Definition 71.14.1. 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. We say $\mathcal{L}$ is relatively ample, or $f$-relatively ample, or ample on $X/Y$, or $f$-ample if $f : X \to Y$ is representable and for every morphism $Z \to Y$ where $Z$ is a scheme, the pullback $\mathcal{L}_ Z$ of $\mathcal{L}$ to $X_ Z = Z \times _ Y X$ is ample on $X_ Z/Z$ as in Morphisms, Definition 29.37.1.
71.14 Relatively ample sheaves
This section is the analogue of Morphisms, Section 29.37 for algebraic spaces. Our definition of a relatively ample invertible sheaf is as follows.
We will almost always reduce questions about relatively ample invertible sheaves to the case of schemes. Thus in this section we have mainly sanity checks.
Lemma 71.14.2. 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. Assume $Y$ is a scheme. The following are equivalent
Proof. This follows from the definitions and Morphisms, Lemma 29.37.9 (which says that being relatively ample for schemes is preserved under base change). $\square$
Lemma 71.14.3. 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. Let $Y' \to Y$ be a morphism of algebraic spaces over $S$. Let $f' : X' \to Y'$ be the base change of $f$ and denote $\mathcal{L}'$ the pullback of $\mathcal{L}$ to $X'$. If $\mathcal{L}$ is $f$-ample, then $\mathcal{L}'$ is $f'$-ample.
Proof. This follows immediately from the definition! (Hint: transitivity of base change.) $\square$
Lemma 71.14.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If there exists an $f$-ample invertible sheaf, then $f$ is representable, quasi-compact, and separated.
Proof. This is clear from the definitions and Morphisms, Lemma 29.37.3. (If in doubt, take a look at the principle of Algebraic Spaces, Lemma 65.5.8.) $\square$
Lemma 71.14.5. Let $V \to U$ be a surjective étale morphism of affine schemes. Let $X$ be an algebraic space over $U$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $Y = V \times _ U X$ and let $\mathcal{N}$ be the pullback of $\mathcal{L}$ to $Y$. The following are equivalent
$\mathcal{L}$ is ample on $X/U$, and
$\mathcal{N}$ is ample on $Y/V$.
Proof. The implication (1) $\Rightarrow $ (2) follows from Lemma 71.14.3. Assume (2). This implies that $Y \to V$ is quasi-compact and separated (Lemma 71.14.4) and $Y$ is a scheme. It follows that the morphism $f : X \to U$ is quasi-compact and separated (Morphisms of Spaces, Lemmas 67.8.8 and 67.4.12). Set $\mathcal{A} = \bigoplus _{d \geq 0} f_*\mathcal{L}^{\otimes d}$. This is a quasi-coherent sheaf of graded $\mathcal{O}_ U$-algebras (Morphisms of Spaces, Lemma 67.11.2). By adjunction we have a map $\psi : f^*\mathcal{A} \to \bigoplus _{d \geq 0} \mathcal{L}^{\otimes d}$. Applying Lemma 71.13.1 we obtain an open subspace $U(\psi ) \subset X$ and a morphism
\[ r_{\mathcal{L}, \psi } : U(\psi ) \to \underline{\text{Proj}}_ U(\mathcal{A}) \]
Since $h : V \to U$ is étale we have $\mathcal{A}|_ V = (Y \to V)_*(\bigoplus _{d \geq 0} \mathcal{N}^{\otimes d})$, see Properties of Spaces, Lemma 66.26.2. It follows that the pullback $\psi '$ of $\psi $ to $Y$ is the adjunction map for the situation $(Y \to V, \mathcal{N})$ as in Morphisms, Lemma 29.37.4 part (5). Since $\mathcal{N}$ is ample on $Y/V$ we conclude from the lemma just cited that $U(\psi ') = Y$ and that $r_{\mathcal{N}, \psi '}$ is an open immersion. Since Lemma 71.13.1 tells us that the formation of $r_{\mathcal{L}, \psi }$ commutes with base change, we conclude that $U(\psi ) = X$ and that we have a commutative diagram
\[ \xymatrix{ Y \ar[r]_-{r'} \ar[d] & \underline{\text{Proj}}_ V(\mathcal{A}|_ V) \ar[d] \ar[r] & V \ar[d] \\ X \ar[r]^-r & \underline{\text{Proj}}_ U(\mathcal{A}) \ar[r] & U } \]
whose squares are fibre products. We conclude that $r$ is an open immersion by Morphisms of Spaces, Lemma 67.12.1. Thus $X$ is a scheme. Then we can apply Morphisms, Lemma 29.37.4 part (5) to conclude that $\mathcal{L}$ is ample on $X/U$. $\square$
Lemma 71.14.6. 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
$\mathcal{L}$ is ample on $X/Y$,
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 $X_ Z/Z$,
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 $X_ Z/Z$,
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 66.13.1. 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 71.14.5. Thus (4) $\Rightarrow $ (3) and the proof is complete. $\square$
Lemma 71.14.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then $f$ is quasi-affine if and only if $\mathcal{O}_ X$ is $f$-relatively ample.
Proof. Follows from the case of schemes, see Morphisms, Lemma 29.37.6. $\square$
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