Lectures onThe Theory of Algebraic Functions

of One Variable

by

M. Deuring

Notes by

C.P. Ramanujam

No part of this book may be reproduced in anyform by print, microfilm or any other means with-out written permission from the Tata Institute ofFundamental Research, Apollo Pier Road, Bom-bay - 1

Tata Institute of Fundamental ResearchBombay

1959

Contents

1 Lecture 1 11 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 12 Ordered Groups . . . . . . . . . . . . . . . . . . . . . . 23 Valuations, Places and Valuation Rings . . . . . . . . . . 3

2 Lecture 2 73 (Contd.) . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Lecture 3 94 The Valuations of Rational Function Field . . . . . . . . 95 Extensions of Places . . . . . . . . . . . . . . . . . . . 10

4 Lecture 4 156 Valuations of Algebraic Function Fields . . . . . . . . . 157 The Degree of a Place . . . . . . . . . . . . . . . . . . . 178 Independence of Valuations . . . . . . . . . . . . . . . . 18

5 Lecture 5 239 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6 Lecture 6 2710 The Space L(U ) . . . . . . . . . . . . . . . . . . . . . 2711 The Principal Divisors . . . . . . . . . . . . . . . . . . 28

7 Lecture 7 3312 The Riemann Theorem . . . . . . . . . . . . . . . . . . 33

iii

iv Contents

13 Repartitions . . . . . . . . . . . . . . . . . . . . . . . . 34

8 Lecture 8 3714 Differentials . . . . . . . . . . . . . . . . . . . . . . . . 3715 The Riemann-Roch theorem . . . . . . . . . . . . . . . 41

9 Lecture 9 4516 Rational Function Fields . . . . . . . . . . . . . . . . . 4517 Function Fields of Degree Two... . . . . . . . . . . . . . 4618 Fields of Genus Zero . . . . . . . . . . . . . . . . . . . 5019 Fields of Genus One . . . . . . . . . . . . . . . . . . . 51

10 Lecture 10 5520 The Greatest Common Divisor of a Class . . . . . . . . 5521 The Zeta Function of Algebraic... . . . . . . . . . . . . . 56

11 Lecture 11 6122 The Infinite Product for (s, K) . . . . . . . . . . . . . . 6123 The Functional Equation . . . . . . . . . . . . . . . . . 6224 L-series . . . . . . . . . . . . . . . . . . . . . . . . . . 64

12 Lecture 12 6725 The Functional Equation for the L-functions . . . . . . . 67

13 Lecture 13 7126 The Components of a Repartition . . . . . . . . . . . . . 71

14 Lecture 14 7527 A Consequence of the Riemann-Roch Theorem . . . . . 75

15 Lecture 15 8128 Classes Modulo F . . . . . . . . . . . . . . . . . . . . 81

16 Lecture 16 8529 Characters Modulo F . . . . . . . . . . . . . . . . . . . 85

Contents v

17 Lecture 17 8930 L-functions Modulo F . . . . . . . . . . . . . . . . . . 8931 The Functional Equations of the L-functions. . . . . . . 89

18 Lecture 18 9732 Extensions of Algebraic Function Fields . . . . . . . . . 97

19 Lecture 19 10333 Application of Galois Theory . . . . . . . . . . . . . . . 103

20 Lecture 20 10934 Divisors in an Extension . . . . . . . . . . . . . . . . . 10935 Ramification . . . . . . . . . . . . . . . . . . . . . . . . 114

21 Lecture 21 11736 Constant Field Extensions . . . . . . . . . . . . . . . . 117

22 Lecture 22 12337 Constant Field Extensions . . . . . . . . . . . . . . . . 123

23 Lecture 23 13338 Genus of a Constant Field Extension . . . . . . . . . . . 13339 The Zeta Function of an Extension . . . . . . . . . . . . 144

Lecture 1

1 Introduction

We shall be dealing in these lectures with the algebraic aspects of the 1theory of algebraic functions of one variable. Since an algebraic func-tion w(z) is defined implicitly by an equation of the form f (z,w) = 0,where f is a polynomial, it is understandable that the study of such func-tions should be possible by algebraic methods. Such methods also havethe advantage that the theory can be developed in the most general set-ting, viz. over an arbitrary field, and not only over field of complexnumbers (the classical case).

Definition . Let k be a field. An algebraic function field K over k isa finitely generated extension over k of transcendence degree at leastequal to one. If the transcendence degree of K/k is r, we say that it is afunction field in r variables.

We shall confine ourselves in these lectures to algebraic functionfields of one variable, and shall refer to them shortly as function fields.

If K/k is a function field, it follows from our definition that thereexists an X in K transcendental over k, such that K/k(X) is a finite al-gebraic extension. If Y is another transcendental element of k, it shouldsatisfy a relation F(X, Y) = 0, where F is a polynomial over K whichdoes not vanish identically. Since Y is transcendental by assumption, 2the polynomial cannot be independent of X. Rearranging in powers ofX, we see that X is algebraic over k(Y). Moreover,

1

2 1. Lecture 1

[K : k(Y)] = [K : k(X, Y)].[k(X, Y) : k(Y)] [K : k(X)].[k(X, Y) : k(Y)] <

and thus Y also satisfies the same conditions as X. Thus, any transcen-dental element of K may be used as a variable in the place of X.

The set of all elements of K algebraic over k forms a subfield kof K, which is called the field of constants of K. Hence forward, weshall always assume, unless otherwise stated, that k = k, i.e., that k isalgebraically closed in K.

2 Ordered GroupsDefinition . A multiplicative(additive) Abelian group W with a binaryrelation < (>) between its elements is said to be an ordered group if

0(1) for , W , one and only one of the relations < , = , < ( > , = > ) holds.

0(2) < , < < ( > , > > )0(3) < , W < ( > , W + > + )

We shall denote the identity (zero) element by 1(0). In this andthe following section, we shall express all our results in multiplicativenotation. > shall mean the same thing as < .

Let W0 be the set { : W < 1}. W0 is seen to be a semi group by0(2) and 0(3). Moreover, W = W0{1}W10 is a disjoint partitioning of3W (where W10 means the set of inverses of elements of W0). Conversely,if an Abelian group W can be partitioned as W0 {1} W10 , where W0is a semi-group, we can introduce an order in W by defining < tomean 1 W0; it is immediately verified that 0(1), 0(2) and 0(3) arefulfilled and that W0 is precisely the set of elements < 1 in this order.

For an Abelian group W , the map n (n any positive integer)is in general only an endomorphism. But if W is ordered, the map is amonomorphism; for if is greater than or less than 1, n also satisfiesthe same inequality.

3. Valuations, Places and Valuation Rings 3

3 Valuations, Places and Valuation RingsWe shall denote the non-zero elements of a field K by K.

Definition. A Valuation of a field K is a mapping v of K onto an orderedmultiplicative (additive) group W (called the group of the valuation orthe valuation group) satisfying the following conditions:V(1) For a, b K, v(ab) = v(a)v(b) (v(ab) = v(a) + v(b)); i.e. v is

homomorphism of the multiplicative group K onto W.V(2) For a, b, a+b K, v(a+b) max(v(a), v(b)) (v(a+b)) min(v(a)

v(b)))V(3) v is non-trivial; i.e., there exists an a K with v(a) , 1(v(a) , 0)

Let us add an element 0() to W satisfying the following(1) 0.0 = .0 = 0. = 0 for every W(+ = + = + = ),(2) > 0 for every W( < ). If we extend a valuation v to the 4

whole of K by defining v(0) = 0 (v(0) = , the new mapping alsosatisfies V(1),V(2) and V(3).The following are simple consequences of our definition.

(a) For a K, v(a) = v(a). To prove this, it is enough by V(1) toprove that v(1) = 1. But v(1). v(1) = v(1) = 1 by V(1), andhence v(1) = 1 by the remark at the end of 2.

(b) If v(a) , v(b), v(a + b) = max(v(a), v(b)). For let v(a) < v(b). Then,v(a + b) max(v(a), v(b)) = v(b) = v(a + b a) max(v(a +b), v(a)) = v(a + b)

(c) Let ai K, (i = 1, . . . n). Then an obvious induction on V(2) givesv(

n1

ai) nmaxi=1

v(ai), and equality holds if v(ai) , v(a j) for i , j.

(d) If ai K, (i = 1, . . . n) such thatn1

ai = 0, then v(ai) = v(a j) forat least one pair of unequal indices i and j. For let ai be such that

4 1. Lecture 1

v(ai) v(ak) for k , i. Then v(ai) = v(n

k=1k,i

ak) nmaxk=1k,i

(v(ak)) = v(a j)

for some j , i, which proves that v(ai) = v(a j).

Let be a field. By (), we shall mean the set of elements of

together with an abstract element with the following properties. + = + = for every

.

. = . = for every

, , 0. + and 0. are not defined.

Definition. A place of a field K is a mapping of K into U() (where5 may be any field ) such that

P(1) (a + b) = (a) + (b).

P(2) (ab) = (a).(b).

P(3) There exist a, b K such that (a) = and (b) , 0 or .P(1)andP(2) are to hold whenever the right sides have a meaning.

From this it follows, taking the b of P(3), that (1)(b) = (b), sothat (1) = 1, and similarly (0) = 0.

Consider the set O of elements a K such that (a) , . Then byP(1), P(2) and P(3), O is a ring which is neither zero nor the whole ofK, and is a homomorphism of this ring into . Since is a field, thekernel of this homomorphism is a prime ideal Y of O

Let b be an element in K which is not in O. We contend that (1b ) =

0. For if this mere not true, we would get 1 = (1) = (b). (1b ) = ,

by P(2). Thus 1b Y , and thus Y is precisely the set of non-units ofO. Since any ideal strictly containing Y should contain a unit, we seethat Y is a maximal ideal and hence the image of O in

is again afield. We shall therefore always assume that is precisely the image ofO by , or that is a mapping onto

U().

The above considerations motivate the

3. Valuations, Places and Valuation Rings 5

Definition . Let K be a field. A valuation ring of K is a proper subringO of K such that if a K, at least one of the elements a 1

ais in O . 6

In particular, we deduce that O contains the unity element. Let Y bethe set of non-units in O . Then Y is a maximal ideal. For, let a O , b Y . If ab < Y , ab would be a unit of O , and hence 1

ab O . This implies

that a 1ab =

1b O , contradicting our assumption that b is a non-unit of

O . Suppose that c is another element of Y . To show that b c Y ,we may assume that neither of them is zero. Since O is a valuation ring,at least one of b

cor

c

b , saybc

, is in O . Hence, bc 1 =

b cc

O . If

b c were not in Y , 1b c O , and hence1

b c b c

c=

1c O ,

contradicting our assumption that c Y . Finally, since every elementoutside Y is a unit of O ,Y is a maximal ideal in O .

Lecture 2

3 (Contd.)In this lecture, we shall establish the equivalence of the concepts of val- 7uation, place and valuation ring.

Two places 1 : K

1 () and 2 : K

2 () are said tobe equivalent if there exists an isomorphism of 1, onto 2 such that2(a) = 1(a) for every a, with the understanding that () = .

This is clearly an equivalence relation, and thus, we can put the setof places of K into equivalence classes. Moreover, equivalent places 1and 2 obviously define the same valuation rings O1 and O2 . Thus, toevery equivalence class of places is associated a unique valuation ring.

Conversely, let O be any valuation ring and Y its maximal ideal.Let

be the quotient field O/Y and the natural homomorphism ofO onto

. It is an easy matter to verify that the map : K

U{}

defined by

(a) =(a) if a O if a < O

is a place, whose equivalence class corresponds to the given valuationring O .

Let v1 and v2 be two valuations of a field K in the ordered groupW1 and W2. We shall denote the unit elements of both the groups by 1, 8since it is not likely to cause confusion. We shall say that v1 and v2 areequivalent if v1(a) > 1 if and only if v2(a) > 1.

Let v1 and v2 be two equivalent valuations. From the definition, itfollows, by taking reciprocals, that v1(a) < 1 if and only if v2(a) < 1,

7

8 2. Lecture 2

and hence (the only case left) v1(a) = 1 if and only if v2(a) = 1. Let be any element of W1. Choose a K such that v1(a) = (this ispossible since v1 is onto W1). Define () = v2(a). The definitionis independent of the choice of a since if b were another element withv1(b) = , then v1(ab1) = 1 so that v2(ab1) = 1, i.e. v2(a) = v2(b).Thus, is a mapping from W1 onto W2 (since v2 is onto W2). It is easyto see that is an order preserving isomorphism of W1 onto W2 and wehave v2(a) = (.v1)(a) for every a K. Thus, we see that the definitionof equivalence of valuations can also be cast into a form similar to thatfor places.

Again, equivalence of valuations is an equivalence relation, and weshall that equivalence classes of valuations of a field K correspond cano-nically and biunivocally to valuation rings of the field K.

Let v be a valuation and O be the set of elements a in K such thatv(a) 1. It is an immediate consequence of the definition that O is aring. Also, if a K, v(a) > 1, then v

(1a

)< 1 and hence 1

a O . Thus, O

is a valuation ring. Also, if v1 and v2 are equivalent, the corresponding9rings are the same.

Suppose conversely that O is a valuation ring in K and Y its maxi-mal ideal. The set difference O Y is the set of units of O and hencea subgroup of the multiplicative group K. Let : K K/O Ybe the natural group homomorphism. Then (O) is obviously a semi-group and the decomposition K/O Y = (O) {1}U (O)1 isdisjoint, Hence, we can introduce an order in the group K/O Y andit is easy to verify that is a valuation on K whose valuation ring isprecisely O .

Summarising, we have

Theorem. The valuations and places of a field K are, upto equivalence,in canonical correspondence with the valuation rings of the field.

Lecture 3

4 The Valuations of Rational Function FieldLet K = k(X) be a rational function field over k ; i.e., K is got by ad- 10joining to k a single transcendental element X over k. We seek for allthe valuations v of K which are trivial on k, that is, v(a) = 1 for everya K. It is easily seen that these are the valuations which correspondto places whose restrictions to k are monomorphic.

We shall henceforward write all our ordered groups additively.Let be a place of K = k(X) onto (). We consider two cases

Case 1. Let (X) = , . Then, the polynomial ring k[X] is containedin O, and Y k[X] is a prime ideal in k[X]. Hence, it should be ofthe form (p(X)), where p(X) is an irreducible polynomial in X. Now, ifr(X) K, it can be written in the form r(X) = (p(X)) g(X)h(X) , where g(X)and h(X) are coprime and prime to p(X). Let us agree to denote theimage in

of an element c in k by c, and that of a polynomial f over k

by f . Then,we clearly have

(r(x)) =

0 if > og()h() if = 0 if < 0

Conversely, suppose p(X) is an irreducible polynomial in k[X] and 11 a root of p(X). The above equations then define a mapping of k(X)onto k() {}, which is a place, as is verified easily. We have thusdetermined all places of k(X) under case 1 (upto equivalence).

9

10 3. Lecture 3

If Z is the additive group of integers with the natural order, the val-uation v associated with the place above is given by v(r(X)) = .

Case 2. Suppose now that (X) = . Then ( 1X

) = 0. Then since K =k(X) = k( 1

X), we see that is determined by an irreducible polynomial

p( 1X

), and since 1X

) = 0, p(Y) should divide Y. Thus, p(Y) must be Y(except for a constant in k), and if

r(X) = a0 + a1x + + anxn

b0 + b1x + + bmxm, an, bm , 0,

(r(X)) = ( 1X )mn

a0xn+

a1xn1

+ + an

b0xm+

b1xm1

+ + bm

=

o if m > nanbm if m = n if m < n

The corresponding valuation with values in Z is given by v(r(X)) =m n = deg r(X), where the degree of a rational function is defined inthe degree of the numerator-the degree of the denominator.

We shall say that a valuation is discrete if the valuation group maybe taken to be Z. We have in particular proved that all valuations of arational function field trivial over the constant field are discrete. We shallextend this result later to all algebraic function fields of one variable.

5 Extensions of Places

Given a field K, a subfield L and a place L of L into

, we wish to12prove in this section that there exists a place K of K into

1, where 1

is a field containing and the restriction of K to L is L. Such a K iscalled an extension of the place L to K. For the proof of this theorem,we require the following

Lemma (Chevalley) . Let K be a field, O a subring and a homomor-phism of O into a field which we assume to be algebraically closed.Let q be any element of K, and O[q] the ring generated by O and q in

5. Extensions of Places 11

K. Then can be extended into a homomorphism of at least one ofthe rings O[q],O[1

q], such that restricted to O coincides with .

Proof. We may assume that is not identically zero. Since the imageof O is contained in a field, the kernel of is a prime ideal Y which isnot the whole ring O . Let O1 = O

{a

b a, b O , b < Y}. Clearly, O1

is a ring with unit, and has a unique extension to O1 as a homomor-phism, give by

(a

b

)=

(a)(b) . The image by is then the quotient field

of (O). We shall denote (a) by a for a O1. Let X and X be indeterminates over O1 and

respectively. can

be extended uniquely to a homomorphism of O1[X] onto [ X] whichtakes X to X by defining

(a0 + a1X + + anXn) = a0 + a1 X + + an Xn.Let U be the ideal O1[X] consisting of all polynomials which van- 13

ish for X = q, and let U be the ideal (U ) in [ X]. We consider threecases.

Case 1. Let U = (0). In this case, we define (q) to be any fixedelement of . is uniquely determined on all other elements of O1[q]by the requirement that it be a ring homomorphism which is an extensionof . In order that it be well defined, it is enough to verify that if anypolynomial over O1 vanishes for q, its image by vanishes for (q).But this is implied by our assumption.

Case 2. Let U , (0),, [ X]. Then U = ( f ( X)), where f is a non-constant polynomial over . Let be any root f ( X) in (there is aroot in since is algebraically closed). Define (q) = . This can beextended uniquely to a homomorphism of O1[q], since the image by ofany polynomial vanishing for q is of the form f ( X) f ( X), and thereforevanishes for X = .Case 3. Suppose U = [ X]. Then the homomorphism clearly cannotbe extended to O1[q]. Suppose now that it cannot be extended to O1[1

q]

12 3. Lecture 3

either. Then if denotes the ideal of all polynomials in O1[X] whichvanish for 1

q, and if is the ideal ()in [ X], we should have =[ X]. Hence, there exist polynomials f (X) = a0 + a1X + + anXn and

b0+b1X++bmXm such that ( f (X)) = (g(X)) = 1, f (q) = g(1q

)= 0.

We may assume that f and g are of minimal degree n and m satisfyingthe required conditions. Let us assume that m n. Then, we have14a0 = b0 = 1, ai = b j = 1 for i, j > 0. Let g0(X) = b0Xm + + bm.Applying the division algorithm to the polynomials bn0 f (X) and g0(X),we obtain

bn0 f (X) = g0(X)Q(X) + R(X), Q(X),R(X) O1[X], deg R < m.

Substituting X = q, we obtain R(q) = 0. Also, acting with , wehave

1 = bn0 f ( X) = g0( X) Q( X) + R( X) = Q( X) Xm + R( X),

and hence, we deduce that Q( X) = 0, R( X) = 1. Thus, R(X) is a polyno-mial with R(q) = 0, R( X) = 1, and deg F(X) < m n, which contradictsour assumption on the minimality of the degree of f (X). Our lemma isthus prove.

We can now prove the

Theorem. Let K be a field and O a subring of K. Let be a homomor-phism of O in an algebraically closed field . Then it can be extendedeither to a homomorphism of K in or to a place of K in (). Inparticular, any place of a subfield of K can be extended to a place of K.Proof. Consider the family of pairs {,O}, where O is a subring ifK containing O and a homomorphism of O in extending on O .The family is non-empty, since it contains (,O). We introduce a partialorder in this family by defining (,O) > (,O) if O O and is an extension of .

The family clearly being inductive, it has a maximal element by15Zorns lemma. Let us denote it by (,O). O is either the whole of K or

5. Extensions of Places 13

a valuation ring of K. For if not, there exists a q K such that neitherq nor

1q

belongs to O . we may then extend to a homomorphism of

at least one of O[q] or O[1q

]in . Since both these rings contain O

strictly, this contradicts the maximality of (,O).If O were not the whole of K, must vanish on every non - unit of

O; for if q were a non-unit and (q) , 0, we may define (1q

)=

1(q)

and extend this to a homomorphism of O[1q

], which again contradicts

the maximality of O . This proves that can be extended to a place ofK by defining it to be outside O .

In particular, a place of a subfield L of K , when considered as ahomomorphism of its valuation ring O and extended to K, gives a placeon K; for if where a homomorphism of the whole of K in , it shouldbe an isomorphism (since the kernel, being a proper ideal in K, shouldbe the zero ideal). But being an extension of , the kernel contains atleast one non-zero element.

Corollary . If K/k is an algebraic function field and X any element ofK transcendental over k, there exists at least one valuation v for whichv(X) > 0.Proof. We have already shown in the previous section that there existsa place Y1 in k(X) such that vY1 (X) > 0. If we extend this place Y1 to aplace Y of K, we clearly have vY (X) > 0.

Lecture 4

6 Valuations of Algebraic Function FieldsIt is our purpose in this paragraph to prove that all valuations of an al- 16gebraic function field K which are trivial on the constant field k arediscrete. Henceforward, when we talk of valuations or places of an al-gebraic function field K, we shall only mean those which are trivial on k.Valuations will always be written additively. We require some lemmas.

Lemma 1. Let K/L be a finite algebraic extension of degree [K : L] = nand let v be a valuation on K with valuation group V. If V denotes thesubgroup of V which is the image of L under v and m the index of V inV, we have m n.

Proof. It is enough to prove that of any n+ 1 elements 1, . . . n+1 of V ,at least two lie in the same coset modulo V.

Choose ai K such that v(ai) = i (i = 1, . . . n + 1). Since there canbe at most n linearly independent elements of K over L, we should have

n+1i=1

liai = 0, li L, not all li being zero .

This implies that v(liai) = y(l ja j) for some i and j, i , j (see Lecture1, 3). Hence, we deduce that

v(li) + v(ai) = v(liai) = v(l ja j) = v(l j) + v(a j),i j = v(ai) v(a j) = v(l j) v(li) V

15

16 4. Lecture 4

and i and j are in the same coset modulo V. Our lemma is proved. 17Let V be an ordered abelian group. We shall say that V is archime-

dean if for any pair of elements , in V with > 0, there correspondsan integer n such that n > . We shall call any valuation with valuegroup archimedean an archimedean valuation.

Lemma 2. An ordered group V is isomorphic to Z if and only if (i) it isarchimedean and (ii) there exists an element > 0 in V such that it isthe least positive element; i.e., > 0 .

Proof. The necessity is evident. Now, let be any element of V .

Then by assumption, there exists a smallest integer n such that n < (n + 1). Thus,

0 n < (n + 1) n = ,

and since is the least positive element, we have = n. The mapping V n Z is clearly an order preserving isomorphism, and thelemma is proved.

Lemma 3. If a subgroup V of finite index of an ordered group V isisomorphic to Z,V is itself isomorphic to Z.

Proof. Let the index be[V : V

]= n. Let , be any two elements of

V , with > 0. Then n and n are in V, n > 0.

Since V is archimedean, there exists an integer m such that mn >n, from which it follows that m > .

Again, consider the set of positive elements in V . Then n are in18V and are positive, and hence contain a least element n (since there isan order preserving isomorphism between V and Z). Clearly, is thenthe least positive element of V .

V is therefore isomorphic to Z, by Lemma 2.We finally have the

Theorem. All valuations of an algebraic function field K are discrete.

7. The Degree of a Place 17

Proof. If X is any transcendental element of K/k, the degree[K : k(X)

]< . Since we know that all valuations of k(X) are discrete, our resultby applying Lemma 1 and Lemma 3.

7 The Degree of a Place

Let Y be a place of an algebraic function field K with constant field konto the field kY . Since Y is an isomorphism when restricted to k,we may assume that kY is an extension of k. We shall moreover assumethat kY is the quotient OY /MY where OY is the ring of the place Yand MY the maximal ideal. We now prove the

Theorem. Let Y be a place of an algebraic function field. Then kY /k isan algebraic extension of finite degree.Proof. Choose an element X , 0 in K such that Y (X) = 0. Then Xshould be transcendental, since Y is trivial on the field of constants.Let

[K : k(X)

]= n < . Let 1, . . . n+1 be any (n + 1) elements of 19

kY . Then, we should have i = Y (ai) for some ai K(i = 1, . . . n + 1).There therefore exist polynomials fi(X) in k[X] such that

n+1i=1

fi(X)ai = 0, not all fi(X) having constant term zero.

Writing fi(X) = li + Xgi(X), we haven+1i=1

liai = Xn+1i=1

aigi(X), and taking

the Y -image,n+1i=1

lii = Y (X)n+1i=1

aigi(Y X) = 0, li k, not all li beingzero.

Thus, we deduce that the degree of kY /k is at most n.The degree fY of kY over k is called the degree of the place Y .

Note that fY is always 1. If the constant field is algebraically closed(e.g. in the case of the complex number field), fY = 1, since kY , beingan algebraic extension of k, should coincide with k.

18 4. Lecture 4

Finally, we shall make a few remarks concerning notation.If Y is a place of an algebraic function field, we shall denote the cor-

responding valuation with values in Z by vY (vY is said to be a normedvaluation at the place Y ). The ring of the place shall be denoted byOY and its unique maximal ideal by Y . (This is not likely to cause anyconfusion).

8 Independence of Valuations

In this section, we shall prove certain extremely useful result on valua-20tions of an arbitrary field K.

Theorem. Let K be an arbitrary field and vi(i = 1, . . . n) a set of valua-tions on K with valuation rings Oi such that Oi 1 O j if i , j. There isthen an element X K such that v1(X) 0, vi(X) < 0 (i = 1, . . . n).

Proof. We shall use induction. If n = 2, since O1 1 O2, there is anX O1, X < O2, and this X satisfies the required conditions. Supposenow that the theorem is true for n 1 instead of n. Then there exists aY K such that

v1(Y) 0, vi(Y) < 0 (i = 2, . . . n 1).

Since O1nsubsetOn, we can find a Z K such that

v1(Z) 0, vn(Z) < 0

Let m be a positive integer. Put X = Y + Zm. Then

v1(Y + Zm) min(v1(Y),mv1(Z)) 0

Now suppose r is one of the integers 2, 3, . . . n. If vr(Z) 0, r cannotbe n, and since vr(Y) < 0, we have

vr(Y + Zm) = vr(Y) < 0.

8. Independence of Valuations 19

If vr(Z) < 0 and vr(Y + Zmr) 0 for some mr, for m > mr we have21vr(Y + Zm) = vr(Y + Zmr + Zm Zmr) = min(vr(Y + Zmr), vr(Zm Zmr)),and vr(Zm Zmr) = vr(Zmr) + vr(1 Zmmr) = mrvr(Z) < 0,

Since vr(1 Zmmr) = vr(1) = 0Thus, vr(Y + Zm) < 0 for large enough m in any case. Hence X

satisfies the required conditions.If we assume that the valuations vi of the theorem are archimedean,

then the hypothesis that Oi 1 O j for i , j can be replaced by the weakerone that the valuations are inequivalent (which simply states that Oi ,O j for i , j).

To prove this, we have only to show that if v and v1 are two archime-dean valuations such that the corresponding valuation rings O and O1satisfy O O1, then v and v1 are equivalent. For, consider an element

a K such that v(a) > 0. Then, v(1a

)< 0, and consequently 1

ais not

in O , and hence not in O1. Thus, v1(1a

)< 0, v1(a) > 0. Conversely,

suppose a K and v1(a) > 0. Then by assumption, v(a) 0. Supposenow that v(a) = 0. Find b K such that v(b) < 0. If n is any positiveinteger, we have v(anb) = nv(a) + v(b) < 0, anb < O .

But since v1 is archimedean and v1(a) > 0, for large enough n wehave

v1(anb) = nv1(a) + v1(b) > 0 , anb O1.22

This contradicts our assumption that O1 O , and thus, v(a) >0. Hence v and v1 are equivalent. Under the assumption that the viarchimedean, we can replace in the theorem above the first inequalityV1(X) 0 even by the strict inequality v1(X) > 0. To prove this letX1 K satisfy v1(X1) 0, vi(X1) < 0, i > 1. Let Y be an element inK with v1(Y) > 0. Then, if X = X1mY , where m is a sufficiently largepositive integer, we have

v1(X) = v1(Xm1 Y) = mv1(X1) + v1(Y) > 0,vi(X) = vi(Xm1 Y) = mvi(X1) + vi(Y) < 0, i = 2, . . . , n.

20 4. Lecture 4

We shall hence forward assume that all valuations considered arearchimedean. To get the strongest form of our theorem, we need twolemmas.

Lemma 1. If vi (i = 1, . . . n) are inequivalent archimedean valuations,and i are elements of the corresponding valuations group, we can findXi(i = 1, . . . n) in K such that vi(Xi 1) > i, v j(Xi) > j, i , jProof. Choose Yi K such that

vi(Yi) > 0, v j(Yi) < 0 for j , i.

Put Xi =1

1 + Ymi. Then, if m is chosen large enough, we have (since the

valuation are archimedean)v j(Xi) = v j(1 + Ymi ) = mv j(Yi) > j, i , j

and vi(Xi 1) = vi(Ymi

1 + Ymi) = mvi(Yi) vi(1 + Ymi ) = mvi(Yi) > i.23

since v i(1 + Ymi ) = 0. A set of valuations vi(i = 1, . . . n) are said to be independent if given

any set of elements ai K and any set of elements i in the respectivevaluation groups of vi, we can find an X K such that

vi(X ai) > i.We then have the following

Lemma 2. Any finite set of inequivalent archimedean valuations areindependent.Proof. Suppose vi(i = 1, . . . n) is a given set of inequivalent archimedeanvaluations. If ai K and i are elements of the valuation group of thevi, put i = i

n

minj=1

vi(a j). Choose Xi as in Lemma 1 for the vi and i.

Put X =n1

aiXi. Then

vi(X ai) = vij,i

a jX j + ai(Xi 1) > i +

n

minj=1

vi(a j) = i,

8. Independence of Valuations 21

and our lemma is proved.

Finally, we have the following theorem, which we shall refer to infuture as the theorem of independence of valuations.

Theorem . If vi(i = 1, . . . n) are inequivalent archimedean valuations,i is an element of the value group of vi for every i, and ai are given 24elements of the field, there exists an element X of the field such that

vi(X ai) = iProof. Choose Y by lemma 2 such that vi(Y ai) > i. Find bi K suchthat vi(bi) = i and an Z K such that vi(Z bi) > i. Then it followsthat vi(Z) = min(vi(Z bi), vi(bi)) = i.

Put X = Y + Z. Then,

vi(X ai) = vi(Z + Y ai) = vi(Z) = i,

and X satisfies the conditions of the theorem.

Corollary. There are an infinity of places of any algebraic function field.Proof. Suppose there are only a finite number of places Y1, . . . ,Yn.

Choose an X such that vYi (X) > 0 (i = 1, . . . n). Then, vYi (X +1) = vYi(1) = 0 for all the places ,Yi, which is impossible since X anconsequently X + 1 is a transcendental element over k.

Lecture 5

9 DivisorsLet K be an algebraic function field with constant field k. We make the 25

Definition . A divisor of K is an element of the free abelian group gen-erated by the set of places of K. The places themselves are called primedivisors.

The group of divisors shall be written multiplicatively. Any ele-ment U of the group of divisors can be written in the form

U =

Y

Y vY (U )

where the product is taken over all prime divisors Y of K, and thevY (U ) are integers, all except a finite number of which are zero. Thedivisor is denoted by n.

We say that a divisor U is integral if vY (U ) 0, for every Y , andthat U divides if U 1 is integral. Thus, U divides if and only ifvY () vY (U ) for every Y .

Two divisors U and are said to be coprime if vY (U , 0 impliesthat vY () = 0.

The degree d(U ) of a divisor U is the integer

d(U ) =Y

fY vY (U ),

where fY is the degree of the place Y . 26

23

24 5. Lecture 5

The map U d(U ) is a homomorphism of the group of divisors into the additive group of integers. The kernel of this homomorphismis the subgroup of divisors of degree zero.

An element X of K is said to be divisible by the divisor U if vY (X) vY (U ) for every Y . Two elements X, Y K are said to be congruentmodulo a divisor U (written X ( mod U )) if X Y is divisible byU .

Let S be a set prime divisors of K. Then we shall denote by (U /S )the set of elements X K such that vY (X) vY (U ) for every Y in S .It is clear that (U /S ) is a vector space over the constant field k, andalso that if U divides , (/S ) (U /S ). Also, if S and S 1 are twosets of prime divisors such that S S 1, then (U /S 1) (U /S )).Finally, (U /S ) = (/S ) if U 1 contains no Y belonging to S witha non-zero exponent.

If S is a set of prime divisor which is fixed in a discussion and U adivisor, we shall denote by U the new divisor got from U by omittingall Y which do not occur in S : U =

Y S

Y vY (U )

Theorem. Let S be a finite set of prime divisors and U , two divisorssuch that U divides . Then,

dimk(U /S )(/S ) = d() d(U) = d(U

1 ).

Proof. By our remark above, we may assume that U = U and = .27Moreover, it is clearly sufficient to prove the theorem when =

U Y , where Y is a prime divisor belonging to S . For, if = U Y1..Ynwe have

dimk(U /S )(/S ) = dimk

(U /S )(U Y1/S )+

dimk(U Y1/S )(U Y1Y2/S ) + + dimk

(UY1Yn1/S(/S )

and d() d(U ) = d(Y1) + d(Y2) + + d(Yn). Hence, we have to prove that if U is a divisor such that all the

prime divisors occurring in it with non-zero exponents are in S , and Y

9. Divisors 25

any prime divisor in S , we have

dimk(U /S )(U Y /S ) = fY = f .

By the theorem on independence of valuations, we may choose anelement u K such that 28

vU (u) = vU (U ) for all U in S .

If X1, X2, . . .X f+1 are any f + 1 elements of (U /S ), the elementsX1u1, . . .X f+1u1 are all in UY . But since the degree of kU = OY /Yover k is f , we have

f+1i=1

aiXiu1 Y , ai k, not all ai being zero .

Hence,f+1i=1

aiXi (U Y /S ), ai k, not all ai being zero, thus

proving that the dimension over k of the quotient (U /S )(U Y /S ) is f .

Now, suppose Y1, . . .Y f are f elements of OY such that they are lin-early independent over k modulo Y . ChooseY1i K such that

vY (Y1i Yi) > 0, vU (Y1i ) 0 for U , Y ,U S .(i = 1, f ).

By the first condition, Y1i Yi( mod Y ), and hence Yi and Y1i de-termine the same element in kY . But by the second condition, the ele-ments uY1i belong to (U /S ), and since Y1, . . . Y f are linearly indepen-dent mod Y , no linear combination of uY11 , . . . uY

1f with coefficients

in k- at least one of which is non-zero-can lie in (U Y /S ). Thus,dimk

(U /S )(U Y /S ) f .

This proves our theorem.

Lecture 6

10 The Space L(U )Let U be any divisor of an algebraic function field K. 29

We shall denote by L(U ) the set of all elements of K which aredivisible by U . Clearly L(U ) = (U /S ) if S is the set of all primedivisors of K, and thus we deduce that L(U ) is a vector space over k andif U divides , L(U ) L(). We now prove the following importantTheorem . For any divisor U , the vector space L(U ) is finite dimen-sional over k. If we denote its dimension by l(U ), and if U divides ,we have

l(U ) + d(U ) l() + d().Proof. Let S be the set of prime divisors occurring in U or .

Then, it easy to see that

L() = L(U ) (/S )Hence, by Noethers isomorphism theorem,

L(U )L() =

L(U )L(U ) (/S )

L(U ) + (/S )L(/S )

L(U /S )(/S ) ,

and therefore, dimkL(U )L() dimk

(U /S )(/S ) = d() d(U )

Now, choose for any integral divisor which is a multiple of U andis not the unit divisor n. Then, L() = (0); for if X were a non-zeroelement of L( ), it cannot be a constant since vY (X) vY () > 0 for

27

28 6. Lecture 6

at least one Y , and it cannot be transcendental over k since vY (X) 30vY () 0 for all Y .

This is not possible. This together with the above inequality provesthat dimk L(U ) = l(U ) < , and our theorem is completely proved.

Since L(N ) clearly contains only the constants, l(N ) = 1.

11 The Principal Divisors

We shall now associate to every non-zero element of K a divisor. Forthis, we need the

Theorem . Let X K. Then there are only a finite number of primedivisors Y with vY (X) , 0.

Proof. If X k, vY (X) = 0 for all Y and the theorem is valid.Hence assume that X is transcendental over k. Let [K : k(X)] = N.

Suppose Y1, . . .Yn are prime divisors for which vYi(X) > 0. Let =n

i=1YivYi (X), and S = {Y1, . . .Yn}.

Then, dimk (N /S )(/S ) = () =n

i=1fYivYi (X). We shall show that this is

at most equal to N.Let in fact Y1, . . . , YN+1 be any (N + 1) elements of (N /S ). Since

[K : k(X)] = N, we should haveN+1j=1

f j(X)Y j = 0, f j(X) k[X], withat least one f j having a non-zero constant term. Writing f j(X) = a j +Xg j(X), the above relation may be rewritten as

N+11

a jY j = XN+1

1g j

(X)Y j, not all a j being zero, and hence

vY

N+1

1a jY j

= vY (X) + vY

N+11

g j(X)Y j vY (X)

31

11. The Principal Divisors 29

This proves thatN+1

1a jY j (/S ), and therefore

n

ni=1

fYivYI (X) = d() = dimk(N /S )(/S ) N,

By considering 1X

instead of X, we deduce that the number of primedivisors Y for which vY (X) < 0 is also finite, and our theorem follows.

The method of defining the divisor (X) corresponding to an elementX K is now dear. We define the numerator zX of X to be the divisorvY (X)>0

Y vY (X) (the product being taken over all Y for which vY (X) >0), the denominator NX of X to be the divisor

vY (X)

30 6. Lecture 6

We then have the

Lemma. If Y is an integral algebraic function of X, and a prime divisorY does not divide NX , it does not divide NY .Proof. Since Y does not divide NX, vY (X) 0, and hence

mvY (Y) = vY (Ym) = vY ( fm1(X)Ym1+ + f0(X))

m1min=0

(vY (Y)) = 0vY (Y),

for some 0 such that 0 0 m 1. This proves that (m 0)vY (Y) 0, vY (Y) 0, and therefore Y does not divide NY . Now, let Y be anyelement of K satisfying the equation

fm(X)Ym + + f0(X) = 0Then, the element Z = fm(X)Y satisfying the equation

Zm + gm1(X)Zm1 + + go(X) = 0,where gk(X) = fk(X) f mk1m (x), and hence Z is an integral function of33X

Suppose then that Y1, . . . YN is a basis of K/k(X). By the aboveremark, we any assume that the Yi are integral functions of X. Theelements XiY j(i = 0, . . . t; j = 1, . . . N) are then linearly independentover k, for any non-negative integer t. By the above lemma, we can findinteger s such that N sX (Y j) are integral divisors. Hence, N s+tX (Xi)(Y j)is an integral divisor for (i = 0, . . . t, j = 1, . . . N), which implies thatXiY j are elements of L(N stX ). Since these are linearly independentand N(t + 1) in number, we obtainN(t + 1) 1(N stX ) l(N ) + d(N ) d(N stX ) = 1 + (s + t)d(Nx),the latter inequality holding because N stX divides N .

Thus, d(NX) Nt + N 1s + t

N as t , which taken togetherwith the opposite inequality we proved earlier shows that d(NX) = N.

It is clearly sufficient to show that d(NX) = N, since the other fol-lows on replacing X by 1

Xand observing that k(X) = k(1/X).

11. The Principal Divisors 31

Corollary 1. If X K, d((X)) = 0. This is clear when X is a constant.If X be a variable, d((X)) = d(zX) d(NX) = N N = 0. Hence we getthe exact sequence

1 k K R0 1,

where 0 is the group of divisor of degree zero and R0 = 0Z

is the groupof divisor classes of degree zero.Corollary 2. Suppose C R is a class of divisors. IF U , are twodivisors of this class, there exists an X K such that U = (X).Hence, d(U ) = d((X)) + d() = d(), and therefore we may define the 34degree d(C) of the class C to be the degree of any one of its divisors.Corollary 3. If X is any transcendental element, there exists an integerQ dependent only on X such that for all integral m, we have

l(N mX ) + d(N mX ) Q.

Proof. We saw in the course of the proof of the theorem that for t 0,

l(N stX ) N(t + 1) = d(NX)(t + 1),

and writing m = s + t, we obtain for m s,

l(N mX ) + d(N mX ) (1 s)d(NX) = Q.

For m < s, since N sX divides Nm

X , we have

l(N mX ) + d(N mX ) l(N sX ) + d(N sX ) Q.

Lecture 7

12 The Riemann Theorem

In the last lecture, we saw that if X is any element of K, the integer 35l(N m) + d(N mX ) remains bounded below as m runs through all in-tegral values. We now prove the following stronger result, known asRiemanns theorem.

Theorem . Let X be any transcendental element of K and (1 g) thelower bound of l(N mX ) + d(N mX ). Then, for any divisor U ,

l(U ) + d(U ) 1 g.Proof. Let U = U1U 12 , where U1 and U2 are integral divisors. Thenclearly U 12 divides U , and we have

l(U ) + D(U ) l(U 12 ) + d(U 12 ).It is therefore enough to prove the inequality with U 12 in the place

of U .

The key to the proof lies in the statement that l()+ d() is unalteredwhen we replace by (Z), where (Z) is a principal divisor. To provethis, consider the map defined on L() by

Y L() YZThis is clearly a k-isomorphism of the vector space L() onto the

vector space L((Z)). This proves that l() = l((Z)), and since wealready know that d() = d((Z)), our statement follows.

33

34 7. Lecture 7

Observe now that for any non-negative integer m, we have l(N mX 36U2) + d(N mX U2) l(N mX ) + d(N mX ) l g. Since X is transcen-dental, d(NX) > 0 and it follows that for large enough m,

l(N mX U2) md(NX d(U2) + l g > 0For such an m therefore, there exists a non-zero element Z in L(N mX

U ). This clearly means that the divisor (Z)N mX U 12 is integral, or thatN mX divides (Z)U 12 . Hence, we deduce that

l(U 12 ) + d(U 12 ) = l((Z)U 12 ) + d((Z)U 12 ) 1(N mX ) + d(N mX ) l g

which proves our theorem.The integer g is called the genus of the field. Since

1 + 0 = l(N ) + d(N ) 1 g,it follows that g is always non-negative. The integer (U 1) = l(U ) +d(U )+ g 1, which is non-negative by the above theorem, is called thedegree of speciality of the divisor U . We say that U is a non-special orspecial divisor according as (U 1) is or is not equal to zero. We shallinterpret (U 1) later. Incidentally, we have proved that if U is anydivisor and X K, the dimensions of the spaces L(U (X)) and L(U )are the same.

This enables us to define the dimension of a divisor classC. Chooseany element U 1 in C and define the dimension N(C) of C to be l(U ).By the remark, this is independent of the choice of U 1 in C.

13 RepartitionsWe now consider the following question, to which we are led naturally37by the theorem of 7. If for every place Y of K, we are given an elementXY of K, can we find an X in K such that vY (XXY ) 0 holds for everyY ? A necessary condition for such an X to exist is that vY (XY ) 0for all but a finite number of Y . For, suppose vY (XY ) < 0 for some Y .Then, since vY (X XY ) 0,vY (X) = vY (X XY + XY ) = min(vY (X XY ), vY (XY )) = vY (XY ) < O,

13. Repartitions 35

and this can hold for at most a finite number of Y . We now make thefollowingDefinition . A repartition C is a mapping Y GY of the set of primedivisors Y of K into the field K such that vY (CY ) o for all but a finitenumber of Y .

We can define the operations of addition and multiplication in thespace X of repartitions in an obvious manner. If C and G are two repar-titions, and a an element of the constant field k,

(C + G )Y = CY + GY , (C G )Y = CY GY , (aC )Y = aCY .The newly defined mappings are immediately verified to be repar-

titions. Thus, X becomes an algebra over the field k. We can imbedthe field K in X by defining for every X K the repartition CX by theequations (CX)Y = X for every Y . The condition for this to be a repar-tition clearly holds, and one can easily verify that this is an isomorphic 38imbedding of K in the algebra X.

We can now extend to repartitions the valuations of the field K bydefining for every place Y ,

vY (C ) = vY (CY ).Clearly, we have the following relations

vY (C G ) = vY (C ) + vY (G )vY (C + G ) min(vY (C ), vY (G )),

and vY (CX) = vY (X)This leads to the notion of the divisibility of a repartition C by a

divisor U . We shall say that C is divisible by U if vY (C ) vY (U )for every Y , and that C and G are congruent modulo U (C G (U ) insymbols) if C G is divisible by U .

The problem posed at the beginning of this article may be restatedin the following generalised form. Given a repartition C and a divisorU , to find an element X of the field such that X C (U ). (The originalproblem is the case U = N ).

If U is a divisor, let us denote by (U ) the vector space (over k) ofall repartitions divisible by U . Then we have the following

36 7. Lecture 7

Theorem . If U and are two divisors such that U divides , then(U ) () and

dimk(U )() = d() d(U ).

Proof. Let S denote the set of prime divisors occurring in either U or39 with a non-zero exponent. Since dimk

(U /S )(/S ) = d() d(U ), it is

enough to set up an isomorphism of (U /S )(/S ) onto the space

(U )() .

If x (U , /S ), define a repartition Gx as follows:

(GX)Y =

X if Y S0 if Y < S

Clearly, GX (U ), and X GX is a k-homomorphism of (U /S )into (U ). The image of an element X (U /S ) lies in () if andonly if vY (X) vY () for every Y S , i.e., if and only if X (/S ).Thus, we have an isomorphism of (U /S )

(/S ) into(U )() . We shall show

that this is onto. Given any repartition C (U ), find X K such that

vY (X C ) vY () for every Y S .

This means that the repartition GX C is an element of (). Also,the above condition implies that for Y S , vY (X) min(vY (C ),vY ()) vY (U ). Thus, X is an element of (U /S ) and its image in(U )() is the coset C + (). Our theorem is thus proved.

Lecture 8

14 Differentials

In this article, we wish to introduce the important notion of a differential 40of an algebraic function field. As a preparation, we prove the

Theorem. If U and are two divisors and U divides , then

dimk(U ) + K() + K = (l() + d()) (l(U ) + d(U ))

and dimkX

(U ) + K = (U1) = l(U ) + d(U ) + g 1.

Proof. We have(U ) + K() + K =

(U ) + (() + K)() + K

(U )(() + K) (U )

But it is easily verified that (() + K) (U ) = () + L(U ).Hence, we obtain

(U ) + K() + K

(U )() + L(U )

(U )/ ()() + L(U )/ ()

(U )/ ()L(U )/L(U ) () =

(U )/ ()L(U )/L()

Thus,

dimk(U ) + K() + K = dimk

(U )() dimk

L(U )L()

37

38 8. Lecture 8

= (d() d(U )) (l(U ) l()) = (l() + d()) (l(U ) + d(U )),

which is the first part of the theorem. 41

Now, choose a divisor L such that

l(L) + d(L) = 1 g.

Putting Y = min(vY (), vY ), and U = Y

Y Y , we see that U

divides both and L. Hence

1 g l(U ) + d(U ) l(L) + d(L) = 1 g,and hence l(U ) + d(U ) = 1 g.

Moreover, we have

dimkX

() + K dimk(U ) + K() + K = l() + d() 1 + g = (

1).

To prove the opposite inequality, suppose C1, . . .Cm are m linearlyindependent elements of X over k module () + K. If we put Y =min

i(vY (Ci), vY ()) and U =

Y

Y Y , clearly all the Ci lie in (U ).We deduce that

m dimk(U ) + K() + K = (l()+d())(l(U )+d(U )) l()+d()1+g,

which proves that dimkX

() + K is finite and (1). The second part

of the theorem is therefore proved.

Definition . A differential is a linear mapping of X into k which van-ishes on some sub space of the form (U + K).

In this case, is said to be divisible by U 1. is said to be of thefirst kind if it is divisible by N

If divides U , clearly (1) + K (U 1) + K, and therefore42every differential divisible by U is also divisible by .

14. Differentials 39

Consider now the set D(U ) of differentials of K which are di-visible by U . This is the dual of the finite dimensional vector spaceX/(U 1)+K , and therefore becomes a vector space over k of dimension

dimk D(U ) = dimk X(U 1) + K = (U ).

If 1 and 2 are two differentials divisible by U1 and U2 respec-tively, their sum is a linear function on X which clearly vanishes on(U 1)+K, where U is (U1,U2). (The greatest common divisor (g.c.d.)of two divisors U1 and U2 is the divisor U =

Y

Y min(vY (U1),vY (U2)).)Thus, 1 + 2 is a differential divisible by U . Similarly, for a differen-tial divisible by U and an element X K, we define the differentialX by

X(C ) = (XC ).X is seen to be divisible by (X)U . We then obtain

(XY) = X(Y)(X + Y) = X + Y

X(1+2) = X1 + X2It follows that the differentials form a vector space over K. We now

prove the

Theorem. If 0 is a non-zero differential, every differential can be writ- 43ten uniquely in the form = X0 for some X K. In other words, thedimension over K of the space of differentials of K is one.Proof. Let 0 be divisible by 10 and by 1. Let U be an integraldivisor, to be chosen suitably later. The two mappings

X0 L(U 10) X00 D(U 1)and X L(U 1) X D(U 1)are clearly k-isomorphisms of L(U 10) and L(U 1) respectively intoD(U 1). Hence, the sum of the dimensions of the images in D(U 1).is

l(U 10) + l(U 1) 2d(U ) d() d(0) + 2 2g,

40 8. Lecture 8

and is therefore > dim D(U 1) = (U 1) = d(U )+g1 if U is chosenso that d(U ) is sufficiently large. With such a choice of U , therefore,we see that the images must have a non-zero intersection in D(U 1).Hence, for some X0 and X different from zero, we must have

X00 = X, = X0X10 = Y0, Y K.

The uniqueness is trivial.

We shall now associate with every differential a divisor. We re-quire a preliminary

Lemma . If a differential is divisible by two divisors U and , it isalso divisible by the least common multiple L of U and (Definition44of l.c.m. obvious).

Proof. Suppose C (L1). Then,

vY (C ) vY (L) = max(vY (U ), vY (z)).

Define two repartitions C 1 and C by the equations

C1Y1= CY ,C

Y= 0 for all C such that vY (U ) vY ()

C1Y= 0,C

Y= CY for all Y such that vY (U ) < vY ().

C 1 and C are by the above definition divisible by U 1 and Y 1 re-spectively, and C = C 1 + C . Hence,

(C ) = (C 1) + (C ) = 0.

Since must vanish on K, must vanish on (L1)+K. Our lemmafollows

Theorem. To any differential , 0, there corresponds a unique divisor() such that is divisible by U if and only if () is divisible by U .

15. The Riemann-Roch theorem 41

Proof. Suppose is divisible by a divisor U . Then, the mapping X L(U 1) X D(N ) is clearly a k-isomorphism of L(U 1) intoD(N ). Hence, we deduce that

l(U 1) dim D(N ) = (N ) = l(N ) + d(N ) + g 1 = g.

On other hand, we have

l(U 1) + d(U 1) = 1 g + (U ) 2 g,

since (U ) 1. Combining these two inequalities, we deduce that

d(U ) 2g 2.

45This proves that the degrees of all divisors dividing a certain differ-

ential are bounded by 2g 2.

Now, choose a divisor () dividing and of maximal degree. If Uwere any another divisor of , the least common multiple of U and() would have degree at least that of (), and would divide by theabove lemma. Hence, we deduce that = () or that U divides ().

The uniqueness of () also follows from this. Our theorem is proved.Corollary 1. If X K, (X) = (X)(). This follows from the easilyverified fact that U divides if and only if (X)U divides X.

This corollary, together with the theorem that the space of differen-tials is one dimensional over K, proves that the divisors of all differen-tials form a class W . This class is called the canonical class

15 The Riemann-Roch theorem

Let C be a class and U any divisor of C. If Ui(i = 1, . . . , n) are elementsof C,UiU 1 = (Xi) are principal divisors. We shall say that the divisorsUi, are linearly independent if Xi(i = 1, . . . , n) are linearly independentover k. This does not depend on the choice of U or of the respectivelyXi of UiU 1, as is easy to verify. We now prove the

42 8. Lecture 8

Lemma. The dimension N(C) of a class C is the maximum number oflinearly independent integral divisors of C. 46Proof. Let U be any divisor of C. Then, the divisors (X1)U , (X2)U , . . .,(Xn)U are linearly independent integral divisors of C if and only ifX1, . . . , Xn are linearly independent elements of L(U 1). Our lemmafollows.

We now prove the celebrated theorem of Riemann-Roch.

Theorem. If C is any divisor class,

N(C) = d(C) g + 1 + N(WC1).

Proof. Let U C. Then,

N(C) = l(U 1) = d(U ) g + 1 + (U ) = d(C) g + 1 + (U ).

But (U ) being the dimension of D(U ) is the maximum number oflinearly independent differentials divisible by U . Hence, it is the max-imum number of linearly independent differentials 1, . . . , n such that(1)U 1, (2)U 1, . . . , (n)U 1 are integral. By the above lemma, weconclude that (U ) = N(WC1), and our theorem is proved.Corollaries E and W shall denote principal and canonical classes re-spectively.

(a) N(E) = l(n) = 1, d(E) = d(N) = 0.1 = N(E) = d(E) g + 1 + N(W) = N(W) = g.g = N(W) = d(W) g + 1 + N(E) = d(W) = 2g 2.

(b) If d(C) < 0, or if d(C) = 0 and C , E, N(C) = 0.If d(C) > 2g 2 or if d(C) = 2g 2 and C , W ,47

N(C) = d(C) g + 1

15. The Riemann-Roch theorem 43

Proof. Suppose N(C) > 0. Then there exists an integral divisor Uin C, and hence d(C) = d(U ) 0, equality holding if and only ifU = N or C = E.

The second part follows immediately on applying that first part tothe divisor WC1.

(c) If W1 is a class and g1 an integer such that

N(C) = d(C) g1 + 1 + N(W1C1),

We must have W = W1 and g = g1.

Proof. Exactly as in (a), we deduce that N(W1) = g1, d(W1) = 2g1 2.Again as in the second part of (b), we deduce that if d(C) > 2g1 2, N(C) = d(C) g1 + 1. Hence, for d(C) > max(2g 2, 2g1 2),

N(C) = d(C) g + 1 = d(C) g1 + 1, g = g1.

Hence N(W1) = g and d(W1) = 2g 2, and it follows from thesecond part of (b) that W = W1.

This shows that the class W and integer g are uniquely determinedby the Riemann-Roch theorem.

Let us give another application of the Riemann-Roch theorem. Weshall say that a divisor U divides a class C if it divides every integraldivisor of C. We then have the following

Theorem. If C is any class and U an integral divisor,

N(C) N(CU ) N(C) + d(U )

The first inequality becomes an equality if and only if U divides theclass CU , and the second if and only if U divides the class WC1.Proof. Since the maximum number of linearly independent integral di- 48visors in CU is clearly greater than or equal to the number of suchdivisors in C, the first part of the inequality follows. Suppose now that

44 8. Lecture 8

equality prevails. Then there exists a maximal set 1, . . . n of linearlyindependent integral divisors in C, such that U 1, . . .U n forms such aset in CU . But since every integral divisor in CU is linear combina-tion of divisors of such a set with coefficients in k (in an obvious sense),every integral divisor of CU is divisible by U .

Now, by the theorem of Riemann-Roch,

N(CU ) = d(C) + d(U ) + 1 g + N(WC1U 1),

and the second inequality together with the condition of equality followsby applying the first to WC1U 1 instead of C.

Corollary. For any class C, N(C) max(0, d(C) + 1).Proof. If N(C) = 0, there is nothing to prove, if N(C) > 0, there existsan integral divisor U in C. Hence we obtain

N(C) = N(EU ) N(E) + d(U ) = d(U ) + 1 = d(C) + 1,

and our corollay is proved.

Lecture 9

16 Rational Function Fields

In this lecture, we shall consider some particular function fields and find 49their canonical class, genus, etc. as illustrations of the general theory.Let us first consider the rational function fields.

Let K = k(X) be a rational function field in one variable over k. Weshall first show that k is precisely the field of constants of K. For lateruse, we formulate this in a more general form.

Lemma. Let K be a purely transcendental extension of a field k. Then kis algebraically closed in K.

Proof. Let (xi)iI be any transcendence basis of K over k such thatK = k(xi). Since any element of K is a rational combination of afinite number of xi, we may assume that I is a finite set of integers(1, . . . n).

We proceed by induction. Assume first that n = 1. Let be anyelement of k(x1) algebraic over k. may be written in the form f (x1)g(x1) ,where f and g are polynomials over k prime to each other.

We then havef (x1) g(x1) = 0

This proves that if were not in k, x1 is algebraic over k(), (since 50the above polynomial for x1 over K() cannot vanish identically ) andhence over k which is a contradiction. Hence is in k.

45

46 9. Lecture 9

Suppose now that the lemma holds for n 1 instead on n. If were an element of k(x1, . . . xn) algebraic over k, it is algebraic overk(x1, . . . xn1). By the first part, it should be the k(x1, . . . xn1), and henceby induction hypothesis in k. Our lemma is proved.

Let us return to the rational function field K. We have already seenthat the prime divisors are (i)NX, the prime divisor corresponding to 1X ,and (ii)Yp(X), the prime divisors corresponding to irreducible polynomi-als p(X) in k[X]. If f (X) is a rational function of X over k having theunique decomposition pe11 (X) penr (X), the principal divisor ( f (X)) isclearly given by

( f (X)) =r

=1Y

ep (X)N deg fX

It follows that the space L(NtX ) for t 0 consists precisely ofall polynomials of degree t, and since there are t + 1 such poly-nomials independent over k, and generating all polynomials of degree t, (1, X, . . . Xt for example), we deduce that

N(NtXE) = t + 1.

But by the corollary to the Riemann - Roch theorem, if d(NtXE) =td(NX) = t > 2g 2, we should have

N(NtXE) = d(NtX E) + 1 g = t + 1 g,

and hence, g = 0. Thus, there are no non-zero differentials of the first51kind.

Since d(N2X E) = 2 = 2g2 < 0, it follows that N(N2X E) = 0 = g.and hence W = n2

XE.

17 Function Fields of Degree Two Over a RationalFunction Field

We start with a lemma which will be useful for later calculations.

17. Function Fields of Degree . . . 47

Lemma. Let K/k be any algebraic function field and X K any tran-scendental element. If R(X) = f1(X)f2(X) is any rational function of X, then

f1 and f2 being prime to each other (R(X)) = z f1z f2 N

deg R(X)X and z f1

and z f2 are prime to each other and both are prime to NX . Moreover,[k(X) : k(R(X))] = max(deg f1, deg f2).Proof. Let f (X) be any polynomial over k of the form

f (X) = ao + a1X + + atXt.If Y is a prime divisor not dividing NX , i.e., if vY (X) 0, we have

vY ( f (X)) min=o

(vY (X)) o). If on the other hand, Y does occur inNX, we have vY (X) < o, and hence

vY ( f (X)) = min=ot

(vY (X)) = tvY (X)

Thus, we see that z f is prime to NX and N f (x) = NtX .

Hence, if R(X) = f1(X)f2(X) , where ( f1, f2) = 1, we have

(R(X)) = z f1z f2 N

deg R(X)X

We assert that z f1 and z f2 are prime to each other. For if not, let z f1 52and z f2 have a prime divisor Y in common. Then vY (X) o. Findpolynomials g1 and g2 such that

f1g1 + f2g2 = 1.Then, 0 = vY (1) = vY ( f1g1 + f2g2)

min(vY ( f1) + vY (g1), vY ( f2) + vY (g2)) > 0,a contradiction. Thus, z f1 and z f2 are prime to each other.To prove the last part of the lemma, we may assume without loss of

generality that deg f1 deg f2 or that deg R(X) 0 (otherwise consider1

R(X)). It then follows that

zR(X) = z f1(X), [k(X) : k(R(X))] = [K : k(R(X))]/[K : k(X)]

48 9. Lecture 9

=d(zR(X))d(NX) =

d(z f1(X))d(NX) =

d(z f1(X))d(NX) = deg f1.

Our lemma is proved.It follows in particular that k(x) = k(R(X)) if and only if max(deg f1,

deg f2) = 1, or R(X) = X + X +

, , , , k and , 0.Now, let k be a field of characteristic different from 2, X a tran-

scendental element over k and K a field of degree two over k(X) whichis not got from an algebraic extension of k (viz., K should not be gotby the adjunction to k(X) of elements algebraic over k). Then k is theconstant field of K, for if it were not, there exists an element of Kwhich is algebraic over k but not in k. cannot lie in k(X), since k is53algebraically closed in k(X). Hence, k(X, ) should be an extension ofdegree at least two over k(X), and should therefore coincide with K. Butthis contradicts our assumption regarding K.

Now, K can be get the adjunctions to k(X) of an element Y whichsatisfies a quadratic equation

Y2 + bY + c = 0, b, c k(X)Completing the square ( note that characteristic k , 2), we get

(Y +

b2

)2+

(c

b4

2)= 0,

and hence k(X, Y) = k(X, Y1) where Y1 = Y + b2

satisfies equation of

the form 12 = R(X),R(X) being a rational function of X. Let R(X) =r

1pe (X), where p(X) are irreducible polynomials in k[X] and e are

integers. Putting e = 2g + , where g are integers and = 0 or

1, and Y = Y1

pggamma(X) , we see that k(X, Y) = k(X, Y

), and Y

satisfies an equation of the form

Y 2 =r1

p (X) = D(X),

17. Function Fields of Degree . . . 49

where D(X) is a polynomial which is a product of different irreduciblepolynomials.

We shall therefore assume without loss in generality that K = k(X, Y)with Y2 = D(X) of the above form. Let us assume that m is the degreeof D.

Now, let be the automorphism of K over k(X) which is not the 54identity. If Z = R1(X) + YR2(X) is any element of K, Z = R1(X) YR2(X). To every prime divisor Y of K, let us associate a prime divisorY by the definition

vY (Z) = vY (1Z)

This can be extended to an automorphism of the group of divisors(see Lecture 19). We shall denote this automorphism again by , andthe image of a divisor U by U . Since X = X, NX = NX.

Suppose now that Z = R1(X) + YR2(X) is any element of L(NtX ) (tany integer). Applying , we deduce that R1(X) YR2(X) should alsobe an element of L(NtX ). Adding, 2R1(X) L(NtX ),R1(X) L(NtX ).If R1(X) = f1(X)g1(X) , where f1 and g1 are coprime polynomials, we have

(R1(X)) =z f1zg1

N deg R1X divisible by NtX , and hence we deduce that zg1 =

n, and g1(X) is a constant. Hence R1(X) is a polynomial of degree t(if t < o,R1(X) = 0).

Also, since both Z and Z are in L(NtX ). This implies as before thatR21 DR

22 is a polynomial of degree 2t. Hence DR

22 is a polynomial

of degree 2t, and since D is square free, R2 is a polynomial of degree t

m

2.

Conversely, by working back, we see that if R1 is a polynomial ofdegree t and R2 a polynomial of degree t = m2 , Z = R1 + YR2

50 9. Lecture 9

L(NtX ) . Hence, we obtain

N(ENtX) = l(NtX ) =

0 if t < 0t + 1 if 0 t m2 1 and m event + 1 if m and 0 t m122t + 2 m2 if m even and t

m2

2t + 2 m+12 if m odd and t m+1

2 .

55Since d(NX) = [K : k(X)] = 2, for t > g1, we have d(ENtX) = 2t >

2g 2 and hence

N(ENtX) = d(NtX) g + 1 = 2t g + 1

Comparing with the above equations, we deduce that g is m2 1 if

m is even and m 12

if m is odd.Thus, we obtain examples of fields of arbitrary genus over any con-

stant field.The canonical class of K is ENg1X . For,

d(Ng1X E) = 2g 2,

and N(Ng1X E) =

g 1 + 1 = g if g > 00 = g if g = 0.

If gg > o, there exists a differential of the first kind with () =Ng1X . Then clearly the differentials , X, . . . , X

g1 are all of the firstkind and are linearly independent over k, and as they are g in number,they form a base over k for all differentials of the first kind.

18 Fields of Genus ZeroWe shall find all fields of genus zero over a constant field k.56

First, notice that any divisor of degree zero of a field of genus zerois a principal divisor. For let C be a class of degree 0. Then sinced(C) > 2 = 2g 2, N(C) = d(C) g + 1 = 1 and therefore C = E.

19. Fields of Genus One 51

Now,

d(W1) = 2 > 2g 2, N(W1) = 2 g + 1 = 3,and therefore there exists three linearly independent integral divisorsU1,U2,U3 in the class W1 (incidentally, this proves there exists inte-gral divisors, and consequently prime divisors of degree at most two).Let

U1U2

= (X). Then clearly NX divides U2, and hence we obtain

[K : k(X)] = d(NX) d(U2) = 2.Thus, any field of genus zero should be either a rational function

field or a quadratic extension of a rational function field. We have thefollowing

Theorem . The necessary and sufficient condition for a field of genuszero to be a rational function field is that it possess a prime divisor ofdegree 1.

Proof. If K = k(X), the prime divisor nX satisfies the requisite condi-tion.

Conversely, let Y be a prime divisor of degree 1 of K. Then,N(Y E) = d(Y ) + 1 = 2, and therefore there are elements X1, X2 in 57K linearly independent over k such that X1Y = U1 and X2Y = U2are integral divisors. Thus if X = X1

X2, (X) = ( X1

X2) = U1

U2, and d(NX)

d(U2) = d(Y ) = 1. But X is not in k, and hence d(NX) = 1 = [K : k(X)],from which it follows that K = k(X). The theorem is proved.

19 Fields of Genus OneLet K/k be an algebraic function field of genus 1. Then, since N(W) =g = 1 and d(W) = 2g 2 = 0, the canonical class W coincides with theprincipal class E.

A function field of genus one which contains at least one prime di-visor of degree one is called an elliptic function field. (The genus beingone does not imply that there exists a prime divisor of degree one. In

52 9. Lecture 9

fact, it can be proved easily that the field R(X, Y), where R is the fieldof real numbers and X, Y transcendental over R and connected by therelation Y2 + X4 + 1 = 0 has every prime divisor of degree two). Let usinvestigate the structure of elliptic function fields.

Let Y be a prime divisor of degree one. Then, since d(Y 2) = 2 >2g 2 = 0, we have l(Y 2) = 2. Let 1, X be a basis of L(Y 2) over k.Since XY 2 is integral, NX divides Y 2.NX cannot be Y , since if it were,we obtain [K : k(X)] = d(NX) = d(Y ) = 1, K = k(X) and hence g = 0.Thus, NX should be equal to Y 2.

Again, since l(Y 3) = 3, we may complete (l, X) to a basis (1, X, Y)of L(Y 3) over k. N should divide Y 3, and since Y is not an element58of L(Y 2), N does not divide Y 2. Thus, N = Y 3.

The denominators of 1, X, Y, X2, XY, X3 and Y2 are respectively N,Y 2, Y 3, Y 4,Y 5,Y 6 and Y 6. Since the first six elements have dif-ferent powers of Y in the denominator, they are linearly independentelements of L(Y 6). But l(Y 6) = 6, and the seventh element, beingin L(Y 6) should therefore be a linear combination of the first six. Wethus obtain

Y2 + XY + Y = 3X3 + 2X2 + 2X2 + 1X + o, , , i k.

Now, if Y where a rational function of X, writing Y = f (X)g(X) , f (X),

g(X) k[X], ( f (X), g(X)) = 1, and substituting in the above equation,we easily deduce that g(X) should be a constant, and that Y should be apolynomial in X of degree 1. But this would mean that Y is divisibleby Y 2, which we have already ruled out. Hence, [k(X, Y) : k(X)] = 2 =d(NX) = [K : k(X)], and consequently, K = k(X, Y).

If the characteristic of k is different from 2, we may as in 16 find aZ such that K = k(X, Y), with

Z2 = f (X)where f (X) is a cubic polynomial in X with non-repeating irreduciblefactors.

A partial converse of the above result is valid. Suppose that K =k(X, Z), where X is transcendental over k and Z satisfies an equation59

19. Fields of Genus One 53

of the form Z2 = f (X), where f (X) is a cubic polynomial which weassume to be irreducible. Let ch(k) , 2. Then, the genus of K is one,by 16. Also, since deg f (X) = 3, Z2 L(N3X ), and N2z divides N3X.But since X is of degree 3 over k(Z), we have d(N2z ) = 2d(Nz) = 2.[K :k(Z)] = 6 = 3.[K : k(X)] = 3d(NX), and therefore Nz = N3X. From this,it is clear that there exists a prime divisor Y with d(Y ) = 1 such thatNz = Y 3, NX = Y 2.

Finally, suppose a field K is of genus greater than 1. Then, N(W) =g > o and d(W) = 2g 2 > o, and hence we deduce the existence ofan integral divisor , N of degree 2g 2. Thus, we have proved thatif g > 1, there always exists prime divisors of degree 2g 2. Theminimal degree of prime divisors for a field of genus one is not known.

Lecture 10

20 The Greatest Common Divisor of a ClassWe wish to find when the greatest common divisor of all integral divisors 60of a class is different from the unit divisor N. We assert that this isimpossible when d(C) 2g. In fact, let U be an integral divisor of theclass C. Then we obtain

d(C) g + 1 = N(C) = N(CU 1) = d(C) d(U ) + 1 g + N(WC1U ),d(U ) = N(WC1U ) max(0, d(WC1U ) + 1)

and since 1 + d(WC1U ) 2g 2 2g + 1 + d(U ) = d(U ) 1, weshould have d(U ) = 0 and U = N.

This is in a sense the best possible result. In fact, if there exists aprime divisor Y of degree 1 in the field, we have d(WY ) = 2g 2, andhence.

N(WY ) = d(W) + d(Y ) g + 1 = g = N(W),which proves that Y divides all integral divisor of the class WY . Nev-ertheless, if g > 0, we can prove that the greatest common divisor of thecanonical class W is N. For, suppose U , N is an integral divisor of W .Then,

dim L(U 1) = N(EU ) = d(U ) + 1 g + N(WU 1)= d(U ) + 1 g + N(W) = d(U ) + 1 2.

Thus, there exists a transcendental element X L(U 1). The divi-sor (X)U is then integral. Also, since N(W) = g > o, we can choose

55

56 10. Lecture 10

a differential of the first kind. Then, (X) = ((X)U ).(()U 1) is 61also integral and therefore X is also first kind. By repetition of theargument, we obtain that Xn is of the first kind for all positive integersn. But since X is transcendental, NX , N , and hence Xn cannot bean integral divisor for large n. This is a contradiction. Our assertion istherefore proved.

For fields of genus g > o, we shall improve the inequality N(C) max(o, d(C) + 1).Lemma. If g > o and d(C) o,C , E, we have

N(C) d(C)Proof. We may obviously assume that N(C) > o. Then there existsan integral divisor U in C, and since C , E,U , N , and therefored(C) = d(U ) > o. Since G > o, U cannot be a divisor of the classW ,and therefore

g = N(W) > N(WU 1) = N(WC1)N(C) = d(C) g + 1 + N(WC1) d(C) g + 1 + g 1 = d(C).Our lemma is proved

21 The Zeta Function of Algebraic Function FieldsOver Finite Constant Fields

In the rest of this lecture and the following two lectures, we shall alwaysassume that k is a finite field of characteristic p > o and with q = ptelements,and that K is an algebraic function field with constant field k.62

If Y is any prime divisor of K, we shall call the number of elementsin the class field kY the norm of Y . Since [kY : k] = d(Y ), we seethat norm of Y (which we shall denote by NY ) is given by

NY = qd(Y )

We may extend this definition to all divisors, by putting

NU =Y

(NY )vY (U ) = qd(U )

21. The Zeta Function of Algebraic Function . . . 57

Clearly we have for two divisors U and

NU = NU .N.

Before introduction the zeta function, we shall prove an important

Lemma . For any positive integer m, the number of prime divisors ofdegree m is finite. The number of classes of degree zero is finite. (Thelatter is called the class number of K and is denoted by h).Proof. To prove the first part of the lemma, choose any transcendentalelement X of K. Let Y be any prime divisor of K with d(Y ) m andwhich does not divide NX. (We may neglect those U which divide NX,since they are finite in number.) Let Y 1 be the restriction of Y to k(X).Y 1 must be a place on k(X). Since Y 1(X) , , there corresponds aunique polynomial (p(X)) which gives rise to Y 1.

Now, since kY kY , we obtain 63

deg(p(X)) = d(Y ) d(Y ) m.

Since the number of polynomials of degree m over a finite field isfinite, and since there the are only a finite number of prime divisors Yof K which divide zp(X) for a fixed p(X), the first part of our theorem isproved.

To prove the second part, choose and fix an integral divisor Uo suchthat d(Uo) g. If C is any class of degree zero, we have

N(CUo) d(C) + d(Uo) g + 1 1,and hence there exists an integral divisor U in CUo such that d(U ) =d(CUo) = d(Uo). But since

d(U ) =

d(Y )vY (U ), d(Y ) 1, vY (U ) 0,and there are only a finite number of Y with d(Y ) d(Uo), there areonly a finite of integral U with d(U ) = d(Uo) and consequently only afinite number of C with d(C) = 0.

The lemma is completely proved.

58 10. Lecture 10

Remark 1. Let denote the least positive integer which is the degree ofa class. since C d(C) is a homomorphism of the group R of divisorclasses into the additive group Z of integers, we see that the degree ofany class of the from where is an integer, and that to any , therecorrespond precisely h classes C()1 , . . . ,C

()h of degree .64

Remark 2. The number of integral divisors in any C is preciselyqN(C) 1

q 1. This is clear if N(C) = 0. If N(C) > 0, let U be any in-

tegral divisor of C. Then all integral divisors of C are of the from (X)U ,where X L(U 1), X , 0. Also (X)U = (Y)U if and only if

(XY

)= N

or X = a, a k. Since the number of non zero elements of L(U 1) isqN(C) 1 and the number of non zero elements of k is q1 our assertionfollows.

Now, let s = + it be a complex variable. For > 1, we define thezeta function of the algebraic function field K by the series

(s, K) =U

1(NU )s , s = + it, > 1

the summation being extended over all integral divisors of the field K.Since

1(NU )s = 1(NU ) , and all the terms of the series are positive

when s is real, the following calculations are valid first for s > 1 andthe for complex s with > 1. In particular, they prove the absoluteconvergence of the series

(s, K)=

C

(number of integral divisors in C). qsd(C), the last summationbeing over all classes,

=1

q 1

C

(qN(C) 1)qsd(C);

writing d(C) = , and noticing that qN(C)1 = 0 if d(C) < 0, the above65

21. The Zeta Function of Algebraic Function . . . 59

expression becomes

1q 1

=o

qsh

l=1qN(Cl

()) h

q 1

=o

qs

Let us now put U = qs. Then |U | = |qs | = q < 1 since > 1,and we can sum the second geometric series. Suppose now that g > o.We may then split the first sum into two parts, the ranging over o

2g 2

and the second over > 2g 2

. (Since d(w) = 2g 2, )

divides 2g 2; or 2g 2

is an integer). In the second summation sinced(C()l ) = > 2g 2, we may substitute N(C

()l ) = g + 1. We

obtain the expression

1q 1

2g2

=o

qsh

l=1qN(C

()l )+

hq 1

>

2g2

qsqg+1h

q 1.

11 qs

= (s, K) = 1q 1

2g2

=o

Uh

l=1qN(C

()l )

+hq1gq 1

(Ug)2g2+1 (Uq)

hq 1

11 U

(1)

If g = o, N(C) = d(C) g + 1 for all C with d(C) > o, and a similarcomputation gives

(S , K) = hqq 1

11 (Uq)

hq 1

11 U

(2)

(1) and (2) may be combined as follows(q 1)(s, K) = F(U) + R(U), (3)

where F(U) is the polynomial 66F(U) =

od(C)2g2

qN(C)Ud(C) (4)

60 10. Lecture 10

and R(U) the rational function

R(U) = hq1g (Uq)max(o,2g2+)

1 (Uq) h

1 U(5)

These formulae provide the analytic continuation of (s, K) to thewhole plane. The only possible poles are the values of s for whichU = 1 or (qU) = 1.

For a rational function field K = k(X), since g = o, h = 1 and = 1,we obtain

(q 1)(s, K) = q1 Uq

1

1 U=

q 1(1 Uq)(1 U)

Lecture 11

22 The Infinite Product for (s, K)Let K be an algebraic function field over a finite constant field k. Then, 67the function of K can be expressed as an infinite product taken over allprime divisors of K.

Theorem. For s = + it, and > 1, we have

(s, K) =Y

11 (NY )s ,

where the product is taken over all prime divisors Y of K. The productis absolutely convergent, and hence does not depend on the order of thefactors.Proof. Since > 1, we have for any integer m > o,

NY m

11 NY s

=

NY m

(1 +

1(NY )s +

1(NY )2s +

),

and since there are only a finite number of factors in the product, eachfactor being an absolutely convergent series, we may multiply these toobtain

NY m

11 (NY )s =

nU m

1(NU )s +

NU >m

1(NU )s ,

where the first summation is over all integral divisors U with NU m,and the second over all integral divisors U which do not contain any

61

62 11. Lecture 11

prime divisor U with NU > m and which satisfy the inequality NU >m.

Hence68

NY m

11 (NY ) s

NU m

1NU

s

=

1NU >m

1(NU )s

NU >m

1(NU )

and letting m , we obtain the asserted equality, since

NU >m( 1NU

),being the remainder of the convergent series for (, K), tends to zeroas m .

The absolute convergence of the product is deduced from the in-equality 1(NY )s +

1(NY )2s +

1(NY ) +1

(NY )2As a corollary to this theorem, we see that (s, K) has no zero for

> 1.

23 The Functional EquationIn the last lecture, we obtained the following formula:

(q 1)(s, K) = F(U) + R(U), where U = qs,F(U) =

od(c)2g2

qN(C)Ud(C)

and R(U) = hq1g (Uq)max(o,2g2+)

1 (Uq) h

1 U.

Substituting for N(C) in F(U) from the theorem of Riemann-Roch,we obtain69

F(U) =

od(C)2g2qd(C)g+1+N(WC

1 )V

d(C)

23. The Functional Equation 63

= q1g

od(C)2g2

(1

qU

)d(WC1)2g+2qN(WC

1)

Writing WC1 = C1, and noticing that C1 runs through the same setof classes as C in the summation, we have

F(U) = qq1U2g2

od(C1)2g2

(1

qU

)d(C1)qN(C

1)= qg1U2g2F

(1

qU

).

We shall prove that a similar functional equation holds for R(U).First suppose that g > o. Then,

R(U) = hg1g (qU)2g2+

1 (qU) h

1 U

= hg1g(

1qU

)2g+21

(1

qU

)+

h.(q. 1Uq )

1 (q. 1Uq

)

= U2g2qg1hq1g

(q 1qU

)2g2+1

(g. 1U q

) h1

(1

Uq

)

= qg1U2g2R(

1qU

).

If g = o, the only divisor class of degree zero is E and hence h = 1.Also, since divides 2g 2 = 2, = 1 or = 2. If = 1, we obtain

R(U) = q1 qU

1

1 U=

q 1(1 U)(1 qU)

= q1U2q 1(

1 1qU) (

1 q. 1qU)

= q1U2R(

1qU

).

If = 2, we get 70

R(U) = q1 (qU)2

11 U2

=

q.(

1qU

)21

(1

qU

)2 + U2

1 (q. 1qU

)2

64 11. Lecture 11

= q1U2R(

1qU

).

Thus, in any case, we have the functional equation

R(U) = qg1U2g2R(

1qU

).

Adding the equation for F(U) and R(U), we see that (s, K) satisfiesthe functional equation

(q 1)(s, K) = qg1qs(22g)(q 1)(1 s, K), or(s, K) = qg1qs(22g)(1 s.K)

We may also rewrite this in the from

qs(q1)(s, K) = q(1s)(g1)(1 s, K).Thus, the function on the left is unaltered by the transformation s

1 s.

24 L-series

We wish to study the L-series associated to characters of the class groupof an algebraic function with a finite constant field.

Definition . A character X of finite order on the class group R is a ho-71momorphism of R into the multiplicative group C of non zero complexnumbers such that there exists in integer N with XN(C) = 1 for all C inR.

X(C) is therefore an Nth root of unity for all C. We may define X onthe group v of divisors by comparing with the natural homomorphismv R, i.e., by putting X(U ) = X(U E) for any divisor U .

The L-function L(s,X, K) associated to a character X (which weshall always assume to be of finite order) is then defined for s = + it, > 1 by the series

L(s,X, K) =U

X(U )(NU )s

24. L-series 65

where the summation is over all integral divisors U of the field. Theabsolute value of the terms of this series is majorised by the correspond-ing term of the series

1(NU ) = (, K) and is therefore absolutely

convergent for > 1. We may prove along exactly the same lines as inthe case of the -function the following product formula:

L(s,X, K) =Y

11 X(Y )(NY )s

where Y runs through all prime divisors and > 1.

Lecture 12

25 The Functional Equation for the L-functions

Let X be a character of finite order on the class group of an algebraic 72field K over a constant field k with q = p f elements. We consider twocases.

Case 1. X when restricted to the subgroup Ro of R of all classes ofdegree zero, is trivial; i.e., X(Co) = 1 for Co Ro.

Let be as before the smallest positive integer which is the degreeof a class, and let C() be a class of degree . Then R is clearly the directproduct of Ro and the cyclic group generated by C(). Set X(C()) =e2i . We then have

L(s,X, K) =U

X(U )(NU )s =

CoRo

n=

U CoC()n

e2inqns

=

U

(NU )(s

2i log q

)=

(s

2i log q

).

Thus, the L-function reduces to a - function in this case. We cantherefore deduce a functional equation for L(s,X, K) from the functionalequation for (s, K). Define the character X conjugate to X by puttingX(C) = X(C). Clearly, X is a character of finite order on the class group

and is trivial on Ro, also XC() = e2i . Then, we obtain the followingrelation for L(s,X, K) by substituting from the functional equation for

67

68 12. Lecture 12

the function.

qs(g1)L(s,X, K) = qs(g1)(s

2i log q

)

= q2(q1)2i log q

(1 s + 2i

log q

)q(1s)(g1)

= X(W)q(1s)(g1)L(1 s, X, K)

since X(W) = (X(C()))2g 2

= e2i

(2g 2)73

Case 2. Suppose now that X when restricted to Ro is not identically 1.Let C1o be a fixed class with X(C1o) , 1. Then,

X(C1o)

CoRo

X(Co) =

CoRo

X(C1oCo) =

CoRo

X(Co),

and therefore CoR

X(Co) =

Again, using the fact that N(C) = 0 if d(C) < 0, we obtain

(q 1)L(s,X, K) =

CoRo

X(Co)

n=0Xn(C())(qN(CoC())n 1)qns

=

CoRo

X(Co)2g2

n=o

Xn(C())(qN(CoC()n ) 1)

+

CoRo

X(Co)

n>2g2

Xn(C())(qng+1 1)qns

The second sum vanishes, since CoRo

X(Co) = 0.Thus, we obtain

(q 1)L(s,X, K) =

od(C)2g2X(C)qN(C)Ud(C)

The coefficient of U2g2 is given by74

25. The Functional Equation for the L-functions 69

d(C)=2g2

X(C)qN(C) = X(W)

CoRo

X(Co)qN(CoW)

= X(W)

Co,EX(Co)qg1 + qg

,

since N(CoW) = d(CoW) g + 1 = g 1 if Co , E and N(W) = g,

= X(W)

CoRo

X(Co)qg1 + (qg qg1)

= (q 1)X(W)qg1 , 0.Thus, L(s,X, K) is a polynomial in U = qs of degree 2g 2 and

leading coefficient qg1X(W)Again, by substituting from the Riemann-Roch theorem, we have

(q 1)L(s,X, K) =

od(C)2g2X(C)qN(C)Ud(C)

=

od(C)2g2

X(C)qd(C)g+1+N(WC1 )Ud(C)

= qg1U2g2X(W)

od(C)2g2X(WC1)qN(WC1)

(1

qU

)d(WC1)

Writing C1 for WC1 and noting that C1 runs through exactly thesame range of summation as does C, we obtain

(q 1)L(s,X, K) = (q 1)qg1U2g2X(W)L(1 s, X, K),

which is again the same functional equation that we got in Case 1. Wehave therefore proved the

Theorem. For any character X of finite order,qs(g1)L(s,X, K) = X(W)q(1s)(g1)L(1 s X, K).

Lecture 13

26 The Components of a Repartition

Our next aim is to introduce the L-functions modulo an integral divisor 75of an algebraic function field over a finite constant filed and to obtaintheir functional equation. We shall develop the necessary results for thisin this and the next two lectures.

Let K be an algebraic function field over an arbitrary (not necessarilyfinite) constant field k. For a repartition C X and a prime divisor Y ,we define the component C Z of C at Z to be the repartition defined by

CZ (U ) =

CZ if U Y0 if Y is a prime divisor other than Y .

The mapping is clearly a k-linear mapping of X into itself. Thisinduces a linear mapping of the dual of X into itself, by which a differ-ential is taken to a liner map Z : X X given by

Y (C ) = (C Z )

Z is called the component of the differential at Y .Z is not, ingeneral, a differential. For X K, we have

(X)Y (C ) = (X)(C Z ) = (XC Z ) = ((XC )Y ) = Z (XC )

We shall now prove a lemma which expresses a differential in termsof its components.

71

72 13. Lecture 13

Lemma. If is a differential and C a repartition, Y (C ) = 0 for all 76but a finite number of prime divisors Y , and we have

(C ) =Y

Y (C ).

Proof. Let U be a divisor which divides the differential . LetY1, . . . ,Yn be the finite number of prime divisors for which eithervYi (U ) , 0 or vYi (C ) < 0. For U , any of the Yi, and any primedivisor Y , we have

vY (C U ) = vY (C U (Y ))

=

vC (YU ) if U = YvY (0) if , Y U

=

vU (C ) if Y = U if Y , U

vY (U ),

and therefore C U (U ), U (C ) = (C U ) = 0.

Also, if we put Y = C n

i=1C Yi , we have for any Y ,

vY (Y ) =

vY (CY ) if Y is not any of theY1vY (0) = if Y is a certain Yi

vY (U ),and therefore Y (U ). Hence,

(C ) = C

ni=1

CYi

+

ni=1

CYi

=

ni=1

CYi

=

ni=1

(C

Yi)=

ni=1

Yi =Y

Y (C ).

Our lemma is proved.We shall now prove another useful

Lemma. Let be a differential and C a prime divisor. Then vY (()) is77the largest integer m such that whenever X K and vY (X) m, wehave Y (X) = 0.

26.