Let $M$ be an $R$-module. I want to show only using the definition of noetherian that if $N$ is a noetherian submodule of $M$ such that $M/N$ is noetherian, then $M$ is noetherian.
I know that if $M_1 \subset M_2 \subset \cdots$ is a chain of submodules in $M$, then since $N$ and $M/N$ are noetherian, the chains $$M_1 \cap N \subset M_2 \cap N \subset \cdots,$$ and $$M_1 + N \subset M_2 + N \subset \cdots$$ stabilize. I feel like this should imply that $M_1 \subset M_2 \subset \cdots$ stabilizes too, but I'm not sure how to show this. Any hints? Thanks.
I think the conditions should be for the ideals chain $\;M_1\subset M_2\subset\ldots\;$ that
$$\begin{cases}M_1\cap N\subset M_2\cap N\subset\ldots\;\text{stabilizes}\\{}\\M_1N/N\subset M_2N/N\subset\ldots\;\text{stabilizes}\end{cases}\;\;\implies\;\;M_1\subset M_2\subset\ldots\;\text{stabilizes}$$
Let us take $\;k\;$ to be the minimal index for which both chains above stabilize: $\;(1)\;\;M_k\cap N=M_{k+1}\cap N\;,\;\;(2)\;\;M_kN/N=M_{k+1}N/N\;$
Let now $\;x\in M_{k+1}\;$ , then we can deduce by (2) that
$$x+n=y+n'\;,\;\;n,n'\in N\;,\;\;y\in M_k\implies x-y=n'-n\in M_{k+1}\cap N$$
since $\;M_k\subset M_{k+1}\;$ , so by (1) we get
$$x-y=t\in M_k\cap N\implies x = y+t\in M_k$$
and we've proved $\;M_{k+1}\subset M_k\;$