## 37.2 Thickenings

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

Definition 37.2.1. Thickenings.

1. We say a scheme $X'$ is a thickening of a scheme $X$ if $X$ is a closed subscheme of $X'$ and the underlying topological spaces are equal.

2. We say a scheme $X'$ is a first order thickening of a scheme $X$ if $X$ is a closed subscheme 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 subscheme $Y$. 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 $S$ be a scheme. We similarly define thickenings over $S$, and morphisms of thickenings over $S$. This means that the schemes $X, X', Y, Y'$ above are schemes over $S$, and that the morphisms $X \to X'$, $Y \to Y'$ and $f' : X' \to Y'$ are morphisms over $S$.

Finite order thickenings. Let $i_ X : X \to X'$ be a thickening. Any local section of the kernel $\mathcal{I} = \mathop{\mathrm{Ker}}(i_ X^\sharp )$ 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

$0 = \mathcal{I}^{n + 1} \subset \mathcal{I}^ n \subset \mathcal{I}^{n - 1} \subset \ldots \subset \mathcal{I} \subset \mathcal{O}_{X'}$

and we see that $X'$ is filtered by closed subspaces

$X = X_1 \subset X_2 \subset \ldots \subset X_ n \subset X_{n + 1} = X'$

such that each pair $X_ i \subset X_{i + 1}$ is a first order thickening over $S$. Using simple induction arguments many results proved for first order thickenings can be rephrased as results on finite order thickenings.

First order thickening are described as follows (see Modules, Lemma 17.27.11).

Lemma 37.2.2. Let $X$ be a scheme over a base $S$. Consider a short exact sequence

$0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_ X \to 0$

of sheaves on $X$ where $\mathcal{A}$ is a sheaf of $f^{-1}\mathcal{O}_ S$-algebras, $\mathcal{A} \to \mathcal{O}_ X$ is a surjection of sheaves of $f^{-1}\mathcal{O}_ S$-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 $X' = (X, \mathcal{A})$ is a scheme and $X \to X'$ is a first order thickening over $S$. Moreover, any first order thickening over $S$ is of this form.

Proof. It is clear that $X'$ is a locally ringed space. Let $U = \mathop{\mathrm{Spec}}(B)$ be an affine open of $X$. Set $A = \Gamma (U, \mathcal{A})$. Note that since $H^1(U, \mathcal{I}) = 0$ (see Cohomology of Schemes, Lemma 30.2.2) the map $A \to B$ is surjective. By assumption the kernel $I = \mathcal{I}(U)$ is an ideal of square zero in the ring $A$. By Schemes, Lemma 26.6.4 there is a canonical morphism of locally ringed spaces

$(U, \mathcal{A}|_ U) \longrightarrow \mathop{\mathrm{Spec}}(A)$

coming from the map $B \to \Gamma (U, \mathcal{A})$. Since this morphism fits into the commutative diagram

$\xymatrix{ (U, \mathcal{O}_ X|_ U) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(B) \ar[d] \\ (U, \mathcal{A}|_ U) \ar[r] & \mathop{\mathrm{Spec}}(A) }$

we see that it is a homeomorphism on underlying topological spaces. Thus to see that it is an isomorphism, it suffices to check it induces an isomorphism on the local rings. For $u \in U$ corresponding to the prime $\mathfrak p \subset A$ we obtain a commutative diagram of short exact sequences

$\xymatrix{ 0 \ar[r] & I_{\mathfrak p} \ar[r] \ar[d] & A_{\mathfrak p} \ar[r] \ar[d] & B_{\mathfrak p} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{I}_ u \ar[r] & \mathcal{A}_ u \ar[r] & \mathcal{O}_{X, u} \ar[r] & 0. }$

The left and right vertical arrows are isomorphisms because $\mathcal{I}$ and $\mathcal{O}_ X$ are quasi-coherent sheaves. Hence also the middle map is an isomorphism. Hence every point of $X' = (X, \mathcal{A})$ has an affine neighbourhood and $X'$ is a scheme as desired. $\square$

Proof. This is a special case of Limits, Proposition 32.11.2. $\square$

Proof for a finite order thickening. Suppose that $X \subset X'$ is a finite order thickening with $X$ affine. Then we may use Serre's criterion to prove $X'$ is affine. More precisely, we will use Cohomology of Schemes, Lemma 30.3.1. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_{X'}$-module. It suffices to show that $H^1(X', \mathcal{F}) = 0$. Denote $i : X \to X'$ the given closed immersion and denote $\mathcal{I} = \mathop{\mathrm{Ker}}(i^\sharp : \mathcal{O}_{X'} \to i_*\mathcal{O}_ X)$. By our discussion of finite order thickenings (following Definition 37.2.1) there exists an $n \geq 0$ and a filtration

$0 = \mathcal{F}_{n + 1} \subset \mathcal{F}_ n \subset \mathcal{F}_{n - 1} \subset \ldots \subset \mathcal{F}_0 = \mathcal{F}$

by quasi-coherent submodules such that $\mathcal{F}_ a/\mathcal{F}_{a + 1}$ is annihilated by $\mathcal{I}$. Namely, we can take $\mathcal{F}_ a = \mathcal{I}^ a\mathcal{F}$. Then $\mathcal{F}_ a/\mathcal{F}_{a + 1} = i_*\mathcal{G}_ a$ for some quasi-coherent $\mathcal{O}_ X$-module $\mathcal{G}_ a$, see Morphisms, Lemma 29.4.1. We obtain

$H^1(X', \mathcal{F}_ a/\mathcal{F}_{a + 1}) = H^1(X', i_*\mathcal{G}_ a) = H^1(X, \mathcal{G}_ a) = 0$

The second equality comes from Cohomology of Schemes, Lemma 30.2.4 and the last equality from Cohomology of Schemes, Lemma 30.2.2. Thus $\mathcal{F}$ has a finite filtration whose successive quotients have vanishing first cohomology and it follows by a simple induction argument that $H^1(X', \mathcal{F}) = 0$. $\square$

Lemma 37.2.4. Let $S \subset S'$ be a thickening of schemes. Let $X' \to S'$ be a morphism and set $X = S \times _{S'} X'$. Then $(X \subset X') \to (S \subset S')$ is a morphism of thickenings. If $S \subset S'$ is a first (resp. finite order) thickening, then $X \subset X'$ is a first (resp. finite order) thickening.

Proof. Omitted. $\square$

Lemma 37.2.5. If $S \subset S'$ and $S' \subset S''$ are thickenings, then so is $S \subset S''$.

Proof. Omitted. $\square$

Lemma 37.2.6. The property of being a thickening is fpqc local. Similarly for first order thickenings.

Proof. The statement means the following: Let $X \to X'$ be a morphism of schemes and let $\{ g_ i : X'_ i \to X'\}$ be an fpqc covering 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, Lemma 35.23.19 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. $\square$

Comment #621 by Anfang on

A typo in the paragraph between definition 36.2.1 and lemma 36.2.2. Do you mean that each pair $X_i \subset X_{i + 1}$ is a first order thickening over $S$ not over $B$ ? There is no $B$ before.

Comment #3848 by Marco on

Hi, I think the indexation after "filtered by closed subspaces" is not correct. After $X_{n-1}$ there is $X_{n+1}$. I think it should be $X_{n}$.

Comment #5486 by Haodong Yao on

In the definition of a first order thickening do we assume the underlying topological spaces are the same so it is in prior a thickening?

Comment #5487 by Matthieu Romagny on

@Haodong Yao In the definition of a first order thickening, the sheaf of ideals defining X has square zero so the closed immersion of X into X' is a homeomorphism.

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