# Computing a sextic twist

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!

• $\mu$ it's more "defined" than "computed": check this answer – Conrado Apr 22 at 11:56
• 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) – Razvan Ursu Apr 22 at 12:34
• 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 – Razvan Ursu Apr 22 at 12:39
• Ok I got it! $\mu^3$ was X and $\mu^2$ was Y – Razvan Ursu Apr 22 at 12:52
• @RazvanUrsu If you have an answer to the question, please submit it (StackExchange doesn't like unanswered questions). – fkraiem Apr 22 at 17:19

$$\mathbb{F}_{p^{2}}/(i^2+1)$$
$$\mathbb{F}_{p^{6}}/(y^3+\xi)$$ 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 $$\mu^3=x$$.