# Mistake in the solution: If $M/N_1,M/N_2$ are noetherian then $M/(N_1\cap N_2)$ is noetherian

Let $$R$$ be a commutative ring with a unit. Let $$M$$ be an $$R$$-module and let $$N_1$$ and $$N_2$$ be sub-modules of $$M$$. Suppose that $$M/N_1$$ and $$M/N_2$$ are noetherian. Show that $$M/N$$ is noetherian where $$N:=N_1\cap N_2$$.

I know that there is an answer for that question here. But I want to ask where is the mistake in my mistaken solution. In my solution I don't use the noethrianess of $$M/N_2$$, which is doesn't sound right:

Let $$S_1\subseteq S_2\subseteq\dots\subseteq M/N$$ be a chain of submodules. Let us define $$A_i:=\{m+N_1:m+N\in S_i\}$$ $$\forall i,A_i$$ is a submodule of $$M/N_1$$. Indeed, if $$m+N_1\in A_i$$ then $$m+N\in S_i\subseteq S_{i+1}\Rightarrow m+N_1\in A_{i+1}$$. Hence, $$A_1\subseteq A_2\subseteq...\subseteq M/N_1$$ is a chain of submodules. $$M/N_1$$ is noetherian. So $$\exists a,\forall a\leq i\leq j, A_i=A_j$$. So $$\forall a\leq i\leq j$$,$$S_i=\{m+N:m+N_1\in A_i\}\\=\{m+N:m+N_1\in A_j\}=S_j$$ Hence $$M/N$$ is noetherian.

You claim that $$S_i=\{m+N:m+N_1\in A_i\}$$, but this has swapped the implication: having $$m+N_1\in A_i$$ does not imply $$m+N\in S_i$$.
The problem is that the classes $$m+N_1$$ are larger than the classes $$m+N$$. Maybe an illuminating example is to take the extreme case where $$N_1=M$$ and $$N_2=0$$. Then $$M/N_1=0$$, so $$A_i=0$$ for all $$i$$.