Lemma 70.14.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{L}$ is ample on $X/Y$,

2. for every scheme $Z$ and every morphism $Z \to Y$ the algebraic space $X_ Z = Z \times _ Y X$ is a scheme and the pullback $\mathcal{L}_ Z$ is ample on $X_ Z/Z$,

3. for every affine scheme $Z$ and every morphism $Z \to Y$ the algebraic space $X_ Z = Z \times _ Y X$ is a scheme and the pullback $\mathcal{L}_ Z$ is ample on $X_ Z/Z$,

4. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that the algebraic space $X_ V = V \times _ Y X$ is a scheme and the pullback $\mathcal{L}_ V$ is ample on $X_ V/V$.

Proof. Parts (1) and (2) are equivalent by definition. The implication (2) $\Rightarrow$ (3) is immediate. If (3) holds and $Z \to Y$ is as in (2), then we see that $X_ Z \to Z$ is affine locally on $Z$ representable. Hence $X_ Z$ is a scheme for example by Properties of Spaces, Lemma 65.13.1. Then it follows that $\mathcal{L}_ Z$ is ample on $X_ Z/Z$ because it holds locally on $Z$ and we can use Morphisms, Lemma 29.37.4. Thus (1), (2), and (3) are equivalent. Clearly these conditions imply (4).

Assume (4). Let $Z \to Y$ be a morphism with $Z$ affine. Then $U = V \times _ Y Z \to Z$ is a surjective étale morphism such that the pullback of $\mathcal{L}_ Z$ by $X_ U \to X_ Z$ is relatively ample on $X_ U/U$. Of course we may replace $U$ by an affine open. It follows that $\mathcal{L}_ Z$ is ample on $X_ Z/Z$ by Lemma 70.14.5. Thus (4) $\Rightarrow$ (3) and the proof is complete. $\square$

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