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I am working through Guide to Elliptic Curve Cryptography, and there is a section which provides the simplified Weierstrass equations and the necessary change of variables depending on the characteristic of the underlying field. One part reads

If the characteristic of $K$ is 2, then there are two cases to consider. If $a_1\neq 0$, then the admissible change of variables $$ (x, y) \to \left(a_1^2x + \frac{a_3}{a_1}, a_1^3y + \frac{a_1^2a_4+a_3^2}{a_1^3}\right)$$ transforms $E$ to the curve $$ y^2+xy=x^3+ax^2+b $$ where a$,b\in K$. Such a curve is said to be non-supersingular and has discriminant $\Delta=b$. If $a_1 = 0$, then the admissible change of variables $$(x,y)\to (x+a_2,y)$$ transforms $E$ to the curve $$y^2+cy=x^3+ax+b$$ where a$,b\in K$. Such a curve is said to be supersingular and has discriminant $\Delta=c^4$.

I have tried to prove/find a proof of the above classification of supersingular curves over a field with characteristic 2, however the closest I got was an exercise in The Arithmetic of Elliptic Curves which stated

Let $K$ be a field of characteristic 2 and let $E/K$ be an elliptic curve defined over $K$. Prove that $E$ is supersingular if and only if $j(E) = 0$.

Cobus Theunissen
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The general Weierstrass equation is $y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$. Now, $j(E)=0$ is equivalent to (using notation from Wikipedia), $c_4=b_2^2-24b_4=0$. Note that since we are in characteristic $2$, this is further equivalent to $b_2=a_1^2+4a_2=a_1^2=0$. That is, $a_1=0$. The change of variables $x\mapsto x+a_2$ makes the quadratic term disappear, giving an equation of the desired form.

Now, using $\Delta=b_6^2=a_3^4\ne 0$, we obtain $a_3\ne0$.

Kenta S
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