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.

  • $\begingroup$ It doesn't really "use" Schur's lemma, it is Schur's lemma. $\endgroup$
    – rschwieb
    Dec 3, 2022 at 22:49

1 Answer 1


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.


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