Isomorphic of elliptic curve in binary fields

Question

Please help me to solve this question:

Suppose that $E1: y^2+y=x^3+x$ and $E2: y^2+y=x^3$ be two elliptic curve over $F_2$,show that $E1$ and $E2$ are not isomorphic over $F_2$ and $F_{2^2}$ and$F_{2^4}$, but over $F_{2^8}$ isomorphic.

Answer 1

You can do it explicitly as follows:

Hint: suppose you have an isomorphism $\varphi: E_2 \rightarrow E_1$ over $\mathbf F_{2^{2^m}}$. It must be given by $$ \varphi(x, y) = (u^2x + a,\, u^3y + bu^2x + c),$$ for suitable $u, a, b, c \in \mathbf F_{2^{2^m}}$. Using the equations of $E_1$ and $E_2$, find a system of equations that $u, a, b$ and $c$ must satisfy. Find the smallest $m$ for which there exist $u, a, b$ and $c$ satisfying those equations.

As a reward, you can also exhibit an explicit isomorphism between them over $\mathbf F_{2^8}$.

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You can also see that $E_1$ and $E_2$ are not isomorphic over $\mathbf F_{2^{2^m}}$ for $m \leq 2$ by computing the cardinalities of $E_1(\mathbf F_{2^{2^m}})$ and $E_2(\mathbf F_{2^{2^m}})$. For $m \leq 2$, they are different, so $E_1$ and $E_2$ are not isogenous over $\mathbf F_{2^{2^m}}$.

Note however that $E_1$ and $E_2$ having the same number of $\mathbf F_{2^{2^m}}$-rational points does not imply that $E_1$ and $E_2$ are isomorphic over $\mathbf F_{2^{2^m}}$.