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.
Best Answer
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 $i$.