Subrings of invariants for actions of finite dimensional Hopf algebras

TL;DR

This survey reviews the structure and properties of invariant rings resulting from actions of finite dimensional Hopf algebras, summarizing key developments and open problems in the field.

Contribution

It provides a comprehensive overview of invariant rings under Hopf algebra actions, highlighting recent advances and unresolved questions.

Findings

Summarizes known results on invariants of Hopf actions

Identifies gaps and open problems in the theory

Connects Hopf invariants to classical invariant theory

Abstract

This is a survey article on the invariant rings of Hopf actions

Equations

$$A^H=\{a\in A\mid ha=\epsilon (h)afor\; allh\in H\}.$$

$$\rho (ab)=\rho (a)\rho (b)for\; alla,b\in A.$$

$$A\otimes H\to A\otimes H,a\otimes h\mapsto \rho (a)\cdot (1\otimes h)$$

$$A^{\mathrm{co}H}=\{a\in A\mid \rho (a)=a\otimes 1\}.$$

$$\rho :A\to A\otimes H^{\ast }\cong {\mathrm{Hom}}_k(H,A),$$

$$(1\#h)(a\#1)=\sum h_{(1)}a\#h_{(2)}fora\in A,h\in H.$$

$$r_P(M)=\frac{{\mathrm{length}}_{A}M/MP}{{\mathrm{length}}_{A}A/P}\in Q.$$

$$A\cong A_1\times \dots \times A_n$$

$$E={⨁}_{P\in \mathrm{Max}A}{F}_P^{m_P/d}whered=\gcd \{m_P\mid P\in \mathrm{Max}A\}.$$

$$(h\cdot f)(x)=\sum h_{(1)}\,f\bigl{(}S(h_{(2)})x\bigr{)},\qquad$$

$$A\otimes H^{\ast }\to A\otimes H^{\ast },a\otimes f\mapsto \rho (a)\cdot (1\otimes f),$$

$$(\mathrm{id}\otimes \Delta ){\rho }^{\prime }=({\rho }^{\prime }\otimes \mathrm{id}){\rho }^{\prime }.$$

$$\gamma :A{\otimes }_{A^H}A\to A\otimes H^{\ast },a\otimes b\mapsto (a\otimes 1)\cdot \rho (b)$$

$$\gamma (a\otimes b)(h)=a(hb)fora,b\in Aandh\in H.$$

$$\gamma (\sum a_i\otimes b_{i}x)=(1\otimes \lambda )\cdot \rho (x)=x\otimes \lambda ,$$

$$\hbox{which means that}\quad\textstyle\sum\limits a_{i}\bigl{(}h(b_{i}x)\bigr{)}=\lambda(h)x\quad\hbox{for all $h\in H$}.$$

$$\textstyle\sum\limits a_{i}f_{i}(x)=\textstyle\sum\limits a_{i}\bigl{(}t(b_{i}x)\bigr{)}=\lambda(t)x=x$$

$$ax=(a\otimes 1)\cdot x,xa=x\cdot \rho (a)fora\in Aandx\in A\otimes H^{\ast }.$$

$$(h\rightharpoonup\xi)(g)=\xi(gh),\qquad(h\rightharpoondown\xi)(g)=\xi\bigl{(}S(h)g\bigr{)}\quad\hbox{for $g,h\in H$ and $\xi\in H^{*}$}.$$

$$h(a\otimes b)=ha\otimes b,h(a\otimes \xi )=\sum h_{(1)}a\otimes (h_{(2)}\rightharpoondown \xi )$$

$$h(a\otimes b)=a\otimes hb,h(a\otimes \xi )=a\otimes (h\rightharpoonup \xi ).$$

$${\mathrm{Hom}}_H(V,A\otimes H^{\ast })\cong {\mathrm{Hom}}_H(V\otimes H,A)$$

$$\rho (a)=\sum (c_i\otimes 1)\rho (a_i)=\gamma (\sum c_i\otimes a_i).$$

$$\sum (g{c}_i\otimes 1)\rho (a_i)=\gamma (\sum g{c}_i\otimes a_i)=\epsilon (g)\rho (a)=\epsilon (g)\sum (c_i\otimes 1)\rho (a_i).$$

$$[A:A^H{]}_l\le nand\text{/}[A:A^H{]}_r\le n.$$

$$a\otimes 1=\textstyle\sum\limits a_{(0)}\otimes S^{-1}(a_{(2)})a_{(1)}=\textstyle\sum\limits\,\bigl{(}1\otimes S^{-1}(a_{(1)})\bigr{)}\cdot\rho(a_{(0)})\in N.$$

$$[A\#H/K:A^H]=[M_m(A^H):A^H]=m^2$$

$$[M_m(A^H):D]=[M_m(A^H):D^{\prime }].$$

$$\alpha :{\mathrm{Hom}}_{\mathcal{A}}(M,N){\otimes }_{B}M\to N$$

$$I_H=\{a\in A\mid Ha\subset I\}=\{a\in A\mid \rho (a)\in I\otimes H^{\ast }\}.$$

Full text

Subrings of invariants for actionsof finite dimensional Hopf algebras

Serge Skryabin

Institute of Mathematics and Mechanics, Kazan Federal University,Kremlevskaya St. 18, 420008 Kazan, Russia E-mail: Serge.Skryabin@kpfu.ru

The classical invariant theory on the threshold of the 20th century considered finite generation of invariants as a key problem. Initially only group actions on polynomial rings in several indeterminates were looked at. In the case of a finite group Emmy Noether’s constructive approach via Galois resolutions yielded certain conclusions also in a more abstract situation. It showed that any commutative ring is integral over the subring of invariants with respect to a finite group of automorphisms. For a finitely generated commutative algebra over a field integrality over a subalgebra is equivalent to finiteness as a module over this subalgebra, while finite generation of the subalgebra is a consequence of these properties. This makes integrality and module-finiteness particularly important in the study of invariants.

Grothendieck and his school made a transition from group actions to actions of group schemes. As it turned out, there are general results on invariants incorporated in the construction of quotients by finite group schemes. These results can be interpreted in terms of coactions of commutative Hopf algebras or, dually, in terms of actions of cocommutative Hopf algebras.

In a different vein many people contributed to research on group actions and Lie algebra actions, and also group gradings, on noncommutative rings. The work on Hopf algebra actions started in the 1980s aimed to unify previously known results in those areas. In spite of the progress made in this study considerable difficulties have been encountered in some questions. Even now the state of knowledge in the general Hopf algebra case has not reached the level recorded in the 1980 manuscript of Susan Montgomery on fixed rings of finite automorphism groups of associative rings [37].

The present paper is intended primarily as a survey of recent work on invariants of Hopf algebra actions. Its highlights are results on integrality of $H$-module PI algebras over central invariants obtained independently by Etingof [24] and Eryashkin [22]. There is a different notion of integrality introduced by Schelter [46] which is suitable for extensions of noncommutative rings. Eryashkin has also proved that an arbitrary $H$-module PI algebra is Schelter integral over the subring of all invariants when the Hopf algebra $H$ is semisimple and cosemisimple [23], thus answering a question of Montgomery [38] in the PI case.

Other recent results on invariants are presented in the author’s two own papers. In [52] it has been proved that, given a semisimple Hopf algebra $H$, all nonzero $H$-stable one-sided ideals of any noetherian $H$-semiprime $H$-module algebra $A$ contain nonzero invariants, and the classical quotient ring of $A$ is obtained by localization at the Ore set of invariant regular elements. We will show that these conclusions are true even when $A$ is not noetherian provided that $A$ has an artinian classical quotient ring. Another article [54] has answered the question of Bergen, Cohen and Fischman [4] on the equality of the left and right dimensions of a skew field over the subfield of invariants. We will also review older results.

In this paper only finite dimensional Hopf algebras over a field will be considered. It should be noted, however, that many results discussed here can be formulated more generally when an arbitrary commutative ring is taken as a base ring and the Hopf algebras are finitely generated projective modules. In fact, this was the setting for several original papers.

1. Terminology and notation

Throughout the whole paper $H$ will stand for a finite dimensional Hopf algebra over a field ${k}$. We denote by $\Delta$, $\varepsilon$, $S$ the comultiplication, the counit and the antipode in either $H$ or the dual Hopf algebra $H^{*}$, depending on the context. For a general information on Hopf algebras and their actions on rings we refer the reader to [1], [38] and other books.

Recall that the categories of $H$-modules and $H$-comodules are monoidal. If $V$ and $W$ are two (left) $H$-modules, then $V\otimes W$ is an $H\otimes H$-module, and $H$ acts on $V\otimes W$ via $\Delta:H\to H\otimes H$. If $V$ and $W$ are two (right) $H$-comodules, then $V\otimes W$ is an $H\otimes H$-comodule, and the coaction of $H$ is obtained by means of the map $H\otimes H\to H$, $\,a\otimes b\mapsto ab$. Here and later $\otimes$ means $\otimes_{k}$ unless the base ring for the tensor product is indicated explicitly.

All algebras and rings are assumed to be associative and unital. An $H$-module algebra is a ${k}$-algebra $A$ equipped with a left $H$-module structure such that the multiplication map $A\otimes A\to A$ is $H$-linear, assuming that $H$ acts on $A\otimes A$ via $\Delta$. If this condition is satisfied, then $H$ acts trivially on the image of ${k}$ in $A$, so that $h1_{A}=\varepsilon(h)1_{A}$ for all $h\in H$ where $1_{A}$ is the unity of $A\,$ [10, Lemma 1.9].

The $H$-invariant elements of an $H$-module algebra $A$ form a subalgebra

[TABLE]

An $H$-comodule algebra is a ${k}$-algebra $A$ equipped with a right $H$-comodule structure such that the multiplication map $A\otimes A\to A$ is a homomorphism of comodules. This condition can be reformulated by saying that the comodule structure map $\rho:A\to A\otimes H$ is multiplicative, i.e.

[TABLE]

Moreover, $\rho$ is then a homomorphism of unital algebras. The fact that $\rho(1)=1\otimes 1$, and so $H$ coacts trivially on the image of ${k}$ in $A$, is easily seen as follows. Clearly $\rho(1)\,x=x$ for all $x$ in the right ideal $I$ of $A\otimes H$ generated by $\rho(A)$. So it suffices to check the equality $I=A\otimes H$, but it does hold because the linear map

[TABLE]

is bijective. In fact the assignment a\otimes h\mapsto({\rm id}\otimes S)\bigl{(}\rho(a)\bigr{)}\cdot(1\otimes h) defines the inverse map. This argument shows also that $\rho$ is an isomorphism of $A$ onto a subalgebra of $A\otimes H$, and $A\otimes H$ is a free $\rho(A)$-module with respect to the action by left multiplications. Similarly, $A\otimes H$ is free over $\rho(A)$ on the right.

With an $H$-comodule algebra one associates its subalgebra consisting of coaction invariants

[TABLE]

As is well-known, the left $H$-module structures are in a bijective correspondence with the right $H^{*}$-comodule structures. This correspondence is compatible with the tensor products of modules and comodules. Therefore each $H$-module algebra is an $H^{*}$-comodule algebra and vice versa. Under the canonical identification of $A\otimes H^{*}$ with $\mathop{\rm Hom}\nolimits_{\,{k}}(H,A)$ the comodule structure on an $H$-module algebra $A$ is given by the map

[TABLE]

In the rest of the paper $A$ is assumed to be an $H$-module algebra. However, sometimes arguments are formulated more naturally in terms of comodule structures. Note, in particular, that $\,A^{H}=A^{\mkern 1.0mu{\rm co}\mkern 1.0muH^{*}}$.

By an ideal we mean a two-sided ideal unless explicitly stated otherwise. Of particular interest are $H$-invariant ideals, i.e. ideals stable under the action of $H$. Several properties of an $H$-module algebra $A$ are defined in terms of the collection of its $H$-stable ideals:

$A$ is $H$-simple if $A\neq 0$ and $A$ has no $H$-stable ideals except the zero ideal and the whole $A$;

$A$ is $H$-prime if $A\neq 0$ and $IJ\neq 0$ for all nonzero $H$-stable ideals $I$ and $J$ of $A$;

$A$ is $H$-semiprime if $A$ contains no nonzero nilpotent $H$-stable ideals.

An $H$-stable ideal $I$ of $A$ is called $H$-prime (respectively $H$-semiprime) if the factor algebra $A/I$ is $H$-prime (respectively $H$-semiprime). For an arbitrary ideal $I$ of $A$ we denote by $I_{H}$ the largest $H$-stable ideal of $A$ contained in $I$. If $I$ is prime (respectively semiprime), then $I_{H}$ is $H$-prime (respectively $H$-semiprime). Conversely, if $I$ is $H$-prime, then $I=P_{H}$ for some prime ideal $P$ of $A$ [8, Lemma 1.5]. An $H$-stable ideal is $H$-semiprime if and only if it is an intersection of $H$-prime ideals [39, Lemma 8.3].

If a ring-theoretic notion is not prefixed by $H$- , then it does not take into account the $H$-module structure. For example, an $H$-module algebra $A$ is PI if $A$ satisfies a polynomial identity as an ordinary algebra.

A left or right $A$-module $M$ is said to be $H$-equivariant if $M$ is equipped with a left $H$-module structure such that the action of $A$ on $M$ comes from an $H$-linear map $A\otimes\nobreak M\to M$ or $M\otimes A\to M$, respectively, assuming that $H$ acts in the tensor products via $\Delta$. Denote by $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ and $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ the categories of $H$-equivariant left and right $A$-modules. Morphisms in these categories are maps which are $A$-linear and $H$-linear simultaneously. Let $\vphantom{{\cal M}}{}_{A}{\cal M}$ and ${\cal M}_{A}$ stand for the categories of left and right $A$-modules.

Similarly, an $A$-bimodule $M$ will be called $H$-equivariant if $M$ is equipped with a left $H$-module structure with respect to which $M$ is an object of both $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ and $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. We denote by $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}_{A}$ the category of $H$-equivariant $A$-bimodules. Note that each $H$-stable ideal of $A$ is an object of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}_{A}$, and any homomorphism of $H$-module algebras $A\to B$ makes $B$ an object of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}_{A}$.

Recall that the smash product algebra $A\#H$ has $A\otimes H$ as its underlying vector space, the canonical maps $A\to A\otimes 1$ and $H\to 1\otimes H$ are isomorphisms of $A$ and $H$ onto subalgebras of $A\#H$, while

[TABLE]

The compatibility of the $A$-module and $H$-module structures requested in the definition of the category $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ means precisely that the two module structures come from a single $A\#H$-module structure. Thus $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ is identified with the category of left $A\#H$-modules.

Let $A^{\rm op}$ be $A$ with the opposite multiplication, and let $H^{\rm cop}$ be $H$ with the opposite comultiplication. Then $A^{\rm op}$ is an $H^{\rm cop}$-module algebra, and $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ may be identified with the category of left $A^{\rm op}\#H^{\rm cop}$-modules.

The algebras $A\#H$ and $A^{H}$ are connected by a Morita context. Several articles [4], [5], [11], [13] derive various information about the invariant ring $A^{H}$ when $A\#H$ is known to be simple, or prime, or semiprime. These results are also discussed in [38]. However, for an arbitrary finite dimensional Hopf algebra $H$ it is quite difficult to understand the ring structure of $A\#H$ in terms of the original algebra $A$. There is still a big gap between what is known in the general case and in the case of a finite group $G$ acting on a ring $R$ where the skew group ring $R*G$ is sufficiently well understood as a normalizing extension of $R$. In our paper we rarely use ring-theoretic properties of $A\#H$ directly. However, equivariant modules are important for many considerations.

Recall that a left (respectively right) integral in $H$ is an element $0\neq t\in H$ such that $ht=\varepsilon(h)t$ (respectively $th=\varepsilon(h)t$) for all $h\in H$. Let $t$ be a left integral. If $V$ is a left $H$-module, then the action of $t$ gives a map ${\hat{t}}:V\to V^{H}$ where $V^{H}$ is the subspace of $H$-invariant elements in $V$. In [13] this map was called a trace by analogy with the terminology used in the case of group actions.

By Maschke’s theorem $H$ is semisimple if and only if $\varepsilon(t)\neq 0$. In this case $t$ acts on $V^{H}$ as a nonzero scalar multiplication. In particular, we have $tV=V^{H}$, i.e. the trace ${\hat{t}}$ is always surjective.

2. Structural properties of $H$-module algebras

In this section we present several results concerned with the structure of $H$-module algebras on which recent work on invariants of Hopf actions is based. Actually most of these results can be formulated for a not necessarily finite dimensional Hopf algebra $H$. Nevertheless it will be assumed in all statements that $\,\dim H<\infty$. With this assumption we do not need to mention any additional restrictions, and also the proofs become considerably simpler.

A key argument used in deriving these results comes from

Theorem 2.1 [49]. Suppose that $A$ is a semilocal $H$-simple $H$-module algebra. Then each object $M\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ is projective in ${\cal M}_{A}$. Moreover, a direct sum of several copies of $M$ is a free $A$-module. A similar conclusion holds in $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$.

There is one application of the freeness properties of $H$-equivariant modules where the $H$-simplicity of the $H$-module algebra is not known beforehand. To deal with this situation one needs the next lemma stated under more technical assumptions about $A$ and $M$ than the previous theorem.

Denote by $\mathop{\rm Max}\nolimits A$ the set of maximal ideals of $A$. If $A$ is semilocal, then its factor algebra by the Jacobson radical is semisimple artinian. This means that the set $\mathop{\rm Max}\nolimits A$ is finite and $A/P$ is simple artinian for each $P\in\mathop{\rm Max}\nolimits A$. An object $M$ of the category $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ is said to be $A$-finite if $M$ is finitely generated as an $A$-module. If $M$ is $A$-finite, then $M/MP$ is an $A$-module of finite length. The rank of $M$ at $P$ is defined as

[TABLE]

Lemma 2.2. Let $A$ be a semilocal $H$-module algebra. Suppose that $M\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ is $A$-finite and there exists $P\in\mathop{\rm Max}\nolimits A$ such that $P$ contains no nonzero $H$-stable ideals of $A$ and $r_{P}(M)\geq r_{Q}(M)$ for all $Q\in\mathop{\rm Max}\nolimits A$. Then $M^{n}$ is a free $A$-module for some integer $n>0$.

For the proof see [49, Lemma 7.5]. This lemma is valid even when $H$ is an infinite dimensional Hopf algebra. However, the assumption $\dim H<\infty$ is needed to deduce the conclusion of Theorem 2.1 for objects $M$ which are not $A$-finite. Since each element of $M$ is contained in an $A$-finite subobject of $M$, a basis for $M^{n}$ over $A$ can be constructed using Zorn’s lemma.

It turns out that Lemma 2.2 and Theorem 2.1 lead to several fundamental facts concerning artinian $H$-module algebras very quickly. Sometimes one has initially less information about an $H$-module algebra, but the left and right artinian conditions can be deduced. For this reason we have to deal with semiprimary algebras. A semilocal ring is called semiprimary if its Jacobson radical is nilpotent.

Lemma 2.3. Let $A$ be a semiprimary $H$-module algebra, and let $K\in\mathop{\rm Max}\nolimits A$. Then the largest $H$-stable ideal $K_{H}$ contained in $K$ is a maximal $H$-stable ideal of $A$.

Proof. Replacing $A$ with the factor algebra $A/K_{H}$, we may assume that $K_{H}=0$, and then we have to prove that $A$ is $H$-simple. First note that $A$ is $H$-prime. Indeed, if $I,J$ are two nonzero $H$-stable ideals of $A$, then both $I\not\subset K$ and $J\not\subset K$, whence $IJ\not\subset K$, and therefore $IJ\neq 0$.

Every semiprimary ring satisfies DCC on finitely generated one-sided ideals. Hence $A$ has a minimal nonzero $H$-stable finitely generated right ideal $M$. If $0\neq x\in M$, then $M=(Hx)A$ since $(Hx)A$ is a nonzero $H$-stable finitely generated right ideal of $A$ contained in $M$. It follows that $M$ is minimal in the set of all nonzero $H$-stable right ideals of $A$. If $I$ is any nonzero $H$-stable ideal of $A$, then $MI$ is an $H$-stable right ideal. Since $MI\subset M$ and $MI\neq 0$ by the $H$-primeness of $A$, we get $MI=M$.

We may view $M$ as an $A$-finite object of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. Pick $P\in\mathop{\rm Max}\nolimits A$ for which $r_{P}(M)$ attains the maximum value. Since $M\neq 0$, we have $r_{P}(M)>0$. This means that $M\neq MP$, but then $M\neq MP_{H}$ too, which is only possible when $P_{H}=0$ by the preceding argument. Thus the hypotheses of Lemma 2.2 are fulfilled, and we deduce that $M^{n}$ is a free $A$-module for some $n>0$. Hence $MI\neq M$ for each ideal $I\neq A$. If $I$ is $H$-stable and $I\neq A$, this entails $I=0$.   $\mathchar 1027\relax$

Theorem 2.4 [55]. Suppose that $A$ is semiprimary and $H$-semiprime. There is an isomorphism of $H$-module algebras

[TABLE]

where $A_{1},\ldots,A_{n}$ are $H$-simple $H$-module algebras. If $A$ has a maximal ideal containing no nonzero $H$-stable ideals of $A$ then $A$ is $H$-simple.

Proof. By Lemma 2.3 the maximal $H$-stable ideals of $A$ are precisely the ideals $K_{H}$ with $K\in\mathop{\rm Max}\nolimits A$. In particular, there are finitely many of them. Let $I_{1},\ldots,I_{n}$ be all the maximal $H$-stable ideals. Then $I_{1}\cap\ldots\cap I_{n}$ is contained in the Jacobson radical $J$ of $A$. Since $J$ is nilpotent and $A$ is $H$-semiprime, we get $I_{1}\cap\ldots\cap I_{n}=0$. But $I_{k}+I_{l}=A$ for each pair of indices $k\neq l$, whence the desired direct product decomposition of $A$ holds with $A_{k}=A/I_{k}$ by the Chinese remainder theorem.   $\mathchar 1027\relax$

Corollary 2.5. Each right coideal subalgebra $B$ of $H^{*}$ is an $H$-simple $H$-module algebra and $H^{*}$ is right and left $B$-free.

Proof. Here $B$ is a subalgebra and a right coideal of $H^{*}$. The restriction of the comultiplication $\Delta$ in $H^{*}$ gives a map $B\to B\otimes H^{*}$ which makes $B$ into an $H^{*}$-comodule algebra. Hence $B$ is also an $H$-module algebra. Put $B^{+}=\mathop{\rm Ker}\varepsilon|_{B}$ where $\varepsilon$ is the counit of $H^{*}$. Then $B^{+}\in\mathop{\rm Max}\nolimits B$ with $B/B^{+}\cong{k}$. Suppose that $I$ is an $H$-stable ideal of $B$. Then $\Delta(I)\subset I\otimes H^{*}$. Recall that $(\varepsilon\otimes{\rm id})\circ\Delta$ is the identity map by the definition of the counit. If $I\subset B^{+}$, we get $x=(\varepsilon\otimes{\rm id})(\Delta x)=0$ for each $x\in I$ since $\varepsilon(I)=0$, and so $I=0$. Thus $B^{+}$ contains no nonzero $H$-stable ideals of $B$. By Theorem 2.4 $B$ is $H$-simple.

Now $H^{*}$ is also an $H$-module algebra, and $B$ is its $H$-stable subalgebra. Hence we may regard $H^{*}$ as an object of both $H\hbox{-}\vphantom{{\cal M}}_{B}{\cal M}$ and $H\hbox{-}\mskip 1.0mu{\cal M}_{B}$. Freeness of $H^{*}$ over $B$ follows from Theorem 2.1.   $\mathchar 1027\relax$

Corollary 2.6. Suppose that $A$ is semiprimary and $H$-semiprime. Then each object $M\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ is projective in ${\cal M}_{A}$.

Proof. The direct product decomposition of $A$ given in Theorem 2.4 implies that $M\cong M_{1}\times\ldots\times M_{n}$ where $M_{i}=M\otimes_{A}A_{i}\in H\hbox{-}\mskip 1.0mu{\cal M}_{A_{i}}$. By Theorem 2.1 $M_{i}$ is projective in ${\cal M}_{A_{i}}$ for each $i$, whence the conclusion.   $\mathchar 1027\relax$

Corollary 2.7. Suppose that $A$ is semiprimary and $H$-semiprime. If $I$ is an $H$-stable right ideal of $A$, then $I=eA$ for some idempotent $e\in A$.

Proof. We may view $A/I$ as an object of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. By Corollary 2.6 $A/I$ is projective in ${\cal M}_{A}$. Hence $I$ is a direct summand of $A$ as a right $A$-module.   $\mathchar 1027\relax$

Theorem 2.8 [55]. Any semiprimary $H$-semiprime algebra $A$ is a quasi-Frobenius ring. In particular, $A$ is left and right artinian.

Proof. By Theorem 2.4 it suffices to consider the case when $A$ is $H$-simple. We are going to apply the general fact that a semiprimary ring is quasi-Frobenius whenever it is left and right selfinjective (see [30, Th. 10]). Let us show that $A$ is left selfinjective. Applying this to the $H^{\rm cop}$-module algebra $A^{\rm op}$, we deduce that $A$ is right selfinjective too, and the conclusion follows.

Take any nonzero injective object $M\in H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$. Then $M$ remains injective in $\vphantom{{\cal M}}{}_{A}{\cal M}$. To see this recall that $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ is identified with the category of left $B$-modules for $B=A\#H$. The forgetful functor $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}\to\vphantom{{\cal M}}_{A}{\cal M}$ is identified with the restriction functor $\vphantom{{\cal M}}{}_{B}{\cal M}\to\vphantom{{\cal M}}_{A}{\cal M}$ that arises from the canonical embedding of $A$ into $B$. The latter functor preserves injectives since it has an exact left adjoint $\,B\,\otimes_{A}\,?\,$.

But $M^{n}$ is a free $A$-module for some $n>0$ by Theorem 2.1. Therefore $A$ is an $\vphantom{{\cal M}}{}_{A}{\cal M}$-direct summand of $M^{n}$. Since $M^{n}$ is injective in $\vphantom{{\cal M}}{}_{A}{\cal M}$, so too is $A$.   $\mathchar 1027\relax$

Thus an $H$-semiprime algebra $A$ is semiprimary if and only if $A$ is left artinian, if and only if $A$ is right artinian, and we will say that $A$ is artinian in this case.

Theorem 2.9. Let $A$ be an $H$-stable subalgebra of an $H$-module algebra $B$. If $A$ is artinian and $H$-simple then $B$ is free as an $A$-module with respect to the action by right (or left) multiplications.

Proof. We may view $B$ as an object of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. Hence Theorem 2.1 applies to $B$. For each $P\in\mathop{\rm Max}\nolimits A$ denote by $F_{P}$ the projective cover in ${\cal M}_{A}$ of a simple right $A/P$-module. These modules $F_{P}$ are indecomposable, and $A\cong\bigoplus_{P\in\mathop{\rm Max}\nolimits A}F_{P}^{\,m_{P}}$ for some multiplicities $m_{P}$. Put

[TABLE]

Then $A\cong E^{d}$. If $M$ is any right $A$-module such that $M^{n}$ is free for some $n>0$, then, by the Krull-Schmidt theorem, $M$ is isomorphic to a direct sum of a family of copies of $E$. Moreover, $M$ is itself free if either $M$ is not finitely generated or $M\cong E^{k}$ with $d$ dividing $k$. If $M$ is in fact an $A$-bimodule, then $M\cong A\otimes_{A}M\cong N^{d}$ where $N=E\otimes_{A}M$. In this case $N$ has to be isomorphic to a direct sum of a family of copies of $E$, whence $M$ is right $A$-free by the previous observation. It remains to apply this for $M=B$.   $\mathchar 1027\relax$

Theorem 2.10 [55]. Suppose that $H$ is semisimple, $A$ is artinian and $H$-semiprime. Then $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ and $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ are semisimple categories. In other words, the smash product algebras $A^{\rm op}\#H^{\rm cop}$ and $A\#H$ are semisimple artinian.

Proof. A key ingredient in the proof is the fact established by Cohen and Fischman [11] according to which a submodule $W$ of a left $A\#H$-module $V$ is a direct summand whenever $W$ is an $A$-module direct summand of $V$. For this one needs only semisimplicity of $H$ but no assumptions about an $H$-module algebra $A$.

In the case of the category $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ a similar argument runs as follows. Let $M,N$ be two objects of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. There is a left $H$-module structure on $\mathop{\rm Hom}\nolimits_{\,{k}}(M,N)$ defined by the rule

[TABLE]

for $h\in H$ with $\Delta h=\sum h_{(1)}\otimes h_{(2)}$, $\,f\in\mathop{\rm Hom}\nolimits_{\,{k}}(M,N)$, $\,x\in M$. It is straightforward to check that $\mathop{\rm Hom}\nolimits_{A}(M,N)$ is stable under this action of $H$ and that a linear map $f\in\mathop{\rm Hom}\nolimits_{\,{k}}(M,N)$ is $H$-invariant if and only if $f$ is $H$-linear. Let $t\in H$ be an integral with $\varepsilon(t)=1$. If $N$ is a subobject of $M$ which splits off as an $A$-module direct summand, then there exists an $A$-linear map $f:M\to N$ such that $f|_{N}={\rm id}$. Now the map $f^{\prime}=t\cdot f$ is $A$-linear and $H$-linear simultaneously, and also $f^{\prime}|_{N}={\rm id}$. Hence $M=N\oplus\mathop{\rm Ker}f^{\prime}$, a direct sum decomposition in $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$.

Under the assumption that $A$ is artinian and $H$-semiprime any subobject $N$ of $M$ is an $A$-module direct summand since the factor object $M/N\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ is projective in ${\cal M}_{A}$ by Corollary 2.6. Hence the previous conclusion holds.   $\mathchar 1027\relax$

The result of Cohen and Fischman mentioned in the proof of Theorem 2.10 means that $A\#H$ is a semisimple extension of $A$ when $H$ is semisimple. In [11] it was used to show that $A\#H$ is semiprime artinian whenever so is $A$. If $A$ is not semiprime but only $H$-semiprime, the same conclusion requires Theorem 2.1 which has been proved much later.

Let $R$ be a ring. A ring $Q$ is said to be a classical right quotient ring of $R$ if

(1) $Q$ contains $R$ as a subring,

(2) all regular elements, i.e. nonzerodivisors, of $R$ are invertible in $Q$, and

(3) each element $q\in Q$ can be written as $q=as^{-1}$ where $a,s\in R$, $\,s$ regular.

Such a ring $Q$ exists if and only if the set of all regular elements of $R$ satisfies the right Ore condition. In this case $Q$ is unique up to isomorphism, and we will denote this ring by $Q(R)$.

If an $H$-module algebra $A$ is not artinian, but $A$ has an artinian classical right quotient ring, the previous results can still be used to derive information about $A$. There are two important cases when this happens:

Theorem 2.11 [55]. If $A$ is right noetherian and $H$-semiprime, then $A$ has a quasi-Frobenius classical right quotient ring.

Theorem 2.12 [22]. If $A$ is PI and $H$-semiprime with finitely many minimal $H$-prime ideals, then $A$ has a quasi-Frobenius classical right quotient ring. In particular, this holds if $A$ is finitely generated, PI, and $H$-semiprime.

In the proof of Theorem 2.11 one first constructs a generalized quotient ring $Q$ using the filter of $H$-stable essential right ideals of $A$. This ring turns out to be semiprimary. In the proof of Theorem 2.12 one starts with the $H$-equivariant Martindale quotient ring $Q$. As we shall see in section 6 it will be a finite module over a central artinian subring. In both cases $H$ acts on $Q$, and $H$-semiprimeness is preserved under passage to $Q$. Hence $Q$ is quasi-Frobenius by Theorem 2.8, and the conclusion that $Q$ is a classical right quotient ring can be deduced from the following ring-theoretic fact:

Proposition 2.13 [51]. Let $R$ be a subring of a quasi-Frobenius ring $Q$. Suppose that ${\cal I}$ is a topologizing filter of right ideals of $R$ with the following properties:**

*(a) *each $I\in{\cal I}$ has zero left and right annihilators in $Q$,

*(b) *for each $q\in Q$ there exists $I\in{\cal I}$ such that $qI\subset R$.

Then each right ideal $I\in{\cal I}$ contains a regular element of $R$, and $Q$ is a classical right quotient ring of $R$.

We say that a family ${\cal F}$ of right ideals in a ring $R$ is a filter if for any pair of right ideals $I,J\in{\cal F}$ there exists $K\in{\cal F}$ such that $K\subset I\cap J$. A filter ${\cal F}$ is topologizing if for each $I\in{\cal F}$ and each $a\in R$ there exists $I^{\prime}\in{\cal F}$ such that $aI^{\prime}\subset I$. The condition that with each $I\in{\cal F}$ all larger right ideals also belong to ${\cal F}$ is often included in the definition of a filter, but omitting it will do no harm.

With small improvements in the proof of [51, Prop. 1.4] the assumption that the filter ${\cal I}$ is topologizing can be actually removed. In the case when $Q$ is semisimple artinian see [52, Prop. 2.3].

Theorem 2.14 [55]. Suppose that $A$ has a right artinian classical right quotient ring $Q$. Then the $H$-module structure on $A$ has a unique extension to $Q$ with respect to which $Q$ becomes an $H$-module algebra.

Proof. We argue in terms of comodule structures. Since the map

[TABLE]

is invertible, $A\otimes H^{*}$ is a free $A$-module with respect to the action of $A$ given by left multiplications by the elements $\rho(a)$. For each regular element $s$ of $A$ it follows that $\rho(s)$ is right regular in $A\otimes H^{*}$, i.e. $\rho(s)x=0$ for $x\in A\otimes H^{*}$ implies $x=0$. Then $\rho(s)$ remains right regular in $Q\otimes H^{*}$, and therefore $\rho(s)$ has to be invertible in the right artinian ring $Q\otimes H^{*}$. This property shows that $\rho:A\to A\otimes H^{*}$ extends to an algebra homomorphism $\rho^{\prime}:Q\to Q\otimes H^{*}$. Now $({\rm id}\otimes\Delta)\rho^{\prime}$ and $(\rho^{\prime}\otimes{\rm id})\rho^{\prime}$ are two algebra homomorphisms $Q\to Q\otimes H^{*}\otimes H^{*}$ which agree on $A$. Hence

[TABLE]

Thus $\rho^{\prime}$ is a structure of an $H^{*}$-comodule algebra extending the given one on $A$.   $\mathchar 1027\relax$

In the situation of Theorem 2.14 we have two subrings of invariants $A^{H}$ and $Q^{H}$. Clearly $A^{H}=A\cap Q^{H}$. Theorem 2.14 is true even without the assumption that $\dim H<\infty$.

The conclusions of Theorems 2.4, 2.8 and 2.11 hold for some classes of infinite dimensional Hopf algebras too. Unfortunately it is still unknown whether the assumption that $H$ has a bijective antipode is sufficient for their validity. However, if $A$ is right artinian and $H$-semiprime, then $A$ is a quasi-Frobenius ring, even when $H$ is an arbitrary infinite dimensional Hopf algebra [50].

3. Module-finiteness over the invariants

In this section we examine those cases where $A$ is known to be a finite $A^{H}$-module. Many results discussed here date back to as early as the 1980s.

The comodule structure $\rho:A\to A\otimes H^{*}$ enables us to define a ${k}$-linear map

[TABLE]

(recall that $A^{H}=\{a\in A\mid\rho(a)=a\otimes 1\}$). Under the canonical identification of $A\otimes H^{*}$ with $\mathop{\rm Hom}\nolimits_{\,{k}}(H,A)$ we have

[TABLE]

The notion of Hopf Galois extensions of algebras is defined in terms of comodule structures (see [38, Ch. 8]). In the case when the Hopf algebra is finitely generated projective as a module over a commutative base ring this notion was introduced by Kreimer and Takeuchi [32].

Since $A$ is an $H$-module algebra by our convention, we say that $A$ is an $H^{*}$-Galois extension of the subalgebra $A^{H}$ if $\gamma$ is bijective. Since $\dim H^{*}<\infty$, it suffices to require surjectivity of $\gamma\,$ [32].

Theorem 3.1 [32]. Suppose that $A$ is an $H^{*}$-Galois extension of $A^{H}$. Then $A$ is a finitely generated projective $A^{H}$-module on the left and on the right.

Proof. As explained in [38, 8.3.1] this conclusion can be proved by verifying the dual basis property which characterizes projective modules. To do this let $t\in H$ be a left integral and $\lambda\in H^{*}$ a right integral such that $\lambda(t)=1$, and let $1\otimes\lambda=\gamma(\,\sum a_{i}\otimes b_{i})$ for some elements $a_{i},b_{i}\in A\,$ ($i=1,\ldots,n$). If $x\in A$, then

[TABLE]

[TABLE]

For each $i$ define a right $A^{H}$-linear map $f_{i}:A\to A^{H}$ by the formula $f_{i}(x)=t(b_{i}x)$. Taking $h=t$ above, we deduce that

[TABLE]

for all $x\in A$. In the proof of the left side version one proceeds similarly replacing $\gamma$ with the map $\,a\otimes b\mapsto\rho(a)\cdot(b\otimes 1)\,$ which is also bijective by [32, Prop. 1.2].   $\mathchar 1027\relax$

Hopf Galois extensions form an important special class of comodule algebras for which more information is available than in general. However, the map $\gamma$ is quite useful, even when $\gamma$ is not bijective. Let us view $A\otimes H^{*}$ as an $A$-bimodule with respect to the left and right actions defined by the rules

[TABLE]

Then $\gamma$ is a homomorphism of $A$-bimodules. In particular, its image is a subbimodule of $A\otimes H^{*}$. Also, $\gamma$ respects certain $H$-module structures. Recall the two natural left actions of $H$ on $H^{*}$ defined by the formulas

[TABLE]

Lemma 3.2. With respect to the $H$-module structures on $A\otimes_{A^{H}}\!A$ and $A\otimes H^{*}$ defined by the formulas

[TABLE]

where $a,b\in A$, $\,h\in H$ and $\xi\in H^{*}$ the map $\gamma$ is a morphism in $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$. On the other hand, $\gamma$ is a morphism in $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ with respect to another pair of $H$-module structures

[TABLE]

Proof. Clearly the $H$-module structures $(*)$ are compatible with the left $A$-module structures, so that $A\otimes_{A^{H}}\!A$ and $A\otimes H^{*}$ become objects of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$. To show that $\gamma$ is $H$-linear with respect to $(*)$ we need only to check that $\rho(b)$ is $H$-invariant for each $b\in A$. In the monoidal category of left $H$-modules the module $H^{*}$ with the action $\rightharpoondown$ is the left dual of $H$ with the action by left multiplications. Therefore

[TABLE]

for each left $H$-module $V$. Taking $V={k}$ with the trivial module structure, we get $(A\otimes H^{*})^{H}\cong\mathop{\rm Hom}\nolimits_{H}(H,A)$. Under this bijection the $H$-linear map $H\to A$, $\,h\mapsto hb$, corresponds to $\rho(b)\in A\otimes H^{*}$. In other words, $(A\otimes H^{*})^{H}=\rho(A)$.

Now $H^{*}$ is an $H$-module algebra with respect to the action $\rightharpoonup$. Hence so too is $A\otimes H^{*}$ with respect to the action given in $(**)$. The map $\rho:A\to A\otimes H^{*}$ is $H$-linear with respect to this action, and so $\rho$ is a homomorphism of $H$-module algebras. In particular, $A\otimes H^{*}\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. Clearly $A\otimes_{A^{H}}\!A\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ too. Finally, the map $\gamma$ is $H$-linear with respect to $(**)$ since the elements in $A\otimes 1$ are $H$-invariant.   $\mathchar 1027\relax$

Lemma 3.3. Put $M=\mathop{\rm Im}\gamma$. Suppose that $M$ is left $A$-free and there exist elements $a_{1},\ldots,a_{n}\in A$ such that $\rho(a_{1}),\ldots,\rho(a_{n})$ form a basis for $M$ over $A\otimes 1$. Then $a_{1},\ldots,a_{n}$ are a basis for $A$ as an $A^{H}$-module with respect to the action by left multiplications. In particular, $A$ is left free of finite rank over $A^{H}$.

Proof. Given $a\in A$, there exist uniquely determined elements $c_{1},\ldots,c_{n}\in A$ such that

[TABLE]

Then $ha=\textstyle\sum\limits c_{i}(ha_{i})$ for all $h\in H$. Taking $h=1$, we get $a=\sum c_{i}a_{i}$. Now let $g\in H$. Since $\rho(a)$ is $H$-invariant and $\gamma$ is $H$-linear with respect to the $H$-module structures $(*)$ considered in Lemma 3.2, we deduce that

[TABLE]

It follows that $gc_{i}=\varepsilon(g)c_{i}$ since $\rho(a_{1}),\ldots,\rho(a_{n})$ are left linearly independent over $A\otimes 1$. Hence $c_{i}\in A^{H}$ for each $i$. On the other hand, if $a=\sum c^{\prime}_{i}a_{i}$ for another collection of elements $c^{\prime}_{1},\ldots,c^{\prime}_{n}\in A^{H}$, then $\rho(a)=\sum\,(c^{\prime}_{i}\otimes 1)\rho(a_{i})$, which entails $c^{\prime}_{i}=c_{i}$ for each $i$.   $\mathchar 1027\relax$

If $A$ is artinian and $H$-simple, then the $A$-bimodule $M=\mathop{\rm Im}\gamma$ is always left (and right) free. Indeed, $M$ may be regarded as an object of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ by Lemma 3.2. Hence $M^{n}$ is left free for some $n>0$ by Theorem 2.1. Moreover, this conclusion holds with $n=1$, as explained in the proof of Theorem 2.9. As a left $A$-module, $M$ is generated by $\rho(A)$, but this does not mean that a basis can be chosen in $\rho(A)$, and so Lemma 3.3 does not apply in general. Here lies the source of possible misbehavior of the subring $A^{H}$.

An example of Björk [7] produces a simple artinian ring $R$ of characteristic 2 which is not a finitely generated module over the subring $R^{G}$ of elements fixed by an automorphism group of order 4. In this example $R^{G}$ is neither left nor right artinian.

The situation becomes much nicer when the $H$-module algebra $A$ has no nontrivial $H$-stable one-sided ideals.

Lemma 3.4. Suppose that $A$ has no nontrivial $H$-stable left (or right) ideals. Then $A^{H}$ is a skew field and $\gamma$ is injective.

Proof. The condition imposed on $A$ means that $A$ is a simple object of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$, i.e. a simple left $A\#H$-module. The fact that $A^{H}$ is a skew field is stated in [4, Lemma 2.1]. It follows from Schur’s lemma since $\,A^{H}\cong(\mathop{\rm End}\nolimits_{A\#H}A)^{\rm op}$.

As explained in Lemma 3.2, $A\otimes_{A^{H}}\!A$ may be regarded as an object of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$. It is a sum of its simple subobjects $A\otimes b$ with $0\neq b\in A$, each isomorphic to $A$. Hence any subobject of $A\otimes_{A^{H}}\!A$ in the category $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ is equal to $A\otimes_{A^{H}}\!V$ for some left vector subspace $V$ of $A$ over $A^{H}$. In particular, this applies to the kernel of $\gamma$. But $\gamma(1\otimes b)=\rho(b)$ for each $b\in A$. This shows that the restriction of $\gamma$ to $1\otimes A$ is injective, and therefore $\,\mathop{\rm Ker}\gamma=0$.   $\mathchar 1027\relax$

Under the hypothesis of Lemma 3.4 the $H$-module algebra $A$ may be regarded as either left or right vector space over the skew field $A^{H}$. Denote by $[A:A^{H}]_{l}$ and $[A:A^{H}]_{r}$ the dimensions of these two vector spaces.

Theorem 3.5 [4]. Suppose that $A$ has no nontrivial $H$-stable left ideals and that $A$ has finite left Goldie rank (i.e., $A$ satisfies ACC on direct sums of left ideals). Then $[A:A^{H}]_{r}\leq n$ where $n$ is the dimension of the image of $H$ in $\mathop{\rm End}\nolimits_{\,{k}}A$.

This result of Bergen, Cohen and Fischman [4, Th. 2.2] was stated for a not necessarily finite dimensional Hopf algebra $H$ with finite dimensional image $\pi(H)$ in $\mathop{\rm End}\nolimits_{\,{k}}A$. In the proof given in [4] an application of Jacobson’s density theorem shows that, whenever $A$ contains $m$ right linearly independent over $A^{H}$ elements, there exists a free left $A$-submodule of rank $m$ in the image $\pi(A\#H)$ of $A\#H$ in $\mathop{\rm End}\nolimits_{\,{k}}A$. Pick $h_{1},\ldots,h_{n}\in H$ whose images give a basis for $\pi(H)$. Then $\pi(A\#H)=\pi(T)$ where $T\subset A\#H$ is the left $A$-submodule generated by $1\#h_{1},\ldots,1\#h_{n}$. It follows that $T$ contains a free $A$-submodule of rank $m$. Since $T$ is a free $A$-module of rank $n$, this entails $m\leq n$ by finiteness of the Goldie rank.

Using the map $\gamma$ we can strengthen the previous theorem. We continue to work under the assumption that $\,\dim H<\infty$. Note, however, that replacing $A\otimes H^{*}$ with $\mathop{\rm Hom}\nolimits_{\,{k}}(H,A)$ in the preceding discussion and modifying all arguments appropriately, the next result can be proved for an infinite dimensional Hopf algebra under the assumption $\,\dim\pi(H)<\infty\,$ used in [4]. In the semilocal case one needs also bijectivity of the antipode.

Theorem 3.6. Suppose that $A$ has no nontrivial $H$-stable left ideals and that either $A$ has finite left Goldie rank or $A$ is semilocal. Let $n=\dim\pi(H)$ where $\pi:H\to\mathop{\rm End}\nolimits_{\,{k}}A$ is the representation afforded by the action of $H$ on $A$. Then

[TABLE]

In particular, $A$ is left and right artinian.

Proof. By Lemma 3.4 $\gamma$ is injective. Thus the $A$-bimodule $M=\mathop{\rm Im}\gamma$ is isomorphic to $A\otimes_{A^{H}}\!A$. Hence $M$ is left free of rank equal to $[A:A^{H}]_{l}$, and $M$ is right free of rank equal to $[A:A^{H}]_{r}$. Note that $\rho(A)\subset A\otimes C$ where $C$ is the subcoalgebra of $H^{*}$ dual to the factor algebra $\pi(H)$ of $H$. Hence $M\subset A\otimes C$ too. Since $\dim C=n$, this shows that $M$, regarded as an $A$-module with respect to the left action, embeds in a free $A$-module of rank $n$. If $A$ has finite left Goldie rank, we get $[A:A^{H}]_{l}\leq n$, while the second inequality is the content of Theorem 3.5.

Suppose further that $A$ is semilocal. Since $C$ is stable under the action $\rightharpoondown$ of $H$ on $H^{*}$, the left $A$-module $A\otimes C$ is an object of $H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ with respect to the $H$-module structure described in $(*)$ of Lemma 3.2. Hence $(A\otimes C)/M\in H\hbox{-}\vphantom{{\cal M}}_{A}{\cal M}$ too. Since $A$ is $H$-simple, Theorem 2.1 shows that $(A\otimes C)/M$ is projective in $\vphantom{{\cal M}}{}_{A}{\cal M}$. Then $M$ is an $\vphantom{{\cal M}}{}_{A}{\cal M}$-direct summand of $A\otimes C$. Denoting by $J$ the Jacobson radical of $A$, we deduce that the free $A/J$-module $M/JM$ of rank equal to $[A:A^{H}]_{l}$ embeds into the free $A/J$-module $A/J\otimes C$ of rank $n$. Since $A/J$ is semisimple artinian, we must have $[A:A^{H}]_{l}\leq n$.

For the remaining part put N=\bigl{(}1\otimes S^{-1}(C)\bigr{)}\cdot\rho(A). Then $N$ is a subobject of $A\otimes H^{*}$ in the category $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$, and $N$ is $A$-free of rank $n$. Let $a\in A$. Writing symbolically $\rho(a)=\sum a_{(0)}\otimes a_{(1)}\in A\otimes C$, we have

[TABLE]

This shows that $A\otimes 1\subset N$. Hence $M=(A\otimes 1)\,\rho(A)\subset N$. Applying Theorem 2.1 to the object $N/M\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$, we deduce that $M$ is an ${\cal M}_{A}$-direct summand of $N$, and passing to quotients modulo $J$, we arrive at $\,[A:A^{H}]_{r}\leq n$.   $\mathchar 1027\relax$

Another old result of Cohen, Fischman and Montgomery determines when the extension $A/A^{H}$ is $H^{*}$-Galois for an $H$-module algebra satisfying the previous assumptions. We will show how the map $\gamma$ can be used in the proof.

Theorem 3.7 [13]. Suppose that $A$ has no nontrivial $H$-stable left ideals and that $A$ has finite left Goldie rank. Then the following conditions are equivalent:**

(1) $[A:A^{H}]_{r}=\dim H$.

($1^{\prime}$) $[A:A^{H}]_{l}=\dim H$.

(2) $A$ is a faithful left $A\#H$-module.

(3) $A\#H$ is a simple algebra.

(4) $A$ is an $H^{*}$-Galois extension of $A^{H}$.

Proof. The map $\gamma$ is an injective homomorphism of $A$-bimodules. Both $A\otimes_{A^{H}}\!A$ and $A\otimes H^{*}$ are free of finite rank as left $A$-modules and as right ones. Since $A$ is left and right artinian by Theorem 3.6, all finitely generated $A$-modules have finite length. Therefore $\gamma$ is bijective if and only if the two bimodules have equal left ranks, and if and only if they have equal right ranks. This shows that $(4)\Leftrightarrow(1^{\prime})\Leftrightarrow(1)$.

Denote by $E$ the endomorphism ring of $A$ as a right vector space over the skew field $A^{H}$. The action of $A\#H$ on $A$ gives a ring homomorphism $\pi:A\#H\to E$. Since $[A:A^{H}]_{r}<\infty$, the ring $E$ is simple artinian. By one of general characterizations of Galois extensions condition (4) holds if and only if $\pi$ is an isomorphism [13, Th. 1.2]. But $\pi$ is surjective by Jacobson’s density theorem since $A$ is a simple left $A\#H$-module. Hence $\pi$ is an isomorphism if and only if $\mathop{\rm Ker}\pi=0$. If $\pi$ is an isomorphism then $A\#H\cong E$ is simple. On the other hand, if $A\#H$ is simple then $\mathop{\rm Ker}\pi=0$. Thus $(4)\Leftrightarrow(2)\Leftrightarrow(3)$.   $\mathchar 1027\relax$

Condition $(1^{\prime})$ has not been included as one of equivalent conditions in [13, Th. 3.3], but it appears in [13, Cor. 3.10] stated for the case when the $H$-module algebra is a division ring. The proof of the equivalence $(1)\Leftrightarrow(2)$ given in [13] is not quite clear since it is based on the equalities

[TABLE]

where $K$ is the kernel of the representation $\pi:A\#H\to\mathop{\rm End}\nolimits_{\,{k}}A$, $\,m=[A:A^{H}]_{r}$, and $M_{m}(A^{H})$ is the ring of $m\times m$ matrices with entries in $A^{H}$. There is a ring isomorphism $A\#H/K\cong M_{m}(A^{H})$, but the natural embedding of $A^{H}$ in $A\#H/K$ may not correspond to the embedding of $A^{H}$ in $M_{m}(A^{H})$ as the subring of diagonal matrices. So one has to deal with two different images $D$, $D^{\prime}$ of the skew field $A^{H}$ in $M_{m}(A^{H})$. Although $D\cong D^{\prime}$, this does not necessarily imply that

[TABLE]

We have given a proof which avoids the difficulty arising here.

If $A$ is semiprimary, then $A$ has no nontrivial $H$-stable left ideals if and only if $A$ has no nontrivial $H$-stable right ideals. Indeed, each of the two conditions imply that $A$ is $H$-simple, but then $A$ is a quasi-Frobenius ring by Theorem 2.8. By a standard property of quasi-Frobenius rings there is a bijective correspondence between left and right ideals of $A$ which assigns to each left ideal its right annihilator in $A$ and to each right ideal its left annihilator in $A$. It is easy to see that $H$-stable left ideals correspond to $H$-stable right ideals.

The question concerning equality of the left and right dimensions of $A$ over $A^{H}$ has been settled only recently. This question was raised in the already mentioned paper of Bergen, Cohen and Fischman for an $H$-module algebra which is a division ring, i.e. a skew field  [4, Question 2.4]. It was motivated by the classical result, due to Jacobson, that such an equality is indeed true in the case of group actions on skew fields.

Theorem 3.8 [54]. Suppose that $A$ is a semiprimary $H$-module algebra without nontrivial $H$-stable one-sided ideals. Then $\,[A\,\mathord{:}\,A^{H}]_{l}=[A\,\mathord{:}\,A^{H}]_{r}$.

Let us outline the proof of this theorem. The desired equality holds precisely when the $A$-bimodule $M=A\otimes_{A^{H}}\!A$ has equal left and right ranks over $A$. By Lemma 3.4 $\gamma$ embeds $M$ into $N=A\otimes H^{*}$. The latter bimodule is also free on each side with the left and right ranks equal to $\dim H$. The conclusion will follow once it is shown that a direct sum of several copies of $M$ is isomorphic to a direct sum of several copies of $N$.

Each $A$-bimodule may be regarded as a right module over the ring ${\cal A}=A^{\rm op}\otimes\nobreak A$. Put ${\cal H}=H^{\rm cop}\otimes H$ and consider $M$ and $N$ as left ${\cal H}$-modules using the two pairs of $H$-module structures described in Lemma 3.2. Note that ${\cal H}$ is a finite dimensional Hopf algebra and ${\cal A}$ is an ${\cal H}$-module algebra. One checks the compatibility condition which makes $M$ and $N$ objects of the category ${\cal H}\hbox{-}\mskip 1.0mu{\cal M}_{\cal A}$.

Put $B=\mathop{\rm End}\nolimits_{\cal A}M$, i.e. $B$ is the endomorphism ring of $M$ as an $A$-bimodule. Then $B$ is an ${\cal H}$-module algebra, and $B$ is semiprimary since $M$ is an $A$-bimodule of finite length. By the assumption on the $H$-stable one-sided ideals of $A$ there cannot exist nontrivial subbimodules of $M$ stable under the two $H$-module structures of Lemma 3.2. In other words, $M$ is a simple object of ${\cal H}\hbox{-}\mskip 1.0mu{\cal M}_{\cal A}$. This implies $H$-semiprimeness of $B$, and an application of Theorem 2.4 leads further to the conclusion that $B$ is $H$-simple.

As a left $A$-module, and therefore as a bimodule, $N$ is generated by $1\otimes H^{*}$. For each $\xi\in H^{*}$ there is a homomorphism of $A$-bimodules $\varphi_{\xi}:M\to N$ sending $a\otimes b\in A\otimes_{A^{H}}\!A\,$ to $(a\otimes\xi)\,\rho(b)$ (note that $(1\otimes\xi)\,\rho(c)=c\otimes\xi$ for all $c\in A^{H}$), and $A\otimes\xi$ is contained in the image of $\varphi_{\xi}$. It follows that $N$, as an $A$-bimodule, is a homomorphic image of $M^{d}$ where $d=\dim H$. This means that the canonical map

[TABLE]

is surjective. The trickiest part is to show that $\mathop{\rm Ker}\alpha=0$. We refer the reader to [54, Th. 3.1] for details.

Thus $N\cong F\otimes_{B}M$ where $F=\mathop{\rm Hom}\nolimits_{\cal A}(M,N)$. Note that $F$ is an ${\cal H}$-equivariant right $B$-module. By Theorem 2.1 $F^{n}$ is a free $B$-module for some $n>0$. Since $M$ and $N$ are ${\cal A}$-finite, we deduce that $F$ is $B$-finite. Hence $F^{n}\cong B^{r}$ in ${\cal M}_{B}$ for some integer $r>0$. This entails $N^{n}\cong F^{n}\otimes_{B}M\cong B^{r}\otimes_{B}M\cong M^{r}$ in ${\cal M}_{\cal A}$, completing the proof.

In some cases the conclusion of Theorem 3.8 is almost obvious. Assume that $A$ is a skew field. If all $A$-bimodule composition factors of $N$ have equal left and right dimensions over $A$, the equality of the left and right dimensions over $A$ will be fulfilled for each subbimodule of $N$. This happens when $H$ is pointed and, more generally, when all simple $H$-comodules have dimension at most 2 over the ground field ${k}$. To see this note that $A\otimes I$ is a subbimodule of $N$ for each right ideal $I$ of $H^{*}$. Taking a composition series of $H^{*}$ as a right module over itself, we get a series of subbimodules of $A\otimes H^{*}$ with factors isomorphic to $A\otimes V$ for various simple right $H^{*}$-modules $V$ where the right action of $A$ on $A\otimes V$ comes from $\rho$. We have $\dim V\leq 2$ for each $V$ by the previous assumption about $H$. The left and right dimensions of $A\otimes V$ over $A$ are both equal to $\dim V$. If $\dim V=1$ then $A\otimes V$ cannot contain any nontrivial subbimodules. If $\dim V=2$ then the $A$-bimodule $A\otimes V$ is either simple or has exactly two composition factors, each of left and right dimension over $A$ equal to 1.

A different approach to finiteness results exploits the trace ${\hat{t}}:A\to A^{H}$ given by the action of a left integral $t\in H$ on $A$. Note that ${\hat{t}}$ is left and right $A^{H}$-linear. In particular, ${\hat{t}}$ is surjective if and only if $\,ta=1\,$ for some $a\in A$.

In terms of comodule structure on $A$ surjectivity of ${\hat{t}}$ is equivalent to the existence of a total integral $H^{*}\to A$, which is a homomorphism of right $H^{*}$-comodules sending $1\in H^{*}$ to $1\in A$ (see [12]). Total integrals were introduced by Doi [19] for comodule algebras over arbitrary Hopf algebras.

Theorem 3.9 [38]. Suppose that $A$ is right noetherian with a surjective trace ${\hat{t}}$. Then $A$ is a noetherian right $A^{H}$-module. In other words, $A^{H}$ is right noetherian and $A$ is right module-finite over $A^{H}$.

This result is stated in [38, Th. 4.4.2]. To prove the theorem Montgomery shows that the lattice of $A^{H}$-submodules of any right $A$-module $V$ embeds into the lattice of submodules of the induced right $A\#H$-module. In particular, the lattice of right $A^{H}$-submodules of $A$ embeds into the lattice of right ideals of $A\#H$. Explicitly, this embedding is obtained by assigning to an $A^{H}$-submodule $U$ of $A$ the right ideal of $A\#H$ generated by $(U\#1)e$ where $e\in A\#H$ is an idempotent such that $A^{H}$ is isomorphic to $e(A\#H)e$.

The same argument shows that $A^{H}$ is right artinian if so is $A$.

The lemma below describes embeddings of lattices of submodules from our point of view on equivariant modules.

Lemma 3.10. Let $M\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$. If the trace map ${\hat{t}}:A\to A^{H}$ is surjective, then:**

*(i) *The lattice of $A^{H}$-submodules of $M^{H}$ embeds into that of $A$-submodules of $M$.

(ii) $M^{H}$ is an artinian $A^{H}$-module whenever $M$ is an artinian $A$-module.

(iii) $M^{H}$ is a noetherian $A^{H}$-module whenever $M$ is a noetherian $A$-module.

Proof. We have $t(vA)=vA^{H}$ for any $v\in M^{H}$ since the map $A\to M$ such that $a\mapsto va$ is $H$-linear. Hence $t(UA)=U$ for each $A^{H}$-submodule $U$ of $M^{H}$. The assignment $U\mapsto UA$ gives therefore the desired embedding of lattices. Assertions (ii) and (iii) are immediate from (i).   $\mathchar 1027\relax$

If $A$ is right noetherian, then each finitely generated right $A$-module is noetherian. Hence, if $M$ is an $A$-finite object of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$, then $M^{H}$ is a noetherian right $A^{H}$-module by Lemma 3.10. To deduce Theorem 3.9 from Lemma 3.10 one needs to find an $A$-finite object $M\in H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ such that $M^{H}\cong A$ as right $A^{H}$-modules. Recall that objects of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ are identified with left $A^{\rm op}\#H^{\rm cop}$-modules. Let $M$ be a cyclic free $A^{\rm op}\#H^{\rm cop}$-module with a free generator $v$. Since the linear map $H\otimes A\to M$ such that $h\otimes a\mapsto h(va)$ is bijective, it follows that $M^{H}=tM=t(vA)$. Hence the assignment $a\mapsto t(va)$ defines a desired right $A^{H}$-linear isomorphism $A\to M^{H}$.

Proposition 3.11. Suppose that $H$ is semisimple and that $A$ is semiprimary and $H$-semiprime. Then:**

*(i) *Each $H$-stable one-sided ideal of $A$ is generated by an $H$-invariant idempotent.

*(ii) *The subring of invariants $A^{H}$ is semisimple artinian.

(iii) $A$ is left and right module-finite over $A^{H}$.

Proof. By Theorem 2.8 $A$ is artinian, and by Theorem 2.10 all objects of $H\hbox{-}\mskip 1.0mu{\cal M}_{A}$ are semisimple. In particular, the latter conclusion applies to $A$. This means that, whenever $I$ is an $H$-stable right ideal of $A$, there exists an $H$-stable right ideal $J$ such that $A=I\oplus J$. Then $I=eA$ for some idempotent $e\in A$. We have $e=p(1)$ where $p$ is the projection of $A$ onto $I$ with kernel $J$. Since $1\in A^{H}$ and $p$ is $H$-linear, it follows that $e\in A^{H}$. This proves assertion (i) for right ideals. Considering the $H^{\rm cop}$-module algebra $A^{\rm op}$, we get (i) for left ideals as well.

Let $U$ be any right ideal of $A^{H}$. Then $UA$ is an $H$-stable right ideal of $A$. By (i) $UA$ is generated by an idempotent $e\in A^{H}$. By Lemma 3.10 applied to $M=A$ the lattice of right ideals of $A^{H}$ embeds into that of right ideals of $A$. Since the right ideal $eA^{H}$ of $A^{H}$ has the same extension to $A$ as $U$, we deduce that $U=eA^{H}$. Thus each right ideal of $A^{H}$ is generated by an idempotent. This yields (ii). Finally, (iii) follows from Theorem 3.9.   $\mathchar 1027\relax$

We have given a self-contained proof. It has been known for a long time that $A^{H}$ is semisimple artinian whenever so is $A\#H$ [13, Th. 3.13]. That $A\#H$ is semisimple artinian, under the hypothesis of Proposition 3.11, has been established in [55] (see Theorem 2.10).

4. Localization at invariants and a Bergman-Isaacs type theorem

In [52] it was shown that for a semisimple Hopf algebra $H$ all right noetherian $H$-module algebras have, loosely speaking, “sufficiently many” $H$-invariant elements. The proofs of these results referred to a few statements from [55] which served as intermediate steps in the process of verifying the existence of artinian classical quotient rings. But in fact only the final conclusion from [55] is needed, and therefore those results of [52] are valid for a larger class of $H$-module algebras. This will be explained below.

Suppose that an $H$-semiprime $H$-module algebra $A$ has a right artinian classical right quotient ring $Q$. By Theorem 2.14 the $H$-module structure extends to $Q$. It is clear then that $Q$ has to be $H$-semiprime since $J\cap A\neq 0$ for each nonzero right ideal $J$ of $Q$. Therefore all results concerning artinian $H$-semiprime algebras apply to $Q$.

A right ideal $I$ of a ring $R$ is called essential if $I$ has nonzero intersection with each nonzero right ideal of $R$.

Lemma 4.1. Suppose that $A$ is $H$-semiprime and $A$ has a right artinian classical right quotient ring $Q$. For a right ideal $I$ of $A$ denote by $I_{H}$ the largest $H$-stable right ideal of $A$ contained in $I$. The following conditions are equivalent:**

(a) $I_{H}$ is an essential right ideal of $A$,

(b) $I$ contains a regular element of $A$.

Proof. Suppose that $I_{H}$ is an essential right ideal of $A$. Then $I_{H}Q$ is an essential right ideal of $Q$. But $I_{H}Q$ is $H$-stable, and therefore this right ideal is generated by an idempotent $e\in Q$ according to Corollary 2.7. Since $eQ\cap(1-e)Q=0$, we must have $e=1$, i.e $I_{H}Q=Q$. Then $1=as^{-1}$ for some $a\in I_{H}$ and a regular element $s\in A$. So $s=a\in I_{H}$, which proves that (a)${}\Rightarrow{}$(b).

Suppose now that $I$ contains a regular element $s$ of $A$. Note that

[TABLE]

We have to prove that $I_{H}$ is an essential right ideal of $A$. Suppose that $I_{H}\cap bA=0$ for some $b\in A$, $\,b\neq 0$. Then $(I\otimes H^{*})\cap\rho(bA)=0$. In particular,

[TABLE]

Since $s\otimes 1$ is a regular element of the ring $A\otimes H^{*}$, it follows that the sum

[TABLE]

is direct, and each summand is a nonzero right $\rho(A)$-submodule of $A\otimes H^{*}$. Consider now $F=A\otimes H^{*}$ as a right $A$-module with respect to the action of $A$ given by right multiplications by the elements $\rho(a)$, $\,a\in A$. We know that this $A$-module is free of rank equal to the dimension of $H$. Hence $F\otimes_{A}Q$ is a finitely generated right $Q$-module containing an infinite direct sum of nonzero submodules. However, this is impossible since $Q$ is right artinian. Thus $I_{H}\cap bA\neq 0$ whenever $b\neq 0$, and so (b)$\Rightarrow$(a).   $\mathchar 1027\relax$

Lemma 4.2. Suppose that $H$ is semisimple, $A$ is $H$-semiprime, and $A$ has a right artinian classical right quotient ring $Q$. Let $I$ be an $H$-stable right ideal of $A$, and let $\,I^{H}=I\cap A^{H}$.

*(i) *If $I^{H}=0$ then $I=0$.

*(ii) *If $I$ is an essential right ideal of $A$ then $Q^{H}I^{H}=Q^{H}$ and $I^{H}Q^{H}=Q^{H}$.

Proof. By Proposition 3.11 $Q^{H}$ is semisimple artinian and $Q$ has finite length as either left or right $Q^{H}$-module.

Consider the $(Q^{H}\!\!,A)$-subbimodule $Q^{H}I$ of $Q$. For each regular element $s$ of $A$ the $Q^{H}$-submodule $Q^{H}Is$ is isomorphic to $Q^{H}I$. Hence these two $Q^{H}$-modules have the same length. Since $Q^{H}Is\subset Q^{H}I$, we get $Q^{H}Is=Q^{H}I$, and therefore $Q^{H}Is^{-1}=Q^{H}I$.

It follows that $Q^{H}I$ is a right ideal of $Q$. Since it is $H$-stable, Proposition 3.11 yields $Q^{H}I=eQ$ for some $e\in Q^{H}$. Let $t\in H$ be an integral. The action of $t$ on $Q$ commutes with the left and right multiplications by $H$-invariant elements. Hence

[TABLE]

If $I^{H}=0$, the above inclusion entails $e=0$, i.e. $I=0$. This proves (i).

Suppose that $I$ is an essential right ideal of $A$. By Lemma 4.1 $I=I_{H}$ contains a regular element of $A$, so that $IQ=\nobreak Q$. Since $Q^{H}I$ is a right ideal of $Q$ containing $I$, we get $Q^{H}I=Q$ as well. The previous argument with $e=1$ shows that $1\in Q^{H}I^{H}$. Thus $Q^{H}=Q^{H}I^{H}$.

Consider now the $H$-stable right ideal $I^{H}Q$ of $Q$. By Theorem 2.10 there exists an $H$-stable right ideal $J$ of $Q$ such that $Q=I^{H}Q\oplus J$. Then $I\cap J$ is an $H$-stable right ideal of $A$ with

[TABLE]

As we have proved already in part (i) this entails $I\cap J=0$. Since $IQ=Q$, any element $y\in J$ can be written as $y=as^{-1}$ where $a\in I$ and $s$ is a regular element of $A$; then $a\in I\cap J$, so that $a=0$ and $y=0$. Therefore $J=0$ and $Q=I^{H}Q$. Hence

[TABLE]

and we are done.   $\mathchar 1027\relax$

Theorem 4.3. Suppose that $H$ is semisimple, $A$ is $H$-semiprime, and $A$ has a right artinian classical right quotient ring $Q$. Denote by $\Sigma$ the set of regular elements of $A^{H}$ and by ${\cal E}$ the set of right ideals of $A$ which satisfy the equivalent conditions (a) and (b) of Lemma 4.1. Then:**

*(i) *The algebra $A^{H}$ is semiprime right Goldie.

(ii) $\Sigma$ is a right Ore subset of regular elements of $A$.

(iii) $Q$ is canonically isomorphic with the right localization of $A$ at $\Sigma$.

*(iv) *The classical right quotient ring of $A^{H}$ is isomorphic with $Q^{H}$.

(v) $I\cap\Sigma\neq\mathchar 1343\relax$ for each right ideal $I\in{\cal E}$.

Proof. Put $I^{H}=I\cap A^{H}$ and ${\cal F}=\{I^{H}\mid I\in{\cal E}\}$. If $I,J\in{\cal E}$ then $I\cap J\in{\cal E}$, whence $I^{H}\cap J^{H}=(I\cap J)^{H}\in{\cal F}$. Therefore ${\cal F}$ is a filter of right ideals of $A^{H}$. If $I\in{\cal E}$ and $a\in A^{H}$, then the right ideal $I_{a}=\{x\in A\mid ax\in I_{H}\}$ is essential and $H$-stable; so $I_{a}\in{\cal E}$ and $I_{a}^{H}\in{\cal F}$. Also, $aI_{a}^{H}\subset I^{H}$ since $aI_{a}\subset I_{H}\subset I$. This shows that ${\cal F}$ is a topologizing filter.

Moreover, ${\cal F}$ satisfies conditions (a) and (b) of Proposition 2.13 with $A$ replaced by $A^{H}$ and $Q$ replaced by $Q^{H}$. Indeed, if $I\in{\cal E}$ then $I^{H}$ has zero left and right annihilators in $Q$ since $I^{H}Q$ and $QI^{H}$ contain the unity 1 by Lemma 4.2. If $q\in Q^{H}$ there exists a regular element $s$ of $A$ such that $qs\in A$. Put $K=sA$. Then $K\in{\cal E}$ by Lemma 4.1 and $qK\subset A$. Hence $K^{H}\in{\cal F}$ and $qK^{H}\subset A^{H}$.

It has been observed already that the ring $Q^{H}$ is semisimple artinian. Now (iv) follows from Proposition 2.13, and (i) is its consequence since a ring $R$ is semiprime right Goldie if and only if $R$ has a semisimple artinian classical right quotient ring.

Given $I\in{\cal E}$, the equality $I^{H}Q^{H}=Q^{H}$ of Lemma 4.2 means that $I^{H}\cap\Sigma\neq\mathchar 1343\relax$, which amounts to (v). For any $q\in Q$ the set

[TABLE]

is a right ideal of $A$ containing a regular element of $A$. By Lemma 4.1 $I\in{\cal E}$. Since $qI\subset A$, assertion (v) shows that $qs\in A$ for some $s\in\Sigma$.

All elements of $\Sigma$ are invertible in $Q^{H}$ and therefore in $Q$. Hence all elements of $\Sigma$ are regular in $A$. Since each element of $Q$ can be written in the form $as^{-1}$ for some $a\in A$ and $s\in\Sigma$, (ii) and (iii) are immediate (see [35, 2.2.4]).   $\mathchar 1027\relax$

In all corollaries below we continue to assume that $H$ is semisimple.

Corollary 4.4. Let $A$ be as in Theorem 4.3. If $s$ is any regular element of $A$, then the right ideal $sA$ contains an $H$-invariant regular element of $A$.

Proof. Since $sA\in{\cal E}$ by Lemma 4.1, we have $sA\cap\Sigma\neq\mathchar 1343\relax$ by Theorem 4.3.   $\mathchar 1027\relax$

Corollary 4.5. Let $A$ and $Q$ be as in Theorem 4.3, and let $I$ be any $H$-stable right ideal of $A$. There exists $x\in I^{H}$ such that $IQ=xQ$.

Proof. Since $IQ$ is an $H$-stable right ideal of $Q$, we have $IQ=eQ$ for some $e\in Q^{H}$ by Proposition 3.11. Now $J=\{a\in A\mid ea\in I\}$ is an $H$-stable right ideal of $A$ containing a regular element of $A$. Hence $J\cap\Sigma\neq\mathchar 1343\relax$ by Corollary 4.4. Pick any $s\in J\cap\Sigma$ and put $x=es$. Then $x\in I\cap Q^{H}=I^{H}$. Since $s$ is invertible in $Q$, we get $eQ=xQ$.   $\mathchar 1027\relax$

Corollary 4.6. All conclusions of Theorem 4.3, as well as Corollaries 4.4, 4.5, hold in each of the following three cases:**

(a) $A$ is semiprime right Goldie,

(b) $A$ is right noetherian and $H$-semiprime,

(c) $A$ is PI and $H$-semiprime with finitely many minimal $H$-prime ideals.

Proof. A right artinian classical right quotient ring $Q$ exists in case (a) by the Goldie theorem, in the other cases by Theorems 2.11, 2.12.   $\mathchar 1027\relax$

Under the assumption that $A\#H$ is semiprime, a short argument given by Bergen and Montgomery [5, Prop. 2.4] shows that $A^{H}$ is semiprime, and that ${\hat{t}}(I)\neq 0$ where ${\hat{t}}:A\to A^{H}$ is the trace map (in particular, $I^{H}\neq 0$) for each nonzero $H$-stable one-sided ideal $I$ of $A$. From this it was further deduced in [5, Lemma 3.4] that, among other things, regular elements of $A^{H}$ are regular in $A$, and that $A^{H}$ is Goldie when so is $A$. If $A,Q,H$ are as in Theorem 4.3, then $Q\#H$ is semisimple artinian by Theorem 2.10; since $Q\#H$ is a classical right quotient ring of $A\#H$, it follows that $A\#H$ is semiprime. This fact was not known at the time when [5] was written.

Several deeper results from [5] use the assumption that $A\#H$ is not only semiprime, but has the ideal intersection property (I I P for short) which means that each nonzero ideal of $A\#H$ has nonzero intersection with $A$. In fact, in the presence of I I P the ring $A\#H$ is semiprime if and only if $A$ is $H$-semiprime. The I I P is satisfied for $X$-outer group actions on semiprime rings and for $X$-outer actions of Lie algebras on prime rings. However, it seems that there are no approaches to analogs of such results for actions of arbitrary finite dimensional or even semisimple Hopf algebras.

It was asked in [5] whether $Q(A)^{H}=Q(A^{H})$ when $A\#H$ is semiprime with I I P. Part (iv) of Theorem 4.3 answers this question, imposing reasonable conditions on $A$ and $H$, but not assuming the I I P. In the case of a finite group $G$ acting on a semiprime ring $R$ without additive $|G|$-torsion the fact that $R^{G}$ is right Goldie if and only if $R$ is right Goldie and the equality $Q(R)^{G}=Q(R^{G})$ were proved by Kharchenko [31]; it was also observed by Montgomery [36] that $Q(R)$ is the localization of $R$ at the Ore set of regular $G$-invariant elements. Analogs of these results for group graded rings are due to Cohen and Rowen [14].

There is a slightly weaker version of Lemma 4.2 for $H$-stable left ideals. The equality $Q^{H}I^{H}=Q^{H}$ in (ii) cannot be proved unless $Q$ is a two-sided quotient ring.

Lemma 4.7. Suppose that $H$ is semisimple, $A$ is $H$-semiprime, and $A$ has a right artinian classical right quotient ring $Q$. Let $I$ be an $H$-stable left ideal of $A$, and let $\,I^{H}=I\cap A^{H}$.

*(i) *If $I^{H}=0$ then $I=0$.

*(ii) *If $QI=Q$ then $I^{H}Q^{H}=Q^{H}$.

Proof. We repeat the steps in the proof of Lemma 4.2 using the two-sided properties of $Q$. First, the $(A,Q^{H})$-subbimodule $IQ^{H}$ is a left ideal of $Q$ since $Q$ has finite length as a right $Q^{H}$-module. Next, $IQ^{H}=Qe$ for some $e\in Q^{H}$ by Proposition 3.11. Applying the integral $t$, we deduce that $e\in I^{H}Q^{H}$.

If $I^{H}=0$, then $e=0$, whence $I=0$. If $QI=Q$, then $IQ^{H}=Q$, and therefore $1\in I^{H}Q^{H}$.   $\mathchar 1027\relax$

Theorem 4.8. Denote by $N$ the prime radical of $A$ and by $N_{H}$ the largest $H$-stable ideal of $A$ contained in $N$. Suppose that $H$ is semisimple, $N$ is nilpotent, and $A/N_{H}$ has a right artinian classical right quotient ring. If $I$ is any $H$-stable one-sided ideal of $A$ such that $I^{H}$ is nilpotent, then $I$ is nilpotent.

Proof. Denote by $\pi$ the canonical surjective homomorphism of $H$-module algebras $A\to A/N_{H}$. Then $\pi(I)$ is an $H$-stable one-sided ideal of $A/N_{H}$ and $\pi$ maps $I^{H}$ onto $\pi(I)^{H}\!$ since $H$ is semisimple. Hence $\pi(I)^{H}$ is nilpotent. The factor algebra $A/N_{H}$ is $H$-semiprime since $N_{H}$ is an $H$-semiprime ideal of $A$. Hence its subring of invariants $(A/N_{H})^{H}$ is semiprime by Theorem 4.3, which entails $\pi(I)^{H}=0$. An application of Lemmas 4.2, 4.7 yields $\pi(I)=0$, i.e. $I\subset N$. So $I$ is nilpotent.   $\mathchar 1027\relax$

Corollary 4.9. Suppose that $H$ is semisimple. The conclusion of Theorem 4.8 holds in each of the following three cases:**

(a) $A$ is left noetherian,

(b) $A$ is right noetherian,

(c) $A$ is finitely generated and PI.

Proof. In all cases $N$ is known to be nilpotent. In cases (b) and (c) the $H$-semiprime factor algebra $A/N_{H}$ has a right artinian classical right quotient ring by Theorems 2.11, 2.12. In case (a) we apply Theorem 4.8 to the right noetherian $H^{\rm cop}$-module algebra $A^{\rm op}$.   $\mathchar 1027\relax$

Let $R$ be a nonunital ring and $G$ a finite group of its automorphisms such that $R$ has no additive $|G|$-torsion. Consider the trace map

[TABLE]

A classical result of Bergman and Isaacs [6] says that $R$ is nilpotent if ${\hat{t}}(R)$ is nilpotent. Moreover, if ${\hat{t}}(R)=0$, then the nilpotency index of $R$ is bounded by a number which depends only on the order $|G|$ of the group $G$, but not on the ring $R$. An easy consequence of this result is that $R^{G}$ is semiprime when so is $R$.

A similar result for group graded rings proved by Cohen and Rowen [14] is even simpler: if $R=\bigoplus_{g\in G}R_{g}$ is a nonunital ring graded by a finite group $G$, and if $R_{1}^{d}=0$ for some integer $d>0$, then $R^{\,d|G|}=0$. In fact one needs only a grading with finite support, while the group $G$ may be infinite.

Theorem 4.8 is different not only in that the conclusion is stated for one-sided ideals of $H$-module algebras satisfying certain conditions, but also because it relies heavily on the $H$-semiprime case. The nilpotency index of $I$ can be bounded only by the nilpotency index of the ideal $N_{H}$, even when $I^{H}=0$.

Bahturin and Linchenko [3] investigated conditions under which one can conclude that $A$ is PI, knowing that $A^{H}$ is PI. They showed that, for a fixed finite dimensional Hopf algebra $H$, in order that each $H$-module algebra $A$ be PI whenever $A^{H}$ is PI it is necessary and sufficient that there exist a natural number $n$ such that $A^{n}=0$ for each nonunital $H$-module algebra $A$ with $A^{H}\cdot A^{H}=0$, and this can happen only if $H$ is semisimple. Several other equivalent conditions are given in [3]. This work of Bahturin and Linchenko elucidates the need for a more precise analog of the Bergman-Isaacs result for Hopf algebra actions.

5. Hopf actions on commutative algebras

Throughout this section we assume that $A$ is a commutative $H$-module algebra. First we are going to recall the algebraic interpretation of the classical result on quotients of affine schemes by actions of finite group schemes.

Given an associative algebra $U$ over a commutative ring $R$ such that $U$ is free of finite rank as an $R$-module the norm $\mathop{\rm Nm}\nolimits_{U/R}(u)\in R$ of an element $u\in U$ is defined as the determinant of the operator $L_{u}\in\mathop{\rm End}\nolimits_{R}U$ of the left multiplication by $u$ in $U$. Considering the polynomial ring $U[t]$, where $t$ is an indeterminate, as an algebra over $R[t]$ we get also the characteristic polynomial

[TABLE]

In particular, $(-1)^{r}\mathop{\rm Nm}\nolimits_{U/R}(u)$ where $r=\mathop{\rm rank}\nolimits_{R}U$ is the coefficient of $t^{0}$ in this polynomial. Passing to localizations of the base ring $R$ these definitions extend to the case where $U$ is not free as an $R$-module, but only projective of finite constant rank.

By the Cayley-Hamilton theorem $P_{U/R}(u,L_{u})=0$ in $\mathop{\rm End}\nolimits_{R}U$. Applying this operator to the identity element $1\in U$, we get $P_{U/R}(u,u)=0$ in $U$, which is a relation of integral dependence of the element $u$ over the ring $R$. The integral dependence of $U$ over $R$ is merely a consequence of module-finiteness. What is important for application to invariants is the fact that the characteristic polynomials enjoy several nice properties. In particular, they are functorial in the sense that, given a homomorphism of commutative rings $\zeta:R\to R^{\prime}$, we have

[TABLE]

where $\zeta^{t}:R[t]\to R^{\prime}[t]$ is the homomorphism extending $\zeta$ and sending $t$ to $t$.

Since $A$ is commutative, the map

[TABLE]

is an isomorphism of $A$ onto a central subalgebra of $A\otimes H^{*}$. So we may regard $A\otimes H^{*}$ as an $A$-algebra via $\iota$. Clearly this algebra is free of rank $d=\dim H$ as an $A$-module. In this way the polynomial $P_{A\otimes H^{*}/A}(u,t)$ is defined for each $u\in A\otimes H^{*}$. Making use of the comodule structure $\rho:A\to A\otimes H^{*}$, we get the polynomial

[TABLE]

for $a\in A$. Suppose that

[TABLE]

Then $\rho(a)^{d}+\sum_{i=0}^{d-1}(c_{i}\otimes 1)\rho(a)^{i}=0$ in $A\otimes H^{*}$. Applying to the left hand side of this equality the algebra homomorphism ${\rm id}\otimes\varepsilon:A\otimes H^{*}\to A$, we get

[TABLE]

If $c_{i}\in A^{H}$ for all $i$, the relation above shows that $a$ is integral over $A^{H}$. On this observation the classical argument reproduced in the following theorem is based:

Theorem 5.1. Suppose that $H$ is cocommutative. Then for each $a\in A$ the characteristic polynomial \,P_{A\otimes H^{*}/A}\bigl{(}\rho(a),t\bigr{)}\, has all coefficients in $A^{H}$. In particular, $A$ is integral over $A^{H}$.

Proof. By the condition imposed on $H$ the dual Hopf algebra $H^{*}$ is commutative, whence $A\otimes H^{*}$ is commutative as well. Since $A^{H}$ is the equalizer of the two algebra homomorphisms $\,\iota,\,\rho:A\to A\otimes H^{*}$, we have to show that

[TABLE]

Note that the commutative diagrams

[TABLE]

where $\iota^{\prime}(x)=x\otimes 1$ for $x\in A\otimes H^{*}$ are cocartesian in the category of commutative $A$-algebras in the sense that each diagram makes $A\otimes H^{*}\otimes H^{*}$ the tensor product of two $A$-algebras given by the respective homomorphisms $A\to A\otimes H^{*}$. Hence $(*)$ can be rewritten as

[TABLE]

by functoriality of the characteristic polynomials. Here $A\otimes H^{*}\otimes H^{*}$ is viewed as an $A\otimes H^{*}$-algebra by means of $\iota^{\prime}$. Next, there is an automorphism $\varphi$ of the algebra $A\otimes H^{*}\otimes H^{*}$ defined by the rule

[TABLE]

Since $\varphi$ acts as the identity on $A\otimes H^{*}\otimes 1$, we have

[TABLE]

for all $y\in A\otimes H^{*}\otimes H^{*}$. If $y=(\iota\otimes{\rm id})\rho(a)$, then $\varphi(y)=({\rm id}\otimes\Delta)\rho(a)=(\rho\otimes{\rm id})\rho(a).$ Hence $(**)$ follows.   $\mathchar 1027\relax$

By passage from $A$ to $A[t]$ the conclusion of Theorem 5.1 can be deduced from the fact that

[TABLE]

The arguments showing this inclusion in the proof of Theorem 5.1 above are tautological, up to different notation and the use of right comodule structures instead of left ones, to those in Mumford’s book on abelian varieties [41, Ch. III, section 12]. Demazure and Gabriel describe quotients by actions of finite group schemes in [18, Ch. III, §2, Cor. 6.1] as a special case of a more general result on quotients of groupoid schemes [18, Ch. III, §2, Th. 3.2].

Thus Theorem 5.1 is a very old result. Somehow it had not been well-known to Hopf algebra theorists for some time in the past. Integrality over invariants for commutative comodule algebras over commutative Hopf algebras was rediscovered by Ferrer Santos in [28]. In the language of module algebras that approach was reformulated by Montgomery [38, §4.2]. It makes use of the characteristic polynomials of endomorphisms of equivariant $A$-modules.

In [38] Montgomery raised the question as to whether $A$ is always integral over $A^{H}$ in the case of an arbitrary finite dimensional Hopf algebra $H$. For pointed Hopf algebras this question was answered shortly afterwards in the affirmative by Artamonov [2] when $A$ is a domain and without any restrictions on $A$ by Totok [56] and Zhu [58] when $\mathop{\rm char}\nolimits{k}>0$. Both [56] and [58] provided counterexamples to integrality in characteristic 0. Zhu also proved that $A$ is integral over $A^{H}$ when $H$ is involutory, i.e. $S^{2}={\rm id}$, and $\mathop{\rm char}\nolimits{k}$ does not divide the dimension of $H$. At that time it remained open what is actually needed for integrality to hold.

The characteristic polynomials have reappeared in a later development:

Theorem 5.2 [48]. If $A$ is $H$-semiprime or, more generally, if there exists a homomorphism of commutative $H$-module algebras $\varphi:A^{\prime}\to A$ such that $A=\varphi(A^{\prime})A^{H}$ and $A^{\prime}$ is $H$-semiprime then for each $a\in A$ the polynomial \,P_{A\otimes H^{*}/A}\bigl{(}\rho(a),t\bigr{)}\, has all coefficients in $A^{H}$. In particular, $A$ is integral over $A^{H}$.

The case when $A$ is $H$-semiprime, which is the main step here, has been subsumed in a recent work of Eryashkin [22] on invariants of $H$-module PI algebras. These results will be discussed in section 7. The original proof of Theorem 5.2 had common elements with the proof of Theorem 7.5, but it didn’t use the Martindale quotient rings.

If $A$ contains nonzero $H$-stable nilpotent ideals, then integrality over invariants may well be lost by the already mentioned examples of Totok and Zhu. There are still two important cases when the $H$-semiprimeness is not needed:

Corollary 5.3. $A$* is integral over $A^{H}$ in each of the following two cases*:**

*(a) *the trace map $A\to A^{H}$ is surjective,

(b) $\mathop{\rm char}\nolimits{k}=p>0$.

Proof. Let $N$ be the largest $H$-stable ideal of $A$ contained in the nil radical of $A$. Since $B=A/N$ is $H$-semiprime, Theorem 5.2 shows that $B$ is integral over $B^{H}$. Let $\pi:A\to B$ be the canonical map.

In case (a) $\pi(A^{H})=B^{H}$. Therefore for each $a\in A$ there exists a polynomial $f\in A^{H}[t]$ with the leading coefficient 1 such that $f(a)\in N$. Then $f(a)^{n}=0$ for some integer $n>0$ since $N$ is nil. Hence $a$ is integral over $A^{H}$.

Suppose that $\mathop{\rm char}\nolimits{k}=p>0$. Put $A^{\prime}=\pi^{-1}(B^{H})$. As in case (a) it is checked that $A$ is integral over $A^{\prime}$. We claim that for each $c\in A^{\prime}$ there exists $n>0$ such that $c^{p^{n}}\in A^{H}$. Indeed, $\rho(c)-c\otimes 1\in N\otimes H^{*}$ since $\pi(c)\in B^{H}$. It follows that $\rho(c)-c\otimes 1$ is nilpotent. Hence

[TABLE]

for sufficiently large $n$, but this means that $c^{p^{n}}\in A^{H}$. Thus $A^{\prime}$ is integral over $A^{H}$, and the final conclusion follows from transitivity of integrality.   $\mathchar 1027\relax$

In [58] Zhu conjectured that $A$ is integral over $A^{H}$ whenever $H$ is involutory. When $\mathop{\rm char}\nolimits{k}=0$ it is known that $H$ is involutory if and only if $H$ is semisimple. In this case the trace $A\to A^{H}$ is surjective. Thus Zhu’s conjecture follows from Corollary 5.3. However, when $\mathop{\rm char}\nolimits{k}>0$ the question of integrality does not depend on any condition on $H$.

As observed by Kalniuk and Tyc [29] the fact that in positive characteristic each commutative $H$-module algebra is integral over the invariants implies a property of $H$ similar to the geometric reductivity known in the theory of algebraic groups. This property was considered in [29] for a not necessarily finite dimensional Hopf algebra $H$, and its main consequence is that, whenever $A$ is a finitely generated commutative $H$-module algebra on which the action of $H$ is locally finite, the algebra of invariants $A^{H}$ is finitely generated. When $\mathop{\rm char}\nolimits{k}>0$ each finite dimensional Hopf algebra is geometrically reductive in this sense [29, Th. 4]. This result can be reformulated as follows:

Theorem 5.4 [29]. Suppose that $\mathop{\rm char}\nolimits{k}=p>0$. If $\varphi:A\to B$ is a surjective homomorphism of commutative $H$-module algebras and $b\in B^{H}$, then $b^{n}\in\varphi(A^{H})$ for some integer $n>0$.

Proof. Consider first the case when $A$ and $B$ are graded, $\varphi$ respects the grading, and $b$ is homogeneous of degree 1. Let $B^{\prime}$ be the subalgebra of $B$ generated by $b$, and let $A^{\prime}=\varphi^{-1}(B^{\prime})$. Since $A^{\prime}$ is integral over $A^{\prime H}$ by Corollary 5.3, $B^{\prime}=\varphi(A^{\prime})$ is integral over $\varphi(A^{\prime H})$. But $\varphi(A^{\prime H})$ is a graded subalgebra of $B^{\prime}$. If $\varphi(A^{\prime H})={k}$, then $b$ is integral over ${k}$, in which case $b$ has to be nilpotent. Otherwise $\varphi(A^{\prime H})$ contains a homogeneous element of positive degree. But each homogeneous component of $B^{\prime}$ is spanned by a power of $b$. Hence $b^{n}\in\varphi(A^{H})$ for some $n$ anyway.

In the general case let $\varphi^{t}:A[t]\to B[t]$ be the extension of $\varphi$ to polynomial rings and extend the action of $H$ to $A[t]$ and $B[t]$ by making $t$ invariant. Then $bt\in B[t]$ is a homogeneous $H$-invariant element of degree 1, and so we are in the situation of the previous case.   $\mathchar 1027\relax$

Integrality of $A$ over $A^{H}$ implies several well-known nice properties of the ring extension $A^{H}\subset A$. In particular, the canonical map $\mathop{\rm Spec}A\to\mathop{\rm Spec}A^{H}$ between the prime spectra is surjective, closed, and satisfies the going-up. However, for deeper conclusions integrality alone is not sufficient, and the characteristic polynomials come into play in an essential way. One application is this:

Theorem 5.5 [48]. Suppose that for each $a\in A$ the characteristic polynomial \,P_{A\otimes H^{*}/A}\bigl{(}\rho(a),t\bigr{)}\, has all coefficients in $A^{H}$. Then the map $\mathop{\rm Spec}A\to\mathop{\rm Spec}A^{H}$ is open, has finite fibers, and satisfies the going-down property.

Theorem 5.5 and its proof generalize the classical results describing properties of the quotient morphism $X\to X/G$ where $X$ is an affine scheme and $X/G$ is its quotient by an action of a finite group scheme.

There are further applications of the technique used in the study of group scheme actions. For each ${p}\in\mathop{\rm Spec}A$ denote by $k({p})$ the residue field of the local ring $A_{p}$. Let $\alpha_{p}:A\to k({p})$ be the canonical ring homomorphism. The composite

[TABLE]

is a homomorphism of $H$-module algebras, assuming that $H$ acts trivially on $k({p})$ and by the left hits $\rightharpoonup$ on $H^{*}$. Hence

[TABLE]

is a commutative right coideal subalgebra of the Hopf algebra $k({p})\otimes H^{*}$ over the field $k({p})$. In [48] $O({p})$ was called the orbital subalgebra associated with ${p}$.

When $H$ is cocommutative and $G$ is the finite group scheme representable by the commutative Hopf algebra $H^{*}$, the algebra $O({p})$ represents the scheme-theoretic $G$-orbit of ${p}$ which is a closed subscheme in the affine scheme $\mathop{\rm Spec}(k({p})\otimes A)$.

Theorem 5.6 [48]. Suppose that $A$ is $H$-semiprime and the function ${p}\mapsto\dim O({p})$ is locally constant on the whole $\mathop{\rm Spec}A$. Then $A$ is a finitely generated projective $A^{H}$-module whose rank at a prime ${q}\in\mathop{\rm Spec}A^{H}$ is equal to $\dim O({p})$ where ${p}$ is any prime ideal of $A$ lying above ${q}$.

Also, the assignment $I\mapsto I\cap A^{H}$ establishes a bijection between the $H$-stable ideals of $A$ and all ideals of $A^{H}$. The inverse correspondence is $J\mapsto JA$.

We will explain briefly the main ideas used in the proof. Given some elements $a_{1},\ldots,a_{n}\in A$, the set $U$ of those prime ideals ${p}$ of $A$ for which $\delta_{p}(a_{1}),\ldots,\delta_{p}(a_{n})$ form a basis of $O({p})$ over $k({p})$ is open in $\mathop{\rm Spec}A$. One can also check that, whenever ${p}$ and ${p}^{\prime}$ are two prime ideals of $A$ with ${p}\cap A^{H}={p}^{\prime}\cap A^{H}$, one has ${p}\in U$ if and only if ${p}^{\prime}\in U$. Then, passing to the localizations $A[s^{-1}]$ of $A$ at suitable elements $s\in A^{H}$, one may assume that there exist $a_{1},\ldots,a_{n}\in A$ such that $\delta_{p}(a_{1}),\ldots,\delta_{p}(a_{n})$ are a basis of $O({p})$ over $k({p})$ for each ${p}\in\mathop{\rm Spec}A$.

The technically complicated part of the proof is to show that the previous assumption implies that $\rho(a_{1}),\ldots,\rho(a_{n})$ form a basis of $(A\otimes 1)\rho(A)\subset A\otimes H^{*}$ over $A$ with respect to the left module structure; once this has been done, the freeness of $A$ over $A^{H}$ follows from Lemma 3.3. Note, however, that, when $A$ is reduced (equivalently, semiprime), there is a general ring-theoretic fact which states that a submodule $M$ of a finite rank free $A$-module $F$ is freely generated by elements $v_{1},\ldots,v_{n}\in M$ provided that for each ${p}\in\mathop{\rm Spec}A$, the image of $k({p})\otimes M$ in $k({p})\otimes F$ has a basis over $k({p})$ consisting of $1\otimes v_{1},\ldots,1\otimes v_{n}$; with

[TABLE]

the desired conclusion is immediate. Since $A$ is assumed to be only $H$-semiprime, one has to overcome several difficulties.

A special case of Theorem 5.6 was given in [47].

Although several fundamental facts of the classical theory generalize to commutative $H$-semiprime algebras, in the case when the Hopf algebra $H$ is not cocommutative it may not admit sufficiently many actions on commutative algebras. The next result has been obtained by Etingof and Walton [25] when either $\mathop{\rm char}\nolimits{k}=0$ or $\mathop{\rm char}\nolimits{k}>0$ and $H$ is also semisimple. Its extension to the case when $H$ is not necessarily semisimple has been given in [53].

Theorem 5.7. Assume ${k}$ to be algebraically closed. Then any action of a finite dimensional cosemisimple Hopf algebra $H$ on a commutative domain $A$ factors through an action of a group algebra, i.e. there exists a Hopf ideal $I$ of $H$ such that $I$ annihilates $A$ and $H/I$ is spanned by grouplike elements.

Etingof and Walton say that the action of $H$ on $A$ is inner faithful if $A$ is not annihilated by any nonzero Hopf ideal of $H$. In [26], [27] they investigated the question of the existence of inner faithful actions on commutative domains for pointed Hopf algebras. Some pointed Hopf algebras admit such actions, while the others do not.

The fact that the annihilator of $A$ in $H$ is often nontrivial had been recognized much earlier. Cohen and Westreich pointed out in [15, Cor. 0.12] that $H$ can act faithfully (in the ordinary sense) on a field $A$ only if $H$ is involutory and all grouplikes of $H^{*}$ lie in the center of $H^{*}$. Here $A$ can be even a domain since then the action of $H$ extends to the quotient field.

All this shows that the class of commutative $H$-module algebras is too narrow when $H$ is not cocommutative, and there is a definite need to study the invariants in the larger class of algebras satisfying a polynomial identity. As yet, not all results known for commutative $H$-module algebras have been extended to the PI case however.

As an extension of the commutative theory in a different direction Cohen and Westreich [16] introduced quantum commutative $H$-module algebras. The commutativity law in these algebras comes from the braiding determined by a quasitriangular structure on $H$. Cohen, Westreich and Zhu proved

Theorem 5.8 [17]. Let $A$ be a quantum commutative $H$-module algebra where $H$ is triangular semisimple and either $\mathop{\rm char}\nolimits{k}=0$ or $\mathop{\rm char}\nolimits{k}>\dim H$. Then $A$ is integral over $A^{H}$ and $A$ is PI.

One may wonder whether the conclusion of this theorem is valid under less stringent restrictions on $H$ and the characteristic of ${k}$ when $A$ is $H$-semiprime.

6. The $H$-equivariant Martindale quotient ring

Here we present results of Eryashkin [22] on quotient rings of $H$-semiprime PI algebras. Generalized Martindale quotient rings can be defined with respect to any filter ${\cal F}$ of ideals of a ring $R$ subject to the conditions that each ideal $I\in\nobreak{\cal F}$ has zero left and right annihilators in $R$ and that $IJ\in{\cal F}$ whenever $I,J\in{\cal F}$. Details of this construction are given, e.g., in [38, §6.4]. If $R$ is prime and ${\cal F}$ is the set of all nonzero ideals of $R$, this construction gives the left, right and symmetric Martindale rings of quotients, as defined in [43, Ch. 3].

Let $A$ be an $H$-module algebra. Denote by ${\cal F}_{H}(A)$ the set of all its $H$-stable ideals with zero left and right annihilators in $A$. If $A$ is $H$-prime, then ${\cal F}_{H}(A)$ consists of all nonzero $H$-stable ideals of $A$. The Martindale quotient rings with respect to this filter were introduced by Cohen [9]. The use of $H$-stable ideals in their definition accounts for the extension of the $H$-module structure to these quotient rings.

We will be concerned only with the $H$-symmetric ring of quotients $Q_{H}(A)$ (see [38, p. 98]). The larger left and right quotient rings are less useful. The ring $Q_{H}(A)$ is characterized by the following properties (cf. [43, Prop. 10.4]):

(1) $Q_{H}(A)$ contains $A$ as a subring;

(2) each $I\in{\cal F}_{H}(A)$ has zero left and right annihilators in $Q_{H}(A)$;

(3) for each $q\in Q_{H}(A)$ there exists $I\in{\cal F}_{H}(A)$ such that $Iq\subset A$ and $qI\subset A$;

(4) given $I\in{\cal F}_{H}(A)$, a left $A$-linear map $f_{l}:I\to A$, and a right $A$-linear map $f_{r}:I\to A$ such that $xf_{r}(y)=f_{l}(x)y$ for all $x,y\in I$, there exists $q\in Q_{H}(A)$ such that $f_{l}(x)=xq$ and $f_{r}(x)=qx$ for all $x\in I$.

Put $Q=Q_{H}(A)$. As explained in [9], $Q$ is an $H$-module algebra, and $A$ embeds in $Q$ as an $H$-stable subalgebra. The centers $Z(A)$, $Z(Q)$ of $A$ and $Q$ are not stable, in general, under the action of $H$. Set

[TABLE]

It follows from (2) and (3) in the characterization of $Q_{H}(A)$ above that each nonzero left or right $A$-submodule of $Q$ has nonzero intersection with $A$. Therefore $Q$ has to be $H$-prime or $H$-semiprime whenever so is $A$.

There are several general properties of the $H$-equivariant Martindale quotient rings of $H$-prime algebras. In particular, the fact that ${Z(Q)^{H}}$ is a field was explicitly stated in Matczuk’s paper [34, Lemma 1.4] which used the right quotient rings, however. This field is called the $H$-extended centroid of $A$.

Lemma 6.1. Suppose that $A$ is $H$-prime. Then ${Z(Q)^{H}}$ is a field. Furthermore, given any $H$-stable ideal $I$ of $Q$ and a morphism $f:I\to Q$ in $H\hbox{-}\mskip 2.0mu\vphantom{{\cal M}}_{Q}{\cal M}_{Q}$, there exists $z\in{Z(Q)^{H}}$ such that $f(x)=zx$ for all $x\in I$.

Proof. Recall that $f$ is a homomorphism of $Q$-bimodules and $H$-modules. Put $I^{\prime}=f^{-1}(A)\cap A$. This is an $H$-stable ideal of $A$. We may assume that $f\neq 0$. Then $I^{\prime}\neq 0$, i.e. $I^{\prime}\in{\cal F}_{H}(A)$. Note that $f(x)y=f(xy)=xf(y)$ for all $x,y\in I$. By (4) there exists $z\in Q$ such that $f(x)=zx=xz$ for all $x\in I^{\prime}$.

If $u\in Q$ is any element, then $uJ\subset A$ for some $J\in{\cal F}_{H}(A)$. Replacing $J$ with $JI^{\prime}$, we will also have $J\subset I^{\prime}$ and $uJ\subset I^{\prime}$. Since $z$ centralizers all elements of $I^{\prime}$, it follows that $(zu-uz)J=0$. Hence $zu=uz$ by (2).

Since $f$ is $H$-linear, we deduce that

[TABLE]

In other words, \bigl{(}hz-\varepsilon(h)z\bigr{)}I^{\prime}=0. Hence $hz=\varepsilon(h)z$ for all $h\in H$, and we conclude that $z\in{Z(Q)^{H}}$.

Define $g:I\to Q$ by the formula $g(x)=f(x)-zx$. Then $g|_{I^{\prime}}=0$. If $u\in I$ and $J\in{\cal F}_{H}(A)$ is such that $J\subset I^{\prime}$ and $uJ\subset I^{\prime}$, then $g(u)J=g(uJ)=0$, yielding $g(u)=0$. Hence $g=0$, and so $f(x)=zx$ for all $x\in I$.

The annihilator of $z$ in $Q$ is an $H$-stable ideal. If this ideal is nonzero, then property (2) entails $z=0$. Hence $f$ has to be injective whenever $z\neq 0$. In this case $f(I)$ is a nonzero $H$-stable ideal of $Q$. Applying the already proved conclusion to the inverse map $f^{-1}:f(I)\to I$, we see that there exists $z^{\prime}\in{Z(Q)^{H}}$ such that $f^{-1}(y)=z^{\prime}y$ for all $y\in f(I)$. Then $(zz^{\prime}-1)I=0$, and it follows that $z^{\prime}=z^{-1}$. But this argument applies to each nonzero element of the commutative ring ${Z(Q)^{H}}$. Thus ${Z(Q)^{H}}$ is a field.   $\mathchar 1027\relax$

Lemma 6.2. Suppose that $A$ is $H$-prime. Let $S$ be any simple algebra whose center contains ${Z(Q)^{H}}$. Consider $S\otimes_{Z(Q)^{H}}Q$ as an $H$-module algebra with respect to the action of $H$ on the second tensorand. If $I$ is a nonzero $H$-stable ideal of this algebra, then $I$ has nonzero intersection with the image of $Q$ in $S\otimes_{Z(Q)^{H}}Q$ under the map $x\mapsto 1\otimes x$.

Proof. Let $n$ be minimal possible for which $I$ contains an element $u\neq 0$ which can be written as $u=a_{1}\otimes b_{1}+\ldots+a_{n}\otimes b_{n}$ with $a_{i}\in S$, $b_{i}\in Q$. Fix such an element and its expression as a sum. Put

[TABLE]

Consider $Q^{n}$ as an object of $H\hbox{-}\mskip 2.0mu\vphantom{{\cal M}}_{Q}{\cal M}_{Q}$ with respect to the natural actions of $H$ and $Q$ on each component. Then $M$ is a subobject of $Q^{n}$ in this category. Note that

[TABLE]

Let $p_{i}:Q^{n}\to Q$, $\,i=1,\ldots,n$, be the projections. Then $J=p_{1}(M)$ is an $H$-stable ideal of $Q$. But $\mathop{\rm Ker}p_{1}|_{M}=0$ since otherwise $I$ would contain a nonzero element written as $\sum_{i=2}^{n}a_{i}\otimes x_{i}$ with less than $n$ summands. Thus $p_{1}|_{M}:M\to J$ is an isomorphism in the category $H\hbox{-}\mskip 2.0mu\vphantom{{\cal M}}_{Q}{\cal M}_{Q}$. Setting $f_{i}=p_{i}\circ(p_{1}|_{M})^{-1}$, we get

[TABLE]

and each map $f_{i}:J\to Q$ is a morphism in $H\hbox{-}\mskip 2.0mu\vphantom{{\cal M}}_{Q}{\cal M}_{Q}$. By Lemma 6.1 there exist $z_{1},\ldots,z_{n}\in{Z(Q)^{H}}$ such that $f_{i}(x)=z_{i}x$ for all $x\in J$. In particular, $b_{i}=z_{i}b_{1}$ for each $i$, and therefore $u=\sum a_{i}\otimes z_{i}b_{1}=(\sum a_{i}z_{i})\otimes b_{1}$.

The minimality of $n$ implies that $n=1$, i.e. $u=a_{1}\otimes b_{1}$. But then $Sa_{1}S\otimes b_{1}\subset I$. Since $S$ is simple, we have $Sa_{1}S=S$. Hence $1\otimes b_{1}\in I$.   $\mathchar 1027\relax$

Lemma 6.3. Suppose that $A$ is $H$-prime, and let $P$ be any prime ideal of $A$ such that $P_{H}=0$. Denote by ${\overline{Q}}$ the symmetric Martindale quotient ring of the prime ring ${\overline{A}}=A/P$. The canonical map $\pi:A\to{\overline{A}}$ extends to a ring homomorphism $Q\to{\overline{Q}}$ which maps the center of $Q$ into the center of ${\overline{Q}}$.

Proof. Let $q\in Q$. There exists $I\in{\cal F}_{H}(A)$ such that $Iq\subset A$ and $qI\subset A$. Since $P_{H}=0$, we have $I\not\subset P$. Hence $\pi(I)$ is a nonzero ideal of ${\overline{A}}$. Note that $Iq(I\cap P)$ is contained in $P$. Applying $\pi$, we get \pi(I)\pi\bigl{(}q(I\cap P)\bigr{)}=0, whence \pi\bigl{(}q(I\cap P)\bigr{)}=0 since ${\overline{A}}$ is prime.

This shows that $q(I\cap P)\subset P$. Similarly $(I\cap P)q\subset P$. Therefore the right and left multiplications by $q$ induce, respectively, a left ${\overline{A}}$-linear map $f_{l}:\pi(I)\to{\overline{A}}$ and a right ${\overline{A}}$-linear map $f_{r}:\pi(I)\to{\overline{A}}$. The pair $(f_{l},f_{r})$ determines an element ${\overline{q}}\in{\overline{Q}}$. It is easy to see that the assignment $q\mapsto{\overline{q}}$ defines a ring homomorphism $Q\to{\overline{Q}}$ whose restriction to $A$ is $\pi$.

Denote this extension of $\pi$ by the same letter $\pi$. If $z\in Z(Q)$, then $\pi(z)$ commutes with all elements of $\pi(A)={\overline{A}}$, but then $\pi(z)$ commutes with all elements of ${\overline{Q}}$.   $\mathchar 1027\relax$

Theorem 6.4 [22]. Suppose that $A$ is PI and $H$-prime. Then the $H$-symmetric quotient ring $Q=Q_{H}(A)$ is an $H$-simple $H$-module algebra of finite dimension over ${Z(Q)^{H}}$. Moreover, $Q={Z(Q)^{H}}A$.

Proof. Take any prime ideal $P$ of $A$ such that $P_{H}=0$. Let $\pi:Q\to{\overline{Q}}$ be the ring homomorphism of Lemma 6.3. Since ${\overline{A}}$ is a prime PI algebra, ${\overline{Q}}$ is the classical quotient ring of ${\overline{A}}$ (see [43, Th. 23.4]). By Posner’s theorem the ring ${\overline{Q}}$ is simple and finite dimensional over its center $Z({\overline{Q}})$. The composite map

[TABLE]

is a homomorphism of $H$-module algebras, assuming the trivial action of $H$ on ${\overline{Q}}$ and the hit action $\rightharpoonup$ on $H^{*}$. It extends to a homomorphism of $H$-module algebras

[TABLE]

Since $({\rm id}\otimes\varepsilon)\circ\varphi=\pi$, we have $\mathop{\rm Ker}\varphi\subset\mathop{\rm Ker}\pi$. It follows that

[TABLE]

But $\mathop{\rm Ker}\varphi$ is an $H$-stable ideal of $Q$. Since $P_{H}=0$, we get $A\cap\mathop{\rm Ker}\varphi=0$, which entails $\mathop{\rm Ker}\varphi=0$. Now $\mathop{\rm Ker}\psi$ is an $H$-stable ideal of $Z({\overline{Q}})\otimes_{{Z(Q)^{H}}}Q$. It has zero intersection with the image of $Q$ by the preceding conclusion, whence $\mathop{\rm Ker}\psi=0$ by Lemma 6.2. Injectivity of $\psi$ entails an upper bound for the dimension

[TABLE]

The $H$-stable subalgebra $A^{\prime}={Z(Q)^{H}}A\subset Q$ is then finite dimensional over ${Z(Q)^{H}}$ too. Also, $A^{\prime}$ is $H$-prime since each nonzero ideal of $A^{\prime}$ has nonzero intersection with $A$. By Theorem 2.4 $A^{\prime}$ is $H$-simple. But for each $q\in Q$ there exists a nonzero $H$-stable ideal $I^{\prime}$ of $A^{\prime}$ such that $qI^{\prime}\subset A^{\prime}$. We must have $1\in I^{\prime}$, and so $q\in A^{\prime}$. Thus $Q=A^{\prime}$.   $\mathchar 1027\relax$

Corollary 6.5. If $A$ is PI and $H$-prime, then $A$ has finitely many minimal prime ideals, and $P_{H}=0$ for each of them.

Proof. Since $Q$ is artinian, it has finitely many maximal ideals. Let $P_{1},\ldots,P_{n}$ be their contractions to $A$. The intersection $\bigcap P_{i}$ is nilpotent since it is contained in the Jacobson radical of $Q$. Hence each prime ideal of $A$ contains $P_{i}$ for some $i$, i.e. all minimal primes are among $P_{1},\ldots,P_{n}$. If $I$ is any $H$-stable ideal of $A$ contained in $P_{i}$, then $IQ$ is an $H$-stable ideal of $Q$ contained in a maximal ideal. It follows that $IQ=0$, and therefore $I=0$.   $\mathchar 1027\relax$

Corollary 6.6. Suppose that $A$ is PI and $H$-prime. If $P$ is any prime ideal of $A$ such that $P_{H}=0$, then the ring homomorphism $\pi:Q\to{\overline{Q}}$ of Lemma 6.3 is surjective.

Proof. We have ${\overline{A}}=\pi(A)\subset\pi(Q)\subset{\overline{Q}}$. If $s$ is any regular element of ${\overline{A}}$, then $s$ is invertible in ${\overline{Q}}$ since ${\overline{Q}}$ is a classical quotient ring of ${\overline{A}}$. But $s\in\pi(Q)$, whence $s$ is a regular element of $\pi(Q)$. Since $\pi(Q)$ is a finite dimensional algebra over a field, it follows that $s^{-1}\in\pi(Q)$. Then ${\overline{Q}}=\pi(Q)$.   $\mathchar 1027\relax$

Corollary 6.7. If $A$ is PI and $H$-simple, then $A$ has finite dimension over its central subfield $Z(A)^{H}$.

Proof. In this case $Q_{H}(A)=A$.   $\mathchar 1027\relax$

Lemma 6.8. Suppose that $K_{1},\ldots,K_{n}$ are minimal $H$-prime ideals of $A$ such that $\bigcap K_{i}=0$. Then $\,Q_{H}(A)\cong Q_{H}(A/K_{1})\times\ldots\times Q_{H}(A/K_{n})$.

Proof. Put $Q=Q_{H}(A)$, $\,A_{i}=A/K_{i}$ and $Q_{i}=Q_{H}(A_{i})$ for each $i$. The canonical map $\pi_{i}:A\to A_{i}$ extends to a homomorphism of $H$-module algebras $Q\to Q_{i}$ as in the proof of Lemma 6.3. The main point here is that $A_{i}$ is $H$-prime and $I\not\subset K_{i}$ for each $I\in{\cal F}_{H}(A)$. In fact $K_{i}$ has nonzero annihilator in $A$ since $K_{i}K^{\prime}_{i}\subset K_{i}\cap K^{\prime}_{i}=0$ where $K^{\prime}_{i}=\bigcap_{j\neq i}K_{j}$. Hence each element of $Q$ gives rise to a left $A_{i}$-linear map $f_{l}:\pi_{i}(I)\to A_{i}$ and a right $A_{i}$-linear map $f_{r}:\pi_{i}(I)\to A_{i}$, and the pair $(f_{l},f_{r})$ determines an element of $Q_{i}$.

Now the collection $\pi_{1},\ldots,\pi_{n}$ gives a homomorphism of $H$-module algebras

[TABLE]

Since $\mathop{\rm Ker}\pi|_{A}=\bigcap K_{i}=0$, it follows that $\mathop{\rm Ker}\pi=0$. It remains to show that $\pi$ is surjective. Suppose $q_{1}\in Q_{1}$. There exists a nonzero $H$-stable ideal $I_{1}$ of $A_{1}$ such that $I_{1}q_{1}\subset A_{1}$ and $q_{1}I_{1}\subset A_{1}$. Take any $H$-stable ideal $J$ of $A$ with the property that $0\neq\pi_{1}(J)\subset I_{1}$. Replacing $J$ with $JK^{\prime}_{1}$ we will also have

[TABLE]

Since $K_{1}\cap K^{\prime}_{1}=0$, there is an isomorphism of $A$-bimodules $K^{\prime}_{1}\cong\pi_{1}(K^{\prime}_{1})$. Define maps $\,f_{l},f_{r}:J\to K^{\prime}_{1}\,$ by the rules

[TABLE]

Put $I=J+\sum_{i\neq 1}K^{\prime}_{i}$. Note that $K^{\prime}_{i}\subset K_{j}$ for $j\neq i$, while $K^{\prime}_{j}\cap K_{j}=0$. Hence the sum $\sum K^{\prime}_{i}$ is direct, and there are extensions of $f_{l},f_{r}$ to maps $I\to A$ vanishing on $K^{\prime}_{i}$ for each $i\neq 1$. Note that $f_{l}$ is left $A$-linear, while $f_{r}$ is right $A$-linear.

Since $\pi_{i}(I)\neq 0$ for each $i$, the left and right annihilators of $I$ in $A$ are contained in each $K_{i}$. Then these annihilators are zero, i.e. $I\in{\cal F}_{H}(A)$. Hence the pair $(f_{l},f_{r})$ determines an element $q\in Q$ such that $\pi_{1}(q)=q_{1}$ and $\pi_{i}(q)=0$ for $i\neq 1$. By symmetry $Q_{1}$ in this argument can be replaced with $Q_{j}$ for any $j$.   $\mathchar 1027\relax$

Corollary 6.9. Suppose that $A$ is PI and $H$-semiprime with finitely many minimal $H$-prime ideals. Then $Q_{H}(A)\cong Q_{1}\times\ldots\times Q_{n}$ where $Q_{1},\ldots,Q_{n}$ are $H$-simple $H$-module algebras with $[Q_{i}:{Z(Q_{i})^{H}}]<\infty$ for each $i$.

Proof. In any $H$-semiprime algebra the intersection of all minimal $H$-prime ideals is zero. So Lemma 6.8 applies. It gives a direct product decomposition of $Q_{H}(A)$ in which each factor $Q_{i}=Q_{H}(A/K_{i})$ is $H$-simple and finite dimensional over ${Z(Q_{i})^{H}}$ by Theorem 6.4.   $\mathchar 1027\relax$

Theorem 6.10. Suppose that $A$ is PI and $H$-semiprime with finitely many minimal $H$-prime ideals. Then $Q_{H}(A)$ is a classical right quotient ring of $A$.

We have made some comments about the proof in the discussion following the statement of Theorem 2.12. Eryashkin proves the conclusion of Theorem 6.10 in the $H$-prime case. If $A$ is PI and $H$-semiprime, and if $K_{1},\ldots,K_{n}$ are all its minimal $H$-prime ideals, then, knowing that $Q_{i}=Q_{H}(A/K_{i})$ is a classical quotient ring of $A/K_{i}$ for each $i$, he concludes that $Q_{1}\times\ldots\times Q_{n}$ is a classical quotient ring of $A$ directly, not using Lemma 6.8.

For a special class of $H$-prime PI algebras Theorems 6.4 and 6.10 were obtained in an earlier article [21].

Lemma 6.11. The set of minimal $H$-prime ideals of $A$ is finite if $A$ is finitely generated and PI. The same conclusion holds if $A$ is either left or right noetherian.

Proof. Under each of these assumptions $A$ has finitely many minimal prime ideals and the prime radical $N$ of $A$ is nilpotent (in the case of a finitely generated PI algebra see [45, Cor. 6.3.36*′*, Th. 6.3.39]). Let $P_{1},\ldots,P_{n}$ be all the minimal primes, and for each $i$ let $K_{i}$ be the largest $H$-stable ideal of $A$ contained in $P_{i}$. Since $\prod K_{i}\subset N$ is nilpotent, any minimal $H$-prime ideal of $A$ has to coincide with one of the $H$-prime ideals $K_{1},\ldots,K_{n}$.   $\mathchar 1027\relax$

7. Integrality of PI algebras over the invariants

If $A$ is a noncommutative $H$-module algebra, it is meaningful to consider integrality of $A$ over invariants in two different senses. One question concerns integrality over central invariants. For this it should be assumed at least that $A$ is integral over its center $Z(A)$. We denote by ${Z(A)^{H}}$ the subalgebra of $Z(A)$ consisting of $H$-invariant central elements.

If $H$ is cocommutative, then $Z(A)$ is stable under the action of $H$, and $Z(A)$ is integral over ${Z(A)^{H}}$ by the classical theory. If $H$ is not cocommutative, the problem becomes highly nontrivial. When $H$ is pointed, or at least when the coradical of $H$ is cocommutative the following result was obtained by Totok:

Theorem 7.1 [57]. The center $Z(A)$ is integral over ${Z(A)^{H}}$, and therefore $A$ is integral over ${Z(A)^{H}}$ if $A$ is integral over $Z(A)$, in each of the following two cases:**

(a) ** $\mathop{\rm char}\nolimits{k}>0$ and $H$ has cocommutative coradical,

(b) ** $\mathop{\rm char}\nolimits{k}=0$, $H$ is pointed, $A$ is reduced, $Z(A)$ is finitely generated.

Making use of the coradical filtration $H_{0}\subset H_{1}\subset\cdots\subset H_{n}=H$, Totok constructs a chain of subalgebras $Z_{0}\supset Z_{1}\supset\cdots\supset Z_{n}$ such that $Z_{0}=Z(A)^{H_{0}}$, and for each $i>0$ the ring $Z_{i-1}$ is integral over $Z_{i}$ and $H_{i}$ acts trivially on $Z_{i}$ in the sense that $hz=\varepsilon(h)z$ for all $h\in H_{i}$ and $z\in Z_{i}$. The conclusion of Theorem 7.1 follows then by transitivity of integrality since $Z(A)$ is integral over $Z_{0}$. This extends the technique applied by Artamonov [2] in the case when $A$ is a commutative domain.

New results on integrality of $A$ over ${Z(A)^{H}}$ for an arbitrary finite dimensional Hopf algebra have appeared very recently. In [24] Etingof observes that the problem admits a bimodule reformulation which can be studied independently of any Hopf algebra theory. In fact ${Z(A)^{H}}$ consists precisely of those elements $a\in A$ for which the left multiplication by $a\otimes 1$ in the algebra $A\otimes H^{*}$ coincides with the right multiplication by $\rho(a)$ or, in other words, the left action of $a$ on $A\otimes H^{*}$ is the same as the right action with respect to the $A$-bimodule structure defined as in section 3. This bimodule has a special property which has been used by Etingof to introduce the notion of Galois bimodules.

An $R$-bimodule $P$ for a ring $R$ is called Galois of rank $d\geq 1$ if $P$ is left and right free of rank $d$ and there is an isomorphism of bimodules $P\otimes_{R}P\cong P^{d}$. Etingof derives a classification of Galois bimodules when $R$ is a semisimple artinian ring module-finite over its center $Z(R)$. Let $R\cong R_{1}\times\ldots\times R_{n}$ where $R_{1},\ldots,R_{n}$ are simple rings. In the process of the classification it is verified that for each Galois bimodule $P$ the ring $R$ is a finite module over the center of $P$ defined as

[TABLE]

Let $\phi_{i}(a)$ be the $R_{i}$-linear endomorphism of $R_{i}\otimes_{R}P$ afforded by the right action of $a\in R$. Now $R_{i}\otimes_{R}P$ is a finite dimensional vector space over the center $Z(R_{i})$ of $R_{i}$, and $\phi_{i}(a)$ may be regarded as a linear transformation of this vector space. So the characteristic polynomial $\chi_{\phi_{i}(a)}\in Z(R_{i})[t]$ makes sense. Let

[TABLE]

be the polynomial whose $i$th component is $\chi_{\phi_{i}(a)}^{m^{2}/m_{i}^{2}}$ where $m_{i}^{2}=[R_{i}:Z(R_{i})]$ and $m$ is the least common multiple of $m_{1},\ldots,m_{n}$. It is shown in [24] that for each central element $a\in Z(R)$ all coefficients of $\chi_{a}$ belong to $Z(P)$. This is a key fact needed for applications to integrality.

Suppose that the $H$-module algebra $A$ has a semisimple artinian classical quotient ring $Q$ which is a finite module over its center $Z(Q)$. Then Etingof’s results discussed in the preceding paragraph apply to the Galois $Q$-bimodule $Q\otimes H^{*}$ whose center is ${Z(Q)^{H}}$. In particular, $Q$ is module-finite over ${Z(Q)^{H}}$ and certain polynomials associated with central elements of $Q$ have coefficients in ${Z(Q)^{H}}$.

However, the ultimate goal is to find conditions ensuring integrality of $A$. Etingof formulates results for comodule algebras, but we stick to the conventions set for the present paper.

Theorem 7.2 [24]. Let $Z$ be a central subalgebra of $A$ whose total quotient ring $Q(Z)$ is a direct product of finitely many fields. Suppose that

(1) $A$ is a finitely generated torsion-free $Z$-module,

(2) $Q(Z)\otimes_{Z}A$ is a semisimple ring with center $Q(Z)$,

*(3) *either $Z$ is integrally closed in $Q(Z)$ or $A$ is a projective $Z$-module.

Then $A$ is integral over $Z\cap A^{H}$.

An $H$-module algebra $A$ is said to be indecomposable if it is not isomorphic to a direct product of two nonzero $H$-module algebras. If $A$ is artinian and $H$-semiprime, this is equivalent, by Theorem 2.4, to the $H$-simplicity of $A$.

Indecomposability of $A$ in the next proposition means that the corresponding Galois $A$-bimodule $P=A\otimes H^{*}$ is connected. In this case $P^{m_{*}^{2}}$ is isomorphic to a multiple of $A\otimes_{Z(P)}A$ by the classification of Galois bimodules. Here $Z(P)=\nobreak{Z(A)^{H}}\!$. Comparing the left ranks of the two bimodules Etingof deduces a divisibility relation involving numeric characteristics of $A$:

Proposition 7.3 [24]. Suppose that $A$ is semisimple artinian, module-finite over $Z(A)$, and indecomposable as an $H$-module algebra. Let $A_{1},\ldots,A_{n}$ be all simple factor rings of $A$. Put

[TABLE]

Then $\sum d_{i}(m_{i}/m_{*})^{2}$ divides the dimension of $H$. In other words, $\,[A:{Z(A)^{H}}]$ divides $\,m_{*}^{2}\,(\dim H)$.

Now we describe a different approach, due to Eryashkin, which makes systematic use of structural properties of $H$-module algebras discussed earlier. The results have been obtained not only for semiprime $H$-module algebras but for $H$-semiprime algebras as well.

The ring homomorphism $A\to A\otimes H^{*}$ given by the assignment $a\mapsto a\otimes 1$ maps the center of $A$ into a central subalgebra of $A\otimes H^{*}$. Therefore $A\otimes H^{*}$ may be regarded as a $Z$-algebra for any central subalgebra $Z$ of $A$. If $A$ is projective of finite constant rank as a $Z$-module, then so too is $A\otimes H^{*}$. As explained in section 5, in this case there are characteristic polynomials for the ring extension $A\otimes H^{*}/Z$. For each $a\in A$

[TABLE]

is the characteristic polynomial of the left multiplication operator by the element $\rho(a)$ in the $Z$-algebra $A\otimes H^{*}$. Alternatively, one could use the characteristic polynomials of the right multiplication operators.

At one point we will need the ring-theoretic fact stated below. For the proof see [22, Prop. 3.1].

Proposition 7.4 [22]. Let $R$ be a ring which has a right artinian classical right quotient ring $Q(R)$. Suppose that $R$ is a finitely generated module over a central subring $Z$ such that $\,\mathop{\rm ann}\nolimits_{R}(z)=\mathop{\rm ann}\nolimits_{Z}(z)R\,$ for each $z\in Z$. Then $Q(R)\cong Q(Z)\otimes_{Z}R$ where $Q(Z)$ is the total quotient ring of $Z$.

Theorem 7.5 [22]. Suppose that $A$ is $H$-semiprime with finitely many minimal $H$-prime ideals. If $A$ is projective of finite constant rank as a module over its center $Z(A)$, then $A$ is integral over ${Z(A)^{H}}$. In fact, for each $a\in A$ the characteristic polynomial \,P_{A\otimes H^{*}\!/Z(A)}\bigl{(}\rho(a),t\bigr{)}\, has all coefficients in ${Z(A)^{H}}$.

Proof. Before we treat the general case let us verify the conclusion of this theorem under additional assumptions about $A$.

Step 1. Suppose that $A$ is $H$-simple.

By Corollary 6.7 ${Z(A)^{H}}$ is a field and $[A:{Z(A)^{H}}]<\infty$. Integrality of $A$ over ${Z(A)^{H}}$ is immediate. We still have to prove the statement about the characteristic polynomials.

Consider the action of $H$ on $A\otimes H^{*}$ defined by the rule $h(a\otimes\xi)=a\otimes(h\rightharpoonup\xi)$ for $h\in H$, $a\in A$ and $\xi\in H^{*}$. Then the map $\rho:A\to A\otimes H^{*}$ is a homomorphism of $H$-module algebras. So too is its extension

[TABLE]

with the action of $H$ on the first algebra defined by the formula $h(z\otimes a)=z\otimes ha$ for $h\in H$, $z\in Z(A)$ and $a\in A$. Put

[TABLE]

We claim that ${\cal B}$ is a free left ${\cal A}$-module with respect to the action afforded by $\psi$. Since the ring $Z(A)$ is artinian and since $\psi$ is a homomorphism of $Z(A)$-algebras, both of which are free modules over $Z(A)$, it suffices to check that for each maximal ideal ${m}$ of $Z(A)$ the ${\cal A}/{m}{\cal A}$-module ${\cal B}/{m}{\cal B}$ is free of rank $r$ where $r$ does not depend on ${m}$. Now

[TABLE]

with $K({m})=Z(A)/{m}$ being a field. Since $A$ is $H$-simple, it follows from Lemma 6.2 that ${\cal A}/{m}{\cal A}$ is $H$-simple too. Hence ${\cal B}/{m}{\cal B}$ is a free ${\cal A}/{m}{\cal A}$-module by Theorem 2.9. The rank $r({m})$ of this free module can be computed as

[TABLE]

where $[A:Z(A)]$ is the rank of $A$ as a $Z(A)$-module. This shows that $r({m})$ has the same value for all ${m}$, as required.

Thus ${\cal B}\cong{\cal A}^{r}$ as an ${\cal A}$-module. Since $\rho(a)=\psi(1\otimes a)$, we deduce that

[TABLE]

which is a polynomial with coefficients in ${Z(A)^{H}}$.

Step 2. Suppose that $A$ is artinian and $H$-semiprime.

By Theorem 2.4 $A=A_{1}\times\ldots\times A_{n}$ where each $A_{i}$ is an $H$-simple $H$-module algebra. Clearly

[TABLE]

Let $\pi_{i}:A\to A_{i}$ be the projection, $\zeta_{i}:Z(A)\to Z(A_{i})$ the restriction of $\pi_{i}$, and $\zeta_{i}^{t}:Z(A)[t]\to Z(A_{i})[t]$ the extension of $\zeta_{i}$ to polynomial rings. Note that

[TABLE]

Since $(\pi_{i}\otimes{\rm id})\rho(a)=\rho(\pi_{i}a)$, we get

[TABLE]

by Step 1. It follows that all coefficients of the polynomial P_{A\otimes H^{*}/Z(A)}\bigl{(}\rho(a),t\bigr{)} are $H$-invariant since they have $H$-invariant images in each $A_{i}$.

It is easy now to complete the proof of Theorem 7.5 in full generality. By Theorem 2.12 $A$ has a right and left artinian classical right quotient ring $Q=Q(A)$ which is an $H$-semiprime $H$-module algebra since so is $A$. Note that $\,\mathop{\rm ann}\nolimits_{A}(z)=\mathop{\rm ann}\nolimits_{Z(A)}(z)A\,$ for each $z\in Z(A)$ since $A$ is a direct summand of a free $Z(A)$-module. By Proposition 7.4 $Q$ is a central localization of $A$. Then the total quotient ring of $Z(A)$ coincides with the center of $Q$, and so $Q\cong Z(Q)\otimes_{Z(A)}A$. From the functorial properties of characteristic polynomials it follows that

[TABLE]

All coefficients of this polynomial lie in ${Z(Q)^{H}}$ by Step 2. Hence they actually lie in $\,Z(A)\cap{Z(Q)^{H}}={Z(A)^{H}}$.   $\mathchar 1027\relax$

All ideas of this proof are taken from [22]. We have used Theorem 2.9 to make some arguments more transparent. Note that Step 1 in the proof yields also the following conclusion:

Corollary 7.6. Suppose that $A$ is $H$-simple and $A$ is a free module of finite rank over its center $Z(A)$. Then the dimension $[Z(A):{Z(A)^{H}}]$ of $Z(A)$ over ${Z(A)^{H}}$ divides the dimension of $H$.

It is not clear to what extent the conditions imposed on $A$ in Theorem 7.5 are optimal. One concern arising here is the finiteness of the set of minimal $H$-primes. An easy extension of Theorem 7.5 is stated below:

Corollary 7.7. Suppose that $A$ is projective of finite constant rank as a module over its center $Z(A)$ and there is a set ${\cal F}$ of $H$-semiprime ideals of $A$ such that

*(1) *each ideal in ${\cal F}$ is an intersection of finitely many $H$-prime ideals,

*(2) *each ideal $I\in{\cal F}$ is generated by $I\cap Z(A)$,

*(3) *for each $I\in{\cal F}$ the image of $Z(A)$ in $A/I$ coincides with the center of $A/I$,

(4) $\bigcap_{I\in{\cal F}}I=0$.

Then for each $a\in A$ the characteristic polynomial \,P_{A\otimes H^{*}/Z(A)}\bigl{(}\rho(a),t\bigr{)}\, has all coefficients in ${Z(A)^{H}}$.

Proof. For each $I\in{\cal F}$ we have $A/I\cong Z(A/I)\otimes_{Z(A)}A$, and this algebra satisfies the hypothesis of Theorem 7.5. Hence all coefficients of P_{A\otimes H^{*}/Z(A)}\bigl{(}\rho(a),t\bigr{)} have $H$-invariant images in $A/I$. But (4) ensures that any element $c\in A$ is $H$-invariant whenever $c+I$ is $H$-invariant in $A/I$ for each $I\in{\cal F}$.   $\mathchar 1027\relax$

The assumptions about $A$ in Corollary 7.7 are admittedly too restrictive. However, they are satisfied when $A$ is commutative and $H$-semiprime. Thus Theorem 5.2 is a special case of Corollary 7.7.

Without projectivity of $A$ over $Z(A)$ the characteristic polynomials for the ring extension $A\otimes H^{*}/Z(A)$ are not defined. One can still exploit finiteness of $Q=Q(A)$ over ${Z(Q)^{H}}$. Recall that, if $A$ is PI and $H$-prime, then $Q$ is $H$-simple and ${Z(Q)^{H}}$ is a field. The next result is based on [22, Prop. 3.3], although it was not stated this way.

Proposition 7.8. Suppose that $A$ is $H$-prime and module-finite over its center $Z(A)$. Let $Q$ be the classical quotient ring of $A$. For each $a\in A$ all coefficients of the characteristic polynomial $\,P_{Q/{Z(Q)^{H}}}(a,t)\,$ are integral over $Z(A)$.

Proof. Since $[Q:{Z(Q)^{H}}]<\infty$, the set $\mathop{\rm Max}\nolimits Q$ of all maximal ideals of $Q$ is finite. For each $M\in\mathop{\rm Max}\nolimits Q$ denote by $\pi_{M}$ the canonical map $Q\to Q/M$. Here $Q/M$ is a simple ring finite dimensional over its center $Z(Q/M)$. Since $Z(A)\subset Z(Q)$, we have \pi_{M}\bigl{(}Z(A)\bigr{)}\subset Z(Q/M).

An arbitrary element $q\in Q$ is integral over $Z(A)$ if and only if $\pi_{M}(q)$ is integral over \pi_{M}\bigl{(}Z(A)\bigr{)} for each $M\in\mathop{\rm Max}\nolimits Q$. Indeed, if the latter property holds, then for each $M$ there exists a polynomial $f_{M}$ in one indeterminate with all coefficients in $Z(A)$ and the leading coefficient 1 such that $f_{M}(q)\in M$. Putting

[TABLE]

we will have $f(q)\in J$ where $J$ stands for the Jacobson radical of $Q$. But $J$ is nilpotent, whence $f(q)^{n}=0$ for some integer $n>0$. Clearly $f^{n}$ is a polynomial with all coefficients in $Z(A)$ and the leading coefficient 1.

We will check that the necessary and sufficient condition of integrality from the preceding paragraph is satisfied for all coefficients of $P_{Q/{Z(Q)^{H}}}(a,t)$. This will yield the final conclusion.

Consider the composite map $\rho_{M}:Q{}\mathrel{\smash{\mathop{\smash{\mathord{-}}\mkern-4.0mu\smash{\mathord{-}}\mkern-4.0mu\mathord{\rightarrow}}\limits^{\rho}}}{}Q\otimes H^{*}{}\mathrel{\smash{\mathop{\smash{\mathord{-}}\mkern-4.0mu\smash{\mathord{-}}\mkern-4.0mu\smash{\mathord{-}}\mkern-4.0mu\smash{\mathord{-}}\mkern-4.0mu\smash{\mathord{-}}\mkern-4.0mu\mathord{\rightarrow}}\limits^{\pi_{M}\otimes{\rm id}}}}{}Q/M\otimes H^{*}$ and its extension

[TABLE]

With the $H$-module structures as in the proof of Theorem 7.5 $\psi$ is a homomorphism of $H$-module algebras, and the first algebra is $H$-simple by Lemma 6.2. It follows, in particular, that $\psi$ is injective. By Theorem 2.9 $\psi$ makes $Q/M\otimes H^{*}$ a free left module over $Z(Q/M)\otimes_{Z(Q)^{H}}Q$. Let $r$ be its rank (it depends on $M$).

Denote by $P_{2}$ and $P_{1}$ the characteristic polynomials of these two rings regarded as finite dimensional algebras over the field $Z(Q/M)$. Since $\rho_{M}(a)=\psi(1\otimes a)$, we get

[TABLE]

Here we identify the field ${Z(Q)^{H}}$ with its image in $Z(Q/M)$ under the map $\pi_{M}$.

Recall that P_{2}\bigl{(}\rho_{M}(a),t\bigr{)} is the characteristic polynomial of the left multiplication operator associated with $\rho_{M}(a)$. But $\rho(a)\in A\otimes H^{*}$, and therefore $\rho_{M}(a)$ lies in the subring $\pi_{M}(A)\otimes H^{*}$ of $Q/M\otimes H^{*}$ which is a finitely generated module over \pi_{M}\bigl{(}Z(A)\bigr{)}. It follows that $\rho_{M}(a)$ satisfies a polynomial relation of integral dependence over \pi_{M}\bigl{(}Z(A)\bigr{)}, whence so too does the corresponding multiplication operator. This means that all eigenvalues of this operator in any algebraic closure of $Z(Q/M)$ are integral over \pi_{M}\bigl{(}Z(A)\bigr{)}. But these eigenvalues are precisely the roots of the characteristic polynomial, i.e. the roots of $P_{Q/{Z(Q)^{H}}}(a,t)$ in view of the equality above. The coefficients of $P_{Q/{Z(Q)^{H}}}(a,t)$ are evaluations of the elementary symmetric functions at the roots of this polynomial. Hence they are integral over \pi_{M}\bigl{(}Z(A)\bigr{)} too.   $\mathchar 1027\relax$

Theorem 7.9 [22]. Suppose that $A$ is $H$-semiprime with finitely many minimal $H$-prime ideals. Let $Q$ be the classical quotient ring of $A$. If $Z(A)$ is integrally closed in $Z(Q)$ and $A$ is a finitely generated $Z(A)$-module, then $A$ is integral over ${Z(A)^{H}}$.

Proof. By Theorem 2.4 $Q\cong Q_{1}\times\ldots\times Q_{n}$ where each $Q_{i}$ is an $H$-simple $H$-module algebra. Let $e_{i}\in Q$ be the $H$-invariant central idempotent whose projection to $Q_{j}$ is 1 for $j=i$ and 0 otherwise. Since $e_{i}\in Z(Q)$ is integral over $Z(A)$, we must have $e_{i}\in Z(A)$. Then

[TABLE]

Clearly $Q_{i}$ is the classical quotient ring of $A_{i}$. It follows that $A_{i}$ is $H$-prime, $A_{i}$ is module-finite over its center $Z(A_{i})$, and $Z(A_{i})$ is integrally closed in $Z(Q_{i})$.

This reduces the proof to the case when $A$ is $H$-prime. But in this case Proposition 7.8 applies. It shows that for each $a\in A$ all coefficients of the characteristic polynomial $P_{Q/{Z(Q)^{H}}}(a,t)$ are in $Z(A)\cap{Z(Q)^{H}}={Z(A)^{H}}$. Since $P_{Q/{Z(Q)^{H}}}(a,a)=0$ by a general property of characteristic polynomials, $a$ is integral over ${Z(A)^{H}}$.   $\mathchar 1027\relax$

The statement of [22, Prop. 3.3] contains the additional assumption that $\mathop{\rm ann}\nolimits_{A}(z)$ is equal to $\mathop{\rm ann}\nolimits_{Z(A)}(z)A$ for each $z\in Z(A)$. The proofs given above show that this assumption is not needed.

In connection with Theorems 7.2, 7.5, 7.9 we are prompted to ask

Question 7.10. Is there any example of an $H$-semiprime algebra $A$ module-finite over its center $Z(A)$ such that $A$ is not integral over ${Z(A)^{H}}$?**

Proposition 7.11 [23]. Suppose that $H$ is semisimple. If $A$ is PI and $A^{H}\subset Z(A)$, then $A$ is integral over $A^{H}$.

Proof. It suffices to consider the case when $A$ is finitely generated. Then integrality of $A$ over $A^{H}$ is equivalent to module-finiteness. There exists a surjective homomorphism of $H$-module algebras $B\to A$ where $B$ has a grading $B=B_{0}\oplus B_{1}\oplus\ldots$ with finite dimensional $H$-stable homogeneous components such that $B_{0}={k}$ and $B_{1}$ generates $B$. For this we can start with the tensor algebra of any finite dimensional $H$-submodule $V\subset A$ which generates $A$ as an algebra. Taking the factor algebra of $B$ by a suitable $H$-stable ideal, we may assume that $B$ is PI. Factoring out another ideal generated by all commutators $xy-yx$ with $x\in B$ and $y\in B^{H}$, we may also assume that $B^{H}\subset Z(B)$.

Since $H$ is semisimple, $B^{H}$ is mapped onto $A^{H}$. Therefore it suffices to show that $B$ is a finite $B^{H}$-module. By the graded Nakayama lemma this holds if and only if $\dim B/B^{H}_{+}B<\infty$ where $B^{H}_{+}=\sum_{i>0}B_{i}^{H}$.

Put $D=B/B^{H}_{+}B$. Note that $B^{H}_{+}B$ is a homogeneous $H$-stable ideal of $B$. Hence $D$ inherits the structure of a graded $H$-module algebra, and $D$ is PI. Since $B^{H}$ maps onto $D^{H}$, we have $D^{H}={k}$. It follows that $D_{+}^{H}=0$ where $D_{+}=\sum_{i>0}D_{i}$.

Let $P$ be any maximal ideal of $D$, and let $P_{H}$ be the largest $H$-stable ideal contained in $P$. By Kaplansky’s theorem the simple algebra $D/P$ is finite dimensional over its center, and, since $D$ is finitely generated, we have $\dim D/P<\infty$ (over ${k}$). Now $P_{H}$ is the kernel of the composite map $D{}\mathrel{\smash{\mathop{\smash{\mathord{-}}\mkern-4.0mu\smash{\mathord{-}}\mkern-4.0mu\mathord{\rightarrow}}\limits^{\rho}}}{}D\otimes H^{*}{}\mathrel{\smash{\mathop{\smash{\mathord{-}}\mkern-4.0mu\mathord{\rightarrow}}\limits}}{}D/P\otimes H^{*}$. It follows that $\dim D/P_{H}<\infty$ too. By Theorem 2.4 $D/P_{H}$ is $H$-simple, which means that $P_{H}$ is a maximal $H$-stable ideal of $D$.

If $P_{H}\not\subset D_{+}$, then $P_{H}+D_{+}=D$. Since all $H$-modules are completely reducible, we deduce that $P^{H}+D_{+}^{H}=D^{H}$ where $P^{H}=P\cap D^{H}$. Hence there exists $d\in D_{+}^{H}$ such that $d\notin P$. This entails $D_{+}^{H}\neq 0$, a contradiction.

Thus $P_{H}\subset D_{+}$ is the only possibility. Since $D_{+}$ is an $H$-stable ideal of $D$, we get $P_{H}=D_{+}$ by maximality of $P_{H}$, but then $P=D_{+}$ too. We conclude that $D$ has a single maximal ideal. Recall that the prime radical of any finitely generated PI algebra is nilpotent by the Braun theorem [45, Th. 6.3.39] and coincides with the intersection of all maximal ideals by the Amitsur-Procesi theorem [45, Th. 6.3.3]. This implies that $D_{+}$ is the prime radical of $D$ and that $D_{+}$ is nilpotent. But then $D_{i}=0$ for sufficiently large $i$. Hence $\dim D<\infty$, as required.   $\mathchar 1027\relax$

The condition $A^{H}\subset Z(A)$ may look artificial, but sometimes it arises very naturally. For instance, this inclusion always holds when $A$ is quantum commutative. In [15] Cohen and Westreich investigated how the condition $A^{H}\subset Z(A)$ affects various properties of an $H$-module algebra, especially in the case when $A$ is an $H^{*}$-Galois extension of $A^{H}$.

In [20] and [21] Eryashkin considered a special class ${\cal A}$ of $H$-module algebras. An $H$-module algebra $A$ belongs to ${\cal A}$ if $A$ has an ideal $I$ such that the factor algebra $A/I$ is commutative and $I$ contains no nonzero $H$-stable ideals of $A$. Such an algebra is PI since it embeds into the algebra $A/I\otimes H^{*}$ which is a finite module over its center. If $z\in A^{H}$, then $\{za-az\mid a\in A\}$ is an $H$-submodule of $A$ contained in the ideal $I$, whence $za=az$ for all $a\in A$ by the conditions imposed on $I$. This shows that $A^{H}\subset Z(A)$.

Starting with an arbitrary left $H$-module $V$ one obtains an $H$-prime algebra in ${\cal A}$ taking $A=T(V)/I^{\prime}_{H}$ where $T(V)$ is the tensor algebra of $V$ and $I^{\prime}_{H}$ is its largest $H$-stable ideal contained in the ideal $I^{\prime}$ generated by all commutators. Here the ideal $I=I^{\prime}/I^{\prime}_{H}$ of $A$ has the property that $A/I$ is the symmetric algebra of $V$. By a careful examination Eryashkin has verified that $A$ is not integral over $A^{H}$ in the case when $\mathop{\rm char}\nolimits{k}=0$, $H$ is the 4-dimensional Hopf algebra described by Sweedler, and $V$ is one of its 2-dimensional indecomposable modules.

In particular, the semisimplicity of $H$ is necessary in Proposition 7.11, even if $A$ is assumed to be $H$-prime. In positive characteristic the previous construction does not give such an example (see Corollary 8.4). This leaves open

Question 7.12. Suppose that $\mathop{\rm char}\nolimits{k}>0$. Is there any example of an $H$-prime PI algebra $A$ such that $A^{H}\subset Z(A)$, but $A$ is not integral over $A^{H}$?**

If $A^{H}\not\subset Z(A)$, then integrality of $A$ over $A^{H}$ should be understood as defined by Schelter [46]. An element $x\in A$ is called Schelter integral over $A^{H}$ if there exists an integer $m>0$ such that $x^{m}$ can be written as a sum of several elements, each of which is a product of elements contained in $A^{H}\cup\{x\}$ with $x$ occurring as a factor in this product less than $m$ times. If all elements of $A$ are Schelter integral over $A^{H}$, then $A$ is said to be Schelter integral over $A^{H}$.

In the 1993 expository lectures Montgomery asked whether Schelter integrality of $A$ over $A^{H}$ holds whenever $H$ is semisimple [38, Question 4.3.1]. By that time the group action case had been settled in full generality by Quinn. If $G$ is a finite automorphism group of a ring $R$ such that $|G|R=R$, then $R$ is Schelter integral over the subring of invariants $R^{G}$. In fact it was proved in [44] that $R$ is fully integral over $R^{G}$, which is a stronger property defined in terms of collections of elements rather than single elements. Quinn also obtained a partial result for Hopf actions:

Theorem 7.13 [44]. Suppose that $H$ is semisimple and the action of $H$ on $A$ is inner. Then each ideal $I$ of $A$ is fully integral, and therefore also Schelter integral, over $I^{H}$ of degree bounded by a function in the dimension of $H$.

The condition that the action is inner means that there exists an invertible element $u\in A\otimes H^{*}$ such that

[TABLE]

In particular, the two nonunital subalgebras $I\otimes 1$ and $\rho(I)$ of $A\otimes H^{*}$ are conjugate by an inner automorphism. It follows that $(I\otimes H^{*})\#H\cong I\otimes\mathop{\rm End}\nolimits_{\,{k}}H$ is fully integral over $\rho(I)\#H$ since $I\otimes\mathop{\rm End}\nolimits_{\,{k}}H$ is known to be fully integral over $I\otimes{k}$ by the Paré-Schelter theorem [42]. Finally, Quinn deduces that $I$ is fully integral over $I^{H}$ using an idempotent $e\in(A\otimes H^{*})\#H$ such that

[TABLE]

as nonunital algebras. In the case of inner action all ideals of $A$ are $H$-stable.

At present it is not known how to extend the previous theorem to arbitrary module algebras for a semisimple Hopf algebra. Special cases of the problem were dealt with in [9], [57]. Eryashkin has succeeded in answering Montgomery’s question in the case of PI algebras:

Theorem 7.14 [23]. Suppose that $H$ is semisimple and cosemisimple. If $A$ is PI, then $A$ is Schelter integral over $A^{H}$.

Proof. The initial idea comes from a paper of Montgomery and Small [40] where a similar problem for group actions on PI rings was considered. By Zorn’s lemma $A$ has an $H$-stable ideal $K$ maximal with respect to the property that all elements of $K$ are Schelter integral over $A^{H}$. It is easy to see that $K$ is $H$-semiprime.

Replacing $A$ with $A/K$, we may assume that $A$ is $H$-semiprime and $A$ has no nonzero $H$-stable ideals which are Schelter integral over $A^{H}$. We have to show that $A=0$. This step requires more effort as compared with the group action case.

Suppose that $A\neq 0$. Since $H$ is cosemisimple, Linchenko and Montgomery tell us that the prime radical of $A$ is $H$-stable [33, Th. 3.5]; hence $A$ is semiprime. Then, by the general PI theory, $A$ has a nonzero ideal $I^{\prime}$ such that for each $x\in I^{\prime}$ the left ideal $Ax$ is contained in a finitely generated $Z(A)$-submodule of $A$. Put $I=HI^{\prime}$. This is a nonzero $H$-stable ideal of $A$. We will show that all elements of $I$ are Schelter integral over $A^{H}$, but this contradicts the assumptions about $A$.

Denote by $C$ the centralizer of $A^{H}$ in $A$. This is an $H$-stable subalgebra of $A$ with $C^{H}\subset Z(C)$ and $Z(A)\subset C$. Let $x\in I$. From the construction of $I$ it follows easily that $Ax$ is contained in a finitely generated $C$-submodule, say $N$, of $A$. Suppose that $N$ is generated as a $C$-module by $e_{1},\ldots,e_{n}$. There exists a finitely generated subalgebra $C_{0}\subset C$ such that $x$ and all elements $e_{i}x$ belong to the $C_{0}$-submodule $N_{0}$ of $N$ generated by $e_{1},\ldots,e_{n}$. Then $N_{0}x\subset N_{0}$.

Without loss of generality we may assume $C_{0}$ to be $H$-stable. Since $C_{0}^{H}\subset Z(C_{0})$, Proposition 7.11 shows that $C_{0}$ is integral, and therefore module-finite, over $C_{0}^{H}$. Hence $N_{0}$ is a finitely generated module over $C_{0}^{H}$. Define $r_{x}\in\mathop{\rm End}\nolimits_{C_{0}}N_{0}$ by the rule $r_{x}(y)=yx$ for $y\in N_{0}$. Since $C_{0}^{H}$ is a commutative ring, the endomorphism $r_{x}$ satisfies a relation

[TABLE]

for some integer $m>0$ and elements $c_{1},\ldots,c_{m}\in C_{0}^{H}$. Applying this operator to $x\in N_{0}$, we deduce that $x^{m+1}+c_{1}x^{m}+\ldots+c_{m}x=0$. It follows that $x$ is Schelter integral over $C_{0}^{H}\subset A^{H}$.   $\mathchar 1027\relax$

8. Comparison with the invariants of the coradical

We continue to assume that $A$ is an $H$-module algebra. If $H$ is pointed with the group $G$ of grouplike elements, it was observed by Artamonov [2] that $A^{H}=A^{G}$ when $\mathop{\rm char}\nolimits{k}=0$ and $A$ is a commutative domain. If $\mathop{\rm char}\nolimits{k}=p>0$, the Hopf algebra is pointed, and $A$ is commutative, then it follows from the results of Totok [56] and Zhu [58] that $z^{p^{s}}\in A^{H}$ for all $z\in A^{G}$ where $s$ is the length of the coradical filtration of $H$.

Etingof and Walton [26] proved the equality ${Z(A)^{H}}=Z(A)^{H_{0}}$ where $H_{0}$ is the coradical of $H$ in the case when $A$ is a prime Azumaya algebra and $\mathop{\rm char}\nolimits{k}=0$. In this section we present stronger conclusions, due to Eryashkin.

Proposition 8.1 [23]. Let $H_{0}\subset H$ be a Hopf subalgebra containing the coradical of $H$. Suppose that $A$ is PI and $H$-simple. Let $K$ be a maximal $H_{0}$-stable ideal of $A$ and $A_{0}=A/K$. Denote by $\nu$ the canonical map $A\to A_{0}$.

*(i) *If $\mathop{\rm char}\nolimits{k}=0$, then {Z(A_{0})^{H_{0}}}=\nu\bigl{(}{Z(A)^{H}}\bigr{)}.

*(ii) *If $\mathop{\rm char}\nolimits{k}=p>0$, then there exists an integer $s\geq 0$ such that z^{p^{s}}\in\nu\bigl{(}{Z(A)^{H}}\bigr{)} for all $z\in{Z(A_{0})^{H_{0}}}$.

Proof. By Corollary 6.7 ${Z(A)^{H}}$ is a field and $[A:{Z(A)^{H}}]<\infty$. Similarly, ${Z(A_{0})^{H_{0}}}$ is a field and $[A_{0}:{Z(A_{0})^{H_{0}}}]<\infty$ since $A_{0}$ is PI and $H_{0}$-simple. Note that $\nu$ maps ${Z(A)^{H}}$ into ${Z(A_{0})^{H_{0}}}$. Define $H$-module structures on

[TABLE]

as in the proof of Theorem 7.5. There is a homomorphism of $H$-module algebras

[TABLE]

where \varphi(a)=(\nu\otimes{\rm id})\bigl{(}\rho(a)\bigr{)}. By Lemma 6.2 ${\cal A}$ is $H$-simple. Hence ${\cal B}$ is a free left ${\cal A}$-module by Theorem 2.9. Let $r$ be its rank. Then

[TABLE]

for all $a\in A$. Here $\nu^{t}:{Z(A)^{H}}[t]\to{Z(A_{0})^{H_{0}}}[t]$ is the homomorphism of polynomial rings induced by the restriction of $\nu$ to ${Z(A)^{H}}$. Thus all coefficients of the polynomial above lie in \nu\bigl{(}{Z(A)^{H}}\bigr{)}.

Now let $z\in{Z(A_{0})^{H_{0}}}$. Pick any $a\in A$ such that $\nu(a)=z$. We have $H_{0}^{*}\cong H^{*}/J$ where $J$ is a Hopf ideal of $H^{*}$. The image of $\rho(a)\in A\otimes H^{*}$ in $A_{0}\otimes H_{0}^{*}$ coincides with $z\otimes 1$ since $z$ is $H_{0}$-invariant. In $A_{0}\otimes H^{*}$ we get then $\varphi(a)-z\otimes 1\in A_{0}\otimes J$. But $J$ is nilpotent since $H_{0}$ contains the coradical of $H$. Hence $\varphi(a)-z\otimes 1$ is nilpotent, and therefore

[TABLE]

where $m=[{\cal B}:{Z(A_{0})^{H_{0}}}]=[A_{0}:{Z(A_{0})^{H_{0}}}](\dim H)$. It follows that

[TABLE]

In particular, mz\in\nu\bigl{(}{Z(A)^{H}}\bigr{)}. If $\mathop{\rm char}\nolimits{k}=0$, this entails z\in\nu\bigl{(}{Z(A)^{H}}\bigr{)}.

Suppose that $\mathop{\rm char}\nolimits{k}=p>0$. Let $p^{s}$ be the largest power of $p$ dividing $m$. Taking $j=p^{s}$, we have ${m\choose j}\not\equiv 0\mskip 12.0mu(\mathop{\rm mod}p)$, whence z^{j}\in\nu\bigl{(}{Z(A)^{H}}\bigr{)}.   $\mathchar 1027\relax$

In [23, Prop. 3.1] it was assumed that $H_{0}$ coincides with the coradical of $H$, but clearly the weaker assumption that $H_{0}$ contains the coradical of $H$ is sufficient.

We have given a slightly different proof. The proof in [23] is based on the embedding of the simple $H$-module algebra $Z(A/M)\otimes_{Z(A)^{H}}A$ into $A/M\otimes H^{*}$ where $M$ is any maximal ideal of $A$ containing $K$. Using this embedding one can see that (ii) holds with $p^{s}$ taken to be the largest power of $p$ dividing the number

[TABLE]

In other words, one obtains a possibly different value of $s$.

Corollary 8.2 [23]. Let $H_{0}\subset H$ be a Hopf subalgebra containing the coradical of $H$. Suppose that $A$ is PI and prime (or at least $H_{0}$-prime). If $\mathop{\rm char}\nolimits{k}=0$, then

[TABLE]

Proof. The quotient ring $Q=Q(A)$ is $H_{0}$-simple, and so Proposition 8.1 applies to the $H$-module algebra $Q$ and its ideal $K=0$.   $\mathchar 1027\relax$

When $\mathop{\rm char}\nolimits{k}>0$ Eryashkin has investigated the relationship between the central invariants for $H$ and for $H_{0}$ in $H$-prime PI algebras. The next result is a consequence of [23, Prop. 3.2]:

Theorem 8.3. Assume that $\mathop{\rm char}\nolimits{k}=p>0$. Let $H_{0}\subset H$ be a Hopf subalgebra containing the coradical of $H$. Suppose that $A$ is PI and $H$-prime. Let $A_{0}=A/P_{0}$ where $P_{0}$ is an $H_{0}$-prime ideal of $A$ containing no nonzero $H$-stable ideals of $A$. If $A_{0}$ is integral over ${Z(A_{0})^{H_{0}}}$, then $A$ is integral over ${Z(A)^{H}}$.

In [23, Th. 3.1] it was assumed additionally that $H_{0}$ is semisimple, which allows one to replace the condition that $A_{0}$ is integral over ${Z(A_{0})^{H_{0}}}$ with two weaker integrality assumptions for intermediate ring extensions.

Corollary 8.4 [21]. Assume that $\mathop{\rm char}\nolimits{k}=p>0$ and $H$ is pointed. If $A$ contains a prime ideal $P$ such that the factor algebra $A/P$ is commutative and $P$ contains no nonzero $H$-stable ideals of $A$, then $A$ is integral over ${Z(A)^{H}}$.

Proof. Here the coradical $H_{0}$ of $H$ is a group algebra ${k}G$. With $P_{0}=\bigcap_{g\in G}gP$ we meet the hypothesis of Theorem 8.3 since $A_{0}=A/P_{0}$ is commutative, and so $A_{0}=Z(A_{0})$ is integral over $A_{0}^{H_{0}}=A_{0}^{G}$ by the classical theory.   $\mathchar 1027\relax$

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