Lemma 66.35.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ is locally of finite type. Then

$\dim (X) \leq \dim (Y) + E$

where $E$ is the supremum of the transcendence degrees of $\xi /f(\xi )$ where $\xi$ runs through the points at which the local ring of $X$ has dimension $0$.

Proof. Immediate consequence of Lemma 66.35.1 and Properties of Spaces, Lemma 65.10.3. $\square$

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