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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – abx
    Apr 12, 2014 at 5:20
  • $\begingroup$ @abx are you sure E_d depends on the choice of point? Maple computed Weierstrass model and maps for the general case without points. $\endgroup$
    – joro
    Apr 12, 2014 at 7:35
  • $\begingroup$ What do you mean? What about the point corresponding to $\infty$ in the Weierstrass model? $\endgroup$
    – abx
    Apr 12, 2014 at 7:43
  • $\begingroup$ @abx I believe the maps in my answer work without distinguished point and checked it. (To make them rational make $a$ square). $\endgroup$
    – joro
    Apr 12, 2014 at 7:54
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    $\begingroup$ 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 $k$. $\endgroup$
    – R.P.
    Apr 12, 2014 at 16:29

3 Answers 3


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.


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


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

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

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

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