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Let $A$ be an algebra over an algebraically closed field $k.$ Recall that if $A$ is a finitely generated module over its center, and if its center is a finitely generated algebra over $k,$ then by the Schur's lemma all simple $A$-modules are finite dimensional over $k.$

Motivated by the above, I would like an example of a $k$-algebra $A,$ such that:

1) $A$ has a simple module of infinitie dimension over $k,$

2) $A$ contains a commutative finitely generated subalgebra over which $A$ is a finitely generated left and right module.

Thanks in advance.

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Is this a homework problem? –  S. Carnahan Jun 19 '10 at 23:36
No, but you could use it as a homework problem if you wish. –  Bedini Jun 20 '10 at 0:25
Is there a reference for the statement in the first paragraph? And are there familiar (noncommutative) infinite dimensional algebras meeting these conditions? The motivation here needs some reinforcement. For me the interesting examples are universal enveloping algebras of finite dimensional Lie algebras in prime characteristic, where Schur's lemma isn't enough to prove finite dimensionality of all simple modules. (Ditto for quantized enveloping algebras or function algebras at a root of unity.) –  Jim Humphreys Jun 20 '10 at 12:51
@Jim: I don't know the exact reference, but one could prove it as follows. Let V be a simple A-module. Without loss of generality we may assume that the annihilator of V is 0. Let Z denote the center of A. It follows that V is f.g. Z-module and any nonzero element of Z acts invertibly on V (Schur's lemma), so Nakayama's lemma implies that Z is a field. But by the assumption Z is a finitely generated algebra over an algebraically closed field k. Therefore Z=k and V is finite over k. –  Bedini Jun 20 '10 at 20:53
I can't follow the last steps in your sketch and would prefer a reference. A textbook version I recall (following Jacobson's original line of proof) treats only universal enveloping algebras over a field of prime characteristic, combining Schur's Lemma with a generalized version of Nakayama's Lemma and the Hilbert Nullstellensatz. Is there a simpler version? (And other natural examples besides enveloping algebras where the same hypotheses are satisfied?) –  Jim Humphreys Jun 20 '10 at 22:46

1 Answer 1

up vote 4 down vote accepted

Doc, this is a stinker. Your condition (2) forces your algebra to be finitely generated PI, and every little hare knows that simple modules over such algebras are finite-dimensional. See 13.4.9 and 13.10.3 of McConnell-Robson...

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Very good! Thanks. –  Bedini Jun 21 '10 at 17:37

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