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.
 A: 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 $P\notin \operatorname{Ass}(N)$. In particular, we have $x\notin N$. 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 $ax\neq0$, 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 $P\notin \operatorname{Ass}(N)$:
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 $a\notin P$ by assumption, so this implies $b\in P$, and indeed $P=(0:n)$, giving us our contradiction.
A: $0\rightarrow L \overset{i}{\rightarrow} M \overset{j}{\rightarrow} N \rightarrow 0$
Let $(0:_{R} m) = P$, $P \in Spec(R)$ and there have non elements $l$ in $L$ such that $(0:_{R} l) = P$ .


*

*If $j(m)=0$ then $m\in Ker(j)=Im(i)$, there exists $l\in L$ such that $i(l)=m$.
So we have $Pi(l)=0 \Rightarrow i(Pl)=0 \Rightarrow Pl=0 \Rightarrow P \subseteq (0:_{R} l)$. Take $t \in ((0:_{R} l \Rightarrow tl=0 \Rightarrow ti(l)=0 $ then $tm=0 \Rightarrow t\in (0:_{R} m)=P$ (contraction)

*If $j(m)\neq 0$; let $j(m)=n \neq 0$. Now, $P(j(m))=j(Pm)=0$ then $P \subseteq( 0:_{R} j(m)) = (0:_{R} n).$ Take $t\in (0:_{R} n)$ then $tn=j(tm)=0$. So, $tm\in KerJ=Imi$. It means there exists $l\in L$ such that $i(l)=tm$. We have $P\subseteq (0:_{R} tm)= (0:_{R} i(l))$. If $t'\in (0:_{R} tm)\notin P$ then $t't\in P$ then $ t \in P$ or $t'\in P$. If $(0:_{R} tm)=(0:_{R} i(l))=P \Rightarrow (0:_{R} l)=P$ (contraction). Deduce that $tm=0\Rightarrow t\in (0:_{R} m)=P$ (done!)
