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which can be written as
s = t 3 + Ats 2 + Bs 3 .
In the following, it will be convenient to write p j
|z for a rational number
z when p j divides the numerator of z . Similarly, we'll write z ≡ 0(mod p j )
in this case. These extended notions of divisibility and congruence satisfy
properties similar to those for the usual notions.
LEMMA 8.3
( x, y ) ∈ E r ifand onlyif p 3 r
|s .If p 3 r
|s ,then p r
|t .
E r ,then p 3 r
divides the denominator of y ,so p 3 r
PROOF
If ( x, y )
s .Then p 3 r
divides the denominator of y . Part (2) of the theorem shows that p 2 r divides
the denominator of x . Therefore, ( x, y ) ∈ E r .
If p 3 r
Conversely, suppose p 3 r
divides the numerator of s =1 /y .
|
|s , then the exact power of p dividing the denominator of y is p 3 k ,
with k ≥ r . Part (2) of the theorem implies that the exact power of p dividing
t = x/y is p k .Since k ≥ r ,wehave p r
|t .
We now continue with the proof of Theorem 8.1.
Let λ r be as in the
statement of the theorem. Note that
p −r x/y =
λ r (
( x, y )) = λ r ( x,
y )=
λ r ( x, y ) .
We now claim that if P 1 + P 2 + P 3 = then
λ r ( P 1 )+ λ r ( P 2 )+ λ r ( P 3 ) 0(mod p 4 r ) .
The proof will also show that if P 1 ,P 2 ∈ E r ,then P 3 ∈ E r (hence E r is a
subgroup). Therefore,
λ r ( P 1 + P 2 )= λ r (
P 3 )=
λ r ( P 3 )= λ r ( P 1 )+ λ r ( P 2 ) ,
so λ r is a homomorphism.
Recall that three points add to if and only if they are collinear (Exercise
2.6). To prove the claim, let P 1 ,P 2 ,P 3 lie on the line
ax + by + d =0
and assume that P 1 ,P 2
E r . Dividing by y yields the s, t line
at + b + ds =0 .
Let P i denote the point P i written in terms of the s, t coordinates. In other
words, if
P i =( x i ,y i ) ,
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