Proof of $M$ Noetherian if and only if all submodules are finitely generated

Solution 1:

The first part is not quite correct: you have assumed $N$ is countably generated.

The second part looks good, though you should make sure you know why the fact that $N$ is finitely generated implies that the sequence is stationary.

Solution 2:

Your first part is strange. I would rephrase the argument as follows: Suppose that $M$ is a Noetherian $A$ - module and there exists a submodule $N$ that is not finitely generated. Now choose $f_1 \in N$. Since $N$ is not finitely generated, $\langle f_1 \rangle \subsetneqq N$ and so we can choose $f_2$ such that $f_2 \in N - \langle f_1 \rangle$. By the same argument as before, we can choose $f_3 \in N - \langle f_1,f_2\rangle$ such that $\langle f_1,f_2,f_3 \rangle \subsetneqq N.$ We now have a chain of submodules of $N$ (and hence of $M$) such that

$$\langle f_1 \rangle \subsetneqq \langle f_1,f_2 \rangle \subsetneqq \ldots \subsetneqq \langle f_1,f_2,f_3 \rangle $$

contradicting the ACC.