# Birationally transforming a quartic elliptic curve

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.)

• See Allan MacLeod's answer to this question math.stackexchange.com/questions/1591990/…. May 14 '16 at 15:27
• On second thought, I can probably edit this instead; that one only did a Weierstrass reduction. May 14 '16 at 15:37
• Are you aware of the transformation of a quartic to a cubic explained in Chapter 4, §3 of Husemoller's book ? May 24 '16 at 15:40
• @user, is that the same as the one Jesper linked to? May 24 '16 at 15:51
• No, but actually similar. Apart from the reduction to a cubic it gives a map from the Jacobi quartic family to the the Legendre cubic family. I suppose one might put it together with known transformations and parametrisations of cubics to get what you want, but I have not tried. May 24 '16 at 15:52

The method explained in Hausemöller'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.

• I see, a Möbius transformation. (I should not be surprised, I guess; such a substitution is customarily done to reduce elliptic integrals to standard forms). I am still in the market for something that would reduce to the Jacobi form $v^2=(1-u^2)(1-mu^2)$ or some other convenient quartic normal form, tho. Thanks for this! May 25 '16 at 17:01
• Should $v = \frac{ad - bc}{(cx + d)^2}y$ (rather than $\frac{ad - bc}{(cx + d)^2}$ as it currently reads)? Dec 20 '16 at 8:34

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}}} }

• Cool, I'll try this out the next time I get to a computer with Maple. Would you happen to be familiar with the algorithm being used behind the scenes? May 25 '16 at 17:28
• The Maple help page says: 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: arXiv.org/abs/math.AG/0203130 May 25 '16 at 17:31
• Okay, the second paper apparently answers the next question I was about to ask. Thank you very much! May 25 '16 at 17:32
• I have to point out that the algorithm of Maple might be wrong, or at least it does not match with eqs. 5.8 and 5.6 in "F-Theory and the Mordell-Weil Group of Elliptically-Fibered Calabi-Yau Threefolds" by D. Morrisson and D. Park. In particular, we can compare the formulas of the comment with these of (eq. 5.8): in the case $f_6 = 0$ we get that $f_9 = f_{Israel}$ and $4 f_{12} = d_{Israel}$. The birational morphism (eq. 5.7 in the paper) is not matched by the birational maps of Maple. In particular, we do not have the term with the strange square roots. Feb 1 at 18:19
• I converted the comment above from an answer. The additional sentence "However, the mapping between the coefficients of the planar quartic, and these of the planar cubic agrees." was cut off in the conversion. Feb 1 at 19:46

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.)

• For my purposes, I'm assuming complex coefficients, so I believe I can get a Weierstrass form at least. Would you happen to know any direct methods for transforming into Jacobi or some other quartic form without having to take a Weierstrass detour? (BTW: I have two of your books; thank you very much for writing them.) May 25 '16 at 19:59
• @J.M. No, sorry, I don't offhand know the direct transformation. Thanks for the kind words about my books. May 25 '16 at 20:33
• I thought $C \to E$ was a 4:1 cover, not a double cover (if $C$ does have a rational point then $C \to E$ is the doubling map, but that has degree $4$). The Jacobi form requires some level-$2$ structure, so in general you don't expect to put $C$ in such a form without extending the field of coefficients. Dec 20 '16 at 6:34
• @NoamD.Elkies I think you're right. Thanks. The curve $C$ is hyperelliptic over $K$, so it has a $K$-rational divisor $D$ of degree 2. This gives a map $$C\to\text{Jac}(C),\quad P\mapsto [2(P)-D].$$ But this map has degree 4. Dec 20 '16 at 13:38

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.