Lemma 29.44.9. Let $f : X \to Y$ be an integral morphism. Then $\dim (X) \leq \dim (Y)$. If $f$ is surjective then $\dim (X) = \dim (Y)$.

Proof. Since the dimension of $X$ and $Y$ is the supremum of the dimensions of the members of an affine open covering, we may assume $Y$ and $X$ are affine. The inequality follows from Algebra, Lemma 10.112.3. The equality then follows from Algebra, Lemmas 10.112.1 and 10.36.22. $\square$

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