Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 247

PROOF

g − 1 φ ( P + T g 1 g 2 )= g − 1 φ g 1 g 2 ( P )

= φ g 2 ( g − 1 P )(by . 19))
= φ ( g − 1 P + T g 2 )(by . 21))
= g − 1

1

φ g 1 ( P + g 1 T g 2 )(by . 19))

1

= g − 1 φ ( P + g 1 T g 2 + T g 1 )(by . 21)) .

1

Applying g 1 then φ − 1 yields

T g 1 g 2 = g 1 T g 2 + T g 1 .

This is the desired relation.

C i ,for i =1 , 2, as above.
We say that the pairs ( C 1 ,φ 1 )and( C 2 ,φ 2 )are eq uivalent if there is a map
θ : C 1 →

Suppose we have curves C i and maps φ i : E →

C 2 defined over Q and a point P 0 ∈ E ( Q ) such that

φ − 1

θφ 1 ( P )= P + P 0 (8.23)

2

for all P E ( Q ). In other words, if we identify C 1 and C 2 with E via φ 1 and

φ 2 ,then θ is simply translation by P 0 .

∈

PROPOSITION 8.33
Thepairs ( C 1 ,φ 1 ) and ( C 2 ,φ 2 ) are equivalent ifand onlyifthe cocycles τ φ 1
and τ φ 2 di er by a coboundary. T hismeansthat there isapoint P 1 ∈

E ( Q )

su ch that

τ φ 1 ( g ) − τ φ 2 ( g )= gP 1 − P 1

for all g ∈ G .

For i =1 , 2, denote τ φ i ( g )= T g ,so
φ i ( P )= φ i ( P + T g )

PROOF

(8.24)

for all P E ( Q ). Suppose the pairs ( C 1 ,φ 1 )and( C 2 ,φ 2 ) a reequivalent,so

there exists θ : C 1 →

∈

C 2 and P 0 as above. For any P ∈ E ( Q ), we have

P + T g + P 0 = φ − 1 θφ 1 ( P + T g )(by . 23))

2

θφ 1 ( P )(by . 24))

= φ − 1

2

2 φ 2 ( φ − 2 θφ 1 ) g ( P ) in e θ g = θ )
=( φ − 2 θφ 1 ) g ( P )+ T g (by (8 . 20))
= g ( φ − 2 θφ 1 )( g − 1 P )+ T g

= φ − 1

(by (8 . 19))

= g ( g − 1 P + P 0 )+ T g

(by (8 . 23))

= P + gP 0 + T g .