Cryptography Reference
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By (8.4),
4 j ( h (2 j P ) 4 h (2 j− 1 P ))
c 1
4 j ,
so the infinite sum converges. Therefore, h ( P ) exists. Since
1
c 1
4 j
= c 1
3 ,
j =1
h ( P )
2 h ( P ) |≤c 1 / 6. It is clear from the definitions that h ( P ) 0
1
we obtain |
for all P .
Proofofpart(3): If h ( P ) ≤ c ,then h ( P ) 2 c + c 3 . There are only finitely
many P satisfying this inequality.
P roof of part (5): We have
1
c 1
4 n |h (2 n P +2 n Q )+ h (2 n P − 2 n Q ) 2 h (2 n P ) 2 h (2 n Q ) |≤
4 n .
Letting n →∞ yields the result.
Proofofpart(4): Since the height depends only on the x -coordinate,
h (
P )= h ( P ).
0. The cases m =0 , 1
are trivial. Letting Q = P in part (5) yields the case m = 2. Assume that we
know the result for m − 1and m .Then
h (( m +1) P )=
Therefore, we may assume m
h (( m
1) P )+2 h ( mP )+2 h ( P ) (by part (5))
=
1) 2 +2 m 2 +2 h ( P )
=( m +1) 2 h ( P ) .
( m
By induction, the result is true for all m .
Proofofpart(6): If mP = ,then m 2 h ( P )= h ( mP )= h ( )=0,so
h ( P ) = 0. Conversely, if h ( P )=0,then h ( mP )= m 2 h ( P ) = 0 for all m .
Since there are only finitely many points of height 0, the set of multiples
of P is finite. Therefore, P is a torsion point. This completes the proof of
Theorem 8.18.
P roof of P roposition 8.19 . It remains to prove Proposition 8.19. It can be
restated as saying that there exist constants c ,c such that
c
2 h ( P )+2 h ( Q )
h ( P + Q )+ h ( P
Q )
(8.6)
2 h ( P )+2 h ( Q )+ c
h ( P + Q )+ h ( P
Q )
(8.7)
for all P, Q . These two inequalities will be proved separately. We'll start with
the second one.
Let the elliptic curve E be given by y 2 = x 3 + Ax + B with A, B ∈ Z .Let
P =( a 1
Q =( a 2
b 1 ,y 1 ) ,
b 2 ,y 2 ) ,
P + Q =( a 3
P− Q =( a 4
b 3 ,y 3 ) ,
b 4 ,y 4 )
 
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