Cryptography Reference
In-Depth Information
C
(
K
), then it is possible to perfo
rm
the following construction using only numbers from
K
rather than from
K
.
This corresponds to the situation in Chapter 2, where we used rational points
to put certain cur
ves
into Weierstrass form. However, we'll content ourselves
with working over
K
.
Corollary 11.19 says that
Choose a point
P
∈
C
(
K
). If
P
∈
(
n
[
P
]) =
n
for all
n
≥
1
.
Since
K ⊆L
([
P
]), which has dimension 1, we have
L
([
P
]) =
K.
Since
(2[
P
]) = 2
>
([
P
]), there exists a function
f
(2[
P
]) having a
double pole at
P
and no other poles. Since
(3[
P
]) = 3
>
(2[
P
]), there exists
a function
g
∈L
(3[
P
]) with a triple pole at
P
and no other poles. Since
functions with different order poles at
P
are linearly independent, we can use
f
and
g
to give bases for several of the spaces
∈L
L
(
n
[
P
]):
([
P
]) = span(1)
L
(2[
P
]) = span(1
,f
)
L
(3[
P
]) = span(1
,f,g
)
L
(4[
P
]) = span(1
,f,g,f
2
)
L
(5[
P
]) = span(1
,f,g,f
2
,fg
)
.
L
We can write down 7 functions in the 6-dimensional space
L
(6[
P
]), namely
1
,f,g,f
2
,fg,f
3
,g
2
.
These must be linearly dependent, so there exist
a
0
,a
1
,a
2
,a
3
,a
4
,a
6
∈
K
with
g
2
+
a
1
fg
+
a
3
g
=
a
0
f
3
+
a
2
f
2
+
a
4
f
+
a
6
.
(11.10)
Note that the coecient of
g
2
must be nonzero, hence can be assumed to
be 1, since the remaining functions have distinct orders of poles at
P
and
are therefore linearly independent. Similarly,
a
0
= 0. By multiplying
f
by a
suitable constant, we may assume that
a
0
=1
.
Let
E
be the elliptic curve defined by
y
2
+
a
1
xy
+
a
3
y
=
x
3
+
a
2
x
2
+
a
4
x
+
a
6
.
We have a map
ψ
:
C
(
K
)
→
E
(
K
)
Q
→
(
f
(
Q
)
,g
(
Q
))
P
→∞.
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