Cryptography Reference
In-Depth Information
Definition 8.2.5
Let
C
1
and
C
2
be
curves over
k
and let
φ
:
C
1
→
C
2
be a non-constant
rational map over
k
.Let
P
∈
C
1
(
k
). The
ramification index
of
φ
at
P
is
v
P
(
φ
∗
(
t
φ
(
P
)
))
e
φ
(
P
)
=
where
t
φ
(
P
)
is a uniformiser on
C
2
at
φ
(
P
). If
e
φ
(
P
)
=
1 for all
P
∈
C
1
(
k
) then
φ
is
unramified
.
We now show that this definition agrees with Definition III.1.5 of Stichtenoth [
529
].
Le
m
ma 8.2.6
Let φ
:
C
1
→
k
∈
C
2
be
a
non-constant morphism of curves over
. Let P
k
=
∈
k
∈ k
C
1
(
)
, Q
φ
(
P
)
C
2
(
)
and f
(
C
2
)
. Then
v
P
(
φ
∗
(
f
))
=
e
φ
(
P
)
v
Q
(
f
)
.
t
Q
h
for some
h
Proof
Let
v
Q
(
f
)
=
n
and write
f
=
∈ k
(
C
2
) such that
h
(
Q
)
=
0.
Then
φ
∗
(
f
)
φ
∗
(
t
Q
)
n
φ
∗
(
h
) and
v
P
(
φ
∗
(
h
)))
0. The result follows since
v
P
(
φ
∗
(
t
Q
)
n
)
=
=
=
nv
P
(
φ
∗
(
t
Q
)).
Exercis
e
8.2.7
Let
φ
:
C
1
→
C
2
be a non-constant rational
m
ap of curves over
k
.Let
∈
k
=
=
∈ k
P
C
1
(
),
Q
φ
(
P
), and suppose
e
φ
(
P
)
1. Show that
t
(
C
2
) is a uniformiser at
Q
if and only if
φ
∗
(
t
) is a uniformiser at
P
.
Exercise 8.2.8
Let
φ
:
C
1
→
C
2
be an isomorphism of curves over
k
. Show that
φ
is
unramified.
The following result is of fundamental importance.
Theorem 8.2.9
Let C
1
and C
2
be curves
o
ver
k
and let φ
:
C
1
→
C
2
be a non-constant
rational map over
k
. Then for all Q
∈
C
2
(
k
)
we have
e
φ
(
P
)
=
deg(
φ
)
.
P
∈
C
1
(
k
):
φ
(
P
)
=
Q
Proof
As mentioned above, one can see this by noting that
φ
∗
(
O
Q
) and
φ
∗
(
k
[
U
]) (for an
open set
U
⊆
C
2
with
Q
∈
U
) are Dedekind domains and studying the splitting of m
Q
in their integral closures in
(
C
1
). For details see any of Proposition 1.10 and 1.11 of
Serre [
488
], Corollary XII.6.3 of Lang [
329
], Proposition I.21 of Lang [
327
], Theorem
III.3.5 of Lorenzini [
355
], Proposition II.6.9 of Hartshorne [
252
], or Theorem III.1.11 of
Stichtenoth [
529
].
k
Corollary 8.2.10
If φ
:
C
1
→
C
2
is a rational map of degree d and Q
∈
C
2
(
k
)
then there
are at most d points P
Q.
Furthermore, if φ is separable then there is an open subset U
∈
C
1
(
k
)
such that φ
(
P
)
=
⊆
C
2
such that for all
∈
U one has
#
φ
−
1
(
Q
)
=
Q
d.
Proof
The first statement is immediate. The second follows by choosing
U
to be the
complement of points corresponding to factors of the discriminant of
(
C
1
)
/φ
∗
(
k
k
(
C
2
)); see
Proposition VII.5.7 of Lorenzini [
355
].
