Divisors - Elliptic Curves: Number Theory and Cryptography - page 378

PROOF Let Q ∈ E [ n ]. Then

ψ ( φQ )=( )

=( ··· ,e n ( φP, φQ ) , ··· ) i e φP = P for P ∈ E ( F q )[ n ])
= ··· ,e n ( P, Q ) φ , ···

··· ,e n ( P, φQ ) , ···

(by part (5) of Theorem 11.7)

=( ··· ,e n ( P, Q ) , ··· ) (since μ n ⊂ F q )

= ψ ( Q ) .

Therefore, ψ (( φ − 1) Q ) = 1, so ( φ − 1) E [ n ] ⊆ Ker ψ . By Lemma 11.30, with
A = B = E [ n ]and C = E ( F q )[ n ], we have # ψ ( E [ n ]) = # E ( F q )[ n ]. Let
Ker ( φ − 1) | E [ n ] denote the kernel of the restriction of φ − 1to E [ n ]. Then

# E ( F q )[ n ]=#Ker( φ − 1) | E [ n ] (since Ker( φ − 1) = E ( F q ))

=# E [ n ] / #(( φ − 1) E [ n ])

(by Lemma 11.29)

≥ # E [ n ] / #(Ker ψ ) (since ( φ − 1) E [ n ] ⊆ Ker ψ )

=# ψ ( E [ n ]) = # E ( F q )[ n ] .

Therefore, we must have equality everywhere. In particular, Ker ψ =( φ −

1) E [ n ].

We can now prove that the Tate-Lichtenbaum pairing is nondegenerate. Let
Q ∈ E ( F q ). Write Q = nR with R ∈ E ( F q ). Suppose that

τ n ( P, Q )= e n ( P, R − φR ) = 1 for all P ∈ E ( F q )[ n ] .

Then R − φR ∈ Ker ψ =( φ − 1) E [ n ]. This means that there exists T ∈ E [ n ]
such that R − φR = φT − T , hence φ ( R + T )= R + T .Sincethepoints
fixed by φ have coordinates in F q ,thisimpliesthat R + T ∈ E ( F q ). Since
Q = nR = n ( R + T ), we have Q ∈ nE ( F q ). Therefore,

τ n : E ( F q )[ n ] × E ( F q ) /n E ( F q ) −→ μ n

is nondegenerate in the second variable. Since the groups E ( F q )[ n ]and
E ( F q ) /n E ( F q ) have the same order (by Lemma 11.29 with α = n ), Lemma
11.27 implies that the pairing is also nondegenerate in the first variable. This
completes the proof of the nondegeneracy of the Tate-Lichtenbaum pairing.

Exercises

11.1 Let E be the elliptic curve y 2 = x 3 − x over Q.

(a) Show that f ( x, y )=( y 4 +1) / ( x 2 +1) 3 has no zeros or poles in

E ( Q ).