The Stacks project

Proof. Let $A' = \text{End}_ A(M)$, so $M$ is a left $A'$-module. Set $A'' = \text{End}_{A'}(M)$ (the bicommutant of $M$). We view $M$ as a right $A''$-module¹. Let $R : A \to A''$ be the natural homomorphism such that $mR(a) = ma$. Then $R$ is injective, since $R(1) = \text{id}_ M$ and $A$ contains no nontrivial two-sided ideal. We claim that $R(M)$ is a right ideal in $A''$. Namely, $R(m)a'' = R(ma'')$ for $a'' \in A''$ and $m$ in $M$, because left multiplication of $M$ by any element $n$ of $M$ represents an element of $A'$, and so $(nm)a'' = n(ma'')$ for all $n$ in $M$. Finally, the product ideal $AM$ is a two-sided ideal, and so $A = AM$. Thus $R(A) = R(A)R(M)$, so that $R(A)$ is a right ideal in $A''$. But $R(A)$ contains the identity element of $A''$, and so $R(A) = A''$. $\square$

Proof. Of course (1) follows from (2) since $A$ is a nonzero $A$-module. For (2), any submodule of minimal (finite) dimension as a $k$-vector space will be simple. There exists a finite dimensional one because a cyclic submodule is one. If $M$ is simple, then $mA \subset M$ is a sub-module, hence we see (3). Any nonzero element of $\text{End}_ A(M)$ is an isomorphism, hence (4) holds. $\square$

Proof. We may choose a simple submodule $M \subset A$ and then the $k$-algebra $K = \text{End}_ A(M)$ is a skew field, see Lemma 11.3.2. By Lemma 11.3.1 we see that $A = \text{End}_ K(M)$. Since $K$ is a skew field and $M$ is finitely generated (since $\dim _ k(M) < \infty $) we see that $M$ is finite free as a left $K$-module. It follows immediately that $A \cong \text{Mat}(n \times n, K^{op})$. $\square$