Elliptic Curves: Notes from Postgraduate Lectures Given in Lausanne 1971 72 (Lecture Notes in Mathematics)

~

cP........

f+ =

div(4=»

~

0 #0 div(f+)

~

-£

establishes an isomorphism (of inverse

f9 ) between the d-dimensional space of holomorphic theta

functions of type (h,a) (Th.(2.10)) and 26

L(~).

- 1.27 -

Let us now indicate how the theta functions are bound with the cohomology of the elliptic curve E, and in particular with the holomorphic line bundles on E. To any L-theta function 9 we associate a holomorphic line bundle F

~ E on E as follows. Let (h,a) be the type of S. We define S a holomorphic action on the trivial line bundle on E = [ , compatible

with the translations in the base by ( z + CAl, a (w) e 7[ h (w) (z +

(z , v)6)

The quotient

E ><

of

[

--+-

J w) v)

•

'E (first proj ection) under this action of

L gives the holomorphic line bundle Fe

~

E . By definition, we could

as well write F(h,a) instead of FS . It is obvious that z

~

(z,S(z))

defines an equivariant section of the trivial line bundle on 1(, hence can be identified with a section of the quotient Fe

~

E . (We say

section instead of meromorphic section.) If S' is another section of F

~ E, S' can be identified with an equivariant section of the S trivial line bundle on E, hence with a theta function of type (h,a).

Thus e'le is elliptic and S' = fS (with an elliptic function f). Thus FS = Fe'

· Moreover, if 8 is a trivial theta function, Fe --+- E

has a holomorphic never vanishing section 8, so that it is holomorphically trivial (hence the denomination of trivial theta function). If 8 and 8' are two theta functions with same divisor, their quotient must be a trivial theta function, so that Fe and F , are holomorphi8 cally equivalent line bundles on E. Again, we might as well note Fd

instead of FS ' and summing up, we see that Fd depends only on the class of the divisor ~ mod P(E). If we assume the basic existence (by no means trivial) theorem of existence of meromorphic sections for all (holomorphic) line bundles over a Riemann surface, it is easy to see that

d

~

Fd defines a bijection between Div(E)/P(E) and the 27

- 1.28 -

set of isomorphism classes of (holomorphic) line bundles over E. This set is naturally a group for the tensor product (corresponding to the usual product of cross sections 9) so that

Fd is a group isomorphism (the line bundles are often called invertible bundles ~ ~

for the reason that among the ho10morphic vector bundles with finite dimensional fibers, they are precisely those which are invertible for the tensor product). Exp1icite1y, the inverse bundle of F(h,a) is F(-h,l/a) as follows from the formula giving the type of 1/9 in function of the type of 9. The theorem of Appe1-Goursat (2.10) may be restated as follows : The holomorphic sections of the line bundle Fd ~ E , when deg(~) ~ 1, make up a complex space of dimension equal to deg(~). (When deg(~) = 0, this dimension is not always 0 indeed, the constants are sections of the trivial line bundle, and so we see that this dimension is one when d is principal. The correction to bring to the formula giving this dimension for small degrees is due to Roch. Riemann only gave the formula for divisors of degree sufficiently high, with an exact bound. With this in view, the Corollary (2.12) should be called Riemann's instead of RiemannRoch' s!) We come finally to the cohomological interpretation of the theta functions and classes of divisors, by introducing the structure sheaf

0

of E (sheaf of germs of holomorphic functions on E), and its

subsheaf of units

17~ (sheaf of germs of nowhere vanishing holomor-

phic functions). Let also

~

denote as usual the normalized exponential

(of period 1) considered as sheaf homomorphism kernel is the constant sheaf

~,

e:

0

~ ~~

.

and we have an exact sequence of

sheaves

28

Its

- 1.29 -

This exact sequence gives rise to the exact cohomology sequence HO (E, II

a:

({)) --+-Ho (E, eJ1--.H I (E ,I) --....H l (E, 0) -.H l (E ,(t'M) -+-H 2 (E ,ll)--..O e U,.."

--=--..

Because e : a:

,y

a: ~

~ a:~

is surjective, we may cut the beginning of the

above sequence and get the shorter exact sequence (2.13) We interpret each term. By Serre's duality HI(E,fP) is complex vector space

HO(E,.i): the

J1l(E) of global holomorphic differentials on E.

These differentials are all proportional to any

non-~ero

of them

(and all invariant under translations, as we shall see later) so that HI(E, 0) ~ Then Hl(E,I) ,.., is a lattice in HI(E,

([·CJ

(!))

~

a: •

(the lattice of "integral"

holomorphic differential forms) which can be identified with L

~

a:

under the preceding isomorphism. The sequence (2.13) starts

0 ~ L ~ a: ~ •••• The last term H2 (E,1l) is easily identified with H2 (L,Z) (group cohomology of L with trivial thus as follows:

action on

,.",

Z because L has no torsion, acts freely on the contractible

space a: with quotient precisely E : Hurewitz Theorem). This group can also be interpreted as space of

Z-bilinear alternate integral forms

on L. These forms are all proportional to the canonical one B : H 2 (L,Z)

~ Z.B· ~

7l

The space HI(E,~~) is the most interesting one. It can be computed by taking a covering of E with open discs U. in ([ (mod L), each

U~

being sufficiently small to avoid pairs z, z+ 4J (w E L - (OJ). Since all the Uo( and their finite intersections

U

~ ... ~

=

n .c, U

t,it'

are

contractible (by convexity), hence cohomologically trivial, the sheaf cohomology HI(E,~~) is already the (sheaf-)~ech cohomology of the nerve of this covering (with variable coefficients in the groups
- 1.30 -

5

on the couple of indices

A l-cocycle

so(fS E: (?><cu..p )

CoC, (3) ~

:

(d,P)

In other words,

(o(,p Soft

~s =

-1

'1) ~ s~1 so(l so(~ =

= se(t1"

s(31

f ¢

of the open intersecting discs U« .

is a l-cochain s such that ~s

when U"'~

on

0, i.e.

1 on

u"Pt = u.nu,..nu, ·

(when this triple intersection

uo(~y

is non-empty). This is a compatibility condition on the transition functions

s.~

defined on the non-empty

.

u~~

A l-cocycle is said to

be a l-coboundary, or more simply to be trivial, when there exists a map

G":

Cl( I--+-

cr(aI)€t1>CCU",) with

5

=~r

C"',13)

:

1---+

rC~)/cr(",) on U

olp

To each l-cocycle s one can associate a holomorphic line bundle Fs

~

E on E as follows. We glue together holomorphically the

a:

trivial line bundles on the Uo( , Pee : U-e x

--+-

U-<

by means of

the slip

::[

(z, v)

(z,s (z)v) 0«(3

u )(: ([

~o(l

Uf!> {\ Uo( Th~

_

Uo(

compatibility conditions on the

s~~

[

n U~ (cocycle condition) ensure

the coherence of these patchings. The line bundle Fs will precisely be holomorphically trivial when s is trivial (existence of a global holomorphic, nowhere vanishing cross-section), so we have an isomorphism

Hl(E,~) ~ s

~

group of holomorphic line bundles /E (Fs~E)

We have already indicated the fact that this group is also isomorphic to the group of classes of divisors Div(E)/P(E) . The co boundary homomorphism

30

- 1.31 -

is given on the line bundles precisely by ~

: F(h,a) ~

(1/2i)ah

d·B

(where d is the degree of any theta function of type (h,a), or equivalentl~ the

degree of any meromorphic section of

F(h,a)~E

).

This justifies in some sense the terminology adopted for "coboundary ah" of

the homomorphism h : L

[ . (More precisely,

~

we should have defined the coboundary of h by C) h (4)1

' lcJz)

=

( 1I

Cc1 h ( w2 ) )

2i) (h (WI ) 4)2 -

If we identify 7l·B with 7l , the coboundary

0 gives what is known

c : Hl(E,~~) ~

under the name of Chern character

class c(F) of a holomorphic line bundle

, but... ! )

F~E

7l : the Chern

is computed (in theory

at least) by taking the degree of the divisor of any meromorphic section of

F~E

. With these interpretations, the exact sequence

(2.13) gives the (isomorphic) exact sequence (2.13)'

o

---+

L -+- [

--+-

deg group of 1ine bundles IE --... Z '--

~

--+-

0

./

or HI (E ,~), or Di v (E) IP (E) The subgroup of H1(E,eJJ consisting of line bundles of Chern class zero (or the subgroup of classes of divisors of degree zero) is thus isomorphic to E = [/L . This is the previously defined (1.24) isomorphism. These line bundles are also called flat line bundles on E and are precisely those which are topologically trivial. So we restate the definition : Pic (E)

"connected component" of Hl(E,~>c) group of holomorphic line bundles over E which are topologically trivial (trivializable!) .

It is important and interesting to note that Hl(E,~~ is also the group of invertible sheaves over E (~odules or O-modules, locally free of rank one), by associating to the bundle

F~E

the sheaf

of germs of its holomorphic cross-sections. The point is that this 31

- 1.32 -

sheaf can be defined without speaking of the line bundle (a real advantage in characteristic, 0

!) as follows. If s EHl(E,~~) is

a coherent system of transition functions, the sheaf by

~s(U)

s is defined

~

= r(U,fs ) = group (or ring) of holomorphic functions on U.

The restrictions are twisted as follows:

f

---+

s~oc

·restr(f) to Uoc(3

Equivalent formulations are, when starting with a divisor d = Ldk(a ), k to put { f:U-..([

d

fez) = (z-a k ) kh(i) in neighbd.

of a k if in U, wi th h hOlomorphiC} , and when starting with a theta function 9 of type (h,a), to put ta(U) = {f:U+L~[ holomorphic, satisfying

f(z+~)=a(~)enh(~)(z+l~)f(z) for Z€U+L} • The sheaf restrictions in these two definitions are the usual restrictions of functions. (2.14) Complement 1. We show why it is sufficient to consider multiplicators of the form Let indeed

F

~

B(W))

exp(A(~)z +

when treating theta functions.

E be any holomorphic line bundle over the elliptic

curve E = ([/L. We use again the basic existence theorem of Riemann surface theory which gives the existence of a meromorphic section s of this bundle. We choose a lifting of this section to the trivial line bundle over the universal covering has the property that (s' Is) (z

+ IU)

-

E = ([ of E. This section 5

(s' Is) (z) is holomorphic

everywhere (entire function) in z for a fixed element w of L. But ~'/s is a section of the canonical (cotangent) line bundle over E.

As the tangent, as well as the cotangent line bundles over E, are trivial, this proves that (s'/s)(z

+

32

~)

-

(s'/s)(z) is constant in z,

- 1.33 -

hence (5' Is)' is an elliptic function. This is what we wanted to prove. (2.15) Complement 2.

It is possible to give more explicitely than

has been done, the possible types of theta functions. Let 9 be a theta function of type (h,a). We have seen that a) h : L

~ ~

is an additive homomorphism,

= 2id·B where d

b) ~h

degdiv(9) ,

and a simple computation shows that a : L ~ ~~satisfies c) a (t.o + w') = (-1) dB(W ,t.t)') a (w) a ( W r) for t..), lAJ' E: L. In particular, fal

: L -+-

TIt.:

is a homomorphism, and this is what

was needed to decompose B = Bo·B red as after (Z.7). As the proof of (2.8) shows, we can write h(z) = (d/S)z + «·z with a complex constant ~.

Moreover, if

win ~L (for

f:

L

new)

denotes the biggest positive integer n such that

tV £L), we see that (_l)dafl(W) a(tJ)

~ [~

~ t - + Re(~tU)

is a homomorphism

the absolute value of which must be of the form for a suitable complex constant

~.

Summing up, the three

above properties a),b), and c) imply that

z + o(·z

h(z)

(diS)

a (c.))

(-1) d-n (c.)) eRe (~c.)) (w)

X

with complex constants

(U = [1 = unit

oC,

(5, and a unitary character ~: L~ V

circle in [). Conversely, all such functions satisfy

the three conditions a),b),c) and hence form the type of a theta function B (with divisor of degree d). (Z .16) In Jacobi t s theory, the variable q

plays as important

.a

= e 1Ci"t where -c =WZllU l

role as ~ itself. If we put t = q2, the

normalized exponential! gives an isomorphism ~

: [I

Z

"-".

[~,

whence also

~ : [I

1l + Z

--r """"'.

[)C I t'll

,

where t~ denotes the multiplicative subgroup of [~ generated by t 33

- 1.34 -

t

Obviously» 1m(1:) > 0

71

=

~t n

: nE

1

~

is equivalent to Itl
defined an elliptic curve as being a quotient of the multiplicative group

Itt < 1

[~

by a discrete subgroup generated by an element t of modulus

(or Itl>l, but not

It(

= 1).

We shall see that this point of

view is more suited to the study of p-adic elliptic curves.

34

- 1.35 3. Variation of the elliptic curve and modular forms

To compare different elliptic curves, we consider two lattices Land L' in [. Then we put E

= [/L , E' = [/L' and compare them by

means of holomorphic maps.

f:

(3.1) Proposition. Let

E ---+- E' be any holomorphic map. If

Cf

is not constant, it is induced by an affine linear transformation Z

...-.. 0( Z

C

: E

Il

+ (3

--+-

z ..-.. z +

and is surj ective. In particular, if we define

E' Q.r Z ....-.

f3

(cons tant map) and T~

~ (translation by

0/ is a homomorphism E ~

p in E'), then E' · (Ii f is not

ep

=

t

= Tp

+ C(3

fez

+ w) -

fez)

0

t

and

constant, it is surjective~

Proof: We choose a lifting, to the universal coverings and note that

.!?L

E' ---... E'

E = E'= [ ,

must be an element of L' for every

W€L. By continuity in z of this expression, we see that this diffe-

rence

is independent of z and by derivation,

L-elliptic function. By (l.l.a) we see that consequently ~(z) Obviously

0(

=

«z +

~

f'

must be an entire

~'must

be constant, and

with two complex constants

must send L into L'

O(·L eL'. When

0<

«and ~ .

f 0, this is a

serious restriction on Land L'. All assertions of the proposition follow from that. (3.2) Corollary. and such that

Ii

is a holomorphic map E

~

~

E' , is injective

0, then it is a group isomorphism, and the

~(O)

lattices L , L' are homothetic. Another proof of this corollary would be as follows. Let a', b' and c'= a'+b'€E'. By (1.22) the divisor (O)+(c')-(a')-(b') is principal, hence of the form div(f') for a function f' on E'. Let f

=

f'of . Its divisor is div(f)

=

(0) + (c) - (a) - (b)

35

- 1.36 -

= ,-l(d), •••. This principal divisor must also

where we have put a

satisfy Abel's condition (apply (I.Z2.d) again), which gives c

=a

+

b. This proves that

f

,-1 is a homomorphism, hence

is an

isomorphism (and is biholomorphic). For

1:

the notation

in the upper half-plane H = {1:' ([: Im(t') >' for the lattice

L~

has a direct basis

7l

L~

+

oj

we introduce

7l·'t'C([ • Every lattice L

homothetic to (at least) one Lt' I ,I4)Z hence is (take 1: = ~Z/ t..)1) • But if we choose a direct basis such that 'flJll is W

minimum among the modules of the non-zero elements then

WE:L

- {OJ , and

CA)Z of module minimum among the elements of L - 7lo41 (so as to

have a direct basis), then we can say much more on by construction, This leads to

J-rJ )

\1: ±

I};;'

(x

imaginary parts,

\xJ

means

= J Rett'»)

Jwz

1. Then also

11:1

and if

~ 1) 2

'1·

+

t

~IJ ~

IWzi

~/~ = ~

by choice of

•

First, <..)2.

1: = x + iy is decomposed in real and

x2

yZ ~

+

Replacing £.)2 by

y2, hence

Wz - tA\

Jx

t

I) ~

Jxl.

This

if necessary, we may

assume (3.3) Again, replacing Re(1:)

'0

(~l' ~Z)

when '--=1 = 1

by

(~2'-~)

if necessary, we may assume

(cf. picture below). We denote by D the subset of the upper half-plane H defined by (3.3) and Re (-r)

H

'0

when 11:1 = 1. Then the

lattices L1:' for

"t'6D are a system

of representants for the classes of homothetic lattices in ([ . It can be checked directly that -f

-.t 2

o

i•

L~

has only

two direct bases satisfying the above conditions when namely

36

(l,~)

and

~~ ~(t/3),i

(-1,-~).

In the

,

- 1.37 case of the square lattice

~ =

i, four direct bases of that kind are

available: (l,i), (i,-l), (-l,-i) and (-i,l). Similarly in the case 1:

= ~(1/3) =

S

(primitive 3~ root of 1), there are six possible

choices of such bases (hexagonal lattice). When we refer to the fundamental region D defined by (3.3), we always implicitely assume that

Re('t)~O

11:1= 1, as indicated just after (3.3).

when

Most functions encountered in Weierstrass' theory (first section) had homogeneity properties in the lattice. For example ~(~Z:AL)

= A,-2 P(Z:L) =

G

2k (A L)

1\-2k 1\

G

2k

and

(L)

= A-12~(L)

IJA ("L)

We introduce slightly different notations. When ~(z:~)

=

0, we put

P(z:LT )

G2k ( ~) = G2k (L~ ) ,

always with Lor =

Im(~)~

A (1: ) = ~ ( L~ )

7l + 7l1: cO: • Since

~

,

never vanishes, we can

define the homogeneous function of degree zero

J(~) = g~(T)/a(~) is homogeneous of degree -4 in the lattice, and we 4 look at this function as function in Im(t') > 0). It is important to (remember g2

60G

note that by uniform convergence of G (T) Zk

for 1:

Im(~)~

£

> 0,

=

L'(m-r + n) -Zk

(k

> 1)

we can evaluate the limit of this expression for

= x + iy (x fixed)

, and y

~ 00,

by taking the limit of each

term in the summation. All terms with m ; 0 tend to zero, so that n -2k

2 ~(2k)

These constants can be computed explicitely (by comparing two expansions of cotg(z), and they can also be linked with the Bernoulli numbers: in particular, they can be seen to be

37

~

0). The homogeneity

- 1.38 -

properties of these functions as function of the lattice give very remarkable properties in the variable (3.4) Proposition. Let

~,~/€H

't'EH

= {z 6(1;

Im(z)

:

7

oj.

and LT,L v be the corresponding

lattices. Then a) L"t' = L"t: if and only if b) L1: / ''''' L-r if and onl y i with a matrix

c(

or + n wi th an integer n ,

t"'

(a 1:

f~'

= (~

~)

E: SL

2

b) / (c 1:

+

(~)

+

d) = 0(' (or)

(integral coefficients

and determinant +1). In this case, the factor of proportionality is

C1: +

d :

(CL +

d)L1:'

= L~

Proof. The first part is easy. For the second, we take any AE[~ such that ALTJ

= L~

· Since

and A must belong to Lt we can write

aT + b

{ wi th 0< = ( ca

A~'

C1:: +

d

shows that there exists another integral matrix of the form

=

~-l A L-r

~.~

and

db) ' an integral matrix. But conversely L1;"'

(1 0

n) 1

~

with

«o~

(n an integer) by the first part. In particular,

must be an invertible integer. This proves det(<<) = tl .

det(~)

A simple calculation shows that

Im(-r') hence det(oc)

=

>0

(ad - bc) 'c~

+

by hypothesis

dl- z lm(1:)

=

det(o<) lct:

+

dl-zlm(t:)

H. Summing up, O(E:SL (Zl) as Z asserted. If conversely 1:' = O«T) with «E:SL ( 7l), the above computaZ tions show also that (C1: + d)L't'1 = L1;: q.e.d. 'r,~' €

(3.5) Corollary. The holomorphic functions

for any

0(=

(ca

b)

G2k ( ex ('1:'))

= ( c 1:

GZk(~)

satisfy

+ d) 2kG 2k ( 1:) , ( k >' l)

d e:SLZ(Zl), and tend to a finite limit when

1:

tends

to infinity on the imaginary axis. This is just a reformulation of the homogeneity properties, in view of the proposition.

38

- 1.39 -

We introduce the following notation : Jacobian of (t'f---.+oO«('t')) at

J (oC , z.)

Thus, if

0(

= (ca ~), J (0<, "t')

(

a1:' +

C'r +

b)' _

d

(ac1:'

+

ad -

~ =7.

ac~

-

,

(DCESL ( 7l)). 2

cb)/(c~ +

d)

2

(C1: + d)-2

hope that this function will never be confused with the modular

~e

invariant J

= g~/ f:.

,especially since they do not depend on the

same variables.) The corollary can be reformulated as

for

't= ~

Hand

0(

E:

J(~(~))

When

0(

't~ ~ +

=

-k

G2k (T) SL 2 ( 7l). Also for example

G2k(<«~))

J(<<,~)

= J(L)

is of the form (~

(k

> 1)

(by homogeneity of degree 0).

~) (nE:Z), it gives an integral translation

n in H, having Jacobian 1, hence all the preceding functions

are periodic of period one (even when the degree of homogeneity is

1

0) and thus can be expanded in a Fourier series of the form

~

L- ane

Z1Cin~

-00

Since the absolute value of e 2'1Cin't

for

t: =

iy purely imaginary, is

e -Z1t'ny , we see that the G

can have no terms with index n negative Zk (they are~ounded when ~ tends to ~ on the imaginary axis). The

same happens for

~(~),

and as consequence,

J(~)

will have only

finitely many non-zero Fourier coefficients of negative index n. The behavior of any holomorphic function with a Fourier expansion (fini tely many an 1 0 for n < 0) for

of the form

z = iy purely imaginary and y with absolute value

land

~ ~

e2~dY

is given by the first term,

if no

= -Inol is negative, and

is the index of the first non-vanishing Fourier coefficient. These properties lead us to a general definition of modular forms. (3.6) Definition. Let k be any integer. A (meromorphic) modular form of weight k is a meromorphic function f on the upper half39

- 1.40 -

plane H, satisfying : J(o<,z) -k fez)

for OCE:SL Z( 7l) (whenever one member is defined!)

a) f(OCCz))

b)

/f(

Z

)1-<

/\y for some (positive) real constant /\,

when y = Im(z)

~

00

•

The second property means that there exists a constant C';)o 0 such that \f(

z)1 <: ceAy

for all y greater than some large yo(depending on

A,

c

and of course f). This condition ensures that the Fourier expansion of f (which exists by a) ) has finitely many non-zero coefficients of negative index. The smallest index n corresponding to a non-zero Fourier coefficient, is called the order of f at infinity, and denoted by ordoo(f) = neo . The orders of f at all points P E: D (fundamental region of H defined by (3.3) ) or at all points P E: H are defined as usual by considering the Laurent (or Taylor) expansion of f around P. ~

For convenience, we put D = DU{oo} , so that we can speak of ordp(f) ~

when PE.D , and f ;

o.

The extent to which our examples GZk exhaust the possibilities of holomorphic modular forms is tackled with a detailed study of the poles and zeros of general modular forms. We have indeed (3.7) Proposition. Let f f 0 be a modular form of weight k. Then ordoo(f) where

~

=

~(1/3)

+

iordi(f)

+

tord~(f)

+

L.ordp(f) = k/6

(primitive third root of 1) and the summation is

extended over the points P E D - {i ,~J

(P f i, t) .

This proposition suggests strongly that we should adopt a new definition for the orders as follows : viCf)

= }ordiCf),

v~Cf)

= ord CCf)/3 and vpCf) = ordpCf)

for all other PE~. The formula could indeed be written then more simply

40

- 1.41 -

As for theta functions, we could show that the modular forms come from meromorphic cross-sections of suitable line bundles over (space of orbits r(z), with t~

r= SL Z( 7l)

r\H

compactified by adding

)

one point at infinity. These line bundles would be the

~-powers

of the canonical one (cotangent bundle), so that a modular form of

u(-J.

weight k comes from a k-differential on r\H

Proposition (3.7)

expresses the degree of their divisors. Proof of (3.7): By hypothesis, n. = ord.(f) zero nor pole when Im(z»

> - co

,

so that f has no

Rand R is large enough. Let DR be the

intersection of the fundamental domain D with is to integrate f'lf on the boundary

~DR

Im(z)~

R. The idea

of DR. In case f has zeros

or poles on the sides, f'lf will have simple poles on the sides, and the usual modification of the contour has to be made. By hypothesis f'lf has no pole on the top horizontal part of

~DR

and if the modified

contour is as shown on the picture (we suppose there that only one pole occurs on the vertical sides, and only one pole occurs on the portion Jzl = 1, -l
(in this direction), we use the

Fourier expansion of f (certainly valid for Im(z)

~

R, because this

region contains no pole of f): fez) = a

n.,

eZ~in_z(l

+

~(e2~iz)),

so that f' (z)/f(z)

G

o

or

If' (z)/f(z)

'/2.

.

27rlnoo + (!J( e

for 41

Im(z)~ ~

· 2 7t:ln 00

27Ciz ) ,

I= 0 ( e - 21r I m( z ) )

. This shows that

- 1.42 -

j[

(f'/f)dz

M

= -2nin~

+

O(e- 2xR )

~ -2xi~

R~~

On the semi-circle (of radius t >0 sufficiently small so as to enclose no pole of f'/f other than i), we get

r

JDE

(f' /f)dz _

-} (21tiord (f))

for

i

£~O.

The arcs BC and PG can be treated similarly and give 2 + (f' I f) (z) dz --+ --627t'i· ord (f) when lBC JpG ~

(r

r )

£,.... 0.

On CD, we suppose that the semi-circle chosen to avoid the pole of f'/f is the transform of that on EF (one of them having radius !), under the mapping z

~

-liz (this mapping transforms EF in DC).

We compute the integral over EF by making the change of variable ~ = -liz, so that z ~~

(f'/f)(z)dz

applies EF --+- DC. Thus (f'/f)(-I/{)d(-I/~) =

(f'/f)(-l/l)t-2d~ = (g'/g)(')d~ with

g(~) =

f(-1/0

= f((~ -~)~) = Z;2kf(~).

because f is supposed

to be a modular form of weight k. Hence (g' I g) (() d~

= [2k/~

+

(f' I f)

(l~d~

Now we gather the two integrals over CD and EF of f'/f :

(r

JCD

1

+

EF

)Cf' /f)Cz)dz

=

(r J

CD

+

r )Cf' /f)dz JrDC 2kd;~ +

i DC

Only the integral of 2k dl over DC remains to be computed, or at 1; least its limit when £~O. For the limit, we can 'rectify" the small semi-circle (of radius £) to put CD on Izi

=

I (this does not change

the value of the integral a bit !) and integrate from i to ~ . This is only the variation of the argument between i and' hence

(r

JeD

+

r ) (f' /f)dz

JEF

-

2k(2n:i/12)

21ri'~

The residue theorem gives on the other hand, the global result ;

(f' /f)dz

=

21[i

L

ordp(f) 42

- 1.43 -

with a summation over the poles P of f'lf enclosed by the contour, i.e. all the poles of f'lf in D, distinct from i and

~.

The compari-

son between the explicit computation and the global result gives the announced formula q.e.d. In the course of the proof, we have used the following result of function theory. Let g (g = f'lf in our case) be an analytic function (in a neighborhood of the origin) having a simple pole at the origin, and let CI ' Cz be two continuously differentiable arcs starting at the origin, having respective tangents d l ' d Z with angle ~. Then the integral of g(z)dz on circles of radii t between CI ' Cz and on circles of radii t between d l ' d Z (both in the positive direction if 0( is positive) have same limit

We are mainly interested in holomorphic modular

forms, so

we define Mk to be the complex vector space of holomorphic modular forms f of weight k with ordoo(f) ":1 0 (thus all vp(f) for p~1} are ~O). We are going to show that all M are finite dimensional, and defining k dk dim[(M k ), we shall show that we have the explicit formula L. d Tk (1 - TZ) -1 (1 - T3) -1 E:. 7l[T]] k k This formal series is called the Poincare series of the graded module M =

ek Mk

·

(3.8) Consequences of (3.7). Let fc:::Mo . Since the constants are in Mo ' f - f(i) will be in Mo . By construction, this function has a zero in i, so that the formula given by the proposition cannot hold. This proves f - f(i)

=

0, hence Mo

= ~

is of dimension 1.

Now if f, fE:M k have all same orders ordp(f) = ordp(r) (PE))), their quotient flf will be in Mo ' hence constant. This shows that a function of M is determined up to a multiplicative constant by its k ~

orders at the P6i}. Obviously Mk = W!when the integer k is negative. 43

- 1.44 -

For k

= 1, there is no possibility to satisfy (3.7) with integral

orders, so that M = {Ol, d = l l lity in (3.7):

ord~(f)

O.

For k = 2, there is only one possibi-

= 1, all other orders being o. This proves

= 1, and all functions in M2 vanish at 4. We know that g2 or G4 is Z a holomorphic modular form of weight two, so that M2 = [g2 = [G 4 and

d

g2 vanishes at { with order one (simple zero)! Similarly for k

1 all other

we have only one possibility to satisfy (3.7) : ord. (f) 1. orders being zero. Hence d

= 1, M3 = [g3 = [G 6 and g3 (or

3

= i.

one simple zero located at z on~

Ag~in

for k

=

3,

G ) has only 6

4, there is only

possibility to satisfy (3.7): ord{(f) = 2, all other orders being 2

so that d 4 1, M4 = [g2 = [G S and in particular, GS ' propor2 tional to G ,has a double zero at ~ (and no other zero). In full: 4 ~ero,

L.'(mz

=

+ n) - S

c(

r:: (m z I

4 + n) - ) 2

( 1m ( z)

> 0) ,

a dream of youth, realized up to a constant (which can be determined explicitely). Finally there is only one possibility to satisfy (3.7) when k

= 5: ordi(f) = ordt(f) = 1 so that d S = 1 and MS = [g2 g 3 = [G 10 .

For larger k's, there are several possibilities. Since vanishes in the upper half-plane, necessarily ord00 ~) 3

tions

~€M6

=

never

1. Other func-

2

in M are g2 and g3. We can quote £ E:M with any prescribed 6 6 simple zero P ; i, ~ in D • It is sufficient to construct Z 332 3 2 g3(P)·gZ - g2(P)·g3 ag 2 + bg 3 3

2

The space M6 is thus of dimension d 6 = 2, spanned by g2 and g3 . But more generall~ we have (3.9) Lemma. Let £:M f

k

~

1-+

[ be the linear map defined by =

~(f)

lim f(iy) y+eo

and 1:1'- : Mk ---... M + be defined by multiplication by the function 6. k 6 Then we have an exact sequence of complex vect~spaces

o

A-

£

--+- Mk

~

M +6 ~ [

and in particular d k + 6

= dk

+

k

1, for k

-. 0 , ~O

•

- 1.45 -

Proof: Because

0, it follows that multiplication by

~;

= Ker(£)CM k +6

tive map. Put M~+6 the composite f

=

£(f)

o

. Since ordoocA)

=

~

is an injec-

1, we see that

{6·) is zero. Conversely, if fE:M k +6 is such that

0, i.e. f vanishes at infinity, it can be divided by

~

(this

function will have the same orders at all P E:D and order one less at 00). Finally, GZ (k+6) does not vanish at infinity (we could also take g;g~ of weight Zoe

3[1 , with

+

C(

and ~ chosen so that ZO( + 3(3 = k+6,

possible since Z and 3 are relatively prime). The exactness gives d k - d k + 6 + 1 = O. From this lemma, and the explicit description of the M and d k

for

0

k

(k < 6, we can check that (1 -

r 2 ) (1

-

r 3)

LdkT

k

k~O

=1

We can also say the essentially equivalent result

1

[k/6 (integral part of k/6) if k51 mod 6 dk

= dima:(M k ) =

l(k/6] +

1 if k ~ 1 mod 6

Here, since these formulas are true for 0 ~ k < 6, they will be true by induction for any k because d k +6 = d k + 1 · We can state the general result. (3.10) Theorem 1. Let M

= E9 Mk be the

~raded algebra, sum of the

spaces M of holomorphic modular forms of weight k (bounded when

k

z

=

iy and y ~ ~). Let on the other hand

a:(x,y] be the algebra over

0:, of polynomials in two indeterminates X,Y with the degree d defined

by the two conditions d(X)

ep : (t[X, YJ is an isomorphism of

= Z and

dey)

= 3. Then

--+ M defined by X ~raded

modular form, bounded when z

t-+

&Z and Y ~ g3

algebras. In other words, every holomorphic

= iy is purely imaginary and

y

~~

is a polynomial in gz and g3 (or a polynomial in G4 and G6 ). Here the degree d has been defined ad hoc, but if a:

[x' ,y ~

is

a polynomiahalgebra in two indeterminates, and if we look at the 4S

- 1.46 -

polynomials subalgebra generated by X

=

X' 2 and y

=

will inheri t the degree d from the natural degree of case, d(Xnym) = 2n

+

y' 3, this subalgebra

a: [X' ,Y ,].

In any

3m by definition. Another way of saying the same

thing would be to consider the graded algebras [[X], d' being defined as the double of the usual degree, and

a: [y] ,

d" being defined as the

triple of the usual degree, and then [[X,Y] = [[X]0 [[Y]with the degree d = d'~ d". The Poincare series of ([ [X] ,d') is obviously 1 + T2 + r 4 + r 6 + = 1/(1 - r 2) because this algebra has one generator Xk in degree d' = 2k and no non-zero element of odd degree. Similarly the Poincare series of (a: [Y] ,d tt ) is (1 - T3 ) -1. I t is then well known (and easy to prove) that the Poicare series of the tensor product of two graded algebras is the product of their Poincare series, so that the Poincare series of ([ [X, Y] ,d) is peT)

= (1 -

T 2 )1(l - T 3 f1.

This proves the theorem completely, but we give the classical proof as well. Proof of Theorem 1. The elements of M for 0 ~ k < 6 have been checked k to be polynomials in gz and g3 Since ~ = g~ - 27g~ is also a polynomial in g2 ' and g3 ' induction applies by the lemma, and shows that any f €M k (any integer k) is a polynomial in g2 and g3 ' so ~ is surjective. We have to emphasize the fact that g2 and g3 satisfy no polynomial relation P(g2,g3) = 0 with 0 t: PE:[[X,Y]. Decomposing P into homogeneous components (for the degree d) shows that we may assume P d-homogeneous to start with. If we had \ ij Lc. ·g2 g 3 = 0 2i+3j=a 1J we could solve e.g. for g2 and get \ , i j L- c .. g2 g 3 j t: 0 1J

(after suitable division by g~g~)

which is impossible, because all monomials in the right hand side contain a positive power of &3 vanishing at z = i, and the left member 46

- 1.47 vanishes only at z = ~ (wi th order i o ). This concludes the proof. Now we call modular function a (meromorphic) modular form of weight

o.

Our example is the modular invariant J

ord.(A) = 1, we have

ord~(J)

3

= g2/~

. Because

= -1, but J is holomorphic in H.

(3.11) Theorem 2. The modular invariant J defines a holomorphic bijection of the fundamental domain D defined in (3.3)(and after) onto the complex line J : D ~~ [ , which is conformal except at z = i (ramification index 2) and at z This function is real on the sides of with the normalizing values

J(~)

=

~ D

(ramification index 3).

and on the imaginary axis,

0, J(i)

= 1, lim J(iy) = ~ . y+oo

Finall~

every modular function is a rational function of J. A

Proof: J has a simple pole at P = 00 E: D, so that the same will be true of any of the functions J - A

any complex value). In the formula

(~

of (3.7), only one compensation is possible, namely vp(J - A) for one (and only one) point P€D. According to

A the

1

following

possibilities occur : ordp(J

- A)

1 with p ~ i, {

A= J(i)

ordi(J -

~)

2 if

-

~)

3 if A= J (~)

ord~(J

(because &3 (i)

0)

0 (because &2({)

0)

1

This proves that J takes once and only once each complex value in the fundamental region

D

(not counting multiplicities). The second part

of the theorem follows from

- = G (z) - = Lr ' (mz- + n) - 2=kG- -(z) G2k (-z) 2k 2k hence J(-z) = J(z). In particular J(z) will be real for -z for z purely imaginary. When Re(z)

=

1 , we have

-z

= z, i.e.

= z - 1 , so that

the periodicity of J shows that it must also be real on the lines Re(z)

= tl. For w = (10 -10) and for z on. the unit circle, we have

-z = w(z) , so that J(z)

=

J(-Z)

J(w(z))

47

=

J(z) must also be real.

- 1.48 -

Since J preserves the orientations, it must apply the part Re(z)

~O

of the fundamental region D onto the closed upper half-plane. From that (and the normalizing conditions), it can be reconstituted globally by means of the symmetry principle of Schwarz. Finally, let f be any (meromorphic) modular function. The conditions on f show that there must exist a meromorphic factorization F f HV{iot) ~ a: u( 00) J ~ ,"~F a: u( 00)

with [u(~)= Wl(a:)

= S2 Riemann sphere. Since any meromorphic function

on the Riemann sphere is a rational function, it follows that F is rational and that f

=

F(J) is a rational function of J.

q.e.d.

The second part of the theorem shows that J gives a conformal representation of D/\ (Re(z) (0) deleted of i , ' still with

J(~)

=

0 , J(i)

=

onto (Im(z) ~ 0) - {O,l}

I (but no conformity at these points).

Then using the symmetry principle of Schwarz on Re(z) and then

J

-1 , +} , •••

Izi = I

0

will give the periodic function J in D + Z.

As the symmetry principle also applies along arcs of be applied to

= -} , ][),

it can

, and inductively, it will eventually lead to

a definition of J (by analytic extension) on H. Although this definition of J may be considered as more elementary than the one we have given (because it does not refer to elliptic curves in any way), it seems far from easy to derive from it the arithmetical properties of

J which we are going to consider later. Also we note that the action of SL 2 (R) on H has a natural extension to JPI (JR) = lRu(ioo) (considered as boundary of H) by fractional linear transformations. The transforms under SL ( 7l) are the rational points ~ em on the real 2 axis, so that a must vanish at all these points (tend to 0 when we

of ioo

approach these points vertically from above)

48

and

the real axis is

t

- 1.49 -

the natural boundary for the analytic extension of

~

(and of J).

(3.lZ) Corollary 1 (Little Picard Theorem). Let f : [

Proof: We may assume a = tion points of J

° and

[ be an

f ((I:)n {a, b} = ¢ , then

entire function. If a f b ~[ are avoided by f f must be constant.

~

b = 1. Let AcH be the set of ramifica-

-1

A

J{O,l}. Then J defines a topological analytical

covering J : H - A --+ [ - {O,l)

(local analytic isomorphism). By the

monodromy principle (G: is simply connected) f : [ be lifted to the covering I F : [

~

~

[ - {O,l} can

H - A. But Liouville's theorem

(applied to exp(iF)) shows that F must be constant (by continuity), and so must f be. More important for us is the following (3.13) Corollary Z. Every non-singular plane (projective) cubic curve yZt = x 3 + Axt Z + Bt 3 , is isomorphic (as analytical group) to an elliptic curve E = [/L , wi th a lattice L C[. Proof. The group law on this cubic is given by the condition that three points have sum zero if and only if they are on a line. If we make the transformation of variable y

~

y =

y/Z , we get an equation

for x,y in the Weierstrass form with gz = -4A and g3 = -4B. For this new equation, the non-singularity condition is gi - Z7g~ ~ O. By the theorem, there exists a z £H such that J(z) define a first lattice L

= gi/~i - z7g~)and we

Lz in ~ . Because J is homogeneous of degree 0 in the lattice, any homothetic lattice L would give the =

same value for J. We choose thus a complex constant ~~Osuch that moreover gZ(AL) = A-4 gZ(L) = gz = -4A. This is possible if A 1 0

A4 = -gZ(L)/(4A)

, and determines ~ up to a power of i

=1=1.

In this

case (A 1 0 implies J ; 0) 2

g3 ().L) = so that

g3(~L)

3

g2(~L) (J(~L)

-

1)/(27JC~L))

=

2

g3 '

= ±g3 · By homogeneity of g3 of degree six in the 49

- 1.50 -

lattice, changing A to iA if necessary, will give the equality. For this lattice L', (1.11) gives an isomorphism E

=

plane cubic curve X : y2 t

=

[/L'

~

X with the

4x 3 - g2xt2 - g3t3. In the case A

= g2

0,

we note that the non-singularity condition implies g3 1 0, so that one (at least) of the lattices

AL~

~E[~)

will do. It is indeed

sufficient to take for A any sixth root of g3(L~)/g3 . (3.14) Corollary 3. Two elliptic curves E

=

[/L and E' = [/L' are

isomorphic (as analytic spaces, or equivalently as analytic groups), if and only if J(E) • J(L) = J(L')

= J(E').

Proof: Indeed, two elliptic curves are isomorphic if and only if their lattices are homothetic. Now there are unique region D with L'VL't and

L'~

L"

~,~'

in the fundamental

. The result follows from the injec-

tivity of J on D. The corollary 2 is the basis of the algebraic study of elliptic curves. For example (3.15) Application. Let X be the plane cubic of equation (g~-27g~rO) 2

Y t = 4x

3

- g2xt

2

g3t

continuous). We look at

3

and take any frE:Aut([) (not necessarily (f

as mapping [2 ----. [2 defined by

(x,y) ~ (x~,yr). Then the image X~ of X X fl [2 is the affine o 0 r part of a plane cubic X , and the two curves X and X~ are isomorphic 3 3 exactly when G" leaves J = g2/(g2 - 27g~) fixed. tr

It is obvious that X

has the equation y2 t

4x 3 - gzxt tr

2

3 - g3CS"'t (non-

singular because X is assumed non-singular). Hence the invariant r of X is Jr where J is the invariant of X. This proves the assertion. r Note that if X is isomorphic to [/L and X to [/L r ' we have in general no simple way of determining Lr in function of L (we mean, no algebraic way of determining Lr in function of L: the connection is given by the transcendental function J).

50

- 1.51 -

This application suggests that the field be chosen to be equal to of E (because

~(J)

~(g2,g3)=>~(J)

might

for a special "model" of the equation

~(J)

is the fixed field of the group of automorphisms

~cAut(~) satisfying Jr = J). It is easy to show that this is true.

g'2 = g'3 = 27J/(J - 1) 323 4x - gxt - &t

E:~(J).

The curve

is isomorphic to E (it has the invariant J = J(E) ). If &3 = 0, any non-zero &2 will do, for instance g2 1, hence an equation y2 t = 4x 3 - xt 2 = 4x(x - }t)(x + }t) for E with coefficients in

=

~

(J = 1 in the case &3 = 0).

~(J)

(3.16) Remark 1. It is easy to check elementarily that G6 vanishes at i and that G vanishes at , . By definition 4 G (i) = (mi + n)-6 = l:'i 6 (-m + ni)-6 6

L.

(m~O

= (-1)3 G6 (i). Hence 2G (i) = O. Similarly since ~3 6 polynomial for 4), we see G (0 = 4

L.

(m/n)'0

(m~

n) -4

L' (2(m I:

Hence (1 - ,2)G4(~)

+

J

+

= 1 and ~2

o

+

~

+

L'z:8 (m{3

+

n(2)-4

n4 2)-4 = L'(2(m

1

+

(minimal

n(-1-4))-4

,2(m-n - n,)-4 = (2G4(~)'

= 0 and G4(~) = O. This method does not give

the orders of these zeros however, nor does it show that these "trivial" zeros are the only ones. (3.17) Remark 2. Let q

=

e 7tit: be Jacobi's variable (cf. proof of (2.10]

and t = q2 = ~('t'). Then ~ ~ t = ~(-r) maps the region -l~Re(~)0 (one - one) onto the punctured unit circle O
~ ~

t.

The condition defining modular forms (3.6.b) means that the corres51

- 1.52 -

ponding function on the punctured circle has at most a pole at the origin. The Laurent expansion of this function will thus be valid in a small punctured circle O
F(t)

=

=

L

a tn

L a e2xinz "~ ... n

=

"~'\>__n

A priori, the Fourier expansions of f are valid in regions r l
t

=

0, as soon as we suppose that tF(t)I

tends to a limit, finite or infinite, for t~O). The variable t

= q2

is better that

~.e:H

for

some

then the lattices

if

t

~,~'£H,

= t

in the sense that if L~

and

t L~,

= ~(1:)

and t'

= ~(--r')

are the same if and only

t •

Automorphic forms have another interpretation in group theory. Let G be the Lie group SL 2 ( m). Then the group of analytic automorphisms of the upper half-plane H is naturally identified with G/(!l), acting by fractional linear transformations. The stabilizer of i is the subgroup of matrices g i

ai ci

+ b + d

d

= (: :) such that

(ai + bled - cil , c i + dl

whence the conditions 2 2 c + = +

lei

'

dl 2 =

1

E:

and

=

(i

ac

+

+

ac

bdJlci

+

+

dl- 2

bd - O.

In G, they characterize the special orthogonal group M = SO(2JR): M ={(

c~s' cos4 sin.-). '

-sln-l

°'''<21t'' J (stabilizer of i)

•

We introduce also the subgroups of G: A

= {(~ ~-1)

: a€ R: },

52

N

= {(~

~):

u€R

J.

H

- 1.53 -

We have -then A.N

b) G

={(ca

dE:: c =

o}

and also obviously G = M·A·N (= N·A·M = G- 1 ). This shows that the action of G on H is transitive : G(i) = NAM(i)

= NA(i)

N(iR»C)

= lR

gM

gM(i)

+

+ iR»C

H.

+

Thus we have an identification (3.18)

G/M ~ H

defined by

t-+

r=

In consequence we can also identify (with r\G/M ~~

r\H by

rgM

= g(i)

SL (1l) )

2

r(z) where z

~

z •

= g(i).

r\G/M appears as set of the lattices of the form Lz Although it is true that through the first identification every In this way,

function f on H can be considered as function over G invariant under M, this is not useful for modular forms. We proceed differently. If £ is a modular form of weight k (holomorphic on H, but no condition

at i 00) , we define the corresponding function F (3.19) and J(g,i)

F(g) = J{g,i)kf(z) where z (ci + d) -2 for g =

(~

for m =( c~s~ Sin'-) .., -sIn'" cos~

(3.20) F(gm~)

:) E: G

,

(cos~-isin~)-2

and then that

= J(gm~,i)kf(gm,(i)) = J(g,i)kJ(m"i)kf(g(l)) =

because

E: M

Ff on the group:

= g(i)E,H ,

With this definition, we see first that J(m"i)

= e2i~

=

gm~(i)

F(g)e

= g(i)

2ik~ =

= F(g)X2k(m~)

,

z, with the even character X2k of M (this

character is the 2k-th power of the basic odd character X of M defined by x(m",J = e i 8-). For a matrix (E: r we have k k-k (3.21) F(¥g) = J(~g,i) · f(t(z)) = J{~g,i) J(r,z) fez) J(g,i)kf{z)

= F(g)

•

This shows that all the lifted functions Ff are functions over

r\G (irrespective of the weight of f), the weight k appearing now S3

- 1.54 -

in the character according to which F transforms along M. If we suppose that f

o

k ' i. e. f is a holomorphic modular form of weight k vanishing at ica ,then the Fourier expansion of f will start E:M

with a term eZ~iz , hence will have the order of magnitude of e

-27Cy

(y = Im(z)) at most If(z))

= O(e- Z1CY )

for y

= Im(z)-"06 ,

and this will in turn give strong decreasing properties of F

=

Ff

along A, which ensure that F will be for example square summable on

r\G (for the invariant measure coming from a Haar measure on G : because G is semi-simple, every Haar measure on G is bi-invariant). 2 Thus the spaces M~ can be embedded in L (r\G), each of them being embedded in an isotypical component for the restriction to M of the right regular representation of G in LZ(r\G). The holomorphy condition on f gives a condition on F which can also be interpreted group theoretically, but this would lead us too far here. Let us treat an example. We take f = G (holomorphic of weight k), and we construct Zk its corresponding function F on the group, which we still denote GZk abusively · First, we note that L'{mz + n) -2k = L.

L

!~l

(clz

+ dt)

L

= .L.!-Zk. = '(2k) Thus, the problem is to lift (cz

(cz

L..

(c,d)=l

(cz

(3.22)

Ek(z)

=

=!

+

d) -Zk

d) -2k .

+

d)-Zk to the group. To simplify,

+

we work with the normalized function Ek(z) = y+~

=

( c , d) = 1

l,~l

satisfies lim Ek(iy)

-2k

(c ,d) =1

GZk(z)/(Z~(Zk))

(which

1). ~

L-

(c,d)=l

(cz

Then we observe that all matrices

+

d)-2

k

1 ~ (~ ~) E

54

r

SL ( 7l) have Z

- 1.55 -

coefficients c,d prime to each other (the determinant condition gives Bezout's condition), and conversely, for each couple (c,d) of integers, prime to each other, Bezout's condition shows the existence of a couple

of integers, say (a,b) such that the matrix (~ Now, it is easy to check that two matrices of

lines if and only if "they are congruent mod if

J and

T'

r

~) has determinant one. have same second

C. = "t (~ J'

have same second line c, d, then

=

~). More precisely

y.. 'r

with a matrix

~ of the form (~ ~) (n a rational integer) or of the formt~ _~) (n as before). This gives in fact a bijection between the couples (c,d)

roO Y € roo\r

of relati vely prime integers, and the classes

.

Using this,

we can write (3.23)

Ek(z)

1 }

L.

(c,d)=l

(cz

+

d) -2k

1

2

L. J(y,Z)k J;,\r

On the group G, we have by definition Ek(g)

J(g,i)kEk(g(i)) =

! If we put

L:

r;.\r

!

J(g,i)kJ(r,g(i))k

L

,:\r

J(l g ,i)k

L(g) = J(g,i) we find the very simple and suggestive formula

Ek(g) = !

(3.24)

L

':o\r

L(1 g )k

showing that on the group, E is obtained in the cheapest way from L k (except for the factor Because L(ng)

!)

so as to be invariant on left under

r.

L(g) for any gE:G and n E:(!I)N, it is indeed sufficient

to take a sum of L (yg) over

6

mod (left)

r fl (±I)N

=

r:

to obtain

a (left-)f -invariant expression. The fact that the hexagonal and square lattices are the only ones admitting more than

two ("canonical") bases as constructed

after (3.3) is equivalent to the fact that

r/(~l)

has only two finite

subgroups of order 3 and 2 respectively, generated by 55

(:1

~) and

- 1.56 -

(~ -~) fixing ,

and i resp. From this and a careful study of the

tesselation of H given by the transforms of the fundamental region D under

r/(tl) one can conclude that this group is the free product

of these two cyclic subgroups SL ( 2

(3.23)

and also that

r

~)/(±l)

= C

2

* C 3

(C

n

= ~/~),

itself is amalgamated product of the subgroups of

order 6 (generated by

e ~))

and of order 4 (generated by l over their intersection of order two (generated by(~l_~)) :

(~ ~l})

(3.23)' (3.24) Proposition. For y

Im(z)

are valid (uniformly in x

Re(z)):

~(az ~(b

+

b : Lz ) ~ -w 2/3 1

2

: L ) .-,.

TC (

z

for a not integer

, (a,b real ),

1 - -) for b not integer

2

sin Kb

( 2 7l") -12A{ z ) ~ ( - z) --+

the following limit formulas

~~

3

,(b real),

1

J(z)~(z) ~ 1/1728 = (12)-3

2

-6

·3

-3

Proof. By definition, we have

pCaz

+

b : L z ) = (az

+

b)-2

+

l::'«az m,n

+

b

+

mz

+

~-2 -(mz

+

n)-2).

The convergence of this series is uniform with respect to Im(z);, t:"7 0, so we compute the limit by taking the limit of each term. When a is not an integer, the terms ((a m. Only remain the terms (mz

+ +

m)z

+

(b

+

n))-2 tend to 0 for every

n)-2 with m = 0 (and n 1 0). The

formula ~ n- 2 = ~2/6 easily gives the first formula. When a = 0, n>O but b is not an integer, only the terms m = 0 give a contribution,

feb : L z ) _b- 2

L

+

L'(b

+

n)-2 - "E.'n- 2

=

(b - n) -2 - 2{(2) = x2 sin -2'J(b - 7(2/ 3

56

- 1.57 -

With the values

~(4)

= 7r4/90 and

'(6) =

6

-n:

/945, we find

= ( 21f) 6/216 •E3

&Z

(2lt) 4/12· E2

!:l

Z (21f) 12/1728 • (E~ _ E ) 3

&3

J

= E3Z/(E 3Z _ E3Z)

But the first coefficients of the t-expansion (Fourier expansion) of EZ and E3 may be computed easily by starting with the equalities ltcot<Z

=

= z-l

ni(t + l)/(t - 1)

+

r='~z - n)-l - n- 1].

and differentiating with respect to z three (resp. five) times both sides. We see then that E2 ( z)

=1

+ 24 0t +

C1 ( t 2 )

• E3 (z)

=

1 - 504 t +

O( t 2 )

(a much stronger result, giving all Fourier coefficients explicitely will be derived in detail later, in (4.1) ). This shows that E3Z - E3Z (3·Z40 + Z·504) t + O(t Z)

This finishes the proof of the proposition. We turn to a closer sudy of the function A of Legendre, and its dependence on the elliptic curve. By definition, if E is an elliptic curve, it has an equation of the form E : y2

and

~

= 4x

3

- &2 X

-

&3

3

4

IT (x i.1

- h( 1w.

a-

1.

))

= (e 3 - eZ)/(e l - e Z) with e i =

possible equation for E of the form : yZ

x(x - l)(x - A)

Ccf. Cl.13) ) with

A'

0 ,1 .

To bring a dependence in z eH, we define explicitely el(z)

P(1

ezCz)

PC1z: Lz )

e 3 (z)

P(lCz+l): Lz )

:

Lz )

and "C z)

Ce 3 Cz) - eZCz))/CelCz) - eZCz)) 57

- 1.58 -

If we take a direct basis all

tc)z

az + b

=

we shall obviously have

=0

, b :: c

a :: d :: 1

~

, tUz of Lz of the form , tell

cz + d

mod Lz , as soon as We define the subgroup f(Z) c r = 5L (1l) Z

4l =

mod Z

=

and ~WZ =

~z

0) mod Z. In other words, (1 (a c ~) - 0 1 kernel of the natural homomorphism of reduction mod Z

by the condition

5 L Z(Z)

---+- SL Z(II ZZ) ,

hence in particular is a normal subgroup of

r

r(Z). is the

with quotient isomorphic

to the non-abelian group of order 6 (necessarily isomorphic to When

~

E'3).

is in f(Z), we can thus write

Hence eZ(z) = J(o< ,z)eZ(O«z)) , and similarly for e l and e • Coming back to 3 property

=

A(O«Z))

We say that

A

for any

A(Z)

0(

A ,

this gives the

E:r(Z) .

is a modular function for the group

f(Z). To have a

full description of the behavior of this function under the fractional linear transformations coming from the full look at a system of representatives of observed that w

o

(1

=

-1 1 0) and n = (0

Z and nw of order 3 in

r

r,

it is sufficient to

mod f(Z). We have already

1 1) are such that w is of order

f/(±l); they will generate such a system of

representatives. Consequently, it is sufficient to connect A(-l/z) and ~

A(z+l)

under

r.

with

A(Z) to have the description of the behavior of

The first matrix corresponds to the change of basis Wz

= -1 ,

or 1~ = z I Z , 1lUZ This gives ~

-1

=z ,

~l

~

~

, so that e 1 and e Z are interchanged. 1 - A(Z)

(-liz) 58

•

- 1.59 -

Under the second transformation, "2 = z

+

1

~

hence e 2 and e 3 are interchanged ~(z + 1) = (e Z - e 3 )/(e l - e 3 )(z) To sum up

(3.25) Proposition. The function

~(Z)/(A(Z)

- 1)

,

•

of Legendre is a modular function

~

r

f(Z) (normal subgroup of

for the group

(z + 1)

of index 6 formed of all

matrices congruent elementwise to the unit matrix mod 2). Under a fractional linear transformation ~

the s ix fun c t ions

t - " , 1/(1

1/'). ,

t

of

r,

A is transformed in one of

- A), 1 - 1/').. = (~- 1)/ A, A/ (A - 1) ,

;\(-l/z) = 1 - A(Z) , A{Z + 1) =

and explicitely

A/(~ -1)

(z) .

To give an explicit relation with J, it is necessary to study the limit of

A

the matrix (~

when Im(z) -+

i)

in some detail. Since

00

contains

acting by a translation of 2 in H, the function

is periodic of period 2. As before we put q = e~iz expansion of"

r(2)

around q

sarily valid for Im(z)

"7

= 0

and the Laurent

gives the Fourier expansion of

0 because

~

(neces-

~

A has no pole in the upper half-

plane H). Using (3.24), we see that ~ 2 1t£ ( 1 - 1/ 3 ) + 'It' / 3

( e 1 - e 2) (z) ~

7(2

for 1m (z)

--+-

00

but we need more. (e 3 - e 2 ) (z) --+ 0 for Im(z) -+- 00 Coming back to the defini tion of e 2 and e by means of the ~ function,

Also

3

and grouping together the corresponding terms in the expansion of

p,

(this cancels the convergence factors and gives an absolutely convergent series), we find

~ (e 3 - eZ)(z) = L-m,n

(~+ ~z

[

~ ~

((m+~)z

m,n

'7(2

[

L

m

+ mz - n)

+ a - n)

-2

-2 -

-

(az + mz - n)

((m+a)z - n)

[cos -2 (m+ D'7(z - sin-2 (m+ D1CZ ]

59

-2]

-2] =

- 1.60 -

where we have used the classical expansion formula

L

1(Z /sinZ'lI'z ..

(z - n)-Z

"€Z Using the duplication formulas for the trigonometric functions, we

l/cosZ~ - 1/sin2~ = -4cosZ~ /sinZZ«

find that the difference

•

We have thus obtained -47tZ

L. cos (Zm+1)7tz/sin Z(2m+1)1t"z m

-81tZL. cos(2m+l)1rz/sinZ(2m+l)"Jrz m~O

Furthermore

I (q +

cosnz

161(Z

q

-1

1= fi
) and sinltz

r. (qZm+1

+ q-2m-1)/(q2m+1 _ q-Zm-l)2

m~O

r. ...

~ qZrn+1(1 + q4rn+2)/(1 _ q4m+2)2

m~O

m~O

q(l + qZ)/(l - q2)Z +

L..

q + O(qZ)

m>l

This gives the desired formula (e

3

- e ) (z) .. 16xZq + O(qZ) , Z

and so

A(z) q-1

(3.26)

=

A{z) e -xiz -.. 16

From this and the relation A(-l/z) )(iy)

~

1 when

gives A(l + iy)

y~O ~ ~

=1

(y real), and when

y~O

~(z

=

24

for Im(z)

-+-

00 •

- A(Z), we infer that + 1)

=

~(Z)/(A(Z)

- 1)

(ibid.).

We come to the explicit relation between

A and

J. Any symmetric

combination of the six functions appearing in (3.25) will be a modular function for

r

(the growth condition for Im{z) -+00 is satisfied by

what we have just seen). We construct the symmetric product of the translates by 1 of these functions : f = (1 + ")(1 + l/~)(l + (l-A))(l + 1/(1-~))(2 - 1/~)(2A-l)/(~-l).

By (3.11), f is a rational function of J. Since f has no pole in H, it must be a polynomial in J, and more precisely since f has a simple 60

- 1.61 -

pole at i

is a linear function in J : f = A

+

B·J • We determine these constants.

The limits q-2 f

e- 2niz f(z) ~ -4.16

q-2 J

e-2~izJ(z) ~ 1/1728

show that B

=

g3(i)

=

=-

_33

=

=

0, e 2 (i)

= 0, and quite generally e 1

-e 3 (i) gives

-1, and finally if e (i) 3

A(i) 0, A(i)

and since J(i) = 1 this proves

= 27 - 27J or

f

= 2- 6 3- 3 ,

27. To determine A, we look at the special z

0 implies e l e 2e 3 (i)

If el(i) A(i)

=

= _2 6

=

+

e2

+

=

e 3 = O.

2. If e 2 (i) = 0, similarly

= 1.

In all cases f(A(i)

A

27. We have obtained

J

1 - f/27.

=

A small computation gives now

=.!. (1

J

(3.27)

-

A2 (1

27

A

e

l

+

e

2

+

e

3

=

0

,

+ ),2)3

_ ~)2.

The relation between J and

th~

e.' s is easily found now. From 1

J

e.2 + e 2 + 2e e {1,2,3j, j k for li J j , k = k J e2 + e2 -2(e 1 e + e 2e + e 3 e l ). Then (3.27) 3 3 2 2

we get e~ 1

hence by summation e 12

+

gives immediately (3.28)

Since g2

-4(e l e 2

+

e e3 2

+

e 3 e 1 ), this expression shows that

A u. = l6(e -e ) 2 (e -e ) 2 (e -e ) 2 l 2 2 3 3 l

IT (e i -e j )

= -16.

J,..

lrJ

(it could,of course, have been derived more simply from it).

61

i.

•

0

- 1.62 -

We draw two possible fundamental regions for f(2) in H.

D(2)

D(2)'

(3.29)

" / /

,,---- " .....

'~.1

\ \1

o

-1

\

The behavior of a near

i~

r(2)-invariant (say meromorphic) function in H, is made by introducing as usual the variable q = e ?tiz . In

the neighbourhood of 0 (in D(2) or D(2)'), we introduce the variable 7Ci z q' e = e -Jrij z' (z' = -l/z near 0 means z near iot ). When z" is in a neighbourhood of 1 (in D(2) or D(2)'), z' = z"-l is near 0, so we take the variable q" = e 1tiz = e- 1Ciz ' = e- 1ti /(z"-l). (3.30) Definition. A modular function for

f(2) is a meromorphic

function f in H satisfying a)

f(~(z))

= fez) for all

~ ~

f(2) (when one member is defined),

b) f has poles at most at the points ioO,O,l with reference to the Laurent expansion in the variables The function

A has

these properties. Indeed,

(uniformly in Re(z)),

'A(z)

~

q,q~q"

A(Z)

~

respectively. 0 when

Im(z).~

1 when z + 0 in D(2), so only the

point 1 deserves verification. But there, we can use ~(-l/(z-l))

= 1/(1 -

A(Z))

We could also look at (3.27) because J has simple poles (in the 62

- 1.63 -

mentioned points) with respect to the variables t = q2 , t' = q,2 and ttl = qtl 2 The orders of a f(2)-modular function are thus · d at a 1 1 · we 11 d e f lne pOlnts

0

f ~ D(2)

D(2)U{ioc),0,1} (if the function

does not vanish identically) , by means of the corresponding Laurent expansions. As in (3.7), integrating f'lf on the boundary

~D(2),

making small semi-circles to avoid the zeros and poles of f and the points

° and 1,

we find

p~

(3.31)

ordp(f)

=

0 •

(This formula is simpler to establish than the corresponding one in

~.7)

because we have no ramification points like we had there in

7;, i , and we take here only the case k=O.) Because

and nowhere that

° in H,

~

is holomorphic

and because A has a simple zero at iaO, we see

A and all the ,,- c (c £

have a simple pole at 1. These ............... functions must have one and only one zero in D(2) to compensate in (3.31). This proves that "

~


gives a bijection

: D(2) ---+-
'6(2) ---.

I IP (
a:

U

{oo}.

Consequently (as in (3.11)), every modular function for rational function in

A.

f(2) is a

We formulate an analogue of (3.11).

(3.32) Proposition. The function

A of

Legendre defines a conformal

bijection of the fundamental region D(2) defined in (3.29) onto
{a, I} . This function is real on the imaginary axis and on the sides

of D( 2) '. The rig ht hal f D( 2)' () (Re ( z)

"7

0)

ismappedon tot he uppe r

half-plane with a continuous-extension to the boundaries. Every r(2)modular function is a rational function in ,.. A couple of diagrams may help to clarify the situation.

63

- 1.64 -

H

1

(3.33)

topological analytical covering ~

f(2)\H

1

ramified covering at t, i

r\H

The fibers of -1 ~ (0)

~

=

!pI (([)

~

xl •

J

!pI (([)

- {O, 1 ,co} 6-sheeted ramified covering

- too}

have 6 elements except

{A(~),

A(W({»

= ~({+l~ has only two elements, each

corresponding to a ramification index 3,

1r:l

(1)

l~(i), ~((i-l)/i), ~(-l/(i-l)~ has only three elements,

each corresponding to a ramification index 2. In fact,

~

is the universal (maximal) covering of pl(([) with

respect to the three ramification conditions index 3 above 0 index 2 above 1 index 2 above

00

•

On the function fields of modular functions for rand f(2) (both have been proved to be purely transcendental fields of degree one), gives the injection

G:(J)

--+ ([(~). The group

r/r(2) ~

~

S3 acts

naturally in ([(A) by field automorphisms over ([ (the group ~3 acts in the purely transcendental extension k(X) by k-automorphisms for any field k). The fixed field is precisely ([(J) (by Luroth's theorem the fixed field in k(X) is purely transcendental over a generator which can be taken X- 2 (1-X)-Z(I-X+X 2)3 of degree 6: compare with (3.27) ). These facts show that

([(~)/([(J)

of group isomorphic to ~3 and we can say that f(2)\H ~

J\H

is a Galois covering of group

r/r(2) ~ ~3

64

is a Galois extension

- 1.65 -

Let us conclude this section by a few formulas, given without proof. (3.34)

A(z)

16 llr(1 q

1

f=

n~l

(3.35)

~(z)

+ q +

2

n

)'

q2n-1

,Cft+tl 1+ 2.L. 1'"

16 n:,O

=

(L.

"EZ

L

'1

(II+-jl' "I-

,,~Z 1

"~1

Also the three singularities {o,l,oo}

of

~

make it possible to

derive an inversion formula z = iF(!,l:l:l-l) F(l,!:l:" ) where F

2Fl is the usual (Gauss') hypergeometric function.

65

- 1.66 -

4. Arithmetical properties of some (elliptic) modular forms

We are going to give explicitely the Fourier coefficients of the modular forms G of weight k, or more precisely of the normaZk lized Ek = GZk/(Z~(Zk)). (4.1) Proposition. The modular form E

k

of weight k (k>l) has

th~

following Fourier expansion Ek(z) where the constant

tk

1 +

L. N


~(z)

)

~l

=

C-I)k C21r )2k/ CZ';C2k)C2k-I)!)

(B k the Bernouilli number of index k), and

br

e Z7tiz

(t

'fk is given by

'fk = C-I)k 4k / Bk

function defined

N

=

GiCN)

~(N)

is the arithmetical

sum of the t-th Eowers d

t of all the

positive divisors d'7 0 of N. Proof ., We start with the classical formula of function theory -1

L

-II

-1

+ (z + n) {(z - n) n )1 (note that the terms nand -n have to be grouped together to secure

1t'cotg7t"Z

z

+

absolute and normal convergence on bounded sets of [ -

~).

By term-

wise derivation, we get C7Ccotgn:z)

I

=

_z-2 -

L

L

{CZ - nf2 + Cz + n f2 )

n~l

-(z + n)-Z

n£71

and by induction (TCcotg-n:Z) (k)

(-l)k.k!

L:

k l (z + nf -

(k

nE:71

•

~l)

But the Fourier expansion of these functions is easily determined " h q W1t

=

e xiz ,an d t = q Z

7C cotg1l"Z

~cos~z

/

"

S1n~z

we get

~(z),

= ~1"( q

+

q-l)/(q _ q-l)

'Jt'i ( t

+

1) / ( t

whence 7t'cotgll'Z

'7l'i - Z1C i / (1 - t )

~i - 2~i(1 66

+ t

+ t

2

+

••• )

•

- 1 ),

- 1.67 -

For the derivation, we note that dt/dz Hence (1t co tg'Jrz) (k)

= - ( Z1tl.) k+ 1 L'\ n ktn n;Jl

, for

k~l

Comparison gives with mz instead of z (t m instead of t)

L.

(mz + n) -k-l

=

(-1) k+l (Zon-i) k+l /k!

n" Zl

nkt mn

(m

f 0) •

n~l

Also, replacing k by Zk - 1

l::

L.

(mz + n)-Zk = (Z~i)Zk/(Zk-l)!

n£Zl

r= n)l

nZk-ltmn

Summing now over m (and treating the case m = 0 separately), we find

Z~(Zk) + (Z7Ci)Zk Z/(Zk-l)!

L L. ... m)l

n~l

because the sums with m and -m give the same result (cf. left handside). The double sum is

L..

nZk-ltmn and its terms may be rearran-

m,n~l

ged at will. We group together the terms with mn = N, a fixed integer N ~ Zk-l N ~ 1, so that the coefficient of t is the sum L- n = o-Zk-l (N) niN with the notations of the proposition. Summing up, we have obtained

with a coefficient

Now we recall that the Bernouilli numbers may be defined by the expansion cotgz = z

-1

1:

Bk22kz2k-1j(2k)!

k~l

The classical Weierstrass product formula sinz = z

IT (1 - L) 2

n~l

n 7[2

= z

IT (n 2 n2

n~l

gives by logarithmic derivation "' cotgz = z -1 + Z L-

z/(z Z-n Z1t Z)

n~l

67

-

z2)j(n

2

i)

- 1.68 -

Comparison gives then

2~(2k)/~2k = Bk 2 2k /(2k)l and hence

4k(-1)k/ Bk as claimed.

'(k

We give the first few values of these numbers B1

'(2) =

r1

B2 = 1/30

1/6 2/6

,

'(4) = 7[4/ 90

7t

= -24 ,

t2

= 240 ,

= 1/42 ,

((6) = ,f /945

B4 = 1/30 x6/3 3 ·S·7

r3

14 = 480

B3

= -504 ,

, ...

, ... , ...

2 (4.2) Application. The relation E2 = E4 (both functions are normalized at iao , hence the proportionality constant is 1) gives the arithmetical relation

of 0"3 and

· Indeed 1 + 240L G"3(n)t n n)l

EZ E

1 - 504

3

E4

r7

=

L

n)l

G"'s (n) t

n

1 + 480 L 0"'7 Cn) t n n~l

hence the explicit relation n-l

~7(n) = f (n) + l20L6"3(m)~3(n-m)

3

m=l

l:

-2k

~

Using (3.23) we may write Ek = (mz + n) with a summation extended over the couple of integers m,n satisfying (m ,n) = 1 , m> 0 , n

>0

if m = 0 ,

hence the relation

(~ valid for Im(z»

(mz + n)

-4) 2 = 1: (mz

+ n)-8

0 (this relation was alluded to in (3.8) ).

Similarly E3E2 = ES gives a relation between r3 ' ~5 ' ~9 · By Theorem (3.10) every Ek is a polynomial in E2 and E3 hence can be computed by means of 0-3 and a"S. 68

r2k - 1

- 1.69 -

(4.3) Formulas.

For reference we give explicite1y some proportionality

constants. With 1

240 L. 0""3 (n) t

+

n

(t

~(z))J

n~l

1 - S04

r:. ()S (n) t n

n~l

we have

60 G = (2~)4·2-2·3-1.E

&2

4

2

g 140 G = (2~)6.2-3.3-3.E 3 6 3

A = (2n)1202-6o3-3(E~

- E;)

J = E3/(E 3 - E2 ) = 22.3-3A-2(1_A)-2(1_~+~a)3

223

and we introduce j

(2 6 .3 3

26 ·3 3 .J

=

(12)3

= 1728)

28A-2(1_~)-2(1_~+Al)3 ( j (t)

=

0

J

j (i)

= 1728 ,

j (ioo)

=

00 .)

(4.4) Theorem 1. The Fourier expansions of (2x)-12~ j

and of

= l728·J have rational integral coefficients (21r) -12

6 (z )

1728J(z)

=

=

L.

n)l

1ft

1:( n) t n

L

+

c(n)t

1:(n)E:ll,

n

(t =

~(z)),

~(l)

= 1,

c(n)E:ll,

n~O

and in particular j is normalized by the condition of having residue +1 at ioo

•

Proof. Let us put E

2

=1

+

240U

=1

+

2 40 3 oS oU and E

= 1 - 23 ·3 2 ·7·Y , with two power series

3

=

1 - S04V

U, Y in t and integral

coefficients. By the above formulas, it is sufficient to show that 4 3 (1 + 2 • 3 • S • U)

= (1

322 6 3 - 2 • 3 • 7 •V) mod 2 •3

in the sense that all coefficients of these series in t satisfy that congruence condition. Hence we have to check that

69

- 1.70 -

or 24 3 2 s.U + 24 3 2 7.V

mod 2 6 3 3

0

=

0 mod 12 (= 22 3). But we observe that for any integer n~l, the number (n-l)n 2 (n+l) = n 4 - n 2 is divisible by 3 and 4 hence by 12, hence the congruence n 4 = n 2 (mod 12) and a

or still

fortiori n

SU + 7V

=

S

=n3

(mod 12). This gives

proves the desired congruence sU + 7V

~3(n)

=0

= ~S(n)

(mod 12) and

(mod 12). Since

~(l)

=

1,

the integrality property of j follows at once. (4.5) Theorem 2. If we introduce the function of Dedekind

",(z)

=

e wiz/12

IT (1

ql/12

_ e21rinz)

n~l

IT (1

_ q2n)

n~l

then we have the following formulas a) (Euler)

,(z) = e7tiz/12

L.

(_1)n q n(3n+l)

nE. 7l

b) (Jacobi) (27() -12A(z)

= ,(z) 24 = t

TT (1

_ tn) 24 ,

n~l

. h th · W1t e usuaI notat1ons q

=

e wi z an d t

=

q2

=

~

() z .

Proof. The basis for this kind of theorem is (3.8) which shows that the dimension of the vector space Mo of modular forms of weight 6 6 vanishing at ioo is dim(M~) = 1, hence M~ is spanned by ~: M~ = ([.~ • We put g ( z)

e lCi z/ 12

L

(-1) nen ( 3n+1) 7l: i z

n€71

= e7tiz/12

L

n£71

exp(lfi3zn 2+ 2nin(!z + 1))

and moreover 8 (z : T) = Hence g{z)

=

L..

nE:ll

exp (7ri zn 2 + 2n'in-c)

(Jacobi's

8 3 ).

e~iz/12 8(3z:1 + lz). We use now the classical summation

formula of Poisson for which we recall the hypothesis. If f is a continuous function on R ,

r=

n E.71 70

f(x

+

n) converges normally

- 1.71 -

on every compact subset of

L

Rand

'fen)

n£ll

Fourier transform of f), then we have For f(x) --

e~izxZ

~

(f denotes the

00

~f(x + n) = L:t(n)eZ~inx .

O ° (Oth ° WI Z = ly, Y ~ 0 a rea 1 parameter ) , t h IS gIves

L. exp (7tiz (x+n) Z) = By analytic continuation for z

L

I <

L. exp (- (xi/ z) n Z

Vi/ z E:

and x = T

H

E

H

+ Z7rinx) .

this gives

exp(lriz(l." + n)Z) = Vi/z 9(-1/z:T) .

Coming back to g : g(-l/z) = e-~i/1Zz 9(-3/z : I - lz) e-7Ci/1Zzr;:;"3i

(Z731 IZ/3i

L. exp('Jt'iz/3(n

+ I -lz)z)

L:: exq(~iz/12Km2 - 2m/z~ = L exp(7rim Zz/1Z + 1t"im/6) m odd m=Zn+l

= fZ/3i y--

m'oad

exp(rrimZz/1Z).cos(rrm/6).

When m is of the form 3m' (m' odd), cos (lfm/6) = cos (mnYZ) = only the terms m m

~

~

o.

Thus

±l (mod 6) contribute in the summation, and since

-m interchanges these two classes, it is sufficient to take

twice the sum over the class m g(-l/z) But cos (7t"k

+

i)

g(-I/z)

YZ/3i·Z =

(-rf €

(z/i

=1

(mod 6)

L. exp(xi(6k+l)Zz/lZ) ·cos(1tk

k€ll

+

L

exp(Jri(36k Z + lZk + l)z/lZ)

r= exp(~i3znZ

+ ~izn) = YZ/i g(z)

We prove now that a similar functional equation holds for Because H is simply connected and choose a branch of 10g'rJ

=

6

/2 • so that

{iii e~iz/1Z

(log,) (z)

!)

7fiz/12 +

~iz/1Z +

~

never vanishes in H, we can

in H :

L

n;,l

10g(1 _ e27rinz)

L: 1:

(_l/m)eZ~inmz

n~l m~l

71

~(z).

- 1.72 -

Hence replacing z by -lIz : log

"1 (-l/z) =

I claim that log ~(-l/z)

=

,(-l/z)

L..

-lli./12z -

=

n ,m;;,l

(l/m)e-ZJrinm/z

!log(z/i)

+

log,(z)

(this will prove

VZ/i ~(z) ), with the principal determination of log in

~ - ilR+ (hence the determination of VZ/i which takes the value +1 at

z

=

i). We have to prove that -1l"i/12z - 'It'iz/12

L.

+

n,m~l

(l/m) (e 2'Jfimnz _ e -21rimn/z)

11 og (/.) Z 1

= 2

By summing first on n (the series is absolutely convergent), using the formula for the sum of a geometric series, we are led to a term e

2Jrimz

e -27tim/z 1 - e -2xim!z

= (i/2)(cotg(xmz)

+

cotg(~m/z))

Remains to prove that the sum -1ti / 1 2z - 1[i z/ 12

+

(

i / 4)

L

n,O

(c 0 t g (7[ nz)

+

co t g (TC n/ z) ) / n

is equal to llog(z/i). To compute the sum, and prove this, we use the following trick of Siegel's. We consider the auxiliary function f(s) = cotg(s).cotg(s/z) It has a double pole at s = 0 and simple poles for s = nx , nnz when 0 !: nE:ll. Since cotg(s) = s-l - s/3 cotg(s/z)

=

zs-l - s/3z

Ress=o ,(s)/(8s)

+

=

•••

•••

,

and also

we get

,

(-~/3

+

- 1/(3z))/8

=

-z/24 - 1/24z

Ress=nlC!:O tfCs)/(8s) = (1/8n1t)cotg(n1C/z) Res s=nXZr~O .(s)/(8s) = T

(1/8n~z)cotg(nxz)·z

This shows that the desiredsum is precisely 2'J{i LResepCs)/8s , with a summation extended over the whole s-p1ane. Let V be a positive number, not an integer and P the parallelogram with vertices Then

.A(-nrp)

f(s)/(Bs)ds

=~p

f(lIn:t)/(Bt)dt

72

±l,~z.

- 1.73 -

tends to the desired sum when bounded on

)p for

."

y~~.

Since t

~ ~(v~t)

is uniformly

and since this function has the limits

~oo

1 on two opposite sides, -Ion the other opposite sides (vertices not included), the limit is easily estimated lim [, \I~eo

ip

If ()/ 5) /

( 85) d 5

=H

JZ JZ +

1

-1

) (d s /

s)

= l(logz - 10g1 + logz - 10g-1) = !log(z/i) with the principal determination of log in

~

- iR+ .

The conclusion of the proof of the theorem is now easy. Indeed, g 24 as well as ~ 24 have Fourier series starting with t = q 2 and are modular forms of weight 6 (indeed SL2(~) is generated by 1 ). This characterizes 6. completely because M0 = (1o -10) and (01 1) 6 More precisely, (g - ~)/g is a modular function vanishing at i~

a:.~.

Since it is holomorphic everywhere on H, it must be identically zero, this proves g =

~

, and finishes the proof of the theorem.

This theorem gives a relatively easy way of computing the first coefficients ~(l)

=

1,

~(n)

L(2) = -24,

~(3)

= 252,

~(4)

= -1472,

Ramanujan conjectured and Mordel1 proved (1920) that moreover (4.6)

L'C(n)n -s n~l

for Re(s»

=

~

(1 _ ~(p)p-s + pll-2s)-1

p prime

13/2. The proof of this has now become a standard applica-

tion of the theory of Hecke operators in modular forms. An equivalent way of saying (4.6) is (4.6) ,

't(mn) = 'C(m) 'ten) when (m,n)

= I

,

m.) 1, n;, 1,

't(pk+l) = 'rep) 't'(pk) - pll,((pk-l) , P prime, k ~ I These Ramanujan coefficients satisfy many famous congruences, e.g. except for a finite number of n, they are all divisible by 691 (although Ramanujan knew that for

n~5000

73

the..:(n) are never divisible

- 1.74 -

by 691, he was lead to conjecture the preceeding result, which was later proved by Watson). Actually, Walfisz proved that the

~(n)

are

2 ·7·691 for nearly all n. Some conjectures divisible by 2 5 ·32·5 remain open. For example, Ramanujan conjectured that 1~(p)I<2pll/2 Deligne could recently show (1969) that this would indeed be the case if the general conjectures of A. Weil on the form of the zeta function of algebraic varieties over finite fields were proved to be true. Lehmer conjectures that

~(n)

~

0 for all n : this has been

proved at least for n <10~. (For all these questions, we refer to J.-P. Serre: Une interpretation des congruences relatives 1 la fonction

T

de Ramanujan, Seminaire Delange-Pisot-Poitou, 1967/68,

n 0 14.)

The Fourier coefficients c(n) of j have also some interesting (and intriguing) arithmetical properties. Their growth is wild c(O) = 744, c(l)

= 196884, c(2) = 21493760,

For example : if n is divisible by 2a , then c(n) is multiple of 2 3a +8 , 3 2b +3 , 3b ,

"

"

"

"

5c ,

"

"

"

"

Sc+l

,

CHAPTER TWO

ELLIPTIC CURVES IN CHARACTERISTIC ZERO

Many properties of (complex) elliptic curves are purely algebraic in the sense that they could also be derived in the field generated over the rationals by the two numbers g2 and g3 ' or perhaps in its alge.raic closure. It is thus sufficient to work in a fixed field of characteristic zero (algebraically closed if needed). Our first aim will be to show that any non-singular (projective) plane cubic can be given in Weierstrass' normal form. The study of the differentials over the curve (and their classification) will also be made purely in algebraic terms. It is true that by Lefschetz' principle the case of the complex field is crucial, and we can always reduce algebraic geometrical problems in characteristic zero to the same problems over [. However, it is not always useful to do that. In particular, when the coefficients &2 ' g3 are rational, it may be very interesting to look at the points of the curve yZ = 4x 3 - gzx - &3 with coordinates in the p-adic fields

~p

(or algebraic extensions of these). It is very strik-

ing how Jacobi's theory of theta functions can be made here if the absolute invariant j is not a p-integer (i.e. j

~ ~

p

). Thus we

conclude the chapter with a brief description of Tate's elliptic curves. We shall assume that the reader is acquainted with the following facts of algebra :

75

- 11.2 -

A[X]

a) If A is a noetherian ring, so is

(Hilbert's basis

theorem), and if A is a unique factorization domain, so is also

A[X] (X

is an indeterminate).

b) Notion of integral element over a ring and simple properties (sum and product of integers are integers, transitivity). c) Notion of valuation (with value group contained in R), and existence of extensions of a given valuation, corresponding to a field extension:)

A certain familiarity with the p-adic fields

~

p

(p a prime) and metric

spaces with an ultra-metric distance, might also help in the last sections, although this is not absolutely required. In the section on analytic p-adic functions, some propositions are proved for their interest and are not used after. Such are ( 4 . 2), (4. 3) and (5. 11) •

v: K)( -+- R = R+ whereas absolute values are considered multiplicatively 1•••l v : K~ ....

*) Valuations are considered additively

R: .

76

- 11.3 -

1. Algebraic varieties and curves

Let k be any field and (Pi) a family of polynomials of k[Xl, ... ,x n ] (where n is a strictly positive fixed integer). For any extension field K of k, we define the set VKCKn as set of common zeros x = (xl' · · · ,x n ) E: Kn of the family (P. ) P.1 (x) = 0 for 1 all indices i. Ifx€V , it is obvious that P(x) = 0 for every K polynomial P of the ideal I generated by the Pi in k[Xl, ..• ,x n ] . By Hilbert's basis theorem (the ring k[Xl, ... ,x n] is noetherian),

the ideal I is generated by a finite number of polynomials, so we may suppose that the family (Pi) is finite to start with. We may then replace k by the field generated over the prime field by the finite number of coefficients occuring in the finite number of polynomials P. , and thus suppose that k is finitely generated over 1

the prime field. If k is of characteristic 0, we are thus reduced to the case of a field k finitely generated over

~,

and all such

fields can be embedded in the field [ of complex numbers ([ is algebraically closed, and its transcendance degree over

~

is not even

denumerable). Thus if the characteristic is 0, we could also have started with k

=[ :

this is Lefschetz t principle. We come back to

the general case, and note that the set V (or V ) may very well k K be empty (e.g. when the ideal I is the whole ring k(Xl, ... ,xn] in which case we say that the ideal is not strict). But when K is algebraically closed

we have

(1.1) Theorem 1 (Hilbert's Nullstellensatz). Let I be a strict ideal of the ring k[X1, .•.

,xn]

• Then there is a common zero x = (Xi)

in a finite algebraic extension K of k. In other words VK is not empty if K is an algebraic closure of k.

77

- 11.4 -

Proof. The proof is made by using essentially the notion of integral element over a ring. First part : we show that if I is a maximal (strict) ideal of k[XI, ...

,xn]

, then the field K

k [X I ' • • • , Xn ] / I

k[xl, ... ,X ] is algebraic over k. This is done by induction over n, n the case n = I being trivial, because k[x] is not a field if x is transcendental over k. We note then that k(x )[X 2 ,· .. ,x n] = k[xl,···,x n ] , l because the right member is a field. By induction hypothesis, the n - I elements x , ... ,x n are algebraic over the field k(x I ) k', 2 hence there is a polynomial P(x l ) E: k[x I ] so that all P(xI)x i for i = 2, ... ,n , are integral over k[x 1]

(this P is a common denominator

for the Xi'S). If now f£.k(xl)Ck[xl, .•. ,x n], a suitable positive integer N

=

N(f) will be such that

pN(XI)·f~k[XJ [P(xI)xZ,· •• ,P(xI)X n ] hence integral over k[x I ] (by transitivity of the notion of integrality) This implies pN f 4::k [xl] , or f

~k [Xl ,P(x I ) -1] and

k(x ) C k [xl' P(x ) I I

-1] ,

Hence the result by transitivity of

and so Xl is algebraic over k the algebraicity. Second part

finally

We

s~lect

a maximal ideal I' containing

I and we look at the homomorphism k [Xl' · · · ,X n] ----... k [Xl' · · · ,X n] /1' = k [Xl' · · · ,x n] = K n By construction x = (xl' ... ,x n ) E:K and P(x) = 0 for all P E: I ' , in particular P(x) = 0 for all PE;IcI' (1.2) Corollary 1. Let PiE:k[XI, •..

,x n]

andKbeanextensionofksuchthat n Pi (x) = 0 for a "point" x = (xl' .•. ,x n ) £K . Then there is a point

y = (Yl' •.• 'Yn) with coordinates in a finite algebraic extension of k such that Pi(y) = 0 (for all indices i). be the ideal of K[XI, •.• ,x n ] generated by I. By hypotheK does not contain I, hence a fortiori I fa I elK. Thus I is strict.

Proof. Let I sis I

K

78

- 11.5 -

(1.3) Corollary 2. Let I be an ideal of k[X , •.• ,xn]and K an algebraic

l n VK(I) cK • Then P E k [Xl' ... ,X n] vanishes on V(I) if and only if a power of P is in I. In particular if I is a closure of k. Put V(I)

prime ideal, P £1 if and only if P vanishes on V(I). Proof (Rabinowitsch). We add an indeterminate Xo and consider the ideal I of k[Xo, ... ,xnJ generated by I and 1 - XoP where P is the polynomial supposed to vanish on V(I). By construction, I has no zero on V(I) hence no zero at all. By the Nullstellensatz, I is not strict and we can write 1 = A (1 - X P)

o

0

+

L

a fini te sum wi th some polynomials P.1

A.P. 1

E:

1

I. Making X0

l/.P in this

formal relation leads to a relation '\

-1

, XI ' · · • , Xn ) Pi (X 1 ' · • · , Xn ) = 1 , and multiplying by the highest power pt of P appearing in the denoL-

Ai (P

minators of the Ai (P

-1

t

,Xl' . · · ,X n ) , we get P

E:

I.

n Let us fix an algebraic closure K of k. An affine variety in K defined over k is by definition a set of the form V(I) for an ideal I of k[Xl, •.. ,xnJ. This variety is said to be irreducible when the corresponding ideal of k[XI, .•. ,Xn]formed of all polynomials vanishing on it is a prime ideal (by definition, a prime ideal is an ideal whose complement is stable under multiplication and contains 1, or equivalently an ideal giving by quotient an integral domain). By the above corollary 2, irreducible (affine) varieties defined over k and prime ideals of polynomial rings over k correspond one-one to each other. This notion is relative to the field k, and we should say k-irreducible instead of irreducible, to be quite precise. When K is an algebraic closure of k and the ideal I K generated by I in K[Xl, ... ,x n ] is still prime, we say that VK(I) = VK(IK)CK n is absolutely irreducible (it is irreducible over any field, containing K or not). 79

- 11.6 -

As before, let k be any field, and let K be a field extension of k. The set of common zeros of a family (F ) of homogeneous polynoi mials (we shall say: forms) of k[Xo' ••• 'X n ] in Kn + l can be identified with a subset of pn(K), the space of lines through the origin in Kn + l (or equivalently the quotient of Kn + l - to} under the equivalence relation given by homotheties). Again this set of zeros depends only on the ideal I generated by the Fi ' and we denote it by VK

=

VK(I)

c: lP n (K)

This is what we call a projective (sub-)variety of Wn(K) defined over the field k. We say that an ideal of a polynomial ring is homogeneous when it can be generated by a family of forms. One sees at once that an ideal of polynomials is homogeneous exactly when the homogeneous components of its members are still in it. Thus, a homogeneous ideal has a set of generators constituted by a finite number of forms (and Lefschetz' principle is still valid for projective varieties in characteristic zero). A hypersurface in pn(K) defined over k is the set of zeros of a single form F with coefficients in k, or equivalently of a principal homogeneous ideal (F) of k[X o , .•• ,X] n ). We shall need (a special case of) a theorem asserting that when K is algebraically closed, a homogeneous ideal wi th m, n genera tors has a zero in pn(K) (such an ideal has always a zero in Kn + 1 , namely the origin, but this trivial zero does not correspond to a point of pn (K) ). In other words, the intersection of m, n hypersurfaces in Wn(K) is not empty if K is algebraically closed. This is a corollary of the dimension theorem, or of elimination theory, but we prove it directly. (1.4) Theorem 2. Let k· be a field, l"m,n two positive integers, and

(F.) .-1 1 1- ,

•••

,m

a set of m forms of k[Xo' ... 'X n ]. Then there is a

fini te algebraic extension K of k and a point 0 , x = (x o 80

' •••

,x n )

E:

Kn+1

- 11.7 -

such that Fi(X) = 0 for i = l, .•• ,m

(if none of the F.1 is constant).

Proof. Let k' = k(Fl(X), ••. ,Fm(X))ck(X0 , ... ,X) n with Fl. (X) = =

Fi(Xo~

..• ,Xn) · The transcendence degree of k' over k is smaller or

equal to m, hence one X.1 at least is still transcendent over k'. Let p = deg k(X o ' ••• ,Xn)/k' has been chosen so that

~

and assume that the numbering

1

Xl' .•. ,X p is a transcendence basis of

k(Xo' ... 'X n ) over k'. We put L = k'(Xl, •.. ,Xp ) and make a diagram. We define a discrete valuation on L

I

k(Xo'···'X n )

trivial on k' by v(X l ) = -1 , v(X Z) • •• = v(X ) = 0 • This defip

finite algebraic

k'(Xl, ... ,X p ) = L purely transcendental of degree p

I

k'

keF.

1

(X)).-l 1-

~

nes v uniquely because if P/Q is a representation of an element of L as

1

rational element over k', then ••• m

v(P/Q) = deg 1 (Q) - degl(P) where deg l denotes the degree in Xl (this

k

is the valuation at infinity associated to Xl). This valuation has an extension to k(Xo' ... 'X n ) which we still denote by v. Let i be such that v(X i ) is minimal, hence v(Xi)~-l. Thus

.

for j = O, ••• ,n v(X·/X.) = v(X j ) - v(X i ) ~ 0 J l. Now the valuation ring Rv V(F)~OJ has a unique { F E: k (X 0' • • • , Xn ) (*)

V(F)~O} , so that Jl= Rv/M v is a field

maximal ideal Mv ={F€R v containing k' (:> k). Also v(F (X IX.

t

01

= - dt

dol

, ... ,X n IX.)) = v(F." (X , ••• ,X )/X. ) l '0 n 1

v ( Xi) ~ d-t

= deg (F.t)

~

1

Let yJo be the class of X./X. in Jl (this is meaningful by (*)), so that J

of Y

=

1

(y.) ~ J1?+1 (remember y. J

1

=

1) is a common zero of all

F,

~

by (**0. The theorem follows by Cor. 1 of Th. 1 applied to the ideal 81

- 11.8 -

generated by I and I - Xi (which has still y as zero). (Alternately, one could also say that if 0 were the only zero of I in Kn + l , with an algebraic closure K of k, then I would contain all X~ , j=O, ... ,n), J

for a sufficiently large integer N by Cor.Z of Th.l, and thus could not vanish on 0 ;: y

E

.n.n+l .)

It is convenient to fix an algebraic closure K of k. As in the affine case, we say that a projective variety (in pn(K)) defined over k is k-irreducible, when its corresponding homogeneous ideal of k[Xo' ... 'X n] is prime, and we say that it is absolutely irreducible when its corresponding homogeneous ideal of K[Xo' ... 'X n] is still prime. Let k[X , .•. ,X] = ffi Ad be the decomposition of this graded o n d)O ring in homogeneous components (the Ad are finite dimensional k-vector spaces). A homogeneous ideal I of this graded ring is precisely an ideal I such that 1= $I d = $(InAd ). By Cor.Z of Th.l, I has 0 as only zero if and only if I d = Ad for some d ~ 0 (hence 1m = Am for all m ~ d). We shall say that a homogeneous ideal I is strict when Id

t:

Ad for all d:) 0 (or equivalently when the corresponding subvarie-

tyof pn(K) is not empty). Th.Z says that an ideal of k[Xo' ... 'X n] generated by m ~n non-constant forms is strict. The only prime homogeneous ideal which is not strict is

d

$

Ad . Let I be a.ny strict,

~l

homogeneous, prime ideal of A = ~Ad ' and let V be the corresponding projective variety in

pn(K). We define the field key) of k-rational

functions on V to be the field of classes of elements F/G with F and Ge:A d for some with the equivalence relation

d~O,

and G¢I d

whenever FIG Z - FZG I e I Because I is a prime ideal, the quotient A/I is an integral domain FI/GI~FZ/GZ

and

82

- 11.9 -

k(Y) c field of fractions of A/I

+

All this is done without reference to a k-basis of Al (only the gradation of A is used), or as we say, without coordinate system (i.e. without the special linear forms X.). But if I is as before, one linear l.

form Yo ¢ II and we may even suppose that Yo is selected among the forms

x.l.

if we want. After renumbering the indeterminates, we may thus

suppose

Xo It. I. We fix now Xo and classify the strict, homogeneous,

prime ideals I which do not contain Xo • With that purpose in mind, we introduce two operations (inverse to each other in a sense), the first one being the homogeneization operation which associates to each polynomial GEo k [Xl' ... ,X n ] the homogeneous polynomial (form) G* in the n + 1 indeterminates Xo, ... ,X n defined by * _ deg(G) (1.5) G (Xo'···'X n ) - Xo G(Xl/Xo' ... 'Xn/X o ) The second one is the dehomogeneization which associates to each form FE: k[X o ' ... 'Xn] the polynomial in the n indeterminates F0 defined by (1.6)

Making Xo = 1 in the definition of G*, it is obvious that (G*)o

G

for every polynomial G. Moreover it is easily seen that (F o )* = F for every form F not divisible by Xo (from this one can see in general that (F )* = X-m·F where m is the largest integer such that F is o 0 divisible by X~ ). These operations are extended to ideals and give a one-one correspondance between (e.g.) strict, prime ideals of k[Xl, ... ,Xn]and strict, prime, homogeneous ideals of k[Xo' ... 'X n ] which do not contain Xo . If we fix again a strict, homogeneous, prime ideal I not containing Xo ' the homomorphism key) cfield of fractions of A/I G

has for kernel precisely the set of G such that G* 83

E.

I and so consists

- 11.10 -

of those G

= (G*) o E: I 0 . In other words, we have an isomorphism

field of fractions of k[XI, .•. ,Xn]/I o ~ key) given by the preceding map. In particular, we could have defined the field of k-rational functions key) on the k-irreducible projective variety V corresponding to the ideal I as being the field of k-rational functions on the k-irreducible affine variety Vo corresponding to the ideal 1 0 (this last field being defined as field of fractions of the integral domain k[Xl, •.. ,Xn]/I o ). The drawback of this method is that it depends a priori on the choice of the hyperplane at infinity Xo = 0 (where Xo t I) · When we work with a finite number of projective varieties, it is always possible to select a linear form Yo (if k is infinite) such that all ideals defining the varieties do not contain Yo (note that it may not be possible to choose Yo among the X.1 here). In other words, it is possible to choose a hyperplane at infinity not containing any member of the finite family. Thus many (local) problems are reduced to affine ones after a suitable choice of hyperplane at infinity. We also note that by Cor.2 of Th.i, the elements of key) have an interpretation as functions defined over non-empty subsets of V

=

VI ' because if FIG is one representation of an element of key), Gt I implies that the set of points of V where G does not vanish is not empty, and on this part of V, F/G defines a function, because F and G are forms of the same degree. Moreover, if Fl/G l is another representation of the same element of k(V) , it will determine the same function in the intersection of the domains where these functions are defined. Conversely, if FI/G1 and FZ/G Z determine the same function on the subset of V where G1G Z does not vanish, they determine the

84

- 11.11 -

same class in key). We do not prove this in general, hence shall refrain from using it (in this general form). We say that an element f

E:

key) is defined at the point P of

V (= V and K is an algebraic closure of k) if f admits a representaK tion FIG with either F(P) 1 0 or G(P) 1 0 (in fact F(P) has no meaning because P is a set of points of the form ~P

o with P0 € Kl1+~\O} and F(P) = (K~)deg(F)F(P ), so that we should write F(P ) 1 0 or o

0

F(P) 1 to}). Then f is said to be defined and finite at the point P when moreover, FIG can be selected (in the class of f) so that G(P) 1

o.

The set of functions f

at P is a ring Rp • For f unambiguously by

FIG

(P)

E:

E:

key) which are defined and finite

Rp c k(V), the value f(P) is defined

= F(Po)/G(P ) o

for a PoE:P. If we assume that

P does not lie on the hyperplane of equation Xo = 0 (which we may after renumbering the indeterminates), and if we identify key) with the field of fractions of R = k[Xl, .•. ,Xn]/I o ' we see that Rp is the subring consisting of fractions FIG with G(P o ) 1 0, where Po (1, ... ) is in the class of P. The set of f £ R such that f(P 0) = 0 is a prime ideal Rlp

r

of R (even a maximal ideal if k

= K) because the quotient

c K. Then Rp is nothing else than the localized of R at

Rp = R = R[(~]-l C field of fractions of R

r

= key)

l'

•

(1.7) Definition. The subring Rp of k(V) consisting of functions defined and finite at PeV K is called local ring of V at P. This definition is justified by the fact that Rp is a local ring in the sense of commutative algebra

Rp

-

R; = {f

= FIG:

F,GER and F(P)

= o}

=1'~

is an ideal (hence the only maximal ideal of Rp ). (1.8) Proposition. The local ring Rp is a noetherian ring.

85

- II.lZ -

Proof. We show that every ideal J p of Rp is finitely generated. Let J = J p n R. Then J is an ideal of the noetherian ring R (a quotient k[Xl, ... ,XnJ/I

of a noetherian ring is noetherian) and so we can o choose a finite family fl, ... ,f N of R-generators of J. I claim that generate J p (over Rp ). Indeed, if fEJ p , we can write f = g/h i with h(P) f 0 and g = L a.f. (a. E:R). Then 111 the f

f

=

r:

(ai/h)f i

with ai/h£R p by definition ·

We turn now to the study of curves. (1.9) Definition. An irreducible variety VClPn(K) (defined over k) is an algebraic curve (over k) if the transcendence degree of key) over k is one. We shall mainly be interested in irreducible plane curves, i.e. curves in lP Z(K) defined by an irreducible form F E: k[X ,Xl' XZ]. o

Plane curves are hypersurfaces inWZ(K), so that we can apply Th.Z. (1.10) Proposition. Let K be an algebraic closure of k and F and G two forms of k[Xo'XI,x ] determining two curves V(F), V(G)ClPZ(K). Z

If F

~

G have no common factor, the intersection V(F)n V(G) is

finite (and not empty). Proof. We assume that neither F nor G is divisible by Xo ,and we prove the assertion of the proposition in the affine piece Xo 1. Thus Fo and Go have no common factor in R = k[Xl,X Z] (notations of (1.6) ) and also

no common factor in k(X l ) [Xz] (Gauss' lemma). As this ring is now principal, we can write aF 0

+

AF o

+

bG o = 1

wi th sui table a, b E: k (Xl )[X ] z and consequently there exists a polynomial 0 t- C(X l ) ~ k[X l ] so that BG o = C f 0 with polynomials A,B

E:k[X l ,X ] 2 The common zeros of Fo and Go must have a first coordinate among the finite set of roots of the polynomial C. Similarly for the second

86

- 11.13 -

coordinates of the intersection points (with another polynomial ~). This proves that the intersection is finite, because V(F) (and V(G) ) have only finitely many points on the line at infinity Xo

O.

(1.11) Corollary 1. Let FE: k [X o ' Xl ' Xz] be a prime form, and V(F) be the corresponding irreducible plane curve. If f

E:

k(V), the set of

points P E: V(F) where f is not defined and fini te f ¢ Rp is fini te. Proof. Let us select a representation AlB of f, with B¢(F). Because F is irreducible, Band F have no common factor and we can apply the proposition, because the set of points where f is not defined and finite is contained in V(B)nV(F). (l.lZ) Definition. Let FEk[X o 'X 1 ,x Z] be an irreducible form, and P £. V(F) c pZ (K). The point P is called regular (or simple, or non-singular) on V(F) when the formal derivatives

(~F/~Xi)(P)

(i = O,l,Z)

do not vanish simultaneously at P (same convention as before regarding the meaning of this). A point which is not regular is called singular. A projective curve is called regular if it has no singular point, and singular if it has at least one singular point. (We shall also say that an affine curve is regular, or non-singular, when the projective curve defined canonically by homogenizing is regular.) Euler's relation gives LXi(dFldXi)

= deg(F)·F

and if P E: V(F) is not on the line at infini ty Xo = 0, we normalize PoE:P by Xo(P o ) = 1 (i.e. Po = (1,·,·) ) and we deduce

But obviously

aFldX.1 = (oF 0 IdX.)* for i = 1,Z , and so, the conditions 1

(dF I~X.)(P )

=

(dF IdX.)*(P ) not both 0 for i

01001

0

= 1,2

are already sufficient for the non-singu1ari ty of a point P €. V(F) not on the line at infinity Xo

=

0 .

87

- 11.14 -

(1.13) Corollary 2. Suppose that the field k is perfect. Then the set of singular points on an irreducible plane curve defined over k is finite. Proof. If F and the

dF/aX i have infinitely many common zeros, they

must have a common factor, and hence (by irreducibility of

must be a multiple of F

dF/~X. 1

F). Because the dF/dX.1 are forms of degree,

deg(F) - 1 , it follows that they vanish identically, and by Euler's relation deg(F)·F =

L.

X.(dF/dX.) -= 0 . 1 1

This is sufficient in characteristic 0, because F defining a curve. In characteristic p

~

=0

is not a form

0, we see that deg(F) must be

a multiple of p, and more precisely, the vanishing of the partial derivatives shows that F depends only of the X~1 . Because k is perfect F = (F')P by taking the pth roots of all coefficients of F, and so F could not be irreducible. We have obtained a contradiction in both cases. (1.14) Examples. We consider the following affine curves.

o

X3

+

y 3 - Xy

=

0

They are irreducible, and have all the origin as singular point. The class f of Y/X is not defined at the origin for any of these curves, nor is l/f defined at the origin for any of them. On the curve of equation Y - X3 = 0 , both £ and l/f are defined at the origin, only f being finite at the origin

3

fe:R(O,O)(Y-X) 88

- 11.15 -

(1.15) Convention. For the rest of the section, K a fixed algebraic closure of k. and V

=

=

k will denote

V (F) Cp 2(K) will be a KK

irreducible (i.e. absolutely irreducible) projective plane curve defined over k by a form F E: k [X o ' Xl' X2 ] . Let X = X denote the set of (discrete) normalized valuations K v of K(V) (i.e. such that vK(V)~ =~) which are trivial on K. This set is going to play the role of "spectrum" of the field K(V) and will be very close to the set of points of V (cf. Cor.l below). We say that a valuation ve:X is centered at a point P = Pv £V if for f E:R p v(f»O is equivalent to f€M p Obviously v is centered at most at one point P because if P ; P' £V there are functions f in RpO Rp ' with

f €.M p but f ¢M p ' (take for f

the quotient of two linear forms gjh where h does not vanish at P,P' and g vanishes at P but not at P'). (1.16) Lemma. Every valuation v E:X is centered at a (unique) point p = PV E.,V , so that we have a well defined mapping VK defined by v ~ ~(v) = Pv Proof. Let X,Y,Z be a permutation of the X.1 such that v(X/Z)

n

X

~

~

° and

v(Y/Z):> 0. Denote by x (resp. y) the class of XjZ (resp. Y/Z) in K(V). If vex - ()

=

° for

all

~ E:

K, we would infer that for every

polynomial P(X)€ K[X] we would also have v(P(x)) = 0, because every polynomial is a product of linear factors in K(algebraically close~. Also v(R(x)) = numerator

° for

every rational function R(X)£K(X) because both

and denominator have valuation 0. But as K(V) is algebraic

over K(x), it would follow that v is trivial: if teK(VJwrite its minimal polynomial t N + alt N-

which implies vet) This

~

=

l

+ •••

=

O. hence Nv(t)

= v(alt N- l + ••• )

0. Thus there is a {E: K such that vex - {) >0.

is unique, because if

~1

were another element of K with 89

- 11.16 -

the same property, we would derive Inf(v(x-~)

,v(x- f'))

>

° contrary to

v(~/_~) = v((x-~)

-

(x-~')) ~

the fact that v is assumed trivial

on K. Now write the equation of the affine part of V in the form

=

Fo(X,Y) and factor Fo(~ ,Y)

L. v(y

c

Fo(~'Y)

11 (Y

- '1i)

- (X -

=

~)A(X,Y)

°,

- ~i) in linear terms. We have

= vex - {)

+ v(A(x,y))

~ vex - 0 > 0

Because x and y have been chosen with yositive valuation, every polynomial in x and y will have positive valuation, and the above relation can only hold if for one i at least v(y -

~i)

>

0. But for

the same reason as above, this strict inequality can hold at most for one

~i

• By construction, the point P

=

( ~, 'Ji ,1)

Pv

is on

the affine curve Va of equation Fo ' and defines a point (still denoted by P) on V. After a translation, we may suppose that this point is at the origin Pv

=

(0,0,1) of the affine coordinate system chosen.

If fE:.R p , then f admits a representation AlB

with B(O,O,l) ;

Because now vex) >0 and v(y) >0, this implies v(Bo(x,y)) = v(f) = v(Ao(x,y)/Bo(x,y)) = v(Ao(x,y)) Ao(O,O) =

° or

>

° is

O.

° and

equivalent to

to f£M p as was to be shown.

We show a kind of converse to this lemma. (1.17) Theorem 3. Let P be a regular point of V. Then there is a

unique valuation v p valuation, f

E:R p

ring and Mp =

ord p E: X which is centered at P. For that

is equivalent to ordp(f)

(~p)

~ 0.

Thus

Rp

is a valuation

is principal with generator any element such that

n M~

ordpC't'p)

1. Also

determined

br ordp(7t~R;)

r)O

lo},

so that the valuation in guestion is

ordp(M~ - Mr 1 ) = r. (K = k, cL end of proaL)

If S = S(V) denotes the finite set (K is perfect,(1.13),(1.15)) of singular points of V, we see that ord : V - S p

~

ord p is inverse of

7t

(rc-v and

on the sets where they are defined). 90

V-1t

~

X defined by

are the identi ty mapping

- 11.17 -

Proof. Let again X,Y,Z be a permutation of the X.1 such that Z(P) , 0 and (dF/dY)(P) ,

° . Normalize Poe:P by Z(Po )

= 1 ,

and identify Po

with a point (~,~) of the affine plane K 2 • After a translation (replacing x -

~

by x' and y -

by y') we may assume that Po = (0,0)

ry

is at the origin of the affine coordinate system defined by X,Y • Thus the affine part Vo of V is defined by the polynomial Fo with no constant term. Supposing that Fo is not proportional to Y (trivial case), we may assume Fo(X,Y) with A(O)

=

YA(Y) - B(X,Y)X

1 (this is no restriction) and

B(X,Y) = Bo(X)

° (because

YBl(X)

+

+ •••

Fo is not divisible by V). We denote also by x and y the classes of X/Z and Y/Z resp. in K(V).

with Bo(X) ,

First step. Obviously Mp = (x,y) is generated (over Rp ) by x and y. We show (under the preceding hypotheses) that Mp is principal and generated by

x , or what amounts to the same, that y is a

rr p

multiple of x in Rp

By (*) we have yA(y)

•

y = xB(x,y)/A(y) e:xR p because A(O) ,

B(x,y)x, hence

=

° implies

l/A(y) E:R p •

Second step. We show more precisely that there is an integer r such that y€xrR;

= M~

-

If on the contrary B(O,O)

Mr

1

= 0,

• This is so with r

=

~

1

1 if B(O,O) "f O.

we proceed as follows: multiplying

by A(y), A(y)B(x,y) = A(y)Bo(x) and writing Bo(X) =

+

Bl(x)A(y)y

+

•••

XB~(X),

A(y)B(x,y) =

= x(B~

x(B~(x)A(y)

+

B1 (x)B(x,y)

+

y( ••• ))

B1Bo + y( ••• )) = x(~o + y( •.• )) with a new polynomial Ero(X) = B~(X) + Bo(X)Bl(X) containing exactly one power of X less than B (X). If Xr - l is the highest power of X +

o

dividing Bo ' we shall get after r-l steps 91

- 11.18 -

A(y)

r-l

B(x,y)

x

=

r-l

~

(Bo(x)

+

y( ••• ))

with fo(O) ~ 0 • This shows that y

= xrC(x,y)/A(y)r

°,

with C(O,O) !

r

hence the assertion (C(x,y) / A(y) E:R p ). )C

Third step. We show now that

n

~ = {O} mlO (this is Krull's theorem valid quite generally for local noetherian rings, but we indicate a direct proof in our particular case). Equivalently we show Rp -

{OJ

=

U

m)O

JCRR"

p p

=

U (~p _ ~+1) p

m~O

If fe:R p - {OJ we choose a representation C/D of f with D(O,O) ! 0. If C(O,O) ! 0 , fER; and we are done. If on the contrary C(O,O)

0 , it is rather obvious that a method similar to that m )( used in the second step will lead to f £ x Rp • =

The conclusion of the proof is now easy. Any normalized valuation (trivial on K) centered at P is such that vex)

=

1 and trivial on

R~ • This shows that there is at most one such valuation. Conversely

we can define for f = xDlg(x,y) with mE.7L ordp(f) = m. As every element f of

K(V)~

and g(x,y)€R; is of this form this gives

the existence of the required valuation, and concludes the proof. We note however, that we have only used the fact that k = K is algebraically closed to be able to suppose that the point under consideration was at the origin. In other words, to have x -

~Ek(V)

and Y -"]Ek(V).

Thus, the proof shows that the result is true if we replace K by the subfield k'

k(~,~) =

k(P) generated by the coordinates of Po

over any field of definition of V (i.e. containing the coefficients of the polynomial F0 (X, Y) defining V). This field k(P) is a fini te algebraic extension of k (independent of the choice of the line at

92

- 11.19 -

infinity, as long as we take the representant Po with one coordinate equal to 1). (1.18) Corollary 1.

Ii

V = VK is an absolutely irreducible non-singular plane curve, n: X ~ V is a bijection. (One could show in general that finite.

7t

is surjective and ~(S{V)) is

In fact X is in one-one correspondence with the normalized

- or any non-singular model - of the curve V.) One should be careful about the fact that even if element v, v{f) >

° (for f E:K{V))

-1 ~(P)

has only one

does not imply f G:M p • Take for example as in (1.14) the curve y2 =x 3 , p at the origin, and f = y/x.

Then there is a unique valuation v centered at P, and for that valuation v{f) Recall now that a subring A of a field F is called valuation ring if for any

°;: x

E:

F, either x or x-I belongs to A. The rings

associated to valuations (by v(f»O) as before, are obviously valuation rings in this sense. But conversely (1.19) Corollary 2. Rp is a valuation ring of K{V) if and only if the point P is regular (and then Rp is the ring of ord p ). Proof. Remains to show that if Rp is a valuation ring, then P is regular. So we suppose that ylx By definition we can write y/x

E:

=

Rp (otherwise interchange y and x) • A(x,y)/B(x,y) with formal polyno-

mials A(X,Y), B(X,Y)E:K[X,Y] and say

B(O~O)

= 1 (we suppose P at

the origin). This gives YB(X,Y) - XA(X,Y) £1 o (F o ). A degree consideration shows that up to a constant factor Fo = YB(X,Y) - XA(X,Y) so that (dFo/3Y)(0,0) = 1 ;:

° and P

(0,0) is regular.

We note explicitely that all f£K(V) are defined at all regular points PEV (possibly infinite at some of them). If f is defined and finite at the regular point P, f(P) is the unique 93

- 11.20 -

element aoof K such that ordp(f f

= ~mod

aJ >0. Equivalently we may write

Mp . From this it follows that f can be expanded in a

formal Taylor series 2

T(fj = a o + alxp + a 2np + ••• -1 where a l is determined by a l = (f - ao)~p mod Mp , and so on inductively. This series converges to f in Rp (or its completed~p) for the (1.20)

topology defined by taking the powers of the maximal ideal Mp as neighbourhoods of 0 (hence all ideals of Rp are neighbourhoods of 0 because Rp is principal). We can also look at this expansion as formal series T(f)

E: K[[1rp]]

• More generally, every f

E:

K(V) can be

expanded in a Laurent series at P, which will be an element of the field of fractions K((np )) of the former ring K[T~] gives the usual

Then ord p

notion of order of a Laurent series.

Suppose now that V is non-singular. Let Div(V) be the free abelian group generated over the set of points of V. Then we can define the divisor of f e: K(V)

by the formal expression L.ordp(f}(P) •

By (1.11) applied to f and l/f , only finitely ordp(f) will be different from 0 which assures div(f)

=r:

ordp(f). (P) £ Div(V) . PE-V (If V were not assumed to be non-singular, we could define the (1.21)

divisor of f on X using the fact that ~(S(V)) is finite, by div(f) =

L

veX

v(f)· (v)

94

E:

Div(X)

.)

- 11.21 -

Another consequence of Th.3 is the possibility of defining purely algebraically the order of contact of a line and the curve V at a regular point PE:V. Let o<X

+ ~Y +

l =

0 be the equation of

an affine line. We can define the order of contact of the curve and this line at the regular point P £V to be the positive integer ordp(<<x

r).

+ ~y +

This order will be 0 exactly when P is not on

the line under consideration, it will be through P. When this order is

~

if the line passes

~l

2 we say that the line is tangent

to the curve at P, and when this order is

~

3 we say that this line

is an inflexion tangent at P, and that P is a flex of V. Every regular point has a unique tangent. To see this, we suppose the point at the origin of the affine coordinate system chosen, and we write the defining equation of the curve in the form Fo(X,Y) = aX terms of degree is

~

bY

+

~2.

2, so that aX

+

G(X,Y) with a polynomial G containing only

+

Then ordp(ax bY

=

+

by) = ordp(-G(x,y)) = ordp(G(x,y))

0 is the equation of a tangent to p (note

that (a,b) f (0,0) because P is regular). In this case ~p

bx - ay is a uniformizing variable (i.e. generator of Mp )' and

if a'X

b'Y

+

get a'x

+

a'X

b'y +

0 was the equation of another tangent line, we would

=

2

= 7t p u

(UE:R p )

b'Y - b 2X2

+

and thus 2abXY - a 2y 2

= 0 mod (F o )

,

hence, by degree consideration, the left hand-side would be equal to Fo (up to a constant in K), and this shows (a' ,b') proportional to (a,b). To find a convenient criterium for the point P to be a flex, we make the further reduction where a we can take

~p

=

0 (so that b

r

0 and

= x as uniformizing variable, as in the proof of

Th.3), and we write more exp1icitely the defining polynomial (1.22) Fo (X,Y) = Y + rX 2 + sXY + ty 2 + G'(X,Y)

95

- 11.22 -

with a polynomial G' containing only terms of degree higher than 3. Then Y = 0 is the equation of the tangent, and ordp(Y)

ordp(rx

2

+

sxy

+

equals 20rd p (x) = 2 as soon as r is a flex precisely when r

=

o.

ty ~

2

+

G'(x,y))

0 (because ordp(Y) ~2). Thus P

We shall transform this condition in

a projective form. Let F be the form defining V and let H(X ,X 1 ,X 2) = det(~2F/aX.OX.) o 1 J This is a form of degree ~ 3(n - 2) if n = deg(F), called the (1.23)

Hessian of F. The plane curve V(H) defined by H(X o 'X 1 ,X 2 ) = 0 is also called the Hessian of V. If the characteristic of the field K is 0, the Hessian form is of degree 3(n - 2) exactly, and we have the following criterium. (1.24) Proposition. Suppose that K is of characteristic 0, and let V(H) be the Hessian of V. Then a regular point P of V is a flex precis,ely when P f:: V ()V(H) is an intersection point of. the Hessian (of V) with V. Proof. We take Z(P) ; 0, is at

~nce mor~

(~F/dY)(P)

the~origin

a permutation X,Y,Z of the Xi such that

; 0, and assume that Po (normalized by Z(Po)=l)

of the coordinate system defined by X,Y. Moreover

we may assume that the tangent line to P is the line Y = 0 and so Fo will be given by (1.22) of which we take the notations. We compute ZH(X,Y,Z) by multiplying by Z the "last" column of H and we add to this column X times the first one and Y times the second one ( this does not change the value of the determinant). Euler's relation in the last column, will give the three terms (n-l)~F/aX

, (n-l)dF/dY , (n-I)()F/dZ ,

because these partial derivatives are forms of degree n-l (we are in

characte~istic

0). We do the same trick with the lines, adding

96

- 11.23 -

X times the first one, Y times the second one, to Z times the last one. We get

F" XX

F"

XY

(n-l)F~

F"

F" YY

(n-l)F y

YX

I

(n-l)F~ (n-l)F~ n(n-l)F

(with notations for the partial derivatives which are better suited to typewri ters! ). If P ~ V we get thus 2r

s

s

2t

a

b

a

b (n-l)2

o

with the notations of (1.22) where we have supposed

a

= 0, b = 1.

This shows that the point Po will be on the Hessian precisely when r = 0 in the case of (1.22), which is a necessary and sufficient condition for P to be a flex. We have used the characteristic assumption to be assured that the partial derivatives are forms of degree n-1 and to conclude r = 0 from 2(n-l)2 r = O. Thus the characteristic assumption can be relaxed Il.'

in special cases. For example, if F is a form of degree 3 (defining a cubic V), the result of the proposition will obviously hold in characteristic p ; 2, 3. (1.25) Corollary. In characteristic 0, a non-singular plane curve of degree > 2 has at least one flex. Proof. Indeed, the Hessian is of degree 3(n-2»0 in that case, and it is sufficient to apply Th.2 (1.4).

97

- 11.24 -

2. Plane cubic curves

Let k be any field, K

=

-

k an algebraic closure of k. A plane

cubic curve defined over k is given by a cubic form (2.1)

3

aoX o

+

(a l X1

+

2

a 2X2)X o

+ •••

+

3

a 6X1

+

2

a 7Xl X2

+

2

a 8 X1 X2

3

+ a9~

with ten coefficients a e:k. Two families (a ) determine the same i i cubic curve in p2(K) whenever they are proportional (with non-zero factor of proportionality). Thus we can identify cubics in

p2(K)

defined over k with elements of p9(k). Some of these cubics are reducible (degenerate in a line plus a conic, or three lines, distinct or not), and some irreducible cubics have singularities. We shall show first, that with mild assumptions, a K-irreducible non-singular such cubic has a very simple equation in a suitable coordinate system. Equivalently, we could say that after a suitable projective transformation of p2(K), the curve of equation (2.1) can be brought on a curve with a much simpler equation. (2.2) Theorem 1. Let V be an irreducible plane cubic defined over the algebraically closed field K of characteristic p f 2. Suppose that V has a flex P. Then there is a coordinate system in which V has an equation of the form y 2Z - F(X,Z) = 0 with a form F of degree three. Conversely, if F is not divisible by Z, this equation defines an irreducible cubic with a flex on the line Z

= o.

Proof. Choose a basis X , Y, Z of linear forms such that the flex has (X,Y,Z) coordinates respectively equal to (0,1,0). Suppose moreover that Z

=

0 is the tangent to V at the flex P. In the

affine coordinate system Y form (1.22) with r

=

=

1, the equation of V will have the

0

Z + sXZ + tZ 2 + A3 (X,Z)

98

0

- II.2S -

with a homogeneous polynomial

~of

degree 3. In projective form, this

equation will be

Zy Z

sXYZ

+

tYZ Z

+

+

A3 (X,Z) = 0 .

Because we assume that the characteristic is not 2 we may define a new form Y'

=

Y

+

Zy,Z

}(sX +

+

tZ) whence an equation

A~(X,Z) = 0

with another homogeneous polynomial A; of degree 3. This is the required form. Conversely, if F is not divisible by Z, the irreducibi1ity amounts to the irreducibility of the polynomial y2 _ F(X) in K[X,y] or in K(X)[Y]

(~auss' lemma)

where F is a polynomial of degree 3. This is obvious because this degree being odd, F cannot be a square. Thus y2 Z - F{X,Z) =

°

is the equation of an irreducible cubic (if F is not divisible by Z), and this cubic has a flex at (X,y,Z) = (0,1,0). We emphasize that if V is K-irreducible and non-singular, and if the characteristic p is not 2 or 3, the existence of a flex follows from (1.25) (and the remark just before it), hence (2.3) Corollary. Every irreducible non-singular cubic in characteristic p r 2,3 has a Weierstrass equation y 2Z = X3 + aXZ 2 + bZ 3 with 4a

5

p

r

+

27b 2

r

0. An irreducible non-singular cubic in characteristic

2 having a flex, has an equation in Legendre form y2 Z = X{X - Z) (X - A Z)

~

r

°, 1

(over the algebraically closed field K). Proof. The first result follows from the remark above combined with the possibility of completing the cube in characteristic r 3 and thus killing the factor X2 in the cubic polynomial in X. The Legendre form follows from a normalization of the three roots of the cubic polynomial. (2.4) Remark. The Weierstrass form in characteristic p 99

r

2,3 can

- 11.26 -

be

attained over any field of definition k of V containing the

coordinates of the flex P. The same remark obviously does not apply to the Legendre form. But we see at once on the Legendre form, that the only irreducible singular cubics having a flex, in characteristic p ; 2 have equations of the form X2 (X - Z) flex at (0,1,0), double point at (0,0,1) X3 flex at (0,1,0), cusp at (0,0,1) Now we turn to the existence of group laws on cubics. Let V be an irreducible plane cubic over the algebraically closed field K of characteristic p ; 2 having a flex P. Let S = S(V) be the set of singular points on V and R(V) = V - S(V) the set of regular points of V. For any two points Q, Q'

E:

R(V), let Q1' be the third

intersection point of V and the line through Q, Q' (this line is the tangent to V if Q we define

Q +

P

and thus Q" = Q if Q = Q'

Q'

=P

). Then

Q' to be the third point of intersection of V and

the line through P and Q" (the symmetric of Q" with respect to

Pl.

Obviously all intersection points involved are regular on V (look at the two singular cubics above, and note that lines through the singular point of V cut V in only one point, and are not tangent to that point). Not less obvious are Q + Q'

P

= Q'

Q

+ Q

P

+

P

P =Q

and if QP denotes the symmetric of Q on V (with respect to P), Q + QP

P

=P

We study this law in the singular cases first. We start with the case of an irreducible cubic with a double point (and p ; 2), hence an equation of the form a) above. The two tangents at the origin (0,0,1) are Y = X and Y

=

-x.

100

In a new system of coordinates

- 11.27 -

admitting these tangents as axis, the equation will have the form X3 - y3 _ XYZ

= 0 • We take the affine coordinate system Y = 1

which contains all regular points on V, and make a picture. y

z

(2.5)

x

x 3 _ y 3 - xy

o

(Z

xz

1)

Z

X3 - 1 (Y = x 2 - l/x

1)

In the last drawing, we see that a line avoiding the singular point is a non-vertical line of equation z = mx

+

h , and this line cuts

the curve in the points of abscissa (mx

+

h)x

1

x

3

- mx

2

- hx - 1

=

0 ,

hence in three points (xi,zi) (i = 1,2,3) with x l x 2x 3 proves that the law + ,where P is the flex x = 1 P

is isomorphic to the multiplicative group projection (x,z)

~

(R(V),

K~under

This

1 Z

=

0 (y

1)

the first

x

p) ---+

K

This result does not use the fact that K is algebraically closed, and is true as soon as the field k contains the coordinates of the flex and of the double point. The picture shows the case K = R. Secondly, we take the singular case corresponding to the equation b) above (case of a cusp). Again, we take the affine coordinate system for which Y

=

1, containing all regular points

of V. The flex is then at the origin of this affine system.

101

- 11.28 -

z

y (2.6)

x y

x Z = X3 (Y = 1)

2 = X3 (Z = 1)

As before, a line avoiding the singular point of V is not vertical in this system, hence has an equation of the form z = m x

+

h ,

cutting the curve in three points of abscissa satisfying x 3 - mx - h = O. The three roots of this equation satisfy xl

x2

+

+

x 3 = 0 (because the equation does not contain a term in x 2 )

hence the first projection (x,z) ---- x gives here an isomorphism

R(V),+

K+

P

where P is the flex x

(additive group of

0, z = 0 (and y

=

K)

,

1). The fact that K is

algebraically closed is not more important here than in the last case of the double point, and the picture illustrates the case K

R.

Finally we come to the case where V is non-singular and the characteristic is neither 2 nor 3. With the choice of flex P, we show that V is a group for the law

+

P

•

Only remains to check the

associativity of this "addition". We choose a coordinate system in which the curve has a Weierstrass' equation, and which puts P on the line at infinity with Z = 0 as tangent. Thus we suppose that the affine equation of the curve is

y2 = X3

+

aX

+

b •

If a line does not go through the chosen flex it is not vertical in this system and has an equation y = mx

+

h , and the intersection

. h ave x-coor d·lnates satls . fylng · pOlnts x 3 - m2x 2 If P.

1

(2.7)

+(a

(xi'Yi) are these three points, we have x3

=

-xl - x 2

+

((Y2 - Yl)/(x 2 - xl))

102

2

- 2rn h) x

+

b

-

0 h 2 =.

- 11.29 -

where (Y2 - Yl)/(x Z - xl) has to be replaced by the slope of the tangent if xl = X z (compare with (1.1.14)). Thus the coordinates of PI

p P2

are rational functions of the coordinates of PI and Pz

'

with coefficients in the prime field. Thus the associativity amounts to the formal verification of the equality of two rational expressions with coefficients in the prime field, or still to the verification that the two polynomials obtained by multiplying the numerator of the first and the denominator of the second on one hand, the denominator of the first and the numerator of the second on the other hand, coincide. Hence the associativity for the sum of three points (distinct from P = (0,1,0); this is sufficient, because this point acts as unit) is reduced to a formal identity between polynomials. In particular, it is sufficient to check it in characteristic

°

(polynomials with rational integral coefficients). But in this case, it follows from the transcendental theory of chapter

I. (Of course

we could have avoided this use of Lefschetz principle had we done more algebraic geometry of curves. The interested reader should read the proof given in Fulton, cf. reference given for this chapterJ The group law has been defined in such a way as to give sum to three points of V on a line. We sum up

103

°

(=

P)

- 11.30 -

(2.8) Theorem 2. Let V be a non-singular, absolutely irreducible plane cubic defined over the field k of characteristic p ; 2,3 , and having a flex P with coordinates in k. Then there is a group law on the set Vk of points of V having coordinates in k, such that P is unit and that three points have zero sum P whenever they lie on a line. (2.9) Corollary 1. The flexes on a non-singular irreducible plane cubic defined over the algebraically closed field K of characteristic p ; 2,3 (once one of them P is fixed) are the points Q£V = V K such that Q Q Q = P , or more briefly 3Q = O.

p p

Proof. Indeed the condition given is equivalent to Q

+

P

Q = QP

because this symmetric QP of Q with respect to P is the inverse of Q for the group law deduced from P. This means that the third intersection point of V and the tangent at Q is Q = (QP)p itself. (2.l0Y Corollary 2. Let V be an irreducible non-singular plane cubic over the algebraically closed field K of characteristic

o.

Then V has nine flexes which make up a subgroup of V (when one of them is chosen as unit element) isomorphic to (71/371)">C: (Z/3Z) . Proof. Had we proved Bezout's theorem, the first part would follow immediately, because the Hessian of a cubi.c is of degree 3 hence has 3x3

=

9 intersection points with the cubic. We proceed differen-

tly (and prove a more precise result: the nine flexes are always distinct). We use Lefschetz' principle and thus suppose that the cubic is defined over k c: ([, and we assume that the algebraic closure

k of k in K is itself embedded in ([. Because the group law is given by rational functions (we suppose that P has coordinates in k, or that the cubic is given in Weierstrass' form) with coefficients in k, the solutions of 3Q = 0 have coordinates in kc([. We are reduced

104

- 11.31 -

to the case K = [, where the situation is easy by the transcendental theory of the first chapter : points of order dividing three on the elliptic curve [/L

are the points (1/3)L/L

~

L/3L , which

form a subgroup of order 9 isomorphic to ~/~)Z as was to be proved. (Z.ll) Remark. The nine flexes on such a cubic curve have a special configuration. Indeed, if PI ' Pz are two distinct flexes, we have 3P l

= 0 , 3P Z = 0 hence 3(-Pl -P Z) = 0 , so that -PI -P Z is a third

flex colinear with PI and Pz

•

To sum up, any line joining two

distinct flexes contains a third (distinct) flex. However, the nine flexes are not on a line, because a line cannot cut the cubic in more than three points. In a suitable coordinate system, these nine points will be contained in an affine portion KZ• We cannot make a picture in R Z of this configuration, because any finite set of points of R Z such that a line joining two distinct of them contains a third point of the set is necessarily on a line (proof?).

(Z.lZ) Example. Let yZ = 4x 3 - gzx - g3 be the (affine) equation of a non-singular plane cubic with gz ' g3

~ ~

. Then the set of

points (x,y) £ ~Z on this curve, form an abelian group with the point at infinity as neutral element (for the group law described before (Z.8) ). It can be shown that this abelian group is finitely generated (Mordell-Weil theorem), hence isomorphic to the direct sum of a finite group and a free group

Zr of rank r ~O. It is still

an open problem to determine this rank r in terms of the cubic.

105

- 11.32 -

Let k be an algebraically closed field, L a function field of one variable over k, i.e. a 0eparable) finitely generated extension of k of transcendental degree one. Then it is well known that there exists a transcendental element xE:L such that L is a finite separable algebraic extension of k. Such an extension has a generator y, so that we can write L = k(x,y). Let F be the minimal polynomial of y over k(x) and multiply it by a suitable polynomial in x so as to get an irreducible polynomial F€k[X,y]. Then, L if V

key)

=

V(F) is the (affine) irreducible variety of zeros of F.

=

We denote by X the set of normalized valuations of L which are trivial on k (hoping that this X will not be confused with the former indeterminate X !). We have a mapping (1.16) 1t

X

:

~

xCv)

V

This mapping is bijective outside the set S(V) of singular points -1

of V. We assume that rr (S(V)) is finite (this could be proved, but we know at least that it is true if V is non-singular). Thus every fE'.L

has a well defined divisor div(f)

=

L.

v(f)· (v)

E:

Div(X)

VE:.X

For every divisor

~ E:

L(~)

=

Di v (X) we define the k-subspace L (~) of L by {fE.L : div(f) ?; -~ }

(remember the convention

o Ei L (~)

div(O)~ -~

for any divisor

~,

so that

).

(2.13) Definition. We say that L is of genus g over k preceding conditions are satisfied and dimkL(~)

Observe that

= deg(~)

h

~

fh

+

1 - g

when the

when

for all

~

E:Div(X) of

deg(~)

defines a k-linear isomorphism of

> 2g

- 2.

L(~)

onto L(i - div(f)). In particular, these spaces have the same dimension and this proves that the degree of the divisors of the form div(f)

106

- 11.33 -

are always zero in a field of genus g.

o.

(2.14) Example 1. The rational field k(x) is of genus

Indeed, in

this case, the set X is in one to one correspondence with the projective line over k, and divisors d over X are thus expressions of the form -d = d00 (~)

+

~ d a (a) with rational integers d a vanishing for

ae:

all but a fini te number of a we observe that f

J---+-

f

E:

k. To compute the dimension of

TT (x - a) d a

L(~),

is a k-linear isomorphism of

a£k L(i) into the vector space of polynomials k[x] , having as image the subspace of polynomials F such that -deg(F)

= ord~(F) ~

or equivalently, such that

-d_ -

~

ae:k

deg(F)~d

polynomials is of dimension d

+

=

da

= -deg(~)

deg(~).

I

This subspace of

1 , so that the conditions of the

definition (2.13) are satisfied with g =

o.

(2.15) Example 2. The function field key) over a singular plane cubic curve V is of genus O. It is sufficient to look at the two special curves defined in (2.4). If the singular cubic has an . . db y ratlona . I functIons, . equatIon y 2 = x 3, . I t can b e parametrIze 3 e.g. y = t and x = t 2 • If the singular cubic has an equation

y2

x 2 (x - 1) , it can also be parametrized by rational functions,

e.g. x

=

1

+

t2 , y

=

t(l

+

t 2 ) • In both cases, key)

k(x,y) ck(t)

is a rational field (LUroth's theorem). It has genus 0 by the example 1 above. (2.16) Example 3. We have proved in (1.2.12) that if L is a lattice in [, then the field of L-elliptic functions is a field of genus one over the complex field.

107

- 11.34 -

0,1 in

The above examples exhaust the possibilities for g the following more precise sense.

(2.17) Theorem 3. Let k be an algebraically closed field, L a function field of one variable over k. If L is of genus 0 over k, k(x) (L is a rational field).

then L is purely transcendental L

If L is of genus one over k, then L is the field of rational functions over a non-singular irreducible plane cubic with a flex. In this case (g = 1), if the characteristic p of k is different from 2, the cubic can be taken in Legendre's form y2 = x ex - 1) (x - A) , and if the characteristic p of k is different from 2 and 3, the cubic can be taken in Weierstrass' form y2 = x 3

+

ax

b . In these

+

forms, the non-singularity amounts to AI: 0,1 and respectively 3 Z 4a + 27h I 0 . Proof. Observe once for all that X is not empty: L contains a rational subfie1d and all valuations from this rational subfield (trivial on k) can be extended to L (this shows more precisely that X is infinite because k is infinite). First part. Assume g = 0, and take any v£X, putting

= (v) for

~

the corresponding divisor on X of degree one. By hypothesis L(Q) is of dimension 1, hence equal to k, and contained in dimension 2. Select any

- L(Q)cL, so that necessarily

xk . By construction vex) = -1 because x ~ L(Q), and consequently v(x n ) = -n for all integers n. By induction on n, one

L(~)

=k

xe:L(~)

of

L(~)

$

sees that L(n~) = k • xk e ... e xnk . I claim that the rational subfield k(x) of L is the whole field L. Take any f in L - k and write its divisor div(f) where

(f)~ ~

div(f)

~

=

(£)0 -

(f)~

-f

-f

(symbolically f (0) - f (-) ),

0 is the polar part of the divisor of f. In particular

- (f)oo and so f E::L( (£)00). This shows that the 2 (n

108

+

1)

- 11.35 -

functions

1 , x , x2

,

...

, x

n

f , xf, x 2 f, are in

L(n~ +

n

+

+

d oO

By hypothesis this last space is of dimension

(f)~).

1 where doo = deg (f)eo . As soon as n ~ d oo

inequali ty n

dIDO

+

1

+

<

2(n

+

there is a strict

'

1). Hence there must be a linear

dependance relation over k between the former functions, and this leads to an equation

L b i xi

L a i xi. f

( ai' b i

E: k ,

0

~i

E; d-o) •

As the xi are linearly independant over k • the coefficients a i i) cannot vanish simultaneously. This proves f = (l:boxi)/{L:aox 1 1 is in k(x) as was to be proved.

=1

Second part. Assume now g Because select L(3~)

L(3~)

L(~)

and

take'~

= (v) with v£X as before.

is of dimension 1 now, we must have -

x£L(2~)

L(~)

This element must have vex)

L(~).

=

= k. So we -2. Since

is of dimension 3 over k we can select YE:L such that k

~

xk

$

yk , and this implies v(y) = -3. From this we

derive

(because v(x 2)

-4)

-S)

$

x 2k $ xyk (v(xy) 2 3 yk ~ x k • xyk • x k

$

yk e x 2k e xyk

L(4~)

k e xk ~ yk e x 2k

L(S~)

k

L(6~)

k e xk xk

k because both v(x 3 )

xk $ yk

$

$

$

$

y2 k

v(y2) = -6 . This shows the existence of a

linear relation over k between the powers in question, that is, a cubic relation between x and y of the form F(x,y) = y2

(2.18)

+

with coefficients in k. replace y

+

! (etX

+~)

equat10n y 2 = JX 3 o

(o(x

+

~)y+ ~x3

+

~x2

+ ex +

"1

0

If the characteristic is not two, we can

by Y and thus kill the term in y, getting the

r:2

+ oX

-

+ Ex +

TJ-

• Th e Legen d re 109

f orm

- 11.36 -

is obtained from there by normalizing x and y in order to get a unitary polynomial in x, and by normalizing its roots as indicated. We have repeatedly said that then A1 0,1 is the non-singulari ty condition, a fact which can be checked in any characteristic p 1 2 (the only possible singular points occur with ordinate y = 0, and their abscissa must be simultaneous roots of the cubic polynomial in x and its derivative). In characteristic p 1 3 , the possibility of killing the term in x 2 is open by completing the cube (and non-singularity condition as claimed if p 1 2,3). Remains to show that the subfield k(x,y) of L is equal to L. As in the first part, we take any f functions

2

L - k , and consider the 3 (n

...

,

x

y

,

xy, x 2y,

, x nY

f

,

xf, x 2f,

,

x

+ 3)~ +

,

,

x

+

1)

n

1

They are all in L( (2n

,

E:

xnf

(f)oo) of dimension 2n

+

3

+

doo (notations

as in the first part). As soon as n is sufficiently large (n :)doo + 1 will do), they will be linearly dependant over k, hence a relation i i i La.x + y Lb.x = f Lc.x I I I

i (otherwise, y€,k(x) : a sheer LC.X 1

with a non-zero polynomial

nonsense in view of the fact that vex) is even and v(y) odd). This gives eventually f = rex)

+

s(x)y E:k(x,y) with r(x), sex) E:k(x)

(and gives once more the fact that y is quadratic over k(x) by taking f = y). Let us just recall that the non-singularity of the curve follows from example 2 above.

q.e.d.

No doubt the reader will have recognized the analogy between the two functions x and y over the curve and the Weierstrass' functions

p and

P'

(the transcendental possibility of taking for y the

derivative of x calls for the factor 4 in yZ 110

= 4x 3

- gzx - g3 ).

- 11.37 -

With the notations of the proof (case g = 1), the divisor called very ample, because Lover k. The divisor

L(3~)

3~

is

contains a set of generators of

itself is called ample because one of its

~

multiples is very ample. This terminology however, adds little to the understanding of the present simple situation. We shall eventually show that conversely, all function fields over non-singular cubics are of genus one. For that, we have to give a purely algebraic proof replacing that of the first chapter using theta functions. The first steps consist in finding substitutes for Liouville's theorem and Abel's condition (I.l.la,c,d). Also, the whole idea behind the use of the subspaces

L(~)

of a function

field L is that they give quite generally a filtration of the k-vector space L wi th fini te dimensional subspaces L(~) cL(~')

: this is why

L(~)

is defined by

implies

(~~ ~'

div(f)~ -~

with a minus

sign !). (2.18) Proposition. Let L be a function field over k. a) L(Q)

=

k , or i'n other words : a function f £L without pole

is constant, f E:k. b) The spaces

L(~)

are finite dimensional for all divisors

and more precisely, if dimkL (~) c)

.!i

~

d

dimkL(~)

+

~~O

~£Div(X),

is a positive divisor of degree d

1.

=

deg(~) +

1 for one positive divisor d f 0, then

L is rational (of genus 0 over k). Proof. a) If f ¢k, define a valuation of kef) by v(f)

=

-1. If w

denotes a normal i zed extens ion of v to L, we have w(f) <: 0 so that f has at least the pole

WE:

X.

The same argument applied to 1/ (f - a)

shows that a non-constant function f £L takes all values a e:k. b) We prove the more precise assertion concerning positive divisors.

111

- 11.38 -

Let

~~O

and

and let us show that

VE:X,

We suppose that

dimL(~ +

(v))~dimL(~)

+

1.

Pv €V is a non-singular point (cf. note after proof) and let Xp be a uniformizing variable at P = Pv ~(v)

(1tp) = Mp CR p , that is v('fp ) = l .

L(~

+

We consider the linear map d +1

(P)) ~ k defined by f ~ (~pp

multiplicity of P in

f)(P) where d p is the

By definition its kernel is precisely

~.

L(~),

and this proves the desired inequality. Using a), induction applies and proves b). c) If dim

L(~)

then dim L((P))

=d

+

1 for one positive divisor

~

, 0 of degree d,.

= 2 by b) for all points P occuring in d (with

multiplicity dp>O). Hence for one such point P L((P)) = k

$

ordp(x) = -1 .

xk,

Consequently L(n(P)) is of maximal dimension n

+

1 for all positive

integers n. Take any y£L, and multiply it by a suitable polynomial in x ih order to make it integral over k[xJ. Then y will have poles only where x has poles, and y

E:

L(N (P)) for a sui tably big integer

N. This shows Y £k(x) and L = k(x) is rational. It is true that we have performed the proof in the non-singular case. More precisely, we have used only the following fact : if v E: X, there exists a plane curve V such that L = key) (a model of L)

and that v is centered at a non-singular point Pv E:. V.

However, we shall only use the results of the proposition in the non-singular case. (2.19) Corollary. If the irreducible plane curve V admits one rational function f€k(V) with only one pole, and a simple one, at the regular point PE;.V, then V is isomorphic to a projective line: key)

=

kef) is a rational field (of genus 0 over k).

This is a particular case of part c) of the above proposition. In particular, f has only one zero and div(f) 112

=

(PI) - (P2) ·

- 11.39 -

(Some readers will probably have noted the analogy between this corollary and the characterization of the sphere 52 = pl([) - e.g. among real compact surfaces - by the existence of a Morse function with only two critical points.) We come back to the case of cubics for Abel's condition (analogous to (I.l.l.d)). Let thus V be an irreducible non-singular plane cubic over

the (algebraically closed) field k of characte-

ristic p f Z. We suppose that the coordinate system has been chosen in such a way that V has the (affine) equation (Z.ZO)

Vo :

Z - (X-e )(X-e )(X-e ) = 0 l 3 Z

Y

with distinct e. eke We could normalize the e. 's so as to have 1

1

e l = 0, e 2 = 1, e 3 = A f 0,1 and work with this equation in Legendre form. However, we prefer to· keep more freedom, because

r

if moreover p

3, we might prefer to work with the Weierstrass'

form (corresponding to the other normalization e l + e + e 3 Z

=

0).

(Z.Zl) Lemma 1. If we denote by ord. the normalized valuation corresponding to the flex P at infinity, we have ord.(x) ord.(y)

=

=

-Z and

-3 .

Proof. We have to remember that x is the class of X/Z (y that of Y/Z) in the function field L

=

key) in our projective definition

of this field. By construction of the coordinate system, P is at the origin of the affine coordinate system defined by Y the line Z

=

=

3 , ordp(class of X/Y)

because the line of equation X gent (distinct from Z

and

1

=

1 and

0 is tangent at the flex P, so that

ordp(class of Z/Y)

ord~(y)

=

=

=

=

1 ,

0 goes through P but is not tan-

0). This shows

ordp(class of Y/Z)

ordp(class of X/Y)

=

-3

= ord~(x)

113

as asserted, -

ord~(y)

= ord~(x)

+

3 .

- 11.40 -

(2.22) Lemma 2. L

=

Proof. The automorphism defined by y

~

o.

k(V) is not of genus ~

of the quadratic extension L of k(x)

-y acts on the set of normalized valuations X

of L and also on the set of points Q £V. Obviously Qr is the point with coordinates x(Q),-y(Q) if Q

r

P and P~ = P is fixed

under this action. Let us take now a function f except at P (f is defined and finite at all Q have the same property and if f We show that f£k[x,y]

= a(x)

€.

r

L which is regular

P). Then f

b(x)y , f

+

~

r

will

= a(x) -b(x)y .

k[V o] the ring of regular functions on

the affine part Vo of V. It is sufficient to show that a(x) and b(x) £ k[x] . We prove it for example for a, by decomposing a in linear terms (k is algebraically closed) : a(x) If one integer

n~

=

TT(x _ ~)n~. I

were strictly negative, a would have

at the intersection points of Vo with the vertical X = apply this to the determination of L( (P))

C

poles ~

. We

k [x ,y]. But a non-constant

polynomial in x has order at infini ty E; -2 (by lemma 1) and a polynomial in x and y containing y has order at infinity This proves L((P))

=

~-3.

o.

k is of dimension 1 so L cannot be of genus

r Q e:.V)

(2.23) Lemma 3. For Q eV o (i.e. P

div(x - x(Q)) = -2(P) d i v (Y - mx - h)

= -

+

3 ( P)

(Q) +

and m,h E:,k, we have +

(Q 1)

(-Q) +

(Q 2)

+

(Q 3)

where the points Qi t:: V are the intersection points of V0 wi th the line of equation Y - mX - h = O. Proof. We have already determined the coefficients of (P) in these divisors in the first lemma (the pole of order 2 of x at P cannot compensate the pole of order 3 of y at the same point). It is tr

obvious tha t x - x (Q) vanishes only a t the two points Q and Q

-Q

=

=

(x(Q),-y(Q)). The multiplicities of these zeros are given by

the order of contact of the vertical X - x(Q) 114

=

0 and V at Q.

P

Q =

- 11.41 -

Because (~/~y)(yZ - (X-e l )(X-e )(X-e 3 )) can vanish at Q only if Z

y(Q)

0, we find that the vertical X -x(Q) = 0 is not tangent to

V at Q (hence the multiplicity is 1) unless y(Q) = 0, i.e. Q

-Q.

In this last case the multiplicity is 2 because these points are not flexes on V. The first formula is thus completely proved. The second one is derived similarly. We note that lemma 3 gives the divisors of all linear functions on V and that their degrees vanish. Also, they satisfy Abel's condi-

p on

tion (when p , 3 so as to have the group law

V). We give

explicitely the special case m = h = 0 with a consequence, in the following lemma. (2.24) Lemma 4. Let Ei(i = 1,2,3) be the three intersection points of Vo with the X-axis of equation Y = 0, so that E.1

-

div(x - e i ) = -2(P) div(y) = -3(P)

+

+

(E l )

(e.1 ,0). Then -

2(E i ) +

(E 2)

+

(E 3 )

The subgroup of V (when moreover p , 2,3) of points Q such that 2Q = 0 is fE o =P= 0, El , E2 , E3 } isomorphic to 71../27i x 71../ 271.. The assumption that the characteristic p is not 3 is not necessary, as we shall presently see. But in characteristic p = 2, we shall see much later (in the third chapter) that there can at most be one non-zero point of order two, and that the case where Eo = P is the only point of order dividing 2 also occurs (supersingular curves). Now we come to a closer study of the subgroup P(X) of principal divisors (of the form div(f) for some f d

= [dQ(Q)

Q~ = [dQ.Q

bL~)

E:Div(X) is any divisor, we put d

of Div(X). If

= deg(~) =

(sum in the group V with respect to

p).

LdQ

Then the

principal divisors d are exactly those for which d = 0 and Qd

115

and

0

- 11.42 -

(2.25) Theorem 4 (Abel). The group of principal divisors on X is a subgroup of Divo(X) , the group of divisors of degree

!.:

V --+ Divo(X)/P(X)

Q

o.

Moreover,

is an isomorphism

(Q) - (P) mod P(X)

~

In particular, the group law on V is independent - up to isomorphismof the choice of the flex P as neutral element. (As the law

+

P

has been shown associative only if p , 2,3 , we have

to make that assumption in the above formulation. However, a more careful analysis shows that we only use the two properties Q

+

p

QP = P

and (Ql

p Q2)

~ Q3

P (Q2 P Q3)

= Ql

= P whenever the

Q. 's are on a line. Thus this theorem gives another proof for the 1

associativity, which works for the characteristics p , 2.) Proof. All linear functions in k [x ,y] have principal divisors in Divo(X) by lemma 3, and they satisfy also the condition Q d Take now any principal divisor dive£)

= L.dQ(Q)

=

o.

E:P(X). Modulo

divisors of linear functions, we have (d - 1) (P) + (Qd)

so that it is sufficient to prove d of the form div(f) = (d - 1) (P)

=

0 and

Qd = 0 for divisors

(Qd) · But Qd ' P implies that f has only one simple zero at Qd (because f having a pole, d-l < 0) which is impossible by (2.19) applied to l/f, L not being of +

genus 0 (lemma 2). Then div(f) = d(P) is possible only with d

0

(loc. cit.) and with fE:k. Now we may consider DivoCX)/PCX) and define

f

as indicated. This map is a homomorphism in virtue of

lemma 3 which shows that !(-Q)

- f(Q)

~(Q 1) +