SEPARABILITY AND HOPF ALGEBRASLARS KADISON AND A.A. STOLIN1. IntroductionOver a �eld of characteristic zero, the separable algebras and the strongly separa-ble algebras coincide with one another and the class of �nite dimensional semisimplealgebras. In this case, separability is the cohomological point of view on semisim-plicity [16]. Strong separability in this case is an additional constraint of symmetryon the separability idempotent, also an interesting point of view [7]. However, itis over a non-perfect �eld F of characteristic p that the three classes of algebrasform a proper chain of inclusions. For example, there is a textbook example of a�nite �eld extension which is not separable but is of course a semisimple algebra[35]. Moreover, the matrix algebraMn(F ) where p divides n (and F need only havecharacteristic p), is separable but not strongly separable.In this paper we survey and study separability and strong separability in itsrelations to Frobenius and Hopf algebras over a commutative ring k. We study theproblem of when Frobenius and Hopf algebras are separable or strongly separable.It turns out that the dual bases tensor, which appears in the study of Frobenius andHopf algebras (cf. [3, 4, 13, 19, 20]), can be used in the construction of a symmetricseparability idempotent (Theorem 4.1). As an application of this result we provethat an involutive, separable Hopf k-algebra is strongly separable (Theorem 5.5).This is generalization of the following well-known result of Kreimer and Larson [22]:if a Hopf algebraH over an algebraically closed �eld F is involutive and semisimple,then the dimensions of simple H-modules are coprime to the characteristic of the�eld. Then we generalize a recent result of Etingof and Gelaki [12], which statesthat if Hopf F -algebra H is semisimple and cosemisimple, then H is involutive. Weshow that with a small condition on 2 2 k, if a Hopf k-algebra is separable andcoseparable, then H is involutive (Theorem 6.1).Our paper is organized as follows. In Section 2 we review some of the basics ofseparable k-algebras and Frobenius k-algebras which are needed. In Section 3, webring together nine conditions for a strongly separable k-algebras in [21, 14, 8, 7, 1]into one theorem (Theorem 3.4). In Section 4 we apply this theorem to Frobeniusk-algebras (Theorem 4.1) and study augmented Frobenius algebras. We prove inTheorem 4.1 that a Frobenius algebra A is strongly separable if and only if thetranspose of its dual bases tensor maps under the multiplication mapping to aninvertible element in A. In Section 5 we study separability of Hopf k-algebrasand apply Theorem 3.4 to involutive separable Hopf k-algebras (Theorem 5.5).In Section 6 we generalize the recent result of Etingof and Gelacki, and obtaininvolutive results for strongly separable Hopf F -algebras under various constraintson the order of the antipode and the size of p.1991 Mathematics Subject Classi�cation. 16W30, 16H05, 16L60 .1

2 LARS KADISON AND A.A. STOLIN2. Preliminaries on Separable and Frobenius AlgebrasLet k denote a commutative ground ring. We refer to k-algebras that are �nitelygenerated and projective over k as being �nite projective algebras.In this section, we set up some notation and recall useful facts for separablealgebras (cf. [9]) and for Frobenius algebras [10, 31, 3, 19]. We also de�ne the traceof an endomorphism of a �nite projective k-module V , its Hattori-Stallings rank(cf. [5]), as well as the standard trace of a �nite projective algebra A.Given an algebra A over k, we make use of several constructions. First, the dualof A is A� := Homk(A; k) and is also �nite projective over k. A� has an A-bimodulestructure given by (afb)(c) := f(bca)(1)for every a; b; c 2 A and f 2 A�. An element f 2 A� is called a trace if af = fafor every a 2 A, and a normalized trace if moreover f(1A) = 1k. A nonzero f 2 A�is said to be nondegenerate if all a 2 A for which fa = 0 are zero, or all a 2 A forwhich af = 0 are zero.Second, the tensor-squareAkA of A has the A-bimodule structure given simplyby a(b c)d = ab cd for every a; b; c; d 2 A. An element Pi xi yi 2 A Ais called symmetric if Pi xi yi = Pi yi xi. An element e 2 A A is calleda Casimir element if ae = ea. The A-A-bimodule epimorphism � : A A ! Agiven on simple tensors by a b 7! ab for every a; b 2 A is referred to as themultiplication mapping. Equivalently, the tensor-square is canonically identi�ed asleft Ae-modules by a b 7! a b with the algebra Ae := AAop, where Aop is theopposite algebra of A and multiplication is given by (a b)(c d) = ac db. Inthis way, � may be viewed as a left Ae-module morphism.A is a separable k-algebra if � is a split A-A-epimorphism. A is a central separablealgebra, or Azumaya algebra, if it is a separable C-algebra where C is its center. IfA is k-separable and K is any intermediate ground ring for A (i.e., there are ringarrows k ! K ! C forming a commuting triangle with the unit map k ! C) thenA is separable K-algebra, since the natural mapping Ak A! AK A pulls backthe splitting map for �.A separability element e 2 AA is the image of 1A under any splitting mapping of�; equivalently, e is a Casimir element such that �(e) = 1. The Casimir condition one 2 Ae is given by ze = �(z)e for all z 2 Ae. If e is symmetric, then ez = e(�0(z)1)where �0(a b) = ba. As a consequence, should a symmetric separability elementexist, it is unique, since given two of these, e and f , we havee = e(�0(f) 1) = ef = �(e)f = f(2)We will see in Section 3 that having a symmetric separability element is equivalentto A being strongly separable, a notion of Kanzaki from 1964 [21].Separability is a transitive notion, in that if A is a separable k-algebra and k is aseparable K-algebra, then A is seen to be a separable K-algebra. For if Pi xi yiis a separability element for k ! A and Pj zj wj is a separability element forK ! k, then it is easily veri�ed that Pi;j xizj K wjyi is a separability elementfor the composite arrow K ! A.If V is a �nite projective k-module, there is a notion of trace of an endomorphismof V , f 2 Endk(V). Let fxig; fgig be a �nite projective base for V . The trace of f

SEPARABILITY AND HOPF ALGEBRAS 3is de�ned to be Tr (f) := nXi=1 gi(f(xi));(3)This de�nition does not depend on the choice of projective base and that Tr (f�g) =Tr (g � f), for under the canonical isomorphismV k V � �=�! Endk(V)given by the mapping v f 7! (w 7! vf(w)) (v; w 2 V; f 2 V �), the trace formsa commutative triangle with the multiplication mapping V V � ! k given byv f 7! f(v). The Hattori-Stallings (HS) rank of V is the trace of the identity:rk(V ) := Tr (IdV):(4)Now any algebra A over k that is �nite projective has a standard trace t : A! kde�ned by t(a) := Tr (�a);(5)where �a(x) := ax is in Endk(A). Of course, t(ab) = t(ba) since Tr is itselfa trace. Note that the standard trace is the so-called \trace of the left regularrepresentation." Note that t(1) = rk(A) and is not necessarily normalizable. If Ais a central separable algebra over k, we will see in Section 3 that t(1) is invertiblein k i� A is strongly separable [14, 8].An algebra A over k is a Frobenius algebra if there exists a k-linear mapping � :A! k, called a Frobenius homomorphism and elements x1; : : : ; xn, y1; : : : ; yn 2 A,called dual bases for A, such that for every a 2 A,Xi �(axi)yi = a(6)or Xi xi�(yia) = a:(7)Either dual bases equation (as we will refer to them) implies the other.It follows from an application of both dual bases equations that Pi xi yi isa Casimir element in A A. As a consequence, it is easy to see that a Frobeniusalgebra A is separable i� there is d 2 A such thatXi xidyi = 1:(8)Since fxig, f�yig is a projective base for the underlying k-module of a Frobeniusalgebra A, it follows that the trace of a k-endomorphism f 2 Endk(A) isTr (f) =Xi �(yif(xi))(9)and the HS rank of Ak isPi �(yixi). For example, let A =Mn(k), � the trace of amatrix, with dual bases given by the matrix units eij ; eji. Then the standard traceTr = n� and the HS rank of A over k is n21k.Another characterization of a Frobenius algebra A is that A is �nite projective,and either AA �= A�A or AA �= AA�. The free generator of A� as an A-module ineither case is a Frobenius homomorphism, and a �nite projective base translatesvia the isomorphism to dual bases.

4 LARS KADISON AND A.A. STOLINIf k is itself a Frobenius algebra over another commutative ring K, then A is aFrobenius algebra also over K. For suppose : k ! K is a Frobenius homomor-phism with dual bases fzjg; fwjg in k. Then it is easy to verify the equations 6and 7 for � � : A ! K and fxizjg; fwjyig. We say then that being Frobenius isa transitive notion. (Transitivity for separability and Frobenius is best formulatedfor noncommutative ring extensions [31, 15, 19]. )The Nakayama automorphism � : A! A may be de�ned by either�(x) =Xi �(xix)yi;(10)for every x 2 A or the equation in A� given byx� = ��(x)(11)for every x 2 A [10, 19]. It follows from either of these equations and the dual basesequations that for every a 2 A,Xi xia yi =Xi xi �(a)yi:(12)If is another Frobenius homomorphism for A, then by a theorem we call thecomparison theorem there is an invertible d 2 A such that = �d in A�. Iffzjg; fwjg are dual bases for , it follows from Equation 7 thatXj zj wj =Xi xi d�1yi:(13)The next proposition shows that a �nite projective, separable extension A=S ofcommutative rings is a special type of symmetric S-algebra ([2, Prop. A.4] with thedi�erent proof sketched in [8]). A symmetric algebra is a Frobenius algebra A suchthat AAA �= AA�A;(14)equivalently, one of the Frobenius homomorphisms is a trace (and the Nakayamaautomorphism is any case an inner automorphism). This proposition will be astepping-stone to proving a similar result for noncommutative separable algebrasin the next section.Proposition 2.1. Suppose A is a commutative ring with subring S such that ASis a �nite projective module. Then A is separable algebra over S if and only ifHomS(A; S) is freely generated as a right A-module by the standard trace t.Proof. Let fyigni=1 and ffigni=1 be a projective basis for AS , and note that t(a) =Pni=1 fi(ayi).(() By hypothesis then there are elements xi 2 A such that xit = fi in A�. ThenPi t(axi)yi = Pi fi(a)yi = a for all a 2 A and t is a Frobenius homomorphismwith dual bases fxig, fyig. It follows that the dual bases tensor e :=Pi xi yi isa Casimir element.We claim that e is moreover a separability element since for all a 2 A we havet(a(1 � Pi xiyi)) = Pi fi(ayi) � Pi fi(yia) = 0. Then by the free generatorassumption on t, Pi xiyi = 1.

SEPARABILITY AND HOPF ALGEBRAS 5()) We may assume that a separability element has the special formPni=1 xiSyi as each x 2 A satis�es x =Pi fi(x)yi. Then, for every a 2 A,Xj t(axj)yj = Xj (Xi fi(axjyi))yj= Xj Xi fi(axj)yiyj= Xj axjyj = a(15)Therefore, ta = 0 implies a =Pi t(xia)yi = 0. In addition,f(x) =Xj t(xxj)f(yj) = t(xXj xjf(yj))for every f 2 A� and x 2 A. Hence, t is free generator of A�.3. Strong separabilityIn this section we bring together what is known about strongly separable al-gebras over a commutative ground ring k [21, 14, 8, 29, 7, 1]. Among the ninecharacterizations we shall consider, strongly separable algebras are characterizedby Demeyer and Hattori as the projective separable algebras with Hattori-Stallingsrank equal to an invertible element in its center [8, 14]. The theory of stronglyseparable algebras introduced by Kanzaki and developed by Demeyer and Hattorishows rather handily that a strongly separable algebra is a symmetric algebra [14].This development led to the question if all projective separable algebras are sym-metric algebras, which was settled in the a�rmative by Endo and Watanabe usingan extended notion of reduced trace for central separable algebras [11].We will need several facts about central separable algebras summarized by theProposition below. The proofs may be found in [2, 9, 30, 19].Proposition 3.1. Suppose A is a central separable algebra with center C. Then:1. For every A-bimodule M , we have an A-bimodule isomorphism� : MA A �=�!M(16) given by �(m a) = ma, where MA := fm 2M jam = mag.2. A is a progenerator Ae-module.3. AC is �nite projective, and Ae is ring isomorphic to EndCA via the mapping� given by (8 a; b; x 2 A) �(a b)(x) = axb:(17)4. There is a C-linear projection � : A! C.Lemma 3.2. Suppose A is a k-algebra with center C. Then A is k-separable ifand only if A is separable over C and C is a separable over k.Proof. (() This follows from transitivity of separability.()) A is C-separable since it is separable over any intermediate ring betweenk and C. In order to show that C is k-separable, it su�ces to show that C is aprojective Ce = CkC-module. By Proposition 3.1, A is projective as a C-module,therefore Ae = Ak Aop is projective as a C k C-module by an easy exercise. ByProposition 3.1, C is a direct summand of A with C-linear projection � : A ! C.

6 LARS KADISON AND A.A. STOLINThen � C � : Ae ! C is also a C-linear projection; whence C is a Ce-directsummand of the projective Ce-module Ae. Hence, C is Ce-projective.Since a central separable C-algebra A is �nite projective over its center C, itsstandard trace t : A ! C is de�ned and t(1) is the Hattori-Stallings rank of theprojective C-module A. The proof of the next proposition will set the pattern formuch of the proof of Theorem 3.4.Proposition 3.3. Suppose A is a separable algebra over its center C. Moreover,suppose its Hattori-Stallings rank c = t(1) is invertible in C. Then there existelements x1; : : : ; xn; y1; : : : ; yn 2 A such that1. Pi xi C yi is a symmetric separability element,2. Pi t(xxi)yi = x,3. Pi xit(yix) = x (8 x 2 A).Proof. Put t1 = c�1t, a C-linear projection and normalized trace from A onto C.By viewing t1 2 HomC(A;A) and surjectivity of the mapping � in Proposition 3.1,we �nd elements x1; : : : ; xn; y1; : : : ; yn 2 A such that t1(a) = Pi xiayi. ThenPi xiyi = t1(1) = 1. Since Im t1 = C, we have a 7! Pi xxiayi = Pi xiayixas mappings of A ! A. Then injectivity of � in Proposition 3.1 implies thatPi xi C yi is a Casimir element, whence a separability element.By the trace property of t1 and again by Proposition 3.1, it follows that 8 a 2 A,Xi xiaC yi =Xi xi C ayi:(18)By applying the transposition C-automorphism on AC A, we get (8 a 2 A)Xi ayi C xi =Xi yi C xia(19)Whence Pi yixxi 2 C for all x 2 A. Recalling that [A;A] denotes the C-linearspace generated by the set f[x; y] := xy � yxj x; y 2 Ag, we note thatA = C � [A;A]as C-modules, since for every x 2 A, x = Pi xiyix = Pi yixxi +Pi[xi; yix] 2C + [A;A]; moreover, C \ [A;A] = f0g since t1jC = IdC and t1([x; y]) = 0 by thetrace property.It follows that 1 =Xi xiyi =Xi yixi +Xi [xi; yi] = 1 + 0;which implies that Pi yixi = 1. Hence, Pi yi C xi is also a separability element,which additionally satis�es Equation 18.

SEPARABILITY AND HOPF ALGEBRAS 7We now complete the proof that e := Pi xi yi is a symmetric separabilityelement. Xi xi �yi = Xi xi(Xj xjyj) yi= Xi;j xiyj xjyi= Xj (Xi xiyi)yj xj= Xj yj xj(20)Now A�AC A �= A� via the multiplication map � in Proposition 3.1. But A�Ais the C-space of (non-normalized) traces. Since A = C � [A;A], A�A is free ofrank 1, being just the scalar multiples of t1 or t. We conclude that A �= A� asA-bimodules, so A is a symmetric algebra, with Frobenius homomorphisms t1 or t(since they di�er by an invertible element).Let fuig, fvig be dual bases for t1, so thatx =Xi t1(xui)vi(21)for every x 2 A, and Pi uivi 2 C. Then ft1uig and fvig is a projective basis forAC . It follows that the HS rank of A as a C-modulec =Xi (t1ui)(vi) = t1(Xi uivi) =Xi uivi:(22)We note moreover that the dual bases tensor of a trace Frobenius homomorphismis a symmetric element in AA, since for every a 2 AXj auj vj = Xi;j vi t1(uiauj)vj= Xi vi uia:(23)Hence e := c�1Pi ui vi is a symmetric separability element. By uniqueness ofsuch an element we have e = e. Since t = ct1, Condition 2 follows from Equation 21by replacing e with e. Condition 3 follows from e being symmetric and t a trace.The proposition is partially summarized by saying that a central separable al-gebra A has a C-linear projection onto its center C given by a 7! Pi xiayi, is asymmetric C-algebra with a dual bases tensor and symmetric separability elementPi xi yi. The data (t; xi; yi) satisfying Conditions 1 to 3 in Proposition 3.3 wetemporarily call a strongly separable base.We de�ne a separable base for a k-algebra A, �nite projective over k, as a k-linear trace f : A ! k together with elements x1; : : : ; xn; y1; : : : ; yn 2 A such thatPni=1 f(axi)yi = a for all a 2 A, and Pni=1 xiyi = 1. In fact, f is necessarily thetrace map t introduced earlier by an easy computation. For example, a commutativeseparable algebra A over k has a separable base by the proof of Proposition 2.1.If (t; xi; yi) is a separability basis for A, then e := Pi xi k yi is a symmetricseparability element by noting the following. First, a computation like Equation 23

8 LARS KADISON AND A.A. STOLINshows that it is symmetric by letting a = 1. It is a Casimir element sinceaXi xi yi =Xj yj xja =Xi xi yiafor every a 2 A. Since Pi xiyi = 1 by de�nition, e is a symmetric separabilityelement. Whence a separable base is in fact a strongly separable base.Theorem 3.4. The nine conditions below on a �nite projective k-algebra A withcenter C are equivalent:1. A is k-separable and the Hattori-Stallings rank of AC is an invertible elementin C;2. The C-algebra A has a separable base, and C is k-separable;3. The standard trace t generates Homk(A; k) as a right A-module;4. The k-algebra A has a separable base;5. There exists an element e =Pi xikyi such thatPi xixkyi =Pi xikxyi,8 x 2 A, and Pi xiyi = 1;6. A is k-separable and there is a C-linear normalized trace map � : A! C;7. A is k-separable and A = C � [A;A] as C-modules;8. A has a symmetric k-separability element.9. For every A-bimodule M , there is a natural k-module isomorphism MA !M=[A;M ], given by m 7! m + [A;M ], where [A;M ] is the k-span of fam �majm 2M;a 2 Ag.A is said to be strongly separable over k if it satis�es any of the nine conditionsabove.Proof. We prove that Conditions 1) 2) 3) 4) 5) 6) 7) 8) 1, and Condition8 ) 9 ) 7.(Condition 1 ) 2.) >From Lemma 3.2, the k-separability of A implies that Ais C-separable and C is k-separable. From Proposition 3.3, the C-algebra A has astrongly separable base, which is a special case of a separable base.(Condition 2 ) 3.) Suppose (t2 : A ! C; xi; yi) is a separable base for theC-algebra A. Then we have seen thatPi xi yi is symmetric separability elementfor A. In particular, A is a central separable algebra, and �nite projective overC by Proposition 3.1. Since C is a C-direct summand in A by Proposition 3.1,it is a k-direct summand, so C is projective over k. Denote its standard trace byt1 : C ! k. >From Proposition 2.1 and the k-separability of C, it follows that C isa symmetric k-algebra with Frobenius homomorphism t1.SincePi xiC yi is a symmetric separability element, we argue as in the proof ofProposition 3.3 to show that A is a symmetric algebra over C: namely, we deducethat A = C � [A;A], from which it follows that A�A �= C and by Proposition 3.1with M := A� that A� �= A as A-bimodules.Since A is a symmetric C-algebra and C is a symmetric k-algebra, it follows thatA is a symmetric k-algebra with trace Frobenius homomorphism t0 = t1 � t2. Thent0 freely generates Homk(A; k) as a right A-module. But it is readily computed (bychoosing projective bases for A and C) that t0 is the standard trace t : A! k.(Condition 3 ) 4.) Let ffig, fyig be a �nite projective basis for A over k. Letxi be elements of A such that fi(x) = t(xxi) for every x 2 A: thenPi t(xxi)yi = x.It follows that t is a nondegenerate trace, and, from Equation 23, with a = 1, thatPi xiyi =Pi yixi.

SEPARABILITY AND HOPF ALGEBRAS 9We need to show that Pi xiyi = 1: this will �nish a proof that (t; xi; yi) is aseparable base for A over k. For every a 2 A,t(a(1�Xi xiyi)) = t(a)� t(aXi xiyi)= Xi fi(ayi)� t(aXi xiyi)= t(a(Xi yixi �Xi xiyi)) = 0:(24)It follows from nondegeneracy of t that 1�Pi xiyi = 0.(Condition 4 ) 5.) Let (t; xi; yi) be a separable base. Then Pi xi k yi is asymmetric separability element for A. It follows that Pi xix k yi =Pj xj xyjfor every x 2 A.(Condition 5 ) 6.) Let �0(a) =Pi xiayi, 8 a 2 A. Then �0 : A! A is C-linear,satis�es �0(xy) = �0(yx) and �0jC = IdC. It follows that C\[A;A] = 0. At the sametime,Pj yjaxj 2 C for every a 2 A: then a =Pi xiyia =Pi yiaxi+Pi[xi; yia] 2C+[A;A], so A = C�[A;A] as C-modules. It follows that the projection � : A! Cfor this decomposition, de�ned by �(a) :=Pj yjaxj , is a C-linear normalized trace.Also, 1 =Pi xiyi =Pi yixi+Pi[xi; yi], where 1;Pi yixi 2 C, whencePi yixi =1. It follows that Pi yi k xi is a separability element for A.(Condition 6 ) 7.) Trivial.(Condition 7) 8.) Since A is k-separable, A is C-separable and C is k-separableby Lemma 3.2. Also by hypothesis, we see that there is a C-linear projection andnormalized trace � of A onto C with kernel [A;A]. By Proposition 3.1, there ise =Pi xi C yi 2 Ae such that �(e) = �.We now argue just like in the �rst stage of the proof of Proposition 3.3 that e isa symmetric separability element.As in the proof of ( 2 ) 3) we see that C is �nite projective over k, so there isa trace t0 : C ! k forming part of a separable basis (t0; uj ; vj) by Proposition 2.1,equation 15. ThenPj ujk vj is a symmetric separability element for C. It followsthat Pi;j xiuj k vjyi is a symmetric separability element for the k-algebra A.(Condition 8 ) 1) First of all, note that A is C-separable, C-�nite projective,and has a symmetric separability element e = Pi xi C yi, since this comes fromthe hypothesis via the canonical epi A k A ! A C A. Let t : A ! C be thestandard trace, and denote the C-rank of A, t(1) = c.Now argue with the element e as in the proof of Proposition 3.3 that we have a C-linear projection and normalized trace �(e) := t1 : A! C, that A = C�[A;A], andA �=! A� as A-bimodules, given by a 7! t1a. ThenPi t1(xi)yic = 1 by Equations 21and 22.(Condition 8 ) 9.) De�ne an inverse to the natural k-linear mapping �M :MA ! M=[A;M ] by M : m+ [A;M ] 7! em, where e is a symmetric separabilityelement and we view M as a left Ae-module. Then M is well-de�ned since e andits twist are Casimir elements, and is an inverse to �M since �(e) = 1. It followsthat MA and M=[A;M ] are naturally isomorphic.(Condition 9 ) 7.) We �rst let M = A. Then the natural mapping �A : AA =C �=! A=[A;A] given by x 7! x+[A;A] has inverse A : A=[A;A]! C. If � : C ! Aand �0 : [A;A] ! A denote the inclusion maps, then � � A is a splitting for the

10 LARS KADISON AND A.A. STOLINcokernel exact sequence of �0,0! [A;A]! A! A=[A;A]! 0:This proves that A = C � [A;A].Now let M = A k A. Set e := M (1 1 + [A;M ]) 2 (A A)A. Since � is anA-bimodule homomorphismM ! A sending 11 into 1, it follows from naturalityof that �(e) = 1.Let us observe a number of corollaries of this theorem. First, it is clear thata separable commutative algebra A which is projective (and automatically �nitelygenerated by a theorem of Villamayor) over k, satis�es any one of several of thenine conditions, and therefore is strongly separable.Corollary 3.5. A separable, projective commutative algebra is strongly separable.Secondly, it follows from Condition 3 that a strongly separable algebra A is asymmetric algebra, since t is necessarily nondegenerate, so A� �= A as A-bimodules.Consequently, the symmetric separability element e = Pi xi C yi belonging toA is an invertible solution of the Yang-Baxter Equation (YBE), when viewed inEndk(A A), by a theorem in Beidar-Fong-Stolin [3]. Now applying Condition 2,Proposition 3.3 and its proof once in the last line below, we have for every a; b 2 Ae(a b)e = Xi;j xiaxj yibyj= Xj Xi bxixjyi yja= bXj t1(xj)yja= c�1b a(25)where c is the HS rank of A over C and t1 = �(e) = c�1t. Thus e�1 = ce and theinner automorphism by e in Autk AA is the permutation solution of the YBE.It follows from Condition 1 that strong separability and separability are indis-tinguishable if k is a characteristic zero �eld.Corollary 3.6. If A is a separable algebra over a �eld k of characteristic zero,then A is strongly separable.Proof. By Wedderburn's theory, we have a ring isomorphism,A �=Mn1(D1)� � � � �Mnt(Dt);(26)where the center of each division ring Di is a �eld Fi �nite separable over k andthe center of A, C �= F1 � � � � � Ft. It follows easily from Equation 4 that the HSrank of A over C is rC(A) = (n21[D1 : F1]; : : : ; n2t [Dt : Ft]) 2 C:(27)But each entry is nonzero, so A is strongly separable.Corollary 3.7. If A is a separable algebra over a �eld k of characteristic p, thenA is strongly separable if and only if each simple A-module M has dimension overZ(EndA(M)) not divisible by p.

SEPARABILITY AND HOPF ALGEBRAS 11Proof. Wedderburn theory gives us the decomposition 26 and consequently the C-rank 27 where C �= F1�� � ��Ft. The hypothesis on simple A-modules is equivalentto each entry in 27 being nonzero in the �eld Fi of characterisic p, since a simpleA-module M is of the form Dnii and Z(EndA(M)) = Fi by Schur theory. It followsthat the HS rank of A is invertible and A is strongly k-separable.By Proposition 3.3, a central separable k-algebra with Hattori-Stallings rankinvertible in k is a strongly separable k-algebra. The reader might enjoy showinghow the following obviously separable algebra A with zero HS rank fails each ofthe nine conditions in the theorem: Let p be a prime, F = Zp and A = Mp(Zp).For example, the central separable F -algebra A does not satisfy A = F � [A;A](Condition 7) since 1 2 [A;A]. Although A has several separability elements, it hasno symmetric separability element by the theorem.We believe that Condition 7 for strong separability in Theorem 3.4 is due toKanzaki [21], as well as Condition 5, Conditions 2, 3 and 4 are due to Demeyer [8],Condition 1 to Hattori and Demeyer, Condition 6 to Hattori [14], and Condition 9 toAguiar [1]. Condition 8 for strong separability is stated by Demeyer [8], consideredin [7] for k = C, and proven in [18]. Onodera's Condition [29] is closely related toConditions 1 and 8 via Equations 22 and 23: it states that A is strongly separablei� it has a trace and Frobenius homomorphism t with dual bases fuig, fvig suchthat Pi uivi is invertible.4. Augmented Frobenius AlgebrasLet k be a commutative ring throughout this section. We �rst consider when aFrobenius algebra is strongly separable.Theorem 4.1. Suppose A is a Frobenius algebra with Frobenius homomorphism �and dual basis fxig; fyig. Then u :=Xi yixiis invertible in A if and only if A is strongly separable. Moreover, the Nakayamaautomorphism is given by �(x) = uxu�1(28)for every x 2 A.Proof. ()) Consider the element e :=Pi yixiu�1 in AA. Then e satis�es theKanzaki Condition 5 for strong separability by choice of u and the fact that thedual bases tensor is Casimir.>From Equation 12 we obtain�(a)e =Xi yi xiau�1;for every a 2 A. Applying the multiplication mapping � to both sides of thisequation, we obtain Equation 28.(() By Demeyer's Condition 4 for strong separability, A has a (stongly) sepa-rable base (t; zj ; wj) where t is the standard trace on A fzjg; fwjg form dual bases

12 LARS KADISON AND A.A. STOLINfor t,Pj zjwj is symmetric, andPj zjwj = 1. Then by the comparison theoremthere exists d 2 A� such that t = �d. Then by Equation 13 we haveXi xi yi =Xj zj dwj :Then u :=Xi yixi = dXj wjzj = dand u is invertible.We �nally note from Equations 6 and 10 that for every x 2 A,�(x) =Xi �(xix)yi =Xi tu�1(zjx)uwj = uxu�1:The implication ( for k a �eld of characteristic zero is equivalent to Beidar-Fong-Stolin's [3, Proposition 4.3].A k-algebra A is said to be an augmented algebra if there is an algebra homomor-phism � : A ! k, called an augmentation. An element t 2 A satisfying ta = �(a)t,8 a 2 A, is called a right integral of A. It is clear that the set of right integrals,denoted by R rA, is a two-sided ideal of A, since for each a 2 A, the element at isalso a right integral. Similarly for the space of left integrals, denoted by RA.Now suppose that A is a Frobenius algebra with augmentation �. We claim thata nontrivial right integral exists in A. Since A� �= A as right A-modules, an elementn 2 A exists such that �n = � where � is a Frobenius homomorphism. Call n theright norm in A with respect to �. Given a 2 A, we compute in A�:�na = (�n)a = �a = �(a)� = �n�(a):By nondegeneracy of �, n satis�es na = n�(a) for every a 2 A.Proposition 4.2. If A is an augmented Frobenius algebra, then the set R rA of rightintegrals is a two-sided ideal which is free cyclic k-summand of A generated by aright norm.Proof. The proof is nearly the same as in [32, Theorem 3], which assumes that Ais also a Hopf algebra. The proof depends on establishing the equation, for everyright integral t, t = �(t)n:(29)The right norm in A is unique up to a unit in k. Similarly the space RA of leftintegrals is a rank one free summand in A, generated by any left norm. If R rA = RA,A is said to be unimodular. In general the spaces of right and left integrals do notcoincide, and one de�nes an augmentation on A that measures the deviation fromunimodularity. In the notation of the proposition and its proof, for every a 2 A,the element an is a right integral since the right norm n is. From Equation 29 oneconcludes that an = �(an)n = (n�)(a)n. The function,m := n� : A! k(30)is called the right modular function, which is an augmentation since 8 a; b 2 A wehave (ab)n = m(ab)n = a(bn) = m(a)m(b)n and n is a free generator of R rA.

SEPARABILITY AND HOPF ALGEBRAS 13We moreover have m � � = �:(31)by [20, Prop. 3.2]. As a consequence, if A is an augmented symmetric algebra, thenA is unimodular.Proposition 4.3. Suppose A is a separable augmented Frobenius algebra with normn and augmentation �. Then �(n) is invertible and A is unimodular.Proof. For any augmented Frobenius algebra (A; �; xi; yi; �) we note the useful iden-tity for the right norm n, n =Xi �(nxi)yi =Xi �(xi)yi:(32)If A is moreover separable, there is d 2 A such thatPi xidyi = 1 by Equation 8.Then by Equation 32 Xi �(xi)�(d)�(yi) = �(d)�(n) = 1:It follows that �(n) is a unit in k. But for every x 2 A, a computation like in [25],namely, �(x)�(n)n = nxn = m(x)n2 = m(x)�(n)n;gives �(x)�(n) = m(x)�(n), whence �(x) = m(x). Thus, A is unimodular.5. Hopf AlgebrasPareigis proved in [32] that every �nite projective Hopf algebra H has a bijectiveantipode S (with inverse denoted by S�1). In addition, the dual Hopf algebra H�has a right Hopf module structure over H with right H-comodule H� the dual ofthe natural H�-module H� and right H-module H� the twist by S of the naturalleft H-module H� [32]. The fundamental theorem of Hopf modules then leads tothe isomorphism, ZH� H �= H�as left H-modules. Whence H is a so-called P -Frobenius algebra where P :=(RH�)�, where RH� is the invertible k-module of left integrals in H� [34]. If P �= k,H is an ordinary Frobenius algebra with Frobenius homomorphism a left norm inH� [24, k = pid]. That RH� �= P �= k is guaranteed if k has Picard group zero (e.g.when k is a �eld, semi-local or a polynomial ring).In this section, k will continue to denote a commutative ring and a Hopf algebraH is always �nite projective as a k-module; moreover, we will assume H is a Frobe-nius algebra with a special Frobenius homomorphism f : H ! k. We require of fthat it be a right norm in the dual Hopf algebra H�, which is no loss of generalitysince S is an anti-automorphism and Sf is a left norm in H�. We will refer to suchan H as simply Hopf algebra in this section.1 It has been shown that also H� andthe quantum double D(H) are Hopf algebras in this sense [20], and more detailedproofs may be found in either of [13] or [19].1These Hopf algebras have been called FH-algebras by Pareigis [33].

14 LARS KADISON AND A.A. STOLINProposition 5.1. Let H be a Hopf algebra with Frobenius homomorphism andright norm f and right norm t in H. Then (f; S�1t2; t1) is a Frobenius system forH.Proof. This follows from a short computation like that in [13, Lemma 1.5] usinga ( f = 1Hf(a) for every a 2 H :X(t) S�1(t2)f(t1a) = X(t);(a)S�1(t3)f(t1a1)t2a2= X(a) f(ta1)a2= f(t)a = a;since ft = �, the counit, and so f(t) = 1.This proposition has two corollaries.Corollary 5.2. A Hopf algebra H is separable if and only if �(t) is invertible.Proof. The forward implication follows from Proposition 4.3. The backward im-plication follows from the proposition, since 1�(t)P(t) S�1(t2) t1 is a separabilityelement.Recall that a Hopf subalgebra of H is a subalgebra K such that S(K) = Kand �(K) � K K. The next corollary is generalizes a proposition in [25] forsemisimple Hopf algebras over a �eld.Corollary 5.3. Suppose k is a local ring. If H is a separable Hopf algebra freeover k and K is a k-free Hopf subalgebra, then K is separable as well.Proof. It follows from [20, Lemma 5.2] that H is free as the natural rightK-module.Let n be a right norm for K. By expanding t in basis for H over K, we �nd � 2 Hsuch that t = �n. Then �(t) = �(�)�(n). Since �(t) is invertible and the non-unitsin k form an ideal, �(n) is invertible. Whence K is separable.We use the notation a ( g := P(a) g(a1)a2 and g * a := Pa1g(a2) for thestandard right and left actions of g 2 H� on a 2 H .Proposition 5.4. Given an Hopf algebra H with right norm and Frobenius homo-morphism f 2 H� and right norm t 2 H, the Nakayama automorphism for f andits inverse are given by:�(a) = S2(a ( m�1) = (S2a)(m�1;(33) ��1(a) = S�2(a ( m) = (S�2a)(m:

SEPARABILITY AND HOPF ALGEBRAS 15Proof. We compute (like in [13, Lemma 1.5] but opposite Nakayama automorphism)using the right modular function m given by at = m(a)t for all a 2 H :S2(��1(a)) = S2(XS�1(t2)f(at1))= X f(at1)S(t2)= X f(a1t1)a2t2S(t3)= X f(a1t)a2= Xm(a1)a2The rest of the proof follows from noting that m is a group-like element in H�,so the convolution-inverse m�1 = m � S and m � S2 = m.Recall that a Hopf algebra is involutive if S2 = Id.Theorem 5.5. Suppose H is an involutive, separable Hopf algebra. Then H isstrongly separable.Proof. We continue with the notation in the two propositions above. By Proposi-tion 4.3, H is unimodular and �(t) is invertible in k by its proof. It follows fromProposition 5.1 that the dual bases tensor is PS�1(t2) t1, which since S = S�1satis�es X(t) t1S�1(t2) = �(t)1Ha unit, whence H is strongly separable by Theorem 4.1.As a special case of the last proposition we have similar results over �elds byKreimer, Larson [22], Beidar-Fong-Stolin [4], and Aguiar [1].We compute the trace of the k-linear automorphism S2 : H ! H in the nextproposition.Proposition 5.6. If H is a Hopf algebra with right norms t 2 H and f 2 H� suchthat f(t) = 1, then Tr (S2) = f(1)�(t)(34)Proof. Since f is a Frobenius homomorphism with dual bases fS�1(t2)g, ft1g, itfollows from Equation 9 that the traceTr (S2) =X(t) f(t1S2(S�1(t2)) =X f(t1S(t2)) = �(t)f(1)as claimed.If H is a �nite dimensional Hopf algebra over a �eld, it follows that the trace ofS2 is nonzero i� H is semisimple and cosemisimple, a result of Larson-Radford [23].We then easily obtain the following generalization from Corollary 5.2, also noted in[4]:Proposition 5.7. If H is a Hopf k-algebra, then Tr (S2) is invertible if and onlyif H is separable and coseparable.

16 LARS KADISON AND A.A. STOLINNow let us assume that H is strongly separable over an algebraically closed �eldk. Then it follows from Theorem 4.1 that u =P(t) t1S�1(t2) is invertible. In thiscase H �= ��Mn�(k), where Mn(k) is the algebra of n� n-matrices with k-entries.Then we have the followingProposition 5.8. Let tr� be the matrix trace on Mn�(k). Thenf(x) =X� n�tr�(xu�1).Proof. First we note that f(xy) = f(yu�1xu) by Theorem 4.1. Then f(xyu) =f(yuu�1xu) = f(yxu) and therefore f(xu) = P� T�tr�(x) for some T� 2 k orf(x) =P� T�tr�(xu�1). We must prove that T� = n�.Since H is a Frobenius algebra with Frobenius homomorphism f , we observeT� 6= 0. Let E�ij be the matrix units in Mn�(k). Then fE�ijg; f(1=T�)E�jiug formdual base with respect to f . Let us introduce u� = Pij(1=T�)E�ijE�jiu. Thenclearly u =P� u� and we have:u =X� (u�) =Xij� (1=T�)E�ijE�jiu =X� (1=T�)n�u�which proves the Proposition.Corollary 5.9. Suppose H is an involutive, semisimple Hopf algebra over an al-gebraically closed �eld k. Then Tr(S2) = dimH and if H is cosemisimple thendimH 6= 0 in k.Proof. Theorem 5.5 and its proof imply that H is strongly separable and u =P(t) t1S�1(t2) = �(t)1H ThenTr (S2) = �(t)f(1) = �(t)X� n�tr�(u�1) =X� n�tr�(1Mn� (k)) =X� n2� = dimHIf H is cosemisimple we already know that Tr (S2) 6= 0 and hence dimH 6= 0.6. Separability and Order of the AntipodeIt was proved recently in [12] that if a Hopf algebraH over a �eld k is semisimpleand cosemisimple, then S2 = Id. Equivalently if H is separable and coseparableover a �eld k, then S2 = Id. Using this result we prove the corresponding statementfor Hopf algebras over rings.Theorem 6.1. Let k be a commutative ring in which 2 is not a zero divisor. LetH be separable and coseparable Hopf algebra which is �nite projective over k. ThenS2 = Id.Proof. First we note that H is unimodular and counimodular. Then it followsfrom [4, Corollary 3.9] or [20, Theorem 4.7] that S4 = Id. Localizing with respectto the set T = f2n; n = 0; 1; ::g we may assume that 2 is invertible in k. ThenH = H+�H� where H� = fh 2 H : S2(h) = �hg, respectively. We have to provethat H� = 0. It su�ces to prove that (H�)m = 0 for any maximal ideal m in H .Since Hm=mHm is separable and coseparable over the �eld k=m, we deduce fromthe main theorem in [12] that (H�)m � mHm and therefore (H�)m � m(H�)m.

SEPARABILITY AND HOPF ALGEBRAS 17The required result follows from the Nakayama Lemma because H� is a directsummand in H .Corollary 6.2. If H is separable and coseparable over a commutative ring k, thenH is strongly separable over k.We believe that strong separability of a Hopf algebra H over a �eld k and thefact that gcd (dimH; char(k)) = 1 imply that H is coseparable. This fact is knownin case of �elds of characteristic 0 simply because strong separability is equivalentto separability in this case [23]. For the remainder of this paper we will assumethat k is algebraically closed, H is strongly separable over k, char(k)) > 2 andgcd (dimH; char(k)) = 1. We continue with the notation we used in Section 5.Lemma 6.3. Let u =P(t) t1S�1(t2). Then tr�(u) = tr�(u�) = n��(t).Proof. We already know from the proof of Proposition 5.8 that fE�ijg; f(1=n�)E�jiugform dual bases with respect to f . Therefore we have:X� Xij (1=n�)E�jiuEij =X� n�1� tr�(u)1Mn� (k) =X(t) S�1(t2)t1 = �(t)1HHence tr�(u) = tr�(u�) = n��(t) as required.Now we recall that �(x) = uxu�1 by Theorem 4.1. On the other hand, �2n = IdHby [13] where n = dimH . Therefore we see that u2n� = q�EMn� (k), where q� 6= 0 2 kand EMn� (k) := 1� is the unit matrix in Mn�(k) (in the sequel we will denotexEMn� (k) := x1� 2Mn�(k) by x�).By considering the Jordan normal form of u� and recalling the assumptiongcd (n; char(k)) = 1, we can assume without loss of generality that u� = a�diag(�ks2n),where a� 6= 0 2 k and �2n is a primitive 2n-root of unity in k. Clearly numbersa� are de�ned up to �k2n and tr�(u�) = a�f�(�2n) where f�(z) =P2n�1k=0 fkzk is apolynomial with non-negative integer coe�cients.Let k0 be the prime sub�eld of k. We see that f�(�2n) 2 k0[�2n]. Let us introducea formal complex conjugation in the following way.Let PN � k0[x] be the set of all polynomials of degree < N . Let � : PN ! PNbe a linear map de�ned by �(xk) = xN�k. Then if q(x) 2 PN and q(�N ) = a 2 k,we de�ne �N (a) := (�q)(�N ).It is clear that if char(k) = 0 then this is the usual complex conjugation.Corollary 6.4. We have the following formula for Tr(S2):Tr(S2) =X� n2� � �(t)a� � �2n(�(t)a� ) =X� f�(�2n)(�2nf�)(�2n):Remark 6.5. We see that in the case char(k) = 0, Tr(S2) 6= 0 as it is a sum ofstrictly positive numbers; whence H is coseparable and S2 = IdH. If we managedto prove that the operator of left multiplication by u on H had eigenvalue �(t) oneach H� :=Mn�(k), then it would follow that Tr(S2) = dimH 6= 0 in k. Howeverwe only know that this is true on H�0 :=M1(k) generated by t 2 H becauseut = �(u)t = tX(t) �(t1)�(S�1(t2)) = tX(t) �(t1�(t2)) = �(t)t

18 LARS KADISON AND A.A. STOLINIt is well-known that if S4 = IdH and char(k) > dimH then semisimplicityimplies cosemisimplicity. The following theorem enables us to improve this result.Theorem 6.6. Let S2m = IdH, where gcd(m; char(k)) = 1. Then there exists adiagonal invertible d 2 H such that S2(x) = dxd�1, dm� = �1�, �2m(tr�(d)) =tr�(d) and Tr(S2) =P�(tr�(d))2.Proof. Let xt 2 H be the transpose of x 2 H . Then �(x) = S(xt) is an auto-morphism of H and therefore there exists an invertible A 2 H such that S(x) =A�1xtA. Further S2(x) = A�1Atx(A�1At)�1 = uxu�1 (since for unimodular Hopfalgebras S2 = � by Proposition 5.4) and we deduce that u = dc where d = A�1Atand c is in the center of H . Clearly we have that S2(x) = dxd�1. We recallthat u� = a�diag(�ksm ) and it follows that d� := dEMn� (k) is a diagonal matrix.On the other hand we have: At = dA and we obtain that A = Atd = dAdor d�1� = A�d�A�1� . Taking into account that um� = am� EMn� (k) we concludethat dm� = bEMn� (k) = d�m� = b�1EMn� (k) for some b 2 k. Then b2 = 1 anddm� = �EMn� (k). The latter implies that entries of d are 2m-roots of unity. Then�2m(tr�(d)) is well-de�ned and it follows that�2m(tr�(d)) = tr�(d�1) = tr�(A�d�A�1� ) = tr�(d):Since u� = c�d� for some c� 2 k and tr�(u) = n��(t) (by Lemma 6.3) we have:Tr(S2) = �(t)X� n�tr�(u�1)= �(t)X� c�1� n�tr�(d�1)= �(t)X� c�1� (�(t))�1tr�(u)tr�(d�1)= X� (tr�(d))2which completes the proof.Corollary 6.7. 1. Let H be strongly separable over a �eld k of characteristicp = 8N � 3 > dimH (this is equivalent to that of p2 62 Fp ). If S16 = IdH,then S2 = IdH.2. Let H be strongly separable over a �eld k of characteristic p = 4N+3 > dimH(this is equivalent to that of p�1 62 Fp ). If S8 = IdH, then S2 = IdH.Proof. We prove the �rst statement. It is su�cient to prove that Tr(S2) 6= 0. Letus consider separately two cases, one where d8� = 1� and the other where d8� = �1�.In the �rst case the eigenvalues of d� are �j8 of multiplicities A�j ; j = 0; 1; ::7.Then dimH� = (P7j=0 A�j )2. We note that �48 = �1 and �8� �38 = �8+ ��18 = p2 62Fp .Then the condition tr�(d) = tr�(d�1) implies that (A1�A5+A3�A7)(�8+�38 ) =2(A6 � A2)�28 and consequently �2(A1 � A5 + A3 � A7)2 = �4(A6 � A2)2. Sincep2 62 Fp we deduce that A�2 = A�6 and A�1 +A�3 �A�5 �A�7 = 0. Thentr�(d) = A�0 �A�4 + (A�1 �A�5 )�8 + (A�3 �A�7 )�38 =A�0 �A�4 + (A�1 �A�5 )(�8 � �38 ) = A�0 �A�4 + (A�1 �A�5 )p2

SEPARABILITY AND HOPF ALGEBRAS 19and (tr�(d))2 = (A�0 �A�4 )2 + 2(A�1 �A�5 )2 + 2p2(A�0 �A�4 )(A�1 �A�5 ) =(A�0 �A�4 )2 + (A�1 �A�5 )2 + (A�3 �A�7 )2 +Xp2; X 2 FpIn the second case the eigenvalues of d� are �2j�116 of multiplicities B�j ; j = 1; :::8.Since �816 = �1 and �216 = �8 we can write that tr�(d) = P4s=1(Bs � Bs+4)�2s�116In this case �16(tr�(d)) = tr�(d) implies that (B1 � B5 + B4 � B8)(�16 + �716) =�(B2 � B6 + B3 � B7)(�316 + �516) and therefore R(1 + �38 ) = S(�8 + �28 ), whereR = B1�B5+B4�B8; S = �(B2�B6+B3�B7). It follows that R(�8+��18 ) = R+Sand hence R = S = B�1 �B�5 +B�4 �B�8 = B�2 �B�6 +B�3 �B�7 = 0Therefore we have obtained the following expression for tr�(d):tr�(d) = (B�1 �B�5 )(�16 � �716) + (B�2 �B�6 )(�316 � �516)Then we compute (tr�(d))2:(tr�(d))2 = 2(B�1 �B�5 )2 + 2(B�2 �B�6 )2 + Yp2 =(B�1 �B�5 )2 + (B�2 �B�6 )2 + (B�3 �B�7 )2 + (B�4 �B�8 )2 + Yp2; Y 2 FpLet us assume that Tr(S2) = x + yp2 = 0. This means that x = y = 0 becausep2 62 Fp . In our case we have:x =X� (A�0 �A�4 )2 + (A�1 �A�5 )2 + (A�3 �A�7 )2 +X� (B�1 �B�5 )2 + (B�2 �B�6 )2 + (B�3 �B�7 )2 + (B�4 �B�8 )2Let us consider x as an element of Z. It is clear that x < dimH < p. On the otherhand x > 0 because we have the one-dimensional component H�0 on which thecorresponding A�00 is 1 and the other multiplicities A�i ; B�j are zeroes. It followsthat Tr(S2) 6= 0 in k.Acknowledgements. The authors thank NorFA of Norway and NFR ofSweden, respectively, for their support of this paper.References[1] M. Aguiar, Symmetrically separable algebras and a right adjoint to the tensorcoproduct of comodules, preprint, Northwestern Univ., 1996.[2] M. Auslander and O. Goldman, The Brauer group of a commutative ring,Trans. A.M.S. 97 (1960), 367{409.[3] K. Beidar, Y. Fong, and A. Stolin, On Frobenius algebras and the quantumYang-Baxter equation, Trans. A.M.S. 349 (1997), 3823{3836.[4] K. Beidar, Y. Fong, and A. Stolin, On antipodes and integrals in Hopf algebraover rings and the quantum Yang-Baxter equation, J. Algebra 194 (1997),36{52.[5] K.S. Brown, Cohomology of Groups, Springer Graduate Texts 87, 1982.[6] A. Connes and H. Moscovici, Hopf algebras, cyclic cohomology and the trans-verse index theorem, IHES preprint, May 1998.[7] J. Cuntz and D. Quillen, Operators on noncommutative di�erential forms andcyclic homology, Universit�at Heidelberg preprint, 1992.

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SEPARABILITY AND HOPF ALGEBRAS 21[35] R. Pierce, Associative Algebras, Grad. Text in Math. 88, Springer, 1982.[36] D. Radford, The order of the antipode of a �nite dimensional Hopf algebra is�nite, Amer. J. Math. 98 (1976), 333{355.[37] H.-J. Schneider, Lectures on Hopf Algebras, Preprint Ciudad University, Cor-doba, Argentina, Trabajos de Matematica, serie B, no. 31, 1995.[38] J.R. Silvester, Introduction to Algebraic K-theory, Chapman and Hall, London,1981.Chalmers University of Technology/G�oteborg University, S-412 96 G�oteborg, Swe-denE-mail address: kadison@math.ntnu.no astolin@math.chalmers.se