Divisors - Elliptic Curves: Number Theory and Cryptography - page 374

Therefore,

div( g/g φ )= n [ R ] − n [ φ ( R )] .

Let P ∈ E ( F q )[ n ]. By Lemma 11.9, we may choose a divisor D P of degree
0 such that sum( D P )= P , such that D P is disjoint from ∞ , Q ,and R ,and
such that φ ( D P )= D P (this means that φ permutes the points in D P ).
Let div( f )= nD P . We assume that f is chosen as in Lemma 11.10, so
f ( φ ( R )) = φ ( f ( R )). In Theorem 11.12, let S = P , T = R − φR , D S = D P ,

[ φR ], F S = f ,and F T = g/g φ .Then

D T =[ R ] −

( g/g φ )( D P )
f ([ R ]

τ n ( P, Q )= e n ( P, R − φ ( R )) =

− [ φ ( R )])

= φ f ( R )
g ( D P )

g ( D P )
f ( R )

= f ( R )
g ( D P )

q− 1

,

since φ raises elements of F q to the q th power. But

f ( R ) n
f ( Q )

f ( ∞ )
f (

= f (div( g )) = g (div( f )) = g ( D P ) n ,

∞ ) n

where the second equality is Weil reciprocity. Therefore,
f ( R )
g ( D P )

n

f ( Q )

f ( ∞ ) f ( ∞ ) n .

=

Raising this to the power ( q − 1) /n yields
τ n ( P, Q )= f ( R )

q− 1

= f ( Q )
f (

( q− 1) /n

f ( ∞ ) q− 1 .

g ( D P ) ∞ )

∈ F q since f φ ) q− 1 But f ( ∞ ) = f and φ ( ∞ )= ∞ . Therefore, f ( ∞ =1.

Define the divisor D Q =[ Q ] − [ ∞ ]. We obtain

τ n ( P, Q )= f ( D Q ) ( q− 1) /n ,

as desired. Since we have shown in the proof of Theorem 11.8 that the value
of τ n is independent of the choice of divisors, this completes the proof.

11.7 Nondegeneracy of the Tate-Lichtenbaum

Pairing

In this section we prove that the Tate-Lichtenbaum pairing is nondegen-
erate. The proof here is partly based on a paper of Schaefer [96]. First, we
make a few remarks on pairings in general.