# Quadratic twist of an elliptic curve given by non-Weierstrass model

Suppose $f(x)$ is a polynomial of degree 4 with integer coefficients and nonzero discriminant. Let $C$ be the hyperelliptic curve of genus 1 defined by $y^2=f(x)$. If we assume that $C$ has a rational point, then $C$ can be given the structure of an elliptic curve $E$. Now let $d$ be a squarefree integer. Thinking of $C$ as just a hyperelliptic curve, it has a quadratic twist $C_d$ defined by $dy^2=f(x)$; this is not necessarily an elliptic curve, as it might have no rational point. Thinking of $C$ as the elliptic curve $E$, it has a quadratic twist $E_d$, which is an elliptic curve. What is the relation between $E_d$ and $C_d$? They are both in some sense the quadratic twist of $C$, but are not the same curve.

• Obviously $E_d$ depends on the choice of a rational point, so you cannot expect a simple relation. If you want to compute $E_d$ you can map your rational point to $\infty$ by a homography in $x$, then you'll have a Weierstrass model. – abx Apr 12 '14 at 5:20
• @abx are you sure E_d depends on the choice of point? Maple computed Weierstrass model and maps for the general case without points. – joro Apr 12 '14 at 7:35
• What do you mean? What about the point corresponding to $\infty$ in the Weierstrass model? – abx Apr 12 '14 at 7:43
• @abx I believe the maps in my answer work without distinguished point and checked it. (To make them rational make $a$ square). – joro Apr 12 '14 at 7:54
• I think joro is right: the isomorphism class of $E_d$ does not depend on the choice of origin on $E$. You see this by simply writing down the obvious $\overline{k}$-isomorphism between $E_d$ and $E_d'$ (which we define as the twist by the same cocycle but wrt some different choice of origin on $E$) and check that it is defined over $k$. – RP_ Apr 12 '14 at 16:29

Let's assume the characteristic of the ground field $k$ is not $2$.

If $C$ is of the form $y^2=f(x)$ with $f$ a separable quartic, and $E$ is the Jacobian of $C$ (hence $E$ is an elliptic curve), then the Jacobian of the twist $C_d$ is $E_d$. (This is not hard to deduce from the construction of the Jacobian of a genus $1$ curve. See e.g. chapter 20 in Cassels's Lectures on Elliptic Curves for an accessible account of this.)

In particular this means that, for any $d$, if $C_d$ has a rational point, then $C_d$ is isomorphic to its own Jacobian, and therefore to $E_d$.

Classical invariant theory gives you the answer given by Rene. I add this one only to show how very straightforward it is. We assume the characteristic of the field is not 2 or 3. In this case every elliptic curve $E$ can be written as $y^2 = 4x^3 - g_2 x - g_3$ and the quadratic twist $E_d$ can be written as $y^2 = 4x^3 - d^2 g_2 x - d^3 g_3$.

The genus one curve $C$ that you describe has a Jacobian $E$ which is an elliptic curve. Given your presentation of $C$, it must represent an element of the 2-Selmer group of $E$. Computing its Jacobian is completed in this paper http://math.arizona.edu/~wmc/Research/JacobianFinal.pdf in section 3.1.

In particular, to $f(x) = a_4x^4 + \dots + a_0$ we have two fundamental invariants $I$ - a quadratic form and $J$- a cubic form, each in $a_4, \dots, a_0$. A change of variables shows that $C_d : dy^2 = f(x)$ can be rewritten as $y^2 = df(x)$. Therefore the Jacobian of $C_d$ is exactly $E_d$.

Maple can compute Weierstrass model without a point.

Here are some experiments for $f(x)=x^4+a$ which are related to congruent numbers.

 with(algcurves):wa:=Weierstrassform(x^4+a-y^2,x,y,u,v):latex(wa);latex(wa);latex(wa);


$$E: {v}^{2}+{u}^{3}-4\,au = 0$$ $$x= 2\,{\frac {{\it RootOf} \left( -a+{{\it \_Z}}^{2} \right) v}{4\,a-{u}^{2}}}$$ $$y= -1/2\,{\frac {2\,{\it RootOf} \left( -a+{{\it \_Z}}^{2} \right) {u}^{2}+8\,{\it RootOf} \left( -a+{{\it \_Z}}^{2} \right) a}{-4\,a+{u}^{2}}}$$

wad:=Weierstrassform(x^4+a-d*y^2,x,y,u,v):latex(wad);latex(wad);latex(wad);


$$E_d: {v}^{2}+{u}^{3}-4\,{d}^{2}au = 0$$ $$x=2\,{\frac {{\it RootOf} \left( {{\it \_Z}}^{2}-da \right) v}{4\,{d}^{2}a-{u}^{2}}}$$ $$y= -1/2\,{\frac {8\,{\it RootOf} \left( {{\it \_Z}}^{2}-da \right) {d}^{2}a+2\,{\it RootOf} \left( {{\it \_Z}}^{2}-da \right) {u}^{2}}{-4\,{d}^{3}a+d{u}^{2}}}$$

Observe that you must twist by $-1$ and change the sign of $u$.

Experiments suggest that if $C$ is of positive rank $E$ has a generator in the number field which maps to rational point.

If $a$ is square, the maps are over the rationals.

For $f(x)=fx^4+ex^3+c x^2+b x+a$. $$E: {u}^{3}+ \left( -1/3\,{c}^{2}-4\,fa+eb \right) u-{\frac {2}{27}}\,{c}^{3}-f{b}^{2}-{e}^{2}a+8/3\,fca+1/3\,bec+{v}^{2} = 0$$ $$E_d: {u}^{3}+ \left( -1/3\,{d}^{2}{c}^{2}-4\,{d}^{2}fa+{d}^{2}eb \right) u-{\frac {2}{27}}\,{d}^{3}{c}^{3}-{d}^{3}{e}^{2}a+1/3\,b{d}^{3}ec-{d}^{3}f{b}^{2}+8/3\,{d}^{3}cfa+{v}^{2} = 0$$

wa:=Weierstrassform(f*x^4+e*x^3+c*x^2+b*x+a-y^2,x,y,u,v):convert(wa,string);convert(wa,string);convert(wa,string);
E: u^3+(-1/3*c^2-4*f*a+e*b)*u-2/27*c^3-f*b^2-e^2*a+8/3*f*c*a+1/3*b*e*c+v^2=0
x=(18*RootOf(-a+_Z^2)*v+3*c*b-18*a*e+9*u*b)/(-9*u^2+36*f*a-c^2-6*u*c)
y=(1296*a^2*f^2*RootOf(-a+_Z^2)+108*f*RootOf(-a+_Z^2)*c*b^2-324*a*f*b*e*RootOf(-a+_Z^2)-216*f*a*c^2*RootOf(-a+_Z^2)+108*a*c*e^2*RootOf(-a+_Z^2)-27*c^2*e*RootOf(-a+_Z^2)*b+5*c^4*RootOf(-a+_Z^2)+(324*f*RootOf(-a+_Z^2)*b^2-648*f*a*c*RootOf(-a+_Z^2)-162*c*b*e*RootOf(-a+_Z^2)+324*e^2*RootOf(-a+_Z^2)*a+42*c^3*RootOf(-a+_Z^2))*u+(9*c^2*b+324*f*b*a-108*c*e*a)*v+(54*c*b-324*a*e)*u*v+81*b*v*u^2+54*u^3*c*RootOf(-a+_Z^2)+(108*c^2*RootOf(-a+_Z^2)-243*e*b*RootOf(-a+_Z^2))*u^2-81*u^4*RootOf(-a+_Z^2))/(c^4+1296*f^2*a^2-72*f*a*c^2+(12*c^3-432*f*c*a)*u+108*u^3*c+(54*c^2-648*f*a)*u^2+81*u^4)