The Stacks project

81.12 Glueing and the Beauville-Laszlo theorem

Let $R \to R'$ be a ring homomorphism and let $f \in R$ be an element such that

\[ 0 \to R \to R_ f \oplus R' \to R'_ f \to 0 \]

is a short exact sequence. This implies that $R/f^ nR \cong R'/f^ nR'$ for all $n$ and $(R \to R', f)$ is a glueing pair in the sense of More on Algebra, Section 15.90. Set $X = \mathop{\mathrm{Spec}}(R)$, $U = \mathop{\mathrm{Spec}}(R_ f)$, $X' = \mathop{\mathrm{Spec}}(R')$ and $U' = \mathop{\mathrm{Spec}}(R'_ f)$. Picture

\[ \xymatrix{ U' \ar[r] \ar[d] & X' \ar[d] \\ U \ar[r] & X } \]

In this situation we can consider the category $\textit{Spaces}(U \leftarrow U' \to X')$ whose objects are commutative diagrams

\[ \xymatrix{ V \ar[d] & V' \ar[l] \ar[d] \ar[r] & Y' \ar[d] \\ U & U' \ar[l] \ar[r] & X' } \]

of algebraic spaces with both squares cartesian and whose morphism are defined in the obvious manner. An object of this category will be denoted $(V, V', Y')$ with arrows suppressed from the notation. There is a functor

81.12.0.1
\begin{equation} \label{spaces-pushouts-equation-beauville-laszlo-glueing-spaces} \textit{Spaces}/X \longrightarrow \textit{Spaces}(U \leftarrow U' \to X') \end{equation}

given by base change: $Y \mapsto (U \times _ X Y, U' \times _ X Y, X' \times _ X Y)$.

We have seen in More on Algebra, Section 15.90 that not every $R$-module $M$ can be recovered from its gluing data. Similarly, the functor (81.12.0.1) won't be fully faithful on the category of all spaces over $X$. In order to single out a suitable subcategory of algebraic spaces over $X$ we need a lemma.

Lemma 81.12.1. Let $(R \to R', f)$ be a glueing pair, see above. Let $Y$ be an algebraic space over $X$. The following are equivalent

  1. there exists an étale covering $\{ Y_ i \to Y\} _{i \in I}$ with $Y_ i$ affine and $\Gamma (Y_ i, \mathcal{O}_{Y_ i})$ glueable as an $R$-module,

  2. for every étale morphism $W \to Y$ with $W$ affine $\Gamma (W, \mathcal{O}_ W)$ is a glueable $R$-module.

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

\[ 0 \to A \to A_ f \oplus A' \to A'_ f \to 0, \quad \text{resp.}\quad 0 \to B \to B_ f \oplus B' \to B'_ f \to 0, \]

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). $\square$

Let $(R \to R', f)$ be a glueing pair, see above. We will say an algebraic space $Y$ over $X = \mathop{\mathrm{Spec}}(R)$ is glueable for $(R \to R', f)$ if the equivalent conditions of Lemma 81.12.1 are satisfied.

Lemma 81.12.2. Let $(R \to R', f)$ be a glueing pair, see above. The functor (81.12.0.1) restricts to an equivalence between the category of affine $Y/X$ which are glueable for $(R \to R', f)$ and the full subcategory of objects $(V, V', Y')$ of $\textit{Spaces}(U \leftarrow U' \to X')$ with $V$, $V'$, $Y'$ affine.

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 15.90.16 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

\[ 0 \to A \to A_1 \oplus A' \to A'_ f \to 0 \]

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. $\square$

Lemma 81.12.3. Let $P$ be one of the following properties of morphisms: “finite”, “closed immersion”, “flat”, “finite type”, “flat and finite presentation”, “étale”. Under the equivalence of Lemma 81.12.2 the morphisms having $P$ correspond to morphisms of triples whose components have $P$.

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 15.90.4 and 15.90.18 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 15.90.2. 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 15.90.4 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. $\square$

Lemma 81.12.4. Let $(R \to R', f)$ be a glueing pair, see above. The functor (81.12.0.1) is faithful on the full subcategory of algebraic spaces $Y/X$ glueable for $(R \to R', f)$.

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. $\square$

Lemma 81.12.5. Let $(R \to R', f)$ be a glueing pair, see above. The functor (81.12.0.1) is fully faithful on the full subcategory of algebraic spaces $Y/X$ which are (a) glueable for $(R \to R', f)$ and (b) have affine diagonal $Y \to Y \times _ X Y$.

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

\[ (U \times _ X W) \times _{U \times _ X Z, a} (U \times _ X Y),\quad (U' \times _ X W) \times _{U' \times _ X Z, c} (U' \times _ X Y), \]

and

\[ (X' \times _ X W) \times _{X' \times _ X Z, b} (X' \times _ X Y) \]

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

\[ (U \times _ X V, U' \times _ X V, X' \times _ X V) \]

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

\[ \xymatrix{ X' \times _ X V \ar[r] \ar[d] & X' \times _ X W \ar[d] \\ X' \times _ X Y \ar[r] & X' \times _ X Z } \]

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. $\square$

Lemma 81.12.6. Let $(R \to R', f)$ be a glueing pair, see above. Any object $(V, V', Y')$ of $\textit{Spaces}(U \leftarrow U' \to X')$ with $V$, $V'$, $Y'$ quasi-affine is isomorphic to the image under the functor (81.12.0.1) of a separated algebraic space $Y$ over $X$.

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

\[ \xymatrix{ & & T_1 \times _ V V' \times _ Y T' \ar[ld] \ar[rd] \\ T_1 \ar[d] & T_1 \times _ V V' \ar[l] \ar[dr] & & V' \times _{Y'} T' \ar[r] \ar[dl] & T' \ar[d] \\ V & & V' \ar[rr] \ar[ll] & & Y' } \]

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

\[ \mathbf{A}^{n'}_{T_1 \times _ V V'} \cong T_1 \times _ V V' \times _{Y'} T' \cong \mathbf{A}^{n_1}_{V' \times _{Y'} T'} \]

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

\[ U \times _ X T \longrightarrow V \quad \text{and}\quad X' \times _ X T \longrightarrow Y' \]

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

\[ ((U \times _ X T) \times _ V (U \times _ X T), (U' \times _ X T) \times _{V'} (U' \times _ X T), (X' \times _ X T) \times _{Y'} (X' \times _ X T)) \]

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

\[ Y = T/W \]

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F9M. Beware of the difference between the letter 'O' and the digit '0'.