Cryptography Reference
In-Depth Information
PROPOSITION 9.25
If
0
<k<
1
,then
K
(
k
)=
I
1
,
1
− k
2
=
I
(1 +
k,
1
−
k
)
.
PROOF
K
(
k
)=
1
0
dt
(1
− t
2
)(1
− k
2
t
2
)
=
π/
2
0
dθ
1
(let
t
=sin
θ
)
k
2
sin
2
θ
−
=
π/
2
0
dθ
cos
2
θ
+(1
− k
2
)sin
2
θ
=
I
(1
,
1
k
2
)
=
I
(1 +
k,
1
− k
)
.
−
The last equation follows from Proposition 9.24, with
a
=1+
k
and
b
=1
−
k
.
Putting everything together, we can now express the periods
ω
1
and
ω
2
in
terms of arithmetic-geometric means.
THEOREM 9.26
Suppose
E
isgiven by
y
2
=4
x
3
− g
2
x − g
3
=4(
x − e
1
)(
x − e
2
)(
x − e
3
)
with real n u m bers
e
1
<e
2
<e
3
.Then
Z
ω
1
+
Z
ω
2
isalattice for
E
,where
πi
ω
1
=
M
(
√
e
3
−
e
1
,
√
e
2
−
e
1
)
π
M
(
√
e
3
− e
1
,
√
e
3
− e
2
)
.
ω
2
=
PROOF
We have, with
k
as in (9.19),
4
ω
2
=
√
e
3
−
e
1
+
√
e
3
−
e
2
K
(
k
)
4
√
e
3
− e
1
+
√
e
3
− e
2
I
(1 +
k,
1
− k
)
.
Use the definition (9.19) of
k
and the relation
cI
(
ca, cb
)=
I
(
a, b
)with
c
=
√
e
3
−
=
e
1
+
√
e
3
−
e
2
2
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