# transform quartic to weierstrass

If we have a quartic surface like $y^2=a(t)x^4+b(t)x^3+...+e(t)$ with a point on it and want to transform it to weierstrass surface (cubic surface) with sage , what can we do? is it possible?

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After a translation in the variable $x$ we can assume that the point on the quartic is $(0,q)$. Then plugging in this point into the equation, we get an equation of the shape: $$y^2 = Ax^4+Bx^3+Cx^2+Dx+q^2\ .$$ For the simplicity of the following formula, let us write $Q=2q$.

The following transformation is given in [Ian Conell, Elliptic Curves Hanbook, page 105, Quartic to Weierstrass] . (Wonderful exposition.) $$x = ( Q(v+q)+Du )/u^2\ ,$$ $$y = ( Q^2(v+q) + Q(Cu^2+Du) - D^2u^2/Q)/u^3\ .$$

(For a sage example with "simpler" coeficients: https://ask.sagemath.org/question/36637/change-of-variable-from-hyperellictic-curve-to-weierstrass-form/

(This is of course a community wiki post... All karma goes to all beautiful books and people in this field.)

more Like that:

sage: t = polygen(QQ, 't')
sage: x, y = polygens(FractionField(t.parent()), 'x,y')
sage: E = Curve(-y**2+x**4+x**3+t*x**2+(1+t)*x+(4+3*t))
sage: Jacobian(E)
Elliptic Curve defined by y^2 = x^3 + (-1/3*t^2-11*t-15)*x + (2/27*t^3-22/3*t^2-6*t+5) over Fraction Field of Univariate Polynomial Ring in t over Rational Field

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