TITLE: (When) does Schur's lemma break?
QUESTION [8 upvotes]: 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?

REPLY [8 votes]: 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.