MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E(\mathbb{F_q})$ - elliptic curve.

$G_1 = E(\mathbb{F_q})[r]$. $|G_1| = r$.

$k$ is minimal such $r | q^k - 1$.

$\pi_q$ - $q$-power Frobenius endomorphism.

$G_2 = E(\mathbb{F_{q^k}})[r] \cap Ker([q] - \pi_q)$.

$\mu_r =$ {$z \in \mathbb{F_{q^k}}: z^r = 1$}.

$f_{s, P} \in \mathbb{F_q}(E)$: div$(f_{s, P}) =s(P) - ([s]P) - (s-1)(O)$.

Let $f_{s, P}^{(q^k - 1)/r}(Q)$ is non-degenerate bilinear pairing from $G_1 \times G_2 \to \mu_r$

Is it true that $r | s$ or $r | (s^k - 1)$ ?

share|cite|improve this question
Is there any relation between $s$ and $q$? What are $P$ and $Q$? – Srilakshmi Mar 20 '13 at 20:02
@Srilakshmi $q$ is fixing number. $s$ is so number that $f_{s,P}^{(q^k-1)/r}(Q)$ is non-degenerate bilinear pairing. $P$ is point from $G_1$ and $Q$ is point from $G_2$ – Alexey Mar 21 '13 at 21:47

The groups $G_1$ and $G_2$ are generated by $P$ and $Q$ respecively. For every $s$, $f_{s,P}$ is the Miller function associated with $s$. In Optimized (lower degree Miller functions) pairing, one imposed condition on $s$ is : $s \equiv q \pmod r$. The integer $k$ is minimal such that $r$ $\mid$ ${q^k-1}$, which implies that $r$ $\mid$ $(s^k-1)$.

share|cite|improve this answer
It is not necessary condition for $s$. For example $s$ can be folowing: $s = q^i($mod$r)$. (…) – Alexey Mar 23 '13 at 13:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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