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Let $C$ be the following curve in $\mathbb{C}^2$. \begin{align} & 11664\, {c_1}^3\, {c_2}^2 + 536544\, {c_1}^3\, c_2 + 6170256\, {c_1}^3 + 67068\, {c_1}^2\, {c_2}^2 + 1542564\, {c_1}^2\, c_2 \\ & + 3085128\, c_1\, {c_2}^2 - 32393844\, c_1\, c_2 + 3085128\, c_1 + 17739486\, {c_2}^2 + 6941538\, c_2 = 0. \end{align} I checked that this curve has genus $1$ using Sage. Therefore it is an elliptic curve. How to change coordinates such that the equation of this curve is of the form $y^2 = f(x)$, where $f$ is some polynomial. Thank you very much.

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First, you need a nonsingular point of the curve. The only singular points are the two points at infinity, so the point $(0,0)$ is safe.

Now, you can use magma to, first, define the affine curve, then its projective closure $CP$, and then call the function EllipticCurve(CP, P); where $P$ is the point $CP![0,0,1]$. It gives the curve in generalized Weierstrass form, and the morphism. Then you can compute other Weierstrass models. The minimal model over $\mathbb Q$ is $$y^2 = x^3 - x^2 + 1018049744x - 672911244106688. $$

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  • $\begingroup$ thank you very much for your answer. I do not have the software magma. Could your method be implemented in Singular or other softwares? $\endgroup$ Commented Aug 2, 2018 at 18:22
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    $\begingroup$ @Jianrong Li You can use the magma online calculator as this computation doesn't take too much time. magma.maths.usyd.edu.au/calc $\endgroup$ Commented Aug 2, 2018 at 19:02
  • $\begingroup$ @FrançoisBrunault, thank you very much. $\endgroup$ Commented Aug 2, 2018 at 19:06
  • $\begingroup$ @Xarles, thank you very much. What is the meaning of $CP![0,0,1]$ in magma? $\endgroup$ Commented Sep 19, 2018 at 17:13
  • $\begingroup$ Is the point with coordinates (0,0,1) seen as a point in the curve CP. $\endgroup$
    – Xarles
    Commented Sep 20, 2018 at 17:14
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You have a form $$ y^2(Ax^3+Bx^3+Cx^2+Dx+E)+y(Fx^3+Gx^2+Hx+I)+(Jx^3+Kx^2+Lx+M)=0$$

Some of the alphabet are zero in your case. You can complete the square in y.

$$y^2f(x)+yg(x)+h(x)=0\rightarrow [2f(x)y+g(x)]^2+4f(x)h(x)-g(x)^2=0$$

Start with analysing the simpler points of your multinomial where $g(x) = 0$ (reduce). This will inform the locus of transition points in the (xy) plane. After this you have at least three options.

The first is to calculate the curvature by transforming into parametric coordinates and locate the critical points.

The second is to regularise beyond the euclidean-normalisable region by extending $y$ to $y+re^{i\theta}$ to test stability and check for splitting.

The third is to examine the discriminant $g(x)^2-4f(x)h(x)$. It is at most degree six. If it is solvable then determine the group structure on permutations of the roots; this will inform the classification of affine structure.

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