## Is a submodule of the sheaf of sections of a smooth vector bundle necessarily finitely generated?

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 2010 at 5:06
I think that your counterexample works in either case. – Martin Brandenburg Oct 12 2010 at 8:23

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).
 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 2010 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 2010 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 2010 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 2010 at 7:51 The union of an infinite strictly ascending chain of submodules can never be finitely generated. – Tom Goodwillie Oct 13 2010 at 10:38