*Edit: I wrote the following question and then immediately realized an answer to it, and moonface gave the same answer in the comments. Namely, $\mathbb C(t)$, the field of rational functions of $\mathbb C$, gives a nice counterexample. Note that it is of dimension $2^{\mathbb N}$.*


The following is one statement of Schur's lemma:



> 
> Let $R$ be an associative unital algebra over $\mathbb C$, and let $M$ be a simple $R$-module. Then ${\rm End}\_RM = \mathbb C$.
> 
> 
> 


My question is: are there extra conditions required on $R$? In particular, how large can $R$ be?


In particular, the statement is true when $\dim\_{\mathbb C}R <\infty$ and also when $R$ is countable-dimensional. But I have been told that the statement fails when $\dim\_{\mathbb C}R$ is sufficiently large.


How large must $\dim\_{\mathbb C}R$ be to break Schur's lemma? I am also looking for an explicit example of Schur's lemma breaking for $\dim\_{\mathbb C}R$ sufficiently large?



(When) does Schur's lemma break?
The idea behind Schur's Lemma is the following. The endomorphism ring of any simple $R$-module is a division ring. On the other hand, a finite dimensional division algebra over an algebraically closed field $k$ must be equal to $k$ (this is because any element generates a finite dimensional subfield over $k$, which must be equal to $k$). 


Thus, when $R$ is an algebra over an algebraically closed field $k$, the endomorphism ring of a finite dimensional simple module is a finite dimensional division algebra over $k$ and hence is equal to $k$. 


On the other hand, let $D$ be any division algebra over a field $k$, which we no longer assume to be algebraically closed. Then $D\_D$ is a simple module, and $\text{End}\_D(D)\cong D$. This allows us to break Schur's Lemma two different ways. If $D$ is infinite-dimensional and $k$ algebraically closed, the endomorphism ring $\text{End}\_D(D)$ will also be infinite dimensional over $k$, hence not isomorphic to $k$. We can easily come up with such $D$, even commutative examples. For instance, $k(x)$ will be an infinite dimensional division algebra over $k$. On the other hand, if $k$ is not algebraically closed, we can take $D$ to be a finite field extension, and we get a $\text{End}\_D(D)\cong D$ not isomorphic to $k$.


