## Schur’s Lemma

Let $R$ be a ring with 1. Prove that if $M_1$ and $M_2$ are irreducible unital left $R$-modules, then any nonzero $R$-module homomorphism $\varphi : M_1 \rightarrow M_2$ is an isomorphism. Deduce Schur’s Lemma: If $M$ is an irreducible unital left $R$-module, then $\mathsf{End}_R(M)$ is a division ring.

Suppose $\varphi : M_1 \rightarrow M_2$ is an $R$-module homomorphism with $M_1$ and $M_2$ irreducible. We showed here that $\mathsf{ker}\ \varphi$ and $\mathsf{im}\ \varphi$ are submodules of $M_1$ and $M_2$, respectively. Since $\varphi$ is not the zero homomorphism, its kernel is not all of $M_1$, and so (since $M_1$ is irreducible) $\mathsf{ker}\ \varphi = 0$. Thus $\varphi$ is injective. Similarly, $\mathsf{im} \varphi$ is not the zero submodule, and so must be $M_2$, hence $\varphi$ is surjective. Thus $\varphi$ is an $R$-module isomorphism.

Now let $M$ be an irreducible unital left $R$-module. Recall that $\mathsf{End}_R(M)$ is a ring under pointwise addition and composition. Now if $\varphi \in \mathsf{End}_R(M)$ is nonzero, then by the above argument it is an isomorphism and so has an inverse $\varphi^{-1}$ which is also in $\mathsf{End}_R(M)$. So $\mathsf{End}_R(M)$ is a division ring. Note that $\mathsf{End}_R(M)$ is not necessarily commutative, so it need not be a field.