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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