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?

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?

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.

and then proceeded to ask:

Is $S$ maximal for this property?

So, is it?