Lemma 66.34.2. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Let $x \in |X|$ and set $y = f(x)$. Assume $f$ and $g$ locally of finite type. Then

- \[ \begin{matrix} \text{relative dimension of }g \circ f\text{ at }x \\ \leq \\ \text{relative dimension of }f\text{ at }x \\ + \\ \text{relative dimension of }g\text{ at }y \end{matrix} \]
equality holds in (1) if for some morphism $\mathop{\mathrm{Spec}}(k) \to Z$ from the spectrum of a field in the class of $g(f(x)) = g(y)$ the morphism $X_ k \to Y_ k$ is flat at $x$, for example if $f$ is flat at $x$,

- \[ \begin{matrix} \text{transcendence degree of }x/g(f(x)) \\ = \\ \text{transcendence degree of }x/f(x) \\ + \\ \text{transcendence degree of }f(x)/g(f(x)) \end{matrix} \]

## Comments (2)

Comment #534 by Kestutis Cesnavicius on

Comment #546 by Johan on