Lemma 67.34.3. Let S be a scheme. Let
\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }
be a fibre product diagram of algebraic spaces over S. Let x' \in |X'|. Set x = g'(x'). Assume f locally of finite type. Then
\begin{matrix} \text{relative dimension of }f\text{ at }x
\\ =
\\ \text{relative dimension of }f'\text{ at }x'
\end{matrix}
we have
\begin{matrix} \text{dimension of local ring of the fibre of }f'\text{ at }x'
\\ -
\\ \text{dimension of local ring of the fibre of }f\text{ at }x
\\ =
\\ \text{transcendence degree of }x/f(x)
\\ -
\\ \text{transcendence degree of }x'/f'(x')
\end{matrix}
and the common value is \geq 0,
given x and y' \in |Y'| mapping to the same y \in |Y| there exists a choice of x' such that the integer in (2) is 0.
Proof.
Choose a surjective étale morphism V \to Y with V a scheme. Choose a surjective étale morphism U \to V \times _ Y X with U a scheme. Choose a surjective étale morphism V' \to V \times _ Y Y' with V' a scheme. Set U' = V' \times _ V U. Then the induced morphism U' \to X' is also surjective and étale (argument omitted). Choose u' \in U' mapping to x'. At this point parts (1) and (2) follow by applying Morphisms, Lemma 29.28.3 to the diagram of schemes involving U', U, V', V and the point u'. To prove (3) first choose v \in V mapping to y. Then using Properties of Spaces, Lemma 66.4.3 we can choose v' \in V' mapping to y' and v and u \in U mapping to x and v. Finally, according to Morphisms, Lemma 29.28.3 we can choose u' \in U' mapping to v' and u such that the integer is zero. Then taking x' \in |X'| the image of u' works.
\square
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