2
$\begingroup$

Let $f: \mathbb{P}^n \to \mathbb{P}^m$ be a rational map. Then there exists $U \subset \mathbb{P}^n$ open so that $f_{|U}$ is a morphism. What can we say about the dimension of $\overline{f(U)}$? We have that $\dim\mathbb{P}^n = \dim U$ and $\dim f(U)= \dim\overline{f(U)}$. Can I conclude that $n=\dim U \geq \dim\overline{f(U)} $ by the surjectvity of $f : U \to f(U)$ ? I think that's not true because $U$ in general is not a projective variety and $f$ need not be a morphism on $\overline{U}$. How can I proceed? Can I find more information assuming that $U \subset \mathbb{A}^n$?

Thanks for the help!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.