I'm currently working on the following exercise:
Exercise: Let $A$ be a noetherian ring and $M$ be an $A$-module, $N$ be an $A$-submodule of $M$. Let $x \in A$. Prove that if $x \not\in \mathfrak{p}$ for any $\mathfrak{p} \in \operatorname{Ass}(M/N)$, then $xM \cap N = xN$.
My attempts 1: Clearly, $xN \subseteq xM \cap N$. So to get the reverse inclusion, I'm trying to localize at every $\mathfrak{p} \in \operatorname{Spec} A$ and see whether the two sides coincides:
- When $\mathfrak{p} \in \operatorname{Ass}(M/N)$, using that $x \not\in \mathfrak{p}$, we see $(xN)_{\mathfrak{p}} = N_{\mathfrak{p}}$ and $(xM \cap N)_{\mathfrak{p}} = M_{\mathfrak{p}} \cap N_{\mathfrak{p}} = N_{\mathfrak{p}}$, as desired.
- But I got stuck on primes not in $\operatorname{Ass}(M/N)$, and have no idea on how to deal with this case. As $\operatorname{Ass}(M/N) \subseteq \operatorname{Supp}(M/N)$, with the equality seldom holds, it seems that we cannot hope the localizations are zero.
My Attempts 2: After searching on this site, I found the post Intersection of ideals equal to their product on noetherian ring is a baby version of this exercise. The main tool is the fact:
Let $I$ and $J$ be ideals of a Noetherian ring $A$. If $JA_P\subseteq IA_P$ for every $P\in \operatorname{Ass}_A(A/I)$, then $J\subseteq I$.
But I also got stuck on this:
- I need a module version of the above fact. But the proof of the above fact (see Associated Prime Ideals in a Noetherian Ring; Exercise 6.4 in Matsumura) relies on the primary decomposition of ideals in noetherian rings. When turn to the modules, we need $M$ to be finitely generated as an $A$-module. But the exercise above does not require this.
- Even if I have the module version of the above fact, we need to do localization at primes in $\operatorname{Ass}(M/xN)$, NOT $\operatorname{Ass}(M/N)$. But we only know $\operatorname{Ass}(M/N) \subseteq \operatorname{Ass}(M/xN)$ by the exact sequence given by the third isomorphism theorem
$$
0 \rightarrow M/N \rightarrow M/xN \rightarrow N/xN \rightarrow 0.
$$
I'm trying to show $\operatorname{Ass}(M/N) = \operatorname{Ass}(M/xN)$, but by writing down the definition carefully, it seems that the inclusion $\supseteq$ is just another way of saying $xM \cap N = xN$, which is the ultimate goal of this exercise.
So finally I got stuck here. Thank you for you all for answering and commenting! 🙂
Best Answer
Let $z\in xM\cap N$. Then $z=xm\in N$, so $\overline{xm}=\overline 0$ in $M/N$. But $x$ is a non-zerodivisor on $M/N$, so $\overline m=\overline 0$. This imples that $m\in N$.