Interlude: Some examples and complements2014-15/3term/ma160c/math160... · Interlude: Some examples and complements 2.1. The Cebotarev density theorem. WehaveseenintheproofofTheorem

COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 31

2. Interlude: Some examples and complements

2.1. The Cebotarev density theorem. We have seen in the proof of Theorem1.3 that for any nontrivial ray class character χ : Hm → C× we could show non-vanishing L(1, χ) �= 0 provided that χ factors through Jm/NL/KJm

L · Pm for somefinite extension L/K. By Theorem 1.2 the subgroup Pm is the norm subgroup ofthe ray class field K(m)/K. Hence we get nonvanishing for all nontrivial χ and thisallows to prove the following strengthening of the the Frobenius density theorem(the analogue of Dirichlet’s theorem on primes in arithmetic progressions for thenumber field K).

Proposition 2.1. For any class c ∈ Hm = Jm/Pm the set of prime ideals p ofK with [p] = c has Dirichlet density 1

|Hm| . In particular, there are infinitely many

primes in each such class.

Proof. We recall equation (6)

logL(s, χ) ∼∑p

χ(p)

Nps

and multiply it with χ(c)−1 and sum over all χ to obtain∑χ

χ(c)−1 logL(s, χ) ∼∑χ

∑p

χ(pc−1)

Nps= |Hm| ·

∑p∈c

1

Nps.

As s→ 1 the functions logL(s, χ) stay finite for χ �= 1 whereas

logL(s, 1) = log ζmK(s) ∼ − log(s− 1).

So we find

lims→1+

∑p∈c

1Nps∑

p1

Nps

= lims→1+

∑p∈c

1Nps

− log(s− 1)=

1

|Hm| .

�

Theorem 2.1. (Cebotarev) Let L/K be a Galois extension with group G and C ⊆ Ga conjugacy class. Then the set of primes p (unramified in L/K) with Frobp ∈ C

has Dirichlet density |C||G| .

Proof. Recall that if P | p and g ∈ G then FrobPg = g FrobP g−1, i.e. all elementsin the conjugacy class C are Frobenius automorphisms provided one of them is. Sowe fix σ ∈ C, denote by Gσ = {g ∈ G|gσg−1 = σ} its centralizer, by E the fixedfield of < σ > and by q a prime of E. Then

Frobp ∈ C ⇔ ∃ P | p,FrobP = σ ⇔ ∃ q | p, NE/Kq = p,Frobq = σ

and since P | q is totally inert

|{q | p, NE/Kq = p,Frobq = σ}| = |{P | p,FrobP = σ}| = [Gσ :< σ >] =|G|c · f

where c := |C| = [G : Gσ] and f = | < σ > |. Hence∑Frobp∈C

1

Nps=

c · f|G|

∑NE/Kq=p,Frobq=σ

1

Nqs=

c · f|G|

∑Frobq=σ

1

Nqs

32 M. FLACH

and

lims→1+

∑Frobp∈C

1Nps

− log(s− 1)=

c · f|G| lim

s→1+

∑Frobq=σ

1Nqs

− log(s− 1)=

c

|G|by Proposition 2.1. �

2.1.1. Some remarks about l-adic representations. In arithmetic geometry one oftenencounters l-adic representations, i.e. continuous homomorphisms

ρ : GK,S → GLn(Ql)

where GK,S is the Galois group of the maximal extension of K unramified outsidea finite set of places S (which we always assume to contain the infinite places andthe places above l). For example, if X is any algebraic variety over K the l-adicetale cohomology spaces

Hi(XK ,Ql)

yield such representations. Another example is the Tate-module Vl(A) of an abelianvariety A/K. If π is an automorphic form on GLn one can sometimes attach anl-adic representation ρπ to π. This process often -but not always- goes throughthe l-adic cohomology of a suitable algebraic variety. Since GK,S is profinite, inparticular compact, ρ(GK,S) is compact hence conjugate to a subgroup of GLn(Zl).So we can assume ρ has image in GLn(Zl) and there is also a reduced representation

ρ : GK,S → GLn(Fl).

One has the following basic observation.

Lemma 2.1. For fixed K, l, S ⊇ {p | l} and n there are only finitely many represen-tations ρ. Hence there is a finite extension L/K through which such representationsfactorize.

Proof. Let Kρ be the fixed field of ρ. Then Kρ/K has degree bounded by |GLn(Fl)|and is unramified outside S, so by Hermite-Minkowski there are only finitely manysuch fields Kρ. For each Kρ there are only finitely many homomorphisms

Gal(Kρ/K)→ GLn(Fl).

�

In many examples one tends to have information about the characteristic poly-nomial

Lp(T, V ) = det(1− Frobp ·T |V ) ∈ Ql[T ]

of a Frobenius element Frobp ∈ GK,S for p /∈ S (since Frobp is well defined up toconjugation its characteristic polynomial is well defined).

Proposition 2.2. Fix K, l, S ⊇ {p | l} and n and let L/K be the extensionfrom Lemma 2.1. Choose primes p1, . . . , pk /∈ S such that the Frobpi

represent allconjugacy classes in Gal(L/K). If V and V ′ are two representations such that

Lpi(T, V ) = Lpi(T, V′)

for i = 1, . . . , k then V and V ′ are equivalent.


Proof. By standard representation theory (over a field of characteristic zero) itsuffices to show that for g ∈ GK,S

tr(ρ(g)) = tr(ρ′(g)).

Let T ⊆ V , T ′ ⊆ V ′ be GK,S-stable lattices and consider the Zl-subalgebra

M ⊆ EndZl(T )× EndZl

(T ′)

generated by {ρ(g)×ρ′(g)|g ∈ GK,S}. It is enough to show tr(ρ(m)) = tr(ρ′(m)) form in a Zl-basis of M and since we already know this identity for m = ρ(Frobpi)×ρ′(Frobpi

) it suffices to show that these elements and their conjugates generate Mover Zl. By Nakayama’s Lemma it suffices to show that they generate M/lM overFl. But the representation ρ × ρ′ factors through G = Gal(L/K) and since theFrobpi

represent all conjugacy classes in G we have

Fl[im((ρ× ρ′)(g Frobpi g−1)] = im((ρ× ρ′)Fl[G]) = M/lM.

�

These ideas often lead to effective criteria to check the equivalence of two l-adicrepresentations. One doesn’t actually compute the field L but works directly withrepresentations. For example one has:

Theorem 2.2. (Serre, Livne) Two irreducible representations

ρi : GQ,{2,3,5} → GL2(Q2)

are equivalent if and only

tr(ρ1(Frobp)) = tr(ρ2(Frobp)); det(ρ1(Frobp)) = det(ρ2(Frobp));

forp ∈ {7, 11, 13, 17, 19, 23, 29, 31, 41, 43, 53, 61, 71, 73}.

This leads to the following application. Let X be the variety of dimension 7defined in P9

Z by the equations

x0 + · · ·+ x9 = 0; x30 + · · ·+ x3

9 = 0.

Then

ζ(X, s) =ζ(s)ζ(s− 1)ζ(s− 2)ζ(s− 3)−84ζ(s− 4)42ζ(s− 5)ζ(s− 6)ζ(s− 7)

L(f, s− 2)

whereζ(X, s) =

∏x ∈ X closed

(1−Nx−s)−1

is the Zeta-function of X, ζ(s) is the Riemann Zeta-function and L(f, s) is theL-function of the elliptic modular form of weight 4, level 10 and trivial character

f(z) = (2− T (3))(η(z)η(2z)η(5z)3η(10z)3

)=

∑n

ane2πinz

where

η(z) = e2πiz24

∞∏n=1

(1− e2πinz)

is Dedekind’s η-function. It is easy to compute ap from this product expansion andone compares those with the traces of Frobenius gotten from counting points on Xfor p ≤ 73.

34 M. FLACH

Another application is to finiteness theorems. For example, if A/K is an abelianvariety of dimension g with good reduction at p, it is known that the characteristicpolynomial of Frobenius on the Tate module of A satisfies

Lp(A, T ) =

2g∏i=1

(T − αi) ∈ Z[T ]

with |αi| =√Np. These are classical results of Artin, Hasse and Weil. So there are

only finitely many possible such polynomials Lp(A, T ) for a given p. These ideaslead to the following beautiful application. First one has

Theorem 2.3. (Faltings) Two abelian varieties A and A′ over a number field Kare isogenous if and only if Vl(A) and Vl(A

′) are isomorphic GK-representations.

Corollary 2.1. For a given K,S and g there are only finitely many isogeny classesof abelian varieties A/K of dimension g with good reduction outside S.

Proof. Pick any prime l and let L be the field of Lemma 2.1 for K, S ∪ {p | l} andn = 2g. By Faltings’ theorem and Prop. 2.2 it suffices note that there are onlyfinitely many possibilities for each Lpi

(A, T ) for i = 1, .., k. �

2.2. Abelian extensions of imaginary quadratic fields and complex multi-plication. Throughout this section K = Q(

√−d) is an imaginary quadratic fieldof discriminant −d. We view K as a subfield of the complex numbers.

We review lattices in the complex plane. For any lattice

Λ = Z · ω1 + Z · ω2

in C one has its j-invariant defined as

j(Λ) = 1728 · g2(Λ)3

g2(Λ)3 − 27 · g3(Λ)2where

g2 = 60 ·G4; g3 = 140 ·G6; G2k =∑

ω∈Λ,ω �=0

1

ω2k

are Eisenstein series of weight 2k. The j-invariant is invariant under homothety,i.e. for all α ∈ C×

j(αΛ) = j(Λ)

and in fact the converse holds as well

j(Λ) = j(Λ′)⇒ ∃α ∈ C× Λ′ = αΛ.

Finally it takes on all complex values hence gives a biholomorphic map

Space of Lattices up to homothety ∼= C.

Some prominent examples:

j(Z[1 +

√−32

]) = 0; j(Z[i]) = 1728.

An endomorphism of a lattice Λ is a complex number α such that αΛ ⊆ Λ. Clearlythe set of all such α forms a subring of C and for most lattices this ring is just Z.When it is bigger than Z we say that Λ has complex multiplication. It is easy tosee that this happens if and only if Λ is homothetic to a lattice

αΛ = Λ′ ⊆ K ⊆ C


in some imaginary quadratic field K. In this case End(Λ) = End(Λ′) is an order Oin K. Moreover, Λ is a fractional O-ideal in K (by definition this just means thatEnd(Λ) = O; so for example if O �= OK then OK wouldn’t be a fractional O-idealbut it is still a finitely generated O-submodule of K).

Lemma 2.2. a) The orders in K are parametrized by f ∈ Z≥1; they are allof the form O = Z+ fOK for a unique such f .

b) The fractional O-ideals form a group under multiplication. The homothetyclasses of fractional O-ideals form a finite abelian group Cl(O) and there isa commutative diagram with exact rows

(26)

0 −−−−→ (OK/f)×

im(O×K)

−−−−→ H(f) −−−−→ H1 −−−−→ 0⏐⏐� ⏐⏐� ‖0 −−−−→ (OK/f)×

im(O×K)(Z/f)×

−−−−→ Cl(O) −−−−→ Cl(OK) −−−−→ 0.

Proof. Part a) is a standard exercise and we leave it as such. The finiteness ofCl(O) can be proven in exactly the same way as the finiteness of the usual classnumber. In fact for any lattice Λ in a number field there is α ∈ Λ \ {0} with

|N(α)| ≤ c ·√| disc(Λ)| = c ·N(Λ) ·

√| disc(O)|

and hence α−1Λ ⊇ O with index bounded by c√| disc(O)|.

Part b) is well known for the maximal order but is more subtle in general becauseO is not a Dedekind ring. It has ”singularities”, i.e. prime ideals p for which Op isnot a regular local ring, and one may view Spec(OK)→ Spec(O) as a ”resolution ofsingularities”. In our case the singular primes are the primes dividing f . Somethingelse happens that is special for quadratic fields: Fractional O-ideals as we havedefined them (O = End(Λ)) are invertible, i.e. there is a fractional ideal Λ′ withΛΛ′ = O. One checks that one can take Λ′ = ΛN(Λ)−1.

Lemma 2.3. Let R be an integral domain with fraction field K.

a) A fractional R-ideal I is invertible if and only if I−1I = R where I−1 ={α ∈ K|αI ⊆ R}.

b) Invertible ideals are locally free, i.e. projective of rank one.c) Two invertible ideals are isomorphic as R-modules if and only if they are

homothetic.d) Invertible ideals form an abelian groups under multiplication. The group of

homothety classes is isomorphic to Pic(R), the Picard group of R.e) Given finitely many maximal ideals m1, . . . ,mn and an invertible ideal I

there is α ∈ K× so that (αI)mi = Rmi for all i.

Proof. If I ′I = R then I ′ ⊆ I−1 hence R = I ′I ⊆ I−1I ⊆ R. This gives a). Writing1 = a1b1 + · · · + anbn with ai ∈ I and bi ∈ I−1, and given a prime ideal p of R,there must be an index i so that aibi ∈ R×p since Rp is a local ring. But thenak = (aibi)

−1biak · ai ∈ Rp · ai for any k since bi ∈ I−1. This gives b). Part c) isclear since any R-isomorphism extends to a K-linear isomorphism which is givenby multiplication with a scalar α ∈ K×. Part d) is clear from the observation thatany abstract invertible R-module P is isomorphic to a submodule of P ⊗R K ∼= K,hence isomorphic to an invertible ideal. For e) one checks that the localization

36 M. FLACH

S−1R where

S =

n⋂i=1

(R \mi) = R \n⋃

i=1

mi

is a semilocal ring with maximal ideals S−1mi. Hence S−1I, being locally free ofrank one, is actually free of rank one and we can take α−1 ∈ S−1I ⊆ Imi

⊆ K tobe a generator. �

We apply the Lemma to R = O and the set of maximal ideals containing (f)which is finite since O/(f) is finite. So given a fractional O-ideal Λ we can replaceit by a homothetic one and assume Λq = Oq for (f) ⊆ q. Moreover Λf has a uniquefactorisation

Λf =∏

p∈Spec(O)\Z(f)

pmp

f

into prime ideals over the Dedekind ring Of = OK,f . But then

Λ =∏

p∈Spec(O)\Z(f)

pmp

since Λq = Oq = pq for q ∈ Z(f). So Cl(O) is a quotient of J(f)K by the subgroup

P(f)O of principal O-ideals with a generator α prime to f , i.e. such that α ∈ O×q for

all q ∈ Z(f). Fix a prime p of OK with q = p ∩ O and set (p) = p ∩ Z. Since

(27) O = {α ∈ OK |∃n ∈ Z α ≡ n mod f}we also have

Oq = {α ∈ OK,p|∃n ∈ Z α ≡ n mod f}.To see this write

α =αp,1

αp,2=

αp,1αp,2

N(αp,2)=

αq,1

αq,2

with αp,i ∈ OK , αp,2 /∈ p. Then αq,2 := N(αp,2) ∈ Z \ (p) and if α is congruent tosome integer modulo f , the same is true for αq,1 := αp,1αp,2. So αq,i ∈ O, αq,2 /∈ q.Hence

(α) ∈ P(f)O ⇔ α ∈

⋂q⊇(f)

O×q =⋂

p⊇(f)

{α ∈ O×K,p|∃n ∈ Z α ≡ n mod f}

={α ∈ K×|∃n ∈ (Z/f)× α ≡ n mod f}.A computation analogous to the computation of Hm then gives the diagram

0 −−−−→ O× −−−−→ {α ∈ K×|∃n ∈ (Z/f)× α ≡ n mod f} −−−−→ P(f)O −−−−→ 0⏐⏐� ⏐⏐�β

⏐⏐�γ

0 −−−−→ O×K −−−−→ {α ∈ K×|(α, f) = 1} −−−−→ P (f) −−−−→ 0.

and hence the diagram (26). �Remark 2.2.1. They key trick to work with invertible O-ideals is to ”move themaway from the conductor (f)”. The group of all fractional O-ideals is not so niceto work with, it for example has torsion elements. In Borevich, Shafarevich Ch. 2.Sec. 7, Ex. 8,9 it is shown that the set of fractional O-ideals Λ with ΛOK = OK

consists of all Λ = Z · f +Z · fω+Z · ξ where ξ ∈ OK = Z+Z ·ω runs through a setof representatives of (OK/f)×/(Z/f)×. So this set is a finite subgroup of the groupof all fractional O-ideals (namely the kernel of the natural map to OK-ideals).


Remark 2.2.2. It is clear that any order O in a number field K has a descriptionsimilar to (27). If f = [OK : O] is the index then fOK ⊆ O and O ⊆ OK/(f) is asubring such that

O = {α ∈ OK |∃ ν ∈ O α ≡ ν mod (f)}.Definition 2.2.1. The ring class field of the order O in K is the class field H(O)associated by class field theory to the quotient H(f) → Cl(O). In particular, H(OK)is the Hilbert class field of K.

Theorem 2.4. Let O = Λ1, . . . ,Λh be a set of representatives for Cl(O). Thenj(O) is an algebraic number of degree h with conjugates j(Λi), and K(j(O)) is thering class field H(O).

Proof. We need a number of facts about elliptic curves which we list without fullproofs.

a) For any lattice Λ ⊆ C the complex manifold E = C/Λ is an algebraic curveof genus one, an elliptic curve. It is the solution set of the equation

y2 = 4x3 − g2(Λ)x− g3(Λ)

in C2 together with a point at infinity. Any complex analytic morphismis algebraic, in particular for an inclusion Λ1 ⊆ Λ2 there is a morphism(isogeny) E1 → E2.

b) Two elliptic curves C/Λ1, C/Λ2 are isomorphic if and only if Λ1 and Λ2 arehomothetic if and only if j(Λ1) = j(Λ2). Hence there is a bijection

Cl(O) ∼= Ell(O); Λ �→ E/Λ

between Cl(O) and the set Ell(O) of isomorphism classes of elliptic curveswith End(E) ∼= O. Curiously, the group structure of Cl(O) has no naturalinterpretation on Ell(O). We view the multiplication

([a], E = C/Λ) �→ [a] ∗ E := C/(a−1Λ)

as a (free and transitive) action of Cl(O) on Ell(O). So one can think ofan integral ideal a as a morphism, in fact an isogeny

E = C/Λ→ C/(a−1Λ) = [a] ∗ E.

c) The set Ell(O) has a natural action of Aut(C) since the map α �→ ασ is abijection End(E) ∼= End(Eσ). Hence if E ∈ Ell(O) then so is Eσ. Since

j(E)σ = j(Eσ)

the finiteness of Ell(O) already implies that j(E) is an algebraic number ofdegree bounded by h. We note that any elliptic curve can be defined overthe field Q(j(E)). In fact the curves

y2 + xy =x3 − 36

j − 1728x− 1

j − 1728j �= 0, 1728

y2 + y =x3 j = 0

y2 =x3 + x j = 1728

have j-invariant j. So from now on we can and will assume that all ourelliptic curves are defined over Q.

38 M. FLACH

d) The set Cl(O) also has a natural action of Aut(C) (of a quite differentnature) since a ⊆ K ⊆ C. Moreover, one has

([a] ∗ E)σ = [a]σ ∗ Eσ.

The proof of this depends on the interpretation of a as an isogeny. Forexample for α ∈ O the action of σ on the endomorphism α : E → E is justthe natural action on α ∈ K ⊆ C if one views C as the tangent space ofE at the origin (which is a purely algebraic construction over any field ofdefinition of E).

e) Since the action of Cl(O) on Ell(O) is simply transitive, we can define amap

F : Gal(Q/Q)→ Cl(O)

by

Eσ = F (σ) ∗ Ewhere E = C/O. If σ ∈ Gal(Q/K) the element F (σ) ∈ Cl(O) only dependson σ, not on E. If [a] ∗ E is another element of Ell(O) we have

([a] ∗ E)σ = [a]σ ∗ Eσ = [a]σ ∗ F (σ) ∗ E = F (σ) ∗ ([a] ∗ E).

Moreover, for σ, τ ∈ Gal(Q/K) we get

F (στ) ∗ E =Eστ = (Eσ)τ = (F (σ) ∗ E)τ

=F (τ) ∗ (F (σ) ∗ E) = (F (σ)F (τ)) ∗ Eand so F is a homomorphism

F : Gal(Q/K)→ Cl(O).

f) The fixed field of the kernel F is simply K(j(E)) since

F (σ) = 1⇔ F (σ) ∗ E = E ⇔ Eσ ∼= E ⇔ j(Eσ) = j(E)σ = j(E).

So K(j(E))/K is an abelian extension of degree bounded by h containingall conjugates of j(E).

g) If m is a modulus for the abelian extension K(j(E))/K we have homomor-phisms

Jm � Gal(K(j(E))/K)F−→ Cl(O).

Then one has the fundamental and rather curious fact that

(28) F ((a,K(j(E))/K)) = [a] ∈ Cl(O)

for all a ∈ Jm. We can assume that f divides m. Then the natural mapJm → Cl(O) is surjective and factors through F by (28). Hence F must besurjective and induces an isomorphism

F : Gal(K(j(E))/K) ∼= Cl(O).

This gives the theorem.

Remark 2.2.3. The homomorphism F is somewhat analogous to the cyclotomiccharacter

Gal(Q/Q)→ Aut(μm) = (Z/mZ)×


in that it provides an inverse of the Artin map via the natural action of the Galoisgroup on a set of objects of algebraic origin ( so that ”Galois conjugate objects” aredefined). A more direct analogue would be the map

Gal(Q/K(j(E)))→ Aut(E[m]) ∼= (O/m)×

for some ideal m (prime to (f)). This map does indeed arise in completely analogousfashion to the map F . Consider the set Ell(OK ,m) of isomorphism classes of pairs(E,P ) where OK

∼= End(E) and P ∈ E(C) is a point whose exact annihilator inOK is the ideal m. In this case the ray class group Hm = Jm/Pm acts freely andtransitively on Ell(OK ,m) by the analogous map

[a] ∗ (C/Λ, P ) = (C/a−1Λ, P ).

Note that if (α) ∈ Pm then there is an isomorphism

(C/(α)−1Λ, P )·α−→ (C/Λ, α · P ) = (C/Λ, P )

since α ≡ 1 mod m. There is an isomorphism of short exact sequences

0 −−−−→ Ell0(m) −−−−→ Ell(OK ,m)π−−−−→ Ell(OK) −−−−→ 0⏐⏐�∼ ⏐⏐�∼ ⏐⏐�∼

0 −−−−→ (OK/m)×

O×K

−−−−→ Hm −−−−→ H1 = Cl(OK) −−−−→ 0

where exactness of the top row has to be understood in the sense of homogeneousspaces, i.e. π is a surjective Hm-equivariant map all of whose fibres are principal

homogeneous spaces under the subgroup (OK/m)×

O×K

of Hm. The steps in the proof of

Theorem 2.4 apply to this situation showing that P ∈ E(Q) and the coordinates ofP generate K(m)/K.

The biggest gap in the above outline is (28) and we add some remarks about itsproof. First we know by Prop. 2.1 that there exists a prime p of OK such that[p] = [a] in Hm which implies

(a,K(j(E))/K) = (p,K(j(E))/K)

and [p] = [a] in Cl(O). We can even assume that p is split in K/Q (recall that thesplit primes always dominate the sum defining Dirichlet density). So it suffices toshow that

F (Frobp) = [p] ∈ Cl(O)

or equivalently, if E = C/Λ then

EFrobp ∼= [p] ∗ E = C/p−1Λ.

This looks like a weird condition mixing analytic and algebraic information. Thekey to analyzing it is to look at the isogeny

φ : E → [p] ∗ Eof degree Np = p and reduce it modulo a prime P | p of K(j(E)). Since we canalways choose p outside a given finite set of primes we can also assume

1) E has good reduction at all P | p.2) K(j(E)/K is unramified at all p | p.3) NK(j(E))/Q(j(Ei)− j(Ek)) is a p-adic unit.

40 M. FLACH

One then shows that the reduction φ is purely inseparable (by computing its tangentmap) and by a general result about elliptic curves over finite fields this implies thatthere is a factorization

EΦ−→ E(Np) ∼−→ [p] ∗ E

where Φ is the relative Frobenius with respect to the Frobenius x �→ xNp on thebase field OK(j(E))/P. For any variety X → Spec(L) over a field of characteristicp, the relative Frobenius Φ with respect to a power F (x) = xq of the absoluteFrobenius is defined via the diagram

X

Φ

��F

��

X(q)

��

�� X

��Spec(L)

F �� Spec(L)

where the square is Cartesian. Hence in our situation E(Np) is nothing but EFrobp .Our last condition on p ensures that the reduction map is injective on j-invariants,so we have

EFrobp ∼= [p] ∗ E ⇒ j(EFrobp) = j([p] ∗ E)

⇒ j(EFrobp) = j([p] ∗ E)

⇒ EFrobp ∼= [p] ∗ E.

�Here are the 13 orders of class number one with their j-invariants as a function

of their discriminant.

-d 3 4 7 8 11 19 43 67 163j 0 2633 −3353 2653 −215 −21533 −2183353 −2153353113 −2183353233293

-d 3 · 22 4 · 22 7 · 22 3 · 32j 243353 2333113 3353173 −2153 · 53

The quickest way to compute these is to use the Fourier expansion of j(τ) (as afunction on the upper half plane) together with the knowledge that they are rationalintegers. In fact j(τ) is always an algebraic integer if τ is imaginary quadratic. Theirminimal polynomials tend to have large coefficients even for small discriminants.Here is one more example

j

(Z[

1 +√−152

]

)= −52515− 85995

1 +√5

2.

CM elliptic curves also illustrate Faltings’ Theorem. For a fixed imaginary qua-dratic field K there are countably many isomorphism classes of elliptic curves withCM by K (and they are all isogenous). In fact, as we have shown, this set is inbijection to

⋃f≥1 Cl(Of ). It turns out that one can always choose elliptic curves

with a given CM j-invariant that have good reduction for all primes larger than afixed bound. Faltings theorem implies that they cannot all be defined over a fixednumber field. Indeed the degrees of the field K(j(E)) go to infinity since the rayclass numbers |Cl(Of )| tend to infinity by the computation (26).


3. The idele class group and idelic class field theory

We have seen that for a global field K the modulus m appears as a parameterfor the tower of ray class fields K(m). The content of class field theory is that theK(m) have an explicit Galois group and that their union is the maximal abelianextension of K (the union of all abelian extensions in a given algebraic closure).This same information can be repackaged in a more canonical construction which atthe same time makes the relation to local class field theory much more transparentand allows to compute the Brauer group of the global field K. This is the role ofthe idele class group. It is a more elegant but slightly more abstract approach toclass field theory since one has to operate with profinite or more general topologicalgroups.

Definition 3.0.2. Let K be a global field. Define the idele group of K as

A×K =⋃S

A×K,S =:∏′

p

K×p

with the direct limit topology where

A×K,S :=∏p∈S

K×p ×

∏p/∈S

O×Kp

has the product topology and S runs through all finite sets of places containing allarchimedean ones. This is a locally compact group by Tychonoff’s theorem.

Remark 3.0.4. As is suggested by the notation A×K is the unit group of the similarlydefined adele ring AK for which we have no use in this course, except for Lemma3.1 below.

Since any element α ∈ K is integral at all but finitely many places we have anatural embedding K× ⊆ A×K . We also have an absolute value map

|(αp)| :=∏p

|αp|p

and K× lies in its kernel by the product formula.

Definition 3.0.3. The idele class group CK of the global field K is the quotientgroup

CK := A×K/K×.

It is a (split) extension

1→ C1K → CK

log | |−−−→ R→ 0

if K is a number field and

1→ C1K → CK

log | |/ log(q)−−−−−−−−→ Z→ 0

if K has characteristic p and Fq is the algebraic closure of Fp in K. Here C1K =

ker(| |).Proposition 3.1. The topology on K× induced from the locally compact topologyof A×K is the discrete topology. Therefore CK is a locally compact abelian group.Moreover, the group C1

K is actually compact.

42 M. FLACH

Proof. It suffices to find a neighborhood of 1 containing no other element of K.Take U =

∏p�∞O×Kp

×∏p|∞B(1, ε) where

B(1, ε) = {x ∈ Kp| |x− 1|p < ε}is the ball of radius ε < 1. Now look at 1 �= α ∈ U ∩K. Then 0 �= α− 1 ∈ OK and|α− 1|p < ε for all archimedean places. But this contradicts the product formula.

Lemma 3.1. There is a constant C > 0 with the following property: If a = (ap) ∈A×K with |a| > C then there exists α ∈ K× such that |α|p ≤ |ap|p.Proof. This is reminiscent of the Minkowski argument in the classical geometryof numbers. We use without proof that K is a discrete subgroup of AK (similarargument to the one just given for A×K) and that AK/K is compact. Let c0 be theHaar measure of AK/K and and c1 that of the set

W = {ξ ∈ AK | |ξp|p ≤ εp}where εp = 1 for non-archimedean p and εp = 1/10 for archimedean p. For C =c0/c1 the set

T = {ξ ∈ AK | |ξp|p ≤ εp|ap|p}has measure

c1∏p

|ap|p > c1C = c0

and therefore there must be a pair of distinct points t1, t2 of T with the same imagein AK/K. Then α := t1 − t2 ∈ K× and |α|p = |t1 − t2|p ≤ |ap|p by the triangleinequality and its ultrametric sharpening. �

For an idele a with |a| > C consider the set

W = {ξ ∈ A×K | |ξp|p ≤ |ap|p}which is compact since it is a product of compact sets. Now for b ∈ A×K with |b| = 1

we also have |b−1a| > C, hence the Lemma gives α ∈ K× with |α|p ≤ |b−1p ap|p, i.e.

αb ∈W . �

3.1. Class field theory using the idele group. In section we reformulate themain theorems of section 1 in terms of the idele group. We start by describing rayclass groups Hm in terms of the idele class group. Given a modulus

m = mfm∞ =∏p

pnp

we define an open subgroup

U(np) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩O×Kp

p finite, np = 0

1 + pnpOKpp finite, np > 0

K2p p real and np = 1

K×p p infinite, np = 0

of K×p for each place p, and then we define the open subgroup

Um =∏p

U(np)


of A×K . To a finite Galois extension L/K we can also associate an open subgroup

NL/K(A×L ) =∏′

p

NLP/KpL×P

where we choose a P | p and note that the local norm subgroup is independent ofthis choice. We remark that this is just the image of the norm map for the finiteflat ring extension AL/AK and this map sends L× to K×. Hence there is also aninduced map

NL/K : CL → CK .

Proposition 3.2. Denoting by Um the image of Um in CK we have

CK/Um = A×K/K× · Um ∼−→ Jm/Pm = Hm

and

CK/NL/K(CL) = A×K/K× ·NL/K(A×L ) ∼= Jm/Pm ·NL/KJmL .

Proof. We have

A×K/Um ∼= Jm ×∏p∈S

K×p /U(np)

where S is the union of the set of primes dividing m and the infinite places. Byweak approximation there is an exact sequence

1→ K×,m → K× →∏p∈S

K×p /U(np)→ 1

where

K×,m = {α ∈ K×|α ≡ 1 mod m}is as in Lemma 1.6. Then

CK/Um ∼= A×K/Um ·K× ∼= (Jm ×∏p∈S

K×p /U(np))/ im(K×)

∼= (Jm × 1)/ im(K×,m)

= Jm/Pm = Hm.

For a finite extension L/K we can choose a modulus m so that

U(np) ⊆ NLP/KpL×P

for all places p and hence that Um ⊆ NL/K(A×L ). Weak approximation gives us theisomorphism (16)

K×/NL/KL× ·K×,m ∼−→∏p∈S

K×p /NLP/Kp

(L×P)

44 M. FLACH

as in the proof of Lemma 1.6. Then we have

CK/NL/K(CL)

∼=A×K/K× ·NL/K(A×L )

∼=⎛⎝Jm ×

∏p∈S

K×p /U(np)

⎞⎠ /

⎛⎝NL/KJm

L ×∏p∈S

NLP/Kp(L×P)/U(np)

⎞⎠ · im(K×)

∼=(Jm × 1)/(NL/KJmL × 1) · im(K×,m ·NL/KL×)

=Jm/(NL/KJmL · Pm).

�Hence we can view the Artin map for a finite abelian extension L/K as a map

ρ = ρL/K : CK → Gal(L/K)

inducing an isomorphism

CK/NL/KCL∼= Gal(L/K).

In particular, viewing K×p as a subgroup of CK , via the embedding

α �→ (1, . . . , 1, αp, 1, . . . ) ∈ A×Kwe get homomorphisms for all primes

ρp : K×p → Gal(L/K)

which clearly coincide with the local Artin map ρp defined in Ma160b if p is unrami-fied in L/K. It is not so clear whether ρp = ρp for all places p. One can now proceedin two ways. Either one shows that ρp takes values in Gal(LP/Kp) ⊆ Gal(L/K)and in fact induces an isomorphism

K×p /NLP/Kp

L×P ∼= Gal(LP/Kp)

for any place p. This is done, for example, in Lang’s Algebraic Number Theory(Thm. 3 in Ch. XI,§4) and this is how local class field theory was discovered in thefirst place (by Hasse). However, the problem still remains to identify this map withρp (and thereby show independence of the choice of K, for example). Perhaps, afterone checks compatibilities with change of fields one can do this but I haven’t seen itanywhere in the literature. Alternatively, one can show that the global Artin map

ρ =∏p

ρp : A×K → Gal(L/K),

defined as the product of local Artin maps, is trivial on K×. Since it is clearlytrivial on NL/K(A×L ) and coincides with ρ on the unramified primes we must haveρ = ρ and therefore ρp = ρp for all primes p.

Theorem 3.1. Let K be a global field. Then one has:

a) (Reciprocity) For every abelian extension L/K the map ρ is trivial on K×.b) (Isomorphism) For every abelian extension L/K the map ρ is surjective

with kernel K×NL/K(A×L ).c) (Existence) For every open subgroup U ⊆ CK of finite index there is a

unique abelian extension L/K with norm subgroup U .d) (Classification) There is an inclusion reversing bijection between abelian

extensions L/K and open subgroups U of CK of finite index.


Proof. The first statement is unfortunately somewhat lengthy to prove, even thoughwe already have the reciprocity theorem in its classical formulation. However,Theorem 1.6 only implies that ρ is trivial on K×,m. We shall give a proof inconjunction with the computation of the cohomology of the idele class group inthe next section. For b) note that ρ is trivial on NL/K(A×L ) by local class field

theory and hence trivial on K×NL/K(A×L ) by a). By Prop. 3.2 this subgroup hasindex [L : K] and ρ is surjective since it restricts to the classical Artin map on theunramified primes. For c) and d) it suffices to note that the groups Um, resp. Um,form a basis of open subgroups of A×K , resp. a basis of open subgroups of finiteindex of CK and apply the classical existence theorem, Theorem 1.7. �

Remark 3.1.1. If K is a global field of characteristic 0 we could replace ”opensubgroup of finite index” by ”open subgroup” in statements c) and d) since anyopen subgroup of CK has finite index.

Remark 3.1.2. Another advantage of the idelic approach is that it easily general-izes to infinite abelian extensions. The inverse limit of the reciprocity maps

ψK : lim←−U

CK/U → lim←−L

Gal(L/K) = Gal(Kab/K)

is injective with image

{σ ∈ Gal(Kab/K|∃n ∈ Z, σ|Fp= Frobn}

if char(K) = p > 0 and it is surjective with kernel the connected component C0K of

the identity in CK if char(K) = 0. One has an isomorphism

C0K∼= R× (S1)r2 × Vr1+r2−1

where

V = lim←−n

S1 ∼= (R× Z)/Z ∼= QD

is the solenoid, the Pontryagin dual of the discrete group Q. Note that the connectedcomponent A×,0

K of the identity in A×K easily computes to

A×,0K

∼=∏

p complex

C× ×∏

p real

R×,>0 ∼= (S1)r2 × Rr1+r2 .

3.2. Galois cohomology of the idele class group. In the next section on classformations we shall give the proofs of Theorem 3.1 a) and b) without using resultsfrom section 1, in particular without any use of ray class L-functions. We shall notreprove c) and d) since the proofs in the idelic setting are really identical to theclassical ones.

In order to verify that the idele class group satisfies the axioms of a class forma-tion axioms we shall compute the Galois cohomology of the idele class group in thissubsection, more precisely just the H0, H1 and H2. This computation goes hand inhand with the determination of the Brauer group of the global field K about whichthe classical approach had nothing to say. It turns out that the Galois cohomologyof the idele class group contains a lot of information which one does not see in theclassical formulation. For a start, Hm or Jm/Pm ·NL/KJm

L are not naturally Galoiscohomology groups whereas CK/NL/KCL is:

46 M. FLACH

Lemma 3.2. For any Galois extension L/K with group G we have

H0(G,CL) = CK

and hence

H0(G,CL) = CK/NL/K(CL).

Proof. The short exact sequence of G-modules

(29) 1→ L× → A×L → CL → 1

induces an exact cohomology sequence

1→ K× → A×K → H0(G,CL)→ 1

since H1(G,L×) = 0 by Hilbert 90. �

The following result is in some sense the analogue of the norm index theorem inthe idelic approach. On the one hand it implies general results on Galois cohomol-ogy of the idele class group which are needed in the proof of Theorem 3.1 a) andb). On the other hand it has a direct proof using purely algebraic techniques, i.e.avoiding the analytic theory of ray class L-functions. At this point we can deduceit rather quickly from Prop. 3.2 and the results of section 1.

Theorem 3.2. If L/K is cyclic with group G = Gal(L/K) then

|Hi(G,CL)| ={[L : K] = |G| i = 0

1 i = 1.

Proof. By Lemma 3.2 and Prop. 3.2 we get

H0(G,CL) = CK/NL/K(CL) ∼= Jm/Pm ·NL/KJmL∼= G

and so it suffices to show that q(CL) = |G|. Consider the open subgroup

A×L,S =∏P/∈S

O×LP×

∏P∈S

L×P

of A×L , where S is a finite G-stable set of places of L containing the archimedeanplaces and those ramified in L/K and large enough so that Cl(OL,S) = 0. Corollary1.6 implies

(30) Hi(G,∏P|pO×LP

) = Hi(Gp,O×LP) = 0

for i = 0, 1 and p /∈ S and since cohomology commutes with products we furtherobtain form Corollary 1.6

q(A×L,S) =∏p∈S

q(Gp, L×P) =

∏p∈S

|Gp|.

Then with m =∏

P∈S P one has

A×L/L× · A×L,S

∼= JmL / im(L×) ∼= Cl(OL,S) = 0

and

CL∼= A×L/L

× ∼= A×L,S/L× ∩ A×L,S = A×L,S/O×L,S .


and therefore

q(CL) = q(A×L,S)/q(O×L,S) =

∏p∈S |Gp|q(O×L,S)

= |G|

by Lemma 1.7. �For the rest of this section we will develop consequences of Theorem 3.2 without

using any other results from class field theory. In other words, Part 1 will onlyenter in this section via Theorem 3.2. Later we will then give an independent proofof Theorem 3.2 making all results independent of Part 1. First, using Theorem 3.2we can compute H1 of the idele class group. It satisfies an analogue of Hilbert’stheorem 90.

Proposition 3.3. For any Galois extension L/K with group G we have

H1(G,CL) = 0.

Proof. Assume first that G is a p-group. Then G has a cyclic quotient of order p.If L′/K denotes the corresponding Galois extension we have an inflation restrictionsequence

0→ H1(Gal(L′/K), CL′)→ H1(G,CL)→ H1(Gal(L/L′), CL)

obtained form the Hochschild-Serre spectral sequence for the group extension

1→ Gal(L/L′)→ Gal(L/K)→ Gal(L′/K)→ 1.

By induction on |G| we deduce H1(G,CL) = 0 from Thm. 3.2. In general letGp ⊆ G be a p-Sylow subgroup. Then the composite map

H1(G,CL)res−−→ H1(Gp, CL) = 0

cor−−→ H1(G,CL)

is multiplication with [G : Gp], hence induces a bijection on the p-primary sub-group H1(G,CL)[p

∞] of H1(G,CL) (recall that this is a |G|-torsion group). SoH1(G,CL)[p

∞] = 0 and since this holds for all p | |G| we get H1(G,CL) = 0. �Our next aim is to compute H2(G,CL) but it is not easy to do this without pass-

ing to the limit and considering the full groupH2(K,CK) = lim−→LH2(Gal(L/K), CL).

Since the system of all finite extensions L/K is filtered, the direct limit

1→ K× → A×K→ CK → 1

of the exact sequence (29) is an exact sequence of discrete GK-modules, and itscohomology in degree 2 is given by the following theorem.

Theorem 3.3. One has a commutative diagram of short exact sequences

0 −−−−→ H2(K, K×) −−−−→ H2(K,A×K) −−−−→ H2(K,CK) −−−−→ 0

‖ ∼=⏐⏐� ∼=

⏐⏐�invK

0 −−−−→ Br(K) −−−−→ ⊕pBr(Kp)

∑p invp−−−−−→ Q/Z −−−−→ 0

where

invp : Br(Kp) ∼=

⎧⎪⎨⎪⎩Q/Z p nonarchimedean12Z/Z p real

0 p complex.

48 M. FLACH

We first need some local as well as semilocal preparations.

Lemma 3.3. For any finite Galois extension L/K with group G and i ∈ Z one has

Hi(G,A×L ) ∼=⊕p

Hi(Gp, L×P)

and for i > 0

Hi(K,A×K) ∼=

⊕p

Hi(Kp, K×p ).

Proof. Since cohomology commutes with products and using Shapiro’s Lemma weget

Hi(G,A×L,S)∼=

∏p/∈S

Hi(Gp,O×LP)×

∏p∈S

Hi(Gp, L×P).

But if S contains all ramified primes the decomposition groups Gp for p /∈ S arecyclic. Hence from (30) and the fact that cohomology of finite groups commuteswith colimits we get

(31) Hi(G,A×L ) = lim−→S

Hi(G,A×L,S) = lim−→S

∏p∈S

Hi(Gp, L×P) =

⊕p

Hi(Gp, L×P).

Now by definition

A×K

=⋃L

A×L = lim−→L

A×L

and Kp :=⋃

L LP = (K)p by Krasner’s Lemma. �

The following Lemma summarizes the statements from local class field theorywe will need.

Lemma 3.4. For any finite Galois extension LP/Kp of local fields with group Gp

there is an isomorphism

invLP/Kp: H2(Gp, L

×P) ∼= 1

[LP : Kp]Z/Z

which is related to the local reciprocity map

ρp = ρLP/Kp: K×

p → Gabp

by the formula

χ(ρp(α)) = invLP/Kp(α ∪ δχ)

where δ : H1(Gp,Q/Z) → H2(Gp,Z) is the connecting homomorphism in the longexact cohomology sequence induced by

0→ Z→ Q→ Q/Z→ 0.

Proof. The existence of the invariant map follow from the determination of theBrauer group of the local field Kp, the key fact being that any central simplealgebra has an unramified splitting field. So there is an isomorphism

invKp: H2(Kp, K

×p ) ∼= H2(Kur

p /Kp, K×p )

vp−→ H2(Kurp /Kp,Z) ∼= Q/Z.


For this isomorphism one proves directly that there is a commutative diagram withexact rows

0 −−−−→ H2(Gp, L×P) −−−−→ H2(Kp, K

×p ) −−−−→ H2(LP, K×

p )

∼⏐⏐�invLP/Kp ∼

⏐⏐�invKp ∼⏐⏐�invLP

0 −−−−→ 1[LP:Kp]

Z/Z −−−−→ Q/Z·[LP:Kp]−−−−−−→ Q/Z.

Setting

γ := inv−1Kp

(1

[LP : Kp]

)∈ H2(Gp, L

×P)

one defines the reciprocity isomorphism

H−2(Gp,Z) = H1(Gp,Z) = Gabp

∪γ−−→ K×p /NLP/Kp

L×P = H0(Gp, L×P).

In other words, setting sα = ρp(α) one has by definition

γ ∪ sα = α

and hence

α ∪ δχ = γ ∪ (sα ∪ δχ).

Since δ commutes with cup products we get sα ∪ δχ = δ(sα ∪ χ) with

sα ∪ χ ∈ H−1(Gp,Q/Z) ∼= 1

[LP : Kp]Z/Z.

For any finite group G, s ∈ Gab and χ ∈ Hom(G,Q/Z) one checks that s∪χ = χ(s)(Serre, Local fields, Appendix to Ch. XI, Lemma 3) and also that δ(r/n) = r underthe boundary map

H−1(Gp,Q/Z) ∼= ker(·n|Q/Z

)=

1

nZ/Z δ−→ Z/nZ = coker (·n|Z) ∼= H0(Gp,Z)

where n = |Gp| = [LP : Kp]. Finally then

invLP/Kp(α ∪ δχ) = invLP/Kp

(u ∪ r) = r/n.

�

Remark 3.2.1. In the next section we shall define a reciprocity map for any classformation of which the formula in Lemma 3.4 is a special case.

Proof. (of Theorem 3.3) We shall write Hi(L/K,M) for Hi(Gal(L/K),M). It isclear that the top sequence is exact at the left because Proposition 3.3 implies

H1(K,CK) = lim−→L

H1(L/K,CL) = 0

and by Lemma 3.4 the bottom sequence is then also exact at the left. It is alsoclear that the top sequence is exact in the middle and the bottom sequence is exactat the right but unfortunately it will not be easy to construct the map invK so thatthe right square commutes. In order to exploit Theorem 3.2 one first looks at theanalogous situation for a finite and then cyclic extension L/K. For finite L/K the

50 M. FLACH

vanishing of H1(L/K,CL) gives the analogous commutative diagram (without thedotted arrow)(32)

0 �� H2(L/K,L×) �� H2(L/K,A×L ) �� H2(L/K,CL) ��

φ

�� 0

0 �� H2(L/K,L×) ��⊕pH2(LP/Kp, L

×P) invL/K �� 1

[L:K]Z/Z �� 0

with rows which are exact at the left and which we use to define the global invariantmap

invL/K(c) :=∑p

invLP/Kp(cp)

as the sum of local invariant maps. Note that invL/K is certainly not alwayssurjective since there are global extensions for which the least common multiple ofall local degrees is a proper divisor of [L : K]. We first record the functoriality ofthe invariant map.

Lemma 3.5. Let L/K be Galois with group G.

a) For K ⊆ K ′ ⊆ L there are commutative diagrams

H2(L/K ′, L×) � � ��

cor

��

H2(L/K ′,A×L )

cor

��

invL/K′�� 1[L:K′]Z/Z

incl

��H2(L/K,L×) � � ��

res

��

H2(L/K,A×L )

res

��

invL/K �� 1[L:K]Z/Z

[K′:K]

��

H2(K ′/K,K ′×)

inf

��

� � �� H2(K ′/K,A×K′)

inf

��

invK′/K�� 1[K′:K]Z/Z

incl

��

0

��

0

��

0

��

where the bottom portion only applies if K ′/K is Galois.b) The columns are exact.c) Moreover, for each α ∈ A×K and χ ∈ H1(G,Q/Z) we have

(33) χ(ρL/K(α))) = invL/K(α ∪ δχ)

where

ρL/K : A×K → Gab

is the global Artin map defined as the product of local ones

ρL/K((αp)) =∏p

ρLP/Kp(αp).

Proof. The commutativity involving the middle and right hand columns followsfrom the corresponding properties of the local invariant map. The commutativityinvolving the middle and left hand columns is just functoriality of res, cor and


inf. The exactness of the left hand column follows from the inflation-restrictionsequence which extends to an exact six term sequence in view of the vanishing

H1(K ′/K,H1(L/K ′, L×)) = 0

by Hilbert 90. Exactness of the right hand column is clear and exactness of themiddle column follows from exactness of the right hand column (for the variouslocal field extensions) and the fact that the invariant map is an isomorphism in thelocal case. Alternatively, it is a consequence of Hilbert 90 for the local fields, justlike the exactness of the left hand column.

To see (33) first note that the global Artin map is well defined since for anyprime p such that αp ∈ O×Kp

and p is unramified in L/K we have ρLP/Kp(αp) = 1.

If χp is the restriction of χ to Gp then αp ∪ δχp is the local component of α ∪ δχand hence

invL/K(α ∪ δχ) =∑p

invLP/Kp(αp ∪ δχp)

=∑p

χp(ρLP/Kp(αp)) = χ(ρL/K(α))).

�

The next theorem verifies that the bottom row in (32) is a complex. It is areciprocity theorem and just like Theorem 1.6 we will eventually reduce its proof tocyclotomic extensions of Q. For such extensions the proof will conclude by a directverification of Theorem 3.1 a).

Theorem 3.4. For finite L/K and any c ∈ H2(L/K,L×) one has

invL/K(c) = 0.

Proof. The key is to construct a diagram of fields

E′ = L′E

L′

��

L

K E

Q

��

where L′/Q is Galois and E/Q is cyclic cyclotomic with certain local properties.By Lemma 3.5 a) the three maps

H2(L/K,L×) inf−−→ H2(L′/K,L′×) cor−−→ H2(L′/Q, L′×) inf−−→ H2(E′/Q, E′×)

52 M. FLACH

do not change invL/K(c). Denoting by c′ ∈ H2(E′/Q, E′×) the image of c, Lemma3.6 applied to S = {p| invp(c′) �= 0} and m a common denominator of the invp(c

′),gives an extension E′ = L′E so that

res(c′) = 0 ∈ H2(E′/E,E′×).

Lemma 3.5 b) then implies that c′ = inf(c′′) for some c′′ ∈ H2(E/Q, E×). ByLemma 3.7 we have invE/Q(c

′′) = 0 and another application of Lemma 3.5 a) givesinvE′/Q(c

′) = invL/K(c) = 0. �

Lemma 3.6. Given a number field L′, finite set of places S and an integer m thenthere exists a cyclic cyclotomic extension E′/L′ so that m divides [E′P : L′p] for allp ∈ S.

Proof. Let E(qr) ⊆ Q(ζqr ) be the cyclic subextension of degree qr−1 for q odd (forq = 2 we leave it as an exercise that there is a totally complex cyclic subextensionE(2r) of Q(ζ2r ) of degree 2r−2). For any prime p (equal to q or not) the localdegree [E(qr)p : Qp] tends to ∞ as r → ∞ since this is true for the local degrees[Q(ζqr )p : Qp] and [Q(ζqr )p : E(qr)p] is bounded by q. If m = qn1

1 · · · qnk

k theextension E = E(qr11 ) · · ·E(qrkk ) will have local degree divisible by m for all p ∈ Sfor ri > ni large enough. By possibly enlarging the ri further we get the sameconclusion for E′/L′ where E′ = EL′ and L′ is any number field. �

Lemma 3.7. For any cyclotomic extension E/Q Theorem 3.1 a) holds, and if E/Qis also cyclic we have invE/Q(c) = 0 for each c ∈ H2(E/Q, E×).

Proof. Let

ρ =∏p

ρp : A×Q → Gal(Q(ζm)/Q) ∼= (Z/mZ)×

be the reciprocity map. It suffices to show that ρ(−1) = 1 and ρ(l) = 1 forevery prime number l. The key ingredient will be Dwork’s theorem saying thatρp(u · pk) = u−1 where u ∈ Z×p and

ρp : Q×p → Gal(Qp(ζpn)/Qp) ∼= (Z/pnZ)×

is the local reciprocity map. For a ∈ (Z/mZ)× ∼= ∏q(Z/q

mqZ)× denote by (a)q its

q-component. If l � m we have

ρp(l) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 p � m, p �= l

l p = l

(l−1)p p | m1 p =∞

and so ρ(l) = l∏

p|m(l−1)p = 1. If l | m we have

ρp(l) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 p � m∏

q|m,q �=l(l)q p = l

(l−1)p p | m, p �= l

1 p =∞


and so ρ(l) =∏

q|m,q �=l(l)q ·∏

p|m,p �=l(l−1)p = 1. Finally

ρp(−1) =

⎧⎪⎨⎪⎩1 p � m

(−1)p p | m−1 p =∞

and so ρ(−1) = (−1)∏p|m(−1)p = 1. If G = Gal(E/Q) is cyclic pick χ ∈

H1(G,Q/Z) so that δχ ∈ H2(G,Z) ∼= Z/|G|Z is a generator. Then

∪δχ : H0(G,M)→ H2(G,M)

is an isomorphism for both M = E× and M = A×E . Writing c = α∪δχ with α ∈ Q×

Lemma 3.5 c) and the first part of this Lemma give

invE/Q(c) = invL/K(α ∪ δχ) = χ(ρ(α)) = 0.

�

We now come back to diagram (32) and assume that L/K is cyclic. Then thefollowing holds:

Step 1. The top row is exact since

H3(G,L×) ∼= H1(G,L×) = 0.

Step 2. There is a map φ so that the diagram commutes since

invL/K(im(H2(G,L×))) = 0

by Theorem 3.4.Step 3. The map invL/K is surjective. Indeed, if L/K is cyclic of prime degree then

we know |CL/NL/KCK | = [L : K], hence there must be a prime p of K with

Gp = G since otherwise NL/K : A×L → A×K would be surjective. The sameconclusion holds if L/K is cyclic of prime power degree by looking at theunique subextension of prime degree. So in this case invL/K is surjectiveand surjectivity for general cyclic L/K follows from Lemma 3.5.

Step. 4 The map φ is an isomorphism since it is surjective by Step 3 and

|H2(L/K,CL)| = |H0(L/K,CL)| = [L : K],

again by Theorem 3.2.

As remarked above, for general finite L/K the rows in (32) need not be exact atthe right. Instead we pass to the direct limit over all cyclic (or even just cyclotomic)extensions L/K and note that

H2(K,A×K) = lim−→

L/K cyclic

H2(L/K,A×L ) =⋃

L/K cyclic

H2(L/K,A×L )

in view of the fact that any element in the local Brauer group has a cyclic (evenunramified) splitting field which in turn is the localization of a suitable global cyclic

54 M. FLACH

(even cyclotomic) extension L/K. In view of the inclusions

0 −−−−→ H2(L, K×) −−−−→ H2(L,A×K)�⏐⏐ �⏐⏐

0 −−−−→ H2(K, K×) −−−−→ H2(K,A×K)�⏐⏐ �⏐⏐

0 −−−−→ H2(L/K,L×) −−−−→ H2(L/K,A×L )�⏐⏐ �⏐⏐0 0

we also obtain (diagram chase)

H2(K, K×) = lim−→L/K cyclic

H2(L/K,L×) =⋃

L/K cyclic

H2(L/K,L×).

This then implies exactness of the bottom row in Theorem 3.3 and concludes thecomputation of the Brauer group of the global field K. One also obtains Q/Z asa subgroup (and hence direct summand) of H2(K,CK) but at this point we stilldon’t know equality of the two groups. For any finite extension L/K one has thecommutative diagram with exact columns

0 −−−−→ Q/ZL −−−−→ H2(L,CK)�⏐⏐[L:K]=res

�⏐⏐res

0 −−−−→ Q/ZKι−−−−→ H2(K,CK)�⏐⏐ �⏐⏐

0 −−−−→ 1[L:K]Z/Z

ιL/K−−−−→ H2(L/K,CL)�⏐⏐ �⏐⏐0 0

where

Q/ZK =⋃

K′/K cyclic

H2(K ′/K,CK′)

is mapped to Q/ZL under the restriction map in view of the commutative diagram

GL −−−−→ Gal(LK ′/L)⏐⏐� ⏐⏐�GK −−−−→ Gal(K ′/K)

and the fact that LK ′/L is again cyclic. A simple diagram chase shows thatι( 1

[L:K]Z/Z) ⊆ H2(L/K,CL) and hence the existence of ιL/K . To show that ι

is an isomorphism it suffices to show that ιL/K is an isomorphism for all L/K.This in turn will follow if we can prove that

(34) |H2(L/K,CL)| ≤ [L : K].


We know equality for cyclic L/K. For a tower K ⊆ K ′ ⊆ L there is an exactinflation-restriction sequence

0→ H2(K ′/K,CK′)→ H2(L/K,CL)→ H2(L/K ′, CL)

in view of the vanishing of H1(L/K ′, CL). Hence by an easy induction we get(34) for all solvable L/K. Finally recall that for any finite group G with Sylowsubgroups Gp, any G-module M and any i ≥ 0

Hi(G,M)resp−−→

∏p||G|

Hi(Gp,M)

is injective. Hence, if Kp denotes the fixed field of the Sylow p-subgroup ofGal(L/K), we get

|H2(L/K,CL)| ≤∏p

|Hi(L/Kp, CL)| ≤∏p

[L : Kp] = [L : K].

This finally completes the proof of Theorem 3.3. �

We summarize our computation of the cohomology of the idele class group inthe following theorem.

Theorem 3.5. For any Galois extension L/K of number fields one has

Hi(L/K,CL) =

⎧⎪⎨⎪⎩CK i = 0

0 i = 11

[L:K]Z/Z i = 2.

For CK = lim−→LCL one has

Hi(K,CK) = lim−→L

Hi(L/K,CL) =

⎧⎪⎨⎪⎩CK i = 0

0 i = 1

Q/Z i = 2.

4. Class formations and duality

Definition 4.0.1. A class formation consists of a profinite group G and a discreteG-module C together with isomorphisms

invU/V : H2(U/V,CV ) ∼= 1

[U : V ]Z/Z

for each pair V � U ⊆ G of open subgroups such that the following hold

a) H1(U/V,CV ) = 0b) If W � U and W ⊆ V the diagram

H2(U/V,CV )inf−−−−→ H2(U/W,CW )

res−−−−→ H2(V/W,CW )

inv

⏐⏐�∼ inv

⏐⏐�∼ inv

⏐⏐�∼1

[U :V ]Z/Z⊆−−−−→ 1

[U :W ]Z/Z·[U :V ]−−−−→ 1

[V :W ]Z/Z

is commutative.

56 M. FLACH

If (G,C) is a class formation then

Hi(G,C) =

⎧⎪⎨⎪⎩CG i = 0

0 i = 11|G|Z/Z := lim−→U

1[G:U ]Z/Z i = 2.

So if for any integer n there is an open subgroup U ⊆ G of index divisible by n wehave

H2(G,C) ∼= Q/Z.

This is the case if there is a surjection G→ Z like in examples a)- e) below.Here are some examples of class formations.

a) (G,C) = (GK , CK) where K is a global field.b) (G,C) = (GK , K×) where K is a local field.

c) (G,C) = (Z,Z)d) (G,C) = (GK , CK) where K/k is an extension of transcendence degree one

over an algebraically closed field k of characteristic zero, i.e. a functionfield of a smooth proper curve X over k, and

CK = lim−→L/Kfinite

HomZ(Pic(XL,Q/Z)

where XL is the smooth proper curve over k with function field L.e) There is class field theory for higher local fields. A 2-local field is a complete

discretely valued field K with residue field a local field, for example K =Qp((T )),Fq((T ))((S)) or

Qp{{T}} ={ ∞∑

i=−∞aiT

i | ai ∈ Qp, infi∈Z

vp(ai) > −∞, limi→−∞

vp(ai) =∞}

which has residue field Fp((T )). Then one has to extend the notion of classformation to allow a complex of GK-modules, and it turns out (G,C•) =(GK , K× ⊗L

Z K×) is a class formation.f) (GS , CS) where S is a set of places of the global field K containing all

archimedean ones, GS = Gal(KS/K) the Galois group of the maximalextension of K unramified outside S and

CS(L) = A×L/L× · US

where US is the compact subgroup

US =∏P/∈S

O×LP×

∏P∈S

{1}.

Note that by (30) we have Hi(G,US) = 0 for i = 0, 1 (and L/K unramifiedoutside S) and in fact for any i ∈ Z since the decomposition groups atP /∈ S are cyclic. By the long exact cohomology sequence induced by

0→ US → CL → CS(L)→ 0

we get isomorphisms Hi(G,CL) ∼= Hi(G,CS(L)) and also H0(G,CS(L)) ∼=CS(K).


We shall now develop a sequence of duality theorems for finite groups which willculminate in the Tate-Nakayama duality for class formations. Throughout we set

A∗ = Hom(A,Q/Z)

for any abelian group A. This is an exact contravariant functor from abelian groupsto abelian groups. If A is finite then A∗ coincides with the Pontryagin dual of A.

Theorem 4.1. Let Γ be a finite group and A a Γ-module. Then for all i ∈ Z thepairing

Hi(Γ, A∗)× H−i−1(Γ, A)∪−→ H−1(Γ,Q/Z) ∼= 1

|Γ|Z/Z ⊆ Q/Z

induces an isomorphism

Hi(Γ, A∗) ∼= H−i−1(Γ, A)∗.

Proof. We first show the statement for i = 0. A homomorphism f : A→ Q/Z is aΓ-homomorphism if and only if f(IΓA) = 0, hence we obtain an isomorphism

H0(Γ, A∗) ∼= H0(Γ, A)∗.

If f ∈ NΓA∗, i.e. f =

∑σ∈Γ σh, then for a ∈ A with NΓa = 0 we have

f(a) =∑

(σh)(a) =∑

h(σ−1a) = h(NΓa) = 0

so we obtain a map(A∗)Γ/NΓA

∗ → (NΓA/IGA)∗.If g : NΓ

A→ Q/Z is a homomorphism that vanishes on IΓA then g can be extendedto a homomorphism g : A → Q/Z which is a Γ-homomorphism since g(IΓA) = 0.So our map is surjective. If f ∈ (A∗)Γ vanishes on NΓA there exists g ∈ (NΓA)

∗

with f(a) = g(NΓa) since NΓ : A/NΓA→ NΓA is an isomorphism. Again g can be

extended to a homomorphism g : A→ Q/Z and then f = NΓg since

(NΓg)(a) =∑σ∈Γ

g(σ−1a) = g(NΓa) = f(a).

For arbitrary i one can use dimension shifting which gives a commutative diagram

Hi(Γ, A∗) × H−i−1(Γ, A)

δi

��

∪ �� H−1(Γ,Q/Z)

(−1)i(i+1)/2

��H0(Γ,Hom(A,Q/Z)i)

δi

��

× H−1(Γ, A−i)∪ �� H−1(Γ,Q/Z)

.

Since Hom(A,Q/Z)i ∼= Hom(A−i,Q/Z) the desired result follows. We recall thatfor any Γ-module A one defines A1 by the exact sequence

0→ A→ A⊗Z Z[Γ]→ A1 → 0

and since the middle term is cohomologically trivial we get an isomorphism

δ : Hi(Γ, A1) ∼= Hi+1(Γ, A).

For p > 0 one defines Ap = (Ap−1)1 and for p < 0 one uses induction and themodule A−1 defined by the exact sequence

0→ A−1 → HomZ(Z[Γ], A)→ A→ 0.

58 M. FLACH

�

Remark 4.0.2. For an arbitrary (discrete) group Γ and (discrete) Z[Γ]-module Aone always has a duality between cohomology and homology

Hi(Γ, A∗) ∼= Hi(Γ, A)∗

for any i ≥ 0. This follows from the Hom-⊗-adjunctionHomZ[Γ](P•,HomZ(A,Q/Z)) ∼= HomZ(P• ⊗Z[Γ] A,Q/Z)

which is a special case of the adjunction

HomS(P,HomR(A,B)) ∼= HomR(P ⊗S A,B)

for a right S-module P , S-R-bimodule A and right R-module B. Theorem 4.1extends this to all i ∈ Z in case Γ is finite.

Proposition 4.1. Let Γ be a finite group and A a Z-free Γ-module. Then for alli ∈ Z the pairing

Hi(Γ,Hom(A,Z))× H−i(Γ, A)∪−→ H0(Γ,Z) ∼= Z/|Γ|Z

induces an isomorphism

Hi(Γ,Hom(A,Z)) ∼= H−i(Γ, A)∗.

Proof. Since A is Z-free we have an exact sequence

0→ Hom(A,Z)→ Hom(A,Q)→ Hom(A,Q/Z)→ 0

and Hom(A,Q) is cohomologically trivial. So we get a commutative diagram

Hi−1(Γ, A∗)

δ

��

× H−i(Γ, A)

id

��

∪ �� H−1(Γ,Q/Z)

δ

��Hi(Γ,Hom(A,Z)) × H−i(Γ, A)

∪ �� H0(Γ,Z)

where the vertical arrows are isomorphisms. �

Remark 4.0.3. One can combine Theorem 4.1 and Prop. 4.1 into an isomorphismin the derived category of abelian groups

RΓ(Γ, RHom(A,Z)) ∼= RHom(RΓ(Γ, A),Q/Z) = RΓ(Γ, A)∗

for any abelian group A. If A is free abelian then RHom(A,Z) ∼= Hom(A,Z) andwe recover Prop. 4.1. If A is arbitrary the exact triangle

RHom(A,Z)→ RHom(A,Q)→ RHom(A,Q/Z)→induces an isomorphism

RΓ(Γ, RHom(A,Q/Z)) ∼= RΓ(Γ, RHom(A,Z))[1]

which implies

Hi(Γ, A∗) ∼= Hi+1(Γ, RHom(A,Z))

and we recover Theorem 4.1.

Theorem 4.2. (Nakayama-Tate) Let Γ = U/V be a layer in a class formation andA a Z[Γ]-module, finite free over Z.


a) If γ ∈ H2(Γ, CV ) is the canonical generator, i.e. the unique element withinvU/V (γ) =

1[U :V ] , then

Hi(Γ, A)∪γ−−→ Hi+2(Γ, A⊗Z CV )

is an isomorphism for all i ∈ Z.b) For all i ∈ Z the cup product

Hi(Γ,Hom(A,CV ))× H2−i(Γ, A)∪−→ H2(Γ, CV ) ∼= 1

[U : V ]Z/Z

induces an isomorphism of finite abelian groups

Hi(Γ,Hom(A,CV )) ∼= H2−i(Γ, A)∗.

Proof. Let

0→ CV → C(γ)→ Z[G]→ Z→ 0

be a Yoneda 2-extension corresponding to

γ ∈ H2(Γ, CV ) = Ext2Z[G](Z, CV ).

The composite map

Z/|Γ|Z = H0(Γ,Z)δ1−→ H1(Γ, IΓ)

δ2−→ H2(Γ, CV )

induced by the short exact sequences

(35) 0→ IΓ → Z[Γ]→ Z→ 0

and

(36) 0→ CV → C(γ)→ IΓ → 0

also coincides with the cup product ∪γ (at least up to sign) by general homologicalalgebra. Since C was a class formation ∪γ is an isomorphism. Since δ1 is alwaysan isomorphism this implies that δ2 is an isomorphism. The long exact sequence

H1(Γ, CV )→ H1(Γ, C(γ))→ H1(Γ, IΓ)δ2−→ H2(Γ, CV )→ H2(Γ, C(γ))→ H2(Γ, IΓ)

induced by (36) together with H1(Γ, CV ) = 0 and H2(Γ, IΓ) ∼= H1(Γ,Z) = 0 thenshows that

H1(Γ, C(γ)) = H2(Γ, C(γ)) = 0.

The same holds for all subgroups, so C(γ) is cohomologically trivial. Statementa) then follows from tensoring (35) and (36) with A, Lemma 4.1 below, and thedescription of the cup product as the composite map

Hi(Γ, A)δ1−→ Hi+1(Γ, A⊗Z IΓ)

δ2−→ Hi+2(Γ, A⊗Z CV ).

Similarly, since A is Z-free applying Hom(A,−) to the exact sequences (35) and(36) yields exact sequences whose middle terms are again cohomologically trivial

60 M. FLACH

by Lemma 4.1 below. This then implies that the maps δ in the diagram

Hi−2(Γ,Hom(A,Z))

δ

��

× H2−i(Γ, A)

id

��

∪ �� H0(Γ,Z)

δ

��Hi−1(Γ,Hom(A, IΓ))

δ

��

× H2−i(Γ, A)

id

��

∪ �� H1(Γ, IΓ)

δ

��Hi(Γ,Hom(A,CV )) × H2−i(Γ, A)

∪ �� H2(Γ, CV )

are isomorphisms and the asserted duality follows from Prop. 4.1. �Lemma 4.1. If X is a cohomologically trivial module over a finite group Γ and Ais any finitely generated Z-free Γ-module then the Z[Γ]-modules HomZ(A,X) andA⊗Z X are cohomologically trivial.

Proof. A Z[Γ]-module is cohomologically trivial if and only if it has finite projectivedimension if and only if it has projective dimension one. So take a resolution

0→ P1 → P2 → X → 0

by Z[Γ]-projective modules. We get induced exact sequences

0→ HomZ(A,P1)→ HomZ(A,P0)→ HomZ(A,X)→ 0

and0→ A⊗Z P1 → A⊗Z P2 → A⊗Z X → 0.

Since A is finitely generated HomZ(A,P ) ∼= HomZ(A,Z)⊗Z P with diagonal actionwhich is well known to be isomorphic to HomZ(A,Z) ⊗Z P with trivial action onthe first factor if P is free, hence to P dimZ A. Similarly, A ⊗Z P ∼= P dimZ A if P isfree. So if P is free then HomZ(A,P ) and A ⊗Z P are again free, hence c.t. Bypassing to direct summands we deduce that HomZ(A,P ) and A⊗ZP are c.t., henceso are HomZ(A,X) and A⊗Z X. �Corollary 4.1. Given any layer Γ = U/V in a class formation there is an isomor-phism

ρ = ρU/V : CU/NΓCV ∼−→ Γab

given byχ(ρ(a)) = invU/V (a ∪ δχ)

for χ ∈ H1(Γ,Q/Z) δ−→ H2(Γ,Z) and a ∈ CV . The map ρ is called the reciprocitymap, or norm residue homomorphism.

Proof. Taking A = Z and i = 0 in Theorem 4.2 gives an isomorphism

ρ : CU/NΓCV = H0(Γ, CV ) ∼= H2(Γ,Z)∗

characterized by the property

invU/V (a ∪ δχ) = ρ(a)(δχ)

and the isomorphism

H2(Γ,Z)∗ ∼= H1(Γ,Q/Z)∗ ∼= Γab

sends the character ψ to the group element σψ ∈ Γab with ψ(δχ) = χ(σψ). We thendefine ρ(a) := σρ(a) and obtain the above formula for χ(ρ(a)). �


Recall from Lemma 3.4 that the reciprocity map for local fields was defined viathis corollary, and this is the only way I know of to define this map in the ramifiedcase. For global fields one has the alternative definition using only the Frobeniusautomorphisms for the unramified places, i.e. the classical Artin map.

The following corollary finally proves Theorem 3.1 a) and b).

Corollary 4.2. Let K be a global field, L/K a finite abelian extension and

ρL/K =∏p

ρp : A×K → Gal(L/K)

the reciprocity map. If α ∈ K× then ρL/K(α) = 1. Moreover, ρL/K induces anisomorphism

A×K/NL/KA×L ·K× = CK/NL/KCL∼= Gal(L/K).

Proof. By Lemma 3.5 c) for each α ∈ A×K and χ ∈ H1(G,Q/Z) we have

χ(ρL/K(α))) = invL/K(α ∪ δχ)

which means that ρL/K coincides with the map ρ of Corollary 4.1. �

Corollary 4.3. For a Galois extension L/K with subfield K ⊆ K ′ ⊆ L there arecommutative diagrams

CK

res

��

Gal(L/K)abρ−1

��

res

��CK′

cor

��

Gal(L/K ′)ab.ρ−1

��

cor

��

Proof. Since ρ−1 is given by ∪γ and res(γ) = γ′ it suffices to show res(x ∪ y) =res(x)∪res(y) and the projection formula cor(x∪res(y)) = cor(x)∪y. By dimensionshifting this reduces to the degree 0 case where we have

cor(x)⊗ y =∑s

s(x)⊗ y =∑s

s(x⊗ y) = cor(x⊗ y).

�

Note that in negative degrees, in particular for i = −2, we have homology groups,so the natural map is the corestriction map whereas the restriction is the ”opposite”or ”Umkehr” map. For i = −2 it is called the transfer.

4.1. Direct proof of Theorem 3.2. In this section we give a proof of Theorem3.2 that does not use results from Part 1 except for some Lemmas which had aself-contained proof. Other Lemmas from Part 1 we will have to reprove but wewill not use any results of section 1.4, the analytic theory of ray class L-functions.

In the above short derivation of Theorem 3.2 we already showed directly, onlyusing Corollary 1.6 and Lemma 1.7, that

(37) q(CL) :=|H0(G,CL)||H1(G,CL)|

= |G| =: n

for a cyclic extension L/K with group G. Hence it suffices to show that |H0(G,CL)|divides n, or that H1(G,CL) = 0 in order to prove Theorem 3.2. By induction on

62 M. FLACH

the number of prime factors of n it furthermore suffices to show this for primedegree n in view of the long exact inflation restriction sequence

0→ H1(Gal(L′/K), CL′)→ H1(G,CL)→ H1(Gal(L/L′), CL)

and the fact that H1 = H1. We can further assume that K contains a primitiven-th root of unity by the following argument. The field K ′ := K(ζn) has degree[K ′ : K] dividing n− 1, hence prime to n, the field L′ := LK ′ is cyclic of degree nover K ′ and G′ := Gal(L′/K ′) is isomorphic to G by restriction. We have maps ofG′-modules

CL → CL′NL′/L−−−−→ CL

whose composite is multiplication by [L′ : L] = [K ′ : K]. Hence the induced map

H1(G′, CL)→ H1(G′, CL′)NK′/K−−−−→ H1(G′, CL)

is multiplication by [K ′ : K] but is also the zero map if we know that H1(G′, CL′) =0. Since H1(G′, CL) is an n-torsion group this implies H1(G′, CL) = H1(G,CL) =0.

Since ζn ∈ K, our field L = K( n√a) is a Kummer extension. We place ourselves

in the situation of Proposition 1.6, making sure that LS,n contains L. So let S bea finite set of places of K such that

• p | ∞ ⇒ p ∈ S• p | n⇒ p ∈ S• {p1, . . . , pk} ⊆ S where 〈p1, . . . , pk〉 = Cl(OK).• vp(a) �= 0⇒ p ∈ S

and define

LS,n := K(

n

√O×K,S

).

Clearly, L ⊆ LS,n since a ∈ O×L,S . By the proof of Prop. 1.6 we also know that

[LS,n : K] = ns

where s = |S|. What we don’t know at this point is whether L or LS,n are classfields. In fact we don’t even know surjectivity of the Artin map since this wasdeduced from the analytic theory. So we reprove here Corollary 1.3.

Lemma 4.2. For any abelian extension L/K and modulus m divisible by all ram-ified primes the Artin map is surjective.

Proof. We again look at the fixed field L′ of the image of the Artin map andconclude that all p � m split completely in L′/K. Let E ⊆ L′ be a subextension sothat E/K is cyclic. Then again all p � m split completely in E/K and in particularK×

p = NEP/Kp(EP)×. So if S = {p | m∞} we have∏′

p/∈SK×

p ×∏p∈S

1 ⊆ NE/KA×E

and by the weak approximation theorem K× surjects onto∏p∈S

K×p /NEP/Kp

(EP)×,

i.e. ∏p/∈S

1×∏p∈S

K×p ⊆ K× ·NE/KA×E .


So we have

H0(E/K,CE) = A×K/(K× ·NE/KA×E) =∏′

p

K×p /(K× ·NE/KA×E) = 1

which together with (37) for the cyclic extension E/K implies E = K. HenceL′ = K, i.e. the Artin map is surjective. �

Using similar ideas we reprove Lemma 1.13, actually a slightly strengthenedversion of it.

Lemma 4.3. Let S′ ⊆ S be a subset also satisfying the above four conditions andput

O×K,S′,n := O×K,S ∩∏p∈S′

(K×p )n.

Then one has O×K,S′,n = (O×K,S)n and therefore by (23)∣∣∣O×K,S/O×K,S′,n

∣∣∣ = ∣∣∣O×K,S/(O×K,S)n∣∣∣ = ns.

Proof. The inclusion O×K,S′,n ⊇ (O×K,S)n is clear. Now take α ∈ O×K,S′,n and

consider the cyclic extension E = K( n√α). Then all p ∈ S′ are split completely in

E/K and all p /∈ S′ are unramified in E/K. So

A×K,S′ :=∏p/∈S′

O×Kp×

∏p∈S′

K×p ⊆ NE/KA×E .

By the fact that S′ contains generators of the ideal class group we have A×K =

K× · A×K,S′ . Therefore

H0(E/K,CE) = A×K/(K× ·NE/KA×E) = K× · A×K,S′/(K× ·NE/KA×E) = 1

which together with (37) for the cyclic extension E/K implies E = K. This meansα ∈ O×K,S ∩ (K×)n = (O×K,S)

n. �

In the situation of Lemma 4.3 we now assume that the primes {p1, . . . , pt} =S \ S′ are chosen so that {Frobp1

, . . . ,Frobpt} is a Z/n-basis of Gal(LS′,n/L) ⊆

Gal(LS′,n/K). This we can do by Lemma 4.2. In particular, the primes pi splitcompletely in L/K and we have t = s′ − 1 and s = s′ + t = 2s′ − 1. We thereforehave

U :=∏p∈S′

(K×p )n ×

∏p∈S\S′

K×p ×

∏p/∈S

O×Kp⊆ NL/K(A×L )

and

H0(G,CL) = A×K/(K× ·NL/K(A×L )) = K× · A×K,S/(K× ·NL/K(A×L ))

has order dividing

[K× · A×K,S : K× · U)] =[A×K,S : U)]

[K× ∩ A×K,S : K× ∩ U ]=

∏p∈S′ [K

×p : (K×

p )n]

[O×K,S : O×K,S′,n]

=n2s′

ns=

n2s′

n2s′−1= n

where we have also used Lemma 1.14 which had a self-contained proof. This finishesour second proof of Theorem 3.2.

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