From: Robin Chapman
Subject: Nilpotent elements in group algebras
Date: Thu, 15 Jul 1999 09:07:41 GMT
Newsgroups: sci.math
Here's a nice little result:
Let K be a field of characteristic zero, and let G be a finite group.
Consider the group algebra A = KG. If x = sum_g a_g g in A
is nilpotent then a_1 = 0.
Here's a proof. Consider the regular representation. In particular
consider the K-endomorphism phi of A given by left multiplication by x,
so that phi(y) = x(y). Then if x is nilpotent, so is phi, and the
trace of phi vanishes. But this trace is |G|a_1, so a_1 = 0. As we might
expect the result holds if K has finite characteristic coprime to |G|.
(But certainly not if its characteristic divides |G|.)
Questions:
(i) Is there an "elementary" proof, in the sense of not using (even
the modest amount used above) representation theory.
(ii) Same as (i) in the special case where x^2 = 0.
(iii) Is the result true when G is an infinite group?
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"They did not have proper palms at home in Exeter."
Peter Carey, _Oscar and Lucinda_
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
==============================================================================
From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: Nilpotent elements in group algebras
Date: 16 Jul 1999 21:18:23 GMT
Newsgroups: sci.math
In article <7mk8co$at8$1@nnrp1.deja.com>,
Robin Chapman writes:
> Here's a nice little result:
>
> Let K be a field of characteristic zero, and let G be a finite group.
> Consider the group algebra A = KG. If x = sum_g a_g g in A
> is nilpotent then a_1 = 0.
>...
> (iii) Is the result true when G is an infinite group?
Yes, there's a proof on p. 47 of D.S. Passman's book _The Algebraic
Structure of Group Rings_. It falls into two parts:
1) In characteristic p, show that nilpotence forces
a_1 + \sum {a_g | g has p-power order} = 0.
2) Suppose now we have a nilpotent element in characteristic 0
with a_1 nonzero. Consider valuations on K where all the nonzero
a_g are in the valuation ring and have nonzero images in the
residue field. Such valuations will occur with residue fields
of infinitely many different prime characteristics. But by
1), that forces a_g to be nonzero for g of infinitely many
different prime power orders, which is impossible.
William C. Waterhouse
Penn State