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Possible Duplicate:
Annihilator of a simple module

Let me ask the same question as before because I still have trouble understanding the problem.

Let $R$ be a finitely generated commutative ring and $C$ an $R$-algebra ($C$ is not necessarily commutative). Assume that $C$ is a finitely generated $R$-module.

If $S$ is a simple $C$-module, then it seems known that the annihilator $I=Ann_{C}(S)$ of $S$ is of the form $I=\mathfrak{m}C$ for some maximal ideal $\mathfrak{m}$ of $R$. Could anyone provide me of a proof?

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marked as duplicate by m_t_, Chris Eagle, rschwieb, LVK, Henning Makholm Aug 20 '12 at 1:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It would be better to edit your old question or comment on Matt E's answer to say why you don't understand it. This is a duplicate. – m_t_ Aug 18 '12 at 8:43

To make explicit a counterexample that Matt E hinted at in the previous question you linked to: let $R$ be a field (I don't know what you mean by "finitely generated commutative ring"), $C=R\times R$ (direct product of rings) which is a finitely generated $R$-algebra, and $S=R$, viewed as $C$-module by the first factor of $R\times R$ (the second component is ignored). This is certainly a simple module, and $I=\mathrm{Ann}_{C}(S)=\{0\}\times R$ is not of the form $I=\mathfrak{m}C$ for any maximal ideal $\mathfrak m$ of $R$ (indeed $\mathfrak m=\{0\}$ would be the only choice, and it fails).

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Thank you for the detail counterexample, Marc. Now that I have a better understanding. I appreciate your assistance. – M. K. Aug 23 '12 at 5:55

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