October 27th, 2010, 10:06 AM  #11 
Newbie Joined: Oct 2010 Posts: 7 Thanks: 0  Re: commutative ring
Oh that's a nice proof. I think you mean...(minus instead of plus) (ab  ba)³ = (ab  ba)(ab(ab  ba)  ba(ba  ab)) = (ab  ba)(0  0) = 0, the hint can be easily proved... but then how do you get ab(ab  ba) = 0 from that?? ab(ab  ba) = 0 doesn't imply ab=0 or (ab  ba) = 0 
October 27th, 2010, 08:45 PM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,467 Thanks: 2038 
No, I meant the plus, not minus. What does the hint give you when applied to the equation mattpi gave? Note that (ab)²(ab  ba) = 0 implies ab(ab  ba) = 0.

October 28th, 2010, 06:15 AM  #13 
Newbie Joined: Oct 2010 Posts: 7 Thanks: 0  Re: commutative ring
Thank you... I gt it.... (finally phew!!) 
October 28th, 2010, 01:49 PM  #14 
Global Moderator Joined: Dec 2006 Posts: 20,467 Thanks: 2038 
Okay. I assume you got ba(a²  b²) = 0, i.e., ba  bab² = 0. Now the key step: multiply by a² to get ba³  bab²a² = 0, and so bababa  bab²a² = 0. Applying the hint again gives ababab  abab²a = 0, i.e., abab(ab  ba) = 0. 
November 3rd, 2010, 09:54 AM  #15 
Newbie Joined: Oct 2010 Posts: 7 Thanks: 0  Re: commutative ring
Hey my prof just pointed out that ab=0 does not imply ba=0..... in any ordinary ring... any light? 
November 3rd, 2010, 10:20 AM  #16 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: commutative ring
If and then 
November 3rd, 2010, 04:31 PM  #17 
Senior Member Joined: Oct 2009 Posts: 105 Thanks: 0  Re: commutative ring
Consider Matrix A = and Matrix B = 
February 2nd, 2013, 09:06 AM  #18 
Global Moderator Joined: Dec 2006 Posts: 20,467 Thanks: 2038 
A ring for which ab=0 implies ba=0 is termed "reversible".


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