Lemma 66.34.3. 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 65.4.3 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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