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Let $k$ be a field of characteristic zero, $R$ a commutative $k$-algebra, and $E$ the monoid of $k$-endomorphisms of $R$. Assume that $e_i \in E$, $1 \leq i \leq 2$, has centralizer $C_{e_i}$ in $E$ ($C_{e_i}$ is of course a sub-monoid of $E$).

Is it possible to know what is the centralizer of $e:=e_1e_2$? Clearly, $C_{e_1} \cap C_{e_2}$ is contained in $C_{e}$, the centralizer of $e$ (but $C_{e}$ may strictly contain $C_{e_1} \cap C_{e_2}$).

(What if we further assume that $e_1e_2=e_2e_1$?).

Any relevant ideas/hints are welcome. Perhaps the first answer in this question is relevant to my question.

Community
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user237522
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1 Answers1

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I doubt there is very much that is interesting you can say about this without adding some strong additional hypotheses. As an illustrative example, suppose $e_1$ is an automorphism of $R$ and $e_2=e_1^{-1}$. Then $e_1e_2=e_2e_1$, but $C_{e_1e_2}$ consists of all of $E$ since $e_1e_2=1$. On the other hand, $C_{e_1}\cap C_{e_2}=C_{e_1}$ will typically be smaller than $E$.

Eric Wofsey
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  • Thank you for your answer. Please, do you know of a 'not too strong' condition which will imply $C_{e_1} \cap C_{e_2}=C_{e_1e_2}$? Also, I am not sure I understand why, in your example, $C_{e_1} \subseteq C_{e_2}$; am I missing something trivial? – user237522 Mar 05 '17 at 01:00
  • I don't know of any such condition, I'm afraid. In my example, if $fe_1=e_1f$ then composing on the right and left with $e_2$ gives $e_2f=fe_2$. – Eric Wofsey Mar 05 '17 at 01:03
  • Thanks again. So I was missing something trivial.. – user237522 Mar 05 '17 at 01:07