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October 27th, 2010, 10:06 AM   #11
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Re: commutative ring

Oh that's a nice proof.

I think you mean...(minus instead of plus)
(ab - ba) = (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
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October 27th, 2010, 08:45 PM   #12
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No, I meant the plus, not minus. What does the hint give you when applied to the equation mattpi gave? Note that (ab)(ab - ba) = 0 implies ab(ab - ba) = 0.
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October 28th, 2010, 06:15 AM   #13
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Re: commutative ring

Thank you...
I gt it....
(finally phew!!)
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October 28th, 2010, 01:49 PM   #14
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Okay. I assume you got ba(a - b) = 0, i.e., ba - bab = 0.
Now the key step: multiply by a to get ba - baba = 0, and so bababa - baba = 0.
Applying the hint again gives ababab - ababa = 0, i.e., abab(ab - ba) = 0.
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November 3rd, 2010, 09:54 AM   #15
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Re: commutative ring

Hey my prof just pointed out that ab=0 does not imply ba=0.....
in any ordinary ring...
any light?
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November 3rd, 2010, 10:20 AM   #16
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Re: commutative ring

If and then
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November 3rd, 2010, 04:31 PM   #17
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Re: commutative ring

Consider Matrix A =

and Matrix B =
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February 2nd, 2013, 09:06 AM   #18
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A ring for which ab=0 implies ba=0 is termed "reversible".
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