Cryptography Reference
In-Depth Information
Since ab
cd
has determinant 1, we have
ab
cd
1
= d −b
.
ca
Let
( m ,n )=( m, n ) ab
cd
=( ma + nc, mb + nd ) .
Then
( m, n )=( m ,n ) d −b
,
ca
so there is a one-to-one correspondence between pairs of integers ( m, n )and
pairs of integers ( m ,n ). Therefore,
G k + b
=( + d ) k
1
( m τ + n ) k
+ d
( m ,n )
=(0 , 0)
=( + d ) k G k ( τ ) .
Since g 2 and g 3 are multiples of G 4 and G 6 ,wehave
g 2 + b
=( + d ) 4 g 2 ( τ ) ,
3 + b
=( + d ) 6 g 3 ( τ ) .
+ d
+ d
Therefore, when we substitute these expressions into the definition of j ,all
the factors ( + d ) cancel.
Let
F
be the subset of z
∈H
such that
1
2
( z ) < 1
π
3
<θ< π
= e
|
z
|≥
1 ,
2 ,
z
for
2 .
Figure 9.3 is a picture of
F
. Since we will need to refer to it several times, we
let
ρ = e 2 πi/ 3 .
PROPOSITION 9.14
Given τ
∈H ,there exists
ab
cd
∈ SL 2 ( Z )
su ch that
+ b
+ d
= z
∈F
.
M oreover, z ∈F isuniquelydeterm ined by τ .
 
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