Lemma 11.3.2. Let $A$ be a $k$-algebra. If $A$ is finite, then

1. $A$ has a simple module,

2. any nonzero module contains a simple submodule,

3. a simple module over $A$ has finite dimension over $k$, and

4. if $M$ is a simple $A$-module, then $\text{End}_ A(M)$ is a skew field.

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. $\square$

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