Mixing
Dimension image of morphism of projective varieties
$begingroup$
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 $dimmathbb{P}^n = dim U$ and $dim f(U)= dimoverline{f(U)}$. Can I conclude that $n=dim U geq dimoverline{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!
algebraic-geometry projective-space dimension-theory
edited Feb 1 at 23:13
KReiser
10.1k21435
asked Feb 1 at 16:22
andresandres
2439
$endgroup$
add a comment |
$begingroup$
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 $dimmathbb{P}^n = dim U$ and $dim f(U)= dimoverline{f(U)}$. Can I conclude that $n=dim U geq dimoverline{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!
algebraic-geometry projective-space dimension-theory
edited Feb 1 at 23:13
KReiser
10.1k21435
asked Feb 1 at 16:22
andresandres
2439
$endgroup$
add a comment |
$begingroup$
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 $dimmathbb{P}^n = dim U$ and $dim f(U)= dimoverline{f(U)}$. Can I conclude that $n=dim U geq dimoverline{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!
algebraic-geometry projective-space dimension-theory
edited Feb 1 at 23:13
KReiser
10.1k21435
asked Feb 1 at 16:22
andresandres
2439
$endgroup$
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 $dimmathbb{P}^n = dim U$ and $dim f(U)= dimoverline{f(U)}$. Can I conclude that $n=dim U geq dimoverline{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!
algebraic-geometry projective-space dimension-theory
algebraic-geometry projective-space dimension-theory
edited Feb 1 at 23:13
KReiser
10.1k21435
asked Feb 1 at 16:22
andresandres
2439
edited Feb 1 at 23:13
KReiser
10.1k21435
asked Feb 1 at 16:22
andresandres
2439
edited Feb 1 at 23:13
KReiser
10.1k21435
edited Feb 1 at 23:13
KReiser
10.1k21435
edited Feb 1 at 23:13
KReiser
10.1k21435
10.1k21435
asked Feb 1 at 16:22
andresandres
2439
asked Feb 1 at 16:22
andresandres
2439
asked Feb 1 at 16:22
andresandres
2439
2439
add a comment |
add a comment |
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