Lemma 106.5.18. If $\mathcal{Z} \hookrightarrow \mathcal{X}$ is a closed immersion of locally Noetherian schemes, and if $z \in |\mathcal{Z}|$ has image $x \in |\mathcal{X}|$, then $\dim _ z (\mathcal{Z}) \leq \dim _ x(\mathcal{X})$.

Proof. Choose a smooth surjective morphism $U\to \mathcal{X}$ whose source is a scheme; the base-changed morphism $V = U\times _{\mathcal{X}} \mathcal{Z} \to \mathcal{Z}$ is then also smooth and surjective, and the projection $V \to U$ is a closed immersion. If $v \in |V|$ maps to $z \in |\mathcal{Z}|$, and if we let $u$ denote the image of $v$ in $|U|$, then clearly $\dim _ v(V) \leq \dim _ u(U)$, while $\dim _ v (V_ z) = \dim _ u(U_ x)$, by Lemma 106.5.9. Thus

$\dim _ z(\mathcal{Z}) = \dim _ v(V) - \dim _ v(V_ z) \leq \dim _ u(U) - \dim _ u(U_ x) = \dim _ x(\mathcal{X}),$

as claimed. $\square$

Comment #7954 by R on

The statement should probably read "locally Noetherian algebraic stacks" instead of "locally Noetherian schemes".

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