## 37.7 Universal first order thickenings

Let $h : Z \to X$ be a morphism of schemes. 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

$\xymatrix{ & Z \ar[ld] & & T \ar[rd] \ar[ll]^ a \\ Z' \ar[rrd] & & & & T' \ar@{..>}[llll]_{a'} \ar[lld]^ b \\ & & X }$

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.

Lemma 37.7.1. Let $h : Z \to X$ be a formally unramified morphism of schemes. There exists a universal first order thickening $Z \subset Z'$ of $Z$ over $X$.

Proof. During this proof we will say $Z \subset Z'$ is a universal first order thickening of $Z$ over $X$ if it satisfies the condition of the lemma. We will construct the universal first order thickening $Z \subset Z'$ over $X$ by glueing, starting with the affine case which is Algebra, Lemma 10.149.1. We begin with some general remarks.

If a universal first order thickening of $Z$ over $X$ exists, then it is unique up to unique isomorphism. Moreover, suppose that $V \subset Z$ and $U \subset X$ are open subschemes such that $h(V) \subset U$. Let $Z \subset Z'$ be a universal first order thickening of $Z$ over $X$. Let $V' \subset Z'$ be the open subscheme such that $V = Z \cap V'$. Then we claim that $V \subset V'$ is the universal first order thickening of $V$ over $U$. Namely, suppose given any diagram

$\xymatrix{ V \ar[d]_ h & T \ar[l]^ a \ar[d] \\ U & T' \ar[l]_ b }$

where $T \subset T'$ is a first order thickening over $U$. By the universal property of $Z'$ we obtain $(a, a') : (T \subset T') \to (Z \subset Z')$. But since we have equality $|T| = |T'|$ of underlying topological spaces we see that $a'(T') \subset V'$. Hence we may think of $(a, a')$ as a morphism of thickenings $(a, a') : (T \subset T') \to (V \subset V')$ over $U$. Uniqueness is clear also. In a completely similar manner one proves that if $h(Z) \subset U$ and $Z \subset Z'$ is a universal first order thickening over $U$, then $Z \subset Z'$ is a universal first order thickening over $X$.

Before we glue affine pieces let us show that the lemma holds if $Z$ and $X$ are affine. Say $X = \mathop{\mathrm{Spec}}(R)$ and $Z = \mathop{\mathrm{Spec}}(S)$. By Algebra, Lemma 10.149.1 there exists a first order thickening $Z \subset Z'$ over $X$ which has the universal property of the lemma for diagrams

$\xymatrix{ Z \ar[d]_ h & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b }$

where $T, T'$ are affine. Given a general diagram we can choose an affine open covering $T' = \bigcup T'_ i$ and we obtain morphisms $a'_ i : T'_ i \to Z'$ over $X$ such that $a'_ i|_{T_ i} = a|_{T_ i}$. By uniqueness we see that $a'_ i$ and $a'_ j$ agree on any affine open of $T'_ i \cap T'_ j$. Hence the morphisms $a'_ i$ glue to a global morphism $a' : T' \to Z'$ over $X$ as desired. Thus the lemma holds if $X$ and $Z$ are affine.

Choose an affine open covering $Z = \bigcup Z_ i$ such that each $Z_ i$ maps into an affine open $U_ i$ of $X$. By Lemma 37.6.5 the morphisms $Z_ i \to U_ i$ are formally unramified. Hence by the affine case we obtain universal first order thickenings $Z_ i \subset Z_ i'$ over $U_ i$. By the general remarks above $Z_ i \subset Z_ i'$ is also a universal first order thickening of $Z_ i$ over $X$. Let $Z'_{i, j} \subset Z'_ i$ be the open subscheme such that $Z_ i \cap Z_ j = Z'_{i, j} \cap Z_ i$. By the general remarks we see that both $Z'_{i, j}$ and $Z'_{j, i}$ are universal first order thickenings of $Z_ i \cap Z_ j$ over $X$. Thus, by the first of our general remarks, we see that there is a canonical isomorphism $\varphi _{ij} : Z'_{i, j} \to Z'_{j, i}$ inducing the identity on $Z_ i \cap Z_ j$. We claim that these morphisms satisfy the cocycle condition of Schemes, Section 26.14. (Verification omitted. Hint: Use that $Z'_{i, j} \cap Z'_{i, k}$ is the universal first order thickening of $Z_ i \cap Z_ j \cap Z_ k$ which determines it up to unique isomorphism by what was said above.) Hence we can use the results of Schemes, Section 26.14 to get a first order thickening $Z \subset Z'$ over $X$ which the property that the open subscheme $Z'_ i \subset Z'$ with $Z_ i = Z'_ i \cap Z$ is a universal first order thickening of $Z_ i$ over $X$.

It turns out that this implies formally that $Z'$ is a universal first order thickening of $Z$ over $X$. Namely, we have the universal property for any diagram

$\xymatrix{ Z \ar[d]_ h & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b }$

where $a(T)$ is contained in some $Z_ i$. Given a general diagram we can choose an open covering $T' = \bigcup T'_ i$ such that $a(T_ i) \subset Z_ i$. We obtain morphisms $a'_ i : T'_ i \to Z'$ over $X$ such that $a'_ i|_{T_ i} = a|_{T_ i}$. We see that $a'_ i$ and $a'_ j$ necessarily agree on $T'_ i \cap T'_ j$ since both $a'_ i|_{T'_ i \cap T'_ j}$ and $a'_ j|_{T'_ i \cap T'_ j}$ are solutions of the problem of mapping into the universal first order thickening $Z'_ i \cap Z'_ j$ of $Z_ i \cap Z_ j$ over $X$. Hence the morphisms $a'_ i$ glue to a global morphism $a' : T' \to Z'$ over $X$ as desired. This finishes the proof. $\square$

Definition 37.7.2. Let $h : Z \to X$ be a formally unramified morphism of schemes.

1. The universal first order thickening of $Z$ over $X$ is the thickening $Z \subset Z'$ constructed in Lemma 37.7.1.

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 $Z$. The following lemma proves that there is no conflict between this definition and the definition in case $Z \to X$ is an immersion.

Lemma 37.7.3. Let $i : Z \to X$ be an immersion of schemes. 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 37.5.1, and

3. the conormal sheaf of $i$ in the sense of Morphisms, Definition 29.31.1 agrees with the conormal sheaf of $i$ in the sense of Definition 37.7.2.

Proof. By Morphisms, Lemmas 29.35.7 and 29.35.8 an immersion is unramified, hence formally unramified by Lemma 37.6.8. The other assertions follow by combining Lemmas 37.5.2 and 37.5.3 and the definitions. $\square$

Lemma 37.7.4. Let $Z \to X$ be a formally unramified morphism of schemes. Then the universal first order thickening $Z'$ is formally unramified over $X$.

Proof. There are two proofs. The first is to show that $\Omega _{Z'/X} = 0$ by working affine locally and applying Algebra, Lemma 10.149.5. Then Lemma 37.6.7 implies what we want. The second is a direct argument as follows.

Let $T \subset T'$ be a first order thickening. Let

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

be a commutative diagram. Consider two morphisms $a, b : T' \to Z'$ fitting into the 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 37.7.5. Consider a commutative diagram of schemes

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

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 Morphisms, Lemma 29.31.3 applied to $(Z \subset Z') \to (W \subset W')$. $\square$

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

be a fibre product diagram in the category of schemes 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 37.7.5 is surjective.

Proof. The morphism $h$ is formally unramified by Lemma 37.6.4. 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 Morphisms, Lemma 29.31.4 for why this implies that the map of conormal sheaves is surjective. $\square$

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

be a fibre product diagram in the category of schemes 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, Lemma 29.25.8. Hence the first statement follows from the description of $W'$ in Lemma 37.7.6. 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 Morphisms, Lemma 29.31.4 for why this implies that the map of conormal sheaves is an isomorphism. $\square$

Lemma 37.7.8. Taking the universal first order thickenings commutes with taking opens. More precisely, let $h : Z \to X$ be a formally unramified morphism of schemes. Let $V \subset Z$, $U \subset X$ be opens such that $h(V) \subset U$. Let $Z'$ be the universal first order thickening of $Z$ over $X$. Then $h|_ V : V \to U$ is formally unramified and the universal first order thickening of $V$ over $U$ is the open subscheme $V' \subset Z'$ such that $V = Z \cap V'$. In particular, $\mathcal{C}_{Z/X}|_ V = \mathcal{C}_{V/U}$.

Proof. The first statement is Lemma 37.6.5. The compatibility of universal thickenings can be deduced from the proof of Lemma 37.7.1, or from Algebra, Lemma 10.149.4 or deduced from Lemma 37.7.7. $\square$

Lemma 37.7.9. Let $h : Z \to X$ be a formally unramified morphism of schemes over $S$. Let $Z \subset Z'$ be the universal first order thickening of $Z$ over $X$ with structure morphism $h' : Z' \to X$. The canonical map

$c_{h'} : (h')^*\Omega _{X/S} \longrightarrow \Omega _{Z'/S}$

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

Proof. The map $c_{h'}$ is the map defined in Morphisms, Lemma 29.32.8. 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 affine case by localization, see Lemma 37.7.8 and Morphisms, Lemma 29.32.3. In this case the result is Algebra, Lemma 10.149.5. $\square$

Lemma 37.7.10. Let $h : Z \to X$ be a formally unramified morphism of schemes over $S$. There is a canonical exact sequence

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

The first arrow is induced by $\text{d}_{Z'/S}$ 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 Morphisms, Lemma 29.32.15. Hence the result follows on applying Lemma 37.7.9. $\square$

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

be a commutative diagram of schemes 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 37.7.5 and the second from Lemma 37.7.10.

Proof. Denote $Z \to Z'$ the universal first order thickening of $Z$ over $X$. Denote $Z \to Z''$ the universal first order thickening of $Z$ over $Y$. By Lemma 37.7.10 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

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

Apply Morphisms, Lemma 29.32.18 to the left triangle to get an exact sequence

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

As $Z''$ is formally unramified over $Y$ (see Lemma 37.7.4) we have $\Omega _{Z'/Z''} = \Omega _{Z/Y}$ (by combining Lemma 37.6.7 and Morphisms, Lemma 29.32.9). Then we have $(i')^*\Omega _{Z'/Y} = i^*\Omega _{X/Y}$ by Lemma 37.7.9. $\square$

Lemma 37.7.12. Let $Z \to Y \to X$ be formally unramified morphisms of schemes.

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 37.7.5 and $i : Z \to Y$ is the first morphism.

Proof. The map $h : Z' \to Y'$ in (1) comes from Lemma 37.7.5. 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 Morphisms, Lemma 29.31.5 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 Morphisms, Lemma 29.31.4 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. $\square$

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