A curve f is called elliptic if the genus is 1. An algebraic function field $C\left(x\right)\left[y\right]/\left(f\right)$ is isomorphic to the field $C\left(\mathrm{x0}\right)\left[\mathrm{y0}\right]/\left(\mathrm{f0}\right)$ where f0 is of the form y0^2 + square-free polynomial in x0 of degree 3 if and only if the curve is elliptic.

For a hyperelliptic curve with genus g there exists a similar normal form: $\mathrm{f0}={\mathrm{y0}}^2+$ a squarefree polynomial in x0 of degree $2g+1$ or $2g+2$.

This procedure computes such normal form f0. It also gives an isomorphism from $C\left(\mathrm{x0}\right)\left[\mathrm{y0}\right]/\left(\mathrm{f0}\right)$ to $C\left(x\right)\left[y\right]/\left(f\right)$ by giving the images of x0 and y0. The inverse isomorphism will also be computed, unless the option `no inverse` is used.

The output is a list of 5 items:

The curve f0

The image of x0 under this isomorphism

The image of y0 under this isomorphism

The image of x under the inverse isomorphism

The image of y under the inverse isomorphism

For a description of the method in the elliptic case see M. van Hoeij, "An algorithm for computing the Weierstrass normal form", ISSAC'95 Proceedings, p. 90-95 (1995). For the hyperelliptic case, see: http://arXiv.org/abs/math.AG/0203130

The analogue of this procedure for curves of genus zero is parametrization.

A regular point $\left[x\,y\,z\right]$ on the curve can be specified as a 6th argument. In some cases this can speed up the computation. In the genus 1 case the option Weierstrass results in a Weierstrass normal form, i.e. $-4{\mathrm{x0}}^3-a\mathrm{x0}+{\mathrm{y0}}^2-b$.

If the curve is not elliptic (which can be verified by computing the genus) nor hyperelliptic (which can be verified with is_hyperelliptic then an error message will be given. If the curve is reducible, which can be checked with evala(AFactor(f)), then the normal form does not exist and Weierstrassform will fail.

$\mathrm{with}\left(\mathrm{algcurves}\right)\:$

$f≔x^4+y^4-2x^3+x^{2}y-2y^3+x^2-xy+y^2\:$

$v≔\mathrm{Weierstrassform}\left(f\,x\,y\,\mathrm{x0}\,\mathrm{y0}\right)$

$v≔\left[{\mathrm{x0}}^3+\frac{2}{3}\mathrm{x0}+\frac{1}{108}+{\mathrm{y0}}^2\,\frac{-3y+2x}{3x}\,-\frac{x^3-2x{y}^2+2y^3-x^2+2xy-2y^2}{2x^2\left(x-1\right)}\,\frac{-162{\mathrm{x0}}^3+324{\mathrm{x0}}^2-162\mathrm{x0}\mathrm{y0}-135\mathrm{x0}+108\mathrm{y0}+156}{162{\mathrm{x0}}^4-432{\mathrm{x0}}^3+432{\mathrm{x0}}^2-192\mathrm{x0}+194}\,\frac{162{\mathrm{x0}}^4-432{\mathrm{x0}}^3+162{\mathrm{x0}}^2\mathrm{y0}+351{\mathrm{x0}}^2-216\mathrm{x0}\mathrm{y0}-246\mathrm{x0}+72\mathrm{y0}+104}{162{\mathrm{x0}}^4-432{\mathrm{x0}}^3+432{\mathrm{x0}}^2-192\mathrm{x0}+194}\right]$

$\mathrm{im1}≔\mathrm{subs}\left(\left\{x=v\left[4\right]\,y=v\left[5\right]\right\}\,f\right)\:$

$\mathrm{evala}\left(\mathrm{subs}\left(\mathrm{y0}=\mathrm{RootOf}\left(v\left[1\right]\,\mathrm{y0}\right)\,\mathrm{im1}\right)\right)$

$0$

$\mathrm{im2}≔\mathrm{subs}\left(\left\{\mathrm{x0}=v\left[2\right]\,\mathrm{y0}=v\left[3\right]\right\}\,v\left[1\right]\right)\:$

$\mathrm{evala}\left(\mathrm{subs}\left(y=\mathrm{RootOf}\left(f\,y\right)\,\mathrm{im2}\right)\right)$

$0$

$\mathrm{Weierstrassform}\left({\left(y^2-1\right)}^2+x{\left(x^2+1\right)}^2\,x\,y\,\mathrm{x0}\,\mathrm{y0}\right)$

$\left[-{\mathrm{x0}}^5+{\mathrm{y0}}^2-\mathrm{x0}-1\,\frac{y^2-1}{x^2+1}\,-y\,-{\mathrm{x0}}^2\,-\mathrm{y0}\right]$