Proposition 1

Let A be a nonzero ring. The following conditions are equivalent:

(i) The A-module $A_s$ is isotypical.

(ii) The ring A is left Artinian, and every two-sided ideal of A is equal to 0 or A.

(iii) The ring A is left Artinian, and there exists a left A-module S that is simple and faithful.

If these conditions are satisfied, then the A-module $A_s$ has finite length and is isotypical of type S, and every simple A-module is isomorphic to S.

Let us prove that (i) implies (ii). Assume that (i) is satisfied. Then the finitely generated A-module $A_s$ is semisimple, hence has finite length and is Artinian (VIII, p. 71, Proposition 10); consequently, the ring A is left Artinian. The endomorphisms of the left A-module $A_s$ are the right multiplications by the elements of A. Since the left A-module $A_s$ is isotypical, it follows from Proposition 6, b) of VIII, p. 86 that the $(A$, A)-bimodule $_sA_d$ is simple. The sub-bimodules of $_sA_d$ are the two-sided ideals of A, so (i) implies (ii).

Let us prove that (ii) implies (iii). The ring A is not reduced to 0; consequently, there exists a simple A-module S. The annihilator of S is a proper two-sided ideal of A. If we assume that (ii) is satisfied, then that annihilator is equal to 0. The A-module S is then faithful, and (ii) implies (iii).

Let us prove that (iii) implies (i). Assume that (iii) is satisfied. Then there exists an integer $m\geqslant 1$ such that $A_s$ is isomorphic to a submodule of $S^m$ (VIII, p. 50, Proposition 5, a)). Since $S^m$ is an isotypical A-module of type S, the same holds for $A_s$ (VIII, p. 61, Proposition 2); hence, (iii) implies (i).

Suppose that conditions (i) through (iii) are satisfied. We have seen above that the A-module $A_s$ has finite length and is isotypical of type S. Every simple left A-module is isomorphic to a quotient of $A_s$, hence to S.