Proposition 5

Let B be a ring, A a simple subring of B, and S a simple left A-module. Then B is a free left A-module of dimension long$_A(B\otimes_AS)$.

Let $r$ be the length of A; the A-module $A_s$ is isomorphic to $S^r$. Now, the left A-module B is isomorphic to $B\otimes_AA_s$ (II, §3, No. 4, p. 249), hence to $(B\otimes_AS)^r$ (II, §3, No. 7, p. 255, Proposition 7). We therefore have long$_A(B) =r$ long$_A(B\otimes_AS)$, and Proposition 5 follows from Proposition 3 of VIII, p. 123.