$M$ Noetherian $\implies $ $M'$ and $M''$ are noetherian.

Question

Let $0\to M'\overset{\alpha }{\to} M\overset{\beta}{\to} M''\to 0$ an exact sequence. I want to show that if $M$ Noetherian, then $M'$ and $M''$ are also Noetherian.

Let $\{M_i'\}$ a chain of $M'$. Since $M_i'\subset \alpha ^{-1}(\alpha (M_i'))$ and that $\alpha (M_i)'$ is stationnary (because $M$ Noetherian), we have the result.

For $M''$ I have problem to conclude since if $\{M_i''\}$ is a chain of $M''$ I can only get $\beta (\beta ^{-1}(M_i''))\subset M_i''$. Any hint ?

Answer 1

The sequence is exact, therefore $\beta $ is surjective, and thus you have the equality $$\beta (\beta ^{-1}(M_i''))=M_i''.$$ Notice that $\alpha $ is injective, and thus you also have the equality $$\alpha ^{-1}(\alpha (M_i'))=M_i'.$$

Answer 2

For any submodule $N \subset M''$, $\beta^{-1}(N)$ is a submodule of the noetherian $M$, in particular finitely generated. Use the surjectivity of $\beta$ to show that $N$ must be finitely generated as well.