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In a comment to an answer to a MO question, in which Bill Dubuque mentioned Jacobson's theorem stating that a ring in which $X^n=X$ is an identity is commutative (theorem which has shown up on MO quite a bit recently, e.g. here), Pierre-Yves Gaillard observed that there is a more general theorem in which $n$ is allowed to be different for each element of the ring, so that in fact we can rephrase the theorem as saying that the set $S=\{X^n-X:n>1\}\subset\mathbb Z[X]$ has the following property:

If $A$ is a ring such that for every $a\in A$ there is an $f\in S$ such that $f(a)=0$, then $A$ is commutative.

Of course, $S\cup (-S)$ also has this property, and even if we construct $S'$ from $S\cup(-S)$ by closing it under the operation of taking divisors in $\mathbb Z[X]$, it also has the same property. Pierre-Yves then proceeded to askasked:

Is $S$ S'$maximal for this property? So, is it? 2 added 2 characters in body In a comment to an answer to a MO question, in which Bill Dubuque mentioned Jacobson's theorem stating that a ring in which$X^n=X$is an identity is commutative (theorem which has shown up on MO quite a bit recently, e.g. here), Pierre-Yves Gaillard observed that there is a more general theorem in which$n$is allowed to be different for each element of the ring, so that in fact we can rephrase the theorem as saying that the set $S={X^n-X:n>1}\subset\mathbb S=\{X^n-X:n>1\}\subset\mathbb Z[X]$ has the following property: If$A$is a ring such that for every$a\in A$there is an$f\in S$such that$f(a)=0$, then$A$is commutative. and then proceeded to ask: Is$S\$ maximal for this property?

So, is it?

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