Cryptography Reference
In-Depth Information
Example
8.1.4
suggests several possible definitions for degree: the first in terms of the
number of preimages of a general point in the image; the second in terms of the degrees
of the polynomials defining the map. A third definition is to recall the injective field
homomorphism
φ
∗
:
1
)
1
). One sees that
φ
∗
(
1
))
(
x
2
)
k
(
A
→ k
(
A
k
(
A
= k
⊆ k
(
x
) and that
(
x
2
)]
[
k
(
x
):
k
=
2. This latter formulation turns out to be a suitable definition for degree.
k
. Let φ
:
C
1
→
Theorem 8.1.5
Let C
1
,C
2
be curves over
C
2
be a non-constant rational
map over
k
. Then
k
(
C
1
)
is a finite algebraic extension of φ
∗
(
k
(
C
2
))
.
Proof
Theorem II.2.4(a) of Silverman [
505
].
Definition 8.1.6
Let
φ
:
C
1
→
C
2
be a non-constant rational map of curves over
k
.The
(
C
1
):
φ
∗
(
degree
of
φ
is [
(
C
2
))].
Let
F
be a field such that
φ
∗
(
k
k
k
(
C
2
))
⊂
F
⊂ k
(
C
1
) and
k
(
C
1
)
/F
is separable and
F/φ
∗
(
(
C
2
)) is purely inseparable (recall the notion of separability from Section
A.6
).
The
separable degree
of
φ
is deg
s
(
φ
)
k
=
[
k
(
C
1
):
F
] and the
inseparable degree
of
φ
is
[
F
:
φ
∗
(
deg
i
(
φ
)
(
C
2
))].
A non-constant rational map of curves is called
separable
(respectively,
inseparable
)
if its inseparable (respectively, separable) degree is 1.
=
k
1
(
1
(
Example 8.1.7
Let
k = F
p
.The
Frobenius map
π
p
:
A
k
)
→ A
k
) is given by
x
p
. Since
1
)
(
x
) and
π
p
(
1
))
(
x
p
) it follows that
(
x
)
/π
p
(
1
))
π
p
(
x
)
=
k
(
A
= k
k
(
A
= k
k
k
(
A
=
(
x
p
) is inseparable of degre
e
p
. Hence, deg
s
(
π
p
)
k
(
x
)
/
k
=
1 and deg(
π
p
)
=
deg
i
(
π
p
)
=
p
.
1
(
Note that
π
p
is one-to-one on
A
F
p
).
1
1
be a non-constant morphism over
Lemma 8.1.8
Let φ
:
A
→ A
k
given by φ
(
x
)
=
a
(
x
)
for some polynomial a
(
x
)
∈ k
[
x
]
. Then
deg(
φ
)
=
deg
x
(
a
(
x
))
.
=
a
(
x
). We have
φ
∗
(
k
A
1
))
= k
⊆ k
Proof
Let
θ
(
(
θ
)
(
x
) and we are required to determine
k
k
k
[
(
x
):
(
θ
)]. We claim the minimal polynomial of
x
over
(
θ
) is given by
F
(
T
)
=
a
(
T
)
−
θ.
First, it is clear that
F
(
x
)
0. Second, it follows from Eisenstein's criteria (see Proposition
III.1.14 of [
529
], Theorem IV.3.1 of [
329
] or Theorem III.6.15 of [
271
]), taking for example
the place (i.e., valuation) at infinity in
=
k
(
θ
), that
F
(
T
) is irreducible. Since deg
T
(
F
(
T
))
=
deg
x
(
a
(
x
)) the result follows.
A
1
→ A
1
be a non-constant rational map over
k
=
Lemma 8.1.9
Let φ
:
given by φ
(
x
)
=
=
{
}
a
(
x
)
/b
(
x
)
where
gcd(
a
(
x
)
,b
(
x
))
1
. Then
deg(
φ
)
max
deg
x
(
a
(
x
))
,
deg
x
(
b
(
x
))
.
a
(
x
)
/b
(
x
) so that
φ
∗
(
1
))
Proof
Let
θ
=
k
(
A
= k
(
θ
)
⊆ k
(
x
). Since
k
(
θ
)
= k
(1
/θ
)wemay
assume deg
x
(
a
(
x
))
≥
deg
x
(
b
(
x
)). If these degrees are equal then one can reduce the degree
of
a
(
x
)byusing
; replacing
θ
by 1
/θ
again we may assume that deg
x
(
a
(
x
))
>
deg
x
(
b
(
x
)). We may also assume that
a
(
x
)
and
b
(
x
) are monic.
k
(
a
(
x
)
/b
(
x
))
= k
((
a
(
x
)
−
cb
(
x
))
/b
(
x
)) for a suitable
c
∈ k
