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. $\square$

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