I'm stuck on a proof I'm reading.
Let $0\to M'\stackrel{\mu}\to M\stackrel{\sigma}\to M''\to 0$ be a sequence of $A$-modules. Then $\operatorname{Ass}(M)\subseteq \operatorname{Ass}(M')\cup \operatorname{Ass}(M'')$.
Let $P\in\operatorname{Ass}(M)$, so $P=\operatorname{ann}(m)$ for some $m\neq 0\in M$.
If $Am\cap im(\mu)=0$, the proof shows $P=\operatorname{ann}(\sigma(m))$, so $P\in \operatorname{Ass}(M'')$.
When $Am\cap im(\mu)\neq 0$ it says there exists $m_0\in Am\cap im(\mu)$ nonzero. Write $m_0=am=\mu(m_0')$ for $m_0'\in M'$. Then $P=\operatorname{ann}(m_0)=\operatorname{ann}(\mu(m_0'))=\operatorname{ann}(m_0')$ since $\mu$ is injective. So $P\in\operatorname{Ass}(M')$.
The only equality above I don't follow is $P=\operatorname{ann}(m_0)$. The containment $P\subseteq\operatorname{ann}(am)=\operatorname{ann}(m_0)$ is clear, since $P=\operatorname{ann}(m)$. But why does the converse hold?
Let's consider the generic example of short exact sequence, that is, $M'$ is a submodule of $M$ and $M''=M/M'$.
So, $P=\operatorname{ann}(m)$, $m\in M$, $m\ne 0$.
There are two cases: $Am\cap M'=0$ or not.
In the first case $P=\operatorname{ann}(\hat m)$, where $\hat m$ is the residue class of $m$ modulo $M'$.
In the second case, let $m_0\in Am\cap M'$, $m_0\ne 0$. Set $m_0=am$. Now you want to show that $P\supseteq\operatorname{ann}(m_0)$ (the other containment being obvious): if $bm_0=0$ then $bam=0$, so $ba\in P$ and since $P$ is prime and $a\notin P$ (why?) we get $b\in P$.