# Faithfully flat morphisms with all fibers complete

Prove or disprove: if $f: X \to Y$ is faithfully flat and each fiber is complete, then $f$ is proper. (I'd especially like to see a counterexample with a morphism of finite type between varieties over algebraically closed fields if one exists.)

(Motivation: in the category of smooth manifolds, it is certainly true that a fiber bundle in which the fibers are compact is proper.)

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In general $f$ needs not be proper.

Consider the finite cover $g: Z=\mathbb A^1\to Y=\mathbb A^1$, $t\mapsto t^2$ (or $t\mapsto t^2-t$ in characteristic $2$), let $X=Z\setminus \{ t=1\}$. Then the restriction $f$ of $g$ to $X$ is faithfully flat, the fibers are proper (because finite), but $f$ is not finite (the corresponding morphism of ring of regular functions is not finite), so it can't be proper (quasi-finite and proper implies finite).

Edit The usual counterexamples are with disconnected fibers (as the above one). But this is essentially the only obstruction:

Let $f : X\to Y$ be a separated morphism of finite type of noetherian schemes. Suppose $f$ is faithfully flat and its fibers are geometrically connected and proper. Then $f$ is proper.

See EGA IV.15.7.10. That $f$ is universally submersive is noted in 15.7.8. In the case of algebraic varieties over an algebraically closed field, the condition is slightly simpler :

Let $f : X\to Y$ be a faithfully flat morphism of algebraic varieties (i.e. scheme of finite type) over an algebraically closed field. Suppose $X$ is separated (e.g. quasi-projective) and that for all closed points $y\in Y$, $f^{-1}(y)$ is connected and proper. Then $f$ is proper.

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Very nice, as usual. You can see what is going on by projecting a conic in the plane from a point outside the conic onto a line not passing through the point.The example is obtained by deleting a point of the conic. – Georges Elencwajg Jan 26 '13 at 8:12