## 76.15 Universal first order thickenings

Let $S$ be a scheme. Let $h : Z \to X$ be a morphism of algebraic spaces over $S$. A universal first order thickening of $Z$ over $X$ is a first order thickening $Z \subset Z'$ over $X$ such that given any first order thickening $T \subset T'$ over $X$ and a solid commutative diagram

76.15.0.1
$$\label{spaces-more-morphisms-equation-universal-first-order-thickening} \vcenter { \xymatrix{ & Z \ar[ld] & & T \ar[rd] \ar[ll]^ a \\ Z' \ar[rrd] & & & & T' \ar@{..>}[llll]_{a'} \ar[lld]^ b \\ & & X } }$$

there exists a unique dotted arrow making the diagram commute. Note that in this situation $(a, a') : (T \subset T') \to (Z \subset Z')$ is a morphism of thickenings over $X$. Thus if a universal first order thickening exists, then it is unique up to unique isomorphism. In general a universal first order thickening does not exist, but if $h$ is formally unramified then it does. Before we prove this, let us show that a universal first order thickening in the category of schemes is a universal first order thickening in the category of algebraic spaces.

Lemma 76.15.1. Let $S$ be a scheme. Let $h : Z \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset Z'$ be a first order thickening over $X$. The following are equivalent

1. $Z \subset Z'$ is a universal first order thickening,

2. for any diagram (76.15.0.1) with $T'$ a scheme a unique dotted arrow exists making the diagram commute, and

3. for any diagram (76.15.0.1) with $T'$ an affine scheme a unique dotted arrow exists making the diagram commute.

Proof. The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) are formal. Assume (3) a assume given an arbitrary diagram (76.15.0.1). Choose a presentation $T' = U'/R'$, see Spaces, Definition 65.9.3. We may assume that $U' = \coprod U'_ i$ is a disjoint union of affines, so $R' = U' \times _{T'} U' = \coprod _{i, j} U'_ i \times _ T' U'_ j$. For each pair $(i, j)$ choose an affine open covering $U'_ i \times _ T' U'_ j = \bigcup _ k R'_{ijk}$. Denote $U_ i, R_{ijk}$ the fibre products with $T$ over $T'$. Then each $U_ i \subset U'_ i$ and $R_{ijk} \subset R'_{ijk}$ is a first order thickening of affine schemes. Denote $a_ i : U_ i \to Z$, resp. $a_{ijk} : R_{ijk} \to Z$ the composition of $a : T \to Z$ with the morphism $U_ i \to T$, resp. $R_{ijk} \to T$. By (3) applied to $a_ i : U_ i \to Z$ we obtain unique morphisms $a'_ i : U'_ i \to Z'$. By (3) applied to $a_{ijk}$ we see that the two compositions $R'_{ijk} \to R'_ i \to Z'$ and $R'_{ijk} \to R'_ j \to Z'$ are equal. Hence $a' = \coprod a'_ i : U' = \coprod U'_ i \to Z'$ descends to the quotient sheaf $T' = U'/R'$ and we win. $\square$

Lemma 76.15.2. Let $S$ be a scheme. Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$. If $Z \subset Z'$ is a universal first order thickening of $Z$ over $Y$ and $Y \to X$ is formally étale, then $Z \subset Z'$ is a universal first order thickening of $Z$ over $X$.

Proof. This is formal. Namely, by Lemma 76.15.1 it suffices to consider solid commutative diagrams (76.15.0.1) with $T'$ an affine scheme. The composition $T \to Z \to Y$ lifts uniquely to $T' \to Y$ as $Y \to X$ is assumed formally étale. Hence the fact that $Z \subset Z'$ is a universal first order thickening over $Y$ produces the desired morphism $a' : T' \to Z'$. $\square$

Lemma 76.15.3. Let $S$ be a scheme. Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$. Assume $Z \to Y$ is étale.

1. If $Y \subset Y'$ is a universal first order thickening of $Y$ over $X$, then the unique étale morphism $Z' \to Y'$ such that $Z = Y \times _{Y'} Z'$ (see Theorem 76.8.1) is a universal first order thickening of $Z$ over $X$.

2. If $Z \to Y$ is surjective and $(Z \subset Z') \to (Y \subset Y')$ is an étale morphism of first order thickenings over $X$ and $Z'$ is a universal first order thickening of $Z$ over $X$, then $Y'$ is a universal first order thickening of $Y$ over $X$.

Proof. Proof of (1). By Lemma 76.15.1 it suffices to consider solid commutative diagrams (76.15.0.1) with $T'$ an affine scheme. The composition $T \to Z \to Y$ lifts uniquely to $T' \to Y'$ as $Y'$ is the universal first order thickening. Then the fact that $Z' \to Y'$ is étale implies (see Lemma 76.13.5) that $T' \to Y'$ lifts to the desired morphism $a' : T' \to Z'$.

Proof of (2). Let $T \subset T'$ be a first order thickening over $X$ and let $a : T \to Y$ be a morphism. Set $W = T \times _ Y Z$ and denote $c : W \to Z$ the projection Let $W' \to T'$ be the unique étale morphism such that $W = T \times _{T'} W'$, see Theorem 76.8.1. Note that $W' \to T'$ is surjective as $Z \to Y$ is surjective. By assumption we obtain a unique morphism $c' : W' \to Z'$ over $X$ restricting to $c$ on $W$. By uniqueness the two restrictions of $c'$ to $W' \times _{T'} W'$ are equal (as the two restrictions of $c$ to $W \times _ T W$ are equal). Hence $c'$ descends to a unique morphism $a' : T' \to Y'$ and we win. $\square$

Lemma 76.15.4. Let $S$ be a scheme. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $S$. There exists a universal first order thickening $Z \subset Z'$ of $Z$ over $X$.

Proof. Choose any commutative diagram

$\xymatrix{ V \ar[d] \ar[r] & U \ar[d] \\ Z \ar[r] & X }$

where $V$ and $U$ are schemes and the vertical arrows are étale. Note that $V \to U$ is a formally unramified morphism of schemes, see Lemma 76.14.2. Combining Lemma 76.15.1 and More on Morphisms, Lemma 37.7.1 we see that a universal first order thickening $V \subset V'$ of $V$ over $U$ exists. By Lemma 76.15.2 part (1) $V'$ is a universal first order thickening of $V$ over $X$.

Fix a scheme $U$ and a surjective étale morphism $U \to X$. The argument above shows that for any $V \to Z$ étale with $V$ a scheme such that $V \to Z \to X$ factors through $U$ a universal first order thickening $V \subset V'$ of $V$ over $X$ exists (but does not depend on the chosen factorization of $V \to X$ through $U$). Now we may choose $V$ such that $V \to Z$ is surjective étale (see Spaces, Lemma 65.11.6). Then $R = V \times _ Z V$ a scheme étale over $Z$ such that $R \to X$ factors through $U$ also. Hence we obtain universal first order thickenings $V \subset V'$ and $R \subset R'$ over $X$. As $V \subset V'$ is a universal first order thickening, the two projections $s, t : R \to V$ lift to morphisms $s', t': R' \to V'$. By Lemma 76.15.3 as $R'$ is the universal first order thickening of $R$ over $X$ these morphisms are étale. Then $(t', s') : R' \to V'$ is an étale equivalence relation and we can set $Z' = V'/R'$. Since $V' \to Z'$ is surjective étale and $v'$ is the universal first order thickening of $V$ over $X$ we conclude from Lemma 76.15.2 part (2) that $Z'$ is a universal first order thickening of $Z$ over $X$. $\square$

Definition 76.15.5. Let $S$ be a scheme. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $S$.

1. The universal first order thickening of $Z$ over $X$ is the thickening $Z \subset Z'$ constructed in Lemma 76.15.4.

2. The conormal sheaf of $Z$ over $X$ is the conormal sheaf of $Z$ in its universal first order thickening $Z'$ over $X$.

We often denote the conormal sheaf $\mathcal{C}_{Z/X}$ in this situation.

Thus we see that there is a short exact sequence of sheaves

$0 \to \mathcal{C}_{Z/X} \to \mathcal{O}_{Z'} \to \mathcal{O}_ Z \to 0$

on $Z_{\acute{e}tale}$ and $\mathcal{C}_{Z/X}$ is a quasi-coherent $\mathcal{O}_ Z$-module. The following lemma proves that there is no conflict between this definition and the definition in case $Z \to X$ is an immersion.

Lemma 76.15.6. Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic spaces over $S$. Then

1. $i$ is formally unramified,

2. the universal first order thickening of $Z$ over $X$ is the first order infinitesimal neighbourhood of $Z$ in $X$ of Definition 76.12.1,

3. the conormal sheaf of $i$ in the sense of Definition 76.5.1 agrees with the conormal sheaf of $i$ in the sense of Definition 76.15.5.

Proof. An immersion of algebraic spaces is by definition a representable morphism. Hence by Morphisms, Lemmas 29.35.7 and 29.35.8 an immersion is unramified (via the abstract principle of Spaces, Lemma 65.5.8). Hence it is formally unramified by Lemma 76.14.7. The other assertions follow by combining Lemmas 76.12.2 and 76.12.3 and the definitions. $\square$

Lemma 76.15.7. Let $S$ be a scheme. Let $Z \to X$ be a formally unramified morphism of algebraic spaces over $S$. Then the universal first order thickening $Z'$ is formally unramified over $X$.

Proof. Let $T \subset T'$ be a first order thickening of affine schemes over $X$. Let

$\xymatrix{ Z' \ar[d] & T \ar[l]^ c \ar[d] \\ X & T' \ar[l] \ar[lu]^{a, b} }$

be a commutative diagram. Set $T_0 = c^{-1}(Z) \subset T$ and $T'_ a = a^{-1}(Z)$ (scheme theoretically). Since $Z'$ is a first order thickening of $Z$, we see that $T'$ is a first order thickening of $T'_ a$. Moreover, since $c = a|_ T$ we see that $T_0 = T \cap T'_ a$ (scheme theoretically). As $T'$ is a first order thickening of $T$ it follows that $T'_ a$ is a first order thickening of $T_0$. Now $a|_{T'_ a}$ and $b|_{T'_ a}$ are morphisms of $T'_ a$ into $Z'$ over $X$ which agree on $T_0$ as morphisms into $Z$. Hence by the universal property of $Z'$ we conclude that $a|_{T'_ a} = b|_{T'_ a}$. Thus $a$ and $b$ are morphism from the first order thickening $T'$ of $T'_ a$ whose restrictions to $T'_ a$ agree as morphisms into $Z$. Thus using the universal property of $Z'$ once more we conclude that $a = b$. In other words, the defining property of a formally unramified morphism holds for $Z' \to X$ as desired. $\square$

Lemma 76.15.8. Let $S$ be a scheme Consider a commutative diagram of algebraic spaces over $S$

$\xymatrix{ Z \ar[r]_ h \ar[d]_ f & X \ar[d]^ g \\ W \ar[r]^{h'} & Y }$

with $h$ and $h'$ formally unramified. Let $Z \subset Z'$ be the universal first order thickening of $Z$ over $X$. Let $W \subset W'$ be the universal first order thickening of $W$ over $Y$. There exists a canonical morphism $(f, f') : (Z, Z') \to (W, W')$ of thickenings over $Y$ which fits into the following commutative diagram

$\xymatrix{ & & & Z' \ar[ld] \ar[d]^{f'} \\ Z \ar[rr] \ar[d]_ f \ar[rrru] & & X \ar[d] & W' \ar[ld] \\ W \ar[rrru]|!{[rr];[rruu]}\hole \ar[rr] & & Y }$

In particular the morphism $(f, f')$ of thickenings induces a morphism of conormal sheaves $f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$.

Proof. The first assertion is clear from the universal property of $W'$. The induced map on conormal sheaves is the map of Lemma 76.5.3 applied to $(Z \subset Z') \to (W \subset W')$. $\square$

Lemma 76.15.9. Let $S$ be a scheme. Let

$\xymatrix{ Z \ar[r]_ h \ar[d]_ f & X \ar[d]^ g \\ W \ar[r]^{h'} & Y }$

be a fibre product diagram of algebraic spaces over $S$ with $h'$ formally unramified. Then $h$ is formally unramified and if $W \subset W'$ is the universal first order thickening of $W$ over $Y$, then $Z = X \times _ Y W \subset X \times _ Y W'$ is the universal first order thickening of $Z$ over $X$. In particular the canonical map $f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ of Lemma 76.15.8 is surjective.

Proof. The morphism $h$ is formally unramified by Lemma 76.14.5. It is clear that $X \times _ Y W'$ is a first order thickening. It is straightforward to check that it has the universal property because $W'$ has the universal property (by mapping properties of fibre products). See Lemma 76.5.5 for why this implies that the map of conormal sheaves is surjective. $\square$

Lemma 76.15.10. Let $S$ be a scheme. Let

$\xymatrix{ Z \ar[r]_ h \ar[d]_ f & X \ar[d]^ g \\ W \ar[r]^{h'} & Y }$

be a fibre product diagram of algebraic spaces over $S$ with $h'$ formally unramified and $g$ flat. In this case the corresponding map $Z' \to W'$ of universal first order thickenings is flat, and $f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ is an isomorphism.

Proof. Flatness is preserved under base change, see Morphisms of Spaces, Lemma 67.30.4. Hence the first statement follows from the description of $W'$ in Lemma 76.15.9. It is clear that $X \times _ Y W'$ is a first order thickening. It is straightforward to check that it has the universal property because $W'$ has the universal property (by mapping properties of fibre products). See Lemma 76.5.5 for why this implies that the map of conormal sheaves is an isomorphism. $\square$

Lemma 76.15.11. Taking the universal first order thickenings commutes with étale localization. More precisely, let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over a base scheme $S$. Let

$\xymatrix{ V \ar[d] \ar[r] & U \ar[d] \\ Z \ar[r] & X }$

be a commutative diagram with étale vertical arrows. Let $Z'$ be the universal first order thickening of $Z$ over $X$. Then $V \to U$ is formally unramified and the universal first order thickening $V'$ of $V$ over $U$ is étale over $Z'$. In particular, $\mathcal{C}_{Z/X}|_ V = \mathcal{C}_{V/U}$.

Proof. The first statement is Lemma 76.14.2. The compatibility of universal first order thickenings is a consequence of Lemmas 76.15.2 and 76.15.3. $\square$

Lemma 76.15.12. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $B$. Let $Z \subset Z'$ be the universal first order thickening of $Z$ over $X$ with structure morphism $h' : Z' \to X$. The canonical map

$\text{d}h' : (h')^*\Omega _{X/B} \to \Omega _{Z'/B}$

induces an isomorphism $h^*\Omega _{X/B} \to \Omega _{Z'/B} \otimes \mathcal{O}_ Z$.

Proof. The map $c_{h'}$ is the map defined in Lemma 76.7.6. If $i : Z \to Z'$ is the given closed immersion, then $i^*c_{h'}$ is a map $h^*\Omega _{X/S} \to \Omega _{Z'/S} \otimes \mathcal{O}_ Z$. Checking that it is an isomorphism reduces to the case of schemes by étale localization, see Lemma 76.15.11 and Lemma 76.7.3. In this case the result is More on Morphisms, Lemma 37.7.9. $\square$

Lemma 76.15.13. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $B$. There is a canonical exact sequence

$\mathcal{C}_{Z/X} \to h^*\Omega _{X/B} \to \Omega _{Z/B} \to 0.$

The first arrow is induced by $\text{d}_{Z'/B}$ where $Z'$ is the universal first order neighbourhood of $Z$ over $X$.

Proof. We know that there is a canonical exact sequence

$\mathcal{C}_{Z/Z'} \to \Omega _{Z'/S} \otimes \mathcal{O}_ Z \to \Omega _{Z/S} \to 0.$

see Lemma 76.7.10. Hence the result follows on applying Lemma 76.15.12. $\square$

Lemma 76.15.14. Let $S$ be a scheme. Let

$\xymatrix{ Z \ar[r]_ i \ar[rd]_ j & X \ar[d] \\ & Y }$

be a commutative diagram of algebraic spaces over $S$ where $i$ and $j$ are formally unramified. Then there is a canonical exact sequence

$\mathcal{C}_{Z/Y} \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/Y} \to 0$

where the first arrow comes from Lemma 76.15.8 and the second from Lemma 76.15.13.

Proof. Since the maps have been defined, checking the sequence is exact reduces to the case of schemes by étale localization, see Lemma 76.15.11 and Lemma 76.7.3. In this case the result is More on Morphisms, Lemma 37.7.11. $\square$

Lemma 76.15.15. Let $S$ be a scheme. Let $Z \to Y \to X$ be formally unramified morphisms of algebraic spaces over $S$.

1. If $Z \subset Z'$ is the universal first order thickening of $Z$ over $X$ and $Y \subset Y'$ is the universal first order thickening of $Y$ over $X$, then there is a morphism $Z' \to Y'$ and $Y \times _{Y'} Z'$ is the universal first order thickening of $Z$ over $Y$.

2. There is a canonical exact sequence

$i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0$

where the maps come from Lemma 76.15.8 and $i : Z \to Y$ is the first morphism.

Proof. The map $h : Z' \to Y'$ in (1) comes from Lemma 76.15.8. The assertion that $Y \times _{Y'} Z'$ is the universal first order thickening of $Z$ over $Y$ is clear from the universal properties of $Z'$ and $Y'$. By Lemma 76.5.6 we have an exact sequence

$(i')^*\mathcal{C}_{Y \times _{Y'} Z'/Z'} \to \mathcal{C}_{Z/Z'} \to \mathcal{C}_{Z/Y \times _{Y'} Z'} \to 0$

where $i' : Z \to Y \times _{Y'} Z'$ is the given morphism. By Lemma 76.5.5 there exists a surjection $h^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{Y \times _{Y'} Z'/Z'}$. Combined with the equalities $\mathcal{C}_{Y/Y'} = \mathcal{C}_{Y/X}$, $\mathcal{C}_{Z/Z'} = \mathcal{C}_{Z/X}$, and $\mathcal{C}_{Z/Y \times _{Y'} Z'} = \mathcal{C}_{Z/Y}$ this proves the lemma. $\square$

Comment #1150 by Matthieu Romagny on

In the initial sentence defining what a universal first order thickening of $Z$ over $X$ is, the morphism $T\subset T'$ is given as a thickening of $X$-schemes so there is a map $T\to X$; then it seems one should require that $a:T\to Z$ is a map of $X$-schemes.

Comment #1171 by on

OK, but we say "solid commutative diagram" which implies that $a$ is a morphism over #$X$. Similarly, $a'$ is forced to be a morphism over $X$. OK?

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