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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.

Arturo Magidin
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francis-jamet
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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
  • I have edited. Thanks for the good wording. – francis-jamet Jun 27 '12 at 19:01
  • @francis-jamet: I've rephrased. The way you wrote it, the plain reading was: "if for each $r\in R$ there exists $p(x)\in E$ such that $p(r)=0$, then $R$ is commutative", but Jacobson's Theorem says that "if there exists $p(x)\in E$ such that for every $r\in R$ we have $p(r)=0$, then $R$ is commutative." – Arturo Magidin Jun 27 '12 at 19:56
  • @ArturoMagidin: No, Jacobson's theorem says: $\forall x \in R, \exists n_x \geq 2, x^{n_x}=x$ implies $R$ is commutative. – francis-jamet Jun 27 '12 at 20:36
  • @francis-jamet: oh, right. I'm thinking of an earlier version of the result, not the more general one. I fixed the statement. By the way, I just found this MO thread which is related. – Arturo Magidin Jun 27 '12 at 20:51

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