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Let $A$ be a ring. Let $E$ be the set of polynomials $\{X^n-X \in \mathbb{Z}[X]|n \in \mathbb{N}^*-\{1\}\}$.

By the theorem of Jacobson, we know that if for each $a\in A$ there is an element of $E$ for which $a$ is a root, then $A$ is commutative.

Is there a characterization of the sets $F \subset \mathbb{Z}[X]$ such that, for all ring $A$, if every element of $A$ is a root of a polynomial in $F$, then $A$ is commutative ?

Thanks in advance.

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In English, it's "polynomial", not "polynom" (though we seem to see the latter a lot in this site). "Annihilated" is the usual term, rather than "annulled". Hope that helps (it may also help in google searches) – Arturo Magidin Jun 27 '12 at 16:54
@rschwieb: How do you mean? The theorem of Jacobson in question is that a ring that satisfies $x^n=x$ (for some fixed $n\gt 1$) is necessarily commutative. – Arturo Magidin Jun 27 '12 at 18:31
@rschwieb: $\mathbb{Z}[x]$ maps into the set of function $R\to R$, and the set of functions $R\to R$ acts on $R$ by evaluation; that's the action in question. That is, the element $p(x)\in\mathbb{Z}[x]$ acts on $R$ by $p(x)\cdot a = p(a)$. – Arturo Magidin Jun 27 '12 at 18:40
@ArturoMagidin Would it be a little less disorienting if it said "every element of $A$ is a root of something in $E$"? I did not recognize the theorem at all, stated this way. – rschwieb Jun 27 '12 at 18:52
@rschwieb: It has been so changed now. – Arturo Magidin Jun 27 '12 at 18:58

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