76.9 Thickenings
The following terminology may not be completely standard, but it is convenient.
Definition 76.9.1. Thickenings. Let S be a scheme.
We say an algebraic space X' is a thickening of an algebraic space X if X is a closed subspace of X' and the associated topological spaces are equal.
We say X' is a first order thickening of X if X is a closed subspace of X' and the quasi-coherent sheaf of ideals \mathcal{I} \subset \mathcal{O}_{X'} defining X has square zero.
Given two thickenings X \subset X' and Y \subset Y' a morphism of thickenings is a morphism f' : X' \to Y' such that f(X) \subset Y, i.e., such that f'|_ X factors through the closed subspace Y. In this situation we set f = f'|_ X : X \to Y and we say that (f, f') : (X \subset X') \to (Y \subset Y') is a morphism of thickenings.
Let B be an algebraic space. We similarly define thickenings over B, and morphisms of thickenings over B. This means that the spaces X, X', Y, Y' above are algebraic spaces endowed with a structure morphism to B, and that the morphisms X \to X', Y \to Y' and f' : X' \to Y' are morphisms over B.
The fundamental equivalence. Note that if X \subset X' is a thickening, then X \to X' is integral and universally bijective. This implies that
76.9.1.1
\begin{equation} \label{spaces-more-morphisms-equation-equivalence-etale-spaces} X_{spaces, {\acute{e}tale}} = X'_{spaces, {\acute{e}tale}} \end{equation}
via the pullback functor, see Theorem 76.8.1. Hence we may think of \mathcal{O}_{X'} as a sheaf on X_{spaces, {\acute{e}tale}}. Thus a canonical equivalence of locally ringed topoi
76.9.1.2
\begin{equation} \label{spaces-more-morphisms-equation-fundamental-equivalence} (\mathop{\mathit{Sh}}\nolimits (X'_{spaces, {\acute{e}tale}}), \mathcal{O}_{X'}) \cong (\mathop{\mathit{Sh}}\nolimits (X_{spaces, {\acute{e}tale}}), \mathcal{O}_{X'}) \end{equation}
Below we will frequently combine this with the fully faithfulness result of Properties of Spaces, Theorem 66.28.4. For example the closed immersion i_ X : X \to X' corresponds to the surjective map i_ X^\sharp : \mathcal{O}_{X'} \to \mathcal{O}_ X.
Let S be a scheme, and let B be an algebraic space over S. Let (f, f') : (X \subset X') \to (Y \subset Y') be a morphism of thickenings over B. Note that the diagram of continuous functors
\xymatrix{ X_{spaces, {\acute{e}tale}} & Y_{spaces, {\acute{e}tale}} \ar[l] \\ X'_{spaces, {\acute{e}tale}} \ar[u] & Y'_{spaces, {\acute{e}tale}} \ar[u] \ar[l] }
is commutative and the vertical arrows are equivalences. Hence f_{spaces, {\acute{e}tale}}, f_{small}, f'_{spaces, {\acute{e}tale}}, and f'_{small} all define the same morphism of topoi. Thus we may think of
(f')^\sharp : f_{spaces, {\acute{e}tale}}^{-1}\mathcal{O}_{Y'} \longrightarrow \mathcal{O}_{X'}
as a map of sheaves of \mathcal{O}_ B-algebras fitting into the commutative diagram
\xymatrix{ f_{spaces, {\acute{e}tale}}^{-1}\mathcal{O}_ Y \ar[r]_-{f^\sharp } \ar[r] & \mathcal{O}_ X \\ f_{spaces, {\acute{e}tale}}^{-1}\mathcal{O}_{Y'} \ar[r]^-{(f')^\sharp } \ar[u]^{i_ Y^\sharp } & \mathcal{O}_{X'} \ar[u]_{i_ X^\sharp } }
Here i_ X : X \to X' and i_ Y : Y \to Y' are the names of the given closed immersions.
Lemma 76.9.2. Let S be a scheme. Let B be an algebraic space over S. Let X \subset X' and Y \subset Y' be thickenings of algebraic spaces over B. Let f : X \to Y be a morphism of algebraic spaces over B. Given any map of \mathcal{O}_ B-algebras
\alpha : f_{spaces, {\acute{e}tale}}^{-1}\mathcal{O}_{Y'} \to \mathcal{O}_{X'}
such that
\xymatrix{ f_{spaces, {\acute{e}tale}}^{-1}\mathcal{O}_ Y \ar[r]_-{f^\sharp } \ar[r] & \mathcal{O}_ X \\ f_{spaces, {\acute{e}tale}}^{-1}\mathcal{O}_{Y'} \ar[r]^-\alpha \ar[u]^{i_ Y^\sharp } & \mathcal{O}_{X'} \ar[u]_{i_ X^\sharp } }
commutes, there exists a unique morphism of (f, f') of thickenings over B such that \alpha = (f')^\sharp .
Proof.
To find f', by Properties of Spaces, Theorem 66.28.4, all we have to do is show that the morphism of ringed topoi
(f_{spaces, {\acute{e}tale}}, \alpha ) : (\mathop{\mathit{Sh}}\nolimits (X_{spaces, {\acute{e}tale}}), \mathcal{O}_{X'}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (Y_{spaces, {\acute{e}tale}}), \mathcal{O}_{Y'})
is a morphism of locally ringed topoi. This follows directly from the definition of morphisms of locally ringed topoi (Modules on Sites, Definition 18.40.9), the fact that (f, f^\sharp ) is a morphism of locally ringed topoi (Properties of Spaces, Lemma 66.28.1), that \alpha fits into the given commutative diagram, and the fact that the kernels of i_ X^\sharp and i_ Y^\sharp are locally nilpotent. Finally, the fact that f' \circ i_ X = i_ Y \circ f follows from the commutativity of the diagram and another application of Properties of Spaces, Theorem 66.28.4. We omit the verification that f' is a morphism over B.
\square
Lemma 76.9.3. Let S be a scheme. Let X \subset X' be a thickening of algebraic spaces over S. For any open subspace U \subset X there exists a unique open subspace U' \subset X' such that U = X \times _{X'} U'.
Proof.
Let U' \to X' be the object of X'_{spaces, {\acute{e}tale}} corresponding to the object U \to X of X_{spaces, {\acute{e}tale}} via (76.9.1.1). The morphism U' \to X' is étale and universally injective, hence an open immersion, see Morphisms of Spaces, Lemma 67.51.2.
\square
Finite order thickenings. Let i_ X : X \to X' be a thickening of algebraic spaces. Any local section of the kernel \mathcal{I} = \mathop{\mathrm{Ker}}(i_ X^\sharp ) \subset \mathcal{O}_{X'} is locally nilpotent. Let us say that X \subset X' is a finite order thickening if the ideal sheaf \mathcal{I} is “globally” nilpotent, i.e., if there exists an n \geq 0 such that \mathcal{I}^{n + 1} = 0. Technically the class of finite order thickenings X \subset X' is much easier to handle than the general case. Namely, in this case we have a filtration
0 \subset \mathcal{I}^ n \subset \mathcal{I}^{n - 1} \subset \ldots \subset \mathcal{I} \subset \mathcal{O}_{X'}
and we see that X' is filtered by closed subspaces
X = X_0 \subset X_1 \subset \ldots \subset X_{n - 1} \subset X_{n + 1} = X'
such that each pair X_ i \subset X_{i + 1} is a first order thickening over B. Using simple induction arguments many results proved for first order thickenings can be rephrased as results on finite order thickenings.
Lemma 76.9.4. Let S be a scheme. Let X \subset X' be a thickening of algebraic spaces over S. Let U be an affine object of X_{spaces, {\acute{e}tale}}. Then
\Gamma (U, \mathcal{O}_{X'}) \to \Gamma (U, \mathcal{O}_ X)
is surjective where we think of \mathcal{O}_{X'} as a sheaf on X_{spaces, {\acute{e}tale}} via (76.9.1.2).
Proof.
Let U' \to X' be the étale morphism of algebraic spaces such that U = X \times _{X'} U', see Theorem 76.8.1. By Limits of Spaces, Lemma 70.15.1 we see that U' is an affine scheme. Hence \Gamma (U, \mathcal{O}_{X'}) = \Gamma (U', \mathcal{O}_{U'}) \to \Gamma (U, \mathcal{O}_ U) is surjective as U \to U' is a closed immersion of affine schemes. Below we give a direct proof for finite order thickenings which is the case most used in practice.
\square
Proof for finite order thickenings.
We may assume that X \subset X' is a first order thickening by the principle explained above. Denote \mathcal{I} the kernel of the surjection \mathcal{O}_{X'} \to \mathcal{O}_ X. As \mathcal{I} is a quasi-coherent \mathcal{O}_{X'}-module and since \mathcal{I}^2 = 0 by the definition of a first order thickening we may apply Morphisms of Spaces, Lemma 67.14.1 to see that \mathcal{I} is a quasi-coherent \mathcal{O}_ X-module. Hence the lemma follows from the long exact cohomology sequence associated to the short exact sequence
0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0
and the fact that H^1_{\acute{e}tale}(U, \mathcal{I}) = 0 as \mathcal{I} is quasi-coherent, see Descent, Proposition 35.9.3 and Cohomology of Schemes, Lemma 30.2.2.
\square
Lemma 76.9.5. Let S be a scheme. Let X \subset X' be a thickening of algebraic spaces over S. If X is (representable by) a scheme, then so is X'.
Proof.
Note that X'_{red} = X_{red}. Hence if X is a scheme, then X'_{red} is a scheme. Thus the result follows from Limits of Spaces, Lemma 70.15.3. Below we give a direct proof for finite order thickenings which is the case most often used in practice.
\square
Proof for finite order thickenings.
It suffices to prove this when X' is a first order thickening of X. By Properties of Spaces, Lemma 66.13.1 there is a largest open subspace of X' which is a scheme. Thus we have to show that every point x of |X'| = |X| is contained in an open subspace of X' which is a scheme. Using Lemma 76.9.3 we may replace X \subset X' by U \subset U' with x \in U and U an affine scheme. Hence we may assume that X is affine. Thus we reduce to the case discussed in the next paragraph.
Assume X \subset X' is a first order thickening where X is an affine scheme. Set A = \Gamma (X, \mathcal{O}_ X) and A' = \Gamma (X', \mathcal{O}_{X'}). By Lemma 76.9.4 the map A \to A' is surjective. The kernel I is an ideal of square zero. By Properties of Spaces, Lemma 66.33.1 we obtain a canonical morphism f : X' \to \mathop{\mathrm{Spec}}(A') which fits into the following commutative diagram
\xymatrix{ X \ar@{=}[d] \ar[r] & X' \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathop{\mathrm{Spec}}(A') }
Because the horizontal arrows are thickenings it is clear that f is universally injective and surjective. Hence it suffices to show that f is étale, since then Morphisms of Spaces, Lemma 67.51.2 will imply that f is an isomorphism.
To prove that f is étale choose an affine scheme U' and an étale morphism U' \to X'. It suffices to show that U' \to X' \to \mathop{\mathrm{Spec}}(A') is étale, see Properties of Spaces, Definition 66.16.2. Write U' = \mathop{\mathrm{Spec}}(B'). Set U = X \times _{X'} U'. Since U is a closed subspace of U', it is a closed subscheme, hence U = \mathop{\mathrm{Spec}}(B) with B' \to B surjective. Denote J = \mathop{\mathrm{Ker}}(B' \to B) and note that J = \Gamma (U, \mathcal{I}) where \mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_{X'} \to \mathcal{O}_ X) on X_{spaces, {\acute{e}tale}} as in the proof of Lemma 76.9.4. The morphism U' \to X' \to \mathop{\mathrm{Spec}}(A') induces a commutative diagram
\xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 }
Now, since \mathcal{I} is a quasi-coherent \mathcal{O}_ X-module we have \mathcal{I} = (\widetilde I)^ a, see Descent, Definition 35.8.2 for notation and Descent, Proposition 35.8.9 for why this is true. Hence we see that J = I \otimes _ A B. Finally, note that A \to B is étale as U \to X is étale as the base change of the étale morphism U' \to X'. We conclude that A' \to B' is étale by Algebra, Lemma 10.143.11.
\square
Lemma 76.9.6. Let S be a scheme. Let X \subset X' be a thickening of algebraic spaces over S. The functor
V' \longmapsto V = X \times _{X'} V'
defines an equivalence of categories X'_{\acute{e}tale}\to X_{\acute{e}tale}.
Proof.
The functor V' \mapsto V defines an equivalence of categories X'_{spaces, {\acute{e}tale}} \to X_{spaces, {\acute{e}tale}}, see Theorem 76.8.1. Thus it suffices to show that V is a scheme if and only if V' is a scheme. This is the content of Lemma 76.9.5.
\square
First order thickening are described as follows.
Lemma 76.9.7. Let S be a scheme. Let f : X \to B be a morphism of algebraic spaces over S. Consider a short exact sequence
0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_ X \to 0
of sheaves on X_{\acute{e}tale} where \mathcal{A} is a sheaf of f^{-1}\mathcal{O}_ B-algebras, \mathcal{A} \to \mathcal{O}_ X is a surjection of sheaves of f^{-1}\mathcal{O}_ B-algebras, and \mathcal{I} is its kernel. If
\mathcal{I} is an ideal of square zero in \mathcal{A}, and
\mathcal{I} is quasi-coherent as an \mathcal{O}_ X-module
then there exists a first order thickening X \subset X' over B and an isomorphism \mathcal{O}_{X'} \to \mathcal{A} of f^{-1}\mathcal{O}_ B-algebras compatible with the surjections to \mathcal{O}_ X.
Proof.
In this proof we redo some of the arguments used in the proofs of Lemmas 76.9.4 and 76.9.5. We first handle the case B = S = \mathop{\mathrm{Spec}}(\mathbf{Z}). Let U be an affine scheme, and let U \to X be étale. Then
0 \to \mathcal{I}(U) \to \mathcal{A}(U) \to \mathcal{O}_ X(U) \to 0
is exact as H^1(U_{\acute{e}tale}, \mathcal{I}) = 0 as \mathcal{I} is quasi-coherent, see Descent, Proposition 35.9.3 and Cohomology of Schemes, Lemma 30.2.2. If V \to U is a morphism of affine objects of X_{spaces, {\acute{e}tale}} then
\mathcal{I}(V) = \mathcal{I}(U) \otimes _{\mathcal{O}_ X(U)} \mathcal{O}_ X(V)
since \mathcal{I} is a quasi-coherent \mathcal{O}_ X-module, see Descent, Proposition 35.8.9. Hence \mathcal{A}(U) \to \mathcal{A}(V) is an étale ring map, see Algebra, Lemma 10.143.11. Hence we see that
U \longmapsto U' = \mathop{\mathrm{Spec}}(\mathcal{A}(U))
is a functor from X_{affine, {\acute{e}tale}} to the category of affine schemes and étale morphisms. In fact, we claim that this functor can be extended to a functor U \mapsto U' on all of X_{\acute{e}tale}. To see this, if U is an object of X_{\acute{e}tale}, note that
0 \to \mathcal{I}|_{U_{Zar}} \to \mathcal{A}|_{U_{Zar}} \to \mathcal{O}_ X|_{U_{Zar}} \to 0
and \mathcal{I}|_{U_{Zar}} is a quasi-coherent sheaf on U, see Descent, Proposition 35.9.4. Hence by More on Morphisms, Lemma 37.2.2 we obtain a first order thickening U \subset U' of schemes such that \mathcal{O}_{U'} is isomorphic to \mathcal{A}|_{U_{Zar}}. It is clear that this construction is compatible with the construction for affines above.
Choose a presentation X = U/R, see Spaces, Definition 65.9.3 so that s, t : R \to U define an étale equivalence relation. Applying the functor above we obtain an étale equivalence relation s', t' : R' \to U' in schemes. Consider the algebraic space X' = U'/R' (see Spaces, Theorem 65.10.5). The morphism X = U/R \to U'/R' = X' is a first order thickening. Consider \mathcal{O}_{X'} viewed as a sheaf on X_{\acute{e}tale}. By construction we have an isomorphism
\gamma : \mathcal{O}_{X'}|_{U_{\acute{e}tale}} \longrightarrow \mathcal{A}|_{U_{\acute{e}tale}}
such that s^{-1}\gamma agrees with t^{-1}\gamma on R_{\acute{e}tale}. Hence by Properties of Spaces, Lemma 66.18.14 this implies that \gamma comes from a unique isomorphism \mathcal{O}_{X'} \to \mathcal{A} as desired.
To handle the case of a general base algebraic space B, we first construct X' as an algebraic space over \mathbf{Z} as above. Then we use the isomorphism \mathcal{O}_{X'} \to \mathcal{A} to define f^{-1}\mathcal{O}_ B \to \mathcal{O}_{X'}. According to Lemma 76.9.2 this defines a morphism X' \to B compatible with the given morphism X \to B and we are done.
\square
Lemma 76.9.8. Let S be a scheme. Let Y \subset Y' be a thickening of algebraic spaces over S. Let X' \to Y' be a morphism and set X = Y \times _{Y'} X'. Then (X \subset X') \to (Y \subset Y') is a morphism of thickenings. If Y \subset Y' is a first (resp. finite order) thickening, then X \subset X' is a first (resp. finite order) thickening.
Proof.
Omitted.
\square
Lemma 76.9.9. Let S be a scheme. If X \subset X' and X' \subset X'' are thickenings of algebraic spaces over S, then so is X \subset X''.
Proof.
Omitted.
\square
Lemma 76.9.10. The property of being a thickening is fpqc local. Similarly for first order thickenings.
Proof.
The statement means the following: Let S be a scheme and let X \to X' be a morphism of algebraic spaces over S. Let \{ g_ i : X'_ i \to X'\} be an fpqc covering of algebraic spaces such that the base change X_ i \to X'_ i is a thickening for all i. Then X \to X' is a thickening. Since the morphisms g_ i are jointly surjective we conclude that X \to X' is surjective. By Descent on Spaces, Lemma 74.11.17 we conclude that X \to X' is a closed immersion. Thus X \to X' is a thickening. We omit the proof in the case of first order thickenings.
\square
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