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.

  • 2
    $\begingroup$ There are no rational parametrization. You need the elliptic functions or similar. $\endgroup$ Jan 5, 2019 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, 2019 at 11:37
  • $\begingroup$ @Yoshi "rational" means the parameters live in $\mathbb C \cup \{ \infty \}$. $\endgroup$
    – Kenny Wong
    Jan 5, 2019 at 11:44
  • $\begingroup$ Sorry but, what's the question? $\endgroup$ Jan 5, 2019 at 11:45
  • $\begingroup$ @KennyWong Oh ok, translation mistake ! Thank you ! $\endgroup$
    – Yoshi
    Jan 5, 2019 at 11:47

1 Answer 1


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}^1\to 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .