# Non-singular elliptic curve parametrization

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.

• There are no rational parametrization. You need the elliptic functions or similar. Jan 5, 2019 at 11:31
• Thank you ! However I didn't said it but i'm not necessarily looking for a rational parametrization Jan 5, 2019 at 11:37
• @Yoshi "rational" means the parameters live in $\mathbb C \cup \{ \infty \}$. Jan 5, 2019 at 11:44
• Sorry but, what's the question? Jan 5, 2019 at 11:45
• @KennyWong Oh ok, translation mistake ! Thank you ! Jan 5, 2019 at 11:47

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.