Schur's Lemma (Representation Theory)/Corollary

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

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

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

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

has the same structure as $k$; it's a field.

Proof
If $f=0$, since $0\in k$ it can be written $f=0 Id_V$.

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 $f=\lambda \operatorname{Id}_V$.

Now we can define $\phi:\operatorname{End}_G(V)\to k$ such that $\phi(\lambda \operatorname{Id}_V)=\lambda$; which is obviously a field isomorphism.