2
$\begingroup$

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 $\mathfrak{q}, \mathfrak{q} \in Supp(M/xM)$?

$\endgroup$
4
  • 7
    $\begingroup$ Not sure I understand. $Ass(M/xM) \subseteq Supp(M/xM)$, right? $\endgroup$ Jun 2, 2010 at 19:45
  • $\begingroup$ 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 $\mathfrak{p} \in Supp(N)$. $\endgroup$
    – ressing
    Jun 2, 2010 at 20:01
  • $\begingroup$ So Long's comment answers the question, right? $\endgroup$ Jun 3, 2010 at 8:24
  • $\begingroup$ @Karl Schwede: It does, yes. I should have been clearer in my comment in response to Hailong's. $\endgroup$
    – ressing
    Jun 3, 2010 at 20:13

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.