§5. Quasi-affine morphisms; quasi-projective morphisms; proper morphisms; projective morphisms

5.1. Quasi-affine morphisms

Definition.

A quasi-affine scheme is a scheme isomorphic to the subscheme induced on a quasi-compact open subset of an affine scheme. We say that a morphism $f:X\to Y$ is quasi-affine, or also that $X$ (considered as a $Y$-prescheme via $f$) is a quasi-affine $Y$-scheme, if there exists a covering $(U_{\alpha })$ of $Y$ by affine opens such that the $f^{-1}(U_{\alpha })$ are quasi-affine schemes.

It is clear that a quasi-affine morphism is separated (I, 5.5.5 and I, 5.5.8) and quasi-compact (I, 6.6.1); every affine morphism is obviously quasi-affine.

Recall that, for any prescheme $X$, setting $A=\Gamma (X,{\mathcal{O}}_X)$, the identity homomorphism $A\to A=\Gamma (X,{\mathcal{O}}_X)$ defines a morphism $X\to \mathrm{Spec}(A)$, called canonical (I, 2.2.4); this is moreover none other than the canonical morphism defined in (4.5.1) in the special case $\mathcal{L}={\mathcal{O}}_X$, once we recall that $\mathrm{Proj}(A[T])$ is canonically identified with $\mathrm{Spec}(A)$ (3.1.7).

Proposition.

Let $X$ be a quasi-compact scheme or a prescheme whose underlying space is Noetherian, and $A$ the ring $\Gamma (X,{\mathcal{O}}_X)$. The following conditions are equivalent:

1. (a) $X$ is a quasi-affine scheme.
2. (b) The canonical morphism $u:X\to \mathrm{Spec}(A)$ is an open immersion.
3. (b') The canonical morphism $u:X\to \mathrm{Spec}(A)$ is a homeomorphism from $X$ onto a subspace of the underlying space of $\mathrm{Spec}(A)$.
4. (c) The ${\mathcal{O}}_X$-module ${\mathcal{O}}_X$ is very ample relative to $u$ (4.4.2).
5. (c') The ${\mathcal{O}}_X$-module ${\mathcal{O}}_X$ is ample (4.5.1).
6. (d) As $f$ ranges over $A$, the $X_f$ form a basis of the topology of $X$.

1. (d') As $f$ ranges over $A$, those of the $X_f$ that are affine form a covering of $X$.
2. (e) Every quasi-coherent ${\mathcal{O}}_X$-module is generated by its sections over $X$.
3. (e') Every quasi-coherent sheaf of ideals of finite type of ${\mathcal{O}}_X$ is generated by its sections over $X$.

Proof. It is clear that (b) implies (a), and (a) implies (c) by virtue of criterion (b) of (4.4.4) applied to the identity morphism (taking into account the remark preceding the statement); on the other hand, (c) implies (c') (4.5.10, (i)), and (c'), (b), and (b') are equivalent by (4.5.2, criteria (b) and (b')). Finally, (c') is equivalent to each of the criteria (d), (d'), (e), (e') by virtue of (4.5.2, criteria (a), (a'), (c)) and (4.5.5, criterion (d'')).

We also observe that, with the preceding notation, those of the $X_f$ that are affine form a basis of the topology of $X$, and that the canonical morphism $u$ is dominant (4.5.2).

Corollary.

Let $X$ be a quasi-compact prescheme. If there exists a morphism $v:X\to Y$ from $X$ into an affine scheme $Y$ that is a homeomorphism of $X$ onto an open subspace of $Y$, then $X$ is quasi-affine.

Proof. Indeed, there exists a family $(g_{\alpha })$ of sections of ${\mathcal{O}}_Y$ over $Y$ such that the $D(g_{\alpha })$ form a basis of the topology of $v(X)$; if $v=(\psi ,\theta )$ and we set $f_{\alpha }=\theta (g_{\alpha })$, then $X_{f_{\alpha }}={\psi }^{-1}(D(g_{\alpha }))$ (I, 2.2.4.1), so the $X_{f_{\alpha }}$ form a basis of the topology of $X$, and criterion (d) of (5.1.2) is satisfied.

Corollary.

If $X$ is a quasi-affine scheme, then every invertible ${\mathcal{O}}_X$-module is very ample (relative to the canonical morphism), and a fortiori ample.

Proof. Indeed, such a module $\mathcal{L}$ is generated by its sections over $X$ (5.1.2, (e)), so $\mathcal{L}\otimes {\mathcal{O}}_X=\mathcal{L}$ is very ample (4.4.8).

Corollary.

Let $X$ be a quasi-compact prescheme. If there exists an invertible ${\mathcal{O}}_X$-module $\mathcal{L}$ such that $\mathcal{L}$ and ${\mathcal{L}}^{-1}$ are ample, then $X$ is a quasi-affine scheme.

Proof. Indeed, ${\mathcal{O}}_X=\mathcal{L}\otimes {\mathcal{L}}^{-1}$ is then ample (4.5.7).

Proposition.

Let $f:X\to Y$ be a quasi-compact morphism. The following conditions are equivalent:

1. (a) $f$ is quasi-affine.
2. (b) The ${\mathcal{O}}_Y$-algebra $f_{\ast }({\mathcal{O}}_X)=\mathcal{A}(X)$ is quasi-coherent, and the canonical morphism $X\to \mathrm{Spec}(\mathcal{A}(X))$ corresponding to the identity homomorphism $\mathcal{A}(X)\to \mathcal{A}(X)$ (1.2.7) is an open immersion.
3. (b') The ${\mathcal{O}}_Y$-algebra $\mathcal{A}(X)$ is quasi-coherent, and the canonical morphism $X\to \mathrm{Spec}(\mathcal{A}(X))$ is a homeomorphism from $X$ onto a subspace of $\mathrm{Spec}(\mathcal{A}(X))$.
4. (c) The ${\mathcal{O}}_X$-module ${\mathcal{O}}_X$ is very ample for $f$.
5. (c') The ${\mathcal{O}}_X$-module ${\mathcal{O}}_X$ is ample for $f$.
6. (d) The morphism $f$ is separated, and for every quasi-coherent ${\mathcal{O}}_X$-module $\mathcal{F}$, the canonical homomorphism $\sigma :f\ast (f_{\ast }(\mathcal{F}))\to \mathcal{F}$ (0, 4.4.3) is surjective.

Furthermore, when $f$ is quasi-affine, every invertible ${\mathcal{O}}_X$-module $\mathcal{L}$ is very ample relative to $f$.

Proof. The equivalence of (a) and (c') follows from the local character on $Y$ of $f$-ampleness (4.6.4), Definition (5.1.1), and the criterion (5.1.2, (c')). The other properties are local on $Y$

and so follow immediately from (5.1.2) and (5.1.4), taking into account the fact that $f_{\ast }(\mathcal{F})$ is quasi-coherent when $f$ is separated (I, 9.2.2, a).

Corollary.

Let $f:X\to Y$ be a quasi-affine morphism. For every open $U$ of $Y$, the restriction $f^{-1}(U)\to U$ of $f$ is quasi-affine.

Corollary.

Let $Y$ be an affine scheme and $f:X\to Y$ a quasi-compact morphism. For $f$ to be quasi-affine, it is necessary and sufficient that $X$ be a quasi-affine scheme.

Proof. This is an immediate consequence of (5.1.6) and (4.6.6).

Corollary.

Let $Y$ be a quasi-compact scheme or a prescheme whose underlying space is Noetherian, and $f:X\to Y$ a morphism of finite type. If $f$ is quasi-affine, then there exists a quasi-coherent ${\mathcal{O}}_Y$-subalgebra $\mathcal{B}$ of $\mathcal{A}(X)=f_{\ast }({\mathcal{O}}_X)$, of finite type (I, 9.6.2), such that the morphism $X\to \mathrm{Spec}(\mathcal{B})$ corresponding to the canonical injection $\mathcal{B}\to \mathcal{A}(X)$ (1.2.7) is an immersion. Furthermore, every quasi-coherent ${\mathcal{O}}_Y$-subalgebra ${\mathcal{B}}^{\prime }$ of $\mathcal{A}(X)$ of finite type that contains $\mathcal{B}$ has the same property.

Proof. Indeed, $\mathcal{A}(X)$ is the inductive limit of its quasi-coherent ${\mathcal{O}}_Y$-subalgebras of finite type (I, 9.6.5); the result is then a particular case of (3.8.4), taking into account the identification of $\mathrm{Spec}(\mathcal{A}(X))$ with $\mathrm{Proj}(\mathcal{A}(X)[T])$ (3.1.7).

Proposition.

1. (i) A quasi-compact morphism $X\to Y$ that is a homeomorphism from the underlying space of $X$ onto a subspace of the underlying space of $Y$ (in particular, a closed immersion) is quasi-affine.
2. (ii) The composition of two quasi-affine morphisms is quasi-affine.
3. (iii) If $f:X\to Y$ is a quasi-affine $S$-morphism, then $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ is a quasi-affine morphism for every base change $S^{\prime }\to S$.
4. (iv) If $f:X\to Y$ and $g:X^{\prime }\to Y^{\prime }$ are two quasi-affine $S$-morphisms, then $f{\times }_{S}g$ is quasi-affine.
5. (v) If $f:X\to Y$ and $g:Y\to Z$ are two morphisms such that $g\circ f$ is quasi-affine, and if $g$ is separated or the underlying space of $X$ is locally Noetherian, then $f$ is quasi-affine.
6. (vi) If $f$ is a quasi-affine morphism, then so is $f_{red}$.

Proof. Taking into account criterion (5.1.6, (c')), (i), (iii), (iv), (v) and (vi) follow immediately from (4.6.13, (i bis), (iii), (iv), (v) and (vi)) respectively. To prove (ii), we may restrict to the case where $Z$ is affine, and then the assertion follows directly from (4.6.13, (ii)) applied to $\mathcal{L}={\mathcal{O}}_X$ and $\mathcal{K}={\mathcal{O}}_Y$.

Remark.

Let $f:X\to Y$, $g:Y\to Z$ be two morphisms such that $X{\times }_{Z}Y$ is locally Noetherian. Then the graph immersion ${\Gamma }_f:X\to X{\times }_{Z}Y$ is quasi-affine, being quasi-compact (I, 6.3.5), and (I, 5.5.12) shows that in (v) the conclusion still holds if we drop the hypothesis that $g$ is separated.

Proposition.

Let $f:X\to Y$ be a quasi-compact morphism, and $g:X^{\prime }\to X$ a quasi-affine morphism. If $\mathcal{L}$ is an ${\mathcal{O}}_X$-module ample for $f$, then $g\ast (\mathcal{L})$ is an ${\mathcal{O}}_{X^{\prime }}$-module ample for $f\circ g$.

Proof. Indeed, since ${\mathcal{O}}_{X^{\prime }}$ is very ample for $g$, and the question is local on $Y$ (4.6.4), it follows from (4.6.13, (ii)) that there exists (for $Y$ affine) an integer $n$ such that

$$g\ast (\mathcal{L}{\otimes }^n)=(g\ast (\mathcal{L})){\otimes }^n$$

is ample for $f\circ g$, so $g\ast (\mathcal{L})$ is ample for $f\circ g$ (4.6.9).

5.2. Serre's criterion

Theorem (Serre's criterion).

Let $X$ be a quasi-compact scheme or a prescheme whose underlying space is Noetherian. The following conditions are equivalent:

1. (a) $X$ is an affine scheme.

2. (b) There exists a family of elements $f_{\alpha }\in A=\Gamma (X,{\mathcal{O}}_X)$ such that the $X_{f_{\alpha }}$ are affine and the ideal generated by the $f_{\alpha }$ in $A$ is equal to $A$.

3. (c) The functor $\Gamma (X,\mathcal{F})$ is exact in $\mathcal{F}$ on the category of quasi-coherent ${\mathcal{O}}_X$-modules; in other words, if

     0 → ℱ' → ℱ → ℱ'' → 0                                                  (*)
   

   is an exact sequence of quasi-coherent ${\mathcal{O}}_X$-modules, then the sequence

     0 → Γ(X, ℱ') → Γ(X, ℱ) → Γ(X, ℱ'') → 0
   

   is also exact.

4. (c') Property (c) holds for every exact sequence (*) of quasi-coherent ${\mathcal{O}}_X$-modules such that $\mathcal{F}$ is isomorphic to an ${\mathcal{O}}_X$-submodule of a finite power ${\mathcal{O}}_X^n$.

5. (d) $H^1(X,\mathcal{F})=0$ for every quasi-coherent ${\mathcal{O}}_X$-module $\mathcal{F}$.

6. (d') $H^1(X,\mathcal{J})=0$ for every quasi-coherent sheaf of ideals $\mathcal{J}$ of ${\mathcal{O}}_X$.

Proof. It is evident that (a) implies (b); furthermore (b) implies that the $X_{f_{\alpha }}$ cover $X$, since by hypothesis the section 1 is a linear combination of the $f_{\alpha }$ and the $D(f_{\alpha })$ then cover $\mathrm{Spec}(A)$. The last assertion of (4.5.2) then implies that $X\to \mathrm{Spec}(A)$ is an isomorphism.

We know that (a) implies (c) (I, 1.3.11), and (c) trivially implies (c'). Let us prove that (c') implies (b). First, (c') implies that, for every closed point $x\in X$ and every open neighbourhood $U$ of $x$, there exists $f\in A$ such that $x\in X_f\subset X-U$. Indeed, let $\mathcal{J}$ (resp. ${\mathcal{J}}^{\prime }$) be the quasi-coherent sheaf of ideals of ${\mathcal{O}}_X$ defining the closed reduced subprescheme of $X$ whose underlying space is $X-U$ (resp. $(X-U)\cup x$) (I, 5.2.1); it is clear that ${\mathcal{J}}^{\prime }\subset \mathcal{J}$ and that ${\mathcal{J}}^{\prime \prime }=\mathcal{J}/{\mathcal{J}}^{\prime }$ is a quasi-coherent ${\mathcal{O}}_X$-module with support $x$ and such that ${\mathcal{J}}_x^{\prime \prime }=\kappa (x)$. Hypothesis (c') applied to the exact sequence $0\to {\mathcal{J}}^{\prime }\to \mathcal{J}\to {\mathcal{J}}^{\prime \prime }\to 0$ shows that $\Gamma (X,\mathcal{J})\to \Gamma (X,{\mathcal{J}}^{\prime \prime })$ is surjective. The section of ${\mathcal{J}}^{\prime \prime }$ whose germ at $x$ is $1_x$ is therefore the image of a section $f\in \Gamma (X,\mathcal{J})\subset \Gamma (X,{\mathcal{O}}_X)$, and by definition $f(x)=1_x$ and $f(y)=0$ on $X-U$, which establishes our assertion. Moreover, if $U$ is affine, then so is $X_f$ (I, 1.3.6), so the union of those $X_f$ that are affine ($f\in A$) is an open set $Z$ containing every closed point of $X$; since $X$ is a quasi-compact Kolmogorov space, we necessarily have $Z=X$ (0, 2.1.3). Because $X$ is quasi-compact, there are finitely many elements $f_i\in A$ ($1\le i\le n$) such that the $X_{f_i}$ are affine and cover $X$. Consider then the homomorphism ${\mathcal{O}}_X^n\to {\mathcal{O}}_X$ defined by the sections $f_i$ (0, 5.1.1); since for every $x\in X$ at least one of the $(f_i{)}_x$ is invertible, this homomorphism is surjective, and we therefore have an exact sequence $0\to \mathcal{R}\to {\mathcal{O}}_X^n\to {\mathcal{O}}_X\to 0$, where $\mathcal{R}$ is a quasi-coherent ${\mathcal{O}}_X$-submodule of ${\mathcal{O}}_X^n$. It then follows

from (c') that the corresponding homomorphism $\Gamma (X,{\mathcal{O}}_X^n)\to \Gamma (X,{\mathcal{O}}_X)$ is surjective, which proves (b).

Finally, (a) implies (d) (I, 5.1.9.2), and (d) trivially implies (d'). It remains to show that (d') implies (c'). If $\mathcal{F}$ is a quasi-coherent ${\mathcal{O}}_X$-submodule of ${\mathcal{O}}_X^n$, the filtration $0\subset {\mathcal{O}}_X\subset {\mathcal{O}}_X^2\subset \cdots \subset {\mathcal{O}}_X^n$ defines on $\mathcal{F}$ a filtration formed by the ${\mathcal{F}}_k=\mathcal{F}\cap {\mathcal{O}}_X^k$ ($0\le k\le n$), which are quasi-coherent ${\mathcal{O}}_X$-modules (I, 4.1.1), and ${\mathcal{F}}_{k+1}/{\mathcal{F}}_k$ is isomorphic to a quasi-coherent ${\mathcal{O}}_X$-submodule of ${\mathcal{O}}_X^{k+1}/{\mathcal{O}}_X^k={\mathcal{O}}_X$, that is, to a quasi-coherent sheaf of ideals of ${\mathcal{O}}_X$. Hypothesis (d') therefore implies $H^1(X,{\mathcal{F}}_{k+1}/{\mathcal{F}}_k)=0$; the exact cohomology sequence $H^1(X,{\mathcal{F}}_k)\to H^1(X,{\mathcal{F}}_{k+1})\to H^1(X,{\mathcal{F}}_{k+1}/{\mathcal{F}}_k)=0$ then allows us to prove by induction on $k$ that $H^1(X,{\mathcal{F}}_k)=0$ for every $k$. Q.E.D.

Remark.

When $X$ is a Noetherian prescheme, in the statements of (c') and (d') one may replace "quasi-coherent" by "coherent". Indeed, in the proof that (c') implies (b), $\mathcal{J}$ and ${\mathcal{J}}^{\prime }$ are then coherent sheaves of ideals, and moreover every quasi-coherent submodule of a coherent module is coherent (I, 6.1.1); whence the conclusion.

Corollary.

Let $f:X\to Y$ be a separated quasi-compact morphism. The following conditions are equivalent:

1. (a) $f$ is an affine morphism.
2. (b) The functor $f_{\ast }$ is exact on the category of quasi-coherent ${\mathcal{O}}_X$-modules.
3. (c) For every quasi-coherent ${\mathcal{O}}_X$-module $\mathcal{F}$, $R^1{f}_{\ast }(\mathcal{F})=0$.
4. (c') For every quasi-coherent sheaf of ideals $\mathcal{J}$ of ${\mathcal{O}}_X$, $R^1{f}_{\ast }(\mathcal{J})=0$.

Proof. All these conditions being local on $Y$, by definition of the functor $R^1{f}_{\ast }$ (T, 3.7.3), we may suppose that $Y$ is affine. If $f$ is affine, then $X$ is affine and property (b) is none other than (I, 1.6.4). Conversely, let us show that (b) implies (a): for every quasi-coherent ${\mathcal{O}}_X$-module $\mathcal{F}$, $f_{\ast }(\mathcal{F})$ is a quasi-coherent ${\mathcal{O}}_Y$-module (I, 9.2.2, a). By hypothesis, the functor $f_{\ast }(\mathcal{F})$ is exact in $\mathcal{F}$, and the functor $\Gamma (Y,\mathcal{G})$ is exact in $\mathcal{G}$ (on the category of quasi-coherent ${\mathcal{O}}_Y$-modules) since $Y$ is affine (I, 1.3.11); hence $\Gamma (Y,f_{\ast }(\mathcal{F}))=\Gamma (X,\mathcal{F})$ is exact in $\mathcal{F}$, which proves our assertion by virtue of (5.2.1, (c)).

If $f$ is affine, $f^{-1}(U)$ is affine for every affine open $U$ of $Y$ (1.3.2), hence $H^1(f^{-1}(U),\mathcal{F})=0$ (5.2.1, (d)), which by definition implies $R^1{f}_{\ast }(\mathcal{F})=0$. Finally, suppose condition (c') is satisfied; the exact sequence of low-degree terms in the Leray spectral sequence (G, II, 4.17.1 and I, 4.5.1) yields in particular the exact sequence

  0 → H¹(Y, f_*(𝒥)) → H¹(X, 𝒥) → H⁰(Y, R¹f_*(𝒥)).

Since $Y$ is affine and $f_{\ast }(\mathcal{J})$ is quasi-coherent (I, 9.2.2, a), $H^1(Y,f_{\ast }(\mathcal{J}))=0$ (5.2.1); hypothesis (c') therefore implies $H^1(X,\mathcal{J})=0$, and we conclude by (5.2.1) that $X$ is an affine scheme.

Corollary.

If $f:X\to Y$ is an affine morphism, then for every quasi-coherent ${\mathcal{O}}_X$-module $\mathcal{F}$, the canonical homomorphism $H^1(Y,f_{\ast }(\mathcal{F}))\to H^1(X,\mathcal{F})$ is bijective.

Proof. Indeed, we have the exact sequence

  0 → H¹(Y, f_*(ℱ)) → H¹(X, ℱ) → H⁰(Y, R¹f_*(ℱ))

of low-degree terms of the Leray spectral sequence, and the conclusion follows from (5.2.2).

Remark.

In Chapter III, §1, we will prove that if $X$ is affine, then $H^i(X,\mathcal{F})=0$ for every $i>0$ and every quasi-coherent ${\mathcal{O}}_X$-module $\mathcal{F}$.

5.3. Quasi-projective morphisms

Definition.

We say that a morphism $f:X\to Y$ is quasi-projective, or that $X$ (considered as a $Y$-prescheme via $f$) is quasi-projective over $Y$, or that $X$ is a quasi-projective $Y$-scheme, if $f$ is of finite type and there exists an invertible ${\mathcal{O}}_X$-module that is $f$-ample.

Note that this notion is not local on $Y$: the counterexamples of Nagata [26] and Hironaka show that, even when $X$ and $Y$ are non-singular algebraic schemes over an algebraically closed field, every point of $Y$ may have an affine neighbourhood $U$ such that $f^{-1}(U)$ is quasi-projective over $U$, without $f$ being quasi-projective.

Note that a quasi-projective morphism is necessarily separated (4.6.1). When $Y$ is quasi-compact, it amounts to the same thing to say that $f$ is quasi-projective or that $f$ is of finite type and there exists an ${\mathcal{O}}_X$-module very ample relative to $f$ (4.6.2 and 4.6.11). Furthermore:

Proposition.

Let $Y$ be a quasi-compact scheme or a prescheme whose underlying space is Noetherian, and let $X$ be a $Y$-prescheme. The following conditions are equivalent:

1. (a) $X$ is a quasi-projective $Y$-scheme.
2. (b) $X$ is of finite type over $Y$, and there exists a quasi-coherent ${\mathcal{O}}_Y$-module $\mathcal{E}$ of finite type such that $X$ is $Y$-isomorphic to a subprescheme of $\mathbb{P}(\mathcal{E})$.
3. (c) $X$ is of finite type over $Y$, and there exists a quasi-coherent graded ${\mathcal{O}}_Y$-algebra $\mathcal{S}$ such that ${\mathcal{S}}_1$ is of finite type and generates $\mathcal{S}$, and such that $X$ is $Y$-isomorphic to a subprescheme induced on an everywhere-dense open subset of $\mathrm{Proj}(\mathcal{S})$.

Proof. This follows immediately from the preceding remark and from (4.4.3), (4.4.6), and (4.4.7).

Note that when $Y$ is a Noetherian prescheme, in conditions (b) and (c) of (5.3.2) we may drop the hypothesis that $X$ is of finite type over $Y$, which is then automatically satisfied (I, 6.3.5).

Corollary.

Let $Y$ be a quasi-compact scheme such that there exists an ample ${\mathcal{O}}_Y$-module $\mathcal{L}$ (4.5.3). For a $Y$-scheme $X$ to be quasi-projective, it is necessary and sufficient that it be of finite type over $Y$ and isomorphic to a $Y$-subscheme of a projective bundle of the form ${\mathbb{P}}_Y^r$.

Proof. Indeed, if $\mathcal{E}$ is a quasi-coherent ${\mathcal{O}}_Y$-module of finite type, then $\mathcal{E}$ is isomorphic to a quotient of an ${\mathcal{O}}_Y$-module ${\mathcal{L}}^{\otimes (-n)}{\otimes }_{{\mathcal{O}}_Y}{\mathcal{O}}_Y^k$ (4.5.5), hence $\mathbb{P}(\mathcal{E})$ is isomorphic to a closed subscheme of ${\mathbb{P}}_Y^{k-1}$ (4.1.2 and 4.1.4).

Proposition.

1. (i) A quasi-affine morphism of finite type (and in particular a quasi-compact immersion, or an affine morphism of finite type) is quasi-projective.
2. (ii) If $f:X\to Y$ and $g:Y\to Z$ are quasi-projective and $Z$ is quasi-compact, then $g\circ f$ is quasi-projective.

1. (iii) If $f:X\to Y$ is a quasi-projective $S$-morphism, then $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ is quasi-projective for every base change $S^{\prime }\to S$.
2. (iv) If $f:X\to Y$ and $g:X^{\prime }\to Y^{\prime }$ are two quasi-projective $S$-morphisms, then $f{\times }_{S}g$ is quasi-projective.
3. (v) If $f:X\to Y$, $g:Y\to Z$ are two morphisms such that $g\circ f$ is quasi-projective, and if $g$ is separated or $X$ locally Noetherian, then $f$ is quasi-projective.
4. (vi) If $f$ is a quasi-projective morphism, then so is $f_{red}$.

Proof. (i) follows from (5.1.6) and (5.1.10, (i)). The other assertions are immediate consequences of Definition (5.3.1), of the properties of morphisms of finite type (I, 6.3.4), and of (4.6.13).

Remark.

Note that $f_{red}$ can be quasi-projective without $f$ being so, even when $Y$ is the spectrum of an algebra of finite rank over $\mathbb{C}$ and $f$ is proper.

Corollary.

If $X$ and $X^{\prime }$ are two quasi-projective $Y$-schemes, then $X\bigsqcup X^{\prime }$ is a quasi-projective $Y$-scheme.

Proof. This follows from (4.6.18).

5.4. Proper morphisms and universally closed morphisms

Definition.

We say that a morphism of preschemes $f:X\to Y$ is proper if it satisfies the following two conditions:

1. (a) $f$ is separated and of finite type.
2. (b) For every prescheme $Y^{\prime }$ and every morphism $Y^{\prime }\to Y$, the projection $f_{(Y^{\prime })}:X{\times }_Y{Y}^{\prime }\to Y^{\prime }$ is a closed morphism (I, 2.2.6).

When this is so, we also say that $X$ (considered as a $Y$-prescheme with structure morphism $f$) is proper over $Y$.

It is immediate that conditions (a) and (b) are local on $Y$. To verify that the image of a closed subset $Z$ of $X{\times }_Y{Y}^{\prime }$ under the projection $q:X{\times }_Y{Y}^{\prime }\to Y^{\prime }$ is closed in $Y^{\prime }$, it suffices to see that $q(Z)\cap U^{\prime }$ is closed in $U^{\prime }$ for every affine open $U^{\prime }$ of $Y^{\prime }$; since $q(Z)\cap U^{\prime }=q(Z\cap q^{-1}(U^{\prime }))$ and $q^{-1}(U^{\prime })$ is identified with $X{\times }_Y{U}^{\prime }$ (I, 4.4.1), we see that to verify condition (b) of Def. (5.4.1) we may restrict to the case where $Y^{\prime }$ is an affine scheme. We will see later (5.6.3) that if $Y$ is locally Noetherian, we may even restrict to verifying (b) when $Y^{\prime }$ is of finite type over $Y$.

It is clear that every proper morphism is closed.

Proposition.

1. (i) A closed immersion is a proper morphism.

2. (ii) The composition of two proper morphisms is proper.

3. (iii) If $X$, $Y$ are $S$-preschemes and $f:X\to Y$ a proper $S$-morphism, then

     f_{(S')} : X_{(S')} → Y_{(S')}
   

   is proper for every base change $S^{\prime }\to S$.

4. (iv) If $f:X\to Y$ and $g:X^{\prime }\to Y^{\prime }$ are two proper $S$-morphisms, then the $S$-morphism $f{\times }_{S}g:X{\times }_S{X}^{\prime }\to Y{\times }_S{Y}^{\prime }$ is proper.

Proof. It suffices to prove (i), (ii), and (iii) (I, 3.5.1). In each of these three cases, the verification of condition (a) of (5.4.1) follows from earlier results (I, 5.5.1 and I, 6.3.4); it remains to verify condition (b). It is immediate in case (i), because if $X\to Y$ is a closed immersion, then so is $X{\times }_Y{Y}^{\prime }\to Y{\times }_Y{Y}^{\prime }=Y^{\prime }$ (I, 4.3.2 and 3.3.3). To prove (ii), consider two proper morphisms $X\to Y$, $Y\to Z$, and a morphism $Z^{\prime }\to Z$. We may write $X{\times }_Z{Z}^{\prime }=X{\times }_Y(Y{\times }_Z{Z}^{\prime })$ (I, 3.3.9.1), and so the projection $X{\times }_Z{Z}^{\prime }\to Z^{\prime }$ factors as $X{\times }_Y(Y{\times }_Z{Z}^{\prime })\to Y{\times }_Z{Z}^{\prime }\to Z^{\prime }$. Taking the initial remark into account, (ii) thus follows from the fact that the composition of two closed morphisms is closed. Finally, for every morphism $S^{\prime }\to S$, $X_{(S^{\prime })}$ is identified with $X{\times }_Y{Y}_{(S^{\prime })}$ (I, 3.3.11); for every morphism $Z\to Y_{(S^{\prime })}$, we may write

  X_{(S')} ×_{Y_{(S')}} Z = (X ×_Y Y_{(S')}) ×_{Y_{(S')}} Z = X ×_Y Z;

$X{\times }_{Y}Z\to Z$ is closed by hypothesis, which proves (iii).

Corollary.

Let $f:X\to Y$, $g:Y\to Z$ be two morphisms such that $g\circ f$ is proper.

1. (i) If $g$ is separated, then $f$ is proper.
2. (ii) If $g$ is separated and of finite type and if $f$ is surjective, then $g$ is proper.

Proof. (i) follows from (5.4.2) by the general procedure (I, 5.5.12). To prove (ii), only condition (b) of Def. (5.4.1) needs to be verified. For every morphism $Z^{\prime }\to Z$, the diagram

                    f × 1_{Z'}
   X ×_Z Z' ─────────────────→ Y ×_Z Z'
         ╲                       │
        p ╲                      │ p'
           ↘                     ↓
                                 Z'

(where $p$ and $p^{\prime }$ are the projections) commutes (I, 3.2.1); furthermore, $f\times 1_{Z^{\prime }}$ is surjective since $f$ is (I, 3.5.2), and $p$ is a closed morphism by hypothesis. Every closed subset $F$ of $Y{\times }_Z{Z}^{\prime }$ is then the image under $f\times 1_{Z^{\prime }}$ of a closed subset $E$ of $X{\times }_Z{Z}^{\prime }$, hence $p^{\prime }(F)=p(E)$ is closed in $Z^{\prime }$ by hypothesis, whence the corollary.

Corollary.

If $X$ is a prescheme proper over $Y$, and $\mathcal{S}$ a quasi-coherent ${\mathcal{O}}_Y$-algebra, then every $Y$-morphism $f:X\to \mathrm{Proj}(\mathcal{S})$ is proper (and a fortiori closed).

Proof. Indeed, the structure morphism $p:\mathrm{Proj}(\mathcal{S})\to Y$ is separated, and $p\circ f$ is proper by hypothesis.

Corollary.

Let $f:X\to Y$ be a separated morphism of finite type. Let $(X_i{)}_{1\le i\le n}$ (resp. $(Y_i{)}_{1\le i\le n}$) be a finite family of closed subpreschemes of $X$ (resp. $Y$), and $j_i$ (resp. $h_i$) the canonical injection $X_i\to X$ (resp. $Y_i\to Y$). Suppose that the underlying space of $X$ is the union of the $X_i$, and that for every $i$ there is a morphism $f_i:X_i\to Y_i$ such that the diagram

              f_i
   X_i ───────────→ Y_i
    │                │
 j_i│                │ h_i
    ↓                ↓
    X ──────────────→ Y
              f

commutes. Then for $f$ to be proper, it is necessary and sufficient that each $f_i$ be proper.

Proof. If $f$ is proper, so is $f\circ j_i$, since $j_i$ is a closed immersion (5.4.2); since $h_i$ is a closed immersion, hence a separated morphism, $f_i$ is proper by (5.4.3). Conversely, suppose that each $f_i$ is proper, and consider the prescheme $Z$ sum of the $X_i$; let $u$ be the morphism $Z\to X$ which reduces to $j_i$ on each $X_i$. The restriction of $f\circ u$ to each $X_i$ equals $f\circ j_i=h_i\circ f_i$, hence is proper since both $h_i$ and $f_i$ are (5.4.2); it then follows immediately from Def. (5.4.1) that $u$ is proper. But since $u$ is surjective by hypothesis, we conclude that $f$ is proper by (5.4.3).

Corollary.

Let $f:X\to Y$ be a separated morphism of finite type; for $f$ to be proper, it is necessary and sufficient that f_red : X_red → Y_red be.

Proof. This is a particular case of (5.4.5) with $n=1$, $X_1=X_{red}$, and $Y_1=Y_{red}$ (I, 5.1.5).

(5.4.7)

If $X$ and $Y$ are Noetherian preschemes and $f:X\to Y$ a separated morphism of finite type, then to verify that $f$ is proper we may reduce to the case of dominant morphisms of integral preschemes. Indeed, let $X_i$ ($1\le i\le n$) be the finitely many irreducible components of $X$ and consider, for each $i$, the unique closed reduced subprescheme of $X$ having $X_i$ as underlying space, which we again denote by $X_i$ (I, 5.2.1). Let $Y_i$ be the unique closed reduced subprescheme of $Y$ having $f(X_i)$ as underlying space. If $g_i$ (resp. $h_i$) is the injection morphism $X_i\to X$ (resp. $Y_i\to Y$), one concludes that $f\circ g_i=h_i\circ f_i$, where $f_i$ is a dominant morphism $X_i\to Y_i$ (I, 5.2.2); we are then in the conditions to apply (5.4.5), and for $f$ to be proper, it is necessary and sufficient that the $f_i$ be.

Corollary.

Let $X$, $Y$ be $S$-preschemes separated and of finite type over $S$, and let $f:X\to Y$ be an $S$-morphism. For $f$ to be proper, it is necessary and sufficient that, for every $S$-prescheme $S^{\prime }$, the morphism $f{\times }_S{1}_{S^{\prime }}:X{\times }_S{S}^{\prime }\to Y{\times }_S{S}^{\prime }$ be closed.

Proof. First note that if $g:X\to S$ and $h:Y\to S$ are the structure morphisms, then by definition $g=h\circ f$, hence $f$ is separated and of finite type (I, 5.5.1 and I, 6.3.4). If $f$ is proper, then so is $f{\times }_S{1}_{S^{\prime }}$ (5.4.2); a fortiori, $f{\times }_S{1}_{S^{\prime }}$ is closed. Conversely, suppose the condition of the statement is satisfied and let $Y^{\prime }$ be a $Y$-prescheme; $Y^{\prime }$ may also be regarded as an $S$-prescheme, and since $Y\to S$ is separated, $X{\times }_Y{Y}^{\prime }$ is identified with a closed subprescheme of $X{\times }_S{Y}^{\prime }$ (I, 5.4.2). In the commutative diagram

                  f × 1_{Y'}
   X ×_Y Y' ─────────────────→ Y ×_Y Y' = Y'
       │                            │
       ↓                            ↓
   X ×_S Y' ─────────────────→ Y ×_S Y'
                  f × 1_{S'}

the vertical arrows are closed immersions; it follows at once that if $f\times 1_{S^{\prime }}$ is a closed morphism, then so is $f\times 1_{Y^{\prime }}$.

Remark.

We will say that a morphism $f:X\to Y$ is universally closed if it satisfies condition (b) of Definition (5.4.1). The reader will observe that

in Propositions (5.4.2) to (5.4.8), one may throughout replace "proper" by "universally closed" without altering the validity of the results (and in the hypotheses of (5.4.3), (5.4.5), (5.4.6) and (5.4.8), one may omit the finiteness conditions).

(5.4.10)

Let $f:X\to Y$ be a morphism of finite type. We will say that a closed subset $Z$ of $X$ is proper over $Y$ (or $Y$-proper, or proper for $f$) if the restriction of $f$ to a closed subprescheme of $X$ with underlying space $Z$ (I, 5.2.1) is proper. Since this restriction is then separated, it follows from (5.4.6) and (I, 5.5.1, (vi)) that the foregoing property does not depend on the closed subprescheme of $X$ having $Z$ as underlying space. If $g:X^{\prime }\to X$ is a proper morphism, then $g^{-1}(Z)$ is a proper subset of $X^{\prime }$: if $T$ is a subprescheme of $X$ having $Z$ as underlying space, it suffices to note that the restriction of $g$ to the closed subprescheme $g^{-1}(T)$ of $X^{\prime }$ is a proper morphism $g^{-1}(T)\to T$ by (5.4.2, (iii)), and then to apply (5.4.2, (ii)). On the other hand, if X'' is a $Y$-scheme of finite type and $u:X\to X^{\prime \prime }$ a $Y$-morphism, then $u(Z)$ is a proper subset of X''; indeed, let us take $T$ to be the closed reduced subprescheme of $X$ having $Z$ as underlying space; the restriction of $f$ to $T$ being proper, so is the restriction of $u$ to $T$ (5.4.3, (i)), hence $u(Z)$ is closed in X''; let T'' be a closed subprescheme of X'' having $u(Z)$ as underlying space (I, 5.2.1), so that $u|T$ factors as

  T ─v→ T'' ─j→ X''

where $j$ is the canonical injection (I, 5.2.2), and $v$ is therefore proper and surjective (5.4.5); if $g$ is the restriction to T'' of the structure morphism $X^{\prime \prime }\to Y$, then $g$ is separated and of finite type, and we have $f|T=g\circ v$; it then follows from (5.4.3, (ii)) that $g$ is proper, whence our assertion.

It follows in particular from these remarks that if $Z$ is a $Y$-proper subset of $X$, then:

1° For every closed subprescheme $X^{\prime }$ of $X$, $Z\cap X^{\prime }$ is a $Y$-proper subset of $X^{\prime }$.

2° If $X$ is a subprescheme of a $Y$-scheme of finite type X'', then $Z$ is also a $Y$-proper subset of X'' (and in particular is closed in X'').

5.5. Projective morphisms

Proposition.

Let $X$ be a $Y$-prescheme. The following conditions are equivalent:

1. (a) $X$ is $Y$-isomorphic to a closed subprescheme of a projective bundle $\mathbb{P}(\mathcal{E})$, where $\mathcal{E}$ is a quasi-coherent ${\mathcal{O}}_Y$-module of finite type.
2. (b) There exists a quasi-coherent graded ${\mathcal{O}}_Y$-algebra $\mathcal{S}$ such that ${\mathcal{S}}_1$ is of finite type and generates $\mathcal{S}$, and such that $X$ is $Y$-isomorphic to $\mathrm{Proj}(\mathcal{S})$.

Proof. Condition (a) implies (b) by virtue of (3.6.2, (ii)): if $\mathcal{J}$ is a quasi-coherent graded sheaf of ideals of $\mathbb{S}(\mathcal{E})$, the quasi-coherent graded ${\mathcal{O}}_Y$-algebra $\mathcal{S}=\mathbb{S}(\mathcal{E})/\mathcal{J}$ is generated by ${\mathcal{S}}_1$, and the latter, the canonical image of $\mathcal{E}$, is an ${\mathcal{O}}_Y$-module of finite type. Condition (b) implies (a) by virtue of (3.6.2) applied to the case where $\mathcal{M}\to {\mathcal{S}}_1$ is the identity map.

Definition.

We say that a $Y$-prescheme $X$ is projective over $Y$, or is a projective $Y$-scheme, if it satisfies the equivalent conditions (a) and (b) of (5.5.1). We say that a morphism $f:X\to Y$ is projective if it makes $X$ a projective $Y$-scheme.

It is clear that if $f:X\to Y$ is projective, then there exists an ${\mathcal{O}}_X$-module very ample relative to $f$ (4.4.2).

Theorem.

1. (i) Every projective morphism is quasi-projective and proper.
2. (ii) Conversely, let $Y$ be a quasi-compact scheme or a prescheme whose underlying space is Noetherian; then every morphism $f:X\to Y$ that is quasi-projective and proper is projective.

Proof. (i) It is clear that if $f:X\to Y$ is projective, it is of finite type and quasi-projective (hence in particular separated); on the other hand, it follows at once from (5.5.1, (b)) and (3.5.3) that if $f$ is projective, so is $f{\times }_Y{1}_{Y^{\prime }}:X{\times }_Y{Y}^{\prime }\to Y^{\prime }$ for every morphism $Y^{\prime }\to Y$. To show that $f$ is universally closed, everything therefore comes down to proving that a projective morphism $f$ is closed. The question being local on $Y$, we may suppose that $Y=\mathrm{Spec}(A)$, hence (5.5.1) $X=\mathrm{Proj}(S)$, where $S$ is a graded $A$-algebra generated by a finite number of elements of $S_1$. For every $y\in Y$, the fibre $f^{-1}(y)$ is identified with $\mathrm{Proj}(S){\times }_Y\mathrm{Spec}(\kappa (y))$ (I, 3.6.1), hence with $\mathrm{Proj}(S{\otimes }_A\kappa (y))$ (2.8.10); consequently $f^{-1}(y)$ is empty if and only if $S{\otimes }_A\kappa (y)$ satisfies condition (TN) (2.7.4), in other words $S_n{\otimes }_A\kappa (y)=0$ for $n$ sufficiently large. But since $(S_n{)}_y$ is an ${\mathcal{O}}_y$-module of finite type, the preceding condition means that $(S_n{)}_y=0$ for $n$ sufficiently large, by Nakayama's lemma. If ${\mathfrak{a}}_n$ is the annihilator in $A$ of the $A$-module $S_n$, the preceding condition also means that ${\mathfrak{a}}_n\not\subset {\mathfrak{j}}_y$ for $n$ sufficiently large (0, 1.7.4). Now, since $S_n{S}_1=S_{n+1}$ by hypothesis, we have ${\mathfrak{a}}_n\subset {\mathfrak{a}}_{n+1}$, and if $\mathfrak{a}$ is the union of the ${\mathfrak{a}}_n$, we see that $f(X)=V(\mathfrak{a})$, which proves that $f(X)$ is closed in $Y$. If now $X^{\prime }$ is an arbitrary closed subset of $X$, there exists a closed subprescheme of $X$ having $X^{\prime }$ as underlying space (I, 5.2.1), and it is clear (5.5.1, (a)) that the morphism $X^{\prime }\to X\to Y$ is projective, so $f(X^{\prime })$ is closed in $Y$.

(ii) The hypothesis on $Y$ and the fact that $f$ is quasi-projective imply the existence of a quasi-coherent ${\mathcal{O}}_Y$-module $\mathcal{E}$ of finite type and of a $Y$-immersion $j:X\to \mathbb{P}(\mathcal{E})$ (5.3.2). But since $f$ is proper, $j$ is closed by (5.4.4), hence $f$ is projective.

Remarks.

(i) Let $f:X\to Y$ be a morphism such that: 1° $f$ is proper; 2° there exists an ${\mathcal{O}}_X$-module $\mathcal{L}$ very ample relative to $f$; 3° the quasi-coherent ${\mathcal{O}}_Y$-module $\mathcal{E}=f_{\ast }(\mathcal{L})$ is of finite type. Then $f$ is a projective morphism: indeed (4.4.4), there is then a $Y$-immersion $r:X\to \mathbb{P}(\mathcal{E})$, and since $f$ is proper, $r$ is a closed immersion (5.4.4). We will see in Chapter III, §3, that when $Y$ is locally Noetherian, condition 3° above is a consequence of the other two, hence conditions 1° and 2° characterize, in this case, the projective morphisms, and if $Y$ is quasi-compact, one may replace condition 2° by the hypothesis of the existence of an ${\mathcal{O}}_X$-module ample for $f$ (4.6.11).

(ii) Let $Y$ be a quasi-compact scheme such that there exists an ample ${\mathcal{O}}_Y$-module. For a $Y$-scheme $X$ to be projective, it is necessary and sufficient that it be $Y$-isomorphic to a closed $Y$-subscheme of a projective bundle of the form ${\mathbb{P}}_Y^r$. The condition is obviously sufficient.

Conversely, if $X$ is projective over $Y$, then it is quasi-projective, so there exists a $Y$-immersion $j$ of $X$ into some ${\mathbb{P}}_Y^r$ (5.3.3) which is closed by (5.4.4) and (5.5.3).

(iii) The argument of (5.5.3) shows that for every prescheme $Y$ and every integer $r\ge 0$, the structure morphism ${\mathbb{P}}_Y^r\to Y$ is surjective, for if we set $\mathcal{S}={\mathbb{S}}_{{\mathcal{O}}_Y}({\mathcal{O}}_Y^{r+1})$, we evidently have ${\mathcal{S}}_y={\mathbb{S}}_{\kappa (y)}(\kappa (y{)}^{r+1})$ (1.7.3), hence $({\mathcal{S}}_n{)}_y\ne 0$ for every $y\in Y$ and every $n\ge 0$.

(iv) It follows from the examples of Nagata [26] that there exist proper morphisms that are not quasi-projective.

Proposition.

1. (i) A closed immersion is a projective morphism.
2. (ii) If $f:X\to Y$ and $g:Y\to Z$ are projective morphisms, and if $Z$ is a quasi-compact scheme or a prescheme whose underlying space is Noetherian, then $g\circ f$ is projective.
3. (iii) If $f:X\to Y$ is a projective $S$-morphism, then $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ is projective for every base change $S^{\prime }\to S$.
4. (iv) If $f:X\to Y$ and $g:X^{\prime }\to Y^{\prime }$ are projective $S$-morphisms, then so is $f{\times }_{S}g$.
5. (v) If $g\circ f$ is a projective morphism and if $g$ is separated, then $f$ is projective.
6. (vi) If $f$ is projective, so is $f_{red}$.

Proof. (i) follows at once from (3.1.7). Here we must prove (iii) and (iv) separately, because of the restriction introduced on $Z$ in (ii) (cf. (I, 3.5.1)). To prove (iii), we reduce to the case $S=Y$ (I, 3.3.11) and the assertion then follows at once from (5.5.1, (b)) and (3.5.3). To prove (iv), we are at once reduced to the case $X=\mathbb{P}(\mathcal{E})$, $X^{\prime }=\mathbb{P}({\mathcal{E}}^{\prime })$, with $\mathcal{E}$ (resp. ${\mathcal{E}}^{\prime }$) a quasi-coherent ${\mathcal{O}}_Y$-module (resp. ${\mathcal{O}}_{Y^{\prime }}$-module) of finite type. Let $p$, $p^{\prime }$ be the canonical projections of $T=Y{\times }_S{Y}^{\prime }$ to $Y$ and $Y^{\prime }$ respectively; by (4.1.3.1) we have $\mathbb{P}(p\ast (\mathcal{E}))=\mathbb{P}(\mathcal{E}){\times }_{Y}T$, $\mathbb{P}(p^{\prime }\ast ({\mathcal{E}}^{\prime }))=\mathbb{P}({\mathcal{E}}^{\prime }){\times }_{Y^{\prime }}T$; whence

  ℙ(p*(𝓔)) ×_T ℙ(p'*(𝓔')) = (ℙ(𝓔) ×_Y T) ×_T (T ×_{Y'} ℙ(𝓔'))
                          = ℙ(𝓔) ×_Y (T ×_{Y'} ℙ(𝓔')) = ℙ(𝓔) ×_S ℙ(𝓔')

upon replacing $T$ by $Y{\times }_S{Y}^{\prime }$ and using (I, 3.3.9.1). Now $p\ast (\mathcal{E})$ and $p^{\prime }\ast ({\mathcal{E}}^{\prime })$ are of finite type over $T$ (0, 5.2.4), hence so is $p\ast (\mathcal{E}){\otimes }_{{\mathcal{O}}_T}{p}^{\prime }\ast ({\mathcal{E}}^{\prime })$; since $\mathbb{P}(p\ast (\mathcal{E})){\times }_T\mathbb{P}(p^{\prime }\ast ({\mathcal{E}}^{\prime }))$ is identified with a closed subprescheme of $\mathbb{P}(p\ast (\mathcal{E}){\otimes }_{{\mathcal{O}}_T}{p}^{\prime }\ast ({\mathcal{E}}^{\prime }))$ (4.3.3), this completes the proof of (iv). To obtain (v) and (vi), we may apply (I, 5.5.13), since every closed subprescheme of a projective $Y$-scheme is a projective $Y$-scheme by (5.5.1, (a)).

It remains to prove (ii); by virtue of the hypothesis on $Z$, this follows from (5.5.3), (5.3.4, (ii)) and (5.4.2, (ii)).

Proposition.

If $X$ and $X^{\prime }$ are two projective $Y$-schemes, then $X\bigsqcup X^{\prime }$ is a projective $Y$-scheme.

Proof. This is an evident consequence of (5.5.2) and (4.3.6).

Proposition.

Let $X$ be a projective $Y$-scheme, and $\mathcal{L}$ a $Y$-ample ${\mathcal{O}}_X$-module; for every section $f$ of $\mathcal{L}$ over $X$, $X_f$ is affine over $Y$.

Proof. The question being local on $Y$, we may suppose $Y=\mathrm{Spec}(A)$; furthermore, $X_{f^{\otimes n}}=X_f$, so by replacing $\mathcal{L}$ with a suitable ${\mathcal{L}}^{\otimes n}$, we may suppose that $\mathcal{L}$ is very ample for the structure morphism $q:X\to Y$ (4.6.11). The canonical homomorphism $\sigma :q\ast (q_{\ast }(\mathcal{L}))\to \mathcal{L}$ is then surjective, and the corresponding morphism

  r = r_{ℒ, σ} : X → P = ℙ(q_*(ℒ))

is an immersion such that $\mathcal{L}=r\ast ({\mathcal{O}}_P(1))$ (4.4.4); furthermore, since $X$ is proper over $Y$, the immersion $r$ is closed (5.4.4). Moreover, by definition $f\in \Gamma (Y,q_{\ast }(\mathcal{L}))$ and $\sigma ♭$ is the identity of $q_{\ast }(\mathcal{L})$; it then follows from formula (3.7.3.1) that $X_f=r^{-1}(D_{+}(f))$. Therefore $X_f$ is a closed subprescheme of the affine scheme $D_{+}(f)$ and is thus an affine scheme.

Taking in particular $Y=X$, one obtains (taking (4.6.13, (i)) into account) the following corollary, whose direct proof is moreover immediate:

Corollary.

Let $X$ be a prescheme and $\mathcal{L}$ an invertible ${\mathcal{O}}_X$-module. For every section $f$ of $\mathcal{L}$ over $X$, $X_f$ is affine over $X$ (and consequently an affine scheme when $X$ is an affine scheme).

5.6. Chow's lemma

Theorem (Chow's lemma).

Let $S$ be a prescheme, $X$ an $S$-scheme of finite type. Suppose that one of the following hypotheses is satisfied:

1. (a) $S$ is Noetherian.
2. (b) $S$ is a quasi-compact scheme, and $X$ has a finite number of irreducible components.

Under these conditions:

1. (i) There exists an $S$-scheme $X^{\prime }$ quasi-projective over $S$ and an $S$-morphism $f:X^{\prime }\to X$ that is projective and surjective.
2. (ii) One can take $X^{\prime }$ and $f$ such that there exists an open $U\subset X$ for which $U^{\prime }=f^{-1}(U)$ is dense in $X^{\prime }$ and the restriction of $f$ to $U^{\prime }$ is an isomorphism $U^{\prime }\stackrel{\sim }{\to }U$.
3. (iii) If $X$ is reduced (resp. irreducible, integral), one may suppose $X^{\prime }$ reduced (resp. irreducible, integral).

Proof. The proof proceeds in several steps.

A) One can first reduce to the case where $X$ is irreducible. Indeed, under hypothesis (a), $X$ is Noetherian, so, in either hypothesis, the irreducible components $X_i$ of $X$ are finite in number. If the theorem is proved for each of the closed reduced subpreschemes of $X$ having the $X_i$ as their underlying spaces, and if $X_i^{\prime }$ and $f_i:X_i^{\prime }\to X_i$ are the corresponding prescheme and morphism, then the prescheme $X^{\prime }$ given by the sum of the $X_i^{\prime }$, and the morphism $f:X^{\prime }\to X$ whose restriction to each $X_i^{\prime }$ is $j_i\circ f_i$ (with $j_i$ the canonical injection $X_i\to X$) will answer the question. It is immediate that $X^{\prime }$ is reduced if the $X_i^{\prime }$ are; on the other hand, (ii) is satisfied by taking for $U$ the union of the sets $U_i\cap \complement ({\bigcup }_{j\ne i}{X}_j)$. Finally, since the $X_i^{\prime }$ are quasi-projective over $S$, so is $X^{\prime }$

(5.3.6); likewise, the morphisms $X_i^{\prime }\to X$ are projective by (5.5.5, (i) and (ii)), hence $f$ is projective (5.5.6), and is obviously surjective by definition.

B) Now suppose that $X$ is irreducible. Since the structure morphism $r:X\to S$ is of finite type, there is a finite covering $(S_i)$ of $S$ by affine opens, and for each $i$ a finite covering $(T_{ij})$ of $r^{-1}(S_i)$ by affine opens, the morphism $T_{ij}\to S_i$ being affine and of finite type, hence quasi-projective (5.3.4, (i)); since in each of the hypotheses (a), (b) the immersion $S_i\to S$ is quasi-compact, it is a quasi-projective morphism (5.3.4, (i)), hence the restriction of $r$ to $T_{ij}$ is a quasi-projective morphism (5.3.4, (ii)). Denote the $T_{ij}$ by $U_k$ ($1\le k\le n$). For each index $k$ there exists an open immersion ${\varphi }_k:U_k\to P_k$, where $P_k$ is projective over $S$ (5.3.2 and 5.5.2). Let $U={\bigcap }_k{U}_k$; since $X$ is irreducible and the $U_k$ non-empty, $U$ is non-empty, hence everywhere dense in $X$; the restrictions to $U$ of the ${\varphi }_k$ define a morphism

  φ : U → P = P_1 ×_S P_2 ×_S ⋯ ×_S P_n

such that the diagrams

              φ
   U ─────────────→ P
   │                │
 j_k│                │ p_k                                                  (5.6.1.1)
   ↓                ↓
   U_k ───────────→ P_k
              φ_k

commute, $j_k$ being the canonical injection $U\to U_k$ and $p_k$ the canonical projection $P\to P_k$. If $j$ is the canonical injection $U\to X$, the morphism $\psi =(j,\varphi {)}_S:U\to X{\times }_{S}P$ is an immersion (I, 5.3.14). Under hypothesis (a), $X{\times }_{S}P$ is locally Noetherian ((3.4.1) and (I, 6.3.7 and 6.3.8)); under hypothesis (b), $X{\times }_{S}P$ is a quasi-compact scheme (I, 5.5.1 and I, 6.6.4); in both cases the closure $X^{\prime }$ in $X{\times }_{S}P$ of the subprescheme $Z$ associated to $\psi$ (and having $\psi (U)$ as underlying space) exists, and $\psi$ factors as

  ψ : U ─ψ'→ X' ─h→ X ×_S P                                                 (5.6.1.2)

where ${\psi }^{\prime }$ is an open immersion and $h$ a closed immersion (I, 9.5.10). Let $q_1:X{\times }_{S}P\to X$ and $q_2:X{\times }_{S}P\to P$ be the canonical projections; we set

  f : X' ─h→ X ×_S P ─q_1→ X,                                               (5.6.1.3)

  g : X' ─h→ X ×_S P ─q_2→ P.                                               (5.6.1.4)

We will see that $X^{\prime }$ and $f$ answer the question.

C) Let us first show that $f$ is projective and surjective, and that the restriction of $f$ to $U^{\prime }=f^{-1}(U)$ is an isomorphism of $U^{\prime }$ onto $U$. Since the $P_k$ are projective over $S$, so is $P$ (5.5.5, (iv)), hence $X{\times }_{S}P$ is projective over $X$ (5.5.5, (iii)), and so is $X^{\prime }$, which is a closed subprescheme of $X{\times }_{S}P$. Furthermore, $f\circ {\psi }^{\prime }=q_1\circ (h\circ {\psi }^{\prime })=q_1\circ \psi =j$, hence $f(X^{\prime })$ contains the open everywhere-dense subset $U$ of $X$; but $f$ is a closed morphism (5.5.3), hence $f(X^{\prime })=X$. Note now that $q_1^{-1}(U)=U{\times }_{S}P$ is induced on an open subset of $X{\times }_{S}P$, and by definition the prescheme $U^{\prime }=h^{-1}(U{\times }_{S}P)$ is induced by $X^{\prime }$ on the open subset $U^{\prime }$; it is therefore the closure with respect

to $U{\times }_{S}P$ of the prescheme $Z$ (I, 9.5.8). But the immersion $\psi$ factors as $U\stackrel{{\Gamma }_{\varphi }}{\to }U{\times }_{S}P\stackrel{j\times 1}{\to }X{\times }_{S}P$, and since $P$ is separated over $S$, the graph morphism ${\Gamma }_{\varphi }$ is a closed immersion (I, 5.4.3), so $Z$ is a closed subprescheme of $U{\times }_{S}P$, whence $U^{\prime }=Z$. Since $\psi$ is an immersion, the restriction of $f$ to $U^{\prime }$ is an isomorphism onto $U$, the inverse of ${\psi }^{\prime }$; finally, by the definition of $X^{\prime }$, $U^{\prime }$ is dense in $X^{\prime }$.

D) Let us now prove that $g$ is an immersion, which will establish that $X^{\prime }$ is quasi-projective over $S$, since $P$ is projective over $S$. Set

  V_k  = φ_k(U_k)     (open subset of P_k)
  W_k  = p_k⁻¹(V_k)   (open subset of P)
  U'_k = f⁻¹(U_k)     (open subset of X')
  U''_k = g⁻¹(W_k)    (open subset of X').

It is clear that the $U_k^{\prime }$ form an open covering of $X^{\prime }$; we will first see that the $U_k^{\prime \prime }$ likewise do, by proving that $U_k^{\prime }\subset U_k^{\prime \prime }$. For this it suffices to show that the diagram

              g|U'_k
   U'_k ─────────────→ P
    │                  │
 f|U'_k                │ p_k                                                (5.6.1.5)
    ↓                  ↓
   U_k ─────────────→ P_k
              φ_k

commutes. Now the prescheme $U_k^{\prime }=h^{-1}(U_k{\times }_{S}P)$ is induced by $X^{\prime }$ on the open subset $U_k^{\prime }$, hence is the closure of $Z=U^{\prime }\subset U_k^{\prime }$ with respect to $U_k^{\prime }$ (I, 9.5.8). To show the commutativity of (5.6.1.5), it therefore suffices (since $P_k$ is an $S$-scheme) to show that composing this diagram with the canonical injection $U^{\prime }\to U_k^{\prime }$ (or equivalently, in view of the isomorphism between $U^{\prime }$ and $U$, with $\psi$), we obtain a commutative diagram (I, 9.5.6). But by definition the diagram one then obtains is precisely (5.6.1.1), whence our assertion.

The $W_k$ therefore form an open covering of $g(X^{\prime })$; to prove that $g$ is an immersion, it suffices to show that each restriction $g|U_k^{\prime \prime }$ is an immersion into $W_k$ (I, 4.2.4). For this, consider the morphism

  u_k : W_k ─p_k→ V_k ─φ_k⁻¹→ U_k → X;

since $X$ is separated over $S$, the graph morphism ${\Gamma }_{u_k}:W_k\to X{\times }_S{W}_k$ is a closed immersion (I, 5.4.3), hence the graph $T_k={\Gamma }_{u_k}(W_k)$ is a closed subprescheme of $X{\times }_S{W}_k$; if we show that this subprescheme majorizes $U^{\prime }$, it will also majorize the subprescheme induced by $X^{\prime }$ on the open subset $U_k^{\prime \prime }$ of $X^{\prime }$, by (I, 9.5.8). Since the restriction of $q_2$ to $T_k$ is an isomorphism onto $W_k$, the restriction of $g$ to $U_k^{\prime \prime }$ will be an immersion into $W_k$, and our assertion will be proved. Let $v_k$ be the canonical injection $U^{\prime }\to X{\times }_S{W}_k$; we must show that there exists a morphism $w_k:U^{\prime }\to W_k$ such that $v_k={\Gamma }_{u_k}\circ w_k$. By the definition of the product, it suffices to prove that $q_1\circ v_k=u_k\circ q_2\circ v_k$ (I, 3.2.1), or, composing on the right

with the isomorphism ${\psi }^{\prime }:U\to U^{\prime }$, that $q_1\circ \psi =u_k\circ q_2\circ \psi$. But since $q_1\circ \psi =j$ and $q_2\circ \psi =\varphi$, our assertion follows from the commutativity of (5.6.1.1), taking into account the definition of $u_k$.

E) It is clear that since $U$, and consequently $U^{\prime }$, is irreducible, so is $X^{\prime }$ in the preceding construction, and the morphism $f$ is therefore birational (I, 2.2.9). If in addition $X$ is reduced, so is $U^{\prime }$, hence $X^{\prime }$ is also reduced (I, 9.5.9). This completes the proof.

Corollary.

Suppose one of the hypotheses (a), (b) of (5.6.1) is satisfied. For $X$ to be proper over $S$, it is necessary and sufficient that there exist a scheme $X^{\prime }$ projective over $S$ and an $S$-morphism surjective $f:X^{\prime }\to X$ (which is then projective by (5.5.5, (v))). When this is so, one may furthermore choose $f$ so that there exists a dense open $U$ of $X$ such that the restriction of $f$ to $f^{-1}(U)$ is an isomorphism $f^{-1}(U)\stackrel{\sim }{\to }U$ and $f^{-1}(U)$ is dense in $X^{\prime }$. If furthermore $X$ is irreducible (resp. reduced), one may suppose that $X^{\prime }$ is too; when $X$ and $X^{\prime }$ are irreducible, $f$ is a birational morphism.

Proof. The condition is sufficient by virtue of (5.5.3) and (5.4.3, (ii)). It is necessary, because with the notation of (5.6.1), if $X$ is proper over $S$, then $X^{\prime }$ is proper over $S$, being projective over $X$ hence proper over $X$ (5.5.3), and our assertion follows from (5.4.2, (ii)); furthermore, since $X^{\prime }$ is quasi-projective over $S$, it is projective over $S$ by virtue of (5.5.3).

Corollary.

Let $S$ be a locally Noetherian prescheme, $X$ an $S$-scheme of finite type over $S$, $f_0:X\to S$ its structure morphism. For $X$ to be proper over $S$, it is necessary and sufficient that for every morphism of finite type $S^{\prime }\to S$, $(f_0{)}_{(S^{\prime })}:X_{(S^{\prime })}\to S^{\prime }$ be a closed morphism. It suffices even for this condition to be satisfied for every $S$-prescheme of the form $S^{\prime }=S{\otimes }_{\mathbb{Z}}\mathbb{Z}[T_1,\cdots ,T_n]$ ($T_i$ indeterminates).

Proof. The condition being obviously necessary, let us show that it is sufficient. The question being local on $S$ and $S^{\prime }$ (5.4.1), we may suppose $S$ and $S^{\prime }$ affine and Noetherian. By Chow's lemma, there exist an $S$-scheme $P$ projective, an immersion $j:X^{\prime }\to P$, and a projective surjective morphism $f:X^{\prime }\to X$, such that the diagram

    X ←─f── X'
    │       │
 f_0│       │ j
    ↓       ↓
    S ←─r── P

commutes. Since $P$ is of finite type over $S$, the first hypothesis implies that the projection $q_2:X{\times }_{S}P\to P$ is a closed morphism. But the immersion $j$ is the composition of $q_2$ and the morphism $f\times 1$ from $X^{\prime }{\times }_{S}P$ to $X{\times }_{S}P$; now $f$, being projective, is proper (5.5.3), hence $f\times 1$ is closed. We conclude that $j$ is a closed immersion, hence proper (5.4.2, (i)). Furthermore, the structure morphism $r:P\to S$ is projective, hence proper (5.5.3), so $f_0\circ f=r\circ j$ is proper (5.4.2, (ii)); finally, since $f$ is surjective, $f_0$ is proper by virtue of (5.4.3).

To prove the proposition using only the second hypothesis, it suffices to show that it implies the first. Now, if $S^{\prime }$ is affine and of finite type over $S=\mathrm{Spec}(A)$,

then $S^{\prime }=\mathrm{Spec}(A[c_1,\cdots ,c_n])$ (I, 6.3.3), and so $S^{\prime }$ is isomorphic to a closed subprescheme of $S^{\prime \prime }=\mathrm{Spec}(A[T_1,\cdots ,T_n])$ ($T_i$ indeterminates). In the commutative diagram

                  1_X × j
   X ×_S S' ─────────────→ X ×_S S''
       │                         │
(f_0)_{(S')}              (f_0)_{(S'')}
       ↓                         ↓
       S' ─────────────────→ S''
                    j

$j$ and $1_X\times j$ are closed immersions (I, 4.3.1) and $(f_0{)}_{(S^{\prime })}$ is closed by hypothesis; so it is also for $(f_0{)}_{(S^{\prime \prime })}$.