Noetherian property for exact sequence Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then $M$ is Noetherian is equivalent to $M'$ and $M''$ are Noetherian.
For the ''$\Leftarrow$'' case:  I guess if we let $(L_n)_{n\geq 1}$ be an ascending chain of submodules of $M$, then $(\alpha ^{-1}(L_n))_{n\geq 1}$ is a chain in $M'$, and $(\beta(L_n))_{n\geq 1}$ is a chain in  $M''$. For large $n$ both these chains are stationary. Then why we know the chain $(L_n)$ is stationary? 
Thanks! 

Let $x\in L_{n+1}$. Then $\beta(x)\in\beta(L_{n+1})$ and let's suppose that $\beta(L_n)=\beta(L_{n+1})$. We have $\beta(x)\in\beta(L_n)$, so there is $y\in L_n$ such that $\beta(x)=\beta(y)$ hence $x-y\in\ker\beta$. From $\ker\beta=\operatorname{im}\alpha$ we get $x-y=\alpha(z)$ for some $z\in M'$. Since $x-y\in L_{n+1}$ we have $z\in\alpha^{-1}(L_{n+1})=\alpha^{-1}(L_{n})$, so $x-y=\alpha(z)\in L_n$ and therefore $x\in L_n$.
