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 :)

Solution 1:

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.

Solution 2:

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.