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
aτ
+
b
=(
cτ
+
d
)
k
1
(
m
τ
+
n
)
k
cτ
+
d
(
m
,n
)
=(0
,
0)
=(
cτ
+
d
)
k
G
k
(
τ
)
.
Since
g
2
and
g
3
are multiples of
G
4
and
G
6
,wehave
g
2
aτ
+
b
=(
cτ
+
d
)
4
g
2
(
τ
)
,
3
aτ
+
b
=(
cτ
+
d
)
6
g
3
(
τ
)
.
cτ
+
d
cτ
+
d
Therefore, when we substitute these expressions into the definition of
j
,all
the factors (
cτ
+
d
) cancel.
Let
F
be the subset of
z
∈H
such that
1
2
≤
(
z
)
<
1
π
3
<θ<
π
=
e
iθ
|
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
aτ
+
b
cτ
+
d
=
z
∈F
.
M oreover,
z ∈F
isuniquelydeterm ined by
τ
.
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