Caltech Algebraic Number Theory 3

7/23/2019 Caltech Algebraic Number Theory 3

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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 : Hm

C we could show non-

vanishing L(1, )= 0 provided that factors through Jm/NL/KJmL Pm for somefinite extension L/K. By Theorem 1.2 the subgroupPm is the norm subgroup ofthe ray class fieldK(m)/K. Hence we get nonvanishing for all nontrivial and thisallows to prove the following strengthening of the the Frobenius density theorem(the analogue of Dirichlets theorem on primes in arithmetic progressions for thenumber fieldK).

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)

log L(s, )p

(p)

Nps

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

(c)1 log L(s, )

p

(pc1)

Nps =|Hm|

pc

1

Nps.

As s1 the functions log L(s, ) stay finite for = 1 whereaslog L(s, 1) = log mK(s) log(s 1).

So we find

lims1+

pc

1Nps

p1Nps

= lims1+

pc

1Nps

log(s 1) = 1

|Hm| .

Theorem 2.1. (Cebotarev) LetL/Kbe a Galois extension with group GandCGa conjugacy class. Then the set of primes p (unramified in L/K) withFrobpChas Dirichlet density |C||G| .

Proof. Recall that ifP|p and gG then FrobPg =g FrobP g1, i.e. all elementsin the conjugacy class Care Frobenius automorphisms provided one of them is. Sowe fix C, denote by G ={g G|gg 1 = } its centralizer, by Ethe fixedfield of< >and by q a prime ofE. Then

FrobpC P|p, FrobP= q|p, NE/Kq= p, Frobq= and since P|q is totally inert

|{q|p, NE/Kq= p, Frobq= }|=|{P|p, FrobP= }|= [G :< >] = |G|

c

f

wherec :=|C|= [G: G] andf=|< >|. HenceFrobpC

1

Nps =

c f|G|

NE/Kq=p,Frobq=

1

Nqs =

c f|G|

Frobq=

1

Nqs


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32 M. FLACH

and

lims1+

FrobpC

1Nps

log(s

1)

= c f

|G

|

lims1+

Frobq=

1Nqs

log(s

1)

= c

|G

|by Proposition 2.1.

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

: GK,SGLn(Ql)whereGK,Sis the Galois group of the maximal extension ofKunramified outsidea finite set of places S(which we always assume to contain the infinite places andthe places above l). For example, ifX 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 abelian

variety A/K. If is an automorphic form on GLn one can sometimes attach anl-adic representation to . This process often -but not always- goes throughthe l-adic cohomology of a suitable algebraic variety. Since GK,S is profinite, inparticular compact,(GK,S) is compact hence conjugate to a subgroup ofGLn(Zl).So we can assumehas image inGLn(Zl) and there is also a reduced representation

: GK,SGLn(Fl).One has the following basic observation.

Lemma 2.1. For fixedK,l,S {p|l} andnthere are only finitely many represen-tations. Hence there is a finite extensionL/Kthrough which such representations

factorize.

Proof. LetKbe the fixed field of . ThenK/Khas degree bounded by | GLn(Fl)|and is unramified outside S, so by Hermite-Minkowski there are only finitely manysuch fieldsK. For eachK there are only finitely many homomorphisms

Gal(K/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 /

Ssuch that theFrobpi represent all

conjugacy classes inGal(L/K). IfV andV are two representations such that

Lpi(T, V) = Lpi(T, V)

for i= 1, . . . , k thenV andV are equivalent.


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

tr((g)) = tr((g)).

LetT V,T V be GK,S-stable lattices and consider the Zl-subalgebraMEndZl(T) EndZl(T)

generated by {(g)(g)|gGK,S}. It is enough to show tr((m)) = tr((m)) formin aZl-basis ofMand since we already know this identity for m = (Frobpi) (Frobpi) it suffices to show that these elements and their conjugates generate MoverZl. By Nakayamas Lemma it suffices to show that they generate M/lM overFl. But the representation factors through G = Gal(L/K) and since theFrobpi represent all conjugacy classes in G we have

Fl[im(( )(g Frobpig1)] = im(( )Fl[G]) =M/lM.

These ideas often lead to effective criteria to check the equivalence of two l-adic

representations. One doesnt actually compute the field L but works directly withrepresentations. For example one has:

Theorem 2.2. (Serre, Livne) Two irreducible representations

i : GQ,{2,3,5}GL2(Q2)are equivalent if and only

tr(1(Frobp)) = tr(2(Frobp)); det(1(Frobp)) = det(2(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 inP9Z by the equations

x0+

+ x9 = 0; x30+

+ x39 = 0.

Then

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

L(f, s 2)where

(X, s) =

x X closed

(1 Nxs)1

is the Zeta-function of X, (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)) (z)(2z)(5z)3(10z)3 = n

ane2inz

where

(z) = e

2iz

24

n=1

(1 e2inz

)

is Dedekinds-function. It is easy to computeap from this product expansion andone compares those with the traces of Frobenius gotten from counting points on Xfor p73.


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34 M. FLACH

Another application is to finiteness theorems. For example, ifA/Kis an abelianvariety of dimensiong with good reduction at p, it is known that the characteristicpolynomial of Frobenius on the Tate module ofA satisfies

Lp(A, T) =2gi=1

(T i) Z[T]

with|i|=

Np. These are classical results of Artin, Hasse and Weil. So there areonly 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 andA over a number fieldKare isogenous if and only ifVl(A) andVl(A

) are isomorphicGK-representations.

Corollary 2.1. For a givenK, Sandg there are only finitely many isogeny classesof abelian varietiesA/K of dimensiong with good reduction outsideS.

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 only

finitely 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 discriminantd. We view Kas a subfield of the complex numbers.

We review lattices in the complex plane. For any lattice

= Z 1+ Z 2inC one has itsj -invariantdefined as

j() = 1728 g2()3

g2()3 27 g3()2where

g2 = 60 G4; g3 = 140 G6; G2k = ,=0

1

2k

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

j() = j()

and in fact the converse holds as well

j() =j() C =.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.

Anendomorphismof a lattice is a complex number such that

. Clearlythe set of all such forms a subring ofC and for most lattices this ring is just Z.When it is bigger thanZ we say that has complex multiplication. It is easy tosee that this happens if and only if is homothetic to a lattice

= K C


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in some imaginary quadratic field K. In this case End() = End() is anorderOin K. Moreover, is a fractionalO-ideal inK (by definition this just means thatEnd() =O; so for example ifO =OK thenOK wouldnt be a fractionalO-idealbut it is still a finitely generatedO-submodule ofK).Lemma 2.2. a) The orders inKare parametrized byf Z1; they are all

of the formO= Z+ fOK for a unique suchf.b) The fractionalO-ideals form a group under multiplication. The homothety

classes of fractionalO-ideals form a finite abelian group Cl(O)and there isa commutative diagram with exact rows

(26)

0 (OK/f)im(OK)

H(f) H1 0 0 (OK/f)

im(OK)(Z/f)

Cl(O) Cl(OK) 0.

Proof. Part a) is a standard exercise and we leave it as such. The finiteness of

Cl(O) can be proven in exactly the same way as the finiteness of the usual classnumber. In fact for any lattice in a number field there is \ {0} with

|N()| c

| disc()|= c N()

| disc(O)|and hence1 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 whichOp 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: FractionalO-ideals as we havedefined them (O = End()) are invertible, i.e. there is a fractional ideal with =O. One checks that one can take =N()1.Lemma 2.3. LetR be an integral domain with fraction fieldK.

a) A fractional R-ideal I is invertible if and only if I1I = R where I1 ={K|IR}.

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 ofR.e) Given finitely many maximal idealsm1, . . . ,mn and an invertible ideal I

there isK so that(I)mi =Rmi for all i.Proof. IfII=R thenI I1 henceR = III1IR. This gives a). Writing1 = a1b1 + +anbn with ai I and bi I1, and given a prime ideal p of R,there must be an index i so that aibi Rp since Rp is a local ring. But thenak = (aibi)

1

biak ai Rp ai for any k since bi I1

. This gives b). Part c) isclear since any R-isomorphism extends to a K-linear isomorphism which is givenby multiplication with a scalar K. Part d) is clear from the observation thatany abstract invertible R-moduleP is isomorphic to a submodule ofPRK=K,hence isomorphic to an invertible ideal. For e) one checks that the localization


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36 M. FLACH

S1Rwhere

S=n

i=1(R \ mi) = R \

n

i=1mi

is a semilocal ring with maximal ideals S1mi. HenceS1I, being locally free ofrank one, is actually free of rank one and we can take 1 S1I Imi K tobe a generator.

We apply the Lemma to R =O and the set of maximal ideals containing (f)which is finite sinceO/(f) is finite. So given a fractionalO-ideal we can replaceit by a homothetic one and assume q=Oqfor (f)q. Moreover fhas a uniquefactorisation

f =

pSpec(O)\Z(f)

pmpf

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

pSpec(O)\Z(f)

pmp

since q =Oq = pq for qZ(f). So Cl(O) is a quotient ofJ(f)K by the subgroupP

(f)O of principalO-ideals with a generator prime tof, i.e. such that Oq for

all qZ(f). Fix a prime p ofOK with q = p O and set (p) = p Z. Since(27) O={ OK|n Zn mod f}we also have

Oq={ OK,p|nZ n modf}.To see this write

=p,1p,2

=p,1p,2N(p,2)

=q,1q,2

with p,i OK, p,2 /p. Thenq,2 := N(p,2)Z \ (p) and if is congruent tosome integer modulof, the same is true forq,1:= p,1p,2. Soq,i O,q,2 /q.Hence

() P(f)O q(f)

Oq = p(f)

{ OK,p|n Zn mod f}

={K|n(Z/f) n mod f}.A computation analogous to the computation ofHm then gives the diagram

0 O {K|n(Z/f) n mod f} P(f)O 0 0 OK {K|(, f) = 1} P(f) 0.and hence the diagram (26).

Remark 2.2.1. They key trick to work with invertibleO-ideals is to move themaway from the conductor(f). The group of all fractionalO-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 fractionalO-ideals with OK =OKconsists of all = Z f+Z f +Z where OK= Z+Z runs through a setof representatives of(OK/f)/(Z/f). So this set is a finite subgroup of the groupof all fractionalO-ideals (namely the kernel of the natural map toOK-ideals).


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Remark 2.2.2. It is clear that any orderOin a number fieldKhas a descriptionsimilar to (27). Iff= [OK :O] is the index thenfOK OandO OK/(f) is asubring such that

O={ OK| O mod (f)}.Definition 2.2.1. Thering class field of the orderO inKis the class fieldH(O)associated by class field theory to the quotientH(f) Cl(O). In particular,H(OK)is the Hilbert class field ofK.

Theorem 2.4. LetO = 1, . . . , h be a set of representatives for Cl(O). Thenj(O) is an algebraic number of degreeh with conjugatesj (i), andK(j(O))is thering class fieldH(O).Proof. We need a number of facts about elliptic curves which we list without fullproofs.

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

y2 = 4x3 g2()x g3()in C2 together with a point at infinity. Any complex analytic morphismis algebraic, in particular for an inclusion 1 2 there is a morphism(isogeny) E1E2.

b) Two elliptic curves C/1,C/2 are isomorphic if and only if 1 and 2 arehomothetic if and only ifj (1) = j(2). Hence there is a bijection

Cl(O)=Ell(O); E/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/)[a] E:= C/(a1)

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/C/(a1) = [a] E.c) The set Ell(O) has a natural action of Aut(C) since the map is a

bijection End(E)=End(E). Hence ifEEll(O) then so is E. Sincej(E) =j(E)

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

y2 + xy=x3 36j 1728x

1

j 1728 j= 0, 1728

y2

+ y=x3

j = 0y2 =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.


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38 M. FLACH

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

([a]

E) = [a]

E.

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

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

F : Gal(Q/Q)Cl(O)by

E =F() EwhereE= C/O. IfGal(Q/K) the elementF()Cl(O) only dependson , not on E. If [a]

Eis another element of Ell(

O) we have

([a] E) = [a] E = [a] F() E= F() ([a] E).Moreover, for, Gal(Q/K) we get

F() E=E = (E) = (F() E)=F() (F() E) = (F()F()) E

and soF is a homomorphism

F: Gal(Q/K)Cl(O).f) The fixed field of the kernel F is simplyK(j(E)) since

F() = 1F() E= EE=Ej(E) = j(E) =j(E).So K(j(E))/Kis an abelian extension of degree bounded by h containingall conjugates ofj (E).

g) Ifm is a modulus for the abelian extension K(j(E))/Kwe have homomor-phisms

Jm Gal(K(j(E))/K) FCl(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 Fmust besurjective and induces an isomorphism

F: Gal(K(j(E))/K)=Cl(O).

This gives the theorem.

Remark 2.2.3. The homomorphism Fis somewhat analogous to the cyclotomiccharacter

Gal(Q/Q)Aut(m) = (Z/mZ)


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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 idealm (prime to(f)). This map does indeed arise in completely analogousfashion to the mapF. Consider the setEll(OK,m)of isomorphism classes of pairs(E, P) whereOK=End(E) and P E(C) is a point whose exact annihilator inOK is the idealm. In this case the ray class group Hm = Jm/Pm acts freely andtransitively onEll(OK,m) by the analogous map

[a] (C/, P) = (C/a1, P).Note that if()Pm then there is an isomorphism

(C/()1, P) (C/, P) = (C/, P)

since1 mod m. There is an isomorphism of short exact sequences0

Ell0(m)

Ell(OK,m)

Ell(

OK)

0

0 (OK/m)OK

Hm H1 = Cl(OK) 0where exactness of the top row has to be understood in the sense of homogeneousspaces, i.e. is a surjectiveHm-equivariant map all of whose fibres are principal

homogeneous spaces under the subgroup (OK/m)

OKofHm. The steps in the proof of

Theorem 2.4 apply to this situation showing thatP E(Q) and the coordinates ofP generateK(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 ofOK such that[p] = [a] inHm 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, ifE= C/ then

EFrobp=[p] E= C/p1.This looks like a weird condition mixing analytic and algebraic information. Thekey to analyzing it is to look at the isogeny

: E[p] E

of degree Np

= p and reduce it modulo a primeP

|p

ofK(j(E)). Since we canalways choose p outside a given finite set of primes we can also assume

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


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40 M. FLACH

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

E

E(Np)

[p] E

where is the relative Frobenius with respect to the Frobenius x xNp on thebase fieldOK(j(E))/P. For any variety X Spec(L) over a field of characteristic

p, the relative Frobenius with respect to a power F(x) = xq of the absoluteFrobenius is defined via the diagram

X

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] Ej(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 functionof 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 ofj () (as afunction on the upper half plane) together with the knowledge that they are rationalintegers. In factj() is always an algebraic integer ifis imaginary quadratic. Theirminimal polynomials tend to have large coefficients even for small discriminants.Here is one more example

j

Z[

1 +152

]

=52515 859951 +

5

2 .

CM elliptic curves also illustrate Faltings Theorem. For a fixed imaginary qua-dratic fieldKthere 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 in

bijection tof1Cl(Of). It turns out that one can always choose elliptic curveswith a given CM j -invariant that have good reduction for all primes larger than a

fixed bound. Faltings theorem implies that they cannot all be defined over a fixednumber field. Indeed the degrees of the fieldK(j(E)) go to infinity since the rayclass numbers| Cl(Of)| tend to infinity by the computation (26).


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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 the

K(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. LetKbe a global field. Define the idele group ofK as

AK=S

AK,S=:p

Kp

with the direct limit topology where

AK,S:=pS

Kpp/S

OKp

has the product topology andSruns through all finite sets of places containing allarchimedean ones. This is a locally compact group by Tychonoffs theorem.

Remark 3.0.4. As is suggested by the notationAKis the unit group of the similarlydefined adele ringAK for which we have no use in this course, except for Lemma3.1 below.

Since any element K is integral at all but finitely many places we have anatural embeddingK AK. We also have an absolute value map

|(p)|:= p|p|p

andK 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 := AK/K

.

It is a (split) extension

1C1KCKlog | | R0

ifKis a number field and

1C1KCKlog | |/ log(q) Z0

ifK has characteristic p andFq is the algebraic closure ofFp inK. Here C1K =

ker(| |).Proposition 3.1. The topology onK induced from the locally compact topologyofAK is the discrete topology. Therefore CK is a locally compact abelian group.Moreover, the group C1K is actually compact.


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42 M. FLACH

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

p OKp

p| B(1, ) where

B(1, ) =

{x

Kp

| |x

1

|p<

}is the ball of radius 0 with the following property: Ifa = (ap)AK with|a|> C then there existsK such that||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 ofAK (similarargument to the one just given for AK) and thatAK/K is compact. Letc0 be theHaar measure ofAK/K and andc1 that of the set

W ={ AK| |p|pp}where p = 1 for non-archimedean p and p = 1/10 for archimedean p. For C =c0/c1 the set

T ={ AK| |p|pp|ap|p}has measure

c1p

|ap|p> c1C= c0

and therefore there must be a pair of distinct points t1, t2 ofTwith the same imagein AK/K. Then := t1 t2 K and||p =|t1 t2|p |ap|p by the triangleinequality and its ultrametric sharpening.

For an idelea with|a|> Cconsider the setW ={ AK| |p|p |ap|p}

which is compact since it is a product of compact sets. Now for bAKwith|b|= 1we also have

|b1a

|> C, hence the Lemma gives

K with

||p

|b1p ap

|p, i.e.

bW. 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) =

OKp pfinite, np= 01 +pnpOKp pfinite, np> 0K2p preal andnp= 1

K

p pinfinite, np= 0ofKp for each place p, and then we define the open subgroup

Um =p

U(np)


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ofAK. To a finite Galois extension L/Kwe can also associate an open subgroup

NL/K(AL ) =

p

NLP/KpLP

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/AKand this map sends L to K. Hence there is also aninduced map

NL/K :CLCK.

Proposition 3.2. Denoting byUm the image ofUm inCKwe have

CK/Um = AK/K

Um Jm/Pm =Hm

and

CK

/NL/K

(CL

) = AK

/K

NL/K

(AL

)=

Jm/Pm

NL/K

JmL

.

Proof. We have

AK/Um=Jm

pS

Kp/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

1K,m K pS

Kp/U(np)1

where

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

CK/Um= AK/Um K=(Jm

pS

Kp/U(np))/ im(K)

=(Jm 1)/ im(K,m)=Jm/Pm =Hm.

For a finite extensionL/Kwe can choose a modulus m so that

U(np)NLP/KpLP

for all placesp and hence thatUm

NL/K(AL ). Weak approximation gives us the

isomorphism (16)

K/NL/KL K,m

pS

Kp/NLP/Kp(LP)


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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

pS

Kp/U(np)

/

NL/KJmL

pS

NLP/Kp(LP)/U(np)

im(K)

=(Jm 1)/(NL/KJmL 1) im(K,m NL/KL)=Jm/(NL/KJ

mL Pm).

Hence we can view the Artin map for a finite abelian extension L/Kas a map

= L/K :CKGal(L/K)inducing an isomorphism

CK/NL/KCL=Gal(L/K).In particular, viewing Kp as a subgroup ofCK, via the embedding

(1, . . . , 1, p, 1, . . . )AKwe get homomorphisms foral lprimes

p: Kp Gal(L/K)

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

Kp/NLP/KpLP

=Gal(LP/Kp)for any place p. This is done, for example, in Langs 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 withp(and thereby show independence of the choice ofK, for example). Perhaps, afterone checks compatibilities with change of fields one can do this but I havent seen itanywhere in the literature. Alternatively, one can show that the global Artin map

=p

p: AKGal(L/K),

defined as the product of local Artin maps, is trivial on K. Since it is clearlytrivial onNL/K(A

L) and coincides with on the unramified primes we must have

= and therefore p= p foral lprimes p.

Theorem 3.1. LetKbe a global field. Then one has:

a) (Reciprocity) For every abelian extensionL/K the map is trivial onK.b) (Isomorphism) For every abelian extension L/K the map is surjective

with kernelK

NL/K(A

L ).c) (Existence) For every open subgroup U CK of finite index there is aunique abelian extensionL/Kwith norm subgroup U.

d) (Classification) There is an inclusion reversing bijection between abelianextensionsL/Kand open subgroupsU ofCKof finite index.


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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 is trivial on K,m. We shall give a proof in

conjunction with the computation of the cohomology of the idele class group inthe next section. For b) note that is trivial on NL/K(AL) by local class field

theory and hence trivial on KNL/K(AL ) by a). By Prop. 3.2 this subgroup has

index [L: K] and 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 ofAK, resp. a basis of open subgroups of finiteindex ofCKand 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 ofCKhas 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

K: limU

CK/U limL

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

is injective with image

{Gal(Kab/K|nZ, |Fp = Frobn}ifchar(K) = p >0 and it is surjective with kernel the connected componentC0K ofthe identity inCK ifchar(K) = 0. One has an isomorphism

C0K= R (S1)r2 Vr1+r21

where

V = limn

S1=(RZ)/Z = QD

is the solenoid, the Pontryagin dual of the discrete group Q. Note that the connectedcomponentA,0K of the identity inA

Keasily 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 theH0, H1 andH2. This computation goes hand in

hand with the determination of the Brauer group of the global field Kabout 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 orJm/Pm NL/KJmL are not naturally Galoiscohomology groups whereasCK/NL/KCL is:


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46 M. FLACH

Lemma 3.2. For any Galois extensionL/Kwith group G we have

H0(G, CL) = CK

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

Proof. The short exact sequence ofG-modules

(29) 1L AL CL1induces an exact cohomology sequence

1K AKH0(G, CL)1sinceH1(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) and

b). 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. IfL/Kis cyclic with group G = Gal(L/K) then

|Hi(G, CL)|=

[L: K] =|G| i= 01 i= 1.

Proof. By Lemma 3.2 and Prop. 3.2 we get

H0(G, CL) =CK/NL/K(CL)=Jm/Pm NL/KJmL=Gand so it suffices to show that q(CL) =|G|. Consider the open subgroup

AL,S=

P/S

OLP

PSL

P

ofAL , where S is a finite G-stable set of places ofL containing the archimedeanplaces and those ramified inL/Kand large enough so that Cl(OL,S) = 0. Corollary1.6 implies

(30) Hi(G,P|p

OLP) = Hi(Gp, OLP) = 0

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

q(AL,S) =pS

q(Gp, LP) =

pS

|Gp|.

Then with m =PSP one hasAL/L

AL,S=JmL / im(L)=Cl(OL,S) = 0and

CL= AL/L= AL,S/L AL,S= AL,S/OL,S.


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and therefore

q(CL) = q(AL,S)/q(OL,S) =

pS |Gp|

q(OL,S)=|G|

by Lemma 1.7.

For the rest of this section we will develop consequences of Theorem 3.2 withoutusing 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 Hilbertstheorem 90.

Proposition 3.3. For any Galois extensionL/Kwith group G we have

H1(G, CL) = 0.

Proof. Assume first that G is a p-group. ThenG has a cyclic quotient of order p.

IfL

/Kdenotes the corresponding Galois extension we have an inflation restrictionsequence

0H1(Gal(L/K), CL)H1(G, CL)H1(Gal(L/L), CL)obtained form the Hochschild-Serre spectral sequence for the group extension

1Gal(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 letGpG be ap-Sylow subgroup. Then the composite map

H1(G, CL) resH1(Gp, CL) = 0 corH1(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 allp| |G| we get H1(G, CL) = 0.

Our next aim is to computeH2

(G, CL) but it is not easy to do this without pass-ing to the limit and considering the full group H2(K, CK) = limLH2(Gal(L/K), CL).

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

1 K AKCK1

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,AK

) H2(K, CK) 0 =

=invK0 Br(K)

pBr(Kp)

pinvp Q/Z 0

where

invp: Br(Kp)=

Q/Z p nonarchimedean12Z/Z p real

0 p complex.


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48 M. FLACH

We first need some local as well as semilocal preparations.

Lemma 3.3. For any finite Galois extensionL/Kwith group G andi Zone has

Hi(G,AL)= p

Hi(Gp, LP)

and for i > 0

Hi(K,AK

)=p

Hi(Kp, Kp).

Proof. Since cohomology commutes with products and using Shapiros Lemma weget

Hi(G,AL,S)=p/S

Hi(Gp, OLP) pS

Hi(Gp, LP).

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,AL) = limS

Hi(G,AL,S) = limS

pS

Hi(Gp, LP) =

p

Hi(Gp, LP).

Now by definition

AK

=L

AL = limL

AL

and Kp:=L LP = (

K)p by Krasners Lemma.

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

Lemma 3.4. For any finite Galois extensionLP/Kp of local fields with group Gpthere is an isomorphism

invLP/Kp :H2(Gp, LP)= 1[LP : Kp]Z/Z

which is related to the local reciprocity map

p= LP/Kp :Kp Gabp

by the formula

(p()) = invLP/Kp( )where: H1(Gp,Q/Z)H2(Gp,Z) is the connecting homomorphism in the longexact cohomology sequence induced by

0Z 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(Kurp /Kp, Kp) vpH2(Kurp /Kp,Z)= Q/Z.


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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 invLP0 1[LP:Kp]Z/Z Q/Z

[LP:Kp] Q/Z.Setting

:= inv1Kp

1

[LP: Kp]

H2(Gp, LP)

one defines the reciprocity isomorphism

H2(Gp,Z) = H1(Gp,Z) = GabpKp/NLP/KpLP = H0(Gp, LP).

In other words, setting s= p() one has by definition

s= and hence

= (s ).Sincecommutes with cup products we get s = (s ) with

s H1(Gp,Q/Z)= 1[LP : Kp]

Z/Z.

For any finite groupG,sGab andHom(G,Q/Z) one checks thats = (s)(Serre, Local fields, Appendix to Ch. XI, Lemma 3) and also that (r/n) = runderthe boundary map

H1(Gp,Q/Z)=ker n|Q/Z = 1

nZ/Z Z/nZ = coker (n|Z)= H0(Gp,Z)

wheren =|Gp|= [LP: Kp]. Finally theninvLP/Kp( ) = 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) = limL

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 invKso 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


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50 M. FLACH

vanishing ofH1(L/K,CL) gives the analogous commutative diagram (without thedotted arrow)(32)

0 H2(L/K, L) H2(L/K,AL) H2(L/K, CL)

0

0 H2(L/K, L) p

H2(LP/Kp, LP)

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 of

the invariant map.Lemma 3.5. LetL/K be Galois with group G.

a) ForKK L there are commutative diagrams

H2(L/K, L)

cor

H2(L/K,AL )

cor

invL/K 1[L:K]Z/Z

incl

H2(L/K,L)

res

H2(L/K,AL)

res

invL/K 1[L:K]Z/Z

[K:K]

H2(K/K,K)

inf

H2(K/K,AK)

inf

invK/K 1[K:K]Z/Z

incl

0

0

0

where the bottom portion only applies ifK/K is Galois.b) The columns are exact.c) Moreover, for eachAK andH1(G,Q/Z) we have

(33) (L/K())) = invL/K( )where

L/K : AKGab

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

L/K((p)) = p

LP/Kp(p).

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


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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 thatp OKp andp is unramified in L/Kwe haveLP/Kp(p) = 1.Ifp is the restriction of to Gp then p p is the local component of and hence

invL/K( ) =p

invLP/Kp(p p)

=p

p(LP/Kp(p)) = (L/K())).

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 ofQ. For such extensions the proof will conclude by a directverification of Theorem 3.1 a).

Theorem 3.4. For finiteL/K and anycH2(L/K,L) one hasinvL/K(c) = 0.

Proof. The key is to construct a diagram of fields

E =LE

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) infH2(L/K,L) corH2(L/Q, L) infH2(E/Q, E)


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52 M. FLACH

do not change invL/K(c). Denoting byc H2(E/Q, E) the image ofc, Lemma

3.6 applied toS={p| invp(c)= 0} and m a common denominator of the invp(c),gives an extensionE =LEso that

res(c) = 0H2(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 fieldL, finite set of placesSand an integerm thenthere exists a cyclic cyclotomic extensionE/L so thatm divides[EP : L

p] for all

pS.Proof. LetE(qr)Q(qr) be the cyclic subextension of degree qr1 forqodd (forq= 2 we leave it as an exercise that there is a totally complex cyclic subextensionE(2r) ofQ(2r) of degree 2r2). 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(qr )p : Qp] and [Q(qr)p : E(qr)p] is bounded by q. If m = qn11 qnkk theextensionE= E(qr11 ) E(qrkk ) will have local degree divisible by m for all pSfor ri > ni large enough. By possibly enlarging the ri further we get the sameconclusion for E/L whereE =EL andL is any number field.

Lemma 3.7. For any cyclotomic extensionE/Q Theorem 3.1 a) holds, and ifE/Qis also cyclic we haveinvE/Q(c) = 0 for eachcH2(E/Q, E).Proof. Let

=p

p: AQGal(Q(m)/Q)=(Z/mZ)

be the reciprocity map. It suffices to show that (1) = 1 and (l) = 1 fo revery prime number l. The key ingredient will be Dworks theorem saying that

p(u pk) = u1 whereuZp andp : Q

pGal(Qp(pn)/Qp)=(Z/pnZ)

is the local reciprocity map. For a(Z/mZ)= q(Z/qmqZ) denote by (a)q itsq-component. Ifl mwe have

p(l) =

1 pm, p=ll p= l

(l1)p p|m1 p=

and so(l) = l

p|m(l

1)p = 1. Ifl|m we have

p(l) =

1 pmq|m,q=l(l)q p= l

(l1)p p|m, p=l1 p=


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and so(l) =q|m,q=l(l)q

p|m,p=l(l

1)p = 1. Finally

p(1) = 1 pm

(1)p p|m1 p=

and so (1 ) = (1)p|m(1)p = 1. If G = Gal(E/Q) is cyclic pick H1(G,Q/Z) so thatH2(G,Z)= Z/|G|Zis a generator. Then

: H0(G, M) H2(G, M)

is an isomorphism for bothM =E andM= AE. Writingc= withQLemma 3.5 c) and the first part of this Lemma give

invE/Q(c) = invL/K( ) = (()) = 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 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, ifL/Kis cyclic of prime degree then

we know|CL/NL/KCK

|= [L: K], hence there must be a primepofKwith

Gp=G since otherwise NL/K :AL AKwould be surjective. The sameconclusion holds ifL/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 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/Kthe rows in (32) need not be exact atthe right. Instead we pass to the direct limit over all cyclic (or even just cyclotomic)extensionsL/Kand note that

H2(K,AK) = limL/Kcyclic

H2(L/K,AL) =

L/KcyclicH2(L/K,AL )

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


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54 M. FLACH

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

0 H2(L, K) H2(L,AK

)

0 H2(K, K) H2(K,A

K)

0 H2(L/K, L) H2(L/K,AL ) 0 0

we also obtain (diagram chase)

H2(K, K) = limL/K cyclic

H2(L/K, L) =

L/Kcyclic

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 obtainsQ/Z asa subgroup (and hence direct summand) ofH2(K, CK) but at this point we stilldont know equality of the two groups. For any finite extensionL/Kone has thecommutative diagram with exact columns

0 Q/ZL H2(L, CK)[L:K]=res res0 Q/ZK H2(K, CK) 0 1[L:K]Z/Z

L/K H2(L/K, CL) 0 0

where

Q/ZK=

K/Kcyclic

H2(K/K,CK)

is mapped toQ/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 that

( 1[L:K]Z/Z) H2(L/K, CL) and hence the existence of L/K. To show that is an isomorphism it suffices to show that L/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].


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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 SylowsubgroupsGp, anyG-moduleMand anyi0

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 extensionL/Kof number fields one has

Hi(L/K, CL) =

CK i= 0

0 i= 11

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

ForCK= limLCL one has

Hi(K, CK) = limL

Hi(L/K, CL) =

CK i= 0

0 i= 1

Q/Z i= 2.

4. Class formations and duality

Definition 4.0.1. Aclass formationconsists of a profinite group G and a discreteG-moduleC together with isomorphisms

invU/V :H2(U/V,CV)= 1

[U :V]Z/Z

for each pairV UG of open subgroups such that the following holda) H1(U/V,CV) = 0b) IfW U andW 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.


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56 M. FLACH

If (G, C) is a class formation then

H

i

(G, C) =

CG i= 0

0 i= 11|G|Z/Z := limU

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

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

H2(G, C)= Q/Z.This is the case if there is a surjection GZ like in examples a)- e) below.

Here are some examples of class formations.

a) (G, C) = (GK, CK) whereKis a global field.b) (G, C) = (GK, K

) whereK is a local field.

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

over an algebraically closed field k of characteristic zero, i.e. a function

field of a smooth proper curve X over k, and

CK= limL/Kfinite

HomZ(Pic(XL,Q/Z)

whereXL 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 Kwith residue field a local field, for example K =Qp((T)),Fq((T))((S)) or

Qp{{T}}=

i=

aiTi | ai Qp, inf

iZvp(ai)>, lim

ivp(ai) =

which has residue fieldFp((T)). Then one has to extend the notion of classformation to allow a complex ofGK-modules, and it turns out (G, C

) =(GK, K LZ K) is a class formation.

f) (GS, CS) where S is a set of places of the global field K containing allarchimedean ones, GS = Gal(KS/K) the Galois group of the maximalextension ofKunramified outside Sand

CS(L) = AL/L

USwhereUS is the compact subgroup

US=P/S

OLP PS

{1}.

Note that by (30) we have Hi(G, US) = 0 for i = 0, 1 (andL/Kunramifiedoutside S) and in fact for any i Z since the decomposition groups atP /

Sare cyclic. By the long exact cohomology sequence induced by

0USCLCS(L)0we get isomorphisms Hi(G, CL)= Hi(G, CS(L)) and alsoH0(G, CS(L))=CS(K).


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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 groupA. This is an exact contravariant functor from abelian groupsto abelian groups. IfA is finite thenA coincides with the Pontryagin dual ofA.

Theorem 4.1. Let be a finite group andA a-module. Then for all iZ thepairing

Hi(, A) Hi1(, A) H1(,Q/Z)= 1||Z/Z Q/Z

induces an isomorphism

Hi(, A)= Hi1(, A).

Proof. We first show the statement for i = 0. A homomorphismf :AQ/Z is a-homomorphism if and only iff(IA) = 0, hence we obtain an isomorphism

H0(, A)=H0(, A).IffNA, i.e. f=

h, then for aA withNa= 0 we have

f(a) =

(h)(a) =

h(1a) = h(Na) = 0

so we obtain a map(A)/NA

(NA/IGA).Ifg : NAQ/Z is a homomorphism that vanishes on IAtheng can be extendedto a homomorphism g :AQ/Z which is a -homomorphism since g (IA) = 0.So our map is surjective. Iff (A) vanishes on NA there exists g (NA)withf(a) = g(Na) sinceN: A/NANAis an isomorphism. Againg can beextended to a homomorphismg : AQ/Z and thenf=Ng since

(Ng)(a) =

g(1a) = g(Na) = f(a).

For arbitraryi one can use dimension shifting which gives a commutative diagram

Hi(, A) Hi1(, A)i

H1(,Q/Z)

(1)i(i+1)/2

H0(, Hom(A,Q/Z)i)

i

H1(, Ai) H1(,Q/Z)

.

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

0AA Z Z[]A10and since the middle term is cohomologically trivial we get an isomorphism

: Hi(, A1)= Hi+1(, A).For p > 0 one defines Ap = (Ap1)1 and for p < 0 one uses induction and themodule A1 defined by the exact sequence

0A1HomZ(Z[], A)A0.


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58 M. FLACH

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

Hi(, A)=Hi(, A)for anyi0. This follows from theHom--adjunction

HomZ[](P, HomZ(A,Q/Z))=HomZ(PZ[]A,Q/Z)which is a special case of the adjunction

HomS(P, HomR(A, B))=HomR(PSA, B)for a right S-module P, S-R-bimodule A and right R-module B. Theorem 4.1extends this to alliZ in case is finite.Proposition 4.1. Let be a finite group andA aZ-free-module. Then for alliZ the pairing

Hi(, Hom(A,Z))

Hi(, A)

H0(,Z)= Z/

|

|Z

induces an isomorphism

Hi(, Hom(A,Z))= Hi(, A).

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

0Hom(A,Z)Hom(A,Q)Hom(A,Q/Z)0and Hom(A,Q) is cohomologically trivial. So we get a commutative diagram

Hi1(, A)

Hi(, A)id

H1(,Q/Z)

Hi(, Hom(A,Z))

Hi(, A)

H0(,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

R(, R Hom(A,Z))=R Hom(R(, A),Q/Z) = R(, A)for any abelian groupA. IfA is free abelian thenR Hom(A,Z)=Hom(A,Z) andwe recover Prop. 4.1. IfA is arbitrary the exact triangle

R Hom(A,Z)R Hom(A,Q)R Hom(A,Q/Z)induces an isomorphism

R(, R Hom(A,Q/Z))= R(, R Hom(A,Z))[1]which implies

Hi(, A)= Hi+1(, R Hom(A,Z))and we recover Theorem 4.1.

Theorem 4.2. (Nakayama-Tate) Let =U/Vbe a layer in a class formation andA aZ[]-module, finite free overZ.


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

1[U:V]

, then

Hi

(, A)

Hi+2

(, A ZCV

)

is an isomorphism for alli Z.b) For alliZ the cup product

Hi(, Hom(A, CV)) H2i(, A) H2(, CV)= 1[U :V]

Z/Z

induces an isomorphism of finite abelian groups

Hi(, Hom(A, CV))= H2i(, A).

Proof. Let

0

CV

C()

Z[G]

Z

0

be a Yoneda 2-extension corresponding to

H2(, CV) = Ext2Z[G](Z, CV).The composite map

Z/||Z = H0(,Z) 1 H1(, I) 2 H2(, CV)induced by the short exact sequences

(35) 0I Z[] Z0and

(36) 0

CV

C()

I

0

also coincides with the cup product(at least up to sign) by general homologicalalgebra. SinceCwas a class formation is an isomorphism. Since1 is alwaysan isomorphism this implies that 2 is an isomorphism. The long exact sequence

H1(, CV)H1(, C())H1(, I) 2H2(, CV)H2(, C())H2(, I)induced by (36) together with H1(, CV) = 0 and H2(, I)=H1(,Z) = 0 thenshows that

H1(, C()) = H2(, C()) = 0.

The same holds for all subgroups, so C() 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(, A) 1 Hi+1(, A ZI) 2 Hi+2(, A ZCV).

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


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60 M. FLACH

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

Hi2(, Hom(A,Z))

H2i(, A)id

H0(,Z)

Hi1(, Hom(A, I))

H2i(, A)id

H1(, I)

Hi(, Hom(A, CV)) H2i(, A) H2(, CV)

are isomorphisms and the asserted duality follows from Prop. 4.1.

Lemma 4.1. IfX is a cohomologically trivial module over a finite group andAis any finitely generatedZ-free -module then theZ[]-modules HomZ(A, X) andA ZXare cohomologically trivial.Proof. A Z[]-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

0P1P2X0byZ[]-projective modules. We get induced exact sequences

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

0A ZP1A ZP2A ZX0.SinceA is finitely generated HomZ(A, P)=HomZ(A,Z) Z Pwith diagonal actionwhich is well known to be isomorphic to HomZ(A,Z) ZPwith trivial action onthe first factor ifP is free, hence to PdimZA. Similarly, A ZP= PdimZA ifP isfree. So if P is free then HomZ(A, P) and AZP are again free, hence c.t. Bypassing to direct summands we deduce that HomZ(A, P) andA Z Pare c.t., henceso are HomZ(A, X) andA ZX.

Corollary 4.1. Given any layer =U/V in a class formation there is an isomor-phism

= U/V :CU/NC

V abgiven by

((a)) = invU/V(a )forH1(,Q/Z) H2(,Z) andaCV. The map is called the reciprocitymap, or norm residue homomorphism.

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

: CU/NCV = H0(, CV)= H2(,Z)

characterized by the property

invU/V(a ) = (a)()and the isomorphism

H2(,Z)=H1(,Q/Z)=absends the character to the group elementab with() = (). We thendefine(a) := (a) and obtain the above formula for ((a)).


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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 Frobenius

automorphisms for the unramified places, i.e. the classical Artin map.The following corollary finally proves Theorem 3.1 a) and b).

Corollary 4.2. LetKbe a global field, L/Ka finite abelian extension and

L/K=p

p : AKGal(L/K)

the reciprocity map. If K then L/K() = 1. Moreover, L/K induces anisomorphism

AK/NL/KAL K =CK/NL/KCL=Gal(L/K).

Proof. By Lemma 3.5 c) for each AK andH1(G,Q/Z) we have(L/K())) = invL/K( )

which means that L/Kcoincides with the map of Corollary 4.1.

Corollary 4.3. For a Galois extensionL/K with subfield K K L there arecommutative diagrams

CK

res

Gal(L/K)ab1

res

CK

cor

Gal(L/K)ab.1

cor

Proof. Since 1 is given by and res() = it suffices to show res(xy) =res(x)res(y) and the projection formula cor(xres(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 fori=2, we have homology groups,so the natural map is the corestriction map whereas the restriction is the oppositeor 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 extensionL/Kwith groupG. Hence it suffices to show that |H0(G, CL)|dividesn, or that H1(G, CL) = 0 in order to prove Theorem 3.2. By induction on


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62 M. FLACH

the number of prime factors of n it furthermore suffices to show this for primedegreen 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 fieldK := K(n) has degree[K :K] dividing n 1, hence prime to n, the field L :=LK is cyclic of degree nover K andG := Gal(L/K) is isomorphic to G by restriction. We have maps ofG-modules

CLCLNL/LCL

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

H1(G, CL)H1(G, CL)NK/KH1(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 ann-torsion group this implies H

1(G, CL) = H1(G, CL) =

0.Sincen

K, our fieldL = 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 letS bea finite set of places ofKsuch that

p| pS p|npS {p1, . . . ,pk} Swherep1, . . . ,pk= Cl(OK). vp(a)= 0pS

and define

LS,n:= K

n

OK,S

.

Clearly, LLS,n sincea OL,S. By the proof of Prop. 1.6 we also know that[LS,n: K] = n

s

where s =|S|. What we dont know at this point is whether L or LS,n are classfields. In fact we dont 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 extensionL/Kand modulusm 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. LetEL be a subextension sothatE/Kis cyclic. Then again all p msplit completely in E /Kand in particularKp =NEP/Kp(EP)

. So ifS={p|m}we havep/S

KppS

1NE/KAE

and by the weak approximation theorem K surjects onto

pS

K

p/NEP/Kp(EP)

,

i.e. p/S

1 pS

Kp K NE/KAE.


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So we have

H0(E/K,CE) = AK/(K

NE/KAE) =

p

Kp/(K NE/KAE) = 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. LetS Sbe a subset also satisfying the above four conditions andput

OK,S,n:=OK,SpS

(Kp)n.

Then one hasOK,S,n= (OK,S)n and therefore by (23)OK,S/OK,S,n

=OK,S/(OK,S)n

=ns.

Proof. The inclusionOK,S,n (OK,S)n is clear. Now take OK,S,n andconsider the cyclic extension E= K( n

). Then all pS are split completely in

E/Kand all p /S are unramified inE /K. SoAK,S :=

p/S

OKp pS

Kp NE/KAE.

By the fact that S contains generators of the ideal class group we have AK =K AK,S . Therefore

H0(E/K,CE) = AK/(K

NE/KAE) = K AK,S/(K NE/KAE) = 1which together with (37) for the cyclic extensionE /K impliesE= K. This means OK,S (K)n = (OK,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 inL/Kand we have t = s 1 and s = s +t = 2s 1. We thereforehave

U :=pS

(Kp)n

pS\S

Kpp/S

OKp NL/K(AL )

and

H0(G, CL) = AK/(K

NL/K(AL)) = K AK,S/(K NL/K(AL))has order dividing

[K AK,S :K U)] =[AK,S: U)]

[K AK,S :K U]=

pS [K

p : (K

p)n]

[OK,S:OK,S,n]

= n2s

ns = n2s

n2s1 =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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