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Let $X$ be a finite-dimensional smooth manifold, $\mathcal C^\infty(X)$ its algebra of smooth functions, $V\to X$ a finite-dimensional smooth vector bundle, and $\Gamma(V)$ the space of smooth sections of $V$. In particular, $\Gamma(V)$ is a $\mathcal C^\infty(X)$-module. I am interested in $\mathcal C^\infty(X)$-submodules $D \subseteq \Gamma(V)$.

Is $D$ necessarily finitely-generated as a $\mathcal C^\infty(X)$-module?

If $X$ is not compact (or maybe even if it is?), then $\mathcal C^\infty(X)$ is not Noetherian. So it is not true that submodules of arbitrary finitely-generated modules are finitely generated. So I expect that the answer to my question is "no", but I'm having trouble coming up with a counterexample.

Actually, what I really want is for $D$ to receive a ($\mathcal C^\infty$-linear) surjection from $\Gamma(W)$ for some finite-dimensional vector bundle $W$. If $X$ is not compact, then I think it is still the case (using partitions of unity) that $\Gamma(W)$ is globally finitely-generated (the idea is to find a cover for which each open intersects only finitely many others in the cover, and then to double up the generators). But if it isn't, the actual question I want to ask is the one with the word "locally" sprinkled in all the necessary places.

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You wrote "sheaf" in the title but the question seems to be concerned with spaces of global sections. – Tom Goodwillie Oct 12 '10 at 5:06
I think that your counterexample works in either case. – Martin Brandenburg Oct 12 '10 at 8:23
up vote 11 down vote accepted

The module of all sections of $V$ that vanish to infinite order at a given point of the manifold will not be finitely generated (unless the bundle has rank zero or the manifold has dimension zero).

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Great. Is it clear how to see this? (Maybe it should be, since it's closely connected to the examples you gave in my previous question.) – Theo Johnson-Freyd Oct 12 '10 at 19:24
Actually I just asserted it without thinking of a proof (but I'm sure it's true and not hard). I'll answer with a question: you say that the ring of smooth functions is not noetherian. If you know that then you know that there are non finitely generated ideals. – Tom Goodwillie Oct 12 '10 at 19:38
So, I'm just thinking out loud. Let $X=\mathbb R$; then the standard example of an ascending chain of ideals is $I_n=\{f\text{ s.t. }f(x)=0\forall x\geq n\}$ for $n\in\mathbb N$. But it's certainly not obvious to me how to turn this particular ascending chain into a particular non-finitely-generated ideal. I mean, presumably each $I_n$ is individually not finitely generated, as its elements vanish to infinite order at $n$. But proving directly that $I_n$ is not f.g. seems no easier nor harder than proving it for your example. – Theo Johnson-Freyd Oct 13 '10 at 7:47
In any case, looking again, I see that Wikipedia does list "All (left) ideals are finitely generated" as a characterization for Noetherianness. Here's what I get for never really learning commutative algebra. C'est la vie --- I'm still young, and have plenty of time to learn. – Theo Johnson-Freyd Oct 13 '10 at 7:51
The union of an infinite strictly ascending chain of submodules can never be finitely generated. – Tom Goodwillie Oct 13 '10 at 10:38

One simple counterexample is to let X be the real line, let V be the trivial line bundle, and consider the submodule of \Gamma(V) of smooth sections with compact support. The same story works for any vector bundle on any non-compact manifold.

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