Proposition 4

Let $M$ be an A-module and $\Phi$ a subset of $\operatorname{Ass}(M)$. Then there exists a submodule $N$ of $M$ such that $\operatorname{Ass}(N) = \operatorname{Ass}(M) - \Phi$ and $\operatorname{Ass}(M/N) = \Phi$.

Let $\mathscr{E}$ be the set of submodules $P$ of $M$ such that $\operatorname{Ass}(P) \subset \operatorname{Ass}(M) - \Phi$. Formula (1) shows that the set $\mathscr{E}$, ordered by inclusion, is inductive; moreover, $\{0\} \in \mathscr{E}$ and hence $\mathscr{E} \neq \varnothing$. Let $N$ be a maximal element of $\mathscr{E}$. Then $\operatorname{Ass}(N) \subset \operatorname{Ass}(M) - \Phi$. We shall see that $\operatorname{Ass}(M/N) \subset \Phi$, which, by Proposition 3, will complete the proof. Let $p \in \operatorname{Ass}(M/N)$; then $M/N$ contains a submodule $F/N$ isomorphic to $A/p$. By Propositions 1 and 3, $\operatorname{Ass}(F) \subset \operatorname{Ass}(N) \cup \{p\}$. Since $N$ is maximal in $\mathscr{E}$, $F \notin \mathscr{E}$ and hence $p \in \Phi$.