IDEMPOTENTS IN GROUP RINGS

by

Zbigniew Marciniak

Dissertation submitted to the Graduate Faculty of the

Virginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

DOCTOR OF PHILOSOPHY

in

Mathematics

APPROVED:

D.R.Farkas, Chairman

E.L.Green R.F.Dickman,Jr.

C.Feustel K.Hannsgen

July, 1982

Blacksburg, Virginia

ACKNOWLEDGEMENTS

First of all, I would like to thank the best teacher

I have ever had: Dr. Daniel Farkas. He was able to share

with me his enthusiasm for mathematics. He succesfully taught

me the most difficult thing: how to progress in mathematics,

as, I hope, this dissertation demonstrates. His talent,

intuition and generosity in sharing his work helped me to

avoid many dead ends in my research. Finally, he was ready

to help in all sorts of every-day problems.

Secondly, I would like to thank all the faculty

members of the Mathematical Department I have met. At least

two people should be mentioned by name: Dr. William Greenberg,

who spent a tremendous amount of energy on arranging my

study at the VPI, organizing my stay here, finding a job

for my wife and many other things, including invaluable

advising. The other is Dr. Raymond Dickman,Jr., whose

teaching skills allowed me to enter his area of research.

Thirdly, I should express my thanks to my wife Danuta.

Over the last two years, while I was studying, she was

taking care of the every-day problems of our family. This

included one especially difficult year of her stay in

Poland, when she had to work, take care of our children

ii

and also stand in food-lines. If not for her devotion,

I would be far from the place where I am now.

Finally, I would like to thank all my friends in

Blacksburg, who, although busy, have helped my wife

while I was away in Wisconsin.

iii

TABLE OF CONTENTS

Acknowledgements •.•.•.•.•...•.•.•.•••.•.•.•.•.•.•.•.• ·•·• ii

Chapter page

I. INTRODUCTION

1. Idempotents .•.•.•.•••.•••.•...•••.•.•••.•.•.•.•.• .• 2

2 • Idempotents in group rings .• .•.•••••.•.•.•.• • 6

3. Results about the trace of an idempotent .••• • •• 9

4. Results about the support of an idempotent .• .... • • 14

II. THE LIFTED TRACE

1 • Witt vectors .•.• . .............................. . •• 1 7

2 • Lifting idempotents ••.•••.••• . . . . . . . . . . . . . . . . . . . . • 28

3 . Invariance of traces ••••••••• . . . . . . . . . . . ........ . • 30

4. Relations among tc(e) for different c. Integrality of tr('€) .•.•.•.•.•.•••.• .......... • • 3 3

5. The lifted trace for finite groups •.•.•.•.•.•••.•. 42

I I I. THE NONVANISHING PROPERTY

1 • Polycyclic-by-finite groups .• .................... • 4 5

2 • Residually finite groups ••••• . . . . . . . . . . . ..... • 4 7

3. Estimates of traces in characteristic zero .• • • 4 9

4. An estimate of traces in characteristic p ••• • • 54

5. Idempotents with small supports .• . •. 56

6. Some combinatorics .•.•••.•.•••.•••••.•.•.•.•.•••.• 66

iv

IV. SUPPORTING SUBGROUPS OF IDEMPOTENTS

1. Classical examples of idempotents

supported on infinite groups ••••.•.•••.•.•••.•.•. 70

2. Idempotents with infinite supporting groups .•.••• 73

3. Supports of central idempotents .•.•.•.•.•.•••.•.• 76

V. LITERATURE CITED•••••••••••••••••••••••••••·•·•·•·•·•81

Vi ta ••••••••••••••••••••••••••••••••••••.•••••••••••••••• 85

v

I. INTRODUCTION

As early as 1940, G.Higman asked in his D.Phil. thesis

whether it was possible to have one-sided units in group

rings, i.e. elements x and y such that xy=l but yx#l.

He was working on units in integral group rings, a topic

essential in the study of the Whitehead group. (He was J.H.C.

Whitehead's student.) The question was important, because

a negative answer guaranteed that all units form a group

[Higman (40aj .

A similar question appeared fourteen years later in

a paper by W.Cockcroft [cockcroft (54)]. He was trying to

prove the famous Whitehead Conjecture which states that a

subcomplex of a two-dimensional, aspherical CW-complex is

aspherical. Equivalently, he was proving that if L is a

non-aspherical complex, then it cannot be made aspherical

by adding 2-cells. If the fundamental group of L is free

LwUe: J

and is aspherical, an easy topological argument

shows that and Jr 2 ( L) Ee 2G n = ZG n for some n,

where G= ~l (L). This is equivalent to the existence of

matrices x and y in M ( ZG) n such that xy=l but yx#l.

Cockcroft showed that this cannot happen with G free and

he asked if other group rings have a similar property.

1

2

Kaplansky was able to prove that xy=l implies yx=l if

x and y belong to k[cl, for any group G and for any

field k of characteristic zero. His approach was to study

the idempotent e=yx. (Notice that 2 e =y(xy)x=y.1.x=e.)

The argument used methods of analysis. Fnr fields of positive

characteristic these methods do not work and the correspon-

ding question is still open. This question motivated a large

part of the work on idempotents presented in this dissertation.

1. Idempotents

We now review the basic facts about idempotents. In

any ring with 1, an element e is called an idempotent if 2 e =e. We always heve the trivial idempotents, namely 0 and 1.

Usually, there exist many others. For instance, if k is a

field, then any projection of kn onto its subspace gives

rise to an idempotent matrix Ee M (k). It is an easy exer-n

cise to see that the space kn decomposes into a direct sum

ImE e Im(l-E) and E acts like on ImE and like 0 on

I (1 E) If h b · f kn the un1'on of bases m - . we c oose as a as1s or

for ImE and Im(l-E), we see that E is similar to a

matrix with IS and O's on the diagonal and O's else-

where. Obviously, the number of 1 I S is equal to the rank

of E. Consequently, rank is the full invariant of E. In

3

particular, rank(E)=O implies E=O.

From these remarks we can conclude that the rank of

E is equal to the matrix trace TrE of E when the

underlying field has characteristic zero. Indeed, it is

obvious for our special basis; but neither rank nor Tr

depend on the choice of a basis. That suggests that Tr is

a powerful tool for the study of idempotent matrices.

Let us keep in mind two properties of the function

Tr: M (k)~k: n it is k-linear and Tr(AB)=Tr(BA) for

any matrices A and B.

To generalize the matrix example we can consider

any ring R with 1 in place of k. For any idempotent

matrix E in M (R) n the direct sum decomposition:

Rn = ImE m Im(l-E)

is still valid, and so ImE is a projective R-module.

Conversely, every finitely generated, projective R-module

can be obtained in this way. This creates a link between

idempotents and algebraic K-theory.

Now, we can generalize even further and instead of

endomorphisms of the free module we can consider

the ring EndR(M) of endomorphisms of an arbitrary R-

module. If M decomposes into a direct sum of R-sub-

modules then the projections

form an (orthogonal) system of idempotents in EndR(M).

4

One can show that M is indecomposable if f EndR (M) has

only trivial idempotents.

When we take M=R, the regular R-module, the sub-

modules become (left) ideals and we are now looking at

decompositions of R into a direct sum of left ideals:

It is easy to see that the components e. l

of 1 in this

decomposition are idempotents. Moreover, I.=Re. for all i.

If the ideals

l l

I. happen to be two-sided, l

the

idempotents e. l

are central: they commute with all elements

of R. This is the situation, for instance, when R is

semisimple. By Wedderburn's Theorem [Curtis-Reiner (81),

Thm.3.22) R can be decomposed into a direct sum of two

sided ideals, which are simple Artinian rings. The

corresponding idempotents are called centrally primitive.

Of course, it may happen that two idempotents give

rise to isomorphic R-modules: Re~ Rf. Equivalent con-

ditions can be spelled out in terms of elements in the ring:

Re ~ Rf if f there exist a and b in R such that

e=ab, f=ba, af=ea=a and fb=be=b ~aplansky (68~

[Passman (77), Lemma 10.1.4]. We call the idempotents e

and f equivalent and write e"""' f.

The existence of a sufficient amount of idempotents

in a ring is important in many situations. For example,

s

* when studying a W -algebra of operators on a Hilbert

space, we need the spectral projections. This leads

naturally to the concept of a von Neumann regular ring

[Goodearl (79~. These rings have plenty of idempotents;

in fact, every finitely generated left ideal is generated

by an idempotent lGoodearl (79), Thm.1.1].

Finally, we can put Cockcroft's observation in

a general context. LEt R be a ring with 1. A left

R-module M is von Neumann finite if it is not iso-

morphic to any proper direct s11mmand of itself. That

is M EB N = M implies N=O. We have

Proposition 1 .1 [Good earl (79), Lemma S .11 A left R-

module M is von Neumann finite iff xy=l for x,yE.EndR(M)

implies yx= 1 •

The ring R itself is von Neumann finite if xy=l

for x,yE.R implies yx = 1 • The name comes from John

* von Neumann who introduced the concept for W -algebras

to characterize those for which the lattice of projections

forms a continous geometry [Goodearl (81)].

Let k be a field of characteristic zero. Here

is a simple proof that the matrix ring M (k) n

is von

Neumann finite. Let AB=l for some matrices A and B .

Then E=l-BA is an idempotent matrix and

TrE = Tr(l) - Tr(BA) = Tr(l) - Tr(AB) = 0.

6

Consequently, rank(E)=O and so E=O, by the first remarks

of the paragraph. Therefore BA= 1 . We shall turn to the

problem of fields of positive characteristic in paragraph 3.

This proof can be repeated in any k-algebra R, as

long as we have a k-linear map t: R---+k which

satisfies t(ab)=t(ba) for a,b f. R and which has the

following "nonvanishing property": t(e)=O implies e=O

for idempotents e <: R. As we shall see in the next

paragraph, group rings constitute examples of k-algebras,

for which such functions can be constructed.

2. Idempotents in group rings

We now consider group rings k[G). A typical

element of this ring will be written as a= L a(x)x

where xEG and only finitely many coe~ficients a(x)E k

are different from zero. We recall two most important

structures associated to an element of this ring: the

support and the trace.

For any element aE k(c] we define the

of a to be [ x f= G a(x)1o1. This subset is

supp(a). It is always finite. Moreover, if a

support

denoted

is a central

element of k[G] then supp(a) is a union of full conjugacy

classes. More precisely, c is a

7

conjugacy class in G, and a €: k c (Passman (77), Lemma 4.1.1]. The subgroup of G generated

by supp(a) will be called the supporting subgroup of a.

(We will typically write (S) for the subgroup generated

by the subset S.)

To find a reasonable definition of the trace on

k[G], we consider the case of a finite group G. The

regular representation of G identifies k(G] with a

subring of the matrix ring M (k), where n n= I GI , by

sending each element a E:. klG1 to the linear transformation

of k(G1 "multiply on the right by a". Consequently, we

can evaluate the matrix trace Tr on an element a=L_a(x)x

of the group ring. Easy calculation shows that Tr(a)=

a(l)·IGI. Now, -1 \GI . Tr still possesses the two basic

properties mentioned in paragraph one. Therefore it seems

reasonable to expect that the function tr:k(G}---? k

defined by tr(a)=a(l) should have trace-like behavior.

(Notice that the definition of tr with the lower case "t"

makes sense even for infinite groups.) One can easily

prove that in any case

i) tr is a k-linear map

and ii) tr(ab)=tr(ba) for all a and b in k[G].

As an immediate corollary, we find that equivalent idempo-

tents have equal traces.

8

Now, there exist other functions from k[Gl to k, satis-

fying conditions i) and ii). To find them, notice

that each such function t is determined by its values

on G (condition i)) and that it must be central:

for all x,yE.G,

by condition ii). Consequently, t must be constant

on each conjugacy class C in G. To each conjugacy

class C in G we can associate a trace function tC'

defined on G as the characteristic function of C:

tc(x) = 1 for x EC and tc(x) = 0 for x Ei G'-C. In

particular, tc(a) =L,{a(x) I x t: c\ for any a€ k (c1 .

Notice that c = { 1} yields the trace tr from which

we started. The preceding discussion shows that tc

satisfies conditions i) and ii) ' and again tc(e)=

tc < f) if e ....._ f, 2 and f 2=f. e =e

If e € k [cl is an idempotent then the numbers

provide a lot of information about e. For instance,

i f G i s f i n i t e , t h e n [ t C ( e )f completely determine

the character of the k(~ -module k (G1 e (Hattori's

Lemma in [Hattori (6S)J, [sass (78), Prop. 5.8]).

The traces tc can be applied to the study of

finitely generated, projective k(G]-modules. For an

idempotent matrix Ef M (klG]) n describing a projective

module P, J.Stallings and A.Hattori constructed

(independently) the rank element

9

which is a formal combination of conjugacy classes.

[Stallings (65)}, LHattori (6S)l . It proves to be an

invariant of P. Its properties were studied further

by H.Bass in [Bass (78A .

The lll - coefficient of rp, equal to tr(TrE),

is known as the Euler characteristic of the projective

module P. It has found a nice application in lFarkas-

Snider (76)].

3. Results about the trace of an idempotent

From the introductory remarks we concluded that

the group ring k[~ is von Neumann finite if the trace

tr has the nonvanishing property. This was proved for

fields of characteristic zero by I.Kaplansky.

Trace Theorem 3 . 1 ( [Kap 1 ans k y ( 6 9 )} ; [Passman ( 7 7) , Th m • 2 . 1 . 8])

Let k be a field of cgaracteristic zero and let e be

an idempotent in k[G), different from 0 or 1. Then tr(e)

is a totally real, algebraic number with the property

that it and all its algebraic conjugates lie strictly

between 0 and l •

Thus, tr(e)=O implies e=O and so k[G) is von Neumann

10

finite. The Trace Theorem also has other interesting

applications. Here we quote two of them.

Corollary 3.2 The only idempotents in the integral group

ring 2[G1 are 0 and 1. Thus 2(G1 is indecomposable as

a left (or right) Z\G1 -module a

Theorem 3.3 (Farkas-Snider (81)1 Assume chark=O and G11.

Then k(GJ is not a primitive ring with a minimal left ideala

From the discussion of the matrix trace Tr it

follows that tr(e)=IGt-.1dim t[G]e for any idempotent

e from the group ring C[G) of a finite group G ( C denotes

complex numbers). Consequently, tr(e) is a rational number.

Kaplansky conjectured, and Zalesskii proved, that this is

true for all groups.

Theorem 3.4 ((Zalesskii (72)] ;lPassman (77),Thm.2.3.S})

Let k 2 be any field and let G be any group. If e =e E klGl

then tr(e) lies in the prime subfield of k •

All known proofs of the Trace Theorem are analytic; by

contrast, the proof of Theorem 3.4 is algebraic and

number theoretical.

Corollary 3.2 is an example of a theorem about the

nonexistence of idempotents. Similar results can be found

in (Coleman (66)] ,[Curtis-Reiner (81)] and [Cliff-Sehgal(77)J

If we consider group rings with field coefficients we have

1 1

the following results:

i) If chark=p>O and G is a p-group then klG1

has only trivial idempotents. This is just

Nakayama's Lemma. The assumption about G is

essential: if G has an element of order q

with p.(q then -1 q-1 e=q ·O+x+ .•. +x ) ,

is a nontrivial idempotent.

ii) If k is arbitrary, G is torsion-free and

either right-orderable or polycyclic-by-finite,

then k [G] has only trivial idempotents. (G right-

orderable (Passman (77) ,Ch.13$2], G polycyclic-by-

finite [Formanek (73)1, ~Cliff (80)) )

The second case is not entirely honest. Here the Zero

Divisor Conjecture has been verified: the group ring of a

torsion-free group has no zero divisors. (G right-orderable

[Passman (77)], G polycyclic-by-finite [Farkas-Snider (76)1,

(Cliff (80)j)

In Kaplansky's fundamental Trace Theorem we assume

that chark=O. In characteristic p>O the nonvanishing part

of the theorem, tr(e)=O implies e=O, still makes sense

but is no longer true.

Example 3.5 Let and e=x + 2 x t: F 2 [ C 3} •

Then 2 e = e I 0 but tr(e)=O.

12

Let E be the 3x3 matrix representing right multiplication

by e. It is easy to calculate that -1

rank 1 E = 2 and TrE=O. 2

In particular, tr ( e) = l C 3\ • Tr E = 0 , so it does not reflect

the fact that rankF E ~ 0. Thus, tr does not have the 2

nonvanishing property here, and so it cannot be used to

verify the von Neumann finiteness problem in characteristic

p. However, we can apply the approach well known in modular

representation theory.

In general, if E is an idempotent matrix over

we can lift E, using Hensel's Lemma [Hasse (81~ in the

F , p

following sense: there exists an idempotent matrix E with A

entries in the ring of p-adic integers 2' , p such that

E = E(mod p). Moreover, rank 2 E = p

if E#O. Notice that the argument

rankF E, and p

proving von

so TrE#O

Neumann

finiteness for matrices over fields of characteristic zero

now adapts to matrices over F . p

We shall follow this pattern in Section II as follows.

If e fi F (G] is an idempotent for any group G, then we lift p

e to an idempotent e E: i- [( G]) so that p

,.. e :!! e(mod p)

( 2 ) ... 1( 2) .. ( 1 Theorem II .. 1 . For instance, e=3 2-x-x (: 2 2 c 3 lifts

e from Example 3.5. Here l (\Gll denotes the ring of p

infinite series [.a(x)x with I a< x) I p 0. The traces

tc(e) E ip depend on e only and do not change within its

equivalence class (Corollary II.3.4). Thus, tc(e) are

invariants of e, written ltC(e). In fact, the results

1 3

we have described are proved with r p replaced by any

field k of characteristic p. The traces ltc then

have values in the ring of Witt vectors W(k). However,

1 t r ( e ) = 1 t { 1 \ ( e ) a 1 w a y s r e mQ)an s w i t h i n t h e c an on i c a 1 ..

copy 2 in W(k) (Theorem II.4.7). p

As a result, we obtain the trace function ltr with ,..

values in the ring z . p The van Neumann finiteness problem

for k[Gl, chark=p > 0, would be settled if we could prove

a p-version of the Trace Theorem: ltr(e)=O implies e=O.

This is the nonvanishing property we study in Section III

of this dissertation. We prove it for polycyclic-by-finite

groups (Theorem III .1 .1). In fact, we obtain there the full

analogue to the Trace Theorem: if e is an idempotent in

k ( GJ different from 0 or 1, then ltr(e) is a rational

number lying strictly between 0 and 1.

In his proof of the Trace Theorem, D.Passman studies

the L2 -norm on C[GJ. In the characteristic p case we

formulate a conjecture about the p-adic norm:

-log lltr(e)I ~ lsupp(e)I. p p

If it is true in general, it clearly implies the non-

vanishing property. If it is true for all finite groups,

it implies the nonvanishing property for residually

finite groups (Proposition III.2.1). In this dissertation

we verify this conjecture for central idempotents (Theorem

III.4.2) and for idempotents in k[G) if the order of G

14

does not exceed 30 (Proposition III.5.4). We also show

that the conjectured estimate is the best possible (Example

III.4.3).

We close this paragraph by mentioning the following

result due to G.Cliff and S.Sehgal, which bridges the

results of Sections III and IV.

Theorem 3.6 Cliff-Sehgal (77) Let

of a polycyclic-by-finite group G

characteristic zero. Let e=[e(x)x

k(G1 be the group ring

over a field k of

be a nontrivial idem-

potent, and write tr(e)=r/s with (r,s)=l. If a prime p

divides s, then there exists 11g~ G of p-power order

with e(g)#O •

This theorem is obviously true for finite groups. If every

idempotent were equivalent to one whose supporting subgroup

was finite, the result would follow. Unfortunately, this is

not the case, as we show in Corollary IV.2.2.

4. Results about the support of an idempotent

We start with a result which describes the nature of

the support for central idempotents.

Theorem 4 .1 ( [Osima (SS)l , (Bovdi-Mihovski (70)], [Burns (70)1,

[Passman (77)1) Let e be a central idempotent in klG1.

Then

and

i) supp(e)

ii) supp(e)

1 5

consists of p-regular elements

is a finite normal subgroup of G •

However, we cannot expect all idempotents to have finite

supporting subgroups.

Theorem 4.2 ((Bovdi-Mihovski (73)} ,tPassman (77), Thm.4.3.91)

Suppose the supporting groups of all idempotents of k(Gl

are finite. Then either all idempotents of klG\ are

central, or G is locally finite •

Section IV starts with a review of two classical

constructions of idempotents with an infinite supporting

group. All idempotents obtained in this way are equivalent

to idempotents supported on a finite group. The constru-

ctions give credence to the well known conjecture that

all idempotents are equivalent to one supported on a finite

group. This idea seems to have sugwated with the fact that

tr(e) is a rational number for eECfG1 and that this

ratio has an "explanation" for finite groups G (compare

with Theorem 3.6). However, the conjecture is not true, as

we show in Corollary IV.2.2.

Finally, we find an estimate on the order of the

supporting subgroup of a central idempotent. It is applied

in the proof of Conjecture III.4.1 for central idempotents.

II. THE LIFTED TRACE

In this section we construct and investigate

invariants of idempotents in k [ G1 for fie 1 d s k of positive

characteristic. We are interested in the lifted traces

constructed for each conjugacy class c in the group G.

These are functions with values in W(k), the ring of Witt

vectors of k.

In the first paragraph we review the construction

and properties of Witt vectors. We construct here the

"completed" group ring with coefficients in W(k), written

W(k) ([ GJ] • In the second paragraph we show idempotents are

1 if t e d f r om k [ G1 to W ( k) fl G l1 . We prove that equiv a 1 en t

idempotents possess equivalent liftings. In the third

paragraph the traces ltc(e), associated with e, are defined

as the corresponding traces of a lifting e of e. We

show that is independent of the way one lifts e and

that is the same for equivalent idempotents.

In paragraph four we prove that certain relations hold

between the traces tc(e) for different

corollary we find that ltr(e)= lt ll} (e)

classes C. As a

is always a

p-adic integer. Finally, in paragraph five we remind the

16

1 7

reader of the explicit formula for ltr(e) in the case

that G is a finite group.

1. Witt vectors

We begin by reviewing the properties of Witt vectors.

The appropriate context is the study of fields with valuations.

Let K be a field. A real valued function \.l :K--+ R+

is called a valuation on K if it has the following pro-

perties:

i) Ix\ = 0 iff x = 0

ii) lxyl \ X\ \ y I

iii) [ X +y\ $ IX l + I yl for all x,yc:K.

By analogy with the complex absolute value, we can make K

into a metric space with the distance d i s t ( x , y ) = I x - y! •

The metric space (K,dist) may be completed to a field,

which has as valuation a natural extension of the original

one [Hasse (80), Ch.8].

All valuations fall into two categories: the

archimedean valuations (including the classical absolute

value) and the non-archimedean ones. A valuation is

non-archimedean provided \ m\ ~ 1 for all integers m.

This implies, in particular, a stronger version of the

18

triangle inequality [Endler (72), Ch.I,$4,Thm.l.141

iv) \x+yl ~ maxt\XI ,1y1\

Sometimes it will be more convenient to work with valuations

in the exponential form: v :K-Rv\.o:i1 , where v (.)=-log I .I. p

Properties i), ii), iv) correspond, then, to:

i)' v(O) =co

ii) '

iv) '

v(xy) = v(x) + v(y)

v(x+y)~ min(v(x),v(y)~

Note that ii) ' * implies that v(K ) is an additive sub-

* group of R, where K = K' lO\

A subclass of non-archimedean valuations will be of

special interest for us. A non-archimedean valuation v is

discrete * if v(K ) is a discrete subgroup of R, i.e.

equal to ~·I! for some ~i=Jt [Endler (72),Ch.I,$4,Thm.4.1}

Let v: K--- Rvloo} be a discrete valuation. We may assume

v is normalized, i.e. * v(K )=2 [Endler (72),Ch.I,$4,Thm.4.4]

We define the valuation ring of v by RV= l x t K ( v ( x) ~

It is a local ring with the unique maximal ideal m = v

{ x ~ Kl v ( x) ~ l } . We call it the valuation ideal .

The quotient k =R /m is the residue field of v . v v v

Here is a basic example. For a fixed prime number

we have a valuation v on the field Q of rational p

numbers: v (pr:1~) =n for pnab_ ~ ~ and Pt ab. p b

completion, we get the field of p-adic numbers

We have R = z , v p the ring of p-adic integers,

After

(Q .~ ) . p p

m = pi v p

OJ .

p

19

and k =Z /pl =f , the Galois field of p elements. v p p p

If char K ; char k then it must be that char K=O v

and char k = p > 0 ~Serre (62) ,Ch.II,$51 • Because p E ZliR v v

is mapped (modulo m ) onto zero, v v(p)~ 1. The integer

e = v(p) is called the ramification index of K. When

e = then we say K is unramified.

The following theorem states that all perfect fields

of positive characteristic can be realized as the residue

field of an (unique) unramified, discrete valuation ring.

Let us recall that a field k, char k = p) 0, is perfect

if the map x.___xP is a bijection of k onto itself.

Theorem 1.1 [Serre (62) ,Ch.II,$5,Thm.31 For every perfect

field k of characteristic p > 0 there exists a unique

field K with a complete discrete valuation which is

unramified and which has residue field k •

The field K is the field of fractions of the corresponding

valuation ring R . The ring v R v

itself can be described

as the ring W(k) of Witt vectors with coordinates in k

(Witt (36)}.

We shall now review the construction of W(k). In

fact, the construction of the ring W(A) can be performed

for any commutative ring A with l [Endler (72) ,Ch.I,$5,

Thm.5.lll ,fserre (62),Ch.II,$6,Thm.6}. However, we will

content ourselves with the assumption that A is a

20

commutative F -algebra with 1. p

In this case the Witt vector ring W(A) consists of

all sequences a.E.A. 1

It is given

the structure of a commutative ring as follows. For each

W (X ,. .. ,X )"' .n o n n ~ 0 consider the Witt polynomial " . n-1

- ~ 1 p .,X)-LpX. defined by W (X ,. • n o n l•O 1 Thus

w = x 0 0

wl = xP + pXl 0

n n-1 w = xP + pXP + .•. + nx

n 0 1 p n

:l(X ,.,X\ o n

Notice that can be reconstructed from w0 ,w 1 , ...

as polynomials with coefficients in l{p- 11. Therefore, if

aEW(A) then the sequence is sometimes referred

to as the list of secondary components of a [Hasse (80),

Ch.10,$4}. Addition and multiplication in W(A) will be

defined as the corresponding operations on the secondary

components: if a,bEW(A) then

W (a+b)=W (a)+ W (b), n n n W (ab)=W (a)·W (b). n n n Here, the right-hand side operations are to be performed

in A"W coordinatewise. More precisely, there exist poly-

nomials S (X ,.,X ,Y ,.,Y ), P (X ,.,X ,Y ,.,Y) n o n o n n o n o n in

l(X , ••• ,X ,Y , ••• ,Y] o n o n such that for all n ~ 0:

W ( S , ... , S ) = W ( X , ... , X ) + W (Y , ... , Y ) n o n n o n n o n

2 1

W (P , ... ,P) = W (X , ..• ,X )·W (Y , ... ,Y) n o n n o n n o n

(Hasse (80), Ch.10,$4], lserre (62), Ch.II,$6,Thm.5].

2[p -l·1 Notice that S , P are defined over ~, not , so n n

we have no trouble applying them to a E W(A). Hence for

a, b~W(A) we can define:

a+ b = (S 0 (a,b),s 1 (a,b) , ... )

a·b = (P 0 (a,b),P 1 (a,b), ... ).

We have S =a + b , s 1 =a 1 + b 1 + p- 1[aP+bP-(a +b )P] 0 0 0 0 0 0 0

' ... P 0 = a 0 b 0 , P 1 =b~a 1 + bia~ + paib 1 , ...

In the ring W(A) the zero element is represented by

(0,0, ... ) and the unit by (1,0, ... ). If A is an integral

domain (of char p) then W(A) is an integral domain of

characteristic zero (Endler (72), Ch.I,$5,Thm.5.111. From

the formulae for S and P it follows that the projection 0 0

on the first coordinate (written ~A) is a ring homomorphism.

On the other hand, the formula r(a)=(a,O, .•. ) gives a

multiplicative (but not additive) function r :A -w(A).

If f :A-B is an F -algebra homomorphism, we define p

W(f):W(A) W(B) by W(f)(a 0 ,a 1 , ... )=(f(a 0 ),f(a 1), ... ).

In this way W(.) becomes a functor from the category of

commutative F -algebras to the category of commutative p

rings of characteristic zero. Moreover, the following

diagrams commute:

W(A) W( f) W(B) 'lt l l 1t

A f B A B

22

Also, if f :A~ B is an imbedding, then we can identify

W(A) as a subring of W(B). Since we always have

we also have W(F ) c:;, W(A). p

F ~ A p

,., The ring W(F ) can p be identified with the ring l

p

of p-adic integers. To see this, choose E.E.z to p be a

primitive th (p-1) root of unity, using Hensel's Lemma

[Hasse (80), Ch.10,$31. Then the set of so-called

- I 2 p - 2t TeichmUller representatives S-lO,l,L,l , ... ,l 1 maps

modulo p onto r . p Denote by 6' :f - s p the inverse

bijection. Then ). :W(F )---ti given by }((a )~ ) = p p n n=o

is the required isomorphism tEndler (72) ,Ch.I .ss1. The shift map V:W(A)--~W(A)

is an additive function whose image VW(A)

is an ideal. Let TC :W(A)--~w (A)=W(A)/VnW(A) n n be the

natural projection. Clearly, the canonical homomorphism

from W(A) to W(A)/ImV is just 'ii: so we use A' 1 and

1t A interchangeably.

The formula F(a 0 ,a 1 , ... )=(a~,af , •.. ) defines a

ring homomorphism (the Frobenius homomorphism) F:W(A)--W(A)

(Serre (62), Ch.II,$6J. Also, F 0 V = V°F is multiplication

by p. In particular, F can be factored

obtaining

For any

F :W (A)-----W (A) n n n

F -algebra homomorphism p

for all n ~ 0.

f:A--+B the diagram:

23

W (A) ___ W_.;(_f-'-)--~ W ( B )

F l l W(A)---W-'-(f~)'-------+W(B)

F

commutes. Indeed: F•W(f)((a.))=F((f(a.)))=((f(a.)p))= i i i ( ( f (a~))) =W ( f) ° F ((a.)) .

i i

Lemma 1.2

Proof: We can reduce to the case A is a domain since

every commutative algebra is the image of a polynomial

algebra. In this case F:W(A)~~~W(A) is an injective

· h h° C ·d Vn(a)·Vm(b) ring omomorp ism. onsi er where a,bEW{A).

Then Fn+m(Vn(a)Vm(b))=Fn+mVn(a)·Fn+mVm(b)=pnFm(a)· pmFn(b)=

pn+mFm (a)· Fn (b) =Fn+mvn+m (Fm (a) Fn (b)). Hence

Vn(a)· Vm(b)=Vn+m(Fm(a) ·Fn(b)) E. Vn+mW(A) •

Corollary 1.3 Let and j=o

W(A) is I-adically complete.

Proof: We have I=VW(A), so Ij~ VjW{A) for all j. Thus .:io co . n Ij C n VJW(A) = (0). For the completeness, let {ak}

j=l j=l be an I-Cauchy sequence in W(A). Then for any N there

exists r(N) such that for all n,m) r(N) holds:

an - am E IN <;; VNW(A). Consequently, the first N coordinates

stabilize: r(N) r(N)+l ai =ai = ... , 0 .:. . .:: - i - N. Then the element

a = ( a ~ ( N ) ) ; = 0 E W ( A ) i s t he I - 1 i m i t o f la k) •

24

For n ~ 2 we have Vn-lW(A) ";> VnW(A). Hence there exist

natural projections

we can reconstruct W(A).

Pro po s i t ion 1 . 4 t Serre ( 6 2 ) , Ch . I I , $ 6} W(A)=lim(W (A), !.A. )a - n In

Finally, asume that A is equal to a field k of

characteristic p. Let K be the field of fractions of W(k).

F o r a E. W ( k ) ' \ O\ 1 e t v(a)=min l i \ a.#0\ EN. We can now l.

define v:K---.. ~u\.°"} by v( O)=o::> and v(a/b)=v(a)-v(b)

for a,b EW(A). It is easy to see that v is a valuation

on K. Moreover R =W(k), m =VW(k) and k = k. Clearly K v v v is unramified, as the ramification index e=v(p·l)=

v (VF ( 1, 0, ... )) =v ( 0, 1, 0, .•. ) = 1.

Let k be a finite field, say f lk\=p =q. Later, we

shall need a more precise description of W(k) and its

field of fractions K. The following statements are contained

in (Koblitz (77), Ch.III,$31 and (Serre (62), Ch.IV,$4,Prop.161

Proposition 1.5

i) K is the unique unramif ied extension of Qp of degree f.

th ii) K = Q ( n , where p l is a primitive (q-1) root of 1.

iii) W(k) can be identified with the ring R of algebraic

integers in Qp(t).

iv) Gal(K/Q )=Z/fZ and it is generated by an automorphism p

F such that F(z)=zp for every (q-l)th root of

unity zEK.

25

v) The above automorphism F preserves W(k)~ K; its

restriction to W(k) coincides with the Frobenius

homomorphism F(a 0 ,a 1 , •.. )=(a~,ar ••.• ) .

vi) The elements 2 q-2 0,1,l.,t. , ...• ~ (mod p)Eok

all distinct.

vii) The ideal

particular

m =VW(k) v coincides with pR;

* * viii) W (k) n n n = R/p R' pR/p R, whence I w <k )\ n

are

in

n n-1 =q -q •

Now we let a group G come into play. Let A be,

again, any commutative F -algebra. For each n ?- 1 we can p

form the group ring W(A)lG). n If n ~ 2 we also have the

homomorphism u. :W (A) tG1 - W l (A) l GJ induced by 1 n n n- the

mapping of coefficients. These homomorphisms form an inverse

system.

Definition 1.6

The elements of W(A) [LGll a re of the form L, a(x)x, where xeG

ixEG \\a(xpl} a(x) E. W(A) and where, for each m E: N, the set m

is finite. Notice that if IGI< co then all the sums are

finite, so we obtain the standard group ring with coefficients

from W(A). Let 1t :W(A) [[G11--~w (A)\:G1 n n denote the

projections associated with the projective limit. From oO •

Corollary 1. 3 it follows that n I l lG11 J= (0) where j = l

I [ [ G11 =ker 7t' 1 . Like in the proof of Corollary 1. 3 we can

26

show that the ring W(A)[(Gj} is I ((G11-complete.

The ring W(A)llGD inherits, through a limiting

process, objects defined for classical group rings. For

example, if cc;; G is a conjugacy class, we have for all

n the trace (see paragraph I .2) tC:Wn(A) [G1---Wn(A).

The diagrams

f-n

obviously commute. Thus we can define tC:W(A)\.\.G1\~w(A)

by tc= lim(W (A) LGl--w (A)). Alternatively, we can - n n

define tc(La(x)x)='2:.la(x) \ x E c) where, if the sum is

infinite, it is understood as the limit of its partial sums

in the VW(A)-topology.

The Frobenius homomorphism F:W(A)~---W(A)

induces F :W (A)lG]---w (A)(G) n n n for all n. Again, the

diagrams F

::r[G]--n-~ wnT~:1

wn-1 (A) (Gi--F __ __. wn-1 (A) lGl n-1

commute and so we have F: W(A)((G11-----+W(A) tlG1\ with

If f:A--B is any F -algebra homomorphism, then p

from the commutativity of the diagrams

27

W (f) (G} n

Wn (Bi ~:1

wn-l (A)[Gl wn-l (f) (Gl wn-l (B) [Gl

we get a ring homomorphism w(f) llG1\ :W(A) Uc\\-w(B) llc\1

Finally, we may insert traces into the above diagrams and

get

W (A) (Cl n

w (f) [cl n

~- \ W n - l ( B) (Cl

t W (A) w ( f) W ( B) n

l· c n t ,. ' (.

wn-1 (A) wn-l(f) wn-1 (B)

It is easy to see that they commute. Thus we have:

W(A) [( Gl1 W(f)(\G1\

W(B) r::J tc l

W(A) w ( f) + W ( B)

28

2. Lifting idempotents

Let A be a commutative F -algebra and let G be p

a group. We say that an idempotent e E: W{A)LlGl1 is a lifting

of the idempotent e E AlGl if -n. 1 (e)=e for the projection

1 from W(A)([Gl1 onto AlG1. In this paragraph we

review the well known fact that each idempotent e G. A\.G\ has

a lifting. (The lifted idempotent e is not uniquely deter-

mined, unless e is a central element of A(Gl.) Finally,

we shall prove that if the idempotents e and f E AtGl are

equivalent then there exist lif tings " e of e and .. f of

such that " e " and f are equivalent.

We start with the existence theorem.

Theorem 2.1 For every idem pot en t e E A \. Gl there ex is ts an

idempotent e E W(A)(l GJl with Jrl (e)=e.

Proof: Because W{A) [[ G}} =lim(W (A) lGl, IA ) it is enough to ,.______ n I n

construct a sequence e EW (A)lG1, n=l,2, .•. n n

i)

ii)

iii)

f'-n(en)=en-1 2 e e n n

e .

such that

f '

We proceed by induction. For n=l we define e 1 to be equal

to et A[G] = wl (A) [G] Let n ~ 2 and assume that we have

already constructed e . Choose any element n stWn+l(A)(G}

2 3 satisfying ?n+l (s)=en. We define en+l=3s -2s . Then

19

i) ( ) Y"n+l en+l = 3e 2 - 2e 3 = 3e - 2e = e n n n n n

ii) 2 en+l- en+l = (3s 2 - 2s 3 ) 2 - (3s 2 - 2s 3 ) =

= (s 2 -s) 2 (2s-3) (2s+l).

Notice that fln+l (s 2-s)=O, so s 2-s E ker fAn+l=VnWn+l (A) [G1.

2 2 2n 1 2 Therefore (s -s) ~ V Wn+l (A) ~G = 0 and so en+l =en+l •

Remark 2.2 General conditions which guarantee that idempo-

tents can be lifted from R/I to R can be found in

[Bourbaki (72), Ch.III,$4.6, Lemma 2]. See also ~Passman

(77), Thm.2.3.71 and (Curtis-Reiner (81), Thm.6.7}

The lifting ~ e is usually not unique. For example,

if A=T 2 , G=<x,y

r 2 lc1 then both

2 3 -1 -1> x =y = 1 , xyx =y

lift e. However, when e EA [Gl is a central idempotent,

the lifting is unique (Dornhof f (72~ .

We show next that equivalent idempotents in A(G1 can

be lifted to equivalent idempotents.

Theorem 2.3 Let 2 e=e and let a , b ~ A l G1 be such that

ab=e, ea=a and be=b. Then there exist a,bf. W(A)(l ell

such that 'ii 1 (a)=a, A ,,. A,. .'\ ..., " ,_,, A ,.. 1f11t,

~ 1 (b)=b, ab=e, ea=a and be=b.

Proof: Let e be a lifting of e and let e = il (e)~ n n

W (A) [G] for n=l ,2, ••.• It suffices to construct elements n

a , b E W (A) ( G1 f or each n s u c h t ha t 1.1 ( a ) =a 1 , ii_ ( b ) = n n n r n n n- ,- n n

b 1 , ab =e , ea =a and b·e =b . n- n n n n n n n n n

30

Let b 1=b. Assume that we have already constructed

an,bnE Wn(A)lG1. Choose s,t E.Wn+l(A)(Gl so that fn+l(s)=an,

f"-n+l (t)=bn. Define and

Then

ii) an+ibn+l=2en+lsten+l-2

en+lsten+l) . Notice that w=e -n+l

v0 wn+l (A)[G1, so w2=o. Therefore

en+l sten+l E ker fn+l =

an+lbn+l=en+l ·

Corollary 2.4 If e ,..,_ f then we can find respective

lif tings with

Proof: If e ,...., f then e=ab, f =ba for some a, b ~ Al Gl . ,. A A

By Theorem 2. 3 we have e=a1. Then f=ba is an idempotent A

equivalent to ... e and '.JI'. (f)=ba=f 1 •

3. Invariance of traces

In this paragraph we prove that the trace tc(e) is

the same for all liftings of an idempotent e E. A{G1.

Consequently, it defines an invariant of e.

Moreover, we show that this invariant has the same value

31

for all idempotents equivalent to e.

Let us recall the following technical result

which plays the crucial role in our argument.

Lemma 3.1 [ C 1 i f f ( 8 0 ) , Lemm a 1 . 2} ; see a 1 so [ 0 k u yam a ( 8 0 )} •

Let R be a ring of prime-power characteristic n p •

r is an integer, r~n, and a 1 , ••. ,amt:.R, then for

we have m ( L.

i=l

r a.) p =

l. r +

The sum on the right is over all s-tuples

If

n-1 s=p

with l~i.::m J

and~E~R,R1 =t ""'(b.c.-c.b.) l b. ,c.€. R}• L i i i 1 1 1

Let us fix a conjugacy class C of elements in a

group G • Firs t we prove that for an idempotent e ~ A l Gl,

the trace tc(e) does not depend on the way we lifted e

to ... e.

Theorem 3.2 If e,e' ~ W(A)(lG1l

1t 1 Ce)= x 1 (e')=e E A(Gl then

are idempotents with

Proof: Let f be the image of e under the map

induced by the map of coefficients r :A---W(A) n

(recall r(a)=(a,O, ... )). We will show that tc(e)=lim tc(f P ~ n-<2>

i . e . tc(e) depends on ~ 1 (e) only. Let h=e-f t w(A)f l c11

Then JC 1 (h)= 'Tt' 1 (e) - 1r 1 r(e) = 0, so hEVW(A)llcll. For

each n ~ 1 let us write, for simplicity of notation, e =1t (e), n n

f = n: (f) and h = 1t (h). From the commuting diagrams n n n n

::Arc1l W (A) (G 1 n

32

tc ------~ W (A) l ~n -------+ W (A)

tc n

we conclude that it will be enough to show that for each n r

we have tc(en) = tc(f~ ) for r sufficiently large. To do

this let us fix n. Notice that we have

pnW (A)(G1 ~ VnFnW (A)(G1= 0. n n

Therefore Lemma 3.1 can be applied with R=W (A)\~, m=2, n

a =f 1 n

where

and For r ~ n we r

= ( f +h ) p = ri n n 1-

get

n-1 s=p , a. =f or h n and ~ ~ ( R, R1 • 1 . n

J Consequently, tc(~)=O and so

tc(en) = L r-n+l

tc((a .... a. )P ). 11 1s

Assume now that r ">, 2n-1. It implies that pr-n+l ~ pn~ n.

Consider one of the products u=a. ••• a . If one of the 11 1 r-n+l s

a. IS is equal to h then u t: vw (AH G1 and so up E-1. n n J

vnw (A)( G1 = 0 . Thus, for r~2n-1, the above sum of n r

traces reduces to tc(f~ ) , as desired •

Now we are able to prove

Theorem 3.3 If e,f f A(G) are equivalent idempotents then

for any liftings e,f fW(A)Clc11.

,. Proof: By Corollary 2.4 we can find liftings e', f' of

33

e,f respectively, so that e' ...._ f'. Thus

as we noticed in the Introduction. On the other hand

~ 1 (e')= ~ 1 (e)=e and ~ 1 (f')= ~1 (f)=f, so by Theorem 3.2

we have and Putting these

equalities together, we obtain tc(e)=tc(f)a

We summarize this paragraph with

Corollary 3.4 Let k be a field of characteristic p) 0.

With each idempotent e E k ( G1 we can assoc i a t e a unique

number ltc(e) E: W(k), which is the same for all idempotents

equivalent to e.

Proof: Define for any lifting e of e •

4 . Relations among t 6 (e) for different C

Integrality of tr(e)

So far we have investigated properties of

a fixed conjugacy class C of G. Now we want to study the

relations between for different conjugacy classes.

At first, we restrict ourselves to the case that A is a

finite field k of characteristic p. Later, using a

specialization argument, we generalize our results to arbitrary

fields of positive characteristic. As a corollary we obtain

that tr(e) is always a p-adic integer, regardless of

34

the field k. Another application will be given in the

section on supports (Lemma IV.~.2).

With the number p we can associate a "Frobenius

map" f :G-- G p defined by f (x)=xp

p for x E. G •

Notice that if two elements x and y are conjugate in

G then 'f'p(x) and ~p(y) are also conjugate. Hence

t.D induces a map T p <P P : ( ~c w he r e t: d en o t es the s e t

of all conjugacy classes of G: clearly qi (C)=C' p

iff

tf (C) ~ C'. The "dynamics" of the ct> -action on C can be p p

described by an oriented graph r , defined as follows. p

The set of vertices of r coincides with (. . Two vertices p

c 1 ,c 2 f '( are connected by an oriented edge: c 1 c2

iff ¢p(C 1 )=C 2 •

< 2 3 Example 4.1 G = x,y \ x =y =l,

(. = lC(l), C(y), C(x)J 2 C(x)=\x,xy,xy ) .

Q cJ

C(x) C(y)

where

-1 -1 > xyx =y ~ s3. 2 C(l)= l11, C(y)={.y,y),

r 3

C(y)

The geometry of r; allows us to distinguish two

complementary subsets of its vertices:

35

i) p-cyclic classes: C is p-cyclic if there exists

n>O with

ii) p-acyclic classes: all others.

4"" (C)=C p

In our example, c (1) and C(y) are the 2-cyclic classes,

and the class C(x) is 2-acyclic.

Proposition 4.2 If the group G is finite then a conjugacy

class C £ G is p-cyclic if and only if it is p-regular.

Proof: Note that permutes the subset G 0

of

elements. Consequently, ...h '±" p preserves the subset

p-regular

consisting of p-regular classes and <:pp\ l: is a permutation 0

of ~ . As such, it decomposes into a product of cycles, i.e. 0

each p-regular class is also p-cyclic.

On the other hand, if C ~ 'e is p-singular, pick x "'- C.

Then p \ o(x), where o(x) denotes the order of x.

n Consequently o( <f (x)) < o(x) for all p

for n ~ 1. Therefore C is p-acyclic •

n)l, i.e.~n(C)#C p

Thus the notion of a p-cyclic class extends the notion

of a p-regular class in finite groups. Let us notice that

the new definition does not make use of the order of an

element; it can be applied, for example, to torsion-free

groups.

Now we prove a special case of the main result of this

paragraph.

36

Theorem 4.3 Let k be a finite field, f \kl = p =q. Let

e EW(k) llG11 be an idempotent. Then

i) tc(e)=o if c is a p-acyclic class,

and ii) t <c)<e)=F<tc(e)) if c is a p-cyclic class. <Pp

Here F denotes the Frobenius homomorphism W(k)~W(k),

as defined in paragraph one.

Proof: Clearly, it will be enough to prove claims i) and

ii) on the W (k) n level. To this end, fix some n. Let

e= r. (~). To simplify notation, let us write R for W (k), n n

m. for for \f p and ~ for ~. p

According to Proposition 1.5, Tl\.=pR, ni..n=O and R is a

ring of characteristic n p • m

n-1 ~ Let s=p and e=L_.e(x.)x. with x.E: G, e(x.)E:R. i=l 1 1 1 1

Further, let~= supp(e)xsupp(e)x ... xsupp(e) (s times).

For ans-tuple X=(x. , ... ,x. )cl.. 1 1 1 s n-1

set X=x.· .. .-x. E.G, 11 ls

ex=e(xi 1 ) ... e(x 1 s) "'R and wx=ei First we prove

Claim: for all pairs of natural numbers b,c satisfying

b~n, O~c<f we have

To see this we apply Lemma 3 .1. For r >,. n we get \ r-n+l_ r-n+l L..., eP · xP

X( l. X w i t h ~ ~ ( R ( G \ , R { G\1 ,

37

Let b,c be any integers satisfying b ~ n, o~ c < f .

Set r=bf+c+n-1. Then for any x " "):. we have r-n+l bf+c b c f -P -P (_q ) p and ex = ex = ex (q=p )

Now, if some e(x.)~'l'Tl b b J

and x. J

occurs in the s-tuple X

then ei '°Tl't.q ~m.n = (0), as b

b n q ) q ~ n. Therefore all _q ..l

eX' With eX T 0, 1 ie in

(see Proposition 1.5 viii)) implies

* n n-1 But I R I = q -q n n-1 ei -q = 1. Hence

for n-1

w = eq x x we have •.• q .. x = wx. In other words,

is a th (q-1) root of 1 in R. We also know from Propo-

sition 1.5 vi) that all are among the images of q-2 l,l, ... ,t, • Consequently, Fn(wx)= w~ (Proposition 1.5 iv).

b n-1 Also, for b~ n we have eq = eq x x and so

c =Li w~ \ i' bf+c (X) ~ c) = Fe (L \.wxl 'fbf+c (X) ~c )

proves the claim.

Let us assume now that C is a p-acyclic class.

We must show that tc(e)=O t-R. Observe that for any

conjugacy class D there exists at most one number d ~ 0

such that In fact, if we had qid 1 (D)=cpdl. (D)=C

with d 1= d 2+b, <b>O then

«;, C = t:f? d '- ( D) = 1' ( 4? d1 ( D) )

~ 4 ( c) '

so C would be p-cyclic. Further, notice that the set

( XI Xi:.lJ <; G is finite, so it is contained in a finite

38

union D 1....,, ••• u D l of conjugacy classes. For each 1 f. i ~ 1

define d. to be the unique n such that c:£n(Di)=C if 1

such n exists and we put d.=O otherwise. 1

Let d =max ld 1 , ..• ,d 1 ~. Take c=O and pick b large

enough, so that bf> d. Then t X \lfbf (X) E:. CJ is empty,

so we have tc(e)=O by our claim.

Let C now be a p-cyclic class. Let

be the cycle with Consider 1

some n} <;. G. Because A is finite and UC. is preserved j=l J 1

t s o 1 a r g e t h a t If t f ( A ) ~ Uc . . j=lJ

by ~. there exists a number

Our claim, applied with b=t, c=O, yields

( *) tc(e) = L, \.wx \ lftf (X) E: c)

The same formula, applied to 4?<c), with b=t, c=l yields

(**) tcp(C) (e) = F ( L ~ wx I lftf+l (X) c ~ (C)~) .

Consider two subsets of G:

and

These two sets are equal. The inclusion s 1 c.;;, s 2 is obvious.

F h 1 S Then lDtf+l(g)" ~(C) and orteconverse, et ge 2 . , ""'-t'

tf l ( lf t f ( g ) ) = If 1- 1 ( '-f t f + 1 ( g ) ) E: lf 1- 1 ( q, ( c ) ) ~ <ii 1 ( c ) = c

(1 is length of the cycle of C).

39

On the other hand, note that

of t. If ~tf {g) EC. then J

1 'ftf(g)E. UC., by definition

j=l J

'fl ( ~ t f ( g) ) E: 'fl ( c . ) ~ J

1 ~ (C.) = C ••

J J

From this and from the previous calculation we obtain C.=C, J

i.e. ~tf(g)E:C and gE.S 1 . Therefore s 1 = s 2 . This,

together with (*) and (**) implies

as desired a

Now we are going to generalize this result for

arbitrary fields k of characteristic p. To this end,

we must develope a specialization argument.

The following lemma is a routine application of the Null-

stellensatz [Passman (77), Lemma 2.2.Sl • Let F denote

the algebraic closure of the Galois field r . p

p

Lemma 4.4 Suppose A is an affine domain over F and

a E: A. If for any

have f (a) = 0

Proposition 4.5

r -algebra homomorphism p

then a = O•

p

f:A-f p

Suppose A is an affine domain over

we

F . p

Let a E W(A) be such that W(f)(a)=O for all F -algebra p

homomorphisms f :A---+ f . p Then a = 0.

Proof: We shall show by induction that for all k

a E vkw (A) •

40

Step one: k=l.

We have the following diagram:

for every f :A-r . p From this diagram we have f Tl (a)= A

'it - W(f) (a)=O r for all f. By Lemma 4.4 we have ~A(a)=O, p

so a E VW(A).

Induction step:

Assume we know that a E: VkW(A). Write a = Vk(b). We claim

that the hypotheses apply to b. Indeed, if f:A~F is p

any r -algebra homomorphism then p v kw ( f) ( b) = w ( f) ( v k ( b) ) = o .

But Vk is a 1-1 map. Therefore W(f)(b)=O and we may

apply step one to b. Hence, b E. VW(A) and

a = Vk(b) ~ VkVW(A) = vk+lW(A) •

Theorem 4.6 Let k be any field, char k = p > 0, and let 2 e = e f k\".G1. If e E W(k)lt G\1 is a lifting of e then

i) tc(e) = 0 if c is p-acyclic

and ii) t<f> (C) (e) = F(tc(e)) if c is p-cyclic p

where F is the Frobenius homomorphism W(k) ~W(k).

Proof: Define A = r [.l__e(x)\ x e:. G~1 ~ k and p

[ 'c <el if c is p-acyclic a =

t~ (C)(e)- F(tC(~)) if c is p-cyclic p

41

Then A is an affine domain since supp(e) is finite.

Also aEW(A). Notice that for any

morphism f :A---+F p

the image f(A)

F -algebra homo-p

is a finite field.

Moreover, W(f)(e) is an idempotent in W(f(A))[tGU and

tc(W(f) (e)) = w(f)(tc(e)). Also FW(f)(tc(e)) =

(see the diagrams in the first paragraph).

Hence, by Theorem 4.3, we have W(f)(a)=O. Then a=O by

Proposition 4.5 a

As an application, we prove

Theorem 4.7 Let k be a field, char k = p> 0, and let

e = e 2 E k [G1. If e E: W(k) U.G\1 is a lifting of e then

tr(~) €: ~ ~ W(k). p

Proof: The class C= \1' is p-cyclic for all p. Moreover

and Theorem 4.6 reads tr(e)=F(tr(e))

where F :W(k)--~W(k) is defined as F(a 0 ,a 1 , ••• )

( p p ) c 1 1"f a 0 ,a 1 , ••.• onsequent y, tr(e)=(a 0 ,a 1 , ••• ) then

a. =a~ and hence a.ET <;k for all i. Therefore 1 1 1 p .

tr(e) E W(F ) = ~ • p p

Using the same method we can prove

Proposition 4.8 Let k be an algebraically closed field

of characteristic p > 0 and let 2 e = e E:. k '\.G'\ • Let

e e: w ( k )l l en be a lifting of e. If Cc; G is a p-cyclic

conjugacy class, which lies in a cycle of length 1, then

42

t cc e) E. w c r Pt ) c; w ck) a

5. The lifted trace for finite groups

In this paragraph we show how to calculate tr(e)

for eEkLG\ without lifting e, if G is finite. The

formula we obtain is well known.

Let \G\<co and char k = p..., O. Consider an

idempotent e €: k \.Gi • Let R = W(k) and let e <:- RlG\ be

any lifting of e. Then we have

R [G1 ~ R ( G 1 ~ ffi R[G\ (1-e).

The natural homomorphism 1\. :RtGl--~ klG1 maps RCG1e

onto k(G'\e and RlG1(1-e) onto kl.G)(l-e). Now,

RlG'\e, Rl.G1 (1-e) are projective R-modules, and hence free,

as R is a local ring [Passman (77), Lemma 10 .4 .14] .

Let m=rankRR(G\e and n=rankRR1i.G1 (1-e). Obviously,

m+n= \GI • If ul, ••• ,um form an R-basis for R ( Gi ~

then ~ (u 1 ), ... , n;(um) span k(G'\ e , i.e. dimkk(G1 e ~ m,

and similarly, dimk k'lG1 (l-e) ~ n. Actually, we have

\GI= m+n ~ dimkk[G\ e + dimkkl:G"\ (l-e) = dimkklG1 = lGl so

m=dimkk(G1e.

43

Because R ( G1 e is R- free , so

K the quotient field of R, and

In conclusion,

Proposition 5.1 If 2 e=e E.klG\, char k = p>O and \Gl<OO

then ltr(e) = 1 ~ 1 · dimkk (G1 e •

I I I . THE NONVANISHING PROPERTY

This section is devoted to the study of the non-

vanishing property of traces. Let R denote either the

group ring k (G1 or the ring of Witt vectors W(k)[l Gl\

We say that the ring R has the nonvanishing property

if for every idempotent e in the ring, tr(e) implies

e = O. Part of the Trace Theorem I.3.1 can be restated

now as follows: if char k = 0 then k (G1 has the

nonvanishing property. On the other hand, as Example I.3.5

shows, r 2 lc~ does not have this property.

In paragraph one we prove the nonvanishing property

f or W ( k ) l l G 11 when char k = p > 0 and G is a poly-

cyclic-by-finite group. In paragraph two we reduce the

corresponding problem for residually finite groups to a

question about idempotents in k [ G1 for finite groups G.

In the third paragraph we review the known estimates of tr

bounding away from zero, if char k = 0. In paragraph four

we formulate a corresponding conjecture for char k = p > 0.

We prove the conjecture here for traces of central idem-

potents. We show by example that the estimate obtained

44

45

is the best possible. In paragraph five we check our

conjecture for groups of order ~ 30. Finally, in

paragraph six we make a combinatorial remark about the

nonvanishing property for Z (( G\\ • p

1. Polycyclic-by-finite groups

Let us recall that a group G is polycyclic-by-

finite if it has a subnormal series

such that for

<f)= G0 <l G 1<1 ••• 4 Gr = G

i=l, •.• ,r the group G./G. l l i-is either

cyclic or finite. It can be shown [Passman (77), Lemma

l0.2.s1 that we can find a subnormal series in G with

all factors G./G. 1 l l-infinite cyclic with the possible

exception of the last one which may be finite. The number

of infinite cyclic factors is an invariant of G, called

the Hirsch number h(G).

In this paragraph we show that for such groups the

lifted trace ltr enjoys all properties guaranteed by

Kaplansky's Trace Theorem in characteristic zero.

Theorem 1.1 Let char k = p > 0 and let G be a poly-

cyclic-by-finite group. For any idempotent e <=- k(G1 we

have i) ltr(e) E tfl

46

ii) 0 ~ ltr(e) ~ l

... ) l l l, lt~(e)=O i~plies e=O

(the nonvanishi~g p~ope~ty).

We need t~o le~rnas.

Le:n:na 1 • 2 fCliff (80), Lemma 21 Let R be a commutative

ring of characteristic n p and let G be a polycyclic-

by-finite group. If 2 e = e E. R l G1 and x E G is of infinite

order then tC(e)=O where C is the conjugacy class

containing x •

Lemma 1.3 (Passman (77), Lemme 10.~.91 An inf init2

polycy~ll~-~y-fii1lre group contains an infinire abelian,

rorsion free, normal subgroup •

?roof of Theorem 1 .1: We proceed by induction on the Hirsch

number h (G). If h(G)=O then G is finite and

where K is the quotient

field of W(k). As charK=O, i)-iii) follow from Theorems

I. 3. 1 and I.3.4. If h(G)~l then G is infinite and

we can apply Lemma 1.3. Let N be the subgroup provided

by this lemma and let 8 :G --~GIN be the natural pro-

jection. Consider e= 6*(e) E. W(k)[(G/Nl\ We have

tr(e) = L, tc(e) = e(l) + 2: t,.(e). C~N Cf~'W ~

Notice that all elemenr:s i.n N' ll\ have infinite order.

On the other hand n p W (k)=O n

(see paragraph II.l).

Therefore we can apply Lemma 1.2 and so

47

and n=l ,2, .... Consequently, tc(e)=O and hence

tr(e)=e(l)=tr(e). But h(G/N) < h(G). Thus tr(e) satisfies

i) and ii).

If moreover tr(e)=O then tr(e)=O, so e=O by

the inductive hypothesis. Therefore e*(e)=O ~ k(G/Nl, as

9*(e): e(mod VW(k)U G/N11 ). Thus e ~ ker 9*= k(G1w(N)

where w ( N) denotes the augmentation ideal of N. Now

we follow Cliff's argument in (Cliff (80)] ao

ideal I n

of a ring R we define I...,= (\In

(I"" t' . Then by n=l

lSehgal (78), Thm.1.3.151

the rank of N is t. But

.....it e ~ ( k {.G1 w ( N) ) c;

...., t k[G1 <-V(N) = 0

2. Residually finite groups

For an <..>n+ 1

and I = ...,t

w(N) =O if

so e = 0 •

Recall that a group G is residually finite if

for any element g E. G ' U\ there exists a finite group H

and a homomorphism ~ :G~H, such that lf(g)~l. Because

we can study idempotents in ktG1 one at a time, we can

restrict ourselves to finitely generated, and hence

countable groups.

of subgroups N. ~ G 1

It is easy to construct, then, a family

for i= 1 • 2, •.• such that

i)

ii)

iii)

N . ..c::j G , 1

Ni+l <; Ni

f't N. =<1> i= l 1

48

tG : N .1 < ..:o 1

We denote by the natural projection G ~ G/N •• 1

Let 2 e=e t. k (G1 and char k = p > 0. Choose any

lifting ,.. e of e . We would like to prove that tr(e)=O

implies e=O. As usual with residually finite groups we

shall translate this problem into a problem about

finite groups.

Observe that the maps q. 1

induce ring homomorphisms

q. : w ( k) ( t G\I --- w ( k HG IN.\ 1 1

. In particular, each q.(e) is 1

an idempotent.

Proposition 2.1 tr(e) = lim tr Ci. (e) 1~00 1

where the limit is taken in the p-adic topology.

Proof: Let us recall (see paragraph II.I) that the

VW(k)-topology is induced by the valuation l I :W(k)~ R

where -n la\ =p for

Pick [> O. Choose n so large that

show that l t r ( e ) - tr q . ( e )l < 1

-n p

-n p <£ • We shall

for i ). i(n)

equivalently, that JC (tr(e)- tr q.(e)) = o. n 1

From the diagram w ( k) tl GI\

tr j W(k)

'i'Cn

1' n w ( k) n

or,

49

we see that we must prove

From another diagram

W(k)ltG11

W(k)[G/N.l 1

tr( 'Jt (e) )=tr 'Ii:" q. (e). n n 1

1C n -------W (k)(Gl n

-q. 1

W ( k) l GIN .1 n 1

we find that 1t q . ( e ) = q . ( TC ( e ) ) . L e t e =J\ ( e ) E: W ( k ) ( GJ. n 1 l n n n

Now we reduced to checking that tr(e)=tr q.(e) 1

for i

sufficiently large.

To this end let S = supp(e). This is a finite subset of

G; hence you can pick i(n) so that for i;l:i(n) we have

Ni(\ Sc; <I'>. For such i we have

tr q. ( e) = 2.. { e ( x) \ x ~ N . (\ s) 1 1

e(l) = tr(e)•

Thus, to prove the nonvanishing property for

w ck) U c1l it is enough to show that the finite group

traces tr Ci.Ce) 1

are bounded away from zero in W(k) by

some constant independent of i. The hope is that this bound

is dependent only on the size of supp(e).

To find a candidate for this constant, we review the known

estimates for traces.

3. Estimates of traces in characteristic zero

Let e Ek ~G1 be an idempotent with char k = 0. Without loss

of generality in further argument we can assume that k = C,

so

the field of complex numbers (see [Passman (77), Thm.2.1.81).

Then, following D.Passman, we have a Hermitian inner

product on ClG1:

(a , b) = L a ( x )·b ( x ) for a=L.a(x)x and b=Lb(x)x. x~G !.<:

= (a,a) 2 • Another norm can be defined I t i n d u c e s a n o rm ll a I\

by setting l a l = L. \a ( x )I They are related by x~G

ll a · b \\ ~ \\a\\ · \ b I [Passman (77), Lemma 2.1.s]

In his proof of Kaplansky's Trace Theorem, Passman obtains

the following estimate: if 2 e =e#O then 2 2 tr(e)~"e" /1e( >0.

However, it is not clear how to make sense o f ll e \l 2 I 1 el 2 i n

p-adic terms. Here the following remark comes into play.

Proposition 3.1 (D.Farkas) Let G be any group and let

e £ C[G1 be an idempotent. Then -1 tr(e)) lsupp(e)I .

(Here \.\ stands for cardinality)

Proof: We define X~ C{.Gl by

__ ~ e ( x ) 0; I e ( x )1

x ( x) l if xE- supp(e)

otherwise

Then \e\ = L \e(x)\ ( e , X.) ~ \\ e \\ · \\ }.\\ by the Cauchy-

Schwarz inequality. But \\'/...i = lsupp(e)( • Thus

l I 2 .,;;: e -2 \\ e ~ l supp ( e )\ and 2 2 -1 t r ( e ) ~ \\ e 11 I t e l ~ \ s u pp ( e ) \ a

In the case of a finite group G we have another

slick proof of the same inequality, due to E.Formanek.

5 1

Proposition 3.2 Let G be a finite group and 2 O#e=e ~ C (G1 •

Then -1 tr(e) '1.\supp(eH .

Proof: Let S be a nonempty subset of the finite group G.

Then for some integer t),IG\ I \S\ one can find sets x 1s,

x 2s, ... ,xtS suctJl.that each such translate of S contains

an element of the group which lies in no other translate.

To see this choose a maximal set y 1s, ... ,ymS of translates

with the desired property. If m < \G\ I \S\ then m

Pick u E G ' U y. S and s E: S. Then i=l 1

-1 (us )S

m uy.S#G. i= 1 1

contains

u while no yiS contains u, contradicting maximality.

Let O#a E C(G1 be any element. Set S = supp(a).

By the previous paragraph there are group elements x 1 , ••• ,xt

such that each of x 1 S, •.• ,xtS contains an element in no

other translate and

are linearly independent; hence dim C[G1 a ~ lGl I ISi .

If a=e E C(G1 is an idempotent then

tr(e) = 1 ~ 1 dim C (G1 e ~ IS\ -l a

We say that an idempotent e is self-adjoint if

* e = e , where ( .L. e(x)x)*= L,;-cx)x- 1 • If the idempotent

e E CtG1 is self-adjoint then the number tr(e) satisfies

the simple relation t r ( e ) = l\ e \\ 2 ~ P a s s m a n ( 7 7 ) , C h a p • 2 • l J • In fact, the above equality is true for the wider class of

essentially self-adjoint idempotents. An idempotent is

essentially self-adjoint if it is equivalent to a self-

52

adjoint idempotent. This definition is motivated by the

observation that equivalent idempotents have equal traces.

It is an interesting problem to determine when all idem-

potents of C[G] are essentially self-adjoint. As a start,

there is

Theorem 3.3 (D.Farkas) The following are equivalent:

i)all idempotents in ~(G1 are essentially self-adjoint,

* ii) e+e -1 is invertible for each idempotent e ~ CtG1, 00 *

l.·i·i·) (' 1) ~ (>.+e-e )n . .11 + L_ - ); +I - - c on v e r g e s in t h e ~ \\ - n o rm t o s om e n=o

element of C[Gl for each e=e 2 E C \:G\ . Here ).=2 \el 2 .

Proof: We prove

i)==} ii): Let

i ) ) i i) ~iii ) ~ i i) ~i ) . 2 e=e t C(G) be essentially self-adjoint.

Then there exists an idempotent f E: C t G1 s u ch t ha t ef=f,

* * * * fe=e and f=f • We have (e+e -l)f=f+(fe) -f=e , so

* * * e E. (e+e -l)CtG]. By symmetry, e E. ( e + e - 1 ) d: lG1 . S inc e

* * (e+e -1) E (e+e -l)CCG1 we see that * 1 E (e+e -l)C(Gl.

* Invertibility on the other side follows bacause e+e -1 is

self-adjoint.

ii)~iii): * Notice first that e+e -1 is invertible if

* and only if 1-e+e is invertible. Indeed,

* 2 * * * * * * * (e+e -1) =ee +e e-e-e +1=(1-e+e )(l+e-e )=(l+e-e )(1-e+e ).

Now, ( C ( G1 , (. , .) ) can be completed. Let L2 (G) denote

the completion. Notice that multiplications by elements

of C ( Gl ~ L 2 ( G) are bounded operators on 2 L (G). We show

53

first that the series of iii) converges in the operator

norm N to * -1 * (1-e+e ) • To see this set q=e-e • We claim

that )+1 > N(). +q). If not, since * q =-q,

2 2 * 2 2 2 2 0+1) ~ N(" +q) =N(()+q) O+q ))=N(). -q )~). +lq\ •

That 2 2 2 2 2 is, ).+2).+1~ }+4\e\ or 2.21e1 +1~41e\,

a contradiction. Therefore, ).+1> N()+q); i.e. N<.?:;\'"!r)

Hence )+q

1- ~+I is invertible and

Of course, 1- ~~g = ~+1

1 * ):+T(l-e+e ).

co (1- ~~g)-1= L <~~s)n

~+1 A+l . n=o

Notice that for any a E'. C(G1 we have

( ) = SU~ ~\~~~ '), \\a l\1 N a , -\, -111- = I\ a II • 1\Xll 0 \IX\\ ,1

< 1 •

~o conclude that our series converges in the II-norm let

us consider its partial sum S • Then m

\\ s m * -1 - (1-e+e ) I\~ N(S * -1 - (1-e+e ) )~ 0 and iii) m

follows.

iii)~ ii): From the argument above we know that our series

converges in the operator norm N to * -1 (1-e+e )

B y o u r as s um p t i o n , i t c on v e r g e s in t h e l\ \\ - n o rm t o s om e

element of C[~ • Thus we must prove that if S ~ x E: C tG1 m

in the a ~ -norm and 2 S ~ y ~ L (G) m in the N-norm then

x = y. By subtracting, we can assume x=O and show that

y=O. It suffices to show that y·g = 0 for all g EG.

(Think of y as an operator.) But

ll y · g 11 ~ ll y · g - s · g ll m

+ \\ S g \\ s N ( y - S ) + ll S ll ~ 0 m m m so, in fact, * -1 ( l - e + e ) E C l. Gl • * Hence e+e -1 is also

54

invertible.

ii) > i) : First note that

* * * (a) (e+e -l)e =e(e+e -1)

* * (b) (e+e -l)e = e ·e

* * ( c) e(e+e -1 ) = ee . Set * -1 f=e(e+e -1) • Clearly, ef=f. Also * * -1 * f =(e+e -1) e =f

by (a). By (c) * * -2 * -1 ee (e+e -1) =e(e+e -1) • Apply (a) to

get f 2= f. Finally, by (b), * * -1 ee (e+e -1) =e. Apply (a)

to get fe=e. Thus e ,...., f and f is self-adjoint a

4. An estimate of traces in characteristic p

In this paragraph we formulate a conjecture about

lltr(e)\ p We verify the conjecture for central idempotents

and show that the estimate obtained is the best possible.

Let k be a field of characteristic p) O. We

denote by v p the exponential valuation on W(k), described

after Proposition II.1.4. The results presented in the

previous paragraph suggest the following

Conjecture 4.1 If O#e=e 2 E k(G1 then v (ltr(e))~ lsuppe\ -1 p

Let we recall that this the type of estimate we wanted for

residually finite groups. Indeed, if e=e 2 E. klG1 and ,.. e is

SS

its lifting, then the above conjecture implies

v (tr q.(e))~ lsupp(q.(e))\ -1 ~ \supp(e)\ -1 for all i. p l l

We used here the fact that the homomcrphisms q . : k ( G1 .- k [ C. IN .1 l l

only decrease the size of the support.

We are able to prove the conjecture for central idempotents.

Theorem 4.2 Let eE kl~ be a central idempotent. Then

v (ltr(e) )~ lsupp(e)\ -1. p

Proof: By Theorem I.4.2 we may assume that G is finite.

Let " e be the lifting of e.Now we use a result which we

shall prove later (Theorem IV.3.1): if G = ( supp ( e ))

then l Cl < lsu pp ( e )I p . Consequently, the highest power

of p which can divide dim KCG1e is lsupp(e)l -1.

(Here K denotes the field of fractions of W(k).) Then

v (1 t r ( e ) ) = v ( t r ( e ) ) = v ( d i m K l G \ e ) -v ( I G I ) =f I supp ( e )l - 1 a p p p p

The estimate above is the best possible. Indeed, we

have

Then 2 e =e.

n n-1 i L e t G = < x \ x 2 - 1 = 1) a n d e = .L x 2 ~ F 2 l G l .

2 3 i=o2n-l_l The elements e, xe,x e,x e, .•. , x e are

Example 4.3

seen to be linearly independent in r 2 Cc1e by comparing

the highest x-term. They also form a basis for the left 2n-l 2 2n-2

ideal. Indeed, we have x · e=~e-(x+x + .•. +x )1e = 2 n-1 e - p(x)e = (1-p(x))e with deg (1-p(x)) < 2 . x

Therefore every element in r 2 (cle can be written as

56

p 1 (x)e with degxpl (x) ~ 2n-l_l yielding an r 2-combi-2n-l_l . n-1

nation of e , x e , •.• , x e. Hence d 1 m F 'f 2 [ Gl e = 2 • 2

Thus v 2 (ltr(e))=v 2 (dimf 2 [G]e)-v 2 ( IGI )= n-1 = lsupp(e)I -1,

by Proposition II.5.1.

5. Idempotents with small supports

In this paragraph we examine idempotents with supports

of cardinality less than or equal to three. It will allow us

to verify Conjecture 4.1 for idempotents in groups of

order ~ 30.

Let C denote the cyclic group of order n. n

Proposition 5.1 If 2 e=e ~ k (C\ and \supp(e)\ =2 then either

i) 1 2 k "I 2 e=-(l+x) where x has order and char 2 -ii) e=ax+a 2 2 where has order 3 and char k 2 and or x x =

a is a cubic root of unity in k •

Proof: Case one: lE.supp(e). Then e=a+bx where a,b Ek

and l#x €.G. We have 2 2 2 2 e =a +2ab·x +bx =a+bx.

As x is neither 1 2 nor x , we have 2ab=b, so 2a=l.

Thus char k "I 2 and 1 a= 2 ~k. Further, after cancellation

of 2ab·x and bx, we get 2 2 2 2 2 a +b x =a, whence x =l and b =~.

Thus we get i).

57

two: e=ax +by where x,y#l.

Then 2 2 2 2 2 e =a x + ab(xy+yx) + b y = ax + by = e.

Because x is not one of 2 xy, yx, x we have 2 x=y and

symmetrically, 2 3 y=x . Thus x =l and 2 2 2 2 a x +2ab+b x=ax+bx •

Consequently, 2ab=O, so char k = 2. Also 2 2 a=b and b=a ,

i . e . 3 a =1. Hence e=ax with 3 a =1 and

We see that all idempotents with tsupp(e)I =2 are

supported on abelian groups. The same is true for idempo-

tents with (supp(e)I =3, with a single exception.

Proposition 5.2 Let e=e 2 e k[~ with \supp(e)\ =3. If the

subgroup <supp(e)) c; G is non-abelian then

i) char k = 2

ii) e=x+y+xy E k [GI where 2 3 = 1 and -1 -1 x =y xyx =y

and iii) (supp(e)) - s 3.

• Proof: Let e=ax + by + CZ with a,b,c€ k, x,y.z E-G ::tnd

x,y,z are all different. First we show that none of x,y,z

can be equal to 1.

For assume (without loss of generality) that x=l. Then 2 e=e

implies 2 2 2 2 2 a+by+cz=a +b y +c z +2aby+2abz+bc(yz+zy).

Let us notice that the term bcyz cannot cancel with any-

thing on the left side of this equation. In fact, yz#y or z

as this would imply z=l (respectively, y=l). Also yz does

not coincide with one of 2 2 l,y ,z ,zy as each possible

equality implies that y and z commute, forcing (supp(e))

58

= (y, z) to be abelian. Thus yz E 2 supp(e )'-supp(e),

a contradiction.

Hence we can assume that e=ax+by+cz where x,y,z#l.

Again, from the equality 2 e=e we have that 2 x,y,z E.supp(e)

in other words each of these letters can be written as a

product of a pair from [x,y,z~. Let us take x for

example. We show now that

x = yz or x = zy.

Because x,y and z are different from 1, so each letter is

a product of two letters different from itself.

Assume that we have 2 x=u for some u ~ {y , z~ • Let v be

the letter different from x and u. Then, by the observation,

v can be expressed as a word in x and u. 2 But x=u •

x,u,v "'- <u) and <supp(e)') is once again abelian.

x=yz or x=zy are the only remaining possibilities.

Of course, the same is true of y and z •

Hence

Thus

Consequently, we have 2 3 possible choices for the system

r: = ulu2 u.#x

l

(*) = vlv2 v.'l-y l

= wlw2 w.'1-z. l

Each such system can be coded by an oriented triangle with

vertices x,y and z. We orient the edge with endpoints u,v

from u to v (u-v) if uv occurs as a right side in

the system.

59

Geometrically, there are only two types possible:

z z

(A) ( B ) /~ x y x y

Notice that two triangles which belong to the same type

differ only by a permutation of vertices. Performing the

same permutation on the system (*),we can always change

our system to one described by (A) or (B). Thus, for

the price of an eventual change in names of our group

elements, we can assume that the relations in G are

described either by (A) or by ( B) •

Now we eliminate the type (A). We have

(A) ) : ~ ::

l z = xy

The relation 2 e=e implies

2 2 2 2 2 2 a x +b y +c z +ab(xy+yx)+ac(xz+zx)+bc(yz+zy)=ax+by+cz

No element of the group which formally appears on the left

side equals x except yz. Indeed:

x .,.

x .,.

2 x ,xy,yx,xz,zx

2 2 y 'z

x 1- zy

(each would imply 1 t::supp(e))

(these were eliminated earlier)

(if x=zy then yz=zy and so (supp(e»

is abelian).

Similar arguments work for y and z. Hence ab=c, ac=b,

60

bc=a and, after cancellation,

2 2 2 2 2 2 a x +b y +c z +cyx+bxz+azy=O

Now, from (A) we have:

2 2 x =xx=x(yz)=(xy)z=zz=z(xy)=(zx)y=yy=y , i . e . 2 2 2 x =y =z

Hence 2 2 2 2 (a +b +c )x +(cyx+bxz+azy)=O.

Again 2 x #yx,xz or zy. Thus the last three terms must

cancel by themselves: yx = xz = zy. Then

2 3 2 zy=(xy)y=xy =x =y x=y(yx)=y(zy) implies y=l, a

contradiction.

Thus we have to deal with case (B) only. It

corresponds to

ix = yz

(B) y = xz

z = xy

As before, x,y,z each appear only once on the left side

of 2 e =e. After cancellation we get

2 2 2 2 2 2 a x +b y +c z +cyx+bzx+azy=O

and ab=c, ac=b and bc=a. When we substitute z=xy into

the first two equations of system (B) we get

x = yxy and x 2 = 1 •

Then we also have 2 2 z =xyxy=x(yxy)=x =l, 2 2 so x =z =l.

However, if we had 2 y =l, it would mean 2 -1 -1 l=z =xyxy=xyx y

Then <supp ( e )) c; ( x , y) would be abelian. 2 Therefore y 11.

Also yx, zx, zy I 1 as each such relation implies that

< supp(e)) is abelian.

61

So and 2 2 a +c =O.

The remaining four terms must also cancel. Now, 2 y is

different from yx, zy, so 2 2 b y =bzx and hence 2 y =zx and

b=l. The last pair gives

cyx = azy, so a=c and yx = zy.

Hence, writing all relations in terms of x and y, we have

x = yxy 2

x = 2 y = xyx

2 yx = xy

We get 3 2 2 y =y y=(xyx)y=x(yxy)=x =l; finally 2 3 x =y =l, -1 2 xyx =y .

Also we have a 2+c 2 =0 and a=c, so char k = 2.

As 2 b=l, so l=b=ac=a . Thus a=l and so a=b=c= 1 E: k.

Therefore e=x+y+xy and

Let e €. i 2 ( S 31 be any 1 if t in g o f e , cons id ere d above .

According to the remark made in Proposition II.5.1,

where M is the transformation obtained by right multi-

plication of e in F 2 ts 3\. It is easy to calculate that ,.. 2 rankF M = 4. Thus tr(e)= 3. In particular

2 < 2= lsupp(e)\ -1. Because Conjecture 4 .1 is valid for

idempotents supported on abelian groups (Theorem 4.2), it

is valid for all idempotents e with [supp(e)l ~ 3.

62

We shall show that this implies that Conjecture 4.1 is

true for all idempotents in k(Gl for \GI ~ 3 0 •

Lemma 5.3 Let G be one of the two nonabelian groups of

order 27. If char k = 2, 2 e = e t k tGl and <supp(e))=G

then [supp(e)! >,. 6.

Proof: The two groups in question are

Q1 ( x,y,z x 3 =y 3 =z 3 =1, [x,y) =z central)

and 3 3 3 > Q 2 = ( x , y , z x = z = 1 , y = z , [ x , y] = z cent r a 1

I n b o t h c a s e s we h av e a n e p i mo r p h i s m TI. : G -----» G I ( z') = C 3 x C 3 •

To simplify the notation we write ~(g)=g for g EG. m

Let us assume that there exists O=#e= L a.g. with i=l l. l.

m ~ 5 and (supp(e)'> =G. Let 1tS={wE:.C 3xc 3l w(z')(I supp(e)1¢}.

Because 2 e =e, for any wE.1lS there are u,v ~'T'S such

that w=uv. Obviously, l~ \r..S\ ~ 5. Write S =supp7'(e), 1t

a subset of 1C' S. It is easy to see by direct calculation

that is invariant under the map 2 <f(x)=x ,

(see also Theorem II.4.6). We shall show now that in all

cases: 11i.Sl=l, .•. ,5 we get a contradiction.

Case l-xSI = 1: Let -x.S={w~. Then - -2 w=w is a product of two

elements in ii.S. Therefore w=l and <'.supp(e))~(z) ~ G, a

contradiction.

Case lit SI =2: Let ~S={w,vJ. If 1 E.11:S, say w=l, then

supp ( e ) ~ < z') v v < z) and < ~ (supp ( e) )) <; < 1 , v'> ~ C 3 x C 3 • If

i ~ irs then the only choice for w is v2 , i.e. ~(supp(e))~

63

~ (v , v 2> 1- C 3 x C 3 . I n b o th c as e s <supp(e~ ~ G, a contra-

diction.

Case l-xSl=3: Now we must have lSn;I ~ l. If then

t he 'f - i n v a r i a n c e i m p 1 i e s Sit' = < 1 ~ . Th e n "( S ' S7t = \. w , ~ a n d

. h - - 2 b f again we ave w=v , so as e ore <supp ( e )) $ G. For

! S'!t \ = 2 we ha v e

t hen (1d s u p p ( e ) )) f (v > • I f w 11 t h e n t h e on 1 y c ho i c e i s

~=v 2 and the same is true. Thus x(supp(eb ~ c3xc 3 , a

contradiction. Finally, if then from the 'f - invar-

iance of s~ it follows that S=S ={l,w,w 2 ~ and 'lt' It

<supp ( e )) 1 G •

Case lx S\ =4: Now \ S7t \ ~ 3 , a s ea c h c o s e t w (z) for

so

w € s, sl( Jt

contains (at 1 ea s t ) two e 1 em en t s of supp ( e) . I f I STI: I = 3

then S , s_ = l w ~ 1 {I\ . C on s e q u en t 1 y , II "

w=u 2 and 1t( supp(e)) ~ <u>'j c3xc 3 , a contradiction. If

I Si<[ =4 then S S f- - 2 - - 21 ~ = ~ =lu,u ,w,w 7 and three of the four

cosets contain precisely one element of supp(e). Without

loss of generality we can assume that u(~ and w < z'> are

among t hem . Le t ~ u z a) = u ( z) r. supp ( e ) and

and the only candidate is

a b so uz ·wz

b a wz ·uz . Let

{ w z b) = w < z') (I s u p p ( e ) •

2 must cancel (e =e)

u=xiyj and

k 1 w=x y . As <u,w) = c3xc 3 , so il-kj~O(mod 3). Now

a b + b a i+k j+l( a+b-jk + a+b-il) Q = UZ • WZ WZ • UZ X y Z Z •

H -jk z-il -- 0. h f ence z + T ere ore il-jk=O, a contradiction.

64

Case l-xSI =5: Then \ S'lt \ = 5 a s we 11 . Now a 11 c o s e t s w <. z)

for wE: '!'CS intersect supp(e) in a single point and the

argument of the previous case can be repeated •

Proposition 5.4

lG1~30. Then

Let 2 e = e <:: k ( G1 , ch a r k = p "> 0

v (ltr(e)) ~ \supp(e)\ -1. p

and

Proof: We have v (ltr(e)) = v (dimkk(G\ e) - v ( tG\ )~ p p p

V - v (\GI) where VG=max { n I pn~ \GI) . G p

If the subgroup <supp ( e )) is abelian or equal to

s 3 then the statement is true by Theorem 4.2 and by the

calculation after Proposition 5.2. This holds for all

lsupp(e) ~ ~ 3. Thus we can assume that \supp(e)\ -1 ~ 3. Now,

as long as V - v ( \GI ) ~ 3 G p

the proposition is established.

Let us consider, first, p=2. The only situation

IGI~ 30 occurs for YG=4 and

2 X IGI. As 'Jc, p =16~1GI we have the following possible orders

of G: 17, 19, 21, 23, 25, 27 and 29. The groups of

orders 17, 19, 23, 25 and 29 are cyclic, so our

proposition is true. All idempotents in groups of order

27 are either supported on an abelian subgroup or have

\supp(e)\ ~ 6. Hence the only case to consider is IGI =21.

This will be done in the next lemma.

This proves the proposition for p=2. The case p 7 3 is

simple. Again, vp(ltr(e))~ vG - v ( IG I ) But this time p

YG~3 for all IG\ ~ 3 0. Hence v (ltr(e))~3, as required p

65

Lemma 5.5 Let e be an idempotent in k\G1, where \GI =21

and char k = 2. Then v (1 tr ( e) ) ~ \supp ( e )\ -1 . p

Proof : We have C 7 ~ < x) -=:::::::\ G

7 3 - 1 r\. G =<x,y\ x =l, y =l, yxy =x I where r=l,2or4.

In particular r yx=x y. Now we proceed as in the proof of

Again,

(*) for any w E. S there are u,v ~ S so that w=uv.

Case \SI =l: By (*), S= 0\ . Thus (supp(e))~<x'> =C 7 and

e has an abelian supporting subgroup.

Case lS\ =2: By ( *) , S={y,y 2 ~ and 2 e= 'f(x)y+ '+'(x)y

...P ,41 f k[ (x)] • By comparison of the coefficients at

in e and 2 e we get 2 r l.t' (xr) ~ (x r ) ~( x ) '+' ( x ) =

r f (x) 'f ( x ) . 'V ( x ) =

r lf( x) . \fl( x ) '+'(x)

We have \ supp ( e ) \ = I sup p If\ + \suppl+'\ = 4. If one of

summands, say !supp~\, is equal to one, then

for

2 1 , y. y

(1)

( 2)

( 3)

the

\supp( '-t'(x) \{'(xr))\ =l -; 3= \supp t¥(x)\ , contradicting (3).

The case \supp'{'\ =l is similar. Hence \supp'fl = \supp~l =2.

From (2) and (3) we get

(**)

If

~(x)= ~(x)·~(xr)=

= '+' ( x). 1.j.1( x r) 2

2 \.V(x)·'f'(xr)·'f'(xr)·\.\J(xr )=

2 'f'(xr).

"' a b '"f' (x)= ~x (l+ ~x ) for 2

a,b#O then from

l+ ~xb = cx.3xa(r +2r)(l+

get

66

It can be easily checked that if b#O then the supports

of the both sides have different cardinalities.

Case IS\ =3: Then y(x)(\supp(e) 2 or y (x) (\ supp(e) must

consist of a single element. Assume that y<x> A supp(e) =

{c:x xa) . Substituting -iJ(x)= a into {t:*) 0( x we get

((2x2ra. 2 a = o< x a ar oCx . cxx .

2 2 Now 1 = o<.3xa(2r+r ) and a(2r+r ).:O(mod 7) • But

r,a 1 O(mod 7); 2+r _ O(mod 7). However, this congruence

is false for r = 1, 2 and 4 a

6. Some combinatorics

Let G be any infinite group and let k be a field

of characteristic p. We have shown in the proof of Theorem

II.3.2

have

( *)

where f

k = F p '

that for

is the

e €. k(G1 and its lifting ~ E: W(k)ll G\1

n tr(e) = lim tr(fp )

n,...oo

"pullback" /'

Let us restrict to

so W(k) = Z • p It is easy to see that the only

we

property of f which is used in the proof of ( *) is that

1C 1 (f)=e or "f projects to e". Consequently, we may

choose any f ~ z Hen p with this property, for instance,

f = L f(x)x with f(x)E Z and 0 ~ f(x) < p for all x ~ G.

67

Th f 11 f n~ "'(G1 and so en, or a n, "" .. t = tr{fn) n

non-negative integer. We know that the sequence

is a

{t .. ) p

converges to tr(e) with respect to l I . Of course, if p

t 1 = tr(f) i 0 then O ~ tr(f) = tr(e)(mod p).

However, in many cases we have t 1 =0 (see Example I.3.5

and Proposition 5.2). In fact, it is not difficult to see

t ha t f o r t h e i d em p o t en t e ~ F 2 [ C t" _ 11 , c on s t r u c t e d i n

Example 4.3, we have

One could ask whether it is possible to have all

t =O for a nonzero idempotent. (Such an example would n

provide a negative solution of the nonvanishing problem).

Of course, if e#O and \Gl<a:> , the Trace Theorem assures

us that some t #0. We are going to show that, in general, n

if Oie=e 2 ~ F [G1 then the sequence of integers p

not consists of zeros only.

Proposition 6.1 Let f€ Z(G) be such that

i) f 2 _ f (mod p)

ii) f= L. f (x)x with Q.::!f(x)<p

Then tr(fn) > 0 for all n ~ \supp(f)\

for

lt \ does n

all x c. G.

Proof: By i) we have f 2=f + pg for some g t 2 tG\ •

Notice that all coefficients g(x) of g are non-negative.

In fact, if we had g(x)~-1 then

f 2 (x) = f(x) + pg(x) ~ f(x) - p < 0.

This is absurd, as f 2 has all coefficients non-negative,

68

like f • Consequently, there are no cancellations between

f and pg, what means supp(f) 2 <; supp(f ). Let supp(f)=

1 ~ i ~ k we have

xi = x o(( i) x ~( i)

for some 1 ~ o((i), ~(i) ~ k. Pick one such pair for

each i.

Le t S be t he f re e s em i gr o u p gene r a t e d by { x 1 , ••• , x kJ

Let 1..f: s- G be the obvous homomorphism. We have that

s is the disjoint union of -Jo

~ S 1 where S consists of \. n 1 n=l n

words of length n. In these terms we must prove that

le lf(S ) n for n ) k.

Consider a sequence of elements in S defined

inductively as follows:

i) a 1 = xl

ii) if a is defined, n ~ 1 ' let a =b x (i.e. n n n s

x is the last letter of a ) . We define s n

a = b nxc.<(s)x ~(s)' n+l

Notice that <.f(a ) = tf( a 1 ) for all n . n

Consider the sequence of last letters in a n

for

n=l,2, ...• Because they all come from a set of cardinality

k, there exist a and n a with the same last letter and

m

0 < n-m \ ~ k. Assume that n < m. Then a =b x n n s and

a m = b x ( )x •.• x x n o< s c 1 cr s We get f(b )·\.f'(x) = n s

. 'f (x ) • s

with r ~ k-1.

'f (a ) n

= ~ (a ) m

69

If we multiply this equality by ~(b )-l on the left and n

by on the right, we obtain

1 = y>(xcX.( )x ..• x ) s c 1 cr

and xol( )x ••. x E. S. s c 1 cr J for some 1 ~ j ~ k •

1 E 'f (S.). Finally, we can take the element J

Therefore

b. E:. S . we J J

have just constructed and blow up its last letter several

times (as in definition of an+l) to obtain elements

b f. S with n n <f(b )=1

n for all n ~ j •

Proposition 6.1 disposes of the simplest way of

constructing an eventual counterexample to the nonvanishing

question. However, it does not solve this problem, because

the traces

powers of p.

t , although nonzero, may be divisible by high n

For example, for e€ r 2 C.c2"-l1 from Example

4. 3 we have t ) 0 for all n >,. k but n 2 n \ t -t-'""' f o r n ~ k - 1 •

IV. SUPPORTING SUBGROUPS OF IDEMPOTENTS

In this section we deal with the supporting subgroups

of idempotents. In paragraph one we show that the two

classical constructions of idempotents with an infinite

supporting group have the same origin: they arise as

conjugates of idempotents supported on finite groups.

In paragraph two we give an example of an idempotent

which is not conjugate (or even equivalent) to any

e "= K (G1

idempotent with a finite supporting group. In paragraph

three we find an estimate on the order of the supporting

subgroup of a central idempotent.

1. Classical examples of idempotents supported on

infinite groups

Here we describe two constructions leading to idem-

potents with infinite supporting groups. We notice that, in

both cases, the idempotents can be obtained in a unified way.

Moreover, each type is conjugate to an idempotent supported

70

7 1

on a finite group.

The two procedures can be briefly described as

follows.

(A) (Rudin-Schneider (64)1 Take an algebraically closed

field k and a finite, nonabelian group H such that

char k 1 lHI . By Wedderburn' s Theorem, k fH1 contains a

subring isomorphic to M2 (k).

Let e = lb g\ and n =(~ ;1 f. M2 (k) <; k (H1 •

Then 2 2 =O, and ne=O. Set G HxZ with e =e, n en=n = Z = (x> • Then f=e+nx Ek lG1 is an idempotent. Also, for

some h ~H, we have hx ~supp(f); hence \<supp(fj>\=oo.

(B) [Passman (77), Thm.4.3.aj Let G be a group, containing

a finitely generated, infinite subgroup. Assume that there

exists a non-central idempotent e~ C~G\. Set H = ~supp(e)) 0

Pick an element x E G 'H with xe:;ex. If xe;lexe set 0

f=e+xe-exe. If xe=exe then it must be that ex;lexe and we

s e t f = e +ex - ex e . I n b o t h ca s es , f i s an id em po t en t , (s u pp ( f )') 2

? H = <.H ,x) and 0

f is not central in ~ \H1. In this way

we can enlarge the supporting group until it becomes

infinite.

Let us notice that in the both examples the new

idempotent f has a "piece" with a special property.

Before we specify this property, let us make precise the

notion of a piece.

72

Definition 1.1 If b"- k~G1 then 01a Ek l.Gl is a piece

of b provided

i) supp(a) f\supp(b-a)=¢, supp(a) usupp(b-a)=supp(b)

and ii) a(x) = b(x) for all x c::. supp(a).

Proposition 1.2 The idempotent f E. k tG1, constructed by

method (A) or (B) has a piece a such that one of the

following conditions holds:

(*)

1 fa = 0

laf = a or ) af = 0

l fa = a

Proof: Case (A): put a = nx. It is a piece of f, as

supp(a) ~ G "{x), supp(e) ~ ex \H and f=e+a. Also

fa=(e+nx)nx=nx=a and af=nx(e+nx)=O.

Case (B): let f=e+xe-exe. Put a=xe-exe. It is a piece

since supp(a) ~ HxH, supp(e) c:;: H and H r. HxH= r:fJ

Also fa=(e+(xe-exe)).(xe-exe)=O and af=(xe-exe)(e+(xe-exe))=

xe-exe=a •

Thus, in both cases, we have f=e+a, where e is

some idempotent and a is a piece of f satisfying the

condition (*). In this circumstance the elements f and

e are conjugate.

Proposition 1. 3 Let f Ek (Gl be an idempotent and let

a E: k [Gl be a piece of f satisfying (*). Then f and f-a

are conjugate in k(G1

73

Proof: Assume that fa=O and af=a. Then

2 a = a·a = (af)a = a(fa) = a·O = O. -1 Hence (1-a)(l+a)=(l+a)·(l-a)=l, i.e. (l+a)=(l-a) . Also

(l-a)f(l-a)- 1=(1-a)f(l+a)=(l-a)(f+fa)=f-af=f-a •

As a conclusion we have that if f is an idempotent and

a is its piece satisfying (*), then e=f-a is also

an idempotent. Obviously, [supp(e)[ = \supp(f)( - \supp(a~ <

\supp(f)\. This suggests the following method of

simplifying idempotents. Find a piece a in f such that

the condition ( *) is fulfilled. Then we get a shorter

idempotent e=f-a with f "-' e. (Thus, we reverse the

procedures (A) and (B)). After several steps of this

kind we must end up with an idempotent which cannot be

further shortened in this way. One could hope that this

"irreducible" idempotent has a finite supporting subgroup.

This hope seems to be encouraged by the fact that tr(e)~ Q

for e E. C lG1 and that this ratio has an "explanation"-

for finite groups G. Unfortunately, no such reduction is

possible, as the next paragraph will show.

2. Idempotents with infinite supporting groups

In this paragraph we show how to construct an

idempotent eE C(G1 such that e is not equivalent to any

74

idempotent supported on a finite group.

Theorem 2 • 1 Let R and s be algebras with 1 over the

field k. Assume that

i) R has an n-generator projective module which

and ii)

is not free

s can be written

S = M (k)xS'. n

as a product of rings

Then sc&>kR has an idempotent which is not equivalent to

any idempotent in

Proof: Let P be a projective non-free R-module, generated

by n elements. Then there exists an epimorphism ~:Rn P

and a splitting map i:P---Rn such that

e= i 0 '1t : Rn ____ Rn is an idempotent. Let

J\• i=idp• Then

EE.M (R) be the n

matrix for e. Of course, E is an idempotent. Because P

is not free, E is not equivalent to any idempotent of the

form [~ ~] where I is the identity matrix. Of course all

idempotents in M ( k) n are equivalent to idempotents of this

type. Therefore E is not equivalent to any idempotent

in M (k). n

Consider

Let us observe that

(E,O) € Mn(R)x(S' ® kR). Evidently (E,O) is an

idempotent in S ®kR, not equivalent to any idempotent in

75

Corollary 2.2 Assume that G is a torsion free group and

that the group ring C [~ has a finitely generated projective

module which is not free. Then, for some finite group F, the

group ring C(FxG1 contains an idempotent not equivalent to

any idempotent whose support lies in a finite subgroup

of FxG.

Proof: Suppose C(G! has a non-free n-generator projective

module. Choose a finite group F which has an irreducible

character of degree n. (The symmetric group on n+l letters

has this property.) Since C(FxG1 C (F1 ®CC CGI (Passman ( 7 7),

Lemma 1.3.41, and C ( F1 = M ( C ) x S ' n there is an idempotent

in C(Fx~ which is not equivalent to any idempotent in

C(F1 But all finite subgroups of FxG lie in F •

J.Lewin has produced in [Lewin (82~ finitely gene-

rated, non-free projective Q [G1 -modules when G is a

torsion free polycyclic-by-finite group which is not

nilpotent and when G is a torsion free one-relator group.

Other examples can be found in CBerridge-Dunwoody (79~ .

Remark 2.3 Notice that we can prove Corollary 2.2 with C

substituted by any algebraically closed field. In case of

positive characteristic p we should choose the group F

s o t h a t p Y \Fl , t o e n s u r e t h e ex i s t e n c e o f t h e d e c om p o s i t i on

of k[~ into a product of matrix rings.

76

3. Supports of central idempotents

Let !Sl denote the cardinality of a (finite)

set S. In this paragraph we prove the following

Theorem 3.1 Let k be a field of characteristic p '> 0.

If e ek[G1 is a central idempotent then

\ <' s u P P < e )>I < [supp ( e )\

p •

We start with a lemma.

Lemma 3.2 Let char k = p> 0 and let e ~ k[G1 be a central

idempotent. If supp(e)=e 1 u e 2 v .•. ven' a union of the

conjugacy classes, then the "Frobenius" map ..f.. defined in "i" p'

II.4, permutes the set S = {e 1 , .•. ,en}.

Proof: By Burns' Theorem (see Theorem I.4.1) we may assume

that the group G is finite. Let e E W(k)t.G\ be the (unique)

lifting of e. Then e N

is central too (see Theorem II.2.2).

Let e = ~ a.C., where i=l 1 1

c.=L°lx Ix 1

~ e .1r ~ W ( k ) \. G \ a n d a . E: W ( k ) • 1 1

Now, e. €: S iff 1

a. 1 ~ VW(k).

Let e.E s. We have te (e)=a.·le.1. Let e.= 1? (e.). 1 i 1 1 J p 1

By Osima's Theorem (see Theorem I.4.1) e. consists of 1

p-regular elements, hence it is p-cyclic (see Proposition

II.4.2). Thus Theorem II.4.6 can be applied:

F (a i )· [ e i \ = F ( t e . ( e) ) = 1

t e. <e) J

a. le .1 J J

77

Obviously (C .l = IC. I 1 J

F(a.)=a .. Since a.<iVW(k), 1 J 1

so

also F(a.)¢.VW(k) 1

and hence a.</: VW(k). Consequently, J

C. E S. Thus ..+.. '±'p preserves the subset S of the collection J

of all p-regular classes, which is itself permuted by 4? . p

Therefore ~p permutes S as well a

The above lemma generalizes Proposition 7.5 in [Bass (78~ ,

where it is shown that {>p permutes the subset

s' = {ci \ PX \ci1) c;, s.

Proof of Theorem 3.1:

Let e ~ k(Gl be a central idempotent and write supp(e) =

.•• v c • By Lemma 3.2, the map ~p permutes the n

action with lengths d 1 , •.• ,dm respectively.

F i x o n e o r b i t , s a y S 1 , a n d c ho o s e C 1 = { x 1 , • . • , x r ( 1 )1 in s 1 . Notice that IC .\

J is constant within an orbit of

as

i . e .

d permutes the p-regular elements. Then~ ~c 1 )=C 1 ,

where d q(l)=p 1

Pick x 1 ' c 1 • After a suitable renumbering we have

x =xq(l) 2 1 '

- q(l) - q(l) - q(l) x3-x2 •• .• ,xt(l)-xt(l)-1' xl-xt(l) so that

q(l)j-1 x.

J

(l)t(l) for j=2,3, •.• ,t(l) and x{ =x 1 .

cp p'

But all elements of c 1 have the same order: all remaining

elements of c 1 fall into similar cycles. Let there be

s (1) such cycles all together. If we choose y 1 , .•. ,ys(l)E:-C 1

78

to be representatives of these cycles then . l

c 1 = tYi(l)J- I l ~i~ s(l), l ~j ~ tO)} .

Choose cycle representatives Y. l.

for each other

orbit s. l.

in a similar way. Let m

Y = U Y .. For i = l l.

each

y E y define A = <. y'> c;- G . The fa mi 1 y y ~Ay}yEY satisfies

the hypothesis of Dietzmann's Lemma [Passman (77), Lemma

4.1.9}: it is closed under conjugation by elements of G.

Hence (A ~ yE: Y) =A· A •.• A where N=lY\. y Y1 Yz YN

Notice that (supp ( e )') = (A I y ~ y') . Indeed , a 11 y E:. Y y

are in supp(e) so the inclusion ? is trivial. On the

other hand, let z Esupp(e). Assume that Z ~CI €- s l • 1

(The case of other subscripts is similar). Then q(l)j-1

z= '1? (x) p

for some

Thus

Therefore

For y~Y. l.

Therefore

and

Now x=yi for some i,j.

1 1 d~j-1)+1 z= ~ (x)=xp =y~ E A

p l. y i

m \<supp(e))\~ fl \A\= \\ n \A \

yE.Y y i=l yf:.Y. y l. d,t ( i) - l

we have already established that yp =l.

n \ A \ ~ ( p ~t ( i ) _ l ) s ( i ) < p d,t ( i ) s ( i ) = p d_r ( i ) y~Y. y

l.

~\ P d;.r ( i ) \ s u p p ( e ) I I l = p • i= l

79

Notice that Theorem II.4.6 is a statement about

arbitrary idempotents. For the purpose of the application

in Lemma 3.2 we could content ourselves with the correspon-

ding result about central idempotents. Then the specializa-

tion argument is not necessary, as we show in the next

proposition.

Prop o s it ion 3 . 3 I f e = .Le ( x) x € k ( G'\ is a cent r a 1 idempotent

and char k = p > O, then all coefficients e(x) lie in

a finite subfield of k.

Proof: By Theorem I.4.1 we can assume that the group G

is finite. Recall Brauer's Theorem [Passman (77), Thm.2.4.61

If K is an algebraically closed field and charK=p) 0

then the number of centrally primitive idempotents in KLG1

is equal to the number of p-regular conjugacy classes in G.

In particular, it is true for K equal to the algebraic

closure of the prime field r p Let f 1 , ••• ,fr be the full

set of centrally primitive idempotents in F [ G1 • p

Now pick an algebraically closed field K containing

the fields k and F • p Then e, as well as

belong to K (G1. We claim that

f 1 , • • • , f r ,

full collection of centrally primitive idempotents in K(G1.

Otherwise, some of the f. would split. Write 1

f. as a 1

sum of centrally primitive idempotents in K(~. From

the orthogonality of f.'s it follows that the idempotents 1

80

from the decomposition of f . l

do not appear in the

decomposition of any other idempotent f .. Thus, if some J

f. decomposed nontrivially, then we would have more than l

r centrally primitive idempotents in K \.G1 , a contra-

diction to Brauer's Theorem.

Write e as a sum of centrally primitive idempo-

tents in K[~ By the previous remark, e is just a sum

of the f. 's, all of which have coefficients in F l p

Consequently, e also has all coefficients in i . Since p

e has finite support, the proposition follows a

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The vita has been removed from the scanned document

IDEMPOTENTS IN GROUP RINGS

by

Zbigniew Marciniak

(ABSTRACT)

The van Neumann finiteness problem for k(G1 is still

open. Kaplansky proved it in characteristic zero. He used the

nonvanishing of the trace: tr(e)=O implies e=O for any

idempotent e f k (G]. Assume now that char k =p > O. Now tr

can vanish on nonzero idempotents. Instead, we study the

lifted trace ltr. For 2 e=e Ek lGl , define ltr(e) by e o)

where e= L. e(x)x lifts e • Here xE.G

with le(x)\ ~ O, where each e(x) p

,.. e is an infinite series

lives in the Witt vector

ring of k. We prove that ltr(e) depends on e only, it is

a p-adic integer and ltr(e)=ltr(f) if f is equivalent

to e. Also ltr(e)f. Q and ltr(e)=O implies e=O if G is

polycyclic-by-finite. We conjecture that -log \ltr(e)\ < p p

2 lsupp(e)l • We prove this for e central and for e=e .: k(G)

with IG\ ~ 3 0. In the last section we give the example of an

idempotent e such that supp(f) is infinite for all f ......._,e.

Finally, we estimate \(s u p p ( e » I f o r c e n t r a 1 i d em po t en t s e .