Genus of a smooth projective curve
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I was trying to prove that the genus of a smooth projective complex curve $F=0$ of degree $d$ is $(d-1)(d-2)/2$.
My attempt was to take the standard projection $$pi:mathbb{P}^2 to mathbb{P}^1$$ $$[x ;y ; z] to [x ; z]$$ which has degree $d$ and then apply Riemann Hurwitz formula.
The problem is that I do not know how to explicitly write the multiplictity of this map at the ramification points(which are the ones for which $dfrac{partial F}{partial y}=0$). In particular, I would like to say that the multiplicity of the map $pi $ in a point $x in mathbb{P}^2$ is $Ileft(x,F cap dfrac{partial F}{partial y}+1right)$, but I'm stuck (the last I is the intersection multiplicity of the two algebraic curves).
algebraic-geometry complex-geometry algebraic-curves riemann-surfaces
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add a comment |
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I was trying to prove that the genus of a smooth projective complex curve $F=0$ of degree $d$ is $(d-1)(d-2)/2$.
My attempt was to take the standard projection $$pi:mathbb{P}^2 to mathbb{P}^1$$ $$[x ;y ; z] to [x ; z]$$ which has degree $d$ and then apply Riemann Hurwitz formula.
The problem is that I do not know how to explicitly write the multiplictity of this map at the ramification points(which are the ones for which $dfrac{partial F}{partial y}=0$). In particular, I would like to say that the multiplicity of the map $pi $ in a point $x in mathbb{P}^2$ is $Ileft(x,F cap dfrac{partial F}{partial y}+1right)$, but I'm stuck (the last I is the intersection multiplicity of the two algebraic curves).
algebraic-geometry complex-geometry algebraic-curves riemann-surfaces
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Are you asking about how to finish this particular proof using the Riemann-Hurwitz formula, or are you interested in alternate ways to prove this statement?
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– KReiser
Jan 16 at 21:00
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I would like to see a proof using Riemann Hurwitz formula
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– Tommaso Scognamiglio
Jan 16 at 21:44
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Is it okay to use adjunction + cohomology ? Then $K_C = mathcal{O}_C(n-3) $. Since the curve is defined by a single equation of degree d it's easy to see that $K_C$ has exactly the same global sections as on $P^2$.
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– aginensky
Jan 16 at 22:52
1
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I think this is basically a duplicated of this question.
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– André 3000
Jan 17 at 4:17
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@André3000 Tthe answer in there is just a broken link, it might be worth it having a written answer.
$endgroup$
– user347489
Jan 17 at 8:31
add a comment |
$begingroup$
I was trying to prove that the genus of a smooth projective complex curve $F=0$ of degree $d$ is $(d-1)(d-2)/2$.
My attempt was to take the standard projection $$pi:mathbb{P}^2 to mathbb{P}^1$$ $$[x ;y ; z] to [x ; z]$$ which has degree $d$ and then apply Riemann Hurwitz formula.
The problem is that I do not know how to explicitly write the multiplictity of this map at the ramification points(which are the ones for which $dfrac{partial F}{partial y}=0$). In particular, I would like to say that the multiplicity of the map $pi $ in a point $x in mathbb{P}^2$ is $Ileft(x,F cap dfrac{partial F}{partial y}+1right)$, but I'm stuck (the last I is the intersection multiplicity of the two algebraic curves).
algebraic-geometry complex-geometry algebraic-curves riemann-surfaces
$endgroup$
I was trying to prove that the genus of a smooth projective complex curve $F=0$ of degree $d$ is $(d-1)(d-2)/2$.
My attempt was to take the standard projection $$pi:mathbb{P}^2 to mathbb{P}^1$$ $$[x ;y ; z] to [x ; z]$$ which has degree $d$ and then apply Riemann Hurwitz formula.
The problem is that I do not know how to explicitly write the multiplictity of this map at the ramification points(which are the ones for which $dfrac{partial F}{partial y}=0$). In particular, I would like to say that the multiplicity of the map $pi $ in a point $x in mathbb{P}^2$ is $Ileft(x,F cap dfrac{partial F}{partial y}+1right)$, but I'm stuck (the last I is the intersection multiplicity of the two algebraic curves).
algebraic-geometry complex-geometry algebraic-curves riemann-surfaces
algebraic-geometry complex-geometry algebraic-curves riemann-surfaces
asked Jan 16 at 18:16
Tommaso ScognamiglioTommaso Scognamiglio
581412
581412
$begingroup$
Are you asking about how to finish this particular proof using the Riemann-Hurwitz formula, or are you interested in alternate ways to prove this statement?
$endgroup$
– KReiser
Jan 16 at 21:00
$begingroup$
I would like to see a proof using Riemann Hurwitz formula
$endgroup$
– Tommaso Scognamiglio
Jan 16 at 21:44
$begingroup$
Is it okay to use adjunction + cohomology ? Then $K_C = mathcal{O}_C(n-3) $. Since the curve is defined by a single equation of degree d it's easy to see that $K_C$ has exactly the same global sections as on $P^2$.
$endgroup$
– aginensky
Jan 16 at 22:52
1
$begingroup$
I think this is basically a duplicated of this question.
$endgroup$
– André 3000
Jan 17 at 4:17
$begingroup$
@André3000 Tthe answer in there is just a broken link, it might be worth it having a written answer.
$endgroup$
– user347489
Jan 17 at 8:31
add a comment |
$begingroup$
Are you asking about how to finish this particular proof using the Riemann-Hurwitz formula, or are you interested in alternate ways to prove this statement?
$endgroup$
– KReiser
Jan 16 at 21:00
$begingroup$
I would like to see a proof using Riemann Hurwitz formula
$endgroup$
– Tommaso Scognamiglio
Jan 16 at 21:44
$begingroup$
Is it okay to use adjunction + cohomology ? Then $K_C = mathcal{O}_C(n-3) $. Since the curve is defined by a single equation of degree d it's easy to see that $K_C$ has exactly the same global sections as on $P^2$.
$endgroup$
– aginensky
Jan 16 at 22:52
1
$begingroup$
I think this is basically a duplicated of this question.
$endgroup$
– André 3000
Jan 17 at 4:17
$begingroup$
@André3000 Tthe answer in there is just a broken link, it might be worth it having a written answer.
$endgroup$
– user347489
Jan 17 at 8:31
$begingroup$
Are you asking about how to finish this particular proof using the Riemann-Hurwitz formula, or are you interested in alternate ways to prove this statement?
$endgroup$
– KReiser
Jan 16 at 21:00
$begingroup$
Are you asking about how to finish this particular proof using the Riemann-Hurwitz formula, or are you interested in alternate ways to prove this statement?
$endgroup$
– KReiser
Jan 16 at 21:00
$begingroup$
I would like to see a proof using Riemann Hurwitz formula
$endgroup$
– Tommaso Scognamiglio
Jan 16 at 21:44
$begingroup$
I would like to see a proof using Riemann Hurwitz formula
$endgroup$
– Tommaso Scognamiglio
Jan 16 at 21:44
$begingroup$
Is it okay to use adjunction + cohomology ? Then $K_C = mathcal{O}_C(n-3) $. Since the curve is defined by a single equation of degree d it's easy to see that $K_C$ has exactly the same global sections as on $P^2$.
$endgroup$
– aginensky
Jan 16 at 22:52
$begingroup$
Is it okay to use adjunction + cohomology ? Then $K_C = mathcal{O}_C(n-3) $. Since the curve is defined by a single equation of degree d it's easy to see that $K_C$ has exactly the same global sections as on $P^2$.
$endgroup$
– aginensky
Jan 16 at 22:52
1
1
$begingroup$
I think this is basically a duplicated of this question.
$endgroup$
– André 3000
Jan 17 at 4:17
$begingroup$
I think this is basically a duplicated of this question.
$endgroup$
– André 3000
Jan 17 at 4:17
$begingroup$
@André3000 Tthe answer in there is just a broken link, it might be worth it having a written answer.
$endgroup$
– user347489
Jan 17 at 8:31
$begingroup$
@André3000 Tthe answer in there is just a broken link, it might be worth it having a written answer.
$endgroup$
– user347489
Jan 17 at 8:31
add a comment |
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$begingroup$
Are you asking about how to finish this particular proof using the Riemann-Hurwitz formula, or are you interested in alternate ways to prove this statement?
$endgroup$
– KReiser
Jan 16 at 21:00
$begingroup$
I would like to see a proof using Riemann Hurwitz formula
$endgroup$
– Tommaso Scognamiglio
Jan 16 at 21:44
$begingroup$
Is it okay to use adjunction + cohomology ? Then $K_C = mathcal{O}_C(n-3) $. Since the curve is defined by a single equation of degree d it's easy to see that $K_C$ has exactly the same global sections as on $P^2$.
$endgroup$
– aginensky
Jan 16 at 22:52
1
$begingroup$
I think this is basically a duplicated of this question.
$endgroup$
– André 3000
Jan 17 at 4:17
$begingroup$
@André3000 Tthe answer in there is just a broken link, it might be worth it having a written answer.
$endgroup$
– user347489
Jan 17 at 8:31