Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 4 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 $M$.

share|improve this answer
    
Thanks a lot. What's your intuition behind your first sentence? –  Stefan Walter 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 at 7:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.