If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.

abstract-algebra ring-theory

Solution 1:

You’re on the right track. From $xy+yx\in Z(R)$ we get $x(xy+yx)=(xy+yx)x$, or, after expanding, $x^2y+xyx=xyx+yx^2$, and hence $x^2y=yx^2$. This shows that every square is in $Z(R)$. Can you see how to finish it from there?

Solution 2:

The answer is Theorem 2 in the paper

http://archive.maths.nuim.ie/staff/sbuckley/Papers/bm_variations.pdf

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