How can associated primes be distributed among submodule and quotient?

Question

Suppose $M$ is a finitely generated module over a Noetherian ring $A$ (commutative, with identity). Then for a submodule $N$, we have the following relation among the sets of associated primes: $$\DeclareMathOperator{\Ass}{Ass}\Ass(N) \subseteq \Ass(M) \subseteq \Ass(N) \cup \Ass(M/N) $$

What sort of converses does this statement have?

1. Which subsets of $\Ass(M)$ appear as $\Ass(N)$?
2. For which subsets $S$ of $\Ass(M)$ can we find a submodule $N$ such that $\Ass(N) = S$ and $\Ass(M/N) = \Ass(M) \setminus S$?

Singleton subsets of $\Ass(M)$ appear as $\Ass(N)$, as in (1), almost by definition. Also, $S = \Ass(M)$ and $S = \varnothing$ work for (2) for trivial reasons. For other $S$'s, I was trying to use something along the lines of take a maximal $N$ such that $\Ass(N) \subseteq S$, and so on. But I could not show that such an $N$ works for (2), or even (1).

For (2), I see that it's enough to show that we can take $S$ to be $\newcommand{\fp}{\mathfrak{p}} \Ass(M) \setminus \{\fp\}$ for each $\fp \in \Ass(M)$. Let $\Ass(M) = \{\fp_1,\dots,\fp_n\}$. Let $N_i$ be a submodule such that $\Ass(N_i) = \Ass(M) \setminus \{\fp_i\}$ and $\Ass(M/N_i) = \{\fp_i\}$ (that is, $N_i$ is $\fp_i$-primary). Then $$N_S = \bigcap _{\fp_i \notin S} N_i$$ works for $S$, since $\Ass(N_S) \subseteq S$ and $\Ass(M/N) \subseteq \Ass(M) \setminus S$.

Answer 1

A proof that (2) works for any subset can be found in Bourbaki's Commutative Algebra, Chapter IV, §1, Proposition 4. The argument is as follows:

Let $\Sigma$ be the collection of submodules $P$ of $M$ such that $\Ass(P) \subseteq S$. Either using Zorn's lemma (so that this works even in the non-Noetherian case) or using that $M$ is Noetherian, let $N$ be maximal in $\Sigma$. It is enough to show that $\Ass(M/N) \subseteq \Ass(M) \setminus S$. Let $\fp \in \Ass(M/N)$. Choose $N \subsetneq F \subseteq M$ such that $F/N \cong A/\fp$. Then $\Ass(F) \subseteq \Ass(N) \cup \{\fp\}$. Since $N$ is maximal in $\Sigma$, $F \notin \Sigma$, and the only way this can happen is $\Ass(F) = \Ass(N) \cup \{\fp\}$ and $\fp \in \Ass(M) \setminus S$.