Divisor on cubic curve $y^2z=x^3-xz^2$
$begingroup$
Let $X$ be the nonsingular cubic curve $y^2z=x^3-xz^2$. Let $P_0=(0,1,0)$. Then the line bundle associated to $3P_0$ is $O_X(1)$. I already know that $3P_0$ is produced by cut the curve with $z=0$. But why this is enough to get it is $O_X(1)$.
algebraic-geometry divisors-algebraic-geometry line-bundles
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
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add a comment |
$begingroup$
Let $X$ be the nonsingular cubic curve $y^2z=x^3-xz^2$. Let $P_0=(0,1,0)$. Then the line bundle associated to $3P_0$ is $O_X(1)$. I already know that $3P_0$ is produced by cut the curve with $z=0$. But why this is enough to get it is $O_X(1)$.
algebraic-geometry divisors-algebraic-geometry line-bundles
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
$endgroup$
$begingroup$
In general you would get that the associated line bundle $mathscr{L}$ is isomorphic to $varphi^*(O_{mathbb{P}^n}(1))$ where $varphi: X to mathbb{P}^n$ is defined by the global sections of $mathscr{L}$. But in this case the divisor $3 P_0$ was used to embed $X$ into projective space in the first place (hence the nice Weierstrass equation), so $varphi$ is just the inclusion map $iota: X to mathbb{P}^2$. I think this means that $mathscr{L} cong iota^*(O_{mathbb{P}^2}(1)) cong O_X(1)$.
$endgroup$
– André 3000
Dec 1 '18 at 20:05
$begingroup$
@André3000 Thank you for your solution. However, I am not quite familiar with the Weierstrass equation. I try to argue it as every divisor in $mathbb{P}^2$ restricts to a divisor on $X$, and $z=0$ is a divisor associated to $O_{mathbb{P}^2}(1)$ (as we have an obvious embedding of this curve to projective space). Thus it should also give a divisor corresponding to $O_X(1)$.
$endgroup$
– Peter Liu
Dec 1 '18 at 23:03
add a comment |
$begingroup$
Let $X$ be the nonsingular cubic curve $y^2z=x^3-xz^2$. Let $P_0=(0,1,0)$. Then the line bundle associated to $3P_0$ is $O_X(1)$. I already know that $3P_0$ is produced by cut the curve with $z=0$. But why this is enough to get it is $O_X(1)$.
algebraic-geometry divisors-algebraic-geometry line-bundles
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
$endgroup$
Let $X$ be the nonsingular cubic curve $y^2z=x^3-xz^2$. Let $P_0=(0,1,0)$. Then the line bundle associated to $3P_0$ is $O_X(1)$. I already know that $3P_0$ is produced by cut the curve with $z=0$. But why this is enough to get it is $O_X(1)$.
algebraic-geometry divisors-algebraic-geometry line-bundles
algebraic-geometry divisors-algebraic-geometry line-bundles
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
asked Nov 30 '18 at 22:21
Peter LiuPeter Liu
305114
305114
$begingroup$
In general you would get that the associated line bundle $mathscr{L}$ is isomorphic to $varphi^*(O_{mathbb{P}^n}(1))$ where $varphi: X to mathbb{P}^n$ is defined by the global sections of $mathscr{L}$. But in this case the divisor $3 P_0$ was used to embed $X$ into projective space in the first place (hence the nice Weierstrass equation), so $varphi$ is just the inclusion map $iota: X to mathbb{P}^2$. I think this means that $mathscr{L} cong iota^*(O_{mathbb{P}^2}(1)) cong O_X(1)$.
$endgroup$
– André 3000
Dec 1 '18 at 20:05
$begingroup$
@André3000 Thank you for your solution. However, I am not quite familiar with the Weierstrass equation. I try to argue it as every divisor in $mathbb{P}^2$ restricts to a divisor on $X$, and $z=0$ is a divisor associated to $O_{mathbb{P}^2}(1)$ (as we have an obvious embedding of this curve to projective space). Thus it should also give a divisor corresponding to $O_X(1)$.
$endgroup$
– Peter Liu
Dec 1 '18 at 23:03
add a comment |
$begingroup$
In general you would get that the associated line bundle $mathscr{L}$ is isomorphic to $varphi^*(O_{mathbb{P}^n}(1))$ where $varphi: X to mathbb{P}^n$ is defined by the global sections of $mathscr{L}$. But in this case the divisor $3 P_0$ was used to embed $X$ into projective space in the first place (hence the nice Weierstrass equation), so $varphi$ is just the inclusion map $iota: X to mathbb{P}^2$. I think this means that $mathscr{L} cong iota^*(O_{mathbb{P}^2}(1)) cong O_X(1)$.
$endgroup$
– André 3000
Dec 1 '18 at 20:05
$begingroup$
@André3000 Thank you for your solution. However, I am not quite familiar with the Weierstrass equation. I try to argue it as every divisor in $mathbb{P}^2$ restricts to a divisor on $X$, and $z=0$ is a divisor associated to $O_{mathbb{P}^2}(1)$ (as we have an obvious embedding of this curve to projective space). Thus it should also give a divisor corresponding to $O_X(1)$.
$endgroup$
– Peter Liu
Dec 1 '18 at 23:03
$begingroup$
In general you would get that the associated line bundle $mathscr{L}$ is isomorphic to $varphi^*(O_{mathbb{P}^n}(1))$ where $varphi: X to mathbb{P}^n$ is defined by the global sections of $mathscr{L}$. But in this case the divisor $3 P_0$ was used to embed $X$ into projective space in the first place (hence the nice Weierstrass equation), so $varphi$ is just the inclusion map $iota: X to mathbb{P}^2$. I think this means that $mathscr{L} cong iota^*(O_{mathbb{P}^2}(1)) cong O_X(1)$.
$endgroup$
– André 3000
Dec 1 '18 at 20:05
$begingroup$
In general you would get that the associated line bundle $mathscr{L}$ is isomorphic to $varphi^*(O_{mathbb{P}^n}(1))$ where $varphi: X to mathbb{P}^n$ is defined by the global sections of $mathscr{L}$. But in this case the divisor $3 P_0$ was used to embed $X$ into projective space in the first place (hence the nice Weierstrass equation), so $varphi$ is just the inclusion map $iota: X to mathbb{P}^2$. I think this means that $mathscr{L} cong iota^*(O_{mathbb{P}^2}(1)) cong O_X(1)$.
$endgroup$
– André 3000
Dec 1 '18 at 20:05
$begingroup$
@André3000 Thank you for your solution. However, I am not quite familiar with the Weierstrass equation. I try to argue it as every divisor in $mathbb{P}^2$ restricts to a divisor on $X$, and $z=0$ is a divisor associated to $O_{mathbb{P}^2}(1)$ (as we have an obvious embedding of this curve to projective space). Thus it should also give a divisor corresponding to $O_X(1)$.
$endgroup$
– Peter Liu
Dec 1 '18 at 23:03
$begingroup$
@André3000 Thank you for your solution. However, I am not quite familiar with the Weierstrass equation. I try to argue it as every divisor in $mathbb{P}^2$ restricts to a divisor on $X$, and $z=0$ is a divisor associated to $O_{mathbb{P}^2}(1)$ (as we have an obvious embedding of this curve to projective space). Thus it should also give a divisor corresponding to $O_X(1)$.
$endgroup$
– Peter Liu
Dec 1 '18 at 23:03
add a comment |
1 Answer
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$begingroup$
Your approach is correct. One nice thing about the divisor and line bundle correspondence is that the divisor is the divisor of any rational section. Sections of $mathcal{O}_{mathbb{P}^2}(1)$ pullback to sections of $mathcal{O}_X(1)$, and the associated divisor of a pulled-back section gives you a divisor $D$ such that $mathcal{O}_X(D)=mathcal{O}_X(1)$. You did this computation and everything works!
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
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1 Answer
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1 Answer
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active
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$begingroup$
Your approach is correct. One nice thing about the divisor and line bundle correspondence is that the divisor is the divisor of any rational section. Sections of $mathcal{O}_{mathbb{P}^2}(1)$ pullback to sections of $mathcal{O}_X(1)$, and the associated divisor of a pulled-back section gives you a divisor $D$ such that $mathcal{O}_X(D)=mathcal{O}_X(1)$. You did this computation and everything works!
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
$endgroup$
add a comment |
$begingroup$
Your approach is correct. One nice thing about the divisor and line bundle correspondence is that the divisor is the divisor of any rational section. Sections of $mathcal{O}_{mathbb{P}^2}(1)$ pullback to sections of $mathcal{O}_X(1)$, and the associated divisor of a pulled-back section gives you a divisor $D$ such that $mathcal{O}_X(D)=mathcal{O}_X(1)$. You did this computation and everything works!
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
$endgroup$
add a comment |
$begingroup$
Your approach is correct. One nice thing about the divisor and line bundle correspondence is that the divisor is the divisor of any rational section. Sections of $mathcal{O}_{mathbb{P}^2}(1)$ pullback to sections of $mathcal{O}_X(1)$, and the associated divisor of a pulled-back section gives you a divisor $D$ such that $mathcal{O}_X(D)=mathcal{O}_X(1)$. You did this computation and everything works!
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
$endgroup$
Your approach is correct. One nice thing about the divisor and line bundle correspondence is that the divisor is the divisor of any rational section. Sections of $mathcal{O}_{mathbb{P}^2}(1)$ pullback to sections of $mathcal{O}_X(1)$, and the associated divisor of a pulled-back section gives you a divisor $D$ such that $mathcal{O}_X(D)=mathcal{O}_X(1)$. You did this computation and everything works!
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
answered Dec 2 '18 at 22:16
Samir CanningSamir Canning
46539
46539
add a comment |
add a comment |
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$begingroup$
In general you would get that the associated line bundle $mathscr{L}$ is isomorphic to $varphi^*(O_{mathbb{P}^n}(1))$ where $varphi: X to mathbb{P}^n$ is defined by the global sections of $mathscr{L}$. But in this case the divisor $3 P_0$ was used to embed $X$ into projective space in the first place (hence the nice Weierstrass equation), so $varphi$ is just the inclusion map $iota: X to mathbb{P}^2$. I think this means that $mathscr{L} cong iota^*(O_{mathbb{P}^2}(1)) cong O_X(1)$.
$endgroup$
– André 3000
Dec 1 '18 at 20:05
$begingroup$
@André3000 Thank you for your solution. However, I am not quite familiar with the Weierstrass equation. I try to argue it as every divisor in $mathbb{P}^2$ restricts to a divisor on $X$, and $z=0$ is a divisor associated to $O_{mathbb{P}^2}(1)$ (as we have an obvious embedding of this curve to projective space). Thus it should also give a divisor corresponding to $O_X(1)$.
$endgroup$
– Peter Liu
Dec 1 '18 at 23:03