Lemma 29.43.16. Let $S$ be a scheme which admits an ample invertible sheaf. Then

1. any projective morphism $X \to S$ is H-projective, and

2. any quasi-projective morphism $X \to S$ is H-quasi-projective.

Proof. The assumptions on $S$ imply that $S$ is quasi-compact and separated, see Properties, Definition 28.26.1 and Lemma 28.26.11 and Constructions, Lemma 27.8.8. Hence Lemma 29.43.12 applies and we see that (1) implies (2). Let $\mathcal{E}$ be a finite type quasi-coherent $\mathcal{O}_ S$-module. By our definition of projective morphisms it suffices to show that $\mathbf{P}(\mathcal{E}) \to S$ is H-projective. If $\mathcal{E}$ is generated by finitely many global sections, then the corresponding surjection $\mathcal{O}_ S^{\oplus n} \to \mathcal{E}$ induces a closed immersion

$\mathbf{P}(\mathcal{E}) \longrightarrow \mathbf{P}(\mathcal{O}_ S^{\oplus n}) = \mathbf{P}^ n_ S$

as desired. In general, let $\mathcal{L}$ be an invertible sheaf on $S$. By Properties, Proposition 28.26.13 there exists an integer $n$ such that $\mathcal{E} \otimes _{\mathcal{O}_ S} \mathcal{L}^{\otimes n}$ is globally generated by finitely many sections. Since $\mathbf{P}(\mathcal{E}) = \mathbf{P}(\mathcal{E} \otimes _{\mathcal{O}_ S} \mathcal{L}^{\otimes n})$ by Constructions, Lemma 27.20.1 this finishes the proof. $\square$

Comment #6646 by Phoebe on

Why does $\textbf{P}(\mathcal{E})=\textbf{P}(\mathcal{E}\otimes_{\mathcal{O}_S}\mathcal{L}^{\otimes n})$ follow from Lemma 27.20.1?

Comment #6647 by Phoebe on

Oh I was being stupid. I get it now. You can delete my comments.

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