## 69.14 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.

Definition 69.14.1. 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.

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 69.14.2. 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

1. $\mathcal{L}$ is ample on $X/Y$ in the sense of Definition 69.14.1, and

2. $X$ is a scheme and $\mathcal{L}$ is ample on $X/Y$ in the sense of Morphisms, Definition 29.37.1.

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 69.14.3. 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 69.14.4. 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 63.5.8.) $\square$

Lemma 69.14.5. 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

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

2. $\mathcal{N}$ is ample on $Y/V$.

Proof. The implication (1) $\Rightarrow$ (2) follows from Lemma 69.14.3. Assume (2). This implies that $Y \to V$ is quasi-compact and separated (Lemma 69.14.4) and $Y$ is a scheme. Then we conclude that $X \to U$ is quasi-compact and separated (Morphisms of Spaces, Lemmas 65.8.8 and 65.4.12). Set $\mathcal{A} = \bigoplus _{d \geq 0} f_*\mathcal{L}^{\otimes d}$. Thus is a quasi-coherent sheaf of graded $\mathcal{O}_ U$-algebras (Morphisms of Spaces, Lemma 65.11.2). By adjunction we have a map $\psi : f^*\mathcal{A} \to \bigoplus _{d \geq 0} \mathcal{L}^{\otimes d}$. Applying Lemma 69.13.1 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 64.26.2. 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 69.13.1 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 65.12.1. 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 69.14.6. 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 64.13.1. 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 69.14.5. Thus (4) $\Rightarrow$ (3) and the proof is complete. $\square$

## Comments (1)

Comment #6271 by Mitchell on

In the proof of Lemma 0D35, when defining $\mathcal{A}$, the notation $f_*$ is used, without having previously noting that $f$ is the morphism $X \to U$. Also, following the definition of $\mathcal{A}$, there is a small typo: "Thus" should be replaced with "This."

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