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?

worksbefore we start worrying about whether it's maximal? I only added $X-9$, I didn't add all $X-n$simultaneously. But my gut instinct is that, if your set is OK, it won't be maximal because there are still plenty of other stupid tricks you can try (square of a linear factor etc). Note however that if someone comes up with an enlargement and then someone else says "OK then is this enlarged set maximal" we could be here all year! – Kevin Buzzard Jul 15 '10 at 19:23