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.

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