Cryptography Reference
In-Depth Information
Proof
Since
σ
(
f
)(
σ
(
P
))
=
σ
(
f
(
P
)) the map
f
→
σ
(
f
) is an isomorphism of local rings
σ
:
O
P,
k
(
C
)
→
O
σ
(
P
)
,
k
(
C
). It also follows that
σ
(m
P
)
=
m
σ
(
P
)
. Since m
P
=
(
t
P
) one has
m
σ
(
P
)
=
(
σ
(
t
P
)), which completes the proof.
We now give an application of uniformisers. It will be used in several later results.
k
→
⊆ P
n
be a rational
Lemma 7.3.6
Let C be a non-singular curve over
and let φ
:
C
Y
map for any projective variety Y. Then φ is a morphism.
Exercise 7.3.7
Prove Lemma
7.3.6
.
7.4 Valuation at a point on a curve
The aim of this section is to define the multiplicity of a zero or pole of a function on a
curve. For background on discrete valuation rings see Chapter 1 of Serre [
488
], Section I.7
of Lang [
327
] or Sections XII.4 and XII.6 of Lang [
329
].
Definition 7.4.1
Let
K
be a field. A
discrete valuation
on
K
is a function
v
:
K
∗
→ Z
such that:
K
∗
,
v
(
fg
)
1. for all
f,g
∈
=
v
(
f
)
+
v
(
g
);
K
∗
such that
f
2. for all
f,g
∈
+
g
=
0,
v
(
f
+
g
)
≥
min
{
v
(
f
)
,v
(
g
)
}
;
K
∗
such that
v
(
f
)
3. there is some
f
∈
=
1 (equivalently,
v
is surjective to
Z
).
Lemma 7.4.2
Let K be a field and v a discrete valuation.
1. v
(1)
=
0
.
K
∗
then v
(1
/f
)
2. If f
∈
=−
v
(
f
)
.
K
∗
:
v
(
f
)
3. R
v
={
f
∈
≥
0
}∪{
0
}
is a ring, called the
valuation ring
.
K
∗
:
v
(
f
)
>
0
4.
m
v
={
f
∈
}
is a maximal ideal in R
v
, called the
maximal ideal
of the
valuation.
5. If f
∈
K is such that f
∈
R
v
then
1
/f
∈
m
v
.
6. R
v
is a local ring.
Exercise 7.4.3
Prove Lemma
7.4.2
.
Lemma 7.4.4
Let C be a curve over
k
and P
∈
C
(
k
)
. For every non-zero function f
∈
m
P,
k
O
P,
k
(
C
)
there is some m
∈ N
such that f
∈
.
Definition 7.4.5
Let
C
be a curve over
k
and
P
∈
C
(
k
). Let m
P
=
m
P,
k
(
C
)beasin
Definition
7.1.1
and define m
P
=
O
P,
k
(
C
). Let
f
∈
O
P,
k
(
C
) be such that
f
=
0 and
∈
m
P
}
define the
order
of
f
at
P
to be
v
P
(
f
)
=
max
{
m
∈ Z
≥
0
:
f
.If
v
P
(
f
)
=
1 then
f
has a
simple zero
at
P
. (We exclude the constant function
f
=
0, though one could define
v
P
(0)
=∞
.)
We stress that
v
P
(
f
) is well-defined. If
f,h
∈
O
P,
k
(
C
) and
f
≡
h
then
f
−
h
=
0in
∈
m
P
then
h
∈
m
P
(and vice versa).
O
P,
k
(
C
). Hence, if
f
