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A morphism of schemes $f:X\to S$ is said to be quasi-compact if for every OPEN quasi-compact subset $K \subset S$ the subset $f^{-1}(K) \subset X$ is also quasi-compact (and open, of course!). The morphism $f:X\to S$ is said to be universally closed if for every morphism $T\to S$ the resulting base-changed morphism $X_T \to T$ is closed. The title question (inspired by topology) is then:

Question 1: If $f:X\to S$ is universally closed, does it follow that $f$ is quasi-compact?

Here is a variant of this question, asking for a stronger conclusion :

Question 2: If $f:X\to S$ is universally closed, does it follow that for every quasi-compact subset $K\subset S$, open or not, $f^{-1}(K)$ is quasi-compact ?

REMARK 1 The converse of Question 1 is false: any morphism between affine schemes is quasi-compact but is not universally closed in general.

REMARK 2 One might wonder whether $f$ proper implies $f$ quasi-compact. The answer is "yes" but for an irrelevant reason: proper is defined as separated, universally closed and of finite type. Since finite type already implies quasi-compact, proper obviously implies quasi-compact.

REMARK 3 In topology "proper" is (or should be !) defined as universally closed; equivalently, closed with quasi-compact fibres. Topologically proper implies that every quasi-compact subset (open or not) of the codomain has quasi-compact inverse image. The converse is not true in general, but it is for locally compact spaces.

REMARK 4 (edited).As BCnrd remarks in his comment below, it is not at all clear that the two questions are equivalent (I had stated they were in the previous version of this post, but I retract that claim ). Also, beware that in topology the notion of quasi-compact continuous map is so weak as to be essentially useless since decent topological spaces, the ones algebraic geometers never use :) , have so few open quasi-compact subsets.

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The equivalence asserted in Remark 4 is unclear, as you impose no quasi-separatedness hypotheses on $f$ nor any constructibility (or pro-constr.) hypotheses on $K$. For example, no reason $K$ should meet an affine open of $S$ in a quasi-compact subset. Can you please edit Remark 4 to justify the equivalence more fully? –  BCnrd May 3 '10 at 20:50
First of all, thanks for reading my post so attentively. I have edited it in order to take your pertinent comments into account, but unfortunately I can't justify the equivalence I had imprudently asserted. So let's consider now that there are two distinct questions. –  Georges Elencwajg May 3 '10 at 23:56
I don't know a single example where one can verify universal closedness in the absence of actually knowing quasi-compactness already, so even if the answers were affirmative I can't imagine how that would be useful. So although I see the motivation based on the analogy with topology, in another sense the questions lack a certain degree of compelling motivation. –  BCnrd May 4 '10 at 2:14
@BCnrd: as you guessed, the topological case motivated the questions. Also, I like to understand the formal implications between concepts when trying to understand them. So my motivation was certainly not to find a practical criterion for universal closedness. But this is personal and it is quite all right with me if others, like you,feel differently about the interest of those questions. –  Georges Elencwajg May 4 '10 at 7:28
@BCnrd: I am pleasantly surprised that you tried your hand at these questions: it feels great to be in such excellent company! Also your re-interpretation of hilberttheorem90's question is very interesting and quite unexpected for me, due to my lamentable knowledge of constructible topology. And I hope hilbertthm90 will forgive me for reading mail addressed to him ... –  Georges Elencwajg May 4 '10 at 8:07

1 Answer 1

up vote 19 down vote accepted

Yes, a universally closed morphism is quasi-compact. (I haven't yet checked whether the same approach answers question 2.)

Proof: Without loss of generality, we may assume that $S=\operatorname{Spec} A$ for some ring $A$, and that $f$ is surjective. Suppose that $f$ is not quasi-compact. We need to show that $f$ is not universally closed.

Write $X = \bigcup_{i \in I} X_i$ where the $X_i$ are affine open subschemes of $X$. Let $T=\operatorname{Spec} A[\{t_i:i \in I\}]$, where the $t_i$ are distinct indeterminates. Let $T_i=D(t_i) \subseteq T$. Let $Z$ be the closed set $(X \times_S T) - \bigcup_{i \in I} (X_i \times_S T_i)$. It suffices to prove that the image $f_T(Z)$ of $Z$ under $f_T \colon X \times_S T \to T$ is not closed.

There exists a point $\mathfrak{p} \in \operatorname{Spec} A$ such that there is no neighborhood $U$ of $\mathfrak{p}$ in $S$ such that $X_U$ is quasi-compact, since otherwise we could cover $S$ with finitely many such $U$ and prove that $X$ itself was quasi-compact. Fix such $\mathfrak{p}$, and let $k$ be its residue field.

First we check that $f_T(Z_k) \ne T_k$. Let $\tau \in T(k)$ be the point such that $t_i(\tau)=1$ for all $i$. Then $\tau \in T_i$ for all $i$, and the fiber of $Z_k \to T_k$ above $\tau$ is isomorphic to $(X - \bigcup_{i \in I} X_i)_k$, which is empty. Thus $\tau \in T_k - f_T(Z_k)$.

If $f_T(Z)$ were closed in $T$, there would exist a polynomial $g \in A[\{t_i:i \in I\}]$ vanishing on $f_T(Z)$ but not at $\tau$. Since $g(\tau) \ne 0$, some coefficient of $g$ would have nonzero image in $k$, and hence be invertible on some neighborhood $U$ of $\mathfrak{p}$. Let $J$ be the finite set of $j \in I$ such that $t_j$ appears in $g$. Since $X_U$ is not quasi-compact, we may choose a point $x \in X - \bigcup_{j \in J} X_j$ lying above some $u \in U$. Since $g$ has a coefficient that is invertible on $U$, we can find a point $P \in T$ lying above $u$ such that $g(P) \ne 0$ and $t_i(P)=0$ for all $i \notin J$. Then $P \notin T_i$ for each $i \notin J$. A point $z$ of $X \times_S T$ mapping to $x \in X$ and to $P \in T$ then belongs to $Z$. But $g(f_T(z))=g(P) \ne 0$, so this contradicts the fact that $g$ vanishes on $f_T(Z)$.

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Bjorn, nice work. To compensate for my earlier pessimism, here is sol'n to "most" of Question 2: if qc $K$ is pro-constructible in $S$ (i.e., loc. on $S$ an intersection of loc. constr. sets) then it is affirmative. Indeed, WLOG $S$ is qc, so by EGA IV_1, 1.9.9, $K$ is qc in the constr. topology of $S$. By 1.9.15(v) and qc of $f$ (Question 1!), $f$ is proper for the constr. topologies. Hence, $f^{-1}(K)$ is qc in $X$ for constr. topology. But that topology has even more open sets, so $f^{-1}(K)$ is qc for usual topology of $X$. So Question 2 remains open only for extremely bizarre $K$. –  BCnrd May 5 '10 at 4:47
Dear Bjorn, I am very happy (but not surprised !) you so brilliantly solved my problem. My heartfelt thanks for taking the time to do so. –  Georges Elencwajg May 5 '10 at 8:38
Thanks also to BCnrd for coming back to my post. I'll definitely have to learn about constructible topology. –  Georges Elencwajg May 5 '10 at 8:48
Bjorn's argument can be simplified somewhat using the simple (topological) fact that closed + quasi-compact fibers => quasi-compact (this is implicit in his argument). In Bjorn's proof we can thus assume that $S$ is a point and that $X$ is not quasi-compact. Another comment (Cor. 8.4 in arXiv:0904.0227v2): If $f:X->S$ is universally closed and separated, then every fiber of $f$ has finite dimension. That is, $f$ is not only quasi-compact but "almost" of finite type (i.e. "almost" proper). –  David Rydh Jul 16 '10 at 8:21

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