Birationally transforming a quartic elliptic curve

Question

Consider the elliptic curve

$$y^2=ax^4+cx^2+dx+f$$

where I assume complex coefficients for the purposes of this question.

I am aware that there are algorithmic methods for birationally transforming a nondegenerate cubic curve into the Weierstrass canonical form (equivalently, deriving a parametrization in terms of Weierstrass elliptic functions).

I want to ask if there are analogous methods for dealing with the quartic curve given above. (Note that I've already performed polynomial depression (removing the cubic term) in advance.) In particular, I want to know if there are birational transformations that can directly convert it into the Jacobi form or some other convenient quartic standard form. (The Weierstrass form has been thoroughly covered here and in some of the answers below.)

(This is effectively a special case of this more general question.)

Answer 1

The method explained in Husemöller's book on elliptic curves is as follows:

Take a general quartic $v^2=f_4(u)=a_ou^4+a_1u^3+a_2u^2+a_3u+a_4$, and let

$$u=\frac{ax+b}{cx+d}\qquad v=\frac{ad-bc}{(cx+d)^2} y$$

Substituting you get:

$$v^2=\frac{(ad-bc)^2}{(cx+d)^4}y^2=f_4\bigg(\frac{ax+b}{cx+d}\bigg)$$

which implies

$$(ad-bc)^2y^2=f_4\bigg(\frac{ax+b}{cx+d}\bigg)(cx+d)^4=\sum_{i=0}^4a_i(ax+b)^{4-i}(cx+d)^i=$$ $$=c^4f_4\bigg(\frac{a}{c}\bigg)x^4+f_3(x)$$

where $f_3(x)$ is a cubic polynomial whose coefficient of $x^3$ is $c^3f'_4(a/c)$. For $a/c$ a simple root of $f_4$ and $ad-bc=1$, this leaves the cubic equation $y^2=f_3(x)$.

From here you can use the tools you mention in the question to take care of the cubic.

Answer 2

This is an elliptic curve. Maple's Weierstrassform function can handle it:

> algcurves:-Weierstrassform(y^2 - a*x^4 - c*x^2 - d*x - f, x, y, u, v);

It returns the normal form in variables $u, v$:

$$u^3+\left(-\frac{c^2}{3}-4af\right)u-ad^2-\frac{2}{27}c^3+\frac{8}{3}afc+v^2$$

as well as the transformation (in both directions). In this case

$$ \eqalign{u&=-{\frac {c{x}^{2}+6\,y\sqrt {f}+3\,dx+6\,f}{3 {x}^{2}}}\cr v &= -{\frac {2\,\sqrt {f}c{x}^{2}+ydx+3\,\sqrt {f}dx+4\,fy+4\,{f}^{3/2}}{{ x}^{3}}}\cr x &= {\frac {18\,\sqrt {f}v+3\,cd+9\,du}{36\,af-{c}^{2}-6\,cu-9\,{u}^{2}}}\cr y &= \sqrt {f}+{\frac {-18\,\sqrt {f}cu+72\,{f}^{3/2}a-6\,\sqrt {f}{c}^{2}- 9\,dv}{-36\,af+{c}^{2}+6\,cu+9\,{u}^{2}}}-108\,{\frac {ad \left( \sqrt {f}cd+3\,\sqrt {f}du+6\,fv \right) }{ \left( -36\,af+{c}^{2}+6\, cu+9\,{u}^{2} \right) ^{2}}} }$$

Answer 3

If you don't have a point defined over your base field $K$, then you can't convert to Weierstrass form with $K$-coefficients using a transformation defined over $K$. (Proof: The Weierstrass equation has a $K$-rational point.) In that situation, it's often convenient to map the curve $C:y^2=ax^4+bx^3+cx^2+dx+e$ to its Jacobian, which is an elliptic curve $E/K$ that admits $C$ as a double cover, with everything defined over $K$. The formula for double cover $C\to E$ with $E$ in Weierstrass form is classical 19th century invariant theory. (See for example Salmon Lessons Introductory to the Modern Higher Algebra 3rd ed., Hodges, Foster, and Co., Cambridge, 1876, pages 187-192.)

Answer 4

For the reduction to Jacobi form, basically you are looking at the orbits of the quartic under Möbius transformation which you can identify (apart from some exceptional cases) from the two invariants $S$ and $T$ of degrees 2 and 3 in $a,c,d,f$. See this MO answer for more details.