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Suppose that $R$ is a ring such that for any $x\in R$ there exists $1<n(x)\in \mathbb{N}$ such that $x^{n(x)}-x\in Z(R)$. Prove that $R$ is commutative or if it is not commutative, then the ideal generated by all additive commutators is null.

I wanted to prove it like Jacobson-Herstein theorem. So if the assertion is true for division rings, from density and subdirect product representation theorems, I can prove it for left primitive rings and next for semiprimitive rings. But I can't prove it for division rings and arbitrary rings from semiprimitive rings.

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    $\begingroup$ Of course, the JH theorem shows that $R/Z(R)$ is commutative. What do you mean by "the ideal generated by all additive commutators is null", if not that it is the 0 ideal (and hence that the ring is commutative)? $\endgroup$
    – LSpice
    Commented Jun 28, 2016 at 20:45
  • $\begingroup$ Suppose that $A=\{ ab-ba : a,b\in R\}$. $I=<x :x\in A>$, the ideal generated by the elements of $A$ or $I= ‎\cap‎ J$ where $A‎\subseteq‎ J$ and $J$ is an ideal of $R$. We should show that $I$ is null. Yes, $R$ is commutative if and only if $I=0$ and in this case $I$ is null. Of course in general we should prove that $I$ is null. $\endgroup$
    – MH.Fakharan
    Commented Jun 28, 2016 at 23:00
  • $\begingroup$ My question is what 'null' means of an ideal (since it doesn't seem to mean "equals the 0 ideal"). EDIT: Maybe it means 'nilpotent'? $\endgroup$
    – LSpice
    Commented Jun 29, 2016 at 3:28
  • $\begingroup$ An ideal $I$ is null, means every element of $I$ is nilpotent. Of course every nilpotent ideal is null. But the converse is not true. $\endgroup$
    – MH.Fakharan
    Commented Jun 29, 2016 at 18:43

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