1 Zp-extensions and ideal class groups.

This chapter will present the theorems of Iwasawa concerning the growth ofClFn

[p∞], where Fn varies over the layers in a Zp-extension of a number fieldF . The main theorem was proved by Iwasawa in the mid 1950s and concernsthe growth of the orders of these groups. However, we will also prove resultsof Iwasawa concerning their group structure. A key ingredient in the proofis to consider the inverse limit

X = XF∞/F = lim←−n

ClFn[p∞]

as a module over the formal power series ring Λ = Zp[[T ]]. The inverse limitX is defined by the norm maps NFm/Fn

for m ≥ n ≥ 0. It turns out to bea finitely generated, torsion Λ-module. We will be able to partially describethe structure of such modules, enough for a proof of the theorem. Finally,in the last section, we discuss the special case where F = Q(µp) for an oddprime p and F∞ = Q(µp∞). Then F∞/F is a Zp-extension. The n-th layer isFn = Q(µpn+1). There is a lot that we can say about the various invariantsand modules introduced in this chapter, a topic which will be continued inlater chapters. We will also discuss the relationship between XF∞/F andthe unramified cohomology groups associated to powers of the cyclotomiccharacter, continuing the topic of section 1.6.

1.1 Introductory remarks about Zp-extensions.

The theorem of Iwasawa alluded to above concerns a certain type of infiniteextension K of a number field F . These infinite extensions were originallyreferred to as “Γ-extensions” by Iwasawa, but later he adopted the moredescriptive term “Zp-extensions.” Let p be a fixed prime. A Galois extensionK/F is called a Zp-extension if the topological group Gal(K/F ) is isomorphicto the additive group Zp of p-adic integers.

Except for the trivial subgroup, all the closed subgroups of Zp have finiteindex. Such a closed subgroup is of the form pnZp for some nonnegativeinteger n and the corresponding quotient group is cyclic of order pn. Thus, ifK/F is a Zp-extension, the finite extensions of F which are contained in Kform a tower F = F0 ⊂ F1 ⊂ · · · ⊂ Fn ⊂ · · · of Galois extensions of F suchthat Gal(Fn/F ) = Z/pnZ for all n. Clearly K =

⋃n≥0 Fn. If one chooses any

1

γo ∈ Gal(K/F ) such that γo|F1is nontrivial, then the infinite cyclic subgroup

generated by γo is dense in Gal(K/F ). We therefore say that Gal(K/F ) is atopologically cyclic group and that the element γo is a topological generatorof Gal(K/F ). We will often use the notation F∞ for a Zp-extension of F .

Let F be any number field and let p be a fixed prime. One importantexample of a Zp-extension of F is quite easy to construct. Let µp∞ denotethe group of p-power roots of unity. The extension F (µp∞)/F is an infiniteGalois extension. At the beginning of section 1.6, we defined a continuoushomomorphism

χ : Gal(F (µp∞)/F )→ Z×p .

This homomorphism is injective. Consequently, Gal(F (µp∞)/F ) is isomor-phic to an infinite closed subgroup of Z×

p . Such a group has a finite torsionsubgroup and the corresponding quotient group will be isomorphic to Zp.Therefore, F (µp∞) contains a unique subfield F∞ such that Gal(F∞/F ) ∼= Zp.We refer to F∞ as the cyclotomic Zp-extension of F . In particular, we will letQ∞ denote the cyclotomic Zp-extension of Q. The cyclotomic Zp-extensionof an arbitrary number field F is then F∞ = FQ∞.

It is easy to show that the primes of F which are ramified in the cyclotomicZp-extension F∞/F are precisely the primes lying over p. For an arbitraryZp-extension, we have the following result.

Proposition 2.1.1. Suppose that F∞/F is a Zp-extension. If v is a primeof F which is ramified in the extension F∞/F , then v lies over p. At leastone such prime must be ramified in F∞/F .

Proof. Let Γ = Gal(F∞/F ). Let Iv denote the inertia subgroup of Gal(F∞/F )corresponding to a prime v of F . If v is ramified in F∞/F , then Iv is non-trivial. Hence Iv must be infinite. If v is an archimedian prime of F , then Ivwould be of order 1 or 2, and so must be trivial. Consequently, archimedianprimes of F split completely in F∞/F . If v is nonarchimedian, but lies over l,where l 6= p, then v is tamely ramified in F∞/F . It is known in general that ifv is tamely ramified in any abelian extension of F , then its ramification indexmust divide N(v)−1, where N(v) denotes the cardinality of the residue fieldfor v. This can be proved either by using properties of ramification groupsor by using local class field theory. (See reference.) Thus, Iv would be finite.Therefore, Iv must be trivial and v must be unramified in F∞/F .

For the final assertion, we just remark that the maximal unramified,abelian extension of F (the Hilbert class field of F ) has finite degree over F .Thus, F∞/F must be ramified for at least one prime. �

2

The existence and ramification properties of Zp-extensions of any numberfield F will be discussed in considerable detail in chapter 3. We will just makea few remarks now. If we take F = Q as the base field, then it is not hardto prove that there is only one Zp-extension, the cyclotomic Zp-extensionQ∞ which was constructed above. To see this, one can use the Kronecker-Weber theorem which asserts that the maximal abelian extension Qab of Qis generated by all the roots of unity. Proposition 2.1.1 then implies thatany Zp-extension of Q must be ramified only at p and therefore contained inQ(µp∞), and so must be Q∞. The cyclotomic Zp-extension of an arbitrarynumber field F is F∞ = FQ∞.

If F is a totally real number field, then it should again be true that thecyclotomic Zp-extension is the only Zp-extension of F . This can be provedif F ⊂ Qab, but is an open question in general (a special case of “Leopoldt’sConjecture”). If F is not totally real, then it turns out that there are infinitelymany distinct Zp-extensions of F . We will discuss this matter in detail inchapter 3. In particular, theorem 3.3 gives a quantitative statement aboutthe existence of Zp-extensions of an arbitrary number field F .

One of the main results to be proved in this chapter is the followingfamous theorem of Iwasawa.

Iwasawa’s Growth Formula. Suppose that F∞ =⋃n≥0 Fn is a Zp-extension

of a number field F . Let hn denote the class number of Fn and let h(p)n = pen

denote the largest power of p dividing hn. Then there exists integers λ, µ,and ν such that en = λn+ µpn + ν for all sufficiently large n.

Iwasawa’s growth formula will be proved in section 2.4, based largely on theresults of section 2.2 and 2.3. The integers λ and µ will be nonnegative. Wewill refer to them as the Iwasawa invariants for F∞/F , often denoting themby λ(F∞/F ) and µ(F∞/F ). Several interpretations of them will be given aswe proceed.

Proposition 1.1.4 implies one very simple special case of Iwasawa’s theo-rem, namely the following useful result.

Proposition 2.1.2. Suppose that F is a number field and that p does notdivide the class number of F . Let F∞ =

⋃n≥0 Fn be a Zp-extension of F

and suppose that only one prime of F is ramified in F∞/F . Then p does notdivide the class number of Fn for any n ≥ 0. Therefore, Iwasawa’s growthformula is valid with λ = µ = ν = 0.

Proof. Suppose that Iv denotes the inertia subgroup of Gal(F∞/F ) for the

3

one ramified prime v. It is clear that v must be totally ramified in F∞/F .Otherwise, F Iv

∞ would be a nontrivial, unramified, cyclic p-extension of F ,contradictict the assumption that p ∤ hF . Hence. for each n ≥ 0, the hy-potheses in proposition 1.1.4 are satisfied for the extension Fn/F . Therefore,the class number of Fn is not divisible by p. �

In particular, this result applies if F has only one prime lying above p andp ∤ hF . For example, p doesn’t divide the class number of Qn, the n-th layerin the cyclotomic Zp-extension Q∞/Q. Also, if we take F = Q(µp), wherep is any odd regular prime, then it follows that the class number of Q(µpn)will not be divisible by p for all n ≥ 1. The class number of Q(µ2n) is 1 forn ≤ 2 and is odd for n > 2, again by proposition 2.1.2.

Now suppose that F∞ =⋃n≥0 Fn is any Zp-extension of F . For every

n ≥ 0, let Ln denote the p-Hilbert class field of Fn. Let L∞ =⋃n≥0 Ln.

Then L∞ is an abelian extension of F∞. Let X = Gal(L∞/F∞). This groupwill arise frequently in this book and will sometimes be denoted by XF∞/F .We then have canonical isomorphisms of topological groups

X ∼= lim←−n

Gal(Ln/Fn) ∼= lim←−n

An

where the inverse limits are defined by the restriction and norm maps

RFm/Fn: Gal(Lm/Fm) −→ Gal(Ln/Fn), NFm/Fn

: Am −→ An

for m ≥ n ≥ 0. The first isomorphism is just a consequence of the definitionof the Galois group for an infinite Galois extension. The second isomorphismis defined by using the inverses of the Artin maps ArtLn/Fn

for n ≥ 0. Thecompatibility of the maps defining the two inverse limits then follows fromthe commutative diagram (4) in the proof of proposition 1.1.1 for the fieldsF = Fn, F

′ = Fm, m ≥ n. The field L∞ could be described more directly asthe maximal, abelian pro-p extension of F∞ which is unramified at all primesof F∞. The adjective ”pro-p” refers to the fact that X = Gal(L∞/F∞) is aprojective limit of finite p-groups. The equivalence of this description andthe one above is not difficult to prove, and is left to the reader. We refer toL∞ as the pro-p Hilbert class field of F∞.

Let γo be a topological generator for Γ = Gal(F∞/F ). For any n ≥ 0, γpn

o

is a topological generator for Γn = Γpn

, the unique subgroup of Γ of indexpn. We have Γn = Gal(F∞/Fn). Now L∞ is a Galois extension of F and we

4

therefore have an exact sequence

1→ X → Gal(L∞/F )→ Γ→ 1

of topological groups. We can then define a continuous action of Γ on Xby inner automorphisms as one normally does for group extensions. Thus,if γ ∈ Γ, let γ be an automorphism of L∞ such that γ|F∞

= γ. One thendefines

xγ = γxγ−1 (1)

for all x ∈ X. Continuity means that the map Γ × X → X defined by(γ, x)→ xγ for all γ ∈ Γ and x ∈ X is continuous. It will be somewhat moreconvenient to use an additive notation for X, and so we will now write γxinstead of xγ. More generally, we can regard X as a module for the groupring Z[Γ] and will write θx if θ ∈ Z[Γ] and x ∈ X. In particular, for anyn ≥ 0, we will denote the element γp

n

o − 1 in this group ring by ωn. Thenωnx corresponds to an element of X which could be written in multiplicativenotation as γo

pn

x(γopn

)−1x−1, a commutator in Gal(L∞/Fn). In fact, we havethe following basic result.

Proposition 2.1.3. For each n ≥ 0, the commutator subgroup of Gal(L∞/Fn)is ωnX. It is a closed subgroup.

Proof. Let Gn = Gal(L∞/Fn) and Γn = Gal(F∞/Fn). We let D(Gn) denotethe commutator subgroup of Gn (as an abstract group). The elements ofωnX are commutators in Gn and so ωnX ⊂ D(Gn). Since X is compact andmultiplication by ωn is continuous, it follows that ωnX is compact and henceclosed. It is clearly a normal subgroup of Gn.

To prove that D(Gn) = ωnX, it is enough to show that the quotientGn/ωnX is abelian. Now γp

n

0 generates a dense, infinite cyclic subgroup Γ′n

of Γn. There is a surjective homomorphism from Gn to Γn. The inverseimage of Γ′

n under that homomorphism is clearly abelian and dense in Gn.It follows that Gn is indeed abelian. �

Remark 2.1.4. We have stated the above proposition for the extensionL∞/F∞. But the proof is obviously more general and would apply wheneverL∞ is an abelian, pro-p extension of F∞ which is Galois over F . Underthat assumption, X = Gal(L∞/F∞) would again have a continuous actionof Γ. Here is an interesting and important example. Let Σ be any subsetof the primes of F . Let Σ∞ be the set of primes of F∞ lying above thosein Σ. Define MΣ

∞ to be the maximal, abelian, pro-p extension of F∞ which

5

is ramified only at the primes in Σ∞. It is easy to verify that MΣ∞ is Galois

over F and so the analogue of proposition 2.1.3 would apply. The field L∞

considered above is the special case where Σ is empty.If Σ contains all the primes of F lying above p, then we have F∞ ⊆ MΣ

∞

according to proposition 2.1.1. For any n ≥ 0, let Σn denote the primes of Fnlying above those in Σ. Let MΣ

n denote the maximal, abelian, pro-p extensionof Fn which is ramified only at the primes in Σn. Then, MΣ

n is the maximalabelian extension of Fn contained in MΣ

∞. If we let XΣ = Gal(MΣ∞/F∞), then

we have

Gal(MΣ∞/M

Σn ) = ωnXΣ, Gal(MΣ

n /F∞) ∼= XΣ

/ωnXΣ

for any n ≥ 0.

As proposition 2.1.2 illustrates, various questions about Zp-extensions,including the proof of the growth formula become simpler under the followinghypothesis about ramification.

RamHyp(1): Exactly one prime of F is ramified in the Zp-extension F∞/Fand this prime is totally ramified.

Under this hypothesis, proposition 2.1.2 already tells us that if p does notdivide the class number of F , then X = 0. The next proposition tells us thatX, together with the action of Γ on it, determines the structure of all thegroups An for n ≥ 0.

Proposition 2.1.5. Suppose that RamHyp(1) is satisfied for the Zp-extensionF∞/F . Then, with the notation as above, we have canonical isomorphisms

X/ωnX ∼= Gal(Ln/Fn) ∼= An

for all n ≥ 0.

Proof. This is a straightforward variation on the proof of proposition 1.1.4.Let v be the unique prime of F which is ramified in F∞/F . Let vn denote theunique prime of Fn lying above v. Then vn is the only prime of Fn ramifiedin the Zp-extension F∞/Fn and it is totally ramified. Let Kn denote themaximal abelian extension of Fn contained in L∞. Proposition 2.1.3 impliesthat we have the isomorphism

X/ωnX → Gal(Kn/F∞)

induced by the restriction map x→ x|Knfor x ∈ X.

6

Let In denote the inertia subgroup of Gal(Kn/Fn) for vn. It is clearthat Ln is the subfield of Kn fixed by In and F∞ is the subfield fixed byGal(Kn/F∞). Since these two subgroups of Gal(Kn/Fn) have trivial inter-section, it follows that Kn = LnF∞. Since vn is totally ramified in F∞/Fn,we have Ln ∩ F∞ = Fn and therefore the restriction map

Gal(Kn/F∞)→ Gal(Ln/Fn)

is indeed an isomorphism. The second isomorphism in the proposition is justthe inverse of the Artin map for the extension Ln/Kn. �

The above proposition reduces the proof of Iwasawa’s formula for a Zp-extension satisfying RamHyp(1) to proving an analogous formula for thegrowth of the quotients X/ωnX of X. We do this in the next two sectionswhere we begin the study of the structure and properties of Γ-modules, atopic that we will return to in Chapter 7. That study will also be the basisfor proving Iwasawa’s growth formula in general.

1.2 The structure of Γ-modules.

We will refer to an abelian, pro-p group X which admits a continuous actionby the group Γ as a Γ-module. This means that there is a homomorphismΓ → Aut(X) such that the map Γ × X → X defined by (γ, x) → γx iscontinuous. Here γ ∈ Γ, x ∈ X, and γx denotes the image of x under theautomorphism given by γ.

Suppose that X is any abelian, pro-p group. Then, for some indexing setI, we have

X = lim←−i

Xi

where Xi is a finite, abelian p-group for each index i ∈ I. It is easy to makeX into a Zp-module. Each Xi is a (Z/ptiZ)-module for some ti > 0. Thecanonical homomorphism Zp → Z/ptiZ makes each Xi into a Zp-module.The projective limit X then inherits the structure of a Zp-module (since themaps defining the projective limit will be Zp-modules homomorphisms). Itis a topological Zp-module in the sense that the map Zp × X → X definedby (z, x) → zx (where z ∈ Zp and x ∈ X) is continuous. Conversely, itis not hard to see that any compact, topological Zp-module is an abelian,pro-p group. One way to prove this is to consider the Pontryagin dual S =Hom(X,Qp/Zp), which is a discrete abelian group and also a topological

7

Zp-module. One then sees that every element of S has finite, p-power order.It follows from this that S is a direct limit of finite Zp-modules Si where ivaries over some indexing set I.

Suppose now that X is an abelian, pro-p group which has a continuousaction of Γ. As above, we have X = lim

←−Xi, where the Xi’s are finite, abelian

p-groups and i varies over an appropiate indexing set I. For each i ∈ I, letYi = ker(X → Xi). Thus, Yi is an open subgroup of X. An easy continuityargument shows that γ(Yi) = Yi for all γ in some subgroup of finite indexin Γ. This means that the orbit of Yi under the action of Γ is finite. HenceYi contains an open subgroup which is Γ-invariant. This implies that we canassume without loss of generality that each Yi is already Γ-invariant and so

X = lim←−i

Xi,

where each Xi is a finite, abelian p-group with a continuous action of Γ. Themaps defining the projective limit are Γ-homomorphisms.

It will be important for us to view X as a module over the ring Λ =Zp[[T ]], the formal power series ring over Zp in the variable T . One seeseasily that Λ is a local ring, m = (p, T ) is its maximal ideal, and Λ is com-plete in its m-adic topology. Also Λ/m = Fp and Λ/mt is finite (of orderpt(t+1)/2) for any Let γo denote a topological generator for Γ, as in section2.1. Roughly speaking, we will make X into a Λ-module by regarding T asthe endomorphism γo − 1.

For each i ∈ I, we have paiXi = 0 for some ai > 0. Also, γo − 1 definesan endomorphism of Xi which has a nontrivial kernel if Xi is nontrivial.Consequently, (γo−1)Xi is a proper subgroup of Xi if Xi 6= 0. It follows that(γo−1)biXi = 0 for some bi > 0. Thus, we can regard Xi as a module over thefinite ring Zp[T ]/(pai , T bi), where we let T act on Xi as the endomorphismγo− 1. However, we obviously have Zp[T ]/(pai , T bi) ∼= Zp[[T ]]/(pai , T bi), andso we can regard Xi as a Λ-module which is annihilated by the ideal (pai , T bi).Taking ti = ai + bi, it is obvious that m

ti ⊂ (pai , T bi) and so each Xi can beregarded as a module over Λ/mti . Regarding theXi’s as Λ-modules, it is clearthat the maps defining the projective limit are Λ-module homomorphisms.Thus X becomes a topological Λ-module. That is, the map Λ × X → Xdefined by (θ, x)→ θx for θ ∈ Λ, x ∈ X is continuous.

Conversely, if X is any compact Λ-module, then X is also a compact Zp-module and hence is an abelian, pro-p group. To make X into a Γ-module,note that Γ can be identified as a topological group with a subgroup of Λ×

8

by the continuous homomorphism γzo −→ (1 + T )z for all z ∈ Zp. Here wedefine

(1 + T )z =∞∑

i=0

(z

i

)T i, where

(z

i

)=

1

i!

i∏

j=1

(z − j + 1).

It is not difficult to prove that the coefficients of the above power series arein Zp. the constant term is 1, and so the power series is indeed invertiblein Λ. Thus, X admits a continuous action of Γ from which the Λ-modulestructure on X arises by letting T act as γo−1, just as above. If X is finitelygenerated as a Λ-module, then it is clear that X is compact and that thequotients X/mnX are finite Λ-modules for all n ≥ 0. In this case, we have

X ∼= lim←−i

X/mnX,

an inverse limit of a sequence of finite Λ-modules.

We proved a version of Nakayama’s lemma in chapter 1, lemma 1.5.3.That proof works in a much more general context. Assume that R is a localring with maximal ideal m, that R is complete in its m-adic topology, andthat R/mt is finite for all t > 0. In particular, k = R/m is a finite field. Letp be its characteristic. Now R = lim

←−R/mt, where the finite rings R/mn have

the discrete topology, and so R is a compact, topological ring. Suppose thatX is a projective limit of finite, abelian groups Xn and that each Xn is amodule over the ring R/mtn for some tn > 0 (which implies that Xn must bea p-group). We can then regard each Xn as an R-module. Assume that themaps defining the projective limit are R-module homomorphisms. Then Xitself becomes an R-module and the map R×X → X defined by (r, x)→ rxis continuous. That is, X is a compact, topological R-module. Conversely,any topological R-module X which is compact arises in the above way.

Proposition 2.2.1. Nakayama’s lemma for compact R-modules. Sup-pose that R and X are as above. Let x1, . . . , xd be a subset of X. Foreach i, 1 ≤ i ≤ d, let xi denote the image of xi under the natural mapX → X/mX. Then x1, . . . , xd is a generating set for the R-module X if andonly if x1, . . . , xd is a generating set for the k-vector space X/mX.

Proof. It is obvious that x1, . . . , xd generate the k-vector space X/mX ifx1, . . . , xd generate the R-module X. Conversely, assume that x1, . . . , xd

9

generate X/mX as a k-vector space. Let Y be the R-submodule of X gener-ated by x1, . . . , xd. Since Y is a continuous image of Rd and R is compact, itfollows that Y is compact. Therefore Y is a closed R-submodule of X. It istherefore enough to prove that Y is dense in X. This follows as before oncewe establish the result in the case where X is finite.

If X is finite, then mtX = 0 for some t > 0. We are assuming that the

image of Y under the canonical homomorphism X → X/mX is all of X/mX.Thus, X = Y + mX. It follows that X = Y + m

iX for all i > 0. Takingi = t, we get Y = X. �

Corollary 2.2.2. Suppose that R and X are as above. Then

1. X = 0 if and only if X = mX.

2. X is a finitely generated R-module if and only if X/mX is finite.

Proof. These statements follow immediately from Nakayama’s lemma. Ofcourse, statement 1 could be proved quite directly by again reducing to thecase where X is finite. Then, on the one hand, m

tX = 0 for some t > 0.But, on the other hand, X = mX ⇒ X = m

tX for all t > 0. It follows thatX = 0. �

We will be primarily interested in the special case where R is ring Λintroduced earlier. Suitable candidates for X are provided by compact Γ-modules. In general, the examples of interest to us will be finitely generatedas Λ-modules.

The ring Λ is Noetherian and has Krull-dimension 2. The maximal idealm has height 2. One simple way to obtain prime ideals of height 1 is askernels of the evaluation homomorphisms. Suppose that α ∈ Qp and hasabsolute value < 1. If f(T ) ∈ Λ, then one can define f(α) ∈ O = Zp[α]since the power series obviously converges to some element of that ring.The ring O is a subring of the ring of integers in Qp(α)—a finite extensionof Qp. The map f(T ) → f(α) defines a surjective ring homomorphismΛ → O and its kernel is a prime ideal of height 1. If α and α′ are twosuch elements of Qp, then it is easy to see that the corresponding evaluationhomomorphisms have the same kernel if and only if α and α′ are conjugateover Qp. Therefore, infinitely many distinct prime ideals arise in this way. Infact, it will become clear later that all but one of the prime ideals of height 1in Λ arise in this way. The exception is the ideal (p). If f(T ) =

∑aiT

i ∈ Λ,

define f(T ) =∑aiT

i ∈ Fp[[T ]], where a denotes the image of a ∈ Zp

10

under the homomorphism Zp → Fp. Then the map Λ → Fp[[T ]] defined by

f(T )→ f(T ) is a surjective ring homomorphism with kernel (p). This makesit clear that (p) is indeed a prime ideal of Λ.

Suppose that X is a finitely generated, torsion Λ-module. If I is any idealof Λ, we will use the notation

X[I] = {x ∈ X | αx = 0 for all α ∈ I}

Let Z =⋃n≥0X[mn] which is a Λ-submodule of X. Since Λ is Noetherian,

Z must be finitely generated and so it follows that Z = X[mt] for some t > 0and that Z is finite. It is clear that any finite Λ-submodule of X is containedin Z and so we refer to Z as the maximal, finite Λ-submodule of X.

Let Y =⋃n≥0X[pn], which is just the Zp-torsion submodule of X. We

will denote Y simply by Xtors in this chapter. Just as above, we see thatY = X[pt] for some t ≥ 0. We have Z ⊂ Y . The quotient X/Y is afinitely generated, torsion Λ-module and is torsion-free as a Zp-module. Iff(T ) =

∑aiT

i is a nonzero element of Λ which annihilates X, then writef(T ) = pmg(T ), where g(T ) ∈ Λ is not divisible by p. It is clear thatg(T ) annihilates X/Y . If X/Y has d generators as a Λ-module, then it isa quotient of the Λ-module Ud, where U = Λ/(g(T )). The following lemmagives the structure of U as a Zp-module.

Lemma 2.2.3. Suppose that g(T ) =∑∞

i=0 biTi ∈ Λ is not divisible by p. Let

l = min{i | bi ∈ Z×p }. Then U = Λ/(g(T )) is a free Zp-module of rank l.

Proof. Let I = (g(T )). It is clear that U = Λ/I is torsion-free as a Zp-module.Otherwise, there would be an element h(T ) ∈ Λ such that ph(T ) ∈ I, but6∈ pI). That is, ph(T ) = g(T )j(T ) where j(T ) is not divisible by p. But thisis not possible since (p) is a prime ideal of Λ.

Now U/pU can be considered as an Fp[[T ]]-module and is isomorphic to

Fp[[T ]]/(g(T )), where, as earlier, g(T ) =∑biT

i. Note that g(T ) is a nonzeroelement of Fp[[T ]] and (g(T )) = (T l). It is clear that U/pU has dimension l asan Fp-vector space and so, by lemma 1.5.3 (Nakayama’s Lemma for compactZp-modules), it follows that U is a finitely generated Zp-module and that lis the minimal number of generators. Since U is torsion-free, it must be freeof rank l. �

We summarize the above observations in the following proposition.

Proposition 2.2.4. Suppose that X is a finitely generated, torsion Λ-module. Then there are uniquely determined Λ-submodules Z and Y of X

11

with the following properties:

a. Z is finite and X/Z has no nonzero, finite Λ-submodules.

b. Y is annihilated by a power of p and X/Y is a free Zp-module of finiterank.

This proposition allows us to define certain important invariants associatedwith X. The Zp-rank of X/Y is obviously equal to dimQp

(V ), where V isthe Qp-vector space X ⊗Zp

Qp. We define

λ(X) = rankZp(X/Y ) = dimQp

(V ). (2)

Then X/Y = Zλ(X)p as a Zp-module. Now Y = X[pt] for some t > 0. (Even

if Y = 0, we will take t > 0 in the following definitions.) For each i such that0 < i ≤ t, the Λ-module X[pi]/X[pi−1] has exponent p and can be consideredas an Fp[[T ]]-module. It will be finitely generated and thus has finite rank.We define

µ(X) =t∑

i=1

rankFp[[T ]]

(X[pi]/X[pi−1]

). (3)

If X is finitely generated as a Zp-module, then µ(X) = 0. To be precise, wehave

µ(X) = 0⇔ Y is finite ⇔ X[p] is finite ⇔ X/pX is finite

In this case, Y and Z coincide. On the other hand, it is also clear thatλ(X) = 0⇔ ptX = 0 for some t > 0

We will refer to λ(X) and µ(X) as the Iwasawa invariants for the Λ-module X. Another invariant which will play an important role will be apolynomial fX(T ) in Zp[T ], which we refer to as the “characteristic polyno-mial” of X. However, this polynomial depends not just on the structure of Xas a Γ-module, but also on the choice of topological generator γo of Γ. (In alater chapter, we will remedy this by redefining the ring Λ in a more intrinsicway, viewing it as the “completed group algebra for Γ over Zp.”) The defi-nition is fX(T ) = pµ(X)gX(T ), where gX(T ) is the monic polynomial whoseroots are precisely the eigenvalues of the linear operator T = γo − 1 actingon the Qp-vector space V defined above. These eigenvalues are in Qp andare counted according to their multiplicities so that gX(T ), and hence fX(T ),

12

has degree equal to λ(X). Since T acts on X topologically nilpotently, it isnot hard to see that the eigenvalues of T have absolute value < 1. Therefore,the nonleading coefficients of gX(T ) are divisible by p.

It will be useful to have finer invariants to describe the structure of Y =Xtors as a Λ-module. For each i ≥ 1, define ri = rankFp[[T ]](X[pi]/X[pi−1]).Assume that µ(X) > 0. This means that r1 > 0. Choose t so that rt > 0,but rt+1 = 0. Note that r1 ≥ · · · ≥ rt. Let r = r1. The finer µ-invariantswill be positive integers µ1, . . . , µr, consisting of r1 − r2 1′s, r2 − r3 2′s, ...,and rt t

′s. With this definition, we have

µ(X) =t∑

i=1

ri =r∑

j=1

µj. (4)

As a simple illustration, suppose that X is a Λ-module and that Y ∼= Λ/pmΛ,where m ≥ 0. We then have r = 1 and µ1 = m. If Y ∼= (Λ/pΛ)m, we thenhave r = m and µ1 = · · · = µm = 1. In both cases, we have µ(X) = m.

Let F∞/F be a Zp-extension and let L∞ be the pro-p Hilbert class fieldof F∞. We will prove that X = Gal(L∞/F∞) is a finitely generated, torsionΛ-module in section 2.5. The integers λ and µ occurring in Iwasawa’s formulaturn out to be precisely the Iwasawa invariants for X: λ = λ(X), µ = µ(X).The polynomial fX(T ) will also be of special interest. In the case whereF/Q is an abelian extension and F∞ is the cyclotomic Zp-extension of F,its roots are related in a certain way to the zeros of the p-adic L-functionsdefined by Kubota and Leopoldt. The precise relationship, which was firstconjectured by Iwasawa in 1969 and proved by Mazur and Wiles in 1979, willbe described in Chapter 8 together with a proof for the special case whereF = Q(µp). Closely related to this is a simple interpretation of the rootsof fX(T ) in terms of the unramified cohomology groups discussed in section1.6. That interpretation will be discussed in the final section of this chapter

If F∞/F satisfies RamHyp(1), then the assertion that X is a finitelygenerated, torsion Λ-module is relatively easy to prove. In this case, wealready know that X/ωnX is finite for all n. In particular, X/TX is finite.The assertion that X is finitely generated then follows immediately from thecorollary to Nakayama’s lemma. The assertion that X is a torsion Λ-modulefollows from the following result.

Proposition 2.2.5. Let X be a finitely generated Λ-module. Then the fol-lowing statements are equivalent:

13

a. There exists an element f(T ) ∈ m such that X/f(T )X is finite.

b. X contains a torsion Zp-submodule Y such that X/Y is finitely gener-ated as a Zp-module.

c. X is a torsion Λ-module.

Proof. We have already proved that (c) ⇒ (b). The converse (b)⇒(c) israther easy. We can assume that X/Y is Zp-torsion free. Since Λ is Noethe-rian, Y is also finitely generated and has bounded exponent. Therefore, itis annihilated by pa for some a. As for X/Y , this is a free Zp-module offinite rank and multiplication by T defines an endomorphism of that mod-ule. If g(x) is the characteristic polynomial of this endomorphism, then g(T )annihilates X/Y . Thus the nonzero element pag(T ) is an annihilator of X,proving (c).

We also have (b) ⇒ (a). To see this, one can simply take g(T ) = T − β,where β ∈ pZp is not an eigenvalue of the endomorphism of X/Y given bymultiplication by T . Then (T − β)(X/Y ) has finite index in X/Y . Supposethat ptY = 0. Then for each i, 0 < i ≤ t, Y [pi]/Y [pi−1] can be consideredas a finitely generated Fp[[T ]]-module. the cokernel of multiplication byT − β is clearly finite. It then follows that (T − β)Y has finite index in Y .Consequently, by the snake lemma, one sees that (T − β)X has finite indexin X.

Finally, we prove that (a) ⇒ (c). If (a) holds and p divides f(T ), thenX/pX is finite too. Hence, by Nakayama’s lemma for the ring Zp, it followsthat X is finitely generated over Zp. Hence (b) is true, and so (c) follows inthis case. Thus, we can now assume that p doesn’t divide f(T ). Then, bylemma 2.2.3, Λ/(f(T )) is a free Zp-module and its Zp-rank l = deg(f(T )) ispositive. Let r = rankΛ(X). The following lemma then completes the proofof proposition 2.2.5. It implies that r = 0 if X/(f(T ))X is finite. �

Lemma 2.2.6. Suppose that X is a finitely generated Λ-module and thatf(T ) ∈ Λ. Let r = rankΛ(X) and l = deg(f(T )). Then

rankZp(X/f(T )X) ≥ rl. (5)

If X is a torsion-free Λ-module, then equality holds.

Proof. To prove (5), we can obviously replace X by the quotient moduleX/XΛ−tors. So we can assume without loss of generality that X is a torsion-free Λ-module. We can also clearly assume that f(T ) is not divisible by p.

14

Now X contains a Λ-submodule Y which is free of rank r over Λ. We havethe exact sequence

0 −→ Y −→ X −→ Z −→ 0

where Z is a finitely generated, torsion Λ-module. The snake lemma thengives an exact sequence

0 −→ Z[f(T )] −→ Y/f(T )Y −→ X/f(T )X −→ Z/f(T )Z −→ 0.

By lemma 2.2.6, Y/f(T )Y is a free Zp-module of rank rl. Let U denote theZp-torsion submodule of Z. Then Z/U is a free Zp-module of finite rank.This implies that Z[f(T )] and Z/f(T )Z have the same (finite) Zp-rank. Thuswe see that rankZp

(X/f(T )X) = rl. �

In a later chapter, we will discuss more precise results concerning thestructure of finitely generated Λ-modules. The results that we are provinghere and in the next section will suffice for the proof of Iwasawa’s growthformula. Some of the later theorems (including a couple in this chapter)involve the following standard ring-theoretic notion. Let R be a Noetherian,local, integral domain. The Krull-dimension of R is then finite. We denoteit by dim(R). If P is any prime ideal of R, then we will denote its height byht(P ). It is related to the Krull-dimension of R/P by the formula:

ht(P ) + dim(R/P ) = dim(R).

Let X be a finitely generated R-module. For any x ∈ X, let Ann(x) denoteits annihilator in R. The primes ideals of R which are associated to X formthe following set.

AssR(X) = {P | P = Ann(x) for some x ∈ X}

This is a finite set. We will say thatX is “pure of dimension d” if dim(R/P ) =d for every prime ideal P ∈ Ass(X). Thus, X is pure if all the prime ideals inAss(X) have the same height h. The dimension will then be d = dim(R)−h.In particular, a torsion-free R-module would be pure of dimension equal todim(R).

If we take R = Λ, then the Krull-dimension is 2. As explained earlier, Λhas one prime ideal of height 2, the maximal ideal m = (p, T ), and infinitelymany prime ideals of height 1. A nonzero, finitely generated, torsion Λ-module X is pure in the following two cases: (a) X is finite and hence pure

15

of dimension 0, (b) X has no nonzero, finite Λ-submodules and hence is pureof dimension 1.

We have been considering Λ-modules which are profinite and thereforecompact as topological groups. It is also quite useful to consider their Pon-tryagin duals. Suppose that X is a pro-p abelian group (i.e. a compactZp-module). Let S = Hom(X,Qp/Zp). One verifies easily that S is ap-primary abelian group and that the usual topology on it (the so-called“compact-open” topology) is just the discrete topology. Also, if S is anyp-primary abelian group with the discrete topology, then we can regard itas a direct limit of finite abelian p-groups. Therefore, its Pontryagin dualX = Hom(S,Qp/Zp) is an abelian, pro-p group. We can regard both S andX as Zp-modules.

If X is a finitely generated Zp-module, we will say that S is a cofinitelygenerated Zp-module. We then define corankZp

(S) = rankZp(X), which we

refer to as the Zp-corank of S. Note that since the Pontryagin dual of Zp isQp/Zp, it follows that any cofinitely generated Zp-module S is isomorphic to(Qp/Zp)

l × T as a Zp-module, where T is finite and l = corankZp(S). Thus,

the maximal divisible subgroup Sdiv of S is isomorphic to (Qp/Zp)l and has

finite index in S. If X is torsion-free as a Zp-module (finitely generated ornot), then S is divisible. The following result is a consequence of Nakayama’slemma for compact Zp-modules, but also can be proved directly.

Proposition 2.2.7. Suppose that S is a discrete, p-primary abelian group.Then S is a cofinitely generated Zp-module if and only if S[p] is finite. If Sis divisible, then corankZp

(S) = dimFp(S[p]).

Suppose that S is a discrete, p-primary abelian group with a continuousaction of Γ. We can translate our earlier results into equivalent statementsabout S. Let X = Hom(S,Qp/Zp), which also admits a continuous actionof Γ. (This is defined on Hom in the usual way, using the given action ofΓ on S and the trivial action on Qp/Zp). Since S is also a Zp-module, it isnot hard to make S into a Λ-module directly (showing first that S is a directlimit of finite Γ-invariant subgroups). Or one can equivalently transfer thestructure of X as a Λ-module to S. Just as above, we say that S is cofinitelygenerated as a Λ-module ifX is finitely generated as a Λ-module, cotorsion asa Λ-module if X is a torsion Λ-module. We refer to rankΛ(X) as corankΛ(S).If S is a cofinitely generated, torsion Λ-module, then we define λ(S) to beλ(X), µ(S) to be µ(X). It is useful to note that if X is torsion-free as aΛ-module, then S is divisible as a Λ-module (i.e. θS = S for every nonzero

16

θ ∈ Λ). Here are the most useful results.

Proposition 2.2.8. Suppose that S is a discrete, p-primary, abelian groupwith a continuous action of Γ. Regard S as a Λ-module. Then

1. S = 0⇐⇒ SΓ = 0⇐⇒ S[m] = 0.

2. S is a cofinitely generated Λ-module ⇐⇒ S[m] is finite .

3. S is a cofinitely generated, cotorsion Λ-module ⇐⇒ S[θ] is finite forsome θ ∈ m.

Proposition 2.2.9. Suppose that S satisfies the assumptions in proposition2.2.8. If SΓ is finite, then S is a cofinitely generated, cotorsion Λ-module.

Proposition 2.2.10. Suppose that S is a cofinitely generated, cotorsionΛ-module. Then Sdiv = (Qp/Zp)

λ as a Zp-module, where λ = λ(S). Thequotient S/Sdiv has bounded exponent.

The proofs are rather easy, based on corollary 2.2.2 , propositions 2.2.4 and2.2.5.

1.3 Growth theorems for quotients of Γ-modules.

Now let X be a finitely generated, torsion Λ-module. It can of course happenthat X/ωnX is infinite for some values of n. This corresponds to the possibil-ity that some of the eigenvalues of γo acting on the vector space V = X⊗Zp

Qp

are pn-th roots of unity. (Of course, the corresponding eigenvectors may bein the vector space V ⊗Qp

Qp obtained by extending scalars to the algebraic

closure Qp of Qp.) The order of such roots of unity is certainly bounded. (Tosee this, note that if pt is the order of some such root of unity, then it is clearthat φ(pt) = pt − pt−1 ≤ λ(X). This gives a bound on t.) In the ring Λ, wecan write ωn =

∏ni=0 φi, where φi = ωi/ωi−1 for i > 0 and φ0 = ω0 = T . For

m ≥ n ≥ 0, define νm,n =∏

n<i≤m φi, which we take to be 1 if m = n. Thus,ωm = ωnνm,n. Note that the roots of the polynomial φi are the numbersζ − 1, where ζ is a primitive pi-th root of unity.

We fix an integer no sufficiently large so that no pt-th root of unity isan eigenvalue of γo acting on V for t > no. Then it is easy to verify thatX/νn,no

X is finite for all n ≥ no. We will prove the following proposition.

17

Proposition 2.3.1. Let X be a finitely generated, torsion Λ-module. Chooseno so that X/νn,no

X is finite for all n ≥ no. Let λ = λ(X), µ = µ(X). Thenwe have

|X/νn,noX| = pλn+µpn+ν

for all n≫ 0, where ν is some integer.

Proof. Suppose that we have an exact sequence of torsion Λ-modules

0 −→ X1 −→ X2 −→ X3 −→ 0

We first show that the validity of the above theorem for X1 and X3 impliesthe validity for X2. It is easy to see that

λ(X2) = λ(X1) + λ(X3), µ(X2) = µ(X1) + µ(X3).

Choose no sufficiently large so that the X2/νn,noX2 is finite for all n ≥ no.

Then the same thing will be true for X1 and X3. The snake lemma gives

0 −→ X1[νn,no] −→ X2[νn,no

] −→ X3[νn,no] −→ X1/νn,no

X1

−→ X2/νn,noX2 −→ X3/νn,no

X3 −→ 0.

Note that each φi is in m and so νn,no∈ m

t for t = n − no + 1. Hence,it is clear that if n ≫ no, then Xi[νn,no

] = Zi for i = 1, 2, 3, where Zidenotes the maximal finite Λ-submodule of Xi. Thus the terms in the firsthalf stabilize for n≫ 0, and therefore the image of the fourth arrow will haveorder |Z1||Z3|/|Z2| = pa, say. This implies that

|X2/νn,noX2| = |X1/νn,no

X1||X3/νn,noX3|p

−a

for n ≫ 0. It is then clear that proving proposition 3 for X = X1 andX = X3 implies it for X = X2.

Based on this last remark and proposition 2.2.4, it is enough to considerthe following three special cases: (a) X is finite, (b) X has exponent p, and(c) X is a free Zp-module of finite rank.

Case (a): This is quite easy. If X is finite, then νn,noannihilates X for

n >> no and so X/νn,noX = X has constant order |X| for such n. The

proposition is valid since λ(X) = µ(X) = 0.

Case (b). If X has exponent p, then one can consider X as a finitely-generated module over Fp[[T ]], which is a principal ideal domain. Therefore,

18

X is isomorphic to Fp[[T ]]r × Z, where r ≥ 0 and Z is the Fp[[T ]-torsionsubmodule of X. Noting that Fp[[T ]]/(T d) is finite (of order pd) for anyinteger d > 0, it follows that Z is finite too. Thus it is enough to justconsider the special case X = Fp[[T ]]. We then have λ(X) = 0, µ(X) = 1.Note that φi has degree pi − pi−1. Hence νn,no

has degree pn − b for n > no,where b is a constant. Also, νn,no

is not divisible by p in Λ. It follows thatX/νn,no

X has order ppn−b for n > no, again verifying the proposition since

λ(X) = 0 and µ(X) = 1.

Case (c). Assume that X is a free Zp-module of rank l. Multiplicationby T defines a Zp-linear endomorphism of X. Let α1, . . . , αl denote theeigenvalues (in Qp) of this endomorphism, counting multiplicity, which arejust the roots of the characteristic polynomial g(t) = det(tI − T ). It is clearthat T acts nilpotently on X/pX, an Fp-vector space of dimension l. Thecharacteristic polynomial for the endomorphism T of X/pX is tl and henceg(t) ≡ tl (mod pZp[t]). This implies the important fact that |αi|p < 1 for1 ≤ i ≤ l.

Thus, if f(T ) ∈ Λ, then f(αi) ∈ Qp is defined Furthermore, multi-plication by f(T ) defines an endomorphism of X which has eigenvaluesf(α1), . . . , f(αl). The determinant of this endomorphism is

∏li=1 f(αi). The

determinant of a matrix over Zp determines the order of the cokernel. To beprecise, if some αi is a root of f(T ), then X/f(T )X is infinite. Otherwise,write

∏li=1 f(αi) = pau, where u ∈ Z×

p . Then |X/f(T )X| = pa.We take f(T ) = νn,no

=∏

no<j≤nφj. The roots of φj(T ) are the numbers

ζ − 1, where ζ varies over the primitive pj-th roots of unity (in Qp). Recallthat ordp(ζ − 1) = 1/(pj − pj−1) for all such roots of unity. Let α be any oneof the αi’s. We have

φj(α) =∏

ζ

(α− (ζ − 1)))

where the product is over all the primitive pj-th roots of unity ζ. Our choiceof no implies that α is not a root of φj for j > no. Obviously, if j is sufficientlylarge (say, j ≥ n1), then ordp(ζ − 1) < ordp(α) for all α ∈ {α1, ..., αl}. Thenumber of factors in the product defining φj(α) is pj − pj−1. Each of thesefactors has valuation equal to ordp(ζ − 1) for j ≥ n1. Thus, for such j, wehave

ordp(φj(α)

)= 1

and so ordp(νn,no

(αi))

= n + ci for n > n1, where ci ∈ Z. It follows that

19

|X/νn,noX| = pln+d for n ≫ 0, where d ∈ Z. Since λ(X) = l and µ(X) = 0,

we have verifies the proposition for case (c).As we already remarked, proposition 2.3.1 follows from these separate

calculations. �

It may be worthwhile to discuss one simple case of Proposition 2.3.1explicitly. Assume that p is odd. Suppose that X = Zp. The action of Γon X would then be given by a continuous homomorphism κ : Γ→ 1 + pZp.Thus κ(γ) = 1+α, where α ∈ pZp. (In fact, α is the eigenvalue of T = γ− 1acting on X.) We assume that the action of Γ is nontrivial. That is, α 6= 0.Suppose that ordp(α) = a. Then X/TX = X/αX is cyclic of order pa.Now 1 + pZp

∼= Zp as a topological group and the image κ(Γ) will be thesubgroup 1 + paZp, which is the unique subgroup of 1 + pZp of index pa−1.The image κ(Γn) will be the unique subgroup of index pn+a−1. That is,ordp(κ(γ

pn

)− 1) = pn+a. It follows that X/ωnX is cyclic of order pn+a. Thisis true for all n ≥ 0. Thus, in the notation of proposition 2.3.1, we haveλ = λ(X) = 1, µ = µ(X) = 0, and ν = a.

Another result which sometimes will be useful is the following proposition.Any integer no satisfying the property in proposition 2.3.1 will also have theproperty that rankZp

(X/ωnX) = rankZp(X/ωno

X) for all n ≥ no. The twoproperties are equivalent. The following result concerns the growth of thetorsion subgroup of X/ωnX.

Proposition 2.3.2. Suppose that X is a finitely generated, torsion Λ-module. Choose no as above. Let λo = rankZp

(X/ωnoX). Then, for n ≫ 0,

we have|(X/ωnX)tors| = p(λ−λo)n+µpn+ν ,

where λ = λ(X), µ = µ(X), and ν is some integer.

Proof. Let Y = ωnoX. Then λ(Y ) = λ − λo. We have ωnX = νn,no

Y forn ≥ no. Also, Y/νn,no

Y is finite for all such n. We have an exact sequence

0→ Y/νn,noY → (X/ωnX)tors → (X/Y )tors → 0

The result follows immediately by applying proposition 2.3.1 to Y . �

One further result concerns the order of quotients of the formX/(T−β)X,where β ∈ pZp.

Proposition 2.3.3. Suppose that X is a finitely generated, torsion Λ-moduleand that fX(β) 6= 0. Then X/(T −β)X is finite. If X has no nonzero, finite

20

Λ-submodules, then

ordp(|X/(T − β)X|

)= ordp

(fX(β)

).

If fX(β) = 0, then X/(T − β)X is infinite.

Proof. We can write fX(T ) = pµ(X)gX(T ), where gX(T ) is a monic polynomialin T . Thus, fX(β) = pµ(X)gX(β). It is clear from the proof of proposition2.3.1 that X/(T − β)X is finite since β is not a root of gX(T ). To calculatethe order of X/(T − β)X, consider the exact sequence

0 −→ Y −→ X −→ X/Y −→ 0

where Y is the Zp-torsion submodule of X. Thus, Y = X[pt] for somet ≥ 0. We assume that X has no nonzero Λ-submodules. If one considersthe maps induced by multiplication by pi−1 on X for i ≥ 1, then one seesthat X[pi]/X[pi−1] is isomorphic to a Λ-submodule of X and hence also hasno nonzero Λ-submodules. This means that each X[pi]/X[pi−1] is free as aFp[[T ]]-module. On the other hand, X/Y is a free Zp-module.

If one then applies the snake lemma as in the proof of proposition 2.3.1,one can reduce the proof to the two cases where either X ∼= Fp[[T ]] or X isa free Zp-module of rank l. One has X/(T − β)X ∼= Fp and µ(X) = 1 inthe first case. The proposition is valid in that case. In the second case, letα1, ..., αl denote the roots of fX(T ) = gX(T ). It is clear that the index of(T − β)X in X is infinite if and only if β is equal to one of the αi’s. If weassume that fX(β) 6= 0, then that index is finite and has the same valuationas the determinant of the operator T − β on X, which is

l∏

i=1

(αi − β) = ± fX(β),

proving the formula in the proposition. �

We will now prove some results about the group-theoretic structure of thequotients of X occurring in propositions 2.3.1 and 2.3.2, starting first withthe case where µ(X) = 0.

Proposition 2.3.4. Let X be a finitely generated, torsion Λ-module. Chooseno so that X/νn,no

X is finite for all n ≥ no. Assume that µ(X) = 0. Then,for all n≫ 0, there is an isomorphism

X/νn,noX ∼=

λ∏

i=1

Z/pn+ciZ× C,

21

where c1, . . . , cλ are certain integers and C is isomorphic to the maximal,finite, Λ-submodule of X.

Proof. As a Zp-module, we have X ∼= U × Z, where U ∼= Zλp . Here Z is a

Λ-submodule of X, but U is just a Zp-submodule. Suppose that t is chosenso that ptZ = 0. Then ptU = ptX is a Λ-submodule of X, and has finiteindex. Hence νn,no

X ⊂ ptU for n≫ no. Note that

X/νn,noX ∼= U/νn,no

X × Z

as a Zp-module. Note that νn,noX is a free Zp-module of rank λ. We will

show below thatνn+1,no

X = pνn,noX (6)

for n ≥ n1, where n1 is chosen in a certain way. Thus, for such n, if U/νn,noX

is isomorphic to a direct product of cyclic groups of orders pa1 , ..., paλ , thenU/νn+1,no

X will be isomorphic to a direct product of cyclic groups of orderspa1+1, ..., paλ+1. The proposition will then follow by induction.

First choose n′o ≥ no so that X ′ = νn′

o,noX ⊂ ptU . Then Γ acts continu-

ously on the quotient group X ′/p2X ′. Choose n1 ≥ n′o so that the subgroup

Γpn1 acts trivially on this quotient. If i > n1, then φi =

∑p−1j=0 γo

jpi−1

acts

on X ′/p2X ′ as multiplication by p. A simple application of Nakayama’slemma (for Zp-modules) implies that φiX

′ = pX ′. Let n ≥ n1. Then, takingi = n+ 1 and multiplying by νn,n′

o, we obtain the identity (6). �

If A is any finite abelian group, its exponent is the smallest positiveinteger m such that A[m] = A, or equivalently, the least common multipleof the orders of elements in A. We will refer to dimFp

(A[p]) as the p-rank ofA. The p-rank of A is also clearly equal to dimFp

(A/pA).

Proposition 2.3.5. Let X be a finitely generated, torsion Λ-module. Chooseno so that X/νn,no

X is finite for all n ≥ no.

1. If λ(X) > 0, then the exponent of X/νn,noX is equal to pn+c for all

n ≫ 0, where c is some integer. If λ(X) = 0, then the exponent ofX/νn,no

X is bounded and becomes constant for n≫ 0.

2. Let r = rankFp[[T ]](X[p]). Then the p-rank of X/νn,noX is equal to

rpn + c for n≫ 0, where c is some constant.

Proof. The first statement follows immediately from proposition 2.3.4 ifµ(X) = 0. Note also that the proof of that proposition shows the following

22

(again under the assumption that µ(X) = 0): Let x ∈ X. Assume thatx 6∈ Z, the maximal finite Λ-submodule of X. Then the image of x inX/νn,no

X has order pn+c for all n≫ 0, where c is a certain integer (dependingon x). We will use this observation in the general case.

In general, let pt denote the exponent of Y = Xtors. Then ptX is a freeZp-module of rank λ(X). The maximal finite Λ-submodule of X/ptX is ofthe form Xo/p

tX for a uniquely determined Λ-submodule Xo of X containingptX. Since p ∤ νn,no

, it is clear that (X/Xo)[νn,no] = 0. This implies that

νn,noX ∩Xo = νn,no

Xo (7)

for any n ≥ no.Let x ∈ X. Assume that ptx 6= 0. Since λ(X) > 0, the image of x in

X/νn,noX has unbounded order as n → ∞. Now ptx ∈ Xo and its images

in X/νn,noX and in Xo/νn,no

Xo have the same order for n ≥ no according to(7). This order is pn+c if n≫ 0, for some c. Thus, the order of the image of xin X/νn,no

X will be pn+c+t for sufficiently large n. The stated result followsimmediately because X is finitely generated as a Λ-module. If x1, ..., xd isa set of generators, then the exponent of X/νn,no

X will be the maximum ofthe orders of the images of x1, ..., xd in that quotient.

We now prove the second statement. One can equivalently define r asrankFp[[T ]](X/pX). This follows from the exact sequence

0→ X[p]→ X → X → X/pX → 0

since the µ-invariant is additive in exact sequences, and so µ(X[p]) = µ(X/pX).Note that

(X/νn,noX)/p(X/νn,no

X) ∼= X/(νn,noX + pX) ∼= (X/pX)/νn,no

(X/pX)

The stated result follows from proposition 2.3.1, since the order of a groupof exponent p determines its dimension over Fp. �

If µ(X) > 0, then the actual structure of the quotients X/νn,noX would

be somewhat more complicated. We will be content to state the followingresult, omitting the proof. If one uses the structure theorem for Λ-modulesto be proved in Chapter 5, then the result is not hard to prove. We usethe following notation. Suppose that {An} and {Bn} are two sequences ofgroups, defined for all n ≥ no, say. We will write {An} ≈ {Bn} if there exists

23

a sequence of group homomorphisms fn : An → Bn defined for n ≥ no suchthat ker(fn) and coker(fn) are finite and of bounded order as n varies.

Proposition 2.3.6. Let X be a finitely generated, torsion Λ-module. Chooseno so that X/νn,no

X is finite for all n ≥ no. Let λ = λ(X) and µ = µ(X). Letµ1, . . . , µr be the finer µ-invariants for X. For n ≥ no, let An = X/νn,no

Xand let Bn = (Z/pnZ)λ ×

∏rj=1(Z/p

µjZ)pn

. Then {An} ≈ {Bn}.

Note that the group Bn in this proposition has order pλn+µpn

. It is alsoworth pointing out that knowing the sequence of groups {An} (up to theequivalence relation ≈) is sufficient to determine the invariants λ(X), µ(X),r, and µ1, . . . , µr.

1.4 Proof of Iwasawa’s growth formula

We will prove Iwasawa’s theorem in complete generality in this section. Ifwe assume RamHyp(1), then the result follows quickly from earlier results.In this case, we know that X/ωnX is finite for all n. As we have alreadypointed out, Nakayama’s lemma and proposition 2.2.5 imply that X mustbe a finitely generated, torsion Λ-module. Now the power of p dividing hnis equal to |X/ωnX| by proposition 2.1.5. The growth for these quantitiesis then given by either proposition 2.3.1 or 2.3.2, and is just as described byIwasawa’s growth formula.

As a first step to the complete proof, we prove the following importantresult.

Proposition 2.4.1. Suppose that F∞/F be a Zp-extension and let L∞ denoteits pro-p-Hilbert class field. Then X = Gal(L∞/F∞) is a finitely generated,torsion Λ-module.

Proof. If a prime v of F is ramified in the Zp-extension F∞/F , then thecorresponding inertia subgroup Iv of Gal(F∞/F ) has finite index in Γ =Gal(F∞/F ). By proposition 1, there are only finitely many such v’s andso the intersection of these inertia subgroups has finite index in Γ. That is,⋂v Iv = Γno

for some no ≥ 0. Every prime of Fnowhich is ramified in F∞/Fno

must be totally ramified. Let t be the number of such primes.Suppose that n ≥ no. Let Kn denote the maximal abelian extension

of Fn contained in L∞. Thus, Gal(L∞/Kn) = ωnX. Let η be any primeof Fn ramified in F∞/Fn. There are t such primes, which we denote by

24

η1, . . . , ηt. Since L∞/F∞ is unramified, only these primes are ramified in theextension Kn/Fn. The corresponding inertia subgroups of Gal(Kn/Fn) areall isomorphic to Zp and they generate the subgroup Gal(Kn/Ln). Thus,Gal(Kn/Ln) is a finitely generated Zp-module which has rank ≤ t. It hasfinite index in Gal(Kn/Fn) since Ln/Fn is a finite extension. Therefore,Gal(Kn/Fn) is also a finitely generated Zp-module with the same rank.

It follows that X/ωnX = Gal(Kn/F∞) is finitely generated as a Zp-module and has rank ≤ t− 1. Nakayama’s Lemma implies that X is finitelygenerated as a Λ-module. On the other hand, lemma 2.2.6 implies that

rankZp(X/ωnX) ≥ rpn

for all n, where r = rankΛ(X). It follows that r = 0. That is, X is torsionas a Λ-module. �

As the above proof might suggest, it is not always true that the quotientsX/ωnX are finite. In a later chapter we will give some examples of thisphenomenon and study it in some detail. However, as noted in proposition2.3.1, if no is sufficiently large, then the quotients X/νn,no

X will be finitefor all n ≥ no. In fact, as we will see in the proof of the next result, onecan take no just as in the proof of proposition 2.4.1. It then turns out that[Ln : Fn]/|X/νn,no

X| becomes constant for n ≫ 0. These facts follow easilyfrom the following proposition, which is somewhat more precise. The proofinvolves keeping careful track of the inertia subgroups of Gal(L∞/Fn) for theprimes over p.

Proposition 2.4.2. Let F∞/F be a Zp-extension. Choose no so that allthe primes of Fno

which are ramified in F∞/Fnoare totally ramified in that

extension. Let X = Gal(L∞/F∞) and Y = Gal(L∞/LnoF∞). Then

X/νn,noY = Gal(Ln/Fn)

for all n ≥ no.

Proof. Let n ≥ no. Let L∗n denote the maximal unramified, extension of Fn

contained in L∞. Then L∗n/Fn is Galois. Obviously, Ln ⊂ L∗

n. But sinceat least one prime of Fn is totally ramified in F∞/Fn, we have L∗

n

⋂F∞ =

Fn. It therefore follows that Gal(L∗n/Fn)

∼= Gal(L∗nF∞/F∞), which is clearly

abelian. Thus, L∗n = Ln.

Let Gn = Gal(L∞/Fn). Let Hn be the smallest closed subgroup of Gn

containing all the inertia subgroups of Gn. The primes of Fn ramified in

25

L∞/Fn are the same as those ramified in F∞/Fn, and we denote them byη1, . . . , ηt. The number t of such primes is independent of n because n ≥ no.Each inertia subgroup for a prime of L∞ above one of the ηi’s is canonicallyisomorphic to Γn (by the restriction map). We have L∗

n = LHn∞ by definition

and therefore Hn = Gal(L∞/Ln).Let Yn = Hn

⋂X. Then Yn = Gal(L∞/LnF∞). In particular, Y (as de-

fined in the proposition) is just Yno. It is also clear that X/Yn = Gal(Ln/Fn).

So it remains just to prove that Yn = νn,noYno

for all n ≥ no.Let R denote the set of primes η of L∞ ramified in the extension L∞/Fno

.For each η ∈ R, let Iη denote the corresponding inertia subgroup of Gno

.Then, as we noted above, Iη = Γno

= Zp. For n ≥ no, the inertia subgroupof Gn for η will be Iη

⋂Gn. This will be the unique subgroup of Iη of index

pn−no , namely Ipm

where we put m = n− no for brevity.Choose a topological generator γno

for Γno. For each η ∈ R, let gη denote

the element of Iη such that gη|F∞= γno

. If η, η′ ∈ R, then gη and gη′ arein the same coset in Gno

/X and so y(η, η′) = gηg−1η′ is in X. Furthermore,

the definitions imply that Y is the smallest closed subgroup of X containingall the y(η, η′)’s, where we allow (η, η′) to vary over R × R. Similarly, itfollows that Yn is the smallest closed subgroup of X containing the elementsyn(η, η

′) = gpm

η g−pm

η′ for (η, η′) ∈ R×R, where m is as above.The rest of this proof will be somewhat clearer if we switch to a mul-

tiplicative notation for Y . Thus, if y ∈ Y and θ ∈ Λ, we will write yθ

in place of θy. We will now do a simple calculation in Gnoto show that

y(η, η′)νn,no = yn(η, η′). This implies that Y νn,no = Yn, from which proposi-

tion 2.4.2 follows.Let a = gη, b = gη′ , y = ab−1 = y(η, η′), and γ = γno

. Note that

νn,no=

∑pm−1i=0 γi. Also, since b|Fno

= γ, we have biyb−i = yγi

for 0 ≤ i < pm.Therefore,

yνn,no =

pm−1∏

i=0

biyb−i = (yb)pm

b−pm

= apm

b−pm

= yn(η, η′)

as we stated above. �

The proof of Iwasawa’s growth formula can now be easily completed.Since Y is a Λ-submodule of X and X/Y is finite, we have λ(Y ) = λ(X) andµ(Y ) = µ(X). Also,

h(p)n = |Gal(Ln/Fn)| = |X/Y ||Y/νn,no

Y |

26

for n ≥ no and so proposition 2.4.2 combined with proposition 2.3.1 (appliedto Y ) establishes the growth formula with λ = λ(X), µ = µ(X).

Remark 2.4.3. Assume that RamHyp(1) holds for the Zp-extension F∞/F .One can then take no = 0 in proposition 2.4.2 and so Y = Gal(L∞/L0F∞).The Galois group G = Gal(L∞/F ) now acts transitively on the set R occur-ring in the proof of proposition 2.4.2. Thus, the inertia groups Iη’s areconjugate in G and the elements gη form a single conjugacy class of G.Therefore, y(η, η′) is a commutator in G. That is, Y ⊂ G′. On the otherhand, G/Y = Gal(L0F∞/F ) is abelian. This implies that G′ ⊂ Y . HenceY = G′ = TX. Therefore, it follows that Yn = νn,0Y = ωnX for all n ≥ 0,which is essentially the content of proposition 2.1.5.

There are a number of interesting and useful consequences of proposition2.4.2 in addition to establishing the growth formula. To state some of these,we introduce the following ramification hypothesis:

RamHyp(2): Every prime of F which is ramified in F∞/F is totally rami-fied.

This simply means that we can take no = 0. It will simplify the statementof the following results. They could be applied to an arbitrary Zp-extensionjust by replacing the base field F by Fno

.

Proposition 2.4.4. Suppose that RamHyp(2) is satisfied for the Zp-extensionF∞/F . Then X/νn,0X is finite for all n ≥ 0. That is, fX(ζ − 1) 6= 0 for allζ ∈ µp∞, except possibly ζ = 1.

Proof. The corresponding statement for Y = Gal(L∞/L0F∞) is part ofproposition 2.4.2. Since [X : Y ] is finite, the first statement in the propositionfollows. The second statement then follows from a previous remark. �

Proposition 2.4.5. Suppose that RamHyp(2) is satisfied for the Zp-extensionF∞/F . Assume that p does not divide the class numbers of F and F1. Thenp does not divide the class number of Fn for any n ≥ 0.

Proof. The assumption implies that X = Y = ν1,0Y . Now ν1,0 ∈ m. There-fore, Nakayama’s lemma implies that Y = 0. Hence X = 0 too. Theconclusion follows from this. �

Remark 2.4.6. If one just assumes that the power of p dividing the classnumbers of F and F1 are equal, in addition to RamHyp(2), then one hasY = ν1,0Y . It again follows that Y = 0. That is, X is finite and the power

27

of p dividing the class number of Fn is equal to |X| for all n ≥ 0. Examplesexist where this actually happens.

If one combines propositions 2.3.4, 2.3.5, and 2.4.4, one obtains informa-tion about the group-theoretic structure of the p-primary subgroup An ofCFn

, summarized in the following proposition. We let λ = λ(X), µ = µ(X),and r = rankFp[[T ]](X[p]).

Proposition 2.4.7. Let F∞/F be a Zp-extension. Then

a. The exponent of An will be pn+c for n≫ 0, where c is some integer.

b. The p-rank of An will be rpn + c, where c is some integer.

c. If µ = 0, then An ∼=∏λ

i=1 Z/pn+ciZ× C for n≫ 0, where c1, ..., cλ arecertain integers and C is a certain finite group.

Proof. Using the notation of proposition 2.4.2, the results concern the struc-ture of the quotients X/νn,no

Y for n ≫ 0. We already have similar resultsfor the quotients Y/νn,no

Y .To prove part a, note that since X is a finitely generated Λ-module, it is

enough to consider the order of the image of x in X/νn,noY , where x is an

element of X which is not of finite order. For some t ≥ 0, we have ptx ∈ Y .As pointed out at the beginning of the proof of proposition 2.3.5, the imageof ptx in Y/νn,no

Y has order pn+c for n≫ 0, where c is some integer. Thus,the image of x in X/νn,no

Y will have order pn+c+t, and a follows.

For part b, note that (X/νn,noY )/p(X/νn,no

Y ) is isomorphic to X/νn,noY ,

where X = X/pX and Y denotes the image of Y under the natural map

X → X. The result then follows from proposition 2.3.1 applied to the Λ-module Y .

The proof of part c is just a slight variation on the proof of proposition2.3.4, using (6) for Y instead of X. �

Remark 2.4.8. The group C occurring in the above proposition is isomor-phic to the maximal, finite Λ-submodule of X, as the proof shows. Also, onecan make the following statement concerning the structure of the An’s with-out the assumption that µ(X) = 0. Let µ1, ..., µr be the finer µ-invariantsfor the Λ-module X, which were defined in section 2.3. Then we have

An ≈ (Z/pnZ)λ ×r∏

j=1

(Z/pµjZ)pn

28

This follows immediately from propositions 2.4.2 and 2.3.6.

1.5 Structure of the ideal class group of F∞.

Suppose that F∞ =⋃n≥0 Fn is a Zp-extension of F . For brevity we will

denote ClFnby Cn. The p-primary subgroup of Cn will be denoted by An.

For m ≥ n ≥ 0, we will denote the norm map NFm/Fn(either from Cm to Cn,

or from Am to An) by Nm,n. We will write Jn,m for JFm/Fn.

The ideal class group of F∞ can be defined as C∞ = lim−→

Cn, where thedirect limit is defined by the maps Jn,m. The natural map Cn → C∞ willbe denoted by Jn,∞. The main object of study in this section will be A∞ =lim−→

An, (defined by the same maps Jn,m, restricted to the An’s). This is thep-primary subgroup of C∞. There is a natural action of Γ on A∞, which wecan then regard as a discrete Γ-module and hence as a discrete Λ-module.Here is one important result. We denote the Iwasawa invariants λ and µoccurring in the growth formula by λ(F∞/F ) and µ(F∞/F ), respectively.

Proposition 2.5.1. The Λ-module A∞ is cofinitely generated, cotorsion, andcopure of dimension 1. In particular, if µ(F∞/F ) = 0, then A∞

∼= (Qp/Zp)λ

as a group, where λ = λ(F∞/F ).

Proof. We will exploit three facts. Let A∗n = Jn,∞(An). Then (1) A∞ =⋃

n≥0A∗n, (2) the groups A∗

n are isomorphic to quotients of X = XF∞/F forn≫ 0, and (3) X is a finitely generated, torsion Λ-module.

The assertion (1) is obviously true and (3) is just the content of proposi-tion 2.4.1. For (2), note that the natural restriction map X → Gal(Ln/Fn)is surjective when n ≫ 0. Also, Gal(Ln/Fn) is canonically isomorphic toAn for all n. Obviously, A∗

n is isomorphic to a quotient of An, and so (2)follows from these remarks. All of these groups have an action of Γ and canbe viewed as Λ-modules. The isomorphisms will be as Λ-modules.

Since X is a finitely generated, torsion Λ-module, proposition 2.2.5 im-plies that X/θX is finite for some θ ∈ m. Since A∗

n is a quotient of X forn≫ 0 (as a Λ-module), A∗

n/θA∗n is then a quotient of X/θX. Also, note that

|A∗n/θA

∗n| = |A

∗n[θ]|, and so we get the following inequality

|A∗n[θ]| ≤ |X/θX|

for all n≫ 0. It follows from this that A∞[θ] is finite. Proposition 2.2.8 thenimplies that A∞ is cofinitely generated and cotorsion as a Λ-module.

29

We will show that A∞ is copure of dimension 1 as a Λ-module. Thismeans that the Pontryagin dual of A∞ has no nonzero, finite Λ-submodules.Such a submodule would correspond to a quotient of A∞. Let A∞ = A∞/B∞

be the maximal, finite Λ-module quotient of A∞. Thus, B∞ is the maximal,Λ-submodule of A∞ which is copure of dimension 1. For each n ≥ 0, de-fine Bn to be the inverse-image of B∞ under the map Jn,∞. Clearly, Bn is

a Gal(Fn/F )-invariant subgroup of An. We denote An/Bn by An. There

are maps Jn,m, Jn,∞, and Nm,n on these quotient groups induced from thecorresponding maps Jn,m, Jn,∞, and Nm,n for m ≥ n ≥ 0. It is clear that

Jn,∞ : An → A∞

is injective for all n and surjective for n≫ 0. Therefore, the maps

Jn,m : An → Am

are isomorphisms for m ≥ n≫ 0. The norm maps

Nm,n : Am → An

will also be isomorphisms for m ≥ n ≫ 0. The surjectivity follows fromproposition 1.1.1. In addition, the identity

Nm,n ◦ Jn,m(a) = apm−n

will hold for all a ∈ An. These observations imply that An is trivial forn≫ 0, and hence for all n. This is because, for m > n, the map a→ ap

m−n

cannot be an isomorphism of An if that group is nontrivial. It follows thatA∞ = 0 and hence A∞ is copure of dimension 1, as claimed.

Finally, if µ(A∞) = 0, then the Pontryagin dual of A∞ will be a finitelygenerated Zp-module. It is pure of dimension 1 as a Λ-module, and so itmust be torsion-free and hence free as a Zp-module. Hence A∞ is indeedcofree as a Zp-module. �

The ingredients in the above proof have some consequences relating thestructure of X = XF∞/F and A∞ as Λ-modules. For example, it follows that

Ann(X) ⊂ Ann(A∞)

To see this, just note that if θ ∈ Ann(X). then θ annihilates the Λ-modulesA∗n for sufficiently large n, and so clearly θ ∈ Ann(A∞). It is also not difficult

30

to see that λ(A∞) ≤ λ(X). Using proposition 2.3.6, one can also verify thatµ(A∞) ≤ µ(X). The following result allows one to turn these inequalitiesinto equalities.

Proposition 2.5.2. Let Z be the maximal, finite Λ-submodule of X. Thenker(Jn,∞) ∼= Z for n≫ 0. In particular, ker(Jn,∞) has bounded order.

Proof. Choose no so that the Zp-extension F∞/Fnosatisfies RamHyp(2).

Let Y = Gal(L∞/LnoF∞) as in Proposition 2.4.2. Then we have a canon-

ical homomorphism αn : X → An which is defined as the composition ofthe restriction map X → Gal(Ln/Fn) with the inverse Artin isomorphismArt−1

Ln/Fn: Gal(Ln/Fn) → An. We then have the following commutative

diagramX −−−→ Anyνm,n

yJn,m

X −−−→ Am

(8)

for m ≥ n ≥ 0. The left vertical arrow is the map X → X defined byx→ νm,nx.

To verify the commutativity of the above diagram, suppose that x ∈ X.The commutative diagram (4) in the proof of proposition 1.1.1 implies that

Nm,n(αm(x)) = αn(x)

According to proposition 1.2.2, we have

Jn,m(αn(x)) = NGal(Fm/Fn)(αm(x)),

where NGal(Fm/Fn) : Am → Am is the norm operator for Gal(Fm/Fn), act-ing on Am. If g is a generator of Gal(Fm/Fn), then the norm operator is∑pm−n−1

i=0 gi.Now an element γ ∈ Γ acts on Am via its restriction γ|Fm

. The mapαm : X → Am then becomes a Γ-homomorphism. We can lift the normoperator NGal(Fm/Fn) on Am to X as follows. Recall that 1 + T ∈ Λ acts onX as γo. Let γn = γp

n

o , which is a topological generator for Gal(F∞/Fn).Then (1 + T )p

n

acts on X as γn. Since g = γn|Fmgenerates Gal(Fm/Fn),

the element∑pm−n−1

i=0 (1 + T )i ∈ Λ is a lifting of NGal(Fm/Fn). This element isωm/ωn = νm,n.

We then have αm(νm,nx) = NGal(Fm/Fn)(αm(x)). This is indeed equal toJn,m(αn(x)), and so diagram (8) is commutative. Now take m ≥ n ≥ no.

31

Define a mapjn,m : X/νn,no

Y → X/νm,noY

by jn,m(x+ νn,noY ) = νm,nx+ νm,no

Y . It follows that

ker(Jn,m) ∼= ker(jn,m) (9)

Since νn,m(νn,noY ) = νm,no

Y , we have

ker(jn,m) = (X[νn,m] + νn,noY )/νn,no

Y ∼= X[νn,m]/(X[νn,m] ∩ νn,noY )

Since m ≥ n ≥ no, the quotient X/νm,nX is finite, and so X[νn,m] ⊂ Z. It isclear that we have X[νn,m] = Z when m≫ n, and therefore

ker(Jn,∞) ∼= Z/(Z ∩ νn,noY ) (10)

The subgroups {νn,noY | n ≥ no} form a base of neighborhoods of 0 in X,

and so it is also clear that Z ∩ νn,noY = 0 for n ≫ no, proving the stated

result. �

Remark 2.5.3. The prime-to-p part of the Cn’s behave quite differently.Suppose that q is a prime, q 6= p. Let Qn denote the q-primary subgroupof Cn. Applying remark 1.2.8 (but for a Galois extension of p-power degreeand for the q-primary subgroup of the class groups, one sees that the mapJn,m : Qn → Qm is injective and that Jn,m(Qn) is a direct factor Qm forany m ≥ n. Thus Q∞, the direct limit of the Qn’s, will be isomorphicto the direct sum of the finite groups Q0, Q1/J0,1(Q0), Q2/J1,2(Q1), .... Thebehavior of the groups Qn/Jn−1,n(Qn−1) is difficult to study. We will describesome results and conjectures about this topic in chapter 4.

1.6 The base field F = Q(µp).

In this section, we will continue to discuss the important example F = Q(µp).Remark 1.2.11, propositions 1.4.5 and 1.4.6, and the discussion in between,and section 1.6 already concern this field. It will be an example which wewill return to periodically throughout this book. We will now illustrate someof the results of this chapter for the base field F .

The cyclotomic Zp-extension of F = Q(µp) is F∞ = Q(µp∞). Earlierresults in this chapter give us some rough, general picture of the structureof X = Gal(L∞/F∞), where L∞ is the pro-p Hilbert class field of F∞. This

32

special case has been studied extensively, both theoretically and computa-tionally, and so we will start with a brief summary of what is now known.We denote the maximal real subfield of F by F+.

I. If p ∤ hF , then X = 0.

II. The µ-invariant µ(X) vanishes.

III. If p ∤ hF+, then X is a free Zp-module.

IV. If p ∤ hF+, then X is a cyclic Gal(F∞/Q)-module.

V. If p < 12, 000, 000, then p ∤ hF and rankZp(X) is equal to the index of

irregularity for p.

The first assertion is an immediate consequence of proposition 2.1.2. Thesecond is a theorem due to B. Ferrero and L. Washington (which is validmore generally for the cyclotomic Zp-extension of any abelian extension F ofQ.) We will give two proofs in Chapter 7. Assuming that result, the thirdassertion just means that the maximal finite Λ-submodule of X is trivial.That is, X is pure of dimension 1. We will justify this statement below.

The fourth result needs some explanation. The field L∞ is a Galois ex-tension of Q. Thus X is a normal subgroup of Gal(L∞/Q) and so admitsa continuous Zp-linear action of G = Gal(F∞/Q). If p ∤ hF+ , then X isa free Zp-module (according to III). The assertion that X is cyclic as aG-module then means that X is spanned as a Zp-module by {gxo | g ∈ G}for some xo ∈ X. Now the structure of X as a Λ-module reflects the actionof Γ = Gal(F∞/F ) on X. The finite group ∆ = Gal(F∞/Q∞) (which is asubgroup of G) also acts on X. Since G is commutative, the actions of Γ and∆ on X commute. That is, the action of ∆ on X is Λ-linear. Therefore, itis natural to consider X as a module for the ring Λ[∆] - the group ring for∆ over Λ. The assertion in IV then means that X is a cyclic Λ[∆]-modulewhen p ∤ hF+ . We will also justify this statement below.

The assertion V is a result of elaborate calculations described in [Buhleret al]. The first such calculations were done in 1967 by Iwasawa and Simsverifying the same assertion for p < 4001. As we proceed, it will becomeclearer how such calculations could be done. It is a conjecture of Vandiverthat h+

F is never divisible by p.

We return now to assertion III. We must explain why the maximal finiteΛ-submodule Z of X is trivial if p ∤ hF+ . By proposition 2.5.2, Z is trivial ifand only if Jn,∞ : An → A∞ is injective for all sufficiently large n. The n-th

33

layer in the Zp-extension F∞/F is Fn = Q(µpn+1). Each of these fields is aCM-field. For m ≥ n ≥ 0, we can apply proposition 1.2.14 to the extensionFm/Fn. Jn,m : An → Am are injective for all m ≥ n ≥ 0. The assumptionsin that proposition are clearly satisfied.

The maximal totally real subfield of Fn, which we denote by F+n , is a cyclic

extension of F+ of degree pn. Only one prime of F+ is ramified in F+n /F

+,namely the unique prime above p, and that prime is totally ramified. Hence,proposition 1.1.4 implies that if p ∤ hF+ , then p ∤ hF+

nfor all n ≥ 0. Therefore,

in the notation of proposition 1.2.14, we have A(ǫ0)n = 0 and hence An = A

(ǫ1)n .

It follows that the maps Jn,m : An → Am are injective. Therefore, the mapsJn,∞ are injective, and we can then conclude that Z is trivial if p ∤ hF+ .

In general, without the assumption that p ∤ hF+ , one can still state thatthe odd ∆-components of X have no nonzero, finite Λ-submodules. In otherwords, for each odd i, the Λ-module X(ωi) is pure of dimension 1. Notethat Z is ∆-invariant. We are asserting that Z(ωi) = 0 if i is odd. if oneexamines the proof of proposition 2.5.2, one sees that the isomorphism is∆-equivariant. Thus, one has an isomorphism

ker(Jn,∞)(ωi) ∼= Z(ωi)

for any i. If i is odd, one can again apply proposition 1.2.14 to conclude thatZ(ωi) is trivial.

Assertion IV is a consequence of proposition 1.4.5 or 1.4.6. Assuming

that p ∤ hF+ , we know that A(ωi)F is cyclic for each i. Since RamHyp(1) holds

for F∞/F , we have a canonical isomorphism X/TX ∼= AF . This isomorphismis ∆-equivariant. We can decompose X as a Zp[∆]-module as follows:

X ∼=

p−2∏

i=0

X(ωi) (11)

We then have an isomorphism

X(ωi)/TX(ωi) ∼= A(ωi)F

for each i. If i is even, then it follows from Nakayama’s lemma thatX(ωi) = 0.If i is odd, Nakayama”s lemma implies that X(ωi) can be generated by oneelement as a Λ-module. Since this is valid for all i, it follows that X is acyclic Λ[∆]-module, as asserted.

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According to assertion III, each ∆-component X(ωi) is a free Zp-module

if p ∤ hF+ . Assertion V then means that rankZp(X(ωi)) ≤ 1, assuming that

p < 12, 000, 000. Thus, for those i’s for which X(ωi) is nontrivial, we haveX(ωi) ∼= Zp and the action of Γ on X(ωi) is by a character 〈χ〉si , where si ∈ Zp.One of the most interesting discoveries of Iwasawa was that this number simust be the zero of a certain analytic function, the Kubota-Leopoldt p-adicL-function Lp(s, ω

j), where j = 1− i. We will prove this in a later chapter.

We now want to explain the relationship of the Λ-module X to the groupsH1unr(Q, Dψ), where ψ is a power of the cyclotomic character χ and Dψ

denotes the Galois module associated to ψ as defined in section 1.6. Thus,we assume that ψ : Gal(F∞/Q) → Z×

p is a continuous homomorphism. For

each i, let fi(T ) denote the characteristic polynomial for X(ωi).

Proposition 2.6.1. Suppose that ψ|∆ = ωi. Then the restriction mapdefines an isomorphism

H1unr(Q, Dψ) −→ HomΓ(X(ωi), Dψ)

If ψ(γo) = 1 + βψ, then the group HomΓ(X(ωi), Dψ) is isomorphic to the

Pontryagin dual of X(ωi)/(T − βψ)X(ωi). The group H1unr(Q, Dψ) is finite if

and only if fi(βψ) 6= 0. We then have

ordp(|H1

unr(Q, Dψ)|)

= ordp(fi(βψ)

).

if i is odd.

Proof. We consider the restriction map in two steps, from GQ to GF andfrom GF to GF∞

. Proposition 1.5.5 implies that we have an isomorphism

H1unr(Q, Dψ) −→ H1

unr(F,Dψ)∆

For the second step, note that H0(F∞, Dψ) = Dψ. The inflation-restrictionsequence for the extension F∞/F becomes

0 −→ H1(Γ, Dψ) −→ H1(F,Dψ) −→ H1(F∞, Dψ)Γ −→ H2(Γ, Dψ)

We will show that H2(Γ, Dψ) = 0 and that H1(Γ, Dψ) is usually also trivial.

For the rest of the proof, and for later arguments, it will be useful tomake some general observations about H i(Γ, A) for any discrete, p-primary

35

abelian group A with a continuous action of Γ. These will be summarized inthe lemma below. Let Γn = Γp

n

for n ≥ 0. By definition, we have

H i(Γ, A) = lim−→n

H i(Γ/Γn, AΓn)

under the natural inflation maps. We will let NΓ/Γndenote the norm map

for the action of the finite cyclic group Γ/Γn on AΓn . The kernel will bedenoted simply by ker(NΓ/Γn

), a subgroup of AΓn . The image is NΓ/Γn(AΓn),

a subgroup of AΓ. First we consider i = 2. We have

H2(Γ/Γn, AΓn)) ∼= AΓ/NΓ/Γn

(AΓn)

If m ≥ n ≥ 0, then the inflation map corresponds to the map

AΓ/NΓ/Γn(AΓn) −→ AΓ/NΓ/Γm

(AΓm)

which is defined by mapping the coset of a ∈ AΓ in the first group to thecoset of NΓn/Γm

(a) in the second group. But NΓn/Γm(a) = pm−na since a is

fixed by Γ. Hence, for each such a and for sufficiently large m, the image istrivial. Thus, the direct limit is trivial. That is, H2(Γ, A) = 0.

Now, if i = 1 and m ≥ n ≥ 0, then the inflation map corresponds to thehomomorphism

ker(NΓ/Γn)/(γo − 1)AΓn −→ ker(NΓ/Γm

)/(γo − 1)AΓm

which is defined by mapping the coset of a ∈ ker(NΓ/Γn) to the coset of

the same a in ker(NΓ/Γm). However, if a ∈ A, then a ∈ AΓn for some n

and, essentially as above, one sees that a ∈ ker(NΓ/Γm) for sufficiently large

m. That is,⋃n≥0

(ker(NΓ/Γn

) = A. On the other hand, it is clear that⋃n≥0

((γo− 1)AΓn

)= (γo− 1)A. We have proved most of the following basic

lemma.

Lemma 2.6.2. If A is a discrete, p-primary Γ-module, then

H1(Γ, A) ∼= A/(γo − 1)A, H2(Γ, A) = 0

Furthermore, if we assume that A is a divisible group, that A[p] is finite, andthat AΓ is also finite, then H1(Γ, A) = 0 too.

To prove the final part of this lemma, note that A ∼= (Qp/Zp)r for some r ≥ 0.

We can regard γo− 1 as an endomorphism of that group. The kernel of that

36

endomorphism is AΓ and, if that kernel is finite, then the image is a divisiblegroup of Zp-corank r, and hence must also be isomorphic to (Qp/Zp)

r. Itfollows easily that (γo − 1)A = A, and therefore that H1(Γ, A) is trivial asstated.

Returning to the proof of proposition 2.6.1, the inflation-restriction se-quence simplifies to

0 −→ Dψ/(γo − 1)Dψ −→ H1(F,Dψ) −→ H1(F∞, Dψ)Γ −→ 0

We can use this to prove that the second step of the restriction map is alsoan isomorphism.

Assume first that ψ|Γ is nontrivial. Then the above lemma implies thatwe have an isomorphism H1(F,Dψ) −→ H1(F∞, Dψ)Γ. Just as in the proofof proposition 1.5.5, we must consider the kernel of the restriction mapH1(Iv, Dψ) → H1(Iη, Dψ), where v is a prime of F , η is a prime of F∞

lying over v, and Iv, Iη are the inertia subgroups of GF and GF∞for a prime

of Q lying over η. However, if v ∤ p, then v is unramified in F∞/F . It followsthat Iv = Iη and the kernel of the restriction map at v is certainly trivial.There is a unique prime v of F lying above p and a unique prime η of F∞

lying over v. Also, v is totally ramified in F∞/F , the inertia subgroup of Γfor v is Γ itself, and hence Iv/Iη can be identified with Γ. It follows thatH1(Iv/Iη, Dψ) = 0. Therefore, the kernel of the restriction map at v is againtrivial. This proves that the restriction map

H1unr(F,Dψ) −→ H1

unr(F∞, Dψ)

is an isomorphism if ψ|Γ is nontrivial.Now assume that ψ|Γ is trivial, i.e., that ψ = ωi for some i. The re-

striction map H1(F,Dψ) −→ H1(F∞, Dψ)Γ has a nontrivial kernel, namelyH1(Γ, Dψ) = Hom(Γ, Dψ). This group is isomorphic to Qp/Zp. However,consider the composite map

H1(Γ, Dψ) −→ H1(Iv/Iη, Dψ) −→ H1(Iv, Dψ)

where v is the prime of F lying above p. The first map is an isomorphism,the second map is injective. Hence, it follows that

ker(H1unr(F,Dψ) −→ H1

unr(F∞, Dψ))

= H1unr(F,Dψ)

⋂H1(Γ, Dψ) = 0

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To show that the cokernel of the map H1unr(F,Dψ) −→ H1

unr(F∞, Dψ)Γ istrivial, we again refer to the proof of proposition 1.5.5. In the diagram(24) and the exact sequence (25), take F ′ = F∞, D = Dψ, and G = Γ.However, ker(bF ′/F ) is now the infinite group H1(Γ, Dψ). Taking into accountthe above remarks about the local restriction map for v ∤ p, it follows thatker(cF ′/F ) = H1(Iv/Iη, Dψ) and that the map ker(bF ′/F ) −→ ker(cF ′/F ) issurjective. Therefore, the map ker(cF ′/F ) −→ ker(aF ′/F ) is the zero-map.Also, since G = Γ, lemma 2.6.2 implies that coker(bF ′/F ) = 0. It then followsthat coker(aF ′/F ) = 0.

Thus, in all cases, the map H1unr(F,Dψ) −→ H1

unr(F∞, Dψ)Γ is an isomor-phism. Combining this with the first step, we obtain the isomorphism

H1unr(Q, Dψ) −→ H1

unr(F∞, Dψ)Gal(F∞/Q)

Now H1unr(F∞, Dψ) = Hom(X,Dψ). Since ∆ acts on Dψ by the character

ψ|∆ = ωi, we have

H1unr(F∞, Dψ)∆ = Hom(X(ωi), Dψ)

It is then clear that H1unr(F∞, Dψ)∆×Γ is isomorphic to HomΓ(X(ωi), Dψ) and

this establishes the first part of proposition 2.5.1.Since γo acts on Dψ as multiplication by 1 + βψ, and this determines

the action of Γ, it follows that any element of HomΓ(X(ωi), Dψ) must factor

through the maximal quotient of X(ωi) on which γo acts in the same way.That quotient is X(ωi)/

(γo − (1 + βψ)

)X(ωi). Conversely, any element of

Hom(X(ωi), Dψ) factoring through that quotient will be Γ-equivariant. Thatis,

HomΓ(X(ωi), Dψ) = Hom(X(ωi)/

(γo − (1 + βψ)

)X(ωi), Dψ

)

which is indeed isomorphic to the Pontryagin dual of X(ωi)/(T − βψ)

)X(ωi)

since Dψ∼= Qp/Zp as a group. This proves the second statement in the

proposition. It is then clear that H1unr(Q, Dψ) is finite if and only if fi(βψ) 6=

0. Furthermore, one can apply proposition 2.3.3 if i is odd because we knowthat X(ωi) has no nonzero, finite Λ-submodules. The final statement followthen follows. �

Remark 2.6.3. It is worth discussing what happens for an arbitrary Zp-extension F∞/F . We consider any number field F and let p be any prime.Let D = Dψ, where ψ ∈ Hom(Γ,Z×

p ). Assume first that ψ has infinite

38

order. Then F∞ = F (D) and DΓ is finite and nontrivial. The cokernel ofthe restriction map

H1(F,D) −→ H1(F∞, D)Γ (12)

is isomorphic to a subgroup of H2(Γ, D). But that group vanishes accordingto lemma 2.6.2 and hence the restriction map is surjective. The kernel isH1(Γ, DGF∞ ). This group also vanishes because DGF∞ = D is divisible. Forthe same reason, the local restriction map H1(F unr

v , D) −→ H1(F unrη , D) is

injective for every ramified prime v in the extension F∞/F , where η is aprime of F∞ lying over v. Therefore, we again get an isomorphism

H1unr(F,D) −→ HomΓ(X,D), (13)

where X = XF∞/F . In particular, if we let β = ψ(γo) − 1, then H1unr(F,D)

is infinite if and only if fX(β) = 0. On the other hand, if ψ has finite order,then corollary 1.5.8 implies that H1

unr(F,D) must be finite. Therefore, sincefX(T ) is a nonzero polynomial, it follows that H1

unr(F,D) is finite for all butfinitely many ψ ∈ Hom(Γ,Z×

p ).If ψ has finite order, then ψ is the trivial character unless p = 2. For

p = 2, the character ψ could have order 1 or 2. It is not difficult to verifythat the kernel of the map (13) is still finite. One uses the fact that if n issufficiently large, then at least one prime of Fn is totally ramified in F∞/Fn.However, the cokernel of (13) can be infinite. In particular, if ψ is trivial(and so β = 0), it is possible to have fX(0) = 0 even though H1

unr(F,D)is finite. As an example to illustrate how this can happen, suppose that Fis an imaginary quadratic field, that p splits in F/Q, and that F∞ is thecyclotomic Zp-extension of F . This situation was discussed briefly followingthe proof of proposition 1.6.4. For s ∈ Zp, let ψs = 〈χ〉s. Thus, ψs hasinfinite order if s 6= 0. Let Ds = Dψs

and βs = ψs(γo)− 1). Thus, βs → 0 ass→ 0 in Zp. For s 6= 0, we have

H1unr(F,Ds) ∼= Hom(X/(T − βs)X,Qp/Zp)

as explained above. However, under the above assumptions on F , the orderof H1

unr(F,Ds) is unbounded as s → 0 in Zp. It follows that fX(βs) → 0 ass→ 0 and therefore fX(0) = 0.

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