Theorem 1

Suppose that the field K is algebraically closed.

a) Let $M_1$ be an $A_1$-module and $M_2$ an $A_2$-module, both simple (resp. semisimple) and finite-dimensional over K. Then $M_1\otimes M_2$ is a simple (resp. semisimple) module over the ring $A_1\otimes A_2$ and is finite-dimensional over K.

b) The mapping from $\mathscr{S}_K(A_1)\times \mathscr{S}_K(A_2)$ to $\mathscr{S}_K(A_1\otimes A_2)$ that sends (cl(S$_1$), cl(S$_2$)) to cl(S$_1\otimes S_2$), where $S_1($resp. $S_2)$ is a simple $A_1$-module (resp. $A_2$-module) that is finite-dimensional over K, is bijective.

To prove part a), it suffices to consider the case when $M_1$ and $M_2$ are simple. Let $M'$ be an A-submodule of $M = M_1\otimes M_2$; it is an $A_1$-submodule of $M_1\otimes M_2$, stable under the set of endomorphisms of the form $1_{M_1}\otimes u$, where $u$ runs through the set of homotheties of the $A_2$-module $M_2$. Since the field K is algebraically closed, Schur’s lemma (VIII, p. 47, Theorem 1) implies that the commutant End$_{A_1}(M_1)$ of $M_1$ is equal to K. By Corollary 2 of VIII, p. 63, the A-submodule $M'$ of $M_1\otimes M_2$ is of the form $M_1\otimes M'_2$, where $M'_2$ is an $A_2$-submodule of $M_2$. We have assumed that $M_2$ is simple; we therefore have $M'_2= 0$ or $M'_2= M_2$, that is, $M'= 0$ or $M'= M$. Consequently, M is simple.

If S is a simple module over $A_1\otimes A_2$ that is finite-dimensional over K, then it follows from Proposition 2 and part a) that S is isomorphic to a module of the form $S_1\otimes S_2$, where $S_1$ (resp. $S_2$) is a simple $A_1$-module (resp. $A_2$-module). Moreover, as an $A_i$-module, S is isotypical of type $S_i$, so the class of $S_i$ only depends on that of S. This proves part b).