## 75.9 Thickenings

The following terminology may not be completely standard, but it is convenient.

Definition 75.9.1. Thickenings. Let $S$ be a scheme.

1. 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.

2. 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.

3. 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.

4. 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

75.9.1.1
$$\label{spaces-more-morphisms-equation-equivalence-etale-spaces} X_{spaces, {\acute{e}tale}} = X'_{spaces, {\acute{e}tale}}$$

via the pullback functor, see Theorem 75.8.1. 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

75.9.1.2
$$\label{spaces-more-morphisms-equation-fundamental-equivalence} (\mathop{\mathit{Sh}}\nolimits (X'_{spaces, {\acute{e}tale}}), \mathcal{O}_{X'}) \cong (\mathop{\mathit{Sh}}\nolimits (X_{spaces, {\acute{e}tale}}), \mathcal{O}_{X'})$$

Below we will frequently combine this with the fully faithfulness result of Properties of Spaces, Theorem 65.28.4. 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 75.9.2. 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 65.28.4, 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 65.28.1), 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 65.28.4. We omit the verification that $f'$ is a morphism over $B$. $\square$

Lemma 75.9.3. 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 (75.9.1.1). The morphism $U' \to X'$ is étale and universally injective, hence an open immersion, see Morphisms of Spaces, Lemma 66.51.2. $\square$

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 75.9.4. 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 (75.9.1.2).

Proof. Let $U' \to X'$ be the étale morphism of algebraic spaces such that $U = X \times _{X'} U'$, see Theorem 75.8.1. By Limits of Spaces, Lemma 69.15.1 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 66.14.1 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 75.9.5. 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 69.15.3. 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 65.13.1 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 75.9.3 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 75.9.4 the map $A \to A'$ is surjective. The kernel $I$ is an ideal of square zero. By Properties of Spaces, Lemma 65.33.1 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 66.51.2 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 65.16.2. 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 75.9.4. 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 75.9.6. 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 75.8.1. Thus it suffices to show that $V$ is a scheme if and only if $V'$ is a scheme. This is the content of Lemma 75.9.5. $\square$

First order thickening are described as follows.

Lemma 75.9.7. 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

1. $\mathcal{I}$ is an ideal of square zero in $\mathcal{A}$, and

2. $\mathcal{I}$ 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 75.9.4 and 75.9.5. 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 64.9.3 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 64.10.5). 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 65.18.14 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 75.9.2 this defines a morphism $X' \to B$ compatible with the given morphism $X \to B$ and we are done. $\square$

Lemma 75.9.8. 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 75.9.9. 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 75.9.10. 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 73.11.17 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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