## 28.26 Ample invertible sheaves

Recall from Modules, Lemma 17.25.10 that given an invertible sheaf $\mathcal{L}$ on a locally ringed space $X$, and given a global section $s$ of $\mathcal{L}$ the set $X_ s = \{ x \in X \mid s \not\in \mathfrak m_ x\mathcal{L}_ x\}$ is open. A general remark is that $X_ s \cap X_{s'} = X_{ss'}$, where $ss'$ denote the section $s \otimes s' \in \Gamma (X, \mathcal{L} \otimes \mathcal{L}')$.

Definition 28.26.1. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. We say $\mathcal{L}$ is ample if

1. $X$ is quasi-compact, and

2. for every $x \in X$ there exists an $n \geq 1$ and $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $x \in X_ s$ and $X_ s$ is affine.

Lemma 28.26.2. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $n \geq 1$. Then $\mathcal{L}$ is ample if and only if $\mathcal{L}^{\otimes n}$ is ample.

Proof. This follows from the fact that $X_{s^ n} = X_ s$. $\square$

Lemma 28.26.3. Let $X$ be a scheme. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. For any closed subscheme $Z \subset X$ the restriction of $\mathcal{L}$ to $Z$ is ample.

Proof. This is clear since a closed subset of a quasi-compact space is quasi-compact and a closed subscheme of an affine scheme is affine (see Schemes, Lemma 26.8.2). $\square$

Lemma 28.26.4. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$. For any affine $U \subset X$ the intersection $U \cap X_ s$ is affine.

Proof. This translates into the following algebra problem. Let $R$ be a ring. Let $N$ be an invertible $R$-module (i.e., locally free of rank 1). Let $s \in N$ be an element. Then $U = \{ \mathfrak p \mid s \not\in \mathfrak p N\}$ is an affine open subset of $\mathop{\mathrm{Spec}}(R)$.

Let $A = \bigoplus _{n \geq 0} A_ n$ be the symmetric algebra of $N$ (which is commutative) and view $s$ as an element of $A_1$. Set $B = A/(s - 1)A$. This is an $R$-algebra whose construction commutes with any base change $R \to R'$. Thus $B' = B \otimes _ R R'$ is the zero ring if $s$ maps to zero in $N' = N \otimes _ R R'$. It follows that if $x \in \mathop{\mathrm{Spec}}(R) \setminus U$, then $B \otimes _ R \kappa (x) = 0$. We conclude that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R)$ factors through $U$ as the fibres over $x \not\in U$ are empty. On the other hand, if $\mathop{\mathrm{Spec}}(R') \subset U$ is an affine open, then $s$ maps to a basis element of $N'$ and we see that $B' = R'[s]/(s - 1) \cong R'$. It follows that $\mathop{\mathrm{Spec}}(B) \to U$ is an isomorphism and $U$ is indeed affine. $\square$

Lemma 28.26.5. Let $X$ be a scheme. Let $\mathcal{L}$ and $\mathcal{M}$ be invertible $\mathcal{O}_ X$-modules. If

1. $\mathcal{L}$ is ample, and

2. the open sets $X_ t$ where $t \in \Gamma (X, \mathcal{M}^{\otimes m})$ for $m > 0$ cover $X$,

then $\mathcal{L} \otimes \mathcal{M}$ is ample.

Proof. We check the conditions of Definition 28.26.1. As $\mathcal{L}$ is ample we see that $X$ is quasi-compact. Let $x \in X$. Choose $n \geq 1$, $m \geq 1$, $s \in \Gamma (X, \mathcal{L}^{\otimes n})$, and $t \in \Gamma (X, \mathcal{M}^{\otimes m})$ such that $x \in X_ s$, $x \in X_ t$ and $X_ s$ is affine. Then $s^ mt^ n \in \Gamma (X, (\mathcal{L} \otimes \mathcal{M})^{\otimes nm})$, $x \in X_{s^ mt^ n}$, and $X_{s^ mt^ n}$ is affine by Lemma 28.26.4. $\square$

Lemma 28.26.6. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume the open sets $X_ s$, where $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ and $n \geq 1$, form a basis for the topology on $X$. Then among those opens, the open sets $X_ s$ which are affine form a basis for the topology on $X$.

Proof. Let $x \in X$. Choose an affine open neighbourhood $\mathop{\mathrm{Spec}}(R) = U \subset X$ of $x$. By assumption, there exists a $n \geq 1$ and a $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $X_ s \subset U$. By Lemma 28.26.4 above the intersection $X_ s = U \cap X_ s$ is affine. Since $U$ can be chosen arbitrarily small we win. $\square$

Lemma 28.26.7. Let $X$ be a scheme and $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume for every point $x$ of $X$ there exists $n \geq 1$ and $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $x \in X_ s$ and $X_ s$ is affine. Then $X$ is separated.

Proof. We show first that $X$ is quasi-separated. By assumption we can find a covering of $X$ by affine opens of the form $X_ s$. By Lemma 28.26.4, the intersection of any two such sets is affine, so Schemes, Lemma 26.21.6 implies that $X$ is quasi-separated.

To show that $X$ is separated, we can use the valuative criterion, Schemes, Lemma 26.22.2. Thus, let $A$ be a valuation ring with fraction field $K$ and consider two morphisms $f, g : \mathop{\mathrm{Spec}}(A) \to X$ such that the two compositions $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(A) \to X$ agree. As $A$ is local, there exists $p, q \ge 1$, $s \in \Gamma (X, \mathcal{L}^{\otimes p})$, and $t \in \Gamma (X, \mathcal{L}^{\otimes q})$ such that $X_ s$ and $X_ t$ are affine, $f(\mathop{\mathrm{Spec}}A) \subseteq X_ s$, and $g(\mathop{\mathrm{Spec}}A) \subseteq X_ t$. We now replace $s$ by $s^ q$, $t$ by $t^ p$, and $\mathcal{L}$ by $\mathcal{L}^{\otimes pq}$. This is harmless as $X_ s = X_{s^ q}$ and $X_ t = X_{t^ p}$, and now $s$ and $t$ are both sections of the same sheaf $\mathcal{L}$.

The quasi-coherent module $f^*\mathcal{L}$ corresponds to an $A$-module $M$ and $g^*\mathcal{L}$ corresponds to an $A$-module $N$ by our classification of quasi-coherent modules over affine schemes (Schemes, Lemma 26.7.4). The $A$-modules $M$ and $N$ are locally free of rank $1$ (Lemma 28.20.1) and as $A$ is local they are free (Algebra, Lemma 10.55.8). Therefore we may identify $M$ and $N$ with $A$-submodules of $M \otimes _ A K$ and $N \otimes _ A K$. The equality $f|_{\mathop{\mathrm{Spec}}(K)} = g|_{\mathop{\mathrm{Spec}}(K)}$ determines an isomorphism $\phi \colon M \otimes _ A K \to N \otimes _ A K$.

Let $x \in M$ and $y \in N$ be the elements corresponding to the pullback of $s$ along $f$ and $g$, respectively. These satisfy $\phi (x \otimes 1) = y \otimes 1$. The image of $f$ is contained in $X_ s$, so $x \not\in \mathfrak {m}_ A M$, that is, $x$ generates $M$. Hence $\phi$ determines an isomorphism of $M$ with the submodule of $N$ generated by $y$. Arguing symmetrically using $t$, $\phi ^{-1}$ determines an isomorphism of $N$ with a submodule of $M$. Consequently $\phi$ restricts to an isomorphism of $M$ and $N$. Since $x$ generates $M$, its image $y$ generates $N$, implying $y \not\in \mathfrak {m}_ A N$. Therefore $g(\mathop{\mathrm{Spec}}(A)) \subseteq X_ s$. Because $X_ s$ is affine, it is separated by Schemes, Lemma 26.21.15, and we conclude $f = g$. $\square$

Lemma 28.26.8. Let $X$ be a scheme. If there exists an ample invertible sheaf on $X$ then $X$ is separated.

Proof. Follows immediately from Lemma 28.26.7 and Definition 28.26.1. $\square$

Lemma 28.26.9. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Set $S = \Gamma _*(X, \mathcal{L})$ as a graded ring. If every point of $X$ is contained in one of the open subschemes $X_ s$, for some $s \in S_{+}$ homogeneous, then there is a canonical morphism of schemes

$f : X \longrightarrow Y = \text{Proj}(S),$

to the homogeneous spectrum of $S$ (see Constructions, Section 27.8). This morphism has the following properties

1. $f^{-1}(D_{+}(s)) = X_ s$ for any $s \in S_{+}$ homogeneous,

2. there are $\mathcal{O}_ X$-module maps $f^*\mathcal{O}_ Y(n) \to \mathcal{L}^{\otimes n}$ compatible with multiplication maps, see Constructions, Equation (27.10.1.1),

3. the composition $S_ n \to \Gamma (Y, \mathcal{O}_ Y(n)) \to \Gamma (X, \mathcal{L}^{\otimes n})$ is the identity map, and

4. for every $x \in X$ there is an integer $d \geq 1$ and an open neighbourhood $U \subset X$ of $x$ such that $f^*\mathcal{O}_ Y(dn)|_ U \to \mathcal{L}^{\otimes dn}|_ U$ is an isomorphism for all $n \in \mathbf{Z}$.

Proof. Denote $\psi : S \to \Gamma _*(X, \mathcal{L})$ the identity map. We are going to use the triple $(U(\psi ), r_{\mathcal{L}, \psi }, \theta )$ of Constructions, Lemma 27.14.1. By assumption the open subscheme $U(\psi )$ of equals $X$. Hence $r_{\mathcal{L}, \psi } : U(\psi ) \to Y$ is defined on all of $X$. We set $f = r_{\mathcal{L}, \psi }$. The maps in part (2) are the components of $\theta$. Part (3) follows from condition (2) in the lemma cited above. Part (1) follows from (3) combined with condition (1) in the lemma cited above. Part (4) follows from the last statement in Constructions, Lemma 27.14.1 since the map $\alpha$ mentioned there is an isomorphism. $\square$

Lemma 28.26.10. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Set $S = \Gamma _*(X, \mathcal{L})$. Assume (a) every point of $X$ is contained in one of the open subschemes $X_ s$, for some $s \in S_{+}$ homogeneous, and (b) $X$ is quasi-compact. Then the canonical morphism of schemes $f : X \longrightarrow \text{Proj}(S)$ of Lemma 28.26.9 above is quasi-compact with dense image.

Proof. To prove $f$ is quasi-compact it suffices to show that $f^{-1}(D_{+}(s))$ is quasi-compact for any $s \in S_{+}$ homogeneous. Write $X = \bigcup _{i = 1, \ldots , n} X_ i$ as a finite union of affine opens. By Lemma 28.26.4 each intersection $X_ s \cap X_ i$ is affine. Hence $X_ s = \bigcup _{i = 1, \ldots , n} X_ s \cap X_ i$ is quasi-compact. Assume that the image of $f$ is not dense to get a contradiction. Then, since the opens $D_+(s)$ with $s \in S_+$ homogeneous form a basis for the topology on $\text{Proj}(S)$, we can find such an $s$ with $D_+(s) \not= \emptyset$ and $f(X) \cap D_+(s) = \emptyset$. By Lemma 28.26.9 this means $X_ s = \emptyset$. By Lemma 28.17.2 this means that a power $s^ n$ is the zero section of $\mathcal{L}^{\otimes n\deg (s)}$. This in turn means that $D_+(s) = \emptyset$ which is the desired contradiction. $\square$

Lemma 28.26.11. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Set $S = \Gamma _*(X, \mathcal{L})$. Assume $\mathcal{L}$ is ample. Then the canonical morphism of schemes $f : X \longrightarrow \text{Proj}(S)$ of Lemma 28.26.9 is an open immersion with dense image.

Proof. By Lemma 28.26.7 we see that $X$ is quasi-separated. Choose finitely many $s_1, \ldots , s_ n \in S_{+}$ homogeneous such that $X_{s_ i}$ are affine, and $X = \bigcup X_{s_ i}$. Say $s_ i$ has degree $d_ i$. The inverse image of $D_{+}(s_ i)$ under $f$ is $X_{s_ i}$, see Lemma 28.26.9. By Lemma 28.17.2 the ring map

$(S^{(d_ i)})_{(s_ i)} = \Gamma (D_{+}(s_ i), \mathcal{O}_{\text{Proj}(S)}) \longrightarrow \Gamma (X_{s_ i}, \mathcal{O}_ X)$

is an isomorphism. Hence $f$ induces an isomorphism $X_{s_ i} \to D_{+}(s_ i)$. Thus $f$ is an isomorphism of $X$ onto the open subscheme $\bigcup _{i = 1, \ldots , n} D_{+}(s_ i)$ of $\text{Proj}(S)$. The image is dense by Lemma 28.26.10. $\square$

Lemma 28.26.12. Let $X$ be a scheme. Let $S$ be a graded ring. Assume $X$ is quasi-compact, and assume there exists an open immersion

$j : X \longrightarrow Y = \text{Proj}(S).$

Then $j^*\mathcal{O}_ Y(d)$ is an invertible ample sheaf for some $d > 0$.

Proof. This is Constructions, Lemma 27.10.6. $\square$

Proposition 28.26.13. Let $X$ be a quasi-compact scheme. Let $\mathcal{L}$ be an invertible sheaf on $X$. Set $S = \Gamma _*(X, \mathcal{L})$. The following are equivalent:

1. $\mathcal{L}$ is ample,

2. the open sets $X_ s$, with $s \in S_{+}$ homogeneous, cover $X$ and the associated morphism $X \to \text{Proj}(S)$ is an open immersion,

3. the open sets $X_ s$, with $s \in S_{+}$ homogeneous, form a basis for the topology of $X$,

4. the open sets $X_ s$, with $s \in S_{+}$ homogeneous, which are affine form a basis for the topology of $X$,

5. for every quasi-coherent sheaf $\mathcal{F}$ on $X$ the sum of the images of the canonical maps

$\Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) \otimes _{\mathbf{Z}} \mathcal{L}^{\otimes -n} \longrightarrow \mathcal{F}$

with $n \geq 1$ equals $\mathcal{F}$,

6. same property as (5) with $\mathcal{F}$ ranging over all quasi-coherent sheaves of ideals,

7. $X$ is quasi-separated and for every quasi-coherent sheaf $\mathcal{F}$ of finite type on $X$ there exists an integer $n_0$ such that $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}$ is globally generated for all $n \geq n_0$,

8. $X$ is quasi-separated and for every quasi-coherent sheaf $\mathcal{F}$ of finite type on $X$ there exist integers $n > 0$, $k \geq 0$ such that $\mathcal{F}$ is a quotient of a direct sum of $k$ copies of $\mathcal{L}^{\otimes - n}$, and

9. same as in (8) with $\mathcal{F}$ ranging over all sheaves of ideals of finite type on $X$.

Proof. Lemma 28.26.11 is (1) $\Rightarrow$ (2). Lemmas 28.26.2 and 28.26.12 provide the implication (1) $\Leftarrow$ (2). The implications (2) $\Rightarrow$ (4) $\Rightarrow$ (3) are clear from Constructions, Section 27.8. Lemma 28.26.6 is (3) $\Rightarrow$ (1). Thus we see that the first 4 conditions are all equivalent.

Assume the equivalent conditions (1) – (4). Note that in particular $X$ is separated (as an open subscheme of the separated scheme $\text{Proj}(S)$). Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Choose $s \in S_{+}$ homogeneous such that $X_ s$ is affine. We claim that any section $m \in \Gamma (X_ s, \mathcal{F})$ is in the image of one of the maps displayed in (5) above. This will imply (5) since these affines $X_ s$ cover $X$. Namely, by Lemma 28.17.2 we may write $m$ as the image of $m' \otimes s^{-n}$ for some $n \geq 1$, some $m' \in \Gamma (X, \mathcal{F} \otimes \mathcal{L}^{\otimes n})$. This proves the claim.

Clearly (5) $\Rightarrow$ (6). Let us assume (6) and prove $\mathcal{L}$ is ample. Pick $x \in X$. Let $U \subset X$ be an affine open which contains $x$. Set $Z = X \setminus U$. We may think of $Z$ as a reduced closed subscheme, see Schemes, Section 26.12. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the quasi-coherent sheaf of ideals corresponding to the closed subscheme $Z$. By assumption (6), there exists an $n \geq 1$ and a section $s \in \Gamma (X, \mathcal{I} \otimes \mathcal{L}^{\otimes n})$ such that $s$ does not vanish at $x$ (more precisely such that $s \not\in \mathfrak m_ x \mathcal{I}_ x \otimes \mathcal{L}_ x^{\otimes n}$). We may think of $s$ as a section of $\mathcal{L}^{\otimes n}$. Since it clearly vanishes along $Z$ we see that $X_ s \subset U$. Hence $X_ s$ is affine, see Lemma 28.26.4. This proves that $\mathcal{L}$ is ample. At this point we have proved that (1) – (6) are equivalent.

Assume the equivalent conditions (1) – (6). In the following we will use the fact that the tensor product of two sheaves of modules which are globally generated is globally generated without further mention (see Modules, Lemma 17.4.3). By (1) we can find elements $s_ i \in S_{d_ i}$ with $d_ i \geq 1$ such that $X = \bigcup _{i = 1, \ldots , n} X_{s_ i}$. Set $d = d_1\ldots d_ n$. It follows that $\mathcal{L}^{\otimes d}$ is globally generated by

$s_1^{d/d_1}, \ldots , s_ n^{d/d_ n}.$

This means that if $\mathcal{L}^{\otimes j}$ is globally generated then so is $\mathcal{L}^{\otimes j + dn}$ for all $n \geq 0$. Fix a $j \in \{ 0, \ldots , d - 1\}$. For any point $x \in X$ there exists an $n \geq 1$ and a global section $s$ of $\mathcal{L}^{j + dn}$ which does not vanish at $x$, as follows from (5) applied to $\mathcal{F} = \mathcal{L}^{\otimes j}$ and ample invertible sheaf $\mathcal{L}^{\otimes d}$. Since $X$ is quasi-compact there we may find a finite list of integers $n_ i$ and global sections $s_ i$ of $\mathcal{L}^{\otimes j + dn_ i}$ which do not vanish at any point of $X$. Since $\mathcal{L}^{\otimes d}$ is globally generated this means that $\mathcal{L}^{\otimes j + dn}$ is globally generated where $n = \max \{ n_ i\}$. Since we proved this for every congruence class mod $d$ we conclude that there exists an $n_0 = n_0(\mathcal{L})$ such that $\mathcal{L}^{\otimes n}$ is globally generated for all $n \geq n_0$. At this point we see that if $\mathcal{F}$ is globally generated then so is $\mathcal{F} \otimes \mathcal{L}^{\otimes n}$ for all $n \geq n_0$.

We continue to assume the equivalent conditions (1) – (6). Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-modules of finite type. Denote $\mathcal{F}_ n \subset \mathcal{F}$ the image of the canonical map of (5). By construction $\mathcal{F}_ n \otimes \mathcal{L}^{\otimes n}$ is globally generated. By (5) we see $\mathcal{F}$ is the sum of the subsheaves $\mathcal{F}_ n$, $n \geq 1$. By Modules, Lemma 17.9.7 we see that $\mathcal{F} = \sum _{n = 1, \ldots , N} \mathcal{F}_ n$ for some $N \geq 1$. It follows that $\mathcal{F} \otimes \mathcal{L}^{\otimes n}$ is globally generated whenever $n \geq N + n_0(\mathcal{L})$ with $n_0(\mathcal{L})$ as above. We conclude that (1) – (6) implies (7).

Assume (7). Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-modules of finite type. By (7) there exists an integer $n \geq 1$ such that the canonical map

$\Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) \otimes _{\mathbf{Z}} \mathcal{L}^{\otimes -n} \longrightarrow \mathcal{F}$

is surjective. Let $I$ be the set of finite subsets of $\Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$ partially ordered by inclusion. Then $I$ is a directed partially ordered set. For $i = \{ s_1, \ldots , s_{r(i)}\}$ let $\mathcal{F}_ i \subset \mathcal{F}$ be the image of the map

$\bigoplus \nolimits _{j = 1, \ldots , r(i)} \mathcal{L}^{\otimes -n} \longrightarrow \mathcal{F}$

which is multiplication by $s_ j$ on the $j$th factor. The surjectivity above implies that $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{F}_ i$. Hence Modules, Lemma 17.9.7 applies and we conclude that $\mathcal{F} = \mathcal{F}_ i$ for some $i$. Hence we have proved (8). In other words, (7) $\Rightarrow$ (8).

The implication (8) $\Rightarrow$ (9) is trivial.

Finally, assume (9). Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. By Lemma 28.22.3 (this is where we use the condition that $X$ be quasi-separated) we see that $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _\alpha I_\alpha$ with each $I_\alpha$ quasi-coherent of finite type. Since by assumption each of the $I_\alpha$ is a quotient of negative tensor powers of $\mathcal{L}$ we conclude the same for $\mathcal{I}$ (but of course without the finiteness or boundedness of the powers). Hence we conclude that (9) implies (6). This ends the proof of the proposition. $\square$

Lemma 28.26.14. Let $X$ be a scheme. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $i : X' \to X$ be a morphism of schemes. Assume at least one of the following conditions holds

1. $i$ is a quasi-compact immersion,

2. $X'$ is quasi-compact and $i$ is an immersion,

3. $i$ is quasi-compact and induces a homeomorphism between $X'$ and $i(X')$,

4. $X'$ is quasi-compact and $i$ induces a homeomorphism between $X'$ and $i(X')$.

Then $i^*\mathcal{L}$ is ample on $X'$.

Proof. Observe that in cases (1) and (3) the scheme $X'$ is quasi-compact as $X$ is quasi-compact by Definition 28.26.1. Thus it suffices to prove (2) and (4). Since (2) is a special case of (4) it suffices to prove (4).

Assume condition (4) holds. For $s \in \Gamma (X, \mathcal{L}^{\otimes d})$ denote $s' = i^*s$ the pullback of $s$ to $X'$. Note that $s'$ is a section of $(i^*\mathcal{L})^{\otimes d}$. By Proposition 28.26.13 the opens $X_ s$, for $s \in \Gamma (X, \mathcal{L}^{\otimes d})$, form a basis for the topology on $X$. Since $X'_{s'} = i^{-1}(X_ s)$ and since $X' \to i(X')$ is a homeomorphism, we conclude the opens $X'_{s'}$ form a basis for the topology of $X'$. Hence $i^*\mathcal{L}$ is ample by Proposition 28.26.13. $\square$

Lemma 28.26.15. Let $S$ be a quasi-separated scheme. Let $X$, $Y$ be schemes over $S$. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module and let $\mathcal{N}$ be an ample invertible $\mathcal{O}_ Y$-module. Then $\mathcal{M} = \text{pr}_1^*\mathcal{L} \otimes _{\mathcal{O}_{X \times _ S Y}} \text{pr}_2^*\mathcal{N}$ is an ample invertible sheaf on $X \times _ S Y$.

Proof. The morphism $i : X \times _ S Y \to X \times Y$ is a quasi-compact immersion, see Schemes, Lemma 26.21.9. On the other hand, $\mathcal{M}$ is the pullback by $i$ of the corresponding invertible module on $X \times Y$. By Lemma 28.26.14 it suffices to prove the lemma for $X \times Y$. We check (1) and (2) of Definition 28.26.1 for $\mathcal{M}$ on $X \times Y$.

Since $X$ and $Y$ are quasi-compact, so is $X \times Y$. Let $z \in X \times Y$ be a point. Let $x \in X$ and $y \in Y$ be the projections. Choose $n > 0$ and $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $X_ s$ is an affine open neighbourhood of $x$. Choose $m > 0$ and $t \in \Gamma (Y, \mathcal{N}^{\otimes m})$ such that $Y_ t$ is an affine open neighbourhood of $y$. Then $r = \text{pr}_1^*s \otimes \text{pr}_2^*t$ is a section of $\mathcal{M}$ with $(X \times Y)_ r = X_ s \times Y_ t$. This is an affine open neighbourhood of $z$ and the proof is complete. $\square$

Comment #8586 by Jinyong An on

                    Let $X= U \cup V$ be a scheme which is disjoint union of two open subschemes.

Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module.

If each $\mathcal{L}|_U$ and $\mathcal{L}|_V$ is ample, then $\mathcal{L}$ is ample?


Comment #8587 by Jinyong An on

Let $X= U \cup V$ be a scheme which is disjoint union of two open subschemes. Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module. If each $\mathcal{L}|_{U}$ and $\mathcal{L}|_V$ is ample, then $\mathcal{L}$ is ample?

Comment #8591 by ZL on

@#8587 If you insist that $U$ and $V$ are disjoint, then your claim is true. Just unravel the definition 28.26.1. Otherwise consider $(X,L)=(\mathbb{P}^1,\mathcal{O}(-1))$. On each standard affine $D_+(x_0)$ and $D_+(x_1)$ the line bundle $L$ is trivial, hence it is ample on each affine. However $L$ has no global section hence it cannot be ample.

Comment #8592 by Jinyong An on

O.K. Can you give more concrete hint? To show that $\mathcal{L}$ is ample, it suffices to show that for every $x\in X$, there exists an $n \ge 1$ and $s \in \Gamma(X, \mathcal{L}^{n}) = \Gamma(U, \mathcal{L}^n) \oplus \Gamma(V, \mathcal{L}^n)$ (Sheaf axiom) such that $x\in X_s$ and $X_s$ is affine. Fix $x\in X = U \cup V$. We may assume that $x\in U$. Since $\mathcal{L}|_U$ is ample, by the definition, there exists an $m \ge 1$ and $s' \in \Gamma(U, \mathcal{L}^m)$ such that $x\in U_{s'}$ and $U_{s'}$ is affine in $U$. From this, how can we find such $n$ and $s \in \Gamma(X, \mathcal{L}^{n})$ (lifting) as above so that $x\in X_s$ and $X_s$ is affine.

What should I catch? What whould be key point to breakthrough this difficulty? :)

Comment #8593 by ZL on

@#8592 You almost get there. Pick another $x'\in V$, then there exists an $s''\in \Gamma(V,\mathcal{L}^{m'})$ such that $x'\in V_{s''}$ is affine. Now set $s\in \Gamma(X,\mathcal{L}^{mm'})$ to be $s| _U=s'^{\otimes m'}$ and $s| _{V}=s''^{\otimes m}$. You get that $X_s=U_ {s'}\coprod V_ {s''}$ is affine.

Comment #8594 by Jinyong An on

@8593 Thanks! I guessed that we may use the affiness of disjoint union of affine schemes but I didn't know how can we emdody it. Good Idea ~ Thank you.

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