# Proper curve over any base is projective?

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?

• 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. – Jason Starr Jul 11 at 10:40
• @JasonStarr does the fibration deform to the non-projective surface? which complex algebraic K3s admit elliptic fibration? – user142965 Jul 11 at 11:48
• 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 $\text{Spec}\ \mathbb{C}$. – Jason Starr Jul 11 at 12:45
• @JasonStarr are there counterexamples if the base is irreducible and admits a morphism to $\mathrm{Spec}\:\mathbb{F}_p$? – user142965 Jul 12 at 9:12