Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}\_R (M,N)$ replaced by $M \otimes\_R N$. So my questions are:


Can we describe $M \otimes\_R N$ explicitly?


In particular, for a fixed $M$, is $N$ such that $M \otimes\_R N \neq 0$ unique up to isomorphism? If not, can we classify such $N$'s in a reasonable way?



Tensor product of simple modules
Sasha's statement is true for any pair of modules.


The center of $R$ is a commutative ring $S$. Since the endomorphisms of a simple module are a division algebra, whose center is a field, the action of $S$ on every simple module factors through some field, so the action of $R$ of course factors through an algebra over that field.


The kenel of a map to a field is a prime ideal $p$, and the map to the field factors through the residue field $k\_p$.


So if we have two finitely-generated modules $M$ and $N$, their annihilators in $S$ are two prime ideals of $S$, $p\_1,p\_2$. If the ideals are distinct, then $S$ annihilates $M \otimes\_R N$ since the action of $S$ factors through $k\_{p\_1} \otimes\_S k\_{p\_2}=0$. The tensor product is zero because one ideal necessarily contains an element $e$ not in the other. In the residue field that element, since it's not in the ideal, has an inverse. Then $1= 1\otimes 1= e^{-1}e\otimes 1=e^{-1}\otimes e=e^{-1} \otimes 0 =0$.


If they are the same ideal, set $R'= R\otimes\_S k\_p$. It is now an algebra over a field. Apply Sasha's statement.


