Proposition 5

Let $A_1$ and $A_2$ be algebras over the commutative field K. Let $M_1$ be a semisimple $A_1$-module that is finite-dimensional over K and $M_2$ a semisimple $A_2$-module. For $i\in \{1,2\}$, denote the commutant of the $A_i$-module $M_i$ by $D_i$ and the center of $D_i$ by $Z_i$. Set $A = A_1\otimes A_2$, M = $M_1\otimes M_2,D = D_1\otimes D_2$, and $Z = Z_1\otimes Z_2$.

a) The commutant of the A-module M can be identified with D, and its center is Z. If the $A_2$-module $M_2$ has finite length, then the A-module M has finite length, the ring D is right and left Artinian, and the ring Z is Artinian.

b) The following properties are equivalent:

(i) The A-module M is semisimple.

(ii) The ring Z is isomorphic to the product of a family of commutative fields.

(iii) The ring Z is reduced.

c) The following properties are equivalent:

(i) The A-module M is isotypical and not reduced to 0.

(ii) The ring Z is a field.

(iii) The ring Z is an integral domain.

By assumption, $M_1$ is finite-dimensional over K. The commutant of M can therefore be identified with D (VIII, p. 213, Lemma 1), and its center is Z (III, §4, No. 4, p. 468, Corollary). Suppose that the $A_2$-module $M_2$ has finite length. The A-module M has finite length by Lemma 2. Since $M_2$ is semisimple and finitely generated, the ring $D_2$ is semisimple (VIII, p. 139, Proposition 6), and its center $Z_2$ is the product of a finite family of commutative fields. Consequently, the $D_2$-module $(D_2)_s$ and the $Z_2$-module $(Z_2)_s$ each have finite length. Furthermore, since $M_1$ is finite-dimensional over K, so are $D_1$ and $Z_1$. By Lemma 2, the module $(D_1\otimes D_2)_s$ has finite length; therefore, the ring $D_1\otimes D_2$ is left Artinian. We prove likewise that the ring $D_1\otimes D_2$ is right Artinian and that the ring $Z_1\otimes Z_2$ is Artinian. Assertion a) follows.

Let us prove b). The center of the commutant of a semisimple module is isomorphic to the product of a family of commutative fields (VIII, p. 87, Proposition 8, a)); this proves that (i) implies (ii). The implication (ii) $\Rightarrow$ (iii) is clear.

Suppose that the ring Z is reduced. We then have $\mathfrak{R}(Z) = 0$ (VIII, p. 215, Theorem 3) and $\mathfrak{R}_A(M) = 0$ (VIII, p. 218, Proposition 4). Since the $A_2$-module $M_2$ is semisimple, there exist a family $(S_i)_{i\in I}$ of simple $A_2$-modules and an isomorphism from $M_2$ to $\bigoplus S_i$. Consequently, the A-module M is isomorphic to $\bigoplus M_1\otimes S_i$. For every $i\in I$, the A-module $M_1\otimes S_i$ is therefore without radical. By a), it has finite length; hence it is semisimple (VIII, p. 153, Proposition 3, b)). The A-module M is then the direct sum of a family of semisimple modules and is consequently semisimple. This proves that (iii) implies (i) and concludes the proof of b).

An A-module is isotypical and nonzero if and only if it is semisimple and the center of its commutant is a field (VIII, p. 87, Proposition 8, b)). So c) follows from b).