Schur's Lemma (Representation Theory)/Corollary

Corollary to Schur's Lemma
Let $\struct {G, \cdot}$ be a finite group.

Let $\struct {V, \phi}$ be a $G$-module.

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

Let:
 * $\map {\mathrm {End}_G} V := \leftset {f: V \to V: f}$ is a homomorphism of $G$-modules$\rightset {}$

Then:
 * $\map {\mathrm {End}_G} V$

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 $k \sqbrk x$.

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:
 * $\map \ker {f - \lambda I_V} \ne \set 0$

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


 * $\paren {\lambda I_V} \circ \paren {\mu I_V} = \paren {\lambda \mu} I_V$


 * $\lambda I_V + \paren {-\mu I_V} = \paren {\lambda - \mu} I_V$

From Subring Test:
 * $\map {\mathrm {End}_G} V$ is a subring of the ring endomorphisms of $V$ as an abelian group.

Let $\phi: \map {\mathrm {End}_G} V \to k$ be defined as:
 * $\map \phi {\lambda I_V} = \lambda$

Then:
 * $\map \phi {\lambda I_V + \mu I_V} = \lambda + \mu = \map \phi {\lambda I_V} + \map \phi {\mu I_V}$
 * $\map \phi {\paren {\lambda I_V} \circ \paren {\mu I_V} } = \lambda \mu = \map \phi {\lambda I_V} \map \phi {\mu I_V}$

Hence $\phi$ is a ring isomorphism.

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