# Locally Noetherian schemes are quasiseparated

I'd like to understand why if $$X$$ is a locally Noetherian scheme, then $$X$$ is quasiseparated. Recall that a scheme is quasiseparated if the intersection of two quasicompact open subsets is quasicompact. This is equivalent to the intersection of any two affine open subsets being a finite union of affine open subsets. Further, locally Noetherian means that $$X$$ can be covered by affine open sets $$\operatorname{Spec}A$$ where $$A$$ is a Noetherian ring.

So suppose that $$U=\operatorname{Spec}A$$ and $$V=\operatorname{Spec}B$$ are two open affine subsets of $$X$$. Further suppose that $$X$$ is covered by $$\{\operatorname{Spec}A_i\}$$, as $$i$$ runs over some index set $$I$$, and where the $$A_i$$ are Noetherian rings. We'd like to show that $$U\cap V$$ can be covered by a finite number of affine open subsets of $$X$$.

Recall that the intersection $$U\cap V=\operatorname{Spec}A\cap\operatorname{Spec}B$$ is a union of open sets that are simulaneously distinguished in both $$\operatorname{Spec}A$$ and $$\operatorname{Spec}B$$. Further, since $$X$$ is covered by $$\{\operatorname{Spec}A_i\}_{i\in I}$$, $$U\cap V$$ is also the union of open sets that are distinguished in some subset of $$\{\operatorname{Spec}A_i\}_{i\in I}$$. But how to show that this cover is finite? Or is there a better way to approach this?

• If $X$ is locally Noetherian, then for any open $\mathrm{Spec}(A) \subset X$ is such that $A$ is a Noetherian ring, so $\mathrm{Spec}(A)$ is a Noetherian topological space, so any open of $\mathrm{Spec}(A)$ is quasi-compact (because of the increasing chain condition for open subsets)
– user598294
Sep 26, 2020 at 18:07
• If you can, see Corollary 3.22 of the book by Görtz & Wedhorn. I am a fan of this book
– user598294
Sep 26, 2020 at 18:10
• @AlexL So then we get that $U\cap V$ is the union of distinguished open sets of each of the $\operatorname{Spec}A_i$'s that intersect $U\cap V$, and that each of these are quasicompact. But I still dont see why there should be a finite cover, since there still might be an infinite number of $\operatorname{Spec}A_i$'s that intersect $U\cap V$ Sep 26, 2020 at 18:33
• It is better if you know that every affine open of a locally Noetherian scheme is the spectrum of a Noetherian ring. That way you are not stuck with just the given covering $\mathrm{Spec}(A_i)$
– user598294
Sep 26, 2020 at 18:52

Let $$U=\mathrm{Spec}(A) \subset X$$ be an open of a locally Noetherian scheme $$X$$. Suppose $$U_i=\mathrm{Spec}(A_i)$$ is an open covering of $$X$$ by spectra of Noetherian rings.

As you said, $$U \cap U_i$$ can be cover by open that are distinguished (or principal) in both $$U$$ and $$U_i$$. In $$U_i$$, a distinguished open is Noetherian, being $$\mathrm{Spec}((A_i)_g)$$ for some $$g \in A_i$$. Thus $$\mathrm{Spec}(A)$$ is cover by (finitely many) distinguished opens $$\mathrm{Spec}(A_f)$$ with $$A_f$$ Noetherian. This implies that $$A$$ is Noetherian (maybe take this as an exercice). But the spectrum of a Noetherian ring is a Noetherian topological space; hence every open of $$\mathrm{Spec}(A)$$ is quasi-compact. In particular for another affine open $$V$$ of $$X$$, $$U \cap V$$ is quasicompact since it is an open of $$U$$

• I hope this is clear enough
– user598294
Sep 26, 2020 at 19:39
• Thanks. But I still don't understand why $\operatorname{Spec}(A)$ is covered by finitely many distinguished opens $\operatorname{Spec}(A_f)$. I understand that each $U\cap U_i$ can be covered by these Noetherian open sets, but I don't understand why there are a finite number of them, let alone a finite number of them covering $\operatorname{Spec}(A)$ . Sep 27, 2020 at 12:59
• So I know that every open subset of a Noetherian space is qc (and so all the $U\cap U_i$ have a finite subcover of distinguished open sets), but there still might be an infinite number of the $U\cap U_i$. Sep 27, 2020 at 13:16
• An affine scheme is always quasi-compact
– user598294
Sep 27, 2020 at 15:14
• of course thanks. Sep 27, 2020 at 15:33