If R is a ring, and has the property that x^3=x for all x in R, then show R is commutative. Generalize for x^k=x, k an integer >3.(adsbygoogle = window.adsbygoogle || []).push({});

I can do this for x^2=x. First use x^2+x^2=x+x=(x+x)^2 = 4x^2 (abusing notation a little) which means x^2+x^2=x+x=0, ie, x=-x for all x in R. Then x^2+y^2=x+y=(x+y)^2 = x^2+xy+yx+y^2, which gives xy+yx=0. But since xy=-xy, we have xy=yx, and R is commutative.

I try to generalize using the same basic idea, but there are a lot of extra terms, and it seems impossible to do anything with R and its elements being so general (ie, I can't cancel anything because I don't know if anything is a zero divisor). Can anyone give me a hint?

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# X^3=x => R is a commutative ring

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