Cryptography Reference
In-Depth Information
Proof.
Let
l
1
:
c
X
X
+
c
Y
Y
+
c
Z
Z
= 0 be the line passing through
P
1
and
P
2
,
where
c
X
,c
Y
,c
Z
∈
K
.
If
P
1
=
P
2
, then we obtain two linear equations in
c
X
,
c
Y
and
c
Z
c
X
X
1
+
c
Y
Y
1
+
c
Z
Z
1
=0
,
c
X
X
2
+
c
Y
Y
2
+
c
Z
Z
2
=0
.
It follows that
c
X
=
Y
1
Z
1
Y
2
Z
2
=
Y
1
Z
2
−
Z
1
Y
2
,
=
Z
1
X
1
Z
2
X
2
c
Y
=
Z
1
X
2
−
X
1
Z
2
,
c
Z
=
X
1
Y
1
X
2
Y
2
=
X
1
Y
2
−
Y
1
X
2
.
Recall that the negative point of
P
3
=(
X
3
:
Y
3
:
Z
3
)is
−P
3
=(
Y
3
:
X
3
:
Z
3
).
Consider the equation
c
X
Y
3
+
c
Y
X
3
+
c
Z
Z
3
=(
Y
1
Z
2
−
Z
1
Y
2
)
Y
3
+(
Z
1
X
2
−
X
1
Z
2
)
X
3
+(
X
1
Y
2
−
Y
1
X
2
)
Z
3
Z
1
Y
2
)(
X
1
Y
2
Z
2
−
Y
1
Z
1
X
2
)+(
Z
1
X
2
−
X
1
Z
2
)(
Y
1
X
2
Z
2
−
X
1
Z
1
Y
2
)
=(
Y
1
Z
2
−
Y
1
X
2
)(
Z
1
X
2
Y
2
−
X
1
Y
1
Z
2
)
+(
X
1
Y
2
−
=0
.
This implies
−
P
3
lies on the line
l
1
. Therefore
div(
l
1
)=(
P
1
)+(
P
2
)+(
−
P
3
)
−
3(
O
)
.
(2)
Let
l
2
be the line passing through
P
3
=(
X
3
:
Y
3
:
Z
3
)and
−
P
3
=(
Y
3
:
X
3
:
Z
3
).
The equation of
l
2
can be easily get that
X
3
Z
3
)(
X
+
Y
)+(
X
3
−
Y
3
)
Z
=0
.
(
Y
3
Z
3
−
Since the identity element is
O
=(
−
1 : 1 : 0), it follows that the point
O
also
lies on the line
l
2
. Therefore
div(
l
2
)=(
P
3
)+(
−
P
3
)
−
2(
O
)
.
(3)
Combine Eq.(2) and Eq.(3), we have
div(
l
1
/l
2
)=(
P
1
)+(
P
2
)
−
(
P
3
)
−
(
O
)
.
In the case of
P
1
=
P
2
.Let
l
1
be the line tangent to
H
(
K
)at
P
1
and it can get
∂E
∂X
1
∂E
∂Y
1
∂E
∂Z
1
that
l
1
:
X
+
Y
+
Z
= 0. Using the similar method, we can also get
div(
l
1
/l
2
)=(
P
1
)+(
P
2
)
−
(
P
3
)
−
(
O
).
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