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Suppose we have a simple left $R$ module $M$, consider $D = End_R(M).$ It is easy see that $D$ is a division ring because, for an element $f \in D$`, the kernel and range of $f$ are submodules of $M$. Since $M$ is simple, it follows that $f$ is an isomorphism. Thus, $D$ is a division ring.

Now I am interested in the converse of this. Suppose $D$ is a division ring, is it true that $M$ is a simple $R$ module? The answer is no, as an example we consider $End_{\mathbb{Z}}(\mathbb{Q})$ is isomorphic to $\mathbb{Q}$ but it is not a simple $\mathbb{Z}$ module.

My question is : what is the condition on $M$ to hold the converse of Schur's lemma?

Infinity_hunter
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    We obviously need $M$ to have no proper submodules that are homomorphic images of $M$ itself. Some kind of a finiteness condition or a chain condition may also be needed to rule out the possibility of surjective endomorphisms with a non-trivial kernel. – Jyrki Lahtonen Dec 04 '22 at 06:38
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    Indecomposability is not enough. In my youth, when I still had a promising future as an algebraist, I encountered indecomposable modules $M$ with three simple composition factors, the socle and the head were isomorphic, so there were non-trivial endomorphisms mapping the head to the socle. – Jyrki Lahtonen Dec 04 '22 at 06:41
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    One usual setting where a converse is considered is, for example, when $R$ is a semisimple ring and say, to be on the safe side, that $M$ is of finite length. Then $M$ is simply the direct sum of finitely many simple modules, and the converse you talk about holds. This is usual, say, when considering representations of a finite group over a field of characteristic $0$. – Sasha Dec 04 '22 at 10:14
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    Such modules are called "bricks", see here for a reference. There is a lot to say about them, little of which would fit in a short answer. – Pierre-Guy Plamondon Dec 04 '22 at 17:07

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