Lemma 42.11.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a morphism. Assume $X$, $Y$ integral and $\dim _\delta (X) = \dim _\delta (Y)$. Then either $f(X)$ is contained in a proper closed subscheme of $Y$, or $f$ is dominant and the extension of function fields $R(X)/R(Y)$ is finite.

**Proof.**
The closure $\overline{f(X)} \subset Y$ is irreducible as $X$ is irreducible (Topology, Lemmas 5.8.2 and 5.8.3). If $\overline{f(X)} \not= Y$, then we are done. If $\overline{f(X)} = Y$, then $f$ is dominant and by Morphisms, Lemma 29.8.6 we see that the generic point $\eta _ Y$ of $Y$ is in the image of $f$. Of course this implies that $f(\eta _ X) = \eta _ Y$, where $\eta _ X \in X$ is the generic point of $X$. Since $\delta (\eta _ X) = \delta (\eta _ Y)$ we see that $R(Y) = \kappa (\eta _ Y) \subset \kappa (\eta _ X) = R(X)$ is an extension of transcendence degree $0$. Hence $R(Y) \subset R(X)$ is a finite extension by Morphisms, Lemma 29.51.7 (which applies by Morphisms, Lemma 29.15.8).
$\square$

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