I'm solving this problem from "Introduction to Commutative Algebra" of Atiyah and Macdonald. Here is the problem:
Let $M$ be an $A$-module and let $N_1, N_2$ be submodules of $M$. If $M/N_1, M/N_2$ are Noetherian, so is $M/(N_1 \cap N_2)$.
I found a solution which states that we have the exact sequence $$0 \rightarrow M/N_1 \rightarrow M/(N_1 \cap N_2) \rightarrow M/N_2 \rightarrow 0$$
so $M/(N_1 \cap N_2)$ is Noetherian if and only if $M/N_1$ and $M/N_2$ are Noetherian.
I can't prove that this sequence is exact but can't find any counter-example for it. Can anyone help me? Thanks so much.
The sequence $0 \to M/N_1 \to M/(N_1 \cap N_2) \to M/N_2 \to 0$ is not exact, in general. A similar sequence that is exact is $0 \to N_2/(N_1 \cap N_2) \to M/(N_1 \cap N_2) \to M/N_2 \to 0$. The first sequence will be exact iff $N_1 + N_2 = M$, in which case the first terms of both sequences are isomorphic. However, the sequence that will solve your problem is the following:
$$0 \to M/(N_1 \cap N_2) \to M/N_1 \oplus M/N_2 \to M/(N_1 + N_2) \to 0$$
If $M/N_1, M/N_2$ are Noetherian, then so is $M/N_1 \oplus M/N_2$, so $M/(N_1 \cap N_2)$ is a submodule of a Noetherian module, hence is Noetherian. For why the sequence is exact, apply the reasoning in this answer.
Incidentally, I would advise you to stop using that particular set of solutions.