Part III Local Fields
Based on lectures by H. C. Johansson
Notes taken by Dexter Chua
Michaelmas 2016
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
The
p
-adic numbers
Q
p
(where
p
is any prime) were invented by Hensel in the late 19th
century, with a view to introduce function-theoretic methods into number theory. They
are formed by completing
Q
with respect to the
p
-adic absolute value
| |
p
, defined
for non-zero
x Q
by
|x|
p
=
p
n
, where
x
=
p
n
a/b
with
a, b, n Z
and
a
and
b
are
coprime to
p
. The
p
-adic absolute value allows one to study congruences modulo all
powers of
p
simultaneously, using analytic methods. The concept of a local field is an
abstraction of the field
Q
p
, and the theory involves an interesting blend of algebra and
analysis. Local fields provide a natural tool to attack many number-theoretic problems,
and they are ubiquitous in modern algebraic number theory and arithmetic geometry.
Topics likely to be covered include:
The p-adic numbers. Local fields and their structure.
Finite extensions, Galois theory and basic ramification theory.
Polynomial equations; Hensel’s Lemma, Newton polygons.
Continuous functions on the p-adic integers, Mahler’s Theorem.
Local class field theory (time permitting).
Pre-requisites
Basic algebra, including Galois theory, and basic concepts from point set topology
and metric spaces. Some prior exposure to number fields might be useful, but is not
essential.
Contents
0 Introduction
1 Basic theory
1.1 Fields
1.2 Rings
1.3 Topological rings
1.4 The p-adic numbers
2 Valued fields
2.1 Hensel’s lemma
2.2 Extension of norms
2.3 Newton polygons
3 Discretely valued fields
3.1 Teichm¨uller lifts
3.2 Witt vectors*
4 Some p-adic analysis
5 Ramification theory for local fields
5.1 Ramification index and inertia degree
5.2 Unramified extensions
5.3 Totally ramified extensions
6 Further ramification theory
6.1 Some filtrations
6.2 Multiple extensions
7 Local class field theory
7.1 Infinite Galois theory
7.2 Unramified extensions and Weil group
7.3 Main theorems of local class field theory
8 Lubin–Tate theory
8.1 Motivating example
8.2 Formal groups
8.3 Lubin–Tate extensions
0 Introduction
What are local fields? Suppose we are interested in some basic number theoretic
problem. Say we have a polynomial
f
(
x
1
, ··· , x
n
)
Z
[
x
1
, ··· , x
n
]. We want to
look for solutions
a Z
n
, or show that there are no solutions at all. We might
try to view this polynomial as a real polynomial, look at its roots, and see if
they are integers. In lucky cases, we might be able to show that there are no
real solutions at all, and conclude that there cannot be any solutions at all.
On the other hand, we can try to look at it modulo some prime
p
. If there
are no solutions mod
p
, then there cannot be any solution. But sometimes
p
is
not enough. We might want to look at it mod
p
2
, or
p
3
, or . . . . One important
application of local fields is that we can package all these information together.
In this course, we are not going to study the number theoretic problems, but
just look at the properties of the local fields for their own sake.
Throughout this course, all rings will be commutative with unity, unless
otherwise specified.
1 Basic theory
We are going to start by making loads of definitions, which you may or may not
have seen before.
1.1 Fields
Definition
(Absolute value)
.
Let
K
be a field. An absolute value on
K
is a
function | · | : K R
0
such that
(i) |x| = 0 iff x = 0;
(ii) |xy| = |x||y| for all x, y K;
(iii) |x + y| |x| + |y|.
Definition (Valued field). A valued field is a field with an absolute value.
Example.
The rationals, reals and complex numbers with the usual absolute
values are absolute values.
Example
(Trivial absolute value)
.
The trivial absolute value on a field
K
is the
absolute value given by
|x| =
(
1 x 6= 0
0 x = 0
.
The only reason we mention the trivial absolute value here is that from
now on, we will assume that the absolute values are not trivial, because trivial
absolute values are boring and break things.
There are some familiar basic properties of the absolute value such as
Proposition. ||x| |y|| |x y|
. Here the outer absolute value on the left
hand side is the usual absolute value of
R
, while the others are the absolute
values of the relevant field.
An absolute value defines a metric d(x, y) = |x y| on K.
Definition
(Equivalence of absolute values)
.
Let
K
be a field, and let
| · |, | · |
0
be absolute values. We say they are equivalent if they induce the same topology.
Proposition.
Let
K
be a field, and
| · |, | · |
0
be absolute values on
K
. Then
the following are equivalent.
(i) | · | and | · |
0
are equivalent
(ii) |x| < 1 implies |x|
0
< 1 for all x K
(iii) There is some s R
>0
such that |x|
s
= |x|
0
for all x K.
Proof.
(i)
(ii) and (iii)
(i) are easy exercises. Assume (ii), and we shall
prove (iii). First observe that since
|x
1
|
=
|x|
1
, we know
|x| >
1 implies
|x|
0
>
1, and hence
|x|
= 1 implies
|x|
0
= 1. To show (iii), we have to show that
the ratio
log |x|
log |x
0
|
is independent of x.
Suppose not. We may assume
log |x|
log |x|
0
<
log |y|
log |y|
0
,
and moreover the logarithms are positive. Then there are
m, n Z
>0
such that
log |x|
log |y|
<
m
n
<
log |x|
0
log |y|
0
.
Then rearranging implies
x
n
y
m
< 1 <
x
n
y
m
0
,
a contradiction.
Exercise.
Let
K
be a valued field. Then equivalent absolute values induce the
same the completion
ˆ
K
of
K
, and
ˆ
K
is a valued field with an absolute value
extending | · |.
In this course, we are not going to be interested in the usual absolute values.
Instead, we are going to consider some really weird ones, namely non-archimedean
ones.
Definition
(Non-archimedean absolute value)
.
An absolute value
| · |
on a field
K
is called non-archimedean if
|x
+
y| max
(
|x|, |y|
). This condition is called
the strong triangle inequality.
An absolute value which isn’t non-archimedean is called archimedean.
Metrics satisfying
d
(
x, z
)
max
(
d
(
x, y
)
, d
(
y, z
)) are often known as ultra-
metrics.
Example. Q, R and C under the usual absolute values are archimedean.
In this course, we will only consider non-archimedean absolute values. Thus,
from now on, unless otherwise mentioned, an absolute value is assumed to be
non-archimedean. The metric is weird!
We start by proving some absurd properties of non-archimedean absolute
values.
Recall that the closed balls are defined by
B(x, r) = {y : |x y| r}.
Proposition.
Let (
K, | · |
) be a non-archimedean valued field, and let
x K
and r R
>0
. Let z B(x, r). Then
B(x, r) = B(z, r).
So closed balls do not have unique “centers”. Every point can be viewed as
the center.
Proof. Let y B(z, r). Then
|x y| = |(x z) + (z y)| max(|x z|, |z y|) r.
So y B(x, r). By symmetry, y B(x, r) implies y B(z, r).
Corollary. Closed balls are open.
Proof. To show that B(x, r) is open, we let z B(x, r). Then we have
{y : |y z| < r} B(z, r) = B(x, r).
So we know the open ball of radius
r
around
z
is contained in
B
(
x, r
). So
B
(
x, r
)
is open.
Norms in non-archimedean valued fields are easy to compute:
Proposition.
Let
K
be a non-archimedean valued field, and
x, y K
. If
|x| > |y|, then |x + y| = |x|.
More generally, if
x
=
P
c=0
x
i
and the non-zero
|x
i
|
are distinct, then
|x| = max |x
i
|.
Proof.
On the one hand, we have
|x
+
y| max{|x|, |y|}
. On the other hand,
we have
|x| = |(x + y) y| max(|x + y|, |y|) = |x + y|,
since we know that we cannot have
|x| |y|
. So we must have
|x|
=
|x
+
y|
.
Convergence is also easy for valued fields.
Proposition. Let K be a valued field.
(i) Let (x
n
) be a sequence in K. If x
n
x
n+1
0, then x
n
is Cauchy.
If we assume further that K is complete, then
(ii)
Let (
x
n
) be a sequence in
K
. If
x
n
x
n+1
0, then a sequence (
x
n
) in
K converges.
(iii) Let
P
n=0
y
n
be a series in K. If y
n
0, then
P
n=0
y
n
converges.
The converses to all these are of course also true, with the usual proofs.
Proof.
(i)
Pick
ε >
0 and
N
such that
|x
n
x
n+1
| < ε
for all
n N
. Then given
m n N , we have
|x
m
x
n
| = |x
m
x
m1
+ x
m1
x
m2
+ ··· x
n
|
max(|x
m
x
m1
|, ··· , |x
n+1
x
n
|)
< ε.
So the sequence is Cauchy.
(ii) Follows from (1) and the definition of completeness.
(iii) Follows from the definition of convergence of a series and (2).
The reason why we care about these weird non-archimedean fields is that
they have very rich algebraic structure. In particular, there is this notion of the
valuation ring.
Definition
(Valuation ring)
.
Let
K
be a valued field. Then the valuation ring
of K is the open subring
O
K
= {x : |x| 1}.
We prove that it is actually a ring
Proposition. Let K be a valued field. Then
O
K
= {x : |x| 1}
is an open subring of
K
. Moreover, for each
r
(0
,
1], the subsets
{x
:
|x| < r}
and {x : |x| r} are open ideals of O
K
. Moreover, O
×
K
= {x : |x| = 1}.
Note that this is very false for usual absolute values. For example, if we take
R with the usual absolute value, we have 1 O
R
, but 1 + 1 6∈ O
R
.
Proof. We know that these sets are open since all balls are open.
To see
O
K
is a subring, we have
|
1
|
=
|
1
|
= 1. So 1
,
1
O
K
. If
x, y O
K
,
then
|x
+
y| max
(
|x|, |y|
)
1. So
x
+
y O
K
. Also,
|xy|
=
|x||y|
1
·
1 = 1.
So xy O
K
.
That the other sets are ideals of O
K
is checked in the same way.
To check the units, we have
x O
×
K
|x|, |x
1
|
1
|x|
=
|x|
1
= 1.
1.2 Rings
Definition
(Integral element)
.
Let
R S
be rings and
s S
. We say
s
is
integral over R if there is some monic f R[x] such that f(s) = 0.
Example. Any r R is integral (take f(x) = x r).
Example.
Take
Z C
. Then
z C
is integral over
Z
if it is an algebraic
integer (by definition of algebraic integer). For example,
2
is an algebraic
integer, but
1
2
is not.
We would like to prove the following characterization of integral elements:
Theorem.
Let
R S
be rings. Then
s
1
, ··· , s
n
S
are all integral iff
R[s
1
, ··· , s
n
] S is a finitely-generated R-module.
Note that
R
[
s
1
, ··· , s
n
] is by definition a finitely-generated
R
-algebra, but
requiring it to be finitely-generated as a module is stronger.
Here one direction is easy. It is not hard to show that if
s
1
, ··· , s
n
are all
integral, then
R
[
s
1
, ··· , s
n
] is finitely-generated. However to show the other
direction, we need to find some clever trick to produce a monic polynomial that
kills the s
i
.
The trick we need is the adjugate matrix we know and love from IA Vectors
and Matrices.
Definition
(Adjoint/Adjugate matrix)
.
Let
A
= (
a
ij
) be an
n × n
matrix with
coefficients in a ring
R
. The adjugate matrix or adjoint matrix
A
= (
a
ij
) of
A
is defined by
a
ij
= (1)
i+j
det(A
ij
),
where
A
ij
is an (
n
1)
×
(
n
1) matrix obtained from
A
by deleting the
i
th
column and the jth row.
As we know from IA, the following property holds for the adjugate matrix:
Proposition.
For any
A
, we have
A
A
=
AA
=
det
(
A
)
I
, where
I
is the
identity matrix.
With this, we can prove our claim:
Proof of theorem. Note that we can construct R[s
1
, ··· , s
n
] by a sequence
R R[s
1
] R[s
1
, s
2
] ··· R[s
1
, ··· , s
n
] S,
and each
s
i
is integral over
R
[
s
1
, ··· , s
n1
]. Since the finite extension of a finite
extension is still finite, it suffices to prove it for the case
n
= 1, and we write
s
for s
1
.
Suppose
f
(
x
)
R
[
x
] is monic such that
f
(
s
) = 0. If
g
(
x
)
R
[
x
], then there
is some
q, r R
[
x
] such that
g
(
x
) =
f
(
x
)
q
(
x
) +
r
(
x
) with
deg r < deg f
. Then
g
(
s
) =
r
(
s
). So any polynomial expression in
s
can be written as a polynomial
expression with degree less than
deg f
. So
R
[
s
] is generated by 1
, s, ··· , s
deg f 1
.
In the other direction, let
t
1
, ··· , t
d
be
R
-module generators of
R
[
s
1
, ··· , s
n
].
We show that in fact any element of
R
[
s
1
, ··· , s
n
] is integral over
R
. Consider
any element b R[s
1
, ··· , s
n
]. Then there is some a
ij
R such that
bt
i
=
d
X
j=1
a
ij
t
j
.
In matrix form, this says
(bI A)t = 0.
We now multiply by (bI A)
to obtain
det(bI A)t
j
= 0
for all j. Now we know 1 R. So 1 =
P
c
j
t
j
for some c
j
R. Then we have
det(bI A) = det(bI A)
X
c
j
t
j
=
X
c
j
(det(bI A)t
j
) = 0.
Since det(bI A) is a monic polynomial in b, it follows that b is integral.
Using this characterization, the following result is obvious:
Corollary.
Let
R S
be rings. If
s
1
, s
2
S
are integral over
R
, then
s
1
+
s
2
and
s
1
s
2
are integral over
R
. In particular, the set
˜
R S
of all elements in
S
integral over R is a ring, known as the integral closure of R in S.
Proof.
If
s
1
, s
2
are integral, then
R
[
s
1
, s
2
] is a finite extension over
R
. Since
s
1
+ s
2
and s
1
s
2
are elements of R[s
1
, s
2
], they are also integral over R.
Definition
(Integrally closed)
.
Given a ring extension
R S
, we say
R
is
integrally closed in S if
˜
R = R.
1.3 Topological rings
Recall that we previously constructed the valuation ring
O
K
. Since the valued
field
K
itself has a topology, the valuation ring inherits a subspace topology.
This is in fact a ring topology.
Definition
(Topological ring)
.
Let
R
be a ring. A topology on
R
is called a
ring topology if addition and multiplication are continuous maps
R × R R
. A
ring with a ring topology is a topological ring.
Example. R
and
C
with the usual topologies and usual ring structures are
topological rings.
Exercise.
Let
K
be a valued field. Then
K
is a topological ring. We can see
this from the fact that the product topology on
K ×K
is induced by the metric
d((x
0
, y
0
), (x
1
, y
1
)) = max(|x
0
x
1
|, |y
0
y
1
|).
Now if we are just randomly given a ring, there is a general way of constructing
a ring topology. The idea is that we pick an ideal
I
and declare its elements to
be small. For example, in a valued ring, we can pick
I
=
{x O
K
:
|x| <
1
}
.
Now if you are not only in
I
, but
I
2
, then you are even smaller. So we have a
hierarchy of small sets
I I
2
I
3
I
4
···
Now to make this a topology on
R
, we say that a subset
U R
is open if every
x U
is contained in some translation of
I
n
(for some
n
). In other words, we
need some y R such that
x y + I
n
U.
But since
I
n
is additively closed, this is equivalent to saying
x
+
I
n
U
. So we
make the following definition:
Definition
(
I
-adically open)
.
Let
R
be a ring and
I R
an ideal. A subset
U R
is called
I
-adically open if for all
x U
, there is some
n
1 such that
x + I
n
U.
Proposition.
The set of all
I
-adically open sets form a topology on
R
, called
the I-adic topology.
Note that the
I
-adic topology isn’t really the kind of topology we are used
to thinking about, just like the topology on a valued field is also very weird.
Instead, it is a “filter” for telling us how small things are.
Proof.
By definition, we have
and
R
are open, and arbitrary unions are clearly
open. If
U, V
are
I
-adically open, and
x U V
, then there are
n, m
such that
x + I
n
U and x + I
m
V . Then x + I
max(m,n)
U V .
Exercise. Check that the I-adic topology is a ring topology.
In the special case where
I
=
xR
, we often call the
I
-adic topology the
x
-adic
topology.
Now we want to tackle the notion of completeness. We will consider the case
of I = xR for motivation, but the actual definition will be completely general.
If we pick the
x
-adic topology, then we are essentially declaring that we take
x to be small. So intuitively, we would expect power series like
a
0
+ a
1
x + a
2
x
2
+ a
3
x
3
+ ···
to “converge”, at least if the
a
i
are “of bounded size”. In general, the
a
i
are
“not too big” if
a
i
x
i
is genuinely a member of
x
i
R
, as opposed to some silly thing
like x
i
.
As in the case of analysis, we would like to think of these infinite series as a
sequence of partial sums
(a
0
, a
0
+ a
1
x, a
0
+ a
1
x + a
2
x
2
, ···)
Now if we denote the limit as
L
, then we can think of this sequence alternatively
as
(L mod I, L mod I
2
, L mod I
3
, ···).
The key property of this sequence is that if we take
L mod I
k
and reduce it mod
I
k1
, then we obtain L mod I
k1
.
In general, suppose we have a sequence
(b
n
R/I
n
)
n=1
.
such that
b
n
mod I
n1
=
b
n1
. Then we want to say that the ring is
I
-adically
complete if every sequence satisfying this property is actually of the form
(L mod I, L mod I
2
, L mod I
3
, ···)
for some
L
. Alternatively, we can take the
I
-adic completion to be the collection
of all such sequences, and then a space is
I
-adically complete it is isomorphic to
its I-adic completion.
To do this, we need to build up some technical machinery. The kind of
sequences we’ve just mentioned is a special case of an inverse limit.
Definition
(Inverse/projective limit)
.
Let
R
1
, R
2
, , ···
be topological rings, with
continuous homomorphisms f
n
: R
n+1
R
n
.
R
1
R
2
R
3
R
4
···
f
1
f
2
f
3
The inverse limit or projective limit of the R
i
is the ring
lim
R
n
=
(
(x
n
)
Y
n
R
n
: f
n
(x
n+1
) = x
n
)
,
with coordinate-wise addition and multiplication, together with the subspace
topology coming from the product topology of
Q
R
n
. This topology is known as
the inverse limit topology.
Proposition. The inverse limit topology is a ring topology.
Proof sketch. We can fit the addition and multiplication maps into diagrams
lim
R
n
× lim
R
n
lim
R
n
Q
R
n
×
Q
R
n
Q
R
n
By the definition of the subspace topology, it suffices to show that the cor-
responding maps on
Q
R
n
are continuous. By the universal property of the
product, it suffices to show that the projects
Q
R
n
×
Q
R
n
R
m
is continuous
for all
m
. But this map can alternatively be obtained by first projecting to
R
m
,
then doing multiplication in
R
m
, and projection is continuous. So the result
follows.
It is easy to see the following universal property of the inverse limit topology:
Proposition.
Giving a continuous ring homomorphism
g
:
S lim
R
n
is the
same as giving a continuous ring homomorphism
g
n
:
S R
n
for each
n
, such
that each of the following diagram commutes:
S R
n
R
n1
g
n
g
n1
f
n1
Definition
(
I
-adic completion)
.
Let
R
be a ring and
I
be an ideal. The
I
-adic
completion of R is the topological ring
lim
R/I
n
,
where
R/I
n
has the discrete topology, and
R/I
n+1
R/I
n
is the quotient map.
There is an evident map
ν : R lim
R/I
n
r 7→ (r mod I
n
)
.
This map is a continuous ring homomorphism if
R
is given the
I
-adic topology.
Definition
(
I
-adically complete)
.
We say that
R
is
I
-adically complete if
ν
is a
bijection.
Exercise. If ν is a bijection, then ν is in fact a homeomorphism.
1.4 The p-adic numbers
For the rest of this course, p is going to be a prime number.
We consider a particular case of valued fields, namely the
p
-adic numbers,
and study some of its basic properties.
Let
x Q
be non-zero. Then by uniqueness of factorization, we can write
x
uniquely as
x = p
n
a
b
,
where a, b, n Z, b > 0 and a, b, p are pairwise coprime.
Definition
(
p
-adic absolute value)
.
The
p
-adic absolute value on
Q
is the
function | · |
p
: Q R
0
given by
|x|
p
=
(
0 x = 0
p
n
x = p
n
a
b
as above
.
Proposition. The p-adic absolute value is an absolute value.
Proof. It is clear that |x|
p
= 0 iff x = 0.
Suppose we have
x = p
n
a
b
, y = p
m
c
d
.
We wlog m n. Then we have
|xy|
p
=
p
n+m
ac
bd
= p
mn
= |x|
p
|y|
p
.
So this is multiplicative. Finally, we have
|x + y|
p
=
p
n
ab + p
mn
cb
bd
p
n
= max(|x|
p
, |y|
p
).
Note that we must have
bd
coprime to
p
, but
ab
+
p
mn
cb
need not be. However,
any extra powers of
p
could only decrease the absolute value, hence the above
result.
Note that if x Z is an integer, then |x|
p
= p
n
iff p
n
|| x (we say p
n
|| x if
p
n
| x and p
n+1
- x).
Definition
(
p
-adic numbers)
.
The
p
-adic numbers
Q
p
is the completion of
Q
with respect to | · |
p
.
Definition (p-adic integers). The valuation ring
Z
p
= {x Q
p
: |x|
p
1}
is the p-adic integers.
Proposition. Z
p
is the closure of Z inside Q
p
.
Proof. If x Z is non-zero, then x = p
n
a with n 0. So |x|
p
1. So Z Z
p
.
We now want to show that Z is dense in Z
p
. We know the set
Z
(p)
= {x Q : |x|
p
1}
is dense inside
Z
p
, essentially by definition. So it suffices to show that
Z
is dense
in Z
(p)
. We let x Z
(p)
\ {0}, say
x = p
n
a
b
, n 0.
It suffices to find x
i
Z such that x
i
1
b
. Then we have p
n
ax
i
x.
Since (
b, p
) = 1, we can find
x
i
, y
i
Z
such that
bx
i
+
p
i
y
i
= 1 for all
i
1.
So
x
i
1
b
p
=
1
b
p
|bx
i
1|
p
= |p
i
y
i
|
p
p
i
0.
So done.
Proposition. The non-zero ideals of Z
p
are p
n
Z
p
for n 0. Moreover,
Z
p
n
Z
=
Z
p
p
n
Z
p
.
Proof.
Let 0
6
=
I Z
p
be an ideal, and pick
x I
such that
|x|
p
is maximal.
This supremum exists and is attained because the possible values of the absolute
values are discrete and bounded above. If
y I
, then by maximality, we have
|y|
p
|x|
p
. So we have
|yx
1
|
p
1. So
yx
1
Z
p
, and this implies that
y
= (
yx
1
)
x xZ
p
. So
I xZ
p
, and we obviously have
xZ
p
I
. So we have
I = xZ
p
.
Now if
x
=
p
n
a
b
, then since
a
b
is invertible in
Z
p
, we have
xZ
p
=
p
n
Z
p
. So
I = p
n
Z
p
.
To show the second part, consider the map
f
n
: Z
Z
p
p
n
Z
p
given by the inclusion map followed by quotienting. Now
p
n
Z
p
=
{x
:
|x|
p
p
n
.
So we have
ker f
n
= {x Z : |x|
p
p
n
} = p
n
Z.
Now since
Z
is dense in
Z
p
, we know the image of
f
n
is dense in
Z
p
/p
n
Z
p
.
But
Z
p
/p
n
Z
p
has the discrete topology. So
f
n
is surjective. So
f
n
induces an
isomorphism Z/p
n
Z
=
Z
p
/p
n
Z
p
.
Corollary. Z
p
is a PID with a unique prime element p (up to units).
This is pretty much the point of the
p
-adic numbers there are a lot of
primes in Z, and by passing on to Z
p
, we are left with just one of them.
Proposition.
The topology on
Z
induced by
| · |
p
is the
p
-adic topology (i.e.
the pZ-adic topology).
Proof.
Let
U Z
. By definition,
U
is open wrt
| · |
p
iff for all
x U
, there is
an n N such that
{y Z : |y x|
p
p
n
} U.
On the other hand,
U
is open in the
p
-adic topology iff for all
x U
, there is
some n 0 such that x + p
n
Z U. But we have
{y Z : |y x|
p
p
n
} = x + p
n
Z.
So done.
Proposition. Z
p
is p-adically complete and is (isomorphic to) the p-adic com-
pletion of Z.
Proof. The second part follows from the first as follows: we have the maps
Z
p
lim
Z
p
/(p
n
Z
p
) lim Z/(p
n
Z)
ν
(f
n
)
n
We know the map induced by (
f
n
)
n
is an isomorphism. So we just have to show
that ν is an isomorphism
To prove the first part, we have
x ker ν
iff
x p
n
Z
p
for all
n
iff
|x|
p
p
n
for all n iff |x|
p
= 0 iff x = 0. So the map is injective.
To show surjectivity, we let
z
n
lim
Z
p
/p
n
Z
p
.
We define a
i
{0, 1, ··· , p 1} recursively such that
x
n
=
n1
X
i=0
a
i
p
i
is the unique representative of
z
n
in the set of integers
{
0
,
1
, ··· , p
n
1
}
. Then
x =
X
i=0
a
i
p
i
exists in
Z
p
and maps to
x x
n
z
n
(
mod p
n
) for all
n
0. So
ν
(
x
) = (
z
n
).
So the map is surjective. So ν is bijective.
Corollary. Every a Z
p
has a unique expansion
a =
X
i=0
a
i
p
i
.
with a
i
{0, ··· , p 1}.
More generally, for any a Q
×
, there is a unique expansion
a =
X
i=n
a
i
p
i
for a
i
{0, ··· , p 1}, a
n
6= 0 and
n = log
p
|a|
p
Z.
Proof.
The second part follows from the first part by multiplying
a
by
p
n
.
Example. We have
1
1 p
= 1 + p + p
2
+ p
3
+ ··· .
2 Valued fields
2.1 Hensel’s lemma
We return to the discussion of general valued fields. We are now going to introduce
an alternative to the absolute value that contains the same information, but is
presented differently.
Definition
(Valuation)
.
Let
K
be a field. A valuation on
K
is a function
v : K R {∞} such that
(i) v(x) = 0 iff x = 0
(ii) v(xy) = v(x) + v(y)
(iii) v(x + y) min{v(x), v(y)}.
Here we use the conventions that r + = and r for all r .
In some sense, this definition is sort-of pointless, since if
v
is a valuation,
then the function
|x| = c
v(x)
for any
c >
1 is a (non-archimedean) absolute value. Conversely, if
| · |
is a
valuation, then
v(x) = log
c
|x|
is a valuation.
Despite this, sometimes people prefer to talk about the valuation rather than
the absolute value, and often this is more natural. As we will later see, in certain
cases, there is a canonical normalization of
v
, but there is no canonical choice
for the absolute value.
Example. For x Q
p
, we define
v
p
(x) = log
p
|x|
p
.
This is a valuation, and if x Z
p
, then v
p
(x) = n iff p
n
|| x.
Example. Let K be a field, and define
k((T )) =
(
X
i=n
a
i
T
i
: a
i
k, n Z
)
.
This is the field of formal Laurent series over k. We define
v
X
a
i
T
i
= min{i : a
i
6= 0}.
Then v is a valuation of k((T)).
Recall that for a valued field K, the valuation ring is given by
O
K
= {x K : |x| 1} = {x K : v(x) 0}.
Since this is a subring of a field, and the absolute value is multiplicative, we
notice that the units in
O
are exactly the elements of absolute value 1. The
remaining elements form an ideal (since the field is non-archimedean), and thus
we have a maximal ideal
m = m
K
= {x K : |x| < 1}
The quotient
k = k
K
= O
K
/m
K
is known as the residue field.
Example. Let K = Q
p
. Then O = Z
p
, and m = pZ
p
. So
k = O/m = Z
p
/pZ
p
=
Z/pZ.
Definition
(Primitive polynomial)
.
If
K
is a valued field and
f
(
x
) =
a
0
+
a
1
x
+
··· + a
n
x
n
K[x] is a polynomial, we say that f is primitive if
max
i
|a
i
| = 1.
In particular, we have f O[x].
The point of a primitive polynomial is that such a polynomial is naturally,
and non-trivially, an element of
k
[
x
]. Moreover, focusing on such polynomials is
not that much of a restriction, since any polynomial is a constant multiple of a
primitive polynomial.
Theorem
(Hensel’s lemma)
.
Let
K
be a complete valued field, and let
f K
[
x
]
be primitive. Put
¯
f = f mod m k[x]. If there is a factorization
¯
f(x) = ¯g(x)
¯
h(x)
with (¯g,
¯
h) = 1, then there is a factorization
f(x) = g(x)h(x)
in O[x] with
¯g = g,
¯
h = h mod m,
with deg g = deg ¯g.
Note that requiring
deg g
=
deg ¯g
is the best we can hope for we cannot
guarantee deg h = deg
¯
h, since we need not have deg f = deg
¯
f.
This is one of the most important results in the course.
Proof.
Let
g
0
, h
0
be arbitrary lifts of
¯g
and
¯
h
to
O
[
x
] with
deg ¯g
=
g
0
and
deg
¯
h = h
0
. Then we have
f = g
0
h
0
mod m.
The idea is to construct a “Taylor expansion” of the desired
g
and
h
term by
term, starting from
g
0
and
h
0
, and using completeness to guarantee convergence.
To proceed, we use our assumption that
¯g,
¯
h
are coprime to find some
a, b O
[
x
]
such that
ag
0
+ bh
0
1 mod m. ()
It is easier to work modulo some element
π
instead of modulo the ideal
m
, since
we are used to doing Taylor expansion that way. Fortunately, since the equations
above involve only finitely many coefficients, we can pick an
π m
with absolute
value large enough (i.e. close enough to 1) such that the above equations hold
with m replaced with π. Thus, we can write
f = g
0
h
0
+ πr
0
, r
0
O[x].
Plugging in (), we get
f = g
0
h
0
+ πr
0
(ag
0
+ bh
0
) + π
2
(something).
If we are lucky enough that
deg r
0
b < deg g
0
, then we group as we learnt in
secondary school to get
f = (g
0
+ πr
0
b)(h
0
+ πr
0
a) + π
2
(something).
We can then set
g
1
= g
0
+ πr
0
b
h
1
= h
0
+ πr
0
a,
and then we can write
f = g
1
h
1
+ π
2
r
1
, r
1
O[x], deg g
1
= deg ¯g. ()
If it is not true that deg r
0
b deg g
0
, we use the division algorithm to write
r
0
b = qg
0
+ p.
Then we have
f = g
0
h
0
+ π((r
0
a + q)g
0
+ ph
0
),
and then proceed as above.
Given the factorization (
), we replace
r
1
by
r
1
(
ag
0
+
bh
0
), and then repeat
the procedure to get a factorization
f g
2
h
2
mod π
3
, deg g
2
= deg ¯g.
Inductively, we constrict g
k
, h
k
such that
f g
k
h
k
mod π
k+1
g
k
g
k1
mod π
k
h
k
h
k1
mod π
k
deg g
k
= deg ¯g
Note that we may drop the terms of
h
k
whose coefficient are in
π
k+1
O
, and the
above equations still hold. Moreover, we can then bound
deg h
k
deg f deg g
k
.
It now remains to set
g = lim
k→∞
g
k
, h = lim
k→∞
h
k
.
Corollary.
Let
f
(
x
) =
a
0
+
a
1
x
+
···
+
a
n
x
n
K
[
x
] where
K
is complete and
a
0
, a
n
6= 0. If f is irreducible, then
|a
| max(|a
0
|, |a
n
|)
for all .
Proof.
By scaling, we can wlog
f
is primitive. We then have to prove that
max
(
|a
0
|, |a
n
|
) = 1. If not, let
r
be minimal such that
|a
r
|
= 1. Then 0
< r < n
.
Moreover, we can write
f(x) x
r
(a
r
+ a
r+1
x + ··· + a
n
x
nr
) mod m.
But then Hensel’s lemma says this lifts to a factorization of
f
, a contradiction.
Corollary (of Hensel’s lemma). Let f O[x] be monic, and K complete. If f
mod m
has a simple root
¯α k
, then
f
has a (unique) simple root
α O
lifting
¯α.
Example.
Consider
x
p1
1
Z
p
[
x
]. We know
x
p1
splits into distinct linear
factors over
F
p
[
x
]. So all roots lift to
Z
p
. So
x
p1
1 splits completely in
Z
p
.
So Z
p
contains all p roots of unity.
Example. Since 2 is a quadratic residue mod 7, we know
2 Q
7
.
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 = |λ|kxk for all λ K and x V .
(iii) kx + yk max{kxk, kyk}.
Note that our norms are also non-Archimedean.
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 K-vector 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 finite-dimensional
K-vector 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 finite-dimensional
K-vector 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
i1
Kx
i+1
··· Kx
n
= span{x
1
, ··· , x
i1
, 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
r1
a
r
x
r1
+ 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
n1
, ··· , a
0
O
K
. Then we
have
|x
n
| > |a
0
+ a
1
x + ··· + a
n1
x
n1
|.
So we know
x
n
+ a
n1
x
n1
+ ··· + a
1
x + a
0
has non-zero norm, and in particular is non-zero. 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 finite-dimensional 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 non-trivial.
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
n1
x
n1
+ 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