Say we have a short exact sequence of modules
$$0 \longrightarrow M' \overset{f}{\longrightarrow} M \overset{g}{\longrightarrow} M'' \longrightarrow 0$$
I want to show that $M$ is Notherian iff $M'$ and $M''$ are.
I'm solid on the proof that if $M$ is Noetherian than $M$ and $M''$ are (since, otherwise, we could push forward or pull back infinite ascending chains of submodules).
For the other direction, I want to try something like this. Assume $M$ is not Noetherian i.e. it has an infinite ascending submodules $N_1 \subset N_2 \subset \cdots$. If $M'$ also has such an infinite ascending of submodules, we are done. Otherwise, $f(M') = ker(g)$ is a submodule of $M$ which is finitely generated. So, we can define:
- $\hat{M} = M/f(M')$
- $\hat{N}_i = N_i/f(M')$
- $\hat{g}:\hat{M} \to M''$ as $\hat{g}(x f(M')) = g(x)$
$\hat{g}$ is a bijection. In addition, the $\hat{N}_i$ are now also an infinite increasing chain of submodules of $\hat{M}$ (perhaps after discarding a finite number of them in the beginning) and, so, we get that the $\hat{g}(\hat{N}_i)$ are an infinite increasing chain of submodules of $M''$ meaning that $M''$ is not Noetherian.
Does this work? I feel like it does but my professor's proof seems more complicated.