Theorem 2

Let $A_1$ and $A_2$ be algebras over the commutative field K; let $S_1$ be a simple $A_1$-module and $S_2$ a simple $A_2$-module. Let $D_1$ and $D_2$ be the respective commutants of $S_1$ and $S_2$. Set $M = S_1\otimes S_2,A = A_1\otimes A_2$, and $D = D_1\otimes D_2$. We view M as a left $(A,D)$-bimodule.

a) The commutant of the A-module M is equal to $D_M$.

b) The mapping $\mathfrak{a}\mapsto \mathfrak{a}M$ is an isomorphism from the set of right ideals of D, ordered by inclusion, to the set of A-submodules of M, ordered by inclusion. The inverse mapping sends a submodule N of M to the ideal of D consisting of the elements $d$ such that $dM\subset N$.

Assertion a) follows from Lemma 1 because a simple module is monogenous.

Let T be the $(A_1,D_2$)-bimodule $S_1\otimes (D_2)_d$. We identify $M = S_1\otimes S_2$ with $T\otimes_{D_2}S_2$ (II, §3, No. 8, p. 258); this identification is compatible with the structures of left modules over the ring $A = A_1\otimes A_2$.

Let N be an A-submodule of M; it is an $A_2$-submodule of $T\otimes_{D_2}S_2$, stable by the endomorphisms of the form $(a_1)_T\otimes 1_{S_2}$ for $a_1$ running through $A_1$. It follows from Corollary 2 of VIII, p. 63 that there exists a unique $(A_1,D_2)$-sub-bimodule V of T such that $N = V\otimes_{D_2}S_2$.

The isomorphism $u$ from $T = S_1\otimes (D_2)_d$ to $((D_1)_d\otimes (D_2)_d)\otimes_{D_1}S_1$ defined by $u(s\otimes d) = 1\otimes d\otimes s$ is $(A_1,D_2$)-linear. We identify these $(A_1,D_2)$-bimodules. A reasoning analogous to that given above proves the existence and uniqueness of a right $(D_1\otimes D_2$)-submodule $\mathfrak{a}$ of $D_1\otimes D_2$ such that $V =\mathfrak{a}\otimes_{D_1}S_1$. In view of the identifications made above, $\mathfrak{a}$ is the unique right ideal of $D = D_1\otimes D_2$ such that $N =\mathfrak{a}M$.

We have just proved that the mapping $\mathfrak{a}\mapsto \mathfrak{a}M$ is bijective; the last assertion follows.