TITLE: On a theorem of Jacobson
QUESTION [20 upvotes]: 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 asked:

Is $S'$ maximal for this property?

So, is it?

REPLY [6 votes]: Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).
EDIT.  Herstein's set may be maximal.  The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained by the condition that if there is $p$-torsion then no $p^2$-torsion.