The Stacks project

Proof. This is true because $s : \mathcal{O}_{U, \overline{u}} \to \mathcal{O}_{R, e(\overline{u})}$ is a flat local homomorphism of Noetherian local rings and hence

by Algebra, Lemma 10.112.7. $\square$

Proof. Suppose we have two smooth neighbourhoods $(U, u)$ and $(U', u')$ of $x$ with $u$ and $u'$ finite type points. After shrinking $U$ and $U'$ we may assume that $u$ and $u'$ are closed points (by definition of finite type points). Then we choose a surjective étale morphism $W \to U \times _\mathcal {X} U'$. Let $W_ u$ be the fibre of $W \to U$ over $u$ and let $W_{u'}$ be the fibre of $W \to U'$ over $u'$. Since $u$ and $u'$ map to the same point of $|\mathcal{X}|$ we see that $W_ u \cap W_{u'}$ is nonempty. Hence we may choose a closed point $w \in W$ mapping to both $u$ and $u'$. This reduces us to the discussion in the next paragraph.

Assume $(U', u') \to (U, u)$ is a smooth morphism of smooth neightbourhoods of $x$ with $u$ and $u'$ closed points. Goal: prove the invariant defined for $(U, u)$ is the same as the invariant defined for $(U', u')$. To see this observe that $\mathcal{O}_{U, u} \to \mathcal{O}_{U', u'}$ is a flat local homomorphism of Noetherian local rings and hence

by Algebra, Lemma 10.112.7. (We omit working through all the steps to relate properties of local rings and their strict henselizations, see More on Algebra, Section 15.45). On the other hand we have

Thus we see that

To prove the lemma it suffices to show that

Observe that this isn't always true (example: if $U'_ u$ is a curve and $u'$ is the generic point of this curve). However, we know that $u'$ is a closed point of the algebraic space $U'_ u$ locally of finite type over $u$. In this case the equality holds because, first $\dim _{(u', u')}(U'_ u \times _ u U'_ u) = 2\dim _{u'}(U'_ u)$ by Varieties, Lemma 33.20.5 and second the agreement of dimension with dimension of local rings in closed points of locally algebraic schemes, see Varieties, Lemma 33.20.3. We omit the translation of these results for schemes into the language of algebraic spaces. $\square$

Proof. Let $s \in S$ be the image of $x_0$. If $\mathcal{O}_{S, s}$ is a G-ring (a condition that is almost always satisfied in practice), then we can prove the lemma as follows. By Lemma 107.2.8, we may find a smooth morphism $U \to \mathcal{X}$, whose source is a scheme, containing a point $u_0 \in U$ of residue field $k$, such that induced morphism $\mathop{\mathrm{Spec}}(k) \to U \to \mathcal{X}$ coincides with $x_0$ and such that $A = \mathcal{O}_{U, u_0}^\wedge $. Write $R = U \times _\mathcal {X} U$. Then we may identify $\mathcal{O}_{R, e(u_0)}^\wedge $ with $B$. Hence the equality follows from the definitions.

In the rest of this proof we explain how to prove the lemma in general, but we urge the reader to skip this.

First let us show that the right hand side is independent of the choice of $(A, B, s, t, c)$. Namely, suppose that $(A', B', s', t', c')$ is a second choice. Since $A$ and $A'$ are versal rings to $\mathcal{X}$ at $x_0$, we can choose, after possibly switching $A$ and $A'$, a formally smooth map $A \to A'$ compatible with the given versal formal objects $\xi $ and $\xi '$ over $A$ and $A'$. Recall that $\widehat{\mathcal{C}}_\Lambda $ has coproducts and that these are given by completed tensor product over $\Lambda $, see Formal Deformation Theory, Lemma 90.4.4. Then $B$ prorepresents the functor of isomorphisms between the two pushforwards of $\xi $ to $A \widehat{\otimes }_\Lambda A$. Similarly for $B'$. We conclude that

It is straightforward to see that

is formally smooth of relative dimension equal to $2$ times the relative dimension of the formally smooth map $A \to A'$. (This follows from general principles, but also because in this particular case $A'$ is a power series ring over $A$ in $r$ variables.) Hence $B \to B'$ is formally smooth of relative dimension $2(\dim (A') - \dim (A))$ as desired.

Next, let $l/k$ be a finite extension. let $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ be the induced point. We claim that the right hand side of the formula is the same for $x_0$ as it is for $x_{l, 0}$. This can be shown by choosing $A \to A'$ as in Lemma 107.2.5 and arguing exactly as in the preceding paragraph. We omit the details.

Finally, arguing as in the proof of Lemma 107.2.10 we can use the compatibilities in the previous two paragraphs to reduce to the case (discussed in the first paragraph) where $A$ is the complete local ring of $U$ at $u_0$ for some scheme smooth over $\mathcal{X}$ and finite type point $u_0$. Details omitted. $\square$