# $𝑅$ a $\mathbb C$- algebra and $𝑀$ a simple $𝑅$-module with a countable basis. Then $\textrm{End}_R(M)\cong \mathbb C.$ [duplicate]

Sсhur's lemma states:

Lemma. Let $$R$$ be a finite-dimensional algebra over an algebraically closed field $$K$$ and let $$M$$ be a simple $$R$$-module. Then $$\textrm{End}_R(M)\cong K.$$

Proof. Since $$M$$ is finitely generated, it's clearly finite-dimensional over $$K.$$ Consider $$\Psi\in\textrm{End}_R(M).$$ It is a $$K$$-linear operator. Let $$\alpha$$ be an eigenvalue of $$\Psi.$$ Then $$\Psi-\alpha\cdot\textrm{Id}$$ has got no inverse. Hence, $$\Psi=\alpha\cdot\textrm{Id}.$$

I want to prove the following:

Proposition. Let $$𝑅$$ be a $$\mathbb C$$- algebra and let $$𝑀$$ be a simple $$𝑅$$-module with a countable basis (over $$\mathbb C$$). Then $$\textrm{End}_R(M)\cong \mathbb C.$$

Here one cannot refer to the finiteness of the dimension. How to get around this difficulty?

• You want Dixmier's lemma, see this post (it's stated for group representations but the argument works fine for modules): math.stackexchange.com/questions/4551157/… Jan 2 at 21:12
• @QiaochuYuan Maybe even vote to close as duplicate?
– Pedro
Jan 4 at 0:12