Assume we have a noncommutative ring $R$ with exactly 2 non-isomorphic simple left modules $S_1$ and $S_2$ (up to isomorphism) and an $R$-bimodule $M$, which switches the simples, i.e. $M\otimes_R S_1=S_2$ and $M\otimes_R S_2=S_1$.

Then we have $Hom_R(S_i,M\otimes_R S_i)=0$ by Schur's lemma ($*$).

Now assume $F$ is an arbitrary left $R$-module of finite length. What can be said about $Hom_R(F,M\otimes_R F)$ with the help of ($*$)?

If $F$ is a quotient of $R$, i.e. we have a surjection $R\rightarrow F \rightarrow 0$, then we get an injection $0\rightarrow Hom_R(F,M\otimes_R F)\rightarrow Hom_R(R,M\otimes_R F)$. Is there a method, using ($*$) to decide whether this is even an isomorphism?

So for example, if $F$ is one of the simples or a direct sum of one of the simples, then $Hom_R(F,M\otimes_R F)=0$ so the map is not an isomorphism. Otherwise both simple modules occur at least once in a Jordan-Hölder composition series. What can be said in this case?

If this question is too broad, the ring $R$ I'm interested in is the following subring of $M_2(A)$: \begin{pmatrix}A &A \\ xA &A \end{pmatrix} where $A$ is a complete regular local ring of dimension 2, and $x\in A$ s.t. $A=\mathbb{C}[[x,y]]$. The $R$-bimodule $M$ is given as the following $R$-submodule in $M_2(Quot(A))$: \begin{pmatrix}A &x^{-1}A \\ A &A \end{pmatrix}

  • $\begingroup$ Is the ring commutative? (as the tag says) Are $S_1$ and $S_2$ the only simple $R$-modules up to isomorphism? $\endgroup$ Jan 11, 2012 at 12:38
  • $\begingroup$ No the ring i'm interested in is not commutative. I gave a description of the ring and changed the tags, thanks for pointing that out. And yes the ring has exactly 2 simple modules up to isomorphism. $\endgroup$
    – TonyS
    Jan 11, 2012 at 19:00

2 Answers 2


The question is too broad in general, but I believe that there is a nice answer for your particular example. The first thing to note is that the given $M$ is a twisted bimodule. Namely, let $\sigma$ be the automorphism of $R$ given by $$\sigma : \begin{pmatrix} r & s \\ xt & u \end{pmatrix} \mapsto \begin{pmatrix} u &t \\ xs &r \end{pmatrix}.$$ Then we can define a bimodule isomorphism $\varphi : {}_1 R_{\sigma} \rightarrow M$ by $$\varphi : \begin{pmatrix} a &b \\ c &d \end{pmatrix} \mapsto \begin{pmatrix} b & x^{-1}a \\ d & c \end{pmatrix}.$$

It follows that tensoring with $M$ can be thought of as the endofunctor on the category of (left) R-modules that is induced by the automorphism $\sigma^{-1}$ ($= \sigma$). Consequently, the modules $F$ and $M \otimes_R F$ have the same structure (in a sense), but with the opposite composition factors. (To contrast with the general situation, the assumption that tensoring with $M$ swaps the simples, alone, does not guarantee that tensoring with $M$ preserves the length of finite-length modules.)

Thus, if you know the structure of $F$, it should not be hard to see if there are any nonzero morphisms from $F$ to $M \otimes_R F$. For example, if $F$ contains a factor of $S_1$ in its top and a factor of $S_2$ in its socle, then $M \otimes_R F$ contains a factor of $S_1$ in its socle, and there will be a nonzero map from $F$ to $M \otimes_R F$. On the other hand, if $F$ is uniserial with composition factors $S_1, S_2, S_1, S_1$ (from the top down), then $\mbox{Hom}_R(F,M\otimes_R F) = 0$.

However, for a finite-length module $F$ over the given $R$, it appears plausible that $\mbox{Hom}_R(F, M \otimes_R F)=0$ if and only if all composition factors of $F$ are isomorphic. I do not have a proof at the moment, but one might try to show that for any $F$ of finite length with $S_1$ in its top, at least one of the three cases occurs: 1) all composition factors of $F$ are isomorphic to $S_1$; 2) the socle of $F$ contains a copy of $S_2$; or 3) the length-$2$ uniserial module with composition factors $S_1, S_2$ is a quotient of $F$ and the length-$2$ uniserial with composition factors $S_2, S_1$ is a submodule of $F$. In cases 2) and 3) there will always be a nonzero map from $F$ to $M \otimes_R F$. (Note: this suggestion is based on my belief that $R$ is isomorphic to the completed path algebra of the quiver with 2 vertices, a loop at each vertex and a pair of arrows connecting the vertices in each direction, modulo the relations that the two paths of length 2 from each vertex to the other are equal. If this is incorrect, then the structure of the finite-length $R$-modules may be more complicated than I envision.)

We can also see that for a cyclic module $F$, in order for the map $\mbox{Hom}_R(F,M\otimes_R F) \rightarrow \mbox{Hom}_R(R,M \otimes_R F)$ to be an isomorphism, there must be an epimorphism $F \rightarrow M \otimes_R F$, which would have to be an isomorphism since these two modules have the same length.


Suppose $R$ is finite dimensional over an algebraically closed field and basic, so that we can suppose it is in fact an admissible quotient of the path algebra of a quiver. Your hypothesis on the simples means then that there are two vertices in the quiver. Let $e_1$ and $e_2$ be the corresponding idempotents, and $S_1$ and $S_2$ the simples.

If $M$ is the bimodule, then as a vector space we have $M=e_1Me_1\oplus e_1Me_2\oplus e_2Me_1\oplus e_2Me_2$. Then $M\otimes_RS_1=e_1Me_1\oplus e_2Me_1\cong S_2$, so that $e_1Me_1=0$ and $e_2Me_1$ is one-dimensional. Likewise, $e_2Me_2=0$ and $e_1Me_2$ is one-dimensional. If follows that $M$ is completely determined by the conditions.

Of course, your example is outside of these hypotheses! :)


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