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.