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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?

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Take $R$ to be a field containing $\mathbb{C}$, and let $R$ act on itself by multiplication. –  moonface May 4 '10 at 22:46
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3 Answers

up vote 5 down vote accepted

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 $k$.

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Just saw that moonface wrote a comment saying the same thing, while I was typing my answer. Sorry! –  Manny Reyes May 4 '10 at 22:54
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k(x) isn't always countably-infinite-dimensional over k. For example if k is uncountable (e.g. the complexes) then by partial fractions the elements 1/(x-lambda) are all lin ind over k, as lambda ranges through the elements of k. –  Kevin Buzzard May 4 '10 at 22:57
    
Oops, thanks you! –  Manny Reyes May 4 '10 at 23:02
    
You're very welcome! In fact this observation is somehow crucial when proving Schur's Lemma for smooth irreducible representations of certain p-adic groups. –  Kevin Buzzard May 5 '10 at 20:08
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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.

  1. 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.

  2. 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.

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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 $k$.

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Re "M is a module... the proof is clear": do you assume that M is a simple module? –  Victor Protsak May 5 '10 at 4:37
    
Oops yes absolutely, thanks! –  Kevin McGerty May 5 '10 at 7:35
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