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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?

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    $\begingroup$ 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) $\endgroup$
    – user598294
    Sep 26, 2020 at 18:07
  • $\begingroup$ If you can, see Corollary 3.22 of the book by Görtz & Wedhorn. I am a fan of this book $\endgroup$
    – user598294
    Sep 26, 2020 at 18:10
  • $\begingroup$ @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 $U\cap V$ $\endgroup$
    – ponchan
    Sep 26, 2020 at 18:33
  • $\begingroup$ 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 $\mathrm{Spec}(A_i)$ $\endgroup$
    – user598294
    Sep 26, 2020 at 18:52

1 Answer 1

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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 $U$

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  • $\begingroup$ I hope this is clear enough $\endgroup$
    – user598294
    Sep 26, 2020 at 19:39
  • $\begingroup$ 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 $\operatorname{Spec}(A)$ . $\endgroup$
    – ponchan
    Sep 27, 2020 at 12:59
  • $\begingroup$ 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 $U\cap U_i$. $\endgroup$
    – ponchan
    Sep 27, 2020 at 13:16
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    $\begingroup$ An affine scheme is always quasi-compact $\endgroup$
    – user598294
    Sep 27, 2020 at 15:14
  • $\begingroup$ of course thanks. $\endgroup$
    – ponchan
    Sep 27, 2020 at 15:33

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