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?



2 Answers 2


This follows immediately from the Five Lemma applied to $$\begin{array}{c} 0 & \longrightarrow & \alpha^{-1}(L_n) & \longrightarrow & L_n & \longrightarrow & \beta(L_n) & \longrightarrow & 0 \\ & & \downarrow && \downarrow && \downarrow & \\ 0 & \longrightarrow & \alpha^{-1}(L_{n+1}) & \longrightarrow & L_{n+1} & \longrightarrow & \beta(L_{n+1}) & \longrightarrow & 0.\end{array}$$

  • 1
    $\begingroup$ Yeah, you are right. $\endgroup$
    – user146507
    Nov 30, 2014 at 22:03

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 $x\in L_n$.

  • $\begingroup$ Interestingly enough, there seems to be only one solution to this problem! $\endgroup$
    – TidSchmod
    Apr 7, 2019 at 21:42

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