Cryptography Reference
In-Depth Information
Lemma 8.5.35
The function φ
∗
of Definition
8.5.34
is
-linear and φ
∗
is injective (
k
=
non-zero) if and only if φ is separable.
k
Proof
The linearity follows since
dx
is
-linear. The second part follows since if
x
is
k
k
(
C
1
)
/φ
∗
k
(
C
2
) and
φ
∗
k
k
(
φ
∗
(
x
)) are
separating for
(
C
2
) and
φ
is separable then
(
C
2
)
/
separable. Hence,
φ
∗
(
x
) is a separating element for
(
C
1
) and
dφ
∗
(
x
)
k
=
0. The reverse
implication is also straightforward.
Lemma 8.5.36
Let φ
:
C
1
→
C
2
be an unramified morphism of curves over
k
and let
k
(
C
2
)
. Then φ
∗
(div(
ω
))
div(
φ
∗
(
ω
))
.
ω
∈
=
∈
k
=
Proof
Let
P
C
1
(
) and
Q
φ
(
P
). Let
t
Q
be a uniformiser at
Q
. Since
φ
is unramified
it follows that
t
P
=
φ
∗
(
t
Q
) is a uniformiser at
P
.Let
f
∈ k
(
C
2
). It suffices to show that
v
P
(
φ
∗
(
df
))
v
Q
(
df
).
Recall from Exercise
8.5.25
that
v
Q
(
df
)
=
=
v
Q
(
∂f/∂t
Q
). If
F
(
x,y
) is a rational function
F
(
t
P
,φ
∗
(
f
))
such that
F
(
t
Q
,f
)
=
0 then 0
=
F
(
t
Q
,f
)
◦
φ
=
F
(
t
Q
◦
φ,f
◦
φ
)
=
=
0.
Hence, by definition,
∂φ
∗
(
f
)
/∂t
P
=−
(
∂F/∂x
)
/
(
∂F/∂y
)
=
∂f/∂t
Q
and so
v
P
(
dφ
∗
(
f
))
=
v
P
(
∂φ
∗
(
f
)
/∂t
P
)
=
v
Q
(
f
).
Corollary 8.5.37
Letφ
:
C
1
→
C
2
be an isomorphism of curves over
k
and letω
∈
k
(
C
2
)
.
deg(div(
φ
∗
(
ω
)))
.
Then
deg(div(
ω
))
=
8.6 Genus zero curves
Theorem 8.6.1
Let C be a curve over
k
(i.e., projective non-singular). The following are
equivalent.
1
.
1. C is birationally equivalent over
k
to
P
k
k
≥
2. The divisor class group of C over
is trivial and
#
C
(
)
2
.
∈
k
≥
3. There is a point P
C
(
)
with
k
(
P
)
2
.
Proof
See Section 8.2 of Fulton [
199
].
Definition 8.6.2
A curve satisfying any of the above equivalent conditions is called a
genus
0 curve
.
Exercise 8.6.3
Write down a curve
C
over a field
k
such that the divisor class group
Pic
0
k
1
.
(
C
)istrivialbut
C
is not birationally equivalent over
k
to
P
Theorem 8.6.4
An elliptic curve does not have genus 0.
Proof
See Lemma 11.3 or Lemma 11.5 of Washington [
560
].
Corollary 8.6.5
Let E be an elliptic curve and P
1
,P
2
∈
E
(
k
)
.IfP
1
=
P
2
then
(
P
1
)
−
(
P
2
)
is not a principal divisor.
