Tensor product of simple modules 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 $M \otimes_R N$...
 A: 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 ... .
