Cryptography Reference
In-Depth Information
and let f,f
∈ k
(
C
)
∗
.
Lemma 7.7.4
Let C be a curve over
k
1.
div(
ff
)
div(
f
)
.
=
div(
f
)
+
2.
div(1
/f
)
=−
div(
f
)
.
≥
P
min
f
)
v
P
(
f
)
,v
P
(
f
)
3.
div(
f
+
{
}
(
P
)
.
4.
div(
f
n
)
=
n
div(
f
)
for n
∈ Z
.
5. Let f
∈ k
(
C
)
and let σ
∈
Gal(
k
/
k
)
. Then
div(
σ
(
f
))
=
σ
(div(
f
))
.
Exercise 7.7.5
Prove Lemma
7.7.4
.
Lemma 7.7.6
With notation as above,
Prin
k
(
C
)
is a subgroup of
Div
k
(
C
)
under addition.
Exercise 7.7.7
Prove Lemma
7.7.6
.
P
1
(
k
Lemma 7.7.8
In
)
, every degree zero divisor is principal.
=
i
=
1
e
i
(
x
i
:
z
i
) where
i
=
1
e
i
=
Proof
Let
D
0. Define
n
zx
i
)
e
i
.
f
(
x,z
)
=
(
xz
i
−
(7.8)
i
=
1
Since
i
=
1
e
i
=
0 it follows that
f
(
x,z
) is a ratio of homogeneous polynomials of the
same degree and therefore a rational function on
1
. Using the uniformisers on
1
P
P
from
Example
7.3.3
one can verify that
v
P
i
(
f
)
=
e
i
when
P
i
=
(
x
i
:
z
i
) and hence that
D
=
div(
f
).
Note that if
D
is defined over
k
then one can show that the function
f
(
x,z
) in equa-
tion (
7.8
) is defined over
k
.
1
) then deg(div(
f
))
Exercise 7.7.9
Prove that if
f
∈ k
(
P
=
0.
Lemma 7.7.10
Le
t E
:
y
2
+
H
(
x
)
y
=
F
(
x
)
be a Weierstrass equation over
k
and let
P
=
(
x
i
,y
i
)
∈
E
(
k
)
be a non-singular point. Then
div(
x
−
x
i
)
=
(
P
)
+
(
ι
(
P
))
−
2(
O
E
)
.
Proof
There are one or two points
P
∈
E
(
k
) with
x
-coordinate equal to
x
i
, namely
P
=
(
x
i
,y
i
) and
ι
(
P
)
=
ι
(
P
)
=
(
x
i
,
−
y
i
−
H
(
x
i
)) (and these are equal if and only if
2
y
i
+
H
(
x
i
)
=
0). By Example
7.3.4
one can take the uniformiser
t
P
=
t
ι
(
P
)
=
(
x
−
x
i
)
unless (
∂E/∂y
)(
P
)
=
2
y
i
+
H
(
x
i
)
=
0, in which case the uniformiser is
t
P
=
(
y
−
y
i
).
In the former case, we have
v
P
(
x
−
x
i
)
=
v
ι
(
P
)
(
x
−
x
i
)
=
1. In the latter case, write
y
i
+
F
(
x
)
=
(
x
−
x
i
)
g
(
x
)
+
F
(
x
i
)
=
(
x
−
x
i
)
g
(
x
)
+
H
(
x
i
)
y
i
and
H
(
x
)
=
(
x
−
x
i
)
a
1
+
H
(
x
i
). Note that
a
1
y
i
−
g
(
x
i
)
=
(
∂E/∂x
)(
P
)
=
0 and so
g
1
(
x
):
=
1
/
(
a
1
y
−
g
(
x
))
∈
O
P
.
Then
y
2
0
=
+
H
(
x
)
y
−
F
(
x
)
y
i
)
2
y
i
+
y
i
−
=
(
y
−
+
2
yy
i
−
(
x
−
x
i
)
a
1
y
+
H
(
x
i
)
y
−
(
x
−
x
i
)
g
(
x
)
−
H
(
x
i
)
y
i
y
i
)
2
=
(
y
−
+
(
x
−
x
i
)(
a
1
y
−
g
(
x
))
+
(
y
−
y
i
)(2
y
i
+
H
(
x
i
))
.
