Show module is Noetherian

Question

Let $R$ be a commutative ring and let $0 \to L \to M \to N \to0$ be an exact sequence of $R$-modules. Prove that if $L$ and $N$ are noetherian, then $M$ is noetherian. I tried considering the pre image of the map $L \to M$ and the image of the map $M \to N$ as they are submodules of $L$ and $N$ respectively, but I couldn't make headway.

Answer 1

Let $S_1\subset S_2 \subset S_3 \subset ... $ be an ascending chain of submodules in $M$.

Now consider the chains $S_1\cap L \subset S_2\cap L \subset ... $ and $f(S_1)\subset f(S_2)\subset ... $ in $L$ and $N$, respectively (here $f:M\rightarrow N$ is the second map in the exact sequence).

As $L$ and $N$ are noetherian, both of these chains stabilize. Now you must show that if $S_i\subset S_j$, $S_i\cap L = S_j\cap L$ and $f(S_i) = f(S_j)$, then $S_i = S_j$.

Answer 2

Let $$0\longrightarrow L\overset{i}\longrightarrow M\overset{p}\longrightarrow N\longrightarrow0$$ be a short exact sequence of $R$-modules. Assume that $L$ and $N$ are Noetherian modules.

Let $(M_n)_{n\geq0}$ be an ascending chain of submodules of $M$.

Since $(p(M_n))_{n\geq0}$ is an ascending chain of submodules of $N$, and $N$ being Noetherian, there exists $n_0\geq0$ such that: $$\tag{$*$}\forall j\geq n_0,\quad p(M_j)=p(M_{n_0}).$$

Similarly, $(i^{-1}(i(L)\cap M_n))_{n\geq0}$ being an ascending chain of submodules of $L$, and $L$ being Noetherian, there exists $n_1\geq0$ such that: $$\forall j\geq n_1,\quad i^{-1}(i(L)\cap M_j)=i^{-1}(i(L)\cap M_{n_1}),$$ which also implies: $$\tag{$**$}\forall j\geq n_1,\quad i(L)\cap M_j=i(L)\cap M_{n_1}.$$

Set $n=\max\{n_0,n_1\}$.

(I guess you got to this point of the proof).

We'll now show that $$\forall j\geq n,\quad M_j=M_n.$$ Let $j\geq n$. We already have the inclusion $M_n\subset M_j$ so let's prove the inclusion $M_j\subset M_n$: let $x\in M_j$. Since $p(x)\in p(M_j)=p(M_n)$ (see Property $(*)$), there exists $y\in M_n$ such that $p(x)=p(y)$, i.e., $p(x-y)=0$, i.e., $x-y\in\ker p$. Now, since we have an exact sequence (namely, $\ker p=\operatorname{Im}i=i(L)$), $x-y\in i(L)$. Now by the inclusion $M_n\subset M_j$ we know that $x-y\in M_j$, hence $x-y\in i(L)\cap M_j=i(L)\cap M_n$ ( see Property $(**)$), hence $x\in M_n+i(L)\cap M_n=M_n$.

Remark. I never used injectivity of the mapping $i$ nor surjectivity of the mapping $p$. Hence the result still holds true for an exact sequence $$L\overset{i}\longrightarrow M\overset{p}\longrightarrow N$$ (and that's why I carefully left the $i$ all the way through the proof).