Cryptography Reference
In-Depth Information
that we do not need to worry about the connection between double roots
in the algebraic calculations and double roots of the corresponding analytic
functions.
Let
y
=
ax
+
b
be the line through
P
1
,P
2
.Let
P
3
=(
x
3
,y
3
) be the third
point of intersection of this line with
E
and let (
x
3
,y
3
)=
P
3
=Φ(
z
3
)with
z
3
∈
C
. The formulas for the group law on
E
show that
y
2
−
y
1
x
2
− x
1
2
1
4
x
3
=
− x
1
− x
2
℘
(
z
2
)
2
℘
(
z
1
)
1
4
−
=
− ℘
(
z
1
)
− ℘
(
z
2
)
.
℘
(
z
2
)
−
℘
(
z
1
)
The function
(
z
)=
℘
(
z
)
− a℘
(
z
)
− b
has zeros at
z
=
z
1
,z
2
,z
3
.Since
(
z
) has a triple pole at 0, and no other
poles, it has three zeros in
F
. Therefore,
div(
)=[
z
1
]+[
z
2
]+[
z
3
]
−
3[0]
.
By Theorem 9.1(4),
z
1
+
z
2
+
z
3
∈
L
. Therefore,
℘
(
z
1
+
z
2
)=
℘
(
−z
3
)=
℘
(
z
3
)=
x
3
.
We obtain
℘
(
z
2
)
2
℘
(
z
1
)
℘
(
z
2
)
− ℘
(
z
1
)
℘
(
z
1
+
z
2
)=
1
−
− ℘
(
z
1
)
− ℘
(
z
2
)
.
(9.7)
4
By continuity, this formula, which we proved with certain values of the
z
i
excluded, now holds for all
z
i
for which it is defined.
We now need to consider the
y
-coordinate. This means that we need to
compute
℘
(
z
1
+
z
2
). We sketch the method (the interested and careful reader
may check the details). Differentiating (9.7) with respect to
z
2
yields an
expression for
℘
(
z
1
+
z
2
)intermsof
x
1
,x
2
,y
1
,y
2
,and
℘
(
z
2
). We need to
express
℘
in terms of
℘
and
℘
. Differentiating (9.5) yields
2
℘
℘
=(12
℘
2
− g
2
)
℘
.
Dividing by
℘
(
z
) (this is all right if
℘
(
z
)
= 0; the other cases are filled in by
continuity) yields
2
℘
(
z
2
)=12
℘
(
z
2
)
2
−
g
2
.
(9.8)
Substituting this into the expression obtained for
℘
(
z
1
+
z
2
) yields an expres-
sion for
℘
(
z
1
+
z
2
)intermsof
℘
(
z
1
)
,℘
(
z
1
)
,℘
(
z
2
)
,℘
(
z
2
). Some algebraic
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