0
$\begingroup$

Is it true that any proper morphism of relative dimension$\leq 1$ is projective (no additional assumptions whatsoever)? Is it true that any such morphism is $H$-projective (https://stacks.math.columbia.edu/tag/01W8)?

Is there a published reference containing a complete proof?

$\endgroup$
  • 4
    $\begingroup$ Welcome new contributor. That is false, and Piotr Achinger explained the issue in a recent comment. Begin with a projective K3 surface that admits an elliptic fibration over $\mathbb{P}^1$. For a very general first order deformation of the surface and its fibration to $\mathbb{P}^1$, the surface is not projective. $\endgroup$ – Jason Starr Jul 11 at 10:40
  • $\begingroup$ @JasonStarr does the fibration deform to the non-projective surface? which complex algebraic K3s admit elliptic fibration? $\endgroup$ – user142965 Jul 11 at 11:48
  • $\begingroup$ Welcome new contributor. The scheme is proper and smooth of relative dimension $2$ over $S=\text{Spec}\ \mathbb{C}[\epsilon]/\langle \epsilon^2 \rangle$, and it admits a proper, flat morphism of relative dimension $1$ to $\mathbb{P}^1_S$. The scheme is not projective over $\text{Spec}\ \mathbb{C}$. $\endgroup$ – Jason Starr Jul 11 at 12:45
  • $\begingroup$ @JasonStarr are there counterexamples if the base is irreducible and admits a morphism to $\mathrm{Spec}\:\mathbb{F}_p$? $\endgroup$ – user142965 Jul 12 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy