HOPF DENSE GALOIS EXTENSIONS OVER A RING

HE Ji-wei, HU Hai-gang

(Department of Mathematics, Hangzhou Normal University, Hangzhou 311121, China)

Abstract: Let R be a commutative domain, let H be a Hopf R-algebra which is a finitely generated free R-module, and let A be an R-algebra which is also a H-comodule algebra.We will say that A/AcoH is a Hopf dense Galois extension if the cokernel of the associated canonical map β :A?AcoH A →A?RH is quotient finite.It is a generalization of Hopf dense Galois extension over a field.This paper shows that a weaker version of Auslander theorem holds for Hopf dense Galois extensions over R.It is also proved that if the algebra A is almost commutative such that gr(A) is a domain, and the canonical map β is strict, then a Hopf dense Galois extension A/AcoH will imply that H is dual to a finite dimensional group algebra over an algebraic closed field containing R.

Keywords: Hopf dense Galois extension; localization; quotient category; filtered algebra

1 Introduction

Motivated by the study of noncommutative isolated singularities,the He-Van Oystaeyen-Zhang introduced in[1]the concept of Hopf dense Galois extensions over a field.Hopf dense Galois extensions provide candidates of noncommutative resolutions of quotient isolated singularities.However,it is usually difficult to see when a Hopf action or coaction on an algebra results a Hopf dense Galois extension.When the algebra A under consideration has a big center, namely, A is finitely generated over its center, then the problem becomes relatively easy[2].Indeed,we may use the mod-p method to reduce the problem to algebras over fields with positive characteristic.For example, if A is a universal enveloping algebra of a finite dimensional Lie algebra, or A is a Weyl algebra over a field of characteristic p > 0, then A is finitely generated over its center.One of the essential parts to use the mod-p method is to find orders of Hopf actions.Hence it is necessary to consider the Hopf (co) actions and Hopf dense Galois extensions over a commutative domain.

In this paper, we introduce the concept of Hopf dense Galois extensions over a commutative domain.The theory involves several torsion theories.We show that Hopf dense Galois extensions work well.In particular, we prove that a weaker version of Auslander theorem holds for Hopf dense Galois extensions over a commutative domain (cf.Theorem 3.7).

Hopf dense Galois extensions depend on the Hopf algebra (co) actions on the algebra under consideration.It was shown in [3, 4]that not too many semisimple Hopf algebras act inner faithfully on a graded commutative domain or Weyl algebras.Let R be a commutative domain of characteristic zero and let k be an algebraically closed field containing R as its subring.Assume that H is a Hopf R-algebra which is a finitely generated free R-module.Suppose that H coacts on an almost commutative algebra A and the coaction preserves the filtration.If A is Hopf dense Galois over the invariant subalgebra AcoH, then H??Rk is isomorphic to a group algebra over k (cf.Theorem 4.10).Theorem 4.10 applies to Hopf algebra coactions on Sridharan enveloping algebras which including universal enveloping algebras of finite dimensional Lie algebras and Weyl algebras.In particular, if H is a finite dimensional Hopf algebra over an algebraically closed field of characteristic zero which acts on a Sridharan enveloping algebra Uf(g) such that the action preserves the filtration of Uf(g) and the associated graded algebra of Uf(g) is a Hopf dense Galois extension on its invariant subalgebra, then H is isomorphic to a group algebra (cf.Corollary 5.12).This result partially generalizes [3, Theorem 4.2].

2 Torsion Theories over a Ring

Let R be a commutative domain.Let Q be the quotient field of R.Given a noetherian R-algebra A, the localizing A ?RQ is a Q-algebra.For simplicity, we write AQfor the Q-algebra A ?RQ.Similarly, if M is a right A-module, then MQ:= M ?RQ is a right AQ-module.We will say that M is R-torsion free if for any x ∈ M,r ∈ R, xr = 0 implies r =0.The localizing functor ? ?RQ induces an exact functor (?)Q:Mod A ?→ Mod AQ.We will frequently use the following properties of localizations.

Lemma 2.1(i) Let M and N be R-modules.Then (M ?RN)QMQ?QNQ.

(ii) Let M be a right A-module and N be a left A-module.Then (M ?AN)QMQ ?AQNQ.

Let us recall some settings in [1].For a right AQ-module M, an element x ∈M is called an AQ-torsion element if xAQis a finite dimensional Q-vector space.Let ΓAQ(M)be the subset of M consisting of all the AQ-torsion elements.Then ΓAQ(M) is a right AQ-submodule of M.If M =ΓAQ(M), then M is called an AQ-torsion module.Let Tor AQbe the full subcategory of Mod AQconsisting of AQ-torsion modules.Then Tor AQis a Serre subcategory of Mod AQ.Denote the quotient category

We refer to the book [5]for the properties of the torsion theory and quotient categories.

Consider the composition of exact functors

Let Tor A be the full subcategory of Mod A consisting of right A-modules M such that F(M)=0.We will say that M is torsion if M ∈ Tor A.Let ? :M → MQbe the localizing map.We have the following easy observation.

Lemma 2.2A right A-module M is in Tor A if and only if for each x ∈ M, ?(x)AQis finite dimensional over Q.

For a right A-module M, let ΓA(M) = {x ∈ M|?(x)AQis finite dimensional}. Then ΓA(M) is a torsion submodule of M.

Lemma 2.3With the above notions, we have

(i) ΓA(M) is the largest torsion submodule of M and M/ΓA(M) is torsion free, which is to say, ΓA(M/ΓA(M))=0.

(ii) ΓA(M)Q= ΓAQ(MQ).

ProofStatement (i) is easy to check.We next prove statement (ii).For x ∈M and an nonzero element s ∈ R, we have (x/s)AQ= ?(x)AQ.It follows that (x/s)AQis finite dimensional if and only if ?(x)AQis finite dimensional.Hence ΓA(M)Q= ΓAQ(MQ).

The subcategory Tor A is a Serre subcategory of Mod A.Denote the quotient category

Then we obtain an exact functor (use the same notation)

As usual conventions, for an object M ∈Mod A, the corresponding object in Q Mod A is denoted by M, and the object in Q Mod AQcorresponding to MQis denoted by MQ.

Let M and N be right A-modules.Assume that M is finitely generated.It is well known

We next show that the above isomorphism may be extended to the quotient categories.

Lemma 2.4Let M be a right A-module.Let L be an AQ-submodule of MQsuch that MQ/L is finite dimensional.Then there is an A-submodule K of M such that MQ/LMQ/KQand M/K is R-torsion free.

ProofLet ? : M → MQbe the localizing map, and K = {m ∈ M|?(m) ∈ L}.Then L=KQ.By the construction, we see that M/K is R-torsion free.

Proposition 2.1Let M be a finitely generated right A-module.For every N ∈Mod A, we have

ProofWe have the following computations

where the first limit runs over all the AQ-submodules L of MQsuch that MQ/L is finite dimensional, and the second limit runs over all the A-submodules K such that MQ/KQ(M/K)Qis finite dimensional.Let T = ΓA(N).Then TQ= ΓAQ(NQ)by Lemma 2.3.Hence we have

3 Hopf Dense Galois Extensions

In this section, R is a noetherian commutative domain.Let Q be its quotient field.An R-module M is said to be quotient-finite if MQis finite dimensional.

Suppose that A is a noetherian R-algebra which is projective as an R-module.Let H be a Hopf R-algebra which is a finitely generated free R-module.Assume that H coacts on A so that A is a right H-comodule algebra through the coaction ρ : A → A ?RH.As the usual convention, we denote AcoH={a ∈ A|ρ(a)=a ? 1} the coinvariant subalgebra of A.

We next extend the concept of Hopf dense Galois extension (cf.[1]) to algebras over a ring.Consider the following map

We call A/AcoHis a Hopf dense Galois extension if the cokernel of β is quotient-finite.Note that if β is an epimorphism, then A/AcoHis a classical Hopf Galois extension (cf.[6, 7]).

Applying the localizing functor (?)Qto the algebra A and the Hopf algebra H, we obtain a finite dimensional Hopf algebra HQand a right HQ-comodule algebra AQ.Note that the coaction of HQon AQis the map ρQ:AQ→ AQ?QHQ.

Lemma 3.5With the notions as above, (AcoH)Q(AQ)coHQ.

ProofLet ? : A → AQand φ : H → HQbe the localizing maps.Applying (?)Qto the inclusion map AcoH→A, we obtain that (AcoH)Qis contained in (AQ)coHQ.On the other hand, assume a ∈ A and ρQ(?(a)) = ?(a)?Q1.Since H is a finitely generated free R-module and R is a noetherian commutative domain, we extend the unit 1 of R to an R-basis h0= 1,h1,...,hnof H.Then we may writeThen ρ(?(a)) =QSince H is free, φ(h),...,φ(h)is a Q-basis of H.Comparing with the0nQassumption ρQ(?(a)) = ?(a)?Q1, we obtain ?(a0) = ?(a) and ?(ai) = 0 for i = 1,...,n.Since A is projective as an R-module, it is R-torsion free, hence ? is injective.It follows that a0=a and ai=0 for i=1,...,n.Hence ρ(a)=a ?R1.

Proposition 3.2Let A be an R-algebra which is projective as an R-module,and let H be an R-Hopf algebra which is R-free.Assume A is a right H-comodule algebra.Then A/AcoHis Hopf dense Galois if and only if AQ/(AQ)coHQis Hopf dense Galois.

ProofApplying (?)Qto the map β :A ?AcoHA → A ?RH, we obtain

by Lemma 3.5.Then the condition that the cokernel of βQis finite dimensional implies both A/AcoHand AQ/(AQ)coHQare Hopf dense Galois.

Since the torsion functor ΓAis left exact,it has right derived functors.Let RiΓA(i ≥ 0)denotes the i-th right derived functor of ΓA.Similarly, we have RiΓAQ.

Lemma 3.6If RiΓA(M)=0 for all i ≤ k, then RiΓAQ(MQ)=0 for all i ≤ k.

ProofLet 0 → M → I0→ I1→ ··· → Ik→ ··· be an injective resolution of M.Since the localizing functor (?)Qpreserves injective modules, it follows that 0 → MQ→is an injective resolution of MQ.Let I?be the complex 0 →I0→ ··· →Ik→ ···.By Lemma 2.3, RiΓAQ(MQ) =Hi(ΓA(I?))Q=RiΓA(M)Q.So RiΓAQ(MQ)=0 for all i ≤ k in case RiΓA(M)=0 for i ≤ k.

Since by assumption H is a finitely generated R-algebra, then similar to equation (2.2),we have an isomorphism of Hopf algebras HomR(H,R) ?RQ ～=HomQ(H ?RQ,Q).Thus we can write

where H?=HomR(H,R) is the dual Hopf algebra of H.

An important feature of Hopf dense Galois extensions over a field is the truth of Auslander theorem (cf.[1, Theorem 3.10]).Note that Theorem 3.10 of [1]is still true if the characteristic is positive.Next result shows that a weaker version of Auslander theorem holds for Hopf dense Galois extensions over a commutative domain.

Theorem 3.7Let A and H be the algebras as in the beginning of this section.Assume further that HQis cosemisimple.If A/AcoHis a Hopf dense Galois extension, and RiΓA(A)=0 for i ≤ 2, then the natural map

is injective.Moreover, for each f ∈ EndAcoH(A), there exist

ProofBy Proposition 3.2, AQ/(AQ)coHQis a Hopf dense Galois extension over the field Q.By Lemma 3.6, RiΓAQ(AQ)=0 for i ≤ 2.Then [1, Theorem 3.10]insures that the natural map

is an isomorphism.By Lemma 3.5,we have(AQ)coHQ=(AcoH)Q.It follows End(AQ)coHQ(AQ)=EndAcoH(A)Q.Moreover,sinceit follows that ξ =ψQ:(A#H?)Q?→EndAcoH(A)Q.We next show that ψ is a monomorphism.Let K =ker ψ and M =coker ψ.Then KQ=ker ξ =0 and MQ=coker ξ =0.Since A is projective over R and H is R-free,it follows that A#H?is projective over R.Since R is a domain, A#H?is R-torsion free.Hence K is R-torsion free, which implies K = 0.Therefore ψ is injective.Moreover, since MQ= 0, it follows that for each f ∈ EndAcoH(A), there is an element 0r ∈ R such that rf lies in the image of ψ, that isHencefor all b ∈ A.

4 Hopf Dense Galois Extensions of Almost Commutative Algebras

In this section,R is a noetherian commutative domain of characteristic zero,and k is an algebraic closed field containing R as a subring. H is an R-Hopf algebra which is a finitely generated free R-module.The filtration of a filtered R-algebra A is an ascending filtration

We call an R-algebra A is almost commutative if A is a filtered R-algebra and the associated graded algebra gr(A)is a graded commutative algebra.Similar to equation(3.1),we will write

for simplicity.

Lemma 4.8Let B =B0⊕ B1⊕ ··· be a graded R-algebra which is a commutative domain and is projective over R.Let ρ : B → B ?RH be a right H-coaction on B which preserves the gradings.If B/BcoHis a Hopf dense Galois extension, thenis isomorphic to a group algebra.

ProofWe write Bk= B ?Rk and Hk= H ?Rk.Applying ? ?Rk to the right coaction ρ : B → B ?RH, we obtain a coaction ρk: Bk→ Bk?kHk.Consider the canonical map βk: Bk?BcoHkBk?→ Bk?kHk.Since B is commutative, βkis indeed an algebra homomorphism, where we view Bk?kHkas the algebra by the usual multiplication of tensor products of algebras.The same proof of Proposition 3.2 shows that Bk/(Bk)coHkis a Hopf dense Galois extension, then the cokernel of βkis finite dimensional over k.Then there is an integer n ≥ 0 such that (⊕i≥nBi)k?kHk? im βk.Since by assumption B is commutative, which implies Bk?BcoHkBkis commutative, and thus im βkis commutative.For g,h ∈ H, taking nonzero elements a,b ∈ (⊕i≥nBi)k, then (a ?kg),(b ?kh) ∈ im βk,which impies (a ?kg)(b ?kh) = (b ?kh)(a ?kg).Then ab ?kgh = ba ?khg = ab ?khg.Since B is a domain and B is projective over R, ab0.Hence we have gh = hg, that is,Hkis commutative.Since H is finitely generated as an R-module, Hkis finite dimensional.Therefore the dual Hopf algebrais cocommutative.Since k is algebraic closed with characteristic zero,is isomorphic to a group algebra.

Let B be a filtered R-algebra.Let M be a filtered right B-module and let N be a left filtered B-module.The tensor product M ?BN has an induced filtration defined by Fn(M ?BN) to be the abelian subgroup of M ?BN generated by elements x ? y for all x ∈ FiM and y ∈ FjN such that i+j = n.There is a graded epimorphism (cf.[8, §6,Chapter I])

where x ∈ FiM Fi?1M, y ∈ FjN Fj?1N andare corresponding elements in the associated graded modules,similarlyis the corresponding element in the graded abelian group associated to M ?BN.

Suppose that there is a right H-coaction ρ :B → B ?RH which preserves the filtration,where the filtration of B ?RH is induced by the filtration of B.Then the induced map

is a right H-coaction on the associated graded algebra gr(B).Then the filtration of B induces a filtration on BcoH.Associated to this filtration,there is a graded algebra gr(BcoH).Then gr(BcoH) is a graded subalgebra of (gr(B))coH.In general, gr(BcoH) is not equal to(gr(B))coH.

Let X and Y be filtered R-modules.An R-module homomorphism f :X →Y is called a strict filtered map [8]if f preserves the filtration and FnY ∩im f =f(FnX) for all n.

Lemma 4.9Keep the notations as above.If B/BcoHis a Hopf dense Galois extension and the canonical map β : B ?BcoHB → B ?RH is strict, then gr(B)/(gr(B))coHis a Hopf dense Galois extension.

ProofLet gr(ρ):gr(B)→ gr(B)?RH be the induced H-coaction on gr(B).Let

be the canonical map associated to gr(ρ).Denote K = coker β.Then K has a natural filtration inherits from B ?RH.Since β is strict, by [8, Theorem 4.2.4, Chapter I], we have an exact sequence

By equation (4.1), the map ?B,B: gr(B)?gr(BcoH)gr(B) ?→ gr(B ?BcoHB) is an epimorphism.Note that gr(B ?RH)=gr(B)?RH, we have the following commutative diagram

where p is an epimorphism induced by the fact that gr(BcoH) is a graded subalgebra of(gr(B))coH.Since ?B,Band p are both epic, we have coker βgr= coker gr(β)gr(K).By assumption, B/BcoHis a Hopf dense Galois extension, thus KQis finite dimensional over Q.Then (gr(K))Q=gr(KQ) is also finite dimensional over Q.Therefore gr(B)/(gr(B))coHis a Hopf dense Galois extension.

Theorem 4.10Let A be an almost commutative R-algebra such that gr(A) is a domain.Assume that A is a right H-comodule algebra such that the right H-coaction preserves the filtration.If A/AcoHis a Hopf dense Galois extension and the canonical map β :A ?AcoHA →A ?RH is strict, thenis isomorphic to a group algebra over k.

ProofAs before, we write Ak= A ?Rk and Hk= H ?Rk.Since A is a filtered R-algebra,Akis also a filtered k-algebra with the obvious induced filtration.Since the right H-coaction preserves the filtration, it induces a right H-coaction on the associated graded algebra gr(A).Applying the functor ? ?Rk to the right H-coaction ρ : A → A ?RH,we obtain that Akis a right Hk-comodule algebra and Ak/(Ak)coHkis a Hopf dense Galois extension.Moreover, since β is strict, the induced canonical map βkis also a strict filtered map.By Lemma 4.9, gr(Ak)/(gr(Ak))coHkis a Hopf dense Galois extension.Since A is almost commutative, gr(A) is a commutative domain.Then gr(Ak) = gr(A) ?Rk is a commutative domain over k.By Lemma 4.8,is a group algebra.

For a filtered algebra A and a filtration preserving right H-coaction, the canonical map

may be not a strict map.Hence the associated graded algebra gr(A) may not be a Hopf dense Galois extension over (gr(A))coH.Some further discussions will be given in the next section.

5 Some Corollaries

In this section, k is an algebraically closed field of characteristic zero.All the algebras and modules in this section are over k.Let H be a finite dimensional Hopf algebra.

The next result is a direct consequence of [2, Proposition 3.6]if A is noetherian and H is semisimple.We give a direct proof and drop the assumptions in [2, Proposition 3.6].

Proposition 5.3Let A be a filtered algebra with an ascending filtration

such that FiA is finite dimensional for all i ≥0.Assume that A is a right H-comodule algebra and the coaction preserving the filtration.If the associated graded algebra gr(A) is a Hopf dense Galois extension over (gr(A))coH, then A/AcoHis a Hopf dense Galois extension.

ProofLet β : A ?AcoHA → A ?kH be the canonical map.Similar to the diagram(4.2), we have the following commutative diagram

Since gr(A)/(gr(A))coHis a Hopf dense Galois extension, which is to say that coker βgris finite dimensional.It follows that there is a positive number n such that for all k ≥n, we have

We claim that β(Fk(A?AcoHA))+Fn?1A?kH =FkA?kH for all k ≥ n.By the commutative diagram (5.1), for every x ∈ FnA ?kH, there is an element y ∈ Fn(A ?AcoHA) such that β(y)+z =x for some z ∈ Fn?1A ?kH.Hence β(Fn(A ?AcoHA))+Fn?1?kH =FnA ?kH.Now assume that β(Fi(A?AcoHA))+Fn?1A?kH =FiA?kH for i ≥ n.By the commutative diagram (5.1), we have β(Fi+1(A ?AcoHA))+FiA ?kH =Fi+1A ?kH.Then

Hence we have dim((A?kH)/im β)≤ dim(Fn?1A?kH)< ∞.Therefore,A/AcoHis a Hopf dense Galois extension.

Combined with Lemma 4.9, we have the following corollary.

Corollary 5.11With the same conditions in Proposition 5.3, if in addition the canonical map β : A ?AcoHA → A ?kH is strict, then A/AcoHis a Hopf dense Galois extension if and only if gr(A)/(gr(A))coHis a Hopf dense Galois extension.

Let g be a finite dimensional Lie algebra, and let f :g × g → k be a 2-cocycle, that is,for every x,y,z ∈g, f(x,x)=0, f(x,[y,z])+f(z,[x,y])+f(y,[z,x])=0.Then a Sridharan enveloping algebra [9]of g is defined to be the associative algebra

where T(g)is the tensor algebra of g over k and I is the ideal of T(g)generated by elements

Assume that {x1,...,xn} is an R-basis of g.Then Uf(g) is a free R-module and it has a basis(cf.[9, Theorem 2.6]).And Uf(g) is a filtered algebra with an ascending filtration defined by

for all k ≥0.The associated graded algebra of Uf(g) is the commutative polynomial ring k[x1,...,xn].

Suppose that there is a right H-coaction ρg:Uf(g)→ Uf(g)?kH which preserves the filtration defined as above.Then the associated graded map gr(ρg):grUf(g)→ grUf(g)?kH is a right H-coaction on grUf(g).

Corollary 5.12With the notions as above, if grUf(g)is a Hopf dense Galois extension over (grUf(g))coH, then Uf(g) is a Hopf dense Galois extension over Uf(g)coHand H?is isomorphic to a group algebra.

ProofThe first part follows from Proposition 5.3.The second part follows from Lemma 4.8 since grUf(g)k[x1,...,xn].

As we know, a right H-coaction on an algebra A is equivalent to a left H-action on A.Moreover, we have AcoH= AH, where AH= {a ∈ A|ha = ε(h)a, for all h ∈ H}.Hence we may rewrite Corollary 5.12 in the Hopf action version.

Corollary 5.13Suppose that H acts on a Sridharan enveloping algebra Uf(g).If the H-action preserves the filtration of Uf(g) and grUf(g) is a right H?-Hopf dense Galois extension over (grUf(g))H, then Uf(g) is a right H?-Hopf dense Galois extension over Uf(g)H, and H is isomorphic to a group algebra.