Lemma 65.34.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. Assume $f$ is locally of finite type. Then we have

$\begin{matrix} \text{relative dimension of }f\text{ at }x \\ = \\ \text{dimension of local ring of the fibre of }f\text{ at }x \\ + \\ \text{transcendence degree of }x/f(x) \end{matrix}$

where the notation is as in Definition 65.33.1.

Proof. This follows immediately from Morphisms, Lemma 29.28.1 applied to $h : U \to V$ and $u \in U$ as in Lemma 65.22.5. $\square$

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