If $R$ is a simple Artinian ring with simple module $M$ then $M$ is finitely generated as an $\mathrm{End}_R(M)$-module. This is usually proven in an indirect way: By Artin-Wedderburn-Theory we may assume that $R = \mathcal{M}_n(D)$ is a matrix ring over a division ring $D$, and $M = D^n$. Then it is obvious that $M$ is finite dimensional over $\mathrm{End}_R(M) \cong D^{op}$.

Is there a more direct / more elegant way to show that $M$ is finitely generated over $\mathrm{End}_R(M)$ which avoids matrix rings? Are there more general conditions on $R$ and $M$ under which the conclusion remains true?


There is the following theorem by Morita [Lang, Algebra - Theorem 7.1]:

Let $R$ be a ring and $M$ an $R$-module. Then $M$ is a generator iff $M$ is balanced and finitely generated projective over $R' = \mathrm{End}_R(M)$.

This generalizes the statement above, as if $R$ is simple Artinian and $M$ is simple, then $R \cong M^{(n)}$ for some $n$, so $M$ is in particular a generator.

Here is an elementary proof of the implication “$M$ is a generator $\implies M$ is f.g. over $R'$”:

Proof. Let $M$ be a right $R$-module, so it can be considered as a left $R'$-module. Being a generator means that $R$ is an epimorphic image of $M^{(n)}$ for some $n$, so there are elements $x_1, \dots, x_n \in M$ and homomorphisms $\varphi_1, \dots, \varphi_n \colon M \to R$ such that $\sum_i \varphi_i x_i = 1$. For any $x \in M$ let $\alpha_x$ be the unique homomorphism $R \to M$ with $\alpha_x(1) = x$. Then for all $x \in M$ we have $$ x = \alpha_x(1) = \alpha_x \left( \sum_i \varphi_i x_i \right) = \sum_i (\alpha_x \varphi_i) x_i. $$ Since $\alpha_x \varphi_i \in R'$ for all $i$, this means that $M$ is generated by $x_1, \dots, x_n$ as an $R'$-module. $\square$


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