The Stacks project

Proof. It is immediate that (2) implies (1). Assume $\{ Y_ i \to Y\} $ is as in (1) and let $W \to Y$ be as in (2). Then $\{ Y_ i \times _ Y W \to W\} _{i \in I}$ is an étale covering, which we may refine by an étale covering $\{ W_ j \to W\} _{j = 1, \ldots , m}$ with $W_ j$ affine (Topologies, Lemma 34.4.4). Thus to finish the proof it suffices to show the following three algebraic statements:

1. if $R \to A \to B$ are ring maps with $A \to B$ étale and $A$ glueable as an $R$-module, then $B$ is glueable as an $R$-module,

2. finite products of glueable $R$-modules are glueable,

3. if $R \to A \to B$ are ring maps with $A \to B$ faithfully étale and $B$ glueable as an $R$-module, then $A$ is glueable as an $R$-module.

Namely, the first of these will imply that $\Gamma (W_ j, \mathcal{O}_{W_ j})$ is a glueable $R$-module, the second will imply that $\prod \Gamma (W_ j, \mathcal{O}_{W_ j})$ is a glueable $R$-module, and the third will imply that $\Gamma (W, \mathcal{O}_ W)$ is a glueable $R$-module.

Consider an étale $R$-algebra homomorphism $A \to B$. Set $A' = A \otimes _ R R'$ and $B' = B \otimes _ R R' = A' \otimes _ A B$. Statements (1) and (3) then follow from the following facts: (a) $A$, resp. $B$ is glueable if and only if the sequence

is exact, (b) the second sequence is equal to the functor $- \otimes _ A B$ applied to the first and (c) (faithful) flatness of $A \to B$. We omit the proof of (2). $\square$

Proof. Let $(V, V', Y')$ be an object of $\textit{Spaces}(U \leftarrow U' \to X')$ with $V$, $V'$, $Y'$ affine. Write $V = \mathop{\mathrm{Spec}}(A_1)$ and $Y' = \mathop{\mathrm{Spec}}(A')$. By our definition of the category $\textit{Spaces}(U \leftarrow U' \to X')$ we find that $V'$ is the spectrum of $A_1 \otimes _{R_ f} R'_ f = A_1 \otimes _ R R'$ and the spectrum of $A'_ f$. Hence we get an isomorphism $\varphi : A'_ f \to A_1 \otimes _ R R'$ of $R'_ f$-algebras. By More on Algebra, Theorem 15.90.17 there exists a unique glueable $R$-module $A$ and isomorphisms $A_ f \to A_1$ and $A \otimes _ R R' \to A'$ of modules compatible with $\varphi $. Since the sequence

is short exact, the multiplications on $A_1$ and $A'$ define a unique $R$-algebra structure on $A$ such that the maps $A \to A_1$ and $A \to A'$ are ring homomorphisms. We omit the verification that this construction defines a quasi-inverse to the functor (81.12.0.1) restricted to the subcategories mentioned in the statement of the lemma. $\square$

Proof. Let $P'$ be one of the following properties of homomorphisms of rings: “finite”, “surjective”, “flat”, “finite type”, “flat and of finite presentation”, “étale”. Translated into algebra, the statement means the following: If $A \to B$ is an $R$-algebra homomorphism and $A$ and $B$ are glueable for $(R \to R', f)$, then $A_ f \to B_ f$ and $A \otimes _ R R' \to B \otimes _ R R'$ have $P'$ if and only if $A \to B$ has $P'$.

By More on Algebra, Lemmas 15.90.5 and 15.90.19 the algebraic statement is true for $P'$ equal to “finite” or “flat”.

If $A_ f \to B_ f$ and $A \otimes _ R R' \to B \otimes _ R R'$ are surjective, then $N = B/A$ is an $R$-module with $N_ f = 0$ and $N \otimes _ R R' = 0$ and hence vanishes by More on Algebra, Lemma 15.90.3. Thus $A \to B$ is surjective.

If $A_ f \to B_ f$ and $A \otimes _ R R' \to B \otimes _ R R'$ are finite type, then we can choose an $A$-algebra homomorphism $A[x_1, \ldots , x_ n] \to B$ such that $A_ f[x_1, \ldots , x_ n] \to B_ f$ and $(A \otimes _ R R')[x_1, \ldots , x_ n] \to B \otimes _ R R'$ are surjective (small detail omitted). We conclude that $A[x_1, \ldots , x_ n] \to B$ is surjective by the previous result. Thus $A \to B$ is of finite type.

If $A_ f \to B_ f$ and $A \otimes _ R R' \to B \otimes _ R R'$ are flat and of finite presentation, then we know that $A \to B$ is flat and of finite type by what we have already shown. Choose a surjection $A[x_1, \ldots , x_ n] \to B$ and denote $I$ the kernel. By flatness of $B$ over $A$ we see that $I_ f$ is the kernel of $A_ f[x_1, \ldots , x_ n] \to B_ f$ and $I \otimes _ R R'$ is the kernel of $A \otimes _ R R'[x_1, \ldots , x_ n] \to B \otimes _ R R'$. Thus $I_ f$ is a finite $A_ f[x_1, \ldots , x_ n]$-module and $I \otimes _ R R'$ is a finite $(A \otimes _ R R')[x_1, \ldots , x_ n]$-module. By More on Algebra, Lemma 15.90.5 applied to $I$ viewed as a module over $A[x_1, \ldots , x_ n]$ we conclude that $I$ is a finitely generated ideal and we conclude $A \to B$ is flat and of finite presentation.

If $A_ f \to B_ f$ and $A \otimes _ R R' \to B \otimes _ R R'$ are étale, then we know that $A \to B$ is flat and of finite presentation by what we have already shown. Since the fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ are isomorphic to fibres of $\mathop{\mathrm{Spec}}(B_ f) \to \mathop{\mathrm{Spec}}(A_ f)$ or $\mathop{\mathrm{Spec}}(B/fB) \to \mathop{\mathrm{Spec}}(A/fA)$, we conclude that $A \to B$ is unramified, see Morphisms, Lemmas 29.35.11 and 29.35.12. We conclude that $A \to B$ is étale by Morphisms, Lemma 29.36.16 for example. $\square$

Proof. Let $f, g : Y \to Z$ be two morphisms of algebraic spaces over $X$ with $Y$ and $Z$ glueable for $(R \to R', f)$ such that $f$ and $g$ are mapped to the same morphism in the category $\textit{Spaces}(U \leftarrow U' \to X')$. We have to show the equalizer $E \to Y$ of $f$ and $g$ is an isomorphism. Working étale locally on $Y$ we may assume $Y$ is an affine scheme. Then $E$ is a scheme and the morphism $E \to Y$ is a monomorphism and locally quasi-finite, see Morphisms of Spaces, Lemma 67.4.1. Moreover, the base change of $E \to Y$ to $U$ and to $X'$ is an isomorphism. As $Y$ is the disjoint union of the affine open $V = U \times _ X Y$ and the affine closed $V(f) \times _ X Y$, we conclude $E$ is the disjoint union of their isomorphic inverse images. It follows in particular that $E$ is quasi-compact. By Zariski's main theorem (More on Morphisms, Lemma 37.43.3) we conclude that $E$ is quasi-affine. Set $B = \Gamma (E, \mathcal{O}_ E)$ and $A = \Gamma (Y, \mathcal{O}_ Y)$ so that we have an $R$-algebra homomorphism $A \to B$. Since $E \to Y$ becomes an isomorphism after base change to $U$ and $X'$ we obtain ring maps $B \to A_ f$ and $B \to A \otimes _ R R'$ agreeing as maps into $A \otimes _ R R'_ f$. Since $A$ is glueable for $(R \to R', f)$ we get a ring map $B \to A$ which is left inverse to the map $A \to B$. The corresponding morphism $Y = \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$ maps into the open subscheme $E \subset \mathop{\mathrm{Spec}}(B)$ pointwise because this is true after base change to $U$ and $X'$. Hence we get a morphism $Y \to E$ over $Y$. Since $E \to Y$ is a monomorhism we conclude $Y \to E$ is an isomorphism as desired. $\square$

Proof. Let $Y, Z$ be two algebraic spaces over $X$ which are both glueable for $(R \to R', f)$ and assume the diagonal of $Z$ is affine. Let $a : U \times _ X Y \to U \times _ X Z$ over $U$ and $b : X' \times _ X Y \to X' \times _ X Z$ over $X'$ be two morphisms of algebraic spaces which induce the same morphism $c : U' \times _ X Y \to U' \times _ X Z$ over $U'$. We want to construct a morphism $f : Y \to Z$ over $X$ which produces the morphisms $a$, $b$ on base change to $U$, $X'$. By the faithfulness of Lemma 81.12.4, it suffices to construct the morphism $f$ étale locally on $Y$ (details omitted). Thus we may and do assume $Y$ is affine.

Let $y \in |Y|$ be a point. If $y$ maps into the open $U \subset X$, then $U \times _ X Y$ is an open of $Y$ on which the morphism $f$ is defined (we can just take $a$). Thus we may assume $y$ maps into the closed subset $V(f)$ of $X$. Since $R/fR = R'/fR'$ there is a unique point $y' \in |X' \times _ X Y|$ mapping to $y$. Denote $z' = b(y') \in |X' \times _ X Z|$ and $z \in |Z|$ the images of $y'$. Choose an étale neighbourhood $(W, w) \to (Z, z)$ with $W$ affine. Observe that

and

form an object of $\textit{Spaces}(U \leftarrow U' \to X')$ with affine parts (this is where we use that $Z$ has affine diagonal). Hence by Lemma 81.12.2 there exists a unique affine scheme $V$ glueable for $(R \to R', f)$ such that

is the triple displayed above. By fully faithfulness for the affine case (Lemma 81.12.2) we get a unique morphisms $V \to W$ and $V \to Y$ agreeing with the first and second projection morphisms over $U$ and $X'$ in the construction above. By Lemma 81.12.3 the morphism $V \to Y$ is étale. To finish the proof, it suffices to show that there is a point $v \in |V|$ mapping to $y$ (because then $f$ is defined on an étale neighbourhood of $y$, namely $V$). There is a unique point $w' \in |X' \times _ X W|$ mapping to $w$. By uniqueness $w'$ is mapped to $z'$ under the map $|X' \times _ X W| \to |X' \times _ X Z|$. Then we consider the cartesian diagram

to see that there is a point $v' \in |X' \times _ X V|$ mapping to $y'$ and $w'$, see Properties of Spaces, Lemma 66.4.3. Of course the image $v$ of $v'$ in $|V|$ maps to $y$ and the proof is complete. $\square$

Proof. Choose $n'$, $T' \to Y'$ and $n_1$, $T_1 \to V$ as in Properties, Lemma 28.18.6. Picture

Observe that $T_1 \times _ V V'$ and $V' \times _{Y'} T'$ are affine (namely the morphisms $V' \to V$ and $V' \to Y'$ are affine as base changes of the affine morphisms $U' \to U$ and $U' \to X'$). By construction we see that

In other words, the affine schemes $\mathbf{A}^{n'}_{T_1}$ and $\mathbf{A}^{n_1}_{T'}$ are part of a triple making an affine object of $\textit{Spaces}(U \leftarrow U' \to X')$. By Lemma 81.12.2 there exists a morphism of affine schemes $T \to X$ and isomorphisms $U \times _ X T \cong \mathbf{A}^{n'}_{T_1}$ and $X' \times _ X T \cong \mathbf{A}^{n_1}_{T'}$ compatible with the isomorphisms displayed above. These isomorphisms produce morphisms

satisfying the property of Properties, Lemma 28.18.6 with $n = n' + n_1$ and moreover define a morphism from the triple $(U \times _ X T, U' \times _ X T, X' \times _ X T)$ to our triple $(V, V', Y')$ in the category $\textit{Spaces}(U \leftarrow U' \to X')$.

By Lemma 81.12.2 there is an affine scheme $W$ whose image in $\textit{Spaces}(U \leftarrow U' \to X')$ is isomorphic to the triple

By fully faithfulness of this construction, we obtain two maps $p_0, p_1 : W \to T$ whose base changes to $U, U', X'$ are the projection morphisms. By Lemma 81.12.3 the morphisms $p_0, p_1$ are flat and of finite presentation and the morphism $(p_0, p_1) : W \to T \times _ X T$ is a closed immersion. In fact, $W \to T \times _ X T$ is an equivalence relation: by the lemmas used above we may check symmetry, reflexivity, and transitivity after base change to $U$ and $X'$, where these are obvious (details omitted). Thus the quotient sheaf

is an algebraic space for example by Bootstrap, Theorem 80.10.1. Since it is clear that $Y/X$ is sent to the triple $(V, V', Y')$. The base change of the diagonal $\Delta : Y \to Y \times _ X Y$ by the quasi-compact surjective flat morphism $T \times _ X T \to Y \times _ X Y$ is the closed immersion $W \to T \times _ X T$. Thus $\Delta $ is a closed immersion by Descent on Spaces, Lemma 74.11.17. Thus the algebraic space $Y$ is separated and the proof is complete. $\square$