The document discusses group rings and zero divisors in group rings. Some key points: - A group ring K[G] is a vector space over a field K with basis G, where elements are finite formal sums of terms with coefficients in K. Multiplication is defined distributively using the group multiplication in G. - If G contains an element of finite order, then K[G] will contain zero divisors. However, if G has no elements of finite order, it is not clear if K[G] contains zero divisors. - The document proves a theorem: K[G] is prime (has no zero divisors) if and only if G has no non-identity finite normal subgroups.

1. WHAT IS A GROUPRING?
D. S. PASSMAN
1. Introduction.Let K be a field.Supposewe aregivensome threeelementset {a, (, y} andwe
areaskedto forma K-vectorspace V withthisset as a basis.Thencertainlywe merelytake V to be
the collectionof all formalsums a ca+ b ,(+ c y with a, b, c E K. In the same way if we were
originallygiven four, five or six element sets we would againhave no difficultyin performingthis
construction.After awhile,however,as the sets get larger,the plussignbecomestediousandat that
pointwe wouldintroducethe E notation.In general,if we aregivena finiteset S, thenthe K-vector
space V withbasisS consistsof allformalsums),2s aa *a withcoefficientsa,aE K.Finally,thereis
no real difficultyin letting S become infinite.We merelyrestrictthe sumsY2a ac to be finite,by
whichwe mean that only finitelymanynonzerocoefficientsaa,can occur.
Of course, additionin V is given by
(E a, ))+ (nba a) = E(aa+ba)b a
and scalarmultiplicationis just
b( , a, a) (ba,,) cta.
Moreover,byidentifying3 E S withtheelement,B'= ab, a E V,whereb, = 1andb",= 0fora ,
we see that V does indeed have this copy of S as a basisand our originalproblemis solved.
Now,howdo we multiplyelementsof V?Certainlythecoefficient-by-coefficientmultiplication
a, .a) 0 & .
a)=(a,) a
isexceedinglyuninterestingandotherthanthatno naturalchoiceseemsto exist.So we arestuck.But
supposefinallythatwe aretold thatS is not justanyset, butratherthatS is in facta multiplicative
group. We would then have a naturalmultiplicationfor the basis elements and by way of the
distributivelaw this could then be extendedto all of V.
Let usnowstartagain.Let K be a fieldandlet G be a multiplicativegroup,notnecessarilyfinite.
Then the group ring K[G] is a K-vector space with basis G and with multiplicationdefined
distributivelyusingthe given multiplicationof G. In other words,for the latterwe have
xE a. x (Eby y) E (axby) (xy)- Ecz z,xeG yEG x,yEG zeG
where
CZ= axby xbx-lz.
xy=z xEG
Certainlytheassociativelawin G guaranteestheassociativityof multiplicationin K[G] so K[G] is a
ringand in fact a K-algebra.It is clearthat these easily definedgroupringsofferratherattractive
objectsof study.Furthermore,as the nameimplies,thisstudyis a meetingplacefor two essentially
differentdisciplinesandindeedthe resultsarefrequentlya ratherniceblendingof grouptheoryand
ringtheory.
Supposefor a momentthat G is finiteso thatK[G] is a finitedimensionalK-algebra.Sincethe
studyof finitedimensionalK-algebras(especiallysemisimpleones overalgebraicallyclosedfields)is
in farbettershapethanthestudyof finitegroups,thegroupringK[G] hashistoricallybeenusedasa
toolof grouptheory.Thisisof coursewhattheordinaryandmodularcharactertheoryisallabout(see
[21forexample).Ontheotherhand,if G isinfinitethenneitherthegrouptheorynortheringtheoryis
particularlyadvancedandwhatbecomesinterestinghereis the interplaybetweenthe two. Ourmain
concernin thisarticlewillbe withinfinitegroupsandin thefollowingwe shalldiscussandprovea few
173

2. 174 D. S. PASSMAN [March
selected results.These results,however,are by no meansrepresentativeof the entirescope of the
subject,butratherwerechosenbecausetheirproofsarebothelegantandelementary.Fewreferences
will be given here, but the interestedreaderis invitedto consultthe book [5] or the more recent
surveys,[4] and [6].
2. Zero divisors.Beforewe go anyfurther,twoobservationsarein order.First,once we identify
the elementsof G witha basisof K[G], thenthe formalsumsandproductsin , ax x areactually
ordinarysumsand products.In particular,we shalldropthe dot in this notation.Second,if H is a
subgroupof G then,since H is a subsetof the basis G, its K-linearspanis clearlyjust K[H]. Thus
K[H] is embeddednaturallyin K[G].
Now let x be a nonidentityelementof G so that G containsthe cyclicgroup(x ). Then K[G]
containsK[(x)] and we brieflyconsiderthe lattergroupring.Supposefirstthat x has finiteorder
n > 1. Then 1,x,_ , xn-' are distinctpowersof x and the equation
(1 -X) (1 + x + * + Xn1) = 1-Xn = O
showsthatK[(x)], andhenceK[G], hasa properdivisorof zero. Onthe otherhand,if x hasinfinite
orderthenallpowersof x aredistinctandK[(x)] consistsof allfinitesumsof theformE aix. Thus
thisgroupringlookssomethinglike the polynomialringK[x] andindeedeveryelementof K[(x)] is
just a polynomialin x dividedby some sufficientlyhighpowerof x. ThusK[(x)] is containedin the
rationalfunctionfield K(x) and is thereforean integraldomain.
Nowwhatwe haveshownaboveis the following.If G hasa nonidentityelementof finiteorder,a
torsionelement,then K[G] has a nontrivialdivisorof zero, butif G hasno nonidentityelementof
finiteorder,then thereare at least no obviouszero divisors.Becauseof this, andwithfranklyvery
littleadditionalsupportingevidence,it wasconjecturedthat G is torsionfreeif andonlyif K[G] has
no zero divisors.Remarkablythis conjecturehas held up for over twenty-fiveyears.
We stillknowverylittleaboutthisproblem.In factall we knowis thatthe conjectureis truefor
somerathersimpleclassesof groupsas,forexample,freegroups,orabeliangroups.Inthelattercase
theproofis even quiteeasy.Thus,supposethat G is torsionfreeabelian,andlet a, P be elementsof
K[G] with a/3= 0. Then, clearly,a and ,Bbelong to K[H] for some finitelygeneratedsubgroup
H C G and,bythefundamentaltheoremof abeliangroups,H isjustthedirectproductof theinfinite
cyclicgroups(x,), (x2), - and (x,). It is then quite easy to see that K[H] is containednaturally
betweenthepolynomialringK[xX,x **, x,] andthe rationalfunctionfieldK(x1,x2,** ,x,). Indeed,
K[H] is just the set of all elements in K(x1,X2,* * , x,) which can be writtenas a polynomialin
xI, x2, , x, divided by some sufficientlyhigh power of (xIx2... x,). Thus K[H] is certainlyan
integraldomainand hence, clearly,either a = 0 or / = 0. Actuallythe best resultto date on this
conjectureconcernssupersolvablegroups.Here for the firsttime nontrivialringtheorycomes into
play but the proof is unfortunatelytoo complicatedto give here.
On a more positive note, there is a variantof the zero divisorproblemwhich we can handle
effectively.A ringR is saidto be primeif aR/ = 0 for a, : E R impliesthata = 0 or : = 0. Clearly
this agreeswith the usualdefinitionin the commutativecase. In addition,thisconceptmakesmore
sense arddis moreimportantfroma ring-theoreticpointof view thanthe verymuchmorestringent
conditionof no zero divisors.Forexample,the matrixringKnis alwaysprime,even thoughit does
have zero divisorsfor n -' 2. The mainresultof interesthere is as follows.
THEOREM I. The groupring K[G] is primeif and only if G has no nonidentityfinite normal
subgroup.
The proofin one directionis quiteeasy.SupposeH is a nonidentityfinitenormalsubgroupof G
andset a = ElhEH h,thesumof thefinitelymanyelementsof H. If h E H thenhH = H so ha = a and
thus
a 2
ha=IHIa.
hEH

3. 1976] WHAT IS A GROUP RING? 175
In particular,if : = IH I1- a then afp= 0. Furthermore,since H is normalin G, we have for all
x E G,Hx = xH,andhenceax = xa. Thereforea commuteswithabasisof K[G], so it iscentraland
we have
aK[G]f = K[G]af = 0.
Finally,since a and,, are clearlynot zero we see that K[G] is not prime.
The conversedirectionis muchmoredifficultandas a starterwe mustaskourselveshow we are
goingto finda finitenormalsubgroupin G. Observethatif H is sucha subgroupandif h E H, then
certainlyh has finiteorder,and furthermore,for all x E G we have
hX= x-1hx E x-1Hx = H.
Thusall G-conjugatesof h arecontainedin H andso certainlythereareonly finitelymanydistinct
ones. This, therefore,leads us to definetwo interestingsubsetsof G, namely,
A(G) = {x E G0 x has only finitelymanyconjugatesin G}
and
Al(G) = {x E G x has only finitely many conjugates
in G and x has finite order}.
At thispointa certainamountof grouptheoryobviouslycomesintoplay.Whilethe proofsareby
no meansdifficultit does seem inappropriateto offer them here. Thereforewe just tabulatethe
necessaryfacts below. See [5, Lemma19.3]for details.
LEMMA1. Let G be a groupand let A(G) and Al(G) be definedas above.Then
(i) A(G) and Al(G) are normalsubgroupsof G.
(ii) A(G): AI(G) and thequotientgroupA(G)IA+(G)is torsionfree abelian.
(iii) I+(0) # (1) if and only if G has a nonidentityfinitenormalsubgroup.
Part(iii)abovecertainlyseemsto answerourfirstquestion.Nowobservethatifx isanyelementof
G, then the G-conjugatesof x are in a naturalone-to-onecorrespondencewith the rightcosets of
CG(x), the centralizer of x. Thus x E A(G) if and only if [G: CG(x)] < x and we shall need the
followingtwo basicpropertiesof subgroupsof finite index ([5, Lemmas1.1 and 1.21).
LEMMA2. Let G be a groupand let H,, H2, ,Hnbe a finitecollectionof subgroupsof G.
(i) If [G: H,] is finitefor all i, then [:G n H,] is finite.
(ii) If G is theset theoreticunionG = U ,Hig, offinitelymanyrightcosetsof thesubgroupsHi,
thenfor some i we have [G :Hi] finite.
We can now proceedwith the remainderof the proofof TheoremI. If a = XaxxE K[G], let us
define the supportof a, Suppa, to be the set of all group elements which occur with nonzero
coefficientin thisexpressionfor a. ThusSuppa is a finitesubsetof G whichis emptypreciselywhen
a = 0. Now supposethatK[G] is not primeandchoose nonzerogroupringelementsa and : with
aK[G]p = 0. If x E Suppa andy E Supp( thencertainly(x-'a)K[G](Py-') = 0, andfurthermore,
1E Suppx-'a, 1E Supp 3y-'.Thuswithoutlossof generality,we mayassumethata and( have1in
theirsupports.
Let us now write a = ao + a, and f = go + f3 where Supp ao, Suppgo CA(G) and Supp a1,
Supp S3QG - A(G). In other words,we have split a and f into two partialsums, the firstone
containingthegroupelementsinA(G) andthe secondone containingtherest.Since1iscontainedin
thesupportsof a and : we see thataoandgo arenonzeroelementsof K[A(G)]. Ourgoalhereis to
prove that aofo = 0. Once this is done the theoremwill follow easily. Let us suppose by way of
contradiction that aofo $ 0. Then clearly ao( = aofo + aofl is not zero since Supp aofo C A(G) and

4. 176 D. S. PASSMAN [March
SuppaoP3CG - A(G) so therecan be no cancellationbetweenthese two summands.Thuswe can
choose z EG to be a fixed element in Suppaof.
We observe now that if u E Suppao then [G:CG(U)] is finite and thus, by Lemma 2 (i),
H= nuueSuppaoCG(U) has finite index in G. Moreover, if h E H, then h centralizes all elements in
Suppao and hence h centralizes ao. Let us now write Suppat = {x1,x2, ,x, } and Suppf =
{yi, Y2, ,ys}.Furthermore,if x, is conjugateto zy7 in G for some i,j, thenwe choosesome fixed
gi,EG with g-' x,g = zy-'.
Nowlet h EH andrecallthataK[G]P = 0. Thensinceahf = 0 andh centralizesao,wehave
0 = h-'ahf = h-'(ao+ al)h . ( = ao( + h-'a,h3.
Thus z occurs in the support of h-'a, hp= - ao( so there must exist i,j with z = h-'xihyj or in other
wordsh-'xih= zy7'. Butthenthissaysthatxi andzy7' areconjugatein G,so bydefinitionof gijwe
have
h-'xih
= zy-' g-' xgij,
andthushg-' centralizesxi.Wehavethereforeshownthatforeachh E H thereexistsanappropriate
i, j with h E CG(xi)gi,and hence
HC U CG(x,)gij.
I,)
Now [G H] is finite,so G = UkHwk is a finiteunionof rightcosetsof H andwe concludefrom
the above that
G = U CG(X,)giWk.
We have thereforeshown that G can be writtenas a finite set theoreticunion of cosets of the
subgroupsCG(xi)andhencewe deducefromLemma2 (ii)thatforsome i, [G CG(xi)] is finite.But
this says that xi E A(G) and this is the requiredcontradiction,since by definitionxi E Suppa1C
G - A(G). Thus aofo = 0.
Wecannowcompletetheproofquitequicklyandeasily.Sinceaoandgfoarenonzeroelementsof
K[A(G)]withao(o = 0 we see thatK[A(G)]hasnontrivialzerodivisors.HenceA(G) cannotpossibly
be torsionfree abelian.On the other hand,accordingto Lemma1 (ii), A(G)/A+(G)is torsionfree
abelianso we must have A+(G)$ (1). Thus finallywe deduce from Lemma 1 (iii) that G has a
nonidentityfinitenormalsubgroupand the theoremis proved.
Inconclusion,we remarkthatTheoremI hasan amusingapplicationto the zerodivisorproblem.
Namely,we canshowthatif G is a torsionfreegroupthen K[G] hasnontrivialzero divisorsif and
only if it has nonzero elements of squarezero. Of course if a EK[G], a X0 with a2 = 0, then
certainlyK[G] haszerodivisors,so thisdirectionis reallytrivial.Intheotherdirectionlet a and( be
nonzeroelementsof K[G] witha(p= 0. SinceG is torsionfree,TheoremI impliesimmediatelythat
K[G] is primeso we have PK[G]aO0. But
(K[G]a .
PK[G]a = PK[G] (a:) K[G]a =0
since ap = 0 and hence we see that every element of (K[G]a has squarezero.
3. Idempotents.Let us returnagainto the groupringof a finitegroupand considerits regular
representation.That is, we view V= K[G] as a K-vector space on which K[G] acts as linear
transformationsby rightmultiplication.In particular,since V is finitedimensionalhere,each choice
of a basisfor V givesriseto a certainmatrixrepresentationforK[G]. Moreprecisely,foreachsuch

5. 1976] WHAT IS A GROUP RING? 177
basisweobtainanappropriatefaithfulhomomorphismr: K[G]-> KnfromK[G] intoKn,theringof
n x n matricesover K. Clearlyn = dimV = IG 1.
Supposefirstthat r correspondsto the naturalbasisfor V,namelyG itself.Thenforeach x E G
rightmultiplicationbyx merelypermutesthe basiselements.Thusr(x) is a permutationmatrix,that
isamatrixof 0'sandl's havingpreciselyone 1ineachrowandcolumn.Moreover,if x# 1,thengxk g
for all g E G so r(x) has only 0 entrieson its maindiagonal.Hence clearlytrace r(x) = 0. On the
otherhand,r(1) is the identitymatrix,so tracer(1) = n = 1.Now matrixtracemapsareK-linear
so for a = >a,x E K[G] we obtainfinally
tracer(a) = E ax tracer(x) = a, IG
Inotherwords,thetraceof r(a) isjustafixedscalarmultipleof a,, theidentitycoefficientof a.
Wearethereforeled,forarbitrarygroupsG,to defineamaptr: K[G] -> K,calledthetrace,by
tr(Iax xy = a,.
Moreover,it seems reasonableto expect thattr shouldhave certaintrace-likebehaviorand that it
should prove to be an interestingobject for study. Indeed, we observe immediatelythat tr is a
K-linearfunctionalon K[G], and furthermore,that for a = E axx,1 = E b,y we have
trag = E axby
xy =I
and this is symmetricin a and : since xy = 1 if and only if yx = 1. Thustr a} = tr (a.
Nowone of themostinterestingquestionsabouttrconcernsidempotentse E K[G] andthevalues
takenon by tre. To see whatwe mightexpectthesevaluesto be, let usagainassumethat G isfinite.
Since the trace of a lineartransformationis independentof the choice of basis,we compute
IG ttre = tracer(e)
by taking a basis more appropriatefor e. Namely we write V as the vector space direct sum
V = Ve+ V(1- e) andthen we choose as a basisfor V the unionof ones for Ve andfor V(1- e).
Since e acts like one on the firstset and like zero on the second, we concludethereforethat
IG Itre = tracer(e) = dim Ve.
Thusifwe areableto dividebyIG tin K(thatis,ifthecharacteristicof K doesnotdivideIG I)then
tre = (dim Ve)/!!G .
In other words, we see that tre is containedin the primesubfieldof K, i.e., the rationalsQ if
charK = 0 or the Galois field GF(p) if charK = p. Furthermore,in characteristic0 since 0_
dim Ve |IGI we have 0?tre - 1
Nowtt turnsoutthatthesetwopropertiesof tre areindeedtrueingeneral,butwe shouldobserve
thatthereis a basicdistinctionbetweenthem.The factthattre is containedin the primesubfieldis
clearlyanalgebraicpropertywhiletheinequality0 c tre _ 1is insomesenseanalyticinnature.With
this basicdichotomyin mind,we now proceedto considersome proofs.
THEOREMII. Let e E K[G] be an idempotent.Then tre is containedin theprimesubfieldof K.
WefirstconsiderfieldsK of characteristicp > 0, andasit turnsout,thepropertyof e we usehere
is eP = e. The reasonfor thisis thatin algebrasover fieldsof characteristicp the pth powermapis
fairlywell behaved.Indeedthe identity(a + b)P= aP+ bPalwaysholdsin commutativeK-algebras
andfor noncommutativeones an appropriategeneralizationexists.To be morepreciselet A be a

6. 178 D. S. PASSMAN [March
K-algebraanddefine[A,AJ, the commutatorsubspaceof A, to be the subspacegeneratedby allLie
products[a,b] = ab - ba witha, b E A. Thentheresultwearealludingto isthefollowingandaproof
can be found in [5, Lemma3.4].
LEMMA3. LetA be an algebraovera field K of characteristicp > 0. If a,, a2, "l am E A and if
n > 0 is a given integerthenthereexistsan elementsb E [A,A with
(a, + a2+ + **+ a.)Pn
= a P1"+ a2Pn + ..+ a
pn
+ b.
Withthisfactwe cannow provethe firstpartof TheoremII. Let e = I a,x be an idempotentin
K[G] andlet S denotethesubsetof Suppe consistingof allthoseelementsof ordera powerof p,the
characteristicof K.Thensince S is finite,thereexistsan appropriatepth power,saypS,withxP'_ 1
forallx E S.Nowlet n be anyintegerwithn ? s andwe applyLemma3 to ePr = e.Thusthereexists
an element y in the commutatorsubspaceof K[GJ with
e =e pn=E (ax) x pn+ Y,
andwe proceedto computethe tracesof bothsidesof thisequation.Observefirstthattry = 0 since
foranya,,BE K[GJwe havetra: = trO3aandhencetr[a, PIJ 0. Also sincen _ s itfollowsthatfor
any x E Suppe we have xP"= I if and only if x E S. Thusclearly
tre = (a.)P
=
(E a,X)
xes xE=S
Nowthisequationholdsforallintegersn ? s andinparticularifwetaken = s ands + 1weobtain
(tre)P -(Eax ) -(E ax) -tr e.
Thereforetre isanelementk E K whichsatisfieskP = k andsinceallsuchelementsk arecontained
in GF(p), the theoremis provedin characteristicp > 0.
We now proceedto considerthe characteristic0 case and here we will need the following(see
[7, AssertionsIV, Y, VI]).
LEMMA4. Let A = Z[ai, a2,, ar] be an integraldomain in characteristic0 which is finitely
generatedas a ringovertheintegersZ andletb E A bean elementnotcontainedintherationalnumbers
Q. Thenthereexistsa maximalidealM of A suchthatF =AIM is a fieldof characteristicp > 0 for
someprimep and such thatthe imageof b in F is notcontainedin GF(p).
We remarkthat this fact is an easy consequenceof the ExtensionTheoremfor Places if b is
transcendentalover Q. But if b is assumedalgebraic,thenits possibleimagesin the fieldsA/M are
greatlyrestricted.Fortunatelyinthiscasewe canapplytheFrobeniusDensityTheorem,a resultfrom
algebraicnumbertheory,to prove the lemma.
Now let K have characteristic0 and let e = E axx be an idempotent in K[GJ. If A =
Z[ax Ix E Suppe], thenA isclearlyanintegraldomainincharacteristic0 andA isfinitelygenerated
asaringovertheintegersZ. Furthermore,e isanidempotentinA [GI, wherethelatteristhesubring
of K[GI consistingof allelementswithcoefficientsin A. Nowlet M be anymaximalidealof A with
F = AIM a field of characteristicp >0. Then underthe naturalhomomorphism
A[G1- *(AIM)[G] = F[G]
theimageof e isanidempotentandhencebythefirstpartof TheoremIItheimageof tre iscontained
in GF(p). The second partof TheoremII now follows immediatelyfromLemma4 with b = tre.
Thuswehaveprovedthefirstconjecturedpropertyof tre,namelythatitisalwayscontainedin the
primesubfieldof K andit remainsto considerthesecondproperty,namelythat0? tre ' 1if K has

7. 1976] WHAT IS A GROUP RING? 179
characteristic0. Here the proofis muchmoretechnicalin nature,it is analyticratherthanalgebraic,
and we shall presentsome basicreductionsand then indicatewhy the proof shouldwork.
Let K be a field of characteristic0 and let e =E aXx be an idempotent in K[G]. If F=
Q(a. Ix E Suppe) thenF is afinitelygeneratedfieldextensionof Q ande is anidempotentinF[GJ.
Furthermore,F canbe embeddedin the complexnumbersC andhenceagaine is an idempotentin
C[G].Thus,asafirstreduction,we mayclearlyassumethatK = C isthefieldof complexnumbers.A
secondminorobservationis thatwe needonlyshowthattre - 0. Thereasonforthisis thatif e is an
idempotent,then so is 1- e and then 1- tre = tr(1- e) >0Oyields tre- 1.
Let us nowconsiderthe complexgroupringC[G].As is to be expectedwhenone combinesthe
structureof groupringswiththe richnessof the complexnumbers,manynice additionalproperties
emerge. For example,if a = E axx and : = E b,x are elementsof C[G] we set
(ae,:)=E abxg
and
where a is the complexconjugateof a and Ia I is its absolutevalue. Then clearly(, ) definesa
Hermitianinnerproducton C[G]withthe groupelementsas anorthonormalbasisandwithjj jjthe
usualassociatednorm.Furthermorelet a* be given by
ar* = aXx-1.
Then the map * is easily seen to satisfy
(a +f)* = a* + a*, (ap)* =P*a*, a** = a,
so that*isaringantiautomorphismof C[G]of order2, thatis, aninvolution.Nowobservethat
(a,,)= tra/3* = trf3*a
and hence we deduceeasily that if y is a thirdelement of C[G], then
(a, 3y) = (ay*,f3) = (13*a,y).
In other words, * is the adjoint map with respect to this inner productfor both right and left
multiplication.
Wecannowuse the abovemachineryto obtainanalternateproofof theassertiontre _ 0 at least
whenG is finite.Let I = eC[G]be the rightidealof C[GJgeneratedby theidempotente andlet I'
be its orthogonalcomplement.Thensince C[G]is a finitedimensionalvectorspace,we knowthat
I + IP= C[G]is a directsumdecomposition.ButIPis notjusta subspaceof C[G],it isinfactalsoa
rightideal.Tosee thislet a E I, f E I' and-yE C[G].ThensinceI isarightidealaRy*E I andhence
(a, 3y) = (ay*, /3)=0.
Thus f3yis orthogonalto all a E I and we concludethat fy E IP.
We have thereforeshownthatI + I' = C[G]is a decompositionof C[G]as a directsumof two
rightidealsandwe let f + f' = 1be thecorrespondingdecompositionof 1.As is wellknown,f andf'
are then both idempotentswith I= fC[G] and I' = f'C[G]. Now f is orthogonalto f'C[G] =
(1-f)C[G] so for all a E C[G] we have
0 =(f,(1- f)a) = ((1- f)*f, a)

8. 180 D. S. PASSMAN [March
and hence (1- f)*f E C[G] =O. Thusf =f*f so
f* =(f*f)* =f*f=f
andwe see thatf is a self-adjointidempotent,a projection.It is nowaneasy matterto completethe
proof.SinceeC[G] = I = fC[G] bothe andf areleftidentitiesforthisidealandhencewehave
tre = trfe = tref = trf.
Furthermoref = f*f so
tre = trf = trf*f=- IfI2=0
and the resultfollows.
It is amusingto observethatthisproofyieldsonly tre 0Oandnot tre E Q,butit does havethe
redeemingvirtueof being extendableto infinitegroups.Of course the finitenessof G was used
cruciallyhere.Infactit wasusedto deducethatI + I' = C[G] sincesuchdecompositionsarenot in
generaltrueforinfinitedimensionalinnerproductspaces.Buttheproofreallyrestsupontheauxiliary
elementf andthereis an alternatecharacterizationof thiselementwhichwe canuse. Let a E I and
considerthe distancebetween a and 1E C[G]. Then by definition
d(a,l)2 = -a-112 =(a - 1,a-1)
and since 1= f +f' and (a - f,f) = O we have
d(a, 1)2= (a -f-f', a-f-f ) a-f 11f2+fff '2.
Thusd(a, 1)f?Lf' andequalityoccursif andonlyif a = f.Inotherwords,f istheuniqueelementof
I whichis closest to 1.
Now let G be an arbitrarygroup,let e be an idempotentin C[G] andset I = eC[G]. Thenwe
definethe distancefrom I to 1 to be
d = infd(a, 1)- inf11a-1 1
SinceC[G] isno longercompleteif G is infinite,thereis no reasonto believethatsomeelementof I
existswhichisclosestto 1.Buttherecertainlyexistsa sequenceof elementsof I whosecorresponding
distancesapproachd. As it turnsout, it is convenient to choose a sequence fl,f2, fn., of
elementsof I with
d2<Ifn -12<d2 + /n4.
Then this sequenceplaysthe role of the auxiliaryelementf of the finitecase and indeed the final
conclusionmirrorsthe originalformulatre = lft112obtainedabove.Namelyaftera certainamountof
workwithinequalitiesandapproximations,whichwewillnotgivehere(see[5,Section22]fordetails),
we deduce finallythat
tre = limlInfI2n0.
This completesthe proof.
4. Semisimplicity.We close with one finalproblem,the semisimplicityproblem.The amusing
thingaboutthisone is thatwhileit hasbeenworkedon forovertwenty-fiveyears,it is onlyrecently
thata viableconjecturehasbeenformulated.Inthissectionwe shalldiscusstheconjecturealongwith
some of the earlywork in characteristic0.
We firstoffersome definitions.If R is a ring,then an R-module V is an additiveabeliangroup
which admits right multiplicationby the elements of R. More precisely, we are given a ring

9. 1976] WHAT IS A GROUP RING? 181
homomorphismR -* End V fromR to theendomorphismringof V ahdbywayof thismapthereisa
naturalactionof R on V whichwe denoteby rightmultiplication.Thusin essence, V is an R-vector
space. Furthermorewe say that V is an irreducibleR-module if there are no R-submodulesof V
otherthan0 or V itself.Forexample,if R = F is a fieldthenthe irreducibleF-modulesareprecisely
the one-dimensionalF-vector spaces.
Nowwhatwe wouldliketo do is studyR in termsof itsirreduciblemodulesbutthereis a natural
obstructionhere.It is quitepossiblethatwe cannoteven tellthe elementsof R apartin thismanner,
thatis, therecouldexisttwodistinctelementsr,s E R suchthatforallirreducibleR-modulesV and
for all v E V we have vr= vs. Of courseif the aboveoccursthen v(r - s) = 0 andhencewe cannot
distinguishthe element r- s from0. Thuswhat is of interesthere is
JR = {rE R IVr= 0 forallirreducibleR-modulesV},
the Jacobsonradicalof R. Thisis easilyseen to be a two sidedidealof R whichcanbe alternately
characterizedas follows (see [3, Theorem1. 2. 3]).
LEMMA5. Let R be a ringwith 1. Then
JR = JrE Rf1 - rsisinvertibleforalls (} R.
Now a ring is said to be semisimpleif its Jacobsonradicalis zero and this then leads to two
problemsin group rings. The first is to characterizethose fields K and groups G with K[G]
semisimpleandthenthe second,moreambitiousproblem,is to characterizein generalthe Jacobson
radicalJK[G]. The initialworkhere concernedfieldsof characteristic0 since for G finiteit was a
classicalfactthatK[G] is semisimple.PresumablyK[G] is alwayssemisimplein characteristic0 and
the firstresulthere on infinitegroupsconcernedthe field C of complexnumbers.
THEOREMIII. Forall groupsG, C[G] is semisimple.
Forourproof,whichcloselymirrorsthe original,we revertto the notationof the previoussection
and we furthermoreintroducean auxiliarynorm on C[G] by definingIa I= E| axI if a = , ajx.
Clearly Ia + I' Ia I+ Ijf and Ia/3 I' a Il I, l. Now let a be a fixed element of JC[GJ. Then for all
complexnumbers4, 1- ;a is invertibleby Lemma5 and we considerthe map
f(;)= tr(1- a)-' .
This is a complexfunctionof the complexvariable ; and we shall show thatf is in fact an entire
functionand we will find its Taylorseries about the origin.
Forconveniencewe set g(;) = (1- ;a)-' so thatf(;) = trg(;). Furthermore,it is clearthat all
g(;) C C[G] commuteand hence we have the basicidentity
g( g(ij) = (1 - {a) -_ (1 - 7a)-'
=[(1 - qa) - (1 - {afl(l - (a)-'(l - -qa)-'
=~(- B)ag(;)g (q).)
We first show that |g(rq)l is bounded in a neighborhoodof ;. Now by the above g(q) =
gso lg(i1)f lg(;)f+f-l fag(;l Ig(7)f and hence
Ig(-q) '{ -,q1 lag(;)l}=g(;) l
In particular,if we choose i1 sufficientlyclose to g thenwe canmakethefactor{.*. } largerthan1/2
and thuswe deduce that q - - impliesIg(iq)IJ?21g(;)I.
Nextwe showthatf(t) is an entirefunction.To do thiswe firstplugthe aboveformulafor g(Qq)
into the righthandside of the basicidentityto obtain
g(;) - g(71)= (; - 7)ag(;) {g(') - (a-71)ag(;)g(71) }

10. 182 D. S. PASSMAN [March
Then we divide this equationby ; - -qand take tracesto obtain
f )0- f( ) - trag(;)2 = _(-7)tr a2g(;)2g(-).
FinallysinceItry y Iweconcludefromtheboundednessof Ig(77)Iin aneighborhoodof ; that
lim = trag()2
Hence f(;) is an entire functionwith f(;) given by f'(;) = trag(;)2.
NowwecomputetheTaylorseriesforf abouttheorigin.Sincef(;) = tr(1- a)-1 whatweclearly
expecthereis thatfor ; smallwe canwrite(1- {a)-1 as the sumof an appropriategeometricseries
and then obtainf by takingtraces.To be more preciseand rigorousset
n
S = E tra.
1=0
Then
f ;)-Sn(;) = tr g(;) - ot
= trg( ){1-(1- a) iai}
trtg(;)gn+ 'an+1
and thus
If(; Sn ()|_||n+1 I a In
+1
I g(;
NowbyourpreviousremarksIg(;) is boundedinaneighborhoodof zeroandhencefor sufficiently
smallwe deduce that
lim sn(') = f(0.n -r
We have thereforeshown that
t(0) E; tra
. =0
is the Taylorseriesexpansionfor f(;) in a neighborhoodof the origin.Furthermore,f is an entire
functionandhencewe can invokea well-knowntheoremfromcomplexanalysis([1,Theorem3, pg.
142])todeducethattheaboveseriesdescribesf(;) andconvergesforall;. Inparticularwehave
limtra n = 0
and this holds for all a E JC[G].
We concludethe proof by showingthat if JC[G] 00 then there exists an element a E JC[G]
which does not satisfy the above. Indeed, suppose is a nonzero element of JC[G] and set
a = f3 */1/3112. Thena E JC[G] sincetheJacobsonradicalis anideal.Furthermorewe havea = a *
and
tra = 11 11-2 tr 3P* = 1 1-2K :112 = 1.
Now the powersof a are also symmetricunder * so for all m ?0 we have
traam+1 = tr a2m( 2m)* 11a2m ?>(tra2m)2.

11. 1976] WHAT IS A GROUP RING? 183
Hence by induction tra"2' =1 for all m -0 and this contradictsthe fact that tra' ->0. Thus
JC[G] = 0 and the resultfollows.
Needless-to-saythis analyticproofboth excitedand annoyedthe algebraistswho rightlyviewed
groupringsarean algebraicsubject.Lateralgebraicproofswereindeedgivenandthe best resultto
datein characteristic0 is asfollows.Let K be a fieldof characteristic0 whichis not algebraicoverthe
rationalsQ.ThenforallgroupsG, K[G] is semisimple.Furthermore,forallthe remainingfieldsthe
semisimplicityquestionis equivalentto that of the rationalgroupring (see[5],Theorem18.3).
Weshouldmentionatthispointthatthereis anamusingployto tryto extendtheaboveargument
to Q[G]. Namely,we againconsiderf(;)= tr(1- ;a)-, thistime as a mapfrom Q to Q, andwe
observe as before that for | small
f()= 2 tra'.
0
Now the righthandside heredescribesan analyticfunctionin a neighborhoodof the originandthis
functionhastheratherstrangepropertythatit takesrationalsto rationals.Thequestionthenis:must
such a function necessarilybe a polynomial?Unfortunatelythis is not the case and a simple
counterexampleis asfollows.Let r,,r2,, rn,... be anenumerationof thenonnegativerationalsand
define
-( d)1 (ri-(2 2 _ 22) ... (r- _2)
n=1 n! (r +) (r2 +1)(r+1)
Thenh iseasilyseento be anentirefunctionwhichtakesQ to Q andwhichisnotapolynomial.
Actually,it now seems that the semisimplicityproblemfor Q[G] will be settled by purelyring
theoreticmeanssincetheresultwillfollowfromanappropriatenoncommutativeanalogof theHilbert
Nullstellensatz.To be more precise,it has been conjecturedthat the Jacobsonradicalof a finitely
generatedalgebrais alwaysa nilideal.Thisconjecture,alongwiththeknownfact([5]Theorem18.5)
thatQ[G] hasno nonzeronilideals,wouldtheneasilyyieldJQ[G] = 0.Thuswhatisreallyofinterest
is the caseof fieldsof characteristicp > 0 andherewe arejustbeginningto guessat whatthe answer
mightbe. Supposefirstthat G is finite.Then K[G] is semisimplehere if and only if p 4 IG 1.Of
course this latterconditiondoes not make sense for infinitegroups,but the equivalentcondition,
namelythat G hasno elementsof orderp, does. None-the-lessthisis not the rightanswer.It is not
truethat JK[G] # 0 if andonly if G has an element of orderp. Whatdoes seem to be trueis that
elementsof orderp do play a role, providedthat they are suitablywell placedin G. To give this
ambiguousideasomemoremeaningwestartbyquotingasimplefact([5]Lemmas16.9and17.6).
LEMMA 6. Leta bean elementof thegroupringK[G]. Thena E JK[G] if andonlyif a E JK[H]
for all finitelygeneratedsubgroupsH of G withH D Suppa.
Thisof coursesaysthatwhatever"wellplaced"meansexactly,itmustsurelydependuponhowthe
particularelementsitsineachfinitelygeneratedsubgroupof G.Nowwe get morespecific.Let H be a
normalsubgroupof G. Then we say that H carriesthe radicalof G if
JK[G] = JK[H] . K[G].
Ourgoalhereis to findan appropriatecarriersubgroupH of G suchthatthe structureof JK[H] is
reasonablywell understood.Admittedlythisis a somewhatvaguestatementbutwe wouldcertainly
insistthatJK[H] beso simpleinnaturethatwecanatleastdecideeasilywhetherornotitiszero.
The firstcandidatefor H is based upon the A' subgroupand Lemma6 above. We define
A+(G) = {x E G Ix E A+(L) forallfinitelygeneratedsubgroupsL of G containingx}.
ThenA+(G)is a characteristicsubgroupof G andin allof theexamplescomputedso farA+(G)does

12. 184 D. S. PASSMAN [March
indeed carrythe radical.Furthermore,the ideal
I = JK[A+(G)]. K[G]
hasbeen studiedin generalandit hasbeen foundto possessmanyof the propertiesexpectedof the
Jacobsonradical.Thusit nowappearsthatA+(G)carriestheradicalbutwe areunfortunatelya long
way from a proof of this fact. Finallywe remarkthat this conjecturehas a nice ring theoretic
interpretation.Namelyit is equivalentto the assertionthatif G is a finitelygeneratedgroup,then
JK[G] is a union of nilpotentideals.
Now giventhe above conjecture,the next naturalstep is to studyA+(G) andJK[A+(G)Iandit
soonbecomesapparentthattheseobjectsarenotasniceaswe hadhopedfor.IndeedA+(G)turnsout
to be just any locally finite group and so the problem of determiningJK[A+(G)] is decidedly
nontrivial.We are thereforefacedwiththe problemof studyinglocallyfinitegroupsin generaland
here a new ingredientcomes into play.
Let G be a locallyfinitegroupso thatbydefinitionallfinitelygeneratedsubgroupsof G arefinite.
Nowif A is sucha finitesubgroupof G we saythatA is locallysubnormalin G if A is subnormalin
all finitegroupsH with A C H C G. In otherwords,we demandthat each such H has a chainof
subgroups.
A = HoC H1C C RH= H
forsome n withH, normalin H,?1.Thenusingtheselocallysubnormalsubgroupsasbuildingblocks
in a certaintechnicalmannerwe can define a new and interestingcharacteristicsubgroupof G
denotedby J(G). Againit turnsoutthatinallcomputedexamplesJ(G) carriestheradicalof G and
that furthermorethe ideal
I = JK[V(G)J. K[G]
possessesin generalmanypropertiesexpectedof the Jacobsonradical.Thusit nowappearsthatif G
is locallyfinitethen Y(G) carriesthe radicalof G.
Thesemisimplicityproblemforgroupsin generalhasthereforebeen splitintotwopieces,namely
the cases of finitely generatedgroupsand of locally finite groups.Moreover,if we combine the
correspondingconjecturesfor each of these cases,thenwhatwe expect,or at leasthope, is thatthe
groupY(A+(G))carriestheradicalof G.Whilethisgroupmayappearto be somewhatcomplicatedin
nature,there are in fact some nice structuretheoremsfor it. Furthermore,andmostimportant,we
have JK[Y(A+(G))]$ 0 if and only if 5(A+(G)) containsan element of orderp.
Added in Proof. Hopefully reference [4] will appear soon in translation in the Journal of Soviet Mathematics.
There is a new monograph in Russian by A. A. Bovdi entitled GroupRings. A second volume is anticipated. This
author is presently working on an expanded and updated version of [5] which will be entitled The Algebraic
Structureof GroupRings. It will be published in two volumes by Marcel Dekker, Inc. There has been some exciting
progress recently on the zero divisor problem, in particular in the case of polycyclic-by-finite groups. This can be
found in the papers by K. A. Brown, On zero divisors in grouprings,to appear in the Journal of the London Math
Society, and by D. R. Farkas and R. L. Snider, K0and Noetherian grouprings,to appear in the Journal of Algebra.
References
1. L. V. Ahlfors, Complex Analysis, McGraw-Hill, New York, 1953.
2. C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras,
Interscience, New York, 1962.
3. I. N. Herstein, Noncommutative Rings, Carus Mathematical Monographs, No. 15, Math. Assoc. Amer.,
1968.
4. A. V. Mihalev and A. E. Zalesskii, Group Rings (in Russian), Modern Problems of Mathematics, Vol. 2,
VINITI, Moscow, 1973.

13. 1976] MATHEMATICAL NOTES 185
5. D. S. Passman, Infinite Group Rings, Marcel Dekker, New York, 1971.
6. , Advances in group rings, Israel J. Math., 19 (1974) 67-107.
7. A. E. Zalesskii, On a problem of Kaplansky, Soviet Math., 13 (1972) 449-452.
MATHEMATICS DEPARTMENT, UNIVERSITY OF WISCONSIN, MADISON, WI 53706.
MATHEMATICAL NOTES
EDITED BY RICHARD A. BRUALDI
Material for this Department should be sent to Richard A. Brualdi, Department of Mathematics, University
of Wisconsin, Madison, WI 53706.
A CONNECTED, LOCALLY CONNECTED,
COUNTABLE HAUSDORFF SPACE
GERHARD X. RITTER
In this MONTHLY,76 (1969) 169-171, A. M. Kirch gave an example of a connected, locally
connectedcountableHausdorffspace.We give another,geometricallymuchsimpler,example.The
examplewe providewillbe a modifiedversionof a connected,but not locallyconnected,countable
Hausdorffspace constructedby R. H. Bing in [1].
LetX = {(a,b) e Qx Q1b > O},whereQdenotestheset of rationalnumbers,andlet 0 be afixed
irrationalnumber.The pointsof ourspacewillbe the elementsof X. Witheach (a,b)E X andeach
? > 0 we associatetwosets L.(a,b) andR. (a, b) whichconsistof allpointsof X in thetrianglesABC
and A'B'C', respectively, where A = (a - b/0,O), B = (a - b/0 + ?,0), C = (a - b/0 + ?/2, 0?/2),
A'= (a + bl0 - ?,0), B' = (a + b10,O),and C'= (a + bl0 - ?12,0?/2). We now define a basic
?-neighborhoodof (a, b) E X by Ne(a, b) = {(a,b)}ULe(a, b) URe(a, b).
Since 0 isirrational,no two pointsof X canlie on a linewithslope 0 or - 0. Thus,for anytwo
distinctpoints of X, we may choose a sufficientlysmall ? > 0 such that the points lie in disjoint
?-neighborhoods.Therefore,X is Hausdorffand, obviously,a countablespace.
(a, b) / / /
L, (a, b) R,.(a, b
FIG. I
Figure 1 shows that the closure of each basic E-neighborhood N. (a,b) consists of the union of four
infinite strips with slope ? 0 emanating from the bases of the triangles R. (a,b) and L.(a, b). Hence
the closures of each pair of basic neighborhoods intersect. Therefore, X is connected but not regular
and, hence, not metrizable.