Lemma 81.7.4. In Situation 81.2.1 let $X, Y/B$ be good and let $f : X \to Y$ be a morphism over $B$. Assume $X$, $Y$ integral and $\dim _\delta (X) = \dim _\delta (Y)$. Then either $f$ factors through a proper closed subspace of $Y$, or $f$ is dominant and the extension of function fields $R(X) / R(Y)$ is finite.
Proof. By Lemma 81.7.1 there is a unique integral closed subspace $Z \subset Y$ such that $f$ factors through a dominant morphism $X \to Z$. Then $Z = Y$ if and only if $\dim _\delta (Z) = \dim _\delta (Y)$. On the other hand, by our construction of dimension functions (see Situation 81.2.1) we have $\dim _\delta (X) = \dim _\delta (Z) + r$ where $r$ the transcendence degree of the extension $R(X)/R(Z)$. Combining this with Spaces over Fields, Lemma 71.5.1 the lemma follows. $\square$
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