# Associated primes in Noetherian rings

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! :)

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 $$m\in N$$.