(When) does Schur's lemma break? 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?
 A: It all depends on what you call "Schur's Lemma". If M is a simple module over a ring R then D=EndRM is always a division ring (think of it as a weak Schur's lemma). The question is, can the endomorphism ring be pinned down more concretely, for example, if R is an algebra over a field k? In the usual Schur's lemma for finite groups, k=C and D=C. More general versions of Schur's lemma assert that D is algebraic over k (so D=k if k is algebraically closed).
If R is an affine (i.e. finitely-generated) commutative algebra over a field k and I is a maximal ideal of R then Hilbert's Nullstellensatz is asserting that R/I is algebraic over k. Since M=R/I is a simple module and R/I=EndRM, you may interpret the statement as a version of Schur's lemma: R/I=EndRM is algebraic over k. It is also known that every prime ideal of R is an intersection of maximal ideals.
Now if R is a noncommutative algebra over k one can ask whether the analogous properties hold: the endomorphism ring of a simple module is algebraic over k ("endomorphism property", implying the usual Schur's lemma when k is algebraically closed) and every prime ideal of R is an intersection of primitive ideals ("R has radical property"). This is true in a range of situations and constitutes noncommutative Nullstellensatz. Duflo proved that the endomorphism property for R[x] implies Nullstellensatz (endomorphism + radical) for R. 


*

*As Kevin pointed out, if the dimension of M over k is smaller than the cardinality of k then the endomorphism property holds for R, and by Duflo, full Nullstellensatz holds for R.

*In general, some extra assumptions are necessary, but Nullstellensatz holds for the Weyl algebra An(k), the universal enveloping algebra U(g) of a finite-dimensional Lie algebra g, and the group algebra k[G] of a polycyclic by finite group. (The first two cases use Quillen's lemma also mentioned by Kevin). 
This is described in chapter 9 of McConnell and Robson, Noncommutative Noetherian rings. 
A: 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 $k$.
A: This is standard stuff: for example, if $A$ is an associative algebra over a field $k$ and $M$ is a simple module over $A$ whose dimension as a $k$-vector space is smaller than the cardinality of $k$, then any element of $\text{End}_k(M)$ is algebraic over $k$ (one just needs to consider the $k$-dimension of $k(\alpha)$ if $\alpha$ is a nonzero endomorphism -- see Kevin Buzzard's comment above for example). On the other hand, there's a nice (short) article of Quillen in Proceedings of the AMS about Schur's Lemma for filtered algebras: he checks that for a filtered algebra $U$ over $k$ whose associated graded is commutative and finitely generated, if $M$ a simple $U$-module, and $\theta \in \text{End}_U(M)$, then one has $\theta$ algebraic over $k$.
