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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)