Irreducible projective varieties and morphisms
$begingroup$
Let $C$ be a smooth and irreducible projective curve, and let $f: X rightarrow mathbb{P}^1(mathbb{C})$ a morphism of varieties: then $f$ is either constant or surjective. I am trying to prove this in an elementary way (I don't know about schemes, and I cannot use the notion of completeness).
I think it should be possible to cover $mathbb{P}^1$ with two affine open sets $U,V$ both isomorphic to $mathbb{A}^1$ and then exploit the fact that the functions from $X$ to $mathbb{A}^1$ have to be constant but I cannot seem how to patch this up and how it relates to surjectivity.
algebraic-geometry
asked Dec 6 '18 at 19:17
KarlKarl
366
$endgroup$
|
show 9 more comments
$begingroup$
Let $C$ be a smooth and irreducible projective curve, and let $f: X rightarrow mathbb{P}^1(mathbb{C})$ a morphism of varieties: then $f$ is either constant or surjective. I am trying to prove this in an elementary way (I don't know about schemes, and I cannot use the notion of completeness).
I think it should be possible to cover $mathbb{P}^1$ with two affine open sets $U,V$ both isomorphic to $mathbb{A}^1$ and then exploit the fact that the functions from $X$ to $mathbb{A}^1$ have to be constant but I cannot seem how to patch this up and how it relates to surjectivity.
algebraic-geometry
asked Dec 6 '18 at 19:17
KarlKarl
366
$endgroup$
2
$begingroup$
You're on the right track. Assume it's not surjective and choose your cover appropriately.
$endgroup$
– user113102
Dec 6 '18 at 19:29
1
$begingroup$
I think that is not that easy. Can you use Riemann-Roch theorem?
$endgroup$
– José Alejandro Aburto Araneda
Dec 6 '18 at 20:55
2
$begingroup$
If $f$ misses a point, WLOG take that point to be the point at infinity, so that $im(f) subset U$. But then $f$ is a map to $mathbb{A}^1$, so must be constant. I'm worried I'm saying something wrong though...been out of the game too long.
$endgroup$
– user113102
Dec 7 '18 at 1:05
2
$begingroup$
@Karl : there is nothing wrong with this argument I think. We started with a map $C to Bbb P^1$. By hypothesis $infty$ is not in the image, so it factors by a map $C to Bbb A^1$. By what you already know this map is constant, so the original map was constant as well.
$endgroup$
– Nicolas Hemelsoet
Dec 7 '18 at 8:12
1
$begingroup$
@Karl That doesn't even make sense. The morphism can't take values on $Bbb{P}^1setminus U$, since $Bbb{P}^1$ is the codomain, not the domain. And by assumption $Bbb{P}^1setminus U$ is not in the image of the morphism, so the morphism doesn't take on that value either.
$endgroup$
– jgon
Dec 8 '18 at 5:57
|
show 9 more comments
$begingroup$
Let $C$ be a smooth and irreducible projective curve, and let $f: X rightarrow mathbb{P}^1(mathbb{C})$ a morphism of varieties: then $f$ is either constant or surjective. I am trying to prove this in an elementary way (I don't know about schemes, and I cannot use the notion of completeness).
I think it should be possible to cover $mathbb{P}^1$ with two affine open sets $U,V$ both isomorphic to $mathbb{A}^1$ and then exploit the fact that the functions from $X$ to $mathbb{A}^1$ have to be constant but I cannot seem how to patch this up and how it relates to surjectivity.
algebraic-geometry
asked Dec 6 '18 at 19:17
KarlKarl
366
$endgroup$
Let $C$ be a smooth and irreducible projective curve, and let $f: X rightarrow mathbb{P}^1(mathbb{C})$ a morphism of varieties: then $f$ is either constant or surjective. I am trying to prove this in an elementary way (I don't know about schemes, and I cannot use the notion of completeness).
I think it should be possible to cover $mathbb{P}^1$ with two affine open sets $U,V$ both isomorphic to $mathbb{A}^1$ and then exploit the fact that the functions from $X$ to $mathbb{A}^1$ have to be constant but I cannot seem how to patch this up and how it relates to surjectivity.
algebraic-geometry
algebraic-geometry
asked Dec 6 '18 at 19:17
KarlKarl
366
asked Dec 6 '18 at 19:17
KarlKarl
366
edited Dec 7 '18 at 1:54
Karl
asked Dec 6 '18 at 19:17
KarlKarl
366
asked Dec 6 '18 at 19:17
KarlKarl
366
asked Dec 6 '18 at 19:17
KarlKarl
366
366
2
$begingroup$
You're on the right track. Assume it's not surjective and choose your cover appropriately.
$endgroup$
– user113102
Dec 6 '18 at 19:29
1
$begingroup$
I think that is not that easy. Can you use Riemann-Roch theorem?
$endgroup$
– José Alejandro Aburto Araneda
Dec 6 '18 at 20:55
2
$begingroup$
If $f$ misses a point, WLOG take that point to be the point at infinity, so that $im(f) subset U$. But then $f$ is a map to $mathbb{A}^1$, so must be constant. I'm worried I'm saying something wrong though...been out of the game too long.
$endgroup$
– user113102
Dec 7 '18 at 1:05
2
$begingroup$
@Karl : there is nothing wrong with this argument I think. We started with a map $C to Bbb P^1$. By hypothesis $infty$ is not in the image, so it factors by a map $C to Bbb A^1$. By what you already know this map is constant, so the original map was constant as well.
$endgroup$
– Nicolas Hemelsoet
Dec 7 '18 at 8:12
1
$begingroup$
@Karl That doesn't even make sense. The morphism can't take values on $Bbb{P}^1setminus U$, since $Bbb{P}^1$ is the codomain, not the domain. And by assumption $Bbb{P}^1setminus U$ is not in the image of the morphism, so the morphism doesn't take on that value either.
$endgroup$
– jgon
Dec 8 '18 at 5:57
|
show 9 more comments
2
$begingroup$
You're on the right track. Assume it's not surjective and choose your cover appropriately.
$endgroup$
– user113102
Dec 6 '18 at 19:29
1
$begingroup$
I think that is not that easy. Can you use Riemann-Roch theorem?
$endgroup$
– José Alejandro Aburto Araneda
Dec 6 '18 at 20:55
2
$begingroup$
If $f$ misses a point, WLOG take that point to be the point at infinity, so that $im(f) subset U$. But then $f$ is a map to $mathbb{A}^1$, so must be constant. I'm worried I'm saying something wrong though...been out of the game too long.
$endgroup$
– user113102
Dec 7 '18 at 1:05
2
$begingroup$
@Karl : there is nothing wrong with this argument I think. We started with a map $C to Bbb P^1$. By hypothesis $infty$ is not in the image, so it factors by a map $C to Bbb A^1$. By what you already know this map is constant, so the original map was constant as well.
$endgroup$
– Nicolas Hemelsoet
Dec 7 '18 at 8:12
1
$begingroup$
@Karl That doesn't even make sense. The morphism can't take values on $Bbb{P}^1setminus U$, since $Bbb{P}^1$ is the codomain, not the domain. And by assumption $Bbb{P}^1setminus U$ is not in the image of the morphism, so the morphism doesn't take on that value either.
$endgroup$
– jgon
Dec 8 '18 at 5:57
2
2
$begingroup$
You're on the right track. Assume it's not surjective and choose your cover appropriately.
$endgroup$
– user113102
Dec 6 '18 at 19:29
$begingroup$
You're on the right track. Assume it's not surjective and choose your cover appropriately.
$endgroup$
– user113102
Dec 6 '18 at 19:29
1
1
$begingroup$
I think that is not that easy. Can you use Riemann-Roch theorem?
$endgroup$
– José Alejandro Aburto Araneda
Dec 6 '18 at 20:55
$begingroup$
I think that is not that easy. Can you use Riemann-Roch theorem?
$endgroup$
– José Alejandro Aburto Araneda
Dec 6 '18 at 20:55
2
2
$begingroup$
If $f$ misses a point, WLOG take that point to be the point at infinity, so that $im(f) subset U$. But then $f$ is a map to $mathbb{A}^1$, so must be constant. I'm worried I'm saying something wrong though...been out of the game too long.
$endgroup$
– user113102
Dec 7 '18 at 1:05
$begingroup$
If $f$ misses a point, WLOG take that point to be the point at infinity, so that $im(f) subset U$. But then $f$ is a map to $mathbb{A}^1$, so must be constant. I'm worried I'm saying something wrong though...been out of the game too long.
$endgroup$
– user113102
Dec 7 '18 at 1:05
2
2
$begingroup$
@Karl : there is nothing wrong with this argument I think. We started with a map $C to Bbb P^1$. By hypothesis $infty$ is not in the image, so it factors by a map $C to Bbb A^1$. By what you already know this map is constant, so the original map was constant as well.
$endgroup$
– Nicolas Hemelsoet
Dec 7 '18 at 8:12
$begingroup$
@Karl : there is nothing wrong with this argument I think. We started with a map $C to Bbb P^1$. By hypothesis $infty$ is not in the image, so it factors by a map $C to Bbb A^1$. By what you already know this map is constant, so the original map was constant as well.
$endgroup$
– Nicolas Hemelsoet
Dec 7 '18 at 8:12
1
1
$begingroup$
@Karl That doesn't even make sense. The morphism can't take values on $Bbb{P}^1setminus U$, since $Bbb{P}^1$ is the codomain, not the domain. And by assumption $Bbb{P}^1setminus U$ is not in the image of the morphism, so the morphism doesn't take on that value either.
$endgroup$
– jgon
Dec 8 '18 at 5:57
$begingroup$
@Karl That doesn't even make sense. The morphism can't take values on $Bbb{P}^1setminus U$, since $Bbb{P}^1$ is the codomain, not the domain. And by assumption $Bbb{P}^1setminus U$ is not in the image of the morphism, so the morphism doesn't take on that value either.
$endgroup$
– jgon
Dec 8 '18 at 5:57
|
show 9 more comments
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2
$begingroup$
You're on the right track. Assume it's not surjective and choose your cover appropriately.
$endgroup$
– user113102
Dec 6 '18 at 19:29
1
$begingroup$
I think that is not that easy. Can you use Riemann-Roch theorem?
$endgroup$
– José Alejandro Aburto Araneda
Dec 6 '18 at 20:55
2
$begingroup$
If $f$ misses a point, WLOG take that point to be the point at infinity, so that $im(f) subset U$. But then $f$ is a map to $mathbb{A}^1$, so must be constant. I'm worried I'm saying something wrong though...been out of the game too long.
$endgroup$
– user113102
Dec 7 '18 at 1:05
2
$begingroup$
@Karl : there is nothing wrong with this argument I think. We started with a map $C to Bbb P^1$. By hypothesis $infty$ is not in the image, so it factors by a map $C to Bbb A^1$. By what you already know this map is constant, so the original map was constant as well.
$endgroup$
– Nicolas Hemelsoet
Dec 7 '18 at 8:12
1
$begingroup$
@Karl That doesn't even make sense. The morphism can't take values on $Bbb{P}^1setminus U$, since $Bbb{P}^1$ is the codomain, not the domain. And by assumption $Bbb{P}^1setminus U$ is not in the image of the morphism, so the morphism doesn't take on that value either.
$endgroup$
– jgon
Dec 8 '18 at 5:57