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This question is motivated by a proof in Bruns' and Herzog's book on "Cohen Macaulay Rings".

Let $\(R,\mathfrak{m}\)$ be a Noetherian local ring, $M \neq 0$ a finitely generated $R$-module. Suppose further that $\mathfrak{p} \in Ass(M)$ and that $x \in \mathfrak{m}$ and also that $x$ is a nonzero divisor on $M$. In the literature this is called an $M$-regular element.

It is clear that $\mathfrak{p}$ consists of zerodivisors of $M/xM$, thus that it is contained in some $\mathfrak{q} \in Ass(M/xM)$. [To see this is suffices to note that the union of the associated primes is the set of zerodivisors, and then apply the prime avoidance theorem.]

Obviously, $\mathfrak{p} \notin Supp(M/xM)$, since $x \notin \mathfrak{p}$. Is it true that for this $\mathfrak{q}, \mathfrak{q} \in Supp(M/xM)$?

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Not sure I understand. $Ass(M/xM) \subseteq Supp(M/xM)$, right? –  Hailong Dao Jun 2 '10 at 19:45
    
Thank you, you're right. $\mathfrak{p} \in Ass(N) \implies 0 \rightarrow A/\mathfrak{p} \rightarrow N$ exact $\implies 0 \rightarrow A/\mathfrak{p} \otimes A_{\mathfrak{p}} \rightarrow N \otimes A_{\mathfrak{p}}$ exact $\implies 0 \rightarrow A_{\mathfrak{p}}/{\mathfrak{p}A_{\mathfrak{p}}} \rightarrow N_{\mathfrak{p}}$ exact. Thus $N_{\mathfrak{p}}$ contains a nonzero submodule and so $\mathfrak{p} \in Supp(N)$. –  ressing Jun 2 '10 at 20:01
    
Do you mean that $\mathfrak{p}$ consists of zerodivisors? of $M/xM$? –  Matthew Morrow Jun 2 '10 at 20:02
    
Indeed. Fixed, thanks. –  ressing Jun 2 '10 at 20:06
    
So Long's comment answers the question, right? –  Karl Schwede Jun 3 '10 at 8:24

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