# Why are the noetherian objects in a category of quasicoherent sheaves just the coherent ones?

The question says it all really. Let $$X$$ be a noetherian scheme. Let $$\mathcal{A}$$ be the category of quasicoherent sheaves on $$X$$. I want to show that an object $$\mathcal{F}$$ in $$\mathcal{A}$$ is noetherian if and only if it is coherent. By noetherian here I mean that it satisfies the ascending chain condition on subsheaves.

The direction coherent $$\Rightarrow$$ noetherian is easy. You just check that any ascending chain becomes constant on stalks by reducing the problem to one of commutative algebra.

My problem is with the other direction. Suppose $$\mathcal{F}$$ is noetherian. Let $$U = \text{Spec }A$$ be a affine open set. I want to show that $$\mathcal{F}(U)$$ is a finitely generated $$A$$-module, or equivalently that it is a noetherian $$A$$-module. My problem is that it's not immediate that just because an ascending chain much become constant globally that it must also become constant on $$U$$. I suspect this has something to do with the fact that you can extend coherent sheaves, but can anyone spell out how I'd prove this?

• Use the equivalence of categories between quasi coherent sheaves on U and A-modules. Use the fact that subsheaves of $\mathcal{F}$ will correspond to submodules of $\mathcal{F}(U)$ – Samir Canning Oct 23 '18 at 1:02
• This is how I proved the forward direction, but it's not clear to me how that proves the reverse direction as well – Luke Oct 23 '18 at 1:26

## 1 Answer

It suffices to show that if $$\mathcal{F}$$ is a noetherian quasicoherent sheaf on $$X$$ then $$\mathcal{F}|_U$$ is a noetherian quasicoherent sheaf on $$U$$, since the equivalence of categories between quasicoherent sheaves on $$U$$ and $$A$$-modules then gives that $$\mathcal{F}(U)$$ is a noetherian $$A$$-module.

So, we prove that $$\mathcal{F}|_U$$ is noetherian in the category of quasicoherent sheaves on $$U$$. Let $$i:U\to X$$ be the inclusion morphism. There is a canonical morphism $$\alpha:\mathcal{F}\to i_*(\mathcal{F}|_U)$$ which is an isomorphism when restricted to $$U$$. Given a quasicoherent subsheaf $$\mathcal{G}\subseteq\mathcal{F}|_U$$, we can consider the quasicoherent subsheaf $$\alpha^{-1}(i_*\mathcal{G})\subseteq\mathcal{F}$$. It is clear that the correspondence $$\mathcal{G}\mapsto \alpha^{-1}(i_*\mathcal{G})$$ preserves the inclusion relation between subsheaves of $$\mathcal{F}|_U$$. Moreover, it preserves strict inclusions, since $$\mathcal{G}$$ can be recovered as the restriction of $$\alpha^{-1}(i_*\mathcal{G})$$ to $$U$$ (since $$\alpha$$ is an isomorphism on $$U$$). Thus any strictly ascending chain of quasicoherent subsheaves of $$\mathcal{F}|_U$$ can be turned into a strictly ascending chain of quasicoherent subsheaves of $$\mathcal{F}$$ by this construction. Since $$\mathcal{F}$$ is noetherian, no such chain can exist, so $$\mathcal{F}|_U$$ is noetherian.

• Thank you, I like this answer a lot. Since asking I came up with a roundabout way of solving it by showing that on a noetherian scheme, the quasicoherent sheaves can be realised as direct limits of their coherent subsheaves, and then using the fact that noetherian objects are compact to show that F must be one of these subsheaves. But that felt like overkill. This is much nicer. – Luke Oct 23 '18 at 2:58