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](http://homepage.sns.it/vistoli/descent.pdf)) 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.



open faithfully flat morphisms are fpqc
[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. 


