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.

Theorem 1 There exists a canonical continuous homomorphism with dense image , called the Artin map, such that
1. (norm and verlagerung functoriality) For any finite separable extension , the following two diagrams commute
2. (existence) Every finite index open subgroup of arises as the kernel of for a unique finite abelian extension (so ).
3. When is a number field, is surjective and consists of divisible elements. In all other cases, is injective. Write (global) or (local) respectively. Then induces an isomorphism between and the Weil group , between and the inertia group .
Theorem 2 (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]).

Definition 1 Let be any group. A left (right) -module is an abelian group equipped with a left (right) action of , i.e., a homomorphism .
Example 1 If is a Galois group of the field extension , then and are both -modules.
Example 2 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.
Remark 1 A -module is the same thing as a -module (in the usual sense when viewing as a ring).
Remark 2 For any commutative ring , one similarly defines the group ring . Then an -module is nothing but an -module, i.e., an -modules equipped with a -action ).
Definition 2 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 .
Definition 3 A -module is called injective (projective) if the functor () is exact.
Remark 3 has enough injectives (in fact, it holds over any ring, c.f. Lang's algebra), i.e., every -module can be embedded into an injective module (using the fact that any divisible module is injective). Any free -module is projective and has enough projective since any -module is the quotient of some free -module.
Definition 4 For , we endow the -module structure on by and the -module structure on by (c.f., Remark 18).
Definition 5 Let be a group homomorphism. This induces a ring homomorphism . We define three functors
1. restriction , viewing a -module as a -module via ;
2. (compact) induction , ;
3. (co-)induction , .
Remark 4 Suppose is right -module and is left -module, then is an abelian group. So is an abelian group when viewing as a right -module. It further becomes a -module when acts on from the left, i.e., .
Remark 5 The -action on is given by .
Remark 6 Suppose is a subgroup of and is a set of coset representatives of . Then and and the -action is given by , where . Under , can also be identified with the set of maps where vanishes almost everywhere: this explains the name "compact induction".
Remark 7 Suppose is a subgroup of and a set of coset representatives of . Then gives and the -action is given by , where . can be identified with all maps via .
Remark 8 Suppose is a subgroup of , we see that there is an -module inclusion . When , since the direct sum is the same as the direct product.
Definition 6 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.
Definition 7 Similarly, is called co-induced. A direct summand of a co-induced module is called relatively injective.
Proposition 1 (Frobenius reciprocity) and . Then
Remark 9 This is the usual adjointness between the restriction of scalars and the extension of scalars of modules.
Remark 10 For , corresponds to the surjective map , . So any -module is canonically a quotient of the induced module .
Remark 11 For , corresponds to the map , .
Remark 12 Notice that (as -modules) given by . So there is no ambiguity to view as the tensor product as -modules or as abelian groups.
Proposition 2 (Frobenius reciprocity) Suppose and . Then
Remark 13 Similarly, we have a canonical injection via . So any -module is canonically injects into the co-induced module . There is also a canonical map given .
Remark 14 Under Frobenius reciprocity, the inclusion comes from the canonical map given by or the canonical map given by So explicitly is given by
Remark 15 When (e.g., is finite), are both left and right adjoint.

02/01/2013

## Group cohomology and homology

Definition 8 Let be a -module. We define the invariants .
Remark 16 Notice (where acts on trivially). So any short exact sequence of -modules gives a left exact sequence of invariants
Definition 9 Let be a -module. The co-invariants is defined to be the largest quotient on which acts trivially, i.e, .
Remark 17 Let be the argumentation ideal, then . Tensoring with gives a right exact sequence , where the image is of the first map is the same as . So and any short exact sequence of -modules gives a right exact sequence of co-invariants
Remark 18 One can check directly (Definition 4 ) that

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

Definition 10 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 .
Remark 19 Recall that the right derived functor can be computed using injective resolutions. Let be an injective resolution -modules of , then is the same as the cohomology of the complex . Similarly, can computed using a projective resolution and taking the cohomology of the complex of co-invariants .

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

Proposition 3
1. and are independent of the choice of the resolutions (up to canonical isomorphisms).
2. and are covariant functors in .
3. and .
4. Any short exact sequence of -modules induces long exact sequences in group cohomology and homology. The connecting maps are natural.
Remark 20 If is injective, then is an injective resolution of , so for . Similarly, if is projective, then for .
Remark 21 The product of two injective modules are injective. So the product of injective resolutions is an injective resolution of the product. Consequently, we have . Similarly, .

## Behavior under induction

Remark 22 Suppose is a group homomorphism. If is injective and is an exact sequence in , then taking (which is exact) gives a exact sequence Hence is injective in . (in general, a functor with exact left adjoint preserves injectives).
Remark 23 Suppose is a subgroup of , then , so is evidently exact (however, this is not true for an arbitrary homomorphism , e.g., for , , which is only left exact).
Remark 24 Using Frobenius reciprocity, we know that Explicitly, is identified with the constant function . When is a subgroup of , is exact and so is an injective resolution of . We can calculate the cohomology In other words, preserves group cohomology. This result is usually known as Shapiro's Lemma.
Remark 25 Similarly, has an exact right adjoint, so it preserves projectives. When is a subgroup of , is exact and . In other words, preserves group homology.
Corollary 1
1. If is relatively injective, then for .
2. If is relatively projective, then for .
Proof Suppose is relatively injective, then for some , where is injective. Thus for . Similarly for the other part. ¡õ
Remark 26 Notice that we can also compute the group cohomology and homology using acyclic resolutions. By the previous proposition, there is canonical acyclic resolution to compute the group cohomology (similarly for group homology), which will be useful in the future.

## 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.

Definition 11 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

Definition 12 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.

Definition 13 We define the an inhomogeneous cochain to be Then So can be identified with inhomogeneous cochains, denoted by .
Remark 27 Notice that forms a cochain complex and its cohomology computes . The boundary map can be computed as
Example 3 A 0-coboundary is of the form . So it verifies that .
Example 4 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):

Example 5 Suppose is a 1-chain, then Hence and it verifies that .
Proposition 4 .
Remark 28 The group cohomology of can be viewed as the cohomology of the classifying space (with coefficient in a local system). The above proposition simply reflects the fact that the first homology group (with -coefficient) is the abelianization of the fundamental group of a space. (In fact, since , we have . Notice is contractible, forms a free resolution of . Hence ).
Proof 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

Definition 14 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 .
Remark 29 When is a subgroup of , there is a natural map . For higher cohomology, we can define it by "dimension-shifting". Namely, let an exact sequence of -modules where is injective, then we can define inductively using the following diagram

02/06/2013

Definition 15 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
Definition 16 Similarly, since homology is a universal attracting -functor, induces a functor On the level of chain complex, this is given by .
Exercise 1 Prove the above map is a map of chain complexes and it induces .
Remark 30 When is a subgroup of , we have a natural map . For higher homology, the corresponding maps again can be defined by "dimension-shifting".
Definition 17 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
Exercise 2 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).
Remark 31 If is a finite index subgroup of . In this case . We further have corestriction on cohomology and similarly restriction on homology
Remark 32 Unwrapping the definition, we can compute the effect of on degree 0: where is identified with , which gets sent to (c.f., Remark 24 and Remark 10).
Definition 18 We define the norm .
Remark 33 By the dimension shifting argument, we know that on cohomology is the unique extension of the norm to higher degrees.

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

Definition 19 We define the conorm , . Then on homology is the unique extension of the norm to higher degrees.
Exercise 3 The restriction on homology is compatible with the isomorphism .

### Inflation and coinflation

Definition 20 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
Example 6 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.
Remark 34 Suppose is a normal subgroup of , is the conjugation by and . Similarly we obtain an automorphism , which is trivial when by the previous example. It follows that acts on and there will be a Hochschild-Serre spectral sequence (more on this later).

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

Suppose is a normal subgroup of and .

Theorem 3 (Inflation-restriction exact sequence)
1. The sequence is exact.
2. If for , then is exact. Moreover, is an isomorphism for .
Remark 35 This can be viewed as part of the spectral sequence mentioned in the previous remark. Nevertheless our direct proof with be fun with computation of cocycles.
Proof
1. 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 .
2. 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
Exercise 4 Prove the analogous theorem on homology. For example, the sequence is exact.

02/11/2013

Proposition 5 on and . In particular, if , then kills .
Proof 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.

Definition 21 Define to be the absolute norm.

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

Definition 22 We define and
Remark 36 These two groups can be thought of as "reduced" cohomology and homology group, which makes the trivial group have all its cohomology and homology groups zero.

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

Proposition 6 Let be a subgroup of and , Then and
Proof 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
Corollary 2 If is relatively injective (equivalently, relatively projective since is finite), then .
Proof 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
Remark 37 Given a short exact sequence of -modules , fits into the following commutative diagram Thus the snake lemma implies the following long exact sequence which "glues" the two long exact sequences of homology and cohomology together.
Definition 23 We define the Tate cohomology group , for , and for . So () form a -functor (infinite in both directions).
Remark 38 The functors and on cohomology and homology extend to Tate cohomology: it suffices to check they extend to and . For example, if then hence descends to . Similarly for the on homology and on cohomology and homology. One can also check directly by diagram chasing that they induce morphisms of -functors.
Remark 39 Suppose is induced, then for all . Similarly for co-induced . This fact makes the dimension shifting argument even easier: we can start anywhere and the induction will extend in both direction (c.f., Remark 10 and Remark 13).

02/13/2013

Proposition 7 . In particular, is killed by (take ).
Proof The same argument as in Proposition 5. ¡õ
Corollary 3 If is a finitely generated abelian group, then is finite.
Proof 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.

Definition 24 We define with the left -module structure (c.f., Definition 4).
Example 7 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 .
Remark 40 If is furthermore a finitely generated free abelian group. Then for any -module , we have as left -modules.
Remark 41 Suppose is free as a -module. Then as left -modules. So is induced and all its Tate cohomology groups are zero. In particular is an isomorphism (c.f., Remark 18).

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.

Proposition 8 is an isomorphism of -functors.
Proof For , this is true by definition. For , this is true since by Remark 41. The remaining cases can be checked by hand. ¡õ
Remark 42 For , is explicitly given by and is given by . so the notion of a homogeneous cochain in negative degree coincides with that of in nonnegative degree (c.f., Example 7).

## 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.

Theorem 4 Suppose is a finite group. Then there exists a system of homomorphisms for all and -modules , , satisfying:
1. These homomorphisms are functorial in and .
2. In degree 0, the homomorphism is induced by and passing the quotient.
3. Suppose is exact and is also exact. Then for and , where is the connecting homomorphism.
4. Suppose is exact and is also exact. Then for and .
Remark 43 The uniqueness immediately follows from these four axioms using dimension shifting. The only trick here is the following. Embedding into the co-induced -module gives an exact sequence of -modules. Notice there is always a splitting as abelian groups sending to . Hence this sequence remains exact when . Furthermore is again co-induced. Thus the dimension shifting argument applies using (c) in the first variable and (d) in the second variable.

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

Proposition 9
1. .
2. .
3. Suppose is a subgroup of , then .
4. Suppose is a subgroup of , .
Proof 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
Remark 44 Viewing as the pushforward and as the pullback induced by the map , (d) is analogous to the projection formula (c.f., Remark 28).

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.

Proposition 10 Suppose there is a family of -module homomorphisms satisfying:
1. .
2. (this gives a chain complex map ).

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

Proof 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.

Definition 25
1. For , we define .
2. For , we define .
3. For , we define
Remark 45 One motivation here is to mimic the construction of the diagonal map as in singular cohomology. It is straightforward but (extremely) tedious to verify thus constructed satisfy the requirements in Proposition 10, we shall not do the verification. Notice there are choices made in this construction and such a family is certainly not unique. You can construct your favorite one too.
Remark 46 Let , be a -cochain and -cochain. Let , be a -chain and -chain. Using the constructed , one can compute explicitly the cup products in all six cases. For example, in the first two cases. and Notice the latter does not make sense for infinite groups : if you have two functions with compact support modulo the -action, the tensor product may not have compact support modulo the diagonal -action. This again explains that the there is a natural cup product structure on group cohomology but not on group homology (as in topology). In contrast, when , is always compact modulo the -action for an infinite group . In fact there is always a cap product In the case of finite groups, the notion of cup product and cap product coincide due to the nice linear duality of -modules.

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 .

Remark 47 Let be a -module. Notice the action of on is determined by the action of , hence and . So we have a nice symmetric expression (Definition 22) In particular, if is induced, then and . Applying this to the free -module , we obtain a complete resolution of , Thus and in particular, since is of period 2. Explicitly,

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

Corollary 4 A short exact sequence of -module gives an exact diagram
Exercise 5 Define . The exact sequence gives a connecting homomorphism Denote the image of under this homomorphism.
1. Prove that is the isomorphism above.
2. Prove that depends on the choice of the generator (it should not surprise you since the complete resolution depends on a choice of ).
Definition 26 Suppose is a finite cyclic group and is a -module, we define the Herbrand quotient of to be if it makes sense.
Remark 48 The Herbrand quotient should be thought of as (a multiplicative version of) the Euler characteristic of . Euler characteristic is additive in short exact sequences. Analogous Herbrand quotient is multiplicative in short exact sequences.
Proposition 11 Suppose is a short exact sequence of -modules. If two of , and exist, then the third also exists and .
Proof It follows easily from the exact hexagon (Corollary 4). ¡õ
Proposition 12 If the -module is finite (as a set), then .
Proof 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
Proposition 13 Suppose there is a -module homomorphism with finite kernel and cokernel. Then (meaning that if one of them exists, then the other also exists).
Proof 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.

Definition 27 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.

Proposition 14 Let be a cyclic group of prime order and be a -module. Suppose exists, then , , all exist and
Remark 49 The second equality should not surprise you since by multiplicativity.
Lemma 1 Suppose is exact and are defined. If Proposition 14 is true for , then it is also true for .
Proof 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.

Lemma 2 If is defined, then there exists a short exact sequence of -modules such that
1. is finitely generated over .
2. is -divisible (i.e., ).

Furthermore, and are both defined.

Proof 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 (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:

Theorem 5 (Tate-Nakayama) Let be a finite group and be -module. . Suppose is a -Sylow subgroup of for any , and
1. (think: Hilbert 90).
2. , where the subscript denotes the restriction of to (think: Brauer group).

Then for any -module such that , is an isomorphism for any and any subgroup of (think: the inverse of the Artin map).

Theorem 6 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)

1. For any , there exists .
2. If is contained in a subgroup of , then there exists such that .
3. For any , , there exists such that .
Example 8 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.
Remark 50 We borrow the terminology from Galois theory in an obvious fashion. For instance, we call fields. We say Galois if and only and only if and write . We say is the composite of if and only if , etc.
Definition 28 A formation is the above data along with such that . Notice that for any , the the stabilizer of is for some .
Example 9 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 .
Remark 51
1. We write for short.
2. For Galois, acts on and .
3. We write , and likewise . Given Galois extensions , with and . Then
1. We have a canonical map compatible with the inclusion .
2. When , this induces the inflation Notice the inflation only makes sense for the cohomology (which does not extend to Tate cohomology in negative degrees).
3. When , this induces the restriction and the corestriction
4. For , we have isomorphisms and , which induces If furthermore , then .
Definition 29 A class formation is a formation along with a homomorphism (called the invariant) for any satisfying the following 2 axioms:

Axiom I: For any Galois, .

This is Hilbert 90 for . Since if and only for all 's, Axiom I equivalent to say that for any cyclic of prime degree, .

Axiom II:

1. For any Galois, is injective and maps isomorphically to . We denote the generator such that and call it the fundamental class.
2. For any , .
Proposition 15 Let be any extension and be a Galois extension containing . Then
1. IF is Galois, then .
2. .
3. .
4. for any .
Proof To check two things in to be equal, only need to check after composing with by the injectivity. ¡õ
Remark 52 Here is my own way to remember the proposition. The elements of the Brauer group of a field correspond to equivalence classes of central simple algebras over . Under this identification, and do not change and keep the same invariant, hence there must be a degree factor to balance the change of the fundamental class. is the operation and multiply the invariant by the degree, hence there is no more degree factor before the fundamental class.
Remark 53 In the sequel we will prove that the formation coming from is indeed a class formation for any complete discretely valued field with quasi-finite residue field, and thus construct the Artin map using the Tate-Nakayama theorem. However, the existence theorem does not follow automatically. In fact, we need the residue field to be furthermore finite in order to prove the existence theorem (in other words, the locally-compactness is the extra input needed).

03/25/2013

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

Theorem 7 For any Galois and , is an isomorphism.
Example 10 When , then since is finite. Therefore .
Example 11 When , using the exact sequence and the fact is cosmologically trivial (it is divisible), we compute that . Hence .
Example 12 When , and . So . This is the expected (inverse of) the Artin map.
Exercise 6 Let be the cocycle representing . Then .
Definition 30 The inverse to the isomorphism , called the (Artin) reciprocity homomorphism, is indeed more useful than itself. We denote it by .
Proposition 16 Let . For , set . Let be the coboundary associated to the exact sequence . Then
Remark 54 It characterizes uniquely by duality. This is more convenient than the previous exercise since is in degree 0.
Proof Write for short. By definition, . So since . Notice that , where .

We claim that for any , . In fact, first we can check that is represented by the homogeneous 1-cycle which maps and maps other things to 0. Then one can compute explicitly that where the last equality identifies the homogeneous cycle with the inhomogeneous cycle.

Suppose , then we compute , then as desired. ¡õ

Proposition 17 (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 ,
Proof 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
Remark 55 Set . Then we can pass to the inverse limit using the last diagram and obtain a reciprocity homomorphism . It again satisfies the first three functoriality (in particular applies to the Artin maps in class field theory).
Remark 56 One can also formulate the existence theorem using class formation. We won't do this here: instead we will discuss Lubin-Tate theory which not only proves the existence theorem but also provide an explicit construction of the local class fields.

## 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.

Lemma 3 For any , .
Proof is filtered by , with successive quotients . Since (normal basis theorem), it remains to prove the following lemma. ¡õ
Lemma 4 Let be a -module filtered by . Suppose is complete and Hausdorff with respect to the topology defined by and for . Then .
Proof 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
Corollary 5 For any , . So is a split exact sequence. Taking direct limits with respect to implies that is a split exact sequence.
Remark 57 Taking gives an split exact sequence . When is finite, (by Hilbert 90 and Herbrand quotient). So (Corollary 6). These hold for a general quasi-finite : this is actually a purely group cohomological consequence of as we shall see next time (Proposition 21).

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 .

Proposition 18 Let . Then .
Remark 58 This can be viewed as a "pro-cyclic" version of .
Proof It follows from and then passing to the inverse limit. ¡õ
Remark 59 The above isomorphism takes a cocycle to . When acts trivially on , we know that , and hence .
Remark 60 .
Corollary 6 .
Proposition 19 Let be a discrete -module. If is divisible or torsion , then .
Proof 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.

Definition 31 Let be a field and . We say is quasi-finite if
1. is perfect,
2. the map is an isomorphism.
Remark 61 Any finite extension of a quasi-finite field is cyclic of the form . Moreover, is also quasi-finite.
Example 13
1. Every finite field is quasi-finite with being the (arithmetic or geometric) Frobenius.
2. 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.
Proposition 20 Let be a quasi-finite field.
1. If is a root of unity, then there exists such that .
2. If , then is surjective.
Proof
1. Let . Then . But contains all the roots of unity.
2. The same argument with using additive Hilbert 90. ¡õ
Proposition 21 If is quasi-finite. Then .
Proof Since is divisible, it follows from Proposition 19that ¡õ
Corollary 7 If is a finite extension, then .
Proof 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

Theorem 8
1. For finite Galois, induces .
2. For finite separable. .
Remark 62 Part (a) follows formally from (b) since is an isomorphism. The real content lies in (b).
Proof (Part (a) of Theorem 8) Write .

When is unramified, and . The result follows from the following commutative diagram

When is totally ramified, and . The result follows from tracing the same diagram by noticing that :

For a general , we can take be the maximal unramified extension of in and reduce to the previous two cases by looking at (totally ramified) and (unramified). ¡õ

03/29/2013

Corollary 8 is split by a finite extension if and only if .
Proof 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
Proof (Part (b) of Theorem 8) By the previous corollary, ¡õ
Remark 63 We have verified all the axioms in the class formation. So we get the Artin maps (Definition 30) isomorphism for any finite Galois. Taking the inverse limit of we obtain It follows that is continuous and has dense image.
Remark 64 These Artin maps satisfy norm and verlagerung functoriality (Remark 55).

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

Theorem 9 (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.
Proposition 22 If is an unramified finite extension. Then .
Proof By Proposition 16, it suffices to compute . Write . Tracing the definition of we find that as wanted. ¡õ
Remark 65 In particular, when is unramified, then Artin map kills . Notice the unramified condition is crucial, otherwise we do not have an identification between and !
Corollary 9 Suppose is finite abelian with . Let be the inertia group. Then .
Proof Let be the maximal unramified subextension of . So by the previous proposition. Hence .

Conversely, let and choose such that . Since . we know that by the previous proposition. Hence . On the other hand, for , So we can choose such that . Let , then and since the Artin map kills the norm. ¡õ

Remark 66 To summarize, we have proved that
1. The Artin is continuous and has dense image and for finite abelian (Remark 63).
2. satisfies the norm and verlagerung functoriality (Remark 64).
3. restricts to (Corollary 9) with dense image. In other words, we have the following commutative diagram Moreover, when is compact (e.g, has finite residue field), is actually surjective.

We are still missing

1. . This part can be done directly, but we shall postpone the discussion due to the time.
2. The existence theorem. Notice this is equivalent to being injective. Because then (the abelianized Weil group) and thus the open subgroups of of finite index correspond to the finite abelian extensions of (Theorem 20). Lubin-Tate theory will construct the inverse of and the injectivity follows. Indeed Lubin-Tate theory does more than that by constructing explicit splitting polynomials for the abelian extensions of .
Remark 67 We really need the fact the residue field is finite. When is not finite, can be neither injective nor surjective.

04/01/2013

## Lubin-Tate theory

Definition 32 Let be a ring. A formal group law over is a power series such that
1. (meaning that ).
2. in .
3. There exists such that .
4. .

We say is commutative if in addition .

Remark 68 (a), (b), (d) implies (c) and the inverse in uniquely determined. So we do not need to worry about the inverse when constructing formal group laws. Notice condition (d) implies that is of the form , i.e., all the higher order terms are crossed terms.
Definition 33 A homomorphism between two formal group laws is given by a power series such that
1. .
2. .
Example 14 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 .
Example 15 The formal additive group is defined to be . Given , .
Example 16 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.

Definition 34 Let .
Lemma 5 For any and . Then there exists a unique such that
1. .
Proof The proof is bases on the successive approximations of the desired result, i.e., the induction on the degree of . Write and for short. We inductively show that the system of equations
1. ,
2. ,

has a unique solution in .

When , we can (and are forced to) take . Suppose we are given . Set , where . Since to fulfill condition (b) we are forced to take Since and is a unit, to show that it suffices to show divides the numerator. It is true because Now taking gives the desired . It is furthermore uniquely determined by this process. ¡õ

Definition 35 Given . We define by
1. .
2. .

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

Remark 69 We will see soon that is indeed a formal group.
Definition 36 Given and . We define such that
1. .
2. .

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

Theorem 10 The following identities hold:
1. .
2. .
3. .
4. .
5. .
6. .
7. .
Proof Apply Lemma 5 repeatedly. ¡õ

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

Definition 37 A formal -module is a (commutative) formal group law together with a homomorphism , denoted by such that .
Corollary 10 Given . There exists a unique formal -module such that and .
Proof 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
Remark 70 Moreover, for , there exists a canonical isomorphism of formal -modules given by .
Example 17 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.
Definition 38 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

Definition 39 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.

Theorem 11
1. is surjective.
2. .
3. .
4. For any , there exists a unique such that for any , .
5. induces an isomorphism and .
6. .
Remark 71 Though the objects are rather complicated, the proof is actually very elementary (using counting argument).
Proof We may assume (Remark 70).
1. 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.
2. 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.
3. It follows from (b) and .
4. The isomorphism in (c) induces an injection . Similarly, .
5. 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.
6. It is immediate since is the constant term of . ¡õ
Remark 72 If is any finite abelian totally ramified extension, the Artin map maps . So for any , there exists such that . Namely, . Define . Then . Conversely, any subextension of is totally ramified. In fact we will see that constructed by Lubin-Tate theory and every maximal totally ramified abelian extension of has this form. (Warning: there is no unique maximal totally ramified extension since the compositum of two totally ramified extensions are not necessarily totally ramified!)

04/05/2013

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

Definition 40 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 .
Theorem 12 The compositum field and the isomorphism are independent of the choice of .
Proof Choose another uniformizer , where . Choose and . Write be the ring of integers of and be the ring of integers of the completion . The following lemma is crucial:
Lemma 6 There exists such that , where such that
1. .
2. .
3. , where acts on coefficient-wisely.
Remark 73 The lemma tells us that is an isomorphism of the formal -modules and but only defined over the larger coefficient ring . That explains why is independent of (though is not).

Assuming the lemma 6, we know that gives an isomorphism of -modules . This isomorphism does not respect the action of , but it does respect the action of . Hence . To show that and are equal, it suffices to show that , since the left hand side does not depend on and all such generate when runs over . Notice that , so it suffices to show that , i.e., acts trivially on Let . Then where Using Lemma 6 , we can now compute that as desired. ¡õ

Corollary 11 and .
Proof Write . Then and we have the Artin map . By Theorem 11 (f), for any and , is a norm from . So acts as the identity on . It follows that and . Therefore and thus .

If , then is totally ramified and hence , where is the inertia group of . But , so we can choose such that and . But then since is an isomorphism, a contradiction. ¡õ

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

Corollary 12 (Existence theorem)
1. induces an isomorphism .
2. 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.