TOPICS IN IWASAWA THEORY

Ralph Greenberg

December 15, 2006

1 Ideal class groups.

The ideal class group of a number field F is defined as the quotient groupClF = FF/PF , where FF denotes the group of fractional ideals of F and PFdenotes the subgroup of principal fractional ideals. It has been an object ofintense study since the nineteenth century. One of the fundamental theoremsof algebraic number theory is that ClF is a finite, abelian group. The factthat it is abelian is obvious, but the finiteness was first proved by Kummerfor the number field F = Q(µp), where p is any prime and µp denotes thegroup of p-th roots of unity.

If F is any number field, we denote the order of ClF by hF , the classnumber of F . If p is a prime, then ClF [p∞] denotes the p-primary subgroup

of ClF and h(p)F denotes its order. Iwasawa’s papers in the 1950s concern the

growth of h(p)Fn

where the Fn’s are a sequence of number fields such that

F = F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...

and Fn is a cyclic extension of F of degree pn for all n ≥ 0. Here p is a fixedprime. One of Iwasawa’s main theorems shows that there is some degree ofregularity in the behavior of h

(p)Fn

. We will discuss this and other theoremsof Iwasawa concerning ClFn [p

∞] in chapter 2. In the first three sections ofthis chapter, we just consider a single finite extension F ′/F . Although someof the results will be more general, the most interesting ones will concernthe case where F ′/F is a cyclic extension, especially a cyclic p-extension.These results will already show some close relationships between ClF [p∞] andClF ′ [p∞] under various hypotheses. Results of this kind were undoubtedlypart of the original inspiration behind Iwasawa’s work.

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The first two sections of this chapter discuss the kernels and images oftwo natural homomorphisms between the ideal class groups ClF and ClF ′ ofthose number fields:

NF ′/F : ClF ′ → ClF , JF ′/F : ClF → ClF ′ (1)

Class field theory provides the main tool for studying NF ′/F . Under rathermild assumptions, one can prove surjectivity. Under more stringent assump-tions, one can obtain some useful results about the kernel. The map JF ′/F ismore difficult to study. It involves the structure of the unit group O×

F ′ as aGalois module.

The third section concerns “genus theory.” which shows the influence ofthe ramified primes on dimFp(ClF ′ [p]) when F ′/F is a cyclic p-extension. Wewon’t attempt to prove the most general or precise results, just enough forcertain applications later on.

The fourth section concerns the so-called “reflection principle.”. We con-sider a number field F containing µp, where p is a prime, and a group ∆of automorphisms of F . It is assumed that p doesn’t divide |∆|. One canregard ClF [p] as a representation space for ∆ over the field Fp. It can bedecomposed as a direct sum of the irreducible representations of ∆ over Fp,each with a certain multiplicity. These irreducible representations occur inpairs in a certain way. The reflection principle shows that the correspondingmultiplicities for each such pair are somehow related. The idea is that onecan study cyclic, unramified extensions of F of degree p by both class fieldtheory and by Kummer theory.

The final topic in this chapter deals with a certain object which can beviewed as a generalization of the ideal class group, or, more precisely, thePontryagin dual of ClF [p∞]. It is defined as the subgroup of a Galois coho-mology group consisting of cocycle classes which are unramified at all primesof F . These groups are certainly closely related to ideal class groups, butover extensions of the field F . It is natural to ask how various results extendto these more general objects. Section 5 will discuss some general propertiesof these groups. Section 6 deals with an important special case associated toone dimensional representations of Gal(F (µp∞)/F ). The behavior of thesegroups is intimately related to classical Iwasawa theory, as we will explainnear the end of chapter 2. These last two sections will also help to makethe transition to the second half of this book, where we describe even morefar-reaching generalizations of the objects studied in the classical theory.

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1.1 The norm map.

Consider an arbitrary finite extension F ′/F of number fields. We first re-call the definition of two basic homomorphisms studied in algebraic numbertheory:

NF ′/F : F ′F → FF , JF ′/F : FF → FF ′

The second map JF ′/F is defined simply by mapping an element I ∈ FF toIOF ′ , the fractional ideal of F ′ generated by I. The map JF ′/F is injective,but not surjective if [F ′ : F ] > 1. To define the first map, it is sufficient todefine NF ′/F (P ′) for every prime ideal P ′ of F ′ since FF ′ is the free abeliangroup on the set of those prime ideals. For any such P ′, let P = P ′∩OF , theprime ideal of OF lying below P ′, and let f(P ′/P ) denote the residue fielddegree [OF ′/P ′ : OF/P ]. Then define

NF ′/F (P ′) = P f(P ′/P ) (2)

If [F ′ : F ] > 1, then the map NF ′/F is neither injective nor surjective.We will also let NF ′/F denote the norm map from F ′ to F as defined in

field theory. One basic result is that if α′ ∈ F ′ and α = NF ′/F (α′), then thecorresponding fractional ideals satisfy

NF ′/F (α′OF ′) = αOF (3)

and, consequently, we have the inclusion NF ′/F (PF ′) ⊂ PF . One can thendefine the map NF ′/F in (1) to be the homomorphism induced by NF ′/F onthe quotient groups ClF ′ and ClF . That is, if c′ ∈ ClF ′ and I ′ is any ideal inc′, then NF ′/F (c′) ∈ ClF is defined to be the class of the ideal NF ′/F (I ′).

The image of the map NF ′/F : ClF ′ → ClF is clearly the subgroup of ClFgenerated by the classes of the ideals P f(P ′/P ) (using the notation above).Most of our arguments in this section will be based on properties of theArtin isomorphism

ArtH/F : ClF → Gal(H/F ),

where H denotes the Hilbert class field of F . The existence of this isomor-phism is a special case of the Artin reciprocity law, and is discussed briefly inan appendix. We will just recall the definition of this map and the propertiesthat we will need.

If P is any prime ideal of F , let σP ∈ Gal(H/F ) denote the Frobeniusautomorphism for P (or, more precisely, for any one of the prime ideals of

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H lying above P ). Then we can define a homomorphism

FrobH/F : FF → Gal(H/F )

by putting FrobH/F (P ) = σP for every prime ideal P of F . As discussed inthe appendix, this map is surjective and its kernel is precisely PF . The Artinisomorphism is then defined by ArtH/F (c) = FrobH/F (I) for every c ∈ ClF ,where I is any element of c.

We first consider the image of the norm map. Surjectivity requires onlya mild assumption about F ′/F .

Proposition 1.1.1. If H ∩ F ′ = F , then NF ′/F : ClF ′ → ClF is surjective.

One immediate consequence is that if F ′ ∩ H = F , then hF divides hF ′ .Another consequence is that the map NF ′/F : ClF ′ [p∞] → ClF [p∞] will besurjective too. One can just assume that [F ′ ∩H : F ] is prime to p for thatassertion to hold.

Proof. The proof depends on the following commutative diagram from classfield theory:

ClF ′ −−−→ Gal(H ′/F ′)yNF ′/F

yRF ′/F

ClF −−−→ Gal(H/F )

(4)

Here H ′ denotes the Hilbert class field of F ′. Note that H ⊂ HF ′ ⊂ H ′. Theright vertical map is the restriction map g → g|H , where g ∈ Gal(H ′/F ′).The horizontal maps are the isomorphisms ArtH′/F ′ and ArtH/F . The leftvertical map is the norm map NF ′/F , as indicated.

The commutativity of (4) follows from the definitions. If P ′ is anyprime ideal of F ′ and P is the prime ideal of F lying below P ′, then letσP ′ ∈ Gal(H ′/F ′) and σP ∈ Gal(H/F ) denote the corresponding Frobeniusautomorphisms. Then, on the one hand, we have the following well-known

property: σP ′|H = σf(P ′/P )P . Comparing this with (2), we see that

FrobH′/F ′(I ′)|H = FrobH/F (NF ′/F (I ′))

for all I ′ ∈ FF ′ . The commutativity of (4) follows from this.The restriction map RF ′/F : Gal(H ′/F ′) → Gal(H/F ) is surjective be-

cause of the hypothesis that H ∩F ′ = F . The surjectivity of the map NF ′/F

then follows from the above commutative diagram. �

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One can give an alternative proof based on the Chebotarev density theo-rem, applied to any finite Galois extension of F containing HF ′. Under theassumptions of the proposition, one can show that if c ∈ ClF , then there ex-ist infinitely many prime ideals P ∈ c such that f(P ′/P ) = 1 for at least oneprime ideal P ′ of F ′ lying above P . If c′ is the class of P ′, then NF ′/F (c′) = c.

Without the assumption that F ′ ∩ H = F , the above proof shows thatNF ′/F (ClF ′) is precisely the inverse image of Gal(H/H ∩ F ′) under the mapArtH/F . Hence, if H ∩F ′ 6= F , then NF ′/F is not surjective. Also, if H ⊂ F ′,then NF ′/F is the zero-map.

Since it will be very useful in the next chapter, we state a corollary forthe p-primary subgroups of the class groups. We will use the notation

AF = ClF [p∞], AF ′ = ClF ′ [p∞] (5)

Let L denote the maximal p-extension of F contained in H, which we willrefer to as the p-Hilbert class field of F . Thus, we have a canonical isomor-phism ArtL/F : AF → Gal(L/F ). The following result is easily deduced fromproposition 1.1.1. Alternatively, there is a diagram just like (4) which givesthe result by the same argument.

Corollary 1.1.2. If L ∩ F ′ = F , then the map NF ′/F : AF ′ −→ AF issurjective.

Suppose that F ′/F is a p-extension and let G = Gal(F ′/F ). Let I1, ..., Irdenote the inertia subgroups of G for all the primes of F ′ which are ramifiedin the extension F ′/F . It is clear that L ∩ F ′ = F if and only if G isgenerated by those inertia subgroups. If one assumes that F ′/F is a cyclicp-extension, then G has a unique maximal subgroup, namely Gp (assumingthat [F ′ : F ] > 1). In that case, F ′ ∩ L = F if and only if Ij 6⊂ Gp for atleast one j, 1 ≤ j ≤ r. But this just means that Ij = G for at least one j,or, equivalently, that there exists at least one prime which is totally ramifiedin F ′/F .

Concerning the kernel of the norm map, diagram (4) shows that ker(NF ′/F )is just the inverse image under ArtH′/F ′ of ker(RF ′/F ), which is obviouslyGal(H ′/HF ′). The following result gives a simple description of this kernelunder certain stringent hypotheses on the extension F ′/F . It is actually aresult in the “genus theory” of cyclic extensions - a topic that we will pursuefurther in section 1.3. If F ′/F is a Galois extension, then we will continue todenote Gal(F ′/F ) by G. There is a natural action of G on ClF ′ , and so we

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can regard ClF ′ as a module for the group ring Z[G]. We will use a multi-plicative notation for the class group in this chapter, and so, if θ ∈ Z[G] andc′ ∈ ClF ′ , we will denote the action of θ on c′ by (c′)θ. Thus, the image ofClF ′ under θ will be denoted by ClθF ′ . We will switch to an additive notationin chapter 2.

Proposition 1.1.3. Suppose that F ′/F is a finite Galois extension andthat G = Gal(F ′/F ) is cyclic. Assume also that at most one prime of F isramified in F ′/F . Then

ker(NF ′/F ) = Clσ−1F ′

where σ denotes a generator of G.

Proof. The kernel of RF ′/F : Gal(H ′/F ′) → Gal(H/F ) is Gal(H ′/HF ′).The extension HF ′/F is clearly abelian. Let K denote the maximal abelianextension of F contained in H ′. Then HF ′ ⊆ K. Under the hypotheses ofthe proposition, we will first show that K = HF ′.

If no prime of F is ramified in F ′/F , then H ′/F is an unramified ex-tension and so we obviously have K = H = HF ′. If there is one prime vof F ramified in F ′/F , let I denote the corresponding inertia subgroup ofGal(K/F ) (which is the same for all primes of K lying above v). Then KI isthe maximal extension of F contained in K which is unramified at that oneprime, and therefore everywhere unramified. Thus we have KI = H and soI = Gal(K/H). But we also have that I ∩Gal(K/F ′) = 1, since K/F ′ is anunramified extension. This means that K = HF ′, as stated.

Now we can describeK in another way. Note thatH ′ is a Galois extensionof F . This is easy to verify just using the definition of the Hilbert class field,and is left to the reader. Let G = Gal(H ′/F ) and N = Gal(H ′/F ′). We thenhave an exact sequence

1→ N → G→ G→ 1

Since N is abelian, there is a natural action of G on N . Let σ be as above,a generator of G. Choose an element σ ∈ G such that σ|F ′ = σ. If η ∈ N ,then σ acts on η as follows: ησ = σησ−1. Considering N as a Z[G]-module(for which we will use an exponential notation), we then have that ησ−1 =

σησ−1η−1. This is a commutator in G, and so we have Nσ−1 ⊂ D(G), where

D(G) denotes the commutator subgroup of G. In fact, we have

D(G) = Nσ−1

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To see this, note that Nσ−1 is a normal subgroup of G. Consider the exactsequence

1→ N/Nσ−1 → G/Nσ−1 → G/N → 1

Clearly, N/Nσ−1 is contained in the center of G/Nσ−1. The quotient group

G/N is cyclic. It follows that G/Nσ−1 is abelian. Hence D(G) ⊂ Nσ−1, andso the two subgroups do coincide.

By definition, K is the subfield of H ′ corresponding to D(G), and so wehave

Gal(H ′/K) = Nσ−1

We have shown before that K = HF ′ and so we also have

Gal(H ′/K) = ker(Gal(H ′/F ′)→ Gal(H/F )

)= ker(RF ′/F )

Since ArtH′/F ′ : ClF ′ → N is a G-equivariant isomorphism, the commutativediagram (4) then implies that the kernel of the norm mapNF ′/F : ClF ′ → ClFis precisely Clσ−1

F ′ , as asserted. �

Now we turn to the important case where F ′/F is a p-extension. Thefollowing result is a consequence of propositions 1.1.1 and 1.1.3 and will be afirst and quite useful step for the results in the next chapter. Its proof pro-vides a simple illustration of some of the ideas which play a role in studyingthe behavior of ideal class groups in Zp-extensions. It asserts that, undercertain stringent assumptions, AF ′ 6= 1⇐⇒ AF 6= 1.

Proposition 1.1.4. Let p be a prime. Suppose that F ′/F is a Galois exten-sion and that G = Gal(F ′/F ) is a cyclic p-group. Assume also that exactlyone prime of F is ramified in F ′/F and that this prime is totally ramified.Then p divides the class number of F ′ if and only if p divides the class numberof F .

Proof. Proposition 1.1.1 implies that hF divides hF ′ . This makes one partof the above proposition obvious: If p divides hF , then p divides hF ′. Toprove the other part, we study the norm map NF ′/F : AF ′ → AF . We knowthat this map is surjective by corollary 1.1.2. Proposition 1.1.3 determinesits kernel because AF ′ is a direct summand of ClF ′ as a Z[G]-module. Thatkernel is Aσ−1

F ′ , where σ again denotes a generator for G. Therefore we obtainan isomorphism

AF ′/Aσ−1F ′ −→ AF (6)

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The other part of the proposition is easily deduced from (6). Assumethat hF ′ is divisible by p. Then the p-group G is acting on the nontrivialp-group AF ′ and therefore the subgroup of elements fixed by the action of Gwill also be nontrivial. That is, if we consider σ− 1 as the endomorphism ofAF ′ defined by mapping a ∈ AF ′ to aσ−1 = σ(a)a−1, then its kernel will benontrivial. It follows that the image Aσ−1

F ′ of that endomorphism is a propersubgroup of AF ′ . Hence, (6) implies that AF must be nontrivial. Therefore,hF is indeed divisible by p. �

Remark 1.1.5. Our proof of the above proposition uses the cyclicity of G.However, it suffices to assume that G is a p-group. This follows easily fromthe case where F ′/F is cyclic of degree p, using the fact that the compositionfactors for a finite p-group are cyclic of order p. The ramification assump-tion for F ′/F implies that the same assumption holds for each intermediateextension. However, one can also give the following direct proof.

Suppose that F ′/F is a p-extension in which exactly one prime is ramified.We assume that this prime is totally ramified in F ′/F . The assertion thatp|hF =⇒ p|hF ′ is again obvious from corollary 1.1.2, and so we just considerthe converse. We will show that if AF ′ 6= 1, then AF 6= 1. Let L′ denotethe p-Hilbert class field of F ′. Clearly, L′/F is Galois, and Gal(L′/F ) is ap-group. Now assume that AF ′ 6= 1 and so [L′ : F ′] > 1. Let P be theunique ramified prime in F ′/F , let Q denote any prime of L′ lying aboveP , and let IQ denote the inertia subgroup of Gal(L′/F ) for Q, which isdetermined by P up to conjugacy. Since e(Q/P ) = [F ′ : F ] < [L′ : F ], IQis a proper subgroup of Gal(L′/F ). Thus there exists a maximal subgroupM of Gal(L′/F ) which contains IQ. Maximal subgroups of a p-group haveindex p and are normal. The nontrivial inertia subgroups of Gal(L′/F ) areconjugate to IQ and hence also contained in M . Therefore, the fixed field(L′)M is an unramified extension of F and Gal((L′)M/F ) ∼= Z/pZ. Thisproves that AF 6= 1.

Remark 1.1.6. Suppose that p = 2 and that the assumptions in proposition1.1.4 are satisfied. If [F ′ : F ] = 2, then the unique prime v of F which isramified in the extension F ′/F could be an infinite prime, which would thenbe totally ramified. In that case, if v′ denotes the prime of F ′ lying above v,then the hypothesis means that Fv = R, F ′

v′ = C, and that F ′ is a quadraticextension of F which is unramified at all other primes of F , finite or infinite.If one assumes only that exactly one finite prime of F is ramified in F ′/F , and

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that this prime is totally ramified, then the argument can easily be adapted toprove the analogous statement for the “strict” class numbers of F and F ′, i.e.that they are either both even or both odd. Recall that the strict class groupClstrF of a number field F is the quotient group FF/P tpF , where P tpF denotes thegroup of principal fractional ideals which are generated by a totally positiveelement of F . (An element α ∈ F is totally positive if its image underevery embedding F → R is positive.) If Hstr denotes the maximal abelianextension of F unramified at all the finite primes of F , then one can use thecorresponding Artin isomorphism ArtHstr/F : ClstrF → Gal(Hstr/F ) to provethe analogues of 1.1.1 - 1.1.4.

Remark 1.1.7. One can extract some information about the structure ofAF ′ in the situation of proposition 1.1.4. As an illustration, let us assume thatF ′/F is cyclic of degree p and that |AF | = p, in addition to the ramificationassumption. For brevity, let τ = σ−1, considered as an endomorphism ofAF ′ .According to (6), AF ′/AτF ′ will then be cyclic of order p. Let |AF ′ | = pm,where m ≥ 1. Then one can easily show that each of the subquotientsAτ

j

F ′/Aτj+1

F ′ will also be cyclic of order p for 0 ≤ j ≤ m − 1 and that Aτm

F ′ istrivial. One can regard τ as an element of the group ring for G over Z orover Fp. In Fp[G], one can easily verify that τ p = 0. Hence, in Z[G], we haveτ p ∈ pZ[G]. It follows that Aτ

p

F ′ ⊆ ApF ′ . This implies that

dimFp

(AF ′ [p]

)= dimFp

(AF ′/ApF ′

)≤ p.

One possibility is that dimFp

(AF ′ [p]

)= 1. Thus AF ′

∼= Z/pmZ. Theautomorphism group of AF ′ is then isomorphic to (Z/pmZ)×. Thus σ(a) = as

for all a ∈ AF ′ , where s is an integer not divisible by p. The order of s modulopm must be 1 or p. Thus sp ≡ 1 (mod pm). Now we have

AτF ′ = As−1F ′ = ApF ′

which means that either m = 1 (and so G acts trivially on AF ′) or m > 1and p||(s− 1). Thus, if m > 1 and p is odd, then it follows that p2||(sp − 1)and therefore m = 2. Hence, for odd p, we must have m ≤ 2. This kind ofargument gives no information if p = 2. Indeed, it is conceivable that AF ′ iscyclic of order 2m, for any m ≥ 1, and that σ(a) = a−1 for a ∈ AF ′ .

Another possibility is that AF ′ is an elementary abelian p-group. Thenm = dimFp(AF ′) and we have 1 ≤ m ≤ p. The endomorphism τ is a nilpotentlinear mapping on the Fp-vector space AF ′ . Suppose that c′ is in AF ′ , but

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not in AτF ′ . Then it is clear that c′ generates AF ′ as a module over the groupring Fp[G]. One then has an isomorphism

AF ′∼= Fp[G]/(τm)

of Fp[G]-modules.One gets a better picture of the possibilities by considering AF ′ as a

module over the group ring Zp[G]. Choosing c′ ∈ AF ′ as above, AF ′ isgenerated by c′, (c′)τ , (c′)τ

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, ... as a group and hence c′ is a generator for AF ′

as a Zp[G]-module. That is, AF ′ is a cyclic Zp[G]-module. Therefore, wehave AF ′

∼= Zp[G]/I, where I is an ideal in Zp[G]. The requirement that[AF ′ : AτF ′ ] = p just imposes a simple condition on I, namely that the imageof I in Zp[G]/(τ)Zp[G] ∼= Zp has index p. There are actually infinitely manysuch ideals and, although one can get strong restrictions on the structure ofAF ′ from this point of view, one cannot bound the index of I. Thus, evenunder the stringent assumptions we have made about AF and F ′/F , therewould seem to be no bound on |AF ′ | in general.

1.2 The map JF ′/F .

The map JF ′/F is induced from the natural homomorphism

JF ′/F : FF → FF ′

defined in section 1. It is obvious that JF ′/F (PF ) ⊂ PF ′ . We obtain ahomomorphism from ClF to ClF ′ as follows. If c ∈ ClF and I is an ideal inc, define JF ′/F (c) to be the ideal class in ClF ′ represented by JF ′/F (I). SinceJF ′/F (PF ) ⊂ PF ′ , the map JF ′/F is well-defined. The map JF ′/F is easilyseen to be injective, but JF ′/F can have a nontrivial kernel and this can bequite difficult to study.

The kernel of JF ′/F has been studied rather extensively in the case whereF ′/F is an unramified abelian extension (i.e., F ′ ⊂ H, where H denotes theHilbert class field of F ). Here are a few of the known results:

1. If F ′/F is a cyclic unramified extension of degree p, then ker(JF ′/F ) isnontrivial.

2. Ker(JH/F ) = ClF .

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3. If F ′/F is any abelian, unramified extension, then |ker(JF ′/F )| is divisibleby [F ′ : F ].

The first result is known as “Hilbert’s Theorem 94.” We will prove this below.The second is the famous “Principal Ideal Theorem.” As one consequence,it is easy to show (using proposition 1.2.1 below) that if L is the p-Hilbertclass field of F , then ker(JL/F ) = AF , the p-primary subgroup of ClF . Thethird result is a generalization of both (1) and (2) proved in 1992 by Suzuki.

There are no really general results for the case where F ′/F is ramified.Later in this section we will give some interesting examples of ramified cyclicextensions F ′/F of degree p such that ker(JF ′/F ) is nontrivial.

Except for proposition 1.2.1, all the results in this section will concern afinite Galois extension F ′/F . We will always let G = Gal(F ′/F ). Our firsttwo results are quite simple, based just on the definitions.

Proposition 1.2.1. Let n = [F ′ : F ]. Then

(NF ′/F ◦ JF ′/F )(c) = cn

for all c ∈ ClF . Consequently, if c ∈ ker(JF ′/F ), then the order of c dividesn. In particular, if (hF , n) = 1, then JF ′/F is injective.

Proof. The proposition follows immediately from the identity

NF ′/F (JF ′/F (I)) = In. (7)

which holds for all I ∈ FF . It suffices to verify this identity if I is a primeideal P of F . But, in that case, using (2), the identity amounts to the familiarfact that ∑

P ′|Pe(P ′/P )f(P ′/P ) = n

where the sum runs over all the prime ideals P ′ of F ′ lying above P , e(P ′/P )denotes the corresponding ramification index, and f(P ′/P ) is the residuefield degree defined before. Alternatively, one can easily verify the identityif I ∈ PF . It then follows for any I since FF is torsion-free and the index[FF : PF ] is finite. �

Proposition 1.2.2. Suppose that F ′/F is a finite Galois extension. Con-sider the mapping NG : ClF ′ → ClF ′ defined by NG(c′) =

∏σ∈G σ(c′). Then

JF ′/F ◦NF ′/F = NG (8)

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In particular, if F ′ ∩H = F , then im(JF ′/F ) = im(NG).

Proof. Suppose that I ′ ∈ FF ′ and let I = NF ′/F (I ′). Then we have

IOF ′ =∏

σ∈Gσ(I ′) (9)

It suffices to verify (9) if I ′ is a prime ideal P ′ of F ′, which is straightforward.Alternatively, one can first consider the case where I ′ ∈ PF ′ . Suppose thatI ′ = α′OF ′ . Then

NF ′/F (α′) =∏

σ∈Gσ(α′)

and the identity (9) then follows from (3). One can prove (9) for any I ′ ∈ FF ′

by using the facts that FF ′ is torsion-free and that PF ′ has finite index.Therefore, if c′ denotes the class of I ′ in ClF ′ , then it follows that the

images of c′ under the maps JF ′/F ◦NF ′/F and NG are the same, namely justthe class of IOF ′ in ClF ′ . Finally, if F ′ ∩ H = F , then proposition 1.1.1shows that NF ′/F (ClF ′) = ClF . It then follows that JF ′/F (ClF ) = NG(ClF ′)as stated. �

If F ′/F is a Galois extension, then ker(JF ′/F ) is somehow related to theway G = Gal(F ′/F ) acts on the unit group O×

F ′ of F ′. This comes from thefollowing observation: Suppose that c ∈ ker(JF ′/F ) and that I is an ideal inc. That is, I ∈ FF and it generates a principal fractional ideal in F ′. Let α′

be a generator for I ′ = IOF ′ . The ideal I ′ is invariant under the action of Gand so it is clear that the map

φ : G→ O×F ′

defined by φ(g) = g(α′)/α′ defines a 1-cocycle on G with values in O×F ′ . Its

cocycle class [φ] is determined by c, the map c → [φ] is a homomorphism,and if [φ] is trivial, then it is easy to see that I ∈ PF and hence c is trivial.Thus, in this way, one defines an injective homomorphism

ker(JF ′/F ) −→ H1(F ′/F,O×F ′).

However, principal ideals I ′ which are invariant under the action of G mayalso arise as products of ramified primes. Such ideals also define a cocycleclass in H1(F ′/F,O×

F ′), exactly as above. These observations are behind the

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following useful result. To simplify the notation, we will identify PF and FFwith their images under the injective homomorphism JF ′/F .

Proposition 1.2.3. Suppose that F ′/F is a finite Galois extension. Wehave an exact sequence

0 −→ ker(JF ′/F ) −→ PGF ′/PF −→ FGF ′/FF −→ ClGF ′/JF ′/F (ClF )

Furthermore, we have isomorphisms

PGF ′/PF ∼= H1(F ′/F,O×F ′), FGF ′/FF ∼=

t∏

i=1

Z/eiZ

where t denotes the number of primes of F which are ramified in F ′/F ande1, ..., et denote the corresponding ramification indices.

Proof. The exact sequence results by applying the snake lemma to the fol-lowing commutative diagram

1 −−−→ PF −−−→ FF −−−→ ClF −−−→ 1yy

yJF ′/F

1 −−−→ PGF ′ −−−→ FGF ′ −−−→ ClGF ′

(10)

The first two vertical maps are injective, and so one obtains an injective map:ker(JF ′/F ) → PGF ′/PF . Explicitly, this map can be defined as follows: If anideal I of F becomes principal in F ′, IOF ′ = (α′), say, then its ideal class(which is in ker(JF ′/F )) is mapped to the coset of (α′) in PGF ′/PF .

Consider the exact sequence 1→ O×F ′ → F ′× → PF ′ → 1, where the map

F ′× −→ PF ′ is defined by mapping an element of F ′× to the principal idealit generates. The maps are G-equivariant and so one obtains the followingexact sequence of Galois cohomology groups (mostly H0’s).

1 −→ O×F −→ F× −→ PGF ′ −→ H1(F ′/F,O×

F ′) −→ 1

To justify the 1 at the end, we use Hilbert’s theorem 90 which states thatH1(F ′/F, (F ′)×) is trivial. The isomorphism PGF ′/PF ∼= H1(F ′/F,O×

F ′) fol-lows immediately because the image of F× in PGF ′ is JF ′/F (PF ) (which we aredenoting by PF ). This isomorphism is just as mentioned before: If (α′) ∈ PGF ′ ,then one maps it to the class of the 1-cocycle g → g(α′)/α′.

13

It remains to show that FGF ′/FF is isomorphic to∏t

i=1 Z/eiZ. To see this,let I ′ be any fractional ideal of F ′. Then I ′ ∈ FGF ′ if and only if the primeideal factorization of I ′ satisfies the following condition: primes which areconjugate under the action of G occur with the same exponent. This meansthat I can be represented uniquely as a product (with integer exponents) ofthe ideals

QP =∏

P ′|PP ′

Note that JF ′/F (P ) = POF ′ = QePP , where eP denotes the ramification index

of P in F ′/F . Thus, we can regard FGF ′ as the free abelian group generatedby the ideals QP , and FF then corresponds to the subgroup generated by theideals QeP

P . It is therefore clear that the quotient group is indeed isomorphicto

∏ti=1 Z/eiZ. �

As we will now show, Dirichlet’s unit theorem has some implicationsconcerning the cohomology group H1(F ′/F,O×

F ′). We will assume that F ′/Fis a cyclic extension. One can regard O×

F ′ as a Z[G]-module. Its structure isdifficult to study. However, the vector space

VO×

F ′

= O×F ′ ⊗Z Q

can be regarded as a Q[G]-module and its structure can be completely deter-mined. In particular, this will allow us to determine the “Herbrand quotient”which is defined by

h(F ′/F,O×F ′) = |H2(F ′/F,O×

F ′)|/|H1(F ′/F,O×

F ′)|

The next result shows that, under certain assumptions, this ratio is 1/[F ′ : F ].

Proposition 1.2.4. Suppose that F ′/F is a cyclic extension of degree n. Ifn is even, assume that the real primes of F are unramified in F ′/F . Then

|H1(F ′/F,O×F ′)| = n|H2(F ′/F,O×

F ′)|

There is a surjective homomorphism O×F

/(O×

F )n −→ H2(F ′/F,O×F ′). In par-

ticular, if F ′/F has prime degree p, then

1 ≤ dimFp(H1(F ′/F,O×

F ′)) ≤ s+ 1

where s = dimFp

(O×F /(O×

F )p).

14

Proof. The proof depends on properties of the Herbrand quotient which wenow recall. We assume that G is a finite cyclic group which acts on a finitelygenerated, abelian group L. Thus, L is a finitely generated Z[G]-module.Then the cohomology groups H i(G,L) for i ≥ 1 are finite. The Herbrandquotient h(G,L) = |H2(G,L)|

/|H1(G,L)| has the following properties:

(1) If L is finite, then h(G,L) = 1.

(2) Suppose that 0 → L1 → L → L2 → 0 is an exact sequence of finitelygenerated Z[G]-modules. Then h(G,L) = h(G,L1)h(G,L2).

Consider the Q-vector space V = L⊗ZQ, which is a representation spaceover G over Q of dimension d = rankZ(L). One can regard L/Ltors as a Z-lattice in V which is invariant under the action of G. One deduces from (1)and (2) that h(G,L) depends only on the isomorphism class of V , not onthe choice of the G-invariant Z-lattice L. This observation together with thefollowing lemma will make it easy to compute h(F ′/F,O×

F ′).

Lemma 1.2.5. Suppose that F ′/F satisfies the assumptions in the aboveproposition. Let W denote the regular representation for G over Q, letWo = WG, which is the trivial representation (of dimension 1), and letW1 = W/Wo. Let r = rankZ(O×

F ). Then

VO×

F ′

∼= W ro ×W r+1

1

as representation spaces for G.

Proof. One fact that we will use is that a cyclic group of order m has a uniquefaithful, irreducible representation defined over Q. Its dimension is φ(m). Asa consequence of this, one can easily see that if G is a finite cyclic group,then a representation space for G over Q is determined (up to isomorphism)by the quantities dimQ(V H), where H varies over all the subgroups of G.

Consider V = VO×

F ′

. If H is any subgroup of G, let E = F ′)H . Then

dimQ(V H) is equal to the rank of the group of units of E since O×E = (O×

F ′)H ,and so O×

E ⊗Q ∼= V H . Let r1 denote the number of real primes of F , r2 thenumber of complex primes of F . Then r = r1 + r2− 1. The real primes of Fare unramified in F ′/F , and hence in E/F . This is obvious if n is odd andis true by assumption if n is even. Thus, if m = [E : F ], then

dimQ(V H) = mr1 +mr2 − 1 = r + (m− 1)(r + 1)

On the other hand, we have dimQ(WHo ) = 1 and dimQ(WH

1 ) = m−1. Hence,(W r

o ×W r+11 )H also has dimension r+(m−1)(r+1), which implies the stated

isomorphism. �

15

The above lemma is valid without the assumption that G is cyclic. If Gis an arbitrary finite group G, then it is still true that a representation spaceV for G over Q is determined by the quantities dimQ(V H). This followsfrom the fact that the isomorphism class of V is determined by its characterwhich, in turn, is determined just by its restrictions to all cyclic subgroupsof G. Those restrictions are determined by the dimQ(V H)’s. It suffices toknow these quantities for all cyclic subgroups H. The proof then proceedsexactly as above.

Returning to the proof of proposition 1.2.4, we can compute h(F ′/F,O×F ′)

by considering any G-invariant Z-lattice in V . According to lemma 1.2.5, wecan choose such a lattice L so that

L ∼= Lro × Lr+11

where Lo = Z, with a trivial action of G, and L1 = IG, the augmentationideal in the group ring Z[G]. The trivial homomorphism G → {1} definesa ring homomorphism Z[G] → Z and IG is defined to be the kernel of thathomomorphism. Now

H1(G,Z) = 0, H2(G,Z) ∼= Z/nZ

and so h(G,Eo) = n. It is obvious that h(G,Z[G]) = 1. The exact sequence

0 −→ IG −→ Z[G] −→ Z −→ 0

shows that h(G,Eo)h(G,E1) = 1, and so we have h(G,E1) = 1/n. This canalso be verified directly.

These observations imply that h(F ′/F,O×F ′) = nr/nr+1 = 1/n, which is

precisely the first statement in the proposition. For the second part, notethat

H2(F ′/F,O×F ′) ∼= O×

F /NF ′/F (O×F ′)

and (O×F ′)n ⊂ NF ′/F (O×

F ′) ⊂ O×F ′ . To prove the final statement, note that if

n = p is prime, then both H1(F ′/F,O×F ′) and H2(F ′/F,O×

F ′) have exponentp, and so their orders determine their dimensions as Fp-vector spaces. Thus,h(F ′/F,O×

F ′) = p implies that dimFp

(H1(F ′/F,O×

F ′))

is bounded above bydimFp

(H2(F ′/F,O×

F ′))

+ 1, which in turn is bounded above by s+ 1. �

Remark 1.2.6. Suppose that F ′/F is any cyclic extension of degree p,where p is an odd prime. Proposition 1.2.1 implies that if c ∈ ker(JF ′/F ),

16

then cp = 1ClF , and so ker(JF ′/F ) is an Fp-vector space. Propositions 1.2.3and 1.2.4 give the following inequality:

dimFp

(ker(JF ′/F )

)≤ s+ 1

where s is as in proposition 1.2.4. Explicitly, we have s = r if µp 6⊂ F ands = r + 1 if µp ⊂ F . Thus, we have a simple bound on |ker(JF ′/F )| just interms of the rank r of the unit group of F .

Now suppose that p = 2, i.e., that F ′/F is any quadratic extension. Theinequality for dimFp

(ker(JF ′/F )

)given above is still valid. It can even be

improved if some of the infinite primes of F are ramified in F ′/F . Supposethat t∞ infinite primes of F are ramified in F ′/F . In lemma 1.2.5, one thenhas V ∼= W r

o ×W r+1−t∞1 and so the Herbrand quotient for the G-module O×

F ′

turns out to beh(F ′/F,O×

F ′) = 2r/2r+1−t∞ = 2t∞−1

Thus,

dimFp

(ker(JF ′/F )

)≤ dimFp

(H1(F ′/F,O×

F ′))≤ s+ 1− t∞

Note that if p = 2, then s = r + 1 = r1 + r2. If F is totally real and F ′

is totally complex, then t∞ = r1 and r2 = 0, in which case one finds thatdimFp

(H1(F ′/F,O×

F ′))≤ 1.

We now discuss other implications of the above propositions in variousspecial cases.

Remark 1.2.7. Assume that F ′/F is an unramified Galois extension. Thenproposition 1.2.3 implies that ker(JF ′/F ) ∼= H1(F ′/F,O×

F ′). If we assume inaddition that F ′/F is a cyclic extension, then proposition 1.2.4 implies thatker(JF ′/F ) has order divisible by n = [F ′ : F ], and hence is nontrivial ifn > 1. Hilbert’s theorem 94 is a consequence. This argument is essentiallythe same as Hilbert’s original proof.

Remark 1.2.8. Consider a Galois extension F ′/F of degree n and a primep such that p ∤ n. As before, we let AF = ClF [p∞], AF ′ = ClF ′ [p∞]. Wewill consider the maps NF ′/F and JF ′/F just on those subgroups. Proposition1.2.1 implies that the composite map

AFJF ′/F−→ AF ′

NF ′/F−→ AF

17

is the isomorphism a → an for a ∈ AF . Therefore, JF ′/F is injective andJF ′/F (AF ) is a direct factor in the Z[G]-module AF ′ isomorphic to AF . Moreprecisely, we have

AF ′∼= JF ′/F (AF )× ker

(NF ′/F : AF ′ → AF

)

as Z[G]-modules. Now G acts trivially on the first factor JF ′/F (AF ). Forthe second factor, we have ker

(NF ′/F : AF ′ → AF

)= ker

(NG : AF ′ → AF ′

),

which we denote by M . It is clear that M satisfies MG = 1, MG = 1. Hencewe have

AF ∼= AGF ′ , (AF ′)G ∼= AF

where the first isomorphism is induced by JF ′/F and the second is inducedby NF ′/F .

Remark 1.2.9. Assume that F ′/F is a cyclic p-extension and that exactlyone prime of F is ramified in F ′/F (which we will assume to be a finiteprime if p = 2). Let σ be a generator of G = Gal(F ′/F ). According to (6)in the proof of proposition 1.1.4, if we regard τ = σ− 1 as an endomorphismof AF ′ , then coker(τ) ∼= AF . Now ker(τ) = AGF ′ has the same order ascoker(τ). Thus, under the above assumptions, AF and AGF ′ have the sameorders. Therefore, if JF ′/F happens to be injective, then we must have AGF ′ =JF ′/F (AF ).

Under the same assumptions, one can instead apply proposition 1.2.3 toprove that equality. We have t = 1 and therefore

dimFp

(H1(F ′/F,O×

F ′))− dimFp

(ker(JF ′/F )

)= 0 or 1

If we make the assumption that JF ′/F is injective, then H1(F ′/F,O×F ′) would

be cyclic of order p, H2(F ′/F,O×F ′) would be trivial, and the map

PGF ′/PF → FGF ′/FF

would be surjective. The last statement means that FGF ′ = FFPGF ′ . Accordingto a result to be proved in the next section (proposition 1.3.4), the vanishingof H2(F ′/F,O×

F ′) implies that every class in ClGF ′ contains an ideal in FGF ′ .Thus, in a different way, we again see that JF ′/F (AF ) = AGF ′ under theassumption that JF ′/F is injective.

The above argument does not require class field theory and gives anotherproof of part of proposition 1.1.4, namely the implication: p | hF ′ ⇒ p | hF .

18

For if p | hF ′ , then AGF ′ will be nontrivial. However, ker(JF ′/F ) ⊆ AF andso, if AF is trivial, then JF ′/F would be injective and hence AGF ′ = JF ′/F (AF )would be trivial too. Therefore, it must be that p|hF .

Remark 1.2.10. This remark should be compared with remark 1.1.7. Incontrast, under certain assumptions, we will obtain a lower bound on A′

F

instead of an upper bound. We will assume that F ′/F is a cyclic extensionof degree p in which at least one prime of F is ramified, that p|hF , andthat the map JF ′/F is injective. Proposition 1.2.2 would then imply thatNG(AF ′) ∼= AF . Now it is obvious that

AGF ′ [p] ⊆ ker(NG

∣∣AF ′

)

and, since p|hF , it is clear that AGF ′ [p] 6= 1. Hence the map NG

∣∣AF ′

has a

nontrivial kernel. It follows that |AF ′| > |AF |. This growth could be either“vertical” or “horizontal”. We will discuss two extreme cases. As in remark1.1.7, we will illustrate the idea by assuming that |AF | = p. Let |AF ′| = pm.As we’ve just explained, we have m ≥ 2.

First assume that AF ′ is a cyclic group. Thus, AF ′∼= Z/pmZ. If p is odd,

then the automorphism group of AF ′ has only one subgroup of order p. Theaction of G may be trivial or through that subgroup. In either case, one findsthat [AF ′ : NG(AF ′)] = p. Hence we must have m = 2 if p is odd. However,for p = 2, we can’t prove anything more than the inequality m ≥ 2.

Now assume that AF ′ is an elementary abelian p-group. Then one wouldhave dimFp(AF ′) ≥ p. To see this, we will use some elementary facts aboutthe group ring Fp[G], where G is a cyclic group of order p. As before, we letτ = σ − 1, where σ is a generator of G. The ideal (τ) of Fp[G] is maximalwith residue field Fp. We have τ p = 0. The distinct proper ideals of Fp[G]are (τ i), where 1 ≤ i ≤ p. It follows that

(τ p−1) = Fp[G]G = (NG), where NG =∑

g∈Gg

Thus, if M is an Fp[G]-module such that dimFp(M) = i < p, then τ i annihi-lates M and hence so does NG. Thus, if NG(M) 6= 0, then dimFp(M) ≥ p.We can just apply this fact to M = AF ′ . Our assumptions imply that|NG(AF ′)| = p.

Remark 1.2.11. We will describe two interesting examples where F ′/F isa ramified cyclic extension of degree p and JF ′/F has a nontrivial kernel, one

19

rather subtle, the other rather straightforward. We take F = Q(µp) andassume that p||hF . It is known that the divisibility p|hF holds for infinitelymany primes p (the so-called “irregular primes). The exact divisibility p||hFprobably holds for infinitely many p’s, but this is not known. The firstirregular prime is p = 37 which does indeed satisfy the assumption. The firstprime for which p2|hF is p = 157. Recall that p is totally ramified in F/Q.Let P denote the unique prime ideal of F lying over p.

Example 1. Consider F ′ = F ( p√p), an extension of F of degree p which is

ramified just at P . By proposition 1.1.1, we know that p|hF ′ . For all primesp < 1000 satisfying the assumption that p||hF , it turns out that p2 ∤ hF ′ . Wecannot explain this here. It is a consequence of a rather difficult calculationdue to McCallum and Sharifi. Actually, it seems reasonable to believe thatthis same statement will be true for any prime p satisfying p||hF . And so,let us assume that we have |AF ′ | = |AF | = p in the rest of this remark. Themap NF ′/F : AF ′ → AF , which is certainly surjective, would then be anisomorphism. As pointed out in remark 1.2.10, it follows that JF ′/F has anon-trivial kernel. In fact, it is clear that ker(JF ′/F ) = AF in this example.

Example 2. This is a much simpler example. Let I be a nonprincipal idealwhose class c ∈ ClF has order p. Then Ip = (α), where α ∈ F×. Nowwe let F ′ = F ( p

√α). Let I ′ = JF ′/F (I). Then I ′ = p

√αOF ′ since both

ideals have the same p-th power. Hence c ∈ ker(JF ′/F ). We again haveker(JF ′/F ) = AF . Note that the field F ′ just defined is a cyclic extension ofF of degree p, but is not uniquely determined by the class c, or even by theideal I. For example, one can choose a different generator αη for the idealIp, where η ∈ O×

F . Under our assumptions, F has only one unramified, cyclicextension of degree p, namely the p-Hilbert class field L of F , and so it isclear that we can obtain a ramified extension F ′/F in this way. One seeseasily that the only prime that can be ramified is P . We will return to thiskind of example in section 4, showing that one can arrange for F ′ = F ( p

√α)

to be Galois over Q. Of course, it is easy to verify that L is Galois over Q.As we will then see, the Galois groups Gal(F ′/Q) and Gal(L/Q) will havedifferent structures, making it obvious that F ′ 6= L.

The final results in this section concern an important special class offields.

Definition 1.2.12. An algebraic extension F of Q is called a CM-field if F istotally complex and contains a totally real subfield F+ such that [F : F+] = 2.

20

The simplest examples of CM-fields are complex, abelian extensions of Q.For example, let m ≥ 3 and let ζm denote a primitive m-th root of unity.Then F = Q(ζm) is a CM-field and F+ = Q(ζm + ζ−1

m ) is its maximal totallyreal subfield. The letters CM stand for “complex multiplication,” referringto the fact that the endomorphism ring of an abelian variety with complexmultiplication is an order in a CM field. In particular, the endomorphism ringof an elliptic curve E is either just Z or an order in an imaginary quadraticfield. In the latter case, we say that E has complex multiplication.

Let ∆ = Gal(F/F+), a group of order 2. Thus, ∆ = {1, δ}, where δdenotes complex conjugation. (To be more precise, δ is the automorphismof F obtained by choosing any embedding F → C and restricting complexconjugation to the image of F .) The group ∆ has two characters: the trivialcharacter ǫ0, and the nontrivial character ǫ1. Let ǫ denote either of thesetwo characters. Suppose that A is an abelian group and that ∆ acts onA. We let A(ǫ) denote the maximal subgroup of A on which ∆ acts by thecharacter ǫ. That is, A(ǫ0) = A∆ and A(ǫ1) is the kernel of the endomorphismN∆ = 1 + δ. If we assume that A is finite and has odd order, then it iseasy to see that we have the direct product decomposition A ∼= A(ǫ0)×A(ǫ1).This is also true just under the assumption that A is a torsion group and hasodd exponent. In general, it is clear that A(ǫ1) ∩ A(ǫ0) = A[2]∆, and this willbe nontrivial precisely when A[2] is nontrivial. Also, it is easy to see that2A ⊆ A(ǫ0) + A(ǫ1).

Suppose that F ′/F is a finite Galois extension and that both F and F ′

are CM-fields. Then one can show that F ′+/F+ is Galois and that F ′ = FF ′

+.Thus Gal(F ′/F ′

+) can be identified with ∆ = Gal(F/F+) and then we have

Gal(F ′/F+) ∼= ∆×GThus, both ∆ and G act on the groups ClF ′ , O×

F ′ , PF ′ , and FF ′ , andthe actions commute with each other. There is also an action of ∆ onH1(F ′/F,O×

F ′). All of the maps and isomorphisms in proposition 1.2.3 are∆-equivariant.

One useful consequence of this is the following result.

Proposition 1.2.13. Suppose that F ′/F is a finite Galois extension, thatboth F and F ′ are CM-fields, and that n = [F ′ : F ] is odd. Let µF ′ denotethe group of roots of unity in F ′. Then

H i(F ′/F,O×F ′)

(ǫ1) ∼= H i(F ′/F, µF ′)

21

for i ≥ 0. If n is even, then there is a homomorphism

H i(F ′/F, µF ′) −→ H i(F ′/F,O×F ′)

(ǫ1)

whose kernel and cokernel are of exponent 2.

Proof. Suppose that A is any abelian group which has an action of thegroup ∆ × G, where G is a finite group and ∆ has order 2. Let ǫ be oneof the two characters of ∆, and ǫ′ the other. For any i ≥ 0, ∆ acts on thecohomology group H i(G,A). This is induced by the action of ∆ on G byinner automorphisms, which is trivial, and by the action of ∆ on A. Clearly,A(ǫ) is a G-invariant subgroup of A. It is also clear that ∆ acts on H i(G,A(ǫ))by the character ǫ. Therefore, we have a map

H i(G,A(ǫ)) −→ H i(G,A)(ǫ). (11)

This map is obviously an isomorphism when i = 0. For i ≥ 1, the kernel is aquotient of H i−1(G,A/A(ǫ)) and the cokernel is a subgroup of H i(G,A/A(ǫ)).Now ∆ acts on A/A(ǫ) by the character ǫ′. Therefore, ∆ acts on the kernel andcokernel of (11) by both ǫ and ǫ′ and therefore those groups have exponent2. But H i(G,A(ǫ)) and H i(G,A)(ǫ) are also killed by |G| if i ≥ 1 and hence(11) will be an isomorphism if G has odd order.

Now take A = O×F ′ . By assumption, G = Gal(F ′/F ) has odd order. Note

that(O×

F ′)(ǫ1) = ker(NF ′/F ′

+: O×

F ′ → O×F ) = µF ′

The first equality is clear by definition. The second follows from the fact thatO×F ′ and O×

F ′

+have the same rank. The statements in the proposition follow

immediately. �

The above proposition allows us to prove that ker(JF ′/F ) ⊂ A(ǫ0)F under

certain hypotheses.

Proposition 1.2.14. Suppose that F ′/F is a finite p-extension, where p isan odd prime, and that both F and F ′ are CM-fields. Suppose that either (i)F does not contain µp or (ii) F ′ = F (µpm) for some integer m. Then themap

A(ǫ1)F → A

(ǫ1)F ′

induced by JF ′/F is injective. Thus, ker(JF ′/F ) ⊆ A(ǫ0)F .

Note that A(ǫ0)F∼= AF+

and that the final conclusion in the proposition impliesthat ker(JF ′/F ) ∼= ker(JF ′

+/F+).

22

Proof. First note that the both the map ker(JF ′/F ) → PGF ′/PF and theisomorphism PGF ′/PF ∼= H1(F ′/F,O×

F ′) are ∆-equivariant. The first mapis injective. Therefore, by proposition 1.2.13, it is enough to show thatH1(F ′/F, µF ′) = 1. In case (i), the p-primary subgroup of µF = µGF ′ istrivial. Since G is a p-group, it follows that the p-primary subgroup of µF ′ istrivial, and hence so is H1(F ′/F, µF ′). In case (ii), we can assume that the p-primary subgroup of µF ′ is µpm . The assumption means that G acts faithfullyon this group. The proposition is then a consequence of the following lemma.

Lemma 1.2.15. Let p be an odd prime. Suppose that G and A are cyclicp-groups and that G acts faithfully on A. Then H i(G,A) = 1. for all i ≥ 1.The statement is true for p = 2 if G is a cyclic 2-group of order at least 4.

Proof. Since G is cyclic, the cohomology is periodic. It is enough to verify thestatement for i = 1, 2. Since A is finite, the Herbrand quotient is trivial, andso it is enough to consider i = 1. Suppose that |G| = pn and that |A| = pm.We can identify A with Z/pmZ and Aut(A) with (Z/pmZ)×, which we regardas a quotient group of Z×

p . That is, any automorphism of A can be realizedas multiplication by a p-adic unit. Let σ be a generator of G. Suppose thatσ acts on A as multiplication by s ∈ Z×

p . Note that s ≡ 1 (mod pZp) if pis odd and s2 ≡ 1 (mod 8Z2) if p = 2. In either case, s is not a root ofunity. The norm map NG on A is multiplication by Φ(s), where Φ(x) is thecyclotomic polynomial 1 + x+ ...+ xp

n−1. If τ denotes the endomorphism ofA defined by σ − 1, then τ acts on A as multiplication by s− 1.

Let a = ordp(Φ(s)

)and b = ordp(s − 1), where ordp denotes the p-adic

valuation, normalized so that ordp(p) = 1. We can assume that n ≥ 1 ifp is odd. By assumption, n ≥ 2 if p = 2. The lemma (for i = 1) assertsthat ker(NG) = im(τ) and this is equivalent to the equality a + b = m. ButΦ(x)(x− 1) = xp

n − 1, and so one must just verify that

ordp(spn − 1) = m

This is true because G acts faithfully on A, which implies that

ordp(spn − 1) ≥ m, ordp(s

pn−1 − 1) < m

But one sees easily that ordp(spn − 1) = ordp(s

pn−1 − 1) + 1 for n ≥ 1 if p isodd and for n ≥ 2 if p = 2. It follows that ordp(s

pn − 1) = m �

23

Remark 1.2.15. We have stated the lemma to include p = 2. In that case,the argument shows that the kernel of the map A

(ǫ1)F → A

(ǫ1)F ′ is of exponent

2. Thus, ∆ acts on that kernel by ǫ0 too.

1.3 Genus theory

Let F ′/F be an arbitrary cyclic extension and let σ be a generator of G =Gal(F ′/F ). The group ClF ′/Clσ−1

F ′ is sometimes called the “genus group” forF ′/F (or the group of “genera”). We will denote it by GF ′/F . The proof ofproposition 1.1.3 shows that GF ′/F

∼= Gal(K/F ′), where K is the maximalabelian extension of F contained in the Hilbert class field H ′ of F ′. Thefield K is often referred to as the “genus field for F ′/F .” Note that if oneassumes that exactly one prime of F is ramified in F ′/F and that this primeis totally ramified, then propositions 1.1.1 and 1.1.3 imply that GF ′/F

∼= ClF .In general, the norm map induces a homomorphism GF ′/F −→ ClF . We

denote its kernel by G(o)F ′/F . Under the assumption that there exists at least

one totally ramified prime for F ′/F , we have an exact sequence

1 −→ G(o)F ′/F −→ GF ′/F −→ ClF −→ 1

Furthermore, one sees easily that every element of G(o)F ′/F has order dividing

[F ′ : F ]. Hence if F ′/F is a p-extension, then G(o)F ′/F is a p-group.

The reader may be familiar with genus theory for quadratic fields, whichhas its roots in the theory of binary quadratic forms developed by Gauss andothers at the beginning of the 19-th century. If F ′ is a quadratic extensionof Q and σ is the nontrivial automorphism of F ′, then one sees easily thatσ(c) = c−1 for c ∈ ClF ′ . Thus, GF ′/Q = ClF ′/Cl2F ′ . Its structure is describedin the following proposition.

Proposition 1.3.1. Suppose that [F ′ : Q] = 2. Let t denote the number offinite primes which are ramified in F ′/Q.

1. If F ′ is an imaginary quadratic field, then GF ′/Q∼= (Z/2Z)t−1.

2. Suppose that F ′ is a real quadratic field.

If −1 ∈ NF ′/Q(F ′×), then GF ′/Q∼= (Z/2Z)t−1.

If −1 6∈ NF ′/Q(F ′×), then GF ′/Q∼= (Z/2Z)t−2.

24

Remark 1.3.2. The two cases which occur in part (2) of this propositioncan be distinguished by using the following fact for a real quadratic field F ′.

Fact. We have −1 ∈ NF ′/Q(F ′×) if and only if every odd prime ℓ ramifiedin F ′/Q satisfies ℓ ≡ 1 (mod 4).

This is a consequence of “Hasse’s Norm Theorem” which states that if F ′/Fis a cyclic extension of number fields and if α ∈ F× is a local norm atall primes of F , then α is a global norm for the extension F ′/F . How-ever, if α ∈ O×

F , there seems to be no simple criterion for predicting whenα ∈ NF ′/F (O×

F ′). In particular, one cannot predict when −1 ∈ NF ′/Q(O×F ′)

(i.e., when H2(F ′/F,O×F ′) = 0). A sufficient condition is t = 1. For then,

proposition 1.2.3 implies that dimF2

(H1(F ′/F,O×

F ′))≤ 1. Proposition 1.2.4

implies that H1(F ′/F,O×F ′) ∼= Z/2Z and that indeed H2(F ′/F,O×

F ′) = 0.

We will prove proposition 1.3.1 later, deducing it from propositions 1.3.4and 1.3.5. The analogous result for cyclic extensions of Q of odd prime degreeis somewhat simpler and we will prove this first. We will give two proofs toillustrate two different approaches to genus theory.

Proposition 1.3.3. Suppose that F ′ is a cyclic extension of Q of degreep, where p is an odd prime. Let t denote the number of primes which areramified in F ′/Q. Then

GF ′/Q∼= (Z/pZ)t−1

Proof. Let K be the genus field for F ′/Q. Suppose that ℓ1, ..., ℓt are theprimes which are ramified in F ′/Q. If ℓ is any one of these primes, letIℓ denote the corresponding inertia subgroup of Gal(K/Q). It is clear thatIℓ∩Gal(K/F ′) = 1 and hence that Iℓ must be cyclic of order p. It is also clearthat Gal(K/Q) is generated by Iℓ1 , ..., Iℓt . This implies that Gal(K/F ′) ∼=(Z/pZ)u for some u ≤ t− 1.

To see that u = t− 1, one explicitly constructs the field K. Using eithersome elementary facts about ramification theory or local class field theory,one can verify that if ℓ is any one of the primes ramified in F ′/Q, then eitherℓ = p or ℓ ≡ 1 (mod p). In both cases, there is a unique cyclic extension ofQ of degree p in which only the prime ℓ is ramified: a subfield of Q(µp2) ifℓ = p, a subfield of Q(µℓ) if ℓ ≡ 1 (mod p). For each ℓi, 1 ≤ i ≤ t, let Ki

denote the field just described.

25

The Kronecker-Weber theorem (which states that every finite abelianextension of Q is contained Q(µm) for some m) implies that F ′ ⊂ K1...Kt.Note also that the inertia subgroup Ii of Gal(K1...Kt/Q) for any one of theℓi’s is of order p and that Ii ∩ Gal(K1...Kt/F

′) is trivial. This implies thatK1...Kt ⊂ K. But it is easy to see that Gal(K1...Kt/F

′) ∼= (Z/pZ)t−1.Comparing this with the inequality u ≤ t−1, it follows that indeed u = t−1and that the field K coincides with the compositum K1...Kt. �

A second proof for proposition 1.3.3 can be given by studying the sub-group of ClF ′ generated by the classes of the primes λ1, ..., λt of F ′ lyingabove ℓ1, ..., ℓt. For 1 ≤ i ≤ t, let ci denote the class of λi. Each of theseclasses has order 1 or p and is invariant under the action of G. One showsthat this subgroup is precisely ClGF ′ and is isomorphic to (Z/pZ)t−1. Thiscan be proved directly, but we will justify it later as an easy consequence ofproposition 1.3.5. Consider the endomorphism τ = σ− 1 of ClF ′ , where σ isa generator of Gal(F ′/Q). Then ker(τ) = ClGF ′ and coker(τ) = GF ′/Q havethe same order. Since τ annihilates GF ′/Q, NG acts on that group simply asmultiplication by p. But NG also annihilates GF ′/Q and so that group is anelementary abelian p-groups and therefore is indeed isomorphic to (Z/pZ)t−1.

Returning to cyclic extensions F ′ of an arbitrary base field F , the sit-uation is somewhat complicated by the fact that O×

F can be infinite. Thegenus group GF ′/F = (ClF ′)G has the same order as (ClF ′)G = ClGF ′ , thesubgroup of G-invariant ideal classes. One obtains information about ClF ′

by studying either of these groups. Both approaches will be useful in laterchapters (where we apply “genus theory” to towers of cyclic extensions). Ourarguments will take the second approach, studying the subgroup ClGF ′ . To

be more precise, we will study the possibly smaller subgroup Cl[G]F ′ consisting

of ideal classes which contain a G-invariant ideal. Obviously, we have

Cl[G]F ′

∼= FGF ′/PGF ′

and so ClGF ′

/Cl

[G]F ′

∼= coker(FGF ′ → ClGF ′

), which is the subject of the next

proposition.

Proposition 1.3.4. Suppose that F ′/F is a finite cyclic extension. Thenthere is an isomorphism

coker(FGF ′ → ClGF ′

) ∼=(O×F ∩NF ′/F (F ′×)

)/NF ′/F (O×

F ′)

26

In particular, if H2(F ′/F,O×F ′) = O×

F /NF ′/F (O×F ′) = 1, then every class in

ClGF ′ contains a G-invariant ideal.

Proof. Consider the exact sequence

1→ PF ′ → FF ′ → ClF ′ → 1

This induces the following exact sequence of cohomology groups

FGF ′ → ClGF ′ → H1(F ′/F,PF ′)→ H1(F ′/F,FF ′)

However, H1(F ′/F,FF ′) = 0. To see this, let FF ′,P denote the group offraction ideals of F ′ generated by the primes lying above P , where P is anyprime ideal of F . Note that FF ′,P is invariant under the action of G and thatFF ′ is isomorphic to a direct sum of these subgroups, For each P , one has anisomorphism FF ′,P

∼= IndGD(Z), where D is the decomposition subgroup of Gfor any one of the prime ideals of F ′ lying above P and Z is given a trivialaction of D. Then by Shapiro’s lemma, one has

H1(F ′/F,FF ′,P ) ∼= H1(D,Z) = Hom(D,Z) = 0

The assertion that H1(F ′/F,FF ′) = 0 follows from this.Therefore, coker(FGF ′ → ClGF ′) ∼= H1(F ′/F,PF ′). Using the injectivity of

the map JF ′/F : PF → PF ′ and the assumption that G is cyclic, generatedby σ, we see that

H1(F ′/F,PF ′) ∼= ker(NF ′/F : PF ′ → PF )/Pσ−1F ′

We also have

ker(NF ′/F : PF ′ → PF )

)=

{(α) ∈ PF ′ | NF ′/F (α) ∈ O×

F

}

and

Pσ−1F ′ =

{(ασ−1) | α ∈ F ′×}

={(α) ∈ PF ′ | NF ′/F (α) ∈ NF ′/F (O×

F ′)}

Hence, it is clear that NF ′/F defines an isomorphism

H1(F ′/F,PF ′)→ (O×F ∩NF ′/F (F ′×))/NF ′/F (O×

F ′),

proving the proposition. �

27

It is sufficient to concentrate on the p-primary subgroups of the ideal classgroups, where p is a fixed prime. As before, we will let AF and AF ′ denotethe p-primary subgroups of ClF and ClF ′ , respectively. According to remark1.2.8, the groups (AF ′)G and AGF ′ are both isomorphic to AF if p ∤ [F ′ : F ].This is valid for an arbitrary Galois extension and ramification plays no role.The isomorphisms are quite simple, given by the maps NF ′/F and JF ′/F . Wewill concentrate in the rest of this section on the case where F ′/F is a cyclicextension whose degree is a power of p, first considering the case of degree p.

Proposition 1.3.5. Suppose that F ′/F is a cyclic extension of degree p.Then

t− s− 1 ≤ dimFp

(G(o)F ′/F

)≤ t− u

where t is the number of distinct prime ideals of F which are ramified inF ′/F , s = dimFp

(O×F /(O×

F )p), and u = min(t, 1).

Proof. We will use the exact sequence in proposition 1.2.3. Note that G actstrivially on all the groups occurring there, that NG annihilates them and actssimply as multiplication by p. Hence those groups are all vector spaces overFp. The image of the map

FGF ′/FF −→ ClGF ′/JF ′/F (ClF )

is Cl[G]F ′ /JF ′/F (ClF ). Thus we have the following relationship between orders

of groups

|Cl[G]F ′ | · |JF ′/F (ClF )|−1 = pt · |H1(F ′/F,O×

F ′)|−1 · |ker(JF ′/F )|

Obviously, |CLF ′ | = |ker(JF ′/F )| · |JF ′/F (ClF )|. Together with proposition1.2.4 (assuming, for p = 2, that the hypothesis there for the infinite primesis satisfied), we then obtain the following formula

|Cl[G]F ′ | = pt−1 · |ClF | · |H2(F ′/F,O×

F ′)|−1

On the other hand, proposition 1.3.4 shows that

|H2(F ′/F,O×F ′)| = pv · |ClGF ′/Cl

[G]F ′ |

where pv = [O×F : O×

F ∩NF ′/F (F ′×)]. Thus, we obtain the formula

|ClGF ′ | = pt−1−v · |ClF |

28

Therefore, |G(o)F ′/F | = pt−1−v. Since v ≥ 0, one immediately obtains the upper

bound on dimFp

(G(o)F ′/F

). To obtain the lower bound, we use the fact that

|H2(F ′/F,O×F ′)| is divisible by pv. That implies that v ≤ s according to

proposition 1.2.4.If p = 2, the assumptions in proposition 1.2.4 may fail to be satisfied.

There may also be some infinite primes of F ramified in F∞. One thenhas to slightly modify the above calculation. If t∞ denotes the numberof such primes, then according to a remark 1.2.6, the Herbrand quotienth(F ′/F,O×

F ′) = 2t∞−1. Hence, in the above argument, we get

|ClGF ′ | = pt−(1−t∞)−v · |ClF | = pt+t∞−1−v

One then has the lower bound t+t∞−s−1 on the p-rank of ClGF ′ , which againimplies the stated result. Note that t + t∞ is the total number of ramifiedprimes in the extension F ′/F . �

We return now to the special cases mentioned at the beginning of thissection, where we take F = Q. The facts that hQ = 1 and the unit groupZ× has order 2 simplify things considerably.

First we complete the alternative proof of proposition 1.3.3. Thus, assumethat F ′ is a cyclic extension of Q of degree p, where p is an odd prime. Obvi-ously, we haveH2(F ′/F,O×

F ′) = 0, and so proposition 1.3.4 shows that ClGF ′ isgenerated by the classes containing ramified primes. Thus ClGF ′

∼= (Z/pZ)u

for some u ≤ t, where t denotes the number of ramified primes in F ′/Q.But the classes of these ramified primes are not independent. One obtainsessentially one nontrivial relationship because PGF ′/PQ

∼= H1(F ′/F,O×F ′) is

a group of order p, as follows from proposition 1.2.3 and the vanishing ofH2(F ′/F,O×

F ′). Hence u = t− 1.Thus we see that ClGF ′

∼= (Z/pZ)t−1. This implies that GF ′/Q = ClF ′/Clσ−1F ′

has order equal to pt−1. Now if we regard ClF ′ as a Z[G]-module, it is clearthat NG ∈ Ann(ClF ′). Also, NG acts as multiplication by p on GF ′/Q. There-fore, GF ′/Q has exponent p and must be indeed be isomorphic to (Z/pZ)t−1.

We now prove proposition 1.3.1. Assume first that F ′ is imaginaryquadratic. Then NF ′/Q(α) > 0 for all α ∈ F ′×, and so proposition 1.3.4implies that ClGF ′ is again generated by the classes of the ramified primes.Also, there is essentially just a single nontrivial relation between those classes

29

because H1(F ′/F,O×F ′) ∼= Z/2Z, which is easily verified directly since O×

F ′ isjust finite. Hence ClGF ′

∼= (Z/2Z)t−1.If F ′ is a real quadratic field and −1 6∈ NF ′/Q(F ′×), then the classes

of the ramified primes generate ClGF ′ , just as in the case of an imaginaryquadratic field. But this time H1(F ′/F,O×

F ′) ∼= (Z/2Z)2 by proposition1.2.3. Therefore, there will be two independent relations between those idealclasses, and so ClGF ′

∼= (Z/2Z)t−2. Note that we must have t ≥ 2, as wepointed out earlier.

If −1 ∈ NF ′/Q(O×F ′), then H2(F ′/F,O×

F ′) = 0 and the argument is thesame as in the case where F ′/Q is cyclic of odd prime degree. But if−1 ∈ NF ′/Q(F ′×), but 6∈ NF ′/Q(O×

F ′), then the classes of the ramified primesgenerate a subgroup of index 2 in ClGF ′ . Also, H2(F ′/F,O×

F ′) ∼= Z/2Z andH1(F ′/F,O×

F ′) ∼= (Z/2Z)2. Thus, this subgroup is isomorphic to (Z/2Z)t−2.It follows that ClGF ′

∼= (Z/2Z)t−1.In all these cases, ClGF ′ = ClF ′ [2] and GF ′/Q = ClF ′/Cl2F ′ are elementary

abelian 2-groups of the same order, and so must be isomorphic, provingproposition 1.3.1 �

Proposition 1.3.4 has some useful consequences if F and F ′ are CM-fields.

Corollary 1.3.6. Suppose that F ′/F is a finite p-extension, where p is anodd prime, and that both F and F ′ are CM-fields. Let ǫ1 be the nontrivialcharacter of ∆ = Gal(F ′/F

′

+) ∼= Gal(F/F+). Suppose that either (i) F doesnot contain µp or (ii) F ′ = F (µpm) for some integer m. Then

H2(F ′/F,O×F ′)

(ǫ1) ∼= H2(F ′/F, µF ′) = 1

and every class in(A

(ǫ1)F ′

)Gcontains a G-invariant ideal.

Proof. The argument is similar to that for proposition 1.2.14. One usesproposition 1.2.13 for i = 2 instead of i = 1. The isomorphism in proposition1.3.4 is ∆-equivariant and hence preserves the ǫ1-components of the groupsin question. In case (i), H2(F ′/F, µF ′) obviously vanishes. In case (ii), onecan apply lemma 1.2.15 to see that that group vanishes. �

Corollary 1.3.7. Suppose that the assumptions in corollary 1.3.6 are satis-fied. Let [F ′ : F ] = pn. Consider the following set of primes of F+:

S = {v | v splits in F/F+ and v is ramified in F ′+/F+}

30

For each v in this set, let pav denote its ramification index in F ′+/F+. Then

(A(ǫ1)F ′ )G

/JF ′/F (A

(ǫ1)F ) ∼=

∏

v∈SZ/pavZ

In particular, if every prime of F+ lying in S is totally ramified in F ′+/F+,

then (A(ǫ1)F ′ )G contains a subgroup isomorphic to (Z/pnZ)|S|.

Proof. We will apply proposition 1.2.3. The homomorphisms and isomor-phisms in that proposition are all ∆-equivariant. Since G = Gal(F ′/F )is a p-group, all the groups occurring there are actually finite p-groups.Since p is odd, the exactness of the sequence and the two isomorphismsare still valid if we take the ǫ1-components of the groups. Note also thatsince ClGF ′/JF ′/F (ClF ) is a p-group, it is isomorphic to AGF ′/JF ′/F (AF ). The

ǫ1-component of that group is (A(ǫ1)F ′ )G/JF ′/F (A

(ǫ1)F ). Corollary 1.3.6 implies

that the map(FGF ′/FF )(ǫ1) −→ (A

(ǫ1)F ′ )G/JF ′/F (A

(ǫ1)F )

is surjective. In fact, it is an isomorphism because H2(F ′/F,O×F ′)(ǫ1) = 1.

Finally, the definition of S implies that for every v ∈ S, there are two primesof F lying above v which are permuted by ∆. They both have ramificationindex pav in F ′/F . Using the final isomorphism in proposition 1.2.3, weobtain

(FGF ′/FF )(ǫ1) ∼=∏

v∈SZ/pavZ

and so the stated isomorphism in the corollary follows. The particular caseis immediate. �

1.4 The reflection principle

Let p be a prime. Suppose that F is a number field which contains µp. Asbefore, let L denote the p-Hilbert class field of F . The idea to be pursuedin this section is that a cyclic unramified extension K of F of degree p isrelated to the ideal class group of F in two different ways. One comes fromclass field theory, the other from Kummer theory. Briefly,

1. Class field theory shows that there is a canonical surjective homomorphismClF → Gal(K/F ). This arises as the composition of the Artin isomorphismClF [p∞] → Gal(L/F ) with the restriction map Gal(L/F ) → Gal(K/F ).

31

Thus Gal(K/F ) can be identified with a certain quotient group of ClF [p∞]of order p.

2. Kummer theory shows that K = F ( p√α), where α ∈ F×. Since K/F is

unramified, it is clear that αOF = Ip, where I ∈ FF . Let c denote the classof I in ClF . Then c has order 1 or p. It is easy to see that the subgroupof ClF generated by c is uniquely determined by the extension K/F . Thisfollows from the fact that the subgroup of F×/(F×)p generated by the cosetα(F×)p is determined by K.

The class c defined in (2) can be trivial. This would be true if and only ifK = F ( p

√η), where η ∈ O×

F . Note also that if α ∈ F× is such that αOF = Ip,then the Kummer extension F ( p

√α)/F can only be ramified at primes of F

dividing p, but is not necessarily unramified.

Suppose that A is a subset of F× containing (F×)p. Kummer theory givesan isomorphism

κA : A/(F×)p −→ Hom(Gal(F (

p√A)/F ), µp

)(12)

Here we let F ( p√A) denote the extension of F generated by { p

√α | α ∈ A}.

The map κA is easy to define. For each α ∈ A, let α′ ∈ F ( p√A) satisfy

α′)p = α. Then one defines a cocyle φ with values in µp by φ(g) = g(α′)/α′

for all g ∈ Gal(F ( p√A)/F ). It is easy to show that the cocyle class [φ] is

determined by the coset of α in A/(F×)p. The Kummer isomorphism κA isdefined by mapping that coset to [φ]. Injectivity is straightforward to verify.Surjectivity is a consequence of Hilbert’s theorem 90.

Suppose that ∆ is a group of automorphisms of F and that A is invariantunder the action of ∆. The extension F ( p

√A) is then a Galois extension

of the fixed field F∆. It follows that ∆ acts (by inner automorphisms) onGal(F ( p

√A)/F ). Now ∆ also acts on µp which is given by a homomor-

phism (or character) ω : ∆ → F×p . Hence one has a natural action of ∆

on Hom(Gal(F ( p

√A)/F ), µp

). One can then verify that the map κA is ∆-

equivariant. In particular, suppose that A/(F×)p is cyclic. The action of∆ on that group is given by a character ψ : ∆ → F×

p . The action of ∆ on

Gal(F ( p√A)/F ) must then be given by the character ϕ = ωψ−1.

There is a theorem of Kummer concerning the class numbers of the CM-field F = Q(µp) and its maximal real subfield F+. We will denote the classnumber of F+ by h+

F , often call the “second factor” in hF . In fact, proposition

32

1.1.1 implies that h+F |hF . The quotient hF/h

+F is called the “first factor” and

is denoted by h−F . By using class number formulas, Kummer showed that ifp|h+

F , then p|h−F . As our first illustration of the reflection principle, we willprove the following more general statement. It concerns a CM-field F andwe will use the same notation h±F .

Proposition 1.4.1. Suppose that F is a CM-field containing µp where p isan odd prime. Let k be the largest integer such that µpk ⊂ F . Assume thatF (µpk+1)/F is ramified for at least one prime of F . If p|h+

F , then p|h−F .

Since F/F+ is ramified at the infinite primes of F , we know that h+F |hF and

so h−F is an integer. One can also state the conclusion this way: p|hF if andonly if p|h−F .

Proof. Let ∆ = Gal(F/F+) and let δ denote its nontrivial element. As insection 1.2, we let ǫ0, ǫ1 denote the two characters of ∆. In the notationdefined above, we have ǫ1 = ω. Since p is odd, we have a direct productdecomposition

ClF [p∞] = ClF [p∞](ǫ1) × ClF [p∞](ǫ0) (13)

Remark 1.2.8 implies that ClF [p∞](ǫ0) ∼= ClF+[p∞] which has order h

(p)F+

, the

power of p dividing h+F . Hence the order of ClF [p∞](ǫ1) is equal to the power of

p dividing h−F . Thus, we must show that ClF [p∞](ǫ0) 6= 1 =⇒ ClF [p∞](ǫ1) 6= 1.Assume that ClF [p∞](ǫ0) 6= 1 . Hence there exists an unramified, cyclic

extension of F+ of degree p. Let K be the compositum of that field andF . Thus, K/F is an unramified, cyclic extension of degree p, and K/F+ isabelian (in fact, cyclic) of degree 2p. We have K = F ( p

√α) for some α ∈ F×.

As mentioned above, α determines an ideal I satisfying Ip = (α) and thecorresponding ideal class c ∈ ClF [p].

Let ∆ = Gal(F/F+) and let δ denote its nontrivial element. As in section1.2, we let ǫ0, ǫ1 denote the two characters of ∆. In the notation definedabove, we have ǫ1 = ω. Taking A to be the subgroup of F× generated byα and (F×)p, so that K = F ( p

√A), note that ∆ acts on Gal(K/F ) by ǫ0.

Therefore, by (12), it follows that ∆ acts on the cyclic group A/(F×)p byǫ1. This means that δ(α) = α−1βp, where β ∈ F×. Therefore, δ(I) = I−1(β)and hence δ(c) = c−1. That is, c ∈ ClF [p](ǫ1) in the notation of section 2. Wemust prove that c 6= 1.

If c = 1, then we would have I = (γ), where γ ∈ F×. Thus, α = γpη,where η ∈ O×

F . We then see that δ(η) = η−1νp, where ν ∈ O×F . Now we nake

33

the following observation: For either character ǫ of ∆, the obvious map

(O×F )(ǫ) −→

(O×F /(O×

F )p)(ǫ)

is surjective. This is easily verified using the facts that |∆| = 2 and that p isodd. As a consequence, we see that η = ζξp, where ξ ∈ O×

F and δ(ζ) = ζ−1.This means that NF/F+

(ζ) = 1 and hence ζ is a root of unity in F . However,

K = F ( p√α) = F ( p

√η) = F ( p

√ζ)

and since this extension is nontrivial, it must be F (µpk+1), a contradictionto our assumption since K/F is unramified. Thus, c 6= 1 and so ClF [p](ǫ1) isindeed nontrivial. �

The argument just given shows more. One can adapt it to obtain the follow-ing inequality

dimFp

(ClF [p](ǫ0)

)≤ dimFp

(ClF [p](ǫ1)

)(14)

under the assumptions of the proposition. Proposition 1.4.2 below will be arefinement of this inequality.

Consider the following situation. Suppose that F is any number fieldcontaining µp and let ∆ be a group of automorphisms of F such that p ∤ |∆|.Let ϕ be any irreducible character of ∆ over Qp, by which we mean thecharacter of an irreducible representation ρ : ∆→ AutQp(Vϕ), where Vϕ is afinite-dimensional vector space over Qp. Let dϕ = dimQp(Vϕ), the degree ofthe character ϕ.

One of the irreducible characters of ∆ is ω, giving the action of ∆ on µp.This description defines a homomorphism ∆ → F×

p , but the reduction mapZ×p → F×

p has a canonical splitting identifying F×p with the group of µp−1 of

(p− 1)-st roots of unity in Z×p . Thus, we can identify ω with a character of

∆ with values in Z×p , the character of a 1-dimensional representation space

Vω. We can make such an identification whenever we have a homomorphism∆→ F×

p .Let ψ be the character of ∆ corresponding to the representation space

Vψ = Hom(Vϕ, Vω),

which is easily seen to be irreducible over Qp. Of course, we also haveVϕ ∼= Hom(Vψ, Vω). We refer to ψ as the ω-dual of ϕ. Note that ϕ and ψ

34

have the same degree. If they are 1-dimensional, then we have the simplerelationship ϕψ = ω.

We will let Irr∆(Qp) denote the set of irreducible characters of ∆ overQp. If ϕ ∈ Irr∆(Qp), the idempotent for ϕ in the group ring Qp[∆] is definedby

eϕ =1

|∆|∑

δ∈∆

ϕ(δ)−1δ (15)

Note that eϕ ∈ Zp[∆] since (p, |∆|) = 1. The group ring Zp[∆] is a directproduct of the ideals generated by the eϕ’s for ϕ ∈ Irr∆(Qp).

Suppose that U is any Zp[∆]-module. Then we have the following decom-position as a direct product of Zp[∆]-submodules:

U =∏

ϕ∈Irr∆(Qp)

U (ϕ) (16)

where U (ϕ) = eϕU . If ϕ is 1-dimensional, then one also has the followingdefinition:

U (ϕ) = eϕU = {a ∈ U | δ(a) = ϕ(δ)a for all δ ∈ ∆} (17)

We refer to (16) as the ∆-decomposition of U and to the submodule U (ϕ) asthe ϕ-component of U .

In particular, suppose that U is an elementary abelian p-group. Wecan regard U as a representation space for ∆ over Fp. Suppose that it isirreducible. Then U = U (ϕ) for a unique ϕ ∈ Irr∆(Qp). Conversely, ifϕ ∈ Irr∆(Qp), one can find a Zp-lattice Tϕ ⊂ Vϕ which is ∆-invariant. ThenU = Tϕ/pTϕ is irreducible and satisfies U = U (ϕ). We denote this space byWϕ. One has dimFp(Wϕ) = dϕ. It is not hard to see that this constructionVϕ Wϕ defines a 1-1 correspondence between the sets of irreducible rep-resentations for ∆ over Qp and over Fp. Also, if V is any finite-dimensionalrepresentation space for ∆ and T is a ∆-invariant Zp-lattice in V , then T/pTis isomorphic to a direct sum of the Wϕ’s, V is isomorphic to a direct sum ofthe Vϕ’s, and the corresponding multiplicities are equal.

The following result is the main theorem of this section. It is an illus-tration of the reflection principle. The structure of O×

F /(O×F )p as a repre-

sentation space for ∆ over Fp plays a role, specifically the Fp-dimension ofthe ϕ-component for an irreducible character ϕ. As we will discuss later, themultiplicity of Wϕ in O×

F /(O×F )p can be determined, in principle, and hence

so can the dimension of (O×F /(O×

F )p)(ϕ).

35

Proposition 1.4.2. Suppose that ∆ is a group of automorphisms of a num-ber field F and that p is a prime not dividing |∆|. Assume that µp ⊂ F .Suppose that ϕ is an irreducible character for ∆ over Qp and that ψ is theω-dual of ϕ. Then

dimFp

(ClF [p](ψ)

)≤ dimFp

(ClF [p](ϕ)

)+ dimFp

((O×

F /(O×F )p)(ϕ)

)

Proof. Let E = F∆. Then ∆ = Gal(F/E). Consider the ψ-componentClF [p∞](ψ) in the ∆-decomposition of ClF [p∞]. The Fp-dimensions of ClF [p](ψ)

and ClF [p∞](ψ)/pClF [p∞](ψ) are equal. Denote this dimension by rψ. TheArtin map for L/F then determines an extension Kψ/F such that Kψ ⊆ L,Kψ/E is a Galois extension, and Gal(Kψ/E) ∼= ClF [p](ψ) as Fp-representationspaces for ∆. Each is isomorphic to a direct sum of rψ/dψ copies of Wψ.

Now Kψ = F ( p√Aϕ) for a certain subgroup Aϕ of F× which contains

(F×)p and is ∆-invariant. The Fp-representation space Aϕ/(F×)p has di-

mension rψ and is a direct sum of rψ/dψ copies of Wϕ. This follows from (12)and is the reason we use the subscript ϕ. The fact that Kψ/F is unramifiedimplies that if α ∈ Aϕ, then αOF = Ip for some fractional ideal I of F . Letc denote the class of I. Of course, if α ∈ (F×)p, then c = 1. Thus, we candefine in this way a homomorphism

Aϕ/(F×)p −→ ClF [p] (18)

which is easily seen to be ∆-equivariant. Thus the image of (18) is a subgroupof ClF [p](ϕ) and hence its Fp-dimension is bounded above by dimFp

(ClF [p](ϕ)

).

The kernel of (18) is of the form Bϕ/(F×)p, where Bϕ is some subgroup

of Aϕ. Suppose that α ∈ Bϕ. Then c = 1 and I = (γ), where γ ∈ F×.Therefore, α = γpη, where η ∈ O×

F . This implies that Bϕ/(F×)p is contained

in the image of the map

(O×F /(O×

F )p)(ϕ) −→

(F×/(F×)p

)(ϕ)(19)

and hence dimFp

(Bϕ/(F

×)p)≤ dimFp

((O×

F /(O×F )p)(ϕ)

). The inequality in

the proposition now follows. �

To deduce the inequality (14) from proposition 1.4.2, one needs just one

additional observation. Just take ψ = ǫ0, ϕ = ǫ1. The group(O×F /(O×

F )p)(ǫ1)

is cyclic and is generated by the coset of a primitive pk-th root of unity ζ.

36

However, since F ( p√ζ) is assumed to be ramified, the image of the coset of ζ

under the map (19) isn’t contained in Aǫ1 and so we get the slightly betterinequality (14).

Another consequence is a theorem proved in the early 1930s, due toScholz, which is for the case p = 3.

Proposition 1.4.3. Let d > 1 be a squarefree integer. Then

dimF3(ClQ(

√d)[3]) ≤ dimF3

(ClQ(√−3d)[3]) ≤ dimF3

(ClQ(√d)[3]) + 1

Proof. In this case, we consider the biquadratic field F = Q(√d,√−3d),

a CM-field with maximal real subfield F+ = Q(√d). The first inequality

follows from proposition 1.4.1. To prove the second inequality, we consider∆ = Gal(F/Q), a group with four characters. The nontrivial charactersare ψ, ϕ, and ω. They factor through the quotient groups of ∆ corre-sponding to the three quadratic fields Eψ = Q(

√−3d), Eϕ = F+, and

Eω = Q(µ3), respectively. We denote the trivial character by ǫ0. Considerthe ∆-decomposition of ClF [3∞]:

ClF [3∞] = ClF [3∞](ψ) × ClF [3∞](ϕ) × ClF [3∞](ω) × ClF [3∞](ǫ0)

One can use remark 1.2.8 to identify each of these components. First of all,ClF [3∞](ǫ0) ∼= ClQ[3∞], which is obviously trivial. Then, similarly, we have

ClF [3∞](ψ) ∼= ClFψ [3∞], ClF [3∞](ϕ) ∼= ClFϕ [3∞], ClF [3∞](ω) ∼= ClFω [3

∞].

The ω-component is also trivial. Proposition 1.4.3 gives an inequality for the3-ranks of the other two ∆-components. The unit groupO×

F has a very simplestructure, namely µ3 and the fundamental unit of Eϕ generate a subgroup ofindex a power of 2. It follows that dimF3

((O×

F /(O×F )3)(ϕ)

)= 1. One obtains

dimF3

(ClF [3](ψ)) ≤ dimF3

(ClF [3](ϕ)) + 1

and the second inequality in the proposition then follows. �

As we mentioned earlier, one can evaluate dimFp

((O×

F /(O×F )p)(ϕ)

)in prin-

ciple. That dimension is determined by the representation space VO×

F=

O×F ⊗Z Qp for ∆ and the torsion subgroup µF of O×

F . For the contributionfrom µF , note that µF/µ

pF is an Fp-vector space of dimension 1 and ∆ acts

37

by ω. Its contribution to the dimension is 1 if ϕ = ω, and 0 otherwise.Now TO×

F= (O×

F /µF )⊗Z Zp is a ∆-invariant Zp-lattice in VO×

Fand therefore

the multiplicity of Wϕ in TO×

F/pTO×

Fis the same as the multiplicity of Vϕ

in VO×

F. Denote that multiplicity by mϕ(VO×

F). Then the Fp-dimension of

(O×F /(O×

F )p)(ϕ) is equal to mϕ(VO×

F)dϕ if ϕ 6= ω, and mϕ(VO×

F)dϕ+1 if ϕ = ω.

In principle, the isomorphism class of VO×

F, and hence the multiplicities of

each of the Vϕ’s in that representation space for ∆, can always be determined.One could take a Galois extension F ′ of Q containing F . Then ∆ is a

subquotient of ∆′ = Gal(F ′/Q) and VO×

F

∼= VGal(F ′/F )

O×

F ′

as representation

spaces for ∆. The action of ∆ on that subspace can be studied in terms ofthe representation space VO×

F ′

for Gal(F ′/Q). Therefore, it seems sufficient

to consider a finite Galois extension of Q and so we will simply assume thatF/Q is Galois. The following theorem is rather well-known. After recallingthe proof, we will describe how to determined the multiplicity of Vϕ in VO×

F.

Proposition 1.4.4. Suppose that F is a finite Galois extension of Q andlet ∆ = Gal(F/Q). Let v be an infinite prime of F and let ∆v denote thedecomposition subgroup of ∆ for v. Suppose that ǫ0 is the trivial characterof ∆v and ϕ0 is the trivial character of G. Let VO×

F= O×

F ⊗Z Qp. Then

VO×

F⊕ Vϕ0

∼= Ind∆∆v

(ǫ0)

as representations spaces for ∆.

Proof. The isomorphism will be proved by showing that the two represen-tations of ∆ have the same character. Both representations can be realizedover Q, but the character is determined by a realization over any field. Theargument depends on the well-known proof of Dirichlet’s unit theorem. Thecompletions Fv of F at the infinite primes v are either all isomorphic to Ror to C. In either case, one defines a log map from F×

v onto R. One thenidentifies O×

F /µF with a Z-lattice in a certain subspace of∏

v|∞ R of codi-

mension 1. This map is ∆-equivariant. The R-vector space∏

v|∞ R is justthe permutation representation determined by the action of ∆ on the setof infinite primes of F . This is isomorphic to Ind∆

∆v(ǫ0), considered as an

R-representation space for ∆. Thus, one has an injective map

O×F ⊗Z R −→ Ind∆

∆v(ǫ0)

and the cokernel is just the trivial representation of ∆ over R. The character

38

of the representation is thus determined. As we mentioned, this is sufficientto prove the stated isomorphism. �

Obviously mϕ0(VO×

F) = 0. If ϕ 6= ϕ0, then mϕ(VO×

F) is just the multi-

plicity of ϕ in Ind∆∆v

(ǫ0). Now Vϕ ⊗Qp Qp may be reducible, a direct sum ofabsolutely irreducible representations spaces, each occurring with a certainmultiplicity. Suppose that ξ is the character of one of the direct summands,a representation space Vξ for ∆ over Qp, and let sξ denote the correspondingmultiplicity. The quantity sξ is the Schur index for ξ over Qp and actuallydepends only on ϕ and not on the choice of ξ. All the characters ξ occurringin ϕ are conjugate over Qp. If mξ

(Ind∆

∆v(ǫ0)

)denotes the multiplicity of Vξ

in Ind∆∆v

(ǫ0), then mϕ(VO×

F) = mξ

(Ind∆

∆v(ǫ0)

)/sξ, assuming that ϕ 6= ϕ0.

One can determine mξ

(Ind∆

∆v(ǫ0)

)by using the Frobenius reciprocity law.

Let dξ = dimQp(Vξ). Note that ∆v has order 1 or 2. Let d+

ξ and d−ξ denote the

multiplicities of ǫ0 and ǫ1 (if ∆v has order 2), respectively, when we regardVξ as a representation space for ∆v. Thus, dξ = d+

ξ + d−ξ . According to the

Frobenius reciprocity law, the multiplicity of ξ in Ind∆∆v

(ǫ0) coincides with

the multiplicity of ǫ0 in ξ|∆v . That is, we have mξ

(Ind∆

∆v(ǫ0)

)= d+

ξ .

If F is totally real, then d+ξ = dξ. If F is a CM-field, then any irreducible

character ξ is either totally even or totally odd, i.e., either d+ξ = dξ or d−ξ = dξ.

In the important special case where ∆ is abelian, and ϕ is an irreduciblecharacter for ∆ over Qp, then each ξ occurring in ϕ is 1-dimensional andoccurs with multiplicity 1. In that case, we have m(ϕ)(VO×

F) = 1 or 0,

depending on whether ϕ is even or odd.

We now consider the field F = Q(µp) and ∆ = Gal(F/Q). The irre-ducible characters of ∆ over Fp are the powers ωi, 0 ≤ i ≤ p− 2.

Proposition 1.4.5. Suppose that p is an odd prime, that 2 ≤ i, j ≤ p− 2,that i is odd, and that i+ j ≡ 1 (mod p− 1). Then

dimFp

(ClF [p](ω

j))≤ dimFp

(ClF [p](ω

i))≤ dimFp

(ClF [p](ω

j))

+ 1.

Also, the ω0 and ω1 components of ClF [p] are trivial.

Proof. First of all, note that i + j ≡ 1 (mod p − 1) implies that ωiωj = ω.Suppose that i is odd, 1 ≤ i ≤ p − 2, and let ψ = ωj, ϕ = ωi. Thenuϕ = 0 if i > 1, uϕ = 1 if i = 1. But when i = 1, one can observe thatF (µp2)/F is a ramified extension. Thus, in all cases, we get the inequality

dimFp

(ClF [p](ω

j))≤ dimFp

(ClF [p](ω

i)).

39

To get the second inequality, take ψ = ωi, ϕ = ωj. The structure ofO×F ⊗Z Qp is rather simple. Each nontrivial even character of ∆ occurs with

multiplicity 1. That is, uϕ = 1 if j 6= 0, uϕ = 0 if j = 0. We get the secondinequality as stated. If j = 0, note that ClF [p](ω

0) = Cl∆F . Remark 1.2.8identifies this group with ClQ[p] which is trivial. Hence ClF [p](ω

1) is trivialtoo. �

We will return frequently to the field F = Q(µp) later in this book.It will be one of our most important examples. Proposition 1.4.5 alreadytouches on one interesting question: what can one say about the dimensionsof the various components in the ∆-decomposition of ClF [p] ? There is animportant criterion for the nontriviality of the ωi-component when i is odd,the Herbrand-Ribet theorem. It involves the Bernoulli numbers Bm whichare defined by the following power series expansion

t

et − 1=

∞∑

m=0

Bmtm

m!

The Bernoulli numbers are nonzero rational numbers for even m > 0. It isknown that Bm is p-integral if m is not divisible by p − 1. We will simplywrite p|Bm when p divides the numerator of Bm.

Herbrand-Ribet Theorem. Suppose that p is an odd prime. Assume thati and j are integers in the range 2 ≤ i, j ≤ p − 2, that i is odd, and thati+ j ≡ 1 (mod p− 1). Let F = Q(µp). Then

ClF [p](ωi) 6= 1 ⇐⇒ p|Bj

We will discuss the proof of this theorem later. It is a refinement of Kummer’sfamous criterion for irregularity:

Kummer’s Criterion. The class number of F = Q(µp) is divisible by p ifand only if p|Bj for at least one even j in the range 2 ≤ j ≤ p− 2.

To explain the connection with the Herbrand-Ribet theorem, suppose firstthat p|hF . Then ClF [p](ω

i) 6= 1 for at least one i in the range 0 ≤ i ≤ p− 2.According to proposition 1.4.5, we know that i 6= 0 or 1 and that i can betaken to be odd. Herbrand proved that p|Bj where j = p− i, an even integer

40

in the range 2 ≤ j ≤ p− 2. Ribet proved that, conversely, if p|Bj for an even

j in the stated range, then ClF [p](ωi) 6= 1 for i = p− j.

As an example, let p = 37, the first irregular prime. Then 37|B32, but37 ∤ Bj for the other even values of j, 2 ≤ j ≤ 34. The Herbrand-Ribettheorem then asserts that ClF [37](ω

5) 6= 1 and that the ωi-component istrivial for the other odd values of i. For an even character ωj, proposition1.4.5 implies that ClF [37](ω

j) = 0 except possibly when j = 32. It turns outthat ClF [37](ω

32) = 0 too. This is so because p ∤ hF+.

In general, it seems reasonable to conjecture that ClF [p∞](ωi) is a cyclic

group, or equivalently, that dimFp

(ClF [p](ω

i))

= 1, for all odd i’s. A sufficient

condition for this to be so is that ClF [p](ωj) = 0 for all even j’s, as proposition

1.4.5 implies. There is no known example where an even component inClF [p] is nontrivial. It may never happen, an assertion often referred to asVandiver’s conjecture. We can state it in an equivalent form as follows:

Vandiver’s Conjecture. Let F = Q(µp). Then p ∤ hF+.

This conjecture has been verified for all p < 16, 000, 000. For all of theseprimes, it turns out that the nontrivial components ClF [p∞](ω

i) are all cyclicof order p.

A useful consequence of Vandiver’s conjecture concerns the structure ofAF = ClF [p∞] as a Zp[∆]-module. It follows immediately from the aboveremarks.

Proposition 1.4.6. Suppose that F = Q(µp). Let ∆ = Gal(F/Q). Assumethat h+

F is not divisible by p. . Then AF is cyclic when considered as a Zp[∆]-module. That is, AF ∼= Zp[∆]/I, where I is a certain ideal in Zp[∆].

We will have a lot to say about this ideal I in later chapters.

We now consider an important variation on proposition 1.4.2, a moreprecise illustration of the reflection principle. Suppose that F is a numberfield, p is a prime, and the hypotheses in proposition 1.4.2 are satisfied. LetM denote the compositum of all abelian, p-extensions of F which are ramifiedonly at primes if F above p. Let X = Gal(M/F ). Note that the p-Hilbertclass field L of F is contained in M . Also, M is obviously a Galois extensionof F∆ and so ∆ acts on X. The next result concerns the ∆-decompositionof X/Xp. Let Sp denotes the set of primes of F lying above p and let OF,Spdenote the ring OF [1

p]. The group of Sp-units of F , which is defined to be

41

O×F,Sp

, will play a role. Also, we let A′F denote the p-ideal class group of that

ring. This is isomorphic to AF/BF , where BF is the subgroup of ideal classesin AF which contain a product of primes in Sp. Note that ∆ acts on bothO×F,Sp

and A′F .

Proposition 1.4.7. Suppose that the assumptions in proposition 1.4.2 aresatisfied. Then

dimFp

((X/Xp)(ψ)

)= dimFp

(A′F [p](ϕ)

)+ dimFp

((O×

F,Sp/(O×

F,Sp)p)(ϕ)

)

.

Proof. We first make the following observation. Suppose α ∈ F×. ThenF ( p√α) ⊂ M if and only if p|ordv(α) for all finite primes v of F such that

v 6∈ Sp. This last condition means that αOF,Sp = Ip, where I is a fractionalideal for OF,Sp . Let A ⊂ F× denote the set of such α’s. Thus the compositum

of all cyclic extensions of F of degree p and unramified outside Sp is F ( p√A).

We haveX/Xp ∼= Gal(F (

p√A)/F ) ∼= Hom(A/(F×)p, µp)

These isomorphisms are ∆-equivariant.Now we also have the following exact sequence

1 −→ O×F,Sp

/(O×F,Sp

)pf−→A/(F×)p

g−→A′F [p] −→ 1

where the map f is induced by the inclusion O×F,Sp→ F× and the map g

is defined as follows. If α ∈ A, then write αOF,Sp = Ip as above. Let c bethe class of I in the class group for OF,Sp . Clearly, c ∈ A′

F [p]. We defineg(a(F×)p

)= c. The fact that g is a well-defined, surjective homomorphism

is easily verified. Also, α represents a coset in the kernel of g if and only ifα = βpη, where η ∈ O×

F,Sp, which proves the exactness.

The maps f and g are obviously ∆-equivariant and the exact sequencesplits because p ∤ |∆|. We have isomorphisms

(X/Xp)(ψ) ∼= Hom(A/(F×)p, µp)(ψ) ∼= Hom

(A/(F×)p)(ϕ), µp)

Now A/(F×)p)(ϕ) ∼=(O×F,Sp

/(O×F,Sp

)p)(ϕ) × A′

F [p](ϕ). Proposition 1.4.6 thenfollows. �

Returning to the case where F = Q(µp) and ∆ = Gal(F/Q), we have thefollowing result. The field M and the Galois group X are as defined above.

42

Corollary 1.4.8. Under the assumptions of proposition 1.4.5, we have

dimFp

((X/Xp)(ωj)

)= dimFp

(ClF [p](ω

i)),

dimFp

((X/Xp)(ωi)

)= dimFp

(ClF [p](ω

j))

+ 1.

Also, dimFp

((X/Xp)(ω1)

)= dimFp

((X/Xp)(ω0)

)= 1.

Proof. This follows easily from proposition 1.4.7. One just notes thatsince the prime of F lying above p is principal, we have AF ∼= A′

F . Also,dimFp

((O×

F,Sp/(O×

F,Sp)p)(ϕ)

)= 1 if ϕ = ωj where j is even or if j = 1. This

dimension is 0 otherwise. �

As an illustration, assume that h(p)F = p in corollary 1.4.8. Thus, ClF [p](ω

i)

is nontrivial for exactly one i. That value of i must be odd. If L denotesthe p-Hilbert class field of F , then L/Q is Galois, Gal(L/F ) is cyclic or or-der p, and ∆ = Gal(F/Q) acts on Gal(L/F ) by the character ωi. Now thecorresponding j is even and ωjωi = ω. The character ωj is the only non-trivial, even character of ∆ for which (X/pX)(ωj) is nontrivial. Furthermore,(X/pX)(ωj) is cyclic of order p. Thus, there is a unique subfield N of Msuch that N/Q is Galois, Gal(N/F ) is cyclic or order p, and ∆ = Gal(F/Q)acts on Gal(N/F ) by the character ωj. Also, N = F ( p

√α), where α is in the

group A defined in the proof of proposition 1.4.7. The coset of α inA/(F×)p

is contained in(A/(F×)p

)(ωi). It follows that the fractional ideal (α) for OF

is of the form Ip where I is a non-principal ideal of FF . This explains aremark that we made before concerning example 2 in Remark 1.2.11.

If we apply a version of Nakayama’s lemma (lemma 1.5.3 to be provedlater), we can say that ωj is the only nontrivial, even character for whichX(ωj) is nontrivial. Furthermore, since (X/pX)(ωj) is cyclic of order p, lemma1.5.3 implies that X(ωj) is either a finite cyclic p-group or isomorphic to Zp.

Although it is quite nontrivial to prove, it turns out that X(ωj) is finite, aconsequence of a proposition to be proved in chapter 3.

1.5 Unramified Galois Cohomology

Let F be a number field and let p be a prime. We will introduce an object inthis section which can be regarded as a generalization of ClF [p∞] (or, to bemore precise, the Pontryagin dual of that group). Let H be the Hilbert class

43

field of F . The Pontryagin dual of Gal(H/F ) is Hom(Gal(H/F ),Q/Z). Thiscan be viewed as a subgroup of the Galois cohomology group H1(GF ,Q/Z),where we are letting GF act trivially on Q/Z. To be precise,

H1(GF ,Q/Z) = Hom(GF ,Q/Z) = Hom(Gal(F ab/F ),Q/Z)

A homomorphism φ : Gal(F ab/F )→ Q/Z factors through Gal(H/F ) if andonly if the restrictions of φ to the inertia subgroups of Gal(F ab/F ) corre-sponding to all the primes of F are trivial. We denote this subgroup byH1unr(GF ,Q/Z). It can be identified with the Pontryagin dual of ClF by

using the Artin isomorphism ArtH/F .We always assume that cocycles or homomorphisms are continuous. The

topology on Q/Z is discrete and so “continuous” means “locally constant.”Since the topology on any Galois group G is compact, cocycles or homomor-phisms from G to a discrete group such as Q/Z will have only finitely manyvalues and will factor through a quotient group G/N where N is an open,normal subgroup of G.

The p-primary subgroup of Q/Z is isomorphic to Qp/Zp, where Zp de-notes the p-adic integers and Qp denotes the fraction field of Zp. Assumethat the action of GF on Qp/Zp is trivial. The p-primary subgroup ofH1unr(GF ,Q/Z) can then be identified with H1

unr(GF ,Qp/Zp). This is iso-morphic to Hom(Gal(L/F ),Qp/Zp), where L denotes the p-Hilbert class fieldof F . We will study a natural generalization, where we replace the trivialGalois module Qp/Zp by any group D ∼= (Qp/Zp)

d which has a continuousaction of GF . We consider D as having the discrete topology. Note that thePontryagin dual Hom(D,Qp/Zp) of D is isomorphic to Zd

p, a free Zp-moduleof rank d. We express this fact by saying that D is a cofree Zp-module andthat its Zp-corank is d.

Given such a D, the subgroup D[pn] (for any n ≥ 0) is isomorphic to(Z/pnZ)d as a group and has a certain action of GF . That action correspondsto a homomorphism GF −→ GLd(Z/p

nZ). We can regard D as the directlimit of these finite Galois modules. We define the “Tate-module” T for Dto be the inverse limit:

T = lim←−n

D[pn]

where the map D[pm] → D[pn] for m ≥ n ≥ 0 is multiplication by pm−n.Then T is a free Zp-module of rank d which has a continuous Zp-linear actionof GF . That action is given by a homomorphism ρD : GF −→ GLd(Zp). We

44

can also define a vector space V = T ⊗Zp Qp. This is a topological vectorspace over Qp of dimension d, and so has the topology of Qd

p. One then hasa continuous, Qp-linear action of GF on V .

If we start instead with such a finite-dimensional Qp-representation spaceV for GF , it is not hard to prove (using continuity and the compactness ofGF ) that V contains a free Zp-module T of rank d = dimQp(V ) which isGF -invariant. One could then take D = V/T as the corresponding discreteGF -module.

In general, suppose that R is a commutative ring and that we have ahomomorphism ρ : GF −→ GLd(R). The kernel will a normal subgroup N

of GF and the fixed field FN

will be a certain Galois extension of F which werefer to as the “the extension cut out by ρ”. For example, if ρD is as describedabove, then we can take R = Zp. We denote the extension cut out by ρD byF (D). Thus, Gal(F (D)/F ) is isomorphic to a subgroup of GLd(Zp), namelyim(ρD). Now ρD is a continuous map, Gal(F (D)/F ) is compact, and henceim(ρD) is a closed subgroup of GLd(Zp). Such a subgroup is known to be a p-adic Lie group and so one could say that F (D)/F is a “p-adic Lie extension”.Of course, it can be an infinite extension. For any n ≥ 0, we can consider therepresention ofGF onD[pn], where we take R = Z/pnZ. This is the reductionmodulo pn of ρD. The field cut out by this representation is a finite extensionof F , denoted by F (D[pn]). Note that F (D) =

⋃n≥0 F (D[pn]). We will use

the notation Ram(D) for the set of primes of F which are ramified in theextension F (D)/F . Thus, if v is a prime of F , then v ∈ Ram(D) if and onlyif the image of the inertia subgroup of GF for a prime of F lying above v isnontrivial. We will say that D is finitely ramified if Ram(D) is a finite set.

Consider the Galois cohomology group H1(GF , D). The subgroup ofH1(GF , D) that is referred to in the title of this section can be defined roughlyas the group of “everywhere unramified cocycle classes” and will be denotedby H1

unr(F,D). We will make this definition more precise.For every prime v of F , finite or infinite, we have the natural embedding

F → Fv, where Fv denotes the v-adic completion of F . This can be extendedto an embedding of F → F v. The choice of this embedding will not be im-portant. We then have the natural restriction maps GFv → GF arising fromthe above embeddings. Let IFv denote the inertia subgroup of GFv . ThusIFv = Gal(F v/F

unrv ), where F unr

v denotes the maximal unramified extensionof Fv. We get homomorphisms

H1(GF , D) −→ H1(GFv , D) −→ H1(IFv , D)

45

for every v. From here on, we will use the customary notation

H1(F, ∗), H1(Fv, ∗), and H1(F unrv , ∗)

instead of H1(GF , ∗), H1(GFv , ∗) and H1(IFv , D), where ∗ is any Galoismodule.

The “unramified Galois cohomology group” for D over F is defined by

H1unr(F,D) = ker

(H1(F,D) −→

∏

v

H1(F unrv , D)

).

where v runs over all the primes of F in the product. That is, if φ : GF −→ Dis a 1-cocycle, then its class [φ] is in H1

unr(F,D) if and only if [φ|IFv ] is trivialin H1(IFv , D) for all v ∈ U . In particular, if S is empty, then U consists ofall primes of F and we denote that group simply by H1

unr(F,D).

It is not true in general that H1unr(F,D) is finite. We will discuss several

examples later to illutrate this. However, we have the following finitenessresult.

Proposition 1.5.1. Assume that D is finitely ramified. Then H1unr(F,D)[p]

is finite.

Proof. The argument involves various simple applications of fundamentaltheorems of group cohomology. This proof will be an opportunity to intro-duce such applications carefully. Later arguments of this kind will be lessdetailed. The proof will be given in three parts.

The map H1(F,D[pn]) −→ H1(F,D)[pn]. First of all, we have an exactsequence 0 → D[pn] → D → D → 0 for any n ≥ 0. The map D → D isgiven by x → pnx for x ∈ D. The kernel of this map is D[pn] and map issurjective because D is a divisible group. For any i ≥ 1, we then have thefollowing part of the corresponding cohomology exact sequence

H i−1(F,D)pn−→H i−1(F,D) −→ H i(F,D[pn]) −→ H i(F,D)

pn−→H i(F,D)

Consequently, the map H i(F,D[pn]) −→ H i(F,D)[pn] must be surjective.Furthermore, we have

ker(H i(F,D[pn]) −→ H i(F,D)

) ∼= H i−1(F,D)/pnH i−1(F,D) (20)

If we take i = 1, then H0(F,D) = DGF is a Zp-submodule of D. The

Pontryagin dual D of D is isomorphic to Zdp; the Pontryagin dual of DGF is

46

a quotient of D. Hence it follows that H0(F,D) ∼= (Qp/Zp)e × C, where C

is finite and 0 ≤ e ≤ d. The subgroup (Qp/Zp)e is the maximal divisible

subgroup H0(F,D)div of H0(F,D) and the corresponding quotient groupH0(F,D)/H0(F,D)div is isomorphic to C. It follows that

H0(F,D)div ⊆ pnH0(F,D) ⊆ H0(F,D)

for all n and that pnH0(F,D) = H0(F,D)div if n is sufficiently large. Tosummarize, if n ≥ 0, then the map

H1(F,D[pn]) −→ H1(F,D)[pn] (21)

is surjective, has finite kernel, and the order of the kernel is bounded inde-pendently of n.

The map H1unr(F,D[pn]) −→ H1

unr(F,D)[pn]. We can use the subscript unreven for finite Galois modules. Consider the map

H1unr(F,D[pn]) −→ H1

unr(F,D)[pn] (22)

We already know that the kernel is finite and of bounded order since that istrue for the kernel of (21). Now we consider the cokernel. We will denotethe images of the following “global-to-local” maps

H1(F,D[pn]) −→∏

v

H1(F unrv , D[pn]), H1(F,D) −→

∏

v

H1(F unrv , D)

by G1(F,D[pn]) and G1(F,D), respectively. By definition, H1unr(F,D[pn])

and H1unr(F,D) are the kernels of those maps. We then have the following

commutative diagram with exact rows. The vertical maps are induced bythe inclusion D[pn] ⊂ D.

0 // H1unr(F,D[pn]) //

αn��

H1(F,D[pn]) //

βn��

G1(F,D[pn]) //

γn

��

0

0 // H1unr(F,D)[pn] // H1(F,D)[pn] // G1(F,D)[pn]

(23)

Note that we can’t say that the last map in the second row is surjective.Now the order of ker(αn) is bounded by the order of ker(βn). For studyingcoker(αn), we apply the snake lemma, obtaining the following useful exactsequence.

ker(γn) −→ coker(αn) −→ coker(βn) (24)

47

But coker(βn) = 0, as pointed out in part 1. We can study ker(γn) factor-by-factor on the entire direct products (over v) which contain G1(F,D[pn])and G1(F,D).

Suppose that v is any prime of F . Consider the map

γn,v : H1(F unrv , D[pn]) −→ H1(F unr

v , D)

An identical argument to the one for (21), applied to F unrv instead of F ,

shows that|ker(γn,v)| ≤ [H0(F unr

v , D) : H0(F unrv , D)div]

and hence is finite and bounded. Also, if v 6∈ Ram(D), then IFv acts triviallyon D, H0(F unr

v , D) = D, a divisible group, and hence the index is then1. That is, |ker(γn,v)| = 1 if v 6∈ Ram(D). Since D is assumed to befinitely ramified, we can conclude that ker(γn) is finite and of bounded order.Therefore, (24) implies that coker(αn) is finite and has bounded order.

Finiteness of H1unr(F,D[p]). To finish the proof, we will take n = 1. It obvi-

ously now suffices to prove that H1unr(F,D[p]) is finite. Let F ′ = F (D[p]), a

finite Galois extension of F . Let G = Gal(F ′/F ). Then we have the followingexact sequence, the first few terms of the inflation-restriction sequence.

0 −→ H1(F ′/F,D[p]) −→ H1(F,D[p]) −→ H1(F ′, D[p])G

Since both G and D[p] are finite, obviously so is H1(F ′/F,D[p]). Theimage of H1

unr(F,D[p]) under the restriction map is clearly contained inH1unr(F

′, D[p]). The kernel is finite and therefore it is enough to provethe finiteness of H1

unr(F′, D[p]). Now GF ′ acts trivially on D[p]. Hence

H1(F ′, D[p]) = Hom(GF ′ , D[p]) and

H1unr(F

′, D[p] = Hom(Gal(L′/F ′), D[p]),

where L′ is the p-Hilbert class field of F ′. The finiteness of H1unr(F,D[p])

follows from the fact that Gal(L′/F ′) ∼= ClF ′ [p∞], which is finite. �

We will refer back to various steps in this proof from time to time. Hereis an important corollary.

Corollary 1.5.2. Assume that D is finitely ramified. Then H1unr(F,D) is a

cofinitely generated Zp-module. Consequently, H1unr(F,D)div ∼= (Qp/Zp)

r forsome r ≥ 0 and H1

unr(F,D)/H1unr(F,D)div is finite.

48

Note that H1unr(F,D) will be isomorphic to the direct sum of its maximal

divisible subgroup and the corresponding finite quotient group.

Proof. Since H1unr(F,D) is a p-primary abelian group, it is a Zp-module.

Consider its Pontryagin dual X = Hom(H1unr(F,D),Qp/Zp

). We must show

that X is a finitely generated Zp-module. This implies that X ∼= Zrp ⊕Xtors

and that Xtors is finite. The final part of the corollary will then follow. ThePontryagin dual of H1

unr(F,D)[p] is X/pX and so proposition 1.5.1 impliesthe finiteness of that group. Furthermore, X is the Pontryagin dual of adiscrete Zp-module and hence will be a compact Zp-module. The followinglemma then completes the proof.

Lemma 1.5.3. (Nakayama’s lemma for compact Zp-modules.) Sup-pose that X is a compact Zp-module. Let x1, . . . , xd be in X. Let x1, . . . , xddenote their images in X/pX. Then x1, . . . , xd is a generating set for X asa Zp-module if and only if x1, ..., xd is a generating set for X/pX as a vectorspace over Fp = Zp/pZp.

Proof. One direction is obvious. For the other direction, where we assumethat x1, . . . , xd generate X/pX, one can give a quick proof that x1, . . . , xdgenerate X by using the compactness of Zp. It is easy to verify this if X isfinite, which we leave to the reader. In general, we have X = lim←−Xn, whereXn = X/pnX. One uses the compactness of X to verify that. Note thatour assumption implies that X/pX is finite. It follows easily that that Xn

is finite for all n. finite. The maps πn : X → Xn are surjective. Hence theinduced maps X/pX → Xn/pXn are also surjective. It follows from the finitecase that the images of x1, . . . , xd under the map πn must generate Xn. LetY be the Zp-submodule submodule of X generated by x1, . . . , xd. Since Y isthe image of Zd

p under a continuous map, it is compact and therefore closed.But πn(Y ) = Xn. This is true for all n and so it follows that Y is also densein X. Hence Y = X, as claimed. �

Remark 1.5.4. One can examine the first two parts of the proof of propo-sition 1.5.1 to determine when the map H1

unr(F,D[pn]) −→ H1unr(F,D)[pn]

is injective and/or surjective. For injectivity, it would suffice that DGF bedivisible. Of course, this could be true simply because DGF = 0. For sur-jectivity, it would suffice that DIFv be divisible for all primes v. For then,ker(γn,v) = 0 for all such v and therefore ker(γn) = 0. Now if v 6∈ Ram(D),then DIFv = D which is a divisible group. However, it could easily happenthat DIFv fails to be divisible for some v ∈ Ram(D). Even if that happens,

49

it is still possible that ker(γn) = 0.

Remark 1.5.5. We will also consider the following object. Suppose thatS is a finite set of primes of F . Let U denote the complement of S. Thendefine

H1U−unr(F,D) = ker

(H1(F,D) −→

∏

v∈UH1(F unr

v , D)).

If S is empty, then this group is H1unr(F,D). The proof of proposition 1.5.1

and its corollary can be applied to this group. One finds that the map

H1U−unr(F,D[p]) −→ H1

U−unr(F,D)[p]

has finite kernel and cokernel. In fact, if one choose S so that Ram(D) ⊂ S,then the cokernel is trivial. Now one can still prove that H1

U−unr(F,D[p]) isfinite. As in the above proof, one can let F ′ = F (D[p]). Let S ′ be the setof primes of F ′ lying above the primes in S. Let U ′ denote its complement.Then the image of H1

U−unr(F,D[p]) under the restriction map is containedin H1

U ′−unr(F′, D[p]), which is a subgroup of Hom(GF ′ , D[p]). If φ′ is in

that subgroup, then φ′ factors through Gal(K ′/F ′), where K ′ is a cyclicextension of F ′ of degree p which is ramified only at primes in S ′. Thediscriminant of such extensions K ′/Q is easily seen to be bounded. One canthen use Hermite’s theorem (which states that there are only finitely manyextensions of Q with a given discriminant) to see that only finitely manysuch extensions K ′ exist. Thus, H1

U ′−unr(F′, D[p]) is finite and therefore so

is H1U−unr(F,D[p]). It follows that H1

U−unr(F,D)[p] is finite and thereforeH1U−unr(F,D) is a cofinitely generated Zp-module.

Here is an important special case. Suppose that S contains Ram(D). LetFS denote the maximal extension of F unramified outside of S. Thus, theaction ofGF onD factors through the quotient group Gal(FS/F ). The inertiasubgroups of GF for primes lying above v ∈ U are all contained in GFS andgenerate a dense subgroup of that group. Furthermore, if φ ∈ H1

U−unr(F,D),then φ|GFS is a homomorphism which is trivial on every inertia subgroup,and is therefore trivial. More precisely, it is clear that

H1U−unr(F,D) = ker

(H1(F,D) −→ H1(FS, D)

) ∼= H1(FS/F,D),

the last isomorphism following from the inflation-restriction sequence. Con-sequently, it follows that H1(FS/F,D) is a cofinitely generated Zp-module.

50

Remark 1.5.6. One can define a generalization of the Pontryagin dual ofthe ideal class group ClF by letting p vary. Consider a compatible systemof p-adic representations V = {Vp} of GF . Thus, (i) each Vp is a Qp-vectorspace of common dimension d, (ii) the action of GF on Vp factors throughGal(FS∪Sp/F ), where S is a fixed finite set of primes of F , and (iii) if v is aprime of F , v 6∈ S, and p is a prime such that v ∤ p, then the characteristicpolynomial for the Frobenius automorphisms for v acting on Vp has coeffi-cients in Q and is independent of the choice of p. For each prime p, choosea Galois-invariant Zp-lattice Tp ⊂ Vp. Then one can consider the Galois-module DV =

⊕p Vp/Tp, which is isomorphic to (Q/Z)d as a group. One can

then define H1unr(F,DV) in the obvious way. One clearly has

H1unr(F,DV) ∼=

⊕

p

H1unr(F, Vp/Tp)

Obviously, H1unr(F,DV) is finite if and only if H1

unr(F, Vp/Tp) is finite for all pand trivial for all but finitely many p. One can say little in general about thefiniteness of this group. However, in remark 1.5.10, we will give an examplewhere H1

unr(F, Vp/Tp) is infinite for all primes p. It is also possible for thisgroup to be finite for all p and nontrivial for an infinite set of p, as we willpoint out in remark 1.6.5. In the special case where each Vp is the trivialrepresentation of GF , H1

unr(F,DV) is the Pontryagin dual of ClF and so isfinite.

The following result concerns Galois theory for H1unr(·, D). By this we

mean that it shows a relationship between an object associated with F andthe G-invariant subobject of an object associated to F ′. One example is thesimple fact that O×

F = (O×F ′)G. However, Galois theory for the unramified

cohomology groups is more subtle.

Proposition 1.5.7. Let F ′/F be a finite Galois extension. Then the kerneland cokernel of the restriction map

H1unr(F,D) −→ H1

unr(F′, D)G

are finite. If p ∤ [F ′ : F ], then the above restriction map is an isomorphism.

Proof. Let n = |G|. First we show that if C is a Z[G]-module such that C[n]and C/nC are both finite, then H i(G,C) is finite for any i ≥ 1. Multiplica-tion by n gives us two obvious exact sequences

0→ C[n]→ C → nC → 0, 0→ nC → C → C/nC → 0

51

The following exact sequences are part of the corresponding cohomologysequences.

H i(G,C[n]) −→ H i(G,C) −→ H i(G, nC),

H i−1(G,C/nC) −→ H i(G, nC) −→ H i(G,C)

Our assumption about C implies that H i(G,C[n]) and H i−1(G,C/nC) areboth finite. The composite map

H i(G,C) −→ H i(G, nC) −→ H i(G,C)

is just multiplication by n on H i(G,C). Both maps have finite kernels andtherefore H i(G,C)[n] is finite. But H i(G,C) is annihilated by |G| = n andtherefore H i(G,C) itself is indeed finite. Note also that if C[n] = 0 andnC = C, then the argument shows that H i(G,C) = 0.

The following exact sequence is part of the inflation-restriction sequence:

0→ H1(F ′/F,D(F ′)) −→ H1(F,D) −→ H1(F ′, D)G −→ H2(F ′/F,D(F ′))

where we letD(F ′) = DGF ′ = H0(F ′, D). Any subgroup C ofD is a cofinitelygenerated Zp-module. It is clear that C[n] and C/nC will be finite. Applyingthe remark at the beginning of the proof to C = D(F ′), it follows thatH1(F ′/F,D(F ′)) and H2(F ′/F,D(F ′)) are both finite. Also, if p ∤ n, thenboth these groups vanish.

Consider the following commutative diagram

0 // H1unr(F,D) //

aF ′/F

��

H1(F,D) //

bF ′/F

��

G1(F,D) //

cF ′/F

��

0

0 // H1unr(F

′, D)G // H1(F ′, D)G // G1(F ′, D)G

(25)

The finiteness of ker(bF ′/F ) implies the finiteness of ker(aF ′/F ). If p ∤ n,then clearly ker(bF ′/F ) = 0. As for the cokernel, the snake lemma gives thefollowing exact sequence

ker(cF ′/F ) −→ coker(aF ′/F ) −→ coker(bF ′/F ) (26)

Now coker(bF ′/F ) is finite and is trivial if p ∤ n. To prove the finiteness, ortriviality, of coker(aF ′/F ), it suffices to prove the finiteness, or triviality, ofker(cF ′/F ).

52

As in the proof of proposition 1.5.1, we study ker(cF ′/F ) factor-by-factor.Suppose v is any prime of F and v′ is a prime of F ′ lying above v. Considerthe restriction map

cv′/v : H1(IFv , D) −→ H1(IFv′ , D) (27)

The kernel is H1(Iv′/v, DIv′ ), where Iv′/v = IFv/IFv′ , a group which can be

identified with the inertia subgroup of G for the prime v′. The kernel of cv′/vfor every v. Furthermore, if v is unramified in F ′/F , then Iv′ = IFv andhence ker(cv′/v) is obviously trivial. Therefore, indeed, ker(cF ′/F ) is finite.Also, if p ∤ n, then p ∤ |Iv′/v| for all v, the kernels of the maps (27) are alltrivial, and hence ker(cF ′/F ) = 0 in that case, as stated. �

Let us now assume that the action of GF on V factors through a finitequotient group ∆ of GF . We will refer to such a V as an Artin representationspace. Let D = V/T , where T is a Galois-invariant Zp-lattice as before. Inthis case, we have the following result.

Corollary 1.5.8. Suppose that V is an Artin representation space. ThenH1unr(F,D) is finite.

Proof. Let F ′ = F (D), the field cut out by ρD, which will be a finite Galoisextension of F . By proposition 1.5.7, it suffices to show that H1

unr(F′, D).

But GF ′ acts trivially on D and hence

H1unr(F

′, D) = Hom(Gal(L′/F ′), D)

where L′ is the p-Hilbert class field of F ′. This group is obviously finite. �

Returning to the trivial Galois module D = Qp/Zp, one can translatesome of the results from the earlier sections rather easily. Consider corollary1.1.2. The surjectivity of the map NF ′/F : AF ′ → AF corresponds to thesurjectivity of the map RF ′/F : Gal(L′/F ′) → Gal(L/F ) which, in turn, isequivalent to the injectivity of the map

H1unr(F,Qp/Zp) −→ H1

unr(F′,Qp/Zp)

Now using the Artin isomorphism ArtL′/F ′ : AF ′ → Gal(L′/F ′), one hasisomorphisms

H1unr(F

′,Qp/Zp)G ∼= HomG(AF ′ ,Qp/Zp) ∼= Hom((AF ′)G,Qp/Zp)

53

If F ′/F is cyclic, then (AF ′)G is the genus group GF ′/F and so we have

GF ′/F∼= H1

unr(F′,Qp/Zp)

G.

If F ′/F satisfies the assumptions of proposition 1.1.3, then the assertionabout ker(NF ′/F ) in that proposition means that the map

H1unr(F,Qp/Zp) −→ H1

unr(F′,Qp/Zp)

G

is surjective. More generally, one can regard genus theory as an assertionabout the cokernel of that map. For, by definition, if F ′/F is a cyclic exten-sion, we have

G(o)F ′/F

∼= coker(H1unr(F,Qp/Zp) −→ H1

unr(F′,Qp/Zp)

G).

In general, one can study coker(H1unr(F,D) −→ H1

unr(F′, D)G

)by using (26).

This involves studying the kernels of the local restriction maps (27), whichis usually rather straightforward, and the image of the global-to-local mapG1(F,D), which is usually a more subtle question.

Next we return to the situation considered in section 1.4. That is, weassume that we have a finite group ∆ of automorphisms of F and that theorder of ∆ is prime to p. Let E = F∆. Suppose that ϕ is an irreduciblecharacter of ∆ over Qp. Thus, Vϕ is an Artin representation space for GE andthe action of GE factors through Gal(F/E). We denote the correspondingD by Dϕ. We then have the following result.

Proposition 1.5.9. Under the above assumptions, there is a canonical iso-morphism

H1unr(E,Dϕ) ∼= Hom∆(ClF [p∞](ϕ), Dϕ)

In particular, if dϕ = 1, then H1unr(E,Dϕ) is isomorphic to the Pontryagin

dual of ClF [p∞](ϕ).

Proof. We apply proposition 1.5.7 to the Galois extension F/E. We thenhave an isomorphism

aF/E : H1unr(E,Dϕ)→ H1

unr(F,Dϕ)∆.

We also have a canonical isomorphism

ArtL/F : Cl[p∞]→ Gal(L/F )

54

This isomorphism commutes with the natural actions of ∆ on Cl[p∞] andon Gal(L/F ) (with ∆ acting on on Gal(L/F ) by inner automorphisms asusual). We can identify H1

unr(F,Dϕ) with Hom(Gal(L/F ), Dϕ) and then weobtain the isomorphisms

H1unr(F,Dϕ)

∆ ∼= Hom∆(Gal(L/F ), Dϕ) ∼= Hom∆(Cl[p∞], Dϕ)

It is clear that a ∆-equivariant homomorphism Cl[p∞] → Dϕ must factorthrough the ϕ-component of Cl[p∞] and this gives the isomorphism in theproposition.

In the special case where ϕ is 1-dimensional, we have

Hom∆(ClF [p∞](ϕ), Dϕ) = Hom(ClF [p∞](ϕ), Dϕ)

which is isomorphic to the Pontryagin dual of ClF [p∞](ϕ) because Dϕ is iso-morphic to Qp/Zp as a group. �

Remark 1.5.10. One can also consider the group of “locally trivial cocycleclasses.” The precise definition is

H1triv(F,D) = ker

(H1(F,D)→

∏

v

H1(Fv, D))

Obviously, H1triv(F,D) is a subgroup of H1

unr(F,D). The two groups can cer-tainly differ. For example, let D = Qp/Zp with a trivial action of GF . ThenH1unr(F,Qp/Zp) = Hom(Gal(L/F ),Qp/Zp) and H1

triv(F,Qp/Zp) correspondsto the subgroup of homomorphisms f : Gal(L/F )→ Qp/Zp which are trivialon every decomposition subgroup of Gal(L/F ). Obviously, ClF is generatedby the ideal classes of the prime ideals P of F and so Gal(H/F ) is generatedby the Frobenius automorphisms σP for those prime ideals. Consequently,the same thing is true for Gal(L/F ). Hence if f(σP ) = 0 for all P , thenf ≡ 0. Thus, H1

triv(F,Qp/Zp) = 0.We will now give an example from the theory of elliptic curves to illustrate

the possibility thatH1triv(F,D) can be infinite. Of course, it would then follow

that H1unr(F,D) is infinite. Suppose that E is an elliptic curve defined over

F . Let p be a prime. For n ≥ 0, let E[pn] denote the elements of E(Q)of order dividing pn. As a group, E[pn] ∼= (Z/pnZ)2 and there is a naturalaction of GF on this group. We will consider

D = E[p∞] =⋃

n≥0

E[pn]

55

which is the p-primary subgroup of E(Q).A famous theorem of Mordell and Weil asserts that the group of F -

rational points E(F ) is a finitely generated abelian group. Let r denoteits rank. Consider the Kummer homomorphism

κ : E(F )⊗Z (Qp/Zp)→ H1(GF , E[p∞])

Note that E(F ) ⊗Z (Qp/Zp) ∼= (Qp/Zp)r. The definition of κ, which we

will now give, is an imitation of classical Kummer theory. (One finds abrief discussion of that in the next section.) Let α = a ⊗ (1/pn + Zp) ∈E(F )⊗Z (Qp/Zp), where a ∈ E(F ). Let b ∈ E(Q) be chosen so that pnb = a.Define a map σ : GF → E[p∞] by σ(g) = g(b)− b for all g ∈ GF . Then σ isa 1-cocycle and we define κ(α) to be the class [σ] in H1(F,E[p∞]) of σ. It isnot hard to verify that κ is injective.

Suppose that v is any prime of F . Consider the local Kummer homomor-phism

κv : E(Fv)⊗Z (Qp/Zp)→ H1(Fv, E[p∞])

This is defined just as above and is again an injective map. We then have acommutative diagram

E(F )⊗Z (Qp/Zp) −−−→ E(Fv)⊗Z (Qp/Zp)yκ

yκv

H1(F,E[p∞]) −−−→ H1(Fv, E[p∞])

(28)

The top horizontal map is induced by the inclusion E(F ) ⊂ E(Fv). Thebottom horizontal map is induced by the restriction map GFv → GF (corre-sponding to a fixed embedding F → F v.

Now it turns out that E(Fv)⊗Z(Qp/Zp) = 0 if v ∤ p. This is a consequenceof the following fact:

Let ℓ be an arbitrary prime. Suppose that Fv is a finite extension of Qℓ andthat E is an elliptic curve defined over Fv. Then E(Fv) ∼= Zn

ℓ × E(Fv)tors,where n = [Fv : Qℓ].

If ℓ 6= p, then Zℓ⊗Z (Qp/Zp) = 0. Since E(Fv)tors is finite, its tensor productwith Qp/Zp is also trivial. However, Zp ⊗Z (Qp/Zp) ∼= Qp/Zp, and so wehave

E(Fv)⊗Z (Qp/Zp) ∼= (Qp/Zp)[Fv :Qp]

56

if v | p. It is then clear from diagram (28) that the kernel of the map

E(F )⊗Z (Qp/Zp) −→∏

v|pE(Fv)⊗Z (Qp/Zp)

is a subgroup of H1triv(Q, E[p∞]). But this kernel is obviously infinite if

r >∑

v|p[Fv : Qp] = [F : Q].

This can certainly happen. For example, one could take F = Q and E to bean elliptic curve whose Mordell-Weil group has rank ≥ 2. If r is that rank,then H1

triv(Q, E[p∞]) contains a subgroup isomorphic to (Qp/Zp)r−1.

This is an example that we alluded to at the end of remark 1.5.6. We canassociate a compatible system V of p-adic representations with E by definingVp(E) = Tp(E)⊗Zp Qp, where Tp(E) is the p-adic Tate-module for E. Then

D = Vp(E)/Tp(E) ∼= E[p∞] and DV ∼= E(Q)tors. The above discussion showsthat H1

unr(F,DV) contains a subgroup isomorphic to (Q/Z)r−1.

1.6 Powers of the cyclotomic character.

Let µp∞ =⋃n≥0 µpn denote the group of p-power roots of unity in Q. The

extension F (µp∞)/F is an infinite Galois extension and one can define acontinuous homomorphism

χ : Gal(F (µp∞)/F )→ Z×p

in the following way. Let g ∈ Gal(F (µp∞)/F ). For every n ≥ 0, there existsan integer un such that g(ζ) = ζun for all ζ ∈ µpn . This integer un is uniquelydetermined modulo pn and is not divisible by p. The definition implies thatun+1 ≡ un (mod pn) and hence that {un} converges to a certain p-adic unitu. We then have g(ζ) = ζu for all ζ ∈ µp∞ . Define χ(g) = u. The statedproperties of χ are easily verified. One refers to χ as the “p-power cyclotomiccharacter,” regarding it often as a character of GF which factors through thequotient group Gal(F (µp∞)/F ).

The character χ gives the action of GF on T = lim←−µpn , which is a freeZp-module of rank 1. The GF -module T with this action is often denoted byZp(1). It is a Galois-invariant Zp-lattice in the vector space V = T ⊗Zp Qp,which is often denoted by Qp(1). The quotient V/T is isomorphic to µp∞ .

57

In this section, we want to discuss one-dimensional representations of GF

over Qp arising from powers of χ. In particular, we will consider the powersχn, where n ∈ Z, but it turns out to be important to consider a more generalclass of representations. The prime p will be assumed to be odd in most ofthe results. We will discuss p = 2 at the very end.

Consider an arbitrary continuous homomorphism

ψ : Gal(F (µp∞)/F )→ Z×p

Then ψ defines a 1-dimensional representation space V for Gal(F (µp∞)/F ).Equivalently, we can regard V as a representation space for GF , where theaction factors through the restriction map GF → Gal(F (µp∞)/F ). Any Zp-lattice T will be GF -invariant and D = V/T will be isomorphic to Qp/Zp

as a group, but with the action of GF given by the character ψ. We willdenote this D by Dψ in the rest of this section. If ψ has finite order, thencorollary 1.5.8 asserts that H1

unr(F,Dψ) is finite. As we will discuss in thenext chapter, it is possible for H1

unr(F,Dψ) to be infinite if ψ has infiniteorder, but this can only happen for finitely many choices of ψ.

Since we are assuming that p is odd, we have a direct product decompo-sition

Z×p∼= µp−1 × (1 + pZp)

Corresponding to this decomposition, we will write χ = ω〈χ〉, where ω is acharacter of order p − 1, the composition of χ with projection to the factorµp−1, and 〈χ〉 is the composition of χ with projection to the factor 1 + pZp.

Assume that χ is surjective. That is, we assume that χ defines an isomor-phism Gal(F (µp∞)/F ) ∼= Z×

p . Then it is not hard to see that an arbitrary ψcan be uniquely expressed in the form

ψ = ωi〈χ〉s

where 0 ≤ i ≤ p − 2 and s ∈ Zp. Thus, in a sense, one can regard ψ as apower of χ. It is actually a limit of integral powers of χ. To explain what wemean, note that the group Hom

(Gal(F (µp∞)/F ),Z×

p

)has a natural topology

on it, the compact-open topology. It is rather simple to describe in this casebecause Gal(F (µp∞)/F ) is compact and contains a dense, cyclic subgroup.If go is a topological generator for Gal(F (µp∞)/F ), then there is a bijection

Hom(Gal(F (µp∞)/F ),Z×

p

)−→ Z×

p

58

defined by sending ψ to ψ(go). The compact-open topology then coincideswith the topology transferred from Z×

p . To see that {χn | n ∈ Z} is a densesubgroup of Hom(Gal(F (µp∞)/F ),Z×

p ), consider ψ = ωi〈χ〉s. One can takea sequence of integers nk such that (i) nk ≡ i (mod p− 1) for all k and (ii)nk → s in Zp as k → ∞. Then χnk → ψ in the compact-open topology onHom

(Gal(F (µp∞)/F ),Z×

p

).

For a fixed i, the following proposition describes a kind of continuity forthe behavior of H1

unr(F,Dψ) as s varies over Zp. In addition to the aboveassumption about χ, we impose a mild ramification assumption.

Proposition 1.6.1. Assume that p is odd, that χ is an isomorphism, andthat every prime of F lying above p is totally ramified in F (µp)/F . Supposethat i is fixed, that 1 ≤ i ≤ p − 2, and that ψ = ωi〈χ〉s for some s ∈ Zp.Then

H1unr(F,Dψ)[pk] ∼= H1

unr(F,Dψ[pk])

for any k ≥ 1. Assume that s1, s2 ∈ Zp satisfy s1 ≡ s2 (mod pk−1). Letψ1 = ωi〈χ〉s1 , ψ2 = ωi〈χ〉s2. Then H1

unr(F,Dψ1)[pk] ∼= H1

unr(F,Dψ2)[pk].

Proof. First note thatGF acts onDψ[p] by the character ωi. The assumptionsimply that this character is nontrivial. Hence H0(F,Dψ) = 0. Also, if v isany prime of F lying above p, then the assumptions imply that ωi|IFv isnontrivial. Therefore, the action of IFv on Dψ[p] is nontrivial and hence wehave H0(F unr

v , Dψ) = 0. If v is an infinite prime, then H0(Fv, Dψ) = 0 if iis odd and H0(Fv, Dψ) = Dψ if i is even. If v is any other prime of F , thenv is unramified in F (µp∞)/F and hence IFv acts trivially on Dψ. Therefore,H0(F unr

v , Dψ) = Dψ. Thus, for all primes v of F , H0(F unrv , Dψ) is a divisible

group. The first part of the proposition then follows from remark 1.5.4,taking n = k. The conditions which imply injectivity and surjectivity aresatisfied.

If s1 ≡ s2 (mod pk−1), then us1 ≡ us2 (mod pk) for all u ∈ 1 + pZp.Therefore, for any g ∈ GF , we have 〈χ〉s1(g) ≡ 〈χ〉s2(g) (mod pk). It followsthat

Dψ1[pk] ∼= Dψ2

[pk]

as (Z/pkZ)-modules with an action of GF . The second part of the propositionthen follows from the first part. �

In the above proposition, one can take k = 1, s1 arbitrary and s2 = 0.Then ψ2 = ωi, the character of an Artin representation. Consequently, wehave the following result.

59

Corollary 1.6.2. Under the assumptions of proposition 1.6.1, we haveH1unr(F,Dψ)[p] ∼= H1

unr(F,Dωi)[p]. Thus, dimFp

(H1unr(F,Dψ)[p]

)depends

only on i.

Now suppose that ψ = χn, where n ∈ Z. The following corollary followsimmediately from proposition 1.6.1.

Corollary 1.6.3. Under the assumptions of proposition 1.6.1, we have

H1unr(F,Dχn1 )[pk] ∼= H1

unr(F,Dχn2 )[pk]

if n1, n2 ∈ Z satisfy the congruence n1 ≡ n2 (mod (p − 1)pk−1) and are notdivisible by p− 1.

The case where ψ = 〈χ〉s, which is excluded in proposition 1.6.1 and theabove corollaries, is actually quite interesting to consider. There can be a dis-continuity at s = 0. If ψ = ψo, which corresponds to s = 0, thenH1

unr(F,Dψo)is the Pontryagin dual of ClF [p∞]. It turns out that H1

unr(F,Dψ) is finite fors close to 0 in Zp, but, in certain cases, its order will be unbounded as s→ 0.

In certain other cases, its order will be bounded, but larger than h(p)F .

We will assume that s 6= 0. Let F∞ = F (Dψ) denote the field cut out byψ. Obviously, F∞ is a subfield of F (µp∞) and Gal(F∞/F ) ∼= im(ψ) which isisomorphic to Zp. We will let Γ denote Gal(F∞/F ). One often refers to F∞as the cyclotomic Zp-extension of F . We will have more to say about it atthe beginning of chapter 2. For simplicity, we will assume that the primesof F lying over p are all totally ramified in F∞/F . It is clear that all othernon-archimedian primes are unramified since the same is true for F (µp∞)/F .Applying the snake lemma to (23) gives us the following extension of (24),where we use the same notation for the maps and take D = Dψ:

ker(βn) −→ ker(γn) −→ coker(αn) −→ coker(βn) (29)

in the notation of those diagrams. We have coker(βn) = 0, as explained inthe first part of the proof of proposition 1.5.1. (See (21).)

Since s 6= 0, H0(F,Dψ) will be a finite cyclic group. Let pu denote itsorder. Note that u ≥ 1. If ordp(s) = k ≥ 0, then u = k + 1. It follows from(20) that ker(βn) will be cyclic and its order will be pmin(n,u). If n = u, thenone can describe this kernel very simply. In that case, GF acts trivially onDψ[pn], H1(F,Dψ[pn]) = Hom(GF , Dψ[pn]), and a 1-cocycle φ is in ker(βn)if and only if φ is the coboundary of an element of D[p2n]. Since the action

60

of GF on D[p2n] factors through Γ = Gal(F∞/F ), any φ ∈ ker(βn) will alsofactor through Γ and hence through Γ/Γp

n. We will denote the fixed field for

Γpn

by Fn, a cyclic extension of F of degree pn. Thus, it follows that

ker(βn) = Hom(Gal(Fn/F ), Dψ[pn]) (30)

if n = ordp(s) + 1. It remains to study ker(γn) and the image of the mapker(βn) −→ ker(γn) in (29). That image is cyclic and not difficult to study.As we will see, it is possible for ker(γn) to be non-cyclic. If that is so,then coker(αn) would obviously be nontrivial and that would imply thatH1unr(F,Dψ) 6= 0.

Since ψ is unramified for all v ∤ p, it follows from the second part of theproof of proposition 1.5.1 that ker(γn,v) = 0 for such v. But for any v|p,we are assuming that the inertia subgroup of Gal(F∞/F ) is the entire groupand hence H0(F unr

v , Dψ) = H0(F,Dψ) will have order pu. In fact, just asexplained above, if n = ordp(s) + 1, then ker(γn,v) will be cyclic of orderpn. Suppose that there are t primes of F lying over p. Then ker(γn) couldpotentially have t cyclic factors of order at least pn. However, the actual sizeof ker(γn) depends on the intersection

im(H1(FSp/F,Dψ[pn])→

∏

v|pH1(F unr

v , D[pn])) ⋂ ∏

v|pker(γn,v)

where Sp denotes the set of primes of F lying above p. We will consider twoextreme cases in the following result.

Proposition 1.6.4. Assume that p is an odd prime and that all primes inSp are totally ramified in F∞/F . Let ψ = 〈χ〉s, where s ∈ Zp, s 6= 0.

(i) If |Sp| = 1 and n = ordp(s) + 1, then coker(αn) = 0. In particular, if

h(p)F = 1, then H1

unr(F,Dψ) = 0.

(ii) If p splits completely in F/Q and n = ordp(s) + 1, then

Hom(Gal(FSp/F ), Dψ[pn])/Hom(Gal(Fn/F ), Dψ[pn]) ∼= H1

unr(F,Dψ)[pn]

In particular, if F has a nontrivial cyclic p-extension which is ramified in Spand not contained in F∞, then H1

unr(F,Dψ) 6= 0.

Proof. If there is just one prime v ∈ Sp, then the map ker(βn) −→ ker(γn) issurjective. This follows from the facts mentioned above: both ker(βn) and

61

ker(γn,v) have order pn, the map is injective since v is totally ramified inFn/F . Hence, in case (i), coker(αn) = 0 and we have an isomorphism

H1unr(F,Dψ)[pn] ∼= H1

unr(F,Dψ[pn])

for n = ordp(s) + 1. In particular, if we make the assumption that p ∤ hF ,then for any s ∈ Zp, it follows that H1

unr(F,Dψ)[pn] = 0 for some n ≥ 1 andhence H1

unr(F,Dψ) = 0.Now assume that p splits completely in F/Q. Then Fv = Qp for all

v ∈ Sp. We will then show that the map

H1(Fv, Dψ[pn]) −→ H1(F unrv , Dψ) (31)

is trivial for n as stated. To see this, note that an element φ ∈ H1(Fv, Dψ[pn])is just a homomorphism of order dividing pn and factors through a cyclic ex-tension of Fv. Now both F unr

v and Fv(µp∞) contain unique cyclic extensionof Fv of degree pn. The intersection of those two extensions is Fv since one isan unramified extension, the other totally ramified. Their compositum con-tains any cyclic extension of Fv of degree dividing pn. This follows from localclass field theory since F×

v /(F×v )p ∼= (Z/pn/Z)2. It follows that φ = φunrφcyc,

where φunr and φcyc are elements of H1(Fv, Dψ[pn]) which factor throughGal(F unr

v )/Fv) and Gal(Fv(µp∞)/Fv), respectively. They are both homomor-phisms of order dividing pn. The earlier argument describing ker(βn) applieswithout change to Fv. It follows that

ker(H1(Fv, Dψ[pn]) −→ H1(Fv, Dψ)

)

contains φcyc. On the other hand, φunr is clearly in the kernel of the restrictionmap H1(Fv, Dψ) −→ H1(F unr

v , Dψ). It follows that the image of φ under themap (31) is indeed trivial, as stated.

Now the inflation map identifies Hom(Gal(FSp/F ), Dψ[pn]) with a cer-tain subgroup of H1(F,Dψ[pn]) = Hom(GF , Dψ[pn]). Under the assumptionthat v splits completely in F/Q, (31) implies that its image under βn is con-tained in H1

unr(F,Dψ)[pn]. Conversely, since ker(γn,v) is trivial for all v ∤ p, ifφ ∈ Hom(GF , Dψ[pn]) and βn(φ) ∈ H1

unr(F,Dψ)[pn], then φ factors throughGal(FSp/F ). It follows that

βn(Hom(Gal(FSp/F ), Dψ[pn])

)= H1

unr(F,Dψ)[pn]

Thus part (ii) of the proposition now follows from (30). �

62

As an example, suppose that F is an imaginary quadratic field and thatp splits in F . Then, it turns out that for any n ≥ 1, FSp contains a subfieldMn such that Gal(Mn/F ) ∼= (Z/pnZ)2. It is not difficult to prove this usingclass field theory. Since the unit group of F is finite, the structure of rayclass groups is relatively easy to study. Therefore, proposition 1.6.4 impliesthat if s 6= 0, but s ≡ 0 (mod pn−1), then H1

unr(F,D〈χ〉s) contains a subgroupof order pn. Thus, the order of H1

unr(F,D〈χ〉s) will be unbounded as s → 0in Zp. We have not justified the assertion that this group is finite if s issufficiently close to 0. That will become easy to verify in chapter 2.

A topic which we will discuss in detail in chapter 3 concerns the specialcase where ψ = χ, the p-power cyclotomic character itself. Classical Kummertheory gives the following important isomorphism:

κ : F× ⊗Z (Qp/Zp)→ H1(F, µp∞)

The map is easily defined. Let α = a ⊗ (1/pn + Zp) ∈ F× ⊗Z (Qp/Zp),

where a ∈ F×. Let b be a pn-th root of a in Q×. Define σ : GF → µp∞

by σ(g) = g(b)/b for all g ∈ GF . Then σ is a 1-cocycle and we define κ(α)to be the class [σ] in H1(F, µp∞) of σ. It is not hard to verify that κ isinjective. The surjectivity is a consequence of Hilbert’s theorem 90, the fact

that H1(F,Q×) = 1, and will also be left to the reader.

Note that H1(F, µp∞) is a very big group. Indeed, it is not hard toshow that F× ⊗Z (Qp/Zp) is a direct sum of a countable number of copiesof Qp/Zp. On the other hand, there is considerable reason to believe thefollowing important conjecture:

Leopoldt’s Conjecture. H1unr(F, µp∞) is a finite group.

The usual way to formulate this conjecture involves the image of the unitgroup of F in the direct product of the completions of F at the primes abovep. The connection with the units of F arises from Kummer theory. If a ∈ O×

F

and n ≥ 1, then one can consider the cocycle class in H1(F, µp∞) associatedto b = pn

√a. This cocycle class is unramified at all primes of F not dividing p.

Whether or not it is unramified at a prime v dividing p is related (althoughnot quite equivalent) to whether or not Fv( pn

√a)/Fv is a ramified extension.

Leopoldt’s conjecture essentially amounts to the assertion that only finitelymany of these cocycle classes are unramified at all v|p.

Suppose now that F = Q, p is an odd prime, and ψ = χn for some n ∈ Zsuch that n 6≡ 0 (mod p − 1). The assumptions in proposition 1.6.1 and its

63

corollaries are satisfied. These groups are trivial if p is a regular prime. Tosee this, it is enough to verify that H1

unr(Q, Dχn)[p] = 0. Choose i so thatn ≡ i (mod p − 1), 1 ≤ i ≤ p − 2. According to corollary 1.6.2, it sufficesto show that H1

unr(Q, Dωi) = 0. But this is true according to proposition1.5.8 since, by assumption, ClQ(µp)[p

∞](ωi) = 0. If n ≡ 0 (mod p − 1), then

H1unr(Q, Dχn)[p] = 0. This follows from part (i) of proposition 1.6.4.

If p is an irregular prime, then ClQ(µp)[p∞](ω

i) 6= 0 for at least one oddvalue of i. Thus, for any such i, it follows that H1

unr(Q, Dχn) 6= 0 for anyinteger n ≡ i (mod p − 1). In a later chapter, we will prove that the groupH1unr(Q, Dχn) is finite for all odd, negative values of n. The order of this

group turns out to be closely related to values of the Riemann zeta functionζ(s). Recall that ζ(s) can be analytically continued to the complex plane(with a simple pole at s = 1). Its functional equation forces ζ(n) = 0 when nis a negative, even integer. It is known that ζ(n) is a nonzero, rational numberif n is negative and odd, closely related to one of the Bernoulli numbers. Tobe precise, if we write n = 1−m, where m is positive and even, then

ζ(n) = −Bm

m.

It is useful to state the congruences

m1 ≡ m2 6≡ 0 (mod p− 1) =⇒ Bm1

m1

≡ Bm2

m2

(mod pZp)

which are known as the Kummer congruences. We have expressed them ascongruences modulo pZp since the quantities Bm/m are p-integral, and hencein Zp, for m 6≡ 0 (mod p− 1). Note that for 0 < m < p, Bm is divisible by pprecisely when Bm/m is divisible by p. Thus, the Herbrand-Ribet theoremcan be stated as follows:

If n is an odd, negative integer, then H1unr(Q, Dχn) 6= 0 if and only if p divides

the numerator of ζ(n).

The following much more precise result is a consequence of a theorem ofMazur and Wiles to be discussed in a later chapter. It is an illustrationof the close connection between values of L-functions and certain objectsdefined by Galois cohomology.

Theorem. Suppose that p is an odd prime, that n is an odd, negative integer,and that n 6≡ 1 (mod p− 1). Then the order of H1

unr(Q, Dχn) is equal to thepower of p dividing ζ(n).

64

In chapter 2, we will show that the groups H1unr(Q, Dχn) can be studied by

an analogue of proposition 1.5.7 for the infinite Galois extension Q(µp∞)/Q.It turns out that the kernel and cokernel are actually trivial. This fact willallow us to relate the structure of the finite groups H1

unr(Q, Dχn), for all n,to the structure of the Galois group of a certain extension of Q(µp∞), theanalogue of the p-Hilbert class field.

For n ≡ 1 (mod p − 1), one has H1unr(Q, Dχn)[p] ∼= H1

unr(Q, Dω)[p] = 0and hence H1

unr(Q, Dχn) = 0. However, assuming n is also negative, thedenominator of ζ(n) is then divisible by p. To be precise, suppose thatn = 1 − m where m > 0 and m ≡ 0 (mod p − 1), then the Clausen-vonStaudt theorem states ordp(Bm) = −1. Thus,

ordp(ζ(n)

)= −1− ordp(m)

and so the power of p dividing the denominator of ζ(n) is unbounded. Infact, restricting to negative n ≡ 1 (mod p− 1), as n→ 1 p-adically, we haveordp

(ζ(n)

)→ −∞. This corresponds to a property of the Kubota-Leopoldt

p-adic L-function Lp(ω0, s), namely that this function has a pole at s = 1.

In light of the above theorem and the continuity properties described incorollary 1.6.3, it is natural to ask whether analogous continuity propertieshold for the powers of p dividing the values of ζ(s) at negative integers. Arefinement of the Kummer congruences stated earlier provides an even moreprecise result. Define

ζp(s) =(1− 1

ps)ζ(s),

which is just the function defined by the Euler product for ζ(s) with the Eulerfactor for p removed, analytically continued to the complex plane (excepts = 1). Thus, if n is a negative, odd integer, then ζp(n) = (1 − p|n|)ζ(n),which is nonzero and divisible by the same poswer of p as ζ(n). Assume thatk ≥ 1. The refined Kummer congruences, stated in terms of values of ζp(s),are

n1 ≡ n2 (mod (p− 1)pk−1) =⇒ ζp(n1) ≡ ζp(n2) (mod pkZp)

where it is assume that n1, n2 are odd, negative integers and n1, n2 6≡ 1(mod p− 1).

Remark 1.6.5. We return to the discussion in remark 1.5.6. Suppose thatn ∈ Z is fixed. We will consider the compatible system V = {Vp} where, for

65

each prime p, Vp is 1-dimensional and GQ acts by χn, where χ is the p-powercyclotomic character. The Mazur-Wiles theorem implies that if n is odd andnegative, then the finite groupH1

unr(Q, Dχn) is trivial for all but finitely manyprimes p. It follows that H1

unr(Q,DV) is finite. This can actually be provedjust using Herbrand’s half of the Herbrand-Ribet theorem, the assertion thatif 0 < m < p, m is even, and p ∤ Bm, then H1

unr(Q, Dω1−m) = 0. In fact, it isnot necessary to assume thatm < p. Since Bm 6= 0, only finitely many primesp can divide Bm. Hence, if n = 1 −m, it follows that H1

unr(Q, Dχn) = 0 ifp ∤ Bm.

The situation is quite different for positive, odd values of n. First of all,apart from the cases discussed above, it is not even known that H1

unr(Q, Dχn)is finite, although this is expected to be so. Furthermore, we have

H1unr(Q, Dχn)[p] ∼= H1

unr(Q, Dωn)[p]

and, if p > n, the Herbrand-Ribet theorem implies that this group is nontriv-ial if and only if p|Bp−n. However, it seems reasonable to conjecture that thisoccurs for an infinite, although very sparse, set of primes p. For example,take n = 3. It turns out that p|Bp−3 for p = 16843 and for p = 2124679.For even values of n, positive or negative, Vandiver’s conjecture implies thatH1unr(Q, Dχn) = 0 for all p.

66