$Ass(M) = Ass(N) \cup Ass(M/N)?$ Suppose $M$ is an $R$ module and $N$ is a submodule of $M.$ I am trying to prove the following: $Ass(M) = Ass(N) \cup Ass(M/N)$. This seemed intuitively true but the proof proved to be difficult. If I have a prime ideal $P \in Ass(M)$ then $P = Ann(m)$ for some $m \in M.$ If $m \in N$ then $P \in Ass(N) \cup Ass(M/N).$ But suppose it isn't. We already have $P \subset Ann(m + N)$ so now I want to show that $Ann(m + N) \subset P.$ How do I go about doing this? If I have an element $x \in R$ such that $xm \in N$ I want to show that $xm = 0.$ If $R$ were a field this would be easy as I could simply multiply the inverse of $x$ and get that $m \in N$ which contradicts the assumption that $m \notin N.$ Should I continue in this approach? 

Disclaimer: I'm not quite good enough to give a nice hint, or point you in the right direction. Here is the argument I am familiar with, but read on only if you want a spoiler

claim:$ Ass(M) \subset Ass(N) \cup Ass(M/N)$.
Let $P \in Ass(M)$. Note that $P$ is associated to $M$ if and only if there is an injection $f:R/P \hookrightarrow M$. Consider the image $\mathrm{Im}(f) \subset M$.
First assume that $\mathrm{Im}(f)$ is disjoint from $N$. Then $\pi:M \to M/N$ restricts to a monomorphism, which we consider as $\pi \circ f:R/P \to M/N$, which shows that $P \in Ass(M/N)$.
Now, assume that $\mathrm{Im}(f)$ intersects $N$ nontrivially, with $x \in \mathrm{Im}(f) \cap N$, but then $ann(x)=P$, since the image of $f$ is isomorphic to $R/P$, so $P$ is an associated prime $Ass(N)$, since it is the anhilator of some element $x \in N$.
