Formal Groups

Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, ZUrich

74 A. FrShlich King's College, London

1968

Formal Groups

Springer-Verlag Berlin. Heidelberg. New York

All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer Verlag. 9 by Springer-Verlag Berlin 9 Heidelberg 1968. Library of Congress Catalog Card Number 68-57940 Printed in Germany. Title No. 3680.

These notes cover the major part of an introductory course on form~1 groups which I gave ~uring the session 1966-67 at King's College London.

They are based on a rough draft by A.S.T. Lue.

I have not included here the last part of the course, on formal complex multiplication end class field theory, as this subject is now accessible in the literature not only in the original paper but also in the Brighton Proceedings.

The literature list on the other

hand includes some papers published since I gave ~

A.F.

course.

CONTENTS CHAPTER I

CHAPTER II

CHAPTER III

PRELIMINARIES 51.

Power series rings ...........................

I

52.

Homomorphisms ................................

16

53.

Formal groups ................................

22

LIE THEORY 51.

The bialgebra of a formal group ..............

29

52.

The Lie algebra of a formal group ............

43

COMMUTATIVE FORMAL GROUPS OF DIMENSION ONE 51.

Generalities .................................

52.

Classification

of formal groups over a

seperably closed field of characteristic 53. CHAPTER IV

p...

Galois cohomology ............................

COMMUTATIVE

51

69 86

FORMAL GROUPS OF DIMENSION ONE

OVER A DISCRETE VALUATION RING 51.

The homomorphisms..., .......................

52.

The group of points of a formal group ........ 1 0 4

53.

Division and rational points .................

119

5~.

The Tare module ..............................

121

Lz~rRE

........................................

96

139

-I-

CHAPTER I. PRELD~NARIES

w

Power Series Rings Let R be a commutative ring.

R[[~,...,X~]

The power series ring

in n indeterminates Xl,...,X n over R is a ring

whose elements are formal power series

iI

in

~(xl,...,x) - ~ ~h"'"in xl """xn with component-wise s~dition and Cauchy multiplication as its operations. Denote by N the set of non-negative integers and let M n

be the set of n-tuples i = (il, o..,in) , with components is 6 N. In other words M n is the set of maps of {l,...,n} into N.

We

define addition and partial order o h M n component-wise, i.e.

i + k = (

+

,,.., i n

i -->k < >

i A _> k A

and

The zero element 0 on M n

for

A = l,,.,,n.

is the n-tuple (0,...,0).

Now we can write

f(7~,...,x)

iI (interpret ~

as X 1

- ~(x) = ~ =

i ...Xn~ ! )" and define

Z is

f -xi n

-2-

(g + f)i = gi + fi'

(g'f)i -- Z

k+j=i

~-

contains R as a subring : identify a ~ R with the power series f, for which fo = a and fi = 0 (the zero of R) when i > 0 (the zero

of Mn). We shall write 9

~-

for the inclusion map.

R[[Xl,...,Xn]] The augmentation

is the ring hcmomorphism with z(f) = fo" R

Note that the dia6ram

R

~[[&,...,xj:; co~tes

9

Note::

we can view t h e f o r m a l power s e r i e s r i n g as t h e s e t

o f maps Mn § R. are not e x p l i c i t l y

If

the particular neede~ we s h a l l

symbols f o r t h e n i n d e t e ~ a t e s s ~

~[[%,...,x]] -- ~.

write

-S-

It is cleat of course that the map

%?_, ;. %~i ieM

i&M

n

n

sets u~ an isomorphism

,[[%,...,x]] ~ R[[Y~,...,~]] com~tible ~ith both z a n d I~-

~ e diagram

Rn_ I ~

-

R

1

=

Rn_I

R

D,

, R

~R,n-i co~mntes, Denote by U(S ) the group of ~nits (invertible elements)of a ring S.

-4-

,PROOF

As U is a functor from rimgs to groups, f 9 U(R n) will imply

~(f) ~ oCt). Let n = I.

If ~(f) = f0 9 U(R) then one cam solve

successively the equations

f0g0 = i,

f0g I + flg 0 = 0,

... ,

f0g r + flgr_ I § ,.. + frg 0 = 0

for the coefficients of the power series gCx) = (fCx)) -I. settles the case n = i.

This

N~r proceed by induction, using Lemma I.

Filtrations of, Abelian ,Groups A filtration v of A is a

Let A denote am abeliam group. map v:A§ which satisfies

(1)

v(o)

(2)

vCx-y) > i n f

I t follows t h a t

=-

, zmvW

,

{vCx), vCy)}

vC-x) = v(x).

(Note : suppose that vCx) = -

Hausdorff f i l t r a t i o n

(-}

(see below)

only if x = O, i.e. that v is a 9

Then by t a k i n g I x I = (~)v(x)

e

-5-

we get a metric space since Ix-Yl

~ote ~ o ,

that v ( ~ )

§ -

Given a f i l t r a t i o n

~ sup (1~1, l y l } ~_ I~1 I%1 ~ 0).

i~lies

v, then f o r m e N, define

I v(x) im)

Am={XeA

A m is a subgroup of A ( = AO) , and A m ~

A

|

+ lyl

=

m~NAm

.

Am+ 1.

Defining

,

we have in fact

A

= {x s A

v is in turn determined h y t h e

Iv(x) ---=) ,

groups Am, for m ~ N, via the

equations v(z) =

sup

x~A

m. m

In fact if we are just given a decreasing sequence {Am} (m ~ N) of subgroups of au abelian group A = A0, then this last equation defines a filtration on A.

LEMMA 2

Suppose A is am S-module for some r i ~

e~e S-mgdules i_~f am_.~donl~ ~

S.

v(sx) ~_ vCx) for. ~

Then the A

x e A, s e So

~aau this is the case, we speak of S-filtrations. A filtration is IIausdorff if A @0

-- {0} .

If {an} is a sequence of elements of A, and n l ~ v ( ~ then we ~rrite lirav am = a. n-~eo

a) = |

For v Hausdorff, a sequence can only have one

-6-

limit.

A seq~ence with a limit is a limit sequence.

A sequence

{an} in A is a Cauchy sequence if

(%+I - %)

n-~=V Ever~ ~ m i t

= o.

sequence is a_ Cauchy sequence.

A filtration v is ~

(or, A is complete under the

filtration v) if it is Hausdorff and if eve~j Cauchy sequence in A has a limit in A. Example (i) If there exists k for which A is complete.

= A k = {0}, then A

The Cauchy sequences are the sequences which are

ultimately constant. @0

Examnle (ii)

A = rl A(k), where A(k) are S-modules, and a(k) k=O denotes the k-th component of a & A. Define

I r = {a e A I a(k) = 0 for .11 k < r},

Pr = {a s A I a(k) = 0 for all k ~ r } ,

v(a) =

inf

n =

a(n)#O

L~5.~. 3

sup a 6 I

r. r

With these definition9, (i) v i s an S-filtration o f

A with the I r as associated subgroups: (ii)

lira a(n) =

v i s a complete filtration, and n + V

a

(a (n) is defined to be the element of A with a(n)(k) = a(k) for k < n, and a(n)(k) = 0 for k > n); (iii) for each r,

A -is - the direct sum , I r + Pr 9

-7-

ExamPle (iii) A is an abelian group, v a filtration on A with associated subgroups A . Denote by ~ m

: A/Am+ 1 § A/Am the

m

natural quotient maps. Consider, in the direct product ~m (A/Am)' the submodule A of elements ~ for which ~m(a(m+l)) = a(m).

The filtration of [-Im(A/Am)

(of. 2nd Ex, page 6) defines a filtration ~ of ~, under which is Hausdorff and complete.

Also, Pm : A § A/Am defines a m

homomorphism A + ~qm (A/Am) whose image is contained in A. therefore a hcmomorphism p : A § A.

L ~

~CpCa))

(ii)

p is injective if and only if A is Hausdorff;

(iii)

p is bijective if and only if A is complete.

h

lira a b nn

vCa), for a 6 A;

If v is a filtration of A, and if A is a ring, then

~Cxy) >_ vCx) V

We have

(i,~

=

This gives

+

vCy) i~ ~ a ~

i~ A A

c An+m. In this case, ~ s o

= lirav an.lirav bn, and w_~es a y v i_~s_a ring filtration.

We leave t h e proofs as exercises. n

If i ~ Mn, define lil = of f to be ord(f) =

~

~.

For f ~ Rn, we define the order

inf k=l fi#0 lil 9 By taking f(k) =

~ . T f'X~ li =k ~

(homogeneous polynomial), we see ord(f) = f(k)#O inf k.

Denote by Rn(k)

t h e R-module of homogeneous polynomials of d e g r e e k in t h e

variables ~ , , ...Xno

Then

Rn

k-O

In Lemma 3, by taking A = Rn, S = R, and ACk) = RnCk) , we have

-8-

" r : (f I q

: o for lil

< r} ,

Pr = {polynomials in XI,...,X n of degree _< r-l} .

PROPOSITION 2 The function ord is ~ complete filtration of R n w i t h a

L

~

L

_

Jl

i

m

m

-

IL

i

~

L ii

~, of R-modules I. associated subgroups Ir, and R n = Ir + P r (direct sum ---

Also, ord(f.g) ~ ord(Z) + ord(g). and the I r are ideals --of I 0 = R n ~ Moreover, lqlp = Ip+q, -------PROOF

By Lemmas 3 dud 4. Note that I I = {f ~ R n i fo = 0} .

Therefore I I = Ker r , and

Rn/I 1 =~ R,

NOTATION : f = g rood dog q means f = g (rood lq), i.e., f-g ~ lq. PROPOSITION 3 I f R is ___~integral $o~-~n, then ord(f.g) = ord(f) + ord(g) an~ R n ---is---anintegr~l domain. PROOF

Verify directly for n = i.

Then by induction on n, using

Suppose mow that J is an ideal of R. %

~he power series f, with

6 J for all i, form an ideal J[[X]] = J [ [ ~ , . . . , X j ]

J[[X]] = Ker ( R m § (RIJ) n } .

of ~ ,

and

If K, J are ideals of R, them

K[[x]].j[[x]] c (K.J)[[x]]. PROPOSITION ~ Le._~tv b_~e_a r i ~ Jp.

filtration of R with associated ideals i

|

'|

V'(T) = inf v(f i) is a ring filtration of R with associated i . . . . n .... ideals %[[X]]. If v i_EsHausdorff/ccmplete then v' i_~sHausdorfflcgmple~e.

~OOF

Then

~,(~) ~_ p c--~ v ( q ) ~_ p for ~ u i < ~ q ~ Jp for ~ u i > f ~ J [Ex]]. P

-9-

%[[x]].q[[x]]c 89 q[[x]]c If V is Hausdorff then n

Jp[[X]] = O, and therefore v" is

Hausdorff. To show that v complete => v" complete, let {f(n)} Cauchy sequence,

Then v'(f(n+l)-f(n)) + -.

v(fi(n+l)-fi(n)) § -.

be

a

V ~"

Therefore, for all i,

Since R is ccmplete under v, then for

each i there exists lira fi(n) = f.. n_~ov

Let f = ~ f.X x.

1

9 1

Given say

1

positive number K, there exists n o such that v'(f(n+l)-f(n)) > K,

f o r a l l n ~ no. and hence

v(fi(n+l)-fi(n)) > K,

f o r a l l n >_no, and f o r a l l i .

v(fi(n) - fi) > K,

f o r a l l n ~ no, and f o r .11 i , and

v'(f(n) - f) > K,

f o r a l l n ~ n O.

So

Therefore

This shc~rs v" i s c o m p l e t e .

f = lim.f(n). n-~. v

THEORy4 1 Jq.

Suppose v i_~s~ring filtration o f R with associated ideg~

Define,

Jq = Jq[~X]] + Jq_l[[X]]Ii § ,. + Jq_r[[X]]I r + . , Iq ; ~Cf) = inf

{n + v'CfCn))} ,

n

(v" is the induced filtration of Prop 4, f(n) denotes the homogeneous component of f of d e g r e e n ) . ~en

(i) v~ i s a_ ring filtration _of _ R n with associated ideals Jq; ,uq .

(iii) if v i s Hausdorff/complete, then v lS Hausdorff/complete, S is a local ring if it has one and only one maxi~-~ ideal w%.

10-

(A ring is local if and only if the non-units form an ideal, which will then be the maximal ideal ~ by the powers ~ COROLLARY 1.

).

We obtain a filtration on S

of ~ .

If R i_~sa locs~l ring then so is R n.

If in addition

R is Hausdorff, so is Rn, and if further R is complete, so i s Rn. The corollary follows from the theorem and the observation that if ~t is the ~ I i~al~[[X]]

§ k

ideal of R then by Prop. i the complement of the of ~

consists of units.

For the proof of the theorem

L~@L% 5

~m

first need a number of lemmas.

If the Ji are ideals Of R, with J q C

Jq-i C

.-- c Jl'

and if K = Jq[[X]] + Jq_l[[X]]l I + ... + lq, then f ~ K if and onl~ if f(A) has coefficients i_~nJq_s , A = O,1,...,q-1. PROOF

The sufficiency is strs/ghtfor~ard.

Then f =

r= 0

gr' where g r 6 J

q-r

[[X]]I . r

For necessity, take f ~ K. For ~ < q-l, f(A) = --

and for r k, for all r,s _9 n O . Hence v'Cfr (q) - fsCq)) > k - q + 1 for all r,s _9 n O 9 For a fixed q,

{fr (~) }

is a Cauchy sequence with respect to v %

Therefore there exists limv. fr (q) = f(q) Prop ~) and

(since v" is complete,

- 12-

v'(fr(q)-

f(q)) > k -

q+Z,

for

rLn o 9

Now, (fr(q+l)) (q) = fr (q) ' and taking limits we obtain (f(q+l))(q) __ f(q).

Therefore there exists a unique power series

f such that f(q) are the terms of f of degree < q. that lim~ r.~v

f

= f.

r

We shoe now

Now

~(f(A) - f) ~ A, and ~(fr CA) - f r ) ~ A, for all r.

Also v'Cf Cz)

-

f

r

CA))

A, for r L nl(A)

(v" Cauchy sequence).

Therefore ~(f C~) - fCA)) ~ A , for r ~ n l ( ~ ) . Hence ~(f - fr ) >_ A, for r >_.nl(~). TH~EO~I 2 PROOF

This completes our proof.

If R is n oetheriau, then so is R n.

It ~ill suffice to establish the theorem for n = l, for then

the general case follows by a trivial induction argument, using

Let J be an ideal of R I.

For any q >.0, J (~ lq = {f E J I ord(f) >.q}

is an ideal of R1, and its image in R

under the map f ~ * f

q

(we here

revert to the notation where f q denotes the coefficient of X q in f) is an ideal Aq of R. Aq+ 1 D Aq.

As f s J • l q implies Xf s

1 we have

The ring R being noetheriau, it follows that we can find

a k >.0, so that A k - Ak+ A for all A >.0. It ~rill suffice to prove that J ~ l k is finitely generated over R1, for J/J/%~. = J + ~ / ~ ,

as an R-submodule of E1/~, is finitely

generated over R, hence over ~ .

Therefore J ~rill then also be

finitely generated over R I. As Ak is finitely generated there is a finite set f(i) (i = l , , . . , s )

- 13

of power series in J /~ ~

(the (i) serving as enumerating index here),

so that the fk (i) generate A k over R. generate J (~ ~

~e contend that the f(i)

over R1.

Let g E J ~ ~ .

We shall construct inductively sequences

(g(i'm)} m (i = 1,...,s) of power series, so that firstly g(i,m) = g(i,m+l)

(mod degree m)

and secondly

s f(i)g(i,m)

g

z ~ i=l

(rood degree re+k).

By the first relation we obtain pc~rer series g(i) = limord g(i,m)

s~ f(i)g(i).

sad by the second one g =

Thus we see that in fact

i=l f(i) generate J ~ IkThe step from m to m+l goes as follows (put g(i,O) = 0 to apply this to the first step') : h = g - ~f(i) g(i,m) lies in

J n ~+=.

~i f(i)

~enoe h ~ = ~ A k ~ = ~ , i,e., h~+= = i=l

ii e R.

Put g(i,m+l) = I.Xm + g(i,m)

Then

1

g " .X f(i) g(i,=+z) = h !

1

: h+= x~+~- X f~(i) ~i ~ + = =- o (=d degree =§ 1

For the rest of this section we suppose R is a complete local ring, with maximal i d e a l s ,

and k = R / ~ .

its image in kn under the epimorphism Rn § ~ The Weierstrasse-order of f,

For f E Rn,

denotes

induced by R § k.

17-ord(f), is defined by

- 14-

w-oral(f)

Then W-oral(f) # ~



=

~ # 0



f has some unit coefficient.

Note t h a t as Ordk(~.~) = Ordk(~ ) + Ordk(E) , also W - o r d ( f . g )

=

~-ord(f) + ~7-ord(g). A di'stimguishe~ ~ fo + flx + f2~

f of ~

is a polynomial of the form

* ''" + fq-iXq-I + Xq' ~here ~11 the fi are in~t~

Note then that N-oral(f) = deg(f).

THEOREM3

(Weierstrasse preparation theorem)

i_~ff ~ R I a~d

~-ord(f) = p < ~, t h e n there exists a unique u ~ U(R I) and unique distinguished polynomial g such that f : u.g.

PROOF

Then of course

~e shall prove by induction on m that (Am) : There exists a

v (m) E U ( ~ ) and a distinguished polynomial g(m), so that

f.

v

This congruence determines v (m) (and hence g(m)) uniquely rood

Ass!~ng (Am) for all m, it follows from the uniqueness part that :

As R I is complete with respect to the filtration ( ~ m [ [ X ] ] }, we obtain in the limit a unit v of ~ , polynomial.

so that f.v = g is a distinguished

Moreover, v is determined uniquely rood ~m~[X]] for all

m, i~e., is unique.

Now multiply through by u = v "I to get the theorem.

To establish (Am) we may work over the residue class ring R / ~TM, i.e., we may suppose t h a t ~

- O.

First for m = i the hypothesis

-

15

-

states that f(X) = X p. u(X), where u(X) is a unit of R I.

But this is

in effect also the assertion. For the induction step write m = r + i.

By the induction

hypothesis there exists a power series @0

v(r)(x) = v(x) =

v.X x i=0

so that, writing f(X) =

i--0

f.~,

we have

:L

v o ~ u(R),

(1)

r

Vofp + ... + v p f 0 = i + up,

Vofs § "'" + V s f o = ~s'

~s s

Up~ ~ ,

(2)

(all s > p).

(3)

This is just the congruence for CAr ) expressed coefficient-wise. By the uniqueness part of (At) the coefficients v~ of v(r+l)(x) must be of the form

v [ = v i + xi'

xi 6 ~ r

~e have to show that the ~. can be chosen so that 1

(2') v&%

+ ... § ~ o

- o, f o r ,~1t ~ ,

p.

(B')

CRe~mber that ~r+l _ ~m _ Ol ). Note t h a t v~ w i l l c e r t ~ n ~ in U(~).

From ( 2 ) , ( 3 ) , ( 2 ' ) and ( 3 ' ) we get the equations

lie

-16-

Xof s + klfs_ I -~ ... + lsf 0 = The Xi are to be chosen in ~ r ,

IJs,

(s >_.p).

and we know that fk ~ ~

for k < p.

Hence we must have

As %

XO% =

- Pp,

Xo~+1

+ ~l~

XoS+

+

15+

= - ~p+1' "'" ' i

"'" +

§

- -

>- o).

* U(R) these equations have unique solutions for Xi in R, and

by induction on k one also sees that Xp+k must lie in r . can solve for the v:. 1

Then we

The uniqueness of the v i nod ~c r and of the X. 1

implies the uniqueness of the vf. 1

,w

Homomorphisms A and B are abelian groups with filtrations v, w respectively.

A continuous homomorphism

e : A § B is a homomorphism of groups

such that, given m e N, there exists A ~ I~ for which (A~)0 c B mHence, if v(a n) § ~, then W(an%) § =. bic0ntin.uous means that

To say that e is

e : A § B is an isomorphism of abelian groups,

mad both e and e-I are continuous. THEOP~4 1 (i) Suppose S i s a_ commutative ring, complete ring filtration v, and R is _a subring o_~fS. S with values vCa i) > i, homomorllhism

there

under...

Given al,,..,a n in

exists a_ uniaue

continuous ring

8 : R n § S (with respect to the order filtration on E n)

which leaves R elemen~rise fixed, and such that X. e 1

= a i.

- 17-

(ii) Explicitly, i_~ff(Xl,,..,~) s Rn, then

li~ fC~)c%,,..,%) - fCxl,,..,x) e . (Here fCq)(Xl,,..,X n) is again the polynomial of degree < q - i t~ich coincides with f rood degree q). (iii) Let T be a c ommutativ9 ring containing R, complete under a ring filtratio_...___.__nnn w, aud with elemauts ~l,...,~n for which w(~ i) L l, s_~othat : Given S and al,...,an as in (i), there exists a_ unique continuous ring homomorphism ~ : T § S with ~i r = ai and leaving R elementwise

.~ze~. Then the continuous hom~mor~hism Rn § T which ~

R elementwise

fixe___~dand maps X i into ~i is a bicontinuous isomorphism.

~oo~

if fCx) ~ ~n' then

fC~+1) Cx) - fC~) Cx) Therefore fCq+l) (al,o..,%) . fCq)

~

lil-q

biX~ , b i E R.

iI im (~U...,%) - [biu...,in e! " " %

and its value under v is at least iI + ,.. + in

vCfC~+!)C%,...,~), {fCq)Cal,,..,~)}

is therefore

fCa)c~,...,%) a

=

q.

Hence

) + | as ~+-.

Cauchy sequence under v.

We put

f C x u . . . , x n) e - !q..~o i = v f(~) C ~ 1 , . . . , % ) . It follows quite easily now that 8 is a continuous ring homomorphism, and the uniqueness of 6 then follows fram continuity.

-

18

-

The proof of (iii) is standard (uniqueness of universal objects). If there is no ambiguity involved, we shall write f(al,...,an)

for liz q_~v

f(~)(~l,...,~).

For a ring S with ring filtration v, define l(S,v) = {s 6 S I v(s) > 0}. Comsider the category YR' whose objects are the pairs S,v as in Theorem i, and whose morphisms are the continuous ring homomorphisms S,v § T,w which maps i(S,v) into !(T,w) " d R D

R' where~R is the

full subcategory with objects Rn,ord (order filtration).

Theorem i

then ss~vs HOm fR(Rn,S ) ~ I(S,v) n, by associating with each e the

el~t

(X1 e ,...,Xe). Co

ider

the

case S -

9 whe

.

the indeterminates of Rm as Y's, to distinguish them from those of Rm, which are still denoted by X's.

Let (r = l,,..,n).

x r e = ~r(Y1,...Jm),

Them

fCx1,,..,x) e - ~ ~

f(~) (glCY),...,~CY)).

shall derive another expression for this elemaut of Rm.

for k 6 M

Write

n kI

~(~)

k

= gl (y)...~n(~)

~ X

g~k

yA

9 k

Since orgy arCY) i 1 we have ordy gk(y) ~ Ik I = d therefore gA = 0 for

I~i

< I~l~

~

it makes sense to define

- 19 -

z(gz,...,@) = z(%,...,%) (Yz,...j=)

=

PROPOSITION I PROOF

~

(

X

k~

yZ 9

f(~)(gl(Y),...,~(Y))= f(gz,...,~) (Yz,...,z=).

Verify for polynomials f.

Then extend to power series f by

continuity. Let f = (fl,,..,fr) be a "vector" of r p~rer series in n indeterminates, and let g = (gl,...,gn) be a "vector" of n power series in m indeterminates.

C%Cgu...,gn),

TJe denote by f o g the vector

... , ~Cgl,---,~))

of r power series in m indeterminates.

With this multiplication the

vectors f = (fl,...,fr) with varying r and n form a category, whose objects are the positive integers, f being viewed as a "map" r ~ > n. In view of the preceding theorem and proposition, a homomorphism % 6 HO~R(Rn,Rm)

determines a vector g8

ge o~ = ge o g~ . categories.

: n + m.

In fact this map 8 ~-~ge

Moreover

is an isomorphism of

In other words we can either use the language of

homomorphisms e or that of vectors of power series. In Rn, consider the ideal I = Ker e, and denote by ~ the image of f under the natural epimorphism I * 1/12 = D(R n) . n

If

n

f =

Y C l ~ + terms of degree 9 2, then f = ~ ciX i. D(R n) is a i=l -i=l free R-module on ~l,,~ ~.~en e : Rn -* Rm is in ~ R ' then I(Rn)e c I(Rm) , and 12(Rn)e c 12(Rm), and so e induces a

homcmorphism D(e) : D(R n) + D(Rm) , of R-modules. Denote by @ R the category of finitely generated free R-modules, or "vector spaces over R".

- 20-

PROPOSITION 2 D is a fu~ct9r: ~ R §162 COROLLARY

!_~fR n is bi.coDtinuously isomorohic to Era, theD n = m.

For, a finitely generated free module over a co~utative ring R has a unique rank. If e is a homomorphism in ~ R

then x.e = ~ CikYk + terms l k=l

m

of degree _> 2, and ~.D(e), =

[ Cik~k . D(8) can be represented in k=l

the matrix form o(e) = (Cik) , and Cik = (~xie/~Yk)y= 0. By Prop. 2 , O defines a map : HO~R(Rn,R m) § HOm~RCD(En),DCR m)). We define a map E in the opposite direction as follc~s, X. onto ~Cigk, then take E(~) : R I

which maps X i onto ~CikY k.

T~O~

2

-

Let

n

§ R

m

to be the homomorphism

We have

e be a c o n t i n u o u s h o m o m o ~ m

'

.

.

.

if $ maps

.

.

.

R n

§ R .

~en

e

n

i_2.Sa_ bicontinuous isomorphism i_~_fs~d only i.f D(8) i~ an ~

of

R-modules. COROLL~

!/_fe is. sur.iectiw then i___ti_~san

PROOF OF COROLLARY isomorphism =>

e surjectlve =>

D(e) surjective =>

D(8)

e isomorphism.

The theorem can be rephrased to read : given fi(Y1,...,Yn), with f.(0,...,O)l = 0, i = l,...,n, then det (~fi/~Yk)Y=0 is a unit if and only if there exist gj(X1,...,Xn) such that fi(gl,...,gn) = X.l (IITVERSE FUNCTION THEOR~). PROOF OF THEOF~ Assume that

~0~eneed only prove the sufficiency of the condition.

~ = D(e) is an isc~orphlsm.

~Trite ~ = E(~).

- 21-

-1 Then D(e @ ~ ) = i. As ~ is an isomorphism, it will suffice to -1 show that 8 ~ ~ is an isomorphism. Without loss of generality we can therefore suppose that D(e) = i.

With this assumption,

X i ~ X.l e mod i2, where XI,... X '

(s gi

are the indetermlnates of R . We construct polynomials n

n

of degree ~ - i so that

x.~ - gi(~)(xe) moa x~, g~ ~+l)(x) ~ gl(~)(x)

rood I~.

By induction on ~ , suppose that

Then X i = gi(~)(xe) Take gi(s

+

X~ ikI=

ck(xe)~ rood I~+ I.

g~)(X) + Ikl- ~

sequence, ~ith limit gi(X), say.

ckXk.

Then {g~)(X)}

is a Cauchy

Also,

x~ = gi(Xe) = gi(X) e . Define ~ by the equations X.T~ = gi(X).

Then x(T~ o e) = gi(x)e = xi,

and so by the uniqueness part of Theorem l, W ~ e = i. 1 = D(~) ~ D(e) = D(W).

As before, there exists X so that X ~ W = I.

Therefore X = X " (~ " e) = e isomorphisms.

Therefore

, and hence ~ and e are inverse

- 22-

Altho!~sh H o ~

(Rn,Rm) is not a group, we can define some

/ %

sort of "filtration" on it by taking

ord(e) = inf (or~(fe) f#O

- or~(f))

inf (oraz(xie) - l). i=l,o..,n ~th

this definition,

ord (e o ~) _> ord(e) + ord(~). ~3.

Form~! Groups In this section ~re take R to be a fixed ring, and all p ~ e r

series are over R. A fo~~F(X,Y)

of dimension n is a system F.(X,Y) of n I

p@rer series in 2n indeterminates X = {X~,o..,X}~a , Y = {Y1 '''''Y-}a satisfying

(1)

;(x,o) = x,

(2)

;(F(x,z),z)

;(o,z) = z; = ;(x,;(Y,z)).

In view of (1), the substitution in (e) makes sense. immediately~r

have F(O,O) = O, and

Fi(X'Y) ~ Xi + Yi mod degree 2. ~[oreover, terms of degree greater than 1 are "mixed", i.e. X's and Y's only occur together. PROPOSITION 1

F is commutatlve if F(X,Y) = F(Y,X).

Given F, there exists aunique i(X) (n p~rer series in

n indeterminates) so that F(X,i(X)) = F(i(X),X) = 0. PROOF

Put gi(X,Y) = X i - Fi(X,y), i = l,...,n,

gi has no constant

-

23

-

term when viewed as a power series in Y.

(~gi/BYk)x=Y=O --(~Fi/~Yk)x=Y=O = - ~ik" By w

Prop. i, the determinant of (~gi/BYk)Y=0 is a unit of

RKKY~l,...,Xn] ].

Apply w

Theorem 2: there exist h i(X,Y), (i = 1,...,n)

such that gi(X,h(X,Y)) = Yi' i.e. X i - Fi(X,h(X,Y)) = Yi' or Fi(X,h(X,Y)) = X i - Y.~, (i = l,...,n).

Put Y = X : Fi(X,h(X,X)) = 0.

i(x) = h(X,X). The proof of the uniqueness of the inverse is a translation of the standard proof of group theory. Suppose n ~ respectively.

that F and G are formal groups of dimensions n and m

A homom0rphism f : F § G is a "vector" f = fl''"" 'fro

of m power series in XI,..~

~rith no constant terms, so that

fCFCxj) ) = GCfCx),fCY)). The homomorphism f determines a homomorphism ef : Rm § Rn, given by ZiG f = fi(X), ~rhere Z i are the indeterminates of R m and X i those of R n.

If f : F § G, g : G § H are homomorphisms of formal groups then

g ~ f : F § H is a homomorphlsm of formal groups. gives the identity homomorphism of F.

Also li(X) = X.

Hence :

PROPOSITION .2. The form~1 ~rou~s and their homo morphlsm form a cate~oznj ~R §

~R

(=~)' and f F-~ 8f defines a con travariant functo_r

(But as f is written on the left, e

on the right ~re still

have %f o g = ef o eg.) Remark : (in ~ R )

A homomorphism f : F § G of formal groups is an isomorphism if and only if ef is an isomorphism (in ~ R ) .

Moreover,

- 94

if f is amy "vector" of n power series with 8f an isomorphism, and if F is a formal group of dimension n, then there is a unique formal group G(= f ~ F ~ "l ) so that f is an isomorphism F +G. THEORF/{ i

(i) Le_~_tF be a f o r ~

grou~ of dimension n, and S,v 6 ~fR"

Then Ho~R(Rn,S)9 becomes a grotrp F(S) under the operation given by

is commutative, then FCs) is abelial. Cii) f

~ e HO~R(S,T) , then (~$) o ~ = Cc~ @ ?)9(~ ~ ~).

(iii) Let G b e ~ fur~er formal group of dimension m, then when f s Hong(G,F) , w_~ehaye

(iv) With the hyDothes.is o f(iii), and if i_~naddition F is commutative, then Hom~(G,F) i_s_sa sub~Toup of the abelian ~oup -

Re~F~:

(i)

Identifying

Ho R( n,S) = i(S,v) n (cf. w

Theorem I), the group operation becomes o~*~ = F(~,8),

m,8 e I(S,v)n. (ii) Again, if we express H o ~ (Rn,Rm) in terms of vectors f of power series we get the group operation

(x) (iii)

-- F C f ( X ) , g C X ) ) .

By the theorem, Ho~R(Rn,R n) is closed under composition

(multiplication) and ~ (addition), with a one-sided distributive law,

- 25-

i,e., it is a near ring. (iv)

The theorem, plus a f ~

is a functor ~ R (v)

X~R§

formal trivialities, tells us that F(S)

groups.

Let ~ a b be the full subcategor$~ of

the commutative formal groups.

~

=~%Rwhose objects are

Then the sets Hom (F,G) for F, G ~

ab

have the structure of abelian groups and the composition of homomorphisms is bilinear.

In particular End~F)

= Ho~(F,F)

is

now a ring. The proof of Theorem 1 is by a straightforward application of the definitions, and the ~:ioms for for~1

groups.

Suppose now that F is a commutative for~]

group of dimension n.

We define a function ~ on the abelian group F(S) = !(S,v) n by

~(~I,...,%) -- inf v(~i). i ~.le

state the following two propositions without proof :

PROPOSiT!~ 3 :

is a filtration of F(S).

(~le have not defined filtrations for non-abelian groups' ) PROPOSIT!OH 4: map Ho%fCG,F)

~,TithF and G as in Theorem i (iv), the composite ~

IIo~RCRn,P~m)

D ~ IIo~R(D(Rn),DCRm))

is a homo-

morphism o_~fgrougs (the compositio.n in Horn (G,F) bein~ ~). For a given prime number p, denote by

=

:R § n

R

n

the

homomorphism which fixes R and ts/~es X. into X p. Then 2 n l a :L o ~ = 7 (2) : x.. ~ x p ands(q) : x ; ~ x p - . Let ~+ denote the I

I

I

I

additive group of R. 'I"I-~,OTRt~..I

2

Let f : F '+ G b_~ea_ homomorphism of for~7 ~

dimensions n and m respectively) and let Of : Rm § R n the

(o_~f

- 26-

corresponding homomorphism of rings.

(i) ~ R

Then D(ef) = 0 i_~faud only i__fff = 0.

(ii) Suppose R + i~so__~fexponent

p (prime).

+ is torsion free.

Then D(ef) = 0 i_~faud onl~ if either f = O, or

ef = ~e_ ~ ~(q), where O(#f) # O__and q > O. PROOF

~[e use the notation 8fi/8~ = fik(~l,..,,Xn);

(3F/~)

(xJ1,...Jn) = F (X,Y); (~Gi/~V~) (U,V) = Gi~ (U,V).

N=r differentiating the equation f'z(F(X'Y) ) = Gi(f(X) ,f(Y) ) ~_th respect to Yk' ~e obtain (chain rule) n

m

; fij CFCxJ));jI~Cx,Y)

j=l

~=l ai~CfC',0,fCY))fz~C~).

Define the matrices

erCx) [email protected]));~ a2FCx,'z) -- C~j~Cx,Y)); a2a(u,v) = (ai~(u,v)). Our equation, for ~11 i a u d k, then gives the matrix equation

~(F(X,Y)).a2F(X J) = a2G(f(X),f(Y)).af(Y). Hence

~(F(X,O)).a2F(X,O) = a2a(f(x),o).af(o) , i.e.,

af(X).a2F(X,O) = a~G(f(x),o).af(o). N~

df(O) = D(ef).

Hence by w

Also, det d2F(0,O) = i i.e., Cn(det d2F(X,O)) = 1.

Prop. !, det d2F(X,O) is a unit, and so d2F(X,0) is aa

-27-

invertlble matrix.

If D(ef) = O, i.e., df(O) = O,

and therefore 3f./BX. = 0 for ~11 i,j. l O this implies f = O.

~en

R + is torsion free

R + is of exponent p this implies

f(X) = g(xP), i.e., e~ = 8 o ~. g induction.

~en

then dr(X) = O,

In the latter case now proceed by

But we must sh~r that e comes from a homomorphism of g

formal groups, i.e., that g(X) is a homomorphism of formal groups.

Now

g(F(P)(xP,YP)) = gCFCX,Y) p) = fCFCX,Y)) = GCf(x),fCY))

=

where F (p) is obtained from F by raising each coefficient to its pth p~er.

}~ have then g(F(P)(x,Y)) = G(g(X),g(Y)).

Since F(P)(x,Y)

is a formal group (the map which sends each element of R into its pth p~er

is an endomorphism of R), g is indeed a homomorphism of formal

g%'OUpS. If

8 = r o =(h), and D(~) # O, then h = ht(8) is called the

height of 8 9

l~e define ht(O) = = .

For f a homomorphism of formal

groups, ht(Sf) = ht(f) is called the height of f.

If f # O, then

h ht(f) = h is the greatest integer so that f is a power series in X p . PROPOS!TION >.

(i) if f, g are homgmorphisms of formal groups an__~df 9 g

is defined , then ht(fe g) L h t ( f ) eo~mtatlve formal o ~ a n d

+ ht(g).

(ii) ~

G is ~

f,g ~ Hom(F,G), then

ht(f~g) ~ inf

{ht(f), ht(g)} .

ht(f) PROOF

(i)

If f is a power series in YP

, and g is a power series in

-

x

p

h

t

28

-

(

g

)

(where Y and X are the corresponding indeterminates) then ht(f) + ht(g)

clearly f e g is a power series in X p ht(f) + ht(g) ~_ ht(fo g).

, and therefore

(NOTE: if our formal groups are of

dimension i, and R is an integral domain, then ht(f) + ht(g) = ht(f 9 g), and the height function is a valuation). (ii) Since ef can be ~itten in the form ~f,

(ht(f)) where

D(~f) # O, then inf {ord Xi~ f} = i. Therefore ~(f) = inf{ord f.} =pht(f). i i The filtration property of ~, established in Prop. 3, n ~ gives pht(~g)

~ i n f { ht(f), pht(g)}

ht(f~g) ~ i n f

, which implies that

{ht(f), ht(g)) . ~Throughout this proof read "power series"

t o mean "vector of power series"]

- 29-

CHAPTER II LIE THEORY w

T h e bialgebra of a f o r ~ 1 group Throughout this chapter R is a fixed commutative ring with

identity and J ~ is the category of R-modules.

For ~I N ~

also M e RN and HomR(M,N) are R-modules. Ne shall need the notions of a coe.lgebra and of a bialgebra over R.

The definitions we shall give are ad~pted to our special

situation.

A coalgebra

is given by an R-module M

{~I, K, s, 8}

and homomorphisms of R-modules

(Z.Z)

(comultiplication),

RM

)M@

= :

7M,

8:M

yR,

so that the foll~ing diagrams commute:

M

(1.2)

.

~

IK

-

.~ M

|

ttM

[I@


kcM

~k (the right--hand sum is in

n

fact a finite sum : u is continuous and hence <X=, u> = 0 for all Ikl sufficiently large).

= k

[ fkCk 9 k

U n is therefore spanned by ~k"

Now

Therefore ~ ck dk = 0 implies k

flck = 0 for all f, and so also ck = 0 for .11 k. k ~" is free on the ~k"

Hence in fact U

n

*

- 33 -

COROLLARY i PROOF

-The -

map f ~-~ sf i s an isomgrghism R n § Ho~(Un,R).

If sf = O, then 0 = [sf,u] = for all u ~ U n.

This

implies f = O, and therefore f~-~sf is in~ective.

Thus

-3~ n " ' is surjective.

f ~-~sf

COROLLARY 2. maps e ~-~ e

k

Th_e_ehomomorphism Homx,(Rn,R m) § Hom~(Um,U n) ~.thich i_~san isomorph"ism.

Thus the functor J ( §

of

Progosit.ion i is an antisomorphism o f .cateaories. PRQOF

Suppose 8

= O.

Then for all f, u,

= O.

0 = = [sfe,u] , which implies Sfo = 0 for all f. an isomorphism, and therefore f8 T h i s proves injectivity.

Therefore But s is

= 0 for all f, which means 8

= O.

If w ~ Ho~CUm,Un) , define 0 6 Ho~(Rn,R m) r

by f8

= ~ 2 . Then e k COROLLARY 3 . The map U n @ RUn

= ~ .

This proves surjectivity.

~ U2n

~iven by

~(kl,...,kn) 0 S(~Z,...,~n) i_s_sa n isom@rphis m ~ PROOF

R-modules.

Obvious. The significance of the last corollary lies in an interpretation

presently to be derived. Let

I = {f~

R

n

I ord

{(z o R~n

(f)

+

9

.

Then

Rn 0 RZ) § ~n 0 Ran}

is an ideal of the ring Rn e RRn .

Let v be the filtration of Rn

corresponding to the p ~ e r s ~q of the ideal ~. v(f @ g) = oral(f) § ord(g).

= -9

Then

Denoting the indeterminates of Rn by

- 34-

X = "" ~l,...,Xn and those of R2n by X',X" = X ~ ,...,~n,~l,..., " ~:" Xn

we

define a homomorphism R n @ RRn + R2n of R-modules by : f(X) @ g(X)

'

" f(X') g(X").

This turns out to be an injective

continuous homomorphism of R-algebras.

In fact the filtration v is

seen to be simply the restriction of the order filtration of R2n. A

Going over to the completion R n @ RRn of R n @ RRn we obtain a bicontinuous isomorphism R n @ HEn = R2n.

Thus we get an isomorphism

URn ~ Homj~(R n @ RRn, R), and the module on the right can be identified with U n @ RUn .

The resulting isomorphism U2n = U n @ RUn

is the inverse of that of the last corollary. The ring multiplication in R

is described by a map : n

R n @ RRn + R n.

As multiplication is continuous and R n is complete,

this map extends uniquely to a continuous homomorphism

Wn : Rn determineduniquely

@ R

7

n'

by the rule

(f(x) e g(X))

n

= f(x)g(x).

A

Identify from now on R n @ RRn = R2n.

In the previously introduced

notation for the indeterminates of R2n ,

h(X',X")

Wn is then given by

= h(X,X).

Let ~n : Rn § R be the augmentation, ~n : R § R n the ring embeddirg.

}[e then have, on identifying U2n = U n Q RUn,

(~iting ~

for ~ R )

- $5-

PRpPOSITIO 3

The o

~n : Un § U2n = U n @ RUn , cn : R §

n P

de f~De on U n the structure of a_ coel~ebr~. The .isomorphism '

~s

rise t_~oa_ bijection

Eom CRn. ) I~,00F

omCo g( ,Un).

For the first assertion we only have to show that ~n' en

and ~n enter into commutative diagrams dual to those postulated for 9~.

the maps

,

~ = ~n ' ~ = r

(see (1.1)-(1.6)).

.*

8 = ;in defining a coalgebra structure

For example (1.2) follo~m from the associ&tive

l~w for the product ~n' (1.3) from the commutative law, and so on. For the second part of the proposition note that a e ~ Homj~(Rn,R m) will actn~.11y lie in Hom~D(Rn,R m) if and only if the diagrams W n

888

R2m

"~

e

)

-36-

8

R

n

,

> R

ra

R

R

R/

e

~

rl

conmmte.

R

]~

The dual diagrams ( s t a r e v e r y t ~ n g and r e v e r s e .11 arrows)

g i v e p r e c i s e l y t h e n e c e s s a r y and s u f f i c i e n t c o n d i t i o n s f o r 8

t o be

a homomorphism o f c o a l g e b r a s . Let {C be t h e c a t e g o r y whose o b j e c t s a r e t h e c o a l g e b r a s {Un' ~n' r

~n } and whose morphisms a r e t h e homomorphisms o f

coalgebras.

The l a s t p r o p o s i t i o n t h e n t e l l s

j ~ § J ~ * y i e l d s an antlsomorphism ~

§ ~

us t h a t t h e f ~ n c t o r of categories.

Let F = F(X' ,X") be a power s e r i e s i n 2n i n d e t e r m i n a t e s x,

-

x"-

v,,

v.

,

w i t h zero c o n s t a n t t e r m .

Let

8F E Hom~ ( R n , ~ n ) be t h e c o r r e s p o n d i n g homomorghism o f power series rings.

Thus f(x)e F = f(F(X',X")).

Then F will be a formal

group i f and o n l y i f t h e f o l l o w i n g diagrams commute :

-3'/-

R

n

Y

(z.7)

/-

R2n @1

n

(~(o,x) :~ x

/R2n @

E

:

F(X,O))

n

R

n

(identii~ring

Rn @ RR = Rn) ,

eF Rn

-'2n

(z.8)

("Associative law"). i | eF R2n

~ R3n

In addition we know of course that eF is a homomorphism of rings, preserving identities.

R

Hence the following two diagrams also co~ute

e F ~ 8F

@ R Rn

^

~ R2n 8 R R2n

(~.9)

Rn

~

:

R2n

R

(~.I0) Bn

~

n

F Rn

> R2n

-38-

In the first diagram the vertical maps are those induced by multiplication, going over from @ to the completed tensor product @. Consider now the dual map

P = eF :

U2n = Un @R Un § Un"

This is an RIlinear multiplication on U (1.7)-(1.10) now show that U

n

and ~Jae duals of diagr~m.~

becomes a bialgebra.

n

E.g. the dual

of (1.8) is the associative law for multiplication and the dual of (1.9) tells us that the comultiplication ~ of U is an algebra n n homomorphism. We can sum up : If F is a formal group of dimension n, then p = eF defines the structure of a bialgebra on U m ( the coalgebra structure being fixed once and for all by ~m' an and ~n ) . Conversely if p : Um @R Un § Un is a map, defining the structure of a bialgebra then in particular p ~ HOmcoalg(U2n,U n) and hence .

p = eF for some unique power series F(X' ,X").

The axioms imposed on

p (as stated above) then imply that F is a formal group.

We have

thus proved the first part of THEOREM 1 ( i ) The map F ~-~ 8F = PF i s a b i j e c t i o n form~ ~

of dimension n onto

on the coalgebra (ii) fo~...1, ~

(iii)

{Un' ~n'

of. th__~es e t o f

the set o f structures o f b.i,algebra

en' ~n } "

The _algebra Um,PF is commutative, if and only if the F i s co~a~-atative. i

I

z_! F

.

.

.

.

]

fprm

res~ectiyely then the iSomorphiSm

of d/mensio=s n an__!m

- 39

HO~

IR n

-

n'

gives rise to a bijection

Ho CF,G)

(Cun,PF), (U ,pG)).

The proof of (ii) is quite analogous to that of (i). For (iii), comaider the vectors f = (fl,...,fm) of m power series in n indeterminates.

We know that these stand in biunique

correspondence under a map f ~ > ef with the e 6 Hom~(Rm,R n) . ,

Going over to the duals we obtain a bisection f ~-* pf = ef e HOmcoalg(Un,U m) . By dualizing the appropriate diagram one sees then that f is a homomorphism F § G precisely when the diagram

PF

U

U n @ RUn

n

Pf @ Pf Um @ RUm

U

-

m

PG commutes, i.e. when pf is a homomorphism (Un,PF) § (Um,PG) of bialgebras. L e t ~ be the category of bialgebras whose underlying *

*

coalgebra is one of the {Un, ~n' The maps F : > Un,PF ,

§

Zn' ~n ) "

We can then sum up

f ~ > pf define an isomorphism

Note that we end up with a covariant functor: ~;e shall now discuss the bialgebra Un,PF in some more detail. If k = (kl,...,kn) ,

A = (AI,...,~ n) are in L!n we define

-40-

=

, and ~.; = zl;...Zn!

\h

/...

.

n

We denote by ~i the element of I~ of the form mi = (O,...,O,l,O,..,O), ~rhlch has 1 in the i-th position and 0 elsa~there. The element Ai of U n is defined by the equation A i = ~

, i.e., 1

<Xi,A i >= I,

<Xj,A.1 9 = o f o r i # ~ ,

.-x-~, A.> = o if 1

Ikl # i .

For the formal group F(X,Y), we introduce the notations FkCX'Y) e Xk + Yk + Bk(X'Y) mod degree 3,

Bk(X'Y) = z0"!" li'j'k XiYj

'

(k = 1,...,n),

Xi'j'ks R.

Then we have PROPOSITION 4 W

! if k=O,

(i) c~(1) = ~o; ~n(~ ) = 0 if k#0.

~+j =k

0

(iii) pF(~O @ u) = PF(U @ ~0 ) = u. (iv) pF(~k @ 6Z) =

s

c.~. , c . ~

Ok+ s +

0 0

0

(~ # o ~ ~). n

(v) PF(Ai,A~)=

~ iJ

~

J

k=l

z,j,k~"

R,

- 41-

n

COROLLARY PROOF

- lj,i,k )A k'

(i) ~nd (iii) are obvious 9 (ii) ~,Tehave

,

%.

xJ,

But by the definition of ~n'

~§

l, i f k = ~ + j O, otherwise.

<X ~ | X j, ~n (~k)> = <X~+J,~k > =

Hence the result, (iv)

Let r = (rl,...,rn). n

F(X,Y) r =

r.

K ~ (X,Y) ~ i=l " i

n

=

=

~

i=l

Then

I".

(xi+Y)~

+ terms of order > Irl

m

Cx + y)r + terms of order

II~, <xr~, pF(~ 9 ~A)~ - ,~yA in F(X.Y) r.

Irl.

is the coefficient of

This is thus 0 when Irl>Ik + A 1 and also when

Irl - Ik § 41 but r ~ ~ + 4.

on the other hand if ~ -- k + 4

this coefficient is clearly (k + &). 4

(v)

>

Finally when r = 0 then

By (iv) we know already that

pF(Ai,Aj) = \

; ~i,j,k n k' % / 6 ~~ .j + k-l

then

-42-

and we have to show that ~i,j,k = Xi,j,k"

"i,j,k = < z ( x J ) %

'

In fact we have

Ai ~ n.>

= J

= ~k (xJ)' ~i e ~j> = ~i,j,k

% ~ + Yk' ~i e ~.> = o = d 3.

A. e

m

Aj> = 0, ~ = e v e r

This completes the proof of the Proposition.

We define T(Rn) to be the submodule of those u ~ U n for vhich = 0 =

.

T(R n) is thus the submodule generated by the A i.

The next

proposition gives an ~n~er characterisation of T(R n) in terms of the coalgebra structure of U n.

m0POSi~O~, ~ ,

ai,~

u ~. u n ~ e

fo~o~

st ~ t ~ n t s

equivalent

(ii)

(iii) (Recall that

~*(u) = u O

r + r ~u,

= c(f) + r

,

r = 60 is always the identity in amy bialgebra PF

structure of Un).

- 43 -

PROOF

(i) => (ii): By Prop. 4 (ii) holds for u = Ai, hence by W

R-linearity of ~

for all u E

T(Rn).

(ii) => (iii)

:

=

=



= ~(g) + ~(f)



.

(iii) => (i) :

If f, g s I then ~(f) = r By linearity = 0.

= 0 and so = 0. Also

= = r

+ r

= 2 , i.e. = O.

Hence = O.

w



The Lie al~ebral of a formal ~rou~ First we list, without proofs, the definitions and results

on Lie algebras to be used. Throughout R is a fixed co~atative ring, and all "algebras" are algebras over R. a Lie algebra ~ Product [a,b] i n ~

For each associative algebra A there exists

(A), which coincides w~th A as a module, the Lie (A) being given in terms of the associative

product ab by

a,b] = ab - ba.

- 44

Each Lie algebra L has an enveloping algebra E(L).

More

precisely E(L) is an associative algebra with identity with an attached homomorphism j : L §

of Lie algebras so that the

map

fl

~f

@ j

(f & Homassoe(E(L),A)) is a bijeetion

(2.1) Note:

Homassoc(ECL),A )

> HomLieCL~CA)).

All associative algebras have identities, and Homasso c is

the set of homomorphisms preserving identities. By (2.1), taking A = R we get from the n1111 a homomorphism of assoclative algebras

: ECL)

§ R.

As E(L) has an identity we also have a homomorphism o : R § E(L). Next if L 1 and L 2 are Lie algebras, then their cartesian set product L I x L 2 has again a Lie algebra structure, and

E(Ti x L2) =~ ECLl)

eR

ECL2).

In particular

E(~ x L) ~ E(~) e R E(~) via

j(~l,~2),

~ j(zl)

e i

+,

e

j(~2).

-

45

-

The diagonal map L § L x L thus gives rise to a homomorphism

D

:

E(L) § E(T) eRE(L)

of associative algebras.

A.

The associative algebra structure on E(L) to~etherwith

the maps D, (of.

w

u, 9

define on E(L) the structure of a bialgebra

for the definition).

To prove this one would only have to verify now the co~utativity of the diagram~ (l.l) - (1.6), and this can be done by going back to the defining property of the enveloping algebra.

For the particular Lie algebrsswhich we shall have to

consider this also follows from the explicit description to be given below. From (2.1) we obtain a map

E : HOmLie(Li,L 2) § HOmBialg(E(Ll),E(L2) ). In fact given a homomorphism

~

: L I + L 2 of Lie algebras there

is one and only one homomorphism E(~) : E(L I) § E(L 2) of associative algebras so that

L1 -

) L2

F,(~) ECh) co~zm~es,

[ ~ECL2)

-46-

In view of the obvious functorial properties of the maps D,

a and ~ associated with each L,

with these. B.

E(m) will in fact commute

In other words

E is a functor frgm Lie algebras t o bialgebras. Now let L be a Lie algebra which as an R-module is free on

say generators ~ " ' ' ' ~ n " C.

Then :

("Poincar~-Birkhoff - Witt Theorem") L § E(L) is in~ective. We shall accordingly vi~v L as embedded in E(L).

~or ~

= (k1,..., ~ ) ~- ~% = ~

...%

(the order of the factors matters')

p_.

Write

(i) (ii) (iii)

c~~ _- ~) Then we have the ~escription

E(L) i_~sthe fr.ee R-module on the dk.

dkd ~ - d ~+~ +

X

a.d j

(k,~ # 0).

D(d i) = 1 g d i + d i g l,

and hence

i+j=k (iv)

~(i) = ~o = i.

~(~) _-

O, k#O i, k=O.

Now we return to the associative algebra U , p F defined in the preceding section, F being a formal group of dimension n.

We

-47

shall write

[']

-

F for the Lie product.

Thus

[ u,v ]F = PF (u'V) - PF (v'u)" In the notation of Ii

w

Prop. 4, we now see that

42.2.)

It follows that the submodule TCR n) of U n generated by the Ai (see w

is closed under ['IF"

In other words T(R n) is a Lie algebra

under ['IF' which we shall denote by LF - the Lie algebra associated with the for~a~ group F. Let f : F + G be a homomorphism of form_~l groups (dim F = n, dim G = m).

The homomorphism ef

: Rm § Rm

maps R

(viewed

as

a

subring) into itself, and maps I~m ) ( = {f e Rml ord f ~_ 2}) into I n)"

Hence the dual homomorphism ef : Un §

~rill map

T(Rn) § T(Rm) 9 Moreover, ef is a homomorphism of associative algebras, i.e., it takes the multiplication PF into PG"

Hence

also

Of u, 8f

G

for u, v @ U . It foll~s that ef gives rise, by restriction, to n a homomorphism Lf : ~ PROPOSITION cate~

I

§ L G of Lie algebras.

He sum up :

L F and Lf define a covariant functor from the

~ of formal groups to_ th_~ecategory o_~fLie algebras ,, and

LF preserves dimensions, i.e., L F is ~ fre___~eR-module on dim F generators.

-48-

Alternative Description: a = (al,...,a n) (ai (

Let R (n) be the module of n-tuples

R).

Let BECx,Y) be t h e homogeneous

quadratic component of Fk(X,Y) (k = l,...,n) (cf. ~(X,Y)

~--"

~Cx,Y)

t

BkCY,X).

w

and write

Define a multiplication on R (n) I

i,e., a map R (n) @ R (n) § R Cn) by

~(a,b) - (A1(a,b),...,%(a,b)). Then, if p is the isomorphism T(Rn) § R (n) of modules given by

pC~.aiAi) = (~,...,an), we get fr= (2.2) p( [u,v]F) -- ~(p(~),p(~) ). Thus R (n) , ~

is a Lie algebra, in fact isomorphic with LF.

For formal groups F and G of d~mensions n and m respectively, we denote by Ai, F and Ai, G the corresponding free module generators of ~

and LG.

If f : F § G is a homomorphism then fik6 R are

defined by

fir

) - ~ fik ~

r

de~ 2)

k--i

m

PROPOSITION 2

Lf(~,F) =

~

fik Ai,G

i=l m

PROOF. Suppose Lf(~,F) = i=l ~

Cik Ai ,G' say.

indeterminates of G by YI'"" ''Ym )

Then (we denote the

- 49 -

Cik =

= -

= fik" COROLLARY 1

The homomorphism f : F § G is an isomorphism of formal

if and onl~ i_~fLf is an isomor,~ism of Lie algebras. PROOF

By I,

COROLLARY 2 PROOF

w Theorem 2. If R + is torsion free then Lf = 0 if and onl~ if f - O.

By I,

w Theorem 2.

For the rest of this section we assume that R is a Q-algebra (Q is the field of rational numbers).

Under this hypothesis we

shall prove that the category of formal groups and the category of Lie algebras which are free E-modules of finite rank are isomorphic. More precisely we have: THE O R ~ i

(i)

Let R b e a Q - ~ .

For each Lie_ algebra L which

is a free module of finite dimension over R, there exists a formal ~rou~ F such that L i s isomorphic to 2 " (ii) (iii) on ~

Hom~(F,G) § HOmLie(~,L G) is abijection. The formal ~rou9s F and G ar.e isomorphic i_/_fand

if the correspondin~ Lie al~ebras L F and L G a~e isomorphic. The proof of Theorem i requires three lemmms.

We take L

-50-

to be a Lie algebra which as module is free on generators ~,... ,dn. The module homomorphism C L : E(L) § U n is defined by the equation

c,..(d~) - kz ~ , where we sh-11 use throughout the description of E(L) given in D. CL is an isomorphism ofmodules.

L~I

E(~) ..

D

U

Moreover, the diagrams

E(~) eR ~(~)

n

>

U

n

n

@~U .t~

n

R

T

~n

/c E(~)

> u n

R COCm~t e. The multiplication E(L) @ E(L) ->E(L) defines through CT. a multiplication qL : Un @ U n § L~4A 2

n~

There exists a_ formal group F fo__~rwhich qL -- PF' i.e.,

qL defines on. U the structure of' a_,bialgebra.

Then also L = LF

-51-

Let now conversely F be a given formal group.

Since LFC~(Un,PF) ,

this inclusion map can be pulled back to a homomorphism : E(~) § LD~3

of algebras (universal property of enveloping algebra).

n : E(L F) § Un,PF i_~san isomorphism.

~.c; = Cn' ~n" ~ = ~

PROOF of Lemma i

Als O

Ca @ a ) . D = w n ~.

and..

Since R + is divisible and by II w

m

Prop. 2, the

k'~ k form a free basis for Un, and so CL is an isomorphism of modules. Also by D a n d II

w

Prop. 4

~.+j=k X

k'

i+j=k

CT '

cai)

e

CT,CdJ)

I i+j=k

=k'

n

z

(6k) = ~n CL(dk)~

By extending linearly to E(L), this proves that the first diagram is commutative.

Similarly for the second diagram.

PROOF of Lemma 2

qL is defined so that

~.C~)

o R ECT.)

Un @Un

-

-

~ ~ ECT.)

qT, -

~,

Un

-52-

is comnutative.

That qL defines a bialgebra structure on U n

is now trivial by Lemma i.

The isomorphism of categories ~

~

of the last section ensures the existence of a formal group F such that t h e Un,qL = Un,PF.

Since CL maps d i onto Ai, L and

are isomorphic under CL as modules, and since CL preserves the Lie product then this is an isomorphism of Lie algebras.

P~90F 9~ Lemma 3 the dkwith

Let E r be the submodule of E ( ~ ) generated by

Ikl ~ r

the ~k with Ikl ~ r.

and let V r be the submodule of U n generated by We shall then prove by induction on r the

assertions

r

When [k[ = r t h e n

(mod Vr_l) , 9 k a

unit of R ;

(Br)

maps E r bijectively onto V r.

As E(LF) is the union of the Er, bijectivity of

~ : E(~) §

U n the union of the V r the

n follows.

By the definition of ~ ,

n(a ~

= n(1) = Co"

n(a~i) = n(ai) = ~i = as. 1

where I%1 : l,

% has a i at the i-th place.

AS for I~l _
_ O.

for f ~ E. is t h e set

For the opposite inclusion consider an

Then a~ E spans J~ over 0~, png E E for some

Now v(png) >_ n, and so by (B),

ht(png) >_ nh, whence by (A)

- 84-

pg=

for some f e E. g ~ E.

o f=

As~

Thus N ~ E ,

f,

is torsion-free, this implies g = f, i,e.,

and hence N = E.

For 4) we first note that it suffices to establish the equation

cent(E)

= Z

P

As Z

c cent(E), it will suffice to show that the Z -rank of P P cent(E) is < 1. If f 6 E, pf 6 cent(E) then f ~ cent(E). Thus cent(E) is a direct summand of E, and therefore its Z -rank P coincides with the dimension over GF(p) of its image cent(E) By Prop9 3 the map E § M is surjective, whence

in the algebra M. cent(E) C

cent(M).

It thus remains to be s h ~ m that the dimension

of centCM) over GF(p) is at most i9 Let a(X) = a~ X ,

h-1

b(x) = X b

.xpj

j=O be two elements of M.

Then o

h-1 (a o b)

(x)-

X

a0bj xp~ .

j=0

h-i (b 9 a)

pj " % b. x p~.

(x) = j=O

For b(X) to lie in the centre of M it is thus necessary that for all a 0 e GF(q) a n d a l l

j = l,

.9

h - i,

pJ

b j ( a 0 - a~ ) = O9

But i f

-

85

-

pJ a 0 is a primitive element of GF(q) then a0 # ~0 and so we must have b. = 0 for these values of j. central element is of the form of a(X). a(X) E cent(M), ~ i l e

(J = i, ..., h - i), In other words a

So now suppose that

bCX) is arbitrary.

If we choose b. = I for

all j we get the equation a 0 = a~, i.e., a 0 ~ GF(p). cent(M) is of dimension

( 1.

Thus in fact

(of course one has equality here).

To prove 5) we first recall the definition of inv ( ~ ) most convenient for our purpose.

There exists an element g o f ~

so that

for all f ~ E

(c)

gfg-l : fp

(mod

) I

is the m a x i E ~

(~ro-sided) ideal of E.

g is of course not unique but

the values v(g) of such elements g form a unique coset mod Z, which is the invariant o f ~

.

One may, by multiplying through by elements of

cent (J~) = 0~, suppose that 0 ~ v ( g )

< 1.

One then has to show that

1 for such a g we have v(g) = ~ . Let then g satisfy (C), and assume that v(g) = K 0 ~ ~ ~h

- 1.

l~e sh~11 sh~r that

~ = 1.

,

From (C) we have,

on multiplying up by g,

N~

we can translate our statements into the language of power

series.

We have a power series g(X) of height ~

g(x) so that for all f(X) ~ E

(mod deg

pK+l)

,

, i.e., with

-86-

(g o f) (x) ~ (f(P) o g) (x)

(mod deg pK+!),

where

f(P) (x) = (f @ f o ... o f) (x) f

~

If f(x) = Zl x + .... , then f~P' (X) = ~ X hence, mod deg p

(p times). + ... , ana

K+I K

(go

f) (x)

(f(P) @

a~

x,

g) (X) ~ af~l X,

K

i.e., a ~1

= a ~1.

As fl can be any element of GF(q) (by Prop. 3)

and as a # 0 it follows that K

=

i.

This completes the proof of the theorem.

w

Galois cohomolo~, Let F and A be topological groups, and suppose A is a F

-group,

so that the elements of r induce automorphisms of A and so that the map

F x A §

is continuous.

For 7 ~ F , a e A we denote by

7a

the image of a under the map defined by Y 9 A cocycle of r in A is a continuous map a : r § A w h i c h satisfies the relation

aC~c~) = a(~c).

"%(~).

We denote the set of cocycles of r in A by zlCr,A)9

Note that

ZI(F,A) is a set with a base point, viz., the trivial cocycle which maps each element of P onto the identity of A. b 9 A, the equation

For a ~ ZI(F,A) and

-87-

= b -1

aC ).

defines a cocycle a1 * zl(r,A).

Yb

~;o cocycles which are related by

such an equation for some b ~ A are said to be associated.

This is

an equivalence relation, and the equivalence classes in zl(F,A) thus defined are called the cohomo!ogy classes.

The set of

cohomology classes is denoted by ~(F,A), which again is a based set with base point the class of the trivial cocycle.

The cocycles

associated with the trivial cocycle are called splitting coc~cles, and they are given by

aC )

=

b-1.

for some b ~ A. Consider now a field k of characteristic p, and let K be a norm~Ll separable extension of k.

Denote by r the Galois group Gal(K/k).

For k I a finite field extension of k in K, we write

Ak! = {Y ~ F I Y

leaves k I fixed elementwise} .

A topology on F is defined by taking as basis of open neighbourhoods of the identity the subgroups Akl for all finite field extensions k I of k in K.

With this topology, a continuous ~ e m e m e ~

map

a : r § A of topological groups has the follc~Ting interpretation : TaP~e y ~ r , and U a neighbourhood of a(Y).

There exists a

finite extension field k I of k so that whenever 8 * F effect on k I as

has the same

Y , then a(6) ~ U.

We state the following two 'le~,~' i~ithout proof.

(K+ denotes

the additive group of K, and K* denotes the multiplicative group

-88-

of the non-zero elements of K).

~sa I

HI( r,K+) -- o.

(This is a consequence of the Nor~1

basis theorem. )

~m

2

(mlbert's

Satz 9 0 ) .

RA(r,K ) - 1.

Let S be the group, with respect to o composition, of power series f(X) defined over K of the form f(X) = flX + f2~

+ ,.. , fl # 0.

A topology on S is defined by the order filtration i.e., by viewing S as a subset of K[[X]].

The action of F on S is defined by

the action of F on the coefficients of the pc~zer series in S.

~lith

this structure we have

PROPOSITION 1

PROOF

HI(F,S) = 1.

Define S (n) = {f(X) s S I f(X) - X (rood deg n+l) }.

is a normal subgroup of S Cexercise for the reader). 1 §

(I) §

§

Then S (n)

The sequence

§

w

where S § K

maps f(X) onto fl' is an exact sequence of

Also, if f(X) f S (n), then f(X) = X + ~ u + l map f(X) ~-~ ~

F -groups.

(rood deg n+2).

The

defines a homomorphism S (n) § K + of r-groups, and

m§

(n+l)§ (n)§ + §

is exact 9 ~,Temust show that, if a ~ Z1 (r,S), then there exists b t S such that a(7) = b -1 ~ A .

- 89-

Then a(y) = [ ar(Y)X r , ar(7) s K. r=l

Take a E ZI(F,S). The map

Y ~ ~alC7) is a cocycle r § K , and hence by Lemma 2

there exists b I 6 K

and define b(1)(X) map

such that blal(Y) = 7b I.

~ a(7)

~ (7b(1)(X))-i

Y ~-~ a(1)(Y) is a cocycle r § s (I).

aCz)(~) =

x

then the map y ; ~ a~ 1)

+ a~ z) ~

,

Take b(1)(X) = blX

= a(1)(7) e S (1).

The

If we write

...

,

i s a c o c y c l e F + K+.

By Le~m~ I there

exists c 2 E K + such that

a~z)(~) = Yc 2 -

Take cCX) = X + c2X2.

c 2.

Then cCx)

o aC1)cy)

This way, we get a Cauchy sequence (b(n)(x)}

b(n)(x) Put b = nl~

bCn) (X).

a(~)

=b

@

(Yc(X)) -1 = a(2)Cy) E S (2) such that

@ aCy) o (YbCn)cx)) - 1 E

S (n).

Then

-I

o

~b.

Let F be a formal group of height h defined over k and fixed once and for all. k

If G is another formal group defined over

and f : F § G is an isomorphism defined over K then

and so Yf : F + G is an isomorphism, a(y) = f - 1

It follows then that

7f is an automorphism of F (defined over K).

Also,

- 90-

a(y6) = f-to

~f

= f-l o ~f o ~f-1 o y6f _ a(v)~ ~ a(~).

Let k I = kCfl,,..,f n) be t h e field obtained by adjoining the first n coefficients of f to k.

Then if Y and ~ have the

same effect on k 1 we have 7f ~ ~f (mod deg n + 1), and hence a(y) = a(6)

(mod deg n + 1).

Thus a is continuous.

Hence

if f : F § G is an isomorphism (over K), then a(7) = f-l~

yf

defines a cocycle of r in AutK(F). Every other isomorphism F § G is of the form f o g for g ~ AutK(F).

Since

,h Cy) _- ( f .

~)-s.

u

o g) _ g-1 o f - ,

. ~.f o ~,g _ g - l o

aC'r) ~ "rg,

then we can associate uniquely with G the cohomology class of

z-I ~ ~f -- a(~c). Suppose now further that G and H are isomorphic formal groups over k and that 9f : F § H

s : G § H is an isomorphism defined over k.

Then

is an isomorphism defined over K and

= f - i o 7f,

since

7A = A .

G and H are therefore associated with the same

class of ~(r,AutK(F~). Denote by ISOK/k(F) the set of k-isomorphic classes of formal groups which become isomorphic to F over K. defined a map ISOK/k(F) § HI(F,AUtK(F)).

THEOP/~4 I

ISOK/k(F) § ~(F,AUtK(F)) i_~sa_bi~ection.

We have then

-91-

PROOF

Suppose G and H are smsociated with the same cohomology

class.

Let f : F § G,

~ : F § H be K-isomorphisms.

Then there

exists g s AutK(F) so that f-i o yf = g-I o A-I ~ YA o yg.

Rearranging we get

o g"

f-l=yz

~ ~gO

~f-l=v(~.

g ~ f-l).

The K-isomorphism A o g o f-i is therefore fixed for all Y s F. It follows that A 9g 9

f-i : G § H is a k-isomorphism.

Thus

the map ISOK/k(F ) § HI(F,AUtK(F)) is an injection. Let a s ZI(F,AUtK(F)).

Since ZI(F,AUtK(F)) c

ZI(p,S),

then by Proposition i there exists f s S such that

a(~)--z - I o

~f

,

for all Y & F.

But if

GCxJ) = ~Cf-iCx),f-iCY)) then f : F § G is an isomorphism of formal groups.

Moreover

~G(X,Y) = f o a(~) F(a(~)-I~ f'l(X), a(~)-I ~ f-1(y)) zFCz'iCx), z-ICY)) - GCx,Y), as a(y) ~ AutK(F).

This being true for all T s F we conclude that

G is defined over k.

It maps onto the cohomology class of a.

Thus we do

have a surjection. Let now I(k,h) be the set of k-isomorphisms classes of form~1 groups of height h.

By

w

Theorem l,

-92-

I(k,h) - IsoK/k(F) when K is a separable closure of k.

We thus obtain a full

classification of formal groups of height h from the theorem. COROLLARy

I f K is a separable closure of k, then

I(k,h) ~ ( r , A u t K ( F ) ) . Note that by w

Let h = I.

Theorem 2 ~e know the group AutK(F).

Then by w

Th. 3 En~(F) = Zp for any K

and hence AutK(F) = Up, the group of p-adic units.

The Galois

group r leaves Z C EndK(F) element~ise fixed, hence leaves the closure Z

P

of Z fixed.

Thus r leaves AutK(F) fixed.

But then

Hl(r,AutK(F)) = Hom(r,AutK(F)) is just the set of continuous homomorphisms r § AutK(F) , i.e. of continuous homomorphisms r+U.

P Now let the field k of definition of F be a finite field

and let K be its algebraic closure.

Then, as a topological

group, the Galois group r is generated by the Frobenius substitution : ~ ~-~ mr , where k = GF(r) (r a power of p).

Moreover, for each

element ~ of Up there is one and only one continuous homomorphism r § Up which takes ~ onto ~ .

Thus we can identify Ham(r,Up) = Up

and ~le end up with a bijection

I(k,1) = !SOK/k(F) ~

~ Up

~henever k is a finite field, and the height of F is i.

-

93

-

~[e can use the preceding Theorem 1 to derive a classification of formal groups of finite height h over a finite field k, which is due to J-P. Serre.

Let F be such a group, fixed once and for all, and write

E = EmdK(F) , K being an algebraic closure of k. Cs

If a e E denote by

its conjugacy class (under inner automorphisms).

Let w be

the normalized p-adic valuation of E @ ~p0~Which__ takes as its set of finite values precisely~, If k has pS elements then ~ i t e

in other words, for f * E, w(f) = ht(f). T

for the set of conjugacy classes S

Cs

of elements with value w(~) = s. Now let G be another formal group of height h defined over k.

Choose an isomorphism

(i) over K.

g : F ~ G

Then

defines an isomorphism EndK(G) = E of ZZ -algebras. Moreover to P ~Tithin an inner automorphism this O is uniquely determined by G. Nine clearly the power series S

t = t(X) = X p is an endomorphism of G, and so

~(G3 = Cs

solely depends on G, and not on the choice of g in (1). THEOREM 2.

(Serre).

The map ~ gives rise to a bijection

I(k,h) m~ T S .

- 94-

S

PROOF r

Let a be the Frobenius automorphism a ~-* ap

of K/k.

As

= GalCK/k) is free profinite on the single generator o it

follows that the map

is a bijection zlcr,u(E)) ~ U(E)

Hence a :

~ a(~) ~ t

(the group of units of E).

is a bijection

(3) zl(r,u(E)) Observe now that if g =

tog=

~ gn Xn then n=l

Og @ t ,

and so when gl # 0 then

(4)

g

-i ~

t " g = c~

C = g

-I

e

a

g

Thus in the map (3) cohomologous cocycles correspond to conjugate elements, i.e. we get a bijection

If now g and e are as in (I) and (2) then, by (4), e(t) = a(a)

~ t, where a is a cocycle corresponding to the isomorphism

class of G under the bijection of Theorem i. through ~(r,U(E)),

i,e.

class of G, and moreover

r

Thus the map ~ factorizes

solely depends on the isomorphism

$(G) ~ T s.

Hence finally r induces a map

I(k,h) § Ts, which factorizes into the product of the bijection of Theorem i and the bijectlon (5) and thus is a bijection. We also note

-95-

PROPOSITION 2.

With e as i_~n(i), (2), End(G) i_~sSsomor~hic to the

subring o_~fE of elements commuting ~rith PROOF byt

e(t)

In EndK(G) the ring Endk(G) is characterized by qm = a, i.e. o

a =

COROLLARY i

m~ Endk(G) is al~ays the maximal order of the O~-a!gebra

i_t_tspans. COROLLARY 2.

There exists a group G defined over k, and of height h

wit__~hEnd(G) = EndK(G), if and only i_~fh divides s. For, the set of values of w on the centre ~

P

of E is the set

of positive multiples of h. COROLLARY 3.

If k is the prime field then End(G) is commutative and

its field of ~uotients i_~s~

ramified of degree h over ~ .

In fact in the algebra 0~ @ Z

Endk(G) = D, the field 0~(t) P 1 has ramification index at least h, as w(t) = ~ w(p). But as a subfield of a central division algebra of rank h 2,

0~(t) is of degree at most h.

Thus in fact h is its degree and ramification index, and moreover ~(t) is then a maximal commutative subfield of D.

-96-

CHAPTER IV. CO~Iu~ATIVE F O ~

GROUPS OF

D~4EI~ION OI~E OVER A DISCRETE VALUATION RI~G

w

~ e homomorphisms. Throughout this chapter we limit our consideration to

co~.,tative formal groups of dimension I. PROPOSITION i

Let L b_~ea field o f characteristic O, and F

a form~1 r o ~

(commutative, of dimension i) over L.

Then

there exists a unique isomorphism AF : F + G a (the additive group) defined over L, so that ZF'(O) = i.

Suppose now that S i_~s

inte~rsl domain w i.th quotient field L, and that F is defined over S.

Th.en s

~ S[[X]],

(We denote the inverse of the isomorphism AF by eF). (Motivation for notation :

If F is the multiplicative group @@

Gm(X,Y) - X + Y + XY, then s

= log(l + X) =

~ (-i) m-I Xn/n, n=l

and _ eF(X)

-- e x

| X

i =

sT.

n=l

.PROOF By II, Corollary i, g : F § G a.

w

Theorem i, Corollary i, or II! w

Theorem 2,

we see that there exists an isomorphism (over L) Also, D(g) - g'(O) # O.

Now D : EndL(G a) § L is an

isomorphism, since the elements of EndL(G a) are the monomials aX. We can therefore find gl ~ EndL(Ga) such that D(g I) = D(g) -I. Thus AF = glg 9F § G a

is an isomorphism with A'F(O) = i.

uniqueness, suppose f, g : F § Ga are isomorphisms with

To show

97

f'(O) = g'(O).

Then f o g-i 6

Therefore f o g

--I

Aut(G a) and D(f ~ g-l) = i.

.

~s the identity on G a and so f = g.

To prove the second p a r t o f the proposition ( ~ i t e

~F = ~)

we differentiate, with respect to Y, the equation

~(F(XJ)) = ~(X) + ~(Y). We obtain

~'(F(XJ)) F2(XJ) = ~,(Y), where F2(X,Y) denotes the derivative of F(X,Y) with respect to Y. Put Y = 0 : ~'(X) F2(X,O) = 1.

From our assumption on F, F2(X,0)

has coefficients in S and leading coefficient 1.

Therefore ~'(X) has

coefficients in S, being the inverse of F2(X,0). COROLLARY i

Ho~L(F,G)

COROLL~RY 2

D : HOmL(F,G) § L i_~s~bijection.

COROLLARY 3

I f F and G are formal groups defined Over S

= eG ~ Endi(Ga)

o

AF.

then D : HomS(F,G ) § S i_~sinjective. PROPOSITION 2

With the same hypothesis as in Prop. i_~ and if

in addition q i s a ~ the set o f ~

intgger, q > i, then HOmL(F,G) i_~s

power series f (with zero constant term) defined

over L so that

f

PR00F

o

F

G

of.

(*)

By Prop. i, Cor. ~ we only have to shc~r that (.), together

- 98 -

We shall

with the equation D(f) = a, determines f uniquely. establish uniqueness of the partial series f(n)

= qx +

. . .

+ q_l

by induction on n. Suppose that

fCn) o

[4F =- [4o ~ f(n)

f(n) o

[q]F" ~ G

(moddeg n),

i,e., that ~ f C n ) _ cx n

(mod

d e g n + i).

Then

r(n+z) ~ [q]F- [~]G ~ f(n+l) ~ c~ + q ( C as D([q]) = q.

Here we must

f

n

~.

c,

,

q-q n

Note that c l e ~ l y

s

q)~

(rood deg n + i),

have "

is a "polynomial" in fl"'" ~s

!~re precisely we see by iteration that there exist polynomi~!~ n(T)' depending on F, G and n so that fn = Cn (fl).

The unique

f satisfying (.) in Proposition 2, with D(f) = a, is thus

o@

~

n=l

~nCa)~.

Suppose from now on that R is a discrete valuation ring with quotient field K of characteristic O, maxims/ ideal ~ residue class fiel~ k of characteristic p # O.

, and

Let v denote the

-99-

valuation on K given by ~ v(p) = I).

(Note:

. (we take v normalized, so that

we are now e l]~ing filtrations whose values

are real, but not necessarily integral. ) l

COROLLA2Y

Let F and G be defined over R.

Then D(HomR(F,G))

is closed in R. PROOF

D : HomK(F,G) § K is a bijectlon by Prop. i.

D-l(a) in @@

HOmK(F,G) has leading c o e f f i c i e n t a, and hence D-l(a) =

~

~n(a)xn.

n=l

R is a closed set (with respect to the valuation topology) in K, and since ~ is a polynomial it is continuous. n

~he elements a E K

for which # (a) e R therefore form a closed subset C n

n

D(H~

= (~n Cn'

of K.

Since

then D(HOmR(F,G)) is closed.

~e denote by ~ the separable closure k. R § k induces a functor ~ R §

(cf. III,

The homomorphism

w

Prop. 1), under

which F : ~ F PROPOSITION 3

I_~f~ is not isomor~%c t_~o~a then Hom2CF,G) § Hom~ (~,~)

i~sinjectlve. PR00 F

Suppose f : F § G is a non-zero homomorphism so that ~ = 0.

Let (~) = ~ o

Then f(X) = r

g(X) where r > 0 and g # 0.

~rg(F(X,y)) = G(~rg(x),

He have

~rg(y))

= wrg(x) + ~rg(y)

(rood ~ r+l R[~X]]).

- i00

-

Hence gCFCx,Y)) - g(X) + g(Y) and

=

§

r

-

-

g ~

Therefore

=

o g = 0.

Since g # O, then [p]~ = O, i . e . , ~ = ~ a (Iii, w In v i ~ of llI, COROLLARY

w

Th.2).

Cor 3 to Prop. i, we have

If Ht(~) # = , Ht(~) # Ht(~), then HomRCF,G) = O.

Suppose n ~

that F and G are formal groups over R, and

~, G are of finite height.

Then we can define three different

filtrations on HomE(F,G) , viz., the filtration induced by the normalized filtration v on R and the injection D : HomR(F,G) § R (again to be denoted by v); the filtration induced by the height filtration ht on Hom~(~,~) and the injection of Prop 3 (again denoted by ht); the p-filtration where the associated subgroups are

{[~1~ Ho~(F,G)} (denotedby Up). Recall now that two filtrations on a group are said to be equivalent if they give rise to the same topology.

Before

stating Theorem l, which gives the relation between v, ht and Up,

we make the following definition.

A filtration w on a free

Zp-module A of finite rank is called a norm if, for some valuation v' of Zp, equivalent to the p-s~ic one,

w(ca) = v'(c) + wCa),

c e Zp, a 9A.

Any t~ro norms are then equivalent. THEORY4 1

Suppose R is complete.

(i) v, ht and up are equi.'.valent

- 101-

f i l t r a t i o n s o_~nHomR(F,G), au~dHOmR(F,G) is complete uuder these filtrations h = Ht(~).

(ii) HomR(F,G) is a free Zp-module of rank < h 2 (iii) (Lubin) End(F) is a

whose quOtient fiel8 has degree over ~ Remark: h.

commutative Zp-order d~'vid/ngh.

One can in fact shoe that the rank of HomR(F,G) divides

See below (Corollary 3 to Theorem 4 in

w

A Zp-order is a Zp-algebra which is free of finite rank as a Zp-module.

Recall that we alrea~ly know End(F) to be an integral

domain. Note that v and ht are in fact valuations on End(F). Hence

~

v(f)

ht[p] F = V(~]F )

CORO~Y

= v(f)

~uswe

Im EndR(F) , h t ( f ) = v ( f ) .

PROOF OF THEOR~ i

have the

Ht(~).

It follows from the Corollary to Prop. 2 that

HOmR(F,G) is complete under the v-topology.

With respect to the

p-adic topology on Z and v-topologies on End(F) and R,

Z

R is a commutative 0/agram of continuous maps.

z § m

CF) to Z P

R

We may therefore

- 102

commutative. (III, w

-

Since HomR(F,G) is a torsion-free En~(F)-module

Prop.2), then HO~R(F,G) is a torsion-free Z -module. P

If g 6 Im {Zp §

, i.e. D(g) ~ Zp, then for f ~ HomR(F,G)

we have

(*~)

vCf o g) = vCf) + vCDCg)),

where v(D(g)) is the p-adic value of D(g). Now consider End(F) and En~(~) with the height filtration, and Z, Zp again with the p-adic topology.

~Te get a diagram of

continuous maps Z P

-L

which is commutative when Z is replaced by Z, hence remains P commutative now. It now follows that C***) --

is a homomorphism of Zp-mOdules.

m

But Kom~(F,G) is a free

Zp-mOdule of rank 0 or rank h 2 (cf. Lemma i, given after this proof).

Since (***) is an embedding (Prop.B), then HomR(F,G)

is a free Zp-module of rank -- inf {v(~), vCB)} . The e l,ements o.f P(F) of finite

order ~orm a subgroup

A(F), the torsion subgroup O f P(F).

(ii) P(F) and A(F) are modules over (iii)

If f : F § G is a homomorphism of formal groups defined

ove~ R then the map P and

F = GaI(K/K).

e ~-~ f(e) is ahomomor2hism P(f) : P(F) + P(G).

A are covariant functors from the cate~olV

c ate~or~ o_~f r-modules.

In particular P(F) and

~ R to the A(F) are modules

over Em~(F), and these endomorphisms commute with PROOF

(i)

If L/K

r .

is finite, and SL the valuation ring of L, then

F(S L) is defined as in I, w

Theorem i and is an abelian group.

We

-

106

-

have then

P(F) = l ~ m F ( S L) = L (ii)

If

UL

F(SL)

y 6 F, then Y F ( % 8 )

= F(Y%YS),

since F is defined

over R and its coefficients are therefore fixed by

7 9

This part

of the proposition is then easily verified. (iii)

If f : F + G is a homomorphism defined over R, then

m

f maps~into

f(;(xj))

itself (since f has zero constant term).

= QCf(x),f(Y)),

Since Yf(~) = fCY~), f commutes ~ t h Remark

Since

y

o

I f F i s the a d d i t i v e group GaD then PCF) i s j u s t ~ w i t h

the ordinary addition.

ACF) = 0.

If F is the multiplicative group Gin, Gm(X,Y) = X + Y + XY, then P(F) is isomorphic to the group U of those units u of ~ for which u - 1 ( m o d e ) .

The isomorphism P(F) §

U is given by m ~ - ~ l + m.

A(F) is isomorphic ~clth the group of pn-th roots of unity for all n. An iso~eny f : F § G is defined to be a non-zero homomorphism defined over R.

Since f' (0) is algebraic over K then f'(0) lies in

some finite extension L of K.

By Prop. l, Cor.2, there exists

g e HOmL(F,G) such that g'(0) = f'(0).

Since f,g ~ Hom~(F,G), then

the same proposition tells us that f = g. L ~

~.

f is thus defined over

IIence every isogeny is defined over some finite extension of R.

From now on all formal groups to be considered are assumed to be of finite

-

107

-

height, unless otherwise mentioned. i

(i)

the map P(f) : P(F) § P(G) is sur~ective;

(ii)

the kernel of P(f) is a finite group of order ht(f).

PROOF

(Lubin,Serre)

Let ~ 6 ~ .

extension S of R.

I@t f : F § G be an isogemy.

Then

THE O R ~

Then f(X) -

is defined over some finite

For the Weierstrass order we have the equation

~-ord(f(X) - ~) = W-ord(f(X)) = pht(f),

and ht(f) is finite by w Theorem (I,

w

Prop. 3.

By the ~eierstrass Preparation

Th.3) therefore, f(x)

-

= u(x).g(x),

where u(X) is an invertible power series and g(X) is a distinguished polynomial: xpht(f)

g(X)

+

O

0

v(%) _~- vCn).

Since ~'F is defined

and aI = i.

~e thus have

Put n =p-Cn) ; then vCn) _ t-(n) v(u) - vCn) ~ pu(n) v(~) - u(n), which tenas to ~ a s n § ~,

provided that v(~) 9 O.

Hence ~F (m)

converges if v(u) > O. Q@

Choose ~ ~ ~ so that v(~) = I/p-l, n=l

e,~e

Bp-I = p.

Then

V(an sn-1) _> ~

- ~(n)

which is 9 0 when v(n) = O.

n-i . v(n) 9 I>-I

--

If v(n) 9 0 we continue

pv(n)-1

- v(n)

p-I

= I + p * p2 + ,,. + p v ( n ) - i . v(n) > _ 0 .

Therefore (S-lo

~ anSn-i xn has coefficients in ~, n-1 Its inverse under composition

AF o 8) (X) =

and leading coefficient i.

@@

(s-I ~ eF ~

=

X bn sn-lxn n=l

is thus also a p ~ e r series ~rith integral coefficients and leading coefficient i.

Hence V(bn Bn'l) > O.

Tahe ~ e Jl/p-l' i.e., such that

- iii-

v(a) 9 I/p-l.

Then

V(bn~=V(bn~-I = V(bn~'l)

(~)n-i ~) as n § ~,

+v((~) n-l) + v ( ~ ) § ~

8

since v(~) > O.

Thus eF(m) converges if m ~ J1/p-l"

V(bn~n) > v(m), if n > 1.

Moreover

Therefore eF(~) = m + ~' ~rhere v(m')>

Hence ~r deduce that, if a ~ J1/p-l' then v(eF(a)) = v(m). if a e J1/p-l' then v(~(m)) = v(m).

v{m).

Similarly,

The maps a ~-~ eF(~) and

~-~ ~(~) thus define inverse bijections J1/p-1 § J1/p-l"

Under

therefore the subgroup of points F(J1/p_ l) becomes isomorphic to the additive group of J1/p-l' and the inverse isomorphism is given by eF.

We have thus established (1) and (iii). Since E + is torsion free, then A(F)C Ker ~ .

i,

F(J1/p-1) for

gF C ~]F (m)) = 0, then by (iii),

Let m & Ker

integer n > 0

~ F (~) = 0.

~.

Since

Therefore m 6 A(F).

Thus in fact Ker ~(F) = A(F). Suppose a E E + 9 Since E+/Jl/p-i p

m

a ~ Jl/p-! for some m.

such that ~F(m) = p a.

is a torsion module, then

Thus by (ill) there exists a ( J!/p-I

But P(F) is divisible (Theorem 2) so there

exists 8 e P(F) such that ~]F(8) = ~

. Since proOF(8) = pma, then

~F (8) = a.

: P(F) § E+ is sur~ective

He have thus sho~m that s

and so that the sequence 0 § A(F) § P(F) § E + § 0 of groups is exact. Since ~

is defined over K this is a sequence of F -modules.

If f ~ ~u~(F), then both A~ ~ f and f'(O) o ~F are homomorphisms F * Ga ~ith derivative f' (0) at O.

They therefore coincide.

From

- 112

-

the commutative diagram

IP(f) I P(F)"

we deduce that P(F) ~ The foll~r

~+

-

is a homomorphism of E n d ( F ) - modules.

theorem is a converse of Theorem i.

that every finite subgroup of

It shc~m

A(F) arises as the kernel of some

isogeny.

THEORI~I I~ (Lubin)

Let r be a finite subgroup o f A(F).

Let L be t h e

fixed field of the stabilizer o f r i n GaI(~/K), and let S denote the integers o f L.

Then ther_~eexist_____~s a formal group G and an

f : F § G, both defined over S, so that

(i)

Ker f = r

(ii)

Ifg

,

: F +H

exists a u n i q u e ~

(we ~ i t e

Ker f for Ker P(f)),

i s an isogeny with Ker g D h

: G §

suc h that g = h

r , then there Q

f.

Ifg

and H

are defined over the integers S I of some finite extension L I o_~fL then so is h.

COROLLARY ,,,m

J

SI,

I

,,

I_f.fthere exists an isogeny F § G defined over some

then there also exists an isogeny G § F defined over S I. OF

PRooF coRo

mY 1

If f

exponent of Ker f is pr.

:

F § G is an isogeny, then suppose the Then Ker f C

Ker ~p]r F"

By Theorem 4

-

113

there exists an isogeny h : G §

-

such that h o f = ~]F' and h

is defined over S1. COROLLARY 2

Either HomsCF,G) = 0, o_rHo~sCF,G) s@~___En~(F)-module

i_~sisomorph~c to a non-zero ideal of Ends(F) , .a~d as an Ends(G)-module i__sisomo~,rph,~ic t~o~nonrzero ~deelo_~fEnds(G). PROOF

Suppose HOms(F,G) # O.

g : G § F over S.

By Cor. 1, there exists an isogeny

The map f ~-~ g 9 f is

an injective homomorphism

Homs(F,G) +Ends(F) of En~(F)-modules, whose image is a non-zero ideal.

Analogously for the map f ~-~ f o g.

CORQLI~. y 3

(1) (ii)

If HOms(F,G) # 0 then the quotient fields of DCEn~CF))

and of DCEndsCG)) coincide;

the rank of Homs(F,G) over Zp i_L th_~erank of End(F)

(and of EndsCa)). PROOF

(ii) fo!l~s immediately from Corollary 2 and from the fact

that any non-zero ideal of an integral dom~A~n !, which is a Z -order, P has the same Z -rauk as I. P For (i) ~rrite ~ and let ~

= D(Ends(F)) , H = HomE(F,G) , T F = En~F(H) ,

be the quotient field of E F (viewed as a subfield of ~).

Define similarly EG, T G and LG.

By Corollary 2, H is isomorphic

to a non-zero ideal of the integral domain EF, and therefore T F is a subring of ~ , c

similarly

containing E F.

Clearly E G C

TF, hence

L G.

For the proof of Theorem 4 we shall need some lemmas. If A is a complete local ring (this always to imply that it is

- 114 -

Hausdorff) then so is A[[X]].

~[e vrite ~u for the maximal ideal

and w for the associated filtration of the latter ring. T ~ A, then we = y L~A

2

If T s

vim, A[[T]I as a subring of A[[X]].

Suppose that X is a root of the polynomial in U

n-i

P(U) -- ~

-

i

Z i=l

pi u

,

with coefficients in A[[T]], an_d s~ that

w(p i ) _ > n -

A[[x]] PROOF

i .

g er te as

l,

"l.

For each non-negative integer m and for i = 0,1,...,n - i,

there are unique elements r

. in A[[TJ], so that in A[[T]][U]

mjl

n-i

~m.

Z rm,i ui

(rood P(u)).

i--O Here rm, i - ~m,~" when m_< n - i, and rn, i = Pi" W(rm, i) _9 m -

i.

For m 9n

Thus, for m_< n,

one easily establishes the same

inequality by induction, using the iteration formulae for the rm, i. o@

}fence if am ~ A, the series

~

Zmrm, i

converges under the

m=o

w-topology and hence

as we were required to show. Reck11 now (of. I,

w

Th.l) that if B is a commutative

ring containing A, complete under some filtration u, and if 8 E B,

- 115

u(8) > 0 then there is a unique continuous homomorphism of rings

e : A[[X]], order § B,u with 8(X) = 8, which leaves A elemen~rise fixed. B = A[[X]], u = w.

Let in particular

Then the resulting 8 is continuous also for

the w-topology. Let now F be a formal group (commutative, of dimension l) defined over A, and let # e *~A (the maximal ideal of A).

Then by

the preceding argument we obtain a continuous automorphism 8r of A[[X]] over A which maps X into F(X,r

Let F(A) be the group

of points, i.e. of elements o f ~ A under the product ~ If r

is the inverse of r in F(A) then

automorphism of

8r .

Hence 8@

e~_1

= F(m,8).

is the inverse

is bicontinuous.

The map r § er

is then an im~ective homomorphism of F(A) into the bicontinuous automorphism group of A[[X]]/A.

Let now r be a finite subgroup of F(A),

and suppose that A is an integral domain. l~.~,[& 3 PROOF

The fixed rinE of r i_~nA[[X]] is A[[T]], where T = ~

r

F(X,r

We consider the T~ierstrass order in U on the power series

ring A[[T]]

[[U]].

We have, with n = card r

W-ord ( ~ F ( U , r

- T) = W-ord(r~F(U,r

r

=

ord F(U,r

,

r

= n,

r

as W-ord F(U,r (I,

w

Th.3)

= I.

Therefore, by the Weierstrass Preparation Theorem

-

116

-

r-] FCu,~) - T = PCU).Q(U), 0 where P(U) is a monic polynomial in U of degree n over A[[T]] and Q(U) is an invertible power series in U.

Clearly the FCx,r

and so in particular X = F(X,O) are roots of

[-I F(U,#) - T, hence

of P(U).

Counting degrees and number of roots we see that

C1)

eCu) =

~

(u - r(x,r

ThUs P(U) satisfies the conditions of Lemma 2, and hence

(2)

A[[X]] is generated by I,X,,..,X n'l

as an A[[T]]-module.

Now let E = quotient field of

A[[X]].

Z 0 = quotient field o f A[[T]], E 1 = fixed field of ~ in E. E"aen (3)

E0 C

El,

and by Galois theory

:

-n.

But by (2), E is generated over E 0 by i, X,..,X n'l.

In view of

(3), (~) it follows firstly that E 0 = El, and secondly that the I,X,... ,Xn-1 are independent over E O. E 0.

Thus P(U) is irreducible over

By (2) therefore AK[x]] is a free A[[T]] -module on 1,X,...,X n-l,

i.e., every element ~ of A[[XI] has a unique representation in the form

-

117

-

n-1

(~)

, =

Suppose n ~

I

a.x',

i=o

~

A[[~]].

a.:

that m is fixed under *

, i.e., in E 1 9

As

E 0 = E 1 and as (5) is the representation of a as an element in terms of a basis over EO, it follows that

A[[x]] n ~l

c

Thus

A[[~]]. The o~site ino~usion ~s t r i ~

PROO____FFOF THEOR~I ~ of L'.

m = a 0 s A[[T]].

Let L' = K(*), and let S' denote the integers

Then f(X) = F] F(X,~) is a power series over S' with

vanishing constant term. Therefore K e r f _~ @ F(m,~) = O.

.

For ~ * *

, f(~) = [~ F(~,~) = O.

Also, if a ~ P(F) and f(~) = O, then

This means that F(m,~) = 0 for some

is the inverse of $ under +. F that Ker f = ~ .

Hence a * ~.

~ ~ ~, sad

Thus we have s h ~ m

Let A = S' [~X]], and define f*(Y) = f(F(X,Y)) ~ A[[Y]]. Then

f*(~) -- ~

r(r(xj),~)

For ~ ~ ~ , f*~(Y) = ~ -- ~

-- ~r(x,r(Y,~)).

F(F(XJ(X,~)),

~)

r(r(x,Y),z(~,~)) - f*(Y).

By Lemma 3, the fixed ring of A[EY]] under ~ is A[[f(Y)]]~

Hence

f*(Y) ~ A[[f(Y)]] = S'[[f(Y),X~] = B[[X]], where B - S'[[f(Y)]]. Consider f**(X) = [7 F(F(X,Y), $). action of ~ on B[[X]] given by X @-- F(X,~).

This is fixed under the We may apply the lemma

again, and deduce that the fixed ring of B[[X]~ under ~ is

-

118

-

B[[f(x)]1 -- s, [[f(x),fcY)]]. NOW we s ~

up : f(F(X,Y) = f*(Y), when considered as a

power series in Y over A = S' [[X]], and we saw that

f*(Y) 6 A[[f(Y)]] = B[[X]], where B = S' ['[f(Y)]]. Thus f(F(X,Y) = f**(X) is an element of B[[f(X)]] = S' [[f(X),f(Y)]].

Hence there exists

G(X,Y) E S' [[X,Y]] so that

(6)

1~ow f,(x) -

f(F(X,Y)) = G(f(X), f(Y)).

[ {F,(x,~) N

F(X,~)} .

~e put X - 0 and observe that

0 s ~, and we get f'(O) = F'(O,O) F-] ~ # O.

Thus, working over L'.

we conclude that

(7)

a-

fo

F o f-1

is a formal group. Let A be the stabilizer of * in Gal(~/K). this is a subgroup of finite index, ~ o s e L.

If ~ ~ A , then f(X)~ =

As , is finite,

fixed field we denoted by

U-~F(X,$~) = F-] F(X,~) = f(X).

f(X) is defined over L, and by (7), so is G.

Thus

Thus finally G is a

formal group defined over S = S' 6~ L and f is an isogeny F § G defined over S so that Ker f = *

.

~Te have thus established (i).

Let now g : F * H be an isogeny with Ker g m

* , defined

over the ring S I of integers in some finite extension ~ may suppose that L ! D

L'.

l.~e see that for

gePCx) = gCFCx,~)) = HCgCx),gCr

of K.

~Te

~ ~

= HCgCX),O) = gCX).

Thus gCX) lies in the fixed ring of # in Sl[[X]] , i.e., gCX) = h(f(X))

-

by Lemma 3, with h(X) s S1 [Ix]]. neither has h(X). w

-

As g(X) has no constant term,

One nc~r verifies easily that h is an isogeny G § H.

Diyision.and Rational Points. ~,R, K, v , ~ , ~, ~

between K and ~, s=

119

{~e~

etc., are as in

w

L is a field

S is the domain of integers of L, i.e,,

[vC~)A0}

.

~=G~C~I~).

Let F be a formal group over R, whose reduction mod is of finite height.

Write

P(F,L) = P(F) ~

L,

A(F,~) = A(F) O L (subgroups of points, and of torsion points in L).

Let moreover

~(F,L) be the subgroup of P(F) of points m which are of finite order rood L, i,e., for which ~ ] F (m) s L for sufficiently large n. Thus ~ (F,L)IP(F,L) is the torsion group of P(F)IP(F,L). TEEOR~4 1

AF gives .rise t_~oa cg~utative diagram with exact

r o w oz h o = = o r ~ % s = 0

§ ACF,L) § PCF,L) ~ L + § ~Cfl,ACF)) § ~(fl,PCF)) § 0

1 0

§ A(F)

COROT.T,~RY 0

o z ~n~(F)-=~dules.

~

I §

O § L+

§

0

get an exact sequence

§ A(F)IA(F,L) § ~(F,L)IP(F,L) § L+llm ~ §

O.

- 120-

PROOF OF ~ PROOF

COROLL~Y

In view of w

Immediate. Theorem 3, we get an exact sequence

0 § HOcn,ACF)) § HOcn,P(F)) § HOcn,~+) § ~Cn,ACF)) § ~C~,PCF))§ But ~C~,~ +) = o.

~e thus get the top roar of the diagram.

that I is given by the restriction of AF. cohomology groups are F~CF)-modules,

~(n,~+).

Note also

It is clear that the

as the operation of En~CF)

on P(F) and on ~+ commutes with the Galois group (~+ is an En~(F)-module via the map End(F) § R § ~).

The proof of the theorem ~-ill be complete

once we have shown that ZF(e) s L + if and only if m e ~

(F,L). Here

we use

~ 1

Let fb_~e an i s o ~ e ~ F

Then f(~) ~ P(GJ) ~ ~

o~

§

~ e ~ n e ~ o v e r S, and ~ ~ P ( F ) .

~

for ~

~ ~ ~ . ~ ~ ~ Ker P(f).

(~ i_~sthe ~fference in_ P(F)). Taking the le~

f o r g r a n t e d a t t h e moment we n o t e t h a t

if

~ ~t (F,L) then ~ ] F (~) ~ L for some n, hence for that n and for all also ~]F(~m ~ m) = O, i,e., us ~ ~ ~ A(F) = Ker ~F"

Thus

mAF(m) = AF(~m) = AF(a).

Conversely,

In other words ~F(a) s L +.

~F(m) & L + implies that ~ . m ~Ker AF for all ~ . But there are only a finite number of elements ~ ~

~.

Hence for all ~ and for some n,

~ s Ker ~]F " This implies that [p]F(e) = ~ ] F ( m )

i.e.,

fp]F(S) ~ L, whence ~ s ~(FtL).

~f(=)~ f(~) - 0 for all < = 9 f(~ ~

=) = 0 for all ~

.

for all m ,

- 121-

SUGGESTION

In the following discussion (Theorem 2 and 3) consider the

particular case when F = G

m

and its relation to Kummer theory.

In the next theorem A c stands for the product of c copies h is the height of F.

of a group A.

ACF)IA(F,L) ( Izp)cI

THEOREM 2

,

:

L*Iz=

C89

=

~(F,L)IP(F,L)

c2

=~ (~/Zp) c,

, c = cI

+

c 2.

Here cI < h, and if the valuation on L i s discrete cI = h. c 2 p .

[P']F (ACF,L)) C J'l/p-1 By w

Thus

~

ICer

Theorem 3, the latter group is null.

A(F,L) C Ker [P]F

and hence is finite.

In other words

Thus in fact cI = h.

Let ~F(X) : ~ % x n. ~e already ~o~ that v(%)~ - v(~). n=l If the valuation of L is discrete, and p is as above then for all 6 P(F,L),

V(s

> inf vC~_ mn) > inf n vCcl) - vCn) > i~ n

~ere

~

= inf

:11.

"

{np - v(n)} > -

n

~ .

I

Thus vCIm A) > t for some

n

integer t, i.e., the fractional ideal

~ t contains Im ~ , and d

80

we have a surjection

,.+/p " (%/z)I:,.. =

Let nc~r @ be a subgroup of A(F,L). isogenies g over S originating from F.

I

Define ~ % = set of

(i.e., g : F § G for some G),

so that Ker P(g) C 0 . ~

= set of a 6 P(F), so that for some ge ~ ~, gCa) s L.

Note that gl(al) ~ L,

~

is a subgroup of P(F). g2(a2) E L.

For suppose gl' g2 6

~

Then the subgroup of A(F) generated by

,

-

123

-

Ker P(gl ) and Ker P(g2 ) is finite, hence of form Ker P(f) where f = fl ~ gl = f2 9 g2" to be defined over S.

As Ker P(f) C ~ we may suppose f, N o w w e see that f t

fl and f2

~ # , say f : F § G.

On the o t h e r h a n d f(al~ a2 ) = fl(gl(al)) ~ f2(g2(a2)) 6 L, as fi(gi(ai)) 6 L. W_~ehave ~cg~nutative diagram with exact rc~m 0 §

r

o § ACFJ)

§

e

PCF,L) § PCF,~)

§

L

§

~(a

(~11§

t~ ( e , P ( r ) )

(Homc = continuous homomorphisms). (ii)

Define for a e ~ r

, ~ efi,

e(a + F Then

Note:

= 0 for all m ~

) ae

= 0 for all a
~

PCF,L), leave_~sL C ~ 0) elementwise fixed.

The last result gives a perfect pairing

Unfortunately this does not in general allow us to determine GalCLC6~@)/L)uniquely. COROLLARY finite, ,

,,,

GslCLC~#)/L) is an Abelian pro p-~oup.

then the ~

that of r .

But we evidently have

o_~fGaI(L(~r

I5" O i_~.s

i_~sfinite and divides

-

124

-

The diagram comes from the diagram

PROOF OF THEORI~{ 3

0 --~ #

§ PCF) §

0

§ P(F) §

§

~

r §

0

§

0

on taking cohomology, provided that we show that

(i)

- H=cCn, on

Cii) i,e.,

and

HOcn,P(F)Ir

a~ a s #

Now if a 6 ~ r g e

r

which is true as ~ acts trivi~11y

),

= ~/r ,

for ~.11 m
h. ~Trite Pm for the surjection T(F) § Ker ~ ] F associated with the inverse limit T(F) = lim Ker ~]F" defined.

M C T(F) and so Pn(M) is

It is the direct product of at most s cyclic subgroups, and

so the number of elements in Pn(M), not in ppn(M) is at most p

ns

(n-l)s

- p

9 Write ~n = Pn (x)"

Pn(M), and not in ppn(M).

Then each element of Fmn lies in

Therefore

card(r%) n o the inequality v(m n) < 1/e p (n'no)h. en be the ramification index of K(an)/K.

On the other hand let

Then certainly en v(a n) >_ l/e,

1/e being the least strictly positive value of v on K.

[K(~n) : K] --> en--> ~ It remains to prove the lemma.

_9 pnh c,

Hence finally

c = p-n~h.

Let ~]F(X) =

~

anXn.

n=! Here aI = p.

Apply I, w Theorem 2 to the ring RIpR mud t/he

reduction of ~]F(X) rood pR. whenever p $

This tells us that v(a n) >__v(p) = i

n, i.e., in particular

-

(~.8)

v(~) .9 z

138

for 0

-