Construction of basis of the group of cyclotomic units of some real abelian extension

Construction of Basis of the group of cyclotomic

units

of some real abelian extensions

Thèse

Azar Salami

Doctorat en mathématiques

Philosophiæ doctor (Ph.D.)

Résumé

Dans cette thèse, nous construisons une base du groupe Ckdes unités cycolotomiques (au

sens de Sinnott) d’une certaine extension abélienne finie k de Q ramifiée exactement sur trois nombres premiers distincts. La première étape consiste en la construction d’une base du goupe Dk des nombres circulaires de k. Par la suite, il sera plus simple d’obtenir une base de Ck.

Abstract

In this thesis, we construct an explicit basis of the group Ckof cyclotomic units of certain

finite abelian extension k of Q ramified at exactly three distinct primes. The first step consists in constructing a basis of the group Dk of circular numbers of k. From there, it is not too

difficult to obtain a basis of Ck. The method is combinatorial in nature. We may visualize

our construction using a three dimensional cuboid formed of | Gal(k/Q)| small cubes, each of these small cubes containing a Galois conjugate of a primitive circular unit of k. The classical norm relations give rise to some identifications on the cuboid. Using these identifications and an Ennola-type relation (a highly non-trivial relation), we manage to construct an explicit basis of Ck.

Contents

Résumé iii Abstract v Contents vii List of Figures ix Acknowledgements xiii Introduction xv Preliminaries xvii

Galois Theory . . . xvii

Dirichlet Characters . . . xviii

Main Theorem . . . xx

1 Basic Definitions and Results 1 1.1 Notations and Consequences . . . 1

1.2 Definition of m . . . 3

1.3 Key Equation . . . 4

2 The Cuboid’s Construction 5 2.1 Notations . . . 5

2.2 Z-basis of Di . . . 6

2.3 Idea for a Z-basis of D . . . 8

2.4 The Cuboid Construction . . . 9

3 The Case of m = 1 19 3.1 a1= a2 = a3= 1. . . 19

3.2 General Case of m = 1. . . 25

4 Ennola Relation 35 4.1 Notations and Restrictions . . . 35

4.2 Computations Toward (4.2) . . . 40

5 General Case 45 5.1 Missing Conjugates and Non-top Layers . . . 47

6 Construction of Abelian Fields 55 6.1 Construction of a Field F . . . 55 6.2 Second Way . . . 58 Conclusion 59 Bibliography 61 viii

List of Figures

2.1 m = 1, a1 = a2= a3 = 1, and r1 = 7, r2 = 5, r3= 4. . . 10 2.2 m = 1, r1= 5, r2 = 4, r3 = 3,and a1= 2, a2= 3, a3= 3. . . 11 2.3 m = 2, a1 = 2, a2 = 3, a3 = 3, and r1 = 5, r2= 4, r3= 3. . . 12 2.4 m = 1, ai = 1, r1 = 7, r2= 5, r3= 4. . . 13 2.5 m = 1, ai = 1, r1 = 7, r2= 5, r3= 4. . . 14 2.6 m = 1, a1 = 2, a2 = 3, a3 = 3, r1 = 5, r2= 4, r3= 3. . . 15 2.7 Staircase-diagonal (2, 0) on a layer . . . 15 2.8 Staircase-diagonals on a layer . . . 16 2.9 Diagonal relation (0, 0). . . 17

2.10 Diagonal relation (0, 0, 0) on a layer. . . 17

3.1 m = 1, ai = 1, r1 = 7, r2= 5, r3= 4. . . 20

3.2 m = 1, ai = 1, r1 = 7, r2= 5, r3= 4. . . 20

3.3 m = 1, ai = 1, r1 = 7, r2= 5, r3= 4. . . 20

3.4 m = 1, ai = 1, r1 = 7, r2= 5, r3= 4. . . 21

3.5 Missing Conjugates. . . 21

3.6 Apply Row Relations. . . 22

3.7 (r2+ r3− 1)Missing Conjugates . . . 22

3.8 (1, r2) = (1, 5); (1, r2− 1) = (1, 4). . . 22

3.9 Two Sets of Missing Conjugates. . . 23

3.10 (1, r2) = (1, 5), (1, r2mod r3) = (1, 1). . . 23

3.11 (1, r2 mod r3) = (1, 1), (1, r2+ (r2mod r3)) = (1, 6). . . 24

3.12 Apply Diagonal relation (1, 6). . . 24

3.13 Apply Column relation (1, 2). . . 24

3.14 Apply Diagonal relation (1, 7). . . 25

3.15 Apply Column relation (1, 3). . . 25

3.16 m = 1, a1 = 2, a2 = 3, a3 = 3, r1 = 5, r2= 4, r3= 3. . . 26

3.17 m = 1, a1 = 2, a2 = 3, a3 = 3, r1 = 5, r2= 4, r3= 3. . . 27

3.18 m = 1, a1 = 2, a2 = 3, a3 = 3, r1 = 5, r2= 4, r3= 3. . . 27

3.19 Missing Conjugates. . . 28

3.20 Missing conjugates on a non-face-layer. . . 29

3.21 Missing conjugates on the Face-layer . . . 29

3.22 (r1a2, 1, 1) = (15, 1, 1). . . 30

3.23 Apply Diagonal relations at all conjugates in rows 2, ..., a2. . . 30

3.24 (r2+ r3− 1)a1 Missing Conjugates. . . 31

3.25 Apply Column Relations. . . 31

3.27 Apply Diagonal Relations. . . 32

3.28 Apply Column Relations. . . 33

3.29 Apply Diagonal Relations. . . 33

3.30 Apply Column Relations. . . 34

4.1 Ennola relation on a layer. . . 43

5.1 Missing Conjugates. . . 48 5.2 Non-top layer. . . 48 5.3 Column Relations. . . 52 5.4 Diagonal Relations. . . 53 5.5 Column Relations. . . 53 x

To Héloïse, my Northern Star through distances and sleepless nights.

Acknowledgements

It is a real pleasure to thank all those who contributed in the realization of this work. First I would like to thank my supervisor, Dr. Hugo Chapdelaine. I am extremely grateful for the opportunity to have worked under his supervision. I cannot thank him enough for supporting and empowering me, while respecting my autonomy. Hugo’s contagious passion instilled in me an even greater love for mathematics.

I would also like to thank Dr. Radan Kucera. Radan’s patience and generosity taught me the essence of being a professor, while his creativity and Socratic spirit pushed me to consistently improve my work and think outside the box.

I would like to thank Dr. Cornelius Greither for his precise yet exhaustive evaluation; it was an honour receiving his insightful comments. I would also like to thank Dr. Claude Levesque for his excellent remarks and for being a role model throughout my years at the department.

Finally, I would like to extend my thanks to the Department of Mathematics and Statistics at Laval University, with special thanks to Sylvie Drolet for her ongoing support.

Introduction

This thesis deals with the group of circular units of an abelian field k introduced by Sinnott in [13]. Even though Sinnott was able to find a formula for its index in the full group of units of k, he has not constructed a basis of this group. In the literature, such a basis can be found only for a few very specific types of abelian fields as follows:

- k is the compositum of real abelian fields of prime power conductor (this seems to be well-known by experts for a long time without publishing; now it can be found as an auxiliary result in [15]),

- k is a cyclotomic field (see [6] and [9]),

- k is the compositum of quadratic fields (but in this case a basis is found for a slightly enlarged group of circular units, see [10]),

- k is the compositum of imaginary abelian fields of prime power conductor (see [11]), - there are exactly two primes ramifying in k/Q (see [2] and [3]),

- [k : Q] = p2_{, where p is an odd prime, and Gal(k/Q) is not cyclic (but here a basis is not}

given explicitly, but only implicitely in two steps: first of the subgroup generated by circular units of all cyclic subfields of k and then of the quotient of the two groups, which turns out to be a vector space over Z/pZ, see [8]).

The aim of this thesis is to construct a basis of Sinnott’s group of circular units for an abelian field k of another type, namely assuming that

1. k is real,

2. there are exactly three primes ramifying in k/Q,

3. the inertia groups of ramified primes in Gal(k/Q) are all cyclic,

4. the Galois group Gal(K/k), where K is the genus field in narrow sense of k, is cyclic as well.

The construction of a basis we are going to explain is a generalization of the approach of R. Kučera made in an unpublished private communication [12], where he gave a basis in a special case (assuming besides our assumptions that we have also r1= r2 = r3 = 1, using the notation

introduced in Section 2.1).

It appears that in our construction a crucial role is played by an Ennola relation among the generators of the group of circular units. The first appearance of a relation of this kind can be found in paper [5] where V. Ennola explains the necessity to work with these relations in the paper [1] of H. Bass. To explain what we have in mind by an Ennola relation we need to be more specific.

The group of circular units of a real abelian field k defined by Sinnott is a Z[G]-module, where G = Gal(k/Q), having plenty of generators, in fact one generator for each non-empty subset of the set of prime numbers ramified in k/Q. When we compute the norm of such a generator going from the subfield of k where the generator is defined to a subfield having less ramified primes, we obtain a relation between the generators which we call a norm relation. By a consequence of these relations we understand a relation which can be obtained by mutliplying these norm relations, more precisely, by any of their multiplicative Z[G]-linear combination. Now an Ennola relation is a valid relation among the generators which is not a consequence of the norm relation. It appears that there are positive integers t ∈ Z such that the tth_power

of this relation is a multiplicative Z[G]-linear combination of norm relations. In Chapter 4 we derive an Ennola relation such that the mth _{(the positive integer m is introduced in Section}

1.2) power of this relation is a multiplicative Z[G]-linear combination of norm relations, and in Chapter 5 we use this relation to construct a basis of the group of circular units if m > 1 (such a basis for m = 1 is obtained already in Chapter 3).

Preliminaries

Galois Theory

In this thesis we assume some knowledge in Galois theory, mainly, chapter 14 in [4]. In this section, we will only state two results from Galois theory that we are going to use over and over.

Proposition 1. Suppose K/F is a Galois extension and L, L0 are two subfields of K containing F . Let H = Gal(K/L) and H0 = Gal(K/L0). Then we have

(i) Gal(K/LL0) = H ∩ H0.

(ii) The subfield invariant by hH, H0i (the smallest subgroup containing H and H0_{), is L ∩ L}0_.

K LL0 L L0 L ∩ L0 F {1} H ∩ H0 H H0 hH ∪ H0i Gal(K/F )

Theorem 1. Suppose K/F is a Galois extension and F0/F is any extension. Then KF0/F0 is a Galois extension, with Galois group

Gal(KF0/F0) ∼= Gal(K/K ∩ F0) isomorphic to a subgroup of Gal(K/F ).

KF0 = K = F0 K ∩ F0 F

Dirichlet Characters

We shall give a brief survey on Dirichlet characters; a reader can find more details in chapter 3of [14].

Definitions

Definition 1. Let n be a positive integer. A Dirichlet character modulo n is a character of the abelian group (Z/nZ)∗. That is, a multiplicative homomorphism of the form

χ : (Z/nZ)∗→ C∗. We call n the modulus of χ.

Suppose χ is a Dirichlet character of modulus n and n|m. Then, by the natural homomorphism φ : (Z/mZ)∗ → (Z/nZ)∗,

define χ0_{= χ ◦ φ. Here χ}0 _{is a Dirichlet character of modulus m. We say χ}0 _{is induced by χ .}

Definition 2. Let fχ be the minimal modulus for the character χ. That is, χ is not induced

by any Dirichlet character of modulus less than f_χ. Then we call f_χ the conductor of χ. A Dirichlet character defined modulo its conductor is called primitive.

Dirichlet Characters and Q(ζn)

The set of all primitive Dirichlet characters form an infinite group [14]. For a fixed integer n the subset of Dirichlet characters whose conductors divide n form a subgroup. In fact using the canonical isomorphism

Gal(Q(ζn)/Q) ∼= (Z/nZ)∗,

the group of Dirichlet characters mod n can be seen as the group of characters of that Galois group.

Definition 3. Let

χ : Gal(Q(ζn)/Q) → C∗,

and let K be the fixed field of the kernel of χ. Then K is the field associated to χ. More generally, let

X be a finite group of Dirichlet characters,

let n = lcm(fχi) where χi runs over X, so X is a subgroup of the group of characters of

Gal(Q(ζn)/Q). Let H =Tχ∈Xker χ, and L be the fixed field of H. Then X is precisely the

set of homomorphisms Gal(L/Q) → C∗; we call L the field associated to X.

The mapping sending a finite group of Dirichlet characters to the corresponding field is an isomorphism between the lattice of all finite subgroups of the group of all Dirichlet characters (with respect to inclusion) to the lattice of all abelian fields (again with respect to inclusion). This implies that if Xi corresponds to Ki for i = 1, 2 then we have

1. X1⊆ X2 ⇔ K1⊆ K2.

2. The group generated by X1 and X2 corresponds to the compositum K1K2.

3. The intersection X1∩ X2 corresponds to K1∩ K2.

4. |X1| = [K1 : Q].

Dirichlet and Ramification Theory

We now show how ramification indices may be computed in terms of characters. Let n =Y

p

pa. Corresponding to the decomposition

(Z/nZ)∗'Y

p

(Z/paZ)∗,

we may write any character χ defined mod n as χ =Y

p

χp

where χp is a character defined mod pa. If X is a group of Dirichlet characters, then we let

Xp= {χp|χ ∈ X}.

Theorem 2. Let X be a group of Dirichlet characters and K its associated field. Let p be a prime number with ramification index e in K. Then e = #(Xp).

Corollary 1. Let χ be a Dirichlet character and K the associated field. Then p ramifies in K ⇔ χ(p) = 0(equivalently p|fχ).

More generally, let L be the field associated with a group X of Dirichlet characters. Then p is unramified in L/Q ⇔ χ(p) 6= 0 for all χ ∈ X.

Theorem 3. Let X be a group of Dirichlet characters, K its associated field. Let Y = {χ ∈ X|χ(p) 6= 0}, Z = {χ ∈ X|χ(p) = 1}.

Then

e = [X : Y ], f = [Y : Z], and g = [Z : 1]

where e, f, gare respectively the ramification index for p in K, the residue class degree, and the number of primes lying above p. In fact

X/Y ' the inertia group, X/Z ' the decomposition group, Y /Zis cyclic of orderf.

Main Theorem

Definition 4 (Genus Field). Let k be an abelian extension of Q. The genus field in the narrow sense of k is the extension of k that is maximal unramified in finite places (it can be ramified at infinite places) and which is an abelian extension of Q.

Theorem 4 (Main Theorem). Let K be the genus field of an abelian extension k in narrow sense. Then

Gal(K/Q) is the direct product of the subgroups Ip where Ip is the inertia group of p, the

product being taken over all ramified prime numbers.

Proof. Let X be the group associated to k, and let L be any abelian field containing k which is unramified over k in all finite places of k. Let Y be the group of Dirichlet characters corresponding to L. Then

X ⊆ Y, Xp⊆ Yp.

But |Xp| = |Yp|which gives Xp = Yp, so Y ⊆ ⊕pXp. Therefore the genus field K of k in the

narrow sense is the field corresponding to ⊕pXp. For a fixed p that ramifies in k/Q, let Fp be

the subfield corresponding to Xp, and Mp be the subfield corresponding to ⊕q6=pXq, where q

runs over all ramified primes different than p. Hence Mp is the maximal subfield of K such

that Mp/Q is unramified at p.

We have ⊕pXp = hXp∪ ⊕qXqicorresponds to K, so FpMp = K.

K = FpMp unram tot ram Fp Mp p unram Q

Then Ip is the inertia group p, i.e., Ip= Gal(K/Mp).

Let M ⊆ K be the fixed field of hIp, p/ni.

Then the union of all subgroups Ip generates a subgroup h∪pIpi = Gal(K/M ). By Galois

theory M = ∩pMp. So M is unramified at all finite places of Q, whereupon M = Q. Thus

h∪_pIpi = Gal(K/Q). Therefore, we have | Gal(K/Q)| = [K : Q] = |X| = |⊕pXp| =Y p |X_p| =Y p ep by Theorem 2 =Y p |I_p|.

Chapter 1

Basic Definitions and Results

1.1

Notations and Consequences

1.1.1 Notations

Whenever i, j or l appear in the same line they are distinct, moreover i, j, l ∈ {1, 2, 3}, and any result (or definition) proven (or defined) with any of these indexes can be proven (or defined) with the others. Let:

- k be a real abelian field ramified exactly at three primes p1, p2, p3.

- K be the genus field of k in the narrow sense. - T := Gal(K/Q).

- Ti be the inertia subgroup of T for pi.

- Ki be the maximal subfield of K ramified only at pi.

- N := Gal(K/k). - G := Gal(k/Q). - n := [K : k] = |N|. - ri := [K : kKi]. - ai:= [(Ki∩ k) : Q]. - πi : T → Ti (projection of T on Ti). - Ni := πi(N ).

1.1.2 Consequences

- T is the direct product T = T1⊕ T2⊕ T3 and

Ti= Gal(K/KjKl) ∼= Gal(Ki/Q)

- ri = [K : kKi]then ri|n.

- πi : T → Ti, then ∀x ∈ T ; x = π1(x) · π2(x) · π3(x).

- πij : T → Ti⊕ Tj, such that πij(x) := πi(x)πj(x) ⇒ ker πij = Tl.

Proposition 2. K = kK_iKj.

Proof. Let ℘i be a prime ideal of KjKl over pi and ˜℘i be a prime ideal of kKjKl above ℘i.

K N ∩Ti unram _℘˜ T ot ram i ⊆ kKjKl unram k ℘i ⊆ KjKl unram KjKl∩ k unram at pi pi∈ Q

Since K is a genus field of k, we have K/k is unramified. Then ˜℘i is unramified in K/kKjKl.

On the other side, ℘i is totally ramified in K/KjKl. Then ˜℘i is totally ramified in K/kKjKl.

So K/kKjKl is at the same time totally ramified and unramified at ˜℘i, then [K : kKjKl] = 1;

therefore K = kKjKl.

Corollary 2. π_ij|_N : N → NiNj is injective.

Proof. Note that ker π_ij|_N = N ∩ Tl, and N ∩ Tl= {1}by Proposition 2.

By the previous hypothesis and results, we draw the following diagrams: 2

K ri kKi n ri Ki n ri k n2 r1r2r3ajal Ki∩ k ai Q K = kKiKj n rl.al n KiKj n k n rl.al KiKj∩ k n rirj.aiaj Q Now we have:

- Recall that [K : kKi] = ri, [K : k] = n, and [Ki∩ k] = ai, then [Ki: Ki∩ k] = [kKi: k] = _rn_i,

therefore |Ti| = [Ki : Q] = _rn_i.ai. - [KiKj : Q] = [Ki : Q].[Kj : Q] = n 2 rirj.aiaj. - [k ∩ KiKj : Q] = _rn irj.aiaj. - [K : Q] = n3 r1r2r3.a1a2a3. - [k : Q] = n2 r1r2r3.a1a2a3. 1.1.3 Assumptions

• We notice from Proposition 2. that N ∼= Gal(KjKl/KjKl∩ k) which is a subgroup of

Gal(KjKl/Q), then N is either cyclic or bi-cyclic. In this thesis, we will assume that N

is cyclic, we set N = hτi.

• We will also assume that the Ti’s are cyclic. For instance, Ti must be cyclic if pi is odd.

1.2

Definition of m

Notice that πi|N : N → Ni is surjective by definition. Then Ni = hπi(τ )i and ker πi =

TjTl= Gal(K/Ki), so | ker πi∩ N | = | Gal(K/kKi)| = ri, then |Ni| = n/ri.

Proposition 3. We have (ri, rj) = 1. Therefore r1r2r3|n.

Recall that N = hτi ⇒ πij(N ) = hπij(τ )i = hπi(τ )πj(τ )i, then the order of πi(τ )πj(τ ) is n,

that is n is the smallest natural number such that πi(τ )nπj(τ )n= 1, then πi(τ )n= πj(τ )n= 1.

Hence n is the smallest positive multiple of both n/ri and n/rj, so n = lcm(n/ri, n/rj), then

(ri, rj) = 1. Therefore r1r2r3|n.

We see that Proposition 3. is a direct consequence of the assumption of the cyclicity of N, moreover, it is going to be present or used almost everywhere in this thesis. In the following we will set

m = n r1r2r3

.

In the next chapters m will play a crucial role in the construction of a basis of the group of circular units C. For instance, we will see that whenever m = 1, the construction of a basis of C is much simpler than in any case of m > 1.

1.3

Key Equation

Proposition 4. Let G = hgi be a cyclic group of order |G| = f . For any positive divisor s of f , there is a unique subgroup L of G of order |L| = s, namely hgf /si. Moreover; any generator of the subgroup L is the (f /s)th power of a suitable generator of G.

Proof. We are going to prove only the second claim of the proposition, which is any generator of the subgroup L is the f/sth _{power of a suitable generator of G.}

We have |g| = f, |L| = s by hypothesis, we write f = sk. Any generator of L is of the form gt where t = ku, (u, s) = 1. By Chinese remainder theorem there is an integer v satisfying v ≡ u (mod s), (v, f) = 1. Then gv is a generator of G and gvk = gt because vk ≡ uk = t (mod f ).

Since τ|Ki generates the subgroup Gal(Ki/k ∩ Ki)of Gal(Ki/Q) of index ai, Proposition

4 allows to choose an element σi∈ Ti (so its restrictions to other Kj and Kl are the identities)

such that σi|Kiis a generator of Gal(Ki/Q) satisfying σ

ai i |Ki = τ |Ki. Therefore since σ a1 1 σ a2 2 σ a3 3

and τ have the same restrictions to each K1, K2, and K3, we get the key equation:

σa1 1 σ a2 2 σ a3 3 = τ . (1.1)

This equation and its restrictions will have a great impact on our work in the next chapters.

Chapter 2

The Cuboid’s Construction

If we manage to construct a Z-basis of the group of circular D of k, it would not be hard for us to get a Z-basis for the group of cyclotomic units C of k. In this chapter, we will set the ground for constructing a Z-basis of D by giving an idea how this Z-basis might look like. We will also give some three dimensional geometrical constructions, in order to be able to make the construction of such Z-basis in the next chapters.

2.1

Notations

Let ζs= e2πi/s for any positive integer s, and f, fij, and fi be the conductor of K,

KiKj, and Ki respectively. So consider:

- η = NQ(ζf)/k(1 − ζf).

- ηij = N_Q(ζfij)/(KiKj∩k)(1 − ζfij).

- ηi = NQ(ζ_fi)/(Ki∩k)(1 − ζfi).

Then we will have D the group of circular numbers of k, and C the group of circular units of k: D = h−1, η, η12, η23, η13, η1, η2, η3iZ[G]. C = h−1, η, η12, η23, η13, η11−σ1, η 1−σ2 2 , η 1−σ3 3 iZ[G].

The group Dij of circular numbers of KiKj∩ k and Di of Ki∩ k are defined similarly:

Dij = h−1, ηij, ηi, ηjiZ[G].

2.2

_{Z-basis of D}

i

As a Z-basis of Di, we take the set Bi of all conjugates of ηi. Since Gal(Ki ∩ k/Q) =

hσi|Ki∩ki is of order ai, which is the Z-rank of Di which is equal to [k ∩ Ki : Q], then:

Bi =η σx

i

i | 0 ≤ x < ai .

Note that if ai= 1 then ηi = pi.

2.2.1 _{Z-basis of D}ij

Proposition 5. Gal(k ∩ K_iKj/Q) = {(σixσ y

j)|k∩KiKj | 0 ≤ x < ai, 0 ≤ y < mrlaj}.

Proof. Recall that [k ∩ K_iKj : Q] = mrlaiaj which is at least equal to

  {(σ x iσ y j)|k∩KiKj; 0 ≤ x < ai, 0 ≤ y < mrlaj}   . Then it is sufficient to prove

Gal(k ∩ KiKj/Q) ⊆ {(σixσ y j)|k∩KiKj | 0 ≤ x < ai, 0 ≤ y < mrlaj}. KiKj (KiKj ∩ k)Ki KiKj∩ k Ki Ki∩ k ai Q

Let ρ ∈ Gal(k ∩ KiKj/Q). From the diagram we see that

Gal(k ∩ KiKj/k ∩ Ki) ∼= Gal(Ki(KiKj∩ k)/Ki),

which is cyclic. Then Gal(k ∩ KiKj/k ∩ Ki) is cyclic generated by σj|k∩KiKj of order [k ∩

KiKj : k ∩ Ki] = mrlaj. Moreover Gal(k ∩ Ki/Q) is cyclic generated by σi|k∩Ki of order

ai. Then ∃ x, 0 ≤ x < ai such that ρ and σxi have the same restriction to k ∩ Ki. So

ρσ_i−x|_k∩KiKj ∈ Gal(k ∩ K_iKj/k ∩ Ki) is of the form σjy|k∩KiKj for a suitable y, 0 ≤ y < mrlaj.

So

ρσ_i−x|_k∩KiKj = σy_j|_k∩KiKj ⇒ ρ = (σx_iσy_j)|k∩KiKj,

as desired.

Now let us define a new set which is:

Bij =η σx iσ y j ij | 0 ≤ x < ai, 0 ≤ y < mrlaj− 1, (x = 0 =⇒ y < (mrl− 1)aj) .

We claim that the set Bij∪ Bi∪ Bj is a Z-basis of Dij. In fact, it is enough to show that

this is a system of generators, because

|Bij ∪ Bi∪ Bj| ≤ (mrl− 1)aj+ (ai− 1)(mrlaj− 1) + ai+ aj = mrlaiaj+ 1,

which is the Z-rank of Dij, which is equal to [k ∩ KiKj : Q] + 1. Actually, it is not hard to

see that the missing generators of Dij are Z-linear combinations of this set.

Indeed, using the relation

mrlaj−1 Y y=0 ησ y j ij = N(KiKj∩k)/(Ki∩k)(ηij) = η 1−Frob(pj)−1 i ,

we see that we can get ησxiσ y j

ij for 0 < x < ai and y = mrlaj − 1as a Z-linear combination of

generators in Bij ∩ Bi. Note that

ησ ai i σ aj j ij = ητij = ηij ⇒ η σai_i ij = η −σ_jaj ij . So ai−1 Y x=0 mrl−1 Y z=0 ησ x iσ zaj j ij = ai−1 Y x=0 mrl−1 Y z=0 ησ x iσ −zai i ij = ai−1 Y y=−(mrl−1)ai ησ y i ij = mrlai−1 Y y=0 ησ y i ij , since ησ_imrlai

ij = ηij. Hence for any c = 0, 1, ..., aj− 1, we have ai−1 Y x=0 mrl−1 Y z=0 ησ x iσ zaj +c j ij = η σc j(1−Frob(pi)−1) j , which gives ησyj

ij for (mrl− 1)aj ≤ y ≤ mrlaj − 1 as a Z-linear combination of generators in

2.3

_{Idea for a Z-basis of D}

We now want to construct a Z-basis B of D of rank [k : Q] + 2 as follows:

Let B =B ∪ Se for a set S of some suitably chosen conjugates of η (to be defined later on) and e

B = B1∪ B2∪ B3∪ B12∪ B23∪ B13.

Remark 1. At the moment we cannot guarantee that the conjugates of eB are linearly inde-pendent, this will be obtained at the very end.

It is clear thatBe generates

D12D23D13= h−1, η12, η23, η13, η1, η2, η3iZ[G].

In order to compute how many conjugates the set S should contain, we assuming the independence of the conjugates ofBe. Then we have:

|S| = ([k : Q] + 2) − | eB|

= m2r1r2r3a1a2a3+ 2 − (mr3a1a2+ 1 + mr2a1a3+ 1 + mr1a2a3+ 1 − a1− a2− a3)

= m2r1r2r3a1a2a3− mr3a1a2− mr2a1a3− mr1a2a3+ a1+ a2+ a3− 1. (2.1)

Remark 2. By (2.1), we see that the set we see that S contains all the conjugates of η −(mr3a1a2+ mr2a1a3+ mr1a2a3− a1− a2− a3+ 1) conjugates. It follows that we need to

get rid of (mr₃a1a2+ mr2a1a3+ mr1a2a3− a1− a2− a3+ 1) conjugates of η. At the moment

we can not decide which conjugates we need to get rid of.

Now, let us find the set of the conjugates of η. Proposition 6. Gal(k/Q) = {(σ1xσ

y

2σz3)|k | 0 ≤ x < mr2r3a1, 0 ≤ y < mr1a2, 0 ≤ z < a3}.

Proof. Recall that [k : Q] = m2r1r2r3a1a2a3 which is at least equal to

|{(σx 1σ

y

2σz3)|k | 0 ≤ x < mr2r3a1, 0 ≤ y < mr1a2, 0 ≤ z < a3}| .

Then it is sufficient to prove Gal(k/Q) ⊆ {(σ1xσ y 2σ z 3)|k| 0 ≤ x < mr2r3a1, 0 ≤ y < mr1a2, 0 ≤ z < a3}. By Proposition 5. we have: Gal(k ∩ K2K3/Q) = {(σy2σ3z)|k∩K2K3 | 0 ≤ y < mr1a2, 0 ≤ z < a3}. 8

Now for any ρ ∈ Gal(k/Q) there are 0 ≤ y < mr1a2; 0 ≤ z < a3 such that ρ and σ2yσz3 have

the same restriction to k ∩ K2K3, so

(ρσ₂−yσ₃−z)|k∈ Gal(k/k ∩ K2K3).

Since Gal(k/k ∩ K2K3)is cyclic generated by σi|kof order [k : k ∩ K2K3] = mr2r3a1, there is

x, 0 ≤ x < mr2r3a1 such that σx1 and ρσ −y 2 σ

−z

3 have the same restriction to k. So

ρ = (σ₁xσ₂yσz₃)|k,

as desired.

Remark 3. By the previous proposition, we conclude that the set {ησx1σ

y

2σz3 | 0 ≤ x < mr₂r₃a₁, 0 ≤ y < mr₁a₂, 0 ≤ z < a₃}

contains all the conjugates of η.

2.4

The Cuboid Construction

To be able to describe the set S or to understand which conjugates it contains (i.e, to be able to decide which conjugates of η we need to get rid of), we need to organize or visualize (in a specific way) the set of all conjugates of η, i.e., the set

{ησ1xσ y

2σ3z | 0 ≤ x < mr_{2r3a1, 0 ≤ y < mr1a2, 0 ≤ z < a3}}.

2.4.1 Construction

We can visualize the set of all conjugates of η being placed in r1× r2r3 boxes (r1 boxes

placed vertically, and r2r3 boxes placed horizontally), each box consisting of ma1× ma2× a3

cubes, each cube containing one conjugate of η. For instance, the conjugate ησx1σ

y 2σz3_,

x = x1ma1+ x2, y = y1ma2+ y2, such that 0 ≤ x1 < r2r3, 0 ≤ y1 < r1, and 0 ≤ x2 < ma1,

0 ≤ y2 < ma2, 0 ≤ z < a3; appears in the cube (x2, y2, z)of the (x1, y1)th box.

So in this chapter when we say (x, y, z), we mean that x is the horizontal coordinate, y is the vertical coordinate and z is the third coordinate placed on the bar perpendicular to the horizontal and vertical directions. Similarly, when we say (x, y) we mean that x is the horizontal coordinate and y is the vertical one. In addition, in this chapter we start counting from 0.

2.4.2 Examples

In the figures below, there is no significant meaning of the yellow and white color except that it helps to visualize the boxes; in particular, it helps us to distinguish the boxes.

1. Let m = 1, a1 = a2 = a3 = 1, and r1 = 7, r2 = 5, r3 = 4, then the conjugates of η

are placed in 7 × 5 · 4 boxes, each box consisting of 1 × 1 × 1 = 1 cube containing one conjugate, (Figure 2.1).

Figure 2.1: m = 1, a1= a2 = a3 = 1, and r1 = 7, r2= 5, r3 = 4.

Consider the conjugate ησ7

1σ22. Since in this case each box consists of one cube, the

conjugate appears in the (7, 2)th _{cube, which in the 8}th_{column and the third row, recall}

that we start counting from 0, and a3 = 1 means we only have one layer. ησ

7 1σ22 is

coloured by blue in Figure 2.1.

2. Let m = 1, r1 = 5, r2 = 4, r3 = 3, and a1 = 2, a2 = 3, a3 = 3; then the conjugates of

η are placed in 5 × 4 · 3 boxes consisting of 2 × 3 × 3 cubes, each cube containing one conjugate.

Figure 2.2: m = 1, r1 = 5, r2= 4, r3 = 3,and a1= 2, a2= 3, a3= 3.

Consider the conjugate ησ0

1σ142 σ30, note that z = 0 means the conjugate is on the face

layer (the facing layer or the first layer); x = 0 means it is in the first column. Then 14 = 4 · 3 + 2, so the conjugate is in the cube (0, 2, 0) of the (0, 4)th box, which the blue coloured conjugate in Figure 2.2.

3. Let m = 2, a1 = 2, a2 = 3, a3 = 3, r1 = 5, r2 = 4, r3 = 3, then the conjugates of η are

Figure 2.3: m = 2, a1 = 2, a2 = 3, a3 = 3, and r1 = 5, r2 = 4, r3 = 3.

Consider the conjugate ησ34

1 σ210. Then 34 = 8 · 4 + 2, 10 = 1 · 6 + 4, so the conjugate is

the cube (2, 4, 0) of the (8, 1)th _{box, which is the blue coloured conjugate in Figure 2.3.}

One more example, consider the conjugate ησ01σ202 σ13. Then 20 = 3·6+2, so the conjugate

is the cube (0, 2, 1) of the (0, 3)th_{box, which is the red coloured conjugate in Figure 2.3.}

Remark 4. We shall define S by explaining which conjugates of η do not belong to S. In this chapter, we only computed the number of the missing conjugates in (2.1). i.e. the number of conjugates which should not be in S, and in chapters 3 and 5 we will locate them in term of our construction.

2.4.3 Norm Relations

In this section, we emphasize more the theoretical aspect of the existence and the distri-bution of some relations. However; in the next chapter, we will give clear examples and figures which make clear what might seem ambiguous in this section.

Remark 5. Recall that (r_i, rj) = 1, so there are dij, dji∈ Z such that dijri+ djirj = 1. Then τmrlridij _{= σ}almrlridij l σ aimrlridij i σ ajmrlridij j by (1.1) = σalmrlridij l σ aimrl(1−djirj) i since σ ajmrirl j = id = σalmrlridij l σ aimrl i . So σaimrl i acts on k as σ −almrlridij l .

Remark 6. Recall that if |b| = s, then hbi = hbfi ⇐⇒ gcd(s, f ) = 1.

We have (ri, rj) = 1, so there are dij, dji ∈ Z such that dijri+ djirj = 1. Then

(rj, dij) = 1 ⇒ (rldij, rj) = 1 ⇒ hσ −almrlridij l i = hσ almri l i since |σalmri l | = rj; so rj−1 X v=0 σ−almrlridijv l = rj−1 X v=0 σalmriv l = rj−1 X v=0 σalmrlridijv l = rj−1 X v=0 σ−almriv l .

The distribution of the conjugates in the subsection 2.4.1, will lead us to detect some relations between the conjugates.

(i) The product of all conjugates in the cubes in the same horizontal row in a given row, can be obtained as a Z-linear combination of conjugates ofBe; we call that a Row Relation; see Figure 3.2 below. For instance, the product of all the conjugates in the first horizontal row is

η1+σ1+σ21+...+σmr2r3a1−11 _{= N}

k/k∩K2K3(η).

Then we can visualize this norm, as Row relation (0, 0) (i.e. Row relation that starts at the conjugate in the first row and in the first column); see Figure 3.2.

Note that Row relation takes place only on one layer. Figure 2.4: m = 1, ai = 1, r1= 7, r2 = 5, r3 = 4.

(ii) The product of all conjugates in the cubes in the same vertical column in any r3 columns

distant by mr2a1 columns consecutively, can also be obtained as a Z-linear combination

of conjugates ofBe; we call that a Column Relation; see Figure 3.3 below. For instance, we have the sum of T2 is

1 + σ2+ σ22+ ... + σa2mr1 −1 2 + σ a2mr1 2 + ... + σ mr1r3a2−1 2 , which is equal to (1 + σ2+ σ22+ ... + σ a2mr1−1 2 )(1 + σ a2mr1 2 + σ 2a2mr1 2 + ... + σ (r3−1)a2mr1 2 ), which is equal to (1 + σ2+ σ22+ ... + σa2mr1 −1 2 ) r3−1 X v=0 σa2mr1v 2 ; by Remark 5, it acts on k as (1 + σ2+ σ22+ ... + σa2mr1 −1 2 ) r3−1 X v=0 σ−a1mr1r2d23v 1 ; by Remark 6 it is equal to (1 + σ2+ σ22+ ... + σa2mr1 −1 2 ) r3−1 X v=0 σa1mr1v 1 . So η(1+σ2+σ22+...+σa2mr1−12 ) Pr3−1 v=0 σa1mr1v1

is the product of the conjugates of the (mr2ba1+ 1)th columns where 0 ≤ b ≤ r3− 1, i.e.,

the product of the conjugates of the first column, by the conjugates of the (mr2a1+ 1)th

column, by the conjugates of the (2mr2a1 + 1)th column,...., by the conjugates of the

(mr2(r3− 1)a1+ 1)th column, which is equal to Nk/k∩K1K3(η). Then we can visualize

this norm as Column relation (0, 0) and (0, 0, 0) in Figures 2.5 and 2.6. Note that Column relation takes place on one layer only.

Figure 2.5: m = 1, ai = 1, r1 = 7, r2= 5, r3 = 4.

Purple color covers all the conjugates belonging to Column relation (0, 0).

Figure 2.6: m = 1, a1 = 2, a2 = 3, a3 = 3, r1= 5, r2= 4, r3 = 3.

Purple color covers all the conjugates to Column relation (0, 0, 0).

Before we define the third relation, let us set a new convention. By a staircase-diagonalwe mean the set of consecutive elements separated by the rectangular distance a1+ a2 where a1is horizontal and a2 is vertical; i.e., the consecutive elements are distant

by ~a1+ ~a2, where ~a1 is a horizontal vector of module a1, and ~a2 is a vertical vector of

module a2; see Figure 2.8.

Note that a staircase-diagonal takes place simultaneously on all layers in the same way.

The numbers 3 in Figure 2.7 consist of all the conjugates of the same staircase-diagonal. Figure 2.7: Staircase-diagonal (2, 0) on a layer

m = 1, a1 = 1, a2 = 1, a3 = 1, r1 = 7, r2 = 5, r3= 4.

The numbers 1 and 2 in Figure 2.8 consist of the conjugates of the staircase-diagonals starting at the conjugates (0, 0, 0) and (1, 0, 0) respectively. We will call the main staircase-diagonal the one starting at (0, 0, 0).

Figure 2.8: Staircase-diagonals on a layer

m = 1, a1= 2, a2= 3, a3= 3, r1 = 5, r2 = 4, r3 = 3.

(iii) Finally the same holds true for the product of all conjugates in any r2staircase-diagonals

parallel to the main staircase-diagonal and distant by mr3a1staircase-diagonals

consecu-tively, called a Diagonal Relation, see Figures 3.4 and 3.18 below. In other words, the distance between any two consecutive conjugates in the same row of the same Diagonal relation is mr3a1.

For instance, we have T3 is equal to

a3−1 X u=0 σ₃u mr1r2−1 X v=0 σa3v 3 = a3−1 X u=0 σ₃u mr1r2−1 X v=0 σ−a3v 3 . (2.2) Which is equal to a3−1 X u=0 σu₃ mr1−1 X v=0 σ−a3v 3 r2−1 X w=0 σ−a3mr1w 3 . (2.3) By Remark 5 it acts on k as a3−1 X u=0 σ₃u mr1−1 X v=0 σa1v 1 σ a2v 2 r2−1 X w=0 σa1mr1r3d32w 1 . (2.4) By Remark 6 it is equal to a3−1 X u=0 σ₃u mr1−1 X v=0 σa1v 1 σa22v r2−1 X w=0 σa1mr3w 1 . (2.5) 16

Now ηPmr1−1

v=0 σ1a1vσa2v2 is the product of the conjugates of the main staircase-diagonal

on the face layer. So ηPmr1−1

v=0 σa1v1 σa2v2 ·σa1mr31 gives the product of the conjugates on the

(a1mr3+ 1)th staircase-diagonal. Then η Pmr1−1

v=0 σa1v1 σ2a2v

Pr2−1

w=0 σa1mr3w1 gives the product

of r2diagonal distant by a1mr3 i.e. it gives the product of all conjugates on the face layer

of Diagonal relation starting at (0, 0, 0) (i.e. the conjugate in the first row, first column, and on the face layer). Consequently, ηPa3−1

u=0 σu3

Pmr1−1

v=0 σ1a1vσ2a2v

Pr2−1

w=0 σa1mr1w1 gives the

product of all the conjugates of Diagonal relation (0, 0, 0), then ηT3 is Diagonal relation

(0, 0, 0).

Note that Diagonal relation takes place on all layers simultaneously and in the same way.

In the following figures, the purpose of the green color is a clearer visualization only, and the number (1) represents a conjugate belonging to the same Diagonal relation.

In the following example we have only one layer since a3 = 1, so the Figure 2.9 gives the

visualization of the Diagonal relation (0, 0).

Figure 2.9: Diagonal relation (0, 0).

m = 1, ai= 1, r1 = 7, r2 = 5, r3 = 4

Figure 2.10: Diagonal relation (0, 0, 0) on a layer.

2.4.4 Basis of C

To finish the proof we must show that all conjugates of η which are not in S can be obtained by linear combinations of conjugates in B =B ∪ Se . Then it is not hard to modify B in order to get a basis of C. Basically, it is enough to change the sets Bi to

B_i0 = {η1−σ

x i

i | 0 < x < ai}.

Chapter 3

The Case of m

= 1

As promised in the previous chapter, we will locate the missing conjugates of the set S in this chapter in the case of m = 1. By using the relations of the subsection 2.4.3 only, we will be able to get back all the missing conjugates, therefore we will be showing that B is a basis for D but only for the case when m = 1. In this chapter, we suppose without the loss of generality, r1 ≥ r2 ≥ r3. On the other hand, we will explore the theory simultaneously with

worked examples which are rich in figures in order to master how the relations work.

3.1

a

1

= a

2

= a

3

= 1.

In this section, we study a very particular case: m = 1 and a1 = a2 = a3 = 1. The

objective of this section is to start with a basic case where the visualization of the relations is simple on one hand; on the other hand, it will be the model we follow while working in more general cases. For the applied example, we set r1 = 7, r2 = 5, and r3 = 4, it is in fact the

same example for which we studied its relations in the subsection 2.4.3.

In order to comply to the usual matrix we are going to slightly change the notations of the conjugates of η in the cuboid.

Notation: The conjugate (i, j) means the conjugate in the ith row and the jth column in the following figures. For instance, we visualize (1, 1) as the conjugate in the first row and in the first column (in the previous notation (0, 0) meant the conjugate in the first row and in the first column). Note that the conjugate (i, j) now is the conjugate ησj−1₁ σ₂i−1 _{(to recall the}

previous notations you can verify section 2.4).

Recall that since a3 = 1, the cuboid consists of one layer. So now the cuboid consists of r1

cubes (placed vertically)× r2r3 cubes (placed horizontally), each cube containing one element

Figure 3.1: m = 1, ai = 1, r1 = 7, r2= 5, r3 = 4.

For instance, the blue coloured cube contains one element (or conjugate).

3.1.1 Relations

Let us recall how the norm relations are in this case: •Row relation consists of one row.

Figure 3.2: m = 1, ai = 1, r1 = 7, r2= 5, r3 = 4.

Orange color covers all the conjugates belonging to Row relation (1, 1).

•Column relation consists of r3 columns. The distance between any two consecutive columns

is r2.

Figure 3.3: m = 1, ai = 1, r1 = 7, r2= 5, r3 = 4.

Purple color covers all the conjugates belonging to Column relation (1, 1).

•Diagonal relation consists of r2 stair-case diagonals. The distance between any two

consec-utive stair-case diagonal is r3.

Figure 3.4: m = 1, ai = 1, r1= 7, r2 = 5, r3 = 4.

Green color covers all the conjugates belonging to Diagonal relation (1, 1).

3.1.2 Missing Conjugates

According to (2.1) and Remark 2, we know that S contains r1r2r3− (r1+ r2+ r3− 2)

conjugates since m = 1, a1 = a2= a3 = 1.. So we need to get rid of r1+ r2+ r3− 2conjugates.

Let us get rid of r1+ r2+ r3− 2 conjugates in the following way:

• Get rid of r₂+ r3− 1conjugates from the first row starting from the upper left corner; (blue

color in Figure 3.5).

• Get rid of r₁− 1conjugates left in the first column; (red color in Figure 3.5).

Figure 3.5: Missing Conjugates.

Now we need to show that we can get back the missing conjugates from the existing conjugates. So we shall add more and more missing conjugates of η step by step as follows:

• We get back the conjugates (2, 1), (3, 1)...(r1, 1) by applying the corresponding Row

Figure 3.6: Apply Row Relations.

(r1, 1) = (7, 1)

3.1.3 (r2 + r3 − 1) Missing Conjugates

So we still have (r2+ r3− 1) missing conjugates; see Figure 3.7.

Figure 3.7: (r2+ r3− 1) Missing Conjugates

• We get back the conjugates (1, r2), (1, r2− 1), (1, r2− 2), ..., (1, r3)by applying the

corre-sponding Column relations; namely, (1, r2), (1, r2− 1), (1, r2− 2), ..., (1, r3) respectively;

see Figure 3.8.

Figure 3.8: (1, r2) = (1, 5); (1, r2− 1) = (1, 4).

Note that Column relations 9, 10 = Column relations 4, 5.

In fact, the missing conjugates now are only in the first row; namely, the conjugates (1, 1), ..., (1, r3− 1)

| {z } first set , and (1, r2+ 1), ..., (1, r2+ r3− 1) | {z } second set

. It is worth to note that the remaining missing conjugates are two different sets of r3− 1 consecutive conjugates; the first set starts from the first left

conjugate of the first row, and the second set starts from (r2+ 1)th conjugate of the first row;

see Figure 3.9. 22

Figure 3.9: Two Sets of Missing Conjugates.

Since we are only working on the first row we will denote the conjugate (1, i) by i; and recall that when we say for example Diagonal relation i, we mean the Diagonal relation starting at i.

• By applying Diagonal relation r2, we get back the conjugate r2 mod r3, see Figure 3.10.

In fact, given that the distance between any two staircase-diagonals of the same Diagonal relation is r3, we get the staircase-diagonal r2+ r3 does not contain a conjugate of the

missing ones.

On the other side, we have r2 mod r3 6= 0since (r2, r3) = 1(i.e. r2 mod r3∈ {1, 2, ..., r3−

1}), we set k1 to be the greatest non-negative integer such that r2 − k1r3 > 0, i.e.,

r2− (k1+ 1)r3 < 0, then r2− k1r3 ∈ {1, 2, ..., r3− 1}, so we have the staircase-diagonal

r2 − k1r3 contains one of the missing conjugates 1,...,r3 − 1; namely, the conjugate

r2mod r3= r2− k1r3.

Note that in the following figures, N denotes the new found conjugate. Figure 3.10: (1, r2) = (1, 5), (1, r2 mod r3) = (1, 1).

• By applying Column relation r2− k1r3, we get the conjugate r2− k1r3+ r2. Recall the

distance between any two consecutive columns of the same Column relation is r2; see

Figure 3.11: (1, r2mod r3) = (1, 1), (1, r2+ (r2 mod r3)) = (1, 6).

Again, let k2 be the greatest non-negative integer such that (r2− k1r3+ r2) − k2r3 > 0.

• By applying Diagonal relation r2− k1r3+ r2, we get the conjugate r2− k1r3+ r2− k2r3;

see Figure 3.12.

If r2− k1r3 = r2− k1r3+ r2− k2r3, then r2− k2r3= 0, which is impossible.

Figure 3.12: Apply Diagonal relation (1, 6).

r2− k1r3+ r2− k2r3 = 2

• By applying Column relation r2− k1r3+ r2− k2r3, we get the conjugate r2− k1r3+ r2−

k2r3+ r2. It’s clear that for any i 6= j ∈ {1, 2, ..., r3− 1}, we have i + r2 6= j + r2; see

Figure 3.13.

Figure 3.13: Apply Column relation (1, 2).

r2− k1r3+ r2− k2r3+ r2= 7

If r2− k1r3+ r2 = (r2− k1r3+ r2− k2r3) + r2= 2r2− (k1+ k2)r3+ r2 then r2− k2r3 = 0,

which is impossible.

Repeat this procedure (Diagonal followed by Column relation) r3− 1times, we get the

r3− 1missing conjugates. If for any i > j of {1, 2, ..., r3− 1}, ir2− (k1+ ... + ki)r3+ r2 =

jr2− (k1+ ... + kj)r3+ r2 then (i − j)r2− (kj+1+ ... + ki)r3 = 0, which is impossible

since i − j ≤ r3− 1.

Even more, if for any i, j ∈ {1, 2, ..., r3− 1}, ir2− (k1+ ... + ki)r3+ r2 = jr2− (k1+ ... +

kj)r3+ r2, then i = j.

Figure 3.14: Apply Diagonal relation (1, 7).

Figure 3.15: Apply Column relation (1, 3).

3.2

General Case of m = 1.

In this section we study the general case of m = 1. For the worked example, we set r1= 5, r2 = 4, r3 = 3, and a1= 2, a2= 3, a3= 3. It is again one of the examples for which we

studied its relations in the subsection 2.4.3.

Notation: The conjugate (i, j, k) means the element in the ith row, the jth column and the kth layer. Note that the conjugate (i, j, k) is the conjugate ησj−11 σ

i−1 2 σ

k−1 3 .

Figure 3.16: m = 1, a1 = 2, a2 = 3, a3 = 3, r1 = 5, r2= 4, r3 = 3.

. Each small cube containing a conjugate; (blue cube contains one conjugate).

3.2.1 Relations

Since Row relation consists of the conjugates of one row in any case, let us recall how the Column and Diagonal relations work in this case:

•Column relation consists of r3 columns, such that the distance between any two consecutive

columns is r2a1.

Figure 3.17: m = 1, a1 = 2, a2 = 3, a3 = 3, r1= 5, r2= 4, r3 = 3.

Purple color covers all the conjugates to Column relation (1, 1, 1).

•Diagonal relation consists of r₁ stair-case diagonals, such that the distance between any two consecutive stair-case diagonals is r3a1.

Figure 3.18: m = 1, a1 = 2, a2 = 3, a3 = 3, r1= 5, r2= 4, r3 = 3.

The 1’s consist of the conjugates belonging to Diagonal relation (1, 1, 1) on a layer.

3.2.2 Missing Conjugates and Face Layer

According to (2.1) and Remark 2., we need to get rid of r1a2a3+ r2a1a3 + r3a1a2− a1−

a2− a3+ 1conjugates.

• Get rid of r₁a2a3− a3 conjugates, precisely, all (i, 1, k) for 2 ≤ i ≤ r1a2, and 1 ≤ k ≤ a3.

The action here takes place on the left side of the cube; (red color in Figure 3.19).

• Get rid of r₃a1a2− a2+ 1 conjugates, precisely, the conjugate (1, 1, 1) and the conjugates

(i, j, 1)for 1 ≤ i ≤ a2, and 2 ≤ j ≤ r3a1; (blue color in Figure 3.19).

• Get rid of r₂a1a3 − a1 conjugates, precisely, the conjugates (1, j, k) for 1 ≤ j ≤ r2a1,

2 ≤ k ≤ a3; and the conjugates (1, j, 1) for r3a1+ 1 ≤ j ≤ r3a1+ r2a1− a1; (orange color

in Figure 3.19).

Figure 3.19: Missing Conjugates.

We need to show that we can get back the missing conjugates from the existing ones. We start by the non-face layers, so we add all the conjugates on any non-face layer as follows:

• Apply Row relation at every missing conjugate in the first column, then apply Column relation at every missing conjugate in the first row. We guarantee that we can get back the missing conjugates of the first row by Column relations since they are r2a1

consecutive ones, which is the distance between any two columns of the same Column relation; see Figure 3.20.

Namely, apply Row relations (2, 1, k)...(r1a2, 1, k), and Column relations (1, 1, k), ..., (1, r2a1, k),

for all 2 ≤ k ≤ a3.

Figure 3.20: Missing conjugates on a non-face-layer.

3.2.3 Face Layer and (r2+ r3 − 1)a1 Missing Conjugates

Figure 3.21: Missing conjugates on the Face-layer

We restrict our work to the face layer where the interesting work takes place. We add all conjugates on the face layer step by step as follows:

Namely, we add the conjugates (a2+ 1, 1, 1), (a2+ 2, 1, 1), ..., (r1a2, 1, 1)by applying the

corresponding Row relations (a2+ 1, 1, 1), (a2+ 2, 1, 1), ..., (r1a2, 1, 1); see Figure 3.22.

Figure 3.22: (r1a2, 1, 1) = (15, 1, 1).

• Apply Diagonal relations at every missing conjugate in the second row to the ath 2 one;

see Figure 3.23.

We add all the conjugates that are not in the first row by Diagonal relations. We guarantee that we can add these conjugates because they fall between the second row and ath

2 row, which are the total of a2− 1 < a2rows and they are spread on exactly r3a1

columns.

Figure 3.23: Apply Diagonal relations at all conjugates in rows 2, ..., a2.

dand e are examples of two different Diagonal relations we apply.

So we still have (r2 + r3− 1)a1 missing conjugates.

Figure 3.24: (r2+ r3− 1)a1 Missing Conjugates.

Therefore we proceed similarly to the subsection 3.1.3.

• We add the conjugates (1, r2a1, 1), (1, r2a1− 1, 1),...,(1, r3a1− a1+ 1, 1) by applying the

corresponding Column relations, see Figure 3.25.

Figure 3.25: Apply Column Relations.

(1, r2a1, 1) = (1, 8, 1), (1, (r2 mod r3)a1, 1) = (1, 2, 1).

We note that similarly to the subsection 3.1.3, the remaining missing conjugates are two different sets of (r3− 1)a1 consecutive conjugates; the first set starts from the first left

conjugate of the first row, and the second set starts from (r2a1+ 1)th conjugate of the

Figure 3.26: Two Sets of Missing Conjugates.

• By applying Diagonal relations (1, r2a1, 1), (1, r2a1− 1, 1),..., (1, r2a1− a1+ 1), we get

the elements (1, (r2 mod r3)a1, 1), (1, (r2 mod r3)a1−1, 1), ..., (1, (r2 mod r3)a1−a1+

1, 1), then followed by the corresponding Column relations, and so on; (Figure 3.27).

So the difference between this case and the case of the subsection 3.1.3 is that, in this case, we start by applying a1 Diagonal relations followed by the corresponding Column relations

instead of applying one Diagonal followed by the corresponding Column relation. For more details reverify the subsection 3.1.3.

Figure 3.27: Apply Diagonal Relations.

(1, r2a1, 1) = (1, 8, 1), (1, r2a1− a1+ 1) = (1, 7, 1).

Figure 3.28: Apply Column Relations.

Figure 3.30: Apply Column Relations.

Chapter 4

Ennola Relation

In this chapter, we will derive a new type of relations, we call it Ennola-type relation, which will be a necessary relation to construct a Z-basis of D as soon as we have m > 1.

4.1

Notations and Restrictions

4.1.1 Notations

Again recall that whenever i, j or l appear in the same line they are distinct, moreover i, j, l ∈ {1, 2, 3}, and any result (or definition) proven (or defined) with any of these indices can be proven with the others; so for any non-negative integer h, let:

- Si(h) =Ph−1_u=0σui ∈ Z[G], so (1 − σi)Si(h) = 1 − σih, (Si(0) = 0). - Ri = Si(ai), Fi =Pm−1_v=0 σa_iiv, ∆i=Pm−1_v=0 vσa_iiv ∈ Z[G], so (1 − σi)Ri= 1 − σ_iai . - Hi=Prjrl −1 v=0 σ aimv i , and Hij = Prj−1 v=0 σ aimv i . For instance, RiFiHi = X ti∈Ti ti. - Γ = F1∆2− F2∆1+ F3∆1.

The Ennola-type relation that we will derive will be the following:

m−1 Y d=0 m−1 Y c=d+1 ησ1a1dσa2c2 R1R2R3H1H21 ∈ hη₁₂, η₁₃, η₂₃i Z[G] (4.1)

It will not be hard to prove (4.1) if we prove ηΓR1R2R3H1H21 _{∈ hη}m

4.1.2 Technicalities

In order to prove (4.2), we start by developing some useful technical equations, especially, when we restrict our work to k. One could skip this section and come back to it when we cite some of its results.

Using the key equation (1.1) we get:

res_K/kσa1 1 σ a2 2 σ a3 3 = τ |k= idk. (4.3)

Remark 7. By Remark 5 there is an integer d₁₃ such that σ−a1mr2

1 acts on k in the same way

as σa2mr1r2d13

2 . Since σ a2m

2 H2 = H2, we have

resK/kσ1−a1mr2H2 = resK/kσ2a2mr1r2d13H2 = resK/kH2.

Proposition 7. For all i 6= j, res_K/kHiHji is the same. For instance,

resK/kH1H21= resK/kH2H12= resK/kH3H13.

Proof. First part. One has

res_K/kH2H12= resK/k r1r3−1 X v=0 σa2mv 2 H12 = res_K/kH21 r3−1 X v=0 σa2mr1v 2 H12 = res_K/kH21H12 r3−1 X v=0 σ−a1mr1r2d23v 1 ; by Remark 5 = res_K/kH21H12 r3−1 X v=0 σa1mr2 1 by Remark 6 = resK/kH21H1.

Second part. Moreover, 36

res_K/kH2H12= resK/kH2 r2−1 X v=0 σa1mv 1 = res_K/kH2 r2−1 X v=0 σ−a1mv 1 by Remark 7 = res_K/kH2 r2−1 X v=0 σa2mv 2 σ a3mv 3 by (4.3) = res_K/k r2−1 X v=0 H2σ3a3mv since σ a2m 2 H2 = H2 = resK/kH2H32.

Using the method of the first part, we get resK/kH2H32= resK/kH3H23; and by the method

of the second part we get resK/kH3H23 = resK/kH3H13. Finally, by the method of the first

part, we get resK/kH3H13= resK/kH1H31. Altogether

resK/kH1H21= resK/kH2H12

= resK/kH2H32

= resK/kH3H23

= resK/kH3H13

= res_K/kH1H31.

Now, let us compute resK/kF3∆1H1H21. Let 0 ≤ hxi < 1 mean the fractional part of a

real number x and [x] = x − hxi mean its integral part. So

res_K/kF3∆1H1H21= resK/kF3∆1H3H13 by Proposition 7.

= resK/k m−1 X c=0 σ−a3c 3 H3 m−1 X b=0 bσa1b 1 H13 since σ3−a3mH3 = H3 = resK/k m−1 X c=0 σa1c 1 σ a2c 2 H3 m−1 X b=0 bσa1b 1 H13 by (4.3) = res_K/k m−1 X b=0 m−1 X c=0 m · hb miσ a1(b+c) 1 H1σa22cH21 note that σ1a1mH1= H1 = resK/k m−1 X c=0 m−1 X d=0 m · hd − c m iσ a1d 1 σ a2c 2 H1H21.

Moreover, resK/kF1∆2H1H21= resK/k m−1 X c=0 σa1c 1 m−1 X b=0 bσa2b 2 H1H21 = res_K/k m−1 X c=0 m−1 X d=0 c · σa1d 1 σ a2c 2 H1H21, and resK/kF2∆1H1H21= resK/k m−1 X c=0 σa2c 2 m−1 X b=0 bσa1b 1 H1H21 = res_K/k m−1 X c=0 m−1 X d=0 d · σa1d 1 σ a2c 2 H1H21. Thus res_K/kΓH1H21= m−1 X c=0 m−1 X d=0 σa1d 1 σ a2c 2 · c − d + m.h d − c m i
H1H21 = − m−1 X c=0 m−1 X d=0 σa1d 1 σ a2c 2 · m · [d−cm ] · H1H21 = m · m−1 X d=0 m−1 X c=d+1 σa1d 1 σ a2c 2 · H1H21. (4.4) Proposition 8. res_K/k∩K iKl(1 − σ ai i )∆iHij = resK/k∩KiKl(FiHij − mHij). Proof. resK/k∩KiKl(1 − σ ai i )∆iHij = resK/k∩KiKl(1 − σ ai i ) m−1 X v=0 vσaiv i Hij = resK/k∩KiKl m−1 X v=0 vσaiv i − m−1 X u=0 uσai(u+1) i ! Hij = resK/k∩KiKl m−1 X v=0 vσaiv i − m X v=1 (v − 1)σaiv i ! Hij = res_K/k∩KiKl m−1 X v=0 vσaiv i − m−1_X v=0 (v − 1)σaiv i + (m − 1)σ aim i + 1  ! Hij = res_K/k∩KiKl m−1 X v=0 σaiv i Hij+ (1 − m)σiaimHij − Hij = resK/k∩KiKlFiHij+ (1 − m).(Hij − 1 + σ aimrj i ) − Hij. 38

Now by Remark 5, we have σaimrj i acts on k as σ −ajmrirjdil j ; since σ −ajmrirjdil j acts trivially on KiKl, we get resK/k∩KiKl(1 − σ ai i )∆iHij = resK/k∩KiKlFiHij + (1 − m).(Hij − 1 + σ −ajmrirjdil j ) − Hij = resK/k∩KiKlFiHij + (1 − m).(Hij − 1 + 1) − Hij = res_K/k∩KiKlFiHij − mHij. Using (1.1) we get ησ ai i σ aj j ij = η τ ij = ηij ⇒ η σai_i ij = η σ_j−aj ij ,

which we use in the following: Proposition 9. ηF_ijiHj = ηF_ijjHj Proof. ησiai ij = η σ−aj_j ij ⇒ η Pm−1 v=0 σaivi ij = η Pm−1 v=0 σ −aj v j ij ⇒ ηFiHj ij = η Pm−1 v=0 σ −aj v j Hj ij ⇒ ηFiHj ij = η Pm−1 v=0 σ aj v j Hj ij since σ ajm j Hj = Hj ⇒ ηFiHj ij = η FjHj ij . Corollary 3. η_ijFiHjl= η_ijFjHjl= ηFiHil ij = η FjHil ij .

Proof. By the previous proposition we have ηFiHj

ij = η FjHj ij , so η FiHjHlj ij = η FjHjHlj ij .

On one hand, by Proposition 7. we get ηFiHjHlj ij = η FiHlHjl ij = η FiHlHil ij , and ηFjHjHlj ij = η FjHlHjl ij = η FjHlHil ij .

On the other hand, ηHl

ij = η rirj ij and ηij ≥ 0. Therefore ηFiHjl ij = η FiHil ij = η FjHjl ij = η FjHil ij .

4.2

Computations Toward (4.2)

In this section, we will prove (4.2); then we will be left with the last small section where we prove (4.1).

By using Proposition 7, we get

ηΓR1R2R3H1H21 _{= η}∆2F1R1R2R3H1H21_η−∆1F2R1R2R3H2H12_η∆1F3R1R2R3H3H13 (4.5)

Note that RiFiHi is the sum of all elements of Ti, so RiFiHi|k is the norm operator with

respect to k/(k ∩ KjKl) (Note that the map Ti → T /N is bijective since Ti ∩ N = {1} by

Corollary 2.), therefore ηΓR1R2R3H1H21 _{= η}∆2R2R3(1−Frob(p1)−1)H21 23 | {z } 1 η−∆1R1R3(1−Frob(p2)−1)H12 13 | {z } 2 η∆1R1R2(1−Frob(p3)−1)H13 12 | {z } 3 . (4.6) For the rest of this section, we will be proving that

1 = η∆2R2R3H21(1−Frob(p1)−1) 23 ≡ η (1−σ2)S2(c32)S2(c12) 2 η −(1−σ3)S3(c23)S3(c13) 3 modulo hη23miZ[G], (4.7) where cij are defined below; then by symmetry we get similar results for 2 and 3 which

give us (4.2).

We start the proof of (4.7) as follows:

Let us define integers 0 ≤ bij < mrirl, 0 ≤ cij < aj by

Frob(p1)−1 = σ₂a2b12+c12σa₃3b13+c13, Frob(p2)−1 = σa11b21+c21σ a3b23+c23 3 , Frob(p3)−1 = σa11b31+c31σ a2b32+c32 2 . Then, η1−Frob(p1)−1 23 = η 1−σ₂c12 23 · η σc12₂ (1−σc13₃ ) 23 · η σ₂c12σ₃c13(1−σa2(b12−b13)₂ ) 23 (4.8)

Let us apply ∆2R2R3H21 and compute the factors of (4.8) successively:

η∆2R2R3H21(1−σ2c12) 23 = η ∆2H21S2(c12)R3(1−σa22 ) 23 = η(F2H21−mH21)S2(c12)R3 23 by Proposition 8. = η(F3H31−mH21)S2(c12)R3 23 by Corollary 3.

From the proof of Proposition 5, we note that FiRiHij acts as the norm operator with

respect to (k ∩ KiKl)/(k ∩ Kl).

Hence, modulo hηm 23iZ[G], we have η∆2R2R3H21(1−σc122 ) 23 ≡ η F3R3H31S2(c12) 23 = η(1−σ2c32)S2(c12) 2 = η(1−σ2)S2(c32)S2(c12) 2 .

For the second factor of (4.8) we obtain similarly ησc122 ∆2R2R3H21(1−σc133 ) 23 = η σ₂c12(1−σa3₃ )∆2H21S3(c13)R2 23 = ησ2c12(1−σ−a22 )∆2H21S3(c13)R2 23 = η−σ2c12−a2(F2H21−mH21)S3(c13)R2 23 . Hence, modulo hηm 23iZ[G], we have ησc122 ∆2R2R3H21(1−σc133 ) 23 ≡ η −σc12−a2₂ F2R2H21S3(c13) 23 = η−σ c12−a2 2 (1−σ3c23)S3(c13) 3 = η−(1−σ3)S3(c23)S3(c13) 3 .

Finally, for the third factor of (4.8) we get ησc122 σc133 ∆2R2R3H21(1−σa2(b12−b13)2 ) 23 = η σ₂c12σ₃c13∆2R2R3H21(1−σ2a2) Pb12−b13+mr1r2r3−1 u=0 σa2u2 23 . But η∆2R2R3H21(1−σ2a2) 23 = η (F2H21−mH21)R2R3 23 ≡ 1 modulo hη m 23iZ[G], because ηF2H21R2R3 23 = N(K2K3∩k)/Q(η23) = 1.

Putting the three previous factors together, we get η∆2R2R3H21(1−Frob(p1)−1) 23 ≡ η (1−σ2)S2(c32)S2(c12) 2 η −(1−σ3)S3(c23)S3(c13) 3 modulo hη23miZ[G].

Using symmetry 1 ↔ 2, we get η∆1R1R3H12(1−Frob(p2)−1) 13 ≡ η (1−σ1)S1(c31)S1(c21) 1 η −(1−σ3)S3(c13)S3(c23) 3 modulo hη13miZ[G].

Now using symmetry 2 ↔ 3, we obtain η∆1R1R2H13(1−Frob(p3)−1) 12 ≡ η (1−σ1)S1(c21)S1(c31) 1 η −(1−σ2)S2(c12)S2(c32) 2 modulo hη12miZ[G].

4.2.1 Ennola Relation

Substituting into (4.6), we arrive to

ηΓR1R2R3H1H21 _{∈ hη}m

12, ηm13, η23miZ[G].

Remark 8. xm = ym⇒ (xy−1)m= 1. So xy−1 is an mth root of unity. If x and y are totally positive, then xy−1 = 1 ⇒ x = y.

By (4.4), we recall that resK/kΓH1H21 is divisible by m. Then by Remark 8., we get

m−1 Y d=0 m−1 Y c=d+1 ησa1d1 σ2a2cR1R2R3H1H21 _{∈ hη} 12, η13, η23iZ[G] (4.9)

We visualize the Ennola relation as the product of all the conjugates located under the staircase-diagonal of all boxes of ma1× ma2× a3 cubes starting from the upper left box. In

fact, we have r1× r2r3 boxes. Note that ηR1R2R3 is the product of a1a2a3 conjugates. Hence

m−1 Y d=0 m−1 Y c=d+1 ησa1d1 σ2a2cR1R2R3 (4.10)

is the product of all conjugates under the main staircase-diagonal of every layer of the (0, 0)th

box by the notation of section 2.4, i.e., the first box starting from the upper left corner. Now multiplying σa1d

1 σ a2c

2 R1R2R3 by H1 in (4.10), it can be seen as repeating the procedure of

(4.10) through r2r3 horizontal boxes, and by H21as repeating the procedure of (4.10) through

r1 vertical boxes; see Figure 4.1 below. (Note that R1R2R3 represents of the product of the

conjugates of one cuboid consisting of a1(horizontal)×a2(vertical)×a3(in depth) conjugates.)

Note that Ennola relation takes place on all layers simultaneously and in the same way.

Figure 4.1: Ennola relation on a layer.

Chapter 5

General Case

In this chapter, we will construct a basis for C in the general case, i.e., for any m ∈ Z+_.

Without loss of generality, we suppose r1 ≥ r2 ≥ r3. On the other side, as we did before,

we will explore the theory simultaneously with a worked example, in order to avoid possible ambiguities. Precisely, when m = 2, r1 = 5, r2 = 4, r3= 3, and a1 = 2, a2 = 3, a3 = 3.

5.0.2 Relations

As described in the subsection 2.4.3, we have Column Relations, Diagonal Relations, and Row Relations. In addition, now we have the Ennola-type Relation. It is clear how Row relation works. So in order to visualize how the other relations work on our worked example, we give a few figures below.

Diagonal relation on a layer.

Ennola relation on a layer.

5.1

Missing Conjugates and Non-top Layers

According to (2.1), we need to get rid of mr1a2a3+ mr2a1a3+ mr3a1a2− a1− a2− a3+ 1

conjugates. We will get rid of them in the following way:

• Get rid of mr₃a1a2− a2+ 1 conjugates, precisely, the conjugates (i, j, 1) for 1 ≤ i ≤ a2, and

2 ≤ j ≤ mr3a1 and the conjugate (1, 1, 1); (blue color in Figure 5.1).

• Get rid of mr2a1a3 − ma1 conjugates, namely, the conjugates (1, j, k) for 1 ≤ j ≤ mr2a1,

2 ≤ k ≤ a3; and the conjugates (1, j, 1) for mr3a1+ 1 ≤ j ≤ mr3a1+ mr2a1− ma1; (orange

color in Figure 3.19).

• Get rid of mr1a2a3− a3 conjugates, namely, the conjugates (i, 1, k) for 2 ≤ i ≤ mr1a2, and

1 ≤ k ≤ a3; (red color in Figure 5.1).

• Get rid of ma1− a1 conjugates, namely, (a2+ 1, j, 1)for a1+ 1 ≤ j ≤ ma1; (purple color in

Figure 5.1: Missing Conjugates.

m = 2, r1= 5, r2 = 4, r3 = 3, a1 = 2, a2= 3, a3= 3.

Figure 5.2: Non-top layer.

We will add more and more missing conjugates of η step by step as follows:

• Apply Row and Column relations, we add all the missing conjugates on any non-top layer; see Figure 5.1.

Non-top layer.

We will restrict our work to the Face layer where the interesting work takes place.

5.2

Face Layer and (r

2

+ r

3

− 1)ma

1

Missing Conjugates

Face Layer

• Apply Row relation on every row from the (a2+ 2)th row till the last row, we add all the

•By applying Ennola relation, we add the conjugate (a2+ 1, 1, 1).

•By applying (4.9)σi

1, we add the conjugates (a₂+ 1, a₁+ i, 1)for 1 ≤ i ≤ ma₁− a₁ (starting

from i = 1, 2..., ma1− a1 respectively), see Figure 5.2.

• By applying Diagonal relation on every missing conjugates from the second row to the ath 2

row, we add all of these missing conjugates, Figure 5.2.

So we still have (r2 + r3− 1)ma1 missing conjugates. Then we proceed similarly

(r2+ r3 − 1)ma1 missing conjugates.

•By applying Column relations from the mr₃a1− ma1+ 1column till the mr2a1 column, we

add all the missing conjugate on these columns.

Figure 5.3: Column Relations.

• By applying Diagonal relations mr2a1, mr2a1 − 1, ..., mr2a1 − mra1 + 1, followed by the

corresponding Column relations and so on. 52

Figure 5.4: Diagonal Relations.

Chapter 6

Construction of Abelian Fields

In this chapter, we show in two different ways that there are infinitely many abelian fields satisfying the assumptions of k. The first way uses the Characters theory, it is long way to reach our goal, but we keep it for pedagogical reasons; while the second way is a straight forward one, it simply uses Galois theory. We start with seven positive integers: m, a1, a2, a3, r1, r2, r3 satisfying (r1, r2) = 1, (r2, r3) = 1, (r1, r3) = 1.

Thanks to Dirichlet theorem on primes in arithmetical progression (chapter 16 in [7]), we know that they are infinitely many primes pi’s for i ∈ {1, 2, 3} satisfying p1 ≡ 1 mod 2mr2r3a1,

p2≡ 1 mod 2mr1r3a2, and p3 ≡ 1 mod 2mr1r2a3.

Let us choose and fix three distinct primes satisfying the previous congruences.

6.1

Construction of a Field F

By fixing three distinct primes p1, p2, p3, we know that there are even Dirichlet characters

χ1, χ2, χ3 of conductors p1, p2, p3 and of orders mr2r3a1, mr1r3a2, mr1r2a3 respectively.

In fact, if p > 2 is a prime then (Z/pZ)∗ _{is a cyclic group of order p − 1 and its group of}

characters is also a cyclic group of order p − 1, so we have a generator χ of this group. If s is a positive integer such that 2sk = p − 1, then χ2 _{is an even character of order} p−1

2 and χ 2k _is

an even character of order s.

We have pi− 1 = 2mrjrlaiki; let χ be a generator of \(Z/piZ)∗, it is of order pi− 1, and let

χi= χ2ki which is even of order p_2ki−1_i = mrjrlai and of conductor pi.