Re: Reduced Tate pairing in supersingular elliptic curves

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Wed, Apr 18, 2018 at 11:13:34AM +0300, Aleksandr Lenin wrote:
> Hello Bill,
> 
> thanks for your answer.
> 
> On 04/18/2018 12:10 AM, Bill Allombert wrote:
> > However E(F_q) is isomorphic to (Z/lZ)^2 with r^6 dividing l,
                                          I meant r^3
> > and p2 is of order r, so p2 can be written as [r].q for some point q,
> > so the Tate pairing (p1,p2) is trivial.
> 
> I do not completely understand the last inference about the triviality
> of the Tate pairing.

E(F_q) is isomorphic to (Z/lZ)^2, so there are two generators g1, g2 of
order l. Since p2 belong to E(F_q), there exist a, b, such that 
p2= a.g1 + b.g2
Now
r.p2 = (r*a).g1 + (r.b)*.g2
since r.p2 = oo then 
r*a = 0 [l]
r*b = 0 [l]
Since r^2 divides l, then
r*a = 0 [r^2]
r*b = 0 [r^2]
so
a=0 [r]
b=0 [r]
So it existe a', b' such that
a= r*a'
b= r*b'
We set q2 = a'.g1+b'.g2
then r.q2 = p2

Now the Tate pairing is bilinear so tate_r(p1,p2) = tate_r(p1,q2)^r
so tate_r(p1,p2)^((q-1)/r) = tate_r(p1,q2)^(q-1) = 1.

Cheers,
Bill