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October 19th, 2010, 10:59 AM   #1
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commutative ring

Let R be a ring. If a ^ 3 = a for all a belongs to R, prove that R is commutative.

I tried this:

(x+x)^3= 2x which gives
6x=0.

(x^2 - x) = (x^2 - x)^3
so we get 3x^2=3x (1)

let S = { 3x : x belongs to R}
then S is a ring with operations of R.
also it is commutative as (3x)^2 = 3x. (using the fact that 6x^2 = 0 and (1))
and ring with all idempotent elements is commutative, hence S is commutative...

but I can't proceed any further...
please guide...
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October 19th, 2010, 11:14 AM   #2
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Re: commutative ring

Work from here and you should easily find that any two elements must commute.
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October 19th, 2010, 04:18 PM   #3
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Re: commutative ring

How do you get the second =?
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October 20th, 2010, 02:30 AM   #4
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Re: commutative ring

Well, so...
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October 20th, 2010, 04:55 AM   #5
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Re: commutative ring

Got it. I would have written ...
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October 20th, 2010, 06:44 PM   #6
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Re: commutative ring

Please give some more idea....
it is only mentioned that it is a ring, it may not have inverse, may not even have identity, and also it may have zero divisors as it may not be integral domain..

so ab(1- a^2*b^2)=0
and ab(1-(ab)^2)=0
leads to nowhere as both the terms may not be zero and yet the product may be...

thank you for the reply
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October 23rd, 2010, 07:27 PM   #7
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Quote:
Originally Posted by mattpi
. . . you should easily find that any two elements must commute.
I'd be interested to see how. I had the impression that proofs of this are a bit awkward to find (although elementary) and need about a dozen steps at least.
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October 24th, 2010, 08:51 AM   #8
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Re: commutative ring

Having looked at this problem I'm no longer sure whether this is so easy to do...

We certainly have for any and so we can deduce that for any pair of elements both nonzero, either at least one is a zero divisor, or Don't know if this helps.
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October 26th, 2010, 05:12 AM   #9
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Re: commutative ring

somebody plz help...
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October 26th, 2010, 04:15 PM   #10
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Various proofs of this result are known, but they leave one wondering whether a slightly shorter proof is possible. So do an exhaustive investigation (before someone else reads this and does it) and get yourself published!

One method is to prove that ab(ab - ba) = 0 (see hint below), then
(ab - ba) = (ab - ba) = (ab - ba)(ab(ab - ba) + ba(ba - ab)) = (ab - ba)(0 + 0) = 0,
so ab = ba.

(Hint: start by showing that ab = 0 implies ba = 0 for any a, b in the ring.)
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