State/prove/explain Schur’s lemma.
Solution
Let \(R\) be a ring and let \(M,N\) be simple left \(R\)-modules.
1. Every nonzero \(R\)-linear map \(f:M\to N\) is an isomorphism.
The kernel \(\ker f\) is an \(R\)-submodule of the simple module \(M\), so \[ \ker f=0 \quad\text{or}\quad \ker f=M. \] Since \(f\ne0\), the second alternative is impossible; hence \(\ker f=0\) and \(f\) is injective.
Likewise, the image \(\operatorname{im}f\) is a nonzero submodule of the simple module \(N\), so \[ \operatorname{im}f=N. \] Thus \(f\) is surjective as well, hence an isomorphism.
2. If \(M\) and \(N\) are nonisomorphic simple modules, then\[ \operatorname{Hom}_R(M,N)=0. \]
Any nonzero homomorphism would be an isomorphism by 1, contradicting \(M\not\cong N\).
3. Every nonzero element of \(\operatorname{End}_R(M)\) is invertible; hence \(\operatorname{End}_R(M)\) is a division ring.
Apply 1 with \(N=M\). Every nonzero endomorphism of \(M\) is an automorphism, which is precisely the statement that every nonzero element of \(\operatorname{End}_R(M)\) is a unit.
4. This is Schur’s lemma.
The usual representation-theoretic formulation is obtained by taking \(R\) to be a group algebra: irreducible representations correspond to simple modules, so a nonzero intertwiner between irreducibles is an isomorphism, and the endomorphism ring of an irreducible representation is a division ring.