The Stacks project

Lemma 11.4.6. Let $A$ be a finite simple $k$-algebra.

  1. There exists exactly one simple $A$-module $M$ up to isomorphism.

  2. Any finite $A$-module is a direct sum of copies of a simple module.

  3. Two finite $A$-modules are isomorphic if and only if they have the same dimension over $k$.

  4. If $A = \text{Mat}(n \times n, K)$ with $K$ a finite skew field extension of $k$, then $M = K^{\oplus n}$ is a simple $A$-module and $\text{End}_ A(M) = K^{op}$.

  5. If $M$ is a simple $A$-module, then $L = \text{End}_ A(M)$ is a skew field finite over $k$ acting on the left on $M$, we have $A = \text{End}_ L(M)$, and the centers of $A$ and $L$ agree. Also $[A : k] [L : k] = \dim _ k(M)^2$.

  6. For a finite $A$-module $N$ the algebra $B = \text{End}_ A(N)$ is a matrix algebra over the skew field $L$ of (5). Moreover $\text{End}_ B(N) = A$.

Proof. By Theorem 11.3.3 we can write $A = \text{Mat}(n \times n, K)$ for some finite skew field extension $K$ of $k$. By Lemma 11.4.5 the category of modules over $A$ is equivalent to the category of modules over $K$. Thus (1), (2), and (3) hold because every module over $K$ is free. Part (4) holds because the equivalence transforms the $K$-module $K$ to $M = K^{\oplus n}$. Using $M = K^{\oplus n}$ in (5) we see that $L = K^{op}$. The statement about the center of $L = K^{op}$ follows from Lemma 11.4.5. The statement about $\text{End}_ L(M)$ follows from the explicit form of $M$. The formula of dimensions is clear. Part (6) follows as $N$ is isomorphic to a direct sum of copies of a simple module. $\square$

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