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Let $M$ be a module and $N$ a submodule of $M$. If $N$ is Noetherian and $M/N$ is Noetherian, so is $M$.

This is usually proven like this: Let $(A_n)$ be an ascending series of submodules of $M$. Then $(A_n\cap N)$ and $((A_n+N)/N)$ are ascending series of submodules of $N$ and $M/N$ respectively, so that they are eventually constant. A little elementary argument shows that, for $A\subset B$ submodules of $M$, $A\cap N=B\cap N$ and $(A+N)/N=(B+N)/N$ together imply $A=B$. So the series $(A_n)$ is eventually constant.

Now suppose we have defined Noetherian as "all submodules are finitely generated" and don't know about the equivalence to the ascending chain condition. There should be a proof of the above statement without the detour via the ascending chain condition, i.e. something like "Let $A$ be a submodule of $M$, $S$ a finite generating set of $A\cap N$ and $T$ a finite generating set of $(A+N)/N$. Then ... is a finite generating set of $A$." I have yet been unable to find such a proof. Can somebody sketch one or link to one?

In case it makes a difference: I am mainly interested in $\mathbf{Z}$-modules.

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up vote 5 down vote accepted

In general, I don't see why any (reasonable) proof shouldn't require a detour through the ACC. In this case, though, one can do so.

Lemma: Let $0 {\to}^{f} M' {\to}^{g} M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $M$ is Noetherian iff $M',M''$ are Noetherian.

Proof: Suppose $M$ is Noetherian. Then $M'$ is Noetherian, as submodules of $M'$ can be identified with submodules of $M$ by the injectivity of $f$, and hence are finitely generated. Similarly, $M''$ is Noetherian, as submodules of $M''$ can be pulled back to submodules of $M$ by the surjectivity of $g$, and since the pull-back is finitely generated, the original submodule is generated by the image of the pull-back generators.

Conversely, suppose $M',M''$ are Noetherian. Let $N$ be a submodule of $M$. It is easily seen that $0 \to N \cap M' \to N \to g(N) \to 0$ is a short exact sequence, and since $N \cap M' \subset M'$, $g(n) \subset M''$, they are both Noetherian. Thus, we obtain the original problem we were trying to show, so it suffices to show $M$ above is finitely generated. Choose generators $m_{1}', \cdots, m_{s}'$ of $M'$ and $m_{1}'', \cdots, m_{r}''$ of $M''$. Choose $m_{1}, \cdots, m_{r}$ such that $g(m_{i}) = m_{i}''$. By the injectivity of $f$, we will assume $M' \subset M$.

We claim that $M$ is generated by $m_{1}, \cdots, m_{r}, m_{1}', \cdots, m_{s}'$.Let $m \in M$. Then $g(m) = a_{1}m_{1}'' + \cdots + a_{r}m_{r}''$. Let $\tilde{m} = a_{1}m_{1} + \cdots a_{r}m_{r}$ (pull back of $g(m)$). Then $g(m - \tilde{m}) = 0$, so $m - \tilde{m} \in \ker(g)$, and by the exactness of the sequence, $m-\tilde{m} \in M'$ (again, remember that we are thinking of $M'$ as being inside $M$). Then write

$m - \tilde{m} = b_{1}m_{1}' + \cdots + b_{s}m_{s}'$ and hence

$m = b_{1}m_{1}' + \cdots + b_{s}m_{s}' + a_{1}m_{1} + \cdots + a_{r}m_{r}$

so $M$ is finitely generated, and by the above observation, any submodule $N$ of $M$ is finitely generated, so that $M$ is Noetherian. $\square$

Corollary: Let $N$ be a submodule of $M$. The sequence $0 \to N \to M \to M/N$ is short exact, and so if $N$ and $M/N$ are Noetherian, so is $M$.

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Thanks a lot. What's your intuition behind your first sentence? – Stefan Mar 11 '12 at 10:36
No specific intuition. It's just that the crux of an argument can rely on a particular equivalent form of a property, and so trying to prove something without use ascending chains of ideals, for example, might make things unnecessarily difficult. – Isaac Solomon Mar 11 '12 at 11:19
This is a really nice proof, reminds me of the Horseshoe lemma. – krey Jan 4 '14 at 7:35

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