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I'm looking at the elliptic curve $C:={\cal Z}(XY^2+ZX^2+YZ^2)$ in the field $k:=\overline{\mathbb{F}_2}$. I want to prove that this curve has 9 inflection points. Since the characteristic of $k$ is 2, I cannot use the Hessian determinant, which is always zero in this case. I have also shown that ${\cal Z}(X)\cap C$, ${\cal Z}(Y)\cap C$ and ${\cal Z}(Z)\cap C$ aren't inflection points. Can anybody help me?

Thanks in advance

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  • $\begingroup$ What happens if you plug in $Z= aX +bY$ and impose that it gives a cube $(\alpha X + \beta Y)^3$ ? Can you solve that for $(\beta:\alpha:1)$ ? $\endgroup$ Jan 3, 2014 at 13:06
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    $\begingroup$ Compute the Hessian determinant as though the equation had integer coefficients, remove the extra factors of two and then reduce. You get the equation $x^3 + y^3 + z^3 + xyz$, cutting out the inflection points on $C$. $\endgroup$
    – M P
    Jan 3, 2014 at 13:25
  • $\begingroup$ @MP: that's neat! does this always work? why? $\endgroup$ Jan 3, 2014 at 15:01
  • $\begingroup$ @MP: In which field extension are the inflection points, found by solving the equation, defined? $\endgroup$
    – user45014
    Jan 3, 2014 at 15:42
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    $\begingroup$ If you just want to know that there are 9 inflection points, why don't you just say that they form a torsor under the group of points of order 3, isomorphic to $(\mathbb{Z}/3)^2$? $\endgroup$
    – abx
    Jan 3, 2014 at 16:47

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