Exposé V. Construction of quotient schemes

by P. Gabriel

The aim of this Exposé is to prove the theorems stated in TDTE III.¹ If $X$ and $T$ are two objects of a category $C$, we write $X(T)$ instead of ${\mathrm{Hom}}_C(T,X)$. Similarly, if $\varphi :Y\to X$ is an arrow (resp. an object $T$) of $C$, then $\varphi (T)$ denotes the map $g\mapsto \varphi \circ g$ from $Y(T)$ to $X(T)$:

        T
       / \
      g   φ ∘ g
     /     \
    Y ─ φ → X,

and $T(\varphi )$ denotes the map $g\mapsto g\circ \varphi$ from $T(X)$ to $T(Y)$:

    Y ─ φ → X
     \     /
      g ∘ φ   g
       \   /
        T.

Finally, if $P$ is a scheme, we write $P$ for the underlying set of $P$.

Exceptionally, in the present Exposé we do not follow the convention stated in IV 4.6.15 on the notation for quotients (loc. cit., top of page 227 of the original), since we wish to give here a construction of quotients which also applies to "pre-equivalence relations"² that are not equivalence relations.

1. $C$-groupoids

a) Let $C$ be a category in which products and fiber products exist. Recall first that a diagram

        d₁       p
   X₁ ⇉ X₀ → Y
        d₀

in $C$ is said to be exact if $p{d}_0=p{d}_1$ and if, for every $T\in C$, $T(p)$ is a bijection of $T(Y)$ onto the subset of $T(X_0)$ consisting of arrows $f:X_0\to T$ such that $f{d}_0=f{d}_1$. One also says that $(Y,p)$ is the cokernel of $(d_0,d_1)$ and writes

(Y, p) = Coker(d₀, d₁).

b) Let, for example, $C$ be the category (Esp.An) of ringed spaces. In this case, there always exists a cokernel $(Y,p)$, which can be described as follows: the underlying topological space of $Y$ is obtained from $X_0$ by identifying the points $d_0(x)$ and $d_1(x)$ and endowing $Y$ with the quotient topology. The canonical map $\pi :X_0\to Y$ together with $d_0,d_1$ then induces a double arrow of sheaves of rings on $Y$:

                       δ₀
    π_∗(O₀) ⇉ π_∗(d_{0∗} O₁) = π_∗(d_{1∗} O₁),
                       δ₁

where $O_i$ is the structure sheaf of $X_i$. We choose for the sheaf of rings on $Y$ the subsheaf of ${\pi }_{\ast }(O_0)$ whose sections $s$ satisfy ${\delta }_0(s)={\delta }_1(s)$. The arrow $p$ is defined in the evident way.

Let³ X₁ ⇉ X₀ be a diagram (with arrows $d_0,d_1$) in (Esp.An) and let $(Y,p)$ be its cokernel. We say that an open set $U$ of $X_0$ is saturated if $d_0^{-1}(U)=d_1^{-1}(U)$, which is equivalent to saying that $U=p^{-1}(p(U))$. In this case, since $Y$ is endowed with the quotient topology, $p(U)$ is an open subset of $Y$.

Lemma 1.1. Let $U$ be a saturated open set of $X$ and $V=p(U)$. If we denote by $U_1$ the open set $d_0^{-1}(U)=d_1^{-1}(U)$ of $X_1$, and by $d_0$, $d_1$, and $p$ the restrictions of $d_0,d_1$ to $U_1$, and of $p$ to $U$, then $(V,p)$ is a cokernel in (Esp.An) of:⁴

         d̃₁      p̃
   U₁ ⇉ U → V.
         d̃₀

The verification is straightforward.

Lemma 1.2. Let X₁ ⇉ X₀ be a diagram in (Sch) (with arrows $d_0,d_1$) and let $(Y,p)$ be its cokernel in (Esp.An).

(i) If $Y$ is a scheme and $p$ a morphism of schemes, then $(Y,p)$ is a cokernel of $(d_0,d_1)$ in (Sch).

(ii) Suppose that every point of $X_0$ possesses a saturated open neighborhood $U$ such that, denoting by $d_0$ and $d_1$ the restrictions of $d_0$ and $d_1$ to $d_0^{-1}(U)=d_1^{-1}(U)$, and by $(Q,q)$ the cokernel of $(d_0,d_1)$ in (Esp.An), the space $Q$ is a scheme and $q$ a morphism of schemes. Then $(Y,p)$ is a cokernel of $(d_0,d_1)$ in (Sch).

(i) is proved in § 4.c); since the proof is short, let us repeat it here. Let $f:X_0\to Z$ be a morphism of schemes such that $f{d}_0=f{d}_1$. By hypothesis, there is a unique morphism of ringed spaces $r:Y\to Z$ such that $f=rp$. It remains to show that, for every $y\in Y$, the homomorphism $O_{r(y)}\to O_y$ induced by $r$ is local. This follows from the fact that $p$ is surjective, so that $y$ is of the form $p(x)$, and from the fact that the homomorphism $O_{f(x)}\to O_x$ induced by $f$ is local.

(ii) follows from (i) and the preceding lemma.

c) In this Exposé we study the existence of $Coker(d_0,d_1)$ when the double arrow $(d_0,d_1)$ is inserted in a richer context; more precisely, let $X_2=X_1{\times }_{d_1,d_0}{X}_1$ denote the fiber product of the diagram

$$X_1\downarrow d_1(\ast )X_0\uparrow d_0{X}_1,$$

and let $d_0^{\prime }$ and $d_2^{\prime }$ be the two canonical projections of $X_2$ onto $X_1$; one then has by definition a Cartesian square

                  d′₀
            X₂ ─────→ X₁
            │          │
        d′₂ │          │ d₁
            ↓          ↓
            X₁ ─────→ X₀.
                  d₀
(0)

Moreover, let us give ourselves a third arrow $d_1^{\prime }:X_2\to X_1$; we say that (d₀, d₁ : X₁ ⇉ X₀, d′₁) is a $C$-groupoid if for every object $T$ of $C$, $X_1(T)$ is the set of arrows of a groupoid $X\ast (T)$ whose set of objects is $X_0(T)$, with source map $d_1(T)$, target map $d_0(T)$, and composition map $d_1^{\prime }(T)$ (one identifies, as usual, $(X_1{\times }_{d_1,d_0}{X}_1)(T)$ with $X_1(T){\times }_{d_1(T),d_0(T)}{X}_1(T)$; we also recall that a groupoid is a category in which every arrow is invertible).⁵

If $\varphi$ is an arrow of the groupoid $X\ast (T)$, the map $f\mapsto \varphi \circ f$ is a bijection from the set of arrows $f$ whose target coincides with the source of $\varphi$ onto the set of arrows having the same target as $\varphi$. One sees easily that this fact can be translated by saying that the square

                  d′₁
            X₂ ─────→ X₁
            │          │
        d′₀ │          │ d₀
            ↓          ↓
            X₁ ─────→ X₀
                  d₀
(1)

is Cartesian.

Similarly, the map $g\mapsto g\circ \varphi$ is a bijection from the set of arrows $g$ of $X\ast (T)$ having source equal to the target of $\varphi$ onto the set of arrows having the same source as $\varphi$. This fact can again be translated by saying that the square

                  d′₁
            X₂ ─────→ X₁
            │          │
        d′₂ │          │ d₁
            ↓          ↓
            X₁ ─────→ X₀
                  d₁
(2)

is Cartesian.

On the other hand, let $s:X_0\to X_1$ be the unique arrow of $C$ such that, for every $T$, $s(T):X_0(T)\to X_1(T)$ associates to every object of $X\ast (T)$ the identity arrow of that object.⁶ The arrow $s$ satisfies the equalities

(3)        d₁ s = id_{X₀},
(3 bis)    d₀ s = id_{X₀}.

Finally, the associativity of the composition maps $d_1^{\prime }(T)$ translates into the commutativity of the diagram

                              d′₁ × X₁
   X₁ ×_{d₁, d₀} X₁ ×_{d₁, d₀} X₁ ─────────→ X₁ ×_{d₁, d₀} X₁
            │                                       │
   X₁ × d′₁ │                                       │ d′₁
            ↓                                       ↓
   X₁ ×_{d₁, d₀} X₁ ──────────d′₁──────────────→ X₁.
(4)

Conversely, the conditions (1), (2), and (4) together with the existence of an arrow $s$ satisfying (3) imply that (X₁ ⇉ X₀, d′₁) is a $C$-groupoid. The condition (3) is harmless; it merely ensures that the map $d_1(T):X_1(T)\to X_0(T)$ is surjective for every $T\in C$. In what follows we shall mostly make use of the Cartesian squares (0), (1) and (2), which we summarize in the diagram

                  d′₁              d₀
            X₂  ────→  X₁  ──────→  X₀
                  d′₀
        d′₂ │           │ d₁
            ↓           ↓
            X₁  ────→  X₀
                  d₀
(0,1,2)

In this diagram the two left-hand squares (i.e. the squares (0) and (2)) are Cartesian; the first row is exact, and $X_2$ is identified with the fiber product $X_1{\times }_{d_0,d_0}{X}_1$.

We use associativity only indirectly, for instance to ensure the existence of an arrow $s$ satisfying (3) and (3 bis), or else to ensure the existence of an arrow

(†)   σ : X₁ → X₁    such that    d₀ σ = d₁    and    d₁ σ = d₀

(one chooses $\sigma$ so that $\sigma (T):X_1(T)\to X_1(T)$ sends every arrow of $X\ast (T)$ to its inverse).⁷

By abuse of language, we shall sometimes call a $C$-groupoid a diagram

        d′₀, d′₁, d′₂      d₀, d₁
   X₂ ⇶ X₁ ⇉ X₀

such that (0), (1) and (2) are Cartesian, (4) is commutative, and there exists $s$ satisfying (3). The object $X_2$ will therefore be allowed to be "a" fiber product of (∗) without being "the" fiber product of (∗).⁸

Terminology. Instead of $C$-groupoid $X\ast$, we shall also speak of the groupoid $X\ast$ with base $X_0$, or of the pre-equivalence relation $X\ast$ in $X_0$.

2. Examples of $C$-groupoids

a) Let $X$ be an object of $C$ and $G$ a $C$-group acting on the left on $X$. We denote by $d_0:G\times X\to X$ the arrow defining the action of $G$ on $X$, by $d_1:G\times X\to X$ the projection of the product onto the second factor, by $\mu :G\times G\to G$ the arrow defining the $C$-group structure of $G$, and finally by $p{r}_{2,3}$ the projection of $G\times G\times X=G\times (G\times X)$ onto the second factor. Then

                  pr_{2,3}              d₁
   G × G × X      ⇉      G × X         ⇉   X
                  μ × X                 d₀
                  G × d₀

is a $C$-groupoid.

b) Let $d_0,d_1:X_1\to X_0$ be an equivalence pair, i.e., if $d_0⊠d_1:X_1\to X_0\times X_0$ is the arrow with components $d_0$ and $d_1$, we suppose that $(d_0⊠d_1)(T)$ is, for every object $T$ of $C$, a bijection of $X_1(T)$ onto the graph of an equivalence relation on $X_0(T)$. The set $X_1(T)$ therefore identifies with the set of pairs $(x,y)$ of elements of $X_0(T)$ such that $x\sim y$; similarly, the set $X_2(T)=(X_1{\times }_{d_1,d_0}{X}_1)(T)$ identifies with the set of triples $(x,y,z)$ of elements of $X_0(T)$ such that $x\sim y$ and $y\sim z$. There is therefore one and only one arrow $d_1^{\prime }:X_2\to X_1$ making the squares (1) and (2) commute: $d_1^{\prime }(T)$ must send $(x,y,z)\in X_2(T)$ to $(x,z)\in X_1(T)$. For this choice of $d_1^{\prime }$, (d₀, d₁ : X₁ ⇉ X₀, d′₁) is a $C$-groupoid.

Conversely, consider a $C$-groupoid $X\ast$ such that $d_0⊠d_1:X_1\to X_0\times X_0$ is a monomorphism. Then $(d_0,d_1)$ is an equivalence pair and $X\ast$ can be reconstructed from $(d_0,d_1)$ as explained a few lines above.⁹

c) If $p:X\to Y$ is any arrow of $C$ and if $p{r}_1$ and $p{r}_2$ are the two projections of $X{\times }_{p,p}X$ onto $X$, then (pr₁, pr₂) : X ×_{p, p} X ⇉ X is an equivalence pair. One says that $p$ is an effective epimorphism if the diagram

                  pr₁           p
   X ×_{p, p} X  ⇉  X  ────→  Y
                  pr₂

is exact, that is, if $(Y,p)=Coker(p{r}_1,p{r}_2)$.

Let, for example, $S$ be a Noetherian scheme and let $C$ be the category of schemes finite over $S$. Let us show that an epimorphism in $C$ is not necessarily effective: take $S$ equal to $\mathrm{Spec}k[T^3,T^5]$, where $k$ is a commutative field, $Y$ equal to $S$, and $X$ equal to $\mathrm{Spec}k[T]$. If $i$ is the inclusion of $B=k[T^3,T^5]$ into $A=k[T]$, take $p$ equal to $\mathrm{Spec}i$. In this case $X{\times }_{p,p}X$ identifies with $\mathrm{Spec}(A{\otimes }_{B}A)$ and $Coker(p{r}_1,p{r}_2)$ with $\mathrm{Spec}{B}^{\prime }$, where $B^{\prime }$ is the subring of $A$ consisting of $a$ such that $a{\otimes }_{B}1=1{\otimes }_{B}a$. Now

T⁷ ⊗_B 1 = (T² T⁵) ⊗_B 1 = T² ⊗_B T⁵ = T² ⊗_B (T³ T²) = T⁵ ⊗_B T² = 1 ⊗_B T⁷.

So $T^7$ belongs to $B^{\prime }$, does not belong to $B$, and $\mathrm{Spec}{B}^{\prime }$ is distinct from $\mathrm{Spec}B$, which yields the counterexample.¹⁰

3. Some sorites on $C$-groupoids

Here, in no particular order, are some remarks used in what follows:

a) Let

        d′₀, d′₁, d′₂      d₀, d₁
   X₂ ⇶ X₁ ⇉ X₀

be a $C$-groupoid and $f_0:Y_0\to X_0$ an arrow of $C$. We shall define a $C$-groupoid with base $Y_0$

        e′₀, e′₁, e′₂      e₀, e₁
   Y₂ ⇶ Y₁ ⇉ Y₀

which we shall call induced by $X\ast$ and $f_0$. One also says that $Y\ast$ is the inverse image of $X\ast$ under the base change $f_0$.

We choose for $Y_1$ the fiber product of the diagram

                 f₁
   Y₁ ─────────→ X₁
   │              │
   │              │ d₀ ⊠ d₁
   ↓              ↓
   Y₀ × Y₀ ────→ X₀ × X₀,
        f₀ × f₀

and for $e_0$ and $e_1$ the arrows obtained by composing the canonical arrow $Y_1\to Y_0\times Y_0$ with the first and second projections of $Y_0\times Y_0$. The morphism $Y_1\to Y_0\times Y_0$ is then $e_0⊠e_1$, and one has $f_0\circ e_i=d_i\circ f_1$ for $i=0,1$, where we have written $f_1$ for the projection of $Y_1$ onto $X_1$.

We set $Y_2=Y_1{\times }_{e_0,e_1}{Y}_1$, cf. 1.c). One can say that the pair $(e_0,e_1)$ is defined in such a way that, for every $T\in C$ and every pair $(y,x)$ of elements of $Y_0(T)$, there is a certain one-to-one correspondence $\psi \mapsto {}_y{\psi }_x$ between the arrows $\psi$ of $X\ast (T)$ with source $f_0(x)$ and target $f_0(y)$ and the arrows ${}_y{\psi }_x$ of $Y\ast (T)$ with source $x$ and target $y$. One therefore determines $e_1^{\prime }:Y_2\to Y_1$ by defining, for every $T\in C$, the composition of arrows of $Y\ast (T)$ by the formula

   _z ψ_y ∘ _y φ_x = _z (ψ ∘ φ)_x.

It is clear that this definition makes each $Y\ast (T)$ into a groupoid.

b) Knowing the $C$-groupoid $X\ast$ and the base change $f_0:Y_0\to X_0$, one can reconstruct the pair (e₀, e₁) : Y₁ ⇉ Y₀ in another way:¹¹ construct $Y_0{\times }_{X_0}{X}_1$, $p{r}_1$ and $p{r}_2$ so that the square

                       pr₂
   Y₀ ×_{X₀} X₁ ─────→ X₁
        │              │
    pr₁ │              │ d₀
        ↓              ↓
        Y₀ ──────────→ X₀
                  f₀

is Cartesian. One then verifies without difficulty, by reduction to the set-theoretic case, that one has the Cartesian square

              e₀ ⊠ f₁
   Y₁ ─────────────→ Y₀ ×_{X₀} X₁
   │                       │
e₁ │                       │ d₁ ∘ pr₂
   ↓                       ↓
   Y₀ ────────────────→ X₀,
              f₀

where $f_1$ denotes the canonical projection of $Y_1=(Y_0\times Y_0){\times }_{(X_0\times X_0)}{X}_1$ onto $X_1$.

c) We shall give two examples of inverse images of a $C$-groupoid. Take $Y_0$ equal to $X_1$, $f_0$ equal to $d_0$. For every object $T$ of $C$, $Y_1(T)$ then identifies with the set of diagrams of the form

        φ
   b ────→ d
   ↑        ↑
   f        g
   │        │
   a        c

of $X\ast (T)$. The source of such a diagram is the arrow $f$, the target is the arrow $g$. These diagrams compose in the evident way.

Now put $Y_0^{\prime }=X_1$, $f_0^{\prime }=d_1$ (we add the primes¹² to avoid any confusion with the preceding example). In this case, $Y_1^{\prime }(T)$ identifies, for every $T\in C$, with the set of diagrams of the form

   b        d
   ↑        ↑
   f        g
   │   ψ   │
   a ────→ c

of the groupoid $X\ast (T)$. The source of such a diagram is $f$, the target is $g$; the composition of these diagrams is evident.

This being so, it is clear that the identity map of $Y_0(T)$ and the map

        φ                       
   b ────→ d           b        d
   ↑        ↑          ↑        ↑
   f        g    ↦     f        g
   │        │          │  g⁻¹φf │
   a        c          a ─────→ c

from $Y_1(T)$ to $Y_1^{\prime }(T)$ define an isomorphism of the groupoid $Y\ast (T)$ onto $Y^{\prime }\ast (T)$. Moreover, this isomorphism depends functorially on $T$, so that the $C$-groupoids $Y\ast$ and $Y^{\prime }\ast$ are isomorphic.¹³

d)

Proposition 3.1. We keep the notations of a) and assume that $f_0$ is an effective and universal epimorphism. Then $Coker(d_0,d_1)$ exists if and only if $Coker(e_0,e_1)$ exists.¹⁴ Moreover, in that case, $f_0$ induces an isomorphism

   Coker(d₀, d₁) ⥲ Coker(e₀, e₁).

Let us first recall that an epimorphism $f_0:Y_0\to X_0$ is said to be universal if, for every Cartesian square

   Y′ ─────→ Y₀
   │         │
f′ │         │ f₀
   ↓         ↓
   X′ ─────→ X₀,

$f^{\prime }$ is an epimorphism. This being so, let us denote by $C(d_0,d_1)$ the covariant functor from $C$ to sets which associates to every $T\in C$ the kernel of the pair T(d₀), T(d₁) : T(X₀) ⇉ T(X₁). We define $C(e_0,e_1)$ similarly. For every $T\in C$, one therefore has a commutative diagram

                           T(d₁)
   C(d₀, d₁)(T) ────→ T(X₀) ⇉ T(X₁)
                           T(d₀)
        │                  │           │
   T(f) │           T(f₀) │           │ T(f₁)
        ↓                  ↓           ↓
                           T(e₁)
   C(e₀, e₁)(T) ────→ T(Y₀) ⇉ T(Y₁),
                           T(e₀)

where $T(f)$ is the injection induced by the injection $T(f_0)$. If we show that $T(f)$ is a surjection for every $T$, we shall have a functorial isomorphism $f:C(d_0,d_1)\stackrel{\sim }{\to }C(e_0,e_1)$, so that the representability of one of these functors will be equivalent to that of the other; this will prove our proposition.

To prove the surjectivity of $T(f)$, consider the diagram

                          f₁
              Y₁ ─────────────→ X₁
            ↗  
        Δ ↗  e₀ │ e₁          d₀ │ d₁
         ↗      ↓                ↓
   Y₀ ×_{X₀} Y₀ ─────→ Y₀ ─────→ X₀,
                   pr₂      f₀
              pr₁

where $\Delta$ is the section of $Y_1\to Y_0\times Y_0$ defined by the morphism $s\circ f_0\circ p{r}_1:Y_0\times Y_0\to X_1$, with $s:X_0\to X_1$ the arrow satisfying equalities (3) and (3 bis) of section 1.

If the arrow $g:Y_0\to T$ is such that $g\circ e_0=g\circ e_1$, then $g\circ e_0\circ \Delta =g\circ e_1\circ \Delta$, so $g\circ p{r}_1=g\circ p{r}_2$. Since $f_0$ is an effective epimorphism, $g$ factors through $f_0$ and an arrow $h:X_0\to T$, that is to say $g=T(f_0)(h)$. It remains to show that $h$ belongs to $C(d_0,d_1)(T)$, i.e. satisfies $h{d}_0=h{d}_1$; now one has

   h d₀ f₁ = h f₀ e₀ = g e₀ = g e₁ = h f₀ e₁ = h d₁ f₁,

whence the desired equality, since $f_1$ is an epimorphism (because $f_0$ is a universal epimorphism).

e) Consider now a scheme $S$ and choose $C$ equal to $(Sch/S)$. The data of a $C$-groupoid

        d′₀, d′₁, d′₂      d₀, d₁
   X₂ ⇶ X₁ ⇉ X₀

allows one to define an equivalence relation on the set $X_0$ underlying the scheme $X_0$: if $x,y\in X_0$, one writes $x\sim y$ when there exists $z\in X_1$ such that $x=d_{1}z$ and $y=d_{0}z$. The reflexivity and symmetry of this relation are evident;¹⁵ let us prove transitivity: if $x\sim y$ and $y\sim z$, there exist $u,v\in X_1$ with $x=d_{1}u$, $y=d_{0}u$, $y=d_{1}v$, $z=d_{0}v$. It follows that $(v,u)$ belongs to the set-theoretic fiber product $X_1{\times }_{d_1,d_0}{X}_1$. Since the canonical map

   X₁ ×_{d₁, d₀} X₁ ⟶ X₁ ×_{d₁, d₀} X₁

from the set underlying the fiber product into the fiber product of the underlying sets is surjective, $(v,u)$ is the image of some $w\in X_2$. One then has $x=d_1{d}_1^{\prime }w$ and $z=d_0{d}_1^{\prime }w$, whence $x\sim z$.

f) We keep the notations of a) and b), $C$ still being $(Sch/S)$. If x, y are points of $Y_0$, we shall see that one has $x\sim y$ if and only if $f_0(x)\sim f_0(y)$ (the inverse image of the equivalence relation defined by a groupoid is the equivalence relation defined by the inverse image of the groupoid).

Indeed, suppose $x\sim y$. There exists therefore $z\in Y_1$ such that $x=e_1(z)$ and $y=e_0(z)$. Since $f_0\circ e_i=d_i\circ f_1$ for $i=0,1$, one then has $f_0(x)=d_1{f}_1(z)$ and $f_0(y)=d_0{f}_1(z)$, whence $f_0(x)\sim f_0(y)$.

Conversely, suppose $f_0(x)\sim f_0(y)$ and let $z\in X_1$ be such that $f_0(y)=d_1(z)$ and $f_0(x)=d_0(z)$. Using the construction and notations of b), there is then a point $t$ of $Y_0{\times }_{X_0}{X}_1$ such that $p{r}_1(t)=x$ and $p{r}_2(t)=z$. Similarly, since $f_0(y)=d_{1}p{r}_2(t)$, there is $s\in Y_1$ such that $y=e_1(s)$ and $(e_0⊠f_1)(s)=t$. One then has $e_0(s)=p{r}_1(e_0⊠f_1)(s)=p{r}_1(t)=x$. Whence $x\sim y$.

4. Passage to the quotient by a finite and flat groupoid (proof of a particular case)

Theorem 4.1. Consider a $(Sch/S)$-groupoid

        d′₀, d′₁, d′₂      d₀, d₁
   X₂ ⇶ X₁ ⇉ X₀.

Suppose the following conditions are satisfied:¹⁶

a) $d_1$ is finite locally free;

b) for every $x\in X_0$, the set $d_0{d}_1^{-1}(x)$ is contained in an affine open of $X_0$.¹⁷

Then:

(i) There exists a cokernel $(Y,p)$ of $(d_0,d_1)$ in $(Sch/S)$; moreover, such a $(Y,p)$ is a cokernel of $(d_0,d_1)$ in the category of all ringed spaces.

(ii) $p$ is integral and open, and $Y$ is affine if $X_0$ is affine.¹⁸

(iii) The morphism $X_1\to X_0{\times }_Y{X}_0$ with components $d_0$ and $d_1$ is surjective.

(iv) If $(d_0,d_1)$ is an equivalence pair, then $X_1\to X_0{\times }_Y{X}_0$ is an isomorphism¹⁹ and $p:X_0\to Y$ is finite locally free.²⁰ Moreover, $(Y,p)$ is a cokernel of $(d_0,d_1)$ in the category of sheaves for the (fppf) topology and, for every base change $Y^{\prime }\to Y$, $Y^{\prime }$ is the cokernel of the groupoid $X\ast {\times }_Y{Y}^{\prime }$ obtained from $X\ast$ by the base change $X_0{\times }_Y{Y}^{\prime }\to X_0$.

In particular, for every base change $S^{\prime }\to S$, $Y^{\prime }=Y{\times }_S{S}^{\prime }$ is the cokernel of the $S^{\prime }$-groupoid $X^{\prime }\ast =X\ast {\times }_S{S}^{\prime }$. So, in this case, "the formation of the quotient commutes with base change".

It evidently follows from (i) that the topological space underlying $Y$ is the quotient of the topological space underlying $X_0$ by the equivalence relation defined by the $(Sch/S)$-groupoid $X\ast$.

We shall first prove this theorem when $X_0$ is affine and $d_1$ is locally free of constant rank $n$. We shall see next how to reduce to this particular case.

In the case in which we have placed ourselves, $X_0$, $X_1$ and $X_2$ are affine. We can therefore suppose that

   X_i = Spec A_i,   d_j = Spec δ_j,   d′_k = Spec δ′_k,

the $A_i$ being commutative rings and the ${\delta }_j$, ${\delta }_k^{\prime }$ ring homomorphisms. One can then replace the diagram (0, 1, 2) by the following

                    δ′₁
            A₂ ⇇ A₁ ⇇ A₀
                    δ′₀         δ₀
   (0,1,2)∗   δ′₂        δ₁
                    δ₁
            A₁ ⇇ A₀,
                    δ₀

where the two left-hand squares are cocartesian.

Let $B$ denote the subring of $A_0$ consisting of those $a\in A_0$ such that ${\delta }_0(a)={\delta }_1(a)$.

a) Let us show that $A_0$ is integral over $B$. If $a$ belongs to $A_0$, let

   P_{δ₁}(T, δ₀(a)) = Tⁿ − σ₁ T^{n−1} + ⋯ + (−1)ⁿ σ_n

be the characteristic polynomial of ${\delta }_0(a)$ when $A_1$ is regarded as an algebra over $A_0$ via the homomorphism ${\delta }_1$ (cf. Bourbaki, Alg. VIII, § 12 and Alg. comm. II, § 5, exercise 9). Since the left-hand squares of $(0,1,2)\ast$ are cocartesian, one has

   δ₀(P_{δ₁}(T, δ₀(a))) = P_{δ′₂}(T, δ′₀ δ₀(a))

and

   δ₁(P_{δ₁}(T, δ₀(a))) = P_{δ′₂}(T, δ′₁ δ₀(a)).

Since ${\delta }_0^{\prime }{\delta }_0={\delta }_1^{\prime }{\delta }_0$, one has

   δ₀(P_{δ₁}(T, δ₀(a))) = δ₁(P_{δ₁}(T, δ₀(a))),

that is, ${\delta }_0({\sigma }_i)={\delta }_1({\sigma }_i)$ for every $i$. Hamilton–Cayley moreover tells us that

   δ₀(a)ⁿ − δ₁(σ₁) δ₀(a)^{n−1} + ⋯ + (−1)ⁿ δ₁(σ_n) = 0.

Since ${\delta }_1({\sigma }_i)={\delta }_0({\sigma }_i)$, one also has

   δ₀(a)ⁿ − δ₀(σ₁) δ₀(a)^{n−1} + ⋯ + (−1)ⁿ δ₀(σ_n) = 0,

whence

   aⁿ − σ₁ a^{n−1} + ⋯ + (−1)ⁿ σ_n = 0,

since there exists a homomorphism $\tau :A_1\to A_0$ such that $\tau {\delta }_0=i{d}_{A_0}$, hence ${\delta }_0$ is injective. It follows that $A_0$ is integral over $B$.

b) Consider now two prime ideals $x$ and $y$ of $A_0$. We shall show that the equality $x\cap B=y\cap B$ entails the existence of a prime ideal $z$ of $A_1$ such that $x=d_0(z)$ and $y=d_1(z)$.

Indeed, if the assertion were not true, $x$ would be distinct from ${\delta }_0^{-1}(t)$ for every prime ideal $t$ of $A_1$ such that ${\delta }_1^{-1}(t)=y$. For such a $t$ one would have ${\delta }_0^{-1}(t)\cap B={\delta }_1^{-1}(t)\cap B=y\cap B=x\cap B$, whence by Cohen–Seidenberg (cf. Bourbaki, Alg. comm. V, § 2, cor. 1 of prop. 1) $x$ would be contained in no ${\delta }_0^{-1}(t)$.²¹ Now there are at most $n$ prime ideals $t$ of $A_1$ such that ${\delta }_1^{-1}(t)=y$ (cf. loc. cit., prop. 3), so, by the "Prime Avoidance Lemma" (loc. cit., II, § 1, prop. 3), there would exist $a\in x$ belonging to no ${\delta }_0^{-1}(t)$. Consequently, ${\delta }_0(a)$ would belong to none of these ideals $t$, and so, by the lemma below, the norm $N_{{\delta }_1}({\delta }_0(a))$ would not belong to $y$ (one computes this norm by regarding $A_1$ as an algebra over $A_0$ via the homomorphism ${\delta }_1$; one has $N_{{\delta }_1}({\delta }_0(a))={\sigma }_n$ with the notations of a)). But, since $(-1{)}^{n-1}{\sigma }_n=a^n+{\sum }_{i=1}^{n-1}(-1{)}^i{\sigma }_i{a}^{n-i}$, this norm belongs to $B\cap x=B\cap y$, whence the contradiction.

Lemma 4.1.1. Let $A\to A^{\prime }$ be a morphism of commutative rings making $A^{\prime }$ into a projective $A$-module of rank $n$. Let $p\in \mathrm{Spec}(A)$, $q_1,\cdots ,q_r$ the elements of $\mathrm{Spec}(A^{\prime })$ above $p$, and $a\in A^{\prime }$. Then $a$ belongs to $q_1\cup \cdots \cup q_r$ if and only if its norm $N(a)$ belongs to $p$.

Indeed, replacing $A$ and $A^{\prime }$ by the localizations $A_p$ and $A_p^{\prime }$, we reduce to the case where $(A,p)$ is local and $A^{\prime }$ is semilocal, with $\mathrm{Spec}(A^{\prime })=q_1,\cdots ,q_r$. In this case, $A^{\prime }$ is a free $A$-module of rank $n$ (cf. Bourbaki, Alg. comm. II, § 3.2, cor. 2 of prop. 5), and $N(a)$ is the determinant of the endomorphism ${\ell }_a:a^{\prime }\mapsto a{a}^{\prime }$ of $A^{\prime }$, so one has the equivalences

   N(a) ∉ p ⟺ N(a) invertible ⟺ ℓ_a invertible ⟺ a ∉ q₁ ∪ ⋯ ∪ q_r.

c) Proof of (i):

Set $Y=\mathrm{Spec}B$ and $p=\mathrm{Spec}i$, where $i$ is the inclusion of $B$ into $A_0$. By a), the morphism $p:X_0\to Y$ is surjective. Let us first show that $(Y,p)$ is a cokernel of $(d_0,d_1)$ in the category of all ringed spaces: it follows indeed from b) that the set underlying $\mathrm{Spec}B$ is obtained from the set underlying $X_0$ by identifying the points $x$ and $y$ such that there exists $z\in X_1$ with $d_{1}z=y$, $d_{0}z=x$. Moreover, since $i$ is integral, $p=\mathrm{Spec}i$ is closed, so $Y$ is endowed with the quotient topology of that of $X_0$. It follows that $p$ is open. Indeed, let $U^{\prime }$ be any open of $X_0$; since $d_1$ is surjective and finite locally free, hence faithfully flat and of finite presentation, and therefore open, the saturation $U=d_1(d_0^{-1}(U^{\prime }))$ of $U^{\prime }$ for the equivalence relation defined by $X\ast$ is open. Then $p(U^{\prime })=p(U)$ is open, since $Y$ is endowed with the quotient topology.

It follows finally from the choice of $B$ and from the fact that $p$, $d_0$ and $d_1$ are affine that the canonical sequence of sheaves of rings

                       p_∗(δ₁)
   O_Y ────→ p_∗(O_{X₀}) ⇉ p_∗(d_{0∗}(O_{X₁})) = p_∗(d_{1∗}(O_{X₁}))
                       p_∗(δ₀)

is exact.

It remains to show that $(Y,p)$ is also the cokernel of $(d_0,d_1)$ in the category of schemes (more generally, in the category of ringed spaces in local rings). Let then $q:X_0\to Z$ be a morphism of schemes such that $q{d}_0=q{d}_1$. By what precedes, there is a unique morphism of ringed spaces $r:Y\to Z$ such that $q=rp$. It remains to show that, for every $y\in Y$, the homomorphism $O_{r(y)}\to O_y$ induced by $r$ is local. This follows from the fact that $p$ is surjective, so that $y$ is of the form $p(x)$, and from the fact that the homomorphism $O_{q(x)}\to O_x$ induced by $q$ is local.

d) Proof of (ii): Follows from a) and c).

e) Proof of (iii):

Recall that one denotes by $P$ the set underlying a scheme $P$, and by $d:P\to Q$ the map induced by a morphism $d:P\to Q$.

Lemma 4.1.2.²² Let $(A,m)$ be a local ring, $k$ its residue field, and $K$ an extension of the field $k$. Then there exists a local and flat $A$-algebra $B$ such that $B/mB$ is $k$-isomorphic to $K$; moreover, one can choose $B$ finite and free over $A$ if $K$ is of finite degree over $k$.

This is proved in EGA 0_III, 10.3.1, where it is moreover shown that one can choose $B$ Noetherian if $A$ is. For the reader's convenience, let us indicate the proof.

Put $A^{\prime }=A[T]$, where $T$ is an indeterminate. If $K=k(T)$, let $p=m{A}^{\prime }$ and $B=A_p^{\prime }$. Then $B/mB\cong k[T{]}_{(0)}=k(T)$, and $B$ is flat over $A^{\prime }$, which is a free $A$-module, so $B$ is flat over $A$.

If $K=k(t)=k[t]$, where $t$ is algebraic over $k$, set $B=A^{\prime }/(F)$, where $F\in A^{\prime }$ is a monic polynomial whose image in k[T] is the minimal polynomial $f$ of $t$ over $k$. Then $B$ is a free $A$-module of finite rank $deg(F)=deg(f)$. In particular, $B$ is integral over $A$, hence every maximal ideal of $B$ contains $m$. Since $B/mB\cong k[T]/(f)\cong K$, it follows that $B$ is local, with maximal ideal mB. This already shows that if $[K:k]<\infty$, one can choose $B$ finite and free over $A$.

In the general case, let $(t_i{)}_{i\in I}$ be a system of generators of $K$ over $k$, and endow $I$ with a well-ordering (i.e., a total order $⩽$ such that every non-empty subset of $I$ has a least element). For every $i\in I$, let $k_i$ (resp. $k_{<i}$) denote the subfield of $K$ generated by the $t_j$ for $j⩽i$ (resp. $j<i$). Adding one element if necessary, we may suppose that $I$ has a greatest element $\xi$, so that $K=k_{\xi }$. Consider the subset $J$ of $I$ consisting of indices $i$ such that there exists an inductive system $(A_j{)}_{j⩽i}$ of local and flat $A$-algebras such that $A_j/m{A}_j\cong k_j$ and $A_j$ is flat over $A_{\ell }$ for every $\ell <j$. Suppose $I-J$ non-empty; let $i$ be its least element and let $A^{\prime }={\lim }_{j<i}{A}_j$. Since tensor product commutes with direct limits, $A^{\prime }$ is flat over $A$ and over each $A_j$ for $j<i$, and one has $A^{\prime }/m{A}^{\prime }\cong A^{\prime }{\otimes }_A(A/m)\cong k_{<i}$. Moreover, $A^{\prime }$ is local, with maximal ideal $m{A}^{\prime }$. Indeed, if $x=f_j(x_j)$ is non-invertible, then $x_j$ is not invertible, hence belongs to the maximal ideal $m{A}_j$ of $A_j$, whence $x\in m{A}^{\prime }$. It then follows from the monogenic case treated above that there exists a local and flat $A^{\prime }$-algebra $A_i$ such that $A_i/m{A}_i\cong k_{<i}(t_i)=k_i$; then $A_i$ is flat over each $A_j$ for $j<i$, and so $i\in J$, contrary to hypothesis. This contradiction shows that $J=I$, and so $A_{\xi }$ answers the question. Lemma 4.1.2 is proved.

Let us now prove 4.1 (iii). Recall that one denotes by $P$ the set underlying a scheme $P$, and by $d:P\to Q$ the map induced by a morphism $d:P\to Q$. One can then translate b) by saying that the map

   d₀ ⊠ d₁ : X₁ ⟶ X₀ ×_Y X₀

with components $d_0$ and $d_1$ is surjective; now this map factors as follows

   X₁ ──d₀⊠d₁──→ X₀ × X₀ ──q──→ X₀ ×_Y X₀,
                       (set-theoretic Y-product)

$q$ being the canonical map; the image of $d_0⊠d_1$ therefore contains all points $v$ of $X_0{\times }_Y{X}_0$ such that $v=q^{-1}(q(v))$. This last condition²³ will be realized in particular if $v$ is rational over $Y$, that is to say, if the residue field $\kappa (v)$ of $v$ identifies with the residue field $\kappa (w)$ of the image $w$ of $v$ in $Y$.

If $v\in X_0{\times }_Y{X}_0$ is not rational over $Y$, let $w$ again be the image of $v$ in $Y$. By lemma 4.1.2, there exists a local ring $C$ of radical $m$ and a local and flat homomorphism $f:O_w\to C$ such that $C/m$ is isomorphic to $\kappa (v)$ as a $\kappa (w)$-algebra. If one sets $Y^{\prime }=\mathrm{Spec}C$ and if $\pi :Y^{\prime }\to Y$ is the morphism induced by $f$, it is clear that the canonical projection of $(X_0{\times }_Y{X}_0){\times }_Y{Y}^{\prime }$ to $X_0{\times }_Y{X}_0$ sends to $v$ a point $v^{\prime }$ of $(X_0{\times }_Y{X}_0){\times }_Y{Y}^{\prime }$ which is rational over $Y^{\prime }$. Since

   (X₀ ×_Y X₀) ×_Y Y′ ≅ (X₀ ×_Y Y′) ×_{Y′} (X₀ ×_Y Y′),

and since the hypotheses of theorem 4.1 and the previous results, in particular point b), remain valid after the base change $\pi :Y^{\prime }\to Y$, then $v^{\prime }$ is the image of an element $u^{\prime }\in X_1{\times }_Y{Y}^{\prime }$ by the morphism deduced from $d_0⊠d_1$ by base change. If $u$ is the image of $u^{\prime }$ in $X_1$, one indeed has $v=(d_0⊠d_1)(u)$.

f) Proof of (iv):²⁴

Lemma 4.1.3. If a monomorphism of schemes $f:T\to Z$ is a finite morphism, it is a closed immersion.

Indeed, covering $Z$ by affine opens $Z_i$ and replacing $f$ by the induced morphisms $f^{-1}(Z_i)\to Z_i$, we reduce ($f$ being finite, hence affine) to the case where $Z=\mathrm{Spec}B$ and $T=\mathrm{Spec}A$. Since $f$ is a monomorphism, the diagonal morphism $T\to T{\times }_{Z}T$ is an isomorphism (EGA I, 5.3.8), i.e., $A{\otimes }_{B}A\to A$ is an isomorphism. Consequently, for every maximal ideal $m$ of $B$, one has an isomorphism

   (A/mA) ⊗_k (A/mA) ≅ (A/mA),

where we have set $k=B/m$. Since $A$ is finite over $B$, $A/mA$ is a $k$-vector space of finite dimension $d$, and the above isomorphism entails $d=0$ or 1, so that the morphism $k=B/m\to A/mA$ is surjective. Hence, by Nakayama's lemma ($A$ being finite over $B$), the morphism $B_m\to A_m$ is surjective. It follows that the morphism of $B$-modules $B\to A$ is surjective (since its cokernel $C$ satisfies $C_m=0$ for every $m$, so is zero). This proves the lemma.

Let us now prove (iv). By hypothesis, $X_0=\mathrm{Spec}{A}_0$, $X_1=\mathrm{Spec}{A}_1$, and, for $i=0,1$, the morphism ${\delta }_i:A_0\to A_1$ makes $A_1$ a finitely generated $A_0$-module; thus, a fortiori, the morphism $A_0{\otimes }_B{A}_0\to A_1$ is finite.

One assumes in addition that $d=d_0⊠d_1:X_1\to X_0{\times }_Y{X}_0$ is a monomorphism; hence, by the preceding lemma, the morphism $A_0{\otimes }_B{A}_0\to A_1$ is surjective.

We shall show that it is an isomorphism (we shall prove along the way that $p:X_0\to Y$ is finite and locally free). It suffices to show that, for every prime ideal $p$ of $B$, the homomorphism $(A_0{)}_p{\otimes }_{B_p}(A_0{)}_p\to (A_1{)}_p$ with components ${\delta }_{0p}$ and ${\delta }_{1p}$ is bijective. In other words, one may suppose $B$ local. It then follows from b) that $(A_0{)}_p$ is semilocal; indeed, if $m$ is a maximal ideal of $(A_0{)}_p$, the other maximal ideals are of the form ${\delta }_0^{-1}(n)$, where $n$ runs over the prime ideals of $A_1$ such that ${\delta }_1^{-1}(n)=m$; the assertion follows from the fact that there are at most $n=[A_1:A_0]$ such prime ideals $n$. Possibly performing a faithfully flat base change,²⁵ one can also suppose that the residue field of $B$ is infinite, so that one can use the following lemma:

Lemma 4.2. Let $B$ be a local ring with infinite residue field, $A$ a semilocal ring, and $i:B\to A$ a homomorphism sending the maximal ideal $n$ of $B$ into the radical $r$ of $A$. Let $M$ be a free $A$-module of rank $n$ and $N$ a $B$-submodule of $M$ that generates $M$ as an $A$-module. Then $N$ contains a basis of $M$ over $A$.

Recall indeed that a sequence $m_1,\cdots ,m_n$ of elements of $M$ is an $A$-basis of $M$ if and only if the canonical images of $m_1,\cdots ,m_n$ in $M/rM$ form a basis of $M/rM$ over $A/r$. One can therefore replace $M$ by $M/rM$, $N$ by $N/(N\cap rM)$, $A$ by $A/r$ and $B$ by $B/n$. In this case the lemma is easy (if $A$ is a product of fields $K_1\times \cdots \times K_r$, one can identify $M$ with the module $K_1^n\times \cdots \times K_r^n$; if $x_j$ is then an element of $N$ whose $j$-th component in $K_1^n\times \cdots \times K_r^n$ is non-zero, show that a certain linear combination $x$ of the $x_j$ with coefficients in $B$ has all components non-zero; then replace $M$ by $M/Ax$ and proceed by induction on $n$).

We apply the preceding lemma in the following situation: $B=B$, $A=A_0$, $i$ is the inclusion of $B$ in $A_0$, $M=A_1$ regarded as an $A_0$-module via the homomorphism ${\delta }_1$, $N={\delta }_0(A_0)$. Indeed, since $d_0⊠d_1:X_1\to X_0{\times }_Y{X}_0$ is a closed immersion, the homomorphism $A_0{\otimes }_B{A}_0\to A_1$ with components ${\delta }_0$ and ${\delta }_1$ is surjective; this means precisely that ${\delta }_0(A_0)$ generates the $A_0$-module $A_1$.

Let then $a_1,\cdots ,a_n$ be elements of $A_0$ such that ${\delta }_0(a_1),\cdots ,{\delta }_0(a_n)$ form a basis of $A_1$ over $A_0$. If we show that $a_1,\cdots ,a_n$ is a basis of $A_0$ over $B$, it will follow that the homomorphism $A_0{\otimes }_B{A}_0\to A_1$ sends the basis $(1\otimes a_i{)}_{1⩽i⩽n}$ to the basis $({\delta }_0(a_i){)}_{1⩽i⩽n}$, hence is bijective. Consequently, if $ϵ:{\mathbb{Z}}^n\to A_0$ is the morphism of abelian groups sending the natural basis of ${\mathbb{Z}}^n$ to $a_1,\cdots ,a_n$, it suffices to prove that the map $B{\otimes }_{\mathbb{Z}}{\mathbb{Z}}^n\to A_0$ with components $i$ and $ϵ$ is bijective.

Now the diagram $(0,1,2)\ast$ considered at the beginning of this proof induces the following commutative diagram:

                    δ′₁                       δ₀
            A₂ ⇇ A₁ ⇇ A₀
                    δ′₀
            │       │ ≅           │
         u₂ │    u₁ │           u₀│
            ↓       ↓             ↓
                    δ₁ ⊗ ℤⁿ      i ⊗ ℤⁿ
            A₁ ⊗_ℤ ℤⁿ ⇇ A₀ ⊗_ℤ ℤⁿ ⇇ B ⊗_ℤ ℤⁿ,
                    δ₀ ⊗ ℤⁿ

where $u_0$, $u_1$ and $u_2$ have respectively as components $i$ and $ϵ$, ${\delta }_1$ and ${\delta }_0ϵ$, ${\delta }_2^{\prime }$ and ${\delta }_0^{\prime }{\delta }_0ϵ$. We know that $u_1$ is an isomorphism. Since the two left-hand squares of $(0,1,2)\ast$ are cocartesian, $u_2$ is bijective. But the two horizontal rows of our diagram are exact, so $u_0$ is bijective.²⁶ This shows that $A_0$ is a $B$-module locally free of rank $n$, and, by the previous reductions, this entails that ${\delta }_0\otimes {\delta }_1:A_0{\otimes }_B{A}_0\to A_1$ is an isomorphism. This completes the proof of theorem 4.1 in the particular case considered ($X_0$ affine and $d_1$ locally free of constant rank $n$).

5. Passage to the quotient by a finite and flat groupoid (general case)

a) Let $U^{(n)}$ be the largest open subset of $X_0$ above which $d_1$ is finite locally free of rank $n$. One knows that $X_0$ is the direct sum of the $U^{(n)}$. It follows on the other hand from the two Cartesian squares

              d′₀                              d′₁
   X₂ ────→ X₁          and       X₂ ────→ X₁
   │         │                     │         │
d′₂│         │ d₁                d′₂│         │ d₁
   ↓         ↓                     ↓         ↓
   X₁ ────→ X₀                     X₁ ────→ X₀
        d₀                                d₁

that the inverse images of $U^{(n)}$ under $d_0$ and $d_1$ both coincide with the largest open subset of $X_1$ above which $d_2^{\prime }$ is locally free of rank $n$;²⁷ one therefore has $d_0^{-1}(U^{(n)})=d_1^{-1}(U^{(n)})$, so that the groupoid $X\ast$ is the direct sum of the groupoids $X{\ast }^{(n)}$ induced by $X\ast$ on the open-and-closed subsets $U^{(n)}$. Consequently, as one sees easily, it suffices to prove theorem 4.1 for each of the $X{\ast }^{(n)}$: one is reduced to the case where $d_1$ is finite locally free of rank $n$.

b) We are now in a position to prove our theorem in the general case.

By a) one may suppose $d_1$ locally free of rank $n$. Let then $(Y,p)$ be a cokernel of $(d_0,d_1)$ in the category of all ringed spaces. The argument at the end of paragraph 4.c) shows that to prove 4.1 (i) it suffices to prove that $Y$ is a scheme and $p:X_0\to Y$ a morphism of schemes. By lemma 1.2, the question is local on $Y$: let $y\in Y$ and let $x\in X_0$ with $p(x)=y$; if $x$ possesses a saturated affine open neighborhood $U$, then $p(U)$ will be an affine open of $Y$ by § 4, and $p|U$ will be a morphism of schemes. It therefore suffices to prove that every $x\in X_0$ possesses a saturated affine open neighborhood $U$. Here is how one proceeds (the proof is taken from SGA 1, VIII, cor. 7.6).

   d₁(d₀⁻¹(x)) ⊂ U = (V_f)′ ⊂ V_f ⊂ V′ ⊂ V ⊂ X₀

      ↑                ↑           ↑
   affine open    special       affine
   special of V   affine open   open
                  of V
                      ↑           ↑
                  largest     largest
                  saturated   saturated
                  open in V_f open in V

By condition b) of 4.1, there exists an affine open $V$ of $X_0$ containing $d_1(d_0^{-1}(x))$;²⁸ if $F=X_0-V$, then $d_1(d_0^{-1}(F))$ is closed since $d_1$ is integral, and $V^{\prime }=X_0-d_1(d_0^{-1}(F))$ is the largest saturated open contained in $V$. Since $V^{\prime }$ is a neighborhood of the finite set $d_1(d_0^{-1}(x))$, there exists a section $f$ of the structure sheaf of $V$ vanishing on $V-V^{\prime }$ and such that $d_1(d_0^{-1}(x))$ is contained in the open $V_f$ of $V$ consisting of points where $f$ does not vanish. We shall show that the largest saturated open $(V_f{)}^{\prime }$ of $V_f$ is affine, and therefore answers the question.

Indeed, let $Z(f)=V^{\prime }-V_f$. Then $d_0^{-1}(Z(f))$ is the set of points of $d_0^{-1}(V^{\prime })=d_1^{-1}(V^{\prime })$ where the image $d_0^{\ast }(f)$ of $f$ under the map induced by $d_0$ vanishes. On the other hand, since $d_1$ induces a locally free morphism of rank $n$ from $d_0^{-1}(V^{\prime })=d_1^{-1}(V^{\prime })$ onto $V^{\prime }$,²⁹ then, by lemma 4.1.1, $d_1(d_0^{-1}(Z(f)))$ is the set of points where the norm $N$ of $d_0^{\ast }(f)$ for the morphism $d_1$ vanishes. It follows that $(V_f{)}^{\prime }=V^{\prime }-d_1(d_0^{-1}(Z(f)))$ is the set of points of $V_f$ where $N$ does not vanish; consequently, $(V_f{)}^{\prime }$ is affine.

This proves 4.1 (i); assertions (ii), (iii), and the first part of (iv) are then clear. Let us finally show the consequences indicated at the end of point (iv) (cf. [Ray67a], th. 1 (iii)).

By hypothesis, the groupoid $X\ast$ comes from an equivalence relation $i:R\to X_0\times X_0$ ($i$ being therefore an immersion, cf. N.D.E. 19), and one has established that $R$ is effective (cf. Exp. IV, 3.3.2) and that $p:X_0\to Y=X_0/R$ is a surjective and finite locally free morphism, hence in particular faithfully flat and of finite presentation.

Consequently, denoting by (M) the family of faithfully flat morphisms locally of finite presentation, $R$ is (M)-effective. Therefore, by Exp. IV, 6.3.3, $(Y,p)$ represents the quotient sheaf of $X_0$ by $R$ for the (fppf) topology, and the assertions concerning base change follow from IV, 3.4.3.1.

Remark 5.1.³⁰ We keep the hypotheses and notations of 4.1, and suppose in addition that $S$ is locally Noetherian and ${\pi }_0:X_0\to S$ is quasi-projective. Let us then show that $\pi :Y\to S$ is quasi-projective.

The above hypotheses imply that $Y\to S$ is of finite type, see the proof of 6.1 (ii). Let $A$ be an invertible $O_{X_0}$-module that is ample for ${\pi }_0$. By EGA II, 6.1.12, $p_{\ast }(A)$ is an invertible $p_{\ast }(O_{X_0})$-module. There therefore exists a covering $(V_i{)}_{i\in I}$ of $Y$ by affine opens such that $A$ is trivial above each of the saturated affine opens $U_i=p^{-1}(V_i)$.