CLASS FIELD THEORY J.S. Milne

Preface.

These^{1}^{2} are the notes for Math 776, University of Michigan, Winter 1997, slightly revised from those handed out during the course. They have been substantially revised and expanded from an earlier version, based on my notes from 1993 (v2.01).

My approach to class field theory in these notes is eclectic. Although it is possible to prove the main theorems in class field theory using neither analysis nor cohomology, there are major theorems that can not even be stated without using one or the other, for example, theorems on densities of primes, or theorems about the cohomology groups associated with number fields. When it sheds additional light, I have not hesitated to include more than one proof of a result.

The heart of the course is the odd numbered chapters. Chapter II, which is on the cohomology of groups, is basic for the rest of the course, but Chapters IV, VI, and VIII are not essential for reading Chapters III, V, and VII. Except for its first section, Chapter I can be skipped by those not interested in explicit local class field theory.

References of the form Math xxx are to course notes available at http://www.math.lsa.umich.edu/∼jmilne.

Please send comments and corrections to me at jmilne@umich.edu.

Books including class field theory Artin, E., Algebraic Numbers and Algebraic Functions, NYU, 1951. (Reprinted by Gordon and Breach, 1967). Artin, E., and Tate, J., Class Field Theory. Notes of a Seminar at Princeton, 1951/52. (Harvard University, Mathematics Department, 1961; Benjamin, 1968; Addison Wesley, 1991). Cassels, J.W.S., and Fröhlich, A., (Eds), Algebraic Number Theory, Academic Press, 1967. Fesenko, I.B., and Vostokov, S.V., Local Fields and their Extensions, AMS, 1993. Goldstein, L.J., Analytic Number Theory, Prentice Hall, 1971. Iwasawa, K., Local Class Field Theory, Oxford, 1986. Iyanaga, S., (ed.), The Theory of Numbers, North-Holland, 1975. Janusz, G.J., Algebraic Number Fields, Academic Press, 1973; Second Edition, AMS, 1996. Koch, H., Number Theory II, Algebraic Number Theory, Springer, 1992. (Encyclopaedia of Mathematical Sciences, Vol. 62, Parshin, A.N., and Shafarevich, I.R., Eds). Lang, S., Algebraic Number Theory, Addison-Wesley, 1970. Neukirch, J., Class Field Theory, Springer, 1985. Neukirch, J., Algebraische Zahlentheorie, Springer, 1992. Serre, J-P., Corps Locaux, Hermann, 1962.

1May 6, 1997; v3.1

2Copyright 1996, 1997, J.S. Milne. You may make one copy of these notes for your own personal use.

Weil, A., Basic Number Theory, Springer, 1967.

The articles of Serre and Tate in Cassels and Fröhlich 1967, and Janusz 1973, have been especially useful in the writing of these notes.

Books including an introduction to class field theory Cohn, H., A Classical Invitation to Algebraic Numbers and Class Fields, Springer, 1978. Cox, D.A., Primes of the Form x^{2} + ny^{2}: Fermat, Class Field Theory, and Complex Multiplication, Wiley, 1989. Marcus, D. A., Number Fields, Springer 1977.

Sources for the history of algebraic number theory and class field theory Edwards, H.M., Fermat’s Last Theorem: A Genetic Introduction to Algebraic Number Theory, Springer, 1977. Ellison, W. and F., Théorie des nombres, in Abrégé d’Histoire des Mathématiques 1700–1900, Vol I, (J. Dieudonné, ed.) Hermann, Paris, 1978, pp 165–334. Herbrand, J., Le Développment Moderne de la Théorie des Corps Algébriques Corps de Classes et Lois de Réciprocité, Gauthier-Villars, Paris, 1936. Smith, H.J.S., Report on the the Theory of Numbers, Reports of the British Association, 1859/1865. (Reprinted by Chelsea, New York, 1965.) Weil, Number Theory: An Approach Through History, Birkhäuser, 1984.

Also: Hasse’s article in Cassels and Fröhlich 1967. Appendix 2 in Iyanaga 1975. Iwasawa’s appendix to the Collected Papers of Teiji Takagi. Tate’s article “Problem 9: The general reciprocity law” in Mathematical Developments Arising from Hilbert’s Problems, AMS, 1976. Weil’s article, Oeuvres 1974c. N. Schappacher, On the history of Hilbert’s twelfth problem (preprint). P. Stevenhagan and H.W. Lenstra, Chebotarëv and his density theorem, Math. Intelligencer, 18.2, 1996, 26–37.

For an introduction to nonabelian class field theory: Arthur’s article in 1980 Seminar on Harmonic Analysis, CMS Conference Proceedings Vol 1, AMS, 1981.

Notations: For a ring R (always with 1), R^{×} denotes the group of invertible elements. We use X =^{df} Y to mean “X is defined to be Y ” or “X = Y by definition”, X ≈ Y to mean that X and Y are isomorphic, and X ≅ Y to mean X and Y are isomorphic with a given (canonical, or unique) isomorphism.

iv

Contents

Introduction 1

Statement of the main problem, 1; Classification of unramified abelian extensions, 2; Classification of ramified abelian extensions, 2; The Artin map, 4; Explicit class field theory, 6; Nonabelian class field theory, 6; Exercises, 7.

Chapter I. Local Class Field Theory 9

1. Statements of the Main Theorems 9 Consequences of Theorems 1.1 and 1.2, 11; Outline of the proof of the main theo-

rems, 13.

2. Lubin-Tate Formal Group Laws 15 Power series, 15; Formal group laws, 16; Lubin-Tate group laws, 19.

3. Construction of the extension K_{π} of K. 22 The local Artin map, 26.

4. The Local Kronecker-Weber Theorem 30 The ramification groups of K_{π,n}/K, 30; Upper numbering on ramification groups,

31; The local Kronecker-Weber theorem, 33; The global Kronecker-Weber theorem, 35; Where did it all come from?, 36; Notes, 37.

5. Appendix: Infinite Galois Theory and Inverse Limits 37 Galois theory for infinite extensions, 38; Inverse limits, 40;

Chapter II. The Cohomology of Groups 41

1. Cohomology 41 The category of G-modules, 41; Induced modules, 42; Injective G-modules, 43;

Definition of the cohomology groups, 44; Shapiro’s lemma, 45; Description of the cohomology groups by means of cochains, 47; The cohomology of L and L^{×}, 49; The cohomology of products, 52; Functorial properties of the cohomology groups, 52; The inflation-restriction exact sequence, 55; Cup-products, 56;

2. Homology; the Tate Groups 57 Definition of the homology groups, 57; The group H_{1}(G, ℤ), 59; The Tate groups,

vi CONTENTS

60; Cup-products, 61; The cohomology of finite cyclic groups, 61; Tate’s Theorem, 64;

3. The Cohomology of Profinite Groups 66 Direct limits, 66; Profinite groups, 67; Notes, 69;

4. Appendix: Some Homological Algebra 69 Some exact sequences, 69; The language of category theory, 70; Injective objects,

71; Right derived functors, 72; Variants, 75; The Ext groups, 75; References, 76; Chapter III. Local Class Field Theory Continued 77 1. Introduction 77 2. The Cohomology of Unramified Extensions 80

The cohomology of the units, 80; The invariant map, 81; Computation of the local Artin map, 83;

3. The Cohomology of Ramified Extensions 85 4. Complements 87

Alternative description of the local Artin map, 87; The Hilbert symbol, 88; Other Topics, 89; Notes, 89; Chapter IV. Brauer Groups 91 1. Simple Algebras; Semisimple Modules 91

Semisimple modules, 91; Simple k-algebras, 93; Modules over simple k-algebras, 95;

2. Definition of the Brauer Group 96 Tensor products of algebras, 96; Centralizers in tensor products, 97; Primordial

elements, 97; Simplicity of tensor products, 98; The Noether-Skolem Theorem, 99; Definition of the Brauer group, 100; Extension of the base field, 101;

3. The Brauer Group and Cohomology 101 Maximal subfields, 102; Central simple algebras and 2-cocycles, 104;

4. The Brauer Groups of Special Fields 107 Finite fields, 107; The real numbers, 108; A nonarchimedean local field, 108;

5. Complements 110 Semisimple algebras, 110; Algebras, cohomology, and group extensions, 110; Brauer

groups and K-theory, 111; Notes, 112; Chapter V. Global Class Field Theory: Statements 113 1. Ray Class Groups 113

Ideals prime to S, 113; Moduli, 114; The ray class group, 115; The Frobenius element, 117;

CONTENTS vii

2. Dirichlet L-Series and the Density of Primes in Arithmetic Progressions 119 3. The Main Theorems in Terms of Ideals 121

The Artin map, 121; The main theorems of global class field theory, 123; The norm limitation theorem, 125; The principal ideal theorem, 126; The Chebotarev Density Theorem, 128; The conductor-discriminant formula, 129; The reciprocity law and power reciprocity, 130; An elementary unsolved problem, 130; Explicit global class field theory: Kronecker’s Jugentraum and Hilbert’s twelfth problem, 131; Notes, 132;

4. Idèles 132 Topological groups, 133; Idèles, 133; Realizing ray class groups as quotients of 𝕀,

135; Characters of ideals and of idèles, 137; Norms of idèles, 139;

5. The Main Theorms in Terms of Idèles 140 Example, 143;

6. Appendix: Review of some Algebraic Number Theory 144 The weak approximation theorem, 144; The decomposition of primes, 145;

Chapter VI. L-Series and the Density of Primes 147 1. Dirichlet series and Euler products 147 2. Convergence Results 149

Dirichlet series, 149; Euler products, 152; Partial zeta functions; the residue formula, 153;

3. Density of the Prime Ideals Splitting in an Extension 155 4. Density of the Prime Ideals in an Arithmetic Progression 157

The Second Inequality, 162; Chapter VII. Global Class Field Theory: Proofs 163 1. Outline 163 2. The Cohomology of the Idèles 164

The norm map on idéles, 168;

3. The Cohomology of the Units 169 4. Cohomology of the Idèle Classes I: the First Inequality 171 5. Cohomology of the Idèle Classes II: The Second Inequality 174 6. The Algebraic Proof of the Second Inequality 175 7. Application to the Brauer Group 180 8. Completion of the Proof of the Reciprocity Law 182 9. The Existence Theorem 184

viii CONTENTS

10. Appendix: Kummer theory 187 Chapter VIII. Complements 191 1. The Local-Global Principle 191

nth Powers, 191; Norms, 192; Quadratic Forms, 192;

2. The Fundamental Exact Sequence and the Fundamental Class 196 The fundamental class, 199; The norm limitation theorem, 200;

3. Higher Reciprocity Laws 200 The power residue symbol, 201; The Hilbert symbol, 203; Application, 206;

4. The Classification of Quadratic Forms over a Number Field 207 Generalities on quadratic forms, 208; The local-global principle, 209; The classifi-

cation of quadratic forms over a local field, 210; Classification of quadratic forms over global fields, 213; Applications, 215;

5. Density Theorems 216 6. Function Fields 218 7. Cohomology of Number Fields 218 8. More on L-series 218

Artin L-series, 218; Hecke L-series, 219; Weil groups and Artin-Hecke L-series, 220; A theorem of Gauss, 220.

Introduction

Class field theory relates the arithmetic of a number field (or local field) to the Galois extensions of the field. For abelian extensions, the theory was developed between roughly 1850 and 1927 by Kronecker, Weber, Hilbert, Takagi, Artin, and others. For nonabelian extensions, serious progress began only about 25 years ago with the work of Langlands. Today, the nonabelian theory is in roughly the state that abelian class field theory was in 100 years ago: there are comprehensive conjectures but few proofs. In this course, we shall be concerned only with abelian class field theory.

Statement of the main problem. For a finite extension L/K of number fields and a finite set S of prime ideals of K containing all those that ramify in L, let Spl_{S}(L/K) be the set of prime ideals 𝔭 of K not in S that split completely in L, i.e., such that 𝔭O_{L} = ∏ 𝔓_{i} with f(𝔓_{i}/𝔭) = 1 for all i. For example, if L = K[α] ≅ K[X]/(f(X)) and S contains the prime ideals dividing the discriminant of f, then Spl_{S}(L/K) is the set of prime ideals 𝔭 such that f(X) splits completely modulo 𝔭. If L/K is Galois, then Spl_{S}(L/K) has^{3} density 1/[L : K], and it follows that Spl_{S}(L/K) determines^{4} L. Thus the problem of classifying the Galois extensions of K ramified only at prime ideals in S becomes that of determining which sets of primes in K arise as Spl_{S}(L/K) for some such L/K.

For quadratic extensions L/ℚ, the answer is provided by the quadratic reciprocity law:

Let m be a positive integer that is either odd or divisible by 4, and let S be a finite set of prime numbers containing all those that divide m. For a subgroup H of (ℤ/mℤ)^{×} of index 2, let

P r(H) = {(p) | p a prime number, p ∉ S, p mod m lies in H}.

Then the sets Spl_{S}(L/ℚ) for L running over the quadratic extensions of ℚ ramified only at primes in S are exactly the sets P r(H).

For example, let p be an odd prime number, and let S be a finite set of prime numbers containing p. Let p^{∗} = (−1)^{\frac{p−1}{2}} p, so that p^{∗} ≡ 1 mod 4. Then ℚ[\sqrt{p^{∗}}]/ℚ is ramified only at p. A prime number q ≠ p splits in ℚ[\sqrt{p^{∗}}] if and only if p^{∗} is a square modulo q, i.e., (\frac{p^{∗}}{q} ) = 1. But if q is odd, then the quadratic reciprocity law says that

(\frac{p}{q}) = (−1)^{\frac{p−1}{2} \frac{q−1}{2}} (\frac{q}{p}) , (\frac{−1}{q}) = (−1)^{\frac{q−1}{2}} ,

and so (\frac{p^{∗}}{q} ) = (\frac{q}{p}). Therefore q splits in ℚ[\sqrt{p^{∗}}] if and only if q is a square modulo p. The same is true of q = 2. The group (ℤ/pℤ)^{×} is cyclic of order p − 1, and so has a unique subgroup H of index 2, namely, that consisting of squares. Therefore, the sets Spl_{S}(ℚ[\sqrt{p^{∗}}]/ℚ), p an odd prime number not in S, are precisely those of the form P r(H) with H the subgroup of (ℤ/pℤ)^{×} of index 2. The proof of the general case is left as an exercise.

3A proof of this, essentially independent of the rest of the course, is given in Chapter VI).

4We are considering only the extensions of K contained in some fixed algebraic closure of K.

By extension, any result identifying the sets Spl_{S}(L/K) for some class of extensions L/K is usually called a reciprocity law, even when no reciprocity is involved.

Classification of unramified abelian extensions. Let I be the group of fractional ideals of K, i.e., the free abelian group generated by the prime ideals, and let i : K^{×} → I be the map sending a ∈ K^{×} to the principal ideal (a). The class group C of K is I/i(K^{×}). To give a subgroup H of C is the same as to give a subgroup H̃ of I containing i(K^{×}).

By a “prime” of K, we mean an equivalence class of nontrivial valuations on K. Thus there is exactly one prime for each prime ideal in O_{K}, for each embedding K ↪ ℝ, and for each conjugate pair of nonreal embeddings K ↪ ℂ. The corresponding primes are called finite, real, and complex respectively. An element of K is said to be positive at the real prime corresponding to an embedding K ↪ ℝ if it maps to a positive element of ℝ. A real prime of K is said to split in an extension L/K if every prime lying over it is real; otherwise it is said to ramify in L.

Let H be a subgroup of the class group C of K. An extension L of K is said to be a class field for H if

(a) L is a finite abelian extension of K;

(b) no prime of K ramifies in L;

(c) the prime ideals of K splitting in L are those in H̃.

The condition (b) means that no prime ideal of K ramifies in L and that no real prime of K has a complex prime lying over it, for example, ℚ[\sqrt{−5}]/ℚ fails the condition at 2, 5, ∞.

Theorem 0.1. A class field exists for each subgroup H of C. It is unique, and C/H ≈ Gal(L/K). Moreover, every extension L/K satisfying (a) and (b) is the class field of some H.

The existence of a class field for the trivial subgroup of C was conjectured by Hilbert in 1897, and proved by his student Furtwängler in 1907—for this reason, this class field is called the Hilbert class field of K. It is the largest abelian extension L of K unramified at all primes of K (including the infinite primes); the prime ideals that split in it are exactly the principal ones, and Gal(L/K) ≈ C. Hilbert also conjectured that every ideal in K becomes principal in the Hilbert class field, and this was proved by Furtwängler in 1930 (Principal Ideal Theorem) after Artin had reduced the proof to an exercise in group theory.

Example 0.2. The class number of ℚ[\sqrt{−5}] is 2 (Math 676, 4.6), and its Hilbert class field is ℚ[\sqrt{−5}, \sqrt{−1}].

Classification of ramified abelian extensions. In order to obtain ramified abelian extensions, we need to consider more general class groups. A modulus 𝔪 is a function

m : {primes of K} → ℤ

such that

(a) m(𝔭) ≥ 0 for all primes 𝔭, and m(𝔭) = 0 for all but finitely many 𝔭;

(b) if 𝔭 is a real prime, then m(𝔭) = 0 or 1; if 𝔭 is complex, then m(𝔭) = 0.

Traditionally, one writes

𝔪 = ∏ 𝔭^{m(𝔭)}.

Let S = S(𝔪) be the set of finite primes for which m(𝔭) > 0, and let I^{S} be the group of S-ideals (free abelian group generated by the prime ideals not in S). Let K_{𝔪} be the set of nonzero elements of K that can be expressed in the form \frac{a}{b} with (a), (b) ∈ I^{S}, and let K_{𝔪,1} be the subgroup of α ∈ K_{𝔪} such that

\{ \begin{matrix}ord_{𝔭}(α − 1) & ≥ m(𝔭) & for all finite 𝔭 with m(𝔭) > 0 \\ α & & is positive at all real 𝔭 with m(𝔭) = 1. \end{matrix}

Let i denote the map sending an element α of K_{𝔪,1} to the principal ideal (α) it defines in I^{S}. The ray class group for the modulus 𝔪 is defined to be

C_{𝔪} = I^{S}/i(K_{𝔪,1}).

Example 0.3. (a) If m(𝔭) = 0 for all 𝔭, then C_{𝔪} is just the usual ideal class

group.

(b) If m(𝔭) = 1 for all real primes and m(𝔭) = 0 otherwise, then K_{𝔪,1} consists of the totally positive elements of K, i.e., those that are positive in every real embedding of K, and C_{𝔪} is called the narrow class group.

(c) Let m be an integer that is either odd or divisible by 4, and let 𝔪 = (m)∞ where ∞ denotes the real prime of ℚ. Then

S = {(p) | p|m};
I^{S} = set of ideals of the form ∏(p)^{r(p)}, gcd(p, m) = 1, r(p) ∈ ℤ;
ℚ_{𝔪} = {b/c | b, c ∈ ℤ, gcd(b, m) = 1 = gcd(c, m)};
ℚ_{𝔪,1} = {a ∈ ℚ_{𝔪} | a > 0, ord_{p}(a − 1) ≥ r if p^{r}|m, r > 0}.

If c ∈ ℤ is relatively prime to m, then there exist d, e ∈ ℤ such that cd+me = 1, and so c becomes a unit in ℤ/mℤ. Each 𝔞 ∈ I^{S} can be represented 𝔞 = (b/c) with b and c positive integers. Therefore there is a well-defined map^{5} map (b/c) ↦ [b][c]^{−1} : I^{S} → (ℤ/mℤ)^{×}, which one can show induces an isomorphism C_{𝔪} → (ℤ/mℤ)^{×}.

To give a subgroup H of C_{𝔪} is the same as to give a subgroup H̃ of I^{S} containing i(K_{𝔪,1}).

Let H be a subgroup of the ray class group C_{𝔪} of K, and let S = S(𝔪). An extension L of K is said to be a class field for H if

(a) L is a finite abelian extension of K;

(b) m(𝔭) = 0 ⟹ 𝔭 does not ramify in L;

(c) the prime ideals not in S that split in L are those in H̃.

Theorem 0.4. A class field exists for each subgroup H of C_{𝔪}. It is unique, and C_{𝔪}/H ≈ Gal(L/K). Moreover, every finite abelian extension L/K is a class field for some 𝔪 and H.

5Here, and elsewhere, I use [x] to denote an equivalence class containing x.

Ray class groups were defined by Weber in 1897. Theorem 0.4 was proved by Takagi in a series of papers published between 1915 and 1922—see especially his talk at the 1920 International Congress, where he states his results almost exactly as we have (Collected Papers, p168, QA3.T341).

The class field L for the trivial subgroup of C_{𝔪} is called the ray class field for 𝔪. It is ramified only at primes dividing 𝔪, and C_{𝔪} ≈ Gal(L/K).

Example 0.5. The field ℚ[\sqrt{6}] has class number 1, and so equals its Hilbert class field. However, its narrow class group has order 2, and indeed ℚ[\sqrt{6}] does possess a quadratic extension, namely, ℚ[\sqrt{−2}, \sqrt{−3}] that is unramified over ℚ at all finite primes (but is ramified at both infinite primes).

For any finite set S of finite primes of K, Takagi’s theorem completely solves the problem of determining the sets Spl_{S}(L/K) for abelian Galois extensions L/K ramified only at primes in S.

Example 0.6. Let d = p_{1}^{∗} ⋯ p_{t}^{∗} where the p_{i} are odd primes and p_{i}^{∗} = (−1)^{\frac{p−1}{2}} p_{i}, as before. The field K = ℚ[\sqrt{d}] is ramified exactly over the primes p_{1}, …, p_{t}. Consider K[\sqrt{p_{i}^{∗}}]. It contains ℚ[\sqrt{p_{1}^{∗} ⋯ p_{i−1}^{∗}p_{i+1}^{∗} ⋯ p_{t}^{∗}}], which is unramified over p_{i}, and so p_{i} can’t be totally ramified in K[\sqrt{p_{i}^{∗}}]. As p_{i} ramifies in K/ℚ, it follows that the prime above p_{i} in K does not ramify in K[\sqrt{p_{i}^{∗}}]. No other prime ramifies, and so K[\sqrt{p_{i}^{∗}}] is unramified over K. From Kummer theory, we find that

L = K[\sqrt{p_{1}^{∗}}, \sqrt{p_{2}^{∗}}, …, \sqrt{p_{t−1}^{∗}}]

has degree 2^{t−1} over K, and Gal(L/K) ≈ (ℤ/2ℤ)^{t−1}. Let C_{∞} be the narrow class group of K. The above construction shows that (C_{∞} : C_{∞}^{2}) ≥ 2^{t−1}. In fact, with a only a little more effort one can prove the following:

Let K be a quadratic extension of ℚ in which t finite primes ramify.

Then (C_{∞} : C_{∞}^{2}) = 2^{t−1}. (Koch 1992, 2.114.)

This result was known to Gauss (by different methods, and in a different language).

In particular, we see that, by using class field theory, it is easy to construct quadratic extensions of ℚ such that (C : C^{2}) is very large. By contrast, as of 1991, no quadratic field was known with (C : C^{3}) > 3^{6}. All methods of constructing elements of order 3 in the class groups of quadratic number fields seem to involve elliptic curves.

The Artin map. Takagi showed that if L is the class field for H, then C_{𝔪}/H is isomorphic to Gal(L/K). Any modern mathematician looking at this result will immediately ask whether there is a natural isomorphism C/H → Gal(L/K). Surprisingly, this question seems not to have occurred to Kronecker, Weber, Hilbert, Takagi et al.. Artin did show that there is a natural isomorphism, but even he did this to obtain another result that interested him.

Let L/K be a finite Galois extension with Galois group G. Recall that the decomposition group D(𝔓) of a prime ideal 𝔓 in O_{L} is the set of σ ∈ G such that σ𝔓 = 𝔓. Any σ ∈ D(𝔓) acts continuously for the 𝔓-adic topology on L, and so extends to the completion L_{𝔓} of L at 𝔓. In this way, we obtain an isomorphism D(𝔓) → Gal(L_{𝔓}/K_{𝔭}), 𝔭 = 𝔓 ∩ K. When 𝔓 is unramified over 𝔭, then the natural map Gal(L_{𝔓}/K_{𝔭}) → Gal(l/k) is an isomorphism. Here l and k are the residue fields at 𝔓 and 𝔭. Therefore, there is a unique element (𝔓, L/K) ∈ D(𝔓) inducing the Frobenius map x ↦ x^{q}, q = (O_{K} : 𝔭), on the residue field. It is the unique σ ∈ Gal(L/K) such that:

(a) σ𝔓 = 𝔓;

(b) for all α ∈ O_{L}, σα ≡ α^{q} mod 𝔓.

If 𝔔 also lies over 𝔭, then 𝔔 = τ𝔓 for some τ ∈ Gal(L/K), and (𝔔, L/K) = τ ∘ (𝔓, L/K) ∘ τ^{−1}. Therefore, to each prime 𝔭 of K that is unramified in L, we have attached a conjugacy class

(𝔭, L/K) =^{df} {(𝔓, L/K) | 𝔓 ∩ O_{K} = 𝔭}

of elements in Gal(L/K). Note that each element in (𝔭, L/K) has order f(𝔓/𝔭).

In an abelian group, conjugacy classes consist of single elements, and so in this case we can regard (𝔭, L/K) as an element of Gal(L/K).

Theorem 0.7. Let L be an abelian extension of K, and let S be the set of finite primes ramifying in L. Then, for some modulus 𝔪 with S(𝔪) = S, the map

𝔭 ↦ (𝔭, L/K) : I^{S} → Gal(L/K)

factors through I^{S}/i(K_{𝔪,1}), and defines an isomorphism C_{𝔪}/H → Gal(L/K) where H is the ray class group corresponding to L.

If the map were not surjective, then there would be a proper extension of K in which every prime of K splits. It is easy to see analytically (much harder algebraically) that no such extension exists. The difficult point to prove is that there exists a modulus 𝔪 such that kernel of the map I^{S} → Gal(L/K) contains i(K_{𝔪,1}) for some 𝔪.

Theorem 0.7 was proved by Artin (published 1927), and the homomorphism C_{𝔪} → Gal(L/K) is called the Artin (or reciprocity) map.

Example 0.8. Let m be a positive integer that is either odd or divisible by 4, and let L = ℚ[ζ] be the field generated by a primitive m^{th} root of 1. Recall (Math 676, Section 6) that there is an isomorphism

(ℤ/mℤ)^{×} → Gal(L/ℚ), [n] ↦ (ζ ↦ ζ^{n}).

Let 𝔓 be a prime of L lying over a prime p not dividing m. Because O_{L}/𝔓 has characteristic p, for any α = ∑ a_{i}ζ^{i}, a_{i} ∈ ℤ,

(∑ a_{i}ζ^{i})^{p} ≡ ∑ a_{i}ζ^{ip} mod 𝔓.

Let σ ∈ Gal(L/ℚ) map ζ to ζ^{p}. Then this congruence shows that α ∈ 𝔓 ⟹ σα ∈ 𝔓, i.e., that σ𝔓 = 𝔓, and that σ acts x ↦ x^{p} on O_{L}/𝔓. Therefore σ = (𝔓, L/ℚ) = (p, L/ℚ).

Let 𝔪 = (m)∞. It follows easily from the above remarks that p ↦ (p, L/ℚ) : I^{S(𝔪)} → Gal(L/ℚ) factors through C_{𝔪} ≅ (ℤ/mℤ)^{×}, and defines an isomorphism C_{𝔪} → Gal(L/ℚ).

Explicit class field theory. Takagi’s theorem states that with every modulus for K there is associated a ray class field L, but it does not explain how to construct L. For the ℚ, the ray class field for 𝔪 = (m)∞ is generated by ζ = e^{2πi/m}. In the 12^{th} of his famous problems, Hilbert asked whether the ray class fields for other number fields can be generated by the special values of explicit holomorphic functions. There are some very beautiful results on the problem, but they remain rather special: in general, we don’t know how to construct class fields explicitly.

Nonabelian class field theory. For abelian extensions, class field theory shows that each set Spl_{S}(L/K) is described by congruence conditions. For nonabelian extensions, this is no longer true, and any description of the sets must be analytic. I briefly explain what is expected to be true. For simplicity, I take K = ℚ.

Let L/ℚ be a finite Galois extension with Galois group G, and let S be the set of finite primes ramifying in L. From each prime (p) ∉ S, we obtain a conjugacy class (p, L/ℚ) of Frobenius elements of G. One group whose conjugacy classes we understand is GL_{n}(ℂ)—this is the theory of Jordan canonical forms—and so it is natural to fix an injective homomorphism ρ : G ↪ GL_{n}(ℂ). Then ρ maps (p, L/ℚ) to a conjugacy class Φ_{p}(ρ) in GL_{n}(ℂ). Note that

Spl_{S}(L/ℚ) = {(p) | Φ_{p}(ρ) = {I}}.

The elements of Φ_{p}(ρ) are diagonalizable, and so Φ_{p}(ρ) is determined by the common characteristic polynomial det(I − Φ_{p}(ρ)T) of its elements. Set

L_{S}(s, ρ) = ∏_{p∉S} \frac{1}{det(I − Φ_{p}(ρ)p^{−s})}, s ∈ ℂ.

For example, if L = ℚ and n = 1, then

L(s, ρ) = ∏_{p} \frac{1}{1 − p^{−s}} = ζ(s).

The product converges to a holomorphic function in some right half plane in ℂ, and an elementary lemma on Dirichlet series implies that the factors \frac{1}{det(I−Φ_{p}(ρ)p^{−s})} are uniquely determined by the analytic function L_{S}(s, ρ). The problem of describing the sets Spl_{S}(L/ℚ) then becomes that of describing the set of analytic functions that arise in this fashion. Langlands has constructed a set of L-series, called automorphic L-series, and conjectures^{6} that each L_{S}(s, ρ) is automorphic, and specifies which automorphic L-series arise in this fashion. Thus, the conjecture answers the original question for all finite Galois extensions of ℚ.

The L-functions L_{S}(s, ρ) were defined by Artin, and so are called Artin Lseries. For n = 1 (so G is abelian) and all K, Artin proved all Artin L-series are automorphic—this was his motivation for proving Theorem 0.7.

For n = 2, Langlands (and Tunnell) have proved the conjecture in some cases. (Their result played a vital role in Wiles’s work on the Taniyama conjecture.)

6Note the similarity to the Taniyama conjecture—in fact, both are special cases of a much more general conjecture.

INTRODUCTION 7

Exercises. 0.1. Complete the proof that the quadratic reciprocity law allows one to describe

the sets Spl_{S}(L/ℚ) with L/ℚ quadratic.

0.2. Prove that ℚ[\sqrt{−5}, \sqrt{−1}] is the Hilbert class field of ℚ[\sqrt{−5}]. 0.3. Prove that the map I^{S} → (ℤ/mℤ)^{×} defined in Example 0.3 has kernel i(ℚ_{𝔪,1}),

and hence induces an isomorphism C_{𝔪} → (ℤ/mℤ)^{×}. 0.4. Prove the statements in Example 0.5. 0.5. Let L = ℚ[\sqrt{−1}, \sqrt{−5}]. Then Gal(L/ℚ) = \{1, σ, τ, στ\} ≈ ℤ/2ℤ × ℤ/2ℤ,

where σ fixes ℚ[\sqrt{−1}], τ fixes ℚ[\sqrt{−5}], and στ fixes ℚ[\sqrt{5}]. (a) Show that only 2, 5, ∞ ramify in L.

(b) Compute (p, L/ℚ) for all p ≠ 2, 5.

(c) Let 𝔪 = (20)∞. Show that p ↦ (p, L/ℚ) defines an isomorphism C_{𝔪}/H →

Gal(L/ℚ) for some H ⊂ C_{𝔪}, and find H. Hint: Show L ⊂ ℚ[ζ] where ζ is a primitive 20^{th} root of 1.

8 CONTENTS

CHAPTER I Local Class Field Theory: Statements of the Main Theorems and Lubin-Tate Extensions

Local class field theory classifies the abelian extensions of a local field. From a different perspective, it describes the local components of the global Artin map.

By a local field, I mean a field K that is locally compact with respect to a nontrivial valuation. Thus it is

(a) a finite extension of ℚ_{p} for some p;

(b) a finite extension of the field of Laurent series 𝔽_{p}((T)) over the field with p elements; or

(c) ℝ or ℂ (archimedean case).

When K is nonarchimedean, the ring of integers (alias, valuation ring) in K is denoted by O_{K} (or A), its maximal ideal by 𝔪_{K} (or just 𝔪), and its group of units by O_{K}^{×} or U_{K}. A generator of 𝔪 is called a prime element of K (rather than the more customary local uniformizing parameter). If π is a prime element of K, then every element of K^{×} can be written uniquely in the form a = uπ^{m} with u ∈ O_{K}^{×} and m =^{df} ord_{K}(a). The residue field k of K has q elements, and its characteristic is p. The normalized valuation on K is defined by |a| = q^{−ord_{K}(a)}.

We fix a separable algebraic closure K^{al} of K. Throughout the chapter “extension of K” will mean “subfield of K^{al} containing K”. Both ord_{K} and | · | have unique extensions to K^{al}.

1. Statements of the Main Theorems

The composite of two finite abelian extensions of K is again a finite abelian extension of K. Therefore the union K^{ab} of all finite abelian extensions of K (in K^{al}) is an infinite abelian extension whose Galois group is the quotient of Gal(K^{al}/K) by the closure of its commutator subgroup. (See the appendix to this chapter for a review of the Galois theory of infinite extensions.)

Let L be a finite unramified extension of K. Then L is Galois over K, and there is a unique element σ ∈ Gal(L/K) such that σα ≡ α^{q} for all α ∈ O_{L}, i.e., such that σ induces the Frobenius automorphism on the residue field of L. This σ is called the Frobenius element of Gal(L/K), and is denoted Frob_{L/K}. It generates Gal(L/K). (See Math 676, 7.41 et seqq.)

Theorem 1.1 (Local Reciprocity Law). For any nonarchimedian local field, there is a unique homomorphism

ϕ_{K} : K^{×} → Gal(K^{ab}/K)

with the following properties:

(a) for any prime element π of K and any finite unramified extension L of K, ϕ_{K}(π)|L = Frob_{L/K};

(b) for any finite abelian extension L of K, Nm_{L/K}(L^{×}) is contained in the kernel of a ↦ ϕ_{K}(a)|L, and ϕ_{K} induces an isomorphism

ϕ_{L/K} : K^{×}/ Nm_{L/K}(L^{×}) → Gal(L/K).

Denote Nm_{L/K}(L^{×}) by Nm(L^{×}). Statement (b) says that, for every finite abelian extension L of K, there is a commutative diagram

K^{×}  ⟶^{ϕ_{K}} Gal(K^{ab}/K)
↓  ↓^{τ↦τ|L}
K^{×}/ Nm(L^{×}) ⟶^{ϕ_{L/K}} Gal(L/K)

with ϕ_{L/K} an isomorphism. I shall refer to ϕ_{K} (and ϕ_{L/K}) as the local Artin map. Other names in use: (local) reciprocity map, norm residue map (symbol), etc..

Theorem 1.2 (Local Existence Theorem). A subgroup N of K^{×} is of the form Nm_{L/K}(L^{×}) for some finite abelian extension L of K if and only if it is of finite index and open.

The proofs of these theorems will occupy most of the rest of this chapter and of chapter 3. First, we note an immediate consequence.

Corollary 1.3. The map L ↦ Nm(L^{×}) is a bijection from the set of finite abelian extensions of K to the set of open subgroups of finite index in K^{×}. Moreover

L_{1} ⊂ L_{2} ⟺ Nm(L_{1}^{×}) ⊃ Nm(L_{2}^{×});
Nm(L_{1} · L_{2}) = Nm(L_{1}) ∩ Nm(L_{2});
Nm(L_{1} ∩ L_{2}) = Nm(L_{1}) · Nm(L_{2}).

Proof. Let K̄ be a finite abelian extension of K. According to the Theorem 1.1, ϕ_{K̄/K} identifies K^{×}/ Nm(K̄^{×}) with Gal(K̄/K) in such a way that a subfield L of K̄ is the fixed field of the subgroup Nm(L^{×})/ Nm(K̄^{×}) of K^{×}/ Nm(K̄^{×}). Moreover, Theorem 1.2 shows that every subgroup H of K^{×} containing Nm(K̄^{×}) is a norm group. Therefore, so far as it concerns subfields of K̄ and subgroups of K^{×} containing Nm(K̄), the corollary is a restatement of the main theorem of Galois theory. The full result is obtained by letting K̄ grow.

Remark 1.4. Corollary 1.3 also holds for archimedean local fields. The abelian extensions of ℝ are ℝ and ℂ, and their norm subgroups are ℝ^{×} and ℝ_{>0}. Let H be a subgroup of finite index in ℝ^{×}. Then H ⊃ ℝ^{×m} for some m, and ℝ^{×m} = ℝ^{×} or ℝ_{>0} according as m is odd or even (apply the intermediate value theorem). Therefore ℝ^{×} and ℝ_{>0} are the only two subgroups of ℝ^{×} of finite index. Moreover, there is a unique isomorphism (local Artin map in the real case)

ℝ^{×}/ℝ_{>0} → Gal(ℂ/ℝ),

namely, that sending r to the identity map or complex conjugation according as r > 0 or r < 0. This verifies the corollary for K = ℝ, and it is even more obvious for K = ℂ.

Remark 1.5. If K has characteristic zero, then every subgroup H of K^{×} of finite index is open. To prove this, observe that a subgroup H of finite index will contain K^{×m} for some m, and that Newton’s lemma (Math 676, 7.23) applied to X^{m} − a shows that any a ∈ O_{K}^{×} such that |1 − a| < |m|^{2} is of the form u^{m} with u ∈ 1 + 𝔪. Therefore H contains an open neighbourhood of 1 in K^{×}, and, since it is a group, this implies that it is open.

If K has characteristic p ≠ 0, then not every subgroup of K^{×} of finite index is open. Note that an open subgroup is also closed (because its complement is a union of cosets of the group, which are also open). Weil 1967, II.3, Proposition 10, shows that 1 + 𝔪 ≈ ∏_{ℕ} ℤ_{p} (product of copies of ℤ_{p} indexed by ℕ), from which it follows that K^{×} has a quotient isomorphic to ∏_{ℕ} 𝔽_{p}. Let a = (a_{n})_{n∈ℕ} ∈ ∏_{ℕ} 𝔽_{p}, and let a(m) = (a(m)_{n})_{n∈ℕ} be such that a(m)_{n} = a_{n} for n ≤ m and a(m)_{n} = 0 otherwise. Then a(m) → a as m → ∞, and so ⊕_{ℕ}𝔽_{p} is dense in ∏_{ℕ} 𝔽_{p}. Therefore any proper subgroup of ∏_{ℕ} ℤ_{p} containing ⊕_{ℕ}ℤ_{p} can not be closed. Because ∏_{ℕ} 𝔽_{p}/ ⊕_{ℕ} 𝔽_{p} is a vector space over 𝔽_{p}, it will have subspaces of finite index, and the inverse image of such a subspace in K^{×} will be a nonclosed subgroup of finite index.

Remark 1.6. The composite of two finite unramified extensions of K is again unramified, and therefore the union K^{un} of all finite unramified extensions of K (in K^{al}) is an unramified extension of K. The residue field k̄ of K^{un} is an algebraic closure of the residue field k of K.

Every automorphism σ of K^{un} fixing K preserves the valuation | · | on K^{un}, and hence induces an automorphism σ̄ of k̄/k. The map σ ↦ σ̄ is an isomorphism Gal(K^{un}/K) → Gal(k̄/k). Therefore, there is a unique element Frob_{K} ∈ Gal(K^{un}/K) inducing the map x ↦ x^{q} on k̄, and the map α ↦ Frob_{K}^{α} : ℤ̂ → Gal(K^{un}/K) is an isomorphism of topological groups. Condition (a) of Theorem 1.1 can be re-stated as:

(a) for any prime element π of K, ϕ(π) acts as Frob_{K} on K^{un}.

Consequences of Theorems 1.1 and 1.2. Assume K possesses a local Artin map, i.e., a homomorphism ϕ_{K} : K^{×} → Gal(K^{ab}/K) with the properties (a) and (b) of Theorem 1.1, and that Theorem 1.2 holds.

As we just remarked, for any prime element π of K, ϕ_{K}(π)|K^{un} = Frob_{K}. If u ∈ U_{K}, then πu is also a prime element, and so

ϕ_{K}(u)|K^{un} = ϕ_{K}(πu)|K^{un} · ϕ_{K}(π^{−1})|K^{un} = id .

Therefore, U_{K} is in the kernel of a ↦ ϕ_{K}(a)|K^{un} : K^{×} → Gal(K^{un}/K), and ϕ_{K}(a)|K^{un} = Frob_{K}^{ord_{K}(a)} for any a ∈ K^{×}. In other words, the map ϕ_{K} : K^{×} → Gal(K^{un}/K) factors into

K^{×} ⟶^{ord_{K}} ℤ ⟶^{n↦Frob_{K}^{n}} Gal(K^{un}/K).

If K_{m} is the unramified extension of K of degree m, then

Nm(K_{m}^{×}) = {a ∈ K^{×} | m|ord_{K}(a)} = U_{K} · π^{mℤ}

because this is the kernel of a ↦ ϕ_{K}(a)|K_{m}.

Choose a prime element π in K. Then any element a ∈ K^{×} can be written uniquely a = u · π^{m}, u ∈ U_{K}, m ∈ ℤ. Thus

K^{×} ≈ U_{K} × ℤ, u · π^{m} ↔ (u, m).

Because U_{K} is both open and closed in K^{×}, this is an isomorphism of topological groups (discrete topology on ℤ). The subgroups 1 + 𝔪^{n}, n ≥ 0, form a fundamental system of neighbourhoods of 1 in K. Clearly therefore, the subgroups (1 + 𝔪^{n}) × mℤ are open of finite index in U_{K}×ℤ, and every open subgroup of finite index contains one of this type. For a finite abelian extension L of K, the smallest f such that Nm(L^{×}) contains 1 + 𝔪^{f} is called the conductor of L/K, except that, when Nm(L^{×}) ⊃ U_{K}, the conductor is said to be 0. Thus L/K is unramified if and only if its conductor is 0, and it is tamely ramified if and only if its conductor is ≤ 1.

The homomorphisms

ϕ_{L/K} : K^{×}/ Nm(L^{×}) → Gal(L/K)

form a projective system as L runs through the finite abelian extensions of K, ordered by inclusion. On passing to the limit, we obtain an isomorphism

ϕ̂_{K} : { \atop (K^{×})} → Gal(K^{ab}/K)

—in particular, there is a one-to-one correspondence between the set of closed subgroups of K̂^{×} and the set of abelian extensions of K. Here K̂^{×} is the completion of K^{×} with respect to the topology for which the norm groups form a fundamental system of neighbourhoods of 1. This topology on K^{×} is called the norm topology. According to Theorem 1.2, the norm groups are the open subgroups of finite index in K^{×}. The norm topology is coarser than the usual topology on K^{×}—for example, U_{K} is not open—but it induces the usual topology on U_{K}. When we complete the terms in the exact sequence

0 → U_{K} → K^{×} ⟶^{ord_{K}} ℤ → 0

with respect to the norm topology, we obtain an exact sequence

0 → U_{K} → K̂^{×} → ℤ̂ → 0.

Here ℤ̂ is completion of ℤ for the topology defined by the subgroups of finite index. Loosely speaking, K̂^{×} can be said to have been obtained from K^{×} by completing ℤ. (See Artin 1951, 9.3, for a detailed discussion of the norm topology on K^{×}.)

The choice of a prime element π determines a decomposition

K̂^{×} = U_{K} · π_{ℤ},

of K̂^{×} into the product of two closed subgroups, and hence (by infinite Galois theory), a decomposition

K^{ab} = K_{π} · K^{un}

where K_{π} is the subfield of K^{ab} fixed by ϕ_{K}(π) and K^{un} is the subfield of K^{ab} fixed by ϕ_{K}(U_{K}). Clearly, K_{π} is the union of all finite abelian extensions L/K (necessarily totally ramified) such that π ∈ Nm(L^{×}).

Aside 1.7. The composite of two totally ramified extensions need not be totally ramified. Consider, for example, ℚ[\sqrt{p}] and ℚ[\sqrt{pq}] where p and q are odd primes and q is not a square mod p. Then p is totally ramifed in each of these extensions, but not in their composite ℚ[\sqrt{p}, \sqrt{q}] because it is unramified in the subfield ℚ[\sqrt{q}]. These statements remain true when ℚ is replaced by ℚ_{p}.

Therefore, in contrast to the situation with abelian and unramified extensions, there is no “largest” totally ramified extension of K in K^{al}; there are only the maximal totally ramified extensions K_{π}, each depending on the choice of π.

Outline of the proof of the main theorems. For m ≥ 1, let K_{m} be the unique unramified extension of K of degree m.

In Section 3 of this chapter, we shall prove:

(∗) For each prime element π of K, there is a totally ramified abelian extension K_{π} = ∪_{n≥1}K_{π,n} of K and a homomorphism

ϕ_{π} : K^{×} → Gal(K_{π} · K^{un}/K)

such that

(a) ϕ_{π}(π)|K^{un} = Frob_{K};

(b) [K_{π,n} : K] = (q − 1)q^{n−1};

(c) for all m and n, ϕ_{π}(a)|K_{π,n} · K_{m} = id for a ∈ (1 + 𝔪^{n})· <π^{m}>.

(d) for all n, π is a norm from K_{π,n}.

Moreover, both K_{π} · K^{un} and ϕ_{π} are independent of the choice of π.

Both the fields K_{π,n} and the homomorphisms ϕ_{π} are explicitly constructed. In Chapter III, we shall prove:

(∗∗) There exists a homomorphism ϕ : K^{×} → Gal(K^{ab}/K) satisfying conditions (a) and (b) of Theorem 1.1.

In the remainder this section, we shall assume these two results, and prove that both Theorem 1.1 and Theorem 1.2 hold, with the added precision that K^{ab} = K_{π} · K^{un} and ϕ = ϕ_{π} for all π.

Let K′ be the subfield K_{π} · K^{un} of K^{ab}, and let ϕ′ = ϕ_{π}—recall that both K′ and ϕ′ are independent of the choice of the prime element π.

Lemma 1.8. For all a ∈ K^{×}, ϕ(a)|K_{π} · K^{un} = ϕ_{π}(a).

Proof. For any prime element π of K, ϕ(π) acts trivially on K_{π,n} because π is a norm from K_{π,n}, and ϕ′(π) acts trivially on K_{π,n} because of condition (∗c) with m = 1 (we may assume that the prime element used in the definition of ϕ′ is π). Since ϕ(π) and ϕ′(π) both act as Frob_{K} on K^{un}, they must agree on K′ = ∪K_{π,n} · K^{un}. But the prime elements of K generate K^{×} as a multiplicative group (a ∈ K^{×} can be written a = uπ^{r} = (uπ)π^{r−1} = π′π^{r−1}), and so this proves the Lemma.

Now fix a prime element π of K, and let

K_{n,m} = K_{π,n} · K_{m},

and

U_{n,m} = (1 + 𝔪^{n})· <π^{m}> .

We are given that ϕ_{π}(a)|K_{n,m} = 1 for all a ∈ U_{n,m}. Hence ϕ(a)|K_{m,n} = 1 for all a ∈ U_{n,m}, and so U_{n,m} ⊂ Nm(K_{n,m}^{×}). But

(K^{×} : U_{n,m}) = (U : 1 + 𝔪^{n})(<π>:<π^{m}>)
= (q − 1)q^{n} · m
= [K_{π,n} : K][K_{m} : K]
= [K_{m,n} : K],

and we are given that ϕ induces an isomorphism

K^{×}/ Nm(K_{n,m}^{×}) → Gal(K_{n,m}/K).

Therefore,

U_{n,m} = Nm(K_{n,m}^{×}).

Lemma 1.9. Let L be a finite extension of K, and assume that Nm(L^{×}) is of finite index in K^{×}. Then Nm(L^{×}) is open in K^{×}.

Proof. Let U_{L} = O_{L}^{×}. Then U_{L} is compact, and so Nm(U_{L}) is closed in K^{×}. Since only units have norms that are units, U_{K}/ Nm(U_{L}) ↪ K^{×}/ Nm(L^{×}). Therefore, Nm(U_{L}) is closed of finite index in U_{K}, and hence is open in U_{K} (and also in K^{×}). Thus Nm(L^{×}) contains an open subgroup of K^{×}, and so is itself open.

Now let L be a finite abelian extension of K. By assumption (∗∗) K^{×}/ Nm(L^{×}) ≈ Gal(L/K), and so Nm(L^{×}) is of finite index in K^{×}. Because it is an open subgroup of finite index in K^{×}, Nm(L^{×}), contains U_{n,m} for some n, m ≥ 0. The map

ϕ : K^{×} → Gal(L · K_{n,m}/K)

is onto and, for a ∈ K^{×},

ϕ(a) fixes the elements of L ⟺ a ∈ Nm(L^{×}),
ϕ(a) fixes the elements of K_{n,m} ⟺ a ∈ Nm(K_{n,m}^{×}) = U_{n,m}.

Because Nm(L^{×}) ⊃ U_{n,m}, this implies that L ⊂ K_{n,m}.

This completes the proof that K^{ab} = K_{π} · K^{un} and that ϕ = ϕ_{π}. To complete the proof of the existence theorem, we have to show that every open subgroup H of K^{×} of finite index is a norm group, but, as we observed above, every such group contains U_{n,m} for some n and m, and U_{n,m} = Nm(K_{n,m}). Let L be the subfield of K_{n,m} fixed by ϕ_{K_{n,m}/K}(H). Then H is the kernel of ϕ : K^{×} → Gal(L/K), and so equals Nm(L^{×}) by (∗∗).

Finally, we prove that there exists at most one homomorphism ϕ : K^{×} → Gal(K^{ab}/K) satisfying the conditions (a) and (b) of Theorem 1.1. Let π be a prime element of K. For all n, π ∈ Nm(K_{π,n}^{×}), and so condition (b) of the theorem implies that ϕ(π) acts as the identity on K_{π,n}. Therefore ϕ(π) acts as the identity on K_{π}, and this shows that ϕ(π) is uniquely determined by the conditions (a) and (b) of the Theorem. Because the prime elements generate the group K^{×}, this shows that ϕ itself is uniquely determined by the conditions.

Exercise 1.10. Use only results from algebraic number theory (e.g., Math 676) to prove that a finite extension L/K of local fields is totally ramified if and only if Nm(L/K) contains a prime element.

The reader not interested in the explicit generation of K^{ab}, nor in the explicit description of the local Artin map ϕ_{K}, can skip the rest of this chapter and go directly to Chapters II and III.

2. Lubin-Tate Formal Group Laws

Power series. Let A be a ring (always commutative with 1). A power series with coefficients in A is an infinite sequence

f = (a_{0}, a_{1}, a_{2}, …), a_{i} ∈ A, i ∈ ℕ.

Addition and multiplication are defined by

(a_{0}, a_{1}, …) + (b_{0}, b_{1}, …) = (a_{0} + b_{0}, a_{1} + b_{1}, …)
(a_{0}, a_{1}, …)(b_{0}, b_{1}, …) = (a_{0}b_{0}, …, ∑_{i+j=k} a_{i}b_{j}, …).

These formulas are easier to remember if we write

f = ∑_{i≥0} a_{i}T^{i}.

The power series with coefficients in A form a commutative ring, which we denote by A[[T]]. Power series can be manipulated in the same way as polynomials, with a few cautions. For example, in general we can not substitute an element c ∈ A into a power series f(T) ∈ A[[T]], because computing f(c) = ∑_{i≥0} a_{i}c^{i} requires us to sum an infinite number of elements of A, which, not being analysts, we can not do. For the same reason, we can not substitute one power series g(T) into a second f(T) if g(T) has a nonzero constant term. However, if the constant term of g(T) is zero, then f(g(T)) is defined—we denote it by f ∘ g.

Lemma 2.1. (a) For all f ∈ A[[T]], g, h ∈ T A[[T]], f ∘ (g ∘ h) = (f ∘ g) ∘ h.

(b) Let f = ∑_{i≥1}^{∞} a_{i}T^{i} ∈ T A[[T]]. There exists a g ∈ T A[[T]] such that f ∘ g = T if and only if a_{1} ≠ 0, in which case g is unique, and has the property that g ∘ f = T.

Proof. (a) In general, (f_{1}f_{2}) ∘ g = (f_{1} ∘ g)(f_{2} ∘ g), and so f^{n} ∘ g = (f ∘ g)^{n}. Therefore, when f = T^{n}, both f ∘ (g ∘ h) and (f ∘ g) ∘ h equal (g ∘ h)^{n}, and when f = ∑ a_{i}T^{i}, both equal ∑ a_{i}(g ∘ h)^{i}.

(b) We seek a g = ∑_{i≥1} b_{i}T^{i} such that

∑_{i≥1} a_{i}g^{i} = T,

i.e., such that

a_{1}b_{1} = 1
a_{1}b_{2} + a_{2}b_{1} = 0
· · · · · · · · ·
a_{1}b_{n} + polynomial in a_{2}, …, a_{n}, b_{1}, …, b_{n−1} = 0
· · · · · · · · ·

The first equation shows that, in order for g to exist, a_{1} must be invertible. Conversely, when a_{1} is invertible, the equations define the b_{i}’s uniquely. Now, because b_{1} is invertible, the same argument shows that there exists an h ∈ T A[[T]] such that g ∘ h = T. But

f = f ∘ T = f ∘ g ∘ h = T ∘ h = h,

and so g ∘ f = T.

Caution: f ∘ (g + h) ≠ f ∘ g + f ∘ h in general.

Power series in several variables can be defined similarly. If f(X_{1}, …, X_{n}) ∈ A[[X_{1}, …, X_{n}]] and g_{1}, g_{2}, …, g_{n} ∈ A[[Y_{1}, …, Y_{m}]], then f(g_{1}, …, g_{n}) is a welldefined element of A[[Y_{1}, …, Y_{m}]] provided that the constant terms of the g_{i} are all zero.

Remark 2.2. Let A be a complete discrete valuation ring, and let 𝔪 be the maximal ideal in A. For any f = ∑_{i≥0} a_{i}T^{i} ∈ A[[T]] and any c ∈ 𝔪, a_{i}c^{i} → 0 as i → ∞. Therefore the series ∑_{i≥0} a_{i}c^{i} converges to an element f(c) ∈ 𝔪.

We often abbreviate “+terms of degree ≥ m” to “+ deg ≥ m”.

Formal group laws. A group is a nonempty set together with a law of composition satisfying the group axioms. A formal group law is a law of composition (without the set) satisfying the group axioms. More precisely:

Definition 2.3. Let A be a commutative ring. A one-parameter commutative formal group law is a power series F ∈ A[[X, Y]] such that

(a) F (X, Y) = X + Y + terms of degree ≥ 2;

(b) F (X, F (Y, Z)) = F (F (X, Y), Z);

(c) there exists a unique i_{F} (X) ∈ XA[[X]] such that F (X, i_{F} (X)) = 0;

(d) F (X, Y) = F (Y, X).

To get an n-parameter group law, replace each of X and Y with sequences of nvariables. Axiom (d) is the commutativity condition. Since we consider no other, we shall refer to one-parameter commutative formal group laws simply as formal group laws. We shall be especially interested in the case that A = O_{K}, the ring of integers in a nonarchimedean local field K.

Remark 2.4. Condition (a) ensures that F (X, Y) has no constant term, and so axiom (b) makes sense: we are comparing finite sums at each degree.

(b) On taking Y = Z = 0 in Axioms (a) and (b), we find that

F (X, 0) = X + deg ≥ 2, F (F (X, 0), 0) = F (X, 0).

Denote the power series F (X, 0) by f(X). The first equality implies that there exists a g such that f ∘ g = X, and the second equality says that f ∘ f = f. On composing the second equality with g we find that f = X. Thus F (X, 0) = X, and similarly F (0, Y) = Y. Hence

F (X, Y) = X + Y + ∑_{\begin{matrix}1≤i<∞ \\ 1≤j<∞ \end{matrix}} a_{i,j}X^{i}Y^{j}.

Exercise 2.5. Let F (X, Y) be a power series such that F (X, 0) = X and F (0, Y) = Y. Show that there is a unique power series G(X) = −X + ∑_{i=2}^{∞} a_{i}X such that F (X, G(X)) = 0. Hence Axiom (c) is redundant.

Let A = O_{K}, and let F = ∑_{i,j}^{∞} a_{ij}X^{i}Y^{j} be a formal group law over O_{K}. For any x, y ∈ 𝔪_{K}, a_{ij}x^{i}y^{j} → 0 as (i, j) → ∞, and so the series

F (x, y) = ∑ a_{ij}x^{i}y^{j}

converges to an element x+_{F} y of 𝔪_{K}. In this way, 𝔪_{K} becomes a commutative group (𝔪_{K}, +_{F}). Similarly, 𝔪_{L} acquires a group structure for any finite extension L of K, and the inclusion (𝔪_{K}, +_{F}) ↪ (𝔪_{L}, +_{F}) is a homomorphism.

Example 2.6. (a) Let F (X, Y) = X + Y. Then +_{F} is the usual addition law on 𝔪_{K}.

(b) Let F (X, Y) = X + Y + XY. The map

a ↦ 1 + a: 𝔪 → 1 + 𝔪

is an isomorphism (𝔪, +_{F}) → (1 + 𝔪, ×). Check:

(a, b)  ⟶ (1 + a, 1 + b)
↓  ↓
a +_{F} b = a + b + ab ⟶ 1 + a + b + ab.

(c) Let E be an elliptic curve over a nonarchimedean local field K, and let

Y^{2}Z + a_{1}XY Z + a_{3}Y Z^{2} = X^{3} + a_{2}X^{2}Z + a_{4}XZ^{2} + a_{6}Z^{3}

be a minimal Weierstrass model of E. Let T = X/Y. On expanding the group law on E as a power series in T_{1}, T_{2}, we obtain a formal group law F_{E}(T_{1}, T_{2}) over O_{K}. See J. Silverman, The Arithmetic of Elliptic Curves, Springer, 1986, Chapter IV.

Definition 2.7. Let F (X, Y) and G(X, Y) be formal group laws. A homomorphism F → G is a power series h ∈ T A[[T]] such that

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

When there exists a homomorphism h′ : G → F such that h ∘ h′ = T = h′ ∘ h, then h is called an isomorphism. A homomorphism h : F → F is called an endomorphism of F.

In the case A = O_{K}, a homomorphism f: F → G defines a homomorphism

a ↦ f(a): (𝔪_{L}, +_{F}) → (𝔪_{L}, +_{G})

for any L ⊃ K.

Example 2.8. Let F = X+Y +XY = (1+X)(1+Y)−1. Then f(T) = (1+T)^{p}−1 is an endomorphism of F, because

F (f(X), f(Y)) = (1 + X)^{p}(1 + Y)^{p} − 1 = f(F (X, Y)).

Note that the following diagram commutes,

𝔪 ⟶^{f} 𝔪
↓^{a↦1+a}  ↓^{a↦1+a}
1 + 𝔪 ⟶^{a↦a^{p}} 1 + 𝔪

i.e., when we identify (𝔪, +_{F}) with (1 + 𝔪, ×), f becomes identified with a ↦ a^{p}.

Let G be a formal group law. For any f, g ∈ T A[[T]], we define

f +_{G} g = G(f(T), g(T)).

Because of the Axioms 2.3a,b,c,d, this composition law makes T A[[T]] into a commutative group. In particular,

f +_{G} (i_{G} ∘ f) = 0.

Lemma 2.9. (a) For any formal group laws F and G, the set Hom(F, G) of homomorphisms from F to G becomes an abelian group with the addition f +_{G} g.

(b) For any formal group law F, the abelian group End(F) of endomorphisms of F becomes a ring with the multiplication f ∘ g.

Proof. Let f and g be homomorphisms F → G, and let h = f +_{G} g. Then

h(F (X, Y)) =^{df} G(f(F (X, Y)), g(F (X, Y)))
= G(G(f(X), f(Y)), G(g(X), g(Y))).

Symbolically (at least), we can write this last power series as

(f(X) +_{G} f(Y)) +_{G} (g(X) +_{G} g(Y)), (∗)

which associativity and commutativity allow us to rewrite as

(f(X) +_{G} g(X)) +_{G} (f(Y) +_{G} g(Y)), (∗∗)

that is, as G(h(X), h(Y)). More formally, the operations that carry (∗) into (∗∗), also carry G(G(f(X), f(Y)), G(g(X), g(Y))) into G(h(X), h(Y)). This proves that h ∈ Hom(F, G). Similarly, one shows that i_{G} ∘ f ∈ Hom(F, G). As 0 ∈ Hom(F, G), this completes the proof that Hom(F, G) is a subgroup of (T A[[T]], +_{G}).

We showed in Lemma 2.1 that f, g ↦ f ∘ g is associative. To show that End(F) is a ring, it remains to observe that, for f, g, h ∈ End(F),

f ∘ (g +_{F} h) =^{df} f(F (g(X), h(Y))) = F ((f ∘ g)(X), (f ∘ h)(Y)) = f ∘ g +_{F} f ∘ h,

and that End(F) has an identity element, namely, T.

Formal group laws are similar to algebraic groups. The main difference is that, because they are defined by power series rather than polynomials, their points must have coordinates “close to 1” in order for products to be defined. There is a very extensive theory of formal group laws—see, for example, M. Hazewinkel, Formal Groups and Applications, Academic Press, 1978.

Lubin-Tate group laws. We now take A = O_{K}, the valuation ring in a nonarchimdean local field K, and we choose a prime element π of A.

Definition 2.10. Let F_{π} be the set of f(X) ∈ A[[X]] such that

(a) f(X) = πX + terms of degree ≥ 2;

(b) f(X) ≡ X^{q} mod π.

Example 2.11. (a) The polynomial f(X) = πX + X^{q} lies in F_{π}.

(b) Take K = ℚ_{p}, π = p; then

f(X) = (1 + X)^{p} − 1 = pX + \binom{p}{2}X^{2} + ⋯ + pX^{p−1} + X^{p} ∈ F_{p}.

Lemma 2.12. Let f, g ∈ F_{π}, and let ϕ_{1}(X_{1}, ⋯, X_{n}) be a linear form with coefficients in A. There is a unique ϕ ∈ A[[X_{1}, …, X_{n}]] such that

\{ \begin{matrix}ϕ(X_{1}, …, X_{n}) & = & ϕ_{1} + terms of degree ≥ 2 \\ f(ϕ(X_{1}, …, X_{n})) & = & ϕ(g(X_{1}), …, g(X_{n})). \end{matrix}

Proof. We prove by induction on r that there is a unique polynomial ϕ_{r}(X_{1}, …, X_{n}) of degree r such that

\{ \begin{matrix}ϕ_{r}(X_{1}, …, X_{n}) & = & ϕ_{1} + terms of degree ≥ 2 \\ f(ϕ_{r}(X_{1}, …, X_{n})) & = & ϕ_{r}(g(X_{1}), …, g(X_{n})) + terms of degree ≥ r + 1. \end{matrix}

The unique candidate for the first polynomial is ϕ_{1} itself. It certainly satisfies the first condition, and, if we write ϕ_{1} = ∑ a_{i}X_{i}, the second says that

π(∑ a_{i}X_{i}) = ∑ a_{i}(πX_{i}) + deg ≥ 2,

which is also true.

Suppose r ≥ 1 and we have defined ϕ_{r}. Because ϕ_{r} is unique, ϕ_{r+1} must equal ϕ_{r} + Q, where Q is a homogeneous polynomial of degree r + 1 in A[X_{1}, …, X_{n}]. We need that

f(ϕ_{r+1}(X_{1}, …, X_{n})) =^{?} ϕ_{r+1}(g(X_{1}), …, g(X_{n})) + deg ≥ r + 2.

The left hand side is

f(ϕ_{r}(X_{1}, …, X_{n})) + πQ(X_{1}, …, X_{n}) + deg ≥ r + 2,

while the right hand side is

ϕ_{r}(g(X_{1}), …, g(X_{n})) + Q(πX_{1}, …, πX_{n}) + deg ≥ r + 2.

As Q is homogeneous of degree r +1, Q(πX_{1}, …) = π^{r+1}Q(πX_{1}, …), and so we need that

(π^{r+1} − π)Q(X_{1}, …, X_{n}) =^{?} f(ϕ_{r}(X_{1}, …, X_{n})) − ϕ_{r}(g(X_{1}), …, g(X_{n})) + deg ≥ r + 2.

Thus Q must be the unique polynomial such that

\frac{f(ϕ_{r}(X_{1}, …, X_{n})) − ϕ_{r}(g(X_{1}), …, g(X_{n}))}{(π^{r} − 1)π} = Q + deg ≥ r + 2.

Note that, because of the simple form that binomial theorem takes in characteristic p,

f ∘ ϕ_{r} − ϕ_{r} ∘ g ≡ ϕ_{r}(X_{1}, …, X_{n})^{q} − ϕ_{r}(X_{1}^{q}, …, X_{n}^{q}) ≡ 0 mod π.

Because π divides f ∘ ϕ_{r} − ϕ_{r} ∘ g and π^{r} − 1 is invertible in A, Q does have coefficients in A, and because ϕ_{r} satisfies the induction hypothesis, it does have degree r + 1.

Having defined the ϕ_{r} for r = 1, 2, … and noted that

ϕ_{r+1} = ϕ_{r} + deg ≥ r + 1,

we can define ϕ to be the unique power series such that

ϕ = ϕ_{r} + deg ≥ r + 1

for all r. Clearly, it has the first of the required properties, and for any r,

f(ϕ(X_{1}, …, X_{n})) = f(ϕ_{r}(X_{1}, …, X_{n})) + deg ≥ r + 1
= ϕ_{r}(g(X_{1}, …, X_{n})) + deg ≥ r + 1
= ϕ(f(X_{1}, …, X_{n})) + deg ≥ r + 1.

Since this holds for all r, ϕ also has the second required property.

Proposition 2.13. For any f ∈ F_{π}, there is a unique formal group law F_{f} with coefficients in A admitting f as an endomorphism.

Proof. According to Lemma 2.12, there is a unique power series F_{f}(X, Y) such that

\{ \begin{matrix}F_{f}(X, Y) & = & X + Y + terms of degree ≥ 2 \\ f(F_{f}(X, Y)) & = & F_{f}(f(X), f(Y)). \end{matrix}

It remains to check that this is a formal group law.

Commutativity: Let G = F_{f}(Y, X). Then

\{ \begin{matrix}G(X, Y) & = & X + Y + terms of degree ≥ 2 \\ f(G(X, Y)) & = & f(F_{f}(Y, X)) = F_{f}(f(Y), f(X)) = G(f(X), f(Y)). \end{matrix}

Since F_{f}(X, Y) is the unique power series with these properties, it follows that G(X, Y) = F_{f}(X, Y).

Associativity: Let G_{1}(X, Y, Z) = F_{f}(X, F_{f}(Y, Z)) and G_{2}(X, Y, Z) = F_{f}(F_{f}(X, Y), Z). Then, for i = 1, 2,

\{ \begin{matrix}G_{i}(X, Y, Z) & = & X + Y + Z + terms of degree ≥ 2 \\ G_{i}(f(X), f(Y), f(Z)) & = & f(G_{i}(X, Y, Z)) \end{matrix}

and again Lemma 2.12 shows that there is only one power series satisfying these conditions.

Example 2.14. Let K = ℚ_{p} and π = p. Then f =^{df} (1 + T)^{p} − 1 ∈ F_{p}, and F =^{df} X + Y + XY admits f as an endomorphism (see 2.8). Therefore, F = F_{f}.

The F_{f}’s are the Lubin-Tate formal group laws. They are exactly the formal group laws that admit an endomorphism reducing mod 𝔪 to the Frobenius T ↦ T^{q} and whose derivative at the origin is a prime element of K.

Proposition 2.15. For f, g ∈ F_{π} and a ∈ A, let [a]_{g,f} be the unique element of A[[T]] such that

\{ \begin{matrix}[a]_{g,f}(T) & = & aT + terms of degree ≥ 2 \\ g ∘ [a]_{g,f} & = & [a]_{g,f} ∘ f. \end{matrix}

Then [a]_{g,f} is a homomorphism F_{f} → F_{g}.

Proof. Let h = [a]_{g,f}—its existence is guaranteed by Lemma 2.12. We have to show that

h(F_{f}(X, Y)) = F_{g}(h(X), h(Y)).

Obviously each = aX + aY + deg ≥ 2. Moreover,

h(F_{f}(f(X), f(Y))) = (h ∘ f)(F_{f}(X, Y)) = g(h(F_{f}(X, Y))),
F_{g}(h(f(X)), h(f(Y))) = F_{g}(g(h(X)), g(h(Y))) = g(F_{g}(h(X), h(Y))),

and we can apply the uniqueness in Lemma 2.12 again.

Proposition 2.16. For any a, b ∈ A,

[a + b]_{g,f} = [a]_{g,f} +_{F_{g}} [b]_{g,f}

and

[ab]_{h,f} = [a]_{h,g} ∘ [b]_{g,f}.

Proof. In each case, the power series on the right satisfies the conditions characterizing the power series on the left.

Corollary 2.17. For f, g ∈ F_{π}, F_{f} ≈ F_{g}.

Proof. For any u ∈ A^{×}, [u]_{f,g} and [u^{−1}]_{g,f} are inverse isomorphisms.

In fact, there is a unique isomorphism h: F_{f} → F_{g} such that h(T) = T + ⋯ and g ∘ h = h ∘ f, namely, [1]_{g,f}.

Corollary 2.18. For each a ∈ A, there is a unique endomorphism [a]_{f}: F_{f}→ F_{f} such that [a]_{f} = aT + deg ≥ 2 and [a]_{f} commutes with f. The map

a ↦ [a]_{f}: A ↪ End(F_{f})

is a ring homomorphism.

Proof. Take [a]_{f} = [a]_{f,f}—it is the unique power series aT + ⋯ commuting with f, and it is an endomorphism of F_{f}. That a ↦ [a]_{f} is a ring homomorphism follows from Lemma 2.16 and the obvious fact that [1]_{f} = T.

Hence the abelian group (𝔪_{L}, +_{F_{f}}) has a natural A-module structure for any finite extension L of K.

Example 2.19. Let K = ℚ_{p} and f = (1+T)^{p} −1 ∈ F_{p}, so that F_{f} = X +Y +XY. For any a ∈ ℤ_{p}, define

(1 + T)^{a} = ∑_{m≥0} \binom{a}{m}T^{m}, \binom{a}{m} = \frac{a(a − 1) ⋯ (a − m + 1)}{m(m − 1) ⋯ 1}.

When a ∈ ℤ, these definitions agree with the usual ones, and if (a_{i})_{i≥1} is a sequence of integers converging to a ∈ ℤ_{p}, then \binom{a_{i}}{m} → \binom{a}{m} as i → ∞. Therefore \binom{a}{m} ∈ ℤ_{p}. I claim that

[a]_{f} = (1 + T)^{a} − 1.

Certainly, (1 + T)^{a} − 1 = aT + ⋯, and

f ∘ ((1 + T)^{a} − 1) = (1 + T)^{ap} − 1 = ((1 + T)^{a} − 1) ∘ f

holds when a is an integer, which (by continuity) implies that it holds for all a ∈ ℤ_{p}.

Under the isomorphism (𝔪, +_{F_{f}}) ⟶^{t↦1+t} (1 + 𝔪, ×), the action of [a]_{f} corresponds to the map sending an element of 1 + 𝔪 to its ath power.

Remark 2.20. (a) Note that [π]_{f} = f, because f satisfies the two defining

conditions.

(b) The homomorphism a ↦ [a]_{f} : A ↦ End(F_{f}) is injective, because a can be recovered as the leading coefficient of [a]_{f}.

(c) The canonical isomorphism [1]_{g,f} : F_{f} → F_{g} commutes with the actions of A on F_{f} and F_{g}, because

[a]_{g} ∘ [1]_{g,f} = [a]_{g,f} = [1]_{g,f} ∘ [a]_{f}.

Summary 2.21. For each f ∈ F_{π}, there exists a unique formal group law F_{f} admitting f as an endomorphism. Moreover, there is a unique A-module structure a ↦ [a]_{f} : A → End(F_{f}) on F_{f} such that

(a) [a]_{f} = aT + deg ≥ 2, all a ∈ A;

(b) [a]_{f} commutes with f.

We have [π]_{f} = f. If g ∈ F_{π}, then F_{f} ≅ F_{g} (by a canonical A-isomorphism).

3. Construction of the extension K_{π} of K.

Again A = O_{K} where K is a nonarchimedean local field with residue field A/𝔪 = k having q (a power of p) elements. We fix a prime element π of K.

According to the discussion in Section 1, there should be a unique totally ramified extension K_{π} of K such that K^{ab} = K_{π} · K^{un} and π is a norm from every finite subextension of K_{π}, namely, the subfield K_{π} of K^{ab} fixed by ϕ(π).

It is easy to construct K^{un}. Let μ_{m} be the set of m^{th} roots of 1 in K^{al}, i.e., μ_{m} is the set of roots of X^{m} − 1. When m is not divisible by p, the discriminant of X^{m} − 1 is a unit in O_{K}, and so the field K[μ_{m}] generated by the elements of μ_{m} is unramified over K; moreover, the residue field of K[μ_{m}] is the splitting field of X^{m} − 1 over k, and so has q^{f} elements where f is the smallest positive integer such that m|p^{f} −1. It follows that K^{un} = ∪_{p∤m}K[μ_{m}]. The Galois group Gal(K^{un}/K) ≅ ℤ̂, and a ∈ ℤ̂ acts K^{un} as follows: for any ζ ∈ μ_{m} and any integer a_{0} sufficiently close to a, a ∗ ζ = ζ^{a_{0}} (= Frob_{K}^{a}(ζ)).

CONSTRUCTION OF THE EXTENSION K_{π} OF K. 23

In the case K = ℚ_{p} and π = p, there is a similar construction for K_{π}, namely, (ℚ_{p})_{p} = ∪ℚ_{p}[μ_{p^{n}}]—we shall prove later that this has the indicated properties. The action

([m], ζ) ↦ ζ^{m} : ℤ/p^{n}ℤ × μ_{p^{n}} → μ_{p^{n}}

makes μ_{p^{n}} into a free ℤ/p^{n}ℤ-module of rank 1. Since ℤ/p^{n}ℤ = ℤ_{p}/p^{n}ℤ_{p}, we can regard μ_{p^{n}} as a cyclic ℤ_{p}-module, isomorphic to ℤ_{p}/(p^{n}). The action of ℤ_{p} on μ_{p^{n}} induces an isomorphism (ℤ_{p}/p^{n}ℤ_{p})^{×} → Gal(ℚ_{p}[μ_{p^{n}}]/ℚ_{p}), and, on passing to limit over all n, we obtain an isomorphism

ℤ_{p}^{×} → Gal((ℚ_{p})_{p}/ℚ_{p}).

Thus, for both the extensions K^{un}/K and (ℚ_{p})_{p}/ℚ_{p}, we have an explicit set of generators for the extension, an explicit description of the Galois group, and an explicit description of the Galois group on the set of generators. Remarkably, the Lubin-Tate groups provide similar results for K_{π}/K for any K and π.

The valuation | · | on K extends uniquely to any subfield L of K^{al} of finite degree over K, and hence to K^{al}. Let f ∈ F_{π}. For any α, β ∈ K^{al} with |α|, |β| < 1 and a ∈ A, the series F_{f}(α, β) and [a]_{f}(α) converge. Therefore, we can define Λ_{f} to be the A-module with:

Λ_{f} = {α ∈ K^{al} | |α| < 1} (as a set),
α +_{Λ_{f}} β = α +_{F_{f}} β = F_{f}(α, β),
a ∗ α = [a]_{f}(α).

We define Λ_{n} to be the submodule of Λ_{f} of elements killed by [π]_{f}^{n}.

Remark 3.1. Recall that [π]_{f}(T) = f(T), and therefore Λ_{n} is the set of roots of

f^{(n)} =^{df} f ∘ f ∘ ⋯ ∘ f (n factors)

in K^{al} with valuation < 1. For simplicity^{1}, we let f be a polynomial πT + a_{2}T^{2} + ⋯ + T^{q}—according to (2.17) it even suffices to take f = πT + T^{q}. Then,

(f ∘ f)(T) =^{df} f(f(T)) = π(πT + ⋯ + T^{q}) + ⋯ + (πT + ⋯ + T^{q})^{q} = π^{2}T + ⋯ + T^{q^{2}},

and

f^{(n)}(T) = π^{n}T + ⋯ + T^{q^{n}}.

From the Newton polygon (see Math 676, 7.35) of f^{(n)}, one sees that its roots all have positive ord_{K}, hence valuation < 1, and so Λ_{n} is the set of all roots of f^{(n)} in K^{al} endowed with the commutative group stucture

α +_{F_{f}} β = F_{f}(α, β) = α + β + ⋯,

and the A-module structure,

[a]_{f}α = aα + · · · .

1The p-adic Weierstrass preparation theorem (Washington, Introduction to Cyclotomic Fields, Springer, 1982, 1997, 7.3) implies that any f ∈ F_{π} factors into f_{1}(T)u(T) where f_{1}(T) is a polynomial πT + ⋯ + aT^{q}, a ≡ 1 mod 𝔪, and u(T) is a unit in A[[T]].

Example 3.2. Take K = ℚ_{p} and f = (T + 1)^{p} − 1 ∈ F_{p}; then

Λ_{n} = {α ∈ ℚ_{p}^{al} | (α − 1)^{p^{n}} = 1} ≅ {ζ | ζ^{p^{n}} = 1} = μ_{p^{n}}.

The addition +_{F} on Λ_{n} corresponds to multiplication on μ_{p^{n}}, and the ℤ_{p}-module structure is as defined before. It follows that Λ_{n} ≈ ℤ/p^{n}ℤ (as a ℤ_{p}-module).

Because A is a principal ideal domain with only one prime element up to conjugates, every finitely generated torsion A-module M decomposes into a direct sum of cyclic modules

M ≈ A/(π^{n_{1}}) ⊕ ⋯ ⊕ A/(π^{n_{r}}), n_{1} ≤ n_{2} ≤ …,

and the sequence n_{1}, …, n_{r} is uniquely determined.

Lemma 3.3. Let M be an A-module, and let M_{n} = Ker(π^{n} : M → M). Assume:

(a) M_{1} has q =^{df} (A : (π)) elements, and

(b) π : M → M is surjective.

Then M_{n} ≈ A/(π^{n}); in particular, it has q^{n} elements.

Proof. We use induction on n. Because A/(π^{n}) has order q^{n}, condition (a) and the structure theorem imply that M_{1} ≈ A/(π). Consider the sequence

0 ⟶ M_{1} ⟶ M_{n} ⟶^{π} M_{n−1} ⟶ 0.

Condition (b) implies that it is exact at M_{n−1}, and is therefore exact. It follows that M_{n} has q^{n} elements. Moreover, M_{n} must be cyclic, because otherwise M_{1} would not be cyclic. Therefore M_{n} is a cyclic A-module of order q^{n}, and every such module is isomorphic to A/(π^{n}).

Proposition 3.4. The A-module Λ_{n} is isomorphic to A/(π^{n}). Hence End_{A}(Λ_{n}) = A/(π^{n}) and Aut_{A}(Λ_{n}) = (A/(π^{n}))^{×}.

Proof. An A-isomorphism h : F_{f} → F_{g} of formal group laws induces an isomorphism of A-modules Λ_{f} → Λ_{g}, and so it does not matter which f ∈ F_{π} we choose. We take f ∈ F_{π} to be a polynomial of the form πT + ⋯ + T^{q}. This is an Eisenstein polynomial (Math 676, 7.46), and so has q distinct roots, each with valuation < 1. Let α ∈ K^{al} have valuation < 1. From the Newton polygon of

f(T) − α = T^{q} + ⋯ + πT − α

we see that its roots have valuation < 1, and so lie in Λ_{f}. Thus, we have verified the hypotheses of the lemma for Λ_{f}, and so Λ_{n} ≈ A/(π^{n}). It follows that the action of A on Λ_{n} induces an isomorphism A/(π^{n}) → End_{A}(Λ_{n}).

Lemma 3.5. Let L be a finite Galois extension of a local field K, with Galois group G. For any F ∈ O_{K}[[X_{1}, ..., X_{n}]] and α_{1}, …, α_{n} ∈ 𝔪_{L},

F (τα_{1}, …, τα_{n}) = τF (α_{1}, …, α_{n}), all τ ∈ G.

Proof. If F is a polynomial, this follows from the fact that τ is a field isomorphism fixing the elements of O_{K}. We know (Math 676, 7.20 et seqq.) that τ preserves the valuation on L, i.e., |τα| = |α| all α ∈ L, and so τ is continuous. Therefore it preserves limits:

lim_{m→∞} α_{m} = L ⟹ lim_{m→∞} τα_{m} = τL.

Let F_{m} be the polynomial of degree m such that F = F_{m} + deg ≥ m + 1. Then

τ(F (α_{1}, …)) = τ( lim_{m→∞} F_{m}(α_{1}, …)) = lim_{m→∞} τF_{m}(α_{1}, …) = lim_{m→∞} F_{m}(τα_{1}, …).

Theorem 3.6. Let K_{π,n} = K[Λ_{n}], the subfield of K^{al} generated over K by the elements of Λ_{n}.

(a) For each n, K_{π,n}/K is totally ramified of degree (q − 1)q^{n−1}.

(b) The action of A on Λ_{n} defines an isomorphism

(A/𝔪^{n})^{×} → Gal(K_{π,n}/K).

In particular, K_{π,n}/K is abelian.

(c) For each n, π is a norm from K_{π,n}.

Proof. Again, we may assume that f ∈ F_{π} is a polynomial of the form πT +⋯+ T^{q}.

(a), (b). Choose a nonzero root π_{1} of f(T) and (inductively) a root π_{n} of f(T) − π_{n−1}. Consider the sequence of fields

K[Λ_{n}] ⊃ K[π_{n}] ⊃^{q} K[π_{n−1}] ⊃^{q} ⋯ ⊃^{q} K[π_{1}] ⊃^{q−1} K.

Each extension is Eisenstein (Math 676, 7.46) with the degree indicated. Therefore K[π_{n}] is totally ramified over K of degree q^{n−1}(q − 1).

Recall that Λ_{n} is the set of roots of f^{(n)} in K^{al}, and so K[Λ_{n}] is the splitting field of f^{(n)}. Therefore Gal(K[Λ_{n}]/K) can be identified with a subgroup of the group of permutations of the set Λ_{n}, but Lemma 3.5 implies that each element of Gal(K[Λ_{n}]/K) acts on Λ_{n} as an A-module isomorphism, and so the image of Gal(K[Λ_{n}]/K) in Sym(Λ_{n}) is contained in

End_{A}(Λ_{n}) = (A/(π^{n}))^{×}.

Hence

(q − 1)q^{n−1} ≥ # Gal(K[Λ_{n}]/K) = [K[Λ_{n}] : K] ≥ [K[π_{n}] : K] = (q − 1)q^{n−1}.

We must have equalities throughout, and so Gal(K[Λ_{n}]/K) ≅ (A/𝔪^{n})^{×} and K[Λ_{n}] = K[π_{n}].

(c) Let f^{[n]}(T) = (f/T) ∘ f ∘ ⋯ ∘ f (n terms), so that

f^{[n]}(T) = π + ⋯ + T^{(q−1)q^{n−1}}.

Then f^{[n]}(π_{n}) = f^{[n−1]}(π_{n−1}) = … = f(π_{1}) = 0. Because f^{[n]} is monic of degree (q − 1)q^{n−1} = [K[π_{n}] : K], it must be the minimum polynomial of π_{n} over K. Therefore,

Nm_{K[Λ_{n}]/K} π_{n} = (−1)^{(q−1)q^{n−2}}π
= π unless q = 2 and n = 1.

In the exceptional case, K[Λ_{1}] = K, and so π is certainly a norm.

Summary 3.7. Let f(T) = πT + ⋯ + T^{q}, and let Λ_{n} be the set of roots of f^{(n)} in K^{al}. Define K_{π,n} = K[Λ_{n}]. Then

K_{π} = ∪K_{π,n}
|
⋮
|
K_{π,n} = K[π_{n}] f(π_{n}) = π_{n−1} 𝔪_{K_{π,n}} = (π_{n})
q  |
⋮  ⋮  ⋮
q  |
K_{π,2} = K[π_{2}] f(π_{2}) = π_{1} 𝔪_{K_{π,2}} = (π_{2})
q  |
K_{π,1} = K[π_{1}] f(π_{1}) = 0 𝔪_{K_{π,1}} = (π_{1}) π_{1} ≠ 0
q − 1 |
K

Moreover, the action

a ∗ λ = [a]_{f}(λ), a ∈ A, λ ∈ Λ_{n},

induces an isomorphism

(A/𝔪^{n})^{×} → Gal(K_{π,n}/K).

On passing to the inverse limit, we obtain an isomorphism

A^{×} → Gal(K_{π}/K).

Example 3.8. Let K = ℚ_{p} and f = (T +1)^{p} −1. For each r, choose a p^{r}th root ζ_{p^{r}} of 1 in such a way that ζ_{p} is primitive and ζ_{p^{r}}^{p} = ζ_{p^{r−1}}. Then π_{r} = ζ_{p^{r}} −1 and (ℚ_{p})_{p,n} = ℚ_{p}[π_{r}] = ℚ_{p}[ζ_{p^{r}}]. Moreover, the isomorphism (ℤ_{p}/(p^{n}))^{×} → Gal(ℚ_{p}[ζ_{p^{r}}]/ℚ_{p}) is the standard one.

The local Artin map. We define a homomorphism

ϕ_{π}: K^{×} → Gal(K^{ab}/K)

as follows. Let a ∈ K^{×}. Because K_{π} ∩ K^{un} = K and K_{π} · K^{un} = K^{ab}, it suffices to describe the actions of ϕ_{π}(a) on K_{π} and K^{un} separately. Let a = uπ^{m}, u ∈ U. We decree that ϕ_{π}(a) acts on K^{un} as Frob^{m}, and that it acts on K_{π} according to the rule

ϕ_{π}(a)(λ) = [u^{−1}]_{f}(λ), all λ ∈ ∪Λ_{n}.

The ^{−1} is inserted so the the following theorem is true.

Theorem 3.9. Both K_{π} · K^{un} and ϕ_{π} are independent of the choice of π.

Recall that K^{al} is not complete (see Math 676, Problems 10, #4); in fact even K^{un} is not complete. We write K̂^{un} for its completion, and B for the valuation ring of K̂^{un} (the valuation | · | on K^{un} extends uniquely to K̂^{un}, and B is the set of elements with value ≤ 1). We write σ for the Frobenius automorphism Frob_{K} of K^{un}/K, and also for its extension to K̂^{un}. For a power series θ[[T]] = ∑ b_{i}T^{i} ∈ B[[T]], (σθ)(T) =^{df} ∑ σb_{i}T^{i}.

Proposition 3.10. Let F_{f} and F_{g} be the formal group laws defined by f ∈ F_{π} and g ∈ F_{ϖ}, where π and ϖ = uπ are two prime elements of K. Then F_{f} and F_{g} become A-isomorphic over B. More precisely, there exists an ε ∈ B^{×} such that σε = εu, and a power series θ(T) ∈ B[[T]] such that:

(a) θ(T) = εT + deg ≥ 2;

(b) σθ = θ ∘ [u]_{f};

(c) θ(F_{f}(X, Y)) = F_{g}(θ(X), θ(Y));

(d) θ ∘ [a]_{f} = [a]_{g} ∘ θ.

The last two conditions say that θ is a homomorphism F_{f} → F_{g} commuting with the actions of A, and the first condition implies that θ is an isomorphism (because ε is a unit).

Lemma 3.11. The homomorphisms

b ↦ σb − b : B → B, b ↦ σb/b : B^{×} → B^{×},

are surjective with kernels A and A^{×} respectively.

Proof. Let R be the valuation ring in K^{un}, and let 𝔫 be its maximal ideal. Then R is a discrete valuation ring, and \underleftarrow{lim} R/𝔫^{n} = B (see (5.7 below). We shall show by induction that the sequence

0 → A/𝔪_{K}^{n} → R/𝔫^{n} ⟶^{σ−1} R/𝔫^{n} → 0 (∗)

is exact. For n = 1, the sequence becomes

0 → k → k̄ ⟶^{x↦x^{q}−x} k̄ → 0.

Here k̄ is the algebraic closure of k. This is obviously exact. Assume that the sequence is exact for n − 1, and consider the diagram

0 → R/𝔫 → R/𝔫^{n} → R/𝔫^{n−1} → 0
↓ ↓ ↓
0 → R/𝔫 → R/𝔫^{n} → R/𝔫^{n−1} → 0

in which all the vertical maps are induced by σ − 1. From the snake lemma, we find that σ − 1 : R/𝔫^{n} → R/𝔫^{n} is surjective and that its kernel has q^{n} elements. As A/𝔫^{n} is contained in the kernel and has q^{n} elements, it must equal the kernel. This shows that (∗) is exact, and, when we pass to the inverse limit over n, the sequence becomes

0 → A → B ⟶^{σ−1} B → 0,

which is therefore exact (see Proposition 5.8 below).

The proofs for A^{×} are similar.

The inverse of a power series h for composition will be denoted h^{−1}. Thus h∘h^{−1} = T = h^{−1} ∘ h.

The proof of the Proposition 3.10 has four steps:

Step 1. Show there exists a θ(T) ∈ B[[T]] satisfying (a) and (b). Step 2. Show that the θ in Step 1 can be chosen so that g = σθ ∘ f ∘ θ^{−1}. Step 3. Show that the power series θ(F_{f}(θ^{−1}(X), θ^{−1}(Y))) has the properties charac-

terizing F_{f}(X, Y), and therefore equals it.

Step 4. Show that the power series θ ∘ [a]_{f} ∘ θ^{−1} has the properties characterizing [a]_{g}, and therefore equals it.

Proof. (of Step 1). Choose an ε ∈ B^{×} such that σε = εu—its existence is ensured by Lemma 3.11. Starting with θ_{1}(T) = εT, we shall construct a sequence of polynomials θ_{r} such that

\begin{aligned}θ_{r}(T) & = & θ_{r−1}(T) + bT^{r}, some b ∈ B, \\ σθ_{r} & = & θ_{r} ∘ [u]_{f} + deg ≥ r + 1. \end{aligned}

Note, that for θ_{1}(T) = εT, the second equation becomes

σεT = ε(uT + ⋯) + deg ≥ 2,

which is true because of our choice of ε. Suppose that θ_{r} has been found, and we wish to find θ_{r+1}(T) = θ_{r}(T) + bT^{r+1}. Write b = aε^{r+1}, a ∈ B. Then the second equation becomes

(σθ_{r})(T) + (σa)(σε)^{r+1}T^{r+1} =^{?} θ_{r}([u]_{f}(T)) + aε^{r+1}(uT)^{r+1} + deg ≥ r + 1.

Thus, we need

(σa − a)(εu)^{r+1} =^{?} c

where c is the coefficient of T^{r+1} in θ_{r} ∘ [u]_{f} − σθ_{r}. We can choose a to be any element of B such that σa − a = c/(εu)^{r+1}.

Proof. (of Step 2). Define

h = σθ ∘ f ∘ θ^{−1} = θ ∘ [u]_{f} ∘ f ∘ θ^{−1} = θ ∘ f ∘ [u]_{f} ∘ θ^{−1}.

Then, because f and [u]_{f} have coefficients in A,

σh = σθ ∘ f ∘ [u]_{f} ∘ σθ^{−1} = σθ ∘ f ∘ θ^{−1} = h.

For the middle equality, we used that [u]_{f} ∘ σθ^{−1} = θ^{−1} which follows from θ ∘ [u]_{f} ∘ σθ^{−1} = T. Because σh = h, it lies in A[[T]]. Moreover,

h(T) = σε · π · ε^{−1}T + ⋯ = ϖT + deg ≥ 2,

and

h(T) ≡ σθ ∘ (θ^{−1})^{q} ≡ σθ(σθ^{−1}(T^{q})) ≡ T^{q} mod 𝔪_{K}.

Therefore, h ∈ F_{ϖ}. Let θ′ = [1]_{g,h} ∘ θ. Then θ′ obviously still satisfies condition (a) of the proposition, and it still satisfies (b) because [1]_{g,h} ∈ A[[T]]. Moreover,

σθ′ ∘ f ∘ θ′^{−1} = [1]_{g,h} ∘ h ∘ [1]_{g,h}^{−1} = g.

The proofs of Steps 3 and 4 are straightforward applications of Lemma 2.12.

Proof. (that K_{π} · K^{un} is independent of π). Let π and ϖ = πu be two prime elements of K. From Proposition 3.10 we find that

(σθ) ∘ [π]_{f} = θ ∘ [u]_{f} ∘ [π_{f}] = θ ∘ [ϖ]_{f} = [ϖ]_{g} ∘ θ,

that is, that

(σθ)(f(T)) = g(θ(T)).

Therefore, for any α ∈ K^{al} (recall that this is the separable algebraic closure of K),

f(α) = 0 ⟹ g(θ(α)) = 0,

and, similarly,

g(α) = 0 ⟹ f(θ^{−1}(α)) = 0.

Therefore θ defines a bijection Λ_{f,1} → Λ_{g,1}, and so

K̂^{un}[Λ_{g,1}] = K̂^{un}[θ(Λ_{f,1})] ⊂ K̂^{un}[Λ_{f,1}] = K̂^{un}[θ^{−1}(Λ_{g,1})] ⊂ K̂^{un}[Λ_{g,1}].

Therefore

K̂^{un}[Λ_{g,1}] = K̂^{un}[Λ_{f,1}].

Now the next lemma shows that

K̂^{un}[Λ_{g,1}] ∩ K^{al} = K^{un}[Λ_{g,1}],  K̂^{un}[Λ_{f,1}] ∩ K^{al} = K^{un}[Λ_{f,1}],

and so

K^{un}[Λ_{g,1}] = K^{un}[Λ_{f,1}].

The argument extends without difficulty to show that

K^{un}[Λ_{g,n}] = K^{un}[Λ_{f,n}]

for all n, and so K^{un} · K_{ϖ} = K^{un} · K_{π}.

Lemma 3.12. Every subfield E of K^{al} containing K is closed (in the topological sense).

Proof. Let H = Gal(K^{al}/E). Then H fixes every element of E and so, by continuity, it fixes every element in the closure of E. By Galois theory, this implies that E equals its closure in K^{al}.

Proof. (that ϕ_{π} is independent of π.) We shall show that, for any two prime elements π and ϖ,

ϕ_{π}(ϖ) = ϕ_{ϖ}(ϖ).

Since π is arbitrary, this implies that for any other prime element π′ of K,

ϕ_{π′}(ϖ) = ϕ_{ϖ}(ϖ) = ϕ_{π}(ϖ).

Since ϖ is also arbitrary, and the prime elements generate the group K^{×}, this implies that ϕ_{π} = ϕ_{π′}.

On K^{un}, both ϕ_{π}(ϖ) and ϕ_{ϖ}(ϖ) induce the Frobenius automorphism. It remains to prove that they have the same effect on K_{ϖ}.

Let θ be an isomorphism F_{f} → F_{g} over K̂^{un} as in Proposition 3.10. It induces an isomorphism Λ_{f,n} → Λ_{g,n} for all n. By definition, ϕ_{ϖ}(ϖ) is the identity on K_{ϖ}, and since K_{ϖ,n} is generated over K by the elements θ(λ) for λ ∈ Λ_{f,n}, it remains to prove that

ϕ_{π}(ϖ)(θ(λ)) = θ(λ), all λ ∈ Λ_{f,n}.

Write ϖ = uπ. Then ϕ_{π}(ϖ) = ϕ_{π}(u) · ϕ_{π}(π) = τσ, say, where

σ = \{ \begin{matrix}Frob_{K} & on K^{un} \\ id & on λ \end{matrix} τ = \{ \begin{matrix}id & on K^{un} \\ [u^{−1}]_{f} & on λ. \end{matrix}

Using that the series θ has coefficients in K̂^{un} and (3.10), we find that

ϕ_{π}(ϖ)(θ(λ)) = τσ(θ(λ)) = (σθ)(τλ) = (σθ)([u^{−1}]_{f}(λ) = θ(λ).

Example 3.13. We describe the local Artin map ϕ_{p} : ℚ_{p}^{×} → Gal(L/ℚ_{p}) in the case L = ℚ_{p}[ζ] where ζ is a primitive n^{th} root of 1.

(i) Suppose n is prime to p. Then L is unramified over ℚ_{p}, with degree equal to the degree of the residue field extension. The residue field is 𝔽_{p^{f}} where p^{f} is the smallest power of p such that n|(p^{f} − 1). The map ϕ_{p} : ℚ_{p}^{×} → Gal(L/ℚ_{p}) sends u · p^{t} to the t^{th} power of the Frobenius element, and its kernel is ℤ_{p}^{×}· <p^{f}>,

(ii) Suppose n is a power p^{r} of p. In this case, L is totally ramified of degree (p − 1)p^{r−1} over K, and L = (ℚ_{p})_{p,n} (see 3.8). The map ϕ_{p} : ℚ_{p}^{×} → Gal(L/ℚ_{p}) can be described as follows: let a = up^{t}, and let u_{0} ∈ ℤ represent the class of u in (ℤ_{p}/p^{r}ℤ_{p})^{×}; then ϕ_{p}(a) sends ζ to ζ^{u^{−1}}. Its kernel is {up^{m} | u ≡ 1 mod p^{r}, m ∈ ℤ}.

(iii) In the general case, write n = m · p^{r} with m prime to p. Then we have

ℚ_{p}[ζ_{n}]
⧸  ⧹
ℚ_{p}[ζ_{p^{r}}]  ℚ_{p}[ζ_{m}].
⧹  ⧸
ℚ_{p}

The map ℚ_{p}^{×}/ Nm(ℚ_{p}[ζ_{n}]^{×}) → Gal(ℚ_{p}[ζ_{n}]/ℚ_{p}) can be described as follows: write a = up^{t}, u ∈ ℤ_{p}^{×}; then a acts on ℚ_{p}[ζ_{m}] by ζ_{m} ↦ ζ_{m}^{t}, and it acts on ℚ_{p}[ζ_{p^{n}}] by

ζ_{p^{n}} ↦ ζ_{p^{n}}^{u_{0}^{−1}} where u_{0} is an integer congruent to u mod p^{r}.

4. The Local Kronecker-Weber Theorem

The main result proved in this section is that K^{ab} = K_{π} · K^{un}. Since this is not needed for the proofs of the main theorems of local class field theory, and is implied by them, this section may be skipped.

The ramification groups of K_{π,n}/K. Let L/K be a finite Galois extension with Galois group G. Recall (Math 676, 7.49 et seqq.) that the i^{th} ramification group is defined to be

G_{i} = {τ ∈ G | ord_{L}(τa − a) ≥ i + 1, all a ∈ O_{L}}.

Moreover, for i ≥ 0,

G_{i} = {τ ∈ G_{0} | ord_{L}(τΠ − Π) ≥ i + 1}

where Π is a prime element for L. Here ord_{L} is the normalized valuation L^{×} ↠ ℤ. Then G/G_{0} = Gal(ℓ/k), and there are inclusions:

(Π ↦ τΠ/Π mod Π) : G_{0}/G_{1} ↪ ℓ^{×}
(Π ↦ (τΠ − Π)/Π^{i+1} mod Π) : G_{i}/G_{i+1} ↪ ℓ

where ℓ ⊃ k are the residue fields of L and K. Thus (G_{0} : G_{1})|q − 1 and (G_{i}: G_{i+1})|q for i ≥ 1. Moreover G_{i} = {1} for i sufficiently large.

Let

U^{(0)} = U = A^{×},
U^{(i)} = 1 + 𝔪^{i}, i ≥ 1.

Then we have a filtration

U/U^{(n)} ⊃ U^{(1)}/U^{(n)} ⊃ ⋯ ⊃ U^{(n)}/U^{(n)} = 0

on A^{×}/(1 + 𝔪^{n}) = U/U^{(n)}.

Proposition 4.1. Under the isomorphism A^{×}/U^{(n)} →^{≈} G of Theorem 3.6, U^{(i)}/U^{(n)} maps onto G_{q^{i}−1}.

Proof. We take f = πT + T^{q}. Certainly G = G_{0}, and U^{(0)}/U^{(n)} maps onto G_{0}. Now take i ≥ 1, and let u ∈ U^{(i)} \ U^{(i+1)}. Then u = 1 + vπ^{i} with v ∈ A^{×}, and

[u]_{f}(π_{n}) = [1]_{f}(π_{n}) + [v]_{f}[π^{i}]_{f}(π_{n}) = π_{n} + [v]_{f}(π_{n−i}) = π_{n} + (unit)π_{n−i}.

For any i ≥ 1, π_{i} = ππ_{i+1}+π_{i+1}^{q} = π_{i+1}^{q}(\frac{ππ_{i+1}}{π_{i+1}^{q}} +1) = π_{i+1}^{q}×unit because ord(\frac{π}{π_{i+1}^{q−1}} ) > 0.

Hence π_{n−i} = π_{n}^{q^{i}} × unit, and

[u]_{f}(π_{n}) − π_{n} = π_{n}^{q^{i}} × unit.

By definition, this means that [u]_{f} ∈ G_{q^{i}−1}, [u]_{f} ∉ G_{q^{i}}. Since this is true for all i, it implies that U^{(i)} maps onto G_{q^{i}−1}.

Hence

G_{0} = G
G_{q−1} = G_{q−2} = ⋯ = G_{1}
G_{q^{2}−1} = G_{q^{2}−2} = ⋯ = G_{q}
⋯
G_{q^{n}−1} = 1

is a complete set of distinct ramification groups for K_{π,n}/K.

Upper numbering on ramification groups. Let L be a finite Galois extension of K with Galois group G. We extend the definition of G_{u} to all real numbers u ≥ −1, by setting

G_{u} = G_{i} where i is the least integer ≥ u.

For u > 0,

G_{u} = {τ ∈ G_{0} | ord_{L}(τΠ − Π) ≥ i + 1}.

Define φ: ℝ_{≥0} → ℝ to be the unique continuous piecewise linear function such that

\{ \begin{matrix}φ(0) & = & 0 \\ φ′(u) & = & (G_{0} : G_{u})^{−1} if u is not an integer. \end{matrix}

Define G^{v} = G_{u} if v = φ(u), i.e., G^{v} = G_{φ^{−1}(v)}.

Example 4.2. Let L = K_{π,n}. Then

(G_{0} : G_{1}) = q − 1, G_{1} = G_{2} = ⋯ = G_{q−1}.

Thus φ′(u) = \frac{1}{q−1} for 0 < u < q − 1, and the first segment of the graph of φ runs from (0, 0) to (q − 1, 1); hence G^{1} = G_{q−1}. Next

(G_{q−1} : G_{q}) = q, G_{q} = G_{q+1} = ⋯ = G_{q^{2}−1}.

Thus φ′(u) = \frac{1}{q(q−1)} for q − 1 < u < q^{2} − 1, and the second segment of the graph of φ runs from (q − 1, 1) to (q^{2} − 1, 2). Thus G^{2} = G_{q^{2}−1}. Continuing in this fashion, we arrive at the following picture:

G_{0} ⊃^{q−1} G_{q−1} ⊃^{q} G_{q^{2}−1} ⊃ ⋯ ⊃ G_{q^{n}−1} = 1
∥  ∥  ∥  ∥
G^{0}  G^{1}  G^{2}  G^{n}

Proposition 4.3. Under the isomorphism A^{×}/U^{(n)} → G,

U^{(i)}/U^{(n)} →^{≈} G^{i}.

Proof. Immediate consequence of (4.1) and (4.2).

The upper numbering is defined so as to be compatible with the passage to the quotient (whereas the lower numbering is compatible with passage to the subgroups).

Proposition 4.4. Consider Galois extensions M ⊃ L ⊃ K, and let G = Gal(M/K) and H = Gal(M/L) (assumed normal) so that G/H = Gal(L/K). Then

(G/H)^{v} = Im(G^{v} → G/H),

i.e., (G/H)^{v} = G^{v}H/H.

Proof. See Serre, 1962, Corps Locaux, IV.3, Pptn 14.

Now consider an infinite Galois extension Ω/K. Using (4.4) we can define a filtration on G = Gal(Ω/K):

τ ∈ G^{v} ⟺ τ ∈ Gal(L/K)^{v}, all L/K finite and Galois L ⊂ Ω.

Definition 4.5. For a finite Galois extension L/K, v is called a jump in the filtration {G^{v}} if, for all ε > 0, G^{v} ≠ G^{v+ε}.

Theorem 4.6 (Hasse-Arf). If L/K is abelian, then the jumps are integers, i.e., if G_{i} ≠ G_{i+1}, then φ(i) ∈ ℤ.

Proof. See Serre, 1962, Corps Locaux, V.7. (The proof is fairly elementary, but complicated. It is does not require that residue fields be finite, but only that the residue field extension be separable.)

Thus, for a finite abelian extension L/K, the filtration on G_{0} = G^{0} is of the form

G^{0} ⊋ G^{i_{1}} ⊋ G^{i_{2}} …. i_{j} ∈ ℕ.

Moreover G^{n} = {1} for n sufficiently large, and (G^{i_{j}}: G^{i_{j+1}}) divides q − 1 or q. For an infinite abelian extension, the same statements hold, except that the filtration need not terminate: we can only say that

∩G^{i} = {1}.

Example 4.7. Let L = K_{π,n}. If G_{i} ≠ G_{i+1}, then i = 0, q − 1, …, q^{n} − 1, and at those points φ takes the values 1, 2, 3, …, n. Thus we have verified the Hasse-Arf theorem for all these extensions, and, because of (4.4), all subextensions.

The local Kronecker-Weber theorem. As usual, K is a local nonarchimedean field, and all extensions of K will be required to be subfields of a fixed separable algebraic closure K^{al} of K.

Theorem 4.8. For any prime element π of K,

K_{π} · K^{un} = K^{ab}.

Lemma 4.9. Let L be an abelian totally ramified extension of K. If L ⊃ K_{π}, then L = K_{π}.

Proof. Let G = Gal(L/K) and H = Gal(L/K_{π}), so that G/H = Gal(K_{π}/K). Consider the diagram (of abelian groups)

1  1  1
↓  ↓  ↓
1 ⟶ G^{n+1} ∩ H ⟶ G^{n+1} ⟶ (G/H)^{n+1} ⟶ 1
↓  ↓  ↓
1 ⟶ G^{n} ∩ H ⟶ G^{n} ⟶ (G/H)^{n} ⟶ 1
↓  ↓  ↓
1 ⟶ \frac{G^{n}∩H}{G^{n+1}∩H} ⟶ \frac{G^{n}}{G^{n+1}} ⟶ \frac{(G/H)^{n}}{(G/H)^{n+1}} ⟶ 1
↓  ↓  ↓
1  1  1

The columns are obviously exact, and Proposition 4.4 shows that the top two rows are exact. Therefore, the third row is exact (by the snake lemma, for example) and so

\begin{aligned}(G^{n}: G^{n+1}) & = & ((G/H)^{n}: (G/H)^{n+1}) & (G^{n} ∩ H: G^{n+1} ∩ H). \\ ≤ q & & = q − 1 or q \end{aligned}

From this we deduce that G^{n} ∩ H = G^{n+1} ∩ H for all n. Thus

G^{n+1} ∩ H = G^{n} ∩ H = ⋯ = G^{0} ∩ H = H,

i.e., H ⊂ G^{n+1} for all n. Since ∩G^{n} = 1, this shows that H = 1.

Lemma 4.10. Every finite unramified extension of K_{π} is contained in K_{π} · K^{un}.

Proof. Let L be an unramified extension of K_{π}. Then L = K_{π} · L′ for some unramified extension L′ of K_{π,n} for some n. Now apply (Math 676, 7.41) to see that L′ = K_{π,n} · L′′ for some unramified extension L′′ of K.

Lemma 4.11. Let L be a finite abelian extension of K of exponent m (i.e., τ^{m} = 1 all τ ∈ Gal(L/K)), and let K_{m} be the unramified extension of K of degree m. Then there exists a totally ramified abelian extension L_{t} of K such that

L ⊂ L_{t} · K_{m} = L · K_{m}.

Proof. For any τ ∈ Gal(LK_{m}/K), τ^{m}|L = 1 = τ^{m}|K_{m}, and so Gal(LK_{m}/K) is still abelian of exponent m. Let τ ∈ Gal(LK_{m}/K) be such that τ|K_{m} is the Frobenius automorphism. Then τ has order m, and so

Gal(L/K) =<τ> ×H (direct product).

for some subgroup H. Let L_{t} = L^{<τ>}; then L_{t} is totally ramified over K and L·K_{m} = L_{t} · K_{m}.

Proof. (of Theorem 4.8). Let L be a finite abelian extension of K; we have to show that L ⊂ K_{π} · K^{un}.

Lemma 4.9 holds with K replaced by K_{π}. If we apply it to the extension L·K_{π}/K_{π} we find that there exists a totally ramified extension L_{t} of K_{π} and an unramified extension L_{u} of K_{π} such that

L · K_{π} ⊂ L_{t} · L_{u} ⊂ (L_{t} · K_{π}) · L_{u}.

Now (4.9) implies that L_{t} ⊂ K_{π} and (4.10) implies that L_{u} ⊂ K_{π} · K^{un}.

Corollary 4.12. Every finite abelian extension of ℚ_{p} is contained in a cyclotomic extension.

Example 4.13. A finite abelian extension L of K need not be of the form L_{t} · L_{u} with L_{t} totally ramified over K and L_{u} unramified over K. Consider:

ℚ_{5}[ζ_{5}, ζ_{624}]
⧸  |  ⧹
ℚ_{5}[ζ_{5}]  L  ℚ_{5}[ζ_{624}].
⧹  |  ⧸
ℚ_{5}

The field ℚ_{5}[ζ_{5}] is totally ramified of degree 4 over ℚ_{5} with Galois group (ℤ/5)^{×}, which is cyclic of order 4. Note that 624 = 5^{4} − 1, and so ℚ_{5}[ζ_{624}] is unramified of degree 4 over ℚ_{5}, and its Galois group is also cyclic. Clearly

Gal(ℚ_{5}[ζ_{3120}]/ℚ_{5}) =<σ> × <τ>

where

\{ \begin{matrix}σ(ζ_{624}) & = & ζ_{624}^{5} \\ σ(ζ_{5}) & = & ζ_{5} \end{matrix} \{ \begin{matrix}τ(ζ_{624}) & = & ζ_{624} \\ τ(ζ_{5}) & = & ζ_{5}^{2} \end{matrix} .

Let L be the fixed field of <σ^{2}τ>. Then L is a cyclic Galois extension of ℚ_{5} of degree 4. Its maximal unramified subfield

L_{u} = L ∩ ℚ_{5}[ζ_{624}] = ℚ_{5}[ζ_{624}]^{<σ^{2}>} = ℚ_{5}[ζ_{624}^{25}]

which has degree 2 over ℚ_{5}. If there existed field a L_{t} such that L = L_{t} · L_{u}, then Gal(L/ℚ_{5}) would be the product of two cyclic groups of order 2, contradicting the fact that it is cyclic.

We recover the fact that K_{π} · K^{un} is independent of π without using Proposition 3.10. However, this proposition is still required to show that ϕ_{π} is independent of π.

Remark 4.14. The original Kronecker-Weber Theorem (proved by Hilbert in 1896 using an analysis of the ramification groups after earlier incomplete proofs by Kronecker and Weber) states that every finite abelian extension of ℚ is contained in a cyclotomic extension. For ℚ_{p} the same statement is called the Local Kronecker-Weber Theorem, and Theorem 4.8 is usually referred to as the Local Kronecker-Weber Theorem for K. It is in fact possible to give an elementary proof of the Local Kronecker-Weber Theorem for ℚ_{p} (see Cassels 1986, Local Fields, Cambridge, p 151).

Remark 4.15. In Chapter III, we shall deduce the Local Kronecker-Weber Theorem from Theorem 1.1 without making use of the Hasse-Arf theorem—this was the original approach of Lubin and Tate. The above proof follows R. Gold, Local class field theory via Lubin-Tate groups, Indiana Univ. Math. Jour., 30, 1981, 795–798. For a proof of the local Kronecker-Weber theorem for local fields of characteristic zero that does not make use of the Hasse-Arf theorem or cohomology, see M. Rosen, Trans. AMS 265 (1981), 599–605. As Iwasawa points out (Iwasawa 1986, p115), once Proposition 4.4 and certain properties of the abelian extensions K_{π,n}/K are taken for granted, then the Local Kronecker-Weber Theorem for K and the Hasse-Arf Theorem are essentially equivalent.

The global Kronecker-Weber theorem. Since it is now so easy, we might as well prove the original Kronecker-Weber theorem.

Theorem 4.16. Every abelian extension of ℚ is contained in a cyclotomic extension.

Lemma 4.17. Let K be a finite Galois extension of ℚ with Galois group G. Then G is generated by the inertia groups of the primes ideals 𝔭 of K that are ramified in the extension K/ℚ.

Proof. Let H be the subgroup of G generated by the inertia groups, and let M be the fixed field of H. Then K^{I(𝔭)} ⊃ M for all prime ideals 𝔭 of K, and so 𝔭 ∩ M is unramified in the extension M/ℚ. Therefore M is an unramified extension of ℚ, and so equals ℚ (by Math 676, 4.8).

Proof. (of the Kronecker-Weber Theorem; following Cassels, 1986, p236). Let K be an abelian extension of ℚ. Let p be a prime number, and let 𝔭 be a prime ideal of O_{K} lying over it. From the local Kronecker-Weber theorem, K_{𝔭} is contained in a cyclotomic extension of ℚ_{p}, say K_{𝔭} = ℚ_{p}[u_{p}, v_{p}] where u_{p} is a p^{s_{p}}th root of 1 and v_{p} is a root of 1 of order prime to p. Note that s_{p} depends only on p (not 𝔭) because Gal(K_{𝔭}/ℚ_{p}) acts transitively on the primes lying over p, and hence on the set of fields K_{𝔭}.

Let L be the cyclotomic extension of ℚ generated by the p^{s_{p}}th roots of 1 for all prime numbers p ramified in K, and let K′ = K · L. Then K′ is again abelian over ℚ, and for any prime 𝔭′ of K′, we have

K′_{𝔭′} ⊂ ℚ_{p}[u_{p}, w_{p}] (∗)

where w_{p} is a root of 1 of order prime to p. Clearly it suffices to prove the theorem with K replaced with K′, and so we can assume K ⊃ L.

We now have

[K: ℚ] ≥ [L: ℚ] = ∏_{p} φ(p^{s_{p}}) (product over p ramifying in K).

Since the Galois group G of K/ℚ is commutative, the inertia group I(𝔭) depends only on the underlying prime p, and so we denote it I(p). From (*) we have

(I(p): 1) ≤ φ(p^{s_{p}})

because I(p) is a quotient of the corresponding group for ℚ_{p}[u_{p}, w_{p}]. By (4.17), G is generated by the groups I(p), and so there is a surjective map ∏ I(p) → G; thus

(G: 1) ≤ ∏_{p}(I(p): 1) ≤ ∏ φ(p^{s_{p}}).

But (G: 1) = [K: ℚ] and so we have equality everywhere, and K = L.

Remark 4.18. At this point, it is not too difficult to complete the proofs of main theorems of global class field theory (see the Introduction) in the case K = ℚ. From the fact that L is contained in a cyclotomic extension we deduce that the Artin map ϕ_{L/ℚ} has a modulus. Now use Dirichlet’s theorem on the density of primes in arithmetic progressions to find that φ_{L/ℚ} is surjective. We know that Nm(I_{L}^{S}) is contained in the kernel, and so the only thing that is lacking at this point is that

(I_{K}^{S}: Nm I_{L}^{S} · i(K_{𝔪,1})) ≤ [L: ℚ].

This is the first inequality, which is not difficult by analytic methods (see Janusz, 1996, IV, 5.6). Once one has that, the existence theorem follows from the fact that ℚ[ζ_{m}] has modulus (m)∞.

Where did it all come from? We have seen that the Lubin-Tate formal group laws provide a remarkably simple solution to an apparently very complicated problem, that of giving explicit generators for the largest abelian extension of a local field and describing how the Galois group acts on them. The only mystery is how anyone ever thought of them. The following speculations may help.

That such a theory might exist was suggested by the theory of complex multiplication of elliptic curves. Here one shows that, for a quadratic imaginary number field K, there exists an elliptic curve E, unique up to isogeny, having O_{K} as its endomorphism ring. For any n, the points of order dividing n on E form a cyclic O_{K}-module, and it is this fact that allows one to prove that adjoining their coordinates gives an abelian extension.

In seeking an analogous theory for local fields, it is natural to replace the algebraic group E by the local analogue, namely, by a formal group law. Thus we seek a formal group law whose endomorphism ring is so large that its torsion points form a cyclic module. The obvious candidate for the endomorphism ring is again the ring of integers O_{K} in K. Initially, it is natural to ask only that the formal group admit an endomorphism corresponding to π, a prime element of O_{K}. Considerations of the heights of formal group laws together with the desire for the torsion points to form a cyclic module suggest that this endomorphism should be given by a power series f(T) such that f(T) = T^{q} mod 𝔪. Moreover, since we truly want the formal group to depend on the choice of π (because the extension K_{π} we wish to construct does), it is natural to require that f(T) = πT + · · ·. Thus, we are led to the set F_{π}, and once we have that, the theory follows naturally.

Notes. The original source for the theory of Lubin-Tate extensions is:

Lubin, J., and Tate, J., Formal complex multiplication in local fields, Ann. of Math., 81 (1965), 380–387.

The theory is also treated by Serre in his article in Cassels and Fröhlich 1967 and in Iwasawa 1986 and Fesenko and Vostokov 1993.

5. Appendix: Infinite Galois Theory and Inverse Limits

We review two topics required for this chapter.

Galois theory for infinite extensions. Fix a field K.

Definition 5.1. A field Ω ⊃ K (not necessarily of finite degree) is said to be Galois over K if

(a) it is algebraic and separable over K, i.e., every element of Ω is a simple root of a polynomial with coefficients in K;

(b) it is normal over K, i.e., every irreducible polynomial with coefficients in K having a root in Ω splits in Ω[X].

Proposition 5.2. A field Ω is Galois over K if and only if it is a union of finite Galois extensions of K.

Proof. Suppose Ω is Galois over K. For any α ∈ Ω, the splitting field in Ω of the minimum polynomial of α over K is a finite Galois extension of K, and Ω is the union of such fields. Conversely, if Ω is a union of finite Galois extensions of K, then it is algebraic and separable over K. Moreover, if f(X) ∈ K[X] has a root in Ω, then it has a root in some finite Galois subextension E of Ω, and therefore splits in E[X].

If Ω is a Galois extension of K, then the Galois group Gal(Ω/K) is defined to be the group of automorphisms of Ω fixing the elements of K, endowed with the topology for which the sets

Gal(Ω/E), Ω ⊃ E ⊃ K, [E: K] < ∞

form a fundamental system of neighbourhoods of 1. This means that two elements of Gal(Ω/K) are close if they agree on some “large” field E, Ω ⊃ E ⊃ K, [E: K] < ∞.

Proposition 5.3. The group Gal(Ω/K) is compact and Hausdorff.

Proof. Consider the map

σ ↦ σ|E : Gal(Ω/K) ⟶ ∏ Gal(E/K)

where the product runs over all E ⊂ Ω with E finite and Galois over K. Proposition 5.2 implies that the map is injective. When we endow each group Gal(E/K) with the discrete topology, and the product with the product topology, then the topology induced on Gal(Ω/K) is the above topology. I claim that the image is closed. The image is equal to

{(σ_{E}) | σ_{E′}|E = σ_{E} whenever E′ ⊃ E}.

Suppose that (σ_{E}) is not in the image. Then there exists a pair of fields E_{2} ⊃ E_{1} such that σ_{E_{2}}|E_{1} ≠ σ_{E_{1}}. Let

U = ∏_{E≠E_{1},E_{2}} Gal(E/K) × {σ_{E_{2}}} × {σ_{E_{1}}}

This is an open neighbourhood of (σ_{E}), and U ∩ Im(Gal(Ω/K)) = ∅.

The topology on Gal(Ω/K) is discrete if and only if Ω is a finite extension of K.

Theorem 5.4. Let Ω be a (possibly infinite) Galois extension of K with Galois group G. Then there is a one-to-one correspondence between the subfields of Ω and the closed subgroups of G. More precisely:

(a) For a subfield E of Ω, H =^{df} Gal(Ω/E) is a closed subgroup of G, and E = Ω^{H}.

(b) If H is a subgroup of G, then Gal(Ω/Ω^{H}) is the closure of H in Gal(Ω/K).

Moreover, the normal closed subgroups of G correspond to the Galois extensions of K, and the open subgroups of G correspond to the finite extensions of K.

Proof. Let E ⊂ Ω be a finite extension of K. Because every K-homomorphism E → Ω extends to Ω, the map σ ↦ σ|Ω : Gal(Ω/K) → Hom_{K}(E, Ω) is surjective, and so induces a bijection

Gal(Ω/K)/ Gal(Ω/E) → Hom_{K}(E, Ω).

This shows that Gal(Ω/E) is of finite index [E : K] in Gal(Ω/E). Because it is closed (by definition), it is also closed (its complement is a finite union of open cosets).

Let E ⊂ Ω be an arbitrary extension of K. Then E = ∪E_{i} where the E_{i} run over the finite extensions of K contained in E. Correspondingly, Gal(Ω/E) = ∩ Gal(Ω/E_{i}), which is therefore closed. Moreover, if α ∈ Ω is not fixed by Gal(Ω/E), then it is not fixed by any Gal(Ω/E_{i}), and so does not lie in E. Thus E = Ω^{Gal(Ω/E)}.

Let H be a subgroup of Gal(Ω/K), and let H′ = Gal(Ω/Ω^{H}). It follows from the Galois theory of finite extensions that, for any open subgroup U of Gal(Ω/K), UH = UH′, and so H̄ = ∩UH = ∩UH′ = H′.

Example 5.5. (a) Endow ℤ with the topology for which the subgroups of finite index form a fundamental system of neighbourhoods, and let ℤ̂ be the completion. Then ℤ̂ = ∏_{ℓ prime} ℤ_{ℓ}, and ℤ̂/mℤ̂ = ℤ/mℤ for every m. Let 𝔽 be the algebraic closure of 𝔽_{q}. There is a canonical isomorphism

ℤ̂ → Gal(𝔽/𝔽_{q})

sending 1 ∈ ℤ̂ to the automorphism x ↦ x^{q}. The extension of 𝔽_{q} of degree m corresponds to the subgroup mℤ̂ of ℤ̂. Let σ be the automorphism of 𝔽/𝔽_{q} such that σ(x) = x^{q}. For α ∈ ℤ̂, define σ^{α} to be the element of Gal(𝔽/𝔽_{q}) such that, for any m, σ^{α}|𝔽_{q^{m}} = σ^{a} where a ∈ ℤ is chosen to be close to α. The above isomorphism sends α to σ^{α}. (For a detailed description of ℤ̂ and the isomorphism ℤ̂ → Gal(𝔽/𝔽_{q}), see Artin 1951, 9.2.)

(b) Let Ω_{p} = ⋃_{r} ℚ[ζ_{p^{r}}], where ζ_{p^{r}} is a primitive p^{r}th root of 1. For u ∈ ℤ_{p}^{×}, write u = a_{0} + a_{1}p + a_{2}p^{2} + ⋯, 0 ≤ a_{i} < p − 1, and define

ζ_{p^{r}}^{u} = ζ_{p^{r}}^{a_{0}+a_{1}p+···a_{s}p^{s}}, any s > r.

This defines an action of ℤ_{p}^{×} on Ω_{p}, and in fact an isomorphism of topological groups

ℤ_{p}^{×} → Gal(Ω_{p}/ℚ).

(c) Let Ω = ⋃ ℚ[ζ_{n}], where n is a primitive nth root of 1. Then

Ω = Ω_{2} · Ω_{3} · Ω_{5} ⋯, Ω_{p} ∩ Ω_{p′} = ℚ, p ≠ p′.

Just as in the case of a finite number of finite Galois extensions, this implies that

Gal(Ω/ℚ) = ∏ Gal(Ω_{p}/ℚ)

(topological product of closed subgroups). Thus there is an isomorphism

ℤ̂^{×} → Gal(Ω/ℚ).

It can be described as follows: if ζ is an n^{th} root of 1 and u ∈ ℤ̂^{×}, then

ζ^{u} = ζ^{m} for any m ∈ ℤ with m ≡ u mod n.

(d) Let Ω_{p} = ∪ℚ[ζ_{p^{r}}], as in (b). Then

ℤ_{p}^{×} ≅ Δ_{p} × C_{p}

where Δ_{p} = (ℤ/(p − 1))^{×} for p ≠ 2 and Δ_{2} = ℤ/2ℤ, and C_{p} ≈ ℤ_{p}. Let Ω′_{p} = Ω_{p}^{Δ_{p}}. Then Gal(Ω′_{p}/ℚ_{p}) ≈ ℤ_{p}. Let

Ω′ = Ω′_{2} · Ω′_{3} ⋯

(composite inside ℚ^{al}). Then Gal(Ω′/ℚ) ≈ ∏ ℤ_{p} ≅ ℤ̂.

(e) For any finite extension K of ℚ, K · Ω′ is a Galois extension of K with Galois group isomorphic to a subgroup of finite index in ℤ̂. But any such subgroup is again isomorphic to ℤ̂ (because it is again the completion of a subgroup of ℤ of finite index). Therefore Gal(K · Ω′/K) ≈ ℤ̂. If we fix an isomorphism, and let K_{m} be the field corresponding to mℤ̂, then we see that:

(a) K_{m} is cyclic of degree m;

(b) K_{m} is cyclotomic, i.e., contained in an extension of K obtained by adjoining roots of 1.

Remark 5.6. In general, there may exist subgroups of finite index in Gal(Ω/K) which are not open^{2} (and hence not closed). For example, K = 𝔽_{p}((T)) has a Galois extension Ω such that Gal(Ω/K) = 𝔽_{p}^{ℕ} (product of an infinite countable number copies of a cyclic group of order p)—see Remark 1.5. Let M be the subgroup of 𝔽_{p}^{ℕ} consisting of the elements (a_{n}) such that a_{n} = 0 for all but finitely many n. Thus M is equal to a direct sum of a countable number of copies of a cyclic group of order p. The closure of M is Gal(Ω/K), and so any subgroup H of finite index in Gal(Ω/K) containing M is not closed (and hence not open).

2My recollection is that Gal(ℚ^{al}/ℚ) has such subgroups, but I don’t know a reference.

Inverse limits. A partially ordered set (I, ≤) is said to be directed if, for any α, β ∈ I, there exists γ ∈ I such that α, β ≤ γ. For example, the set of positive integers ordered by divisibility, m ≤ n ⟺ m|n, is directed.

An inverse system (or projective system) of sets is a family (S_{α})_{α∈I} of sets indexed by a directed set I together with, for each pair α ≤ β, a map φ_{α,β} : S_{β} → S_{α} such that

(a) for all α ∈ I, φ_{α,α} is the identity map;

(b) for all α ≤ β ≤ γ in I, φ_{α,β} ∘ φ_{β,γ} = φ_{α,γ}.

A set S together with, for each α ∈ I, a map φ_{α} : S → S_{α} such that φ_{α} = φ_{α,β} ∘ φ_{β} is said to be the inverse limit (or projective limit) of the inverse system (S_{α}), (φ_{α,β}) if it satisfies the obvious universal property.

Every inverse system of sets, groups, or rings has an inverse limit. For example, ℤ̂ = \underleftarrow{lim} ℤ/mℤ.

Example 5.7. The completion R̂ of a discrete valuation ring R is ≅ \underleftarrow{lim} R/𝔪^{n}, where 𝔪 is the maximal ideal in R.

The profinite completion of a group G is defined to be \underleftarrow{lim} G/H where H runs through the normal subgroups of finite index in G.

Proposition 5.8. The inverse limit of an inverse system of exact sequences of finite abelian groups is again exact.

Remark 5.9. The Galois group Gal(Ω/K) is the projective limit of the groups Gal(E/K) where E runs over the subfields of Ω that are finite and Galois over K. A topological group that is a projective limit of finite groups is called a profinite group. They are precisely the compact totally disconnected topological groups. (A topological space is totally disconnected if its connected components are the one-point subsets.) See Serre, Cohomologie Galoisienne, Springer, 1964, or Shatz, Profinite Groups, Arithmetic, and Geometry, Princeton, 1972.

CHAPTER II The Cohomology of Groups

We take a respite from number theory and do some homological algebra. In an appendix to the chapter, we review the general theory of derived functors.

1. Cohomology

The category of G-modules. Let G be a group. A G-module is an abelian group M together with a map

(g, m) ↦ gm: G × M ⟶ M

such that, for all g, g′ ∈ G, m, m′ ∈ M,

(a) g(m + m′) = gm + gm′;

(b) (gg′)(m) = g(g′m), 1m = m.

Equivalently, a G-module is an abelian group M together with a homomorphism of groups G → Aut(M).

A homomorphism of G-modules (or a G-homomorphism) is a map α: M → N such that

(a) α(m + m′) = α(m) + α(m′) (i.e., α is a homomorphism of abelian groups);

(b) α(gm) = g(α(m)) for all g ∈ G, m ∈ M.

We write Hom_{G}(M, N) for the set of G-homomorphisms M → N.

Remark 1.1. The group algebra ℤ[G] of G is the free abelian group with basis the elements of G and with the multiplication provided by the group law on G. Thus the elements of ℤ[G] are the finite sums

∑ n_{g}g, n_{g} ∈ ℤ, g ∈ G,

and

(∑ n_{i}g_{i})(∑ n′_{j}g_{j}) = ∑ n_{i}n′_{j}(g_{i}g_{j}).

To endow an abelian group M with a G-module structure is the same as to endow it with a ℤ[G]-module structure. Thus the category Mod_{G} of G-modules can be identified with the category of modules over the ring ℤ[G]. In particular, Mod_{G} is an abelian category.

If M and N are G-modules, then the set Hom(M, N) of homomorphisms φ : M → N (regarded only as abelian groups) becomes a G-module with the structures

(φ + φ′)(m) = φ(m) + φ′(m)
(gφ)(m) = g(φ(g^{−1}m)).

Induced modules. Let G be a group and H a subgroup. For an H-module M, we define Ind_{H}^{G}(M) to be the set of maps^{1} φ : G → M such that φ(hg) = hφ(g) for all h ∈ H. Then Ind_{H}^{G}(M) becomes a G-module with the operations

(φ + φ′)(x) = φ(x) + φ′(x)
(gφ)(x) = φ(xg).

A homomorphism α : M → M′ of H-modules defines a homomorphism

φ ↦ α ∘ φ : Ind_{H}^{G}(M) → Ind_{H}^{G}(M′)

of G-modules.

Lemma 1.2. (a) For any G-module M and H-module N,

Hom_{G}(M, Ind_{H}^{G}(N)) ≅ Hom_{H}(M, N).

(b) The functor

Ind_{H}^{G} : Mod_{H} → Mod_{G}

is exact.

Proof. (a) Given a G-homomorphism α : M → Ind_{H}^{G}(N), we define β(m) = α(m)(1_{G}), where 1_{G} is the identity element in G. The various definitions show that, for any g ∈ G,

β(gm) = (α(gm))(1_{G}) = (g(α(m)))(1_{G}) = α(m)(g).

Because α(m) ∈ Ind_{H}^{G}(M), when g ∈ H, α(m)(g) = g(α(m)(1_{G})) = g(β(m)). Therefore, β is an H-homomorphism M → N.

Conversely, given an H-homomorphism β : M → N, we define α to be the map M → Ind_{H}^{G}(M) such that α(m)(g) = β(gm). Then α is a G-homomorphism.

Since the maps α ↦ β and β ↦ α are inverse, both are isomorphisms.

(b) Given an exact sequence

0 → M → N → P → 0,

we have to prove that

0 → Ind_{H}^{G} M → Ind_{H}^{G} N → Ind_{H}^{G} P → 0

is exact. This is obvious except at the last position. Let S be a set of right coset representatives for H in G, so that G = ∪_{s∈S}Hs, and let φ ∈ Ind_{H}^{G}(P). For each s ∈ S, choose an n(s) ∈ N mapping to φ(s) in P, and define φ̃(hs) = h · n(s). Then φ̃ ∈ Ind_{H}^{G}(N) and maps to φ.

1The φ’s are not required to be homomorphisms.

When H = {1}, an H-module is just an abelian group. In this case, we drop the H from the notation Ind_{H}^{G}. Thus

Ind^{G}(M_{0}) = {φ : G → M_{0}} (maps, not necessarily homomorphisms)
= Hom(ℤ[G], M_{0}) (homomorphisms of abelian groups).

A G-module is said to be induced if it isomorphic to a Ind^{G}(M_{0}) for some abelian group M_{0}.

Remark 1.3. Let G be a finite group. Then a G-module M is induced if and only if there exists an abelian group M_{0} ⊂ M such that

M = ⊕_{g∈G}gM_{0} (direct sum of abeliangroups).

Also, there is an isomorphism of G-modules

φ ↦ ∑_{g∈G} g ⊗ φ(g^{−1}) : Ind^{G}(M_{0}) → ℤ[G] ⊗_{ℤ} M_{0}.

Here ℤ[G] ⊗_{ℤ} M_{0} is endowed with the G-structure such that

g(z ⊗ m) = gz ⊗ m.

Let H be a subgroup of G. An induced G-module is also induced when considered as an H-module: let S be a set of right coset representatives for H in G, so that G = ∪_{s∈S}Hs; if M = ⊕_{g∈G}gM_{0}, then M = ⊕_{h∈H}hM_{1} with M_{1} = ⊕_{s∈S}sM_{0}.

Let M be a G-module, and let M_{0} be M regarded as an abelian group. Then

π : Ind^{G}(M_{0}) → M, φ ↦ ∑_{g∈G} gφ(g^{−1})

is a surjective homomorphism of G-modules. It corresponds to the map

ℤ[G] ⊗ M_{0} → M, (∑ n_{g}g) ⊗ m ↦ ∑ n_{g}gm.

Remark 1.4. Let M and N be G-modules. Then the rule

g(m ⊗ n) = gm ⊗ gn

defines a G-module structure on M ⊗_{ℤ}N. Let M_{0} be M regarded as an abelian group. Then ℤ[G] ⊗_{ℤ} M = ℤ[G] ⊗_{ℤ} M_{0} as abelian groups, but their G-module structures do not correspond. However, one checks easily that the map ℤ[G] ⊗_{ℤ} M_{0} → ℤ[G] ⊗_{ℤ} M sending g ⊗ m to g ⊗ gm is an isomorphism of G-modules.

Injective G-modules. A G-module I is said to be injective if every Ghomomorphism from a submodule of a G-module extends to the whole module, or, equivalently, if Hom_{G}(·, I) is an exact functor.

Proposition 1.5. The category Mod_{G} has enough injectives, i.e., every G-module M can be embedded into an injective G-module, M ↪ I.

Proof. When G = {1}, so that Mod_{G} is the category of abelian groups, this is proved in the Appendix (Proposition 4.3). Now let M be a G-module, and let M_{0} be M regarded as an abelian group. We can embed M_{0} into an injective abelian group, say, M_{0} ↪ I. On applying the functor Ind^{G}, we obtain an inclusion Ind^{G}(M_{0}) ↪ Ind^{G}(I) of G-modules. There is an inclusion of G-modules

M ↪ Ind^{G}(M_{0}), m ↦ (function g ↦ gm).

On composing these maps, we obtain an injective homomorphism M ↪ Ind^{G}(I), and so it remains to show that Ind^{G}(I) is an injective G-module, but this follows from the fact that Ind^{G} has an exact left adjoint, namely, the forgetful functor (see Proposition 4.5).

Definition of the cohomology groups. For a G-module M, define

M^{G} = {m ∈ M | gm = m all g ∈ G}.

The functor

M ↦ M^{G}: Mod_{G} ⟶ Ab

is left exact, i.e., if

0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

is exact, then

0 ⟶ M^{′G} ⟶ M^{G} ⟶ M^{′′G}

is exact. Since the category of G-modules has enough injectives, we can apply the theory of derived functors (see the appendix to this chapter) to this situation.

Let M be a G-module, and choose an injective resolution

0 ⟶ M ⟶ I^{0} ⟶^{d^{0}} I^{1} ⟶^{d^{1}} I^{2} ⟶^{d^{2}} ⋯

of M. The complex

0 ⟶^{d^{−1}} (I^{0})^{G} ⟶^{d^{0}} (I^{1})^{G} ⟶ ⋯ ⟶^{d^{r−1}} (I^{r})^{G} ⟶^{d^{r}} (I^{r+1})^{G} ⟶ ⋯

need no longer be exact, and we define the r^{th} cohomology group of G with coefficients in M to be

H^{r}(G, M) = \frac{Ker(d^{r})}{Im(d^{r−1})}.

These groups have the following basic properties.

1.6. The zeroth group H^{0}(G, M) = M^{G}, because

0 ⟶ M^{G} ⟶ I^{0G} ⟶^{d^{0}} I^{1G}

is exact, and H^{0}(G, M) =^{df} \frac{Ker(d^{0})}{Im(d^{−1})} = Ker(d^{0}).

1.7. If I is an injective G-module, then H^{r}(G, I) = 0 for all r > 0, because 0 → I → I → 0 → ⋯ is an injective resolution of I.

1.8. For any homomorphism α : M → N of G-modules and any injective resolutions M → I^{·} and N → J^{·}, α extends to a map of complexes

\begin{aligned}M & → & I^{·} \\ ↓ α & & ↓ α̃ \\ N & → & J^{·} \end{aligned} ,

and the homomorphisms

H^{r}(α̃) : H^{r}(I^{·}) → H^{r}(J^{·})

are independent of the choice of α̃. On applying this statement to the identity map id : M → M, we find that the groups H^{r}(G, M) are well-defined up to a canonical isomorphism. The general statement then implies that M ↦ H^{r}(G, M) is a functor from the category of G-modules to the category of abelian groups.

1.9. A short exact sequence

0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

of G-modules gives rise to a long exact sequence

0 ⟶ H^{0}(G, M′) ⟶ ⋯ ⟶ H^{r}(G, M) ⟶ H^{r}(G, M′′) ⟶^{δ^{r}} H^{r+1}(G, M′) ⟶ ⋯

Moreover, the association

short exact sequence ↦ long exact sequence

is functorial, i.e., a morphism of short exact sequences induces a morphism of long exact sequences.

Remark 1.10. The family of functors (H^{r}(G, ·))_{r≥0} and coboundary maps δ^{r} are uniquely determined by the properties (1.6, 1.8, 1.9).

Shapiro’s lemma. Let M be a G-module, and regard ℤ as a G-module with the trivial action: gm = m for all g ∈ G, m ∈ ℤ. A homomorphism α : ℤ → M is uniquely determined by α(1), and m ∈ M is the image of 1 under a G-homomorphism ℤ → M if and only if it is fixed by G. Therefore

Hom_{G}(ℤ, M) ≅ M^{G}.

Proposition 1.11 (Shapiro’s Lemma). ^{2} Let H be a subgroup of G. For any H-module N, there is a canonical isomorphism

H^{r}(G, Ind_{H}^{G}(N)) ⟶ H^{r}(H, N),

all r ≥ 0.

Proof. For r = 0, the isomorphism is the composite

N^{H} ≅ Hom_{H}(ℤ, N) ≅^{1.2} Hom_{G}(ℤ, Ind_{H}^{G}(N)) ≅ Ind_{H}^{G}(N)^{G}.

Now choose an injective resolution N → I^{·} of N. On applying the functor Ind_{H}^{G}, we obtain an injective resolution Ind_{H}^{G}(N) ⟶ Ind_{H}^{G}(I^{·}) of the G-module Ind_{H}^{G}(N), (because Ind_{H}^{G} is exact (1.2) and preserves injectives (proof of 1.5). Hence

H^{r}(G, Ind_{H}^{G}(N)) = H^{r}((Ind_{H}^{G}(I^{·}))^{G}) = H^{r}(I^{.H}) = H^{r}(H, N).

Corollary 1.12. If M is an induced G-module, then H^{r}(G, M) = 0 for r > 0.

Proof. If M = Ind^{G}(M_{0}), then

H^{r}(G, M) = H^{r}({1}, M_{0}) = 0 for r > 0.

2For the origin of this elementary but very useful result see Weil, Oeuvres I, pp 577-578.

Remark 1.13. Consider an exact sequence

0 → M → J → N → 0

of G-modules. If H^{r}(G, J) = 0 for all r > 0, then the cohomology sequence becomes

0 → M^{G} → J^{G} → N^{G} → H^{1}(G, M) → 0

and

H^{r}(G, N) ⟶^{≈} H^{r+1}(G, M), r ≥ 1.

For example, let M be a G-module, and let M_{∗} be the induced module Ind^{G}(M_{0}) where M_{0} is M regarded as an abelian group. As we have already noted, M can be identified with the G-submodule of M_{∗} consisting of maps of the form g ↦ gm, m ∈ M, and we let M_{†} = M_{∗}/M. On applying the above remark to the sequence

0 → M → M_{∗} → M_{†} → 0

we find that

H^{r}(G, M_{†}) ≅ H^{r+1}(G, M), all r ≥ 1.

More generally, if

0 → M → J^{1} → ⋯ → J^{s} → N → 0

is exact and H^{r}(G, J^{i}) = 0 for all r, i > 0, then there are canonical isomorphisms

H^{r}(G, N) ⟶^{≈} H^{r+s}(G, M), all r ≥ 1.

To prove this, break the sequence up into short exact sequences

0 → M → J^{1} → N^{1} → 0
0 → N^{1} → J^{2} → N^{2} → 0
⋯
0 → N^{s−1} → J^{s} → N → 0

and note that we have isomorphisms

H^{r}(G, N) ≅ H^{r+1}(G, N^{s−1}) ≅ H^{r+2}(G, N^{s−2}) ≅ ⋯

Remark 1.14. Let

0 ⟶ M ⟶^{ε} J^{0} ⟶^{d^{0}} J^{1} ⟶^{d^{1}} J^{2} ⟶ ⋯

be an exact sequence such that H^{s}(G, J^{r}) = 0 for all s > 0 and all r. Then

H^{r}(G, M) = H^{r}(J^{·G}).

This remark applies to any resolution of M by induced modules.

Description of the cohomology groups by means of cochains. Let M be a G-module. Let P_{r}, r ≥ 0, be the free ℤ-module with basis the (r + 1)-tuples (g_{0}, …, g_{r}) of elements of G, endowed the action of G such that

g(g_{0}, …, g_{r}) = (gg_{0}, …, gg_{r}).

Note that P_{r} is also free as a ℤ[G]-module, with basis {(1, g_{1}, …, g_{r}) | g_{i} ∈ G}. Define a homomorphism d_{r} : P_{r} → P_{r−1} by the rule

d_{r}(g_{0}, …, g_{r}) = ∑_{i=0}^{r}(−1)^{i}(g_{0}, …, ĝ_{i}, …, g_{r})

where the symbol ˆ· means that · is omitted. Let P_{·} be

⋯ → P_{r} ⟶^{d_{r}} P_{r−1} → ⋯ → P_{0}

One checks easily that d_{r−1} ∘ d_{r} = 0, and so this is a complex. Let ε be the map P_{0} → ℤ sending each basis element to 1.

Lemma 1.15. The complex P_{·} ⟶^{ε} ℤ → 0 is exact.

Proof. Choose an element o ∈ G, and define k_{r} : P_{r} → P_{r+1} by

k_{r}(g_{0}, …, g_{r}) = (o, g_{0}, …, g_{r}).

One checks easily that d_{r+1} ∘ k_{r} + k_{r−1} ∘ d_{r} = 1. Hence, if d_{r}(x) = 0, then x = d_{r+1}(k_{r}(x)).

Proposition 1.16. For any G-module M,

H^{r}(G, M) ≅ H^{r}(Hom_{G}(P_{·}, M)).

Proof. This follows from the fact that P_{·} → M is a projective resolution of M— see Example 4.14.

An element of Hom(P_{r}, M) can be identified with a function φ : G^{r+1} → M, and φ is fixed by G if and only if

φ(gg_{0}, …, gg_{r}) = g(φ(g_{0}, …, g_{r})) all g, g_{0}, …, g_{r} ∈ G.

Thus Hom_{G}(P_{r}, M) can be identified with the set C̃^{r}(G, M) of φ’s satisfying this condition. Such φ are called homogeneous r-cochains of G with values in M. The boundary map d̃^{r} : C̃^{r}(G, M) → C̃^{r+1}(G, M) induced by d_{r} is

(d̃^{r}φ)(g_{0}, …, g_{r+1}) = ∑(−1)^{i}φ(g_{0}, …, ĝ_{i}, …, g_{r+1}).

Proposition 1.16 says that

H^{r}(G, M) ≅ \frac{Ker(d̃^{r})}{Im(d̃^{r−1})}.

A homogenous cochain φ : G^{r+1} → M is determined by its values on the elements (1, g_{1}, g_{1}g_{2}, …, g_{1} … g_{r}). We are therefore led to introduce the group C^{r}(G, M) of inhomogeneous r-cochains of G with values in M consisting of all maps φ : G^{r} → M. We set G^{0} = {1}, so that C^{0}(G, M) = M. Define

d^{r}: C^{r}(G, M) ⟶ C^{r+1}(G, M),

by (d^{r}φ)(g_{1}, ⋯, g_{r+1}) =

g_{1}φ(g_{2}, …, g_{r+1}) + ∑_{j=1}^{r}(−1)^{j}φ(g_{1}, …, g_{j}g_{j+1}, …, g_{r+1}) + (−1)^{r+1}φ(g_{1}, …, g_{r}).

Define

Z^{r}(G, M) = Ker(d^{r}) (group of r-cocycles)

and

B^{r}(G, M) = Im(d^{r−1}) (group of r-coboundaries).

Proposition 1.17. The sequence of maps

C^{0}(G, M) ⟶^{d^{0}} C^{1}(G, M) ⟶^{d^{1}} ⋯ ⟶^{d^{r−1}} C^{r}(G, M) ⟶^{d^{r}} C^{r+1}(G, M) ⟶ ⋯

is a complex, i.e., d^{r} ∘ d^{r−1} = 0, and there is a canonical isomorphism

H^{r}(G, M) ≅ \frac{Z^{r}(G, M)}{B^{r}(G, M)}.

Proof. For φ ∈ C̃^{r}(G, M), define

φ′(g_{1}, …, g_{r}) = φ(1, g_{1}, g_{1}g_{2}, …, g_{1} ⋯ g_{r}).

Then φ ↦ φ′ is a bijection C̃^{r}(G, M) → C^{r}(G, M) transforming the boundary maps in C̃^{·}(G, M) into the boundary maps in C^{·}(G, M).

Example 1.18. (a) A map φ: G → M is a crossed homomorphism if φ(στ) = σφ(τ) + φ(σ). For example, for any m ∈ M, the map σ ↦ σm − m is a crossed homomorphism—called a principal crossed homomorphism. According to the proposition

H^{1}(G, M) = \frac{\{crossed homomorphisms G → M\}}{\{principal crossed homomorphisms\}}.

If the action of G on M is trivial (i.e, σm = m all σ ∈ G, m ∈ M), then a crossed homomorphism is a homomorphism, and zero is the only principal crossed homomorphisms. Thus, in this case,

H^{1}(G, M) = Hom(G, M).

(b) Let M be an abelian group (with the law of composition written as multiplication). An extension of G by M is an exact sequence

1 ⟶ M ⟶ E ⟶^{π} G ⟶ 1.

If s(σ) ∈ E maps to σ ∈ G, then we set

σm = s(σ) · m · s(σ)^{−1}, m ∈ M.

Because M is commutative, σm depends only on σ, and so this defines an action of G on M. Note that

s(σ) · m = σm · s(σ), all σ ∈ G, m ∈ M.

Choose a section s to π, i.e., a map (not necessarily a homomorphism) s: G → E such that π ∘ s = id. Then s(σ)s(σ′) and s(σσ′) both map σσ′ ∈ G, and so they differ by an element φ(σ, σ′) ∈ M:

s(σ)s(σ′) = φ(σ, σ′) · s(σσ′).

From

s(σ)(s(σ′)s(σ′′)) = (s(σ)s(σ′))s(σ′′)

we deduce that

[σφ(σ′, σ′′)]φ(σ, σ′σ′′) = φ(σ, σ′)φ(σσ′, σ′′),

i.e., that φ ∈ Z^{2}(G, M). If s is replaced by a different section, φ is replaced by a cohomologous cocycle, and so the class of φ in H^{2}(G, M) is independent of the choice of s. Every such φ arises from an extension, and so H^{2}(G, M) classifies the isomorphism classes of extensions of G by M with a fixed action of G on M.

Example 1.19. Let G be a cyclic group of order m, and let M be a G-module. Define Nm_{G}: M → M to be m ↦ ∑_{σ∈G} σm. Choose a generator σ for G. I claim that φ ↦ φ(σ) defines an isomorphism

H^{1}(G, M) ⟶ Ker(Nm_{G})/(σ − 1)M.

Let φ be a crossed homomorphism. Observe that

φ(1) = φ(1 × 1) = 1φ(1) + φ(1) = 2φ(1)

and so φ(1) = 0. Hence,

0 = φ(σ^{m}) = φ(σ^{m−1}σ) = σφ(σ^{m−1})+φ(σ) = σ^{2}φ(σ^{m−2})+σφ(σ)+φ(σ) = Nm_{G} φ(σ),

and so φ(σ) ∈ Ker(Nm_{G}). Finally note that

φ is principal ⟺ φ(σ) = σm − m some m ⟺ φ(σ) ∈ (σ − 1)M.

Remark 1.20. Let

0 → M → N → P → 0

be an exact sequence of G-modules. The boundary map

δ^{r} : H^{r}(G, P) → H^{r+1}(G, M)

has the following description: let γ ∈ H^{r}(G, P) be represented by the r-cocycle φ : G^{r} → P; because N maps onto P, there exists an r-cochain φ̃ : G^{r} → N lifting φ; because dφ = 0, dφ̃ takes values in M—it is the cocycle representing δ^{r}γ.

The cohomology of L and L^{×}. Let L be a finite Galois extension of the field K, and let G = Gal(L/K). Then both L (regarded as a group under addition) and L^{×} are G-modules.

Proposition 1.21. Let L/K be a finite Galois extension with Galois group G. Then H^{1}(G, L^{×}) = 0.

Proof. Let φ: G → L^{×} be a crossed homomorphism, i.e.,

φ(τσ) = τφ(σ) · φ(τ).

For a ∈ L^{×}, let

b = ∑_{σ∈G} φ(σ) · σa.

Suppose b ≠ 0. Then

τb = ∑_{σ} τφ(σ) · τσa = ∑_{σ} φ(τ)^{−1}φ(τσ)τσa = φ(τ)^{−1}b.

Hence

φ(τ) = b/τb = τ(b^{−1})/b^{−1},

which shows that φ is principal.

It remains to show that there is an a for which b ≠ 0. Recall (Math 594f, 5.13):

(Dedekind’s theorem on the independence of characters.) Let L be a field and H a group; then any finite set {f_{i}} of distinct homomorphisms H → L^{×} is linearly independent over L, i.e.,

∑ a_{i}f_{i}(α) = 0 all α ∈ H ⟹ a_{1} = a_{2} = ⋯ = a_{n} = 0.

When we apply this with H = L^{×} and ∑_{σ∈G} φ(σ)σ, we find that there exists an a such that b ≠ 0.

Corollary 1.22. Let L/K be a cyclic extension, and let σ generate Gal(L/K). If Nm_{L/K} a = 1, then a is of the form \frac{σb}{b}.

Proof. We have 1 = H^{1}(G, L^{×}) = Ker(Nm_{G})/(σ − 1)L^{×}.

Corollary 1.22 occurs as Satz 90 of Hilbert’s book, Die Theorie der algebraischen Zahlkörper, 1894/95, and Theorem 1.21 is Emmy Noether’s generalization. Both are usually referred to as Hilbert’s Theorem 90.

Proposition 1.23. Let L/K be a finite Galois extension with Galois group G. Then H^{r}(G, L) = 0 for all r > 0.

Proof. The Normal Basis Theorem (see below) states that L ≈ K[G] as a Gmodule. But K[G] = Ind_{{1}}^{G}K, and so H^{r}(G, L) = H^{r}({1}, K) = 1 for r > 0.

Lemma 1.24 (Normal Basis Theorem). Let L/K be a finite Galois extension (of arbitrary fields) with Galois group G. Then there exists an α ∈ L such that {σα | σ ∈ G} is a basis^{3} for L as a K-vector space.

Proof. We give two proofs, the first of which assumes that K is infinite and the second that G is cyclic. Since every finite Galois extension of a finite field is cyclic, this covers all cases.

Assume that K is infinite. This has the consequence that, for f(X_{1}, …, X_{m}) ∈ K[X_{1}, …, X_{m}],

f(a_{1}, …, a_{m}) = 0 all a_{1}, …, a_{m} ∈ K ⟹ f(X_{1}, …, X_{m}) = 0.

3A basis of this form is said to normal.

This can be proved by induction. For m = 1 it follows from the fact that a nonzero polynomial in one variable has only finitely many roots. For m > 1, write

f = ∑ c_{i}(X_{1}, …, X_{m−1})X_{m}^{i}.

When we substitute values a_{1}, …, a_{m−1} for the X_{1}, …, X_{m−1}, the resulting polynomial in X_{m} has infinitely many roots, and therefore each of its coefficients is zero, i.e., c_{i}(a_{1}, …, a_{m−1}) = 0. Since this holds for all (a_{1}, …, a_{m−1}), the induction hypothesis shows that c_{i}(X_{1}, …, X_{m−1}) is zero.

Now number the elements of G as σ_{1}, …, σ_{m} (with σ_{1} = 1).

Let f(X_{1}, …, X_{m}) ∈ K[X_{1}, …, X_{m}] have the property that

f(σ_{1}α, …, σ_{m}α) = 0

for all α ∈ L. For a basis α_{1}, …, α_{m} of L over K, let

g(Y_{1}, …, Y_{m}) = f(∑_{i=1}^{m} Y_{i}σ_{1}α_{i}, ∑_{i=1}^{m} Y_{i}σ_{2}α_{i}, …).

The hypothesis on f implies that g(a_{1}, …, a_{m}) = 0 for all a_{i} ∈ K, and so g = 0. But the matrix (σ_{i}α_{j}) is invertible (see Math 676, 2.25). Since g is obtained from f by an invertible linear change of variables, f can be obtained from g by the inverse linear change of variables. Therefore it also is zero.

Write X_{i} = X(σ_{i}), and let A = (X(σ_{i}σ_{j})), i.e., A is the m × m matrix having X_{k} in the (i, j)^{th} place if σ_{i}σ_{j} = σ_{k}. Let f(X_{1}, …, X_{m}) = det(A). Then f(1, 0, …, 0) is the determinant of a matrix having exactly one 1 in each row and each column and its remaining entries 0. Hence the rows of the matrix are a permutation of the rows of the identity matrix, and so its determinant is ±1. Hence f is not identically zero, and so there exists an α ∈ L^{×} such that f(σ_{1}α, …, σ_{m}α) (= det(σ_{i}σ_{j}α)) is nonzero. Now suppose

∑_{j=1}^{m} a_{j}σ_{j}α = 0

for some a_{j} ∈ K. On applying σ_{1}, …, σ_{m} successively, we obtain a system of mequations

∑ a_{j}σ_{i}σ_{j}α = 0

in the m “unknowns” a_{j}. Because this system of equations is nonsingular, the a_{j}’s are zero. We have shown that the σ_{j}α’s are linearly independent over K. This completes the proof of the lemma in the case that K is infinite.

Now assume that G is cyclic generated, say, by an element σ_{0} of order n. We regard σ_{0} as an endomorphism of L considered as a K-vector space. The characteristic polynomial P (X) of σ_{0} is the monic polynomial in K[X] of least degree such that P (σ_{0}) = 0 (as an endomorphism of L). It has the property that it divides every polynomial Q(X) ∈ K[X] such that Q(σ_{0}) = 0. Since σ_{0}^{n} = 1, P (X) divides X^{n} − 1. On the other hand, Dedekind’s theorem on the independence of characters (see above) implies that id, σ_{0}, …, σ_{0}^{n−1} are linearly independent over K, and so deg P (X) > n − 1. We conclude that P (X) = X^{n} − 1. Therefore, as a K[X]-module with X acting as σ_{0}, L is isomorphic to K[X]/(X^{n} − 1). For any generator α of L as a K[X]-module, α, σ_{0}α, …, σ_{0}α^{n−1} is a K-basis for L.

The cohomology of products. A product M = ∏ M_{i} of G-modules becomes a G-module under the diagonal action:

σ(…, m_{i}, …) = (…, σm_{i}, …).

Proposition 1.25. For any G-modules M_{i},

H^{r}(G, ∏ M_{i}) = ∏ H^{r}(G, M_{i}).

Proof. A product of exact sequences of abelian groups is again exact. From this it follows that a product I = ∏ I_{i} of injective G-modules is again injective, because

Hom_{G}(·, I) ≅ Π_{i} Hom_{G}(·, I_{i})

is exact. Let M_{i} → I_{i}^{·} be an injective resolution of M_{i}. Then ∏ M_{i} ⟶ ∏ I_{i}^{·} is an injective resolution of ∏ M_{i}, and

H^{r}(G, ∏ M_{i}) = H^{r}((∏ I_{i}^{·})^{G}) = H^{r}(∏(I_{i}^{·G})) = ∏ H^{r}(I_{i}^{·G}) = ∏ H^{r}(G, M_{i}).

Remark 1.26. The formation of inverse limits of arbitrary abelian groups is not exact. Therefore, in general, one can not expect cohomology to commute with inverse limits.

Functorial properties of the cohomology groups. Let M and M′ respectively be G and G′ modules. Homomorphisms

α: G′ → G, β: M → M′

are said to be compatible if

β(α(g)m) = g(β(m)).

Then (α, β) defines a homomorphism of complexes

C^{·}(G, M) ⟶ C^{·}(G′, M′), φ ↦ β ∘ φ ∘ α^{r},

and hence homomorphisms

H^{r}(G, M) ⟶ H^{r}(G′, M′).

Example 1.27. (a) Let H be a subgroup of G. For any H-module M, the map

β : Ind_{H}^{G}(M) → M, β(φ) = φ(1),

is compatible with the inclusion H ↪ G, and the induced homorphism

H^{r}(G, Ind_{H}^{G}(M)) → H^{r}(H, M)

is the isomorphism in Shapiro’s Lemma.

(b) Let H be a subgroup of G. Let α be the inclusion H ↪ G, and let β be the identity map on a G-module M. In this case, we obtain restriction homomorphisms

Res: H^{r}(G, M) ⟶ H^{r}(H, M).

They can also be constructed as follows: let M → Ind_{H}^{G}(M) be the homomorphism sending m to the map φ_{m} with φ_{m}(g) = gm; the restriction map is the composite of the homomorphism it defines on cohomology with the isomorphism in Shapiro’s Lemma,

H^{r}(G, M) → H^{r}(G, Ind_{H}^{G}(M)) ⟶^{≈} H^{r}(H, M).

(c) Let H be a normal subgroup of G, let α be the quotient map G → G/H, and let β be the inclusion M^{H} ↪ M. In this case, we obtain the inflation homomorphisms

Inf : H^{r}(G/H, M^{H}) ⟶ H^{r}(G, M).

(d) For any g_{0} ∈ G, the homomorphisms α: G → G, σ ↦ g_{0}σg_{0}^{−1}, and β: M → M, m ↦ g_{0}^{−1}m, are compatible. I claim that the homomorphisms

H^{r}(G, M) ⟶ H^{r}(G, M).

they define are each the identity map. For r = 0, the homomorphism is

m ↦ g^{−1}m: M^{G} ⟶ M^{G},

which is obviously the identity. Let r > 0, and suppose the statement is known for r − 1. From the sequence

0 → M → M_{∗} → M_{†} → 0, M_{∗} = Ind^{G}(M_{0}),

in (1.13) we obtain a diagram

H^{r−1}(G, M_{∗}) → H^{r−1}(G, M_{†}) → H^{r}(G, M) → 0
↓ ↓ ↓
H^{r−1}(G, M_{∗}) → H^{r−1}(G, M_{†}) → H^{r}(G, M) → 0.

The 0s at right result from the fact that M_{∗} is an induced module. The vertical maps are those defined by the pair (α, β) (for the different modules). One checks easily that the diagram commutes. By induction, the middle vertical map is the identity, which implies that the third is also.

Remark 1.28. (a) The method of proof in (d) is called dimension shifting.

(b) Let H be a normal subgroup of G. For any G-module M, the recipe in (d) gives an action of G on H^{r}(H, M), which the above result shows to factor through G/H.

Example 1.29. We shall need one more functorial map of cohomology groups. Let H be a subgroup of finite index of G, and let S be a set of left coset representatives for H in G, G = ∪_{s∈S}sH. Let M be a G-module. For any m ∈ M^{H},

Nm_{G/H} m =^{df} ∑_{s∈S} sm

is independent of the choice of S, and is fixed by G. Thus Nm_{G/H} is a homomorphism

M^{H} ⟶ M^{G}.

This can be extended to a corestriction homomorphism

Cor: H^{r}(H, M) ⟶ H^{r}(G, M)

for all r as follows: for any G-module M, there is a canonical homomorphism of G-modules

φ ↦ ∑ sφ(s^{−1}) : Ind_{H}^{G} M → M;

the map on cohomology which it defines, when composed with the isomorphism in Shapiro’s lemma, gives Cor,

H^{r}(H, M) ⟶^{≈} H^{r}(G, Ind_{H}^{G} M) ⟶ H^{r}(G, M).

Proposition 1.30. Let H be a subgroup of G of finite index. The composite

Cor ∘ Res: H^{r}(G, M) ⟶ H^{r}(G, M)

is multiplication by (G: H).

Proof. The map Cor ∘ Res is the map on cohomology defined by the composite of

M → Ind_{H}^{G}(M) → M, m ↦ φ_{m} ↦ ∑_{s} sφ_{m}(s^{−1}) = ∑_{s} m = (G : H)m.

Corollary 1.31. If (G: 1) = m, then mH^{r}(G, M) = 0 for r > 0.

Proof. According to the proposition, multiplication by m factors through H^{r}({1}, M) = 0,

H^{r}(G, M) ⟶^{Res} H^{r}({1}, M) ⟶^{Cor} H^{r}(G, M).

Corollary 1.32. If G is finite and M is finitely generated as an abelian group, then H^{r}(G, M) is finite.

Proof. From the description of H^{r}(G, M) as the group of cocycles modulo coboundaries, it is clear that H^{r}(G, M) is finitely generated as an abelian group, and we have just seen that it is killed by (G: 1). Therefore it is finite.

For any abelian group A and prime p, the p-primary component A(p) of A is the subgroup consisting of all elements killed by a power of p.

Corollary 1.33. Let G be a finite group, and let G_{p} be its Sylow p-subgroup. For any G-module M, the restriction map

Res : H^{r}(G, M) → H^{r}(G_{p}, M)

is injective on the p-primary component of H^{r}(G, M).

Proof. By definition, (G : H) is not divisible by p. The composite

Cor ∘ Res : H^{r}(G, M) → H^{r}(G_{p}, M) → H^{r}(G, M)

is multiplication by (G : H), and so is injective on the p-primary component of H^{r}(G, M).

The inflation-restriction exact sequence.

Proposition 1.34. Let H be a normal subgroup of G, and let M be a G-module. Let r be an integer > 0. If H^{i}(H, M) = 0 for all i with 0 < i < r, then the sequence

0 → H^{r}(G/H, M^{H}) ⟶^{Inf} H^{r}(G, M) ⟶^{Res} H^{r}(H, M)

is exact.

Proof. We first prove this for r = 1—in this case, the hypothesis on H^{r}(H, M) is vacuous. Let φ : G → M be a crossed homomorphism whose restriction to H is principal, say, φ(h) = hm_{0} − m_{0}. Define φ′(g) = φ(g) − (gm_{0} − m_{0}). Then φ′ represents the same class in H^{1}(G, M), but now φ′(h) = 0 for all h ∈ H, and so φ′ comes by “inflation” from a crossed homomorphism G/H → M^{H}. This shows that the sequence is exact at H^{1}(G, M). The exactness at H^{1}(G/H, M^{H}) is even easier to prove.

Now assume that r > 1 and that the statement is true for r −1. Consider the exact sequence (1.13)

0 → M → M_{∗} → M_{†} → 0.

Then

H^{i}(H, M_{†}) ≅ H^{i+1}(H, M), i > 0,

and so H^{i}(H, M_{†}) = 0 for 0 < i ≤ r − 1. By induction, the sequence

0 → H^{r−1}(G/H, M_{†}^{H}) ⟶^{Inf} H^{r−1}(G, M_{†}) ⟶^{Res} H^{r−1}(H, M_{†})

is exact, and this is isomorphic to the sequence

0 → H^{r}(G/H, M^{H}) ⟶^{Inf} H^{r}(G, M) ⟶^{Res} H^{r}(H, M).

Remark 1.35. For the experts, the Hochschild-Serre (alias Lyndon) spectral sequence has the form

H^{r}(G/H, H^{s}(H, M)) ⟹ H^{r+s}(G, M).

It defines a filtration on each of the groups H^{r}(G, M), and expresses each quotient of the filtration in terms of the cohomologies of G/H and H. It implies the above proposition. Without any hypotheses, one can show that there is an exact sequence

0 ⟶ H^{1}(G/H, M^{H}) ⟶^{Inf} H^{1}(G, M) ⟶^{Res} H^{1}(H, M)^{G/H} ⟶

H^{2}(G/H, M^{H}) ⟶ H^{2}(G, M)^{∗} ⟶ H^{1}(G/H, H^{1}(H, M))

where H^{2}(G, M)^{∗} = Ker(Res: H^{2}(G, M) → H^{2}(H, M)).

Example 1.36. Let Ω/K and L/K be Galois extensions, with L ⊂ Ω. Then H =^{df} Gal(Ω/L) is a normal subgroup of G =^{df} Gal(Ω/K). According to Proposition 1.21, H^{1}(H, Ω^{×}) = 0, and so there is an exact sequence

0 ⟶ H^{2}(G/H, L^{×}) ⟶ H^{2}(G, Ω^{×}) ⟶ H^{2}(H, Ω^{×}).

A direct proof (not involving dimension shifting) that this sequence is exact can be found in Artin 1951, 6.4.

Cup-products. Let G be a group. For G-modules M and N, we write M ⊗ N for M ⊗_{ℤ} N, regarded as a G-module with

g(m ⊗ n) = gm ⊗ gn, g ∈ G, m ∈ M, n ∈ N.

Proposition 1.37. There exists one and only one family of bi-additive pairings

(m, n) ↦ m ∪ n : H^{r}(G, M) × H^{s}(G, N) ⟶ H^{r+s}(G, M ⊗ N),

defined for all G-modules M, N and all integers r, s ≥ 0, satisfying the following conditions:

(a) these maps become morphisms of functors when the two sides are regarded as covariant bifunctors on (M, N);

(b) for r = s = 0, the pairing is

(m, n) ↦ m ⊗ n: M^{G} ⊗ N^{G} ⟶ (M ⊗ N)^{G};

(c) if 0 → M′ → M → M′′ → 0 is an exact sequence of G-modules such that

0 → M′ ⊗ N → M ⊗ N → M′′ ⊗ N → 0

is exact, then

(δm′′) ∪ n = δ(m′′ ∪ n), m′′ ∈ H^{r}(G, M′′), n ∈ H^{s}(G, N).

Here δ denotes the connecting homomorphism H^{r}(G, M′′) ⟶ H^{r+1}(G, M′) or H^{r+s}(G, M′′ ⊗ N) ⟶ H^{r+s+1}(G, M′ ⊗ N).

(d) if 0 → N′ → N → N′′ → 0 is an exact sequence of G-modules such that

0 → M ⊗ N′ → M ⊗ N → M ⊗ N′′ → 0

is exact, then

m ∪ δn′′ = (−1)^{r}δ(m ∪ n′′), m ∈ H^{r}(G, M), n′′ ∈ H^{s}(G, N′′).

Proof. The uniqueness is proved using dimension shifting. For the existence, one proves that the pairing defined as follows has the required properties: let m ∈ H^{r}(G, M) and n ∈ H^{s}(G, N) be represented by the cocycles φ and ψ; then m ∪ n is represented by the cocycle

(g_{1}, …, g_{r+s}) ↦ φ(g_{1}, …, g_{r}) ⊗ g_{1} ⋯ g_{r}ψ(g_{r+1}, ⋯, g_{r+s}).

Proposition 1.38. The following formulas hold:

(a) (x ∪ y) ∪ z = x ∪ (y ∪ z) (in H^{r+s+t}(G, M ⊗ N ⊗ P));

(b) x ∪ y = (−1)^{rs}y ∪ x if x ∈ H^{r}(G, M), y ∈ H^{s}(G, N);

(c) Res(x ∪ y) = Res(x) ∪ Res(y);

(d) Cor(x ∪ Res y) = Cor(x) ∪ y.

Proof. In each case, one verifies the formula when x, y, … have degree 0, and then uses dimension shifting to deduce the general case. For example, the proof of (d) is written out in detail in Cassels and Fröhlich, 1967, IV, Proposition 9.

A pairing

x, y ↦ (x, y) : M × N → P

such that

g(x, y) = (gx, gy), all g ∈ G,

defines a homomorphism of G-modules

M ⊗ N → P,

and hence maps

H^{r}(G, M ⊗ N) → H^{r}(G, P).

The composites of the cup-product pairings with these maps, namely, the pairings

H^{r}(G, M) × H^{s}(G, N) → H^{r+s}(G, P),

will also be referred to as cup-product pairings.

2. Homology; the Tate Groups

Definition of the homology groups. For a G-module M, let M_{G} be the largest quotient of M on which G acts trivially. Thus M_{G} is the quotient of M by the subgroup generated by

{gm − m | g ∈ G, m ∈ M}.

Note that this is the dual notion to M^{G}, which is the largest subobject of M on which G acts trivially. The definition of the cohomology groups dualizes to give us homology groups.

In detail, the functor

M ↦ M_{G}: Mod_{G} ⟶ Ab

is right exact, i.e., if

0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

is exact, then

M′_{G} ⟶ M_{G} ⟶ M′′_{G} ⟶ 0

is exact.

An object P of an abelian category is projective if for any object N and quotient object M, every morphism P → M lifts to a morphism P → N. For example, any free ℤ[G]-module is projective as a ℤ[G]- (equivalently, G-) module.

Let M be a G-module. Let (m_{i})_{i∈I} be a family of generators for M as a ℤ[G]- module, and let ℤ[G]^{(I)} be a direct sum of copies of ℤ[G] indexed by I. The map ∑_{i∈I} γ_{i} ↦ ∑ γ_{i}m_{i} is a surjective G-homomorphism ℤ[G]^{(I)} → M. This shows that every G-module is a quotient of a free G-module, i.e., that Mod_{G} has enough projectives.

Let M be a G-module, and choose a projective resolution

⋯ ⟶ P_{2} ⟶^{d_{2}} P_{1} ⟶^{d_{1}} P_{0} ⟶ M ⟶ 0

of M. The complex

⋯ ⟶ (P_{2})_{G} ⟶^{d_{2}} (P_{1})_{G} ⟶ (P_{0})_{G} ⟶ 0

need no longer be exact, and we set

H_{r}(G, M) = \frac{Ker(d_{r})}{Im(d_{r+1})}.

These groups have the following basic properties.

2.1. H_{0}(G, M) = M_{G}, because

P_{1G} → P_{0G} → M_{G} → 0

is exact.

2.2. If P is a projective G-module, then H_{r}(G, P) = 0 for all r > 0, because ⋯ → 0 → P → P → 0 is a projective resolution of P.

2.3. Let P_{·} → M and Q_{·} → N be projective resolutions of G-modules M and N. Then any homomorphism α : M → N of G-modules extends to a morphism of complexes

\begin{aligned}P_{·} & → & M \\ ↓ α̃ & & ↓ α \\ Q_{·} & → & N \end{aligned} ,

and the homomorphisms

H_{r}(α̃) : H_{r}(P_{·}) → H_{r}(Q_{·})

are independent of the choice of α̃. When we apply this to the identity map id : M → M, we see that the groups H_{r}(G, M) are well-defined up to a canonical isomorphism. The general statement then implies that M ↦ H_{r}(G, M) is a functor from the category of G-modules to the category of abelian groups.

2.4. A short exact sequence

0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

of G-modules gives rise to a long exact sequence

⋯ ⟶ H_{r}(G, M) ⟶ H_{r}(G, M′′) ⟶ H_{r−1}(G, M′) ⟶ ⋯ ⟶ H_{0}(G, M′′) → 0.

Moreover, the association

short exact sequence ↦ long exact sequence

is functorial, i.e., a morphism of short exact sequences induces a morphism of long exact sequences.

Remark 2.5. The properties (2.1–2.4) determine the functors H^{r}(G, ·).

Just as in the case of cohomology, it is possible to give an explicit description of H_{r}(G, M) as the quotient of a group of r-cycles by a subgroup of r-boundaries—see the references later in this chapter.

The group H_{1}(G, ℤ). Using only the properties of the homology groups listed above, we shall compute H_{1}(G, ℤ).

The augmentation map is

ℤ[G] ⟶ ℤ, ∑ n_{g}g ↦ ∑ n_{g}.

Its kernel is called the augmentation ideal I_{G}. Clearly I_{G} is a free ℤ-submodule of ℤ[G] with basis {g − 1 | g ∈ G}, and

M/I_{G}M = M_{G} =^{df} H_{0}(G, M).

Consider the exact sequence:

0 ⟶ I_{G} ⟶ ℤ[G] ⟶ ℤ ⟶ 0.

The G-module ℤ[G] is projective (because it is free, hence projective, as a ℤ[G]- module), and so H_{1}(G, ℤ[G]) = 0. Therefore we obtain an exact sequence of homology groups

0 ⟶ H_{1}(G, ℤ) ⟶ I_{G}/I_{G}^{2} ⟶ ℤ[G]/I_{G}ℤ[G] ⟶ ℤ ⟶ 0.

The middle map is induced by the inclusion I_{G} ↪ ℤ[G], and so is zero. Therefore the sequence shows that

H_{1}(G, ℤ) ⟶^{≈} I_{G}/I_{G}^{2} (∗)

and

ℤ[G]_{G} = ℤ

i.e., ℤ is the largest quotient of ℤ[G] on which G acts trivially. Note that I_{G}^{2} =^{df} I_{G} · I_{G} is the ℤ-submodule of M generated by elements of the form

(g − 1)(g′ − 1), g, g′ ∈ G.

Lemma 2.6. Let G^{c} be the commutator subgroup of G, so that G/G^{c} is the largest abelian quotient G^{ab} of G. Then the map g ↦ (g − 1) + I_{G}^{2} induces an isomorphism

\frac{G}{G^{c}} ⟶ \frac{I_{G}}{I_{G}^{2}} .

Proof. Consider the map

g ↦ (g − 1) + I_{G}: G ⟶ I_{G}/I_{G}^{2}.

This is a homomorphism because

gg′ − 1 = (g − 1)(g′ − 1) + (g − 1) + (g′ − 1) ≡ (g − 1) + (g′ − 1) mod I_{G}^{2}.

Since I_{G}/I_{G}^{2} is commutative, the map factors through G^{ab}. To prove it is an isomorphism, we construct an inverse. Recall that I_{G} is freely generated by the elements g − 1. Consider the homomorphism I_{G} ⟶ G^{ab} mapping g − 1 to the class of g. From the identity

(g − 1)(g′ − 1) = (gg′ − 1) − (g − 1) − (g′ − 1)

we see that (g − 1)(g′ − 1) maps to 1. Since I_{G}^{2} is generated by elements of this form, this shows that the map factors through I_{G}/I_{G}^{2}. The two maps we have constructed are inverse.

Proposition 2.7. There is a canonical isomorphism

H_{1}(G, ℤ) ⟶ G^{ab}.

Proof. Combine the isomorphism (∗) with that in the lemma.

Remark 2.8. For any group G, there exists a topological space BG, called the classifying space of G, such that G = π_{1}(BG) and H_{r}(BG, ℤ) = H_{r}(G, ℤ) for all r (J. Rosenberg, Algebraic K-Theory and Its Applications, Springer, 1994, 5.1.16, 5.1.27). Therefore the proposition simply states that H_{1}(BG, ℤ) is the maximal abelian quotient of the fundamental group of BG.

The Tate groups. For the remainder of this section, we require G to be finite. For any G-module M, the norm map Nm_{G}: M → M is defined to be

m ↦ ∑_{g∈G} gm.

Let g′ ∈ G. As g runs through the elements of G, so also do gg′ and g′g, and so

Nm_{G}(g′m) = Nm_{G}(m) = g′(Nm_{G}(m)).

Hence

I_{G}M ⊂ Ker Nm_{G}, Im(Nm_{G}) ⊂ M^{G}.

As H_{0}(G, M) = M/I_{G}M and H^{0}(G, M) = M^{G}, it follows that Nm_{G} defines a homomorphism

Nm_{G} : H_{0}(G, M) ⟶ H^{0}(G, M).

For any short exact sequence of G-modules

0 ⟶ M′ ⟶ M ⟶ M′′ ⟶ 0

we get a diagram

H_{1}(G, M′′) ⟶ H_{0}(G, M′) ⟶ H_{0}(G, M) ⟶ H_{0}(G, M′′) ⟶  0
↓^{Nm}_{G}  ↓^{Nm}_{G}  ↓^{Nm}_{G}
0  ⟶ H^{0}(G, M′) ⟶ H^{0}(G, M) ⟶ H^{0}(G, M′′) ⟶ H^{1}(G, M′) ⟶ ⋯

On applying the extended snake lemma (4.1) to the middle part of the diagram, we get a (very) long exact sequence

⋯ ⟶ H_{T}^{r} (G, M′) ⟶ H_{T}^{r} (G, M) ⟶ H_{T}^{r} (G, M) ⟶^{δ} H_{T}^{r+1}(G, M) ⟶ ⋯ −∞ < r < ∞.

where

H_{T}^{r} (G, M) =^{df} \{ \begin{matrix}H^{r}(G, M) & r > 0 \\ M^{G}/ Nm_{G}(M) & r = 0 \\ Ker(N_{G})/I_{G}M & r = −1 \\ H_{−r−1}(G, M) & r < −1. \end{matrix}

Since it causes no ambiguity, we often omit the subscript T when r ≠ 0.

Most of the results we proved for the groups H^{r}(G, M) with r ≥ 0 extend to all r. For example, Shapiro’s lemma and its consequences are true, and the maps we defined in (1.27) and (1.29) extend to all the Tate cohomology groups. Specifically, there are canonical homomorphisms:

Res: H_{T}^{r} (G, M) ⟶ H_{T}^{r} (H, M) (H a subgroup of G);

Cor: H_{T}^{r} (H, M) ⟶ H_{T}^{r} (G, M) (H a subgroup of G);

Inf: H_{T}^{r} (G/H, M^{H}) ⟶ H_{T}^{r} (G, M) (H a normal subgroup of G). Moreover, there is a natural action of G/H on M^{H}.

The composite Res ∘ Cor is still multiplication by (G: H). As H_{T}^{r} ({1}, M) = 0 for all r, the argument in the proof of (1.30) shows that H_{T}^{r} (G, M) is killed by m for all r.

Note that H_{T}^{−2}(G, ℤ) = H_{1}(G, ℤ) ≅ G/G^{c}.

Proposition 2.9. Let H be a subgroup of G.

(a) The map Cor : H^{−2}(H, ℤ) → H^{−2}(G, ℤ) corresponds to the map H/H^{c} → G/G^{c} induced by the inclusion H ↪ G.

(b) The map Res : H^{−2}(G, ℤ) → H^{−2}(H, ℤ) corresponds to the Verlagerung map G/G^{c} → H/H^{c}.

Proof. See E. Weiss, Cohomology of Groups, Academic Press, 1969, 3-5.

There is a uniform explicit description of all the groups H^{r}(G, M). In fact, there is an explicit complex L. of G-modules (infinite in both directions) such that

H_{T}^{r} (G, M) = H^{r}(Hom_{G}(L., M)).

For r > 0 this leads directly to the description we gave above of H^{r}(G, M) in terms of inhomogeneous cocycles and coboundaries.

Cup-products. When the group G is finite, the cup-products extend in a unique way to all the cohomology groups, and have the same list of properties.

The cohomology of finite cyclic groups. We first compute the cohomology of ℚ, ℤ, and ℚ/ℤ, each regarded as a G-module with the trivial action.

Lemma 2.10. For any finite group G,

(a) H_{T}^{r} (G, ℚ) = 0 all r (ℚ regarded as a G-module with the trivial action);

(b) H_{T}^{0}(G, ℤ) = ℤ/(G : 1)ℤ and H^{1}(G, ℤ) = 0;

(c) there is a canonical isomorphism

Hom(G, ℚ/ℤ) ⟶ H^{2}(G, ℤ).

Proof. (a) The group ℚ is uniquely divisible, i.e., for all integers m, multiplication by m: ℚ → ℚ is an isomorphism. Therefore the map H^{r}(m) : H_{T}^{r} (G, ℚ) ⟶ H_{T}^{r} (G, ℚ), which is also multiplication by m, is an isomorphism. Now take m = (G: 1). Then multiplication by m on H_{T}^{r} (G, M) is both zero (see 1.30) and an isomorphism—this is possible only if H_{T}^{r} (G, M) = 0.

(b) Because G acts trivially on ℤ, ℤ^{G} = ℤ and the norm map is multiplication by (G : 1). Hence H_{T}^{0}(G, ℤ) = ℤ/(G : 1)ℤ. Moreover, H^{1}(G, ℤ) = Hom(G, ℤ) (see 1.18), but, because ℤ is torsion-free, there are no nonzero such homomorphisms.

(c) The cohomology sequence of

0 ⟶ ℤ ⟶ ℚ ⟶ ℚ/ℤ ⟶ 0

is an exact sequence

H^{1}(G, ℚ) ⟶ H^{1}(G, ℚ/ℤ) ⟶ H^{2}(G, ℤ) ⟶ H^{2}(G, ℚ).
∥  ∥  ∥
0  Hom(G, ℚ/ℤ)  0

Let G be a finite cyclic group of order m, with generator σ. Then, by definition,

H_{T}^{0}(G, M) = \frac{Ker(σ − 1)}{Im(Nm_{G})}

and

H_{T}^{−1}(G, M) = H_{0}(G, M) = \frac{Ker(Nm_{G})}{Im(σ − 1)}.

Note that H_{T}^{−1}(G, M) = H_{T}^{1}(G, M) (see 1.19). In fact, the cohomology groups are periodic with period 2.

Proposition 2.11. Let G be a cyclic group of finite order, and let M be a Gmodule. Then, for all r, there exists an isomorphism

H_{T}^{r} (G, M) ⟶^{≈} H_{T}^{r+2}(G, M)

depending only on the choice of a generator for G.

Proof. Let σ generate G. Then the sequence

0 → ℤ ⟶^{m↦m1_{G}} ℤ[G] ⟶^{σ−1} ℤ[G] ⟶^{σ^{i}↦1} ℤ → 0

is exact. Because the groups in the sequence and the kernel I_{G} of ℤ[G] → ℤ are free ℤ-modules, the sequence remains exact after it is tensored with M. Thus

0 → M → ℤ[G] ⊗_{ℤ} M → ℤ[G] ⊗_{ℤ} M → M → 0

is an exact sequence of G-modules. Recall (1.4) that ℤ[G]⊗_{ℤ}M ≈ ℤ[G]⊗_{ℤ}M_{0}, where M_{0} is M regarded as abelian group, and so H^{r}(G, ℤ[G]⊗_{ℤ}M) = 0 for all r. Therefore (see 1.13), the sequence defines isomorphisms

H_{T}^{r} (G, M) ⟶^{≈} H_{T}^{r+2}(G, M)

for all r.

Remark 2.12. Let γ be the element of H^{2}(G, ℤ) corresponding under the isomorphism H^{2}(G, ℤ) ≅ Hom(G, ℚ/ℤ) to the map sending the chosen generator σ of G to 1/m. Then the map H^{r}(G, M) → H^{r+2}(G, M) is x ↦ x ∪ γ.

Let G be a finite cyclic group, and let M be a G-module. When the cohomology groups H^{r}(G, M) are finite, we define the Herbrand quotient

h(M) = \frac{#H_{T}^{0}(G, M)}{#H_{T}^{1}(G, M)}.

Proposition 2.13. Let 0 → M′ → M → M′′ → 0 be an exact sequence of Gmodules. If any two of the Herbrand quotients h(M′), h(M), h(M′′) are defined, then so also is the third, and

h(M) = h(M′)h(M′′).

Proof. We can truncate the (very) long exact cohomology sequence as follows,

0 ⟶ K ⟶ H^{0}(M′) ⟶ H^{0}(M) ⟶ H^{0}(M′′) ⟶ H^{1}(M′) ⟶ H^{1}(M) ⟶ H^{1}(M′′) ⟶ K′ ⟶ 0

where

K = Coker(H^{−1}(M) ⟶ H^{−1}(M′′)) ≈ Coker(H^{1}(M) ⟶ H^{1}(M′′)) ≈ K′.

The first statement is now obvious, and the second follows from the next lemma.

Lemma 2.14. Let

0 ⟶ A_{0} ⟶ A_{1} ⟶ ⋯ ⟶ A_{r} ⟶ 0

be a finite sequence of finite groups. Then

\frac{#A_{0}#A_{2} ⋯}{#A_{1}#A_{3} ⋯} = 1.

Proof. For a short exact sequence, that is, r = 3, this is obvious, but any exact sequence can be broken up into short exact sequences,

0 ⟶ A_{0} ⟶ A_{1} ⟶ C_{1} ⟶ 0
0 ⟶ C_{1} ⟶ A_{2} ⟶ C_{2} ⟶ 0
⋯
0 ⟶ C_{r−1} ⟶ A_{r−1} ⟶ A_{r} ⟶ 0.

Here C_{i} = Coker(A_{i−1} → A_{i}) = Ker(A_{i+1} → A_{i+2}). From these sequences we find that

1 = \frac{#A_{0}#C_{1}}{#A_{1}} = \frac{#A_{0}#A_{2}}{#A_{1}#C_{2}} = · · · .

Proposition 2.15. If M is a finite module, then h(M) = 1.

Proof. Consider the exact sequences

0 ⟶ M^{G} ⟶ M ⟶^{g−1} M ⟶ M_{G} ⟶ 0

and

0 ⟶ H^{−1}(M) ⟶ M_{G} ⟶^{Nm_{G}} M^{G} ⟶ H^{0}(M) ⟶ 0

where g is any generator of G. From the first sequence we find that M^{G} and M_{G} have the same order, and then from the second that H^{−1}(M) and H^{0}(M) have the same order.

Corollary 2.16. Let α: M → N be a homomorphism of G-modules with finite kernel and cokernel. If either h(M) or h(N) is defined, then so also is the other, and they are equal.

Proof. Suppose h(N) is defined, and consider the exact sequences:

0 ⟶ α(M) ⟶ N ⟶ Coker(α) ⟶ 0
0 ⟶ Ker(α) ⟶ M ⟶ α(M) ⟶ 0.

From the first sequence, we find that h(αM) is defined and equals h(N), and from the second sequence we find that h(M) is defined and equals h(αM).

Tate’s Theorem. For the remainder of this section, all cohomology groups will be the Tate groups, and so we drop the subscript T.

Theorem 2.17. Let G be a finite group, and let M be a G-module. If

H^{1}(H, M) = 0 = H^{2}(H, M)

for all subgroups H of G, then H^{r}(G, M) = 0 for all r ∈ ℤ.

Proof. If G is cyclic, this follows from the periodicity of the cohomology.

Assume now that G is solvable. We shall prove the theorem in this case by induction on the order of G.

Because G is solvable, it contains a proper normal subgroup H such that G/H is cyclic. Because H has order less than that of G, and the pair (H, M) satisfies the hypotheses of the theorem, H^{r}(H, M) = 0 for all r. Therefore (see 1.34), we have exact sequences

0 → H^{r}(G/H, M^{H}) → H^{r}(G, M) → H^{r}(H, M) (∗)

for all r ≥ 1. Because H^{1}(G, M) = 0 = H^{2}(G, M), H^{1}(G/H, M^{H}) = H^{2}(G/H, M^{H}) = 0, and because G/H is cyclic, this implies that H^{r}(G/H, M^{H}) = 0 for all r. Therefore, the sequences (∗) show that H^{r}(G, M) = 0 for all r > 0. We next show that H^{0}(G, M) = 0. Let x ∈ M^{G}. Because H^{0}(G/H, M^{H}) = 0, there exists a y ∈ M^{H} such that Nm_{G/H}(y) = x, and because H^{0}(H, M) = 0, there exists a z ∈ M such that Nm_{H}(z) = x. Now

Nm_{G}(z) = (Nm_{G/H} ∘ Nm_{H})(z) = x.

Thus, we now know that H^{r}(G, M) = 0 for all r ≥ 0.

To proceed further, we use the exact sequence (1.13)

0 → M → M_{∗} → M_{†} → 0.

Because M_{∗} is induced as an H-module (see 1.3), H^{r}(H, M_{∗}) = 0 for all r and all subgroups H of G, and so

H^{r}(H, M) = H^{r−1}(H, M_{†})

for all r and all H. Therefore, M_{†} satisfies the hypotheses of the theorem, and so (by what we have proved) H^{r}(G, M_{†}) = 0 for r ≥ 0. In particular,

0 = H^{0}(G, M_{†}) = H^{−1}(G, M).

The argument, when repeated, gives that H^{−2}(G, M) = 0, etc.. This proves the theorem when G is solvable.

Now consider the case of an arbitrary finite group G. If G and M satisfy the hypotheses of the theorem, so also do G_{p} and M where G_{p} is a Sylow p-subgroup. Therefore H^{r}(G_{p}, M) = 0 for all r and p, and so (see 1.33), the p-primary component of H^{r}(G, M) is zero for all r and p. This implies that H^{r}(G, M) = 0 for all r.

Theorem 2.18. Let G be a finite group and let C be a G-module. Suppose that for all subgroups H of G (including H = G),

(a) H^{1}(H, C) = 0, and

(b) H^{2}(H, C) is a cyclic group of order equal to (H : 1).

Then, for all r, there is an isomorphism

H^{r}(G, ℤ) ⟶ H^{r+2}(G, C)

depending only on the choice of a generator for H^{2}(G, C).

Proof. Choose a generator γ for H^{2}(G, C). Because Cor ∘ Res = (G : H), Res(γ) generates H^{r}(H, C) for any subgroup H of G.

Let φ be a cocycle representing γ. Define C(φ) to be the direct sum of C with the free abelian group having as basis symbols x_{σ}, one for each σ ∈ G, σ ≠ 1, and extend the action of G on C to an action on C(φ) by setting

σx_{τ} = x_{στ} − x_{σ} + φ(σ, τ).

The symbol “x_{1}” is to be interpreted as φ(1, 1). This does define an action of G on C(φ) because

ρσx_{τ} = x_{ρστ} − x_{ρσ} + φ(ρσ, τ),

whereas

ρ(σx_{τ}) = ρ(x_{στ} − x_{σ} + φ(σ, τ))
= x_{ρστ} − x_{ρ} + φ(ρ, στ) − (x_{ρσ} − x_{ρ} + φ(ρ, σ)) + ρφ(σ, τ).

These agree because φ satisfies the cocycle condition

ρφ(σ, τ) + φ(ρ, στ) = φ(ρσ, τ) + φ(ρ, σ).

Note that φ is the coboundary of the 1-cochain σ ↦ x_{σ}, and so γ maps to zero in H^{2}(G, C(φ)). For this reason, C(φ) is called the splitting module for γ.

We shall first show that the hypotheses imply that

H^{1}(H, C(φ)) = 0 = H^{2}(H, C(φ))

for all subgroups H of G.

Recall that we have an exact sequence

0 → I_{G} → ℤ[G] → ℤ → 0

where I_{G} is the free abelian group with basis the elements σ−1, σ ∈ G, σ ≠ 1 Because ℤ[G] is induced, H^{r}(H, ℤ[G]) = 0 for all r, and so

\begin{aligned}H^{1}(H, I_{G}) & ≅ H^{0}(H, ℤ) ≅ ℤ/(H : 1)ℤ, \\ H^{2}(H, I_{G}) & ≅ H^{1}(H, ℤ) = 0 \end{aligned}

Define α to be the additive map C(φ) → ℤ[G] such that

α(c) = 0 for all c ∈ C(φ)
α(x_{σ}) = σ − 1.

Clearly,

0 → C → C(φ) ⟶^{α} I_{G} → 0

is an exact sequence of G-modules. Its cohomology sequence reads

0 → H^{1}(H, C(φ)) → H^{1}(H, I_{G}) → H^{2}(H, C) ⟶^{0} H^{2}(H, C(φ)) → 0

The zeros at the ends use that H^{1}(H, C) = 0 and H^{2}(H, I_{G}) = 0. The map H^{2}(H, C) → H^{2}(H, C(φ)) is zero because H^{2}(H, C) is generated by Res(γ), and this maps to the restriction of the image of γ in H^{2}(G, C(φ)), which is zero. Therefore, H^{1}(H, I_{G}) → H^{2}(H, C) is onto, and hence is an isomorphism (because the two groups have the same order). Its kernel and cokernel, namely, H^{1}(H, C(φ)) and H^{2}(H, C(φ)), are therefore both zero

We deduce from the Theorem 2.17 that H^{r}(H, C(φ)) = 0 for all r. On splicing the two short exact sequences together, we obtain an exact sequence

0 → C → C(φ) → ℤ[G] → ℤ → 0

with the property that H^{r}(G, C(φ)) = 0 = H^{r}(G, ℤ[G]) for all r. Therefore, the double boundary map is an isomorphism (1.13)

H^{r}(G, ℤ) → H^{r+2}(G, C).

Remark 2.19. If M is a G-module such that Tor_{1}^{ℤ}(M, C) = 0, for example, if either M or C is torsion-free as a ℤ-module, then one can tensor the above 4-term sequence with M and obtain isomorphisms

H^{r}(G, M) → H^{r+2}(G, M ⊗ C).

Remark 2.20. The map H^{r}(G, ℤ) → H^{r+2}(G, C) is cup-product with the chosen element γ ∈ H^{2}(G, C).

Example 2.21. Let K be a local field. We shall prove that for any finite Galois extension L of K with Galois group G, H^{2}(G, L^{×}) is cyclic of order [L: K] with a canonical generator u_{L/K}, called the (local) fundamental class. Moreover, in a tower of fields L ⊃ K′ ⊃ K, Res(u_{L/K}) = u_{L/K′}. Since we know H^{1}(G, L^{×}) = 0 (Hilbert’s Theorem 90), Tate’s theorem shows that cup-product with u_{L/K} is an isomorphism

G^{ab} = H^{−2}(G, ℤ) ⟶ H^{0}(G, L^{×}) = K^{×}/ Nm L^{×}.

The reverse map is the local Artin map. The global Artin map can be obtained by a similar argument.

3. The Cohomology of Profinite Groups

Direct limits. A partially ordered set (I, ≤) is said to be directed if for any two elements i and j of I, there exists a k such that i, j ≤ k. Suppose that for every element i of a directed set (I, ≤) we have a set A_{i}, and for every inequality i ≤ j we have a map α_{ji} : A_{i} → A_{j}. If

(a) α_{ii} = id for all i ∈ I, and

(b) α_{kj} ∘ α_{ji} = α_{ki} whenever i ≤ j ≤ k,

then the family (A_{i}, α_{ji}) is called a direct system. On the disjoint union ∐ A_{i} of the A_{i}, introduce the equivalence relation under which a_{i} ∈ A_{i} is related to a_{j} ∈ A_{j} if and only if α_{ki}(a_{i}) = α_{kj}(a_{j}) for some k ≥ i, j. The corresponding quotient set is called the direct limit of the A_{i} (relative to the α_{ji}):

A = \underrightarrow{lim} A_{i}.

There is for each i a canonical map

α_{i} : A_{i} ⟶ A,

possessing the following properties:

(a) α_{i} = α_{j} ∘ α_{ji} for j ≥ i;

(b) α_{i}(a_{i}) = α_{j}(a_{j}) ⟺ α_{ki}(a_{i}) = α_{kj}(a_{j}) for some k ≥ i, j;

(c) A = ⋃ α_{i}(A_{i}).

The system (A, α_{i}) has the following universal property: let T be a set and let (β_{i}), β_{i} : A_{i} → T, be a family of maps such that β_{i} = β_{j} ∘ α_{ji} for i ≤ j; then there exists a unique map β : A → T such that β_{i} = β ∘ α_{i} for all i.

If the A_{i} are abelian groups and the α_{ij} are homomorphisms, then A has a unique structure of an abelian group for which the α_{i} are homomorphisms.

Lemma 3.1. For any direct system of exact sequences

A_{i} ⟶ B_{i} ⟶ C_{i},

the sequence

\underrightarrow{lim} A_{i} ⟶ \underrightarrow{lim} B_{i} ⟶ \underrightarrow{lim} C_{i}

is again exact. Therefore the formation of direct limits commutes with passage to cohomology in complexes.

Proof. Exercise for the reader.

Profinite groups. Let G be a profinite group. This means that G is a compact^{4} topological group for which the open normal subgroups form a fundamental system of neighbourhoods of 1. Note that every open subgroup is of finite index (because its cosets cover G). For example, a finite group with the discrete topology is profinite, and every discrete profinite group is finite. A Galois group G = Gal(L/K) is a profinite group—the open subgroups are exactly those fixing a finite extension of K contained in L—and every profinite group occurs as a Galois group. For a profinite group, we use the topology to modify our notion of cohomology group.

First, we consider only those G-modules for which the map

G × M ⟶ M

is continuous when M is endowed with the discrete topology, i.e., the topology in which every subset is open. Equivalent conditions:

(a) M = ∪M^{H}, H runs through the open subgroups of G;

(b) the stabilizer in G of any element of M is open.

A module satisfying these conditions is called a discrete G-module.

The discrete G-modules form an abelian category with enough injectives, and so we can define cohomology groups H_{cts}^{r}(G, M) by taking injective resolutions, just as before. Moreover, the groups can be calculated using continuous cocycles: let C_{cts}^{r}(G, M) be the group of continuous maps G^{r} → M, and define d^{r}: C_{cts}^{r}(G, M) → C_{cts}^{r+1}(G, M) as before; then

H_{cts}^{r}(G, M) = \frac{Z_{cts}^{r}(G, M)}{B_{cts}^{r}(G, M)}

where Z_{cts}^{r}(G, M) = Ker(d^{r}) and B_{cts}^{r}(G, M) = Im(d^{r−1}).

4Following Bourbaki, I require compact spaces to be Hausdorff.

Let φ : G^{r} → M be a continuous r-cochain. Then φ(G^{r}) is compact, but M is discrete, and so φ(G^{r}) is finite, and hence is contained in M^{H_{0}} for some open normal subgroup H_{0} of G. The inverse image φ^{−1}(m) of each point m of φ(G^{r}) is open, and so contains a translate of H(m)^{r} for some open normal subgroup H(m) of G. Let H_{1} = ∩_{m∈φ(G^{r})}H(m). This is again an open subgroup of G, and φ factors through (G/H_{1})^{r}. Let H = H_{0} ∩ H_{1}. Then φ arises by inflation from an r-cocycle on G/H with values in M^{H}.

Proposition 3.2. The maps Inf : H^{r}(G/H, M^{H}) → H_{cts}^{r}(G, M) realize H_{cts}^{r}(G, M) as the direct limit of the groups H^{r}(G/H, M^{H}) as H runs through the open normal subgroups H of G:

\underrightarrow{lim} H^{r}(G/H, M^{H}) = H_{cts}^{r}(G, M).

Explicitly, the statement means that each element of H_{cts}^{r}(G, M) arises by inflation from some group H^{r}(G/H, M^{H}) and if a ∈ H^{r}(G/H, M^{H}) and a′ ∈ H^{r}(G/H′, M^{H′}) map to the same element in H_{cts}^{r}(G, M), then they map to the same element in H^{r}(G/H′′, M^{H′′}) for some open subgroup H′′ ⊂ H ⋂ H′.

Proof. The above argument shows that

C_{cts}^{r}(G, M) = \underrightarrow{lim} C^{r}(G/H, M^{H}).

The proposition then follows from the fact that passage to the direct limit commutes with the formation of kernels and cokernels, and hence with the formation of cohomology.

Most of the theory concerning the cohomology groups H^{r}(G, M) for r ≥ 0 continues to hold for the groups defined by continuous cochains. For example, if H is a closed subgroup of G, there are maps Inf, Res, and Cor (the last requires H to be of finite index), and there are cup-product maps.

In future, all cohomology groups will be defined using continuous cochains (and the subscript cts will be dropped). In practice, this will mean that either G is an infinite profinite group, and it matters that we take continuous cochains, or G is finite, in which case it doesn’t (and the groups are defined for all r ∈ ℤ).

Proposition 3.3. Let G be a profinite group, and let M be a discrete G-module. If M = \underrightarrow{lim} M_{i} where M_{i} ⊂ M, then H^{r}(G, M) = \underrightarrow{lim} H^{r}(G, M_{i}).

Proof. Because G is compact and M is discrete, the image of any r-cochain f : G^{r} ⟶ M is finite. Since the M_{i} form a directed system of submodules of M (i.e., given M_{i} and M_{j}, there is an M_{k} containing both of them) and M = ⋃ M_{i}, any finite subset of M is contained in an M_{i}. It follows that C^{r}(G, M) = \underrightarrow{lim} C^{r}(G, M_{i}), and so Lemma 3.1 shows that

H^{r}(C^{·}(G, M)) = \underrightarrow{lim} H^{r}(C^{·}(G, M_{i})).

Aside 3.4. (For the experts.) Let G be a profinite group, let M(G) be the category of all G-modules, and C(G) the category of discrete G-modules. Then C(G) is a full subcategory of M(G). Moreover, there is a functor

M ↦ M^{∗} =^{df} ⋃_{H open in G} M^{H}: M(G) ⟶ C(G).

Clearly,

Hom_{G}(M, N^{∗}) = Hom_{G}(M, N)

for M a discrete G-module. The inclusion functor i: C(G) → M(G) is exact, but doesn’t preserve injectives—hence H^{r}(G, M) ≠ H^{r}(G, iM) in general. On the other hand, M ↦ M^{∗} preserves injectives, but is only left exact—hence C(M) has enough injectives. Again H^{r}(G, M) ≠ H^{r}(G, M^{∗}) (however, there is a spectral sequence …).

Notes. In the mid-1930s, Hurewicz showed that the homology groups of an “aspherical space” X depend only on the fundamental group π of the space. Thus one could think of the homology groups H_{r}(X, ℤ) of the space as being the homology groups H_{r}(π, ℤ) of the group π. It was only in the mid-1940s that Hopf, Eckmann, Eilenberg, MacLane, Freudenthal and others gave purely algebraic definitions of the homology and cohomology groups of a group G. It was then found that H^{1} coincided with the group of crossed homomorphisms modulo principal crossed homomorphisms, and H^{2} with the group of equivalence classes of “factor sets”, which had been introduced much earlier (e.g., I. Schur, Über die Darstellung der endlichen… , 1904; O. Schreier, Über die Erweiterungen von Gruppen, 1926; R. Brauer, Über Zusammenhänge…, 1926). For more on the history, see S. MacLane, Origins of the cohomology of groups, Enseign. Math. (2) 24, 1978, 1–29.

Our proof of Tate’s theorem follows Tate’s original proof (Ann. of Math., 56 (1952), 294–297). At that time, there was no published account of the Tate groups, and so Tate proved his theorem only for r ≥ 0, but ended with an enigmatic promise to extend the result to negative r and to recover the Artin map. The construction of the Tate cohomology groups was first published in Cartan and Eilenberg, Princeton Univ. Press, 1956, following Tate’s ideas.

The best source for the above material is Serre 1961, Part 3. There is a somewhat abbreviated version of the same material in Cassels and Fröhlich 1967, Chapter IV. See also Iyanaga 1975, Chapter I, and E. Weiss, Cohomology of Groups, Academic Press, 1969. For the cohomology of profinite groups, see J.-P. Serre, Cohomologie Galoisienne, Springer, 1964, and S. Shatz, Profinite Groups, Arithmetic, and Geometry, Princeton University Press, 1972.

4. Appendix: Some Homological Algebra

Some exact sequences. ^{5}

5Based on F. Lemmermeyer, The Snake Lemma.

Lemma 4.1 (The Extended Snake Lemma). Every commutative diagram of abelian groups

A ⟶^{f} B ⟶^{g} C ⟶ 0
↓^{α}  ↓^{β}  ↓^{γ}
0 ⟶ A′ ⟶^{f′} B′ ⟶^{g′} C′

with exact rows gives rise to an exact sequence

0 → Ker f → Ker α → Ker β → Ker γ → Coker α → Coker β → Coker γ → Coker g′ → 0.

Proof. Except for the first and last terms, this standard. A small diagram chase shows that Ker f ⊂ Ker(α), from which exactness at Ker α follows. The proof of exactness at Coker γ is similarly straightforward.

Lemma 4.2 (Kernel-Cokernel Lemma). Every pair of homomorphisms A ⟶^{f} B ⟶^{g} C of abelian groups gives rise to an exact sequence

0 → Ker f → Ker g ∘ f ⟶^{f} Ker g → Coker f → Coker g ∘ f → Coker g → 0.

Proof. Apply the extended snake lemma to

A ⟶^{f} B ⟶ Coker f ⟶ 0
↓^{g∘f}  ↓^{g}  ↓
0 ⟶ C ⟶^{id} C ⟶ 0

The language of category theory. A category C consists of a nonempty class ob(C) of objects, a set Hom(A, B) for each pair of objects A, B (called the set of morphisms from A to B), and a map

(α, β) ↦ β ∘ α : Hom(A, B) × Hom(B, C) → Hom(A, C)

for each triple of objects A, B, C, satisfying the following conditions:

(a) composition of morphisms is associative;

(b) for each object A, Hom(A, A) has an element id_{A} that is a left and right identity for composition.

It is understood that the sets Hom(A, B) are disjoint, so that a morphism determines its source and target.

A covariant functor F : C → D is a “map” that with each object A of C associates an object F (A) of D and with each morphism α : A → B a morphism F (α) : F (A) → F (B) such that F (α ∘ β) = F (α) ∘ F (β) and F (id_{A}) = F (id_{F (A)}).

A functor F : C → D is left adjoint to the functor G : D → C if

Hom_{D}(F (A), B) ≈ Hom_{C}(A, G(B))

functorially.

If the sets Hom(A, B) are endowed with the structures of abelian groups in such a way that the composition maps are bi-additive, and every finite collection of objects in C has a direct sum, then C (together with the structures) is called an additive category. To say that A and B admit a direct sum means that there is an object A⊕B in C and maps i_{A} : A → A ⊕ B, i_{B} : B → A ⊕ B, p_{A} : A ⊕ B → A, p_{B} : A ⊕ B → B such that:

p_{A} ∘ i_{A} = id_{A} p_{B} ∘ i_{B} = id_{B}
p_{A} ∘ i_{B} = 0 p_{B} ∘ i_{A} = 0
i_{A}p_{A} + i_{B}p_{B} = 1_{A⊕B}.

Let C be an additive category. A sequence

0 → A → B ⟶^{α} C

is exact if the sequence of abelian groups

0 → Hom(T, A) → Hom(T, B) → Hom(T, C)

is exact for all objects T. A sequence

A ⟶^{β} B → C → 0

is exact if the sequence of abelian groups

0 → Hom(C, T) → Hom(B, T) → Hom(A, T)

is exact for all objects T. When the first sequence is exact, A is called the kernel of α, and when the second is exact, C is called the cokernel of β.

Let C be an additive category in which every morphism has both a kernel and a cokernel. Let α : A → B be morphism. The kernel of the cokernel B → C of α is called the image of α, and the cokernel of the kernel of α is called the coimage of α. There is canonical map from the coimage of α to the image of α, and if this is always an isomorphism, then C is called an abelian category.

Functors between additive categories will be assumed to be additive, i.e., such that the maps Hom(A, B) → Hom(F (A), F (B)) are homomorphisms of abelian groups. Such a functor is exact if it maps exact sequences to exact sequences.

For example, for any ring R, the category of R-modules is an abelian category, and, for any topological space X, the category of sheaves of abelian groups on X is an abelian category.

In the remainder of this section, C will be an abelian category. The reader will lose little (so far as this course is concerned) by taking C to be the category of modules over a ring, for example, the category of modules over a group ring ℤ[G].

Injective objects. Let C be an abelian category. An object I of C is injective if Hom(·, I) is an exact functor, i.e., if

0 → A → B → C → 0

exact in C implies that

0 → Hom(C, I) → Hom(B, I) → Hom(A, I) → 0

is exact. This sequence is automatically exact except at Hom(A, I), and so to say that I is injective means that every homomorphism A → I extends to B.

Proposition 4.3. The category of abelian groups has enough injectives.

Lemma 4.4. A module M over a principal ideal domain R is injective if and only if it is divisible, i.e., the map x ↦ rx is surjective for all r ∈ R, r ≠ 0.

Proof. Let m ∈ M, and consider the map x ↦ xm : R → M. The map x ↦ rx : R → R is injective, and any extension of x ↦ xm to R will send 1 to an element m′ such that rm′ = m. Therefore, if M is injective, it is divisible.

Conversely, suppose that M is divisible, and consider a homomorphism α : A → M where A is a submodule of B. On applying Zorn’s lemma to the set of pairs (A′, α′) where A′ is a submodule of B containing A and α′ extends α to A′, we obtain a maximal such pair (A_{1}, α_{1}). If A_{1} ≠ B, there exists a b ∈ B \ A_{1}, and we define I = {r ∈ R | rb ∈ A_{1}}. Because M is divisible, the map r ↦ α_{1}(rb) : I → M extends to R, but this implies that α_{1} extends to A_{1} + Rb, which contradicts the maximality of (A_{1}, α_{1}).

The lemma shows that ℚ/ℤ is an injective ℤ-module. For an arbitrary ℤ-module M, define M^{∨} = Hom(M, ℚ/ℤ). Check that the canonical map M → M^{∨}^{∨} is injective, and that M^{∨}^{∨} is divisible.

Proposition 4.5. Any functor F : C → D that admits an exact left adjoint preserves injectives.

Proof. Let F′ be an exact left adjoint to F. For any injective object I in C, the functor Hom_{D}(·, F (I)) is isomorphic to the functor Hom_{C}(F′(·), I), which is exact because it is the composite of two exact functors, namely, F′ and Hom_{C}(·, I).

Right derived functors. Let C be an abelian category with enough injectives, and let F : C → D be a left exact functor from C to a second abelian category. Thus, a short exact sequence

0 → M′ → M → M′′ → 0

in C gives rise to an exact sequence

0 → F (M′) → F (M) → F (M′′)

in D. The theory of derived functors provides a natural extension of this last sequence to a long exact sequence.

Let M be an object of C. A resolution of M is a long exact sequence

0 → M → I^{0} → I^{1} → ⋯ → I^{r} → · · · .

If the I^{r}’s are injective objects of C, then it is called an injective resolution. We sometimes abbreviate this complex to M → I^{·}.

Lemma 4.6. An injective resolution M → I^{·} of M exists, and if M → J^{·} is a second injective resolution, then there is a homomorphism from M → I^{·} to M → J^{·}, i.e., there is a commutative diagram:

0 ⟶ M ⟶ I^{0} ⟶ I^{1} ⟶ ⋯
∥  ↓  ↓
0 ⟶ M ⟶ J^{0} ⟶ J^{1} ⟶ ⋯

Proof. By assumption, there exists an injective morphism

0 ⟶ M ⟶ I^{0}

with I^{0} injective. Let B^{1} be the cokernel of the map. Then we know that there is an inclusion

0 ⟶ B^{1} ⟶ I^{1}

with I^{1} injective. Now

0 ⟶ M ⟶ I^{0} ⟶ I^{1}

is exact. Let B^{2} = Coker(B^{1} → I^{1}), and continue in this fashion.

A morphism of resolutions can be constructed step by step, using the definition of an injective object.

Let M → I^{·} be an injective resolution of M. On applying the functor F, we obtain a complex

F (I^{·}) : F (I^{0}) → F (I^{1}) → ⋯ → F (I^{r}) ⟶^{d^{r}} F (I^{r+1}) → ⋯

which may no longer be exact. Define

(R^{r}F)(M) = H^{r}(F (I^{·})) =^{df} \frac{Ker(d^{r})}{Im(d^{r−1})}.

Proposition 4.7. Let α : M → N be a morphism of objects of C. For any injective resolutions M → I^{·} and N → J^{·}, there exists a morphism α^{·} : I^{·} → J^{·} making

\begin{matrix}M & ⟶ & I^{·} \\ ↓ & & ↓ \\ N & ⟶ & J^{·} \end{matrix}

commute. The morphism H^{r}(I^{·}) → H^{r}(J^{·}) is independent of the choice of α^{·}.

We discuss the proof below.

The proposition (applied to the identity map M → M) implies that the objects (R^{r}F)(M) are well-defined up to a canonical isomorphism. Moreover, a morphism α : M → N gives rise to a well-defined morphism (R^{r}F)(M) → (R^{r}F)(N), and, in fact, the R^{r}F are functors. They are called the right derived functors of F.

Example 4.8. Because F is left exact,

0 → F (M) → F (I^{0}) ⟶^{d^{0}} F (I^{1})

is exact. Therefore,

(R^{0}F)(M) =^{df} Ker(d^{0}) = F (M).

In other words, R^{0}F = F.

The next two lemmas prove something a little more precise than the proposition.

Lemma 4.9. Let M → I^{·} and N → J^{·} be resolutions of objects M and N of C. If N → J^{·} is an injective resolution, then any morphism α : M → N extends to morphism

\begin{matrix}M & ⟶ & I^{·} \\ ↓ & & ↓ \\ N & ⟶ & J^{·} \end{matrix}

of complexes.

Proof. Bucur and Deleanu 1968, 7.5.

Two morphisms α^{·}, β^{·} : I^{·} → J^{·} of complexes are said to be homotopic if there exists a family of morphisms k^{r} : I^{r} → J^{r−1} such that

α^{r} − β^{r} = d^{r−1} ∘ k^{r} + k^{r+1} ∘ d^{r}

for all r.

Note that, for any x ∈ Z^{r}(I^{·}) =^{df} Ker(d^{r}),

α^{r}(x) − β^{r}(x) = d^{r−1}(k^{r}(x)) ∈ B^{r}(I^{·}).

Therefore α^{r}(x) and β^{r}(x) have the same image in H^{r}(J^{·}), and so homotopic morphisms define the same morphism on cohomology.

Lemma 4.10. Let M → I^{·} be a resolution of M, and let N → J^{·} be an injective resolution N. Any two extensions α^{·} and β^{·} of morphisms M → N to I^{·} → J^{·} are homotopic.

Proof. Ibid..

Proposition 4.11. A short exact sequence

0 → A → B → C → 0

in C gives rise to a long exact sequence

0 → F (A) → F (B) → F (C) → R^{1}F (A) → ⋯
⋯ → R^{r}F (A) → R^{r}F (B) → R^{r}F (C) → ⋯

and the association of the long exact sequence to the short exact sequence is functorial.

The second condition means that a commutative diagram

0 → A → B → C → 0
↓ ↓ ↓
0 → A′ → B′ → C′ → 0

gives rise to a commutative diagram

⋯ → R^{r−1}F (C) → R^{r}F (A) → R^{r}F (B) → R^{r}F (C) → ⋯
↓ ↓ ↓ ↓
⋯ → R^{r−1}F (C′) → R^{r}F (A′) → R^{r}F (B′) → R^{r}F (C′) → ⋯

For the proof of the proposition, see Bucur and Deleanu 1968, 7.6.

Remark 4.12. The right derived functors of F are characterized by the three properties:

(a) R^{0}F = F;

(b) R^{r}F (I) = 0, r > 0, I injective;

(c) the property in (4.11).

Variants. By reversing the directions of some of the arrows, one obtains variants of some of the above definitions, for example, projective objects, left derived functors, etc.

The Ext groups. Let C be an abelian category.

Let A ∈ C. If C has enough injectives, then we can define the right derived functors of the left exact functor Hom(A, ·). Denote the rth right derived functor by Ext^{r}(A, ·). To compute Ext^{r}(A, B), we choose an injective resolution B → I^{·} of B, and set

Ext^{r}(A, B) = H^{r}(Hom(A, I^{·})).

Let B ∈ C. If C has enough projectives, then we can define the right derived functors of the left exact contravariant functor Hom(·, B). Denote the rth right derived functor by Ext^{r}(·, B). To compute Ext^{r}(A, B), we choose a projective resolution P^{·} → A of A, and we set

Ext^{r}(A, B) = H^{r}(Hom(P^{·}, B).

Proposition 4.13. If C has enough injectives and enough projectives, then the two definitions of Ext^{r}(A, B) coincide.

Proof. We define the Ext^{r} using projectives, and prove that they have the properties characterizing the right derived functors of Hom(A, ·).

First, certainly Ext^{0}(A, B) = Hom(A, B).

To say that I is injective means that Hom(·, I) is exact. Therefore Hom(P^{·}, I) is exact, and so

Ext^{r}(A, I) =^{df} H^{r}(Hom(P^{·}, I)) = 0.

Finally, if

0 → B′ → B → B′′ → 0

is exact, then because P^{·} → A is a resolution of A by projectives, the sequence of complexes

0 → Hom(P^{·}, B′) → Hom(P^{·}, B) → Hom(P^{·}, B′′) → 0

is exact. By a standard procedure, we get out of this a long exact sequence

⋯ → H^{r}(Hom(P^{·}, B′)) → H^{r}(Hom(P^{·}, B)) → H^{r}(Hom(P^{·}, B′′) → · · · .

Example 4.14. Let G be a group. Then Mod_{G} has both enough injectives and enough projectives. For any G-module M, Hom_{G}(ℤ, M) = M^{G}, and so the functors Hom(ℤ, ·) and H^{0}(G, ·) agree. Hence, so also do their right derived functors:

Ext_{G}^{r}(ℤ, M) = H^{r}(G, M).

The last proposition allows us to compute these groups by choosing a projective resolution P^{·} → ℤ of ℤ and setting

H^{r}(G, M) = H^{r}(Hom_{G}(P^{·}, M)).

Remark 4.15. It would shorten the exposition in this chapter a little by adopting the last formula as the definition of H^{r}(G, M)—this is the approach taken in Chapter IV of Cassels and Fröhlich, 1967. However, it is not the natural definition.

References. For the general notion of derived functors, see Chapter 7 of Bucur, I., and Deleanu, A., Introduction to Categories and Functors, Wiley, 1968,

or Chapter 4 of Hilton, P.J., and Stammbach, U., A Course in Homological Algebra, Springer, 1971.

CHAPTER III Local Class Field Theory: Cohomology and Completion of the Proofs

We develop enough of the cohomological approach to local class field theory to complete the proofs of the main theorems. Throughout this chapter, “local field” will mean “nonarchimedean local field”. As before, K^{al} denotes an algebraic closure of K (or separable algebraic closure in the case that K has characteristic p ≠ 0), and “extension of K” means “subfield of K^{al} containing K”. All cohomology groups will computed using continuous cochains (see II.3).

1. Introduction

Recall that, after Chapter I, to complete the proof of the main theorems of class field theory, it remains to show that, for any local field K, there exists a homomorphism (local Artin map)

ϕ_{K} : K^{×} → Gal(K^{al}/K)

with the following properties:

(a) for any prime element π of K, ϕ_{K}(π)|K^{un} = Frob_{K};

(b) for any finite abelian extension L of K, Nm_{L/K}(L^{×}) is contained in the kernel of a ↦ ϕ_{K}(a)|L, and ϕ_{K} induces an isomorphism

ϕ_{L/K} : K^{×}/ Nm_{L/K}(L^{×}) → Gal(L/K).

For a Galois extension of fields L/K (possibly infinite), set

H^{2}(L/K) = H^{2}(Gal(L/K), L^{×}).

In the next chapter we shall see that H^{2}(L/K) has an explicit interpretation as the relative Brauer group of L/K. Because of Hilbert’s Theorem 90, there is an exact sequence

0 → H^{2}(L/K) ⟶^{Inf} H^{2}(E/K) ⟶^{Res} H^{2}(E/L)

for any tower of Galois extensions E ⊃ L ⊃ K (II.1.36).

We shall prove:

Theorem 1.1. For any local field K, there exists a canonical isomorphism

inv_{K} : H^{2}(K^{al}/K) → ℚ/ℤ.

Moreover, if [L : K] = n < ∞, then the diagram

0 ⟶ H^{2}(L/K) ⟶ H^{2}(K^{al}/K) ⟶^{Res} H^{2}(K^{al}/L)
↓^{inv}_{K}  ↓^{inv}_{L}
0 ⟶ \frac{1}{n}ℤ/ℤ ⟶  ℚ/ℤ  ⟶^{n} ℚ/ℤ

commutes, and therefore defines an isomorphism

inv_{L/K} : H^{2}(L/K) → \frac{1}{n}ℤ/ℤ.

The proof of Theorem 1.1 will occupy the next two sections. In the remainder of this section, we shall explain how it (together with Proposition 1.6 below) implies the existence of a local Artin map with the required properties.

Thus, assume Theorem 1.1. The element u_{L/K} of H^{2}(L/K) corresponding to \frac{1}{n} mod ℤ under the isomorphism inv_{L/K} is called the fundamental class of the extension L/K.

Lemma 1.2. Let E ⊃ L ⊃ K be a tower of finite Galois extensions. Then

\begin{aligned}Res(u_{E/K}) & = & u_{E/L}, \\ Inf(u_{L/K}) & = & [E : L]u_{E/K}. \end{aligned}

Proof. Consider

H^{2}(K^{al}/K) ⟶^{Res} H^{2}(K^{al}/L) ⟶^{Res} H^{2}(K^{al}/E)
↓^{inv}_{K}  ↓^{inv}_{L}  ↓^{inv}_{E}
ℚ/ℤ ⟶^{[L:K]} ℚ/ℤ ⟶^{[E:L]} ℚ/ℤ.

The vertical maps are isomorphisms. On applying the kernel-cokernel lemma to the rows, we obtain a commutative diagram:

0 ⟶ H^{2}(L/K) ⟶^{Inf} H^{2}(E/K) ⟶^{Res} H^{2}(E/L)
↓^{inv}^{L/K}  ↓^{inv}^{E/K}  ↓^{inv}^{E/L}
0 ⟶ \frac{1}{[L:K]}ℤ/ℤ ⟶^{id} \frac{1}{[E:K]}ℤ/ℤ ⟶^{[L:K]} \frac{1}{[E:L]}ℤ/ℤ.

The commutativity of the two squares implies the two equalities.

Now let L/K be a finite Galois extension of local fields with Galois group G. For any subgroup H of G,

(a) H^{1}(H, L^{×}) = 0 (Hilbert’s theorem 90);

(b) H^{2}(H, L^{×}) is cyclic of order (H : 1), with a canonical generator u_{L/L^{H}} (= Res(u_{L/K})).

The pair (G, L^{×}) thus satisfies the hypotheses of Tate’s theorem (II.2.18).

Corollary 1.3. Let L/K be a finite Galois extension of local fields with Galois group G. For all r, there exists a canonical isomorphism

H_{T}^{r} (G, ℤ) → H_{T}^{r+2}(G, L^{×}),

namely, cup product with u_{L/K}. When r = 2, this becomes an isomorphism

G^{ab} ≅ K^{×}/ Nm_{L/K}(L^{×}).

Corollary 1.4 (Norm Limitation Theorem). Let E be a finite Galois extension of K, and let L be the largest abelian extension of K contained in E; then Nm_{E/K}(E^{×}) = Nm_{L/K}(L^{×}).

Proof. Because Nm_{E/K} = Nm_{E/L} ∘ Nm_{L/K}, certainly Nm_{E/K}(E^{×}) ⊂ Nm_{L/K}(L^{×}). However, Gal(L/K) = Gal(E/K)^{ab}, and so the preceding corollary shows that the norm groups have the same index in K^{×}. This implies that they are equal.

Local class field classifies the abelian extensions of a local field by means of the norm groups of the fields. The corollary shows that this approach fails to classify the nonabelian extensions.

For a finite abelian extension L/K, we define the local Artin map

φ_{L/K} : K^{×}/ Nm(L^{×}) → Gal(L/K)

to be the inverse of the isomorphism in Corollary 1.3.

Proposition 1.5. If E ⊃ L ⊃ K is a tower of finite abelian extensions of K, then φ_{E/K}(a)|L = φ_{L/K}(a) for all a ∈ K.

Proof. This can be checked directly from the definition of the local Artin maps, using that Inf(u_{L/K}) = [E : L]u_{E/K}.

Define φ_{K} : K^{×} → Gal(K^{ab}/K) to be the homomorphism such that, for every finite abelian extension L/K, φ_{K}(a)|L = φ_{L/K}(a) all a ∈ K^{×}—Proposition 1.5 shows that this definition makes sense. By its very definition, φ_{K} satisfies condition (b) to be the local Artin map, and the next proposition implies that it also satisfies condition (a).

Proposition 1.6. When L/K is unramified, φ_{L/K} maps every prime element of K to Frob_{L/K}.

This will be proved in the next section.

Thus, to complete the proofs of the main theorems of local class field theory, it remains to prove Theorem 1.1 and Proposition 1.6.

Remark 1.7. To a finite Galois extension L/K of local fields, we have attached the groups K^{×}/ Nm(L^{×}) and H^{2}(L/K). When L/K is cyclic, they are canonically (up to the choice of generator for G = Gal(L/K)) isomorphic, but not otherwise. The first group is always isomorphic to G^{ab}, and the second is always cyclic of order [L : K]. Thus, when G is abelian but not cyclic, the two groups have the same order but are not isomorphic, and when G is nonabelian, they have different orders.

2. The Cohomology of Unramified Extensions

The cohomology of the units.

Proposition 2.1. Let L/K be a finite unramified extension with Galois group G, and let U_{L} be the group of units in L. Then

H_{T}^{r} (G, U_{L}) = 0, all r.

Let π be a prime element of K. Then π is also a prime element in L, and

L^{×} ≅ U_{L} × π^{ℤ}.

Therefore, H^{r}(G, U_{L}) is a direct summand of H^{r}(G, L^{×}) (see II.1.25). Since H^{1}(G, L^{×}) = 0 (by Hilbert’s theorem 90), this shows that H^{1}(G, U_{L}) = 0. Because G is cyclic, to complete the proof of the theorem, it suffices (by II.2.11) to show that H_{T}^{0}(G, U_{L}) = 0. This is accomplished by the next proposition.

Proposition 2.2. For any finite unramified extension L/K, the norm map Nm_{L/K} : U_{L} → U_{K} is surjective.

Let ℓ and k be the residue fields of L and K. The action of G on O_{L} identifies G with Gal(ℓ/k).

Lemma 2.3. For m > 0, let U_{L}^{(m)} = 1 + 𝔪_{L}^{m}. Then

U_{L}/U_{L}^{(1)} ⟶^{≈} ℓ^{×}
U_{L}/U_{L}^{(m)} ⟶^{≈} ℓ

as G-modules.

Proof. Let π be a prime element of K. It remains prime in L, and

U_{L}^{(m)} = {1 + aπ^{m} | a ∈ O_{L}}.

The maps

u ↦ u mod 𝔪_{L} : U_{L} → ℓ^{×}
1 + aπ^{m} ↦ a mod 𝔪_{L} : U_{L}^{(m)} → ℓ

induce the required isomorphisms.

Lemma 2.4. For all r, H_{T}^{r} (G, ℓ^{×}) = 0. In particular, the norm map ℓ^{×} → k^{×} is surjective.

Proof. By Hilbert’s Theorem 90, H^{1}(G, ℓ^{×}) = 0, and because ℓ^{×} is finite, its Herbrand quotient h(ℓ^{×}) = 1 (see II.2.15)). Therefore H^{2}(G, ℓ^{×}) = 0, and this implies that all the groups are zero (see II.2.11).

Lemma 2.5. The group H_{T}^{r} (G, ℓ) = 0 for all r. In particular, the trace map ℓ → k is surjective.

Proof. See Proposition II.1.23.

We now prove Proposition 2.2. Consider u ∈ U_{K}. Because the norm map ℓ^{×} → k^{×} is surjective, there exists a v_{0} ∈ U_{L} such that Nm(v_{0}) ≡ u mod U_{K}^{(1)}. Because the trace map ℓ → k is surjective, so also is the norm map U_{L}^{(1)}/U_{L}^{(2)} → U_{K}^{(1)}/U_{K}^{(2)}, and so there exists a v_{1} ∈ U_{L}^{(1)} such that Nm(v_{1}) ≡ u/ Nm(v_{0}) mod U_{K}^{(2)}. Continuing in this fashion, we obtain a sequence of elements v_{0}, v_{1}, v_{2}, v_{3}, …, v_{i} ∈ U_{K}^{(i)}, such that u/ Nm(v_{0} ⋯ v_{i}) ∈ U_{K}^{(i+1)}. Let v = lim_{m→∞} ∏_{j=1}^{m} v_{j}. Then u/ Nm(v) ∈ ∩U_{K}^{(i)} = {1}.

Remark 2.6. Let L be an infinite unramified extension of K. Then L is Galois over K, and, for r ≥ 0,

H^{r}(Gal(L/K), U_{L}) = \underrightarrow{lim}_{K′}H^{r}(Gal(K′/K), U_{K′})

where the limit is over the finite extensions K′ of K contained in L. Therefore

H^{r}(Gal(L/K), U_{L}) = 0

for all r > 0 (continuous cohomology).

The invariant map. Let L be an unramified extension of K (possibly infinite), and let G = Gal(L/K).

From the cohomology sequence of G-modules

0 ⟶ U_{L} ⟶ L^{×} ⟶^{ord_{L}} ℤ ⟶ 0,

we obtain (using 2.6) an isomorphism

H^{1}(G, L^{×}) ⟶ H^{1}(G, ℤ).

From the cohomology sequence

0 ⟶ ℤ ⟶ ℚ ⟶ ℚ/ℤ ⟶ 0

(trivial G-actions) we obtain an isomorphism

H^{1}(G, ℚ/ℤ) ⟶^{δ} H^{2}(G, ℤ).

Recall that

H^{1}(G, ℚ/ℤ) = Hom_{cts}(G, ℚ/ℤ).

If [L : K] = n < ∞, then G is a cyclic group of order n with generator Frob_{L/K}, and the map

f ↦ f(Frob_{L/K}) : Hom(G, ℚ/ℤ) → ℚ/ℤ

is an isomorphism from Hom(G, ℚ/ℤ) onto the unique cyclic subgroup \frac{1}{n}ℤ/ℤ of ℚ/ℤ of order n; if [L : K] = ∞, then G is generated in the topological sense by Frob_{L/K}, i.e., G is the closure of the group {Frob_{L/K}^{i} | i ∈ ℤ}, and f ↦ f(Frob_{L/K}) is an isomorphism of Hom_{cts}(G, ℚ/ℤ) onto an infinite subgroup of ℚ/ℤ (in fact, the whole of ℚ/ℤ if L = K^{un}, because the image contains \frac{1}{n}ℤ/ℤ for all n).

The composite of

H^{2}(L/K) ⟶^{≈} H^{2}(G, ℤ) ⟵_{≈}^{δ} H^{1}(G, ℚ/ℤ) = Hom(G, ℚ/ℤ) ⟶^{f↦f(σ)} ℚ/ℤ

is called the invariant map

inv_{L/K} : H^{2}(L/K) → ℚ/ℤ.

Now consider a tower of field extensions

E ⊃ L ⊃ K

with both E and L unramified (hence Galois) over K. Then

H^{2}(L/K) ⟶^{inv_{L/K}} ℚ/ℤ
↓ Inf  ∥
H^{2}(E/K) ⟶^{inv_{E/K}} ℚ/ℤ

commutes, because all the maps in the definition of inv are compatible with Inf. In particular, there is a canonical isomorphism

inv_{K} : H^{2}(K^{un}/K) ⟶ ℚ/ℤ,

with the property that, for any L ⊂ K^{un} of finite degree n over K, inv_{K} induces an isomorphism

inv_{L/K} : H^{2}(L/K) ⟶ \frac{1}{[L:K]}ℤ/ℤ.

We next need to prove that the diagram in Theorem 1.1 commutes with K^{al} replaced by K^{un}. In fact, we shall prove a more general result.

Proposition 2.7. Let L be a finite extension of K of degree n, and let K^{un} and L^{un} be the largest unramified extensions of K and L. Then the following diagram commutes:

H^{2}(K^{un}/K) ⟶^{Res} H^{2}(L^{un}/L)
↓^{inv}_{K}  ↓^{inv}_{L}
ℚ/ℤ  ⟶^{n} ℚ/ℤ.

Proof. To obtain the largest unramified extension of a local field, we only have to adjoin the mth roots of 1 for m not divisible by the residue characteristic. Therefore, L^{un} = L · K^{un}, and so

τ ↦ τ|K^{un} : Gal(L^{un}/L) ⟶ Gal(K^{un}/K)

is injective. The map denoted Res in the above diagram is that defined by the compatible homomorphisms:

Gal(K^{un}/K) ⟵ Gal(L^{un}/L)
K^{un×}  ⟶  L^{un×}.

Let Γ_{K} = Gal(K^{un}/K) and Γ_{L} = Gal(L^{un}/L). Consider the diagram:

H^{2}(K^{un}/K) ⟶^{≈} H^{2}(Γ_{K}, ℤ) ⟵_{≈}^{δ} H^{1}(Γ_{K}, ℚ/ℤ) ⟶ ℚ/ℤ
↓^{Res}  ↓^{e Res}  ↓^{e Res}  ↓^{fe}
H^{2}(L^{un}/L) ⟶^{≈} H^{2}(Γ_{L}, ℤ) ⟵_{≈}^{δ} H^{1}(Γ_{L}, ℚ/ℤ) ⟶ ℚ/ℤ.

Here e and f are the ramification index and residue class degree for L/K. The left hand square is obtained from the commutative square

K^{un×} ⟶^{ord_{K}} ℤ
↓  ↓^{e}
L^{un×} ⟶^{ord_{L}} ℤ.

The second square expresses the fact that the restriction map commutes with the boundary map. Apart from the factor “e”, the third square is

Hom(Γ_{K}, ℚ/ℤ) ⟶^{φ↦φ(Frob_{K})} ℚ/ℤ
↓^{φ↦φ|Γ_{L}}  ↓^{f}
Hom(Γ_{L}, ℚ/ℤ) ⟶^{φ↦φ(Frob_{L})} ℚ/ℤ.

The Frobenius elements are determined by the fact that they induce x ↦ x^{q} and x ↦ x^{q^{f}} on the residue fields, where q = #k and q^{f} = #ℓ. Hence Frob_{L} |K = Frob_{K}^{f}. It is now clear that the square commutes, and since n = ef, this proves the proposition.

Computation of the local Artin map. Let L be a finite unramified extension of K with Galois group G, and let n = [L : K]. From the last subsection, we know that (G, L^{×}) satisfies the hypotheses of Tate’s theorem (II.2.18), and so we have a canonical isomorphism

H^{−2}(G, ℤ) → H^{0}(G, L^{×})
∥  ∥
G  K^{×}/ Nm(L^{×}).

We now compute this map explicitly.

If π is a prime element of L, then every element of L^{×} can be written uniquely in the form α = uπ^{m}, u ∈ U, m ∈ ℤ; thus

L^{×} = U × π^{ℤ} ≅ U × ℤ.

Since L is unramified over K, we can choose π ∈ K. Then τα = τ(uπ^{m}) = (τu)π^{m} for τ ∈ Gal(L/K), and so the above decomposition is a decomposition of G-modules (G acting trivially on π^{ℤ} ≈ ℤ).

Lemma 2.8. For any group G and G-modules M, N,

H^{r}(G, M ⊕ N) = H^{r}(G, M) ⊕ H^{r}(G, N).

Proof. This was proved in Proposition II.1.25. Alternatively, to say that a module P is a direct sum of modules M and N means that certain maps, and relations between the maps, exist (see II.4). These maps and relations persist when we apply the additive functor H^{r}(G, ·).

Thus

H^{r}(G, L^{×}) = H^{r}(G, U_{L}) ⊕ H^{r}(G, π^{ℤ}).

Choose a generator σ of G (e.g., the Frobenius generator), and let

f ∈ H^{1}(G, ℚ/ℤ) = Hom(G, ℚ/ℤ)

be the element such that f(σ^{i}) = \frac{i}{n} mod ℤ for all i. It generates H^{1}(G, ℚ/ℤ). From the exact sequence

0 → ℤ → ℚ → ℚ/ℤ → 0

and the fact that H^{r}(G, ℚ) = 0 for all r, we obtain an isomorphism

δ : H^{1}(G, ℚ/ℤ) → H^{2}(G, ℤ).

According to the description of δ in (II.1.20, to construct δf, we first choose a lifting of f to 1-cochain f̃: G → ℚ. We take f̃ to be the map σ^{i} ↦ \frac{i}{m} where 0 ≤ i < m − 1. Then

df̃(σ^{i}, σ^{j}) = σ^{i}f̃(σ^{j}) − f̃(σ^{i+j}) + f̃(σ^{i}) = \{ \begin{matrix}0 & if & i + j ≤ n − 1 \\ 1 & if & i + j > n − 1. \end{matrix}

When we identify ℤ with the subgroup π^{ℤ} of L^{×}, we find that the fundamental class u_{L/K} ∈ H^{2}(G, L^{×}) is represented by the cocycle:

φ(σ^{i}, σ^{j}) = \{ \begin{matrix}1 & if & i + j ≤ n − 1 \\ π & if & i + j > n − 1 \end{matrix}

From the exact sequences

0 → I → ℤ[G] → ℤ → 0
0 → L^{×} → L^{×}(φ) → I → 0

(see the proof of II.2.18) we obtain boundary maps

H^{−2}(G, ℤ) → H^{−1}(G, I)
H^{−1}(G, I) → H^{0}(G, L^{×}),

which are isomorphisms because ℤ[G] and L^{×}(φ) have trivial cohomology. Here L^{×}(φ) is the splitting module L^{×} ⊕ ⨁_{σ∈G,σ≠1} ℤx_{σ} of φ.

Finally, H^{−2}(G, ℤ) =^{df} H_{1}(G, ℤ) ≅ G (II.2.7).

Proposition 2.9. Under the composite

\begin{matrix}H^{−2}(G, ℤ) & → & H^{0}(G, L^{×}) \\ ∥ & & ∥ \\ G & & K^{×}/ Nm_{G} L^{×} \end{matrix} ,

of the above maps, σ maps to π mod Nm(L^{×}).

Note that, because H^{0}(G, U_{L}) = 0, U_{L} ⊂ Nm_{L/K}(L^{×}), and the class of π mod Nm(L^{×}) is independent of the prime element π. On the other hand, L^{×}(φ) and the map depend on the choice of the generator σ for G.

Proof. From the construction of the isomorphism H^{−2}(G, ℤ) ≅ G, we see that the image of σ under the boundary map H^{−2}(G, ℤ) → H^{−1}(G, I_{G}) ⊂ I_{G}/I_{G}^{2} is represented by σ − 1.

The boundary map H^{−1}(G, I_{G}) → H^{0}(G, L^{×}) is that given by the snake lemma from the diagram (we write I for I_{G}):

H^{−1}(G, I)
↓
(L^{×})_{G} ⟶ L^{×}(φ)_{G} ⟶ (I)_{G} ⟶ 0
↓  ↓  ↓
0 ⟶ L^{×G}  ⟶ L^{×}(φ)^{G} ⟶  I^{G}
↓
H^{0}(G, L^{×}).

The vertical maps connecting the rows are Nm_{G} = ∑_{i=0}^{n−1} σ^{i}. The element (σ − 1) + I^{2} is the image of x_{σ} + I · L^{×}(φ) in L^{×}(φ)_{G}, and Nm_{G}(x_{σ} + I · L^{×}(φ)) is the sum of the elements:

\begin{aligned}x_{σ} & = & x_{σ} \\ σx_{σ} & = & x_{σ^{2}} − x_{σ} + φ(σ, σ) \\ σ^{2}x_{σ} & = & x_{σ^{3}} − x_{σ^{2}} + φ(σ, σ^{2}) \\ ⋯ & & ⋯ \\ σ^{n−1}x_{σ} & = & ‘x_{1}’ − x_{σ^{n−1}} + φ(σ, σ^{n−1}) \end{aligned} .

On summing these, remembering that ‘x_{σ}’= φ(1, 1) = 1 and that + on the factor L^{×} of L(φ) is actually ·, we find that

Nm_{G}(x_{σ}) = ∏_{i=1}^{n−1} φ(σ, σ^{i}) = π.

This completes the proof.

Remark 2.10. The above proof of Proposition 2.9, using Tate’s original definition of the isomorphism H^{r}(G, ℤ) → H^{r+2}(G, C), is simpler than that found in other references, which uses the description of the map in terms of cup products.

3. The Cohomology of Ramified Extensions

Lemma 3.1. If L/K is Galois of finite degree n, then H^{2}(L/K) has a subgroup of order n.

Proof. Consider the diagram

0 ⟶ Ker(Res) ⟶ H^{2}(K^{un}/K) ⟶^{Res} H^{2}(L^{un}/L)
↓  ↓^{Res}  ↓^{Res}
0 ⟶ H^{2}(L/K) ⟶ H^{2}(K^{al}/K) ⟶^{Res} H^{2}(K^{al}/L)

Since the two restriction maps are injective, so also is the first vertical map, but (2.7) shows that the kernel of the restriction map on the top row is \frac{1}{n}ℤ/ℤ.

To complete the proof of Theorem 1.1, it suffices to prove that the map \frac{1}{n}ℤ/ℤ ↪ H^{2}(L/K) is an isomorphism (see the last paragraph of this section). There are two different approaches to proving this. In the next chapter on Brauer groups, we shall show that H^{2}(K^{al}/K^{un}) = 0; this implies that the restriction maps in the diagram are isomorphisms, and hence also that the first vertical map is an isomorphism. The second proof, which we now present, shows that #H^{2}(L/K) ≤ n.

Lemma 3.2. Let L be a finite Galois extension of K with Galois group G. Then there exists an open subgroup V of O_{L} stable under G such that H^{r}(G, V) = 0 all r > 0.

Proof. Let {x_{τ} | τ ∈ G} be a normal basis for L over K (see II.1.24). The x_{τ} have a common denominator d in O_{K} (see Math 676, 2.4). After replacing each x_{τ} with d · x_{τ}, we may suppose that they lie in O_{L}. Take V = ∑ O_{L}x_{τ}. Then

V ≅ O_{L}[G] = Ind^{G} O_{L},

and so H^{r}(G, V) = 0 for all r > 0.

Lemma 3.3. Let L, K, and G, be as in the last lemma. Then there exists an open subgroup V of U_{L} stable under G such that H^{r}(G, V) = 0 for all r > 0.

Proof. (Assume charK = 0.) The power series

e^{x} = 1 + x + ⋯ + x^{n}/n! + ⋯

converges for ord(x) > ord(p)/(p − 1). It defines an isomorphism of an open neighbourhood of 0 in L onto an open neighbourhood of 1 in L^{×}, with inverse mapping log(x) = (x−1)−(x−1)^{2}/2+(x−1)^{3}/3−· · ·. Clearly both maps commute with the actions of G. If V′ is an open neighbourhood of 0 as in (3.2), then π^{M}V′ will have the same properties, and we can take V = exp(π^{M}V′) with M chosen to be sufficiently large that the exponential function is defined on π^{M}V′.

Lemma 3.4. Let L/K be a cyclic extension of degree n; then h(U_{L}) = 1 and h(L^{×}) = n.

Proof. Let V be an open subgroup of U_{L} with H^{r}(G, V) = 0 for all r. Because U_{L} is compact, the group U_{L}/V is finite, and so h(U_{L}) = h(V) = 1 (see II.2.16). Finally h(L^{×}) = h(U) · h(ℤ) = h(ℤ), and

h(ℤ) = #H_{T}^{0}(G, ℤ)/#H^{1}(G, ℤ) = #H_{T}^{0}(G, ℤ) = #(ℤ/nℤ) = n.

Lemma 3.5. Let L be a finite Galois extension of K of order n, then H^{2}(L/K) has order n.

Proof. We know that the order of H^{2}(L/K) is divisible by n, and that it equals n when L/K is cyclic. We prove the lemma by induction on [L: K]. Because the group Gal(L/K) is solvable (see Math 676, 7.50), there will be a Galois extension K′/K with L ⫌ K′ ⫌ K. From the exact sequence

0 ⟶ H^{2}(K′/K) ⟶ H^{2}(L/K) ⟶ H^{2}(L/K′)

we find that

#H^{2}(L/K) ≤ #H^{2}(K′/K)#H^{2}(L/K′) = n.

We now complete the proof of Theorem 1.1. From the diagram in the proof of (3.1) we see that, for any finite Galois extension L of K, the subgroup H^{2}(L/K) of H^{2}(K^{al}/K) is contained in H^{2}(K^{un}/K). Since H^{2}(K^{al}/K) = ⋃ H^{2}(L/K), this proves that the inflation map H^{2}(K^{un}/K) → H^{2}(K^{al}/K) is an isomorphism. Therefore, the invariant map inv_{K} : H^{2}(K^{un}/K) → ℚ/ℤ defines an isomorphism H^{2}(K^{al}/K) → ℚ/ℤ. It follows from (3.1) that this homomorphism has the property required for Theorem 1.1. Moreover, Proposition 2.9 (with σ taken to be the Frobenius generator) shows that the homomorphism has the properties required for Proposition 1.6. This completes the proofs of the main theorems of local class field theory.

4. Complements

Alternative description of the local Artin map. Let L/K be a finite abelian extension with Galois group G, and let u_{L/K} ∈ H^{2}(G, L^{×}) be the fundamental class. The local Artin map ϕ_{L/K} is the inverse to the isomorphism

x ↦ x ∪ u_{L/K} : H_{T}^{−2}(G, ℤ) → H_{T}^{0}(G, L^{×}).

This definition is very difficult to work with because cup-products involving both homology and cohomology groups have no very convenient description. Instead, we re-interprete the map purely in terms of cohomology groups. Consider the cupproduct pairing

H^{0}(G, L^{×})×H^{2}(G, ℤ) ⟶ H^{2}(G, L^{×}) ⟶^{inv_{L/K}} ℚ/ℤ.

Given an element a ∈ H^{0}(G, L^{×}) = K^{×} and a class c ∈ H^{2}(G, ℤ) represented by a cocycle f: G × G → ℤ, the cup-product class a ∪ c is represented by the cocycle (σ, τ) ↦ a^{f(σ,τ)}. Recall also that we have an isomorphism

Hom(G, ℚ/ℤ) = H^{1}(G, ℚ/ℤ) ⟶^{δ} H^{2}(G, ℤ).

Proposition 4.1. For any χ ∈ Hom(G, ℚ/ℤ) and a ∈ K^{×},

χ(ϕ_{L/K}(a)) = inv_{K}(a ∪ δχ).

Proof. Omitted. (See Serre, Local Fields, “Annexe” to Chapter XI, and his article in Cassels-Fröhlich, p140.)

Using this, we can get another proof of Proposition 1.6.

Lemma 4.2. If L/K is unramified, ϕ_{L/K} sends a ∈ K^{×} to Frob^{ord_{K}a}.

Proof. Recall that inv_{K} is defined to be the composite

H^{2}(G, L^{×}) ⟶^{ord} H^{2}(G, ℤ) ⟵^{δ} Hom(G, ℚ/ℤ) ⟶^{χ↦χ(σ)} ℚ/ℤ.

Because of the functoriality of cup-products

ord(a ∪ δχ) = ord(a) ∪ δχ, a ∈ K^{×}, χ ∈ Hom(G, ℚ/ℤ)

where, on the left ord denotes the map on H^{2} induced by ord_{L}: L^{×} ↠ ℤ, and on the right it is the map itself. Let a ∈ H^{0}(G, L^{×}) = K^{×}, and let m = ord_{L}(a). For any χ ∈ Hom(G, ℚ/ℤ), in the above diagram,

a ∪ δχ ↦ ord(a) ∪ δχ ↦ mχ ↦ χ(σ^{m}), σ = Frob,

i.e., inv_{K}(a ∪ δχ) = χ(σ^{m}). On combining this with the formula in (4.1) we find that

χ(ϕ(α)) = χ(σ^{ord(α)})

for all χ ∈ Hom(G, ℚ/ℤ), and so ϕ(α) = σ^{ord(α)}.

For any character χ of G, we get a character a ↦ inv_{K}(a∪δχ) of K^{×}. By duality we get a map K^{×} → G. Proposition 4.1 shows that this map is ϕ_{L/K} : K^{×} → Gal(L/K).

The Hilbert symbol. Let a, b ∈ ℚ_{p}^{×} with p a prime number or infinity (we let ℚ_{∞} = ℝ). The Hilbert symbol of a and b relative to ℚ_{p} is defined by

(a, b)_{p} = \{ \begin{matrix}1 & if z^{2} = ax^{2} + by^{2} has a nonzero solution in ℚ_{p} \\ −1 & otherwise. \end{matrix}

Then a, b ↦ (a, b) is a nondegenerate bilinear pairing

ℚ_{p}^{×}/ℚ_{p}^{×2} × ℚ_{p}^{×}/ℚ_{p}^{×2} → {±1}.

It has many interesting properties, the most profound being the product formula (proved by Hilbert). For fixed a, b ∈ ℚ, (a, b)_{p} = 1 for all but finitely many primes p, and

∏_{p≤∞}(a, b)_{p} = 1.

This proof of the product formula uses the quadratic reciprocity law. The product formula has an interpretation in terms of Brauer groups (see the next chapter), and has implications for the theory of quadratic forms over ℚ (see Chapter VIII).

We can define Hilbert symbols for any local field K. Let μ_{n} be the group of nth roots of 1 in K^{al} (if the characteristic of K is p ≠ 0, then n should not be divisible by p). Let G = Gal(K^{al}/K). There is an exact sequence of G-modules,

1 → μ_{n} → K^{al×} ⟶^{n} K^{al×} → 1,

and hence an exact cohomology sequence

K^{×} ⟶^{n} K^{×} → H^{1}(G, μ_{n}) → 0 → 0 → H^{2}(G, μ_{n}) → H^{2}(K^{al}/K) → H^{2}(K^{al}/K).

Thus:

H^{1}(G, μ_{n}) ≅ K^{×}/K^{×n}
H^{2}(G, μ_{n}) ≅ \frac{1}{n}ℤ/ℤ ≅ ℤ/nℤ.

There is a cup-product pairing

H^{1}(G, μ_{n}) × H^{1}(G, μ_{n}) → H^{2}(G, μ_{n} ⊗ μ_{n}).

In the case that K contains an nth root ζ of 1, the cup-product pairing

H^{2}(G, μ_{n}) × H^{0}(G, μ_{n}) → H^{2}(G, μ_{n} ⊗ μ_{n})

defines an isomorphism

H^{2}(G, μ_{n}) ⊗ μ_{n} → H^{2}(G, μ_{n} ⊗ μ_{n}),

and hence an isomorphism

H^{2}(G, μ_{n} ⊗ μ_{n}) → H^{2}(G, μ_{n}) ⊗ μ_{n} → (ℤ/nℤ) ⊗ μ_{n} = μ_{n}.

Therefore, in this case, the pairing becomes

a, b ↦ (a, b) : K^{×}/K^{×n} × K^{×}/K^{×n} → μ_{n}.

This pairing is again called the Hilbert symbol. Class field theory shows that it has most of the properties of the first pairing defined above.

Proposition 4.3. The Hilbert symbol is related to the local Artin map by the formula

ϕ_{K}(b)(a^{\frac{1}{n}}) = (a, b)a^{\frac{1}{n}} .

Note that Galois theory tells us that, for any τ ∈ Gal(K[a^{\frac{1}{n}}]/K), τa^{\frac{1}{n}} = ζa^{\frac{1}{n}} for some nth root of one ζ (remember, we are assuming that K contains the nth roots of 1), and so the point of the formula is that roots of 1 are the same. The proof of the formula is an exercise in cup-products, starting from Proposition 4.1.

Other Topics. At this point, it would not be difficult to give a proof of the Local Existence Theorem (I.1.2) that is independent of Chapter II—see Serre 1962, 11.5, 14.6.

The reader who is interested in understanding the (conjectural) statement of nonabelian local class field theory should look first at Chapter VII of C.J. Moreno, Advanced Analytic Number Theory, Part I: Ramification Theoretic Methods, AMS, 1983.

Notes. It follows from Krasner’s lemma (see Math 676, 7.51) that every finite abelian extension of local fields arises by completing a finite abelian extension of global fields. In the 1930’s Hasse was able to deduce the main theorems of local class field theory from those of global class field theory.

From the modern perspective, this seems a strange way to do things. In the 1940’s, in his algebraic approach to class field theory Chevalley developed local class field theory directly, and used it in the construction of global class field theory. F. K. Schmidt also showed that local class field theory can be constructed independently of global class field theory.

At that time, there was no good description of the local Artin map, and nor was there an explicit way of constructing the maximal abelian extension of a local field (except for ℚ_{p} of course).

In 1958 Dwork gave an explicit description of the local Artin map, which is reproduced in Serre, Local Fields, but it was not very pleasant.

In 1965 Lubin and Tate introduced the Lubin-Tate formal group laws, and gave an explicit construction of K^{ab} and an explicit description of the local Artin map. However, they made use of the existence of the local Artin map (our Theorem I.1.1) in their proofs.

In 1981 Gold and Rosen independently gave “elementary” proofs that K^{ab} = K_{π} · K^{un}. In his book (Local Class Field Theory, 1986), Iwasawa develops the whole of local class field theory from the Lubin-Tate perspective, and also gives explicit formulas (due to de Shalit and Wiles) for the Hilbert symbols etc..

Other noncohomological approaches can be found in (Hazewinkel, Local class field theory is easy, Adv. Math. 18 (1975), 148–181), (Neukirch, Class Field Theory, Springer, 1986), and (Fesenko and Vostokov, Local Fields and Their Extensions: A constructive approach, AMS, 1993).

The disadvantage of the noncohomological approaches is, naturally, that they provide no information about the cohomology groups of local fields, which have important applications to other topics, for example elliptic curves.

In this chapter, we have largely followed Serre 1962 and Serre’s article in Cassels and Fröhlich 1967.

CHAPTER IV Brauer Groups

In this chapter, we define the Brauer group of a field, and show that it provides a concrete interpretation of the cohomology group H^{2}(K^{al}/K). Besides clarifying the class field theory, Brauer groups have many applications, for example, to the representation theory of finite groups and to the classification of semisimple algebraic groups over nonalgebraically closed fields.

Throughout the chapter, k will be a field, and all vector spaces over k will be finite dimensional.

Terminology for k-algebras. By a k-algebra we mean a ring A containing k in its centre and finite dimensional as a k-vector space. We do not assume A to be commutative. For example, A could be the ring M_{n}(k) of n × n matrices over k. A k-subalgebra of a k-algebra is a subring containing k. A homomorphism φ : A → B of k-algebras is a homomorphism of rings with the property that φ(a) = a for all a ∈ k. The opposite A^{opp} of a k-algebra A is the algebra with the same underlying set and addition, but with multiplication · defined by α · β = βα. If A is a k-subalgebra of a k-algebra E, then the centralizer of A in E, sometimes denoted C(A) or C_{E}(A), is the set of γ ∈ E such that γα = αγ for all α ∈ A. It is again a k-subalgebra of B. For example, C_{A}(A) is the centre of A. Let e_{1}, …, e_{n} be a basis for A as a k-vector space. Then

e_{i}e_{j} = ∑_{l} a_{ij}^{l}e_{l}

for some a_{ij}^{l} ∈ k, called the structure constants of A relative to the basis (e_{i})_{i}. Once a basis has been chosen, the algebra A is uniquely determined by its structure constants.

1. Simple Algebras; Semisimple Modules

Semisimple modules. By an A-module, we mean a finitely generated left Amodule V. In particular, this means that 1v = v for all v ∈ V. Such a V is also finite-dimensional when considered as a k-vector space, and so to give an A-module is the same as to give a (finite-dimensional) vector space over k together with a homomorphism of k-algebras A → End_{k}(V), i.e., a representation of A on V. The module is said to be faithful if this homomorphism is injective, i.e., if ax = 0 for all x ∈ V implies a = 0.

An A-module V is said to be simple if it is nonzero and contains no A-submodules apart from the obvious two, namely, V and 0, and it is said to be semisimple if it can be written as a direct sum of simple A-modules. It is indecomposable if it can not be written as a direct sum of two nonzero A-modules. Thus a simple module is semisimple, and an indecomposable module is semisimple if and only if it is simple. Some authors use “irreducible” and “completely reducible” for “simple” and “semisimple” respectively.

Example 1.1. Let V = k^{2}, and let A = k[α] for some α ∈ M_{2}(k). The Asubmodules of V are the k-subspaces stable under α.

If α = ( \begin{matrix}a & 1 \\ 0 & a \end{matrix} ), then \{( \begin{matrix}∗ \\ 0 \end{matrix} )\} is an A-submodule of V. In fact, it is the only

nontrivial submodule, and so V is indecomposable, but not semisimple.

If α = ( \begin{matrix}a & 0 \\ 0 & b \end{matrix} ), a ≠ b, then the only lines stable under α are L_{1} =^{df} \{( \begin{matrix}∗ \\ 0 \end{matrix} )\} and

L_{2} =^{df} \{( \begin{matrix}0 \\ ∗ \end{matrix} )\}. Since V = L_{1} ⊕ L_{2} (as an A-module), it is semisimple.

If α = ( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ), then V again decomposes as the direct sum of two lines, but the

decomposition is no longer unique.

Finally, if A = M_{2}(k) then V is a simple A-module: there are no subspaces of k^{2} stable under M_{2}(k) apart from 0 and V.

Theorem 1.2. Any A-module V admits a filtration

V ⊃ V_{1} ⊃ ⋯ ⊃ V_{r} = 0

whose quotients V_{i}/V_{i+1} are simple A-modules. If

V ⊃ V′_{1} ⊃ ⋯ ⊃ V′_{r′} = 0

is a second such filtration, then r = r′ and there is a bijection i ↦ i′ such that V_{i}/V_{i+1} ≈ V′_{i′}/V′_{i′+1} for all i.

Proof. If V is simple, then V ⊃ 0 is such a filtration. Otherwise, V contains a submodule W, V ≠ W ≠ 0, and we can apply the same argument to V/W and to W. This procedure terminates after finitely many steps because V is a finite dimensional k-vector space.

The uniqueness statement can be proved exactly as in the Jordan-Hölder theorem (in fact, the statement is a special case of the Jordan-Hölder theorem with operators) (see, for example, Math 594g, Section 6.4).

For example, the simple summands S_{i} occurring in a decomposition V = ⊕S_{i} of a semisimple A-module are unique up to isomorphism and renumbering.

Proposition 1.3. Let V be a semisimple A-module, say V = ⊕_{i=1}^{r}S_{i} with the S_{i} simple. For any submodule W of V, there is a subset I of {1, …, r} such that V = W ⊕ ⊕_{i∈I}S_{i}.

Proof. For a subset I of {1, …, r}, define S_{I} = ⊕_{i∈I}S_{i}. Let I be maximal among the subsets of {1, …, r} such that W ∩ S_{I} = 0. I claim that W + S_{I} equals V. To prove this, it suffices to show that each S_{i} is contained in W + S_{I}. Because S_{i} is simple, S_{i} ∩ (W + S_{I}) equals S_{i} or 0. In the first case, S_{i} ⊂ W + S_{I}, and in the second W ∩ (S_{I} + S_{i}) = 0, which contradicts the definition of I. Therefore V = W + S_{I}, and because W ∩ S_{I} = 0, the sum is direct.

Corollary 1.4. Every submodule and every quotient module of a semisimple module is semisimple.

Proof. The proposition shows that V/W ≈ S_{I} and that W ≈ V/S_{I} ≈ S_{I′} where I′ is the complement of I in {1, 2, …, r}.

Simple k-algebras. A k-algebra A is said to be simple if it contains no two-sided ideals except for the obvious two, namely, 0 and A. We shall make frequent use of the following observation:

The kernel of a homomorphism f : A → B of k-algebras is an ideal in A not containing 1; therefore, if A is simple, then f is injective.

Example 1.5. Consider the matrix algebra M_{n}(k). For A, B ∈ M_{n}(k), the jth column (A · B)_{j} of A · B is A · B_{j} where B_{j} is the jth column of B. Therefore, a given matrix B,

B_{j} = 0 ⟹ (A · B)_{j} = 0
B_{j} ≠ 0 ⟹ (A · B)_{j} arbitrary.

It follows that the left ideals of M_{n}(k) are the sets of the form L(I) where I is a subset of {1, 2, …, n} and L(I) is the set of matrices whose jth columns are zero for j ∉ I. In particular, the minimal left ideals are the sets L({j_{0}}). Similar statements hold for the right ideals, from which it follows that any nonzero two-sided ideal in M_{n}(k) is the whole ring: M_{n}(k) is a simple k-algebra.

Example 1.6. A k-algebra A is said to be a division algebra if every nonzero element a of A has an inverse, i.e., there exists a b such that ab = 1 = ba. Thus a division algebra satisfies all the axioms to be a field except commutativity (and for this reason is sometimes called a skew field). Clearly, a division algebra has no nonzero proper ideals, left, right, or two-sided, and so is simple.

If D is a division algebra, then, as in (1.5), the left ideals in M_{n}(D) are the sets of the form L(I) and M_{n}(D) is simple.

Example 1.7. For a, b ∈ k^{×}, let H(a, b) be the k-algebra with basis 1, i, j, κ (as a k-vector space) and with the multiplication table:

\begin{matrix}1 & i & j & κ \\ i & a & k & aj \\ j & −κ & b & −bi \\ κ & −aj & bi & −ab \end{matrix}

i.e., i^{2} = a, j^{2} = b, ij = κ = −ji ( ⟹ iκ = iij = aj etc.). Then H(a, b) is a kalgebra, called a quaternion algebra over k. For example, if k = ℝ, then H(−1, −1) is the usual quaternion algebra. One can show (see 5.1) that H(a, b) is either a division algebra or it is isomorphic to M_{2}(k). In particular, it is simple.

Remark 1.8. Much of linear algebra does not require that the field be commutative. For example, the usual arguments show that a finitely generated module V over a division algebra D has a basis, and that all bases have the same number n of elements—n is called the dimension of V. Let V be a D-module of dimension 1, so that V = De_{0} for some e_{0} ∈ V. A D-linear map φ : V → V is determined by its value at e_{0}, φ(e_{0}) = ae_{0}, and φ ↦ a is a bijection End_{D}(V) → D. If φ(e_{0}) = ae_{0} and ψ(e_{0}) = be_{0}, then

(φ ∘ ψ)(e_{0}) =^{df} φ(ψ(e_{0})) = φ(be_{0}) = b(φ(e_{0})) = bae_{0},

and so φ ↦ a is an isomorphism of k-algebras End_{D}(V) → D^{opp}. Similarly, if V is a D-module of dimension n, then the choice of a basis for V determines an isomorphism of k-algebras End_{D}(V) → M_{n}(D^{opp}).

Theorem 1.9. Any simple k-algebra is isomorphic to M_{n}(D) for some n and some division k-algebra D.

Proof. Choose a simple A-module S, for example, any minimal left ideal of A. Then A acts faithfully on S, because the kernel of A → End_{k}(S) will be a two-sided ideal of A not containing 1, and hence is 0.

Thus, we can regard A as a subalgebra of E =^{df} End_{k}(S), the ring of k-linear map S → S. Note that the centralizer C_{E}(A) of A in E is equal to End_{A}(S), the ring of A-linear maps S → S, and that S can be regarded as a C_{E}(A)-module. We shall complete the proof of the theorem by proving:

(a) D =^{df} C_{E}(A) is a division algebra;

(b) A is the centralizer of D in E, i.e., A = End_{D}(S).

It will then follow from the Remark 1.8 that End_{D}(S) ≈ M_{r}(D^{opp})

Lemma 1.10 (Schur’s Lemma). For any k-algebra A and any simple A-module S, End_{A}(S) is a division algebra.

Proof. Let γ be an A-linear map S → S. Then Ker(γ) is an A-submodule of S, and so it is either S or 0. In the first case, γ is zero, and in the second it is an isomorphism, i.e., it has an inverse that is also A-linear.

Theorem 1.11 (Double Centralizer Theorem). Let A be a k-algebra, and let V be a faithful semisimple A-module. Then C(C(A)) = A (centralizers taken in End_{k}(V)).

Example 1.12. Let V = k^{n}. In the following, centralizers will be taken in End_{k}(V) = M_{n}(k).

(a) Let A = k acting by left multiplication on V = k^{n}. Then C(A) = M_{n}(k), and C(C(A)) is the centre of M_{n}(k). To see this, let α = (a_{ij}) lie in the centre of M_{n}(k). Let e_{ij} be the matrix with 1 in the (i, j)th position and zeros elsewhere, so that

e_{ij}e_{lm} = \{ \begin{matrix}e_{im} & if j = l \\ 0 & if j ≠ l. \end{matrix}

Then α = ∑_{i,j} a_{ij}e_{ij}, and so αe_{lm} = ∑_{i} a_{il}e_{im} and e_{lm}α = ∑_{j} a_{mj}e_{lj}. If αe_{lm} = e_{lm}α, then a_{il} = 0 for i ≠ l, a_{mj} = 0 for j ≠ m, and a_{ll} = a_{mm}. It follows that the centre of M_{n}(k) is k (identified with the set of scalar matrices).

(b) Let A be the set of diagonal matrices in M_{n}(k), and let V = k^{n}. Then C(A) = A, and so C(C(A)) = C(A) = A.

Proof. In the situation of the proposition, let D = C(A) and let B = C(D). Clearly A ⊂ B, and the reverse inclusion follows from the next lemma when we take v_{1}, …, v_{n} to generate V as a k-vector space.

Lemma 1.13. Under the hypotheses of the proposition, for any v_{1}, …, v_{n} ∈ V and b ∈ B, there exists an a ∈ A such that

av_{1} = bv_{1}, av_{2} = bv_{2}, …, av_{n} = bv_{n}.

Proof. We first prove this for n = 1. Note that Av_{1} is an A-submodule of V, and so (see 1.3) there exists an A-submodule W of V such that V = Av_{1} ⊕ W. Let π : V → V be the map (av_{1}, w) ↦ (av_{1}, 0) (projection onto Av_{1}). It is A-linear, hence lies in D, and has the property that π(v) = v if and only if v ∈ Av_{1}. Now

π(bv_{1}) = b(πv_{1}) = bv_{1},

and so bv_{1} ∈ Av_{1}.

We now prove the general case. Let W be the direct sum of n copies of V with A acting diagonally, i.e.,

a(x_{1}, …, x_{n}) = (ax_{1}, …, ax_{n}), a ∈ A, x_{i} ∈ V.

Then W is again a semisimple A-module. The centralizer of A in End_{k}(W) consists of the matrices (γ_{ij})_{1≤i,j≤n}, γ_{ij} ∈ End_{k}(V), such that (γ_{ij}α) = (αγ_{ij}) for all α ∈ A, i.e., such that γ_{ij} ∈ D. In other words, the centralizer of A in End_{k}(A) is M_{n}(D). An argument as in the above example, using the matrices e_{ij}(δ) with δ in the ijth position and zeros elsewhere, shows that the centralizer of M_{n}(D) in End_{k}(W) consists of the diagonal matrices

( \begin{matrix}β & 0 & ⋯ & 0 \\ 0 & β & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & β \end{matrix} )

with β ∈ B. We now apply the case n = 1 of the lemma to A, W, b, and the vector (v_{1}, …, v_{n}) to complete the proof.

Modules over simple k-algebras. When we regard A as a k-vector space and let A act on it by left multiplication, then we obtain a homomorphism A → End_{k−linear}(A)—this is called the regular representation of A.

Proposition 1.14. For any simple k-algebra A, the simple submodules of A (regarded as a left A-module) are the minimal left ideals in A; any two such ideals are isomorphic as left A-modules, and A is a direct sum of its minimal left ideals.

Proof. After Theorem 1.9, we may assume that A = M_{n}(D) for some division algebra D. An A-submodule M of A is a left ideal, and M will be simple if and only if it is minimal. We saw in (1.6) that the minimal left ideals in M_{n}(D) are those of the form L({j_{0}}). Clearly A = ⊕_{1≤j≤n}L({j}) and each L({j}) is isomorphic to D^{n} with its natural action of M_{n}(D).

Theorem 1.15. Let A be a simple k-algebra, and let S be a simple A-module. Then any A-module V is isomorphic to a direct sum of copies of S. In particular, it is semisimple.

Proof. Let S_{0} be a minimal left ideal of A. The proposition shows that, as a left A-module, A ≈ S_{0}^{n} for some n.

Let e_{1}, …, e_{r} be a set of generators for V as an A-module. The map

(a_{1}, …, a_{r}) ↦ ∑ a_{i}e_{i}

realizes V as a quotient of the A-module A^{r}, and hence as a quotient of S_{0}^{nr}. Now Proposition 1.3 shows that V ≈ S_{0}^{m} for some m.

Corollary 1.16. If A is a simple k-algebra, then any two simple A-modules are isomorphic, and any two A-modules having the same dimension over k are isomorphic.

2. Definition of the Brauer Group

Tensor products of algebras. Let A and B be k-algebras, and let A ⊗_{k} B be the tensor product of A and B as k-vector spaces. There is a unique k-bilinear multiplication on A ⊗_{k} B such that

(a ⊗ b)(a′ ⊗ b′) = aa′ ⊗ bb′, a, a′ ∈ A, b, b′ ∈ B.

When we identify k with k·(1⊗1) ⊂ A⊗_{k}B, then A⊗_{k}B becomes a k-algebra. If (e_{i})_{i} and (f_{j})_{j} are bases of A and B as k-vector spaces, then (e_{i}⊗f_{j})_{i,j} is a basis for A⊗_{k} B, and the structure constants for A ⊗_{k} B can be obtained from those of A and B by an obvious formula. We shall use that tensor products are commutative and associative in the sense that, for any two k-algebras A, B, there is a unique isomorphism

A ⊗_{k} B → B ⊗_{k} A

sending a ⊗ b to b ⊗ a, and for any three k-algebras A, B, C, there is a unique isomorphism

A ⊗_{k} (B ⊗_{k} C) → (A ⊗_{k} B) ⊗_{k} C

sending a ⊗ (b ⊗ c) to (a ⊗ b) ⊗ c.

Example 2.1. For any k-algebra A, A ⊗_{k} M_{n}(k) ≅ M_{n}(A). To see this, note that a ring B containing a subring R is isomorphic to M_{n}(R) if and only if it admits a basis (e_{ij})_{1≤i,j≤n} as a left R-module such that

e_{ij}e_{lm} = \{ \begin{matrix}e_{im} & if j = l \\ 0 & if j ≠ l. \end{matrix}

If (e_{ij}) is the standard basis for M_{n}(k), then (1⊗e_{ij}) is an A-basis for A⊗M_{n}(k) with the correct property. More generally, A ⊗_{k} M_{n}(A′) ≅ M_{n}(A ⊗_{k} A′) for any k-algebras A and A′.

Example 2.2. For any m, n, M_{m}(k) ⊗ M_{n}(k) ≅ M_{mn}(k). To see this, note that according to the preceding example, M_{m}(k) ⊗_{k} M_{n}(k) ≅ M_{m}(M_{n}(k)), and an m × mmatrix whose entries are n × n-matrices is an mn × mn-matrix (delete the inner parentheses). Alternatively, let (e_{ij}) and (f_{lm}) be standard bases for M_{m}(k) and M_{n}(k), and check that (e_{ij} ⊗ f_{lm}) has the correct multiplication properties.

Centralizers in tensor products.

Proposition 2.3. Let A and A′ be k-algebras, with subalgebras B and B′, and let C(B) and C(B′) respectively be the centralizers of B and B′ in A and A′. Then the centralizer of B ⊗_{k} B′ in A ⊗_{k} A′ is C(B) ⊗_{k} C(B′).

Proof. Certainly C(B ⊗_{k} B′) ⊃ C(B) ⊗_{k} C(B′). Let (f_{i})_{i} be a basis for A′ as a k-vector space. Then (1 ⊗ f_{i})_{i} is a basis for A ⊗_{k} A′ as an A-module, and so an element α of A ⊗_{k} A′ can be written uniquely in the form α = ∑_{i} α_{i} ⊗ f_{i}, α_{i} ∈ A. Let β ∈ B. Then α commutes with β ⊗ 1 if and only if βα_{i} = α_{i}β for all i. Therefore, the centralizer of B ⊗ 1 in A ⊗ A′ is C(B) ⊗ A′. Similarly, the centralizer of 1 ⊗ B′ in C(B) ⊗ A′ is C(B) ⊗ C(B′), which therefore contains C(B ⊗ B′).

In particular, the centre of the tensor product of two k-algebras is the tensor product of their centres: Z(A ⊗_{k} B) = Z(A) ⊗_{k} Z(B).

Corollary 2.4. The centre of a simple k-algebra is a field.

Proof. Obviously, the centre of a division algebra is a field, but Wedderburn’s theorem (1.9) shows that every simple k-algebra is isomorphic to M_{n}(D) for some division algebra D. Now M_{n}(D) ≅ M_{n}(k) ⊗_{k} D, and so Z(M_{n}(D)) ≅ k ⊗_{k} Z(D) ≅ Z(D).

A k-algebra A is said to be central if its centre is k, and a k-algebra that is both central and simple is said to be central simple. The corollary shows that every simple k-algebra is central simple over a finite extension of k.

Primordial elements. Before continuing, it will be useful to review a little linear algebra from the second edition of Bourbaki’s Algebra.

Let V be a k-vector space, and let (e_{i})_{i∈I} be a basis for V. Any v ∈ V can be written uniquely v = ∑ a_{i}e_{i}, and we define J(v) to be the set of i ∈ I such that a_{i} ≠ 0. Let W be a subspace of V. An element w ∈ W is said to be primordial (relative to V and the basis (e_{i})_{i∈I}) if

(a) J(w) is minimal among the sets J(w′) for w′ a nonzero elements of W, and (b) in the expression w = ∑ b_{i}x_{i}, at least one b_{i} = 1.

Proposition 2.5. (a) Let w ∈ W be primordial, and let w′ ∈ V. Then J(w′) ⊂ J(w) if and only if w′ = cw for some c ∈ k, in which case w′ = 0 or J(w′) = J(w).

(b) The set of primordial elements of W generates it.

Proof. (a) Let w =∑_{i∈J(w)} a_{i}e_{i} and w′ = ∑_{i∈J(w′)} a′_{i}e_{i} be elements of W with J(w′) ⊂ J(w). We may assume w′ ≠ 0, so that J(w′) ≠ ∅. Choose an i_{0} ∈ J(w′), and set c = a_{i_{0}} · a_{i_{0}}^{′−1}. Then J(w − cw′) ⊂ J(w) and i_{0} ∉ J(w − cw′), and so w − cw′ = 0.

(b) Let w = ∑_{i∈J(w)} a_{i}e_{i} ∈ W. We shall use induction on the number n of elements in J(w) to prove that w lies in the subspace generated by the primordial elements. For n = 0 the statement is obvious, and so suppose n > 0. Among the nonzero elements w′ of W with J(w′) ⊂ J(w), choose one w_{0} such that J(w_{0}) has the fewest elements. After replacing it by a scalar multiple, we may assume that w_{0} is primordial, say w_{0} = ∑ b_{i}e_{i} with b_{i_{0}} = 1. Then w − a_{i_{0}}w_{0} ∈ W but J(w − a_{i_{0}}w_{0}) has at most n − 1 elements. By induction, w −a_{i_{0}}w_{0} is a linear combination of primordial elements, and so therefore is w.

The results (and proofs) of this section do not require V to be finite-dimensional over k, and they do not require k to be commutative, i.e., k can be a division algebra.

Simplicity of tensor products.

Proposition 2.6. The tensor product of two simple k-algebras, at least one of which is central, is again simple.

Proof. After (1.9), we may suppose that one of the algebras is M_{n}(D) where D is a division algebra with centre k. Let A be the second simple k-algebra. If A ⊗_{k} D is simple, say, A ⊗_{k} D ≈ M_{m}(D′) with D′ a division algebra, then

A ⊗_{k} M_{n}(D) ≅ M_{n}(A ⊗_{k} D) ≈ M_{n}(M_{m}(D′)) ≅ M_{mn}(D′),

which is simple. Thus the proposition follows from the next lemma.

Lemma 2.7. Let A be a k-algebra, and let D be a division algebra with centre k. Then any two-sided ideal 𝔄 in A ⊗ D is generated (as a left vector space over D) by

𝔞 =^{df} 𝔄 ∩ (A ⊗ 1).

Proof. We make A ⊗_{k} D into a left D-module by the rule:

δ(α ⊗ δ′) = α ⊗ δδ′, α ∈ A, δ, δ′ ∈ D.

The ideal 𝔄 of A ⊗_{k} D is, in particular, a D-submodule of A ⊗_{k} D.

Let (e_{i}) be a basis for A as a k-vector space. Then (e_{i} ⊗ 1) is a basis for A ⊗_{k} D as a left D-vector space. Let α ∈ 𝔄 be primordial with respect to this basis, say

α = ∑_{i∈J(α)} δ_{i}(e_{i} ⊗ 1) = ∑_{i∈J(α)} e_{i} ⊗ δ_{i}.

For any nonzero δ ∈ D, αδ ∈ 𝔄, and αδ = ∑ e_{i}⊗δ_{i}δ = ∑_{i∈I}(δ_{i}δ)(e_{i}⊗1). In particular, J(δα) = J(α), and so αδ = δ′α for some δ′ ∈ D (Proposition 2.5a). As some δ_{i} = 1, this implies that δ = δ′, and so each δ_{i} commutes with every δ ∈ D. Hence δ_{i} lies in the centre k of D, and α ∈ A ⊗ 1. We have shown that every primordial element of 𝔄 is in A ⊗ 1, which completes the proof because 𝔄 is generated (as a D-module) by its primordial elements.

Corollary 2.8. The tensor product of two central simple k-algebras is again central simple.

Proof. Combine Proposition 2.3 with Proposition 2.6.

Let A be a central simple algebra over k, and let V denote A regarded as a k-vector space. Then left multiplication makes V into a left A-module, and right multiplication makes it into a right A-module, or, what is the same thing, a left A^{opp}-module. These actions identify A and A^{opp} with commuting subalgebras of End_{k}(V). From the universality of the tensor product, we obtain a homomorphism

a ⊗ a′ ↦ aa′ : A ⊗_{k} A^{opp} → End_{k}(V).

As A ⊗_{k} A^{opp} is simple and the kernel of the homomorphism does not contain 1, it is injective. On counting degrees, we find that

[A ⊗_{k} A^{opp} : k] = [A : k]^{2} = n^{2} = [End_{k}(V) : k],

and so the homomorphism is an isomorphism. We have shown:

Corollary 2.9. For any central simple k-algebra A,

A ⊗_{k} A^{opp} ≅ End_{k}(A) ≈ M_{n}(k), n = [A : k].

The Noether-Skolem Theorem.

Theorem 2.10 (Noether-Skolem). Let f, g : A → B be homomorphisms from the k-algebra A to the k-algebra B. If A is simple, and B is central and simple over k, then there exists an invertible element b ∈ B such that f(a) = b · g(a) · b^{−1} for all a ∈ A.

Proof. If B = M_{n}(k) = End_{k}(k^{n}), then the homomorphisms define actions of A on k^{n}—let V_{f} and V_{g} denote k^{n} with the actions defined by f and g. According to (1.16), two A-modules with the same dimension are isomorphic, but an isomorphism b : V_{g} → V_{f} is an element of M_{n}(k) such that f(a) · b = b · g(a) for all a ∈ A.

In the general case, we consider the homomorphisms

f ⊗ 1, g ⊗ 1 : A ⊗_{k} B^{opp} → B ⊗_{k} B^{opp}.

Because B ⊗_{k} B^{opp} is a matrix algebra over k, the first part of the proof shows that there exists a b ∈ B ⊗_{k} B^{opp} such that

(f ⊗ 1)(a ⊗ b′) = b · (g ⊗ 1)(a ⊗ b′) · b^{−1}

for all a ∈ A, b′ ∈ B^{opp}. On taking a = 1 in this equation, we find that (1 ⊗ b′) = b · (1 ⊗ b′) · b^{−1} for all b′ ∈ B^{opp}. Therefore (see 2.3), b ∈ C_{B⊗_{k}B^{opp}}(k ⊗ B^{opp}) = B ⊗_{k} k, i.e., b = b_{0} ⊗ 1 with b_{0} ∈ B. On taking b′ = 1 in the equation, we find that

f(a) ⊗ 1 = (b_{0} · g(a) · b_{0}^{−1}) ⊗ 1

for all a ∈ A, and so b_{0} is the element sought.

Corollary 2.11. Let A be a central simple algebra over k, and let B_{1} and B_{2} be simple k-subalgebras of A. Any isomorphism f : B_{1} → B_{2} is induced by an inner automorphism of A, i.e., there exists an invertible a ∈ A such that f(b) = aba^{−1} for all b ∈ B_{1}.

Proof. This is the special case of the theorem in which g is the identity map B_{1} → B_{1}.

Corollary 2.12. All automorphisms of a central simple k-algebra are inner.

For example, the automorphism group of M_{n}(k) is PGL_{n}(k) =^{df} GL_{n}(k)/k^{×}I_{n}.

Definition of the Brauer group. Let A and B be central simple algebras over k. We say that A and B are similar, A ∼ B, if A ⊗_{k} M_{n}(k) ≈ B ⊗_{k} M_{m}(k) for some m and n. This is an equivalence relation: it is obviously reflexive and symmetric, and (2.2) implies that it is transitive. Define Br(k) to be the set of similarity classes of central simple algebras over k, and write [A] for the similarity class of A. For classes [A] and [B], define

[A][B] = [A ⊗_{k} B].

This is well-defined (i.e., if A ∼ A′ and B ∼ B′, then A ⊗_{k} B ∼ A′ ⊗_{k} B′), and the associativity and commutativity of tensor products show that it is associative and commutative. For any n, [M_{n}(k)] is an identity element, and because A ⊗_{k} A^{opp} ≈ M_{n}(k) (see 2.9) [A] has [A^{opp}] as inverse. Therefore Br(k) is an abelian group, called the Brauer group of k.

Example 2.13. (a) If k is algebraically closed, then Br(k) = 0, for let D be a central division algebra over k. We have to show that D = k. Let α ∈ D, and let k[α] be the subalgebra of D generated by k and α. Then k[α] is a commutative field of finite degree over k (because it is an integral domain of finite degree over k). Hence k[α] = k, and α ∈ k. Since α was arbitrary, this shows that D = k.

(b) The Brauer group of ℝ is cyclic of order 2. Its elements are represented by k and the usual quaternion algebra. (See Section 4 below.)

(c) The Brauer group of a finite field is zero (Theorem 4.1).

(d) The Brauer group of a nonarchimedean local field is canonically isomorphic to ℚ/ℤ (Proposition 4.3).

(e) If K is a number field, then there is an exact sequence

0 → Br(K) → ⊕_{v} Br(K_{v}) → ℚ/ℤ → 0.

The sum is over all the primes of K (including the infinite primes). This statement is of the same depth as that of the main theorems of class field theory. (See Chapter VIII).

Remark 2.14. (a) Wedderburn’s theorem (1.9) shows that every central simple algebra over k is isomorphic to M_{n}(D) for some central division algebra D. Moreover, D is the opposite algebra of End_{A}(S) for any simple A-module S. Because any two simple A-algebras are isomorphic (see 1.16), this shows that D is uniquely determined by A (even by the similarity class of A) up to isomorphism. Therefore, each similarity class is represented by a central division algebra, and two central division algebras representing the same similarity class are isomorphic.

(b) We should verify^{1} that the similarity classes form a set, and not merely a class. For each n > 0, consider the families (a_{ij}^{l})_{1≤i,j,l≤n} that are structure constants for central division algebras over k. Clearly, these families form a set, each family defines a central division algebra over k, and these division algebras contain a set of representatives for the Brauer group of k.

1I once heard Brauer, who normally had a gentle manner, deliver a tirade against “modern”

mathematicians who ignored the distinction between sets and classes. As he pointed out, if you ignore the distinction, then you obtain a contradiction (Russell’s paradox), and once you have one contradiction in your system, you can prove everything.

Extension of the base field.

Proposition 2.15. Let A be a central simple algebra over k, and let K be a field containing k (not necessarily of finite degree over k). Then A⊗_{k} K is a central simple algebra over K.

Proof. The same argument as in the proof of Proposition 2.3 shows that the centre of A ⊗_{k} K is k ⊗_{k} K = K (the argument does not require K to have finite degree over k). Also, the proof of Lemma 2.7 does not use that D is finite-dimensional over k. Therefore, when A is a division algebra, any two-sided ideal in A ⊗_{k} K is generated as an A-module by its intersection with K, and therefore is 0 or A ⊗_{k} K. A general A ≈ M_{n}(D), and

A⊗_{k} K ≈ M_{n}(D)⊗_{k} K ≈ M_{n}(k)⊗_{k} (D ⊗_{k} K) ≈ M_{n}(D ⊗_{k} K) ≈ M_{n}(K)⊗_{K} (D ⊗_{k} K)

which is simple.

Corollary 2.16. For any central simple algebra A over k, [A : k] is a square.

Proof. Clearly [A : k] = [A ⊗_{k} k^{al} : k^{al}], and A ⊗_{k} k^{al} ≈ M_{n}(k^{al}) for some n.

Let L be a field containing k (not necessarily of finite degree). Then

M_{n}(k) ⊗ L ≅ M_{n}(L),

and

(A ⊗_{k} L) ⊗_{L} (A′ ⊗_{k} L) = A ⊗_{k} (L ⊗_{L} (A′ ⊗_{k} L)) = (A ⊗_{k} A′) ⊗_{k} L.

Therefore the map A ↦ A ⊗_{k} L defines a homomorphism

Br(k) → Br(L).

We denote the kernel of this homomorphism by Br(L/k)—it consists of the similarity classes represented by central simple k-algebras A such that the L-algebra A ⊗_{k} L is a matrix algebra.

A central simple algebra (or its class in Br(k)) is said to be split by L, and L is called a splitting field for A, if A ⊗_{k} L is a matrix algebra over L. Thus Br(L/k) consists of the elements of Br(k) split by L.

Proposition 2.17. For any field k, Br(k) = ∪ Br(K/k), where K runs over the finite extensions of K contained in some fixed algebraic closure k^{al}.

Proof. Let A be a central simple algebra over k. Then A ⊗_{k} k^{al} ≈ M_{n}(k^{al}), i.e., there exists a basis (e_{ij})_{1≤i,j≤n} for A ⊗_{k} k^{al} such that e_{ij}e_{lm} = δ_{jl}e_{im} for all i, j, l, m. Because A ⊗_{k} k^{al} = ∪_{[K:k]<∞}A ⊗_{k} K, the e_{ij} ∈ A ⊗_{k} K for some K, and it follows that A ⊗_{k} K ≈ M_{n}(K).

3. The Brauer Group and Cohomology

For a Galois extension L/k of fields, let H^{2}(L/k) = H^{2}(Gal(L/k), L^{×}). We shall show that there is a natural isomorphism H^{2}(L/k) ≈ Br(L/k), but first we need to investigate the maximal subfields of a central simple algebra.

Maximal subfields. We need a variant of the double centralizer theorem.

Theorem 3.1. Let A be a central simple algebra over k, and let B be a simple subalgebra over A. Then the centralizer C = C(B) of B in A is simple, and B is the centralizer of C. Moreover,

[B : k][C : k] = [A : k].

Proof. Let V denote B regarded as a k-vector space. Then B and B^{opp} act on V, by right and left multiplication respectively, and each is the centralizer of the other (see 1.8).

Consider the simple algebra A ⊗_{k} End_{k}(V). Proposition 2.3 shows that the centralizer of B ⊗ 1 in this algebra is C ⊗ End_{k}(V) and that of 1 ⊗ B is A ⊗ B^{opp}. On applying the Noether-Skolem theorem to the two embeddings b ↦ b ⊗ 1, 1 ⊗ b of B into A ⊗_{k} End_{k}(V), we obtain an invertible element u of this k-algebra such that b ⊗ 1 = u(1 ⊗ b)u^{−1} for all b ∈ B. Clearly then

u · C(B ⊗ 1) · u^{−1} = C(1 ⊗ B)

(centralizers in A ⊗_{k} End_{k}(V)), which shows that these centralizers are isomorphic. Therefore C ⊗_{k} End_{k}(V) is simple because A ⊗_{k} B^{opp} is simple (see 2.6), and this implies that C itself is simple because, for any ideal 𝔞 of C, 𝔞 ⊗_{k} End_{k}(V) is an ideal in C ⊗_{k} End_{k}(V). As End_{k}(V) has degree [B : k]^{2} over k,

[C ⊗_{k} End(V) : k] = [C : k][B : k]^{2},

and obviously

[A ⊗ B^{opp} : k] = [A : k][B : k].

On comparing these equalities, we find that

[A : k] = [B : k][C : k].

If B′ denotes the centralizer of C in A, then B′ ⊃ B. But after the above, [A : k] = [C : k][B′ : k]; so [B : k] = [B′ : k] and B = B′.

Remark 3.2. In the case that A = End_{k}(V) for V a k-vector space, Theorem 3.1 follows from Theorem 1.11 because V will be a faithful semisimple B-module. This observation can be used to give an alternative proof of the theorem, because A becomes of this form after a finite extension of the base field (see 2.17).

Corollary 3.3. If in the statement of the theorem, B has centre k, then so also does C, and the canonical homomorphism B ⊗_{k} C → A is an isomorphism.

Proof. The centres of B and C both equal B ∩ C, and so B central implies C central. Therefore the k-algebra B ⊗_{k} C is central simple, which implies that B⊗_{k}C → A is injective. It is surjective because the algebras have the same dimension over k.

Corollary 3.4. Let A be a central simple algebra over k, and let L be a subfield of A containing k. The following are equivalent:

(a) L equals its centralizer in A;

(b) [A : k] = [L : k]^{2};

(c) L is a maximal commutative k-subalgebra of A.

Proof. Because L is commutative, it is contained in its centralizer C(L), but

[A : k] = [L : k][C(L) : k],

and so C(L) = L if and only if [A : k] = [L : k]^{2}.

The equivalence of (b) with (c) follows from the observation that any commutative k-subalgebra of A containing L is contained in C(L).

Corollary 3.5. The maximal subfields containing k of a central division k-algebra D are exactly those with degree \sqrt{[D : k]} over k.

Proof. Any commutative k-subalgebra of D is an integral domain of finite degree over k, and hence is a field.

Corollary 3.6. Let A be a central simple algebra over k, and let L be a field of finite degree over k. Then L splits A if and only if there exists an algebra B similar to A containing L and such that [B : L] = [L : k]^{2}.

Proof. Suppose L splits A. Then L also splits A^{opp}, say, A^{opp} ⊗_{k} L = End_{L}(V). This equality states that A^{opp} ⊗_{k} L is the centralizer of L in End_{L}(V), and so L is the centralizer of A^{opp} ⊗_{k} L in End_{L}(V) (see 3.1). Let B be the centralizer of A^{opp} in End_{k}(V). I claim that B satisfies the required conditions. Certainly, B ⊃ L, and Corollary 3.3 shows that B is central simple and that B ⊗_{k} A^{opp} ≅ End_{k}(V). On tensoring both sides with A and using that A ⊗_{k} A^{opp} is a matrix algebra, we find that B ∼ A.

For the converse, it suffices to show that L splits B. Because L is commutative, L = L^{opp} ⊂ B^{opp}, and because [L : k] = \sqrt{[B : k}], L is equal to its centralizer in B^{opp}. Therefore the centralizer of 1⊗L in B ⊗_{k} B^{opp} is B ⊗_{k} L. When we identify B ⊗_{k} B^{opp} with End_{k}(B) (endomorphisms of B as a k-vector space—see 2.9), the centralizer of L becomes identified with End_{L}(B) (endomorphisms as an L-vector space). This completes the proof.

Corollary 3.7. Let D be a central division algebra of degree n^{2} over k, and let L be a field of degree n over k. Then L splits D if and only if L can be embedded in D (i.e., there exists a homomorphism of k-algebras L → D).

Proof. If L splits D, then there is a central simple algebra B over k containing L, similar to D, and of degree [L : k]^{2}. But B ∼ D implies B ≈ M_{m}(D) for some m (see Remark 2.14a), and the condition on the degrees implies that m = 1.

Proposition 3.8. Any division algebra over k contains a maximal subfield separable over k.

Proof. We omit the proof, because we are mainly interested in fields of characteristic zero and finite fields, for which the problem doesn’t arise.

Corollary 3.9. For any field k, Br(k) = ∪ Br(L/k) where L/k runs over the finite Galois extensions of k contained in a fixed (separable) algebraic closure of k.

Proof. The proposition shows that every element of Br(k) is split by a finite separable extension, and therefore by a finite Galois extension.

Central simple algebras and 2-cocycles. First, we need a remark that should have been made in Chapter II. Let G be a group, and let M be a G-module. For m ∈ M, let φ_{m} : G → M be the constant map σ ↦ φ_{m}. Then

(dφ_{m})(σ, τ) = σm − m + m = σm.

In particular, (dφ_{m})(1, 1) = m. Therefore, every class in H^{2}(G, M) is represented by a 2-cocycle φ with φ(1, 1) = 0. Such a 2-cocycle is said to be normalized.

Fix a finite Galois extension L of k, and let G = Gal(L/k). Define A(L/k) to be the class of central simple algebras A over k containing L and of degree [A : k] = [L : k]^{2} (hence, L equals its centralizer in A).

Fix an A ∈ A(L/k). For any σ ∈ G, Corollary 2.11 of the Noether-Skolem theorem shows that there exists an element e_{σ} ∈ A such that

σa = e_{σ}ae_{σ}^{−1} for all a ∈ L (∗).

Moreover, e_{σ} is determined by (∗) up to multiplication by an element of L^{×}, because if f_{σ} has the same property, then f_{σ}^{−1}e_{σ} centralizes L. Note that (∗) can be written as

e_{σ} · a = σa · e_{σ} for all a ∈ L (∗′)

which says that moving e_{σ} past a ∈ L replaces it with σa. Clearly e_{σ}e_{τ} has the property (∗) for στ, and so

e_{σ}e_{τ} = φ(σ, τ)e_{στ} (∗∗)

for some φ(σ, τ) ∈ L^{×}. Note that

e_{ρ}(e_{σ}e_{τ}) = e_{ρ}(φ(σ, τ)e_{στ}) = ρφ(σ, τ) · φ(ρ, στ) · e_{ρστ}

and

(e_{ρ}e_{σ})e_{τ} = φ(ρ, σ)e_{ρσ}e_{τ} = φ(ρ, σ)φ(ρσ, τ) · e_{ρστ}.

Therefore the associative law implies that φ is a 2-cocycle. It is even a normalized 2-cocycle if we choose e_{1} = 1. A different choice of e_{σ}’s leads to a cohomologous 2-cocycle, and so we have a well-defined map A ↦ γ(A) : A(L/k) → H^{2}(L/k).

Theorem 3.10. The map A ↦ γ(A) defines a bijection

A(L/k)/≈ → H^{2}(L/k).

We first need a lemma.

Lemma 3.11. Let A ∈ A(L/k), and define e_{σ} to satisfy (∗). Then the set (e_{σ})_{σ∈L} is a basis for A as a left vector space over L.

Proof. Note that

dim_{L}(A) = dim_{k}(A)/ dim_{k}(L) = n,

and so it suffices to show that the e_{σ} are linearly independent. Suppose not, and let (e_{σ})_{σ∈J} be a maximal linearly independent set. If τ ∉ J, then

e_{τ} = ∑ a_{σ}σ

for some a_{σ} ∈ L. Let a ∈ L. When we compute e_{τ}a in two different ways,

e_{τ}a = τa · e_{τ} = ∑_{σ∈J} τa · a_{σ}e_{σ},

e_{τ}a = ∑_{σ∈J} a_{σ}e_{σ}a = ∑_{σ∈J} a_{σ} · σa · e_{σ}

we find that τa · a_{σ} = σa · a_{σ} for all σ ∈ J. For at least one σ ∈ J, a_{σ} ≠ 0, and then the equation shows that τ = σ, contradicting the fact that τ ∉ J. Therefore J = G.

Now A is uniquely determined by the following properties: A ⊃ L; (e_{σ})_{σ∈G} is a basis for A as an L-vector space; multiplication in A satisfies the equation (∗) and (∗∗).

Let A′ ∈ A(L/k) and suppose that γ(A) = γ(A′). The condition implies that we can choose bases (e_{σ}) and (e′_{σ}) for A and A′ satisfying (∗) and (∗∗) with the same 2-cocycle φ. The map ∑ a_{σ}e_{σ} ↦ ∑ a_{σ}e′_{σ} : A → A′ is an isomorphism of k-algebras.

Next suppose that A and A′ are isomorphic elements of A(L/k). The Noether-Skolem theorem allows us to choose the isomorphism f : A → A′ so that f(L) = L and f|L is the identity map. If e_{σ} satisfies condition (∗∗) for A, then f(e_{σ}) satisfies (∗∗) for A′. With the choices (e_{σ}) and (f(e_{σ})), A and A′ define the same cocycle.

These remarks show that the map A ↦ γ(A) defines a injection

A(L/k)/≈ → H^{2}(L/k).

To show that the map is surjective, we construct an inverse.

Let φ : G × G → L^{×} be a normalized 2-cocycle. Define A(φ) to be the L-vector space with basis (e_{σ})_{σ∈G} endowed with the multiplication given by (∗) and (∗∗). Then e_{1} is an identity element for multiplication, and the cocycle condition (exactly) shows that

e_{ρ}(e_{σ}e_{τ}) = (e_{ρ}e_{σ})e_{τ}.

It follows that A(φ) is a k-algebra. We identify L with the subfield Le_{1} of A(φ).

Lemma 3.12. The algebra A(φ) is central simple over K.

Proof. Let α = ∑ a_{σ}e_{σ} centralize L, and let a ∈ L. On comparing aα = ∑ aa_{σ} ·e_{σ} with αa = ∑ a_{σ}(σa) · e_{σ}, we find that a_{σ} = 0 for σ ≠ 1, and so α = a_{1}e_{1} ∈ L. Therefore, the centralizer of L in A(φ) is L.

Let α lie in the centre of A(φ). Then α centralizes L, and so α ∈ L, say α = ae_{1}, a ∈ L. On comparing e_{σ} · α = (σa)e_{σ} with α · e_{σ} = ae_{σ}, we see that α ∈ k. Thus A(φ) is central.

Let 𝔄 be a two-sided ideal in A(φ); in particular, 𝔄 is an L-subspace of A(φ). If 𝔄 contains one element e_{σ}, then (**) shows that it contains all, and so equals A(φ). Suppose 𝔄 ≠ 0, and let α = ∑ a_{σ}e_{σ} be a primordial element of 𝔄, with say a_{σ_{0}} = 1. If a_{σ_{1}} ≠ 0, σ_{1} ≠ σ_{0}, then for any a ∈ L,

(σ_{1}a) · α − α · a = ∑ a_{σ}(σ_{1}a − σa)e_{σ} ∈ 𝔄.

If a is chosen so that σ_{1}a ≠ σ_{0}a, then this element is nonzero but has fewer nonzero coefficients than α, contradicting its primordality. Therefore, α = e_{σ_{0}}, and we have shown that 𝔄 = A(φ).

Let φ and φ′ be cohomologous 2-cocycles, say,

a(σ) · σa(τ) · φ′(σ, τ) = a(στ) · φ(σ, τ)

for some map a : G → L^{×}. One checks immediately that the L-linear map A(φ) → A(φ′) sending e_{σ} to a(σ)e′_{σ} is an isomorphism of k-algebras. Therefore φ ↦ A(φ) defines a map H^{2}(L/K) → A(L/k)/≈, which is clearly inverse to A ↦ γ(A). This completes the proof of Theorem 3.10.

The algebras A(φ) are called crossed-product algebra. Before group cohomology existed, 2-cocycles φ : G × G → L^{×} were called factor sets.

Theorem 3.13. For any finite Galois extension L/k, the map φ ↦ [A(φ)] defines an isomorphism of abelian groups H^{2}(L/k) → Br(L/k).

To show that this map is bijective, it suffices (after Theorem 3.13) to show that the map A ↦ [A] : A(L/k)/≈ → Br(L/k) is bijective.

If A and A′ are similar central simple algebras over k, then (see 2.14) there exists a central division algebra D such that A ∼ D ∼ A′, say, A ≈ M_{n}(D), A′ ≈ M_{n′}(D). But if [A : k] = [A′ : k], then n = n′, and so A ≈ A′. This proves that the map A(L/K)/≈ → Br(L/k) is injective, and 3.6 proves that it is surjective.

Lemma 3.14. For any two 2-cocycles φ and φ′, A(φ + φ′) ∼ A(φ) ⊗_{k} A(φ′).

Proof. The proof is a little messy because we have to recognize A(φ)⊗_{k}A(φ′), not as a crossed-product algebra, but as matrix algebra over a crossed-product algebra. I merely sketch the proof (see Blanchard, 1972, p94–95, or Farb and Dennis, 1993, p126–128 for the details).

Set A = A(φ), B = A(φ′), and C = A(φ + φ′). Regard A and B as left L-modules (using left multiplication), and define

V = A ⊗_{L} B.

Concretely, V is the largest quotient space of A ⊗_{k} B such that

ℓa ⊗_{L} b = a ⊗_{L} ℓb

holds for all a ∈ A, b ∈ B, ℓ ∈ L.

The k-vector space V has a unique right A ⊗_{k} B-module structure such that

(a′ ⊗_{L} b′)(a ⊗_{k} b) = a′a ⊗_{L} b′b, all a′, a ∈ A, b′, b ∈ B,

and a unique left C-module structure such that

(ℓe′′_{σ})(a ⊗_{L} b) = ℓe_{σ}a ⊗_{L} e′_{σ}b, all ℓ ∈ L, σ ∈ G, a ∈ A, b ∈ B.

Here (e_{σ}), (e′_{σ}), and (e′′_{σ}) are the standard bases for A = A(φ), B = A(φ′), and C = A(φ + φ′) respectively.

The two actions commute, and so the right action of A ⊗_{k} B on V defines a homomorphism of k-algebras

f : (A ⊗_{k} B)^{opp} → End_{C}(V).

This homomorphism is injective because A ⊗_{k} B (and hence its opposite) is simple. Since both (A ⊗_{k} B)^{opp} and End_{C}(V) have degree n^{4} over k, where n = [L : k], f is an isomorphism. As we noted in (1.16), any two modules over a simple ring of the same k-dimension are isomorphic, and it follows that V ≈ C^{n} as a C-module. Hence

End_{C}(V) ≈ End_{C}(C^{n}) = M_{n}(C^{opp}),

and on composing this isomorphism with f we obtain an isomorphism of k-algebras

(A ⊗_{k} B)^{opp} → M_{n}(C)^{opp}.

The same map can be interpreted as an isomorphism

A ⊗_{k} B → M_{n}(C).

Corollary 3.15. For any separable algebraic closure k^{al} of k, there is a canonical isomorphism Br(k) → H^{2}(k^{al}/k).

Proof. For any tower of fields E ⊃ L ⊃ k with E and L finite and Galois over k, the diagram

H^{2}(L/k) ⟶^{Inf} H^{2}(E/k)
↓  ↓
Br(L/k) ↪ Br(E/k)

commutes (the vertical maps send φ to [A(φ)]. Now use that

Br(k) = ∪ Br(L/k) (see 3.9),

and

H^{2}(k^{al}/k) = ∪H^{2}(L/k)

where both unions run over the finite Galois extensions L of k contained in k^{al}.

Corollary 3.16. For any field k, Br(k) is torsion, and for any finite extension L/k, Br(L/k) is killed by [L : k].

Proof. The same statements are true for the cohomology groups.

4. The Brauer Groups of Special Fields

The results of the last section allow us to interpret the results of Chapter III as statements concerning the Brauer group of a field. In this section, we shall derive the same results independently of Chapter III (but not quite of Chapter II).

Finite fields. Let k be a finite field. We saw in Chapter III (see III.2.3) that, for any finite extension L of k, H^{2}(L/k) = 0, and hence Br(k) = 0. The following is a more direct proof of this fact.

Theorem 4.1 (Wedderburn). Every finite division algebra is commutative.

Proof. Let D be a finite division algebra with centre k, and let [D : k] = n^{2}. Every element of D is contained in a subfield k[α] of D, and hence in a maximal subfield. Every maximal subfield of D has q^{n} elements. They are therefore isomorphic, and hence conjugate (Noether-Skolem). Therefore, for any maximal subfield L, D^{×} = ∪αL^{×}α^{−1}, but a finite group can not equal the union of the conjugates of a proper subgroup (the union of the conjugates has too few elements), and so D = L.

The real numbers. Let G = Gal(ℂ/ℝ) = {1, σ}. Then

H^{2}(ℂ/ℝ) ≈ H_{T}^{0}(G, ℂ^{×}) = ℝ^{×}/ Nm_{G}(ℂ^{×}) = {±}

and so Br(ℂ/ℝ) is a cyclic group of order 2. The nonzero element of H^{2}(ℂ/ℝ) is represented by the 2-cocycle φ : G × G → ℂ^{×},

φ(ρ, τ) = \{ \begin{matrix}−1 & if ρ = σ = τ \\ 1 & otherwise \end{matrix} .

Let H be the usual quaternion algebra over ℝ. Then the ℂ-linear map A(φ) → H sending x_{σ} to j is an isomorphism of ℝ-algebras. It follows that every central simple algebra over ℝ is isomorphic either to a matrix algebra over ℝ or to a matrix algebra over H.

A nonarchimedean local field. Let K be a nonarchimedean local field.

Let K be a local field, and let D be a central division algebra over K. Let n^{2} = [D : K].

For any subfield L of D containing K, the valuation | · | has a unique extension to L. Since any element α of D is contained in such a subfield of D, for example, in K[α], the valuation | · | has a unique extension to D. It is possible to verify that | · | is a nonarchimedean valuation on D in the obvious sense, i.e.,

(a) |α| = 0 ⟺ α = 0;

(b) for all α, β ∈ D, |αβ| = |α||β|;

(c) for all α, β ∈ D, |α + β| ≤ max{|α|, |β|}.

Let q be the number of elements in the residue field k of K, and define ord(α) for α ∈ D by the formula:

|α| = (1/q)^{ord(α)}.

Then ord extends the additive valuation ord_{K} on K (normalized to map K^{×} onto ℤ) to D. For any subfield L of D containing K, [L : K] ≤ n, and so ord(L^{×}) ⊂ n^{−1}ℤ. Hence also ord(D^{×}) ⊂ n^{−1}ℤ.

Let

O_{D} = {α ∈ D | ord(α) ≥ 0}
𝔓 = {α ∈ D | ord(α) > 0}.

Then O_{D} is a subring in D, called the ring of integers. For any subfield L of D containing K, O_{D} ∩ L = O_{L}, and so O_{D} consists precisely of the elements of D that are integral over O_{K}. Moreover 𝔓 is a maximal 2-sided ideal in O_{D} (obviously), and the powers of it are the only 2-sided ideals in D (the proof is the same as in the commutative case). Hence 𝔓^{e} = 𝔭O_{D} for some e. Then ord(D^{×}) = e^{−1}ℤ, and therefore e ≤ n.

Clearly, the elements of O_{D} not in 𝔓 are units. Therefore d =_{df} O_{D}/𝔓 is again a division algebra, and hence a field. Let f be its degree over k. Write d = k[a]. We can lift a to an element α of O_{D}. Because [K[α] : K] ≤ n, we have f ≤ n.

The same argument as in the commutative case shows that n^{2} = ef, namely, O_{D} is a free O_{K}-module of some rank m. Because O_{D} ⊗_{O_{K}} K = D, m = n^{2}. Moreover, because O_{D} ⊗_{O_{K}} k = O_{D}/𝔭O_{D}, it also is free of dimension of n^{2} over k. Now consider the filtration of k-vector spaces

O_{D} ⊃ 𝔓 ⊃ 𝔓^{2} ⊃ ⋯ ⊃ 𝔓^{e} = 𝔭O_{D}.

From our definition of f, O_{D}/𝔓 = d has dimension f as a k-vector space, and the successive quotients are one-dimensional vector spaces over d. Hence O_{D}/𝔭O_{D} has dimension ef over k, and so ef = n^{2}.

Because e ≤ n, f ≤ n, the equality ef = n^{2} implies that e = f = n. In particular, every central division algebra ≠ K is ramified. Again write d = k[a], and lift a to an element α ∈ D. Then K[α] is a field with residue field d, and so [K[α] : K] ≥ [d : k] = n. Therefore K[α] has degree n over K and is unramified. It is a maximal subfield, and hence splits D. We have shown that every element of Br(K) is split by an unramified extension, i.e., Br(K) is equal to its subgroup Br(K^{un}/K).

We next define the map

inv_{K} : Br(K) ⟶ ℚ/ℤ.

An element of Br(K) represented by a central division algebra D over K (unique up to isomorphism). According to what we have just proved, there is a maximal subfield L of D that is unramified over K. Let σ be the Frobenius automorphism of L. According to the Noether-Skolem theorem, there is an element α ∈ D such that σx = αxα^{−1} for all x ∈ L. If α′ also has this property, then α′ = cα for some c ∈ L, and so

ord(α′) = ord(c) + ord(α) ≡ ord(α) mod ℤ.

We define

inv_{K}(D) = ord(α) mod ℤ.

It depends only on the isomorphism class of D.

Example 4.2. Let L be the unramified extension of K of degree n, and let σ be the Frobenius automorphism of L/K, so that G =^{df} Gal(L/K) = {σ^{i} | 0 ≤ i ≤ n − 1}. Let φ be the 2-cocycle

φ(σ^{i}, σ^{j}) = \{ \begin{matrix}1 & if & i + j ≤ n − 1 \\ π & if & i + j > n − 1, \end{matrix}

where π is a prime element of K (see the discussion preceding III.2.9). The crossedproduct algebra A(φ) equals ⊕_{0≤i≤n−1}Le_{i} with the multiplication determined by

e_{i} · a = σ^{i}a · e_{i} all a ∈ L,

and

e_{i}e_{j} = \{ \begin{matrix}e_{i+j} & if & i + j ≤ n − 1 \\ πe_{i+j−n} & if & i + j > n − 1, \end{matrix}

We identify L with a subfield of A(φ) by identifying e_{0} with 1. Because e_{1}ae_{1}^{−1} = σa for a ∈ L, we can use e_{1} to compute the invariant of A(φ). According to the above rules, e_{1}^{n} = e_{n−1}e_{1} = πe_{0} = π. Hence

inv_{K}(A(φ)) = ord(e_{1}) = \frac{1}{n}ord(e_{1}^{n}) = \frac{1}{n}ord(π) = \frac{1}{n},

as expected.

Proposition 4.3. The map inv_{K} : Br(K) → ℚ/ℤ just defined is a bijection.

Proof. Let L be the unramified extension of K of degree n (contained in a fixed algebraic closure K^{al} of K), and let ℓ/k be the corresponding extension of residue fields. Because the norm maps ℓ → k, ℓ^{×} → k^{×} are surjective, and U_{L} has a filtration whose quotients are ℓ^{×} or ℓ one finds that the norm map U_{L} → U_{K} is surjective (see III.2.2). Therefore, H_{T}^{0}(G, U_{L}) = 0, and (because the cohomology of cyclic groups is periodic) this implies that H^{2}(G, U_{L}) = 0. As L^{×} = U_{L} × π^{ℤ} for any prime element π of K,

H^{2}(L/K) = H^{2}(G, π^{ℤ}).

Consideration of the cohomology sequence of

0 → ℤ → ℚ → ℚ/ℤ → 0

shows that H^{2}(G, π^{ℤ}) is cyclic of order n and is generated by the class of the cocycle φ considered in the last example (see the discussion preceding III.2.9). Therefore, Br(L/K) is cyclic of order n, and it is generated by [A(φ)]. It now follows that inv_{K} : Br(K^{un}/K) → ℚ/ℤ is an isomorphism, and we saw above that Br(K^{un}/K) = Br(K^{al}/K).

Remark 4.4. (a) The calculation in Example 4.2 shows that the invariant map

defined in this chapter agrees with that in the preceding chapter.

(b) A calculation as in Example 4.2 shows that inv_{K}(A(φ^{i}) = \frac{i}{n} mod ℤ. I claim that if i is relatively prime to n, then A(φ^{i}) is a division algebra. If not, then A(φ^{i}) ∼ M_{r}(D) for a division algebra D of degree m^{2} some m < n, and inv_{K}(A(φ^{i})) = inv_{K}(D) ∈ \frac{1}{m}ℤ/ℤ, which is a contradiction. It follows that each division algebra over K is isomorphic to exactly one division algebra of the form A(φ^{i}) for some n ≥ 1 and some i relatively prime to n. In particular, for a division algebra D, the order of [D] in Br(K) is \sqrt{[D : K]}.

(c) Let D be a division algebra of degree n^{2} over K. Because the map Br(K) → Br(L) multiplies the invariant by [L : K] (Theorem III.1.1), D is split by every extension L of K of degree n. Therefore (3.7), every such L can be embedded into D. Hence every irreducible polynomial in K[X] of degree n has a root in D.

5. Complements

Semisimple algebras. A k-algebra A is said to be semisimple if every A-module is semisimple. Theorem 1.15 proves that a simple k-algebra is semisimple, and (yet another) theorem of Wedderburn shows that every semisimple k-algebra is a product of simple k-algebras. For any finite group G, the group algebra k[G] is semisimple provided (G : 1) is not divisible by the characteristic of k. For more on semisimple k-algebras, see the references below.

Algebras, cohomology, and group extensions. Let A be a central simple algebra of degree n^{2} over k, and assume that A contains a field L of degree n over k (e.g., A a division algebra). Let E be the set of invertible elements α ∈ A such that αLα^{−1} = L. Then each α ∈ E defines an element x ↦ αxα^{−1} of Gal(L/K), and the Noether-Skolem theorem implies that every element of Gal(L/K) arises from an α ∈ E. Because [L : k] = \sqrt{[A : k]}, the centralizer of L is L itself, and so the sequence

1 → L^{×} → E^{×} → Gal(L/K) → 1

is exact. It is not difficult to show that the map sending A to this sequence defines an isomorphism from A(L/K) to the set of isomorphism classes of extensions of Gal(L/K) by L^{×}, and hence to H^{2}(L/k) (see II.1.18). See Serre 1950/51.

Brauer groups and K-theory. Let k be a field containing a primitive nth root ζ of 1. To any elements a, b ∈ k^{×}, one attaches the k-algebra A(a, b; ζ) having generators i and j and relations

i^{n} = a, j^{n} = b, ij = ζji.

It is a central simple algebra over k.

The (Milnor) K-group K_{2}F of a field F is the quotient of F^{×} ⊗_{ℤ}F^{×} by the abelian group generated by the elements of the form u ⊗ (1 − u) with u an element of F^{×} such that 1 − u ∈ F^{×}. Thus K_{2}F has as generators pairs {a, b}, one for each pair of elements in F^{×}, and relations

{ab, c} = {a, c} + {b, c}
{a, bc} = {a, b} + {a, c}
{u, 1 − u} = 0.

It is known that these relations imply that

{u, v} = {v, u}^{−1}
{u, −u} = 1

(see J. Rosenberg, Algebraic K-Theory and Its Applications, Springer, 1994, p214).

It is not difficult to show that the A(a, b; ζ), considered as elements of Br(k) satisfy these relations, and so there is a well-defined homomorphism

K_{2}k → Br(k).

Remarkably, it has been proved (Theorem of Merkuryev-Suslin, early 1980s) that this map defines an isomorphism from K_{2}k/nK_{2}k onto the subgroup of Br(k) of elements killed by n, and so we have an explicit description of Br(k)_{n} in terms of generators and relations. This theorem is discussed in the book (Kersten, I., Brauergruppen von Körpern, Vieweg, 1990).

Exercise 5.1. Let F be a field of characteristic ≠ 2, and define the quaternion algebra H(a, b) as in (1.7). Thus H(a, b) has basis 1, i, j, k and i^{2} = a, j^{2} = b, ij = −ji. It is a central simple algebra over F.

(a) Show that every 4-dimensional central simple algebra over k is isomorphic to H(a, b) for some a, b ∈ F^{×}.

(b) According to Wedderburn’s theorem, either H(a, b) ≈ M_{2}(F) or H(a, b) is a division algebra. Show that the first case occurs if and only if w^{2} − a^{2}x^{2} − b^{2}y^{2} + abz^{2} has a nontrivial zero in K. (Hint: for α = w + xi + yj + zk, let ᾱ = w − xi − yj − zk, and note that αᾱ = w^{2} − a^{2}x^{2} − ⋯)

(c) Show that H(1, 1) ≈ M_{2}(F). (Hint: consider the matrices e_{12} + e_{21} and e_{11} − e_{22}.)

(d) Show that H(a, b) ≈ H(ax^{2}, by^{2}) any x, y ∈ F^{×}.

(e) Show that H(a, b) ⊗_{F} L is the quaternion algebra defined by a, b ∈ L^{×}.

(f) Verify that H(a, b) is in fact central simple over F.

(g) Show that H(a, 1 − a) ≈ M_{2}(F), provided a, 1 − a ∈ F^{×}.

(h) Show that H(1, b) ≈ H(a, −a) ≈ M_{2}(F) (Hint: consider j + k and i + j.)

(i) Show that H(a, b) ≈ H(a, b)^{opp}.

(j) Show that H(a, b) ≈ M_{2}(k) if and only if a ∈ Nm(F [\sqrt{b}]).

(k) Show that the map {a, b} ↦ [H(a, b)] : K_{2}F → Br(F) is well-defined.

Notes. Brauer groups were introduced and studied by R. Brauer, E. Noether, A. Albert, H. Hasse, and others, starting in the nineteen-twenties. The classic accounts are: Deuring, M., Algebren, Springer, 1935. Artin, E., Nesbitt, C., and Thrall, R., Rings with Minimum Condition, University of Michigan. Press, 1944.

Apart from the quaint terminology (e.g., Kronecker products for tensor products), the latter is still an excellent book.

Other books include: Blanchard, A., Les Corps Non Commutatifs, PUP, 1972.

This gives a concise elementary treatment of material in the chapter. Herstein, I., Noncommutative Rings, Carus, 1968. Farb, B., and Dennis, R.K., Noncommutative Algebra, Springer, 1993.

These books include the Brauer group, but also cover much more (but no number theory). The second has lots of exercises.

Much of this chapter is based on: Serre, J-P., Applications algébriques de la cohomologie des groupes, I, II, Séminaire Henri Cartan, 1950/51.

CHAPTER V Global Class Field Theory: Statements of the Main Theorems

La théorie du corps de classes a une réputation de difficulté qui est en partie justifiée. Mais il faut faire une distinction: il n’est peut-être pas en effet dans la science de théorie où tout à la fois les démonstrations soient aussi ardues, et les résultats d’une aussi parfaite simplicité et d’une aussi grande puissance.

J. Herbrand, 1936, p2.

In this chapter, we state and explain the theorems of global class field theory. The main theorems will be proved Chapter VII.

Throughout this chapter, K will be a number field, although most of the results hold also for finite extensions of 𝔽_{p}(T).

Recall that for a number field K, we define a prime of K to be an equivalence class of nontrivial valuations of K. There are two types of primes: the finite primes, which can be identified with the prime ideals of O_{K}, and the infinite primes. A real infinite prime can be identified with an embedding of K into ℝ, and a complex infinite prime can be identified with a conjugate pair of embeddings of K into ℂ. We use 𝔭 or v to denote a prime, finite or infinite. We use S denote a finite set of primes of K, and also the set of primes of a finite extension L of K lying over K. The set of infinite primes is denoted by S_{∞}.

The completion of K at a prime 𝔭 (resp. v) is denoted by K_{𝔭} (resp. K_{v}), and the inclusion K ↪ K_{𝔭} (resp. K ↪ K_{v}) is denoted a ↦ a_{𝔭} (resp. a ↦ a_{v}).

1. Ray Class Groups

Ideals prime to S. Let I = I_{K} be the group of fractional ideals in K. For a finite set S of primes of K, we define I^{S} to be the subgroup of I generated by the prime ideals not in S. Each element 𝔞 of I^{S} factors uniquely as

𝔞 = 𝔭_{1}^{n_{1}} ⋯ 𝔭_{s}^{n_{s}}, 𝔭_{i} ∉ S, n_{i} ∈ ℤ,

and so I^{S} can be identified with the free abelian group generated by the prime ideals not in S. Define

K^{S} = {a ∈ K^{×} | (a) ∈ I^{S}} = {a ∈ K^{×} | ord_{𝔭}(a) = 0 all finite 𝔭 ∈ S}.

There is a natural map i : K^{S} → I^{S} sending an element a of K^{S} to the ideal aO_{K}. For example, if K = ℚ and S is the set of prime numbers dividing n, then I^{S} can be identified with

{r/s | r, s ∈ ℤ, r, s > 0, gcd(r, n) = 1 = gcd(s, n)}

and

ℚ^{S} = {r/s | r, s ∈ ℤ, gcd(r, n) = 1 = gcd(s, n)}.

In this case, the natural map ℚ^{S} → I^{S} is surjective with kernel {±1}.

Lemma 1.1. For any finite set S of prime ideals in O_{K}, the sequence

0 → U_{K} → K^{S} → I^{S} → C → 0

is exact. (Here U_{K} = O_{K}^{×} and C is the full ideal class group I/i(K^{×}).)

Proof. We first show that every ideal class C is represented by an ideal in I^{S}. Let 𝔞 = 𝔟𝔠^{−1} with 𝔟 and 𝔠 integral ideals. For any nonzero c ∈ 𝔠, 𝔠|(c), and so 𝔞(c) = 𝔟(c)𝔠^{−1} is an integral ideal. Therefore, C will be represented by an integral ideal 𝔞. Write 𝔞 = ∏_{𝔭∈S} 𝔭^{n(𝔭)}𝔟 where 𝔟 ∈ I^{S}. For each 𝔭 ∈ S, choose a π_{𝔭} ∈ 𝔭 \ 𝔭^{2}, so that ord_{𝔭}(π_{𝔭}) = 1. By the Chinese Remainder Theorem, there exists an a ∈ O_{K} such that

a ≡ π_{𝔭}^{n(𝔭)} mod 𝔭^{n(𝔭)+1}

for all 𝔭 ∈ S. These congruences imply that ord_{𝔭}(a) = n(𝔭) for all 𝔭 ∈ S, and so (a) = ∏_{𝔭∈S} 𝔭^{n(𝔭)}𝔟′ with 𝔟′ ∈ I^{S}. Now a^{−1}𝔞 ∈ I^{S} and represents the same class as 𝔞 in C.

We have shown that I^{S} → C is surjective. If 𝔞 ∈ I^{S} maps to zero in C, then 𝔞 = (α) for some α ∈ K^{S}, and α is uniquely determined up to a unit.

Remark 1.2. In fact, every class in C is represented by an integral ideal 𝔞 in I^{S}: suppose the class is represented by 𝔞 ∈ I^{S}; write 𝔞 = 𝔟𝔠^{−1} with 𝔟 and 𝔠 integral ideals in I^{S}, choose a nonzero c ∈ 𝔠 ∩ K^{S} (exists by the Chinese remainder theorem), and note that c𝔞 is integral.

Moduli.

Definition 1.3. A modulus for K is a function

m : {primes of K} → ℤ

such that

(a) m(𝔭) ≥ 0 for all primes 𝔭, and m(𝔭) = 0 for all but finitely many 𝔭;

(b) if 𝔭 is real, then m(𝔭) = 0 or 1;

(c) if 𝔭 is complex, then m(𝔭) = 0.

Traditionally, one writes

𝔪 = ∏_{𝔭}𝔭^{m(𝔭)}.

A modulus 𝔪 = ∏ 𝔭^{m(𝔭)} is said to divide a modulus 𝔫 = ∏_{𝔭} 𝔭^{n(𝔭)} if m(𝔭) ≤ n(𝔭) for all 𝔭. In particular, a prime 𝔭 divides a modulus 𝔪 if and only if m(𝔭) > 0.

A modulus 𝔪 can be written

𝔪 = 𝔪_{∞}𝔪_{0}

where 𝔪_{∞} is a product of real primes and 𝔪_{0} is product of positive powers of prime ideals, and hence can be identified with an ideal in O_{K}.

RAY CLASS GROUPS 115

The ray class group. For a modulus 𝔪, define K_{𝔪,1} to be the set of a ∈ K^{×} such that

\{ \begin{matrix}ord_{𝔭}(a − 1) & ≥ & m(𝔭) all finite 𝔭 dividing 𝔪 \\ a_{𝔭} & > & 0 all real 𝔭 dividing 𝔪. \end{matrix}

Note that ord_{𝔭}(a − 1) ≥ m(𝔭) ⟺ π^{m(𝔭)}|(a_{𝔭} − 1) ⟺ a ↦ 1 in (O_{𝔭}/𝔭^{m(𝔭)})^{×} = (Ô_{𝔭}/𝔭̂^{m(𝔭)})^{×} where π is a prime element in the completion K_{𝔭} at 𝔭. Let

S(𝔪) = {primes dividing 𝔪}.

For any a ∈ K_{𝔪,1} and prime ideal 𝔭 dividing 𝔪, ord_{𝔭}(a − 1) > 0 = ord_{𝔭}(1), and so

ord_{𝔭}(a) = ord_{𝔭}((a − 1) + 1) = 0.

Therefore, for any a ∈ K_{𝔪,1}, the ideal i(a) =^{df} (a) lies in I^{S(𝔪)}. The quotient

C_{𝔪} = I^{S(𝔪)}/i(K_{𝔪,1})

is called the ray class group modulo 𝔪.

Example 1.4. The expression 𝔪 = (2)^{3} · (17)^{2} · (19) · ∞ is a modulus for ℚ with 𝔪_{0} = (2)^{3} · (17)^{2} · (19) and 𝔪_{∞} = ∞ (here ∞ denotes the unique infinite prime of ℚ). Moreover, ℚ_{𝔪,1} consists of the positive rational numbers a such that

\{ \begin{matrix}ord_{2}(a − 1) & ≥ & 3 \\ ord_{17}(a − 1) & ≥ & 2 \\ ord_{19}(a − 1) & ≥ & 1 \end{matrix} .

The condition at 2 says that a is the quotient of two odd integers, a = b/c, and that the image of bc^{−1} in (ℤ/8ℤ)^{×} is 1. The other conditions can be expressed similarly.

Theorem 1.5. For any modulus 𝔪 of K, there is an exact sequence

0 → U/U_{𝔪,1} → K_{𝔪}/K_{𝔪,1} → C_{𝔪} → C → 0

and canonical isomorphisms

K_{𝔪}/K_{𝔪,1} ≅ ∏ ; \begin{matrix}𝔭 real \\ 𝔭|𝔪 \end{matrix} \{±\} × ∏ ; \begin{matrix}𝔭 finite \\ 𝔭|𝔪 \end{matrix} (O_{K}/𝔭^{m(𝔭)})^{×} ≅ ∏ ; \begin{matrix}𝔭 real \\ 𝔭|𝔪 \end{matrix} \{±\} × (O_{K}/𝔪_{0})^{×},

where

K_{𝔪} = K^{S(𝔪)} = {α ∈ K^{×} | ord_{𝔭}(α) = 0 for all 𝔭|𝔪_{0}}
U = O_{K}^{×}, the group of units in K,
U_{𝔪,1} = U ∩ K_{𝔪,1}.

Therefore, C_{𝔪} is a finite group of order

h_{𝔪} = h · (U : U_{𝔪,1})^{−1} · 2^{r_{0}} · ℕ(𝔪_{0}) · ∏_{𝔭|𝔪_{0}}(1 − \frac{1}{ℕ𝔭})

where r_{0} is the number of real primes dividing 𝔪 and h is the class number of K (order of C).

Proof. The inclusion I^{S(𝔪)} → I defines a homomorphism C_{𝔪} → C. Consider the pair of maps

K_{𝔪,1} ⟶^{f} K_{𝔪} ⟶^{g} I^{S(𝔪)}.

According to the Lemma 1.1, the kernel and cokernel of g are U and C respectively. The cokernel of g ∘ f is C_{𝔪} (by definition) and its kernel is K_{𝔪,1} ∩ U = U_{𝔪,1}. Finally, f is injective. Therefore, the kernel-cokernel sequence (see II.4.2) of the pair of maps is

0 → U_{𝔪,1} → U → K_{𝔪}/K_{𝔪,1} → C_{𝔪} → C → 0.

We next prove that K_{𝔪} is canonically isomorphic to the given groups.

Lemma 1.6. Let S be a finite set of prime ideals of K. Then any element α ∈ K^{S} can be written α = a/b with a, b ∈ O_{K} ∩ K^{S}.

Proof. Because α ∈ K^{S}, (α) = 𝔞/𝔟 with 𝔞, 𝔟 integral ideals in I^{S}. Clearly 𝔞 and 𝔟 represent the same element C of the ideal class group, and according to Remark 1.2 we can choose an integral ideal 𝔠 in I^{S} to represent C^{−1}. Now (α) = 𝔞𝔠/𝔟𝔠 = (a)/(b) for some a, b ∈ O_{K} ∩ K^{S}.

Let 𝔭 be a prime dividing 𝔪. If 𝔭 is real, we map α ∈ K_{𝔪} to the sign of α_{𝔭} (recall that a real prime is an embedding K ↪ ℝ, and that α_{𝔭} denotes the image of α under the embedding). If 𝔭 is finite, i.e., it is a prime ideal in O_{K}, then we map α ∈ K_{𝔪} to [a][b]^{−1} ∈ (O_{K}/𝔭^{m(𝔭)})^{×} where a, b are as in the lemma. As a and b are relatively prime to 𝔭, their classes [a] and [b] in O_{K}/𝔭^{m(𝔭)} are invertible, and so this makes sense. The weak approximation theorem (6.3) shows that the map K_{𝔪} → ∏{±}×∏(O_{K}/𝔭^{m(𝔭)})^{×} is surjective, and its kernel is obviously K_{𝔪,1}.

The Chinese Remainder Theorem shows that there is an isomorphism of rings

O_{K}/𝔪_{0} ≅ ∏_{𝔭|𝔪}O_{K}/𝔭^{m(𝔭)},

and hence an isomorphism of groups

(O_{K}/𝔪_{0})^{×} ≅ ∏(O_{K}/𝔭^{m(𝔭)})^{×}.

This completes proof of the isomorphisms. It remains to compute the orders of the groups. Note that O_{K}/𝔭^{m} is a local ring with maximal ideal 𝔭/𝔭^{m} (because its ideals correspond to the ideals of O_{K} containing 𝔭^{m}), and so its units are the elements not in 𝔭/𝔭^{m}. The filtration

(O_{K}/𝔭^{m})^{×} ⊃ (1 + 𝔭)/𝔭^{m} ⊃ ⋯ ⊃ (1 + 𝔭^{m−1})/𝔭^{m} ⊃ 0

has quotients isomorphic to

k^{×}, k, …, k, k =^{df} O_{K}/𝔭,

and so (O_{K}/𝔭^{m})^{×} has order (q − 1)q^{m−1}, q = (O_{K} : 𝔭) =^{df} ℕ𝔭. This shows that

(C_{𝔪} : 1) = (C : 1) · (K_{𝔪} : K_{𝔪,1}) · (U_{𝔪} : U_{𝔪,1})^{−1}

is equal to the expression in the statement of the theorem. □

Example 1.7. (a) If 𝔪 = 1, then C_{𝔪} = C.

(b) When 𝔪 is the product of the real primes, C_{𝔪} is the narrow class group and there is an exact sequence

0 → U/U_{+} → K^{×}/K_{+} → C_{𝔪} → C → 1

where K_{+} is the group of totally positive elements (i.e., positive under all real embeddings) and U_{+} is the group of all totally positive units. Moreover, K^{×}/K_{+} ≅ ∏_{𝔭 real}{±}, and so the kernel of C_{𝔪} → C is the set of possible signs modulo those arising from units.

For ℚ, the narrow class group is trivial. For ℚ[\sqrt{d}], d > 0, there are two real primes, and U = {±ε^{m} | m ∈ ℤ} ≈ (ℤ/2ℤ) × ℤ where ε is a fundamental unit. Let ε̄ be the complex conjugate of ε. Then h_{𝔪} = h or 2h according as ε and ε̄ have the same or different signs. Note that Nm(ε) = +1 if the signs are the same and −1 if they differ. For small values of d we have

\begin{matrix}d & h & ε & Nm(ε) \\ 2 & 1 & 1 + \sqrt{2} & −1 \\ 3 & 1 & 2 + \sqrt{3} & 1 \\ 5 & 1 & (1 + \sqrt{5})/2 & −1 \\ 6 & 1 & 5 + 2\sqrt{6} & 1 \end{matrix}

Therefore, ℚ[\sqrt{3}] and ℚ[\sqrt{6}] have class number 1 but narrow class number 2, whereas for ℚ[\sqrt{2}] and ℚ[\sqrt{5}] both class numbers are 1.

(c) For the field ℚ and the modulus (m), the sequence becomes

0 → {±1} → (ℤ/mℤ)^{×} → C_{𝔪} → 0.

For the modulus ∞(m), the sequence becomes

0 → {±1} → {±} × (ℤ/mℤ)^{×} → C_{𝔪} → 0.

Here −1 maps to (−, [−1]), and the subgroup (ℤ/mℤ)^{×} of the product maps isomorphically onto the quotient C_{𝔪}.

The Frobenius element. Let K be a number field, and let L be a finite Galois extension of K with group G. Let 𝔭 be an ideal of K, and let 𝔓 be an ideal of L lying over it. The decomposition group D(𝔓) (or G(𝔓)) is defined to be

{τ ∈ G | τ𝔓 = 𝔓}.

Equivalently, it is the set of elements of G that act continuously for the 𝔓-adic topology, and so extend by continuity to an automorphism of the completion L_{𝔓}. In this way we obtain an isomorphism

D(𝔓) ⟶ Gal(L_{𝔓}/K_{𝔭}).

Assume 𝔓 is unramified over 𝔭. Then the action of Gal(L_{𝔓}/K_{𝔭}) on O_{L} induces an isomorphism

Gal(L_{𝔓}/K_{𝔭}) ⟶ Gal(l/k)

where l and k are the residue fields. Pictorially:

𝔓  L — L_{𝔓} — l
f | | | D(𝔓)
𝔓_{D} L^{D(𝔓)} — K_{𝔭} — k
g  | ⧸
𝔭  K

The group Gal(l/k) is cyclic with a canonical generator, namely, the Frobenius element x ↦ x^{q} where q is the number of elements of k. Hence D(𝔓) is cyclic, and the generator of D(𝔓) corresponding to the Frobenius element in Gal(l/k) is called the Frobenius element (𝔓, L/K) at 𝔓. It is the unique element σ of Gal(L/K) satisfying the following two conditions:

(a) σ ∈ D(𝔓), i.e., σ𝔓 = 𝔓;

(b) for all α ∈ O_{L}, σα ≡ α^{q} mod 𝔓, where q is the number of elements the residue field O_{K}/𝔭, 𝔭 = 𝔓 ∩ K.

We now list the basic properties of (𝔓, L/K).

1.8. Let τ𝔓 be a second prime dividing 𝔭. Then D(τ𝔓) = τD(𝔓)τ^{−1}, and

(τ𝔓, L/K) = τ(𝔓, L/K)τ^{−1}.

Proof. If ρ ∈ D(𝔓), then

τρτ^{−1}(τ𝔓) = τρ𝔓 = τ𝔓,

and so τρτ^{−1} ∈ D(τ𝔓). Thus τD(𝔓)τ^{−1} ⊂ D(τ𝔓), and since they have the same order, they must be equal.

Let α ∈ O_{L} and let σ = (𝔓, L/K); then

τστ^{−1}(α) = τ((τ^{−1}α)^{q} + a), some a ∈ 𝔓, and
τ((τ^{−1}α)^{q} + a) = α^{q} + τa ≡ α^{q}modτ𝔓.

As G acts transitively on the primes dividing 𝔭, this implies that

{(𝔓, L/K) | 𝔓|𝔭}

is a conjugacy class in G, which we denote (𝔭, L/K). When L/K is abelian, (𝔭, L/K) contains a single element, and we regard it as an element of Gal(L/K) (rather than a set consisting of a single element).

1.9. Consider a tower of fields

\begin{matrix}M & 𝔔 \\ | \\ L & 𝔓 \\ | \\ K & 𝔭 \end{matrix}

and assume that 𝔔 is unramified over 𝔭; then

(𝔔, M/L) = (𝔔, M/K)^{f(𝔓/𝔭)}.

L-SERIES 119

Proof. Let k(𝔔) ⊃ k(𝔓) ⊃ k(𝔭) be the tower of residue fields. Then f(𝔓/𝔭) =^{df} [k(𝔓) : k(𝔭)], and the Frobenius element in Gal(k(𝔔)/k(𝔓)) is the f(𝔓/𝔭)^{th} power of the Frobenius element in Gal(k(𝔔)/k(𝔭)). The rest is straightforward.

1.10. In ( 1.9), assume that L is Galois over K; then

(𝔔, M/K)|L = (𝔓, L/K).

Proof. Clearly (𝔔, M/K)|L satisfies the conditions characterizing (𝔓, L/K).

Let L_{1} and L_{2} be Galois extensions of K contained in some field Ω, and let M = L_{1} · L_{2}. Then M is Galois over K, and there is an injective homomorphism

σ ↦ (σ|L_{1}, σ|L_{2}): Gal(M/K) → Gal(L_{1}/K)) × Gal(L_{2}/K).

1.11. Let 𝔔 be a prime ideal of O_{M}, and let 𝔓_{i} = 𝔔 ∩ O_{L_{i}}. Under the above map,

(𝔔, M/K) ↦ (𝔓_{1}, L_{1}/K) × (𝔓_{2}, L_{2}/K).

Proof. Apply (1.10).

Note that 𝔭 splits completely in L if and only if (𝔓, L/K) = 1 for one (hence all) primes 𝔓 lying over it. Hence, in the situation of (1.11), 𝔭 splits completely in M if and only if it splits completely in L_{1} and L_{2}.

2. Dirichlet L-Series and the Density of Primes in Arithmetic Progressions

We begin by briefly reviewing the elementary theory of Dirichlet L-series (see, for example, J.-P. Serre, Cours d’Arithmétique, PUP, 1970, Chapter VI).

Let m be an integer. A Dirichlet character modulo m is a homomorphism χ : (ℤ/mℤ)^{×} → ℂ^{×}. Because (ℤ/mℤ)^{×} is finite, χ([n]) is a root of 1 for all n. A Dirichlet character modulo m can be regarded as a multiplicative function on the set of integers prime to m whose value at n depends only on n mod m. Often one extends χ to a function on all the integers by setting χ(n) = 0 when gcd(m, n) ≠ 1. The Dirichlet character modulo m that takes the value 1 for all integers prime to m is called the principal Dirichlet character χ_{0}.

To a Dirichlet character χ modulo m, one attaches a Dirichlet series

L(s, χ) = ∏_{p∤m} \frac{1}{1 − χ(p)p^{−s}} = ∑_{n>0} χ(n)/n^{s}.

Both expressions converge for s a complex number with ℜ(s) > 1—their equality is the analytic expression of the unique factorization. Note that L(s, χ_{0}) differs from the Riemann zeta function ζ(s) only in that it is missing the factors \frac{1}{1−p^{−s}} for p dividing m.

Theorem 2.1. (a) The zeta function ζ(s) extends to a meromorphic function on the half-plane ℜ(s) > 0, and

ζ(s) = \frac{1}{s − 1} + φ(s)

where φ(s) is holomorphic for ℜ(s) > 0.

(b) If χ ≠ χ_{0}, then the series for L(s, χ) converges for ℜ(s) > 0 and L(1, χ) ≠ 0.

Proof. Serre, ibid. Propositions 10, 12, Théor̀eme 1.

On applying log to the equality in (a), one finds that

∑ 1/p^{s} ∼ log \frac{1}{1 − s} as s ↓ 1.

By this we mean that the quotient ^{\frac{∑ 1/p^{s}}{− log(1−s)}} converges to 1 as s approaches 1 through real numbers > 1. This result makes reasonable the definition that a set T of primes has Dirichlet density δ if

∑_{p∈T} 1/p^{s} ∼ δ log \frac{1}{1 − s} as s ↓ 1.

Define f_{χ}(s) = ∑_{p∤m} χ(p)/p^{s}. Then (2.1b) shows that, for χ ≠ χ_{0}, f_{χ}(s) is bounded near s = 1. An elementary argument (Serre, ibid., Lemme 9) shows that, for any a prime to m,

∑_{p≡a mod m} 1/p^{s} = \frac{1}{φ(m)} ∑_{χ} χ(a)^{−1}f_{χ}(s),

where φ(m) = #(ℤ/mℤ)^{×} and the sum is over all Dirichlet characters modulo m.

Theorem 2.2. For any a prime to m, the primes in the arithmetic progression

…, a − 2m, a − m, a, a + m, a + 2m, …

have Dirichlet density 1/φ(m).

Proof. For χ ≠ χ_{0}, f_{χ}(s) remains bounded near s = 1, and so

∑_{p≡a mod m} 1/p^{s} ∼ \frac{1}{φ(m)}χ_{0}(a)^{−1}f_{χ_{0}}(s) ∼ \frac{1}{φ(m)} log \frac{1}{1 − s} as s ↓ 1.

Corollary 2.3. For any m, the set of primes splitting in the cyclotomic field ℚ[ζ_{m}] has Dirichlet density 1/φ(m).

Proof. A prime ideal (p) splits in ℚ[ζ_{m}] if and only if p ≡ 1 mod m.

We now explain how the above results generalize to arbitrary number fields. Proofs will be given in Chapter VI.

Let K be a number field, and let 𝔪 be modulus for K. A Dirichlet (or Weber) character modulo 𝔪 is a homomorphism χ : C_{𝔪} → ℂ^{×}—again, its values are roots of 1. Alternatively, a Dirichlet character is a multiplicative function I^{S} → ℂ^{×} that is zero on i(K_{𝔪,1}) for some modulus 𝔪 with S(𝔪) = S. The principal Dirichlet character modulo 𝔪 is the function χ_{0} : C_{𝔪} → ℂ^{×} taking only the value 1.

To a Dirichlet character χ modulo 𝔪, one attaches a Dirichlet series

L(s, χ) = ∏_{𝔭∤𝔪} \frac{1}{1 − χ(𝔭)ℕ𝔭^{−s}} = ∑_{(𝔞,𝔪_{0})=O_{K}} χ(𝔞)/ℕ𝔞^{s}.

The product is over the prime ideals relatively prime to 𝔪_{0}, and the sum is over ideals in O_{K} relatively prime to 𝔪_{0}. Again, both expressions converge for ℜ(s) > 1, and their equality is the analytic expression of the unique factorization of ideals. The L-series L(s, χ_{0}) differs by only a finite number of factors from the Dedekind zeta function

ζ_{K}(s) =^{df} ∏_{𝔭} \frac{1}{1 − χ(𝔭)ℕ𝔭^{−s}} = ∑_{𝔞⊂O_{K}} χ(𝔞)/ℕ𝔞^{s}.

Theorem 2.4. (a) The zeta function ζ_{K}(s) extends to a meromorphic function on the half-plane ℜ(s) > 0, and

ζ_{K}(s) ∼ \frac{2^{r_{1}}(2π)^{r_{2}}Reg(K)}{w_{K}|Δ_{K/ℚ}|^{1/2}}h_{K} \frac{1}{s − 1} as s ↓ 1

where r_{1} and r_{2} are the numbers of real and complex primes of K respectively, Reg(K) is the regulator of K (see Math 676), w_{K} is the number of roots of 1 in K, Δ_{K/ℚ} is the discriminant of K/ℚ, and h_{K} is the class number.

(b) If χ ≠ χ_{0}, then the series for L(s, χ) converges for ℜ(s) > 0 and L(1, χ) ≠ 0.

The proof of (b) uses the Existence Theorem (see 3.6).

Again, on applying log to the equality in (a), one finds that

∑_{𝔭}1/ℕ𝔭^{s} ∼ log \frac{1}{1 − s} as s ↓ 1,

and one says that a set T of prime ideals in T has Dirichlet density δ if

∑_{𝔭∈T} 1/ℕ𝔭^{s} ∼ δ log \frac{1}{1 − s} as s ↓ 1.

A similar argument to that in the previous case proves:

Theorem 2.5. For any ideal 𝔞 relatively prime to 𝔪_{0}, the prime ideals in O_{K} whose class in C_{𝔪} is [𝔞] have Dirichlet density 1/h_{𝔪}.

The analysis in the proofs of Theorems 2.4 and 2.5 is the same as in the case K = ℚ, but the number theory is much more difficult.

3. The Main Theorems in Terms of Ideals

The Artin map. Let L/K be an abelian extension Galois group G. Recall that, for a prime ideal 𝔭 of K that is unramified in L, there is a Frobenius automorphism σ = (𝔭, L/K) of L uniquely determined by the following condition: for any prime ideal 𝔓 of L lying over 𝔭, σ𝔓 = 𝔓, and σα ≡ α^{ℕ𝔭} mod 𝔓.

For any finite set S of primes of K containing all primes that ramify in L, we have a homomorphism

ψ_{L/K} : I^{S} → Gal(L/K), 𝔭_{1}^{n_{1}} ⋯ 𝔭_{t}^{n_{t}} ↦ ∏(𝔭_{i}, L/K)^{n_{i}}

called the global Artin map (or reciprocity map).

Example 3.1. Let K = ℚ[\sqrt{m}] where m is a square-free integer. The set S of finite primes ramifying in K consists of the primes dividing m if m ≡ 1 mod 4 and the primes dividing m together with 2 otherwise. Identify Gal(K/ℚ) with {±1}. The Artin map is the homomorphism determined by

p ↦ (\frac{m}{p}) : I^{S} → Gal(K/ℚ)

where (\frac{m}{p}) is the quadratic residue symbol.

Example 3.2. Let L = ℚ[ζ_{n}] where n is a primitive n^{th} root of 1. Assume that n is odd or divisible by 4 (so that the primes ramifying in L are precisely the primes dividing n). The map sending an integer m prime to n to the automorphism ζ ↦ ζ^{m} of L is an isomorphism Gal(L/ℚ) ⟶^{≈} (ℤ/nℤ)^{×}. For p not dividing n, (p, L/K) = [p] (see 0.8). If r and s are positive integers prime to n, then r/s defines a class [r/s] = [r][s]^{−1} ∈ (ℤ/nℤ)^{×}, and the Artin map is the composite of

I^{S} ⟶^{(r/s)↦[r/s]} (ℤ/nℤ)^{×} ⟶^{[m]↦(ζ↦ζ^{m})} Gal(L/ℚ).

Recall (Math 676, p63) that for any finite extension of number fields L/K, the norm map Nm_{L/K}: I_{L} → I_{K} from the group of fractional ideals of L to the similar group for K, is the unique homomorphism such that, for any prime ideal 𝔓 of L, Nm_{L/K}(𝔓) = 𝔭^{f(𝔓/𝔭)} where 𝔭 = 𝔓 ∩ O_{K}. For any α ∈ L, Nm_{L/K}((α)) = (Nm_{L/K} α).

Proposition 3.3. Let L be an abelian extension of K, and let K′ be any intermediate field: L ⊃ K′ ⊃ K. Then the following diagram commutes:

I_{K′}^{S} ⟶^{ϕ_{L/K′}} Gal(L/K′)
↓^{Nm}  ↓^{inclusion}
I_{K}^{S} ⟶^{ϕ_{L/K}} Gal(L/K).

Here S is any finite set of prime ideals of K containing all those that ramify in L, and also the set of primes of K′ lying over a prime in S.

Proof. Let 𝔭′ be any prime ideal of K′ lying over a prime ideal 𝔭 of K not in S. Then Nm_{L/K}(𝔭′) = 𝔭^{f(𝔭′/𝔭)}, and we have to show that ψ_{L/K′}(𝔭′) = ψ_{L/K}(𝔭^{f(𝔭′/𝔭)}), i.e., that (𝔓, L/K′) = (𝔓, L/K)^{f(𝔭′/𝔭)} for any prime ideal 𝔓 of L lying over 𝔭. But this was proved in (1.9).

Corollary 3.4. For any abelian extension L of K,

Nm_{L/K}(I_{L}^{S}) ⊂ Ker(ψ_{L/K} : I^{S} → Gal(L/K)).

Proof. Take K′ = L in the above diagram.

Thus the Artin map induces a homomorphism

ψ_{L/K} : I_{K}^{S}/ Nm(I_{L}^{S}) → Gal(L/K)

whenever L/K is an abelian extension. Note that I^{S}/ Nm(I_{L}^{S}) is still an infinite group, and so ψ_{L/K} can not be injective.

The main theorems of global class field theory. Let S be a finite set of primes of K. We shall say that a homomorphism ψ : I^{S} → G admits a modulus if there exists a modulus 𝔪 with S(𝔪) ⊂ S such that ψ(i(K_{𝔪,1})) = 0. Thus ψ admits a modulus if and only if it factors through C_{𝔪} for some 𝔪 with S(𝔪) ⊂ S.

Theorem 3.5 (Reciprocity Law). Let L be a finite abelian extension of K, and let S be the set of primes of K ramifying in L. Then the Artin map ψ : I^{S} → Gal(L/K) admits a modulus 𝔪 with S(𝔪) = S, and it defines an isomorphism

I_{K}^{S(𝔪)}/i(K_{𝔪,1}) · Nm(I_{L}^{S(𝔪)}) → Gal(L/K).

A modulus as in the statement of the theorem is called a defining modulus for L. Note that the theorem does not imply that K has even a single nontrivial abelian

extension. Write I_{K}^{𝔪} for group of S(𝔪)-ideals in K, and I_{L}^{𝔪} for the group of S(𝔪)′- ideals in L, where S(𝔪)′ contains the primes of L lying over a prime in S. Call a subgroup H of I_{K}^{𝔪} a congruence subgroup modulo 𝔪 if

I_{K}^{𝔪} ⊃ H ⊃ i(K_{𝔪,1}).

Theorem 3.6 (Existence Theorem). For any congruence subgroup H modulo 𝔪, there exists an abelian extension L/K such that H = i(K_{𝔪,1}) · Nm_{L/K}(I_{L}^{𝔪}).

Note that, for H and L as in the theorem, the Artin map ψ_{L/K} induces an isomorphism

I^{S(𝔪)}/H → Gal(L/K).

In particular, for each modulus 𝔪 there is a field L_{𝔪}, called the ray class field modulo 𝔪 such that the Artin map defines an isomorphism C_{𝔪} → Gal(L_{𝔪}/K). For a field L ⊂ L_{𝔪}, set Nm(C_{L,𝔪}) = i(K_{𝔪,1}) · Nm(I_{L}^{𝔪}) mod i(K_{𝔪,1}).

Corollary 3.7. Fix a modulus 𝔪. Then the map L ↦ Nm(C_{L,𝔪}) is a bijection from the set of abelian extensions of K contained in L_{𝔪} to the set of subgroups of C_{𝔪}. Moreover,

L_{1} ⊂ L_{2} ⟺ Nm(C_{L_{1},𝔪}) ⊃ Nm(C_{L_{2},𝔪});
Nm(C_{L_{1}·L_{2},𝔪}) = Nm(C_{L_{1},𝔪}) ∩ Nm(C_{L_{2},𝔪});
Nm(C_{L_{1}∩L_{2},𝔪}) = Nm(C_{L_{1},𝔪}) · Nm(C_{L_{2},𝔪}).

In Section 5 below, we shall restate Theorems 3.5 and 3.6 in terms of idèles, and in Chapter VII we prove the restated theorems.

As we discuss below, there is a rather simple analytic proof that the Artin map is surjective. Thus the difficulty in proving the Reciprocity Law is in showing that the Artin map admits a conductor and that

(I^{S(𝔪)} : i(K_{𝔪,1}) · Nm(I_{L}^{S(𝔪)}) = [L : K].

To prove the Existence Theorem we must construct a ray class field for each modulus. Unfortunately, we don’t know how to construct the ray class field directly. Rather we construct enough extensions to force the theorem to be true.

Remark 3.8. Let L/K be an abelian extension with Galois group G. According to the Reciprocity Law, there is a modulus 𝔪 with support the set of primes of K ramifying in L such that the Artin map ψ_{L/K} : I^{S(𝔪)} → G takes the value 1 on i(K_{𝔪,1}). Consider the map in Theorem 1.5

(O_{K}/𝔭^{m(𝔭)})^{×} ↪ K_{𝔪}/K_{𝔪,1} ⟶^{i} C_{𝔪} ⟶^{ψ_{L/K}} G.

Clearly, there will be a smallest integer f(𝔭) ≤ m(𝔭) such that this map factors through (O_{K}/𝔭^{f(𝔭)})^{×}. The modulus 𝔣(L/K) = 𝔪_{∞} ∏ 𝔭^{f(𝔭)} is then the smallest modulus such that ψ_{L/K} factors through C_{𝔣}—it is called the conductor of L/K.^{1} The conductor 𝔣(L/K) is divisible exactly by the primes ramifying in L.

The subfields of the ray class field L_{𝔪} containing K are those with conductor 𝔣|𝔪. Every abelian extension of K is contained in L_{𝔪} for some 𝔪.

Example 3.9. The ray class group for the modulus 𝔪 = 1 is the ideal class group, and the corresponding ray class field is the Hilbert class field; it is the maximal abelian extension of K that is unramified at all primes (i.e., such that each finite prime is unramified in the usual sense, and each real prime remains real). For example, the Hilbert class field of ℚ is ℚ itself (because ℚ has class number 1). The Hilbert class field of ℚ[\sqrt{−5}] is ℚ[\sqrt{−1}, \sqrt{5}]—both 2 and 5 ramify in ℚ[\sqrt{−5}], but only 2 ramifies in ℚ[\sqrt{−1}] and only 5 ramifies in ℚ[\sqrt{5}], from which it follows that the primes of ℚ[\sqrt{−5}] dividing 2 and 5 do not ramify ℚ[\sqrt{−1}, \sqrt{5}].

Example 3.10. Let m be a positive integer which is odd or divisible by 4. The ray class field for (m) is ℚ[ζ_{m} + ζ̄_{m}], and the ray class field for ∞(m) is ℚ[ζ_{m}]. Thus the Reciprocity Law implies the Kronecker-Weber theorem: every abelian extension of ℚ has conductor dividing ∞(m) for some m, and therefore is contained in a cyclotomic field.

Example 3.11. Let d be a square-free integer. We compute the conductor of K = ℚ[\sqrt{d}] by finding the smallest integer m such that ℚ[\sqrt{d}] ⊂ ℚ[ζ_{m}].

First, consider an odd prime p. Then Gal(ℚ[ζ_{p}]/ℚ) ≅ (ℤ/pℤ)^{×} is cyclic of order p − 1, and so has a unique quotient group of order 2. Therefore, ℚ[ζ_{p}] contains a unique quadratic field, which because it can only be ramified at p, must equal ℚ[\sqrt{p^{∗}}] where p^{∗} = (−1)^{\frac{p−1}{2}} p (the sign is chosen so that p^{∗} ≡ 1 mod 4).

Second, note that ζ_{8} = (1 + i)/\sqrt{2}, and so ζ_{8} + ζ̄_{8} = \sqrt{2}. Therefore ℚ[\sqrt{2}] ⊂ ℚ[ζ_{8}] (in fact, ℚ[\sqrt{2}] is the largest real subfield of ℚ[ζ_{8}], and ℚ[ζ_{8}] = ℚ[i, \sqrt{2}]).

Let n be the product of the odd primes dividing d (so d = ±n or ±2n). I claim that

ℚ[\sqrt{d}] ⊂ ℚ[ζ_{n}] if d ≡ 1 mod 4,
ℚ[\sqrt{d}] ⊂ ℚ[ζ_{4n}] if d ≡ 3 mod 4,
ℚ[\sqrt{d}] ⊂ ℚ[ζ_{8n}] if d ≡ 2 mod 4

and that, in each case, this is the smallest cyclotomic field containing ℚ[\sqrt{d}]. For example, note that d = p_{1} … p_{r}, d ≡ 1 mod 4, implies that d = p_{1}^{∗} ⋯ p_{r}^{∗}, and so ℚ[\sqrt{d}] ⊂ ℚ[ζ_{n}]. Also note that if d is even, then ℚ[\sqrt{d}] is not contained in ℚ[ζ_{4n}] because otherwise ℚ[ζ_{4n}] would contain i, \sqrt{d}, \sqrt{d/2}, and hence would contain i, \sqrt{2}, and ζ_{8}.

1Führer in German—in Germany in the 1930s, conversations in public on class field theory could be hazardous.

We conclude that the conductor of ℚ[\sqrt{d}] is |Δ_{K/ℚ}| or ∞|Δ_{K/ℚ}| depending on whether d > 0 or d < 0—here Δ_{K/ℚ} is the discriminant of K/ℚ.

Exercise 3.12. Compute the conductor of ℚ[\sqrt{d}]/ℚ by applying the quadratic reciprocity law to find the smallest 𝔪 such that i(ℚ_{𝔪,1}) is in the kernel of the Artin map (cf. Exercise 0.1).

The field L corresponding to a congruence subgroup H is called the class field of H, whence the name of the subject. Note that for a prime 𝔭 of K not dividing the conductor of L/K, the residue class degree f(𝔓/𝔭) for a prime lying over 𝔭 is the order of 𝔭 in I^{𝔪}/H (because this is the order of the Frobenius element in Gal(k(𝔓)/k(𝔭))). Thus we have obtained a classification of the abelian extensions of K in terms of the ideal structure of K, and for each abelian extension we know the decomposition laws of the primes in K.

Exercise 3.13. Verify the last row in the following table:

Discriminant  −15  −20  −23  −24  −31
Class number  2  2  3  2  3
Hilbert class field X^{2} + 3 X^{2} + 1 X^{3} − X − 1 X^{2} + 3 X^{3} + X − 1

The first row lists the discriminants of the first five imaginary quadratic fields with class number not equal to 1, the second row lists their class numbers, and the final row lists the minimum polynomial of a generator of the Hilbert class field. (Note that for a totally imaginary field, the class number and the narrow class number coincide.)

Exercise 3.14. This exercise explains what happens when we ignore a finite set S of prime ideals of K. Let 𝔪 be a modulus of K with S(𝔪) ∩ S = ∅, and let H be a subgroup of I^{S∪S(𝔪)} containing i(K_{𝔪,1}). Define an extension L of K to be an S-class field for H if

(a) L is a finite abelian extension of K, and the prime ideals in S split completely in L;

(b) 𝔪(𝔭) = 0 ⟹ 𝔭 does not ramify in L;

(c) the prime ideals not in S ∪ S(𝔪) that split in L are precisely those in H.

Prove that an S-class field L exists for each group H as above, that it is unique, and that I^{S∪S(𝔪)}/H ≅ Gal(L/K); moreover, every field L satisfying (a) is the S-class field for some H.

Hint: Show I^{S∪S(𝔪)}/i(K_{𝔪,1}) ≅ I^{S(𝔪)}/ <S> ·i(K_{𝔪,1}), where <S> is the subgroup of I^{S(𝔪)} generated by the primes in S.

The norm limitation theorem. In our classification of the abelian extensions of K, we attach to L the group H = i(K_{𝔪,1}) · Nm(I_{L}^{S(𝔪)}) for 𝔪 a modulus sufficiently large to be a defining modulus (and then (I^{S(𝔪)} : H) = [L : K]). One might hope that something similar works for nonabelian extensions, but the following theorem shows that it does not.

Theorem 3.15 (Norm Limitation Theorem). Let L be a Galois extension of K, and let L′/K be the maximal abelian subextension of L/K. For any defining modulus 𝔪 for L′/K,

i(K_{𝔪,1}) · Nm_{L/K}(I_{L}^{S(𝔪)}) = i(K_{𝔪,1}) · Nm_{L′/K}(I_{L′}^{S(𝔪)}).

This indicates that, for a nonabelian extension L/K, Spl(L/K) is not described by congruence conditions.

The principal ideal theorem. The following theorem was conjectured by Hilbert about 1900.

Theorem 3.16. Every ideal in K becomes principal in the Hilbert class field of K.

I explain the idea of the proof. Recall that for a group G, the commutator (or derived) subgroup G′ of G is the subgroup generated by the commutators ghg^{−1}h^{−1}, g, h ∈ G. The quotient G^{ab} = G/G′ is abelian, and it is the largest abelian quotient of G. If L is a Galois extension of K with Galois group G, then L^{G′} is an abelian extension of K with Galois group G^{ab}, and it is the largest abelian extension of K contained in L.

Suppose we have fields

L ⊃ K′ ⊃ K

with L Galois over K (not necessarily abelian). For any finite set of primes S of K, 𝔞 ↦ 𝔞O_{K′} is a homomorphism I_{K}^{S} → I_{K′}^{S}. Consider:

I_{K}^{S} ⟶^{ψ_{L/K}} Gal(L/K)^{ab}
↓^{can.}  ↓^{?}
I_{K′}^{S} ⟶^{ψ_{L/K′}} Gal(L/K′)^{ab}.

What is the map “?” making the diagram commute?

Before describing it, we need to explain a construction in group theory. Let H be a group of finite index in a group G, and write G as a disjoint union of cosets,

G = Hg_{1} ∪ Hg_{2} ∪ … ∪ Hg_{n}.

For g ∈ G, set φ(g) = g_{i} if g ∈ Hg_{i}, and define

V (g) = ∏_{i=1}^{n} g_{i}gφ(g_{i}g)^{−1} mod H′

where H′ is the commutator subgroup of H.

Proposition 3.17. The map g ↦ V (g) is a homomorphism G → H/H′, and it is independent of the choice of the coset representatives g_{i}.

Proof. The verification is straightforward—see, for example, M. Hall, The Theory of Groups, Macmillan, 1959, 14.2.1.

Thus, whenever we have a group G and a subgroup H of finite index, we have a well-defined homomorphism

V : G^{ab} → H^{ab},

called the Verlagerung (or transfer) map.

In the situation of the above diagram, Gal(L/K′) is a subgroup of Gal(L/K), and hence the Verlagerung is a homomorphism

V : Gal(L/K)^{ab} → Gal(L/K′)^{ab}.

Emil Artin showed that this is the map making the above diagram commute (cf. II.2.9b).

Consider the fields

K′′ ⊃ K′ ⊃ K

where K′ is the Hilbert class field of K, and K′′ is the Hilbert class field of K′. Then

(a) K′′ is normal over K (because any conjugate of K′′ is again an abelian unramified extension of K′, and hence is contained in K′′);

(b) K′ is the maximal abelian extension of K contained in K′′ (because every abelian extension of K contained in K′′ is unramified over K, and hence is contained in K′).

From (b) we find that Gal(K′′/K)^{ab} = Gal(K′/K). Therefore, when L = K′′, the diagram becomes

C_{K} ⟶^{≈} Gal(K′/K)
↓^{can.}  ↓^{V}
C_{K′} ⟶^{≈} Gal(K′′/K′),

where C_{K} and C_{K′} are the class groups of K and K′. Let G = Gal(K′′/K) and let H = Gal(K′′/K′). Because of (b), H is the commutator subgroup of G. The next theorem (which was conjectured by Emil Artin) shows that V is zero in this situation, and hence that the canonical map C_{K} → C_{K′} is zero, i.e., that every ideal of K becomes principal in K′.

Theorem 3.18. Let G be a finite group, and let H be its commutator subgroup; then

V : G^{ab} ⟶ H^{ab}

is zero.

Proof. This is a theorem in group theory. It was proved by Furtwängler in 1930. For a simple proof, see: Witt, Proc. International Congress of Mathematicians, Amsterdam, 1954, Vol 2, pp71–73.

Remark 3.19. It is in fact easy to see that there exists an extension L of K of degree dividing the class number h of K such that every ideal in K becomes principal in L: write the class group of K as a direct sum of cyclic groups; choose a generator 𝔞_{i} for each summand, and let h_{i} be the order of 𝔞_{i} in the class group; write 𝔞_{i}^{h_{i}} = (a_{i}), and define L to be the field obtained from K by adjoining an h_{i}th root of a_{i} for each i.

However, a field constructed in this fashion will not usually be the Hilbert class field of K—it need not even be Galois over K. There may exist fields of degree < h over K in which every ideal in K becomes principal.

Remark 3.20. The principal ideal theorem does not, of course, imply that every ideal in the Hilbert class field K′ of K is principal, because not every ideal of K′ is in the image of the homomorphism I_{K} → I_{K′}. One can form the Hilbert class field K′′ of K′, and so on, to obtain a tower

K ⊂ K′ ⊂ K′′ ⊂ ⋯ ⊂ K^{(n)} ⊂ ⋯

in which K^{(n+1)} is the Hilbert class field of K^{(n)}. Note that the same argument that proved K′′ is Galois over K shows that K^{(n)} is Galois over K. The class field tower problem (stated by Hasse in 1925) asks whether this tower is always finite, and so terminates in a field with class number one. The answer was shown to be negative by Golod and Shafarevich in 1964 (see Roquette’s article in Cassels and Fröhlich 1967). For example, ℚ[\sqrt{−2.3.5.7.11.13}] has infinite class field. In fact, ℚ[\sqrt{d}] has an infinite class field tower whenever d has more than 8 prime factors.

The Chebotarev Density Theorem. Let L be a Galois extension of K with Galois group G. Recall that, for any prime ideal 𝔭 of K unramified in L,

(𝔭, L/K) =^{df} {(𝔓, L/K) | 𝔓|𝔭}

is a conjugacy class in G.

Theorem 3.21 (Chebotarev Density Theorem). Let L/K be a finite extension of number fields with Galois group G, and let C be a conjugacy class in G. Then the set of prime ideals of K such that (𝔭, L/K) = C has density #C/#G in the set of all prime ideals of K. In particular, if G is abelian, then, for a fixed τ ∈ G, the set of prime ideals 𝔭 of K with (𝔭, L/K) = τ has density (G : 1).

Proof. For an abelian extension L/K this follows from Theorem 2.5 and Theorem 3.5: the latter says that the map 𝔭 ↦ (𝔭, L/K) induces a surjective homomorphism C_{𝔪} → Gal(L/K) for some modulus 𝔪, and the former says that the primes are equidistributed among the classes in C_{𝔪}. The nonabelian case is derived from the abelian case by an ingenious argument—see Chapter VIII.

Corollary 3.22. If a polynomial f(X) ∈ K[X] splits into linear factors modulo 𝔭 for all but finitely prime ideals 𝔭 in K, then it splits in K[X].

Proof. Apply the theorem to the splitting field of f(X).

For a finite extension L/K of number fields and a finite set S of primes of K, let Spl_{S}(L/K) be the set of primes of K not in S that split in L.

Theorem 3.23. If L and M are Galois extensions of K, then

L ⊂ M ⟺ Spl_{S}(L/K) ⊃ Spl_{S}(M/K).

Hence

L = M ⟺ Spl_{S}(L/K) = Spl_{S}(M/K).

Proof. As a consequence of (1.11),

Spl_{S}(LM/K) = Spl_{S}(L/K) ∩ Spl_{S}(M/K).

Hence

Spl_{S}(L/K) ⊃ Spl_{S}(M/K) ⟹ Spl_{S}(LM/K) = Spl_{S}(M/K),

which, by the Chebotarev density theorem, implies

[LM : K] = [M : K],

and so L ⊂ M. The reverse implication is obvious.

Remark 3.24. (a) Theorem 3.23 is not true without the Galois assumption (see Cassels and Fröhlich 1967, p363).

(b) In the statement of Theorem 3.23, S can be replaced by any set of primes of density 0.

(c) Let f(X) be an irreducible polynomial in K[X]. If f(X) has a root modulo 𝔭 for almost all prime ideals 𝔭, then f(X) has a root in K (ibid. p363, 6.2).

The conductor-discriminant formula. Two Dirichlet characters χ : I^{S} → ℂ^{×} and χ′ : I^{S′} → ℂ^{×} are said to be cotrained if they agree on I^{S′′} for some S′′ ⊃ S ⋃ S′. This is an equivalence relation. In each equivalence class, there is a unique χ with smallest S—such a χ is said to be primitive.

Let χ_{1} be the primitive character equivalent to χ. The smallest modulus 𝔪 such that χ_{1} is zero on K_{𝔪,1} is called the conductor of 𝔣(χ) of χ. Set 𝔣(χ) = 𝔣_{∞}(χ)𝔣_{0}(χ) where 𝔣_{∞} and 𝔣_{0} are respectively divisible only by infinite primes and finite primes.

Theorem 3.25 (Führerdiskriminantenproduktformel). For any finite abelian extension L/K of number fields with Galois group G,

disc(L/K) = ∏_{χ∈G^{∨}} 𝔣_{0}(χ ∘ ψ_{L/K}), G^{∨} =^{df} Hom(G, ℂ^{×});

𝔣(L/K) = lcm𝔣(χ ∘ ψ_{L/K}).

Clearly ∩ Ker(χ : G → ℂ^{×}) = 0, from which the second statement follows. We omit the proof of the first—it is really a statement about local fields.

Remark 3.26. Let H be the kernel of the character χ. Then 𝔣(χ) = 𝔣(L^{H}/K). For example, if χ is injective, then 𝔣(χ) = 𝔣(L/K).

Example 3.27. Let L = ℚ[\sqrt{d}]. Then G ≈ ℤ/2ℤ, and there are only two characters G → ℂ^{×}, one χ_{0} trivial and the other χ_{1} injective. Therefore, the theorem says that Δ_{K/ℚ} = 𝔣_{0}(χ_{1}) = 𝔣_{0}(K/ℚ), as we showed in Example 3.11.

Example 3.28. Let L = ℚ[ζ_{p}], p an odd prime. Then G ≅ (ℤ/pℤ)^{×}, which is cyclic of order p − 1. It therefore has p − 2 nontrivial characters. If χ is nontrivial, then 𝔣(χ)|∞(p), but 𝔣(χ) = 1 or ∞ is impossible (because it would imply χ is trivial). Therefore, 𝔣(χ) = (p) or ∞(p) if χ is nontrivial, and so the conductor-discriminant formula shows (correctly) that Δ_{L/ℚ} = ±p^{p−2}.

Exercise 3.29. Verify the conductor-discriminant formula for the extension ℚ[ζ_{p^{2}}]/ℚ.

Remark 3.30. Let L/K be a finite extension of number fields with Galois group G (not necessarily abelian). To a representation ρ : G → GL(V) of G on a finitedimensional vector space V, Artin attaches a Dirichlet L-series L(s, ρ) (see the introduction). The analytic properties of these Artin L-series are still not fully understood.

When G is commutative, the Artin map I^{S} → G identifies characters of G with Dirichlet characters, and hence Artin L-series with Dirichlet L-series. This was Artin’s motivation for seeking the map (not, as seems natural today, in order to construct a canonical isomorphism between the groups C_{𝔪} and G, already known to be abstractly isomorphic).

A part of Langlands’s philosophy is a vast generalization of this correspondence between Dirichlet L-series and abelian Artin L-series.

The reciprocity law and power reciprocity. Assume K contains a primitive nth root of 1, and let a ∈ K. If \sqrt[n]{a} is one root of X^{n} − a, then the remaining roots are of the form ζ \sqrt[n]{a} where ζ is an nth root of 1. Therefore L =^{df} K[\sqrt[n]{a}] is Galois over K, and σ \sqrt[n]{a} = ζ \sqrt[n]{a} for some nth root ζ of 1.

If 𝔭 is a prime ideal of K that is relatively prime to n and a, then 𝔭 is unramified in L, and we can define an nth root (\frac{a}{𝔭})_{n} of 1 by the formula

(𝔭, L/K)(\sqrt[n]{a}) = (\frac{a}{𝔭})_{n} \sqrt[n]{a}.

One can show that

(\frac{a}{𝔭})_{n} = 1 ⟺ a is an n^{th} power modulo 𝔭,

and so (\frac{a}{𝔭})_{n} generalizes the quadratic residue symbol. For this reason (\frac{a}{𝔭})_{n} is called the power residue symbol. Artin’s reciprocity law implies all known reciprocity laws for these symbols, and so, as Artin pointed out, it can be viewed as a generalization of them to fields without roots of unity. We shall explain this in Chapter VIII.

An elementary unsolved problem. Let K be a number field, and let S be a nonempty finite set of prime ideals of K. Does there exist for every prime number p not divisible by any prime in S a sequence of fields …, L_{n}, L_{n+1}, … such that

(a) L_{n} is unramified outside the primes of S;

(b) p^{n}|[L_{n} : K]?

A key case, for which the answer is unknown, is K = ℚ and S = {ℓ}.

More explicitly (and slightly harder), fix a prime ℓ in ℚ. Does there exist for every prime p ≠ ℓ a sequence of monic irreducible polynomials f_{n}(X) ∈ ℤ[X] such that

(a) disc(f_{n}(X)) is not divisible by any prime other than ℓ;

(b) p^{n}| deg f_{n}(X).

See (Milne, Arithmetic Duality Theorems, Academic Press, 1986, p60/61) to find where this problem turned up. It is certainly very difficult.

Explicit global class field theory: Kronecker’s Jugentraum and Hilbert’s twelfth problem. Unlike local class field theory, global class field theory does not (in general) provide an explicit construction of the abelian extensions of a number field K.

Gauss knew that the cyclotomic extensions of ℚ are abelian. Towards the end of the 1840s Kronecker had the idea that the cyclotomic fields, and their subfields, exhaust the abelian extensions of ℚ, and furthermore, that every abelian extension of a quadratic imaginary number field E is contained in the extension given by adjoining to E roots of 1 and certain special values of the modular function j. Many years later he was to refer to this idea as the most cherished dream of his youth (mein liebster Jugendtraum).

More precisely, Kronecker’s dream^{2} is that every abelian extension of ℚ is contained in the field obtained by adjoining to ℚ all values of the function e^{2πiz} for z ∈ ℚ^{×}, and that every abelian extension of an imaginary quadratic field K is contained in the field obtained by adjoining to K all values of the function j(z) for z ∈ K^{×}.

Later Hilbert took this up as the twelfth of his famous problems: for any number field K, find functions that play the same role for K that the exponential function plays for ℚ and the modular function j plays for a quadratic imaginary field ^{3}, Proc. Symp. Pure Math. XXVIII, Part 1, AMS, 1976. The first part of Kronecker’s dream, that every abelian extension of ℚ is a subfield of a cyclotomic extension, was proved by Weber (1886, 1899, 1907, 1911) and Hilbert (1896).

The statement above of “Kronecker’s dream” is not quite correct. Let K be an imaginary quadratic field. We can write O_{K} = ℤ + ℤτ with ℑ(τ) > 0. It is known that K[j(τ)] is the Hilbert class field of K. Now adjoin to K all roots of unity and all values j(τ) with τ ∈ K, ℑ(τ) > 0. The resulting field K′ is abelian over K, and [K^{ab} : K′] is product of groups of order 2. To get the whole of K^{ab}, it is necessary to adjoin special values of other elliptic functions. These statements were partially proved Weber (1908) and Feuter (1914), and completely proved by Takagi (1920).

From the modern point of view, special values of elliptic modular functions are related to the arithmetic of elliptic curves with complex multiplication, and it is results about the latter that allow one to prove that the former generate abelian extension of a quadratic imaginary field.

Beginning with the work of Taniyama, Shimura, and Weil in the late fifties, the theory of elliptic curves and elliptic modular curves has been generalized to higher dimensions. In this theory, an elliptic curve with complex multiplication by an imaginary quadratic field is replaced by an abelian variety with complex multiplication by a “CM-field”, that is, a quadratic totally imaginary extension K of a totally real field F, and an elliptic modular function by an automorphic function.

Philosophically, one expects that (with the exception of ℚ), one can not obtain abelian extensions of totally real fields by adjoining special values of automorphic functions. However, it is known that, roughly speaking, one does obtain the largest possible abelian extension of a CM-field K consistent with this restriction.

More precisely, let K be a CM-field and let F be the largest totally real subfield of K. Then G =^{df} Gal(ℚ^{al}/K) is a subgroup of index 2 in G′ =^{df} Gal(ℚ^{al}/F), and the corresponding Verlagerung is a homomorphism V : G^{′ab} → G^{ab}. In this case, V has a very simple description.

2For a careful account of Kronecker’s idea and work on it, see Schappacher, N., On the history of Hilbert’s twelfth problem, I, Paris 1900—Zürich 1932: The comedy of errors (preprint).

3See pp18-20 of Mathematical Developments arising from Hilbert’s Problems

Theorem 3.31. Let K be a CM-field, and let F be the totally real subfield of K with [K : F] = 2. Let H be the image of the Verlagerung map

Gal(F^{ab}/F) ⟶ Gal(K^{ab}/K).

Then the extension of K obtained by adjoining the special values of all automorphic functions (defined on canonical models of Shimura varieties with rational weight) is (K^{ab})^{H} · ℚ^{ab}.

Proof. See Wafa Wei’s thesis (Michigan 1993).

Notes. The relation between congruence groups and abelian extensions of K was known before Artin defined his map. It emerged only slowly over roughly the period 1870–1920. The main contributors were Kronecker, Weber, Hilbert, and Takagi. Chebotarev proved his theorem in (1926) (a less precise result had been proved much earlier by Frobenius), and Artin defined his map and proved it gave an isomorphism in 1927. (Earlier, it had been known that I^{𝔪}/H ≈ Gal(L/K), but no canonical isomorphism was known.) The fact that analysis, in the form of Chebotarev’s (or Frobenius’s) theorem was required to prove the main theorems, which are purely algebraic in form, was regarded as a defect, and in 1940, after much effort, Chevalley succeeded in giving a purely algebraic proof of the main theorems. (The difficult point is proving that if L/K is an abelian extension of number fields of prime degree p, then at least one prime of K does not split or ramify in L.) He also introduced idèles, which make it possible to state class field theory directly for infinite extensions. Group cohomology (at least 2-cocycles etc.) had been used implicitly in class field theory from the 1920s, but it was used systematically by Nakayama, Hochschild, and Tate in the 1950s. In 1951/52 in a very influential seminar, Artin and Tate gave a purely algebraic and very cohomological treatment of class field theory. Since then there have been important improvements in our understanding of local class field theory (mainly due to Lubin and Tate). Nonabelian class field theory is a part of Langlands’s philosophy, which is a vast interlocking series of conjectures, and some progress has been made, especially in the local case and the function field case (Drin’feld). A fairly satisfactory abelian class field theory for more general fields (fields of finite transcendence degree over ℚ or 𝔽_{p}) has been created by Bloch, Kato, Saito, and others. It uses algebraic K-theory (see W. Raskind, Abelian class field theory of arithmetic schemes, Proc. Symp. Pure Math, AMS, Vol 58.1 (1995), 85–187).

4. Idèles

Theorems 3.5 and 3.6 show that, for any number field K, there is a canonical isomorphism \underleftarrow{lim}_{𝔪}C_{𝔪} → Gal(K^{ab}/K). Rather than studying \underleftarrow{lim}_{𝔪}C_{𝔪} directly, it turns out to be more natural to introduce another group that has it as a quotient— this is the idèle class group.

Topological groups. A group G with a topology is called a topological group if the maps

g, g′ ↦ gg′: G × G → G, g ↦ g^{−1} : G → G

are continuous. The translation map

g ↦ ag: G → G

is then a homeomorphism.

In general, to determine a topology on a set we have to give a fundamental system of neighbourhoods of each point, i.e., a set of neighbourhoods of the point such that every neighbourhood contains one in the set. Because the translation map is a homeomorphism, the topology on a topological group is determined by a fundamental system of neighbourhoods of 1.

We shall need to make use of various generalities concerning topological groups, which can be found in many books. Fortunately, we shall only need quite elementary things.

Let (X_{i})_{i} be a (possibly infinite) family of topological spaces. The product topology on ∏ X_{i} is that for which the sets of the form ∏ U_{i}, U_{i} open in X for all i and equal to X_{i} for all but finitely many i, form a basis. Tychonoff’s theorem says that a product of compact spaces is compact. However, an infinite product of locally compact spaces will not in general be locally compact: if V_{i} is a compact neighbourhood of x_{i} in X_{i} for all i, then ∏ V_{i} will be compact, but it will not be a neigbourhood of (x_{i}) unless V_{i} = X_{i} for all but finitely many i.

Idèles. We now often write v for a prime of K. Then:

| · |_{v} = the normalized valuation for v (for which the product formula holds),
K_{v} = the completion of K at v,
𝔭_{v} = the corresponding prime ideal in O_{K}, (when v is finite),
O_{v} = the ring of integers in K_{v},
U_{v} = O_{v}^{×}
𝔭̂_{v} = the completion of 𝔭_{v} = maximal ideal in O_{v}.

Recall that, for all v, K_{v} is locally compact—in fact, O_{v} is a compact neighbourhood of 0. Similarly K_{v}^{×} is locally compact; in fact

1 + 𝔭̂_{v} ⊃ 1 + 𝔭̂_{v}^{2} ⊃ 1 + 𝔭̂^{3} ⊃ ⋯

is a fundamental system of neighbourhoods of 1 consisting of open compact subgroups.

We want to combine all the groups K_{v}^{×} into one big topological space, but ∏ K_{v}^{×} is not locally compact. Instead we define the group of idèles to be

𝕀_{K} = {(a_{v}) ∈ ∏ K_{v}^{×} | a_{v} ∈ O_{v}^{×} for all but finitely many v}.

For any finite set S of primes that includes all infinite primes, let

𝕀_{S} = ∏_{v∈S} K_{v}^{×} × ∏_{v∉S} O_{v}^{×}

with the product topology. The first factor is a finite product of locally compact spaces, and so is locally compact, and the second factor is a product of compact spaces, and so is compact (by Tychonoff). Hence 𝕀_{S} is locally compact. Note that

𝕀 = ⋃ 𝕀_{S}.

We want to endow 𝕀_{K} with a topology such that each 𝕀_{S} is open in 𝕀 and inherits the product topology. We do this by decreeing that a basis for the open sets consists of the sets of the form

∏_{v} U_{v} = \{ \begin{matrix}U_{v} & open in K_{v}^{×} for all v; \\ U_{v} & = O_{v}^{×} for almost all v. \end{matrix}

An intersection of two sets of this form contains a set of this form, and so they do form a basis for a topology. It is clear that the topology does have the property we want, and moreover that it endows 𝕀 with the structure of a topological group. The following sets form a fundamental system of neighbourhoods of 1: for each finite set of primes S ⊃ S_{∞} and ε > 0, define

U(S, ε) = {(a_{v}) | |a_{v} − 1|_{v} < ε, v ∈ S, |a_{v}|_{v} = 1, all v ∉ S}.

4.1. There is a canonical surjective homomorphism id

(a_{v}) ↦ ∏_{v finite} p_{v}^{ord_{p}(a_{v})} : 𝕀_{K} → I_{K}

whose kernel is 𝕀_{S_{∞}}.

We can think of the idèles as an enlargement of the ideals: it includes factors for the infinite primes, and it includes the units at the finite primes. Note that 𝕀_{K}/𝕀_{S_{∞}} ≈ ℤ^{(ℕ)} (direct sum of countably many copies of ℤ with the discrete topology), but that ∏ K_{v}^{×}/𝕀_{S_{∞}} ≈ ℤ^{ℕ} (product of countably many copies of ℤ, which is itself uncountable).

4.2. There is a canonical injective (diagonal) homomorphism

a ↦ (a, a, a, …): K^{×} → 𝕀_{K}.

I claim that the image is discrete. Because we have groups, it suffices to prove that 1 ∈ K^{×} is open in the induced topology. Let U = U(S, ε) with S any finite set containing S_{∞} and 1 > ε > 0. For any a ∈ K^{×} ∩ U,

\{ \begin{matrix}|a − 1|_{v} < ε & for all v ∈ S \\ |a|_{v} = 1 & for all v ∉ S. \end{matrix}

The second condition implies that

|a − 1|_{v} ≤ max(|a|_{v}, | − 1|_{v}) ≤ 1.

Therefore, if a ∈ K^{×} ∩ U, then ∏_{v} |a − 1|_{v} < ε^{#S} < 1, which contradicts the product formula unless a = 1.

The quotient C = 𝕀/K^{×} is called the idèle class group of K. It maps onto the ideal class group of K. It is not compact (see (4.4) below).

4.3. There is a canonical injective homomorphism

a ↦ (1, …, 1, a, 1, … 1): K_{v}^{×} ⟶ 𝕀

(a in the v^{th} place). The topology induced on K_{v}^{×} is its natural topology, because

U(S, ε) ∩ K_{v}^{×} = \{ \begin{matrix}\{a | |a − 1|_{v} < ε\} & v & ∈ S \\ \{a | |a|_{v} = 1\} & v & ∉ S \end{matrix}

and such sets form a fundamental system of neighbourhoods of 1 in K_{v}^{×}.

4.4. There is canonical surjective homomorphism

a = (a_{v}) ↦ c(a) = ∏ |a_{v}|_{v} : 𝕀 → ℝ_{>0}.

The image of a is called the content of a. Define

𝕀^{1} = Ker(c) = {a ∈ 𝕀 | c(a) = 1}.

Note that, because of the product formula, K^{×} ⊂ 𝕀^{1}. The quotient 𝕀/K^{×} can’t be compact because it maps surjectively onto ℝ_{>0}, but one can prove that 𝕀^{1}/K^{×} is compact.

Aside 4.5. Define 𝕀_{f} the same way as 𝕀, except using only the finite primes. We call 𝕀_{f} the group of finite idèles. We have

∏_{v finite} O_{v}^{×} ⊂ 𝕀_{f} ⊂ ∏_{v finite} K_{v}^{×}.

The subgroup ∏ O_{v}^{×} is open and compact in 𝕀_{f}, and 𝕀_{f}/ ∏ O_{v}^{×} = I (the group of ideals of K).

Again there is a diagonal embedding of K^{×} into 𝕀_{f}, but this time the induced topology on K^{×} has the following description: U_{K} =^{df} O_{K}^{×} is open, and a fundamental system of neighbourhoods of 1 is formed by the subgroups of U_{K} of finite index (nontrivial theorem). In particular, K^{×} is a discrete subgroup of 𝕀_{f} ⟺ U_{K} is finite ⟺ K = ℚ or an imaginary quadratic field.

Realizing ray class groups as quotients of 𝕀. We have seen that the class group C_{K} = I/i(K^{×}) can be realized as the quotient of 𝕀. We want to show the same for C_{𝔪}.

Let 𝔪 be a modulus. For 𝔭|𝔪, set

W_{𝔪}(𝔭) = \{ \begin{matrix}ℝ_{>0} & 𝔭 real \\ 1 + 𝔭̂^{m(𝔭)} & 𝔭 finite. \end{matrix}

Thus, in each case, W_{𝔪}(𝔭) is a neighbourhood of 1 in K_{𝔭}^{×}, and

K_{𝔪,1} = K^{×} ∩ ∏_{𝔭|𝔪}W_{𝔪}(𝔭).

Define

𝕀_{𝔪} = (∏_{𝔭∤𝔪}K_{𝔭}^{×} × ∏_{𝔭|𝔪}W_{𝔪}(𝔭)) ⋂ 𝕀
W_{𝔪} = ∏ ; \begin{matrix}𝔭 ∤ 𝔪 \\ 𝔭 infinite \end{matrix} K_{𝔭}^{×} × ∏_{𝔭|𝔪}W_{𝔪}(𝔭) × ∏ ; \begin{matrix}𝔭 ∤ 𝔪 \\ 𝔭 finite \end{matrix} U_{𝔭}.

Thus

(a_{𝔭}) ∈ 𝕀_{𝔪} ⟺ a_{𝔭} ∈ W_{𝔪}(𝔭) for all 𝔭|𝔪,
(a_{𝔭}) ∈ W_{𝔪} ⟺ (a_{𝔭}) ∈ 𝕀_{𝔪} and a_{𝔭} is a unit for all finite 𝔭 ∤ 𝔪.

Note that 𝕀_{𝔪} ∩ K^{×} = K_{𝔪,1} (intersection inside 𝕀).

Proposition 4.6. Let 𝔪 be a modulus of K.

(a) The map id : 𝕀_{𝔪} → I^{S(𝔪)} defines an isomorphism

𝕀_{𝔪}/K_{𝔪,1} · W_{𝔪} →^{≈} C_{𝔪}.

(b) The inclusion 𝕀_{𝔪} ↪ 𝕀 defines an isomorphism:

𝕀_{𝔪}/K_{𝔪,1} → 𝕀/K^{×}.

Proof. (a) Consider the pair of maps

K_{𝔪,1} → 𝕀_{𝔪} ⟶^{id} I^{S(𝔪)}.

The first map is injective, and the second is surjective with kernel W_{𝔪}, and so the kernel-cokernel sequence (II.4.2) of the pair of maps is

W_{𝔪} → 𝕀_{𝔪}/K_{𝔪,1} → C_{𝔪} → 1.

The proves (a) of the proposition.

(b) The kernel of 𝕀_{𝔪} → 𝕀/K^{×} is K^{×} ∩ 𝕀_{𝔪} (intersection in 𝕀) which, we just saw, is K_{𝔪,1}. Hence the inclusion defines an injection

𝕀_{𝔪}/K_{𝔪,1} ↪ 𝕀/K^{×}.

For the surjectivity, we apply the weak approximation theorem (Theorem 6.3 below). Let S = S(𝔪) and let a = (a_{v}) ∈ 𝕀. If we choose b ∈ K to be very close to a_{v} in K_{v}^{×} for all v ∈ S, then a_{v}/b will be close to 1 in K_{v}^{×} for all v ∈ S; in fact, we can choose b so that a_{v}/b ∈ W_{𝔪}(𝔭) for all v ∈ S. For example, for a real prime v in S, we need only choose b to have the same sign as a_{v} in K_{v}. Then a/b ∈ 𝕀_{𝔪}, and it maps to a in 𝕀/K^{×}.

Characters of ideals and of idèles. Let S ⊃ S_{∞} be a finite set of primes of K, and let G be a finite abelian group. A homomorphism

ψ: I^{S} → G

is said to admit a modulus if there exists a modulus 𝔪 with support in S such that ψ(i(K_{𝔪,1})) = 1. For example, for any abelian extension L/K, the Artin map

I^{S} → Gal(L/K)

admits a modulus.

Proposition 4.7. If ψ : I^{S} → G admits a modulus, then there exists a unique homomorphism ϕ : 𝕀 → G such that

(a) ϕ is continuous (G with the discrete topology)

(b) ϕ(K^{×}) = 1;

(c) ϕ(a) = ψ(id(a)), all a ∈ 𝕀^{S} =^{df} {a | a_{v} = 1 all v ∈ S}.

Moreover, every continuous homomorphism ϕ : 𝕀 ⟶ G satisfying (b) arises from a ψ.

Proof. Because ψ is admits a modulus 𝔪, it factors through I^{𝔪}/i(K_{𝔪,1}) = C_{𝔪}. Hence we have the diagram:

I^{𝔪} ⟶  C_{𝔪}  ⟶^{ψ} G
↑^{≈}
𝕀_{𝔪}/K_{𝔪,1} ⟶ 𝕀_{𝔪}/K_{𝔪,1}W_{𝔪}
↓^{≈}
𝕀 ⟶ 𝕀/K^{×}.

The isomorphisms are those in Proposition 4.6, and the remaining unnamed maps are quotient maps. Define ϕ to be the composite 𝕀 → G. It certainly has properties (a) and (b), and it also has the property that

ϕ(a) = ψ(id(a)) for all a ∈ 𝕀_{𝔪},

and so, a fortiori, it has property (c).

To prove that the map is uniquely determined by (a), (b), and (c), it suffices to prove that 𝕀^{S}K^{×} is dense in 𝕀, but this follows from the weak approximation theorem (Theorem 6.3): let a ∈ 𝕀; choose b ∈ K^{×} to be very close to a_{v} for v ∈ S, and let a′ be the element of 𝕀^{S} such that a′_{v}b = a_{v} for all v ∉ S. Then a′b ∈ 𝕀^{S} · K and is close to a in 𝕀.

For the converse, let ϕ : 𝕀 → G be a continuous map. The kernel contains an open neighbourhood of 1, and so U(S, ε) ⊂ Ker(ϕ) for some S and ε. Consider an infinite prime v. The restriction of ϕ to K_{v}^{×} is a continuous map ℝ^{×} → G or ℂ^{×} → G. Clearly, the connected component of K_{v}^{×} containing 1, namely, ℝ_{>0} or ℂ^{×}, maps to 1, and so is in the kernel. On combining these remarks, we see that the kernel of ϕ contains W_{𝔪} for some 𝔪.

Now we can use the diagram at the start of the proof again. We are given a homomorphism ϕ : 𝕀/K^{×} → G, which we can “restrict” to a homomorphism 𝕀_{𝔪}/K_{𝔪,1} → G.

This homomorphism is trivial on W_{𝔪}, and hence factors through 𝕀_{𝔪}/K_{𝔪,1}W_{𝔪}. The homomorphism can now be transferred to C_{𝔪}, and composed with 𝕀 ↠ C_{𝔪}. This is the ψ we are looking for.

Remark 4.8. Let G be a commutative topological group. Define a homomorphism ψ : I^{S} → G to be admissible if for every neighbourhood N of 1 in G, there exists a modulus 𝔪 such that ψ(i(K_{𝔪,1})) ⊂ N. Then every admissible homomorphism ψ defines a homomorphism ϕ : 𝕀 → G satisfying conditions (a), (b), (c) of the proposition. Moreover, if G is complete and has “no small subgroups” i.e., there exists a neighbourhood of 1 containing no nontrivial subgroup, then every continuous homomorphism ϕ : 𝕀 → G satisfying (b) arises from an admissible ψ. The proof is the same as that of the proposition (see Proposition 4.1 of Tate’s article in Cassels and Fröhlich 1967).

The circle group G = {z ∈ ℂ | |z| = 1} is complete and has no small subgroups. The admissible ψ : I^{S} → G, and the corresponding ϕ, are called Hecke characters.

Remark 4.9. Given ψ we chose an 𝔪, and then showed how to construct ϕ. In practice, it is more usually more convenient to identify ϕ directly from knowing that it satisifies the conditions (a), (b), (c). For this, the following observations are useful.

(a) Let a = (a_{v}) be an idèle such that a_{v} = 1 for all finite primes and a_{v} > 0 for all real primes; then ϕ(a) = 1. To see this, note that the topology induced on ∏_{v|∞} K_{v}^{×} as a subgroup of 𝕀 is its natural topology. Therefore, the restriction of ϕ to it is trivial on the connected component containing 1.

(b) Let a = (a_{v}) be an idèle such that a_{v} = 1 for all v ∈ S and a_{v} is a unit for all v ∉ S; then ϕ(a) = 1. In fact, this follows directly from condition (c).

(c) If a is “close to 1”, ϕ(a) = 1. In fact, this follows directly from condition (a) in view of the fact that G has the discrete topology.

(d) On combining (a), (b), (c), we find that if a = (a_{v}) is such that

a_{v} > 0 when v is real;

a_{v} is “close to 1” when v ∈ S is finite; a_{v} is a unit when v ∉ S

the ψ(a) = 1. In fact, (a) and (b) say that we can multiply a with idèles of certain types without changing the value ψ(a). Clearly, if a_{v} is close to 1 for the finite v in S, we can multiply it by such idèles to make it close to 1.

Example 4.10. Let L = ℚ[ζ_{p}], and let ψ be the Artin map

I^{S} → (ℤ/pℤ)^{×} → Gal(L/ℚ), S = {p, ∞}.

Recall that first map sends the ideal (uniquely) represented by (r/s), r, s > 0, (p, r) = 1 = (p, s), to [r][s]^{−1}, and that the second sends [m] to the automorphism ζ ↦ ζ^{m}. Overall, for any prime number ℓ ≠ p, the map sends (ℓ) to the Frobenius automorphism at ℓ, ζ ↦ ζ^{ℓ}. Let ϕ : 𝕀 → Gal(L/ℚ) be the homomorphism corresponding to ψ as in the theorem. We wish to determine ϕ explicitly.

Let a = (a_{∞}, a_{2}, …, a_{p}, …, a_{ℓ}, …) be an idèle of ℚ. If a_{∞} = 1 = a_{p}, then ϕ(a) = ϕ(id(a)). Thus ϕ(a) = ζ_{p}^{m} where m = ∏ ℓ^{ord_{ℓ}(a_{ℓ})}.

Consider p = (1, …, 1, p, 1, …) (p in the p-position). Then

p/p = (p^{−1}, …, p^{−1}, 1, p^{−1}, …).

According to (d) of the above Remark, ϕ(p/p) = 1, and so

ϕ(p) = ϕ(p/p)ϕ(p) = 1.

In this, p denotes both the element p ∈ ℚ_{p} and the principal idèle (p, p, …). Now consider a = (1, …, 1, u, 1, …), u ∈ ℤ_{p}^{×}, u in the p-position. Write

u^{−1} = a_{0} + a_{1}p + ⋯ + a_{s}p^{s} + ⋯, 0 ≤ a_{i} < p, a_{i} ∈ ℤ,

and let c = a_{0} + ⋯ + a_{s}p^{s} ∈ ℤ. Then uc ∈ 1 + p^{s+1}ℤ_{p}, i.e., for large s it is “close to 1”. Write

ac = (c, c, …, c, 1^{ℓ|c}, c, …, u^{p} c, c, …)(1, …, 1, c^{ℓ|c}, 1, …).

The first factor is ac except that we have moved the components at the primes ℓ dividing c to the second factor. For large s, ϕ(first factor) = 1 by (d) of the above remark. The second factor lies in 𝕀^{S}, and the description we have of ϕ|𝕀^{S} shows that ϕ(second factor) maps ζ to ζ^{c}. In conclusion,

ϕ(a)(ζ) = ζ^{c} = ζ^{u^{−1}}.

Because ϕ is a homomorphism, this completes the explicit description of it.

Remark 4.11. The map π_{𝔪} : 𝕀 → C_{𝔪} is the unique continuous homomorphism such that

(a) π_{𝔪}(K^{×}) = 1;

(b) π_{𝔪}(a) = id(a) for all a ∈ 𝕀^{S(𝔪)}.

If 𝔪|𝔪′, then the composite of π_{𝔪′} with the canonical homomorphism C_{𝔪′} → C_{𝔪} satisfies the conditions characterizing π_{𝔪}. Therefore, the π_{𝔪} combine to give a continuous homomorphism π : 𝕀 → \underleftarrow{lim} C_{𝔪}. We wish to determine the kernel and image of this map.

Because each map π_{𝔪} : 𝕀 → C_{𝔪} is onto, the image is dense. In fact, 𝕀^{1} → C_{𝔪} is onto, and so π_{𝔪}(𝕀^{1}) is dense. But π_{𝔪}(𝕀^{1}) is compact, because π_{𝔪} factors through the compact group 𝕀^{1}/K^{×}, and therefore is complete. This shows that π is onto.

Let 𝕀_{∞}^{+} be the set of idèles a such that a_{v} = 1 if v is finite and a_{v} > 0 if v is real. Thus 𝕀_{∞}^{+} is isomorphic to the identity component of (K ⊗_{ℚ} ℝ)^{×} = ∏_{v|∞} K_{v}^{×}. The kernel of 𝕀 → \underleftarrow{lim} C_{𝔪} contains 𝕀_{∞}^{+} · K^{×}, and hence its closure. In fact, it equals it.

Norms of idèles. Let L be a finite extension of the number field K, let v be a prime of K. Recall from (Math 676, 8.2) that there is a canonical isomorphism

L ⊗_{K} K_{v} → ∏_{w|v} L_{w}.

It follows (ibid. 8.3) that for any α ∈ L,

Nm_{L/K} α = ∏_{w|v} Nm_{L_{w}/K_{v}} α (equality in K_{v}).

For an idèle a = (a_{w}) ∈ 𝕀_{L}, define Nm_{L/K}(a) to be the idèle b ∈ 𝕀_{K} with b_{v} = ∏_{w|v} Nm_{L_{w}/K_{v}} a_{w}. The preceding remark shows that the left hand square in the following diagram commutes, and it is easy to see that the right hand square commutes:

L^{×} ⟶ 𝕀_{L} ⟶^{id} I_{L}
↓^{Nm}^{L/K}  ↓^{Nm}^{L/K}  ↓^{Nm}^{L/K}
K^{×} ⟶ 𝕀_{K} ⟶^{id} I_{K}.

Thus we get a commutative diagram:

C_{L} ⟶ C_{L}
↓^{Nm}^{L/K}  ↓^{Nm}^{L/K}
C_{K} ⟶ C_{K}

(C_{K} = idèle class group 𝕀/K^{×}; C_{K} =ideal class group I/i(K^{×})).

Proposition 4.12. If L/K is a finite extension of local fields of characteristic zero, then

(a) Nm_{L/K}(L^{×}) = ℝ_{>0} (case K = ℝ, L = ℂ);

(b) Nm_{L/K}(L^{×}) ⊃ 1 + 𝔭_{K}^{m} for some m (case K is nonarchimedean);

(c) Nm_{L/K}(L^{×}) ⊃ O_{K}^{×} (case K is nonarchimedean and L/K is unramified).

Proof. Statement (a) is obvious. For (b), see (I.1.9), and for (c), see III.2.2.

Corollary 4.13. Let L/K be a finite extension of number fields. Then Nm_{L/K} 𝕀_{L} ⊃ W_{𝔪} for some modulus 𝔪.

5. The Main Theorms in Terms of Idèles

The statement of the main theorems of class field theory in terms of ideals is very explicit and, for many purposes, it is the most useful one. However, it has some disadvantages. One has to fix a modulus 𝔪, and then the theory describes only the abelian extensions whose conductor divides 𝔪. In particular, it provides no description of the infinite abelian extensions of K. The statement of the main theorems in terms of idèles allows one to consider infinite abelian extensions, or, what amounts to the same thing, all finite abelian extensions simultaneously. It also makes transparent the relation between the local and global Artin maps.

Let L be a finite abelian extension of K. Let v be a prime of K, and let w be a prime of L lying over v. Recall that the decomposition group D(w) of w is the subgroup

D(w) = {σ ∈ Gal(L/K) | σw = w}.

Its elements extend uniquely to automorphisms of L_{w}/K_{v}, and D(w) ≅ Gal(L_{w}/K_{v}). Local class field theory provides us with a homomorphism (the local Artin map)

ϕ_{v} : K_{v} → D(w) ⊂ G.

Lemma 5.1. The subgroup D(w) of G and the map ϕ_{v} are independent of the choice of the prime w|v.

Proof. Any other prime lying over v is of the form σw for some σ ∈ G, and σ : L → L extends by continuity to a homomorphism σ : L_{w} → L_{σw} fixing K_{v}. We have

D(σw) = σD(w)σ^{−1},

which equals D(w) because G is commutative.

Let Ω and Ω′ be maximal abelian extensions of K_{v} containing L_{w} and L_{σw} respectively. From Chapter III, we obtain local Artin maps ϕ_{v} : K^{×} → Gal(Ω/K_{v}) and ϕ′_{v} : K^{×} → Gal(Ω′/K_{v}). The choice of an isomorphism σ̃ : Ω → Ω′ determines an isomorphism

ρ ↦ σ̃ ∘ ρ ∘ σ̃^{−1} : Gal(Ω/K_{v}) → Gal(Ω′/K_{v})

which is independent of σ̃. Moreover, its composite with ϕ_{v} is ϕ′_{v} (because it satisfies the conditions characterizing ϕ′_{v}).

Proposition 5.2. There exists a unique continuous homomorphism ϕ_{K} : 𝕀 → Gal(K^{ab}/K) with the following property: for any L ⊂ K^{ab} finite over K and any prime w of L lying over a prime v of K, the diagram

K_{v}  ⟶^{ϕ_{v}}  Gal(L_{w}/K_{v})
↓  ↓
𝕀_{K} ⟶^{a↦ϕ_{K}(a)|L} Gal(L/K)

commutes.

Proof. Let a ∈ 𝕀, and let L ⊂ K^{ab} be finite over K. If a_{v} ∈ U_{v} and L_{w}/K_{v} is unramified, the ϕ_{v}(a_{v}) = 1 (see III.1). Therefore, ϕ_{v}(a_{v}) = 1 except for finitely many v’s, and so we can define

ϕ_{L/K}(a) = ∏_{v} ϕ_{v}(a_{v}).

(product inside Gal(L/K)). Clearly, ϕ_{L/K} is the unique homomorphism making the above diagram commute.

If L′ ⊃ L, then the properties of the local Artin maps show that ϕ_{L′/K}(a)|L = ϕ_{L/K}(a). Therefore there exists a unique homomorphism ϕ : 𝕀 → Gal(K^{ab}/K) such that ϕ(a)|L = ϕ_{L/K}(a) for all L ⊂ K^{ab}, L finite over K.

Again, the properties of the local Artin maps show that, for any fields K ⊂ K′ ⊂ L ⊂ K^{ab} with L finite over K,

𝕀_{K′}^{S} ⟶^{ϕ_{L/K′}} Gal(L/K′)
↓^{Nm}  ↓^{inclusion}
𝕀_{K}^{S} ⟶^{ϕ_{L/K}} Gal(L/K)

commutes. On taking K′ = L, we find that Nm_{L/K}(𝕀_{L}) is contained in the kernel of ϕ_{L/K}. In particular, the kernel of ϕ_{L/K} contains an open subgroup of 𝕀_{L} (Corollary 4.13), and this implies that ϕ_{K} is continuous.

Theorem 5.3 (Reciprocity Law). The homomorphism ϕ_{K} : 𝕀_{K} → Gal(K^{ab}/K) has the following properties:

(a) ϕ_{K}(K^{×}) = 1;

(b) for every finite abelian extension L of K, ϕ_{K} defines an isomorphism

ϕ_{L/K} : 𝕀_{K}/(K^{×} · Nm(𝕀_{L})) → Gal(L/K).

We saw in the proof of the proposition that ϕ_{L/K}(Nm(𝕀_{L})) = 1, and so (assuming (a) of the theorem) we see that ϕ_{L/K} does factor through 𝕀_{K}/K^{×} · Nm(𝕀_{L}). Part (b) can also be stated as: ϕ defines an isomorphism

ϕ_{L/K} : C_{K}/ Nm(C_{L}) → Gal(L/K).

Example 5.4. Statement (a) of the theorem says that, for any b ∈ K^{×}, ∏ ϕ_{v}(b) = 1. On applying this to the extension K[a^{\frac{1}{n}}]/K under the assumption that K contains a primitive nth root of 1, one obtains the product formula for the Hilbert symbol:

∏_{v}(a, b)_{v} = 1.

See (III.4.3).

Theorem 5.5 (Existence Theorem). Fix an algebraic closure K^{al} of K; for every open subgroup N ⊂ C_{K} of finite index, there exists a unique abelian extension L of K contained in K^{al} such that Nm_{L/K} C_{L} = N.

The subgroups N open and of finite index in C_{K} are called the norm groups, and the abelian extension L of K such that Nm(C_{L}) = N, i.e., such that N = Ker(ϕ_{L/K}), is called the class field of K belonging to N.

As stated, the Existence Theorem is valid for all global fields. In the number field case, all subgroups of finite index in C_{K} are open.

Corollary 5.6. The map L ↦ Nm(C_{L}) is a bijection from the set of finite abelian extensions of K to the set of open subgroups of finite index in C_{K}. Moreover,

L_{1} ⊂ L_{2} ⟺ Nm(C_{L_{1}}) ⊃ Nm(C_{L_{2}});
Nm(C_{L_{1}·L_{2}}) = Nm(C_{L_{1}}) ∩ Nm(C_{L_{2}});
Nm(C_{L_{1}∩L_{2}}) = Nm(C_{L_{1}}) · Nm(C_{L_{2}}).

Remark 5.7. (a) In the number field case, the map

ϕ_{K} : 𝕀_{K} → Gal(K^{ab}/K).

is surjective. For an infinite prime v of K, write K_{v}^{+} for the connected component of K_{v}^{×} containing 1; thus K_{v}^{+} ≈ ℂ^{×} or ℝ_{>0} according as v is complex or real. Clearly ∏_{v|∞} K_{v}^{+} ⊂ Ker(ϕ_{K}). By definition K^{×} ⊂ Ker(ϕ_{K}), and so K^{×} · (∏_{v|∞} K_{v}^{+}) ⊂ Ker(ϕ_{K}). But ϕ_{K} is a continuous homomorphism and Gal(K^{ab}/K) is Hausdorff, and so the kernel is a closed subgroup. Thus Ker(ϕ_{K}) contains the closure of K^{×} · (∏_{v|∞} K_{v}^{+}). It is a theorem that this is precisely the kernel. The image of the closure of K^{×} · (∏_{v|∞} K_{v}^{+}) in C_{K} is the connected component of C_{K} containing 1.

(b) In the function field case, the Artin map ϕ_{K} : 𝕀_{K}/K^{×} → Gal(K^{ab}/K) is injective, but it is not surjective (its image is dense).

Remark 5.8. Assume that the global Artin map ϕ : 𝕀 → Gal(K^{ab}/K) contains K^{×} in its kernel. Then, for every finite abelian extension L/K, ϕ_{L/K} : 𝕀 → Gal(L/K) arises (as in Proposition 4.7) from a homomorphism ψ : I^{S} → Gal(L/K) admitting a modulus. Moreover, because ϕ is the product of the local Artin maps, ψ must be the ideal-theoretic global Artin map (which therefore admits a modulus). It is a straightforward exercise to derive Theorems 3.5 and 3.6 from their idèlic counterparts, Theorems 5.3 and 5.5. We shall prove Theorems 5.3 and 5.5 in Chapter VII.

Example.

Lemma 5.9. The map

(r, t, (u_{p})) ↦ (rt, ru_{2}, ru_{3}, ru_{5}, …) : ℚ^{×} × ℝ_{>0} × ∏ ℤ_{p}^{×} ⟶ 𝕀_{ℚ}

is an isomorphism of topological groups (ℚ^{×} with the discrete topology).

Proof. Any idèle a = (a_{∞}, a_{2}, …, a_{p}, …) can be written

a = a(t, u_{2}, u_{3}, u_{5}, …), a ∈ ℚ^{×}, t ∈ ℝ_{>0}, u_{p} ∈ ℤ_{p}^{×};

—take a = (sign(a_{∞})) ∏ p^{ord_{p}(a_{p})}, t = a_{∞}/a, u_{p} = a_{p}/a. Moreover, the expression is unique because the only positive rational number that is a p-adic unit for all p is 1.

The subsets

{1} × U × ∏_{p finite} U_{p}

with U, U_{p} open neighbourhoods of 1 in ℝ^{×}, ℚ_{p}^{×}, and U_{p}^{×} = ℤ_{p}^{×} for all but finitely many p’s, form a fundamental system of neighbourhoods 1 on the left, and also on the right.

Thus there is a canonical isomorphism of topological groups

C_{ℚ} → ℝ_{>0} × ∏_{p finite} ℤ_{p}^{×} = ℝ_{>0} × ℤ̂^{×}.

Let

ℚ^{cyc} = ⋃ ℚ[ζ_{n}].

In this case, the global reciprocity map is

ϕ : 𝕀_{ℚ} → ℤ̂^{×} → Gal(ℚ^{cyc}/ℚ)

where 𝕀_{ℚ} → ℤ̂^{×} is the above projection map, and ℤ̂^{×} → Gal(ℚ^{cyc}/ℚ) is the canonical isomorphism (see I.5.5c).

Exercise 5.10. Show that every subgroup of finite index in C_{K} is open (K a number field).

Hint: Use that every subgroup of K_{v}^{×} of finite index is open.

6. Appendix: Review of some Algebraic Number Theory

The weak approximation theorem. The theorem in this subsection should have been proved in Math 676. Recall that a valuation on a field K is homomorphism a ↦ |a| : K^{×} → ℝ_{>0} such that |a + b| ≤ |a| + |b| for all a, b ∈ K^{×}. We extend it to K by setting |0| = 0. A valuation is trivial if |a| = 1 for all a ≠ 0. Two nontrivial valuations | · |_{1} and | · |_{2} are equivalent if |a|_{1} < 1 implies |a|_{2} < 1, in which case | · |_{2} = | · |_{1}^{r} for some r ∈ ℝ_{>0} (Math 676, 7.8). The statements in this section continue to hold if we replace “valuation” with “positive power of a valuation” (which, in the archimedean case, may fail to satisfy the triangle rule).

Lemma 6.1. If |·|_{1}, |·|_{2}, …, |·|_{n} are nontrivial inequivalent valuations of K, then there is an element a ∈ K such that

\{ \begin{matrix}|a|_{1} & > & 1 \\ |a|_{i} & < & 1, i ≠ 1. \end{matrix}

Proof. First let n = 2. Because | |_{1} and | |_{2} are inequivalent, there are elements b and c such that

\{ \begin{matrix}|b|_{1} & < & 1, |b|_{2} ≥ 1 \\ |c|_{1} & ≥ & 1, |c|_{2} < 1. \end{matrix}

Now a = \frac{c}{b} has the required properties.

We proceed by induction assuming that the lemma is true for n − 1 valuations. There exist elements b, c such that

\{ \begin{matrix}|b|_{1} & > & 1, |b|_{i} < 1, i = 2, 3, …, n − 1 \\ |c|_{1} & < & 1, |c|_{n} > 1 \end{matrix}

If |b|_{n} ≤ 1, then a = cb^{r} works for sufficiently large r. If |b|_{n} > 1, then a_{r} = \frac{cb^{r}}{1+b^{r}} works for sufficiently large r, because \frac{b^{r}}{1+b^{r}} converges to 0 or 1 according as |b| < 1 or |b| > 1.

Lemma 6.2. In the situation of the last lemma, there exists an element of K that is close to 1 for | · |_{1} and close to 0 for | · |_{i}, i = 2, … n.

Proof. Choose a as in (6.1), and consider a_{r} = \frac{a^{r}}{1+a^{r}}. Then

|a_{r} − 1|_{1} = \frac{1}{|1 + a^{r}|_{1}} ≤ \frac{1}{|a|_{1}^{r} − 1} → 0

as r → ∞. For i ≥ 2,

|a_{r}|_{i} = \frac{|a|_{i}^{r}}{|1 + a|_{i}^{r}} ≤ \frac{|a|_{i}^{r}}{1 − |a|_{i}^{r}} → 0

as r → 0.

Theorem 6.3. Let | · |_{1}, | · |_{2}, …, | · |_{n} be nontrivial inequivalent valuations of a field K, and let a_{1}, …, a_{n} be elements of K. For any ε > 0, there is an element a ∈ K such that |a − a_{i}|_{i} < ε for all i.

Proof. Choose b_{i}, i = 1, …, n, close to 1 for | |_{i} and close to 0 for | |_{j}, j ≠ i. Then

a = a_{1}b_{1} + ⋯ + a_{n}b_{n}

works.

Let K_{i} be the completion of K for | · |_{i}. The statement of the theorem also holds with a_{i} in K_{i} (rather than K)—choose a′_{i} ∈ K very close to a_{i} and a ∈ K very close to each a′_{i}. Thus K (embedded diagonally) is dense in ∏ K_{i}.

The theorem shows that there can be no finite product formula. More precisely:

Corollary 6.4. Let | · |_{1}, | · |_{2}, …, | · |_{n} be nontrivial inequivalent valuations on a field K. If

|a|_{1}^{r_{1}} ⋯ |a|_{n}^{r_{n}} = 1, r_{i} ∈ ℝ,

for all a ∈ K^{×}, then r_{i} = 0 for all i.

Proof. If any r_{i} ≠ 0, an a for which |a|_{i} is sufficiently large and the |a|_{j}, j ≠ i, are sufficiently small provides a contradiction.

The reader should compare the Weak Approximation Theorem with what the Chinese Remainder Theorem gives.

Exercise 6.5. Let | · |_{1}, …, | · |_{n} be the valuations on a number field K corresponding to distinct prime ideals 𝔭_{i}, and let a_{1}, …, a_{n} be elements of K. Let d be a common denominator for the a_{i} (so that da_{i} ∈ O_{K}). Show that, for any ε > 0, there is an element a ∈ K such that |a − a_{i}|_{i} < ε for i = 1, …, n and |a| ≤ 1/|d| for all valuations | · | corresponding to prime ideals other than the 𝔭_{i}.

Hint: Apply the Chinese Remainder Theorem to the da_{i}.

Notes: The Weak Approximation Theorem first occurs in: Artin, E., and Whaples, G., Axiomatic characterization of fields by the product formula for valuations, Bull. AMS, 51, 1945, pp. 469–492. Our account follows the original.

The decomposition of primes. Recall the following theorems from Math 676.

Theorem 6.6. Let A be a Dedekind domain with field of fractions K, and let L be a finite separable extension of K of degree n. Let B be the integral closure of A in K.

(a) Let 𝔭 be a prime ideal of A and write

𝔭B = ∏_{i=1}^{g} 𝔓_{i}^{e}, f_{i} =^{df} [B/𝔓_{i} : A/𝔭_{i}].

Then

∑_{i=1}^{g} e_{i}f_{i} = n.

If L is Galois over K, then Gal(L/K) acts transitively on the 𝔓_{i}, and the numbers e_{i}, f_{i} are independent of i: n = efg.

(b) A prime 𝔭 ramifies in L (i.e., at least one of the e_{i} is > 1) if and only if 𝔭| disc(B/A).

Note that, for a Galois extension, except for the finitely many primes that ramify, the type of the decomposition of 𝔭B into prime factors is described by the single number f.

Theorem 6.7. Let K and A be as in the preceding theorem, and let f(X) be a monic polynomial in A[X]. Let L = K[α] with α a root of f(X), and let B be the integral closure of A in L. The following conditions on 𝔭 are equivalent:

(a) 𝔭 does not divide disc(f(X));

(b) 𝔭 does not ramify in B and A_{𝔭}[α] is the integral closure of A_{𝔭} in L (here A_{𝔭} = {a/b | b ∉ 𝔭});

(c) there is a factorization

f(X) ≡ f_{1}(X) ⋯ f_{g}(X) mod 𝔭

with the f_{i} distinct, monic, and irreducible modulo 𝔭.

When these conditions hold, the factorization of 𝔭 into prime ideals in L is

𝔭B = (𝔭, f_{1}(α)) ⋯ (𝔭, f_{g}(α)).

Thus class field theory is really about polynomials in one variable with coefficients in a number field and their roots: abelian extensions of K correspond to monic irreducible polynomials f(X) ∈ K[X] such that the permutations of the roots of f(X) that give a field automorphism form an abelian group; Theorem 6.7 shows that the factorization of all but finitely many prime ideals of K in an abelian extension L corresponds to the factorization of a polynomial over a finite field.

CHAPTER VI L-Series and the Density of Primes

Euler used the Riemann zeta function in rudimentary form to prove that there are infinitely many prime numbers. In order to prove that the primes are equally distributed among the different arithmetic progressions modulo m, Dirichlet attached L-series (regarded as functions of a real variable) to a character of (ℤ/mℤ)^{×}. Riemann initiated the study of the Riemann zeta function as a function of a complex variable. In this section, we shall (following Weber) extend Dirichlet methods to the study of the distribution of the prime ideals among the classes in a ray class group. Except for the definition of the ray class group, this chapter is independent of the preceding chapters.

In this chapter, we shall need to use a little complex analysis. Recall that the power series 1 + z + \frac{z^{2}}{2!} + ⋯ converges for all z ∈ ℂ to a holomorphic function, which is denoted e^{z}. For any positive real number n and complex number z, n^{z} is defined to be e^{(log n)z} where log is the natural log (function ℝ_{>0} → ℝ inverse to e^{r}).

1. Dirichlet series and Euler products

A Dirichlet series is a series of the form

f(s) = ∑_{n≥1} \frac{a(n)}{n^{s}} a(n) ∈ ℂ, s = σ + it ∈ ℂ.

An Euler product belonging to a number field K is a product of the form

g(s) = ∏_{𝔭} \frac{1}{(1 − θ_{1}(𝔭)ℕ𝔭^{−s}) ⋯ (1 − θ_{d}(𝔭)ℕ𝔭^{−s})}, θ_{i}(𝔭) ∈ ℂ, s ∈ ℂ,

in which 𝔭 runs over all but finitely many of the prime ideals of O_{K}.

Example 1.1. (a) The Riemann zeta function is

ζ(s) = ∑_{n≥1} \frac{1}{n^{s}} = ∏_{p} \frac{1}{1 − p^{−s}} .

It is known that the behaviour of ζ(s), especially in the critical strip 0 ≤ ℜ(s) ≤ 1, is related to the distribution of the prime numbers.

(b) The Dedekind zeta function. For any number field K,

ζ_{K}(s) = ∑_{𝔞≥0} \frac{1}{ℕ𝔞^{s}} = ∏_{𝔭} \frac{1}{1 − ℕ𝔭^{−s}} .

148 VI. L-SERIES AND THE DENSITY OF PRIMES

Here ℕ𝔞 = (O_{K} : 𝔞). The sum is over the integral ideals in O_{K}, and the product is over the prime ideals in O_{K}.

(c) A Dirichlet character is^{1} a homomorphism

χ : I^{𝔪} ⟶ ℂ^{×}

whose kernel contains i(K_{𝔪,1}) for some modulus 𝔪, i.e., χ is a character of the ray class group C_{𝔪}. For such a character, the corresponding Dirichlet L-series is

L(s, χ) = ∑_{𝔞⊂O_{K}, (𝔪,𝔞)=1} \frac{χ(𝔞)}{ℕ𝔞^{s}} = ∏_{(𝔪,𝔭)=1} \frac{1}{1 − χ(𝔭)ℕ𝔭^{−s}} .

(d) A Hecke character (or Grössen character) is a continuous homomorphism

ψ : 𝕀_{K}/K^{×} ⟶ ℂ^{×}

with image in the unit circle. If it is 1 on the identity components of 𝕀_{K} at the infinite primes, then it factors through C_{𝔪} for some 𝔪, and is a Dirichlet character; conversely, a Dirichlet character defines a Hecke character with discrete image. A Hecke character will take the value 1 on some set ∏_{v∉S} U_{v} (S a finite set of primes containing the infinite primes), and the corresponding Hecke L-series is

L_{S}(s, ψ) = ∏_{v∉S} \frac{1}{1 − ψ(π_{v})ℕ𝔭_{v}^{−s}}

where π_{v} is an idèle with a prime element in the v-position and 1 elsewhere.

(e) Let L be a finite Galois extension of K with Galois group G. Let V be a finite dimensional vector space over ℂ and let

ρ : G → GL(V)

be a homomorphism of G into the group of linear automorphisms of V. We refer to ρ as a (finite-dimensional) representation of G. The trace of ρ is the map sending σ to the trace of the automorphism ρ(σ) of V. For σ ∈ G, let

P_{σ}(T) =^{df} det(1 − ρ(σ)T | V) = ∏_{i=1}^{dim V} (1 − a_{i}(σ)T), a_{i} ∈ ℂ,

be the characteristic polynomial of ρ(σ). Because P_{σ}(T) depends only on the conjugacy class of σ, for any prime 𝔭 of K unramified in L, we can define P_{𝔭}(T) to be the characteristic polynomial of (𝔓, L/K) for any prime 𝔓 of L dividing 𝔭. The Artin L-series attached to ρ is

L(s, ρ) = ∏_{𝔭} \frac{1}{P_{𝔭}(ℕ𝔭^{−s})} = ∏_{𝔭} \frac{1}{(1 − a_{1}(𝔭)ℕ𝔭^{−s}) ⋯ (1 − a_{dim V} (𝔭)ℕ𝔭^{−s})}

(product over all unramified primes of K; a_{i}(𝔭) = a_{i}((𝔓, L/K))).

1In the case K = ℚ and 𝔪 = ∞(m), so that C_{𝔪} = (ℤ/mℤ)^{×}, these characters and L-series were introduced by Dirichlet. For arbitary ray class groups, they were introduced by Weber. Some authors restrict the terms “Dirichlet character” and “Dirichlet L-series” to the case ℚ and refer to the more general objects as “Weber characters” and “Weber L-series”. Dirichlet used L to denote his L-functions, and the letter has been used ever since.

2. Convergence Results

We study the elementary analytic properties of Dirichlet series and Euler products.

Dirichlet series.

Proposition 2.1. Let

f(s) = ∑_{n≥1} \frac{a(n)}{n^{s}}.

Write S(x) = ∑_{n≤x} a(n), and suppose there exist positive constants a and b such that |S(x)| ≤ ax^{b} for all large x. Then the series f(s) converges uniformly for s in

D(b, δ, ε) = \{ℜ(s) ≥ b + δ, | arg(s − b)| ≤ \frac{π}{2} − ε\}

for all δ, ε > 0, and it converges to an analytic function on the half plane ℜ(s) > b.

Proof. Since every point s with ℜ(s) > b has a neighbourhood of the form D(b, δ, ε), the second part of the statement follows from the first. To prove the first, we use Cauchy’s criterion for uniform convergence. For large integers n_{1} < n_{2},

| ∑_{n=n_{1}}^{n_{2}} \frac{a(n)}{n^{s}} | = |∑_{n_{1}}^{n_{2}} \frac{s(n) − s(n − 1)}{n^{s}} |

= |∑_{n_{1}}^{n_{2}} \frac{s(n)}{n^{s}} − ∑_{n_{1}−1}^{n_{2}−1} \frac{s(n)}{(n + 1)^{s}} |

= |\frac{s(n_{2})}{n_{2}^{s}} − \frac{s(n_{1} − 1)}{n_{1}^{s}} + ∑_{n_{1}}^{n_{2}−1} s(n) (\frac{1}{n^{s}} − \frac{1}{(n + 1)^{s}} )|

≤ \frac{|s(n_{2})|}{n_{2}^{σ}} + \frac{|s(n_{1} − 1)|}{n_{1}^{σ}} + ∑_{n_{1}}^{n_{2}−1} |s(n)| |s ∫_{n}^{n+1} \frac{dt}{t^{s+1}} |

≤ \frac{a}{n_{2}^{σ−b}} + \frac{a}{n_{1}^{σ−b}} + ∑_{n_{1}}^{n_{2}−1} |s|an^{b} |∫_{n}^{n+1} \frac{dt}{t^{s+1}} |

≤ \frac{2a}{n_{1}^{σ−b}} + ∑_{n_{1}}^{n_{2}−1} |s|a |∫_{n}^{n+1} \frac{t^{b}dt}{t^{s+1}} |
≤ \frac{2a}{n_{1}^{σ−b}} + |s|a ∫_{n_{1}}^{∞} \frac{dt}{t^{σ+1−b}}
≤ \frac{2a}{n_{1}^{σ−b}} − \frac{|s|a}{σ − b} \frac{1}{t^{σ−b}} |_{n_{1}}^{∞}

≤ \frac{2a}{n_{1}^{σ−b}} + \frac{|s|a}{(σ − b)n_{1}^{σ−b}} .

But for s ∈ D(b, δ, ε),

\frac{|s|}{σ − b} = \frac{|s − b + b|}{σ − b} ≤ \frac{|s − b|}{σ − b} + \frac{b}{σ − b} = \frac{1}{cos θ} + \frac{b}{σ − b} ≤ \frac{1}{cos θ} + \frac{b}{δ}

with θ = arg(s − b). Now because |θ| ≤ \frac{π}{2} − ε, \frac{1}{cos θ} is bounded by some number M, and so

| ∑_{n=n_{1}}^{n_{2}} \frac{a(n)}{n^{s}} | ≤ \frac{2a}{n_{1}^{σ−b}} + \frac{(M + \frac{b}{δ})a}{n_{1}^{σ−b}}.

The right hand side of this equation tends to zero as n_{1} → ∞, and so we can apply Cauchy’s criterion to deduce the uniform convergence of f(s).

Remark 2.2. (a) For the Dirichlet series ζ(s), S(x) is [x], and so the series of ζ(s) converges for ℜ(s) > 1. For ζ_{K}(s), S(x) is the number of integral ideals in K with numerical norm ≤ x. It is obvious that S(x) is finite, but in fact (see 2.8) S(x) ≤ Cx. Therefore the series for ζ_{K} (and for L(s, χ)) converge for ℜ(s) > 1. (It is also possible to show directly that the Euler products converge for ℜ(s) > 1, which implies that the Dirichlet series converge. See Fröhlich and Taylor, Algebraic Number Theory, CUP, 1991, VIII.2.2.)

(b) Let f(s) = ∑ \frac{a(n)}{n^{s}} be a Dirichlet series with a(n) ≥ 0. If f(s) converges for all s with ℜ(s) > b, but does not converge on the half-plane {s | ℜ(s) > b − ε} for any ε > 0, then f(s) → ∞ as s → 1 through real numbers > 1. i.e., the domain of convergence of f(s) is limited by a singularity of f situated on the real axis. (See Serre, Cours..., 1970, III.2.3.) For example, the series for ζ(s) does not converge on any half-plane ℜ(s) > 1 − ε, ε > 0, and, as we shall see, ζ(s) does have a pole at s = 1.

Lemma 2.3. The zeta function ζ(s) has an analytic continuation to a meromorphic function on ℜ(s) > 0 with its only (possible) pole at s = 1.

Proof. Define

ζ_{2}(s) = 1 − \frac{1}{2^{s}} + \frac{1}{3^{s}} − \frac{1}{4^{s}} + ⋯

For this Dirichlet series, S(x) = 0 or 1, and so ζ_{2}(s) is analytic for s > 0. Note that

(1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + ⋯) − 2(\frac{1}{2^{s}} + \frac{1}{4^{s}} + \frac{1}{6^{s}} + ⋯) = 1 − \frac{1}{2^{s}} + \frac{1}{3^{s}} − \frac{1}{4^{s}} + ⋯,

that is,

ζ(s) − \frac{2}{2^{s}} ζ(s) = ζ_{2}(s),

or

ζ(s) = \frac{ζ_{2}(s)}{1 − 2^{1−s}} .

Thus ζ(s) is analytic for ℜ(s) > 0 except possibly for poles where 2^{s−1} = 1. But

2^{s−1} = 1 ⟺ e^{(log 2)(s−1)} = 1 ⟺ (log 2)(s − 1) = 2kπi,

and so ζ(s) is analytic except possibly at

s = 1 + \frac{2kπi}{log 2}, k ∈ ℤ.

In fact, the only possible pole is s = 1. To see this, define

ζ_{3}(s) = 1 + \frac{1}{2^{s}} − \frac{2}{3^{s}} + \frac{1}{4^{s}} + \frac{1}{5^{s}} − \frac{2}{6^{s}} + ⋯

and observe (as for ζ_{2}(s)) that ζ_{3}(s) is analytic for s > 0, and

ζ(s) = \frac{ζ_{3}(s)}{1 − 3^{1−s}} .

Hence ζ(s) is analytic for s > 0, except possibly for poles at

s = 1 + \frac{2kπi}{log 3}.

Thus, at a pole for ζ(s), we must have

\frac{2kπi}{log 2} = \frac{2k′πi}{log 3},

or

2^{k′} = 3^{k}, k, k′ ∈ ℤ.

Because of unique factorization, this is possible only if k = 0 = k′.

Lemma 2.4. For s real and s > 1,

\frac{1}{s − 1} ≤ ζ(s) ≤ 1 + \frac{1}{s − 1}.

Hence ζ(s) has a simple pole at s = 1 with residue 1, i.e.,

ζ(s) = \frac{1}{s − 1} + function holomorphic near 1.

Proof. Fix an s > 1, s real. By examining the graph of y = x^{−s}, one finds that

∫_{1}^{∞} x^{−s}dx ≤ ζ(s) ≤ 1 + ∫_{1}^{∞} x^{−s}dx.

But

∫_{1}^{∞} x^{−s}dx = \frac{x^{1−s}}{1 − s}|_{1}^{∞} = \frac{1}{s − 1},

which gives the inequalities. Because ζ(s) is meromorphic near s = 1,

ζ(s) = \frac{c}{(s − 1)^{m}} + \frac{g(s)}{(s − 1)^{m−1}}

near s = 1 for some m ∈ ℕ, c ∈ ℂ, and g(s) holomorphic near s = 1. The inequalities imply that m = 1 and c = 1.

Proposition 2.5. Let f(s) be a Dirichlet series for which there exist real constants C and b, b < 1, such that

|S(n) − a_{0}n| ≤ Cn^{b}.

Then f(s) extends to a meromorphic function on ℜ(s) > b with a simple pole at s = 1 with residue a_{0}, i.e., near s = 1

f(s) = \frac{a_{0}}{s − 1} + holomorphic function

near s = 1.

Proof. For the Dirichlet series f(s)−a_{0}ζ(s), we have |S(n)| ≤ Cn^{b}, and therefore f(s) − a_{0}ζ(s) converges for ℜ(s) > b.

Euler products. Recall that an infinite product ∏_{n=1}^{∞} 1 + b_{n}, b_{n} ∈ ℂ, b_{n} ≠ −1, is said to converge if the sequence of partial products

Π_{m} = ∏_{n=1}^{m} 1 + b_{n}

converges to a nonzero value. Moreover, the product is said to converge absolutely if ∏_{n=1}^{∞} 1+|b_{n}| converges. It is a standard fact that ∏_{n=1}^{∞} 1+b_{n} converges if it converges absolutely, in which case, any reordering of the product converges (absolutely) to the same value.

Lemma 2.6. The product ∏_{1}^{∞} 1 + b_{n} converges absolutely if and only if the series ∑ b_{n} converges absolutely.

Proof. We may suppose that b_{n} ≥ 0 for all n. Then both Π_{m} =_{df} ∏_{1}^{m} 1 + b_{n} and Σ_{m} =_{df} ∑_{1}^{m}b_{n} are monotonically increasing sequences. Since Π_{m} ≥ Σ_{m}, it is clear that Σ_{m} converges if Π_{m} does. For the converse, note that

e^{Σ_{m}} = ∏_{i=1}^{m} e^{b_{i}} ≥ ∏_{i=1}^{m}(1 + b_{i}) = Π_{m}

and so, if the sequence Σ_{m} converges, then the sequence Π_{m} is bounded above, and therefore also converges.

Recall that a product of finite sums, say,

(∑_{i=1}^{l} a_{i})(∑_{i=1}^{m} b_{i})(∑_{i=1}^{n} c_{i})

is a sum

∑_{\begin{matrix}1≤i≤l \\ 1≤j≤m \\ 1≤k≤n \end{matrix}} a_{i}b_{j}c_{k}

of products, each of which contains exactly one term from each sum. Also that

\frac{1}{1 − t} = 1 + t + t^{2} + ⋯, |t| < 1.

Hence (formally at least),

∏_{p} \frac{1}{1 − p^{−s}} = (1 + 2^{−s} + (2^{2})^{−s} + ⋯)(1 + 3^{−s} + (3^{2})^{−s} + ⋯)(1 + 5^{−s} + (5^{2})^{−s} + ⋯) ⋯
= ∑ n^{−s}

because each positive integer can be written as a product of powers of primes in exactly one way. This identity is sometimes referred to as the analytic form of unique factorization. We now prove a more general result.

Proposition 2.7. Let χ be a Dirichlet character of a number field K. For all s with ℜ(s) > 1, the Euler product ∏_{𝔭∤𝔪} \frac{1}{1−χ(𝔭)ℕ𝔭^{−s}} converges to L(s, χ).

Proof. For ℜ(s) > 1,

\frac{1}{1 − χ(𝔭)ℕ𝔭^{−s}} = 1 + \frac{χ(𝔭)}{ℕ𝔭^{s}} + \frac{χ(𝔭^{2})}{(ℕ𝔭^{2})^{s}} + ⋯

Now

∏_{\begin{matrix}(𝔭,𝔪)=1 \\ ℕ𝔭≤t_{0} \end{matrix}} \frac{1}{1 − χ(𝔭)ℕ𝔭^{−s}} = ∑ \frac{χ(𝔞)}{ℕ(𝔞)^{−s}}

where the second sum runs over all integral ideals expressible as a product of prime ideals with numerical norm ≤ t_{0}. As t_{0} → ∞, the right hand side converges (absolutely) to L(s, χ). Therefore the infinite product converges, and its value is L(s, χ).

Partial zeta functions; the residue formula. Let K be a number field, let 𝔪 be a modulus. For any class 𝔨 in C_{𝔪} =^{df} I^{𝔪}/i(K_{𝔪,1}), we define the partial zeta function

ζ(s, 𝔨) = ∑_{𝔞≥0, 𝔞∈𝔨} \frac{1}{ℕ𝔞^{s}} (sum over the integral ideals in 𝔨).

Note that for any character χ of C_{𝔪},

L(s, χ) = ∑_{𝔨∈C_{𝔪}} χ(𝔨)ζ(s, 𝔨).

In particular,

ζ_{K}(s) = ∑_{𝔨∈C_{𝔪}} ζ(s, 𝔨).

Therefore, knowledge of the ζ(s, 𝔨) will provide us with information about L(s, χ) and ζ_{K}(s).

Let

S(x, 𝔨) = #{𝔞 ∈ 𝔨 | 𝔞 integral ℕ𝔞 ≤ x} ,

i.e., it is the S(x) for the Dirichlet series ζ(s, 𝔨). Recall from Math 676 that there is a homomorphism

ℓ : U ⟶ ℝ^{r+s}, u ↦ (log |σ_{1}(u)|, …, 2 log |σ_{r+s}(u)|)

whose kernel is the torsion subgroup of U and whose image is an r + s − 1 dimensional lattice. The regulator reg(K) is defined to be the volume of a fundamental parallelopiped for this lattice. Let U_{𝔪,1} = U ∩ K_{𝔪,1}. Then U_{𝔪,1} has finite index in U, and we define reg(𝔪) to be the volume of the fundamental parallelopiped for ℓ(U_{𝔪,1}). Thus reg(𝔪) = reg(K)(U : U(𝔪)).

Proposition 2.8. For all x ≥ 1,

|S(x, 𝔨) − g_{𝔪}x| ≤ Cx^{1− \frac{1}{d}} , g_{𝔪} = \frac{2^{r}(2π)^{s}reg(𝔪)}{w_{𝔪}ℕ(𝔪)|Δ_{K/ℚ}|^{\frac{1}{2}}} , d = [K : ℚ],

where

r = number of real primes,
s = number of complex primes,
w_{𝔪} = number of roots of 1 in K_{𝔪,1},
ℕ(𝔪) = ℕ(𝔪_{0})2^{r_{0}},
r_{0} = number of real primes in 𝔪, and
Δ_{K/ℚ} = discriminant of K/ℚ.

Proof. First show that there is an integral ideal 𝔟_{0} ∈ 𝔨^{−1}. Then for any 𝔞 ∈ 𝔨, 𝔞 integral, 𝔞𝔟_{0} = (α), some α ∈ O_{K}. Now S(x, 𝔨) is the number of principal ideals (α) such that α ∈ 𝔟_{0} ∩ K_{𝔪,1} with | Nm(α)| ≤ xℕ(𝔟_{0}). Now count. The techniques are similar to those in the proof of the unit theorem. For the details, see Lang 1970, VI.3, Theorem 3. (A slightly weaker result is proved in Janusz 1996, IV.2.11).

Corollary 2.9. The partial zeta function ζ(s, 𝔨) is analytic for ℜ(s) > 1 − \frac{1}{d} except for a simple pole at s = 1, where it has residue g_{𝔪}.

Proof. Apply Proposition 2.5.

Note that g_{𝔪} does not depend on 𝔨.

Lemma 2.10. If A is a finite abelian group, and χ: A → ℂ^{×} is a nontrivial character (i.e., homomorphism not mapping every element to 1), then

∑_{a∈A} χ(a) = 0.

Proof. Because χ is nontrivial, there is a b ∈ A such that χ(b) ≠ 1. But

∑_{a∈A} χ(a) = ∑_{a∈A} χ(ab) = (∑_{a} χ(a))χ(b),

and so

(χ(b) − 1) ∑_{a} χ(a) = 0,

which implies that ∑_{a} χ(a) = 0.

Corollary 2.11. If χ is not the trivial chaaracter, then L(s, χ) is analytic for ℜ(s) > 1 − \frac{1}{d}.

Proof. Near s = 1,

L(s, χ) = ∑_{𝔨∈C_{𝔪}} χ(𝔨) · ζ(s, 𝔨) = \frac{∑_{𝔨∈C_{𝔪}} χ(𝔨)g_{𝔪}}{s − 1} + holomorphic function,

and the lemma shows that the first term is zero.

Later we shall see that L(1, χ) ≠ 0.

Corollary 2.12. The Dedekind zeta function ζ_{K}(s) is analytic for ℜ(s) > 1 − \frac{1}{d} except for a simple pole at s = 1, where it has residue

\frac{2^{r}(2π)^{s}reg(K)}{w_{K}|Δ|^{\frac{1}{2}}}h_{K}

Proof. Recall that ζ_{K}(s) = ∑_{𝔨∈C_{K}} ζ(s, 𝔨).

Example 2.13. (a) For K = ℚ, the last formula becomes 1 = \frac{2}{2}.

(b) For K = ℚ[\sqrt{d}], the formula becomes

lim_{s→1}(s − 1)ζ(s) = \{ \begin{matrix}\frac{2 log(u)}{Δ^{\frac{1}{2}}}h_{K}, & u > 1 a fundamental unit , d > 0 \\ _{\frac{2π}{w_{K}|Δ|^{\frac{1}{2}}}} h_{K}, & d < 0. \end{matrix}

It is possible to find a closed formula for the expression on the left, and this leads to a very simple expression for the class number. Recall that the Artin map for K/ℚ can be regarded as a character χ: I^{S} → {±1} where S is the set of primes that ramify. Rather than a map on ideals, we regard it as a map on positive integers, and we extend it to all positive integers by setting χ(m) = 0 if m is divisible by a prime that ramifies in K. Thus χ is now the multiplicative map on the set of positive integers taking the values

χ(p) = \{ \begin{matrix}1 & if p splits in K \\ −1 & if p remains prime in K \\ 0 & if p ramifies in K. \end{matrix}

For a quadratic imaginary field with discriminant < −4, the formula becomes

h_{K} = \frac{1}{2 − χ(2)} ∑_{\begin{matrix}(x,Δ)=1 \\ 0<x<|Δ|/2 \end{matrix}} χ(x).

For example, if K = ℚ[\sqrt{−5}], then |Δ| = 20, and

h = \frac{1}{2 − 0}(χ(1) + χ(3) + χ(7) + χ(9)) = \frac{4}{2} = 2,

because 2 ramifies, and

−5 ≡ 1 ≡ 1^{2} mod 3, −5 ≡ 2 ≡ 3^{2} mod 7.

See Borevich and Shafarevich, 1966, Chapter 5, Section 4, for more details.

3. Density of the Prime Ideals Splitting in an Extension

For a set T of prime ideals of K, we define ζ_{K,T} (s) = ∏_{𝔭∈T} \frac{1}{1−ℕ𝔭^{−s}}. If some positive integral power ζ_{K,T} (s)^{n} of ζ_{K,T} (s) extends to a meromorphic function on a neighbourhood of 1 having a pole of order m at 1, then we say^{2} that T has polar density δ(T) = m/n.

Proposition 3.1. (a) The set of all prime ideals of K has polar density 1.

(b) The polar density of any set (having one) is ≥ 0.

(c) Suppose that T is the disjoint union of T_{1} and T_{2}. If any two of T, T_{1}, T_{2} have polar densities, then so also does the third, and δ(T) = δ(T_{1}) + δ(T_{2}).

2Following Marcus, Number Fields, Springer 1977, p188.

(d) If T ⊂ T′, then δ(T) ≤ δ(T′) (when both are defined).

Proof. (a) We know that ζ_{K}(s) itself extends to a neighbourhood of 1, and has a simple pole at 1.

(b) To say that T has negative density means that ζ_{K,T} (s) is holomorphic in a neighbourhood of s = 1, and is zero there. But ζ_{K,T} (1) = ∏_{p∈T} \frac{1}{1−p^{−1}} > 0.

(c) Clearly,

ζ_{K,T} (s) = ζ_{K,T_{1}}(s) · ζ_{K,T_{2}}(s).

Suppose, for example, that ζ_{K,T} (s)^{m} and ζ_{K,T_{1}}(s)^{m_{1}} extend to meromorphic functions in neighbourhoods of 1, with poles of order n and n_{1} at 1. Then ζ_{K,T_{2}}(s)^{mm_{1}} = ζ_{K,T} (s)^{mm_{1}}/ζ_{K,T_{1}}(s)^{mm_{1}} extends to a meromorphic function in a neighbourhood of 1, and has a pole of order m_{1}n − mn_{1} at 1. Therefore

δ(T_{2}) = \frac{m_{1}n}{mm_{1}} − \frac{mn_{1}}{mm_{1}} = δ(T) − δ(T_{2}).

(d) Combine (c) with (b).

Proposition 3.2. If T contains no primes for which ℕ𝔭 is a prime (in ℤ), then δ(T) = 0.

Proof. For 𝔭 ∈ T, ℕ𝔭 = p^{f} with f ≥ 2. Moreover, for a given p, there are at most [K : ℚ] primes of K lying over p. Therefore ζ_{K,T} (s) can be decomposed into a product ∏_{i=1}^{d} g_{i}(s) of d infinite products over the prime numbers each factor of a g_{i}(s) being 1 or of the form \frac{1}{1−p^{f}} with f ≥ 2. For each i, g_{i}(1) ≤ ∑_{n>0} n^{−2} = ζ(2). Therefore g_{i}(s) is holomorphic at 1.

Corollary 3.3. Let T_{1} and T_{2} be sets of prime ideals in K. If the sets differ only by primes for which ℕ𝔭 is not prime and one has a polar density, then so does the other, and the densities are equal.

Theorem 3.4. Let L be a finite extension of K, and let M be its Galois closure. Then the set of prime ideals of K that split completely in L has density 1/[M : K].

Proof. A prime ideal 𝔭 of K splits completely in L if and only if it splits completely in M. Therefore, it suffices to prove the theorem under the assumption that L is Galois over K. Let S be the set of prime ideals of K that split completely in L, and let T be the set of prime ideals of L lying over a prime ideal in S. Corresponding to each 𝔭 in S, there are exactly [L : K] prime ideals 𝔓 in T, and for each of them Nm_{L/K} 𝔓 = 𝔭, and so ℕ𝔓 = ℕ𝔭. Therefore, ζ_{K,S}(s) = ζ_{L,T} (s)^{[L:K]}. But T contains every prime ideal of L that is unramified in L/K for which ℕ𝔓 = p. Therefore T differs from the set of all prime ideals in L by a set of polar density 0, and so T has density 1. This implies that ζ_{K,S}(s) has the property signifying that S has density 1/[L : K].

Corollary 3.5. If f(X) ∈ K[X] splits into linear factors modulo 𝔭 for all but finitely many prime ideals 𝔭, then f splits into linear factors in K.

Proof. Apply the theorem to the splitting field of f.

Corollary 3.6. For Galois extensions L and M of a number field K,

L ⊂ M ⟺ Spl(L) ⊃ Spl(M).

Hence

L = M ⟺ Spl(L) = Spl(M),

and

L ↦ Spl(L)

is an injection from the set of finite Galois extensions of K (contained in some fixed algebraic closure) to the set of subsets of {𝔭 ⊂ O_{K}}.

Proof. See the proof of (V.3.23).

Example 3.7. Let f(X) be an irreducible polynomial of degree 3. The density of the set of primes 𝔭 for which f(X) splits modulo 𝔭 is 1/3 or 1/6 depending on whether f(X) has Galois group C_{3} or S_{3}.

Corollary 3.8. For any abelian extension L/K and any finite set S ⊃ S_{∞} of primes of K including those that ramify in L, the Artin map ψ_{L/K} : I^{S} → Gal(L/K) is surjective.

Proof. Let H be the image of ψ_{L/K}. For all 𝔭 ∉ S, (𝔭, L^{H}/K) = (𝔭, L/K)|L^{H} = 1, which implies that 𝔭 splits in L^{H}. Hence all but finitely many prime ideals of K split in L^{H}, which implies that [L^{H} : K] = 1.

4. Density of the Prime Ideals in an Arithmetic Progression

Let f(s) and g(s) be two functions defined (at least) for s > 1 and real. We write

f(s) ∼ g(s) as s ↓ 1

if f(s) − g(s) is bounded for

1 < s < 1 + ε, s real, some ε > 0.

Note that

f(s) ∼ δ log \frac{1}{s − 1} as s ↓ 1

implies

lim_{s↓1} \frac{f(s)}{log \frac{1}{s−1}} = δ.

When f(s) and g(s) are functions holomorphic in a neighbourhood of s = 1 except possibly for poles at s = 1,

f(s) ∼ g(s) as s ↓ 1

if and only if f(s) and g(s) differ by a function that is holomorphic on a neighbourhood of 1.

Let T be a set of primes of K. If there exists a δ such that

∑_{𝔭∈T} \frac{1}{ℕ𝔭^{s}} ∼ δ log \frac{1}{s − 1} ass ↓ 1,

then we say that T has Dirichlet density δ.

If the limit

lim_{x→∞} \frac{number of 𝔭 ∈ T with ℕ𝔭 ≤ x}{number of 𝔭 with ℕ𝔭 ≤ x}

exists, then we call it the natural density of T.

Proposition 4.1. (a) If the polar density exists, then so also does the Dirichlet

density, and the two are equal.

(b) If the natural density exists, then so also does the Dirichlet density, and the two are equal.

Proof. (a) If T has polar density m/n, then

ζ_{K,T} (s)^{n} = \frac{a}{(s − 1)^{m}} + \frac{g(s)}{(s − 1)^{m−1}}

where g(s) is holomorphic near s = 1. Moreover, a > 0 because ζ_{K,T} (s) > 0 for s > 1 and real. On taking logs, we find that

n ∑_{T} \frac{1}{ℕ𝔭^{s}} ∼ m log \frac{1}{s − 1} as s ↓ 1,

which shows that T has Dirichlet density m/n.

(b) See Goldstein 1971, p252.

Remark 4.2. (a) A set T may have a Dirichlet density without having a natural density. For example, let T be the set of prime numbers whose leading digit (in the decimal system) is 1. Then T does not have a natural density, but its Dirichlet density is log_{10}(2) = ·3010300 … (statement in Serre 1970, Cours..., VI.4.5). Thus it is a stronger statement to say that a set of primes has natural density δ than that it has Dirichlet density. All of the sets whose densities we compute in this course will also have natural densities, but we do not prove that.

(b) By definition, polar densities are rational numbers. Therefore any set having a natural density that is not rational will not have a polar density.

Recall that the exponential function

e^{z} = ∑ \frac{z^{n}}{n!} = e^{x}(cos y + i sin y), z = x + iy,

defines an isomorphism from

{z ∈ ℂ | −π < ℑ(z) < π}

onto the complement of the negative real axis

{z ∈ ℝ | z ≤ 0}

in ℂ whose inverse is, by definition, the (principal branch of) the logarithm function log. With this definition

log z = log |z| + i arg z

where the log on the right is the function defined in elementary calculus courses and

−π < arg z < π.

With this definition

log \frac{1}{1 − z} = z + \frac{z^{2}}{2} + \frac{z^{3}}{3} + ⋯, |z| < 1.

Lemma 4.3. Let u_{1}, u_{2}, … be a sequence of real numbers ≥ 2 such that

f(s) =^{df} ∏_{j=1}^{∞} \frac{1}{1 − u_{j}^{−s}}

is uniformly convergent on each region D(1, δ, ε), δ, ε > 0. Then

log f(s) ∼ ∑ \frac{1}{u_{j}^{s}} ass ↓ 1.

Proof. We have

log f(s) = ∑_{j=1}^{∞} log \frac{1}{1 − u_{j}^{−s}}

= ∑_{j} ∑_{m=1}^{∞} \frac{1}{mu_{j}^{sm}}

= ∑_{j} \frac{1}{u_{j}^{s}} + ∑_{j} ∑_{m=2}^{∞} \frac{1}{mu_{j}^{sm}}
= ∑_{j} \frac{1}{u_{j}^{s}} + g(s),

where

|g(s)| ≤ ∑_{j=1}^{∞} ∑_{m=2}^{∞} |\frac{1}{mu_{j}^{sm}} | = ∑_{j=1}^{∞} ∑_{m=2}^{∞} \frac{1}{mu_{j}^{mσ}} , σ = ℜ(s).

Estimate the inner sum by using (u ≥ 2, σ > 1)

∑_{m=2}^{∞} \frac{1}{mu^{mσ}} ≤ ∑_{m=2}^{∞} \frac{1}{2} (\frac{1}{u^{σ}} )^{m} = \frac{1}{2} \{\frac{1}{1 − u^{−σ}} − u^{−σ} − 1\} = \frac{1}{2} \frac{u^{−2σ}}{1 − u^{−σ}} < \frac{1}{u^{2σ}} .

Hence

|g(s)| ≤ f(2σ).

Because f(s) is holomorphic for ℜ(s) > 1, f(2s) is holomorphic for ℜ(s) > \frac{1}{2}, and so g(σ) is bounded as σ ↓ 1.

Proposition 4.4. (a) The set of all prime ideal of K has Dirichlet density 1.

(b) The Dirichlet density of any set (having one) is ≥ 0.

(c) If T is finite, then δ(T) = 0.

(d) Suppose that T is the disjoint union of T_{1} and T_{2}. If any two of δ(T_{1}), δ(T_{2}), δ(T) are defined, so is the third, and δ(T) = δ(T_{1}) + δ(T_{2}).

(e) If T ⊂ T′, then δ(T) ≤ δ(T′) (assuming both are defined).

Proof. (a) The set of all primes ideals even has polar density 1.

(b) For s > 0 real, \frac{1}{ℕ𝔭^{s}} > 0, and for s = 1 + ε, log \frac{1}{s−1} = − log ε, which is positive for 0 < ε < 1.

(c) When T is finite, ∑_{𝔭∈T} \frac{1}{ℕ𝔭^{s}} is holomorphic for all s and hence bounded near any point.

(d) Clearly

∑_{𝔭∈T} \frac{1}{ℕ𝔭^{s}} = ∑_{𝔭∈T_{1}} \frac{1}{ℕ𝔭^{s}} + ∑_{𝔭∈T_{2}} \frac{1}{ℕ𝔭^{s}} ℜ(s) > 1.

Therefore, if, for example,

∑_{𝔭∈T_{1}} \frac{1}{ℕ𝔭^{s}} ∼ δ_{1} log \frac{1}{s − 1},  ∑_{𝔭∈T_{2}} \frac{1}{ℕ𝔭^{s}} ∼ δ_{2} log \frac{1}{s − 1},

then

∑_{𝔭∈T} \frac{1}{ℕ𝔭^{s}} ∼ (δ_{1} + δ_{2}) log \frac{1}{s − 1}.

(e) If both δ(T) and δ(T′) exist, then so also does δ(T′ \ T), and

δ(T′) − δ(T) =^{(c)} δ(T′ \ T) ≥^{(a)} 0.

Proposition 4.5. Let T be the set of prime ideals of K having degree 1 over ℚ, i.e., such that the residue class degree f(𝔭/p) = 1. Then δ(T) = 1.

Proof. The complement of T has polar density 1 (Proposition 3.2)

Corollary 4.6. Let T be as in the Proposition. For any set S of primes of K having a Dirichlet density

δ(T ∩ S) = δ(S).

Proof. The complement T′ of T has density 0, and it follows easily that δ(S∩T′) = 0. Because S is the disjoint union of S ∩ T and S ∩ T′, this implies that δ(S ∩ T) is defined and equals δ(S).

Lemma 4.7. Let A be a finite abelian group, and let a ∈ A. Then

∑_{χ∈A^{∨}} χ(a) = .

Here A^{∨} is the group of characters of A, i.e., A^{∨} = Hom(A, ℂ^{×}).

Proof. If a = 1, then χ(a) = 1 for all χ, and so the statement follows from the fact that A^{∨} has the same number of elements as A (it is in fact noncanonically isomorphic to A). If a ≠ 1, there is a character χ_{1} such that χ_{1}(a) ≠ 1. Then

∑_{χ∈A^{∨}} χ(a) = ∑_{χ∈A^{∨}}(χ_{1}χ)(a) = ∑_{χ∈A^{∨}} χ_{1}(a)χ(a) = χ_{1}(a) ∑_{χ∈A^{∨}} χ(a).

Since χ_{1}(a) ≠ 1, this implies that ∑_{χ∈A^{∨}} χ(a) = 0.

Alternatively, identify A with A^{∨}^{∨} by means of the isomorphism

a ↦ (χ ↦ χ(a)): A ⟶ (A^{∨})^{∨},

and apply (2.10).

Theorem 4.8. Let 𝔪 be a modulus for K, and let H be a congruence subgroup for 𝔪:

I^{𝔪} ⊃ H ⊃ i(K_{𝔪,1}).

Then

δ(\{𝔭 ∈ H\}) = \{ \begin{matrix}1/(I^{S(𝔪)} : H) & if L(1, χ) is nonzero for all characters χ ≠ χ_{0} of I^{S(𝔪)}/H; \\ 0 & otherwise. \end{matrix}

Proof. Let χ be a character of I^{𝔪} trivial on H, and let

L(s, χ) = ∏_{𝔭∤𝔪} \frac{1}{1 − χ(𝔭)ℕ𝔭^{−s}} .

Then the argument in the proof of (4.3) shows that

log L(s, χ) − ∑_{𝔭∤𝔪} \frac{χ(𝔭)}{ℕ𝔭^{s}}

is holomorphic for ℜ(s) > \frac{1}{2}. In particular,

log(L(s, χ)) ∼ ∑_{𝔭∤𝔪} \frac{χ(𝔭)}{ℕ𝔭^{s}} as s ↓ 1.

But (see 4.7)

∑_{χ} χ(𝔭) = \{ \begin{matrix}h & 𝔭 ∈ H \\ 0 & 𝔭 ∉ H, \end{matrix}

and so, on summing over all χ, we find that

∑_{χ} log L(s, χ) ∼ h ∑_{𝔭∈H} \frac{1}{ℕ𝔭^{s}} as s ↓ 1.

If χ ≠ χ_{0}, then L(s, χ) is holomorphic near s = 1, say L(s, χ) = (s − 1)^{m(χ)}g(s) where m(χ) ≥ 0 and g(1) ≠ 0. Thus

log L(s, χ) ∼ m(χ) log(s − 1) = −m(χ) log \frac{1}{s − 1}.

If χ = χ_{0}, then L(s, χ) = ζ_{K}(s)/ ∏_{𝔭|𝔪} \frac{1}{1−ℕ𝔭^{−s}}, and so

log L(s, χ_{0}) ∼ log ζ_{K}(s) ∼ log \frac{1}{s − 1} as s ↓ 1.

On combining these statements, we find that

h ∑_{𝔭∈H} ℕ𝔭^{−s} ∼ (1 − ∑_{χ≠χ_{0}} m(χ)) log \frac{1}{s − 1},

and hence

δ(\{𝔭 ∈ H\}) = \frac{1 − ∑_{χ≠χ_{0}} m(χ)}{h}.

This shows that δ(\{𝔭 ∈ H\}) = \frac{1}{h} if L(1, χ) ≠ 0 for all χ ≠ χ_{0}, and δ(\{𝔭 ∈ H\}) = 0 otherwise (and at most one L(s, χ) can have a zero at s = 1, and it can only be a simple zero).

The Second Inequality.

Theorem 4.9. For any Galois extension L of K and modulus 𝔪 of K,

(I^{S(𝔪)} : i(K_{𝔪,1}) · Nm(I_{L}^{S(𝔪)})) ≤ [L : K].

Proof. Let H = Nm_{L/K} I_{L}^{𝔪} ·K^{×}. From Theorem 4.8, we know that δ({𝔭 ∈ H}) = 1/(I^{S(𝔪)} : H) or 0, and that the first case holds exactly when, for all nontrivial characters χ of I^{S}/H, L(1, χ) ≠ 0.

If 𝔭 splits in L, i.e., f(𝔓/𝔭) = 1 for all 𝔓|𝔭, then 𝔭 is the norm of any prime ideal of O_{L} lying over it, and so {𝔭 ∈ H} contains the set of prime ideals splitting in L. Hence, Theorem 3.4 shows that

δ({𝔭 ∈ H}) ≥ [L : K]^{−1} ≠ 0.

On combining the two statements we find

(a) δ({𝔭 ∈ H}) ≠ 0;

(b) that for all nontrivial characters χ of I^{S}/H, L(1, χ) ≠ 0;

(c) (I^{S} : H) = δ({𝔭 ∈ H})^{−1} ≤ [L : K].

Corollary 4.10. Let χ be a nontrivial character of C_{𝔪}, and suppose that there is a Galois extension L of K such that Nm_{L/K} C_{𝔪,L} ⊂ Ker(χ). Then L(1, χ) ≠ 0.

Proof. This was shown in the course of the proof of the theorem.

The Reciprocity Law (Theorem V.3.5) implies that the hypothesis of the corollary holds for all χ. It is possible to prove that L(1, χ) ≠ 0 without using class field theory, but, at this point we prefer to return to class field theory. We shall complete the proof of the Chebotarev Theorem in Chapter VIII.

CHAPTER VII Global Class Field Theory: Proofs of the Main Theorems

In this chapter we prove the main theorems of global class field theory, namely, the Reciprocity Law and the Existence Theorem (Theorems V.5.3, V.5.5), following the method of Tate’s article in Cassels and Fröhlich 1967 (see also Artin and Tate 1951/52). Throughout, we work with idèles rather than ideals.

This chapter is independent of Chapter VI, except that Theorem VI.4.9 can be used to replace Section 6. We shall need to refer to Chapter V only for the definitions of the idèle class group and the the definition of the global Artin map ϕ : 𝕀 → Gal(L/K) as the “product” of the local Artin maps (Section 5). On the other hand, we shall make frequent use of the results in Chapters II and III.

1. Outline

Let L/K be a finite Galois extension of number fields with Galois group G. The idèle class group C_{L} =^{df} 𝕀_{L}/L^{×} plays the same role for global class field theory that the multiplicative group L^{×} plays for local class field theory. In particular, when L/K is abelian, we shall prove that there is a isomorphism

ϕ : C_{K}/ Nm_{L/K}(C_{L}) → G

whose local components are the local Artin maps, i.e., such that for any prime v of K and prime w of L lying over it, the following diagram commutes,

K_{v}^{×} ⟶^{ϕ_{v}} Gal(L_{w}/K_{v})
↓  ↓
𝕀_{K} ⟶^{ϕ} Gal(L/K)

where ϕ_{v} is the local Artin map of Chapters II and IV.

According to Tate’s theorem II.2.18, to obtain such an isomorphism, it suffices to prove that, for every finite Galois extension L/K with Galois group G,

(a) H^{1}(G, C_{L}) = 0;

(b) H^{2}(G, C_{L}) is cyclic of order [L : K] with a canonical generator u_{L/K};

(c) if E ⊃ L ⊃ K are two finite Galois extensions of K, then Res(u_{E/K}) = u_{L/K}.

The isomorphism ϕ_{L/K} is then the inverse of that defined by u_{L/K},

H_{T}^{−2}(G, ℤ) → H_{T}^{0}(G, C_{L}).

Once the fundamental class u_{L/K} has been shown to be compatible with the local fundamental classes, ϕ_{L/K} will be a product of the local Artin maps.

In fact, we adopt a slightly different approach. We shall define the global Artin map C_{K} → Gal(L/K) to be the “product” of the local Artin maps, and we shall use results slightly weaker than (a) and (b) to deduce that it has the correct properties.

In Section 2, we express the cohomology of the idéles in terms of the cohomology of the local fields,

H^{0}(G, 𝕀_{L}) = 𝕀_{K}; H_{T}^{r} (G, 𝕀_{L}) = ⊕_{v}H_{T}^{r} (G^{v}, L^{v×})

(sum over the primes v of K; for some choice of a prime w|v, G^{v} is the decomposition group of w and L^{v} = L_{w}). After computing the Herbrand quotient of the group of S-units in Section 3, we prove the first inequality,

For any cyclic extension L/K, (C_{K} : Nm_{L/K} C_{L}) ≥ [L : K].

in Section 4. We also prove in Section 4 that, for any abelian extension, the Galois group is generated by the Frobenius elements. In Section 5 we state the theorem,

For any Galois extension L/K of number fields,

(a) (C_{K} : Nm_{L/K} C_{L}) ≤ [L : K] (second inequality);

(b) H^{1}(G, C_{L}) = 1;

(c) H^{2}(G, C_{L}) has order ≤ [L : K].

and we prove it using Theorem VI.4.9. In the following section, we give a different proof of the theorem that avoids the use complex analysis.

After some preliminaries on Brauer groups, in Section 7 we complete the proof of the reciprocity law by showing that, for any abelian extension L/K, K^{×} is contained in the kernel of ϕ_{L/K} : 𝕀_{K} → Gal(L/K). Because we already know that Nm_{L/K}(𝕀_{L}) is contained in the kernel of ϕ_{L/K} and that ϕ_{L/K} is surjective (because Gal(L/K) is generated by the Frobenius elements), the second inequality implies that ϕ_{L/K} is an isomorphism.

We prove the Existence Theorem in Section 9 by showing the every (open) subgroup of finite index in C_{K} contains the norm group of some subextension of the extension obtained by first adjoining a root of unity to K and then making a Kummer extension.

To some extent, the cyclic cyclotomic extensions of K play the same role as the unramified extensions of a local field. For example, a key point in the last step of the proof of the Reciprocity Law is that every element of Br(K) is split by a cyclic cyclotomic extension.

2. The Cohomology of the Idèles

Let L/K be a finite Galois extension of number fields with Galois group G. Recall that σ ∈ G acts on the primes w of L lying over a fixed prime v of K according to the rule |σa|_{σw} = |a|_{w}. Therefore σ is an isomorphism of valued fields

(L, | |_{w}) ⟶ (L, | |_{σw}),

and so extends to the completions: there is a commutative diagram

L_{w} ⟶^{σ} L_{σw}
↑^{i_{w}}  ↑^{i_{σw}}
L ⟶^{σ} L.

Fix a prime v of K, and let w_{0} be a prime of L lying over v. The map σ ↦ σw_{0} defines a bijection

G/G_{w_{0}} ⟶ {w|v},

where G_{w_{0}} is the decomposition group of w_{0}.

We wish to extend the action of G on L to an action of G on ∏_{w|v} L_{w}. Recall (Math 676, 8.2) that the map

a ⊗ b ↦ i_{w}(a)b : L ⊗_{K} K_{v} ⟶ ∏_{w|v} L_{w}

is an isomorphism. The group G acts on L ⊗_{K} K_{v} through its action on L, and we use the isomorphism to transfer this action to ∏_{w|v} L_{w}. Thus,

(a) the elements of G acts continuously on ∏_{w|v} L_{w};

(b) all elements of the form (a, …, a), a ∈ K_{v}, are fixed by G;

(c) for any a ∈ L, σ(…, i_{w}(a), …) = (…, i_{w}(σa), …).

These conditions determine the action uniquely.

Lemma 2.1. For σ ∈ G and α = (α(w)) ∈ ∏_{w|v} L_{w},

(σα)(w) = σ(α(σ^{−1}w)). (∗)

Proof. The rule (∗) does define a continuous action of G on ∏ L_{w}, and so it suffices to check that it satisfies (b) and (c). Condition (b) is obvious. For (c), let α(w) = i_{w}(a), a ∈ L. Then (by (∗))

(σα)(w) = σ(i_{σ^{−1}w}(a)).

When we replace w with σ^{−1}w in the commutative diagram above, we obtain the formula σ ∘ i_{σ^{−1}w} = i_{w} ∘ σ. Therefore

(σα)(w) = i_{w}(σa),

as required.

In more down-to-earth terms, (σα)(σw) = σ(α(w)): if α has the element a in the w-position, then σα has the element σa in the σw-position.

Note that the action of G on ∏_{w|v} L_{w} induces an action of G on the subsets ∏_{w|v} L_{w}^{×} and ∏_{w|v} U_{w} of ∏_{w|v} L_{w}.

Proposition 2.2. Choose a w_{0}|v, and let G_{w_{0}} be its decomposition group. For α ∈ ∏_{w|v} L_{w} and σ ∈ G, define f_{α}(σ) = σ(α(σ^{−1}w_{0})). Then f_{α} ∈ Ind_{G_{w_{0}}}^{G}(L_{w_{0}}), and the map

α ↦ f_{α} : ∏_{w|v} L_{w} → Ind_{G_{w_{0}}}^{G}(L_{w_{0}})

is an isomorphism of G-modules. Similar statements hold with L_{w} replaced with L_{w}^{×} and with U_{w}.

Proof. Recall (II.1) that

Ind_{G_{w_{0}}}^{G}(L_{w_{0}}) = {f : G → L_{w_{0}} | f(ρσ) = ρf(σ), all ρ ∈ G_{w_{0}}}

and that τ ∈ G acts on f ∈ Ind_{G_{w_{0}}}^{G}(L_{w_{0}}) according to the rule (τf)(σ) = f(στ). For ρ ∈ G_{w_{0}},

f_{α}(ρσ) = ρσ(α(σ^{−1}ρ^{−1}w_{0})) = ρσ(α(σ^{−1}w_{0})) = ρf_{α}(σ),

and so f_{α} ∈ Ind_{G_{w_{0}}}^{G}(L_{w_{0}}). Moreover,

(τf_{α})(σ) = f_{α}(στ) = στ(α(τ^{−1}σ^{−1}w_{0})) = σ(τα)(σ^{−1}w_{0}) = f_{τα}(σ),

and so α ↦ f_{α} is a homomorphism of G-modules ∏_{w|v} L_{w} → Ind_{G_{w_{0}}}^{G}(L_{w_{0}}). Given f ∈ Ind_{G_{w_{0}}}^{G}(L_{w_{0}}), set

α_{f}(w) = σ(f(σ^{−1})), w = σw_{0}.

Then f ↦ α_{f} is an inverse to α ↦ f_{α}.

Proposition 2.3. For all r,

H^{r}(G, ∏_{w|v} L_{w}^{×}) ≅ H^{r}(G_{w_{0}}, L_{w_{0}}^{×}).

In particular,

H^{0}(G, ∏_{w|v} L_{w}^{×}) ≅ K_{v}^{×}.

Similar statements hold with L_{w}^{×} replaced with U_{w}.

Proof. We have

H^{r}(G, ∏_{w|v} L_{w}^{×}) = H^{r}(G, Ind_{G_{w_{0}}}^{G}L_{w_{0}}^{×}) = H^{r}(G_{w_{0}}, L_{w_{0}}^{×})

by Shapiro’s lemma (II.1.11).

Remark 2.4. The group H^{r}(G_{w_{0}}, L_{w_{0}}^{×}) is independent of the prime w_{0} dividing v up to a canonical isomorphism, for let w be a second such prime. Then we can write w = σw_{0}, and we have a compatible pair of isomorphisms

τ ↦ στσ^{−1} : G_{w_{0}} → G_{w}, x ↦ σ^{−1}x : L_{w} → L_{w_{0}},

and hence isomorphisms

H^{r}(G_{w}, L_{w}^{×}) ⟶ H^{r}(G_{w_{0}}, L_{w_{0}}^{×})

for each r (see II.1).

If w = σ′w_{0}, then σ′ = στ with τ ∈ G_{w_{0}}. The maps defined by σ and σ′ differ by the automorphism of H^{r}(G_{w_{0}}, L_{w_{0}}^{×}) defined by τ, which is the identity map (II.1.27d). Therefore H^{r}(G_{w_{0}}, L_{w_{0}}^{×}) and H^{r}(G_{w}, L_{w}^{×}) are canonically isomorphic. This suggests the following notation: choose a prime w|v and set,

G^{v} = G_{w}, L^{v} = L_{w}, U^{v} = U_{w}.

These objects are defined only up to noncanonical isomorphisms, but H^{r}(G^{v}, L^{v×}) and H^{r}(G^{v}, U^{v}) are defined up to canonical isomorphisms.

We endow 𝕀_{L} with the action of G such that the inclusions

∏_{w|v} L_{w}^{×} → 𝕀_{L}

are G-homomorphisms. Thus if α has a_{w} in the w-position, then σα has σa_{w} in the σw-position.

Proposition 2.5. (a) The map 𝕀_{K} ↪ 𝕀_{L} induces an isomorphism 𝕀_{K} → 𝕀_{L}^{G}.

(b) For all r ≥ 0, H_{T}^{r} (G, 𝕀_{L}) = ⊕_{v}H_{T}^{r} (G^{v}, L^{v×}).

Proof. (a) Clearly α = (a_{w}) is fixed by G if and only if each subfamily (a_{w})_{w|v} is fixed by G. But (a_{w})_{w|v} is fixed by G only if a_{w} is independent of w and lies in K_{v}^{×}.

(b) For each finite set S of primes of K, let

𝕀_{L,S} = ∏_{v∈S}(∏_{w|v} L_{w}^{×}) × ∏_{v∉S}(∏_{w|v} U_{w}).

Then 𝕀_{L,S} is stable under the action of G, and 𝕀_{L} is the directed union of the 𝕀_{L,S} as S runs over the finite sets of primes of K containing all infinite primes and all primes that ramify in L. Hence (see II.3.3),

H^{r}(G, 𝕀_{L}) = \underrightarrow{lim} H^{r}(G, 𝕀_{L,S}).

On applying (II.1.25) and (2.3), we find that

H^{r}(G, 𝕀_{L,S}) = ∏_{v∈S} H^{r}(G, ∏_{w|v} L_{w}^{×}) × ∏_{v∉S} H^{r}(G, ∏_{w|v} U_{w})
= ∏_{v∈S} H^{r}(G^{v}, L^{v}^{×}) × ∏_{v∉S} H^{r}(G^{v}, U^{v}).

Because of (III.2.1), the second factor is zero when r > 0, and so

H^{r}(G, 𝕀_{L}) = \underrightarrow{lim}_{S}H^{r}(G, 𝕀_{L,S}) = \underrightarrow{lim}_{S} ⊕_{v∈S} H^{r}(G^{v}, L^{v×}) = ⊕_{all v}H^{r}(G^{v}, L^{v×}).

The same argument works for r ≤ 0 when one uses the Tate groups.

Corollary 2.6. (a) H^{1}(G, 𝕀_{L}) = 0;

(b) H^{2}(G, 𝕀_{L}) ≈ ⊕_{v} (\frac{1}{n_{v}} ℤ/ℤ), where n_{v} = [L^{v} : K_{v}].

Proof. (a) Apply Hilbert’s theorem 90.

(b) From Theorem III.1.1 we know that the invariant map gives an isomorphism

H^{2}(G^{v}, L^{v×}) ⟶^{≈} \frac{1}{n_{v}} ℤ/ℤ.

Proposition 2.7. Let S be a finite set of primes of K, and let T be the set of primes of L lying over primes in S. If L/K is cyclic, then the Herbrand quotient

h(𝕀_{L,T}) = ∏_{v∈S} n_{v}, n_{v} = [L^{v} : K_{v}].

Proof. We have

𝕀_{L,T} = ( ∏_{v∈S}(∏_{w|v} L_{w}^{×})) × ( ∏_{v∉S}(∏_{w|v} U_{w})) .

The Tate cohomology of the second factor is zero, and so the Herbrand quotient

h(G, 𝕀_{L,T}) = ∏_{v∈S} h(G, ∏_{w|v} L_{w}^{×}) = ∏_{v∈S} h(G^{v}, L^{v×}) = ∏_{v∈S} #H^{2}(G^{v}, L^{v×}) = ∏_{v∈S} n_{v}.

The norm map on idéles. Let L/K be a finite Galois extension of number fields. As for any G-module, there is a norm map

x ↦ ∏_{σ∈G} σx : 𝕀_{L} → 𝕀_{L}^{G} = 𝕀_{K}.

We need to examine this map. Recall (Math 676, 8.3) that there is a commutative diagram:

L^{×} ⟶ ∏_{w|v} L_{w}^{×}
↓^{Nm}^{L/K}  ↓^{(a_{w})↦∏ Nm_{Lw/K_{v}} a_{w}}
K^{×} ⟶ K_{v}^{×}.

For any w, Nm L_{w}^{×} is open in K_{v}^{×} (for example, because it is of finite index), and for any unramified w, the norm map sends U_{w} onto U_{v} (see III.2.2). The image of the right hand vertical map in the diagram is just Nm L_{w}^{×} for any w|v (because any two L_{w}’s are K_{v}-isomorphic). We denote it by Nm L^{v×}.

Let S ⊃ S_{∞} be a finite set of primes of K containing those that ramify, and let T be the set of primes lying over a prime of S. The above remarks show that

Nm_{L/K} 𝕀_{L,T} = ∏_{v∈S} V_{v} × ∏_{v∉S} U_{v}

where V_{v} is an open subgroup of finite index K_{v}^{×}. This is an open subgroup in 𝕀_{K,S} and 𝕀_{K,S} is open in 𝕀_{K}. We have proved:

Proposition 2.8. For any finite Galois extension L/K of number fields, Nm_{L/K} 𝕀_{L} contains an open subgroup of 𝕀_{K} and therefore is itself open.

Consider

0 ⟶ L^{×} ⟶ 𝕀_{L} ⟶ C_{L} ⟶ 0
↓^{Nm}^{L/K}  ↓^{Nm}^{L/K}  ↓
0 ⟶ K^{×} ⟶ 𝕀_{K} ⟶ C_{K} ⟶ 0.

The left hand square commutes, and so the norm map 𝕀_{L} → 𝕀_{K} induces a norm map C_{L} → C_{K}. From the snake lemma, we find that the quotient map 𝕀_{K} → C_{K} induces an isomorphism 𝕀_{K}/K^{×} · Nm(𝕀_{L}) → C_{K}.

3. The Cohomology of the Units

Let L/K be a finite extension of number fields with Galois group G. Let S ⊃ S_{∞} be a finite set of primes of K, and let T be the set of primes of L lying over a prime of K in S. Because T is stable under the action of G, the group of T -units

U(T) =^{df} {α ∈ L | ord_{w}(α) = 0 all w ∉ T }

is also stable under G.

Proposition 3.1. In the above situation, assume that G is cyclic. Then the Herbrand quotient h(U(T)) is defined, and satisfies

n · h(U(T)) = ∏_{v∈S} n_{v}

where n = [L : K] and n_{v} = [L^{v} : K_{v}].

We first show that any two G-stable full lattices^{1} in the same real vector space have the same Herbrand quotient. Then we construct two such lattices, one with Herbrand quotient n · h(U(T)) and the other with Herbrand quotient ∏ n_{v}.

Lemma 3.2. Let G be a finite group, and let k be an infinite field. Let M and N be k[G]-modules that are of finite dimension when regarded as k-vector spaces. If M ⊗_{k} Ω and N ⊗_{k} Ω are isomorphic as Ω[G] modules for some field Ω ⊃ k, then they are already isomorphic as k[G]-modules.

Proof. First note that if V is the space of solutions for a system of homogeneous linear equations over k, then the solution space for the same system of equations over Ω admits a basis with coordinates in k. In fact, the standard algorithm for finding a basis for the solution space yields the same result when carried out over k or Ω.

A k-linear map α : M → N is a G-homomorphism if α(σm) = σα(m) all m ∈ M, σ ∈ G. Once bases have been chosen for M and N, giving a k-linear map α : M → N is the same as giving a matrix A, and the condition that α be a G-homomorphism takes the form A · B(σ) = C(σ) · A for certain matrices B(σ) and C(σ). This is a linear condition on the coefficients of A, and so the remark shows that there are k[G]-homomorphisms α_{1}, …, α_{r} : M → N that form an Ω-basis for the space of Ω[G]-homomorphisms M ⊗_{k} Ω → N ⊗_{k} Ω.

Because M ⊗_{k} Ω and N ⊗_{k} Ω are isomorphic as Ω[G]-modules, there exist a_{1}, …, a_{r} ∈ Ω such that ∑ a_{i}α_{i} is an isomorphism, and hence has nonzero determinant. But det(∑ a_{i}α_{i}) is a polynomial in the a_{i} with coefficients in k, and the preceding sentence shows that not all of its coefficients are zero. As k is infinite, there exist a_{i}’s in k such that ∑ a_{i}α_{i} has nonzero determinant (see the proof of the Normal Basis Theorem II.1.24), and hence is a k[G]-isomorphism M → N.

Remark 3.3. (For the experts). It is possible to give an alternative proof of the lemma (at least when k has characteristic zero). The group H of automorphisms of M as a k[G]-module is a product of groups of the form GL_{d}(D), D a division algebra over k. The functor of isomorphisms M → N is a principal homogeneous space for H (nonempty, because there exists an isomorphism over some field containing k), and hence defines an element of H^{1}(k, H). Now a generalization of Hilbert’s theorem 90 shows that H^{1}(k, H) = 1.

1Recall that a subgroup M of a real vector space V is called a full lattice if M is the ℤ-submodule generated by a basis for V; equivalently, if the canonical map ℝ ⊗_{ℤ} M → V is an isomorphism. The definition of a full lattice in a ℚ-vector space is similar.

Lemma 3.4. Let G be a finite cyclic group, and let M and N be G-modules that are finitely generated as ℤ-modules, and such that M ⊗_{ℤ} ℚ and N ⊗_{ℤ} ℚ are isomorphic as G-modules. If either h(M) or h(N) is defined, so also is the other, and the two are equal.

Proof. After (II.2.16), we may assume that M and N are torsion free. Choose an isomorphism

α : M ⊗ ℚ ⟶ N ⊗ ℚ.

Then α(M) and N are lattices in the same ℚ-vector space, and so α(M) ⊂ n^{−1}N for some n ∈ ℕ (express the elements of a basis for α(M) in terms of a bases for N, and let n be a common denominator for the coefficients). After replacing α with nα, we have that α(M) ⊂ N. Now we have an exact sequence

0 ⟶ M ⟶^{α} N ⟶ N/α(M) ⟶ 0

with N/α(M) finite, and we can apply (II.2.16) again to deduce that h(M) = h(N).

Lemma 3.5. Let G be a finite cyclic group, and let V be a real vector space on which G acts linearly (i.e., V is an ℝ[G]-module). Let M and N be two G-stable full lattices in V. If either h(M) or h(N) is defined, then so is the other, and they are equal.

Proof. Because M and N are full lattices in V, the canonical maps

M ⊗_{ℤ} ℝ → V, N ⊗_{ℤ} ℝ → V

are isomorphisms. These maps are G-homomorphisms, and therefore (3.2) M ⊗_{ℤ}ℚ ≈ N ⊗_{ℤ} ℚ as ℚ[G]-modules, and we can apply Lemma 3.4.

We now complete the proof of the Theorem. Let V be a product of copies of ℝ indexed by the elements of T, i.e.,

V = Hom(T, ℝ).

We let G act on V according to the rule:

(σf)(w) = f(σ^{−1}w), σ ∈ G, w ∈ T.

The first lattice in V is N =^{df} Hom(T, ℤ). For each v ∈ S, choose a w lying over v, and let G^{v} be the decomposition group of w. The sets G^{v} · w, v ∈ S, are the orbits of G acting on T. In particular, T is a disjoint union of these sets, and so

Hom(T, ℤ) = ⊕_{v} Hom(G/G^{v}, ℤ).

But Hom(G/G^{v}, ℤ) = Ind_{G^{v}}^{G}(ℤ) (ℤ regarded as a trivial G^{v}-module). Therefore,

h(G, N) = ∏_{v} h(G, Ind_{G^{v}}^{G}(ℤ)) = ∏ h(G^{v}, ℤ) = ∏ n_{v}.

We now define the second lattice. Let λ : U(T) → V be the map a ↦ (…, log |a|_{w}, …), and let M^{0} to be the image of λ. Note that λ commutes with the action of G. The kernel of λ consists of the elements a of L^{×} such that |a|_{w} = 1 for all w (including the infinite primes). These are the roots of 1 in L, and so h(U(T)) = h(M^{0}). The product formula shows M^{0} is contained in the subspace

V^{0} : ∑ x_{w} = 0,

of V, and the proof of the T -unit theorem shows that M^{0} is a lattice in V^{0} (cf. Math 676, 5.7). The vector e = (1, 1, …, 1) is stable under G, and we define M = M^{0} +ℤe. Then M ⊗_{ℤ} ℝ = V^{0} + ℝe = V, and so M is a lattice in V. Moreover,

h(M) = h(M^{0}) · h(ℤ) = h(M^{0}) · n.

This completes the proof Proposition 3.1.

4. Cohomology of the Idèle Classes I: the First Inequality

Let L/K be a finite Galois extension of number fields with Galois group G. We have a commutative diagram of G-modules with exact rows,

0 ⟶ K^{×} ⟶ 𝕀_{K} ⟶ C_{K} ⟶ 0
↓  ↓  ↓
0 ⟶ L^{×} ⟶ 𝕀_{L} ⟶ C_{L} ⟶ 0,

That the rows are exact is the definition of the idèle class groups. The vertical arrows in the left hand square are the natural inclusions. The square therefore commutes, which shows that the right hand vertical arrow exists.

Lemma 4.1. The canonical map C_{K} → C_{L} induces an isomorphism

C_{K} ⟶ C_{L}^{G} = H^{0}(G, C_{L}).

Proof. From the bottom row of the above diagram, we obtain a cohomology sequence

0 ⟶ H^{0}(G, L^{×}) ⟶ H^{0}(G, 𝕀_{L}) ⟶ H^{0}(G, C_{L}) ⟶ H^{1}(G, L^{×})
∥  ∥  ∥  ∥
L^{×G}  𝕀_{L}^{×G}  C_{L}^{G}  0

which can be identified with

0 ⟶ K^{×} ⟶ 𝕀_{K} ⟶ C_{K} ⟶ 0.

The ideal class group of a number field is finite, and it is generated by the classes of prime ideals. Therefore, it is generated by a finite number of prime ideals.

Lemma 4.2. Let K be a number field, and let S ⊃ S_{∞} be a finite set of primes of K containing a set of generators for the ideal class group of K. Then

𝕀_{K} = K^{×} · 𝕀_{S}.

Proof. The condition that S contains a set of generators for the ideal class group of K means that every fractional ideal 𝔞 can be written

𝔞 = 𝔟 · (c)

with 𝔟 in the group generated by the prime ideals in S and c ∈ K^{×}. Therefore, 𝔞 = (c) in the quotient group I^{S} = I/ <S>, and so I^{S}/i(K^{×}) = 0.

For any finite set S ⊃ S_{∞} of primes of K, the natural map 𝕀 → I^{S} defines an isomorphism 𝕀/𝕀_{S} ⟶^{≈} I^{S}. On dividing out by K^{×} on both sides, we find that 𝕀/K^{×} · 𝕀_{S} ≅ I^{S}/i(K^{×}) ≅ 0.

Recall that we want to prove that for any abelian extension L/K, C_{K}/ Nm_{L/K} C_{L} ≅ Gal(L/K) and that for any Galois extension H^{1}(G, C_{L}) = 1. For a cyclic extension, the two statements imply that the Herbrand quotient

h(C_{L}) =^{df} \frac{(C_{K} : Nm C_{L})}{#H^{1}(G, C_{L})} = [L : K].

As a first step, we verify this equality.

Theorem 4.3. For any finite cyclic extension L/K of number fields, h(C_{L}) = [L : K].

Proof. Let S any finite set of primes of K such that:

(a) S ⊃ S_{∞}, the set of infinite primes of K;

(b) S contains all primes that ramify in L;

(c) S contains the prime ideals 𝔓 ∩ O_{K} for a set of generators 𝔓 of the ideal class group of L.

Let T be the set of primes of L lying over a prime in S. Condition (c) implies that 𝕀_{L} = 𝕀_{L,T} · L^{×}, and so

C_{L} =^{df} 𝕀_{L}/L^{×} = L^{×} · 𝕀_{L,T} /L^{×} ≅ 𝕀_{L,T} /L^{×} ∩ 𝕀_{T} .

Note that

L^{×} ∩ 𝕀_{T} = {α ∈ L | ord_{w}(α) = 0, ∀w ∉ T } = U(T),

and so

h(C_{L}) = h(𝕀_{L,T})/h(U(T)).

The theorem now follows from Proposition 2.7 and Proposition 3.1.

Corollary 4.4 (First inequality). If L/K is cyclic of degree n, then

(𝕀_{K} : K^{×} · Nm(𝕀_{L})) ≥ n.

Proof. Since h(C_{L}) = n, its numerator must be ≥ n.

We now give some application of the First Inequality.

Lemma 4.5. Let L be a finite solvable extension of K (i.e., a finite Galois extension with solvable Galois group). If there exists a subgroup D of 𝕀_{K} such that

(a) D ⊂ Nm_{L/K} 𝕀_{L}; and

(b) K^{×} · D is dense in 𝕀_{K}

then L = K.

Proof. If L ≠ K, then the exists a subfield K′ of L that is cyclic over K and ≠ K. Then

D ⊂ Nm_{L/K}(𝕀_{L}) = Nm_{K′/K}(Nm_{L/K} 𝕀_{L})) ⊂ Nm_{K′/K}(𝕀_{K′}).

Therefore, K^{×} · Nm_{K′/K} 𝕀_{K′} is dense in 𝕀_{K}. Because it is a union of translates of Nm_{K′/K} 𝕀_{K′}, it is open (2.8), and because it is a subgroup of 𝕀_{K}, it is also closed. Therefore, K^{×} · Nm_{K′/K} 𝕀_{K′} = 𝕀_{K}, and the first inequality implies that K′ = K.

Proposition 4.6. Let L be a finite solvable extension of K. If L ≠ K, then there are infinitely many primes of K that do not split completely in L.

Proof. Suppose there are only finitely many, and let S ⊃ S_{∞} be a finite set of primes of K including all those that do not split completely. We shall apply the lemma with

D = 𝕀^{S} =^{df} {(a_{v}) | a_{v} = 1 for all v ∈ S}.

For w|v ∉ S, L_{w} = K_{v}, and so clearly D ⊂ Nm(𝕀_{L}). Let a = (a_{v}) ∈ 𝕀. By the Weak Approximation Theorem (V.6.3), there is an element b ∈ K^{×} that is very close to a_{v} in K_{v} for all v ∈ S. Choose a′ to be the element of 𝕀^{S} such that the v component of ba′ is equal to a_{v} for all v ∉ S. Then ba′ is close to a in 𝕀_{K}. Hence K^{×} · D is dense in 𝕀_{K}.

Proposition 4.7. For any finite solvable extension L/K with Galois group G, and any finite set of prime ideals T of L including those that ramify from K, the Frobenius elements (𝔓, L/K) for 𝔓 ∉ T generate G.

Proof. Let H be the subgroup generated by the Frobenius elements at the 𝔓 ∉ T, and let E = L^{H}. Recall (V.1.10) that

(𝔓, E/K) = (𝔓, L/K)|_{E},

which is the identity map. Therefore all primes 𝔭 ∉ S split in E, which shows that E = K. By the main theorem of Galois theory, this implies that H = G.

Corollary 4.8. For any abelian extension L/K and finite set of primes S ⊃ S_{∞} of K including the primes that ramify in L, the map

𝔭 ↦ (𝔭, L/K) : I^{S} ⟶ Gal(L/K)

is surjective. (Recall that I^{S} is the group of fractional ideals generated by prime ideals not in S.)

Proof. The image contains the Frobenius elements (𝔓, L/K) for all 𝔓|𝔭 ∈ S, and these generate Gal(L/K).

Remark 4.9. Of course, Proposition 4.6 is much weaker than the result available using complex analysis—see Theorem VI.3.4—but it suffices for the proof of the Reciprocity Law.

5. Cohomology of the Idèle Classes II: The Second Inequality

Theorem 5.1. Let L/K be a Galois extension of number fields with Galois group G. Then

(a) (second inequality) the index (𝕀_{K} : K^{×} · Nm(𝕀_{L})) is finite, and divides [L : K];

(b) the group H^{1}(G, C_{L}) = 0;

(c) the group H^{2}(G, C_{L}) is finite, and its order divides [L : K].

Lemma 5.2. If G is cyclic, then statements (a), (b), and (c) of the theorem are equivalent (and (𝕀_{K} : K^{×} · Nm(𝕀_{L})) = (H^{2}(G, C_{L}) : 1) = [L : K]).

Proof. Without restriction on G,

𝕀_{K}/K^{×} · Nm_{L/K}(𝕀_{L}) ≅ C_{K}/ Nm_{L/K}(C_{L}) = H_{T}^{0}(G, C_{L}).

If G is cyclic, its cohomology is periodic, and so H_{T}^{0}(G, C_{L}) ≈ H^{2}(G, C_{L}). This proves that (a) and (c) are equivalent. Theorem 4.3 states that the Herbrand quotient h(C_{L}) = [L : K], and so each of (a) and (c) is equivalent to (b).

Lemma 5.3. It suffices to prove the theorem in the case that G is a p-group, p prime.

Proof. Recall (II.1.33), that if H is the Sylow p-subgroup of G then, for any G-module M, the maps

Res : H_{T}^{r} (G, M) → H_{T}^{r} (H, M)

are injective on the p-primary components. Therefore, if the theorem holds for L/L^{H}, then p does not divide the order of H_{T}^{1}(G, C_{L}) and the power of p dividing the orders H_{T}^{0}(G, C_{L}) and H_{T}^{2}(G, C_{L}) is less than the power of p dividing [L : K]. On applying this for all p, we obtain the lemma.

Lemma 5.4. It suffices to prove the theorem in the case that G is a cyclic group of prime order p.

Proof. After the last lemma, we may assume that G is a p-group. We shall prove the theorem for G by induction on (G : 1). Because G is a p-group, it has a normal subgroup H of index p (see Math 594g, 4.15). Consider the exact sequence (II.1.34)

0 ⟶ H^{1}(G/H, C_{K′}) ⟶^{Inf} H^{1}(G, C_{L}) ⟶^{Res} H^{1}(H, C_{L})

where K′ = L^{H}. By assumption H^{1}(G/H, C_{K′}) = 0 and by induction H^{1}(H, C_{L}) = 0. Therefore H^{1}(G, C_{L}) = 0—statement (b) is true.

Because H^{1}(H, C_{L}) = 0, the sequence

0 ⟶ H^{2}(G/H, C_{K′}) ⟶ H^{2}(G, C_{L}) ⟶ H^{2}(H, C_{L})

is exact, from which it follows that statement (c) is true.

Finally, note that

(C_{K} : Nm_{L/K}(C_{L})) = (C_{K} : Nm_{K′/K}(C_{K′}))(Nm_{K′/K}(C_{K′}) : Nm_{L/K}(C_{L})),

which divides p[L : K′] because Nm_{K′/K} defines a surjection

C_{K′}/ Nm_{L/K′}(C_{L}) → Nm_{K′/K}(C_{K′})/ Nm_{L/K}(C_{L}).

It therefore remains to prove that the Second Inequality holds for a cyclic extension of characteristic p, but in (VI.4.9 we proved that the Second Inequality holds for all finite Galois extensions. (For the translation between the idealic and the idèlic form of the statement, see Proposition V.4.6). In the next section, we give an algebraic proof of the Second Inequality, independent of Chapter VI.

Remark 5.5. To a finite Galois extension L/K of number fields, we have attached the group C_{K}/ Nm(C_{L}) and H^{2}(G, C_{L}). When L/K is cyclic, they are canonically (up to a choice of a generator for G) isomorphic, but not otherwise. The first group is always isomorphic to G^{ab}, and the second is always cyclic of order [L : K]. Thus, when G is abelian but not cyclic, the two groups have the same order but are not isomorphic, and when G is nonabelian, they have different orders.

6. The Algebraic Proof of the Second Inequality

We shall prove the Second Inequality in the case that L/K is cyclic of prime degree p.

Lemma 6.1. It suffices to prove the Second Inequality in the case that K contains a p^{th} root of 1.

Proof. Let ζ be a primitive p^{th} root of 1 (in some fixed algebraic closure of K containing L), and let K′ = K[ζ] and L′ = K′ · L = L[ζ]. Then [K′ : K] = m|p − 1, and so is relatively prime to p. Hence K′ ∩ L = K, and we have the picture:

\begin{matrix}L & —^{m} & L′ \\ |p & & |p \\ K & —^{m} & K′. \end{matrix}

The map

Gal(L′/K) ⟶ Gal(L/K) × Gal(K′/K)

is an isomorphism. Consider the diagram:

C_{L} ⟶^{Nm_{L/K}} C_{K} ⟶ C_{K}/ Nm C_{L} ⟶ 0
↓^{i_{L}}  ↓^{i_{K}}  ↓
C_{L′} ⟶^{Nm_{L′/K′}} C_{K′} ⟶ C_{K′}/ Nm C_{L′} ⟶ 0
↓^{Nm}_{L′/L}  ↓^{Nm}_{K′/K}  ↓
C_{L} ⟶^{Nm_{L/K}} C_{K} ⟶ C_{K}/ Nm C_{L} ⟶ 0

Here i_{L} and i_{K} are the maps induced by the inclusions 𝕀_{L} ↪ 𝕀_{L′} and 𝕀_{K} ↪ 𝕀_{K′}, Nm_{L/K} and Nm_{L′/K′} are the maps

x ↦ ∑ σx, σ ∈ Gal(L/K) = Gal(L′/K′),

and Nm_{L′/L} and Nm_{K′/K} are the maps

x ↦ ∑ σx, σ ∈ Gal(L′/L) = Gal(K′/K).

Clearly the squares at left commute, and this implies that the rest of the diagram exists. The composites

Nm_{L′/L} ∘i_{L} and Nm_{K′/K} ∘i_{K}

are both multiplication by m. Therefore the composite

C_{K}/ Nm C_{L} ⟶ C_{K′}/ Nm C_{L′} ⟶ C_{K}/ Nm C_{L}

is also multiplication by m, and hence is an isomorphism (clearly, p^{th} powers in C_{K} are norms, and so C_{K}/ Nm C_{L} is killed by p). In particular, the second map is surjective, and so

(C_{K} : Nm C_{L}) divides (C_{K′} : Nm C_{L′}),

which by assumption, divides p.

We shall prove the Second Inequality in the case the K contains a primitive pth root of 1, and L is a finite abelian extension of K of exponent p with Galois group G. Let [L : K] = p^{r}, so that G ≈ (ℤ/pℤ)^{r}. By Kummer theory (see the appendix to this chapter),

L = K(a1^{\frac{1}{p}} , …, ar^{\frac{1}{p}}).

Let S be a finite set of primes of K such that

(a) S contains the infinite primes;

(b) S contains all divisors of p;

(c) S is so large that all a_{i} are S-units.

(d) S contains a set of generators for the ideal class group of K, and so 𝕀_{K} = 𝕀_{K,S} · K^{×} (see 4.2).

Note that, by (10.5), (b) and (c) imply that S contains all primes that ramify in L.

As usual, we write U(S) for the group of S-units, i.e., the group of elements of K^{×} that are units for all primes outside S. Recall that the unit theorem says that

U(S) ≈ ℤ^{s−1} × U(S)_{tors}, s = #S,

and U(S)_{tors} is a finite cyclic group. In our case, the order of U(S)_{tors} is divisible by p (because it contains μ_{p}), and so

U(S)/U(S)^{p} ≈ (ℤ/pℤ)^{s}.

Let M = K[U(S)^{\frac{1}{p}}]. This is the Kummer extension corresponding to the group

U(S) · K^{×p}/K^{×p} ≈ U(S)/U(S) ∩ K^{×p} = U(S)/U(S)^{p} ≈ (ℤ/pℤ)^{s}.

We therefore have extensions

M ⊃^{p^{t}} L ⊃^{p^{r}} K, r + t = s.

Lemma 6.2. There exists a set of primes T of K, disjoint from S, such that {(𝔭_{v}, M/K) | v ∈ T } is a basis for Gal(M/L) (regarded as an 𝔽_{p}-vector space).

Proof. Note that if w′|v ∉ S, then M_{w′} is an unramified extension of K_{v}. Hence Gal(M_{w}/K_{v}) is cyclic, and it has exponent p (because it is a subgroup of Gal(M/K)). Therefore it is either cyclic of order p or trivial. In particular, if M_{w′} ≠ L_{w}, then L_{w} = K_{v}.

According to (4.7), there is a finite set {w_{1}, …, w_{t}} of primes of L, none lying over a prime in S, such that the Frobenius elements (𝔭_{w_{i}}, M/L) form a basis for Gal(M/L). Let v_{i} be the prime of K lying under w_{i}. Then, according to the above observation, L_{w_{i}} = K_{v_{i}}, and therefore (𝔭_{w_{i}}, M/L) = (𝔭_{v_{i}}, M/K). We can take T to be {v_{1}, …, v_{t}}.

Note that the order of T is t where p^{t} = [M : L], and that for any w|v ∈ T, L_{w} = K_{v}.

Lemma 6.3. With the above notations,

L^{×p} ∩ U(S) = {a ∈ U(S) | a ∈ K_{v}^{×p}, all v ∈ T }.

Proof. ⊂: If a ∈ LHS, then it is in U(S) and it becomes a p^{th} power in L. Therefore it is a p^{th} power in L_{w}^{×} for all w, but if w|v ∈ T, then L_{w} = K_{v}, and so it is a p^{th} power in K_{v}.

⊃: If a ∈ U(S), then a^{\frac{1}{p}} ∈ M. If further a is a p^{th} power in K_{v} for v ∈ T, then a^{\frac{1}{p}} is fixed by (𝔭_{v}, M/K). Since these Frobenius elements generate Gal(M/L), a^{\frac{1}{p}} lies in L, and so a ∈ L^{p}.

Lemma 6.4. The subgroup

E = ∏_{v∈S} K_{v}^{×p} × ∏_{v∈T} K_{v}^{×} × ∏_{v∉S∪T} U_{v}

of 𝕀_{K} is contained in Nm_{L/K}(𝕀_{L}).

Proof. Let a = (a_{v}) ∈ E. We have to show that each component a_{v} of a is a norm.

v ∈ S: From local class field theory, we know that

K_{v}^{×}/ Nm L_{w}^{×} ⟶^{≈} Gal(L_{w}/K_{v}).

Because the second group is killed by p, so also must be the first group, which means that K_{v}^{×p} ⊂ Nm L_{w}^{×}.

v ∈ T: Here L_{w} = K_{v}, and so every element of K_{v} is a norm from L_{w}.

v ∉ S ∪ T: Because L_{w} is unramified over K_{v}, the norm map U_{w} → U_{v} is surjective (see III.2.2)

Now

(C_{K} : Nm C_{L}) = (𝕀_{K} : K^{×} · Nm 𝕀_{L}),

which divides (𝕀_{K} : K^{×}E), and so it remains to show that

(𝕀_{K} : K^{×}E)|p^{r}.

But 𝕀_{K} = K^{×} · 𝕀_{S} = K^{×} · 𝕀_{S ⋃ T}, and so

(𝕀_{K} : K^{×}E) = (K^{×}𝕀_{S∪T} : K^{×}E).

Lemma 6.5. Let A, B, and C be subgroups of some abelian group, and assume that A ⊃ B. Then

(AC : BC)(A ∩ C : B ∩ C) = (A : B)

in the sense that, if two of the indexes are finite, so is the third, and the equality holds.

Proof. In the following commutative diagram, the columns and the top two rows are obviously exact, and it follows (from the snake lemma for example) that the bottom row is exact. This implies the statement.

0  0  0
↓  ↓  ↓
0 ⟶  B ∩ C  ⟶ B ⟶ BC/C ⟶ 0
↓  ↓  ↓
0 ⟶  A ∩ C  ⟶ A ⟶ AC/C ⟶ 0
↓  ↓  ↓
0 ⟶ A ∩ C/B ∩ C ⟶ A/B ⟶ AC/BC ⟶ 0
↓  ↓  ↓
0  0  0

On applying the lemma with A = 𝕀_{S∪T}, B = E, and C = K^{×} we find that

(𝕀_{K} : K^{×}E) = \frac{(𝕀_{S∪T} : E)}{(U(S ∪ T) : K^{×} ∩ E)}.

Lemma 6.6. With the above notations:

(𝕀_{S∪T} : E) = p^{2s}.

Lemma 6.7. With the above notations:

(U(S ∪ T) : K^{×} ∩ E) = p^{s+t}.

Since r + t = s, this will prove the boxed formula.

Proof. (of 6.6). Obviously (𝕀_{S∪T} : E) = ∏_{v∈S}(K_{v}^{×} : K_{v}^{×p}). Since there are s primes in S and K contains p distinct p^{th} roots of 1, the next proposition shows that

(𝕀_{S∪T} : E) = \frac{p^{2s}}{∏_{v∈S} ∥p∥_{v}} .

By assumption, S contains all the primes for which ∥n∥_{v} ≠ 1, and so

∏_{v∈S} ∥n∥_{v} = ∏_{all v} ∥n∥_{v},

which equals 1 by the product formula.

Proposition 6.8. Let K be a local field of characteristic zero, and let U be the group of units in K. Then

(U : U^{n}) = \frac{(μ_{n} : 1)}{|n|}, (K^{×} : K^{×n}) = n\frac{(μ_{n} : 1)}{|n|}

where μ_{n} is the group of n^{th} roots of 1 in K^{×}.

Proof. For an abelian group M, we write

h_{n}(M) = (M : nM)/(M_{n} : 1), M_{n} = {x ∈ M | nx = 0}.

Then h_{n}(M) is the Herbrand quotient of M regarded as a ℤ/nℤ-module with trivial action, and so we may apply the results in II.2.

As we saw in the proof of (III.3.3), the exponential map defines an isomorphism from a subgroup of finite index in O_{K} onto a subgroup of finite index in U. Therefore

h_{n}(U) = h_{n}(O_{K}) = (O_{K} : nO_{K}) =^{df} |n|^{−1}.

Hence

(U : U^{n}) = \frac{(U_{n} : 1)}{|n|}

and U_{n} = μ_{n}. Since K^{×} ≈ U × ℤ, we have

(K^{×} : K^{×n}) = (U : U^{n})(ℤ : nℤ) = \frac{#μ_{n}}{∥n∥} n.

Proof. (of 6.7.) Clearly K^{×} ∩ E ⊃ U(S ∪ T)^{p}. It follows from the unit theorem (as before) that (U(S ∪ T) : U(S ∪ T)^{p}) = p^{s+t}, and so it remains to prove that

K^{×} ∩ E ⊂ U(S ∪ T)^{p}.

This is accomplished by the next two lemmas (the first shows that the second may be applied to prove the inclusion).

Lemma 6.9. With the above hypotheses, the obvious map

U(S) ⟶ ∏_{v∈T} U_{v}/U_{v}^{p}

is surjective.

Proof. Let H be the kernel of the map. To prove that the map is surjective, we shall show that

(U(S) : H) = ∏_{v∈T}(U_{v} : U_{v}^{p}).

Because T is disjoint from S, ∥p∥_{v} = 1 for all v ∈ T, and so (6.8) shows that the right hand side is p^{t}. On the other hand, by Lemma 6.3, H = U(S) ∩ L^{×p}, and so

U(S)/H = U(S)/U(S) ∩ L^{×p} ≅ U(S) · L^{×p}/L^{×p}.

This last group corresponds by Kummer theory (see 10.3) to the extension M/L, and hence has order [M : L] = p^{t}.

Proposition 6.10. Let K be a number field containing a primitive n^{th} root of 1. Let S be a set primes containing the infinite primes, the divisors of n, and a set of representatives of the ideal class group of K. Let T be a set of primes disjoint from S and such that

U(S) ⟶ ∏_{v∈T} U_{v}/U_{v}^{n}

is surjective. Suppose that b ∈ K^{×} is an n^{th} power in K_{v} for all v ∈ S and a unit outside S ∪ T. Then b ∈ K^{×n}.

Proof. Let L = K[b^{\frac{1}{n}}]—we have to show that L = K. Put

D = ∏_{v∈S} K_{v}^{×} × ∏_{v∈T} U_{v}^{n} × ∏_{v∉S∪T} U_{v}.

By Lemma 4.5, in order to show that L = K, it suffices to shows that

(a) D ⊂ Nm_{L/K} 𝕀_{L}, and

(b) D · K^{×} = 𝕀_{K}.

(a) Let d = (d_{v}) ∈ D. We have to check that d_{v} is a norm from K_{v}[b^{\frac{1}{n}}] for all v. v ∈ S: In this case K_{v}[b^{\frac{1}{n}}] = K_{v}, and so every element of K_{v} is a norm. v ∈ T: By local class field theory, the index (K_{v}^{×} : Nm K_{v}[b^{\frac{1}{n}}]^{×}) is equal to the

degree [K_{v}[b^{\frac{1}{n}}] : K_{v}], which divides n. Hence every n^{th} power in K_{v} is a norm.

v ∉ S ∪ T: Because nb is a unit at v, the field K_{v}[b^{\frac{1}{n}}] is unramified over K_{v}, and hence every unit is a norm.

(b) Obviously 𝕀_{S}/D = ∏_{v∈S} U_{v}/U_{v}^{n}, and by hypothesis U(S) → ∏_{v∈S} U_{v}/U_{v}^{n} is surjective. Hence 𝕀_{S} = D · U(S), and therefore

𝕀_{K} = 𝕀_{S} · K^{×} = D · U(S) · K^{×} = D · K^{×}.

This completes the proof of Theorem 5.1 (the Second Inequality).

7. Application to the Brauer Group

Readers, especially those who skipped Chapter IV, may interprete the notation Br(L/K) as shorthand for H^{2}(Gal(L/K), L^{×}) and Br(K) as shorthand for H^{2}(Gal(K^{al}/K), K^{al×}).

Theorem 7.1. For any Galois extension L/K of number fields (possibly infinite), the canonical map

Br(L/K) → ⊕_{v} Br(L^{v}/K_{v})

is injective.

Proof. Assume initially that L/K is a finite Galois extension with Galois group G. Because H^{1}(G, C_{L}) = 0, the cohomology sequence of

0 ⟶ L^{×} ⟶ 𝕀_{L} ⟶ C_{L} ⟶ 0

is

0 ⟶ H^{2}(G, L^{×}) ⟶ H^{2}(G, 𝕀_{L}) ⟶ · · · .

But

H^{2}(G, L^{×}) = Br(L/K)

and (see 2.5)

H^{2}(G, 𝕀_{L}) = ⊕H^{2}(G^{v}, L^{v×}) = ⊕ Br(L^{v}/K_{v}),

and so this proves the theorem in this case. To obtain the theorem for an infinite extension, pass to the limit over the finite Galois subextensions.

An extension L/K of fields is said to be cyclotomic if L ⊂ K[ζ] for some root ζ of 1. The next proposition will play a role in the proof of the global reciprocity law

Proposition 7.2. For any β ∈ Br(K), there exists a cyclic cyclotomic extension L of K such that β maps to zero in Br(L).

Proof. The theorem shows that β is determined by its images in Br(K_{v}), and hence by the invariants inv_{v}(β_{v}) ∈ ℚ/ℤ (see Theorem III.1.1). For any finite extension L of K and prime w|v of L, inv_{w}(β|L) = [L_{w} : K_{v}] · inv_{v}(β) (ibid.), and so we have to find a cyclic cyclotomic extension L/K such that

[L^{v} : K_{v}] · inv_{v}(β_{v}) = 0 mod ℤ

for all v. Note that, because Br(L/K) maps into the direct sum of the local Brauer groups, inv_{v}(β_{v}) = 0 for almost all v. Hence there exists an integer m such that m inv_{v}(β_{v}) = 0 for all v. The existence of an L with the correct properties is ensured by the next lemma.

Lemma 7.3. Given a number field K, a finite set S of finite primes of K, and an integer m > 0, there exists a totally complex cyclic cyclotomic extension L of K such that m|[L^{v} : K_{v}] for all v ∈ S.

Proof. It suffices to prove this for ℚ and m·[K : ℚ]. Hence we can simply assume K = ℚ.

Let ℓ be a prime, and let ζ be a primitive ℓ^{r}th root of 1 with r > 2. Then Gal(ℚ[ζ]/ℚ) ≅ (ℤ/ℓ^{r}ℤ)^{×}, and

(ℤ/ℓ^{r}ℤ) ≈ \{ \begin{matrix}Δ × C(ℓ^{r−2}) & ℓ odd \\ Δ × C(2^{r−3}) & ℓ = 2 \end{matrix}

where Δ is of order ℓ − 1 and 2 in the two cases and C(t) denotes a cyclic group of order t (Serre 1970, Cours..., II.3.2). Therefore L(ℓ^{r}) =^{df} ℚ[ζ]^{Δ} is a cyclic cyclotomic extension of ℚ of degree ℓ^{r−2} or ℓ^{r−3}.

Next consider ℚ_{p}[ζ]. If p = ℓ, then ℚ[ζ] is totally ramified over p, and so [ℚ_{p}[ζ] : ℚ_{p}] = [ℚ[ζ] : ℚ] = φ(ℓ^{r}). If p ≠ ℓ, then ℚ[ζ] is totally unramified over p, and [ℚ_{p}[ζ] : ℚ_{p}] is the smallest integer t such that ℓ^{r}|p^{t} − 1. In either case, we see that [ℚ_{p}[ζ] : ℚ_{p}] → ∞ as r → ∞. Thus, for any p, [L(ℓ^{r})^{p} : ℚ_{p}] is a power of ℓ that tends to ∞ as r tends to ∞.

A product of cyclic groups of distinct prime power orders is again cyclic. Therefore, for distinct primes ℓ_{1}, …, ℓ_{s}, L = L(ℓ_{1}^{r_{1}}) ⋯ L(ℓ_{s}^{r_{s}}) will be cyclic, and clearly, by choosing ℓ_{1}^{r_{1}} … ℓ_{s}^{r_{s}} to be sufficiently large, we can ensure that the local degrees m|[L^{p} : ℚ_{p}] are divisible by m for all p ∈ S.

In more concrete terms, the two results say that:

If a central simple algebra over K splits over K_{v} for all v, then it splits over K.

and

Every central simple algebra over K splits over a cyclic cyclotomic extension of K.

8. Completion of the Proof of the Reciprocity Law

Recall that, for a finite abelian extension L/K of number fields with Galois group G, we have defined a homomorphism ϕ_{L/K} : 𝕀_{K} → G such that ϕ_{L/K}(a) = ∏_{v} ϕ_{v}(a_{v}).

Theorem 8.1. (a) Let L/K be a finite abelian extension of number fields.

Then ϕ_{L/K} takes the value 1 on the principal idèles K^{×} ⊂ 𝕀_{K}.

(b) Let L/K be a finite Galois extension of number fields. Then ∑ inv_{v}(α) = 0 for all α ∈ Br(L/K).

Before proving this theorem, we explain why (a) implies the Reciprocity Law for L/K. Statement (a) says that ϕ_{L/K} : 𝕀_{K} → Gal(L/K) contains K^{×} in its kernel. We know already that it contains Nm_{L/K}(𝕀_{L}) in its kernel^{2}, and therefore it defines a homomorphism

𝕀_{K}/K^{×} · Nm_{L/K} 𝕀_{L} → Gal(L/K) (∗).

For any finite prime v of K unramified in L, ϕ_{L/K} maps the idèle with a prime element in v-position to the Frobenius element (𝔭_{v}, L/K), and so (4.7) shows that ϕ_{L/K} is surjective. On the other hand, the Second Inequality (5.1) states that

(𝕀_{K} : K^{×} · Nm_{L/K} 𝕀_{L}) ≤ [L : K]

and so the homomorphism (∗) is an isomorphism.

Example 8.2. We verify (8.1a) for the extension ℚ[ζ_{m}]/ℚ, where ζ_{m} a primitive mth root of 1. We identify Gal(ℚ[ζ_{m}]/ℚ) with (ℤ/mℤ)^{×}. Thus, for n an integer relatively prime to m, [n] denotes the automorphism of ℚ[ζ_{m}] sending ζ_{m} to ζ_{m}^{n}. It suffices to show that ϕ(a)|ℚ[ζ_{ℓ^{r}}] = 1 for all ℓ|m. Thus, we may assume that m = ℓ^{r}, m ≠ 2.

The homomorphism ϕ_{∞} : ℝ^{×}/ Nm(ℂ^{×}) → Gal(ℚ[ζ_{m}]/ℚ) sends any negative real number to complex conjugation. Therefore ϕ_{∞}(a) = [sign(a)].

Let a = up^{s} ∈ ℚ_{p}^{×}. If p ≠ ℓ, then p is unramified in ℚ[ζ_{m}], and ϕ_{p}(a) acts as the sth power of the Frobenius at p:

ϕ_{p}(up^{s}) = [p^{s}].

The prime ℓ is totally ramified in ℚ[ζ_{m}] and

ϕ_{ℓ}(a) = [u^{−1}]

(see I.3.13)

2Because this is true locally.

It suffices to show that ϕ(a) = 1 in the three cases: a = −1, a = ℓ, a = a prime q ≠ ℓ. We have:

ϕ_{p}(−1) = \{ \begin{matrix}[−1] & if & p = ∞ \\ [−1] & if & p = ℓ \\ [1] & if & p ≠ ℓ, ∞. \end{matrix}

ϕ_{p}(ℓ) = \{ \begin{matrix}[1] & if & p = ℓ \\ [1] & if & p ≠ ℓ \end{matrix}

ϕ_{p}(q) = \{ \begin{matrix}[q] & if & p = q \\ [q^{−1}] & if & p = ℓ \\ [1] & if & p ≠ ℓ, q. \end{matrix}

In each case, ∏ ϕ_{p}(a) = 1.

Remark 8.3. In Example V.4.10, we showed that the homomorphism ϕ : 𝕀_{ℚ} → Gal(ℚ[ζ_{m}]/ℚ) attached to the Artin map ϕ : C_{∞(m)} → Gal(ℚ[ζ_{m}]/ℚ) has local components equal to the local Artin maps. Since, by definition, ϕ(ℚ^{×}) = 1, this gives an alternative proof of (8.1a) in the case ℚ[ζ_{m}]/ℚ.

Lemma 8.4. (a) If ( 8.1a) holds for L/K, then it holds for any subextension.

(b) If ( 8.1a) holds for L/K, then it holds for L · K′/K′ for any number field K′ ⊃ K.

Proof. (a) Suppose L ⊃ K′ ⊃ K. Then ϕ_{K′/K} is the composite of ϕ_{L/K} and the restriction map Gal(L/K) → Gal(K′/K) (because this is true for the local Artin maps).

(b) Let L′ = L · K′. For each prime w of K′, we have a commutative diagram

K_{w}^{′×} ⟶^{ϕ_{w}} Gal(L^{′w}/K′_{w})
↓^{Nm}  ↓^{inclusion}
K_{v}^{×} ⟶^{ϕ_{v}} Gal(L^{v}/K_{v}).

On combining them, we get a commutative diagram:

𝕀′_{K} ⟶^{ϕ_{L′/K′}} Gal(L′/K′)
↓^{Nm}  ↓^{inclusion}
𝕀_{K} ⟶^{ϕ_{L/K}} Gal(L/K).

Because the norm map on idèles carries K^{×} into ℚ^{×}, we see that this lemma follows from the previous one.

From the example and the lemma, we find that (8.1a) holds for all cyclotomic extensions^{3} of a number field K.

We next need to relate the two statements in Theorem 8.1.

Lemma 8.5. Let L/K be an abelian extension of number fields. If ( 8.1b) holds for L/K, then so also does ( 8.1a). Conversely, if L/K is cyclic and ( 8.1a) holds for L/K, then so also does ( 8.1b).

3An extension L/K is said to be cyclotomic if L ⊂ K[ζ] for some root ζ of 1.

Proof. Let χ ∈ Hom(G, ℚ/ℤ). We can regard χ as an element of H^{1}(G, ℚ/ℤ), and then its image under the boundary map arising from the sequence

0 → ℤ → ℚ → ℚ/ℤ → 0

is an element δχ ∈ H^{2}(G, ℤ). Consider the diagram:

K^{×}  ⟶  I_{K}  ⟶^{ϕ_{L/K}} G
↓^{∪δχ}  ↓^{∪δχ}  ↓^{χ}
H^{2}(G, L^{×}) ⟶ H^{2}(G, I_{L}) ⟶ Q/Z.

The first two vertical arrows are cup-product by δχ: if δχ is represented by the 2cocycle n_{σ,τ} then the image of x is represented by the 2-cocycle σ, τ ↦ x^{n_{σ,τ}}. Clearly, the left-hand square commutes. The right-hand vertical map is χ itself. That the right hand square commutes follows from (III.4.1). Now assume 8.1b is true for L/K. Then χ(ϕ_{L/K}(a)) = 0 for all characters χ of G, and so ϕ_{L/K}(a) itself is zero. Conversely, when G is cyclic, we can choose χ to be injective, and then (a) implies (b).

Since we know 8.1a for cyclotomic extensions, it follows that we know 8.1b for cyclic cyclotomic extensions. Moreover, we will have proved the whole theorem once we have proved (b) of the theorem. Thus, the next result completes the proof of the theorem

Lemma 8.6. If ( 8.1b) is true for cyclic cyclotomic extensions, then it is true for all finite Galois extensions.

Proof. Let β ∈ Br(K). We are given that, if β ∈ Br(L/K) for some cyclic cyclotomic extension L/K, then ∑ inv_{v}(β) = 0, where β_{v} is the image of β in Br(K_{v}), but Proposition 7.2 says that every β in Br(K) lies in Br(L/K) for some cyclic cyclotomic extension L/K.

9. The Existence Theorem

In this section we prove the Existence Theorem: every open subgroup of finite index in the idèle class group is a norm group. A large part of the proof can be extracted from Section 6. However, at the cost of some repetition, I give a proof independent of Section 6 (except for some elementary statements).

Lemma 9.1. If U is a norm group, and V ⊃ U, then V also is a norm group.

Proof. Suppose U = Nm C_{L}. According to the Reciprocity Law, the Artin map defines an isomorphism

ϕ : C_{K}/U ⟶ Gal(L/K).

If M is the fixed field of ϕ(V), then ϕ defines an isomorphism

C_{K}/V ⟶ Gal(M/K),

but, according to the Reciprocity Law, the kernel of ϕ : C_{K} → Gal(M/K) is Nm_{M/K} C_{M}.

It is obvious from its factorization into primes, that a rational number a is an nth power if and only if it is an nth power in ℝ and in ℚ_{p} for all primes p|a. The proof of the analogous statement for number fields requires the Reciprocity Law (or complex analysis).

Proposition 9.2. Let K be a number field containing a primitive nth root of 1, and let S ⊃ S_{∞} be a finite set of primes of K containing all those dividing n and enough primes to generate the class group of K. Any a ∈ K^{×} such that

a is an nth power in K_{v} for all v ∈ S;
a is a unit in K_{v} for all v ∉ S.

is an nth power in K.

Proof. Let L = K[a^{1/n}]—because ζ_{n} ∈ K, this is an abelian extension of K. For any prime v ∈ S, X^{n} − a splits completely in K_{v}[X], and so L_{w} = K_{v} for all w|v. Hence the norm map L_{w}^{×} → K_{v}^{×} is onto. On the other hand, L is unramified over K at any prime v ∉ S, and so the norm map Nm_{L/K} : U_{w} → U_{v} is onto. Therefore, Nm_{L/K}(𝕀_{L}) ⊃ 𝕀_{S}, and so

K^{×} · Nm_{L/K}(𝕀_{K}) ⊃ K^{×} · 𝕀_{S} = 𝕀_{K}.

The Reciprocity Law now shows that L = K, and so a is an nth power in K.

Lemma 9.3 (Key case of the Existence Theorem). Let K be a number field containing a primitive p^{th} root of 1 (p prime). Then every open subgroup V of C_{K} such that C_{K}/V is a finite group killed by p is a norm group.

Proof. Let S ⊃ S_{∞} be a finite set of primes of K containing the infinite primes, those dividing p, and enough primes so that 𝕀_{K} = K^{×} · 𝕀_{S}. Let L be the extension of K corresponding by Kummer theory to the group U(S) · K^{×p}, i.e., L = K[U(S)^{\frac{1}{p}}], and let

E = ∏_{v∈S} K_{v}^{×p} × ∏_{v∉S} U_{v}.

We shall prove that K^{×} · E = K^{×} · Nm(𝕀_{L}) by verifying that

(a) E ⊂ Nm(𝕀_{L});

(b) (𝕀_{K} : K^{×} · E) = p^{s} = (𝕀_{K} : K^{×} · Nm_{L/K}(𝕀_{L})).

For any prime v of K and prime w of L lying over it, the local Artin map is an isomorphism

K_{v}^{×}/ Nm(L_{w}^{×}) → Gal(L_{w}/K_{v}).

Because L/K is has exponent p, Nm(L_{w}^{×}) ⊃ K_{v}^{×p}.

For any prime v ∈ S, L is unramified over K and v, and so the norm map U_{w} → U_{v} is onto.

On combining the statements in the last two paragraphs, we obtain (a).

From the Reciprocity Law,

(𝕀_{K} : K^{×} · Nm(𝕀_{L})) = [L : K],

and from Kummer theory,

[L : K] = (U(S) · K^{×p} : K^{×p}).

But

U(S) · K^{×p}/K^{×p} ≈ U(S)/U(S) ∩ K^{×p}.

If a^{p} ∈ U(S), then a ∈ U(S), and so U(S) ∩ K^{×p} = U(S)^{p}. Now, by the Dirichlet Unit Theorem (Math 676, 5.9),

U(S) ≈ U(S)_{torsion} ⊕ ℤ^{s−1},

and U(S)_{torsion} is the group of roots of 1 in K, which is a cyclic group whose order is divisible by p. Hence (U(S) : U(S)^{p}) = p^{s}.

On the other hand,

(𝕀_{K} : K^{×} · E) = (𝕀_{S} · K^{×} : E · K^{×}),

which, by (6.5), equals

(𝕀_{S} : E)/(𝕀_{S} ∩ K^{×} : E ∩ K^{×}).

Therefore (see 6.8),

(𝕀_{S} : E) = ∏_{v∈S}(K_{v}^{×} : K_{v}^{×p}) = ∏_{v∈S} \frac{p}{|p|_{v}} p = p^{2s}.

Here, we have used that K contains a primitive pth root of 1 and that S contains all v for which |p|_{v} ≠ 1, and so ∏_{v∈S} |p|_{v} = ∏_{all v} |p|_{v} = 1 by the product formula. It follows that K^{×} · E = K^{×} · Nm 𝕀_{L}.

Now let V be an open subgroup of C_{K} such that C_{K}/V is killed by p, and let U be the inverse image of V in 𝕀_{K}. Then U is open in 𝕀_{K} and so there is a finite set of primes S such that U ⊃ ∏_{v∈S} 1 × ∏_{v∉S} U_{v}. Moreover, 𝕀_{K}/U has exponent p, and so U ⊃ 𝕀_{K}^{p}. Hence U ⊃ E · K^{×}, and because E · K^{×}/K^{×} is a norm group, so also must be U/K^{×} = V.

For simplicity, in the proof of the next lemma, we assume the Norm Limitation Theorem, which is not proved until the next chapter. For a proof avoiding that theorem, see p202 of Tate’s article in Cassels and Fröhlich, 1967.

Lemma 9.4. Let U be an open subgroup of finite index in C_{K}. If there exists a finite extension K′/K such that Nm_{K′/K}^{−1}(U) is a norm group, then so also is U.

Proof. Write U′ for Nm_{K′/K}^{−1}(U), and let L be the abelian extension of K′ with Nm C_{L} = U′. If M is the maximum abelian subextension of L/K, then we have

Nm_{M/K} C_{M} = Nm_{L/K} C_{L} = Nm_{K′/K} U′ ⊂ U

and we can apply Lemma 9.1.

Theorem 9.5. Every subgroup U of finite index in C_{K} is a norm group.

Proof. We prove this by induction on the index of U. If the index is 1, then there is nothing to prove. Otherwise, there exists a prime p dividing (C_{K} : U). After (9.4) we may assume that K contains a p^{th} root of 1. Choose a subgroup U_{1} of C_{K} containing U and of index p in C_{K}. After (9.3), there exists an abelian extension K′ of K such that Nm_{K′/K} 𝕀_{K′} = U_{1}; moreover K′ is cyclic of degree p over K. Put U′ = Nm_{K′/K}^{−1} U. The map

Nm_{K′/K} : C_{K′} ⟶ C_{K}/U

has image U_{1}/U and kernel U′. Hence

(C_{K′} : U′) = (C_{K} : U)/p

and so, by induction, U′ is a norm group. Now we can apply (9.4) to deduce that U is a norm group.

10. Appendix: Kummer theory

Throughout this subsection, K is a field containing a primitive n^{th} root of 1, ζ. In particular, K either has characteristic 0 or characteristic p not dividing n. Write μ_{n} for the group of n^{th} roots of 1 in K. Then μ_{n} is a cyclic subgroup of K^{×} of order n with generator ζ.

Consider a field L = K[α] generated by an element α whose n^{th} power is in K. Then α is a root of X^{n}−a, and the remaining roots are the elements ζ^{i}α, 1 ≤ i ≤ n−1. Since these are all in L, L is a Galois extension of K, with Galois group G say. For any σ ∈ G, σα is also a root of X^{n}−a, and so σα = ζ^{i}α for some i. Hence σα/α ∈ μ_{n}. The map

σ ↦ σα/α : G ⟶ μ_{n}

doesn’t change when α is replaced by a conjugate, and it follows that the map is a homomorphism: \frac{στα}{α} = \frac{σ(τα)}{τα} \frac{τα}{α}. Because α generates L/K, the map is injective. If it is not surjective, then G maps into a subgroup μ_{d} of μ_{n}, some d|n, d < n. In this case, (σα/α)^{d} = 1, i.e., σα^{d} = α^{d}, for all σ ∈ G, and so α^{d} ∈ K. Thus the map is surjective if n is the smallest positive integer such that α^{n} ∈ K. We have proved the first part of the following statement.

Proposition 10.1. Let L = K[α] where α^{n} ∈ K and no smaller power of α is in K. Then L is a Galois extension of K with cyclic Galois group of order n. Conversely, if L is cyclic extension of K of degree n, then L = K[α] for some α with α^{n} ∈ K.

Proof. It remains to prove the second statement. Let σ generate G and let ζ generate μ_{n}. It suffices to find an element α ∈ L^{×} such that σα = ζ^{−1}α, for then α^{n} ∈ K, and α^{n} is the smallest power of α that lies in K. According to the Normal Basis Theorem (II.1.24), there exists an element γ ∈ L such that {γ, σγ, …, σ^{n−1}γ} is a basis for L/K as a K-vector space. Form the sum

α = ∑ ζ^{i}σ^{i}γ.

Then α ≠ 0 because the σ^{i}γ are linearly independent and the ζ^{i} ∈ K, and σα = ζ^{−1}α.

Proposition 10.2. Two cyclic extensions K[a^{\frac{1}{n}}] and K[b^{\frac{1}{n}}] of K of degree n are equal if and only if a = b^{r}c^{n} for some r ∈ ℤ relatively prime to n and some c ∈ K^{×}, i.e., if and only if a and b generate the same subgroup of K^{×}/K^{×n}.

Proof. Only the “only if” part requires proof. We are given that K[α] = K[β] with α^{n} = a and β^{n} = b. Let σ be the generator of the Galois group with σα = ζα, and let σβ = ζ^{i}β, (i, n) = 1. We can write

β = ∑_{j=0}^{n−1} c_{j}α^{j}, c_{j} ∈ K,

and then

σβ = ∑_{j=0}^{n−1} c_{j}ζ^{j}α^{j}.

On comparing this with σβ = ζ^{i}β, we find that ζ^{i}c_{j} = ζ^{j}c_{j} for all j. Hence c_{j} = 0 for j ≠ i, and therefore β = c_{i}α^{i}.

The last two results give us a complete classification of the cyclic extensions of K of degree n (recall that we are assuming K contains a primitive n^{th} root of 1). It is not difficult to extend this to a classification of all abelian extensions of exponent n. (We say that a group G has exponent n if σ^{n} = 1 for all σ ∈ G. A finite abelian group of exponent n is isomorphic to a subgroup of (ℤ/nℤ)^{r} for some r.)

Let L/K be a finite Galois extension with Galois group G. From the exact sequence

1 ⟶ μ_{n} ⟶ L^{×} ⟶^{x↦x^{n}} L^{×n} ⟶ 1

we obtain a cohomology sequence

1 ⟶ μ_{n} ⟶ K^{×} ⟶^{x↦x^{n}} K^{×} ∩ L^{×n} ⟶ H^{1}(G, μ_{n}) ⟶ 1.

The 1 at the right is because of Hilbert’s Theorem 90. Thus we obtain an isomorphism

K^{×} ∩ L^{×n}/K^{×n} ⟶ Hom(G, μ_{n}).

This map can be described as follows: let a be an element of K^{×} that becomes an n^{th} power in L, say a = α^{n}; then a maps to the homomorphism σ ↦ \frac{σα}{α}. If G is abelian of exponent n, then

# Hom(G, μ_{n}) = (G : 1).

Theorem 10.3. The map

L ↦ K^{×} ∩ L^{×n}/K^{×n}

defines a one-to-one correspondence between the finite abelian extensions of K of exponent n contained in some fixed algebraic closure Ω of K and the finite subgroups B of K^{×}/K^{×n}. The extension corresponding to B is K[B^{\frac{1}{n}}], the smallest subfield of Ω containing K and an n^{th} root of each element of B. If L ↔ B, then [L : K] = (B : K^{×n}).

Proof. For any finite Galois extension L of K, define B(L) = K^{×} ∩ L^{×n}. Then L ⊃ K[B(L)^{\frac{1}{n}}], and for any group B containing K^{×n} as a subgroup of finite index, B(K[B^{\frac{1}{n}}]) ⊃ B. Therefore,

[L : K] ≥ [K[B(L)^{\frac{1}{n}}] : K] = (B(K[B(L)^{\frac{1}{n}}]) : K^{×n}) ≥ (B(L) : K^{×n}).

If L/K is abelian of exponent n, then [L : K] = (B(L) : K^{×n}), and so equalities hold throughout: L = K[B(L)^{\frac{1}{n}}].

Next consider a group B containing K^{×n} as a subgroup of finite index, and let L = K[B^{\frac{1}{n}}]. Then L is a composite of the extensions K[a^{\frac{1}{n}}] for a running through a set of generators for B/K^{×n}, and so it is a finite abelian extension of exponent n. Therefore

a ↦ (σ ↦ \frac{σa^{\frac{1}{n}}}{a}) : B(L)/K^{×n} → Hom(G, μ_{n}), G = Gal(L/K),

is an isomorphism. This map sends B/K^{×n} isomorphically onto the subgroup Hom(G/H, μ_{n}) of Hom(G, μ_{n}) where H consists of the σ ∈ G such that σa^{\frac{1}{n}} /a = 1 for all a ∈ B. But such a σ fixes all a^{\frac{1}{n}} for a ∈ B, and therefore is the identity automorphism on L = K[B^{\frac{1}{n}}]. This shows that B(L) = B, and hence L ↦ B(L) and B ↦ K[B^{\frac{1}{n}}] are inverse bijections.

Example 10.4. (a) The quadratic extensions of ℝ are in one-to-one correspondence with the subgroups of ℝ^{×}/ℝ^{×2} = {±1}.

(b) The finite abelian extensions of ℚ of exponent 2 are in one-to-one correspondence with the finite subgroups of

ℚ^{×}/ℚ^{×2} ≈ {±1} × ℤ/2ℤ × ℤ/2ℤ × ⋯

(copies of ℤ/2ℤ indexed by the prime numbers).

After this excursion into algebra, we return to some number theory.

Proposition 10.5. Let K be a number field (containing a primitive n^{th} root of 1, as always in this subsection), and let L = K[a1^{\frac{1}{n}} , ⋯, an^{\frac{1}{n}}]. Then L is unramified at a finite prime v of K if na_{i} is a unit in K_{v} for all i.

Proof. We need only consider a cyclic extension K[a^{\frac{1}{n}}], which (see 10.1) we can assume to have degree n. Let α = a^{\frac{1}{n}} and f = X^{n} − a. Then

disc f = ± Nm_{L/K} f′(α) = ± Nm_{L/K} nα^{n−1} = ±n^{n}a^{n−1}.

Thus if 𝔭_{v} does not divide na, it does not divide disc f, and, a fortiori, it does not divide disc(O_{L}/O_{K}). (See Math 676, 2.33, 2.22.)

Remark 10.6. Determining the ramification in K[a^{\frac{1}{n}}]/K at prime 𝔭 dividing n can be quite complicated. The following summarizes what happens when n = p, a prime, and K contains a primitive pth root ζ of 1. Let π = ζ − 1, and let a ∈ K^{×} be relatively prime to p and not a pth power in K.

(a) A prime 𝔭|p splits completely in the extension K[a^{\frac{1}{p}}]/K if and only if X^{p} ≡ a mod pπ𝔭 has a nonzero solution in O_{K} (a is then said to be hyperprimary at 𝔭).

(b) A prime 𝔭|p is unramified in the extension K[a^{\frac{1}{p}}]/K if and only if X^{p} ≡ a mod pπ has a nonzero solution in O_{K} (a is then said to be primary at 𝔭).

(c) The extension K[a^{\frac{1}{p}}]/K is unramified at all finite primes of K if and only if a is primary and (a) = 𝔞^{p} for some ideal 𝔞.

For hints of proofs of these statements, see L. Washington, Introduction to Cyclotomic Fields, Springer, 1982/1997, Exercise 9.3, and Cassels and Fröhlich 1967, Exercise 2.12, p353. In the case K = ℚ[ζ], see also J. Cassels, Local Fields, Cambridge, 1986, pp 139–140, and A. Fröhlich and M. Taylor, Algebraic Number Theory, Cambridge, 1991, III.3.11.

190 VII. GLOBAL CLASS FIELD THEORY: PROOFS

CHAPTER VIII Complements

In this Chapter, we add some complements to the theory of class fields, and we give some applications. In particular, we extend all results proved in Serre, Cours..., 1970, Chapters I–IV, VI to arbitrary number fields.

1. The Local-Global Principle

The local-global (or Hasse) principle asks whether a statement is true over a number field K whenever it is true over each of the completions of K. In this section, we give three cases where class field theory allows us to prove that the principle holds.

nth Powers.

Theorem 1.1. Let K be a number field containing a primitive nth root of 1, and let S be a finite set of primes of K. An element a of K^{×} is an nth power in K if and only if it is an nth power in K_{v} for all primes v.

Proof. Only the sufficiency needs to be proved. Consider L = K[a^{\frac{1}{n}}]. This is an abelian extension of K, and a prime v splits in L if and only if a is an nth power in K_{v}. Therefore, every prime not in S splits in L, and Proposition VII.4.6 implies that L = K.

Remark 1.2. (a) If we use (VI.3.4) rather than (VII.4.6), we obtain the stronger result: under the hypothesis of the theorem, a is an nth power in K if it an nth power in K_{v} for all v in a set of density > 1/2.

(b) In 1933, Grünwald proved Theorem 1.1 without the assumption that K contain a primitive nth root of 1 (Grünwald’s theorem). In 1942, Whaples gave another proof of Grünwald’s theorem. Then in 1948, Wang gave a counterexample to Grünwald’s theorem (see the exercise below), and later gave a proof of a corrected theorem (Grünwald-Wang theorem). The correct theorem states:

Let K be a number field, and let n be a positive integer. Let a ∈ K^{×}, and suppose that a ∈ K_{v}^{×n} for all but finitely many primes v. Then at least one of the following is true:

(a) a ∈ K^{×n};

(b) n = 2^{t}n′ for some odd n′, Gal(K[ζ_{2^{t}}]/K) is not cyclic, and a^{2} ∈

K^{×n}.

See Cassels, Local Fields, 1968, Exercise on p248, and Artin and Tate 1951/52, p96.

Example 1.3. Show that 16 is an 8th power in ℝ and ℚ_{p} for all odd p (but not in ℚ). (Hint: Show that ℚ[ζ_{8}] = ℚ[i, \sqrt{2}] is unramified at all odd p, and deduce that, for p odd, ℚ_{p} contains at least one of 1 + i, \sqrt{2}, or \sqrt{−2}.)

Norms.

Theorem 1.4. Let L/K be a cyclic extension of number fields, and let a ∈ K^{×}. Then the image of a in K_{v} is a norm from L^{v} for all but finitely many v, and if it is a norm for all v, then it is a norm in K.

Proof. According to Theorem VII.5.1, H^{1}(G, C_{L}) = 0. Because of the periodicity of the cohomology of cyclic groups, this implies that H_{T}^{−1}(G, C_{L}) = 0. Therefore, from the cohomology sequence of

1 → L^{×} → 𝕀_{L} → C_{L} → 0

we find that

H_{T}^{0}(G, L^{×}) → H_{T}^{0}(G, 𝕀_{L})

is injective. But (see VII.2.5), this is

K^{×}/ Nm(L^{×}) → ⊕_{v}K_{v}^{×}/ Nm(L^{v×}).

Remark 1.5. The proof fails for noncyclic extension, and, in fact, the statement is not true for noncyclic extensions. For example, 2 is a local norm from ℚ[\sqrt{13}, \sqrt{17}] at all primes but is not a global norm.

Quadratic Forms. Recall that a quadratic form on a vector space V over a field k is a map Q : V → k such that

(a) Q(av) = a^{2}Q(v);

(b) B(v, w) =^{df} Q(v + w) − Q(v) − Q(w) is a bilinear form on V.

The quadratic form Q is said to be nondegenerate if its associated bilinear form B is nondegenerate. Let c ∈ k. A nondegenerate quadratic form Q is said to represent c if there exists a nonzero v ∈ V such that Q(v) = c.

Lemma 1.6. If a nondegenerate quadratic form Q represents 0, then it represents all c ∈ k.

Proof. Note that, for t ∈ k,

Q(tv + w) = t^{2}Q(v) + tB(v, w) + Q(w).

If v_{0} is a nonzero vector such that Q(v_{0}) = 0, then, because B is nondegenerate, there exists a vector w_{0} such that B(v_{0}, w_{0}) ≠ 0. As t runs through all values of k, so also does Q(tv_{0} + w_{0}) = tB(v_{0}, w_{0}) + Q(w_{0}).

When k has characteristic ≠ 2, there exists a basis {e_{1}, …, e_{n}} for V such that B(e_{i}, e_{j}) = 0 for i ≠ j. Then

Q(∑ x_{i}e_{j}) = ∑_{i=1}^{n} a_{i}x_{i}^{2}, n = dim V.

Henceforth, we shall write the quadratic form as

q(X_{1}, …, X_{n}) = a_{1}X_{1}^{2} + ⋯ + a_{n}X_{n}^{2},

and keep in mind that an invertible change of variables will change none of our statements.

Lemma 1.7. A nondegenerate quadratic form q(X_{1}, …, X_{n}) represents a if and only if r =^{df} q − aY^{2} represents 0.

Proof. If q(x_{1}, …, x_{n}) = a, the r(x_{1}, …, x_{n}, 1) = 0. Conversely, suppose r(x_{1}, …, x_{n}, y) = 0. If y = 0, then q represents 0 and hence represents every element in k. If y ≠ 0, then q(\frac{x_{1}}{y} , ⋯, \frac{x_{n}}{y}) = q(x_{1}, …, x_{n})/y^{2} = a.

Theorem 1.8. Let q be a nondegenerate quadratic form in n variables with coefficients in a number field K.

(a) If n ≥ 3, then q represents 0 in K_{v} for all but finitely many v.

(b) The form q represents 0 in K if it represents 0 in K_{v} for all v.

Before beginning the proof, we note a consequence.

Corollary 1.9. Let c ∈ K. A nondegenerate quadratic form q with coefficients in K represents c in K if and only if it represents c in K_{v} for all v.

Proof. Let r = q − cY^{2}. Then r represents 0 if and only if q represents c.

We begin the proof with a purely algebraic result.

Proposition 1.10. Let k be a field of characteristic ≠ 2.

(a) The form q = X^{2} does not represent 0.

(b) The form q = X^{2} − aY^{2} represents 0 if and only if a is a square.

(c) The form q = X^{2} − aY^{2} − bZ^{2} represents 0 if and only if b is a norm from the field k[\sqrt{a}].

(d) The form q = X^{2} − bY^{2} − cZ^{2} + acT^{2} represents 0 in k if and only if c, as an element of k[\sqrt{ab}], is a norm from k[\sqrt{a}, \sqrt{b}].

Proof. (a) This is obvious.

(b) According to Lemma 1.7, X^{2} − aY^{2} represents 0 if and only if X^{2} represents a.

(c) According to 1.7, X^{2} −aY^{2} −bZ^{2} represents 0 if and only if X^{2} −aY^{2} represents b, i.e., if and only if b is a norm from k[\sqrt{a}].

(d) Clearly,

q(x, y, z, t) = 0 ⟺ c = \frac{Nm_{k[\sqrt{b}]/k}(x + \sqrt{b}y)}{Nm_{k[\sqrt{a}]/k}(z + \sqrt{a}t)}.

Because the inverse of a norm is also a norm, this shows that q represents zero if and only if c is the product of norm from k[\sqrt{a}] and a norm from k[\sqrt{b}]. Thus (d) follows from the next lemma.

Lemma 1.11. Let k be a field of characteristic ≠ 2. An element c ∈ k^{×} is the product of a norm from k[\sqrt{a}] and a norm from k[\sqrt{b}] if and only if, as an element of k[\sqrt{ab}], it is a norm from L = k[\sqrt{a}, \sqrt{b}].

Proof. We leave the degenerate cases, in which one of a, b, or ab is a square in k to the reader. Thus, we may suppose that Gal(k[\sqrt{a}, \sqrt{b}]/k) = \{1, σ, τ, στ\} where each of σ, τ, and στ is of order 2, and fix respectively \sqrt{a}, \sqrt{b}, and \sqrt{ab}. The first condition asserts,

(∗) There exist x, y ∈ k[\sqrt{a}, \sqrt{b}] such that σx = x, τy = y, and xy ·
στ(xy) = c.

and the second asserts,

(∗∗) There exists z ∈ k[\sqrt{a}, \sqrt{b}] such that z · στ(z) = c.

Clearly, (∗) ⟹ (∗∗). For the converse, note that

z · σz = Nm_{k[\sqrt{a},\sqrt{b}]/k[\sqrt{a}]} z ∈ k[\sqrt{a}].

Moreover,

Nm_{k[\sqrt{a}]/k}(z · σz) = z · σz · τz · στz ∈ k.

As z · στz = c ∈ k, this implies that σz · τz ∈ k, and so

σz · τz = σ(σz · τz) = z · στz = c.

Therefore,

Nm_{k[\sqrt{a}]/k}(z · σz) = c^{2}.

Now Hilbert’s theorem 90 (II.1.22), applied to z · σz/c ∈ k[\sqrt{a}], shows that there exists an x ∈ k[\sqrt{a}]^{×} such that τx/x = z · σz/c. Let y = στz/x. Then

τy = σz/τx = c/z · x = z · στz/z · x = y

(use: definition of y; definition of x; definition of z; definition of y) and

xy · στ(xy) = στz · στ(στz) = στz · z = c

(use: definition of y; (στ)^{2} = 1; definition of z) as required.

Proof of (a) of the Theorem. If q = q_{1}(X_{1}, …, X_{m}) + q_{2}(X_{m+1}, …, X_{n}) and q_{1} represents zero, then so also does q. Therefore, it suffices to prove (1.8a) for a quadratic form in 3 variables. After multiplying q by a nonzero scalar, we may suppose q = X^{2} − aY^{2} − bZ^{2}, and for such a quadratic form, the statement follows from (1.10) and Theorem 1.4. □

Proof of (b) of the Theorem We prove this by induction on the number n of variables. When n = 1, there is nothing to prove, because the hypothesis is never fulfilled. When n = 2, then, after multiplying q by nonzero scalar, we may suppose that

q = X^{2} − aY^{2}, and for such a quadratic form, the statement follows from (1.10) and Theorem 1.1.

When n = 3, 4 the statement follows in a similar fashion from (1.10c,d) and Theorem 1.1.

Before proving the general case, we make some elementary observations.

(a) A nondegenerate quadratic form q_{1}(X_{1}, …, X_{m}) − q_{2}(X_{m+1}, …, X_{n}) represents 0 in a field k if and only if there is a c ∈ k such that both q_{1} and q_{2} represent c.

(b) If q represents c in k^{×}, then q represents every element in the coset c · k^{×2}.

(c) The subgroup K_{v}^{×2} of K_{v}^{×} is open. When v is real or complex, this is obvious. When v is nonarchimedean, Newton’s Lemma (Math 676, 7.23) shows that 1 can be refined to a root of X^{2} − a for any a with |1 − a|_{v} < |2|_{v}^{2}.

On combining (b) and (c), we see that a quadratic form q with coefficients in K_{v} represents the elements in a nonempty open subset of K_{v}^{×}.

Assume now that n ≥ 5 and that Theorem 1.4b has been proved for n − 1. Let

q(X_{1}, …, X_{n}) = aX_{1}^{2} + bX_{2}^{2} − r(X_{3}, …, X_{n}), n − 2 ≥ 3.

From (a) of the theorem, we know that, except for v in a certain finite set S, R represents 0 in K_{v}. Let v ∈ S. Because q represents 0 in K_{v}, there exists an element c_{v} ∈ K_{v}^{×} that is represented by both aX_{1}^{2} + bX_{2}^{2} and r, i.e., there exist x_{i}(v) ∈ K_{v} such that

ax_{1}(v)^{2} + bx_{2}(v)^{2} = c_{v} = r(x_{3}(v), …, x_{n}(v)).

Now apply the weak approximation theorem, to find elements x_{1}, x_{2} ∈ K that are close to x_{1}(v), x_{2}(v) in K_{v} for all v ∈ S. Then

c =^{df} ax_{1}^{2} + bx_{2}^{2}

will be close to c_{v} for each v ∈ S; in particular, we may suppose that c/c_{v} ∈ K_{v}^{×2} for all v ∈ S.

Consider the quadratic form cY^{2} − r. It represents 0 in K_{v} for v ∉ S because r represents zero in K_{v}, and it represents 0 in K_{v} for v ∈ S because r represents c in K_{v}. By induction, cY^{2} − r represents zero in K. It follows that q represents 0 in K because each of aX_{1}^{2} + bX_{2}^{2} and r represents c in K. □

Proposition 1.12. A nondegenerate quadratic form q in 4 variables over a finite extension K of ℚ_{p} represents every nonzero element of K^{×}.

Proof. If q represents 0, then it represents every element of K. We assume the contrary. After multiplying q by a nonzero element of K, we may suppose that

q = X^{2} − bY^{2} − cZ^{2} + acT^{2}.

Because q does not represent 0 in K, neither b nor a is a square.

If K[\sqrt{a}] ≠ K[\sqrt{b}], then (by local class field theory, Theorem I.1.1), Nm(K[\sqrt{a}]^{×}) and Nm(K[\sqrt{b}]^{×}) are distinct subgroups of index 2 in K^{×2}, and therefore K^{×} = Nm(K[\sqrt{a}]^{×}) · Nm(K[\sqrt{b}]^{×}). Since the inverse of a norm is also a norm, this means that we can write c as

c = \frac{x^{2} − by^{2}}{z^{2} − at^{2}} ,

some x, y, z, t ∈ K. On multiplying out, we find that q represents 0, contradicting our assumption. Therefore K[\sqrt{a}] = K[\sqrt{b}], and a = b × (square) (see VII.10.1). The square may be absorbed into the T^{2}, and so we may write

q = X^{2} − bY^{2} − cZ^{2} + bcT^{2}.

Consider the quaternion algebra H(b, c) (see IV.5.1). For

α = x + yi + zj + tk

we define

ᾱ = x − yi − zj − tk

so that

Nm(α) =^{df} αᾱ = x^{2} − by^{2} − cz^{2} + bct^{2}.

The map α ↦ Nm(α) : H(b, c)^{×} → K^{×} is a homomorphism, which we must show to be surjective.

For any α ∈ H(b, c),

P_{α}(X) =^{df} (X − α)(X − ᾱ) = X^{2} − (α + ᾱ)X + Nm(α) ∈ K[X].

Therefore, P_{α}(X) is the characteristic polynomial of α in the extension K[α]/K. In particular,

Nm(α) = Nm_{K[α]/K}(α).

Now (see IV.4.4), H(b, c) contains copies of every quadratic extension of K, for example, the unramified quadratic extension of K and a totally ramified quadratic extension of K. Therefore Nm(H(b, c)^{×}) contains the norm groups of these two distinct quadratic extensions, and so (as above) equals K^{×}.

Corollary 1.13. Every nondegenerate quadratic form q in ≥ 5 variables over a finite extension of ℚ_{p} represents 0.

Proof. We can write q = r(X_{1}, …, X_{4}) − aX_{5}^{2} + q′(X_{6}, …), where r is a nondegenerate quadratic form in 4 variables and a ≠ 0. Then r represents a and so q represents 0.

Corollary 1.14. A nondegenerate quadratic form q in ≥ 5 variables over a number field K represents 0 if and only if it represents 0 in every real completion of K.

Proof. Combine (1.13) with (1.8).

Remark 1.15. The proof of Proposition 1.12 also for K = ℝ down to the last step: the only quadratic extension of ℝ is ℂ, and so

Nm(ℍ(b, c)^{×}) = Nm(ℂ^{×}) = ℝ_{>0}.

It shows therefore, that a nondegenerate form in 4 variables over ℝ that does not represent zero represents all strictly positive real numbers.

2. The Fundamental Exact Sequence and the Fundamental Class

For a Galois extension L/K, we write

Br(L/K) = H^{2}(Gal(L/K), L^{×}),

and for a Galois extension L/K of number fields, we write

H^{2}(L/K) = H^{2}(Gal(L/K), C_{L}).

Because H^{1}(G, L^{×}) = 0 (see II.1.21) and H^{1}(G, C_{L}) = 0 (see VII.5.1), for any tower of Galois extensions E ⊃ L ⊃ K, we get exact sequences

0 → Br(L/K) → Br(E/K) → Br(E/L)

THE FUNDAMENTAL EXACT SEQUENCE AND THE FUNDAMENTAL CLASS 197

and

0 → H^{2}(L/K) → H^{2}(E/K) → H^{2}(E/L).

On passing to the direct limit over larger fields E ⊂ K^{al}, we obtain exact sequences

0 → Br(L/K) → Br(K) → Br(L)

and

0 → H^{2}(L/K) → H^{2}(K^{al}/K) → H^{2}(K^{al}/L).

Thus, we can regard Br(L/K) as the subgroup of Br(K) of elements split by L, and similarly for H^{2}(L/K).

Let L/K be a Galois extension of number fields of finite degree n, and consider the diagram:

H^{2}(L/K)
↗
0 → Br(L/K) → ⊕ Br(L^{v}/K_{v})
↘
ℚ/ℤ

The top row is part of the cohomology sequence of

0 → L^{×} → 𝕀_{L} → C_{L} → 0.

The zero at left comes from the fact that H^{1}(G, C_{L}) = 0. The top row is exact, but the map ⊕_{v} Br(L^{v}/K_{v}) → H^{2}(L/K) will not in general be surjective—we denote its image by H^{2}(L/K)′.

Recall (III.1.1), that for each prime v, we have a homomorphism

inv_{v} : Br(K_{v}) → ℚ/ℤ.

If v is nonarchimedean, it is an isomorphism of Br(K_{v}) onto ℚ/ℤ, and if v is real, it is an isomorphism of Br(K_{v}) onto \frac{1}{2}ℤ/ℤ. Moreover, if L_{w}/K_{v} has degree n_{v}, then inv_{w}(β) = n_{v} · inv_{v}(β), and so inv_{v} defines an isomorphism

inv_{v} : Br(L_{w}/K_{v}) → \frac{1}{n_{v}} ℤ/ℤ.

The southeast arrow in the diagram is

Σ : ⊕ Br(L^{v}/K_{v}) → ℚ/ℤ, (β_{v}) ↦ ∑ inv_{v}(β_{v}).

The image of inv_{v} is the cyclic subgroup of order n_{v} in the cyclic group \frac{1}{n}ℤ/ℤ, and therefore the image of Σ is the cyclic subgroup \frac{1}{n_{0}} ℤ/ℤ, where n_{0} = lcm(n_{v}).

According to Theorem VII.5.1, the order of H^{2}(L/K) divides n. According to Theorem VII.8.1, the bottom row of the diagram is a complex, and so the maps in the diagram induce a surjective homomorphism

H^{2}(L/K)′ → \frac{1}{n_{0}} ℤ/ℤ.

The lcm n_{0} of the local degrees always divides n, but need not equal it (see Example 2.5 below). Suppose, however, that the extension L/K has the property that n = n_{0}. Then:

(a) The map H^{2}(L/K)′ → \frac{1}{n_{0}} ℤ/ℤ is an isomorphism. (It is surjective, and (H^{2}(L/K)′ : 1) ≤ (H^{2}(L/K) : 1) ≤ n.)

(b) H^{2}(L/K)′ = H^{2}(L/K), and each has order n.

(c) The bottom row is an exact sequence

0 → Br(L/K) → ⊕_{v} Br(L^{v}/K_{v}) ⟶^{Σ} \frac{1}{n}ℤ/ℤ → 0

(because it is isomorphic to the top row).

Lemma 2.1. If L/K is cyclic, then n = n_{0}.

Proof. Let S ⊃ S_{∞} be a set of primes of K including all those that ramify in L. For v ∉ S, (𝔭_{v}, L/K) is an element of Gal(L/K) of order n_{v} (= f_{v}), and so the image of the Artin map I^{S} → Gal(L/K) has order n_{0} =^{df} lcm(n_{v}). According to , the Artin map is onto, which implies that n_{0} = n. [Using complex analysis, one can show more, namely, that for all v in a set of density φ(n)/n, L^{v}/K_{v} is cyclic of order n: let 𝔪 be the modulus of L/K, and let 𝔞 be an ideal in I^{S} such that (𝔞, L/K) generates Gal(L/K); then the set of prime ideals 𝔭 ≡ 𝔞 in C_{𝔪} has density 1/n.]

Let ℚ^{c} be the infinite cyclic cyclotomic extension of ℚ defined in (I.5.5d) (see also (VII.7.3)), and let Ω = ℚ^{c} · K. For every n, Ω contains a unique cyclic extension of Ω_{n} of degree n. The preceding lemma and remarks show that

0 → Br(Ω_{n}/K) → ⊕ Br(Ω_{n}^{v}/K_{v}) ⟶^{Σ} \frac{1}{n}ℤ/ℤ → 0

is exact. On passing to the direct limit (actually, directed union) over all n, we obtain an exact sequence

0 → Br(Ω/K) → ⊕_{v} Br(Ω^{v}/K_{v}) → ℚ/ℤ → 0.

Theorem 2.2. For any number field K, the sequence

0 ⟶ Br(K) ⟶ ⊕_{v} Br(K_{v}) ⟶^{Σ} ℚ/ℤ ⟶ 0

is exact.

Proof. According to Proposition VII.7.2, Br(Ω/K) = Br(K). Moreover, Br(Ω^{v}/K_{v}) = Br(K_{v}) because, in the nonarchimedean case, [Ω_{n}^{v} : K_{v}] → ∞ as n → ∞ (see VII.7.3).

The sequence in the theorem is called the fundamental exact sequence (of global class field theory).

Corollary 2.3. For any finite extension L/K, the sequence

0 ⟶ Br(L/K) ⟶ ⊕_{v} Br(L^{v}/K_{v}) ⟶^{Σ} \frac{1}{n_{0}} ℤ/ℤ ⟶ 0, n_{0} = lcm(n_{v}),

is exact.

Proof. Apply the snake lemma to the diagram obtained by mapping the fundamental exact sequence for K to that for L.

Example 2.4. (a) For a finite cyclic extension of number fields L/K, the fundamental exact sequence becomes identified with

0 → K^{×}/ Nm(L^{×}) → ⊕K_{v}^{×}/ Nm(L^{v×}) → \frac{1}{n}ℤ/ℤ → 0, n = [L : K].

(b) Let D be a division algebra over a number field K, and let i_{v} = inv_{v}(D ⊗_{K} K_{v}). Then: i_{v} = 0 for all but finitely many v; i_{v} = 0 if v is complex; i_{v} ∈ \frac{1}{2}ℤ/ℤ if v is real; and ∑ i_{v} ≡ 0 mod v. The family (i_{v}) determines the isomorphism class of D, and any family (i_{v}) satisfying the conditions is the family of invariants of the division algebra. Clearly, the order of the class of D in Br(K) is the least common denominator n of the i_{v}. One can also prove that [D : K] = n^{2}. For example, to give a quaternion algebra over ℚ is the same as to give a set of primes of ℚ having an even finite number of elements.

Example 2.5. Let L = ℚ[\sqrt{13}, \sqrt{17}]. Clearly n = 4, but I claim that n_{v} = 1 or 2 for all v. Because both 13 and 17 are congruent to 1 modulo 4, 2 is unramified in L. Therefore, for w|p, p ≠ 13, 17, L_{w} is an unramfied extension of ℚ_{p}. In particular, its Galois group is cyclic. Since it is a subgroup of Gal(L/ℚ), it is killed by 2, and therefore has order 1 or 2. On the other hand, (\frac{17}{13}) = 1 (obviously) and (\frac{13}{17}) =

(\frac{17}{13}) = 1. Hence, 17 is a square modulo 13, and Hensel’s lemma implies that it is a square in ℚ_{13}. Similarly, 13 is a square in ℚ_{17}.

The fundamental class. It follows from the above discussion that there is an isomorphism

inv_{K} : H^{2}(Ω/K) → ℚ/ℤ

uniquely characterized by having the property that the composite

⊕_{v} Br(K_{v}) ⟶ H^{2}(Ω/K) ⟶^{inv_{K}} ℚ/ℤ

is (β_{v}) ↦ ∑ inv_{v}(β_{v}).

Lemma 2.6. For any finite extension L/K of number fields and γ ∈ H^{2}(Ω/K), inv_{L}(γ) = n inv_{K}(γ), n = [L : K].

Proof. Use that the sum of the local degrees is the global degree.

Therefore, for any L/K finite and Galois, we obtain an isomorphism

inv_{L/K} : H^{2}(L/K) → \frac{1}{n}ℤ/ℤ.

On passing to the direct limit over all L ⊂ K^{al}, we obtain an isomorphism

inv_{K} : H^{2}(K^{al}/K) → ℚ/ℤ.

Theorem 2.7. For every finite Galois extension L/K of number fields, H^{2}(L/K) is cyclic of order n = [L : K] having a canonical generator u_{L/K}.

Proof. Take u_{L/K} to be the element such that inv_{L/K}(u_{L/K}) = \frac{1}{n} mod ℤ.

The generator u_{L/K} of H^{2}(L/K) is called the fundamental class. One shows as in the local case (see III.1.2) that for any tower E ⊃ L ⊃ K of finite Galois extensions,

Res(u_{E/K}) = u_{E/L}
Inf(u_{L/K}) = [E : L]u_{E/K}.

Therefore, one may apply Tate’s theorem (II.2.18) to obtain an isomorphism

Gal(L/K)^{ab} → C_{K}/ Nm(C_{L}).

That this is inverse to the global Artin map ϕ_{L/K} defined in the last chapter follows from the fact that the global fundamental classes are compatible with the local fundamental classes.

The norm limitation theorem.

Theorem 2.8. Let E be a finite extension of K (not necessarily Galois), and let M be the maximal subextension of E such that M/K is an abelian Galois extension. Then

Nm_{E/K} C_{E} = Nm_{M/K} C_{M}.

Proof. Let L be a Galois extension of K containing E, and let G = Gal(L/K) and H = Gal(L/E). Consider the commutative diagram

H_{T}^{−2}(H, ℤ) ⟶^{≈} H_{T}^{0}(H, C_{L})
↓^{Cor}  ↓^{Cor}
H_{T}^{−2}(G, ℤ) ⟶^{≈} H_{T}^{0}(G, C_{L})

in which the horizontal arrows are cup-product with the fundamental classes. This can be identified with the commutative diagram:

H^{ab} ⟶^{≈} C_{E}/ Nm_{L/E} C_{L}
↓  ↓^{Nm}^{E/K}
G^{ab} ⟶^{≈} C_{K}/ Nm_{L/K} C_{L}

Hence the cokernel of H^{ab} → G^{ab} is isomorphic to C_{K}/ Nm_{E/K}(C_{E}). But the cokernel is equal to Gal(M/K), which is isomorphic to C_{K}/ Nm_{M/K}(C_{M}). Since Nm(C_{M}) ⊃ Nm(C_{E}), the two groups must be equal.

3. Higher Reciprocity Laws

For an odd prime p and integer a not divisible by p, one defines (Legendre symbol, quadratic residue symbol)

(\frac{a}{p}) = \{ \begin{matrix}1 & if a is a square modulo p \\ −1 & otherwise. \end{matrix}

The group 𝔽_{p}^{×} is cyclic of order p − 1 with −1 as its unique element of order 2. Therefore, for u ∈ 𝔽_{p}^{×}, u^{\frac{p−1}{2}} is 1 or −1 according as u is a square or not, and so (\frac{a}{p}) is the unique square root of 1 such that

(\frac{a}{p}) ≡ a^{\frac{p−1}{2}} mod p.

The quadratic reciprocity law says that, for odd primes p and q,

(\frac{p}{q}) (\frac{q}{p}) = (−1)^{\frac{p−1}{2} \frac{q−1}{2}} .

The supplement to the quadratic reciprocity law says that

(\frac{−1}{p}) = (−1)^{\frac{p−1}{2}} , (\frac{2}{p}) = (−1)^{\frac{n^{2}−1}{8}} .

For α a Gaussian integer (i.e., element of ℤ[i]) and π an odd Gaussian prime (i.e., prime element of ℤ[i] not dividing 2), Gauss defined (\frac{α}{π}) (quartic residue symbol) to be the unique 4th root of 1 such that

(\frac{α}{π}) ≡ α^{\frac{ℕπ−1}{4}} mod π

and proved a quartic reciprocity law for these symbols. Later Eisenstein proved a cubic reciprocity law. Emil Artin remarked that his theorem (V.3.5) implied all possible such reciprocity laws, and therefore can be considered as a “reciprocity law for fields not containing an nth root of 1”. In the remainder of this section, we explain this remark.

The power residue symbol. Let K be a number field containing a primitive nth root of 1. For any finite set a, b, … of elements of K, we define S(a, b, …) to be the set of prime ideals of K such that ord_{𝔭}(n) ≠ 0, or ord_{𝔭}(a) ≠ 0, or ord_{𝔭}(b) ≠ 0,... . In particular, S itself consists only of the divisors of n.

Recall that the discriminant of X^{n} − 1 is divisible only by the primes dividing n. Therefore X^{n} − 1 has n distinct roots in 𝔽_{p}^{al} for any p ∤ n, and the map

ζ ↦ ζ mod 𝔭 : μ_{n}(K) → μ_{n}(O_{K}/𝔭)

is bijective for any prime ideal 𝔭 ∤ n. For such a prime 𝔭, let q = ℕ𝔭 =^{df} (O_{K} : 𝔭). Then 𝔽_{q}^{×} is cyclic of order q − 1, and so n|q − 1 and u^{\frac{q−1}{n}} ∈ μ_{n} ⊂ 𝔽_{q}^{×}.

For a ∈ K^{×} and 𝔭 ∈ S(a), define (\frac{a}{𝔭}) to be the unique nth root of 1 such that

(\frac{a}{𝔭}) ≡ a^{\frac{ℕ𝔭−1}{n}} mod 𝔭.

3.1. For any a, b ∈ K^{×} and 𝔭 ∈ S(a, b),

(\frac{ab}{𝔭}) = (\frac{a}{𝔭}) (\frac{b}{𝔭}) .

This is obvious from the definition.

3.2. For a ∈ K^{×} and 𝔭 ∈ S(a), the following are equivalent:

(a) (\frac{a}{𝔭}) = 1;

(b) a becomes an nth power in O_{K}/𝔭;

(c) a becomes an nth power in K_{𝔭}.

The equivalence of (a) and (b) follows from the exactness of

1 → 𝔽_{q}^{×n} → 𝔽_{q}^{×} ⟶^{x↦x^{q−1/n}} μ_{n} → 1, q = ℕ𝔭.

If X^{n} − a has a solution modulo 𝔭, then Hensel’s lemma (Math 676, 7.24) shows that it has a solution in K_{𝔭}. Conversely, if a = α^{n}, α ∈ K_{𝔭}, then ord_{𝔭}(α) = \frac{1}{n}ord_{𝔭}(a) = 0, and so α ∈ O_{K_{𝔭}}. The map O_{K} → O_{K_{𝔭}}/𝔭 is surjective, and so there is an α_{0} ∈ O_{K} mapping to α modulo 𝔭.

We extend the mapping 𝔭 ↦ (\frac{a}{𝔭}) to I^{S(a)} by linearity: thus, for 𝔟 = ∏ 𝔭_{i}^{r_{i}} ∈ I^{S(a)},

(\frac{a}{𝔟}) = ∏ (\frac{a}{𝔭_{i}} )^{r_{i}} .

We abbreviate (\frac{a}{(b)}) to (\frac{a}{b}).

For an abelian extension L/K in which the primes in S′ do not ramify, ψ_{L/K} : I^{S} → Gal(L/K) denotes the Artin map (see Chapter V).

3.3. For any a ∈ K^{×} and 𝔟 ∈ I^{S(a)},

ψ_{K[a^{\frac{1}{n}}]/K}(𝔟)(a^{\frac{1}{n}}) = (\frac{a}{𝔟}) a^{\frac{1}{n}} .

From Galois theory, we know that there is an nth root ζ(𝔟) of 1 such that ψ(𝔟)(a^{\frac{1}{n}}) = ζ(𝔟) · a^{\frac{1}{n}} and that the map 𝔟 ↦ ζ(𝔟) is a homomorphism. Therefore, it suffices to prove the equality with 𝔟 = 𝔭, a prime ideal. By definition,

ψ(𝔭)(x) ≡ x^{ℕ𝔭} mod 𝔭.

From

ψ(𝔭)(a^{\frac{1}{n}}) = ζ(𝔭) · a^{\frac{1}{n}}

we find that

ζ(𝔭) · a^{\frac{1}{n}} ≡ x^{\frac{ℕ𝔭}{n}} mod 𝔭,

from which it follows that ζ(𝔭) = (\frac{a}{𝔭}).

3.4. Let a ∈ O_{K}, and let 𝔟 be an integral ideal in I^{S(a)}. If a′ ∈ O_{K}, a′ ≡ a mod 𝔟, then 𝔟 ∈ I^{S(a′)} and

(\frac{a}{𝔟}) = (\frac{a′}{𝔟} ) .

For any prime ideal 𝔭 dividing 𝔟, a′ ≡ a mod 𝔭, and so (\frac{a}{𝔭}) = (\frac{a′}{𝔭}).

The Artin Reciprocity Law allows us to prove a similar, but weaker, result for (\frac{a}{𝔟}) regarded as a function of 𝔟.

3.5. Let a ∈ K^{×}. There exists a modulus 𝔪 with support in S(a) such that (\frac{a}{𝔟}) depends only on the class of 𝔟 in the ray class group C_{𝔪}.

According to Proposition VII.10.5, S(a) contains all primes ramifying in K[a^{\frac{1}{n}}]. Therefore, Artin’s Reciprocity Law (V.3.5) shows that there exists a modulus 𝔪 with support in S(a) such that ψ(𝔟) depends only on the class of 𝔟 in the ray class group C_{𝔪}.

The Hilbert symbol. Let K_{v} be a local field containing a primitive nth root of 1. The Hilbert symbol is a pairing

a, b ↦ (a, b)_{v} : K^{×}/K^{×n} × K^{×}/K^{×n} → μ_{n}

where μ_{n} is the group of nth roots of 1 in K_{v}. Probably the most natural way of defining this as the cup-product map

H^{1}(G, μ_{n}) × H^{1}(G, μ_{n}) → H^{2}(G, μ_{n} ⊗ μ_{n}), G = Gal(K^{al}/K),

followed by the isomorphism

H^{2}(G, μ_{n} ⊗ μ_{n}) = H^{2}(G, μ_{n}) ⊗ μ_{n} → μ_{n}

defined by the invariant map inv_{v}. However, in the spirit of the 1920s and 1930s, I’ll define it in terms of central simple algebras.

Recall (IV.5) that for any a, b ∈ K_{v}^{×}, we define A(a, b; ζ) to be the K_{v}-algebra with generators elements i, j and relations

i^{n} = a, j^{n} = b, ij = ζji.

It is a central simple algebra of degree n over K_{v}. In the case that n = 2, A(a, b; −1) is the quaternion algebra H(a, b). We define

(a, b)_{v} = ζ^{−n·inv_{v}([A(a,b;ζ)])}

where [A(a, b; ζ)] is the class of A(a, b; ζ) in Br(K_{v}). Because A(a, b; ζ) is split by a field of degree n (in fact, by any maximal subfield, for example, ℚ[i]), its invariant is an element of \frac{1}{n}ℤ/ℤ, and hence n · inv_{v}([A(a, b; ζ)]) is an element of ℤ/nℤ. Clearly the isomorphism class of A(a, b; ζ) depends only on a and b as elements of K_{v}^{×}/K_{v}^{×n}, and so we do have a pairing

K_{v}^{×}/K_{v}^{×n} × K_{v}^{×}/K_{v}^{×n} → μ_{n}.

However, it is not obvious from this perspective that the pairing is bilinear.

Example 3.6. Consider the case K_{v} = ℚ_{p}, p an odd prime, and n = 2. Then (a, b)_{p} = ±1, and

(a, b)_{p} = 1 ⟺ H(a, b) ≈ M_{2}(K_{v});
⟺ X^{2} − aY^{2} − bZ^{2} + abT^{2} represents 0 in K_{v};
⟺ b is a norm from K[\sqrt{a}];
⟺ X^{2} − aY^{2} − bZ^{2} represents 0 in K_{v}.

To prove the equivalences use (respectively) that: a quaternion algebra has invariant \frac{1}{2} if and only if it is a division algebra; Exercise IV.5.1; Proposition 1.10d; Proposition 1.10c. The last condition shows that our definition of the Hilbert symbol agrees with that, for example, in Serre, Cours..., 1970, III.

3.7. For any a, b,

A(b, a; ζ) ≈ A(a, b; ζ^{−1}) ≈ A(a, b; ζ)^{opp}.

Therefore

(b, a)_{v} = (a, b)_{v}^{−1}.

By definition A(b, a; ζ) is the K_{v}-algebra with generators i′, j′ and relations i^{′n} = b, j^{′n} = a, and i′j′ = ζi′j′. The map i′ ↦ j, j′ ↦ i is an isomorphism A(b, a; ζ) → A(a, b; ζ^{−1}). The map i ↦ i, j ↦ j is an isomorphism A(a, b; ζ)^{opp} → A(a, b; ζ^{−1}).

3.8. Let a, b ∈ K^{×}. For any v ∈ S(a), (a, b)_{v} = (\frac{a}{𝔭_{v}} )^{ord_{v}(b)}.

For simplicity, we assume that A(a, b; ζ) is a division algebra. Recall (IV.4) that, to compute the invariant of a central division algebra D over a local field K_{v}, we

(a) choose a maximal unramified field L ⊂ D;

(b) find an element β ∈ D such that α ↦ βαβ^{−1} is the Frobenius automorphism of L (such an α exists by the Noether-Skolem Theorem);

(c) set inv_{v}([D]) = ord_{v}(α).

We apply this with L = K_{v}[i] = K_{v}[a^{\frac{1}{n}}]. Note that, because v ∈ S(a), this extension is unramified. Let (\frac{a}{𝔭_{v}} ) = ζ^{r}, so that (𝔭, L/K_{v})(i) = ζ^{r}i. Since jij^{−1} = ζ^{−1}i, we see that we can take β = j^{−r}. Then β^{n} = b^{−r}, and so ord_{v}(β) = −\frac{r}{n}ord_{v}(b). Hence

(a, b)_{v} =^{df} ζ^{−n inv_{v}(A(a,b;ζ))} = ζ^{r·ord_{v}(b)} = (\frac{a}{𝔭_{v}} )^{ord_{v}(b)} .

Remark 3.9. In fact,

(a, b)_{v} = \frac{ϕ_{v}(b)(a^{\frac{1}{n}})}{a^{\frac{1}{n}}}

for all a, b, v. See III.4.3.

3.10. For a, b ∈ K^{×},

∏_{v}(a, b)_{v} = 1.

In the course of proving the Reciprocity Law, we showed that, for any β ∈ Br(K), ∑ inv_{v}(β) = 0. In particular, ∑ inv_{v}(A(a, b; ζ)) = 0, and this implies the formula.

For a, b ∈ K^{×}, define

(\frac{a}{b}) = ∏_{v∉S(a)} (\frac{a}{v})^{ord_{v}(b)} = (\frac{a}{(b)^{S(a)}} )

where (b)^{S(a)} is the ideal in I^{S(a)} generated by b. The symbol (\frac{a}{b}) is multiplicative in

b, but (\frac{aa′}{b} ) = (\frac{a}{b}) (\frac{a′}{b} ) will not always hold unless S(b) ∩ S(a, a′) = S.

Theorem 3.11 (Power Reciprocity Law). Let a and b be elements of K^{×} such that S(a) ∩ S(b) = S (for example, a and b could be relatively prime). Then

(\frac{a}{b}) (\frac{b}{a})^{−1} = ∏_{v∈S}(b, a)_{v}.

Moreover, if S(c) = S, then

(\frac{c}{b}) = ∏_{v∈S}(c, b)_{v}.

Proof. Let S′(a) = S(a) \ S and S′(b) = S(b) \ S. Our assumption is that S′(a) and S′(b) are disjoint. Then

(\frac{a}{b}) = ∏_{v∈S′(b)} (\frac{a}{𝔭_{v}} )^{ord_{v}(b)} = ∏_{v∈S′(b)}(a, b)_{v}

and

(\frac{b}{a}) = ∏_{v∈S′(a)} (\frac{b}{𝔭_{v}} )^{ord_{v}(a)} = ∏_{v∈S′(a)}(b, a)_{v}.

Therefore

(\frac{a}{b}) (\frac{b}{a})^{−1} = ∏_{v∈S′(a)∪S′(b)}(a, b)_{v}.

For v ∉ S ∪ S′(a) ∪ S′(b), (a, b)_{v} = 0 (by 3.8 for example), and so the product formula shows that

∏_{v∈S′(a)∪S′(b)}(a, b)_{v} × ∏_{v∈S}(a, b)_{v} = 1.

This completes the proof of the first equality, and the second is obvious.

To obtain a completely explicit formula, it remains to compute the Hilbert symbol for the v ∈ S. For the infinite primes, this is easy: if v is complex, then (a, b)_{v} = 1 always, and if v is real, then

(a, b)_{v} = 1 ⟺ X^{2} − aY^{2} − bZ^{2} represents 0 ⟺ a > 0 or b > 0.

For K = ℚ and n = 2,

(u2^{r}, v2^{s})_{2} = (−1)^{\frac{u−1}{2} \frac{v−1}{2} +r \frac{v^{2}−1}{8} +s \frac{u^{2}−1}{8}}

where u and v are 2-adic units, and the exponent is to be interpreted modulo 2. For an elementary proof of this, see Serre, Cours..., 1970, III.1.2. On applying this formula successively to the pairs (p, q) with p and q odd primes, (2, p) with p an odd prime, and to (−1, p) with p an odd prime, one obtains the classical quadratic reciprocity law (including the supplements).

For p an odd prime and K = ℚ[ζ] with ζ a primitive pth root of 1, one can make the Hilbert symbol (a, b)_{p} completely explicit. Recall that p is totally ramified in K and (p) = (π)^{p−1} where π = 1 − ζ. Let K_{π} denote the completion of K at (π), and let U_{i} denote the group of units in K_{π} congruent to 1 mod π^{i}. We have a filtration

O_{K_{π}}^{×} ⊃ U_{1} ⊃ U_{2} ⊃ ⋯ ⊃ U_{p+1} ⊃ · · · .

If u ∈ U_{p+1}, then u is a pth power in K_{π} (see VII.10.6a). From this, one can deduce that K_{π}^{×}/K_{π}^{×p} is freely generated (as an 𝔽_{p}-vector space) by the elements

π, ζ, 1 − π^{2}, …, 1 − π^{p}.

Let η_{i} = 1 − π^{i}, i ≥ 1 (e.g., η_{1} = ζ).

Proposition 3.12. The Hilbert pairing

a, b ↦ (a, b)_{π} : K_{π}^{×} × K_{π}^{×} → μ_{p}

is the unique skew-symmetric pairing satisfying

(a) (η_{i}, η_{j})_{π} = (η_{i}, η_{i+j})_{π}(η_{i+j}, η_{j})_{π}(η_{i+j}, π)_{π}^{−j} for all i, j ≥ 1;

(b) (η_{i}, π)_{π} = \{ \begin{matrix}1 & if & 1 ≤ i ≤ p − 1 \\ ζ & if & i = p. \end{matrix}

(c) (·, ·)_{π} = 1 on U_{i} × U_{j} if i + j ≥ p + 1.

For hints, see Cassels and Fröhlich 1967, p354.

Example 3.13. (Cubic reciprocity law; Eisenstein). Let p = 3, so that K = ℚ[ζ], ζ = \frac{−1+\sqrt{3}}{2}, and π = −ζ\sqrt{3}. Then O_{K} = ℤ[ζ], and every nonzero element of O_{K} can be written in the form ζ^{i}π^{j}a with a ≡ ±1 mod 3O_{K}. In this case, the reciprocity law becomes:

(\frac{a}{b}) = (\frac{b}{a})

if a and b are relatively prime and congruent to ±1 mod 3O_{K}, and

\{ \begin{matrix}(\frac{ζ}{a}) & = & ζ^{−m−n} \\ (\frac{π}{a} ) & = & ζ^{m} \end{matrix}

if a = ±(1 + 3(m + nζ)).

Note that, if a ∈ ℤ, then a ≡ ±1 mod 3O_{K} is automatic.

Exercise 3.14. Let p ∈ ℤ be a prime congruent to 1 modulo 3 (so that 𝔽_{p} contains the cube roots of 1). Show that 2 is a cube modulo p if and only if p is of the form x^{2} + 27y^{2}, x, y ∈ ℤ.

Application. Fix an odd prime p and a primitive pth root ζ of 1. If x, y, z are integers such that x^{p} + y^{p} = z^{p}, then

∏_{i=0}^{p−1}(x + ζ^{i}y) = z^{p}.

We may suppose that x, y, z have no common factor. If p ∤ xyz, then the elements x + ζ^{i}y of ℤ[ζ] are relatively prime (Math 676, 6.9). Therefore, each generates an ideal that is a pth power, and the same is true of

α = \frac{x + ζy}{x + y} = 1 − \frac{yπ}{x + y} , π = 1 − ζ.

Hence (\frac{α}{β}) = 1 for all β ∈ ℤ[ζ] relatively prime to α.

Theorem 3.15. Let x, y, z be relative prime positive integers such that p ∤ xyz and x^{p} + y^{p} = z^{p}. For any prime q dividing xyz, q^{p−1} ≡ 1 mod p^{2}.

THE CLASSIFICATION OF QUADRATIC FORMS OVER A NUMBER FIELD 207

Proof. In this case, the Power Reciprocity Law becomes

(\frac{β}{α}) (\frac{α}{β})^{−1} = ζ^{Tr_{ℚ[ζ]/ℚ}(η)}

where η = \frac{β−1}{p} \frac{α−1}{π}. We apply this equation with β = q^{p−1}. Without loss of generality, we may assume that q|y, so that α ≡ 1 mod q and (\frac{α}{q}) = 1. Moreover,

Tr(η) = \frac{q^{p−1} − 1}{p} Tr(\frac{α − 1}{π}),

but

Tr \frac{α − 1}{π} = Tr −\frac{yπ}{x + y} = −\frac{y}{x + y} (p − 1),

which is not divisible by p. Therefore \frac{q^{p−1}−1}{p} is divisible by p.

Corollary 3.16 (Wieferich’s Condition). If X^{p}+Y^{p} = Z^{p} admits a solution x, y, z with x, y, z positive integers none of which is divisible by p, then 2^{p−1} ≡ 1 mod p^{2}.

Proof. If x^{p} + y^{p} = z^{p}, then at least one of x, y, or z must be even.

A similar argument (with a different β) proves Mirimanoff’s condition: 3^{p−1} ≢ 1 mod p^{2}.

The only primes < 3 × 10^{9} satisfying Wieferich’s condition are 1093 and 3511, and they fail Mirimanoff’s condition. Thus this proves the first case of Fermat’s last theorem for p < 3 × 10^{9}.

Notes: Theorem 3.15 was proved by Furtwängler (see Hasse 1970, II, 22); see also Koch 1992, II.6.3, and Herbrand 1936, p47. Class field theory also allows one to simplify the proof of Kummer’s second criterion when the second case of Fermat’s theorem holds (J. Herbrand, Sur les classes des corps circulaires, J. Math. Pures Appl., IX. Sér. 11, 417–441.

4. The Classification of Quadratic Forms over a Number Field

Earlier we showed that a nondegenerate quadratic form over a number field represents 0 in the field if and only if it represents zero in every completion of the field. In this section, we completely classify the quadratic forms over a number field. Specifically, we shall:

(a) Show that two quadratic forms over a number field K are equivalent if and only if they are equivalent over every completion of K.

(b) Give a complete list of invariants for the quadratic forms over a local field.

(c) Determine which families of local invariants arise from a global quadratic form.

Generalities on quadratic forms. In this subsection, k is an arbitrary field of characteristic ≠ 2. Let (V, Q) be a quadratic space over k with corresponding bilinear form B, and let U_{1} and U_{2} be subspaces of V. If every element of V can be written uniquely in the form v = u_{1}+u_{2} with u_{1} ∈ U_{1} and u_{2} ∈ U_{2}, then we write V = U_{1}⊕U_{2}. If, addition, B(u_{1}, u_{2}) = 0 for all u_{1} ∈ U_{1} and u_{2} ∈ U_{2}, then we write V = U_{1} ⊥ U_{2}. For any subspace U of V,

U^{⊥} = {v ∈ V | B(u, v) = 0 for all u ∈ U}.

If Q|U is nondegenerate, then V = U ⊥ U^{⊥}.

Let (V, Q) and (V′, Q′) be quadratic spaces over k. A morphism s : (V, Q) → (V′, Q′) is a linear map s : V → V′ of k-vector spaces such that Q′(s(v)) = Q(v) for all v ∈ V. A morphism is an isomorphism if it admits an inverse that is also morphism. An isomorphism (V, Q) → (V′, Q′) will also be called an isometry.

Proposition 4.1. Let (V, Q) be a quadratic space. If Q represents a ∈ k^{×}, then there exists an e ∈ V with Q(e) = a and a subspace U of V such that V = U ⊥ k · e.

Proof. Because Q represents e, there does exist an e ∈ V such that Q(e) = a, and we can take U to be the orthogonal complement of k · e.

Let (V, Q) be a quadratic space. For any y ∈ V with Q(y) ≠ 0, we define the symmetry with respect to y (or with respect to the line k · y) to be the map

τ_{y}(x) = x − \frac{2B(x, y)}{Q(y)}y.

Note that τ_{y} is a morphism (V, Q) → (V, Q) and that τ_{y}∘τ_{y} id, and so τ_{y} is an isometry. It reverses every vector in the line k · y, and leaves every vector in the hyperplane (k · y)^{⊥} fixed. It is therefore reflection in the hyperplane (k · y)^{⊥}.

Proposition 4.2. Let U and W be isometric subspaces of a quadratic space (V, Q) and assume that Q|U is nondegenerate. Then U^{⊥} and W^{⊥} are isometric.

Proof. We prove this by induction on the dimension of U. Suppose first that U and W are lines, say U = ku and W = kw. Then Q(u) ≠ 0, Q(w) ≠ 0, and we may suppose that Q(u) = Q(w). From

Q(u + w) + Q(u − w) = 2Q(u) + 2Q(w) = 4Q(u)

we see that at least one of Q(u + w) or Q(u − w) is nonzero, and, after replacing w with −w if necessary, we may suppose that it is the latter. Therefore the symmetry τ_{u−w} is defined:

τ_{u−w}x = x − \frac{2B(x, u − w)}{Q(u − w)}(u − w).

Then τ_{u−w}(u) = w, because

Q(u − w) = Q(u) + Q(w) − 2B(u, w) = 2Q(u) − 2B(u, w) = 2B(u, u − w),

and so τ_{u−w} maps U^{⊥} isometrically onto W^{⊥}.

Thus, we may suppose that dim U ≥ 2, and so admits a nontrivial decomposition U = U_{1} ⊥ U_{2}. Because W ≈ U, there is a decomposition W = W_{1} ⊥ W_{2} with U_{1} ≈ W_{1} and U_{2} ≈ W_{2}. Note that Q|U_{1} will be nondegenerate, and that U_{1}^{⊥} = U_{2} ⊥ U^{⊥}.

The induction hypothesis implies that U_{2} ⊥ U^{⊥} is isometric to W_{2} ⊥ W^{⊥}, and the choice of an isometry defines a decomposition U_{2} ⊥ U^{⊥} = X ⊥ Y with X ≈ W_{2} and Y ≈ W^{⊥}. But then X ≈ U_{2}, and the induction hypothesis shows that Y ≈ U^{⊥}. Hence W^{⊥} ≈ U^{⊥}.

On choosing a basis e_{i} for a quadratic space (V, Q), we obtain a quadratic form

q(X_{1}, …, X_{n}) = ∑ a_{ij}X_{i}X_{j}, a_{ij} = B(e_{i}, e_{j}).

Conversely, a quadratic form q defines a quadratic space (k^{n}, q).

Two quadratic forms q and q′ are said to be equivalent, q ∼ q′, if they define isomorphic quadratic spaces, i.e., if one can be obtained from the other by an invertible change of variables. If q and q′ are quadratic forms in distinct sets of variables, then we denote q + q′ by q ⊥ q′; then (k^{m+n}, q ⊥ q′) = (k^{m}, q) ⊥ (k^{n}, q′).

From Proposition 4.2 we find that:

Let q = r ⊥ s and q′ = r′ ⊥ s′ be two quadratic forms, and assume that r is nondegenerate. If q ∼ q′ and r ∼ r′, then s ∼ s′.

From Proposition 4.1 we find that

A nondegenerate quadratic form q in n variables represents a if and only if q ∼ r ⊥ aZ^{2} where r is a quadratic form in n − 1 variables.

The rank of a quadratic space (V, Q) is defined to be the rank of the matrix (B(e_{i}, e_{j})) for some basis e_{i} of V. The rank of a quadratic form q is the rank of the corresponding quadratic space. When q is written in diagonal form, q = a_{1}X_{1}^{2} + ⋯ + a_{r}X_{r}^{2}, then the rank of q is the number of nonzero coefficients a_{i}, i.e., the number of variables actually occurring in q.

The local-global principle.

Theorem 4.3 (Hasse-Minkowski). Let q and q′ be quadratic forms over a number field K. If q and q′ become equivalent over K_{v} for all primes v, then q and q′ are equivalent over K.

Proof. We may suppose that q and q′ are nondegenerate. We use induction on the common rank n of q and q′. If n = 0, both forms are zero, and there is nothing to prove. Otherwise, there exists an a ∈ K^{×} represented by q. Then q(X_{1}, …, X_{n})−aZ^{2} represents 0 in K, and hence in K_{v} for all v. On applying Theorem 1.8, to q′ − aZ^{2}, we find that q′ represents a in K. Therefore, q ∼ q_{1} ⊥ aZ^{2} and q′ ∼ q′_{1} ⊥ aZ^{2} for some quadratic forms q_{1} and q_{2} of rank n − 1. Now (4.2) shows that q_{1} ∼ q_{2} over K_{v} for all v, and so (by induction) they are equivalent over K. This implies that q and q′ are equivalent over K.

Remark 4.4. Let (V, Q) be a quadratic space over a field k, and let O be its group of isometries. Theorem 4.3 says that

H^{1}(K, O) → ∏_{v} H^{1}(K_{v}, O)

is injective.

The classification of quadratic forms over a local field. The archimedean case. Any quadratic form over ℂ (as for any algebraically closed field of characteristic ≠ 2) is equivalent to a unique quadratic form

X_{1}^{2} + ⋯ + X_{n}^{2}.

Thus two quadratic forms over ℂ are equivalent if and only if they have the same rank n.

According to Sylvester’s theorem, a quadratic form q over ℝ is equivalent to a unique quadratic form

X_{1}^{2} + ⋯ + X_{r}^{2} − X_{r+1}^{2} − ⋯ − X_{r+t}^{2}.

The number t of −1s is the index of negativity. Thus, two quadratic forms over ℝ are equivalent if and only if they have the same rank n and the same index of negativity t.

The nonarchimedean case. Let K be a local field. Recall that the Hilbert symbol (·, ·) can defined for a, b ∈ K^{×} by

(a, b) = \{ \begin{matrix}1 & ⟺ X^{2} − aY^{2} − bZ^{2} represents 0 & ⟺ aY^{2} + bZ^{2} represents 1 \\ −1 & otherwise. \end{matrix}

Lemma 4.5. The Hilbert symbol has the following properties:

(a) it is bi-multiplicative and (ac^{2}, bd^{2}) = (a, b) for all a, b, c, d ∈ K^{×};

(b) for any nonsquare a ∈ K^{×}, there exists a b ∈ K^{×} such that (a, b) = −1;

(c) (b, a) = (a, b)^{−1} = (a, b);

(d) (a, −a) = (1, a) = 1.

Proof. Obviously, (a, b) does not change when a or b is multiplied by a square. Also, (c) is obvious.

Note that (a, b) = 1 if and only if b is a norm from K[\sqrt{a}]. From local class field theory, we know that if a is not a square in K, then Nm(K[\sqrt{a}]^{×}) is a subgroup of index 2 in K^{×}, and therefore b ↦ (a, b) is an isomorphism K^{×}/ Nm(K[\sqrt{a}]^{×}) → \{±1\}. This completes the proof of (a) and (b). Finally, aX^{2}−aY^{2} = a(X^{2}−Y^{2}), and X^{2}−Y^{2} represents a^{−1} because it represents 0.

If q ∼ a_{1}X_{1}^{2} + ⋯ + a_{n}X_{n}^{2} with a_{1}, …, a_{n} ∈ K^{×}, then we set

n(q) = n
d(q) = a_{1} ⋯ a_{n} (in K_{v}^{×}/K_{v}^{×2})
S(q) = ∏_{1≤i≤j≤n}(a_{i}, a_{j}) = ∏_{1≤i≤n}(a_{i}, d_{i}) (in {±1})

where d_{i} = a_{1} … a_{i}. Thus n(q) is the rank of q and d(q) is the discriminant of q. Both depend only on the equivalence class of q. We shall prove that the same is true of S(q). It is called the Hasse invariant of q.

Remark 4.6. Serre, Cours..., 1970, defines

ε(q) = ∏_{1≤i<j≤n}(a_{i}, a_{j}) = ∏(a_{i}, d_{i−1}).

Note that

S(q) = ε(q) ∏_{i=1}^{n}(a_{i}, a_{i}) = ε(q) ∏_{i=1}^{n}(−1, a_{i})(−a_{i}, a_{i}) = ε(q)(−1, d(q)).

Thus the knowledge of (d(q), S(q)) is equivalent to the knowledge of (d(q), ε(q)).

Proposition 4.7. The element S(q) depends only on the equivalence class of q.

Proof. It suffices to prove that ε(q) depends only on the equivalence class of q. When q has rank 1, there is nothing to prove: ε(q) = 1 (empty product) for all q.

Next suppose that q ∼ aX^{2} + bY^{2} ∼ a′X^{2} + b′Y^{2}. Because they are equivalent, either both aX^{2} + bY^{2} and a′X^{2} + b′Y^{2} represent 1 or neither represents 1, and so (a, b) = (a′, b′).

Next suppose that n > 2 and that

q ∼ a_{1}X_{1}^{2} + ⋯ + a_{i}X_{i}^{2} + a_{i+1}X_{i+1}^{2} + ⋯ ∼ a′_{1}X_{1}^{2} + ⋯ + a′_{i}X_{i}^{2} + a′_{i+1}X_{i+1}^{2} + ⋯

with a_{j} = a′_{j} except possibly for j = i, i + 1. We then have to prove that

(a_{i}, d_{i−1})(a_{i+1}, d_{i}) = (a′_{i}, d′_{i−1})(a′_{i+1}, d′_{i}).

But

(a_{i}, d_{i−1})(a_{i+1}, d_{i}) = (a_{i}, d_{i−1})(a_{i+1}, d_{i−1})(a_{i+1}, a_{i}) = (a_{i}a_{i+1}, d_{i−1})(a_{i}, a_{i+1})

and a_{i}a_{i+1} differs from a′_{i}a′_{i+1} by a square, and so it remains to show that (a_{i}, a_{i+1}) = (a′_{i}, a′_{i+1}). According to Proposition 4.2, a_{i}X_{i}^{2} + a_{i+1}X_{i+1}^{2} ∼ a′_{i}X_{i}^{2} + a′_{i+1}X_{i+1}^{2}, and we already shown that this implies (a_{i}, a_{i+1}) = (a′_{i}, a′_{i+1}).

The following elementary lemma now completes the proof.

Lemma 4.8. Let B and B′ be orthogonal bases for a nondegenerate quadratic space (V, Q). Then there exists a chain of orthogonal bases B_{1}, B_{2}, …, B_{m} such that B_{1} = B and B_{m} = B′, and each B_{i} is obtained from B_{i−1} by altering at most two adjacent elements.

Proof. See O. O’Meara, Introduction to Quadratic Forms, Springer, 1963, Lemma 58.1.

Proposition 4.9. Let q be a nondegenerate quadratic form in n variables over a nonarchimedean local field K, and let a ∈ K^{×}. Then q represents a if and only if

(a) n = 1 and a = d(q) (in K^{×}/K^{×2});

(b) n = 2 and (a, −d)(−1, d) = S(q) (equivalently, (a, −d) = ε(q));

(c) n = 3 and either a ≠ −d(q) (modulo squares) or a = −d(q) (modulo squares) and (−1, −1) = S(q);

(d) n ≥ 4.

Proof. (a) Clearly dX^{2} represents a if and only if a = d (in K^{×}/K^{×2}).

(b) Let q = bX^{2} + cY^{2}. Clearly bX^{2} + cY^{2} represents a if and only if abX^{2} + acY^{2} represents 1, i.e., if and only if (ab, ac) = 1. But

(ab, ac) = (a, a)(a, b)(a, c)(b, c) = (a, −1)(a, d(q))(b, c) = (a, −d(q)) · ε(q)

and so the condition is that

ε(q) = (a, −d(q)).

(c) Let q = a_{1}X_{1}^{2} + a_{2}X_{2}^{2} + a_{3}X_{3}^{2}. Then q represents a if and only if there exists an e ∈ K^{×} for which the equations

a_{1}X_{1}^{2} + a_{2}X_{2}^{2} = e = a_{3}X_{3}^{2} − aX_{4}^{2}

have solutions. According to (b), this will be so if and only if

(e, −a_{1}a_{2}) = (a_{1}, a_{2}), (e, a_{3}a) = (a_{3}, −a). (∗)

Consider two linear forms f, g : V → 𝔽_{2} on an 𝔽_{2}-vector space V of dimension ≥ 2. The simultaneous linear equations f(x) = ε_{1}, f(x) = ε_{2} will have a solution unless they are inconsistent, i.e., unless f = 0 and ε_{1} = −1; or g = 0 and ε_{2} = −1; or f = g and ε_{1} = −ε_{2}.

When we apply this observation to the linear forms (·, −a_{1}a_{2}), (·, a_{3}a) : K^{×}/K^{×2} → {±1}, we find that there will exist an e satisfying (∗) unless −a_{1}a_{2} = a_{3}a (in K^{×}/K^{×2}) and (a_{1}, a_{2}) = −(a_{3}, a). The first equality says that a = −d(q) (mod squares), and (when a = −d(q)) the second says that (−1, −1) = S(q).

(d) In this case, q(X_{1}, …, X_{n}) − aZ^{2} has rank ≥ 5, and therefore represents 0 (see 1.13).

Theorem 4.10. Two quadratic forms over a nonarchimedean local field are equivalent if and only if they have the same rank, the same discriminant, and the same Hasse invariant.

Proof. We showed in Proposition 4.7 that equivalent forms have the same invariants. For the converse, we use induction on the common rank n of the two forms q and q′. Two quadratic forms of rank 1 are obviously equivalent if they have the same discriminant, and so we may suppose n > 1. From Proposition 4.9, we see that q and q′ represent the same elements in K^{×}. In particular, there is an a ∈ K_{v}^{×} that is represented by both q and q′. Thus,

q ∼ q_{1} + aZ^{2}, q′ ∼ q′_{1} + aZ^{2}

with q_{1} and q′_{1} quadratic forms of rank n − 1. Now

d(q) = a · d(q_{1}), S(q) = (a, d(q_{1})) · S(q_{1})

and similarly for q′ and q′_{1}. Therefore, q_{1} and q′_{1} have the same invariants, and the induction hypothesis shows that q_{1} ∼ q′_{1}.

Proposition 4.11. Let q be a quadratic form of rank n over a nonarchimedean local field K.

(a) If n = 1, then S(q) = (−1, d).

(b) If n = 2, then d(q) = −1 (mod squares) implies S(q) = (−1, −1).

Apart from these constraints, every triple n ≥ 1, d ∈ K^{×}/K^{×2}, s = ±1, occurs as the set of invariants of a quadratic form over K.

Proof. Case n = 1. Then q = dX^{2}, and S(q) = (d, d) = (−1, d).

Case n = 2. For q = aX^{2} + bY^{2}, S(q) = (a, a)(b, d), and so

d = −1 ⟹ S(q) = (−1, a)(−1, b) = (−1, d) = (−1, −1).

Conversely, the form X^{2} − Y^{2} has d = −1 and S = (−1, −1).

Now suppose d ≠ −1 and s are given. We seek an a ∈ K^{×} such that q =^{df} aX^{2}+adY^{2} has S(q) = s. But

S(q) = (a, a)(ad, d) = (a, −1)(a, d)(d, d) = (a, −d)(d, d).

Because −d ≠ 1 (in K^{×}/K^{×2}), we can choose a so that (a, −d) = s · (d, d).

Case n = 3. Choose an a ∈ K^{×} such that a ≠ −d in K^{×}/K^{×2}. Because of the condition on a, there exists a quadratic form q_{1} of rank 2 such that

d = d(q_{1})a, s = S(q_{1})(a, d).

Take q = q_{1} + aZ^{2}.

Case n ≥ 4. There exists a quadratic form with the shape

q_{1}(X_{1}, X_{2}, X_{3}) + X_{4}^{2} + ⋯ + X_{n}^{2}

having the required invariants.

Generalities. We define the Hasse invariant for a quadratic form q over ℝ or ℂ by the same formula as in the nonarchimedean case. For ℂ, S(q) = 1 always, and for ℝ, S(q) = (−1)^{t(t+1)/2} where t is the index of negativity (because (−1, −1) = −1). Note that in the second case, d(q) = (−1)^{t} (in ℝ^{×}/ℝ^{×2}), and that d(q) and S(q) determine t when r ≤ 3 but not for r > 3.

We say that a system (n, d, s, …), n ∈ ℕ, d ∈ K^{×}/K^{×2}, s ∈ {±1},... is realizable there exists a quadratic form q having n(q) = n, d(q) = d, S(q) = s, . . ..

(a) For a nonarchimedean local field K, (n, d, s) is realizable provided s = 1 when n = 1 and s = (−1, −1) when n = 2 and d = −1.

(b) For ℝ, (n, d, s, t) is realizable provided 0 ≤ t ≤ r, d = (−1)^{t}, s = (−1)^{t(t+1)/2}.

Classification of quadratic forms over global fields.

Theorem 4.12. Let n ∈ ℕ, and suppose that for each prime v of the number field K there is given a nondegenerate quadratic form q(v) of rank n over K_{v}. Then there exists a quadratic form q_{0} over K such that (q_{0})_{v} ∼ q_{v} for every v if and only if

(a) there exists a d_{0} ∈ K^{×} such that d_{0} ≡ d(q_{v}) mod K_{v}^{×}^{2} for all v;

(b) S(q(v)) = 1 for almost all v and ∏_{v} S(q(v)) = 1.

The conditions are obviously necessary. In view of Proposition 4.11 and the following remarks, we can restate the theorem as follows. Suppose given:

• an n ≥ 1 and a d_{0} ∈ K^{×}/K^{×2};

• for each prime v, finite or real, an s_{v} ∈ {±1};

• for each real prime v, an integer t_{v}.

Then, there exists a quadratic form q over K of rank n, discriminant d_{0}, Hasse invariant S_{v}(q) = s_{v} for all v, and index of negativity t_{v}(q) = t_{v} for all real v if and only if

(a) s_{v} = 1 for all but finitely many v and ∏_{v} s_{v} = 1;

(b) if n = 1, then s_{v} = (−1, d)_{v}; if n = 2, then either d ≠ −1 in K_{v}^{×}/K_{v}^{×2} or s_{v} = (−1, −1)_{v}; (c) for all real v, 0 ≤ t_{v} ≤ n, d_{v} = (−1)^{t_{v}} (modulo squares), and s_{v} =

(−1)^{t_{v}(t_{v}+1)/2}.

Proof. In the case n = 1, q_{v} = d(q_{v})X^{2}, and we can take q_{0} = d_{0}X^{2} where d_{0} is the element of K^{×} whose existence is guaranteed (a).

The key case is n = 2, and for that we need the following lemma, whose proof requires class field theory.

Lemma 4.13. Let T be finite set of real or finite primes of K, and let b ∈ K^{×}. If T has an even number of elements and b does not become a square in K_{v}^{×} for any v ∈ T, then there exists an a ∈ K^{×} such that

(a, b)_{v} = \{ \begin{matrix}−1 & for v ∈ T \\ 1 & otherwise. \end{matrix}

Proof. (Following Tate, 1976, 5.2). Let L be the composite of all abelian extensions of K of exponent 2, and let G = Gal(L/K). By class field theory,

G ≅ C_{K}/2C_{K} ≅ 𝕀/K^{×} · 𝕀^{2}.

The cohomology sequence of

0 → μ_{2} → L^{×} ⟶^{x↦x^{2}} L^{×2} → 0

is an exact sequence

K^{×} ⟶^{x↦x^{2}} K^{×} ∩ L^{×2} → Hom_{conts}(G, μ_{2}) → 0.

Every element of K^{×} becomes a square in L, and so we have an isomorphism

K^{×}/K^{×2} → Hom_{conts}(𝕀/K^{×} · 𝕀^{2}, μ_{2}).

This map sends a ∈ K^{×} to the continuous homomorphism

(c_{v}) ↦ ∏(a, c_{v})_{v}

(because of the relation between the Hilbert symbol and the local Artin map). Thus, finding a is equivalent to finding a homomorphism f : 𝕀/𝕀^{2} → μ_{2} such that

(a) f = 1 on principal idèles;

(b) f(1, …, 1, i_{v}(b), 1, …, 1) = \{ \begin{matrix}−1 & for v ∈ T \\ 1 & otherwise. \end{matrix}

where i_{v} is the inclusion K ↪ K_{v}. For each v, let B_{v} be the 𝔽_{2}-subspace of K_{v}^{×}/K_{v}^{×2} generated by i_{v}(b), and let B = ∏ B_{v} ⊂ 𝕀/𝕀^{2}. Because i_{v}(b) is not a square for v ∈ T, there exists a linear form (automatically continuous) f_{1} : B → μ_{2} satisfying condition (b), and f_{1} will extend to a continuous linear form on 𝕀/𝕀^{2} satisfying (a) if and only if f_{1} takes the value 1 on every principal idéle in B. The value of f_{1} on the principal idéle of b is ∏_{v∈T} −1, which is 1 because of our assumption that T contains an even number of elements. Let c ∈ K^{×} be such that its idéle lies in B. Then, for every v, i_{v}(c) = 1 or i_{v}(b) in K_{v}^{×}/K_{v}^{×2}. Therefore, i_{v}(c) becomes a square in K_{v}[\sqrt{b}] for all v, which (by 1.1) implies that c is a square in K[\sqrt{b}]. Hence c = 1 or b in K^{×}/K^{×2}, and so f_{1} takes the value 1 on its idèle.

We now prove the case n = 2 of the theorem. We are given quadratic forms q(v) = a(v)X^{2} + a(v)d(v)Y^{2} for all v, and we seek q = aX^{2} + ad_{0}Y^{2} such that q ∼ q(v) over K_{v} for all v. Thus, we seek an a ∈ K^{×} such that

S_{v}(q) =^{df} (a, a)_{v}(ad_{0}, d_{0})_{v} = S(q(v))

for all v. Now

(a, a)_{v}(ad_{0}, d_{0})_{v} = (a, −1)_{v}(a, d_{0})_{v}(−1, d_{0})_{v} = (a, −d_{0})_{v}(−1, d_{0})_{v}.

We apply the lemma with T equal to the set of primes for which S(q(v))(−1, d_{0})_{v} = −1 and with b = −d_{0}. The set T is finite because of condition (b) of the theorem, and it has an even number of elements because

∏_{v} S(q(v))(−1, d_{0})_{v} = ∏_{v} S(q(v)) · ∏_{v}(−1, d_{0})_{v} = 1 × 1 = 1.

Moreover,

S(q(v)) · (−1, d_{0})_{v} = (a(v), −d(v))_{v}

and so −i_{v}(d_{0}) = −d(v) ≠ 1 in K_{v}^{×}/K_{v}^{×2} when v ∈ T. Thus the lemma gives us the required element a.

We next prove the case n = 3. For a form q = q_{1} + aZ^{2}, a ∈ K^{×},

n(q_{1}) = n(q) − 1, d(q_{1}) = a · d(q), S_{v}(q_{1}) = (a, d(q))_{v} · S_{v}(q).

We seek an a for which the invariants (2, a · d_{v}, (a, d_{v})_{v} · s_{v}) are realizable for all v, i.e., such that i_{v}(a)d_{v} = −1 ⟹ (a, d_{v})_{v} · s_{v} = 1. Let T = {v | s_{v} ≠ 1}—by hypothesis, it is a finite set. By the weak approximation theorem, there exists an a ∈ K^{×} such that, for all v ∈ T, i_{v}(a)d_{v} ≠ −1. Now, for v ∈ T, d(q_{1}) ≠ −1, and so (2, d_{v}, s_{v}) is realizable. For v ∉ T, (a, d_{v}) · s_{v} = (a, d_{v}), and i_{v}(a)d_{v} = −1 implies (a, d_{v})_{v} = (−d_{v}, d_{v})_{v} = 1. Hence, for such an a, there exists a quadratic form q_{1} of rank 2 such that q_{1} + aZ^{2} has the required invariants.

We prove the case n ≥ 4 by induction. If t_{v} < n for all n, we can find a quadratic form with shape q_{1}(X) + Z^{2} with the correct local invariants. If no t_{v} = 0, then we can find a quadratic form with shape q_{1}(X) − Z^{2} with the correct invariants. In the general case, we use the weak approximation theorem to find an element a that is positive at the real primes where t_{v} < n and negative at the real primes where t_{v} = 0. Then the induction hypothesis allows us to find a q_{1}(X) such that q_{1}(X) + aZ^{2} has the correct invariants.

Applications.

Proposition 4.14 (Gauss). A positive integer n is a sum of three squares if and only if it is not of the form 4^{a}(8b − 1) with a, b ∈ ℤ.

Proof. Apply the above theory to the quadratic form X_{1}^{2} + X_{2}^{2} + X_{3}^{2} − aZ^{2}—see Serre, Cours..., 1970, Chap. IV.

5. Density Theorems

Throughout this section, K is a number field.

Theorem 5.1. For any modulus 𝔪 of K and any nontrivial Dirichlet character χ : C_{𝔪} → ℂ^{×}, L(1, χ) ≠ 0.

Proof. As we noted at the end of Chapter VI, this follows from the proof of Theorem VI.4.9 once one has the Reciprocity Law.

Theorem 5.2. Let 𝔪 be a modulus for K, and let H be a congruence subgroup for 𝔪: I^{𝔪} ⊃ H ⊃ i(K_{𝔪,1}). For any class 𝔨 ∈ I^{𝔪}/H, the set of prime ideals in 𝔨 has Dirichlet density 1/(I^{𝔪} : H).

Proof. Combine Theorem 5.1 with Theorem VI.4.8.

Corollary 5.3. Let L/K be a finite abelian extension with Galois group G, and let σ ∈ G. Then the set of prime ideals 𝔭 in K that are unramified in L and for which (𝔭, L/K) = σ has Dirichlet density \frac{1}{[L:K]}.

Proof. The Reciprocity Law V.3.5 says that the Artin map defines an isomorphism I^{𝔪}/H → Gal(L/K) for some modulus 𝔪 and some H ⊃ i(K_{𝔪,1}), and we can apply the theorem to the inverse image of σ in I^{𝔪}/H.

Theorem 5.4 (Chebotarev). Let L be a finite Galois extension of the number field K with Galois group G, and let C be a subset of G stable under conjugation, i.e., such that

x ∈ C, τ ∈ G ⟹ τxτ^{−1} ∈ C.

Let

T = {𝔭 | 𝔭 unramified in L, (𝔭, L/K) ⊂ C}.

Then T has Dirichlet density

δ(T) = \frac{number of elements in C}{number of elements in G}.

Proof. It suffices to prove this in the case that C is the conjugacy class of a single element σ,

C = {τστ^{−1} | τ ∈ G}.

Let σ have order f, and let M = L^{<σ>}. Then L is a cyclic extension of M of degree f, and therefore the Artin map gives an isomorphism

C_{𝔪}/H ⟶< σ > .

for some modulus 𝔪 of M and for H = M^{×} · Nm_{L/M} C_{L,𝔪}. We use the notations

𝔓|𝔮|𝔭

for primes 𝔓 of O_{L}, 𝔮 of O_{M}, and 𝔭 of O_{K}. Let d = [L: K] = (G: 1), and let c be the order of C. We have to show that

δ(T) = \frac{c}{d}.

In the proof, we ignore the (finitely many) prime ideals that are not prime to 𝔪.

Let

T_{M,σ} = {𝔮 ⊂ O_{M} | (𝔮, L/M) = σ, f(𝔮/𝔭) = 1}.

The Chebotarev density theorem for abelian extensions (5.3 shows that the set of primes satisfying the first condition in the definition of T_{M,σ} has density \frac{1}{f}, and it follows (see 4.5) that T_{M,σ} itself has density \frac{1}{f}.

Let

T_{L,σ} = {𝔓 ⊂ O_{L} | (𝔓, L/K) = σ}.

We shall show:

(a) the map 𝔓 ↦ 𝔮 = 𝔓 ∩ O_{M} defines a bijection T_{L,σ} → T_{M,σ};

(b) the map 𝔓 ↦ 𝔭 = 𝔓 ∩ O_{K} : T_{L,σ} → T sends exactly \frac{d}{cf} primes of T_{L,σ} to each prime of T.

On combining these statements, we find that the map 𝔮 ↦ 𝔭 = 𝔮∩O_{K} defines a \frac{d}{cf} : 1 map T_{M,σ} → T. For such a 𝔮, Nm_{M/K} 𝔮 = 𝔭 and so ℕ𝔮 = ℕ𝔭; hence

∑_{𝔭∈T} \frac{1}{ℕ𝔭^{s}} = \frac{cf}{d} ∑_{𝔮∈T_{M} ,σ} \frac{1}{ℕ𝔮^{s}} ∼ \frac{cf}{d} \frac{1}{f} log \frac{1}{s − 1} = \frac{c}{d} log \frac{1}{s − 1}

as required. It remains to prove (a) and (b).

Let 𝔓 ∈ T_{L,σ}, and let 𝔮 = 𝔓 ∩ O_{M} and 𝔭 = 𝔓 ∩ O_{K}. Then the Galois group of L_{𝔓}/K_{𝔭} is generated by σ, but σ fixes M_{𝔮}, and so M_{𝔮} = K_{𝔭}. Therefore f(𝔮/𝔭) = 1, which shows that 𝔮 ∈ T_{M,σ}, and so we have a map

𝔓 ↦ 𝔮 =_{df} 𝔓 ∩ O_{M}: T_{L,σ} ⟶ T_{M,σ}.

This is injective because f(𝔓/𝔮) = f(𝔮/𝔭)^{−1}f(𝔓/𝔭) = 1 × f, and so 𝔓 is the only prime of L lying over 𝔮. It is surjective because, for any prime 𝔓 lying over 𝔮 ∈ T_{M,σ},

(𝔓, L/K) = (𝔓, L/K)^{f(𝔮/𝔭)} = (𝔓, L/M) = σ

(first condition for 𝔮 to lie in T_{M,σ}), and so 𝔓 ∈ T_{L,σ}. This proves (a).

Fix a 𝔭_{0} ∈ T, and let 𝔓_{0} ∈ T_{L,σ} lie over 𝔭. Then, for τ ∈ G,

(τ𝔓_{0}, L/K) = τ(𝔓_{0}, L/K)τ^{−1}

and so

τ(𝔓_{0}, L/K)τ^{−1} = σ ⟺ τ ∈ C_{G}(σ)

(centralizer of σ in G). Therefore the map τ ↦ τ𝔓_{0} is a bijection

C_{G}(σ)/G(𝔓_{0}) ⟶ {𝔓 ∈ T_{L,σ} | 𝔓 ∩ O_{K} = 𝔭_{0}}

The decomposition group G(𝔓_{0}) equals <σ >, which has order f, and C_{G}(σ) has order \frac{d}{c} because there is a bijection

τ ↦ τστ^{−1} : G/C_{G}(σ) ⟶ C.

Therefore (C_{G}(σ) : G(𝔓_{0})) = \frac{d}{cf}, and we have shown that, for each 𝔭 ∈ T, there are exactly \frac{d}{cf} primes 𝔓 ∈ T_{L,σ} lying over 𝔭. This proves (b) and completes the proof of the theorem.

Remark 5.5. For effective forms of the Chebotarev density theorem, see Lagarias and Odlysko (Algebraic Number Fields, Ed. Fröhlich, 1977). Let L be a finite Galois extension of K, and let

π_{C}(x) = #{𝔭|(𝔭, L/K) = C, ℕ𝔭 ≤ x}.

Then

π_{C}(x) = \frac{c}{d} \frac{x}{log x} + specific error term.

6. Function Fields

We should also include the class field theory of function fields (finite extensions of 𝔽_{p}(T) for some p). For this, one can either mimic proofs in the number field case (see Artin and Tate 1951/52) or (better) one can base the proofs on Tsen’s theorem (see J. Milne, Arithmetic Duality Theorems, Academic Press, 1986, Appendix to Chapter I).

7. Cohomology of Number Fields

We should also include proofs of the theorems of Nakayama and Tate (e.g., J. Tate, The cohomology groups of tori in finite Galois extensions of number fields, Nagoya Math. J., 27, 1966, 709–719) and Poitou and Tate (e.g., J. Milne, ibid., Chapter I).

8. More on L-series

Let χ be a Dirichlet L-series. Then there exist constants A(χ) ≥ 0, a(χ), b(χ) ∈ ℂ, such that

Φ(s, χ) =_{df} A(χ)^{s}Γ(\frac{s}{2})^{a(χ)}Γ(\frac{s + 1}{2})^{b(χ)}L(s, χ)

satisfies the functional equation

Φ(s, χ) = W (χ)Φ(1 − s, χ̄) |W (χ)| = 1.

See W. Narkiewicz, Elementary and Analytic Theory of Numbers, PWN, 1974.

Artin L-series. Let L be a finite Galois extension of K with Galois group G. Let V be a finite dimensional vector space over ℂ and let

ρ: G → Aut_{ℂ}(V)

be a homomorphism of G into the group of linear automorphisms of V. We refer to ρ as a (finite-dimensional) representation of G over ℂ. The trace of ρ is the map

σ ↦ χ(σ) = Tr(ρ(σ)).

(Recall that the trace of an m × m matrix (a_{ij}) is ∑ a_{ii}, and the trace of an endomorphism is the trace of its matrix relative to any basis.) For σ ∈ G, let

P_{σ}(T) = det(1 − ρ(σ)T | V) = ∏_{i=1}^{dim V} (1 − a_{i}T), a_{i} ∈ ℂ,

be the characteristic polynomial of ρ(σ). Note that P_{σ}(T) depends only on the conjugacy class of σ, and so for any prime 𝔭 of K unramified in L, we can define

P_{𝔭}(T) = P_{σ}(T), σ = (𝔓, L/K) some 𝔓|𝔭.

MORE ON L-SERIES 219

For such a 𝔭, let

L_{𝔭}(s, ρ) = \frac{1}{P_{𝔭}(ℕ𝔭^{−s})}

and let

L(s, ρ) = ∏ L_{𝔭}(s, ρ).

For example, if L/K is abelian, then the representation is diagonalizable^{1}

ρ ≈ χ_{1} ⊕ ⋯ ⊕ χ_{m},

where each χ_{i} is a homomorphism G → ℂ^{×}. When composed with the Artin map

C_{𝔪} ⟶ G,

χ_{i} becomes a Dirichlet character, and so the Artin L-series becomes identified with a product of Dirichlet L-series. This was the original reason Artin defined his map.

One can show that if (V, ρ) = Ind_{H}^{G}(V_{0}, ρ_{0}), then

L(s, ρ) = L(s, ρ_{0}).

To handle more general Artin L-series, Artin proved that every character of a finite group G is a linear combination (over ℚ) of induced characters from cyclic subgroups. Hence

L(s, ρ) = ∏(Dirichlet L-series)^{r_{i}}, r_{i} ∈ ℚ.

Later Brauer proved a stronger theorem that allows one to show that

L(s, ρ) = ∏(Dirichlet L-series)^{r_{i}}, r_{i} ∈ ℤ.

Artin conjectured that, provided ρ does not contain the trivial representation, L(s, ρ) extends to a holomorphic function on the whole complex plane. The last formula implies that this is true if the r_{i} are all positive integers. Little progress was made in this conjecture until Langlands succeeded in proving it in many cases where V has dimension 2 (see R. Langlands, Base Change for GL(2), Princeton, 1980).

Hecke L-series. A Hecke (or Grössen) character is a continuous homomorphism from 𝕀 into the unit circle in ℂ^{×} such that ψ(K^{×}) = 1 and, for some finite set S, ψ is 1 on a set {(a_{v}) | a_{v} = 1 for v ∈ S, a_{v} =unit for all v}

Example 8.1. Let D ∈ ℤ, cube-free, and let ζ be primitive cube root of 1. If p ≡ 2 mod 3, then p remains prime in ℚ[ζ], and we set ψ(p) = 1. If p ≡ 1 mod 3, then p = ππ̄ in ℚ[ζ], and we choose π to be ≡ 1 mod 3. Then π = \frac{1}{2}(a + 3b\sqrt{−3}) and 4p = 4ππ̄ = a^{2} + 27b^{2}. Now there exists a Hecke character such that ψ(p) = 1 for all odd p ≢ 1 mod 3 and ψ(p) = \frac{π}{\sqrt{p}} (\frac{D}{π̄} ).

ρ(g) = ( \begin{matrix}χ_{1}(g) & 0 & ⋯ & 0 \\ 0 & χ_{2}(g) & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & χ_{m}(g) \end{matrix} ) , g ∈ G.

1By this we mean that, relative to suitable basis for V,

For such a character, we define

L(s, ψ) = ∏_{v∉S} \frac{1}{1 − ψ(π_{v})ℕ𝔭_{v}^{−s}}

where π_{v} is an idèle having a prime element in the v position and 1 elsewhere. The basic analytic properties of Hecke L-series (meromorphic continuation, functional equation etc.) are well understood (e.g., J. Tate, Fourier analysis in number fields and Hecke’s zeta function, thesis, 1950; reprinted in Cassels and Fröhlich 1967).

Weil groups and Artin-Hecke L-series. For this topic, see J. Tate, Number theoretic background, pp 3–26, in: Automorphic Forms, Representations, and L-Functions, AMS, 1979.

A theorem of Gauss. Having begun the course with theorem first proved by Gauss, namely, the quadratic reciprocity law, it seems appropriate to end it with another theorem of his.

Consider the elliptic curve E : X^{3} + Y^{3} + Z^{3} = 0. Let N_{p} be the number of points on E with coordinates in 𝔽_{p}. Gauss showed:

(a) if p ≢ 1 mod 3, then N_{p} = p + 1;

(b) if p ≡ 1 mod 3, then N_{p} = A, where A is the unique integer ≡ 1 mod 3 for which 4p = A^{2} + 27B^{2}.

See J. Silverman and J. Tate, Rational Points on Elliptic Curves, Springer, 1992, IV.2.

Gauss’s theorem implies that the Weil conjecture for E/𝔽_{p}, namely, that

|N_{p} − p − 1| < 2\sqrt{p}.

It also implies the Taniyama conjecture for E/ℚ, because it shows that the L-series L(s, E) equals L(s − \frac{1}{2}, ψ) where ψ is the Hecke character in the above example associated with D = −1.

He wrote to me that algebraic number theory was the most beautiful topic he had ever come across and that the sole consolation in his misery was his lecturing on class field theory.... This was indeed the kind of mathematics he had admired most: the main results are of great scope, of great aesthetic beauty, but the proofs are technically extremely hard.

A. Borel, in: Current Trends in Mathematics and Physics: A Tribute to Harish-Chandra, Editor S.D. Adhikari, Narosa Publishing House, New Dehli, Madras, 1995, p213.

Index

abelian category 71 additive category 71 admissible 138 admit a modulus 123, 137 algebra, k- 91 algebra, opposite 91 Artin L-series 6 Artin map 5 Artin map, global 121 Artin map, local 79 augmentation ideal 59 augmentation map 59 automorphic L-series 6 Brauer group 100 category 70 central simple 97 central 97 centralizer 91 character, Dirichlet 148 character, Hecke 138, 148, 219 class field 2, 3, 125, 142 classifying space 60 coboundaries 48 cochain, homogeneous 47 cochain, inhomogeneous 47 cocycles 48 cohomology group 44 coimage 71 cokernel 71 commutator subgroup 126 conductor 12, 124, 129 congruence subgroup 123 content 135 continuous cocycle 67 corestriction homomorphism 53 cotrained 129 covariant functor 70 crossed homomorphism 48 crossed-product algebra 106 cyclotomic 181 defining modulus 123 density, Dirichlet 120, 121, 157 density, natural 158 density, polar 155 derived functor, right 73 dimension shifting 53 dimension 94 Dirchlet character, principal 119 direct limit 66 direct system 66 directed 66 Dirichlet character 119 discrete G-module 67 divide, a modulus 114 division algebra 93 endomorphism 17 Euler product 147 exact 71 exponent 188 extension, of groups 48 factor sets 106 faithful 91 formal group 16 Frobenius element 9, 118 functor, contravariant 75 fundamental class 78, 200 fundamental class, local 66 fundamental exact sequence 198 Galois group 37 Galois 37 G-homomorphism 41 G-module 41 group algebra 41 group, norm 142 Hasse invariant 210 Hasse principle 191 Herbrand quotient 62 Hilbert class field 2 Hilbert symbol 88, 89, 203 homomorphism 17 homotopic 74 hyperprimary 189 idele class group 134 ideles, finite 135 ideles 133 image 71 indecomposable 92 index of negativity 210 induced 43 inflation homomorphism 53 injective 43, 71, 72 isometry 208 isomorphism 17 jump 32 kernel 71 K-group 111 lattice, full 169 left adjoint 70 Legendre symbol 200 local Artin map 10, 11 local-global principle 191 L-series, Artin 130, 148 L-series, Dirichlet 148 L-series, Hecke 148 Lubin-Tate formal group 21 modulus 2, 114 morphisms 70 narrow class group 3 nonabelian 89 nondegenerate 192 norm map 60 norm topology 12 normalized, 2-cocycle 104 positive 2 power residue symbol 130 power series 15 p-primary component 54 primary 189 prime 113 prime element 9 prime, finite 113 prime, real 113 primitive 129 primordial 97 principal crossed homomorphism 48 profinite group 40 projective 57 quadratic form 192 quartic residue symbol 201 quaternion algebra 93 ramify 2 rank 209 ray class field 3, 4, 115, 123 realizable 213 represent 192 representation, regular 95 representation 91 resolution 72 restriction homomorphism 52 ring of integers, division algebra 108 semisimple, algebra 110 semisimple 91, 92 series, Dirichlet 147 similar 100 simple 91 simple 93 skew field 93 split 2, 101 splitting field 101 splitting module 65 structure constants 91 symmetry 208 topological group 133 totally disconnected 40 totally positive 3 Verlagerung 127 zeta function, Dedekind 121, 147 zeta function, partial 153 zeta function, Riemann 147