Lemma 2

Let D be a field and V a right vector space of finite dimension $r\geqslant 1$ over D. Then V is a simple module over the ring E = End$_D(V)$, and its commutant is equal to $D_V$. The ring E is simple, and its left length is equal to $r$.

We know that V is a simple E-module (VIII, p. 45, Example 3) and that its commutant is equal to $D_V$ (VIII, p. 82, Corollary 1). Let $(x_i)_{1\leqslant i\leqslant r}$ be a basis of V over the field D. The mapping $u\mapsto (u(x_i))_{1\leqslant i\leqslant r}$ is an isomorphism from the E-module $E_s$ to the E-module $V^r$; consequently, the E-module $E_s$ is isotypical of length $r$, so the ring E is simple.

Let us now prove Theorem 1. Recall (II, §10, No. 7, p. 349) that the ring $\mathbf{M}_r(D)$ can be identified with the endomorphism ring of the right D-vector space $D^r_d$; moreover, every right vector space of finite dimension $r$ over a field D is isomorphic to $D^r_d$ (II, §7, No. 1, p. 292). Theorem 1 therefore follows from Lemmas 1 and 2.