# Noetherian assumption for coherent sheaves — equivalence of categories

Using Hartshorne's definition of coherence, we should still have that an $\mathcal{O}_X$-module $\mathscr{F}$ is coherent if and only if for every affine open $U=\operatorname{Spec}(A)$ of $X$, there exists a finitely generated $A$-module $M$ such that $\mathscr{F}|_U=\widetilde{M}$, $\textbf{even for$X$not noetherian}$, because if $(f_1,\ldots,f_r)=A$ and each $M_{f_i}$ is a finite $A_{f_i}$-module, then $M$ is a finite $A$-module even if $A$ is not noetherian (see for example Stacks Project).

But then why does the equivalence of categories between finitely generated $A$-modules and coherent $\mathcal{O}_{\operatorname{Spec}(A)}$-modules fail without the noetherian assumption? The functor $M\mapsto \widetilde{M}$ is fully faithful and essentially surjective.

## 1 Answer

The stacks project tag you link to is irrelevant to this question. Of course if you simply take the Hartshorne definition of a coherent sheaf even in the non-Noetherian case, you will get the equivalence with finitely generated $A$-modules (by definition).

The main point is that if $A$ is not noetherian, then the subcategory of $QCoh(Spec(A))$ (the category of quasi-coherent sheaves) given by the sheaves $\tilde{M}$, where $M$ is finitely generated, is not abelian. The problem is that kernels will not be finitely generated.

Once you know this you understand that for the correct definition of a coherent sheaf you need to fix this, which leads you to the definition that you find on Wikipedia for example. Note that this definition might exclude $\tilde{A}$ as a coherent sheaf.

Edit: A ring $A$ for which the structure sheaf is a coherent sheaf on $Spec(A)$ is called a coherent ring. See here.

There are very important rings that are not noetherian but coherent. For example the ring of holomorphic functions on a Stein manifold, which leads to Oka's theorem saying that the structure sheaf of a complex manifold is coherent (note that this is a very hard theorem).

Moreover I think that the ring of smooth functions on a real manifold is not coherent. Unfortunately I cannot find a reference. If anyone finds a reference (or a proof) for this, I would be happy to see it.

• I don't think that $\mathcal{O}_{\mathrm{Spec}(A)}=\tilde{A}$ is excluded as a coherent sheaf, regardless which definition you take. This sheaf has global and local trivial finite presentation. – user213008 Jan 4 '18 at 12:35
• For $\tilde{A}$ to be coherent you also need that finitely generated ideals of $A$ are finitely presented. This is not always true. Of course in my answer one gets the impression that $\tilde{A}$ is always excluded in the non-noetherian case. This is not true, and I will edit my answer – Rieux Jan 4 '18 at 13:11
• Thanks, you are right. – user213008 Jan 4 '18 at 13:53