# Endomorphism ring of a module

1. Jan 30, 2012

### Bleys

Let R be a finite dimensional C-algebra (C=Complex numbers) and S a simple R-module. Why does it follow that $End_{R}(S)$ is also finite dimensional (as C-vector spaces, I'm guessing)? I'm not really sure how to construct a basis for it using one of S, and there's probably another reason for it (is end(S) embedded in S or something?)

2. Jan 30, 2012

### morphism

Is S finitely generated over R (hence C)?

3. Jan 30, 2012

### Bleys

yes; I have this result
"For a finite dimensional C-algebra R, there are only finitely many isomorphism classes of simple R-modules and they are finite dimensional"

4. Jan 30, 2012

### morphism

Perfect. So S is a finite-dimensional C-vector space and End_R(S) is a subspace of End_C(S), hence is finite-dimensional over C. In fact, you can use Schur's lemma to show that End_R(S) is one-dimensional.

5. Jan 30, 2012

### Bleys

ah of course, I didn't think of the fact End_{R}(S) is a subspace of End_{C}(S).
I'm actually going through Schur's Lemma's proof to show End_R(S) is isomorphic to C but this was the detail I wasn't understanding.

Thank you, morphism!