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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 $$ y^2+xy=x^3+ax+b. $$ 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.

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.

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 $$ y^2+xy=x^3+ax+b. $$ 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 $$ y^2+xy=x^3+(a+D)x+b. $$ (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.

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 $$ y^2+xy=x^3+ax+b. $$ 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.

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 $$ y^2+xy=x^3+ax+b. $$ 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 $$ y^2+xy=x^3+(a+D)x+b. $$ (See The Arithmetic of Elliptic Curves, Springer, Exercise A.2.)

The situation for $j(E)=0$ is much more complicated, because $\text{Aut}(E)$ is the twisted product of $\mathbb Z/3\mathbb Z$ and the quaternion group. So $\text{Aut}(E)$ has a lot of elements of order 2. Offhand, I don't know a reference that describes all of the possible quadratic twists.

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 $$ y^2+xy=x^3+ax+b. $$ 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 $$ y^2+xy=x^3+(a+D)x+b. $$ (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.