Quadratic twist of an elliptic curve given by Weierstrass model [closed]

Question

What is the equation of quadratic twist of Weierstrass curve over prime field GF(p) if p mode 4 = 1 or 3 if we want to have isomorphism over GF(p)? What is the equation for binary field ? Thanks.

Answer 1

This is the answer to your question for "binary" fields, i.e., fields of characteristic 2. So let $k$ have characteristic 2 and let $E/k$ be an elliptic curve. If $j(E)\ne0$, then there is a Weierstrass equation for $E/k$ of the form $$ y^2+xy=x^3+ax^2+b. $$ Further, $\text{Aut}(E)=\{\pm1\}$, so there is a unique quadratic twist corresponding to the unique quadratic extension $K/k$. Precisely, let $K=k(\alpha)$, where $\alpha$ is a root of the polynomial $X^2-X-D$ for some $D\in k$. Then the associated twist of $E$ is $$ y^2+xy=x^3+(a+D)x^2+b. $$ (See The Arithmetic of Elliptic Curves, Springer, Exercise A.2.)

The situation for $j(E)=0$ is that $\text{Aut}(E)$ is the twisted product of $\mathbb Z/3\mathbb Z$ and the quaternion group. So $\text{Aut}(E)$ again has a unique element of order 2, leading to a quadratic twist, but it will have a lot of quartic twists. Offhand, I don't know a reference that describes all of the possible twists.

Answer 2

Let $k$ be a field of characteristic not equal to $2$ and let $$C: y^2 = f(x)$$ be an elliptic curve over $k$. Then a quadratic twist of $k$ is a curve of the form $$C_d: dy^2 = f(x)$$ for $d \in k^*$. The isomorphism classes of the $C_d$ over $k$ are given by the group $k^*/k^{*2}$.

Now assume that $k = \mathbb{F}_p$ with $p \neq 2$. Then $k^{*}/k^{*2} \cong \{\pm 1\}$, given by the quadratic residue. If $p \equiv 3\bmod 4$, then $-1$ is a quadratic non-residue, hence $C_{-1}$ is the unique (up to isomorphism) non-trivial quadratic twist of $C$. If $p \equiv 1\bmod 4$, however, there is no canonical choice of representative for the group $k^*/k^{*2}$. Consequently, there is no "uniform" way to write down an equation for the non-trivial quadratic twist.