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Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. My questions are:

How can we describe $M \otimes_R N$ explicitly? Well, I guess that it is a quotient of $R$ by a sum of a left and a right ideal, but it seems somehow unsatisfactory...

Is $N$ such that $M \otimes_R N \neq 0$ uniquely determined by $M$, up to isomorphism? If not, can we classify such $N$'s in a reasonable way?

Generally, I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}_R (M,N)$ replaced by $M \otimes_R N$...

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    Ah, sorry, I misinterpreted the meaning of "unique."2012-08-22

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Here are some things that you probably know, but just to say something ... :

Suppose that $R$ is a $k$-algebra for a field $k$, and that $M$ is finite dimensional over $k$. Then $M$ is simple as a right $R$-module if and only if $Hom_k(M,k)$ (with the transpose $R$-action) is simple as a left $R$-module. Then $M\otimes_k N$ is a $k$-vector space, and since it is the universal recipient of a $R$-bilinear pairing from $M\times N$, it is non-zero if and only if there is a $R$-linear pairing $M \times N \to k$, if and only if there is a non-zero $R$-module homomorphism $N \to Hom_k(M,k)$. If $N$ is simple, such a non-zero homomorphism has to be an isomorphism, and so we see that $M\otimes N \neq 0$ for simple $M$ and $N$ if and only if $N = Hom_k(M,k)$.

Of course my assumptions imply that $M\otimes_R N = Hom_R(Hom_k(M,k),N),$ and so this reduces to the "Hom" case that is motivating your whole question.

But maybe it suggests a way of thinking which might work in a more general setting (trying to replace $k$ by some kind of "minimal" quotient of $M\otimes_R N$). I didn't actually succeed yet in saying anything more general, though ... .