Continuing a simple proof involving $\operatorname{Ass}(M)$. > Let $M$ be an $R$-module and $N$ be an $R$-submodule of $M$. The question is to prove $$\operatorname{Ass}(M)\subseteq \operatorname{Ass}(N) \cup \operatorname{Ass}(M/N).$$ I know there are some proofs for this problem and they are all kind of simple. But I believe there exists a more simple proof just using definition because if I remember correctly, I made one. Here's what I do: Let $P\in \operatorname{Ass}(M)$. Then $\exists\, x\in M\setminus \\{0\\}$ s.t. $P=(0:x)$. If $x\in N$ then $P\in \operatorname{Ass}(N)$, else $x\notin N$. Can anyone continuing this proof as simply as possible? Thank you for all your help.

You're on the right track! Throughout this proof, I am going to write $\bar x$ to denote the residue of $x\in M$ in $M/N$.

As you say, if $P\in \operatorname{Ass}(M)$ then there is some nonzero $x\in M$ such that $P=(0:x)$. Now, suppose $P\
otin \operatorname{Ass}(N)$. In particular, we have $x\
otin N$. Now, we would like to show that $P=(0:\bar x)=(N:x)$, so that $P\in \operatorname{Ass}(M/N)$.

Clearly, $P\subseteq(N:x)$. For the reverse inclusion, suppose $a\in(N:x)$. Then $ax\in N$; if $ax=0$, then $a\in P$ and we are done.

We now show that if $ax\
eq0$, this creates a contradiction. If this is the case, let $n=ax\in N$. Then we show that $P=(0:n)$, contradicting our assumption that $P\
otin \operatorname{Ass}(N)$:

Again, we clearly have $P\subseteq(0:n)$; for the reverse, suppose $b\in(0:n)$. Then $0=bn=(ab) x$, which tells us $ab\in P$. But $P$ is prime and $a\
otin P$ by assumption, so this implies $b\in P$, and indeed $P=(0:n)$, giving us our contradiction.