Cryptography Reference
In-Depth Information
+
√
2
,
1
−
√
2)
,Q
E
xa
mple
7.
6.7
Let
C
:
x
2
y
2
+
=
6 over
Q
and let
P
=
(1
=
(1
−
√
2
,
1
+
√
2)
(
√
2))
∈
C
(
Q
⊆
C
(
Q
). Define
D
=
(
P
)
+
(
Q
)
.
(
√
2)
/
It is
s
ufficient
t
o consider
σ
(
D
)for
σ
∈
Gal(
Q
Q
). The only non-trivial element is
σ
(
√
2)
=−
√
2
and one sees that
σ
(
P
)
=
Q
and
σ
(
Q
)
=
P
. Hence,
σ
(
D
)
=
D
for all
(
√
2)
/
σ
∈
Gal(
Q
Q
) and
D
is defined over
Q
. Note that
C
(
Q
)
= ∅
, so this example shows
it is possible to have Div
k
(
C
)
={
0
}
even if
C
(
k
)
= ∅
.
7.7 Principal divisors
This section contains an important and rather difficult result, namely that the number of
poles of a function on a curve (counted according to multiplicity) is finite and equal to the
number of zeroes (counted according to multiplicity). The finiteness condition is essential
to be able to represent the poles and zeroes of a function as a divisor. The other condition
is required to show that the set of all divisors of functions is a subgroup of Div
0
k
(
C
).
In this chapter, finite poles and finite zeroes is only proved for plane curves and
deg(div(
f
))
0 is proved only for elliptic curves. The general results are given in Sec-
tion
8.3
in the next chapter.
=
(
C
)
∗
. Then f has finitely many poles
Theorem 7.7.1
Let C be a curve over
k
and f
∈ k
and zeroes.
2
where
F
is irreducible.
By Exercise
5.2.30
there are only finitely many points at infinity, so we can restrict to the
affine case
C
Proof
(Special case of plane curves.) Let
C
=
V
(
F
(
x,y,z
))
⊆ P
=
V
(
F
(
x,y
)).
Let
f
=
f
1
(
x,y
)
/f
2
(
x,y
) with
f
1
,f
2
∈ k
[
x,y
]. Then
f
is regular whenever
f
2
(
P
)
=
0
so the poles of
f
are contained in
C
V
(
f
2
). Without loss of generality
f
2
(
x,y
) contains
monomials featuring
x
. The resultant
R
x
(
f
2
(
x,y
)
,F
(
x,y
)) is a polynomial in
y
with a
finite number of roots; hence,
C
∩
V
(
f
2
) is finite.
To show there are finitely many zeroes write
f
∩
=
f
1
/f
2
. The zeroes of
f
are contained
∩
∪
in
C
(
V
(
f
1
)
V
(
f
2
)) and the argument above applies.
Definition 7.7.2
Let
f
∈ k
(
C
)
∗
and define the
divisor of a function
(this is a divisor by
Theorem
7.7.1
)
div(
f
)
=
v
P
(
f
)(
P
)
.
P
∈
C
(
k
)
The divisor of a function is also called a
principal divisor
. Note that some authors write
div(
f
)as(
f
). Let
(
C
)
∗
}
Prin
k
(
C
)
={
div(
f
):
f
∈ k
.
Exercise 7.7.3
Show that the zero element of Div
k
(
C
) lies in Prin
k
(
C
).
