Chapter 10 Algebraic Function Fields

CHAPTER 10 ALGEBRAIC FUNCTION FIELDS

Definitions

10.10

Let that

L

F(L)

and

Introduction

be a lattice in

@.

We have seen in Chapter 7

is a field extension of

as follows. p

which can be described

@

is transcendental over

finite algebraic extension of

@(?).

C

and

F(L)

is a

This will serve as a model

for the definition to follow. Let

K

be a subfield of a field

F.

KcF

will be called

an algebraic function field of one variable, or merely an algebraic function field, if there exists over

K,

such that

F

x

E

F,

transcendental

is an algebraic extension of

K(x)

of

finite degree. Example.

@ c

F(L)

is an algebraic function field.

The theory of algebraic function fields is a large one. We will give only a very brief sketch of it in this chapter, giving frequent references to the literature.

Hopefully we will

write enough to motivate what has been done when

K

is IR

or

C, and thus make Part I11 of this monograph more accessible to the reader not knowledgeable in these matters. The standard reference to this subject is Chevalley's concise and extensive Introduction to the theory of algebraic functions of one variable [16]. A more classical reference to the state of the art c. 1900 is Hensel and Landsberg's Theorie 189

Norman L. Alling

190

der - algebraischen Funktionen einer Veriabeln 1341. Let

10.11

(1)

Let

E

XI

KcF

F

be an algebraic function field (10.10).

be transcendental over

K;

Indeed, since the transcendence degree of definition, 1, x

is algebraic over

[F:K(x)], we see that

[F:K(x,x')]

then [F:K(x')] <

F

over

K(x');

thus

thus


K

m.

is, by

[F:K(x')l

< m l

establishing (1). Thus the definition of an algebraic function field does not depend on the choice of Let

K

be called a field of constants and

of functions (2)

Let

then

It

of f

{a

F :a

E

function field. 2 E x a m p l e 1.

algebraic over F,

and

KI;

EcF

is an algebraic

IRc@(x)

KcF.

is an algebraic function field.

The

@.

will be called a complex (resp. real) algebraic

function field if Let

the field --

is called the field of constants of

field of constants of it is

(3)

F

KcF.

is a subfield of

KcF

x.

FcF1

K = @ (resp. K=IR). be a finite algebraic extension; then

KcF1

is an algebraic function field. We have seen in 18.13 that a compact Riemann surface

x

may be associated with a complex algebraic function field, using the "cut and paste method". structing

X.

Many choices were made in s o con-

That these arbitrary choices do not influence

the final surface

X

is true, but was not manifest in 58.13.

To motivate the way we will presently construct

sider the following example.

x,

let us con-

Algebraic Function Fields Let a:

Example 2.

over

(4)

where

z ( 2 )

If

Oa

Let

=F(C);

thus the

corresponds to an object of analysis

with some algebraic subobject of

C

is transcendental

Our problem is initially to associate a point

and geometry, 1. of

(z)

We saw in Theorem 8.30 that a:

@.

algebraic object @ ( z )

u

c@

191

f

E C ( Z ) :

is regular at

~ ( 2 ) .

ul.

is a proper subring of ~ ( z ) that contains C I

Clearly

-

f E ~ : ( z ) U0

such that for all

I

l/f

E

0,.

Subrings of this

sort occur elsewhere in algebra; they are called valuation rings. (5)

A proper subring

f t - F-L) implies

l / f E

L) L)

Historical note:

of

F

that .contains K

such that

is called a valuation ring of

KcF.

in this form the idea of a valuation

ring goes back to Krull's classic Allgemeine Bewertungstheorie 1461 I written c. 1931. (6)

Let

RiemKF

{ L ) : L)

is a valuation ring of ,Kc F).

We will think of this set as the formal Riemann surface of K

c

F. It is not difficult to show that

Rie%C

( z ) = {Ou:u

E

11.

In order to work with valuation rings we must know more about them.

10.12 (7

E

Let

KcF

be an algebraic function field and let

RiemKF.

(1)

Let

ul

U(L))

(or U M(O)

and let

for short) be the group of units of (or M

for short) be

0 - U.

It may be useful to refer back to Example 2 in 510.11. Clearly

03;

U

thus

(OD) M((7,)

= {f

E

a: ( 2 ) : f has neither a zero or a pole at

is the maximal ideal of

0,.

Further

192 f

E

Norman L. Alling f(a) E C

Oa ->

M(O,).

as kernel

M

af

fc M

M;

for a moment

that

then it is in

that b a f = l ,

Now let g e M;

(3)

f-g

f E U,

Assume that

is a valuation ring

f

f

M;

by (2) is in

f/g

proving (3).

is an ideal in

is zero then clearly

g

are non-zero.

g

g/f

or

f/g

or

and

generality we may assume that

0,

which is absurd; proving

then

+ 1 U.~ If

Indeed, note that

of

such

M.

E

(3) holds.

0, is

af, which must be in

Thus there exists b e 0

U.

showing that

(2)

M

a E 0; then

and let

M.

E

Assume not in

0.

is the maximal ideal of

Let

Proof.

having

These facts suggest general results.

Theorem.

(2)

is a C-linear homomorphism onto d:

is in E

(1.

0

Without loss of

f - g = g (f/g

(1.

Since

- 1) ,

which

From ( 2 ) and (3) we see that

0. Since 0 - M = U,M

is the maximal ideal

proving the Theorem. Clearly

is a subgroup of

K*(:K-{Ol)

U,

thus

K n M = (01.

FO

Clearly K

maps injectively, under the canonical homomorphism

pO,

is called the residue -class field of

onto a subfield of

to a map

let

(4)

U

F*

h

f(O)

Let

E pO(f)

vO

(or v F*/U:

pO

(or p

F

by sending any

Po

EFO u

{a}.

for short) extend

fcF-0

So extended

to po

00,

a

is the

h

E

of course

onto

Let

0.

f6 F

-

is

Let

(5)

FO.

0. Given

place of

F.

of all of

p

point not in

of

O/M

GO

FO.

-

a subgroup of

F*.

for short) be the canonical homomorphism (or G

for short).

193

Algebraic Function Fields The Abelian group

0.

is called the -value group of

G

It will

be written additively, and can be ordered as follows:

(6)

let

v ( 0 nF*).

G'Z

0 = v (1)E G+ ,

Clearly

and clearly

-

What is - at first glance given

(7)

gc

then

then

G,

g

or

f E F*

f

a valuation ring

or

-g

is, accordingly, in

in

G',

-g

0: i.e. ,

and so

g = 0, For

f

and

g

Let

G.

v(f)

0

l/f

is in

G

Assume that

+. l/f

and

h

(=

let

E G

0

g. Since

(10.11:5); hence

g

and

so that

a,b E U

are in

=

0.

Thus

-g fa f

is

g

or

are both and

b/€

is a unit

g,h

if

g-h

is in

linearly) ordered Abelian group.

be an element not in E

If both,

G'.

establishing (7).

is a totally

g

is in

such that

then there exist units

are in

all

a bit surprising is that

g = 0.

Indeed, there exists

G

is closed under addition.

G+

G, ordered to be greater than

G+;

then

Let

m

g

for

v(0) E m .

(8)

For all

a,b

(9)

Further, if

E

F, v(a

k

b) Lmin(v(a) , v ( b ) ) .

v(a) # v(b)

equality holds in (8).

Indeed, without l o s s of generality we may assume that v(a) Iv(b).

If

b#O;

v(b/a),O,

then

b=O

then (8) and (9) hold. andso

clearly (8) holds and, since

b/aEOnF*.

-1 E U , v(a)

=

Assume that If

a + b = 0 then

w (b). Assume that

aib#O.

v(a+b) =v(a(l+b/a)) =v(a) +v(li-b/a) 'v(a),

ing (8).

Assume now that

hence

b/a

E

M.

As a consequence

v (a 2 b) = v (a) + v

The map

v(a) < v(b).

v

(1 f b/a) of

F*

=v

(a),

onto

lkb/a

Then E

U,

v(b/a) > 0

provand

and so

showing that (9) holds. G

is known as the valuation

Norman L. Alling

194

with

associated

0.

Let us return to

E x a m p l e 2 (510.11

f E Q : ( z ) (=F(C)); let

with

{n E Z: n 2 0 )

0,,

and

Let

A

C

thus

vu

RE

f

Q:

u

at

onto

(z)*

2

vU(Uu n @ ( z ) * ) =

Further

is the valuation associated

is its value group.

be an integral domain and let

~

and let

C

E

be the order of

U ( O u ) nq: ( z ) * .

;

Z

quotient field. element

u

Let

is a homomorphism of

having as its kernel Z+(-

vu(f)

vu

Clearly

(86.41).

continued).

R

Let

denote its

A

be a field extension of

is called integral

A

A.

An

if there exists a

monic polynomial

... + mn-l

(10) m(x) E m 0 + m1 x +

as a root.

A

xn-l + xn E A[x]

is integrally closed in

is integral over

A

Lemma.

Each

Proof.

Let

is in 0

a

E

E

if each

is integrally closed in

be integral over

be a monic polynomial with coefficients in a root. hence

Assume, for a moment, that v(an)

<

v(an-l) <

that

. . . < v(a) < 0.

a{O;

F.

Let m(x)

I).

that has

I)

then

(10)

as

a

u(a) < 0

and

Since 1 nmn-l a r nv(a) ,min(v(mo)

... + w (a), . . . , v (mn-l)+ (n-l)v (a)) 2 (n-l)v (a);

a n = - m o - m1 c1-

v (m,)

a~ A

c1

A.

RiemKF

F

A

which has

which is

absurd, proving the Lemma.

10.13

Valuation theory can also be used to good effect

in number theory.

Let

a prime ideal in A. (1)

Let Pe

A

be an integral domain and let P

Let

A

A p - {a/bE A: a E A {a/b

E

A: a E P

and

be the field of quotients of and b

E

bEA-PI

A -PI;

be A.

and let

then

Ap, which is

Algebraic Function Fields called the ring

A

localized at P,

contains A, and Pe Pe n A = P . Further (2)

Pe

Indeed, give

- Pe.

Lemma.

Let

p

E

- Pel f = a/b with a,b E A - P; thus Hence Ap - Pe = u (Ap), establishing (2).

A

is a prime ideal in

in

A

be a unique factorization domain and let

Let

A.

Ap

A.

Let

P ZAP;

f e A-A,.

Then

f=a/b,

with

thus

plb.

Hence

a

is not in

P,

a

Since

and so

then

P

A.

is a valuation ring of

having no common irreducible factors.

b e P;

such that

Ap.

denote an irreducible element in

Proof.

Ap

that

f E Ap

Ap

f-l = b/a

A

is a subring of

is a prime ideal in

is the maximal ideal in

195

and

b

fpAp, b/aEAp,

proving the Lemma. We can apply this lemma to the following examples. E x a m p l e 1.

0 - z P (PI numbers.

is

Let

p

be a prime number; then by the Lemma,

is a valuation ring of the field Q1

of rational

It is easy to see that the residue class field of

Z

PI

the field having

p

elements.

easy to see that the value group of E x a m p l e 2.

Let

K

reducible polynomial in

Up

is

Further it is Z.

be a field and let K[x]

.

m(x)

be an ir-

By the lemma,

0 - 0

:K[x] is a valuation ring in the field K(x). m (x) (m(x)1 It is again easy to see the residue class field of 0 is

K-isomorphic to

K[x]/(m(x))

and that the value group is

Z.

That these two examples seem so similar, one drawn from number theory and the other from algebraic function theory, might be superficial; however it is not.

The development and

interrelations between these two fields lies outside the scope

Norman L. Alling

196

of this monograph; however before leaving this fascinating topic in comparative mathematics, let us point out an essential difference between these two topics. It is not difficult to show that given

Example 1 ( c o n t . ) .

any valuation ring number

p

see that

such that 2

Q, then there exists a unique prime

of

I)

o=I)

P'

Further it is not difficult to

is the intersection of all valuation subrings of

Q. Example 2

(cont.)

om 5 K [l/x] (l/x)

.

K[l/x]

is a subring of

is a valuation subring of

valuation ring at

(Note: if

co.

K(x),

I)

E

that is

0=0

(m(x))

*

the

here defined

It is not difficult

o = om

RiemKK(x), then either

there exists an irreducible polynomial m(x)

and

called

om

K = @ , then

and as defined in (10.11:4) are identical.) to show that given any

K(x)

in

K[x]

Clearly the intersection of all

0

E

or

such RiemKK(x)

K. Historical note.

The names of Hansel and Hasse are

associated with early applications of valuation theoretic methods to algebraic number theory.

See Endler's excellent

monograph on valuation theory [21] for additional details. Finally let us consider an example that will prove useful in Part I11 of this monograph. Example

3.

In studying R i e w ( x )

Example 2 above, that we must study maximal ideal of IR[x]

.

Given

m

E

we have seen, in

SpecmIR[xl, the set of

Specm lR[x] ,

it is, of course,

a principal ideal generated by an irreducible monic polynomial m(x) x-r,

in IR[x]. for some

m(x)

is of one of two types:

r E ~ R ; or it is

it is either

2

x + b x + c , with

Thus it is natural to associate SpecmIR[x]

with

b 2 - 4 c < 0.

Algebraic Function Fields

Q;+ E { z

(3)

c c : Im(z) > 03.

Hence it is natural to associate

with c + u

RieW(x)

{a}.

Extensions

10.20

Let

197

be an algebraic function field and let

KcF

be a finite algebraic extension of degree

n.

FcF

(As a general

reference on valuations and extensions see e.g., [ 7 0 , vol.11, Chapter VI].

This theory is not trivial.

Most of the results

in this section will be stated, but not proved.

Proofs may be

found e.g., in [ 7 0 , ~01.111. ) Let (1)

U

0E X

8 be in RiemK $:?

0 : ~ ( 8 )-

and let

8

nF;

then

E Riem F.

K

5

will be said to be under

6

and

extension theorem (see e.g. [16, p . 6 1 )

0. The place has many consequences.

One of them is the following: (2)

given Let

0 E X,

M

there exists

8 2

such that

E

0 and let

be the maximal ideal of

8;

maximal ideal of

then

k

n

0 = M.

~ ( 8 =) 0. be the

Then the following dia-

gram is commutative and row exact:

O ->

2

->

8

->

U / M ->

0

0 ->

M ->

0

->

U / M ->

0

A unique K-isomorphism

g

(3)

of

I

0, into ?J/G

t

*

of

-

U/M,

the residue class field

the residue class field of

fined so that ( 3 ) is commutative.

fication. (4)

Let

g

8,

serve as an identi-

Then

[a/i: O / M l

:f,

is finite.

It is called the relative degree of

-

can be de-

0 over

0.

198

Norman L. Alling

fi

Let

-

U.

(5)

-

0.

valuation associated with group of

0=

U n

0 ->

i j L > F*

0 ->

1 2

-’ GV

F* ->

U->

->

0

.

G - > O

of

Let

h

is of finite index

into

G

can be defined

serve as an identification.

Then

e

be the value

ih 13

making (5) commutative.

G

Let

U.

The following diagram is commutative and row exact:

A unique group isomorphism h

(6)

8 and let 5 be the

be the group of units of

e

in

is called ramification index of

-

G.

8

over

0.

As a corollary of (6) we have the following

(7)

is isomorphic to

2.

then

is finite. We have noted (Example 2 (10.13))

G

is isomorphic to

2.

and let

F - K(x);

x

[F:F]

be transcendental over

K

Indeed, let

that

Using (6) we see that (7) holds.

Let us consider some examples to illustrate these

10.21

phenomena. E x a m p l e 1.

U FlR[x] is IR

and let (XI and that of

of

is 2 .

-

f

Example 2 .

0 E@ [ z ]

(XI

10.22

8

8

F -lR(x)

a :Q: [XI

over

and

EC(x).

Let

-

The residue class field of 0 (XI is @ ; thus the relative degree (10.20:4)

-

K-Q:,

Let

and let

(10.20:6) of

K-IR,

Let

0 EQ: [ z

2

F

I

is

Assume now that

- @ ( Z ) ~

(,2)

-

and let

2 F - @ ( z ).

Let

then the ramification index

2.

-

FcF

is a separable extension.

Algebraic Function Fields

-

(This will - of course

be assured if

K

199

is of characteristic

zero, which it will be in all the central applications in this monograph.)

over

0. For

over

0

1 1 1k'

and let

e

0;

(1)

f

O1,...,Ok

E

-

X

aj

be the relative degree of

j

be the ramification index of

j

be the points

aj

over

... + ekfk =n.

e1f1 +

then

let

-

*

0 E X , let

Given any

Theorem.

(It takes a fair amount of space and care to prove this. A proof may be found as Theorem 20 [ 7 0 , vol.11, pp.60-611.

See

also the remark in the penultimate paragraph of 170, vol.11,

-

P.631 1 There are many applications of this beautiful theorem. Let

E x a m p l e 1.

K -@.

numbers and let L

elle2, and

-U E X-

Let

0.

be over

is

@.

Since

of

6

is also

or 2 points.

Finally

E

E

f =l.

n-l(o)

0 :

(10.11:4). 0

Over each

*-'((I) consists of 1

contains 1 point if and only if

Let

K=IR,

F EIR(x)

0

E

e=l=f.

X-

and let

F :@ (x) (as in

3 x 5 { 0 E X : the residue class field of

Let

nl. ax = m[xI (x-r):r c I R 1

each point

and let

Hence

u

ax there is only one point in

f=2.

C

00.

Example 1, 110.21).

0

u

(x,y), where

Clearly the residue class field of

thus

@;

Example 2.

is

Let

F SQ:

is algebraically closed the residue class field

Q:

u=el,e2,e3, or

0

be distinct complex

F -a: (x) and let

Let

y = 4(x-el) (x-e2)(x-e,).

e3

ax

'&[[l/xl over

1.

For each

0, since n = 2

and

there exist two points, since for

Norman L. Alling

200

The

10.30

d complex a l g e b r a i c f u n c t i o n

Riemann s u r f a c e

field

Let C c F and 10.11)

.

be a complex algebraic function field (10.10

X

Let

5

Riem F

C

ically closed,each 0 E X has C h

F O = Z,

(10.12); thus C

Since C

(10.11:6).

is algebra-

as its residue class field

the Riemann sphere ( ( 1 0 . 1 2 ) and ( 8 . 3 0 ) ) .

has, of course, a topology on it which makes it a compact

surface.

0 E X ->

(1)

Let

fc F

gives rise to a map

?(0) E

C

(10.12:4).

have the weak topology making (1) continuous, for a1 L

X

f E F. (2)

Each

Chevalley [16, p.133 ff.] has proved the following. X

is a compact space.

As remarked before ( 1 0 . 2 0 : 7 ) , thus there exists

2;

fE F

the value group of

such that

0

is

vO(f) = 1.

Chevalley has also shown that (3)

f

is a local uniformizer at

0; thus X

is a compact

surface. Finally Chevalley has shown that the Atlas so defined is analytic ( 8 . 2 0 ) (4)

X

;

thus

becomes a compact Riemann surface X

is a meromorphic function on

X;

and

such that each

f E F ->

2

E

F(X)

is

a surjective @-isomorphism. Remark.

structing X

In the author's opinion,Chevalley's method of confrom C c F

is a great improvement over the "cut

and paste method" described in 58.13.

4 theorem

10.40

Let

C

of

coeguivalence

be the category whose objects are complex alge-

Algebraic Function Fields

201

braic function fieldsand whose morphisms are @-linear isomorphisms of one such object into another.

S

Let

be the category

whose objects are compact (connected) Riemann surfaces and whose morphisms are analytic surjections of one such object onto (See,e.g., 1501 for the revelent definitions in cate-

another.

gory theory.) let

f

Let

@

cF

and let

of generality, we may take @

@ cF

and

cF

of the morphism

f.

Let

into

[F : F].

Riem F

5

X

and

@

n

(Riem f) ( ? ) E

then

Riem f

@

F

Since

C F is a finite

is called the degree

-

,..

X :Riem F ;

c

S.

a

Given

2 n F;

Riem@

Lemma 1.

-

F.

then,

are compact Riemann surfaces:

is an analytic map of

a:

and

Then, without l o s s

and let

each is an object in the category

(1)

C

be objects in

F.

to be a subfield of

n: X

Let

as we noted in 510.30, X

,

F

-

cF

are algebraic function fields

algebraic extension.

i.e.

F

be a @-isomorphism of

@

2 onto

X.

let

E

Hence

is a contravariant function of

c

into

S. Y -> a

Now let S.

object in

Y.

tions on

Let

for details.) onto

g

is an

be a triple of object, morphism, and

F(Y)

be the field of a11 meromorphic func-

Using the fundamental existence theorem on Riemann

surfaces, F(Y),

Y

Y

is a proper extension of

Let

g

E

F(Y)

then

-@;

(2)

and

is an analytic map of

It may be shown that there exists

C.

n - to - 1 map of

Y

onto

Z

Then it is not difficult to show that F(Y)

f

(See e.g. , [ 5 8 1

@.

F(V)

are objects in

F(a) (h) ha,

to show that

for all

h

6

c. F(Y)

nc N

such that

(counting rami€ication). [F(Y):

@

(g)]

=n;

thus

Let

.

Then it is not difficult

202

Norman L. Alling

Theorem of c o e q u i v a l e n c e .

Riem

and

Q:

s

is a contravariant functor of

F

Lemma 2 .

into

c.

The contravariant functions

establish a coequivalence of the categories

F

c

S.

and

(See [5, pp.95-1041 for details.) f is a triple of object, morphism, and Thus if F1 > F2

C;

object in F1

(3)

L.

If

X 2 ->

in

S,

(4)

then

@ F(Rie% F1)

a

X1

F (Riem@f )

> F(Riemc F 2 ) = @ F 2 .

is a triple of object, morphism, and object

then X 2 = Riem F ( X 2 )

Rie%F (a)

Let

Theorem.

x :C/L

X':@/L'.

and

let F '

> Riem F(X1) =

c

c

L

and Let

L'

F

be lattices in

@.

Let

F(L) (or equivalently F ( X ) )

(or equivalently F ( X ' ) ) .

F(L')

xl. and

The following are

equivalent: L

and

L'

are equivalent lattices ( 6 . 2 3 ) .

(ii) X

and

X'

are analytically equivalent.

(iii) F

and

F'

are @-isomorphic.

(i)

(iv) J (L) = J (L') Proof.

.

By Theorem 8.45, (i) and (ii) are equivalent.

Theorem 10.40, (ii) and (iii) are equivalent.

By

By Theorem 9.29,

(ii) and (iv) are equivalent; proving the Theorem.

The Riemann-Roch Theorem

10.50

Let

X

be a compact Riemann surface, and let

the underlying topological space.

Since

X

X

denote

is a compact

orientable surface,it is homeomorphic to a sphere to which

g

Algebraic Function Fields handles have been adjoined. X

called its genus.

g

203

is a topological invariant of

(See e.g., [51] as a general reference on

the topology of surfaces.)

Let

a

be a divisor on

X.

(See

58.60 for a definition.)

(1)

Let

L(a)

then

L(a)

is a complex subspace of

(2)

L(a)

is of finite complex dimension

Riemann's

(3)

L(a)1'

(4)

Let

{f

E

F(X): (f) >a]; F(X)

. L(a).

Theorem

- g - deg(a) .

i(a),

the index - of specialty of

be the non-negative integer

a,

be defined to

-1 + L (a) + deg (a) + g;

then we have

the following. The Riemann-Roch

Theorem.

For all

a

E

divX,

the follow-

ing holds :

(5)

O=l-.l(a) -deg(a) -g+i(a). The Serre D u a l i t y Theorem.

For all

aEdivX

c U (X) (-a), where

(6)

i (a) = dim

(7)

U(X) (-a) :(6

(Note: D(X)

E

D(X):

(6) > -a}.

was defined in 1 8 . 2 3 . )

The literature on the Riemann-Roch and the Serre Duality Theorems is extensive.

Nevertheless, Gunning's excellent

Lectures 0" Riemann surface [29] would serve to give complete proofs of these beautiful and important results.

10.51

The Riemann-Roch theorem can be stated and proved

for any algebraic function field 113.)

KcF.

In so doing a non-negative integer

the genus of

KcF.

(See e . g . , g

[16, Chapter

emerges, called

One of the great accomplishments of contem-

204

Norman L. Alling

porary algebraic geometry is that it has been able to carry over much algebraic and analytic geometry, that was first discovered and investigated over

@,

to arbitrary ground fields.