Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 42 (2001), No. 2, 395-400.

Actions of Hopf Algebras on Fully Bounded Noetherian Rings

Thomas Guédénon

21, Avenue de Versailles 75016 Paris, France e-mail: guedenon@caramail.com

Abstract. Let k be a commutative ring, H a finitely generated projective Hopf algebra over k and R a k-algebra which is a left H-module algebra. Assume that for every H-invariant left ideal I of R and every x + I ∈ (R/I)^{H} there exists s ∈ R^{H}, such that s − x ∈ I. The main result of the paper is that R is left FBN if and only if R is left Noetherian and R^{H} is left FBN. This result generalizes [4, Theorem 8] and [6, Theorem 2.3].

0. Introduction

A ring A is left bounded if every essential left ideal of A contains a nonzero two-sided ideal. The ring A is left fully bounded if for every prime ideal P of A, A/P is left bounded. We say that A is left FBN if it is left Noetherian and left fully bounded. The best known class of left FBN rings are left Noetherian P.I. rings. Right FBN rings are defined in a symmetric fashion.

Let k be a field, G a finite group and R an associative unitary k-algebra which is also a right G-module. Assume that the following condition holds: (⋆) For every G-invariant right ideal I of R and every x + I ∈ (R/I)^{G}, there exists r ∈ R^{G},

such that r − x ∈ I. Then J. J. Garcia and A. Del Rio [6, Theorem 2.3] have shown that R is right FBN if and only if R is right Noetherian and R^{G} is right FBN.

If there exists an r ∈ R such that tr(r) = 1 (this is the case if |G|, the order of G is invertible in R), then condition (⋆) holds. So [6, Theorem 2.3] gives a positive answer to Fisher and Osterburg’s question for right Noetherian rings [4, Question 7 page 367].

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2001 Heldermann Verlag

Let k be a commutative ring, H a finitely generated projective Hopf algebra over k and R a right Noetherian left H-module algebra with an element of trace 1. Then Dǎscǎlescu, Kelarev and Torrecillas proved that R is right FBN if and only if the subalgebra of invariants R^{H} is right FBN [4, Theorem 8]. This result generalizes partially [6, Theorem 2.3], since an example in [6] shows that condition (⋆) doesn’t imply that R has an element of trace 1. Our aim is to generalize [4, Theorem 8] and [6, Theorem 2.3] in the case where the action comes from a finitely generated projective Hopf algebra over k.

Throughout the paper, k is a commutative ring, H is a Hopf k-algebra with comultiplication Δ, counit ϵ, and antipode s and R is an H-module algebra, i.e. an associative unitary k-algebra which is also a left H-module such that h.(ab) = ∑_{h}(h_{1}.a)(h_{2}.b) for all h ∈ H and a, b ∈ R. We denote by R#H the associated smash product. The expression rh means r#h. The multiplication in R#H is defined by the rule (ah)(bg) = ∑_{h} a(h_{1}.b)(h_{2}g).

The group algebra kG of a finite group G is a finite-dimensional cocommutative Hopf algebra and R#kG is the usual skew group algebra R#G.

For further informations about Hopf algebras and the ring R#H, the reader is referred to [1, 8, 13].

In the remainder of the paper, all modules are left modules. An R-module M which is an H-module such that h.(am) = ∑_{h}(h_{1}.a)(h_{2}.m) is an R#H-module. Conversely, if M is an R#H-module, M may be thought of as an R-module with an action of H such that the above formula holds. It is clear that R is an R#H-module defined by (ah).b = a(h.b); a, b ∈ R. If M is an H-module, denote by M^{H} = {m ∈ M | h.m = ϵ(h)m ∀h ∈ H} the subspace of invariant elements of M. Clearly, R^{H} is a subring of R called the fixed subring of R (or the subring of invariants of R). The elements of R^{H} commute with H. If P is an R#H-module, P^{H} is an R^{H}-module with trivial H-action.

From now on, H is a finitely-generated projective k-module. Let us denote by x_{1}, x_{2}, …, x_{n} a generator set for H. We know from [7, Proposition 1.1] that H has a nonzero left integral and that the antipode s is a bijective antimorphism of algebras and an antimorphism of coalgebras. Also, R#H is finitely generated R-free module with generators x_{1}, x_{2}, …, x_{n}. If R is left Noetherian, then clearly so is R#H.

The main result of this article states that if for every H-invariant left ideal I of R and every x + I ∈ (R/I)^{H} there exists s ∈ R^{H} such that s − x ∈ I, then R is left FBN if and only if R is left Noetherian and R^{H} is left FBN. The main tool to prove this result is the basic fact that R has a canonical structure of R#H-module such that Hom_{R#H} (R, R) is isomorphic to R^{H}. We use the same techniques as in [6].

1. Preliminary results

We recall briefly some basic definitions. Let A be a ring, P and M two A-modules. We say that M is

− finitely P -generated if there exists an epimorphism P^{(I)} → M for some finite set I; − P -faithful if Hom_{A}(P, M′) ≠ 0 , for every nonzero submodule M′ of M.

If M is finitely generated, clearly M is finitely A-generated.

For every subset X of M (resp. of Hom_{A}(P, M )), we set

l_{A}(X) = {a ∈ A | am = 0 for all m ∈ M} (resp. l_{P} (X) = ∩_{f∈X}Kerf).

Let A be a ring. An A-module M is said to be quasi-projective if for every submodule N of M and every homomorphism f : M → M/N there is an endomorphism g : M → M such that p ∘ g = f where p : M → M/N is the canonical epimorphism. If R is finitely generated as R^{H}-module and if M is a finitely generated R-module, then M is a finitely generated R-faithful R^{H}-module. A subset I of R is H-invariant if H.I ⊆ I. Clearly, the H-invariant left ideals of R are just the R#H-submodules of R. If I is an H-invariant two-sided ideal of R, then R/I is an H-module algebra.

The left integral space of H is defined by

∫_{H} = {t ∈ H | ht = ϵ(h)t for all h ∈ H}.

We always fix an element 0 ≠ t ∈ ∫_{H}. Let M be an R#H-module. If m ∈ M, the Hsubmodule Hm of M is a finitely generated k-submodule of M containing m. More precisely, Hm is generated over k by the x_{i}m.

Lemma 1.1. An R#H-module is finitely generated as R#H-module if and only if it is finitely generated as R-module.

Proof. Let M be an R#H-module finitely generated as R#H-module. For every m ∈ M, (R#H)m = R(Hm) = ∑ R(x_{i}m). So M is generated as R-module by the x_{i}m_{j}; 1 ≤ i ≤ n, 1 ≤ j ≤ l; where m_{1}, m_{2}, …, m_{l} ∈ M is a generator set for M as R#H-module. □

The following lemma is the analogue of Nǎstǎsescu and Dǎscǎlescu’s result [9] used in the proof of [6, Theorem 2.3].

Lemma 1.2. If R is left FBN, then so is R#H.

Proof. By Lemma 1.1, R#H is finitely R-generated. By [6, Corollary 1.9], R is an FBN left R-module. Let M be a finitely generated R#H-module. Then M is a finitely generated R#H-faithful R-module. Consider the subset M = Hom_{R#H} (R#H , M ) of Hom_{R}(R#H , M ). By [6, Corollary 1.8], there exists a finite subset F of M such that l_{R#H}(M) = l_{R#H}(F). Since R#H is left Noetherian, the result follows from [6, Theorem 1.2]. □

2. The main results

We continue with the preceding notations. The map t̃: R → R given by t̃(r) = t.r is an R^{H}-bimodule morphism with values in R^{H}. Consider the following two conditions: (C_{1}) For every H-invariant left ideal I of R and every x + I ∈ (R/I)^{H}, there exists s ∈ R^{H},

such that s − x ∈ I. (C_{2}) There exists an r ∈ R, such that t̃(r) = 1.

Lemma 2.1. (C_{2}) ⇒ (C_{1}).

Proof. Let r ∈ R such that t̃(r) = 1, I be an H-invariant left ideal of R and x + I ∈ (R/I)^{H}. Then t̃(rx) − x = t̃(rx) − t̃(r)x = t.(rx) − (t.r)x = ∑_{t}(t_{1}.r)(t_{2}.x − ϵ(t_{2})x) ∈ I. Since t̃(rx) ∈ R^{H}, the result follows. □

An example in [6] shows that (C_{1}) doesn’t imply (C_{2}).

Lemma 2.2. The following statements are equivalent.

(a) R is R#H-quasi-projective.

(b) Condition (C_{1}) is satisfied.

(c) For every H-invariant left ideal I of R, (R/I)^{H} = (R^{H} + I)/I.

Proof. The equivalence (b) ⇔ (c) is obvious. (a)⇒(b) Let I be an H-invariant left ideal of R and x+I ∈ (R/I)^{H}. Then right multiplication by x+I is an R#H- morphism f : R → R/I. Let π : R → R/I be the canonical epimorphism. Since R is R#H-quasi-projective, there exists g ∈ Hom_{R#H} (R, R) such that π ∘ g = f. Take s = g(1), then s ∈ R^{H} and s − x ∈ I. (b)⇒(a) Let f : R → R/I be an R#H-morphism, where I is an R#H-submodule of R. Then I is an H-invariant left ideal of R and f(1) + I ∈ (R/I)^{H}. Let s ∈ R^{H}, such that f(1)+I = s+I and g : R → R be the right multiplication by s map. Then g ∈ Hom_{R#H} (R, R) and if we denote by π the canonical epimorphism R → R/I, then π ∘ g = f. □

Lemma 2.3. Let M be an R#H-module.

(a) The map f ↦ f(1) defines an isomorphism of R^{H}-modules between Hom_{R#H} (R, M ) and M^{H}.

(b) End_{R#H}(R) is isomorphic to R^{H}.

(c) R is R^{H}-isomorphic to (R#H)^{H}, where R#H is considered as left R#H-module via left multiplication.

Proof. (a) and (b) follow from [11, Corollary 3.5] and [12, Definition 3.1].

(c) The map R → tR; r ↦ tr is an R^{H}-isomorphism. By [8, Proof of Theorem 8.3.3 page 139], tR = (R#H)^{H}. Note that in [8, 11, 12], k is a field but there is no problem with k being now only a commutative ring. □

We can now state the main theorem of the paper.

Theorem 2.4. Assume condition (C_{1}) holds. Then the following statements are equivalent:

(a) R is left FBN.

(b) R is left Noetherian and R^{H} is left FBN.

Proof. By assumption and Lemma 2.2, R is R#H-quasi-projective. (a)⇒(b) Assume that R is left FBN. By Lemma 1.2 , R#H is left FBN too and, by [6, Corollary 1.9], R is FBN as R#H-module. Now [6, Theorem 1.7] and Lemma 2.3 (b) imply that R^{H} ≃ End_{R#H}(R) is left FBN.

(b)⇒(a) Since R is R#H-quasi-projective and R#H is left Noetherian, Lemma 2.3 and [2, Corollary 4.11] imply that R is a Noetherian R^{H}-module. So R is finitely generated as R^{H}- module. Let M be a finitely generated R-module. Then M is a finitely generated R-faithful R^{H}-module. Consider the subset M = Hom_{R}(R, M ) of Hom_{R^{H}} (R, M ). Since R^{H} is left FBN, there exists a finite subset F of M such that l_{R}(F) = l_{R}(M) (see [6, Corollary 1.8]). Since R is left Noetherian, the result follows from [6, Theorem 1.2]. □

Corollary 2.5. Assume that 1 ∈ t̃(R). Then the following statements are equivalent.

(a) R is left FBN.

(b) R is left Noetherian and R^{H} is left FBN.

Proof. By Lemma 2.1, condition (C_{1}) is satisfied. □

We close the paper by the following remark:

Remark 2.6. If the map t̃ is surjective, 1 ∈ t̃(R). If k is a field, then H is a finitedimensional Hopf algebra over k. If H is semisimple, then the map t̃ is surjective [8, page 55]. If H has a finite global dimension, H is semisimple [3, Corollary 1.7]. If k has characteristic 0, then H is semisimple if and only if s is involutive [10, Theorem 5.4]. If H is cocommutative or commutative, then s is involutive.

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400 T. Guédénon: Actions of Hopf Algebras on Fully Bounded Noetherian Rings

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Received February 2, 2000