Lemma 10.72.8. Let $(R, \mathfrak m)$ be a local Noetherian ring and $M$ a finite $R$-module. Let $x \in \mathfrak m$, $\mathfrak p \in \text{Ass}(M)$, and $\mathfrak q$ minimal over $\mathfrak p + (x)$. Then $\mathfrak q \in \text{Ass}(M/x^ nM)$ for some $n \geq 1$.

Proof. Pick a submodule $N \subset M$ with $N \cong R/\mathfrak p$. By the Artin-Rees lemma (Lemma 10.51.2) we can pick $n > 0$ such that $N \cap x^ nM \subset xN$. Let $\overline{N} \subset M/x^ nM$ be the image of $N \to M \to M/x^ nM$. By Lemma 10.63.3 it suffices to show $\mathfrak q \in \text{Ass}(\overline{N})$. By our choice of $n$ there is a surjection $\overline{N} \to N/xN = R/\mathfrak p + (x)$ and hence $\mathfrak q$ is in the support of $\overline{N}$. Since $\overline{N}$ is annihilated by $x^ n$ and $\mathfrak p$ we see that $\mathfrak q$ is minimal among the primes in the support of $\overline{N}$. Thus $\mathfrak q$ is an associated prime of $\overline{N}$ by Lemma 10.63.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).