K-Theory of Finite Groups and Orders

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

Author: Richard G. Swan | Graham E. Evans

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Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, Z0rich

149 Richard G. Swan University of Chicago, Chicago / IL / USA Notes by

E. Graham Evans University of Chicago, Chicago / IL / USA

K-Theory of Finite Groups and Orders

$ Springer-Verlag Berlin. Heidelberg. New York 1970

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin. Heidelberg 197(XLibrary of Congress Catalog Card Number 75-133576 Printed in Germany. Title No. 3306 ~ck:

Julim Beitz, W(finheim/B~gstr.

Introduction This set of notes is a sequel to my previous notes in this series [SKI. of Chicago.

The notes are from a course given at the University No pretense of completeness

is made.

A great deal of

additional material may be found in Bass' book [BK] which gives a remarkably complete account of algebraic K-theory. notes, however,

The present

contain a number of recent results of Jacobinski

[JS and Roiter [RS.

An excellent survey of the theory of orders

with detailed references may be found in Reiner's article

[RS].

I would like to thank E. Graham Evans who compiled, and wrote up these notes.

edited,

Thanks are also due to Professor S.

Mac Lane who suggested editing these notes for publication, to Mrs. M. Benson who typed them.

and

T A B L E OF C O N T E N T S

Chapter

1.

Introduction ...............................

Chapter

2.

Frobenius

Chapter

5-

Finiteness

Chapter

4.

Chapter

i

Functors .........................

13

Theorems ........................

37

Results

on K 0 and G O . . . . . . . . . . . . . . . . . . . . . . .

54

5.

Maximal

Orders

o

83

Chapter

6.

Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

104

Chapter

7.

K 0 of a M a x i m a l

Order ......................

126

Chapter

8.

K 1 and G 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

140

Chapter

9.

Cancellation

167

Appendix

O

o

o

o

o

o

e

e

~

e

o

e

o

e

e

e

e

e

o

e

e

o

e

e

o

e

Theorems ......................

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

References

205

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

233

of S y m b o l s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

235

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

236

List

Chapter One: Introduction Let R be a ring.

Then Ko(R) is the abelian group given by

generators [P] where P is a finitely generated projective R module, with relations KP] = [P'S + [P"S whenever O -4PP' - ~ D P - ~ P P "

-~-O

is an exact sequence of finitely generated projective R modules. K 0 is a covariant functor from rings to abelian groups.

If

f: R -~PR', then Ko(f'~ Ko(R) -*mKo(R') by [P] - ~ [ R ' ~ P ] .

If R

is left noetherian, then Go(R) is the abelian group with generators [M] where M is a finitely generated left R module with relations [M] = [M'] + [M"] whenever O -o-M' --b-M - ~ M "

-~O

quence of finitely generated left R modules.

G O is not a functor

since the tensor product R ' ~

will preserve all the relations only

when R' is flat as a right R module. given by [ P ] ~ [ P ] .

is an exact se-

There is a map Ko(R) - ~ G o ( R )

This map is called the Cartan map.

It is

natural with respect to maps of rings R -~-R' such that R' a flat right R module.

If R is left artinian then Go(R) is free abelian

on [S1] , ..., [Sn] where S i are the distinct classes of simple R modules.

Ko(R) is free abelian on [Ii], ..., [I r] where I i are the

distinct classes of indecomposable projective R modules. The map Ko(R) -~-Go(R) gives a matrix (aij) where aij = the number of times Sj occurs in a composition series for Iio Definition.

A ring R is left regular if

l)

R is left noetherian, and

2)

every finitely generated left R module has a finite

resolution by finitely generated projective left R modules.

-

If R is commutative,

2 -

we omit the adjective

"left".

We recall the following theorem whose proof is well known [SKI and is essentially the same as that of Theorem 1.2 below. Theorem 1.1.

If R is left regular,

then the Cartan map is an iso-

morphism. Examples

(Bass-Murty).

nor an isomorphism

The Cartan map is neither a monomorphism

in general.

abelian group, and Z ~ i s

a finitely generated

its integral group ring, then G o ( Z ~ )

finitely generated but K o ( Z ~ ) If ~ i s

If ~ i s

is

is not finitely generated in general

a finite group, then Ko(Z ~ )

= Z~finite

group but G o ( Z ~ )

has rank greater than 1 in general. Definition.

Let R be a commutative ring and A an R algebra.

G~(A) is the abelian group with generators

[M] where M is a left A

module which is finitely generated and projective with relations O -~M'

as an R module,

[M] ~ [M'] + [M"] for each A exact sequence

-~-M -~-M" -~-0 of A modules which are finitely generated

as R modules. Theorem 1.2.

Let R be a regular commutative

ring and A an R alge-

bra which is finitely generated and projective Then the map f: G~(A) -I-Go(A)

as an R module.

given by [M]-'~q~[M]

is an isomor-

phism. Remark.

The examples of main concern are integral group rings of

finite groups. Proof.

First we need a few preliminary results.

-3-

Lemma 1.~.

(Schanuel~

Let R be a ring and 0 -~PB - - P

and O - - ~ B '

-~PP' - ~ A - ~ - O

with P and P' projective. Proof.

Consider

be two exact sequences Then P ~ B '

-I~A -~-0

of R modules

is isomorphic

to P ' ~ B .

the diagram 0

0

B

= B

0 -~PB' -qP X - ~ D P

-~0

II 0 -~P B' -~p P' ~ A

0

where X is the pullback. isomorphic Corollary

to P ~ B ' 1.#.

0

Then since P and P' are projective

and to P' ~ B o

Let R be a ring and

0 - ~ B n -~PPn-1 - ~ ' ' "

-~'Po -~PA -~PO and

0 -qpB~ - ~ Pro-1 -~" "'" -~'Po - ~ A

-~PO be exact sequences

modules with Pi and Pi' all projective. Bn~P' n-l(~Pn-2 Proof. Remark

~

" . " . is isomorphic

(I. Kaplansky). dimension

Corollary of a module

to B'n ~ Pn-1 ~ P ' n-2 ~ argument

1.4 enables (without

projective

By Corollary

resolution.

" ." "

from lemma 1.3.

one to define the

reference

to Ext) as

the first n such that the kernel of any n step projective tion is projective.

of R

Then

This follows by an easy induction

homological

X is

resolu-

1.4, we get the same n for any

-4-

We return to theorem 1.2. for f: G~(A) -~-G0(A). ators.

We want to define an inverse, g,

It clearly is enough to define g on gener-

Let M be a finitely generated A module.

Then M is a

finitely generated R module and has finite homological dimension over R.

Say hdRM = n.

Let 0 -~-B n -W-Pn_ I - ~ ... -~-Pj -~DM ~w-O

be ariA exact sequence with all the Pi finitely generated projective A modules.

Then all the Pi are finitely generated projective

R modules since A is a finitely generated projective R module. Hence B n is a finitely generated projective R module. g([M]) = [P0 ] - [P1 ] + ... (-1)n[Bn ].

We must show that g is well

defined on generators and preserves relations. i)

, _~pp,

0 -~B m

m-i -~P "'" -~'P0 - ~ M

We define

Let

-q~0

be an A exact sequence with P~ finitely generated projective A modules with m ~ n. 2)

Let

0 -~PB m - ~ Pm-i - ~ "'" -~'Pn -~DBn -~PO

be an exact sequence with all P finitely generated projective A modules. 5)

Then

0-~'Bn-~Pm_

I -~P... - ~ P P n - ~ - P n _ l

is an exact sequence.

--~-... - ~ P o - ~ P N - ~ - 0

Apply corollary 1.4 to I) and 5).

Then we

want to show that 2) implies that [B m] - [Pm_l ] + ... ± [Pn ] $ [B n] = 0 in G~(A).

That is we must

-5-

show that if 0 -*-Mp ~

... -~-M 0 -~-0 is an A exact sequence of

finitely generated projective R modules

t h e n ~]( -~1 )-i [ ~ Mi

= O.

We prove this by induction on n by making two A exact sequences O - ~ - M k - ~ - . . . 4)

-~-M2-~PX-~-O

and

0 -~PX - ~ M I - ~ M 0 -~DO

and noting that 4) implies that the A module X is a finitely generated projective R module since $) splits as a sequence of R modules. Now given two resolutions

of M, we can assume they have the

same length by the result just proved.

Corollary 1.4 now shows

that g is well defined on generators. Let 0 -~-M' -~-M -~PM" - ~ 0 nitely generated A modules. step resolutions modules.

be an A exact sequence of fi-

Pick n ~ h d R M '

, hdRM , hdRM".

Pick n

of M' and M" by finitely generated projective A

Combine to get an n step resolution of M (see Cartan-

Eilenberg, p. 80) giving the following commutative diagram of A modules with all rows and columns exact: 0

0

0 -i.p, - ~

ker h ,

0 -~M' 0

-~-

m

0 -rap @

M 0

-~PO

.

-~DN' --~PO 0

-6-

where all PI, P[ are finitely generated projective A modules.

Then

g([M]) = g([M']) + g([M"]) since g([N]) can be defined from any projective resolution of length greater than or equal to hdRN where N is a finitely generated A module. Hence g gives a well defined map g: Go(A) -~ G~(A).

g is

clearly the inverse of f. Using theorem 1.2 we will investigate the functorality of G O and G~.

Let R be a regular commutative ring and h: R -~-R' a h a

momorphism of commutative rings.

Let ~ b e

any finite group.

Then

we have a homomorphism h ~ of group rings induced by h, h': R ~ - ~ ' R ' ~ .

R'~

-- induces a map K o ( R ~ )

-~-Ko(R'~).

We also have the diagram

GO(R [ )

a (RT[)

GO(R']]" )

k

G

(R'][)

where f is an isomorphism and k is defined by k([M]) = [ R ' ~

M]

which preserves relations since we are only considering modules which are projective over R, and, hence, all short exact sequences split over R. f'kg: G o ( R ~ )

If g is the inverse of f, then -m-Go(R'~).

Therefore, Go(R ~ ) is a functor on the

category of regular commutative rings.

The main use of this will

be for the cannonical projection Z -a~Z/pZ.

-7-

If M and N are finitely generated R~modules, then M ~ N is a finitely generated R ~ m o d u l e where p ( m ~ n ) all p 6 ~ .

= (pm~pn)

for

If M', M, and M" are R ~ m o d u l e s which are finitely

generated and projective as R modules, if N is amy R~module, and if the R~sequence 0 ~ M ' sequence 0 - - ~ Z ' ~

-~M

N -~Z~

-~PM" -m~O is exact then the R ~

N -~-M"~

N -~0

is exact.

If N is

also a finitely generated projective R module then so are M ' ~ N , M"~N,

and M ~ N . G~(R~)~

Hence there is a well defined map G~(R~) ~

by [ N ] ~ [ N ] ~ [ M ~ N ] .

hN:

)

o -~.~,

~N

in M~N

G ~ ( R ~ ) given

If we fix N then we have

) given by EMI -~N"

-~0

CM NI.

i s e x a c t then hN=hN,,+h~,.Since

the ~ a c t i o n

i s the d i a g o n a l t h i s product i s a s s o c i a t i v e and commuta-

tive.

The R ~ m o d u l e R with trivial ~ a c t i o n is clearly the iden-

tity.

Hence we have proved:

Theorem 1.5. Then G ~ ( R ~ )

Let R be a commutative ring and ~ a

finite group.

is an associative commutative ring with identity.

Corollary 1~6.

If in the above R is regular, then Go(R~-) is an

associative commutative ring with 1. By the same proof, we can show Corollar ~ 1. 7.

With the nota$ion of theorem 1.5 K o ( R ~ ) is an

associative commutative ring without 1 (unless ~ = wise R is not projective over R ~ ) .

{1) since other-

-8-

If R - ~ - R '

is a homomorphism of rings such that R' is a

finitely generated R module then [ M ] ~ [ M ] map GO(R') - ~ G o ( R ) .

gives a well defined

Let R be a commutative noetherian

ring, A

an R a l g e b r a w h i c h is finitely generated as an R module, and S a multiplicatively closed set of R containing 1 but not containing O. Then A S is an algebra over R S and M ~ R s ~ = defined map Go(A) -@mGo(As).

M, induces a well

If M is a finitely generated A module

such that M S = 0 then there exists an s • S such that sM = O. Hence M can be thought of as an A/(s) module.

Hence the map

Go(A ) -~PGo(As) ~nnihilates the images of all Go(A/(s)) for s • S. The following two theorems describe in full detail the map Go(A) -~-Go(As). Theorem 1.6.

The proofs may be found in [SK, Chapter 5].

Let R be a commutative noetherian

ring, A an R alge-

bra finitely generated as an R module, S a multiplicatively closed set containing 1 but not O.

Then the sequence

Go(A/(s)) -~-Go(A) -~-Go(A S) -~-0 is exact. s~ S Theorem 1. 7. With the same notation as the above the sequence Go(A/PiA) -~PGo(A) ~ G o ( A S) -~-0 is exact where {pi ) is the i~I set of all prime ideals of R with pi ~ S ~ ~. Let k be a field and A a k algebra which is finitely generated as a k module then Go(A) is free abelian on IS1] , ..., ISn] where the S i represent all the isomorphism classes of simple A modules. n

If M is a finitely generated A module, then

[El = ~ a i [ S

i] a i ~ Z in Go(A).

We want to recover the a i from M.

-9-

We could try characters, characteristic

of k.

but they only yield information up to the

Pick a ~ A.

a(m) = am for all m ~ M.

Then a: M -Ip M is given by

We define ~ a ( M )

= characteristic poly-

A

nomial of a on M.

Then ~ a ( M ) ~ k[X].

If 0 -~PN' -~PM -~pM" -~PO

is an exact sequence of finitely generated A modules, then ~a(M)

= ~a(M')

~a(M").

For, pick a basis of M' ~

to a basis for M, which projects to a basis for M". determinant

involved gives the result.

For, if dimkM = n ~ a ( M )

= Xn + ....

M and extend Computing the

If M # 0 then ~ a ( M )

~ O.

If M = O, define ~ a ( M )

Now k[x] - (0) = k[x] + is contained in k(x) - (0) = k(x)*.

= 1.

Define

(~a: Go(A) -~D k(X)* to be the homomorphism sending addition in Go(A) to multiplication Theorem 1.8 (Brauer).

in k(X)* such that [M] ~-%-~@a(M). If x ~ Go(A) and ~ a ( X )

= 1 for all a ~ A,

then x = O. Remark.

We can even pick out a finite number of a ~ A such that

it is enough to check for them. Proof.

x = ~

k i[S i] = E mi[S i ] - E n i[S i] where each isomormi, O ni~O

phism class [S i] occurs only once in the sum. i.

mi Let M = m~ oSi and

Then x = [M] - IN] where M and N are semisimple A

ni>,O modules. Hence,

~a(X)

= 1 = ~a(M)~a(N)

-I implies that ~ a ( M )

= ~a(N).

it is enough to show that a semisimple A module is deter-

mined up to isomorphism by the values ~ a ( M )

a ~ A.

-

lO

Let R be the radical of A. simple.

Then RM = 0 since M is semi-

Thus, H is an A/R = ~ module and ~ is semislmple since A

has finite dimension over k. are simple algebras. identity of B i.

the~ai(M)

Hence A = B 1 x...x B n where the B i

A acts on S i through B i.

determine M. ~ 1

S i.

Lift the e i to a i in A.

We claim

a i acts on Sj in the same way as e i.

Say

Then a i acts like 0 on the first summand

like 1 in the second.

Hence

~ai(M)

Thus we can recover the m i from ~ a i ( H ) oz the

Let e i be the

Then the e i are primative central idempotents and

e i acts on Sj as (~ j if' ii ~f j

and

-

= X r (X-l) mi(dimkSi).

and H from the collection

(M).

Theorem 1.$.

Let R be a regular commutative domain, A an R alge-

bra which is a finitely generated projective R module, m a maximal ideal of R, and K the quotient field of R. f: G o ( K ~ A ) G~(A) ~

There is a unique map

-~-G0(A/mA) which makes the following diagram commute: Go(A) ~

Go(K~A) ." f G O (AlmA )

Remarks.

The hypothesis insures that all maps exist since

G~(A) -@-G0(A) is an isomorphism,

f is given by a matrix which is

called Brauer's matrix of decomposition. map Go(A) - ~ G 0 ( K ~ A )

By theorem 1.6 or 1.7 the

is onto and hence f, if it exists, is umiqu~

-

Proof.

Let S = R - (0).

ll

-

Then K = R S and K ~ A

= AS .

Let M be a

finitely generated A module such that sM = O for some s E S. [M]~O or 1°7°

If

in Go(A/mA) for all such M, then f exists by theorem 1.6 Let l) 0 - ~ P n - ~ P . . .

-~pP0 --~DM-~PO be an A exact se-

quence with all the Pi finitely generated and projective over R. Then [M~ a G0(A) corresponds to ~

(-1)liP i] ~ G~(A).

Since

A/mA = Am/mmA m we can factor the map G0(A) -~PG0(A/mA) as Go(A) --~G0(A m) -~-G0(A/mA).

That is we can assume R is local.

Thus all the Pi are free over R. Localizing the sequence l) at S we get 0 -~PPhs - ~ ' ' "

-~PPos -~PMS -@m0 and M S = O.

(-1)i[Pi~ = 0 in G0(As). ~a(Pis

)(-1)i = 1.

Therefore,

Hence for each a ~ A,

Since Pi are all free over R, we pick a basis A

for each and obtain matrices for a on the Pi which are the same matrices as for a on the^Pis. teristic polynomial Pi a - ~ P i -~ ~a(Pi) i even

=

Therefore, F i = ~a(Pis) = characis in R[xS.

~ ~a(Pi). i odd

Reducing mod m we have ~

Therefore

Hence by theorem 1.8 ~ Fi = ~ F i. i even i odd

(-1)i[Pi/mPiS E Go(A/mA).

The bases for

Pi reduced mod m give bases for Pi/mPi since R is local with maximal ideal m.

Let a ~ A and reduce mod m to get ~ ~ ~ = A/mA.

Then

A

the matrix for the characteristic polynomial for ~ , ~ ( P i / m P i ) , ~a(Pi/mPi) reduced mod m. phism.

But reducing mod m is a ring homomor-

Hence i ~odd~ ( P i / m P i )

= i ~even~(Pi/mPi)

since

is

-

Fi/mF i = i odd Go(A/mA).

~ Fi/mF i. i even

12

-

Thus [M] ~ ~

(-l)i[Pi/mPi ] = 0 in

Hence the map f exists.

Theorem I.I0.

Let R be a commutative local domain (not necessarily

noetherian) with quotient field K and maximal ideal m.

Let A be an

R algebra finitely generated and projective as an R module. t~at the Cartan map ~ :

Assume

Ko(A/mA) -a~Go(A/mA) is a monomorphism.

Let P and Q be finitely generated projective A modules such that K~P

is isomorphic to K ~ Q

Remarks.

over K ~ A .

Then P is isomorphic to Q.

We will show later that the Cartan map is a monomorphism

for R ~ w h e r e

~is

a finite group.

This theorem is false even for

ideals of Dedekind rings if R is not local. Proof.

P and Q are finitely generated free R modules.

PI' "''' Pn and ql' "''' qn where n = d i m K K ~ P

Pick bases

= dimKK~Q.

For

A

each a ~ A we have a: P -~-P and a: Q -~-Q yielding matrices A and B with respect to the bases (pi) and (qi). characteristic polynomials of A and B. isomorphic f = g.

Let f and g be the

Since K ~ P

P = P/mP is a module over A/mA.

module ~ is free on (pi), the images of (pi).

and K ~ Q

are

As a k = R/m

As before the char-

A

acteristic polynomial of a on P is f with coefficients reduced mod m.

Hence for every ~ ~ A/mA the characteristic polynomials for

and Q are the same. Go(A/mA>.

Thus, by theorem 1.8 [~] - [Q] = 0 in

That is if X ~ G~(A) goes to 0 in G o < K ~ A )

to 0 in Go(A/mA). a monomorphism.

Thus ~ ( [ ~ ]

- [Q]) = 0 in G0(A/mA).

Hence [~] - [Q] = 0 in K0(A/mA).

then X goes But

~

Hence P / m P ~ F

is

-

is isomorphic to Q / m Q ~ F module.

13

-

where F is a finitely generated free A/mA

A/mA is artinian.

Therefore, the Krull-Schmidt theorem

applies and we cancel the F's getting P/mP isomorphic to Q/mQ. Thus we have shown K ~ P isomorphic to Q/mQ.

isomorphic to K ~ Q

implies P/mP is

We examine the diagram p..,. P/mP

f

• -~Q

f' L.~ Q/mQ

0

0

where f' is any isomorphism of ~ with Q. jective and maps onto Q/mQ.

f is onto by Nakayama's lemma.

projective and hence f splits.

Q is

Now N = ker f is finitely generated

since it is a direct slmmand of Q.

Since f splits, N/mN = ker f'=O,

and hence N is O by Ns~ayama's lemma. and P and Q are isomorphic.

f exists since P is pro-

Hence f is an isomorphism

More quickly, we note that P and Q

are projective covers for ~ and Q which are isomorphic and hence P and Q are isomorphic.

Chapter Two:

Fr0benlus Funct0rs

In this chapter we discuss Go(R~ ) and K o ( R ~

as functors

in ~. I

Let ~ b e the inclusion.

a finite group, ~ Let ~--

~

~'gi

t

a subgroup, and i: ~

-~

gi ~ ~ be the decomposition of

-

into disjoint cosets. gl' "''' gn"

14

-

Then R ~ is a free R ~

We have G o ( R ~ ) ~ I

module with base

G0(Rl0given by iJ([M]) = [M].

This is functorial for category of finite groups and inclusion maps.

Since R ~ i s i, Go(R~ ) ~ G o ( R ~

a free R ~

module, it is flat and hence we have

given by i,[M] = [ R ~

M].

This was defined

by Frobenius long before tensor products were known.

It was called

induction or forming the induced representation. We assume R is a regular commutative ring. Go(R~ ) = G~(R~ ) are all rings. i: ~

-~P~

Then

For any map of finite groups

i* is defined and is a ring homomorphism where we use

the ring structure on G R ( R ~ ) and push it down to Go(R~ ) isomorphism.

Even without regularity i*: GoR(R~ ) -~-GR(R~) is a

ring homomorphism. homomorphism.

via the

However i,: G R ( R ~ ) -~-GR(R~) is not a ring

For i,([M]) = [ R ~

M] and R ~

M is isomor-

phic to

II M as an R module. Hence i, does not preserve the 1 identity as a r-nk count shows. But i, does give a covariant func-

tor from finite groups and monomorphisms to abelian groups. Theorem 2.1.

(Frobenius)

Let R be a commutative ring, i: ~

-~-~

an inclusion of finite groups, x ~ GR(R~ ), and y ~ GR(R~).

Then

i,(i'(x)y) - xi,(y). Remarks. >

This says i~,: GoR(R~) -a,-G~(R~) is a homomorphism of a

> mo u e

-

of i x.

15

-

The formula is similar to the relation between homology and

cohomology. Proof.

We prove this by proving the following:

Theorem 2.2. and N an

With the same notation as above let M be an R ~ m o d u l e module

M~(R~N)

Then

is isomorphic

where M is considered as an R ~

to

module on the left

side. Proof.

We define a map f on generators which will be an isomor-

phism.

Let p ~

f(p~(m~n)) q g ~.

(m~n)

~ R~(~R~(M~N).

= pm~(p~n),

Then q ( f ( p ~ ( m O n ) )

Hence f is an R ~ m a p . and f ( p i ( q ) ~ m ~ n ) f(p~qm~n).

= qpm~(qp~n)

Let q t ~ ,

= f(q(p~

then p i ( q ) @ m ~ n

~ pi(q)m~pi(q)~n

= p@

Let

(m~n))). qm~n

= p(qm)~(p~qn) in

We define an inverse g to f by ~ p~(p-lm~n).

R we need it is over ~ . pi(q)~(i(q)-ip-lm~n) all q ~ .

f is automatically an R map.

Hence f is well defined over action of R ~

R~R~(M~N). g(m~(p~n))

We define

This is clearly multilinear over

g(m~(pi(q)~n)) ~ p~(p-lm~qn)

g and f are clearly inverses.

~ g(m~(p~qn))

for

This completes theorems

2.1 and 2.2. Definition (Lam). l)

A Frobenius functor is

a contravariant functor F:

finite groups and monomorphisms

commutative rings ~ith 1 and homomorphisms preserving 1.

-

2)

For each i: ~

-~an

i.: F ( ~ ) -~mF(~)

16

-

inclusion of finite groups we are given

such that i~ makes F into a covariant functor

from finite groups and monomorphisms to abelian groups,

i, is

additive, (ij)~ = i,j~, and 1. ~ i. 5)

i: ~

-~P~.

Let x 6 F ( ~

and y J F ( ~ ) .

Then

i,(i*(x)y) = xi,(y). Definition.

A morphism ~

between two Frobenius functors F and G

is defined by giving a ring homomorphism ~ each finite group ~ s u c h

that if i: ~

: F ( ~ ) -~mG(~)

for

- ~ - ~ i s a monomorphism of

finite groups then the diagrams i,

---*- F(V )

G(TF ) i*

i,

C O ~ U t e. For example for the Frobenius functors GR(R~) and GR'(R'~ ), if R --I~R' is a homomorphism of commutative rings then :

GR(R~) -~pGR'(R'~)

given b y ~

([M]) = [ R ' ~ M ]

is a

morphism. If R is not regular, then Go(R~ ) fails to be a Frobenius functor since it is not a ring.

If i: ~ - ~ - ~ i s a monomorphism of i* Go(R ~ ) Note that finite groups, then we have Go(R ~ ) ~

Ko(R~ ) is a ring but without a unit.

If [P] ~

Ko(R~ ) then

-

[P] ~ K o ( R ~ ) over R ~ .

17

-

since R ~ is a finitely generated projective module

Hence we have i" and i, in this case also.

and Ko(R ~ ) are modules over GR(R~).

Both Go(R ~ )

For if M is a finitely gener-

ated R ~ m o d u l e projective over R and N is a finitely generated R ~ module, then M ~ N O -~M'

is a finitely generated R~module.

-~PM -~PM" - ~ 0

If

is an exact sequence of finitely generated

R ~ m odules projective over R, then 0 -~-M' ~ N

-~-N~N

-~-M" ~ N

quence splits over R.

-4PO is exact since the first se-

If 0 -~-N' -~tN -~mN, -~-0 is an exact se-

quence of finitely generated R~modules, then O -~M~N'

-@DM~N

-~-M ~ ' - ~ 0

tive (and hence flat) R module.

is exact since M is a projecThus KM] ~ IN] = K M ~ N S

well defined product making G o ( R ~ )

into a GR(R~)module.

gives a This

suggests the following Definition

(Lam).

Let F be a Frobenius functor.

Then a Frobenius

module M over F is defined by giving, for each finite group ~, an F(~)

module M ( ~ )

and, for each inclusion of groups i: ~

two maps i*: M ( ~ ) - ~ - M ( ~ ) and i,: M ( ~ ) l)

-~b~,

-~-M(~ ) such that

i. and i* are functorial and additive, i.e., M with i, is an

additive covariant functor and M with i* is an additive contravariant functor. 2)

If x ~ F ( ~ )

and y a M ( ~ ) ,

then i*(xy) = i~(x)i*(y), and

3)

if x a F ( ~ )

and y ~ M ( ~ ) ,

then i,(i~(x)y) = xi,(y), and if

x i F(~), y E M(V),

then i.(xi ~(y)) = i~(x)y.

-

Examples.

18

-

Go(R ~ ) is a Frobenius module over G ~ ( R ~ ).

a Frobenius functor, then F ( ~ ) Definition.

If F ( ~ ) is

is a Frobenius module over F ( ~ ) .

If M and N are Frobenius modules over the Frobenius

functor F then a m orphism of Frobenius modules is defined by givin~ for each ~, a morphism f : M ( ~ ) - ~ - N ( ~ ) of F ( ~ ) modules which is natural with respect to i* and i,. Frobenius modules and morphisms over a Frobenius functor F form an abelian category as the reader can easily check. A more important example of a Frobenius module is K0(R ~ ) over G~(R~).

i. and i* are defined as usual.

M is a finitely generated R ~ m o d u l e

projective over R and P a

finitely generated projective R ~ m o d u l e , generated projective R ~ m o d u l e . multiplication as before. over R, M ~ P

We need to check that if

then M ~ P

is a finitely

Then we will have a well defined

Since M and P are finitely generated

will be finitely generated over R (and hence over R ~ .

Hence we only need the following: Proposition 2.~.

If M is an R ~ m o d u l e

which is projective

(resp.

free) as an R module and P is a projective (free) R ~ m o d u l e , M~P Proof.

then

is a projective (free) Rl~module. If N is free as an R module with base m i and P is free as

an R ~ m o d u l e with base pj, it is easy to see that M ~ P an R ~ m o d u l e

with base m i ~ P j .

an R ~ m o d u l e

Q such that P ~ Q

module N such that M ~ N

is free as

In the general case, we can find = F is a free R ~ m o d u l e

= G is a free R module.

and an R

We let ~ a c t

-

19

-

trivially on N making N an R ~ module. G~F

= (M~P)

@

(M~Q)

jective since G ~ F

~ (N~P)

Then

~ (N~Q).

Ko(R ~ ) - ~ GR(R~ ) given

and GR(R~ ) --~Go(R ~ ) given by [ M ~

morphisms of Frobenius modules. Ko(R~ ) -~-Go(R ~ )

is pro-

is free.

Thus Ko(R ~ ) is a Frobenius functor. by C P I A ~ E P 3

Hence M ~ P

EM~ are both

Hence the Cartan map

is also, so the kernel of the Cartan map is a

Frobenius functor. There are two basic types of morphisms of the Frobenius modules we have developed so far. forgetting.

We cover them both in the next proposition.

Proposition 2.4. rings. i)

The first is tensoring, the second

Let R -q~R' be a homomorphism of commutative

Then

CMS~CR'~ITM3

induces a morphism of Frobenius modules over J.LI!

GR(R~ ) in the cases ,

2)

a)

GR(R~ ) - ~ G R ' (R'~),

b)

Go(R~ ) -qPGo(R'~ ) when R' is a flat R module, and

EZ~-

EZ3 induces a morphism of Frobbnius modules over G ~ ( R ~

in thb cases

a)

G~(R~)~

GoR'(R'~) when R' is a finitely generated pro-

jective R module, b)

Go(R~ )~r- Go(R'~ ) when R' is a finitely generated projective R module, and

- 20 -

c) Proof.

Ko(ell" ),,,- Ko(e'll" ). This is routine for the most part.

facts which are necessary.

We only list a few

We need to show that the maps are mor-

phisms of modules for each R' and R and that the maps are natural with respect to both i *

l)

and i, induced from the inclusion ~

For example, we need that

~M~NS

commutes.

But this is true since R ~ R ~ fore, ( R ' ~ R ~ ) ~ , N 2)

-~.

is isomorphic to R' ~ M ,

and, there-

is isomorphic to M ~ N .

Here we need that

[N] ~

INS

commutes.

But it does because R' ~ M ~ , N

is isomorphic to M ~ N .

Thus,

all maps in l) and 2) are defined with respect to GR(R~ ). Next we check naturality. be G R, GO,

or K O-

First for i* the maps

Let ~

-~D~be

an inclusion and let

-

(~)

~

21

-

~(R,I[ )

i*~

$i*

sends

which commutes. For i= we note that the maps

sends M so w. ~.ed that

ule.

~

~

~

is isomorphic ~o ~ ~ M

~s ~

~o~

-R ! RII We could prove this by checking generators and relations or

via an explicit map using the fact that R ' ~ , M I|

M

M which is isomorphic to R ~ M .

is isomorphic to

But this was Frobenius'

fit original construction for the tensor product.

Don_._~e.

The reason for this work is to aid calculation of Frobenius functors. functor F.

Suppose we are given a finite group ~ a n d In general F ( ~ )

for ~ c y c l i c however. tion on F ( ~ )

is hard to calculate.

a Frobenius We may know it

So we want a method for obtaining informa-

from {F(~)} where ~

runs over the cyclic subgroups

of ~. More generally, let ~ functor, and M a module.

be a class of groups, F a Frobenius

-

22

-

Definitions.

TF If B is a group and A a subgroup of B then we say A has ex-Loonent dividing n in B (B/A has expln) if x ~ B implies nx ~ A. A typical theorem (to be proved later) is the following due to E. Artin: Theorem 2_~..

Let ~ b e

characteristic

a finite group of order n, K a field of

0 (so K 0 = G O = GR), and ~

subgroups of ~, then Go(K ~ )/Go(K ~ )~

be the class of cyclic

has expln.

We will use the machinery of Frobenius functors to extend Artin' s Theorem. Lemma 2.6.

Let M be a Frobenius module over the Frobenius functor

F then a)

M(~)~

is a sub F ( ~ ) m o d u l e

of M(~),

b)

F(~)~

is an ideal of F(~), and

c)

F(~)/F(~)~

is a ring and M ( W ) / N ( ~ ) ~

is a module over it.

Proof. a)

M(~)~

is closed under addition by definition and under multi-

plication by elements of F ( ~ ) b y

the definition of Frobenius modul~

a) implies b) by taking M to be F.

-

c)

25

-

We only need to check that F ( ~ ) ~

That is, we must show F ( ~ ) ~ ators.

Let x 6 F ~

M(~)~

and m g N(~),

annihilates N ( ~ ) / M ( ~ ) ~ .

M(~)~.

We check it on gener-

Then Done.

(i,x),m = i,(x,i*m)6 M ( V ~ . Lemma 2.7.

Let R be a ring and N a module over R.

If there exists

an integer n such that nx = 0 for all x ~ R, then n annihilates N. Proof.

nx = n(lx) = (nl)x = Ox = O.

Corollary 2~8.

If F ( ~ ) / F ( ~ ) ~

C o r o l l a s 2.~.

Let ~ be an algebra over R.

G~(R~)-~G~(A~)

has expln so does M ( V ) / M ( ~ ) ~ . Then R -~-A induces

and makes G~(A~) a G~(R~)module.

G~(R11")/G~(RT~)~ has expln, SO does G~(A~) / G ~ ( A ~ ) ~

So if .

If we could prove Artin's theorem for Z we would have it for any ring, since any ring is cannonically an algebra over Z. theorem is true for Z with n replaced by n 2.

We prove this later.

We can dualize the above theory using i*. over the Frobenius functor F and ~

Let M be a module

be a class of groups.

Definition.

M(~) ~

Lemma 2.10.

a)

M(~) ~

is a F(~)submodule of M(~),

b)

M(~) ~

is a module over F ( V ) / F ( V ) ~ .

c)

If F ( ~ ) / F ( ~ ) ~

Proof.

= ~=~

ker[N(~) i - ~ p M ( ~ ) ] .

a) and c) are clear.

Artin's

has expln so does M(~) ~ .

-

b)

24

-

Observe that we only need F ( ~ ) ~ .

x ~ F ( ~ ) with ~ 6 ~ But i*y = 0.

M(~) ~

= 0.

If

and y e M(~) e , then (i,x).y = i,(x.i~y).

Hence (i,x),y ~ 0.

Now we state an extension of Artin's theorem. Theorem 2.11.

Let ~ b e

a finite group, ~

a class of subgroups, R

a Dedekind ring, and K its quotient field.

If Go(K~)/G0(K~) ~

has

exp In then a)

if p is a nonzero prime ideal of R and k = R/p, then

Go(k~)/Go(k~) ~

has expln, and

b) Q0(R }/G0(RT[} Corollary 2.12.

has expln2

Let ~ b e

a finite group of order n and C

be the

class of cyclic subgroups of ~, then a)

if R is an algebra of characteristic p, p ~ 0, then Go(R~)/G0(R~) G

b)

has expln, and

if R is any ring, then GR(R~)/GoR(R~)~

has expln 2.

Proof of corollary is easy from Artin's theorem, theorem 2.11 applied to Z and corollary 2.9. These are not the best possible results.

Lam [ L ] has im-

proved on them by improving the exponent in Artin's theorem. Proof of Theorem 2.11.

By Theorem 1.9, there is a map

Go(R~)-a-Go(R/p~) which is clearly a morphism of Frobenius functore since G0(R~ ) -a~Go(KK) and Go(K~) -amGo(R/p~) are. Go(R/p~) is a Frobenius module over Go(K~). Corollary 2.8.

Therefore

This proves (a) by

-

25

-

For (b), we are given x E Go(R~)and must show n2x ~ Go(R~) c • We examine the diagram

II GO(R/P~) _a~ GO(R~) ~-~ GO(K~) --~ 0 pCR

p~o

~

~.J o

pCR

p~O We have f(nx) = n f ( x ) a

Go(K~) ~ .

If Go(R~) ~

were onto then there would exist

y ~ Go(R~) ~

-~-Go(K~) ~

with f(y) = f(nx).

ness nx-y = g(z) for some z ~

Thus f(nx-y) = O.

~ Go(R/p~). pCR

Hence by exact-

Now, by (a),

p#O n~ ~

~R%(~/pW)C.

Thus n2x -- ~y + g(nz) ~ d n2x ~ % ( R ] D ~

as

P

p/O needed.

It remains to check that Go(R~) ~

-~-Go(K~) ~

is onto.

is enough to show we can get a set of generators for Go(K~) ~ Pick y ~ Go(K ~ ) mapping onto it. Corollar~ 2.12.

where ~

.

Let ~ b e

Done.

a finite group of order n and ~

class of cyclic groups, then Go(Z~)/Go(Z~) ~ C o r o l l a ~ 2.1~.

the

has expln 2.

a finite group of order n, ~

of cyclic groups and R any ring, then GR(R~)/G~(R~)~

K0(R]]')/Ko(R'~') ~ has expln 2.

•

Then there exists a z ~ Go(R~)

Clearly i,(z) maps onto i.(y). Let ~ b e

It

the class and

-

26

-

Before we can prove 12tin's Theorem we need some preliminary remarks on trace and class function. Let K be a field of characteristic K algebra,

and H a finite dimensional A module.

let a: M - ~ D H H.

O, A a finite dimensional

denote the function a(m) = am.

Pick a ~ A and

XN(a) = trace of a on

The trace is K-linear in a, that is, if a, b 6 A and r, s ~ K

then XN(ra + sb) = rXM(a) + sXM(b). If 0 -~-M' -~PH -~PM" -~bO is an exact sequence of finite dimensional A module,

then we can pick a base of N" and lift its ele-

ments to elements of H.

The resulting set, together with a base

for M' will be a base for M.

With respect to that base the matrix

o)

for a on H looks like a onM'

on

Trace is diagonal X(a):

sum, therefore,

~"

XH(a)

= XH,(a)

+ XH.(a).

Thus,

Go(A) --I~K when we think of the trace as a function of the

module.

Since X M ~ HomK(A,K)

and X M = XM, + XM. we can regard X as

a map from Go(A ) to HomK(A,K). Theorem 2.14.

If K is a field of characteristic

dimensional K-algebra, Remark.

then X: Go(A) -~PHOmK(A,K)

0 and A a finite is injective.

Go(A) is free abelian on [Si] where S i are representatives

of the isomorphism classes

of simple modules.

Since HomK(A,K)

divisible group for char K = O, X cannot be onto. characteristic

p # O, X cannot be injective.

is a

Also, if K has

-

Proof.

Pick x 8 G0(A ) .

27

Then x = ~ m i [ S

and N are semisimple A modules. are ~ modules.

-

i] = [M] - IN] where M

Let ~ = A/rad(A).

Then M and N

Thus we need to show that if M and N are semisimple

A module and X M = XN, then N is isomorphic to N. ni M = ~(l~Si).

If we can determine the n i we are done.

r A =i=lllA i

where the A i are simple rings and the S i are simple A i modules. Let e i be the identity of A i. idempotents, delta.

ei: Sj ~ S j

Hence, ej: M ~ M

we put the Sj's first.

Then the e i are primitive central

is ~i

where ~ij is the Kronecher

has a matrix which looks like (~ Hence the trace is njdim Sj.

Pick a i 6 A mapping onto e i ~ ~. since a i acts on M through e i. uniquely.

Then XN(ai) = n i dim S i

dim S i # 0 so we can solve for n i

Note that we need characteristic 0 here.

recover the n i from M.

~) if

Thus X:G0(A) -~-HOmK(A,K)

Thus we can

is one-one. Done.

Let A = K ~ w h e r e from ~ t o

K}.

~is

a finite group.

X: G o ( K ~ ) - ~ - { K

HomK(A,K) . (function

valued functions on ~}.

We make the

latter a ring by (fg)(s) = f(s)g(s) for all s ~ . Theorem 2.1 ~. Proof.

X is a ring homomorphism.

Let M and N be K ~ m o d u l e s

XM~N(S).

and s ~ .

We need to calculate

We need the trace of s: M Q N - ~ M ~ N .

s(mQn)

sm~sn.

If m i i~ a base for M and nj a base for N then M @ N

mi@

If sm i

nj.

~

aiKm K and snj

E

bjenj, then

= has

-

28

-

s ( m i ~ n j) = s m i ~ s n j = k , ~ a i k b j ~ ( m k ~ n ~ ) .

So the trace is XM(S)XN(S).

Next we check that the map preserves the identity. the unit of Go(K~).

sl = 1 for all s ~ ~.

[K] is

Thus XK(S) = 1. Done.

Thus X identifies Go(K~)with a subring of K valued functions on ~. Now let F(~, K) be the ring of K valued functions on ~. i: ~

~

~

are

If

homomorphisms of groups, we clearly get a commuta-

tive diagram o(KTF ) .......

Go( ]f )

l •

?(IT,K) ....... i" f

~-

~

F(T~ ,K) fi

If i is a monomorphism (which we can assume is an inclusion), then we have ~ = U s i ~

where the s i are coset representatives.

Then K ~ is free as a right K ~ K~K~N

= ~ ( s i ~ M).

That is, every element is uniquely expres-

sible in the form ~ s i ~ m t(~si@m Ubsi~

i) = ~ t s i @ m

module with s i as a base and

i

mi~

N.

If t ~ ~, then

i is not in the correct form.

~ t ~ = ~ and the union is disjoint.

However,

Hence there is a permu-

tation of the cosets induced by t such that t s i ~

= st(i)~.

That

-

29

-

is , there are elements Pt,i ~ ~' such that ts i = st(i)Pt,i. fore, ~ t s i ~ m

i = ~st(i)Pt,i~mi

= ~ s t ( i ) e P t , i m i.

al, ..., an be a base for M over K. Then K ~ K ~ M (si~aj).

If p ~ ' ,

Let

has a base

then paj = ~R(p)j,ka k with R(p) a K. k

st(i) e P t , i a j = k~St(i)~R(Pt,i)j,kak

There-

Thus,

= k ~ ( P t , i ) j , k ( S t ( i ) ~ a k )"

That is, the coefficient of s i ~ a j in above is 0 if t(i) ~ i and R(Pt,i)~, j if t(i) = i.

x

(t)

K~@--M

=

Thus we can compute the trace by

2~

R(pt,i

i such that t(i) = i

=

j,j

~ ~ (trace of Pt,i on M). i such that j t(i) = i

That is,

Xi*(M)(t) =i s~uchthatXN(Pt'i)" t(i) = i t(1) = i if and only if ts i = siPt, i with Pt,i ~ ~' which happens if and only if si-ltsi -- Pt,i ~ ' XM(p) if p ~ ' XM(P) -- ( O if p ~ '

.

If we define

, then Xi,(M)(t) = ~, XM(s-lts).

s~

-

Lemma 2.16.

30

-

Let K be any field, ~ a n y

finite group and M a finite

K~module, then XH(t) = XH(sts -I) for all s,t ~ . Proof.

Consider the commutative diagram where the columns are

isomorphisms

Therefore, trace t: N -~-H is the same as trace sts-l: M ~ H . Done. Now we return to ~ = U s i ~ ' .

Pick p G ~ ' ,

then

(si-ltsi) = X(p-lsi-ltsi p) since one of the products is in ~ only if the other is.

Hence X(s

where I~'I is the order of ~ .

L) = ~ p ~

if and

,X(p-lsi-ltsi p)

Finally

xi.<M>. I III

Definition. for some t ~

A function f: ~ - a P K is a class function if s = p t p -I implies f(s) = f(t).

Let Fa(~,K) be the set of such

functions. Generalizing the formula for Xi,(M ) just given, we can define i,: F a ( ~ , K ) -%~Fa(~,K ) such that the diagram

-

Theorem 2 . 1 ~ s,t ~

Q~

%(K]1')

GO( V)

Fa( ~ ,K) ~

Fa(g,K)

commutes.

Let M be a Q ~ m o d u l e where ~ i s

a finite group.

If

generate the same subgroup (Ws~ ~ ~t'), then XM(s) = XN(t).

Remark. Proof.

31-

This is false if

Q is replaced by ¢, the complex numbers.

It is enough to consider the case ~

. ~s, = 4t~.

module has trace equal to XM restricted to ~

(XM~,).

M as a There-

fore, it is enough to consider the case of ~ c y c l i c of order n. Q~ ~(x)

Q[x]/(xn-l) where x A ~ g e n e r a t o r

~ ~ (x-r i) G Q[x]. primitive n-th roots of 1

Since the ~

Then xn-i =

of ~.

Let

~ (x-r i) = d~/n~d(X~ all n-th roots of 1

are relatively prime (having no common factor x-ri) ,

we have Q ~ Q [ x ] / ( ~ d ( X ) ) . We use the fact that ~d(X) are irreducible over Q and note that the remainder of this proof works for any field such that all ~(x)

are irreducible.

Q]I" Q xS/( n-l

]]'Kd

where K d is Q with a primitive d-th root r d of 1 adjoined. are fields.

All K d

We only need to check the theorem for simple modules.

The simple modules are just K d.

~acts

on K d through x by

-

32

-

A

rd: Kd -~DK d giving the usual trace TQ(rd)/Q(r d) ~ Q and TQ(rd)/Q = ~

conjugate8 of r d.

which generator of ~ w a s tive d-th roots of 1.

Hence TQ(rd)/Q is independent of

picked since the conjugates are all prima-

Therefore X(s) = X(t).

Now let K be any field of characteristic O. that for any K ~ m o d u l e of unity.

M and any s ~

Done. First we show

that XM(s) is a sum of roots

Let L be a field containing K.

Then X L ~ M ( S )

= XM(S)

because if e i is a K base for M then l ~ e

i is a L base for L ~ M .

Then s has the same matrix for both L ~ M

and M and hence the

characters are the same.

Thus we can assume K is algebraically

closed to compute characters.

K ~s~

Then xn-1 =

n = KEx3/(xn-1) = ~ K r i a n d

~ (x-ri). all n-th roots of 1,r i

s acts on Kri by multiplication by

ri • Theorem 2.18.

Let R be an integrally closed domain of characteris-

tic O, X be a trace of a K ~ m o d u l e where K is the quotient field of R.

Then X(s) ~ R for all s ~ ~.

Proof. R.

X(S) is a sum of roots of unity and hence is integral over

X(s) ~ K by definition.

But R integrally closed so X(s) ~ R. Done.

Definition. and

~t~

Let ~ b e

a group and s,t ~ .

are conjugate subgroups.

Write s ~ Q t

if

~s~

-

33

-

We can s1~mmarize what we have done so far as follows: is a character of a Q ~ m o d u l e and X ( s ) t

then l)

s ~Qt

If X

implies X(s) ~ X(t)

Z all s t

Theorem 2.1~.

(Artin).

be a function on ~ s u c h

Let ~ b e

a finite group of order n and f

that

a)

f(s) ~ Z for all s i W a n d

b)

S~Qt

implies f(s) = f(t).

Then nf = ~

aviv,(l

) where ~

runs over the cyclic subgroups of

w

~, i

is the character of the trivial representation,

Remark.

and av L Z.

If ~ is the class of cyclic groups, apply theorem 2.19 to

the function f which is identically i and get nf 4. G o ( Q ~ ) ~ a n d hence that G o ( Q ~ / G o ( Q ~ 6 has expln.

That is, this will complete

the proof of theorem 2.5. Proof. i)

It is enough to prove this for the case where f is the

characteristic function of a Q class that is ~[ = UK i where each K i is the equivalence class of the equivalence relation ~'~ Q.

~

fi(s) =

i

S ~K i

0

s ~K i

.

Then any f =

E

f(si)f i where s

Let runs over

i

a set of representatives of the classes. 2)

i~:

-a~is

an inclusion of finite groups and g is a

function on ~ ' of the form g = ~

aviv.(1

) over ~

~cyclic

and

-

= E

av(iiv,)(l

3a

-

) is of the same form.

We finish the proof by an induction on the order of ~.

Let

f be the characteristic function of the Q-class of s ~ ~. Case I.

s ~ ~does

not generate ~.

tion the theorem is true for ~ .

g(x)

=

f:

eraes

if x does not generate ~

aviv (1

).

i,(n'g)(t) =

= < s~.

By induc-

Now the function

tion of the Q class of s in ~ . n'g(x) -- ~

~

is the characteristic func-

If n' = I~'I,

then we know

By 2) i,(n'g) has the required form.

~

~ n'g(ptp-1) p~ll

= ~ g(ptp -1) p~.Ir i if ptp "l generates "I]" 0 if not

p aT = the number of p for which ptp -1 generates If t is not Q equivalent to s this number is O. (t ~

= ~asa "l~

= a ~ a -I.

Now ptp -1 generates

If t ~

-

p k so #S ~ k

be as in (4) and choose an

+ 1.

z # O,

z = ZlW 1 +

s - s' with s, s ' g [A:

~z]~

C(2~) n.

S.

i/k + ~ - i ~

By

S is >~-(k + l)n > [2%" I] so Subtracting them gives

+ ZnW n where each z, has the form

By (5), Izil ~ Z~.

Since A ~

Ix: A z] = [A: A z ] [ A so ~ / k ~

...

Rllxl ~ ~).

Therefore the number of

there are two such x's congruent mod I. I,

CN n.

Choose an integer k so

Let S = ( x ~

x = XlW 1 + ... + XnW n with all x i E

z E

N for all i

is finite by (1) and (2) since it is a

torsion module and finitely generated. that k n ~

Using (5),

I]-l~ 1 + ~-i.

I ~z,

Therefore

we have

C(2~)nk -n

u~ k )

but ~..~ 1 + k ~ -I

Therefore [I: A z] _~ D where

D = C(2 + 2 ~ - i ) n is independent of I.

We need one more lemma in

order to apply this. Lemma 3,13.

Let R satisfy (1), (2), (~) and (5).

Given D ~ ,

there is an r E R, r ~ 0 such that r annihilates every R-module C with # ( C ) ~

O.

-54-

Proof.

By (1), C ~ ~ R / R d i.

d i # O.

Since

Now d -- ~ d i a n n i h i l a t e s

(4) implies that #R ~ ~ ,

C and

Idl = ~

(5) there are a finite number of elements bl, every d with

Idl ~

D has the form d = ~ b i ,

Idil = #C ~ ..., b k G

~

all

D.

By

R such that

a unit of R.

Clearly

r = b, b 2 ... b k will do. Now if I is any left ideal of A , [I" ~ z ]

~

D.

Lemma 3.13. But

~/~r

Therefore

we have found a z ~ A

[Iz-l: ~ ] = [I: A z] ~

Then r ( I z - 1 / A ) is finite of order

~).

A -- Q[x]/(x2),

J ~r. .

This

for J. Let

and M n -- Z ~ Z where x acts by the

These are all distinct

show that we can recover n from M n. vectors.

~ ~

Irln where n is the rank of A

Theorem 3.9 is false if A is not semisimple.

L -- Z[x]/(x2), matrix (0

Let r be as in

= 0 so I ~ J = rIz -1 and

allows only a finite number of possibilities Example.

D.

so

Then xu ~ nv and xv -- O.

for n ~ 0.

We only need to

Let u and v be the two basic Hence ker(x)

x: M n -m~ M n is Zv

^

and the image

(x) ~ Znv.

recover n from M n.

Chapter Four:

Theorem 4.1. Zassenhaus

Therefore,

Note that QM n ~

Results

ker x/im x -- Z/nZ and we can A for all n by

u ~-~l, v ~

1-x" n

on K 0 and G O

Let R be a Dedekind ring for which the Jordan-

theorem holds, K the quotient field of R, and ~ a

group of order prime to char K. Go(R ~ ) - ~ G o ( K ~ ) is finite.

Then the kernel of

finite

-

Proof.

(Rim)

55

-

We have an exact sequence

o -~ Oo(RIT) - ~ Ko(~lT) -,- KO(KIT) where Co(R~) is defined to be the kernel. Co(R~) is finite by Theorem 3.8.

Let X be the kernel of G0(R~)-q-G0(K~).

We get

a diagram

0 --~ Co(R1]') ---~ Ko(R'TI" ) ~

0 ~

X ~

Ko(KTF )

Go(R'11") - ~ Go(K'II" ) --~ 0

where the right hand map is an isomorphism.

Since Go(R~) is a

finitely generated abelian group, X will be finite if it is torsion. Hence it is enough to show its generators have finite order.

We

have an exact sequence

II Go(e/p "IT) --'-- Go(Z~IT)--~ Go(KIT) ~ o p#O pC R

where the

le~t

Ko(R/p~) order s.

hand map has image X.

-~-Go(R/p~)

The map

has finite coker by Theorem 2.20, say of

Pick [M] ~ G o ( R / p ~ ) .

s[M] lifts to K o ( R / p ~ )

will have finite order if and only if s[M] does. enough to show that the image of K 0 ( R / p ~ ) Let M be a projective R / p ~

module.

direct summand of a free R / p ~

module.

Therefore, it is

-~-Go(R~)is

We claim pd

and [M]

RV

torsion.

M _~-1.

M is a

Hence, it is enough to show

-

PdR

R/p~

1.

R/p~

56

-

= R/p~R~.

The sequence

0 -~-p -~-R -~-R/p -~-0 is exact. serves exactness since R ~ i s

Tensoring with R ~ o v e r

a free R module.

0-~p~R~-~-R~R~-~-R/p~R~-~-O R~R~are modules.

Hence pd R

R/p~

Hence

is exact.

projective R ~ m o d u l e s

But p ~ R ~ a n d

since p and R are projective R

1 as claimed.

Next we claim that if M

is a finitely generated torsion module over R with pd R [Z] lies in the image of C o ( R ~ ) - ~ - G 0 ( R ~

Since pd I M ~

0 -~P-~-F

-~M

-q-O of R ~ m o d u l e

with F finitely generated free Then [M] = [F] - [P] lifts to

If [F] - [P] goes to 0 in Ko(K~) then IF] - [PJ ~ Co(R ~ )

as desired.

Tensoring over R with K which is flat over R we get

0 -~- K ~ P But K ~ M

will complete

1 we can find an exact sequence

and P finitely generated projective. Ko(R~

M ~ 1 then

Since Co(R~) is finite,

this will show that IN] has finite order in G o ( R ~ ) a n d the proof.

R pre-

-~ K~F

-~- K ~ M

-~-0

exact.

= 0 since M is torsion and, hence, IF] - [P] = 0 in

Ko(K~) as desired. As an application of this we give a generalization of the Herbrand quotient. e.g., ~ c y c l i c

Suppose ~ h a s

and q = 2.

periodic cohomology of period q,

If M is a finitely generated Z ~ m o d u l e ,

let hi(~, M) be the order of Hi(~, M). Q(M) = ~ hi(~, M) (-1)i. i=l then the exact

Let

If 0 -~-M' -~-M -~-M" -~-0,

is exact,

-

57

-

cohomology sequence shows that Q(N) ~ Q(M')Q(M").

Therefore Q de-

fines a map Q: G o ( Z ~ ) - a - Q * + , the multiplicative group of rational numbers~0.

Since Q*+ is torsion free and

X = ker[G0(Z~)-m~Go(K~)]

is finite, Q(X) = O.

through a map Q: G o ( K ~ ) - ~ - Q Q(Z) = Q ( [ K ~ M ] )

,+

= Q(O) = 1.

.

Therefore Q factors

If M is finite, Q ~ M

If ~ i s

= 0 so

cyclic, q = 2, Q(M) is the

Herbrand quotient h2/l(M ) and we have reproved Herbrand's result h2/l(M) ~ 1 for finite M. Theorem 4,2.

Let R be a Dedekind ring of characteristic 0 with

quotient field K and let ~ b e

a finite group of order n.

If no

prime dividing n is a unit in R and P is a finitely generated

project~e R~[ mod~e, then K~)RP is free over K]~. S

Proof.

We will prove this by computing characters.

Then X F ~ s

. If g ~ g

then g: K ~ - ~ - K g g i v e n

trace equal to the number of t i ~ ~

IO .

by g(t) ~ gt has

such that gt = t.

if g ~ i

×K~(g)=

. We need to sn

show that the character of K ~ P

So

if g ~

Hence ×?(g) =

if g ~ I

Let F = I~K~.

if g = ~

is equal to ×F for some F.

Since

Xp(1) = dimK(P ) we will need to show that n divides the rank of P. Lemma 4.~.

If k is a field of characteristic p, ~ i s

group, and I is the kernel of the augmentation

a finite p

~: k ~ - ~ k ,

then I

is nilpotent. Proof.

If G is a group and N a normal subgroup let J be the kernel

of the map RG -~-RG/N.

As a left ideal J is generated by all n - 1

-

with n ~ N.

58

-

As an R module J is generated by all xn - x with x E G

and n ~ N.

Since ~ i s

a p group there exists 1 # x ~

central and of order p.

Let N = ~ x ~ .

which is

We have a commutative

diagram with exact row 0 -~J

-~

.--',-0

k]T-~-kT[/~x~

'k/ k which shows J C I ~.

Hence

(I

. We use induction on the number of elements in

/J~

= 0 or equivalently I m C

enough to show that J is nilpotent. x n - 1 = (x - l) (something) Therefore,

jn = k ~ ( x

Hence jn = k ~ ( x

- i) n.

Thus, J = k ~ ( x -

i).

p.

- i) p = k ~ ( x p - i) = 0

This completes

the proof.

Corollar2~.4.

Under the above assumptions k ~ i s

Corollar2 4.~.

Under the above assumptions

projective k ~ m c d u l e s

it is

- 1)... but ( x - l) is central.

So JP = k ~ ( x

since k has characteristic

Therefore

But J is generated by

for all n.

- 1)k~(x

J.

local.

all finitely generated

are free.

Propositiqn 4.~.

(Nakayama)

acteristic O, ~ a

finite group of order n, and P a finitely gener-

ated projective R ~ m o d u l e .

Let R be a Dedekind domain of char-

If no prime dividing n is a unit in R

then n divides rankRP. Proof.

Let Pln and ~ p b e

a Sylow p subgroup of ~.

finitely generated projective

R~pmodule.

Then P is a

Hence if the theorem is

- 59 -

true for p groups it is true for finite groups.

Let pln ~ pa

then p is not a unit.

Hence there is a prime ideal ~ C R such that

p ~.

is a projective R/~Wmodule.

P/~P = R / ~ P

P/~P is free over R / f ~ b y an R module P ~ A l ~ P/~P = A1/~A 1 ~ dimR~P/~P

module. M/~

... ~ As/~A s = R / ~

... ~ R/~.

As

Then

Therefore,

This completes the proof of proposition ~.6.

Let R be a ring, ~ a

Then _H~ =

I dlmR~P/~P.

... ~ A s where A i are ideals.

= rankRP.

Definitiqn.

corollary 4.5 so ~ I

Therefore,

finite group, and M be a R ~

(x g Mlsx = x for all s ~ )

= M/iM = M/((s - l)M)

and

That is, M / ~ i s

obtained from M by

sE~'"

identifying x and sx for all s i ~ . Let N ~ R ~ b e

the element N = ~

s.

If t & ~ ,

tN = N = Nt.

s all

If x ~ M, then Nx = ~ s x Therefore, N: M - ~ - N ~ C factors M N _ ~ M ~

and tNx = Nx. M.

Thus Nx ~ M ~ f o r

N(s - l)x = (Ns - N)x = O.

all x L M. Therefore,

"

M/W Lemma 4. 7 .

If M is projective then the induced map N: M / ~ - ~ M

~ is

an isomorphism. Proof.

There exists M' such that M ~

M~® M,~

M' = F is free.

Clearly

F~ and M / ~ ® M'/W ~ F/V ~ d if ~: F/W ~ FV is an iso-

morphism then ~[: M/]T-'~M ]]"is also.

F~

II (R~)~ and

F/~ = II(R~/~).

Hence we only need to prove the lemma for M = R~.

R~/~

~ R where the map R ~ / ~ - ~ - R

= R~=

R~I

is given by the

-

60

-

augmont~tioo If ~ ass ~ C ~ ~a~o

~ o S,to

~hont E a ~ Eass ~here~o~o

~a~s~ .~e~e~

~

O~ ~ o

That is, a s = a for all s ~ 11"o Hence, (R~) ~

map N: R ~ / ~

--i,D(R~) ~

clearly sends 1 ~ R ~ T

a ~ o ~o~ ~ = R.N.

Now the

= R to N ~ (R~) ~

and is an isomorphism. Lemma 4.8°

If 11"is a finite solvable group of order n, R is a

Dedekind ring of characteristic 0 such that no prime dividing n is a unit in R, and P is a finitely generated projective R ~ m o d u l e , then rankRP = n rankRP~. Proof.

Lemma 4.7 allows us to use P / ~ for P~.

We proceed by in-

duction on I ~ l l)

If ~ i s

of ~o

not of prime order, there is a proper normal subgroup N

rankRP -- INlrankR(P/N) by inductive hypothesis.

(R~RNR~)(~R ~ P = R~RNR~= R(~/N)~R

R('~'/N)(~R.I1.P. The

R V / ( m - 1 ) R ~ for all n ~ N. P is projective over R~/N,

=

I tT/N I rankR(P/'N')

last step is true because Therefore, R ~ R N ( R ~ ( ~ R

P) =

-

by the inductive hypothesis.

61

-

Hence rankRP = I ~ [ rankRP~ for I ~ I

not prime. 2) ~

is cyclic of prime order p.

y G R with p & y.

Then there is a prime ideal

We reduce mod y.

since change of rings is transitive. R/y~module.

Then (P/~)/y(P/~)

= (P/yP)/~

Now (P/yP)/~ is a projective

By corollary 4.5) (P/yP)/~is a free R/y~module.

Therefore, rankRP/~-- dimR/y(P/~)/y(P/~) rankRP = dimR/y(P/yP).

= dimR/y(P/yP)/~while

Let k -- R/y and

r r r P/yP = ~ k~. T h i s h a s dim p r and (PIyP)ITF= LL k ~ / T F = I I k h a s 1 1 1 dimension r. This completes the proof of lemma 4.8. Now we return to the proof of theorem 4.2.

R is a characteristic 0 Dedekind ring,

a finite group of order n, and no prime dividing n is a unit in R. P is a finitely generated projective R ~ m o d u l e . rankRP -- ns for some integer s. generators.

characters.

Let F be a free R ~ m o d u l e on s

Then rankRF = rankRP.

phic over K ~ t o

K ~ R F.

We know that

We need to show K ~ R F is isomor-

It is enough to show they have the same

Each element of ~ g e n e r a t e s

a cyclic subgroup of ~ .

Thus if K ~ R P is isomorphic to K ~ R F as a K~' module for every cyclic ~ ' C ~ , then they have the same characters. assume~

is cyclic.

Now K ~ i s

semisimple.

Thus, we can

Let SI, o.., S r be

representatives of all the isomorphism classes of simple modules, K~RP

r ni = l~ l~ Si and K ~ R F

n i = m i for all i.

r mi = l~_ ll~ S i.

We only need to show

-

62

-

By Schur's lemma, if S and T are non isomorphic K~modules, then Hom

(S, T) = 0 while Hom

(S, S) # 0.

If

KV r ni M = l~ l~ Si, then Horn

KV

(Si, M ) =

ni II Horn

1

KV

(Si, Si)-

Hence,

dimKHOmK (Si,M) I

.

Thus M is K~isomorphic

to N if and only

dimKHomK~(Si,S i)

if HOmK~ (Si, M ) and HOmK~ (Si, N) have the same dimension over K for all i.

Thus it is enough to show that dimEHornK~Si, K ~ R P )

depends onl 7 on r a n k R P f o r all i.

We define a ~ a c t i o n all g ~ ~, f ~ HomK(S , M).

on HomE(S , N) by g(f)(s) = gf(g-ls) for It is clear that

HOmK~ (S, M) = HomK(S , M ) ~ w i t h respect to this action. S* = HomK(S , K).

Then S * ~

map sending f ~ m

to the function g with g(s) = f(s)m.

K ~ isomorphism. pf~pm~h:

Let p ~ ~.

M is isomorphic to HomK(S , M) by the This is a

Then p(f ~ m) = pf ~ pm and

S -~-N is given by h(s) = p(f)(s)pm = pf(p-ls)pm =

f(p-ls)pm since ~ a c t s

trivially on K.

of p on g = image of f ~ m. f(p-ls)pm.

Let

Next we compute the action

p(g)(s) = pg(p-ls) = p(f(p-ls)m) =

Hence the map is a ~ m a p .

Thus, HOmE~ (S, N) -- (S* ~ M ) ~ a n d

by lemma 4.7 Horn (S, M)

- 63 -

is isomorphic to ( S * ~ M ) / ~ . K~module,

therefore there exists a finitely generated R ~ m o d u l e

T C S* such that K ~ T the R ~ m o d u l e K~.

S* is a finitely generated projective

is isomorphic to S*.

We need only let T be

generated by a finite set of generators for S* over

To compute the rank of H O m K ~ (S, K O R P ) w e

(S*~)K(~RP)/I[

dim K K ~

= (K@

(T ~ e ) / ~

so is T @ P .

T ~K~P)/I[

= (K(~)T~P)/[[.

= rank(T ~ P ) / ~ .

Since P is projective over R~,

Therefore r a n k ( T ~ P ) / ~ =

~ rank T rank P. n from r~nW P.

note that

1 rank(T@

Hence we can compute the dimKHomK~ (Si, P) solely

This completes the proof of theorem 4.2.

Proposition 4.$.

Let R be a Dedekind ring with quotient field K,

a finite group of order n and L an R order of K~. Remark.

If characteristic

Since R ~ i s

P) =

Then n L C

R~.

of K does not divide n, then L C ~IR~""

a noetherian R module, this gives an easy proof that

there exist maximal orders in K~. Proof.

We examine the trace map tr: K ~ - ~ P K

Where x(a) = xa. Lemma 4.10. Proof.

given by tr(x) = tr(x)

We need some information on the image of tr.

If L is an R order, then tr: L -~-R.

It is enough to check this locally s i n c e ~ R

Lp is a free R xw i ~ Lp.

module.

Let (wi} be a basis for Lp.

(w i} is also a basis for K ~ o v e r

write xw i = ~

aijw j with aij ~ K.

implies aii E Rp.

K.

= R.

Then

Then

To compute trace we

Then tr(x) = ~

Thus tr(x) g Rp as desired.

P

aii.

xw i ~ Lp

-

64

-

To complete the proof we compute trace using the s ~ basis.

The coefficient of s in ts is 0 if t ~ 1 for t t ~

efficient of s in ls is 1.

and co-

Thus tr(t) = 0 if t I 1 and tr(1) = n.

Trace is additive, therefore tr( ~

want to show nk i R~.

as a

asS) = na 1.

Let k = ~ a s S

Given k i L we

the na I = tr(k) t R.

If t ~ ,

then t-lk t L, t-lk = ~ a s t - l s , and so tr(t-lk) = na t ~ R for every t ~.

Thus, nk = ~ a t t Let ~ b e

E R~as

desired.

a finite abelian group of order n and R a Dedekind

ring with quotient field K whose characteristic does not divide n. Pick x ~ K ~ w h i c h

is integral over R~.

generated R ~ ( a n d

hence also R) module.

an R order.

Then R ~ E x ]

Therefore, R ~ [ x ]

By the previous proposition, A C

integral closure of R ~ i n

KW.

~R~.

Then, by the above,

hence is finitely generated over R.

is a finitely

Thus,

~

Let ~C

= A

P

is

be the

~R~and

is an R order of K W .

But any order has to be finitely generated and integral over R~. Therefore, of K~.

~

is a maximal order and even the unique maximal order

Since R ~ i s

integral over R, ~

is also the integral clo-

sure of R in K ~ . r K ~ = i = ~ L i where the L i are fields.

Let e i be the unit of L i-

The e i satisfy the equation x 2 - x and hence are integral over R and, thus, in ~ .

1 = ~ e i and eie j = 0 if i ~ j so

-

= ~ i x ... X V r

where V i

65

-

= V e i.

the integral closure of R in L i.

Since P i

The C i

= Li~

C,

C i is

are all Dedekind rings.

Now K ~ - - ~ L i is onto and L i is generated over K by elements the images of the s E ~

in L i.

sn = 1 implies ~ ns = i.

generated over K by its n-th roots of i. temio extension of K.

So L i is

That is L i is a cyclo-

Now [x ~ Lilxn = i) is a cyclic group which

is generated by one root of 1 say ~ . cyclic of order n, then K ~ = is K ( ~ ) where ~

~ s'

Then L : K ( ~ ) .

K[x]/(x n - 1).

If ~ i s

One of the fields L i

is a primitive n-th root of 1.

Our remarks thus

yield the following result which does not involve group rings explicitly. Corollar?f ~.ll. and L : K ( ~ )

Let R be a Dedekind ring, K its quotient field,

with ~

a primitive n-th root of 1.

integral closure of R in L. Remark.

Then n R ' C

We might hope that R' = R [ ~ ].

Corollar~ #.12.

Let R' be the

R[ ~ ]. This is false in general.

With the above notation, if p is a prime ideal of

R that does m~t contain n, then p is unramified in R'. eI er Proof. Let R'p = P1 "'" Pr " We want to show all the e i are 1. r e. Now R'/R'p = ~ R ~ P i l . If we could show R'/R'p is semisimple then 1 we would be done because commutative semisimple rings have no nilpotent elements. Nothing is changed by localizing at p. n is a unit.

Hence we can assume

Then, by corollary #.ll, R' = R[ ~ ] so

-

R'/pR'

=

R[~]/pR[~].

66

-

This is a quotient of R/p~.

characteristic of R/p is prime to I ~ I , R / p ~ i s

Since the

semisimple and thus

so is R'/pR'. Theorem 4.1~.

Let R be a Dedekind ring of characteristic O, K its

quotient field and p ~ z a prime such that l)

p is not a unit in R and

2)

p is unramified in R (i.e., (~) = P1 ..... Pr where the Pi all

distinct). Let

~

be a primitive pn root of 1 (for any n), then

i)

[K(~): K] = ~ ( p n )

2)

the integral closure R' of R in K( I ) is R[ I ] , and

3)

PI' "''' Pr are totally ramified in R', i.e. , R'Pi ~ A~i~(pn).

Corollar2 $.14.

= pn-l(p _ I),

Let R be a Dedekind ring of characteristic O, K

its quotient field, n ~ Z.

If for all primes p ~ Z dividing n we

have l)

p is not a unit in R and

2)

p is unramified in R.

Then if l)

I

is a primitive nth root of l, we have K]

=

2)

the integral closure of R in K( I ) is R[ I ], and

3)

p~R

is ramified if and only if n g p.

Proof of the corollary from the theorem, KCK(r)

CK~)

n = pVm where p2m.

where r is a primitive mth root of i.

on n we can go from K to K(r).

By induction

To get from K(r) to K(~) we only

-

need adjoin since

a primitive

67

p V - t h root

-

of i.

p is not a unit

it is not in R and p does not ramify

since p~m.

in R[rS

This proves

the corollary. Proof

of Theore m

in R.

Hence

4.1~.

ordp(p)

If p ~ P, then p ~ p2 since p is u n r a m i f i e d Let R ' P = ~

= 1 for all PIP.

el

@

1

"''"

er r

r and fl = [ R ' / ~ i : characteristic 3) asserts

R/PS.

Then i~'eif i = KK(z):

K3

O, there

is no inseparability).

(since R has

that r = f = 1 or that R'P =

P

~e

where

e = CK(z):

KI.

~ primative p n - t h root of i

pn(1) = II so p = I

(i

) LR'.

~ primitive pn-th of 1

root

Let z and w = z k be two primitive

pn-th

1 - w = (1 - z k) = (1 - z)(1 + ... + zk-1).

of 1.

Therefore,

1 - wll - z.

Thus

p = V(1-

~ ) as above,

we have p = (unit)

(1 - ~ ) ~ ( P n )

root

= 1 - ~ .

(P) = ( ~ ) _ ( p n ) tain el. ordpx

Examining = ord(~Ix.

of 1. ~l

Hence,

e1

Let

~

(1 - z).

Then

1 - zll - w.

Similarly

a fixed p n - t h p r i m i t i v e

1 - w = (unit)

roots

Since for

Then

above we have for x L R we

ob-

-

68

-

eI ord~ = ord@iP - ~ ( p n ) o ~ @ l ( ~ ) tion.

So e I = ~ ( p n ) o r d @ l ~

But ordp(p) = i by assump-

and e l ~

(pn).

But

r

±=~elf I = [K(z): KS ~ ~ ( p n ) . [K(z): K] = ~ ( p n )

and so R'P = ~ l ~ ( P n )

prime divisor of R'P. ord~i~

Thus r = i, e I = i, and

= I so ( ~ )

where ~ i

This proves (I) and (9). =~i

"''~r where~i

We also see that

is as in (3).

Finally we want to show that R' = R[z]. pnR' C R[z].

By corollary @.ii

Since we have seen that f = I f o r ~ i

R ' / ~ i = R/P i where ~ i

is the unique

over P, we have

is the unique prime over Pi for

i = i, ..., r.

If x • R', there exists Yi ~ R such that

x - Yimod ~ i "

By the Chinese Remainder Theorem on PI' "''' Pr in

R there exists y g R such that y "--Yi mod Pi i = i, ..., r. r

Thus

r

y - x " i=l ~ i

= iTTlei. = < ~ )"

Consider R'/R[~ ] as a R[~ ] module.

Since for any x C R', there is y a R with x - y ~

R' we know that

N

~'IR[ 11 ] = '~ (R'/R[~]) . . . . . for all N ~ i . Thus R ' / R [ ~ ]

But

~%@(pn) = (unit)p so

= pt(R'/R[~]).

p n(R'/R[~]) = R ' / R [ ~ Corollary 4.1~.

~

(RIR['~ ]) ~ ~ ( p n ) t = (unit)p t.

Since pnR' C R[~] and

we have R'/R[~] = 0 and R' = RIll.

Let R be a Dedekind ring of characteristic 0 and

K its quotient field.

Let n ~ Z such that for all primes pln, p is

-

69

-

not a unit in R and is unramified in R. n-th root of i.

Then [K(~):K]

in

is R l.

Proof.

Let n = Pl

eI

e "'" Pr n.

primitive piei-th root of 1.

= ~(n)

Let

S

Let

~

be a primitive

and the integral c l o s u r e ~ R

=

~

i "'"

Adjoin the ~ i

~

r where

i is a

one at a time and

apply Theorem $.13. Let ~ b e

a finite abelian group.

examine the integral closure ~ ,

of Z ~ i n

Now we are going to Q~.

Now Q ~ = ~ A

A i = Q ( ~ i ) is a cyclotonic extension of Q for all i. = ~i'

where ~ i

we project Q ~ o n t o

= Z[ ~

i where

Thus

is the ring of integers of Q ( ~ i ).

Ai, the image of Z ~ i s

~Go(~i). l

If

Then

[M] L Oo(Z]T). Theore ~ 4.16. Proof.

The map G O ( ~ )

G0(Z~) = GoZ(Z~).

free Z~module. GO(~)

-~Go(Z~)

sending [M] to [M], is onto.

Let M be a finitely generated torsion

We must show [M] is in the image of

-~PGo(Z~).

Consider the Q ~ m o d u l e W = Q ~ M . Z

Choose a

composition series W = Wt:~bWt_l~ Let M i = M # ~ W i. M = M t ~ ~t-I ~

torsion free.

... ~ W I ~

WO = 0

for W.

Then .... ~ z = o Mi/Zi_ 1 C. Wi/Wi_ 1 and Mi/Mi_ 1 is

Since [M] = ~ [ M i / M i _ l ] , it will suffice to consider

-

70

-

the modules Mi/Mi, i.e., we can assume W is simple. Q~=

~ A i as above.

Then W will be a module over one Ai, the re-

maining Aj acting trivially. its image in A i. for j ~ i.

Therefore Z ~ a c t s

Regard N as a ~

Then [N] ~ G o ( ~ )

Theorem $.17.

Write

If ~ i s

on N through ~i,

module with ~ j

acting trivially

maps into [M] in Go(Z~).

abelian of order pn with p a prime, then

G O ( ~ ) --I~Go(Z~ ) is a isomorphism. Proof.

pn~

Z~by

corollary 4.11.

Then Z S = Z[~] and Z S ~ =

(Z~) S = ~S"

Let S = (1, p, p 2

p3

...).

Consider the commutative

diagram with exact rows [M]

Qo( /pr")

[M]

o(Z]IVpzg)

ao(Zg)

Go(1"s)

o

ao((z]1") s)

o

where the middle map is onto by theorem 4.16. that G o ( r / p p )

--Ib-Go(~) is the zero map.

= P 1 x ... x P r

where ~ i

It is enough to show Now

is the ring of integers in a oyclo-

tomic field Q ( ~ i ) where ~ i is a primitive pV-th root of 1 with v ~ n.

We want to show that any r module M which is annihilated by

p is zero in G O ( ~ ) . ~/p~-modules,

Since [MS = ~ n i [ M i ]

where the N i are simple

it is enough to show that if S is a simple ~

mod-

ule annihilated by p then [S] = 0 in G O ( ~ ) . If S simple then S is a non trivial ~ i only.

module for one i

Thus it is enough to show that if R = Z [ ~ ] where

E

is a

-

primitive

71

-

pV-th root of 1 and S is a simple R module with pS = 0

then [S] = 0 in Go(R). exact sequence

cyclic

so we have an

0 -a.p -aPR -~-S -~-0 where p ~ P since pS = O.

That is P occurs

in the factorization

By theorem 4.15 P = (1 - 4 ) .

Now S is certainly

(p) is totally ramified.

Therefore

ideals are isomorphic

of (p).

P is a principal

to R since

(P) = p ~ ( p V )

ideal.

R is a domain.

and

But principal Thus

[S] = [R] - [P] = 0 as desired. Example. GO(~)

There exists a finite cyclic group ~ s u c h

-J~Go(Z ~ ) is not an isomorphism.

proof that there are cyclotomic a principal

ideal domain.

fields

that the map

We shall assume without Q(~)

such that Z[ ~]

Note that Z[ ~ ] with

~

is not

a primitive

25rd root of 1 is an example. Lemma #.18.

Let k be an integer,

and p a prime not dividing Proof.

Let/~

~ok(X)

= ~(x-~

k.

be a primitive

a primitive

Then Pl ~ D k ( ~

is a primitive

pkth root of 1 with

mod pk such that

pth root of i, Pl % ( i ) ,

kth root of l,

).

r) where r runs over the classes

where r runs over the classes ~k

~

~

of ~ ( I )

(r, pk) = 1.

and ~ ( i )

occur in % k ( ~ )

Then

mod pk such that

where s runs over the classes mod p with (s, p) = i. can show all the factors

= ~.

Now

= ~(i-~

ks)

Thus if we

we are done.

That is if s is prime to p we must show there is an r prime to pk with r - p = ks.

Consider

all r = p + ks where s is prime to p.

-

72

-

Then (r, pk) = i for let q be a prime such that qlpk and qlr. p = q then Plr implies plks =~ Plk but p ~ k by assumption. q ~ p then qlk and qlr so qlP.

Thus (r, pk) = 1.

If If

Next we claim if

we pick different classes s mcd p which are prime to p, then we get different classes mod pk. pl(s I - s2). pl~pk(~)

If p + ks I

~ p + ks 2

mod pk then

Thus every factor of ~p(1) occurs in ~ p k ~ )

and

as desired.

Returning to the example we pick k such that Z [ ~ ] is not a P.I.D. where

~

non principal

is a primitive kth root of 1.

ideal a which we can choose prime to k.

a = P1 .... Pn with Pi prime. least one is not principal.

x.

Then G O ( ~ )

Call it P.

note

~

1 and above.

= Z[~] ~

Let n = pk and let ~ b e --~Go(Z~) x Z[ ~ ] x

=~P.

Say

Then Pi are all prime to k and at

P must contain a prime p ~ Z. where p ~ k.

Then there exists a

Thus R/P has characteristic p cyclic of order n generated by

is not a monomorphism. ... where ~

To see this,

is a primitive n-th root of

Let M = Z[ ~ ]/P where P is the prime constructed

As a Z ~ m o d u l e

x acts on N as

~ (i.e., xm = S

m).

If

rn~ N, then (~n(~)m = ~ n ( ~ ) m = (algebraic integer)p.m = O. Let N = N as a module over Z V / ( ~ n ( X ) )

= Z[/~] (considered

as a factor in ~ )

this makes sense since ~n(X)M = O.

IN] - [M] ~ G O ( ~ )

and has image 0 in Go(Z~).

P is not principal and [N] I [M] in G O ( ~ ) ferent components. phism.

Therefore,

GO(~)

Then

But [M] ~ O since

since they lie in dif-

-~-G0(Z ~ ) is not an isomor-

-

73

-

We will now compute Ko(Z~) for ~ c y c l i c of prime order p by using a theorem of Reiner which classifies all finitely generated torsion free Z ~- modules.

Later we will give a different proof

using a theorem of Serre. If ~ i s

cyclic of prime order p, then Q ~ =

is a primitive p-th root of I.

Q ~ Q ( ~ ) where

The maximal order ~

where R = Z[~ ] is the ring of integers of Q ( ~ ).

is Z ~ R

As above, every

Z or R-module is a W-module through the projections Z ~ R-4P Z or R and so is also a Z~module through the inclusion Z ~ C

~

(or more

directly, through the projection Z~--~DZ or R). We now single out 3 types of Z~- modules M. Type I Type II

Type III

N = Z (with trivial ~ - a c t i o n ) M =~,

a non-zero ideal of R (Z~acts through

Those M for which there is a non-split extension O -4~~M

~Z

-~0

where A

is of Type II (and

Z of Type I). We have just shown that G O ( Z ~ ) = G O ( ~ ) Z~(Z~Co(R))

= Go(Z)~Go(R)

where Co(R) is the class group of R.

=

If M is a

finitely generated Z~-module, we define the ideal class cl(M) of M to be the component of [M] in the summand Co(R). 0 -~PM' - ~ - N - ~ M '

Clearly

-qPO exact implies cl(M) = cl(M')cl(M") where we

consider Co(R) as a multiplicative group. n c I ( M I ~ . . o ~ M n) = ~cl(Mi). I

Therefore

If M is of type I, cl(M) = O while

-

74

-

if M has type II or III cl(M) is the ideal class of the ideal occurring in the definition of these types. Theorem 4.19

(Reiner).

Let ~ b e

cyclic of prime order p.

every torsion free (over Z) finitely generated Z ~ - m o d u l e . direct sum M = M I Q . . .

Then M is a

Q M n where each M i is of type I, II, or III.

Let rl, r2, r 3 be the number of i with M i of type I, II, or III respectively.

Then rl, r2, r 3 depend only on M (and not of the

choice of the decomposition M = M I ~ those numbers.

Write ri(M) for

Then M is determined up to isomorphism by

rl(M) , r2(M) , r3(M) , and cl(M). ri E

... Q M n ) .

Any set (rl, r2, r3, k),

Z, k ~ Co(R) is realized by a module M of the above type pro-

vided r i ~

0 for i = l, 2, 5 and provided that k is trivial if

r 2 = r 3 = O.

Finally M is projective if and only if

rl( ) = r2(N)

= o.

Corollar~ 4.20.

If ~ i s

cyclic of prime order p, every finitely

generated projectiv e module P over Z ~ i s type III.

a direct sum of modules of

Furthermore P is determined up to isomorphism by

rkP = pr3(P) and cl(P).

If n 6

Z, n ~ 0 ,

pln and if k ~ C0(R) ,

there is a finitely generated projective module P over Z ~ w i t h rkP = n, cl(P) = k . Corollar~ 4.2~.

If ~ i s

K0(Z~) = Z ~ C 0 ( R ) ~:

, G0(Z~) = Z ~

K0(Z~)-P-G0(Z~)is

is a monomorphism.

cyclic of prime order p, Z~C0(R

) and the Cartan map

given by (n, k) ~-~-(n, n, k).

Therefore

-

Proof.

Glearly~(KPS)

P of type iIIo ~KP]

= ~KQ]

75

-

= (r3(P) , r3(P) , k) since this is true for

If x E

ker~

write x = [PS - KQ].

so r3(P) = r3(Q) , cl(P) ~ cl(Q).

Since

ri(P) = O = ri(Q) for i = l, 2, the theorem shows P ~ Since im ~

Then

Q so x = O.

is a group and contains all (n, n, k) for n ~ O,

k E Co(R ) we see that Ko(Z~)=

((n, n, k) I n E

C o r o l l a ~ 4.22.

Z, k E

C0(R)}~

Z(~)C0(R).

If ~ is cyclic of prime order and P, Q are

finitely generated projective Z ~ - m o d u l e s with [P] = [Q] in Ko(Z~) , then P ~ Q. Z~-module

and P, Q are finitely generated torsion free Z ~ -

modules, then X ~ P ~ Proof.

More generally if X is any finitely generated

X~)Q

Q.

If T is the torsion submodule of X and Y = X/T, factoring

out torsion shows Y ~ P ~ cl(Y@P) Remark.

implies P ~

Y~Q.

Since ri(Y ~ P) = ri(Y) + ri(P) ,

~ cl(Y)cl(P) we see that ri(P) = ri(Q) , cl(P) = cl(Q). This result does not extend to all finite groups ~ .

fact if ~ i s

the generalized quoternion group of order 32, there

is a projective Z ~

P

In

module P such that P ~

Z~

but

zV. The proof of Reiner's theorem wi&l be based on the following

well known lemma. Lemma 4.2~.

Let R be a Dedekind ring, y a prime ideal of R and A

a finitely generated torsion free R-module. morphism of A/yA with det @~

= l, then

@~

If @~

is an auto-

lifts to an

-

76

-

automorphism @ of A i.e., @ is such that A

@~

A/yA Proof.

A

A/yA

Let A = A l ~ . . .

~ A n with A i of rank 1.

A/yA = A1/YAl( ~ . . . ( ~ A n / Y A n. be a generator.

commutes.

Now A i / Y A i ~

R/y.

Then Let v i E Ai/YA i

Then Vl, ..., v n is a base for A/yA.

With respect

to this base, we can identify Aut(A/yA) = GLn(R/y) and we have ~

SLn(R/y).

Since R/y is a field, SLn(R/y) is generated by

elementary matrices in those of the form 1 + teij where eij has 1 in position ij and other entries O. elementary.

If each ~ i

let @ = @l "'" @r" automorphisms

g(vi)

tvj.

Let

~=

~l

°'" ~ r

with ~ i

lifts to an automorphism @i of A we can

Thus it is sufficient to lift all elementary

= 1 + teij.

Let g: Ai/YA i ~ A j / y A j

by

Since A i is projective we can lift g to f: A i ~ A j

so

ii ij Ai/YA i

~-Aj/yAj g

commutes.

Let ~

be the endomorphism of A = ~ A k sending A i to Aj

-

77

-

by f and sending all other A k to O. (I

-

~)(I

+T)

= (I + ~ ) ( I

-~)

Since i ~ j,

= i.

~ =

0 so

Thus @ = 1 + ~

is an

automorphism which clearly lifts ~ . Corollary

4.2@.

Let A = A I ~

Let R be a Dedekind

... ~ ) A n be an R-module

ated and torsion

free of rank i.

module with S i ~

R/y for all i.

then m ~ n

and there

Factor

~

m and

A/yA w i t h ~

kernel

.

base Let

~:

~(@Ai)

Then

(vi) = s i.

@ of A such that

= 0 for i > m . Let s i generate

S i-

Let (wj) be a base for the

{vi, wj) is a base for A/yA.

We can get a new

{ui) for A/yA by choosing one generator u i for each Ai/YA i~

be the automorphism

CVl, V2,

..., Vm, Wl,

det

~-

A.

Now ~ ( @ A i )

= 1.

of A/yA sending Ul,

= @~

(A i) =

@~

to an automorphism

@ of

~ ( A i / Y A i) = R/yv i for i4. m, and

This @ clearly has the required properties.

We now turn to the proof of Reiner's Let N = ~ , ~

..., u n to

..., W h where we choose c 6 R/y - (0) so that

By the lemma we can lift

R/Ywi_ m for i • m.

~.

... ~ ) S m be an R/y-

A -~-S is an epimorphism,

into A ~-a~A/yA ~ - ~ S .

Choose v i ~ of~

Let S = S I G If

ideal of R.

with each A i finitely gener-

is an automorphism

~ ( @ A i) ~ S i for i ~ Proof.

ring and y a prime

theorem.

= 1 + x + ... + x p-1 = ~p(X) 6 Z~.

Z~/(N) = R and Z ~ / I = Z where

Let x generate Then

I = (x - l) is the augmentation

ideal.

-

Let M ~ (x

(m~

78

-

Mlxm = m) and M N ~ ( m ~

- 1)N -- 0, we have

(x - 1)M = I M C

M I N m = 0). M N and N M ~ M

A = M/M N is a module over Z ~ / I ~ Z and B ~ M/M ~ I s Z ~ / ( N ) = R.

Since ~.

Therefore

a module over

These modules are torsion free (over Z) since if

h • Z, h # O, m E M, and hm ~ M ~ t h e n (x - 1)m -- 0 i.e., m ~

M ~.

(x - l)hm = 0 and so

The same argument works for M/M N.

Con-

sequently B is also torsion free over R.

In fact if r G

then r divides

Thus if rb = O, b G B

its norm n G Z and n # 0.

R, r # 0

then nb = 0 so b = 0. Now let C = M / ( M ~ + J = (N) + I.

MN).

This is a module over Z ~ / J

Since N maps to p G Z and I maps to y = ( ~

we have Z/pZ • Z ~ / J ~

R/y.

We also note that M N ~

where

- l) in R

M ~ = 0 because

if m is an element of this we have xm = m so 0 = Nm -- pm but M is torsion free. Consider the diagram

n-~A B ~ C s

where the maps are the canonical cocartesian M ~

A

~

B

(a pushout). ~

C

quotient maps.

T h i s is equiyalent ~

0.

to the exactness of

Since ker(p,

this sequence is also exact on the left.

This is clearly

q) = M N ~ M

In other words

~

~, is

- 79 -

Thus we can recover M as the pullback

also cartesian.

of the

diagram A

(.)

~r

B~C Choose decompositions C = C l@ B2 ~

A ~ AI~...~A

... @ C c where C i ~

... ~ B b ~

Z~/J

R and B l £ ~ o r

a where A i ~ and B = B l ~

a non-zero

corollary 4.24 to s (using Z ~ / J ~

iS

Replacing

c and s(Bi)

Applying

R/y) shows there is an automors(@B i) = O for

the B i by the @Bi, we can assume s(B i) = C i if

= O for i ~ c .

Similarly,

can alter the A i so r(A i) = C i for i ~ Therefore

... @ B b where

ideal of R.

phism @ of B such that s(@B i) = C i for i ~ c , i ~ c.

Z,

the diagram

using Z ~ / J ~

Z/pZ, we

c and r(A i) = 0 for i ~ c.

* is the direct

sum of diagrams

of the

form Ai (i)

~ 0 ~

0 , (2)

0

Bi ~

and so M is the direct pullbacks Consider

~,and

Ai (,)

0

sum of the pullbacks

$ Bi ~

CI

of such diagrams.

The

of (1) and (2) are A i and B i which are of Types I and II. a diagram

of type

(3)

-

80

P--

-

z z/pz

where P is the pullback. and q.

Since r and s are epimorphisms

Also ker p = ker s = b say and ker q = k e r r

fore we have exact sequences O - - ~ b 0 -~-Z -~-P ~

~

split epimorphism II.

-a-O.

-~-P ~ Z

so are p

Z Z.

There-

-aDO,

Therefore P is of type IIl unless p is a

in which case P ~

b~)Z

is a sum of types I and

This proves the first assertion of the theorem.

We remark

that, in fact, p csnnot be a split epimorphism.

If h: Z -~-P

splits p then h ( Z ) C

~

P~[ so q h ( Z ) ~

~ t ~[.

But ~

= 0 since

aC~][implies

(x - l)a = O, Na = O, so pa = 0 but X ~

is torsion

free so a = O.

Thus qh = 0 so r = rph = sqh = O, an obvious

contradiction. The proof of uniqueness depends on the following lemma.

If

R is a ring, U(R) denotes the group of units of R. L e m m a 4.20. Proof.

U(R) ~ U ( R / y )

R/y ~ Z ~ / J ~ Z/pZ.

is onto. If r is an integer prime to p, so that

r rood p is a unit of Z/pZ, then 1 + x + ... + x r - l c r mod p.

Its image in R is

~

= 1 + ~ + ... + E r - 1

into the element of R/y corresponding (1 - ~ ) ~

= 1 - ~ r.

to r.

Z~

maps into

and ~

maps

Now

Since ~ r is also a primitive p-th root of

i, (i - ~ r ) I(l - ~ ) so ~

is a unit of R.

-

Lemma 4.26.

Let

P ~

Z

q~

of ~

-

be cartesian with r and s epimor-

~r s

phisms.

81

z/pz

Then the isomorphism class of P depends only on the class (and not on r and s).

Furthermore P is a projective

Z~-

module. Proof.

We may take r(1) as the generator of Z/pZ so r is the

canonical

quotient map.

~ ~ / y ~

The map s factors as

R/y ~ Z / p Z .

Two different maps s, s' will

correspond to two different @, @'. R/y, @' = u@. ping onto u.

By the previous

These differ by a unit u of

lemma, there is a unit ~

Therefore @' = @ ~ .

of R map-

This gives an isomorphism

of

diagrams

Z/pZ

!d =~ Z/pZ

~---,~,.., Since isomorphic diagrams have isomorphic pullbacks,

this proves

the first statement. If we perform the construction on M = Z ~

we see that the diagram

at the beginning of the proof

-

82

-

z]F~ R ~

is cartesian.

z

Z/pZ

Given any ideal class we can choose an ideal ~

this class such that R / ~ @

has order n prime to p.

maps onto Z/pZ under the map R -~-Z/pZ.

in

Therefore

This gives us a subdiagram

Z

V with s onto. nRC~,

z/pz

Its pullback P is a submodule of Z~.

we have n Z ~ C

P is projective.

P.

Since

Since p ~ n, Rim's lemma implies that

By the first part we get all possible isomorphism

classes of P's in this way. We can now show that N is determined by the rl, r2, r3, and clM for the decomposition obtained above. i ~ 1.

R for

Therefore, the modules obtained will all be isomorphic to

Z, R, or Z ~

except for the first one.

with ideal ~ ~ ~=

We choose B i ~

Bl°

This will be of type III

= B 1 unless r 3 = 0 when it will be of type II and Therefore the modules occurring are determined by

rl, r2, r 3 and the class o f ~

which is clearly cl(M).

We must now show that rl, r2, r3, and clM are uniquely determined by M.

Now Go(Z~) ~

Z QZQCo(R

) and under this isomor-

phism, [M] corresponds to (r I + r3, r 2 + r3, cl(Z)).

Consider

-

H@(~[, N) = M ~ / N "M.

83

-

If M is of type I we see that H @ ( ~ , M) = Z/pZ.

If M is of type II, M ~ =

0 so H @ ( ~ , M) = O.

of type III by Lemma 4 . 7

The same is true for M

since such M are projective.

r I is determined by ~ o ( ~ , N).

Therefore

Since r I + r3, r 2 + r 3 are deter-

mined by N so are rl, r2, and r 3. We have seen above that there is a module P of type III w i t h any given ideal class.

Therefore we can realize any rl, r2, r 3 ~ O

and class k provided k = 0 if r 2 = 0 = r 3. Finally we have seen that all P of type III are projective while Z is not.

If~5

III with class ~rt..

is a non-zero

We have shown above that there is an exact

sequence 0 -~-Z - ~ - P - ~ - A ~

-~-0.

this sequence would split.

Therefore

If~M.

of P and hence would be projective. jective.

ideal of R, let P be of type

were projective

Z would be a direct s!,mmand Therefore ~

cannot be pro-

This completes the proof of Reiner's theorem.

Chapter ~.

Maximal Orders

Definition.

Let R be a Dedekind ring, K its quotient

a semisimple

separable K algebra.

ring ~

Remark.

over Z~[,

field, and A

Then an R order of A is a sub-

of A which is a finitely generated R module such that

Let N be a finitely generated torsion free R module such

that KM ~ A.

Let

~

= (x ~ A I x M C M ) .

Then A

is a finitely

-

84-

generated R module since the map ~ E~

-~PHOmR(M , M) given by sending

to f ~ HomR(M , M) where f(x) = ~x.

Embeds

HomR(M , M) which is finitely generated over R. nitely generated,

Clearly ~

r # 0 in R such that r a M C fore K ~

= A.

Therefore

Proposition 5.1.

A is semisimple.

rings.

Let A

Hence, ~

A

If a ~ A we can find an There-

is an order called the left order of M.

Say A = A 1 x ,,. x A n with the A i simple

be an R order of A.

is an R order of Aio

Let A i

Then A C

be the image of A

under

A 1 x .., x A n and each A i

Hence A 1 x ,o. x A n is an R order of A,

by A 1 x ... x ~ n o is maximal for all i,

~

integral closure of R in Z.

Hence we can assume A is simple.

Then R ' A

Hence we can replace ~

We

is maximal if and only if

Let Z be the center of A (which is separable over K),

module.

is fi-

M (since M is finitely generated.

the projection of A to A i.

~iCAi

in

Any order can be embedded in a maximal one.

Proof.

replace A

is a ring.

A

by R ' ~

Let R' be the

is a finitely generated R and assume K is the center

of A. Before we finish the proof we need a few lemmas. Lemma 5.2.

Let A be a central simple K algebra and L a field con-

taining K, then L ~ K A Proof.

is a central simple L algebra.

Let e i be a base of L over K.

L~KZ(A)

clear (where Z(B) denotes the center of B). ~e i~a (~e i~a

i a Z(L~KA). i).

Let a a A .

Hence ~ e i ~ a a i

C

is

Let

Then ( l ~ a ) ( ~ e

= ~e i~aia.

Z(L~KA)

i~a

i) =

But e i is a base.

-

85

-

Therefore, aa i ~ aia for all a L A. ~e i~

Hence a i • Z(A) and

ai • L@KZ(A). Next we show L ~ K A

is simple.

A is simple.

Hence

A ~ Mn(D) , the n by n matrices over a division ring D. to show L ~ K D simple.

It is enough

is simple since a matrix ring over a simple ring is

Let 0 # B C L ~ K D

be a proper 2 sided ideal.

Pick a non zero x in B such that the representation x =~ s e i ~ d i ± .

has the smallest number of non zero di's where e i

is a base as above. is in B. (l~d)

Say dj # O.

Then ( l @ d j l ) x

Hence we can assume dj ~ 1.

= ~ ei ~ d ~ l d i ia S Pick any d # O in D. Then

x (l~d

-1) ~ ~ e i ~ d d i d - 1 is in B and has jth term i~ S Thus ( l ~ d ) x ( l ~ d -1) - x = 0 since it has fewer non

ej~l.

zero terms.

Therefore d i ~ Z(D) = K for all i and x ~ L ~ K Z ( D )

Thus x is a unit contradicting the properness of B.

= L.

Hence L ~ K A

is simple.

Done with Lemma 5.2.

Lemma ~.~.

If F is a separably closed field, D a division ring

with center F, and D finite dimensional over F, then D = F. Proof.

Suppose not.

Pick x £ D - F.

inseparable field extension of F.

Then F(x) is a proper purely

Say IF(x): F] = pn where

p = char(F).

[D: F] ~ [D: F(x)]KF(x).F].

xpn

~

F.

Let F be the algebraic closure of F.

Plm since pl[D: F]. Apn~

Hence pl[D: F].

Let l ~ x

F.I since ( l ~ x ) pn • F.

correspond to A E

~

F~FD Mm(F).

Also

~ Mm(F) and Now

Hence A pn ~ rI = spnI for some

-

s~

F.

86

-

(A - sl)P n = A p n - s p n I = 0 and A - sI is n i l p o t e n t .

Hence

Thus tr(A - el) = 0 and so tr(A)

~ ~r(1)

T h e s e A g e n e r a t e M m ( F ) as an F module. has trace O.

Definition.

This

The c h a r a c t e r i s t i c

polynomial

teristic p o l y n o m i a l

= Mn(E) by l e m m a 5.3.

the charac-

polynomials. closure

of F, t h e n

n N o w M n ( E ) : l ~ S w h e r e S is a row of

polynomial

polynomial

of Mn(E) w i t h E ~ F A .

into t h e m s e l v e s .

H e n c e if

of a, t h e n f = gn w h e r e g is the

of alS for a r o w S.

F i x an i s o m o r p h i s m

We will show that the c o e f f i c i e n t s

The c o e f f i c i e n t s

p h i s m of E.

over F and a ~ A.

it is c l e a r that a and l ~ a

= x a sends these rows

f is the c h a r a c t e r i s t i c

in F.

5.3.

of a is, by d e f i n i t i o n ,

G i v e n A let E be the s e p a r a b l e

characteristic

Done w i t h lemma simple a l g e b r a

is field extension,

l e m g t h n and a(x)

in Mm(F~

of a: A - ~ - A g i v e n by x ~ - ~ . x a .

have the same c h a r a c t e r i s t i c

E~FA

Thus e v e r y t h i n g

is a c o n t r a d i c t i o n .

Let A be a central

If E ~ F

= sm z 0 since plm.

of f c e r t a i n l y are.

of g are

Let s be an F a u t o m o r -

S i n c e f = gn and f = fs ~ (gS)n we see that gn ~ (gS)n

and so gS = (unit)g.

All the p o l y n o m i a l s

unit

Thus the c o e f f i c i e n t s

~ 1 and g = gS.

are monic.

H e n c e the

of g are fixed b y s and

hence are in F. Definition.

W i t h the above n o t a t i o n g is the r e d u c e d c h a r a c t e r i s -

tic p o l 2 n o m i a l

of a.

If g = X TM - t ( a ) x m-1 + ... + (-1)mn(a)

then

-

87

-

t(a) is the reduced trace of a and n(a) is the reduced norm of a. Clearly t is F linear and n is multiplicative. Definition.

Let Wl, ..., w n be a~ F base for A.

Then

A ( w I, ..., w n) = It(wlwj)l.

If W~ = ~ a i j w j is another base, then It(w!w~)l i 0 = laij I It(wj k) II ak~l" ~(Wl,

Hence if one base has

..., w n) ~ O, then every base does.

Pick the base, eli , of matrix units. ~

t(eijek~) =

j ~ k or i ~ ~ j = k and i s

Lemma ~.~.

w=)

Thus (t(eijek~)) is a permu-

.

Proof. Rp~%

~p

be an R order of A and a ~ A ,

equation of a has all coefficients

R = ~Rp =

discrete valuation ring.

characteristic

Then

in R.

~

is a free R module with base

equation, f, of a is in R[x]. equation.

g has coefficients

cient of g with ordpbm~. O,

in R.

Clearly

Hence we can assume R is a

Now wia = ~ a i j w j with aij L R.

the reduced characteristic Therefore,

then the reduced

where p runs over all the primes of R.

is an order of A over Rp.

Wl, ..°, w n.

Thus

o.

Let A

characteristic

~ Nn(E).

Then

tatlon matrix and has a non-zero determinant. I, ...,

But E ~ F A

Therefore, the But f = gn where g is

R is integrally closed. (If b m is the highest coeffi-

then the coefficient of x nm in f has

-

ordp

~ O.

88

-

This is impossible).

Done with lemma 5.@.

Now we complete the proof of proposition 5.1. Pick a K base for A in w~ = ~ a i j w j with alj ~ K. ~akjt(wiwj).

But

A,

Wl, ..., w n.

Let

Then t(wiw ~) = t ( ~ a k j w i w j) =

~ (Wl, ..., w n) ~ O.

Hence the matrix

(t(wiwj)) is nonsingular so we can pick aij such that t(wiw ~) is any given matrix over K. then x = ~ t ( w i x ) w ~.

Pick these so that t(wiw~) = ~ i j "

If x is in any order containing

hence containing the wi) , t(wix) ~ R by Lemma 5.@. containing A R module. modules.

is contained i n ~ w ~

But ~ R w ~

A

(and

Hence any order

which is a finitely generated

has the ascending chain condition on R sub-

Thus we can find a maximal order containing A

desired. Definition.

x ~ Aj

as

Done with proposition 5.1. Let A be a central simple K algebra where K is the

quotient field of a Dedekind ring R and let ~

be an order in A.

By a (fractional) ideal of A we mean a finitely generated A submodule N of A with KN = A. fine M-I = (x t A I N x M C M ) . (Y E A I y N C N )

Note that M -1 = (x ~ AIMx C ~ )

is an order containing

and so N x M C N ~ = Theorem 5.~.

It is called integral if I C ~ .

MxC

A

A

and hence equals

since ~,

.

For every l~ft ideal Z of a maximal order

have NM -1 = ~ .

De-

~,

we

-

Proof.

If

~

A

89

-

and x ~ M -I then Nx ~ (

x ~ L M -1 and so M -1 is a right ideal of zero two sided ideal. Definition.

If

A

~

A

~

C

A.

Hence

.

Hence ME -1 is a non-

To finish the proof we need several lemmas.

is an R order of A then a 2 sided ideal P C

A

is called prime if for any 2 sided ideals S, T with ST C P we have sC

P or T C P. Let A

Note that we only consider ideals with KP = A.

= A/P

then in A ,

a_~b = 0 implies a = 0 or b = 0

where a and b are 2 sided ideals of ~ .

A

over R since K ~ R A

has D.C.C.

of A

is O.

= 0.

Therefore, ~

is a torsion module

In fact, the radical N is a 2 sided ideal.

N m = 0 we have N = 0 by the above property of semisimple.

The radical of

If

A

=

which is impossible. sided ideal of A

1 x ... x Hence A

A

.

Since

Hence, A

n with n ~ l, then A

is simple.

is

2 = 0

Thus P is a maximal 2-

if and only if P is prime,

i.e., P is prime if

and only if L/P is simple. Lemma ~.6. ideals of Proof.

Any 2 sided ideal M C A ~

.

Suppose not.

ated over R.

contains a product of prime

A

is noetherian

since it is finitely gener-

Hence there is an ideal M C ~

counterexample.

Clearly M cannot be prime.

and T 2 sided ideals of A

such that ST C M .

which is a maximal Hence,

Now (S + M)(T + M ) C M

so we can assume that S and T properly contain M. contain product of primes. Lemma ~-7STCA

A

•

But then S and T

Thus, ST C M does also.

Let S and T be (fractional)

, then T s C

there exist S

2 sided ideals of A

Done. with

-

Proof.

(TS)T = T ( S T ) C

of T.

This

i)~

90

T A = T.

and so is A

-

Therefore, T S C

since ~

the left order

is maximal.

Hence TS C

A.

Done. Lemma 5.8.

Let M C A

be a proper 2 sided ideal of ~ ,

then

M-Ix A. Proof. S

=

R

We first note that KM ~ A implies M f'IR # (0}. -

{O}.

Then ( M g ~ R ) S = M s ~ R

S ~ A~K

Let

~ K.

Now M is contained in P, a maximal 2 sided ideal by Zorn's lemma (or since ~ O ~ u • PP~R. r is minimal.

is noetherian).We know P ~ R

Let P1 .... P r C U A C P

This exists by lemma 5.6.

Pi"

A

Since P is prime, it

since u is central. }.

~

where u-lB and PC are 2 sided ideals

Therefore, Hence

M -I = ( x l M x ~ A ) I ~ P cu-lB C P - 1 C

Changing notation

where B and C are the products of the remaining

Hence ( u - l B ) ( P C ) C A

p-1 = ( x l P x C A

Pick

where the Pi are prime and

contains some Pi and, hence, is equal to it. we have B P C ~ u

# (0}.

-I. .

(PC)(u-IB)C~

cu-lBCp

-1.

Thus, if M -I =

But u is central.

duct of fewer than r primes.

by lemma 5.7.

But

But A

, then

P-IcA

Thus C B C u ~

so

and is a pro-

This contradicts the choice of r. Done.

Now we finish the proof of Theorem 5.5. Let B = MN -1. Since ~ 3

Then B is a 2 sided

BB -1 = MM-1B -1 we have M - 1 B - l ~

right order of M -1 ~ ~ B =~

since

by the previous lemma.

~

~

ideal contained in M -1.

Therefore, B-1C

is a maximal order.

Therefore, Done.

-

Corollary ~.lO.

Let A

91

-

be a maximal order and M a 2 sided

ideal, then M is projective as a left or right

A

module.

A(This

will be considerably improved in Theorem 5.12.) Proof.

We have N-1M ~ ~

since M is a right ideal by Theorem 5.5

(applied to the opposite ring of ~

i.e., A

with x o y = yx).

But the definition of M -1 is independent of left or right. M-1M ~ ~

for 2 sided ideals.

Hence

Pick m i E M and n i ~ M -1 such that

r ~ , him i = 1. i=l

We map A ~ A

i(x) ~ (xnl, ..., Xnr) and

~ ... e r times A

~

e ... ~ ) A

by

-~-~M by r

J(kl, ..., k r) = klm I + ... + krm r. r x ( ~ nim i) = x. i~l

Then ji(x) ~

~ shim i i~l

Hence M is projective as a left module since it is

a direct summand of

A

~

... ~

~

•

Similarly N is projective as

a right module.

Done.

Theorem 5.11.

Let

A

be a maximal order in a central simple alge-

bra A over K.

Then the 2 sided (fractional) ideals form a free

abellan group with generators the prime ideals of A Proof.

O)

plication.

The 2 sided ideals clearly form a monoid under multiThe unit is

ME -1 = M-IM = ~ l)

•

~.

This monoid is a group since

for all M in it.

Let M C ~

be a 2 sided ideal.

is the product of primes of

~.

We first show that M

Suppose not.

Let M be a maximal

-

counter example. Also M ~ p - 1 M

Now M ~ P

for some prime P so P - I M ~

= pp-1 ~ A

the maximality of M.

.

~

p-ip = A

•

= MM-1 = p-1NN-1 = p-1

Thus P-1M is a product of primes by

But N = P(p-1N).

If N is a fractional

primes and their inverses. uMC~

-

since M = P-1M would imply

and so P ~ P A

2)

92

ideal, then N is the product of

In fact, there exists u ~ R

since M is finitely generated.

proper 2 sided ideal.

Hence, B = P1

Let B = uM. Pr' u ~

.o.•

u ~ O with

Then B is a

~ P~ . . •.

P'S

and M ° p -I .... p -iPl .... Pr" 3) P' 3 P ' P P'PC

The group is abelian.

= p(p-ip,p)

Let P and P' be primes.

so either P G P '

P so p - i p , p ~ p - i p

~

)

or P - I P ' P C P ' .

If P ~ P '

then P

(Note that P' since

primes are maximal 2 sided ideals and so P and P' commute. P M P', then P - I P ' P C . P ' P'P = PP'. 4)

so P ' P ~ P P ' .

If

By symmetry P P ' C P ' P

i.e.,

Thus all generators commute. The group is free abelian.

Ul pUrp '-vl P '-vs = ~ PI "'" r i "'" s

prime we have Pi C

Let

be a relation with all ui, v j ~ O .

uI ur vI p,Vs Then P1 .... Pr = P1 "'" s C.P~.

Therefore,

P~ and so Pj = P~ for some i.

uI ui-1 p~-l gives P1 "'" Pi .... ~

Then

,Vl-1 v 2 P1 P2 ....

u i we see that all relations are trivial•

since P is

Multiplying by By induction on Done.

-

Theorem 5.12.

Let

A

93

-

be a maximal order in a central simple K

algebra and M be a finitely generated torsion free left

A

module.

Then M is projective. Proof.

Consider K ~ R M

as an A module.

A module is projective.

Therefore

Since A is simple,

(K~RM)~(K~R

every

N) = A m for

some integer m and some finitely generated torsion free module N.

If M ~ N

is projective,

for M, we can assume K ~ R M

isomorphic

to N.

Hence, using M ~ N

is free and even that K ~ R M

i.e., M C A m and KM = A n . an r # 0 in R with r N C

then M is.

~ Am ,

Since M is finitely generated,

A m C A m.

there is

Since M torsion free, rE is

Hence we can assume that M ~

KN = K A m = A m .

Therefore,

~m/M

Thus, considering

the resolution 0 - ~ M

A m and

is a torsion module over R. -~D # ~ m _ ~ D ~

m/M -~PO,

it

is enough to show the following lemma. Lemma ~-l~.

If B is an A

module which is finitely generated and

torsion over R, then the projective dim over ~ Proof.

B has a composition

series, B = B o ~ B I ~

Consider 0 -qPB 1 -~DB -~-B/B 1 -q~O. pd A

B/Bl~- l, then pd A

assume B is a simple generated,

~

B~d-1. module.

If pd A

Bl~

Hence, r A

...~B

1.

t = O.

1 and

Thus, by induction on t we can Since B is torsion,

there is some r # 0 in R with rB = O.

2 sided ideal.

of B is ~

and finitely

Now r A

vI v ~ P1 ... Ps s with Pi primes.

is a By the

simplicity of B, if I is a 2 sided ideal either IB = 0 or IB = B. Since

(r~)B

= O, there must exist a prime P ~ Pi with PB = O.

Therefore, B = B/PB and B is an

~/Pmodule.

so B is a direct s~mmand of a free enough to show that pd A O--~P-~P

A

-4~ A / P - ~

But

A/P-module.

( A/P)..~I.

A/P

is simple

Hence, it is

But

0 is exact and P is projective by

corollary 5.10. CorollaxT ~.i@. algebra A.

Let ~

be a maximal order in a central simple K

Then every finitely generated A

pd A M ~ - I .

Proof.

Done.

Thus A

-module M has

is left (and right) regular so

Let M be a finitely generated module.

O -~P-~-F-@pM-~-O Theorem 5.12. Problem:

with F free.

Let

Then P is projective by

The last part follows from Theorem i.I.

Is corollary 5.1@ true for non finitely generated modules?

Theorem ~.i~.

Let A be a central simple K algebra with ~

maximal orders.

Then the categories of left ~

and left ~

and modules

are equivalent and this equivalence restricts to one between the subcategories of finitely generated modules.

A ~

module is tor-

sion free (resp. torsion) if and only if the corresponding ~

-

module is. Proof. B =

This is again a special case of the Morita theorems.

~.

a left ~

This is a finitely generated R submodule of A which is and right ~

and right ~ C ~

H.

Let

ideal.

ideal.

Let C = B -I.

If M is a left ~

If N is a left ~

Then C is a left

module send it to

module send it to B ~ .

Then the

-

95

composite sends M to B ~ C ~ A to ~ @ ~

M = M.

an isomorphism.

M which we claim is isomorphic to

We only need to show that B ~ C The mapping clearly is onto.

we see that the kernel of B ~ projective right ~ sion free. B~Y

C ~

A

C)~

Therefore, B ~ is O.

~

is torsion free.

Similarly C ~

Now B is a

M.

~

= C s is tor-

Hence the kernel of

B is isomorphic to ~ . The functors C ~

give the required equivalence.

(resp torsion) so is C ~ r~

-

module Y such that

(Y(~ C) = ( B ~ Y ) ~

the composition is also the identity. B ~ ~

Applying K ~ R

is

module since it is finitely generated and tor-

Then ( B ~

B~C -~PBC

-~-BC = ~

is torsion.

Thus there exists a right ~

= ~s.

sion free.

-

Hence,

~,

If M is finitely generated

If M is torsion free, then for all

R, r ~ O we have 0 -~-M r--~-M. Since C is projective over ~

we have 0 - ~ C ~ Theorem ~.16.

~M -~C~ Let p C R

a unique prime P of ~

~M

so C ~

~M

is torsion free.

be a prime ideal # O. such that P ~ p ~

,

.

Then there exists

As always ~

is a

maximal order in a central simple K-algebra. Proof.

Let ~

R = lim R/pR. ated R-module.

= lim~/pn~gthe completion of ~ Then R ~ R ~ If p ~

= ~

since

at p and let

is a finitely gener-

vI v = P1 ... Ps s, then ( p ~ ) n

nv I nv s = P1 "'" Ps

-

since the primes commute.

96

-

/pnA

Thus,

= "~" ~ / P nvi b y the i--1

Chinese R e m a i n d e r Theorem w h i c h holds since 2 sided ideals commute. nv i Let el(n) be the identity of ~ e i = (... - ~ e i ( n + l )

-~ei(n)

/Pi

.

-*- . . . ) £ m

Then and is an idempotent.

S

• hus

A

-- U

Let K be the quotient field of R.

A e i-

Then

i=l S

K ~

A

A

=i=ll~K ~ R ^ ( A e i )

lemma 5.2.

= K~KA

is central simple over K by

Thus s = 1 and there is only one prime over each prime

of R.

Done.

P r o p o s i t i o n 5.17.

Let R be a (not n e c e s s a r i l y commutative ring)

and let A be a nilpotent 2 sided ideal, then if e g R/A is an idempotent, there exists an idempotent e' ~ R w i t h image e in R/A. Proof.

By induction on n with A n = 0, we can assume A 2 = 0 i.e.,

lift e to R/A 2 and from there to R. e and let a = x 2 - x.

Let x g R be any preimage of

Then a ~ A since e is idempotent.

y = (x - a) 2 = x 2 - 2ax = x + a - 2ax.

Let

Then y is the desired ele-

ment since y2 = x 2 + 2x(a - 2ax) = x + a + 2ax - ~ a ( x + a) = x + a - 2ax = y. C o r o l l a r y 5.18. p and

A

Let R be a complete local ring w i t h maximal ideal

an R algebra w h i c h is finitely g e n e r a t e d as an R module.

If e is an idempotent of mapping onto it.

~/p

~,

there is an idempotent e' of A

-

Proof.

~

is isomorphic to lim

But p A / p n A

97

-

A/pnA

is a nilpotent ideal of

since R is complete.

~ /pn A

5.17 we can lift e to an element en in each liftings are compatible. Corollary ~.l 9. p and /~

~/pn~

morphism mod P A

Then A

and let x be its image in

A .

since

is local.

A /P ~ •

Then x is

A/p~

Hence x is a unit if

Next let J be the Jacobson radical

is finite dimensional over R/p, J is

Hence ~ is a unit if and only if its image ~ ~ A / J

By assumption ~

has no idempotents except 0 and 1.

Hence,

~/J

has no idempotents except 0 and 1 by corollary 5.18.

~/J

is a division ring, and by the above results, every element

of

~

of

~.

- J is a unit of A

Theorem 5.20.

•

Let R be a complete local ring and

~

an R algebra.

modules finitely

generated over R. By SK, Theorem 2.11

Thus

Thus J is the unique maximal ideal

Then the Krull-Schmidt Theorem holds for ~

Proof.

.

then it is an epimorphism by Nakayama's lemma.

The converse is obvious.

nilpotent. is.

/pn~

For if multiplication by x is an epi-

Thus x has right and left inverses if ~ does.

A/p

lim ~

an R algebra which is a finitely generated R module and

a unit if and only if x is.

of

such that the

Let R be a complete local ring with maximal ideal

Let x ~ A

is.

Hence by Prop.

Thus, we lift e to (en ) ~

which has no idempotents except 0 and 1. Proof.

•

we need to prove two things.

-

l)

98

-

Every finitely generated

composable

A

2)

modules,

A

module is a sum of inde-

and

If M is an indecomposable

E = Hom A

module,

then

(M, M) is local.

i)

follows since finitely generated

A

modules are hoe-

therian. 2)

EC

HomR(M , M) which is finitely generated over R.

Hence E is finitely generated over R. 0 and 1 since an idempotent M = ker e ~ i m

e.

Corollary 5.21.

E has no idempotents

endomorphism

except

e gives a decomposition

Hence, E is local by corollary 5.19. With the above hypothesis K 0 ( A

ian group generated

) is a free abel-

(freely) by the classes of the indecomposable

projective modules. Definition.

Let A and B be R modules and f: A -~-B a homomorphism.

Then f is an essential epimorphism if f is an epimorphlsm and if X is any R module and g: X -~-A is any homomorphism an epimorphism. Definition.

Then g is an epimorphism.

A projective cover of an R module M is a projective R

module P with an essential Remark.

If f: P - 4 P M

eplmorphism f: P - a P M .

and f': P' -q~M are two projective

then P and P' are isomorphic. @: P - ~ - P ' phism.

such that fg is

with f'@ = f.

Since f is essential, isomorphisms.

~

Since P is projective we can find

Since f' is essential,

Since P' is projective

covers,

there is a ~ :

is an epimorphism

@ is an epimor-

P' -~-P so @ ~

so @ and

~

are

= id.

-

Lemma ~.22.

99

-

Let R be a (not necessarily commutative) ring, A a

nilpotent 2 sided ideal, and M an R module.

Then p: M - ~ P M / A M

is

an essential epimorphism. Proof.

Let g: X - ~ M

g(X) = N.

be a map with pg an epimorphism.

Since pg is an epimorphism M = N + AM.

M/N = A(M/N) = A2(M/N) . . . . .

An(M/N) = O.

Let

Hence

Thus, M = N and g is

an epimorphlsm. Lemma 5.23.

Let A be a nilpotent 2 sided ideal of a ring R and Q

a finitely generated projective R/A module. tive cover over R.

Then Q has a projec-

In fact, Q is isomorphic to P/AP where P is a

finitely generated projective R module. Proof.

There exists a Q' such that Q ~ Q '

idempotent matrix e: F - ~ - F 0-~-Mn(A)

with Q = e(F).

Mn(A) is a nilpotent idempotent of Mn(R).

matrices with entries in B).

ideal of Mn(R). Let P = im(e'). Tensoring R n

R/A gives (R/A) n ~ ( R / A ) n Proposition ~.2~.

The sequence

-q~ Mn(R ) -q~Mn(RIA ) -~-0

is exact when Mn(B) = ( n x n

P = Rn/(1 - e')R n.

= (R/A) n = F and an

Hence, e lifts to e', an Now Q = F/(1 - e)F and

(1-e')t-Rn ~ P

~

Clearly

P/PA ~

~

0 with

0 so P/PA ~ Q.

Let A be a nilpotent ideal of a ring R.

Then

there is a 1 - 1 correspondence between the isomorphism classes of finitely generated projective R modules and finitely generated

-

i00

-

projective R/A modules giving by passing from P to P/AP and from Q to a projective

cover of Q as an R module.

duces an isomorphism Proof.

This correspondence

in-

of Ko(R) with Ko(R/A).

Immediate.

Remark.

The same theorem is true for the case in which A is not

nilpotent but R is complete with respect to A, i.e., R = lim R/A n . We lift idempotents Corollar~ ~ 2 4 A .

by the argument of corollary 5.18.

Let R be a complete local ring and A

algebra finitely generated as an R-module. finite number of indecomposable KO(A

projective

an R-

Then there are only a A

modules.

Hence

) is finitely generated.

Proof.

If J is the Jacobson radical of A

then A

is complete

with respect to J. Proposition

5.25.

If R is an artinian ring, then every finitely

generated R module has a projective Proof.

Let E = radical(R),

cover.

r_n = 0 for some n since R is artinian.

Let M be a finitely generated R module. sential epimorphism

and R/E is semisimple.

jective as an R/E module. P-~-M/~N morphism.

Then M - b - M / E M

Hence, M/EM has a projective

as an R module P is projective Hence,

Therefore,

M/EM is procover

and M -~-M/EM is an epi-

there exists an f: P - ~ - M

making

P

N -~-----~,.N/r_M

W

is an es-

commute.

-

Then f: P - ~ M

is a projective

iO1

cover.

note that if fg is an epimorphism Since

~

Corollar~

is essential,

~.26.

Let R be an artinian

Ko(R)

Theorem quotient

~fg.

ring.

projectives.

out by the radical

If A is a separable

modules with KM isomorphic

semisimple

valuation

1.10~

A

ring R,

with the A i simple.

Then

ideal of R.

A

A/pA

G O and K 0 preserve

of R in K'.

well as over R.

Now pR'

cover.

over K, the

is a maximal torsion

free

to N.

In

to show that the Caftan Let A = A I x ... x A n

~ A 1 x ... x A n with = ~

consider the case A simple. gral closure

A

generated

it will suffice

is a monomorphism.

Then

is

is principal.

map K O ( A ) - ~ - G O ( A )

order in A i.

algebra

to KN, then M is isomorphic

every ideal of

By Theorem

and Go(R)

Go(R)

and taking projective

order over R in A, and M and N finitely

Proof.

Then Ko(R)

and these bases are in 1 - 1 correspondence

field of a discrete

particular

so is g.

by the Caftan map).

is free on indecomposable

~.27.

Therefore

f must be an epimorphism.

(but not necessarily

free on simple modules by factoring

To see that f is essential

so is

itself

are isomorphic Proof.

-

A i a maximal

~ i / P A i where p is the maximal finite

sums.

Hence it is enough to

Let K' be its center and R' the inte-

Then

~

is a maximal

is an ideal of R'.

Say

order over R' as

-

vI vi

"'" Pr

A

with the Pi primes of R'.

unique P in A

~ /Pi

vi

containing pi A

maximal 2 sided ideal so ( A /Pi ~/Pi ^

Then

~/p~

and it remains to check that the Cartan map is a mono-

morphism for the factors

fore

-

vr

Say pR' = Pl ~A/Pi

102

vi

~

A

. vi

•

By Theorem 5.16, there is a

Thus ~

~ /Pi

vi

A

has a unique

)/radical is simple.

There-

has only one simple module S and so only one in-

decomposable projective P. by CS~ and K o ( A / p i

vi

Thus GO( ~ / p i V i ~

) = Z generated by

~ ) = z generated by CP~.

The Cartan map is

multiplication by n, the number of times S occurs in a composition series for P. Corollary.

This map is clearly a monomorphism.

Theorem 5.27 remains true if R is any semilocal Dede-

kind ring. Proof.

If pl, ..., Pn are the primes of R.

crete valuation ring and Theorem 5.28 below.

Then Rpi is a dis-

A pi is a maximal order over Rpi by

Since KMpi = KM isomorphic to KNpi = KN,

Theorem 5.27 implies Mpi is isomorphic to Npi say by fPi~ i.e., M and N have the same genus.

By Roiters lemma we can find

0 -~-M -~-N -~-X -~-0 where the annihilator of X is prime to Pl' "''~ Pr"

Therefore X = O.

-

Theorem ~.28.

io5

-

Let R be a commutative noetherian ring, and

A

a

maximal order over R in a separable K algebra A where K is the quotient field of R. Then

Let S C. R be a multiplicatively closed set.

A S is a maximal order over R S in A.

Proof.

Suppose not.

Let

A s ~ ~

where ~

Then there exists r ~ R such that F -- r p ~ is a 2 sided ideal Of O~(f) ~ A ~f

A

~ .

~ ~F ~F

Pick ~ ~

-- r ~ C

~

since it is contained in that s ~ f_ ~ A Hence s ~ and

•

~.

Then f = F

S" A.

A

A).

Therefore, O~(f) = A

Then

since

Then ~ F C F and

f is a finitely generated R module Therefore, there exists s ~ S such

and s ~ f ~ sF C F . O~(f)_ = ~

A S"

Let O~(f) = ( x l x f ~ f and x ~

and is an order over R.

is maximal.

is an R S order in A.

and so ~

Therefore, s ~ f = ms~ C

AS.

cAlF

Hence ~

= f. -- A s'

A S is a maximal R S order as desired.

Corollar 7 ~,25.

Let A be a separable semisimple K algebra where K

is the quotient field of a noetherian ring R. in A over R.

Then ~

Let ~

be an order

is a maximal order if and only if A

is a

maximal order over R Proo f .

Embed

~

for every maximal ideal p of R. P in ~ a maximal order. Then A p C ~ p .

P

A p are maximal orders, then

Ap

= ~p

for all p and so A

The converse is a special case of Theorem 5.28.

If all = ~.

-

Chapter 6.

104-

Orders

In this chapter we present results of Jacobinski and Roiter concerning arbitrary orders.

Let A be a separable semisimple K

algebra where K is the quotient field of a Dedekind ring R. A

be any R order in A and let U ~

be a maximal order.

there exists an r # 0 in R with r ~ C R with r ~ p, then r ~ p a Ap~ ing r.

~p.

A p C

A.

Then

If p is a prime ideal of

~ p and r is a unit at p.

Hence,

There are only a finite number of primes of R contain-

Thus, except for those primes,

Lemma 6.1.

Let

~ p is a maximal order.

Let M be a finitely generated torsion free

B -~-C -~-0 an exact sequence of homomorphism.

A

~

modules and f: M - ~ - C

module, a

Then rf lifts to a map, g, from M to B making the

diagram B

g"

M ---gK~ c

0 Corollary 6. 2 .

r Ext~ (M, X) = 0 for all A

Proof of corollary from the exact with I an injective 0 -~-HomA(H,X) exact.

commute.

A

lemma. module.

Let 0 -a-X -~-I - a ~ Y - ~ - O be Then

-a-Hom A (H,l) -4~HomA(H,Y)

Pick u ~ E x t , ( M ,

X).

modules X.

-~Ext~(M,X)

Lift u to f ~ H o m A ( M , Y).

-a~0 is Then rf

io5-

-

is a lifting of ru and rf comes from some g i H o m ~ (M, I) by the lemma. Remark.

Hence,

rf goes to zero in E x t ~ (M, X).

Actually we have shown more.

then a Ext~(M, X) = 0 for n ~ 0.

Hence ru = O.

If ~ = {r £ R I r ~ C

~

},

This follows by exactly the same

method. Proof of the lemma. mapping onto M.

Let F be a finitely generated free module

Let ~: M - ~ - M

be given by multiplication by r.

Since F is free there is a map, h, making h

F g

M

r ~M

O If we can find a g': M - ~ - F will lift rf.

c ommut e.

making the diagram commute,

Hence we can assume C = M and B is a finitely gen-

erated free module mapping onto C. done by Theorem 5.12.

If we were over ~

To take advantage

An

vn . t

!

•

5

0

"

we would be

of this, we tensor with

and obtain the diagram

0

then hg'

gl!

-

~

M/torsion is ~

projective and since pj is onto, there is a

g" making the diagram commute. in r ~ n ~

~ n.

106-

The image of rg" will be contained

Hence rg"ji will lift r to a map M - ~ ' A n .

check that p'(rg"ji)

To

= r we can apply ji which is injective since

M is torsion free. We recall the following Definition.

Let M and N be finitely generated torsion free

modules where A

is an order over R.

same genus,

N), if Mp is isomorphic to Np over A p

(Z~

primes p of R.

Then we say M and N have the for all

This terminology will only be applied to finitely

generated torsion free A-modules. Lemma 6.~. p, ~

Let R be a discrete valuation ring with maximal ideal

an R order, M and N finitely generated A

modules, R the

A

completion of R at p, ~

= A~RR

, M = M~RR

, and N = N ~ R R .

Then M is isomorphic to N if and only if M is isomorphic to N. Proof.

Let g: M -~-N be an isomorphism.

M and N are finitely presented. morphic to Hom A (M, N). such that l ~ f

M/pM

Hence, R ~ R H O m

A(M,

N) is iso-

Thus, there exists an f ~ Hom A (M, N)

~ g mod p.

N/pN

R is a flat R module and

Hence g - ( l ~ f ) :

MM^/p^

M-4~pN.

Consider

( 1 @ f > "~- N/pN^ g

-

107

-

Since g is an isomorphism and l ~ f is onto.

But ( l ~ f ) "

~ g mod p, we see that ( l ~ f ) *

~ f~ under the natural identification of

M/pM with M/pM and N/pN with N/pN. onto by Nakayama's lemma. torsion.

Thus f* is onto.

Hence f is

Since rank N = rank M the kernel is

But M is torsion free.

Hence f is an isomorphism.

The converse is obvious. We return now to an order A

over a Dedekind ring R, in a

semisimple K-algebra. Lemma 6.4.

If M, M', N, and N' are ~

and torsion free with N ~ N ~ M ' Proof.

~N'

modules, finitely generated and N ~ M '

then N N N '

By lemma 6.3 it is enough to check at the completion.

But

the Krull-Schmidt theorem holds for orders over complete local rings. Remark.

In ch~cking whether M ~ N

we can restrict our attention to

a finite set of primes of R by the following method. maximal order containing r~ (

~

.

A

.

Let ~

be a

Then there is an r ~ 0 in R with

Let P be a finite non empty set of primes of R includ-

ing all the primes containing r. Lemma 6.5.

Two finitely generated torsion free ~

modules M and

N will have the same genus if and only if Np is isomorphic to Mp for all p ~ P. Proof. to K ~ R N

If K is the quotient field of R, then K ~ R M since E is non empty.

is isomorphic

Now Mp and Np p $ P are finitely

generated torsion free modules over a maximal Rp order ~ p = ~ p

-

which become isomorphic isomorphic

to Np.

108-

over K.

Therefore,

by theorem 5.27, Mp is

Hence M and N are isomorphic

R if they are so at all p ~

at every prime of

P.

Recall now the following definition. Definition.

Let R be a commutative domain.

Then the Jordan-

Zassenhaus Theorem holds for R if for every R algebra

A

which is

finitely generated and torsion free as an R module with K ~ R A semisimple

and separable

for every integer n ~ O phism classes of

A

over K (K is the quotient field of R) and there are only a finite number of isomor-

modules which are finitely generated and

torsion free of rank ~_ n. Lemma 6.6

(Roiter).

Let R be a Dedekind ring such that the Jordan

Zassenhaus Theorem holds for R. simple separable K algebra A. and N be two A

modules

Let

~

be an order in a semi-

Let 0 # ~ be an ideal of R.

of the same genus.

exact sequence 0 - ~ - M - ~ - N - ~ - U - ~ - O

Let H

Then there exists an

where U is torsion free with

annihilator prime to ~ and with U = IIU i where the U i are simple modules and the annihilators

of the U i are pairwise relatively

prime. Proof.

Let n = rank M.

Look at all A

modules M i of rank n

which are finitely generated and torsion free and all exact sequences 0 -~-M i f-~Mj -q~Sf - ~ 0 hence,

its annihilator

where the Sf is simple, and,

is a prime ideal p.

gers i, j one of the following happens

For each pair of inte-

-

l)

109

-

Infinitely many primes of R are obtainable as annihi-

lators of Sf for differing f. 2)

Only finitely many primes can occur as annihilators

of

Sf. Since there are only finitely many pairs i, j, only finitely many occur in case 2).

Let ~ be the product of all the primes that can

occur in 2) for any i, j.

Apply Roiter's theorem to the ideal

b to obtain an exact sequence O - ~ M annihilator prime to ~ b.

Let X = X O ~

tion series for X and N i = j-l(xi). N

=

No~NI~...

Consider N o ~ N

~Nm_l~Nm 1.

~N

~X-4~O

... ~ X m = 0 be a composi-

Then

= M has all quotients Ni/Ni+ 1 simple.

For some i, j N O isomorphic

morphic to M i and 0 ~ N

to Mj and N 1 iso-

1 -~-N O -m. S -~-0 is exact with the anni-

lator of S prime to ~ b. applies to (i, j).

where X has

Hence, by the choice of ~

, case l)

Hence we can modify the inclusion N l ~

NO

so that No/N 1 is simple with annihilator any of an infinite number of primes,

Doing this at each stage we get a new embedding of M

into N with 0 ~ U = Uo~UI~...

N -~-N -m-U - ~ 0

exact and

~ U n = 0 a composition series with the annihila-

tors of Ui_l/U i pairwise relatively prime.

Then U is the direct

sum of its primary components which are simple phic to the Ui_l/U i. Lemma 6.2.

A

modules isomor-

This is what we wanted.

(Roiter).

R order in a semisimple

Let R be any Dedekind ring. separable K algebra A, ~

Let

~

be an

a maximal order

-

containing

A,

r # 0 in R with r ~ C

which is the direct sum of U1, ~nnihilators

Proof.

~

.

Let U be a

~

module

..., U r with U i simple and with the

of the U i pairwise relatively prime and prime to (r).

Let M be a faithful generated.

ii0-

~

module which is torsion free and finitely

Then there is epimorphism f: M - ~ - U .

It suffices to show there exists an epimorphism

fi: M -~-U i.

If so, take f = (fi): M ~ U

at any p ~ R ,

at most one Uip will be # O.

Therefore f is locally

and thus i ss an epimorphism.

Hence we can assume U

an epimorphism

is simple with annihilator p with r ~ p. mal order.

Since M is a faithful

A

i = U.

Now ~ p

module, K ~ M

If we localize

= ~p

is a maxi-

is a faithful

s

A module.

Therefore II KN -- (KN) s -- A ( ~ W for big enough s since 1 A is semisimple. Thus by Theorem 5.27, l"Isp ~- Ap t~ Np for some ~

module N C W with KN = W.

Hence, A p is a direct summand of M~.

Now Up = U since all elements of R - p act as units on U. have an epimorphism

~

-~-U.

P

So we

In the sequence of maps

s

iII Mp -~- ~ p hence,

-b~ U -w. 0

one composite Mp -~- U must be non zero and,

onto since U is simple.

we have a non zero image.

Thus in the composite M -~-Mp -~-U

Again the simplicity of U implies the

map is onto. Theorem 6.8

Done. (Roiter).

Let

A

be an R-order in a semisimple

separable K algebra A where K is the quotient

field of R and R is

-

iii

-

a Dedekind ring satisfying the Jordan-Zassenhaus Theorem. H, N1, and N 2 be torsion free finitely generated A Nl~

N 2 and M faithful.

that M O N I ~ Proof. A

modules with

Then there exists an M' with M ' ~

with r P ~

A

•

Let ~ = (r) in lemma 6.6.

exact sequence 0 -*bN 1 ~ N ditions of lemma 6.7.

2 -~-U-~-0

Then we get an

where U satisfies the con-

If r £ p, then Up = O.

By the following lemma we have N l ~ M ~ (Roiter). O

~

A

0 ~ C

containing

This lemma gives us an exact sequence

0 -~-H' - ~ - M - ~ - U - ~ - O .

Lemma 6. 9

M such

H' ( ~ N 2.

There exists r # 0 in R and a maximal order ~

lemma 6.5.

Let

Hence M ' ~

M by

M' ( ~ N 2.

Let ~

B

~ D

~

W

~

~ W

O

--~0

be short exact sequences with B and D finitely generated and torsion free over all y such that Proof.

A

. Ay

Assume W is a torsion module and Wy ~ 0 for is not a maximal order.

Form the pullback diagram 0

0

A-~A 0 -~-C - ~ - P - ~ - B O-~-C

-~-0

-~D-~W-~-O 0

0

Then A G D ~

B~C.

-

where P = ((b, d) lj(b) = k ( d ) ) C 0 -*-A-~-P-~-D-a~O would be isomorphic

112

-

B OD.

If the sequences

and 0 -aPC ~ P - @ - B

-~-0 both split, then P

to both A(~)D and C(~)B which is what we want.

If there were maps f: D - ~ - B

and g: B-I- D such that jf = k and

kg = j then the sequences would split for D - - i ~ ! ~ B ~ D

would

have image contained in P and would split the sequence 0

-a~A

-i-p

-~-D

--~0.

Thus we must find a map f: D -~-B which

makes D s

B --4. W - - a ~ O commute.

This is possible

is not a maximal Y order we have W = O. If A is a maximal order, D is projective. Y Y Y All modules here are finitely generated so Hom localizes. Since Hom~(D,

B) - ~ - H o m ~ ( D ,

locally.

If

~

W) is locally onto, it is onto and we are

done. Theorem 6.8 has the following corollary. Corollary

(Jacobinski).

Jordan-Zassenhaus

Let R be a Dedekind ring satisfying the

Theorem.

separable K-algebra A.

Let

A

Then K O ( A )

be an R-order in a semisimple and G O ( A )

are finitely

generated. Proof. KO(A)

This was done for G O ( A )

in Chapter 2.

The argument for

is given more generally in theorem 6.10 below.

-

Definition. generated

Let A ~

ll~

-

be an order and M a torsion free finitely

module.

D M is the abelian group with generators

[N]

for all N such that N is a direct snmmand of M s for some s and with relations [N I] + KN2] ~ [ N I ~ N 2 S . Remark.

It is easy to see that in DM [N l] = [N2] only if there

exists an s with Ms O N 1 isomorphic to M S ( ~ N 2 •

The argument is

the same as that for K 0 ( A ) = D ~ . Jacobinski generalizes corollary 6.9 as follows. Theorem 6.10.

If R satisfies the Jordan-Zassenhaus

Theorem bhen

DN is finitely generated. As always R is Dedekind and

A

is an order in a semisimple

separable K-algebra. Proof.

Let ~ D ~

be a maximal order.

Let r ~ 0 in R with r ~

and let p be a finite nonempty set of primes of R containing all p ~(r).

Then ~ p

= ~p

is a maximal order if p ~ P.

NOW/p

the

completion of M at p is a torsion free finitely generated

P Define DM -~ II D ~ by sending [N] to ([Np]). D~ is a pEP p p finitely generated free abelian group by the Krull Schmidt Theorem.

module.

In fact, the only indecomposable modules that can occur in a direct summand of M~ are the ones that occur in Mp. DM i s

finitely

generated.

Let

[N l ]

-

Hence, the image of

[N 2] b e i n

A A [Nip] - [N2p] = 0 in DN~p for all p L P.

the

kernel.

Then

~ Thus, Nip is isomorphic

-

A to N 2

114

-

for all p ~ P by the Krull Schmidt Theorem.

Hence by lemma

P 6.5 N l ~

N 2.

Let I be the annihilator

is still semi-simple in A1, and M is a DE remains

and separable.

A 1 module.

As a

the same if computed over

assume M is a faithful with M ' ~ M

and M ~ N

A

module.

1 isomorphic

[N l] - IN 2] = [M'] - [M].

in A of KN.

~I ~l A

Then A 1 = A/I

= ~/I~A

is an order

module M is faithful. or

A 1.

Hence, we can

Apply theorem 6.8 to obtain M' to M ' O N

2.

In D M we have

By the Jordan-Zassenhaus

hypothesis

there are only finitely many such M' up to isomorphism.

Thus the

kernel is finite. Remark.

Done.

This also shows that if N I N

N 2 then [N l] - [N 2] has

finite order since [N l] - [N 2] will be in the kernel of DE ~

TT D~ • pip p

Theorem 6.11.

This can be strengthened

Let R be a Dedekind ring satisfying the Jordan-

Zassenhaus Theorem. A

Let M and N be torsion free finitely generated

modules with N ~ N .

isomorphic

as follows.

Then there exists an s such that M s is

to N s.

The converse

is clear by completing and applying the Krull

Schmidt Theorem. Proof.

We can assume M is faithful by the construction used in

the proof of theorem 6.10. tors of M and N are the same Lemma 3.1 ).

Then N is faithful since the annihila(we can embed N in M by Roiter's

By Theorem 6.8, we can find M' with M Q M ~

M' Q N .

-

Therefore EM3 Mr~

-

KN~ ~ D N.

Thus EMks ~ ENk~ so

Mr Q N k for all large r.

we can let s = kt. now have M t ~ N

-

By the remark preceding Theorem 6.11,

EN~ has order k " o@. Mk ~

ll5

If we can show (Mk) t ~ (Nk) t

Therefore we replace M and N by M k and N k.

~ Nt(~N

We

for all large t by the isomorphism just

proved. In other words it will suffice to prove the theorem under the additional hypothesis that N t ~ This implies M t ~ Mt~M

N ~ Mt ~N

for all t ~ t O .

N r = (M t ( ~ N ) ( ~ N r-I = M(~ (M t O

Nr-l) = ... ~

r for all r ~ i, t S t O . Let NI, ..., N h represent all isomorphism classes of modules

Ni~

N.

Note h ~ O @

by the Jordan-Zassenhaus

soms+n we have M n ( ~ N i ~ Choose n o ~

MnoM,

If for

choose one such n and call it n i-

n i for i = i, ..., h.

By Theorem 6.8, if m ~ I , im.

Theorem.

If t ~ to, then M t + m + l ~

we have N m ~ N

Mt QMm+I8

~MmONim

M t ~ N m+l • M t ~ M

for some m~N

i . m

By the choice of no, we have

Mno+l

~ MnO~Ni

. m m-n 0

start with m ~ no, then N m+l = M m ~ N i m

= M

As always R is a Dedekind ring and

~

Therefore if we ~MnO~Nim~

Mm+l"

an R-order in a

semisimple separable K-algebra. Theorem 6.12.

(Jacobinski).

primes such that if p ~ P then

Let P be a finite non empty set of ~p

is a maximal order.

If M and N

- 116 are finitely generated torsion free A

modules such that Mp is a

direct s1~mmand of Np for all p ~ P, then there exists M' such that M'~

M and M' is a direct summand of N.

Corollary 6.15.

If M is locally a direct summand of N, then there

exists M' such that M ' ~ Proof of theorem. i~

v~

M and M' is a direct summand of N.

Let P = (Pl' "''' Pn )"

A i n we have Mpv ~

~ Npv ~

For each v with

A Npv such that Jviv is an

isomorphism. Now HOmApv(NPv , Npv) = ~p_ (~ Hom#~

(Np, Np) = Hom~ Pv

( % , Npv ) Pv

v

since M and N are finitely presented and Rpv is flat over Rqv. Hence, there exist i'v g Hom A pv(NPv , Npv) and Jv g HOmApv(Npv,MPv) such that i'v ~ iv mod Pv and j,~. = Jv mod Pv"

Hence Jvlv

ji mod Pv' so by Nakayama's lemma, Jvlv ''' is onto. finite rank and is torsion free. all v.

But Mp has

Hence ~'i' is an isomorphism for ~V V

Now we can forget the completions.

We have

i

Pv

Pv

Pv

with composition an isomorphism.

Now

H°mLpv(MPv' Npv) = H°mL(M' N)pv since M and N are finitely presented.

Say i v

i ~ / s v and Jv

Jv / v w i t h

" ~ HomL(N , M). iv' • HOmL(N , N), and Jv

sv

tv ~ R

Pv'

We can replace iv and Jv

-

117

-

by i'v and Jv" and the composition will still be an isomorphism.

By

the Chinese Remainder Theorem there are i L HOmL(M , N) and j ~ HomL(N , M) such that i ~ iv mod Pv and j ~ Jv mod Pv for all v. Then ji: Mpv --~-M

is onto for all v by Nakayama's Lemma and Pv

hence an isomorphism since Mpv is torsion free finitely generated. We claim M' = j(N) will do.

We need to show that M ~

is a split epimorphism and that j ( N ) ~ M . is an isomorphism.

For p ~ P, j(N)p -~-Mp

Hence M and J(N) have the same genus Lemma 6.5.

If p • P then Np -~- j(N)p = Mp splits by construction. then

j(N)

If p ~ P,

a

is a maximal order and hence j(N)p is projective since P it is finitely generated torsion free. Thus Np - ~ j ( N ) p splits for all p.

The proof of Theorem 6.12 and its corollary will be

completed with the proof of the following lemma. Lemma 6.14. and f: C - ~ D

If C and D are finitely generated modules over a is locally a split epimorphism then f is a split

epimorphism. Proof.

f is an epimorphism since the coker of f is locally 0.

We

have the exact sequence 0 -~-B -~-C f-~D -*-0. A splitting will be a k: D -~-C with fk = 1D.

If the map

(i, f): H o m ~ (D, C) -~-Hom a (D, D) is onto, there will be such a k.

Now D is finitely generated and hence finitely presented.

Thus

-

Hom~

118

-

localizes and hence, the coker of Hom~ (D, C) - ~ - H o m ~ (D, D)

is locally 0 and hence 0.

So Hom A (D, C) -~-Hom A (D, D) is onto

and a splitting exists. Corollary 6.15.

Let R satisfy the Jordan-Zassenhaus Theorem.

N and N satisfy the hypothesis of Theorem 6.12.

Let

If every simple A

module S which occurs in KN occurs strictly more times in KN than in KE, then N is a direct summand of N. Proof.

By Theorem 6.12 (which we assume applies), N = M' ~ P

M'~

M.

P'~

P with M' ~ P

If P were faithful we could apply Theorem 6.8 to get some

mand of N. A/~

with

~A

~ M~P'

then N ~ M ~ P '

and H would be a sum-

Let a be the annihilator in A of KN.

We pass to

= A 1. All modules concerned are A 1-modules since

KM~KH'CKN.

The condition on simple submodules of KN implies

KP is faithful and hence P is faithful. Corollary 6.16. N

If H ~ N ,

then M is indecomposable if and only if

is.

Proof.

Immediate from the theorem.

Definition.

The genus of M, GM, is ( N I N O N ) .

Addition of genera is well defined.

A genus is called in-

decomposible if any. one of its elements is (and therefore all are by corollary 6.16). Theorem 6.17

(Jacobinsky).

Let

~

be an R order in a semisimple

separable K-algebra where R is a Dedeklnd ring satisfying the Jordan-Zassenhaus Theorem.

Let M be a torsion free finitely

-

generated

~

module.

I19-

Then there are only finitely many indecom-

posible modules N such that N is a direct summand of M s for some s. Corollar~ 6.18.

If

A

and R are as above, there are only finitely

many indecomposible projectives. Proof of the theorem. ideals such that A p

Let ~

be a finite nonempty set of prime

is maximal if p ~ ~ .

quence 0 -,P T -b.D M - I ~ D

4

where T = (IN I] - IN 2]INIP ~"- ^

all p £ 6 3 ) = (IN I] - [ N 2 S I N I ~ N2). s11mmand of M s for some s). ~

Let~=

N

Lemma 6.1~. ~

is finitely generated.

We first prove the theorem

Let N be an indecomposible

~N".

this shows that G N G genera.

summand of M s .

in.

By

since G N = GN, + GN,,

If S is a finite set of generators for ~ , S so N belongs to one of a finite set of

Hence the rank of indecomposibles

Jordan-Zassenhaus

(GNIN is a direct

•

corollary 6.16, GN is also indecomposible implies N ~ N '

for

is a monoid with addition defined by

GN ÷ GN, =GN

from the lemma.

We have an exact se-

is bounded and the

Theorem implies there are only a finite number

of indecomposibles. Proof of Lemma 6.15 .

We map ~

to DM/T by G N ~ E N S

is well defined, additive and one-one.

2~

MS~N'~N

This

In fact IN] = [N'] mod T

implies that IN3 - [N'] = 3NS - IN2] where N l ~ some s, M S ~ N ( ~ ) N

mod T.

I so N ~ N

'.

N2 .

Thus for

However, it is not

true that a submonoid of a finitely generated abelian group is

-

finitely generated.

120

-

For example, the submonoid of Z ~

Z given by

[(m, n) l(m, n) = (0, O) or m ~ 0 and n ~ O} is not finitely generated. We need to characterize the image more precisely.

Now D ^

Mp

is free abelian on [PI ], ..., CP s] where PI' "''' Ps are the indecomposibles occuring in a direct sum decomposition ni

s

Mp = i~ 1 Pi

"

We define an isomorphism of D~

with Z e

... Q Z

P (s times) by [ N ] ~ ( f l ( [ N ] ) ~

..., re(IN])) where fi([N]) ~ the

number of times Pi occurs in the decomposition of N into a direct sum of indecomposibles.

We extend the map fi to D N by ~i-~DZ. The fipj all anihilate T since

T is a finite group.

Hence we have functionsfip j defined on DM/T.

Identifying G with its image in DM/T we claim tha~ if x g DM/T , then x I G if and only if all f i p j ( X ) ~ 0 . fipj(X) $ 0 fipj(x) $0. y=

y

IN ]~

EPp]

If x • G, then all

clearly since x = [N] for some N. Say x = [N] - [N'].

- [~].

In D~

Now x ~ y

Suppose all t D~p where

there does exist a module P

such that

P

Pp = j pfi( )

By the Krull-Schmidt Theorem,

-

Np is isomorphic

to N'p e

there exists an N" ~

121

-

Pp for all p ~ P.

Hence by Theorem 6.12,

N' and a Q such that N = N" (~Q.

Now

x = [N] - IN'] = [N"] - IN'] + [Q] = [Q] rood T since N" ~ N ' plies

IN"] - [N'] C

T.

Hence x ~ [ G Q ] G ~

im-

as claimed.

Lemma 6.19 now follows from the following generalization of a classical result. Theorem 6.20.

(Gordon's Lemma).

Let A be a finitely generated

commutative monoid and let fi: A - ~ - Z i = l, ..., n be a finite collection of homomorphisms.

Then G = (x & A l f i ( x ) ~

0 for all i)

is a finitely generated monoid. We note that a finitely generated abelian group A is also finitely generated as a monoid.

If a i generate A as a group,

then

all ± ai.generate A as a monoid. Remark.

Let ~

f: Z ~ Z

~

be an irrational real number.

~Rby f((m, n)) = m - n ~ .

Define

Then G = (xTf(x) ~ O) is

not a finitely generated monoid.

This shows that we c~uuot re-

place Z by IR or even by Z + Z ~ C

~ in this theorem.

m - n ~ O

if and only if m - nr ~ 0

Since

for all rational r ~

~

,

the same G can be defined by an infinite number of maps into Z. Thus the restriction to a finite number of fi is essential. example G = ((x, y) G

Z~

Zlx ~ 0 and y ~ 0 }

replace the condition fi(x) ~ 0 by fi(x) ~ Proof.

shows that we cannot O.

We use induction on n, the number of fi"

(x C A l f i ( x ) ~

0 for l~. i ~

The

If G 1 =

n - l) then G 1 is finitely generated

-

by the induction

hypothesis

122

-

and G = (x @ G l l f n ( x ) $ O } .

it will suffice to do the case n = 1. Let al,

..., a r generate

A.

Then f ( a ) $ O or X o ~ O

Z, x i ~

of A has the form

O°

Let Pi = f(ai)"

if and only if the x i s a t i s f Y ~ i f ( a

where we set x 0 = ~ i x i Let H be the set of (x0,

conditions

We write f = fl o

Any element

a = xla I + ... + Xra r where all x i ~

xi$0

i) =

$0

. ..., x r) ~

for O _~- i ~ r

Therefore

Z r+l satisfying

and x 0 = ~ i x i n

.

the

Then H is a

monoid under addition and clearly G = (l~xiail ~ x 0 with (Xo,

..., x n) G

H).

Therefore

the map

~:

H -~-G by

~@(Xo,

..., x n) = xla I + ... + Xra r is a monoid homomorphism

onto.

Thus it will suffice

and is

to show that H is a finitely generated

monoid. We relabel x0, Xl, Wl,

..., x n as Ul,

.o., Ua, v l, ..., Vb,

..., w c in some order so that the equation x 0 = ~ i x i

takes

the form rlu I + ... + raU a = SlV 1 + ..o + SbV b where all r i ~ 0 and all s j , Define

0. elements

all other coordinates

hi, hij G

0 and by letting hij have u i = sj, vj = r i

and all other coordinates r lhl = E x i . 0

Then

H by letting h i have w i = 1 and

zero.

lh + h'l

If h = (x0,

= lhl + lh'l.

..., x n ) G

H, let

If h has w i ~ 0 for

-

123-

some i, we can write h = h i + h' where but some u i ~ N'min(ri)~ where

lh'l~

lhl.

If all w i = 0

N (an integer to be determined) then

~riui

= ~,sjvj ~ ( ~ s j )

= ( ~ s j ) - l ( m i n ri).

max(vj) so some vj ~ @ ~ N

Choose N so that N and L

greater than all ri, sj (and also ~ 1).

N are

Then we will have

u i ~ sj, vj ~ r i so we can write h = hij + h' where

lh'l ~

lhl.

By induction on lhl we see that H is generated by the hi, hij and those h G

H with all w i = 0 and all ui_~ N.

are only a finite number of such h since u i ~ vj~

But there

N for all i implies

~,sjvj = ~ r i u i _~ ( ~ r i ) N .

Theorem 6.21

(Roiter).

Let

a

be an order over a Dedekind ring

R in a separable semisimple algebra over K, the quotient field of R.

Let R satisfy the Jordan-Zassenhaus Theorem and assume that

R/p is finite for each nonzero prime p of R.

Then there exists an

integer N such that every genus of A - m o d u l e s

contains ~. N iso-

morphism classes. Note that the hypothesis on R/p holds if R is the ring of integers of an algebraic number field or a finite extension of k[t] with finite k. Proof.

Let G be a genus and let A = A1, A2, ..., A n be

representatives of all the isomorphism classes of G. be an element of R such that r ~ C ~ order.

By lemma 6.6, we can

where

~D~

Let r # 0 is a maximal

-

124-

embed each A i in A such that A/A i is a direct sum of simple modules whose orders are relatively prime and prime to r, and such that if i # j the orders of A/A i and A/Aj are relatively prime. Let B = f % A i in A and let

U = A/B,

Then

U

is a direct sum of

simple modules of relatively prime orders and orders relatively prime to r. eplmorphism

In fact A/B - . ~ A / A j

is clearly injective and is an

since this is true locally.

epimorphism A

-~"

U.

Since A

By Lemma 6.7 there is an

is projective

there exists a map

- ~ A making the diagram

A 0 -~B

--~A

--~U

---~0

0 Let C be the image of A

-~m A.

Then rk C ~

commute. rk ~

and B + C = A.

We want such a C which is pure in A (with respect to R). D = (KC)~

A in KA.

Then rk D = rk C, D ~ C ,

torsion free, and rk D ~ rk A . A i + D~

B + D = A.

Let

B + D = A, A/D is

Since A i ~ ) B , we have

Therefore A i + D = A so A i -~- A/D is onto.

Consider the following commutative diagram with exact rows and columns :

-

0 ~ A i ~ D

~ A

0 ~

-~B

B~D

0

125

-

A/D

i ~

~A/D

0

0

--~0

0

We can recover A i from fi by forming the pushout. need to calculate

the number

If h: B - ~ - A i ~ D replacing

of possible

pushouts°

, we can get a new commutative

gi by gi + h and fi by fi + hi.

yield the same pushout° (up to isomorphism)

Therefore

Ai~

D) ~

diagram by

Thus fi and fi + hj

the number

of possible

is d. the order of the cokernel

(j, I): Hom#~(B,

Hence we

pushouts

of

H o m A ( B ~ ' ~ D , Ai~'~ D)-

We claim the image of (j, i) contains

r Hom(B~

D, Ait'~ D).

Since A/D is torsion free, we can apply lemma 6.1 to the diagram A/D

B ~ A / D e and get m: A/D -~- B with em = rlA/D°

~ 0

Now e(rl B - me) = 0 so we

can write rl B - me = jk where k: B - ~ - B ~ D . jkj = rj - mej = rj but j is a monomorphism fore,

if f: B ~

Now so kj = r l B ~ D.

D -~-Ait-X D, then fk: B - ~ - A i ~

There-

D and (fk)j = rf.

-

126-

This proves our assertion. number of possible pushouts Hom A ( B ~ D , since R/(r)

is less than or equal to the order of

Aif~ D)/r H o m ~ (Bf~ D, Aigl D). is finite and Hom A ( B ~

ated over R.

Now B ~

D~

D and A ± ~

rk(E) are bounded by rk(D) ~ Theorem, B~D

rk A .

D, A i ~

This order is finite D) is finitely gener-

D CD.

Hence r k ( B g ~ D )

and

Hence by the Jordan-Zassenhaus

there are only a finite number N 1 of possible pairs of

and A i ~

Hom(B~

It follows from this that the

D.

Let N 2 be the maximum of the orders of

D, Aif~ D)/r Hom(Bf~ D, A i ~ D )

over all such pairs.

Then

the number of possible A i is less than or equal to NIN 2 for every genus.

Done.

Chapter 7:

K 0 of a Maximal Order

In this chapter we will compute K 0 for maximal orders in algebras over global fields.

Let R be a Dedekind ring with quo-

tient field K, A a semisimple

separable K-algebra and A

mal R-order in A. = ~l

a maxi-

Then A = A 1 x ... x A r with A i simple and

x ... x A r.

Let A i have center K i.

Then h i is a max-

imal Ri-order in A i where R i is the integral closure of R in K iTherefore KO( A ) = ~ K o ( A i ) .

We are thus reduced to looking at

A . so it will suffice to consider the case where A is central l simple over K. Let S be a simple A module. Then there exists a prime ideal p C

R such that pS z O.

By Theorem 5.16 there is a

-

unique 2 sided prime P of A e ~ O.

Then

~/P

127

-

containing p ~

and p A

is a simple algebra over R/p.

pes = 0 implies PS = 0 so S is a

~/P-module.

= pe some

Since S is simple, Since

~ /P is

simple, there is exactly one simple S with pS ~ 0 for any p ~ Hence the simple ~

R.

modules are in 1-1 correspondence with the

non-zero prime ideals of R.

Let I R be the group of fractional

ideals of R.

This is free abelian on the non-zero primes p of R.

We map K 0 ( ~ )

~

K0(A) by [ M ] ~ [ K ~ R M ] .

For any finitely generated torsion

a

This is clearly onto. module C we have pd~ C ~ 1

by Lemma 5.13 so there is a projective resolution 0 -4pp, -~ P-~- C -4-0.

By Schanuel's lemma, [P] - [P'] G K O ( ~ )

is independent of the choice of the resolution. T~

Hence if we let

be the category of finitely generated torsion modules over ~

there is a well defined map Ko(T ~ ) -D~ K O ( ~ ) KPS - [P'].

The sequence Ko(T ~ ) - ~ - K O ( ~ )

In fact, given C E

TA

we have K ~ R C

Conversely if [P] - [P'] ~ By Theorem 5.27, P ~ P ' O-4~P'

~

P-~C

sending [C] to

-0-Ko(A) -J-0 is exact.

= 0 so K ~ R P ~

0 in Ko(A) , then K ~ R P

K~RP'• K~RP'.

so by Roiter's lemma, we can find

-~-0 with C a torsion module.

Ko(A) is isomorphic to Z.

Since A is simple,

Also K0(T ~ ) is free on the simple

~

modules.

Hence we can define an isomorphism Ko(T ~ )~ IR by

[S]~p

where p is the unique non zero prime with pS = 0 above.

Let C G T ~ . intrinsic way.

,

-

We want to describe the image of [C] in I R in a more Let 0(C), the order ideal of C,be the ideal of R

-

128

-

defined by O(C) = ~ P i

where there exists a series

of C,

... ~ C 0 with Ci/Ci_ 1 isomorphic

C = Cn~Cn_l~

This is well defined by the Jordan-HSlder on the R-module exact,

structure

of C.

then we can construct

from such series for C and C'. Hence we have O: Ko(T ~ ) ~ phism.

but dimR/pS

-4mC -~-C" -~-0 is series for C (over R)

This shows that O(C) = O(C')O(C").

I R.

This is not the above isomor-

Since R/p is a field,

is not necessarily

matrices

dimR/pS

= n.

1.

For example

0(S) = pdimR/pS if

~

= Mn(R) , the

over R, then a simple module S with pS = 0 has We want to show this also happens

case provided

R/p is finite.

Theorem 7.1.

Let R be a Dedekind

A

only

If S is simple and p C R is the unique prime with pS = O,

then S is an R/p module.

n~n

to R/Pi.

theorem and depends

If 0 - ~ C '

a composition

of R submodules

be a maximal R-order

ring with quotient

in a central

dimKA = n 2, R/p is finite,

in the general

field K.

simple K-algebra

and S is a simple

A

A.

Let

If

module with

pS = O, then dimR/pS_ = n. Proof. A/pp~p module,

l)

Localize

= A/pA, nothing

a discrete

changes

A p is a maximal

= R/p.

order over Rp~

Since S is a simple A / P A -

and so we can assume R is local and hence

ring.

Next we show that we can assume R is complete.

be the completion

of R and K be the quotient

A

= K~KA~A

Then

and Rp/ppRp

valuation

2)

at p.

A

~ R~.

Then A / p ~

Let

field of R.

^

= ~ /p~

and R/pR = R/p.

-

We need to show that ~

129-

is a maximal order.

The remaining condi-

tions are clearly preserved by completion. Lemma 7.2.

Let R be a discrete valuation ring with quotient field

K.

be a maximal R-order in a semisimple, separable K-

Let ~

algebra. X

Then the c o m p l e t i o n ~

=

is a maximal order over R in

A

Proof.

Let ~ ~

be a maximal R-order in A.

We want to show

A

= A •

This is accomplished by

Lemma 7-7-

Let R be a discrete valuation ring, R its completion,

K be the quotient field of R, K the quotient field of R. a finite dimensional vector space over K, V = K ~ K a finitely generated R submodule of V. (1)

N = M~

(2)

If rk M = rk V, then RN = M.

Remark. ~

Since ~

is maximal, ~

= ~

A and (2) implies

Let el, ..., en be a basis for V over K (and hence also Let ml, ..., m r be a basis for M over R.

m i = ~ a i j e j with aij ~ K. ord gi $

O.

ord( E g i ~ i j ) $ i m = ~

Then

This proves lemma 7.2 since ~q'~ A is an order by (1) and

for V over K).

SO

V, and let M be

V is a finitely generated R module, and

A ~.

Proof.

Let V be

If m & M, then m = ~ g i m i

Thus m = ~ g i ~ i j e j min(ord@~i~) = s.

tjej, then ord t

Then

with gi ~ ~

and This shows that if m G M and

$ s for all j.

Let n E N = M ~ V .

Then

-

n

150

-

tjej with tj ~ K and we have just shown that oral(tj) $ s.

~

Hence N ~

~R~Sej

which is a finitely generated R module.

is any element of R with R ~ =

p).

(Here

Hence N is a finitely gener-

ated module since R is noetherian. We now prove part (2). basis for V over K. verse

(hij) say.

dense in M. tj L K.

Here we have r = n so the m i form a

Hence (aij) is an invertible matrix with in-

Thus ej = ~, bjkm k.

Let m a M.

We want to show that N is

Then m = Z g i m i

= Zgiaijej

We approximate the tj by t~ t K.

since tj' ~ K.

= ~tjej

Then x = ~ t ~ e j

with

~ V

But ~ t ~ ej = ~ g ~ aije j = ~ g ~ m i and the gi' =

~,bkit ~ are close to the gi = ~ b k i t k "

Hence if the t~ are picked J

close enough all the ord g~ will be positive since ord g i ~ O . Hence all the gi'

R.

Thus x = ~ g ~ m i £ M.

Hence x L M ~

V =

N.

Since x can be made arbitrarily close to m by taking the t~ close to t i we can approximate m i N by elements of N and so N is dense in M. Hence in the proof of Theorem 7.1 we can assume R is complete local discrete valuation ring. of Theorem 7.1. r~

We now proceed with the proof

By Wedderburn theory we know that A = Mr(D) , the

r matrices over a division ring D.

Lemma 7.~.

If ~

mal order of A.

is a maximal order of D, then Z r ( ~ )

is a maxi-

- 151 -

Proof.

Let A

7)

(d a D1 R.

be a maximal order containing M r ( ~ ) and let ~ @ s & A ).

Also A 0 D ~ ,

of D containing ~

D and is finitely generated over

Then A 0

A OA Oc

A 0 and i g

so A 0 = ~ .

~0"

Hence ~ O

Let (tij) ~ A

is an order

with tij g D.

Then

z A°

where the first matrix

has a 1 in the k-th column and 1st row

and the last has a 1 in the 1st column m-th row. all i and j.

Hence Mr( p ) = A -

Thus tij ~ ~

for

This proves Lemma 7.4.

We now observe that it will suffice to prove Theorem 6.1 for a single maximal order A P = A'~,

in A.

If A' is another and

then Th. 5.15 shows that the functor S%--%- P ~ A

gives an equivalence between the categories of modules.

modules and ~

Since R is local, P ~ A

module by Theorem 5.27 so P ~

as a

S ~ S as an R-module and

our assertion follows immediately. We can now assume A Mr(~IPV)

-

Since simplicity can be defined in terms of the category,

simple modules will correspond. right A

A

S

= mr(~).

Now Mr(~)/PMr(]'1 ) =

and factoring out the radical gives Mr< ~IE)-

If S' is

-

a simple

~/~

module,

132

-

then S = (

"

Is i ~ S') is a simple

sT Mr(~/P) for ~ ,

module

and dimR/pS = r dimR/pS'.

then dimR/pS'

= n/r,

to prove the theorem is a division A/~

so dimR/pS

= n.

in the case A = D.

it is enough

DVR.

This implies that

To see this note that A / P

so ~ / P

= M S (division

S ~ 1.

If not there is an idempotent

Since R is complete,

Hence,

ring) by Wedderburn

theory.

e t A/E

is simple and We want to show

with e # O, 1.

we can lift e to an idempotent

e' ~

Since e ~ O, 1 we have e' # O, 1 but this is impossible e' ~ A, a division finite module ~S~

ring.

Hence

over R/p.

and so ~ / ~

is a 2 sided ideal of ~

(by Wedderburn's

and p ~

~ a r+l.

~a

Mapping

It is clearly ta r L ~ r + l

= ~ = ~ ~/P

onto.

~

~/p~

= pe for some e.

Assume

ring and is a so is

Now p ~

Since R is

= ~a~

this for some r.

so pr+l = ~ a ~ a

_pr/pr+l by l ~ a

ring.

P ~p20

ef where f = d i m R / p ~ / P =

Therefore

r = ~aa r =

if and only if But we can cancel the

t = ua ~ ~ a

... ~ pe ~ P A

dimR/pS.

Since

r gives an isomorphism.

we have ta r = ua r+l for such t.

as ~ ~

since

theorem).

Since ta r = 0 in pr/pr+l

a r since A is a division filter

.

Let P = ~ a.

We claim that pr = ~ a r. is 2-sided,

is a division

A

Since R/p is finite by hypothesis,

is a field

local, ~ is principal.

~/~

is true

That is, we can assume A

ring and R is a complete

is a division ring.

If the theorem

.

= P.

We

Thus d i m R / p ~ / p A ~

-

133

-

We want to show f = n. n2 = dimR/pA/p and 2) f ~ - n . ~

~ = ef.

Hence it is enough to show that l) e ~

For 1), let p = R ~ .

= ~ a e we have ~ =

of ~ .

Now K ( a ) ~

We know that

Since P = a

ua e and a e = v ~

Also

index K(a)/K) ~

(residue class degree)

which is a finitely generated R module.

u is integral

over R.

Similarly v is integral over R.

Let R' be the integral closure of R in K(a). uv = 1.

=

Note there is only one prime over p since R is com-

Also R [ u ] C A

Therefore,

We claim that

index of K(a)/K is greater than or equal to e.

But (ramification

plete.

and con-

[K(a): K] ~_. n since every element

of A satisfies an equation of degree ~ n over K.

[K(a): K].

= Ee =

where u and v are units

A is a field since K(a) is commutative

tained in a division ring.

the ramification

a and p ~

n

Since ~ =

ua e and a e = v ~ ,

Hence the ramification

Then u, v ~ R' with

we have o r d ~ =

index of K(a) over K is

be

e ord a in R'. and e ~ n as

claimed. To prove f ~ implies that a / ~ hence

~/P

n we need the fact that R/p is finite. is a finite separable

= (R/p)(b).

(Note:

were algebraically closed.)

This

field extension of R/p and

this would also be true if R/p

There exists a g ¢ ~

with

b ~ g mod ~ and such that g satisfies a monic equation of degree n with coefficients

in R, i.e., its minimal equation.

Therefore,

b

satisfies the equation obtained from this by reduction mod p and so

-

f = [A/~:

R/p]

as desired. Hence

= deg b ~ n .

-

Hence e, f ~ n and ef = n 2 so f = n

This completes the proof of Theorem 7.1. O(S) = pn.

Thus if C is a torsion

O(C) is an n-th power in I R. Definition.

134

~

(Use a composition

module then

series for C.)

Let C be a finitely generated torsion ~

module.

the reduced order ideal of C is defined to be ~ r ( C ) If S is simple and pS = O, then r: Ko(T~) ~ I R

= p.

is the desired isomorphism.

Ko(T A) ~ K o ( A ) reduces to I R - ~ Z and Ko(T~)

~r(C)

~Ko(A)

KO(~)

~

Z -~

is isomorphic

to I R.

~

Then

= O(C) 1/n. Therefore

The exact sequence

0

0 since Ko(A) is isomorphic

to

We must now determine the kernel

of I R

Lemma 7-~-

The kernel of the map Ko(T ~) ~ K o ( A )

is generated by

all [C] such that there exists a finitely generated free module F and an exact sequence 0 - ~ - F - ~ - F - b - C Proof.

Clearly all such [C] belong to the kernel.

x £ Ko(TA) go to 0 in K O ( ~ ). C, D G

~

TA •

Let O - ~ - P - ~ - F - m - D - ~ O

O. Let

be exact w i t h F finitely

Since D is torsion,

R such that sD = O.

Hence s F C

D' ~

-

Write x = [C] - [D] for some

generated and free.

0 ~

~

there exists an s # 0 in

P and we have an exact sequence

F/sF --m- D ~ 0

.

-

Thus x = [ C ~ D ' ]

- [D~D']

135

-

and [ D ~ D ' ]

= [F/sF].

the expression x = [C] - [D] for any x I Ko(T ~ ) has the form F/sF. of the specified [CS since

Therefore

elements.

If

in

we can assume D

-~PD -I-0 so [DS is one

x goes to 0 in K O ( ~ ) then so does

[D] clearly does.

Thus we are reduced KO(~)

0 -~-F - ~ F

Therefore,

to showing

then there is a finitely

quence 0 -~- F-~- F -D-C -~-0. 0 ~P-w-F K O ( ~ ) but

generated

free F and an exact se-

Choose an exact sequence

-~-C -*-0 with F free.

Then

[C] = 0 so KF] - [P] = O.

such that F' ~ P

that if [C] goes to 0 in

is isomorphic

[C] = IF] - [P] in

Therefore,

to F' ~ F .

there exists F'

Hence there is an exact

sequence 0 ~

F~F'

~

F(~F'

~

C ~

0

as desired. Let

[C] E Ker(Ko(T ~ ) ~

0 -~-F f-~F -~-C ~ 0 f ~ End(F)

be an exact sequence

It is sufficient

is a free R-module.

and so for Ms(A)

over K.

Then

Clearly f is

if and only if f is a unit in M s ( A ) ~ M s (

the order ideal of M s ( A ) / f Z s ( A )

so ~

as in the lemma.

= MS( A ) where F is free on s generators.

a monomorphism

norm of f.

K O ( ~ )) and let

~ ).

Now,

is (N(f)) where N(f) is the usual

to check this locally.

Assume R local

Let (el) be a base for M s ( ~ ) Let fe i = ~ a i j e

j.

Then N(f)

over R = det(aij).

-

136

-

Since R has been localized, (aij) is equivalent to a diagonal matrix diag(dl, ..., dr).

Thus M i ( A ) / f M s ( A )

~ II R/d i) as an

R-module so its order ideal is (dl, ..., dr).

Now s

Ms(A)

= F~...

~F

(s times).

Therefore, N s ( ~ ) / f M s ( A )

= ~F/fF i

and so s

s

(N(f)) = o(jl1 F/fF) = O(Jl1 C) = O(C) s =~r(C) ns. If n(f) is the reduced norm of f, then N(f) = n(f) ns since dimRZs(A ) = n2s 2.

Hence~r(C)

= n(f).

Thus we have an exact

sequence O -~-X -~-I R - ~ K o ( ~ ) -~DKo(A ) - ~ 0 where X is generated by all (n(f)) where f A Ms(A ) and f is a unit of Ms(A). Let a be a unit in Ms(A). such that ra • M n ( ~ ) . Ms(A).

Then there exists an r # 0 in R

Now r.I ~ M s ( A )

Thus n(ra) and n(r) are in X.

(n(a) = n(ra)/n(r) • X.

and r.I is a unit in

Since X is a group,

Thus, X is generated by all (n(f)) where

f is a unit in Ms(A) as claimed. Let Ms(A)* = GLs(A) be the group of units of Ms(A).

Define

GLs(A) -~- I R by a~-~-(n(a)) and let Hs be the image of this. Then we have an exact sequence

E

Hs -~I R -~Ko(~)

-~ Ko(A) -~0

.

-

137

-

Later we will show this is a part of an exact sequence of K-theory. We must now determine i.e., an algebraic

the He.

Assume that K is a global field,

number field or function field of dimension 1

over a finite field Our problem is to find the image of n: Ms(A)* ~ K ~(A)

~.

Since

is a central simple algebra over K, it will suffice to de-

termine the image of n: B* -~-K * for any central simple K-algebra B.

If y is a prime

By = K y S K

(i.e., a valuation)

B denote the completions

Definition.

of K, let Ky and

of K and B at y.

We say that y is unramified

in B if and only if

By ~ Mn(Ky), a matrix algebra over Ky. In particular,

if y is a real archimedian prime,

and By = Nn(l~ or M m ( ~

then K

where ~ denotes the quaternions.

y is ramified if and only if B ~ M m ( ~ .

Such a

The following classical

result answers our question. Theorem 7.6.

(Hasse-Schilling

norm theorem).

field and B a central simple K-algebra. n: B* - ~ K * consists

Let K be a global

Then the image of

of those a ~ K* which are positive at each

real archimedian prime which is ramified in B. Corollary 7~7.

If char K I O, n: B* -~-K*

is onto.

To say that a is positive at y means that a corresponds a positive real number under the unique isomorphism K y ~ It is easy to see that the condition is necessary. a commutative diagram

to

R. We have

-

B"

158

~

K

-

*

It will suffice to show that the image of Mm(~B)* ~ P ~

con-

sists of the positive reals.

But n is continuous and Mm( ~ "

connected topological group.

Our assertion clearly follows from

this.

To see that ~ ( ~ ) "

is connected let A ~ M m ( ~ ) * .

A to diagonal form D = diag(dl, colnmn operations.

Reduce

..., dm) by elementary row and

Then A and D are connected by a path in Mm(~)*.

For example, if we replace the i-th row r i by r i + ~ r j , we can instead replace r i by r i + t ~ r j ,

~

• H,

0 ~ t d~_l. This gives a

path between the initial and final values. connected.

is a

Now H~* = ~ 4 _

Let di(t) be a path from d i to 1 in ~'.

(0) is

Then

D(t) = diag(di(t)) gives a path from D to I. The sufficiency of the condition will be proved in Chapter 9. Recall now the following definition used in class field theory. Definition.

If ~ =

Yl "'" Yr is a formal product of real archi-

median primes, the ray m o d ~

is defined to be S ~ =

((a)la~O

at Yl' "''' Yr )' where (a) = Ra. Since y ramifies in Ms(A) if and only if it ramifies in A, Theorem 7.6 shows that for each s, the image Hs of Ms(A)" in IR is precisely S ~

where ~

is the (formal) product of all real archi-

median primes ramified in A.

- 159-

7.8.

Theorem let ~

Let R be a Dedekind ring with quotient field K and

be a maximal R-order in a central simple K-algebra A.

Assume K is a global field and let ~

be the product of all real

archimedian primes of K which ramify im A.

Then we have an exact

sequence 0 ~IR/S

~

~

KO(A) ~Ko(A)

~

0

so Ko(A ) • m O ZR/S The class group of R is Cl(R) = IR/P R where PR = ((a)laeK*}. If char K / O, then IR/S ~

= Cl(R).

If char K = 0, we have an

exact sequence 0 ~PR/S~

--IR/S %

~CI(R)--0

.

For real archimedian y, define Ky* -b. ~/2Z~ by Ky* ~ R~*-b-R~*/R~ *+. Let D -- ~ / 2 ~

over the y I E

K* ~y,..~FJY* ~ y I ~ T tion theorem. of X in PR"

z~/2~, -- D.

This is onto by the approxima-

If X is the kernel of this, then S ~

is the image

Now K* -~'PR is onto and its kernel is clearly U(R),

the group of units of R. D/im(U(R) -~P D). U(R)

and define K* -~PD by

Therefore we have P R / S ~

Thus we have an exact sequence

--a~D ~ I R I S

R~

~CI(R)

~ 0

.

= K*/U(R)X =

-

140

-

Since D and CI(R) are finite (by the Jordan-Zassenhaus theorem), so is I R / S R .

Chapter 8:

(Of course this was already known by Theorem 3.8 .)

K 1 and G 1

If R is any ring then GLn(R) is, by definition, the group of all invertible n x

n matrices over R. 0 A

GLn+l(R) by sending A to (

0 ). 0...0 1

We embed GLn(R) in

Let GL(R) = lim GLn(R).

This is the group of infinite invertible matrices of the form

l

©

IA 1 1

o

where A is finite. Definition.

KI(R) = GL(R)/[GL(R), GL(R)].

We refer to SK Ch 15 for elementary properties of this and in particular for the fact that [GL(R), GL(R)] = E(R), the subgroup of GL(R) generated by all elementary matrices of the form i + reij , r E R, i ~ j. The main object of this chapter is to give a partial calculation of K l ( A )

for an order A .

Before doing this, we will

generalize some of the results of Chapter 7.

-

141

-

Let R be a Dedekind ring with quotient field K. an R-order in a semisimple separable K-algebra A. we determined the kernel of KO( ~ ) ~ K o ( A ) maximal order.

M~N.

In Chapter 7

~ K~N

is any order, does not imply

Let P be a finite non-empty set of primes such that A y

a maximal order for y ~ P. M~N.

be

for the case of a

In the more general case where A

this argument will not work because K ~ M

Let ~

Then M y ~

Therefore we consider K O ( A )

is

Ny for all y @ P implies ~

K O ( A v) instead of pEP

KO( A ) --~ Ko(A)More generally, let M be a finitely generated torsion free -module.

As in Chapter 6, we let D M be the abelian group gener-

ated by all IN] where N is a direct summand of M S for some s, with relations [ N I ~ N

2] = [NI]+ IN2].

Thus K O ( A )

= D~ .

If we l e t ~

We will

determine the kernel o f ~ M ~

II D~ . YtP y

be the full

subcategory of all A - m o d u l e s

which are direct summands of M s for

some s, we can describe D M as K o ( D M , ~ ) . As in Chapter 6, let J be the annihilator of KM in A and let i = A/J,

A

= ~/A~

J.

Then M is a faithful ~ -module.

S i n c e ~ M and so D M is the same whether we use

~

or A ,

we can

assume without loss of generality that M is a faithful ~ -module. Let P be a finite non-empty set of primes of R such that is a maximal order for y ~ P.

Let T ~ ,p be the category of fi-

nitely generated torsion ~ -modules with no y torsion for y e P.

Y

-

If C ~ T ~ , p

-~-0 with NI, N 2 E ~ N ,

= CNlJ - K N 2 S ~

Lemma 8.1.

-

and if there is an exact sequence

0 -~-N 2 -~-N I - ~ C Y(C)

142

(i)

then

D M is well defined by Lemma 6.9.

If C ~ T A ,p, there is an epimorphism M s -~-C ~ - 0 for some s.

(2)

If 0 -~-N 2 -~-N 1 -~DC -~D0 with C G T A , p N1 ~ ~ M '

then N 2 G ~ M "

(3)

If N I G ~

M and N 2 ~

(4)

If O - ~ - C '

-~-C - ~ C "

~(C) = ~(C') homomorphism Proof.

If C is simple,

implies (1) for C.

~C"

+ ~(C")

so ~

gives a well defined

~: K0(T ~,p) -~-D z. We now use in-

It will suffice to consider an exact ~0

and show that (1) for C' and C"

More generally given epimorphisms

f': N' ~s-C', f": N" - ~ C " , f: N" - ~ C

M.

-~-0 in T A ,p, then

(1) is true by Lemma 6.7.

duction on the length of C. sequence 0 -~-C' ~ C

N1, then N 2 G ~

and

with N', N" E ~ M, we can lift f" to

such that N t!

C -J-~C" commutes.

~

0

This argument was used in the proof of Lemma 6.9.

It

-

143

-

is now trivial to show that (if', f): N' ~ N " phlsm.

This completes

-IPC is an epimor-

the proof of (1).

We also get a diagram

0

0

0

o

Ni

Nx

Ni

0 ~

N' ~ N '

~N"

4

4

0 ~C'

~

C

0

~

N" ~ 0

4, ~ C "

0

~ 0

0

If we have (2), the modules in the top row lie i n ~ M. top row splits. Y

P,

We claim the

By Lemma 6.14 we can check this locally.

For

A y is maximal and so N~

so the top row becomes

is projective. If y ~ P, Cy = 0 Y isomorphic to the middle row locally at y.

Since the top row splits,

[N l] = [N~] + [N~] in D E and we see

immediately that (4) is satisfied. If we localize the sequence N2y~

Nly since Cy = O.

implies

in (2) at any y ~ P, we get

By Lemma 6.5, N 2 ~

N 1.

Therefore

(3)

(2). It remains to prove

Since N l ~

(3).

We use the argument of Theorem 6.8.

N 2 we can find an exact sequence 0 -~-N 1 -~-N 2 -hJ-C -I~O

with C ~ T ~ , p

by Lemma 6.6

O - ~ - P - m - M s -~-C -~-0.

.

By (1) we can find

By Lemma 6.9, N 2 ~ P

• Nl~MS.

Since N I

is a summand of M t for some t, we see that N 2 is a summand of M s+t.

Lemma

8.2.

Ko(T ~

,p) ~ D

M~

TT D~ is Y~

generated by all

[C]

If C E T ~ , p ,

then as above N l ~

is

with C 6 T ~ ,p such that there is an exact

sequence 0 - ~ M s -~-M s -~-C ~ 0 Proof.

exact and ker ~

y

for some s.

0 --q~N 2 ~ N

1 ~ C

N 2 so [N l] - [N 2] ~ 0

~ 0

with N1, N 2 ~ D M

in D~ . Y

A

versely that x = [N l] - [N 2] ~

in DMy for y E P.

by Lemma 6.5 and Theorem 5.20.

Suppose conThen N l ~

N2

As above we find

0 -m-N 2 -4~N I -,-C -~-O with C ~

T~,p

using Lemma 6.6

•

There-

fore x ~ ~ ( C ) . If O -~-M s --b-Ms -~-C -~-0 then clearly ~(x)

= 0 where x = [C] - [D] ~ Ko(T ~ ,p).

0 ~m-N - ~ - M s -~-D -~-0.

:

y would imply ~ C

[c~o']

0 -~-M s ~ erators.

- [DQo,]

P.

(If ~

y some y ~

is the ~nnihilator of D, P but D

:

[cOD']

- [MS/rMS].

= 0.)

Therefore,

Now ~ut

M s - ~ MS/rM s -~-O so [MS/rM s] is one of our chosen genAlso this shows 0 = ~ ( x )

We now claim that

0 -~-N -~-M s -~-C -~-0. N~M

= ~[C ~D'].

~ [C] = 0 implies the existence of a se-

quence 0 -~-M s -~-M s -~-C -~-0.

M s ~M t~

Let

By Lemma 8.1 (1), find

Y we have 0 -~-D' -~-MS/rM s -~-D -~-0 for some D'. x

= O.

Since Dy ~ 0 for y ~ P, there is an r # O

in R with rD = O, y~r for y ~ C

~(C)

In fact, by Lemma 8.1 (1), find

Now ~ [ C ]

= [M s ] - IN] = 0 so

t for some t and 0 - ~ - N ( ~ M t -~-M s ( ~ M t -~-C -~-0.

This proves the lemma.

-

Now let S = R - U yGP only if C S = O. automorphism

y.

145

-

Then C

= 0 for all y E P if ~ d Y

If 0 -~-N s f--4-Ms -~-C -~-0 we see that f induces s of M S.

Conversely,

choose t ~ S and f : M S ~

M s so~=

if ~

is an automorphism

f/t.Since

of M~,

(ker f)s = O, ker f is

a torsion module but M is torsion free.

Thus f is a monomorphism.

Let C = ckr f, 0 -~-M s -m-N s -~-C - ~ 0 .

Then C S = 0 so C ~ Tp, A .

Also M S / t M S ~ d(~)

T A , P.

Define d: A U t A s ( M ~ ) - - ~

= KC] - [MS/tMsS.

K o ( T A , P) by

We claim that d is a well defined group

homomorphism. To prove this we use the following well known result. Lemma 8,3. O-~-ker

If A f - - ~ b B - ~ C

f-~-ker

to these compositions If~

o

~d

~ f'/t'

gf ~

then t'f = tf'.

Sply

Lena

8.5

= g/u, Lemma 8.5 on fg and tu shows

+ d(p

>. Then E n d A s ( ~ )

= Ms(E) so

(M~) = GLs(E ) so we have a homomorphism S

It is trivial

ckr g ~ 0 .

also to t't = tt' to see that d is well

Let E = E n d A s ( M S ) . Aut~

exact sequence

gf -b-ker g -~-ckr f - ~ c k r

If we also have ~

defined.

then we have ~

d: GLs(E) -~-Ko(T

). A,P

to see that

GLs(E)

I

"--.°

GLs+I(E)

---d-a~K0(T A,p)

-

146

-

commutes, so we have d: GL(E) -~-K0(T A ,p)o tive, this induces Pr°p°si~i°n 8~4. able K-algebra A. such that A y

~:

Since K 0 is commuta-

El(E) -~-K0(T A ,p).

Let A

be an order over R in a semisimple separ-

Let P be a finite non-empty set of primes of R

is a maximal order for y ~ P.

generated ~ -module and E = End A s(MS).

Let N be a finitely

Then we have an exact

sequence

KI(E) ~Ko(TA,P) ~DM -a'lT D~ . FGP

y

This is immediate from the above results.

An exact sequence

of this type was considered by Heller and Reiner for ordinary Ktheory.

The generalization to D M is due to Jacobinski.

In Chap-

ter 9 we will apply this result in a special case where E is semilocal (i.e., E/J is artinian where J is the Jacobson radical of E). In this case we can replace Kl(E ) by U(E) the group of units of E using Proposition 8 ~

(Bass).

If E is a semilocal ring, then U(E)

GLI(E) --~-Kl(E) is an epimorphism. Proof.

A more general result is proved in SK, Theorem 13.5.

We

will give the proof for the semilocal case again since it is very simple compared to the general case. Bass 8K Lemma 11.8, i f ~ then there is an a E ~

By a very useful lemma of

is a right ideal of E and Ex + ~ t such that x + a = u is a unit of E.

= E,

-

Suppose (aij) E

GLn(E).

147-

Let x s all and ~

= al2E + ... + alnE.

We get a unlt u = all + al2Y 2 + ... + alnY n.

Thus by elementary

row and col!,mn operations we reduce all other entries in the flrs~ row and col~m~ to O. diagonal matrix. can reduce

Repeating, we eventually reduce (aij) to a

Since all (~

x1 ) to ( "i (aij

~-l) ~

)~

E(E) by SK Lemma 13.3 we

GLI(E)"

"l We now recall Bass' categorical definition of Kl(R).

We

consider pairs (P, f) where P is a finitely generated projective R-module and f is an R-automorphism of P°

A map

g: (P, f) ~ ( P ' ,

such that the diagram

f') is a map f: P - e ~ P ' p---~p, f~

,~f'

P ---~P'

commutes.

Let Kl(R) be the abelian group generated by all such [(P, f)S with the relations

i)

[(P, z)] = E(P', f')S + [(P"

f")S if there are maps

i: (P', f') --~-(P, f) and j: (P, f) -~-(P", f") with 0 -~-P' i~.p ~ p , , 0-4~(p,, 2)

-~-0

exact, i.e.,

f') -~-(P, f) -~-(P", f") -~-0 is exact• [(P, fg)] = [(P, f)] + [(P, g)] where f and g are both

automorphisms of P.

-

TheoremS.6.

148-

KI(R) under the definition given is naturally isomor-

phic to the Whitehead group GL(R)/[GL(R), GL(R)] = GL(R)/E(R) (the definition of Kl(R) we have used previously). This is one of the standard facts of K-theory.

See SK Chap-

ter 13. As we shall see, the ability to switch back and forth between these definitions will be crucial in many proofs. If R is a commutative ring then Ko(R) is made into a ring by setting [P][Q] = [ P ~ R Q ] ,

and Kl(R) is a Ko(R) module by

[P][(Q, g)] = [ ( P ~ R Q, l ~ g ) ] .

If R is commutative and A

are R algebras then we can define K O ( A ) the same map, [P][(Q, g)] = [ ( p ~ R Q ,

x Kl(~)

and

-m'KI(~RU

) by

l~g)].

If R is any ring and R' = Nn(R) then Kl(R) = KI(R' ) since Mk(R') = Nkn(R) and hence GLk(R' ) = GLkn(R ). If A is a simple ring, then A = Mn(D) for some division ring D and Kl(A ) = Kl(D). field K.

Let A be a central simple algebra over a

Then the reduced norm n: Mr(A) -w-K gives a homomorphism

GLn(A) = Mn(A)* - ~ K * . fact, if L ~

The diagram

K splits A then

G L~+ n(A)~K* GL 1 ( A ) ~

commutes.

In

-

Mr(A)* ~

(L@

K+

l#e

-

KMr(A)) * = M r ( L ~ K A ) *

~

G L r ( ~ (~ KA) ~

K*

commutes by the definition of n, and GLn+I(LQKA)~det clearly commutes.

Thus n defines n: KI(A) -~PK*.

The image of

this is known when K is an algebraic number field by the norm theorem.

S. S. Wang has shown in the algebraic number field case

the kernel is 0 (BK V, Theorem 9.7). We now give a partial extension of this to arbitrary fields. A map f: A -~-B of abelian groups is called an isomorphism mod torsion if ker f and ckr f are torsion groups or equivalently if l~f:

Q~A

Theorem 8. 7 .

~-~Q~

B.

Let K be any field and A a central simple K algebra.

Then n: Kl(A) -~-K* is an isomorphism mod torsion.

In fact ker f

and ckr f are ~n~ihilated by m where m 2 = dimKA. Proof. x ~ x A.

The composite map K* C m where dimKA = m 2.

Then

A* = GLl(A) ~ K l ( A )

For let L ~ K

-~K*

is just

be a splitting field for

-

15o-

"L* r

K •

A* ~ g , -

(L~)KA)*

= Mm(L)"

KI(A) ~

KI(L~ ) K A)

n~

~det

K*

'~-- L*

commutes.

But x ~ K* goes to xI in Mm(L)* which goes to x m ~ K +~

L +.

Hence x ~ x

Kl(A) ~ K *

m as desired.

Kl(Mm(L)) = Kl(L) = L'.

Thus the cokernel of

plete the proof.

split A.

Then K l ( L ~ K A )

Hence if x ~ Ker(Kl(A) ~ K * )

x ~ Ker(Kl(A) - 4 ~ K l ( L ~ K A ) ) .

=

then

Thus the following lemma will com-

We remark that if A = Ms(D) w i t h D

ring, we may let L be a maximal subfield of D°

a division

Thus [L: K]IM.

Let L be a finite extension of a field K of degree r.

Then the kernel of Kl(A) - ~ K l ( L ~ K A ) Proof.

But

is annihilated by m.

For the kernel let L ~ K

Lemma 8.8.

L*.

Let [(P, f)] & K l ( L ~ ) K A )

projective L ~ K A

has exponent dividing r.

where P is a finitely generated

module and f is an automorphism.

L~KA

is a

free finite&y generated A module since L is free and finitely generated over K. Hence, P is a finitely generated projective A module and f is an A automorphism, so we can define [(P, f)] ~ Kl(A).

This

-

151

yields a well defined map K I ( L ~ K A ) El(A) - I b K I ( L ~ K A ) L~K

-~Kl(A)

-

--~-KI(A).

The composite

is multiplication by r = EL: KS since

r ~ P as an A-module. I

P~

Thus if x & kerCKl(A) - ~ - K I ( L ~ K A ) S

then rx = 0 in KI(A)

as desired. Corollar~ 8. 9 .

If A is a semisimple algebra with center Z then the

kernel and cokernel of Kl(A) - - ~ K l ( Z ) bounded exponent.

are torsion groups of

(Kl(A) = Kl(A l) x ... x KI(A n) where A =

A1 x -.. x An.) We will now show that a similar result holds for orders. Let R be a Dedekind ring with quotient field K, A a separable semisimple K algebra with center C, and A R.

Then we have maps KI( A ) -~-KI(A ) ~ C * .

gral closure of R in C. contained in R'*.

For everything is well behaved on products so we

We can replace ~

R'-order.

Let R' be the inte-

Then we claim the image of K I ( A ) in C* is

can assume that A is simple. ring.

an order in A over

Thus C is a field and R' a Dedeklnd

by the bigger R' A , so we assume ~

For any s, M s ( A ) is an R' order.

Thus n: M s ( ~ )

is an -*-R'.

We can now state the main result which slightly generalizes a result of Bass. Theorem 8.10.

Let R be a Dedekind ring with quotient field K, A a

separable semisimple K algebra, A

an R order in A, C the center of

-

152

-

A and R' the integral closure of R in C.

Then n: Kl(L) -~-R '~ is

an isomorphism mod torsion if l)

R/~

is finite for every non zero ideal ~

, and

2)

If L/K is a finite separable extension and R 1 is the integral

closure of R in L, then the class group of R 1 is a torsion group. Remark.

This implies rk K l ( A )

= rk R ~ for such orders.

We recall the definitions of the relative K 1. ring and I a 2-sided ideal of R. ker[GL(R) ~

GL(R/I)].

Let R be any

Then GL~R~ I N =

We let ~

be the smallest normal sub-

group of E(R) containing all I + qeij

i / j

q L I and let

KI(R , I) = GL(R, I)/E(R, I). Proposition 8.11. ideal of ~

Under the hypothesis of Theorem 8.10, if I is an

with K I - - A ,

then K I ( ~ ,

I)-~KI(~)

is an isomorphism

mod torsion. Proof.

GL(~/I)

GLn(A/I)

=~.~ G L n ( ~ / I ). But A / I n=l is finite for each n and G L ( ~ / I )

Also E ( A )

= [GL(A),

GL(~)]

and E ( ~ ,

is finite hence is locally finite.

I) = [ G L ( A ) ,

GL(~,

I)].

See SK, Theorem 15.2 and Theorem 15.1. Hence it is enough to show the following lemma: Lemma 8.12.

Let G be a group and N a normal subgroup with G/N lo-

cally finite.

Then N/[G, N] -~-G/[G, G] is an isomorphism mod

torsion. Proof.

(Assuming results from cohomology of groups. ) O ~N

-~G

-~-G/N -'~0

is exact.

- 153-

Letting Hn(G) = Hn(G ~ Z) we get an exact sequence H2(G )

-~H2(GIN) -~-Ho(GIN , Hl(N)) -IbHl(G) --~HI(GIN) -~-0.

But

Hl(N) = N/IN, N], Ho(G/N , Hl(N)) = N/[G, N], and Hl(G) = G/[G, G] so H2(G ) -IPH2(G/N) - ~ N / [ G ,

N] -IPG/[G, G] -4PG/N[G, G] -~-0.

Hence it is enough to show that HI(G/N) and H2(G/N) are torsion. But G/N is locally finite.

Hence G/N = lim H i with H i finite.

Therefore, Hn(G/N) = lim Hn(H i) and each Hn(H i) is finite. HI(G/N ) and H2(G/N) are torsion.

Thus,

This completes the proof of

proposition 8.11. Keeping the above hypothesis. A then so is A ~

~.

If ~

and ~

KI(A~ ~) -~KI(A) and

If

are R orders in KI(A~

~ ) -~-

K I ( ~ ) are both isomorphisms mod torsion, the diagram

KI(A)~---- K l ( A ~

R

for A

~

Kl( ~ ) shows that Theorem 8.10 holds

~

if and only if it holds for ~ .

Lemma 8.1~.

Let G be a group and N a normal subgroup such that G/N

is locally finite.

Proof.

~)

[G~ N ] C

~

N t'~ [G, G]/EG, N I t

Then [G, G]/[G, N] is locally finite.

[G~ G ] C

[G, G].

But

ker[N/[G, N] ~ G / [ G ,

torsion and abelian by Lemma 8.12.

G]] and therefore is

Hence N f'~ [G, G]/[G, N] is

- 154 -

locally finite.

Now [G, G S / N ~ [ G ,

G/N which is locally finite. cally finite. Lemma 8.14.

G] is the image of [G, G] in

Hence [G, G]/Nf~ [G, G] is also lo-

Apply the following lemma. If B is a group and A a normal subgroup with A and

B/A locally finite, then B is locally finite. Proof.

Let C be a finitely generated subgroup of B.

image C / C ~ Thus C g ~ A ated.

Then the

A of C in B/A is finitely generated and hence finite. has finite index in C so C ~ A

is also finitely gener-

Thus C t'~ A is finite since A is locally finite.

Propositio n 8.1~.

With the hypothesis of Theorem 8.10, if A C

are both R orders of A then K l ( A ) - 4 ~ K l ( ~ )

is an isomorphism mod

torsion. Proof.

Pick r # 0 in R with r ~

ideal of both A both ~

and ~

and ~ . then G L ( A ,

then I = r ~

is a 2-sided

We claim that if J is a 2-sided ideal of J) = G L ( U ,

J).

In fact, G L n ( A ,

J) is

the group of invertible matrices of the form 1 + Q where Q has entries in J.

Thus it depends only on J and not on A •

We have a commutative diagram

Kl( A, J) - - - ~ K I (

KI(~)

P, J)

~.~--K l ( P )

By Proposition 8.11 the columns are isomorphism mod torsion. Hence the bottom will be an isomorphism mod torsion if the top is.

- 155 -

As above, pick r # 0 in R with r ~ C ~ J = 12 = r 2 ~ . E(~, AB)~

For any ideals A, B of ~ w e

[E(~, A), E ( ~ ,

, let I = r ~

and

have

B)] since 1 + ab eij =

K1 + aeik , 1 + bekj]. Hence E ( ~ , But G L ( ~ ,

J)~

I) = G L ( ~ ,

[GL(A, I), G L ( A ,

[E(~, I).

I)]~

I), E ( ~ ,

I)]~

[GL(~, I), G L ( ~

Hence [GL(~, I), G L ( ~ ,

[GL(A), GL( A ,

I)l

I)] =

I)] = E ( A ,

I)C

E(A).

The top row of * is GL(~,

J)IE(A,

J) - - ~ G L ( ~ ,

J)IE(U,

J) .

This map is onto since the GL's are the same and the kernel is E(~,

J)/E(A,

J).

calculation above.

But E ( ~ ,

J)/E(A,

J)C

E(A)/E(A,

J) by the

Hence, it is enough to show that

E(~ )/E(~ , J) is locally finite. Let G = G L ( A )

and N = G L ( ~ , J).

locally finite since ~ / J

is finite.

[G, G]/[G, N] is locally finite. [G, N] = E ( A ,

J).

Then G/N C GL(~/J)

is

Hence, by lemma 8.13,

But [G, G] = E(A ) and

Hence E ( ~ ) / E ( ~ ,

J) is locally finite as

desired. This completes the proof of Proposition 8.15. Now we return to the proof of Theorem 8.10. mal order ~ 3

~

.

Then K l ( A ) - ~ K l ( ~ )

torsion by Proposition 8.15.

We pick a maxi-


Hence K l ( ~ ) - - ~ R ' * will be an

-

i~6

isomorphism mod torsion if K I ( ~ )

-

-~R'*

is.

Everything is now well behaved on products since if A = A I x ... x Ar, then ~ A is simple and ~

= ~l

x ... x ~ r "

is a maximal order.

Hence we can assume

Now C is a field and R' is

a Dedekind ring finitely generated as an R-module. which is an order over R, we see that A A.

= R'~

Since ~

R'

is an R' order in

The hypotheses on R in Theorem 8.10 are inherited by R'.

Therefore we can replace R' by R and assume that A is central simple over K (and so R' = R). First we show the map is an epimorphism mod torsion. examine the maps R* ~ K l ( A ) K* ~ K l ( A x to xm where m 2 = dimKA.

~R"

.

We

The bottom composite sends

) ~ K * Hence, so does the top.

Thus

ckr[Kl(A ) -~-R*] has exponent dividing m. To show the map is a monomorphism mod torsion we begin with Lemma 8.16.

Let R be a Dedekind ring, ~

an R algebra which is

finitely generated as an R module, and R' a commutative ring which is finitely generated and projective as an R module. Kl(~ ) -~KI(R'~R~) Proof.

Then the map

is a monomorphism mod torsion.

If P is a finitely generated projective R ' ~ R A

then P is also a finitely generated projective KI(R'~R~)-~Kl(~) well defined.

given by [(P, f ) ] ~

A

module.

module, The map

[(P, f)] is clearly

The composite K l ( ~ ) -%- KI(R' ~ R

~ ) - ~ K l ( A ) is

-

given by [(P, f ) ] " ~ a ~ [ ( R ' ~ R P

157-

, l~f)],

multiplication by JR'] t Ko(R ).

which is nothing but

We claim there is a non zero in-

teger n and an x ~ Ko(R ) such that x[R'] = n[R]. y~

Kl(~)

goes to 0 in K I ( R ' ~ R A )

n[R]y = ny = 0 in K l ( a ) . Kl(A)

-a~KI(R'~RA).

Then, if

, then x[R']y = 0 so

Thus n annihilates the kernel of Now R' = F ~ a

as an R module where F is

a finitely generated free R-module and a is an ideal. [F~-~.

Then x[R'] = [ F ( ~ ' I ] [ F @ a ]

= [(F(~F)]

= [(F~F)]

CF.(~-I]

+ [a-l~)~]

+ [F][~)~

-I]

[(F(~F)]

+ [F][R~R]

claimed.

We use here the fact ~hat ~ ) ~ - i ~

Let x = + [F~a]

+

+ JR] :

+ [R] : n[R] for some non zero integer n as R~R,

and

Let L be a finite separable field extension of K such that L~KA L.

= Ms(L) for some s.

Let R' be the integral closure of R in

Then R' satisfies the hypothesis

of lemma 8.16.

The diagram

) R*

Kl( ' ® R A ) -- ~-~ R'*

But K I ( A ) - ~ ' K I ( R ' ~

R A)

commutes.

is a monomorphism mod torsion.

Thus, Kl( A ) -%-R* will be a monomorphism mod torsion if KI(R'~RA]

-~pR'*

is.

Thus, we can assume that A = Mn(K) for

-

some n. A

158-

By the remark just before Lemma 8.15 we can also assume

= Mn(R)-

Now Kl(Nn(R))

have already observed.

is naturally isomorphic to Kl(R) as we

This isomorphism,

given bY M k ( M n ( R ) ) ~

Mnk(R) clearly preserves n = det: K 1 -@bR*.

Thus it is enough to

prove the theorem for det: Kl(R) -@DR ~. By SK Theorem 15.5 the map GL2(R) to Kl(R) is onto.

Let

x i Kl(R) go to 1 in R*. a We claim we can pick (c ~) ~

GL2(R) representing x such that

a separable field extension L of K contains a root of the characteristic polynomial of (c

)"

If char(K) # 2, this is automatic.

If char(K) = 2, then the characteristic polynomial is separable unless a + d = 0.

If a + d = O, then we modify (~ ~) by the elem-

entary matrix (~ ~) which does not change the image in El(R). (~ ~)(~ ~) = (~

a+b~. d+c ~

polynomial unless c = 0.

The latter has separable characteristic If c = 0 and b # 0, then

(~ 0 a a b 1)(O ~) = (0 b+d ) has a separable characteristic polynomial. b = c = O, then a -1 = d since det(~ ~) = det x ~ 1. -1 (~

~) ~ E(R) by

SK, Lemma 15.5.

If

But

a0 Thus (0 a -1) is

equivalent to the identity which has its characteristic roots in K. Of course x = 1 in this case. in L.

Let R' be the integral closure of R

Then lemma 8.16 applies to the map Kl(R) ~ K I ( R ' ) .

Since

159

-

-

KI(R) -'--~KI(R') the diagram

det~

~det

R*

commutes, we can assume K con-

' "~----R'*

rains a root of the characteristic polynomial of (~ ~). Thus x I = (~ ~) has an eigenvalue in K. Kl@K

We break K 2 up as

2 where K 1 is generated by u with XlU = ~ u ,

is stable under x I and Xl-1. and X1-1.

~ ~ K and hence

Then P = R 2 f ~ K 1 is stable under x 1

Passing to the categorical definition of K 1 we have a

commutative diagram of exact sequences 0 ~ P

~ R

0 ~ P

~

2 ~

Q --~-0

Q ~ 0

where f and g are induced from x I and are isomorphisms. x = [(R 2, Xl)S = [(P, f)] + [(Q, g)S. free finitely generated.

Then

P and Q are rank 1 torsion

Hence P is isomorphic to ~ and Q to

where a and ~ are ideals and f and g must be multiplication by units r and s of R. x goes to 1 in K*. [(2,

~)]

- [(~, r-l) ] ~

Therefore, rs = 1.

-[(~, ~)]-

Thus

Hence,

x = ([~]

where Kl(R) is considered as a K0(R) module.

- [b])([R,

~])

But [al-[b] ~ Co(R),

the class group of R, which is a torsion group by hypothesis.

-

Hence,

x ~ image(Co(R)

- ~ K I ( R ) ) given by z ~ z

this image is a torsion group. pletes

the proof of Theorem

Corollary tegers,

8.17.

(Bass).

A a semisimple

But

order.

This com-

numbers,

Z the in-

a Z order in A.

Let q be

8.10.

Let Q be the rational

Q algebra,

of simple components

[(R, r)].

Hence has finite

the number of simple components number

160-

and A

of A = A 1 x ... x Aq, and r the

of ~ Q A

where ~ i s

the real numbers.

Then rank K l ( ~ ) = r - q. Proof.

Let C be the center of A.

be the ring of integers

Then C = K 1 x ... x Kq.

of K i and R = R 1 x ... x Rq.

Theorem 8.10 rank K I ( ~ ) =i=~rank R~.

By Dirchlet's

(W, 0M), the rank R'~ is one less than the number 1 primes

of K i.

ErkR~

of archimidean

primes

= ~ ( n i - l) = ~ n i - q.

ents of ~ Q A Corollar~

unit theorem

of archimidean

of K i is n 1.

~or

Thus

But the number of simple compon-

i is n i.

8.18.

Let G be a finite group.

number of irreducible irreducible

Then by

Now I R ~ Q K i = L 1 x ... x Lni with each L v ~

and the number

Let R i

rational

real representations representations

of G.

Then rank Kl(ZG)

= the

of G -- the number of

-

Remark.

If ~

161

-

is an order over ~ , a theorem of Siegel shows that

GLn(A ) is finitely generated for all n.

By SK, Theorem 13.5,

GL2(A)

Thus K l ( ~ )

-~PKl(A)

generated.

is finitely generated.

is finitely

(See BK X, Theorem 3.2.)

We will now obtain a result similar to Theorem 8.10 for G l ( A ). Definition.

Let A

be a left noetherian ring.

Then G I ( ~ )

is the

abelian group defined by generators [(M, f)] where f is an automorphism of the finitely generated left a

module M with the relations

(1) for every exact sequence 0 - ~ ( M ' ,

f') -a~(M, f) -~-(Z", f")-~-O

we have [(M, f)] = [(M', f')] + [(M", f")] and

(2) [(M, fg)] = [(M, f)] + [(N, g)]. The definition is just like that of K I except for the choice of the modules M.

Thus there is a natural map

~ : KI(A) --'~GI(A) by ~[(P, f)] : [(P, f ) ] . Theorem 8.1~. 2 sided ideal.

Let A

be a left noetherian ring and ! a nilpotent

Then we have an isomorphism G l ( ~ / I )

given by [(M, f ) ] " ~ - E ( M ,

f)].

module M can be regarded as a a

~-~-Gl(~ )

(This makes sense since each A / I module.)

-

162

-

We claim the map GI( A ) - ~ G I ( A / I ) [(M, f ) ] ~ ~ [ ( M i / M i + l ,

given by

f)] is the inverse of G l ( ~ / I )

~Gl(A

).

By the Jordan-Holder-Zassenhaus theorem any two such filtrations will have a common refinement.

Since the Jordan-Holder-

Zassenhaus construction uses just sums and intersections, the refinement will still be stable under f and f-1. will be preserved.

Hence l) and 2)

Therefore, to prove the map is well defined it

is enough to show that adding one extra term to the filtration does not change the image.

Say M i ~

N ~ M i + 1.

0 -~-N/Mi+ 1 -IPMi/Mi+ 1 -m-Mi/N - ~ 0 [(N/Mi+l, f)] + [(MI/N , f)].

Then

is exact and [(Mi/Mi+ l, f)] =

Thus the map is well defined.

If O --~(M', f') -~P(M, f) - ~ ( M " ,

f") -~PO is exact, take a

filtration for M' and N" and paste together to get one for N. that filtration shows the relations (1) are preserved.

use M ~ I ~ I ~

. . . D InM = O, f o r [(M, f g ) ] ,

r(M, f ) ]

Then

For (2),

and [(Z, g ) ] .

This shows relations (2) are preserved. Theorem 8.20 (Lam).

Let R be a Dedekind ring with quotient field K

and A be a separable semisimple K algebra such that l)

R/p is finite for every non-zero prime ideal p C

R, and

2)

If L is a finite separable field extension of K and R' is the

integral closure of R in L, then the class group of R' is a torsion group.

Then l)

~: Kl(A )

If ~

is an R order in A, the map

- ~ - G l ( ~ ) is an isomorphism mod torsion,

- 163

2)

If ~

-

is an R order in A containing A ,

then the

map G I ( ~ ) -~PGI( A ) is an isomorphism mod torsion, 5)

If C is the center of A and R' is the integral

closure of R in C, then the composite Gl( a ) -~-Gl(A) = Kl(A) n - ~ R ' *

is an isomorphism mod torsion, and 4)

R-orders a orders A Proof.

If Kl( A ) is finitely generated for all maximal

of A, then G l ( ~ )

is finitely generated for all R

in A. If S = R - (0}, then localizing at S is an exact functor

and gives a map G I ( ~ ) -~-GI( A S ) = GI(A).

Since A is semisimple,

every module is projective and so Gl(A) = Kl(A).

We have a commu-

tative diagram

KI(A ) ~

)~

R'"

C

GI(#~)

al(A)

C*

where C is the center of A and R' is the integral closure of R in C. By Theorem 8.10 the map Kl( ~ ) -~-R'" is an isomorphism mod torsion. Hence ker(Kl(A ) -O--Gl( A )) is torsion. Let A C r~

= I cA

~

with ~

a maximal order.

Choose r ~ 0 in R with

then we have maps G l(~/I) - ~ G I ( ~ )

and

-

Sl(~) -~ Gl(A).

1 6 4 -

We claim G I ( ~ ) ~ G l ( ~ / I )

-%-GI(~)

is onto.

By lemma 8.22 below, GI(~ ) is generated by the [(M, f)] where M is torsion free as an R module.

For such an (M, f), M C

KM = ~ R

M

and f induces an automorphism of KM and hence also of ~ M C K M since p M is stable under f and f-1. sequence of ~ I~M

= r~ ~M

We examine the short exact

modules 0 -a~M-~b ~ M _a~ ~ M / M - ~ 0 . = rUMC-AM

Now

= M, s o V M / M is snnihilated by I.

duces automorphism f' and f" on ~ M

f in-

and ~ M / M respectively and

[(M, f)] = [(VM, f')] - [(~M/M, f")] which is in the image of a

GI(G) ~ G I ( L / I ) r~

I ~ A.

A = A/a.

as claimed. Thus A

Now ~

= ~ / I is finite since

has a radical ~ which is nilpotent.

Then GI( ~ ) = Gl( A ) by Theorem 8.19.

simple, G I ( A )

= K I ( A ) and furthermore,

Hence KI(A ) is finite since ~

is.

Since A

Let

Thus, if ~

A' be any order between ~

G I ( ~ ) -~ G I ( A ' ) ~

GI(A)

is a maximal order

and ~ .

is finitely generated if K I ( ~ ) Now if (5) is true, and ~

is.

is

Then we have has a fi-

Since maximal orders

are hereditary, SK, Th. 16.11 shows that K I ( ~ )

diagram

GI(A/I)

so the map GI(A') -4~ G I ( A )

nite cokernel since G I ( ~ ) -4~GI(~) does.

GI(A)

is semi-

A * -a~Kl(A ) is onto.

the map G I ( ~ ) --~PGI(~ ) has a finite cokernel since finite.

Let

= GI(~).

Thus

This establishes 4).

~ , we have a commutative

-

165

-

oz(P') R,.

Gz(A)/ and the maps G I ( N ) - ~ R ' * torsion.

and GI(A ) -@-R'* are isomorphism mod

Hence so is G l ( ~ )

-4~GI(~).

Hence 5) implies 2).

Also, the map K l ( A ) -~.R '~ is an isomorphism mod torsion by Theorem 8.10.

If (5) holds the map Gl( A ) -~-R'* is an isomor-

phism mod torsion. torsion.

Hence K l ( A )

-I~GI(A)


Thus (5) implies (1).

Therefore we only have to prove (5). to verify (3) for maximal orders.

If ~ C ~ ,

We claim it is enough we have the commu-

tative diagram

R,~

GI(A) and G I ( ~ ) -~-GI(A) for ~ , then G I ( C )

is an epimorphism nod torsion. -~-Gl(A)

the diagram commutes.

If (5) is true

is a monomorphism mod torsion simce

Hence G l ( ~ ) -~DGl(A)

and G l ( ~ ) -~DR'*

are both isomorphisms mod torsion and thus, Gl( ~ ) ~ R Hence we can assume A Kl(~)

= ~

is a maximal order.

= GI( ~ ) since maximal orders are hereditary.

'~ is also. Then But

-

166

-

K I ( ~ ) - ~ R '~ is an isomorphism mod torsion by Theorem 8.10. Proposition 8.21. A

Let R be a Dedekind ring with quotient field K,

be a semisimple K algebra, and ~

an R order in A.

Let A =

A 1 x ... x An be the decomposition of A into simple components and let ~ i be the image of ~

in A i so that ~ C

Then the map G l ( ~ ) - ~ G l ( a ) Proof.

~

= A 1 x ... x A n .

is onto.

By Lemma 8.22, GI(A ) is generated by [(M, f)] with M a

torsion free finitely generated A Now K ~ R M

is an A module.

module. Hence K ~ R M

= NI~...

~N n

where each N i is an A i module and the N i are fully invariant. filter K ~ R M

by K ~ R M

V i = Ni~Ni+l~... Mi = M~V

= Vl~

~ N r.

V2~

and [(Mlf)S = ~[(Mi/Mi+l,

Let

This is an automorphism of N i since M i

is clearly stable under f and f - 1

Now M = M l ~

M2~

... ~ M r ~ 0

Ti)] where ~i is the map induced on

Hence it is enough to show that all [(Mi/Mi+l, ~i)]

are in the image of Gl(~). i = j.

where

Then Vi/Vi+ 1 is isomorphic to N i.

i and fi = f/m " i

Mi/Mi+ 1 by f.

... D V r D O

We

Now Aj ~nnihilates Vi/Vi+ 1 unless

Hence Vi/Vi+ 1 is an A i module and a

through the projection a the projection ~

- ~ . a i.

-~- ~i"

Thus ~

acts on Mi/Mi+ 1

acts on Mi/Zi+ 1 through

Hence [(Zi/Zi+l, Ti)] is in G I ( ~ ) and

the map G I ( ~ ) -~-GI(~ ) is onto as desired.

-

167

-

The following result follows from SK, Chapter 16, but we include a proof for the reader's convenience. Lemma 8.22.

Let R be a Dedekind ring and A

generated and torsion free as an R-module.

an R-algebra finitely Then Gl( ~ ) is gener-

ated by the [(N, f)] where M is finitely generated and torsion free as an R-module. Proof.

Let [(M, f)] be any generator of G I ( ~ ) .

be an epimorphism with Q projective over A . consider P = Q ~ Q ~ Q P = Q@Q

~

(f0 of-i ) (~)

M~M

of M ~ M .

and (~ ~).

projective.

--4~M. -~M.

Let 7 :

Q-~M

Let P = Q O Q

and

This factors as Lift f to the automorphism

This is a product of elementary automorphisms These lift to automorphisms

of Q e

Q since Q is

Thus we can lift f to an automorphism g of P.

The

sequence 0 -~-(N, b) -b-(p, g) -a-(M, f) -~-0 shows that [(M, f)] = [(P, g)] - [(N, b)] and N, P are torsion free. Remark.

With a bit more work we can show Gl(A ) is K 1 of the

category of finitely generated torsion free A - m o d u l e s .

See

SK, Corollary 16.21.

Chapter 9:

Cancellation Theorems

In this chapter we will be concerned with the following problem.

If X ~ M

~ X~N,

is M • N?

As always, we will consider

-

1 6 8 -

finitely generated torsion free ~ -modules where A

is an order

over a Dedekind ring R in a semisimple separable K-algebra A, K being the quotient field of R. due to Bass and Jacobinski.

We will give two partial answers

Jacobinski's result depends on a deep

theorem of Eichler and most of this chapter will be concerned with the proof of Eichler's theorem. In SK, Chapter 12 we gave Bass' original proof of his theorem.

Here we will give a different proof which shows more

clearly the relation between the results of Bass and Jacobinski. The approach here is much more specialized since it depends on the fact that A

is an order.

projective here.

We do not require the modules to be

However, Dress [DR] has shown that the original

proofs of Serre and Bass can be modified to avoid this assumption. We will illustrate our approach by first proving Dress' extension of Serre's theorem for the case of orders. The hypotheses on R, K, A, A

stated above will be in force

throughout this chapter and will not be repeated. finite non empty set of primes such that ~ y Theorem 9.1. modules.

Let P be a

is maximal for y ~ P.

Let M and N be torsion free finitely generated ~ -

If % ~ %

is a direct snmmand of My for all y ~

P, then

N is a direct summand of M. Proof.

By Theorem 6.12, M = N I ~ M '

where N l ~

N 1.

Since the

Krull-Schmidt theorem holds over ~ y, Ny is a direct snmmand of M~j for y ~

P.

By Theorem 6.12 again, M' = N 2 ~ P

where N 2 ~

N.

By

-

169-

Roiter's lemma we can embed N 1 and N 2 in N so that N/Nl~ N/N 2 have relatively prime orders.

Thus N 1 + N 2 = N so we have an exact

sequence 0 -~-Q-a-NI~)N For any y either (N/ N 1 )y

=

2 -*~N -abO .

0 or (N/ N 2 )y

=

O.

or N2y = Ny so the sequence splits locally. splits and so N 1 O N 2 =

N~

Q.

Therefore Nly = Ny By Lemma 6.14 it

But M = N 1 ~ ) N 2 ~ ) P = N ~ ) Q O P .

We now turn to the cancellation problem. X~)M

• X~N

A-modules.

where X, M, N are finitely generated torsion free In both theorems to be proved we must assume X locally

a direct summand of M n for some n.

Note that this is true in B ass'

original theorem in which X is projective By Theorem 6.12, of M n.

We have

this implies X ~ - X '

By Lemma 8.1 (5), X G D ~ s o

some M s = X ~ ) Y . Since X ~ ) M ~

where X' is a direct summand X itself is a direct summand of

We conclude that M S ~ ) M

X~N,

and M has a free summand.

~ MS~N

so N G

we have [M] -- IN] = 0 in DN-

Let P be a finite non empty set of primes such that a maximal

D

order for y ~ P.

For Jacobinski's

is Y theorem we will also

have to assume that certain other primes lie in P.

A

By Roiter's

lemma we can embed N in M such that C = M/N has C (Clearly N ~ M by Lemma 6.4.) of Proposition 8.5. Ko(T A ,p) -~pD M. ~

= 0 for y ~ P. Y Thus C 6 T ~ ,p using the notation

Now [C] ~-a-[M]

- [N] = 0 under the map

By Proposition 8.5 there is an

Kl(End~s(MS))

such that ~

~

[C].

-

170

-

Now S = R - ~ y so R S is semilocal having as maximal yt P ideals the YS for y G P.

By the next lemma, E + E n d A s ( M S) is

semilocal since E is finitely generated as an R S module. Lemma 9.2.

If R is a semilocal commutative ring and E is an R-

algebra finitely generated as an R-module, then E is semilocal. Proof.

By definition, this means that E/J is artinian where J is

the Jacobson radical of E. ideals of R and ~ E/~E

= ~E/~iE

= ~l ~

Let ~ l '

"'''~n

"''~n

be the maximal

its Jacobson radical.

by the Chinese Remainder Theorem and E / ~ i E

a finite dimensional algebra over a field R / ~ will suffice to show t h a t ~ that ~

annihilates all simple ~

module and k ~ S at ~ i

E C J.

# 0 then ~ S

i.

Then is

Therefore it

For this we need only show

-modules.

= S so ~ i S

If S is a simple A = S for all i.

and use Nakayama's lemma to conclude S ~ i

= O.

Localize Thus

S = O, a contradiction. Since E is semilocal, the map U(E) = GLl(E) -~PKl(E) is onto by Proposition 8.5. The map Kl(E) ~ K o ( T ~

Therefore we can choose ~

~ U(E) = Aut A S ( M S ) .

A ,p) was constructed as follows.

If

GL A (E), write @~ = f/s where s ~ S and f ~ Z n ( E n d A ( M ) ) .

We

get 0 -~-H n f--~-Nn -~-D -~-0 and 0 -4-H n s-~Mn -~-E -~-0 and the image o f ~

in K0(T A p) is [D] - [El.

[D] - [ES = [C] so [D] = [ C ~ E ] .

In our case, n = 1 and also

Therefore D and C ~ E

have

- 17i -

c o m p o s i t i o n series w i t h isomorphic factors. 0 ~ N

~ M

~ C

~ 0

Now we have

and 0 ~ M - ~ M - ~ B - E

Therefore

~O.

A

if we define H b y 0 ~ N 0 -m-C -~pH - m - E - ~ - O

~-~bM --g- H ~ 0

we have

so H and D have isomorphic c o m p o s i t i o n fac-

tors and 0 - b - M f - - ~ M - ~ - D -b-O, 0 -~-N ~ M

~-H-~-O.

We can n o w reduce to the case where H = D and is simple by using the f o l l o w i n g lemma. Lemma 9,3.

If D is a torsion ~

module with D

= 0 for y ~ P, we Y can find a c o m p o s i t i o n series for D in w h i c h the c o m p o s i t i o n factors appear in any p r e - a s s i g n e d order.

Proof.

If D = D l ~ D

2 and the lemma holds for D 1 and D2, then it

clearly holds for D.

Since D is the sum of its y - p r i m a r y comport-

ents, we can assume D is y-primary. regard D as a module over A y. are reduced to the case where 1 x ... x A s

where K A i

Therefore D = b

Since A y is a maximal order, we A

is maximal.

In this case ~ =

= Ai' A = A 1 x ... x A s with

Now D = D 1 x ... x D s where D i is a ~ i module. to consider each D i.

Thus we can replace ~

words we need only do the case where A simple algebra A. of A then R ' ~

and we can

Ai

simple.

It will suffice

by ~ i"

In other

is a maximal order in a

If R' is the integral closure of R in the center

is an order c o n t a i n i n g A

(Note that all our alge-

bras are assumed separable so R' is a finitely generated R-algebra). Thus A

= R'A

so A

is an order over R'.

Since R does not occur

-

172

-

in the statement of the lemma we can replace R by R' so we can assume A is central simple over K.

Repeating the first part of the

argument we reduce to the case where R is a DVR° maximal

ideal.

Let y be its

By Theorem 5.16, there is a unique prime ideal (=

maximal 2 sided ideal) P of A

with y ~ C

P and P e G

y~

for some eo

If S is a simple ~ -module then yS = O (If yS = S then S = O by Nakayama's

lemma).

Therefore PS = O so S is a

~/P-module.

But

/P is simple since P is a maximal 2-sided ideal and so has only one simple module up to isomorphism. for A

Therefore the same is true

and any composition series will do. We now return to our sequences 0 --~-M ~ M

0 ~ N

~ M

~--~--H ~ 0 .

Since D and H have the same composition

factors,

Lemma 9.3 gives us composition series

D = DO~

D l~

Di/Di+ l ~

... ~ D h = 0 and H = H O ~ H l ~

Hi/Hi+ 1.

S i = Di/Di+ 1.

~---~-D --~-0 and

... ~ H h = O with

Let M i = ~ - l ( D i ) , N i = ~ - l ( H i ) , and set

Then we have 0 ~Mi+

1 ~ M

i ~ S

i ~ 0

0 ---b-Ni+ 1 ~ N

i ~ S

i ~ O

and M 0 = M, N O = M, M h ~

M, N h ~

N.

This leads at once to the following result° Proposition 9.4.

Let M be a finitely generated

module with the following property:

torsion free A

There is a finite set P of

-

primes of R such that if M ' ~

173

-

M and S is a simple A -module with

Sy = 0 for all y 6 P~ then any two epimorphisms M' -~DS have isomorphic kernels.

Let X be a finitely generated ~ -module which is

a local direct s~mmand of M n for some n.

Then X ~ M

~ X(~N

im-

plies M ~ N. Proof.

We can enlarge P without spoiling the hypothesis and so

assume that ~ Y is maximal for y ~ P. applies. N i~

Since (Si)y = 0 for y ~ P, we have M i ~

Ni+ 1.

Therefore M i ~

hypothesis implies N i + l ~ M ~Mh~

The construction above now

M, N i ~ Ni+ 1.

M for all i. But N 0 = M = N 0.

Mi+ 1 and If M i ~

Ni, the

Therefore

N h ~ N. We must now prove the hypothesis of Proposition 9.4 holds

under certain conditions.

We do this by showing that if

f, g: M' -~-S are epimorphisms then g = f@ for some automorphism @ of M'.

This clearly implies ker f ~ ker g.

proposition $.~. maximal for y $ P.

Let P be a finite set of primes such that ~ y

is

Let M be a finitely generated torsion free ~ -

module such that M = M l ( ~ M 2 where M 1 and M 2 are f a i t h f u l ~ modules.

If S is a simple ~ - m o d u l e and f, g: M - ~ - S

are epimor-

phisms, then there is an automorphism @ of M with g = f@. Proof.

The relation "g = f@ for some automorphism @" is clearly an

equivalence relation.

We will show any epimorphism f: M - ~ - S

equivalent to one particular f0: M - ~ S "

is

By Lemma 6.7 there are

-

epimorphisms ~ i : Ml~N2

-~-S.

N1 -~-S, ~ 2 : N 2

If f = ( ~ , ~ ) :

of @~ and ~ , say ~ ,

=

-~-S.

Ml~M

Let fo = (~l'

~ 2 ):

2 -~-S is an epimorphism one

must be an epimorphism since S is simple.

By Lemma 9.6 below, if ~ : where ~ :

175

M 2 -b-M 1 i.e.,

M2 ~ S

is any map we can write

MIL~- ~-

Z2

commutes.

~ =@(

This is clearly

possible locally since y ~ P implies Sy = 0 while y ~ P implies that ~ y

is maximal and hence M2y is Ay-projective.

be g i ~ e ~ by M1 ® . ~2

"

~h~

f~ ~ ( ~ ,

,~

+ 0~)

Let @: M ~ M

-- (Or, #

+ ~)-

l'q2

Therefore we can replace ~

with ~ 2 using

~ = W 2 - ~.

Since

~ 2 is an eplmorphism we can apply the same argument again to replace ~

by ~l'

i.e., we get f@@l = (~l' ~ 2 ) = fo"

same argument works if ~ Lemma ~ .

and not ~

Let A f-~ C ~ -

each y, there is a map ~ y

Clearly the

is an epimorphism.

B with A finitely presented. making the diagram

Ay

By C

commute, then there is a map ~

Y making

If for

-

175

A

-

~

B

C commute. In other words, the d i a g r ~

can be filled in if it c ~

be

filled in locally. Proof. H o m A, (_ A

We must show f lies in the image of B) 1 - - ~ H o m ~ ( A ,

HomA(A ' m)

localizes.

of (1, g) in P.

C).

Since A is finitely presented,

Let P = H o m ~ ( A ,

C) and let Q be the image

Then for each y, f/1 ~ ~ C

Py.

Then t(sf - q) = 0 so ts f ~ Q for some ts $ y.

(r

E R~rf G Q) then ~ = R so I G ~

is an ideal and ~

¢

Write f/l = q/s. If X9% =

y for all y.

Thus

and f t Q.

Combining Proposition 9.@ and 9.5 we get the following result which generalizes Bass' theorem for the case of orders.

A

closely related result was proved by Dress [DR] for much more general algebras. Theorem 9.7.

Let M be a finitely generated torsion free A - m o d u l e

such that M = M l ~ M

2 where M 1 and M 2 are faithful A -modules.

Let

X be a finitely generated A - m o d u l e which is a direct summand of M n for some n.

Then X ~ M

Proof.

M, then by C o r o l l a ~

If M' ~

• X~N

implies M ~ N. 6.13 we have M' = M ~ M ~

where M!I ~ Mi and hence M!l is faithful.

By Proposition 9.6 we see

that the hypothesis of Proposition 9.5 is satisfied.

-

176

-

In the case considered by Bass, mand of M. since A C

Thus M = A ~ M M 1 locally.

~2

is a local direct sum-

1 by Theorem 9.1 and M 1 is faithful

Also X is assumed projective

so X is a

direct summand of A n and hence of M n. We now turn to Jacobinski's global field,

theorem.

For this, K must be a

i.e., an algebraic number field or a function field

of dimension 1 over a finite field. algebra over K, we say that a prime

If A is a central simple (= valuation)

fled in A if the completion Ay is isomorphic we say y is ramified.

of K is unrami-

to Mn(Ky).

Otherwise

A real archimedian prime y is ramified in

A if and only if Ay ~ M m ( ~ )

where H~ is the algebra of quat ernions.

If R is a Dedekind ring with quotient field K, we say a prime of K comes from R if it is non-archimedian

and its valuation

ring is Rp for some prime ideal P of R. Definition.

Let K be a global field and let R be a Dedekind ring

with quotient field R.

Let A be a simple separable K-algebra with

center C and let R' be the integral closure of R in C. satisfies Eichler's (1)

condition

We say A

(relative to R) if either

There is a prime of K which is unramified

in A and

which does not come from R, or (2)

char K = 0 and dimcA # 4.

If A is a semisimple

separabl~ K-algebra then A = A 1 x ... x A s

where the A i are simple. if and only if all A i do.

We say A satisfies Eichler's condition

-

Remark.

177

-

Suppose K is an algebraic number field and R is the ring

of integers of K.

Then A satisfies Eichler's condition if and only

if no A i is a totally definite quaternion algebra.

(A totally

definite quaternion algebra is a simple algebra such that Ay for every archimedian prime of the center of A). Definition. K-algebra,

If R and K are as above, A is a semisimple separable a

is an R-order in A, and M is a finitely generated

torsion free A - m o d u l e ,

we say that M satisfies Eichler's

condition

if the K-algebra EndK(KM) does. If ~ mod ~

is an ideal of R and f, g: M ~ N

if f - g sends ~ into ~ N C N .

M and @ ~ 1 m o d ~

0(8-i

_ I)M :

(i

If @ is an automorphism of

, then also @-l ~ 1 mod ~

- @)M ~ ,

we write f m g

M so ( @ - i _ I ) M C

because

@-I(L~t,M)

:

We now state the following generalized version of Eichler's theorem. Theorem 9.8.

Assume that K is a global field.

generated torsion free A tion.

module which satisfies Eichler's condi-

Then there is a finite set of primes P with the following

property:

If S is a simple a -module with yS = 0 for some y ~ P

and f, g: M -~-S are epimorphisms, automorphism ~ g = f@.

Let M be a finitely

then either g = @~f for some

of S or there is an automorphism @ of M with

Furthermore,

in the second case,

if yS = 0 and ~

ideal of R prime to y, we may choose @ so that @ m 1 mod ~ .

is any

-

As a consequence tion theorem.

178-

of this we obtain Jacobinski's

cancella-

(The last part of Theorem 9.8 is not needed for

this.) Theorem g . 9

(Jacobinski).

Assume that K is a global field.

Let M

be a finitely generated torsion free A -module which satisfies Eichler's condition.

Let X be a finitely generated

is a local direct summand of M n for some n.

~

module which

Then X Q M

~X(~N

implies M ~ N. Proof.

This is an immediate consequence

Theorem 9.8.

There are only a finite number of M ' ~

Jordan Zassenhaus theorem. gives us some P.

M by the

If we take the union of all these we get a finite

As an application

M.

of this, let R be the ring of integers of

an algebraic number field K and let ~ an indecomposable if K ~

9.5 and

For each of these M', Theorem 9.8

set which will do for all M' ~

Therefore

of Proposition

be a finite group.

finitely generated R ~ - m o d u l e ,

If P is

then KP • K ~ .

has no simple component which is a totally defin-

ite quaternion algebra,

Theorem 9.9 shows that F ~ P

~ F~Q

implies P ~ Q for finitely generated projective E ~ - m o d u l e s . will certainly be so if K is not totally real. real and K ~

has a bad component,

The image of ~ over

R~.

in ~

If K is totally

we will have H ~

= ~x

....

will be a finite subgroup of H~ which spans

The only finite subgroups

the generalized

This

of

~H are the cyclic groups,

quaternion groups Qn and 3 exceptional

groups

-

179

T, O, I of order 24, 48, and 120. since

H{ is non commutative.

-

The cyclic groups cannot span

Therefore

if ~

has no quotient

iso-

morphlc to Qn' T, O, or I then cancellation holds for finitely generated R ~ - m o d u l e s .

This will be the case if ~ i s

simple, or of odd order,

abelian or

etc.

We now turn to the proof of Theorem 9.8.

Our first object

is to reduce to the case of a maximal order in a central simple algebra.

This is the case treated by Eichler.

The reduction does

not depend on the fact that K is a global field so we state it a bit more generally. Lemma 9.10. R~

Let ~

be a class of Dedekind ringssuch that if

has quotient field K, L/K is a finite separable extension

and R' is the integral closure of R in L, then R' ~ ~

.

If the conclusion of Theorem 9.7 holds whenever R ~ is central simple over K, and A for any order in a separable

RE'

is a maximal order, then it holds

semlsimple

algebra over K for any

.

Proof.

We first reduce to the case of a maximal order.

be maximal.

Assume the theorem holds for ~ .

it also holds for A .

KM.

M does.

Since K ~ M

Choose r ~ R, r # 0 so r ~ C

= KM,

Let S be a simple ~

~M

Let

~

We will show that ~ •

assume all y with r ~ y lle in P and that P works for ~ ~Ma

, A

We can and

satisfies Eichler's condition if

module with Sy = O for y G

is a unique y with yS = 0 (i.e., Sy # O) and y #

P.

P.

There

We can regard

-

S as a A y - m o d u l e .

But A y

garded as a p - m o d u l e .

~ ~y

~ My and

eplmorphism Hom~(~M,

f':

~M

--~- ( ~ M ) y

~M

-~-S.

S) ~

since y ~

P so s can also be re-

If f: M -q~ S is an epimorphism,

factor f as M -4PMy -~-Sy = S. (UM)y

180-

Since

~y

= Ay,

we have

-- My -~-Sy -~-S extends

In fact the extension

Hom A (M, S) (by restriction)

we can

f to a

is unique

since

is an isomorphism

(check it locally). Let f, g: M -~-S be epimorphisms. f' , g' : ~ M not,

- ~ S.

let ~

prime to y. @ of ~ M

Aut S, then g = ~ f .

for ~

be any given ideal of R prime By the assumption

on ~ ,

~ ~M

M.

since r ~ C

The same argument automorphism

In fact,

of M.

~

applies

to y.

(and so @-l .= 1 mod r ~ ) .

if x 6 M, then @(x) - x g r ~ .

Thus @ ( M ) ~

the case where

~

M.

C

Thus @IM is an

is a maximal

Then

and S = S 1 x ... x Ss, M -- N 1 x ... x M s .

~

~ ~l

is maximal.

Let

x ... x ~ s

N i - ~ S i.

of M i with the required properties, ... xl on N.

order.

Since S is simple,

one S i ~ 0 and we are looking at epimorphisms

simple and ~

~M

M and @IN ~ 1 mod ~ .

to @-l so @ - l ( M ) C

A = A 1 x ... x A s with all A i simple.

lx °.o xlx@xlx

is also

It clearly has the required properties.

We now consider

an automorphism

Then r ~

If

there is a ~ - a u t o m o r p h i s m

with g' -- f'@ and @ "-- 1 rood r ~

We claim @ ( M ) C ~ M

If g' = ~ f '

Extend them to

only

If @ is

use

T h e r e f o r e we can assume that A is

Let L be the center

of A and let R' be

-

181

-

be the integral closure of R in L. we see that ~ of A

~ R' ~

so ~

and the separability

R' ~

As usual,

considering

A C

is an R' order (using the maximality of L over K).

By our hypothesis

and the theorem holds for ~

as an R'-algebra.

Let

be the required set of primes of R'.

Set P = ( R ~ y ' I y '

E P').

S is a simple A yE

y' C

and let ~

C

P'.

Let f, g: M - ~ - S

R prime to y where yS = O.

R' with y'S ~ O.

have y' + R ' ~

P' If

module with Sy = 0 for y ~ P, then S = yS for

P so S = y'S for y ' ~

g #~f

R'A

= R'.

Clearly y = R ~ y ' . By the hypothesis,

be ~ -epimorphisms There is a unique

Since y + ~

= R, we

we can find an automor-

phism @ of M with g = f@, @ ~ 1 mod R ' ~ .

Thus

(@

Of course we must check

-

1)M ~ R ' ~ M

~ M

so @ m 1 m o d e .

that M satisfies Eichler's

condition with respect to R' and ~

but

this is obvious. We next show that Theorem 9.8 is a consequence

of the fol-

lowing result. Proposition 9.11.

Let R be a Dedekind ring whose quotient field K

is a global field. condition.

Let A

Let A be a simple K-algebra be an order in A over R.

set of primes P of R such that if

~

satisfying Eichler's

Then there is a finite

is any non-zero

ideal of R

prime to all y ~ P and f(x) = x n + alx + ... + (-1) n E dimKA ~ n 2, then there is a unit Furthermore, can choose ~

if ~ with ~

~ G~

is any non-zero ~ 1 mod ~ .

with f ( ~ )

REx] where

~ 0 sod ~

ideal of R prime to ~

, we

.

-

182

-

The following well known result about maximal orders will be useful.

Let R be a Dedekind ring with quotient field K and let A

be a central simple K-algebra. and

~

are conjugate

proposition 9.12.

if p

If A ,

p

are orders in A we say

~ a A a -1 for some unit a of A.

If R satisfies the Jordan Zassenhaus

theorem

there are only a finite number of conjugacy classes of maximal orders in A.

If R is local, there is only one class.

Proof.

Fix one maximal order A .

M = ~

.

This is a left

Theorem 5.5, MM -1 = ~ • ~'~,

~:

a ~ U(A).

module and right ~ -module.

If N' ~

M -~-M'.

as a right A-module.

is any other one, let By

If P ' is another maximal order and M' =

then M'M '-I = ~ ' .

an isomorphism

a ~ a -1.

P

If P

M as a right ~

Apply K ~ R

Thus K ~ R ~ :

Therefore Z' = ~ ( M )

module, choose

- and get K ~ R ~ :

x~ax

A ~ A

for all x @ A where

= aM so ~ ' ~ aM(aM) -I = aMM-la -1 =

Thus the number of conjugacy classes of maximal orders is

the number of isomorphism

classes of right A -ideals.

satisfies the Jordan Zassenhaus

theorem this is finite.

If R If R is

local, there is only one class by Theorem 5.27. Corollary 9,13.

If A is unramified

at y C R, then A / y ~

=

Mn(R/y) where n 2 = dimKA. Proof.

Since ^ y is unramified , A Y ~

M n ( Ky ) .

By Theorem 5.28 and

A

Lemma 7.2, A y is a maximal

order in Ay.

By Lemma 7.4, so is

A

Mn(y) so^Ay A /yA

~ h y/YAy

Nn(Ry) by Proposition 9.12. ~ Mn(Ry/YRy) ~ Mn(R/Y).

Thus

-

183

-

The following result will not be needed here but we include it for completeness. Corollary 9°14.

Let A

be a maximal order and M a finitely gener-

ated torsion free A - m o d u l e .

Then

~

= E n d A (M) is a maximal order

in B = EndA(KM). Proof.

Let A = A 1 x ... x A s .

Then A

= A I x ... x

As,

M = M 1 x ... x M s and we reduce immediately to the case where M is simple.

If R' is the integral closure of R in the center of A (=

center of B), then maximal orders over R and R' are the same so we can assume A (and hence B) is central simple.

By Corollary 5.29

we can assume R is a DVR.

Let A = Mr(D) where D is a division

ring.

order of D.

Let ~

be a maximal

order of A by Lemma 7.4. can assume

A

=

s.

Thus

By Proposition 9.12,

Mr(~.).~_~ Let N = ~

KN is a simple A-module,

~

~

is a maximal Mr(~)

so we

as a left A - m o d u l e .

s Theorem 5.27 shows that M ~ ~

p ~ End A (M) ~ M s ( E n d A N ) .

a maximal order.

Then M r ( ~ )

But E n d A ( N ) ~

Since

N for some

N°

which is

By Lemma 7.4 so is Ms(End ~ (N)).

To prove Theorem 9.8 from P~oposition 9.11, we require the following lemma. Lemma 9.15.

Let k be a field, S a simple Mm(k)-module,

nitely generated Mm(k)-module. of S with h = oCg.

M a fi-

S~ppose there is no automorphism

Let f(x) bei a polynomial

over k with no

-

root in k.

Let @: M - ~ - M

r Let M = ~ i

Proof.

since h # ~ g

of M with f(@) = O.

of M and there is an automorphism

S.

Then End(M)

= Mr(End S) ~ Mr(k).

S) = k r = V say.

for a n y o ~ g

End(S)

= k.

-aol(ar @r-1 + ... + a l) is an inverse

then for each v E V, v@ = ~ v

with

= arxr + ... + a O.

We claim there is an

independent k.

over k.

f(~) Let

= ~2

so @ = ~ I .

= 0 but this is impossible ~-l

be an automorphism

Then g ~ @ ~ - l

= e@,~-i

+ ~2v2

Since f(@) = O, we have ~ E k.

Thus e must exist.

of V over k sending

e to g and e@ to h.

9.8 for an order

in a central

A

simple

A.

Let P be a finite given by Proposition

set of primes

of R including

9.11 and all those ramified

A / y ~ ~ Mr(R/y ) by Corollary free finitely

generated

=

= h.

We now prove Theorem K-algebra

since

If not,

If Vl, v 2 ~ V and

v -- v I + v 2 we have Vl@ = ~ lVl , Vl@ = ~ 2 v 2 , v@ = ~ l V l ~l

over k

Thus

for @.

~ ~

= Mr(k)

independent

Let f(x)

e in V such that e and e@ are linearly

Thus

of

This acts

Now @ E End(M)

Note that g and h are linearly

Since f has no root in k, a O = f(O) ~ O.

~ ( v I + v2).

~

= h.

from the right on Hom(M, and g, h E V.

-

be an automorphism

Then @ is an automorphism M such that g- ~ @ ~ - l

184

9.13.

~-module,

in A.

all those If y ~ P,

Now suppose M is a torsion S is a simple ~ -module,

and

-

g, h: M - b - S

are epimorphisms such that h # @~g for any automor-

phism @~ of S. M-~-M/yM

185-

Suppose yS = 0 for y ~ P.

= M =~S.

= EndA(M)C

We will apply Proposition 9.12 to the order

B = EndA(KN).

Let dimKB = m 2.

Then g, h factor as

Clearly B is central simple over K.

By Lemma 9.16 below we can find a polynomial

over k = R/y of the form ~(x) = x m + al xm-I + ... + (-i) m with no root in k.

Lift ~ to f(x) = x m + alxm-1 + ... + (-1) m E R[x].

Applying Proposition 9.11 we get a unit ~ and with ~ y.

Let ~

~(~)

= 0.

morphlsm ~ lift ~

~ 1 mod~

where ~

6

bQ the automorphism of M = M/yM induced b y e . Lemma 9.15 now shows that h = g ~ of M.

We claim that if ~

Then

-1 for some auto-

-1 to an automorphism @ of N.

phism ~ i 6

E n d ~ (M).

Since ~ i :

Nakayama's lemma shows that ~ i : hence an isomorphism. and M i = ~ l M

Since M is

~ Aut M to an endomor-

M -~-M induces

~i:

N/yM ~ M/yM,

My -~-My is an epimorphism and

Thus ~

Let ~

My -~-My is defined.

Since M

be the ideal chosen in Theorem 9.8.

prime to y we have ~

We now let t

L,~: M "~P M.

Aut M.

lie in My we can find some r ~ R - y such that

for all i.

Since ~ i s y.

~ 0 mody

is properly chosen, we can

projective over A , we can lift each ~ i

a~

with ~ ( ~ )

is any chosen ideal of R prime to

There are only a finite number of ~ ~

rMi~M

~

Since ~ - i

y.

= (ra) so that ~ - 1 = (~

Choose a 6 ~ = 1 + ra~

- i)~'[-I.

where

We can also

with

-

write ~ - i

= i + ra~.

The same argument

- l ~ i -1 so @-i: M -~-M. -- ~ i ~ '

there

Thus

applies

to @-l =

Thus @ is an automorphism

reducing sod y shows that @ ~ i

of N.

-- ~ i ~

or

soh=

Thus Lemma $.16..

all maps as endomorphisms

-1 = 1 + ~ i a ~ ( r ~ i - 1 ) .

@: M -~-M and @ - 1 mod(a).

Since @ ~ i

-

Now, considering

of My, we have @ ~ ~ i ~ i

~i ~

186

If k is any finite field and m ~ I is an integer,

is a polynomial

then

f(x) = x m + a I x n-I + ... + (-l)m G k[x] with

no root in k. Proof. q

m-i

(x

.

Let q = Ikl. If f has a root

The number of f(x) of the required ~

form is

k, then f(x) =

~ ) ( x m-I + b, x m-2 + ... + bm_l) where - ~ b m _ 1 = (-I) m.

-

bm_ 1 # 0 and bm_ 1 determines

~.

such f(x) is ~__ qm-2(q _ i )
I but any prime P ~ I must contain I' or I". Finally note that an increasing chain of submodules

corresponds

of M

to a sequence of epimorphisms M - ~ - M 1 -~-M 2 -~- . . . .

If M is noetherian such a sequence will eventually consist of isomorphisms. morphism.

In particular,

any epimorphism M - ~ M

will be an iso-

- 208-

3.

~

.

Let A

be a ring and J a 2-sided ideal of ~ . ^

T h e n we d e f i n e

the completion

of ~

at J to be

M is a A-module, we define M = lim M/ . d~jn~ complete at J if ~ is a

-~

~

We say

= ~lim ] ~ / j n o

~ (resp M) is

(resp M -~-M) is an isomorphism.

A-module. ^

In general, M ~

is not flat

over A.

@A M, M ~

If

Clearly

M is not exact,

A

and A

The f o l l o w i n g

theorem enables

us to

establish these properties in certain cases. Theorem A2 (Artin-Rees)°

Assume J is generated by a finite number

of central elements of ~ . and M' a submodule.

Let M be a noetherian left ~ -module

Then there is a k with JnMf~ M' =

jn-k(jkMf~ M') for n ~ k ° Proof. ~0 ~

The Rees ring ~ ~l @

~2 ~''"

of ~

is defined to be ~

where ~ 0 = ~ l ~ n

=

= jn and the multiplica-

tion ~ i ~< P j -~" ~i+j is given by the ordinary multiplication ji x

jj .~_ji+j.

Define a surjective map A [ t l, ..., tr] ~

sending t i to Ji E ~ l Since ~ @ A M

= J where 3i E J t~ ~ ( ~ ) generate J.

is a quotient of A Itl, ..., tr] @ A M it is noether-

Jan over A [tI, ---, tr] and so over P . P

~ by

-- A ~ J ~ j2 ~ -.., we have ~ @ A M

N ~ M' Q (JM{3 Z') ~ ( J 2 M ~ M') O . . . .

Since = M~

JM(~ J 2 M ~

....

Let

This is a ~-submodule of

- 209--

~AM M' ~

and so is finitely generated.

(JMt'~ M') ~

JnM~

M' = ~ n M ~

J(JiMN

M,)C

Theorem A~.

... ~

(jkM~

Therefore it is generated by

M') = N k say.

Now if n ~ k ,

k ~ N k = ~, jn-i(jiMf~ M') = jn-k(jkM~] M') since i=O

ji+lMAM,. Assume J is generated by a finite number of central

elements of ~ .

If 0 -*-M' -~-M -~-M" -~-0 is exact and M is

noetherian then 0 -~-M' -~-M - p M " If also A

-~-0 is exact.

is left noetherian,

then A ~ M - ~ - M

is an iso-

A

morphism for finitely generated M and ~

is flat as a right

~

-

module. Proof.

The sequence 0 - ~ - M ' / M ' ~

exact.

Since J n M ' c

M'r'~ J n M c

llm M'/M't'~ jnM = M'. ment.

JnM -~-M/jnM -=-M"/jnM" -~-0 is

jn-kM' by Theorem A2, we have

The lemma below now gives the first state-

The s e c o n d f o l l o w s

a s i n Lemma A1.

The f i r s t

two p a r t s

show

A

that

A I~ A -

preserves

exactness

for

finitely

generated

modules.

Any short exact sequence is a direct limit of finitely generated ones and A ~ A Lemma ALl-. L e t

preserves direct limits. (0 - - - ~

---~B n -.,PC n ---~0) be a n i n v e r s e

system of

short exact sequences of modules indexed by the positive integers. If all the maps An+ 1 -~-A n are onto, then

0 ~

lim A n ~ l i m

Bn ~ l i m

Cn ~ 0

-

210

--

is exact. This is standard and easy. By applying this lemma to 0 _~.j/jn _~. ~ / j n _~.~/j -~-0 we see that, without any hypothesis, 0 -*-J - * - ~ A generally, 0 -m. jn -4~ ~ _~.~/jn _~. 0 and 0 -*-JnM-~-M-~-M/JnM-~-O.

-*-~ /J --~-0. More

However it is not clear that

jn = ~n or jnM = ~n~. Corollary AS.

If A

is left noetherian and satisfies the hypothe-

sis of Theorem A3, and M is finitely generated, then

jn~ = ~n~ = ~n~ and ~ / ~ n ~ Proof.

M/JM = A / J @ & M

=

M/JM = M/J~ so JM = JM.

M/~nM.

A/~@XA®^M

= A/~®~

Use induction on n.

: ~/~

but

Also applying

A

A ®~-

to ~° -~ J -

0 -~'AJ

-~'A

^ -~ A/5 -.-~o gives

-b- ~ / J

-4-0 so J = ~ J and so is finitely

generated. Coroll~

A6.

If A

if 0 -~-M' - ~ - M - 4 ~ M "

satisfies the hypothesis of Theorem A3, and -~-0 with M noetherian, then M is complete

if and only if M' and M" are.

Therefore if A

is also left noe-

therian and complete, then so are all finitely generated left A-modules. This is immediate from the 5-1emma. We say a right A -module M is faithfully flat if a sequence is exact if and only if its image under M ~ A

-- is exact.

This is

-

211-

so if and only if M is flat and M ~ N

= 0 implies N = O.

It is

sufficient to consider cyclic modules N. Corollary A7.

If ~

is left noetherian and satisfies the hypothea sis of Theorem A3, then A is faithfully flat as a right ~ - m o d u l e if and only if J is contained in the Jacobson radical of A .

particular,

this is so if

A

In

is complete. A

Proof.

If N is finitely generated and A

N/JN = O.

If J C

J a c . r a d ( A ) then Nakayama's

If J is not contained M ~J.

~ A N = 0 then N/JN

in J a c . r a d ( A ) ,

Let S = A / M .

lemma implies N = O.

there is a maximal left ideal

Then JS = S so s/jns ~ 0 and

~ ~S

= S = O

while S # O. The hypothesis (R a commutative J = ~ I.

of Theorem A3 will hold if ~

ring),

IC

R is a finitely generated ideal, and

We can always reduce to this case by taking R = ~ ( A ) .

If M is a ~ -module,

then JnM = InM so the completions

respect to I and J are the same. plete R-module,

4.

is an R algebra

then ~

Dedekind Rings.

In particular,

if ~

of M with is a com-

is complete and so J C J a c . r a d . ( ~ ) .

A Dedekind ring is a commutative

main in which all ideals are projective.

integral do-

The obvious example is a

principal ideal domain

(PID) for which all ideals are free.

is Dedekind so is R S.

A PID with only one non zero prime ideal is

called a discrete valuation ring DVR. Dedekind ring is a DVR.

If R

We will show that a local

If R is a DVR, let (p) be the unique

-

maximal ideal. sup{n{pnJr}.

212

-

The associated v a l u a t i o n is defined b y ord(r) = This satisfies ord(xy) = ord x + ord y~ ord(x + y)

min(ord x, ord y), and ord x = ~ - x note R is n o e t h e r i a n must stop.

(as a PID) and so (x) ~

To see that last, (p-lx) ~

(p-2x)~

...

We extend ord to the quotient field K of R by

ord(x/y) = ord x - ord y. Definition.

= 0.

The above properties still hold.

If R is an integral domain with quotient field K, a

fractional ideal of R is an R-submodule I of K with I ~ 0 and sICR

for some s ~ 0 in K.

Define 1-1 = (x ~ K { x I C R } .

1-1 is a fractional ideal and I I - 1 C ii -1 = R.

R.

Then

We say I is invertible if

If I is finitely generated then (Is)-l = (I-1)S for any

SCR. The fractional ideals form a monoid u n d e r (I~ S) --~-IJ and the invertible ones form a group. The following theorem compares the various possible definitions of Dedekind rings. Theorem AS.

Let R be a commutative integral domain.

The following

conditions are equivalent. (1)

R is a Dedekind ring.

(2)

Every non zero ideal of R is invertible,

i.e., the

fractional ideals form a group. (3)

R is noetherian,

integrally closed~ and every non-zero

prime ideal is maximal. (4)

R is n o e t h e r i a n and R

is a DVR for all maximal ideals Y

Y.

-

(5)

Ry is a DVR for all maximal

there are only a finite number

ideals

ideals y and if a ~ 0, a ~

of maximal

(6)

Every ideal is a product

(7)

Every non-zero

ideals containing

ideal is uniquely

g((xi))

a product

R = I1-1 so 1 = ~ a i b 1

i with a i ~

n n Let f: I -~'II R by f(x) = (xb i) and g: ~ 1 1

= ~ , xia i.

generated.

of prime

n

If I is invertible,

I -I.

Then gf = id so I is projective

Conversely,

and finitely

if we have such f, g (even with infinite n)

form x ~-b~xb with b ~ K.

Thus Jan.

But a i G

(i)~(2) Since R

(1) Y

one generator

I, b i E

1-1, ~ , aib i = 1 so I is invertible.

~-(4).

is local,

We have just seen that R is noether-

all ideals are free.

since for more generators

is projective

two conditions

has the

Since f(1) = (b i) only a finite number

O, a non trivial relation. Clearly I ~

I,

R -~-I by

then g has the given form and so does f since a map I - ~ R

of b i ~ O.

a.

of prime ideals.

(up to order),

Proof. b i~

213-

a, b,

Thus Ry is a PID. for a l l ~ .

can be verified

locally.

There can be only ... we have ba - ab = (4)

(4) = = ~ ( 3 ) .

=~(1). The last

They clearly hold for a

DVR.

(4) over it.

Since all primes

are maximal,

the y ~ R a

are minimal

-

(~

~(4~.

Let I I C

I2C

214

....

We can assume I 1 ~ 0.

is noetherian we can find n so Iny = In+ly . . . . mumber of y with I i C Y .

Since Ry

for the finite

For the other y, Ily . . . . .

Ry.

There-

fore I n = In+ 1 . . . . . (4) ~ P ' ( 7 ) .

If I # O, define ordyI = ordya where Iy = Rya. If ordyI ordyI ~ 0 then I C y . Let J = fly , a finite product." Then ni J = I since they are locally equal. is unique.

Clearly (7) =mD-(6).

then n i = ordyiI

The fact that (6) =~D-(2) does

not seem to come up in practice, to verify than (6).

If I ~ ~ Y l

the other conditions being easier

A proof will be found in Zariski-Samuel,

Commutative Algebra, volume I. It remains to show that (3) = ~ - ( 4 ) , satisfying

(3) is a DVR.

Let ~ =

i.e., that a local ring

(a): b = (r ~ Rlrb ~ (a)) be

maximal among proper ideals of the form (a): b° but

z ~ ,

then (a): xb ~ ~

xb E (a) so x ~ .

+ Ry ~ ,

This shows that ~

If xz ~

so (a): xb = R. is prime.

but Thus

By (3) the

only primes are 0 and y, the unique maximal ideal so y ~ ~ (a): b.

Now x = b/a E R otherwise

xy = R or x y ~ y .

(a): b ~ R.

cija j.

so

In the second case x would be integral over R

since y is a finitely generated R-module xa i

But x y ~ R

Then

x ~ R, a contradiction.

(let y = (al, ..., an),

Ix ~ i j - Ci~l ~ O).ance R is ~ntegrally closed, Thus xy ~ R and y ~ (p) is principal with

-

p = i/x. ~

If ~gW~ is any ideal,

~p-l~p-2~t~

p-n~ and ~

215-

the chain

..° C R must stop, say at p - n ~ .

~ R it lies in (p) and p - n - l ~ G

R.

If

Therefore p - n ~

= R

= (pn). The next theorem gives many examples of Dedekind rings.

Theorem A 9.

Let R be a Dedekind ring with quotient field K.

L be a finite field extension of K.

Let R' be the integral closure

of R in L.

Then R' is a Dedekind ring and KR' = L.

separable,

R' will be a finitely generated R-module.

Proof.

Let

If L/K is

Since R S = K, S = R - (0) we have KR' = R~ = L, the inte-

gral closure of K in L.

By taking the integral closure in two

steps it will suffice to do the separable case and the purely inseparable case. A8 (3).

Since R' is integrally closed, we consider Theorem

Let P' be a prime of R' and P = R C%P'.

tegral over R/P. is maximal.

Then R'/P'

is in-

If P ~ O, R/P is a field hence so is R'/P'

and P'

We show P ~ 0 if P' # 0.

Let x g P'

x n + alxn-i + ... + a n = 0 where the a i g a n ~ O.

Clearly a n E

x ~ O.

Let

R and n is minimal.

Then

R g%P'.

We now assume L/K is separable and prove the last statement, which shows also that R' is noetherian. for L over K with w i ~

R' (Note KR'

Let Wl,

..., w n be a base

= L since integral closure

n

localizes).

The map L -~'IIK by x ~ ( t r ( w i x ) ) 1

is a monomorphism

-

216--

and hence an isomorphism since tr(wix) = 0 all i =~-tr(yx)

= 0 all

y =~-tr(z) = 0 all z E L (a field) =~.L/K is not separable. n wi' E

L so tr(w~w~) ~ d =

tr(xwi).

~.lj"

If x G L, x = ~1

Choose

ciwi' where c i =

If u ~ R', all conjugates of u are integral over R.

Hence so is tr(u) E

K so tr(u) E

R.

Thus x ~ R' implies c i ~

R

n so R' C

~ 1

Rwl, a finitely generated R-module.

We now consider the p u r e l y inseparable case. Theorem A8 (5).

If char K = p, there is a q = pn so x & L = ~ x q G K

Suppose first that R is a DVR. We have R ' ~

O.

If y E

so y / x m is a unit of R'.

q ~ R~ord

xq ~ 0 .

~ ~ . Let x G

R'

R', then ord y q = m ord x q some m

From this it is trivial that R' is a DVR.

if P' is a non-zero prime of R', then P = R A P '

(if a ~ P', a q ~ P). R~.

Consider L* ° - - q - ~ K *

K = R so x ~ R ' - ~ x

have least ord x q >

In general,

We use

# 0

Since Rp = R S w i t h S = R - P is a DVR, so is

Thus so is R~,, a localization of R~.

F i n a l l y let a ~ 0 in R'.

In fact Rg, = R~.

If a 6 P', then a q ~ P = R ~ % P ' .

There

are only a finite n u m b e r of possible P but P determines P' = P~,~% R' = R'(% max. ideal of R~. We now make a few general remarks on r e l a t i v e l y prime ideals.

Here R can be any ring, not n e c e s s a r i l y commutative.

consider only 2-sided ideals. prime if I + J = R.

We say I and J are relatively

One useful consequence of this is that

We

-

I~J

= IJ + JI.

Clearly the right side always lies in I ~ J .

I + J = R, let i + j = 1. If R is commutative,

If x ~ I g~J,

this reduces to I ~

I, J' are relatively prime, in R/I is the product R/I.

217-

In particular,

If

then x = xi + xj ~ JI + IJ. J = IJ = JI.

If I, J and

so are I, JJ' since the image of JJ'

of the images of J and J' both of which are Im , jn are relatively prime for all n, m.

Note that I and J are relatively prime if and only if no maximal (2-sided)

ideal contains both I and J.

Theorem A10

(Chinese Remainder Theorem).

Let I1,

..., I n be pairwise

M-~-I~

M/IiM is onto. In particular , R / I I ~

Proof.

Fix i.

Let M be any R-module.

relatively prime.

Then

... f'~In = TI R/I i.

For each j let xj + yj = 1 with xj E I i ,

Then yj ~ 1 mod Ii, yj m 0 mod Ij. mod Ii, a i ~ 0 mod Ij for j ~ i.

Let a i = TT y j-

yj E I j .

Then a i ~ 1

If m i E M let m = ~ , aim i.

Then

m ~ m i mod I i. Corollary All. Proof.

R/I~

If I and J are relatively prime, J ~R/I~

R/J.

then I/If'~ J ~

The inverse image of O ~

R/~

R/J is

clearly I/I ~'~ J° We now return to Dedekind rings. tional ideals of R w r i t e ~

~ ~

and say ~

same class if there is an x G K with ~ and only i f ~

~

If ~

as R modules.

and and ~

= x~.

In fact f: ~

~

are frachave the

Then~ -~- ~ ,

~

~-

if

-

applying K ~ - ~O~

218-

shows that f(a) = xa for some x E

if and only i f ~

= ~

(with ~

K.

Note that

= V~'-I~c

R).

Thus

two ideals are relatively prime if and only if their prime factorizations have no common y. Theorem AI2. ideals.

Let R be a Dedekind ring.

Then there is ~

C

Let ~

R such that ~ ~

and ~

C

and ~

R be is prime

to ~f. Proof.

,~i

For each y ~

By Theorem

AIO we can find a ~ - i

generator~ylby

= Ryay for some ayG~-~

so a may

mod y~-l.

Thus a

Nakayama's lemma since it generates

~i/yy~l.

Let ~

Corollar~ A13.

Let ~

Dedekind ring R. Proof.

= (~-l)y

~ a~C

R.

and ~ C

If y O

~

then t y

= ay~y

R be any non zero ideals of a

Then ~ / ~ , ~

R/~

.

This depends only on the isomorphism class of ~ .

Theorem A12 we can assume ~ &

and ~

By

are relatively prime.

Apply

Corollary All. Corollar 2 Al@. = (a,

If a E ~

, a ~ O.

Then there is b E ~

b).

Let b generate

the cyclic

module

~/a~.

Then

=

If all R / ~ N~

= IR/~I.

=

are finite f o r ~

f 0 we can define

Considering R ~ 3 ~ ' ~

and using

with

-

Corollary AI3 we see that N ( ~ ) If J

= N(~)N(~).

R is the category of finitely generated torsion mod-

ules over R, then K O ( ~ R )

is free abelian with base ([R/y]).

IR is the group of fractional with base (y}.

Let ~ :

IR ~

K0(~R)

Modules over Dedekind rings.

by ~ ( y )

= [R/m

If A

i.e., all left ideals are projective,

= [R/y].

] for ~ C

Then

R.

is a left hereditary ring, then every projective module

is a direct sum of ideals and every submodule ule is projective.

If

ideals of R, I R is free abelian

Corollary AI5 shows that ~ ( ~ )

~.

219-

of a projective mod-

The proof may be found in Cartan-Eilenberg.

We consider here the finitely generated case. Theorem Al~.


Then every finitely gener-

ated torsion free R-module M ~ 0 is projective where F is free and ~

is an ideal.

and M ~

F ~

The rank of M and class of

are unique. Proof.

Let M be such a module.

F = R n is free.

Then K ~ R M

Replace M by sM, s / 0 so M ~ F .

be projection on the last coordinate Then 0 -~-M' -~,M - ~ M = M' ~ .

= Kn = K~RF

-~-0.

of F.

Since ~

then ~

~

= p(M)C

R.

is projective,

a direct sum of ideals and hence projective.

Lemma A16.

Let p: F -,-R

Let ~

By induction on rk M = d i m K K S R M

M = Rn-l~

where

we see that M is

To get

we apply the following lemma. If ~ ~

~

and ~ R ~ ~

are non-zero .

ideals of a Dedeklnd ring R

- 220 -

Proof.

We can assume ~

+ ~

= R and ~ f ~

0 -~-~~

and ~

~

-~" ~ O

are relatively

= ~ .

~

The sequence

-~-~99. + ~

Finally we show the uniqueness of rk F is clear. ~(M ~N)

If

~(M)

this shows that

~

Since

of a projective

be free and note A ( M ) ~ R R For n >/2, K(~) A n ( ~ , )

A(~)

= RO~%

is a non-zero

algebra

of ~ .

on M, then

of a free module

module

is free,

is projective

summand

That

(let M ~ ) N

of ~ ( M ) ~

= 0 so ~ n ( ~ )

~(N)).

= O.

Thus

and A n ( z ) ~ . consider

the situation

of Theorem A9 with

Then R' is torsion free of rank n = [L: K]. prime of R, Theorem

(R/y) n.

Let R'y = Pl ei

R'/R'y

~

: ~n(K~,)

eI R'/yR' ~

of the class

is a direct

As an application, L/K separable.

-~-0 splits.

is the exterior

= ~(M)~R~(N)"

prime so

= UR'/Pi

A15 and Corollary e "'" Ps s.

factors R'/p i by Corollary

A15.

A13 show that

By Theorem AlO,

e. Also R'/p i i has a composition

.

If y

series with e i

If we let fi = [R'/Pi:

R/y],

this

s

shows that ~ elf i = n. 1 We now determine KO(R)

and Go(R).

These are equal since R

is regular.

Let I R be the group of fractional

the subgroup

of principal

is, by definition,

cl(R)

group of isomorphism tive modules

ideals Rx, x 6 K*. = IR/P R.

classes

ideals

The class group of R

We also define Pic R to be the

of finitely generated

with multiplication

of R and PR

[P][Q]

= [P~RQ].

rank 1 projec-

-

Theorem A17.

2 2 1 -

If R is Dedekind, Go(R) = Ko(R) = ~ @ c l ( R )

and

Pic R = cl(R). Proof.

If M is a finitely generated projective module of rank n,

define det M = An(M). det(M~N)

Using A ( M ~ N )

= det(M) ~ d e t ( N )

det: Ko(R) - ~ P i c

R.

= A (M)~A

so we have a homomorphism

Since every finitely generated projective

rank 1 module P is isomorphic to an ideal ( P C that Pic(R) = cl(R). ~ R

~

(N) we see that

K~RP

~ K) we see

The multiplications agree since

-~-R is a monomorphism with i m a g e ~ b ~ .

(Apply K ~ - -

to show that the kernel is torsion and thus O s i n c e ~ projective.) ~cl(R)

Now rk, det: Ko(R ) -~- ~ x cl(R).

-~Ko(R)

by (n, [a]) ~-~-n[R] + [ ~ ]

~R

~

is

The map is clearly an

inverse by Theorem A15. The following theorem of Steinitz and Chevalley classifies all finitely generated R-modules. Theorem A18.


erated torsion free R-modules. NI(~ ... ~ N m ~ N m + l ~ . . .

Let M C N be finitely gen-

Then we can write N =

Q Nn, M = Ml ~

... ~ Z m where the M i and

N i are isomorphic to ideals of R, M i = ~ i N i

2 1 C ~t2C "'" C~tm' ~ l ~

where

~i

G

R and

o.

Any finitely generated R-module A has the form A = Rn ~ ) ~

~)R/~

I(~ ... (~) R / ~

is torsion) where 0 i ~ i C -

m (or M = R/~9~ 1 (~) ... ~ ) R / ~

~, 2c

...C~m.

The ideals

m if M

- 222 -

~i'

"''' ~ m '

determined Proof.

the integer n, and the class o f ~

by M.

If A is an R-module

is finitely generated, A = P~T.

let T be the torsion submodule.

A/T ~ P is projective

This shows the uniqueness

the second part since T = R / ~ l the existence

~i:

Q...

of such a decomposition

be the snnihilator phism ~ :

A -~-R/~.

of A.

= ~Yi

since it is locally so.

where ~

is prime to y.

Then ~

= O then y n - l A C y n A

of ~ .

will suffice

Therefore

~ ynM.

B C U y r/ y 1

We must now show

when A = T is torsion.

ni

"

Let

If we find epimorphisms

annihilates

is an epi-

ynA.

yn~.

If

so ~ yn-lA = 0 contradicting

the

of B : A/ynA is yn.

It

Write B = F/N where F is a

In KN take M = y-nN.

Thus B = F / N O

@i: B -*-R/y n be the i-th coordinate

ule) of R/y n.

m.

in

Since y n F C N

By Theorem A15 and Corollary A13,

r M/N • U R / y n where r = rk N. 1

then @ i B C y / y

~R/~

the annihilator

Then KN = KF.

we have M ~ F ~ N

of n and the class of A ~

Let y = Yi' n = hi, ~ =

to find B -~-R/y n -~-0.

free R-module.

by Theorem A15 so

A -~p T~R/y i n.i = R / ~

= (~i):

morphism

yn-l(A/ynA)

If A

We will show that there is an epimor-

Let A ~

A -~'R/Yi ni , then ~

choice

are uniquely

n the unique maximal

r M/N = U R / y n. 1

projection.

ideal

If @i is not onto

(= unique proper

If this is so for all i, then

n and so yn_lB = O, a contradiction.

Let

submod-

- 223 -

We now have an epimorphism R/A-module

and R / h i s

Thus A = R / ~ ~ C R/~

~' 10

~A'.

2 ~

A -~-R/~.

Since A is an

a free module over this ring, ~ If

~'

and A' = R / ~ ' R/~

~:

is the annihilator of A' then

OA".

... • R / ~

splits.

Repeating,

n @A

we get A =

(n+l) with ~ l

~ ~2

C

....

n

Since A is noetherian the increasing sequence ~ l ' R / ~ i must stop. To prove the uniqueness we can use induction on n, observe that

~l

is the annihilator

of A and that A' in A = R / ~ I ( ~ A '

unique by the Krull Schmidt theorem. Fitting invariants ordy~

Alternatively we can use the

of A or we can observe that we can calculate

i from a knowledge

of dimR/yynA/yn+lA

We now prove the first part. ent proof of the existence erated module. T = Torsion

for all

n.

This gives a slightly differ-

of a decomposition

for a finitely gen-

We use induction on the rank of N.

(A), P = A/T.

Thus N = N' ~ N "

is

If A = N/M, let

Then T = N'/M for some N' and N/N'

where N" • P.

= P.

By Theorem A15, N" = N m + l ( ~ ... O N n

whil~ M C N'.

Thus we are done by induction unless N" = O so that

A is torsion.

In this case let ~

be the annihilator

of A.

what we have shown above, there is an epimorphism A - ~ - R / ~ . Since N is projective by Theorem A15 there is a map ~ : making the diagram

JJ N

w

A -~-

R

I.

R/~

N ~R

By

- 224 -

commute where

V(r)

R/~I~

O.

so " ~ #

have

~(~)

If ~

-- ~

Let ~

= O. ker ~

# ~

= t

~

~(~),

= ~.

~/~

=

~ •

(~)C

~

M1

looks like ~

Now ~

= ida.

~(~) R/~

=

= R/~.

since R / O ~

M so ~I ~

=

~(Z)

with non-

is noetherian.

= ~ ,

we have ~ :

splits M -~- ~

-~0.

=

Let

=

M', M l c

N l, M ' C

~ C ~

under ~

so M I = ~ N

N' in the required form.

If

so we can find ~ :

= ~

ker ~ ,

=~.

But

-*-R/~

is projective

M' so all ideals of N' C

~ N ) C

by Corollary A13.

Since ~ ( M )

N'

Then ~ ( ~ )

Since M has image 0 in A we

N = Ml~

= ~

we can put M' C

NC

~(~),

Then N = N l ( ~ N ' ,

~

~.

This is impossible

-4-N with ~

~N'

= ~(N).

we would get an epimorphism R / ~

Thus ~

N1

= ~(M)C

then ~ / ~ and

trivial kernel.

(~)C~

Let ~

Since ~I~N C M we have ~

~CtC~C~ ~

= r mod ~ .

N'.

Also M l c

N1

1.

By induction

Since ~ N

C M we have

M' will contain ~ .

As in the case of abelian groups,

there is an alternative

approach to the second part of Theorem A18 where we first decompose a torsion module into its primary component. y-primary component Definition.

as a module over Ry.

Let M be a torsion R-module.

of Z is Z (y) = ( x C

We then regard the

The y-primary component

Mlynx = 0 for some n).

We say M is y-primary

if M : z (y).

Lemma Al$.

If M is y-primary

then M -~-My is an isomorphism.

M is torsion and M (y) = O, then M then M (y) ~ M y .

Y

= O.

Therefore

If

if M is torsion

-

Proof.

Let M be y-primary.

nihilator of x. Thus

(yn

yn and s.

but ( y n

t$

] let ~ be the anY and sx = 0 so s ~ for some s ~

s) = R since no maximal ideal contains

Suppose now M is torsion and M (y) = O.

rx = 0 for some r E R. yn~x

If x E ker[M--~-M

Then y n c ~

s)C~

225-

= O.

Let (r) = ~ y n

Since M (y) = O, ~ x

= O.

y, tx = 0 implies x/s = 0 in My.

where ~

Theorem A20. Proof.

6.

(M(Y))y ~

y.

Then

Then

Since there is some t @ ~ Finally, we localize

0 -~-M (y) -~-M -~-M/M (y) -~-0 at y and note that This gives M (y) ~

Let x G M.

(M/M(Y)) (y) = O.

My.

Let M be a torsion R-module.

Then M = ~ M (y). Y The map ~IM (y) -b-M is locally an isomorphism by Lemma A19.

Global fields.

A global field is either an algebraic number

field (finite extension of ~ )

or an extension of a finite field

which is finitely generated and of transcendence degree 1.

We want

to classify all Dedekind rings whose quotient field is a global field. Proposition A21.

Let K be an algebraic number field and let R be

the ring of integers of K, i.e., the integral closure of ~ i n

K.

If R' is a Dedekind ring with quotient field K, then R' : R S for some multiplicatively Proof.

closed subset S of R.

Since R is finitely generated as a ~ - module

(Theorem A9),

the Jordan-Zassenhaus

theorem applies and shows that the class

group cl(R) is finite

(only the very last part of the proof of the

,

- 226-

Jordan-Zassenhaus

theorem

proof of the finiteness integrally

closed

is needed.

This is just the classical

of class number).

Now ~ C R '

(Theorem AS) so R C R'.

lemma which makes no assumption

and R' is

We now use the following

on K and does not even require R'

to be Dedekind. Lemma A22.

Let R be a Dedekind

that cl(R) is a torsion group.

ring with quotient

If R' is any subring

ing R, then R' -- R S where S -- U ( R ' ) ~ Proof. r~

Clearly R S C

R' and U ( R ' ) ~

R, s E S implies

assume that U ( R ' ) ~

r6

U(R')

fi

in R so Rx -- fly i

in R and are relatively an h with ~ h ~

u~

U(R).

Therefore Remark. prime U(R')~

xE~

C

Thus ~ h

U y i el

,

(b)=

and ~

lie

is torsion we can find

= (c), ~ h

= (d),

Those

ideals

are so we can write

= d-lE

R'.

But d G R C

= (d) -- R so ~

R' so

= R and Rx = ~ .

R.

ideal of R with y n ~

# y because

:

Let

h-1 = Rcd -1 so x h = ucd -1 where

This is false without

R = U(R).

R by R S we can

say, where ~

and ~

u-lgx h + k

R = U(R).

Now (a)

Since cl(R)

prime since ~

Therefore

d E U(R')~

_-~-i

Let ~ h

Now Rx h ~ ~ h ~

Replacing

We must show that R' = R.

prime.

R ~ ~h.

are relatively gc + k d = 1.

ei-fi

of K contain-

R S -- U(R S) since r/s ~ U(R'),

so r ~ S.

R = U(R).

Assume

R.

x ~ R' and write x = a/b where a, b ~ R. fly i

field K.

the hypothesis

R for n # O.

In fact if x ~ U ( R ' ) ~

R~_~--

R~.

on cl(R).

Let R' ~ ~ y - n .

Let y be a Then

R then ord~#&x ~ 0 for

If x ~ R, then ordyx ~ -n ~ 0 so

- 227-

(x) = y-n and y n ~

R.

For a treatment

of the general case see

w § 4-5. We now turn to the case where K has characteristic Proposition A2~.

Let K be a global field over a finite field k.

Let R' be a Dedekind ring w i t h quotient field K. x ~ R' such that K/k(x) is a finite separable the integral closure of k[x] Proof.

For the existence

nitely generated

Then there is an

extension.

x i must be transendental

R.

of x we only need to know that K is fiand R is any set generating

Since K is finitely generated,

..., x n ~

If R is

in K then R' ~ R S where S = U ( R ' ) ~

over k, k is perfect,

K as a field over k. finite set Xl,

p # O.

R such that K = k(Xl, over k so K/k(x)

we can find a

..., Xn).

is algebraic.

Some x Choose

such an x such that the largest possible number of xj are separable over k(x). Xr+l,

Suppose Xl,

..., x n are not.

..., x r are separable

Let y = Xr+ 1 and let h(Y) = 0 be the mini-

mal equation of y over k(x).

Clearing denominators

that h(Y) = f(x, Y) where f(X, Y ) ~ y is not separable

over k(x) and

we can assume

k[X, Y] is irreducible.

over k(x), we have f(x, Y) = fl(X, YP).

Since If we

had f(X, Y) = f2(X p, YP) we could write f(X, Y) = g(X, Y)P using the fact that k is perfect to extract a p-th root of each coefficient of f.

This would imply g(x, y) = O, an equation of degree

smaller than h. is irreducible k(g).

Thus f does not have the stated form.

If f(X, y)

over k(y), this implies that x is separable over

But then so are Xl,

..., x r so we have a contradiction.

- 228 °

Suppose f(X, y) = p(X, y)q(X, y). (otherwise

it is separable

Since y is transcendental

over k (which is perfect)).

over k

Gauss'

lemma shows that f(X, Y) factors in k[X, Y] which is again a contradiction. The rest of the proof is just like that of Proposition A21. Combining Proposition A21 and A23 we get Corollar F A24.

If R is a Dedekind ring whose quotient field is a

global field then R satisfies

the Jordan-Zassenhaus

theorem.

We also remark that any integrally closed domain whose quotient field is a global field is a Dedekind ring.

In fact the

above proofs used only this property of R'. Remark.

If K is a global field over a finite field k, the field of

constants k' of K is, by definition, K.

the algebraic

closure of k in

This is also finite since a subfield of a finitely generated

field extension is also finitely generated. from Proposition A23 since k [ x ] C

k'[x]~

We can also see this

R so k'[x]

is finitely

generated as a k~x]-module. We conclude by giving some alternative proofs of the strong approximation

theorem

(SAT) Theorem 9. 21

.

The following standard

result shows a connection between the SAT and the theory of Dedekind rings. Theorem A2~.

Let K be any field.

Then there is a i-i correspon-

dence between Dedekind rings R with quotient field K and sets E of valuations

on K such that

- 229 -

(1)

Every valuation y E E is discrete.

(2)

If x E

(3)

E satisfies the SAT.

K, x ~ O, then IXly = i except for a finite number of y.

If R corresponds to E, then R = ~ @ where @y is the valuation yGE y ring of y, and E is the set of valuations ordy given by the DVR's Ry, y i O . Proof. (2).

If R is given, then E = (ordy) clearly satisfies

(1) and

(3) follows from the Chinese Remainder Theorem (CRT) if we

observe that E satisfies the SAT if and only if given Yl' Y2 ~ E, Yl # Y2 and E > 0 we can find @ ( ~ and ICily ~ 1 for the other y 6 E. the CRT (Theorem AlO).

the CRT.

K with l@~ly I

K-Theory of Finite Groups and Orders

K-Theory of Finite Groups and Orders

K-Theory of Finite Groups and Orders

Whitehead groups of finite groups

Classes of Finite Groups

Classes of finite Groups

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Moonshine of finite groups

Finite Groups

Enumeration of Finite Groups

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Injectors of finite groups

Atlas of Finite Groups

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REPRESENTATION THEORY OF FINITE GROUPS

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K-Theory of Finite Groups and Orders