Is a $R \otimes S$-module of the form $V \otimes W$ with $V$ simple $R$-module and $W$ simple $S$-module a simple module?

Question

Let $R$ and $S$ be $\mathbb{C}$-algebras. Show that $X$ is a simple $R \otimes S$-module if and only if it is isomorphic to a $R \otimes S$-module of the form $V \otimes W$ with $V$ simple $R$-module and $W$ simple $S$-module.

Starting with the backwards direction, I know that if we assume the semisimplicity of $R$ and $S$, I can show that the endomorphism ring of $V \otimes W$ is a division ring (it is isomorphic to $\mathbb{C}$), since we can also show $R \otimes S$ is finite dimensional and semisimple, that would imply that $V \otimes W$ is simple.

But now I would like to show both directions $without$ assuming the semisimplicty of $R$ and $S$, and I'm lost.

Answer 1

(I assume all tensor products are taken over $\mathbb{C}$ here. Also, this is not a complete answer.)

This is false if $R$ and $S$ can be infinite-dimensional. Take $R = S = \mathbb{C}(t)$; then the only tensor product of simple modules available is the free module $R \otimes S$ of rank $1$, but $R \otimes S$ is not a field (e.g. it admits a natural surjective map to $\mathbb{C}(t)$ given by multiplication which has nontrivial kernel, so it has a nonzero proper ideal) so it is not simple as a module over itself.

For a more complicated counterexample involving the Weyl algebra see this math.SE answer. According to that question it should be true if we assume that $X$ is finite-dimensional but I'm not sure how to prove that.

If $R$ and $S$ are finite-dimensional, then a simple $R \otimes S$-module is a module over the quotient $(R \otimes S)/J(R \otimes S)$ by the Jacobson radical, and the tensor product $V \otimes W$ of a simple $R$-module and a simple $S$-module is a module over the tensor product $R/J(R) \otimes S/J(S)$. There's a natural map

$$R/J(R) \otimes S/J(S) \to (R \otimes S)/J(R \otimes S)$$

and it would suffice to prove that this is an isomorphism, because then we'd be reduced to the semisimple case. I think this is true (we will need to use the fact that $\mathbb{C}$ is algebraically closed, or at least that it's perfect; this is false over a non-perfect field) but I also don't know how to prove this off the top of my head.

Answer 2

Let's begin with some generalities. Let $K$ be any field, $R$, $S$ two $K$-algebras, and $V$ (resp. $W$) a simple right module over $R$ (resp. $S$). As Qiaochu noted, $V\otimes_KW$ need not be simple if $V$ and $W$ are not finite dimensional, so let's restrict to that case. Set $D=\mathrm{End}_R(V)$ and $E=\mathrm{End}_S(W)$, division algebras acting on the left on $V$ and $W$ respectively.

Passing to $R/\mathrm{Ann}(V)$ we may assume that $R$ is a finite dimensional simple $K$-algebra. Applying Morita theory, we may assume that $R=D$ (no opposite needed since endomorphisms act on the left), and $V=R$ as right modules. Similarly for $S$. Thus $V\otimes_KW$ is just the regular representation for $D\otimes_KE$.

Let $Z(D)$ be the centre of $D$, and set $L:=Z(D)\otimes_KZ(E)$. Then $D\otimes_KE\cong D\otimes_{Z(D)}L\otimes_{Z(E)}E$. The properties of this are all controlled by the properties of $L$. In particular, $V\otimes_KW$ is simple if and only if $L$ is a field.

To see this, suppose $L$ is a field. Then $D$ is a central simple algebra (CSA) over $Z(D)$, so $D\otimes_{Z(D)}L$ is a CSA over $L$. Similarly for $L\otimes_{Z(E)}E$, and so their tensor product over $L$ is again a CSA over $L$.

If $L$ is a proper product of rings, then so too is $D\otimes_KE$, and hence $V\otimes_KW$ has proper summands. If $L$ has nilpotent elements, then so too does $D\otimes_KE$, and $V\otimes_KW$ has a proper submodule.

Finally, assume $K$ is algebraically closed. Then $D=K=E$, so $L=K$, and the tensor product $V\otimes_KW$ is simple.