The Stacks project

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

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). $\square$

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 $V = \{ \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 $x \in \mathop{\mathrm{Spec}}(R) \setminus V$, then $B \otimes _ R \kappa (x) = 0$. We conclude that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R)$ factors through $V$ as the fibres over $x \not\in V$ are empty. On the other hand, if $\mathop{\mathrm{Spec}}(R') \subset V$ 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 $\mathop{\mathrm{Spec}}(B) \to V$ is an isomorphism and $V$ is indeed affine. $\square$

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. $\square$

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. $\square$

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 $f = g$. $\square$

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

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. $\square$

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. $\square$

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

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. $\square$

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

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

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

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

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. $\square$

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 $X'_{s'} = i^{-1}(X_ s)$ by Modules, Remark 17.25.11 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. $\square$

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. $\square$