III Local Fields (Full)

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

-adic numbers

(where

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

with respect to the

-adic absolute value

| − |

, defined

for non-zero

x ∈ Q

|x|

−n

, where

a/b

with

a, b, n ∈ Z

and

are

coprime to

. The

-adic absolute value allows one to study congruences modulo all

powers of

simultaneously, using analytic methods. The concept of a local field is an

abstraction of the field

, 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

(

, ··· , x

)

∈ Z

[

, ··· , x

]. We want to

look for solutions

a ∈ Z

, 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

. If there

are no solutions mod

, then there cannot be any solution. But sometimes

not enough. We might want to look at it mod

, or

, 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

be a field. An absolute value on

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

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

, 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

be a field, and let

| · |, | · |

be absolute values. We say they are equivalent if they induce the same topology.

Proposition.

Let

be a field, and

| · |, | · |

be absolute values on

. Then

the following are equivalent.

(i) | · | and | · |

are equivalent

(ii) |x| < 1 implies |x|

< 1 for all x ∈ K

(iii) There is some s ∈ R

such that |x|

= |x|

for all x ∈ K.

Proof.

(i)

⇒

(ii) and (iii)

⇒

(i) are easy exercises. Assume (ii), and we shall

prove (iii). First observe that since

−1

|x|

−1

, we know

|x| >

1 implies

|x|

1, and hence

|x|

= 1 implies

|x|

= 1. To show (iii), we have to show that

the ratio

log |x|

log |x

is independent of x.

Suppose not. We may assume

log |x|

log |y|

and moreover the logarithms are positive. Then there are

m, n ∈ Z

such that

log |x|

log |y|

log |x|

log |y|

Then rearranging implies



< 1 <



a contradiction.

Exercise.

Let

be a valued field. Then equivalent absolute values induce the

same the completion

, and

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

is called non-archimedean if

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

(

x, z

)

≤ max

(

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

. 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

around

is contained in

(

x, r

). So

(

x, r

)

is open.

Norms in non-archimedean valued fields are easy to compute:

Proposition.

Let

be a non-archimedean valued field, and

x, y ∈ K

. If

|x| > |y|, then |x + y| = |x|.

More generally, if

∞

c=0

and the non-zero

are distinct, then

|x| = max |x

Proof.

On the one hand, we have

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|

Convergence is also easy for valued fields.

Proposition. Let K be a valued field.

(i) Let (x

) be a sequence in K. If x

− x

n+1

→ 0, then x

is Cauchy.

If we assume further that K is complete, then

(ii)

Let (

) be a sequence in

. If

− x

n+1

→

0, then a sequence (

) in

K converges.

(iii) Let

∞

n=0

be a series in K. If y

→ 0, then

∞

n=0

converges.

The converses to all these are of course also true, with the usual proofs.

Proof.

(i)

Pick

ε >

0 and

such that

− x

n+1

| < ε

for all

n ≥ N

. Then given

m ≥ n ≥ N , we have

− x

| = |x

− x

m−1

+ x

m−1

− x

m−2

+ ··· −x

≤ max(|x

− x

m−1

|, ··· , |x

n+1

− x

< 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

be a valued field. Then the valuation ring

of K is the open subring

= {x : |x| ≤ 1}.

We prove that it is actually a ring

Proposition. Let K be a valued field. Then

= {x : |x| ≤ 1}

is an open subring of

. Moreover, for each

r ∈

1], the subsets

|x| < r}

and {x : |x| ≤ r} are open ideals of O

. Moreover, O

= {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

, but 1 + 1 6∈ O

Proof. We know that these sets are open since all balls are open.

To see

is a subring, we have

|−

= 1. So 1

, −

∈ O

. If

x, y ∈ O

then

y| ≤ max

(

|x|, |y|

)

≤

1. So

y ∈ O

. Also,

|xy|

|x||y| ≤

1 = 1.

So xy ∈ O

That the other sets are ideals of O

is checked in the same way.

To check the units, we have

x ∈ O

⇔ |x|, |x

−1

| ≤

⇔ |x|

|x|

−1

= 1.

1.2 Rings

Definition

(Integral element)

Let

R ⊆ S

be rings and

s ∈ S

. We say

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

if it is an algebraic

integer (by definition of algebraic integer). For example,

√

is an algebraic

integer, but

√

is not.

We would like to prove the following characterization of integral elements:

Theorem.

Let

R ⊆ S

be rings. Then

, ··· , s

∈ S

are all integral iff

R[s

, ··· , s

] ⊆ S is a finitely-generated R-module.

Note that

[

, ··· , s

] is by definition a finitely-generated

-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

are all

integral, then

[

, ··· , s

] 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

The trick we need is the adjugate matrix we know and love from IA Vectors

and Matrices.

Definition

(Adjoint/Adjugate matrix)

Let

= (

) be an

n × n

matrix with

coefficients in a ring

. The adjugate matrix or adjoint matrix

∗

= (

∗

) of

is defined by

∗

= (−1)

i+j

det(A

where

is an (

n −

(

n −

1) matrix obtained from

by deleting the

column and the jth row.

As we know from IA, the following property holds for the adjugate matrix:

Proposition.

For any

, we have

∗

det

(

)

, where

is the

identity matrix.

With this, we can prove our claim:

Proof of theorem. Note that we can construct R[s

, ··· , s

] by a sequence

R ⊆ R[s

] ⊆ R[s

, s

] ⊆ ··· ⊆ R[s

, ··· , s

] ⊆ S,

and each

is integral over

[

, ··· , s

n−1

]. Since the finite extension of a finite

extension is still finite, it suffices to prove it for the case

= 1, and we write

for s

Suppose

(

)

∈ R

[

] is monic such that

(

) = 0. If

(

)

∈ R

[

], then there

is some

q, r ∈ R

[

] such that

(

) =

(

)

(

) +

(

) with

deg r < deg f

. Then

(

) =

(

). So any polynomial expression in

can be written as a polynomial

expression with degree less than

deg f

. So

[

] is generated by 1

, s, ··· , s

deg f −1

In the other direction, let

, ··· , t

-module generators of

[

, ··· , s

We show that in fact any element of

[

, ··· , s

] is integral over

. Consider

any element b ∈ R[s

, ··· , s

]. Then there is some a

∈ R such that

j=1

In matrix form, this says

(bI − A)t = 0.

We now multiply by (bI − A)

∗

to obtain

det(bI − A)t

= 0

for all j. Now we know 1 ∈ R. So 1 =

for some c

∈ R. Then we have

det(bI − A) = det(bI − A)

(det(bI − A)t

) = 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

∈ S

are integral over

, then

and

are integral over

. In particular, the set

R ⊆ S

of all elements in

integral over R is a ring, known as the integral closure of R in S.

Proof.

, s

are integral, then

[

, s

] is a finite extension over

. Since

+ s

and s

are elements of R[s

, s

], they are also integral over R.

Definition

(Integrally closed)

Given a ring extension

R ⊆ S

, we say

integrally closed in S if

R = R.

1.3 Topological rings

Recall that we previously constructed the valuation ring

. Since the valued

field

itself has a topology, the valuation ring inherits a subspace topology.

This is in fact a ring topology.

Definition

(Topological ring)

Let

be a ring. A topology on

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

with the usual topologies and usual ring structures are

topological rings.

Exercise.

Let

be a valued field. Then

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

, y

), (x

, y

)) = max(|x

− x

|, |y

− y

|).

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

and declare its elements to

be small. For example, in a valued ring, we can pick

{x ∈ O

|x| <

}

Now if you are not only in

, but

, then you are even smaller. So we have a

hierarchy of small sets

I ⊇ I

⊇ I

⊇ ···

Now to make this a topology on

, we say that a subset

U ⊆ R

is open if every

x ∈ U

is contained in some translation of

(for some

). In other words, we

need some y ∈ R such that

x ∈ y + I

⊆ U.

But since

is additively closed, this is equivalent to saying

⊆ U

. So we

make the following definition:

Definition

(

-adically open)

Let

be a ring and

I ⊆ R

an ideal. A subset

U ⊆ R

is called

-adically open if for all

x ∈ U

, there is some

n ≥

1 such that

x + I

⊆ U.

Proposition.

The set of all

-adically open sets form a topology on

, called

the I-adic topology.

Note that the

-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

are open, and arbitrary unions are clearly

open. If

U, V

are

-adically open, and

x ∈ U ∩ V

, then there are

n, m

such that

x + I

⊆ U and x + I

⊆ 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

, we often call the

-adic topology the

-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

-adic topology, then we are essentially declaring that we take

x to be small. So intuitively, we would expect power series like

+ a

x + a

+ a

+ ···

to “converge”, at least if the

are “of bounded size”. In general, the

are

“not too big” if

is genuinely a member of

, 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

+ a

x, a

+ a

x + a

, ···)

Now if we denote the limit as

, then we can think of this sequence alternatively

(L mod I, L mod I

, L mod I

, ···).

The key property of this sequence is that if we take

L mod I

and reduce it mod

k−1

, then we obtain L mod I

k−1

In general, suppose we have a sequence

∈ R/I

)

∞

n=1

such that

mod I

n−1

. Then we want to say that the ring is

-adically

complete if every sequence satisfying this property is actually of the form

(L mod I, L mod I

, L mod I

, ···)

for some

. Alternatively, we can take the

-adic completion to be the collection

of all such sequences, and then a space is

-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

, , ···

be topological rings, with

continuous homomorphisms f

: R

n+1

→ R

···

The inverse limit or projective limit of the R

is the ring

lim

←−

(

) ∈

: f

n+1

) = x

)

with coordinate-wise addition and multiplication, together with the subspace

topology coming from the product topology of

. 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

←−

× lim

←−

lim

←−

By the definition of the subspace topology, it suffices to show that the cor-

responding maps on

are continuous. By the universal property of the

product, it suffices to show that the projects

→ R

is continuous

for all

. But this map can alternatively be obtained by first projecting to

then doing multiplication in

, 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

S → lim

←−

is the

same as giving a continuous ring homomorphism

S → R

for each

, such

that each of the following diagram commutes:

S R

n−1

Definition

(

-adic completion)

Let

be a ring and

be an ideal. The

-adic

completion of R is the topological ring

lim

←−

R/I

where

R/I

has the discrete topology, and

R/I

n+1

→ R/I

is the quotient map.

There is an evident map

ν : R → lim

←−

R/I

r 7→ (r mod I

)

This map is a continuous ring homomorphism if

is given the

-adic topology.

Definition

(

-adically complete)

We say that

-adically complete if

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

-adic numbers,

and study some of its basic properties.

Let

x ∈ Q

be non-zero. Then by uniqueness of factorization, we can write

uniquely as

x = p

where a, b, n ∈ Z, b > 0 and a, b, p are pairwise coprime.

Definition

(

-adic absolute value)

The

-adic absolute value on

is the

function | · |

: Q → R

≥0

given by

|x|

(

0 x = 0

−n

x = p

as above

Proposition. The p-adic absolute value is an absolute value.

Proof. It is clear that |x|

= 0 iff x = 0.

Suppose we have

x = p

, y = p

We wlog m ≥ n. Then we have

|xy|



n+m



= p

−m−n

= |x|

|y|

So this is multiplicative. Finally, we have

|x + y|



ab + p

m−n



≤ p

−n

= max(|x|

, |y|

Note that we must have

coprime to

, but

m−n

need not be. However,

any extra powers of

could only decrease the absolute value, hence the above

result.

Note that if x ∈ Z is an integer, then |x|

= p

−n

iff p

|| x (we say p

|| x if

| x and p

n+1

- x).

Definition

(

-adic numbers)

The

-adic numbers

is the completion of

with respect to | · |

Definition (p-adic integers). The valuation ring

= {x ∈ Q

: |x|

≤ 1}

is the p-adic integers.

Proposition. Z

is the closure of Z inside Q

Proof. If x ∈ Z is non-zero, then x = p

a with n ≥ 0. So |x|

≤ 1. So Z ⊆ Z

We now want to show that Z is dense in Z

. We know the set

(p)

= {x ∈ Q : |x|

≤ 1}

is dense inside

, essentially by definition. So it suffices to show that

is dense

in Z

(p)

. We let x ∈ Z

(p)

\ {0}, say

x = p

, n ≥ 0.

It suffices to find x

∈ Z such that x

→

. Then we have p

→ x.

Since (

b, p

) = 1, we can find

, y

∈ Z

such that

= 1 for all

i ≥



−



|bx

− 1|

= |p

≤ p

−i

→ 0.

So done.

Proposition. The non-zero ideals of Z

are p

for n ≥ 0. Moreover,

∼

Proof.

Let 0

I ⊆ Z

be an ideal, and pick

x ∈ I

such that

|x|

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|

≤ |x|

. So we have

|yx

−1

≤

1. So

−1

∈ Z

, and this implies that

= (

−1

)

x ∈ xZ

. So

I ⊆ xZ

, and we obviously have

⊆ I

. So we have

I = xZ

Now if

, then since

is invertible in

, we have

. So

I = p

To show the second part, consider the map

: Z →

given by the inclusion map followed by quotienting. Now

|x|

≤ p

−n

So we have

ker f

= {x ∈ Z : |x|

≤ p

−n

} = p

Now since

is dense in

, we know the image of

is dense in

But

has the discrete topology. So

is surjective. So

induces an

isomorphism Z/p

∼

Corollary. Z

is a PID with a unique prime element p (up to units).

This is pretty much the point of the

-adic numbers — there are a lot of

primes in Z, and by passing on to Z

, we are left with just one of them.

Proposition.

The topology on

induced by

| · |

is the

-adic topology (i.e.

the pZ-adic topology).

Proof.

Let

U ⊆ Z

. By definition,

is open wrt

| · |

iff for all

x ∈ U

, there is

an n ∈ N such that

{y ∈ Z : |y − x|

≤ p

−n

} ⊆ U.

On the other hand,

is open in the

-adic topology iff for all

x ∈ U

, there is

some n ≥ 0 such that x + p

Z ⊆ U. But we have

{y ∈ Z : |y − x|

≤ p

−n

} = x + p

So done.

Proposition. Z

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

lim

←−

/(p

) lim Z/(p

)

We know the map induced by (

)

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

for all

iff

|x|

≤ p

−n

for all n iff |x|

= 0 iff x = 0. So the map is injective.

To show surjectivity, we let

∈ lim

←−

We define a

∈ {0, 1, ··· , p −1} recursively such that

n−1

i=0

is the unique representative of

in the set of integers

{

, ··· , p

−

}

. Then

x =

∞

i=0

exists in

and maps to

x ≡ x

≡ z

(

mod p

) for all

n ≥

0. So

(

) = (

So the map is surjective. So ν is bijective.

Corollary. Every a ∈ Z

has a unique expansion

a =

∞

i=0

with a

∈ {0, ··· , p − 1}.

More generally, for any a ∈ Q

, there is a unique expansion

a =

∞

i=n

for a

∈ {0, ··· , p − 1}, a

6= 0 and

n = −log

|a|

∈ Z.

Proof.

The second part follows from the first part by multiplying

−n

Example. We have

1 − p

= 1 + p + p

+ p

+ ··· .

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

be a field. A valuation on

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

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

|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

, but there is no canonical choice

for the absolute value.

Example. For x ∈ Q

, we define

(x) = −log

|x|

This is a valuation, and if x ∈ Z

, then v

(x) = n iff p

|| x.

Example. Let K be a field, and define

k((T )) =

(

∞

i=n

: a

∈ k, n ∈ Z

)

This is the field of formal Laurent series over k. We define





= min{i : a

6= 0}.

Then v is a valuation of k((T)).

Recall that for a valued field K, the valuation ring is given by

= {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

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

= {x ∈ K : |x| < 1}

The quotient

k = k

= O

is known as the residue field.

Example. Let K = Q

. Then O = Z

, and m = pZ

. So

k = O/m = Z

/pZ

∼

Z/pZ.

Definition

(Primitive polynomial)

is a valued field and

(

) =

··· + a

∈ K[x] is a polynomial, we say that f is primitive if

max

| = 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

[

]. 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

be a complete valued field, and let

f ∈ K

[

]

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

This is one of the most important results in the course.

Proof.

Let

, h

be arbitrary lifts of

¯g

and

[

] with

deg ¯g

and

deg

h = h

. Then we have

f = g

mod m.

The idea is to construct a “Taylor expansion” of the desired

and

term by

term, starting from

and

, and using completeness to guarantee convergence.

To proceed, we use our assumption that

¯g,

are coprime to find some

a, b ∈ O

[

]

such that

+ bh

≡ 1 mod m. (†)

It is easier to work modulo some element

instead of modulo the ideal

, 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

+ πr

, r

∈ O[x].

Plugging in (†), we get

f = g

+ πr

(ag

+ bh

) + π

(something).

If we are lucky enough that

deg r

b < deg g

, then we group as we learnt in

secondary school to get

f = (g

+ πr

b)(h

+ πr

a) + π

(something).

We can then set

= g

+ πr

= h

+ πr

and then we can write

f = g

+ π

, r

∈ O[x], deg g

= deg ¯g. (∗)

If it is not true that deg r

b ≤ deg g

, we use the division algorithm to write

b = qg

+ p.

Then we have

f = g

+ π((r

a + q)g

+ ph

and then proceed as above.

Given the factorization (

∗

), we replace

(

), and then repeat

the procedure to get a factorization

f ≡ g

mod π

, deg g

= deg ¯g.

Inductively, we constrict g

, h

such that

f ≡ g

mod π

k+1

≡ g

k−1

mod π

≡ h

k−1

mod π

deg g

= deg ¯g

Note that we may drop the terms of

whose coefficient are in

k+1

, and the

above equations still hold. Moreover, we can then bound

deg h

≤ deg f −deg g

It now remains to set

g = lim

k→∞

, h = lim

k→∞

Corollary.

Let

(

) =

···

∈ K

[

] where

is complete and

, a

6= 0. If f is irreducible, then



| ≤ max(|a

|, |a

for all .

Proof.

By scaling, we can wlog

is primitive. We then have to prove that

max

(

|, |a

) = 1. If not, let

be minimal such that

= 1. Then 0

< r < n

Moreover, we can write

f(x) ≡ x

+ a

r+1

x + ··· + a

n−r

) mod m.

But then Hensel’s lemma says this lifts to a factorization of

, 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

has a (unique) simple root

α ∈ O

lifting

¯α.

Example.

Consider

p−1

−

∈ Z

[

]. We know

p−1

splits into distinct linear

factors over

[

]. So all roots lift to

. So

p−1

−

1 splits completely in

So Z

contains all p roots of unity.

Example. Since 2 is a quadratic residue mod 7, we know

√

2 ∈ Q

2.2 Extension of norms

The main goal of this section is to prove the following theorem:

Theorem.

Let

be a complete valued field, and let

L/K

be a finite extension.

Then the absolute value on

has a unique extension to an absolute value on

given by

|α|

L/K

(α)|,

where

= [

] and

L/K

is the field norm. Moreover,

is complete with

respect to this absolute value.

Corollary.

Let

be complete and

M/K

be an algebraic extension of

. 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

be a complete valued field and

L/K

a finite extension. If

σ ∈ Aut(L/K), then |σ(α)|

= |α|

Proof.

We check that

α 7→ |σ

(

)

is also an absolute value on

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

. Instead, we will look at norms of

as a

-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

-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

be a valued field and

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

be norms on

. Then

two norms are equivalent if they induce the same topology on

, i.e. there are

C, D > 0 such that

C kxk ≤ kxk

≤ D kxk

for all x ∈ V .

One of the most convenient norms we will work with is the max norm:

Example

(Max norm)

Let

be a complete valued field, and

a finite-

dimensional K-vector space. Let x

, ··· , x

be a basis of V . Then if

x =

then

kxk

max

= max

defines a norm on V .

Proposition.

Let

be a complete valued field, and

a finite-dimensional

K-vector space. Then V is complete under the max norm.

Proof.

Given a Cauchy sequence in

under the max norm, take the limit of each

coordinate to get the limit of the sequence, using the fact that

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

be a complete valued field, and

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

(kx

k). Then we have

kxk =



≤ max (|a

|kx

k) ≤ (max |a

|)D = kxk

max

We find

by induction on

. If

= 1, then

kxk

|kxk

kxk

max

k. So C = kx

k works.

For n ≥ 2, we let

= Kx

⊕ ··· ⊕Kx

i−1

⊕ Kx

i+1

⊕ ··· ⊕Kx

= span{x

, ··· , x

i−1

, x

i+1

, ··· , x

By the induction hypothesis, each

is complete with respect to (the restriction

of) k·k. So in particular V

is closed in V . So we know that the union

[

i=1

+ V

is also closed. By construction, this does not contain 0. So there is some

C >

such that if x ∈

i=1

+ V

, then kxk ≥ C. We claim that

C kxk

max

≤ kxk.

Indeed, take x =

∈ V . Let r be such that

| = max

(|a

|) = kxk

max

Then

kxk

−1

max

kxk =



−1



+ ··· +

r−1

+ x

r+1

+ ··· +



≥ C,

since the last vector is an element of x

+ V

Before we can prove our theorem, we note the following two easy lemmas:

Lemma.

Let

be a valued field. Then the valuation ring

is integrally

closed in K.

Proof.

Let

x ∈ K

and

|x| >

1. Suppose we have

n−1

, ··· , a

∈ O

. Then we

have

| > |a

+ a

x + ··· + a

n−1

So we know

+ a

n−1

+ ··· + a

x + a

has non-zero norm, and in particular is non-zero. So

is not integral over

So O

is integrally closed.

Lemma.

Let

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

| ≤

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

be a complete valued field, and let

L/K

be a finite extension.

Then the absolute value on

has a unique extension to an absolute value on

given by

|α|



L/K

(α)



where

= [

] and

L/K

is the field norm. Moreover,

is complete with

respect to this absolute value.

Proof.

For uniqueness and completeness, if

|·|

is an absolute value on

, then

it is in particular a

-norm on

as a finite-dimensional vector space. So we

know L is complete with respect to |·|

|·|

is another absolute value extending

|·|

, then we know

|·|

and

|·|

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 >

such that

|·|

. But the two norms agree on

, 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/K

(α)



is a norm.

(i) If |α|

= 0, then N

L/K

(α) = 0. This is true iff α = 0.

(ii)

The multiplicativity of

|α|

and follows from the multiplicativity of

L/K

|·| and

√

·.

To show the strong triangle inequality, it suffices to show that

|α|

≤

1 implies

|α + 1|

≤ 1.

Recall that

= {α ∈ L : |α|

≤ 1} = {α ∈ L : N

L/K

(α) ∈ O

We claim that

is the integral closure of

. This implies what we

want, since the integral closure is closed under addition (and 1 is in the integral

closure).

Let

α ∈ O

. We may assume

α 6

= 0, since that case is trivial. Let the

minimal polynomial of α over K be

f(x) = a

+ a

x + ··· + a

n−1

+ x

∈ K[x].

We need to show that

∈ O

for all

. In other words,

| ≤

1 for all

. This

is easy for a

, since

L/K

(α) = ±a

and hence |a

| ≤ 1.

By the corollary of Hensel’s lemma, for each i, we have

| ≤ max(|a

|, 1)

By general properties of the field norm, there is some

m ∈ Z

≥1

such that

L/K

(α) = ±a

. So we have

| ≤ max





L/K

(α)

1/m



, 1



= 1.

So f ∈ O

[x]. So α is integral over O

On the other hand, suppose

is integral over

. Let

K/K

be an algebraic

closure of K. Note that

L/K

(α) =

σ:L→

σ(α)

for some

d ∈ Z

≥1

, and each

(

) is integral over

, since

is (apply

to the

minimal polynomial). This implies that

L/K

(

) is integral over

(and lies

in K). So N

L/K

(α) ∈ O

since O

is integrally closed in K.

Corollary

(of the proof)

Let

be a complete valued field, and

L/K

a finite

extension. We equip

with

| · |

extending

| · |

. Then

is the integral

closure of O

in L.

2.3 Newton polygons

We are going to have a small digression to Newton polygons. We will not make

use of them in this course, but it is a cute visual devices that tell us about roots

of polynomials. It is very annoying to write down a formal definition, so we first

look at some examples. We will work with valuations rather than the absolute

value.

Example. Consider the valued field (Q

, v

), and the polynomial

+ p

− p

+ pt + p

We then plot the coefficients for each power of

, and then draw a “convex

polygon” so that all points lie on or above it:

power of t

valuation of coefficient

1 2 3 40

Example. Consider (Q

, v

) with the polynomial

+ 5t

t +

Here there is no t

term, so we simply don’t draw anything.

power of t

valuation of coefficient

1 2 3 40

−1

We now go to come up with a formal definition.

Definition (Lower convex set). We say a set S ⊆ R

is lower convex if

(i) Whenever (x, y) ∈ S, then (x, z) ∈ S for all z ≥ y.

(ii) S is convex.

Definition

(Lower convex hull)

Given any set of points

T ⊆ R

, there is a

minimal lower convex set

S ⊇ T

(by the intersection of all lower convex sets

containing

– this is a non-empty definition because

satisfies the property).

This is known as the lower convex hull of the points.

Example.

The lower convex hull of the points (0

0) is

given by the region denoted below:

Definition

(Newton polygon)

Let

(

) =

···

∈ K

[

], where

(

K, v

) is a valued field. Then the Newton polygon of

is the lower convex hull

of {(i, v(a

)) : i = 0, ··· , n, a

6= 0}.

This is the formal definition, so in our first example, the Newton polygon

really should be the shaded area shown above, but most of the time, we only

care about the lower line.

Definition

(Break points)

Given a polynomial, the points (

i, v

(

)) lying on

the boundary of the Newton polygon are known as the break points.

Definition

(Line segment)

Given a polynomial, the line segment between two

adjacent break points is a line segment.

Definition

(Multiplicity/length)

The length or multiplicity of a line segment

is the horizontal length.

Definition (Slope). The slope of a line segment is its slope.

Example. Consider again (Q

, v

) with the polynomial

+ 5t

t +

power of t

valuation of coefficient

The middle segment has length 2 and slope 1/2.

Example. In the following Newton polygon:

The second line segment has length 3 and slope −

It turns out the Newton polygon tells us something about the roots of the

polynomial.

Theorem. Let K be complete valued field, and v the valuation on K. We let

f(x) = a

+ a

x + ··· + a

∈ K[x].

Let

be the splitting field of

over

, equipped with the unique extension

of v.

If (

r, v

(

))

→

(

s, v

(

)) is a line segment of the Newton polygon of

with

slope −m ∈ R, then f has precisely s − r roots of valuation m.

Note that by lower convexity, there can be at most one line segment for each

slope. So this theorem makes sense.

Proof.

Dividing by

only shifts the polygon vertically, so we may wlog

= 1.

We number the roots of f such that

w(α

) = ··· = w(α

) = m

w(α

) = ··· = w(α

) = m

w(α

) = ··· = w(α

) = m

t+1

where we have

< m

< ··· < m

t+1

Then we know

v(a

) = v(1) = 0

v(a

n−1

) = w





≥ min

w(α

) = m

v(a

n−2

) = w





≥ min

i6=j

w(α

) = 2m

v(a

n−s

) = w





6=...6=i

...α





= min w(α

···α

) = s

It is important that in the last one, we have equality, not an inequality, because

there is one term in the sum whose valuation is less than all the others.

We can then continue to get

v(α

n−s

−1

) ≥ min w(α

···α

) = s

+ m

until we reach

v(α

n−s

−s

) = s

+ (s

− s

We keep going on.

We draw the Newton polygon.

(n, 0)

(n − s

, s

)

(n − s

− s

, s

+ (s

− s

)

···

We don’t know where exactly the other points are, but the inequalities imply

that the (i, v(a

)) are above the lines drawn. So this is the Newton polygon.

Counting from the right, the first line segment has length

n −

(

n − s

) =

and slope

0 − s

n − (n − s

)

= −m

In general, the

th segment has length (

n − s

k−1

)

−

(

n − s

) =

− s

k−1

, and

slope



k−2

i=1

i+1

− s

i+1



−



k−1

i=1

i+1

− s

i+1



− s

k−1

−(s

− s

k−1

− s

k−1

= −m

and the others follow similarly.

Corollary.

is irreducible, then the Newton polygon has a single line segment.

Proof.

We need to show that all roots have the same valuation. Let

α, β

be in

the splitting field

. Then there is some

σ ∈ Aut

(

L/K

) such that

(

) =

Then w(α) = w(σ(α)) = β. So done.

Note that Eisenstein’s criterion is a (partial) converse to this!

3 Discretely valued fields

We are now going to further specialize. While a valued field already has some

nice properties, we can’t really say much more about them without knowing

much about their valuations.

Recall our previous two examples of valued fields:

and

((

)). The

valuations had the special property that they take values in

. Such fields are

known as discretely valued fields.

Definition (Discretely valued field). Let K be a valued field with valuation v.

We say

is a discretely valued field (DVF) if

(

)

⊆ R

is a discrete subgroup

of R, i.e. v(k

) is infinite cyclic.

Note that we do not require the image to be exactly

Z ⊆ R

. So we allow

scaled versions of the valuation. This is useful because the property of mapping

into

is not preserved under field extensions in general, as we will later see. We

will call those that do land in Z normalized valuations.

Definition

(Normalized valuation)

Let

be a DVF. The normalized valuation

is the unique valuation on

in the given equivalence class of valuations

whose image is Z.

Note that the normalized valuation does not give us a preferred choice of

absolute value, since to obtain an absolute value, we still have to arbitrarily pick

the base c > 1 to define |x| = c

−v(x)

Definition

(Uniformizer)

Let

be a discrete valued field. We say

π ∈ K

uniformizer if

(

)

0 and

(

) generates

(

) (iff

(

) has minimal positive

valuation).

So with a normalized valuation, we have v

(π) = 1.

Example.

The usual valuation on

is normalized, and so is the usual valuation

on k((T )). p is a uniformizer for Q

and T is a uniformizer for k((T )).

The kinds of fields we will be interested are local fields. The definition we

have here might seem rather ad hoc. This is just one of the many equivalent

characterizations of a local field, and the one we pick here is the easiest to state.

Definition

(Local field)

A local field is a complete discretely valued field with

a finite residue field.

Example. Q

and

with

are both discretely valued fields, and

is a local

field. p is a uniformizer.

Example. The Laurent series field k((T)) with valuation





= inf{n : a

6= 0}

is a discrete valued field, and is a local field if and only if

is finite field, as the

residue field is exactly k. We have

k((T ))

= k[[T ]] =

(

∞

n=0

: a

∈ k

)

Here T is a uniformizer.

These discretely valued field are pretty much like the p-adic numbers.

Proposition.

Let

be a discretely valued field with uniformizer

. Let

S ⊆ O

be a set of coset representatives of O

= k

containing 0. Then

(i) The non-zero ideals of O

are π

for n ≥ 0.

(ii)

The ring

is a PID with unique prime

(up to units), and

πO

(iii)

The topology on

induced by the absolute value is the

-adic topology.

(iv) If K is complete, then O

is π-adically complete.

(v) If K is complete, then any x ∈ K can be written uniquely as

x =

∞

n−∞

where a

∈ S, and

|x| = |π|

−inf{n:a

6=0}

(vi)

The completion

is also discretely valued and

is a uniformizer, and

moreover the natural map

∼

is an isomorphism.

Proof. The same as for Q

and Z

, with π instead of p.

Proposition.

Let

be a discretely valued field. Then

is a local field iff

is compact.

Proof.

is compact, then

−n

is compact for all

n ≥

0 (where

is the

uniformizer), and in particular complete. So

K =

∞

[

n≥0

−n

is complete, as this is an increasing union, and Cauchy sequences are bounded.

Also, we know the quotient map

→ k

is continuous when

is given the

discrete topology, by definition of the

-adic topology. So

is compact and

discrete, hence finite.

In the other direction, if

is local, then we know

/π

is finite for

all

n ≥

0 (by induction and finiteness of

). We let (

) be a sequence in

Then by finiteness of

/πO

, there is a subsequence (

1,i

) which is constant

modulo

. We keep going, choosing a subsequence (

n+1,i

) of (

) such that

(

n+1,i

) is constant modulo

n+1

. Then (

i,i

)

∞

i=1

converges, since it is Cauchy as

− x

| ≤ |π|

for j ≤ i. So O

is sequentially compact, hence compact.

Now the valuation ring

inherits a valuation from

, and this gives it a

structure of a discrete valuation ring. We will define a discrete valuation ring in

a funny way, but there are many equivalent definitions that we will not list.

Definition

(Discrete valuation ring)

A ring

is called a discrete valuation

ring (DVR) if it is a PID with a unique prime element up to units.

Proposition. R is a DVR iff R

∼

for some DVF K.

Proof.

We have already seen that valuation rings of discrete valuation fields are

DVRs. In the other direction, let

be a DVR, and

a prime. Let

x ∈ R \ {

}

Then we can find a unique unit

u ∈ R

and

n ∈ Z

≥0

such that

(say,

by unique factorization of PIDs). We define

v(x) =

(

n x 6= 0

∞ x = 0

This is then a discrete valuation of

. This extends uniquely to the field of

fractions K. It remains to show that R = O

. First note that

K = R





This is since any non-zero element in





looks like

u, u ∈ R

, n ∈ Z

, and

is already invertible. So it must be the field of fractions. Then we have

v(π

u) = n ∈ Z

≥0

⇐⇒ π

u ∈ R.

So we have R = O

Now recall our two “standard” examples of valued fields —

((

)) and

. Both of their residue fields are

, and in particular has characteristic

However,

((

)) itself is also of characteristic

, while

has characteristic 0.

It would thus be helpful to split these into two different cases:

Definition

(Equal and mixed characteristic)

Let

be a valued field with

residue field k

. Then K has equal characteristic if

char K = char k

Otherwise, we have K has mixed characteristic.

has mixed characteristic, then necessarily

char K

= 0, and

char k

Example. Q

has mixed characteristic, since

char Q

= 0 but

char k

Z/pZ = p.

We will also need the following definition:

Definition

(Perfect ring)

Let

be a ring of characteristic

. We say

perfect if the Frobenius map

x 7→ x

is an automorphism of

, i.e. every element

of R has a pth root.

Fact.

Let

be a field of characteristic

. Then

is perfect if and only if every

finite extension of F is separable.

Example. F

is perfect for every q = p

3.1 Teichm¨uller lifts

Take our favorite discretely valued ring

. This is

-adically complete, so we

can write each element as

x = a

+ a

p + a

+ ··· ,

where each

is in

{

, ··· , p −

}

. The reason this works is that 0

, ··· , p −

1 are coset representatives of the ring

/pZ

∼

Z/pZ

. While these coset

representatives might feel like a “natural” thing to do in this context, this is

because we have implicitly identified with

/pZ

∼

Z/pZ

as a particular subset

Z ⊆ Z

. However, this identification respects effectively no algebraic structure

at all. For example, we cannot multiplying the cosets simply by multiplying the

representatives as elements of

, because, say, (

p −

−

+ 1, which is

not 1. So this is actually quite bad, at least theoretically.

It turns out that we can actually construct “natural” lifts in a very general

scenario.

Theorem.

Let

be a ring, and let

x ∈ R

. Assume that

-adically

complete and that

R/xR

is perfect of characteristic

. Then there is a unique

map [−] : R/xR → R such that

[a] ≡ a mod x

and

[ab] = [a][b].

for all

a, b ∈ R/xR

. Moreover, if

has characteristic

, then [

−

] is a ring

homomorphism.

Definition

(Teichm¨uller map)

The map [

−

] :

R/xR → R

is called the Te-

ichm uller map. [x] is called the Teichm¨uller lift or representative of x.

The idea of the proof is as follows: suppose we have an

a ∈ R/xR

. If we

randomly picked a lift

, then chances are it would be a pretty “bad” choice,

since any two such choices can differ by a multiple of x.

Suppose we instead lifted a

th root of

, and then take the

th power

of it. We claim that this is a better way of picking a lift. Suppose we have picked

two lifts of a

−1

, say, α

and α

. Then α

= xc + α

for some c. So we have

(α

)

− α

= α

+ pxc + O(x

) − α

= pxc + O(x

where we abuse notation and write

(

) to mean terms that are multiples of

We now recall that

R/xR

has characteristic

, so

p ∈ xR

. Thus in fact

pxc = O(x

). So we have

(α

)

− α

= O(x

So while the lift is still arbitrary, any two arbitrary choices can differ by at most

. Alternatively, our lift is now a well-defined element of R/x

We can, of course, do better. We can lift the p

th root of a to R, then take

the

th power of it. Now any two lifts can differ by at most

(

). More

generally, we can try to lift the

th root of

, then take the

th power of

it. We keep picking a higher and higher

, take the limit, and hopefully get

something useful out!

To prove this result, we will need the following messy lemma:

Lemma.

Let

be a ring with

x ∈ R

such that

R/xR

has characteristic

. Let

α, β ∈ R be such that

α = β mod x

(†)

Then we have

= β

mod x

k+1

Proof.

It is left as an exercise to modify the proof to work for

= 2 (it is actually

easier). So suppose p is odd. We take the pth power of (†) to obtain

− β

p−1

i=1





p−i

∈ x

p(k+1)

We can now write

p−1

i=1

(−1)





p−i

p−1

i=1

(−1)





(αβ)



p−2i

− β

p−2i



= p(α − β)(something).

Now since

R/xR

has characteristic

, we know

p ∈ xR

. By assumption, we know

α − β ∈ x

k+1

R. So this whole mess is in x

k+2

R, and we are done.

Proof of theorem.

Let

a ∈ R/xR

. For each

, there is a unique

−n

∈ R/xR

We lift this arbitrarily to some α

∈ R such that

≡ a

−n

mod x.

We define

= α

The claim is that

[a] = lim

n→∞

exists and is independent of the choices.

Note that if the limit exists no matter how we choose the

, then it

must be independent of the choices. Indeed, if we had choices

and

then

, β

, ···

is also a respectable choice of lifts, and thus must

converge. So β

and β

must have the same limit.

Since the ring is

-adically complete and is discretely valued, to show the

limit exists, it suffices to show that β

n+1

− β

→ 0 x-adically. Indeed, we have

n+1

− β

= (α

n+1

)

− α

We now notice that

n+1

≡ (a

−n−1

)

= a

−n

≡ α

mod x.

So by applying the previous the lemma many times, we obtain

(α

n+1

)

≡ α

mod x

n+1

So β

n+1

− β

∈ x

n+1

R. So lim β

exists.

To see [a] = a mod x, we just have to note that

lim

n→∞

≡ lim

n→∞

−n

)

= lim a = a mod x.

(here we are using the fact that the map

R → R/xR

is continuous when

given the x-adic topology and R/xR is given the discrete topology)

The remaining properties then follow trivially from the uniqueness of the

above limit.

For multiplicativity, if we have another element

b ∈ R/xR

, with

∈ R

lifting b

−n

for all n, then α

lifts (ab)

−n

. So

[ab] = lim α

= lim α

lim γ

= [a][b].

If R has characteristic p, then α

+ γ

lifts a

−n

+ b

−n

= (a + b)

−n

. So

[a + b] = lim(α

+ γ

)

= lim α

+ lim γ

= [a] + [b].

Since 1 is a lift of 1 and 0 is a lift of 0, it follows that this is a ring homomorphism.

Finally, to show uniqueness, suppose

R/xR → R

is a map with these

properties. Then we note that

(

−n

)

≡ a

−n

mod x

, and is thus a valid choice

of α

. So we have

[a] = lim

n→∞

φ(a

−n

)

= lim φ(a) = φ(a).

Example. Let R = Z

and x = p. Then [−] : F

→ Z

satisfies

[x]

p−1

= [x

p−1

] = [1] = 1.

So the image of [

] must be the unique

p −

1th root of unity lifting

(recall we

proved their existence via Hensel’s lemma).

When proving theorems about these rings, the Teichm¨uller lifts would be

very handy and natural things to use. However, when we want to do actual

computations, there is absolutely no reason why these would be easier!

As an application, we can prove the following characterization of equal

characteristic complete DVF’s.

Theorem.

Let

be a complete discretely valued field of equal characteristic

and assume that k

is perfect. Then K

∼

((T )).

Proof.

Let

be a complete DVF. Since every DVF the field of fractions of

its valuation ring, it suffices to prove that

∼

[[

]]. We know

has

characteristic

. So [

−

] :

→ O

is an injective ring homomorphism. We

choose a uniformizer π ∈ O

, and define

[[T ]] → O

∞

n=0

7→

∞

n=0

]π

Then this is a ring homomorphism since [

−

] is. The bijectivity follows from

property (v) in our list of properties of complete DVF’s.

Corollary.

Let

be a local field of equal characteristic

. Then

∼

for

some q a power of p, and K

∼

((T )).

3.2 Witt vectors*

We are now going to look at the mixed characteristic analogue of this result. We

want something that allows us to go from characteristic

to characteristic 0.

This is known as Witt vectors, which is non-examinable.

We start with the notion of a strict

-ring. Roughly this is a ring that satisfies

all the good properties whose name has the word “p” in it.

Definition

(Strict

-ring)

Let

be a ring. A is called a strict

-ring if it is

p-torsion free, p-adically complete, and A/pA is a perfect ring.

Note that a strict

-ring in particular satisfies the conditions for the Te-

ichm¨uller lift to exist, for x = p.

Example. Z

is a strict p-ring.

The next example we are going to construct is more complicated. This is in

some sense a generalization of the usual polynomial rings

[

, ··· , x

], or more

generally,

Z[x

| i ∈ I],

for

possibly infinite. To construct the “free” strict

-ring, after adding all these

variables

, to make it a strict

-ring, we also need to add their

th roots, and

the p

th roots etc, and then take the p-adic completion, and hope for the best.

Example. Let X = {x

: i ∈ I} be a set. Let

B = Z[x

−∞

| i ∈ I] =

∞

[

n=0

Z[x

−n

| i ∈ I].

Here the union on the right is taken by treating

Z[x

| i ∈ I] ⊆ Z[x

−1

| i ∈ I] ⊆ ···

in the natural way.

We let

be the

-adic completion of

. We claim that

is a strict

-ring

and A/pA

∼

−∞

| i ∈ I].

Indeed, we see that

-torsion free. By Exercise 13 on Sheet 1, we know

A is p-adically complete and torsion free. Moreover,

A/pA

∼

B/pB

∼

−∞

| i ∈ I],

which is perfect since every element has a p-th root.

If A is a strict p-ring, then we know that we have a Teichm¨uller map

[−] : A/pA → A,

Lemma.

Let

be a strict

-ring. Then any element of

can be written

uniquely as

a =

∞

n=0

for a unique a

∈ A/pA.

Proof. We recursively construct the a

= a (mod p)

≡ p

−1

(a − [a

]) (mod p)

Lemma.

Let

and

be strict

-rings and let

A/pA → B/pB

be a ring

homomorphism. Then there is a unique homomorphism

A → B

such that

f = F mod p, given by





[f(a

)]p

Proof sketch.

We define

by the given formula and check that it works. First of

all, by the formula,

-adically continuous, and the key thing is to check that

it is additive (which is slightly messy). Multiplicativity then follows formally

from the continuity and additivity.

To show uniqueness, suppose that we have some

lifting

. Then

(

) =

So ψ is p-adically continuous. So it suffices to show that ψ([a]) = [ψ(a)].

We take α

∈ A lifting a

−n

∈ A/pA. Then ψ(α

) lifts f(a)

−n

. So

ψ([a]) = lim ψ(α

−n

) = lim ψ(α

)

−n

= [f(a)].

So done.

There is a generalization of this result:

Proposition.

Let

be a strict

-ring and

be a ring with an element

such that

-adically complete and

B/xB

is perfect of characteristic

. If

A/pA → B/xB

is a ring homomorphism. Then there exists a unique ring

homomorphism

A → B

with

F mod x

, i.e. the following diagram

commutes:

A B

A/pA B/xB

Indeed, the conditions on

are sufficient for Teichm¨uller lifts to exist, and

we can at least write down the previous formula, then painfully check it works.

We can now state the main theorem about strict p-rings.

Theorem.

Let

be a perfect ring. Then there is a unique (up to isomorphism)

strict p-ring W (B) called the Witt vectors of R such that W (R)/pW (R)

∼

Moreover, for any other perfect ring

, the reduction mod

map gives a

bijection

Hom

Ring

(W (R), W (R

)) Hom

Ring

(R, R

)

∼

Proof sketch.

(

) and

(

) are such strict

-rings, then the second part

follows from the previous lemma. Indeed, if

is a strict

-ring with

C/pC

∼

(

)

/pW

(

), then the isomorphism

¯α

(

)

/pW

(

)

→ C/pC

and its

inverse

¯α

−1

have unique lifts

(

)

→ C

and

−1

C → W

(

), and these

are inverses by uniqueness of lifts.

To show existence, let R be a perfect ring. We form

−∞

| r ∈ R] → R

7→ r

Then we know that the

-adic completion of

[

−∞

| r ∈ R

], written

, is a

strict p-ring with

A/pA

∼

−∞

| r ∈ R].

We write

I = ker(F

−∞

| r ∈ R] → R).

Then define

J =

(

∞

n=0

∈ A : a

∈ I for all n

)

This turns out to be an ideal.

J A R

0 I A/pA R 0

We put

(

) =

A/J

. We can then painfully check that this has all the required

properties. For example, if

x =

∞

n=0

∈ A,

and

px =

∞

n=0

n+1

∈ J,

then by definition of

, we know [

]

∈ I

. So

x ∈ J

. So

(

)

-torsion

free. By a similar calculation, one checks that

∞

n=0

W (R) = {0}.

This implies that

(

) injects to its

-adic completion. Using that

-adically

complete, one checks the surjectivity by hand.

Also, we have

W (R)

pW (R)

∼

J + pA

But we know

J + pA =

(

| a

∈ I

)

So we have

W (R)

pW (R)

∼

−∞

| r ∈ R]

∼

So we know that W (R) is a strict p-ring.

Example. W

(

) =

, since

satisfies all the properties

(

) is supposed

to satisfy.

Proposition.

A complete DVR

of mixed characteristic with perfect residue

field and such that p is a uniformizer is the same as a strict p-ring A such that

A/pA is a field.

Proof.

Let

be a complete DVR such that

is a uniformizer and

A/pA

perfect. Then

-torsion free, as

is an integral domain of characteristic 0.

Since it is also p-adically complete, it is a strict p-ring.

Conversely, if

is a strict

-ring, and

A/pA

is a field, then we have

⊆

A \ pA, and we claim that A

= A \ pA. Let

x =

∞

n=0

with

= 0, i.e.

x 6∈ pA

. We want to show that

is a unit. Since

A/pA

is a

field, we can multiply by [

−1

], so we may wlog

= 1. Then

= 1

− py

for

some y ∈ A. So we can invert this with a geometric series

−1

∞

n=0

is a unit. Now, looking at Teichm¨uller expansions and factoring out multiple

, any non-zero element

can be written as

for a unique

n ≥ Z

≥0

and

u ∈ A

. Then we have

v(z) =

(

n z 6= 0

∞ z = 0

is a discrete valuation on A.

Definition

(Absolute ramification index)

Let

be a DVR with mixed charac-

teristic

with normalized valuation

. The integer

(

) is called the absolute

ramification index of R.

Corollary.

Let

be a complete DVR of mixed characteristic with absolute

ramification index 1 and perfect residue field k. Then R

∼

W (k).

Proof.

Having absolute ramification index 1 is the same as saying

is a uni-

formizer. So

is a strict

-ring with

R/pR

∼

. By uniqueness of the Witt

vector, we know R

∼

W (k).

Theorem.

Let

be a complete DVR of mixed characteristic

with a perfect

residue field k and uniformizer π. Then R is finite over W (k).

Proof.

We need to first exhibit

(

) as a subring of

. We know that

k → k

lifts to a homomorphism

(

)

→ R

. The kernel is a prime ideal because

an integral domain. So it is either 0 or

(

). But

has characteristic 0. So it

can’t be pW(k). So this must be an injection.

Let e be the absolute ramification index of R. We want to prove that

R =

e−1

i=0

W (k).

Looking at valuations, one sees that 1

, π, π, ··· , π

e−1

are linearly independent

over W (k). So we can form

M =

e−1

i=0

W (k) ⊆ R.

We consider R/pR. Looking at Teichm¨uller expansions

∞

n=0

]π

≡

e−1

n=0

]π

mod pR,

we see that 1

, π, ··· , π

e−1

generate

R/pR

(

)-modules (all the Teichm¨uller

lifts live in W (k)). Therefore R = M + pR. We iterate to get

R = M + p(M + pR) = M + p

r = ··· = M + p

for all

m ≥

1. So

is dense in

. But

is also

-adically complete, hence

closed in R. So M = R.

The important statement to take away is

Corollary.

Let

be a mixed characteristic local field. Then

is a finite

extension of Q

Proof.

Let

be the residue field of

. Then

is finite over

(

) by the

previous theorem. So it suffices to show that

(

) is finite over

(

) =

Again the inclusion

⊆ F

gives an injection

(

)

→ W

(

). Write

and let x

, ··· , x

∈ W (F

) be lifts of an F

-bases of F

.. Then we have

W (F

) =

i=1

+ pW (F

and then argue as in the end of the previous theorem to get

W (F

) =

i=1

4 Some p-adic analysis

We are now going to do some fun things that is not really related to the course.

In “normal” analysis, the applied mathematicians hold the belief that every

function can be written as a power series

f(x) =

∞

n=0

When we move on to

-adic numbers, we do not get such a power series expansion.

However, we obtain an analogous result using binomial coefficients.

Before that, we have a quick look at our familiar functions

exp

and

log

, which

we shall continue to define as a power series:

exp(x) =

∞

n=0

, log(1 + x) =

∞

n=1

(−1)

n−1

The domain will no longer be all of the field. Instead, we have the following

result:

Proposition.

Let

be a complete valued field with an absolute value

| · |

and

assume that

K ⊇ Q

and

| · |

restricts to the usual

-adic norm on

. Then

exp

(

) converges for

|x| < p

−1/(p−1)

and

log

(1 +

) converges for

|x| <

1, and

then define continuous maps

exp : {x ∈ K : |x| < p

−1/(p−1)

} → O

log : {1 + x ∈ K : |x| < 1} → K.

Proof.

We let

−log

| · |

be a valuation extending

. Then we have the

dumb estimate

v(n) ≤ log

Then we have





≥ n · v(x) − log

n → ∞

if v(x) > 0. So log converges.

For exp, we have

v(n!) =

n − s

(n)

p − 1

where s

(n) is the sum of the p-adic digits of n. Then we have





≥ n · v(x) −

p − 1

= n ·



v(x) −

p − 1



→ ∞

if v(x) > 1/(p −1). Since v





≥ 0, this lands in O

For the continuity, we just use uniform convergence as in the real case.

What we really want to talk about is binomial coefficients. Let

n ≥

1. Then

we know that





x(x − 1) ···(x − n + 1)

is a polynomial in

, and so defines a continuous function

→ Q

x 7→





When n = 0, we set





= 1 for all x ∈ Z

We know





∈ Z

x ∈ Z

≥0

. So by density of

≥0

⊆ Z

, we must have





∈ Z

for all x ∈ Z

We will eventually want to prove the following result:

Theorem

(Mahler’s theorem)

Let

→ Q

be any continuous function.

Then there is a unique sequence (a

)

n≥0

with a

∈ Q

and a

→ 0 such that

f(x) =

∞

n=0





and moreover

sup

x∈Z

|f(x)| = max

k∈N

We write

(

, Q

) for the set of continuous functions

→ Q

as usual.

This is a Q

vector space as usual, with

(λf + µg)(x) = λf(x) + µg(x)

for all λ, µ ∈ Q

and f, g ∈ C(Z

, Q

) and x ∈ Z

If f ∈ C(Z

, Q

), we set

kfk = sup

x∈Z

|f(x)|

Since

is compact, we know that

is bounded. So the supremum exists and

is attained.

Proposition.

The norm

k · k

defined above is in fact a (non-archimedean)

norm, and that C(Z

, Q

) is complete under this norm.

Let

denote the set of sequences (

)

∞

n=0

such that

→

0. This is

a Q

-vector space with a norm

k(a

)k = max

n∈N

and

is complete. So what Mahler’s theorem gives us is an isometric isomor-

phism between c

and C(Z

, Q

We define

∆ : C(Z

, Q

) → C(Z

, Q

)

∆f(x) = f (x + 1) − f(x).

By induction, we have

∆

f(x) =

i=0

(−1)





f(x + n −i).

Note that ∆ is a linear operator on C(Z

, Q

), and moreover

|∆f(x)|

= |f(x + 1) −f(x)|

≤ kfk.

So we have

k∆fk ≤ kfk.

In other words, we have

k∆k ≤ 1.

Definition

(Mahler coefficient)

Let

f ∈ C

(

, Q

). Then

th-Mahler coeffi-

cient a

(f) ∈ Q

is defined by the formula

(f) = ∆

(f)(0) =

i=0

(−1)





f(n − i).

We will eventually show that these are the

’s that appear in Mahler’s

theorem. The first thing to prove is that these coefficients do tend to 0. We

already know that they don’t go up, so we just have to show that they always

eventually go down.

Lemma. Let f ∈ C(Z

, Q

). Then there exists some k ≥ 1 such that

k∆

fk ≤

kfk.

Proof.

= 0, there is nothing to prove. So we will wlog

kfk

= 1 by scaling

(this is possible since the norm is attained at some

, so we can just divide by

f(x

)). We want to find some k such that

∆

f(x) ≡ 0 mod p

for all x. To do so, we use the explicit formula

∆

f(x) =

i=0

(−1)





f(x + p

− i) ≡ f(x + p

) − f(x) (mod p)

because the binomial coefficients





are divisible by

for

i 6

= 0

, p

. Note that

we do have a negative sign in front of

(

) because (

−

1 as long as

odd, and 1 = −1 if p = 2.

Now

is compact. So

is uniformly continuous. So there is some

such

that

|x − y|

≤ p

−k

implies

(

)

− f

(

)

≤ p

−1

for all

x, y ∈ Z

. So take this

k, and we’re done.

We can now prove that the Mahler’s coefficients tend to 0.

Proposition.

The map

f 7→

(

))

∞

n=0

defines an injective norm-decreasing

linear map C(Z

, Q

) → c

Proof. First we prove that a

(f) → 0. We know that

(f)k

≤ k∆

fk.

So it suffices to show that

∆

fk →

0. Since

∆

k ≤

1, we know

∆

monotonically decreasing. So it suffices to find a subsequence that tends to 0.

To do so, we simply apply the lemma repeatedly to get k

, k

, ··· such that



∆

+...+k



≤

kfk.

This gives the desired sequence.

Note that

(f)|

≤ k∆

k ≤ kfk.

So we know

k(a

(f))

k = max |a

(f)|

≤ kfk.

So the map is norm-decreasing. Linearity follows from linearity of ∆. To finish,

we have to prove injectivity.

Suppose a

(f) = 0 for all n ≥ 0. Then

(f) = f (0) = 0,

and by induction,we have that

f(n) = ∆

f(0) = a

(f) = 0.

for all

n ≥

0. So

is constantly zero on

≥0

. By continuity, it must be zero

everywhere on Z

We are almost at Mahler’s theorem. We have found some coefficients already,

and we want to see that it works. We start by proving a small, familiar, lemma.

Lemma. We have







n − 1





x + 1



for all n ∈ Z

≥1

and x ∈ Z

Proof.

It is well known that this is true when

x ∈ Z

≥n

. Since the expressions

are polynomials in

, them agreeing on infinitely many values implies that they

are indeed the same.

Proposition. Let a = (a

)

∞

n=0

∈ c

. We define f

: Z

→ Q

(x) =

∞

n=0





This defines a norm-decreasing linear map

→ C

(

, Q

). Moreover

(

) =

for all n ≥ 0.

Proof. Linearity is clear. Norm-decreasing follows from

(x)| =









≤ sup









≤ sup

= ka

where we used the fact that





∈ Z

, hence









≤ 1.

Taking the supremum, we know that

k ≤ kak.

For the last statement, for all k ∈ Z

≥0

, we define

(k)

= (a

, a

k+1

, a

k+1

, ···).

Then we have

∆f

(x) = f

(x + 1) − f

(x)

∞

n=1



x + 1



−





∞

n=1



n − 1



∞

n=0

n+1





= f

(

(x)

Iterating, we have

∆

= f

(k)

So we have

) = ∆

(0) = f

(n)

(0) = a

Summing up, we now have maps

C(Z

, Q

) c

with

F (f) = (a

(f))

G(a) = f

We now that

is injective and norm-decreasing, and

is norm-decreasing

and

F G

. It then follows formally that

and the maps are norm-

preserving.

Lemma.

Suppose

V, W

are normed spaces, and

V → W

W → V

are

maps such that

is injective and norm-decreasing, and

is norm-decreasing

and F G = id

. Then GF = id

and F and G are norm-preserving.

Proof. Let v ∈ V . Then

F (v −GF v) = F v − F GF v = (F − F )v = 0.

Since F is injective, we have

v = GF v.

Also, we have

kvk ≥ kF vk ≥ kGF vk = kvk.

So we have equality throughout. Similarly, we have kvk = kGvk.

This finishes the proof Mahler’s theorem, and also finishes this section on

p-adic analysis.

5 Ramification theory for local fields

From now on, the characteristic of the residue field of any local field will be

denoted p, unless stated otherwise.

5.1 Ramification index and inertia degree

Suppose we have an extension

L/K

of local fields. Then since

⊆ m

, and

⊆ O

, we obtain an injection

→

= k

So we also get an extension of residue fields

. The question we want to ask

is how much of the extension is “due to” the extension of residue fields

and how much is “due to” other things happening.

It turns out these are characterized by the following two numbers:

Definition

(Inertia degree)

Let

L/K

be a finite extension of local fields. The

inertia degree of L/K is

L/K

= [k

: k

Definition

(Ramification index)

Let

L/K

be a finite extension of local fields,

and let

be the normalized valuation of

and

a uniformizer of

. The

integer

L/K

= v

(π

)

is the ramification index of L/K.

The goal of the section is to show the following result:

Theorem. Let L/K be a finite extension. Then

[L : K] = e

L/K

We then have two extreme cases of ramification:

Definition

(Unramified extension)

Let

L/K

be a finite extension of local fields.

We say L/K is unramified if e

L/K

= 1, i.e. f

L/K

= [L : K].

Definition

(Totally ramified extension)

Let

L/K

be a finite extension of local

fields. We say L/K is totally ramified if f

L/K

= 1, i.e. e

L/K

= [L : K].

In the next section we will, amongst many things, show that every extension

of local fields can be written as an unramified extension followed by a totally

ramified extension.

Recall the following: let

be a PID and

a finitely-generated

-module.

Assume that

is torsion-free. Then there is a unique integer

n ≥

0 such that

∼

. We say

has rank

. Moreover, if

N ⊆ M

is a submodule, then

finitely-generated, so N

∼

for some m ≤ n.

Proposition.

Let

be a local field, and

L/K

a finite extension of degree

Then

is a finitely-generated and free

module of rank

, and

an extension of degree ≤ n.

Moreover, L is also a local field.

Proof.

Choose a

-basis

, ··· , α

. Let

k · k

denote the maximum norm

on L.



i=1



= max

i=1,...,n

as before. Again, we know that

k · k

is equivalent to the extended norm

| · |

L as K-norms. So we can find r > s > 0 such that

M = {x ∈ L : kxk ≤ s} ⊆ O

⊆ N = {x ∈ L : kxk ≤ r}.

Increasing r and decreasing s if necessary, we wlog r = |a| and s = |b| for some

a, b ∈ K.

Then we can write

M =

i=1

bα

⊆ O

⊆ N =

i=1

aα

We know that

is finitely generated and free of rank

over

, and so is

So O

must be finitely generated and free of rank n over O

Since m

= m

∩ O

, we have a natural injection

→

= k

Since

is generated over

elements, we know that

is generated by

n elements over k

, so it has rank at most n.

To see that

is a local field, we know that

is finite and

is finite,

is finite. It is complete under the norm because it is a finite-dimensional

vector space over a complete field.

Finally, to see that the valuation is discrete, suppose we have a normalized

valuation on K, and w the unique extension of v

to L. Then we have

w(α) =

L/K

(α)).

So we have

w(L

) ⊆

v(K

) =

So it is discrete.

Note that we cannot just pick an arbitrary basis of

L/K

and scale it to give

a basis of

. For example,

(

√

)

has basis 1

√

, but

√

and cannot be scaled to 1 by an element in Q

Even if such a scaled basis exists, it doesn’t necessarily give a basis of the

integral rings. For example,

(

√

−1

)

has a

-basis 1

1 + 3

√

−1

and

|1 + 3

√

−1| = 1, but

√

−1 6∈ Z

+ Z

(1 + 3

√

−1).

So this is not a basis of O

(

√

−1)

over Z

Theorem. Let L/K be a finite extension. Then

[L : K] = e

L/K

and there is some α ∈ O

such that O

= O

[α].

Proof.

We will be lazy and write

L/K

and

L/K

. We first note that

is separable, so there is some

¯α ∈ k

such that

(

¯α

) by the

primitive element theorem. Let

f(x) ∈ k

[x]

be the minimal polynomial of

¯α

over

and let

f ∈ O

[

] be a monic lift of

with deg f = deg

We first claim that there is some

α ∈ O

lifting

¯α

such that

(

)) = 1

(note that it is always

≥

1). To see this, we just take any lift

. If

(

)) = 1,

then we are happy and set

. If it doesn’t work, we set

, where

is the uniformizer of L.

Then we have

f(α) = f (β + π

) = f(β) + f

(β)π

+ bπ

for some

b ∈ O

, by Taylor expansion around

. Since

(

))

≥

2 and

(

)) = 0 (since

is separable, we know

(

) does not vanish when we

reduce mod m), we know v

(f(α)) = 1. So f(α) is a uniformizer of L.

We now claim that the elements

for

= 0

, ··· , f −

1 and

= 0

, ··· , e −

are an O

-basis of O

. Suppose we have

i,j

= 0

for some a

∈ K not all 0. We put

f−1

i=0

We know that 1

, α, ··· , α

f−1

are linearly independent over

since their re-

ductions are linearly independent over

. So there must be some

such that

6= 0.

The next claim is that if

= 0, then

e | v

(

). We let

be an index for

which |a

| is maximal. Then we have

−1

f−1

i=0

−1

Now note that by assumption, the coefficients on the right have absolute value

≤ 1, and is 1 when i = k. So we know that

−1

6≡ 0 mod π

because 1, ¯α, ··· , ¯α

f−1

are linearly independent. So we have

−1

) = 0.

So we must have

) = v

) + v

−1

) ∈ v

) = ev

) = eZ.

Now we write

e−1

j=0

= 0.

= 0, then we have

(

) =

(

) +

j ∈ j

. So no two non-zero

terms in

e−1

j=0

have the same valuation. This implies that

e−1

j=0

= 0,

which is a contradiction.

We now want to prove that

i,j

We let

M =

i,j

and put

N =

f−1

i=0

Then we have

M = N + πN + π

N + ··· + π

e−1

We are now going to use the fact that 1

, ¯α, ··· , ¯α

f−1

span

over

. So we

must have that O

= N + πO

. We iterate this to obtain

= N + π(N + O

)

= N + πN + π

= ···

= N + πN + π

N + ··· + π

e−1

N + π

= M + π

using the fact that

and

have the same valuation, and thus they differ by

a unit in O

. Iterating this again, we have

= M + π

for all

n ≥

1. So

is dense in

. But

is the closed unit ball in the subspace

i,j

Kα

⊆ l

with respect to the maximum norm with respect to the given basis. So it must

be complete, and thus M = O

Finally, since

(

)

is a polynomial in

, we know that

[α].

Corollary. If M/L/K is a tower of finite extensions of local fields, then

M/K

= f

L/K

M/L

M/K

= e

L/K

M/L

Proof.

The multiplicativity of

M/K

follows from the tower law for the residue

fields, and the multiplicativity of

M/K

follows from the tower law for the local

fields and that f

M/K

= [M : K].

5.2 Unramified extensions

Unramified extensions are easy to classify, since they just correspond to extensions

of the residue field.

Theorem.

Let

be a local field. For every finite extension

/k

, there is a

unique (up to isomorphism) finite unramified extension

L/K

with

∼



over

. Moreover, L/K is Galois with

Gal(L/K)

∼

Gal(/k

Proof.

We start with existence. Let

¯α

be a primitive element of

/k

with

minimal polynomial

f ∈ k

[

]. Take a monic lift

f ∈ O

[

] of

such that

deg f

deg

. Note that since

is irreducible, we know

is irreducible. So we

can take L = K(α), where α is a root of f (i.e. L = K[x]/f). Then we have

[L : K] = deg f = deg(

f) = [ : k

Moreover,

contains a root of

, namely the reduction

. So there is an

embedding  → k

, sending ¯α to the reduction of α. So we have

: k

] ≥ [ : k

] = [L : K].

So L/K must be unramified and k

∼

 over k

Uniqueness and the Galois property follow from the following lemma:

Lemma.

Let

L/K

be a finite unramified extension of local fields and let

M/K

be a finite extension. Then there is a natural bijection

Hom

K - Alg

(L, M) ←→ Hom

- Alg

, k

)

given in one direction by restriction followed by reduction.

Proof.

By the uniqueness of extended absolute values, any

-algebra homomor-

phism

L → M

is an isometry for the extended absolute values. In particular,

we have

(

)

⊆ O

and

(

)

⊆ m

. So we get an induced

-algebra

homomorphism ¯ϕ : k

→ k

So we obtain a map

Hom

K-Alg

(L, M) → Hom

-Alg

, k

)

To see this is bijective, we take a primitive element

¯α ∈ k

over

, and take a

minimal polynomial

f ∈ k

[

]. We take a monic lift of

[

], and

α ∈ O

the unique root of

which lifts

¯α

, which exists by Hensel’s lemma. Then by

counting dimensions, the fact that the extension is unramified tells us that

= k

(¯α), L = K(α).

So we can construct the following diagram:

ϕ Hom

K-Alg

(L, M) Hom

-Alg

, k

) ¯ϕ

ϕ(α) {x ∈ M : f (x) = 0} {¯x ∈ k

f(¯x) = 0} ¯ϕ(¯α)

∼

reduction

∼

reduction

But the bottom map is a bijection by Hensel’s lemma. So done.

Alternatively, given a map

¯ϕ

→ k

, we can lift it to the map

L → M

given by



]π



[ ¯ϕ(a

)]π

using the fact that

is a uniformizer in

since the extension is unramified.

So we get an explicit inverse.

Proof of theorem (continued).

To finish off the proof of the theorem, we just

note that an isomorphism

¯ϕ

∼

over

between unramified extensions.

Then

¯ϕ

lifts to a

-embedding

L → M

and [

] = [

] implies that

ϕ is an isomorphism.

To see that the extension is Galois, we just notice that

|Aut

(L)| = |Aut

)| = [k

: k

] = [L : K].

L/K

is Galois. Moreover, the map

Aut

(

)

→ Aut

(

) is really a

homomorphism, hence an isomorphism.

Proposition.

Let

be a local field, and

L/K

a finite unramified extension,

and

M/K

finite. Say

L, M

are subfields of some fixed algebraic closure

Then

LM/M

is unramified. Moreover, any subextension of

L/K

is unramified

over K. If M/K is unramified as well, then LM/K is unramified.

Proof.

Let

¯α

be a primitive element of

, and

f ∈ k

[

] a minimal polyno-

mial of

¯α

, and

f ∈ O

[

] a monic lift of

, and

α ∈ O

a unique lift of

lifting

¯α. Then L = K(α). So LM = M(α).

Let

¯g

be the minimal polynomial of

¯α

over

. Then

¯g |

. By Hensel’s

lemma, we can factorize

[

], where

is monic and lifts

¯g

. Then

(

) = 0 and

is irreducible in

[

]. So

is the minimal polynomial of

over

M. So we know that

[LM : M] = deg g = deg ¯g ≤ [k

: k

] ≤ [LM : M ].

So we have equality throughout and LM/M is unramified.

The second assertion follows from the multiplicativity of

L/K

, as does the

third.

Corollary.

Let

be a local field, and

L/K

finite. Then there is a unique

maximal subfield

K ⊆ T ⊆ L

such that

T/K

is unramified. Moreover, [

] =

L/K

Proof.

Let

T/K

be the unique unramified extension with residue field extension

. Then

→ k

lifts to a

-embedding

T → L

. Identifying

with its image, we know

[T : K] = f

L/K

Now if

is any other unramified extension, then

is an unramified extension

over K, so

[T : K] ≤ [T T

: K] ≤ f

L/K

= [T : K].

So we have equality throughout, and T

⊆ T . So this is maximal.

5.3 Totally ramified extensions

We now quickly look at totally ramified extensions. Recall the following irre-

ducibility criterion:

Theorem

(Eisenstein criterion)

Let

be a local field, and

(

) =

n−1

···

∈ O

[

]. Let

be the uniformizer of

. If

n−1

, ··· , a

and π

- a

, then f is irreducible.

Proof.

Left as an exercise. You’ve probably seen this already in a much more

general context, but in this case there is a neat proof using Newton polygons.

We will need to use the following characterization of the ramification index:

Proposition.

Let

L/K

be an extension of local fields, and

be the normalized

valuation. Let

be the unique extension of

. Then the ramification

index e

L/K

is given by

−1

L/K

= w(π

) = min{w(x) : x ∈ m

Proof.

We know

and

differ by a constant. To figure out what this is, we

have

1 = w(π

) = e

−1

L/K

(π

So for any x ∈ L, we have

w(x) = e

−1

L/K

(x).

In particular, putting x = π

, we have

w(π

) = e

−1

L/K

(π

) = e

−1

L/K

The equality

w(π

) = min{w(x) : x ∈ m

is trivially true because the minimum is attained by π

Definition

(Eisenstein polynomial)

A polynomial

(

)

∈ O

[

] satisfying the

assumptions of Eisenstein’s criterion is called an Eisenstein polynomial.

We can now state the proposition:

Proposition.

Let

L/K

be a totally ramified extension of local fields. Then

L = K(π

) and the minimal polynomial of π

over K is Eisenstein.

Conversely, if

(

) and the minimal polynomial of

over

is Eisenstein,

then L/K is totally ramified and α is a uniformizer of L.

Proof.

Let

= [

be the valuation of

, and

the unique extension

to L. Then

[K(π

) : K]

−1

≤ e

−1

K(π

)/K

= min

x∈m

K(π

)

w(c) ≤

where the last inequality follows from the fact that π

∈ m

L(π

)

But we also know that

[K(π

) : K] ≤ [L : K].

So we know that L = K(π

Now let

(

) =

n−1

···

∈ O

[

] be the minimal polynomial

of π

/K. Then we have

= −(a

+ a

+ ··· + a

n−1

So we have

1 = w(π

) = w(a

+ a

+ ··· + a

n−1

) = min

i=0,...,n−1



) +



This implies that v

) ≥ 1 for all i, and v

) = 1. So it is Eisenstein.

For the converse, if K = K(α) and n = [L : K], take

g(x) = x

+ b

n−1

+ .. + b

∈ O

[x]

be the minimal polynomial of

. So all roots have the same valuation. So we

have

1 = w(b

) = n · w(α).

So we have w(α) =

. So we have

−1

L/K

= min

x∈m

w(x) ≤

= [L : K]

−1

So [L : K] = e

L/K

= n. So L/K is totally ramified and α is a uniformizer.

In fact, more is true. We have

[

], since every element in

can

be written as

i≥0

where a

is a lift of an element in k

= k

, which can be chosen to be in O

6 Further ramification theory

6.1 Some filtrations

If we have a field

, then we have a unit group

. We would like to

come up with a filtration of subgroups of the unit group, namely a sequence

··· ⊆ U

(2)

⊆ U

(1)

⊆ U

(0)

= U

of subgroups that tells us how close a unit is to being 1. The further down we

are in the chain, the closer we are to being 1.

Similarly, given a field extension

L/K

, we want a filtration on the Galois

group (the indexing is conventional)

··· ⊆ G

(L/K) ⊆ G

−1

(L/K) = Gal(L/K).

This time, the filtration tells us how close the automorphisms are to being the

identity map.

The key thing about these filtrations is that we can figure out information

about the quotients

(s)

(s+1)

and

(

L/K

)

s+1

(

L/K

), which is often easier.

Later, we might be able to patch these up to get more useful information about

and Gal(L/K).

We start with the filtration of the unit group.

Definition (Higher unit groups). We define the higher unit groups to be

(s)

= U

(s)

= 1 + π

We also put

= U

(0)

= U

(0)

= O

The quotients of these units groups are surprisingly simple:

Proposition. We have

(1)

∼

, ·),

(s)

(s+1)

∼

, +).

for s ≥ 1.

Proof.

We have a surjective homomorphism

→ k

which is just reduction

mod

, and the kernel is just things that are 1 modulo

, i.e.

(1)

. So this

gives the first part.

For the second part, we define a surjection U

(s)

→ k

given by

1 + π

x 7→ x mod π

This is a group homomorphism because

(1 + π

x)(1 + π

y) = 1 + π

(x + y + π

xy),

and this gets mapped to

x + y + π

x + y

∼

x + y mod π

Then almost by definition, the kernel is U

(s+1)

The next thing to consider is a filtration of the Galois group.

Definition (Higher ramification group). Let L/K be a finite Galois extension

of local fields, and v

the normalized valuation of L.

Let s ∈ R

≥−1

. We define the sth ramification group by

(L/K) = {σ ∈ Gal(L/K) : v

(σ(x) − x) ≥ s + 1 for all x ∈ O

So if you belong to

for a large

, then you move things less. Note that we

could have defined these only for

s ∈ Z

≥−1

, but allowing fractional indices will

be helpful in the future.

Now since σ(x) −x ∈ O

for all x ∈ O

, we know

−1

(L/K) = Gal(L/K).

We next consider the case of G

(L/K). This is, by definition

(L/K) = {σ ∈ Gal(L/K) : v

(σ(x) − x) ≥ 1 for all x ∈ O

}

= {σ ∈ Gal(L/K) : σ(x) ≡ x mod m for all x ∈ O

In other words, these are all the automorphisms that reduce to the identity when

we reduce it to Gal(k

Definition

(Inertia group)

Let

L/K

be a finite Galois extension of local fields.

Then the inertia group of L/K is the kernel of the natural homomorphism

Gal(L/K) → Gal(k

)

given by reduction. We write this as

I(L/K) = G

(L/K).

Proposition.

Let

L/K

be a finite Galois extension of local fields. Then the

homomorphism

Gal(L/K) → Gal(k

)

given by reduction is surjective.

Proof.

Let

T/K

be maximal unramified subextension. Then by Galois theory,

the map

Gal

(

L/K

)

→ Gal

(

T/K

) is a surjection. Moreover, we know that

= k

. So we have a commutative diagram

Gal(L/K) Gal(k

)

Gal(T/K) Gal(k

∼

So the map Gal(L/K) → Gal(k

) is surjective.

Then the inertia group is trivial iff

L/K

is unramified. The field

sometimes called the inertia field.

Lemma.

Let

L/K

be a finite Galois extension of local fields, and let

σ ∈ I

(

L/K

Then σ([x]) = [x] for all x.

More generally, let

x ∈ k

and

σ ∈ Gal

(

L/K

) with image

¯σ ∈ Gal

(

Then we have

[¯σ(x)] = σ([x]).

Proof. Consider the map k

→ O

given by

f : x 7→ σ

−1

([¯σ(x)]).

This is multiplicative, because every term is multiplicative, and

−1

([¯σ(x)]) ≡ x mod π

So this map f has to be the Teichm¨uller lift by uniqueness.

That’s all we’re going to say about the inertia group. We now consider the

general properties of this filtration.

Proposition.

Let

L/K

be a finite Galois extension of local fields, and

the

normalized valuation of

. Let

be the uniformizer of

. Then

s+1

(

L/K

) is

a normal subgroup of G

(L/K) for s ∈ Z

≥0

, and the map

(L/K)

s+1

(L/K)

→

(s)

(s+1)

given by

σ 7→

σ(π

)

is a well-defined injective group homomorphism, independent of the choice of

Proof. We define the map

φ : G

(L/K) →

(s)

(s+1)

σ 7→ σ(π

)/π

We want to show that this has kernel G

s+1

(L/K).

First we show it is well-defined. If σ ∈ G

(L/K), we know

σ(π

) = π

+ π

s+1

for some x ∈ O

. So we know

σ(π

)

= 1 + π

x ∈ U

(s)

So it has the right image. To see this is independent of the choice of

, we let

u ∈ O

. Then σ(u) = u + π

s+1

y for some y ∈ O

Since any other uniformizer must be of the form π

u, we can compute

σ(π

(π

+ π

s+1

)(u + π

s+1

= (1 + π

x)(1 + π

s+1

−1

≡ 1π

x (mod U

s+1

So they represent the same element in in U

(s)

(s+1)

To see this is a group homomorphism, we know

φ(στ) =

σ(τ(π

))

σ(τ(π

))

τ(π

)

τ(π

)

= φ(σ)φ(t),

using the fact that τ(π

) is also a uniformizer.

Finally, we have to show that ker φ = G

s+1

(L/K). We write down

ker φ = {σ ∈ G

(L/K) : v

(σ(π

) − π

) ≥ s + 2}.

On the other hand, we have

s+1

(L/K) = {σ ∈ G

(L/K) : v

(σ(z) − z) ≥ s + 2 for all z ∈ O

So we trivially have

s+1

(

L/K

)

⊆ ker φ

. To show the converse, let

x ∈ O

and

write

x =

∞

n=0

]π

Take σ ∈ ker φ ⊆ G

(L/K) ⊆ I(L/K). Then we have

σ(π

) = π

+ π

s+2

y, y ∈ O

Then by the previous lemma, we know

σ(x) − x =

∞

n=1

] ((σ(π

))

− π

)

∞

n=1

]



(π

+ π

s+2

− π



= π

s+2

(things).

So we know v

(σ(x) − x) ≥ s + 2.

Corollary. Gal(L/K) is solvable.

Proof. Note that

(L/K) = {id}.

So (

(

L/K

))

s∈Z

≥−1

is a subnormal series of

Gal

(

L/K

), and all quotients are

abelian, because they embed into

(s)

(s+1)

∼

(

+) (and

−

1 can be checked

separately).

Thus if

L/K

is a finite extension of local fields, then we have, for

s ≥

injections

(L/K)

s+1

(L/K)

→ k

Since k

is a p-group, it follows that

(L/K)|

s+1

(L/K)|

is a pth power. So it follows that for any t, the quotient

(L/K)|

is also a

th power. However, we know that the intersection of all

(

L/K

)

{id}

, and also

Gal

(

L/K

) is finite. So for sufficiently large

, we know that

(L/K)| = 1. So we conclude that

Proposition. G

(L/K) is always a p-group.

We now use the injection

(L/K)

→ k

and the fact that

has order prime to

. So

(

L/K

) must be the Sylow

p-subgroup of G

(L/K). Since it is normal, it must be the unique p-subgroup.

Definition

(Wild inertia group and tame quotient)

. G

(

L/K

) is called the wild

inertia group, and the quotient G

(L/K)/G

(L/K) is the tame quotient.

6.2 Multiple extensions

Suppose we have tower

M/L/K

of finite extensions of local fields. How do the

ramification groups of the different extensions relate? We first do the easy case.

Proposition.

Let

M/L/K

be finite extensions of local fields, and

M/K

Galois.

Then

(M/K) ∩ Gal(M/L) = G

(M/L).

Proof. We have

(M/K) = {σ ∈ Gal(M/L) : v

(σx − x) ≥ s + 1} = G

(M/K) ∩ Gal(M/L).

This is trivial, because the definition uses the valuation

of the bigger field

all the time. What’s more difficult and interesting is quotients, namely going

from M/K to L/K.

We want to prove the following theorem:

Theorem

(Herbrand’s theorem)

Let

M/L/K

be finite extensions of local fields

with M/K and L/K Galois. Then there is some function η

M/L

such that

(L/K)

∼

(M/K)

(M/L)

for all s, where t = η

M/L

(s).

To better understand the situation, it helps to provide an alternative charac-

terization of the Galois group.

Definition (i

L/K

). We define

L/K

(σ) = min

x∈O

(σ(x) − x).

It is then immediate that

(L/K) = {σ ∈ Gal(L/K) : i

L/K

(σ) ≥ s + 1}.

This is not very helpful. We now claim that we can compute

L/K

using the

following formula:

Proposition.

Let

L/K

be a finite Galois extension of local fields, and pick

α ∈ O

such that O

= O

[α]. Then

L/K

(σ) = v

(σ(α) − α).

Proof.

Fix a

. It is clear that

L/K

(

)

≤ v

(

)

− α

). Conversely, for any

x ∈ O

, we can find a polynomial g ∈ O

[t] such that

x = g(α) =

where b

∈ O

. In particular, b

is fixed by σ.

Then we have

(σ(x) − x) = v

(σg(α) − g(α))

= v

i=1

(σ(α)

− α

)

≥ v

(σ(α) − α),

using the fact that σ(α) − α | σ(α)

− α

for all i. So done.

Now if

M/L/K

are finite Galois extensions of local fields, then

[

]

implies O

= O

[α]. So for σ ∈ Gal(M/L), we have

M/L

(σ) = i

M/K

(σ).

Going in the other direction is more complicated.

Proposition.

Let

M/L/K

be a finite extension of local fields, such that

M/K

and L/K are Galois. Then for σ ∈ Gal(L/K), we have

L/K

(σ) = e

−1

M/L

τ∈Gal(M/K)

τ|

=σ

M/K

(τ).

Here

M/L

is just to account for the difference between

and

. So the

real content is that the value of

L/K

(

) is the sum of the values of

M/K

(

) for

all τ that restrict to σ.

Proof.

= 1, then both sides are infinite by convention, and equality holds.

So we assume

σ 6

= 1. Let

[

] and

[

], where

α ∈ O

and

β ∈ O

. Then we have

M/L

L/K

(σ) = e

M/L

(σβ − β) = v

(σβ − β).

Now if τ ∈ Gal(M/K), then

M/K

(τ) = v

(τα − α)

Now fix a τ such that τ|

= σ. We set H = Gal(M/L). Then we have

∈Gal(M/K),τ

=σ

M/K

(τ

) =

g∈H

(τg(α) − α) = v





g∈H

(τg(α) − α)





We let

b = σ(β) − β = τ(β) − β

and

a =

g∈H

(τg(α) − α).

We want to prove that v

(b) = v

(a). We will prove that a | b and b | a.

We start with a general observation about elements in

. Given

z ∈ O

we can write

z =

i=1

, z

∈ O

Then we know

τ(z) − z =

i=1

(τ(β)

− β

)

is divisible by τ(β) − β = b.

Now let

(

)

∈ O

[

] be the minimal polynomial of

over

. Then explicitly,

we have

F (x) =

g∈H

(x − g(α)).

Then we have

(τF )(x) =

g∈H

(x − τg(α)),

where

τF

is obtained from

by applying

to all coefficients of

. Then all

coefficients of

τF −F

are of the form

(

)

−z

for some

z ∈ O

. So it is divisible

by b. So b divides every value of this polynomial, and in particular

b | (τF − F )(α) =

g∈H

(α − g(α)) = ±a,

So b | a.

In other direction, we pick

f ∈ O

[

] such that

(

) =

. Then

(

)

−β

= 0.

This implies that the polynomial

(

)

− β

divides the minimal polynomial of

in O

[x]. So we have

f(x) − β = F (x)h(x)

for some h ∈ O

[x].

Then noting that f has coefficients in O

, we have

(f − τβ)(x) = (τf −τb)(x) = (τF )(x)(τh)(x).

Finally, set x = α. Then

−b = β − τβ = ±a(τh)(α).

So a | b.

Now that we understand how the

L/K

behave when we take field extensions,

we should be able to understand how the ramification groups behave!

We now write down the right choice of η

L/K

: [−1, ∞) → [−1, ∞):

L/K

(s) =

−1

L/K

σ∈G

min(i

L/K

(σ), s + 1)

− 1.

Theorem

(Herbrand’s theorem)

Let

M/L/K

be a finite extension of local

fields with M/K and L/K Galois. We set

H = Gal(M/L), t = η

M/L

(s).

Then we have

(M/K)H

= G

(L/K).

By some isomorphism theorem, and the fact that

H ∩ G

(

M/K

) =

(

M/L

this is equivalent to saying

(L/K)

∼

(M/K)

(M/L)

Proof.

Let

Gal

(

M/K

). Fix a

σ ∈ Gal

(

L/K

). We let

τ ∈ Gal

(

M/K

) be an

extension of σ to M that maximizes i

M/K

, i.e.

M/K

(τ) ≥ i

M/K

(τg)

for all g ∈ H. This is possible since H is finite.

We claim that

L/K

(σ) − 1 = η

M/L

M/K

(τ) − 1).

If this were true, then we would have

σ ∈

(M/K)H

⇔ τ ∈ G

(M/K)

⇔ i

M/K

(τ) − 1 ≥ s

Since η

M/L

is strictly increasing, we have

⇔ η

M/L

M/K

(τ) − 1) ≥ η

M/L

(s) = t

⇔ i

L/K

(σ) − 1 ≥ t

⇔ σ ∈ G

(L/K),

and we are done.

To prove the claim, we now use our known expressions for

L/K

(

) and

M/L

M/K

(τ) − 1) to rewrite it as

−1

M/L

g∈H

M/K

(τg) = e

−1

M/L

g∈H

min(i

M/L

(g), i

M/K

(τ)).

We then make the stronger claim

M/K

(τg) = min(i

M/L

(g), i

M/K

(τ)).

We first note that

M/K

(τg) = v

(τg(α) − α)

= v

(τg(α) − g(α) + g(α) − α)

≥ min(v

(τg(α) − g(α)), v

(g(α) − α))

= min(i

M/K

(τ), i

M/K

(g))

We cannot conclude our (stronger) claim yet, since we have a

≥

in the middle.

We now have to split into two cases.

(i)

M/K

(

)

≥ i

M/K

(

), then the above shows that

M/K

(

τg

)

≥ i

M/K

(

But we also know that it is bounded above by

. So

M/K

(

τg

) =

M/K

(

So our claim holds.

(ii)

M/K

(

)

< i

M/K

(

), then the above inequality is in fact an equality as

the two terms have different valuations. So our claim also holds.

So done.

We now prove an alternative characterization of the function

L/K

, using a

funny integral.

Proposition. Write G = Gal(L/K). Then

L/K

(s) =

(L/K) : G

(L/K))

When −1 ≤ x < 0, our convention is that

(L/K) : G

(L/K))

= (G

(L/K) : G

(L/K)),

which is just equal to 1 when −1 < x < 0. So

L/K

(s) = s if − 1 ≤ s ≤ 0.

Proof.

We denote the RHS by

(

). It is clear that both

L/K

(

) and

(

) are

piecewise linear and the break points are integers (since

L/K

(

) is always an

integer). So to see they are the same, we see that they agree at a point, and

that they have equal derivatives. We have

L/K

(0) =

|{σ ∈ G : i

L/K

(σ) ≥ 1}|

L/K

− 1 = 0 = θ(0),

since the numerator is the size of the inertia group.

If s ∈ [−1, ∞) \ Z, then

L/K

(s) = e

−1

L/K

(|{σ ∈ G : i

L/K

(σ) ≥ s + 1}|)

(L/K)|

(L/K) : G

(L/K))

= θ

(s).

So done.

We now tidy up the proof by inventing a different numbering of the ramifica-

tion groups. Recall that

L/K

: [−1, ∞) → [−1, ∞)

is continuous, strictly increasing, and

L/K

(−1) = −1, η

L/K

(s) → ∞ as s → ∞.

So this is invertible. We set

Notation.

L/K

= η

−1

L/K

Definition

(Upper numbering)

Let

L/K

be a Galois extension of local fields.

Then the upper numbering of the ramification groups of L/K is defined by

(L/K) = G

L/K

(t)

(L/K)

for t ∈ [−1, ∞). The original number is called the lower numbering.

To rephrase our previous theorem using the upper numbering, we need a

little lemma:

Lemma.

Let

M/L/K

be a finite extension of local fields, and

M/K

and

L/K

be Galois. Then

M/K

= η

L/K

◦ η

M/L

Hence

M/K

= ψ

M/L

◦ ψ

L/K

Proof.

Let

s ∈

[

−

, ∞

), and let

M/L

(

), and

Gal

(

M/L

). By Her-

brand’s theorem, we know

(L/K)

∼

(M/K)H

∼

(M/K)

H ∩ G

(M/K)

(M/L)

Thus by multiplicativity of the inertia degree, we have

(M/K)|

M/K

(L/K)|

L/K

(M/L)|

M/L

By the fundamental theorem of calculus, we know that whenever the derivatives

make sense, we have

M/K

(s) =

(M/K)|

M/K

So putting this in, we know

M/K

(s) = η

L/K

(t)η

M/L

(s) = (η

L/K

◦ η

M/L

)

(s).

Since

M/K

and

L/K

◦ η

M/L

agree at 0 (they both take value 0), we know that

the functions must agree everywhere. So done.

Corollary.

Let

M/L/K

be finite Galois extensions of local fields, and

Gal(M/L). Let t ∈ [−1, ∞). Then

(M/K)H

= G

(L/K).

Proof. Put s = η

L/K

(t). Then by Herbrand’s theorem, we have

(M/K)H

M/K

(t)

(M/K)H

∼

M/L

(ψ

M/K

(t))

(L/K)

= G

(L/K)

= G

(L/K).

This upper numbering might seem like an unwieldy beast that was invented

just so that our theorem looks nice. However, it turns out that often the upper

numberings are rather natural, as we could see in the example below:

Example.

Consider

a primitive

th root of unity, and

(

). The

minimal polynomial of ζ

is the p

th cyclotomic polynomial

(x) = x

n−1

(p−1)

+ x

n−1

(p−2)

+ ··· + 1.

It is an exercise on the example sheet to show that this is indeed irreducible.

K/Q

is a Galois extension of degree

n−1

(

p −

1). Moreover, it is totally

ramified by question 6 on example sheet 2, with uniformizer

π = ζ

− 1

is a uniformizer. So we know

= Z

[ζ

− 1] = Z

[ζ

We then have an isomorphism





→ Gal(L/Q

)

obtained by sending m → σ

, where

(ζ

) = ζ

We have

K/Q

(σ

) = v

(σ

(ζ

) − ζ

)

= v

(ζ

− ζ

)

= v

(ζ

m−1

− 1)

since

is a unit. If

= 1, then this thing is infinity. If it is not 1, then

m−1

is a primitive

n−k

th root of unity for the maximal

such that

| m −

1. So

by Q6 on example sheet 2, we have

(ζ

m−1

− 1) =

n−1

(p − 1)

n−k−1

(p − 1)

= p

Thus we have

(ζ

m−1

− 1) ≥ p

⇔ m ≡ 1 mod p

It then follows that for

≥ s + 1 ≥ p

k−1

+ 1,

we have

(K/Q

)

∼

{m ∈ (Z/p

)

: m ≡ 1 mod p

Now m ≡ 1 mod p

iff σ

(ζ

) = ζ

. So in fact

(K/Q

)

∼

Gal(K/Q

(ζ

)).

Finally, when s ≥ p

− 1, we have

(K/Q

) = 1.

We claim that

K/Q

− 1) = k.

So we have

(K/Q

) = Gal(K/Q

(ζ

)).

This actually looks much nicer!

To actually compute

K/Q

, we have notice that the function we integrate to

get η looks something like this (not to scale):

p−1

p(p−1)

(p−1)

p − 1

− 1 p

− 1

The jumps in the lower numbering are at

−

1 for

= 1

, ··· , n −

1. So we have

K/Q

− 1) = (p −1)

p − 1

+ ((p

− 1) −(p −1))

p(p − 1)

+ ··· + ((p

− 1) −(p

k−1

− 1))

k−1

(p − 1)

= k.

7 Local class field theory

Local class field theory is the study of abelian extensions of local fields, i.e. a

Galois extension whose Galois group is abelian.

7.1 Infinite Galois theory

It turns out that the best way of formulating this theory is to not only use

finite extensions, but infinite extensions as well. So we need to begin with some

infinite Galois theory. We will mostly just state the relevant results instead of

proving them, because this is not a course on Galois theory.

In this section, we will work with any field.

Definition

(Separable and normal extensions)

Let

L/K

be an algebraic exten-

sion of fields. We say that

L/K

is separable if, for every

α ∈ L

, the minimal

polynomial

∈ K

[

] is separable. We say

L/K

is normal if

splits in

for

every α ∈ L.

Definition

(Galois extension)

Let

L/K

be an algebraic extension of fields.

Then it is Galois if it is normal and separable. If so, we write

Gal(L/K) = Aut

(L).

These are all the same definitions as in the finite case.

In finite Galois theory, the subgroups of

Gal

(

L/K

) match up with the

intermediate extensions, but this is no longer true in the infinite case. The

Galois group has too many subgroups. To fix this, we need to give

Gal

(

L/K

) a

topology, and talk about closed subgroups.

Definition

(Krull topology)

Let

M/K

be a Galois extension. We define the

Krull topology on M/K by the basis

{Gal(M/L) : L/K is finite}.

More explicitly, we say that

U ⊆ Gal

(

M/K

) is open if for every

σ ∈ U

, we can

find a finite subextension L/K of M/K such that σ Gal(M/L) ⊆ U .

Note that any open subgroup of a topological group is automatically closed,

but the converse does not hold.

Note that when

M/K

is finite, then the Krull topology is discrete, since we

can just take the finite subextension to be M itself.

Proposition.

Let

M/K

be a Galois extension. Then

Gal

(

M/K

) is compact

and Hausdorff, and if

U ⊆ Gal

(

M/K

) is an open subset such that 1

∈ U

, then

there is an open normal subgroup N ⊆ Gal(M/K) such that N ⊆ U.

Groups with properties in this proposition are known as profinite groups.

Proof. We will not prove the first part.

For the last part, note that by definition, there is a finite subextension of

M/K

such that

Gal

(

M/L

)

⊆ U

. We then let

be the Galois closure of

over

. Then

Gal

(

M/L

)

⊆ Gal

(

M/L

)

⊆ U

, and

Gal

(

M/L

) is open and normal.

Recall that we previously defined the inverse limit of a sequence rings. More

generally, we can define such an inverse limit for any sufficiently nice poset

of things. Here we are going to do it for topological groups (for those doing

Category Theory, this is the filtered limit of topological groups).

Definition

(Directed system)

Let

be a set with a partial order. We say that

is a directed system if for all

i, j ∈ I

, there is some

k ∈ I

such that

i ≤ k

and

j ≤ k.

Example. Any total order is a directed system.

Example. N with divisibility | as the partial order is a directed system.

Definition

(Inverse limit)

Let

be a directed system. An inverse system (of

topological groups) indexed by

is a collection of topological groups

for each

i ∈ I and continuous homomorphisms

: G

→ G

for all i, j ∈ I such that i ≤ j, such that

= id

and

= f

◦ f

whenever i ≤ j ≤ k.

We define the inverse limit on the system (G

, f

) to be

lim

←−

i∈I

(

) ∈

i∈I

: f

) = g

for all i ≤ j

)

⊆

i∈I

which is a group under coordinate-wise multiplication and a topological space

under the subspace topology of the product topology on

i∈I

. This makes

lim

←−

i∈I

into a topological group.

Proposition.

Let

M/K

be a Galois extension. The set

of finite Galois

subextensions

L/K

is a directed system under inclusion. If

L, L

∈ I

and

L ⊆ L

then we have a restriction map

·|

: Gal(L

/K) → Gal(L/K).

Then (Gal(L/K), ·|

) is an inverse system, and the map

Gal(M/K) → lim

←−

i∈I

Gal(L/K)

σ 7→ (σ|

)

i∈I

is an isomorphism of topological groups.

We now state the main theorem of Galois theory.

Theorem

(Fundamental theorem of Galois theory)

Let

M/K

be a Galois exten-

sion. Then the map

L 7→ Gal

(

M/L

) defines a bijection between subextensions

L/K

M/K

and closed subgroups of

Gal

(

M/K

), with inverse given by sending

H 7→ M

, the fixed field of H.

Moreover,

L/K

is finite if and only if

Gal

(

M/L

) is open, and

L/K

is Galois

iff Gal(M/L) is normal, and then

Gal(L/K)

Gal(M/L)

→ Gal(L/K)

is an isomorphism of topological groups.

Proof.

This follows easily from the fundamental theorem for finite field extensions.

We will only show that

Gal

(

M/L

) is closed and leave the rest as an exercise. We

can write

L =

[

⊆L

/K finite

Then we have

Gal(M/L) =

⊆L

/K finite

Gal(M/L

and each Gal(M/L

) is open, hence closed. So the whole thing is closed.

7.2 Unramified extensions and Weil group

We first define what it means for an infinite extension to be unramified or totally

ramified. To do so, we unexcitingly patch up the definitions for finite cases.

Definition

(Unramified extension)

Let

be a local field, and

M/K

be alge-

braic. Then

M/K

is unramified if

L/K

is unramified for every finite subextension

L/K of M/K.

Note that since the extension is not necessarily finite, in general

will not

be a local field, since chances are its residue field would be infinite.

Definition

(Totally ramified extension)

Let

be a local field, and

M/K

algebraic. Then

M/K

is totally ramified if

L/K

is totally ramified for every

finite subextension L/K of M/K.

Proposition.

Let

M/K

be an unramified extension of local fields. Then

M/K

is Galois, and

Gal(M/K)

∼

Gal(k

)

via the reduction map.

Proof.

Every finite subextension of

M/K

is unramified, so in particular is Galois.

M/K

is Galois (because normality and separability is checked for each

element). Then we have a commutative diagram

Gal(M/K) Gal(k

)

lim

←−

L/K

Gal(L/K) lim

←−

L/K

Gal(k

)

reduction

∼ ∼

reduction

∼

The left hand map is an isomorphism by (infinite) Galois theory, and since all

finite subextensions of

are of the form

by our finite theory, we

know the right-hand map is an isomorphism. The bottom map is an isomorphism

since it is an isomorphism in each component. So the top map must be an

isomorphism.

Since the compositor of unramified extensions is unramified, it follows that

any algebraic extension M/K has a maximal unramified subextension

T = T

M/K

/K.

In particular, every field K has a maximal unramified extension K

We now try to understand unramified extensions. For a finite unramified

extension L/K, we have an isomorphism

Gal(L/K) Gal(k

)

∼

By general field theory, we know that

Gal

(

) is a cyclic group generated by

Frob

L/K

: x 7→ x

where

is the size of

. So by the isomorphism, we obtain a generator

of Gal(L/K).

Definition

(Arithmetic Frobenius)

Let

L/K

be a finite unramified extension

of local fields, the (arithmetic) Frobenius of

L/K

is the lift of

Frob

L/K

∈

Gal(k

) under the isomorphism Gal(L/K)

∼

Gal(k

There is also a geometric Frobenius, which is its inverse, but we will not use

that in this course.

We know

Frob

is compatible in towers, i.e. if

M/L/K

is a tower of finite

unramified extension of local fields, then

Frob

M/K

Frob

L/K

, since they both

reduce to the map

x 7→ x

Gal

(

), and the map between

Gal

(

)

and Gal(L/K) is a bijection.

So if M/K is an arbitrary unramified extension, then we have an element

(Frob

L/K

) ∈ lim

←−

L/K

Gal(L/K)

∼

Gal(M/K).

So we get an element

Frob

M/K

∈ Gal

(

M/K

). By tracing through the proof of

Gal(M/K)

∼

Gal(k

), we see that this is the unique lift of x 7→ x

Note that while for finite unramified extensions

M/K

, the Galois group is

generated by the Frobenius, this is not necessarily the case when the extension

is infinite. However, powers of the Frobenius are the only things we want to

think about, so we make the following definition:

Definition

(Weil group)

Let

be a local field and

M/K

be Galois. Let

M/K

be the maximal unramified subextension of

M/K

. The Weil group

of M/K is

W (M/K) = {σ ∈ Gal(M/K) : σ|

= Frob

T/K

for some n ∈ Z}.

We define a topology on

(

M/K

) by saying that

is open iff there is a finite

extension L/T such that σ Gal(L/T ) ⊆ U.

In particular, if M/K is unramified, then W(M/K) = Frob

T/K

It is helpful to put these groups into a diagram of topological groups to see

what is going on.

Gal(M/T ) W (M/K) Frob

T/K

Gal(M/T ) Gal(M/K) Gal(T /K)

Here we put the discrete topology on the subgroup generated by the Frobenius.

The topology of

(

M/K

) is then chosen so that all these maps are continuous

homomorphisms of groups.

In many ways, the Weil group works rather like the Galois group.

Proposition.

Let

be a local field, and

M/K

Galois. Then

(

M/K

) is dense

Gal

(

M/K

). Equivalently, for any finite Galois subextension

L/K

M/K

the restriction map W(M/K) → Gal(L/K) is surjective.

If L/K is a finite subextension of M/K, then

W (M/L) = W (M/K) ∩ Gal(M/L).

If L/K is also Galois, then

W (M/K)

W (M/L)

∼

Gal(L/K)

via restriction.

Proof.

We first prove density. To see that density is equivalent to

(

M/K

)

→

Gal

(

L/K

) being surjective for all finite subextension

L/K

, note that by the

topology on

Gal

(

M/K

), we know density is equivalent to saying that

(

M/K

)

hits every coset of

Gal

(

M/L

), which means that

(

M/K

)

→ Gal

(

L/K

) is

surjective.

Let

L/K

be a subextension. We let

M/K

. Then

L/K

T ∩ L

. Then

we have a diagram

Gal(M/T ) W (M/K) Frob

T/K

Gal(L/T ∩ L) Gal(L/K) Gal(T ∩ L/K)

Here the surjectivity of the left vertical arrow comes from field theory, and the

right hand vertical map is surjective because

T ∩ L/K

is finite and hence the

Galois group is generated by the Frobenius. Since the top and bottom rows are

short exact sequences (top by definition, bottom by Galois theory), by diagram

chasing (half of the five lemma), we get surjectivity in the middle.

To prove the second part, we again let

L/K

be a finite subextension. Then

L · T

M/K

⊆ T

M/L

. We then have maps

Frob

M/K

Gal(T

M/K

/K) Gal(k

)

Frob

M/L

Gal(T

M/L

/L) Gal(k

)

∼

So the left hand vertical map is an inclusion. So we know

Frob

M/L

= Frob

M/K

∩ Gal(T

M/L

/L).

Now if σ ∈ Gal(M/L), then we have

σ ∈ W (M/L) ⇔ σ|

M/L

∈ Frob

M/L

⇔ σ|

M/K

∈ Frob

M/K

⇔ σ ∈ W (M/K).

So this gives the second part.

Now

L/K

is Galois as well. Then

Gal

(

M/L

) is normal in

Gal

(

M/K

). So

W (M/L) is normal in W(M/K) by the second part. Then we can compute

W (M/K)

W (M/L)

W (M/K)

W (M/K) ∩ Gal(M/L)

∼

W (M/K) Gal(M/L)

Gal(M/L)

Gal(M/K)

Gal(M/L)

∼

Gal(L/K).

The only non-trivial part in this chain is the assertion that

(

M/K

)

Gal

(

M/L

) =

Gal

(

M/K

), i.e. that

(

M/K

) hits every coset of

Gal

(

M/L

), which is what

density tells us.

7.3 Main theorems of local class field theory

We now come to the main theorems of local class field theory.

Definition

(Abelian extension)

Let

be a local field. A Galois extension

L/K is abelian if Gal(L/K) is abelian.

We will fix an algebraic closure

, and all algebraic extensions we will

consider will be taken to be subextensions of

K/K

. We let

sep

be the separable

closure of K inside

M/K

and

M/K

are Galois extensions, then

LM/K

is Galois, and the map

given by restriction

Gal(LM/K) → Gal(L/K) × Gal(M/K).

is an injection. In particular, if

L/K

and

M/K

are both abelian, then so is

LM/K. This implies that there is a maximal abelian extension K

Finally, note that we know an example of an abelian extension, namely the

maximal unramified extension

sep

⊆ K

, and we put

Frob

Theorem

(Local Artin reciprocity)

There exists a unique topological isomor-

phism

Art

: K

→ W (K

/K)

characterized by the properties

(i) Art

(π

= Frob

, where π

is any uniformizer.

(ii) We have

Art

L/K

(x))|

= id

for all L/K finite abelian and x ∈ L

Moreover, if

M/K

is finite, then for all

x ∈ M

, we know

Art

(

) is an

automorphism of

, and restricts to an automorphisms of

. Then

we have

Art

(x)|

= Art

M/K

(x)).

Moreover, Art

induces an isomorphism

M/K

)

→ Gal



M ∩ K



To simplify this, we will write

(

L/K

) =

L/K

(

) for

L/K

finite. From

this theorem, we can deduce a lot of more precise statements.

Corollary. Let L/K be finite. Then N(L/K) = N((L ∩K

)/K), and

: N(L/K)) ≤ [L : K]

with equality iff L/K is abelian.

Proof.

To see this, we let

L ∩ K

. Applying the isomorphism twice gives

N(L/K)

∼

Gal(M/K)

∼

N(M/K)

Since

(

L/K

)

⊆ N

(

M/K

), and [

]

≥

[

] =

|Gal

(

M/K

)

, we are

done.

The theorem tells us if we have a finite abelian extension

M/K

, then we

obtain an open finite-index subgroup

M/K

(

)

≤ K

. Conversely, if we are

given an open finite index subgroup of

, we might ask if there is an abelian

extension of

whose norm group is corresponds to this subgroup. The following

theorem tells us this is the case:

Theorem. Let K be a local field. Then there is an isomorphism of posets



open finite index

subgroups of K

 

finite abelian

extensions of L/K



H (K

)

Art

(H)

N(L/K) L/K

In particular, for L/K and M/K finite abelian extensions, we have

N(LM/K) = N(L/K) ∩ N(M/K),

N(L ∩ M/K) = N(L/K)N (M/K).

While proving this requires quite a bit of work, a small part of it follows from

local Artin reciprocity:

Theorem.

Let

L/K

be a finite extension, and

M/K

abelian. Then

(

L/K

)

⊆

N(M/K) iff M ⊆ L.

Proof.

By the previous theorem, we may wlog

L/K

abelian by replacing with

L ∩ K

. The ⇐ direction is clear by the last part of Artin reciprocity.

For the other direction, we assume that we have

(

L/K

)

⊆ N

(

M/K

), and

let

σ ∈ Gal

(

). We want to show that

σ|

. This would then imply

that M is a subfield of L by Galois theory.

We know

(

) is dense in

Gal

(

). So it suffices to show this for

σ ∈ W (K

/L). Then we have

W (K

/L)

∼

Art

(N(L/K)) ⊆ Art

(N(M/K)).

So we can find

x ∈ M

such that

Art

(

M/K

(

)). So

σ|

by local

Artin reciprocity.

Side note: Why is this called “class field theory”? Usually, we call the field

corresponding to the subgroup

the class field of

. Historically, the first type

of theorems like this are proved for number fields. The groups that appear on

the left would be different, but in some cases, they are the class group of the

number field.

8 Lubin–Tate theory

For the rest of the course, we will indicate how one can explicitly construct the

field K

and the map Art

There are many ways we can approach local class field theory. The approach

we use, using Lubin–Tate theory, is the most accessible one. Another possible

approach is via Galois cohomology. This, however, relies on more advanced

machinery, namely Galois cohomology.

8.1 Motivating example

We will work out the details of local Artin reciprocity in the case of

as a

motivating example for the proof we are going to come up with later. Here we

will need the results of local class field theory to justify our claims, but this is

not circular since this is not really part of the proof.

Lemma. Let L/K be a finite abelian extension. Then we have

L/K

= (O

: N

L/K

)).

Proof.

Pick

x ∈ L

, and

the valuation on

extending

, and

= [

Then by construction of w, we know

L/K

(x)) = nw(x) = f

L/K

(x).

So we have a surjection

N(L/K)

L/K

The kernel of this map is equal to

N(L/K)

∼

∩ N (L/K)

L/K

)

So by local class field theory, we know

n = (K

: N(L/K)) = f

L/K

: N

L/K

)),

and this implies what we want.

Corollary.

Let

L/K

be a finite abelian extension. Then

L/K

is unramified if

and only if N

L/K

) = O

Now we fix a uniformizer

. Then we have a topological group isomorphism

∼

hπ

i × O

Since we know that the finite abelian extensions correspond exactly to finite

index subgroups of

by taking the norm groups, we want to understand

subgroups of K

. Now consider the subgroups of K

of the form

hπ

i × U

(n)

We know these form a basis of the topology of

, so it follows that finite-index

open subgroups must contain one of these guys. So we can find the maximal

abelian extension as the union of all fields corresponding to these guys.

Since we know that

(

LM/K

) =

(

L/K

)

∩ N

(

M/K

), it suffices to further

specialize to the cases

hπ

i × U

(n)

and

hπ

i × O

separately. The second case is easy, because this corresponds to an unramified

extension by the above corollary, and unramified extensions are completely

characterized by the extension of the residue field. Note that the norm group

and the extension are both independent of the choice of uniformizer. The

extensions corresponding to the first case are much more difficult to construct,

and they depend on the choice of

. We will get them from Lubin–Tate theory.

Lemma.

Let

be a local field, and let

be the extension corresponding

to hπ

i × O

. Let

L =

[

Then we have

= K

Lemma. We have isomorphisms

W (K

/K)

∼

W (K

L/K)

∼

W (K

/K) × Gal(L/K)

∼

Frob

× Gal(L/K)

Proof.

The first isomorphism follows from the previous lemma. The second

follows from the fact that

∩ L

is totally ramified. The last

isomorphism follows from the fact that

trivially, and then by

definition W (K

ur/K

)

∼

Frob

Example. We consider the special case of K = Q

and π

= p. We let

= Q

(ζ

where

is the primitive

th root of unity. Then by question 6 on example

sheet 2, we know this is a field with norm group

N(Q

(ζ

)/Q

) = hpi × (1 + p

) = hpi × U

(n)

and thus this is a totally ramified extension of Q

We put

(ζ

∞

) =

∞

[

n=1

(ζ

Then again this is totally ramified extension, since it is the nested union of

totally ramified extensions.

Then we have

Gal(Q

(ζ

∞

)/Q

)

∼

lim

←−

Gal(Q

(ζ

)/Q

)

= lim

←−

(Z/p

= Z

Note that we are a bit sloppy in this deduction. While we know that it is true

that

∼

lim

←−

(

Z/p

)

, the inverse limit depends not only on the groups

(

Z/p

)

themselves, but also on the maps we use to connect the groups together.

Fortunately, from the discussion below, we will see that the maps

Gal(Q

(ζ

)/Q

) → Gal(Q

(ζ

n−1

)/Q

)

indeed correspond to the usual restriction maps

(Z/p

→ (Z/p

n−1

It is a fact that this is the inverse of the Artin map of

restricted to

Note that we have

(

∞

)

) =

Gal

(

∞

)

) because its maximal

unramified subextension is trivial.

We can trace through the above chains of isomorphisms to figure out what

the Artin map does. Let m = Z

. Then we can write

m = a

+ a

p + ··· ,

where a

∈ {0, ··· , p − 1} and a

6= 0. Now for each n, we know

m ≡ a

+ a

p + ··· + a

n−1

mod p

By the usual isomorphism Gal(Q

(ζ

)/Q

)

∼

Z/p

Z, we know m acts as

7→ ζ

p+...+a

n−1

“=” ζ

(

), where we abuse notation because taking

to powers of

greater

than

gives 1. It can also be interpreted as (1 +

)

, where

−

1 is

a uniformizer, which makes sense using binomial expansion.

So the above isomorphisms tells us that

Art

restricted to

acts on

(ζ

∞

) as

Art

(m)(ζ

) ≡ σ

−1

(ζ

) = ζ

−1

The full Artin map can then be read off from the following diagram:

W (Q

)

hpi × Z

W (Q

) × Gal(Q

(ζ

∞

)/Q

)

∼

Art

restriction

∼

where the bottom map sends

, mi 7→ (Frob

, σ

−1

In fact, we have

Theorem (Local Kronecker-Weber theorem).

[

n∈Z

≥1

(ζ

[

n∈Z

≥1

(n,p)=1

(ζ

Not a proof.

We will comment on the proof of the generalized version later.

Remark.

There is another normalization of the Artin map which sends a

uniformizer to the geometric Frobenius, defined to be the inverse of the arithmetic

Frobenius. With this convention, Art

(m)|

(ζ

∞

)

is σ

We can define higher ramification groups for general Galois extensions.

Definition

(Higher ramification groups)

Let

be a local field and

L/K

Galois.

We define, for s ∈ R

≥−1

(M/K) = {σ ∈ Gal(M/K) : σ|

∈ G

(L/K) for all finite

Galois subextension M/K}.

This definition makes sense, because the upper number behaves well when

we take quotients. This is one of the advantages of upper numbering. Note that

we can write the ramification group as the inverse limit

(M/K)

∼

lim

←−

L/K

(L/K),

as in the case of the Galois group.

Example.

Going back to the case of

. We write

for the unramified

extension of degree n of Q

. By question 11 of example sheet 3, we know that

(ζ

)/Q

) =











Gal(Q

(ζ

)/Q

) s = −1

Gal(Q

(ζ

)/Q

) −1 < s ≤ 0

Gal(Q

(ζ

)/ζ

) k − 1 < s ≤ k ≤ m − 1

1 s > m − 1

which corresponds to











hpi × U

(0)

i × U

(m)

s = −1

i × U

(0)

i × U

(m)

−1 < s ≤ 0

i × U

(k)

i × U

(m)

k − 1 < s ≤ k ≤ m − 1

1 s > m − 1

under the Artin map.

By taking the limit as n, m → ∞, we get

Theorem. We have

) = Art

(1 + p

) = Art

(k)

where k is chosen such that k − 1 < s ≤ k, k ∈ Z

≥0

Corollary. If L/Q

is a finite abelian extension, then

(L/Q

) = Art



N(L/Q

)(1 + p

)

N(L/Q

)



where n − 1 < s ≤ n.

Here Art

induces an isomorphism

N(L/Q

)

→ Gal(L/Q

So it follows that

L ⊆ Q

(

) for some

if and only if

(

L/Q

) = 1 for all

s > m − 1.

8.2 Formal groups

The proof of local Artin reciprocity will be done by constructing the analogous

versions of

for an arbitrary local field, and then proving that it works. To

do so, we will need the notion of a formal group. The idea of a formal group is

that a formal group is a rule that specifies how we should multiply two elements

via a power series over a ring

. Then if we have a complete

-module, then

the formal group will turn the

-module into an actual group. There is then a

natural notion of a formal module, which is a formal group

with an

-action.

At the end, we will pick

. The idea is then that we can fix an

algebraic closure

, and then a formal

-module will turn

into an actual

-module. Then if we adjoin the right elements of

, then we obtain

an extension of

with a natural

action, and we can hope that this restricts

to field automorphisms when we restrict to the unit group.

Notation. Let R be a ring. We write

R[[x

, ··· , x

]] =







,...,k

∈Z

≥0

,...,k

···x

: a

,...,k

∈ R







for the ring of formal power series in n variables over R.

Definition

(Formal group)

A (one-dimensional, commutative) formal group

over R is a power series F (X, Y ) ∈ R[X, Y ] such that

(i) F (X, Y ) ≡ X + Y mod (X

, XY, Y

)

(ii) Commutativity: F (X, Y ) = F (Y, X)

(iii) Associativity: F (X, F (Y, Z)) = F (F (X, Y ), Z).

This is most naturally understood from the point of view of algebraic geometry,

as a generalization of the Lie algebra over a Lie group. Instead of talking about

the tangent space of a group (the “first-order neighbourhood”), we talk about its

infinitesimal (formal) neighbourhood, which contains all higher-order information.

A lot of the seemingly-arbitrary compatibility conditions we later impose have

such geometric motivation that we unfortunately cannot go into.

Example.

is a formal group over

, where

is a complete valued field,

then

(

x, y

) converges for all

x, y ∈ m

. So

becomes a (semi)group under

the multiplication

(x, y) 7→ F (x, y) ∈ m

Example. We can define

(X, Y ) = X + Y.

This is called the formal additive group.

Similarly, we can have

(X, Y ) = X + Y + XY.

This is called the formal multiplicative group. Note that

X + Y + XY = (1 + X)(1 + Y ) − 1.

So if

is a complete valued field, then

bijects with 1 +

by sending

x 7→

1 +

, and the rule sending (

x, y

)

∈ m

7→ x

xy ∈ m

is just the

usual multiplication in 1 + m

transported to m

via the bijection above.

We can think of this as looking at the group in a neighbourhood of the

identity 1.

Note that we called this a formal group, rather than a formal semi-group. It

turns out that the existence of identity and inverses is automatic.

Lemma. Let R be a ring and F a formal group over R. Then

F (X, 0) = X.

Also, there exists a power series i(X) ∈ X · R[[X]] such that

F (X, i(X)) = 0.

Proof. See example sheet 4.

The next thing to do is to define homomorphisms of formal groups.

Definition

(Homomorphism of formal groups)

Let

be a ring, and

F, G

formal groups over

. A homomorphism

F → G

is an element

f ∈ R

[[

]]

such that f (X) ≡ 0 mod X and

f(F (X, Y )) = G(f(X), f(Y )).

The endomorphisms

F → F

form a ring

End

(

) with addition +

given by

(f +

g)(x) = F (f(x), g(x)).

and multiplication is given by composition.

We can now define a formal module in the usual way, plus some compatibility

conditions.

Definition

(Formal module)

Let

be a ring. A formal

-module is a formal

group

over

with a ring homomorphism

R → End

(

), written,

a 7→

[

]

such that

[a]

(X) = aX mod X

Those were all general definitions. We now restrict to the case we really care

about. Let K be a local field, and q = |k

|. We let π ∈ O

be a uniformizer.

Definition

(Lubin–Tate module)

A Lubin–Tate module over

with respect

to π is a formal O

-module F such that

[π]

(X) ≡ X

mod π.

We can think of this condition of saying “uniformizer corresponds to the

Frobenius”.

Example.

The formal group

is a Lubin–Tate

module with respect to

given by the following formula: if a ∈ Z

, then we define

[a]

(X) = (1 + X)

− 1 =

∞

n=1





The conditions

(1 + X)

− 1 ≡ aX mod X

and

(1 + X)

− 1 ≡ X

mod p

are clear.

We also have to check that

a 7→

[

]

is a ring homomorphism. This follows

from the identities

((1 + X)

)

= (1 + X)

, (1 + X)

(1 + X)

= (1 + X)

which are on the second example sheet.

The objective of the remainder of the section is to show that all Lubin–Tate

modules are isomorphic.

Definition

(Lubin–Tate series)

A Lubin–Tate series for

is a power series

e(X) ∈ O

[[X]] such that

e(X) ≡ πX mod X

, e(X) ≡ X

mod π.

We denote the set of Lubin–Tate series for π by E

Now by definition, if

is a Lubin–Tate

module for

, then [

]

is a

Lubin–Tate series for π.

Definition

(Lubin–Tate polynomial)

A Lubin–Tate polynomial is a polynomial

of the form

+ π(a

q−1

+ ··· + a

) + πX

with u ∈ U

(1)

, and a

q−1

, ··· , a

∈ O

In particular, these are Lubin–Tate series.

Example. X

+ πX is a Lubin–Tate polynomial.

Example.

and

, then (1 +

)

−

1 is a Lubin–Tate polynomial.

The result that allows us to prove that all Lubin–Tate modules are isomorphic

is the following general result:

Lemma. Let e

, e

∈ E

and take a linear form

L(x

, ··· , x

) =

i=1

, a

∈ O

Then there is a unique power series

(

, ··· , x

)

∈ O

[[

, ··· , x

]] such that

F (x

, ··· , x

) ≡ L(x

, ··· , x

) mod (x

, ··· , x

)

and

(F (x

, ··· , x

)) = F (e

), e

), ··· , e

)).

For reasons of time, we will not prove this. We just build

by successive

approximation, which is not terribly enlightening.

Corollary.

Let

e ∈ E

be a Lubin–Tate series. Then there are unique power

series F

(X, Y ) ∈ O

[[X, Y ]] such that

(X, Y ) ≡ X + Y mod (X + Y )

e(F

(X, Y )) = F

(e(X), e(Y ))

Corollary.

Let

, e

∈ E

be Lubin–Tate series and

a ∈ O

. Then there exists

a unique power series [a]

∈ O

[[X]] such that

[a]

(X) ≡ aX mod X

([a]

(X)) = [a]

(X)).

To simplify notation, if e

= e

= e, we just write [a]

= [a]

e,e

We now state the theorem that classifies all Lubin–Tate modules in terms of

Lubin–Tate series.

Theorem.

The Lubin–Tate

modules for

are precisely the series

for

e ∈ E

with formal O

-module structure given by

a 7→ [a]

Moreover, if

, e

∈ E

and

a ∈ O

, then [

]

is a homomorphism from

→ F

If a ∈ O

, then it is an isomorphism with inverse [a

−1

]

So in some sense, there is only one Lubin–Tate module.

Proof sketch.

is a Lubin–Tate

-module for

, then

= [

]

∈ E

definition, and

satisfies the properties that characterize the series

. So

F = F

by uniqueness.

For the remaining parts, one has to verify the following for all

e, e

, e

∈ E

and a, b ∈ O

(i) F

(X, Y ) = F

(Y, X).

(ii) F

(X, F

(Y, Z)) = F

(X, Y ), Z).

(iii) [a]

(X, Y )) = F

([a]

(X), [a]

(Y )).

(iv) [ab]

(X) = [a]

([b]

(X)).

(v) [a + b]

(X) = [a]

(X) + [b]

(X).

(vi) [π]

(X) = e(X).

The proof is just repeating the word “uniqueness” ten times.

8.3 Lubin–Tate extensions

We now use the Lubin–Tate modules to do things. As before, we fixed an

algebraic closure

K of K. We let

m = m

be the maximal ideal in O

Proposition.

is a formal

-module, then

becomes a (genuine)

module under the operations +

and ·

x +

y = F (x, y)

a · x = [a]

(x)

for all x, y ∈

m and a ∈ O

We denote this

This isn’t exactly immediate, because

need not be complete. However,

this is not a problem as each multiplication given by

only involves finitely

many things (namely two of them).

Proof.

x, y ∈

, then

(

x, y

) is a series in

(

x, y

)

⊆

. Since

(

x, y

) is

a finite extension, we know it is complete. Since the terms in the sum have

absolute value

1 and

→

0, we know it converges to an element in

K(x,y)

⊆

The rest then essentially follows from definition.

To prove local class field theory, we want to find elements with an

(n)

action for each

, or equivalently elements with an

(n)

action. Note that

the first quotient is a quotient of groups, while the second quotient is a quotient

of a ring by an ideal. So it is natural to consider the following elements:

Definition

(

-division points)

Let

be a Lubin–Tate

-module for

. Let

n ≥ 1. The group F (n) of π

-division points of F is defined to be

F (n) = {x ∈

| [π

]

x = 0} = ker([π

]

This is a group under the operation given by F , and is indeed an O

module.

Example. Let F =

, K = Q

and π = p. Then for x ∈

, we have

· x = (1 + x)

− 1.

So we know

(n) = {ζ

− 1 | i = 0, 1, ··· , p

− 1},

where ζ

∈

is the primitive p

th root of unity.

(n) generates Q

(ζ

To prove this does what we want, we need the following lemma:

Lemma. Let e(X) = X

+ πX. We let

(X) = (e ◦ ··· ◦ e)

| {z }

n times

(X).

Then f

has no repeated roots. Here we take f

to be the identity function.

Proof.

Let

x ∈

. We claim that if

(

)

| <

1 for

= 0

, ··· , n −

1, then

(X) 6= 0.

We proceed by induction on n.

(i) When n = 1, we assume |x| < 1. Then

(x) = e

(x) = qx

q−1

+ π = π



1 +

q−1



6= 0,

since we know

has absolute value

≤

1 (

vanishes in

, so

q/π

lives in

), and x

q−1

has absolute value < 1.

(ii) in the induction step, we have

n+1

(x) = (qf

(x)

q−1

+ π)f

(x) = π



1 +

(x)

q−1



(x).

By induction hypothesis, we know

(

)

= 0, and by assumption

(

)

| <

1. So the same argument works.

We now prove the lemma. We assume that

(

) = 0. We want to show that

(x)| < 1 for all i = 0, ··· , n −1. By induction, we have

(x) = x

+ πg

(x)

for some

(

)

∈ O

[

]. It follows that if

(

) = 0, then

|x| <

1. So

(

)

| <

for all i. So f

(x) 6= 0.

The point of the lemma is to prove the following proposition:

Proposition. F

(

) is a free

/π

module of rank 1. In particular, it has

elements.

Proof. By definition, we know

· F (n) = 0.

So F (n) is indeed an O

/π

-module.

To prove that it is free of rank 1, we note that all Lubin–Tate modules

for

are isomorphic. This implies that all the honest

modules

(

) are

isomorphic. We choose

, where

πX

. Then

(

) consists

of the roots of the polynomial

(

), which is of degree

and has no

repeated roots. So

(

)

. To show that it is actually the right thing, if

∈ F (n) \ F (n −1), then we have a homomorphism

→ F (n)

given by

A 7→ a · λ

. Its kernel is

by our choice of

. By counting, we

get an O

-module isomorphism

→ F (n)

as desired.

Corollary. We have isomorphisms

∼

End

(F (n))

(n)

∼

Aut

(F (n)).

Given a Lubin–Tate O

-module F for π, we consider

n,π

= L

= K(F(n)),

which is the field of

division points of

. From the inclusions

(

)

⊆ F

(

+1)

for all n, we obtain a corresponding inclusion of fields

⊆ L

n+1

The field

depends only in

, and not on

. To see this, we let

be another

Lubin–Tate O

-module, and let f : F → G be an isomorphism. Then

G(n) = f(F (n)) ⊆ K(F(n))

since the coefficients of f lie in K. So we know

K(G(n)) ⊆ K(F (n)).

By symmetry, we must have equality.

Theorem. L

is a totally ramified abelian extension of degree

n−1

(

q −

with Galois group

Gal(L

/K)

∼

Aut

(F (n))

∼

(n)

Explicitly, for any σ ∈ Gal(L

/K), there is a unique u ∈ U

(n)

such that

σ(λ) = [u]

(λ)

for all λ ∈ F (n). Under this isomorphism, for m ≥ n, we have

Gal(L

)

∼

(n)

(m)

Moreover, if F = F

, where

e(X) = X

+ π(a

q−1

+ ··· + a

) + πX,

and λ

∈ F (n) \ F (n −1), then λ

is a uniformizer of L

and

(X) =

(X)

n−1

(X)

= X

n−1

(q−1)

+ ··· + π

is the minimal polynomial of λ

. In particular,

(−λ

) = π.

Proof. Consider a Lubin–Tate polynomial

e(X) = x

+ π(a

q−1

+ ··· + a

) + πX.

We set F = F

. Then

(X) =

(X)

n−1

(X)

= (e

n−1

(X))

q−1

+ π(a

n−1

(X)

q−2

+ ··· + a

n−1

(X)) + π

is an Eisenstein polynomial of degree

n−1

(

q −

1) by starting at it long enough.

So if

∈ F

(

)

\ F

(

n −

1), then

is a root of

(

), so

(

)

is totally

ramified of degree q

n−1

(q − 1), and λ

is a uniformizer, and

K(λ

)/K

(−λ

) = π

as the norm is just the constant coefficient of the minimal polynomial.

Now let

σ ∈ Gal

(

). Then

induces a permutation of

(

), as these

are the roots of e

(X), which is in fact O

-linear, i.e.

σ(x) +

σ(y) = F(σ(x), σ(y)) = σ(F (x, y)) = σ(x +

σ(a · x) = σ([a]

(x)) = [a]

(σ(x)) = a · σ(x)

for all x, y ∈ m

and a ∈ O

So we have an injection of groups

Gal(L

/K) → Aut

(F (n)) =

(n)

But we know



(n)



= q

n−1

(q − 1) = [K(λ

) : K] ≤ [L

: K] = |Gal(L

/K)|.

So we must have equality throughout, the above map is an isomorphism, and

K(λ

) = L

It is clear from the construction of the isomorphism that for

m ≥ n

, the

diagram

Gal(L

/K) U

(m)

Gal(L

/K) U

(n)

∼

restriction

quotient

∼

commutes. So the isomorphism

Gal(L

)

∼

(m)

(n)

follows by looking at the kernels.

Example. In the case where K = Q

and π = p, recall that

(n) = {ζ

− 1 | i = 0, ··· , p

n−1

− 1},

where ζ

is the principal p

th root of unity. The theorem then gives

Gal(Q

(ζ

)/Q

)

∼

(Z/p

)

given by if a ∈ Z

≥0

and (a, p) = 1, then

(ζ

− 1) = [a]

(n)

(ζ

− 1) = (1 + (ζ

− 1))

− 1 = ζ

− 1.

This agrees with the isomorphism we previously constructed.

Back to the general situation, setting

∞

[

n=1

we know L

∞

/K is Galois, and we have isomorphisms

Gal(L

∞

/K) lim

←−

Gal(L

/K) lim

←−

(n)

∼

σ (σ|

)

∼ ∼

This map will be the inverse of the Artin map restricted to L

∞

To complete the proof of Artin reciprocity, we need to use the following

theorem without proof:

Theorem (Generalized local Kronecker-Weber theorem). We have

= K

∞

(for any π).

Comments on the proof.

One can prove this from the Hasse-Arf theorem, which

states that in an abelian extension, the jumps in the upper ramification groups

occur only at integer values. This, together with the calculation of ramification

groups done later, easily implies the theorem. Essentially,

∞

maxed out all

possible jumps of the upper ramification groups. However, the Hasse-Arf theorem

is difficult to prove.

Another approach is to prove the existence of the Artin map using other

techniques (e.g. Galois cohomology). Consideration of the norm group (cf. the

next theorem) then implies the theorem. The content of this section then

becomes an explicit construction of a certain family of abelian extensions.

We can characterize the norm group by

Theorem. We have

N(L

/K) = hπi × U

(n)

Comments on the proof.

This can be done by defining Coleman operators, which

are power series representations of the norm. Alternatively, assuming the

description of the local Artin map given below and local Artin reciprocity,

(n)

is in the kernel of

Art|

, so

hπi × U

(n)

⊆ N

(

). The result follows by

comparing order.

We can then construct the Artin map as follows:

Theorem.

Let

be a local field. Then we have an isomorphism

Art

→

W (K

/K) given by the composition

W (K

/K)

hπi× U

Frob

× Gal(L

∞

/K)

∼

Art

∼

where the bottom map is given by (π

, u) 7→ (Frob

, σ

−1

), where

(λ) = [u]

(λ)

for all λ ∈

∞

n=1

F (n).

The inverse shows up in the proof to make sure the map defined above is

independent of the choice of uniformizer. We will not prove this, nor that the

map obtained has the desired properties. Instead, we will end the course by

computing the higher ramification groups of these extensions.

Theorem. We have

/K) =











Gal(L

/K) −1 ≤ s ≤ 0

Gal(L

) q

k−1

− 1 < s ≤ q

− 1, 1 ≤ k ≤ n − 1

1 s > q

n−1

Proof. The case for −1 ≤ s ≤ 0 is clear.

For 0 ≤ s ≤ 1 (which we may wlog is actually 1), we know that

Gal(L

)

∼

(k)

(n)

under the isomorphism

Gal

(

)

∼

(n)

. On the other hand, we know

/K) is the Sylow p-subgroup of Gal(L

/K). So we must have

/K)

∼

(1)

(n)

So we know that

(

) =

Gal

(

). Thus we know that

(

) =

Gal(L

/K) for 0 < s ≤ 1.

We now let σ = σ

∈ G

/K) and u ∈ U

(1)

(n)

. We write

u = 1 + επ

for some

ε ∈ U

and some

(

)

≥

1. Since

is not the identity, we know

k < n. We claim that

(σ) = v

(σ(λ) − λ) = q

Indeed, we let

λ ∈ F

(

)

\ F

(

n −

1), where

is a choice of Lubin–Tate module

for π. Then λ is a uniformizer of L

and O

= O

[λ]. We can compute

(λ) = [u]

(λ)

= [1 + επ

]

(λ)

= F (λ, [επ

]

(λ))

Now we can write

[επ

]

(λ) = [ε]

([π

]

(λ)) ∈ F (n − k) \ F (n − k − 1),

since [

]

is invertible, and applying [

n−k

]

to [

]

(

) kills it, but applying

[π

n−k−1

]

gives [π

n−1

]

, which does not kill.

So we know [

επ

]

(

) is a uniformizer of

n−k

. Since

n−k

is totally

ramified of degree q

, we can find ε

∈ O

such that

[επ

]

(λ) = ε

Recall that F (X, 0) = X and F (0, Y ) = Y . So we can write

F (X, Y ) = X + Y + XY G(X, Y ),

where G(X, Y ) ∈ O

[[X, Y ]]. So we have

σ(λ) − λ = F (λ, [επ

]

(λ)) − λ

= F (λ, ε

) − λ

= λ + ε

+ ε

G(λ, ε

) − λ

= ε

+ ε

G(λ, ε

In terms of valuation, the first term is the dominating term, and

(σ) = v

(σ(λ) − λ) = q

So we know

(σ

) ≥ s + 1 ⇔ q

k(u)

− 1 ≥ s.

So we know

/K) = {σ

∈ G

/K) : q

k(u)

− 1 ≥ s} = Gal(L

where q

k−1

− 1 < s ≤ q

− 1 for k = 1, ··· , n − 1, and 1 if s > q

n−1

= 1.

Corollary. We have

/K) =











Gal(L

/K) −1 ≤ t ≤ 0

Gal(L

) k − 1 < t ≤ k, k = 1, ··· , n − 1

1 t > n −1

In other words, we have

/K) =

(

Gal(L

dte

) −1 ≤ t ≤ n − 1

1 t > n − 1

where we set L

= K.

Once again, the numbering is a bit more civilized in the upper numbering.

Proof. We have to compute the integral of

/K) : G

/K)

We again plot this out

q−1

q(q−1)

(q−1)

q − 1

− 1 q

− 1

So by the same computation as the ones we did last time, we find that

(s) =











s −1 ≤ s ≤ 0

(k − 1) +

s−(q

k−1

−1)

k−1

(q−1)

k−1

− 1 ≤ s ≤ q

− 1, k = 1, ··· , n − 1

(n − 1) +

s−(q

n−1

−1)

n−1

(q−1)

s > q

n−1

− 1.

Inverting this, we find that











t −1 ≤ t ≤ 0

dte−1

(q − 1)(t − (dte − 1)) + q

dte−1

− 1 1 < t ≤ n − 1

n−1

(q − 1)(t − (n − 1)) + q

n−1

− 1 t > n − 1

Then we have

/K) = G

ψ(L

/K)(t)

/K),

which gives the desired by the previous theorem.

So we know that

Art

−1

/K)) =

(

dte

(n)

−1 ≤ t ≤ n

1 t ≥ n

Corollary. When t > −1, we have

ab/K

) = Gal(K

dte

and

Art

−1

/K)) = U

(dte)

Proof.

Recall the following fact from the examples class: If

L/K

is finite un-

ramified and

M/K

is finite totally ramified, then

LM/L

is totally ramified, and

Gal(LM/L)

∼

Gal(M/K) by restriction, and

(LM/K)

∼

(M/K).

via this isomorphism (for t > −1).

Now let

be the unramified extension of degree

. By the lemma and

the previous corollary, we have

/K)

∼

/K) =

(

Gal(L

dte

) −1 < t ≤ n

1 t ≥ n

(

Gal(K

dte

) −1 < t ≤ n

1 t ≥ n

So we have

/K) = G

∞

/K)

= lim

←−

m,n

/K)

= lim

←−

m,n

n≥dte

Gal(K

dte

)

= Gal(K

∞

dte

)

= Gal(K

dte

and

Art

−1

(Gal(K

dte

)) = Art

−1







lim

←−

m,n

n≥dte

Gal(K

dte

)







= lim

←−

m,n

n≥dte

Art

−1



Gal(K

dte

)



= lim

←−

m,n

n≥dte

(dte)

(n)

= U

dte

Corollary.

Let

M/K

be a finite abelian extension. Then we have an isomor-

phism

Art

N(M/K)

∼

Gal(M/K).

Moreover, for t > −1, we have

(M/K) = Art

(dte)

N(M/K)

Proof. We have

(M/K) =

/K)G(K

/M )

G(K

/M )

= Art

(dte)

N(M/K)