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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 asked:

Is $S'$ maximal for this property?

So, is it?

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@Litt, I guess you are taking minimal in cardinality. But minimal in what sense, such that it implies commutative? x^2-x=0 is enough. Notice that putting more elements in the set doesn't necessarily make things better to get the commutativity because then some a's can satisfy equations from S and some others satisfy the new equations. (or all satisfy the new equations). –  ABC Jul 15 '10 at 16:56
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Dear Mariano: Thanks for mentioning this question. I think one should ask if $S\cup(-S)$ [and not $S$] is maximal for the property in question. –  Pierre-Yves Gaillard Jul 15 '10 at 16:57
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$S$ isn't remotely maximal, as far as I can see. For example any divisor of $X^n-X$ for any $n$ can be added to it, as if $P(X)=0$ for $P$ some divisor of $X^n-X$ then $X^n-X=0$. Moreover, if you have a ring in which every element other than the number 9 satisfies $X^n=X$ for some $n$, then 9 will also satisfy this, because $3^n=3$ implies $9^n=9$. So you can also add $X-9$ to $S$. And so on and so on... –  Kevin Buzzard Jul 15 '10 at 18:17
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@Pierre: can you prove that your set works before 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
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It still might be true though! I certainly don't know a counterexample (as you probably guessed---because if I did I would have played it instantly!). I am not optimistic about finding a "nice" maximal set though. I think the first thing to do is to read the proof and to see what's really going on, and to go from there. –  Kevin Buzzard Jul 15 '10 at 20:38

1 Answer 1

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.

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I think I found the reference: Herstein, I. N. The structure of a certain class of rings. Amer. J. Math. 75, (1953). 864-871. --- Herstein even proves this: If, for every $a$ a ring $R$, there exists a polynomial $p_a(t)$ (with integral coefficients) such that $a^2p_a(a)-a$ is central, then $R$ is commutative. –  Pierre-Yves Gaillard Jul 15 '10 at 22:56
    
Yes, I should have explained that the full result only requires central rather than zero, but the latter is enough to answer the question. I never proved the full Jacobson theorem but I did work out the $f(x) = x^n - x$ problem some time ago (using the universal ring generated by the required relations) and if I remember correctly, the key conditions are that $f(x)=x$ mod $p$ and the form of $f$ rules out nilpotent elements. These two conditions easily force a direct sum decomposition into finite fields and this concludes the proof. Herstein apparently showed this is enough in general. –  T.. Jul 15 '10 at 23:07
    
I hope the following is correct. Let $A$ be the set of $f(X)\in\mathbb Z[X]$ with constant term $\pm1$, let $XA$ be the set of the $Xf(X)$, $f(X)$ in $A$. Let $R$ be a ring and $C$ its center. Herstein's Theorem says that if for any $r$ in $R$ there is an $f(X)$ in $XA$ such that $f(r)$ is in $C$, then $R=C$. It implies trivially the following. Put $B:=\{X-n\ |\ n\in\mathbb Z\}$. If for any $r$ in $R$ there is an $f(X)$ in $XA\cup A$ such that $f(r)=0$, then $R=C$. If for any $r$ in $R$ there is an $f(X)$ in $XA\cup B$ such that $f(r)$ is in $C$, then $R=C$. –  Pierre-Yves Gaillard Jul 16 '10 at 5:41
    
@T.: Herstein's set isn't maximal because you can add $X-1$. –  Kevin Buzzard Jul 16 '10 at 5:43
    
If you allow the zero ring as a ring, yes. But the $\pm X-n$ and the ones from A in Pierre-Yves' comment seem to be all you can add to Herstein's set. – T. 2 mins ago –  T.. Jul 16 '10 at 6:47

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