Let $M$ be an $R$-module for a ring $R$. Let $N$ be a submodule of $M$. I read that one can prove that $M$ is Noetherian if and only if $N$ and $M/N$ are Noetherian using these two results:
1) If $M$ is finitely generated and $A$ is a submodule, then $M/A$ is finitely generated.
2) If, for a submodule $A$, $A$ and $M/A$ are finitely generated, then $M$ is finitely generated.
I see how to prove the original result based on ascending chains but not how to do it using these results... Could someone enlight me here? It would be very helpful as I'm getting confused.
Let's prove that, if $N$ and $M/N$ are noetherian, then $M$ is noetherian. We need to see that every submodule of $M$ is finitely generated.
Consider a submodule $L$ of $M$. Then $L\cap N$ is a submodule of $N$, hence finitely generated. Also $(L+N)/N$, as a submodule of $M/N$ is finitely generated. Consider the exact sequence $$ 0\to L\cap N\to L\to L/(L\cap N)\to 0 $$ Since $$ L/(L\cap N)\cong (L+N)/N $$ is finitely generated, also $L$ is finitely generated because of fact 2.
For the converse, assume $M$ is noetherian and that $N$ is a submodule of $M$. Then $N$ is noetherian, because every submodule of $N$ is also a submodule of $M$. For $M/N$ recall that a submodule of $M/N$ is of the form $L/N$, where $L$ is a submodule of $M$ and $N\subseteq L$. Then $L/N$ is finitely generated because of fact 1.