Is a rational map in which fibers are generically singletons necessarily birational?

Asked Oct 31 '13 at 16:24

Active Feb 16 '14 at 23:33

Viewed 220 times

This is a very basic question about algebraic geometry which for some reason I am having trouble thinking about clearly.

Let $X,Y$ be irreducible varieties over an algebraically closed field $k$ (in case this is ambiguous, definition below). Let $f$ be a rational map from $X$ to $Y$ that is regular on an open subset $U\subset X$. If the fibers of $f$ are generically singletons, i.e. if there is an open subset $V\subset Y$ such that if $y\in V$, there is exactly one $x\in U$ with $f(x)=y$, does it follow that $f$ is birational? In other words, is the inverse map $g:V\rightarrow U$ defined by $g(y)=f^{-1}(y)$ necessarily a rational map?

My intuition says yes, but I'm having trouble thinking about how I know. Can you help me?

Thank you in advance.

Definition of variety I am working with: topological space equipped with a sheaf of functions to $k$ that is locally isomorphic as a locally ringed space to Zariski-closed subsets of affine space.

Addendum 11/1/13: It's pointed out in the comments that the Frobenius map gives a counterexample in characteristic $p$; but Cantlog asserts that the answer is positive in characteristic zero. What's the proof?

edited Feb 16 '14 at 23:33

asked Oct 31 '13 at 16:24

Ben Blum-Smith

20,766

1

What about $\mathbb{A}^1$ to itself defined by $x \to x^p$, where the ground field is $\overline{\mathbb{F}_p}$? – Oct 31 '13 at 16:28
@Sanchez Good point. – Bruno Joyal Oct 31 '13 at 16:40
2

However the answer is positive in characteristic $0$. – Cantlog Oct 31 '13 at 21:26
2

You should also suppose $V$ is non-empty. Then in characteristic $0$, prove that the field extension $K(X)/K(Y)$ has degree $1$. Search in SE for "number of points in a fiber". – Cantlog Nov 01 '13 at 18:33

Is a rational map in which fibers are generically singletons necessarily birational?

0 Answers0