# A reference book for Schur's lemma

We want to find the reference book for some version of the schur's lemma which covers the following result

Let A be an assoiative algebra over {\mathbb C} with countable basis, then any central element acts on any simple A-module as a scalar.

Thanks

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Why is it relevant that the algebra has a countable basis? – André Henriques Jun 25 '12 at 18:58
@André: Take $A = M = \mathbb{C}(x)$. Every element is central, but only a few act as scalars! – Evan Jenkins Jun 25 '12 at 19:32
I'd bet it must be also in Dixmier's Universal eneveloping algebras. – Vít Tuček Jun 27 '12 at 12:52

This is (an immediate consequence of) Lemma 2.1.3(b) in Chriss-Ginzburg.

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Doc, if you want to be anal with your references, you should quote Amitsur, A. S. Algebras over infinite fields. Proc. Amer. Math. Soc. 7 (1956), 35–48.

Otherwise, this is a well-known fact and you can just refer to it as "Amitsur's Trick" or "Noncommutative Nullstellensatz". Chapter 9 of McConnell-Robson-Noncommutative-Noetherian-Rings is devoted entirely to this property and its finer variations.

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From Wikipedia:

• David S. Dummit, Richard M. Foote. Abstract Algebra. 2nd ed., pg. 337.
• Lam, Tsit-Yuen (2001), A First Course in Noncommutative Rings, Berlin, New York: Springer-Verlag, ISBN 978-0-387-95325-0

• William Fulton, Joe Harris. Representation Theory.
• William Arveson. An Invitation to C*-Algebras.

I hope that helps you a bit.

P.s: I wanted to post this in a comment, but the comment button isn't available for some reason.

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Thanks, i only have the book of Lam in hand. However i don't find it yet. – r_l Jun 25 '12 at 19:11
Johan, just so you know, you need at least 50 reputation in order to leave comments on posts that are not your own. I imagine you will get there pretty soon, though! – B R Jun 25 '12 at 19:27
@BR Aha! Well, You are right. I just passed that limit (-; – jmc Jun 25 '12 at 19:38

This is also in Bourbaki's Algebra 8 (most recent edition), Section 3, number 2, Example, page 43. This reference is online if you have access to SpringerLink.

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