# Endomorphism ring between simple module is a division ring

I'm curious about a statement I read on Wikipedia about endomorphism between simple $$R$$-modules. The statement goes as follows:

In general, if $$R$$ is a ring and $$S$$ is a simple module over $$R$$, then, by Schur’s lemma, the endomorphism ring of $$S$$ is a division ring.

In particular when $$R$$ is a commutative ring, is there a more simple way to prove this statement without using Schur’s lemma?

Any help is greatly appreciated.

• It doesn't really "use" Schur's lemma, it is Schur's lemma. Dec 3, 2022 at 22:49

Any simple left $$R$$-module is of the form $$R/\mathfrak{m}$$, where $$\mathfrak{m}$$ is a maximal left ideal. If $$R$$ is commutative, then $$\mathfrak{m}$$ is a two-sided ideal, so that $$R/\mathfrak{m}$$ is a ring (even a field). Any $$R$$-linear map $$R/\mathfrak{m} \to R/\mathfrak{m}$$ is in fact $$R/\mathfrak{m}$$-linear, so that we obtain
$$\mathrm{End}_R(R/\mathfrak{m})=\mathrm{End}_{R/\mathfrak{m}}(R/\mathfrak{m})=R/\mathfrak{m}$$, which is a field, hence a division ring.