Schur's lemma

Let $R$ be a ring with unity

Statement

Let $E$ and $F$ be a simple $R$-module. Any nonzero homomorphism $f:E\to F$ is an isomorphism.

Proof

Since $E$ is a simple $R$-module, kernel of $f$ is $0$ or $E$. Since it is nonzero homomorphism, $ker(f)=0$. Therefore, $E{\cong }_{mod}Im(f)$. Similarly, $F$ is a simple $R$-module, image of $f$ is $0$ or $F$. But $f$ is nonzero homomorphism, $Im(f)=F$. Therefore, $E$ and $F$ are isomorphic.

Application

Theorem

Let $E$ be a simple $R$-module. Then $En{d}_R(E)$ is a skew field.

$\varphi \in En{d}_R(E)$ is a trivial homomorphism or isomorphism. Multiplication of isomorphisms are also isomorphism. Therefore $En{d}_R(E)$ is a division ring.

By the Shcur’s lemma,