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

1. $\begin{matrix} \text{relative dimension of }g \circ f\text{ at }x \\ \leq \\ \text{relative dimension of }f\text{ at }x \\ + \\ \text{relative dimension of }g\text{ at }y \end{matrix}$
2. 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 $x$,

3. $\begin{matrix} \text{transcendence degree of }x/g(f(x)) \\ = \\ \text{transcendence degree of }x/f(x) \\ + \\ \text{transcendence degree of }f(x)/g(f(x)) \end{matrix}$

Proof. Choose a diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z }$

with $U, V, W$ schemes and vertical arrows étale and surjective. (See Spaces, Lemma 64.11.6.) Choose $u \in U$ mapping to $x$. Set $v, w$ equal to the images of $u$ in $V, W$. Apply Morphisms, Lemma 29.28.2 to the top row and the points $u, v, w$. Details omitted. $\square$

Comment #534 by Kestutis Cesnavicius on

'The' --> 'the'

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