Prove if $S \subseteq Ass_N(M)$ then their exist a submodule $N$ such that $Ass_R(N)=S$.

Question

$R$ is a commutative ring and $M$ is a $R$-module.

I am trying to prove,

If $S \subseteq Ass_R(M)$ then their exist a submodule $N$ such that $Ass_R(N)=S$.

I want use Zorn lemma to prove this problem. Let $N'$ be an element of a collection of submodules such that $Ass_R(N') \subseteq S$. This collection of submoules is partially ordered. Must exist a maximal element by Zorn's lemma...

Answer 1

I have just realised how to prove this statement. It will be great if someone can verify.

Let $N$ be a maximal element in the collection of submodules $M'$ such that $Ass_R(M') \subseteq S$ due to Zorn's lemma.

I claim that $Ass_R(N) =M$. If it was not true then exists $m \in M$ such that $Rm \cong R/P$ where $P$ is a prime ideal, thus $N \subset N+Rm$.For any $n \in N$, $Rn \cong R/I$ meaning that $rn=0$ iff $r \in I$.

Concluding that $r(n+m)=0$ when $r \in I\cdot P$. So from this statement we can see that $R(n+m) \cong R/I'$. We see that $I'$ is not a prime ideal. So $Ass_R(N+Rm) \subseteq S$ and $N $ is a submodule of $N+Rm$ which contradicts our assumption $m \notin N$, so $Ass_R(N)=S$.