The Stacks project

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

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

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

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. $\square$

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. $\square$

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

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. $\square$

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 $(Z \subset Z') \to (W \subset W')$. $\square$

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. $\square$

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. $\square$

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. $\square$

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. $\square$

Proof. We know that there is a canonical exact sequence

see Morphisms, Lemma 29.32.15. Hence the result follows on applying Lemma 37.7.9. $\square$

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

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

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. $\square$

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

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. $\square$