These are my live-TeXed notes for the course Math 223a: Algebraic Number Theory taught by Joe Rabinoff at Harvard, Fall 2012.

09/05/2012

## Introduction

This is a one-year course on class field theory — one huge piece of intellectual work in the 20th century.

Recall that a global field is either a finite extension of (characteristic 0) or a field of rational functions on a projective curve over a field of characteristic (i.e., finite extensions of ). A local field is either a finite extension of (characteristic 0) or a finite extension of (and sometimes we also include and as local fields) . The major goal of class field theory is to describe all abelian extensions of local and global fields (an abelian extension means a Galois extension with an abelian Galois group). Suppose is the maximal abelian extension of , then , the topological abelianization of the absolute Galois group . Moreover, there is a bijection between abelian extensions of and closed subgroups of . So we would like to understand the structure of .

We also would like to know information about ramification of abelian extensions. For example, does have a degree 3 extension ramified only over 5? This can be nicely answered by class field theory. Class field theory also allows us to classify infinite abelian extensions via studying the topological group . The course will start with lots of topological groups in the first week and one may be impressed by how seemingly unrelated to number theory at first glimpse.

Here some useful applications of class field theory.

Example 1 (Primes in arithmetic progressions) The famous Dirichlet theorem says that for an integer , the primes is equidistributed in . Notice that there is a canonical isomorphism between and and this isomorphism sends the Frobenius element associated to any to . So the classical theorem of Dirichlet can be viewed as a special case of the following Chebotarev's density theorem.
Theorem 1 Let be a finite Galois extension. Then the Frobenius elements (conjugacy classes) for primes of are equidistributed on the conjugate classes of .

Chebotarev's density theorem is proved via reducing to the case of cyclic extensions using a nice counting argument and then applying class field theory (cf. 34).

Example 2 (Artin -functions) An Artin representation is a continuous representation where is a number field and is a finite complex vector space. One can attach an Artin -function to each Artin representation. When is one-dimensional, an Artin representation is simply a character of , which must factor through . So a one-dimensional Artin representation is nothing but a character of . By continuity, this character factors through , where is a finite abelian extension of .

Weber generalized the Dirichlet -functions to Weber -functions over any number fields. He proved that a Weber -function has analytic continuation to the whole and satisfies a functional equation. Class field theory then tells us that the Weber -functions are exactly the one-dimensional Artin -functions.

Example 3 Let be a smooth projective connected curve over a finite field and be the function field of . Then class field theory classifies all abelian covers of . In particular, this gives the abelianization of the etale fundamental group of . The proof of Weil conjecture II uses it in an essential way.
Example 4 (Cohomology of ) In the second semester we will study the Tate global and local duality, Brauer groups and introduce all the cohomological machinery in order to prove class field theory.

Now we briefly turn to the main statements of class field theory. Class field theory gives Artin maps (in the global case) and and the kernel and image of the Artin maps can be described. The crucial thing is that the source of the Artin maps are intrinsic to the field (doesn't involve ). Moreover the Artin maps satisfy the local-global compatibility: the diagram commutes. In other words, class field theory is functorial in . For a finite abelian extension , the Artin map induces the relative Artin maps and . They are surjective and the kernel is exactly the norm subgroups. We can furthermore read ramification data from the relative Artin maps. In the local case is exactly the inertia group and is exactly the -th ramification group of . In the global case, the ramification data can be extracted from the local-global compatibility. We will spend most of the first semester to state the class field theory and draw important consequences from it and devote the second semester to the proofs.

Here are a few words about the proofs of class field theory. The classical approach is to do the global case first, using cyclotomic extensions, Kummer extensions and Artin-Schreier extensions (in characteristic ) to fill up the absolute Galois group, and then derive the local case from it. The cohomological approach is to establish local class field theory using group cohomology and then "glue" the local Artin maps to obtain the global Artin maps. One of the advantage of the cohomological approach is that the local-global compatibility comes from the construction. We will take this approach in the second semester.

Finally we may also talk about explicit class field theory, i.e., finding explicit construction (e.g. as splitting field of polynomials) of abelian extensions. This is a highly open problem in general with several known cases:

1. When , , the union of all cyclotomic extensions of . So the polynomials exhaust all abelian extensions of . This the most satisfactory case.
2. When is an imaginary quadratic field. The CM theory of elliptic curves assert that can be obtained essentially by adjoining all the torsion points on an elliptic curve with complex multiplication by .
3. When is a global function field, there is a theory of Drinfeld modules to obtain most abelian extensions of (apart from some ramification restriction).
4. When is a nonarchimedean local field, Lubin-Tate theory tells that can be obtained by adjoining all torsion points of the Lubin-Tate formal groups.

Somehow adjoining torsion points of a group law is possibly the only known way to construct explicit class fields.

09/07/2012

## Global fields

Today and next Monday we will review the basic notions we learned from Math 129, taking this opportunity to set up the notations.

Definition 1 A number field is a finite extension of . It is an abstract field and have many embeddings into the complex numbers (we do not specify one).
Definition 2 A global function field is a finite (separable) extension of . It is a fact that the algebraic closure of in a global function field is a finite field, called the field of constants. Equivalently, is the field of rational functions on a smooth projective geometrically connected curve over , unique up to isomorphism. The geometrically connectedness ensures that is the field of constants.
Remark 1 If is a finite homomorphism of function fields, then we have a corresponding map of curves. In particular, a finite homomorphism gives a map .
Definition 3 A global field is either a number field or a global function field.
Definition 4 Let be a global field. A place of is an equivalent class of nontrivial absolute values on . Two absolute values and are equivalent if and only if they induce the same topology on under the metric , if and only if for . The set of all places of is denoted by . Suppose , then we have the completion of with respect to any absolute values corresponding to .
Definition 5 Let be an absolute value.
• is called non-archimedean if
• is called archimedean if , when , .

It can be shown that these two properties are mutually exclusive.

Remark 2 If is a function field, then any absolute value on is non-archimedean. If is a number field, let be the number of real embeddings and be the number of complex embeddings (in conjugate pairs). Then there are exactly archimedean places.
Example 5 When , there is only one archimedean place given by the usual absolute value and . For any prime number , the -adic absolute value is defined on the generators of by In this way we have a bijection between the set of primes of and all non-archimedean places of . The completion with respect to is denoted by .
Remark 3 Let is a homomorphism of global fields, then the map given by restriction is a surjection with finite fibers. For , we have if and only ( lies above ). In this case the homomorphism extends uniquely to a continuous map .
Definition 6 Suppose is an archimedean place of , then there exists an absolute value such that for any . By the Gelfand-Mazur theorem, is either or , so the absolute value is the usual absolute value on or . The normalized absolute value is often defined as (when is real) and (when is complex). This normalization simplifies many statements like the product formula.
Definition 7 Suppose is a non-archimedean place of . Let be a representative of . Then is discrete (Ostrowski's theorem), hence is equal to for some . We normalized the valuation such that . This valuation is intrinsic to and is called the normalized valuation. It extends uniquely to .
Definition 8 We denote , the valuation ring of ; , the maximal ideal of and , the residue field of (a finite field). We define the normalized -adic absolute value to be . It is equivalent to absolute value we started with.
Theorem 2 (Product formula) For , for almost all places and
Remark 4 In fact, the global fields can be characterized intrinsically using the product formula.
Theorem 3 (Artin-Whaples, 1945) Let be any field and be the set of places of . Then is a global ield if and only if:
1. There exists for all places such that the product formula holds.
2. The completion is locally compact for any .

Notice that the second condition excludes the fields like .

Definition 9 Let be a nonempty finite set of places of containing the set of archimedean places. The ring of -integer is . This is a Dedekind domain. For , we denote , and .
Theorem 4 Every maximal ideal of is of the form for .
Example 6 For , . For any prime , .
Definition 10 Let be a number field. is called the ring of integers of . It is the integral closure of in .
Remark 5 Analogous to the case of , we have a bijection between the maximal ideals of and the non-archimedean places of .
Remark 6 Let be a global field. Let be the constant field of and be the associated curve. We have a bijection between the places of and closed points of . Any place is given by sending a rational function to the order of zero or pole at a fixed closed point. For a place , the residue field is then the residue field of the corresponding closed point . When is empty, is the constant field . When is nonempty, is the ring of regular functions on . In particular, .
Example 7 Let . Then taking the usual degree of rational function gives exactly the valuation associated to the closed point . Moreover, . A homomorphism gives a map . We know that and is the integral closure of in .

09/10/2012

Definition 11 Let be a global field and be a finite set of places of containing . Then is a Dedekind domain. We denote the group of fractional ideals of by . Then . Any generates the principal fractional ideal . Let be the subgroup of principal ideals. We define the -class group of to be .
Remark 7 When is a number field, a finite group and when , we call the class group of . In geometrical language, . Since is finite (we will prove this fact using ideles), by killing the finitely many nontrivial ideal classes we know that for sufficient large , i.e., is a PID.
Remark 8 When is a global function field, . When , . The degree map of divisors induces a surjection , thus is not finite. Nevertheless, is always finite.

## Extensions of global fields

Let be a finite extension of global fields and be a finite set of places of containing . Let be the preimage of . Then natural map makes a finite projective -module of rank .

Fix a place , we factorize . Then the 's correspond to the places of restricting to .

Definition 12 Denote . The number is called the ramification index of , denoted by . The number is called the residue degree of . We have .
Definition 13 is called unramified at if for all 's, totally split at if and inert at if and .
Remark 9 When is Galois, the 's and 's are the same, we write and . So .

Now fix an archimedean place . Then is either 1 or 2.

Definition 14 is called unramified if and ramified otherwise. Similarly we have .
Remark 10 For any place , we have an isomorphism given by sending and to the diagonal element under the embeddings and .

Suppose is Galois with , then acts on given by . This action is transitive on the fibers of .

Definition 15 The decomposition group of is defined to be the stabilizer of in . We have . In particular, when is abelian the 's for all coincide, we simply denoted it by .
Remark 11 For , acts on by isometries with respect to . So we obtain a map . Its composition with the restriction map is the identity map, thus . A simple counting shows that , hence is also Galois.
Definition 16 When is archimedean and is ramified, we denote , where is the complex conjugation. These complex conjugations are related by .
Definition 17 When is non-archimedean, acts on the residue extension and gives a map . This map is surjective and the kernel is called the inertia subgroup, denoted by . Similarly we have . Counting shows that . So is unramified if and only if , if and only if .
Definition 18 Suppose is unramified. The generator of the cyclic group is called the Frobenius, denoted by (you may think it as the analogue of the complex conjugation . Similarly we have . In particular, when is abelian we have a unique Frobenius attached to , denoted by .
Remark 12 If are finite Galois extension over . Suppose , , . Then natural maps , and are surjections and maps to (when it makes sense). So the construction of the Frobenius element is functorial in the field extensions of .

## Valued fields

Definition 19 A valued field is a pair , where is a field and is a nontrivial absolute value on . We endow with the metric . We say that is complete if is complete under this metric.
Remark 13 Suppose is a complete valued field and is a finite extension. Then there exists a unique extension of to , given by . Moreover, is a complete valued field. In general, when is complete and is any algebraic extension, there is a unique extension of to . But may not be complete.
Definition 20 A valued field extension is a map , i.e., a homomorphism such that .
Theorem 5 (Gelfand-Mazur) If is a complete archimedean field. Then is isomorphic to either or .
Definition 21 Let be a non-archimedean valued field. The ring is called the ring of integers. It is a valuation ring with valuation group contained in . Denote its fraction field by . We similarly define the maximal ideal and the residue field .
Definition 22 We say is discretely valued if is discrete (equivalently, equal to for some ). Any element such that is called a uniformizer. Let be the unique valuation such that .

The famous Hensel's lemma holds for any valued fields (but the proof in this generality is different from the discrete valued case).

Theorem 6 (Hensel's lemma) Suppose is monic and such that and . Then there exists such that and .

09/12/2012

Definition 23 Let be a complete discretely valued field and be a complete discretely valued field extension of . Let be a uniformizer of . We define to be the ramification index of and the residue degree. We say that is unramified if and is separable (so is separable), or equivalently, is etale. We say that is totally ramified if . Notice that and are multiplicative in towers . So is unramified (resp., totally ramified) if and only if and are unramified (resp. totally ramified).
Remark 14 The functor restricts to an equivalence of categories between the category of finite unramified extensions and the category of finite separable extensions . In particular we have , thus is Galois if and only if is Galois. In particular, fixing a finite unramified extensions, the the separable closure of in corresponds to , the maximal unramified sub-extension of . contains all other unramified sub-extensions of .
Remark 15 If is a finite totally ramified extension, then and the minimal polynomial of is Eisenstein, i.e., it is of the form , where and . When , we can choose such that the minimal polynomial is , i.e, is the uniformizer of .
Definition 24 We say is tamely ramified if and is separable, and is wildly ramified otherwise. In particular, every unramified extension is tamely ramified.
Remark 16 If is tamely ramified, then and are both tamely ramified. Fixing an , there exists a maximal tamely ramified sub-extension . In particular, and the residue field of is the separable closure of in . The ramification index is then the prime-to- part of .

## Local fields

Definition 25 A local field is a locally compact complete valued field . So and are locally compact Hausdorff topological group.
Remark 17 In fact the completeness can be deduced from the locally compactness.
Remark 18 When is archimedean, then is either or by Gelfand-Mazur's theorem.
Remark 19 When is non-archimedean complete field. Then the requirement that is locally compact holds if and only is discretely valued with finite residue field. When , must be a finite extension of . When , the residue field embeds into and choosing a uniformizer of gives an isomorphism .
Remark 20 Let be a finite extension of non-archimedean local fields with . Then for any , So , where is the absolute value of extending . In other words, the normalized absolute value is not functorial with respect to inclusion of fields. On the other hand, for , Thus the normalized absolute value is functorial with respect to the norm. The similar statement is true for archimedean local fields.
Remark 21 Let be a non-archimedean local field. Then we have an exact sequence Any choice of a uniformizer of splits this exact sequence and gives a decomposition . For archimedean local fields, we have the decompositions and .
Definition 26 Let be a non-archimedean local field and be a finite Galois extension with . acts on by isometries, so preserves for any and the action induces a map . We define to be the kernel of the map , called higher ramification groups (in the lower numbering). The higher ramification groups give a filtration
Definition 27 is called the inertia group. We have , . In particular, and is totally ramified.
Definition 28 The fixed field of is the maximal tamely ramified sub-extension and is the prime-to- part of . is called the tame inertia group and is called the wild inertia group. is cyclic of order prime to and is the unique -Sylow subgroup of .
Remark 22 For , is an elementary -group. It follows that is solvable.
Definition 29 For we define a function , where , and for . Then is a piecewise linear continuous increasing convex-up function. We define to be the inverse function of and define the higher ramification groups (in the upper numbering) . Then and , and for . For more details, see Chapter 4 of Serre's local fields.
Remark 23 Supoose is a sub-extension of , then the restriction map maps to , where . So the upper numbering behaves better than the lower numbering under extensions: Herbrand's theorem says that for any , we identify , where is the fixed field of , then .

## Topological groups

Definition 30 A continuous map of topological space is called proper if is compact for any compact, open if is open for any open, closed if is closed for any closed.
Exercise 1
1. A continuous map between locally compact Hausdorff spaces is proper if and only if it is closed with compact fiber.
2. If and are proper and is Hausdorff, then is proper.
3. If and are open, then is open.
Remark 24 The last statement is not true when "open" is replaced by "closed": this is one place where the properness can help us.

09/14/2012

Definition 31 A topological group is a group with a topology under which the multiplication and the addition are continuous.

The following basic properties of topological groups are easy exercises.

Exercise 2 Let be a topological group.
1. If is a subgroup, the the closure is also a subgroup. If is normal, so is .
2. Any open subgroup of is closed.
3. Any finite index closed subgroup of is open.
4. An open subgroup of a compact group has finite index.
5. is Hausdorff if and only if is closed.
Definition 32 For any subgroup , we endow the coset space the quotient topology, i.e., is open if and only if is open. By definition, if is continuous homomorphism, then is continuous.
Remark 25 If is furthermore normal, then is a topological group. If is normal and closed, then is Hausdorff. If locally compact Hausdorff and is normal and closed, then is also locally compact Hausdorff. In this case, is proper if and only if it is closed and is compact.
Exercise 3
1. Let be a surjective continuous homomorphism. Then is a quotient map if and only if is open.
2. Let be a closed continuous homomorphism of Hausdorff topological groups, then is open if and only if is open.
3. Let be a surjective continuous homomorphism and . Suppose is a continuous homomorphism such that . Then is a homeomorphism and is a quotient map. In particular, when is abelian, we have an isomorphism of topological groups .
Definition 33 Let be a locally compact Hausdorff topological group. A Haar measure on is a nonzero Radon measure on such that for any and any measurable subset of .
Theorem 7 (Haar) The Haar measure exists and is unique up to scalar multiplication.
Remark 26 Recall that the Borel -algebra on is the smallest set of subsets of that containing all the closed sets and open sets and is closed under countable union and intersection. A Haar measure takes a Borel set in the Borel -algebra to , called the measure of , satisfying
1. is finite if is compact.
2. if is open.
3. is additive with respect to countable disjoint union.
Remark 27 If is a nice (e.g. continuous with compact support) function, then the integral makes sense. In particular, for the characteristic function of . The left-invariance of the Haar measure implies that
Example 8 The Lebesgue measure on satisfies and is a Haar measure. The standard measure on is a Haar measure.
Example 9 Let be any local field and be a Haar measure on . For , set , then is also a Haar measure. It follows from Haar's theorem that , where is a constant.
Lemma 1 .
Proof The archimedean case is obvious. Suppose is non-archimedean. Since is compact open, we know that is a positive real number. Thus it is enough to show that , Replacing by , we may assume that . It easy to see (via the filtration ) that . Hence is a disjoint union of cosets of . Therefore using the left-invariance of . ¡õ
Corollary 1 Let be local field and be a Haar measure. Then is a Haar measure on , i.e., is left-invariant.
Remark 28 Let be an exact sequence of locally compact Hausdorff topological groups, where is endowed with the quotient topology and is endowed with the subspace topology. Let and be Haar measures on and respectively. Then there exists a unique Haar measure on such that in other words, is a measurable function on , independent on the choice of a lift of .

## Profinite groups

Definition 34 A profinite group is a (filtered) inverse limit of finite groups . We endow with the product topology, which makes it a compact and Hausdorff topological group. Then is a closed subspace, hence is also compact and Hausdorff and becomes a topological group under the subspace topology (equivalently, the weakest topology such that the projections are continuous). Moreover, if then there exists a projection , hence and are disjoint open and closed subsets, we conclude that is totally disconnected.

In fact, we can characterize the profinite groups topologically as follows.

Theorem 8 A topological group is profinite if and only if it is compact and totally disconnected. In this case, , where runs over all open normal subgroups of
Remark 29 So any open subgroup of a profinite group has finite index. Any closed subgroup is compact and totally disconnected, hence is also profinite. If is furthermore normal, then is also profinite.
Remark 30 If is a continuous homomorphism where is profinite and is Hausdorff. Then is closed (since is compact). Hence is open if and only if is open. If is surjective, then is a quotient map and is also profinite.
Remark 31 If is compact and Hausdorff, then weaker notion of totally connectedness that subsets all have one point, is equivalent to the stronger notion that every 2 points are separated by disjoint clopen subsets.
Definition 35 Let be any group. Then the profinite completion of is defined to be , where runs over all normal subgroup of finite index of . is profinite and the natural map has dense image. Any homomorphism of , for profinite, factors through .
Example 10 Let be a field, then is a profinite group. Hence any homomorphism factors as .
Example 11 .
Example 12 For any non-archimedean local field , and are profinite.
Example 13 If are profinite, then the product is also profinite.

09/17/2012

## Infinite Galois theory

Definition 36 Let be a field. We say is separably closed if there is no finite separable extension of . We define the separable closure of to be an algebraic field extension of which is separably closed. Any two separable closure are isomorphic.
Definition 37 A separable extension is Galois if any of the following equivalent conditions hold
1. Any two embeddings from to has the same image.
2. The image of any embedding is stable under .
3. The minimal polynomial of any splits completely in .
4. is the union of 's, where every is finite Galois.

In this case, the Galois group is .

Definition 38 The absolute Galois group of is defined to be .
Remark 32 If is Galois and is any Galois finite subextension, then we have a surjection . Since , we know that . In particular, any Galois group is profinite. We call the profinite topology on a Galois group the Krull topology.
Theorem 9 (infinite Galois theory) Suppose is Galois. Then there is a bijection between subextension and closed subgroups of under the Krull topology given by and . Moreover, is Galois if and only if is normal. In this case, we have as topological groups.
Definition 39 We say is abelian if it is Galois with abelian Galois group . In this case, it is easy to see that any subextension of is abelian.
Definition 40 Let be a topological group. Then is a normal topological subgroup. The topological abelianization of is defined to be the Hausdorff topological group , i.e., the maximal Hausdorff abelian quotient of .
Remark 33 Suppose is Galois with . Let be the fixed field of . Then is Galois with Galois group . Suppose is abelian, then the kernel of is closed and contains . Thus , i.e., is the maximal abelian extension of . When , we write this maximal abelian extension as .
Remark 34 Suppose and and are Galois extensions. Let be a -algebra homomorphism. Then the restriction gives a continuous map . In particular, a continuous map depending on the embedding . This continuous map is determined up to conjugation by an element of . By passing to the abelianization, the map is unique. So the association is functorial with respect to and is intrinsic to . In view of the Langlands program, this explains the study of the nonabelian structure of via representation theory, which is insensitive to conjugation. The geometrical situation is analogous: the (etale) fundamental group depends on the choice of the base point, hence is determined only up to conjugation. On the other hand, its abelianization (the first homology group) is intrinsic.

Now let be a global field and be a (possibly infinite) Galois extension.

Lemma 2 For any , there exists a place of such that and acts transitively on such .
Proof The finite case is known. Assume that is infinite. We have a -algebra homomorphism induced by . There is a unique absolute value on extending and restricting to gives a . Now suppose and are two such places of . Then is nonempty. So the inverse limit with respect to finite Galois extensions is again nonempty. ¡õ
Definition 41 The decomposition group at of is defined to be the stabilizer . This is a closed subgroup of .
Definition 42 Suppose is non-archimedean. Let be the the residue field of representing . This is an algebraic extension of . Then acts on and give a continuous homomorphism . This map at each finite level is surjective, hence the image is dense. But is compact, so this map is actually surjective. The kernel of this map is called the inertia group. It is a closed subgroup of and is equal to . We say is unramified in if for any . In this case, by surjectivity, we have a Frobenius element whose image is . When is abelian, this doesn't depend on and we have a Frobenius element .
Remark 35 Any acts by isometries on , hence we have a homomorphism . But in general, is not algebraic (and Galois groups don't make sense). For example, when , is not algebraic over . In other words, is not complete.
Lemma 3 Define . Then , where runs over all finite separable extensions of . Moreover, is Galois and the composition is an isomorphism of topological groups, with all maps being bijective.
Proof Let , there exists such that . By Krasner's lemma, for , . Set . Then . Now since , where runs over finite Galois subextensions, we know that any is Galois, hence is Galois. The composition map is an inverse limit of isomorphisms, hence is a topological isomorphism. The injectivity of follows from the fact that is dense in . ¡õ

09/19/2012

Theorem 10 Suppose . Then .
Proof By definition, . Let be a monic irreducible separable polynomial and be a root of . Let with coefficients close to and the same degree as . Then is small. On the other hand, write , where . Then can be made as small as possible. Now Krasner's lemma implies that . Comparison of degrees shows that . So is separable and irreducible over , hence over . We conclude that , thus . ¡õ
Corollary 2 .
Remark 36 The inverse isomorphism can be described as follows, by extending continuously, we obtain , then restricting gives . The composition maps isomorphically to .
Remark 37 Suppose is Galois. Then the surjection carries (resp. ) surjectively to (resp. ), where and is a place of .
Remark 38 Since maps surjectively to , we know that maps surjectively to . Is this injective? The injectivity simply means that every abelian extension of a local field is obtained as the completion of an abelian extension of some global field (the same statement without the adjective "abelian" follows from the previous corollary). The answer is yes, but not obvious (we will need the power of class field theory).
Exercise 4 Let be a finite separable extension of local fields. Then
1. is a homeomorphism onto a closed subgroup of .
2. For , is continuous, proper, and it is open if or .
3. The norm map is continuous, proper, and it is open if or .
Remark 39 In fact the following general theorem holds, which is much harder to prove.
Theorem 11 is always open.
Proof is proper, hence is closed. It is enough it show that is open is by Exercise 3. Since we already know that the image is closed, it suffices to show that is of finite index. This follows from the norm-index formula. In fact, homological methods can show that (and equality holds if and only is abelian). ¡õ

The unramified case is relatively simpler.

Lemma 4 If is unramified, then . Hence .
Proof is clear. Since the image is closed in , it suffices to show that the image is dense. Any acts by isometry on , . So , where the last equality holds because is unramified. The norm map induces a map , which is coincides with the norm map on the residue fields. By a counting argument we know that the norm map on finite fields is surjective. The norm map also induces a map , which coincides with the trace map, so it is nonzero and -linear (trace is always surjective for finite separable extension). This concludes that the norm map has dense image. ¡õ

Definition 43 Let be a family of topological spaces, be open subset defined for almost all . We define and endow with the topology given by the base of open subsets The topological space is called the restricted topological product of with respect to . Notice that this topology is different from the subspace topology induced from the product topology.

The following lemma is easy to check.

Lemma 5 Let be a finite set of indices containing all 's such that is not defined. Then is open in and the subspace topology on is the product topology on .

The following proposition partially explains the reason of introducing the restricted product.

Proposition 1 If 's are locally compact Hausdorff and the 's are compact, then is locally compact Hausdorff.
Proof Notice that is locally compact: it is a product of a locally compact Hausdorff space and a compact space. Then result then follows from the fact that . ¡õ
Definition 44 Let be a global field. The ring of ideles is defined to be the restricted product of with respect to for non-archimedean. It is a subring of and is a locally compact Hausdorff topological ring.
Remark 40 We have a natural diagonal embedding . It is well-defined since for almost all . It is a topological ring homomorphism so we can regard as a subring of .
Example 14 Since , we have . The similar identification works for general number fields when replacing by . We also have .
Exercise 5 Suppose is a finite extension. Then we have an diagonal embedding for , and hence an embedding .
1. is a homeomorphism onto a closed subring of .
2. The natural map is an isomorphism of topological rings, where is given by the product topology viewing as copies of .
3. as topological groups.
Theorem 12 is a discrete closed subgroup and the quotient is compact.

09/21/2012

Remark 41 Noticing the discreteness of the embedding and the compactness of the quotient, the exact sequence can be thought as an analogy to the exact sequence Moreover, one can show that in fact is the Pontryagin dual of like the familiar fact that is the Pontryagin dual of .
Proof Using Exercise 5, we reduce to the case or since any global field is a finite extension of one of them. We shall do the case , the function field case is similar.

For discreteness of , we need to find a neighborhood of 0 in such that . The open neighborhood suffices since the only units of are . Since every discrete subgroup of a Hausdorff topological group is closed, we know that is also closed in . In particular, the quotient is Hausdorff. To prove the compactness of this quotient, it suffices to show that the natural continuous map is surjective. Let . Since is dense in , we can choose such that and for almost all . Let , then for any . Now choose such that . Then satisfying that , hence maps to in . ¡õ

We will omit the tedious measure-theoretic check of the following the lemma.

Lemma 6 There exists a Haar measure on such that where is the normalized Haar measure on such that .
Remark 42 From the exact sequence of locally compact Hausdorff groups we can construct another canonical measure on . Let be the Haar measure on with total measure 1 and be the counting measure on . By the general construction, there exists on such that The above two Haar measures differs by a scalar and it is an interesting question to figure out this scalar.

## Ideles

Definition 45 The group of ideles is defined to be . The embeddings defined by gives an embedding . The quotient is the called the idele class group of .
Exercise 6 Let be a topological ring. We have an embedding given by . Then
1. is a topological group with respect to the subspace topology coming from .
2. When , is the restricted product of with respect to . Hence is a locally compact Hausdorff topological group.
3. is not a topological group under the subspace topology coming from . But is continuous under the above topology.
4. When is finite and separable, is a homeomorphism onto a closed subgroup.
Definition 46 The idelic norm is defined to be the homomorphism given by . The unit ideles is defined to be the subgroup .
Exercise 7
1. is continuous. Hence is a closed subgroup.
2. When is a number field, is surjective and open, hence is a quotient map.

When is a global function field, the image of can be described as follows.

Theorem 13 Let be a global function field with constant field and . Then .
Proof We will use the following lemma, to be proved when we discuss the Chebotarev's density theorem (Corollary 10).
Lemma 7 If is finite separable of global fields and almost all places of split completely in , then .

Obviously , hence the image is of the form . We need to show that . Notice that . There fore . Hence . If is the degree extension of , then for any . Hence . Write . Then is finite separable. Now shows that there are exactly places over , i.e., splits completely. By the lemma, we know that and . ¡õ

Remark 43 Geometrically the theorem means that there exists a degree 1 divisor on the curve with . In fact, suppose is a closed point corresponding to . Then , hence . So for any divisor , . (Question: is it true over an arbitrary constant field)?
Lemma 8 For , then multiplication by scales by .
Proof It follows from the case of local fields (Lemma 1) and the Lemma 6. ¡õ
Corollary 3 is a Haar measure on .

From this we can obtain a slick proof of the product formula.

09/24/2012

Proof (Theorem 2) Consider the measure in Remark 42. Suppose , then Since is invariant under multiplication by , this integral is equal to Hence . ¡õ
Remark 44 When is a number field, we have an exact sequence Sending the positive real numbers back to an archimedean place of gives a (non-canonical) isomorphism of topological groups . Similarly, as topological groups. We will show that is compact, so this can be seen as the analogy to .
Remark 45 When is a global function field, we have a similar (non-canonical splitting) and , where is compact.
Example 15 Let . Then the fractional ideal where determined up to sign. Hence . But implies that . Replacing by if neccesary, there exists a unique such that as well. Hence is a fundamental domain for . Therefore as topological groups. Hence as topological groups. Now also embeds into , hence as topological groups (this can also be seen directly by extracting all -powers of ).

The classical Minkowski's theorem says that for a compact convex and symmetric around 0 region , implies that there exists a nonzero such that . The following is a reformulation.

Theorem 14 (Minkowski) Let be a -dimensional -vector space and be a lattice. Suppose is a Haar measure on constructed from the counting measure on and the volume one measure on . If is compact convex and symmetric around 0, then implies that .

The idea of the proof of the following adelic version is essentially the same as the classical version.

Theorem 15 (Adelic Minkowski) Let be a global field and . Then is compact. There exists depending only on such that if , then .
Proof Let be the Haar measure on as before. Let Then for any , for any . Then is compact and contains a neighborhood of 0. So , which is an intrinsic quantity of . Let . We claim that this works. Suppose with . Then .

We claim that the natural map has some fiber with at least two elements. Write be the characteristic function of . Then So if every fiber has size at most one, then the above integral is less than , a contradiction to .

Hence there exists such that . Write and , where . Then by the definition of . Therefore . ¡õ

Theorem 16 (Strong approximation) Let and . Then the diagonal embedding has dense image.
Remark 46 In contrast, if we keep all the places, the image is discrete as in Theorem 12.
Remark 47 Concretely the strong approximation means the following: let and be a finite set of places not containing . Then there exists such that for all and for and . For comparison, the weak approximation says that for a finite set of places. Then for , there exists such that for any (without restriction on other places).
Proof We claim that there exists such that . This follows from the fact that is open (Exercise 3), and is compact.

For such a , let , and . Then for

• , we choose such that .
• and , we let .
• , we choose such that has , where is the constant in the Adelic Minkowski's Theorem 15.

By Adelic Minkowski's Theorem 15, there exists . Write such that and . So , then . Then , where and

• for any by construction.
• for and . ¡õ

09/26/2012

Lemma 9 is closed in and the subspace topology on from coincides with the subspace topology from .
Proof First we show that is closed in . For , the products will eventually be decreasing. So is well-defined. Suppose , then . There are two cases:
1. . Let be a finite set of places such that and for all . For and , we let and Then is an open neighborhood of in . For small enough, .
2. . Let be a finite set of places such that
1. for .
2. for . This implies that if , and , then .
3. .

For , let be a small open neighborhood of small enough such that for . Then is a neighborhood of in . Let . If for some , then . If for all , then . Hence . This concludes that is closed in .

Now is continuous and has closed image. So we need to show that any neighborhood of in contains for some a neighborhood of in . By homogeneity, we may assume . The basic open neighborhood of in is of the form , where and is an open neighborhood of 1 in . Shrinking the 's we may assume for any . We claim that works, i.e., . If for some , we have , then , thus a contradiction. ¡õ

Theorem 17 Let be a global field. Then is a discrete closed subgroup and is compact.
Proof The assertion that is discrete and closed follows form the case of (Theorem 12) since the topology on is finer than . It remains to prove the compactness of . By the previous lemma, if is compact, then is compact in . So it suffices to show that there is a surjection for some compact. Let be as in the Adelic Minkowski's Theorem 15 and choose such that . We claim that the compact set works. Let , then . It follows from the Adelic Minkowski's Theorem 15 that there exists , i.e., for any . Now , hence . This concludes the surjectivity of . ¡õ

## Classical finiteness theorems

Definition 47 Let be a finite set of places. We denote an open subring of . Similarly, we denote an open subgroup of . We have and .

Recall that the -class group (Definition 11) is , the fractional ideals of quotient by the principal ideals. We have a natural surjection with kernel and . Hence we have an isomorphism

Theorem 18 (Finiteness of the class groups and finite generation of units) Let be a nonempty finite set of places.
1. .
2. is a finitely generated abelian group of rank .
Proof Write . Then we have an exact sequence Notice that the first inclusion has open image, hence the quotient is discrete. But is compact, hence is compact and is finite.

Now the natural homomorphism has cokernel , which is always finite: in the number field case, both are ; in the global function field case, both are nontrivial subgroup of . It follows that is finite.

It remains to prove that is finitely generated. Consider the log map given by , where . Then maps into the hyperplane . One can easily (and classically) show that is discrete in (only finitely many polynomial with bounded integral coefficients) and the kernel is the roots of unity in , which is finite. This shows that is finitely generated with rank at most . To show is of full rank in , it suffices to show is compact (hence of full rank) as the image is isomorphic to . This follows from the surjection and the compactness of . ¡õ

09/28/2012

## Idele class groups

Let be a finite extension of global fields. We know that as topological rings. In particular, is a closed embedding (i.e., a homeomorphism onto a closed subgroup). Hence is also a closed embedding.

Proposition 2 The natural map is a closed embedding.
Proof For any , . Hence . When is a global function field, we have two exact sequences Since is compact, the first vertical map is a closed embedding. The last vertical map is also a closed embedding. Now the middle term is infinite union of these closed embeddings, hence is also a closed embedding. The number field case is similar: The same argument shows that the middle map is a closed embedding. ¡õ
Exercise 8
1. For any , is a closed embedding.
2. If is a finite set of places and . Then is not a closed embedding.

Notice that is a finite -module, we obtain a norm map , compatible with the norm . Using , we know that for any , the norm map is also compatible with the local norm . Moreover, for any .

Theorem 19 is continuous, open and proper.
Proof It is continuous since each local norm is continuous and (hence the inverse image of a basic open subset is open). For the properness, we use the splitting and . Then is the norm on the compact factors and identity on or , so it is a product of two proper maps, hence is proper. To prove the openness, we use the fact that local norms are open (Theorem 11) and the local norm is surjective for unramified local extensions (Lemma 4) to conclude that the image of a basic open subset is open. ¡õ
Exercise 9 The map is continuous and proper.

Our next goal is to describe the connected component of 1 in ideles class group (which turns out to be exactly the kernel of the global Artin map by class field theory).

First suppose is a number field. Then the connected component of 1 in is where and are the numbers of real and complex places of . It is divisible, i.e., is surjective for any . Let be its closure in . Then is a closed connected divisible subgroup (the divisibility follows from the fact that is proper, thus closed).

Lemma 10 Suppose is a number field. Then is the connected component of 1 in and is profinite. Moreover, is the set of all divisible elements in .
Proof Notice that is finite and , we know that the natural map has finite cokernel. Let be the image of this last map. Since is profinite, we know that the image is also a profinite group (Remark 30). Since is of finite index, it is also open in . Combining the fact that is compact and totally disconnected, we find that is also compact and totally disconnected, thus is profinite.

Let be the connected component of 1, then is killed under the map to by totally disconnectedness, therefore . But is already connected, this shows that . Every divisible element maps to 1 in since profinite group has no nontrivial divisible elements, so must lie in . But is already divisible, hence it consists of all divisible elements of . ¡õ

Remark 48 If (i.e., or an imaginary quadratic field), then is a closed embedding (Exercise 8), hence or . In general we have (for a proof, see Artin-Tate Ch. IX) Notice that is the inverse limit of -fold covers of (called a solenoid).

Now consider the global function field case.

Lemma 11 Suppose is a global function field. Then is totally disconnected and has no nontrivial divisible elements.
Proof Then is an open neighborhood of 1 in , hence is an open neighborhood of 1 in . Hence is totally disconnected, thus is profinite. Hence has no divisible elements. ¡õ

## Cyclotomic extensions

Let be any field and . Let be a primitive -th root of unity. Then , the splitting field of , is separable, hence is Galois. Let . Then any is determined by its action on , thus we obtain a injection This in particular shows that is abelian. The map is functorial, i.e., for any field extension , we have a commutative diagram

Definition 48 For , and . So forms a filtered directed system. We define the maximal cyclotomic extension to be .
Remark 49 Combining all 's gives an injection The right hand side is simply when .

10/01/2012

Example 16 When or , the Kronecker-Weber theorem says that . However, this is not true for general number fields. For example, for , the extension is abelian over but is not even Galois over , hence cannot be contained in a cyclotomic extension of since is abelian over for any .
Example 17 Suppose , where . Suppose and is the order of . Since , we know that . Therefore and is a degree extension of . We have , and is given by .
Example 18 Suppose and , then and sends the complex conjugation to as .
Example 19 Suppose is a non-archimedean local field. If , then is unramified. Indeed, if is any finite extension, Hensel's lemma implies that if and only if . Hence is the unramified extension with residue field . We have as the case of finite fields.
Exercise 10 Suppose is a global function field, then . However, it is not clear whether this is the maximal abelian extension (indeed, not in general).
Example 20 Let be a number field and . Then is ramified at most over finite places and ramified at the real places if . For , , where , due to the compatibility of with respect to the inclusion , and the decomposition group maps to . For , maps to for . The question remaining is what takes for (which can be answered by Artin reciprocity law).

Now consider the case , we have . is ramified at if and only if (and is even for ). In fact, is totally ramified in since . Write , then is the compositum field of (where is unramified) and (where is totally ramified). By Chinese remainder theorem, we know that hence these two fields are linear disjoint. Comparing ramification index shows that the prime above of is totally ramified in and the prime above in is unramified in . So is the maximal unramified subextension at in . Hence maps isomorphic to under and given by . Therefore maps isomorphically to under .

Proposition 3 Let be a finite subextension. Then there exists a unique smallest (the notation comes from German word Führer) such that . Moreover, if and only if is ramified in .
Proof Since is finite, there exists an such that . Since , the gcd of all such 's is the smallest . If , then is unramified in . Conversely, if is unramified in , we write , then the restriction map on inertia groups is . Hence is contained in the fixed field of of , i.e., . ¡õ
Remark 50
1. is roughly the Artin conductor of (after adding a possible factor at ), which "controls" the abelian extension of .
2. and the maximal totally real subextension are examples of ray class fields, which form a cofinial system of finite abelian extensions of .

## Artin maps

Recall the following commutative diagram We know that for , maps surjectively to the inertia group and maps surjectively to the wild inertia subgroup, i.e., the -Sylow subgroup of . For , the element (with 1 at the place ) maps to , which is equal to .

Definition 49 Let be a finite abelian extension . Define the Artin map by sending to and as the restriction . It is a continuous surjection. It follows that maps surjectively to the inertia group and maps to .

10/03/2012

Lemma 12 , . When is unramified is , (opposite to the usual Artin map), where is the image of .
Proof We already know that . Without loss of generality we may assume . Suppose is unramified, then maps to , i.e., . For arbitrary , we write . Then . Because any prime is unramified in , we also know that . Hence We know that This completes the proof. ¡õ
Remark 51 In general, to compute the image of an idele under the Artin map, one needs scale it by a global element (via strong approximation) to obtain an idele with units at all ramified places (these are mapped into the inertia groups) and compute the product of the images of components at the unramified places (i.e., powers of the Frobenius elements).
Definition 50 Now for any with infinite degree, makes sense by taking the inverse limit over finite over . is obviously continuous by construction. Since is compact and the image is dense (surjective on every finite , we find the Artin map is surjective.
Remark 52 Write and . Then Under the isomorphism , corresponds to the subgroup of the idele class group . These form a cofinal system of subgroups of . Moreover, for any , is an open subgroup.

The following proposition summarizes easy properties of the Artin map . We will see how they generalize for any global field.

Proposition 4
1. is a bijection between the finite subextension of and open subgroups (of finite index) of . Indeed, any open subgroup contains some , hence is a subgroup of , hence corresponds to a finite extension .
2. is continuous (and surjective onto ).
3. , the connected component of .

Now let us turn to the local case .

Suppose , then is unramified over . Recall that sending to , so , where is the order of in . On the other hand, is totally ramified of degree . Hence is an isomorphism. Hence for general , we have and the inertia subgroup .

Taking the inverse limit of the exact sequence we obtain two isomorphic exact sequences

Definition 51 We define the Artin map sending This is a continuous map with dense image and maps surjective onto the inertia group.
Remark 53 The image of consists of which induces an integral power of on . is continuous but not homeomorphism for the subspace topology on . Indeed is not open in (in other words, the subspace topology is not discrete).

10/05/2012

Remark 54 Suppose , we define the Artin map by composing with the natural quotient map . In particular, when is finite, is surjective since the cosets of are open. Moreover, is the inertia group and is the wild inertia group.

The following important proposition is left as an exercise.

Proposition 5 (Local-global compatibility) We have the following commutative diagram
Remark 55 We have similar (trivial) local-global compatibility for and .

## Weil groups

Let be a nonarchimedean local field (resp. a global function field) and be the residue field (resp. the constant field). The residue field (resp. constant field) of is . Sow we have a continuous surjection . Denote its kernel by (which is the inertia group in the local case). As a group, the Weil group is simply , i.e., the elements in which induces integral powers of Frobenius on . However, as we have seen in Remark 53, may not be open in under the subspace topology from . But there exists a finer topology on such that is open in . under this finer topology, (in the local case) or (in the global function field case) will be isomorphisms of topological groups.

More precisely, suppose we have a short exact sequence of profinite groups and .

Lemma 13 There exists a unique topology on such that is open in , has the subspace topology under and any splitting of induces an isomorphism of topological groups.
Proof See the handouts for details. ¡õ

Under this topology on the Weil group , is continuous and has dense image (but is not a homeomorphism onto its image). Moreover, the topology is compatible under abelianization as in following proposition.

Proposition 6 Let be the abelianization of and : Then with the abelianization as topological groups (but not for ).
Remark 56 Notice that the abelianization of topological groups usually does not preserve injectivity!

The following theorem allows us to use Weil groups as a replacement of to classify finite abelian extensions of local or global function fields.

Theorem 20 The maps and gives a bijection between open subgroups of and open finite index subgroup of . For any an open subgroup, is isomorphic to as discrete coset spaces.
Exercise 11 This bijection extends to a bijection between closed subgroups of such that or has finite index in , and closed subgroups of . This will allows us to partially classify infinite abelian extensions of local fields and global function fields.

## Statement of global class field theory

Theorem 21 (Global class field theory) Let be a global field. There exists a continuous homomorphism, called the Artin map, with dense image, satisfying:
1. For any finite abelian and . The composition satisfies:
1. When is nonarchimedean, it kills if and only if is unramified in , in which case it sends any uniformizer to .
2. When is archimedean, it kills the connected component of . When is unramified, it kills all of ; when is ramified, it sends to the complex conjugation.
2. (Existence theorem) Every finite index open subgroup arises from a finite extension as the kernel of . is called the class field to the subgroup .
Remark 57 Since is the inverse limit of for finite abelian extensions , we know that is determined by the maps . Moreover, is surjective since has dense image.
Lemma 14 Property (a) in Theorem 21 uniquely characterizes .
Proof Observe that if is a finite set of places of . Then is dense in by the weak approximation (Remark 47). If is a finite abelian extension, we let contain and all the ramified places. Then is determined by (a), hence by continuity is determined by (a). ¡õ
Remark 58 In particular, is the opposite to the one defined in Definition 50.

Assuming Theorem 21, let us prove the following "real version" of the existence theorem (and please hope for the "real real version").

Theorem 22 (Existence Theorem) is an inclusion reverse bijection between finite abelian extensions and open subgroups of finite index in .
Proof It suffices to show the injectivity. Suppose and are finite abelian extensions such that . Let and be the corresponding open subgroup in , then . Hence . Suppose , then there is an neighborhood since and are both open and closed. But is dense, this would contradict . We conclude that , thus . ¡õ

10/10/2012

Suppose is a number field and be the connected component of . We have shown that is profinite (Lemma 10). is profinite, thus contains no divisible elements, hence . But the image of is dense, we know that is surjective in this case. On the other hand, is profinite implies that the intersection of all open subgroups is trivial. We find that the intersection of all open subgroups (of finite index) of is , as any subgroup of finite index contains divisible elements . Therefore by the existence theorem. Namely, we have shown

Theorem 23 Suppose is a number field, then induces an isomorphism of topological groups .

We thus deduce a stronger version of existence theorem for number fields.

Corollary 4 The map is a bijection between all abelian extensions and closed subgroups of such that . Under this bijection, is called the class field of .
Remark 59 Suppose is open. Write , then we have an isomorphism using the splitting . But is compact, hence any open subgroup of is of finite index. This is false for a global function field, e.g., is open and has infinite quotient is .

Now suppose is a global function field. We have an isomorphism of topological groups , a product of a discrete group and a profinite group. Then the intersection of all open subgroup is trivial, hence is injective.

But is not surjective, indeed we claim that if , then , an integral power of the Frobenius. In fact, for a finite extension , then is finite abelian and unramified everywhere and (Theorem 13, or the base change of an etale map is etale). It suffices to show that . For , lifts , By definition, as is unramified everywhere and . In other words, we have proved the following diagram commutes

It follows that . We claim that is actually an isomorphism. We already know it is injective, so it suffices to show the surjectivity. Choose such that . Then there exists a unique continuous homomorphism sending 1 to by the universal property of profinite groups. Hence as topological groups. Under this identification, is simply . Let . Then is a closed subgroup and , hence the closure of in is . Since the image of is dense, it follows that , which proves the surjectivity.

We have proved

Proposition 7 induces vertical isomorphisms of topological groups So can viewed as the Weil group and is the profinite completion of .
Corollary 5 gives a bijection between all abelian extensions such that the constant field extension is either finite or equal to , and closed subgroups of .
Proof By Exercise 11. ¡õ

10/12/2012

## Norm and Verlagerung functoriality

Let be a global field and be any finite separable extension. gives us a canonical map . We also have the compatibility of local and global norms:

Proposition 8 (Norm functoriality) The follows diagram commutes:
Proof We only need to show the diagram for is finite abelian and (which finite abelian over ): Suppose is a finite set of places containing:
• such that is ramified in
• such that is ramified in
• ramified in
• .

Since is dense in by weak approximation. It suffices to show that for any . Let and . For the latter equality, notice that and since is unramified. It remains to prove the first equality. Let be a uniformizer. Then since is unramified. We compute On the other hand, . Since both side of the first equality are the unique element of which fixes and reduces to , they are actually equal. ¡õ

Corollary 6 Suppose is finite abelian, then . In other words,.
Proof Take in the previous theorem. ¡õ
Theorem 24 (Global norm index inequality) Suppose is a global field and is a finite and separable extension. Then .
Proof We will give an easy analytic proof later (cf. Exercise 17). In fact, the cohomological proof will even give a division relation. ¡õ
Remark 60 If particular, we conclude that when is finite abelian we actually have an equality, as the maps surjectively onto in this case.
Theorem 25 (Existence theorem III) Suppose is a global field and is a finite and separable extension. Then if and only if is abelian, in which case .
Proof It remains to prove the "only if" direction, which is left as an exercise (hint: the left-hand-side is always . ¡õ

Now let us briefly turn to the verlagerung functoriality of the Artin map.

Definition 52 Let be a group and be a subgroup of finite index. For any , let be the smallest such that . This only depends the choice of in , a finite double coset. Write be the coset representatives. We define the verlagerung (or transfer) . Then is a group homomorphism. In terms of group cohomology, this is the restriction map and functorial in .

Suppose is a field and is finite separable. Then is of finite index (depending on the choice of an isomorphisms ). We then obtain the verlagerung , a continuous map of the topological abelianization of and , which does not depend on the choice of .

Proposition 9 Suppose is a global field and is a finite separable extension. The following diagram commutes:
Proof Suppose is finite Galois and . It suffices to show at the finite level. The remaining check will be an easy calculation which we leave as an exercise. ¡õ
Remark 61 Notice the nice interchange of position between and in the previous two propositions.

## Statement of local class field theory

Theorem 26 (Local class field theory) Let be a nonarchimedean local field. There exists a canonical injective homomorphism satisfying:
1. Namely, the Artin map identifies with the Weil group
2. If is finite separable, then we have norm functoriality and verlagerung functoriality for and :
3. If is finite abelian, then is surjective with , (the -th ramification group in the upper numbering, ) and (the inertia group).
4. (Existence theorem) Every open subgroup of finite index of is of the form for some finite abelian.

10/15/2012

Remark 62 The statements in the previous theorem are not independent. For example, given the local norm index equality, the norm functoriality implies the first part of (c). Also (a) implies the second part of (c) concerning . Moreover, (a) and (c) implies (d) by Theorem 20.
Remark 63 This theorem does not (at least, not obviously) uniquely characterize . But if is unramified, then is uniquely determined by (a).
Remark 64 Of course, there are obvious Artin maps for archimedean local fields: and

Analogously, the local existence theorem will follow from the local norm index inequality.

Theorem 27 (Local norm index inequality) Suppose is a local field and is a finite and separable extension. Then .
Theorem 28 (Existence Theorem) Suppose is a local field and is a finite and separable extension. Then if and only if is abelian, in which case, .

The local-global compatibility will follow from defining the global Artin map via "gluing" local Artin maps.

Theorem 29 (Local-global compatibility) Suppose is a global field and . Then we have a commutative diagram If is finite abelian and , then is also finite abelian and we have a corresponding commutative diagram at finite level.

We can then derive part of (a) in global class field theory (Theorem 21) using the local-global compatibility at archimedean places.

Corollary 7 Suppose is a number field and is a real place. If is finite abelian, then is unramified in if and only if kills all . Otherwise kills and is the complex conjugation.

## Ray class fields and conductors

Now let us discuss the classical formulation of class field theory in terms of ideal classes.

Definition 53 Let be a global field. A modulus of is a formal product , where and for almost all ; for a real place, ; a complex place, . We denote by the -component of . We write or if . The modulus is used to keep track of the ramification of the places of in some sense.
Definition 54 Let , we write if
• is nonarchimedean and , then ;
• is nonarchimedean and , then ;
• is archimedean and , then ;
• no requirement for archimedean and .

In other words, should go to under the Artin map.

Definition 55 Let , we say if and only if for any . This is a multiplicative condition, so it makes sense to say that whenever .
Example 21 For , each modulus is of the form or , where . It follows easily that an idele if and only the finite part (which can be thought of the usual congruence relation ), and if we further have .
Definition 56 Suppose is a modulus, we define , namely It is an open subgroup of and if and only if . We know that forms a cofinal filtered system of open subgroups in . This generalizes Proposition 4.
Definition 57 Define
• be the free abelian group of fractional ideals of generated by .
• be the image of under the natural map .
• the ray class group .
Remark 65 When is a number field and , we recover that , and . If is the product of all real places, then , is generated by totally positive elements and is called the narrow class group.
Example 22 For and , we have . We have a surjection and the kernel is exactly . Therefore . Similarly, when , we find that .

10/17/2012

Proposition 10 There are isomorphisms
Proof We define (it properly contains ) and . Then we have an injection It is actually an isomorphism by weak approximation. Now define the homomorphism which is obviously a surjection with kernel . Taking quotient by the image of , we obtain that . ¡õ
Remark 66 The second isomorphism is a bit hard to write down explicitly since the weak approximation is involved in the proof.
Remark 67 When is a number field, is finite since any open subgroup of is of finite index. But when is a global function field is infinite, since and the quotient contains at least .
Definition 58 The ray class field of the modulus is the class field corresponding to . When is a number field, is finite abelian. When is a global function field, contains since the image of has trivial image when projected to .
Remark 68 When is a number field, the Artin map induces an isomorphism . When is a global function field, the Artin map induces a map with dense image. In either case, for any , .
Exercise 12 When , we have and and the following diagram commutes
Remark 69 Notice that forms a filtered cofinal system of open subgroups of . Hence forms a cofinal system of abelian extensions of , in which every finite abelian extension of is contained in some . In particular, the classical Kronecker-Weber theorem for follows from the previous exercise.

We summarize the easy properties of ray class fields as follows.

Proposition 11
1. is unramified at all .
2. If , then .
Definition 59 Suppose is a finite abelian extension. We say a modulus of is admissible for if . The gcd of all admissible modulus of is called the conductor of . It follows that .
Example 23 When . is essentially the same as Proposition 3, except that if and only if is totally real.
Lemma 15 is admissible for if and only if .
Proof The "if" direction follows from the fact that . For the "only if" direction, suppose , then the local-global compatibility implies that for nonarchimedean (and a similar thing for archimedean places), hence by the definition of idelic norm, . Therefore . ¡õ
Remark 70 In some literature, the notion of admissible modulus is defined via this lemma.
Proposition 12 is ramified in if and only if .
Proof Let , where . If is ramified, the , hence , hence , which implies that . Now suppose is unramified, then for any , . Then we can find a modulus such that and . It follows that by the previous lemma and the definition of . ¡õ

## Ideal-theoretic formulation of global class field theory

Definition 60 Let be a finite abelian extension and be a modulus which is divisible by all ramified places. We define the Artin symbol
Exercise 13 is admissible if and only if . (This is the way Artin originally introduced the notion of admissible moduli. The existence of admissible moduli is truly surprising and is the key difficulty of class field theory!)

10/19/2012

Definition 61 Suppose is a finite extension and is a modulus of . We denote by the free abelian group generated by prime ideals of whose restriction to are coprime to . The usual ideal norm restricts to a group homomorphism . We denote the image of by .
Exercise 14 Let be a finite extension. Suppose is finite abelian and . Then we have a commutative diagram where is a modulus of divisible by the primes of ramified in , is a modulus of divisible by primes ramified in or restrict to primes of ramified in .
Remark 71 Suppose is finite abelian and is divisible by primes ramified primes. Then it follows from the previous exercise (norm functoriality) that kills . is surjective by weak approximation and surjectivity of (which is completely nontrivial without class field theory!). Suppose further that is admissible for , then also kills by Exercise 13.
Theorem 30 (Class field theory in ideal-theoretic language) Let be a number field and be a finite abelian extension.
1. (Reciprocity law) There exists an admissible modulus for and induces an isomorphism .
2. (Existence theorem) For any modulus , there exists a unique finite abelian extension for which is admissible and (i.e., ).
Remark 72 One can show that this classical version will imply the idele-theoretic version of global class field theory.
Remark 73 An inconvenience in idele-theoretic language is that one has to constantly change the moduli when dealing with several field extensions.

## Weber L-functions

Before stepping into the cohomological proof of class field theory, we will discuss various applications of class field theory in the following several weeks. From now on we will assume is a number field for simplicity (though some results are also valid for function fields). Write the degree .

Definition 62 We defined the Dedekind zeta function where is the absolute norm.

We omit the proof of the simple analytic property.

Proposition 13 is analytic on except a simple pole at .
Definition 63 Let be a modulus of and be an ideal class. We define

A similar analytic property holds for too.

Theorem 31 is analytic on except a simple pole at . The residue at depends on the modulus but not on ideal class .

Recall that for a finite abelian group, we have the notion of Pontryakin dual consisting of characters of , and the following elementary properties holds:

Proposition 14
1. canonically.
2. If , then
3. If , then
Definition 64 The Weber -function for a character is defined to be It follows from the previous theorem that is analytic on except a possible simple pole at . When , it is actually analytic at by the previous proposition, since all the residues of at are the same.
Example 24 Consider and . Then is simply the Riemann zeta function. A character is the same thing as a Dirichlet character . Then is simply a Dirichlet -function, where we extend on by letting whenever .

Now class field theory easily imply the following result on special values of Weber -functions.

Theorem 32 if .
Proof By global class field theory, there exists a class field such that . So can be viewed as a character of . Let be a modulus of divisible exactly by the primes of restricting to . Then as a special case of the lemma below () The result then follows since each of the -functions and has a simple pole at . ¡õ

10/22/2012

Lemma 16 Let be a finite abelian extension. Suppose is an admissible modulus for and is exactly divisible by primes of restricting to primes dividing . Then where runs over all characters .
Proof This equality actually holds at the level of local Euler factors. Say , then is unramified and , we claim that Write , then . The right hand side becomes Write for short. Taking the logarithms of both sides, it reduces to show that Notice that and has order in . Then the character of the Pontryagin dual of is nontrivial if and only if . Therefore the right-hand-side is simply which coincides with the left-hand-side. ¡õ
Remark 74 The above lemma can be viewed as a special case in the more general setting of Galois representations. Consider a Galois representation , where is a -dimensional complex vector space. Since has no small subgroup, we know that factors through a finite extension . We define the Artin -function of to be It converges on and satisfy the following nice (and easy) properties:
1. .
2. for a finite extension and .
Example 25 Consider a one-dimensional Galois representation . Then factors through a finite cyclic extension , where . Let be the conductor of , then induces a character . By definition, since if and only if is unramified in . In other words, Weber -functions can be viewed as the same thing as one-dimensional Artin -functions via class field theory. Notice Weber -functions involves geometry of numbers (ideal classes, etc.), which are crucial for establishing the analytic properties. On the other hand, the functoriality properties of Artin -functions give handy ways to establish non-vanishing results of special values. This picture motivates the Langlands program of the study of general Artin -functions via relating Galois representations and automorphic representations and class field theory can be viewed as the Langlands program for .
Example 26 Suppose is a finite abelian extension with . Then (notice the left-hand-side is the regular representation of ). Thus we have This is analogous to the previous lemma except that more local Euler factors are involved here.

General Artin -functions can be reduced to one-dimensional Artin -functions via Brauer's induction theorem.

Theorem 33 (Brauer) Suppose is a finite group and . Then there exists , subgroups of and characters such that as virtual representations.
Corollary 8 With the same notation,
Remark 75 In particular, we obtain a meromorphic continuation for Artin -functions from the meromorphic continuation of Weber -functions (but the Artin conjecture for holomorphy when is still open in general!).

## Chebotarev's density theorem

Definition 65 Let be a number field of degree . Let be a set of finite primes of . We define the natural density of to be the limit (if it exists) We define the Dirichlet density to be
Remark 76 The natural density is the "strongest notion" of density in the following sense: when exists, also exists and . We will mainly talk about the Dirichlet density since they are more convenient for analytic arguments.

10/24/2012

Remark 77 For holomorphic in a neighborhood of 1 except possibly at 1, we write if is holomorphic at 1. It may be helpful to notice that .Then Since is analytic for . We see that Also observe that has a simple at , so is analytic and non-vanishing at 1. Hence Because is an equivalence relation, it follows that and consequently

The following basic properties of the (Dirichlet) density follows easily from definition.

Proposition 15
1. If has a density, then .
2. If is finite, then .
3. If is the disjoint union of and and two of , , have density, then so does the third one and .
4. If and both have density, then .
5. If has a density and ,then .
6. If has density and is the complement of , then has density and .
Exercise 15 Suppose is a number field. Denote the degree of a prime by . Prove that has density 1. (Not to be confused with the density of of split primes of in , which will be shown to be ).
Proposition 16 Let be a modulus and . Then .
Proof Notice that On the other hand, since is analytic at 1 whenever , we know that Expanding the last sum gives The desired result then follows. ¡õ
Remark 78 The non-vanishing result of can be proved without class field theory, which we leave as an exercise (cf. Exercise 17).
Example 27 When , , we have and an ideal class is given by simply given by a residue class modulo . From the previous proposition we recover the classical theorem of Dirichlet on primes in progressions: . Namely, the primes are equidistributed modulo .
Theorem 34 (Chebotarev's density) Suppose is a Galois (but not necessarily abelian) extension of global fields with and . Let and be the conjugacy class of in (with ). Then has density . In other words, the Frobenius conjugacy classes are equidistributed in .
Proof Let us first show the case that is abelian. Let be the conductor of . Let be the kernel of . Then and , where is any preimage of . Since contains exactly ideal classes, we know that as needed. This is the only step where the usage of class field theory is crucial.

10/26/2012

For the general case, we let be the fixed field of under the cyclic group generated by . Then is a cyclic extension with Galois group and we can reduce the previous case as follows. Let be the primes of unramified in and . By the abelian case, we know that . Also, let be the primes in such that , then by Exercise 15. We claim that for any , . Assuming this claim, we know that as desired.

It remains to show the claim. Let be the set of primes of such that is unramified in and . For , write and . Since acts trivially on , it also acts trivially on . Therefore , i.e. . Since , we know that is the unique prime of over . On the other hand, given , let and be a prime of over . Since has order , we know that . Hence is the unique prime of over and . Thus . In this way we have exhibited a bijection between and .

Now let and a prime of over . We can choose . Then is the orbit of under the centralizer of . So which proves the claim. ¡õ

Remark 79 The corresponding Chebotarev's density theorem for global function fields can be proved using Riemann hypothesis for curves.

## Split primes

Definition 66 Let be a global field and be a finite separable extension. We define For , we also define .
Corollary 9 Suppose is Galois and . Then and have density .
Proof Use Chebotarev's Density Theorem 34 for . ¡õ
Remark 80 Notice this special case doesn't use class field theory since it only involves the second part of the proof of Chebotarev's density theorem:
Exercise 16
1. Suppose are finite separable. Then .
2. Suppose is finite separable and is its Galois closure. Then . In particular, if and only if is Galois.
Corollary 10 If has density 1, then .
Proof Apply the previous exercise to the Galois closure of . ¡õ
Remark 81 We used this corollary in Theorem 13. But as remarked above, it doesn't depend on class field theory and there is no danger of circular reasoning.

We can now prove the following results without using class field theory.

Exercise 17
1. for .
2. (Global norm index inequality) for a finite extension of number fields.
Theorem 35 Suppose are finite Galois, . Then the following are equivalent:
1. .
2. .
3. .
Proof (a) implies (c) and (c) implies (b) are obvious. For (b) implies (a), notice that . Therefore and . ¡õ
Remark 82 The statement is obviously false without the Galois assumption.
Corollary 11 Suppose and are finite Galois, . Then if and only if . In other words, a Galois extension is determined by the set of split primes.
Remark 83 Which subsets of are of the form for Galois? The answer for general case is still unknown. But when is abelian, we can answer this question using class field theory as follows. Suppose is the conductor of . Let be the kernel of . Then splits completely if and only if , if and only if is an element of the subgroup .
Remark 84 The last condition can be thought as a congruence condition. For example, splits completely in if and only if . If corresponds to the subgroup , then splits completely in if and only if .

## Hilbert class fields

Definition 67 Let be a number field. The Hilbert class field of is defined to be the ray class field for . Then and is unramified at every place (including archimedean places). The narrow Hilbert class field of is defined to be the ray class field , where is the product of all real places of . Then (the narrow class group) and is unramified at all finite places. Notice that and surjectis onto .
Lemma 17 (resp. ) is the maximal abelian extension of which is unramified everywhere (resp. at all finite places).
Proof If is unramified everywhere, then by Proposition 12. Hence . Similarly for . ¡õ

10/31/2012

Remark 85 Denote by the totally possibly elements of . By weak approximation, . So surjects onto the kernel of quotient map .
Proposition 17 Suppose is finite extension of number fields and is totally ramified at some . Then .
Proof Since is abelian and unramified everywhere, we know that . So it suffices to show that and are linearly disjoint over (i.e. . Suppose with the minimal polynomial of . Let be a monic polynomial dividing over . Since is Galois, splits into linear factors over , which implies also splits over . Hence , which is equal to by the assumption that is totally ramified at some place. Hence and is irreducible over as needed. ¡õ
Example 28 Consider , . We know that using the previous proposition. When is a prime, the famous criterion of Kummer asserts that divides (indeed equivalent to that divides ) if and only if divides the numerator of some Bernoulli number . Such a prime is called irregular.

## Artin's principal ideal theorem

Theorem 36 (Principal Ideal Theorem) Let be a number field and be its Hilbert class field. Then any ideal becomes principal in .

Using class field theory, it will reduce to the following purely group-theoretic theorem (we omit the proof).

Theorem 37 (Furtwangler) Let be a finite group and . Then is trivial.
Proof (Proof of the Principal Ideal Theorem) Let be the Hilbert class field of . Then is Galois since is intrinsic to . Notice that is the maximal abelian subextension of . Therefore and , where . We have the following commutative diagram By the previous theorem, we know that the , hence is principal. ¡õ

## Class field towers

The principal ideal theorem motivates the following construction. Let be a number field and be the Hilbert class field of . The we obtain the Golod-Shafarevich tower Does stabilize (i.e., )? The answer in general is no.

Exercise 18 The following conditions are equivalent:
1. .
2. for some .
3. There exists an number field such that and embeds into .
Remark 86 Clearly (the maximal unramified extension of ). But this is usually not an equality (e.g., is always solvable).

The analysis on this tower may be easier if we consider a single prime at one time. Let be the maximal unramified abelian -extension of . We obtain a tower We now state a theorem of Golod-Shafarevich (for the proofs, cf. Cassels-Frolich Ch. IX).

Theorem 38 (Golod-Shafarevich) Let be a number field of degree . If , then where is the -rank of a finite group of .

On the other hand, Brumer's theorem provides a lower bound on .

Theorem 39 (Brumer) Suppose is Galois of degree . Let be the number of primes such that for any above . Then
Corollary 12 Suppose is Galois of degree . If , then .
Example 29 When , it follows from the previous corollary that a quadratic field has infinite Golod-Shafarevich tower whenever the number of ramified primes is at least 8, e.g., where has at least 8 different prime factors. In particular, there are infinite such quadratic fields.
Remark 87 In general, for any , there exists infinitely many number fields of degree with infinite Golod-Shafarevich towers. For this, one needs the non-Galois version of Brumer's theorem.

## Hilbert class fields of global function fields

Now suppose is a global function field. Then is the maximal abelian unramified extension of . In particular, and is infinite. To get better situation, we may ask what is the maximal unramified extension of with constant field . As class fields with constant fields corresponds to subgroups that surjects onto under , we know that a maximal unramified extension with constant field corresponds to a minimal subgroup such that there exists an element with . However, there may be more than one such minimal subgroup. The set of such subgroups forms a principal homogeneous space under (in geometrical terms, it is simply ). So there are such groups and hence there are maximal abelian unramified extension of with constant field .

11/02/2012

Nevertheless, the following construction gives a maximal unramified extension which is canonical in some sense. Choose with . Then modulo is independent on the choice of and the subgroup is canonically defined. Let be the class field of . We then have an exact sequence So is everywhere unramified of degree with constant field the degree extension of .

We end the discussion by applying a similar idea to prove a useful proposition concerning -adic characters of the Weil group of a global function field.

Definition 68 Let be a Hausdorff group. A continuous character is called unramified if the class field corresponding to is unramified at .
Proposition 18 Suppose is a global function field of characteristic and is a finite extension of for some . Let be a continuous homomorphism unramified outside a finite set . Then there exists and continuous of finite order such that .
Proof Since is abelian, factors through . Pick such that . Let and . Notice that by construction. But , it suffices to show that is finite as is finite. But and by assumption for , it suffices to show that is finite. But has a finite index pro- group and has a finite index pro- group. Because there is no nontrivial map from a pro- group to a pro- group, the image must be finite. ¡õ

## Grunwald-Wang theorem

Proposition 19 Let be a global field. Suppose and contains the -th roots of unity. If is an -th power in for almost all , then itself is an -th power in .
Proof Let be an -th root of and be its minimal polynomial. So . If is an -th power in , then splits completely in as contains the -th roots of unity. Since is separable, it follows that splits completely in . Now the split primes has density 1, hence . ¡õ
Exercise 19
1. 16 is an 8-th power in and in for , but not in (hence not in , which is obvious).
2. Let , then 16 is an 8-th power in for any place but not an 8-th power in .
Theorem 40 (Grunwald-Wang) Let be a global field. If is a -th power in for almost all , then is an -th power in , except potentially if is a number field and is not cyclic, where with odd.
Remark 88 is not cyclic, which explains the previous exercise.
Remark 89 Grunwald proved and published the theorem without noticing the exceptional cases (and even a second proof due to Whaples was published and reviewed by Chevalley later!). Wang found a counter-example and proved the right version of the theorem in his thesis. The mistake occurred in one of Grunwald's lemmas where the case fails and is essentially due to the sloppy usage of the notation whose meaning depends on a choice of -th root of unity. For this reason Artin-Tate strongly urged that the notation should never be used unless contains the -th roots of unity!

The proof only uses the result on split primes (Corollary 10), so it is not really an application of class field theory.

11/05/2012

Proof It suffices to treat the case that is a power of a prime. In fact, suppose with coprime, then by the Euclidean algorithm we can write . Suppose , then . We can further assume that . If and such that . Then is purely inseparable over . But is separable for any , which implies that and .

First consider the case that is cyclic of -power order. We know that splits into linear factors over by the previous proposition. Now look at the factorization over and choose a root of in for each . If for some , then for some and splits in (notice that is always abelian). Since is a cyclic of -power order, its subfields are totally ordered. Hence there exists an such that for any . Therefore splits in for almost all . By Corollary 10, we know that and has a linear factor over , i.e., .

The case and the follows since is always cyclic of 2-power order, due to the assumption in the number field case and the fact that every finite extension of the constant field is cyclic extension in the global function field case. It remains to prove the case odd and . Notice that the extension is always cyclic of -power order and we can apply the first case to find such that . Let , then is an -th power, where is the degree of . Because is coprime to , we know that itself is an -th power. ¡õ

Let us analyze the exceptional case in more detail. It is remarkable that we can write down the exceptional cases completely. In general when the local-global principle fails, it is quite rare that the failure can be completely classified.

Suppose is not cyclic. We choose a -th root of unity such that , and . Write .

Lemma 18
1. .
2. .
3. is cyclic of 2-power order.
Proof Corresponding to the decomposition of , we have the decomposition of . The lemma follows immediately from this decomposition. ¡õ

Let be the unique integer such that and .

Lemma 19 is cyclic for any if and only if .
Proof If is cyclic, then it contains a unique quadratic extension as . We find that , hence . If , then is cyclic for all by the previous lemma. For , is also cyclic because . ¡õ
Remark 90 Suppose we are in the exceptional case, then the previous lemma implies that . Then is a degree 4 abelian extension of , which is the compositum of the three quadratic extensions , and . In particular, , and are all non-squares in .
Definition 69 Let be a finite set. We define .
Remark 91 Write . Suppose and , then we know that by the Grunwald-Wang theorem, hence but by the Euclidean algorithm. Moreover, because is cyclic, the Grunwald-Wang theorem implies that .

The following key lemma characterizes all the counter examples.

Lemma 20 If and , then , where .
Proof is always cyclic, so by the Grunwald-Wang theorem there exists such that . Let be the complex conjugation. We compute So is a -th root of unity. But , hence is a -root of unity. So we can write for some . Define an element for some to be determined. Then If is even, then we choose so that . Then and (as ), which contradicts our assumption. If is odd, we choose so that . Replacing by , we may assume that . Notice that , we compute If , then , hence . So , where Notice that . Hence . Also , hence the lemma follows. ¡õ

11/07/2012

We can actually show that . If not, then . Thus for some -th root of unity . But , which contradicts our assumption. Therefore if then is the disjoint union of and .

We have shown that in order to ensure to be non-cyclic, we necessarily have , are non-squares in . We shall show this obstruction always fails with the further condition (d) as in the following strong version of Grunwald-Wang theorem.

Theorem 41 (Strong Grunwald-Wang) except the following condition holds:
1. is a number field.
2. .
3. , where .
4. contains the set of places over 2 such that are non-squares in . In the exceptional case, .
Proof Suppose is a number field and , then (a)-(c) holds and . We have to show that (d) holds, i.e., for , then . If . Then we deduce that by Euclidean algorithm. Hence there exits and . If , then , we know . Otherwise, , hence .

For the other direction, we must show that if (a)-(d) hold, then . It is equivalent to showing that for any . Let . Condition (b) implies that is a degree four extension with Galois group . Moreover, is unramified away from 2 by construction. If is a palce of above . Then is quadratic. Consequently one of is a square in . If and , the one of is a square in by the assumption (d). In either case,

1. if or is a square in . Then . Therefore .
2. if , then , hence . We conclude that . ¡õ

11/30/2012

## Outline of the proof

### Group cohomology

Suppose is any group. From an exact sequence of -modules, taking -invariants gives an exact sequence But the last map is not necessarily surjective, e.g., consider and the Kummer sequence The group cohomology functor , extends the above left exact sequence to the expected long exact sequence.

Dually, we can also take the -coinvariants (the largest quotient on which acts trivially) which does not preserve injectivity and the group homology functor fills in the corresponding long exact sequences.

When is a finite group, Tate defines the norm map . It maps into and factors through . Then we can connect both long exact sequence in cohomology and homology as Write , and . Then the five lemma gives a long exact sequence in Tate group cohomology in both directions.

When is cyclic, we further have , i.e., the Tate cohomology groups are periodic of period 2.

Example 30 Let be finite. Then the Hilbert 90 theorem asserts that . In particular, when is cyclic, , which implies that if and only if (the classical formulation of Hilbert 90).
Example 31 .
Example 32 and dually , where is endowed with the trivial -action.

### Class formations

Class formations tell you all the group cohomology input in order to derive all the statements in class field theory (e.g., Artin maps).

Let be a local or global field. Suppose and is a -module. We say is a continuous -module if (e.g., and . Fix such an , for any Galois extension , we define and .

Suppose is Galois and , we have an inflation map given by the precompostion by the natural surjection on cocycles.

Suppose , we have a restriction map given by the precompostion by the natural inclusion on cocycles.

Definition 70 A class formation is the data of plus an invariant map for any finite and finite Galois satisfying
1. for any finite Galois.
2. is injective and induces an isomorphism . Moreover, . Equivalently, there exists a generator (called the fundamental class) of order such that (which formally implies that ).

Using pure group cohomology, we will prove the following main theorem.

Theorem 42 Cup product with gives isomorphisms . In particular, when and is abelian, we obtain that . We define the Artin map to be the inverse of this isomorphism.

### Local class field theory

The main reference will be Serre's local fields and Galois cohomology. We set , then . The first axiom in class formation is simply Hilbert 90 and the second axiom amounts to proving that . Let (for local fields , but this als0 works for general complete discretely valued fields with quasi-finite residue fields, however, the existence does not work, the local compactness is still needed). For any field , we write .

Theorem 43 We have an exact sequence where is the residue field of .

The only non-formal part is the following

Proposition 20 (in other words, any central simple algebra over is split after an unramified base change.
Theorem 44 When is finite, (in other words, there are no noncommutative central division algebra over a finite field).
Corollary 13 When is local, .

The class formation machinery then proves local class field theory except the existence theorem. Proving the latter boils down to constructing enough abelian extensions using cyclotomic, Kummer and Artin-Schrier extensions (together with pure topological arguments).

### Global class field theory

The main reference will be Cassels-Frolich and Artin-Tate. In global class field theory, the idele class group plays the role of in local class field theory. The first axiom in class formation is again Hilbert 90. For the second axiom, we will prove is the the product of local Artin maps and it kills all the global elements. By the exact sequence , the Brauer group of a global field fits into As we will see, the proof of the existence theorem (norm index inequality) interplays with the proof of Artin maps (class formation): first proved for cyclotomic extensions and then in general.