# Sextic twist over BN elliptic curves

I am struggling to understand how to perform a sextic twist over a BN elliptic curve. This is what I understood so far:

Let's consider a BN elliptic curve: $$E: y^2=x^3+b$$ And let's consider a point $$Q \in E(F_{p^{12}})$$. What I would like to do is to "easily" go from/to $$Q \in E(F_{p^{12}})$$ to/from $$Q' \in E'(F_{p^{2}})$$, where $$Q'$$ is the sextic twist defined as $$E': y^2=x^3+b/\xi$$ In fact I can use this isomorphism (I guess) to move from/towards $$E$$ and $$E'$$: $$\psi: E'(F_{p^2}) \rightarrow E(F_{p^{12}})$$ $$\psi((x',y')) = (\xi^{1/3}x',\xi^{1/2}y')$$ $$\psi^{-1}(x,y) = \left(\frac{x}{\xi^{1/3}},\frac{y}{\xi^{1/2}}\right)$$ My questions are:

1. Is the isomorphism I wrote correct?
2. Is $$\xi$$ an integer number in $$F_p$$?
3. Can I choose any value I want for $$\xi$$? I guess that I should choose it so that $$\xi^{1/3}$$ and $$\xi^{1/2}$$ are integers, but I'm not sure...

1. The map $$\psi(x',y')$$ is correct, but it is a homomorphism and not an isomorphism. The two curves have different numbers of points and so there can be no isomorphism between them. One can check that the map is a well-defined as for $$x',y',\xi\in\mathbb F_{p^2}\Rightarrow \xi^{1/3}x',\xi^{1/2}y'\in\mathbb F_{p^{12}}$$ and $$y'^2=x'^2+\frac b\xi\Rightarrow (\xi^{1/2}y')^2=(\xi^{1/3}x')^3+b.$$ The homomorphism can be tediously checked by working through the addition formulae (note that the slope formula will scale by a factor of $$\xi^{1/6}$$. The inverse $$\psi^{-1}$$ function only lands on $$E'(\mathbb F_{p^2})$$ if $$x/\xi^{1/3}$$ and $$y/y^{1/2}$$ are both in $$\mathbb F_{p^2}$$ which will not be the case for a general point of $$E(\mathbb F_{p^{12}})$$. It will however be correct for any point on the subgroup of $$E(\mathbb F_{p^{12}})$$ that is the image of $$\psi$$.
2. and 3. For any value $$\xi$$ in $$\mathbb F_{p^2}$$ the map is defined. However, you should choose $$\xi$$ that is neither a square (in particular not an element of $$\mathbb F_p$$) nor a cube in $$\mathbb F_{p^2}$$ otherwise the image of $$\psi$$ will lie in $$\mathbb F_{p^6}$$ or $$\mathbb F_{p^4}$$ respectively and pairings between points in the image of $$\psi$$ will all be equal to 1.
• Worth noting that the map can be clearly defined over the algebraic closure, where it is clearly an isomorphism.. (already over $F_{p^{12}}$). Oct 19, 2021 at 12:54
• @Fractalice Good point, I've added some ground field notation to make this explicit. Formally $$\psi: E(\overline\mathbb F_p)\to E'(\overline \mathbb F_p)$$ and $$\psi^{-1}:E'(\overline\mathbb F_p)\to E(\overline\mathbb F_p)$$ are both isomorphisms. For cryptographic purposes though we will always want to restrict to something finite. Oct 19, 2021 at 18:45