Cryptography Reference
In-Depth Information
The line through
P
3
=(
x
3
,y
3
)and
−
P
3
is
x
−
x
3
= 0. The divisor of the
function
x
−
x
3
is
div(
x − x
3
)=[
P
3
]+[
−P
3
]
−
2[
∞
]
.
(11.1)
Therefore,
div
ax
+
by
+
c
x
=div(
ax
+
by
+
c
)
−
div(
x−x
3
)=[
P
1
]+[
P
2
]
−
[
−P
3
]
−
[
∞
]
.
−
x
3
Since
P
1
+
P
2
=
−P
3
on
E
, this may be rewritten as
]+div
ax
+
by
+
c
x − x
3
.
[
P
1
]+[
P
2
]=[
P
1
+
P
2
]+[
∞
The following important result is the analogue of Theorem 9.6.
THEOREM 11.2
Let
E
be an elliptic curve. Let
D
be a divisor on
E
with
deg(
D
)=0
.Then
there isafunction
f
on
E
with
div(
f
)=
D
ifand onlyif
sum(
D
)=
∞.
PROOF
We have just shown that a sum [
P
1
]+[
P
2
] can be replaced by
[
P
1
+
P
2
]+[
∞
] plus the divisor of a function, call it
g
. Note also that
sum(div(
g
)) =
P
1
+
P
2
−
(
P
1
+
P
2
)
−∞
=
∞.
Equation (11.1) shows that [
P
1
]+[
P
2
]equals2[
∞
] plus the divisor of a function
when
P
1
+
P
2
=
. Therefore, the sum of all the terms in
D
with positive
coe
cients equals a single symbol [
P
] plus a multiple of [
∞
] plus the divisor
of a function. A similar result holds for the sum of the terms with negative
coecients. Therefore, there are points
P
and
Q
on
E
, a function
g
1
,andan
integer
n
such that
∞
D
=[
P
]
−
[
Q
]+
n
[
∞
]+div(
g
1
)
.
Also, since
g
1
is the quotient of products of functions
g
with sum(div(
g
)) =
∞
,
we have
sum(div(
g
1
)) =
∞.
Since deg(div(
g
1
)) = 0 by Proposition 11.1, we have
0=deg(
D
)=1
−
1+
n
+0=
n.
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