Cryptography Reference
In-Depth Information
1
1
x
2
Example 8.2.11
Consider
φ
:
A
→ A
given by
φ
(
x
)
=
as in Example
8.1.4
.This
1
1
(
x
2
/z
2
extends to the morphism
φ
:
P
→ P
given by
φ
((
x
:
z
))
=
: 1), which is re
gu
-
(
√
a
:
(1 : 0) via the equivalent formula (1 :
z
2
/x
2
). One has
φ
−
1
((
a
:1))
lar at
∞=
={
−
√
a
:1)
,
φ
−
1
((0 : 1))
and
φ
−
1
((1 : 0))
1)
,
(
}
={
(0 : 1)
}
={
(1 : 0)
}
. At a point
Q
=
(
a
:1)
with
a
=
0 one has uniformiser
t
Q
=
x/z
−
a
and
−
√
a
)(
x/z
+
√
a
)
.
φ
∗
(
t
Q
)
x
2
/z
2
=
−
a
=
(
x/z
(
√
a
: 1) one has
φ
(
P
)
Writing
P
=
=
Q
and
e
φ
(
P
)
=
1. However, one can verify that
e
φ
((0 : 1))
=
e
φ
((1 : 0))
=
2.
Lemma 8.2.12
Let
φ
:
C
1
→
C
2
and ψ
:
C
2
→
C
3
be non-constant morphisms of curves
over
k
. Let P
∈
C
1
(
k
)
. Then e
ψ
◦
φ
(
P
)
=
e
φ
(
P
)
e
ψ
(
φ
(
P
))
.
Exercise 8.2.13
Prove Lemma
8.2.12
.
Exercise 8.2.14
Let
φ
:
C
1
→
C
2
be defined over
k
.Let
P
∈
C
1
(
k
) and let
σ
∈
Gal(
k
/
k
).
Show that
e
φ
(
σ
(
P
))
=
e
φ
(
P
).
8.3 Maps on divisor classes
We can now define some important maps on divisors that will be used in several proofs
later. In particular, this will enable an elegant proof of Theorem
7.7.11
for general curves.
Definition 8.3.1
Let
φ
:
C
1
→
C
2
be a non-constant morphism over
k
. Define the
pullback
φ
∗
:Div
k
(
C
2
)
→
Div
(
C
1
)
k
=
P
∈
φ
−
1
(
Q
)
e
φ
(
P
)(
P
) and extend
φ
∗
to Div
) define
φ
∗
(
Q
)
as follows. For
Q
∈
C
2
(
k
(
C
2
)
k
by linearity, i.e.
=
φ
∗
n
Q
φ
∗
(
Q
)
.
n
Q
(
Q
)
Q
∈
C
2
(
k
)
Q
∈
C
2
(
k
)
(
C
1
) are not varieties, it does not make sense to ask
whether
φ
∗
is a rational map or morphism.
Note that, since Div
(
C
2
) and Div
k
k
1
1
given by
φ
(
x
)
x
2
.Let
D
Example 8.3.2
Consider
φ
:
A
→ A
=
=
(0)
+
(1) be a divisor
1
. Then
φ
∗
(
D
)
on
A
=
2(0)
+
(1)
+
(
−
1).
Let
φ
:
C
1
→
C
2
be a non-constant morphism over
k
and let
P
∈
C
2
(
k
). Then the divisor
φ
∗
(
P
) is also called the
conorm
of
P
with respect to
(
C
1
)
/φ
∗
(
k
k
(
C
2
)) (see Definition III.1.8
of Stichtenoth [
529
]).
(
C
)
∗
be a non-constant rational
Lemma 8.3.3
Let C be a curve over
k
and let f
∈ k
1
by φ
function. Define the rational map φ
:
C
→ P
=
(
f
:1)
(in future we will write f
φ
∗
((0 : 1)
instead of φ). Then φ is a morphism and
div(
f
)
=
−
(1 : 0))
.
