Divisor on cubic curve $y^2z=x^3-xz^2$












1












$begingroup$


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







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$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












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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!






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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!






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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!






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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!






        share|cite|improve this answer









        $endgroup$



        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!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 22:16









        Samir CanningSamir Canning

        46539




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