Divisors - Elliptic Curves: Number Theory and Cryptography - page 359

Let

D (0 , 3) =[(0 , 3)] − [ ∞ ] , D (5 , 1) =[(3 , 6)] − [(6 , 1)] .

The second divisor was obtained by adding R =(6 , 1) to (5 , 1) to obtain
(3 , 6) = (5 , 1) + (6 , 1). A calculation (see Section 11.1) shows that

div 4 x

=3 D (5 , 1) .

− y +1

div( y − 3) = 3 D (0 , 3) ,

5 x − y − 1

Therefore, we take

(5 , 1) = 4 x

− y +1

f (0 , 3) = y − 3 ,

1 . 5 x − y −

We have

f (0 , 3) D (5 , 1) = f (0 , 3) (3 , 6) = 6 − 3

1 − 3 ≡

2(mod .

f (0 , 3) (6 , 1)

Similarly,

f (5 , 1) ( D (0 , 3) )=4

(to evaluate f (5 , 1) ( ∞ ), see below). Therefore,

e 3 ((0 , 3) , (5 , 1)) = 4

2 ≡ 2(mod .
The number 2 is a cube root of unity, since 2 3 ≡ 1(mod7).

There are several ways to evaluate f (5 , 1) ( ∞ ). The intuitive way is to observe
that y has a pole of order 3 at ∞ while x has a pole of order 2. Therefore,
the terms −y in the numerator and denominator dominate as ( x, y ) →∞ ,so
the ratio goes to 1. Another way is to change to homogeneous form and use
projective coordinates:

f (5 , 1) ( x : y : z )= 4 x − y + z

5 x − y + z .

Then

f (5 , 1) ( ∞ )= f (5 , 1) (0:1:0)=1 .

The Tate-Lichtenbaum pairing can be calculated as

n = f ( Q + R )

P, Q

f ( R )

for appropriate f (depending on P )and R (as long as there are enough points
in E ( F q ) to choose R = P,−Q, P − Q,∞ ) . We can express the Weil pairing
intermsofthispairing:

e n ( S, T )= T,S n

n , S, T