The exercise 5.7 in chapter V of J. Silverman, The Arithmetic of Elliptic Curves, is following:

Let $E/K$ be an elliptic curve with char($K$)=2. Then E is supersingular if and only if j(E)=0.

I have that E/K is given by two ways :

$E : y^2+xy=x^3+a_2x^2+a_6$, with $j(E)=\frac{1}{a_6}$ or $E : y^2+a_3y=x^3+a_4x+a_6$ with $j(E)=0$.

How to prove that the first is not supersingular? i.e is ordinary?


1 Answer 1


The curve $$E : y^2+xy=x^3+a_2x^2+a_6$$ contains the point $P=(0,\sqrt{a_6})$. Of course, this point may live over a quadratic extension of $K$, but that's okay. You can check using the addition law that $$ [2]P = \mathcal{O}, $$ and hence $E(\overline K)$ has a non-trivial $2$-torsion point. Therefore $E$ is ordinary.

  • 1
    $\begingroup$ Thank you for your answer. I'm honored to talk you. $\endgroup$
    – WHERE 234
    May 29, 2023 at 5:46

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