Show the direct sum of noetherian R-modules is noetherian. I have been struggling to understand the proof of how the direct sum of noetherian R-modules is noetherian. I have been working with a proof through induction and the exact sequence for the 2nd step of induction.
The sequence being:
$0 \rightarrow M_1 \rightarrow M_1 \oplus M_2 \rightarrow M_2 \rightarrow 0$
where we have: $m \mapsto (m,0)$ for the inclusion: $M_1 \rightarrow M_1 \oplus M_2$ and $(m,n) \mapsto m$ for the surjection: $M_1 \oplus M_2 \rightarrow M_2$.
Where I have trouble is with how this proves $M_1 \oplus M_2$ to be noetherian. I know it involves some sort of intersections, but I can't seem to understand where these intersections are defined from and how they preserve the noetherian properties from the images/pre-images.
In advance; thank you :)
 A: There is a result that says the following:
Let $0 \to L \stackrel{f}{\longrightarrow} M \stackrel{g}{\longrightarrow} N \to 0$ be an exact sequence of $A$-modules and $A$-homomorphisms. Show that $L$ and $N$ are Noetherian $A$-modules if and only if $A$-module $M$ is Noetherian.
To prove this result take an ascending sequence of $M$ submodules
$$
M_1\subseteq M_2 \subseteq \cdots
$$
So
$$
g(M_1)\subseteq g(M_2) \subseteq \cdots
$$
$$
f^{-1}(M_1)\subseteq f^{-1}(M_2) \subseteq \cdots
$$
are sequences of submodules in $N$ and $L$, respectively. Since $N$ and $L$ are notherian, such sequences park, hence
$$
g(M_k)=g(M_t), \ and \ f^{-1}(M_k)=f^{-1}(M_t)\ for\ all\ t\geq k  
$$
Under these conditions it can be shown that $M_k=M_t$ for all $t\geq k$.
Another slightly simpler way is as follows:
the exact sequence being $Ker(g)=Im(f)$ and furthermore $f$ is injecting and $g$ is surjective. Soon,
$$
\dfrac{M}{ker(g)}=\dfrac{M}{Im(f)}\cong N
$$
and
$$
L\cong Im(f).
$$
Then, $\dfrac{M}{Im(f)}$ and $Im(f)$ is Noetherian, what completes the test.
A: You are missing the fact that given an exact sequence of $R$-modules $$0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0,$$
$M$ is Noetherian if and only if $M_1$ and $M_2$ are Noetherian. You can find a proof here: Question about direct sum of Noetherian modules is Noetherian.
