Ring of endomorphism Let $M$ be a simple $R$-module. My goal is to show that End$_R(M)$ is a field. I have proven a version of Schur's Lemma. That is, I have proven to myself that a homomorphism $\phi: M \rightarrow M'$ of simple $R$ -modules is either an isomorphism or the zero-morphism.
This version of Schur's Lemma immediately implies that End$_R(M)$ is a division ring. My problem arises with proving that End$_R(M)$ is commutative. I don't immediately see why End$_R(M)$ is commutative, and I was not given any information about $R$.
 A: With no further hypotheses this is not true; Schur's lemma tells you that $\text{End}_R(M)$ is a division ring (and slightly more precisely it's a division algebra over $Z(R)$) and that's all. Moreover every noncommutative division ring occurs, e.g. if $R$ itself is a division ring $D$ and $M = D$ is the regular representation, in which case $\text{End}_D(D) \cong D^{op}$.
It is true if $R$ is commutative; for this you can show that we have $M \cong R/m$ where $m$ is a maximal ideal, and then show that $\text{End}_R(R/m) \cong R/m$. It is worth thinking in some detail about the case that $R = k[x]$ for $k$ a field and $m = (f(x))$ is the maximal ideal generated by an irreducible polynomial.
A: You're exactly right that $\text{End}_R(M)$ is (in general) only a division ring. However, we can use the fact that $R$ is commutative to learn more.
In particular, if $M$ is a simple $R$ module, then $M \cong R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$.
But can you show that every endomorphism of $R / I$ looks like multiplication by $r$ for some $r \in R$? (Hint: if $\varphi$ is your endomorphism, look at $\varphi(1)$).
Then there is a surjection $R \to \text{End}_R(M)$. What is its kernel? Do you see why this shows $\text{End}_R(M)$ is a field?

I hope this helps ^_^
