# Frobenius Map on BN Curve sextic twist?

I'm new to cryptography and I am working on R-ate Pairings on a BN Curve:

$y^2=x^3+b$ with $b$=5

with its M-type twist:

$y^2=x^3+b\beta$, and $\beta^2 = -2$.

The base finite field characteristic is P

Now, according to the paper, there is a Frobenius endomorphism

$\pi^p:(x,y)\longmapsto(x^p,y^p)$

So, for a given point:

$Q(x,y)$ on $y^2=x^3+5\beta$,

Its Frobenius map $Q`(x^p,y^p)$ should also have $(y^p)^2=(x^p)^3+5\beta$.

My calculation shows the results is $(y^p)^2=(x^p)^3-5\beta$

So where am I going wrong?

Reference of the paper (in Chinese):

## Translation of algorithm image:

$\pi_q$ as Frobenius endomorphism, $\pi_q: E \rightarrow E$, $\pi_q(x,y) = (x^q,y^q)$

$\pi_{q^2}: E \rightarrow E$, $\pi_{q^2}(x,y) = (x^{q^2},y^{q^2})$

R-ate pairing ("pair's calculation"):

Input: $P \in \mathbb{E(\mathbb{F_p}})[r], Q \in \mathbb{E'(\mathbb{F_{p^2}})}[r]$, $a = 6t+2$

Output: $R_a(Q,P)$

a) Let $a = \sum_{i=0}^{L-1}a_i 2^i, a_{L-1}=1$

b) Set $T=Q, f=1$

c) With $i$ from $L-2$ down until 0, execute:

• c.1) calculate $f = f^2 \cdot g_{T,T}(P), T=[2]T$

• c.2) if $a_i = 1$, compute $f = f\cdot g_{T,Q}(P), T = T+Q$

d) Compute $Q_1 = \pi_q(Q), Q_2=\pi_{q^2}(Q)$

e) compute $f = f\cdot g_{T,Q_1}(P), T = T+Q_1$

f) compute $f = f\cdot g_{T,-Q_2}(P), T = T-Q_2$

g) compute $f = f^{(q^{12}-1)/r}$;

h) output $f$.

Edit:

In step d) $Q_1$ is a point after applying the Frobenius map on $Q$. Then in step e) there is $T=T+Q_1$ Confusion: According my calculation above, $Q_1$ should not be on the same curve where $T$ is. Is it correct to calculate $T=T+Q_1$? Or is my calculation wrong?

• Comments are not for extended discussion; this conversation has been moved to chat. – e-sushi Dec 13 '17 at 7:46

In step d) $Q_1$ is a point after applying the Frobenius map on $Q$. Then in step e) there is $T=T+Q_1$. Confusion: According to my calculation above, $Q_1$ should not be on the same curve where $T$ is. Is it correct to calculate $T=T+Q_1$? Or is my calculation wrong?
The Frobenius map $\pi$ is a mapping from $\mathbb{E}′(\mathbb{F}_q)$ to itself. $T$ is a point on this curve, and since $\pi:\mathbb{E}′(\mathbb{F}_q) \rightarrow \mathbb{E}′(\mathbb{F}_q)$, $Q_1$ is also on the curve and so the sum is defined.