + a 2 X 2 == 0
in 1vhich Xo = 0, namely {(O, -ta 2 , tal)}.
PROJECTIVE PLANES
So far we have spoken only of the affine plane. By a point of the projective
plane over K one means a class of triples (tx o , tx v tx 2 ), where xO' Xl' X 2 are in
K, but are not all zero, and t runs over K - O. The triple (x o , Xv x 2 ), abbrevi-
ated X, is called a representative of the point; for t *- 0, (tx o , tx v tx 2 ), abbrevi-
ated tx, is a representative of the same point. The point with representative x
is denoted P x; we sometimes also refer, though loosely, to the point x.
The triple (x o , Xv x 2 ) is also called a set of coordinates for the point P x.
By a line of the projective plane over K one means the set of points P x such that
aoxo + a1x I + a 2 x 2 == 0, where a o , a v a 2 are in K and are not all zero.
The equation aoXo + alX 1 + a 2 X 2 == 0 is called an equation of the line;
taoXo + talX I + ta 2 X 2 == 0, for any t in K - 0, is also an equation of the
same line, and one proves that there are no others. Therefore one speaks,

8 HOMOGENEOUS COORDINATES
though with a slight inexactitude, of aoXo + a1X 1 + a 2 X 2 == 0 as being
the equation of the line. The triple of coefficients (a o , a v a 2 ) == a is called a
representative of the line; then (ta o , tal' ta 2 ), t E K - 0, is also a representative
of the same line, and there are no other representatives for the line. Notation-
ally, we refer to the line La. The point Px is said to be on the line La if Px
is one of the points of the line La' or, in other words, if
aoXo + alx t + a 2 x 2 == o.
The notion made up of the set of points, the set of lines, and the relation
on just defined, is called the projective (number) plane over K and is denoted
PNK.
It is usual to write the triples which represent lines as row triples and
those which represent points as column triples; for typographical reasons,
however, we will also write the point representatives in a row if no confusion
is anticipated.
THE LINE AT INFINITY
It is clear that there must be a relation between the affine and projective
planes.
Start with the projective plane PNK and subtract the line Xo == 0:
every point not on Xo == 0 has a set of coordinates of the form (1, x, y).
Then (1, x, y)  (x, y) is clearly a one-to-one correspondence between the
points of PNK not on Xo == 0 and the points of ANK. Moreover the points
of the projective line aoXo + alX 1 + a 2 X 2 == 0, with a 1 =F 0 or a 2 =F 0,
make up, with just one exception [namely, the point (0, -a 2 , at)], the points
of the affine line a o + alX + a 2 Y == o. Thus the correspondence
(1, x, y)  (x, y) induces a one-to-one correspondence between the lines of
PNK, with the exception of Xo == 0, and the lines of ANK, the line
aoXo + alX l + a 2 X 2 == 0 corresponding to the line a o + alX + a 2 Y == o.
Because of these correspondences, one says that the plane AN K is part of
the plane PNK; one also says that aoXo + alX l + a 2 X 2 == 0 (with a l =F 0
or a 2 =F 0) and a o + at X + a 2 Y == 0 are the same line.
Of course, one can start with AN K and add on the nlissing line Xo == 0
to get the projective plane.
From the point of view of the projective plane, any line is very much
like any other: the line Xl == 0, for example, is just as good, or bad, as the
line Xo == O. But from the point of view of the affine plane obtained by
removing Xo == 0, the line Xo == 0 is special: it is then referred to as
the line at infinity. Points on Xo == 0 are referred to as the points at infinity,
while the points in AN K itself are said to be at finite distance (though no
concept of distance is involved).
From the projective plane one can subtract any of the lines Xo == 0,
Xl == 0, or X 2 == 0 to obtain an affine plane. In this way three different

PROJECTIVE PLANES OVER AN EXTENSION FIELD OF K 9
X2=0: Y=O
Eo: (1,0,0) El: (0,1,0)
Fig. 2.1
(though isomorphic) ANK's are obtained. (Two affine planes are said to be
isomorphic if the points and lines of one plane can be put in one-to-one
correspondence with the respective points and lines of the other in such way
that a point and a line of one plane are incident if and only if the correspond-
ing point and line of the other plane are incident.) The projective plane can
be thought of as these three affine planes appropriately patched together.
To have a diagram of the projective plane in relation to the affine plane,
one draws a triangle as in Fig. 2.1. The line at infinity Xo == 0 is brought
nearer, where one can see it. The X-axis, X 2 == 0 or Y == 0, is drawn
horizontally; the Y-axis, Xl == 0 or X == 0, is drawn nearly, but not quite
vertically, since angle, and in particular right angle, is not a concept of affine
or of projective geometry; moreover the events taking place near Eo are
just like those taking place near E I , or E 2 , or, indeed, any other point.
The point Eo: (1, 0, 0) is the origin (0, 0) of the affine plane. The point
E 1 : (0, 1, 0) is the point at infinity on the X-axis, and E 2 : (0, 0, 1) is the point
at infinity on the Y-axis. Lines through E 2 are called vertical, those through
E 1 , horizontal. The parallel but nonvertical lines Y == mX + b l and
Y == mX + b 2 meet at infinity in the point (0, 1, m).
PROJECTIVE PLANES OVER AN EXTENSION FIELD OF K
Let L be an extension field of K, Le., a field containing K. (Consider,
for example, K == real field, L == complex field.) Then, obviously, ANK
is contained in ANL; in symbols, ANK c ANL. If in PNL we con-
sider the points and lines having coordinates in K, then these points and
lines form a plane isomorphic to P N K, and in this sense one can say
PNK c PN L. Note, however, that a point in PNK may have coordinates
not in J{; for example, P : (1,1,1) is in PNK, but P also has coordinates
(t, t, t) for any tEL - 0, and t may not be in K.
Exercise. By normalized coordinates for a point in a P N K one means coordinates
in which at least one entry is 1. Every point has at least one and at most three
sets of normalized coordinates. If K c L, then a point in P N L is in P N K if
and only if its normalized coordinates are in K.

Chapter 3
ALGEBRAIC CURVES*
A point of the projective plane is represented by a triple (x o , Xl' X 2 ), some
Xi =F 0, and equally well by (pxo, px l , px 2 ) for any p =F 0: the point is the
collection of triples (pxo, pXv px 2 ). If F (X o , Xl' X 2 ) E C[X o , Xv X 2 ] is a
polynomial that vanishes at (x o , Xv x 2 ), then we shall want F also to vanish
at (pxo, pXv px 2 ); otherwise we could not speak of F vanishing or not vanish-
ing at the point represented by (x o , Xv x 2 ). For this it will be sufficient (and,
in fact, also necessary) to assume that F is homogeneous, Le., that all its
terms have the same total degree.
Then let F(X o , Xv X 2 ) E C[X o , Xl' X 2 ] be a nonconstant homogeneous
polynomial. The set of points (xo, Xv x 2 ) such that F (x o , Xv x 2 ) = 0 is called
an algebraic curve (in the projective plane).
N ow we ask the following question: Does every pair of algebraic curves
(in the projective plane) have points in common?
FIRST CASE: One of the curves is a line. One then has
n
r : F(X o , Xv X 2 ) = 0
L : aoXo + alX I + a 2 X 2 = 0
(n = degree of F),
(some ai =F 0).
We may suppose that a 2 =F o. To solve siInultaneously, we eliminate X 2 ,
getting the equation F(Xo, Xl' -(a O /a 2 )X O - (a l /a 2 )X I ) = o. Consider
the polynomial
( ao a l )
F Xo, Xl' - - xo - - Xl = H(X o , Xl).
a 2 a 2
Two cases can occur: either H is the polynomial zero or H is different from
zero. In the first case, every point of the line is a point of r, i.e., L cr.
* In the previous chapter we have allowed an arbitrary field K to intervene, but
in the present and subsequent chapters, until further notice is given, we again
take K = C, the complex field: the discussion takes place in the projective
number plane over the complex numbers.
10

THE EQUATIONS OF A CURVE 11
In the second case, H is a homogeneous polynomial of degree n; then
H(X o , XI)/X is a polynomial in XI/X o and as such can be factored over 0
into linear factors. Multiplying by X, we see that H(X o , Xl) can be written
in the form
n
H(X o , Xl) == II (x&i)X 1 - Xi)Xo).
i=l
The linear factor X)XI - x.)Xo gives rise to the (complex) solution
(xg>, x&i» or, rather, to the class {(px&i), pxi i »)}, p =F 0, which, however,
counts only as one solution. Hence H(X o , Xl) == 0 has at most n and at
least one solution (H may have multiple factors). Each of these solutions
gives us a point of intersection of rand L. Thus:
Theorem 3.1. E1,ery curve r defined by an equation F == 0 of degree
n cuts every line L in at least one point and, unless it contains L, in at
most n points.
Corollary 3.2. If every point of L s on r, then F s divisible by
aoXo + alX I + a 2 X 2 .
This follows upon observing that H(X o , Xl) is the remainder upon divi-
sion of F(X o , Xv X 2 ), regarded as a polynomial in X 2 , by
a l a o
X 2 + - + - Xo.
a 2 a 2
Exercise. Establish Theorem 1.4 for projective curves.
The following result is a generalization of the last corollary, and its proof
,vill be given later on in the chapter.
Theorem 3.3. Let F(Xo, Xl' X 2 ) be a (homogeneous) irreducible poly-
nomial. If G is a (homogeneous) polynomial that vanishes at every point of
F == 0, then G is divisible by F in C[X o , Xv X 2 ].
THE EQUATIONS OF A CURVE
An algebraic curve in the projective plane is given by a polynomial equation:
F (Xo, Xl' X 2 ) == 0, but the polynomial F is not unique; for example, the
same curve is also given by F2(XO' Xv X 2 ) == O. However, one can describe
all the possibilities for the equations of a curve r.
We recall some facts about a polynomial ring K[ Y I , . . . , Y n ] in n
letters over a field K. Bya unit in K[ Y v . . . , Y n ] we mean any element e
for which there is anf in the ring with ef == 1. Since deg uv == deg u + deg v
(where deg is the abbreviation for degree) for any elements u, v, we find that
deg e + deg f == 0; since deg u > 0 for every element u, we find that

12 ALGEBRAIC CURVES
deg e = 0, Le., e E X (and e i=- 0). Conversely, if e E X and e i=- 0, then
since K is a field, there is an f E K such that ef = 1. Thus:
3.4. An element e in X[ Y 1 , . . . , Y n ] is a unit in K[ Y v · · · , Y n ] if
and only if e E X and e i=- O.
An element is called reducible if it can be factored into two nonunits;
otherwise it is called irreducible. If u is factored: u = vw, with neither v
nor w a unit then deg v > 0, deg w > 0, and since
deg u = deg v + deg w,
we have deg v < deg u, deg w < deg u. This shows by induction that any
element of u in X[ Y v . . . , Y n] can be factored into irreducible factors.
Now suppose we have factored u in two ways:
u = Cfl · · · fs,
,
u = C Yl · . · Yt,
where c and c' are units and the fi and Yj are irreducible nonunits. Then it
can be shown that s = t and that for some rearrangement of the subscripts,
Yi = cifi, where C i is a unit. Since a unit and its inverse can always be
introduced into a factorization to give another, the above factorizations are
regarded as essentially the same. These facts are summed in the following
theorem, which we will not prove.
Theorem 3.5. A polynomial ring in any number of letters over a field
is a unique factorization domain.
Polynomials f, Y such that f = cy, where c is a unit, are called associates.
Given a complete factorization u = Cfl · · · fs of an element u, Le., a factoriza-
tion with c a unit and the fi irreducible nonunits, it is convenient to group
the associates among the fi together, to write each fi in a group as a unit
times some one element of the group, and to let c absorb the new units thus
arising. In this way, the complete factorization is re"'Titten as
U - cy TI . . . Y T t
- / 1 t ,
where c is a unit, the Yi are irreducible nonunits, and Yi, Y; are nonassociates
if i i=- j. The r i are positive integers and Yi is said to occur with multiplicity
ri as a factor of u. Sometimes for notational reasons it is convenient to
allow some r i to be zero, but of course if r l = 0, say, then Yl does not occur
as a factor of u.
We have already used above that deg fy = deg f + deg y. By the
subdegree of f one means the degree of the lowest degree terms in f. It is
also true, and easy to prove, that subdeg fy = subdeg f + subdeg y. Now
if u = fy is homogeneous, then deg u = subdeg u, from which
subdeg f + subdeg g = deg f + deg g,

THE EQUATIONS O:E' A CURVE 13
and since subdeg f < deg f and subdeg g < deg g, we conclude that
subdeg f = deg f and subdeg g = deg g. In other words:
3.6. If a homogeneous polynomial u is factored, then its factors are also
homogeneous.
Now let r be an algebraic curve and let F = 0, G = 0 be equations for
r. Write
F = C F T 1FT 2 · · · FTm
12m'
G = d Gi1G2 · · · Gn,
where c, d are units, the F i , G k are irreducible nonunits, F i is not the associate
of F; for i "* j, and G i is not the associate of G; for i "* j. By Theorem 3.3,
since G vanishes wherever FI does, G is divisible by F I , G = FIB. Factoring
H completely and applying the unique factorization theorem, we see that
FI is the associate of some G;, and each F i is the associate of some G;.
Similarly, each G i is the associate of some F;. Thus m = n, and writing each
G i as a unit times an F;, absorbing the new units in d, and rearranging the
subscripts, we see that the second factorization above can be rewritten:
G = d Fi 1 . · · F;:a. Thus we have the following theorem.
Theorem 3.7. If r is given by F = 0, where F = c F1 · · · F,
c is a unit, the F i are irreducible nonunits, F i and F; are nonassociates
for i "* j, and the r i are> 0, then every other equation of r is of the form
G = 0 with G = d Fi 1 · · · F;:a, where d is a unit and the Si are> o.
Taking the r i all as small as possible, namely, r i = 1, we see:
Corollary 3.8. The equations of r of minimal degree are all of the form
c F 1 F 2 . · · Fm = o. Thus disregarding the inessential factor c, we can
say that the equation of r of minimal degree is essentially unique.
The equation of r of minimal degree will sometimes be referred to as
the equation of r .
Definition 3.9. A curve r is said to be reducible if it is the proper union
of two other (algebraic) curves: in symbols r = r 1 U r 2' r "* r l'
r "* r 2. If r is not reducible, it is called irreducible.
Theorem 3.10. A curve r is irreducible if and only if its m'inimal
equation F = 0 is such that F is irreducible.
Proof. Let r be given by F 1 · · · Fs = 0, F i an irreducible nonunit, F i not
an associate of F; for i "* j, so that F = FI · · · Fs is irreducible if and
only if s = 1. Suppose first that s > 1. Let r I be the curve F 1 = 0, r 2
the curve F 2 . · · Fs = o. Since FI is not divisible by F 2 , by Theorem 3.3 F 1
cannot vanish over F 2 = 0 and a fortiori cannot vanish over r 2; therefore

14 ALGEBRAIC CURVES
r 2 is not contained in r v in symbols: r 2 q: r 1. Similarly one proves
r 1 q: r 2 . Therefore r == r 1 U r 2 with r * rv r * r 2 , whence r is
reducible.
Now suppose that s == 1, and let r == r 1 U r 2 . Let G 1 == 0 by the
minimal equation for r 1. Since F == F 1 vanishes over r v F 1 is divisible
by each irreducible factor of G 1 and hence by G 1 . Thus F == hG 1 ; and since
F 1 is irreducible, h must be a constant and r 1 is given by F I == 0, i.e.,
r == r 1. Hence r is irreducible.
Thus a curve is irreducible if and only if it is given by an equation F == 0
with F irreducible (or, as one says, by an irreducible equation).
Theorem 3.11. Let r l' r 2 be irreducible curves with equations F 1 == 0,
F 2 == 0, where Fv F 2 are irreducible. Then r 1 == r 2 if and only if F I
and F 2 are associates.
Proof. If Fv F 2 are associates, then clearly r 1 == r 2. Conversely, if
r I == r 2' then F 1 == 0 and F 2 == 0 are minimal equations of r l' and F 1
and F 2 are associates by 3.8.
Theorem 3.12. Let r be a curve with F == 0 as a (not necessarily
minimal) equation and let  be an irreducible curve with minimal equation
G == O. Then  is contained in r if and only if F is divisible by G, i.e.,
if and only if G is one of the irreducible factors of F. The irreducible
curve  is contained in a union r 1 U r 2 u · .. u r s of curves r i if
and only if it is contained in one of the r i.
Proof. If F is divisible by G [in symbols, F -- O(G)], then F vanishes
wherever G does, so  c r. Conversely, if  c r, then F vanishes where-
ever G does, and since G is irreducible, F - O(G), by 3.3. As for the second
part of the theorem, if  is contained in some r i, then of course  is contained
in r 1 U · · · u rs. Conversely, if r i is given by F i == 0, then r 1 U · · · u rs
is given by F 1 · · · Fs == 0, and if  c r 1 U . · · u r s' then G is a factor
of F 1 · · · Fs, by the first part of the theorem, and hence is a factor of one of
the Fh whence  is contained in one of the r i .
Corollary 3.12a. If r,  are irreducible curves and  c r, then  == r.
Proof. If F == 0, G == 0 are the irreducible equations for 1"", , then from
 c r we have already seen that F - O(G). Since F is irreducible, this
imples that F and G are associates, whence  == r.
Theorem 3.13. A curve r can be written as the union of irreducible
curves in one and only one way.
Proof. If F 1 F 2 . · · Fs == 0 is the minimal equation of r (with the F i
irreducible nonunits) and r i : F i == 0, then the r i are irreducible and
distinct and r == r I U · · · u r s. To prove the uniqueness, let
r == r I U · · · u r s == l U · · . U t,

INTERSECTIONS 15
where the i are irreducible and distinct. Each i is contained in a r j (by
3.12), hence is equal to r j (by 3.12a). Similarly, each r i is a j. Hence
the set {r v . . . , r s} equals the set {1' . . . , t}. Q.E.D.
The distinct irreducible curves r v . . . , r s whose union is
r-r U ... u r
- 1 s'
are called the components of r.
Let r be a curve and F = 0 an equation for r, not necessarily a minimal
equation: F = c F 1 1. · · FJ:. Many proofs proceed by taking an equation
F = 0 for r, and the proofs in no way make use of whether or not F is
minimal. In this way, although the statement of a theorem may be about a
curve r, the theorem proved is actually more general than the original
statement. To have a terminology for such a situation we introduce the
notion of a cycle. A cycle is the notion made up of a number of irreduc-
ible curves r v . . . , r m (m > 0) and associated positive multiplicities
rv . . . , r m: notationally, we write r 1 r 1 + · · · + r m r m for this cycle
(or 0, if m = 0). As equation for this cycle we take (by definition)
c Fl · · · F';: = o. Thus every equation F = 0, F E C[X, Y] - 0, is the
equation for one and only one cycle; and every cycle has one and essentially
only one equation.
For a curve r we usually tacitly take its minimal equation,. in effect, we
regard a curve as a cycle in which the multiplicities r i are = 1.
Although many of our theorems will really hold for cycles, we will
continue to use the word "curve"; and we will do this even when we intend
to apply the theorem to cycles.
Remark. The cycles we have defined are so-called pure I-dimensional cycles.
One may define a pure O-dimensional cycle as a notion made up of points
PI' . . . , Pn with associated multiplicities. More generally, one may allow
both points and curves in a cycle; when this is done, the pure I-dimensional
cycles are called divisorial cycles (or divisors). The associated multiplicities
r l' . . . , r m may also be allowed to be negative, and if this is done, one calls
a divisorial cycle in which the ri are> 0 (or, for notational reasons, > 0)
effective, and otherwise one calls it virtual. As our cycles throughout are
the so-called effective divisorial cycles, we may refer to these more briefly
simply as cycles.
INTERSECTIONS
Theorem 3.3, the proof of which was postponed, can be reformulated and
strengthened as follows:
Theorem 3.14. Let F(X o , Xl' X 2 ) be an irreducible homogeneous
polynomial with F  C. If G(Xo, Xl' X 2 ) is a homogeneous polynomial not
divisible by F, then G vanishes at only a finite number of points of F = o.

16 ALGEBRAIC CURVES
Proof. Since C[X o , Xl' X 2 ] is a unique factorization domain, we can speak
of the greatest common divisor of F and G in C[X o , Xl' X 2 ], and we have
G.C.D. of F, G = 1. Let us consider now the ring C(X o , X I )[X 2 ], i.e., the
ring of polynomials in X 2 with coefficients in the field C(X o , Xl) (= set of
quotients f(Xo, X l )/ g(X o , Xl)' where f, g E C[X o , Xl]). In C(X o , X l )[X 2 ] the
G.C.D. of F and G is still 1. In fact, suppose that F and G have a nonunit
common factor k(Xo, Xl' X 2 ) in C(X o , X I )[X 2 ]; since h is a nonunit,
degx 1 h > o. Then F = uk, G = vk. Shifting the denominator of h to u
and v, we may suppose that h E C[X o , Xl' X 2 ]. Writing u, v as quotients
of polynomials with a common denominator d(Xo, Xl) E C[X o , Xl] and
multiplying through by d, we get dF = Uh, dG = Vh, where now U, V E
O[X o , Xl' X 2 ]. The irreducible factors of k involving X 2 cannot go into
(i.e., factor) d, so they go into F; similarly they must go into G. This
contradicts that the G.C.D. of F, Gin C[X o , Xl' X 2 ] is 1.
The ring C(X o , X I )[X 2 ] is a polynomial ring in one letter over a field, and
we recall that in such a ring, the G.C.D. of two elements F, G can be written
in the form uF + vG, where u, v are in the ring. Hence we have
A' ( X X X ) B' ( X X X )
1 = 0' l' 2 G ( X X X ) + 0' l' 2 F ( X X X )
L ( X X ) 0' l' 2 M ( X X ) 0' l' 2'
0' 1 0' 1
where A', B', L, M are polynomials (not necessarily homogeneous). Reducing
to a common denominator d (X o , Xl) we can write
d(Xo, Xl) = A(X o , Xl' X 2 )G(X O ' Xl' X 2 ) + B(X o , Xl' X 2 )F(X o , Xl' X 2 ).
By retaining on both sides only the terms of a given degree, one can assume
here that d, A, B are homogeneous (and d ::F 0). Since' d (X o , Xl) = 0
represents a finite number of lines through the point (0, 0,1), it follows from
the last equality that the common points of F = 0 and G = 0 are on those
lines. See Fig. 3.1. Reasoning in the same way with respect to the other
variables one obtains that the points of intersection of F = 0 and G = 0
are among the points in which three finite sets of lines cut themselves
(see Fig. 3.1) and hence F = 0 and G = 0 intersect in only a finite number
of points.
(0,0, 1)
Fig. 3.1

AFFINE ALGEBRAIC CURVES 17
We give now another proof of a slightly weaker statement, namely,
that ifG vanishes wherever F does, then G - O(F), given that Fis irreducible.
In fact, suppose G =1= O(F). As in the previous proof one shows that
d(Xo, Xl) = A(X o , Xv X 2 )G(X O ' Xv X 2 ) + B(X o , Xv X 2 )F(X o , Xl' X 2 ),
with d, A, B honl0geneous. We may suppose that degx F > O. Let
2
F = ao(X o , XI)Xr + al(X O ' XI)Xr- 1 + · · · , a o -=I=- O. Since d -=I=- 0 as be-
fore, the polynomial aod vanishes only for a finite number of ratios xo: Xl.
Let xo, Xl be such that ao(x o , xl)d (xo, Xl) -=I=- o. Then one solves F (xo, Xv x 2 ) = 0
to obtain xo: Xl: x 2 such that F (x o , Xv x 2 ) = o. By hypothesis G(xo, Xl' X 2 ) = 0
also and hence
d (xo, Xl) = A(x o , Xv x 2 )G(X O ' Xl' X 2 ) + B(x o , Xv x 2 )F (X o , Xl' X 2 ) = 0,
which is a contradiction.
The same argument holds for any number of variables, both with
Xo, Xv . . . , X n in projective space as well as with Y v Y 2 , . . . , Y n in
affine space. In nonhomogeneous or affine terms, the theorem says:
Theorem 3.15. Let G, FE C[Y v . . . , Y n ] with F irreducible. If G
vanishes wherever F does, then G == O(F) (in C[Y v . . . , Y n ]).
Exercises
1. One says that G( Y l' . . . , Y n) vanishes almost everywhere on
F( Y l' · · · , Y n) = 0
if G vanishes everywhere on F = 0 except perhaps where another poly-
nomial H vanishes on F = 0, provided H does not vanish everywhere on
F = o. Show that if F is irreducible and G vanishes almost everywhere on
F = 0, then G == O(F).
2. In 3.15, if one drops the hypothesis that F is irreducible, one can still con-
clude that a power of G is divisible by F.
AFFINE ALGEBRAIC CURVES
By a (plane) affine (algebraic) curve one means (as we have already said in
Chapter 1) the set of points (x, y) with x, y in C that satisfy an equation
F(X, Y) = 0, where F is in C[X, Y] but not in C.
Let F*(Xo, Xv X 2 ) = 0 be a curve in the projective plane. Then F*
is a homogeneous polynomial, of degree m, say, and we can write it in the
form:
F* = aoX + al(X 1 , X2)X--1 + a 2 (X 1 , X2)X-2 + · · · ,
where ai(X V X 2 ) is homogeneous of degree i in Xv X 2 . Now assume that the
curve is not the line Xo = o. Then F* is not a power, nor a constant times

18 ALGEBRAIC CURVES
a power, of Xo; hence ai -=I=- 0 for some i > O. Let as "* 0, but
a s + 1 == a s + 2 == · · · == O.
Then
F*(I, X, Y) == a o + al(X, Y) + a 2 (X, Y) + · · · + as(X, Y),
from which one sees that F*(I, X, Y) tt C. Clearly the point (1, x, y) on
F*(Xo, Xl' X 2 ) == 0 yields the point (x, y) on F*(I, X, Y) == 0, and con-
versely. Thus, taking Xo == 0 as line at infinity of the associated affine
plane, we have:
3.16. The points at finite distance on a projective curve different from loo
are the points of an affine curve. If F*(Xo, Xv X 2 ) == 0 is an equation for
the projective curve, then F*(I, X, Y) == 0 is an equation for the affine
curve.
The converse of 3.16 also holds. Starting with the affine curve
F(X, Y) == 0, where
F(X, Y) == a o + al(X, Y) + a 2 (X, Y) + · · · + as(X, Y)
and ai(X, Y) is the sum of the terms in F of degree i, one merely writes
F*(Xo, Xv X 2 ) == aoX + al(X V X 2 )X O -l + · · · + as(X v X 2 )
to obtain a projective curve, F*(Xo, Xv X 2 ) == 0, whose affine part is the
given curve F(X, Y) == o. Note that
( Xl X2 )
F*(X o , Xl' X 2 ) == XF -, - .
Xo Xo
If F (X, Y) is of degree 8, we call
XF ( Xl , X 2 ) == F*(X o , Xv X 2 )
Xo. Xo
the standard (or canonical) homogeneous polynomial associated with F; it is
of degree s and is not divisible by XO' If F*(X o , Xv X 2 ) is homogeneous of
degree s and is not divisible by Xo, we call F*(I, X, Y) the standard non-
homogeneous polynomial associated with F*; it is of degree s. Thus:
3.17. The standard (or canonical) operations just dfined set up a one-to-one
correspondence between the homogeneous polynomials in Xo, Xv X 2 not
divisible by Xo and the nonhomogeneous polynomials in X, Y. Correspond-
ing polynomials have the same degree.
A simple straightforward argument shows:
3.18. Let the homogeneous polynomial F*(Xo, Xv X 2 ) be ¥= O(X o ). Then
F * is irreducible if and only if F (1, X, Y) is.

AFFINE ALGEBRAIC CURVES 19
Definition. Let r* be a curve in the projective plane not having
Xo === 0 as a component. Then the curve r consisting of the points not
on Xo === 0 is called the affine representative of r* (relative to Xo === 0 as
line at infinity).
Theorem 3.19. Let ri : Fi === 0, r: F == 0 be two curves not having
Xo === 0 as component. Let r v r 2 be the associated affine representatives.
Then ri === ri if (and only if) r 1 === r 2.
Proof. Clearly XoFi === 0 and XoF; === 0 are the same locus. Hence XoFi
and XoF have the same irreducible factors. Since Fi, Fi are neither
- O(X o ), also Fi and Fi have the same irreducible factors. Hence ri === ri.
Definition. Let r* be a curve in the projective plane not having
Xo === 0 as component. If r* has r as affine representative, then (on
the basis of 3.19) we call r* the projective completion of r.
Theorem 3.20. Let r v . . . , r s be affine curves having rr,..., r:
as projective completions. Then the projective completion of r 1 u · · · u rs
is r i u · · · u r:.
Proof. ri u · · · u r: is a projective curve meeting Xo === 0 in only a
finite number of points and having r 1 u · · . u r s as its affine part,
whence the assertion follows.
The notions of reducibility and irreducibility of affine curves are defined
analogously to the projective notions: an affine curve r is reducible if it
can be written in the form r === r 1 U r 2 , r * rv r * r 2 , rv r 2 affine
curves; and irreducible otherwise.
3.21. A n affine curve r is reducible if and only if its pro.iective completion
r* is.
Proof. Suppose that r === r 1 U r 2 with r * r 1 and r * r 2. Then for
the projective completions we have (by 3.19) r* * ri and r* * ri.
Hence (by 3.20) r* === ri u r is reducible. The converse is proved in the
same way.
Theorem 3.22. Let r v . . . , r s be distinct irreducible affine curves
and rr: Fi === 0, i === 1, . . . , s, their projective completions. By 3.21
the ri are irreducible, and the Fi are taken irreducible. Let Fi(X, Y) ===
Fi(l, X, Y). (By 3.18 the F i are irreducible.) Then all equations for
r 1 u. .. u rs are of the form FlF2. · · Fs === 0 with the r i > o.
Proof. Let F (X, Y) === 0 be an equation for r 1 U... u r s. Let F* be
the standard homogeneous polynomial associated with F. Then F* === 0 is
an equation of ri u.. · u ri. The ri are distinct, hence F* is of the
form F* === Fi r1 . · · Firs with the ri > O. Hence F === Fl. · · Fs. Q.E.D.

20 ALGEBRAIC CURVES
Defining cycle in the affine plane as the notion made up of a finite number
of distinct irreducible affine curves r v . . . , r s and associated positive
multiplicities rl'...' rs and taking Fl... F8 = 0 as the equation of
r1r 1 + · · · + rsrs one obtains:
Corollary 3.24. There is a one-to-one correspondence between the cycles
in the affine plane and those in the projective plane not containing Xo = o.
The one-to-one correspondence between the cycles is obtained through
the one-to-one correspondence of their equations.
The last few theorems and definitions are simply translations into affine
terms of previous projective theorems and definitions. Such translations,
and the reverse translations, are usually (though not always) a straightforward
and simple matter. Usually, then, we will make and use these translations
without much ado.
Exercises
1. Sketch the curve XY = 1 in the real affine plane. Sketch its projective
completion X 1 X 2 - X = 0 in the real projective plane (especially near
the line at infinity).
2. Find the points in which (the projective completion of)
X3 - 6X2Y + 11XY2 - 6Y3 + 1 = 0
meets the line at infinity, loo.
3. Sketch (the projective completion of) Y 2 - X 3 = 0 in the real projective
plane. Do the same for Y - X3 = o.
4. All the curves X 2 + Y 2 + aX + bY + c = 0 go through the same pair
of points I, J on the Hne at infinity. (These are the so-called circular points
at infinity.)
5. Let F(X o , Xl' X 2 ) be homogeneous. Show that if F(l, X, Y) = 0, then
F(X o , Xl' X 2 ) = o.
6. Let F(X o , Xl' X 2 ), G(X o , X 19 X 2 ) be homogeneous polynomials with
deg F > deg G Assume that F(l, X, Y) = G(l, X, Y). Then
F(X o , Xl' X 2 ) = X G(X o ' Xl' X 2 ),
where d = deg F - deg G.
7. a) Show that
Xu X l2
X 21 X 22
is irreducible in C[X n , . . · , X 22 ].
b) Show that
I Xu X l2 XiS
X 21 X 22 X 23
X S1 X 32 X 33
is irreducible in C[X 11' · · · , Xss].

Chapter 4,
RESULTANTS
Consider two polynomials
f = aoX m + a1X m - 1 + · · · + am,
g - b Xn + b Xn-l + · · · + b
- 0 1 n'
with indeterminate coefficients a = (a o , . . . , am), b = (b o , . . . , b n ). Let
us suppose that for certain special values a = a, b = b of the coefficients,
these polynomials,! and g, have a root X in common, that is:
fIoxm + a 1 xm-l + · · · + am = 0,
Doxn + D 1 xn-l + · · · + b n = O.
Multiplying the first equation successively by X n-l, X n-2, . . . ,land the
second by xm-l, xm-2, . . . , 1, we obtain the equalities
d o xm+n-l + d 1 xm+n-2 + · · · · · · · · · + a m xn-l
= 0,
a o xm+n-2 + · · · · · · · · · + a m _ 1 xn-l + a m xn-2 = 0,
doxm + a 1 xm-l +.......... + am = 0,
6oX m + n - 1 + D 1 xm+n-2 + · · · · · · · . . + b n xm-l = 0,
boX m+n-2 +. · · · · · · · · · · · · · · · · · · · + bnX m-2 = 0,
.......................................................... .
DOxn + b 1 xn-l +.......... + b n = o.
Considering the homogeneous linear system with a and b as coefficients
21

22 RESULTANTS
obtained by replacing in these equalities the distinct powers of X by in-
determinates, we see that the system thus obtained has a nontrivial solution,
namely, Xm+n-l, xm+n-2, . . . , 1. Hence the determinant of the system
a o
am
a o
am
R(j, g) ==
b o
a o
am
b n
b o
b n
b o
b n
is equal to O. This determinant is called the resultant ofJ and g, and one has
proved that:
4.1. A necessary condition for! and g to have a common root is that the
resultant be zero.
[Multiply the left-hand sides of the above equations respectively by the
corresponding cofactors of the last column of the above determinant and sum:
by familiar theorems on determinants, the coefficients of xm+n-l, . . . , X
in the sum will be zero and the constant term will be R(J, g). Moreover, the
sum is a linear combination of xn-lJ(x), Xm-1g(X), . . . , g(X). Hence one
obtains:
R(J, g) == A(X)!(X) + E(X)g(X),
and, for indeterminate a, b, X,
4.1.1. R(f, g) == A(X)f(X) + B(X)g(X), where A(X), B(X) are
polynomials of degrees n - 1 and m - 1 respectively.]
4.2. If R(!, g) == 0 and deg! == m or deg g == n, then! == 0 and g = 0
have a common solution.
Proof. j == 0 and g == 0 have a common solution if and only if J and g
have a common factor h of positive degree. Suppose that J == c(X)h,
g == d(X)h with deg h > o. Then d! == cg with deg c < deg! and c =P 0
(also, of course, deg d < deg g and d -=I=- 0). Conversely, if d! == cg with
deg c < deg J and c * 0, then J, g must have a common factor of positive
degree. In fact, we factor both sides completely, cancel the factors from c,
and are still left with a factor of!, which goes into g. Thus!, g have a common
factor if and only if there are polynomials c * 0, d -=I=- 0, deg c < m,
deg d < n such that cJ == dg. Writing
[(doXn-l + · · · + dn_1)(aOXm + · · · + am)
== (coXm-l + · · · + cm_1)(boxn + · · · + Dn)],

RESULTANTS 23
expanding both sides, and comparing coefficients, we get the equations
aod o = boc o ,
a 1 d o + a O d 1 = blc O + bOc l ,
a 2 d o + a 1 d l + a O d 2 = b 2 c O + b 1 c l + b O c 2 , etc.
These are m + n homogeneous linear equations in m + n unknowns c i , die
They have a solution with not all c i = 0, d j = 0 if and only if th&
determinant
a o
a l a o
a 2 a l a o
b o
b 1 b o
b 2 b 1 b o
is not equal to o. Note that if there is a solution with not all d j = 0, then
also not all C i = 0, as cg = df *- O. Interchanging the rows and columns,
which does not change the value of the determinant, we get R(f, g). Thus,
assuming deg f = m or deg g = n, f, g have a common solution if and only
if R(f, g) = O. Q.E.D.
Taken together, 4.1 and 4.2 can be stated as follows:
4.2.1. If a o *- 0 or b o *- 0, then a necessary and sufficient condition for
I and g to have a common zero is that R(/, g) = o.
Exercise. Using the resultant of a polynomial and its derivative, find the
condition on the coefficients for a polynomial of degree 2 to have a multiple root.
Do the same for a polynomial of degree 3.
If one makes the previous polynomials homogeneous, we have:
f* = aoXr + a 1 X r- 1X O + · · · + amX o ,
g* = boX + b1X-lXo + · · . + bnX,
and in this case if a o = b o = 0, then one has the common solution (0, 1) and
thus we have:
4.3. A necessary and sufficient condition for 1* and g* to have a common
solution ( #(0, 0)) is that R(/*, g*) = o.
We mention the following generalization, of which we omit the proof:
4.4. Consider afinite set of polynomial equations in the letters a, b, . . . , f, X
1) {Fi(a, b, . . . , f, X) = O}, a = (a o , . . . , am), b = (b o , . . . , b n ) etc.,
homogeneous in the ai' in the b j , etc. Then there exists a finite set of
polynomial equations 2) {Rj(a, b, . . . ,f) = O} homogeneous in the ai'
in the bi' etc., such that if a, b,...,J, X is a solution of (1) with

24 RESULTANTS
x *- (O, 0, . . . ,0) then ii, b, . . . ,J is a solution of (2), and if ii, b,...,J
is a solution of (2), then there exists an X *- (O, 0, . . . , 0) such that
ii, b, . . . , J, X is a solution of (I).
This statement holds not only for C, but for any algebraically closed field.
And, just as in the case of two polynomials in X = (X o , Xl' X 2 ), one can
construct a (so-called resultant) system {Ria,. . . , f) = O} in a uniform
way depending only on the number of F i , their degrees, and the number of
variables in X = (X o , Xl' . . . , X p ). The construction involves only
operations of addition, subtraction, and multiplication. Similar statements.
hold for systems {F i = O} containing several sets X = (X o , . . . , X p ),
Y = (yo, . . . , Y q), etc., of unknowns. One merely eliminates the {sets of}
unknowns X, Y, etc., successively.
The following two observations will be useful to us later:
1) The resultant R{f, g) is homogeneous in the a i of degree nand
homogeneous in the b j of degree m.
2) The concept of weight. One defines the weight of ai as i and the weight
of b j as j. Every constant *- 0 is given the weight zero; to zero is associated
every weight. The weight of a term ca il · · · ai,' b jl ... bit, c E C, c *- 0, is
by definition i l + · · · + is + jl + · · · + jt, in other words, the sum of
the weights of the factors. The weight of a sum of (unlike) terms different
from zero is the maximum of the weights of the terms. Such a sum is called
isobaric if all the terms have the same weight.
Theorem 4.5. R{f, g) is isobaric of weight mn.
Proof. Let au be the element in the ith row and jth column of R{f, g),
i = 1, . . . , n; it has weight j - i. Let b uv be the element in the (n + u)th
row and vth column of R{f, g), u = 1, . . . , m; it has weight v - u. Any
term in the expansion of the determinant will have weight
n m
 (j - i) +  (v - u),
i=l u=l
where j, v together cover the set 1, 2, . . . , m + n, and hence the term has
weight
(m + n)(m + n + 1)
2
m{m + 1)
2
n{n + 1)
2
whence R{f, g) is isobaric. Considering the main diagonal, one sees that the
weight is mn (or one can simplify the above expression).
Theorem 4.6. If F = 0 and G = 0 are two (projective) curves with
deg F = m and deg G = n respectively that intersect in only a finite
number of points, then they intersect in at least one and in at most mn points.

RESULTANTS 25
Proof. We consider first the case (a) in which at least one of the curves does
not pass through (0, 0, 1), and no pair of intersections of F == 0 and G == 0
are collinear with (0, 0, 1). We write
F(X o , Xv X 2 ) == aoX + al(X O ' XI)X-l + · · · ,
G(Xo, Xv X 2 ) == boX + bl(X O ' XI)X-l + · · ·
The resultant of F and G with respect to X 2 is
R X2 (F, G) == P(a o , aI' . . . , b o , b v · . .).
As at least one of the curves does not pass through (0, 0, 1), a o or b o =1= 0, so
we can apply 4.2. Now
P == H (X O , Xl)'
where H is homogeneous of degree mn, since the degrees of the ai and of the
b j are i andj respectively, that is, equal their weights, and it has been shown
that R is isobaric of weight mn. Moreover, H (X O , Xl) =1= 0, as in the con-
trary case for any pair of numbers Xo, Xl there would exist an X 2 such that
F(X o , Xv X 2 ) == 0 and G(Xo, Xv X 2 ) == 0 so that the two curves would have
infinitely many points in common. Every solution Xo: Xl of H (X O , Xl) == 0
gives rise to an intersection (X o , Xl' X 2 ) of F == 0 and G == 0, and to only
one, since the point (X o , Xv X 2 ) is on the line Xo: Xl == Xo: Xl' which
passes through (0, 0, 1). As H (X O , Xl) == 0 has at least one solution, the
first part of the theorem is proved. On the other hand, H (X o , Xl) == 0
has at most mn solutions, and each solution yields just one point common
to the two curves. Hence the two curves have in common at most mn
points.
The ren1aining case [case (b)] will be treated in the next chapter.
Remark. In forming R(J, g), the polynomials I, g are formally assigned
degrees m and n; the assigned (or formal) degrees may be greater than the
actual degrees. The reason for this is that we think of R(J, g) as arising from
R(f, g), where f, g have indeterminate coefficients, by specialization of these
coefficients; one or both of the leading coefficients may specialize to zero.
It is usually to be understood from context what the assigned degrees of!
and g are.
Remark. In developing the determinantal expression for R(f, g), we have
written the polynomials f, g in descending powers of X. We could just as
well have written them in ascending powers. This would have led to a
determinant ==R(g, f) == (-I)mnR(f, g). Below we shall frequently write
f, g in ascending powers and neglect the (_I)mn sign. As we are invariably
concerned with the divisibility properties of the resultant, this negligence
will not be serious.

26 RESULTANTS
Exercises
1. Find the polynomial f (X ) of degree 4 such that the abscissas of the points of
intersection of X2 + y2 = 1 and of Y2 - 2XY - X2 = 0 are the roots
of f(X).
2. Write the polynomial f (X ) of Exercise I-the polynomial giving the abscissas
of the points of intersection of X2 + Y2 = 1 and of Y2 - 2XY - X 2 = 0
-as a linear combination of X2 + Y2 - 1 and of Y2 - 2XY - X 2
(with polynomial multipHers).
3. Let F(X o , Xl' X 2 ) = 0, G(Xo, Xl' X 2 ) = 0 be two curves having only a finite
number of points in common. Then R X2 (F, G) = 0 if and only if both curves
pass through E 2 : (0, 0, 1).
4. Let f = aoXm + alXm + · · ., g = boxn + b l xn-1 + . . · , where the ai'
b; are indeterminates. Let Ri(f, g) be the resultant obtained by formally
assigning to g the degree n + i (and to f the degree m). Then aoRo, R 1 , R 2 , . . .
all have the same irreducible factors.
5. Find the error in the following proof of 4.2: From 4.1.1, AJ = -By with
deg E < m,degA < n. Ifdegj = m,say,Le.,a i:- O,thenfromAj = -By
one sees that some factor of j must go into g.

Chapter 5
LINEAR TRANSFORMA TIONS
Let A be a 3 X 3 matrix with entries from C and let the determinant of A
be different from zero: IA I =1= O. Consider the transformation
T : x -+ Ax.
Assume that x =1= (0, 0, 0). (Strictly speaking, we should write
() ,
and more generally
(= '
but for typographical reasons we also ,vrite x = (x o , Xv x 2 ) when there is no
likelihood of confusion.) Then Ax =1= 0 as Ax = 0 implies that
A-lAx = A-IO = 0,
whence x = O. The transformation T is defined on the set of triples
(x o , Xv x 2 ), but from it one obtains a transformation of points: namely,
the point P x with representative x is transformed into the point with repre-
sentative Ax. One must prove that the representative px leads to the same
image. This is immediate: px -+ A(px) = pAx, which also represents PAx.
The transformation obtained in this way will be denoted LA. Such a trans-
formation is called a homogeneous nonsingular linear transformation.
The transformation LA is univalent (i.e., it transforms distinct points into
distinct points) and it is onto (Le., every point of the plane is an image of some
point under LA). It transforms straight lines into straight lines.
Associated with T we consider a transformation on the set {F (X)} of
homogeneous polynomials according to the rule:
F(X) -+ F(A-IX) = F'(X).
27

28 LINEAR TRANSFORMATIONS
We agree to call 0 homogeneous of any degree. Then we have the following
properties:
1) If F is homogeneous of degree n, then F' is also homogeneous of
degree n.
2) If F = 0, then F' = o.
3) If F' = 0, then F = o. [One applies (2) to the transformation
x  A-Ix.]
4) The value of F at x is equal to the value of F' at x' = Ax. Hence
the locus of F = 0 is transformed (under LA) into the locus of F' = O.
5) If F is reducible, then also F' is reducible.
6) If F' is reducible, then also F is reducible. [One applies (5) to the
transformation x  A-Ix.] Hence if F is irreducible, then also F' is
irreducible.
7) An equation for a curve r of minimal degree is transformed into an
equation of minimal degree for the transformed curve r'.
8) If F and G are associates, then F' and G' are associates and conversely.
Thus if F and G are not associates, then F' and G' are not associates.
Definition. The degree of an equation of minimal degree for a curve r
is called the order of r .
9) The order of r is equal to the order of r'.
The assertion (9) can be expressed by saying that the order of r is a
projective invariant. The second part of (4) is expressed by saying that the
concept of an algebraic curve is a projectively invariantive concept. According
to (6), the irreducibility of an algebraic curve is a projectively invariantive
concept.
5.1 (Fundamental theorem of linear transformations). Let Ao, A), A 2 , As
be four points no three of which are collinear,. let Bo, Bv B 2 , Bs be another
four points no three of which are collinear. Then there exists one and only
one homogeneous nonsingular linear transformation sending Ai into
B i , i = 0, 1, 2, 3. (We omit the proof.)
The rule LA = L pA is immediate. The converse is also true: if LA = L B ,
then B = pA for some p.
Exercise. Let
Ao: G) , AI: (i) , AI: G) ,
Bo:G)' BI:G)' BI:(i),
A3 : G) ,
B3: G) ·
No three of the Ai and no three of the Bj are collinear. Find a matrix A such
that LA(A i ) = B i , i = 0, 1, 2, 3.

COORDINATE SYSTEMS 29
COORDINATE SYSTEMS
Let Ao, Av A 2 , Aa be four points no three of which are collinear. Let LA
be such that E i is sent into Ai' i == 0, 1, 2, 3 under LA. Here Eo: (1,0, 0),
El : (0, 1, 0), E 2 : (0, 0, 1), Ea: (1, 1, 1). Given a point Px we assign new
coordinates to it according to the rule: new coordinates of P == coordinates
of LA-1(P). That is, designating the original coordinates of P as x and
its new coordinates as x', one obtains
x' == A -IX.
One will say that Ao, AI' A 2 , Aa is a system of coordinates: Ao, Av A 2 are
called the first, second, and third vertices, respectively, and Aa is called the
unit point.
Let a curve r be given in the old (or original) coordinate system by
F (X) == O. Then one sees that in the new coordinate system r is given by
F(AX) == O.
The algebraic formulas that we use in considering transformations and
in considering changes of coordinates are the same, but the point of view is
different. In the latter case, we speak of an alias transformation and in the
fornler, of an alibi transformation. The basic facts are the same in either
case, but sometimes it is convenient to adopt one point of view, and some-
times the other.
In particular, to study r in coordinate system AoAIA2Aa is the same as
to study LA-1(r) in the coordinate system EoElE2Ea. In more technical
terms, this assertion can be formulated as follows:
The equation of r in the coordinate system AoAIA2Aa is also the equation
of L A-l(r) in the original coordinate system.
The preceding are preliminaries to case (b) of the last theorem of Chapter 4.
We proceed as follows.
The idea is to transform r,  by means of a transformation LA into
curves r A' A that will be in permissible position, i.e., into curves for which
the hypothesis of case (a) holds (see the first sentence of the proof of Theorem
4.6). The intersections Pv . . . , Ps of r and  are transformed into the
intersections Pt, . . . , P1 of r A and A. The order is a projective invari-
ant; thus ord r A == m, ord A == n. Hence applying case (a) to r A and
A' we obtain 1 < s < mn. It is, then, only a matter of establishing the
existence of an adequate transformation LA.
Here we change point of view and we look for a coordinate system in
which r,  are in permissible position. Let l,C' k == 1, . . . , s(s + 1 )/2 be
the lines PiP j , i, j == 1, . . . , s, i =1= j. Let A 2 be a point outside of
r u  U li U . . · U ls(s + 1)/2. Let Ao, Al' Aa be three points such that
no three of Ao, Av A 2 , Aa are collinear. Then r and  are in permissible
position relative to the coordinate system AoAIA2Aa. This completes the
proof.

30 LINEAR TRANSFORMATIONS
TRANSFORMATIONS IN THE AFFINE PLANE
As we frequently pass back and forth between the projective and the affine
planes, it will be well to have the equations of a homogeneous nonsingular
linear transformation
LA : (x o , Xv x 2 )  (x, x, x),
X =  a..x.
  l1 "
in affine terms. We take Xo = 0 as line at infinity and confine our attention
to points Px such that Px and its transform Px' are at finite distance. Then
P x has the affine coordinates (YI' Y2) = (xl/X O ' X2/XO) and P x' has the affine
coordinates (Y, Y) = (x/x, x/x). Thus I
Yi
aioXO + ail Xl + a i2 x 2
aooxo + aOlx l + a 02 x 2
aiO + ailYI + a i2 Y2
,
aoo + aOIYI + a i2 Y2
i = 1, 2. As we have already said, we are not considering points Px on the
line aooXo + aOIX I + a 02 X 2 = O. If a 01 or a 02 =1= 0, then this line has the
affine equation aoo + a OI Y I + a 02 Y 2 = O. Otherwise it is the line at infinity,
and the transformation takes the simpler form:
Y = b lO + bllYI + b 12 Y2,
Y = b 20 + b 2l Yl + b 22 Y2'
where b ii = aii/aOO. Here
1
b lO
b 20
o
b ll
b 21
o
b l2
b 22
=--t-O
3 -r- ,
aOO
whence b II b 22 - b 2I b I2 =1= O. Hence one can solve for (YI' Y2) in terms of
(y, Y); the transformation is onto the affine plane. These transformations
are called (nonsingular) affine linear transformations. They are the trans-
formations induced by the nonsingular homogeneous linear transformations
of the associated projective plane that leave the line at infinity invariant.
Exercise. Let Bo, B 1 , B 2 be three points on the irreducible conic aooX + . . . = o.
By a Hnear transformation taking Bo, B 1 , B 2 into Eo, E 1 , E 2 , the equation of the
transformed conic takes the form C 2 X O X 1 + C O X 1 X 2 + C 1 X 2 X O = 0 with
COC1C2 i:- o. The linear transformation xi = XdCi (Le., x  x') transforms this
conic into XOX1 + X 1 X 2 + X 2 X O = o. Hence any irreducible conic can be sent
into any other by a linear transformation taking any three given points of the
first into any three given points of the second and, moreover, this can be done
in just one way.

Chapter 6
SIMPLE POINTS AND SINGULAR POINTS
Consider the intersection of a curve r having F (X o , Xv X 2 ) == 0 as equation
of minimal degree with a straight line L. Let Xo == (xg, x, xg), Xl == (X5, xL x)
be two points of L. An arbitrary point of L will then be of the form
AoXo + AIX I . This will lie on r if and only if F(AoXO + A 1 X I ) == O. Thus to
find the intersections ofr and L one solves the equation F (AoxO + A 1 x l ) == o.
It is possible that F(AoxO + Alx l ) is the element 0 in C[A o ' AI]. This
will be so if and only if every point of L is on r. Let us consider the case that
L is not contained in r.
Let
k
F (A XO + A Xl )  - H ( A A ) - c II ( A ,(i) - A '(i» ) 8i
° 1 - 0' I - 011.1 111.0'
i=1
with Ai): Ah i ) *- Aj): A) for i *- j, c E C, c *- O.
Definition. The number Si is called the intersection multiplicity of r
and L at the point Pi determined by Ah i ): Ai), Le., the point A)XO + Ai)XI.
A similar definition can be given for a cycle F == O.
We shall show later that the intersection multiplicity of rand L at Pi
is a projective invariant and does not depend on the choice of XO and Xl
[but only on L (and r)].
We suppose now that the point XO is taken to be on the curve
r. Then XO == 1 · XO + 0 · Xl and corresponds to the ratio Al : Ao == 0: 1, and
hence to the factor A o . 0 - Al · 1 == -AI of F(AoxO + Alx l ). Let us
place Ao == 1, Al == A. (This is similar to the passage from the projective
plane to the affine plane explained in Chapter 3.) Then we have:
6.1. The intersection multiplicity of L with r at XO is equal to the maximum
power of A that can be factored from F (XO + AXI).
Moreover we have
F(xg + AX5, x + AxL xg + Ax) == F(xO) + A(H o (xO)x5 + H1(XO)X
+ H2(XO)X) + A2(. · .) + .. ·
31

32 SIMPLE POINTS AND SINGULAR POINTS
Two cases may present themselves:
a) (Ho(xO), HI(xO), H 2 (xO)) == (0,0,0),
b) the remaining case.
In case (a) every line L through XO cuts r at XO with multiplicity > 2.
In case (b) L intersects r at XO with multiplicity> 1 if and only if
Ho(xO)x + HI(XO)X + H2(XO)X == 0, Le., if and only if Xl is on the line
Ho(xO)X o + HI(xO)X I + H 2 (xO)X 2 == O.
In case (a) one says that XO is a singular point of r and in case (b) that
it is simple. The line mentioned in case (b) is called the tangent to r at xo.
One will see below that these concepts are projective invariants.
Exercises
1. Find the intersections of 25X - X - X = 0 and the line joining
Po: (1,7,1), PI: (1, -1,7).
2. If F has two (not not necessarily distinct) factors that vanish at x O , then the
intersection multipHcity of the cycle F = 0 with any Hne through X O is
greater than 1.
3. Let r be the cycle having X + Xf + X = 0 as equation, and let
(X o , Xl' x 2 ) be a (Le., any) point on it. Show that (x o ' Xl' X 2 ) is simple on r.
Hence the cycle r is an irreducible curve. (See the remark at the end of
Chapter 1.) Write the equation of the tangent to r at (X o , Xl' x 2 ).
4. Do Exercise 3 for the cycle X + X + X + XOX1 = o.
We now study the nature of a curve r at (or, as one also says, near) a
point P of the curve. Since any point is in an affine plane (for some choice
of the line at infinity), we make an affine study: this has at least the advan-
tage that there is one less variable to deal with.
E2
Let F(X o , Xv X 2 ) == 0 be the given curve and Mo the point we wish to
study; here F(X) == 0 is an equation of minimal degree for r. We may
suppose that Mo == Eo == (1,0,0), that is, we take Mo as first vertex of a
coordinate system (see Fig. 6.1). Let EI == (0, 1, 0), E 2 == (0, 0, 1) and
A == (0, 1, m). Let L be the line EoA: its equation is X 2 - mX 1 == O.
Passing to affine coordinates, the affine part of the curve r is given by the
equation F(I, X, Y) == 0; we write F(I, X, Y) == F(X, Y). The affine

SIMPLE POINTS AND SINGULAR POINTS 33
equation of L in Y - mX = O. From 6.1, the intersection multiplicity of
EoA with r at Eo is the maximum power of A that can be factored from
F(1 + A · 0, 0 + A · 1,0 + A · m) = F(I, A, Am). Thus:
The intersection multiplicity of rand L at (0, 0) is the maximum power
of X that can be factored from F(X, mX).
Notation. The intersection multiplicity of rand L at P is denoted by
i(r, L; P).
Exercise. Let r,  be curves without common component meeting at a point
P and let L be a line through P. Show that
i(r u , L; P) = i(r, L; P) + i(, L; P).
We write
F(X, Y) = a oo -1- a10X + a01Y + a 20 X2 + a11XY + a 02 Y2 + · · ·
The point (0, 0) is on r if and only if a OO = o. We suppose this to be the case.
Corresponding to the alternatives (a) and (b) above, we assert:
6.1.1. a) If a 10 = a Ol = 0, then every line through Eo cuts r at Eo with
multiplicity > 2.
b) If a 10 or a Ol is not zero, then every line through Eo with just one exception
cuts r at Eo with multiplicity 1. This exception, the tangent (by the defini-
tion already given), has as equation: a10X + a01Y = o.
In fact, substituting Y = mX in F(X, Y), one obtains in case (a)
F(X, mX) = a 20 X2 + a 11 mX2 + a 02 m 2 X2 + · · · ,
Le., at least X2 can be factored out. This takes care of every line through
(0, 0) except the vertical line X = o. The result for this line follows by
interchanging the roles of X and Y (i.e., by what has already been proved).
In case (b), if, say, a Ol -=F 0, then one obtains
F(X, mX) = (a 10 + 1na 01 )X + X2(. · .)
and X but no higher power can be factored out, except in the case that
a 10 + ma Ol = 0, i.e., if and only if m = -a10/a 01 ' This takes care of every
line except the vertical line X = 0; here the intersection multiplicity is
given by the maximum power of Y that can be factored out of F (0, Y), which
one sees is 1. This gives Y = -(a10/a01)X or a10X + a01Y = 0 as the
tangent. If a Ol = 0, then a 10 -=F 0, and the desired result follows by an inter-
change of the roles of X and Y.
In case (b), i.e., when the point is simple, one can say, speaking without
precision, that the tangent approximates the curve near the given point.
We may write our polynomial F(X, Y) as
F(X, Y) = Fr(X, Y) + F r + 1 (X, Y) + · · · ,
Fr(X, Y) =1= 0,

34 SIMPLE POINTS AND SINGULAR POINTS
with Fi(X, Y) homogeneous of degree i and with r > 1. The point is simple
if and only if r = 1. For r > lone has
8
F ( X Y ) - c II ( m(i)X - m(i)Y ) 8 i
r' - 0 l'
i=l
where CEO and m):mi) -=F m):mj) for i -=F j. In this case, i.e., when the
point (0,0) is singular, every line through this point cuts r at the point with
multiplicity r except for the lines
m) X - mi) Y = 0 (i = 1, 2, . . . , s),
which are called the tangents to r at (0, 0). In this case one says that (0, 0)
is of multiplicity exactly r. One says that the point is ordinary if every Si = 1.
Examples
1) The curve Y2 - X2 - xa = 0 passes through (0, 0). The poly-
nomial Y 2 - X 2 - X a, which determines it, is irreducible. In this case
r = 2 and every line through (0,0) intersects the curve at (0, 0) with multi-
plicity 2 except for the tangents, Le., the lines given by the factors of
Y2 - X2 = 0, or Y - X = 0 and Y + X = o. This is shown in Fig. 6.2.
This type of singular point is called ordinary double point. The curves
Y2 - X2 + xa = 0 and Y2 + X2 + xa = 0 also have at the origin
ordinary double points.
"-
"-
"-
"-
"-
"-
"-

Fig. 6.2
2) Let r have at the origin a nonordinary double point. Then at the
origin there is just one tangent to r; let this be the X-axis. Let F = 0 be
the minimal equation for r. Then
F = Y2 + aaoxa + a 21 X2Y + a 12 XY2 + aoaya + · · ·
The origin is called an ordinary cusp if the tangent cuts the curve there with
multiplicity 3. Necessary and sufficient for this is that aao -=F o. Thus
Y2 = X a has a cusp at (0, 0). The appearance is illustrated in Fig. 6.3.
Not all nonordinary double points are ordinary cusps. For example, the
origin is not an ordinary cusp of Y2 - X4 - X5 = O. The appearance
here is shown in Fig. 6.4.

SIMPLE POINTS AND SINGULAR POINTS 35
Fig. 6.3 Fig. 6.4
Remark. The above does not yet say what is essential about a cusp. Later
we shall define branch, and a cusp will be defined as a double point at which
there is only one branch. If the double point is a cusp at the origin, then
(in an appropriate coordinate system) for points near the origin, y is given
by a power series in X 1 / 2 ; whereas if it is not a cusp, then the points near the
origin are given by two power series in x, Le., y = Pi (x), i = 1, 2.
Theorem 6.2. If r and  are two irreducible curves, then every common
point of r and  is a singular point of the curve r u .
Proof. We may suppose that the common point is the origin. If F = 0 and
G = 0 are the (minimal) equations for r and , then F · G = 0 is the
minimal equation for r u  (we are supposing that r =1= ). Since subdeg
FG = subdeg F + subdeg G > 2, the origin is singular for FG = O. Q.E.D.
Remark. One can define simple and singular points for cycles. The theorem
corresponding to Theorem 6.2 would then say: If r and  are cycles and P
is a point common to r and , then P is singular for the cycle r + .
Corollary 6.3. Every point of a curve that belongs to two distinct com-
ponents of it is singular.
Let F(X, Y) E C[X, Y], (a, b) a point. Then
F(X, Y) = ! CijXiYi = ! cil a + (X - a)]i [b + (Y - b)]i
= ! cijaib i + ! ciiiai-1bi(X - a) + ! ciijaib i - 1 ( Y - b)
+ terms of degree > 2 in X - a, Y - b
of of
= F(a, b)+ - (X - a) + - (Y - b)
ax (a,b) a Y (a,b)
+ higher-degree terms.
Here
of
ax (a,b)
is the partial derivative of F(X, Y) with respect to X evaluated at (a, b).

36 SIMPLE POINTS AND SINGULAR POINTS
(In writing of loX, we do not have to enter into the notions of the calculus,
but merely work formally, using the familiar symbols and rules.)
If (a, b) is on the curve, then F(a, b) = 0 (and conversely). Then writing
X' = X - a, Y' = Y - b, the equation
of X' + oF I Y' + . . . = 0
ax (a,b) a YI(a,b)
is simply the equation of F = 0 in a new coordinate system having (a, b)
as origin . Thus:
Theorem 6.4. The singular points of F(X, Y) = 0 are the common
points of F(X, Y) = 0, of loX = 0, and of laY = o. At a simple
point (a, b), the tangent is given by
of of
(X - a) + - (Y - b) = o.
ax (a,b) a Y (a,b)
Corollary 6.5. A curve r has at most a finite number of singularities.
Proof. First suppose that r : F = 0 is irreducible. F involves at least one-
of the letters X, Y, say Y. Then of laY =1= 0, hence deg (oFloY) < deg F.
Hence of laY cannot be divisible by F, so by 3.14 (in nonhomogeneous
form), of /0 Y can vanish only at a finite number of points of r, which must
include all the singularities of r (at finite distance-and at infinity r only
has a finite number of points). If r is reducible, the corollary still holds, as:
the singularities of r are the singularities of its individual components plus
points of intersection of pairs of components.
Corollary 6.6. Let r: F(X, Y) = 0 be a curve and P: (a, b) a point
thereon. Then
=0
a Y (a,b)
if and only if either P is singular for r or it is simple and the tangent to r
at P is vertical.
of
Proof. If P is singular, then
=0
a Y (a,b)
by the theorem. If P is simple, then the tangent at P is
of of
(X - a) + - (Y - b) = O.
ax (a,b) a Y (a,b)
This line is vertical if and only if
of
of
= O.
a Y (a,b)

SIMPLE POINTS AND SINGULAR POINTS 37
Corollary 6.7. Let r be a curve of order n, L a line not a component of r.
Then L cuts r in at most n points, and this number is reached for some L.
Proof. We already have the first assertion (see 3.1). For the second, take
the coordinate system so that E 2 : (0, 0, 1), the point at infinity on the Y-axis,
does not lie on r. Through E 2 take a line not passing through any singularity
of r and not tangent to r: to see that this can be done, consider first the
case that r: F(X, Y) = 0 is irreducible. Then F(X, Y) involves Y:
otherwise F(X, Y) would be a polynomial G(X) in X alone, in fact of the
first degree, since G(X) is irreducible; F (X, Y) = 0 would be a vertical line,
therefore a curve through E 2 , and this is not so. Hence of laY =1= o. Now
of laY =1= O(F), as deg (oFjoY) < deg F. Hence of laY = 0 at only a
finite number of points PI' . . . , Ps of r. Any other point of r at finite
distance is simple with nonvertical tangent. Take, then, a line through E 2
not through P v . . . , P sand =1= L 00 . Such a line will satisfy the require-
ments. If r = r 1 u · · · u r s with the r i irreducible, one takes a line
through E 2 not tangent to any r i, through none of the singularities of the
r i , and through no intersection of a r i and a r;, i =1= j. Such a line L will
meet r in n points each counted simply, hence in n distinct points.
Exercises
1. Show that the polynomials Ho, HI' H 2 on p. 31 are 8F/8X o , 8F/8Xl' of/oX 2
respectively. Hence in homogeneous terms, the singularities of
F(X o , Xl' X 2 ) = 0
are given by the simultaneous solutions of F = 0, 8F/8Xo = 0, 8F/8X l = 0,
8F/8X 2 = o.
2. Let F(X o , Xl' X 2 ) be homogeneous of degree n. Show that
Xo 8F/ 8X o + Xl 8F/8X 1 + X 2 8F/8X 2 = nF.
(This is known as Euler's Formula. ) Use this identity and the affine criterion
for singularities (Theorem 6.4) to get the projective criterion given in the
last exercise.
3. Let F(X o , Xl' X 2 ) be homogeneous of degree 2. Then F can be written in
the form
(X o , Xl' X 2 )
( a o o aOl a 02 ) ( Xo )
a 10 all au Xl
a 20 au a 2 2 X 2
with aii = aji (all i, j). Show that F(X) = 0 has a singularity, hence is
reducible, if and only if det (aii) = o.
4. Find the singularities (in the complex projective plane) of
a) X3 + Y3 = 3XY (Folium of Descartes),
b) (X2 + y2)(X - 1)2 = X2 (Conchoid of Nicomedes),
c) (X2 + Y2)3 = 4X 2 Y2 (Four-leafed rosette),
and sketch the curves (in the real projective plane).

38 SIMPLE POINTS AND SINGULAR POINTS
5. Sketch the curve Y 2 + X 2 + X3 = 0 in the real affine plane, especially
near the origin.
6. Let Xo, Xl' X 2 be indeterminates over C and let x 2 = X/Xo.
a) Show that if f(Xo, Xl' X 2 ) E C[X o , Xl' X 2 ] vanishes (X o , Xl' x 2 ), then
f == O(X - X O X 2 ) in C[X o , Xl' X 2 ].
b) Show that C[X o , Xl' x 2 ] (= set of elements in the field C(X o ' Xl) that
can be written as polynomials in Xo, Xl' x 2 with coefficients in C) is not a
unique factorizat.ion domain.
THE POLAR
Definition 6.8. Let r: F(X o , Xl' X 2 ) = 0 be a curve (or cycle) and
let P : (a o , aI' a 2 ) be a point. If
of of of
a- + a- + a--I-O
o ax 1 oX 2 ax -r- ,
012
then by the polar of P relative to r one means the cycle
of of of
a o X + a l - X + a 2 X = o.
a 0 a 1 a 2
Theorem 6.9. The notion of a polar is projective. In more detail, if
X' = AX,whereAisa3 X 3nonsingularmatrixandF'(X') = F(A-IX'),
then
, of' of
! a 1 oX = ! ai ax. '
 t
where a' = Aa.
Proof. Let deg F = n. By a computation like the one leading to 6.4 (or
by the directly ensuing computational rules)
of
F(X + a) = F(X) + ! a i ax. + terms of degree < n - 2
t
and
of'
F'(X' + a') = F'(X') + ! ai oX'. + terms of degree < n - 2 .
t
Since F'(X' + a') = F(X + a), the terms of degree n - 1 (in X or in X')
on the two sides must be the same, i.e., ! ai(oF loX i ) = ! ai(oF'loXi).
Q.E.D.
Let us take a coordinate system in such way that P gets the coordinates
(0,0, 1), Le., P = E 2 . Then the polar cycle, if defined, is of loX 2 = O. The
polar cycle will not be defined if and only if F does not involve X 2 , or, expressed
invariantively, if and only if F = 0 consists of a number of lines through P.

SOME COMPUTATIONAL RULES 39
From
of(X o , Xv X 2 ) of(l, X, Y)
-
oX 2 X=(l,X,Y) 0 Y
one sees that the affine part of the polar of G(X, Y) = 0 with respect to
E 2: (0, 0, 1) is oG /0 Y = o.
Exercises
1. If P is on r, then the polar of P (if defined) passes through P.
2. If P is simple on r, and  is the polar of P, then  has P as simple point and
r and  have the same tangent at P.
Theorem 6.10. The polar of P with respect to r (assuming it is defined)
passes through the singular points of r, through the points of contact of the
tangents from P to r, and through no other point of r.
Proof. We may suppose that P is at E 2 . Let Q be any other point of
F(X o , Xt, X 2 ) = O. We may suppose that Q is at finite distance: let
Q = (a, b). Then Q lies on the polar if and only if
= o.
( a,b )
By 6.6, then, Q is either singular or is a point of contact of a tangent from P.
This takes care of every point Q of r except P. If P is not on r, there is
nothing to prove; and if P: (a o , at, a 2 ) is on r, then P is on the polar
I ai of /oX i = 0 by Euler's Formula, which says that I Xi of /oX i = nF.
The proof is complete.
of(I, X, Y)
oY
SOME COMPUTATIONAL RULES
We summarize some rules familiar from calculus. Throughout we deal
only with polynomials; as a consequence these rules are really merely com-
putational rules. To emphasize this, we start from the following definition:
Definition. Let f(X) E O[X] and let h be a new indeterminate. Then
f(X + h) E O[X, h]:
f(X + h) = fo(X) + fl(X)h + f2(X)h 2 + · · · , fi(X) E O[X],
and by definition the derivative of f(X) is f'(X) = fl(X), Le., the
coefficient of h in f(X + h).
Placing h = 0, note that f(X + 0) = fo(X), so that we write:
f(X + h) = f(X) + f'(X)h + h2(. · .).
a) (f + g)' = f' + g', b) (fg)' = f'g + fg',
c) c' = 0, for CEO, d) X' = 1.

40 SIMPLE POINTS AND SINGULAR POINTS
Proof. Let us prove (b); the others are proved similarly:
f(X + h) = f(X) + f'(X)h + h2(. . .),
g(X + h) = g(X) + g'(X)h + h2(. · .),
hence
f(X + h)g(X + h) = f(X)g(X) + [f'(X)g(X) + f(X)g'(X)]h + h2(. · .),
and the coefficient of h is the familiar expression. Q.E.D.
For
e) (Xn)' = nXn-l
the proof is by induction, using (b). Hence
f) (! CiXi)' = ! iCiXi-l.
Now for the function of a function rule:
g) f(g(X))' = f'(g(X)) · g'(X).
Proof
f(X + h) = f(X) + f'(X)h + h2(. · .).
Therefore,
f(u + v) = f(u) + f'(u)v + v 2 (. · .) (u, V E C[X]),
g(X + h) = g(X) + g'(X)h + h2(. · .)
= g(X) + v = u + v,
f(g(X + h)) = f(g(X)) + f'(g(X))[g'(X)h + h2(. . .)]
+ [g'(X)h + h2(. . .)]2(. · .)
= j(g(X)) + f'(g(X))g'(X)h + h2(. · .).
If instead of C we have any other coefficient domain, say C[ Y], where Y
is a second indeterminate, then the above rules and proofs continue to hold.
In the stated case, the derivative is denoted a/ox. Similarly, looking at
C[X, Y] in the form C[X][Y], one defines a/ay.
a of 0 2 F
- - is denoted
ax ay axay ·
Since
0 2 a
(  C..XiYi ) = - (  J .C_.Xiyi-l ) =  i1c..Xi-lyi-l
axa y £., U oX £., 'l1 k 'J 'l1 ,

SOME COMPUTATIONAL R1JLES 41
and similarly for 8 2 /ayaX, we have:
a 2 a 2
h) -
8XaY - ayaX '
in other words, the operators a/ax and a/ay commute.
Now for Taylor's Theorem:
h 2 h 3
i) f(X + h) == f(X) + f'(X)h + f"(X) - + f"'(X) - + · · ·
2! 3!
Proof. f(X + h) == f(X) + f'(X)h + f2(X)h 2 + h 3 (. · .). Applying the func-
tion of a function rule on the variable h, we get
f'(X + h) · 1 == f'(X) + 2f2(X)h + h2(. · .),
whence from the definition
f"(X) == (f'(X))' == 2f2(X),
showing that Taylor's Theorem is correct so far as the coefficient of h 2 is
concerned. Now we make an induction, assuming that
h 2 h i - 1
f(X + h) == f(X) + f'(X)h + f"(X) - + · · · + f(1:-1)(X) .
2! (-1)!
+ fi(X)h i + hi+l(. · .) ,
for some i and every polynomialf(X). Applying the function of a function
rule, we have
hi-2
f'(X + h) · 1 == f'(X) + f"(X)h + · · · + f(i-l)(X) .
( - 2) !
+ i fi(X )hi-l + hi(. · .),
and applying the induction assumption tof'(X), we find
i fi(X) == (f'(X))(i-l)/(i - I)!,
and the induction is complete.
Writing Df for f', D2f for f", etc., we can write Taylor's Theorem
symbolically in the form
h 2 D2
f(X + h) == (1 + hD + - + · · .) f(X) .
2!
Recalling from calculus that for any number x
x 2
e == 1 + x + 2! + · · · ,

42 SIMPLE POINTS AND SINGULAR POINTS
we also write the above (more symbolically) in the form
(*)
f(X + h) = ehDf(X).
For several variables, let X abbreviate (Xl' . . . , X n ) and h abbreviate
(h l , . . . , h n ), where the Xi and h j are 2n indeterminates or letters, and let
f(X) = f(Xv . . . , X n ) by a polynomial in C[Xl' . . . , Xn]. Let
Dl = a/ax l , . . · , Dn = a/ax n .
Then
f(X I + hv X 2 + h 2 , . . . , X n + h n )
= eh1Dtj(Xl' X 2 + h 2 , Xs + hs, . . . , X n + h n )
= e h1D1 ( e h2D2 (f(Xl' X 2 , Xs + hs, . . . , X n + h n ))
and, since (hI DI)i( (h 2 D2)if) = (hi hDi D)f, this
= (e h1D1 · e h2D2 )f(Xl' X 2 , Xs + hs, . . . , X n + h n ),
where
h 2 D 2
e lt1D1 = 1 + h D +  + · · ·
I I 2 !
and
h 2 D2
e h2D2 = 1 + h D +  + · · ·
2 2 2 !
are to be multiplied out according to the usual rules for working with numbers.
Now one knows that
eX · eV = e X + v ,
when x and yare numbers, and one checks directly that also
eh1D1 . eh2D2 = e h1D1 +h 2 D 2 .
Continuing, then, our previous computation, we find
f(X l + hv . . . , X n + h n ) = e h1D1 + .. . +hnDnf(Xl' . . . , X n ).
An abbreviation for hlDI + · · · + hnDn is hD, so we obtain Taylor's
Theorem for several variables in the form
(**)
f(X + h) = ehDf(X).
Exercise. Let r : F(X) = 0 be a cycle with deg F = n and let a be a point.
Then F(X + a) = F(X) + F 1 (X, a) + F 2 (X, a) + . . . + F n _ 1 (X, a) + F(a),
where F i is homogeneous of degree n - i in X (and homogeneous of degree
i in a). If F i, F i+1 define cycles (Le., are not = 0), then F i+1 = 0 is the polar of
F i = 0 with respect to a.

Chapter 7
BEZOUT'S THEOREM
In Chapter 5 we said that we would prove the following results:
Theorem 7.1. The intersection multiplicity of r and L at Pi (a point
of intersection of rand L) is independent of the points x o , Xl used to
determine L.
Theorem 7.2. The intersection multiplicity of r and L at Pi is a pro-
jective invariant.
However we still postpone the proof for reasons that will shortly become
clear (see the remark at the end of the chapter).
Now suppose we have two arbitrary curves (or cycles) instead of a curve
and a line. Let these be
r : F(X o , Xv X 2 ) = 0
and
 : G(Xo, Xv X 2 ) = 0
of orders m and n and without common components. Let PI' . . . , Ps be
their points of intersection.
Problem. To define the intersection multiplicity
i(r,; Pi)
of r and  at P; in such a way that
8
! i(r,; Pi) = mn.
j=l
Suppose first that r and  are in permissible position with respect to
the coordinate system Eo, E I , E 2 , E 3 ; that is, at least one of the curves does
not pass through E 2 and no two points of intersection of r and  are collinear
w1:th E 2 . Let Pi = (XiO' XiI' X i2 ). Then
R X2 (F(X), G(X)) = cII(XIXio - XoXil)8 i ,
43
CEO, C # o.

44 BEZOUT'S THEOREM
We can then propose:
i(r,; Pj) = Sj.
Observe that deg x2 F < deg F, deg x2 G < deg G and that equality
holds at least once, since r,  are in permissible position. In forming
Rx (F, G) we agree to give to F and G formally the degrees m and n in X 2 .
2
If r and  are not in permissible position, then one considers a linear
transformation LA such that if LA : r  r A'   A' and Pj  PI, then
r A and A are in permissible position. Then one can define
i(r,; Pj) = i(r A , A; P1).
It is clear that for this definition to be correct we shall have to prove that if
L B is a linear transformation sending r and  into curves r B , B in per-
missible position, then
(*) i(r A , A; PI) = i(r B , B; Pf).
Once this is proved, we shall have:
Theorem 7.3 (Bezout's Theorem). Two curves of orders m and n without
common components meet in mn points, counting their multiplicities.
By a complete set of intersections of r and  one means the system
PI' P 2' . . . , P mn of intersections, taken in any order, where a point P of
multiplicity r occurs r times in the system.
The equality (*) amounts to saying that the proposed definition of
intersection multiplicity is a projective invariant. This in turn could be said
to be the main point of Bezout's Theorem.
To prove the equality we will introduce a general linear transformation
Lv. This is determined by the matrix
u = ( ::: :: ::: ) ,
u 20 U 21 U 22
where u oo , . . . , U 22 are nine indeterminates over O. In order to prove the
desired equality (*) we will prove:
( * * ) i( r A' A; Pf) = i( r u' u; Pj).
Now we will work over the field 0 1 = O(u oo , . . . , U 22 ) = algebraic closure
of C(u oo ,..., U 22 ), Le., 0 1 is a]gebraic over O(u oo ,..., U 22 ) and C 1 is
algebraically closed. t All that has been proved up to now for 0 is valid for
0 1 . We will designate by PNO the projective number plane over 0 and
similarly PNC 1 . One has that PNO c PN0 1 . Let us suppose that F = 0
is a curve in PNO. In PNC 1 , F = 0 defines a curve that contains the points
of F = 0 in PNO. To continue the proof of (**), we first prove the following
theorem.

BEZOUT'S THEOREM 45
Theorem 7.4. Let rand Ll be curves in PNO without common com-
ponents and let r, Ll be given by F = 0, G = 0, F, G E O[X o , Xl' X 2 ].
Then the intersections of F = 0 and G = 0 in PNO I are the same as the
intersections of F = 0 and G = 0 in PNO.
We may take the curves r, Ll in permissible position with respect to
EoElE2Ea and suppose moreover that no intersection is on EIE2 (the line
at infinity). Using the resultant R X2 (F, G), we see that the points of inter-
section of F = 0 and G = 0, whether regarded in PNO or in PNO I , are
on the complex lines through E 2 given by the factorization over 0 of
R X2 (F, G) = H(X o , Xl) into linear factors. A similar statement is true for
R X1 (F, G). Since the intersections of these two finite sets of lines are points
with complex coordinates, the theorem is proved.
We return to the linear transformation Lu. Let ru be the curve
F (U -IX) = 0 and Llu the curve G( U -IX) = O. Let PI' be the transform
of Pi. We now assert that r u and Llu are in permissible position. In fact,
if (0, 0, 1) satisfied F (U -IX) = 0, then, specializing U to A (Le., substituting
au for uii)' we would obtain that (0, 0, 1) is on F(A-lX) = 0 for every A,
which is surely not so. Analogously, one shows that PI' PI' does not pass
through E 2 .
Since the PI' are the points of intersection of r u and Llu, we have
Rx (F(U-lX), G(U-lX)) = c'IT[Xl(UOoXiO + UOIXil + UOi2)
2 t
- XO(UloXiO + UIIXil + UIi2)] i
with c' in 0 1 . Observe first that c' E O(u oo , . . . , U 22 ): one sees this by
comparing the coefficient of any power product of Xo, Xl occurring on the
right with the coefficient of the same power product on the left. We write
then c' = c(U)/d(U), where c, d are polynomials having no common factors
(other than units, of course). Next observe that each factor of the product IT
is irreducible in O[ Xo, Xl' u oo ,..., U 22 ]; for since it is homogeneous of
degree 1 in Xo, Xl and homogeneous of degree 1 in u oo , . . . , U 22 ' it could,
at worst, factor into a polynomial in Xo, Xl alone times a polynomial in
u oo , . . . , U 22 alone; but then the coefficients of Xo and of Xl would differ
only by a factor in 0, in particular, would involve the same Uij, and this is
not so. Finally, observe that on the left only powers of the determinant
I U I occur in the denominator (since this is true for the entries of U-l).
Cross multiplying and using the unique factorization theorem, one sees that
d(u) is a factor of f U fe, for some e. Hence c' can be written as cl(U)/1 U fe,
C l a polynomial.
Now we specialize U  A. Observe that the operations of forming the
resultant and of specializing taken in either order give the same result.
(This observation depends on our agreement to give F and G formal degrees
in X 2 equal, respectively, to their total degrees in Xo, Xl' X 2 . See p. 44.)

46 BEZOUT'S THEOREM
Then we get
Rx.(F(A-IX), G(A-IX)) = CI! II[XIX -XoXiJi,
where (xto, xii, x) are the coordinates of Pf. Here xlxto - Xrii and
xlXto - XoXA are nonassociates for i =1= j because r A' Ll A are in permissible
position. Hence
i(r u, Llu; Pf) = i(r A' Ll A ; Pf),
and the proof is complete.
The main object of the preceding proof was to establish the projective
invariance of the intersection multiplicity, but it shows quite generally that
on specializing U to A, the i(r u, Llu; PI') factors of R X2 ( F (U-lX), G(U-IX))
corresponding to PI' go over into that number of factors of R x2 (F(A-IX),
G(A -IX)) corresponding to Pt; the curves r A' Ll A need not be in permissible
position, though E 2 is not to lie on both of them. Hence we get:
Theorem 7.5. Assume that E 2 : (0, 0, 1) does not lie on both r : F = 0
and Ll : G = 0 and that rand Ll are without common component. Then the
multiplicity with which XlX o - XoXl occurs in R X2 (F, G) is the sum of the
intersection multiplicities along the line XIX o - XoXl = o.
Exercises
1. Let
f(Y) = aoym + a l ym-1 + · ..,
g (Y) = Y - b.
Show that in this case R(f, g) = ::I:: f (b).
2. Let r have at P an ordinary cusp, and let  have a simple point there.
Assume that r and  have a common tangent at P. Show that i(r, ; P) = 3
(and not more!).
3. Prove that c( U) = cl U 1'\ c E C (see p. 45).
Now we have a definition of intersection multiplicity (at a point) of two
curves rand Ll and, in particular, for Ll = L, of a curve r and a line L.
We already had a definition for the latter case, but we see that the two
definitions coincide in that case: for let
r: F(X o , Xv X 2 ) = 0,
L : X 2 - mX l = o.
According to one definition, i(r, L; Eo) equals the maximum power of Xl
that can be factored from F (X O , Xl' mX l ); according to the other, it equals
the maximum power of Xl factorable from R X2 (F(X o , Xl' X 2 ), X 2 - mX I ).

BEZOUT'S THEOREM 47
Hence by Exercise 1 above, the two definitions coincide. Since all of our
results concerning singularities follow from the theorem that the intersection
multiplicity of F(X, Y) = 0 and Y = mX at Eo is the maximum power of
X that can be factored from F(X, mX) and since we have obtained this
result starting from the second definition of intersection multiplicity we may
now simply abandon the first definition and then no longer need to prove
Theorems 7.1 and 7.2.
Exercise. Prove Theorems 7.1 and 7.2.

Chapter 8
THE BASIC INEQUALITY
Theorem 8.1. Let r and  be two curves without common components,
each containing P as a simple point. Then i(r, Ll; P) > 1 and equality
holds if and only if the tangents to rand Ll at P are distinct.
Let F(X, Y) = 0 and G(X, Y) = 0 be the minimal equations of r
and Ll. We may suppose that rand Ll are in permissible position and that
P = (0, 0). We write
F(X, Y) = alOX + a01Y + ... ,
G(X, Y) = blOX + b01Y + · · · ,
a 10 X + · .. a 01 +. · ·
a 10 X + · · ·
Ry(F, G) =
b 10 X + · .. b 01 +... ...
blOX + · .. ...
........................
Expanding the determinant by the first two columns according to Laplace's
rule, t we find that one of the terms will be
X a 10 + · .. a +...
b + ... b 01 + . .. · cofactor.
10 01
The other four terms, as one sees, have X2 as factor. Thus one obtains
a a
X 10 01. cofactor + X 2(. . .),
b lO b Ol
whence the first point, Le., i(r, Ll; P) > 1.
If the tangents are equal, then a10b 01 - a01b 10 = 0, and then
i(r, Ll; P) > 1.
If the tangents are distinct, then a10b 01 - a01b 10 i= O. ""7 e shall show that
the cofactor mentioned above is not divisible by X, Le., that evaluated at
48

THE BASIC INEQUALITY 49
0, it is i= o. The cofactor evaluated at 0 is the resultant of the polynomials
F(O, Y)/Y, G(O, Y)/Y:
R ( F(O, Y) G(O, Y) ) t
Y , Y .
Now F(O, Y) = 0 gives the points of intersection of r with X = 0; the
same is true for G(O, Y) = o. Thus F(O, Y)/Y and G(O, Y)/Y have no root
in common except possibly 0, because rand Ll are in permissible position;
but at least one of F(O, Y)/Y, G(O, Y)/Y does not have 0 as a root (since at
least one of the tangents is not vertical). Hence
( F(O, Y) G(O, Y) )
R , #0.
Y Y
Exercise. Let r have an ordinary double point at P and let P be simple for
. Show that i(r, Ll; P) > 2 and = 2 if and only if the tangent at P to Ll is
distinct from the tangents to r at P.
Generalizing Theorem 8.1, we have:
Theorem 8.2. Let rand Ll be curves without common components and
let the point P be an exactly r-fold point for r and an exactly s-fold point
for Ll. Then
i(r, Ll; P) > rs,
and inequality holds if and only if rand Ll have a common tangent at P.
Remark. If one fills out the determinant
a o a l
a o a l
am
am
b n
with the letters a_. l , a_ 2 , . . . , b_ l , b_ 2 , . . . in the form
a o a l .
b o b l byt
b o b l b n
b o b l
am
a o a l a 2 am a m + l a m + n - l
a_I a o a l a m + n - 2
a_ 2 a_I a o am+n-a
b o b l b 2 b
m+n-l
b_ l b o b l b m + n - 2

50 THE BASIC INEQUALITY
then, just as in the proof of 4.5, one sees that this last deterIninant is isobaric
of weight mn. The same is true if we assign a i the weight m - i and b;
the weight n - j; see the remark at the end of Chapter 4.
We will prove 8.2 for r == 2, s == 3, though the method is the same in
the general case. We may suppose that r and  are in permissible position
and that P == (0, 0). We write
F == a 20 X2 + a1lXY + a 02 Y2 + · · · ,
G == b 30 X3 + b 21 X2Y + b 12 XY2 + b 03 Y3 + · · ·
Then
Ry(F, G) ==
a 20 X 2 + . .. all X +... a 02 +. .. a 03 +... a 04 + · · ·
a 20 X 2 + · .. all X +... a 02 +... a 03 + · · ·
a 20 X 2 + · .. all X + · .. a 02 + · · ·
b X 3 + · .. b X2 + . .. b X + ... b + · .. b + ... ...
30 21 12 03 04
b X 3 -L... b X2 + ... b X + ... b + ... ...
30 I 21 12 03
We make a Laplace expansion by the first r + s (==2 + 3) columns. One
of the (r + s) X (r + s) determinants is formed from the first s rows of
the a's and the first r rows of the b's, namely,
a X2 + · · · a ll X + . . . a 02 + . . . a 03 + . . . a + ...
20 04
a X2 + · · · a1lX + . . . a 02 + . . . a + ...
20 03
a 20 X 2 + · · · aX+... a 02 + · · · .
11
b X3+... b 2l X 2 + · · · b 12 X + . . . b 03 + . . . b + ...
30 04
b X3+... b X2 + · · · b 12 X + · · · b + ...
30 21 03
If we give the entry a 20 X2 + · · · the weight 2, a 11 X + · · · the weight 1,
etc., as in the initial remark, we see that weight is < subdegree for each
entry and hence that the weight (== rs) of the determinant is < the subdegree
of the determinant. So rs is < the subdegree of the determinant. Moreover,
if weight is < subdegree of some entry, then weight < subdegree of any term
containing that entry as a factor. For example,
weight (a 03 + · · .) == -1 < 0 < subdegree (a 03 + · · .).
Hence any term in the expansion of the determinant containing a 03 + · · ·
as a factor has subdegree > rs; and similarly for the other terms. Hence
the above determinant is equal to
a X2 + · · · a 11 X + . . . a 02 + . . . 0 0
20
a X2 + · · · a 1l X + . . . a 02 + . . . 0
20
a X2 + · · · aX+... a + ... + X7 (. ·
20 11 02
b X3 + · · · b X2 + · · · b 12 X + . . . b 03 + . . . 0
30 20
b X3+... b X2 + · · · b 12 X + · · · b + ...
30 21 03

THE BASIC INEQUALITY 51
In this determinant the term cX 6 comes from the lowest-degree ternlS of
the various entries. Hence the coefficient of X 6 is
a 20 all a 02
a 20 all a 02
a 20 all a 02 ,
b ao b 21 b 12 boa
b ao b 21 b 12 boa
which is R(R n G s ), and which vanishes if and only if the homogeneous
polynomials Fn G s have a common factor.
We have to examine the other subdeterminants of the first r + 8
columns. Let us take 8 rows of the a's not the first 8, say rows iI' i 2 , . . . , is,
with i 1 < i 2 < · . . < is. Then 1 < iv 2 < i 2 , . . . , 8 < is and inequality
holds at least once. The elements of row ij have sub degree > subdegree of
the corresponding elements of row j, and inequality holds at least once.
Hence the subdegree of any such sub determinant is >r8.
We still have to consider taking 8 :f:: k rows of a's and r =f k rows of
b's, say the first 8 + 1 of the a's and the first r - 1 of the b's. We recall
that we give to zero every degree. Giving the appropriate degrees to the
zeros, we find that the sub degrees for row 8 + 1 of the a's are:
. . . , r, r - 1, . . . , 1.
The subdegrees for row r of the b's are:
...,8,8 -1,...,0.
Hence here too, we replace elements by elements having greater subdegree,
and hence the subdeterminant is - o(Xrs+I). The same is true for the re-
maining subdeterminants.
Resuming, we have
Ry(F, G) = R(F n Gs)xrs · cofactor + Xrs+l(. · .).
In every case, then, i(r,; P) > r8. The cofactor evaluated at X = 0 is
R(F(O, y)/yr, G(O, y)/ys).
Since r,  are in permissible position, F(O, y)/yr and G(O, y)/ys have no
common roots other than o. Choosing the coordinate system conveniently,
we may suppose that the Y-axis is not tangent either to r or to  at
P = (0,0). In that case, F(O, Y) is exactly divisible by yr and G(O, Y) by
Ys, whence neither F(O, y)/yr nor G(O, y)/ys has 0 as a root, and hence
R(F(O, y)/yr, G(O, y)/ys) * o.
Consequently i( r,; P) = r8 if and only if R(F n G s ) =1= 0, Le., if and only if
r and  have no common tangent at P.

52 THE BASIC INEQUALITY
Theorem 8.3. If r, Ll are two cycles having no common component with
a third cycle E, then
i(r + Ll, E; P) = i(r, E; P) + i(Ll, E; P).
This theorem is an immediate consequence of the following:
Theorem 8.4. Iff, g, h, are polynomials in Y with indeterminate coefficients,
then
R(f(Y)g(Y), h(Y)) = R(f(Y), h(Y)). R(g(Y), h(Y)).
Thus let
f(Y) = aoym + · · · ,
g( Y) = b o Y n + · · · ,
h( Y) = coYP +...,
We shall need first:
ai' b j , Ck indeterminates.
Lemma 8.5. R(f, g) is irreducible in C[a o , . . . , b o , . . .].
We first prove 8.4 using 8.5. Letf*, g*, h* be, respectively, the homo-
geneous polynomials corresponding to f, g, h [then, for example, ,,-e have
R{f*, h*) = R(f, h)]. Applying 4.3, we find that if for certain values
a o , . . . , 6 0 , . . . , Co' . · ·
the resultant R(f, h) vanishes, then also R(fg, h) vanishes, since if f*, h* have
a common solution [#(0, 0)], then so do f*g*, h*. Hence by Theorem 3.15,
in C[ a o , . . . , b o , . . . , co' . . .]
R(fg, h) == O(R(f, h)).
Also
R(fg, h) == O( R(g, h))
and since C[ a o , . . . , b o , . . . , co' . . .] is a unique factorization domain and
since R(f, h), R(g, h) (which are distinct) are irreducible by 8.5, one has
R{fg, h) == O(R{f, h)R(g, h)).
Then let
R(fg, h) = H R(f, h)R(g, h).
The degree of R{f, h) in the C k is m and that of R(g, h) in the C k is n. Moreover
the degree of R(fg, h) in the C k is m + n. Hence H cannot involve the C k .
A similar argument holds for the ai' b j . Hence H is a constant; and moreover

THE BASIC INEQUALITY 53
this constant is 1, but we omit the proof as this fact is not vital for the
moment.
Finally we prove 8.5. Let R{f, g) = P · Q · R · . .. be a complete
factorization of R{f, g). Making an induction on deg f + deg g, after
checking 8.5 for deg f = 1 and deg g = 1, we assume (as inductive hypo-
thesis) that R{f, gl) is irreducible, where gi = b i yn-l + · .. Since a o
and R{f, gl) are irreducible, and R{f, g)lbo=O = aoR{f, gl)' we must have
Plbo=O = a o · const
or = R{f, gl) · const
or = aoR{f, gl) · const
or = const (const =1= 0),
and the same may be said of Qlbo=O' . .. Hence the complete factorization
of R{f, g) has the form
(I) R{f, g) = [a o + b o {. · .)][R{f, gl) + b o {. · .)]
X [const + b o {. · .)] · · · [const + b o {. · .)],
or possibly
(2) R{f, g) = [aoR{f, gl) + b o {. · .)]
X [const + b o {. · .)] · · · [const + b o {. · .)].
Since R{f, g) is homogeneous in the bi' the same is true for each factor and
hence the coefficient of b o in each factor [const + b o {. · .)] must be zero.
In other words, except for units (=constants =1= 0), one cannot have such
factors. Now in case (2) one sees that R{f, g) is irreducible (since there is only
one factor in the complete factorization). In case (I), as argued before, the
coefficient of b o in [a o + b o {. · .)] is zero. Hence R{f, g) has a o as factor.
But this is certainly not so as R{f, g) lao=o =1= o. [This last point already
occurred previously in the argument, when we used R{f, g)lbo=O =1= 0.]
Exercises
1. Let r be a curve, F(X o , Xl' X 2 ) = 0 its minimal equation, and a = (a o , aI' a 2 )
a point. The cycle
of of of
 : a o - + a 1 - + a2 - = 0
oX o oX l oX 2
is the polar of a with respect to r (see 6.8). By 6.10  (if it is defined) meets
r in only a finite number of points. Show that if r has as singularities only
ordinary double points, d in number, and ordinary cusps, k in number, then
if a is not on r,  intersects r at its singularities with a total multiplicity
of 2d + 3k at least. Hence if a is not on any tangent to r at a singular point,
then from a one can draw at most n(n - 1) - 2d - 3k tangents to r.

54 THE BASIC INEQUALI'I'Y
2. Let
f(X) = ao(X - X 1 )(X - X 2 ) · · · (X - X m ),
g(X) = bo(X - Y 1 )(X - Y 2 ). . · (X - Y n ),
where a o , b o , X l' . . . , Y n are indeterminates. Show that
R(f(X), g(X)) = abrr(Xi - Y j ) = arrg(Xi).
i,j i
Following this up, one could get another proof that
R(f1f2' g) = R(f1' g) · R(f2' g).

Chapter 9
CUBICS
Definition. By an ordinary point of inflection of a curve one means a
simple point P of the curve such that the tangent to the curve at P
has intersection multiplicity 3 with the curve at P.
If the intersection multiplicity is 2, then the simple point P is called
ordinary.
The following theorem is immediate:
Theorem 9.1. Every simple point of an irreducible cubic is either an
ordinary simple point or an ordinary point of inflection.
Lemma 9.2. Let P be a simple point of curve r and let L be the tangent
to r at P. Let K be a curve. Then if one of the numbers i(K, r; P),
i(K, L; P) is less than i(r, L; P), so is the other, and th; two are equal.
This lemma holds for curves r of arbitrary degree, but for the moment
we will prove it only for cubics. There are three cases:
a) i(r, L; P) = 2, Le., P is an ordinary simple point;
b) i(r, L; P) = 3, Le., P is an ordinary point of inflection;
c) i(r, L; P) = 00, whereby we mean that L is a component of r.
Case (c) is trivial.
Take P: (1, 0, 0), L: Y = 0, and r : F = 0, K: G = 0 in permissible
position. By 4.1.1 R(F, G) = AF + BG: placing Y = 0, one sees that
i(K, r; P) > min {i(r, L; P), i(K, L; P)},
so that if i(K, L; P) > i(r, L; P), then i(K, r; P) > i(r, L; P). It
thus remains to show that
i(K, L; P) < i(r, L; P) implies i(K, r; P) = i(K, L; P).
Consider first case (b). By assumption we have
i(K, L; P) < i(r, L; P) = 3.
55

56 CUBICS
Then i(K, L; P) == 1 or 2, so K can have at most a double point at P, and
inthatcasei(K,L; P) == 2. IfKhasasimplepointatP,theni(K,L; P) == 2
or I according as L is the tangent to K at P or not. Let us now suppose,
first, that P is a double point of K. Then L is not a tangent to K at P,
whence i(K, r; P) == 2, by Theorem 8.2. Next suppose that P is simple
for K and that L is not tangent to Kat P. Then i(K, r; P) == 1, again by
8.2. There remains the case that P is a simple point of K and that L is the
tangent to K at P. Taking P: (1, 0, 0) and L: Y == 0, we have
K: Y + dX 2 + · · · == 0
(d =F 0),
r : Y + aX Y + bY 2 + cX 3 + · · · = 0
(c =1= 0).
One then computes directly, by means of the resultant, that i(K, r; P) = 2.
(One makes a Laplace expansion by the first two columns of
dX 2 + · · · 1 +... ... ...
dX 2 + · .. 1 + · · ·
cX 3 + · · · 1 +... ... ...
eX 3 + · .. 1 + · · ·
as in the proof of 8.1.)
Finally, in case (a), i(K, L; P) < i(r, L; P) = 2. Therefore
i(K, L; P) = I, so P is simple on K and L is not tangent to Kat P. Hence
i(K, r; P) == I.
Exercise. Establish Lemma 9.2 for curves r : Y - cX P = 0 [and P: (1, 0, 0)].
Theorem 9.3. Let a curve r of order n be cut by a line L, not a com-
ponent of r, in the points PI' . . . , Pn with Pi simple for r, i = 1, . . . , n.
(It is understood that each Pj occurs i(r, L; Pj) times in Pv . . . , Pn.)
If the mn intersections of r with a curve K of order m contain P v . . . , P n'
then the remaining (m - l)n intersections are the points of intersection
with r of a curve K' of order m - 1.
Since the proof uses Lemma 9.2, for the moment the theorem will hold
only for cubics. We continue to use the notation introduced in the proof of
9.2; in particular, the coordinate system is taken so that L has the equation
X 2 = o.
From the hypothesis it follows that every root of F (X o , Xl' 0) is a root
of G(Xo, Xv 0), and moreover with a multiplicity at least as great, whence
G(Xo, Xv 0) - o( F (X o , Xl' 0)),
that is,
G(Xo, Xl' 0) = H(X o , X1)F(X o , Xv 0),

CUBICS 57
where H(X o , Xl) is homogeneous of degree m - n, if m > n, and zero if
m < n. Hence
G(Xo, Xv X 2 ) === H(X o , XI)F(X o , Xl' X 2 ) + X 2 G l (X O ' Xl' X 2 ).
This equality shows that G === 0 and X 2 G l === 0 meet r in the same points;
but we have to count with multiplicity. If m < n, then H === 0 and
G === X 2 G I ; since this case is trivial, we suppose that m > n. Taking a
permissible coordinate system for G === 0 and F === 0 (without changing
X 2 === 0), we note that the system is also permissible for X 2 G I === 0 and
F === 0, since the same intersections are involved. To complete the proof we
will use:
Lemma 9.4. Letf, g, h be polynomials in Y with indeterminate coefficients.
Then R(f, g + hf) is divisible by R(f, g).
Supposing that the lemma is known, we take f === F(X o , Xv X 2 ) and
g === G(X O 'X V X 2 ) with X 2 === Y and find R(F,X 2 G l ) === R(F,G - HF) ==
o (R(F, G)). Since both resultants have degree mn,
R(F, X 2 G t ) === const · R(F, G).
Hence the intersections of G === 0 and F === 0 are the same, including multi-
plicities, as those of X 2 G l === 0 and F === 0, and since deg G l === m - 1, the
proof is complete.
Proof of the lemma. R(f(Y), g(Y) + h( Y)f(Y)) vanishes (for special values
of the coefficients ai' hi of f, g) whenever R(f( Y), g( Y)) does. Since R(f, g)
is irreducible by 8.5, the first resultant is divisible by the second.
Theorem 9.5. An irreducible cubic has at most one singularity.
The proof is trivial.
Notation. Let S, T be sets. By S X T, the so-called cartesian product of S
and T, one means the set of (ordered) pairs (P, Q) with P in Sand Q in T.
Fig. 9.1
Theorem 9.6. Let r be an irreducible cubic without singularities and
o a point on r. Let A and B be arbitrary on r. We define an operation
r X r -+ r as follows (see Fig. 9.1): the line determined by A B (or the

58 CUBICS
tangent in the case A = B) cuts r in a third point R. The line OR cuts r
in a third point which we will call A + B. Then r with this operation
becomes a commutative group.
Fig. 9.2
Commutativity, the existence of a neutral element (namely, 0), and the
solvability of the equation P + X = S are all immediate. All that remains
is to establish associativity.
Examining Fig. 9.2, one sees that the problem of proving that
(A + B) + C = A + (B + C)
comes to proving that A, B + C, and Q are collinear.
Consider the cubic
K = ABP + C(A + B)Q + O(B + C)R.
The intersection of K and r (counting multiplicities) consists of nine points:
K · r = {A, B, P, C, A + B, Q, 0, B + C, R}.
Now the points C, B, R are collinear. These are three of the nine mentioned
points. Hence by Theorem 9.3, the remaining six, Le.,
A,
P,
A + B,
Q,
0,
B+C
are cut out on r by a conic. Of these six, 0, A + B, P are collinear. Hence,
again by Theorem 9.3, A, B + c, Q are collinear. Q.E.D.
In the case that the cubic r has a singular point D one can define a sum
in exactly the same way on the set r - {D}, and r - {D} becomes an
abelian group.
Exercises
1. Let 0, 1, 00 be three points on an irreducible conic r. Then r - {oo} can
be made into a commutative group by defining X + Y as in Fig. 9.3;
and r - {O, oo} can be made into a commutative group by defining X · Y
as in Fig. 9.4.

CUBICS 59
y
x
x
Fig. 9.3
Fig. 9.4
2. Let 0 be the zero of the group G on a nonsingular cubic r. If A, B, 0 are
collinear (Le., the complete set of intersections of some line with r), then
A + B + C = K, where K is the third point of intersection with r of the
tangent at O. Conversely, if A + B + C = K, then A, B, C are collinear.
The points of inflection of r are given by the solutions of 3X = K.
3. Let G, r be as in the previous exercise. Then there is a point X on r such that
0, X, 2X are collinear, and hence r contains at least one point of inflection
(namely X).
4. There is a line through any two given points in the plane, a conic through
any five, a cubic through any nine, and a quartic through any fourteen.
Theorem 9.7. Let 0 be the zero of the group G on a nonsingular cubic r
and let H == 0, where deg H == m, cut out Pv..., Pam on r. Then
P l + · · · + Pam == mK, where K is the third point of intersection with
r on the tangent at o.
Proof. The proof is by induction on 1n and is checked immediately from the
definition of G for m == 1. Let P l P 2 cut r in a third point R and let PaP4
cut r in S. Let RS cut r in T;
R + S + T == K.
Then Pv P 2 , . . . , Pam' R, S, T are cut out by an (m + 1)-ic (i.e., a curve of
order m + 1). Since PI' P 2 , R are collinear, Pa, P 4 , . . . , Pam' S, T are cut out
by an m-ic; and since P a, P 4' S are collinear, P 5' P 6' . . . , Pam' T are cut out
by an (m - 1)-ic. Hence P 5 + P 6 + · · · + Pam + T == (m - I)K; using
P 1 + P 2 + R == K and P a + P 4 + S == K, we get
(PI + P 2 + R) + (Pa + P 4 + S) + (P 5 + P 6 + · · · + Pam + T)
== (m + I)K.
Now using R + S + T == K, we get the desired result.
The converse of 9.7 is proved similarly.

60 CUBICS
RATIONAL FUNCTIONS ON AN IRREDUCIBLE CURVE
Let r be an irreducible affine curve, F(X, Y) = 0 its minimal equation.
By a rational function on r we mean, at least roughly speaking, a function
given by a rational expression a(x, y)Jb(x, y), where (x, y) varies over rand
a(X, Y), b(X, Y) are polynomials in O[X, Y]. If the expression is to be
meaningful for even one point (x, y), we must have b(X, Y) ¥= O(F(X, Y)),
and we impose this condition; the function is then defined at all but perhaps
a finite number of points, namely, the points where b(X, Y) === 0 intersects
F (X, Y) = O. However, we wish to allow ourselves to rewrite the expression
a(x, y)Jb(x, y) taking into account relations ,vhich hold for a point (x, y) in
general position on r. For example, let r be the curve Y = X2, and consider
the expression yJx. This expression is undefined at (0, 0), but if we write
yJx = x, which holds for a point in general position, we will be able to
say that the rational function yJx is defined at (0, 0), and that its value there
is O.
In order, then, to define rational function with appropriate precision, we
first define a representative of a rational function on r as a function given by a
rational expression a(x, y)Jb(x, y) where (x, y) varies over rand
a(X, Y), b(X, Y) E C[X, Y]
with b(X, Y) ¥= O(F); this function is defined at the points (x, y) of r for
which b(x, y) * 0, and nowhere else. Two representatives aJb, cJd are called
equivalent if they are equal wherever both are defined, Le., for points (x, y)
on r for which b(x, y)d (x, y) if o. Note that aJb and cJd are equivalent if and
only if a(X, Y)d(X, Y) - b(X, Y)c(X, Y) = O(F); in fact, if aJb and cJd
are equivalent, then a(x, y)d (x, y) - b(x, y)c(x, y) = 0 except at a finite nunl-
ber of points, so a(X, Y)d(X, Y) - b(X, Y)c(X, Y) - O(F). Conversely,
if this congruence holds, then a(x, y)d (x, y) - b(x, y)c(x, y) = 0 for all
points of r. The same argument shows that aJb and cJd are equivalent if
they are equal at infinitely many points of r. It follows that if aJb is equivalent
to cJd and cJd is equivalent to eJ f, then aJb is equivalent to eJ f, for all three
will be equal where bdf * o. We then define a rational function on r as an
equivalence class of representations of a rational function. The value of a
rational function r at any point (x, y) is the value of its representatives at
(x, y). Since none of the representatives may be defined at (x, y), it is possible
for r not to be defined at (x, y); for example, if r: Y - X2 = 0, then xJy
is not defined at (0, 0). Rational functions are added, subtracted, multiplied,
and divided via their representatives.
The situation for an irreducible projective curve r with minimal equation
F (X o , Xv X 2 ) = 0 is quite the same except that in the definition of a repre-
sentative aJb one requires a, b to be homogeneous of the same degree.
Now let r, Ll be two irreducible affine curves; the curves need not be
distinct or even in the same plane, in fact, for notational purposes it is

RATIONAL FUNCTIONS ON AN mREDUCIBLE CURVE 61
convenient to think of them as being in distinct planes and then write their
minimal equations as F(X, Y) = 0, G(U, V) = O. We now wish to define
a rational function for the variable (x, y; u, v), where (x, y) is on rand
(u, v) on d; or, in other words, we want to define a rational function on
r X d. We need:
Lemma 9.8. Let r: F(X, Y) = 0, d: G(U, V) = 0 be irreducible
curves. If H 1 (X, Y; U, V) · H 2 (X, Y; U, V) vanishes on r X d,
where HI' H 2 are polynomials, then HI or H 2 vanishes on r X d.
Proof. Assume that neither HI nor H 2 vanishes on r X d. Then there is a
point (x, y; u, v) in r X d where HI does not vanish. Thus there is a point
(u, v) on d such that H1(X, Y; u, v) does not vanish on r. Similarly, there
is a point (u', v') on d such that H 2 (X, Y; u', v') does not vanish on r.
Hence HI(X, Y; u, v)H 2 (X, Y; u', v') does not vanish on r. Hence there
is a point (x, y) on r such that HI(x, y; u, v) · H 2 (x, y; u', v') =1= o.
Then HI(x, y; u, v) =1= 0, so H 1 (x, y; U, V) does not vanish on d, and
for a similar reason H 2 (x, y; U, V) does not vanish on d. Hence it follows that
H1(x, y; U, V) · H 2 (x, y; U, V) does not vanish on d. Hence there is a
point (u, v) on d such that HI(x, y; u, v) · H 2 (x, y; u, v) =1= o. This
contradicts the assumption that HIH2 vanishes on r X d.
The definition of a rational function on r X d can now be given: one
proceeds very much as with the definition of a rational function on r.
A representative of a rational function is a function given by an expression
a(x, y; u, v)/b(x, y; u, v), where a, b are polynomials and b does not vanish
everywhere on r X d. Two representatives a/b, e/d are said to be equivalent
if ad - be vanishes over r X d. The proof that if a/b and c/d are equivalent
and if c/d and e/ f are equivalent, then a/b and e/ f are equivalent is slightly
different, since a polynomial b may vanish at infinitely many points of
r X d without vanishing everywhere: assuming ad - be and cf - de
vanish on r X d, one finds that adf - bde (=f(ad - be) + b(cf - de))
vanishes on r X d, and since d does not, af - be does, by the lemma.
Now we define a rational function on r X d as an equivalence class of
representatives.
If a/b "-' a' /b' (Le., a/b is equivalent to a' /b') and e/d "-' e' /d', then
ac/bd "-' a' e' /b' d'. For the proof, one first shows ac/bd "-' ae' /bd' and then
00' /bd' "-' a' e' /b' d', whence ac/bd "-' a' e' /b' d'. Hence the product of rational
functions can be defined via their representatives. Similarly for sum,
difference, and quotient.
A similar situation obtains for irreducible projective curves r, d, given
by F(X o , Xl' X 2 ) = 0, G(U o , U I , U 2 ) = o. Here a, b should be homogeneous
of the same degree in Xo, Xl' X 2 and homogeneous of the same degree in
U o , U 1 , U 2 .
In the following theorem we take r = d.

62 CUBICS
Theorem 9.9. Let r be an irreducible affine curve with F(X, Y) = 0
as minimal equation. Let (Xl' Yl)' (X 2 , Y2) be points on r. Then the rational
function given by (Y2 - YI)/(X 2 - Xl) (which, of course, is the slope of
the line joining the points if Xl =1= x 2 ) is still defined for (xl' Yl) = (X 2 , Y2)
provided that (Xl' Yl) is a simple point of r with nonvertical tangent;
and, in that case, the value of the function at (Xl' Yl) is the slope of the tangent
to r at (xl' Yl).
Proof. By the computational rules given in Chapter 6,
o = F (X 2 , Y2) - F (Xl' Yl) = (F (X 2 , Y2) - F (Xl' Y2))
+ (F(x 1 , Y2) - F(x v Yl))
of of
- (x 2 - Xl) ax + (x 2 - XI)2[. .J + (Y2 -Yl) ay
(Xl,Y2) (XttYl)
+ (Y2 - Yl)2[. .],
where the [. · J stand for the polynomial expressions in Xl' YI, X 2 , Y2. Then
BF
+ (x 2 - x 1 )[. .]
ax
(Xl,Y2)
of
Y2 - Yl
X 2 - Xl
BY + (Y2 - Yl)[. -J
(XbYl)
The theorem follows from this way of rewriting (Y2 - Yl)/(X 2 - Xl).
Let r, d be irreducible affine curves. By an obvious generalization of
rational functions, we say that a point R (varying in an affine plane) is a
rational function of points P E d, QEd if the coordinates of R are rational
functions of the coordinates of P and of Q. We frequently write R: (s, t),
where s, t are representatives of the functions in question.
The situation for projective curves is quite the same. Here P: (X o , Xl' X 2 ),
Q : (u o , ul' u 2 ), R: (ro, r l , r 2 ), and ro, r l , r 2 are to be rational functions
of P, Q, at least one of which does not = 0, Le., does not vanish every..
where on r X d. We say that R = R(P, Q) is defined at P, Q if for some
r i that does not vanish on r X d, the functions rO/ri' rl/ri, r2/ri are all
defined at P, Q. [If this happens for two values of i, say i = 1,2, then the
points obtained are the same. In fact, one first observes that if g and hare
rational functions defined at P, Q, then so is gh, and the value of gh at P, Q
is the product of the values of g and h at P, Q. Then the values of rl/r2 and
r 2/r I are both =1= 0, as the product of these values = 1. The assertion now follows
upon observing that (r k /r l )/(r 2 /r l ) = rk/r2 and that (rk/r l ) I p,ij/(r 2 /r l ) /p,'Q
= (rk/ r 2) /i\ij.]
By a complete set of representatives for a function r ,ve mean a set of
representatives of r such that at least one of them is defined at any point
(of PNC or of PNC X PNO) at which r is defined.

RATIONAL FUNCTIONS ON AN ffiREDUCIBLE CURVE 63
Let WI' . . . , wN be indeterminates over 0 and let 0 1 == O(w l , . · . , WN).
We wish to define a rational function on r, or on r X d, over 0 1 . The in-
dependent variables will be the same as before, but the values of the functions
may involve the 'Wi. The procedure is quite the same as before once we have
the following:
Lemma 9.10. Let F be an irreducible polynomial in C[X, Y]. Then F
remains irreducible in Ol[X, Y].
Proof. This can also be formulated: if HI' H 2 E Cl[X, Y] and HlH2 == O(F)
in Ol[X, Y], then HI or H 2 == O(F) in Cl[X, Y]. The Hi may have denomina-
tors involving the Wi' but this is of no importance, and we may suppose the
Hi are in O[w l , . . . , wN; X, Y]; we consider HI' H 2 as polynomials in the
Wi with coefficients in O[X, Y]. If HlH2 == O(F), then
HI(W l , . . . , W n ; X, y) · H 2 (W l , . . . , W n ; X, y) == 0
for every point (x, y) (in ANO) on F == o. Then Hl(wl' . . . , W n ; x, y) = 0
or H 2 (W l , . . . , W n ; x, y) == o. Thus for any (x, y) on F = 0, all the co-
efficients of HI or all those of H 2 vanish. Hence the product al(X, Y)a 2 (X, Y)
of any coefficient a l of HI and any coefficient a 2 of H 2 vanishes everywhere on
F == 0, whence a l a 2 == O(F). If HI =1= O(F), then some coefficient a l of
HI is =1= O(F), whence every coefficient of H 2 is == O(F). Q.E.D.
Theorem 9.11. Let r be a cubic curve in the projective plane free of
singularities. Let PI' P 2 be points on r and let P a be the third intersection
of P l P 2 with r. (As usual, we understand P l P 2 to be the tangent at PI if
PI == P 2 ). Then Pa is an everywhere (on r X r) defined rational function
of PI' P 2 .
Proof. Let Pi: (x&i), xi), xi»), i == 1, 2, 3. Let Xo = 0 be taken as the
line at infinity; and consider for a moment the case that the Pi are at finite
distance. Let (1, xi)/x), x)/x») == (1, Xi' Yi) be the normalized coordinates
of the Pi. Assume, too, that P l P 2 is not vertical and even, at first, that
PI =1= P 2 . Then the slope of P l P 2 is m == (Y2 - Yl)/(X 2 - Xl)' and the
equation of P l P 2 is Y - Yl == m(X - Xl). Let F(X, Y) == 0 be the
minimal equation for the affine part of r. Eliminating Y, we find a cubic
in X whose roots are Xl' X 2 , Xa: let this equation be
aoxa + a l X2 + a 2 X + aa = 0,
where the ai are polynomial expressions in m, xl' Yl. Here a o * 0, since the
equation has 3 roots, and -al/a O == Xl + X 2 + xa. Solving for xa, we find
that Xa is a rational function of Xl' Yl' X 2 , Y2; the same now holds for Ya from
Ya == m(x a - Xl) + Yl. Thus Pa, for the case considered, is a rational func-
tion of PI' P 2 . Moreover, by 9.9, Pa is still defined for PI = P 2 , provided
the tangent at PI is not vertical.

64 CUBICS
The possibility that PIP 2 is vertical and the possibility that PI' P 2' or
Pa (or some two or all three of them) are at infinity require further considera-
tion.
As in Bezout's Theorem (7.3), we introduce nine indeterminates
U oo , . . . , U 22 over C and consider the linear transformation Lu. Let
pV . (y (i) Y (i) Y (i» ) where Y (i) = Ux(i) be the transform of p. As ex p lained
. 0' l' 2 , , Z.
for 7.3, pf PY does not pass through E 2 : (0, 0, 1); a similar argument shows
that pf cannot be at infinity (if it were, then Pf would be, too, for any
nonsingular matrix A with entries from C, and this is clearly absurd). If
r : F (X) = 0, then the Pf lie on F (U -IX) = 0, and by the considerations
given just above, Pf: (1, y3)/y&3), y&3)/y&3») is an everywhere (on r X r)
defined rational function of PI' P 2 . Note that we need only two representa-
tions of pf to bring this definition to expression: one for the case that
PI =/:; P 2 , in which one writes the slope of PfPf in the obvious (Le., universal)
manner, and one for the case that PI = P 2 , in which one writes the slope as
in 9.9. Applying U-l to these representations, we get two representations
(Ao, AI' A 2 ), (Bo, B l , B 2 ) of Pa: (X&3), X3), x&3»). Here Ai and B i are repre-
sentatives of the same function, i = 1, 2, 3. The proof would be complete,
except that the Ai, Bj involve the U pq . To complete the proof, we first
establish:
Lemma 9.12. Let r be a rational function on r X d (say) over
0 1 = O(w l , . . . , wN) and assume that r(P, Q) EO whenever defined. Let
r l' . . . , r (J be a complete set of representatives of r,. each r i can be written as
Ik tkaki
ri = ,
Ik tkb ki
where the t k are power products of the Wj and the aki, b ki are polynomial
expressions in P, Q (P E r, QEd). Oonsider all the aki/bki with b ki :f=. O.
Then these form a set of representatives for a function r' on r X d over
C defined wherever r is; and r'(P, Q) = r(P, Q) wherever r is defined.
Proof. Consider one of the aki/bki, say all/b ll . We say that
all(I tkb kl ) - bll(I tka kl ) = 0
on r X d. In fact, for any point P, Q on r X , if I tkbkl(P, Q) = 0, then
bll(P, Q) = 0 and, in this case, the equality follows; if I tkbkl(P, Q) :f=. 0,
then there is a number f E C such that f I tkbkl(P, Q) - I tkakl(P, Q) = 0,
whence fbll(P, Q) - all(P, Q) = 0, and again the equality follows. Thus
all/b ll is a representative of r. As this holds for all the akdbki, they are all
equivalent and represent a function r' over C. If, now, r is defined at P, Q,
then I tkbki(P, Q) :f=. 0 for some i, say i = 1. Then for some Ie, bkl(P, Q) :f=. 0,
say Ie = 1. Then all/b ll is defined at P, Q, and the proof is complete.

ABELIAN VARIETIES 65
To complete the proof of 9.11, note that Ao, AI' A 2 , and not merely their
ratios, are defined for PI -=1= P 2 and have values not all == o. Let ak),
k == 1, . . . , gi, be the complete set of representatives of Ai given by the
lemma. Then the gog1g2 representations (a&k o ), akl), ak2»), where k i runs from
1 to gi, are over C; and for any PI' P 2 in r X r with PI -=1= P 2 , at least
one of these yields P a. These glg2ga representations, together with a similar
set derived from Bo, B 1 , B 2 , yield a point (r 0' r l' r 2) which is everywhere
defined on r X r and which for any PI' P 2 has the value Pa.
Remark. Let r be given by an irreducible equation F (X) == o. As in 7.3,
F == 0 defines a curve r* in PNO v where 0 1 is the algebraic closure of
C 1 == C(U oo , . . . , U 22 ). To define rational functions on r* we must know
that it is irreducible. It is a property of any irreducible polynomial in
C[X] to remain irreducible over any extension field K ofC (Le., in K[X]);
this is its so-called absolute irreducibility. In the case of a curve r : F == 0
without singularities, a simple proof of this can be given by observing that r
cannot acquire singularities by being regarded over an (algebraically closed)
extension field of C. We could, then, have defined rational functions on r*
for our cubic F == 0; but, in fact, this was not necessary.
ABELIAN VARIETIES
Let r be a cubic curve without singularities in the projective plane and let a
line cut r in the points P v P 2' P a. Then, as we have just seen, P a is an every-
where (on r X r) defined rational function of Pv P 2 . From this one sees
that in the group defined on r one has:
a) in P + Q == R, R is an everywhere (on r X r) defined rational
function of P, Q, and
b) in P + Q == 0 (the neutral element), Q is an everywhere (on r)
defined rational function of P.
We will now make some observations with the aim of relating the above
example to a branch of algebraic geometry that has attracted much attention,
namely, abelian varieties. We need some definitions.
An affine variety is the locus of a point satisfying a finite number of
simultaneous polynomial equations: F1(Y V ... ,Y n ) == 0,..., FS(Y 1 ,... ,Y n )
== 0, F i E C[Y 1 , . . . , Y n ] (where F i E C is allowed). A projective
variety is similarly defined by homogeneous polynomials Gi(Xo,..., X n ).
A variety V (affine or projective) is called red'ucible if V == VI U V 2 ,
V -=1= VI' V -=1= V 2 , where VI' V 2 are varieties (affine or projective
according as V is affine or projective). A variety is called irreducible if it is
not reducible.
It would be convenient to introduce two more terms: abstract variety
and complete variety. We will not give the general definitions, but will limit
ourselves to some special cases.

66 CUBICS
By a special abstract variety we mean an irreducible variety (projective
or affine) minus a proper subvariety. Thus every special abstract variety
has the form V - F with V, F varieties and V irreducible. By a special
complete variety we mean an irreducible projective variety. Thus every special
complete variety is a special abstract variety of the form V - F with V
projective and F empty.
By a special group variety V we mean a special abstract variety that is
also a group with the conditions:
a) in P · Q = R, R is an everywhere (on V X V) defined rational func-
tion of P, Q, and
b) in p. Q = E (the neutral element), Q is an everywhere (on V)
defined rational function of P.
The main point of this definition is that the group operations are not to
be taken arbitrarily, but must be defined algebraically.
A special group variety need not be commutative. For example, let us
consider the points (u oo , U 01 , U IO , U II ) in projective 3-space such that
UOOU II - UIOU OI :f=. 0; these form a special abstract variety. It can be made
into a group according to the rule.
(U OO ' U OI , U IO , U II ) · (V OO ' VOl' VIO' VII) = (WOO, WOl' WIO' W II ),
where
( WOO WOI ) = ( U OO UOI ) . ( V OO VOl )
W IO W II U IO U II V IO VII
(ordinary matrix multiplication).
A nonsingular cubic made into a group IS a special complete group
variety.
There is a known theorem assuring us that a special complete group
variety is necessarily commutative, and because of this the special complete
group varieties are called special abelian varieties. Thus it was no accident
that in our example of the cubic there was commutativity. In fact, it is
impossible to make a nonsingular cubic into a noncommutative group
variety.

Chapter 10
CUBICS (CONTINUED)
POINTS OF INFLECTION
Let r: F(X O ' Xl' X 2 ) = 0 be an irreducible curve of order n. Our first
question is: Does r necessarily have a point of inflection (Le., a simple
point whose tangent to r cuts r there with multiplicity > 3) 1
If r is a line, then every point of r is a point of inflection; if r is a
conic, then no point is a point of inflection: hence we will assume ord r > 3.
If x, yare two points, then (as in Chapter 6)
F(x + AY) = F(x) + ( I f)F Yi ) A +  ( I f)2F YiY; ) A2 + ...
oX i 2 ! OXiOXj
(Here OFjOXi is a standard notation for oFjoXilx=x; and similarly for
02FjoXiOXj.) Ifxisonr,thenF(x) = 0; ifitissimple,thenI (OFjOXi)X i =
o is a line, the tangent line. If y is on this line, then I (oFjOXi)Yi = 0; if
the tangent line is to cut with multiplicity > 3, then also
02F
I 0 0 YiYi = 0,
Xi Xj
i.e., y lies on the locus
02F
I XiX; = o.
OX-OX,
1. ,
As this is to hold for each point of I(oF/oXi)X i = 0, we find:
Theorem 10.1. The simple point x of r : F = 0 is a point of inflection
of r if and only if
0 2 F ( oF )
I -- XiX j == 0 I - Xi .
OX.X. OX.
1. , 1.
In particular, therefore, if x is a point of inflection then the conic
02F
I x.x. = 0
ox-ox. 1. ,
1. ,
67

68 CUBICS (CONTINUED)
must be reducible. The condition for this is that
det
fJ2F
fJx .fJx .
t ,
- O .
- ,
or, in other words:
Corollary 10.2. A point x of r is a point of inflection only if it lies on
the curve H == 0, where
fJ2F fJ2F fJ2F
--
fJX2 fJXOfJX I fJX o fJX 2
0
H= fJ2F fJ2F fJ2F
fJXlfJX o fJX 2 fJX l fJX 2
1
fJ2F fJ2F fJ2F
fJX 2 fJX O fJX 2 fJX I fJX 2
2
We will see in a moment that not every point of r is a point of inflection.
Hence the above determinant is not the polynomial 0, and the use of the word
"curve" is justified. Hence also not all fJ2 F/fJXifJXj can vanish over r, which
justifies the word "conic" in our reference to the locus
fJ2F
! XiX; == O.
fJx .fJx .
Z J
Corollary 10.3. Oonversely, if x is a simple point of r on H == 0 (the
so-called Hessian of r), then x is a point of inflect1:on of r.
Proof. Multiplying the first column of H by Xo and adding to it Xl times
the second and X 2 times the third, and applying Euler's formula, one finds
fJF fJ2F fJ2F
(n - 1)- fJXOfJX I fJX O fJX 2
fJX o
fJF fJ2F fJ2F
XoH== (n - 1)- fJX2 fJX l fJX 2 .
fJX I 1
fJ2F fJ2F fJ2F
(n - 1)- fJX 2 fJX I fJX 2
fJX 2 2
Applying a similar operation to the rows of this determinant, placing
Xo == 1, Xl == X, X 2 == Y, and writing F for F(I, X, Y) and H for
H(I, X, Y), one finds H == 0 as the affine equation of the Hessian, where
n(n - 1)1"
H == (n - l)fJ1"/fJX
(n - l)fJF/fJY
(n - l)fJF/fJX
fJ 2 1"/fJX2
fJ21"/fJ YfJX
(n - l)fJ1"/fJY
fJ2 1"/fJXfJ Y
fJ21"/fJ Y2

POINTS OF INFLECTION 69
Now assume that the origin (0, 0) is a simple point but not an inflection.
Let the tangent at the origin be Y = o. Then r is given by F = 0 with
F = Y + eX 2 + · · · , c :f=. o. Then
o 0 n-l
H (O, 0) =
o
2c
- -2(n - 1)2 c :f=. 0,
n-l
so that the Hessian does not pass through the origin.
Remark. In assuming that the simple point is at the origin, we are using
the fact that the Hessian curve is a projective notion: this should be checked.
We are also using Theorem 7.2 (see the exercise on p. 47).
As the order of the Hessian (more precisely: the cycle H = 0) IS
3(n - 2), we have:
Theorem 10.4. A nonsingular curve of order n > 3 has at least one
point of inflection.
Let r be a nonsingular cubic. Take the coordinate system in such a way
that E 2 : (0, 0, 1) is on r. With reference to Xo = 0 as line at infinity, the
equation of r takes the form a(X) Y2 + b(X) Y + c(X) = 0, where a, b, c
are polynomials without common factor; the degree in Y is 2, since some
vertical lines cut r in two points and none cuts in more than 2 (at finite
distance). Taking (0,0,1) to be a point of inflection with Xo = 0 as tangent,
every vertical line cuts r at finite distance in two points (counting with
multiplicity), and hence a(X) = 0 can have no root, Le., a(X) = const.
Since i(r, loo; E 2 ) = 3, one finds that deg b(X) < 1. Making the linear
transformation X' = X, Y' = Y + b(X)/2a, one eliminates the Y'-term:
the equation of r then has the form Y2 = g(X). Here g(X) is a cubic,
since r is a cubic. Thus r is given by Y2 - d (X - a)(X - b)(X - c) = O.
Replacing Y by Y'V d, we may suppose that d = 1. The roots a, b, c of g
are distinct, since if a = b, for example, then r has a multiple point at
(a, 0). Moreover, by taking a new coordinate system, keeping (0, 0, 1) and
(0, 1, 0) as vertices, but taking (a, 0) as a vertex and the unit point collinear
with E 2 and (b, 0), the equation takes the form Y2 = X (X - I)(X - c).
Hence:
Theorem 10.5. With an appropriate choice of the coordinate system,
every nonsingular cubic can be put in the form Y2 = X (X - 1)(X - c),
c :f=. o. (See Fig. 10.1.)
Let Pa be the point (a, 0), for arbitrary a. Then c gives the cross-ratio
of Po, Pv Pc, P 00 in some order. Since E 2 P O ' E 2 P V E 2 P C , E 2 P 00 are the four
tangents from E 2 to r, c can be described directly in terms of r, namely, as
the cross-ratio of these tangents at E 2 . In the above construction (of the

70 CUBICS (CONTINUED)
\
y2 = X (X - 1) (X - c)
Fig. 10.1
equation Y2 = X(X - I)(X - c)), the roles of Pa, P b , Pc can be arbitrarily
interchanged. For the six possibilities, one gets the cross-ratios c, lie, 1 - c,
1/(1 - c), -c/(1 - c), (1 - c) I -c, which are in general distinct. Thus in
general one gets six different standard equations for r.
Let S be the system of values c, l/c,..., (1 - e)/ -c. These are
defined in terms of a selected point of inflection, but we will show that S
depends only on r and not on the point of inflection.
Theorem 10.6. The Hessian of r cuts r at an ordinary point of inflection
with multiplicity 1.
The theorem is true for r of arbitrary order n > 3, but we will prove
it for the present only for n = 3.
Proof. Taking a flex at (0, 0) with Y = 0 as tangent (and applying 10.5
with change of notation), the equation of r can be written:
Y - X(X - Y)(X - cY) = O.
In the Hessian determinant H , dropping terms divisible by Y or by X2
we get
o
o
2
o
-6X
- -24X
,
2
i.e., H(X, Y) contains the term -24X. Hence H = 0 has a simple point
at the origin and Y = 0 is not the tangent. Hence the intersection multi-
plicity is 1 as stated.
Corollary 10.7. r has only a finite number of points of inflection;
H  O(F).
Corollary 10.8. A nonsingular cubic has exactly nine points of inflection.

POINTS OF INFLECTION 71
Theorem 10.9. Let r be a nonsingular cubic. Then any two points of
inflection are collinear with a third. For any two points of inflection
A, B, there is a linear transformation taking r into itself and A into B.
Hence the cross-ratios of the tangents at A are the same as those at B.
Proof. Let A be placed at E 2 and let r have the equation
Y2 = X(X - I)(X - c).
This curve is symmetric in the X-axis (i.e., is sent into itself by the linear
transformation X' = X, Y' = -Y). The points Po, PI' Pc are not points
of inflection, since the tangents there go through E 2 . Hence B is not on the
X -axis. Hence the symmetric point, B', is also a point of inflection; and
BB' passes through E 2 . This proves the first point. Now let A, B be two
given points of inflection and let C be the third point of inflection collinear
with A and B. Take 0 at E 2 . Then A, B become symmetric points of
inflection on r, and the transformation X' = X, Y' = -Y interchanges
A and B and sends r into itself.
Corollary 10.10. The system S = Sr depends only on r, and not on
the point of inflection by means of which it is defined.
Theorem 10.11. Two nonsingular cubics r, r' are projectively equivalent
if and only if Sr = Sr'.
Proof. If a linear transformation takes r into r', it takes the points of
inflection of r into those of r', and the tangents to r from a point of in-
flection P into the tangents to r' from the corresponding point P'; hence
Sr = Sr'. Conversely, let Sr = Sr'. If c is an element of Sr, then r can
be given the equation Y2 = X(X - I)(X - e); and for the same reason,
r' can be given the same equation. Hence rand r' are projectively
equivalent.
If c is the cross-ratio of four lines taken in some order, then the cross-ratios
for various orders are c, lie, 1 - c, 1/(1 - c), 1 - l/c, 1 - 1/(1 - c), all
of which can be obtained from c by a succession of substitutions of the form
c -+ lie and c -+ 1 - e. Since
A = 4(1 - e + C 2 )3f(c + 1)2(2c - 1)2(c - 2)2
is invariant under these substitutions, the six cross-ratios yield the same value
of A. Conversely, given A, the above equality yields a sixth-degree equation,
which can have at most six roots. Except for the cases 1 - c + c 2 = 0 and
(c + 1)(2c - l)(c - 2) = 0, which may be handled separately, the values
c, l/c, etc., are distinct, hence any other cross-ratio c yields a different value
of A. In the case of a nonsingular cubic r, then, Sr determines and is

72 CUBICS (CONTINUED)
uniquely determined by A: this number is called the modulus of A. Theorem
10.11 can also be phrased:
10.11.1. Two nonsingular cubics r, r', are projectively equivalent if
and only if they have the same modulus.
Exercise. Let r be a nonsingular cubic and define a group on it with 0 at a
point of inflection. Show that the nine points of inflection are 0, X, - X, y,
- Y, -X - Y, X + Y, X - Y, -X + Y, the nine solutions of 3Z = o.
Call these points AI' A 2 , Aa, B 1 , C 1 , C 2 , B 2 , C a , Ba (in the given order). Take a
coordinate system in such way that AI: (1, 1), A 2 : (1, -1), B 1 : (-1, 1),
B 2 : (-1, -1). One then finds that C 2 : (0,0), Aa: (1, r.J.), Ba: (-1, -r.J.), C 1 : (fJ, 1),
C 2 : ( - fJ, -1) with r.J. = V -3 , fJ = - V - 3. Thus the nine points are deter-
mined, up to a projective transformation, independently of r.
Standard forms for cubics with a singularity are found by methods
similar to the one used for a nonsingular cubic. If r has a cusp, place it at
E 2 with Xo = 0 as tangent. Then r takes the form Y = g(X), where g(X)
is a cubic. Taking some point on r as Eo and the tangent there as the X-axis,
we see that r takes the form Y = aX2(X - c). Here c = 0 is possible,
but taking Eo to be not a point of inflection, we have c :f=. o. By appropriate
choice of the unit point, we may suppose that c = 1. Replacing Y by
Y' = Y/a, we have a = 1. Therefore all cubics with a cusp are projectively
equivalent. Hence:
Theore"" 10.12. Every cubic with a cusp is projectively equivalent to
Y2 = X3.
Exercise. If a cubic has a cusp, then it has a point of inflection.
Now let r have an ordinary double point. Place it at E 2 with Xo = 0,
Xl = 0 as the tangents. Then r takes the form X O X 1 X 2 + a cubic form in
Xo, Xl (as one sees by referring r to X 2 = 0 as line at infinity and then
restoring homogeneous coordinates). Thus
r : X O X 1 X 2 + aXg + bXgX I + cXoXi + dX = O.
Replacing X 2 + bX o + cX I by X 2 (and then d by b), we get
r : X O X 1 X 2 + aXg + bX = o.
Here a :f=. 0 and b :f=. 0, as r is irreducible. Applying the transformation
X = {/ a o Xo, X = {/b Xl' X = X 2 , we may achieve a = 1, b = 1.
Hence:
Theorem 10.13. All irreducible cubics with an ordinary double point
are projectively equivalent.

CUBICS THROUGH EIGHT POINTS 73
Exercise. Obtain another proof of this therom as follows: Show that the
Hessian of a cubic r with an ordinary double point has intersection multiplicity
6 with r there. Hence r has a point of inflection. Place a point of inflection
at E 2 , etc.
CUBICS THROUGH EIGHT POINTS
Every conic has an equation of the form
aooX + aOIXOX I + a 02 X O X 2 + allXi + a I2 X I X 2 + a22X = o.
There are six coefficients here. If we ask that this conic pass through a
point (X o , Xl' X 2 ), we impose the condition
aooxg + · · · = o.
Regarding the au as unknowns, we find that this is a linear condition on the
au. From the theory of linear equations, it is known that any five such
conditions can be satisfied nontrivially. Hence:
Theorem 10.14. Through any five points in the plane one can pass at
least one conic.
As the condition aooX + · · · = 0 and the condition
A( aooX + · · .) = 0,
A -::/:= 0 (A E C),
are the same, we associate the condition with the point (x, XOx I , . . . , x)
in projective 5-space. Then we use the terminology of this 5-space also for
the conditions; e.g., we say that one condition is (linearly) dependent on
others, that several conditions are (linearly) independent, etc.
Let PI' P 2 , . . . , Ps be five points and let Lv L 2 , . . . , Ls be the
corresponding conditions on a conic to pass through them. Then there will
be only one conic through the points if and only if L I , . . . , Ls are independent:
this is a direct translation from a familiar theorem on homogeneous linear
equations (namely, that r linearly independent equations in n unknowns has
n - r, but not more, linearly independent solutions). Now if Lv . . . , Ls
are independent, then none is a consequence of the others (i.e., by definition, any
four have a solution which is not a solution of the fifth-this will be so
because any four have 6 - 4 = 2 linearly independent solutions), and there
will be a conic through any four of PI' . . . , Ps not passing through the
fifth. Conversely, if through every four there is a conic not through the
fifth, then the conditions are independent. If four of the points are on a line
m, then any conic through three of them contains the line m and hence the
fourth point; the four (also the five) points impose dependent conditions on a
conic to contain them, and there will be infinitely many conics through the
five points. On the other hand, if no four of PI' . . . , Ps are collinear, then

74 CUBICS (CONTINUED)
one can pass a conic (even a reducible one) through any four of them not through
the fifth,. the five points impose independent conditions,. hence one can pass only
one conic through the five points.
(We say that a linear condition L is a consequence of linear conditions
L 1 , . . . , Lr if every solution of L 1 , . . . , Lr is a solution of L. By a theorem
from the theory of linear equations, this will be so if and only if L is a linear
com bina tion of Lv . . . , Lr.)
The same ideas apply to cubics. The general cubic equation has ten
coefficients. Hence:
Theorem 10.15. A cubic can be made to pass through any nine points.
Lemma 10.16. Let PI'...' Ps be eight distinct points and let
L 1 , . . . , Ls be the conditions imposed on a cubic to pass through them.
If no four of the Pi are collinear and no seven co-conical, then Lv . . . , Ls
are independent.
Proof. We show that a cubic can be put through any seven of the Pi' say
Pv . . . , P 7 , without passing through the eighth. On PIPS there is at most
one other Pi' hence there is a Pi' say P 2 , such that P 1 P 2 does not pass through
Ps. A similar argument shows that there is a third line PI Pi' say PIP a,
not through Ps. Now pass a conic O 2 through P 4 , . . . , P 7 and P 2 , and a
conic Oa through P 4 ,. . . , P 7 and Pa. Then not both O 2 and Oa can pass
through Ps. If they did, then O 2 * Oa; otherwise the points P 2 , . . . , Ps
would be co-conical. Then the distinct conics O 2 , Oa pass through P 4' . . . , P s'
whence some four of P 4' . . . , P s are collinear, which is excluded by hypo-
thesis. Thus O 2 or Oa does not pass through Ps, say O 2 does not. Then O 2
and the line PIP a form a cubic through Pv . . . , P 7 but not through Ps.
Theorem 10.17. Let two cubics meet in just nine (distinct) points,
PI' . . . , P 9. Then any cubic through any eight of the Pi also passes
through the ninth.
Remark. Let Lv . . . , Lg be the conditions imposed on a cubic to contain
PI' . . . , P g. Since the conditions Lv..., Lg have two solutions,
L 1 , . . . , Lg are dependent, and one of them must depend on the others.
Hence all the cubics through some eight of the Pi also pass through the ninth.
This doesn't quite prove the theorem.
Proof of the theorem. No four of the Pi are collinear and no seven co-conical,
since otherwise the two cubics would have a common component. Hence
any eight of the conditions L i are independent, so the ninth must depend
on them.
A slightly deeper statement is the following.

THE MODULUS OF A NONSINGULAR CUBIC 75
Theorem 10.18. All cubics through any eight given (distinct) points,
P v . . · , P s, also pass through a ninth. Here the ninth point may be
"infinitely near" one of the Pi' say PI' whereby we mean that there is a
line m through PI such that i(r, m; PI) > 1 for every cubic r through
P v . . · , P s.
Proof. If four of the Pi lie on a line g, then any cubic through the Pi contains
g as a component, and any point of g (i=-P i , i == 1, . . . , 8) can be taken as
the ninth point. If no four Pi are collinear but some seven lie on a conic C,
then C is irreducible, since otherwise some four of the Pi would lie on a
component of C. Then any cubic through the Pi contains C as a component,
and any point of C (i=-P i , i == 1, . . . , 8) can be taken as the ninth point.
We may assume, then, that no four of the Pi are collinear and that no seven
are co-conical. Now take two cubics r v r 2 through P v . . . , P s and suppose
they meet in a ninth point P 9 . Since, by Lemma 10.16, the conditions
L 1 , . . . , Ls are independent, but conditions Lv..., L9 are dependent,
L9 depends on L],. . . , Ls, and all cubics through P v . . . , P s also pass
through P 9 . However, two cubics through PI' . . . , Ps may meet just in
those points, therefore with multiplicity > 1 at one of the points, say PI;
in fact, since the cubics meet just in the points P v . . . , P s, they have no
common component, and the intersection multiplicity will be 1 at seven of the
points and 2 at the eighth (PI). Thus at least one of r l' r 2 has a simple
point at PI' say r l does, with m as tangent; then r 2 must have at PI either
a double point or a simple point with m as tangent. Place PI at (0, 0) with
m as X-axis and write the general cubic in the form
boo + b 10 X + b OI Y + · · · == o.
The condition that a cubic pass through (0, 0) is that boo == 0, which is
a linear condition; the condition that it pass through (0, 0), be simple there,
and have m as tangent (or be a double point) is boo == 0 and b iO == 0, i.e.,
two linear conditions. The condition boo == 0 is L 1 ; designate the condition
b 10 == 0 by L9. Then Lv..., L9 are dependent, since they have two
solutions, and as L 1 , . . . , Ls are independent, L9 is a consequence of them.
Hence any cubic through PI' . . . , Ps either has a simple point at PI with
m as tangent or has a double point there. This completes the proof.
THE MODULUS OF A NONSINGULAR CUBIC
Above (see 10.5 through 10.11) we classified the nonsingular cubics from a
projective point of view. Later (see Chapter 20, Example 6) we shall want
to do this from a so-called birational point of view and shall need a result of
Salmon on the nonsingular cubics.

76 CUBICS (CONTINUED)
Let r : F = 0 be a cycle and P a point. If F (X o , Xl' X 2 ) is irreducible
(or even if it merely has no multiple factors) and if P is not on r (so that the
polar is certainly defined), then the polar  meets F = 0 in n(n - 1) points
Q, where n = degree of F. From P, then, one can draw n(n - 1) tangents
PQ to F = o. These may not be distinct, but if we count PQ i times, where
i = i(r,; Q), we get the full number n(n - 1) of tangents.
If r : F = 0 is an irreducible nonsingular cubic, then from P one can
draw six tangents. For orientation, consider the rema if P approaches
a point A of the curve, two of the six tangents coalesce with the tangent at
A; thus one can draw four tangents from A to r, not counting the tangent
at A.
We now make this remark precise. One checks the following statements:
a) The polar  of A passes through A. The intersection multiplicity
i( r, ; A) is 2 if and only if A is not a point of inflection, in which
case it is 3.
b) IfQ * A is on r and on , then i(r, ; Q) = 1.
Exercise. Establish the following statements:
a') rand Ll both have simple points at A and have a common tangent there
and nowhere else.
b') Ll splits into two Hnes, one of which will be the tangent at A, if and only
if A is a point of inflection of r.
Establish (a) and (b).
Thus if A is not a point of inflection, one can draw precisely four distinct
tangents from A to r, not counting the tangent at A. If A is a point of inflection,
then one can draw precisely four, this time counting the tangent at A.
Note that no two of the intersections Qv Q2 of r and  can be collinear
with A, as otherwise QIQ2 would intersect r more than three times.
Appropriately interpreted, we have:
From every point A of r one can draw precisely four tangents to r.
Let P, Q be points on a nonsingular cubic r and consider the construc-
tion illustrated in Fig. 10.2. Through P draw a line, cutting r again in A
and A'; let QA cut r in the third point B, and QA' in the third point B'.
Then we say:
The lines BB' pass through a fixed point R on r.
Proof. Setting up a group on r, let 0 be the zero and K the third intersection
with r of the tangent at 0; by taking 0 to be a point of inflection, we may
also arrange to have K = o. Now by 9.7 we have
P + A + A' = K,
Q + A' + B' = K,
Q + A + B = K,
R + B + B' = K,

THE MODULUS OF A NONSINGULAR CUBIC 77
Fig. 10.2
from which one deduces P - Q = Q - R, whence R = 2Q - P is fixed.
Moreover,
P and R can be taken arbitrarily on r.
That is, we are to show that R + P = 2Q can be solved for Q. Write this
in the form: D + 2Q = O. This is equivalent to (D + K) + 2Q = K.
This comes to saying that D + K, Q, Q are collinear, i.e., that Q is the point
of contact with a tangent to r from D + K; we know that such tangents
exist. Q.E.D.
The above construction obviously sets up a one-to-one correspondence
between the lines on P and those on R; moreover A = A' if and only if
B = B', so that the tangents from P to r correspond to those from R to r.
We will prove that this correspondence is a projectivity, and then we shall
have:
Theorem 10.19. There is a projectivity between the pencils at P and R
such that the four tangents from P to r correspond to the four tangents
from R to r. Hence the cross-ratio of the four tangents from P (taken
in some order) is the same as the cross-ratio of the four tangents from R
(taken in some order).
We first give a rough indication of why the above correspondence is a
projectivity. Take a coordinate t for the lines of the pencil of center P, Le.,
let lo = 0, II = 0 be two lines through P and write the lines through P
in the form lo + tl l = o. Let t' be similarly defined for the pencil at R.
Now we can find the intersections of lo + tl l = 0 with r: F (X, Y) = 0;
eliminating Y, say, from F(X, Y) = 0 and lo(X, Y) + tll(X, Y) = 0, we
would get a cubic in X, yielding the abscissas of P, A, A' (see Fig. 10.2).
Since P is known, we have one of the roots of the cubic, and so have only to
solve a quadratic to get the other roots; this will introduce a square root
of an expression in t into the computations. Now we can compute the
coordinates of B and of B' (we only have to solve linear equations for this),

78 CUBICS (CONTINUED)
and then can write the equation for BB'. In this way we can write t' in
terms of t; a square root may enter the expression for t' in terms of t, but if
so, we eliminate it, and get a polynomial relation G(t, t') = O. The poly-
nomial G(T, T') may have multiple factors, but we keep each irreducible
factor only to the first power: in this way we get an equivalent condition
on (t, t'), and may suppose that G(T, T') is free of multiple factors. Consider
the curve G(T, T') = O. For each value of t, we get just one value of t'.
The values of t' are obtained from G(t, T') = O. Let G(T, T') be of degree
n in T'. Then, with the exception of a finite number of values of t, G(t, T')
is still of degree n in T'. Now G(t, T') = 0 may have multiple roots: this
will happen if and only if the line T - t = 0 is tangent to G(T, T') = 0 or
passes through one of its singularities (at finite distance). Thus avoiding a
finite number of further values of t, we get n distinct values for t'. Thus
n = 1. Similarly one argues that G(T, T') is of degree 1 in T. Thus
G(T, T') = 0 is of the form a + bT + cT' + dTT' = o. Then
-a - bt
t' =
c + dt ·
Here ad - bc =1= 0, otherwise t' would be constant. Hence the correspondence
is a projectivity.
The above argument, though quite suggestive, does contain a slippery
point or two, since the nonhomogeneous treatment does not take into account
all the elements of the situation. We give a precise proof. The proof will
show quite generally that if a one-to-one correspondence between the lines of
two pencils can be described in an algebro-geometric manner, then the correspon-
dence is a projectivity, though we stick, at subsidiary points, to our present
situation with a cubic.
We use homogeneous coordinates. Let P, Q, R, A, A', B, B' have
coordinates (Po, PI' P2)' (qo, qv q2)' etc. Let lo(Xo, Xl' X 2 ), ll(X o , Xv X 2 ) be
two linear forms yielding two (distinct) lines through P, so that every
line through P has an equation of the form tolo + tIll = O. Similarly let
l = 0, l = 0 be lines through R, so that every line through R has an
equation of the form tl + tl = O. Let (to, t l ) be the coordinates of the line
P AA' (i.e., P AA' is given by tolo + tIll = 0) and let (t, t) be the coordinates
of RBB'. Let r be given by F(X o , Xv X 2 ) = o. Now we write equations
mirroring the geometric facts:
(1)
F (a o , aI' a 2 ) = 0,
F (a, a, a) = 0,
(2)
tolo(a o , aI' a 2 ) + t l l 1 (a O ' aI' a 2 ) = 0,
tolo(a, a, a) + tIll (a, a, a) = O.
These equations say that A and A' are on the intersection of rand

THE MODULUS OF A NONSINGULAR CUBIC 79
tol o + tIll = 0, but they do not say that (P, A, A') is a complete set of inter-
sections of r and the line; for example, Eqs. (1) and (2) are both satisfied
by the triple (P, A, A), or even by (P, P, P). For a reason that will be
apparent in a moment, we do not wish to write inequalities. We proceed
as follows. Let a coordinate system be taken in such way that E 2 : (0, 0, 1)
and EI : (0, 1, 0) are not on r and such that P, Q, R are not on EIE2. Then
R X2 (F, tol o + tIll) = 0 yields the ratios Po: PI' a o : a v ab: a of the points
P, A, A' as solutions (see 7.5). Therefore we wish to say that this resultant is
const (POX I - PIXO)(aOX I - alXo)(abXI - aXo). Let u o , U I , . . . be the
various coefficients of R X : 2 = P(Xo, Xl) and let v o , Vv . . . be the correspond-
ing coefficients of Q(Xo, Xl) = (POX I - PIXO)(aOX I - alXo)(abXI - aXo).
We do not say that (u o , U v . . .) and (v o , Vv . . .) are equal, but only that
they are proportional. This is expressed by:
(3)
U o u l
= 0,
V o VI
U o U 2
= 0, etc.
V o V 2
The Ui are homogeneous and of the same degree (= deg F) in (to, t l ), and the
Vi are homogeneous of a degree 1 in a and in a'. Consider also
R X1 (F, tol o + tIll) = P'(X o , X 2 )
and
Q'(X o , X 2 ) = (P O X 2 - P2 X O)(a O X 2 - a 2 X o )(ab X 2 - a;X o ).
Let (ub, u,...) and (vb, v,...) be the coefficients and consider the
conditions:
(4)
U' u'
o 1
= 0,
U' u'
o 2
= 0, etc.
V' v'
o 1
Of course, we also have
V' v'
o 2
(5)
F (Po, PI' P2) = o.
Throughout we are speaking of points (a o , a v a 2 ), (Po, Pv P2)' etc., Le.,
not all the entries are zero; and the same goes for (to, t l ) and (tb, t).
The conditions (1) through (5), all equalities, say that (P, A, A') is a
complete set of intersections of rand tol o + tIll = O. Conditions (1), (2), (3),
and (5) do not quite do it, as one sees by considering the case that
tol o + tIll = 0 passes through E 2.
Write, now,
(6)
(7)
F(b) = 0,
tblb(b) + tl(b) = 0,
F(b') = 0,
tblb(b') + t l (b') = 0,
and, introducing coordinates t, t for the pencil at Q, write conditions (8)

80 CUBICS (CONTINUED)
and (9), like (3) and (4), saying that (Q, A, B) is a complete set of intersections
of line BA with r; similarly write conditions (10), (11) for line A'B' and
conditions (12), (13) for line BB'.
Statements (1) through (4) and (6) through (13) are a precise counterpart
of the construction leading to the correspondence P A -+ RB. [Statement (5)
and corresponding statements for Q, R need not be made.]
Now by the result quoted in (4.4), one can eliminate any of the unknowns
a, b, etc. In this way we can successively eliminate a, a', b, b', t". This yields
a condition
(14)
G(t o , t 1 ; t, t) = o.
This condition is equivalent to conditions (1) through (13): any solution
t, a, . . . , t' of (1) through (13) yields a solution (t, t') to (14); and for any
solution (t, t') of (14), there exist a,..., b' satisfying (1) through (13).
Thus the correspondence P A -+ RB is given by (14).
The polynomial G(T, T') is homogeneous in T = (To, T 1 ) and in
T' = (T, T). By 3.6, the irreducible factors of G(T, T') will also be homo-
geneous in T and in T'. Keeping each irreducible factor only to the first power,
we may suppose that G is without multiple factors.
Also, G is not divisible by To; otherwise for (to, t 1 ) = (0, 1), (14) would
have infinitely many solutions (t, t). Hence
deg T G(I, T; T, T) = deg(To,Tl)G(To, T 1 ; T, T).
Similarly,
deg T , G(I, T; 1, T') = deg(T,Tl)G(To' T 1 ; T, T).
By considerations similar to those in the last part of Chapter 3, one also sees
that G(I, T; 1, T') is without multiple factors. Moreover, any solution of
(15)
G(I, t; 1, t') = 0
yields a solution of (14). Hence by an argument already made in the pre-
liminary remarks, G(I, T; 1, T') is of degree 1 in T and of degree 1 in T'.
A similar statement can be made for G(T, T') and G(T, T') is of the form
a ToT + b TTl + c ToT + d TIT. Then (15) is
a + bt + et' + dtt' = 0,
and, as before, ad - be -::/:= O. Hence (14) yields a projectivity. Q.E.D.
Remark. The difficulty mentioned concerning the lack of generality in
proceeding nonhomogeneously can also be overcome by a device employed
in Chapter 9, as follows. One first writes the equation F = 0 for r homo-
geneously, Le., F = F(X o , Xv X 2 ), and one takes homogeneous coordinates
(to, t 1 ), (t, t) for the pencils at P and at R. Then one subjects r to a general
linear transformation. In the transformed situation one considers only

THE MODULUS OF A NONSINGULAR CUBIC 81
transformed elements, i.e., points and lines in the original plane P NO. The
transform of r and any transformed line will be in permissible position.
Let pU be the transform of P; then (to, t 1 ) are also coordinates in the pencil
at pu. We wish to pass to the nonhomogeneous coordinate tl/t o , but have to
face the exceptional case to == O. To nleet this one also subjects to, t 1 (and
similarly t, t) to a general linear transformation:
( t ) == ( v oo VOl ) ( to ) ;
t 1 V IO VII t l
and now tri =1= 0 for any transformed line. Now one dehomogenizes and uses
the argument given at the beginning. The result is that corresponding lines
(to, t 1 ), (t, t) satisfy a relation G(To, T l ; T, T) == 0 of the desired form,
except that G may have coefficients in an extension field 0 1 of O. To meet
this new point, one writes G in the form G == coG o + · · · + csG s , where
C i E 0 1 and the G i have coefficients in O. If a C i can be written linearly in
terms of the others with coefficients inO, one can eliminate it; in this way, we
may suppose that the C i are linearly independent over O. But then corre-
sponding lines (to, t 1 ), (t, t) satisfy all of the G i .

Chapter 11
EXPANSION AT A SIMPLE POINT
We start with some motivating remarks.
Let r : F (X, Y) == 0 be an irreducible curve having a simple point at
the origin with Y == 0 as the tangent. Then F(X, Y) == cY - G(X, Y),
c E C, and subd G > 2. Without loss of generality, we assume that c == 1.
Then any point on r satisfies
(1)
Y == G(X, Y).
If we substitute this expression for Y into the right-hand side, we get
(2)
Y == G(X, G(X, Y)),
which is also, clearly, satisfied by any point on r. Now in (1), Y appears
on the right only in terms of degree > 2, and hence in (2), Y appears on the
right only in terms of degree > 3. And we can repeat the process, getting
relations Y == Hn, which are satisfied by all points of r and in which Y
appears on the right only in terms of degree > n + 1. For example, starting
with Y == X3 + XY, we first get Y == X3 + X(X3 + XY), l.e.,
Y == X3 + X4 + X2Y; after another step we get
Y == X3 + X4 + X2(X3 + X4 + X2Y),
Le., Y == X3 + X4 + X5 + X6 + X4Y, etc.
Since Y - G(X, G(X, Y)) vanishes over r, we have
Y - G(X, G(X, Y)) == 0 (Y - G(X, Y)),
Le., Y - G(X, G(X, Y)) == (Y - G(X, Y)) · H(X, Y). Here H(X, Y) ==
1 + · · ., since Y occurs on the left, and hence must occur on the
right. The curve (2), then, consists of r plus, perhaps, a component not
passing through the origin. Thus (2) has exactly the same appearance
as (1) near the origin. And the same can be said for the various curves
Y == Hn obtained by repeating the process.
82

FORMAL POWER SERIES 83
Now roughly speaking, if Ixl and I yl are small, we can neglect higher-
degree terms relative to lower ones. For example, we could say of the curve
Y == X 3 + X4 + X5 + X6 + X4 Y by neglecting the terms of degree > 5,
that it is approximated by the curve Y == X 3 + X4 near the origin. If
we make the substitution mentioned, we get
Y == X3 + X4 + X5 + X6 + X4 · (X3 + X4 + X5 + X6 + X4Y),
and neglecting the terms of degree > 9, we could say that the original curve
(Y == X3 + XY) is approximated by Y == X3 + X4 + · · · + X8. But
without neglecting anything, if we repeated the process infinitely many
times and could use some kind of limiting process, we would say that
r: Y == X3 + X2Y is given near the origin by Y == X3 + X4 + · · ·
Surely, working with complex numbers, we would conjecture that
X3 + X4 + · · · is convergent near X == 0 and that Y == X3 + X4 + · · ·
gives the points of Y == X 3 + X 2 Y near the origin, and the conjecture
would be correct.
If we could replace Y == X3 + X2Y by Y == X3 + X4 + · .. in
investigating phenomena near the origin, it would be quite helpful. For
example, in studying intersection multiplicity, the resultant of G(X, Y) and
Y - X3 - X4 - · · · is easy to handle-it is simply G(X, X3 + X4 + · · .)
-the unwieldy determinants would be eliminated.
The above suggests that the proper tool for a study near the origin is
the power series-we should work with power series rather than polynomials.
It turns out that we need not be concerned with questions of convergence,
but can work formally with the power series. How this is done, we now
explain.
FORMAL POWER SERIES
Let K be a field. By K[ [X, Y]] we designate the set of objects of the form
F == a OO + alOX + aOIY + a 20 X2 + a1lXY + a 02 Y2 + .. ., au E K,
among which are defined the operations of sum and product according to the
rules:
(a OO + alOX + aOIY + ...) + (boo + blOX + bOlY + ...)
== (a oo + boo) + (a lO + blO)X + (a OI + bOl) Y + · · · ,
(a oo + alOX + aOIY + · · .) · (boo + blOX + bOIY + · · .)
== aoob oo + (aOOb lO + aloboo)X + · · ·
With these two operations K[ [X, Y]] becomes a commutative ring whose
elements are called formal power series.
It is customary to write the power series in the form
F == Fo + FI + F 2 + · · .,

84 EXPANSION AT A SIMPLE POINT
where F i is a homogeneous polynomial in X, Y of degree i; (also, if an
F i = 0, it may be left out of the above expression). Starting from
this convention, we define (Fo + FI + . · .) + (Go + G I + · ..) as
(Fo + Go) + (F I + G I ) + · .. and (Fo + FI + · · .) · (Go + G I + · · .)
as FoGo + (FOG I + FIG O ) + · .. If one has
F = Fr + Fr+l + · · · ,
Fr -#- 0,
then one calls r the subdegree or order of F. It is immediate that
ord (F + G) > min {ord F, ord G}
and
ord FG = ord F + ord G.
Taking into account this last property, one sees easily that K[ [X, Y]] is an
integral domain. One designates its quotient field by K( (X, Y)).
If one has a sequence of power series F i such that ord F i  00 as
i  00, one may define
00
2 F i E K[ [X, Y]]
i=l
as follows: Let
F i = F iO + Fil + · · · ,
with F ii homogeneous of degree j. Then we place
00
2 F i = (2 F iO ) + (2 F il ) + · · · ,
i=O
which makes sense since each sum 2i(F ii ) is finite by the hypothesis that
ord F i  00 as i  00.
Remark. If F = Fo + FI + · · ., then F = 2to F i in the sense just
defined.
One also has the properties
2 F i + 2 G i = 2(F i + G i ),
F 2 G i = 2FGi.
Theorem 11.1. The units of K[ [X, Y]] are the power series of order zero.
Proof. If FG = 1, then ord F + ord G = ord 1 = 0, whence orO. F = O.
Conversely, let
F = aoo + alOX + aOIY + · · · = a oo (1 - G),
where aoo #- 0 and ord G > O. Now 1 - G has 2i':O Gi as inverse, as one
easily proves using the properties mentioned above. Hence F has a oo 1 2i':O Gi
as Inverse.

FORMAL POWER SERIES 85
These definitions can be extended to any number of indeterminates; In
particular we consider K[ [t]]. Its elements #- 0 are of the form
F = a ta. + a ta.+l + · · ·
a. a.+l
(aa. *- 0, (l > 0)
and may be written as
F = ta.E(t),
where E(t) is a unit in K[ [t]] and (l > o.
Theorem 11.2. K[ [t]] is a unique factorization domain. The element t
is the only irreducible element (except for associates).
In fact, any element is of the form ta.E(t), E(t) a unit, and ifit is irreducible,
then (l = 1; moreover tE(t) and t are associates. Thus any complete
factorization of F has exactly (l = ord F factors of the form tEi(t), or of the
form t if we replace tEi(t) by an associate. Thus the only complete factoriza-
tion of ta.E(t) is t · . . . · t · E(t).
It is also true that K[ [X, Y]] is a unique factorization domain, but since
we will not use this property, we omit the proof. What is immediate is that
K[[X]][Y] and K((X))[Y] are unique factorization domains (according
to the known theorem that if R is a unique factorization domain, then so is
the ring R[X], X an indeterminate).
Let us examine the form of the (nonzero) elements of K( (t)). These are
F tlJE (t )
_ = 1 = tlJ-VE(t)
(J t V E 2 (t) ,
where E1(t), E 2 (t), E(t) are units. Placing ft - v = A and distributing t).
through E(t), one writes the above elements as a).t). + a).+lt).+l + · · · ,
a i E K, a). #- 0, A an integer. Thus K( (t)) - 0 is the set of elements of this
form. (For more than one indeterminate, the situation is not quite so simple.)
Theorem 11.3. Let F(X, Y) = 0 be the equation of a curve r having
the origin (0, 0) = 0 as simple point and whose tangent at 0 is not vertical.
Then there is one and only one power series
c1X + C 2 X2 + · · ·
such that
F(X, c1X + C 2 X2 + · · .) = o.
Proof. One has, from the hypothesis,
F (X, Y) = Y + a10X + · · · ,
of/oY = 1 + G(X, Y)
(where subd G > 0).

86 EXPANSION AT A SIMPLE POINT
We will first prove the uniqueness. Suppose that
c1X + · · · + Ci_lXi-1 + Ci Xi + · · · ,
c1X + · · · + Ci_IXi-1 + diXi + · · ·
satisfy the condition of the theorem. Expanding and taking into account that
of
== 1 + G(X, c i + · · · + Ci_lXi-I),
oY Y=c1X+".+Ci_1Xi-1
we have
o == F(X, c1X + ... + CiXi + ...)
== F(X, clX + · · · + Ci_lXi-l)
+ (1 + G(X, clX + · · · + Ci_IXi-l) )(CiXi + · · .)
+ terms of deg > i.
Writing the same for F(X, clX + · · · + Ci_IXi-1 + diXi + · ..) and
comparing the terms in Xi, we get C i == d i , which proves the uniqueness.
The existence will be proved by inductively constructing polynomials
c1X + · · · + CiXi such that ord F(X, clX + · · · + CiXi) > i. For i ==
1, it suffices to take C I == -alOe Let us suppose that clX + · · · + Ci_IXi-1
has been constructed. We will calculate C i . We write
F ( X C X + · · · + c. X i-I + c.X i ) == F ( X C X + · · · C. Xi-I )
, I t-l t , I t-l
+ [1 + G(X, C1X + ... + Ci_IXi-I)]CiXi + terms ofdeg > i
== (dXi + · · .) + CiXi + terms of deg > i.
We take Ci == -d. With this construction we assure that
F(X, clX + · . . + CiXi + · · .) == o.
In fact,
F(X, clX + · · · + Ci_IXi-1 + CiXi + · · -)
== F(X, clX + · · · + Ci_IXi-l) + Xi(.. .).
By the construction, F(X, c1X + · · · + Ci_IXi-l) is of order> 1, from
which ord F (X, c1X + · . .) > i, and since this holds for all i, one must have
F(X, clX + · · .) == o. Q.E.D.
Remark. Thus far, K is an arbitrary field. If K == C, then clX + C 2 X2 +
· .. has a positive radius of convergence. From the same construction
one sees that if K == k == the real field, the C i are real. As a corollary one
obtains: A simple point of a real curve is not isolated.

FORMAL POWER SERIES 87
As Y = c1X + C 2 X2 + · · · is a solution in K[[X]J of F(X, Y) = 0,
we have that
F(X, Y) = (Y - c1X - C 2 X2 - · · .)F1(X, Y), F1(X, Y) E k[[X]][Y].
Moreover F1(0, 0) =F 0, since 0 is simple on F = 0; that is, ord Fl = o.
Hence ord F(X, 0) = ord (c1X + · · .). Thus we have proved:
Theorem 11.4. i(r, (Y = 0); 0) = ord (c1X + C 2 X2 + · · .).
This can be generalized as follows:
Theorem 11.5. Let r: F(X, Y) = 0 and : G(X, Y) = 0 be two
curves without common component and let r have a simple point at the
origin with nonvertical tangent, so that (by 11.3) there is a unique power
series c1X + C 2 X2 + · · · such that F (X, c1X + · · .) = o. Then
i ( r . 0 ) - ord G ( X c X + · · . )
,,- , 1 .
We will suppose that r and  are in permissible position, although the
theorem holds without this hypothesis. This will suffice for studying
i(r, ; 0) as we first put r,  in permissible position and then apply the
theorem.
Proof
Ry(G(X, Y), F(X, Y)) = Ry(G, (Y - c1X - · · .) F 1 )
= Ry(G, (Y - c1X - .. .)) · Ry(G, F 1 )
= G(X, c1X + · · .) · Ry(G, F 1 ).
Thus it remains to show that Ry(G, F 1 ) =1= O(X), or that R(G, F 1 ) Ix=o =F 0,
or that R(G(O, Y), F1(0, Y)) =F o. Now F(O, Y) = YF1(0, Y) and hence
it will suffice to prove that
R(G(O, Y), F(O, Y)/Y) =1= 0,
which follows from the fact that G(O, Y) and F (0, Y)/ Y have no common
root (because of the permissible position of r and ) and the fact that
Y = 0 is not a root of F (0, Y)/ Y (because r has a simple point at 0 with
nonvertical tangent).
With this criterion we can give a proof of Lemma 9.2 of Chapter 9.
Take a permissible coordinate system for K: G(X, Y) = 0, r:
F(X, Y) = 0 in such way that P is the origin and L is t.he X-axis. The
case K = L is easily checked and we assume that K =F L. Then
G(X, Y) = dXIJ + · · · + Y (. · .),
d =F O.

88 EXPANSION AT A SIMPLE POINT
Let
Y = eX" + · · · , v > 0, e *- 0,
be the series satisfying F(X, Y) = O. Then
i(K, L; 0) = fl, i(r, L; 0) = v,
i(K, r; 0) = ord [dXIl + · · · + (eX" + · · .)(.. e)].
From this one sees at once that
if fl < v,
if ft > v,
This is Lemma 9.2.
then i(K, r; 0) = ft,
then i(K, r; 0) > v.

Chapter 12
BRANCHES
Definition 12.1. A branch representation is a pair of elements ((t), 1')(t))
in C( (t)) not both in C, or, in other words, a point of ANC( (t)), the
affine number plane over O((t)), that is not in ANO, the affine number
plane over O.
The branch representation ((t), 1')(t)) is given the homogeneous co-
ordinates (1, , 1]) and (p, p, P1]), where P E O((t)), P =F o. More generally,
one defines a branch representative (in the projective plane) as a point of
PNO((t)), the projective number plane over O((t)), that is not in PNO, the
projective number plane over O. A branch representation will, then, have
coordinates (o(t), l(t), 2(t)) and (po' Pl' P2)' where o(t), l(t), 2(t), p(t) E
C( (t)); and the P(t)i(t) will not all be in 0 for any p(t) #- o.
If (o' v 2) is a branch representation, we may multiply the i by a
power tV of t, and so obtain coordinates for the branch with order > 0 in t;
then dividing by an appropriate power of t, we can get coordinates at least
one of which has order o. Thus every branch representation has coordinates
of the form (a o + · · · , a l + · · · , a 2 + · · .), where a i E 0, (a o , a v a 2 ) E P NO,
and the omitted terms are of positive order; such coordinates will be called
special. If (o' v 2) are special, then (po' pv P2) are also special if and only
if p is a unit in O[[tJ]. Assuming that (o' l' 2) is special, we call the point
(o(O), l(O), 2(0)) the center of the branch representation.
If one of the i is 1 and ord i > 0 for every i, then the coordinates are
called normalized. Every branch representation has at least one and at most
three sets of normalized coordinates.
If (o(t), l(t), 2(t)) are special coordinates of a branch representation
centered at finite distance (relative to Xo == 0 as loo)' then o(O) #- 0 ando (t)
is a unit. Thus the branch representation has affine coordinates (t), 1](t)
with ord  > 0, ord 1] > 0; its center is ((O), 1](0)).
Definition 12.2. A branch representation (t), 1](t) is called imprimitive
or not primitive if both are power series in a power series 'T of order> 1.
89

90 BRANCHES
In the contrary case, the representation is called primitive. A branch
representation (o, 1' o) is called primitive if its normalized coordinates
yield a primitive branch representation in the sense just defined.
If ((t), 17(t)) is a branch representation and , 17 can both be written as
power series in a power series T of order 'V > 1, (t) = 1(T), 17(t) = 17l(T),
then one says that ((t), 17(t)) is of redundancy at least v. The reason for this
terminology is that we regard 1 (T), 171 (T) as simpler expressions for the branch
representation. Technically one allows the value v = 1 also. We leave it
as an exercise to show that there is an integer v > 1 such that (t), 17(t) can
be written in terms of a power series T of order 'V but not in terms of a power
series of order >'1'; if (t) = 1(T), 17(t) = 17l(T), then (I(T), 17l(T)) is primitive.
The redundancy of (t), 17(t) is by definition this number v. A branch represen-
tation is primitive if and only if its redundancy is 1.
If in a branch representation (t), 17(t) we substitute for t the power
series T = ct + · · · , c *- 0, of order 1, then in general, of course, we get
a different branch representation (ct + · · .), 17(ct + · · .); but we wish
to regard this as merely a different expression for the same thing. This leads
us to study the substitutions of order 1.
Theorem 12.3. Every substitution t  T of order 1 yields an auto-
morphism of C[[t]] over C, and conversely.
Proof. Let T = ct V + · · · E C[ [t]], v > 1, c *- O. Consider the substitution
(j : t  T, Le., f(t)  f(T) for f(t) E C[ [t]]. If f(t) *- 0, then
f = d t s + d t s + 1 + · · ·
s s+1 ,
8 > 0, d s *- 0;
and
f(T} = ds(ct V + · · .)s + d s + 1 (ct V + · · .)s+1 + · · · = dsct VS + · · . ,
and as dsc *- 0, we have f(T} *- O. Therefore the substitution (j : t  T is
an isomorphism ofC[[t]] onto G[[T]] c C[[t]]. We will show that if v = I,
then (j is an automorphism. To show this we have to show that
C[[T]] = C[[t]];
and for this it will be sufficient to show that t E C[[T]]. If (j sends c 1 t +
c 2 t 2 + · · · into t, then
c 1 (ct + · · .) + C 2 (ct + · · .) 2 + · · · = t,
and conversely. Therefore it is sufficient (and necessary) to find C v c 2 , . . .
such that
c 1 (ct + · · .) + C 2 (ct + . · .) 2 + · · · = t.
Comparing the coefficients of t, t 2 , . . . on both sides, we see that the above

BRANCHES 91
equality is equivalent to the following infinite sequence of equalities:
C 1 C = 1,
C 2 C 2 + c 1 (. .) = 0,
c a c 3 + c 2 (. .) + c 1 (. .) = 0,
.
Ci Ci + C i - 1 (. .) + · · · + c 1 (. .) = 0,
where the omitted expressions are built up from the coefficients of ct + · · ·
These equalities introduce c 1 ' c 2 , . . . successfully, and as the coefficient of
C i in the ith equality, the one that introduces C i , is *- 0, we can solve for the
c i successively.
For the converse, let (J be an automorphism of the stated kind. (J trans-
forms units into units (every automorphism of a ring does this). Hence (J
transforms elements of order 0 into elements of order 0, and (J transforms
nonunits into nonunits (every automorphism of a ring does this). Hence (J
transforms elements of positive order into elements of positive order. (J
transforms a product of exactly v irreducible elements into a product of
exactly v irreducible elements (every automorphism of a ring does this).
Hence (J transforms elements of order 'V into elements of order v. In particular,
(J : t -+ 'T = ct + · · · , c*-O (c E C). Hence
m m
(J : ! Citi -+ ! Ci'T i ,
i=O i=O
because (J is an automorphism over C. However, we still have to prove
00 00
(J : ! Citi -+ ! Ci'T i ,
i=O i=O
as, in general, the definition of an automorphism says nothing about infinite
sums. One reasons as follows:
00 m
! Ci ti = ! cit i + Rm,
i=O i=O
where Rm E C[[t]] has order> m. Hence
(J ( . Cit i ) = (J ( .! Cit i ) + (J(Rm)
=o =o
m
= ! Ci'T i + an element of order > m
i=O
00
= ! Ci'T i + an element of order > m.
1=0

92 BRANCHES
Hence
a ( .! Cl i )
'1,=0
00
- ! C{T i = an element of order> m for arbitrary m,
i=O
whence
a ( .! Cit i )
'1,=0
00
= ! C{T i .
i=O
Definition 12.4. On the basis of Theorem 12.3 we define two branch
representations to be equivalent if one is obtained from the other by a
substitution of order 1 [so that, in particular, if (, 'Y)) is equivalent to
(', 'Y)'), then (', 'Y)') is equivalent to (, 'Y))]. A branch is a class of equiva-
lent primitive branch representations.
The center of a branch is the center of anyone of its associated special
representa tions.
Finally, let us examine the action of a transformation LA on a branch.
Let A = (ai1)' aij E C. It is immediate that:
a) L A transforms branch representations into branch representations, and,
moreover, primitive ones into primitive ones and nonprimitive ones
into nonprimitive ones. Thus the notion of a primitive branch represen-
tation is proj ecti ve.
b) LA also transforms equivalent branch representations into equivalent
branch representations. Hence LA transforms branches into branches,
and the notion of a branch is projective.
Theorem 12.5. The center of a branch of a curve r is a point of r.
(By a branch of r one means, of course, a branch whose representatives
satisfy the equation of r.)
Let G(Xo, Xv X 2 ) = 0 be the equation of r and let (o(t), l(t), 2(t))
be a special representation of the given branch. Then
G((o(t), l(t), 2(t)) = o.
Placing t = 0, one finds that G( o(o), 1(0), 2(0)) = O. Q.E.D.
We have already encountered branch representations (in Theorem 11.3).
Thus if r : F (X, Y) = 0 has a simple point at (0, 0) with nonvertical tangent,
then there is a power series c1X + C 2 X2 + · · · such that
F(X, c1X + C 2 X 2 + · · .) = O.
Writing  = t, 'Y) = c1t + c 2 t 2 + · · ., one sees that (, 'YJ) is a branch
representation of r centered at the origin. This can be rephrased as part
of the following theorem.

BRANCHES 93
Theorem 12.6. There is one and only one branch (of r) centered at a
simple point of a curve r.
Proof. With the notation just used,  = t, 'Yj = cit + c 2 t 2 + · .. is ob-
viously primitive and yields a branch of r centered at the origin. Conversely,
let (t), 'Yj(t) be a primitive branch representation centered at (0, 0), so that
ord  > 0, ord 'Yj > 0, and let F (, 'Yj) = O. Since
F(X, clX + C 2 X2 + ...) = 0,
one has the factorization F(X, Y) = (Y - clX - C 2 X2 - · · .)H(X, Y)
in C[[X]][Y], where H(X, Y) has a nonzero constant term, since F contains
linear terms. In C[ [t]] we can write
F (, 'Yj) = ('Yj - CI - C22 - · · · )H(, 'Yj) = o.
Here H(, 'Yj) is a unit in C[[t]] and one concludes that 'Yj = CI + C22 + · · ·
Then  must be of order 1 (as otherwise (, 'Yj) would not be primitive). The
substitution a: t   is a substitution of order 1, hence is an automorphism,
of C[[t]] over C; its inverse, a-I, is also an automorphism, hence produced
by a substitution of order 1. Then a-I carries (, 'Yj) into (t, clt + c 2 t 2
+ · · .), Le., into the branch representation considered at the beginning of
the proof. Thus this gives the only branch of r centered at the origin.
Let y be a branch and let (o(t), I(t), 2(t)) be special coordinates for y;
by the definition of branch, the representation (o' v 2) is primitive. Let d
be a curve, or more generally a cycle, of equation G(Xo, Xv X 2 ) = o. Let
P be the center of y.
Definition 12.7. i(d, y; P) = ord t G(o(t), l(t), 2(t)).
If (po' Pl' P2) are also special, then P is a unit and
ord t G(o' l' 2) = ord t G(po' pv P2).
To compute i(d, y; P), then, one may suppose that (o' I' 2) are normalized
coordinates. The order does not change under substitutions of order 1, and
thus i(d, y; P) is seen to depend only on y, not on the representation. Note,
also, that the definition is of a projective character. In fact,
LA:G(X) = 0  G(A-IX) = 0,
and
LA: ()  A ( ) .
Moreover G(X)lx= = G(A-IX)lx=A()' whence the order is preserved.
If G(o' I' 2) = 0, we place i(d, y; P) = 00. We also write
ord y G(Xo, Xl' X 2 ) for ord t G(o' I' 2). A simple reformulation of 11.5 gives:
Theorem 12.8. Let y be the branch of a curve r centered at a simple point
P of r and let d be another curve. Then
i(d, y; P) = i(d, r; P).

94 BRANCHES
More generally, the following theorem connects i(d, y; P) with our
previous intersection multiplicity for curves at a point:
Let r : F (X o , Xv X 2 ) = 0 be an irreducible curve, d : G(Xo, Xl' X 2 ) = 0,
a curve not having r as a component, and P a point of r. Then
i(d, r; P) = ! i(d, y; P),
y
where y runs over the set of branches of r having center at P.
We shall prove this theorem in Chapter 21, and meanwhile make no use
of it.
Let y be a branch, of center P. By the multiplicity r of y (at P) one
means the minimum intersection multiplicity with y of lines l through
P: r = min les {i(l, y; P)}, where S is the set of lines through P. Assuming
without loss of generality that P = 0 = (0, 0) and that y is given by
 = ct A + · · · , 'YJ = dtl-' + · · · , cd =1= 0 (A. > 0, ft > 0), one sees that
r = min {A., ft}.
If A < ft, then the line Y = 0 and no other cuts y with multiplicity> r;
similarly if ft < A.. If A. = ft, then the line dX - cY = 0 and no other
cuts y with multiplicity> r. This exceptional line is called the tangent of y.
(If the tangent meets y with multiplicity r + s, then (r, s) is called the
characteristic of y. This can be useful in classifying branches, but we shall
not have occasion to refer to it.)
If the multiplicity r is 1, the branch is called linear. Thus the branches
(of r) centered at simple points of a curve r are linear.
Theorem 12.9. Two distinct irreducible curves r,  cannot have a
common branch.
Proof. The proof of Theorem 7.4 shows that even if one allows coordinates
from a field larger than C, nevertheless the common points of r,  must have
coordinates in C. As a branch representation by definition is not in PNO
(or ANC), no branch representation can satisfy the equations of distinct
irreducible curves.
Corollary 12.10. If (, 'YJ) is a branch and  = 0, then the branch is a
branch of X = 0 and of no other irreducible curve.
Remark. Having explained the relation between branch representation and
branch, we will now use the expression "the branch (, 'YJ)" instead of "the
branch having (, 'YJ) as branch representation."
So far, we still do not know whether there are any branches centered at
a singular point of a curve. To analyze a singular point, which we may suppose

BRANCHES 95
to be the origin (0, 0), we are going to apply transformations of the form
(1)
X' = X,
Y' = Y/X,
so-called locally quadratic transformations. The point (0, 0) is called the
center of the transformation (1).
We start with some orienting remarks. The transformation (1) is, of
course, not defined for x = 0, or at least if x = 0 and y = 0 (if x = 0 and
y =f. 0, we could consider the transformed point to be E 2 , the point at infinity
on the Y-axis). Nevertheless, (1) has a definite action near (0, 0). Suppose
that we let a point (x, y) approach (0, 0) along the line Y = mX; then the
transformed point moves along the line Y' = m (or, dropping the prime,
Y = m). We would expect that a curve passing through the origin and
having Y = mX as tangent would be transformed into a curve through the
point (0, m) (and this is correct). Roughly speaking, we can say that there
is a one-to-one correspondence between the directions about the origin and the
points of the Y-axis.
Let (t), 'Yj(t) be a branch centered at the origin and assume that  =f. o.
Then (1) is defined at (, 'Yj), the transform being (, 'Yj / ) . We call the branch
(, 'YJ/) the transform of the branch (, 'Yj) under (1). We can make the content
of our orienting remarks formal by:
Theorem 12.11. The locally quadratic transformation (1) sets up a
one-to-one correspondence between the branches centered at the origin and
having nonvertical tangents and the branches centered at finite distance on
the Y-axis.
The proof is rather immediate, and we omit it.
Let
F(X, Y) = Fr(X, Y) + Fr+1(X, Y) + ... ,
Fr =f. 0,
and let r : F = 0 be a curve (or cycle) not necessarily through the origin.
Applying (1), consider
F(X, XY') = Fr(X, XY') + Fr+1(X, XY') + · · ·
= xr[F r (l, Y') + XF r + 1 (1, Y') + · · .].
One sees that F(X, XY') is exactly divisible by xr in O[X,Y']. We define
r': F(X, xy)/xr = 0 as the transform of r: F = 0 under (1). Note:
12.11.1. If r is irreducible ana. not the line X = 0, then the transform r'
of r is irreducible.
Proof. We are given that F(X, Y) is irreducible and not divisible by X
and are to prove that F'(X, Y') = F(X, XY')/xr is irreducible (in
C[X, Y']). At any rate, F'(X, Y') is not divisible by a (nonconstant)

96 BRANCHES
polynomial h(X); for if it were, replacing Y' by Y IX and multiplying by a
power of X, we would find that XPF(X, Y) is divisible by h(X) in O[X, Y].
Thus h(X) = cX(J, CEO. As F' (X, Y') is not divisible by X, (J = 0, proving
the point mentioned. Hence if F'(X, Y') is reducible, it has two factors of
positive degree in Y'. Replacing Y' by Y IX and multiplying by a powet;
of X, we would find that XPF(X, Y) has two factors of positive degree in Y,
and this is not so.
As a corollary we have:
12.11.2. Let r v . . . , r s be irreducible and distinct from X = o. Then
the transform of m 1 r 1 + · · · + msr s is m 1 r + · · · + msr;, where r'
is the transform of r i.
Theorem 12.12. Let r be a curve not having X = 0 as tangent at (0, 0).
Let r' be the transform of r under (1). Then there is a one-to-one corre-
spondence between the branches of r centered at (0, 0) and the branches of r'
centered at finite distance on the Y-axis.
Proof. Neither r nor r' has X = 0 as component, so that all branches in
question have first coordinate =1= 0 (see 12.10). Moreover if (, 'YJ) satisfies
F (X, Y) = 0, then (, 'YJ/) satisfies F (X, XY) = 0 and F (X, XY)lxr = 0,
and conversely; so (1) sets up a one-to-one transformation between the
branches of i-' and those of r'. Observe now that r has no branch (, 'YJ) at
the origin with vertical tangent, for if so, then ord  > ord 'YJ > 0; the
assumption that X = 0 is not a tangent to Fr + Fr+l + · · · = 0 can be
expressed by saying that Fr has a term yr with nonzero coefficient:
Fr(X, Y) = coxr + c1xr-lY + ... + cryr, c r =1= o. Then
ord Fr(' 'YJ) = ord 'YJr = r · ord 'YJ;
and
ord Fr+l(' 'YJ) > (r + 1) ord 'YJ, etc.
Hence ord F (, 'YJ) = r · ord 'YJ. This is impossible, since F (, 'YJ) = 0 and
ord F(, 'YJ) = 00. In view of Theorem 12.11, the proof is complete.
Theorem 12.13. At an ordinary r-fold point 0 of r : F (X, Y) = 0 there
are centered exactly r branches (of r). These are linear, and their tangents
are the same as those of r at o.
Proof. We may suppose that 0 = (0, 0) and that
r
F(X, Y) = II (Y - miX) + ...
i=l
Then r', the transform of r under (1), is given by F' = 0, where
r
F' = II (Y - mi) + X(.. .).
i=l

BRANCHES 97
Then X == 0 cuts r' in the points (0, m l ), . . . , (0, m r ), at each with multi-
plicity 1, since the m i are distinct; hence each of these points is simple for
r'. Thus r' has exactly r branches centered at finite distance on X == 0;
therefore r has exactly r branches centered at (0, 0). According to 11.3 or
12.6, the branch of r' at (0, m i ) is given by X == t, Y == m i + ... The
corresponding branch of r is X == t, Y == m/ + · · .; its tangent is
Y - m'iX == o. The proof is now complete.
Using locally quadratic transformations we can prove:
Theorem 12.14. At every point P of a curve r: F(X, Y) == 0 there is
centered at least one and at most a finite number of branches (of r).
Proof. We may suppose F to be irreducible, as every branch of r annihilates
some irreducible factor of F. We may also suppose that P == (0,0) and that
r does not have X == 0 as tangent at P. Let P be r-fold for r, so that
p"1 == IIi = 1 (Y - miX) + · · · , where the mi are not necessarily distinct.
Applying (1), we find that the transformed curve r' is given by
r
F' == IT (Y - m i ) + X(.. .).
i=l
Then (as one sees by placing X == 0 in F'):
12.14.1. The sum of the intersection multiplicities of X == 0 with r'
at the points (0, mi) is r.
If at least two of the m i are distinct, then each of the points (0, m i )
of r' is of multiplicity less than r. In such a case we could apply induction
on r (we have the theorem for r == 1); by induction, there would be at least
one branch and at most a finite number centered at each point (0, m i ) and
so, by 12.12, there would be at least one and at most a finite number of
branches of r centered at P. But we have to consider the possibility that
all the m i are equal, or, what is the same, that P is r-fold with only one
tangent: F == (Y - mlX)r + . · · , r > 1. In that case, we apply (1) to
get F'(X, Y) == (Y' - m1t + X(.. .); we make the translation X" == X,
Y" == ' - m l to get Y"r + X(. · -) == 0 as the equation of r', and then
apply the argument again. But Y"r + X(...) == 0 may also have an
r-fold point at the origin with only one tangent, .so we are faced again with
the same situation. Note, however, that if such an r-fold point occurs, then
at any rate the tangent at the origin is not X == 0 (since a term Y"r is present),
and hence we do not have to make preliminary changes in the coordinate
system before applying (1). Rearranging, we get (Y" - m 2 Xt + · · · == 0
as an equation of r'. Applying a transformation of type (1), we get
(Y'" - m 2 )r + X (. · .) == 0 as the equation of the transformed curve r (2);
then we make a translation to (0, m 2 ), etc. The question then is: Can we
get in this way an infinite sequence of r-fold points?

98 BRANCHES
The answer is no, so we finally get a reduction of r; for suppose the answer
to be yes: Then define m 3 , m4'... by means of r(2), r(3),... just as
m l and m 2 were defined by rand r'. Write F(X, Y) in the form
(*) F(X, Y) == I cabXa(Y - mlX - m 2 X2 - · .. - miXi)b + Xh,pi(X),
b>O
Cab E C, Pi E C[X], Pi(O) =1= 0; this can be done in one and only one way.
Then a + b > r: for replace Y - mIX by Y l in F and consider the terms
in which a + b is minimal; the ab- term yields Xa Y, which cannot cancel
from another term. Since F(X, Y I + mlX) == 0 has an r-fold point at
(0, 0), we must have a + b > r. Similarly hi > r. Then
F'(X, Y') == I CabXa+b-r(y' - m i - m 2 X - .. · - miXi-I)b
b>O
+ Xhi-rPi(X),
which is just the form (*) for F'. Hence h i ( F) == h i - l (F') + r. Repeating
the argument, we see that hi(F) == h i - 2 (F") + 2r; in this way one finds that
hi(F) > ire Hence hi  00 as i  00.
Then F (X, mIX + m 2 X2 + · · .) == 0; for substituting Y == Itl miXi
in (*), one finds ord x F(X, mIX + m 2 X2 + · · .) > min {i + 1, hi} for
every i. Hence ord F(X, mIX + · · .) == 00, F(X, mIX + · · .) == 0 and
F(X, Y) == (Y - mIX - m 2 X2 - .. .)G(X, Y),
where G E C[[X]][Y]. We have, then, a branch y of r centered at (0, 0),
namely,  == t, 'YJ == mit + · .. The first point of 12.14 is therefore proved,
but for the second we establish:
12.15. The sequence of locally quadratic transformations and translations
occurring in the above proof cannot yield a sequence of r-fold points with
r > 1.
Proof. Since r is irreducible, the branch  == t, 'YJ == mit · · · is not a branch
of d: 8F j8Y == 0 (by 12.9). Let d': 8F'j8Y' == 0 (so that d' has the same
relation to r' as d to r). Now
8F(X, Y) j xr-I == 8F(X, XY') . 8Y' j x r-I
8Y y=Xy' 8Y' 8Y
= o, (r F(X, XYI)) ,
whence
i(d, y; 0) == i(d', y'; 0') + r - 1,
where 0, 0' are the centers of y, y'. Hence, if r > 1,
i(d, y; 0) > i(d', y'; 0') > O.

BRANCHES 99
Repeating the argument, we get an infinitely decreasing sequence of positive
integers, which is impossible.
Corollary 12.15.1. Let r,  have no common component through P,
or equivalently, no common branch centered at P. Then!y i(, y; P),
where y runs over the branches of r centered at P, is a finite number.
We write !y i(, y; P) = j(, r; P) and can refer to this temporarily
as the j-multiplicity of  and r at P. As already remarked
j(, r; P) = i(, r; P).
Remark. For the moment, we do not know thatj(, r; P) = j(r,; P).
Notation. Instead of i(, y; P) we also write ord y F, where : F = O.
The above device is the basis of a general inductive method: if one wishes
to prove something about the branches of r at (0, 0), one applies a locally
quadratic transformation (1) to r; by induction one assumes the desired
properties of the branches of r' at the transformed points of (0, 0), i.e., the
points (0, mi), and deduces the desired result for r.
In this connection we mention the following as an exercise: Let y be a
branch of r at (0, 0). Then, first,
{ 8F 8F }
It = min ord y -, ord)I -
8X oY
IS invariant under transformations X' = aX + bY, Y' = eX + dY,
ad - be =1= 0 (hence the axes can be rearranged without disturbing ft).
Second, assuming that X = 0 is not the tangent of y, if y' is the branch of
r' corresponding to y under (1) and
{ 8F' 8F' }
11.' = min ordJ' -, ordJ' - ,
r r oX' r 8Y'
then ft' < -(r - 1) ord X + ft, where (0, 0) is r-fold for r. Thus ft is
reduced if r > 1.
We have already remarked that
(*)
i(r,; P) = j(r,; P).
As a consequence, j(r,; P) has the following properties:
12.15.2. a) j(r,; P) is a projective invariant.
b) j( Y = 0, F(X, Y) = 0; (0,0)) = ord x F(X, 0) (i.e., for a line
l, j (l, A; P) = i( l, ; P) ) .
c) j(r,; P) = j(, r; P).
d) Let r,  have r- and s-fold points, respectively, at P. Then
j(r,; P) > rs and equality holds if and only if r,  do not have
a common tangent at P.

100 BRANCHES
e) Bezout's Theorem holds for j(r, d; P), Le.,
!j(r, d; P) == ord r · ord d,
p
where r, dare assun1ed to have only a finite number of points in
common and the !p is over these common points.
However, we are going to prove these separately. This will involve some
duplication of effort, but we prefer this as we do not wish to make the
equivalence (*) central in our exposition. Indeed, once we have properties
(a) through (e) we could really dispense with i(r, d; P) altogether (though,
because of existing literature, one would still want to prove the equivalence).
a) This we already have; see the paragraph following 12.7.
b) Because of (a) we may suppose that the Y-axis is not tangent to
F(X, Y) == 0 at (0,0). Then F(X, Y) == IIi=l(Y - miX) +..., where we
have dismissed the trivial case r == 0; we also have (b) for r == 1, since we
know (by 12.8) that (*) holds for P a simple point of d; so we suppose
r > 1. We apply the transformation X' == X, Y' == Y IX. The cycle
: F(X, Y) == 0 goes over into d' : F'(X, Y') == 0, where
F'(X, Y') == F(X, XY')/xr.
The various branches y of d centered at (0, 0) go over into the varIOUS
branches y' of Ll' centered at the points Pi : (0, mi). Then
j(Y == 0, F(X, Y) == 0; (0,0)) ===! ord" Y ==! ord y ' X +! ord y ' Y',
y y' y'
and if we could apply (b) to d' at the points Pi' then (using 12.14.1) this
would
== r + ord x F'(X, 0) == ord x F(X, 0).
Now if at least two of the mi are distinct, then the points Pi are ri-fold for
Ll' with r < r, and we can apply induction on r. Even if there is only one
point Pi' if r' < r we can do this. If there is only one Pi and r' === r, we
apply the 10cally quadratic transformation X" === X', Y" == (Y' - mt)/X,
etc.; eventually r is reduced. We can make an induction on the (mini-
mum) number of steps needed to reduce r. The proof is thus complete.
c) To prove (c) we first establish:
Theorem 12.16. Let r : F == 0, d: G == 0 be curves (or cycles) having
r- and s-fold points at P : (0, 0) and assume that neither r nor d is tangent
there to X === O. Let r', d' be their transforms under the locally quadratic
transformation X' === X, Y' == Y I X, and let P, . . . , P; be the points
of intersection of r' and d' on the Y-axis at finite distance. Then
j (r, d; P) === r s + ! = 1 j (r', d'; Pi) .

IMPRIMITIVE BRANCH REPRESENTATIONS 101
Proof
j(r,; P) === ! ord y F(X, Y) === ! ord y ' XrF'(X, Y')
y y'
=== r ! ord y ' X + ! ord y ' F'(X, Y')
y' y'
t
=== r8 + !j(r', '; Pi).
i=l
This proves the theorem. As for (c), using the inductive scheme of (b), we
may apply (c) to r', ' at the Pi. Then
j(r, ; P) === rs + !j(r', '; Pi) === rs + ! j(', r'; Pi) === .j(, 1-'; P),
Pi Pi
and the proof of (c) is complete. (The transformations employed separate
the branches of r + : in the first step of the induction r and  do not
meet at (0, 0), and both sides of (c) obviously ===0.)
d) If F === IIi=l(Y - miX) + · · ., G === IIi=l(Y - niX) + .. ., then
the branches of F === 0 and G === 0 go over into the branches of F' === 0 and
G' === 0 centered at the points (0, mi), (0, nj). The point (d) is now a corollary
of 12.16.
The proof of (e) is postponed to Chapter 14.
1M PRIMITIVE BRANCH REPRESENTATIONS
Let x(t), y(t) be a branch representation of redundancy v > 1. Then we can
rewrite x(t), y(t) in terms of a power series T === ct V + · · · :
x(t) === X 1 (T),
y(t) === Yl(T),
and X 1 (T), Yl(T) will be primitive. In this way we can get a branch; but T
is not unique, and we should make sure that any such procedure leads to the
same branch.
Theorem 12.17. Let x(t), y(t) be a branch representation, not necessarily
primitive. Let T be an element of O{x(t), y(t)) of minimal positive order, g.
Then any element' of O{x(t), y(t)) has order divisible by g,. and' can be
expanded into a power series in T (with coefficients in 0). Hence if x(t),
y(t) is primitive, there is an element in C{ x(t), y(t)) of order 1.
Proof. Let' be of order ft and write # in the form # === qg + r, 0 < r < g.
Then '/T q is in C( x(t), y(t)) and has order r, whence r === 0 (from the minimal
condition on T) and the first point is proved. We have: '/T q === c + '1'
where CEO, C -=F 0, and '1 is a power series in t of positive order. The element
'1 === ('/T q ) - C, so '1 E C{x(t), y(t)). Applying the argument again, we have
'1/Tq1 == C 1 + '2 with ql > o. Now we have , === CT q + C 1 T q + Q1 + T q1 q1'2'
and repeating we expand' into a power series in T. This proves the second

102 BRANCHES
point. In particular, x(t) and y(t) can be expanded in terms of T, so if x(t),
y(t) is primitive T must be of order 1, and the proof is complete.
Corollary 12.18. Let x(t), y(t) be of redundancy v > 1 and let
x(t) == X1(T 1 ) == X 2 (T2)'
y(t) == Yl(Tl) == Y2(T2)'
where Tl, T2 are power series of order v in t. Then Tl is a power series of
order 1 in T2.
Proof. Let T be an element of least order (in Tl) in O(X 1 (Tl)' Yl(Tl)). Since
X1(Tl)' Yl(Tl) is primitive, T is of order 1 in TI : T == CTI + · .. By 12.3, Tl
is a power series in T. Now T is a rational function of xl' Yv hence also, of
course, of X 2 (T2)' Y2(T2)' so T is a power series in T2, of positive order, in fact,
of order 1, since T and T 2 are both of order v in t. Hence T 1 is a power series
in T2; since, similarly, T2 is a power series in TV each is of order 1 in the other.
The corollary establishes the point mentioned at the beginning of this
section. We should still show:
12.19. Equivalent branch representations have the same redundancy, and
the associated primitive branch representations are equivalent.
Since the proof is now straightforward, we omit further details.
Exercises
1. Show that Y2 = X3 has just one branch centered at the origin and that
y2 = X4 + X 5 has just two.
2. Find the first few terms in the expansions for the two branches of
XY - X2 + y3 = 0
centered at the origin.
3. Examine the action of X' = X and y I = y / X on the branches centered
on Zoo-
4. How many branches does y3 - X3y + X5 = 0 have (centered) at (0, O)?
Find the intersection multiplicity of Y = 0 with each of them.
5. I.Jet x(t), y(t) be a branch representation, and let x(t) = x'(7"), y(t) = Y'(7"),
7" = ctJ.l + . . . , c i:- 0, ft > O. Assume x'(7"), Y'(7") is primitive. Then the
redundancy of x(t), y(t) is ft.
6. Show that an automorphism of C((t))/C induces an automorphism of
C[[t]]/C.

Chapter 13
GENERIC POINTS
By a point we will now mean any pair (x, y) of elements x, y belonging to an
arbitrary field containing C. (We have, indeed, already encountered such
points, namely, the branch representations.) One will say that a point is
constant, if both its coordinates are in C, and variable otherwise. By a curve
we still mean t:be set of points satisfying a polynomial with coefficients in C.
Everything that has been said up to now holds also for points in the new sense
(with perhaps slight changes in the proofs).
Definition. A point (x, y) is called a generic or general point of the curve
r, of minimal equation F (X, Y) = 0, if G(x, y) = 0 with
G(X, Y) E C[X, Y]
implies G(a, b) = 0 for every point (a, b) of r. In other words, (x, y) is
generic if G(x, y) = 0 implies G == O(F).
Theorem 13.1. Reducible curves do not have generic points.
(Hence when we speak of a generic point of a curve, it is tacitly under-
stood that the curve is irreducible.)
Proof. Let r u Ll (r q: Ll, Ll q: r) be a reducible curve. Let F = 0 be
an equation (not necessarily minimal) for r, and G = 0 an equation for Ll.
Then FG = 0 is an equation for r u d; here we use that the coordinates
of any point (a, b) are in an extension field ofC, as from F(a, b) · G(a, b) = 0
we want to conclude that F(a, b) = 0 or G(a, b) = o. Now suppose r u Ll
has a generic point (x, y). Then F(x, y) = 0 or G(x, y) = 0, say F(x, y) = O.
Then by definition F(X, Y) vanishes over r u Ll, so that Ll is part of
F (X, Y) = 0, Le., of r; this contradicts the hypothesis.
Theore,,,,13.2. Every variable point (x, y) of an irreducible curve r is
a generic point of the curve.
103

104 GENERIC POINTS
Proof. Let F(X, Y) = 0 be the minimal equation for r and suppose that
G(x,y) = ObutthatG(X, Y) ¥= O(F(X, Y)),GEC[X, Y]. Asinaprevious
theorem, one shows that G = 0 and F = 0, considered as curves in ANC(x, y),
cut in only a finite number of points, all of which are in ANC. Hence (x, y)
is in ANC, which contradicts the hypothesis.
Definition. Two points (x, y), (xv Y1) are said to be isomorphic over C
if there is an isomorphism of C[ x, y] over C sending x into Xl and y into Yl.
Theorem 13.3. Any two generic points (x, y), (xv Yl) of a curve rare
isomorphic over c.
Proof. We want to define a mapping a of C[x, y] into C[x v YI] by placing
a(G(x, y)) = G(x v YI)' where G is a polynomial in C[X, Y] but we have to
check that if G(x, y) = H(x, y), then G(x v Y1) = H(xl' Y1) (G, H E C[X, Y]).
'fhis is immediate, since G(x, y) - H(x, y) = 0 implies
G(x l , Y1) - H(x 1 , Y1) = 0
[because (x, y) is generic and (xl' Y1) is on r]. Hence we have a mapping, a;
a( G 1 (x, y) + G 2 (x, y)) = a( G 1 (x, y)) + a( G 2 (x, y)) is immediate, and so is
a( GI(x, y) · G 2 (x, y)) = a( G 1 (x, y)) · a( G 2 (x, y)). Hence a is a homomor-
phism. a is obviously onto C[xl' Y1]' and if G(x 1 , YI) = 0, then G(x, y) = 0,
as (xv YI) is generic and (x, y) is on r. Thus a is an isomorphism. Q.E.D.
From 13.2 one deduces easily the following corollary.
Corollary 13.4. An irreducible curve r has infinitely many generic
points.
In fact, assuming without loss of generality that r is not a vertical Jine,
if F(X, Y) = 0 is the minimal equation for r, then n = degyF
> o. Let F = ao(x)yn + al(X)Yn-1 + ... + an(X), ai E C[X]; here
ao(X) =F O. Taking an indeterminate x over C (there are infinitely many such)
one solves the equation F (x, Y) == 0 [in the algebraic closure of k(x)] to
obtain a nonconstant point (x, y) of r.
Definition. Let r : F = 0 be an irreducible curve and (x, y), a generic
point of r. The field  = C(x, y) is called the field of rational functions
on r. By 13.3, this is uniquely determined except for (or up to) an
isomorphism over C. (Rational functions on r were defined in Chapter 9:
with the present definition, we adopt a different point of view.)
We recall that if K and L are fields with K eLand  E L is algebraic
over K, Le., satisfies a nontrivial polynomial relation over K, and if fJ E L
is algebraic over K, then  ::!: fJ,  . fJ and /fJ (if fJ -#- 0) are algebraic over K.
Thus, for example, if L is obtained from K by the adjunction of a finite
number of quantities-in symbols: L = K (l' . . . , s)-and if each

RATIONAL MAPPINGS 105
(Xi (i == 1, . . . , s) is algebraic over K, then every element of L is algebraic
over K. If K and L are fields with K c L, then L is called an extension field
of K; if every element of L is algebraic over K, then L is said to be algebraic
over K, or to be an algebraic extension field of K. If K c L c M are fields
with L algebraic over K, and if fX is an element of M algebraic over L, then,
as one knows, fX is algebraic over K (so-called "transitivity of algebraic
dependence") . t
If L == K (fXl' . . . , fXs), then L is called a finite extension (or, frequently,
a finitely generated extension) of K. t If each fXi (i == 1, . . . , s) is algebraic
over K, then L is a finite algebraic extension of K.
Let K and L be fields with K c L. If elements fXl' . . . , fXs of L
satisfy a nontrivial polynomial relation over K, then fXv . . . , fXs are said
to be algebraically dependent over K, and otherwise, to be algebraically in-
dependent over K; an infinite system fXl, fX 2 , . . . is called algebraically inde-
pendent over K if every finite subsystem is algebraically independent over
K, and otherwise algebraically dependent over K. If fXv fX 2 , . . . are algebraic-
ally independent over K and L is algebraic over K (fX 1 , fX 2 , . . .), then
fXl' fX 2 , . . . is called a transcendency basis of Lover K. If fX 1 , . . . , fXs is a
transcendency basis of Lover K, it is known that any s + 1 or more elements
of L are algebraically dependent over K. In particular, if PI' . . . , Pt
is another transcendency basis of Lover K, then t < s. By the same argu-
ment, s < t, whence s == t: thus any two (finite) transcendency bases of L
over K have the same number of elements, and this number (assuming such
bases exist) is called the degree of transcendency of Lover K. (The case of
infinite degree of transcendency also occurs.)
Note that the field C(x, y) of rational functions on an irreducible curve
r is a finite extension of C of degree of transcendency lover C. In particular,
then, we have:
Theorem 13.5. Any two elements , 'YJ of  == C(x, y) are algebraically
dependent over C.
Corollary 13.6. Every point (, 'YJ) in AN - ANC is a generic point
of some curve.
RATIONAL MAPPINGS
Speaking without precision, we say that a rational mapping is a transforma-
tion
X' == f(X, Y)
h(X, Y)'
Y' == g(X, Y 1
h(X, Y)
(f, g, h E C[X, Y]),
determined by rational functions. As h(X, Y) may vanish at certain points,
the formula above does not define images for such points. We shall speak

106 GENERIC POINTS
more of this in a moment. Moreover, if we wish the domain of the transforma-
tion to be nonempty and to be contained in r, a given irreducible curve, we
will want h(X, Y) ¥= 0 (F(X, Y)), where F == 0 is the equation of r.
In that case, h(a, b) is zero at only a finite number of points of r. We can
then define, for (a, b) in rand h(a, b) =F 0,
,1(a, b) === (f(a, b)/h(a, b), g(a, b)/h(a, b)).
In particular, A is defined at every generic point of r and transforms the
generic point (x, y) into a point (x', y') E AN. We give, then, the following
defini tion :
Definition. A rational mapping A of an irreducible curve r is a concept
made up of a generic point (x, y) of r and a pair of elements
x', y' E  == C(x, y).
One then has
, f(x, y)
x == ,
h(x, y)
, g(x, y)
y == ,
h(x, y)
where f, g, h are polynomials with coefficients in C, and h does not vanish
everywhere on r (since h(x, y) =F 0). From these formulas one can then
define, for (a, b) E r,
,1(a, b) == (f(a, b)/h(a, b), g(a, b)/h(a, b)),
provided that (a, b) does not annihilate h.
Remark. The expressions for x', y' as rational functions of x, yare not unique,
but nonetheless ,1(a, b) does not depend on the form of the expressions.
Proof. Let, for example, f(x, y)/h(x, y) == fl(X, y)/h1(x, y). Then
f(X, Y)h1(X, Y) - fl(X, Y)h(X, Y)
is a polynomial that vanishes at the generic point (x, y) of r, whence (by the
definition of generic point) f(a, b)h1(a, b) - fl(a, b)h(a, b) == o. Assuming that
h(a, b)h1(a, b) =F 0, one concludes that f(a, b)/h(a, b) == fl(a, b)/h 1 (a, b).
One will say that A is defined at (a, b) if for some expression of x', y'
one has h(a, b) =F o. If there is no such expression, one will say that A is
not defined at (a, b).
Examples. Let r be the curve Y - X2 == 0 and let (x, y) be a generic
point for it. Let A be determined by the pair x', y', where x' == y, y' == y/x.
Then ,1(0, 0) is defined, as, although % is not defined, y/x == x, so that
x', y' can be written in the form x' == y, y' == x, whence ,1(0, 0) == (0, 0).
On the other hand, consider the transformation A determined by y' == x/y,
x' == y. One cannot rewrite x/y in the form g(x, y)/h(x, y) with h(O, 0) -#- o.

RATIONAL MAPPINGS 107
For from h(x, y)x - yg(x, y) == 0 one concludes that
h(X, Y)X - Yg(X, Y) == 0 (Y - X2),
which is impossible, as the terms of least degree in h(X, Y)X - Y g(X, Y)
contain a term cX (c E C, c -=1= 0), but that cannot be the case for a multiple
of Y - X 2.
The above discussion is easily extended to the projective plane. First
one defines a generic point 01 a curve r in the projective plane over C to be a
point P: (X o , Xl' x 2 ), with entries from an extension field of C such that if
G(Xo, Xl' X 2 ) is a homogeneous polynomial in C[X o , Xl' X 2 ] that vanishes
at P, then G vanishes at every point of r-this amounts to saying that
G - O(F), where F == 0 is the minimal (homogeneous) equation for r.
As in the affine plane, a curve has a generic point if and only if it is irreducible.
If Xi -=1= 0, then C(XO/X i ' Xl/Xi' X 2 /X i ) == C( P) is called the field of the point P;
it does not depend on i. IfC(P) == C, then P is called constant, and otherwise
variable. As before, a point of an irreducible curve is generic if and only if
it is variable.
A rational mapping A of r is a concept made up of a generic point
P: (x o , Xv x 2 ) of r and a point P': (x, x, x;) such that C(P') c C(P).
One can then write
X : x : x; == fo(X o , Xl' X 2 ) : fl(X O ' Xv X 2 ) : f2(X O ' Xl' x 2 ),
where 10' Iv f2 are three forms of the same degree. If a == (a o , a v a 2 ) is a
point of r, and 10' 11' f2 do not all vanish at a, then one defines A(a) as
(fo(a), fl(a), f2(a)): the fi are not unique, but if
fo(x) : fl(x) : 12(X) == go(x) : gl(x) : g2(X),
where the gi are also forms of like degree, and if go, gl' g2 also do not all
vanish at a, then fo(a) : fl(a) : f2(a) == go(a) : gl(a) : g2(a). In this way, A is
defined at all but a finite number of points of r.
Examples. In the example given above in which r : Y - X 2 == 0 and A is
determined by x' == y, y' == X / y, A is not defined at (0, 0); however, in
the projective plane, we may write
1 : x' : y' == y : y2 : X == X2 : yx2 : X == X : yx : 1 == 1 · X : y · X : 1 2
== 10(1, x, y) : fl(l, x, y) : f2(1, x, y),
where fo == XOXl' fl == X l X 2 , f2 == X. Then A(I, 0, 0) == (0, 0, 1). In
other words, A is not defined at (0, 0) in the affine plane, but it is defined at
that point, Le., at (1, 0, 0), in the projective plane. For an example of a
rational transformation in the projective plane not everywhere defined (on a
given curve), consider the curve r: Y 2 - X 2 - X 3 == 0 and the transforma-

108 GENERIC POINTS
tion A determined by x' == x, y' == ylx. As an exercise, show that A is not
defined at (1, 0, 0).
Going back to the affine case, we see that if x', y' E C(x, y), two cases
present themselves:
1) In the case x', y' E C, all the points of r are transformed into the
same point, namely,. (x', y'). We will call this case trivial.
2) Not both x', y' are in C. Then d.t. C(x', y')/C == 1 [since
C(x', y') c C(x, y),
we have d. t. C1 (x', y') 10 < 1, and since not both x' , y' are in C, and hence not
algebraic overC, one has d.t. C(x', y')IC > 1]. Thus (x', y') is a generic point
of some curve r', called the image curve (of r under A). This case will be
called nontrivial, and we usually tacitly assume we are in this case.
From the definitions, we obtain:
Theorem 13.7. All transformed points (of r under A) are on the image
curve.
Theorem 13.8. A rational transformation T of r is always defined at a
generic point of r and (in the nontrivial case) the image of a generic point
of r is a generic point of the image curve r'.
The proofs are immediate.
As a special case, we can take a branch representation of r as a generic
point of r. This gives rise to:
Theorem 13.9. The image of a branch representation of r under a
rational transformation is a branch representation of the image curve r' (in
the nontrivial case).
The proof is immediate.
Remark. The image of a primitive branch representation need not be
primitive. Still, equivalent branch representations yield equivalent branch
representations, so that a given branch of r gives rise (under A) to a definite
branch y' of r' (see 12.19). We will say that y' is the image of y under A.
Let A be defined by a pair x', y' E C(x, y). We define a transformation
A* as follows: if (a, b) is a point of rand y is a branch of r with center (a, b),
we place A*(a, b) == (a', b') where (a', b') is the center of the image y' of y
(see the previous remark). Here A* need not be a so-called univalued trans-
formation, as r may have several branches of center (a, b); thus we should
define A*(a, b) more precisely as {(a', b') I (a', b') == the center of the branch
y' corresponding to a branch y of r of center (a, b)}. It will also not be
univalent, in general. The center of the transformed branch y' of y need
not be at finite distance, but if we consider the discussion to be taking place

RATIONAL MAPPINGS 109
in the projective plane, then A * will be defined at every point of r, since there
is at least one branch of r with given center (on r).
Let (a o , aI' a 2 ) be a point of r and assume that A is defined at (a o , a v a 2 ).
Then we can write
1 : x': y' == fo(l, x, y) :fl(l, x, y) :f2(1, x, y),
where fv f2' fs are forms of the same degree and fi(a O ' a v a 2 ) =F 0 for some i.
Let Xo == a o + t(. · .), Xl == a 1 + t(. · .), x 2 == a 2 + t(. · .) be a special
branch representation of r. Placing t == 0 in f(x o , Xv x 2 ), one sees that y'
has center (fo(a o , aI' a 2 ), fl(a O ' aI' a 2 ), f2(a O ' a v a 2 )). In other words, A*
extends A, but has the advantage of being defined everywhere on r. (Hence note
that A cannot be defined at a point where A* is multivalued.)
Definition. A rational transformation r : r  r', determined by the
generic points (x, y) and (x', y') of rand r', is called birational if
C(x', y') == C(x, y).
In this case, the generic point (x', y') of r' and the pair (x, y) E C(x', y')
yield a rational transformation f : r'  r.
Theorem 13.10. If T is defined at P and f is defined at T(P), then
f(T(P)) == P.
Hence f is called the inverse of r. One writes f == T- 1 .
Proof of the theorem. For simplicity we will assume that P and P' are at
finite distance relative to the lines at infinity, Zoo and Z'oo, of the planes of r
and r'. For complete generality, the proof should be written in terms of
homogeneous coordinates; or we may observe that our assumption is made
without loss of generality, since one can always take Zoo and ['00 not to be
through any given finite number of points, namely, P, T(P), and f(r(P)).
(Whether rand r' are in the same plane or not makes no difference.)
Then let P == (a, b) and let r, 8, t E C[X, Y] be such that
, r(x, y)
8 == ,
t(x, y)
, s(x, y)
y == ,
t(x, y)
t(a, b) -=1= 0,
let P' == r(P) == (a', b'), and let r', s', t' E C[X, Y] be such that
r' (x', y')
x== ,
t' (x', y')
s '(x', y')
y == t' (x', y') ,
t'(a', b') =F o.
Now
r'(r(x, y)/t(x, y), s(x, y)/t(x, y))
x== ,
t'(r(x, y)/t(x, y), s(x, y)jt(x, y))
y = . . .

110 GENERIC POINTS
These are equalities, though not (in general) polynomial equalities; if they
were, then by the definition of generic point, the same equalities would hold
with (x, y) replaced by (a, b). Therefore we argue as follows:
( ) [ ( r(x, y) s(x, y) ) ( r(x, y) s(x, y) )]
t(x, y) h xt' , - r' , = 0
t(x, y) t(x, y) t(x, y) t(x, y)
is a polynomial equality for some h, hence it holds for (x, y) replaced by
(a, b). Since t(a, b) =1= 0 and t'(a', b') =1= 0 we can in turn deduce
r'(r(a, b)/t(a, b), s(a, b)/t(a, b))
a= ,
t'(r(a, b)/t(a, b), s(a, b)/t(a, b))
that is,
r'(a', b')
a=
t'(a', b') ,
and similarly
s'(a', b')
b = .
t'(a', b')
Since T is defined at every generic point P of rand T(P) is generic for
r', we have:
Corollary 13.11. A birational transformation T of a curve r sets up a
one-to-one correspondence between the generic points of r and the generic
points of the image curve r'.
Let, in particular, P:((t), r;(t)), be a branch representation of r, not
necessarily primitive, say of redundancy A. Then T(P) = P' is a branch
representation of r'; let it be of redundancy fl. It is immediate that fl > A.;
but for the same reason, A > fl. In particular, if P is primitive, then so is
T( P) . Hence:
Corollary 13.12. A birational transformation T of a curve r sets up a
one-to-one correspondence between the branches of r and the branches of
the image curve r'. A primitive branch representation corresponds under
T to a primitive branch representation.
Definition. If  is a field of rational functions on a curve (or if it is an
extension of C of the form  = C(x, y) of degree of transcendency 1
over C), then by a model of  one means the notion made up of a point
(, r;) such that  = C(, r;) together with the curve having (, r;) as
generic point. Notation: (r; (, r;)).
Remark. Let r: Y = 0, (x, 0), a generic point of r,  = C(x). Note that
(x + a, 0), a E C, is also a generic point of r and that  = C(x + a).

PLACES 111
Hence  has infinitely many models in which r: Y = 0 is the specified
curve.
Let (r; (, 17))' (r'; (', 17')) be two models of. Then the associated
points (, 17), (', 17') determine a birational transformation of r with r'.
Hence:
Theorem 13.13. Any two models of a given field  are birationally
equivalent in a canonical (i.e., standard) way.
Definition. A property of the field  is called a birationally invariantive
property of anyone of its models.
PLACES
Let (r; (, 17)) be a model of a field , and let P = (1(t), 171(t)) be a branch
representation of r, not necessarily primitive. Then we know (by 13.3)
that (, 1]) and (1' 171) are isomorphic points, and hence we have an isomor-
phism (J over C of  into C((t)): (J sends any r(, 17) E  into r(l' 171). If
A : t -+ r is a substitution of order 1, then (P)A [or A(P), depending on choice
of notation] is equivalent to P, and the corresponding isomorphism is aAr
( = a followed by A). In a diagram:
( C( 1(t), 171(t))
 = C(, 7}) a l A
aA
C(l(r), 171(r)).
This situation leads to the following:
Definition. By a representation of a place we mean an isomorphism (J
over C of the field  in to the field C ( (t) ) . One says that the place repre-
sentation (J is primitive if the images of a pair of generators , 17 of 
yield a primitive branch representation; one checks that whether such
is the case does not depend on the generators chosen. Two place repre-
sentations a and (J' are called equivalent if there exists a substitution A
of order 1 in C( (t)) such that a' = aA. A place is an equivalence class
of primitive place representations.
A place is merely the field-theoretic or birational counterpart of the
projective notion of a branch. The following two theorems explicate this
remark.
Theorem 13.14. The places of  are in one-to-one correspondence
(in standard fashion) with the branches of any given model of.
Proof. Let (r; (x, y)) be a model of. We have seen how, from a primitive
branch representation x1(t), Y1(t), to get a primitive place representation (J,
and it is equally clear how to get a primitive branch representation from a

112 GENERIC POINTS
primitive place repesentation a, namely, take a(x), a(y); in this way the
branch representation and place representation give rise to each other.
Moreover equivalent branch representations correspond to equivalent place
representations, so the branches and places also are in one-to-one correspon-
dence.
Theorem 13.15. Let (r; (x, y)), (r'; (x', y')) be two models of, and
let T be the birational transformation sending (x, y) into (x', y'). Then
(as has been seen) T induces a one-to-one correspondence between the branches
of r and those of r'; and two branches correspond if and only if they
correspond to the same place of .
Proof. Let the place  send (x, y) into (x(t), y(t)) and (x', y') into (x'(t), y'(t))
-for simplicity we say place rather than place representation. In a diagram:
(x, y)  (x(t), y(t))
71 Y
(x', y')  (x'(t), y'(t)).
Let x' = r(x, y){s(x, y). Applying, we find that
x'(t) = r(x(t), y(t) ){s(x(t), y(t)),
and similarly for y'. This shows that T(X(t), y(t)) = (x'(t), y'(t)), justifying
the fourth arrow in the above diagram, and proving one part of the theorem.
A similar argument gives the converse.

Chapter 14
ZEROS AND POLES
Let r: F(X, Y) = 0 be an irreducible curve with (x, y) as generic point.
Let  be a place of the field L = C(x, y) and 7": L -+ C( (t)) a primitive
representation of the place. Let  be an element of. We define:
ord  = ord t 7"().
One checks that this does not depend on the choice of the (primitive) repre-
sentation 7" of .
Let y be the branch of r associated with the place. By the center of
 on r we mean the center of y. As there is a one-to-one correspondence
between the branches of r and the places of , and as there are only a finite
number of branches of r with given center, there are only a finite number of
places of L with given center on r.
We define:
ord y  = ord .
We say that  is a zero of  if ord  > 0 and that  is a pole of  if
ord  < o. If  is a zero of , it is said to be a zero of multiplicity ord 
if it is a pole of , then it is said to be a pole of multiplicity -ord .
From here on, in this section, let' stand for an element of L - C.
Theorem 14.1. '(EL - C) has only a finite number of zeros. Thenumber
of zeros of " counted with multiplicity, is [ : C()].t
Proof. Since' is transcendental over C, L is a finite algebraic extension of
C(,) and [L : C(,)] is a finite number. We now make use of the Theorem of
the Primitive Element:
If L is a finite algebraic extension of K (of characteristic 0), then L is a
simple extension of K.
Because of this, there exists an 'Y) E L such that  = C(', 'Y)). Thus'
is the first coordinate of a generic point of a model of. Notationally it is
113

114 ZEROS AND POLES
convenient to write , = x, 'Y) = y (in other words, we look at , as being the
first coordinate of a generic point of a model r of L).
Let F(X, Y) = co(x)ym + · · · + cm(X) be the irreducible poly-
nomial satisfied by x, y over C. Then
co(x)m-lF(X, Y) = (co(x)y)m + cl(X)(cO(x)y)m-l + · · ·
= Zm + cl(X)Zm-l + · · ·
is irreducible in C[X, Z] and is satisfied by (x, co(x)y). Since co(x)y generates
Lover C(x), we may replace y by co(x)y. In other words, we may suppose
that co(X) = 1.
Let r be the model of L = C(x, y) having (x, y) as generic point. We
are concerned with the branches Y of r such that ord y x > o. These will
have their centers on the Y-axis, but a priori possibly at infinity (this can
happen, but not if co(X) = 1). Hence we first show that if ord y x > 0,
then the center of Y is at finite distance. If ord y y > 0, then the center of y
is at (I, (x), (y)) It=o, which is at finite distance. Thus if the center is at
infinity, then ord y y < O. But then
ord y (ym + c](x)ym-l + · · .) = ord y ym < O.
This is a contradiction, as ym + cl(x)ym-l + .. · = 0 and ord y 0 = +00.
Let Yv . . . , i's be the branches for which ord y x > 0, i.e., the branches
ofrwithcenteronthe Y-axis (and at finite distance). Theni(X = O,Yj; Pj)
= intersection multiplicity of X = 0 with Yi (at its center Pj); and
Lj i(X = 0, Yj; Pj) = number of intersections of X = 0 with r at finite
distance. On the one hand, this is the number of zeros of x; on the other,
it is m = [L : C(x)]. Q.E.D.
Corollary 14.2. 'also has only a finite number of poles. The number,
counted with multiplicity, is [L : C(')], i.e., number of zeros of , = number
of poles.
This can also be written:
14.2.1
! ord , = 0,

where the sum is taken over all places. (This also holds, obviously, for
, E C, , #- 0.)
Proof. The poles of , are the zeros of 1/'. Hence there are only a finite
number of them, and this number, counting with multiplicity, is
[L : C(I/')] = [L : C(,)]. The rest of the corollary is immediate.
Theorem 14.3. Bezout's Theorem for j(r, Ll; P). Let
r: F(X o , Xv X 2 ) = 0,
Ll : G(Xo, Xl' X 2 ) = 0

DIFFERENTIALS 115
be cycleAS of orders m and n having only a finite number of points P in
common. Then
!j(r, Ll; P) = mn.
p
Proof. We first prove the theorem for d irreducible; there will be no
difficulty in extending the result to arbitrary cycles. By the projective
invariance of j(r, Ll; P) (see 12.15.2a), we may suppose Ll is not the line
Xo = o. Then let d be irreducible with (1, x, y) as generic point. Let 
be an arbitrary place of L = G(x, y) and let (xo(t), x 1 (t), x 2 (t)) be special
coordinates for the branch of  corresponding to . Then
ord t F (x o , Xl' x 2 ) = ord t xW + ord F (1, X, y).
Summing over all places (and using Corollary 14.2 and 12.15.2b), we get
!pj(r, d; P) = m times total intersection multiplicity of Xo = 0 with
d = mn. Q.E.D.
DIFFERENTIALS
Bya derivation D of a field K into itself one means a mapping a  a' = D(a)
of K into itself such that
(a + b)' = a' + b'
and
(ab)' = a'b + ab'.
From
0' = (0 + 0)' = 0' + 0'
one finds that
0' = 0,
and from
(1)' = (1 · 1)' = l' · 1 + 1 · l'
one finds that
l' = o.
From
a' = (b · a/b)' = b'(a/b) + (a/b)'b
one finds that
(a/b)' = (a'b - ab')/b 2 .
If R is an integral domain with quotient field K, one defines a derivation
D of R into K by means of the same two initial rules given above. One can
then extend D (uniquely) to a derivation of K by placing
(a/b)' = (a'b - ab')/b 2 ,
where a, bare arbitra.ry elements of R with b =I=- o. One must check that

116 ZEROS AND POLES
ajb = cjd implies (alb)' = (cld)'. For rE R, r i= 0, one has
(ralrb)' = ((ra)'(rb) - ra(rb)')j(rb)2,
by definition, and this reduces at once to (a'b - ab')lb 2 . Then one obtains
(a/b)' = (adlbd)' = (bcjbd)' = (cld)'. Q.E.D.
For the polynomial ring K[X, Y] one has the familiar derivation 0loX
given by
°
-  C..Xiyj =  iC..Xi-lyj
oX k 'l1 k 'l1
(cij E K) ,
and also the derivation 0loY.
Let K and L be fields with K c L. A derivation D of L is said to be
over K if Dc = 0 for every c E K. Let D be a derivation of Lover K and
let 0, q; E L. Then D(Oiq;j) = iOi-lq;i DO + jOiq;i-l Dq;, from which one finds
the familiar rule:
of of
D(F(O, q;)) = _ 0 DO + - Dq; ,
° oq;
whereFEK[X, Y]; here of 100 means of loX evaluated at X = 0, Y = vY.
By a simple exercise, the last formula also holds for F a quotient fig of
polynomials, provided, of course, that g(O, q;) i= o.
After these preliminary generalities, we consider the more special fields
with which we are concerned. In C( (t)) the mapping defined by
(d Idt)! cl i = ! it i - 1
(c i E C)
is a derivation, as one easily checks. Now let dt be a new indeterminate and
consider the field C( (t) )(dt). If ° E C( (t)), we define the differential dO as
dO = ((d Idt)O) · dt.
Now let q; E C((t)) - C. Then we define
dOjdq; = dO divided by dq;
[according to the operations in C( (t) )(dt)]. From
d d d
d . (0102) = 0 1 - O 2 + O 2 d Ol
t dt t
one gets, by multiplying by dt,
d (Ol ( 2 ) = 0ld0 2 + 02 dO V
and a similar rule for addition; then dividing by dq;, one gets
d (0 1 ( 2 ) _ ° d0 2 ° dOl
df{! - 1 df{! + 2 df{! ,

DIFFERENTIALS 117
which (together with a similar rule for addition) shows that fJ -+ dfJldq; is a
derivation ofC( (t)) (which coincides with the above derivation d Idt for q; = t).
This derivation is denoted: d /dgJ. We have already established the rule
d 0 of dO of dgJ
dt F ( , cp) = ao dt + acp dt ;
and hence
of of
dF (0, gJ) === _ 0 dO + - dgJ ,
o ogJ
where F is any elementf/g, f, g E C[X, Y], provided that g(O, gJ) =1= o.
Let  be a place of  and 'T:  -+ C( (t)) a representation of. Let
,  E  (with   C). Let  === 'T(),  === 'T(), and let f(, ) === 0 be the
irreducible polynomial relation satisfied by , , over O. Then from
f(, ) === 0 one finds that
(of/o)d + (of/o)d === 0,
that is,
d/d === -(of/o)/(of/o);
[of/o =1= 0 asf(X, Y) is irreducible and involves Xl Let us define
d/d === -(of/o)/(of/o).
The mapping  -+ d/d' is obviously a derivation, since   d/d is. It
does not, however, depend on , as one sees from the definition (which is
independent of ).
Exercise. Show that d/d' = 0 only if  is a constant, i.e.,  E C. Show that
d/d' = C E C if and only if  = C, + d, d E C.
Let d be a new indeterminate. A differential is defined to be an expres-
sion of the form x d', x E , Le., an element of the linear space  · d'. The
differential d of an element  E  is defined by d = (d/d')d' (such differ-
en tials are called exact).
If l is another element of  - C, then dl/d =1= 0, and if one identifies
the "new" indeterminate d'l with (d'l/d')d', then d as defined through
dl is the same as d defined through d'. In other words, the notion of a
differential is independent of the element  E  - C used in the definition.
Remark. The above definition of differential, though sufficient, is awkward,
in that it singles out an element  for special attention. This can be avoided
as follows. One introduces for every  E , including elements of 0, an in-
determinate d' and considers the (infinite-dimensional) linear space t

118 ZEROS AND POLES
 · d'l +  · d'2 + · · · , where '1' '2' . . . range over ; then one subjects
these symbols to the relations
of of
8{1 d{1 + 8{2 d{2 + · · · = 0
and no others, where F (Xl' X 2 , . . .) is a rational function over C satisfied
by '1' '2'... The resulting (one-dimensional) linear space is the desired
space of differentials.
Let 7" be a primitive representation of the place , and let 7":  -.... ,
, -.... ,. Then we define
d
ord d = ord t -;
dt
one checks that this is independent of the representation 7" of . A place 
where ord d > 0 is called a zero of d of order = ord d; a place  where
ord d < 0 is called a pole of d of order = -ord d.
Lemma 14.4. Let x E  - C. Then ord dx = 0 except at a finite
number of places. Alternatively, dx has only a finite number of zeros and
poles.
Proof. Let y be a primitive element for  over C(x), so that  = C(x, y),
and consider the curve r: F(X, Y) = 0 having (x, y) as generic point.
Since x  c, r is not a vertical line. We examine dx at places centered at
simple points at finite distance with nonvertical tangent: this is sufficient,
since only a finite number of places are dismissed in this way. At any such
point (a, b), however, the expansion for x is x = a + ct + · · ., c i= 0
(since X - a = 0 intersects r at (a, b) with multiplicity 1) and ord dx = o.
Q.E.D.
Hence we may consider the sum ! ord d', as this sum is really a
finite sum, since all the terms with a finite number of exceptions are zero.
Theorem 14.4.1. For any elements " " E  - c,
! ord d' = ! ord d".
 
Proof. d" fd' =1= 0 and ! ord d" - ! ord d' = ! ord (d" fd') = 0
by 14.1.
THE GENUS
Definition. Let p be the number defined by
! ord d' = 2p - 2.
q't

THE GENUS 119
This number (which will turn out to be a nonnegative integer) is called
the genus of  or of any model of. The genus of a curve, is, there-
fore, a birational invariant.
To show that the genus is a nonnegative integer, we will prove:
Theorem 14.5. Let r be an irreducible curve and  its field of rational
functions. Then  has a (plane) model with oniy ordinary singularities.
In other words, r can be transformed birationally into a curve with only
ordinary singularities.
Theorem 14.6. If P l' . . . , P s are the singular points of a curve r of
order n and all these points are ordinary, and if Pi is r ilold, then
(n - 1) (n - 2) ri(r i - 1)
p= -!.
2 2
Corollary 14.7. P is an integer.
Theorem 14.8. p is nonnegative.
Proof of 14.5. Our principal tool will be "transformations" of the type
(1)
X = I/X o ,
X = I/X I ,
x; = I/X 2 .
These are the so-called standard quadratic transformations. (There is one and
only one such transformation T for each coordinate system. ) We can introduce
an arbitrary factor p =1= 0 into (1) and rewrite the equations of transformation
as
(2)
X = p/X o ,
X = pIX I ,
x; = p1X 2 .
Taking p = X o X 1 X 2 , we get
(3)
X = X I X 2 '
X = X 2 X O ,
x; = XOX I .
From (1) .we see that the "transformation" T is its own inverse. Together
with T let us consider the mapping
(4)
A : F(X o , Xv X 2 )  F(X I X 2 , X 2 X O , XOX I )
defined on the set of homogeneous polynomials in C[X o , Xv X 2 ]. Note that
(5) A: F(X I X 2 , X 2 X O ' XoXI)  F(X 2 X O XoXv X O X 1 X I X 2 , X I X 2 X 2 X O )
= (XoXIx 2 )nF(X o , Xv X 2 ),
where n = deg F. If F(X I X 2 , X 2 X O , XOX 1 ) has only Xo, Xl' X 2 as irre-
ducible factors, then A(F(X I X 2 , X 2 X O ' XoXI)) has only Xo, Xv X 2 as
irreducible factors (since A: Xo  X I X 2 , etc.). Hence if F(X o , Xv X 2 ) has
an irreducible factor different from Xo, Xl' X 2 , then (5) shows that the same
is true for A(F(Xo, Xl' X 2 )). In the same way one sees that if F(X o , Xl' X 2 )

120 ZEROS AND POLES
has two irreducible factors distinct from Xo, Xv X 2 , then the same can be
said for F(X 1 X 2 , X 2 X o , XOX I ). Hence, if F(X o , Xl' X 2 ) has only one
irreducible factor different from Xo, Xv X 2 , then the same can be said for
F(X I X 2 , X 2 X O ' XOX I ).
The "transformation" T gives rise to a true transformation on the set
of points of the plane not on the axes (the lines Xi = 0), and moreover it is a
univalent transformation of this set onto itself; it is its own inverse, or, what
is the same, its square is the identity. We will designate this transformation
also by T.
Let r : F(X o , Xl' X 2 ) = 0 be the given irreducible curve. We suppose
r is not an axis, as there is nothing to prove in that case. Taking Xo = 0
as line at infinity, let (x, y) be a generic point of r, with L = C(x, y).
The point (x, y) has (1, x, y) as homogeneous coordinates. As r is not
Xl = 0 nor X 2 = 0, one has x * 0 and y =1= o. Thus T is defined at (1, x, y),
and its image is (1, l/x, l/y). Let us define a birational transformation of r
by means of this point.
Let r' be the transformed curve. As F(X 1 X 2 , X 2 X O ' XOX 1 ) vanishes at
the generic point of r', the irreducible homogeneous polynomial F' (X 0' Xv X 2)
that defines r' is a factor of F(X I X 2 , X 2 X O ' XOX I ). Hence to get
F'(X o , Xl' X 2 ) one simply suppresses the superfluous factors Xi from
F(X I X 2 , X 2 X O ' XOX I ).
Let us suppose that r has at Eo: (1,0,0) a point of multiplicity (exactly)
r. Then
(6) F(X o , Xv X 2 ) == Fr(Xv X2)x-r
+ Fr+1(X v X2)x-r-l + ...,
F r =1= 0,
(7) F(X I X 2 , X 2 X O ' XOX I ) = XFr(X2' X I )(x 1 x 2 )n-1' +...,
from which one concludes that F (X IX 2' . . .) is exactly divisible by X.
Supposing that r has a point of multiplicity exactly ri at E i , i = 0, 1, 2,
one sees that
(8)
F'(X o , Xv X 2 ) = F(X I X 2 , X 2 X O ' XOXl)/XOXlX2,
i.e., r' is given by the irreducible equation F'(X o , Xv X 2 ) == o.
From (8) one sees that the order of r' is given by
(9)
n' == 2n - r l - r 2 - ra.
Let P be a point of multiplicity r of r. Take P = Eo, i.e., as first vertex
of a coordinate system. (See Fig. 14.1.) Through P == Eo passes a line not
tangent to r at P and cutting r in n - r further distinct points. On this
line take a point El such that El ft r. Similarly take another line through

THE GENUS 121
E2
Fig. 14.1
Eo satisfying the same conditions and in a similar way take E 2 on it. More-
over we can, and do, take El and E 2 so that EIE2 cuts r in n distinct points
(n == order of r). The equation of r is F == 0 with
(10) F(X o , Xl' X 2 ) == Fr(X 1 , X2)x-r + · .. + Fn(Xv X 2 ).
The transformed curve is F' == 0 with F' == F(X 1 X 2 , .. .)/X, and hence
(11) F'(X o , Xv X 2 ) == F r (X 2 , X I )(x 1 x 2 )n-r + · · · + F n (X 2 , XI)x-r.
From the construction, Fn(X I , X 2 ) has n distinct factors (which give the
intersections of r with Xo == 0) and none of these is Xl or X 2 . The same can
then be said for F n (X 2 , Xl)' whence, from (11), r' has at Eo an ordinary
singular point of multiplicity n. A similar argument shows that r' has at
each of El and E 2 an ordinary (n - r)-fold point. As r' has an n-fold point
at Eo and an (n - r)-fold point at E I , and as the order of r' is 2n - r
[by (9)], the line EoEI cuts r' only at Eo and at EI. Similarly for EoE2.
The intersection of r' with Xo == 0 is given, according to (11), by
F r (X 2 , X I )(X I X 2 )n-r == 0, or, not counting EI and E 2 , by F r (X 2 , Xl) == o.
If r has at P == Eo an ordinary r-fold point, then its transform has r simple
points on loo outside of EI and E 2 ; nevertheless, we are not interested in
this simplification as, in view of the statement of the theorem, the ordinary
singularities are no problem. Similarly, the appearance of new ordinary
singularities at El and E 2 does not bother us.
We therefore put a nonordinary singularity at Eo, with the object of
simplifying it by means of T. Then Fr(X 1 , X 2 ) does not have r distinct
factors, and r' cuts loo in less than r points, not counting E I , E 2 , say at
U I , . . . , U t (these will be given by F r (X 2 , Xl) == 0).
In a moment we will prove that an s-fold point of r not on an axis is
transformed into an s-fold point of r', and if the point is ordinary, then so is
its transform. In other words, in trying to simplify the singularity at P, the
curve does not acquire complications elsewhere. Hence, if Fr has at least
two factors, then t > 1, and each of the points U i is of multiplicity less than
r, since !; i(r', loo; U j ) == r. Hence we obtain a simplification. The principal

122 ZEROS AND POLES
problem is when t = 1, that is, when the r-fold point P is transformed into a
point of the same multiplicity (possibly nonordinary). Before considering
this difficulty, we will take care of the point mentioned at the beginning of
this paragraph.
Let 1': (1, a + ct r + · · · , b + dt r + · · .) be a primitive branch
representation with center outside the axes, Le., a i= 0, b i= O. Let r be
such that c i= 0 or d i= O. Then every line through (a, b) with one exception,
namely, d (X - a) - c( Y - b) = 0, intersects I' at (a, b) with multiplicity r.
We say that I' is of multiplicity r and that the exceptional line is its tangent
(up till now we have not used the condition a i= 0, b i= 0). Under T, I'
is transformed into
1" : (1, (l{a)(1 - cIa · t r - · · .), (1/b)(1 - dlb · t r - · · .)),
which is also primitive and of multiplicity r. Let P be an s-fold point of r
not on an axis and let 1'1' . . . , Yt ,be the branches of r with center P and
with multiplicities S1' . . . , St, respectively. Then by 12.15.2(b)
i(l, r; P) = !i(l, Yj; P)
j
for every line l through P, whence s = ! 8j. Let T (P) = P'. The branches
of r with center P are in one-to-one correspondence (under T) with the
branches of r' with center P'. Hence P' is of multiplicity s.
The lines l for which i(l, r; P) > s are the tangents to r at P, and the
lines l for which !j i(l, Yj; P) > s are the tangents to the Yj. Hence the
tangents to r at P are the same as the tangents of the branches of r at P.
If P is ordinary, then there are s tangents to r at P, hence at least s branches
with center at P, Le., t > s. As moreover t < s, one obtains t = s, Le., at
an ordinary s-fold point there are exactly s branches each of order 1 (a branch
of order 1 is called linear). From the representation of 1" above, one checks
easily that branches with distinct tangents transform into branches with
distinct tangents. Hence, if P is ordinary, then P' will have s branches with
distinct tangents, whence P' will also be ordinary. Thus T does not disturb
the multiplicity or character-of being ordinary or not-of points not on the
axes.
Related to this last point, though not needed for the present, is the
following:
Lemma 14.9. Let Ll: G(Xo, Xv X 2 ) = 0 be a curve (or cycle) that cuts
r in only a finite nU1nber of points and let Ll' : G'(XO' Xv X 2 ) = 0 be the
transform of Ll. Let I' be a branch of r with center P and let 1" be the
corresponding branch of r' with center P'. Assume that P is on no axis.
Then
i(d', 1"; P') = i(d, 1'; P).

THE GENUS 123
Proof. Let
Xo = a o + a 1 t + · · · ,
Xl = b o + blt + · · · ,
X 2 = Co + c 1 t + · · ·
be special coordinates of y (Le., a o , b o , Co are not all = 0) and let (x o , Xv x 2 )
be a primitive branch representation of y. Then P = (a o , b o , co), and hence
aoboc o -=I o. Consequently, l/xo, l/x l , l/x2 are special coordinates for y'.
Not counting the factors X h ' is given by G(X 1 X 2 , X 2 X o , XOX 1 ) = 0, and
as neither P nor P' is on an axis, it remains to prove that
ord t G(I/X 1 X 2 , l/x2 x o, I/XOX1) = ord t G(xo, Xv x 2 ).
This, however, follows from the observation that
G(I!x l x 2 , l!x 2 x o , I/XoX1) = (l!xoXl x 2)mG(x o , Xv x 2 )
and that ord t xoX1X2 = o.
We consider now the essential difficulty previously mentioned.
By Theorem 14.6, which we will not use here, the genus of a curve having
only ordinary singularities is given by
(n - l)(n - 2) ri(ri -1)
p= 2 -,! 2
and by Theorem 14.8, one has p > o. In the case that r has arbitrary
singularities P l' . . . , P s of multiplicities r l' . . . , r s respectively, one also
considers the number
(n - l)(n - 2) ri(r i - 1)
'TT= -,!,
2 2
"\vhich is called the apparent genU8. Lemma 14.10 will show that 'TT > o.
Let us now compare the apparent genera 'TT(r) and 'TT(r') of rand r'.
According to our construction, we suppose that P 1 = Eo is a point of multi-
plicity r l = r of r and that the Pj, of multiplicity rj for r, are not on the
axes for j = 2, . . . , 8. Then r' will have an rrfold point at Pj, j = 2, . . . 8,
an ordinary n-fold point at Eo, and an ordinary (n - r)-fold point at each
of E l , E 2 , and possibly other singularities in E 1 E 2 , but otherwise only simple
points. Suppose for a moment that all these points, including Pv are ordinary.
Then
'TT(r) = p(r)
and
'TT( r') = p(r').
Moreover, since p is a birational invariant, p(r) = p(r'), and hence also
'TT(r) = 'TT(r'). As P 1 (if it is ordinary) is transformed into simple points
(so that r' has no singularities on Zoo except the ordinary ones at E 1 and E 2 ),

124 ZEROS AND POLES
one will have:
(12)
(n - l)(n - 2) _ r(r - 1) _ i ri(ri - 1)
2 2 i=2 2
= (2n - r - 1)(2n - r - 2) _ i ri(r i - 1)
2 i=2 2
2(n - r)(n - r - 1)
2
n(n - 1)
2
that is,
(13)
(n - l)(n -=- 2)
2
r( r - 1)
2
(2n - r - 1)(2n - r - 2)
2
2(n - r)(n - r - 1)
2
n(n - 1)
2
But this is a simple identity in nand r, which one checks easily, and it is
independent of any hypothesis that one may make about r.
If PI is an ordinary singularity, then the right-hand side of (12) gives
7T(r'), whence 7T(r) == 7T(r'). If PI is not ordinary, and the corresponding
points U l' . . . , U t of r' are not simple, then to compute 7T(r') one will
still have to subtract the contributions from the U i from the right-hand side
of (12), or, in other words, 7T(r) > 7T(r'). However, 7T(r') > 0 (as we shall
see), so that this kind of reduction can occur only a finite number of times
(in applying successively transformations of the type T). Hence we will
eventually resolve all the singularities into ordinary ones. Q.E.D.
To complete the proof of 14.5 it remains to show that the apparent genus,
of any curve, is > o. This is done in the following lemma.
Lemma 14.10. Let the irreducible curve r of order n have PI' . . . , P,
as singularities, and let Pi be r rfold for r. Then the apparent genus
1T(r) = (n - l)(n - 2) _ I ri(r i - 1)
2 2
is > o.
Remark. If each r i == 2, then I ri(ri - 1)/2 is the number of double points
of r. For general values of r i , it is convenient to count an rrfold point as
ri(r i - 1)/2 double points. Then the lemma says that r can have no more
than (n - l)(n - 2)/2 double points.
Proof of the lemma. With the object of motivating the proof, we ask why
a cubic cannot have more than (3 - 1)(3 - 2)/2 == 1 double points. The
answer is that ifit had two distinct double points, the line joining them would

THE GENUS 125
cut the cubic in 4 points, and this is not possible. Why can a quartic have at
most (4 - 1)(4 - 2)/2 == 3 double points 1 The answer is that if it had 4,
then a conic passed through these and one other point of the quartic
would cut the quartic in 4 . 2 + 1 points, which is impossible.
In both cases we have used curves of order n - 2 having an (ri - I)-fold
point at each r rfold point of r (so-called adjoints of order n - 2; see Chapter
17) . In the general case, we will also use curves of order n - 2.
In the second example analyzed, we had to know that a conic can be
made to pass through five points. The reason is: the general equation of
degree 2 has 6 coefficients, and these can be subjected to 6 - 1 == 5 linear
homogeneous conditions (yielding a nontrivial solution). In general, a
homogeneous polynomialcP(Xo, Xv X 2 ) of degree m has the same number of
coefficients as (1 X, Y), and this has 1 constant term, 2 linear terms, 3
terms of degree 2, . . ., rn + 1 terms of degree m, so altogether it has
(m + l)(m + 2)/2 terms. Hence we will be able to subject a cPn-2 to
(n - l)n/2 - 1 conditions.
Now we require cPn-2 to have at Pi an (r i - I)-fold point. Assume Pi
is at (0, 0); this involves a linear transformation, which transforms cPn-2 == 0
into cP-2 == 0, but the conditions on the coefficients of cP-2 can be translated
at once into conditions on the coefficients of cP-2. Thus the assumption
Pi : (0, 0) is made without loss of generality. The condition that cPn-2 == 0
pass through (0, 0) is that its constant term be zero, and the condition that
it pass doubly through (0, 0) is that the linear terms have zero coefficients.
Similarly, by imposing 1 + 2 + . · . + (r i - 1) == (r i - l)rd2 conditions
on cPn-2' we can make cPn-2 == 0 have an (ri - I)-fold point at Pi. Therefore if
(1)
( n - l ) n r. ( r. - 1 )
_1>t'l
2 - £., 2 '
then the conditions at the Pi can be met, and moreover cPn-2 == 0 can be
made to pass through
(n - l)n _ 1 _ ! ri(ri - 1)
2 2
other points. Thus cPn-2 == 0 would meet r in at least
(n - l)n ri(r i - 1)
ri(ri-l)+-2--! 2 -1
ri(r i - 1) (n - l)(n - 2) (n - l)(n - 2) (n - l)n
==!---- + +-1
2 2 2 2
== -'1T + n(n - 2)
points. This would be a contradiction if '1T < o.

126 ZEROS AND POLES
Thus it remains to prove (1). Let r: F(X, Y) == 0 and assume that no
singularity is on Xo == 0, the line at infinity, and that no tangent at a singular
point is vertical. Then one sees easily that of /0 Y == 0 has at Pi an
(ri - I)-fold point; we also may suppose r involves Y and of/oY #- O.
Since of/oY == 0 is of order n - 1, we obtain: n(n - 1) > !ri(r i - 1).
'We can take E 2 : (0, 1,0) to be on a tangent at a simple point of r, and hence
get n(n - 1) > ! ri(r i - 1) + 1. Since n(n - 1) and! ri(r i - 1) are both
even, we get: n(n - 1) > !ri(r i - 1) + 2, which proves (1). The proof
of 14.10, and thereby of 14.5, is now complete.
Theorem 14.6 can be rephrased as follows: If r has only ordinary
singularities, then the genus equals the apparent genus. To prove this, we
compute the genus in terms of projective features of r. First we need:
Lemma 14.11. Let P be a simple point of r (r, irreducible, and not a
vertical line, but otherwise arbitrary) and let the tangent at P be vertical. Let
x, y be a generic point of r : F (X, Y) == o. Then ord p dx == ord p oFjo Y.
First proof. Let Pbeplacedat (0, 0). ThenF(X, Y) == X + cY1' + terms
involving X2, X Y, or yr+l (c #- 0), where r == i(r, X == 0; P). The branch
of r at P is given by x == -cyr + · · ., hence ord p y == 1; then
dx == (-rcyr-l + ...) dy and ord p dx == r - 1. From of/oY == rcyr-l +
terms involving X + terms involving yr, one sees that also
ord p of joY == r - 1.
Second proof. We have (oF joy) dy + (oF lox) dx == o. Now note that
ord p dy == 0 and ord p of jox == 0, whence ord p dx == ord p of joy. Q.E.D.
Proof of 14.6. Let r be of order n and have only ordinary singularities.
Place r so that it cuts Xo == 0, the line at infinity, in n distinct points, none
at E 2 : (0, 0, 1) or E 1 : (0, 1,0), and so no tangent at a singularity is vertical.
Let (x, y) be a generic point of r. Then by definition 2p - 2 == ! ord dx ==
the sum of the orders of zeros of dx - the sum of the orders of the poles of
dx . We are going to show:
a) dx has poles only at the branches centered on Xo == 0 and ord dx ==
-2 at each of these;
b) dx has zeros only at simple points P of r having vertical tangent
and ord p dx == ord p oFjoY at such points;
c) at a singular r-fold point of r,
i ( : = 0, r; p) = r(r - 1) .
From these points there will follow
of
n(n -1) == !ri(r i -1) + !ord p -,
p oY

INFINITELY NEAR POINTS 127
where P runs over the simple points of r with vertical tangent (both sides
give the total intersection multiplicity of of /0 Y === 0 and F === 0). Hence
sum of the orders of the zeros of dx === n(n - 1) - ! ri(r i - 1),
sum of the orders of the poles of dx === -2n,
and
2p - 2 === n(n - 1) - !ri(r i - 1) - 2n,
whence
(n - l)(n - 2) ri(r i - 1)
p=== 2 -! 2 .
To prove (a), note that (1, x, y) and (1/x, 1, ylx) are homogeneous co-
ordinates of (x, y), and hence that (1/x, ylx) is a generic point of r referred
to Xl === 0 as the line at infinity. Hence the intersections of r with Xo === 0
are the zeros of l/x, Le., the poles of x. At each of these intersections, then,
ord x === -1; letting t === l/x, we find that dx === -(llt 2 ) dt, whence
ord dx === -2 at each of the branches.
To prove (b), let y be a branch of r with center at a point P: (a, b) at
finite distance. If X - a === 0 is not the tangent to y, and in particular
this will be the case if y is centered at a singular point, then x - a has order
1 at y. Placing x - a == t, one has dx === dt and ord y dx === O. This, together
with 14.11, proves (b).
To prove (c), observe that of 10 Y === 0 has an (r i - I)-fold point at an
rrfold of F === 0; and that the curves have no common tangent at such a
point.
The proof of 14.6 is now complete. In view of this, Theorem 14.8 is
just a special case of 14.10.
INFINITELY NEAR POINTS
A standard quadratic transformation transforms the branches centered at
Eo into branches centered on a line l. The events at Eo are mirrored in the
events along l. One thinks of the neighborhood of Eo as a little closed circuit
about Eo, and this is blown up, by means of the transformation, into the
line l (which is also a closed circuit). Whatever takes place along l we then
say takes place in the (first) infinitely near neighborhood of Eo. For example,
if r has an r-fold point at Eo and r', the transformed curve, has an rl-fold
point and an r 2 -fold point on l (not at EI or E 2 , the axes having been appro-
priately arranged), then we say r has an r1-fold point and an r 2 -fold point
in the (first) neighborhood of Eo. By applying a standard quadratic trans-
formation (or a locally quadratic transformation) with a point P on l playing
the role of Eo, one defines a second infinitely near neighborhood of E, and
so on, for third, fourth, etc., infinitely near neighborhoods of Eo.

128 ZEROS AND POLES
Although the foregoing description does not constitute a satisfactory
definition, it is nevertheless true that the notion of infinitely near neighbor-
hoods can be given a precise meaning. The main difficulty is whether the
neighborhood points depend on the transformations by means of which they
have been described, and it is a fact that they do not. As, however, we merely
want to use a suggestive phraseology and do not really need formal or
rigorous results in this regard, we do not enter into the details.
The apparent genus 7T(r) == (n - l)(n - 2)/2 - ! ri(r i - 1)/2 merely
takes into account the explicit multiple points; the apparent genus 7T(r')
of the transformed curve r' also takes into account the multiple points of r
infinitely near, and in the first neighborhood, of the explicit ones, etc. One
could, then, phrase the following theorem:
If the irreducible curve r has the multiple points Pv..., Ps with
multiplicity rv . . . , rs, including the infinitely near points, then in every case
the genus is given by p(r) == (n - l)(n - 2)/2 - ! ri(r i - 1)/2.
As another example, let r, Ll be distinct irreducible curves meeting in
Pv . . . , Pt with Pi rrfold for rand srfold for Ll. Let us call ris i the
apparent intersection multiplicity for r, Ll at Pi and b == mn - ! ris i the
intersection defect (m == deg r, n == deg Ll). Arrange r u Ll appropriately
with PI at Eo, apply a standard quadratic transformation, let r', Ll' be the
transformed curves, and let !' be the contribution to the apparent multi-
plicity of r' and Ll' along E 1 E 2 outside of E 1 and E 2 . Then
b' == (2m - r 1 )(2n - S1) - mn - 2(m - r 1 )(n - S1) - ! risi - !'
i2
== b - !'.
Thus the intersection defect diminishes unless !' == 0, i.e., unless r, Ll have
no common tangent at Pt. Since this can happen only a finite number of
times, we have:
Theorem 14.12. Let r, Ll be distinct irreducible curves. Then the
resolution of singularities into ordinary ones also holds for r u Ll.
Informally we may say:
The total intersection multiplicity of rand Ll (Le., mn) is given in every
case by ! ris i , provided that the infinitely near points are counted.

Chapter 15
NOETHERIAN CONDITIONS
Let r:F(X, Y) == 0 and :G(X, Y) == 0 be two curves (or cycles) with-
out common components. Let r (1  == Pv . . . , Ps. Given an arbitrary
polynomial H (X, Y) we are interested in establishing conditions for H
to be in (F, G), i.e., to be of the form H == AF + BG, where A, B are poly-
nomials. An obvious necessary condition is that H vanish at Pv . . . , Ps.
This condition is also sufficient in the case that the intersection multiplicity
of F == 0 and G == 0 at each common point is 1. We will not stop to prove
this here, as it will follow from a later theorem.
Let P be an arbitrary point. We define
{ AF + BG }
qp == H IH == d ' d(P) -=I 0, for some d, A, BE C[X, Y] ,
Le., as the set of polynomials that can be written in the stated form. Clearly
qp is an ideal in C[X, Y], Le., if HI and H 2 are in qp, then so is HI + H 2 ,
and if H is in qp and A is in C[X, Y], then AH is in qp.
We will use the following lemma, which is a consequence of Hilbert's
N ullstellensa tz. t
Lemma 15.1. If {d(X(X, Y)} is a collection of polyrwmials without a
common zero, then the ideal generated by the d(X(X, Y) is (1); or in other
words, 1 is a linear combination of the d(X with polynomial coefficients.
Definition. One says that the polynomial H satisfies the Noetherian
conditions for F, G at P if H E qp.
Theorem 15.2. If H satisfies Noether's conditions for F, G at every
point P, then H is in (F, G).
Proof. For each P one has a polynomial d p such that dp(P) -=I 0 and
d p · H E (F, G). By 15.1, the ideal generated by the d p is 1, whence
1 == !Apd p , and as (!Apdp)H E (F, G), one has HE (F, G).
129

130 NOETHERIAN CONDITIONS
Theorem 15.3. qp = (1) except for P = Pv . . . , p. fl ; and qp. -=I (1).

Proof. Let P -=I Pi' i = 1, . . . , s. Then either F or G, say F, does not
vanish at P. Writing
1 = F /F,
we see that 1 E qp. Conversely, suppose that 1 E qp (for some point P).
Then one can write
1 = (AF + BG)/d,
with d(P) * o. Asd = AF + BG and F and G vanish at Pi' i = 1, . . . , s,
we must have d (Pi) = o. Hence P -=I Pi. Q.E.D.
Corollary 15.4 (Lasker-Noether)
( F, G) = q p () · · · () q 1 .
1 8
Corollary 15.5. Let PI: (0, 0). Then all suffic1:ently high powers of X,
of Y, and of the ideal (X, Y) are in qp .
1
Proof. Ry(F, G) = AF + BG = XP(c o + c1X + · · .), C i E 0, Co -=I o.
As PI: (0,0) is in both F = 0 and G = 0, one has p > o. Hence
XP = (AF + BG)/(c o + · · .) E qp 1 . Simila.rly ya E qP 1 for some G. Let
T = P + G - 1. Then (X, Y)T c qp 1 . For in a power product Xiyj with
i + j = 'T, one must have i > p or j > a (i < p - 1 andj < a-I implies
i + j < p + G - 2) whence Xiyj E qp 1 .
Theorem 15.6. Let PI: (0, 0). A necessary and sufficient condition
for H E qP 1 is that there exist A, B E O[ [X, Y] ] such that H = AF + BG.
Proof. If H E qp 1 , then H = (AF + BG)/d with Av Bv d E O[X, Y] and
d = d oo + dlOX + dOl Y + · · · ,
with d oo -=I 0, since d(P 1 ) -=I o. Hence d is a unit in O[[X, YJ], and
A = AId-I, B = Bld- l are the sought elements in O[[X, Y]].
Conversely, let H = AF + BG, A, BE O[[X, Y]]. Let us write
A = Ar + A;, B = Br + B;, where An Br are the sums of the terms of
degree < r in A, B, respectively. Then
H = ArF + BrG + A;F + B;G = Ar F + Br G + T r + v
where Tr+l contains only terms of degree> r. Since Hand Ar F + BrG
are polynomials, their difference, T r + v is also a polynomia1. By 15.5, for r
sufficiently great, Tr+1 E q p . Moreover ArF + BrG E q p . Hence H E qp .
1 1 1
Q.E.D.

NOETHERIAN CONDITIONS 131
Consider the ring C[[X]][Y]. Let
r
I = II (Y - miX - .. .),
i=l
with rni -=I m; for i -=I j,
G, H be three elements in that ring. One has:
Theorem 15.7. If for i = 1, . . . , r,
ord x H(X, miX + .. .) > ord x G(X, miX + · ..) + r - 1,
then
H = Af + BG
(A, BE C[[X]][Y])
and moreover A and B can be taken 80 that subd B > r - 1.
Proof. We prove the theorem by induction on r. For r = 1,
ord x H(X, mIX + ...) > ord x G(X, mIX + · · .),
whence
H(X, mIX + · · .) = B(X)G(X, mIX + .. .),
and
H(X, Y) - B(X)G(X, Y) = A(X, Y)(Y - mIX - · · .),
which proves the case r = 1.
We apply the induction to
r
11 = II (Y - miX - .. .),
i=2
G l = XG,
which, as one checks, satisfy the required conditions. rrhus H = Alfl +
BlXG with subd Bl > r - 2. Since mi -=I m i for i =I=- 1, one now has
ord x A 1 (X, mlX + · · .) + r - 1
= ord x (H(X, mIX + · ..) - Bl(X, mIX + · · .)XG(X, mlX + · · .))
> ord x G(X, mIX + · · .) + r - 1,
using that both ord x H(X, mIX + · ..) and ord x Bl(X, mIX + · · .)X
G(X, mIX + · ..) are > ord x G(X, mIX + · · .) + r - 1. Hence one
obtains
ord x Al(X, m 1 X + · ..) > ord x G(X, mlX + · · .),
whence
AI(X, Y) = (. · .)(Y - mIX - · · .) + (. · .)G(X, Y).
Substituting Al in H = Aifl + BIXG, one obtains the desired expression
for H (with subd B > r - 1).

132 NOETHERIAN CONDITIONS
Let r: F (X, Y) == 0 be a curve having at 0 == (0, 0) an ordinary
r-fold point. We take the coordinate system in such a way that X == 0 is
not a tangent to r at O. Then
r
F(X, Y) == II (Y - miX) + ...,
i=l
with mi =1= mj for i -=I j. Then from 12.13 one has:
Theorem 15.8. r has r branches at 0,. these are of the form
x == t,
y == mit + · · ·
From algebra it is known that if R is a unique factorization domain, then
so is R[ Y], the ring of polynomials in an indeterminate Y with coefficients in
R, and that the units in R[Y] are those in R. We apply this to R == o [[X]] .
Hence if d (X) is a unit, then d (0) -=I 0, and Y - miX - · .. and
Y - mjX - · .. are not associates in C [[X]] [Y] if mi -=I mj. Now
F(X, miX + ...) == 0, so F(X, Y) is divisible by Y - miX -... in
C[[X]][Y], and as the Y - miX - · · · are nonassociates, we have:
15.8.1. F (X, Y) == IIi=l( Y - miX - · · .) · g(X, Y), g E C[[X]J[ Y],
and g(O, 0) -=I 0, as subd F == r.
Now we can state:
Corollary 15.9. Let F(X, Y) == 0 be as before and let Yv..., Yr
be the branches centered at PI == (0, 0). If : G(X, Y) == 0 and
E : H (X, Y) == 0 are two curves such that
i(E, Yj; PI) > i(, Yj; PI) + r - 1,
j == 1, . . . , r,
then
H E qp 1 .
Proof. We have F(X, Y) == IIi=l(Y - miX -- .. .)g(X, Y), g(O, 0) -=I O.
We place IIi=l(Y - miX - · · .) == f, whence F == fy. Let Yi be the branch
X == t, Y == mjt + · .. Then
i( , Y j  PI) == ord t G( t, mjt + · · .) == ord x G( X, miX + · · .),
and similarly for E and H. From the hypothesis, by 15.7, one obtains
H == Af + BG,
A, BE O[[X, Y]J.
As g is a unit in C[[X]], H == Ag-IF + BG and HE qP 1 by 15.6.
Let {y} be the set of all branches of an irreducible curve r. A formal sum
! nyY with integral ny and with ny =I=- 0 only for a finite number of branches
Y is called a divisor of r. Defining! myY + ! nyY to be ! (my + ny)Y' the

NOETHERIAN CONDITIONS 133
set of divisors becomes an abelian group. We say that! nyY is effective if
ny > 0 for every y; and we write! nyY > O. We say D > D' if D -
D' > o. If a divisor is not effective, it is called virtual.
Let r be a curve with only ordinary singularities; let these be
PI' . . . , Ps, and let Pi be rr fold . Let Yo, j == 1, . . . , r h be the branches of
r centered at Pi' Then the divisor
D == ! (ri - I)Yii
i,j
is called the double-point divisor of r.
If r and  are two curves, we denote by  · r the divisor
. r == !i(, y; P)y,
y
where y runs over all the branches of r. (This is the intersection of  with
r with the "points" of intersection counted ,vith the "right" multiplicity.
We write "points," though we really mean branches. If r were without
singularities, there would be a one-to-one correspondence between the points
of r and the branches of r; and one could speak of points instead of branches
without any fear of confusion. However, if r has an ordinary singularity,
there will be several branches centered at the same point, and one will have
to make the distinction. The reason we continue to say "points," instead
of using the more exact word "branches," is that we think of the branches as
"accidentally" coming together at a singularity.)
As before, let r : F == 0 be a curve with only ordinary singularities and
with (minimal) equation F == o. Let  : G(X, Y) == 0 and E : H (X, Y) == 0
be two curves (or cycles); and assume that r and  are without common
component.
Theorem 15.10 (The "AF + BG Theorem"). If E · r > D +  · r,
then H == AF + BG u,ith A, B E C[X, Y].
Proof. Let y be a branch of r with center at point P of multiplicity r > 1
for r. From the hypothesis one has
i(E, y; P) > i(, y; P) + r - 1.
By Corollary 15.9 one then has that H belongs to qp. Then, by Theorem 15.2,
H belongs to (F, G). Q.E.D.
We shall need the projective formulation of this last theorem. Let
r : F*(Xo, Xv X 2 ) == 0 be the given curve (always with only ordinary
singularities). Let : G*(Xo, Xv X 2 ) == 0, E: H*(X o , Xl' X 2 ) == O. Then
one has:
Theorem 15.11. If E. r > D + . r, then H* == A*F* + B*G*
with A* and B* homogeneous polynomials in C[X o , Xv X 2 ].

134 NOETHERIAN CONDITIONS
Proof. The theorem is of a projective character, and hence we may suppose
thatandrdonotmeetonXo = o. (Hence, in particular, G*(O,X v X 2 ) =1= 0
and F*(O, Xv X 2 ) =1= 0.)
From the previous theorem,
H*(I, X, Y) = A(X, Y)F*(I, X, Y) + B(X, Y)G*(I, X, Y).
Multiplying by a power of Xo, we obtain
(1) XH*(Xo' Xl' X 2 ) = Ai(X o , Xv X 2 )F*(X o , Xv X 2 )
+ Bi(X o , Xv X 2 )G*(X O ' Xl' X 2 ),
where Ai, Bi are homogeneous and p > 0 is possible: let us suppose that
p > o. Placing Xo = 0 we obtain
Ai(O, Xl' X 2 )F*(0, Xv X 2 ) + Bi(O, Xv X 2 )G*(0, Xl' X 2 ) = O.
As F*(O, Xl' X 2 ) and G*(O, Xv X 2 ) have no common factors, one obtains
Ai(O, Xl' X 2 ) = A:(X v X 2 )G*(0, Xl' X 2 ),
whence
Ai(X o , Xl' X 2 ) = A:(X v X 2 )G*(X O ' Xv X 2 ) + XoA:(X o , Xl' X 2 ),
with A:, A: homogeneous. Substituting this expression for Ai in (1) and
rearranging the terms, we obtain
(2) XH*(Xo' Xv X 2 ) = XoA:(X o , Xl' X 2 )F*(X o , Xl' X 2 )
+ B:(X o , Xv X 2 )G*(X O ' Xl' X 2 ).
Placing Xo = 0, we see that B:(O, Xl' X 2 ) = 0, whence
B:(X o , Xv X 2 ) = XoB:(X o , Xl' X 2 ).
Canceling Xo, we obtain
(3) X-lH*(Xo' Xv X 2 ) = A:(X o , Xv X 2 )F*(X o , Xl' X 2 )
+ B:(X o , Xv X2)G*(XO' Xl' X 2 )'.
Repeating the process, we reduce p to 0, whence the theorem follows.

Chapter 16
LINEAR SERIES
Let L = C(X, y) be the field of rational functions of some curve, i.e., a field
C(x, y) of degree of transcendency lover C. Let {} be the set of all places
of L. Consider the formal sums ! n, where n = 0 except for a finite
number of places. The set of such formal sums forms an abelian group under
the rule for group addition given by: ! m  + ! n  = !(m + n).
The formal sums ! n  are called divisors: as there is a one-to-one
correspondence between the places of L and the branches of any model of
L, the divisors of this chapter are essentially the same as those of the last.
One defines effective and virtual divisors as before.
Examples of divisors are:
1) ()o = set of zeros of  ( E L - C). One has ()o > 0, Le., every
nonconstant function has at least one zero. ()o is called the divisor of zeros
of .
2) () 00 = set of poles of  ( E L - C). Also () 00 > o.
One also defines ()o, () 00 for  E C,  * 0, namely, by placing
()o = () 00 = the zero of the group of divisors.
Definition. If A and B are two divisors, we say that A and Bare
equivalent, in symbols: A - B, if A - B = ()o - ()oo for some
 E L.
This relation is a so-called equivalence relation, i.e.,
(r) A == A for every A,
(s) A == B implies B == A,
(t) A == Band B - 0 implies A == C.
For example, to prove (t), we have for some  and 'YJ that
A - B = ()o - ()oo = ! (ord )

and
B - C = ('YJ)o - ('YJ) 00 = ! (ord 'YJ).

135

136 LINEAR SERIES
Adding we get A - C == ! (ord 1J)'13 == (1])0 - (1]) 00' whence A == C.
The properties (r) and (s) are proved similarly.
Remark. It is not true in general that (1])o == ()o + (1])0.
Theorem 16.1. If A, B, 0 are divisors, then A == B implies
A + 0 == B + C.
Definition. ord (! n '-P) == ! n.
Theorem 16.2. If A and B are equivalent divisors, then ord A == ord B.
Proof. Let A - B == ()o - () 00. If   0, then ord ()o == [L : C()] and
ord () 00 == [L : C()]; if  E 0, then A == B. Thus ord (A - B) == 0, i.e."
ord A == ord B.
Remark. For a proof we need only (14.2.1), which says that ord ()o ==
ord ()oo' and not the somewhat deeper statement of 14.1.
Let xo, Xl' . . . , x r E L, not all zero, and let B be a fixed divisor. Consider
the divisor
(*) Ac == (coXo + · · · + crxr)o - (coxo + · · · + crxr)oo + B
(c i E 0, ! CiXi -=1= 0).
Definition. The set L of divisors Ac for all permissible c == (co, . . . , c r )
is called a linear series.
We may, and generally will, suppose that XO' . . . , X r are linearly
independent over C. Then the condition on the C i is simply that they be not
all zero.
Theorem 16.3. All the elements of a linear series L are equivalent to B
and hence are equivalent to each other.
The common order of the elements of a linear series is called the order
of the linear series.
Theorem 16.4. For any  ::I=- 0
Ac == (coxo + · · · + crxr)o - (coxo + · · · + crxr) 00
- ()o + ()oo + B.
Corollary 16.5. If B == B', then the elements of L can be represented
in the form
(CoXo + . · · + CrXr)o - (CoXo + . · · + CrX r ) 00 + B',
for some Xo, . . . , Xr.

LINEAR SERIES 137
In fact, if B == B', then
B - B' = (')0 - ()oo
and
B' = -()o + ()oo + B
for some .
Theorem 16.6. Suppose that the (typical element of the) linear series L
is also given by
A = (doYo + · · · + dsYs)o - (doYo + · · · + dsYs)oo + B',
and suppose that the Yi are linearly independent over C, and suppose the
same for the Xi of (*). Then r = s and for some ,
Yi = ! Ci.1 X j,
ICiil -=1= o.
Proof. Since (Yi)O - (Yi)oo + B' E L, we can write
(Yi)O - (Yi)oo + B' = (! CijXi)O - (! CijXj) 00 + B, B == Ac == B'.
Write B - B' = ()o - ()oo. Then
(Yi)O - (Yi) 00 = (! C i .1 X j)o - (! CijXj) 00 + ()o - () 00
= (! CijX.1)O - (! CijX j ') 00.
As there is no cancellation in (Yi)O - (Yi)oo nor in (! Ci.1Xj)O - (! CijXj)oo'
one obtains
(Yi)O = (! CijXj)O'
(Yi) 00 = (! CijXj)oo
and hence Yi/! CijX j ' has no zeros. Hence Yi = (const) · ! CijXj, or, absorbing
the constant into the Cij, we can write Yi = ! CijX j '. From this there follows
s < r. Similarly r < s, Le., r = s.
Corollary 16.7. The integer r depends only on the seres L, not on
the representation.
The integer r is called the dimension of L.
If d = AC, A E C, then Ad = Ac. Conversely, if Ad = Ac, Le.,
(! CiXi)O - (! CiXi) 00 + B = (! dixi)O - (! dix i ) 00 + B, then (! CiXi)O =
(! dixi)O and! dix i = (const) · ! CiXi. Assuming the Xi linearly independent
over C, we conclude that d = AC. Thus the set of divisors Ac of L are
put in one-to-one correspondence with the points of projective r-space (overC),
the divisor Ac corresponding to the point with coordinates C = (co, C v . . . , c r ).
A change of representation (*) changes C by a homogeneous nonsingular
linear transformation (still assuming that the Xi are linearly independent over

138 LINEAR SERIES
C). In fact, using the notation of Theorem 16.6, and given that A == Ac,
Le., that
(doYo + · · · + drYr)o - (doYo + · · · + drYr) 00 + B'
= (coXo + · · · + crxr)o - (coXo + · · · + crxr)oo + B
and
B - B' = (')0 - (')00'
from Yi == ,! CijXj one obtains
(! dicijXj}o - (! dicijxj) 00 == (! CiXi}O - (! CiXi) 00'
i,j
so that c:J = (const) ! cijd i ; and the Cj and d i are related in the way stated.
If C == (Cij) and CT = C "transpose" == the matrix obtained from C
by interchanging the rows a.nd columns, then, considering c, d as column
vectors, we see that C == A · CTd, and d == p(C T )-lC.
We can say, then, that L is a projective space, the divisors in L are
the points of that space. A representation (*) of a divisor in L gives a set
of coordinates for the divisor (or "point"). A change from XO' . . . , x,
to any other linearly independent set Yo' . . . , Yr (with a change from B to
B') merely gives coordinates to the divisor in a new coordinate system.
Definition. By a subseries of a linear series L one means the set of
elements of L corresponding to a subspace of the space representing L.
Theorem 16.8. A subseries of a linear series is a linear series.
Proof. Let Ac == (coXo + · · · + crxr)o - (coXo + · · · + crxr)oo + B be a
representation of L with Xo,..., X r linearly independent over C. The
elements of L are in one-to-one correspondence with the points of a pro-
jective r-space pr: Ac  P: (c). Let PS be an s-dimensional subspace of
pro Then ps is generated by s + 1 linearly independent points:
Co == (coo, . . · , cor), . . . , C s == (c so , . . . ,C Sr ). The subseries corresponding to
ps is by definition the set of divisors
8 8
(1) ((! AiCiO}X O + · · · + (! AiCir}Xr}o - (. · .)00 + B.
i=O i=O
Let Yi == ciOX O + · · · + CirXr. Then (1) can be rewritten as
(2)
8 8
(! AiYi)O - (! AiYi) 00 + B,
i=O i=O
tha t is, as the divisors of a linear series.
One says that the representation (*) is normalized if ! CiXi == 1 for some
c i E C.
One may suppose (though generally we will not) that the representation
is normalized. In fact, as we have seen, we may replace the Xi by Xi' with

LINEAR SERIES 139
 arbitrary ( -=1= 0). Let Xj -=1= 0 and take  = l/xj. Then one of the Xi'
namely xi' is 1; and thus 1 is a linear combination of the Xi.
Theorem 16.9. The divisor (*) is normalized if and only if B belongs
to L.
Proof. Suppose that (*) is normalized. Take the Ci so that I CiXi = 1. Then
Ac = (1)0 - (1)00 + B = B,
i.e., BEL. Conversely, if BEL, then B = Ac for some c = (co, . . . , c r ).
For such a c,
(I CiXi)O - (I CiXi) 00 = 0,
i.e., I CiXi has no zeros or poles, whence I CiX i = const = d, and
I (ci/d)x i = 1.
Theorem 16.10. If Land L' are two linear series and an element of L
is equivalent to one of L', then L and L' are both contained in a linear series.
Proof. Let
L: Ac = (I CiXi)O - (I CiXi) 00 + B,
L': A = (IdjYi)o - (IdjYi)oo + B'.
We may replace B' by any divisor equivalent to it, in particular B. Hence
we may assume B = B'. Then the linear series
LO: (ICiXi + IdjYi)o - (ICiX i + I diYj) 00 + B
contains both Land L'.
TheorenJ, 16.11. One can add (or subtract) an arbitrary divisor C to
the elements of a linear series to obtain another linear series of the same
dimension.
The proof is immediate.
Theorem 16.12. The multiplicity with which a place appears in the
divisors of a linear series L is bounded above and below.
Proof. The divisor B in (I CiXi)O - (I CiX i ) 00 + B does not influence the
question of boundedness, so we may (and do) omit B. If '13 is not a pole of
any Xi' then 0 is a lower bound. In general, let ord Xi = -Vi. Then
ord (I CiX i ) > min { -Vi},
which gives a lower bound. As for an upper bound, observe that a place
occurs with negative multiplicity in (I CiXi)O - (I CiX i ) 00 only if it is a pole of
some Xi. Each Xi has only a finite number of poles. Let '131' . . . , '138 be

140 LINEAR SERIES
the poles of all the Xi. Then by the first part of the proof applied to
v · . . , s, - ord (CiXi) 00 is bounded below and hence ord (CiXi)O
(= ord (CiXi) (0) is bounded above. If  is the given place, then the multi-
plicity with which  occurs in (CiXi}O - (CiXi) 00 is equal to or less than
the upper bound of ord (CiXi)O.
Theorem 16.13. Let  be a place and 'V the minimum multiplicity with
which  occurs in the elements of L. Then the elements of L in which 
occurs with multiplicity > 'V + 1 is a subseries of L of dimension =
dim L - 1.
Proof. Let Ac = (CiXi}O - (CiXi) 00 + B be a representation of L. Take
BEL and in such way that  occurs in B with multiplicity 'V. Then 
is not a pole of any Xi [for otherwise  would occur with multiplicity less
than 'V in some (xi)o - (Xi) 00 + B]. Moreover (for a similar reason)  is not
a zero of all the Xi. Let  :   C( (t)) be a representation of the place .
Then Xi -+ a iO + ailt + · .. with not all a iO = o. The elements of L in
which  occurs with multiplicity > 'V + 1 correspond to the points
c = (co, . . . , c r ) for which  CiaiO = 0, which proves the theorem, as this
is a homogeneous linear condition on the c i .
Corollary 16.14. The set of divisors of L in which a given place  occurs
with multiplicity > It is a subseries of L.
Proof. If ft < 'V, then the set is L itseJf. If ft = 'V + 1, then the coroUary
coincides with the theorem. If ft = 'V + 2, one applies the theorem to the
sub series of elernents in which  occurs with multiplicity > 'V + 1, etc.
Corollary 16.15. If 1'...' s is an arbitrary collection of places
and ftv . . . , Its are integers, then the elements of L in which each i
occurs with multiplicity > Iti form a subseries of L.
Theorem 16.16. The effective elements in a linear series form a subseries.
The notation for a linear system of effective divisors is g, in which n is
the order and r the dimension.
Proof of the theorem. If Ac = (CiXi)O - (CiXi) 00 + B, there are only a
finite number of places in which the elements Ac can have negative order
(namely, the places occurring in B and the poles of the Xi). The theorem now
follows from the last corollary, taking 1' . . . , s as this set of places and
J.ti = 0, i = 1, . . . , s.
Theorem 16.17. In a g, r < n.
Proof. If r = 0, then obviously r < n. Let r > o. Take a place  that
does not occur in every element of g. Then the divisors in g in which 

LINEAR SERIES 141
occurs form a g-l. Subtracting  from each of the divisors mentioned, one
obtains a g-=-l; applying induction on r, one gets r - 1 < n - 1.
Theorem 16.18. Let g and g, be such that an element of one of them
is equivalent to an element of the other (and hence n == n'). Then both
are contained in a common linear series of effective divisors.
Proof. From 16.10 it follows that g and g, are both contained in a linear
series L. Let L' be the set of all effective divisors in L. Then L' is a linear
series (by 16.16) and g, g, c L'.
Theorem 16.19. Let 0 be a divisor. Then the set of effective divisors
equivalent to 0 is a linear series, called a complete linear series.
Notation. The linear series of effective divisors equivalent to 0 is denoted 101.
Proof of the theorem. It may be that there are no effective divisors equivalent
to 0, and then the set in question forms the empty linear series. Otherwise,
we may (and will) suppose that 0 is effective, because there exists an effective
D, and the divisors equivalent to D are the same as those equivalent to O.
Then 0 is contained in some g, for 0 itself forms a g. If such a g is properly
contained in a g, then s > r; since s < n, there is a g containing 0 and
not contained properly in a g. Every effective divisor D equivalent to 0
must be in this g, otherwise there is a g containing D and g, i.e., properly
containing g. This proves the theorem.
At the beginning we could have announced various other rules for working
with equivalences of divisors. Thus
A == B
A == B
implies
-A == -B,
and
O - D
im plies
A + 0 - B + D.
The proofs are immediate.
Let I A I, I B I be two complete linear series (with A and B not necessarily
effective). Let Al - A, Bl - B. Then Al + BI - A + B, whence
IAl + Bli == IA + BI. This linear series is called the sum of IA I and IBI
and is denoted II A I + I B II. Similarly one defines the difference II A I - I B II.
Theorem 16.20. Let A and B be effective. Then /lAI - IB/I consists
of the divisors of IA I that contain B, minus B.
Proof. Let 0 E II A I - I B /I. Then 0 - A - B; 0 + B == A; 0 + B ==
Al E IA I; 0 == Al - B. Conversely, if 0 == Al - B with Al E IA I
and Al > B, thenOE IIAII -IBII == IIAI -IBII.
Thedifferentialsoftheformdx (x E L - 0) are called exact. Bydefinition,
every differential is of the form  dx,  E L (recall that dy == (dyfdx) dx).
Let  -=1= o. Then ( dx)o - ( dx) 00 == (dx)o - (dx) 00 + ()o - () 00' whence

142 LINEAR SERIES
all the divisors (dx)o - ( dx) 00 belong to the same equivalence class of
divisors, called the canonical class. Its order, directly from the definition of
p, is 2p - 2. One also considers the complete linear series I (dx)o - (dx) co I,
which is called the canonical series.
Exercises
1. Let g be a linear series having a fixed place  (hence n > 1). Show that the
elements of g minus  form a g-l.
2. Let g, *g be two linear series of equivalent sets having no element in
common. Show that the two series are contained in a g:, but in no g:.
3. Let g be a Hnear series having no fixed place and let B E g. Show that there
exists a B' E g having no place in common with B.

Chapter 17
LINEAR SYSTEMS
Lemma17.1. Let r:F(Xo, Xl' X 2 ) = 0 be an irreducible curve with
(1, x, y) as generic point and d: G(Xo, Xv X 2 ) = 0 a cycle of order m
such that G =1= O(F). Then
(G(I, x, y))o - (G(I, x, y))oo = d · r - m(loo · r).
Proof. Let  be an arbitrary place, (x o , Xv x 2 ) special coordinates for the
branch of r corresponding to  (that is, ord Xi > 0 for all i, and ord Xi = 0
for some i), and let (xo, Xv x 2 ) be primitive. Then
ord G(xo, Xt, x 2 ) = ord (x o ) + ord G(I, X, y).
Multiplying by  and summing over all , one obtains
! (ord G(xo, Xv X2)) = ! (ord xo) + ! (ord G(I, X, y)),
  
whence
d · r = m(loo · r) + (G(I, X, y))o - (G(I, X, y)) 00.
Corollary 17.2. G = 0 cuts r in a divisor equivalent to m(loo · r).
Corollary 17.3. Let G 1 = 0 also be a cycle of order m and G 1 =1= O(F).
Then G = 0 and G 1 = 0 cut r in equivalent divisors.
Let qJl = 0, . . . , qJr = 0 be curves (or cycles, rather) of the same order
m. The totality of curves of the form C 1 qJl + · · · + crqJr = 0, C i E C, is.
called a linear system of curves. It is said to be generated by
qJl = 0, . . . , qJr = O.
If one of the qJi, say qJr, is a linear combination of the other qJi with coefficients.
in C, then it can be eliminated; i.e., qJ1 = 0, . . . , qJr-1 = 0 will generate
the same system.
143

144 LINEAR SYSTEMS
Theorem 17.4. The members of a linear system of curves (of some given
order m) which do not contain r cut r in the members of a linear series,
and, except for a fixed divisor, conversely.
The system is then said to cut out the series.
Proof. For the converse, let L be a linear series and let
Ac == (! CiX i ) 0 - (! CiX i ) 00 + B
be a typical element of L. Then
Ac == (! ClPi(l, x, y)/d (1, x, y))o - (! ClPi(l, x, y)/d (1, x, y)) 00 + B
== (! ClPi(l, x, y))o - (! ClPi(l, x, y)) 00 + B'
== <Dc · r - m(loo · r) + B',
where the ({Ji are homogeneous, of like degree m, and <Dc IS gIven by
! Ci({Ji(X O , Xv X 2 ) == o.
For the direct part, one reverses the above steps.
Observe that the linear independence of the Xi corresponds to the Jinear
independence of the ({Ji(X O ' Xv X 2 ) mod F. Given a linear system
<l>c : !i=l Ci({Ji(X O ' Xv X 2 ) == 0, if one of the ({Ji, say ({Jv is linearly dependent
on the others mod F, we can eliminate it; i.e., the system !i=2 Ci({Ji == 0 cuts
out the same series on r as does !i=l Ci({Ji == o. Thus we can suppose that
the ({Ji are linearly independent mod F. (If m < ord r, then the ({Ji are
linearly independent mod F if and only if they are linearly independent.)
From here on we will suppose that r has only ordinary singularities and
we designate by D the double-point div1:sor.
Definition. d is called an adjoint of r, and is said to be adjoint to r,
if d · r > D. Then d · r == D + E with E > o.
One says that the adjoint d cuts out E on r (though it really cuts out
E + D) . One says that d passes through F if 0 < F < E.
Theorem 17.5. d is an adjoint (of r) if and only if d has at least an
(r - I)-fold point at every r-fold point of r.
Proof. The if is immediate: if P is an r-fold point of rand y is a branch with
center P, then i(d, y; P) > r - 1.
Let us suppose now that d has only an (r - 2)-fold point at a point P
of multiplicity r for r. Let Yv . . . , Yr be the branches of r centered at
P. Then the tangents at P to the Yi are distinct (the tangent line to a branch
was defined in the proof of Theorem 14.9; in the present case, these are
simply the tangents to r at P). Then the tangent of some Yi, say Yv is
distinct from the tangents to d at P. Then i(d, Yl; P) == r - 2, and d

LINEAR SYSTEMS 145
is not an adjoint. The same argument holds if Ll has at P an s-fold point with
s < r - 2.
Theorem 17.6. The ad joints of a given order cut out on r a complete
linear series.
Proof. Let <I> be of order m and let
<I>.r=D+E
(E effective)
and let E' - E, E' also effective. We must show that there exists a <1>'
of order m such that <1>' · r = D + E'. As E == E', by 17.4 there exist
'Y, 'Y' such that
'Y · r = E + F,
'Y' · r = E' + F.
Here F need not be effective (if E and E' do not have places in common,
then F will be effective, as 'Y, 'Y' cut out effective divisors on r, and any
negative part of F would have to cancel in E and in E'). Nevertheless, we
can suppose F is effective. In fact, let F = Fl - F 2 , Fv F 2 effective. Let
F 2 = l + · · · + t, let L i be a line through i' and let 0 = LI + · · · + Lt.
Then (0 + 'Y) · r = E + FI - F 2 + F 2 + F 3 , with F3 effective. More-
over, (0 + 'Y') · r = E' + FI - F 2 + F 2 + F 3 , that is, we can suppose F
is effective. We have
(<I> + 'Y') · r = D + E + E' + F > D + 'Y · r.
Let
<I> : gy(Xo, Xv X 2 ) = 0, 'Y : 1p(X o , Xv X 2 ) = 0, 'Y' : 1p'(X o , Xl' X 2 ) = 0;
then by Theorem 15.10
gy1p' = AF + gy'1p
[with A, gy' homogeneous, of course; here gy' -=F 0 as gy1p' =I=- O(F)]. Hence
(<I> + 'Y') · r = (<I>' + 'Y) · r = D + E + E' + F,
where <1>' : gy'(XO' Xl' X 2 ) = 0, and hence
<1>' · r = D + E'.
Theorem 17.7. Every complete series is cut out, within a fixed divisor,
by the adjoints of some order m.
Proof. Let I E I be the linear series, EEl E I. We can make an adjoint pass
through E; for example, we can take lines passing through the center of
the branches of E and r - 1 lines passing through each of the points of
multiplicity r of gy; the curve made up of all these lines will be an adjoint
and will pass through E. Let <I> be such an adjoint and let m be its order:
<I>.r=D+E+F
(F effective).

146 LINEAR SYSTEMS
Then the complete linear series IE + F I is cut out by the adjoints of order
m. The elements of IE + F I that contain F are cut out by the adjoints of
order m that pass through F. These elements, minus F, form the complete
series lEI (see 16.20); that is, lEI is cut out by the adjoints of order m that
pass through F.
Definition. The adjoints of order n - 3 (where n == order r) are
called canonical adjoints.
By a computation of the kind occurring in Chapter 14, there are at least
p linearly independent canonical adjoints.
Theorem 17.8. The canonical series I (dx)o - (dx) 00 I is cut out by the
canonical adjoints.
Proof. Let r: F (X, Y) == 0, d: of /0 Y = 0, E, a canonical adjoint, D,
the double-point divisor, (x, y), a generic point of r. Using the notation of
the proof of Theorem 14.6, we have
d · r == D + (dx)o
and
(dx) 00 = -2(l00 · r).
On the other hand, (E + 2loo) · r - d · r, whence
E · r - D + (dx)o - (dxL.
Thus any canonical adjoint cuts out a set equivalent to (dx)o - (dx)oo. Since
the adjoints of a given order cut out a complete series, the proof is complete,
at least if p > 0 and canonical adjoints exist. If p == 0, then the canonical
series, which is of order 2p - 2, must be empty; there can be no canonical
adjoints, and the proof is complete in this case also.

Chapter 18
THE THEOREM OF RIEMANN-ROCH
Let r be a model of  = O(x, y) of order m with only ordinary singularities.
In the calculations below we suppose that m > 2; later we will check the
results separately for m = 1 and m = 2. We also wish to speak of adjoints
of order m - 3, and we apparently have a difficulty for m = 3, as there are
no curves of order m - 3 = O. This is not a real difficulty, however, as we
merely have to use the word cycles instead of the imprecise word curves,
and there is one and only one cycle of order 0, namely, the one defined by
1 = O.
We use I W I to denote the canonical series, i.e., the series determined
by the equivalence class of (dx)o - (dx) 00. This is the series cut out on r
by the canonical adjoints, Le., the adjoints of order m - 3. We use W to
denote an element of I WI, or, more generally, an element equivalent to
(dx)o - (dx) 00.
Theorem 18.1 (Theorem of Riemann).
r > n - p.
I n a complete
r
seres gn'
Theorem 18.2 (Theorem of Riemann-Roch). In a complete sere.'
101 = g with ° effective
r = n - p + i,
where
i = dim IIWI - 1011 + 1,
or, what is the same, i = maximum number of linearly independent
canonical adjoints through a given element ° of 101.
(For the definition of II W I - 1011 see the paragraph preceding 16.20.
One sees, then, that the two definitions of i just given are equivalent and that
the second depends only on 101 and not on 0.)
147

148 THE THEOREM OF RIEMANN-ROCH
In Chapter 14 we showed that the maximum number of linearly indepen-
dent adjoints of order m - 2 is
(1)
(m - l)rn/2 - Z ri(r i - 1)/2 + 8,
with 8 > 0;
this number is equal to p - 1 + m + 8. These cut out divisors of order
m(m - 2) - Z ri(r i - 1) == 2p - 2 + m. Thus the adjoint <D m - 2 's cut
out a complete gp-;J+mm+ E . A priori 8 may possibly be greater than 0 as the
Z ri(r i - 1)/2 conditions imposed on a <Dm-2 to be adjoint may be linearly
dependent. However, i(gp-32++mm+E) == 0, as the canonical series is of order
2p - 2, and consequently no canonical adjoint can cut out a divisor of
order 2p - 2 + m. By the theorem of Riemann-Roch it folJows that 8 = O.
For the present, though, we have to allow for the possibility that 8 may be
greater than o.
As a reformulation of the assertion 8 == 0, we get:
Corollary 18.3. The Z ri(r i - 1)/2 conditions imposed on a <Dm-2 to
make it adjoint are linearly independent.
By a calculation analogous to the one leading to (1), one finds that the
maximum number of linearly independent canonical adjoints is:
(2) (m - l)(m - 2)/2 - zri(r i - 1)/2 + 8 = P + 8,
(Here again we shall see that 8 == 0.)
First we prove a special case of Riemann's theorem.
Lemma 18.4. In a complete series g if n == p + 1, then r > o.
with 8 > O.
Proof. As m > 2, we have 2p + m - 2 > p + 1. By (1), the adjoints
<Dm-2 cut out a g;i::E == L. Let 0 E g (n == p + 1). The elements of L
that contain 0, minus 0, form a g;3t+mm+(pi)1)+El, with 8 1 > o. As
the dimension (== -3 + m + 8 + 8 1 ) of this series is > 0, there exists a
DEL such that D > O. Consider now the elements of L that contain
D - 0 minus D - 0 These form a g P-2+m+E-(2 p -2+m-(p+l»+E 2 8 > 0
, . p+l , 2 - ,
that contains O. As the dimension of this series is > 1, a fortiori r > 1.
Corollary 18.5. Lemma 18.4 also holds for n > p + 1.
Proof. Let 0 + E E g, 0, E effective, ord 0 = p + 1. By the lemma,
dim 101 > 1, whence there exists an effective divisor 0' - 0 with 0' "* O.
Then 0' + E - C + E and 0' + E "* C + E; this implies r > 1.
To prove the theorem of Riemann-Roch, we need the so-called Reduction
Theorem.
Theorem 18.6 (Reduction Theorem). If  is a place with a simple
point P of r as center and there exists a canonical adjoint through 0 not
through C + , then  is a fixed place of 1 0 +  I. (0 >  is allowed,
as the places are being counted with multiplicity.)

EXTENSION OF THE THEOREM OF RIEMANN-ROCH 149
Proof. For some adjoint <Pm-a, <Pm-a · r = D + 0 + E, where D is the
double-point divisor, E is effective, and E does not contain. Through 
draw a line l cutting r in m distinct places , 2'...' m. Then
(<Pm-a + l) · r = D + 0 + E +  + 2 + · · · + m. The adjoints<I>m_2
cut out 10 + E +  + 2 + · · · + ml; and the <Pm-2 through E +
2 + · · · + m cut out 1 0 +  I, Le.,
1 0 + I = 11 0 + E +  + 2 + ... + ml - IE + 2 + ... + mll.
These <Pm-2 cut l in m - 1 = (m - 2) + 1 points and hence contain l as a
component. Hence each passes through , and as  is not in E + 2 +
... + m'  is a fixed place of 10 + I.
Proof of the theorem of Riemann-Roch. The proof is by induction on i.
First we consider i = 0, and here the proof is by induction on r. The first
case, then, is r = 0 (and i = 0). By Corollary 18.5 of Lemma 18.4,
n < p + 1. By (2), the canonical adjoints cut out a gp-!ie, whence a canon-
ical adjoint can be made to pass through any p - 1 places. Since i = 0,
we must have n > p. Hence n = p, Le., r = n - p + i.
Suppose now r > 0 (and i = 0). Let  be a place with a simple point
as center and not a fixed place of the given g. Then by Theorem 16.13, the
elements ofg that contain ,minus,formag-=-\. Nowi(g-=-\)also = 0,
as otherwise there would be a canonical adjoint through an element 0 of
g-=-, and this would not pass through 0 +  [as i(g) = 0]; hence, by the
Reduction Theorem,  would be a fixed point of g, a contradiction. By
induction, r - 1 = n - 1 - p, whence r = n - p.
Suppose now i > o. Let 0 E g. We take a canonical curve through 0
and let P be a simple point of r not on this curve. Then, the place of
center P, is fixed for 10 + I. Hence this is a g+l. Now i(IO + /) =
i(I01) - 1. Hence r = n + 1 - P + (i - 1) = n - p + i. Q.E.D.
EXTENSION AND REFORMULATION
OF THE THEOREM OF RIEMANN-ROCH
Let us write the theorem in the form
(1)
dim 101 = ordO -p + 1 + dim IW -01.
This was proved under the hypothesis that 101 contains at least one effective
divisor (or r > 0). Let us show that (1) holds also when 101 contains no
effective divisor (r = -1).
If dim I W - 01 > 0, then applying (1) to W - 0 we obtain:
(2) dim I W - 0 I = ord (W - 0) - p + 1 + dim I W - (W - 0) I
and reordering (2) obtain (1): hence we can suppose that neither 101 nor

150 THE THEOREM OF RIEMANN-ROCH
I W - 0 I contain effective divisors. In that case we have to prove
(3)
ord 0 = p - 1
or, what is the same,
(4)
ord (W - C) = p - 1.
Suppose that ord 0 > p - 1. Let 0 = A - B with A and B effective.
Then dim 10 + BI > ord 0 + ord B - p > p + ord B - p. Hence we
can impose ord B conditions on the elements of 10 + BI to obtain an
A' E IB + 01 with A' > B. Then A' - B == 0 and A' - B is effective,
which is a contradiction.
If ord 0 < p - 1, then ord (W - 0) > p - 1, and applying the same
argument to W - 0, we again reach a contradiction. Thus (1) holds for
arbitrary o.
The theorem of Riemann-Roch has also been given the following formu-
lation. Let 0 be an arbitrary divisor, possibly virtual, and let xo, xl' . . . , X r
be Jinearly independent elements of L such that
(!CiXi)O - (!CiXi)oo (=! (ord !CiXi)) > -0

for all Ci E O. Then
Ac = (! CiXi)O - (! CiX i ) 00 + 0
is a linear series of effective divisors; hence r is bounded (namely,
r < ord 0). Let 2(0) be the linear space of functions x (over the field 0
of complex numbers) such that ! (ord x) > -0. This space is of finite
dimension with
dim 2(0) = 1 + dim 101.
Then the reformulation we mentioned says that
dim 2(0) = ord 0 - p + 1 + dim 2(W - 0),
which, as one sees immediately, is equivalent to the previous form (1).
Exercises
In all these exercises,  = C(x, y) is the field of rational functions on a curve r
of genus p.
1. If  is a simple transcendental extension of C [Le.,  = C(t)], then r is
birationally equivalent to a Hne and  has a gt.
2. If P = 0, then  has a gt.
3. If  has a gt (necessarily complete), then there exists an element x E :E
having exactly one zero and exactly one pole. Hence  is a simple transcen-
dental extension of C.
4. Let r be a conic. The lines of the plane cut out on r a g.

EXTENSION OF THE THEOREM OF RIEMANN-ROCH 151
5. Let r be a cubic without singularities. The lines of the plane cut out on
r a g, which is complete.
6. Let r be as in Exercise 5. Then any pair of "points" (Le., branches) of r
determine a complete g. The Hnes through a "point" P of r cut out on r
(not counting P) a complete g. The g's belonging to distinct "points"
P are distinct (in other words, if A, B, 0, Dare "points" of r, then
A + B == 0 + D if and only if the residual intersection of AB with
r = residual intersection of OD with r).
7. Show that Y2 = X(X - l)(X - 2) is irreducible and of genus 1, that
Y2 = X(X - l)(X - 2)(X - 3) is irreducible and of genus 1, that
y 2 = X (X - 1 )(X - 2)(X - 3 )(X - 4) is irreducible and of genus 2, etc.
8. If r (or :E) is of genus 1, then it has a complete g. Hence there is a function
 E :E with exactly two zeros and exactly two poles. As a consequence :E
has a model with the equation
a o (X)Y2 + al(X)Y + a 2 (X) = O.
By means of (elementary) birational transformations this can be brought
to the form
Y2 = X(X - c1)(X - c 2 ),
C I C2(C 1 - c 2 )  o.
9. If r is a quartic with one ordinary double point as sole singularity, then
p = 2. This curve cannot be transformed birationally into a (plane) curve
without singularities.
10. The field :E has exactly one g;!2' namely, the canonical series.
11. If r: F = 0, of degree m, has only ordinary singularities and n > m, then
the (maximum) number of cPn Hnearly independent mod F is
(n + l)(n + 2)/2 - (n - m + l)(n - m + 2)/2.
The! r i(r i-I )/2 conditions imposed on a cPn to make it adjoint are Hnearly
independent for every n.
12. Let 0 be an arbitrary divisor (not necessarily effective). Then
i = 1 + dim IW - 01
is the (maximum) number of linearly independent differentials  dx such that
! (ord (dx)) ·  > O.

This follows from the observations at the end of the chapter according to
which 1 + dim ICI is the (maximum) number of linearly independent
functions with
!(ord ).  > -0.

Hence show that there exist exactly p Hnearly independent differentials
without poles (= differentials of the first kind). There are no differentials
with exactly one pole. The number of Hnearly independent differentials with
exactly two assigned poles, or exactly one with multipHcity two, is p + 1.
The number of Hnearly independent differentials with exactly one more pole
is p + 2; the number with exactly s more poles is p + 1 + s.

Chapter 19
THE QUESTION OF THE CHARACTERISTIC
Until now we have worked over the field C of complex numbers, of which,
to be sure, we have used only two essential properties: (1) C is algebraically
closed, and (2) C has characteristic zero. We have also used a subsidiary
property, namely that C is infinite; but this is a property of any algebraically
closed field. Hence, all of our results hold for fields having properties (1)
and (2).
We will now indicate how to eliminate the condition on the characteristic.
In the case of characteristic 0, if r is a curve of order nand E is an r-fold
point of r, r > 0, then there is a line through E cutting r outside E in n - r
sinlple points. This turns out no longer to be true for arbitrary characteristic.
Another difficulty is that the Theorem of the Primitive Element, which was
used in the proof of Theorem 14.1, needs a more precise formulation for
arbitrary characteristic. But let us consider our results seriatim.
There is hardly occasion for comment until we come to Theorem 6.4
and Corollary 6.5. In these there occur 8F 18X, 8F 18 Y. One can say that
wherever a derivative occurs, a scrutiny of the proof is called for: for in
the case of characteristic p > 0, in taking a derivative we may get zero where
we might, without due caution, expect something not equal to zero (because
that is what happens for p == 0). Thus for n > 0, 8ynl8Y == nyn-l,
which is not zero if the characteristic is zero, but is zero if the characteristic
p is positive and n == O(p).
The proof of 6.4 can be maintained as it stands; but 6.5 calls for comment.
There we said that F (X, Y) involves either X or Y, say Y, and then
8F 18 Y -=F 0; this is no longer true for p > 0, since all exponents of Y
occurring in F(X, Y) may be divisible by p. But (recalling that
(a + b)P == a P + b p in a field of characteristic p) we can now say that either
X or Y occurs with exponent ¥= O(p) (otherwise F, assumed irreducible,
would be a pth power), say Y, and then, again, 8F 18 Y -=F 0, and the proof
proceeds as before.
152

THE QUESTION OF THE CHARACTERISTIC 153
Corollary 6.6 continues to hold. However, we may note that, unlike
the case p == 0, if P > 0, then it may happen that the tangent at every simple
point of a curve r may be vertical (pass through E 2 ). For example, this is so
for the curve X - Yp == o. Or we may put the point thus: it may be
that every line through a point E 2 is tangent to r. However, this can happen
for at most one position of E 2 . For let E, E' be distinct and let P be a simple
point of r not on E, E' (where we have dismissed the trivial case that
r == EE'), then PE, PE' cannot both be tangent at P. Say PE' is not.
Take E' as point at infinity on the Y-axis, and let F (X, Y) == 0, F irreducible,
be the equation of r. Then (by 6.6) 8F 18Y #- 0, and a vertical tangent
can only occur for points satisfying F == 0 and 8F 18Y = 0, i.e., for
only a finite number of points. On the basis of this argument, one gets
6.7.
As for the polar, the definition, the proof of invariance in Theorem 6.9,
and the characterization in Theorem 6.10 remain. As before, if r consists
of a number of lines through E, the polar of E is not defined; but it is also
undefined if every line through E is a tangent. The computational rules
(a) through (h) all hold, but Taylor's Theorem fails for arbitrary character-
istic, as might be expected from the appearance of the n! in the denominators;
we have not used Taylor's Theorem in our general developments, however,
but only in some illustrative material of Chapter 10, so this will cause us
little inconvenience.
Chapters 7 and 8 call for no comment. Chapters 9 and 10 take up cubics,
partly in anticipation of our general results for curves. Generally speaking,
in dealing with cubics one will assume p #- 2, P #- 3 (for conics, p #- 2),
and we are not concerned with removing these restrictions. Still, we note that
all the results of Chapter 9 (in particular 9.6, which establishes a group on a
nonsingular cubic) hold for arbitrary characteristic. Since the material of
Chapter 10 is brought in largely for illustrative purposes, RO that we may
as well assume p == 0, we will not go over the results in detail and merely
remark that, in speaking of points of inflection, we assume p #- 2 (in writing
1 8 2 F
_ 2 , 1 8 .8 . YiYi
. Xl x]
as the coefficient of A 2 ), and that 10.7 assumes p #- 2, P #- 3.
Chapter 11 holds for arbitrary characteristic, with no changes.
In Chapter 12, the characteristic plays no role up to 12.15 and in par-
ticular 12.6, which says that at any simple point there is centered one and only
one branch, and 12.14, which says that at any point (of the given curve r)
there is centered at least one and at most a finite number of branches,
continue to hold. In the proof of 12.15, 8F 18 Y intervenes, but if this is
#- 0, the proof continues to hold; this condition can be obtained by a suitable
choice of E 2 (avoiding at most one point; then 8F/8Y #- 0, and also

154 THE QUESTION OF THE CHARACTERISTIC
of'lo Y' =F 0, etc., without further rearranging of the axes). The character-
istic does not intervene in the remaining results of Chapter 12 (except for
point (e), Bezout's Theorem for j, which was postponed to Chapter 14).
Chapter 13 requires no comment.
In Chapter 14, Theorem 14.1 causes difficulty, and to overcome it we
must recall some definitions and facts from field theory.
Let K, L be fields with K c L. An element (X in L is separably algebraic
over K if in the polynomial F (X) E K[ X] of least degree satisfied by (X
not all the exponents (of terms with nonzero coefficient) are - O(p). If
(x, f3 E L are separably algebraic over K, then so are (X :t: f3, (X · f3, and (XI f3
(if f3 =F 0). If every element of L is separably algebraic over K, then L is said
to be separably algebraic over K. If L == K ((Xv . . . , (Xm) and for each i, (Xi
is separably algebraic over K ((Xv. . . , (Xi-I) then L is 8eparably algebraic
over K. The Theorem of the Primitive Element continues to hold for Lover
K provided L is separably algebraic over K. A transcendency basis Xl' . . .
of Lover K such that L is separably algebraic over K (xv. . .) is called a
separating transcendency basis. Let K be algebraically closed. Then: if
L ==.K ((Xv. . . , (Xm), then L has a separating transcendency basis, and,
moreover one consisting of some of the (Xi. If a separating transcendency basis
contains only a single element x, X is called a separating element.
For the proof of 14.1 we need two lemmas.
Lemma 19.1. Let K be of characteristic p > 0 and let X be transcendental
over K. Let x 1 / P be a pth root of X (in an algebraic closure of K (X), say) and
let x l / pe be a root of order pe. Then
[K (x l / P ) : K (x)] == p
and
[ K (x 1 / pe ) : K (x)] == pe.
Proof. x l / P satisfies the equation ZP - x == 0, and it is thus a matter of
showing that ZP - x is irreducible over K (x). For this, it is sufficient to
show that ZP - x is irreducible in K[x, Z]. Suppose on the contrary that
ZP - x == (zml + · · · + a l (x))(Zm 2 + · · · + a 2 (x)), where
a'l(x), a 2 (x) E K[ x]
and
m l > 0, m 2 > o.
Then a l (x)a 2 (x) == -x, so a l ¥=- O(x) or a 2 ¥= O(x) in K[x], say a l ¥=- O(x).
Then a (0) =F o. Now ZP - x == (Z - xl/P)P so the only root of ZP - x
is Xl/po Hence Zml + · · · + al(x), which ca,n have no root other than
x 1 / P , has x l / P as a root. Then (xllp)m 1 + · · · + al(x) == o. Replacing xllP
by 0, one finds al(O) == 0, a contradiction. Hence [K(xI/P):K(x)] ==p;
and for the same reason, [K (x l / p2 ) : K (x I/P )] == p. 1'hen [K (x 11p2 ) : K (x)] == p2,
and similarly [K (x 1 / pe ) : K (x)] == pe.
Lemma 19.2. Let K be algebraically closed, of characterist1.c p > o.
Let x be transcendental over K and let u be in an extension field of K (x)

THE QUESTION OF THE CHARACTERISTIC 155
and algebraic over K(x), satisfying an equation F(x, U) === 0, where F
is irreducible over K (x) and can be 'lvritten as a polynomial in Up 6 but not
as a polynomial in UP6+1. Th.en K(x, u) contains a root of order pB
of x.
Proof. Let F (x, U) === G(x, UP 6 ), and let deg v G(x, V) === d, so that
deg u F(x, U) === dpe and [K(x, u): K(x)] === dpe. We may suppose
F E K[ x, U] and write G === ! CijX i ( UP 6 )j . Working in an algebraic closure
of K (x), we see that u satisfies (Gl/P6 ===) ! Cijllp6(Xl/p6)iUj === O. This equation,
which is of degree d, is the minimal equation for u over K (x 1 / p6 ), because if u
satisfied an equation H(x llp6 , U) === 0 of degree d' < d, then u would satisfy
H P 6(x llp6 , U) = 0, which is an equation over K(x) of degree d'e P < de'P.
Hence [K (x 1 / p6 , u) : K (x 1 / pe )] === d and (from 19.1)
e
[K(x 1 / P , 'u): K(x)] === dpe.
Hence
K (x, u) = K (x 1 / pe , u),
so l/x P6 E K (x, u). Q.E.D.
Now for the proof of (14.1) in the case p > 0:
Proof. Let  = K (u, v) be our field of aJgebraic functions of one variable.
Let x E  be transcendental over K. If there were an element y E  such
that  === K (x, y), then the previous proof of (14.1) (for p === 0) would appJy
here; for example, if x were a separating element. In general, consider the
field K (x, u) and let e be the integer introduced in Lemma 19.2, so that
x llP6 E K (x, 'u). Let e' be the corresponding integer defined for x and v.
Without loss of generality, we may assume e > e' (if e' > e, we interchange
the roles of u and v). Then x llP6 E K (u, v), a.nd u, v are separable over
K (x 1 / P6 ). By the Theorem of the Primitive Element,
K (x 1 / p6 , U, v) = ]( (x 1 / P6 , 0).
1'hus  === K (x llp6 , 0). By the remark at the beginning of the paragraph,
x llP6 has [: K (x 1 / pe )] zeros (counted with multiplicity). Each zero of
x 1 / P6 counts pe times as often as zero of x. Hence x has pe · [L : K (x llpe ) 1 =
[ : K (x)] zeros. Q.E.D.
We still mention the following:
Theorem 19.3. Let  be a field of algebraic functions of one variable
over an algebraically closed ground field K (i.e.,  is finite over K and d.t.
/K = 1) and let x E  be transcendental over K. Then there is ayE 
such that  = K (x, y). (A similar theorem for d.t. L/K > 1 is false.)

156 THE QUESTION OF THE CHARACTERISTIC
Proof. Let u be a separating element for /K. By the Theorem of the
Primitive Element, there is a v such that  == K (u, v). In the notation of
the previous proof, u pB and v pB ' are separable over K (x), so by the Theorem
of the Primitive Element, K (x, u pB , v pB ') == K (x, fJ) for some fJ.
We now show that u E K (x, u pB , x l / pB ); and similarly v E K (x, v pB ', Xl/pB').
We have that xl/ pB satisfies the equation ZpB - x == 0 over K (x, u pB ), and
claim this equation is irreducible over that field; for suppose
ZpB _ x = (Zm 1 + .. .)(Zm 2 + · · .), 0 < m 1 < pee
Working over the algebraic closure of K(x, u pB ), we see that
ZpB _ x = (Z _ Xl/ pB )p8,
so that Zml + · .. and Zm2 + · · · must each be a power of Z - xl/fl.
Let pY == G.C.D. (m 1 , pe); g < e as m 1 < pee Then pY == qm l + rpe, q, r
integers; and x 1 / pB - g = (xml/pB)qxr E K (x, u pB ). The equation of least degree
for x 1 / pB - g over K (x) is ZpB-U - X == 0, which contradicts the separabiJity
of K (x, u pB ) over K (x). So now we have that [K (x, u pB , XI/pB) : K (x, u pB )] == pe;
and, recalling the notation [K (x, u pB ) : K (x)] == d, we have
[K (x, u pB , xl/ pB ) : K (x)] == dpe.
Since K (x, u pB , xl/pB) C K (x l / pB , U) == K (x, u) (by 19.2) and
K[ (x, u) : K (x)] == dpe,
we have
K (x, u) = K (u pB , XI/pB),
whence u E K (x, u pB , XI/pB).
From this there follows: u, v E K (x, fJ, XI/pB), or  == K (x, fJ, XI/pB).
Recall that the Theorem of the Primitive Element says that if in the field
L(rJ.., f3), rJ.. and f3 are separable over the field L, then L(rJ.., f3) has a single
generator over L. We now quote a well-known and somewhat deeper asser-
tion which says that the same conclusion holds if one of rJ.., (3 is separable over
L. Since fJ is separable over K (x), we conclude that  has a single generator
over K (x). Q.E.D.
Remark. For the developments of Chapters 14 to 18, we do not so much
need 14.1, which tells the number of zeros of a function, as merely 14.2.1,
which says that the number of zeros == number of poles. A rather simple
proof of this could be based on Bezout's Theorem for j(r,; P), therefore on
quite elementary results provided we knew that i(r,; P) == j(r,; P). As,
however, we did not wish to make this equivalence central in our exposition,
we preferred to prove 14.1; Theorem 14.3, Bezout's Theorem for j(r,; P),
is then made to rest on 14.1. If 14.1 for p > 0 had caused us more difficulty,
we would have taken the way just indicated to get the results of Chapters 14
to 18 for arbitrary characteristic.

THE QUESTION OF THE CHARACTERISTIC 157
To continue with Chapter 14, differentials are defined as before. Note
that d' == 0 for , == '" C i .t ij (c i . "* 0) if and only if all the ij are == O(p).
k J J
Let  be a place and (x, y) -+ (x, fj) == ((x), (y)) in a primitive place
representation ((x, y) a generic point of r). Then dx, dfj cannot both be
zero, as  (x),  ( y) are not both po\ver series in t p . Moreover if x is a separating
element, then dx * 0; for (oF jox)dx + (oF jofj)dfj == 0, and dx == 0 implies
of jofj == 0 and of Joy == 0, which is not so. Thus if , is a separating element
of K (x, y)jK, then we can define the derivation d jd' as before.
Because d' * 0 for a separating element " we frequently let a separating
element take over the role of the (arbitrary) element of the characteristic 0
theory. Thus 14.4 continues to hold if x is separating. Similarly, the genus
is defined as before, by the formula ! ord d' == 2 P - 2, except that here
we require' to be a separating element. Again by 14.1 (or 14.2.1), we see
that p does not depend on ,.
We now conle to 14.5, the resolution of singularities. The first main point
is that the apparent genus is nonnegative: the proof is the same as before,
with slight nlodifications of the kind already encountered. Let, then, r
have a (nonordinary) singularity and place it at Eo: (1,0,0). Take the axes
EoEv EoE2 to be different from the tangents to r at Eo, take E 1 , E 2 not on
r and such that EIE2 passes through no singularity. Applying the transfor-
mation 'T: X == IjXo, X == IjX 1 , X; == IjX 2 , we find that the curve r
goes over into a curve r' . We need to know that an s-fold point P of r
not on the axes goes over into an s-fold point P' of r', and that if P is ordinary,
then so is P'. We can establish this as before, using branches, or it can be
done by a straightforward computation making no use of the characteristic.
Now unless the point at Eo is resolved into simple points, the apparent genus
(by the argument of Chapter 14) diminishes. As this can happen only a finite
number of times, we may suppose that none of our transformations reduces
the apparent genus, as none of them will raise it.
Let A be an r-fold point of r (to be placed at Eo). It may be that for
every point P of r, AP cuts r at P with multiplicity> 1; but we have seen
that this can happen for at most one point A. Supposing A does not have this
undesirable property (7T), we place it at Eo, take EoEl and EoE2 so that they
each cut r in n - r further distinct simple points (n == ord r), and take
EIE2 so as to cut r in n distinct simple points. Applying the transformation
'T, we see that A is resolved into simple points on r', and three ordinary
singularities are introduced. In this way we can resolve all the nonordinary
singularities except, perhaps, one.
We suppose, then, that r has only one nonordinary singularity A and that
A has the property (7T). We place A at Eo and take Ev E 2 suitably. (See
Fig. 19.1.) Then Ev E 2 do not have the property (7T) for r. The transformed
curve r' has a (new) ordinary n-fold point at Eo and (new) nonordinary
(n - r)-fold points at E 1 , E 2 . The points Ev E 2 also do not have the property

158 THE QUESTION OF THE CHARACTERISTIC
n-r
E2
Eo
n-r
El
Fig. 19.1
(1T) for r'; for there exists a line through E 2 , not an axis, cutting r in n
distinct points. This line is transformed by 'T into a line through E 2 cutting
r' in n points outside of E 2 (and with multiplicity n - r at E 2 ). Since r'
is of order 2n - r, these n points are simple for r', and therefore E 2 does
not have the property (7T) for r'; the same is true for El. Since this is so,
we can resolve the (n - r)-fold point of r' at E 2 into simple points without
introducing non ordinary singularities. Moreover we do it in such way that
El continues not to have property (7T) for the transformed curve r": we
keep E 2 as a vertex and take new vertices E, E such that EE2' EE2 cut
r' in n distinct simple points outside of E 2 ; E, E should not be on r'; EE
should cut r' in 2n - r distinct simple points, none at E, E; El should not
lie on the new axes EE2' EE2' EE; and EIE should cut r' in n distinct
simple points outside of El. This last condition assures us that the El will
also not have the property (7T) for the transformed curve r". Applying
one further transformation, the singularity at El is resolved and the trans-
formed curve rIll has only ordinary singularities.
For 14.6, Lemma 14.11 was needed. We gave two proofs, the second of
which holds for arbitrary characteristic. Theorem 14.6 then holds for
arbitrary characteristic.
Chapter 15 never refers to a derivative, or to the Theorem of the Primitive
Element; its arguments hold for arbitrary characteristic. So do those of
Chapters 16 and 17 (though these refer back to 14.1).
The canonical class is defined as before except that in speaking of differ-
entials 'YJ d, we always suppose  to he separating.
In Chapter 18, in the Reduction Theorem, we drew a line l through the
(simple) center P of a place , cutting the curve r in m - 1 other distinct
simple points. In characteristic p > 0, we cannot be snre of being able
to do this, though we can make the other intersections fall at simpJe point.
A reference to Lemma 9.2 (proof at the end of Chapter 11) shows, however,
that the previous argument for the Reduction Theorem holds, with a minor

THE QUESTION OF THE CHARACTERISTIC 159
revision, also for arbitrary characteristic. The other results of Chapter 16
require no comment.
To sum up: All oj' our birational results, including the Theorem of Rie1nann-
Roch, hold for arbitrary fields  = K (xv . . . , Xs) of algebraic functions of one
variable (Le., degree of transcendency of  over K is 1) over an algebraically
closed ground field K. (The restriction to plane curves is made without loss
of generality, since every such field  has a pair of generators over K.)

Chapter 20
LINEAR SERIES AND RA TIONAL MAPPINGS;
SPACE CURVES
The definition of algebraic curves in higher-dimensional spaces is best given
in the context of the definition of arbitrary algebraic varieties (see Chapter
9), but in order to come directly to an important connection between linear
series and rational mappings, we give an ad hoc definition. This bypasses a
difficulty which is overcome in Chapter 23.
Let K denote an arbitrary algebraically closed ground field.
A point in affine n-space, AN Kn, is merely an n-tuple (a v ..., an)
witha i E K, and we define branch, as before, as a point in AN K( (t))n - AN Kn.
Similar definitions hold for projective space. The center of a branch is also
defined as before; for example, if Xl = a l + · · · , . . . , X n = an + · · · ,
then its center is (a v . . . , an).
Let P: (x o , . . . , x n ) be a point in projective n-space, where the Xi
are contained in an extension field of K. Then for some i, Xi =F o. By K(P),
the field of the point, one means K (Xo/x i , . . . , xn/x i ) ; if also Xj =F 0, then
K (Xo/x i , . . . , Xn/Xi) = K (xo/Xj, . . . , xn/Xj).
Let  be a finite extension of K of degree of transcendency 1, so that
(by previously mentioned theorems)  is of the form  = K (x, y). Let
P: (x o , . . . , x n ) be a point and assume that the Xi E; moreover, assume
that K (P) = .
Definition. By an irreducible algebraic curve r in n-space, one means the
set of centers of the branches ((xo)' . . . , (xn)) as  varies over the
places of. These branches are called the branches of r.
Note that this definition agrees with our previous definition for plane
curves.
Theorem 20.1. Every irreducible curve r contains infinitely many points.
Proof. Let r be determined by P: (x o , . . . , x n ) and let  = K(P). We
may assume Xo =F 0; let Yi = xdxo. Dismissing the places of  where any
160

LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES 161
Yi has a pole, we still have left infinitely many places  of. For any
such place, (1, (YI)' . . . , (Yn)) has center at finite distance. If
(Yi) == ai + · · ., this center is (a l ,..., an). One of the Yi, say YI,
is transcendental over K. Then Yl - a l has  as a zero. Since Yl - a l has
only a finite number of zeros, only a finite number of places can be centered at
(aI' . . .). As any finite number of points can be the centers for only a finite
number of places, r must have infinitely many points (at finite distance,
even). Q.E.D.
By definition, a hypersurface is given by a polynomial equation:
F(X v ..., X n ) == 0, FE C[X 1 ,..., X n ], F  C; a hyperplane is given by
a linear equation: a o + alX 1 + · · · + anX-n == 0; similar definitions hold
in projective space.
Most of Chapter 3 can be repeated for hypersurfaces. In particular, two
hypersurfaces F == 0 and G == 0 have the same locus (Le., are the same) if
and only if F and G have the same irreducible factors; if l : Fl == 0, . . . ,
s : F's == 0 are irreducible hypersurfaces, then we say that Fl · · · F. == 0
is the equation of the cycle rll + · · · + rss.
If y is a branch of center A and (xo(t), . . . , xn(t)) are special coordinates
for y [i.e., the representation is primitive, ord t xi(t) > 0 for all i, and
ord xi(t) == 0 for some i] and : G(Xo, . . . , X n ) == 0 is a hypersurface,
then by definition
?:(, y; A) = ord t G(xo(t), . . . , xn(t)).
The branch is said to be linear if i(H, y; A) == 1 for some hyperplane H.
Theorem 20.2. Let the curve r be deter1nined by P: (x o , . . . , x n )
and let  : G == 0 be a hypersurface. Then  ::::> r if and only if
G(xo, . . . , x n ) == o.
Proof. The if part is immediate. For the converse, assume G(xo, . . . , x n ) * o.
Notationally, let Xo * 0, Xi/XO == Yi. Then G(I, YI ..., Yn) * 0, and
G(I, Yv . . . , Yn) can have only a finite number of zeros. If A : (1, a v . . . , an)
is a point of rand YI == a l + · · · , Yn == an + · · · is a branch y of r
centered at A, and if G(I, a v . . . , an) == 0, then y, or the corresponding
place, rather, is a zero of G(l, Yv . . . , Yn). Hence  can pass through only
a finite number of points of r at finite distance. As r has infinitely many
points at finite distance (see the Proof of 20.1), the proof is complete.
By a homogeneous nonsingular linear transformation, any point can
be moved to any other. Hence without loss of generality we may suppose
r has a point on none of the hyperplanes Xi == o. Then Xi * 0, all i, and
we get:
Corollary 20.3. If  :p r, then  meets r in only a finite number of
poi nts.

162 LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES
Proof. Assume Xi -=1= 0, all .i. We saw in the above proof that  Ineets r
in only a finite number of point.s at finite distance relative to Xo = o.
The same holds for any Xi = 0, and as every point is at finite distance for
some Xi = 0, the proof is complete.
Theoren 20.4. Let y be a branch of r centered at A and assume that
 : G = 0 does not contain r. Then i(, y; A) is a .finite number (i.e.,
if y: (xo(t), . . . , xn(t)), then G(x(t)) -=1= 0).
Proof. Let y arise from the place . Then  maps K (x o , . . . , x n ) iso-
morphically onto K(xo(t), . . . , xn(t)), with Xi mapping into Xi(t). Thus the
xi(t} satisfy the same polynomial relations over K as do the Xi. As G(x) -=1= 0,
also G(x(t) -=1= o. Q.E.D.
With a slight change in the wording and proof of 17.1 we get:
Theorem 20.5. Let r be an irreducible curve, determined by (1, Xl' . . . , x n )
and : G(Xo, . . . , X n ) = 0 a hypersurface of order m (Le., deg G = m)
such that  :p r. Then (with H 00 : Xo = 0)
(G(I, Xl'... ,Xn))o - (G(I, XV... , X n )) 00 = . r - m(Hoo. r).
By ord ( ,. r) one means the total intersection multiplicity of  with r.
Hence:
Corollary 20.6. ord ( · r) = ord  · ord (H 00 · r), and
ord (H · r) = ord (H 00 · r)
for any hyperplane not containing r.
This number, ord (H · r), is called the order of r. Hence:
Corollary 20.7 (Bezout's Theore1n). ord ( · r) = ord  · ord r.
Parallel to 17.2, 17.3, and 17.4, and with similar proofs, we have:
20.8.  cuts r in a divisor equivalent to m(H 00 · r).
20.9. Let 1 ::p r be another hypersurface of order m. Then l. r
is equivalent to  · r.
20.10. A linear system of hypersurfaces (of some given order m) cuts
r in a linear series, and, except for a fixed divisor, conversely.
Relative to 20.10, observe that if f{Jo = 0, . . . , f{Js = 0 are hyper-
surfaces, it is possible for  Cif{Ji = 0 to contain r even if none of the f{Ji = 0
do. In the case of plane curves, we spoke of linear independence mod F
(where r was given by F = 0), and we need a parallel concept here. We say
that f{Jo, . . . , f{Js are linearly dependent mod r if  Cif{Ji vanishes over r for
some C i E K not all zero; or, what is the same, if  Cif{Ji(X O ' . . . , x n ) = 0,
where r is determined by (xo, . . . , x n ). One can then eliminate one of the
Pi' and in this way suppose the f{Ji are linearly independent mod r.

LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES 163
Birational transformation can be defined as before, namely, as a notion
made up of two "points" P, Q with K(P) == K (Q) ==  (d.t. /K == 1), and
if K (P) == K(Q) == , then the curve defined by Pwill be birationallyequivalent
to the one defined by Q. Points A and B (with coordinates in K) correspond
if they are the centers of corresponding branches (and branches correspond
if they arise from the same place of I;).
Since every field K (P) of degree of transcendency lover K has a pair
of generators, we have:
Theorem 20.11. Every irreducible curve is birationally equivalent to
a plane curve.
By a simple point of a curve r one means a point which is the center of
just one branch, and that linear, of r.
Theorem 20.12. Every point of an irreducible curve r is the center of
only a finite number of branches (of r), and r has only a finite number
of singular (i.e., nonsimple) points.
Proof. Let A be a given point of r . We may suppose A is at finite distance.
If r is defined by (x o , . . . , x n ), then Xo =1= 0, and we may suppose Xo == 1.
If A : (1, al' . . . , an), the branches centered at A are given by expansions
Xl == a l + · · · , . . . , X n == an + · · · , where the omitted terms have positive
order . Now some Xi' say Xl' is not in K. In all these branches, ord (Xl - a l ) > 0;
the corresponding places are zeros of Xl - a v and there are only a finite
number of them. This proves the first point. Hence there are only a finite
number of branches centered on H ex>, and we dismiss these.
Let y be one of the branches centered at A. One of the Xi' say Xv is
separating; let Xl == a l + ct V + · · · , c =1= o. If v > 1, then  is a zero
of dx l , and this is so only for a finite number of places. Hence all but a finite
number of branches of r are linear. Dismissing the nonlinear branches, we
may suppose that the branches centered at A are linear.
For the remaining point, let (x, y) be a generic point of a plane curve
r' birationally equivalent to r. Let X == f(x v . . . , xn)/d (Xl' . . . , x n ),
y == g(xl' . . . , xn)/d (Xl' . . . , X n ), f, g, d polynomials. Since d (Xl' . . . , X n ) =F 0,
the hypersurface d (Xv. . . , X n ) == 0 meets r in only a finite number of
points. We dismiss all the places centered at these points. Any branch
centered at A : (aI' . . . , an) goes over into (Le., corresponds to) a branch of
r' centered at
A': (f(al' . . . , an)/d(a l , . . . , an), g(a l , . . . , an)/d(a l , . . . , an)).
If this point is singular for r', we dismiss the branch spoken about; in this
way we dismiss only another finite number of branches. Then there is only
one branch of r' centered at A', and hence only one of r centered at A.
Q.E.D.

164 LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES
Rational transformations are also defined as before. Here we have
"points" P, Q with K (P) ==  and K (Q) == ' c. In the nontrivial
case, d.t. ' IK == 1; in the trivial case ' == K (and r is transformed into
a point). We usually tacitly suppose we are in the nontrivial case. If 
is a place of , given, therefore, by an isomorphic mapping T of I; into
K((t)), then T induces an isomorphic mapping T' of' into K((t)), thereby
giving rise to a place ' of '; one says that  lies over '. Similarly, if
the points P, Q define curves r, r', then a branch of r gives rise to a branch
of r'; the center of the first branch is transformed into the center of the
second, by definition.
To facilitate the discussion of rational transformations (or mappings),
we first consider a purely field-theoretic question.
Let ' be a place of' and  a place of  over it. Let  be given by T
and let T induce T'. Then even if T is primitive, which let us assume, T' need
not be; let v be its redundancy. If  E ' - K has  as a zero, then it also
obviously has ' as a zero, and conversely. It is also immediate that
ord  == v ord,; thus v == ord /ord,  for every  E ' - K.
Now consider the linear series Im' I, which is a g in '. Then
r > m - p, p == genus of'. For sufficiently large m, then, r > 1, so there
exists an effective divisor B - m', B =J= m'. If B has ' as component,
cancel ' as often as possible. We may suppose, then, that ' is not a
component of B. Hence there exists an element x such that
(x o ) - (x) 00 == m' - B.
Therefore x has ' and no other place as a zero. Usually m > 1, in fact,
m == number of zeros of x in ' == [' : K (x)]. In , however, x has
[L: K (x)] zeros. Therefore:
20.13. Over every place ' of' there lies at least one and at most a finite
number of places of .
Let v . . . , s be the places of  lying over '. Let Vi be the redun-
dancy encountered at i' so that ordi x == Vi ord, x. Then we say that
to ' corresponds VI I + · · · + VSS, i.e., we count the i with multiplicity.
Now:
[: K (x)] == ! ord. x == ! Vi ord, x == (! Vi) · [': K (x)],
m Z .
1Ji 
whence
! Vi == [ : '].
i
Thus:
Theorem 20.14. To each place ' of' there correspond [I; : '] places
of  (counted with multiplicity).
In a rational transformation, multiplicity is counted in a similar way.
Hence we get the following corollary.

LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES 165
Corollary 20.15. A rational transformation from r to r' is birational
if and only if to each place of r' there corresponds just one place of r;
and, also, if and only if to some place of r' there corresponds just one
place of r.
Exercise. Using differentials, show that if  is of genus 0, then so is L'; assume
K of characteristic o.
Theorem 20.16. Let r be defined by P: (1, Xl'...' x n ). Then for
some pair of elements YI == a&l) + all)x I + · · · +' al)xn' Y2 == a&2) +
al 2 )x I + . . . + a2)xn' aj) E K, the curve r' defined by Q: (1, Yv Y2) is
birationally equivalent to r.
Proof. Let '-13 be a place of I; == K (P) centered at a simple point A of r.
By avoiding the poles of the Xi' we may suppose A is at finite distance
(Le., not on Xo == 0). Then for some a)l) E K, the hyperplane
H{l): a&l) + all)X I + · · · + aI)Xn == 0
passes through A and ord (a&l) + all)x l + · · · + al)xn) == 1 in the place
centered at A. We take Yl == a&l) + all)x I + · · · + al)xn. The element
YI has only a finite number of zeros; let
H{2): a&2) + al 2 )X I + · · · + a2)Xn == 0
be a hyperplane through A but not through the other zeros of YI. We take
Y2 == a&2) + al 2 )x I + · · · + a2)xn. Let I;' == K (Yl' Y2) and let '-13 lie
over '-13'. Then '-13' is a common zero of YI and Y2' and '-13 is the only place
lying over '-13'; moreover the multiplicity is 1, because already from the fact
that ord Yl == lone sees that a representation T of '-13 induces one of '
of redundancy 1. By 20.15, the proof is complete.
By taking H(1) and H(2) as above and in addition ap), j == 3, . . . , n,
i == 1, . . ., n such that det laj) I "* 0 (i, j == 1, . . . , n), one may subject
n-space to an affine transformation (or to a change of coordinates). We
may then assume that the transformation in 20.16 is then given by Yl == Xl'
Y2 == x 2 . The transformation (a l , . . . , an) --+ (aI' a 2 ) is a projection. Thus
we can say that the rational transformation of 20.16 determined by (P, Q) is
(induced by) a projection.
The transformation (a v ..., an) --+ (al' a 2 ) is induced by the trans-
formation in projective space given by (a o , al' a 2 , . . . , an) --+ (a o , aI' a 2 ),
which is defined at points not on the linear space Ln-3 given by Xo == 0,
Xl == 0, X 2 == 0, and which in fact is the projection from Ln-3 as center
to the plane given by X 3 == 0, . . . , X n == 0, Le., the join of (a o , aI' a 2 , . . . , an)
and Ln-3 is an Ln-2 meeting the plane X 3 == 0, . . . , X n == 0 in (a o , a l , a 2 ,
0, . . . , 0), or, omitting the last coordinates as superfluous, in (a o , a v a 2 ). It is to
be noted that r may intersect the center, Ln-3, and if so, the transform, or
transforms, of the points of intersection are defined; not in a simple-minded
fashion, to be sure, but via the branches centered at those points.

166 LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES
To summarize: Every irreducible curve can be transformed by means of
a projection into a birationally equivalent plane curve.
Let P: (x o , . . ., x n ) and Q: (Yo, . . . , Ym), Xi' Yi E  = K(P), deter-
mine curves r, r' and consider the rational transformation T determined
by P, Q. Together with the transformation we can consider the linear series
of typical element
m m
Ac = (! CiYi)O - (! CiYi) 00 + B,
i=O i=O
where B is a fixed divisor. Conversely, the typical element of a linear series
on a curve r can be written in the above form, thus determining a point
Q : (Yo, . . . , Ym) and a rational transformation of r. Hence there is an
obvious connection between linear series and rational transformations.
Our object is to explicate this relation a bit.
The curve r' lies in projective m-space, 8m, but if Yo, Yv . . . , Ym are
linearly dependent over K, say Co Yo + · · · + CmYm = 0, then the hyper-
plane H: Co Yo + · · · + cmY m = 0 passes through the center of every
place of I;' = K (Q) on r', i.e., r' lies in H. Conversely, if r' lies in H, then
coYo + · · · + CmYm has infinitely many zeros, and hence
CoYo + · · · + CmYm = o.
The significance of our usual assumption when discussing linear series, that
Yo, . . . , Ym are linearly independent, is thus clear: this assumption holds
if and only if r' is contained in no proper subspace of 8m.
If to the divisorsA c = (! CiYi)O - (! CiYi) 00 + Bis added a fixed divisor
0, a new series results, but the associated rational transformation is the same.
Hence we will assume that the series {Ac} in question contains only effective
divisors and is without fixed point.
To describe the transform under the rational transformation T, it will
be helpful to make the following purely field-theoretic definition. Let'
be a place of I;' and let VI l + · · · + V s  s be the divisor of I; corresponding
to ' (this was introduced just prior to Theorem 20.14).
Definition 20.17. The set of divisors VI1 + · · · + vss obtained as
' varies over the places of I;' is called an involution (of I;), of order
V = VI + · . · + 'V S. The notation for an involution of order V is y V.
If one uses the letter ' also to designate the corresponding branch on
r', and similarly for l' . . . , s and r, then one uses the same terminology
for r and the series {Ac}.
Definition 20.18. If the place  of I; lies over the place ' of I;', we
write T() = ' (though there is no T given); and similarly for divisors:
T(! ftii) = ! ftiT(i). Similar terminology is used for a rational
transformation T.

LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES 167
Theorem 20.19. Let x E ' and let (x)o, (x) denote the divisor of zeros
of x relative to  and to I;'. Then T( (x)o) == [ : '] · (x).
Proof. Let (x) == ftl + · · · + fts; (i -=1= j for i -=1= j). Let Viiij
correspond to j. Then ordii x == Vii ordj x == Vijftj. Hence
(x)o == ftI(! Vilil) + · · · + fts(! ViSiS)
and
T( (x)o) == ftl(! Vii ) + · · · + fts(! Vis;)
== [ : '](ftl + · · · + fts). Q.E.D.
Corollary 20.20. If (x) ==  + · · · + ;, then (x)o is the sum of
the divisors corresponding to the i, hence, in particular, is made up of the
divisors of an involution.
Let, now,
(*)
Ac == (! CiYi)O - (! CiYi) 00 + B
be the typical element of our series. We may assume (*) is normalized:
this involves replacing the (Yi) by a multiple (PYi), but this does not change
r'. Then B is a member of the given linear series. Let l' . . . , s be
the places of B. The condition that Ac contain i is a linear condition
! AijCj == 0 on the Cj (see 16.13); this condition is not trivial, Le., not all the
Aii == 0, since i is not fixed for {Ac}. Thus the polynomial! AijC j is not
zero, and hence neither is IIi=l (!AijC j ). Just as we proved that there are
points not on a given plane curve, so we can prove that there are points not
on the hypersurface !i=l (! AijC j ) == O. If c is such a point, then Ac goes
through none of the i. Let c be chosen this way. Then Ac is the set of
zeros and B the set of poles of ! C i Y i ; B is therefore also the set of zeros of a
function. By 20.20, B is made up of elements of -an involution. Similarly
any (! CiYi)O and (! CiYi) 00 are made up of elements of the same involution.
Hence:
The elements of {Ac} are made up of elements of an involution.
Every series {Ac} or rational transformation 'T defines an associated
involution, y({Ac}) or Y(T), and throughout we have been speaking of this
involution. If y({A.c}) == Yv and v > 1, then {Ac} is said to be made up
of or composed of or composite with the involution y v; otherwise {Ac} is said
to be simple.
Clearly, T is birational if and only if the corresponding series is simple.
(Proof. I; == I;' if and only if [ : I;'] == 1.)
Applying T to (*) we get
T(Ac) == v[(! CiYi) - (! CiYi)'oo] + T(B),
where V == [I; : ']. Each place in T(B) occurs with a multiplicity that is a

168 LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES
multiple of v, and similarly for T(A e ); hence we can divide by v to get
(**)
1 , , 1
- T(Ae) == C: CiYi)O - C: CiYi) 00 + - T(B) .
v v
The sets (l/v)T(A e ) clearly form a linear series on r'. If  + · · · + ;
is an element of this series, then the sum of the divisors corresponding to the
 is an ...4 e ; conversely, every Ae is obtained this way. The series {(I/v)T(A e )}
has no fixed point; for suppose it did and ' were such a fixed point. Let
v . . . , s be the places over '. By a proof given a moment ago, there
is an Ae passing through none of the i' and T(Ae) does not go through ',
a contradiction. Thus (**) is without fixed points; of course, it is also effec-
tive. The hyperplanes of 8m cut out
(! CiYi) - (! CiYi)'oo + D;
as this is without fixed points and effective, the multiplicity A with which
any place  enters D is -mine {ord (!CiYi)}. Since (l/v)T(B) can be
described in the same way, D == (l/v)T(B). Thus (**) is the series of hyperplane
sections of r'. In particular:
20.21. If a linear series on r defines a birational transformation, then
the series is given on the transform r' by the hyperplane sections.
20.22. If a gr;: has no fixed point and is composed of an involution, then
for every place  of I;, every divisor of gr;: containing  also contains
another place 2' where 2 depends only on  (2 ==  is allowed, i.e.,
the divisor gr;: is to contain  with multiplicity > 2); and conversely.
Proof. Let g': be composed of the involution Yv, v > 1. Let  + 2 +
· · · + v be an element of the Yv. Then every divisor of gr;: containing  con-
tains each of2' . . . , v. This proves the direct part. For the converse, assume
that gr;: is simple, so that it defines a birational transformation; and gr;: is cut
out on the transform r' by the hyperplanes of 8m. Let  be a simple point of
r'. Then for every point 2 of r' there is a hyperplane through  not
through 2 (for 2 == , this means that there is a hyperplane cutting out
 + · · · but not 2'l3 + · · · on r'). This proves the converse.
This proof also contains a criterion for the transform of  on r' (i.e.,
the center of the place  on r') to be simple. We state it as:
20.23. Let gr;: be the series cut out on r' by the hyperplane sections. Then
 is simple for r' if and only if the subseries of gr;: containing , minus
, does not have  as fixed point.
Exercise. Let gr; have no fixed point and be composed of the involution Yv.
Then with the exception of a finite number of places, every place  belongs to just
one member of Yv and this member is the common part of t.he elements of g':

LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CtJ"'"RVES 169
containing. In general, the common part will be made up of several elements
of the YV.
An easy consequence of the last two theorems is:
20.24. Every complete g with n > 2p on a curve r of genus p defines
a birationally equivalent curve r' free of singularities.
Proof. The g can have no fixed point, otherwise we would get a complete
g-l (by subtracting a fixed point); then r == n - p and r == n - 1 - P
(since n - 1 > 2p - 2, the g-l is nonspecial); this is a contradiction.
Now let  be any point on r. Then we get a complete g=-l from the subseries
containing. This series can also not have a fixed point, otherwise we would
get a complete g-=-, leading to a contradiction as before. The proof is now
complete.
Examples
1) Let r be an irreducible conic (in a plane). The lines through some
given point on r cut out a g. The transform r' is a one-dimensional
projective space (i.e., a line). The transformation is birational.
2) Let r be an irreducible cubic with a double point. The lines through
the double point cut out a g. The transform is a line. The transformation
is birational.
3) Let r be an irreducible plane quartic with just two ordinary double
points, so that p(r) =- 1. The conics through the double points cut out a
g:. This series has no fixed points, otherwise r would have a gg, hence would
be of genus o. If  is any place, the elements of g: containing , minus ,
form a g:. This also cannot have a fixed point. Hence the transformation
defined by g is birational. The transform r' is a 3-space quartic of genus 1
without singularities.
4) There is also a 3-space quartic of genus 0 without singularities, but
it is harder to get. Let g be the complete series on a curve r of genus 0,
determined by any four points on r. From it, we will select a suitable g:.
The g defines a curve r' of order 4 in 4-space, and this is without singularities,
as one sees by arguments already given. The selection of the g: amounts to
projecting r' birationally into a 3-space curve without singularities; the
same technique can be used on any n-space curve free of singularities.
Let, then,
4 4
Ac == (! CiXi)O - (! CiX i ) 00 + B
i=O i=O
be the typical element of the g. We may take B to be any element of the
g, and we take it to consist of 4 distinct places. Then! CiX i == 1 for some c,
so we may assume Xo == 1. Let B == 1 + · · · + 4. Since each Ac is

170 LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES
effective, Xi can have i as pole to at most the multiplicity 1, and as the Ac
have no fixed point, each i is a pole of some Xi. The requirement that
C1X I + · · · + C 4 X 4 have j as a pole is a polynomial (in fact, linear) in-
equality condition on the C i . Place these conditions (for each j, j == 1, . . . , 4)
on (Ck),..., Ck»), k == 1, . . . , 4, and also impose the condition det
ICk)1 -=1= o. These conditions can be met-the various polynomials (none
of which is zero) define hypersurfaces in a 16-space, and one takes a
point on none of them. Hence we may assume that Xv . . . , x 4 each ha,ve
v . . . , 4 as poles. One of the Xi' say Xv is a separating element of
L == K (Xl' . . . , x 4 ) (if K == C, any Xi will do). Hence dX I has only a finite
number of zeros. Let these be (1), . . . , (S). If  is any other place and
 -=1= j, j == 1, . . . , 4, then Xl has an expansion Xl == a l + bIt + · · · ,
b l -=1= 0, at. (The number a v defined for any  and any Xl not having a
pole at  is called the residue of Xl at .) Thus the gl determined by 1,
Xl' i.e., the set of divisors of the form (co + CIX1)O - (co + Clx l ) 00 + B, is
such that some element of it contains  only once. As B belongs to this
gl, the same can be said for  == j, j == 1, . . . ,4.
Now consider the (i). These have distinct simple centers at finite
distance on r', the curve defined by the g:. Let Xi == res'.p(i) Xj + · · · be
the expansion of Xj at (i) [where res(i) Xj stands for the residue of Xj at
(i)]. Then some hyperplane !1=1 Cj(X j - res(i) Xj) == 0 cuts r' at (i)
with multiplicity 1. By the argument on inequalities in the previous para-
graph, we may suppose this occurs simultaneously at every (i). We may
suppose! CiX i == X 2 . Then (1, Xl' x 2 ) determines a g such that every place
 occurs with multiplicity 1 in some of its divisors. Moreover, as in 20.16,
we may suppose K (xl' X 2 ) == L; or in other words that the g defines a
birational transformation. The transform r" is a plane quartic having only
linear branches; the g is cut out by the lines of the plane, in particular,
B is cut out by loo : Xo == o.
Let ', ", . . . be the places centered at singular points of r". They
have distinct simple centers at finite distance on r' (on r', B is cut out by
the hyperplane at infinity). There is a hyperplane !1=1 ci(X j - res, Xj) == 0
cutting r' at ' with multiplicity 1 and not passing through '13", . . . ;
by appropriate choice of the C i , we can suppose that the same is true relative
to each of ', ", .... We may write !CiX i == Xa. Thus for each of
', ",... there is a divisor (co + caxa)o - (co + caxa)oo + B containing
it just once and the others not at all. Then the series of typical element
(co + CIX I + C 2 X 2 + caxa)o - (co + C1X 1 + C 2 X 2 + caxa) 00 + B yields the
desired g:.
5) Hyperelliptic and nonhyperelliptic curves. A curve of genus 1 is called
elliptic. Any such curve carries a g and hence is birationally equivalent to a
plane cubic without singularities. A curve of genus p > 1 whose canonical
series gK;.2 2 is composite with an involution is called hyperelliptic. If a curve

LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES 171
has a complete g, and p > 1, in particular, if p == 2, then it is hyperelliptic.
In fact, let 0 == l + 2 E g. The canonical series I W I has no fixed point,
so requiring aWE I W I to contain l (or 2) imposes one linear condition.
If these W do not automatically contain 2 (respectively, l)' then requiring
W E I W I to contain 0 imposes two linear conditions. In that case,
dim I W - 01 == p - 1 - 2, and i(O) == p - 2. Applying Riemann-Roch
to the g, we get 1 == 2 - P + (p - 2), a contradiction. Hence I W I is
made up of the elements of g. As p - 1 arbitrary places can be assigned to
aWE I WI, I W I is composed of the g.
Conversely, let p > 1 and let 1 W I be composite with an involution Yv.
I W I defines a rational mapping and (from the general discussion one
sees that) the hyperplanes of the space containing the transformed
curve r' cut out a gr2;2)/V. Since in a g, r < n, we have v < 2; and
v == 2, since I W I is not simple. Thus I W I is composed of a Y2. The
question now is whether the elements of this Y 2 are the elements of a g:
the answer is yes. As we have just seen, r' has a g:=L hence is a rational curve,
Le., its genus == o. Hence its field I;' has a single generator,' == K (x). Let
o ==  + · · · +  -1 E g = t. Suppose x has nonnegative order at .
Then  is a zero of x - res x, and, moreover, a simple zero, since
[I;' : K (x)] == 1. Over  in I; lie just two places (or one with multiplicity 2),
and these are the places in I; where x has res x as residue. A similar situa-
tion holds if  is a pole of x. Thus the Y2 is just the series whose typical
element is (co + c1x)o - (co + c1x) 00 + (x) 00.
Consider now the nonhyperelliptic curves with p > 3. Let r be such a
curve. Its canonical series gG p -!l defines a birational transform r' of r. If
p-1 p-1
Ac == (! CiXi)O - (! CiX i ) 00 + B
i=O i=O
is the typical element of the gP-!l' then xo, . . . , x p - 1 is the general point
of r'. The curve r' lies in a (p - I)-space and moreover belongs to it, i.e.,
cannot lie in a proper subspace, as the Xi are linearly independent. The
hyperplanes cut out the canonical series on r'. Conversely, let r" be a curve
birationally equivalent to r, of order 2p - 2, and belonging to a (p - 1)-
space. Then the hyperplanes cut out a gf P -!2 on r". As p - 1 > (2p - 2) -
p, this series is special; hence, obviously, this is the canonical series. We
are going to show that r' and r" are projectively equivalent.
The typical element of the canonical series can be written in the form
(! CiXi)O - (! CiX i ) 00 + B,
where (x o , Xv . . . , x p - 1 ) deternlines r'; we may suppose BEl W 1 and that
Xo == 1. Similarly, it can be written in the form
(! diYi)o - (! diYi) 00 + B,

172 LINEAR SERIES AND RATIONAL MAPPINGS; SPACE CURVES
where (Yo, YV..., YP-l) determines r"; we may suppose that Yo = 1.
Then
(Yj)O - (Yj) 00 + B = (! Cijxi)o - (! CijX i ) 00 + B
for some Cij, whence
Y . == const  c..x.
1  II l'
and similarly the Xi can be expressed in terms of the Yj' Thus
(1, Xv . . . , x p - 1 ),
(1, Yl' . . . , YP-l)
are obtained from each other via a nonsingular linear homogeneous trans-
formation. The same can then be said of r' and r". Thus:
Every nonhyperelliptic curve of genus p > 3 is birationally equivalent
to one and, up to projective equivalence, only one curve of order 2p - 2 belonging
to (p - 1 )-space.
This theorem can be used as a basis for classifying the nonhyperelliptic
curves of genus p > 3.
6) We come now to the birational classification of plane cubics promised
in Chapter 10: we prove Salmon's Theorem that two plane cubics r, r'
without singularities are birationally equivalent if and only if they are pro-
jectively equivalent, or if and only if they have the same modulus. The direct
part is trivial. For the converse, let r, r' be birationally equivalent and let 
be the field of rational functions on r and on r'. Consider a g of I;. It is
given by means of a pencil on a point A of r; and similarly by a pencil on
a point B of r'. Via the g we can set up a one-to-one correspondence between
the lines of the two pencils. It is fairly evident (though it could be spelled
out further) that this correspondence is of an algebro-geometric character.
Hence by a theorem of Chapter 10, the correspondence is a projectivity.
In this correspondence the tangents from A to r and those from B to r'
correspond to the element of g of the form 2X. Hence the tangents from A
correspond to the tangents at B; and since the correspondence is a projec-
tivity, the cross ratios, i.e., the moduli, are equal. Q.E.D.

Chapter 21
ANALYTIC BRANCHES
In this chapter, K denotes an arbitrary algebraically closed ground field.
According to a previous remark, the tool for the local study of a curve
is the power series. Hence, to study the curve F (X, Y) == 0 at (0, 0), say,
we ought to study F as an element of the ring X[ [X, Y]]. If instead of
starting with a polynomial F, we start with an arbitrary element F in
X[ [X, Y]], we come to the so-called theory of analytic branches.
In the case K == C, an element F E C[[X, Y]] n1ay be convergent near
(0,0) and F(X, Y) == 0 represents an actual locus. In the more general case,
the only point (a, b) E ANK that we can substitute into F(X, Y) is (0,0),
and there is no locus to speak about.
We can, however, substitute a branch representation (x(t), y(t)} into
F(X, Y) == !;:ooociiXiyjprovidedord x(t) > Oandord y(t) > o. Moreover,
if a is an automorphism of X[ [t]] over K and a: (x(t), y(t)}  (x'(t), y'(t)),
then !bn Ciixiyi  !bn Ciix'iy'i, and, by an argument occurring in the
proof of 12.3, !;:ooo Ciixiyj  !;:ooo CijX'iy'j. Hence if F(x(t), y(t)} == 0, then
F(x'(t), y'(t)} == 0, and we can say that the branch x((t), y(t)} annihilates
F(X, Y).
Definition. By an analytic branch [centered at (0, 0)] one means a
branch [centered at (0, 0)] in the sense of Definition 12.4. We shall see
(in 21.15.1) that any such branch annihilates some F (and hence, also,
some irreducible nonunit) in K[ [X, Y]].
Shortly we shall prove that there is one and only one branch annihilating
a given irreducible nonunit FE X[ [X, Y]], and that if Fl and F 2 are ir-
reducible nonunits annihilated by a given branch, then Fl and F 2 are
associates. Hence we could equally well define an analytic branch as an
irreducible nonunit in K[ [X, Y]], or, rather, the equivalence class of such
a nonunit under the equivalence: Fl f'"'oo..J F 2 if Fl == unit times F 2 .
173

174 ANALYTIC BRANCHES
Just as for curves, we will want to make transformations of the form
X = aX' + bY', Y = cX' + dY', ad - bc "* 0, to avoid special positions
of the coordinate axes. For example, if F = Fr + Fr+l + · · · , we can,
by a transformation of the kind mentioned, assume that Fr t=. O(X).
Moreover we have:
Theorem 21.1 ( Weierstrass's Preparation Theorem). If
P(X l' . . . , X n ) E K[ [X l' . . · , Xn]]
and
P(O, . . . , 0, X n ) = cvX + Cv+lX+ 1 + · · · 9
C v "* 0,
then there exists a unique unit E in K[ [Xv. . . , Xn]] such that
(1) EP = X + A1(X V . . . , Xn_l)X-l + · · · + Av(Xv . . . , X n - 1 ),
with Ai(X I ,..., X n - 1 ) E K[ [Xl' . . . , X n - l ]] and Ai(O, ..., 0) = 0
(i = 1, . . . , v).
An element of K[ [Xv. . . , Xn]J containing a term in X n alone is said
to be regular in Xn. Thus P is regular. The notation is such that X is the
lowest power of X n occurring in P. A polynomial having the form of the
right-hand side of (1), Le., Ai(O, . . . , 0) = 0, is called special.
This theorem follows from a more general one:
21.2 (Weierstrass's Formula). IfQ and P are elements of
K[ [X, . . . , Xn]]
and P is regular in X n ,
P( 0, . . . , 0, X n ) = cvX + c v + 1 Xj + 1 + · · · ,
C v "* 0,
then there exist elements A, B in K[ [Xl' . . . , Xn]] such that
(WF)
Q - AP = B,
where B involves no power of X n of degree > v. Moreover, A and Bare
unque.
The theorem follows from the formula by taking Q = X. Then A must
be a unit, as otherwise X could not cancel out on the left.
Proof of the formula. The proof is by induction on n. The formula is easily
checked for n = 1, and we suppose that n > 1. If qi, ai' Pi' b i are the co-
efficients of Xi (not X!), then (WF) is equivalent to the following infinite

ANALYTIC BRANCHES 175
set of equalities:
90 - aopo == b o ,
91 - aOPl - alPo == b 1 ,
.
qi - aOPi - · . · - ai-lPl - aiPo == bi'
.
On Po(X 2 , . . . , X n ) we have the same hypothesis as on P, namely,
Po(O, . . . , 0, X n ) == cvX + · · . , and on b i we have the same requirement
as on B. By induction on n, ai' b i are (successively) uniquely determined
and exist; hence the same can be said for A and B. Q.E.D.
Corollary 21.3. If P E K[ [Xv. . . , X l1 ]] and
P(O, . . . , 0, X n ) == cvXv + c v + 1 xv+ 1 + · · ·
with
C v i= 0,
and if A' P is in K[ [Xv. . . , Xn-1J][X n ] is of degree < v, where A'
is assumed to be in K[ [Xv. . . , XnJ], then A' == o.
Proof. Take Q == O. We have 0 - A' P == B' and 0 - o. P == 0, so by the
uniqueness of A and B in (WF), A' == O.
Corollary 21.4. If P is a special polynomial in K[ [Xl' . . . , Xn]] of
degree v and A' P == B' E K[ [Xv. . . , Xn-1J][X n ] is of degree v, then
A' (assumed in K[ [Xv. . . , XnJ]) is in K[ [Xv. . . , X n - 1 ]].
Proof. Let B' == CX + · · · , then CX - A' P == CX - B' is of degree
< v. Take Q == CX. Then Q - AP = B is satisfied by A == A' and
B == CX - B' and also by A == C and B == Q - CPo By the uniqueness
of A, B in (WF), A' == C.
Given an elenlent F(X, Y) in K[ [X, Y]], or several such elements
F 1 , . . . , Fs, by applying a transformation (or substitution) X == aX' + bY',
Y == cX' + dY', ad - bc i= 0, we can suppose without loss of generality,
that F, or Fv . . . , Fs, are regular in Y. Then by Weierstrass's Theorem we
can replace F, or Fv . . . , Fs, by associates in K[ [X]][ Y]. We can then,
in effect, work in K[ [X]][Y]. As K[ [X]][Y] is a unique factorization
domain (it is a polynomial ring over a UFD), we could perhaps manage
without the following theorem.
Theorem 21.5. K[ [Xv. . . , Xn]] is a unique factorization domain.
Proof. The proof is by induction on n, so inductively we have that.
K[ [X l' . . . , X n - 1 ]][ XnJ is a unique factorization domain.

176 ANALYTIC BRANCHES
If F == G l · · · G m , then subd F == ! subd G i . From this it follows
that any nonunit F has a factorization into irreducible nonunits.
Now let ElFl... Fs == E 2 G l · · · G t be two complete factorizations
of F in X[ [Xv. . . , Xn]J, where E l , E 2 are units and the F i and G j are
irreducible nonunits.
1) Applying a homogeneous nonsingular linear transformation X == AX',
we may suppose F i , G j are regular in Xn.
2) Applying Weierstrass's Theorem (and absorbing the extra units into
El and E 2 ), we may suppose the F i , G j are special in Xn.
3) If F is special in X n and irreducible in X[ [X I' . . . , X n ]], then
F is irreducible in X[ [Xl' . . . , X n _ I ]][ Xn].
Proof. Suppose F == G l G 2 with deg x G i < deg x F. Since F is irreducible
n n
in K[ [Xl' . . . , X n ]], one of the G i is a unit in that ring, say G v is. Then
G l l F == G 2 , and by 21.3, G 2 == O. This is impossible.
4) Multiplying both sides of EIFI · · · Fs == E 2 G I · · · G t by Ell, we
maysupposeE I == 1. By21.3, we cannot have deg Fl . · . Fs < deg G I · . · G h
and from E:;l FI · . · Fs == G I · · . G t , we cannot have deg FI . · · Fs >
deg G l ... G t . Hencethedegreesareequal,andby21.4,E 2 == 1. Nowapplying
unique factorization in K[ [Xl'. . . , Xn-l]][X n ], wegets == t,and that, for some
reordering of the su bscripts, F i and G i are associates in X[ [X v . . . , X n - l ]][ X n ].
As the units in this ring are also units in X[ [Xl' . . . , X n ]], we see that F i
and G i are also associates in K[ [Xv. . . , Xn]J. The proof is complete.
Consider now the ring K[ [X]][ Y], and first, for a moment, the case
K == C. To work in C[ [X]][Y] means to work about the Y-axis: this is
clear if one considers that from an element f(X, Y) in C[ [X]J[Y] one can
obtain a geometric locus f(X, Y) == 0, that is, the set of points (x, y) such
that f(x, y) == 0, and that, in general, the coefficients of f (as a polynomial
in Y with coefficients C[ [X] J will be convergent only for small values of
Ixl. However, as with C[[X, Y]], we do not wish, nor need, to treat questions
of convergence, but can work formally in C[[X]][Y] (and in K[ [X]][Y]).
Moreover, we shall make no substitution (X, Y)  (a, b) E ANC in
f{X, Y) E C[[X]][Y] except for the points (a, b) with a == o.
We have seen in Chapter 11 that if the curve F(X, Y) == 0 has a simple
point at (0, 0) with nonvertical tangent, then F(X, Y) has a factor Y-
clX - C 2 X2 - · . . in C[[X]][Y]. Likewise, if F(X, Y) == 0 hasasimplepoint
at (0, b) with nonvertical tangent, then F has a factor of the form.
Y - b - d X - d X2 - · · · ·
12,
so that if F == 0 cuts the Y-axis in at least two simple points with nonverticaI
tangents, then F will have at least two factors in C[[X]][Y]. We may
conjecture that this phenomenon does not depend on the special hypotheses
mentioned: if one intersects the curve F == 0, assumed to pass through

ANALYTIC BRANCHES 177
(0, 0) and (0, b) with b =I- 0, with a thin parallel strip about the Y-axis, the
part of F == 0 cut out decomposes into at least two parts, and one may
conjecture that to this geometrical decomposition there corresponds a
decomposition of F(X, Y) in C[[X]][Y].
More generally, we can make the same conjecture for an arbitrary f in
K[ [X]][Y] (f(O, Y) =I- 0). Thus in particular we can conjecture that if
f(O, Y) has at least two distinct roots, then f(X, Y) is reducible in
K[ [X]][Y]. This is a correct theorem. Setting aside a trivial difficulty, we
win assume thatf(X, Y) is monic (in Y). Then we obtain:
21.6 (Hensel's Lemma). Let f(X, Y) E K[ [X]][Y] be monic of degree
n in Y and such that j(O, Y) has at least two roots. Then j(X, Y) is
reducible in K[ [X]][Y]. Moreover, if f(O, Y) == go(Y)ho(Y) are prime
to each other and are of degrees rand n - r respectively, then
f == g(X, Y)h(X, Y)
with g, h E K[ [X]][Y] of degrees rand n - r n Y and with
g( 0, Y) == g o( Y) and h( 0, Y) == ho( Y ) .
Proof. Let us write
g(X, Y) == go(Y) + gl(Y)X + ... + gs(Y)XS + ... ,
h(X, Y) == ho(Y) + hl(Y)X + ... + hs(Y)XS + ...,
f(X, Y) == fo(Y) + f1(Y)X + ... + fs(Y)XS + ... ,
with deg go == r, deg ho == n - r, deg fo == n (since f is monic) and for
i > 0, deg gi < r, deg hi < n - r, deg fi < n; here the gi, hi are unknown.
The relation f == gh is equivalent to the conditions:
fo( Y) == go( Y )ho( Y ),
fl(Y) == go(Y)h 1 (Y) + g1(Y)h o (Y),
Since go( Y), ho( Y) are prime to each other, we can write
1 == (. .) go + (. · )ho,
whence
f1 == h 1 g o + g1 h O.
However, we require that deg gi < r, deg h 1 < n - r. If this condition
does not already obtain, we can achieve it by writing gi == q go + g, with
deg g < r and substituting in the last equality to obtain
f1 == hI go + (qgo + g)ho == (hI + qh o ) go + gho.
That is, one has that the coefficient of ho is of degree < r; now one has
deg ho == n - r, hence deg gho < n, and as deg go == r, one obtains

178 ANALYTIC BRANCHES
deg (hI + qh o ) < n - r. This shows that one can suppose deg gl > r,
deg hi < n - r.
The general case proceeds by induction, that is, supposing that gi, hi are
already known for i < s (with the required limitation on their degrees),
one writes
fs == gohs + glh s - l + · · · + gS-lh l + gsho,
or, rather,
fs - glh s - l - · · · - gS-lh l == gohs + gsho,
which can be solved for g s' hs since go' ho are relatively prime; then we lower
the degrees of gs, hs just as we lowered those of gl' hI. Thus the lemma is
proved.
The methods of Chapter 12 now apply with scarcely a change. One
applies the local quadratic transformation
X' == X,
Y'==Y/X
to elements in K[[X]][Y]. Starting from an
fEK[[X]][Y], f==fr+fr+l+...'
we may have to apply a preparatory transformation X == aX' + bY',
Y == cX' + dY', ad - bc =F 0, to obtain the assumptionfr ¥= O(X). This
leads into K[ [X, Y]], but then we return to K[ [X]][Y] via Weierstrass's
Preparation Theorem. We go over the results briefly.
First, we can refer to a nonunit Fin K[ [X, Y]] having no multiple factor
as an analytic curve and, more generally to any F E K[ [X, Y]], F =F 0, as
an analytic cycle (and can say F == 0 is its equation). We say that the center
(0, 0) is r-fold if F == Fr + Fr+l + · · ., Fr =F 0, and that it is simple if
r == 1. The factors of F r yield (by definition) the tangents to F == 0 at its
center. All these notions are invariant under transformations
X' == aX' + bY',
Y == cX' + dY',
ad - bc =F o.
(One understands that F and EF, where E is a unit, refer to the same analytic
curve.)
Corresponding to 12.6 we have:
21.7. If FE K[ [X, Y]] and subd F == 1, then there is one and only
one branch (x(t), y(t)) such that F(x(t), y(t)) == o.
The proof is the same. The existence goes back to 11.3 and shows that if
F == Y - clX - · · · , then the branch is given by (t, clt + · · .).
Definition 12.7 can be repeated: if (x(t), y(t)) is a primitive representation
of y, and  : G == 0 is an analytic cycle, then, by definition,
i(, y; 0) == ord t G( x(t), y(t)).

ANALYTIC BRANCHES 179
The multiplicity of a branch at its center has already been defined
(see Chapter 12).
Corresponding to 12.9 we have:
21.8. Two distinct irreducible analytic curves r: F === 0, : G === 0
cannot have a common branch.
Proof. Applying a transformation X === aX' + bY', Y === cX' + d Y',
ad - bc ::j=. 0, we may suppose F, G regular in Y, and then, by the Preparation
Theorem, that they are special. By point (3) in the proof of Theorem 21.5
F, G remain irreducible in K[ [X]][Y]. Then they also remain irreducible
in K((X))[Y]. Now, by hypothesis, F, G are not associates in K[ [X, Y]].
Then they are also not associates in K[ [X]][Y] and in K((X))[Y]. Their
G.C.D. in K( (X))[ Y] is, then, 1, so 1 === aFl + bF 2 , a, b E K( (X))[ Y].
Clearing a possible denominator, we get XP === AF + BG, A, B E K[ [X]][ Y].
Now assuming (x(t), y(t)) is a common branch, and substituting, we find
x P === 0 and x === o. As F is special, one obtains y === o. This contradicts
the definition of a branch.
Corollary 21.9. If (, 'YJ) is a branch and  === 0, then the branch is a
branch of X === 0 and of no other irreducible analytic curve.
Theorem 21.10. Let r be an analytic cycle centered at 0 === (0, 0) not
containing X === 0 as a component. Then r has a unique special equation
F === 0, and F is reducible in K[ [X] J[ Y] if and only if it is reducible in
K[ [X, Y]].
Proof. This merely summarizes points made in 21.1, 21.3, 21.4, and 21.5
(point 3).
Preparatory to considering the locally quadratic transformation X' === X,
Y' === Y / X, we need some definitions. By an analytic cycle along the Y-axis
we mean the notion made up of a finite number of irreducible analytic cycles
centered on the Y-axis, together with associated multiplicities. The notation
is mlr l + .. · + msrs. (We could include centers at infinity, but we
exclude this for simplicity.) If F E K[ [X]][ Y] is monic and irreducible,
then by Hensel's Lemma F (0, Y) has only one root, m, and F === 0 defines
an irreducible analytic cycle centered at (0, m) different from X === 0; if
m === 0, then F is special, and in general F is special relative to Y 1 === Y - m.
Conversely, every irreducible analytic cycle centered on X === 0 but different
from it is given by a unique, monic, irreducible F E K[ [X]][ Y]. If r v . . . , r s
are such cycles with monic irreducible equations F 1 === 0, . . . , F s === 0,
then, by definition, Fl... F";s === 0 is the (monic) equation for
m1r l + · · · + msrs. Conversely, every monic FE X[ [X]][Y] has a
unique factorization into monic irreducible factors: F == Fl · · · Fm,.
and defines the cycle mlr 1 + · · · + msrs. [If F === EFl · · · Ft is a

180 ANALYTIC BRANCHES
complete factorization in K[[X]J[Y], where E is a unit and the F i are
nonunits, then the F i need not be monic; but if e i is the leading coefficient
of F i , then ei is a unit, E == ell... etl, and F == (ellF 1 )... (etlFt) is
a complete factorization into monic irreducible polynomials.]
Consider now the locally quadratic transformation
(1)
X' - X
- ,
Y' == Y/X.
Let r: F(X, Y) == 0 be an analytic cycle centered at (0,0). We define the
transform only in the case that X == 0 is not a tangent to r (the other case is
excluded for simplicity). To define it, we replace F == 0 by the special
equation of r, i.e., we assume F special. (We insist on this, because the units
do not transform well.) Then we define the transform r' of r by
r' : F (X, X Y')/xr == 0, where (0, 0) is r-fold for r.
Theorems 12.11.1, 12.11.2, 12.12, 12.13 continue to hold for analytic
curves r centered at (0, 0) and not tangent to X == o. The proofs are the
same, too, if in them one takes F == 0 to be the special equation of r.
Corresponding to 12.14 we have:
21.11. Let r: F(X, Y) == 0 be an analytic curve centered at (0, 0).
Then there is at least one and at most a finite number of branches (x(t), y(t))
that satisfy F.
Proof. The proof is much like that of 12.14, but we must note one new point.
Let F == Fr(X, Y) + · .. After a transformation X' == aX + bY,
Y' == cX + dY, we may suppose Fr =1= O(X), and then by Weierstrass's
Theorem, we get yr + a1(X)yr-l + · · · + ar(X) == 0 as the special
equation for r. We may assume r is irreducible. Then
F'(X, Y') == F(X, XY')/xr
is also irreducible. By Hensel's Lemma, F'(O, Y') has only one root, m,
and after a translation Y" == Y' - m, we may assume m == o. We have
F'(X, Y') == Y'r + b1(X)Y'r-l + ... If r is not reduced, then F' is
already special and r' is not tangent to X == o. It is this point we wish to
note, as in the proof of 12.14 we do not make preparatory transformations
(except translations) before applying the locally quadratic transformation
the second, third, etc., time.
Corresponding to 12.15 but with a different proof, we have:
21.12. The sequence of locally quadratic transformations and translations
occurring in the proof of 12.11 cannot yield a sequence of r-fold points for
r > 1.
Proof. Let r be irreducible, not tangent to X == 0, and with F == 0 as special
equation. As in the proof of 12.15, if the sequence of locally quadratic

ANALYTIC BRANCHES 181
transformations does not yield a reduction in r, then we find
F(X, Y) = (Y - mIX - m 2 X2 - · · .)G(X, Y).
As F is irreducible and monic, we conclude that
F(X, Y) = Y - mIX - m 2 X2 - · · · ,
whence.r = 1. Q.E.D.
N ow we can prove:
Theorem 21.13. An irreducible analytic cycle r has only one branch.
Proof. Let (0,0) be r-fold for r. If r = 1, then 21.7 gives the result. For
r > 1, after rearranging the axes and preparing the equation F = 0 of r,
we apply the locally quadratic transformation (1). If there is a reduction in
r, then we are through by an induction on r. If not, after a translation we
make another locally quadratic transformation. Eventually r is reduced
(by 21.12) and the proof is complete.
Let F(X, Y) = ye + al(X)ye-1 + ... + ae(X) E K[ [X]J[Y] be
monic and irreducible. By Hensel's Lemma, F(O, Y) = 0 has only one root,
m, and we may suppose m = o. Let x = ct d + · · ., y = · · ., c i= 0,
be the branch of F = o. We are going to prove that d = e.
Lemma 21.14. If (x(t), y(t)) is a primitive branch representation, then
the field K(x(t), y(t)) contains an element of order 1. (An element of order 1
is called a uniformizing parameter.)
Proof. Let'T be an element of K(x(t), y(t)) of least positive order, g: then
g = 1. In fact, every element' in K(x, y) c K((t)) must have order h
divisible by g; for suppose h = qg + r, 0 < r < g. Then '/T'l has order
r < g, contradicting the minimality of g. It follows that' can be expanded
into a power series in'T. For say h = qg. Then' and'T q have equal order, and
, - C'T q has order hI > h for some c E K. Repeating the argument, we see
that there is an integer ql > q and a c 1 E K such that' - cT'l - clT'll is of
order h 2 > hI' etc. Thus x and y can be expanded into power series in 'T.
Since (x(t), y(t)) is primitive, order'T = g = 1. Q.E.D.
Lemma 21.15. Let x = ct d + · · · , C =1= o. Then
K[ [x]][t] = K[ [t]]
and
[K((x))(t): K((x))] = d.
Proof. Note that 1, t,..., t d - I , x, tx,..., td-Ix, x 2 , . .. are of orders
0, 1, 2, . . . in t. Hence any element in X[ [t]J can be written (uniquely) as
an (infinite) linear combination of them. This proves K[ [t]J c K[ [x]][t].
The opposite inclusion is immediate. To prove the second point, consider
the element X - CTd - · .. in X[ [X, T]J. It is obviously irreducible
(its subdegree is 1). By Weierstrass's Theorem, there is a unit E(X, T) such

182 ANALYTIC BRANCHES
that E. (X - CTd - · · .) == Td + a1(X)Td-l + · · · , and this element is
irreducible in X[ [X]][T] (by the proof of 21.5) and hence also irreducible in
K{(X))[T]. Since E(x, t) =F 0, x(t), t satisfy Td + a1(X)Td-l + · · · = 0
and t satisfies Td + a1(x(t)) Td-l + · · · = o. Thus t satisfies an irreducible
equation of degree d over K((x(t))), whence [K{(x))(t): K((x))] = d.
Remark. Concerning the last sentence, note that ! CiX i = 0 implies all
Ci = 0, so that one computes with x(t) just as with X, Le., K( (x(t))) 
K( (X)). In fact, if some Ci =F 0, and j is the least subscript for which
Ci =F 0, then! CiXi has order j ord x, which is impossible, as ord 0 = + 00.
Corollary 21.15.1. If x(t), y(t) is a primitive branch representation of a
branch centered at (0, 0) and x(t) =F 0, then y is algebraic over K( (x)).
Proof. By 21.14, there is an element T of order 1 in K(x, y). With a change
in notation, we nlay suppose t E K(x, y). If t E K(x), then t E K( (x)),
K{ (t)) == K( (x)), and y E X( (x)). If t ft K(x), then from t E K(x, y) one sees
that y is algebraic over K(x, t), hence also over K( (x) )(t), whence by the
lemma, y is algebraic over K( (x)).
Theorem 21.16. Let ye+a1(X)ye-l+...EK[[X]][Y] be monic
and irreducible and let x = ct d + · · ., y = y( t), c =F 0, be its branch,
then d == e.
Proof. y(t) E K[ [t]] == K[[x(t)]][t] and [K((x(t)))(t): K((x(t)))] = d. Hence
y satisfies an equation of degree d over K( (x)). As all such equations
must be multiples of the irreducible equation ye + a1(x) ye-l + · · · = 0,
we have e < d. For the converse, by 21.14, there is an element T of order 1 in
K(x(t), y(t)). We replace t by T; so we may assume t E K(x, y). Then t
satisfies an equation over K((x)) of degree [K((x))(y): K((x))] = e,
whence d < e. Hence d == e. Q.E.D.
Theorem 21.17. Let Kbe of characteristic 0, let F = Yd + a1(X)Yd-l +
· · · E K[ [X]][Y] be monic and irreducible, and let x(t), y(t) be its
branch. Then F( x(t), Y) splits completely into linear factors over K( (t))
[=K( (x) )(y)]. [In other words, F (X, Y) = 0 is a so-called normal
equation over K( (X)).]
Proof. Let x = ct d + · · ., c =F o. Placing,-d = ct d + · · ., we want to
sol ve for T as a power series in t (of order 1); the case d == 1 is trivial, and
we suppose d > 1. Dividing by t d , we have to solve (T/t)d == C + · · ·
As Zd - c has at least two roots (in fact it has d distinct roots), the element
Zd - (c + · ..) is reducible in K[ [t]][Z], by Hensel's Lemma. Applying
the lemma repeatedly, one finds that Zd - (c + · · .) factors into d linear

ANALYTIC BRANCHES 183
factors in K[ [t]][ Z]. Hence c + · .. has a dth root {/c + · .. and
d -
T=yCt+...
Let (] be the substitution t  T. By 12.3, the set of substitutions of order
1 is the same as the set of automorphisms of X[ [t]] over K; as the inverse
of an automorphism is also an automorphism, there is a substitution of order 1
sending T into t and therefore transforming x(t) = ct d + · · · , Y = y(t), into
x = t d , Y = · .. In short, we may suppose x = t d .
Let
(1)
y = !Ci/ i ;
(Ci -# 0),
;
that is, we write only the terms with nonzero coefficient. Then
(2)
G.C.D. (d, . . . , ii' . . .) = 1,
for if the G.C.D. were =e > 1, we could rewrite (x(t), y(t)) in terms of t e ,
contradicting the primitive character of (x, y).
Consider for a moment the case K = C, the complex field. In the
Argand diagram of C, the dth roots of 1 are represented by points on the
unit circle, that is, they are the numbers 1, cos 21T/d + i sin 21T/d,
cos 2 · 21T/d + i sin 2 · 21T/d, . . . , cos (d - 1) 21T/d + i sin (d - 1) 21T/d. Let
, = cos 21T/d + i sin 21T/d. Then' is a primitive dth root of 1, Le., a dth
root of 1 such that' -# 1, '2 -# 1, . . . , ,d-l -# 1.
Now it is known that any algebraically closed field of characteristic 0
has a primitive dth root of 1, and with this remark we drop the assumption
K = C.
Consider the d substitutions
(3)
t  'it,
i = 0, . . . , d - 1.
They all send t d into t d , and they send y(t) into Yi(t), i = 0, . . . , d - 1.
As the substitutions are automorphisms of K[ [t]] over K, the Yi(t) are roots
of F (t d , Y). If we can show that the Yi are distinct, then they are the d roots
of F(td, Y), and the proof will be complete.
The expression for Y A is:
(4)
Y A = ! c ij ' Ai;t i ;,
where the Ci. are the Ci. in (1). If YA = YP,' then 'Ai; = 'P,i;, whence
J J
'<A-P,)i; = 1 and therefore (A - ft)ii == O(d). Let e = G.C.D. {.. ., i j , .. .}.
Then (A - ft)e == O(d). Since G.C.D. (e, d) = 1, we conclude that
A - ft - O(d), hence A = ft, since 0 < A < d - 1 and 0 < ft < d - 1.
The proof is complete.
Remark. If the characteristic p =F 0, then t d + · · · need not have a dth
root t + · .. For example, t P + t p + 1 has no pth root t + · · · , as every pth
power must have its exponents divisible by p.

184 ANALYTIC BRANCHES
Theorem 21.17 is convenient, but we only need a somewhat weaker
theorem, which holds for arbitrary characteristic.
Theorem 21.18. Let K be again of arbitrary characteristic, and let
F == Yd + a 1 (X)Yd-l + · · · E K[ [X]][Y] be monic. Then for some
x == x(v) E K[ [v]], F(x(v), Y) splits completely into linear factors in
K[ [v]][ Y ] .
Proof. Taking a branch x(t), y(t) of F == 0, we see that we can split off at
least one linear factor from F(x(t), Y). Let
F(x(t), Y) == (Y - y(t)) · F1(x(t), Y).
Now by the same argument (if d > 1) there is a t == t(u) E K[ [u]] such that
F1(x(t(u)), Y) splits off a linear factor. Thus from F(x(t(u)), Y) we can
split off two linear factors, etc., until we have finished.
The following theorem is the basis for proving that the intersection
multiplicities i and j are the same.
Theorem 21.19. Let F == Yd + a1(X)Yd-l + ... be monic and
irreducible and let r: F == 0 be centered at (0, 0), (so F is special); let
Ll : (x(t), y(t)) be its branch. Let y : G == 0 be an arbitrary analytic cycle
centered at (0, 0) == O. Then
i(, y; 0) == ord x Ry(G(X, Y) F(X, Y)),
that is,
ord t G(x(t), y(t)) == ord x Ry(G(X, Y), F(X, Y)).
Proof. For simplicity we first give the proof for characteristic 0, though the
proof in the general case is nearly the same.
We know, then, that F(x(t), Y) == IIf=l (Y - Yi(t)) and that there are
automorphisms of K[ [t]] over K [namely, those in (3)] taking x into x
and y into Yi. Hence
ord t G(x(t), Yi(t)) == ord t G(x(t), y(t)).
Now
d d
Ry( G(x, Y), II (Y - Yi)) == II G( x(t), Yi(t))
i=l i=l
whence
ord t Ry(G(x, Y), F(x, Y)) == d ord t G(x, y).
Recalling that ord t x == d, we see that ord t (. .) == d ord x (. .), whence the
theorem follows (for characteristic 0).
In the general case, we have to write F(x(u), Y) == IIil=l (Y - Yi(U)),
where (x(u), y(u)) and (x(u), Yi(U)) need not be primitive. Bya substitution
t i == CiUfJi + · · ., (x(u), Yi(U)) goes over into the primitive X(ti)' y(ti).

ANALYTIC BRANCHES 185
By 21.16, ord t . X(ti) = d; whence ord u x(u) = Pid. Thus all the redun-
.
dancies Pi are equal: Pi = p.
Now (xi(t i ), yi(t i )) and (xj(t j ), yj(t j )) are primitive branch representations
for the branch of r. The substitution t i -+ t j replaces (X(ti)' yi(t i )) by the
primitive branch representation (xi(t j ), yi(t j )), which also obviously represents
the branch of r. Hence the substitution t i -+ t j followed by an automorphism
of K[ [t j ]] over K takes (xi(t i ), yi(t i )) into (xj(t j ), yj(t j )). Then
ord ti G( xi(t i ), y(ti)) = ord tJ G( xj(t j ), yj(t j )),
whence
ord u G( x(u), Yi(U)) = ord u G( x(u), Yj(u)).
The rest of the proof is as before.
Corollary 21.20. Let r,  be two plane algebraic curves without common
component and let P be a point. Then the intersection multiplicity
i( Ll, r; P) as defined by means of resultants is the same as the j-intersection
multiplicity j(Ll, r; P) defined by means of the branches of r at P.
Proof. If P is not on both curves, then i = 0 and j = 0; so we suppose P
is on r and on Ll. We take a coordinate system so that P = (0, 0), the point
at infinity of the Y-axis is not on r or on , and so that r, Ll do not meet
on the Y-axis except at P. Then i(, r; P) = ord x R(G, F), where
r: F = 0, Ll: G = o. As r does not pass through E 2 : (0,0, 1), F(X, Y)
is monic in Y. Let F = FIF2 · · · Fs be the complete factorization of F
in K[ [X]][ Y] into monic irr"educible elements. The F i = 0 give the branches
')Ii of r centered on the Y-axis. Of these, let Fl = 0, . . . , Fk = 0 be
centered at (0, 0), the others not centered at (0, 0). Note that Ry(G, F i ) =
Pi(X) is a unit in K[ [X]] for i > k, otherwise G(O, Y) and Fi(O, Y) would
have a common root. Now
8 8
i(Ll, r; P) = ord x Ry(G, F) = ord x Ry(G, II F i ) = ord x II Ry(G, F i )
i=l i=l
8 k
= I ord x Ry(G, F i ) = Iord x Ry(G, F i )
i=l i=l
k
= I i(, ')Ii; P) = j(, r; P). Q.E.D.
i=l

Chapter 22
NOETHERIAN CONDITIONS AND
INTERSECTION MULTIPLICITY
Let R = K[X, Y], where K is an arbitrary algebraically closed ground field.
Let M = {d Id E R, d (0, 0) =I=- O}. Let R1'JIJ = {f jd If E R, d E M}; one
checks that RM is a ring and a unique factorization domain; it is called the
local ring of the plane at the point 0 = (0, 0).
Let P be a point, which we may assume is the origin, and let
r : F(X, Y) = 0, d: G(X, Y) = O. We retain the assumption of Chapter 15
that rand d are without common component, though in working at or near
a point P only the components through P matter, and we could assume merely
that rand d have no common component containing P.
We will be working in Ry and in S = X[ [X, Y]]. Every element in
RM can, as we have seen, be expanded into a power series in X and Y:
RM c S. Let qp = {H IH E R, H = (AF + BG)/d for some A, BE R,
dE M}, thus qp = RM(F, G) f1 R. We have seen (in 15.5) that all power
products of sufficiently high degree are in qp. Hence for every h E RlJJ
there is a P(X, Y) E Rsuch thath - P E X[ [X, Y]Jqp c X[ [X, Y]J(F, G).
Let h = hl/d, hi E R, d E M. Then h l - dP E S · (F, G). By 15.6,
hi - dP E qp, whence h - P E RJ1(F, G). Thus every h E R.I.'! is congruent
to a polynomial mod RJ1(F, G). Moreover, we can limit the degree of P,
as high-degree power products are in RJ1(F, G). Hence the elements of
RJ mod R./."l(F, G) form a finite-dimensional linear space over X, or, what is
the same, a finite X-module. We denote this space by R.. 7J1 /R J1 (F, G) and its
dimension by p(F, G; P) or p(r, d; P). We call p(F, G; P) the number
of Noetherian conditions for rand d at P.
It is to be noted that p is a projective invariant. If H*, F*, G* are the
standard homogeneous polynomials associated with H, F, G, one can write
the condition
H = (AF + BG)jd, d(P) # 0,
in the form
H* = (Ai F* + BiG*}Jd*,
186

NOETHERIAN CONDITIONS AND INTERSECTION MULTIPLICITY 187
where Ac, Bt, dt are homogeneous and di(P) =I=- O. The invariance of the
number of Noetherian conditions follows at once from this remark.
Our object is to show that p(r, Ll; P) == i(r, Ll; P).
Let F == fl · · · fs, G == Yl · · · Yt be complete factorizations in K[X, Y].
The following theorem shows that we may restrict ourselves to irreducible
r,Ll:
Theorem 22.1. #(fl · · · fs, Yl · · · Yt; P) == !i,j P(fi, Yj; P).
Proof. If anfi is in M, then R.,:w:(fi, Yj) == R M , and P(fi, Yj) == o. We may
therefore suppose that all fi' Yj vanish at P.
The elements of Rl(F, Yl... Yi) mod RM(F, Yl... YiYi+l) form a
linear space, RM(F, Yl... Yi)/RM(F, Yl... YiYi+l)' over K; bases for
RM/RM(F, Yl)' RM(F, Yl)/RM(F, YlY2)' . . . , RM(F, gv . . . , Yt-l)/RM(F, G)
together form a basis for RM/RM(F, G) as one easily checks. We wish to
show that dim RM(F, Yl · · · Yi)/RM(F, Yl · · · Yi+l) == #(F, Yi+l; P). Let
CXl, · · · , CXk be a basis for RM/RM(F, Yi+l)' where k === p(F, Yi+l; P). Then
cxlgl · . . Yi, . . . , CX1Bl · · · Yi are in R:Nl(F, Yl · · · Yi) and are linearly inde-
pendentmodR1J;l(F, Yl... Yi+l); forif!cjcxjYl... Yi === AF + BYI ... YiYi+l,
then AF ---: O(y} · · · Yi), whence A == O(YI · · · Yi), since RJ.Y: is a unique
factorization domain. Canceling Yl · · · Yi, we get! CjCXj == A' F + BYi+l'
whence the Cj == o. This shows that
dim RM(F, Yl · · · Yi)fRM(F, gl · · · Yi+l) > p(F, Yi+l; P).
For the converse, let 0 == AF + BYI · · · Yi E RlI(F, gl · · · Yi). Then
B == AlF + BlYi+l + ! CjCXj, whence
o - ! CjCXjYl · · · Yi E R 11 1.(F, Yl · · · Yi+l)'
showing the opposite inequality. Hence p(F, G; P) === ! p(F, Yi; P). The
same argument shows that #(F, Yi; P) == !; p(fj, Yi; P), whence the proof
is complete.
Theorem 22.2. Let r : F == 0, Ll: G == 0 have an ordinary r-fold point
and an ordinary s-fold point at P and let them have no common tanyent
at P. Then p(r, ; P) === rs.
Proof. Letting P == (0, 0) and adjusting the axes, we shall have
r
F(X, Y) === II (Y - miX - · · .)Fl(X, Y)
i=l
and
8
G(X, Y) === II (Y - niX - · · .)G1(X, Y)
i=l
in K[ [X]][Y], with F1Gl(0, 0) =I=- o.

188 NOETHERIAN CONDITIONS AND INTERSECTION MULTIPLICITY
These are obviously complete factorizations in K[ [X, Y]]. Now one can
define for S = K[ [X, Y]] a function il analogous to the function ft for RM
and prove for it a theorem analogous to Theorem 22.1. In fact, the proof
would be the same, but for it one would have to know that K[ [X, Y]] is a
unique factorization domain. We can bypass this however, because of our
special situation: we have fi = Y - miX - · · · , Yi = Y - niX - · · · ,
and we only need to know that AF == O(Y1 · · · Yi) implies A == O(Y1 · · · gi)
in the case that the Yj are of the form gj = Y - njX - · · · (see the above
proof). To prove this observe that
AFly=n1x+...=O and FIY=n1x+...=I=O,
whence A I y = nIX +. .. = 0 and A = A' · (Y - n 1 X.. .), A' E K[ [X, Y]].
Similarly A' = A"(Y - n 2 X - · · .), etc. We have then:
il(f1 · · · fr, Y1 · · · gs) = ! il(fi, Yj).
Now each il(fi, Yj) = 1; for write fi = aX + bY, Yi = cX + dY, and
note that a, b, c, d and ad - bc are units in K[ [X, Y]], whence
S · (fi, Yj) = S · (X, Y).
Hence il(f1 · · · fr, Y1 · · · Ys) = rs, so there exist (Xv . . . , (Xrs E S, which we
may suppose to be polynomials, that are linearly independent mod S · (F, G).
They are a fortiori such mod R1J;l(F, G). Now let (X E R M ; then there exist
Ci E K such that (X - ! Ci(Xi E l:j · (F, G). Using Theorem 15.6, one concludes
that ex - !Ci(Xi E R M . (F, G). Thus the (Xi form a basis for Ry/R.tu:(F, G)
and ft(F, G; P) = rs. Q.E.D.
The idea for proving ft(r, ; P) = i(r,; P) is contained in:
Theorem 22.3. Let ft'(I\ Ll; P) be an inteyral-valued function of P
and of cycles r,  without common component, and let:
i) ft'(r,; P) = rs if P is an ordinary r-fold point of r, an ordinary
s-fold point of Ll, and r, Ll have no common tanyent at P.
ii) If P is not on the axes and r', ', P' are the transforms of r, , P
under a standard quadratic transformation, then
u' ( r . P ) = ' ( r''. P' )
r " ft".
iii) Bezout's Theorem holds for ft', i.e., !pft'(r,; P) summed over the
intersections of r and  = deg r · deg .
Then ft'(r, Ll; P) = i(r,; P).
Proof. Let P be a given point. We apply standard quadratic transformations,
always keeping P and its transforms off the axes. In this way, ft'(r,; P)
does not change value [by (ii)]. Also i(r, Ll; P) = i(r', Ll'; P'), as one
checks directly using branches. Then by an argument just like the one leading
to Theorem 14.5, we may suppose that every point Q of r u  except P

NOETHERIAN CONDITIONS AND INTERSECTION MULTIPLICITY 189
is an ordinary singularity (or simple point) of r u. By (i), ft' = i at
every point except possibly P, hence by (iii), ft' = i also at P. Q.E.D.
Now we prove (ii) for ft.
Theorem 22.4. ft(r, ; P) == ft(r', Ll'; P').
Proof. Let R === K[X, Y] and let M === {d Id E R, d(a, b) #- O}, where
P === (a, b), and let RM === {fld If E R, d EM}. This is the ring mentioned
previously in connection with the Noetherian conditions, except that before
we assumed without loss of generality that P === (0, 0), whereas here we
assume a#-O and b #- O. Similarly we define R M , relative to the point
P'. Note that X and Yare units in RM and in R M ,.
Under the standard quadratic transformation T, (1, a, b) goes into
(1, Ifa, lib). Hence the nonhomogeneous equations for Tare
T: X' === If X, Y' === IfY.
Letf(X, Y) E Rand letf*(Xo, Xt, X 2 ) be the associated standard form,
sothatf*(I,X, Y) ===f(X, Y)anddegf=== degf*. UnderT,f*(X o , Xl' X 2 )
goes over into f*(X t X 2 , X 2 X o , XOXt), and except for factors Xi'
f*(X t X 2 , . . .) === 0 gives the transform of f*(Xo, . . .) = o. In non-
homogeneous terms, f*(XY, Y, X) === 0 does this. If we neglect powers of
X and of Y, we can write this asf*(I, l/ X , l/Y) === 0, or f(lfX, IfY) = O.
With T we consider a mapping, also denoted T, of K(X, Y) into itself,
namely, T : r( X, Y)  r( 1 I X, 1 I Y ). If f (X, Y) E K[ X, Y], then f (1 I X, 1 I Y)
is a polynomial divided by a power product in X and Y, and hence
T(f) E R M ,. Let now d E M. Then d(X, Y) does not vanish at P, so
d (1/X, II Y) does not vanish at P'. Hence T sends R It[ into R lU': T(R lJ;/) c R M'.
Similarly T(R M ,) C Rlf. Hence T(T(R M ,)) C T(R M ), Le., R M , C T(R M );
so R M , === T(R1f).
Let r:F(X, Y) === 0, :G(X, Y) === 0, r':F'(X, Y) === 0, and ':
G'(X, Y) === o. Then except for a power product in X and Y as a factor,
F'(X, Y) = F(I/X, l/Y). Hence T(RM(F, G)) C RM,(F', G'). An argu-
ment like the one in the last paragraph now shows that
T( R J1 (F, G)) === R\l,(F', G').
Elements OCt, . . . , OC s linearly dependent mod R.."l(F, G) obviously are
sent into elements of R J l' linearly dependent mod RJ.ll,(F', G'), but also
linear independent ones are sent into linearly independent ones; for if
T( OCt), . . . , T( oc s ) were linearly dependent, then T2( OCt), . . . , T2( oc s ), Le.,
OC v . . . , OC s , would be linearly dependent. Hence the spaces R M I R M( F, G)
and RM,IRM,(F', G') have the same dimension. Q.E.D.
Now we prove (iii), Bezout's Theorem, for ft.

190 NOETHERIAN CONDITIONS AND INTERSECTION MULTIPLICITY
Lemma 22.5. Letfr(X, Y), gs(X, Y) be forms of degree r, s without com-
monfactor. Thenpr+s-1 cR. (Ir, gs), where R = K[X, Y], p = (X, Y);
and for any f = fr + fr+1 + . · · + fm and any g = gs + · · · + gn,
RM · pr+s-1 C RM · (f, g).
Proof. With U = Y/X, fr(l, U), gs(l, U) are without common factor.
Hence 1 = A(U)fr(l, U) + B(U)gs(l, U), A, BE K[U]. Moreover, (as in
the proof of Hensel's Lemma) we may suppose deg B < r, deg A < 8.
Rendering this relation homogeneous, we conclude that Xr+s-1 E (fT' gs).
Making the affine transformation Y' = Y, X' = X + cY, one concludes
that also (X + cy)r+s-1 E (fr, gs) for every c E K.
N ow for r < 1 and s < lone checks the first part of the lemma directly,
and we make an induction on r + s. Let, say, s > 1 and factor gs : gs =
l]l2 · . . ls, where the li are linear. By an affine transformation we may assume
is = Y. By induction, pr+s-2 C (ir, l1l2 · · · lS-1)' whence
Y p r+s--2 c (f l ... l Y ) - (f g )
r, 1 s-1 - r, S.
As also Yr+s-1 E (fT' gs), we get pr+s-1 C (Ir, gs).
For the second part (which we do not need in the following), we have
pr+s-1 C (f, g, pr+s),
sInce
x yr+s-l- = AS-l fr + Bt-l gs E (I, g, pr+s);
hence
pr+s = p(pr+s-1) C (pf, pg, pr+s+1) C (I, g, pr+s+1),
whence
pr+s-l C (f, g, pr+s+1),
and repeating,
pr+s-1 C (f, g, pr+s+i),
for all i.
Since some power of X and some power of Y is in Rl(/, g), we get
R M pr+s-1 c Rll(/, g). Q.E.D.
Theorem 22.6. Bezout's Theorem holds for p, i.e., !pp(F, G; P) = mn.
Proof. First we show that !pp(F, G; P) gives the (maximum) number of
polynomials linearly independent mod (F, G). With the notation of 15.4,
( F, G) = q p f1 · · · f1 q p .
1 8
Let M i = {d Id E R, d(P i ) -=1= O} and let R i = R11.' so that
t
p(F, G; Pi) === dim Rd Ri(F, G),
and let k i be this number. Then there are polynomials (Xi), . . . , (Xki that
t
are linearly independent over K mod qp.. Let hi be a polynomial that vanishes
i

NOETHERIAN CONDITIONS AND INTERSECTION MULTIPLICITY 191
at all the Pi except Pi; it is easy to construct an hi of this kind. Now hi is
a unit in R i , so we may replace the r:J.)i) by hir:J.)i), and hence may suppose the
r:J.)i) to be in every qp , k =F i; we do this. Then the  p(F, G; Pi) elements
r:J.i), are linearly independent mod (F, G). For suppose
 c(i)r:J.(i) E ( F G ) == q n. .. n q
k:J '3 ' PI P 8 .
All the r:J.)i), i > 1, are in qp l , by construction. Hence
 C(1) r:J.(1) E q
k J '3 PI'
whence the CJ1) == 0; and similarly all the cJi) == o. Conversely, let r:J. E R.
Then there exist C)i) E K such that, for each i, r:J. - i cJi)r:J.)i) E qp.. Then
r:J. -  i,i cJi)r:J.i) E qP 1 n... n qp s == (F, G). Thus we have proed that
the linear space R/R(F, G) of polynomials mod (F, G) forms a finite-dimen-
sional space of dimension  p(F, G; P).
Now we show that, if F == 0 and G == 0 do not meet on loo, then there are
mn but no more polynomials linearly independent mod (F, G). Let
F == Frn + F m-l + · · · + Fr, G == G n + G n - l + · · · + G s . The assump-
tion on loo (which is made without loss of generality) comes to saying that
F m, G n have no common factor. Then by 22.5, for p == (X, Y),
pm+n-l C (F m, G n ).
If hv is a form of degree 'V > m + n - 1, then
hv == Av_m F m + Bv-nGn
== Av_m F + Bv_n G + Av_m(F m - F) + Bv-n(G n - G)
== Av_mF + Bv_mG + a polynomial of degree < 'V - 1,
where Av-m, Bv-n are forms of the indicated degrees. Hence all polynomials
are alrea{1y linear combinations mod (F, G) of polynomials of degree
< m + n - 2.
The polynomials of degree < m + n - 2 form a linear space !f of
dimension (m + n - 1)(m + n)/2. We build a basis for this space starting
from a basis r:J.v . . . , r:J.k of the space K of polynomials of this limited degree
in (F, G), and complete this to a basis of the whole space. Thus we should
find
rnn == (m + n - l)(m + n)/2 - k.
To computek,letH == AF + BGE (F,G),withA == Ap + Ap-l + ...,
B == B(J + B(J-l + · · · (A p =F 0 and B(J =F 0) and deg H < m + n - 2.
If p + m > m + n - 2, then p + m == (] + nand ApFm + BqG n == o.
Then Ap == q;G n , B q == -q;F m and
H == (A - cpG)F + B + cpF)G == A'F + B'G
with deg A' < deg A and deg B' < deg B. Thus we may suppose

192 NOETHERIAN CONDITIONS AND INTERSECTION MULTIPLICITY
deg A < n - 2 and deg B < m - 2. Allowing A and B to range over the
power products of these limited degrees, we find
k < (m - l)m/2 + (n - l)n/2.
By the argument of the first part of the paragraph, if deg A < n - 2 and
deg B < m - 2, then AF + BG = 0 implies A = 0 and B = o. Hence
the opposite inequality is proved and the desired value of k is obtained.
Q.E.D.
Remark. A similar argument shows that if Fr, G s are without common factor,
then p(F, G; P) = rs, and thus one gets another proof of 22.2.
Summing, we have:
Theorem 22.7. p(I\; P) = i(r,; P).

Chapter 23
SPECIALIZA TIONS; SPACE CURVES
In Chapter 20 we gave an ad hoc definition of an irreducible space curve,
namely, letting P: (x o , Xv . . . , x n ) be a point and L = K (P) a field of
algebraic functions of 1 variable over K (and assuming the Xi in L), we
defined an irreducible curve as the locus of centers of the branches
('13 (x o ), . . . , '13 (x n )) as '13 varied over the places of L / K . [Assuming Xo i=- 0
and writing xilx o = Yi, i = 1, . . . , n, we find that the affine part of the
curve relative to Xo = 0 as hyperplane at infinity is the locus of centers of
('13(YI)' . . . , '13(Yn)) as '13 varies over the places for which ord '13(Yi) > 0,
i = 1, . . . , no] In this chapter we will define irreducible algebraic variety
and then define an irreducible curve as an irreducible algebraic variety of
dimension 1.
The ground field K is an arbitrary algebraically closed field.
Definition. Let Xv . . . , X n be elements of an extension field of K and
let u 1 , . . . , Un be elements of an extension field of Ko Then we say
that (u l , . . . , un) is a specialization of (Xl' . . . , x n ) over K if every
polynomial F (Xl' 0 . . , X n ) E K[ Xl' . . . , Xn] that vanishes at
(Xl' . . . , x n ) also vanishes at (u l , . . . , un).
Notation. (xv..., x n )  (u t , . . 0 , un), or simply (xl' . 0 . , x n )  (u v . . . , un).
K
We "rill be especially concerned with the points (u v . . . , un) in AN Kn.
Definition. A locus in ANKn is said to be an irreducible (affine) alge-
braic variety if it is the locus of specializations (in AN Kn, over K) of a
point (Xl' . . . , x n ) in ANL n , where L is a field containing K. If dot.
K (xv 0 . . , xn)/k = r, then the variety is said to be of dimension r.
If r = 1, the variety is called a curve 0
The problem, of course, is to see that the definitions for a curve given
above and in Chapter 20 agree.
Let r be a curve defined, as in this chapter, by the point (Xl' . . . , x n )
and let r' be the curve defined, as in Chapter 20 (see first paragraph
193

194 SPECIALIZATIONS; SPACE CURVES
above), by the same point. Let '13 be a place of K (xv. . . , xn)/K such
that ord '13(x i ) > 0, i === 1, . . . , n. Let 'l3(x i ) === a i + bit + · · · Let
F(X I , . . . , X n ) E K[X 1 , . . . , Xn] vanish at (xv..., x n ). Since  is an
isomorphism over K, F also vanishes at (a l + · · · , . . . , an + · · .).
Placing t === 0, one finds F (a v . . . , an) === O. Therefore
(Xl' . · . , x n )  (a v . . . an)
K
and (aI' . . . , an) is a point of r; in other words, r' cr. We will show the
converse, r c r', also. This can be formulated as follows:
Theorem 23.1. Every point of r is the center of a branch
((XI)' . . . , (xn)).
For the proof we need a strengthened version of the Theorem of the
Primitive Element, which we will prove here.
Theorem 23.2 (Theorem of the Primitive Element (Strong Form)). Let
F be a field containing in:finitely many elements and let (1., fJ be in an
extension field of F. If (1., fJ are separably algebraic over F (or merely
algebraic if F is of characteristic 0), then for some A E F, (1. + AfJ is a
primitive element for F((1., fJ) over F; that is, F((1. + AfJ) === F((1., fJ).
Proof. Let A be an indeterminate over F ((1., fJ) . We remark first that if
G(X) E F[X] is irreducible in F[X], then it remains irreducible in F(A)[X]:
for if G(X) === U(A, X)V(A, X)/d(A), where U, V E F[A, X] and
d (A) E F[ A], then taking a A E F such that d (A) i=- O-here we use the
infinitude of F-we obtain the factorization G(X) === U(A, X)V(A, X)/d(A).
Hence (1., fJ remain separable over F (A), in fact, they satisfy the same ir-
reducible equations over F (A) as over F. Then w === (1. + AfJ is separable
over F(A). Let G(A; W) === 0 be the irreducible equation for w over F(A).
Then, because of the separability of w, 8G/aW i=- 0, and since
aG
deg w aw < deg JV G,
one has
aG
i=- O.
aw W=w
Now
G(A, (1. + AfJ) === 0,
whence, taking a/aA by a familiar computational rule (see the methods in
Chapter 6), one obtains
aG(A, W)
aA
ff = IX + AfJ
+ 8G(A, W) . {J = 0 .
aw W=cx+Atl

SPECIALIZATIONS; SPACE CURVES 195
One can "solve" this for f3, finding f3 E F (A)((X + A(3). Hence one can
wri te
(J = H(A, at + AfJ)
d l ( A) ,
where H(A, W) E F[A, W] and dl(A) E F[A]. Taking a A E F such that
dl(A) *- 0, one finds
(J = H(J-, at + J-{J)
d l (A) ,
so f3 E F ((X + A(3); then also (X E F ((X + A(3) and F((X, (3) == F ((X + AfJ).
Q.E.D.
To prove 23.1, recall that one of Xl'...' x n is separating for
K (xv. . . , xn)/K, say Xl is (in the case of characteristic 0 any Xi trans-
cendental over K will do). Let (aI' . . . , an) be the given point of r. Apply-
ing a translation: X == Xi - ai' of space, we may replace the Xi by Xi - ai'
or, in other words, we may suppose the a i == o. Let l' . . . , lc be the
places where Xl has residue 0; we know they are finite in number. Consider
the element C 2 X 2 + · · · + CnX n , C i E K. By 23.2, for appropriate Ci E K,
C 2 X 2 + . . · + CnX n is a primitive element of K (xv. . . , xn)/K (Xl); more-
over, by a proof analogous to the proof of 23.2, employing the form
A 2 x 2 + · · · + Anxn (instead of (X + A(3), one can find a polynomial
d(A 2 , . . . , An) [analogous to the dl(A) of that proof] such that
C 2 X 2 + · · · + cnx n will be a primitive element provided d (c 2 , . . . , cn) :1= o.
Now consider the plane curve d having (xv C 2 X 2 + · · · + cnx n ) as generic
point. From the theory of plane curves we know that d is the locus of the
specializations of its general point (thus for plane curves we have the result
we are trying to prove more generally for space curves). Now
(Xl' C 2 X 2 + · · · + CnX n ) -+ (0, 0),
K
sInce
(Xl' X 2 , . . . , X n ) -+ (0, 0, . . . , 0).
K
Hence at least one of the i' i == 1, . . . , k is centered at (0,0) on d; hence
C 2 X 2 + · · · + cnx n has residue 0 in at least one of the i' i == 1, . . . , k.
We may suppose, without loss of generality, that x 2 , . . . , X n are linearly
independent over K; for suppose, say, that X n were a linear combination of
X 2 , . . . , x n - l . Employing an induction on n, we would have a place  of
K(x 1 , . . . , xn_1)/K, i.e., of K(x v . . . , xn)/K, in which Xv . . . , x n - l all
have residue zero; but then X n , as a linear combination X 2 , . . . , X n - 1 ,
would also have residue 0 in , and  would be the place sought. Hence we
assume X 2 , . . . , X n to be linearly independent over K.

196 SPECIALIZATIONS; SPACE CURVES
W . t . (i) (i). 1 1 h th
rl lng c 2 x 2 + · · · + CnX n ,  == ,..., n - , we can c oose e
C(i) such that
J
C(1)
2
Cl) I
*0
C(n -1)
2
. C(n-1)
n
and in this way get n - 1 elements of the form C 2 X 2 + · · · + CnX n linearly
independent over K. Moreover by the same argument, letting i == 1, . . . , N,
we can arrange to have any n - 1 of N elements to be linearly independent
over K (by requiring a certain product of determinants not to vanish).
We also impose the conditions d (ci), . . . , c») * 0, so that we always
get primitive elements. Now let N == (k + l)(n - 1). Then for some n - 1
values of the index i, C&i)x 2 + · · · + cnx) will have residue 0 in the same
j. Solving for X 2 , . . . , X n in terms of these linear combinations, one sees
that X 2 , . . . , X n (as well as Xl) also have residue 0 in that place. This place
will, then, be centered at (0, . . . , 0). Q.E.D.
To sum up: The definitions of irreducible algebraic curves of Chapter 20
and of the present chapter are equivalent. To be sure, we have only considered
affine varieties and affine curves, but as the main difficulty has been met, we
will not go further into this point.
SPECIALIZATIONS AND RESIDUE MAPS ("PlACES")
Definition. Let F be an arbitrary field. By a projective field one means
the notion made up of F and an extra symbol 00 together with the
following rules:
a · 00 == 00 · a == 00,
for a E F (Le., a * (0),
for a * 0 (a E F or a == (0).
a + 00 == 00 + a == 00,
(It is understood that 00 + 00, 00 · 0 and 0 · 00 are undefined. These are)
roughly speaking the rules one encounters in the theory of limits in calculus.
Notation. (F, (0). If (F, (0) and (F', (0) are two projective fields, then by
a morphis1n
T: (F, (0) -+ (F', (0),
one means a mapping a -+ a' of (F, (0) into (F', (0) satisfying the following
rules:
1) 0 -+ 0, 1 -+ 1, 00 -+ 00,
2) a + b -+ a' + b', whenever both sides are defined,
3) a · b -+ a' · b', whenever both sides are defined.

SPECIALIZATIONS AND RESIDUE MAPS ("PLACES") 197
Let F" be the set of images under T contained in F'. If T( a), T( b) E F",
then T(a + b) = T(a) + T(b) E F' and hence T(a) + T(b) = F"; similarly,
if T(a) E F" - 0, then 1 = T(a · l/a) = T(a) · T(lja), so T(I/a) = I/T(a) E F'
and hence I/T(a) E F". Then one proves easily that F" is a field. Hence we
will tacitly assume that in a morphism (F, (0)  (F', (0), the mapping is
onto (F', (0) .
Examples
1) Let t be an indeterminate over K and let F = K(t). Any element in
F can be written as f(t)/g(t), f, g E K[t], G.C.D. (f, g) = 1. Let a E K. If
g(a) #- 0, place f(t)/ g(t)  f(a)/ g(a); if g(a) = 0, [then f(a) #- 0 and] place
f(t)/g(t)  00. This is a morphism: (F, (0) --+ (K, (0).
2) Let u = l/t, so that K (u) = K (t) = F. Placing u  0 as in the
first example, one gets another morphism (F, (0)  (K, (0). One can show
that these are all the morphisms of (F, (0) into (K, (0) over K, Le., which send
every element of K into itself.
3) Let L be a field of algebraic functions of 1 variable over K. Let 
be a place of L/K. If x E L has nonnegative order, and x = a + · · · , we
have defined res x = a; if ord x < 0, we define res x = 00. Then one
checks that x --+ res x is a morphism of (L, (0) into (K, (0) over K. 'Ve call
this mapping a residue map. Thus we have:
Theorem 23.3. Every place  of L/K gives rise via the residue map,
res, to a morphism over K: (L, (0)  (K, (0). Or, stated otherwise, every
residue map is a morphism over K.
Let L be a field of algebraic functions of 1 variable over K (i.e., L is
finite over K, L = K (xv. . . , x n ), and d.t. L/K = 1). We pose the problem
of finding all morphisms (L, (0)  (L', (0) over K of L.
Since the morphism is to be over K, L' must contain K.
Lemma 23.4. If in a morphism x  00, then x#-o and l/x -+ 0 in
the morphism.
Proof. x cannot be zero, as 0 --+ 0, not 0 --+ 00. Now x · l/x = 1. Therefore
x'. (l/x)' = 1', i.e., 00. (l/x)' = 1, unless 00. (l/x)' is not defined. Now
whenever 00. a is defined, 00. a = 00. Hence 00. (l/x)' is not defined.
Hence (1/x), = O. Q.E.D.
Theorem 23.5. W ith  as before, in any morphism (L, (0)  (', (0)
over K either L' = K or the morphism maps L isomorphically onto L'.
The latter morphisms are called trivial. (Recall our convention that
morphisms are onto.)
Proof. Let x E L be transcendental over K, and let x  x'. If x' = 00 or is
algebraic over K, hence in K, for every x, then we are in the first case:

198 SPECIALIZATIONS; SPACE CURVES
L' == K. Otherwise let x', for some x, be transcendental over K. Let y # 0
be another element of L. Then y satisfies an equation
ao(X)ym + al(x)ym-l + · · · + am(x) == 0,
where ai(X) E K[X] and ao(x) .:/= O. Hence also
ao(x) + a l (x)(l/y) + · · · + a m (x)(l/y)m == o.
Then II y  0 is impossible as (ao(x))' == ao(x') i= O. Hence y  00 is
excluded. Thus no element of L maps into 00. Hence for any elements
a, bEL, (a + b)' == a' + b' and (a · b)' == a'b'. Thus the mapping is an
isomorphism on L unless a' == 0 for some a # o. Now if a i= 0, then
a · l/a == 1, a' · (1/a)' is defined, and a' · (1/a)' == l' == 1, whence a' == 0 is
impossible. The proof is complete.
Theorem 23.6. With L as before, every nontrivial morphism of LIK is
the residue map of some place.
Proof. By 20.24, L has a model r free of singularities. Let this be defined
by means of the point P: (xo, . . . , x n ); some Xi i= 0, and we may suppose
Xo == 1. Let T be the given morphism. Not all T(X i ) == 0 or 00, as
T(XO) == T(I) == 1. Let T(X i ) == 00 for s subscripts. We say that we may
assume s == O. For if T(X I ) == 00, say, we define r by means of
P : (1/xv 1, x2/xv . . . , xnlxl).
If T(X j ) # 00, then T(xjlx l ) == T(X j · l/xl) == T(X j ) T(l/xI) == 0, so that more
finite quantities show up for the new form of the coordinates than for the
old (0 and 1 show up in the first two places); Le., there will now be at most
s - 1 indices at which T is infinite. In this way s can be reduced to O.
Supposing s == 0, i.e., that T(X I ), . . . , T(X n ) are all finite, we note that
(xv..., x n )  (T(X I ), ..., T(X n )), so that Q : (T(X I ), ..., T(X n )) is on r.
K
We call this point center of the morphism T. We know that there is a place 
centered at Q; it is then immediate that the morphism res is centered there.
Now we will show that there is only one morphism centered at Q: this will
prove that T == res.
If possible, let T' be another morphism centered at Q. Then there is a
y E L such that T(Y) # T'(y). We want to show that we may suppose
T(Y), T'(y) are both finite. Not both == 00; suppose T'(y) == 00. There are
two cases: (i) T(Y) # 0, (ii) T(Y) == O. If T(Y) # 0, then T(l/y) # 00; also
T(11 y) # 0, for T( Y . II y) == T( y) · T(11 y) is defined and == 1. Then
T(11 y) # T'(11 y) == 0, and we replace y by II y. If T( y) == 0 [and T'(y) == 00],
then replacing y by 1 + y, we are in the first case. Hence in every case there
is a y with T(Y), T'(y) finite and unequal.
Now consider the curve  defined by the point (Xl' . . . , X n , y). Then
( T(X I ), . . . , T(X n ), T( y) ) and (T' (Xl)' . . . , T' (x n ), T' (y)) are distinct points on

VALUATIONS AND MORPHISMS ("PLACES") 199
. Let,' be places centered at these points of. Then  i= ', since
T(Y) i= T'(y). But these places both have the center (T(X 1 ), . . . , T(X n )} ==
(T'(X 1 ), . . . , T'(X n )} == Q on r. This is impossible, as Q is simple for r. The
proof is complete.
Remark. In the literature, morphisms are usually called places. Thus 23.6
justifies (if any justification is needed) our using the term for a different,
though associated, concept.
VALUATIONS AND MORPHISMS ("PLACES")
In a moment, we will define the notion of a valuation of a field L over a base
field F. When we have done so it will be obvious in the case F == K and L
a field of algebraic functions of 1 variable over K, that for any place , ord
yields a valuation; moreover, in this way one gets all the valuations of
LIK. Hence so far as the theory of algebraic curves is concerned, we do not
really need this extra term. Since, however, the concept is of frequent
occurrence, we establish the relation between valuations and places.
By an ordered abelian group (under +) one means the notion made up
of an abelian group and a transitive relation «) satisfying the rule:
(*)
a<b=>a+b<b+c
for all a, b, c in the group; and for every a in the group, precisely one of the
relations a == 0, a < 0, 0 < a holds.
The additive group of integers orderedin the usual way is an example.
For another example, consider the set of ordered pairs (a, b) of integers;
by defining (a, b) + (c, d) as (a + c, b + d) one gets an abelian group, and
by placing (a, b) < (c, d) if a < c or if a == c and b < d, one gets an ordered
a belian group.
If a < band c < d, then a + c < b + d. In fact, a + c < b + c <
b + d by two applications of (*), and a + c < b + d follo,vs by transitivity.
Definition. By a valuat.ion of a field L one means a mapping v of L - 0
onto an ordered abelian group r satisfying the following rules:
1) v(ab) == v(a) + v(b),
2) v(a + b) > min {v(a), v(b)} (a + b # 0)
for all a, b in L - 0 (with the stated exceptions).
To avoid exceptions one introduces a symbol + 00 "W"ith the rules
ex < + 00 and ex + (+ (0) === (+ (0) + ex < + 00 for all ex E r, and
+00 + (+00) === +00. Then (1) and (2) hold for all elements in.
From v(l) === v(12) === v(l) + v(I), one sees that v(l) === O. Then
o == v(l) === v( (-1)2) == v( -1) + v( -1). In an ordered group, one proves
easily that x + x === 0 implies x === o. Hence v( -1) == O. Therefore
v(a) == v( -a) for all a E L.

200 SPECIALIZA TIONS ; SPACE CURVES
In (2), if v(a) # v(b), then v(a + b) = min {v(a), v(b)}. In fact, suppose
without loss of generality that v(a) > v(b). Werev(a + b) > min {v(a), v(b)},
i.e., v(a + b) > v(b), one would find
v(b) = v(a + b - a) > min {v(a + b), v( -a)} > v(b),
a contradiction.
If F is a field contained in L, the valuation v is said to be over F if
v(a) = 0 for all a E F - O.
As already mentioned, in the case F = K and L a field of algebraic
functions of 1 variable over K, ord yields a valuation for every place '13.
Theorem 23.7. Let v be a valuation of a field L. Then
 = {alv(a) > O}
is a ring, called the valuation ring of v. It has the property that for any
a E L - 0, a E  or IJa E. Conversely, given a ring  having this
property, there is a valuation of L of which  is the ring. (Hence such rings
are called valuation rings.)
Proof. That  is a ring follows at once from points (1) and (2) of the defini-
tion of a valuation. Now 0 = v(l) = v(a · IJa) = v(a) + v(lja); then
v(a) < 0 and v(lja) < 0 is impossible, as together these imply 0 < 0:
this proves the second point.
Before proving the converse, we show how to describe the group r of
the valuation directly in terms of , up to an isomorphism. Let us write
a '"" b for a, bEL - 0 if v(a) = v(b); this is obviously an equivalence rela-
tion. We can eliminate v from this description by saying a '"" b if ajb and bJa
are in , or if ajb is a unit in. Let Ka = {xix E L, x '"" a}. Note that
Ka = Kb if and only if a '"" b; in what follows we make some constructions
with the sets Ka (a E L) and though these constructions will be given in
terms of the subscripts a, it will be important (though not difficult) to check
that the constructions are independent of the subscript chosen.
Define, then, Ka + Kb as Kab. We also place Ka > Kb if v(a) > v(b),
or, eliminating v, if ajb is a nonunit in. Then the set {Ka} has been made
into an ordered group, and it is not difficult to show that v(a)  Ka is an
order-preserving isomorphism of r and the group just constructed.
Now for the converse, we define a '"" b, if aJb is a unit in , and check
that this is an equivalence relation; then place Ka = {x la '"" x},
Ka + Kb = K ab , and Ka > Kb if ajb is a nonunit in. Then the set {Ka}
has been made into an ordered group. (The zero of this group is KI') One
checks that the mapping a  Ka is a valuation. In this way one completes
the proof.
Let v be a valuation of L and let r be its group. Let 'T : r  r' be an
order-preserving isomorphism of r and an ordered group r'. Then

VALUATIONS AND MORPHISMS ("PLACES") 201
a  T{v(a)) is obviously also a valuation; but because of the trivial nature
of this construction, the valuations are regarded as the same. We adopt this
point of view also. By doing so we can say: A valuation can be described
(in standard fashion) in terms of its ring.
Theorem 23.8. Let L be a field. Every morphism T : (L, (0)  (L', (0)
gives rise to a valuation in a standard way, namely, the set of elements in
L mapping into L' form a valuation ring, and this yields the valuation.
Conversely, every valuation ring  yields a morphism in a standard way,
and from the morphism one recovers (as in the first part of the theorem) the
valuation ring.
Proof. One checks directly from the definition of a morphism that the
elements of L mapping into L' form a ring,. If a ELand a f# , i.e.,
T(a) = 00, then T(I/a) = 0, so l/a E. This shows that  is a valuation
rIng.
Before proving the converse, we show how the set of images in L'
(i.e., L' according to our agreement) can be described, up to an isomorphism,
in terms of .
Let us write a f'.J b (for a, b E ) if T(a) = T(b); this is obviously an
equivalence relation. Now T(a) = T(b) if and only if T(a - b) = 0; in fact
[having checked that T( -b) = -T(b)], we have T(a - b) = T(a) - T(b),
from which the assertion follows. Now T(X) = 0 if and only if x is a nonunit
in ; in fact, if x is a nonunit, then T(I/x) = 00, as otherwise
1 = T(X · l/x) = T(X) · T(I/x) = 0;
i.e., l/x f#. In this way we can eliminate T from the definition of a '"" b:
we have a f'.J b if and only if a - b is a nonunit in .
Now define Ka = {xix E , x f'.J a}. Then Ka - T(a) is obviously a one-
to-one correspondence between the set {Ka} and L'. Defining Ka + Kb =
Ka+b and Ka. K b = K ab , we see that this correspondence becomes an
isomorphism.
We make still one more comment and formulate a lemma before proving
the converse part of the theorem. Namely, if  is the valuation ring arising
from a morphism T, then the nonunits in  form an ideal. In fact, the non-
units are just the elements x E  such that T(X) = 0; if T(X 1 ) = 0 and
T(X 2 ) = 0 and a E, then T(X 1 + x 2 ) = 0 and T(ax 1 ) = 0, and this proves
the assertion. Moreover, this property of  already follows from the defining
property of a valuation ring, as stated in the following:
Lemma 23.9. Let  be a valuation ring. Then the nonunits of  form
an ideal '13. [Since every other ideal (i=- ) in  must be contained in '13,
'13 is the sole maximal ideal in .]
Proof. Let x, y be nonunits in : to show that x + y is a nonunit. We may
exclude as trivial the case that x = 0 or y = o. Now x/y or y/x is in , say

202 SPECIALIZATIONS; SPACE CURVES
yJx E. Suppose I/(x + y) E. Then I/(x + y) = (IJx)J(1 + yJx) E 
and l/x E , a contradiction.
Now for the converse part of 23.7, starting from the valuation ring 
(of quotient field L) we define a 1"-1 b, if a - b is a nonunit in . It follo\vs
from the lemma that this is an equivalence relation. Let
Ka = {xix E , x 1"-1 a},
and define Ka + K b = Ka+b' Ka · K b = Kab. Then one checks that the set
{Ka} is thereby made into a field L', and, moreover, placing T(a) = Ka for
a E  and T(a) = 00 for a tt , that T: (L, (0)  (L', (0) is a morphism
having  as associated valuation ring. Q.E.D.
Theorem 23.7 shows that there is a standard one-to-one correspondence
between the valuations of a field L and its morphisms [or the morphisms of
(L, (0), rather]. Moreover it is not difficult to show that if there is a base field
F C L, then a valuation over F gives rise to a morphism over F, and'vice
versa. Thus to find an valuations of L/ F, it is sufficient to find all morphisms
of LJ F. In the case F = K and L a field of algebraic functions of 1 variable
over K, we know all the morphisms, and hence we know all the valuations.
The morphisms all arise as residue maps of places. Let  be a place of LIK.
Then its valuation ring  consists of the elements having finite residue, or,
stated differently, of the elements of nonnegative order. Thus the ring of
the valuation ord is the same as the ring of the residue map res. Hence
we have:
Theorem 23.10. In the case L is afield of algebraic functions of one variable
over the base field K, every valuation of LJK is of the form ord for some
place  of LIK.

Chapter 24
INFINITELY NEAR POINTS
In Chapter 14 we gave a rough explanation of the notion of infinitely near
points sufficient for our purpose. In order not to leave a loose end, we give
a rigorous definition. We follow, somewhat, the exposition of B. L. van der
Waerden (Indicationes Mathematicae, vol. 12, North Holland Publishing
Company, Amsterdam, 1950, pp. 401-410). Unlike van der Waerden, who
works at a simple point of an arbitrary d-dimensional variety, we consider
only the plane: the main point can already be seen in this case.
Let P be a point, say the origin. The points infinitely near P will be
defined by means of the cycles, or divisors, through P and the branches
centered at P. We first define a point PI infinitely near to P and in the fir8t
neighborhood. The idea is to define PI by means of the divisors that we "know"
pass through Pl' Thus we come to the following definition.
Definition AI. By a point PI infinitely near and in the first neighbor-
hood of P one means the set of divisors D having with some branch fJ
of center P an intersection multiplicity > rs, where r === multiplicity
of D at P and s === multiplicity of fJ at P (==min {ord p x, ord p y}). A
divisor D is said to pass through PI if D E Pl'
In the definitions we may allow all (i.e., analytic) branches, or we may
restrict ourselves to branches of algebraic curves.
Each branch fJ uniquely determines a PI; but different branches may
determine the same Pl'
By analogy, P itself is the set of all divisors through P.
Recall that i(D, fJ; P) > rs. As i(D, fJ; P) > rs if and only if the tan-
gent to fJ is a tangent to D at P, we see that for every line l through P there
is one and only one point PI infinitely near and in the first neighborhood of
P, namely, the set D of divisors having l as a tangent at P.
Definition BI. A branch y centered at P is said to pass through PI
if i(D, y; P) > r(D)s(y) for every D in Pv where r(D), s(y) are the
multiplicities of D, y at P.
203

204 INFINITELY NEAR POINTS
Thus y passes through PI if and only if it has the same tangent as fJ,
a branch defining Pl. Hence also:
Theorem C 1 - If a branch y passes through Pv then it determines the
same infinitely near point Pl.
Definition D 1 - The multiplicity 8 1 === (fJ)Pl of a branch at PI is the
minimum excess of i(D, fJ; P) over r8 for all D E Pl.
Clearly 8 1 > 0 if and only if fJ passes through Pl.
Definition E 1 - The nlultiplicity r 1 === (D) 1\ of a divisor at PI is the
minimum excess of i(D, fJ; P) over r8 for all fJ passing through Pl.
Clearly r 1 > 0 if and only if D passes through Pl.
We will also prove:
Theorem Fp i(D, fJ; P) > r8 + r 1 8 1 , where r, 8, rv 8 1 are the multipli-
cities of D, fJ at P, Pl.
Taking F 1 for granted, we define a point P 2 infinitely near to P and
8ucce88or to PI using the divisors and branches through Pl.
Definition A 2 - By a second neighborhood point P 2 infinitely near P
and successor to PI one means the set of divisors D through PI
having ,,"ith some branch fJ through PI an intersection multiplicity
> r8 + r 1 8 1 (see F 1 ). These divisors are said to pass through P 2 .
Definition B 2 - A branch y through PI is said to pass through P 2 if
i(D, y; P) > r(D)8(Y) + r 1 (D)8 1 (y) for every D through P 2 .
Theorem C 2 - If a branch y passes through P 2 , then it determines the
same successor P 2 of Pl.
We postpone the proof of O 2 (and F 2 ) for a moment.
Definition D 2 - The rnultiplicity 8 2 === (fJ)p of a branch fJ (through PI)
2
at P 2 is the minimum excess of i(D, fJ; P) over r8 + r 1 8 1 for all D E P 2 .
Clearly 8 2 > 0 if and only if fJ passes through P 2.
Definition E 2 - The multiplicity r 2 === (D) P2 of a divisor D (through PI)
at P 2 is the minimum excess of i(D, fJ; P) over r8 + r 1 8 1 for all fJ passing
through P 2 .
Clearly r 2 > 0 if and only if D passes through P 2 .
Theorem F 2 - i(D, fJ; P) > r8 + r 1 8 1 + r 2 8 2 .
Taking O 2 and F 2 for granted, we set up Definitions Aa, Ba, Theorem 0a,
Definitions Da, Ea, Theorem Fa, etc.

INFINITELY NEAR POINTS 205
According to the above, a point Pi in the ith neighborhood of P is a
notion made up of the divisors through Pi and the predecessors P, PI' . . · , P i - 1
of Pi. However, we will prove:
Theorem Gie Given an infinitely near point Pi' the set of divisors
through Pi itself determines the integer i and the predecessors
P , PI' · · · , Pi-I.
In other words, an infinitely near point is determined by, and may simply
be identified with, the divisors through it.
We have postponed the proofs as they use locally quadratic transforma-
tions and we wish to emphasize that the definitions make no reference to any
transformation.
To facilitate the proofs, we will also prove:
Theoren Hie Let D : F = 0 be a divisor passing through P , PI' . . . , Pi
with multiplicities r, r 1 , . . . , r i (we may allow r i = 0). Then if P(X, Y)
is a polynomial of sufficiently high subdegree, the divisor F + P = 0
also passes through P, PI' . . . , Pi with the same multiplicities.
Similarly, if fJ: (x(t), y(t)) passes through P, PI'...' Pi with multi-
plicities 8, 8 1 , . . . , 8i, and the primitive representation x(t), y(t) is
modified by terms of sufficiently high degree (keeping primitivity), then
the modified branch also passes through P, PI' . . . , Pi with the same
multiplicities.
In the proofs we assume PI is determined by the X -axis and consider
the locally quadratic transformation X' = X, Y' = Y /X. Recall from
Chapter 12 that the transform of F(X, Y) = 0 is F(X', X'Y')/X'r = 0,
where r is the multiplicity of F = 0 at P: (0, 0). If F == O(X), the com-
ponents X = 0 are lost in the transformation; otherwise, Le., if F =I=- O(X),
then F can be recovered from the transform. Conversely, every divisor
F' = 0 not having X' = 0 as component arises from a divisor F = 0,
its counter-transform. The transformation sets up a one-to-one correspon-
dence between the divisors F(X, Y) = 0 through P and tangent to the
X-axis and not having X = 0 as _component and the divisors F'(X', Y') = 0
through P': (0, 0) not having X' = 0 as component. More generally, we
will prove:
Theorem 'ie Let fJ be a branch centered at P and tangent to the
X-axis; let Pv P 2 , . . . , Pi be the initial successive infinitely near
points on fJ; and let 8, 8 1 , . . . , 8i be the multiplicities of fJ at
P, P 1 , . . . , Pi. Let fJ' be the transform of fJ under X' = X, Y' = Y/X,
let P', p, . . . , P;-1 be the center and the initial i-I successive in-
finitely near points on fJ', and let 8', s{, . . . , 8;-1 be the multiplicities of
fJ' at P', P, · . . , Pi-I. Then 8i-I = 8i. The transformation sets up a

206 INFINITELY NEAR POINTS
one-to-one correspondence of the divisors D through Pi and not having
X === 0 as component and the divisors D' through P-l not having
X' === 0 as component; if D passes through Pi with multiplicity ri,
then D' passes through P;-l with this same multiplicity. If P, PI' . . . , Pi
are the initial successive points on another branch y, then P', P{, . . . ,
P-l are also the initial points on the transform y' of y.
Now for the proofs, and first. that of F 1 (we already have C 1 ). We may
assume D and fJ are tangent to the X-axis, otherwise r 1 === 0 or Sl === 0 and Fl
is clear. By a simple calculation one has i(D, fJ; P) === rs + i(D', fJ'; P'),
where D' and fJ' are the transforms of D and fJ. Let r', s' be the multiplicities
at P'. Then i(D, fJ; P) > rs + r's'. Take D' to have a simple point at P'
with tangent different from the tangent of fJ' and not having X' === 0 as
component and let D be its counter-transform. Then i(D, (J; P) === rs + s'.
Hence s' is the minimum (positive) excess, Le., s' === SI. In a similar way, one
sees that r' === r l . Hence i(D, (J; P) > rs + rls l . This proves F l , and II as
well.
In G I , we are given the set S of divisors of some PI' Le., all the divisors
through P having a given line as tangent at P. These divisors have only P
in common, hence cannot determine a point infinitely near to any point ::j::P.
Inthepreviousparagraphwesawthatini(D,{J; P) > rs + rls V the equality
sometimes holds (for given fJ); this shows that the set of divisors through a
P 2 is always a proper subset of the set through Pl. Hence S can only belong
to a point in the first neighborhood of P, Le., i === 1. Q.E.D.
In HI' let D : F === 0 have multiplicities r, r 1 at P, Pl. Then
D': F(X', X'Y')/X'T === 0
has multiplicity r l at P' (by II). Hence the first part of HI holds for
subd P > r + r l . Similarly, if
fJ: x( t) === t S + · · · , y( t) === ct s + s ' + · · ·
(c =1= 0, s' > 0),
then the second part holds if x( t), y( t) are modified in terms of degree
>s + s'.
In the induction, we first prove C i (i > 1). This is of some importance,
as otherwise one has meticulously to distinguish the branches {J which
define a Pi from the branches y which go through Pi; but with this exception,
the order of the proofs is immaterial.
Let, then, {J define the successive point P, P v . . . , Pi; and let y pass
through them. By induction assumption, P, PI' . . . , P i - l is the initial set
of successive points on y. Let Qi be the successor of Pi-Ion y. In C i we are
given: Pi c Qi, and are to prove Pi === Qi. Let {J', y' be the transforms of
{J, y under X' === X, Y' === Y/X. By I i - v {J', y' have the same initial suc-
cessive points P', P{,. . . , P-2. Let P-l' Q-l be the successors of P-2

INFINITELY NEAR POINTS 207
on fJ', y'; we prove: P -1 = Q -1. Let D' E P -1 and first suppose D' does
not have X' = 0 as component. Let r l , . . . , r i - l be its multiplicities at
P', P{, . . . , P-2. As D' E P-2' by I i - l its counter-transform D has multi-
plicities rv . . . , ri-l at Pv . . . , P i - l (and let r be its multiplicity at P).
Let fJ, y have multiplicities s, Sv . . . , Si-V t, t l , . . . , t i - l at P, Pl' · . . , Pi-l'
and hence multiplicities Sl, . . . , Si-V t l , . . . , t i - l at P', P{, . . . , P-2 (by
I i - l ). We have i(D', fJ') > rls l + · · · + ri-lsi-l, hence
i(D, fJ) > rs + rls l + · · · + ri-lsi-l.
Hence D E Pi and therefore D E Qi. Hence
i(D, y) > rt + · · · + r i - I t i - 1
and
i(D', y') > rlt l + · · · + ri-lti-l'
whence D' E Q;-1. Now assume D': F' = 0, D' E P-1' does have X' = 0
as component. By adding on high-degree terms to the factors X' in F', one
gets a divisor D": F" = 0 not having X' = 0 as component. Moreover, by
H i - l , one does it in such way that D" and D' have the same multiplicities at
P', P{, . . . , P;-2; and we may also suppose D", D' have the same inter-
section multiplicities with f3' and with y'. Hence D" E P-1 and D" E Q-v
by the first case, whence D' E Q; -1. Hence P -1 C Q -1' so by induction
(C i - l ), P-1 = Q-1. By arguments just given we also have part of Ii,
namely, that the transformation sets up a one-to-one correspondence of the
divisors D through Pi not having X = 0 as component and the divisors D'
through P-l not having X' = 0 as component. Hence one concludes that
Pi and Qi coincide except perhaps in the divisors having X = 0 as conl-
ponent. To conclude, one may argue that D : XF = 0 is in Pi if and only if
F = 0 is in Pi (which is easily proved); or one may vary D as above D'
was varied. Thus Pi = Qi. Q.E.D.
Now we prove Ii' We already have the stated correspondence between
certain divisors through Pi and certain ones through P-1. This shows that
s;-1 < Si; however, nontransformed divisors, i.e., divisors having X' = 0
as component enter into the definition of S-1' so a priori s-1 < Si might hold.
Let, then, D' E P-1 yield the minimum (positive) excess
i(D', fJ') - rls l - · · · - ri-lsi-l = S;_1.
If D' has X' = 0 as a component, we vary D' using H i - l (as in the proof of
C i ) to get a transformed divisor D" E P-1 yielding the same minimum.
Hence s-1 = Si' The assertion that the multiplicity of D at Pi is the same
as that of D' at P;-1 is proved similarly (using the second part of Hi_vas
the branch x' = 0, y' = t enters into the definition of the multiplicity of D'
at P-l)' The remaining part of Ii has already been established.

208 INFINITELY NEAR POINTS
Now F i is immediate. We have i(D, fJ) = rs + i(D', fJ') and
i{D', fJ') > r 1 s 1 + · · · + risi by induction, whence F i follows.
Also, Hi is immediate. Let D': F' = 0 be the transform of D: F = 0
and let h be so large that F' = 0 and F' + P' = 0 win have the same
multiplicities at P', P{, . . . , P-l provided subd P' > h. Then F = 0 and
F + P = 0 will have the same multiplicities at P, Pv..., Pi if
subd P > h + r. The second part of Hi is proved similarly.
As G i did not enter the inductive scheme, the proofs ofO i , F i , Hi, and Ii
are complete.
As for G i , let Pi be a given infinitely near point, say the ith along a branch
fJ, and assume that the divisors through Pi are the same as those through Qj,
an infinitely near point, jth along a branch y. Let P, Q be the centers of
fJ, y with P at the origin. Let F = 0 be a divisor through Pi. If Q i=- P,
we can modify F by high-degree terms to get a divisor F + P = 0 through
Pi but not through Q; a contradiction. So P = Q. Now we prove that
fJ, y have the same tangent. Let the tangent fJ be the X-axis and take the
Y-axis not tangent to y. Let the tangent y be Y = eX. Apply the trans-
formation X' = X, Y' = Y IX. Then y goes over into a branch y' of center
(0, e). The transforms of the divisors through Pi therefore pass through
(0, e) and through P-l' the transform of Pi. If e i=- 0, we easily modify a
divisor through P-1 and (0, e) to get a transformed divisor through Pi-l
and not through (0, c), a contradiction. Therefore e = 0 and y also has the
X -axis as tangent. Let Qj -1 be the (j - 1 )th successive point along y'.
Then P-1 and Qj-1 coincide in their transformed members. By an argument
of type already given (using H), Pi-l and Qj-l coincide also in their non-
transformed members; i.e., P--l = Qj-1. By induction on i we conclude
i-I = j - 1, whence i = j. The rest of the proof follows similarly by
induction.
Noether's Theorem (the last theorem in Chapter 14) and its generalization
to a divisor and branch [Le., i(P, fJ) = ! risiJ are immediate.
We would also want to show that infinitely near points transform as
expected under a standard quadratic transformation; but this is secondary to
the main point, namely, to give the definitions wi thou t using transforma t ions.

NOTES
Page 1. By a ring R one means a set in which two operations + and. are defined
and in which one computes according to some of the familiar rules for addition
and multiplication. More precisely, for + we have the following rules: a + b
is in R for every a, b in R; a + b = b + a for every a, b in R; (a + b) + c
= a + (b + c) for every a, b, c in R; there is an element 0 in R, called zero,
such that a + 0 = a for every a; every equation a + x = b, where a, b are in
R, has a unique solution x. (These rules, especially the last, allow one to intro-
duce subtraction, and the solution to a + x = b is then denoted x = b - a.)
For . we ha ve the rules: a. b is in R for every a, b in R; a. b = b · a for
every a, b in R; (a. b) . c = a . (b . c) for every a, b, c in R; there is an element
1, called identity (or one), such that a . 1 = a for every a in R.
There is also a rule connecting + and., namely: a. (b + c) = (a . b) + (a · c)
for every a, b, c in R.
It is also possible to have a . b = 0 with a i:- 0 and b i:- O. If this does not
happen and if 1 i:- 0, then the ring is called an integral domain.
A field is a ring R with 1 i:- 0 in which every equation a . x = b, a, b in R,
a i:- 0, has a unique solution. (One can then introduce division, and the solution
to a . x = b is denoted x = b/a.)
If R is a ring, then by a polynomial in one letter (or indeterminate) X over
R one means an expression of the form ao + alX + a 2 X2 + . . ., where the ai
are in R and all but a finite number of the ai equal zero. It is understood that
two such expressions, ao + a l X + a 2 X2 + . .. and b o + b l X + b 2 X2 + . · .,
are the same or equal if and only if ai = b i for every i. The set of polynomials
in one letter X over R is denoted R[X]. If one defines + and. in R[X] in the
expected way, namely,
(a o + a 1 X + a 2 X2 + · . .) + (b o + blX + b 2 X 2 + . . .)
= (ao + b o ) + (at + bl)X + (a 2 + b 2 )X2 + · · .,
and
(a o + alX + a 2 X 2 + . . .). (b o + b l X + b 2 X2 + . . .)
= (aob o ) + (aOb l + atbo)X + (a o b 2 + alb l + a 2 b o )X2 + . · .,
then R[X] becomes a ring.
209

210 NOTES
In a similar way one defines a polynomial ring over R in two or more letters.
In writing down a o + a l X + a 2 X2 + . . ., it is usual to omit some or all the
terms aiXi for which ai = 0, as this can be done without confusion. In this
way the polynonlial a o + 0 . X + 0 . X 2 + . . . is identified with the element a o .
Page 44. Let S be a ring and let R be a subset of S which is a ring under the
operations of S, Le., let R be a sub ring of S; in particular, we suppose the 1 of S
to be in R. Let a l ,..., am E S. The set of elements in S of the form
"" Ci l . . . i ail. . . a m im , Ci ... i E R, Le., the set of elements which can be written
k m 1 m
as polynomial expressions in aI' . . . , am with coefficients in R, form a subring
of S, denoted R[a l ,. . . , am]. This ring is said to arise by ring adjunction of
aI' · · · , am to R. It is the smallest sub ring of S containing R and aI' . . . , am.
If Rand S are fields, the set of elements of the form I Ci] . . . imai m . . . a:r/
I d il . .. i m a : 1 . · . am forms a subfield of S, denoted R(a l ,... , am). This field
is said to arise by field adjunction of aI' . . . , am to S. It is the smallest subfield
of S containing aI' . . . , am.
Let R be an integral domain. The quotient field of R is obtained by formally
forrning the expressions a/b, with a, b in Rand b i:- 0, and subjecting these to
the familiar rules for working with quotients in a field. The construction imitates
the one that leads from the ring of integers to the rational numbers.
On page 44 we have nine in determinates U 00' . . . , U 22 over C. In the first
note, we have explained how to construct C[ U 00' . . . , U 22]. The field
C( U 00, . . . , U 22) is its quotient field.
Now let K and L be fields with K contained in L: K c L. An element (or
quantity) r.J. in L is said to be algebraic over K if there exist elements co' c l ' . . . , C n
in K with Co i:- 0 such that cor.J.n + c 1 r.J.n-l + . . . + C n = 0; in other words, if
r.J. satisfies a nontrivial polynomial relation over K. If r.J., f3 are in L and algebraic
over K, then from algebra it is known that r.J. =f:: fJ, r.J. . fJ, and r.J./f3 (if fJ i:- 0) are
also algebraic over K. r.t'he field L is said to be algebraic over K if every element
in L is algebraic over K; L is then also. called an algebraic extension of K. The
field K is called algebraically closed if it has no proper algebraic extensions, Le.,
if L algebraic over K implies L = K. By an algebraic closure of a field K one
means an algebraically closed algebraic extension field of K. It is known that
every field has an algebraic closure.
Page 48. Laplace's expansion. Let A = (aij) be an n X n matrix with ai; in
the ith row and jth column. Let i'l' . . . , i r be r of the numbers from 1 to n;
and let jl' . . . , jr be r of the numbers froln 1 to n. Denote by A(i 1 , . . · , i r ;
jl' · · . , jr) the r X r determinant whose entry in the pth row and qth column
is ai j. If i 1 < . . . < i r and jl < . . . < jr, then A (iI' . . . , i r ; jl' . . . , jr) is called
p q
a minor. It is, of course, made up from the elements in rows iI' . . . , i r and
columns jl' . . . , jr of A. The minor made up from the other n - r rows and the
other n - r columns is called the complementary minor; this complementary
'minor times (-1 )il + . . . + i r + i l + . . . + i r is called the cofactor of A (iI' . . . , i r ;
jl'... ,jr)' and is denoted by A'(i l ,..., i r ; jl'... ,jr). Now fix i 1 ,..., i r with
it < . . · < ire According to a theorem of Laplace,
det A = I A(i l ,..., i r ; jl'... ,jr)A'(i l ,..., i r ; jl'... ,jr)'
,vhere one sums over all choices of jl' . . . , jr with jl < · . . < jr. This way of

NOTES 211
computing det A is called the Laplacian expansion of det A by the rows
iI' . . . , ire For r = 1, one gets the familiar expansion by a specified row.
Page 49. Let deg(x,Y) F = m, deg(x,Y) G = n. Since F = 0 or G = 0 does not
pass through (1,0,0), deg ]?(O, Y) = m or deg G(O, Y) = n. It is convenient
(and possible, though not necessary) to arrange the coordinate system so that
neither }? = 0 nor G = 0 passes through (1,0, 0); and then deg G(O, Y) = rn
and deg G(O, Y) = n. Thus the determinant giving R{ F(O, V), G(O, Y)) is
m X n, and in fact is = R(F(X, V), G(X, Y))lx=o. Moreover, F(O, Y) == O(Y)
and G(O, Y) == O( Y); and one sees that R{ F(O, Y)/ Y, G(O, Y)/ V), is the stated
cofactor.
Page 105. * If K and L are two fields with ]( eLand if there are elements
Xl' . . . , X m in L linearly independent over K such that
L = K . Xl + K . x 2 -+ . . . + K . X m ,
then one says that the degree of Lover K is 'in, and one writes [L : K] = m. If
Yl' · · · , Ys is another system of elements linearly independent over K and with
L = K . Yl + . . . + K . Ys, then from linear algebra it is known that m = s;
so the degree is uniquely determined. In the terminology of linear algebra, L is
a finite-dimensional linear (or vector) space over K of dimension m. (See note
to page 117 below.)
If L = K(r.J. 1 , . . . , r.J. n ) and each r.J.i is algebraic over K, then L is a finite-
dimensional Unear space over K. Let us prove this; and first for n = 1. Let
(Xl = r.J..
Since r.J. is algebraic over K, there is a polynomialf(X) in K[X] - 0 such that
f(r.J.) = o. Of all such polynomials take one of least degree and denote it by f(X).
Note that f(X) is irreducible in K[X]. Let now h(X) E K[X] be such that
h(r.J.) i:- o. We are going to show that l/h(r.J.) can be written as a polynomial in r.J..
We cannot have h(X) == O{f(X)), as otherwise h(r.J.) = 0; and asf(X) is irreducible,
we have G.O.D. (f, h) = 1. Hence we can write 1 = A(X)f(X) + B(X)h(X) for
some A, B in K[X]. Then 1 = B(r.J.)h(r.J.) and l/h(r.J.) = B(r.J.). This shows that
K(r.J.) = K[cx.], where K[r.J.] denotes ring adjunction to K, Le., the set of polynomials
in r.J. with coefficients in K; for any element of K (ex.) can be written in the form
g(r.J.)/h(r.J.) with g(X), h(X) E K[X]; and then one rewrites l/h(r.J.) as a poly-
nonlial in r.J.. Now if f(X) = coX r + c l xr-l + . .., Ci E K, CO i:- 0, then any
element of K[r.J.] can be written as a linear combination of 1, r.J., . . . , r.J.r-l. In
fact, let g(r.J.) E K[r.J.], g E K[X), and write g = gf + s, where g, s E K[X] and
s = 0 or deg s < r. Then g(r.J.) = s(r.J.), as desired. Moreover, 1, r.J., . . . , r.J.r-l are
linearly independent over K (otherwise r.J. would satisfy a polynomial of degree
less than r), so here [K(r.J.) :K] = r.
Now for arbitrary n, applying the case n = 1, we can write K(r.J. 1 , . . . , r.J. n )
= K(r.J. l , . . . , r.J. n - 1 )[r.J. n ]; applying it again, we have
K(r.J. 1 , · · · , r.J. n - 1 )[r.J. n ] = K(r.J. 1 , · · · , r.J. n -2)[r.J. n - 1 ][r.J. n ];
and finally K (r.J. 1 , . . . , r.J. n ) = K[ r.J. l , . . . , r.J. n ]. Thus the elements of K (r.J. 1 , . . . , r.J. n )
can be written as polynomials in r.J.l' . . . , r.J. n ; and moreover of limited degree;
i.e., the elements of L can be written as a linear combination of certain power-
products of the r.J.i. The proof is complete.

212 NOTES
f>age 105.** Here we are deviating from the standard terminology. What we
call a finite extension is usually called a finitely generated extension; and what
is usually called a finite extension, we call a finite algebraic extension. rrhe usual
term "finite extension" is bad, as it refers to the larger field as a vector space
(rather than a field). Our terminology has the field structure alone in view. We
think our terminology is better than the standard one, but do not doubt that it
may cause additional confusion. Perhaps the best way to avoid the confusion
(temporarily) would be to drop the term "finite extension" altogether and on
the one hand speak of finitely generated extensions and on the other of finite
algebraic extensions.
Page 113. If L = K(r.J. 1 , . . . , r.J. n ) is an algebraic extension of K, then the degree
of Lover K, [L: K], has been defined on page 211. If L is a simple extension of
K, Le., of the form L = .K(r.J.), then [L:K] is the degree of an equation over K
of least degree satisfied by r.J.; this was proved on page 211.
Page 117. A linear (or vector) space over the field k is the notion made up of an
additive abelian group V and a mapping of k X V into V subject to the following
rules: c(r.J. + fJ) = Cr.J. + ('fJ, (c + c')r.J. = Cr.J. + c'r.J., «('c')r.J. = c(c'r.J.), 1r.J. = r.J.. Here
c, c' are elements of k, 1 is the identity of k, and , {J are elements of V; and Cr.J.
designates the image of (c, r.J.) in the mapping.
Page 129. Let F, p\, . . . , F q be polynomials in C[X 1 ,..., X n ]. Hilbert's
Nullstellensatz says that if F vanishes at every common zero (Xl'...' X n ) of
F l' . . . , F q (where the Xi are in C), then some power of F is a linear combination
of F l' . . . , F q with coefficients from C[X l' · · · , Xn]:
FP-AF + ... + A F
- 1 1 q q.
We have encountered special cases of this theorem. :B'or a proof, see
o. Zariski and P. Samuel, Communicative ..Algebra, vol. 2, page 164.
Page 152. Let k be an arbitrary field; 1, the identity of k. "orm the sums
1 + 1, 1 + 1 + 1, . .. These may all differ from zero, in which case k is said
to be of characteristic zero. Otherwise there is a smallest positive integer p such
that the p-fold sum 1 + 1 + . · . + 1 equals zero. Then the field is said to be
of characteristic p.

INDEX
abelian variety, 65
absolute irreducibility, 65
adjoint, 144
adjunction, 210
AF + BG Theorem, 133
algebraic, 210
algebraic curve, 2, 3, 10, 17, 160
algebraically closed, 44, 210 (note), 152,
159
algebraically (in)dependent, 105
analytic curve, 178
analytic cycle, 1 78
along the Y-axis, 179
ANK (affine number plane over K),
6
ANKn, 160
apparent genus, 123
apparent intersection multiplicity, 128
associates, 12
basic inequality, 48, 49
Bezout's Theorem, 43, 100
for j, 114
for It', 188, 190
birational invariant, 111
birationally equivalent, 109, 163
branch, 35, 89-102, 160
analytic, 173
linear, 94, 122
branch representation, 89
canonical adjoints, 146
canonical class, 142, 158
Cartesian product, 57
center, of a branch, 92, 160
of a branch representation, 89
of a locally quadratic transforma-
tion, 95
of a morphism, 198
of a place, 113
characteristic, 152
of a branch, 94
complete factorization, 12
complete Hnear series, 141
complex number field, 2, 10
component, 15
composite series, 167
Conchoid of Nicomedes, 37
conic, 30
consequence, 73, 74
constant point, 103
coordinate system, 29
corresponding points, branches, 163
cross-ratio, 69
cubic, 55-66, 172
cusp, 34
cut out a series, 144
cycle, 15
degree, assigned or formal, 25
of transcendency, 105
derivation, 115
derivative, 39
difference, of linear series, 141
differential, 115, 116, 141
exact, 117, 141
dimension, of a linear series, 137
213

214 INDEX
divisor, effective or virtual, 15, 133, 135
double point, 34
double-point divisor, 133
effective divisor, 15, 133, 135
elliptic curve, 170
equation(s), of an analytic cycle, 178
of a curve, 13
irreducible, 14
minimal, of a curve, 13
equivalent branch representations, 92
equivalent representations of a rational
function, 60
Euler's Formula, 37, 39
extension field, 105
extensions, algebraic, 105
finite, 105
finitely generated, 105
sinlple, 212
field, of a point, 160
of rational functions, 104
field adjunction, 210
folium of Descartes, 37
formal power series, 83
Four-leafed rosette, 37
function, of a function, 40
rational, 60, 104
g, 140
generic point, 103
genus, 118
Hensel's Lemma, 177
Hessian, 68
Hilbert's Nullstellensatz, 129
homogeneous coordinates, 6
homogeneous polynomial, 10
horizontal, 9
hyperelliptic, 170
hyperplane, 161
hypersurface, 161
ideal, 129
image curve, 108
imprimitive branch representation, 89,
101
infinitely near points, 127, 203-208
integral domain, 209
intersection defect, 128
intersection of two curves, 43-47
invariant(s), 28, 32, 38, 44, 92
involution, 166
irreducible algebraic variety, 193
irreducible curve, 13, 193
irreducible element (polynomial), 1, 12
irreducible equation, 14
isobaric, 24
isomorphic ANK's, 9
isomorphic points, 104
Laplace, 48
Lasker, 130
lie over, 164
line, 2, 7
at infinity, 8
linear branch, 94, 122
Hnear condition, 73
Hnear series, 135-142, 160-172
linear space, 212
linear system of curves, 143
Hnearly dependent mod r, 162
locally quadratic transformation, 95
mapping; see transformation
minor, 210
model, 110
module, 186
modulus, 72, 75, 172
morphism, 196, 199
multiplicity, of a branch (at its center),
94, 122
of a curve and a branch, 93
intersection, of a curve and a hyper-
surface, 161
of a curve and line at a point, 31,
33
of an analytic cycle and a branch,
178
of two curves at a point, 43-47,
186-192
of an irreducible factor, 12
of a point of a curve, 34
of a pole, 113
of a zero, 113

Noether, 130
Noetherian conditions, 129-134, 186-
192
normalized coordinates of a point, 9, 89
normalized representation of a series,
138
notation; see the list on page viii
onto, 3, 27
order, of a curve, 28
of a differential, 118
of a divisor, 136, 162
of a function, 113
of a linear series, 136
of a power series, 84
ordered abelian group, 199
ordinary cusp, 34, 35
ordinary double point, 34
ordinary multiple point, 34
ordinary point of inflection, 55
ordinary simple point, 55
pencil, 77
permissible position, 29, 43
place, 111, 196, 199
plane, affine, 4, 6
projective, 4, 6, 7
over an extension field, 9
PNK (projective number plane over
K),8
point, constant, 103
generic, 103
variable, 103
points, circular, 20
at finite distance, 8
infinitely near, 127, 203-208
in n-space, 160
at infinity, 8, 9
isomorphic, 104
of inflection, 55, 67
singular, 32
simple, 32, 178
polar, 38
pole, of a differential, 118
of a function, 113
primiti ve branch representation, 89,
90
INDEX 215
primitive element, 113, 154, 194
primitive place representation, 111
projective completion, of an affine
curve, 19
projective field, 196
quotient field, 210
rational function, 60, 104
rational mapping, 106, 107, 164
Reduction Theorem, 148
redundancy, 90, 102
regular polynomial, in x n ' 174
residue, 170, 197
residue map, 196
resultant, 122
r-fold point, 34, 178
Riemann-Roch, 147, 150
ring adjunction, 210
separably algebraic, 154
separating, 154
simple point, 32, 178
simple series, 167
singular point, 32
space curve(s), 160-172
special coordinates, of a branch repre-
sentation, 89
special polynomial, 174
specialization, 193, 196
tangent, 32, 34, 122, 178
Taylor's Theorem, 41, 42
transcendency basis, 105
separating, 154
transformation(s-), in the affine plane,
30
alias, 29
alibi, 29
birational, 109
fundamental theorem of linear, 28
general linear, 44
homogeneous nonsingular linear, 27
inverse, 109
locally quadratic, 95
rational, 106, 107, 164
standard quadratic, 119
univalued, 108

216 INDEX
transformed, 81
trivial, 7
uniformizing parameter, 181
unique factorization, 12, 132
unit, 11
unit point, of a coordinate system, 29
univalent, 3, 27
valuation, 199
over F, 200
valuation ring, 200
variable point, 103
variety, a beHan, 65
abstract, 65
affine, 65
complete, 65
irred uci ble, 65
projective, 65
reducible, 65
special a beHan, 66
special abstract, 66
special complete, 66
special group, 66
vector space, 212
vertical, 9
vertices, of a coordinate system, 29
virtual divisor, 15, 133, 135
Weierstrass's Formula, 174
Weierstrass's Preparation Theorem,
174
weight, 24
zero, of a differential, 118
of a function, 113