direct sum of noetherian modules is noetherian

Question

We say that an $A$-module $M$ is Noetherian if all of its submodules are finitely generated. Having that definition in mind can anyone give me some hints to prove that if $M$ and $N$ are Noetherian modules then $M \oplus_A N$ must be Noetherian?

The problem I'm having is that an arbitrary submodule $Q$ of $M \oplus_A N$ is not necessarly of the form $Q = \bar{M} \oplus \bar{N}$ where $\bar{M}$ and $\bar{N}$ are submodules of $M$ and $N$ respectively. So I am out of ideas, and will be very happy to recieve some help.

Answer 1

Hint: first show that if $P \subseteq Q$ are $A$-modules, then $P, Q/P$ Noetherian implies $Q$ Noetherian.

Answer 2

Let $L_1\subseteq L_2\subseteq\cdots$ be an ascending chain of sub-modules of $M\oplus N$. Then $L_1\cap N\subseteq L_2\cap N\subseteq\cdots$ is stationary, so for some $k$ we have $L_k\cap N=L_{k+1}\cap N=\cdots$.

Similarly, $(L_1+N)/N\subseteq(L_2+N)/N\subseteq \cdots$ will be stationary, so for some $\ell$ we have $L_\ell+N=L_{\ell+1}+N=\cdots$.

Now for any $n\ge\max\{k,\ell\}$ let $x\in L_{n+1}\setminus L_{n}$. Since $L_n+N=L_{n+1}+N$, we have $x\in N$. However, then $x\in (L_{n+1}\cap N)\setminus (L_n\cap N)$, which is impossible.