The term “quasi-projective scheme” has not yet been defined. A possible definition could be a scheme which has an ample invertible sheaf. However, if $X$ is a scheme over a base scheme $S$, then we say that $X$ is quasi-projective over $S$ if the morphism $X \to S$ is quasi-projective (Morphisms, Definition 29.40.1). Since the identity morphism of any scheme is quasi-projective, we see that a scheme quasi-projective over $S$ doesn't necessarily have an ample invertible sheaf. For this reason it seems better to leave the term “quasi-projective scheme” undefined.
Lemma 37.49.1. Let $S$ be a scheme which has an ample invertible sheaf. Let $f : X \to S$ be a morphism of schemes. The following are equivalent
$X \to S$ is quasi-projective,
$X \to S$ is H-quasi-projective,
there exists a quasi-compact open immersion $X \to X'$ of schemes over $S$ with $X' \to S$ projective,
$X \to S$ is of finite type and $X$ has an ample invertible sheaf, and
$X \to S$ is of finite type and there exists an $f$-very ample invertible sheaf.
Proof. The implication (2) $\Rightarrow $ (1) is Morphisms, Lemma 29.40.5. The implication (1) $\Rightarrow $ (2) is Morphisms, Lemma 29.43.16. The implication (2) $\Rightarrow $ (3) is Morphisms, Lemma 29.43.11
Assume $X \subset X'$ is as in (3). In particular $X \to S$ is of finite type. By Morphisms, Lemma 29.43.11 the morphism $X \to S$ is H-projective. Thus there exists a quasi-compact immersion $i : X \to \mathbf{P}^ n_ S$. Hence $\mathcal{L} = i^*\mathcal{O}_{\mathbf{P}^ n_ S}(1)$ is $f$-very ample. As $X \to S$ is quasi-compact we conclude from Morphisms, Lemma 29.38.2 that $\mathcal{L}$ is $f$-ample. Thus $X \to S$ is quasi-projective by definition.
The implication (4) $\Rightarrow $ (2) is Morphisms, Lemma 29.39.3.
Assume the equivalent conditions (1), (2), (3) hold. Choose an immersion $i : X \to \mathbf{P}^ n_ S$ over $S$. Let $\mathcal{L}$ be an ample invertible sheaf on $S$. To finish the proof we will show that $\mathcal{N} = f^*\mathcal{L} \otimes _{\mathcal{O}_ X} i^*\mathcal{O}_{\mathbf{P}^ n_ S}(1)$ is ample on $X$. By Properties, Lemma 28.26.14 we reduce to the case $X = \mathbf{P}^ n_ S$. Let $s \in \Gamma (S, \mathcal{L}^{\otimes d})$ be a section such that the corresponding open $S_ s$ is affine. Say $S_ s = \mathop{\mathrm{Spec}}(A)$. Recall that $\mathbf{P}^ n_ S$ is the projective bundle associated to $\mathcal{O}_ S T_0 \oplus \ldots \oplus \mathcal{O}_ S T_ n$, see Constructions, Lemma 27.21.5 and its proof. Let $s_ i \in \Gamma (\mathbf{P}^ n_ S, \mathcal{O}(1))$ be the global section corresponding to the section $T_ i$ of $\mathcal{O}_ S T_0 \oplus \ldots \oplus \mathcal{O}_ S T_ n$. Then we see that $X_{f^*s \otimes s_ i^{\otimes n}}$ is affine because it is equal to $\mathop{\mathrm{Spec}}(A[T_0/T_ i, \ldots , T_ n/T_ i])$. This proves that $\mathcal{N}$ is ample by definition.
The equivalence of (1) and (5) follows from Morphisms, Lemmas 29.38.2 and 29.39.5. $\square$
Lemma 37.49.2. Let $S$ be a scheme which has an ample invertible sheaf. Let $\text{QP}_ S$ be the full subcategory of the category of schemes over $S$ satisfying the equivalent conditions of Lemma 37.49.1.
if $S' \to S$ is a morphism of schemes and $S'$ has an ample invertible sheaf, then base change determines a functor $\text{QP}_ S \to \text{QP}_{S'}$,
if $X \in \text{QP}_ S$ and $Y \in \text{QP}_ X$, then $Y \in \text{QP}_ S$,
the category $\text{QP}_ S$ is closed under fibre products,
the category $\text{QP}_ S$ is closed under finite disjoint unions,
if $X \to S$ is projective, then $X \in \text{QP}_ S$,
if $X \to S$ is quasi-affine of finite type, then $X$ is in $\text{QP}_ S$,
if $X \to S$ is quasi-finite and separated, then $X \in \text{QP}_ S$,
if $X \to S$ is a quasi-compact immersion, then $X \in \text{QP}_ S$,
add more here.
Proof. Part (1) follows from Morphisms, Lemma 29.40.2.
Part (2) follows from the fourth characterization of Lemma 37.49.1.
If $X \to S$ and $Y \to S$ are quasi-projective, then $X \times _ S Y \to Y$ is quasi-projective by Morphisms, Lemma 29.40.2. Hence (3) follows from (2).
If $X = Y \amalg Z$ is a disjoint union of schemes and $\mathcal{L}$ is an invertible $\mathcal{O}_ X$-module such that $\mathcal{L}|_ Y$ and $\mathcal{L}|_ Z$ are ample, then $\mathcal{L}$ is ample (details omitted). Thus part (4) follows from the fourth characterization of Lemma 37.49.1.
Part (5) follows from Morphisms, Lemma 29.43.10.
Part (6) follows from Morphisms, Lemma 29.40.7.
Part (7) follows from part (6) and Lemma 37.43.2.
Part (8) follows from part (7) and Morphisms, Lemma 29.20.16. $\square$
The following lemma doesn't really belong in this section, but there does not seem to be a good spot for it anywhere else.
Lemma 37.49.3. Let $X$ be a quasi-affine scheme. Let $f : U \to X$ be an integral morphism. Then $U$ is quasi-affine and the diagram is cartesian.
\[ \xymatrix{ U \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U)) \ar[d] \\ X \ar[r] & \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X)) } \]
Proof. The scheme $U$ is quasi-affine because integral morphisms are affine, affine morphisms are quasi-affine, a scheme is quasi-affine if and only if the structure morphism to $\mathop{\mathrm{Spec}}(\mathbf{Z})$ is quasi-affine, and compositions of quasi-affine morphisms are quasi-affine. The first two statements follow immediately from the definition and the third is Morphisms, Lemma 29.13.4. Set $U' = X \times _{\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))} \mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ and consider the extended diagram
The morphism $j$ is closed by Morphisms, Lemma 29.41.7 combined with the fact that an integral morphism is universally closed (Morphisms, Lemma 29.44.7) and the fact that the vertical arrows are in the diagram are separated. On the other hand, $j$ is open because the horizontal arrows in the diagram of the lemma are open by Properties, Lemma 28.18.4. Thus $j$ identifies $U$ with an open and closed subscheme of $U'$. If $U \not= U'$ then $U$ isn't dense in $U'$ and a fortiori not dense in the spectrum of $\Gamma (U, \mathcal{O}_ U)$. However, the scheme theoretic image of $U$ in $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ is $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ because any ideal in $\Gamma (U, \mathcal{O}_ U)$ cutting out a closed subscheme through which $U$ factors would have to be zero. Hence $U$ is dense in $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ for example by Morphisms, Lemma 29.6.3. Thus $U = U'$ and we win. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)