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 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, 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 ... 
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  
Global Moderator Joined: Dec 2006 Posts: 20,484 Thanks: 2041  Quote:
 
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 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. 
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: 20,484 Thanks: 2041 
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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