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 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: 20,931 Thanks: 2205 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: 20,931 Thanks: 2205 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 $ab=0$ and $(ba)^3=ba$ then $ba=bababa=b(ab)^2a=0.$
 November 3rd, 2010, 04:31 PM #17 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: commutative ring Consider Matrix A = $\begin{pmatrix}0 &1 \\ 0 &-1 \end{pmatrix}$ and Matrix B = $\begin{pmatrix}1 &1 \\ 1 &1 \end{pmatrix}$
 February 2nd, 2013, 09:06 AM #18 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 A ring for which ab=0 implies ba=0 is termed "reversible".

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r be a ring in which x^4=x for all x belongs to r prove that r is commutative

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