Non-singular elliptic curve parametrization












0












$begingroup$


It is known that some singular elliptic curves can be expressed with parametric equations.



For example :
$y^2=x^3$ can be parametrized with $x=t^2$ and $y=t^3$
$y^2=x^3+x^2$ can be parametrized with $x=t^2-1$ and $y=t^3-t$ [source]



But, is there any parametric equations for some non-singular elliptic curves with a graph looking like :
simple elliptic curve

And if yes, what would be the correct way to obtain it.



We could assume that it can be represented by : $y^2 = f(x)$
where $f(x)=x^3+ax^2+bx+c$ have only one simple real root.






algebraic-geometry elliptic-curves parametrization elliptic-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    There are no rational parametrization. You need the elliptic functions or similar.
    $endgroup$
    – user10354138
    Jan 5 at 11:31










  • $begingroup$
    Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization
    $endgroup$
    – Yoshi
    Jan 5 at 11:37












  • $begingroup$
    @Yoshi "rational" means the parameters live in $mathbb C cup { infty }$.
    $endgroup$
    – Kenny Wong
    Jan 5 at 11:44












  • $begingroup$
    Sorry but, what's the question?
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 5 at 11:45










  • $begingroup$
    @KennyWong Oh ok, translation mistake ! Thank you !
    $endgroup$
    – Yoshi
    Jan 5 at 11:47


















0












$begingroup$


It is known that some singular elliptic curves can be expressed with parametric equations.



For example :
$y^2=x^3$ can be parametrized with $x=t^2$ and $y=t^3$
$y^2=x^3+x^2$ can be parametrized with $x=t^2-1$ and $y=t^3-t$ [source]



But, is there any parametric equations for some non-singular elliptic curves with a graph looking like :
simple elliptic curve

And if yes, what would be the correct way to obtain it.



We could assume that it can be represented by : $y^2 = f(x)$
where $f(x)=x^3+ax^2+bx+c$ have only one simple real root.






algebraic-geometry elliptic-curves parametrization elliptic-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    There are no rational parametrization. You need the elliptic functions or similar.
    $endgroup$
    – user10354138
    Jan 5 at 11:31










  • $begingroup$
    Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization
    $endgroup$
    – Yoshi
    Jan 5 at 11:37












  • $begingroup$
    @Yoshi "rational" means the parameters live in $mathbb C cup { infty }$.
    $endgroup$
    – Kenny Wong
    Jan 5 at 11:44












  • $begingroup$
    Sorry but, what's the question?
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 5 at 11:45










  • $begingroup$
    @KennyWong Oh ok, translation mistake ! Thank you !
    $endgroup$
    – Yoshi
    Jan 5 at 11:47
















0












0








0





$begingroup$


It is known that some singular elliptic curves can be expressed with parametric equations.



For example :
$y^2=x^3$ can be parametrized with $x=t^2$ and $y=t^3$
$y^2=x^3+x^2$ can be parametrized with $x=t^2-1$ and $y=t^3-t$ [source]



But, is there any parametric equations for some non-singular elliptic curves with a graph looking like :
simple elliptic curve

And if yes, what would be the correct way to obtain it.



We could assume that it can be represented by : $y^2 = f(x)$
where $f(x)=x^3+ax^2+bx+c$ have only one simple real root.






algebraic-geometry elliptic-curves parametrization elliptic-equations







share|cite|improve this question











$endgroup$




It is known that some singular elliptic curves can be expressed with parametric equations.



For example :
$y^2=x^3$ can be parametrized with $x=t^2$ and $y=t^3$
$y^2=x^3+x^2$ can be parametrized with $x=t^2-1$ and $y=t^3-t$ [source]



But, is there any parametric equations for some non-singular elliptic curves with a graph looking like :
simple elliptic curve

And if yes, what would be the correct way to obtain it.



We could assume that it can be represented by : $y^2 = f(x)$
where $f(x)=x^3+ax^2+bx+c$ have only one simple real root.






algebraic-geometry elliptic-curves parametrization elliptic-equations




algebraic-geometry elliptic-curves parametrization elliptic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 18:33







Yoshi

















asked Jan 5 at 11:19









YoshiYoshi

133




133








  • 2




    $begingroup$
    There are no rational parametrization. You need the elliptic functions or similar.
    $endgroup$
    – user10354138
    Jan 5 at 11:31










  • $begingroup$
    Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization
    $endgroup$
    – Yoshi
    Jan 5 at 11:37












  • $begingroup$
    @Yoshi "rational" means the parameters live in $mathbb C cup { infty }$.
    $endgroup$
    – Kenny Wong
    Jan 5 at 11:44












  • $begingroup$
    Sorry but, what's the question?
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 5 at 11:45










  • $begingroup$
    @KennyWong Oh ok, translation mistake ! Thank you !
    $endgroup$
    – Yoshi
    Jan 5 at 11:47
















  • 2




    $begingroup$
    There are no rational parametrization. You need the elliptic functions or similar.
    $endgroup$
    – user10354138
    Jan 5 at 11:31










  • $begingroup$
    Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization
    $endgroup$
    – Yoshi
    Jan 5 at 11:37












  • $begingroup$
    @Yoshi "rational" means the parameters live in $mathbb C cup { infty }$.
    $endgroup$
    – Kenny Wong
    Jan 5 at 11:44












  • $begingroup$
    Sorry but, what's the question?
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 5 at 11:45










  • $begingroup$
    @KennyWong Oh ok, translation mistake ! Thank you !
    $endgroup$
    – Yoshi
    Jan 5 at 11:47










2




2




$begingroup$
There are no rational parametrization. You need the elliptic functions or similar.
$endgroup$
– user10354138
Jan 5 at 11:31




$begingroup$
There are no rational parametrization. You need the elliptic functions or similar.
$endgroup$
– user10354138
Jan 5 at 11:31












$begingroup$
Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization
$endgroup$
– Yoshi
Jan 5 at 11:37






$begingroup$
Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization
$endgroup$
– Yoshi
Jan 5 at 11:37














$begingroup$
@Yoshi "rational" means the parameters live in $mathbb C cup { infty }$.
$endgroup$
– Kenny Wong
Jan 5 at 11:44






$begingroup$
@Yoshi "rational" means the parameters live in $mathbb C cup { infty }$.
$endgroup$
– Kenny Wong
Jan 5 at 11:44














$begingroup$
Sorry but, what's the question?
$endgroup$
– José Alejandro Aburto Araneda
Jan 5 at 11:45




$begingroup$
Sorry but, what's the question?
$endgroup$
– José Alejandro Aburto Araneda
Jan 5 at 11:45












$begingroup$
@KennyWong Oh ok, translation mistake ! Thank you !
$endgroup$
– Yoshi
Jan 5 at 11:47






$begingroup$
@KennyWong Oh ok, translation mistake ! Thank you !
$endgroup$
– Yoshi
Jan 5 at 11:47












1 Answer
1






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oldest

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3












$begingroup$

First, there are not "singular elliptic curves", you want to say "singular curves given by a Weierstrass equation".



Now, if you want to do such parametrization (through polynomials) to a curve given by a Weierstrass equation, if the curve is non-singular, that can not be done. If your curve $C$ is defined over $mathbb{P}^2$ over $mathbb{C}$, you will have a regular (holomorphic) map
begin{align}
phi:mathbb{P}^1to C.
end{align}
By the Riemann-Hurwitz Theorem, as $phi$ is surjective, the genus of the codomain must be less or equal to the genus of the domain. But, the genus of $C$ is $1$, and the genus of $mathbb{P}^1$ is zero.






share|cite|improve this answer









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    $begingroup$

    First, there are not "singular elliptic curves", you want to say "singular curves given by a Weierstrass equation".



    Now, if you want to do such parametrization (through polynomials) to a curve given by a Weierstrass equation, if the curve is non-singular, that can not be done. If your curve $C$ is defined over $mathbb{P}^2$ over $mathbb{C}$, you will have a regular (holomorphic) map
    begin{align}
    phi:mathbb{P}^1to C.
    end{align}
    By the Riemann-Hurwitz Theorem, as $phi$ is surjective, the genus of the codomain must be less or equal to the genus of the domain. But, the genus of $C$ is $1$, and the genus of $mathbb{P}^1$ is zero.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      First, there are not "singular elliptic curves", you want to say "singular curves given by a Weierstrass equation".



      Now, if you want to do such parametrization (through polynomials) to a curve given by a Weierstrass equation, if the curve is non-singular, that can not be done. If your curve $C$ is defined over $mathbb{P}^2$ over $mathbb{C}$, you will have a regular (holomorphic) map
      begin{align}
      phi:mathbb{P}^1to C.
      end{align}
      By the Riemann-Hurwitz Theorem, as $phi$ is surjective, the genus of the codomain must be less or equal to the genus of the domain. But, the genus of $C$ is $1$, and the genus of $mathbb{P}^1$ is zero.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        First, there are not "singular elliptic curves", you want to say "singular curves given by a Weierstrass equation".



        Now, if you want to do such parametrization (through polynomials) to a curve given by a Weierstrass equation, if the curve is non-singular, that can not be done. If your curve $C$ is defined over $mathbb{P}^2$ over $mathbb{C}$, you will have a regular (holomorphic) map
        begin{align}
        phi:mathbb{P}^1to C.
        end{align}
        By the Riemann-Hurwitz Theorem, as $phi$ is surjective, the genus of the codomain must be less or equal to the genus of the domain. But, the genus of $C$ is $1$, and the genus of $mathbb{P}^1$ is zero.






        share|cite|improve this answer









        $endgroup$



        First, there are not "singular elliptic curves", you want to say "singular curves given by a Weierstrass equation".



        Now, if you want to do such parametrization (through polynomials) to a curve given by a Weierstrass equation, if the curve is non-singular, that can not be done. If your curve $C$ is defined over $mathbb{P}^2$ over $mathbb{C}$, you will have a regular (holomorphic) map
        begin{align}
        phi:mathbb{P}^1to C.
        end{align}
        By the Riemann-Hurwitz Theorem, as $phi$ is surjective, the genus of the codomain must be less or equal to the genus of the domain. But, the genus of $C$ is $1$, and the genus of $mathbb{P}^1$ is zero.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 12:05









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110






























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