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

zerodivisors? of $M/xM$? – Matthew Morrow Jun 2 '10 at 20:02