Frobenius Modules Associated to Algebra Automorphisms

Abstract

Here, we study Frobenius bimodules associated with a pair of automorphisms of an algebra and discuss their basic properties. In particular, some equivalent conditions for a finite-dimensional bimodule are proved to be Frobenius and some isomorphisms between Ext-groups and Tor-groups of Frobenius modules over finite dimensional algebras are established.

1. Introduction

The study of Frobenius algebras and Frobenius extensions has a long history. It has long been well known that Frobenius algebras and extensions receive extensive applications; for instance, they are related to Hopf algebras [1,2], topological quantum field theory [3], Yang–Baxter equations [4], representation and homology theory [5,6,7,8,9], Lie theory [10,11], etc. Recently, Frobenius extensions have found applications in matrix theory and invariant theory [12,13,14].

Let R and S be rings and let ${}_{R}M_S$ be a bimodule. Assume that M is projective both as a left R-module and as a right S-module. If M satisfies certain self-dual properties (Definition 2.1 in [15]), then it is call a Frobenius bimodule. Now, assume R is a ring and that S is a subring of R. It was proved in [15] that the ring extension $R/S$ is a Frobenius extension if and only if R, viewed as an R-S-bimodule, is a Frobenius module. Many other properties of ring extensions may be determined by Frobenius bimodules; for example, separable Frobenius extensions are determined by Frobenius bimodules, and two rings are separable equivalent if and only if they are linked by a Frobenius biseparable bimodule [16]. More properties and applications of Frobenius bimodules may be found in [14,15,16,17]. Note that Frobenius bimodules are assumed to be projective as both left modules and right modules; however, many examples show that if we drop the assumption of the projectiveness in the definition of a Frobenius bimodule and keep the self-dual property, the resulting bimodules continue to possess many properties similar to those of Frobenius bimodules.

In this paper, we provide a modified definition of Frobenius bimodules over a single algebra. Let A be an algebra and let ${}_{A}M_A$ be a finite-dimensional bimodule. If M admits a nondegenerate bilinear form which is balanced associated to a pair of automorphisms of A, then we say that M is a Frobenius module (more precisely, see Definition 2). Note that we drop the assumption that M is projective as a left or right A-module. Such Frobenius modules exist extensively; indeed, as is shown in Theorem 3, every finite-dimensional A-bimodule is a direct summand of a Frobenius module. Because a Frobenius module is not necessary projective, it has many nontrivial homological properties.

The rest of this paper is organized as follows.

In Section 2, we provide a precise definition of Frobenius modules associated with a pair of automorphisms of a given algebra, then discuss the basic properties of Frobenius modules. In particular, we provide a criterion condition for a bimodule to be Frobenius (Theorem 4) and prove that the Nakayama automorphism of a Frobenius module is a bimodule homomorphism (Proposition 4 and Corollary 3).

In Section 3, we focus on the homological properties of Frobenius modules over finite-dimensional algebras. The main results of this paper are as follows.

Theorem 1

(=Corollary 4). Let A be a finite dimensional algebra and let $(\varsigma ,\sigma )$ be a pair of automorphisms of A. Assume that M is a $(\varsigma ,\sigma )$-Frobenius module; then, we have isomorphisms

$${Ext}_A^n(M,M)\cong {Ext}_{A^{\circ }}^n(M,M)$$

for all $n\ge 0$, where ${Ext}_A^n(M,M)$ is the Ext-group of the left A-module ${}_{A}M$ and ${Ext}_{A^{\circ }}^n(M,M)$ is the Ext-group of the right A-module $M_A$.

The above theorem shows that the Ext-group of a Frobenius module is left–right symmetric, which is a consequence of a more general result (Theorem 5).

Theorem 2

(=Theorems 6 and 8). Let A and M be the same as in Theorem A and let ${}_{A}X$ and $Y_A$ be finitely generated A-modules; then, we have the following isomorphisms:

(i)

${Tor}_n^A{(X,M)}^{*}\cong {Ext}_{A^{\circ }}^n(X,M^{\sigma })$ for $n\ge 0$;

(ii)

${\underline{Hom}}_A(Y,M)\cong {Tor}_1^A(Tr\left(Y\right),M)$.

In the above theorem, ${\underline{Hom}}_A(Y,M)$ is the stable Hom-set and $Tr\left(Y\right)$ is the Auslander–Retein translation of $Y_A$ (see the main text above Theorem 6). Isomorphism (ii) in the above theorem may be viewed as a new explanation of Auslander–Reiten duality for Frobenius modules.

Throughout this paper, $\mathbb{k}$ is a field with characteristic zero and all algebras and modules considered are over the field $\mathbb{k}$. Letting V be a vector space, we write $V^{*}={Hom}_{\mathbb{k}}(V,\mathbb{k})$.

2. Frobenius Modules Associated with Algebra Automorphisms

Let A be an algebra and let $M_A$ be a right A-module. For $\sigma \in Aut\left(A\right)$, we write $M^{\sigma }$ for the A-module whose right A-action is twisted by $\sigma$. Below, to avoid possible confusion, we use ⋄ to denote the right A-action twisted by $\sigma$, that is,

$$x\diamond a=x\sigma \left(a\right)$$

for $x\in M$ and $a\in A$. Similarly, if ${}_{A}N$ is a left A-module, then ${}^{\sigma }N$ denotes the left A-module obtained from ${}_{A}N$ with the left A-action twisted by the automorphism $\sigma$.

Definition 1.

Let σ be an automorphism of A and let ${}_{A}M_A$ be an A-bimodule.

(i)

A bilinear form $\langle -,-\rangle :M\times M⟶\mathbb{k}$ is said to be σ-inner-balanced if it satisfies the following condition: for all $a\in A$, $x,y\in M$,

$$\langle xa,y\rangle =\langle x,\sigma \left(a\right)y\rangle .$$

(ii)

A bilinear form $\langle -,-\rangle :M\times M⟶\mathbb{k}$ is said to be σ-outer-balanced if it satisfies the following condition: for all $a\in A$, $x,y\in M$,

$$\langle ax,y\rangle =\langle x,y\sigma \left(a\right)\rangle .$$

(iii)

A bilinear form $\langle -,-\rangle :M\times M⟶\mathbb{k}$ is nondegenerate if $\langle x,y\rangle =0$ for all $y\in M$ implies that $x=0$.

A bimodule with a nondegerate balanced bilinear form has nice dual properties.

Proposition 1.

Let A be an algebra and let ${}_{A}M_A$ be a finite-dimensional A-bimodule. The following are equivalent:

(i)

There is a nondegerate σ-inner-balanced bilinear form $\langle -,-\rangle :M\times M⟶\mathbb{k}$;

(ii)

There is an isomorphism of right A-modules $f:M^{{\sigma }^{-1}}⟶M^{*}$;

(iii)

There is an isomorphism of left A-modules $g:{}^{\sigma }M⟶M^{*}$.

Proof. 

(i) ⟹ (ii). We define a linear map $f:M⟶M^{*}$ by setting $f\left(x\right)=\langle x,-\rangle$ for all $x\in M$. For $a\in A$ and $y\in M$, we have

$$f\left(xa\right)\left(y\right)=\langle xa,y\rangle =\langle x,\sigma \left(a\right)y\rangle =f\left(x\right)\left(\sigma \right(a\left)y\right)=\left(f\right(x)·\sigma (a\left)\right)\left(y\right),$$

where $f\left(x\right)·\sigma \left(a\right)$ is the right A-module action on $M^{*}$; therefore,

$$f\left(x{\sigma }^{-1}\left(a\right)\right)=f\left(x\right)·\sigma \left({\sigma }^{-1}\left(a\right)\right)=f\left(x\right)·a.$$

Hence $f:M^{{\sigma }^{-1}}⟶M^{*}$ is a right A-module homomorphism. As the bilinear form is nondegenerate, it follows that f is injective. Because M is finite-dimensional, we have $dim\left(M\right)=dim\left(M^{*}\right)$; hence, f is indeed an isomorphism.

(ii) ⟹ (i). Define a bilinear map $\langle -,-\rangle :M\times M⟶\mathbb{k}$ by setting $\langle x,y\rangle =f\left(x\right)\left(y\right)$. Because $f:M^{{\sigma }^{-1}}⟶M^{*}$ is a right A-module homomorphism, we have

$$\begin{array}{c}\hfill \langle xa,y\rangle =f\left(xa\right)\left(y\right)=f(x\diamond \sigma (a\left)\right)\left(y\right)=\left(f\right(x)·\sigma (a\left)\right)\left(y\right)\\ \hfill =f\left(x\right)\left(\sigma \right(a\left)y\right)=\langle x,\sigma \left(a\right)y\rangle .\end{array}$$

The injectivity of f implies that $\langle -,-\rangle$ is non-degenerated.

(ii) ⟹ (iii). Taking the vector space dual of the right A-module isomorphism f, we obtain an isomorphism of left A-modules $f^{*}:{\left(M^{*}\right)}^{*}⟶{\left(M^{{\sigma }^{-1}}\right)}^{*}$. Note that ${\left(M^{{\sigma }^{-1}}\right)}^{*}={}^{{\sigma }^{-1}}\left(M^{*}\right)$. Let $\tau :M⟶{\left(M^{*}\right)}^{*}$ be the valuation map, that is, $\tau \left(x\right)\left(\alpha \right)=\alpha \left(x\right)$ for all $x\in M$ and $\alpha \in M^{*}$. Now, for $a\in A$ we have $\tau \left(ax\right)\left(\alpha \right)=\alpha \left(ax\right)=(\alpha ·a)\left(x\right)=\tau \left(x\right)(\alpha ·a)=(a·\tau (x\left)\right)\left(\alpha \right)$. Therefore, $\tau$ is an isomorphism of left A-module isomorphism. Setting $g=f^{*}\circ \tau$, g is indeed an isomorphism of left A-modules from ${}^{\sigma }M$ to $M^{*}$.

(iii) ⟹ (ii). This case is similar to the previous case. □

The proof of the above proposition shows that the nondegeneracy of the bilinear form defined in Definition 1(iii) is symmetric.

Corollary 1.

If $\langle -,-\rangle$ is a nondegenerate σ-inner balanced bilinear form defined on M, then $\langle x,y\rangle =0$ for all $x\in M$ implies that $y=0$.

Proof. 

From the proof of Proposition 1, $f=\langle x,-\rangle :M^{{\sigma }^{-1}}⟶M^{*}$ is an isomorphism of right A-modules and $g=f^{*}\circ \tau :{}^{\sigma }M⟶M^{*}$ is an isomorphism of left A-modules. For $y\in M$, we have $g\left(y\right)=f^{*}\left(\tau \left(y\right)\right)=\langle x,y\rangle$. If $\langle x,y\rangle =0$ for all $x\in M$, then $g\left(y\right)=0$. Because g is an isomorphism, it follows that $y=0$. □

Similar to the above proposition, we have the following results for outer-balanced bilinear forms.

Proposition 2.

Let A be an algebra and let ${}_{A}M_A$ be a finite-dimensional A-bimodule. The following are equivalent:

(i)

There is a nondegerate σ-outer-balanced bilinear form $\langle -,-\rangle :M\times M⟶\mathbb{k}$;

(ii)

There is an isomorphism of right A-modules $f:M^{\sigma }⟶M^{*}$;

(iii)

There is an isomorphism of left A-modules $g:{}^{{\sigma }^{-1}}M⟶M^{*}$.

Proof. 

We only show the following two directions, as the others are similar to the proof of Proposition 1.

(i) ⟹ (iii). Similar to the proof of Proposition 1, set $g\left(x\right)=\langle x,-\rangle$ for all $x\in M$. To avoid possible confusion, we use ⋄ to denote the left A-action on ${}^{{\sigma }^{-1}}M$, that is, $a\diamond x={\sigma }^{-1}\left(a\right)x$ for $a\in A$ and $x\in M$. We have $g(a\diamond x)\left(y\right)=\langle {\sigma }^{-1}\left(a\right)x,y\rangle =\langle x,ya\rangle =(a·g\left(x\right))\left(y\right)$ for $y\in M$; therefore, g is a left A-module homomorphism. The injectivity follows from similar arguments as those in Proposition 1.

(iii) ⟹ (ii). Note that ${\left({}^{{\sigma }^{-1}}M\right)}^{*}\cong {\left(M^{*}\right)}^{{\sigma }^{-1}}$. Taking the vector dual of the map g, we obtain $g^{*}:{\left(M^{*}\right)}^{*}⟶{\left({}^{{\sigma }^{-1}}M\right)}^{*}\cong {\left(M^{*}\right)}^{{\sigma }^{-1}}$. As in the proof of the Proposition 1, the valuation map $\tau :M⟶{\left(M^{*}\right)}^{*}$ is a right A-module isomorphism. We obtain an isomorphism $f:=g^{*}\circ \tau :M⟶{\left(M^{*}\right)}^{{\sigma }^{-1}}$. Note that f is indeed an isomorphism $M^{\sigma }⟶M^{*}$. □

In view of the propositions above, we make the following definition of Frobenius modules.

Definition 2.

Let A be an algebra and let ${}_{A}M_A$ be a finite-dimensional A-bimodule. Assume that $(\varsigma ,\sigma )$ is a pair of automorphisms of A.

(i)

If there is a nondegenerate bilinear form

$$\langle -,-\rangle :M\times M⟶\mathbb{k}$$

which is both ς-inner-balanced and σ-outer-balanced, then we call M a $(\varsigma ,\sigma )$-Frobenius module, or simply, a Frobenius module.

If the automorphisms $\varsigma =\sigma =id$, then we call M a balanced Frobenius module.

(ii)

If there is an $id$-inner-balanced nondegenerate bilinear form

$$\langle -,-\rangle :M\times M⟶\mathbb{k}$$

such that $\langle x,y\rangle =\langle y,x\rangle$ for all $x,y\in M$, then we call M a symmetric module.

Remark 1.

In [15], Kadison introduced the notion of a Frobenius bimodule ${}_{R}M_S$ over rings R and S, where M is assumed to be projective both as a left R-module and as a right S-module (see Definition 2.1 in [15]). We drop these assumptions so that the homological properties of a Frobenius module are not trivial.

Next, we provide an example of a Frobenius module.

Example 1.

Let $A=\mathbb{k}\langle x,y\rangle /(xy+yx)$ be a skew polynomial algebra, define an automorphism σ on A by setting $\sigma \left(x\right)=-x$ and $\sigma \left(y\right)=-y$, and let $M=A/(x^{2}A+y^{2}A)$. Then, M is an A-bimodule. Note that M has a basis $\{1,x,y,xy\}$. Defining a bilinear form $\langle -,-\rangle :M\times M\to \mathbb{k}$ by

$$\langle k_0+k_{1}x+k_{2}y+k_{3}xy,l_0+l_{1}x+l_{2}y+l_{3}xy\rangle =k_0{l}_3+k_3{l}_0+k_1{l}_2+k_2{l}_1,$$

where $k_0,\dots ,k_3,l_0,\dots ,l_3\in \mathbb{k}$, it is easy to check that M is an $(id,\sigma )$-Frobenius A-module.

Condition (ii) in Definition 2 is stronger than the condition of balanced Frobenius algebras. Indeed, we have the following proposition.

Proposition 3.

If ${}_{A}M_A$ is a symmetric module, then it is a balanced Frobenius module.

Proof. 

For any $a\in A$, $x,y\in M$, we have $\langle ax,y\rangle =\langle y,ax\rangle =\langle ya,x\rangle =\langle x,ya\rangle$. Hence, M is a balanced Frobenius module. □

Remark 2.

If A is a Frobenius algebra, then it is a Frobenius module when viewed as an A-bimodule. Indeed, from the definition of a Frobenius algebra, there is a nondegenerate bilinear form $\langle -,-\rangle :A\times A⟶\mathbb{k}$ which is $id$-inner-balanced. Assume σ is the Nakayama automorphism of A; then, for all $a,b,c\in A$ we have $\langle ab,c\rangle =\langle a,bc\rangle =\langle bc,\sigma \left(a\right)\rangle =\langle b,c\sigma \left(a\right)\rangle$. Hence, the bilinear form is σ-outer-balanced and ${}_{A}A_A$ is a $(id,\sigma )$-Frobenius module.

The next result shows that Frobenius modules exist extensively. Indeed, every finite dimensional bimodule can be viewed as a direct summand of a Frobenius module.

Theorem 3.

Let A be an algebra and let ${}_{A}M_A$ be a finite dimensional A-bimodule. Let σ be an automorphism of A and set $T_{\sigma }\left(M\right):={}^{\sigma }M^{{\sigma }^{-1}}\oplus M^{*}$. Then, $T_{\sigma }\left(M\right)$ is a $({\sigma }^{-1},\sigma )$-Frobenius module.

In particular, $T\left(M\right):=M\oplus M^{*}$ is a symmetric module.

Proof. 

We define a bilinear map $\langle -,-\rangle :T_{\sigma }\left(M\right)\times T_{\sigma }\left(M\right)⟶\mathbb{k}$ by setting

$$\langle (x,\alpha ),(y,\beta )\rangle =\alpha \left(y\right)+\beta \left(x\right)$$

for all $x,y\in M,\alpha ,\beta \in M^{*}$. Now, for $a\in A$ we have

$$\begin{array}{ccc}\hfill \langle (x,\alpha )·a,(y,\beta )\rangle & =& \langle (x\diamond a,\alpha ·a),(y,\beta )\rangle \hfill \\ & =& (\alpha ·a)\left(y\right)+\beta \left(x{\sigma }^{-1}\left(a\right)\right)\hfill \\ & =& \alpha \left(ay\right)+({\sigma }^{-1}\left(a\right)·\beta )\left(x\right)\hfill \\ & =& \langle (x,\alpha ),({\sigma }^{-1}\left(a\right)\diamond y,{\sigma }^{-1}\left(a\right)·\beta )\rangle \hfill \\ & =& \langle (x,\alpha ),{\sigma }^{-1}\left(a\right)·(y,\beta )\rangle ,\hfill \end{array}$$

and similarly, we have

$$\langle a·(x,\alpha ),(y,\beta )\rangle =\langle (x,\alpha ),(y,\beta )·\sigma \left(a\right)\rangle .$$

The nondegeneracy of the bilinear form is easy to see. Hence, $T_{\sigma }\left(M\right)$ is a $({\sigma }^{-1},\sigma )$-Frobenius module. □

Propositions 1 and 2 imply the following criteria in order for a bimodule to be Frobenius.

Theorem 4.

Let A be an algebra and let ${}_{A}M_A$ be a finite-dimensional A-bimodule. Suppose that $(\varsigma ,\sigma )$ is a pair of automorphisms of A. Then, the following are equivalent:

(i)

M is a $(\varsigma ,\sigma )$-Frobenius;

(ii)

There is an A-bimodule isomorphism ${}^{{\sigma }^{-1}}M^{{\varsigma }^{-1}}\cong M^{*}$;

(iii)

There is an A-bimodule isomorphism ${}^{\varsigma }M^{\sigma }\cong M^{*}$.

Proof. 

(i) ⟹ (ii). As was shown in Proposition 1, the map $f:M^{{\varsigma }^{-1}}⟶M^{*}$ is an isomorphism of right A-modules where $f\left(x\right)=\langle x,-\rangle$. Proposition 2 shows that f is indeed an isomorphism of left A-modules $f:{}^{{\sigma }^{-1}}M⟶M^{*}$. Hence, f is an A-bimodule isomorphism.

(ii) ⟹ (i). This is similar to the proof of Proposition 1.

(ii) ⟺ (iii). This is obtained by taking the vector space dual. □

The following is an immediate consequence of the above theorem.

Corollary 2.

If ${}_{A}M_A$ is a $(\varsigma ,\sigma )$-Frobenius module, then $M^{*}$ is a $(\sigma ,\varsigma )$-Frobenius module.

Let ${}_{A}M_A$ be a $(\varsigma ,\sigma )$-Frobenius module. Similar to Frobenius algebras, there is a Nakayama automorphism of M. Indeed, from Corollary 1, for an element $x\in M$ there is a unique element $x^{\prime }\in M$ such that $\langle x,-\rangle =\langle -,x^{\prime }\rangle$ in $M^{*}$, which induces a linear map $\mathfrak{n}:M\to M$ such that

$$\langle x,y\rangle =\langle y,\mathfrak{n}\left(x\right)\rangle$$

for all $x,y\in M$. Indeed, from Proposition 1 we have $\mathfrak{n}=g^{-1}\circ f$, where f and g are isomorphisms in Proposition 1; hence, $\mathfrak{n}$ is a linear automorphism. We call $\mathfrak{n}$ the Nakayama automorphism of M.

It is clear that ${}_{A}M_A$ is a symmetric module if and only if the Nakayama automorphism of M is the identity map.

Proposition 4.

Let ${}_{A}M_A$ be a $(\varsigma ,\sigma )$-Frobenius module. The Nakayama automorphism $\mathfrak{n}$ is an A-bimodule isomorphism

$$\mathfrak{n}:{}^{\varsigma \sigma }M^{\sigma \varsigma }⟶M.$$

Proof. 

For $x,y\in M$ and $a\in A$, we have

$$\langle y,\mathfrak{n}\left(xa\right)\rangle =\langle xa,y\rangle =\langle x,\varsigma \left(a\right)y\rangle =\langle \varsigma \left(a\right)y,\mathfrak{n}\left(x\right)\rangle =\langle y,\mathfrak{n}\left(x\right)\sigma \left(\varsigma \right(a\left)\right)\rangle ;$$

therefore,

$$\mathfrak{n}\left(xa\right)=\mathfrak{n}\left(x\right)\sigma \varsigma \left(a\right).$$

Similarly,

$$\langle y,\mathfrak{n}\left(ax\right)\rangle =\langle ax,y\rangle =\langle x,y\sigma \left(a\right)\rangle =\langle y\sigma \left(a\right),\mathfrak{n}\left(x\right)\rangle =\langle y,\varsigma \sigma \left(a\right)\mathfrak{n}\left(x\right)\rangle ;$$

therefore,

$$\mathfrak{n}\left(ax\right)=\varsigma \sigma \left(a\right)\mathfrak{n}\left(x\right).$$

Hence, the result follows. □

The above proposition implies the following result.

Corollary 3.

Let σ be an automorphism of A. If ${}_{A}M_A$ is a $(\sigma ,{\sigma }^{-1})$-Frobenius module, then the Nakayama automorphism is an A-bimodule automorphism of ${}_{A}M_A$.

3. Homological Properties of Frobenius Modules over Finite-Dimensional Algebras

In this section, we always assume that A is a finite-dimensional algebra. We write $A^{\circ }$ for the opposite algebra of A. Then, a right A-module can be viewed as a left $A^{\circ }$-module. If X and Z are left A-modules, then we write ${Hom}_A(X,Z)$ and ${Ext}_A^n(X,Z)$ for the Hom-set and extension groups of X and Z, respectively, while if X and Z are right A-modules, then we write ${Hom}_{A^{\circ }}(X,Z)$ and ${Ext}_{A^{\circ }}^n(X,Z)$ for the Hom-set and extension groups of X and Z.

Let $(\varsigma ,\sigma )$ be a pair of automorphisms of A and let ${}_{A}M_A$ be a $(\varsigma ,\sigma )$-Frobenius module. Then, take a projective resolution of the left A-module ${}_{A}M$:

$$\cdots ⟶P_n⟶\cdots ⟶P_1⟶P_0⟶M⟶0,$$

(1)

where $P_n$ is a finitely generated projective left A-module for all $n\ge 0$. Twisting the left A-actions on the modules in the above sequence, we obtain a projective resolution of ${}^{\sigma }M$:

$$\cdots ⟶{}^{\sigma }P_n⟶\cdots ⟶{}^{\sigma }P_1⟶{}^{\sigma }P_0⟶{}^{\sigma }M⟶0.$$

(2)

Taking the vector space dual of sequence (1), we obtain the following exact sequence:

$$0⟶M^{*}⟶P_0^{*}⟶P_1^{*}⟶\cdots ⟶P_n^{*}⟶\cdots .$$

(3)

Because $P_n$ is a projective left A-module for every $n\ge 0$, it follows that $P_n^{*}$ is an injective right A-module for every $n\ge 0$. Therefore, the exact sequence (3) is an injective resolution of the right module $M^{*}$. From Theorem 4, we have the A-bimodule isomorphism ${}^{\varsigma }M^{\sigma }\cong M^{*}$. Hence, the exact sequence (3) is an injective resolution of the right A-module $M^{\sigma }$. Then, we have the following injective resolution of the right A-module M:

$$0⟶M⟶{\left(P_0^{*}\right)}^{{\sigma }^{-1}}⟶{\left(P_1^{*}\right)}^{{\sigma }^{-1}}⟶\cdots ⟶{\left(P_n^{*}\right)}^{{\sigma }^{-1}}⟶\cdots ,$$

(4)

or equivalently,

$$0⟶M⟶{\left({}^{{\sigma }^{-1}}P_0\right)}^{*}⟶{\left({}^{{\sigma }^{-1}}P_1\right)}^{*}⟶\cdots ⟶{\left({}^{{\sigma }^{-1}}P_n\right)}^{*}⟶\cdots .$$

(5)

Now, let ${}_{A}Y$ be a finite-dimensional left A-module and consider the right A-module ${\left({}^{{\sigma }^{-1}}Y\right)}^{*}$. Applying the functor ${Hom}_{A^{\circ }}({\left({}^{{\sigma }^{-1}}Y\right)}^{*},-)$ to the injective resolution (5), we obtain the following complex:

$$0⟶{Hom}_{A^{\circ }}({\left({}^{{\sigma }^{-1}}Y\right)}^{*},{\left({}^{{\sigma }^{-1}}P_0\right)}^{*})⟶\cdots ⟶{Hom}_{A^{\circ }}({\left({}^{{\sigma }^{-1}}Y\right)}^{*},{\left({}^{{\sigma }^{-1}}P_n\right)}^{*})⟶\cdots .$$

(6)

Taking the n-th cohomology of complex (6), we obtain the extension group ${Ext}_{A^{\circ }}^n({\left({}^{{\sigma }^{-1}}Y\right)}^{*},M)$.

Notice that we have

$${Hom}_{A^{\circ }}({\left({}^{\theta }X\right)}^{*},{\left({}^{\theta }Z\right)}^{*})\cong {Hom}_A(Z,X)$$

for any finite-dimensional left A-modules $X,Z$ and any automorphism $\theta$ of A. The sequence (6) is equivalent to the following complex:

$$0⟶{Hom}_A(P_0,Y)⟶{Hom}_A(P_1,Y)⟶\cdots ⟶{Hom}_A(P_n,Y)⟶\cdots ,$$

(7)

which is exactly the complex obtained from projective resolution (1) by applying the functor ${Hom}_A(-,Y)$. The n-th cohomology of complex (7) is the extension group ${Ext}_A^n(M,Y)$. Therefore, we have

$${Ext}_A^n(M,Y)\cong {Ext}_{A^{\circ }}^n({\left({}^{{\sigma }^{-1}}Y\right)}^{*},M)$$

for all $n\ge 0$.

Notice that from Theorem 4 we also have an A-bimodule isomorphism ${}^{{\sigma }^{-1}}M^{{\varsigma }^{-1}}$. Replacing the isomorphism ${\sigma }^{-1}$ in sequences (5) and (6) with $\varsigma$, we finally obtain the isomorphism

$${Ext}_A^n(M,Y)\cong {Ext}_{A^{\circ }}^n({\left({}^{\varsigma }Y\right)}^{*},M)$$

for all $n\ge 0$.

Summarizing the above narratives, we obtain the following result.

Theorem 5.

Let ${}_{A}M_A$ be a $(\varsigma ,\sigma )$-Frobenius module and let ${}_{A}Y$ be a finite-dimensional module. For each $n\ge 0$, we have

$${Ext}_A^n(M,Y)\cong {Ext}_{A^{\circ }}^n({\left({}^{{\sigma }^{-1}}Y\right)}^{*},M)\cong {Ext}_{A^{\circ }}^n({\left({}^{\varsigma }Y\right)}^{*},M).$$

The above theorem implies that the Ext-groups of a Frobenius module are left–right symmetric.

Corollary 4.

Let ${}_{A}M_A$ be a $(\varsigma ,\sigma )$-Frobenius module. For each $n\ge 0$, we have

$${Ext}_A^n(M,M)\cong {Ext}_{A^{\circ }}^n(M,M).$$

Proof. 

Note that as a right A-module, from Theorem 4 we have

$$M\cong {\left(M^{*}\right)}^{{\sigma }^{-1}}\cong {\left({}^{{\sigma }^{-1}}M\right)}^{*}.$$

The result follows from Theorem 5 by setting $Y=M$. □

Next, let $X_A$ be a right A-module. Applying the functor ${(X{\otimes }_A-)}^{*}$ to the projective resolution (1) of M, we obtain the following complex:

$$0⟶{\left(X{\otimes }_A{P}_0\right)}^{*}⟶{\left(X{\otimes }_A{P}_1\right)}^{*}⟶\cdots ⟶{\left(X{\otimes }_A{P}_n\right)}^{*}⟶\cdots ,$$

(8)

the n-th cohomology of which is equal to ${Tor}_n^A{(X,M)}^{*}$.

Note that this complex is equivalent to the following complex:

$$0⟶{Hom}_{A^{\circ }}(X,{\left(P_0\right)}^{*})⟶{Hom}_{A^{\circ }}(X,{\left(P_1\right)}^{*})⟶\cdots ⟶{Hom}_{A^{\circ }}(X,{\left(P_n\right)}^{*})⟶\cdots .$$

(9)

From sequence (4), we can see that complex (9) is indeed obtained by applying the functor ${Hom}_{A^{\circ }}(X,-)$ to the injective resolution of $M^{\sigma }$. Hence, the n-th cohomology of complex (9) is the extension group ${Ext}_{A^{\circ }}^n(X,M^{\sigma })$.

Summarizing the above narratives, we obtain the following Tor–Ext translation.

Theorem 6.

Let ${}_{A}M_A$ be a $(\varsigma ,\sigma )$-Frobenius module and let $X_A$ be a right A-module. Then, we have the following isomorphisms:

$${Tor}_n^A{(X,M)}^{*}\cong {Ext}_{A^{\circ }}^n(X,M^{\sigma })$$

for all $n\ge 0$.

Let ${}_{A}Y$ be a finitely generated left A-module. There is an Auslander–Reiten transpose $Tr\left(Y\right)$ of Y (for instance, see [6]) which is defined as follows. Let

$$P_1\stackrel{d}{⟶}{P}_0⟶Y⟶0$$

be a minimal projective resolution of Y; applying the functor ${Hom}_A(-,A)$ to the left A-module homomorphism $P_1\stackrel{d}{⟶}{P}_0$, we denote the result of the right A-module homomorphism by ${Hom}_A(d,A)$, that is, we have

$${Hom}_A(d,A):{Hom}_A(P_0,A)⟶{Hom}_A(P_1,A).$$

The Auslander–Reiten transpose is defined to be the right A-module:

$$Tr\left(Y\right):=coker{Hom}_A(d,A).$$

Let ${}_{A}X$ be another finitely generated left A-module. Let $P(Y,X)$ be the subspace of ${Hom}_A(Y,X)$ consisting of homomorphisms f which factors through a projective module, that is, there is a projective module P such that $f=gh$ where $h:Y\to P$, $g:P\to X$. Now, we write

$${\underline{Hom}}_A(Y,X):={Hom}_A(Y,X)/P(Y,X).$$

The isomorphism in the next theorem is called the Auslander–Reiten duality.

Theorem 7

([18]). Let ${}_{A}X$ and ${}_{A}Y$ be finitely generated left A-modules. There is an isomorphism

$${\underline{Hom}}_A{(Y,X)}^{*}\cong {Ext}_A^1(X,Tr{\left(Y\right)}^{*}).$$

Lemma 1.

Let ${}_{A}Y$ be a finitely generated left A-module. Assume that θ is an automorphism of A. We have $Tr\left({}^{\theta }Y\right)\cong Tr{\left(Y\right)}^{\theta }$.

Proof. 

Let $P_1\stackrel{d}{⟶}{P}_0⟶Y⟶0$ be a minimal projective resolution of Y; then, ${}^{\theta }P_1\stackrel{{}^{\theta }d}{⟶}{}^{\theta }P_0⟶{}^{\theta }Y⟶0$ is a minimal projective resolution of ${}^{\theta }Y$, where the map ${}^{\theta }d$ is indeed the same as d. Applying the functor ${Hom}_A(-,A)$, we have the following right A-module homomorphism:

$${Hom}_A({}^{\theta }d,A):{Hom}_A({}^{\theta }P_0,A)⟶{Hom}_A({}^{\theta }P_1,A).$$

(10)

Because ${}^{{\theta }^{-1}}A\cong A^{\theta }$ as A-bimodules, we have the following right A-module isomorphisms:

$${Hom}_A({}^{\theta }P_0,A)\cong {Hom}_A(P_0,{}^{{\theta }^{-1}}A)\cong {Hom}_A(P_0,A^{\theta })\cong {Hom}_A{(P_0,A)}^{\theta }.$$

Then, the right A-module homomorphism in (10) is equivalent to the following map:

$${Hom}_A{(d,A)}^{\theta }:{Hom}_A{(P_0,A)}^{\theta }⟶{Hom}_A{(P_1,A)}^{\theta }.$$

(11)

Hence,

$$Tr\left({}^{\theta }Y\right)=coker{Hom}_A({}^{\theta }d,A)\cong coker{Hom}_A{(d,A)}^{\theta }=Tr{\left(Y\right)}^{\theta }.$$

 □

Remark. We end this note with the following explanation of Auslander–Reiten duality for Frobenius modules.

Theorem 8.

Let $(\varsigma ,\sigma )$ be a pair of automorphisms of A. Assume that ${}_{A}M_A$ is a $(\varsigma ,\sigma )$-Frobenius module and that ${}_{A}Y$ is a finitely generated left A-module. Then, we have the following isomorphism:

$${\underline{Hom}}_A(Y,M)\cong {Tor}_1^A(Tr\left(Y\right),M).$$

Proof. 

From the Auslander–Reiten duality, we have

$${\underline{Hom}}_A{(Y,M)}^{*}\cong {Ext}_A^1(M,Tr{\left(Y\right)}^{*}).$$

From Theorem 5,

$${Ext}_A^1(M,Tr{\left(Y\right)}^{*})\cong {Ext}_{A^{\circ }}^1({\left({}^{{\sigma }^{-1}}\left(Tr{\left(Y\right)}^{*}\right)\right)}^{*},M).$$

Now, per Lemma 1,

$${\left({}^{{\sigma }^{-1}}\left(Tr{\left(Y\right)}^{*}\right)\right)}^{*}\cong {\left(Tr{\left(Y\right)}^{{\sigma }^{-1}}\right)}^{*}{)}^{*}\cong Tr{\left(Y\right)}^{{\sigma }^{-1}}\cong Tr\left({}^{{\sigma }^{-1}}Y\right);$$

therefore, we have

$${\underline{Hom}}_A{(Y,M)}^{*}\cong {Ext}_{A^{\circ }}^1(Tr\left({}^{{\sigma }^{-1}}Y\right),M).$$

(12)

Because the functor ${(-)}^{\sigma }$ is an auto-equivalence of the Abelian category of right A-modules, we obtain the following isomorphisms:

$${Ext}_{A^{\circ }}^1(Tr\left({}^{{\sigma }^{-1}}Y\right),M)\cong {Ext}_{A^{\circ }}^1(Tr{\left({}^{{\sigma }^{-1}}Y\right)}^{\sigma },M^{\sigma })\cong {Ext}_{A^{\circ }}^1(Tr\left(Y\right),M^{\sigma }).$$

(13)

From Theorem 6, we have the isomorphism

$${Ext}_{A^{\circ }}^1(Tr\left(Y\right),M^{\sigma })\cong {Tor}_1^A{(Tr\left(Y\right),M)}^{*}.$$

(14)

Combining isomorphisms (12)–(14), we obtain

$${\underline{Hom}}_A{(Y,M)}^{*}\cong {Tor}_1^A{(Tr\left(Y\right),M)}^{*}.$$

Hence, the result follows. □

Conclusions

In this short note, we have introduced the notion of Frobenius modules over a single algebra, which is a modification of the concept of Frobenius bimodules in the literature. Because a Frobenius module in our sense is not necessary projective as a left module or a right module, it enjoys many nontrivial homological properties. We have proved some symmetric properties of Ext-groups and Tor-groups of Frobenius modules. The following questions deserve further consideration:

(a)

Is the differential graded structure on the co-chain complex of a Frobenius module still left–right symmetric?

(b)

Does a Frobenius module have certain duality properties between Hochschild cohomology and Hochschild homology, say, Poincare duality?

(c)

Does a Frobenius module relate to certain weak versions of Frobenius extensions?