2Valued fields
III Local Fields
2.2 Extension of norms
The main goal of this section is to prove the following theorem:
Theorem.
Let
K
be a complete valued field, and let
L/K
be a finite extension.
Then the absolute value on
K
has a unique extension to an absolute value on
L
,
given by
α
L
=
n
q
N
L/K
(α),
where
n
= [
L
:
K
] and
N
L/K
is the field norm. Moreover,
L
is complete with
respect to this absolute value.
Corollary.
Let
K
be complete and
M/K
be an algebraic extension of
K
. Then
 ·  extends uniquely to an absolute value on M.
This is since any algebraic extension is the union of finite extensions, and
uniqueness means we can patch the absolute values together.
Corollary.
Let
K
be a complete valued field and
L/K
a finite extension. If
σ ∈ Aut(L/K), then σ(α)
L
= α
L
.
Proof.
We check that
α 7→ σ
(
α
)

L
is also an absolute value on
L
extending the
absolute value on K. So the result follows from uniqueness.
Before we can prove the theorem, we need some preliminaries. Given a finite
extension
L/K
, we would like to consider something more general than a field
norm on
L
. Instead, we will look at norms of
L
as a
K
vector space. There
are less axioms to check, so naturally there will be more choices for the norm.
However, just as in the case of
R
vector spaces, we can show that all choices of
norms are equivalent. So to prove things about the extended field norm, often
we can just pick a convenient vector space norm, prove things about it, then
apply equivalence.
Definition
(Norm on vector space)
.
Let
K
be a valued field and
V
a vector
space over K. A norm on V is a function k·k : V → R
≥0
such that
(i) kxk = 0 iff x = 0.
(ii) kλk = λkxk for all λ ∈ K and x ∈ V .
(iii) kx + yk ≤ max{kxk, kyk}.
Note that our norms are also nonArchimedean.
Definition
(Equivalence of norms)
.
Let
k·k
and
k·k
0
be norms on
V
. Then
two norms are equivalent if they induce the same topology on
V
, i.e. there are
C, D > 0 such that
C kxk ≤ kxk
0
≤ D kxk
for all x ∈ V .
One of the most convenient norms we will work with is the max norm:
Example
(Max norm)
.
Let
K
be a complete valued field, and
V
a finite
dimensional Kvector space. Let x
1
, ··· , x
n
be a basis of V . Then if
x =
X
a
i
x
i
,
then
kxk
max
= max
i
a
i

defines a norm on V .
Proposition.
Let
K
be a complete valued field, and
V
a finitedimensional
Kvector space. Then V is complete under the max norm.
Proof.
Given a Cauchy sequence in
V
under the max norm, take the limit of each
coordinate to get the limit of the sequence, using the fact that
K
is complete.
That was remarkably easy. We can now immediately transfer this to all other
norms we can think of by showing all norms are equivalent.
Proposition.
Let
K
be a complete valued field, and
V
a finitedimensional
Kvector space. Then any norm k·k on V is equivalent to k·k
max
.
Corollary. V is complete with respect to any norm.
Proof. Let k·k be a norm. We need to find C, D > 0 such that
C kxk
max
≤ kxk ≤ D kxk
max
.
We set D = max
i
(kx
i
k). Then we have
kxk =
X
a
i
x
i
≤ max (a
i
kx
i
k) ≤ (max a
i
)D = kxk
max
D.
We find
C
by induction on
n
. If
n
= 1, then
kxk
=
ka
1
x
1
k
=
a
1
kxk
=
kxk
max
kx
1
k. So C = kx
1
k works.
For n ≥ 2, we let
V
i
= Kx
1
⊕ ··· ⊕Kx
i−1
⊕ Kx
i+1
⊕ ··· ⊕Kx
n
= span{x
1
, ··· , x
i−1
, x
i+1
, ··· , x
n
}.
By the induction hypothesis, each
V
i
is complete with respect to (the restriction
of) k·k. So in particular V
i
is closed in V . So we know that the union
n
[
i=1
x
i
+ V
i
is also closed. By construction, this does not contain 0. So there is some
C >
0
such that if x ∈
S
n
i=1
x
i
+ V
i
, then kxk ≥ C. We claim that
C kxk
max
≤ kxk.
Indeed, take x =
P
a
i
x
i
∈ V . Let r be such that
a
r
 = max
i
(a
i
) = kxk
max
.
Then
kxk
−1
max
kxk =
a
−1
r
x
=
a
1
a
r
x
1
+ ··· +
a
r−1
a
r
x
r−1
+ x
r
+
a
r+1
a
r
x
r+1
+ ··· +
a
n
a
r
x
n
≥ C,
since the last vector is an element of x
r
+ V
r
.
Before we can prove our theorem, we note the following two easy lemmas:
Lemma.
Let
K
be a valued field. Then the valuation ring
O
K
is integrally
closed in K.
Proof.
Let
x ∈ K
and
x >
1. Suppose we have
a
n−1
, ··· , a
0
∈ O
K
. Then we
have
x
n
 > a
0
+ a
1
x + ···+ a
n−1
x
n−1
.
So we know
x
n
+ a
n−1
x
n−1
+ ··· + a
1
x + a
0
has nonzero norm, and in particular is nonzero. So
x
is not integral over
O
K
.
So O
K
is integrally closed.
Lemma.
Let
L
be a field and
 · 
a function that satisfies all axioms of an
absolute value but the strong triangle inequality. Then
 · 
is an absolute value
iff α ≤ 1 implies α + 1 ≤ 1.
Proof.
It is clear that if
 · 
is an absolute value, then
α ≤
1 implies
α
+ 1
 ≤
1.
Conversely, if this holds, and
x ≤ y
, then
x/y ≤
1. So
x/y
+ 1
 ≤
1. So
x + y ≤ y. So x + y ≤ max{x, y}.
Finally, we get to prove our theorem.
Theorem.
Let
K
be a complete valued field, and let
L/K
be a finite extension.
Then the absolute value on
K
has a unique extension to an absolute value on
L
,
given by
α
L
=
n
q
N
L/K
(α)
,
where
n
= [
L
:
K
] and
N
L/K
is the field norm. Moreover,
L
is complete with
respect to this absolute value.
Proof.
For uniqueness and completeness, if
·
L
is an absolute value on
L
, then
it is in particular a
K
norm on
L
as a finitedimensional vector space. So we
know L is complete with respect to ·
L
.
If
·
0
L
is another absolute value extending
·
, then we know
·
L
and
·
0
L
are equivalent in the sense of inducing the same topology. But then from one of
the early exercises, when field norms are equivalent, then we can find some
s >
0
such that
·
s
L
=
·
0
L
. But the two norms agree on
K
, and they are nontrivial.
So we must have s = 1. So the norms are equal.
To show existence, we have to prove that
α
L
=
n
q
N
L/K
(α)
is a norm.
(i) If α
L
= 0, then N
L/K
(α) = 0. This is true iff α = 0.
(ii)
The multiplicativity of
α
and follows from the multiplicativity of
N
L/K
,
· and
n
√
·.
To show the strong triangle inequality, it suffices to show that
α
L
≤
1 implies
α + 1
L
≤ 1.
Recall that
O
L
= {α ∈ L : α
L
≤ 1} = {α ∈ L : N
L/K
(α) ∈ O
K
}.
We claim that
O
L
is the integral closure of
O
K
in
L
. This implies what we
want, since the integral closure is closed under addition (and 1 is in the integral
closure).
Let
α ∈ O
L
. We may assume
α 6
= 0, since that case is trivial. Let the
minimal polynomial of α over K be
f(x) = a
0
+ a
1
x + ···+ a
n−1
x
n−1
+ x
n
∈ K[x].
We need to show that
a
i
∈ O
K
for all
i
. In other words,
a
i
 ≤
1 for all
i
. This
is easy for a
0
, since
N
L/K
(α) = ±a
m
0
,
and hence a
0
 ≤ 1.
By the corollary of Hensel’s lemma, for each i, we have
a
i
 ≤ max(a
0
, 1)
By general properties of the field norm, there is some
m ∈ Z
≥1
such that
N
L/K
(α) = ±a
m
0
. So we have
a
i
 ≤ max
N
L/K
(α)
1/m
, 1
= 1.
So f ∈ O
K
[x]. So α is integral over O
K
.
On the other hand, suppose
α
is integral over
O
K
. Let
¯
K/K
be an algebraic
closure of K. Note that
N
L/K
(α) =
Y
σ:L,→
¯
K
σ(α)
!
d
,
for some
d ∈ Z
≥1
, and each
σ
(
α
) is integral over
O
K
, since
α
is (apply
σ
to the
minimal polynomial). This implies that
N
L/K
(
α
) is integral over
O
K
(and lies
in K). So N
L/K
(α) ∈ O
K
since O
K
is integrally closed in K.
Corollary
(of the proof)
.
Let
K
be a complete valued field, and
L/K
a finite
extension. We equip
L
with
 · 
L
extending
 · 
on
K
. Then
O
L
is the integral
closure of O
K
in L.