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 $f = 0 I_V$.

Let $f$ be an automorphism

We have that $k$ is algebraically closed.

Therefore the characteristic polynomial of $f$ is complete reducible in $k \left[{x}\right]$.

Hence $f$ has all eigenvalue in $k$.

Let $\lambda \in k$ be an eigenvalue of $f$.

Consider the endomorphism:
 * $f - \lambda I_V: V \to V$

Because $\lambda$ is an eigenvalue:
 * $\ker \left({f - \lambda I_V}\right) \ne \left\{ {0}\right\}$

From Schur's Lemma:
 * $f = \lambda I_V$


 * $ \left({\lambda I_V}\right) \circ \left({\mu I_V}\right) = \left({\lambda \mu}\right) I_V$


 * $\lambda I_V + \left({-\mu I_V}\right) = \left({\lambda - \mu}\right) I_V$

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

Let $\phi: \operatorname{End}_G \left({V}\right) \to k$ be defined as:
 * $\phi \left({\lambda I_V}\right) = \lambda$

Then:
 * $\phi \left({\lambda I_V + \mu I_V}\right) = \lambda + \mu = \phi \left({\lambda I_V}\right) + \phi \left({\mu I_V}\right)$
 * $\phi \left({\left({\lambda I_V}\right) \circ \left({\mu I_V}\right)}\right) = \lambda \mu = \phi \left({\lambda I_V}\right) \phi \left({\mu I_V}\right)$

Hence $\phi$ is a ring isomorphism.

But since $k$ is a field it is a field isomorphism.