These are my (slightly reorganized) live-TeXed notes for the course *Math 223b: Algebraic Number Theory* taught by Joe Rabinoff at Harvard, Spring 2012. This course is a continuation of Math 223a. Please let me know if you notice any errors or have any comments!

01/28/2013

##
Overview

Suppose is a global field or a local field. Denote or respectively. Recall that the main theorems of class field theory states the following.

(local-global compatibility)
Suppose

is a place of a global field

, then the following diagram commutes

In particular, if

is finite abelian , then

kills

if and only if

is unramified, in which case it sends

to

. This determines

by continuity.

We will prove the main theorems of local and global class field theory in the first part of this semester (as sketched at the end of last semester). In the remaining part, our additional topics may include central simple algebras and quaternion algebras, Lubin-Tate formal groups (explicit local class field theory), CM elliptic curves (explicit class field theory for imaginary quadratic fields), (possibly) Drinfeld modules (explicit class field theory for global function fields), local and global Tate duality theorems (c.f., Serre's *Galois cohomology* [1]) and (possibly) Langlands classification of representations of tori.

01/30/2013

##
-modules

We will start with basics on group cohomology (c.f., Serre's *local fields* [2]).

Let

be any group. A

*left (right) -module* is an abelian group

equipped with a left (right) action of

, i.e., a homomorphism

.

If

is a Galois group of the field extension

, then

and

are both

-modules.

Denote the free abelian group on the elements of

by

(called the

*group ring*). Under the left (right) action of

on

by

(

),

becomes a left (right)

-module.

A

*-module homomorphism* is a

-module homomorphism. The category of

-modules, denoted by

, is an abelian category. For

, the set of

-module homomorphisms between them is denoted by

.

A

-module

is called

*injective* (projective) if the functor

(

) is exact.

For

, we endow the

-module structure on

by

and the

-module structure on

by

(c.f., Remark

18).

For

and

any abelian group, we have

. We say

is

*induced* if

is isomorphic to

as a

-module. A direct summand of an induced module is called

*relatively projective*.

Similarly,

is called

*co-induced*. A direct summand of a co-induced module is called

*relatively injective*.

(Frobenius reciprocity)
and

. Then

(Frobenius reciprocity)
Suppose

and

. Then

02/01/2013

##
Group cohomology and homology

Let

be a

-module. We define the

*invariants* .

Let

be a

-module. The

*co-invariants* is defined to be the largest quotient on which

acts trivially, i.e,

.

The group cohomology (resp. homology) of measures the failure of the right (resp. left) exactness of taking -invariants (resp. co-invariants).

The

*group cohomology* functor

is defined to be the right derived functor of

. The

*group homology* functor

is defined to be the left derived functor of

.

The group cohomology (homology) is a cohomological (homological) functor satisfying the following basic properties.

- and are independent of the choice of the resolutions (up to canonical isomorphisms).
- and are covariant functors in .
- and .
- Any short exact sequence of -modules induces long exact sequences in group cohomology and homology. The connecting maps are natural.

##
Behavior under induction

- If is relatively injective, then for .
- If is relatively projective, then for .

Suppose

is relatively injective, then

for some

, where

is injective. Thus

for

. Similarly for the other part.

¡õ
##
Cochains and chains

Notice that , so is simply another name for . Similarly, is simply the other name for . It is a general principle that one can compute these groups via resolutions in *both* variables. In our cases, we can start with a projective resolution of the trivial -module . Finding such an explicit resolution will give us an concrete description of cohomology and homology groups in terms of cochains and chains.

Let

. Then the diagonal morphism makes

a

-module. Define the boundary map

by

and

be the degree map. The complex

is exact and

is free (hence projective). This projective resolution of

is called the

*standard resolution*.

Thus . Concretely, we have Such an element is called a *homogeneous cochain*. The boundary map is concretely

A homogeneous cochain

is called a

*cocycle* if

and a

*coboundary* if

for some

. Then

is the quotient group of the

-cocycles by the

-coboundaries.

02/04/2013

Notice that is uniquely determined on its values at . But inhomogeneous cochains are defined slightly differently as follows to make the boundary maps look nicer.

We define the an inhomogeneous cochain to be

Then

So

can be identified with

*inhomogeneous cochains*, denoted by

.

A 0-coboundary is of the form

. So it verifies that

.

A 1-coboundary is of the form

. A function

such that

(i.e.

) is called a

*twisted homomorphism*. So we can identify set of 1-coboundaries as the set of twisted homomorphisms. In particular, when

acts on

trivially,

(there is no "twisting").

Analogously, homology can be computed using homogeneous and inhomogeneous chains. We start with a *right* -module projective resolution of , where with the right -action Choose the coset representatives of , Then and can be identified with functions such that , and has finite support modulo (which is the same as ). Such an element is called a *homogeneous chain*. Similarly we can define inhomogeneous chains and the chain complex of inhomogeneous chains computes . We have a similar formula for boundary maps (adding one index instead of deleting):

Suppose

is a 1-chain, then

Hence

and it verifies that

.

.

Let

be the argumentation ideal of

. Then

. The long exact sequence of group homology implies that

is exact. Notice that

(as

is free) and

. Hence

. On the other hand,

In fact,

is easily seen to be a group homomorphism, hence factors through

. Conversely,

is a

-module freely generated on

. The map

and

gives the inverse.

¡õ
##
Change of groups

###
Restriction and corestriction

Let

be a homomorphism. Then

is also a

-functor (i.e. a short exact sequence gives a long exact sequence). Since

is universal

*repelling* (derived functors are universal

-functors), we obtain a morphism of

-functors

Alternatively, we even have a morphism on the level of cochains

via composing with

.

02/06/2013

Here is a slightly more general construction. Suppose

and

. We say

is a homomorphism

*compatible with * if

for any

, equivalently,

is a

-module homomorphism. Composing

with

gives a functor

. When

is a subgroup of

, we call this functor the

*restriction functor*, denoted by

Similarly, since homology is a universal

*attracting* -functor,

induces a functor

On the level of chain complex, this is given by

.

Prove the above map is a map of chain complexes and it induces

.

Similarly, suppose

is a homomorphism compatible with

, then composing with

gives a functor

. When

is a subgroup of

, we call this functor the

*corestriction functor*, denoted by

Use Shapiro's Lemma (Remark

24) and the natural map

(Remark

13) to construct the restriction functor

. Do the same for the corestriction functor using the natural map

(Remark

10).

We define the

*norm* .

Similarly, we can compute the effect of on degree 0 : where is sent to , which is identified with .

We define the

*conorm* ,

. Then

on homology is the unique extension of the norm

to higher degrees.

The restriction on homology

is compatible with the isomorphism

.

###
Inflation and coinflation

Suppose

is a

*normal* subgroup of

. Then

is again a

-module and

acts trivially on it, hence

is a

-module. Suppose

is compatible with the quotient map

. Then

and

induce the

*inflation functor* and

*coinflation functor*
Suppose

and

, for some

. Then

is compatible with

and they induce an automorphism

. We claim that this map is the identity. In fact, since this is a morphism of

-functors, we only need to check it on the degree 0 part ("dimension-shifting" argument), in which case

is the identity.

##
(Co)inflation-(co)restriction exact sequence

Suppose is a normal subgroup of and .

(Inflation-restriction exact sequence)
- The sequence is exact.
- If for , then is exact. Moreover, is an isomorphism for .

- First we show the injectivity. Say is a cocycles such that is a coboundary. We need to show that it self is a coboundary. Suppose . Then . But , we find that , hence for any , i.e. . It follows that is a coboundary. Next let us show that composition is zero. Suppose is a cocycle. Then which is clearly zero. Finally let us show the exactness in the middle. Suppose is a coboundary in . Then there exists such that for any . Subtracting by the coboundary , we may assume . Now for any , we know that , i.e. factors through . On the other hand, since is normal, for some , hence , namely, . Thus factors through .
- Induction on . Suppose . Choose injective such that . Notice so . It follows that is an injective -module as preserves injectives (Remark 22). Since we assume , we obtain a short exact sequence By the assumption for . We obtain the following diagram The induction hypothesis implies the exactness of the first row, hence the second row is exact as desired.
¡õ

Prove the analogous theorem on homology. For example, the sequence

is exact.

02/11/2013

on

and

. In particular, if

, then

kills

.

By dimension shifting, it suffices to check on degree zero, e.g., for cohomology, the composition is given by

.

¡õ
##
Tate cohomology

From now on we assume that is a finite group. A phenomenon unique to finite groups is that the group homology can be also understood as group cohomology — the Tate cohomology as we shall define.

Define

to be the

*absolute norm*.

Let . Notice that descends to a map . Hence we obtain a map which is functorial in .

We define

and

One can check directly that Shapiro's lemma (Remark 24) also holds for these two groups:

Let

be a subgroup of

and

, Then

and

Let us prove the first identity (similarly for the second). By Shapiro's lemma (Remark

24), there is an isomorphism

induced by

. By the definition of

, it remains to show that

. On the one hand, for

, we have

On the other hand, for any

, if we let

then

Thus

as desired.

¡õ
If

is relatively injective (equivalently, relatively projective since

is finite), then

.

We may assume

is (co-)induced since

and

commutes with direct sum. Then the result follows from the previous proposition (c.f., Corollary

1).

¡õ
We define the

*Tate cohomology group* , for

,

and

for

. So

(

) form a

-functor (infinite in both directions).

02/13/2013

. In particular,

is killed by

(take

).

The same argument as in Proposition

5.

¡õ
If

is a finitely generated abelian group, then

is finite.

Notice that if

is finitely generated and

is finite, then

is a finitely generated abelian group, hence

is finitely generated. But

is also torsion by the previous proposition.

¡õ
##
Tate cohomology via complete resolution

A more conceptual way to understand the Tate cohomology is via complete resolution. For a finite group , the existence of a complete resolution boils down to the nice duality properties of -modules (c.f., Cassels-Frohlich [3] and Brown [4]).

Let us start with a discussion of the linear duality of -modules. Let be a finite group and be a finitely generated left -module.

We define

with the left

-module structure

(c.f., Definition

4).

Taking

, then

has a dual basis

, where

. The action of

on

is given by

Hence

as left

-modules. Consequently, if

is a free

-module, then so is

.

Now we are in position to construct the *complete resolution* (i.e. a -module resolution of which extends in both directions). Let be a resolution by finite free -modules (e.g. the standard resolution). Taking dual gives another resolution by finite free -modules Write for , then gluing the above two resolutions together gives a complete resolution

As we expected, the complete resolution computes all Tate cohomology groups.

is an isomorphism of

-functors.

For

, this is true by definition. For

, this is true since

by Remark

41. The remaining cases

can be checked by hand.

¡õ
##
Cup products

We first give an axiomatic description of cup products, its resemblance to the usual cup product (e.g., in singular cohomology) should not surprise you too much.

Some basic properties of cup products are in order before we prove the existence.

- .
- .
- Suppose is a subgroup of , then .
- Suppose is a subgroup of , .

It suffices to check on degree 0 by dimension shifting. (a), (b) and (c) then follow immediately. For (d), it reduces to the fact that

, for

and

.

¡õ
02/15/2013

To prove the existence of cup products, we will construct a family of -module homomorphisms for a complete resolution of as in the following proposition.

Suppose there is a family of

-module homomorphisms

satisfying:

- .
- (this gives a chain complex map ).

Then such a family of -module homomorphisms is enough to construct the cup product in Theorem 4.

First, suppose

,

. We define

Then one can check that

It follows easily that the

thus defined indeed descends to the level of Tate cohomology.

Second, we need to check this construction actually satisfies the axioms in Theorem 4. This construct is certainly functorial in and and Axiom (b) follows from Requirement (a) in Proposition 10. For Axiom (c), since 's are free, the diagram has exact rows. Now one can check that by diagram-chasing the connecting homomorphisms. Similarly for Axiom (d).
¡õ

Now we use the standard complete resolution to construct a family of (see the previous section). There are six cases depending on the signs of the degrees.

- For , we define .
- For , we define .
- For , we define

02/20/2013

##
Cohomology of finite cyclic groups

The general strategy of computing cohomology of a finite group is by reducing to computing the cohomology of its Sylow -groups. By filtering these -groups, the problem reduces to the case of finite cyclic -groups. In this section we shall discuss the general theory of cohomology of finite cyclic groups.

Suppose is a finite cyclic group of order . Then its argumentation ideal is generated by as a -module, where is any generator of .

The long exact sequence of Tate cohomology thus retracts to an *exact hexagon* due to the above periodicity.

A short exact sequence of

-module

gives an exact diagram

Suppose

is a finite cyclic group and

is a

-module, we define the

*Herbrand quotient* of

to be

if it makes sense.

Suppose

is a short exact sequence of

-modules. If two of

,

and

exist, then the third also exists and

.

It follows easily from the exact hexagon (Corollary

4).

¡õ
If the

-module

is finite (as a set), then

.

Since the number of elements is always multiplicative in exact sequence, looking at the exact sequence

we know that

. The same trick applies to the exact sequence

and gives the desired result.

¡õ
Suppose there is a

-module homomorphism

with finite kernel and cokernel. Then

(meaning that if one of them exists, then the other also exists).

Applying the previous two propositions to the two short exact sequences

implies what we want.

¡õ
Now suppose is cyclic of prime order . The structure theory of -modules in this case is not so complicated and we are going to classify them completely.

The

*trivial Herbrand quotient* of

is defined to be the Herbrand quotient of

regarded trivial

-action, i.e.,

if it exists (notice if

acts trivially on

, then

and

).

The goal in remaining of the this section is to prove the following formula of the Herbrand quotient.

Let

be a cyclic group of prime order

and

be a

-module. Suppose

exists, then

,

,

all exist and

Suppose

is exact and

are defined. If Proposition

14 is true for

, then it is also true for

.

Notice

and

are both defined. It remains to show that

and

are defined. The long exact sequence in group cohomology gives

where

is the image of

in

. By the assumption that

is defined, we know

is finite, so

is also finite and

by Proposition

12. Hence

by multiplicativity. The same argument for

.

¡õ
To prove Proposition 14, we can filter the -module and apply the above lemma.

Since

is finite, there exists

finitely generated over

such that

. Replacing

by

(still finitely generated over

since

is finite), we obtain a

-submodule

such that

. Write

. Applying the snake lemma to multiplication by

gives a short exact sequences

Since

is surjective by construction, we know that

, i.e.,

is

-divisible. Since

is finitely generated over

, we know that

and

are both finite and

is defined, so

is also defined.

¡õ
(Proof of Proposition 14)
By the previous two lemmas, we may assume

is finitely generated over

or

is

-divisible.

First assume is finitely generated over . Suppose and are two -modules finitely generated over and as -modules. Let . Then is a -stable lattice in . For , the inclusion has finite cokernel. Hence and for . In other words, we have proved that if is a finite dimensional -module, and is a -stable lattice, then and do not depend on the choice of . By Proposition 12, and do not depend on the choice of either. So we reduce to the case of finite dimensional -representations by replacing with .

02/22/2013

By filtering the finite dimensional -representation, it suffices to prove the formula when is simple. As a ring and . So any -module is a direct product of a -module and a -module. In particular, has exactly two simple modules: (the trivial representation) and (the representation of dimension ). For the first case (with a lattice inside it), then formula is obvious. For the second case, we identify , where via . Let be a -stable lattice inside . Then The shape of the desired formula essentially boils down to this computation.

It remains to treat the case where is -divisible. From the exact sequence and the fact and , the snake lemma tells us that and . In particular, is an isomorphism and is an isomorphism by functoriality. But kills by Proposition 7, hence and . We now have reduced to the case with . Recall that there is a duality between such discrete torsion abelian groups and finitely generated free -modules given by taking the Pontryagin dual . This duality further hold on the -module level. One can check that and dualizes to . Moreover, In particular, So it remains to prove the formula for a finitely generated free -module. The exactly same argument for finitely generated -modules works using the fact that , where has degree over .
¡õ

03/15/2013

The notes are incomplete at this point because I was out of town for AWS 2013. Meanwhile we did two essential inputs for the proof of class field theory: the study of cohomology of finite groups which leads to the proof of Tate-Nakayama (c.f., Chap VII-IX in [1]), and a detailed analysis of ramification and norm which allows one show that , the maximal unramified extension of a complete discretely valued field with perfect residue field, has universally trivial Brauer group (c.f., Chap IV-V, X in [1]). We summarize these two main results:

Suppose

is a complete discretely valued field with perfect residue field. Then

.

##
Class formations

The Tate-Nakayama lemma provides cohomological testing hypotheses for class field theory. These cohomological data can be formalized as *class formation*. We will first state in full generality purely in group cohomological terms, then specialize to Galois cohomology.

Let be any group and be a nonempty collection of finite index subgroups of . Assume that if and only if . We make the following hypotheses (formal properties that an infinite Galois group must satisfy)

- For any , there exists .
- If is contained in a subgroup of , then there exists such that .
- For any , , there exists such that .

Let

be a field and

be any Galois extension of

. Then

and

and

satisfy the previous hypotheses.

is the most interesting case for class field theory.

A

*formation* is the above data

along with

such that

. Notice that for any

, the the stabilizer of

is

for some

.

For

a local field, we are interested in the formation coming from

. Similarly, for

a number field, we are interested in the class formation coming from

.

To check two things in

to be equal, only need to check after composing with

by the injectivity.

¡õ
03/25/2013

Take , , and in the Tate-Nakayama Theorem 5, we obtain

For any

Galois and

,

is an isomorphism.

When

, then

since

is finite. Therefore

.

When

, using the exact sequence

and the fact

is cosmologically trivial (it is divisible), we compute that

. Hence

.

When

,

and

. So

. This is the expected (inverse of) the Artin map.

Let

be the cocycle representing

. Then

.

The inverse to the isomorphism

, called the (Artin)

*reciprocity homomorphism*, is indeed more useful than

itself. We denote it by

.

Let

. For

, set

. Let

be the coboundary associated to the exact sequence

. Then

(Functoriality)
Let

, where

are Galois. Then we have norm and verlagerung functoriality for the reciprocity homomorphism

For

, we have the follow commutative diagram

Also the compatibility with respect to field extensions

,

The norm functoriality follows from that

The verlagerung functoriality follows from that

The third diagram follows from

(c.f., Proposition

15). For the last diagram, we use the previous proposition. Let

and

. Then by the previous proposition

which finishes the proof.

¡õ
##
Brauer group of a complete discretely valued field

Let be a complete discretely valued with perfect residue field (not necessarily characteristic ). We proved that (Theorem 6). It follows that . So every element of is split by a finite unramified Galois extension (the Galois condition is needed when is not finite). We would like to relate and . Let be the residue field of and write and write . Then is an exact sequence of -modules, split by a choice of uniformizer (this exists because is unramified!). Hence for any , is a split exact sequence.

For any

,

.

is filtered by

, with successive quotients

. Since

(normal basis theorem), it remains to prove the following lemma.

¡õ
Let

be a

-module filtered by

. Suppose

is complete and Hausdorff with respect to the topology defined by

and

for

. Then

.

We omit the proof. The idea is that using completeness we can add up coboundaries valued in the filtration

to get a coboundary valued in

.

¡õ
For any

,

. So

is a split exact sequence. Taking direct limits with respect to

implies that

is a split exact sequence.

03/27/2013

##
Cohomology of

Recall that every open subgroup of is of the form for some . A discrete -module is none other than an abelian group equipped with an automorphism (the action of ) such that , i.e., every element of is fixed by some power of .

Let

. Then

.

It follows from

and then passing to the inverse limit.

¡õ
.

Let

be a discrete

-module. If

is divisible or torsion , then

.

First suppose

is finite. Then

. For

, we claim that

is induced by multiplication by

,

Recall the isomorphism between degree 0 and degree 2 is by cup product with

, where

is the coboundary of

(Exercise

5). So

because

,

and

. Our claim follows. So taking

implies that

for any

, hence

.

When is torsion, we can write as the union of finite -submodules. Then we are done by taking the direct limit.

When is divisible, we have , the taking the cohomology of the exact sequence shows that is injective on for any . But any cohomology group is torsion and we are done again.
¡õ

Our next goal is to show that any quasi-finite field has trivial Brauer group.

Let

be a field and

. We say

is

*quasi-finite* if

- is perfect,
- the map is an isomorphism.

- Every finite field is quasi-finite with being the (arithmetic or geometric) Frobenius.
- Let be an algebraically closed field of characteristic 0 and . Then is the field of Puiseux series. Choose a compatible system of primitive -th root of unity and let such that . Then . Hence is quasi-finite.

- Let . Then . But contains all the roots of unity.
- The same argument with using additive Hilbert 90.
¡õ

If

is quasi-finite. Then

.

Since

is divisible, it follows from Proposition

19that

¡õ
If

is a finite extension, then

.

Write

. Then

.

¡õ
##
Local class field theory

We can now finally construct the local class formation for a complete discretely valued field with quasi-finite residue field . Write . Notice that any finite extension of also satisfies the same hypothesis. Let be as in Remark 57. We would like to check the two axioms of the class formation is satisfied. Axiom I is simply Hilbert 90. Axiom II is the following

- For finite Galois, induces .
- For finite separable. .

03/29/2013

is split by a finite extension

if and only if

.

By definition,

is split by

if and only if

. Since

is an isomorphism, this is is equivalent to

by Part (b) of Theorem

8, if and only if

since

is an isomorphism.

¡õ
(Part (b) of Theorem 8)
By the previous corollary,

¡õ
In particular, the local norm index equality (which we treated as a black box last semester) follows:

(local norm index inequality)
Suppose

is finite Galois, then

is a finite index subgroup of

and

. This becomes an equality if and only if

is abelian.

If

is an unramified finite extension. Then

.

By Proposition

16, it suffices to compute

. Write

. Tracing the definition of

we find that

as wanted.

¡õ
Suppose

is finite abelian with

. Let

be the inertia group. Then

.

04/01/2013

##
Lubin-Tate theory

A

*homomorphism* between two formal group laws is given by a power series

such that

- .
- .

Let

be the ring of integers of a local field

. Suppose

. If

is a formal group law over

, then it gives

a structure of a group by simply evaluating

for

.

The

*formal additive group* is defined to be

. Given

,

.

The

*formal multiplicative group* is defined to be

. Then

gives an isomorphism

.

The more interesting formal groups are those not so easily described in terms of explicit power series.

From now on let be a local field with uniformizer and with elements.

Let

.

For any

and

. Then there exists a unique

such that

- .

Given

. We define

by

- .
- .

Such a power series exists and is unique by Lemma 5.

Given

and

. We define

such that

- .
- .

Again such a power series exists and is unique by Lemma 5.

Apply Lemma

5 repeatedly.

¡õ
The importance of this theorem lies in the following definition (intuitively, a formal group with "extra endomorphism").

A

*formal -module* is a (commutative) formal group law

together with a homomorphism

, denoted by

such that

.

Given

. There exists a unique formal

-module

such that

and

.

works by the previous theorem. The only thing needs to check is that

. In fact, it again follows from the uniqueness of Lemma

5 applied to

and

.

¡õ
When

,

, we can take

. Then

and

. The

-torsion of

is exactly the

-th roots of unity and adjoining all of them gives us a maximal totally ramified extension of

! This picture will generalize to any local field and is the main content of Lubin-Tate theory.

Now fix a

. For a valued field extension

, we define

endowed with the structure of a

-module: for

,

, we set

and

. It is easy to check that when

is a finite Galois extension,

is an

-module on which

acts by

-linear maps. Moreover,

acts on

by

-linear maps.

04/03/2013

For

, we define

Define

and

. These only depend on the choice of

and

and do not depend on the choice of

. In fact, since

and

are isomorphic as

-module by

(Remark

70) and thus there is an induced isomorphism

as both

and

-modules. Also let

and

.

The main theorem for today is the following.

- is surjective.
- .
- .
- For any , there exists a unique such that for any , .
- induces an isomorphism and .
- .

We may assume

(Remark

70).

- For any . By definition, is the same as being a root of . We must show that has a root in . In fact, the Newton polygon of is strictly above the -axis, hence all the roots of has positive valuation.
- When , is a is a -vector space. But has exactly elements, so is 1-dimensional -vector space. In general, we assume by induction that there are elements , such that induces an isomorphism and furthermore for . Now look at the sequence It is exact: the injectivity and the exactness in the middle are obvious; the surjectivity follows from the (a). Now we are done by induction if we choose any preimage of under . In fact, induces an injection by construction and it must be an isomorphism by counting.
- It follows from (b) and .
- The isomorphism in (c) induces an injection . Similarly, .
- It suffices to show the surjectivity of the maps in (d). By injectivity, we only need to check that . By definition is the splitting field of the polynomial , which a polynomial of degree . Notice that . We define . Then . Now and has constant coefficient , thus is an Eisenstein polynomial. It follows that is irreducible and in particular as wanted. As a byproduct, we have also shown that is
*totally ramified*.
- It is immediate since is the constant term of .
¡õ

04/05/2013

Today we are going to remove the choice of the uniformizer from the whole picture, by adding the maximal unramified extension .

Define

by sending

to

, where

is the isomorphism in Theorem

11 (d). Namely,

under the identification

. We will soon see this naturally defined homomorphism

agrees with the Artin map

.

The compositum field

and the isomorphism

are independent of the choice of

.

and

.

As Remark 66, the existence theorem then follows from the isomorphism .

(Existence theorem)
- induces an isomorphism .
- Any open finite index subgroup of has the form for some finite abelian extension .

#### References

[1]Jean-Pierre Serre, Galois Cohomology, Springer, 2001.

[2]Jean-Pierre Serre, Local Fields (Graduate Texts in Mathematics), Springer, 1980.

[3]John William Scott Cassels and Albrecht Frohlich, Algebraic Number Theory, London Mathematical Society, 2010.

[4]Kenneth S. Brown, Cohomology of Groups (Graduate Texts in Mathematics, No. 87), Springer, 1982.