Proposition 4

Let A be a simple ring and M a nonzero A-module. Denote the endomorphism ring of the A-module M by B, and view M as a left B-module.

a) The mapping $a\mapsto a_M$ is an isomorphism from A to the endomorphism ring of the B-module M.

b) Suppose that M has finite length as an A-module. Then the ring B is simple, and we have

(2) long$_A(M) =$ long(B) and long$_B(M) =$ long(A).

The A-module M is generating by Proposition 2 of VIII, p. 122. By definition, we have $B = A'_M$, and assertion a) therefore follows from Theorem 2 of VIII, p. 82.

Suppose that M is an A-module of finite length. Choose a simple A-module S, and let D be the opposite field of the endomorphism ring of S. By Lemma 3, the A-module M is isomorphic to a module of the form $S\otimes_DV$, where V is a left vector space over the field D. The vector space V is finite-dimensional. By Theorem 3 of VIII, p. 64, the ring B is isomorphic to End$_D(V)$; by Lemma 2 (VIII, p. 121), the ring B is therefore simple. Taking Remark 1 of VIII, p. 63 into account, we obtain the equalities

long(B) = dim$_D(V) =$ long$_A(M)$.

By Remarks 1 of VIII, p. 63 and 3 of VIII, p. 121, we have the relations

long$_B(M) =$ long$_{End_D(V)}(S\otimes_DV) =$ dim$_D(S) =$ long(A),

which gives the last relation.