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Let $(x,y) \in E'_{\mathbb{F}_{p^2}}$ be a point of the sextic twist.

I am currently trying to compute:

$\psi : (x, y) \leftarrow (\mu^2x,\mu^3y)$ with $\mu \in \mathbb{F}_{p^{12}}$ the root of $(Y^2 - \xi)$ with $\xi = 1+i \in \mathbb{F}_{p^{2}}$.

From the original curve over $\mathbb{F}_{p^{12}}$ (that is $y² = x³ + 2$) I found the twisted curve over $\mathbb{F}_{p^{2}}$ (that is $y² = x³ + (1-i)$) so I am currently able to easily find points in the twisted curve. However I don't really understand how I am supposed to compute $\mu$ in order to compute $\psi$ and get my point in $\mathbb{F}_{p^{12}}$.

Does anyone have any idea of how I can find $\mu$? I mean is there an algorithm that computes $\mu$? I know that $\mu^6=1+i$ but how do I compute $\mu$?

Thank you very much!

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  • $\begingroup$ $\mu$ it's more "defined" than "computed": check this answer $\endgroup$ – Conrado Apr 22 at 11:56
  • $\begingroup$ What happens when you build $\mathbb{F}_{p^{12}}$ differently? I for one I applied the following towering: Fp12 = Fp6/(X^2-xi) Fp6 = Fp2/(Y^3-xi) Fp2 = Fp(i^2+1) $\endgroup$ – Razvan Ursu Apr 22 at 12:34
  • $\begingroup$ I tried setting $\mu$ to X in the extension from Fp6 to Fp12 and when I apply the transformation and check if the point is on the curve ... it is not $\endgroup$ – Razvan Ursu Apr 22 at 12:39
  • $\begingroup$ Ok I got it! $\mu^3$ was X and $\mu^2$ was Y $\endgroup$ – Razvan Ursu Apr 22 at 12:52
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    $\begingroup$ @RazvanUrsu If you have an answer to the question, please submit it (StackExchange doesn't like unanswered questions). $\endgroup$ – fkraiem Apr 22 at 17:19
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The correct answer depends on the way the towering is done (if any towering was done). I used the following towering:

$\mathbb{F}_{p^{2}}/(i^2+1)$

$\mathbb{F}_{p^{6}}/(y^3+\xi)$ with $\xi \in \mathbb{F}_{p^{2}}$

$\mathbb{F}_{p^{12}}/(x^2+\xi)$ with $\xi \in \mathbb{F}_{p^{2}}$

For this towering, $\mu^2=y$ and $\mu^3=x$.

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