TITLE: Quadratic twist of an elliptic curve given by non-Weierstrass model
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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 $E_d$.