# Any subset of associated primes of a module M is $\text{Ass}_R(N)$ for some sub-module $N$ of $M$.

Given an $$R$$-module $$M$$ ($$R$$ is commutative) and a $$S\subseteq {\rm Ass}_R(M)$$, I want to show that there is a sub-module $$N$$ of $$M$$ s.t.$${\rm Ass}_R(N)=S$$.

So far, I've used Zorn's Lemma on the set of sub-modules $$N'$$ such that $${\rm Ass}_R(N')\subseteq S$$, to show that this set contains a maximal element $$T$$. I don't see why $${\rm Ass}_R(T)$$ could not be a proper subset of $$S$$.

I presume that $$R$$ is commutative with unity.

I will use the following theorem:

If $$N$$ is a submodule of $$M$$ then $$\newcommand{\as}{\text{Ass}} \as(N)\subseteq {\as (M) \subseteq {\as (N ) \cup {\as (M / N)}}}$$.

Suppose $$\as(T)\subsetneq S$$. Then there is a prime ideal $$P\in S\setminus\as(T)$$ such that $$P=\newcommand{\an}{\text{Ann}}\an(x)$$ for some $$x\in M\setminus T$$. Consider the submodule $$T+Rx$$. Then $$\as(T+Rx)\subseteq \as(T)\cup\as((T+Rx)/T)=\as(T)\cup\as(Rx/(T\cap Rx)).$$ If $$\as(Rx/(T\cap Rx))=\emptyset$$, then $$\as(T+Rx)\subseteq S$$, a contradiction to the maximality of $$T$$. So assume $$\as(Rx/(T\cap Rx))\neq \emptyset$$. Let $$Q=\an(\bar y)$$ be a prime ideal for some $$y\in Rx$$. Clearly, $$P\subseteq Q$$. Is $$Q\subseteq P$$? Yes!

Suppose $$a\in Q$$ but not in $$P$$. Then $$a\bar y=0\neq ax$$. Let $$y=rx$$. Then $$a\bar y=0$$ implies $$arx\in Rx\cap T$$. In particular, $$arx\in T$$.

I claim that $$\an(x)=\an(arx)$$.

Clearly $$\an(x)\subseteq\an(arx)$$. Now, let $$z\in \an(arx)$$. So $$zarx=0$$ and therefore $$zar\in\an(x)=P$$. Since $$P$$ is prime and $$a\notin P$$, we have $$zr\in P$$. So either $$z\in P$$ or $$r\in P$$; but if $$r$$ were in $$P$$, then that would imply $$y=0$$ and hence $$Q=(1)$$, a contradiction.

So $$\an(x)=\an(arx)$$. But $$arx\in T$$. Therefore $$P=\an(x)\in\as(T)$$, a contradiction to the choice of $$P$$. So $$a\in P$$ and hence $$Q\subseteq P$$.

So, we have $$P=Q$$. Therefore $$\as(T+Rx)\subseteq \as(T)\cup\{P\}\subseteq S$$, again a contradiction.