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!


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.