The Stacks project

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. $\square$