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?

  • 2
    $\begingroup$ 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/… $\endgroup$ Jan 2 at 21:12
  • $\begingroup$ @QiaochuYuan Maybe even vote to close as duplicate? $\endgroup$
    – Pedro
    Jan 4 at 0:12