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Let $f : X \to Y$ be an open faithfully flat morphism of schemes. In the text I'm reading (Angelo Vistoli's notes on descent) it is claimed that then every point $x \in X$ admits an open neighorhood $U$ such that $f(U)$ is open and the morphism $U \to f(U)$ is quasi-compact. The latter is one of the possible definitons of a fpqc morphism. However, I don't understand at all why $U \to f(U)$ should be quasi-compact.

This is needed to prove that the fpqc topology is finer than the fppf topology.

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  • $\begingroup$ Well, if you take for $U$ any open affine neighbourhood of $x$ then $f(U)$ is open because $f$ is, and $U\to f(U)$ is quasi-compact because $U$ is... $\endgroup$
    – Matthieu Romagny
    Commented Sep 18, 2010 at 11:46
  • $\begingroup$ I don't understand. There are morphisms $f : X \to Y$, which are not quasi-compact, although $X$ is affine and $Y$ is quasi-compact. For example, take some affine $X$ which has an open subset $U$ which is not quasi-compact, and glue two copies of $X$ along $U$ to get $Y$. Then the, say first, inclusion $X \to Y$ is not quasi-compact. $\endgroup$
    – Martin Brandenburg
    Commented Sep 18, 2010 at 12:03

1 Answer 1

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[Edit] (add some details).

Replacing $Y$ with an affine open neighborhood $V$ of $f(x)$ (and $X$ with $f^{-1}(V)$), one can suppose that $Y$ is affine. Cover $X$ by affine open subsets {$U_i$}$_i$. As $Y$ is quasi-compact, a finite number of the $f(U_i)$ cover $Y$. If necessarily, we can add one more $U_i$ so $x$ belong to one of these $U_i$'s. The union $U$ of these (finitely many) $U_i$ is quasi-compact, and we have $f(U)=Y$, $x\in U$. The morphism $f|_U : U\to Y$ is a morphism from a quasi-compact scheme to an affine scheme, so it is quasi-compact because for any affine open subset $V$ of $Y$, $(f|_U)^{-1}(V)\cap U_i= V\times_Y U_i$ is affine.

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  • $\begingroup$ hopefully this answers 1,2; though it's mostly just a re-hash of Liu's answer 1.if $x \not \in U$ then as Liu says just add pick a open affine $U_0$ containing $x$ and replace $U$ with $U \cup U_0$. 2.Any q-compact open $V$ in $Y$ is covered by a finite number of affine opens $V_i$ so $f|_U^{-1}(V)$is covered by the finitely many $f|_U^{-1}(V_i)$ so it suffices to show $f|_U^{-1}(V_i)$ is q-compact; Liu shows $f|_U^{-1}(V_i)$ is covered by a finite number of affines hence quasi-compact. $\endgroup$
    – solbap
    Commented Sep 18, 2010 at 20:54

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