Hello,
Kristian Gjøsteen wrote:
> Andreas Meinenhof <a.meinenhof77_nospam@gmx.de> wrote:
>
>>So let e: G_1 x G_1 -> G_2 be an admissible bilinear pairing. Let G_1
>>and G_2 be 2 groups of prime order q. Let Q be a generator of G_1.
>>First, 2 basic questions:
>>(i) Is the pairing symmetric? So is the following correct, P \in G_1:
>>e(P,Q)=e(Q,P)?
>
>
> Yes, the pairing is symmetric.
Is it hard to prove that?
>>(ii) Let a,b,c \in Z_q and Q_a =aQ, Q_b =bQ then we have:
>>e(P,Q_a)^c * e(P,Q_b)^c
>>=e(P,Q)^{ac} * e(P,Q)^{bc}
>>=e(P,Q)^{ac+bc}
>>=e(P,Q)^{c(a+b)}
>>=e(P,(a+b)Q)^{c}
>>=e(P,Q_a*Q_b)^{c}
>>I suppose this is correct. Is my suggestion correct?
>
>
> You should write Q_a+Q_b, not Q_a*Q_b!
They wrote this (see: http://crypto.stanford.edu/~pgolle/papers/nim.pdf
pp. 10): e(\prod y_j, P)^{x_i}, where P is some point in G_1 and y_j =
x_j*Q. Maybe there is a mistake in the paper. At least, I think so.
Andreas
Received on Thu Sep 29 21:52:26 2005