# Examples of non-projective morphisms with projective fibres

Let $X\to S$ be a morphism of noetherian schemes such that, for all $s$ in $S$, the morphism $X_s\to$ Spec $k(s)$ is projective.

Then it doesn't follow that $X\to S$ is projective in general. In fact, if $X\to S$ is an open immersion then every fibre $X_s\to$ Spec $k(s)$ is projective (some of the fibres are presumable empty, and I am using that $\emptyset \to S$ is a projective morphism). So we at least need something like surjectivity.

Thus, assume $X\to S$ is surjective. Then, still it doesn't follow that $X\to S$ is proper as one can find non-projective proper threefolds $X$ over $\mathbf C$ that fibre over the projective line $S=\mathbf P^1_{\mathbf C}$ whose singularities on the fibres aren't too bad. This morphism $X\to S$ has projective fibres, but $X$ is not projective. Thus $X\to S$ is not projective.

I'm wondering what one can say about flat surjective morphisms $X\to S$ whose fibres are projective (resp. proper). Under which conditions is the morphism $X\to S$ projective (resp. proper)?

• Tautologically, existence of a relatively ample line bundle. What else do you expect? – abx Dec 8 '13 at 13:04
• @abx: the case of an open immersion is an obvious counterexample. – Laurent Moret-Bailly Dec 9 '13 at 8:22
• @Kuipers: if $U\subset X$ is open (but not closed) then $U\coprod X\to X$ is surjective with projective fibers, but not proper. – Laurent Moret-Bailly Dec 9 '13 at 8:25

Let $$f : X\to Y$$ be a separated morphism of finite type, with $$Y$$ Noetherian. Suppose $$f$$ is faithfully flat and its fibers are geometrically connected and proper. Then $$f$$ is proper. See EGA IV.15.7.10 and 15.7.8. If you do not require fibers to be connected, counterexamples are not hard to find.