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Message from discussion x^3=x implies ring is commutative
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Bill Dubuque  
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 More options Jul 12 2005, 2:52 pm
Newsgroups: sci.math
From: Bill Dubuque <w...@nestle.csail.mit.edu>
Date: 12 Jul 2005 14:52:19 -0400
Local: Tues, Jul 12 2005 2:52 pm
Subject: Re: x^3=x implies ring is commutative
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themadhatter...@yahoo.com wrote

> If R is a ring with unity and x^3 = x for all elements
> then this ring is commutative.

1. ab = 0 => ba = 0  via  ba = (ba)^3 = b ab ab a = 0

2. cc = c => c central [i.e. xc = cx  for all x]

   Proof: c(x-cx)=0 so (x-cx)c=0 by 1, so  xc = cxc
          (x-xc)c=0 so c(x-xc)=0 by 1, so  cx = cxc

3. xx central  via  c = xx  in  2.

4. cc = 2c => c central: c = ccc = 2cc central by 3.

5. x + xx central  via  c = x + xx  in 4.

6. x = (x + xx) - xx  central  via  3, 5.  QED

This is a special case of a famous theorem of Jacobson
which implies that a ring is commutative if for every
element x there is an integer n > 1 such that x^n = x.

The above is from my post in the thread below
which you might also find informative
http://groups-beta.google.com/group/sci.math/browse_frm/thread/5a0072...

-Bill Dubuque


 
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