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I've been reading Vakil's notes on algebraic geometry (on my own -- this is not part of a class), and I'm stuck on one problem (number 6.1.H). It goes as follows.

Let $X$ be a scheme. Prove that $X$ is quasicompact and quasiseparated if and only if $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.

It's not hard to show one direction, namely that if $X$ is quasicompact and quasiseparated then it has a cover of the indicated form. It's also not hard to prove that if $X$ has a cover of the indicated form, then $X$ is quasicompact. I'm having difficulty with the "quasiseparated" part.

Thank you very much for any help.

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    I wonder when algebraic geometers are going to get on with the program and drop the quasis... :)2010-11-11
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    I think the key think to keep in mind is that the intersection $U \cap V$ for any is the pre-image of $U \times V$ in the diagonal $X \to X \times X$. (Here "pre-image" can also be interpreted in the categorical sense, i.e. via pull-backs; it's also the usual set theory as these are open inclusions). So if the diagonal is compact, then the pull-back of $U \times V$ for $U,V$ q.c. (q.c. being preserved by products) is q.c. as well.2010-11-11

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first we try to solve the easiest case :

suppose $X=U_1 \cup U_2$ and we know that $U_1$ and $U_2$ are both affine and their intersection is the union of finite affine open sets like $U_1 \cap U_2 = V_1 \cup V_2 \cup ... \cup V_n$ .

now suppose that specA is affine in X. i'm going to prove that specA $\cap$ U1 is the union of finite affine open sets.

for each point $p \in specA \cap U_1$ consider a distinguished open set inside $specA \cap U_1$ of specA. and for each $p \in specA-(specA \cap U_1)$ consider again a distinguished open set of specA inside $specA \cap U_2$. now all these distinguished open set provide a covering of specA and as the latter is quasicompact therefore we get a cover of specA by some $D(f_1) \cup ... \cup D(f_k)=specA$ where the $D(f_i)$'s are the distinguished open sets. now if $specA \cap U_1$ is covered by those distinguished open sets which were inside $specA \cap U_1$ then we'll be done. otherwise we have that some of the $D(f_i)$ like for example $D(f_1)$ is one of those distinguished which were in $specA \cap U_2$. well in this case consider the intersections $D(f_1) \cap V_j$ which is a subset of $(specA \cap U_2) \cap U_1$ and as $U_2$ is affine and therefore quasiseperated so $D(f_1) \cap V_j$ is the union of finite open affine set(which will be in $specA \cap U_1$ and also note that as $U_1 \cap U_2 = V_1 \cup V_2 \cup ... \cup V_n$ we get $\cup ($D(f_1) \cap V_j$) = D(f_1) \cap (U_2 \cap U_1) = D(f_1) \cap U_1$). Do the same now for every $D(f_i)$ that u find in $specA \cap U_2$ and at last u will get a finite affine open cover of $specA \cap U_1$.

now u can easily see that if $specB$ and $specA$ are affine open subsets of X then their intersection is a finite union of affine open sets( first intersect specA with U1 and U2 and apply whay we proved and do the same for the intersections of specB with U1 and U2 and use the fact that U1 and U2 are quasiseperated as they are affine and ...).

as I'm very lazy and i won't write it down, u can see yourself that we can easily do the same argument for the general case where X=$U_1 \cup ... \cup U_n$. just follow exactly the same path. first prove that specA $\cap U_i$ is a finite union of affine open sets and then do the same reasoning in the easiest case for proving that $specB$ and $specA$ are affine open subsets of X then their intersection is a finite union of affine open sets.

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    Sloppiness aside, this answer gives the correct strategy to prove the claim with the tools available at this point in the book.2015-02-25
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I also wanted a topological proof, since it's nice to know what the proofs are in each language and knowing what the proof is when translated in another language (e.g. quasi-separatedness via diagonal morphisms) usually tells something a different way, which is not quite as enlightening. I cleaned up moji's proof (because I am not lazy ;)!)

A scheme $X$ is qcqs (short for quasi-compact and quasi-separated) if and only if there exists a finite open affine cover $\{U_1,\cdots,U_n\}$ such that each intersection $U_i \cap U_j$ admits a finite open affine cover $\{V_{ij1},\cdots,V_{ijk_{ij}}\}$ (where $k_{ij} \in \mathbb N$ depends on $i$ and $j$).

Proof : ($\Rightarrow$) Pick a finite open affine cover $\{U_1,\cdots,U_n\}$ of $X$ by quasi-compactness. Affine schemes are qcqs, so the intersections $U_i \cap U_j$ are quasi-compact and therefore admit a finite open affine cover $\{V_{ij1},\cdots,V_{ijk_{ij}}\}$.

($\Leftarrow$) Let $U \subseteq X$ be a quasi-compact open subset. We claim that for each $\alpha=1,\cdots,n$, $U \cap U_{\alpha}$ is quasi-compact. It suffices to deal with the case of $\alpha=1$. Because $U$ is a scheme, its topology admits a basis consisting of quasi-compact open neighborhoods (take a finite open affine cover and the basis of distinguished open subsets of each of those affines). Write $$ U = \bigcup_{j=1}^n U \cap U_j = \bigcup_{j=1}^n \bigcup_{\ell \in L_j} W_{j\ell} $$ where $W_{j\ell} \subseteq U \cap U_j$ is a quasi-compact open subset. Since $U$ is quasi-compact, choose finite subsets $M_1 \subseteq L_1, \cdots, M_n \subseteq L_n$ such that the above equality still holds. Intersecting this with $U_1$, we get $$ U \cap U_1 = \bigcup_{j=1}^n \bigcup_{\ell \in M_j} W_{j\ell} \cap U_1. $$ Pick $j > 1$ and $\ell \in M_j$, so that for any $1 \le k \le k_{1j}$, the open subsets $V_{1jk}, W_{j\ell} \subseteq U_j$ are quasi-compact. Because $U_j$ is quasi-separated, $V_{1jk} \cap W_{j\ell}$ is quasi-compact. This means that $$ U \cap U_1 = \bigcup_{j=1}^n \bigcup_{\ell \in M_j} W_{j\ell} \cap U_1 \overset{(!)}= \bigcup_{j=1}^n \bigcup_{\ell \in M_j} W_{j\ell} \cap U_1 \cap U_j = \bigcup_{j=1}^n \bigcup_{\ell \in M_j} \bigcup_{k=1}^{k_{1j}} W_{j\ell} \cap V_{1jk} $$ is quasi-compact. (The $(!)$ is because $W_{j\ell} \subseteq U_j$ for each $j$. This seemed to be the cause of many incorrect edits to my proof.)

With this lemma in hand, if $U, U' \subseteq X$ are quasi-compact, then for $i=1,\cdots,n$, we see that $U \cap U' \cap U_i = (U \cap U_i) \cap (U' \cap U_i)$ is quasi-compact by the quasi-separatedness of $U_i$ and the quasi-compactness of $U \cap U_i$ and $U' \cap U_i$, so $X$ is quasi-separated.

Hope that helps,

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    Guys, I know the indices in this proof are somewhat tricky, but stop making edits unless you've written it down on paper completely at least once. I must have reverted at least 3 incorrect edits so far and it's kind of annoying.2018-02-13
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Hint: Starting from your open affine cover of $X$, construct an open affine cover of $X \times X$, and then show that the preimage of each member of that open affine cover is a union of finitely many affines.

Another hint, more directly relevant to Vakil's definition of quasi-separated: I am implicitly cheating somewhat, because I am still secretely thinking about quasi-separated as meaning that the diagonal map is quasi-compact. If I do that, then the exercise at hand translates into the following: show that if a morphism $f$ has the property that the target has an open affine cover, the preimage of each member of which is quasi-compact, then $f$ is quasi-compact (i.e. each quasi-compact in the target has quasi-compact preimage).

So you should certainly remind yourself how to prove this fact, because the tools used there will be the same tools you need in your present exercise. (Even though you may not be working explicitly with the formulation in terms of the diagonal morphism, the underlying argument will have to be essentially the same.)

E.g. first convince yourself that you only have to check that the intersection of any two open affines is quasi-compact.

Also, an important point in all these kinds of arguments: if $f: X \to Y$ with $X$ and $Y$ affine, say $X =$ Spec $B$ and $Y =$ Spec $A$, and $a \in A$, so that Spec $A_a$ is a distiguished open, then the inverse image of Spec $A_a$ is a distinguished open affine in $X$ (it is Spec $B_a$). This a fundamental tool: it is the one way that you have to construct arbitrarily small open affine neighbourhoods in $Y$ whose preimages in $X$ are again affine.

So you will want to use the idea of the preceding paragraph, but now "inverse image" will be replaced by "intersection". (When you take inverse images of product neighbourhoods along the diagonal map, you are computing intersections, so this replacement makes sense; but again, you won't have to think about the diagonal map explicitly if you dont' want to.)

That's probably enough of a rambling hint for now. If you have those ideas in mind and put everything you know together, you hopefully can find your way to a complete proof.

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    It looks like you are using the definition of quasiseparated found in wikipedia (ie a scheme is quasiseparated if the diagonal morphism is quasicompact). However, that is not how Vakil defines it at this point in the notes (the table of contents promises that 11.1 will have "quasiseparatedness done properly", but that isn't posted yet). Instead, Vakil says that a scheme is quasicompact if the intersection of any two quasicompact open subsets of it is quasicompact. Is there a nice way to translate your suggestion into that language?2010-11-11
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    @T_P: Dear T_P, Ah, I didn't check Vakil's definition, and you are right, I was using the definition in terms of diagonals. I'll try to add another hint that is more relevant to the definition Vakil uses.2010-11-11
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    Thank you for your help so far (and all the help you have given newbies to algebraic geometry like me). I look forward to your promised hint.2010-11-11
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    I am stuck on the same problem and came here. But I cannot follow this hint. I understood that we only have to check that the intersection of any two open affines is quasi-compact. At the problem 6.1.H in Vakil's note, we do not know closed schemes nor products of scheme. It would be appreciated if somebody writes an answer readable by beginner at that level.2012-11-25
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    @Tom: Dear Tom, The second hint, beginning in the second paragraph with "Another hint, ... ", does not use products or closed subschemes. Rather, it points out that the problem is similar to the following one: "If $f: X \to Y$ is a morphism s.t. $Y$ has a cover by open affines such that $f^{-1}$ of each member of the cover is quasicompact, then $f$ is quasicompact, i.e. $f^{-1}$ of *any* open affine is quasicompact." In your case, you are given a particular cover by finitely many open affines with quasi-compact intersections, and you have to show that ...2012-11-25
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    ... the intersection of *any* two open affines is quasicompact. Hopefully you see the analogy: in both problems, you have to pass from a particular cover by open affine with some additional property, to deducing that same property for all open affines. As I wrote in the answer, you should remined yourself how to prove this kind of fact (which I think Vakil has treated at this point in his text). The rest of the answer is a discussion of some of the main techniques involved in these sorts of arguments; if you already are familiar with them, you can ignore it. In any even, if you apply ...2012-11-25
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    ... these techniques you should be able to solve your problem. Regards,2012-11-25
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    @Matt E Thank you for explaining. But I haven’t yet got a clear picture from your hint, though it might be a good hint for people who once read Hartshorne. In fact, I haven’t yet understood even the definition of morphisms between schemes(ringed spaces, too), which will be given 27 pages later in his notes. As far as he put this question as an exercise in that page, I think Vakil has a not-so-difficult solution which can be written clearly only with words define between page 1 and 1422012-11-26
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    @Tom: Dear Tom, Gregor Bruns suggested in his answer to your recent question that you look ahead to section 6.3, and use some of the techniques introduced there. My hint is essentially the same; what Vakil calls the "affine communication lemma" (or more precisely its proof) and related ideas are what I had in mind. Regards,2012-11-26
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    @Matt E: Thank you for the kind answer!2014-06-19