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I have given the following elliptic curve over $\mathbb{F}_9 = \mathbb{F}_3[\zeta_8]$ (where $\min_{\mathbb{F}_3}(\zeta_8) = x^2 - x - 1$):

$$ E/\mathbb{F}_9: y^2 = \zeta_8 x^3 + x. $$ Now I would like to know how to bring this equation in Weierstrass form, i.e. in the form $y^2 = x^3 + Ax + B$ for some $A,B \in \mathbb{F}_9$. I thought about doing a substitution $y = \zeta_{16} y'$ for some element $\zeta_{16}$ with $\zeta_{16}^2 = \zeta_8$, so I could divide by $\zeta_8$ on both sides. However, I don't think that $\zeta_{16}$ is in $\mathbb{F}_9$.

Could you please help me here?

Ribbity
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    In characteristics two or three we cannot always bring an elliptic curve into the short Weierstrass form. Observe that you are in characteristic three, so it is not necessarily possible to depress the cubic.This time it works like charm though. – Jyrki Lahtonen Mar 22 '21 at 04:52

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Let $\zeta = \zeta_8$. Multiplying both sides of $y^2 = \zeta x^3 + x$ by $\zeta^2$ yields $$ (\zeta y)^2 = \zeta^2 y^2 = \zeta^3 x^3 + \zeta^2 x = (\zeta x)^3 + \zeta (\zeta x) \, . $$ Letting $\tilde{x} = \zeta x$ and $\tilde{y} = \zeta y$, then $$ \tilde{y}^2 = \tilde{x}^3 + \zeta \tilde{x} \, . $$

Viktor Vaughn
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  • Thank you! Is there an instructive or intuitive way of seeing how to choose the substitutions as you did? Or is that mere trial and error? – Ribbity Mar 22 '21 at 17:12
  • @Ribbity The only change of variables preserving a short Weierstrass equation is $x = u^2 x'$ and $y = u^3 y'$ for some $u \in \overline{k}$, where $k$ is the base field. (Cf., Silverman's AEC, III.1.) The equation you gave was almost already in short Weierstrass form, so I figured a scaling would work, and I just needed to "absorb" the $\zeta$ into the $x$ variable. In general one can use Nagell's algorithm to put a cubic plane curve in Weierstrass form---see here or here for more. – Viktor Vaughn Mar 22 '21 at 23:13