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 October 19th, 2010, 10:59 AM #1 Newbie   Joined: Oct 2010 Posts: 7 Thanks: 0 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...
 October 19th, 2010, 11:14 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: commutative ring $ab=a^3b^3=(ab)^3.$ Work from here and you should easily find that any two elements must commute.
 October 19th, 2010, 04:18 PM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: commutative ring How do you get the second =?
 October 20th, 2010, 02:30 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: commutative ring Well, $(ab)^1=(ab)^3,$ so...
 October 20th, 2010, 04:55 AM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: commutative ring Got it. I would have written $(ab)^3=ab=a^3b^3$...
 October 20th, 2010, 06:44 PM #6 Newbie   Joined: Oct 2010 Posts: 7 Thanks: 0 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
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.

 October 24th, 2010, 08:51 AM #8 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: commutative ring Having looked at this problem I'm no longer sure whether this is so easy to do... We certainly have $a(a^2-b^2)b=0=b(a^2-b^2)a$ for any $a,b,$ and so we can deduce that for any pair of elements $a,b,$ both nonzero, either at least one is a zero divisor, or $a^2=b^2.$ Don't know if this helps.
 October 26th, 2010, 05:12 AM #9 Newbie   Joined: Oct 2010 Posts: 7 Thanks: 0 Re: commutative ring somebody plz help...
 October 26th, 2010, 04:15 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 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)³ = (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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### r be a ring in which x^4=x for all x belongs to r prove that r is commutative

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