# Dominant rational maps

The University I go to doesn't have any courses in (classical) Algebraic Geometry so I am trying to learn myself. I am fairly comfortable with the content I have covered so far aside from a so called "easy" results which I fail to understand. So my question:

Why is the concept of a rational map between two varieties being dominant well-defined. Specifically let $X$ and $Y$ be varieties; let $U$ and $V$ be open sets of $X$; and let $(f,U)$ and $(g,V)$ be equivalent rational maps from $X$ to $Y$. Why is the image of $f$ dense in $Y$ if and only if the image of $g$ is dense in $Y$?

Thanks for any help.

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Please give some context: what is the definition of rational maps you are asking about , what are the domain and codomain of these maps, do you consider varieties or schemes and what book are you studying from? – Georges Elencwajg Aug 13 '12 at 12:17
OK Georges, will do. – M Davolo Aug 13 '12 at 13:23
After reading the answers, the key fact for me is that a non-empty open subset of an irreducible variety is dense. Hence $U$ and $V$ cannot be very different, indeed $(f,U\cap V)$ still has dense image. – Jack Schmidt Aug 13 '12 at 15:24

This is a purely topological fact:

First we have $f(U \cap V) = g(U \cap V) \subset g(V)$, as $(f,U)$ and $(g,V)$ are equivalent. If $f(U)$ is dense in $Y$ we get

$Y=\overline{f(U)}=\overline{f(\overline{U \cap V})} = \overline{f(U \cap V)} \subset \overline{g(V)}$

where the third equality follows from continuity of $f$. Note also that I take the closure of $U \cap V$ in $U$, so that $\overline{U \cap V}=U$. Hence $g(V)$ is dense in $Y$ as well.

EDIT: In view of M Turgeons answer I should add that my varieties are always irreducible. I used this fact to conclude that $\overline{U \cap V}=U$, as $U \cap V$ is dense in $X$ (it is a non-empty open subset of an irreducible space).

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Dear Nils, You may be interested in my comment to M Turgeon's answer. Regards, – Matt E Aug 13 '12 at 17:24
Thanks very much. – M Davolo Aug 13 '12 at 19:41
Dear @MattE: thank you for the heads up. Good to see some experts discussing foundational matters! – Nils Matthes Aug 14 '12 at 12:57

The reason why it is well-defined is because the definition of a dominant rational map is the following:

Definition: A rational map $(f,U)$ is dominant is there exists a rational map $(g,V)$ equivalent to $(f,U)$ such that the image of $g$ is dense in its codomain.

The key point here is that I could take a smaller open set as the domain for my rational map in such a way that the image is not dense, but still get a dominant rational map. (Added: As MattE mentionned in the comments, the correct definition of rational map in the reducible case makes sure that this never happens.)

Added: For my last paragraph, this all depends on your definition of variety. If you assume (as is often the case in algebraic geometry, but not in the theory of algebraic groups) that a variety is irreducible, then we have the following:

Lemma: Let $(f,U)$ and $(g,V)$ be two equivalent rational maps, where we denote the codomain by $Y$. Then $f(U)$ is dense in $Y$ if and only if $g(V)$ is.

Proof: We have the following general fact from topology: For any subset $O\subseteq U$, we have $f(\overline{O})\subseteq \overline{f(O)}$. But if $O$ is open and non-empty, it is dense in $U$, and so $\overline{O}=U$.

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@JackSchmidt See my edit. In the reducible case, we could take a (reducible) variety with two connected components and concoct an example easily, I think. – M Turgeon Aug 13 '12 at 15:19
Thanks! Yes, the reducible case is clear. Take f to be the identity, and g the identity on a component. – Jack Schmidt Aug 13 '12 at 15:20
Dear M Turgeon, What definition of rational map on a reducible variety are you using that makes your third paragraph ("The key point here ...") true? In any definition of rational map I can imagine, the domain of definition would be required to be a dense open subset of the variety in question, and the intersection of any two such is again dense. (See e.g. math.columbia.edu/~dejong/wordpress/?p=1049 .) In short, in the definitions of rational map that I know, one representative will be dominant if and only if every representative is. Regards, – Matt E Aug 13 '12 at 17:24
@MattE I think the problem here stems from the fact that conflicting definitions of variety are used in different fields, i.e. sometimes the irreducibility condition is assumed, sometimes it is not. Hence, the definition with which I am familiar only assumes the domain to be a non-empty subset of the variety - which of course is automatically dense when the variety is irreducible. Thank you for clearing this up. – M Turgeon Aug 13 '12 at 17:52
Thanks very much. – M Davolo Aug 13 '12 at 19:42

Thanks for your help Nils and M Turgeon. What I didn't realize was that if $f: X \to Y$ is a continuous map of topological spaces, and $U$ is a subset of $X$, then $f(\overline{U})$ is contained in $\overline{f(U)}$. Is the following a correct proof of this fact?

Since $\overline{f(U)}$ is closed in $Y$, by continuity $f^{-1}(\overline{f(U)})$ is closed in $X$. Since also, $U\subseteq f^{-1}(\overline{f(U)})$, it follows that $\overline{U}\subseteq f^{-1}(\overline{f(U)})$ i.e. $f(\overline{U})\subseteq \overline{f(U)}$.

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Yes, this is a valid proof. – M Turgeon Aug 13 '12 at 19:52
Cheers. Was just checking as my topology is poor :( – M Davolo Aug 13 '12 at 19:55
Dear @MDavolo: glad we could help you! And have fun with your studies in classical algebraic geometry! – Nils Matthes Aug 14 '12 at 12:59