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 29.8.5 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). $\square$

Perhaps the first sentence of the proof could be followed by a reference to 0379 and 004W?

Note that (24.47.4) applies because by (24.16.8), the morphism $f$ is locally of finite type.

Under the assumptions of (7.4) of this chapter, $\delta(\eta_X) = \delta(\eta_Y)$ is clear, but why does this hold in general?

Comment #125 by on

Added these clarifications to the proof.

Not sure what you are asking in the final question. Can you clarify?

Never mind, I just thought we were assuming dim(X) = dim(Y) instead of equality of the \delta-dimensions.

In order to talk about the field extension $R(Y) \hookrightarrow R(X)$, we need to show that the homomorphism is injective; I couldn't find such a statement in the Stacks project, so feel free to use the following proof.

Proposition: If $X = \mathrm{Spec}(A)$ and $Y = \mathrm{Spec}(B)$ are affine schemes, a morphism $f : X \to Y$ is dominant if and only if the kernel of the corresponding homomorphism $\varphi : B \to A$ is contained in the nilradical of $B$.

Proof: Recall that there is an equality for ideals $I \subset A$; in particular, for the ideal $I = (0)$ one obtains if and only if $f$ is dominant. This is equivalent to the ideal $\varphi^{-1}(0) = \ker(\varphi)$ being contained in every prime ideal, i.e. in the intersection of all the prime ideals; but this ideal is precisely the nilradical of $A$.

Proposition: If $X$ and $Y$ are integral schemes and the morphism $f : X \to Y$ is dominant, then the induced homomorphism $R(Y) \to R(X)$ of fields of rational functions is injective.

Proof: Let $U = \mathrm{Spec}(A)$ and $V = \mathrm{Spec}(B)$ be (nonempty) affine open subsets of $X$ and $Y$, respectively, such that $f(U) \subset V$; let $\varphi : B \to A$ be the homomorphism corresponding to the restriction $f|_U : U \to V$. Recall that one can identify canonically the fields of rational functions of $X$ and $Y$ with the fields of fractions of $A$ and $B$, respectively, so it is sufficient to show that the induced homomorphism $\mathrm{Frac}(B) \to \mathrm{Frac}(A)$ is injective. It is clear that the restriction of $f$ to $U$ is still dominant, since $U$ and $V$ must contain the generic points $\eta_X \in X$ and $\eta_Y \in Y$, respectively (recall that a morphism of integral schemes is dominant if and only if it maps generic point to generic point). By the above proposition it follows that the homomorphism $\varphi : B \to A$ is injective, and therefore so is the induced homomorphism $\mathrm{Frac}(B) \to \mathrm{Frac}(A)$ of fraction fields.

Well, I am assuming that the canonical homomorphisms $R(Y) \to R(X)$ and $\mathrm{Frac}(B) \to \mathrm{Frac}(A)$ are compatible, but surely this shouldn't be a problem.

Comment #133 by on

@#132: The first proposition is Lemma Tag 10.30.5. For the second proposition I have two remarks: (1) we should define what we mean by a dominant rational map, and (2) show that we can define the composition $f \circ g$ of rational morphisms, provided that g is dominant with irreducible source. Namely, then we get a ring map R(Y) --> R(X) which will be injective as any ring map between fields is injective (by our conventions that a ring map sends 1 to 1).

Ah, that does seem simpler.

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