Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 248

Therefore,

T g − T g = τ φ 1 ( g ) − τ φ 2 ( g )= gP 0 − P 0 .

Conversely, suppose there exists P 1 such that

τ φ 1 ( g ) − τ φ 2 ( g )= gP 1 − P 1 . (8.25)

Define θ : C 1 → C 2 by

θ ( Q )= φ 2 ( φ − 1

( Q )+ P 1 ) .

1

Clearly, θ satisfies (8.23). We need to show that θ is defined over Q . f

Q ∈ C ( Q ), then

θ g ( Q )= gθ ( g − 1 Q )(by . 19))
= gφ 2 φ − 1 ( g − 1 Q )+ P 1
= φ 2 (( φ 1 ) − 1 ( Q )+ gP 1 )
= φ 2 ( φ − 1

φ 2 )(( φ 1 ) − 1 ( Q )+ gP 1 )

2

= φ 2 ( φ 1 ) − 1 ( Q )+ gP 1 + T g (by (8 . 24))
= φ 2 φ − 1

T g ( g )+ gP 1 + T g (by (8 . 22)) ( Q ) −

1

= φ 2 ( φ − 1

( Q )+ P 1 )(by . 25))

1

= θ ( Q ) .

Therefore, θ is defined over Q , so the pairs ( C 1 ,φ 1 )and( C 2 ,φ 2 )areequivalent.

Proposition 8.33 says that we have a map

H 1 ( G, E ( Q )) . equivalence classes of pairs ( C, φ ) →

It can be shown that this is a bijection (see [109]). The most important

property for us is the following.

PROPOSITION 8.34
Let τ φ correspond tothe pair ( C, φ ) .Then τ φ ∈ B ( G, E ( Q )) (= cobou n daries)
ifand onlyif C has a rationalpoint(that is, a pointwithcoordinates in Q ).

PROOF Let P ∈ E ( Q ). Then

gP + T g = φ − 1 φ g ( gP )= φ − 1 ( gφ ( P ))

and

P = φ − 1 ( φ ( P )) .

Therefore,

T g = P − gP ⇐⇒ gφ ( P )= φ ( P ) .