Zero Divisors in Hopf Algebras

9 Pages·2015·0.163 MB·English

by  Caleb McWhorter

Zero Divisors in Hopf Algebras
CalebMcWhorter
Spring2015
1 Historical Background
The development of group rings was slow. The study of group rings began with Arthur Cayley (study-
ing CS ) in about 1854. Group rings then appeared in some works in the late 1800s in complex algebras.
3
However,itwasnottillaftertheturnofthecenturythatgroupringswereusedinrepresentationtheoryto
answerrealquestions. Finitedimensionalcomplexringsoccurnaturallyinthestudyofrepresentationsof
finitegroups,i.e. Maschke’sTheorem(ifGisafinitegroupandkisafieldsuchthat|G|·1 (cid:54)=0,thenkGis
k
asemisimplek-algebra)andSchur’sLemma(ifLisasimpleR-modulethenEnd Lisadivisionring).Nev-
R
ertheless, the first monogrammed work on group rings was not until “The Algebraic Structure of Group
Rings” by Donald S. Passman in 1977. Much of what is known about group rings came when studying
semisimplerings. Rickartshowedin1950thatifK = C,thenKGissemisimpleregardlessofthegroupG.
∼
First,wewillneeddefineagroupring. Oneinitialopenquestion(nowsolved)wasdoesRG = RH imply
thatG ∼= H? ThiscouldnotalwaysbethecasegiventhatifR =Cand|G| < ∞,thenCGisisomorphictoa
directproductofmatrixrings(asweshallsee)soforanytwononisomorphicabeliangroupsG,Hoforder
n,wehaveCG ∼=Cn ∼=CH. Infact,Hertweckshowedin2001thattherearetwononisomorphicgroupsof
order225·972havingnonisomorphicgrouprings.
2 Group Rings
Definition2.1(GroupRing). Let R bearingwithidentityand G agroup. Agroupring, denoted RG, isaring
consistingofsumsoffinitesupport
∑
a g
g
g∈G
wherea ∈ Randg ∈ G. Theoperationsaredefinedasfollows:
g
∑ ∑ ∑
a g+ b g = (a +b )g
g g g g
g∈G g∈G g∈G
(cid:32) (cid:33)(cid:32) (cid:33)
∑ ∑ ∑
a g b g = c g
g g g
g∈G g∈G g∈G
wherecg = ∑h∈Gagbg−1h. Thisringisunitalwithidentity1R1G. ThisgroupringisaleftR-moduleunderaction
∑ ∑
r· a g = (ra )g
g g
g∈G g∈G
1
Thegroupringisthenafree R-modulewithbasisconsistingofcopiesofelementsof G withrank|G|. If Risafield
(oftendenotedk),kGisavectorspaceoverk,withacanonicalbasisconsistingoftheelementsofG. IfGisfinite,then
kGisafinitedimensionalk-algebra. TheringkGisoftencalledthegroupalgebra.
Remark 2.2. Multiplication in a group ring might seem strange but it is exactly what is required when
we force (a g)(b h) = a b gh. One could simply define this to be the product. One could also define
g h g h
(cid:16) (cid:17)(cid:16) (cid:17)
∑ a g ∑ b g = ∑ c g,wherec = ∑ a b . Onenaturalseestheequivalenceviagh = u
g∈G g g∈G g u∈G u u gh=u g h
ifandonlyifg = uh−1orviceversagh = uifandonlyifh = g−1u.
Definition 2.3 (Support Group). For a function f ∈ RG, the support of f, denoted Supp(f), consisting of the
finitesubsetofpoints x ∈ G forwhich f(x) (cid:54)= 0. Thesupportgroupof f isthesmallestsubgroupof G containing
Supp(f).
Theidentityofagroupringis1 ·1 . Ingeneral, RG isnotcommutative. Infact, RG iscommutative
R G
if and only if both R and G are commutative.The group ring RG is a ring extension of R as we have a
ringhomomorphism R → RG givenbyr (cid:55)→ r·1 . So RG isbydefinitionan R-algebra. Finally, themap
G
G → RGgivenbyg (cid:55)→1 ·gisagroupembeddingofGintothegroupofunitsofRG. Furthermore,these
R
typesofmapscanbenaturallyextended.
Proposition2.4. Ifϕ : S → Risaringhomomorphism,thenϕextendsuniquelytoahomomorphismofgrouprings
ϕ : SG → RGgivenby
(cid:32) (cid:33)
∑ ∑
ϕ a g = ϕ(a )g
g g
g∈G g∈G
If ϕ isinjective, thensotoois ϕ. Inthiscase, SG canbeinterpretedasasubringof RG as SG canonicallyembeds
intoRG.
Proposition2.5. If ϕ : H → G isagrouphomomorphism, then ϕ extendsuniquelytoahomomorphismofgroup
rings ϕ : RH → RGgivenby
(cid:32) (cid:33)
∑ ∑
ϕ a g = a ϕ(g)
g g
g∈G g∈G
ThisshowsthatifH ≤ G,thenRHcanbecanonicallyembeddedintoRG.
Proposition2.6. Let R beacommutativering, G agroup, and A an R-algebra. If ϕ : G → I(A), where I(A) is
the set of invertible elements of A, be a group homomorphism. Then ϕ induces a unique R-algebra homomorphism
ϕ : RG → Asuchthat ϕ(g) = ϕ(g)forallg ∈ G.
TherearealsoprojectionmapsoftheR-moduleRGontosubmodulesRH,whereH ≤ G.Theprojection
map is π : RG → RH mapping the basis elements g ∈ G\H to 0. That is for a ∈ RG given by a =
H
∑ a g,wehave
g∈G g
∑
π (a) = a g
H g
g∈H
Proposition 2.7. Let H ≤ G be groups and R be a ring. The projection map π : RG → RH is an R-module
H
homomorphismsuchthatfora ∈ RG,b ∈ RH,wehave
π (ab) = π (a)b
H H
π (ba) = bπ (a)
H H
2
Example 2.8. Let k be a field and let G = (cid:104)g(cid:105) be a cyclic group of order n. We know k[x] is a PID. Take
the map ϕ : k[x] → kG given by x (cid:55)→ g is a surjective ring homomorphism. It is clear that xn −1 ∈
kerϕ. Furthermore, if p(x) ∈ kerϕ, then we know p(x) = q(x)(xn−1)+r(x) for q(x),r(x) ∈ k[x] with
degr(x) < n so that kerϕ is the PID generated by xn−1. Therefore by the First Isomorphism Theorem,
kG ∼= k[x]/(xn−1).
Example2.9. BytheArtin-WedderburnTheorem, anygroupringover k isisomorphicasa k-algebratoa
directproductofmatrixringsoverafinitedimensionaldivisionalgebraD. Then
i
n
kG ∼= ∏M (D)
ni i
i=1
Remark2.10. Asareminder,theArtin-WedderburnTheoremstatesthatifRisafinitedimensionalsemisim-
plealgebra,thenRisisomorphictoaproductofmatrixalgebrasoverdivisionrings.
Example2.11. Asaspecificexampleofthepreviousexample,ifkisanalgebraicallyclosedfield,thepoly-
nomialxn−1factorscompletelyintoirreduciblefactors
n
xn−1= (x−ζ )(x−ζ )···(x−ζ ) = ∏x−ζ
1 2 n i
i=1
Furthermore,ifnisaprimenumberandk =C,thentherootsofunityζ ,ζ ,··· ,ζ arealldistinct. IfM is
1 2 n i
themaximalidealofC[x]generatedbythepolynomialx−ζ ,thenbytheChineseRemainderTheorem
i
n n
CG ∼=C[x]/∏M ∼= ∏C[x]/M ∼=Cn
i i
i=1 i=1
Example2.12. IfR =CandG =Z,weknowthatCZ∼=C[x,x−1],theringofformalsums∑n∈Zanxnwith
finitesupport.
Example 2.13. Given any linear group representation ρ : G → GL(V), where V is a vector space over a
fieldk. ThisgrouprepresentationhasakG-modulestructureonV andviceversa.
Agroupring RG canbeidentifiedwiththefinitelysupported R-valuedfunctionson G. Suppose x ∈
RG. Then x = ∑ a G. Wecantheninterpret xasafunction x : G → Rgivenby x(g) = a . Thegroup
g∈G g g
ringRGcanthenbegivenadditionoperations
(x+y)(g) = x(g)+y(g)
forallx,y ∈ RGandg ∈ G. Multiplicationissimilar
(x∗y)(g) = ∑ x(gh−1)y(h)
h∈G
forallx,y ∈ RGandg ∈ G,where∗isjustordinaryconvolution.
For any nontrivial group, a group ring can never be simple as they always contain a proper nonzero
ideal.
Definition 2.14 (Augmentation Ideal). Let G be a nontrivial group and let E be the trivial group. Consider the
homomorphism ϕ : G → E. IfRisaring,itisclearthatRE ∼= R. Theextendedringhomomorphism ϕ : RG → R
iscalledtheaugmentationmap. Thekernelofthismapiscalledtheaugmentationideal:
(cid:40) (cid:41)
∑ ∑
kerϕ = a g | a =0
g g
g∈G g∈G
Thisidealisdenotedω(RG).
3
Notice that if R is a field then the augmentation ideal of RG is maximal as it is the kernel of a ring
homomorphismintoafield.
3 Idempotents and Zero Divisors
Justtomakeclearthedefinitionforgrouprings.
Proposition 3.1. An element x = ∑ a g ∈ RG is in the center of RG if and only if the coefficients a are
g∈G g g
containedinthecenterofRandag = ahgh−1 forallg,h ∈ G.
ThezerodivisorconjectureofKaplanskyiscloselyrelatedtoanotherconjectureasthetwoconceptsare
similar. Wewilldiscussthemsimultaneously. However,wefirstgivesomedefinitions.
Definition3.2(ZeroDivisor). Anelement y ∈ R isazerodivisorifthereisanonzeroelement x ∈ R suchthat
xy = 0. Anelementisatrivialzerodivisorify = 0. Leftandrightzerodivisorsaredefinedintheobviousway. If
theringRiscommutative,allthedefinitionsareequivalent.
Example3.3. Ifn = pqforsomeprimes p,q,then p,qarezerodivisorsinZ/nZ.
Example3.4. Wehavetwoexamplesofzerodivisorsin M (R)
2
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
1 1 1 1 0 0
=
2 2 −1 −1 0 0
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
1 0 0 0 0 0
=
0 0 0 1 0 0
Definition3.5(Idempotent). Anelement e ∈ R isanidempotentif e2 = e. Amap e : R → R isanidempotent
mappingife2 = e. Anidempotentistrivialife =0ore =1.
Example3.6. Anmatrixwith0’sand1’salongitsdiagonalsandzeroselsewhereisanidempotentmatrix
in M (R).
n
Example3.7. If M = N⊕LisanR-module,thentheprojectionmapsπ andπ areidempotentmaps.
N L
Remark3.8. Ifeisanidempotent,thenalltheconjugatesofebyaunitareidempotents. Thatis,ifeisan
idempotentandx ∈ Risaunit,thenxex−1 isanidempotent. Furthermore,ifeisanidempotentthen1−e
isanidempotent. Noticealsothatanilpotentelementisalsoazerodivisor.
Observe that nontrivial idempotents e ∈ R imply the existence of zero divisors as e(1−e) = 0 =
(1−e)e. However,theexistenceofidempotentsismorestringentthanrequiringzerodivisorsasZ/4Zhas
nontrivialzerodivisorsbutnonontrivialidempotentelements.
Conjecture1(Kaplanksy). LetkbeafieldandGatorsionfreegroup. ThenkGcontainsnonontrivialidempotents.
ItisclearthatifRcontainsanidempotentelements,sotoodoesRG. Furthermore,itisnecessarythatG
betorsionfreeforotherwiseitwouldhaveafinitesubgroupHandthen
1 ∑
x = h
|H|
h∈H
wouldbeanidempotent(sethat Hh = H forallh ∈ H). ThisconjecturehasbeenconfirmedbyFormanek
for noetherian groups (groups which satisfy the ascending chain condition) in the case where k has char-
acteristic 0. Furthermore, if R is an integral domain and G is torsion free abelian then RG is an integral
domainsoRGcannotcontainnontrivialidempotents(forotherwiseitwouldcontainzerodivisors).
4
Proposition3.9. IfRisanintegraldomainandGisatorsionfreeabeliangroup,thenRGisanintegraldomain.
Proof: IfH ≤ G,wehavetheinclusionRH (cid:44)→ RG. Ifx ∈ RGisatorsionelement,thereisa0(cid:54)= y ∈ RG
suchthatxy =0. LetHbethesubgroupofGgeneratedbythesupportsofx,yandformthesubgroupring
RH. Thenx,y ∈ RH,whereHisfinitelygenerated. ItthensufficestoconsiderthecasewhereGisafinitely
generatedtorsionfreeabeliangroup.
Let G = Zn and let Q = Frac(R) be the field of fractions. Consider the field of rational functions
Q(x ,··· ,x ). Theelementsof G aren-tuplesofintegers. Mapanyn-tupleto xmi···xmn. Thisisagroup
1 n 1 n
homomorphism,whichthenextendstoaringhomomorphismRG (cid:44)→ Q(x ,··· ,x ). ButQ(x ,··· ,x )is
1 n 1 n
afield.
However,thereisastrongerconjecture,alsoformulatedbyKaplanksy.
Conjecture2(Kaplanksy). LetkbeafieldandGatorsionfreegroup. ThenkGhasnonontrivialzerodivisors.
It is clear that R need be a domain as R ⊆ RG and that G need be torsion free for G ⊆ RG. If G
containedanytorsionelementgofordern >2,thenx =1−gwouldbeatorsionelementinRGbyletting
y =1+g+···+gn−1,whichisnonzerosothatwehave
0= gn−1= (1−g)(1+g+···+gn−1) = xy
This conjecture has been shown for several classes of groups. For example in 1976, Brown, Farkas,
Sniderconfirmedtheconjectureforpolycyclic-by-finitegroupsoverfieldsofcharacteristic0.
Definition3.10(PolycyclicGroups). Apolycyclicgroupisasolvablegroupwithcyclicfactors. Thatis,wehavea
chainofnormalsubgroups
1= G (cid:69)G (cid:69)G ···(cid:69)G = G
0 1 2 n
suchthatGi+1/Gi iscyclicfori = 0,1,··· ,n−1. Agroupiscalledapolycyclic-by-finteifitcontainsapolycyclic
normalsubgroupHoffiniteindex.
Itisalsoknownthatif k isafieldand G isatorsionfreegroup,then kG hasnonontrivialcentralzero
divisors.Theresulthasalsobeenshownforfiniteconjugacyclassgroups(FCC-groups)anduniqueproduct
groups(up-groups). However, theconjecturehasnotevenbeenshownforanytorsionfreegroup G over
F .Therearesomeresultsthatgiveequivalenciesfortheconjecture,butnoneofwhichareofanyparticular
2
use.
Proposition3.11. LetkbeafinitefieldandGbeatorsionfreegroup. IfkGisadomainthenCGisadomain.
Proposition3.12. LetGbeatorsionfreegroup. Thefollowingareequivalent:
(i) Foranyfieldk,kGisadomain.
(ii) Foranyfinitefieldk,kGisadomain.
Remark 3.13. There is actually a stronger conjecture which would imply the two given above: Let k be
a field and G a torsion free group. Then kG contains no nontrivial unit. This is known as the Kadison-
Kaplanksyconjecture. IfonecanshowthisconjectureholdsforthereducedC∗-algebra,thentheconjecture
holds and we obtain the above conjectures as corollaries. The Kadison-Kaplanksy conjecture was shown
to hold in 1997 by Higson and Kasparov for groups a-(T)-menable groups or those with the Haagerup
property;thatis,groupsadmittingametricallyproperaffineisometricactiononHilbertspaces. Examples
of such groups are amenable, free coxeter, countable subgroups of GL (k), or any discrete subgroups of
2
SO(n,1)orSU(m,1).
5
4 Hopf Algebras
Definition4.1(k-algebra). Ak-algebraisak-vectorspacewithtwolinearmaps
m : A⊗ A → Aandu : k → A
k
calledmultiplicationandunit,respectively,suchthatthefollowingdiagramsarecommutative
A⊗A⊗A m⊗1 A⊗A A⊗A
u⊗1 1⊗u
1⊗m m k⊗A A⊗k
A⊗A m A A
Therightdiagramgivestheidentityelementin Abysetting1 = u(1 ).
A k
Definition4.2(TwistMap). Forany k-spacesV,W, thetwistmap τ : V⊗W → W⊗V isgivenby v⊗w (cid:55)→
w⊗v.
Definition4.3(Coalgebra). Ak-coalgebra(withcounit)isak-vectorspace, A,withtwok-linearmaps
∆ : A → A⊗Aand(cid:101) : A → k
calledco-multiplication/co-productandcounit,respectively,suchthatthefollowingdiagramscommute
A⊗A⊗A ∆⊗1 A⊗A A⊗A
(cid:101)⊗1 1⊗(cid:101)
1⊗∆ ∆ k⊗A A⊗k
1⊗− −⊗1
∆
A⊗A A A
Thisandthepreviousdefinitiongivethat∆isinjectiveandmissurjective.
Definition4.4(Bialgebra). Abialgebra A isa k-vectorspace, A = (A,m,u,∆,(cid:101)), where (A,m,u) isanalgebra
and(A,∆,(cid:101))isacoalgebrasuchthatitsatisfiestheequivalentconditions.
(i) ∆,(cid:101)arealgebrahomomorphisms
(ii) m,uarecoalgebrahomomorphisms.
Example 4.5. If G is any group, let B = kG be the group ring. Then B is a bialgebra via ∆g = g×g and
(cid:101)(g) =1forallg ∈ G.
Example4.6. If gisanyk-Liealgebraand B = U(g)isitsuniversalenvelopingalgebra,then Bbecomesa
bialgebravia∆x = x⊗1+1⊗xand(cid:101)(x) =0forallx ∈ g.
6
Definition4.7(BialgebraMorphism). Abialgebramorphismisamorphismwhichisbothanalgebraandcoalgebra
morphism.
Remark4.8. If A,Barek-algebras,then A⊗Bisak-algebrawithoperation(a⊗b)(c⊗d) = ac⊗db. That
is,
A⊗B⊗A⊗B1⊗−τ→⊗1 A⊗A⊗B⊗Bm−A⊗→mB A⊗B
a
whereτ : B⊗A → A⊗Bistheflipb⊗a −→ ⊗b. TheunituA⊗B of A⊗Bisgivenby
k ∼= k⊗k u−A⊗→uB A⊗B
Likewise,ifC,Darecoalgebras,then⊗Dwith∆C⊗D givenby
C⊗D ∆−C⊗→∆D C⊗C⊗D⊗D1⊗−τ→⊗1C⊗D⊗C⊗D
andcounit
C⊗D (cid:101)−C⊗→(cid:101)D k⊗k ∼= k
Inparticular,thisappliesfor A = BandC = D.
Definition4.9. LetAbeak-algebra.ThefinitedualofAisA◦ = {f ∈ A∗ | f(I) =0forsomeideal I of Asuchthat dimA/I <
∞}, where A∗ = Hom (A,k) is the linear dual of V. Note that A,A∗ determine a nondegenerate bilinear form
k
(cid:104), (cid:105) : A∗⊗A → kvia(cid:104)f,a(cid:105) = f(a).
Proposition4.10. If Aisanalgebra,then A◦ isacoalgebrawithcomultiplication∆ = m∗ andcounit(cid:101) = u∗. If A
iscommutative,then A◦ iscocommutative.
Remark4.11. A◦ isthelargestsubspaceV of A∗ suchthatm∗(V) ⊆V⊗V.
Definition 4.12 (Convolution Product). Let C be a coalgebra and A an algebra. Then Hom (C,A) becomes an
k
algebraundertheconvolutionproduct
(f ∗g)(c) = m◦(f ⊗g)(∆c)
forall f,g ∈ Hom (C,A),c ∈ C. TheunitelementinHom (C,A) = u(cid:101). Onedefinesthetwistconvolution, or
k k
anti-convolution,omHom (C,A)by
k
(f ×g)(c) = m◦(f ⊗g)(τ◦∆(c))
Definition4.13(HopfAlgebra). Let A = (A,m,u,∆,(cid:101))beabialgebra. ThenalinearendomorphismSfrom Ato
Aisanantipodefor Aifthefollowingdiagramcommutes
A⊗A m A m A⊗A
1⊗s u(cid:101) s⊗1
A⊗A A A⊗A
∆ ∆
Thatisforalla ∈ A,(cid:101)(a) = ∑a S(a ) = ∑S(a )a . AHopfalgebraisabialgebrawithanantipode. Equivalently,
1 2 1 2
thereisanelementS ∈Hom (H,H)whichisaninverseto1 underconvolution.
k H
Example4.14. LetGbeanygroupandkafield. ThenkGisaHopfalgebrabydefiningS(g) = g−1 forall
g ∈ G. Since S is linear, we know 1 = (cid:101)(g) = gg−1 = g−1g for all g ∈ G. We define ∆(g) = g⊗g and
(cid:101)(g) =1forallg ∈ G.
7
Remark4.15. Inparticular,thisexampleshowsthataproofofthezerodivisorconjectureforHopfalgebras
mayverywellprovetheKaplanskyZeroDivisorConjecture.
Proposition4.16. LetHbeaHopfalgebrawithantipodeS. Then
(i) Sisananti-algebramorphism. Thatis,
S(hk) = S(k)S(h)forallh,k ∈ H; andS(1) =1
(ii) Sisananti-coalgebramorphism;thatis,
∆◦S = τ◦(S⊗S)◦∆and(cid:101)◦S = (cid:101)
Definition4.17(Group-LikeElements). LetAbeanycoalgebra.Thena ∈ Aisagroup-likeelementif∆a = a⊗a
and(cid:101)(a) =1.
Remark4.18. OnecanseethenomenclatureforagrouplikeelementbyexaminingExample4.5andExam-
ple4.14
ThestructureofHopfalgebrasissimilartothatofgroupringsledPeterLinnelltoconjectureavariant
oftheKaplanskyZeroDivisorConjectureforHopfAlgebras.
Conjecture 3 (Linnell’s Conjecture). Let H be a Hopf algebra and G be a torsion free subgroup of its group-like
elements. Thenifα ∈ kGandβ ∈ Harenonzerotheproductαβisnonzero.
IfGisaninfinitecyclicgroupgeneratedbyxandkisafield. ConsidertheHopfalgebra
H = (cid:104)g,h,x | gh =1,gx = qxg(cid:105) = (cid:77)kGxi
i≥0
where 0 (cid:54)= q ∈ k\{ζn}, ζ is a root of unity, n ∈ Z, and g,h ∈ G are group-like elements. We note that
comultiplication∆,counit(cid:101),andantipodeSaredefinedas
∆(g) = g⊗g
∆(h) = h⊗h
∆(x) = g⊗x+x⊗1
(cid:101)(g) =1
(cid:101)(h) =1
(cid:101)(x) =0
S(g) = h
S(h) = g
S(x) = −hx
ItisclearthatSisinvertibleas
S2(x) = S(−hx) = −S(h)S(x) = −gS(x) = −g(−hx) = ghx =1x = x
Furthermore,S2(g) = S(h) = gandS2(h) = S(g) = h. Itisclearthat H isofinfiniteorderas0 (cid:54)= q ∈ kis
notarootofunity.
8
Proposition4.19. Let G = Zandkisafieldand H istheHopfAlgebra H = (cid:76) kGxi. Ifα ∈ kG and β ∈ H
i≥0
arenonzero,thenαβ (cid:54)=0.
Proof: Supposethatα ∈ kGand β ∈ H arenonzero. Bydefinitionandassumption, β = ∑ α xi,where
i i
the α arenotallzeroi ∈ Z. Againbydefinition, α = ∑ a gj, wherethe a ∈ k arenotallzerofor j ∈ Z.
i j j j
Then αβ = ∑ αα xi. Soitsufficestoshowthat αα (cid:54)= 0forsomei ∈ Zastheproductistakenpointwise.
i i i
Butasαi ∈ kG,wehaveαi = ∑ki akigki,whereki ∈Z. Thenforsomei ∈Z,
(cid:32) (cid:33)(cid:32) (cid:33)
αα = ∑a gj ∑a gki
i j ki
j ki
is the product of two nonzero elements of kG, which is never zero (we showed this in Proposition 3.9).
Hence,αα (cid:54)=0forsomei ∈Z. Hence,αβhasatleastonenonzerocoefficient.
i
Roman andLinnell have managed toshow that CG0 (the Hopfalgebra of C linear maps α : CG → C
such that kerα contains an ideal I of finite co-dimension d in CG) has no nontrivial zero divisors for any
infinite finitely generated group G as well as a special case for Hopf algebras of representative functions
overcompactgroups.
References
[1] S.Montgomery,HopfAlgebrasandtheirActionsonRings.AmericanMathematicalSociety,RhodeIsland,
1992.
[2] A. Roman, Peter Linnell, Zero Divisors of Hopf Algebras: A Generalization of the Classical Zero-
DivisorConjecture.May2012.
[3] D.S.Passman,TheAlgebraicStructureofGroupRings.JohnWiley&Sons,NY.1977.
[4] D.S.Passman,InfiniteGroupRings,MarcelDekker,NY,1971.
[5] B.Sagan,TheSymmetricGroup.Springer-Verlag.NY,2001.
[6] Bartosz Malman. Zero-Divisors and Idempotents in Group Rings. Master’s Thesis. Lund University,
2014.
[7] KenBrown.HopfAlgebras.http://www.maths.gla.ac.uk/~kab/Hopf%20lects%201-8.pdf
9