# Question on a property of $\mathrm{Ass}(M)$ for modules over noetherian rings

I got stuck reading a proof of the following lemma (Lemma 0.19 in this file):

Lemma Suppose that $M$ is a module over a commutative noetherian ring $R$ and let $m\neq 0 \in M$. Let $S$ be a multiplicative set in $R$ such that $S\cap \mathrm{Ann}(m) = \emptyset$. Then there exists $\mathcal{P} \in \mathrm{Ass}(M)$ such that $\mathrm{Ann}(m)\subseteq \mathcal{P}$ and $S\cap \mathcal{P} = \emptyset$.

The proof goes like this:

Use Zorn's lemma to get a maximal ideal $\mathcal{Q}$ with respect to the properties $\mathrm{Ann}(m)\subseteq \mathcal{Q}$ and $S\cap \mathcal{Q} = \emptyset$ and prove that it is a prime ideal.

Use that $R$ is noetherian to get a maximal ideal $\mathcal{P}$ among the ideals $\mathcal{P} \subseteq \mathcal{Q}$ such that $\mathcal{P} = \mathrm{Ann}(rm)$ for some $r\in R$ with $rm \neq 0$.

Now we want to prove that $\mathcal{P}$ is prime but here is where I got stuck. If $r_1 r_2 \in \mathcal{P}$ and $r_2 \notin \mathcal{P}$ then $r_2 r m \neq 0$ and $\mathcal{P} \subseteq \mathcal{P} + (r_1) \subseteq \mathrm{Ann}(r_2 r m) \subseteq \mathcal{Q}$. I don't see why the last inclusion, $\mathrm{Ann}(r_2 r m) \subseteq \mathcal{Q}$ holds.

Thanks!

• When you write ‘$\text{Ass}(M)$’, do you mean the set of associated primes of $M$? – Berrick Caleb Fillmore Apr 19 '15 at 7:06
• Yes, that's what I mean. – Pipicito Apr 19 '15 at 16:02

Suppose $\mathrm{Ann}(r_2rm)\not\subseteq Q$. Then there is $a\in \mathrm{Ann}(r_2rm)$, $a\notin Q$. Now $\mathrm{Ann}(arm)\subseteq Q$: $b\in\mathrm{Ann}(arm)\Rightarrow b(arm)=0\Rightarrow (ba)rm=0\Rightarrow ba\in P\Rightarrow ba\in Q\Rightarrow b\in Q$. Since $P\subseteq\mathrm{Ann}(arm)$ and $P$ is maximal with some properties that $\mathrm{Ann}(arm)$ also satisfies we must have $P=\mathrm{Ann}(arm)$. In particular, $r_2\in P$, a contradiction.