The compositum of wild extensions of localfields of prime degree

Ilaria Del Corso and Roberto Dvornicich

.

Abstract

In this paper we present a general view of the totally and wildly ramified exten-sions of degree p of a p-adic field K. Our method consists in deducing the propertiesof the set of all extensions of degree p of K from the study of the compositum CK(p)of all its elements. We show that in fact CK(p) is the maximal abelian extension ofexponent p of F = F (K), where F is the compositum of all cyclic extensions of K ofdegree dividing p− 1. By our method, it is fairly simple to recover the distributionof the extensions of K of degree p (and also of their isomorphism classes) accordingto their discriminant.

2000 MSC: 11S15; 11S20.

1 Introduction

Let p be a prime number, and let K be a finite extension of Qp. For each pair (e, f) ofpositive integers, the set EK(e, f) of extensions of K with ramification index e and inertialdegree f is finite, and the problem arises of characterizing this set as explicitly as possible.If p - e, namely in the tamely ramified case, then EK(e, f) is easy to describe, and, in thegeneral case, the main difficulty is to deal with the totally and wildly ramified part ofthese extensions.

By examining all possible Eisenstein polynomials, Krasner [4]-[8] obtained an explicitformula for the total number of elements in EK(e, f). Later on, Serre [11], with a similarmethod, was able to refine Krasner’s result by counting the elements in EK(e, f) withgiven discriminant for all possible discriminants; in the same paper, he also showed thatthe distribution of the extensions in EK(e, f) according to their discriminant satisfies a“mass formula” of general type.

The method of examining Eisenstein polynomials, however, seems unsuitable for deter-mining the number IK(e, f) of isomorphism classes of fields in EK(e, f). In the general

1

case, no explicit formula for IK(e, f) is known. Recently, Hou and Keating [3] were ableto find general formulas for IK(e, f) when p2 - e and, under some additional assumptionson e and f , also when p2||e.In this paper we present a general and explicit view of the totally and wildly ramifiedextensions of degree p of a p-adic field K. Our method consists in deducing the propertiesof the set of all extensions of degree p of K from the study of the compositum CK(p) ofall its elements. We shall show that in fact CK(p) is the maximal abelian extension ofexponent p of F = F (K), where F is the compositum of all cyclic extensions of K ofdegree dividing p− 1 (Theorem 1).

In virtue of Theorem 1, we shall endow the Galois group G := Gal(CK(p)/K) with a veryrich structure, which is extremely useful for studying the properties of all extensions ofK of degree p. To start with, we shall prove that G is isomorphic to a semidirect productH oG, where H := Gal(CK(p)/F ) and G := Gal(F/K), so G acts on H via conjugation.This action gives to H a structure of G-module, and it is not hard to recognize that theG-submodules of H correspond, via Galois theory, to the fields L such that F ⊆ L ⊆ CK

and L is normal over K. Also, H is a vector space over Fp and decomposes as a directsum of G-submodules of dimension 1. To determine such a decomposition is the same asdetermining a basis ofH made up with “eigenvectors” under the action of G, i.e., elementsλ ∈ H such that gλ is a multiple of λ for all g ∈ G. We shall find a basis of eigenvectorsof H as follows. First we shall notice that F ∗/F ∗p, a group which is isomorphic to H, hasalso a natural structure of G-module and hence has a decomposition into G-submodulesof dimension 1. Secondly, we shall see that the G-modules H and F ∗/F ∗p are dual to eachother, so that a suitable isomorphism between F ∗/F ∗p and H sends a basis of eigenvectorsfor F ∗/F ∗p into a basis of eigenvectors for H. Thirdly, we shall use the description ofF ∗/F ∗p made in [2] to find a basis of eigenvectors for F ∗/F ∗p which corresponds to a basisof eigenvectors for H most convenient for our purposes.Once we have such a basis, we know all G-submodules of H, hence all normal extensions Lof K such that F ⊆ L ⊆ CK(p). In particular, we know one by one all normal extensionsL of K such that [L : F ] = p. We shall show that there is a bijection between the set L ofsuch extensions and the set IK(p) of isomorphism classes of the extensions of K of degreep, thus retrieving some results of [3] very easily. Also, since in fact the field F containsthe p-th roots of unity, each L ∈ L is of type L = F ( p

√a) for some a ∈ F , and this makes

the computation of Disc(L/K) easy too. The properties of the discriminant in towersof extensions then allow us to reconstruct immediately the distribution of the extensionsof K of degree p (and also of their isomorphism classes) according to their discriminant(Section 6).While studying the structure of the G-module H, we shall also obtain some more infor-mation on the Galois group G and on the extensions of K of degree p. Along with thenotion of “eigenvector”, we can think of a notion of “eigenvalue”. In our case, since G isgenerated by two elements, σ and τ , the eigenvalue relative to an eigenvector λ can bethought as the pair (s, t) such that σλ = λs and τλ = λt. The set of eigenvectors of H

2

with a given eigenvalue (s, t) is a G-submodule, that we shall denote by Hst. Clearly Hdecomposes uniquely as a direct sum of its G-submodules Hst, and we shall determine thedimensions of these submodules as well. We shall see that these dimensions determinehow many extensions E of K of degree p are such that their normal closure intersects Fin a given cyclic extension of K.As a further application, we shall show that the basis of eigenvectors chosen allows us todetermine the discriminant of CK(p)/K and the ramification groups of its Galois groupG.

As mentioned above, the results of this paper regard the extensions of prime degree p only.We intend to show in a forthcoming paper the applications of our method to extensionsof degree a higher power of p, as well as its connections to extensions of degree d = d0p

n,where (d0, p) = 1 and n is fixed.

2 Notation and statement of the main theorem

Throughout the paper, p will be a fixed prime number. For a finite extension K of Qp,we shall denote by eK and fK the ramification index and the inertial degree of K/Qp,respectively, and by nK = eKfK = [K : Qp] the absolute degree of K.Also, we shall denote by OK the ring of integers of K, by πK a uniformizer of K (i.e., agenerator of the maximal ideal of OK), and by vK the valuation of K normalized so thatvK(πK) = 1. The residue field of K will be indicated by K and its cardinality by qK = pfK .If x ∈ OK, its projection to K will be denoted by x.The symbol ζn will indicate a generic primitive nth root of unity, except for the casen = p − 1, where we shall choose once and for all a fixed primitive (p − 1)-th root ofunity ζp−1 in Qp, and we shall let r be the integer with 1 ≤ r ≤ p− 1 such that ζp−1 ≡ r(mod p). The set of Teichmuller representatives of K (i.e., the union of the (qK − 1)-throots of unity in K and zero) will be denoted by T (K).We shall use the standard notations NK′/K, DK′/K, Disc(K′/K) for the norm, the differ-ent and the discriminant of K′/K, respectively, and the notation Ui(K) (i ≥ 1) for themultiplicative subgroups of (OK)∗ defined by Ui(K) = {x ∈ OK |x ≡ 1 (mod πi

KOK)}.Finally, for a real number v, the shall use the standard notations bvc and dve for thelargest integer ≤ v and the smallest integer ≥ v, respectively.

From now on, K will be a given finite extension of Qp. We shall denote by

E = EK(p) = {E ⊇ K | [E : K] = p}

the set of all extensions of K of degree p within a fixed algebraic closure of K, by CK(p)the compositum of all extensions E ∈ E , and by IK(p) the number of isomorphism classesover K of the elements E ∈ E .

3

First of all, we shall need to determine the field CK(p). Let F = FK be the compositumof all cyclic extensions of K of degree dividing p− 1, and let AF be the maximal abelianextension of F of exponent p. We shall prove the following theorem (Section 5).

Theorem 1 The compositum of all extensions of K of degree p coincides with AF .

Clearly, all extensions in the tower K ⊆ F ⊆ AF are normal, so we can define

G := Gal(F/K), H := Gal(AF/F ), G := Gal(AF/K) .

In the following, we shall describe many properties of CK(p) by studying the naturalstructure of H as a G-module. We shall see that from these properties one can getdetailed information on the single elements E ∈ E and, in particular, an explicit formulafor IK(p).

3 The G-modules F ∗/F ∗p and Gal(AF/F )

Proposition 1 F = K(ζqp−1K −1, π), where πp−1 = πK . Moreover, G := Gal(F/K) ∼=

〈σ〉 × 〈τ〉 ∼= (Z/(p− 1)Z)2, where σ and τ are defined by

σ(ζqp−1K −1) = ζqp−1

K −1, σ(π) = ζp−1π;

τ(ζqp−1K −1) = ζqK

qp−1K −1

, τ(π) = π.(1)

Proof. The unramified extensionK(ζqp−1K −1)/K and the totally ramified extensionK(π)/K

are cyclic of degree p− 1, hence K(ζqp−1K −1, π) ⊆ F . Let K ′/K be any cyclic extension of

degree d|p − 1, and let f ′, e′ be its inertial degree and its ramification index (f ′e′ = d).Then K ′ = K(ζ

qf ′K−1

, π′), where π′ is a root of a polynomial of type Xe′ − uπK , where u

is a root of unity of K(ζqf ′K−1

). Since the e′-th roots of u belong to K(ζqp−1K −1), it follows

that K ′ ⊆ K(ζqp−1K −1, π), whence F = K(ζqp−1

K −1, π).

The statement about the Galois group is trivial, since clearly the extensions K(ζqp−1K −1)/K

and K(π)/K are cyclic and linearly disjoint. 2

Remark 1 Since the degree of K(ζp) over K divides p−1, we have K(ζp) ⊆ F . We shalldenote by k, h the integers such that 1 ≤ k, h ≤ p− 1 and

σ(ζp) = ζkp , τ(ζp) = ζh

p . (2)

Proposition 2 G ∼= H oG, where G acts on H by conjugation,

4

Proof. Observe that G has H as a normal subgroup with quotient isomorphic to G, henceacting on G by conjugation. Since (|H|, |G|) = 1, the Schur-Zassenhaus theorem (see [10,9.12]) yields G ∼= H oG. 2

The action of G on H by conjugation makes H a G-module. On the other hand, alsoF ∗/F ∗p is a G-module with the action of G induced by the action on F . We shall seein the next proposition that they are canonically dual to each other. In fact, the Galoisaction makes also the group 〈ζp〉 of the p-th roots of unity a G-module, and we can give toHom(H, 〈ζp〉) the usual structure of G-module attached to the group of homomorphismsbetween twoG-modules, namely, we can define theG-action on Hom(H, 〈ζp〉) by g(χ)(λ) =g(χ(g−1λ)).

Proposition 3 F ∗/F ∗p ∼= Hom(H, 〈ζp〉) as a G-module. In particular, the group H isisomorphic to (Z/pZ)nF +2.

Proof. For a ∈ F ∗, denote by a the class of a in F ∗/F ∗p. By Kummer theory, the map

ψ : F ∗/F ∗p → Hom(H, 〈ζp〉) defined by ψ(a) = χa, where χa(λ) = λ( p√a)p√a

, is well-definedand is a group-isomorphism. Regarding the action of G, extend all g ∈ G to elementsof Gal(AF/K) and denote again by g these extensions. Observe that, for any a ∈ F ∗

and for any choice of p√a and p

√g(a), the ratio p

√g(a)/g( p

√a) is a p-th root of unity, and

therefore is left fixed by all λ ∈ H. We have

χg(a)(λ) =λ( p

√g(a) )

p√g(a)

=λg( p

√a)

g( p√a)

= g

(g−1λg( p

√a)

p√a

)= g(χa(g

−1λg)) = (g(χa))(λ)

and hence ψ(g(a)) = gψ(a).As to the second statement, we note that H is a finite abelian group of exponent p andthat Hom(H, 〈ζp〉) is the group of its characters, hence they are isomorphic.

From the first part of the proposition, it follows that H ∼= F ∗/F ∗p. Now, since F containsthe p-th roots of unity, we get F ∗/F ∗p ∼= (Z/pZ)nF +2 (see [9, Ch. 5, Prop. 5.8 and Thm.5.7]).

2

We now recall some elementary facts of the theory of group representations, which wewant to apply to the G-modules H and F ∗/F ∗p.

Proposition 4 Let V be vector space over Fp endowed with a structure of G-module.

(i) If V ′ ⊆ V is a G-submodule then V ∼= V ′ ⊕ V/V ′ as a G-module;

(ii) V decomposes as a direct sum of G-submodules of dimension 1;

(iii) for m,n ∈ F∗p, let Vmn = {v ∈ V |σv = mv, τv = nv}. Then V =⊕

m,n∈F∗pVmn.

5

Proof. We observe that a G-module structure on V is the same as a representation ofG in V . In particular, each G-submodule V ′ of V admits a complementary G-submoduleV ′. Clearly, V/V ′ ∼= V ′ and (i) follows.As for (ii), note that G is abelian of exponent p−1 and Fp contains the (p−1)-th roots ofunity; in this case Schur’s lemma implies that every irreducible representation of G intoan Fp-vector space has degree 1.For statement (iii), observe that a subspace 〈v〉 of V of dimension 1 is a G-submodule ifand only if there exist elements m,n ∈ F∗p such that

σv = mv, τv = nv.

In other words, every G-submodule of dimension 1 of V is contained in some Vmn. Group-ing together all G-submodules of an irreducible decomposition of V contained in the sameVmn, we obtain (iii). 2

In analogy with the notation of last proposition, writing the action of G explicitly, wedefine, for m,n ∈ F∗p,

(F ∗/F ∗p)mn := {w ∈ F ∗/F ∗p |σw = wm, τw = wn}

and, for s, t ∈ F∗p,Hst = {λ ∈ H |σλσ−1 = λs, τλτ−1 = λt} .

With this notation, we have F ∗/F ∗p =⊕

m.n∈F∗p(F ∗/F ∗p)mn and H =

⊕s,t∈F∗p

Hst.

Now, letH = 〈λ1〉⊕· · ·⊕〈λnF +2〉 be a decomposition ofH intoG-submodules of dimension1, and associate to each λi the homomorphism χi defined by χi(λi) = ζp, χi(λj) =1 for j 6= i. Extending this definition to all elements of H by linearity, this gives agroup-isomorphism Ψ : H → Hom(H, 〈ζp〉). This isomorphism, combined with the G-isomorphism of Proposition 3, allows us to relate the G-modules Hst and (F ∗/F ∗p)mn.

Proposition 5 For s, t ∈ F∗p, let s∗, t∗ ∈ F∗p be defined by s∗ = ks−1, t∗ = ht−1. ThenΨ(Hst) = Hom(H, 〈ζp〉)s∗t∗ .

Proof. Let H = 〈λ1〉 ⊕ · · · ⊕ 〈λnF +2〉 be a decomposition of H into G-submodules ofdimension 1. If σλiσ

−1 = λsi , then (σχi)(λi) = σ(χi(λ

s−1

i )) = σ(χi(λi))s−1

= ζks−1

p

and (σχi)(λj) = 1 for j 6= i, whence σχi = χks−1

i . Similarly, one shows that, ifτλiτ

−1 = λti, then τχi = χht−1

i , whence Ψ(Hst) ⊆ Hom(H, 〈ζp〉)s∗t∗ . On the other hand,the map (s, t) 7→ (s∗, t∗) is a bijection of (F∗p)2 into itself, and hence

∑s,t∈F∗p dimFp Hst =∑

s,t∈F∗p dimFp Hom(H, 〈ζp〉)s∗t∗ = nF + 2, so there is equality everywhere.2

6

4 The dimensions of the G-modules Hst

In this section we shall determine the dimension of Hst for all s, t ∈ F∗p (Corollary 3).In view of Propositions 3 and 5, this is equivalent to determining the dimensions of theG-modules (F ∗/F ∗p)mn for all m,n ∈ F∗p..We first observe that F ∗/F ∗p ∼= 〈π〉/〈πp〉 × U1(F )/U1(F )p, where π is the uniformizer ofF defined in Proposition 1, and it is easy to see that the canonical isomorphism is alsoa G-module isomorphism. From now on we shall write Uj for Uj(F ) (j ≥ 1); settingW (0) = 〈π〉/〈πp〉, we have F ∗/F ∗p ∼= W (0) ⊕ U1/U

p1 as G-modules.

Now our first step is to choose a suitable Fp-basis for U1/Up1 ; actually, we shall take

the projection of the Zp-basis of U1 given in [2, Ch. 1 Prop. 6.4], and we shall provesome features of this basis which will be crucial in the following. We note that the resultsconcerning this basis hold for a generic Galois extension F of K (provided that F containsζp) and for its Galois group G.

Consider the filtration of G-modules

U1/Up1 ⊇ U2U

p1 /U

p1 ⊇ · · · ⊇ UpeK+1U

p1 /U

p1 = 0 ,

where the last equality holds since UpeK+1 ⊆ Up1 (see [2, Ch. 1, Sec. 5.8, Cor. 2]).

By Proposition 4, the G-module Uj+1Up1 /U

p1 has a complement W (j) in UjU

p1 /U

p1 and we

clearly have W (j) ∼= UjUp1 /Uj+1U

p1 . It follows that

U1/Up1∼=

peK⊕j=1

W (j) (3)

as G-modules. Let θ0, θ1 be the the unique elements of T (F ) such that

p ≡ θ0πeF (mod πeF +1OF ), ζp ≡ 1 + θ1π

eK (mod πeK+1OF ). (4)

It is easy to check that θ0 = −θp−11 , and that the linear map ψ : F → F defined by

ψ(ε) = εp + θ0ε has a kernel of dimension 1 over Fp, and hence ψ(F ) is a linear subspaceof F of codimension 1 (see [2, Ch. 1] for details).Let α1 . . . , αfF

, β ∈ T (F ) be such that {α1, . . . , αfF} is a basis of F and β is a generator

of F /ψ(F ).

Proposition 6 For 1 ≤ j ≤ peK , a set of representatives for a basis of UjUp1 /U

p1 is given

byXj = {1 + αiπ

l | 1 ≤ i ≤ fF , j ≤ l < peK , (l, p) = 1} ∪ {1 + βπpeK}. (5)

Proof. We prove the proposition by induction on j. The case j = 1 is an immediateconsequence of [2, Ch. 1, Prop. 6.4], since, by Nakayama’s Lemma, a minimal set ofgenerators of U1 as Zp-module is the same as a basis of the vector space U1/U

p1 over

Fp∼= Zp/pZp.

7

Lemma 1 If 1 ≤ j < peK and (j, p) = 1 then Uj ∩ Uj+1Up1 = Uj+1.

Proof. The inclusion ⊇ is trivial, so let x = yzp, where x ∈ Uj, y ∈ Uj+1 and z ∈ U1:we have to show that zp ∈ Uj+1. So we may assume that z 6= 1 and that z ≡ 1 + θπµ

(mod Uµ+1) for some positive integer µ and for some θ ∈ T − {0}. By the binomialtheorem, zp ≡ 1 + θpπµp (mod πmin{µp+1,(p−1)eK+µ}OK).If µp+ 1 > (p− 1)eK + µ, then µ ≥ eK , whence zp ∈ UpeK

⊆ Uj+1.If µp + 1 ≤ (p − 1)eK + µ, then zp ∈ Uµp \ Uµp+1. Now, from zp = xy−1 ∈ Uj it followsµp ≥ j and, since (j, p) = 1, µp ≥ j + 1.

2

Assume now that the statement of the Proposition is true for some j < peK . Then

Yj ={y({mil},meK

) =( ∏

1≤i≤fF

∏j≤l<peK

(l,p)=1

(1 + αiπl)mil

)· (1 + βπeK )meK |mil,meK

∈ Fp

}

is a set of representatives of all elements of UjUp1 /U

p1 . If (j, p) = 1,

y({mil},meK) ≡

fF∏i=1

(1 + αiπj)mij ≡ 1 +

fF∑i=1

mijαiπj (mod Uj+1U

p1 /U

p1 ) .

By Lemma 1, y({mil},meK) ∈ Uj+1U

p1 if and only if 1 +

∑fF

i=1mijαiπj ∈ Uj+1. In turn,

this is equivalent to∑fF

i=1mijαi = 0 and to mij = 0 for 1 ≤ i ≤ fF . This means thatXj+1 is a set of representatives for a basis of Uj+1U

p1 /U

p1 .

Finally, if p|j, then Xj ⊆ Uj+1, therefore Xj = Xj+1 and UjUp1 /U

p1 = Uj+1U

p1 /U

p1 . 2

Corollary 1 (i) For 1 ≤ j < peK and (j, p) = 1, W (j) ∼= Uj/Uj+1 as a G-module,hence dimFp W

(j) = fF ;

(ii) for 1 ≤ j < eK , W (pj) = 0;

(iii) dimFp W(peK) = 1 and W (peK) = 〈1 + βπpeK mod Up

1 〉.

Proof. (i): The projection Uj/Uj+1 onto UjUp1 /Uj+1U

p1∼= W (j) is a G-module homo-

morphism; since, by Proposition 6, dimFp UjUp1 /Uj+1U

p1 = fK = dimFp Uj/Uj+1, it is a

G-module isomorphism.(ii): We have shown at the end of the proof of Proposition 6 that UpjU

p1 /U

p1 = Upj+1U

p1 /U

p1 ,

hence the quotient W (pj) is zero.(iii): This is the case j = peK of Proposition 6. 2

8

We now return to our specific case when F is the compositum of all extensions of K ofdegree dividing p − 1. Proposition 4 (iii) applied to the G-module W (j) gives W (j) =⊕m.n∈F∗p

W(j)mn, where W

(j)mn := {w ∈ W (j) |σw = wm, τw = wn}. Now, by (3), it follows that

F ∗/F ∗p ∼=⊕peK

j=0 W(j), hence

(F ∗/F ∗p)mn =

peK⊕j=0

W (j)mn . (6)

Our next step is to determine dimFp W(j)mn for all j,m, n.

Lemma 2 With the notation of Remark 1, let k, h be the integers such that 1 ≤ k, h ≤p− 1 and σ(ζp) = ζk

p , τ(ζp) = ζhp . Then{k ≡ reK (mod p)

h ≡ θqK−11 (mod p).

Proof. By equation (4) we have ζp − 1 ≡ θ1πeK (mod πeK+1OF ). Applying σ and τ we

get

ζkp − 1 = σ(ζp − 1) ≡ σ(θ1π

eK ) ≡ θ1ζeKp−1π

eK ≡ reK (ζp − 1) (mod πeK+1OF )

andζhp − 1 = τ(ζp − 1) ≡ τ(θ1π

eK ) ≡ θqK1 πeK (mod πeK+1OF ) .

On the other hand, for each integer l ≥ 1

ζ lp − 1 = (ζp − 1)(ζ l−1

p + ζ l−2p + · · ·+ 1) ≡ l(ζp − 1) (mod πeK+1OF ),

whence k ≡ reK (mod πOF ) and h ≡ θqK−11 (mod πOF ). Now since k and h are integers,

k ≡ rek (mod p) and h ≡ θqK−11 (mod p). 2

Lemma 3 For all ε ∈ F we have εqK − hε ∈ ψ(F ).

Proof. We prove by induction on l that, for 1 ≤ l ≤ fF and for each ε ∈ F , εpl ≡θ

(pl−1)qF /pl−1

1 ε (mod ψ(F )). For l = 1 we have

εp ≡ −θ0ε ≡ θp−11 ε ≡ θ

(p−1)qF

1 ε (mod ψ(F )) .

Assuming the statement true for l, we have

εpl+1 ≡ (εpl

)p ≡ θ(p−1)qF

1 εpl ≡ (θ(p−1)qF /pl

1 ε)pl

≡ θ(pl−1)qF /pl−1

1 θ(p−1)qF /pl

1 ε ≡ θ(pl+1−1)qF /pl

1 ε (mod ψ(F )) .

It follows that εqK ≡ θ(qK−1)pfF−fK+1

1 ε (mod ψ(F )). Now, by lemma 2, θqK−11 ≡ h (mod p),

whence εqK ≡ hpfF−fK+1ε ≡ hε (mod ψ(F )). 2

9

Proposition 7 (i) dimFp W(0)11 = dimFp W

(0) = 1.

(ii) Let j be an integer with 1 ≤ j < peK and (j, p) = 1. Then

dimFp W(j)mn =

{fK if m ≡ rj (mod p)

0 otherwise.

(iii) dimFp W(peK)kh = dimFp W

(peK) = 1.

Proof. (i) We have σ(π) = ζp−1π = ζpp−1π and τ(π) = π.

(ii) By Corollary 1, a set of representatives for W (j) is given by {1 + θπj | θ ∈ T}. Also,by Lemma 1, two elements of Uj represent the same element of W (j) if and only if theyare congruent modulo Uj+1. We have

σ(1 + θπj) = 1 + ζjp−1θπ

j ≡ 1 + rjθπj ≡ (1 + θπj)rj

(mod Uj+1)

andτ(1 + θπj) = 1 + θqKπj ≡ (1 + θπj)n (mod Uj+1) ⇔ θqK = nθ .

Now, any non-zero solution of the equation XqK = nX is also a solution of X(p−1)(qK−1) =1, and since (p− 1)(qK − 1) | qp−1

K − 1 = qF − 1, it belongs to F . It follows that, for eachn with 1 ≤ n ≤ p− 1, the qK distinct solutions of the equation XqK = nX form a vectorspace over Fp of dimension fK .

(iii) By Corollary 1, we need only to show that 1 + βπpeK mod Up1 ∈ W

(peK)kh . We have

σ(1 + βπpeK ) = 1 + ζpeKp−1βπ

peK ≡ 1 + rpeKβπpeK

≡ (1 + βπpeK )rpeK (mod UpeK+1).

By Lemma 2, rpeK ≡ kp ≡ k (mod p), whence

σ(1 + βπpeK ) ≡ (1 + βπpeK )k (mod UpeK+1).

Since UpeK+1 ⊆ Up1 , the last congruence holds a fortiori modulo Up

1 . Moreover, by Propo-sition 6, two elements 1 + β1π

peK , 1 + β2πeK of UpeK

represent the same element of W peK

if and only if β1 − β2 ∈ ψ(F ). Hence

τ(1 + βπpeK ) = 1 + βqKπpeK ≡ (1 + βπpeK )n (mod Up1 )

⇔ βqK − nβ ∈ ψ(F ).

By Lemma 3, βqK − nβ ≡ (h − n)β (mod ψ(F )). Since β is a generator of the groupF /ψ(F ), (h− n)β ∈ ψ(F ) if and only if n = h. 2

10

Corollary 2dimFp(F

∗/F ∗p)mn = nK + δ11mn + δkh

mn,

where δijmn = 1 if (m,n) = (i, j) and δij

mn = 0 otherwise.

Proof. For 1 ≤ µ ≤ eK , in each interval p(µ−1) < j < pµ there is exactly one j for whichm ≡ rj (mod p). Since there are eK such intervals, the contribution to dimFp(F

∗/F ∗p)mm

given by Proposition 7 amounts to fKeK = nK . The contributions δ11mn and δkh

mn are givenby Proposition 7 (i) and (iii), respectively. 2

Corollary 3 dimFp Hst = nK + δkhst + δ11

st .

Proof. Since (k∗, h∗) = (1, 1) and (1∗, 1∗) = (k, h), the corollary follows from Corollary2 and Proposition 3. 2

Remark 2 We note that (k, h) = (1, 1) if and only if K(ζp) = K. In fact, (k, h) = (1, 1)if and only if ζp is left fixed by G, i.e., if and only if ζp ∈ K.

5 Proof of Theorem 1

Since AF/F is abelian, each field L such that F ⊆ L ⊆ AF is normal over F . Regardingthe normality of L over K, we have the following

Lemma 4 A field L such that F ⊆ L ⊆ AF is normal over K if and only if Gal(AF/L)is a G-submodule of H.

Proof. Let HL = Gal(AF/L). By Galois theory, F ⊆ L if and only if HL ⊆ H andL is normal over K if and only if HL is a normal subgroup of G. Since H is abelian,each element of H normalizes HL, hence HL is a normal subgroup of H if and only ifgHLg

−1 ⊆ HL for all g ∈ G, i.e., if and only if HL is a G-submodule of H. 2

For each pair (s, t), the subgroup Hst has an unique G-submodule complement, namely

Hst =⊕

(s′,t′) 6=(s,t)

Hs′t′ .

Let Lst be the fixed field of Hst; since Hst is a G-submodule of H, Lst is normal over K.By Galois theory, Gal(Lst/F ) ∼= Hst and Gal(Lst/K) ∼= Hst oG. Moreover, the canonicalisomorphisms between these groups are also G-module isomorphisms.Let Gst be the subgroup of G of elements which act trivially on Gal(Lst/F ), and let Fst

be the subfield of F fixed by Gst. Note that

Gst = {σxτ y | sxty = 1} .

11

Proposition 8 Fst is a cyclic extension of K of degree dividing p−1. Conversely, if F ′ isa cyclic extension of K of degree dividing p−1, then F ′ = Fst for some s, t. In particular,F =

∏s,t∈F∗p

Fst.

Proof. The map G → F∗p sending σxτ y to sxty is a homomorphism with kernel Gst andwhose image is isomorphic to G/Gst

∼= Gal(Fst/K).Conversely, let F ′/K be cyclic of degree d | p− 1. Projecting Gal(F/K) onto Gal(F ′/K)and embedding the image in F∗p we get a homomorphism ϕ : G → (Z/pZ)∗ with kernelGal(F/F ′). If ϕ(σ) = s and ϕ(τ) = t, then Gal(F/F ′) = Gst and F ′ = Fst. 2

Now consider extensions L/F such that L ⊆ AF and [L : F ] = p. By Lemma 4, L isnormal over K if and only if L is the fixed field of a G-submodule of H, i.e., if and onlyif L ⊆ Lst for some s, t. Hence, letting

Lst = {L ⊆ Lst | [L : F ] = p} , L =⋃

s,t∈F∗p

Lst , (7)

we have that L is precisely the set of abelian extensions of F of degree p which are normalover K. Moreover, if L ∈ Lst and Gal(L/F ) = 〈λL〉, then Gal(L/K) ∼= 〈λL〉 o G, wherethe action of G is induced by the action on Hst, namely, σλLσ

−1 = λsL, τλLτ

−1 = λtL.

Proposition 9 Each element L ∈ L contains exactly one isomorphism class over K ofelements of E .

Proof. By the Schur-Zassenhaus Theorem [10, 9.12], the subgroups of index p of Gal(L/K)are conjugate to each other, hence the corresponding fixed fields, i.e., the subfields of L/Kof degree p over K, are conjugate over K. Moreover, since L/K is a normal extension, itcontains the whole isomorphism class over K of any of its subextensions. 2

Since Gal(L/K) ∼= 〈λL〉oG, we can identify G and Gst with subgroups of Gal(L/K). Wedenote by LG and LGst the subfields of L fixed by G and Gst, respectively. A priori, thedefinition of these fields depends on the isomorphism between Gal(L/K) and 〈λL〉 o G;however, the last proposition ensures that LG is well defined up to isomorphism over K,and the next proposition shows that LGst is well defined.

Proposition 10 LGst is the normal closure of LG over K. Moreover, L = LGF = LGstFand LGst = LGFst.

Proof. Let E ′ be the normal closure of LG over K. Since LGst/K is Galois, then LG ⊆E ′ ⊆ LGst and G′ = Gal(L/E ′) ⊇ Gst. On the other hand, G′ is the largest subgroup ofG which is normal in Gal(L/K), hence the subgroup 〈λ〉oG′ is in fact a direct product;it follows that G′ commutes with λ, i.e., G′ ⊆ Gst.

12

As to the last statements, observe that ([L : LG], [L : F ]) = (|G|, p) = 1, henceGal(L/LG) ∩ Gal(L/F ) = {id}. It follows that L = LGF and, a fortiori, L = LGstF .Similarly, Gal(LGst/LG) ∩Gal(LGst/Fst) = {id} and LGst = LGFst.

2

Proposition 11 If E ∈ E , then EF ∈ L.

Proof. Let E ∈ E and let E ′ be its normal closure over K. The Galois group Gal(E ′/K)is a transitive solvable subgroup of the permutation group on p elements, hence it iscontained in the normalizer Np of a p-cycle (see [1, Ch.3, Thm. 7]). The group Np isisomorphic to a semidirect product Z/pZoZ/(p−1)Z, hence Gal(E ′/K) ∼= Z/pZoZ/dZfor some d | p − 1. It follows that E ′ = EF ′ is a normal extension of degree p of a fieldF ′, where F ′ is a cyclic extension of degree d of K. In particular, EF = E ′F is a normalextension of F of degree p and it is normal over K, whence it belongs to L. 2

We can now conclude the proof of Theorem 1. The inclusion CK(p) ⊆ AF follows fromProposition 11, so it remains to prove that AF ⊆ CK(p). By Proposition 5, for all (s, t)we have dimFp Hst > 0, so there exists L ∈ Lst. As remarked before, LG ∈ E , hence,by Proposition 10, L = LGF and the normal closure LGst of LG over K is LGFst. SinceLGst ⊂ CK(p), we have that Fst is contained in CK(p) for all s, t, whence, by Proposition8, F ⊂ CK(p). This implies that CK(p) contains LGF = L for each L ∈ L.Now, from the definitions of Lst and L it is clear that AF =

∏s,t∈F∗p

Lst =∏

L∈LL ⊆ CK(p).

2

Remark 3 We already observed (see Remark 1) that K(ζp) is a cyclic extension of Kof degree dividing p − 1, hence, by Proposition 8, K(ζp) = Fst for some s, t. It is easyto check that actually K(ζp) = Fkh. In fact, Fkh is the fixed field of Gkh, and kxhy ≡ 1(mod p) if and only if σxτ y(ζp) = ζkxhy

p = ζp.

Remark 4 Proposition 8 says that the map which assigns to each element (s, t) ∈ (F∗p)2

the field Fst is surjective on the set of cyclic extensions of K of degree dividing p − 1.However, this map is not a bijection. In fact, it is an easy exercise to show that Fs′t′ ⊆ Fst

if and only if there exists an integer a such that (s′, t′) = (sa, ta), and that Fs′t′ = Fst ifand only if a can be chosen coprime to p− 1.Also, by the same arguments as in the conclusion of the preceding proof, one can showthat

p−1∏a=1

∏L∈Lsata

LG = AFst .

13

6 Applications

6.1 Isomorphism classes of extensions of degree p

Looking at the structure of CK(p) one can reconstruct the formula for the number ofisomorphism classes IK(p) of the extensions of degree p of K. This formula is not new(see [3]), but it is a very simple consequence of our results.

Proposition 12

IK(p) =

{(p− 1)(pnK − 1) + 2pnK if ζp 6∈ K;

(p− 1)(pnK − 1) + pnK+1 + pnK if ζp ∈ K.

Proof. By Proposition 9 there is a one to one correspondence between L and the iso-morphism classes of the elements of E , hence IK(p) = |L| =

∑s,t∈F∗p |Lst|. One obtains

the formula in the proposition by using Corollary 3 and by observing that a vector spaceof dimension n over Fp has exactly (pn − 1)/(p− 1) subspaces of codimension 1. 2

6.2 Extensions of degree p with given discriminant

The number of extensions of degree p with given discriminant of a local field K is well-known (see for instance [7] and [11]), but again it can be derived directly from the resultsof Section 5. In fact, since F contains the p-th roots of unity, each abelian extension Lof F of degree p is of type F ( p

√a) for some element a ∈ F and therefore determining

Disc(L/F ) is an easy task. Using the relations between the discriminants in towers ofextensions, we immediately obtain the corresponding discriminants of the fields E ∈ E .

In the following we use the notation of Proposition 6.

Lemma 5 The unramified extension of F of degree p is L0 = F ( p√

1 + βπpeK ).

Proof. Let b be a p-th root of 1 + βπpeK . Then (b− 1)/πeK is a root of the polynomial

f(X) = Xp +

p−2∑i=1

(p

i

)1

πieKXp−i +

p

πeFX − β.

The reduction of f(X) modulo π is

f(X) = Xp + θ0X − β.

Since θ0 6= 0, the polynomial f(X) has distinct roots, whence L0 = F ((b − 1)/πeK ) isunramified over F . 2

14

Lemma 6 Let L/F be an abelian extension of F of degree p of type L = F ( p√ul), where

ul ∈ Ul \ Ul+1, 1 ≤ l < peK , and (l, p) = 1. Then Disc(L/F ) = (π(p−1)(peK−l+1)).

Proof. Let Gal(L/F ) = 〈λL〉 and let γ be a p-th root of ul; clearly γ ∈ Ul(L). Now letl′ be an integer such that ll′ ≡ 1 (mod p). Then πL = (γ − 1)l′/π(ll′−1)/p is a uniformizerof L, hence

DL/F =

p−1∏i=1

(λiL(πL)− πL) =

1

π(p−1)(ll′−1)/p

p−1∏i=1

((ζ ipγ − 1)l′ − (γ − 1)l′)

=1

π(p−1)(ll′−1)/p

p−1∏i=1

(ζ ip − 1)γ

l′−1∑j=0

(ζ ipγ − 1)j(γ − 1)l′−1−j.

Taking valuations, we have vL(π) = p, vL((ζ ip−1)γ) = vL(ζ i

p−1) = eL/(p−1) for all i and

vL((ζ ipγ−1)j(γ−1)l′−1−j) = l(l′−1) for all j. Regarding the last sum, let θ ∈ T (L)−{0}

be such that γ ≡ 1 + θπlL (mod πl+1

L OL); we get

l′−1∑j=0

(ζ ipγ − 1)j(γ − 1)l′−1−j ≡ (

l′−1∑j=0

ζ ijp )θl′−1π

l(l′−1)L (mod π

l(l′−1)+1L OL)

and, since∑l′−1

j=0 ζijp = (ζ il′

p − 1)/(ζ ip − 1) is a unit in OL, the valuation of this sum is

exactly l(l′ − 1). Summing up, we obtain

vL(DL/F ) = −(p− 1)(ll′ − 1) + eL + (p− 1)l(l′ − 1) = eL − (p− 1)(l − 1)

and, taking norms,

vF (Disc(L/F )) = eL − (p− 1)(l − 1) = (p− 1)(peK − l + 1) .

2

Lemma 7 Let L = F ( p√πu), where u ∈ U1. Then Disc(L/F ) = (π)(p−1)(peK+1).

Proof. Let πL be a p-th root of πu. Then πL is a uniformizer of L and hence Disc(L/F )is equal to the ideal generated by the discriminant of the polynomial Xp − πu. 2

Proposition 13 LetNF,L(D) be the number of fields L ∈ L such that Disc(L/F ) = (π)D.Then

(i) NF,L(0) = 1;

(ii) if 1 ≤ l < peK and (l, p) = 1, then

NF,L((p− 1)(peK − l + 1)) =

qeK−b l

pcK − q

eK−b lpc−1

K if l 6≡ eK (mod p− 1) ;

2(qeK−b l

pcK − q

eK−b lpc−1

K ) if l ≡ eK (mod p− 1) .

15

(iii) NF,L((p− 1)(peK + 1)) =

{qeKK if ζp 6∈ K ;

pqeKK if ζp ∈ K .

Also, NF,L(D) = 0 for all D not occurring in the preceding list.

Proof. By Kummer theory, the abelian extensions of F of degree p are in one to onecorrespondence with the cyclic subgroups of F ∗/F ∗p of order p, hence they are in p − 1to one correspondence with the set

X = {πiu | 0 ≤ i ≤ p− 1, u ∈ X ′, πiu 6∈ F ∗p} , (8)

where X ′ ⊂ U1 is a set of representatives of U1/Up1 . Denote by [πiu] the class of πiu

modulo F ∗p. By Proposition 5, an abelian extension F (p√πiu)/F belongs to L if and only

if [πiu] ∈ ∪m,n∈F∗p(F∗/F ∗p)mn, so we are left to count the number of elements πiu ∈ X

such that [πiu] ∈ ∪m,n∈F∗p(F∗/F ∗p)mn and F (

p√πiu)/F has a prescribed discriminant. For

each m,n ∈ F∗p, let Xmn be the subset of X of non-trivial representatives of (F ∗/F ∗p)mn =peK⊕j=0

W(j)mn .

Consider first the case i = 0, namely, L = F ( p√u) for u ∈ X ′∩ (∪m,n∈F∗pXmn). In this case

Disc(L/F ) is given by Lemmas 5 and 6, and depends on l = max{j |u ∈ Uj} only.

If l = peK , then, by Propositions 6 and 7, 〈[u]〉 = 〈[1 + βπpeK ]〉 = W (peK) = W(peK)kh ,

so F ( p√u) belongs to L and in fact, by Lemma 5, it is the unramified extension of F of

degree p. This establishes the first case of the proposition.

Now let 1 ≤ l < peK ; for each m,n ∈ F∗p, we must determine the cardinality of the set(Ul \ Ul+1) ∩Xmn. By Proposition 6 we have

#(Ul \ Ul+1) ∩Xmn = #

peK⊕j=l

W (j)mn −#

peK⊕j=l+1

W (j)mn . (9)

This cardinality is clearly 0 whenever W(l)mn = 0, hence, by Proposition 7, we need only

consider the case when m = rl. In this case, for j < peK , the space W(j)

rln= 0 unless

rj = rl, and W(peK)

rln= 0 unless (rl, n) = (k, h) (hence l ≡ eK (mod p− 1)). Summarizing,

for 1 ≤ l < peK and (l, p) = 1 we have

#(Ul \ Ul+1) ∩Xrln =

qeK−b l

pcK − q

eK−b lpc−1

K if (rl, n) 6= (k, h)

p(qeK−b l

pcK − q

eK−b lpc−1

K ) if (rl, n) = (k, h) .(10)

Applying the formula

NF,L((p− 1)(peK − l + 1)) =1

p− 1

∑n∈F∗p

# ((Ul \ Ul+1) ∩Xrln)

16

we get the second case of the proposition.

Finally, consider the case i 6= 0, namely, L = F (πiu) for some i ∈ {1, . . . , p − 1} andsome u ∈ X ′. The subgroup of F ∗/F ∗p generated by [πiu] contains exactly one elementof type [πu′], so we may just count the elements πu where u ∈ X ′ and L = F (πu) ∈ L.By Proposition 7, L ∈ L if and only if [πu] ∈ (F ∗/F ∗p)11. Again we have to distinguishtwo cases. If ζp 6∈ K, i.e., if (1, 1) 6= (k, h), then the dimension of the Fp-vector space(F ∗/F ∗p)11 is nK + 1, hence the number of fields of this type is qeK

K . If ζp ∈ K, i.e., if(1, 1) = (k, h), then the preceding dimensions must be increased by 1, so the number offields of this type is pqeK

K . By Lemma 7, this establishes case (iii) of the proposition andconcludes the proof.

2

Lemma 8 Let L ∈ L be such that Disc(L/F ) = (π)DL , and let E ∈ E , E ⊂ L. Then

Disc(E/K) =

{(πK)DL/(p−1)+p−2 if DL > 0;

1 if DL = 0.

Proof. By the well-known formula on discriminants in tower of extensions, we have

Disc(L/K) = (Disc(E/K))[L:E]NE/K(Disc(L/E)) = (Disc(F/K))[L:F ]NF/K(Disc(L/F )) .

The lemma follows by observing that, since L/E and F/K are tamely ramified with bothinertial degree and ramification index equal to p− 1, we have Disc(L/E) = (πE)(p−1)(p−2)

and Disc(F/K) = (πK)(p−1)(p−2). 2

Proposition 14 LetNK(δ) be the number of fields E ∈ E such that Disc(E/K) = (πK)δ.Then

(i) NK(0) = 1;

(ii) if 1 ≤ l < peK and (l, p) = 1, then

NK(p(eK + 1)− l − 1) = p(qeK−b l

pc

K − qeK−b l

p]c−1

K ) ;

(iii) NK(p(eK + 1)− 1) = pqeK .

Also, NK(δ) = 0 for all δ not occurring in the preceding list.

Proof. By Proposition 9, each L ∈ L contains exactly one isomorphism class over K ofelements of E . This isomorphism class has one element if E = LG is normal over K and pelements otherwise. By Proposition 10, LG is normal over K if and only if L ⊆ L11, and,by Proposition 5, this is true if and only if L is of type F (

p√πiu) where [πiu] ∈ (F ∗/F ∗p)kh.

17

So the proof of the proposition will follow directly by the examination of cases made inthe proof of Proposition 13 and by Lemma 8.

Case (i) is clear, since it corresponds to the unique unramified extension of K of degreep.

If 1 ≤ l < peK and (l, p) = 1, then for each n ∈ F∗p there are exactly pp−1

(qeK−[ l

p]

K −qeK−[ l

p]−1

K )

fields E ∈ E contained in Lrln, independently of whether l ≡ eK (mod p − 1) or not.Summing up over n ∈ F∗p, we obtain (ii).

Similarly, the number of fields E ∈ E contained in a field L ∈ L of type L = F ( p√πu)

with u ∈ X ′ is pqeK , independently of whether (1, 1) = (k, h) or not, thus proving (iii).2

6.3 The ramification groups and the discriminant of CK/K

As usual, for i ≥ −1, we shall denote by Gi the i-th ramification group of G = Gal(CK/K),namely

Gi = {g ∈ G | vCK(g(x)− x) ≥ i+ 1 ∀x ∈ OCK

} .

By standard theory, G−1 = G, G0 is the inertia subgroup of G, and G1 is the p-Sylowsubgroup of G0. Let H0 be the subgroup of H fixing the unramified extension L0 of F ofdegree p. Then clearly H0

∼= (Z/pZ)nF +1 and G0 is the subgroup of G generated by H0

and σ, hence G0∼= H0 o Z/(p − 1)Z. Moreover, Gi

∼= Hi for all i ≥ 1, so we are left todetermine the higher ramification groups of CK/F .

For all real numbers v ≥ −1, it is easy to describe the lower numbering and the uppernumbering ramification groups with index v of all abelian extensions L/F of degree p. Forevery such L, let HL = Gal(CK/L), so that Gal(L/K) = H/HL. If Disc(L/F ) = (π)DL ,then

(H/HL)v = (H/HL)v =

{Z/pZ for v ≤ DL

p−1− 1

0 for v > DL

p−1− 1 .

(11)

By classical theory, this allows us to reconstruct the ramification groups of H. In fact,for all v, we have the relation (H/HL)v = HvHL/HL (see [12, Ch. IV, Prop. 14]), hence

Hv ⊆ HL ⇐⇒ v >DL

p− 1− 1 . (12)

Since HL runs over all subgroups of index p of H as L runs over all abelian extension ofdegree p of F , it follows that

Hv =⋂

DLp−1

<v+1

HL . (13)

18

Lemma 9

dimFp Hv =

nF + 2 for v = −1;

nF + 1 for − 1 < v ≤ 1;

nF + 1−(dve −

⌈vp

⌉)fF for 1 < v ≤ peK ;

0 for v > peK .

Proof. By (13), for all v we have to determine the dimension of⋂

DLp−1

<v+1HL. Via

Galois theory, this group corresponds to CHv

K =∏(v) L, where the product is taken over

all abelian extensions L/F such that DL < (p− 1)(v+1). The case v = −1 is trivial and,by Lemmas 5, 6 and 7, we have:

(i) if −1 < v ≤ 1, then CHv

K = L0, where L0 is the unramified extension of F of degreep;

(ii) if v > 1 and L = F (p√πiu) then DL < (p − 1)(v + 1) if and only if [πiu] ∈⊕

l>peK−v

W (l). In other words, CHv

K corresponds, via Kummer theory, to the subgroup

UpeK+1−dve]Up1 /U

p1 =

⊕l>peK−v

W (l) of F ∗/F ∗p.

Using Corollary 1 and the relation #Hv · [CHv

K : F ] = [CK : H] = pnF +2 we obtain thelemma.

2

In particular, there are (p − 1)eK + 1 non-negative integers v such that Hv 6= Hv+1,namely, the integers in the set {1 ≤ v < peK | v 6≡ 0 (mod p)} ∪ {peK}. Now recall (see[12, Ch. IV.3]) that the relation between the lower numbering and the upper numberingramification groups is given by Hu = Hϕ(u), where ϕ : [1,∞] → [1,∞] is the bijectivefunction defined by

ϕ(u) =1

h0

(h1 + · · ·+ hbuc + (u− buc)hbuc+1) and hi = #Hi.

It follows that there are exactly (p−1)eK +1 non-negative integers i such that Hi 6= Hi+1.Denote these integers by s0, s1, . . . s(p−1)eK

, with s0 < s1 < · · · < s(p−1)eK. By simple

calculations, we get

Proposition 15 For i = 0, 1, . . . , (p− 1)eK − 1 we have

si =i∑

ν=0

pνfF +i∑

ν=1ν≡0 (mod p−1)

pνfF ,

19

s(p−1)eK= s(p−1)eK−1 + pnF

and dimFp Hsi= nF + 1− ifF , for all i. Moreover, for si−1 < u ≤ si,

Hu = Hsi=

{H i+d i+1

p−1e if i < (p− 1)eK ;

HpeK if i = (p− 1)eK .

and Gu = Hu for u > 0.

We conclude by determining the discriminant of CK/K.

Proposition 16

vK(Disc(CK/K)) = p(p− 1)

[(eK + 1)pnF +2 − pnF +1 − pnF − pnF − 1

pfF − 1− pnF − 1

p(p−1)fF − 1

].

Proof. By substituting the cardinalities of the ramification groups given by Proposition15 into the well-known formula

vCK(DCK/K) =

∞∑i=0

(|Gi| − 1),

we get

vCK(DCK/K) = [(p− 1)pnF +1 − 1] +

∑0≤i<(p−1)eK

pifF (pnF +1−ifF − 1)

+∑

0<i<eK

pi(p−1)fF (pnF +1−i(p−1)fF − 1) + pnF (p− 1).

Rearranging and observing that NCK/K(πCK) = (πK)p(p−1) we get the formula in the

statement. 2

References

[1] E, Artin, Galois Theory, Notre Dame Mathematical Lectures, No.2, University ofNotre Dame Press, London, 1942.

[2] I.B. Fesenko and S.V. Vostokov, Local fields and their extensions, Second Edi-tion, Translation of Mathematical Monographs, Vol. 121, American MathematicalSociety, 2002.

20

[3] X. Hou and K. Keating, Enumeration of isomorphism classes of extensions ofp-adic fields, J. Number Theory 104 (2004), no. 1, 14-61.

[4] M. Krasner, Nombre des extensions d’un degre donne d’un corps p-adique: enoncedes resultats et preliminaires de la demonstration (espace des polynomes, tranforma-tions T ), C.R. Acad. Sci. Paris 254 (1962), 3470-3472.

[5] M. Krasner, Nombre des extensions d’un degre donne d’un corps p-adique: suitede la demonstration, C.R. Acad. Sci. Paris 255 (1962), 224-226.

[6] M. Krasner, Nombre des extensions de degre donne d’un corps p-adique: les condi-

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[7] M. Krasner, Nombre des extensions de degre donne d’un corps p-adique: calcul deNn

k,j,s; demonstration du theoreme 1, C.R. Acad. Sci. Paris 255 (1962), 2342-2344.

[8] M. Krasner, Nombre des extensions de degre donne d’un corps p-adique:complements au theoreme 1 dans le cas non p-adique; demonstration du theoreme 2,C.R. Acad. Sci. Paris 255 (1962), 3095-3097.

[9] W. Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers, 2nd Edi-tion, Springer Verlag, 1990.

[10] D.J.S. Robinson, A course in the theory of groups, 2nd Ed., Springer Verlag GTM80, 1995.

[11] J.-P. Serre, Une “formule de masse” pour les extensions totalement ramifiees dedegre donne d’un corps local, C.R. Acad. Sci. Paris Ser A-B 286 (1978), no. 22,A1031–A1036.

[12] J.P. Serre, Local Fields, Springer-Verlag, New-York (1979).

Authors’ address:Dipartimento di Matematicavia Buonarroti, 256127 PISA (Italy)

E-mail: delcorso@dm.unipi.itdvornic@dm.unipi.it

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