The Stacks project

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). $\square$

Comments (7)

Comment #119 by Adeel Ahmad Khan on

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

Comment #122 by Adeel Ahmad Khan on

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

Under the assumptions of (7.4) of this chapter, 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?

Comment #131 by Adeel on

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

Comment #132 by Adeel Ahmad Khan on

In order to talk about the field extension , 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 and are affine schemes, a morphism is dominant if and only if the kernel of the corresponding homomorphism is contained in the nilradical of .

Proof: Recall that there is an equality for ideals ; in particular, for the ideal one obtains if and only if is dominant. This is equivalent to the ideal being contained in every prime ideal, i.e. in the intersection of all the prime ideals; but this ideal is precisely the nilradical of .

Proposition: If and are integral schemes and the morphism is dominant, then the induced homomorphism of fields of rational functions is injective.

Proof: Let and be (nonempty) affine open subsets of and , respectively, such that ; let be the homomorphism corresponding to the restriction . Recall that one can identify canonically the fields of rational functions of and with the fields of fractions of and , respectively, so it is sufficient to show that the induced homomorphism is injective. It is clear that the restriction of to is still dominant, since and must contain the generic points and , 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 is injective, and therefore so is the induced homomorphism of fraction fields.

Well, I am assuming that the canonical homomorphisms and 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 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).

Comment #137 by Adeel Ahmad Khan on

Ah, that does seem simpler.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02R1. Beware of the difference between the letter 'O' and the digit '0'.