Schur's Lemma (Representation Theory)/Corollary

Corollary to Schur's Lemma
Let $\left({G, \cdot}\right)$ be a finite group.

Let $\left({V, \phi}\right)$ be a $G$-module.

Let the underlying field $k$ of $V$ be algebraically closed.

Let:
 * $\operatorname{End}_G \left({V}\right) := \left\{{f: V \to V:\ f }\right.$ is a homomorphism of $G$-modules$\left.\right\}$

Then:
 * $ \operatorname{End}_G \left({V}\right)$

is a field, with the same structure as $k$.

Proof
Denote the identity mapping on $V$ as $I_V: V \to V$.

If $f=0$, since $0\in k$ it can be written $f = 0 \operatorname{Id}_V$.

Now if $f$ is an automorphism, the characteristic polynomial of $f$ is complete reducible in $k[x]$ because $k$ is algebraically closed; hence $f$ has all eigenvalue in $k$.

We take $\lambda \in k$ an eigenvalue of $f$ and consider the endomorphism $f-\lambda \operatorname{Id}_V: V \to V$.

Since $\ker(f-\lambda \operatorname{Id}_V)\ne\{0\}$ because $\lambda$ is an eigenvalue, it follows using Schur's Lemma that $f=\lambda \operatorname{Id}_V$.


 * $(\lambda\operatorname{Id}_V)\circ (\mu\operatorname{Id}_V) =(\lambda\mu)\operatorname{Id}_V$
 * $\lambda\operatorname{Id}_V+(-\mu\operatorname{Id})=(\lambda-\mu)\operatorname{Id}_V$

From Subring Test $\operatorname{End}_G(V)$ is a subring of the ring endomorphisms of $V$ as an abelian group.

Now we can define $\phi:\operatorname{End}_G(V)\to k$ such that $\phi(\lambda \operatorname{Id}_V)=\lambda$.


 * $\phi(\lambda\operatorname{Id}_V+\mu\operatorname{Id}_V)=\lambda+\mu=\phi(\lambda\operatorname{Id}_V)+\phi(\mu\operatorname{Id}_V)$
 * $\phi((\lambda\operatorname{Id}_V)\circ(\mu\operatorname{Id}_V))=\lambda\mu=\phi(\lambda\operatorname{Id}_V)\phi(\mu\operatorname{Id}_V)$

Hence $\phi$ is a ring isomorphism, but since $k$ is a field it is a field isomorphism.