Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a ring with unit element $1$ such that $(ab)^2=(ba)^2$ for all $a,b$ in $R$. If in $R$, $2x=0$ implies $x=0$, how do I show that R is commutative?

Is there any general approach to attack this kind of problem?

share|improve this question
    
Some solution is given as Theorem 4 here. It was among the first google hits for "ab^2=ba^2" commutative. [By this I do not suggest that googling should be considered the general approach for such problems ;-)] –  Martin Sleziak Nov 18 '11 at 11:44
add comment

1 Answer

up vote 5 down vote accepted

Here is the proof in Martin's link (so this question doesn't go unanswered and we don't have to link out):

Let $F(a,b) = (ab)^2 - (ba)^2 = abab - baba = 0 $. Then routine verification shows $ F(1+x, 1+y) - F(1+x,y) - F(x,1+y) + F(x,y) = -2yx+ 2xy $ so $ 2(xy-yx) = 0 $, which implies $xy=yx.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.