0
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ $\mu$ it's more "defined" than "computed": check this answer $\endgroup$
    – Conrado
    Apr 22, 2019 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) $\endgroup$ Apr 22, 2019 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 $\endgroup$ Apr 22, 2019 at 12:39
  • $\begingroup$ Ok I got it! $\mu^3$ was X and $\mu^2$ was Y $\endgroup$ Apr 22, 2019 at 12:52
  • 1
    $\begingroup$ @RazvanUrsu If you have an answer to the question, please submit it (StackExchange doesn't like unanswered questions). $\endgroup$
    – fkraiem
    Apr 22, 2019 at 17:19

1 Answer 1

1
$\begingroup$

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 $\mu^3=x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.