vtechworks.lib.vt.edu€¦ · On the Units and the Structure of the 3-Sylow Subgroups of the Ideal Class Groups of Pure Bicubic Fields and their Normal Closures Alberto Pablo Chalmeta

On the Units and the Structure of the 3-Sylow Subgroups of theIdeal Class Groups of Pure Bicubic Fields and their Normal

Closures

Alberto Pablo Chalmeta

Dissertation submitted to the faculty of theVirginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

Doctor of PhilosophyIn

Mathematics

Dr. Charles Parry (Chair)Dr. Ezra Brown

Dr. Edward GreenDr. Peter Linnell

Dr. Michael Renardy

September 29, 2006Blacksburg, Virginia

Keywords: Bicubic Fields, Normal Closure, Class Number, Invariants, Ideal Class Group

On the Units and the Structure of the 3-Sylow Subgroups of the IdealClass Groups of Pure Bicubic Fields and their Normal Closures

Alberto Pablo Chalmeta

ABSTRACT

Let Q( 3√m) and Q( 3

√m, 3

√n), where m and n are cube free rational integers, be called a cubic and a

bicubic field respectively. The number theoretic invariants for the cubic fields and their normal closuresare well known. Some work has been done on the units, classnumbers and other invariants of the bicubicfields and their normal closures by Parry but no method is available for calculating those invariants. Thisdissertation provides an algorithm for calculating the number theoretic invariants of the bicubic fields andtheir normal closure. Among these invariants are the discriminant, an integral basis, a set of fundamentalunits, the class number and the rank of the 3-class group.

Acknowledgements

I want to thank my advisor, Charles Parry. He has been encouraging and patient throughout the entiredissertation process. Without him I would have not been able to finish this document. He has been a kind,wise and patient mentor and I appreciate him greatly.

I want to thank the other members of my committee of their efforts in reviewing this work. I want tothank all the teachers I have had at Virginia Tech who have guided me through my graduate program.

I also want to thank my wife Sandi for never giving up on me and my son Ben for being my inspiration.

iii

Contents

1 Statement of Problem 1

2 Notation 2

3 Integral Basis for Ki, K and L 4

4 Unit Group of Ki 174.1 Types of Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Calculation of B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Unit group for Ki . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Unit group of K 195.1 Units in K from Type I fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.2 Example Type I units in K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.3 Units in K from Type IV fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.4 Example Type IV units in K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.5 Units in K from Type III Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.6 Cube Root Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.7 Example Type III units in K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6 Unit Group of L 326.1 Criteria for Units in L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326.2 Units in L from Type I Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 666.3 Example Type I units in L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696.4 Units in L from Type III Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.5 Example Type III units in L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716.6 Basis for the Unit Group of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

7 Rank of the Class Group of K and L 767.1 Class numbers of L and all its subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 767.2 Calculation of the cubic Hilbert symbol for divisors of 3 . . . . . . . . . . . . . . . . . . . . 777.3 Calculation of NB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 817.4 Calculation of ND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 827.5 Rank of the 3-Class Group Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Bibliography 86

A Units of Cubic Fields and their Normal Closures 87

B Some invariants of K and L where mi ≤ 500 for all i 104

iv

List of Tables

A.1 Units of ki = Q( 3√M) and Ki = ki(ζ) where M < 495 . . . . . . . . . . . . . . . . . . . . . 88

A.2 Units of ki and Ki not on Table A.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

B.1 Unit Basis, Class Numbers and Rank of the 3-Class Group for K and L . . . . . . . . . . . 105

v

Chapter 1

Statement of Problem

A field K = Q( 3√m1, 3

√m2) where m1 and m2 are positive integers and [K : Q] = 9 will be called a

bicubic field. The objective of this dissertation is to compute the number theoretic invariants of these

fields and their normal closure. Among these invariants are the discriminant, an integral basis, a set of

fundamental units, the class number and the rank of the 3-class group. The discriminant and integral basis

are determined in chapter 3. The determination of a set of fundamental units and class numbers of these

fields requires a knowledge of the same invariants for the pure cubic subfields and their normal closure.

Williams, et al [12] describe a method for determining the fundamental units of a cubic field using Vornoi’s

algorithm. Once this is known the class number of a cubic field can be determined by estimating the zeta

function. Barrucand and Cohn in [1] and [2] give a method for determining a set of fundamental units and

class numbers of the normal closure of a cubic field, knowing these invariants for the cubic field. Parry [9]

describes relationships between a set of fundamental units and the class number of a bicubic field and its

cubic subfields, as well as similar relations for the normal closure of a bicubic field. Using ideas from this

article we develop algorithms for determining a set of fundamental units and the class number of a bicubic

field and its normal closure in chapters 4 - 6. In chapter 7, we would like to determine the rank of the

3-class group of a bicubic field and its normal closure. Using methods of Gerth [5] we are able to do this

when the bicubic field has a cubic subfield with class number relatively prime to 3.

1

Chapter 2

Notation

The following notation will be used throughout this article.

ζ = e2πi/3

ω = e2πi/9

ki = Q( 3√mi) where i = 1, 2, 3, 4

K = Q( 3√m1, 3

√m2)

Ki = Q(ζ, 3√mi) where i = 1, 2, 3, 4

L = Q(ζ, 3√m1, 3

√m2)

Mi = Q(ω, 3√mi) where i = 1, 2, 3, 4

OF : the ring of integers of a field F .

εi: the fundamental unit of ki where i = 1, 2, 3, 4

ui: unit of Ki such that {εi, ui} form a fundamental set of units of Ki

G = Gal(L/Q): Galois group of L/Q

σi: nontrivial element of G that fixes Ki for i = 1, 2, 3, 4

τ : nontrivial element of G that fixes K

NM/F : norm function for the field extension M/F

Bi: the unique primitive integer of Ki such that εi = Bi

Bσi

where σ = σ1 for i �= 1 and σ = σ2 for i = 1.

Ai: the unique primitive integer of Ki such that Bi = Ai

Aσi

(Defined only when NKi/k(Bi) = 1) or Ai is

the unique primitive integer of Mi such that ωjBi = Ai

Aσij = 1 or 2 (Defined when NKi/k(Bi) = ζ or ζ2)

H, h: class number of L, K respectively.

Hi, hi: class number of Ki, ki respectively.

E: group of units of L.

2

Chapter 2. Notation 3

ε: subgroup of units of E generated by the units of K1, K2, K3, K4.

ε0: subgroup of units of ε generated by the units of k1, k2, k3, k4 and their congugates.

εi: subgroup of units of E generated by the units of K, Kσi , Ki.

e: group of units of K.

e0 = e ∩ ε0: subgroup of e generated by the units of k1, k2, k3, k4.

Ui (i = 1, 2, 3,4): group of units of Ki.

ui (i = 1, 2, 3,4): subgroup of Ui generated by the units of ki and its conjugates.

Chapter 3

Integral Basis for Ki, K and L

Let M be an an algebraic number field of degree n over Q and let OM be the rings of algebraic integers for

M . A field basis B = {β1, β2, . . . , βn} for M/Q is an integral basis if every α ∈ OL is uniquely representable

in the form a1β1 + a2β2 + · · ·+ anβn with an ∈ Z. It is well known [8] that for any field basis consisting of

integers A = {α1, α2, . . . , αn} and and integral basis B = {β1, β2, . . . , βn} of a number ring OM that the

discriminants are related by

disc(A) = N2disc(B)

where N ∈ Z.

Any two integral bases have the same discriminant and this discriminant can be considered an invariant

of the ring OM so δM/Q = disc(B). We will use this to find integral bases for the rings of integers of

L = Q(ζ, 3√m1, 3

√m2) and all its subfields. If we can find a basis which has the same discriminant as the

ring then we have found an integral basis. Marcus [8] provides a basis for the ring of integers of the pure

cubic field Q( 3√m) where m = ab2:

1, 3√m,

3√m2

b if m �≡ ±1 (mod 9)

1, 3√m,

(b2±b2 3√mi+ 3√mi2)

3b if m ≡ ±1 (mod 9).

The theorems in this section will provide an integral basis for the rings of integers of Ki, K and L.

Theorem 3.1 Let mi ∈ Z with mi = ab2 where a and b are relatively prime and square free. Let OKi be

the ring of integers of Ki = Q( 3√mi, ζ). Then an integral basis for OKi is

A1 if mi ≡ 2, 4, 5 or 7 (mod 9)

A2 if mi ≡ 0 (mod 3)

A3 if mi ≡ ±1 (mod 9)

,

4

Chapter 3. Integral Basis for Ki, K and L 5

The Ai’s are defined as follows:

A1 = {1, ζ, 3√mi, (1 − ζ) (1∓2 3√mi+ 3√mi

2/b)3 , ζ 3

√mi,

1−ζ±4 3√mi±2ζ 3√mi+ 3√mi2/b+2ζ 3√mi

2/b3 }

where ± or ∓ correspond as

top sign if mi ≡ 4 or 7 (mod 9)

bottom sign if mi ≡ 2 or 5 (mod 9)

A2 = {1, ζ, 3√mi, (1 − ζ)

3√mi2

3b , ζ 3√mi,

3√mi2+2ζ 3√mi

2

3b }

A3 = {1, ζ, 3√mi,

1± 3√mi+ 3√mi2/b

3 , 13 (2 + ζ ± 3

√mi ∓ ζ 3

√mi), ζ

(1± 3√mi+ 3√mi

2/b3

)}

where ± or ∓ correspond as mi ≡ ±1 (mod 9).

Proof: To show that the Ai’s are integral bases for the ring of integers we will use the relationship

between the different, ∆Ki/Q, and the discriminant, δKi/Q, which is given by result P. in section 13.2 of

Ribenboim [10]:

NKi/Q(∆Ki/Q) = δKi/Q.

To calculate the different we need to know which primes are ramified from Q to Ki. Then we know from

result O. in section 13.2 of [10] that ∆Ki/Q =∏QsQ where Q is any nonzero prime ideal of OKi . Let eQ

be the ramification index of Q in Ki/Q then sQ ≥ eQ − 1. Moreover, sQ = eQ − 1 if and only if q = Q⋂

Z

does not divide the ramification index eQ.

We begin by looking at those primes p which are ramified in Ki but are not 3. If p ≡ 2 (mod 3) then

in ki we can factor (p) = p3 with e = 3 and f = 1. In Ki, (p) = P 3 with e = 3 and f = 2. If we look at

norms we get that NKi/Q(P ) = p2. The relation on the different P 3−1 | ∆Ki/Q follows from the formula

and applying norms we get that NKi/Q(P ) | δKi/Q =⇒ p4 | δKi/Q

We can perform a similar analysis on p ≡ 1 (mod 3). In ki we can factor (p) = p3 with e = 3 and

f = 1 and in Ki it factors as (p) = (P1P2)3 with e = 3 and f = 1. If we look at norms we get that

NKi/Q(P1) = NKi/Q(P2) = p. The relation on the different (P1P2)3−1 | ∆Ki/Q is clear and applying norms

we get that NKi/Q(P 21P

22 ) | δKi/Q =⇒ p4 || δKi/Q.

For the prime 3 we need to consider 3 different cases. If mi ≡ ±1 (mod 9) then in ki we can factor

(3) = p21p2. In Ki (3) = (P1P2P3)2 with e = 2 and f = 1. If we look at norms we get that NKi/Q(Pi) = 3

for i = 1, 2, and 3. Since the ramification index is relatively prime to the characteristic of OKi/Pj we have

that (P1P2P3)2−1 | ∆Ki/Q and applying norms we get that NKi/Q(P1P2P3) | δKi/Q =⇒ 33 | δKi/Q. Since

disc(A1) = 33∏q4i , where the qi’s are the ramified primes relatively prime to 3, then A1 is an integral

basis for Ki.


If mi ≡ 2, 4, 5 or 7 (mod 9) then in ki we can factor (3) = p3 and in Ki as (3) = P 6 with e = 6,

f = 1 and NKi/Q(P ) = 3. Since the ramification index from Ki to Q is not relatively prime to the

characteristic of OKi/P we will look at intermediate fields. From ki to Ki we can factor (p) = P 2

and the ramification index is relatively prime to the characteristic so (P )2−1 | ∆Ki/ki. We know that

∆ki/Q = (3A2), where A = ( 3√ab2,

3√a2b), so we can use the product formula for differents to get that

∆Ki/Q = ∆Ki/ki· ∆ki/Q =⇒ P · 3A2 | ∆Ki/Q. Applying norms we get that NKi/Q(P · 3A2) | δKi/Q =⇒

37 | δKi/Q. Since disc(A2) = 37∏q4i and then A2 is an integral basis for Ki.

If mi ≡ 0 (mod 3) then the factorization of 3 in the fields is the same as the previous case. The

only thing which changes is ∆ki/Q. In this case we can always choose mi such that mi = 3m∗ where

3 � m∗ because if 9 | mi then we can choose mi = m2i /27. We then get that ∆ki/Q = (3A2), where

A = ( 3√

3ab2, 3√

9a2b), and when we use the product formula ∆Ki/Q = ∆Ki/ki· ∆ki/Q we get P · 3A2 |

∆Ki/Q =⇒ NKi/Q(P · 3A2) | δKi/Q =⇒ 3 · 36 · 34 = 311 | δKi/Q. Since disc(A3) = 311∏q4i then A3 is an

integral basis for Ki.

Theorem 3.2 Let m1,m2 ∈ Z with mi = aib2i where ai and bi are relatively prime and square free. Let

OK be the ring of integers of K = Q( 3√m1, 3

√m2). Then an integral basis for OK is

A1 if m1 ≡ 1 (mod 9) and m2 ≡ ±1 (mod 9)

A2 if m1 ≡ 1 (mod 9) and m2 ≡ 2, 4, 5, or 7 (mod 9)

A3 if m1 ≡ −1 (mod 9) and m2 ≡ 2, 4, 5, or 7 (mod 9)

A4 if m1 ≡ ±1 (mod 9) and m2 ≡ 0 (mod 3)

A5 if m1 ≡ 2, 4, 5, or 7 (mod 9) and m2 ≡ 0 (mod 3).

Let d1 = GCD(a1, a2), d2 = GCD(a1, b2), d3 = GCD(b1, a2), and d4 = GCD(b1, b2) and the Ai’s are

defined as follows:

A1 = {1, 3√m1,

13

(1 + 3

√m1 +

3√m12

b1

), 3√m2,

13

(1 ± 3

√m2 +

3√m22

b2

), 1

3

(1 − 3

√m1 ∓ 3

√m2 ± 3√m1m2

d2d3d4

),

13

(3√m2 +

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4

), 1

3

(3√m1 +

3√m1m2

d2d3d4+

3√

m1m22

b2d1d3d4

),

19

(1 + 3

√m1 +

3√m12

b1+ 3

√m2 +

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4+

3√m22

b2+

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3

)}

where ± or ∓ correspond as m2 ≡ ±1 (mod 9).

A2 = {1, 3√m1,

13

(1 + 3

√m1 +

3√m12

b1

), 3√m2,

3√m22

b2, 1

3

(3√m2 +

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4

),

3√

m1m22

b2d1d3d4, 1

3

(3√m2

2

b2+

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3

), α}


where α =

13

(2 − 2 3

√m1 − 2 3

√m2 −

3√

m21m2

b1d1d2d4−

3√

m1m22

b2d1d3d4−

3√

m21m

22

b1b2d1d2d3

)if m2 ≡ 4 or 7 (mod 9)

13

(2 − 2 3

√m1 − 3

√m2 +

3√

m21m2

b1d1d2d4−

3√

m1m22

b2d1d3d4−

3√

m21m

22

b1b2d1d2d3

)if m2 ≡ 2 or 5 (mod 9)

A3 = {1, 3√m1,

13

(1 − 3

√m1 +

3√m12

b1

), 3√m2,

3√m22

b2, 1

3

(3√m2 − 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4

),

3√

m1m22

b2d1d3d4, 1

3

(3√m2

2

b2−

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3

), α}

where α =

13

(− 3√m1 − 3√m1

2

b1+ 3

√m2 +

3√m1m2

d2d3d4− 3√m2

2

b2+

3√

m21m

22

b1b2d1d2d3

)if m2 ≡ 4 or 7 (mod 9)

13

(3√m1 +

3√m12

b1+ 3

√m2 +

3√m1m2d2d3d4

+3√m2

2

b2−

3√

m21m

22

b1b2d1d2d3

)if m2 ≡ 2 or 5 (mod 9)

A4 = {1, 3√m1,

13

(1 ± 3

√m1 +

3√m12

b1

), 3√m2,

3√m22

b2,

3√m1m2

d2d3d4, 1

3

(3√m2 ± 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4

),

13

(∓ 3√m2

2

b2+

3√

m1m22

b2d1d3d4

), 1

3

(∓

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3

)}

where ± or ∓ correspond as

top sign if m1 ≡ 1 (mod 9)

bottom sign if m1 ≡ 8 (mod 9)

A5 = {1, 3√m1,

3√m12

b1, 3√m2,

3√m22

b2,

3√m1m2

d2d3d4, 1

3

(3√m2

2

b2±

3√

m1m22

b2d1d3d4

),

13

(1 ± 2 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4

), 1

3

(3√m2

2

b2± 2 3

√m1m2

2b2d1d3d4

+3√

m21m

22

b1b2d1d2d3

)}

where ± corresponds as

+ if m2 ≡ 2 or 5 (mod 9)

− if m2 ≡ 4 or 7 (mod 9).

Proof:

Note: If neither m1 nor m2 ≡ 0 (mod 3) then at least one of m1 or m2 ≡ 1 or 8 (mod 9). If both m1

and m2 ≡ 8 (mod 9) then, without loss of generality, we can choose m1 ≡ m1m2 ≡ 1 (mod 9).

We will first show that all the terms in the basis elements are integers. Certainly 3√m2

i = 3√a2i b

4i =

bi3√a2i bi, so

3√

m2i

biis an algebraic integer in the ring of integers of K for i = 1, 2. Similarly 3

√m1m2 =

3√a1b21a2b22 and d2 = GCD(a1, b2), d3 = GCD(b1, a2), and d4 = GCD(b1, b2) are all perfect cubes so

3√m1m2

d2d3d4is an algebraic integer of K. We can similarly show that for

3√

m21m2

b1d1d2d4,

3√

m1m22

b2d1d3d4and

3√

m21m

22

b1b2d1d2d3the

denominators are those elements that are perfect cubes in their respective products and hence the fractions

are algebraic integers of K.

To show that the Ai’s are integral bases for the ring of integers we will use the relationship between


the different, ∆K/Q, and the discriminant, δK/Q, which is given by result P. in section 13.2 of [10]:

NK/Q(∆K/Q) = δK/Q.

To calculate the different we only need to know which primes are ramified from Q to K. Then we know

from result O. in section 13.2 of [10] that ∆K/Q =∏QsQ where Q is any nonzero prime ideal of OK . Let

eQ be the ramification index of Q in K/Q then sQ ≥ eQ − 1. Moreover, sQ = eQ − 1 if and only if the

characteristic of OK/Q does not divide the ramification index eQ.

This last statement tells us for any prime Q � 3 that sQ = eQ − 1 because eQ = 3j , j = 0 or 1 as Q is

unramified or ramified respectively. Then for those primes that are unramified QeQ−1 = Q1−1 = 1. For

those primes Q ∈ OK that are ramified there will be exactly one of the cubic subfields ki where NK/ki(Q)

is unramified. Without loss of generality let that field be k1 and we will consider the three cases. If q ≡ 2

(mod 3) is a rational prime then in k1 it splits as q = q1q2 and in K it ramifies as q = (Q1Q2)3 where the

degree of Q1 is 2 and of Q2 is 1. Then QeQj

−1

j = Q3−1j = Q2

j for j = 1 and 2 where NK/Q(Q1) = q2 and

NK/Q(Q2) = q, thus q6 | δK/Q. If q ≡ 1 (mod 3) then either q stays prime in k1 or splits completely. If q

stays prime then q = Q3 in K and QeQ−1 = Q3−1 = Q2 where NK/Q(Q) = q3. If q splits completely in

k1 then q = (Q1Q2Q3)3 in K and QeQj

−1

j = Q3−1j = Q2

j where NK/Q(Qj) = q for j = 1, 2 and 3. Thus

q6 | δK/Q for the last two cases as well.

For Q = 3 the problem is more complicated and we will look at different cases.

Case 1: m1 and m2 ≡ ±1 (mod 9).

In this case m3 and m4 ≡ ±1 (mod 9) as well. In ki, 3 factors as (3) = p2ia

pib for i = 1, 2, 3, 4 and in

K as (3) = P 21P

22P

23P

24P5. Since the ramification indices 2 and 1 are relatively prime to the characteristic

of OK/Pj then

P 2−11 P 2−1

2 P 2−13 P 2−1

4 P 1−15 | ∆K/Q =⇒ P1P2P3P4 | ∆K/Q

and NK/Q(Pi) = 3 for i = 1, 2, 3, 4. Using the norm relationship between the different and the discriminant

gives us that

NK/Q(P1P2P3P4) | NK/Q(∆K/Q) =⇒ 34 | δK/Q.

Then the lower bound on the discriminant is δK/Q ≥ 34∏q6i where the qi’s are the primes that ramify in

K but are not 3. Since disc(A1) = 34∏q6i then A1 must be an integral basis for OK .

Case 2: m1 ≡ ±1 (mod 9) and m2 ≡ 2, 4, 5, or 7 (mod 9).

The factorization of 3 in the real subfields is different for k1 and k2 in this case. In k1, (3) = p21p2 and in

k2, (3) = p3. In K we have (3) = P 61P

32 where P1 lies over p1, P2 lies over p2, p = P 2

1P2 and NK/Q(Pi) = 3.


It is well known that 3 || ∆k2/Q. Since p = P 21P2 in K and the ramification index is relatively prime to 3

then we have that

P 2−11 p1−1

2 || ∆K/k2 =⇒ P1 || ∆K/k2 .

Using the multiplicative property of the different we get that 3P1 || ∆K/Q. Using the norm relationship

between the different and the discriminant gives us that

NK/Q(3P1) || NK/Q(∆K/Q) =⇒ 310 || δK/Q.

Since disc(A2) = disc(A3) = 310∏q6i then A2 and A3 are integral bases for OK when m1 ≡ 1 (mod 9) or

m1 ≡ 8 (mod 9) respectively.

Case 3: m1 ≡ ±1 (mod 9) and m2 ≡ 0 (mod 3).

In this case we can factor (3) in k1 as (3) = p21p2, in k2 as (3) = p3 and in K as (3) = P 6

1P32 . From

k1 to K, (pi) = P 3i for i = 1, 2 and from k2 to K, (p) = P 2

1P2 with NK/Q(P1) = 3. We will calculate the

different of K/k2 since 3 is relatively prime to the ramification index of p. Again for k2/Q we know that

3 3√

32 || ∆k2/Q and for K/k2 we have that P 2−1

1 p1−12 || ∆K/k2 =⇒ P1 || ∆K/k2 so using the multiplicative

property we get

NK/Q(3 3√

32P1) || NK/Q(∆K/Q) =⇒ 316 || δK/Q.

Since disc(A4) = 316∏q6i then A4 is an integral basis for OK .

Case 4: m1 ≡ 2, 4, 5, or 7 (mod 9) and m2 ≡ 0 (mod 3).

Since 3 is totally ramified in k1, k2 and K then (3) = p3i in ki for i = 1, 2 and (3) = P 9 in K. For K/k2,

p2 = P 3 and for K/k1, p1 = P 3 so 3 is wildly ramified in both cases. Here we have that NK/Q(P ) = 3.

Using k2 as the intermediate field we will calculate: ∆K/Q = ∆K/k2 · ∆k2/Q. We already know that

∆k2/Q = 3 3√m2

2 and since 3 | m2 then we have that 3 3√

32 | ∆k2/Q. For K/k2 we know that p2 is wildly

ramified so at least P 3 | ∆K/k2 . We get the relationship:

P 3 · 3 3√

32 | ∆K/k2 · ∆k2/Q.

The norm relationship between the different and the discriminant gives us that

NK/Q(P 3 · 3 3√

32) | NK/Q(∆K/Q) =⇒ 318 | δK/Q

and the discriminant of the basis A5 is 318∏q6i so A5 must be an integral basis for OK .

Theorem 3.3 Let m1 and m2 be as in Theorem 3.2. Let OL be the ring of integers of L = Q(ζ, 3√m1, 3

√m2)

then an integral basis for OL is


A1 if m1 ≡ 1 (mod 9) and m2 ≡ ±1 (mod 9)

A2 if m1 ≡ 1 (mod 9) and m2 ≡ 2, 4, 5, or 7 (mod 9)

A3 if m1 ≡ −1 (mod 9) and m2 ≡ 2, 4, 5, or 7 (mod 9)

A4 if m1 ≡ ±1 (mod 9) and m2 ≡ 0 (mod 3)

A5 if m1 ≡ 2 or 5 (mod 9) and m2 ≡ 0 (mod 3)

A6 if m1 ≡ 4 or 7 (mod 9) and m2 ≡ 0 (mod 3)

where d1 = GCD(a1, a2), d2 = GCD(a1, b2), d3 = GCD(b1, a2), and d4 = GCD(b1, b2) and the Ai’s are as

follows:

A1 = {1, 3√m1,

13 (1 + 3

√m1 +

3√m12

b1), 3√m2,

1± 3√m2+ 3√m22/b2

3 , 13 (1 − 3

√m1 ∓ 3

√m2 ± 3√m1m2

d2d3d4),

13 ( 3√m2 +

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4), 1

3 ( 3√m1 +

3√m1m2

d2d3d4+

3√

m1m22

b2d1d3d4),

19 (1 + 3

√m1 +

3√m12

b1+ 3

√m2 +

3√m1m2d2d3d4

+3√

m21m2

b1d1d2d4+

3√m22

b2+

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3)

ζ, 13 (1 − ζ)(1 − 3

√m1), 1

3ζ(1 + 3√m1 +

3√m12

b1), 1

3 (1 − ζ)(1 ∓ 3√m2), ζ 1

3 (1 ± 3√m2 +

3√m22

b2),

ζ 13 (1 − 3

√m1 ∓ 3

√m2 ± 3√m1m2

d2d3d4), 1

9 (1 − ζ)(1 + 3√m1 +

3√m12

b1∓ 3

√m2 ∓ 3√m1m2

d2d3d4∓

3√

m21m2

b1d1d2d4),

19 (1 − ζ)(1 − 3

√m1 ± 3

√m2 ∓ 3√m1m2

d2d3d4+

3√m22

b2−

3√

m1m22

b1d1d3d4),

19ζ(1 + 3

√m1 +

3√m12

b1± 3

√m2 ± 3√m1m2

d2d3d4± 3

√m2

1m2

b1d1d2d4) +

3√m22

b2+

3√

m1m22

b1d1d3d4) +

3√

m21m

22

b1b2d1d2d3)}

where ± or ∓ correspond as m2 ≡ ±1 (mod 9).

A2 = {1, 3√m1,

13 (1+ 3

√m1 +

3√m12

b1), 3√m2,

3√m22

b2, 1

3 (∓ 3√m1± 3√m1

2

b1− 3√m2 +

3√m1m2

d2d3d4∓ 3√m2

2

b2±

3√

m1m22

b2d1d3d4),

13 ( 3√m2 +

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),

3√

m1m22

b2d1d3d4, 1

3 (3√m2

2

b2+

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3), ζ, 1

3 (1−ζ− 3√m1 +ζ 3

√m1),

ζ 13 (1 + 3

√m1 +

3√m12

b1), ζ 3

√m2,

13 (1 − ζ)(1 ± 3

√m2 +

3√m22

b2), ζ 1

3 ( 3√m2 +

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),

ζ 13 (∓ 3

√m1 ± 3√m1

2

b1− 3

√m2 +

3√m1m2

d2d3d4∓ 3√m2

2

b2± 3

√m1m2

2b2d1d3d4

, α, β}

where ± and ∓ corresponds as

top sign if m2 ≡ 4 or 7 (mod 9)

bottom sign if m2 ≡ 2 or 5 (mod 9)

and α =

13 (1 − ζ + 2 3

√m1 − 2ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1+ 2

3√m1m2

d2d3d4− 2ζ

3√m1m2

d2d3d4+ 2

3√

m21m2

b1d1d2d4−

2ζ3√

m21m2

b1d1d2d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3) if m2 ≡ 4, 7 (mod 9)

13 (1 − ζ − 3

√m1 + ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1+

3√m1m2

d2d3d4− ζ

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4− ζ

3√

m21m2

b1d1d2d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3) if m2 ≡ 2, 5 (mod 9)


and β =

19 (1 − ζ + 3

√m1 − ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1+ 3

√m2 − ζ 3

√m2 +

3√m1m2

d2d3d4− ζ

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4−

ζ3√

m21m2

b1d1d2d4+

3√m22

b2− ζ

3√m22

b2+

3√

m1m22

b2d1d3d4− ζ

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3)

if m2 ≡ 4, 7 (mod 9)

19 (1 − ζ − 3

√m1 + ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1− 3

√m2 + ζ 3

√m2 − 3√m1m2

d2d3d4+ ζ

3√m1m2

d2d3d4− 3

√m2

1m2

b1d1d2d4+

ζ3√

m21m2

b1d1d2d4+

3√m22

b2− ζ

3√m22

b2+

3√

m1m22

b2d1d3d4− ζ

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3)

if m2 ≡ 2, 5 (mod 9)

.

A3 = {1, 3√m1,

13 (1− 3

√m1 +

3√m12

b1), 3√m2,

3√m22

b2, 1

3 (− 3√m1− 3√m1

2

b1± 3√m2± 3√m1m2

d2d3d4+

3√m22

b2+

3√

m1m22

b2d1d3d4,

13 ( 3√m2− 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),

3√

m1m22

b2d1d3d4, 1

3 (3√m2

2

b2−

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3), ζ, 1

3 (2+ζ+2 3√m1 +ζ 3

√m1),

ζ 13 (1 − 3

√m1 +

3√m12

b1), ζ 3

√m2,

13 (1 − ζ)(1 ± 3

√m2 +

3√m22

b2), ζ 1

3 ( 3√m2 − 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),

ζ 13 (− 3

√m1 − 3√m1

2

b1± 3

√m2 ± 3√m1m2

d2d3d4+

3√m22

b2+

3√

m1m22

b2d1d3d4, α, β}

where ± and ∓ corresponds as

top sign if m2 ≡ 4 or 7 (mod 9)

bottom sign if m2 ≡ 2 or 5 (mod 9)

and α =

13 (1 − ζ − 2 3

√m1 + 2ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1− 2

3√m1m2

d2d3d4+ 2ζ

3√m1m2

d2d3d4+ 2

3√

m21m2

b1d1d2d4+

2ζ3√

m21m2

b1d1d2d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3) if m2 ≡ 4, 7 (mod 9)

13 (1 − ζ + 3

√m1 − ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1− 3√m1m2

d2d3d4+ ζ

3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4−

ζ3√

m21m2

b1d1d2d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3) if m2 ≡ 2, 5 (mod 9).

and β =

19 (2 − 2ζ + 3

√m1 − ζ 3

√m1 + 2

3√m12

b1+ 2ζ

3√m12

b1+ 2 3

√m2 − 2ζ 3

√m2 +

3√m1m2

d2d3d4− ζ

3√m1m2

d2d3d4+ 2

3√

m21m2

b1d1d2d4−

2ζ3√

m21m2

b1d1d2d4+ 2

3√m22

b2− 2ζ

3√m22

b2+

3√

m1m22

b2d1d3d4− ζ

3√

m1m22

b2d1d3d4+ 2

3√

m21m

22

b1b2d1d2d3) − 2ζ

3√

m21m

22

b1b2d1d2d3)

if m2 ≡ 4, 7 (mod 9)

19 (1 − ζ + 5 3

√m1 − 5ζ 3

√m1 +

3√m12

b1− ζ

3√m12

b1− 3

√m2 + ζ 3

√m2 − 5

3√m1m2

d2d3d4+ 5ζ

3√m1m2

d2d3d4−

3√

m21m2

b1d1d2d4+

ζ3√

m21m2

b1d1d2d4+

3√m22

b2− ζ

3√m22

b2+ 5

3√

m1m22

b2d1d3d4− 5ζ

3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3) − ζ

3√

m21m

22

b1b2d1d2d3)

if m2 ≡ 2, 5 (mod 9)

.

A4 = {1, 3√m1,

13 (1 ± 3

√m1 +

3√m12

b1), 3√m2,

3√m22

b2,

3√m1m2d2d3d4

, 13 ( 3√m2 ± 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),


13 (∓ 3√m2

2

b2+

3√

m1m22

b2d1d3d4), 1

3 (∓3√

m1m22

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3), ζ, ζ 3

√m2,

13 (1−w)

3√m22

b2, 1

3 (1−w∓ 3√m1±ζ 3

√m1),

ζ 13 (1 ± 3

√m1 +

3√m12

b1), ζ

3√m1m2

d2d3d4, ζ 1

3 ( 3√m2 ± 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),

19 (1 − ζ)( 3

√m2 ∓ 2

3√

m1m22

b2d1d3d4+

3√

m21m2

b1d1d2d4), α}

where α =

19 (10

3√m22

b2− 10ζ

3√m22

b2+

3√

m1m22

b2d1d3d4− ζ

3√

m1m22

b2d1d3d4− 2

3√

m21m2

b1d1d2d4+ 2ζ

3√

m21m2

b1d1d2d4)

if m1 ≡ 1 (mod 9)

19 (17

3√m22

b2− 17ζ

3√m22

b2+

3√

m1m22

b2d1d3d4− ζ

3√

m1m22

b2d1d3d4+ 2

3√

m21m2

b1d1d2d4− 2ζ

3√

m21m2

b1d1d2d4)

if m1 ≡ −1 (mod 9)

.

A5 = {1, 3√m1,

13 (1 − ζ)(1 + 2 3

√m1 +

3√m12

b1), 3√m2,

13 (1 − ζ)

3√m22

b2, 1

3 (1 − ζ)( 3√m2 +

3√m1m2

d2d3d4),

13 (

3√m22

b2+

3√

m1m22

b2d1d3d4), 1

3 ( 3√m2 + 2 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4), 1

3 (3√m2

2

b2+ 2 3

√m1m2

2b2d1d3d4

+3√

m21m

22

b1b2d1d2d3), ζ, ζ 3

√m2,

13 (

3√m22

b2+ 2ζ 3√m2

2

b2), ζ 3

√m1,

13 (1 + 2 3

√m1 +

3√m12

b1+ ζ(2 + 4 3

√m1 + 2 3√m1

2

b1)),

13 ( 3√m2 +

3√m1m2

d2d3d4+ ζ(2 3

√m2 + 2 3√m1m2

d2d3d4)), 1

3 (ζ(3√m2

2

b2+

3√

m1m22

b2d1d3d4)),

13 (ζ( 3

√m2 + 2 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4)), α}

where α =

13 (6 + 3ζ + 6 3

√m1 + 3ζ 3

√m1 + 6 3

√m2 + 3ζ 3

√m2 + 5

3√m22

b2+ ζ

3√m22

b2+

3√

m1m22

b2d1d3d4+ 2ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22

b1b2d1d2d3) if m2 ≡ 3 (mod 9) and m1 ≡ 2 (mod 9)

13 (6 + 3ζ + 6 3

√m1 + 3ζ 3

√m1 + 3 3

√m2 + 6ζ 3

√m2 + 5

3√m22

b2+ ζ

3√m22

b2+

3√

m1m22

b2d1d3d4+ 2ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


13 (3 + 6ζ + 3 3

√m1 + 6ζ 3

√m1 + 6 3

√m2 + 3ζ 3

√m2 + 5

3√m22

b2+ ζ

3√m22

b2+

3√

m1m22

b2d1d3d4+ 2ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


13 (3 + 6ζ + 3 3

√m1 + 6ζ 3

√m1 + 3 3

√m2 + 6ζ 3

√m2 + 5

3√m22

b2+ ζ

3√m22

b2+

3√

m1m22

b2d1d3d4+ 2ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


.

and A6 = {1, 3√m1,

3√m12

b1, 3√m2,

3√m22

b2,

3√m1m2

d2d3d4, 1

3 (3√m2

2

b2+

3√

m1m22

b2d1d3d4), 1

3 ( 3√m2 − 2 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4),

13 (

3√m22

b2− 2 3√m1 3√m2

2

b2d1d3d4+

3√

m21m

22

b1b2d1d2d3), ζ, ζ 3

√m2,

13 (1 − ζ)

3√m22

b2, ζ 3

√m1,

13 (1 − ζ)(1 − 2 3

√m1 +

3√m12

b1), 1

3 (1 − ζ)(− 3√m2 +

3√m1m2

d2d3d4), 1

3 (ζ)( 3√m2

2 −3√

m1m22

b2d1d3d4),

13ζ( 3

√m2 − 2 3√m1m2

d2d3d4+

3√

m21m2

b1d1d2d4), α}


were α =

13 (6 + 3ζ + 3 3

√m1 + 6ζ 3

√m1 + 6 3

√m2 + 3ζ 3

√m2 + 5

3√m22

b2+ 4ζ

3√m22

b2+ 8

3√

m1m22

b2d1d3d4+ 4ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


13 (6 + 3ζ + 3 3

√m1 + 6ζ 3

√m1 + 3 3

√m2 + 6ζ 3

√m2 + 5

3√m22

b2+ 4ζ

3√m22

b2+ 8

3√

m1m22

b2d1d3d4+ 4ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


13 (3 + 6ζ + 6 3

√m1 + 3ζ 3

√m1 + 3 3

√m2 + 6ζ 3

√m2 + 5

3√m22

b2+ 4ζ

3√m22

b2+ 8

3√

m1m22

b2d1d3d4+ 4ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


13 (3 + 6ζ + 6 3

√m1 + 3ζ 3

√m1 + 6 3

√m2 + 3ζ 3

√m2 + 5

3√m22

b2+ 4ζ

3√m22

b2+ 8

3√

m1m22

b2d1d3d4+ 4ζ

3√

m1m22

b2d1d3d4+

53√

m21m

22

b1b2d1d2d3) + ζ

3√

m21m

22


Proof: The proof follows much the same as the basis for K. We will determine the discriminant of the

number field L/Q by first finding the different ∆L/Q and then applying NL/Q(∆L/Q) = δL/Q.

We will define our subfields in the usual way: L = Q(ζ, 3√m1, 3

√m2), Ki = Q(ζ, 3

√mi), ki = Q( 3

√mi)

and k = Q(ζ) where mi is cube free for i = 1, 2, 3 and 4, and then we will look at the factorization of all

ramified primes in L. We can write

δL/Q = 3a∏

pbl (3.1)

where pl �= 3. We begin with those primes p �= 3 and we will show that b = 12 by considering the different

cases based on the congruence of p (mod 3). For some mi we know that p � mi so without loss of generality

we will always choose that mi to be m1

Case 1: p ≡ 2 (mod 3)

We can factor p in each of the subfields as follows:

in k (p) = π where f = 2 and e = 1,

in k1 (p) = p1p2 where f1 = 2 and f2=1 and e = 1 ,

in K1 (p) = P1P2P3 where fj = 2 for j = 1, 2, 3 and e = 1,

in L (p) = (P1P2P3)3 where fj = 2 for j = 1, 2, 3 and e = 3,

where NL/Q(Pj) = p2 for j = 1, 2, 3. In this case the characteristic p does not divide ep = 3 so we have

that

(P1P2P3)3−1 || ∆L/Q =⇒ NL/Q[(P1P2P3)2] = p12 || δL/Q.

For p ≡ 1 (mod 3) we have to consider two possibilities, either p stays prime in k1 or p splits completely.

Case 2: p ≡ 1 (mod 3) and stays prime in k1.



in k1 (p) = p where f = 3 and e = 1,

in K1 (p) = P1P2 where fj = 3 for j = 1, 2 and e = 1,

in L (p) = (P1P2)3 where fj = 3 for j = 1, 2 and e = 3,

where NL/Q(Pj) = p3 for j = 1, 2. In this case p does not divide ep = 3 so we have that

(P1P2)3−1 || ∆L/Q =⇒ NL/Q[(P1P2)2] = p12 || δL/Q.

Case 3: p ≡ 1 (mod 3) and splits completely in k1.


in k1 (p) = p1p2p3 where fj = 1 and ei = 1 for j = 1, 2, 3 ,

in K1 (p) = P1P2P3P4P5P6 where fj = 1 and ej = 1 for j = 1, 2, . . . , 6

in L (p) = (P1P2 . . .P6)3 where fj = 1 and ei = 3 for j = 1, 2, . . . , 6 ,

where NL/Q(Pj) = p for j = 1, 2, . . . , 6. In this case p does not divide ep = 3 so we have that

(P1P2P3P4P5P6)3−1 || ∆L/Q =⇒ NL/Q[(P1P2P3P4P5P6)2] = p12 || δL/Q.

In each case we have that p12 || δL/Q so b = 12 in equation (3.1).

For p = 3 we have to consider 4 different cases based on the congruences of m1 and m2 (mod 9). In

all these cases we will let qi be those primes that ramify in L but are different from 3.

Case 1: m1 and m2 ≡ ±1 (mod 9).

In this case m3 and m4 ≡ ±1 (mod 9) as well. We can factor (3) in each of the subfields as follows:

in ki (3) = p21p2 where fj = 1 for j = 1, 2,

in Ki (3) = (P1P2P3)2 where fj = 1 for j = 1, 2,

in L (3) = (P1P2 . . .P9)2 where fj = 1 and ei = 2 for j = 1, 2, . . . , 9 ,

where NL/Q(Pj) = 3 for j = 1, 2, . . . , 9. In this case 3 does not divide ep = 2 so we have that

(P1P2 . . .P9)2−1 || ∆L/Q =⇒ NL/Q[(P1P2 . . .P9)] = 39 || δL/Q

Since disc(A1) = 39∏q12i then A1 must be an integral basis for OL.

Case 2: m1 ≡ ±1 (mod 9) and m2 = 2, 4, 5, or 7 (mod 9).

We can factor (3) in each of the subfields as follows:


in k2 (3) = p3 where f = 1 and e = 3,

in K2 (3) = (P )6 where f = 1 and e = 3,

in L (3) = (P1P2P3)6 where f = 1 and e = 6,

where NL/Q(Pi) = 3. In this case 3 is wildly ramified in L and 3 does not divide m2 so we can apply the

theorem to get a lower bound for the discriminant:

(P1P2P3)6 | ∆L/Q =⇒ NL/Q[(P1P2P3)6] = 318 | δL/Q

Since disc(A2) = disc(A3) = 318∏q12i then A2 and A3 must be integral bases for OL.

Case 3: m1 ≡ ±1 (mod 9) and m2 = 0 (mod 3).


in k1 (3) = p21p2 where fj = 1 for j = 1, 2 and e = 1,

in K1 (3) = (P1P2P3)2 where f = 1 and e = 2,

in k2 (3) = p3 where f = 1 and e = 3,

in K2 (3) = (P )6 where f = 1 and e = 3,

in L (3) = (P1P2P3)6 where f = 1 and e = 6,

where NL/Q(Pi) = 3, NL/K2(P ) = P 3 and NL/Q(P ) = 33. In this case 3 is wildly ramified in L so we will

use the product rule using K2 as the intermediate field to get ∆L/K2 ·∆K2/Q = ∆L/Q. The prime P of K2

is unramified in L so ∆L/K2 is relatively prime to 3 and in the proof of Theorem 3.1 it was shown that

p11 || ∆K2/Q so p11 || ∆L/Q. Then the norm relation implies that

NL/Q[P 11] = 333 | δL/Q

and since disc(A4) = 333∏q12i then A4 must be an integral basis for OL.

Case 4: m1 ≡ 2, 4, 5, or 7 (mod 9) and m2 = 0 (mod 3).


in k1 (3) = p3 where f = 1 and e = 3,

in K (3) = P 9 where f = 1 and e = 9,

in L (3) = (P)18 where f = 1 and e = 18,

in L (P ) = P2 where f = 1 and e = 2,

and NL/Q(P) = 3; NL/Q(P ) = 32 and NL/Q(Q) = 3.

We know from the proof of Theorem 3.2 that P 18 || ∆K/Q and from K to L we know that P | ∆L/K .


By the product rule we get

P · P 18 | ∆L/K · ∆K/Q =⇒ P37 | ∆L/Q =⇒ NL/Q[P37] = 337 | δL/Q

and since disc(A5) = disc(A6) = 337∏q12i then A5 and A6 must be integral bases for OL.

Chapter 4

Unit Group of Ki

4.1 Types of Cubic Fields

Let ki = Q( 3√mi) be a pure cubic field with unit group < εi >. Vornoi’s algorithm [12] will quickly generate

the element εi. In [1] Barrucand and Cohn give a classification of the pure cubic fields and their normal

closures. In [9] Parry showed that Type II fields do not exist and consequently simplified the definitions

of the remaining Types. Since Nki/Q(εi) = 1 we can write εi = Bi

Bσ2i

, where Bi is a primitive integer in Ki,

then the fields ki and Ki are of

Type I if N(Bi) = 1

Type III if N(Bi) is not a unit

Type IV if N(Bi) = ζa where a = 1 or 2.

An integer Bi of Ki is said to be primitive if it is not divisible by any integer of k other than roots of unity.

To find the Type of the subfield it is necessary to calculate the element Bi. Since the calculation of Bi

is the same in all the fields we will drop the subscript and simply write B, ε and σ.

4.2 Calculation of B

Let ki = Q( 3√mi) and let ε be the fundamental unit for ki then it is known from Hilbert’s Theorem 90 that

there exists B, a unique (up to multiplication by roots of unity in k) primitive integer of Ki, such that

ε = BBσ . To find B we start with a solution of the form β = 1 + ε+ εσε which gives us β = Bα where α ∈ k

and B ∈ Ki is primitive. In Barrucand and Cohn [1] it is shown that NKi/Q(β) = [3 + tr(ε) + tr(1/ε)]3

17

Chapter 4. Unit Group of Ki 18

where tr is the trace for ki/Q. Since β = Bα then b31 = NKi/Q(β) = NKi/Q(B)(Nk/Q(α))3 and hence

NKi/Q(B) is a cube in Z. Let b31 = brbk where br is divisible only by ramified primes and bk is divisible

by no ramified primes. We know that B is divisible only by ramified primes so bk = NKi/Q(α1), where

α1 | α. To find α1 we use the gcd algorithm for the third cyclotomic field on NKi/k(β) = NKi/k(B)(α3)

and bk. We can now easily find B∗ = βα1

with B∗ ∈ Ki. At this point B∗ will be divisible by units of Ki

and may be divisible by ramified primes. If NKi/k(B∗) = 1, ζ, ζ2 and the field Ki is either Type I or IV

and B = B∗ is a unit of Ki. If B∗ is not a unit then more reduction may be needed.

Let γ = NKi/k(B∗) and write γ = 3a0πa11 πa2

2 · · ·πavv p

av+1v+1 · · · pan

n where πj for 1 ≤ j ≤ v is a prime

divisor in k of a rational prime congruent to 1 modulo 3 and pj ≡ 2 (mod 3) for v + 1 ≤ j ≤ n. It was

shown above that NKi/Q(B) is a cube in Z. Let p ≡ 2 (mod 3) with pa || NKi/Q(B∗) and P the prime

divisor of p in Ki. Since NKi/Q(P ) = p2 if follows that 6 | a. Suppose that P b || B∗ then p2b || NKi/Q(B∗)

so 2b = a = 6c =⇒ b = 3c and so pc = P 3c || B∗. Hence all primes pj ≡ 2 (mod 3) can be removed from

B∗ by dividing B∗ by paj/3j .

The same is true for 3 since a prime divisor of 3 will only divide B if m �≡ ±1 (mod 9). Hence 3 has

only one prime divisor P3 in Ki and (1 − ζ)2 = (3) = P 63 . As above if P b

3 || B∗ then b is divisible by 3 so

(1 − ζ)b/3 | B∗ and can be removed.

For a prime p ≡ 1 (mod 3) let p = ππ in k then (π) = P 3 and (π) = P 3 for distinct primes P and

P of Ki. If P b || B∗ then π�b/3� | B∗ and can be removed. Hence we can assume that the powers of π

and π dividing NKi/k(B∗) are both less than or equal to 2. Since the power of p dividing NKi/Q(B∗) is a

multiple of 3 then if either π or π divides NKi/Q(B∗) that norm must be exactly divisible by ππ2 or π2π.

Now B∗ is primitive so B = B∗.

Now there are 3 cases. If B =v∏

j=1

πaj

j πbj

j where aj + bj = 3 and aj , bj �= 0 then Ki is a Type III field.

If NKi/k(B) = ζ or ζ2 then Ki is Type IV and if NKi/k(B) = 1 then Ki is Type I.

4.3 Unit group for Ki

If Ki is Type III then the only units in Ki are those units in the real subfield ki and their conjugates so

the unit group of K has basis {εi, εσi }.

If Ki is Type I or IV then we can find Bi as above where NKi/k(Bi) = 1, ζ, or ζ2 so Bi is a unit in

Ki which is not a product of the the units of the subfields. Let ui = Bi then the unit group of Ki has the

basis {εi, ui}. Moreover εi

εσ2i

= u3i

NKi/k(ui)so u3

i = ζa εi

εσ2i

for some a = 0, 1 or 2.

Chapter 5

Unit group of K

Let K = Q( 3√m1, 3

√m2) and e be the unit group of K. Then K has 4 fundamental units, e1, e2, e3, e4,

and a basis for e can be chosen in one of four possible ways as described in Theorem VI of [9]. These bases

will distinguish K into 4 Kinds.

(1) (e : e0) = 27 and e1 = ε1, e32 = εa1

1 ε2, e33 = εb11 ε3, e

34 = εc11 ε4

(2) (e : e0) = 9 and e1 = ε1, e2 = ε2, e33 = εa1

1 εa22 ε3, e

34 = εb11 εb22 ε4

(3) (e : e0) = 3 and e1 = ε1, e2 = ε2, e3 = ε3, e34 = εa1

1 εa22 εa3

3 ε4

(4) (e : e0) = 1 and e1 = ε1, e2 = ε2, e3 = ε3, e4 = ε4

To find a basis for the unit group of K we will be able to use the different Types of subfields to

”construct” the basis elements. Theorem VI of [9] tells us the form of the basis elements based on the Kind

of K and Theorem XII and XIII tell us the conditions required for each Kind to occur. The units in K are

dependent on the Type of the subfields and thus we need to look at three cases based on those Types.

To find a new unit in K from Type I fields is the easiest of the three cases. By [9] Theorem IX we know

that for a Type I field ki that we can write the fundamental unit εi = A3i

riwith Ai ∈ ki, ri ∈ Z and ri | 9m2

i .

Since Bi ∈ Ki where εi = Bi

Bσ2i

is a unit of norm 1 we can calculate Ai by using Hilbert’s Theorem 90 we

can write Bi = Ai

Aσi

as we did for the units of Ki. Using this method we see that

εi =Bi

Bσ2

i

=A3

i

(AiAσ2

i Aσi )

=A3

i

N(Ai)

which implies that ri = N(Ai).

Using Barracund and Cohn’s [1] terminology we will define a principal factor of the discriminant ∆ki

to be an element αi ∈ Z such that αi | ∆ki and there exists A ∈ ki such that Nki/Q(A) = αi. For brevity

we will refer to the principal factors of the discriminant simply as ”principal divisors”.

19

Chapter 5. Unit group of K 20

In section 7 of [1] Barrucand and Cohn show that a basis for the principal divisors of ki = Q( 3√mi) can

be constructed from the elements N(Ai) from εi = A3i

N(Ai). Let mi = ab2 with (a, b) = 1 for square-free a

and b then principal divisors for are all of the form

1, ab2, a2b, d, dab2, da2b, d2, d2ab2, d2a2b.

where d = N(Ai).

5.1 Units in K from Type I fields

We will look for products of the principal divisors of the form αa11 αa2

2 αa33 αa4

4 = αp with 0 ≤ ai ≤ 2 whereA3

i

αi= εi such that αp = m3 where m ∈ Z. Then

A = Aa11 Aa2

2 Aa33 Aa4

4 with 0 ≤ ai ≤ 2, i = 1, 2, 3, 4

then

εa11 εa2

2 εa33 εa4

4 =A3

αp=(A

m

)3

. (5.1)

This gives us a new unit in K, e = 3√εa11 εa2

2 εa33 εa4

4 and we choose the largest subscript i with ai �= 0 such

that ai = 1.

It will be advantageous in the calculation of the units of L to choose the unit basis in K such that

whenever possible al = 0, for some l = 1, 2, 3, or 4 the class number of kl is relatively prime to 3. Choose a

maximal independent set elements from the list of all solutions to equation (5.1); this will provide at most

3 elements depending on the kind of K. Since e has four elements we can choose the remaining non-cube

elements as described at the beginning of this chapter.

5.2 Example Type I units in K

Example 1 K = Q( 3√

2, 3√

5)

The cubic subfields k1 = Q( 3√

2), k2 = Q( 3√

5), k3 = Q( 3√

10), and k4 = Q( 3√

20) are all Type I so for

each we can calculate Ai, and the principal divisor αi where εi = A3i

αi.

A1 = 1 + 3√

2, α1 = 3 and ε1 = 1 + 3√

2 + 3√

4

A2 = 4 + 2 3√

5 + 3√

25, α2 = 9 and ε2 = 41 + 24 3√

5 + 14 3√

25

A3 = 13 (4 + 3

√10 + 3

√100), α3 = 2 and ε3 = 1

3 (23 + 11 3√

10 + 5 3√

100)

A4 = 2 + 3√

20 + 3√

50, α4 = 18 and ε4 = 11 + 4 3√

20 + 3 3√

50


Then to find the units in K we look for products of the α’s that are cubes in K.

α1α2 = 33 so e1 = 3√ε1ε2 = A1A2/3 = 1

3 (4 + 4 3√

2 + 2 3√

5 + 2 3√

10 + 3√

25 + 3√

50)

α3 = ( 3√

2)3 so e2 = 3√ε3 = A3/

3√

2 = 13 (2 3

√4 + 3

√5 + 3

√50)

α1α4 = (3 3√

2)3 so e3 = 3√ε1ε4 = A1A4/(3 3

√2) = 1

3 (2 + 3√

4 + 3√

10 + 3√

20 + 3√

25 + 3√

50)

[e : e0] = 33 and e =< ε1, e1, e2, e3 >.


10, 3√

42)

The cubic subfields k1 = Q( 3√

10), k2 = Q( 3√

42), and k3 = Q( 3√

420) are all Type I so for each we can

calculate Ai, and the principal divisor αi and k4 = Q( 3√

525) is Type III.

A1 = 13 (4 + 3

√10 + 3

√100), α1 = 2 and ε1 = 1

3 (23 + 11 3√

10 + 5 3√

100)

A2 = 49 + 14 3√

42 + 4 3√

1764, α2 = 49 and ε2 = 21169 + 6090 3√

42 + 1752 3√

1764

A3 = 12 + 15 3√

420 + 4 3√

22050), α3 = 28 and ε3 = 13 (453610 + 60570 3

√420 + 16176 3

√22050)

Then to find the units in K we look for products of the α’s that are cubes in K.

α1α2α3 = (14)3 so e1 = 3√ε1ε2ε3 = A1A2A3/14 = 1

3 (4208+1952 3√

10+908 3√

100+1213 3√

42+562 3√

420+

1042 3√

525 + 116 3√

1764 + 108 3√

2205 + 50 3√

22050)

No other products produce a cube so [e : e0] = 3 and e =< ε1, ε2, ε4, e1 >.

5.3 Units in K from Type IV fields

If ki is a Type IV field then εi = BBσ2 where N(B) = ζk with k = 1 or 2. We would like to apply Hilbert’s

Theorem 90 as we did in the case for the Type I fields but we need an element of norm 1. We can do this

by adjoining the 9th roots of unity to Ki.

Let ω be a root of x6 + x3 + 1 = 0, Mi = Ki(ω) and F = Q(ω) where ω is chosen so ω3 = ζ, then

NMi/F (ωjB) = ωjB · ωjBσ · ωjBσ2= ω3jNKi/k(B) = ζjζk

If N(B) = ζk then choose j = 3 − k then NMi/F (ωjB) = 1. Then Hilbert’s Theorem 90 can be applied

and ωjB = AAσ where A ∈ Mi is unique up to multiplication by an element α ∈ Q(ω).

Let A0 = 1+ωjB+(ωjB)σ(ωjB) then ωjB = A0Aσ

0and it can be shown that NMi/Ki

(A0) = εε′(3−tr(ε))

where tr is the trace from ki to Q and NKi/k [εε′(3 − tr(ε))] = (3 − tr(ε))3 = NMi/k(A0) = a3m3 with

gcd(a,mi) = 1 and gcd(a,m) = 1. Similarly we can calculate that a0 = NMi/F (A0) = 6+tr(ε)+3tr(B)ω2+

3tr(B · Bσ)ω4 and NF/k [a0] = (3 − tr(ε))3.

Now (A0) = A1A2 where A1 is an ideal of F and A2 is an ideal of Mi that is divisible by no nontrivial

ideals of F . Since hF = 1 then A1 = (At1) is principal and hence A2 = (At2 ) is principal and we can


write A0 = At1At2 as integers. We will say that At2 is primitive. Note that since At1 ∈ F then At1 = Aσt1

and ωjB = A0Aσ

0= At1At2

Aσt1

Aσt2

= At2Aσ

t2so (At2) =

(Aσ

t2

). Since At2 is primitive it can only be divisible by prime

ideals of R that are ramified over F with exponent at most 2. To calculate At2 we use the GCD function

in the 9th cyclotomic field on NMi/F (A0) and NMi/k(A0)/m3 to find At1 . Then At2 = A0/At1 and At2 is

divisible by no prime ideals of F and

εi =B

Bσ2 =ωjB

ωjBσ2 =A3

t2

NMi/F (At2).

Lemma 5.1 Let A be a primitive integer of Mi such that ε = A3/NMi/F (A) and P be a prime ideal of F

with P a || NMi/F (A) and P⋂

Q = p ≡ 1 (mod 9) then P a || NMi/F (A) where P = P τ .

Proof: (We will abbreviate NMi/F (α) = N(α)) Since ε = ε then A3/N(A) = (Aτ )3/N(Aτ ) and we can

rearrange the expression to be (A/Aτ )3 = N(A)/N(Aτ ). Since 3√N(A)/N(Aτ ) is in Mi, Kummer Theory

says that N(A)/N(Aτ ) = mjiβ

3 for some β ∈ F and j = 0, 1, or 2. So we have that N(A) = N(Aτ )mjiβ

3.

Suppose that P b || N(A) then

a ≡ b + jc (mod 3) and b ≡ a + jc (mod 3)

where pc || mi. Adding the two congruences together gives

a + b ≡ a + b + 2jc (mod 3) =⇒ 0 ≡ 2jc (mod 3)

so

a ≡ b (mod 3).

and since 0 ≤ a, b ≤ 2 then a = b

Theorem 5.2 Let A be a primitive integer of Mi such that εi = A3

NMi/F (A) and let R be the maximal real

subfield of F . Then (A) is an ideal of Ri = R( 3√mi). If p �= 3 is a prime that divides NMi/Q(A) then

(p) = P1P2P3 in R and Pj = p3j in Ri. Moreover, subscripts can be assigned so that p2

1p2 || (A) and

p3 � (A).

Proof: Since ki is Type IV, p ≡ ±1 (mod 9) so p splits completely in R. Suppose P is a prime

ideal of Mi lying over p with Pa || (A). If p ≡ 8 (mod 9) then P = p is a prime ideal of Ri so assume

p ≡ 1 (mod 9). Then P a = NMi/F (Pa) || NMi/F (A), by Lemma 5.1 P a || NMi/F (A), hence Pa || (A).

Since p ≡ 1 (mod 9) and P �= P then (PP)a = (p)a || (A) and hence (A) is an ideal of Ri.

Let a0 = NMi/F (A) and p �= 3 be a prime divisor of mi with (p) = P1P2P3 in R, p | NF/Q(a0) and

εi ≡ a + b 3√mi + c 3

√mi

2 (mod 3) where a, b, c ∈ Z. Then the constant term of a0, which we know from


above to be 6 + tr(εi), must be divisible by p so we have that 6 + 3a ≡ 0 (mod p) =⇒ a ≡ −2 (mod p).

Since p | NF/Q(a0) then we have that P b11 P b2

2 P b33 | a0 with b1 + b2 + b3 ≡ 0 (mod 3) since N(a0) is

a cube. If none of the bi’s are equal to zero then we have that p | a0 but we know from above that

Nki/Q(εi) = 1 ≡ a3 (mod p) so (−2)3 ≡ 1 (mod p) =⇒ 9 ≡ 0 (mod p). Clearly this is impossible for

p �= 3 and since A is primitive all the exponents on Pi’s must be less than 3 so for some choice of subscripts

p21p2 || (A).

We know from [9] Theorem XI that if e3 = εa1εb2ε

c3 with 1 ≤ a, b, c ≤ 2 has a solution in K then

k1, k2, k3 are Type IV fields and if e3 = εa1εb2ε

c3ε

d4 with 1 ≤ a, b, c, d ≤ 2 has a solution in K then exactly

three of k1, k2, k3, k4 are Type IV fields and the remaining field is Type I. We will consider cases which

fit these criteria.

To begin we present the field diagram for the Type IV field calculations:

K(ω)

Mi = ki(ω) K(ω + ω8)

F = Q(ω)

k = Q(ζ)

Ri = ki(ω + ω8) K

R = Q(ω + ω8) ki = Q( 3√mi)

Q

For each ki, i = 1, 2, 3, 4 that is Type IV choose Ai to be a primitive integer satisfying the conditions

of Theorem 5.2. If ki is Type I choose Ai to be a primitive integer of ki such that εi = A3i

N(Ai)and if ki

is Type III choose Ai = 1. We will also define ai = N(Ai) where N(Ai) = NMi/F (Ai) for ki Type IV

and N(Ai) = NKi/k(Ai) otherwise. Since at least three of the fields are of Type IV we know that for p

a prime and p | mi then either p ≡ ±1 (mod 9) or p = 3 so the prime factors of m1m2 can be written

3, p1, p2, · · · , pt. Hence each pj factors as Pj1Pj2Pj3 where the Pjk’s are prime ideals of R.


Lemma 5.3 Let k1, k2, k3 be Type IV fields, k4 Type I or IV and suppose

A3e = εb11 εb22 εb33 εb44 (5.2)

has a solution Ae ∈ K where 1 ≤ bi ≤ 2 for i = 1, 2, 3 and 0 ≤ b4 ≤ 2. If

εb11 εb22 εb33 =A3

0

a0(5.3)

where a0 ∈ R and A0 ∈ M4 then ab44 a0 = ml

4α3 for some l = 0, 1, or 2 and α ∈ R.

Proof: Combining equations (5.2) and (5.3) gives (A0Ae

)3 = a0ε−b44 . It follows from [9] Theorem XI that

if k4 is Type IV then b4 = 0 and if k4 is Type I then ε4 = A34

a4. So (A0

Ae)3 = a0( a4

A34)b4 =⇒ (A

b44 A0

Ae)3 = a0a

b44

so Ab44 A0

Ae= 3√a0a

b44 is a cube root of an element of R. Then a0a

b44 = ml

4α3 for some α ∈ F and l = 0, 1,

or 2. Since α3 ∈ R and [F : R] = 2 then one of α, ζα or ζ2α is in R and hence the equation a0ab44 = ml

4α3

has a solution with α ∈ R and the theorem is proved.

Let NK(ω+ω8)/R4 = N , P be a prime divisor lying over 3 in R and e1 = ω + ω8 and e2 = ω4 + ω5 be

fundamental units of R. If we can find a product A0 = Ab11 Ab2

2 Ab33 such that, for each prime pj �= 3 that

divides m1m2 with pj as above, the prime ideal factorization of N(A0) has the form

(t∏

j=1

Pcj1j1

Pcj2j2

Pcj3j3

)P c

and cj1 ≡ cj2 ≡ cj3 (mod 3) then let a0 = N(A0). By Theorem 5.2 N(Ai) ∈ R for i = 1, 2, 3 so a0 ∈ R.

Furthermore if a0ab44 = ml

4α3 as described in Lemma 5.3 and suppose that pj | a0a

b44 where paj exactly

divides a4 and pbj exactly divides m4 then

N(Ae) = N(Ab11 Ab2

2 Ab33 Ab4

4 ) = aoab44 = ml

4α3

and the exponents on the Pjk’s are (cj1 , cj2 , cj3) + b4(a, a, a) ≡ l(b, b, b) + (0, 0, 0) (mod 3) =⇒ cjk

≡l · b − b4 · a (mod 3) for each k. So if there exists a solution to (5.2) then there has to exist at least one

choice of bi’s so that cj1 ≡ cj2 ≡ cj3 ≡ cj (mod 3) for all j = 1, . . . , t. In addition Lemma 5.3 shows that

if (5.2) has a solution in K then the bi’s can be chosen so that c ≡ 0 (mod 3). We would like to choose

0 ≤ b4 ≤ 2 so that we can find

εb11 εb22 εb33 εb44 =(Ab1

1 Ab22 Ab3

3 Ab44 )3

a0ab44

=

(A0A

b44

α 3√ml

4

)3

.

At this point we have that a0ab44 = neγ3

1 for some n ∈ Z, γ1 ∈ R and e a unit of R. We will show that

unless nml

4is a power of 3 times a cube of a rational number then (5.2) will have no solution for this choice

of b =< b1, b2, b3, b4 >.

Assume that (5.2) has a solution for this choice of b then ml4α

3 = neγ31 for some γ1 ∈ R and e a unit

of R of the form e = ±eu1ev2, where {e1, e2} form a fundamental set of units of R, so

e = eu1ev2 =

ml4

n

(α

γ1

)3

. (5.4)


Now let ϕ be the element of Gal(F/Q) with ϕ(ω) = ω4 then ϕ fixes k = Q(ω3) and N(e1) = e1eϕ1 e

ϕ2

1 = 1

and eϕ1 = e2. By Hilbert’s Theorem 90 we can find β1 ∈ F , in fact β1 = 1 + e1 + e1e2, such that e1 = β1βϕ1

,

NF/k(β1) = 9 ande1e2

=NF/k(β1)

β31

=9β3

1

so e2 = e1β31

9 . Now we can rewrite (5.4) and obtain

e = eu1

(e1β

31

9

)v

= eu+v1

(βv1 )3

9v=

ml4

n

(α

γ1

)3

so eu+v1 = qγ3

2 where q ∈ Q and γ2 ∈ R. This says that

(e1e2

)u+v

=(γ2

γϕ2

)3

=(

9β3

1

)u+v

or

9u+v =(γ2β

u+v1

γϕ2

)3

is a cube of an element of R. But this is false unless u + v ≡ 0 (mod 3) and thus v ≡ 2u (mod 3). Hence

e = (e1e22)u = sγ33 for some s ∈ Q, γ3 ∈ R. But e1e22 = 9β3

2 for some β2 ∈ R so ml4

n

(αγ1

)3

= e = (e1e22)u =

9uβ3u2 which can be simplified to ml

49un =

(β2γ1α

)3

= γ34 for some γ4 ∈ R. Since γ3

4 ∈ Q then γ4 ∈ Q so

ml4 = 9unγ3

4 . Thus n differs from ml4 by a power of 3 times a rational cube, so n = 3rml

4γ30 for r ∈ Z and

γ0 ∈ Q. Thus for a fixed vector b either n satisfies this condition or no solution exists for this value of

b. Thus we assume this condition holds and since 3 differs from a unit of R by a cube in R we have that

a0ab44 = neγ3

1 = 3rml4(γ0γ1)3e so

εb11 εb22 εb33 εb44 =

(A0A

b44

a0ab44

)3

=

(A0A

b44

3√ml

4

)3

· e · γ3

for some unit e of R and γ ∈ R.

If e is a cube of a unit in R, we are done. Suppose e = (e1e22)u and either 3√

3 is in K or k4 is Type I and

3 is a principal divisor in k4 so that 3 = α3εi4 for α ∈ k4 and i =0, 1, 2. In the first case e1e22 =

(3√

9β2

)3is a cube in K(ω + ω8) and in the second case e1e

22 =

(α2β2

)3ε2i4 which gives us that e1e22ε

i4 = (α0)3 for

some α0 ∈ K(ω + ω8). In either case we get a solution to

A3e = εb11 εb22 εb33 εb44

for some integer b4. In the case where neither 3√

3 is in K nor is 3 a principal divisor, Lemma 5.3 shows

that no solution exists.


5.4 Example Type IV units in K


3, 3√

17)

Then the cubic subfields are k1 = Q( 3√

3), k2 = Q( 3√

17), and k4 = Q( 3√

153) are all Type IV so for each

we calculate Bi ∈ Ki and Ai ∈ Mi and k3 = Q( 3√

51) is Type I.

B1 = −1 + 3√

3 + ζ 3√

3 − 3√

32/3 − 2ζ 3

√32/3

A1 = 73/3−17 3√

3/3−7 3√

32/3+(−21+3 3

√32)ω+(47/3+17 3

√3/3−4 3

√32)ω2+(47/3−15 3

√3+5 3

√32/3)ω3+

(−21 + 37 3√

3/3)ω4 + (73/3 − 28 3√

3/3 − 5 3√

32/3)ω5

B2 = 13

[−7 + 13 3

√17 − 4 3

√17

2+ w(28 + 2 3

√17 − 5 3

√17

2)]

A2 = −10− 3√

17/3+5 3√

172/3+(43/3−7 3

√17

2/3)ω+(31/3+ 3

√17/3+5 3

√17

2/3)ω2+(−31/3+11 3

√17/3)ω3+

(43/3 − 17 3√

17/3)ω4 + (−10 + 4 3√

17)ω5

B4 = 28 + 19ζ − 5 3√

153/3 − 16/3ζ 3√

153 − 2 3√

867 + ζ 3√

867

A4 = −1403/3 + 98 3√

153/3 + 98 3√

867/3 + (400 − 131/3 3√

867)ω + (−898/3 − 98/3 3√

153 + 51 3√

867)ω2 +

(−898/3 + 91 3√

153 − 55/3 3√

867)ω3 + (400 − 78 3√

153)ω4 + (−1403/3 + 175/3 3√

153 + 55/3 3√

867)ω5

Let A0 = A1A2A4∗e61∗e13

23√17

2 then N(A0) = e2e23. We can calculate e21e2 =

[3√33 (−ω + ω2 + ω4 + 2ω5)

]3then

A = A0 ∗3√33 (−ω + ω2 + ω4 + 2ω5) and to get a unit in K we multiply A by its complex conjugate to get

e1 = AAτ = 3 − 4 3√

3/3 − 7 3√

32/3 − 3

√17 + 4 3

√51/3 − 3

√867/3 + 3

√2601/3 and e1 = 3

√ε1ε2ε4.

It turns out that in this case since k3 is Type I with A3 = 1513 − 408 3√

51 − 408ζ 3√

51 + 110ζ 3√

512,

α3 = ( 3√

172)3 and ε3 = 107846641 + 29081484 3

√51 + 7841994 3

√51

2then e2 = 3

√ε3 = A3

( 3√17)2= 110 3

√9 +

89 3√

17 + 24 3√

867.

Then [e : e0] = 32 and the basis for e can be chosen < 3√ε1ε2ε4, 3

√ε3, ε1, ε2 >.


17, 3√

19)

Then the cubic subfields are k1 = Q( 3√

17), k3 = Q( 3√

323), and k4 = Q( 3√

5491) are all Type IV so for

each we calculate Bi ∈ Ki and Ai ∈ Mi and k2 = Q( 3√

19) is Type III.

B1 = 13

[−7 + 13 3

√17 − 4 3

√17

2+ ζ(28 + 2 3

√17 − 5 3

√17

2)]

A1 = 13

[−30 − 3

√17 + 5 3

√17

2+ (43 − 7 3

√17

2)ω + (−31 + 3

√17 + 5 3

√17

2)ω2 + (−31 + 11 3

√17)ω3

+(43 − 17 3√

17)ω4 + (−30 + 12 3√

17)ω5]

where NK1/k(B1) = −ζ2 and ωB1 = A1A

σ21


B3 = −52177936089095795/3− 43275181296929777ζ/3 + 1297548413554459 3√

323/3

+2534923169868269ζ 3√

323 + 306419189646737 3√

3232 − 189113586167644ζ 3

√323

2/3

A3 = −5482098754152 + 5482098754152ω2 + 7800620169767ω3/3 + 6788827068112ω4

+24246916432223ω5/3 + 3√

323(−1136915770715/3− 989450119403ω− 1177970515295ω2

−1177970515295ω3/3 − 989450119403ω4 − 1136915770715ω5/3) + 3√

3232(515056454665/3

+144209232164ω+ 55233961812ω2 + 349354569229ω3/3 + 349354569229ω5/3)

where NK3/k(B3) = ζ2 and ωB3 = A3A

σ23

B4 = −57735118463347/3 + 828970907077ζ/3 + 3319562986196 3√

5491/3 + 3272574786271ζ 3√

5491/3 −45278001347 3

√6137/3 − 3198743037625ζ 3

√6137/3

A4 = −12674882642/3 + 106264169 3√

5491/3 + 196633197 3√

6137 + 12674882642ω/3

−612180730ω 3√

5491/3 − 102396542ω 3√

6137/3 + 18498534619ω2/3 − 1010381267ω2 3√

6137/3

−10800158675ω3/3 + 239481633ω3 3√

5491 − 102396542ω3 3√

6137/3 + 624907989ω4

−239481633ω4 3√

5491 + 196633197ω4 3√

6137 + 18498534619ω5/3 − 1048544449/3ω5 3√

5491

where NK4/k(B4) = ζ and ω2B4 = A4A

σ24

The prime divisors of 17 in R are p17a = 3 + ω − ω2 − ω5, p17b = pϕ17a = 3 + ω4 + ω5 and p17c =

pϕ2

17a = 3 − ω + ω2 − ω4 so A0 = A1A3A43√

19 ∗ p217ap17b ∗ e71e63 and NK(ω)/M2(A0) = ω3(172 · 19)3. Then

A = A0172·19 and e1 = AAτ = 10070766629−5253124558 3

√17/3−2250284651 3

√17

2/3+11216410025 3

√19/3−

2009801662 3√

17 · 19/3 − 859315960 3√

172 · 19/3 + 1342037855 3√

192 − 274047206 3

√17 · 192 −

348851731 3√

172 · 192/3 where e1 = 3√

(ε1ε3ε4). Here ε1 = 18 − 7 3√

17, ε3 = −4167355395831946/3 +

12496683414448621 3√

17 · 19/3 − 1732828610414410 3√

17 · 192/3, and

ε4 = 20614589130834 + 1027551896964 3√

172 · 19 − 2116111936999 3√

17 · 192.

Then [e : e0] = 3 and the basis for e can be chosen < 3√ε1ε3ε4, ε1, ε2, ε3 >.

5.5 Units in K from Type III Fields

In order to get a unit in e that is not in e0 and involves units from Type III subfields, it is shown in [9]

Theorem XII that K is not Kind 1. The same theorem shows that K can be of Kind 2 if exactly three of

the subfields are Type III and that K can be of Kind 3 if at least 3 are of Type III. It can be shown that

the first case is not possible.


Definition. If α and β are in k we shall say α and β are equivalent and write α ∼ β if α = β or

α = β.

Theorem 5.4 If k1, k2 and k3 are all of Type III and k4 is of Type I then K is not of Kind 2. Furthermore,

if k1 and k2 are of Type III such that N(B1) ∼ N(B2) and k3 and k4 are of Type I then K is also not of

Kind 2.

Proof: Suppose K is of Kind 2 and ki = Q( 3√mi) for i = 1, 2, 3, 4 and k1, k2 and k3 are all of Type

III and k4 is of Type I. Then since k1, k2 and k3 are all of Type III then 3√εa1ε

b2ε

c3 ∈ K with 1 ≤ a, b, c ≤ 2

and since k4 is of Type I then 3√ε4 ∈ K. Thus K = k4( 3

√ε4) and only prime divisors of 3 can ramify from

k4 to K. All other primes that ramify in K must also be ramified in k4 over Q so m4 must be divisible

by all ramified primes, except possibly 3. We know that m1, m2 and m3 must all have a common prime

divisor p ≡ 1 (mod 3) by Corollary III of Theorem X of [9]. But any prime that divides m1, m2 and m3

can not divide m4 which is a contradiction. Hence K is not of Kind 2.

Suppose K is of Kind 2 and k1 and k2 are of Type III such that N(B1) ∼ N(B2) and k3 and k4 are of

Type I, then 3√ε3, 3

√ε4 ∈ K. Thus K = k3( 3

√ε3) = k4( 3

√ε4) and only prime divisors of 3 can ramify from

k3 to K or from k4 to K. As above m3 and m4 must be divisible by all ramified primes, except possibly

3. Since N(B1) ∼ N(B2) then there exists a prime p �= 3 such that p | m1 and p | m2 so p divides exactly

one of m3 or m4 but this is not possible unless p = 3. Thus K is not of Kind 2 and the theorem is proved.

Let k1, k2, k3 by Type III fields and k4 be of any Type. We know from Corollaries III and IV of

Theorem X of [9] that to find a unit in K from Type III fields we will look for units of the form 3√εa1ε

b2ε

c3ε

d4

with 1 ≤ a, b, c ≤ 2 and 0 ≤ d ≤ 2. Corollary IV of [9] had a minor notational error which we correct here

as Theorem 5.5. First we will note that the significance of Remark A from [9] is that if we wish to replace

εi with ε2i in any part of [9] Theorem X, its corollaries or Theorem 5.5 then we should replace N(Bi) with

N(Bi).

Theorem 5.5 If k1 is Type III and e3 = εa1εb2ε

c3ε

d4 has a solution e in K where 1 ≤ a, b, c, d ≤ 2 then at least

three of the fields k1, k2, k3 and k4 are of Type III. If k4 is not of Type III then N(B4) = ζt with t = 0, 1 or

2 and{N(B1) = ζs·tN(B2) or N(B1) = ζs·tN(B2)

}and

{N(B1) = ζ2s·tN(B3) or N(B1) = ζ2s·tN(B3)

}where s = 1 or 2 and m1, m2, m3 must have a common prime divisor p ≡ 1 (mod 3).

Proof: The proof of the first statement of the Theorem is correct in [9] so we will only prove the last part

of the Theorem. To do so we will first look at the case where all the exponents are 1 and then see how the

exponents on the εi’s effect the relations between the norms. Assume that e3 = ε1ε2ε3ε4 and that k4 is not of


Type III then N(B4) = ζt with t = 0, 1 or 2. We know from the proof of the first statement that in this case

N(B1) = ζ2tN(B2) and N(B1) = ζtN(B3). Now suppose that e3 = ε21ε2ε3ε4 then from Remark A we can

replace N(B1) with N(B1) in the previous case which gives N(B1) = ζ2tN(B2) =⇒ N(B1) = ζtN(B2)

and N(B1) = ζtN(B3) =⇒ N(B1) = ζ2tN(B3). Suppose that e3 = ε1ε22ε3ε4 then N(B1) = ζ2tN(B2)

and N(B1) = ζtN(B3). Suppose that e3 = ε1ε2ε23ε4 then N(B1) = ζ2tN(B2) and N(B1) = ζtN(B3).

Suppose that e3 = ε1ε2ε3ε24 then we replace N(B4) = ζt with N(B4) = ζ2t so we get N(B1) = ζtN(B2)

and N(B1) = ζ2tN(B3). By combining these cases we get all possible exponents on the εi’s and it is clear

that if the exponent on ζ is s · t for the N(B2) term it will be 2s · t on the N(B3) term. That m1,m2 and

m3 have a common prime divisor is shown in [9].

To find a unit in K we first find a product, A, of powers of three Bi’s and/or Bi’s for i =1, 2, 3

such that the norm of A is a cube of some element α ∈ k. We will apply Hilbert’s Theorem 90 to

A/α. We know from the corollaries to Theorem X in [9] and Theorem 5.5 that we need to have that

N(B1) ∼ ζuN(B2) ∼ ζvN(B3) with 0 ≤ u, v ≤ 2. We can find the exponents by considering the cases

in the proof of the Theorem 5.5. Suppose that e3 = ε1ε2ε3εd4 where 0 ≤ d ≤ 2 has a solution e in K.

Then N(B1B2B3) = N(B1)N(B2)N(B3) = N(B1)ζtN(B1)ζ2tN(B1) = N(B1)3. Then if A0 = B1B2Bτ3

then NL/K4(A0) = α3 where α = N(B1) ∈ k. Similarly if e3 = ε21ε2ε3εd4 where 0 ≤ d ≤ 2 has a solution

e in K then N(B21B2B

τ3 ) = N(B1)2N(B2)N(B3) = N(B1)2ζtN(B1)ζ2tN(B1) = (N(B1)N(B1))2. Since

N(B1)N(B1) is the cube of a rational integer then for A0 = B21B2B

τ3 we have that NL/K4(A0) = α3 for

some α ∈ Z. A similar product can be found for the other 2 cases so let A = A0α then NL/K4(A) = 1.

Let σ = σ4, since NL/K4(A) = 1 we can use Hilbert’s Theorem 90 to find an element E0 ∈ K such that

A = E0Eσ

0. Without loss of generality let A = B1B2B3

α and ε1ε2ε3 = B1B2B3(B1B2B3)σ then

ε1ε2ε3 =A

Aσ=

Eσ2

0 E0

(Eσ0 )2

=N(E0)(Eσ

0 )3. (5.5)

Taking the complex conjugate of (5.5) and multiplying we obtain

(ε1ε2ε3)2 =N(E0)N(E0)

(Eσ0E

στ0 )3

where ρ = N(E0)N(E0) ∈ k4. To find a unit in K we need ρ to be a cube times a unit in k4 so we want

ρεd4 = β3 with β ∈ K. By Kummer Theory this is true if and only if ρεd4ml1 = γ3 where γ3 ∈ k4 and

0 ≤ l ≤ 2. Since γ ∈ K4 and [K4 : k4] = 2, ζiγ ∈ k4 for some i = 0, 1, 2. Use the cube root function

(section 5.6) to test if ρεd4ml1 is a cube in k4 for d, l ∈ {0, 1, 2}. If there is a solution for β then

εd4(ε1ε2ε3)2 =ρεd4

(Eσ0 E

στ0 )3

=β3ml

1

(Eσ0E

στ0 )3

=

(β 3√ml

1

(Eσ0E

στ0 )

)3

= E3

and E3 = εd11 εd2

2 εd33 εd4

4 where 0 ≤ d4 ≤ 2.


5.6 Cube Root Function

Given an integer β0 ∈ K1 we would like to be able to solve the equation α30 = β0 for some α0 an integer of

K1 if such a solution exists. If m1 = ab2 where a and b are relatively prime and square free and β0 and α0

are both integers of K1 then we can express them as α0 = a1+a2ζ+a33√m1+a4ζ 3

√m1+a5

3√m2

1 +a6ζ3√m2

1

and 33b3β0 = b1 + b2ζ + b33√m1 + b4ζ 3

√m1 + b5

3√m2

1 + b6ζ3√m2

1 where aj, bj ∈ Z for all j. The advantage

of multiplying by 33b3 is that this will eliminate all the denominators in β0 and then solutions with integer

coefficients can be found. Since α0 has 6 unknowns then we will need 6 equations. We can find these

equations by conjugating with the elements of Gal(K1/Q) to get β1 = βσ0 , β2 = βσ2

0 , β3 = βτ0 , β4 = βστ

0

and β5 = βσ2τ0 and similarly α1, . . . , α5 are the conjugates of α0. To solve the equation α3

0 = β0 we replace

ζ and 3√m1 with a numerical approximation in each βj and αj and we get a set of 6 equations of the form

αj = 3b 3√βj = Cj + Dj

√−1, j = 1, . . . , 6, Cj , Dj ∈ R.

When a solution for α0 is produced this way, the coefficients will have real values. To find integer solutions

we see if there is a solution to the equation where the coefficients are close to integers. It is easy to check

if the solution is correct by verifying α30 = 33b3β0. If a solution exists in Z then

(α03b

)3 = β0.

5.7 Example Type III units in K


7, 3√

19)

Then the cubic subfields, k1 = Q( 3√

7), k2 = Q( 3√

19), k3 = Q( 3√

133), and k4 = Q( 3√

931) are all Type

III so for each εi we calculate Bi ∈ Ki with εi = Bi

Bσi

. The prime 7 factors as (3 + 2ζ)(3 + 2ζ2) = π7π7 and

19 factors as (5 + 2ζ)(5 + 2ζ2) = π19π19 in k. Let σ = σ1 and N = NKi/k for 1 ≤ i ≤ 4 then

B1 = 13 (7 + 14ζ + 3

√7 − 4ζ 3

√7 − 2 3

√49 − ζ 3

√49) and N(B1) = ζ2π2

7π7

B2 = 13 (19 − 3 3

√19 − 5ζ 3

√19 − 3

√361 + 2ζ 3

√361) and N(B2) = −ζπ2

19π19

B3 = 13 (18297 + 5871ζ − 2436 3

√133 − 3585ζ 3

√133 − 225 3

√17689 + 477ζ 3

√17689) and N(B3) = π2

19π19

B4 = 13 (−855 − 114ζ + 75 3

√931 + 87ζ 3

√931 + 9 3

√2527 − 54ζ 3

√2527) and N(B4) = π2

19π19

Let A0 = B2B3B4, then NL/K1(A0) = π619π19

3 = α3 and A = A0α = (945269 + 253083ζ + 493886 3

√7 +

132373ζ 3√

7 + 258362 3√

49 + 69331ζ 3√

49 − 259218 3√

19 − 354292ζ 3√

19 − 135610 3√

133 − 185186ζ 3√

133 −70822 3

√931 − 96758ζ 3

√931 − 35564 3

√361 + 97126ζ 3

√361 − 18628 3

√2527 + 50782ζ 3

√2527 − 9712 3

√17689 +

26566ζ 3√

17689)/3 with NL/K1(A) = 1.

Using Hilbert’s Theorem 90 on A we get that A = E0Eσ

0where E0 = 229816/3+ 40064 3

√7 + 20948 3

√49 +

86182 3√

19/3 + 15007 3√

133 + 7849 3√

931 + 32290 3√

361/3 + 5628 3√

2527 + 2940 3√

17689 and NK/k1(E0) =


(9 + 7 3√

7 + 4 3√

49)3 = β3. The new unit in K is then E = E0β = 6226/3 + 3242 3

√7/3 + 1694 3

√49/3 +

2323 3√

19/3 + 1220 3√

133/3 + 635 3√

931/3 + 871 3√

361/3 + 455 3√

2527/3 + 239 3√

17689/3 and E3 = ε2ε3ε4.

Then [e : e0] = 3 and the basis for e can be chosen < 3√ε2ε3ε4, ε1, ε2, ε3 >.

Chapter 6

Unit Group of L

Once we know the basis for the group of units of K we would like to be able to determine the basis for the

group of units of L. The units in L are dependent on the Kind of K and the Types of the cubic subfields.

In this section we will provide some criteria for when there can be units in L that are not products of the

units of the subfields and what their form will be. We will also give a method for computing the unit basis

of L.

6.1 Criteria for Units in L

For this section we will define a new equivalence relation. This equivalence relation says that the units in

L are equivalent up to multiplication by a cube of an element in L.

Definition 1 : If e1 and e2 are units in L we shall say e1 and e2 are equivalent and write e1 ≈ e2 if and

only if e1 = ζfe3e2 for some e ∈ L and f = 0, 1, or 2.

Using this equivalence relation we can define some new relationships for the units of Type I and IV

fields. These relationships are summarized below with f = 0, 1, or 2.

For i = 1 let σ = σ2

For i > 1 let σ = σ1

εσ2

1 = ε′′ ≈ ε2(ε′)2

εσ1 = ε′1 = u31ε

−21 ζf ≈ ε1.

εσ2 = ε′2 = u32ε

−22 ζf ≈ ε2.

32

Chapter 6. Unit Group of L 33

εσ3 = ε′3 = u33ε

−23 ζf ≈ ε3.

εσ4 = ε′4 = u34ε

−24 ζf ≈ ε4.

uσi = ε−1

i ui

uσ2

i = εiu−2i ≈ εiui

uτi = ui = εiu

−1i

(ε′)τ = ε′′

uu = u2

ε = 3√ε(ε′′)2

Lemma 6.1 Let K1 be of Type I or IV and ζ be a cube root of unity, possibly 1. If ε1 and u1 form a

fundamental system of units for K1 then the equation

e3 = ζiεa1ub1

has no solutions e ∈ L when b ≡ 1 or 2 (mod 3).

Proof: Assume e3 = ζiεa1ub1 has a solution e ∈ L with b ≡ 1 or 2 (mod 3). Then (e1−τ )3 = ζ−i(u1

u1)b =

ζ−i 3√ε1(ε′′1 )2

b. Then by changing notation we may assume e3 = ζi 3

√ε1(ε′′1 )2. Thus if L = K1( 3

√n) for an

integer n then

nj = α3ζi 3

√ε1(ε′′1 )2 where j = 1 or 2 and α ∈ K1.

By taking complex conjugates and multiplying we get

(nj)1+τ = n2j = (α1+τ )3 3

√ε21(ε′1)2(ε′′1 )2 = (α1+τ )3,

but then 3√n ∈ K1 which is a contradiction. Hence the equation has no solution.

Lemma 6.2 Let K1 be a Type III field then the equation

e3 = ζaεb1(ε′1)c

has no solution in L unless a ≡ b ≡ c ≡ 0(mod 3).

Proof: Suppose the equation e3 = ζaεb1(ε′1)c had a solution with not all of a, b, c ≡ 0 (mod 3). If b and

c ≡ 0 (mod 3) then the 9th roots of unity would be in L so it must be that one of b or c is not divisible by

3. Assume without loss of generality that b is not divisible by 3. Then

(e1−σ2τ )3 = ζ2a

(ε1ε′′1

)b

,

but by Theorem X of [9] this equation can have no solution when K1 is Type III, contradiction. Thus

a ≡ b ≡ c ≡ 0 (mod 3).


Lemma 6.3 Let K1 and K2 be of Type I and 3√ε1, 3

√ε2 ∈ K. If the equation

e3 = ζi 3√ε1

a 3√ε′1

b3√ε2

c 3√ε′2

d

has a solution in L then none of a, b, c, and d are divisible by 3 and b ≡ 2a, d ≡ 2c (mod 3)

Proof: We know from Corollary I of Theorem III of [9] that E3 ⊂ ε. Then, since e ∈ E, we know that

e3 = ζiεr1us1ε

t2u

v2 (6.1)

and by hypothesis e3 = ζi 3√ε1

a 3√ε′1

b3√ε2

c 3√ε′2

dso

e3 = ζi 3√ε1

au−b

13√ε1

b 3√ε2

cu−d

23√ε2

d = ζi 3√ε1

a+bu−b

13√ε2

c+du−d

2 . (6.2)

Then equations (6.1) and (6.2) must be equal to each other so the exponents must be equivalent mod 3,

thus a + b = 3r and c + d = 3t =⇒ b ≡ 2a (mod 3) and d ≡ 2c (mod 3) and the result is proved.

Lemma 6.4 Let K1 and K2 be of Type I or IV. If e3 = e1e2 has a solution in L but not in K where

ei ∈ Ki and ei not a cube in Ki for i = 1 or 2, then 3√ε1, 3

√ε2 ∈ K.

Proof: Note that for e1 ∈ K1 we can write e1 = ζiua1ε

b1 for some integers 0 ≤ a, b ≤ 2 (not both zero).

Now

(e1−σ2)3 = e1−σ21 =

ua1ε

a+b1

ua1ε

′1b

= εa1(ε21ε′′1)b = εa1u

3b1 .

Thus the equation E3 = εa1 has a solution in L and then 3√ε1 ∈ L so 3

√ε1 ∈ K.

Assume a = 0. We can write e2 = ζiuc2ε

d2 for some integers 0 ≤ c, d ≤ 2 (not both zero). If c �= 0 then

by considering e1−σ1 we see that 3√ε2 ∈ K.

If c = 0 then e3 = ζiεb1εd2 so (e1+τ )3 = ε2b1 ε2d2 . Hence e3 = e1e2 has a solution in K, contradictory

to hypothesis so c �= 0 and we have e3 = ζiεb1uc2ε

d2 = ζiεb1u

c2

3√ε2

3d. Moving all cubes to the left of

the equation we see that E3 = ζiεb1uc2 has a solution in L. Conjugating with τ and multiplying we get

(E1+τ )3 = ε2b1 εc2 = ε2b1 3√ε2

3c so 3√ε1 ∈ K. Thus 3

√ε1, 3

√ε2 ∈ K

Theorem 6.5 Under the hypotheses of Lemma 6.4, either e3 = ζtu1u2 or e3 = ζtu1u22 with t = 0, 1 or 2

has a solution in L.

Proof: Since 3√ε1 and 3

√ε2 are in K then e3 = ζtua

1ub2 with 0 ≤ a,b ≤ 2 and not both zero. Lemma

6.1 shows that neither can be zero, so the theorem follows.

There is a special ideal in the integers of L, called the different of OL (with respect to OKi) which is

divisible by exactly those primes which are ramified over OL (see [8]). The term ”different” comes from


the idea of the derivative: If L = Ki( 3√εi) then f(x) = x3 − εi is the irreducible polynomial in Ki[x] for

the extension and its derivative is f ′(x) = 3x2. The number different of εi for the extension L/Ki is then

f ′( 3√εi) = 3 3

√ε2i and the different for L/Ki is the GCD of all the number differents of integers in L.

Lemma 6.6 If 3√ε1, 3

√ε2, 3

√ε3 ∈ K then 3

√3 ∈ K

Proof: Since 3√ε1, 3

√ε2, 3

√ε3 ∈ K then L = Q( 3

√m1, 3

√m2, ζ) = Ki( 3

√εi) for i = 1, 2, 3. The only

primes that ramify are those that divide the different of L over Ki. Since the different of L over Ki divides

3 then only prime divisors of 3 can ramify in the extension L/Ki.

Suppose p|m1, p � m2 and p �= 3 then P must ramify from K2 to L where P is a prime divisor of p in

K2 but as shown only divisors of 3 can ramify from Ki to L so P does not exist. Thus p|m1 and p|m2 and

hence m1 and m2 have the same prime divisors, except 3. A similar argument will show that m3 will have

the same prime divisors, except 3, as m1 and m2. Since any prime divides exactly three of m1, m2, m3

and m4 or none of them, then p � m4 so 3√

3 ∈ K.

Lemma 6.7 If K1, K2 and K3 are Type IV and e = 3√ε1εa2ε

b3 ∈ K then we can write e = B

Bσ4 with B ∈ L

and NL/K4 [B] is not a unit in K4.

Proof: Since K1,K2,K3 are Type IV 1 ≤ a, b ≤ 2.

Case 1: K4 Type I or IV

Suppose N [B] ≈ {unit in K4} ≈ εc4ud4 then using the usual technique we can get e

eσ24

= B3

N [B] . So we would

be able to get a solution to

E3 ≈ e

eσ24εc4u

d4 ≈ 3

√(ε1ε′1

)(ε2ε′2

)a(ε3ε′′3

)b

εc4ud4 ≈ u2

1u2a2 ub

3εc4u

d4. (6.3)

Then if we conjugate with σ1

(E3)σ1 ≈ u21(u2a

2 )σ1(ub3)σ1 (ε′4)c(ud

4)σ1 ≈ u21ε

−2a2 (u2a

2 )ε−b3 (ub

3)(ε4)cε−d4 (ud

4)

and take the quotient then we get (E

Eσ1

)3

≈ ε2a2 εb3εd4.

Thus there is no solution unless K4 is of Type IV.

If K4 is of Type IV we show in Case (3A) of Theorem 6.9 and in Corollary 6.10.2 that N(B) is not a

unit.

Case 2: Suppose K4 is Type III. Then N [B] ≈ {unit in K4} ≈ εc4(ε′4)d and equation (6.3) becomes:

E3 ≈ u21u

2a2 ub

3εc4(ε′4)d.


Then if we conjugate with σ2 (Note: σ2 = σ21 on K4.)

(E3)σ2 ≈ u21ε

−21 (u2a

2 )ε−b3 (ub

3)(ε′′4 )c(ε4)d ≈ u21ε

−21 (u2a

2 )ε−b3 (ub

3)(ε4)2c+d(ε′4)c

and take the quotient (E

Eσ2

)3

≈ ε21εb3

εc4ε′4d

(ε′4)cε2c+d4

≈ ε21εb3ε

−(c+d)4 ε′4

d−c.

By Theorem X of [9], c + d ≡ 0 (mod 3) and d − c ≡ 0 (mod 3) but (E3) = ε21εb3 has no solution so then

N[B] �≈ {unit in K4}.

Lemma 6.8 If K1, K2 and K3 are Type III and e = 3√ε1εa2ε

b3 ∈ K then we can write e = B

Bσ4 with B ∈ L

and NL/K4 [B] not a unit in K4.

Proof: Since K1,K2,K3 are Type III 1 ≤ a, b ≤ 2.

Case 1: K4 Type I or IV

Suppose N [B] ≈ {unit in K4} ≈ εc4ud4. Following a similar procedure to the proof of Lemma 6.7 we can

find a solution to:

E3 ≈ 3

√(ε1ε′1

)(ε2ε′2

)a(ε3ε′′3

)b

εc4ud4.

As before we conjugate with σ1

(E3)σ1 ≈ 3

√(ε1ε′1

)(ε′2ε′′2

)a(ε′3ε3

)b

(ε′4)c(u′4)d

and take the quotient to get

(E

Eσ1

)3

≈ 3

√√√√( ε2ε′′2ε′2

2

)a(ε23ε′′3ε′3

)bεc4u

d4

ε′4cu′4

d≈ (ε′2)2aεb3ε

d4.

This has no solution by Corollary III to Theorem X of [9].

Case 2: Suppose K4 is Type III. Then N [B] ≈ {unit in K4} ≈ εc4(ε′4)d and with the same procedure we

get that: (E

Eσ1

)3

≈ (ε′2)2aεb3εc+d4 (ε′4)2d−c

Which has no solution by Corollary III of Theorem X of [9].

We would like to be able to identify when we can find units in L which are not products of units in

the subfields. In [9] there are criteria outlined for when a unit in L, which is not a product of units in the

subfields, can exist for Type III fields and the following theorem will more specifically define the criteria

for all Types of subfields.


Theorem 6.9 Let E0 = ε4∏

i=1

εi then [E : E0] = 3b∗ where b∗ ≤ 2. Furthermore b∗ can not be 2 when K is

Kind 2. Moreover, if K is Kind 3 or 4 and b∗ = 2 then at least three of the cubic subfields are Type III or

3√εi ∈ K for some i.

Proof: The group of units generated by E0 = ε4∏

i=1

εi represent all the units in L that are products of

units in the subfields of L. Let e1, e2, · · · , e8 be a basis for E0.

We are concerned with the number of solutions to an equation of the form e3 =n∏

i=1

ei, where ei ∈ E0

and 1 ≤ n ≤ 8. The basis E0 is dependent on the Type of the four real subfields so it is necessary to

consider the possible cases. Theorem IV of [9] gives us 4 possible choices for the basis (e : e0) (the Kind of

K) and Theorem XII of [9] tells us the subcases for each Kind.

Let {E1, E2, · · · , E8} be a basis for E. We know from Corollary I of Theorem III in [9] that E3 ⊂ ε

and E3 ⊂ εi for i = 1, 2, 3, 4 so we can write

(e1, e2, · · · , e8) = (E1, E2, · · · , E8)A where A is an 8x8 matrix (the exponents for the Ei’s). We can

put A in upper triangular form with elementary row operations that act as a change of basis for E. We

can assume that A is in upper triangular form with diagonal entries either 1 or 3.

A =

a11 a12 a13 · · · a18

0 a22 a23 · · · ...

0 0 a33 · · · ......

... 0. . . a78

0 · · · · · · 0 a88

[E : E0] = det(A) = 3b∗ where b∗ ≤ 8

We need to solve the system of 8 equations but many of our entries are 0. In all cases e1 = ε1 and

e2 = ε′1 or u1 as K1 is Type III or not. Our first equation is e1 = Ea111 but Theorem V in [9] shows us that

this can only have a solution when a11 = 1 because no noncube of a unit of K1 is in E3. Similarly a22 = 1

so we know that E1 = e1 and E2 = e2.

Note: Once we know that a22 = 1 we can let a12 = 0 by row reduction.

We now proceed to consider the individual cases. For the basis E0 we will always assume that the basis

elements are are in the order {e1, e2, · · · , e8}.

Case 1: Kind 1: (e : e0)= 27

All fields are Type I or IV: E0 =< ε1, u1, 3√εa1ε2,

3√εa1ε

′2,

3√εb1ε3,

3√εc1ε4,

3√εb1ε

′3,

3√εc1ε

′4, >

We already have the first two columns of A from above:


A =

1 0 a13 · · · a18

0 1 a23 · · · ...

0 0 a33 · · · ......

... 0. . . a78

0 · · · · · · 0 a88

For convenience of notation when we are looking at an equation of the form ei = ea1i1 ea2i

2 · · · ea(j−1)ij−1 E

aji

i

we will drop the second subscript and write ei = ea11 ea2

2 · · · eaj−1j−1 E

aj

i . Also, when we do calculations we will

be looking at the equation as Eaj

i = . . . but we will continue to write the exponents as positive.

Suppose a33 = 3. Then

E33 = ea1

1 ea22 e3 = εa1

1 ua21

3√εa1ε2

but Lemma 6.4 says that if the equation has a solution then 3√ε1 ∈ K but it is not. So a33 = 1 and E3 = e3.

A similar argument shows that a44 = 1 and E4 = e4.


E35 = ea1

1 ea22 ea3

3 ea44 e5 = εa1

1 ua21

3√εa1ε2

a3 3√εa1ε

′2

a4 3√εb1ε3. (6.4)

Multiplying (6.4) by its complex conjugate gives

(E5E5)3 = ε2a11 ua2

1 ua21

3√εa1ε2

2a3 3√εa1ε

′2

a4 3√εa1ε

′′2

a4 3√εb1ε3

2

= ε2a1+a21

3√εa1ε2

2a3 3√ε2a1 ε′2ε′′2

a4 3√εb1ε3

2(ui = εiu

−1i )

= ε2a1+a21

3√εa1ε2

2a3−a4 3√ε3a1 ε2ε′2ε

′′2

a4 3√εb1ε3

2

= ε2a1+a2+a·a41

3√εa1ε2

2a3−a4 3√εb1ε3

2

All exponents must be divisible by 3 but 2 is clearly not, therefore a55 �= 3 and so a55 = 1 and E5 = e5.

Suppose a66 = 3, then

E36 = εa1

1 ua21

3√εa1ε2

a3 3√εa1ε

′2

a4 3√εb1ε3

a53√εc1ε4. (6.5)

Multiplying (6.5) by its complex conjugate gives (E6E6)3 = ε2a1+a2+a·a41

3√εa1ε2

2a3−a4 3√εb1ε3

2a5 3√εb1ε4

2. All

exponents must be divisible by 3, therefore a66 �= 3. So a66 = 1 and E6 = e6.

So A =

1 0 0 · · · a18

0 1 0 · · · ...

0 0. . . · · · ...

...... 0 a77 a78

0 · · · · · · 0 a88

and det(A) = 3b∗ where b∗ ≤ 2.

Case 2: Kind 2: (e : e0)= 9


(A) All fields are Type I or IV: E0 =< ε1, u1, ε2, u2,3√εc11 εc22 ε3,

3√εc11 ε′2

c2ε′3,3√εb11 εb22 ε4,

3√εb11 ε′2

b2ε′4 >

It is clear from Lemma 6.4 that a11 = a22 = a33 = a44 = 1 so suppose a55 = 3. Then E35 =

εa11 ua2

1 εa32 ua4

23√εc11 εc22 ε3 and by taking the complex conjugates and multiplying we get

(E5E5)3 = ε2a1+a21 ε2a3+a4

23

√εc11 εc22 ε3

2

.

All exponents must be divisible by 3 but 2 is clearly not, therefore a55 �= 3 and so a55 = 1 and E5 = e5.


E36 = εa1

1 ua21 εa3

2 ua42

3√εc11 εc22 ε3

a53√εc11 ε′2

c2ε′3 (6.6)

and multiplying by the complex conjugate gives

(E6E6)3 = ε2a1+a21 ε2a3+a4

23√εc11 εc22 ε3

2a5 3√εc11 ε′2

c2ε′33√εc11 ε′′2

c2ε′′3

= ε2a1+a21 ε2a3+a4

23√εc11 εc22 ε3

2a5−1 3

√ε3c11 (ε2ε′2ε

′′2)c2(ε3ε′3ε

′′3)

= ε2a1+a2+c11 ε2a3+a4

23√εc11 εc22 ε3

2a5−1

=⇒ 2a5 − 1 ≡ 0 (mod 3) 2a1 + a2 + c1 ≡ 0 (mod 3) and 2a3 + a4 ≡ 0 (mod 3)

so a5 ≡ 2 (mod 3) a2 ≡ a1 + 2c1 (mod 3) a3 ≡ a4 (mod 3).

Note: 3√εc11 εc22 ε3

2 3√εc11 ε′2

c2ε′3 = 3√ε3c11 (ε22ε

′2)c2(ε23ε

′3) = εc11 uc2

2 u3.

So E36 = εa1

1 ua1+2c11 εa3

2 ua32 εc11 uc2

2 u3 and conjugating with σ1 and dividing gives

(E1−σ16 )3 =

εa32 ua3

2 uc22 u3

ε′2a3u′2

a3u′2c2u′3

≈ εa3+c22 ε3

From the basis we know that e3 = εc11 εc22 ε3 ∈ K so we can take the quotient and rename to get

E∗3 = ε−c11 εa3

2

where all exponents must be divisible by 3 so c1 ≡ a3 ≡ 0 (mod 3). Then (6.6) becomes

E36 = εa1

1 ua11 uc2

2 u3 (6.7)

and by conjugating with σ2 and dividing we get (E1−σ26 )3 = ε

a11 u

a11 u

c22 u3

ε′1a1u′

1a1u

c22 u′

3≈ εa1

1 ε3. We know that e3 =

εc22 ε3 ∈ K so we can take the quotient and rename to get (E∗)3 = εa11 ε−c2

2 where all exponents must be

divisible by 3 so a1 ≡ c2 ≡ 0 (mod 3). Then (6.7) becomes

E36 = εa5

3 u3

and this equation can have no solution by Lemma 6.1 so a66 = 1 and E6 = e6.



E37 = ea1

1 ea22 ea3

3 ea44 ea5

5 ea66

3√εb11 εb22 ε4. (6.8)

Multiplying (6.8) by its complex conjugate gives (E7E7)3 = e2a1+a2+a61 e2a3+a4

3 e2a5−a65

3√εb11 εb22 ε4

2

. All

exponents must be divisible by 3 so we get a contradiction. Thus a77 = 1 and E7 = e7.

So A =

1 0 0 · · · a18

0 1 0 · · · ...

0 0. . . · · · ...

...... 0 1 a78

0 · · · · · · 0 a88

and det(A) = 3b∗ where b∗ ≤ 1.

(B) Three Type I and one Type III: E0 =< ε1, u1,3√εb1ε2,

3√εb1ε

′2,

3√εc1ε3,

3√εc1ε

′3, ε4, ε

′4 >

Here we let k4 be Type III.

Clearly we can take a11 = a22 = a33 = a44 = a55 = a66 = 1. Suppose a77 = 3. Then e0 =

εa11 ua2

13√εb1ε2

a3 3√εb1ε

′2

a43√εc1ε3

a5 3√εc1ε

′3

a6 and E37 = e0ε4, so (E7E7)3 = e0e0ε

24. All exponents must be

divisible by 3 so a77 = 1 and E7 = e7. The matrix A is the same as in case(2A) so det(A) = 3b∗ where

b∗ ≤ 1.

(C) Two Type I and two Type III:

In this case k1 and k2 are Type III.

Suppose N(B1) � N(B2): E0 =< ε1, ε′1, ε2, ε

′2, 3√ε3, u3, 3

√ε4, u4 >

We can take a11 = a22 = a33 = a44 = a55 = a66 = 1. Suppose a77 = 3. Let e0 = εa11 ε′1

a2εa32 ε′2

a4 3√ε3

a5ua63

and E37 = e0 3

√ε4, so (E7E7)3 = e0e0 3

√ε4

2. All exponents must be divisible by 3 so a77 = 1 and E7 = e7.

Suppose a88 = 3. Then E38 = εa1

1 ε′1a2εa3

2 ε′2a4 3√ε3

a5ua63

3√ε4

a7u4 and

(E8E8)3 = ε2a1−a21 ε2a3−a4

23√ε3

2a5+3a6 3√ε4

2a7+3.

All the exponents must be divisible by 3 so the following relations are clear

a2 ≡ 2a1 (mod 3), a4 ≡ 2a3 (mod 3), a5 ≡ a7 ≡ 0 (mod 3)

and E38 = εa1

1 ε′12a1εa3

2 ε′22a3ua6

3 u4 so det(A) = 3b∗ where b∗ ≤ 1.

Suppose N(B1) ∼ N(B2): This case does not exist by Theorem 5.4.

Case 3: Kind 3: (e : e0) = 3

(A) All fields are Type I or IV: E0 =< ε1, u1, ε2, u2, ε3, u3,3√εc11 εc22 εc33 ε4,

3√εc11 ε′2

c2ε′3c3ε′4 >


We know that a11 = a22 = a33 = a44 = 1 by previous work and a55 = 1 by a similar proof to Case 2.


E36 = εa1

1 ua21 εa3

2 ua42 εa5

3 u3. (6.9)

Conjugating (6.9) with σ1 and dividing gives (E1−σ16 )3 = ε

a32 u

a42 ε

a53 u3

ε′2a3u′

2a3 ε′3

a5u′3≈ εa4

2 ε3. Once again we need all

exponents divisible by 3 but the exponent on ε3 is not. This is a contradiction so a66 = 1 and E6 = e6.


E37 = εa1

1 ua21 εa3

2 ua42 εa5

3 ua63

3

√εc11 εc22 εc33 ε4

and

(E7E7)3 = ε2a1+a21 ε2a3+a4

2 ε2a5+a63

3

√εc11 εc22 εc33 ε4

2

.

All exponents must be divisible by 3 so we have a contradiction. Thus a77 = 1 and E7 = e7.


E38 = g1g2g3

3

√εc11 εc22 εc33 ε4

a73

√εc11 ε′2

c2ε′3c3ε′4

where gi = εa2i−1i ua2i

i and gigi = ε2a2i−1+a2i

i and

(E8E8)3 = g1g1g2g2g3g33√εc11 εc22 εc33 ε4

2a7 3√εc11 ε′2

c2ε′3c3ε′4

3√εc11 ε′′2

c2ε′′3c3ε′′4

= ε2a1+a2+c11 ε2a3+a4

2 ε2a5+a63

3√εc11 εc22 εc33 ε4

2a7−1

2a7 −1 ≡ 0 (mod 3)

a7 ≡ 2 (mod 3)

a3 ≡ a4 (mod 3)

a5 ≡ a6 (mod 3)

2a1 + a2 + c1 ≡ 0 (mod 3)

So E38 = g1g2g3

3

√ε3c11 (ε22ε

′2)c2(ε23ε

′3)c3(ε24ε

′4) and, since 3

√ε2i ε

′i = ui for i = 2, 3, 4, then

E38 = g1g2g3ε

c11 uc2

2 uc33 u4. Also, if gi = εsi

i uti

i then the quotient after conjugating with σ1 is gi

gσ1i

= gi

g′i≈ εti

i

for i = 2, 3. Then

E38 = εa1+c1

1 ua1+2c11 εa3

2 ua3+c22 εa5

3 ua5+c33 u4

and conjugating with σ1 and dividing gives

(E1−σ18 )3 =

εa32 ua3+c2

2 εa53 ua5+c3

3 u4

ε′2a3u′2

a3+c2ε′3a5u′3

a5+c3u′4≈ εa3+c2

2 εa5+c33 ε4.

Since e3 = εc11 εc22 εc33 ε4 ∈ e we divide both sides by e3 to and rename to get E∗3 = ε−c11 εa3

2 εa53 , where

all exponents must be divisible by 3 so c1 ≡ a3 ≡ a5 ≡ 0 (mod 3). The equation reduces to E38 =


εa11 ua1

1 uc22 uc3

3 u4 and by conjugating with σ2 and dividing we get

(E1−σ28 )3 =

εa11 ua1

1 uc22 uc3

3 u4

ε′1a1u′1

a1u′2c2u′3

c3u′4≈ εa1

1 εc33 ε4.

As above we get that c3 ≡ a1 ≡ 0 (mod 3).

If we conjugate with σ3 we get that c2 ≡ 0 ( mod 3) so E38 = u4. Lemma 6.1 says this can have no solutions

so a88 = 1 and E8 = e8.

So A =

1 0 0 · · · 0

0 1 0 · · · ...

0 0. . . · · · ...

...... 0 1 0

0 · · · · · · 0 1

and det(A) = 3b∗ where b∗ = 0.

(B) One Type III: E0 =< ε1, ε′1, ε2, u2, ε3, u3,

3√εc22 εc33 ε4,

3√ε′2

c2ε′3c3ε′4 >

We know that a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1 from the previous case so suppose a88 = 3.

Then

E38 = εa1

1 ε′1a2εa3

2 ua42 εa5

3 ua63

3

√εc22 εc33 ε4

a73√ε′2

c2ε′3c3ε′4 (6.10)

and (E8E8)3 = ε2a1−a21 ε2a3+a4

2 ε2a5+a63

3√εc22 εc33 ε4

2a7−1so the exponents must all be divisible by 3 and

a2 ≡ 2a1 (mod 3), a4 ≡ a3 (mod 3), a6 ≡ a5 (mod 3) and a7 ≡ 2 (mod 3)

Then equation (6.10) becomes E38 = εa1

1 ε′12a1εa3

2 ua32 εa5

3 ua53

3√εc22 εc33 ε4

2 3√ε′2

c2ε′3c3ε′4 and

(E1−σ18 )3 =

εa32 ua3

2 εa53 ua5

33√εc22 εc33 ε4

2 3√ε′2

c2ε′3c3ε′4

ε′2a3u′2

a3ε′3a5u′3

a5 3√ε′2

c2ε′3c3ε′4

23√ε′′2

c2ε′′3c3ε′′4

≈ εa3+c22 εa5+c3

3 ε4.

Since e3 = εc22 εc33 ε4 ∈ e we divide both sides by e3 to and rename to get E∗3 = εa32 εa5

3 , where all exponents

must be divisible by 3 so a3 ≡ a5 ≡ 0 (mod 3). Equation (6.10) reduces to

E38 = εa1

1 ε′12a1 3

√εc22 εc33 ε4

23√ε′2

c2ε′3c3ε′4

and by conjugating with σ2 and dividing we get

(E1−σ28 )3 ≈ εc33 ε4,

which has a solution if c2 ≡ 0 (mod 3). If we conjugate with σ3 we see that c3 ≡ 0 (mod 3) so E38 =

εa11 ε′1

2a1u4. Then det(A) = 3b∗ where b∗ ≤ 1.

(C) Two Type III and two Type I or IV:



′2, ε3, u3, 3

√εa3ε4, u4 >

In this case k1 and k2 are Type III, k4 is a Type I field and k3 is Type I if a �= 0.

We know from [9] Theorem X that E3 = εa11 ε′1

a2εa32 ε′2

a4 has no solution in L so a11 = a22 = a33 = a44 =

1. Suppose that a55 = 3. Then

E35 = εa1

1 ε′1a2εa3

2 ε′2a4ε3

and if we multiply E35 by it’s complex conjugate we get

(E5E5

)3= ε2a1−a2

1 ε2a3−a42 ε23.

All exponents must be divisible by 3 so this equation has no solution and a55 = 1.

We will number the basis elements of E0 as e1 to e8 and we can find some relations between the

conjugates.

e1e1 = e21, e2e2 = ε′1ε′′1 = ε−1

1 = e−11 , e3e3 = e23, e4e4 = ε′2ε

′′2 = ε−1

2 = e−13 ,

e5e5 = e25, e6e6 = u3u3 = ε3 = e5, e7e7 = e27, e8e8 = u4u4 = ε4 = e37e−a5


E36 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

3 u3

and

(E6E6)3 = ε2a1−a21 ε2a3−a4

2 ε2a5+13 .

All exponents must be divisible by 3 so we get the following relations

a2 ≡ 2a1 (mod 3), a4 ≡ 2a3 (mod 3), a5 ≡ 1 (mod 3).

Then

E36 = εa1

1 ε′12a1εa3

2 ε′22a3ε3u3.

We can also find some relations by conjugating with σ1.

e1−σ13 = ε2

ε′2= e3e

−14 , e1−σ1

4 = ε′2ε2”

= e3e24, e1−σ1

5 = ε3ε′3

= e36, e1−σ16 = ε3 = e5,

e1−σ17 = 3

√εa3ε4

ε′3aε′4

= ua3u4 = ea6e8

Then E1−σ16 = ε3a3

2 ε′23a3u3

3ε3, which has no solution, so a66 = 1.


E37 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

3 ua63

3√εa3ε4

and

(E7E7)3 = ε2a1−a21 ε2a3−a4

2 ε2a5+a63

3√εa3ε4

2.


All exponents must be divisible by 3 so a77 = 1. Suppose a88 = 3. Then

E38 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

3 ua63

3√εa3ε4

a7u4

and

(E8E8)3 = ε2a1−a21 ε2a3−a4

2 ε2a5+a6−a3

3√εa3ε4

2a7+3.

All exponents must be divisible by 3 so the following relations are clear

a2 ≡ 2a1 (mod 3), a4 ≡ 2a3 (mod 3), a6 ≡ a5 + a (mod 3) a7 ≡ 0 (mod 3).

Now for brevity we change to the basis notation ei and we have that

E38 = ea1

1 e2a12 ea3

3 e2a34 ea5

5 ea5+a6 e8

and we can conjugate with 1 − σ1 to get

(E1−σ18 )3 = e3a3

3 e3a34 e3a5

6 ea55 e37.

Then moving all cubes to the left and renaming we get (E∗8 )3 = ea5

5 = εa53 so a5 ≡ 0 (mod 3). Now

E38 = ea1

1 e2a12 ea3

3 e2a34 ea6e8 so if we consider

(E1−σ28 )3 = e3a1

1 e3a12 e−2a

5 e3a56 e37

then by moving the cubes to the left and renaming we get (E∗8 )3 = e−2a

5 = ε−2a3 so a ≡ 0 (mod 3). So

E38 = εa1

1 ε′12a1εa3

2 ε′22a3u4 and b∗ ≤ 1.

Suppose N(B1) ∼ N(B2): E0 =< ε1, ε′1, ε2, ε

′2, ε3, u3, 3

√εa3ε4, u4 >

In this case k1 and k2 are Type III, k4 is a Type I field and k3 is Type I if a �= 0.

We know from [9] Theorem X that e3 = ζa(ε1ε2/ε′1ε′′2) has a solution e ∈ L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′′2, ε3, u3, 3

√εa3ε4, u4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]≤ 3, thus[

E : E0

]=[E : E1

] [E1 : E0

]≤ 32.

We know from above that a11 = a22 = a33 = 1. Suppose that a44 = 3. Then E34 = εa1

1 ε′1a2εa3

23

√ε1ε2ε′1ε

′′2

and conjugating with σ1 gives:

(E1−σ14 )3 =

εa3+12

ε′2a3

which has no solution by [9] Theorem X.

Suppose that a55 = 3. Then E35 = εa1

1 ε′1a2εa3

23

√ε1ε2ε′1ε

′′2

a4ε3 and multiplying by the complex conjugate gives

(E5E5)3 = ε2a1−a2+a41 ε2a3+a4

2 ε23


which has no solution because all exponents must be divisible by 3 so a55 = 1.


1 ε′1a2εa3

23

√ε1ε2ε′1ε

′′2

a4εa53 u3. Multiplying by the complex conjugate

gives

(E6E6)3 = ε2a1−a2+a41 ε2a3+a4

2 ε2a5+13

which provides the following relations:

a2 ≡ 2a1 + a3 (mod 3), a4 ≡ a3 (mod 3), a5 ≡ 1 (mod 3).

Now E36 = εa1

1 ε′12a1+a3εa3

23

√ε1ε2ε′1ε

′′2

a3ε3u3 and we conjugate with σ1 to get

(Eσ16 )3 = εa1

1 ε′12a1+a3ε′2

a3 3

√ε1ε′2ε′1ε2

a3

ε′3u′3.

Taking the quotient gives

(E1−σ16 )3 = ε2a3

2 ε′2−a3ε3u

33

and all exponents must be divisible by 3 because 3√ε3 �∈ E1 so a66 = 1.


1 ε′1a2εa3

23

√ε1ε2ε′1ε

′′2

a4εa53 ua6

33√εa3ε4. Multiplying by the complex conju-

gate gives

(E7E7)3 = ε2a1−a2+a41 ε2a3+a4

2 ε2a5+a63

3√εa3ε4

2

which has no solution since the exponents must be divisible by 3. Thus a77 = 1.


1 ε′1a2εa3

23

√ε1ε2ε′1ε

′′2

a4εa53 ua6

33√εa3ε4

a7u4. Multiplying by the complex

conjugate gives

(E8E8)3 = ε2a1−a2+a41 ε2a3+a4

2 ε2a5+a6−a3

3√εa3ε4

2a7+3

which provides the following relations:

a2 ≡ 2a1 + a3 (mod 3), a4 ≡ a3 (mod 3), a6 ≡ a5 + a (mod 3) a7 ≡ 0 (mod 3).

Now E38 = εa1

1 ε′12a1+a3εa3

23

√ε1ε2ε′1ε

′′2

a3εa53 ua5+a

3 u4 and we conjugate with σ1 to get

(Eσ18 )3 = εa1

1 ε′12a1+a3ε′2

a3 3

√ε1ε′2ε′1ε2

a3

ε′a53 u′a5+a

3 u′4. Taking the quotient and simplifying gives

(E1−σ18 )3 = εa3

1 ε′1−a3ε4a3

2 εa53

so a3 ≡ a5 ≡ 0 (mod 3).

We are left with E38 = εa1

1 ε′12a1ua

3u4 and if we conjugate with σ3 and take the quotient and simplify

then

(E1−σ38 )3 = 3

√εa3ε4

−3εa3


so a ≡ 0 (mod 3). Thus E38 = εa1

1 ε′12a1u4 and b∗ ≤ 2.

(D) Three Type III and k4 Type I or IV:

Suppose N(B1) ∼ ζc1N(B2) ∼ ζc2N(B3):

There are two possibilities for this case. There can be a unit in K of the form 3√ε4 from a Type I field

or the three Type III fields produce 3√ε1ε2ε3. The first case doesn’t occur because if the principal divisor

of k4 is d then d | 9m24 and K = Q( 3

√d, 3

√m4). Since the norms of the Bi’s are similar they must all share

a common prime divisor p ≡ 1 (mod 3) but p must also divide d which implies that p | m4 which is not

possible.

Consider the case where the unit in K comes from the Type III fields. Then

E0 =< ε1, ε′1, ε2, ε

′2, 3√ε1ε2ε3, ε

′3, ε4, u4 >

We know from [9] Theorem X that e3 = ζa(ε1ε2/ε′1ε′′2) and e3 = ζc(ε1ε3/ε′1ε

′3) both have solutions e ∈ L

so b∗ is at least 2. Let E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε3, 3

√ε1ε2ε′1ε

′′2, 3

√ε1ε3ε′1ε

′3, ε4, u4 >. Then E0 ⊂ E1 and we will show

that[E : E1

]= 1, thus

[E : E0

]=[E : E1

] [E1 : E0

]= 32.

The same reasoning as in Case 3 (B) gives us that a11 = a22 = a33 = a44 = 1. We will again number

the basis elements of E1 as e1 to e8 and we can find some relations between the conjugates.

e1e1 = e21, e2e2 = ε′1ε′′1 = ε−11 = e−1

1 , e4e4 = e24,

e3e3 = e23, e5e5 = 3

√ε21ε

22

ε′1ε′′1 ε′2ε

′′2

= ε1ε2 = e1e3, e6e6 = e−11 e−1

3 e34,

e7e7 = e27, e8e8 = u4u4 = u4u−14 ε4 = e7

Suppose a55 = 3. Then E35 = ea1

1 ea22 ea3

3 ea44 e5 and taking the product with the complex conjugate gives:

(E5E5)3 = e2a11 e−a2

1 e2a33 (e4)2a4e1e3

= e2a1−a2+11 e2a3

3 e2a44 .

All exponents must be divisible by 3 so we get the following relations

a2 ≡ 2a1 + 1 (mod 3), a3 ≡ a4 ≡ 0 (mod 3).

So E35 = ea1

1 e2a1+12 e5 = εa1

1 ε′12a1+1

3

√ε1ε2ε′1ε

′′2

which can have no solution in L since E3 ⊂ ε so a55 = 1. Similarly

a66 = 1.


1 ea22 ea3

3 ea44 ea5

5 ea66 e7 and taking the product with the complex conjugate

gives:

(E7E7)3 = e2a11 e−a2

1 e2a33 (e4)2a4(e1e3)a5(e34e

−13 )a6e27


= ε2a1−a2+a51 ε2a3+a5−a6

23√ε1ε2ε3

2a4+3a6ε24.

Since all exponents must be divisible by 3 so a77 = 1.

Suppose that a88 = 3 then E38 = ea1

1 ea22 ea3

3 ea44 ea5

5 ea66 ea7

7 e8. Multiplying by the complex conjugate gives:

(E8E8)3 = e2a11 e−a2

1 e2a33 (e4)2a4(e1e3)a5(e34e

−13 )a6e2a7

7 e8

= ε2a1−a2+a51 ε2a3+a5−a6

23√ε1ε2ε3

2a4+3a6ε2a7+14 .

Then the following relations are clear

a4 ≡ 0 (mod 3), a2 ≡ 2a1 + a5 (mod 3), a6 ≡ 2a3 + a5 (mod 3), a7 ≡ 1 (mod 3)

and E38 = ea1

1 e2a1+a52 ea3

3 ea55 e2a3+a5

6 e7e8. We will now conjugate with σ1 and we find the following relations

e1−σ13 = ε2

ε′2= e1e

−12 e33e

−15 , e1−σ1

5 = ε2 = e3, e1−σ16 = ε′3

−1 = e2e3e−34 e36,

e1−σ17 = ε4

ε′4= u3

4 = e38, e1−σ18 = ε4 = e7.

Now (E1−σ18 )3 =

(e1e

−12 e33e

−35

)a5ea53

(e2e3e

−34 e36

)2a3+a5e7e

38 and we can move all the cubes to the left

at rename to get

(E∗8 )3 = ea5

1 e2a32 e2a3+2a5

3 e7

= εa51 ε′1

2a3ε2a3+2a52 ε4.

All exponents must be divisible by 3 so a88 = 1 and b∗ = 1.

Suppose N(B1)N(B2) ∼ ζaN(B3): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, 3√ε4, u4 >

In this case we have a solution in K from the Type I field.

We know from [9] Theorem X that e3a = ζa ε1ε2ε3ε′1ε

′2ε

′′3

has a solution ea in L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, ε

′2, ε3, 3

√ε1ε2ε3ε′1ε

′2ε

′′3, 3√ε4, u4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]≤ 3, thus[

E : E0

]=[E : E1

] [E1 : E0

]= 3b∗ where b∗ ≤ 2.

The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = 1. Suppose

a66 = 3, then

E36 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

33

√ε1ε2ε3ε′1ε′2ε′′3

.

Taking complex conjugates and multiplying we get

(E6E6)3 = ε2a11 (ε′1ε

′′1)a2ε2a3

2 (ε′2ε′′2)a4ε2a5

33

√ε21ε

22ε

23

ε′1ε′′1 ε′2ε

′′2 ε′3ε

′′3

= ε2a1−a2−11 ε2a3−a4−1

2 ε2a5+13 .


Then the following relations are clear

a2 ≡ 2a1 + 1 (mod 3), a4 ≡ 2a3 + 1 (mod 3), a5 ≡ 1 (mod 3)

and E36 = εa1

1 ε′12a1+1

εa32 ε′2

2a3+1ε3 3


′2ε

′′3

. We will now conjugate with σ1 and we find the following relations

e1−σ13 = ε2

ε′2= e3e

−14 , e1−σ1

4 = ε2ε′22 = e3e

24, e1−σ1

5 = ε23ε′′3 = e1e

−12 e3e

−14 e35e

−36 , e1−σ1

6 = ε3ε′2

= u−14 e5.

Then (E1−σ16 )3 = (e3e−1

4 )a3(e3e24)2a3+1(e1e−12 e3e

−14 e35e

−36 )e−1

4 e5 and by moving all cubes to the left and

renaming we get

(E∗6 )3 = ε1ε

′1ε

22ε3

which has no solution so a66 = 1.

Suppose a77 = 3, then E37 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

33


′2ε

′′3

a63√ε4. Taking complex conjugates and multiply-

ing we get

(E7E7)3 = ε2a11 (ε′1ε

′′1)a2ε2a3

2 (ε′2ε′′2)a4ε2a5

33

√ε21ε

22ε

23

ε′1ε′′1 ε′2ε

′′2 ε′3ε

′′3

a63√ε4

2

= ε2a1−a2+a61 ε2a3−a4+a6

2 ε2a5+a63

3√ε4

2.

This can have no solution in K unless all exponents are divisible by 3 so a77 �= 3 which implies that a77 = 1.

Suppose a88 = 3. Then a similar calculation results in

E38 = εa1

1 ε′12a1εa3

2 ε′22a3u4

so det(A) = 3b∗ where b∗ ≤ 2.

Suppose N(B1) ∼ ζaN(B2) � N(B3): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, 3√ε4, u4 >

We know from [9] Theorem X that e3 = ζa(ε1ε2)/(ε′1ε′′2) has solutions e in L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′′2, ε3, ε

′3, 3√ε4, u4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]≤ 3, thus[

E : E0

]=[E : E1

] [E1 : E0

]= 3b∗ where b∗ ≤ 2.

The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = a66 = 1. We

will again number the basis elements e1 to e8 and we can find some relations between the conjugates.

e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e1e3, e5e5 = e25, e6e6 = e−1

5 , e7e7 = e27


1 ea22 ea3

3 ea44 ea5

5 ea66 e7 and multiplying by the conjugate gives:

(E7E7)3 = e2a11 e−a2

1 e2a33 (e1e3)a4e2a5

5 e−a65 e27

=e2a1−a2+a41 e2a3+a4

3 e2a5−a65 e27


=ε2a1−a2+a41 ε2a3+a4

2 ε2a5−a63

3√ε4

2

All exponents must be divisible by 3 so so a77 �= 3 which implies that a77 = 1.

Suppose a88 = 3. Then a similar calculation results in

E38 = εa1

1 ε′12a1εa5

3 ε′32a5u4


Suppose N(B1) � N(B2) � N(B3) � N(B1) and N(B1)N(B2) � N(B3): Then

E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, 3√ε4, u4 >

The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1.

Suppose a88 = 3. Then a similar calculation as above results in

E38 = εa1

1 ε′12a1εa3

2 ε′22a3εa5

3 ε′32a5u4


(E)Four Type III:

We can have at most 3 Type three fields of similar norm so the proof of this case is the same as the

proof of case(3D) Three Type III fields.

Case 4: Kind 4: (e : e0) = 1

(A) All fields are Type I or IV: E0 =< ε1, u1, ε2, u2, ε3, u3, ε4, u4 >

We know that a11 = a22 = a33 = a44 = a55 = a66 = 1 by a similar proof to Case 3 so suppose a77 = 3.

Then E37 = εa1

1 ua21 εa3

2 ua42 εa5

3 ua63 ε4 and multiplying by the complex conjugate gives:

(E7E7)3 = ε2a11 (u1u1)a2ε2a3

2 (u2u2)a4ε2a53 (u3u3)a6ε24 ≈ ε2a1+a2

1 ε2a3+a42 ε2a5+a6

3 ε24.

This clearly has no solution because K is Kind 4 all exponents must be divisible by 3 so a77 = 1.

Suppose a88 = 3, then E38 = εa1

1 ua21 εa3

2 ua42 εa5

3 ua63 εa7

4 u4 and conjugating with σ1 and dividing we get

(E1−σ18 )3 =

(ε2ε′2

)a3 (u2

u′2

)a4 ( ε3ε′3

)a5 (u3

u′3

)a6 (ε4ε′4

)a7 (u4

u′4

)≈ εa4

2 εa63 ε4.

Again all the exponents must be divisible by three hence a88 �= 3 =⇒ a88 = 1. As with the other cases

we then have det(A) = 1 and b∗ = 0.

(B) One Type III: E0 =< ε1, ε′1, ε2, u2, ε3, u3, ε4, u4 >

The proof for this case is identical to the previous one.


(C) Two Type III and two Type I or IV:

Suppose N(B1) ∼ ζaN(B2): E0 =< ε1, ε′1, ε2, ε

′2, ε3, u3, ε4, u4 >

We know from [9] Theorem X that e3 = ζa(ε1ε2/ε′1ε′2) has a solutions e in L so b∗ is at least 1.

Let E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, u3, ε4, u4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1, thus[

E : E0

]=[E : E1

] [E1 : E0

]= 31.

The proof follows much as the previous ones. The same reasoning as in the previous case gives us that

a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1. We will again number the basis elements e1 to e8 and find

some relations between the conjugates.

e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e1e3, e5e5 = e25, e6e6 = e5, e7e7 = e27, e8e8 = e7


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7

7 e8 and multiplying by the complex conjugate:

(E8E8)3 = e2a1−a2+a41 e2a3+a4

3 e2a5+a65 e2a7+1

7 = ε2a1−a2+a41 ε2a3+a4

2 ε2a5+a63 ε2a7+1

4

For this to have a solution in K we need

2a1 − a2 + a4 ≡ 0 (mod 3), 2a3 + a4 ≡ 0 (mod 3), 2a5 + a6 ≡ 0 (mod 3), 2a7 + 1 ≡ 0 (mod 3).

So E38 = ε2a2+a3

1 ε′1a2εa3

23

√ε1ε2ε′1ε

′2

a3εa53 ua5

3 ε4u4 and conjugating with σ1 and taking the quotient:

(E1−σ18 )3 =

(ε2ε′2

)a3

3

√ε′′2ε2ε′2

2

a3 (ε3ε′3

)a5 (u3

u′3

)a5 ε4ε′4

u4

u′4=(ε2ε′2

)a3

(ε′2)−a3ε3a53 u−3a5

3 εa53 ε34u

−34 ε4.

Then renaming and moving cubes to the left gives

(E∗)3 = εa32 ε′2

a3εa53 ε4.

Since a11 = . . . = a77 = 1 all the exponents must be divisible by 3. Thus a88 �= 3 =⇒ a88 = 1 and b∗ = 1.


′2, ε3, u3, ε4, u4 >

The proof follows much as the previous ones. The same reasoning as in the previous case gives us that

a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1. We will again number the basis elements e1 to e8 and find

some relations between the conjugates.

e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e3−1, e5e5 = e25, e6e6 = e5, e7e7 = e27, e8e8 = e7


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7

7 e8 and multiplying by the complex conjugate gives:

(E8E8)3 = e2a1−a21 e2a3−a4

3 e2a5+a65 e2a7+1

7

For this to have a solution in K we need

2a1 − a2 ≡ 0 (mod 3)


2a3 − a4 ≡ 0 (mod 3)

2a5 + a6 ≡ 0 (mod 3)

2a7 + 1 ≡ 0 (mod 3)

So we can rewrite E38 = εa1

1 ε′12a1εa3

2 ε′22a3εa5

3 ua53 ε4u4, conjugating with σ1 and take the quotient to get:

(E1−σ18 )3 = ε

a32 ε′2

2a3 εa53 u

a53 ε4u4

ε′2a3 ε′′2

2a3 ε′3a5u′

3a5 ε′4u

′4

= (ε′′2 )−3a3ε3a53 u−3a5

3 εa53 ε34u

−34 ε4 ≈ εa5

3 ε4.

This can have no solution unless all exponents are divisible by 3, contradiction. Thus a88 �= 3 =⇒ a88 = 1.

As with the other cases we then have det(A) = 1 and b∗ = 0.

(D) Three Type III and one Type I or IV:

Suppose N(B1) � N(B2) � N(B3) � N(B1) and N(B1)N(B2) � N(B3): then

E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, u4 >


We will again number the basis elements e1 to e8 and find some relations between the conjugates.

e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e−1

3 , e5e5 = e25, e6e6 = e−15 , e7e7 = e27, e8e8 = e7


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7


(E8E8)3 = e2a1−a21 e2a3−a4

3 e2a5−a65 e2a7+1

7

So the following relations on the exponents are clear: a2 ≡ 2a1 (mod 3), a4 ≡ 2a3 (mod 3), a6 ≡2a5 (mod 3), and a7 ≡ 0 (mod 3) and the equation can be rewritten:

E38 = ea1

1 e2a12 ea3

3 e2a34 ea5

5 e2a56 e7e8.

We will conjugate with σ1 to get some relations.

e1−σ13 = e3e

−14 , e1−σ1

4 = e3e24, e1−σ1

5 = e5e−16 , e1−σ1

6 = e5e26, e1−σ1

7 = e37e−38 , e1−σ1

8 = e7

Then (E1−σ18 )3 = (e3e−1

4 )a3ea33 (e5e−1

6 )a5(e5e26)2a5e37e−38 e7. Combining like terms, moving all cubes to the

left, and renaming gives (E′8)3 = e7 = u4 and this clearly has no solution hence a88 �= 3 =⇒ a88 = 1. As

with the other cases we then have det(A) = 1 and b∗ = 0.

Suppose N(B1) ∼ ζaN(B2) � N(B3): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, u4 >

We know from [9] Theorem X that e3a = ζa(ε1ε2/ε′1ε′′2) has solutions ea in L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′′2, ε3, ε

′3, ε4, u4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1, thus

[E : E0

]=[

E : E1

] [E1 : E0

]= 31.




e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e1e3, e5e5 = e25, e6e6 = e5, e7e7 = e27, e8e8 = e7


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7


(E8E8)3 = e2a1−a2+a41 e2a3+a4

3 e2a5−a65 e2a7+1

7 . So the following relations on the exponents are apparent:

a2 ≡ 2a1 + a4 (mod 3), a4 ≡ a3 (mod 3), a6 ≡ 2a5 (mod 3), and a7 ≡ 1 (mod 3) and the equation can

be rewritten:

E38 = ea1

1 e2a1+a32 ea3

3 ea34 ea5

5 e2a56 e7e8.

Conjugating with σ1 gives the following relations:

e1−σ13 = e−1

1 e2e33e

−34 , e1−σ1

4 = e−11 e2e

−13 e34, e1−σ1

5 = e5e−16 , e1−σ1

6 = e5e26, e1−σ1

7 = e37e38, e1−σ1

8 = e7

Thus (E1−σ18 )3 = (e−1

1 e2e33e

−34 )a3(e−1

1 e2e−13 e34)a3(e5e−1

6 )a5(e5e26)2a5e37e38e7.

Combining like terms, moving the cubes to the left and renaming (E1−σ18 )3

(E∗8 )3 = ε′1

2a3 ε′33a5 ε4

ε2a31 ε

a32 ε

3a53

This can only have solutions for Type III fields with similar norm hence a88 �= 3 =⇒ a88 = 1 and b∗ = 1.

Suppose N(B1) ∼ ζaN(B2) ∼ ζcN(B3): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, u4 >

We know from [9] Theorem X that e3a = ζa(ε1ε2/ε′1ε′2) and e3c = ζc(ε1ε3/ε′1ε

′3) both have solutions ea

and ec in L so b∗ is at least 2. Let E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, 3

√ε1ε3ε′1ε

′3, ε4, u4 >. Then E0 ⊂ E1 and we will

show that[E : E1

]= 1, thus

[E : E0

]=[E : E1

] [E1 : E0

]= 32



e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e1e3, e5e5 = e25, e6e6 = e1e5, e7e7 = e27, e8e8 = e7

Suppose a88 = 3, then E38 = ea1

1 ea22 ea3

3 ea44 ea5

5 ea66 ea7


(E8E8)3 = e2a1−a2+a4+a61 e2a3+a4

3 e2a5+a65 e2a7+1

7

We get the following relations on the exponents: a2 ≡ 2a1 + a4 + a6 (mod 3), a4 ≡ a3 (mod 3),

a6 ≡ a5 (mod 3), and a7 ≡ 1 (mod 3) and the equation becomes E38 = ea1

1 e2a1+a3+a52 ea3

3 ea34 ea5

5 ea56 e27e8.

We will again take conjugates with σ1.

e1−σ13 = e−1

1 e2e33e

−34 , e1−σ1

4 = e−11 e2e

−13 e34, e1−σ1

5 = e−11 e2e

36, e1−σ1

6 = e−11 e2e

−15 e36, e1−σ1

7 = e37e−38 ,

e1−σ18 = e7

Taking the quotient gives (E1−σ18 )3 = (e−1

1 e2e33e

−34 )a3(e−1

1 e2e−13 e34)a3(e−1

1 e2e36)a5(e−1

1 e2e−15 e36)a5e37e

−38 e7.

And combining like terms, moving the cubes to the left and renaming (E1−σ18 )3:

(E∗8 )3 = e−2a3−2a5

1 e2a3+2a52 e−a3

3 e−a55 e7 = (ε′1)

2a3+2a5 ε4

ε2a3+2a51 ε

a32 ε

a53


This can have no solution because the exponent on ε4 is not divisible by 3 hence a88 �= 3 =⇒ a88 = 1 so

b∗ = 2.

Suppose N(B1)N(B2) ∼ ζaN(B3): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, u4 >


′2ε

′′3

has a solution ea in L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, ε

′2, ε3, 3


′2ε

′′3, ε4, u4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1, thus[

E : E0

]=[E : E1

] [E1 : E0

]= 31



e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e−1

3 , e5e5 = e25, e6e6 = e1e3e5, e7e7 = e27, e8e8 = e7


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7

7 e8 and multiplication with the complex conjugate

gives:

(E8E8)3 = e2a1−a2+a61 e2a3−a4+a6

3 e2a5+a65 e2a7+1

7

We get the following relations on the exponents: a2 ≡ 2a1 + a6 (mod 3), a4 ≡ 2a3 + a6 (mod 3),

a6 ≡ a5 (mod 3), and a7 ≡ 1 (mod 3) and the equation can be rewritten:

E38 = ea1

1 e2a1+a52 ea3

3 e2a3+a54 ea5

5 ea56 e7e8.


e1−σ13 = e3e

−14 , e1−σ1

4 = e3e24, e1−σ1

5 = e1e−12 e3e

−14 e35e

−36 , e1−σ1

6 = e−14 e5, e1−σ1

7 = e37e−38 ,

e1−σ18 = e7.

Taking the quotient gives (E1−σ18 )3 = (e3e−1

4 )a3(e3e24)2a3+a5(e−14 e5)a5(e1e−1

2 e3e−14 e35e

−36 )a5e37e

−38 e7 and

combining like terms, moving the cubes to the left and renaming (E1−σ18 )3 = (E′

8)3

(E′8)3 = ea5

1 e−a52 e3a3+2a5

3 e3a34 ea5

5 e7 =(ε1)a5ε2a5

2 εa53 ε4

ε′1a5 .

This can have no solution because the exponent on ε4 is not divisible by 3, hence a88 �= 3 =⇒ a88 = 1 so

b∗ = 1.

(E) Four Type III:

Suppose no norms or products of norms are similar: E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >


We will again find the following relations between the conjugates: ε2i−1ε′2i−1 = ε22i−1 and ε2iε

′2i = ε−1

2i−1.

Suppose a88 = 3. Then E38 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

3 ε′3a6εa7

4 ε′4 and multiplying by the complex conjugate


gives:

(E8E8)3 = ε2a1−a21 ε2a3−a4

2 ε2a5−a63 ε2a7−1

4

The following relations on the exponents are apparent: a2 ≡ 2a1 (mod 3), a4 ≡ 2a3 (mod 3), a6 ≡2a5 (mod 3), and a7 ≡ 2 (mod 3) and the equation can be rewritten:

E38 = εa1

1 ε′12a1εa3

2 ε′22a3εa5

3 ε′32a5ε24ε

′4.

We know that this can have no solutions in L by [9] because no product of the norms is a cube hence b∗ = 0

Suppose N(B1) ∼ ζaN(B2) and no other norms or products of norms are similar:

E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >

We know from [9] Theorem X that e3a = ζa(ε1ε2/ε′1ε′2) has solutions ea in L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, ε

′3, ε4, ε

′4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1, thus

[E : E0

]=[

E : E1

] [E1 : E0

]= 31.




5 , e7e7 = e27, e8e8 = e−17


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7


(E8E8)3 = e2a1−a2+a41 e2a3+a4

3 e2a5−a65 e2a7−1

7 = ε2a1−a2+a41 ε2a3+a4

2 ε2a5−a63 ε2a7−1

4 .

The following relations on the exponents are apparent: a2 ≡ 2a1 + a4 (mod 3), a4 ≡ a3 (mod 3), a6 ≡2a5 (mod 3), and a7 ≡ 2 (mod 3) and the equation can be rewritten:

E38 = ea1

1 e2a1+a32 ea3

3 ea34 ea5

5 e2a56 e27e8.

Conjugating with σ1 gives the following relations:

e1−σ13 = e−1

1 e2e34, e1−σ1

4 = e−11 e2e

−13 e34, e1−σ1

5 = e5e−16 , e1−σ1

6 = e5e26, e1−σ1

7 = e7e−18 , e1−σ1

8 = e7e28.

Thus (E1−σ18 )3 = (e−1

1 e2e34)a3(e−1

1 e2e−13 e34)a3(e5e−1

6 )a5(e5e26)2a5(e7e−18 )2e7e28. Combining like terms, mov-

ing the cubes to the left and renaming (E1−σ18 )3 gives

(E∗8 )3 =

ε′12a3

ε2a31 εa3

2

and all exponents must be divisible by 3 so a3 ≡ 0 (mod 3). Now

E38 = (ε1ε′1

2)a1(ε3ε′32)a5ε24ε

′4


which has no solution in L unless N(B1) ∼ ζaN(B2) ∼ ζcN(B4) hence a88 �= 3 =⇒ a88 = 1 and b∗ = 1.

Suppose N(B1) ∼ ζaN(B2) ∼ ζcN(B3): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >

Any three of the norms can be similar in this case but the fourth will always be different.

We know from [9] Theorem X that e3a = ζa(ε1ε2/ε′1ε′2) and e3c = ζc(ε1ε3/ε′1ε

′3) both have solutions ea

and ec in L so b∗ is at least 2. Let E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, 3

√ε1ε3ε′1ε

′3, ε4, ε

′4 >. Then E0 ⊂ E1 and we will

show that[E : E1

]= 1, thus

[E : E0

]=[E : E1

] [E1 : E0

]= 32

This case is similar to Case 3 with three Type III fields so we know that a11 = a22 = a33 = a44 =

a55 = a66 = a77 = 1. We will again number the basis elements e1 to e8 and find some relations between

the conjugates.

eiei = e2i for i = 1, 3, 5, 7, e2e2 = e−11 , e4e4 = e1e3, e6e6 = e1e5, e8e8 = e−1

7


1 ea22 ea3

3 ea44 ea5

5 ea66 ea7


(E8E8)3 = e2a1−a2+a4+a61 e2a3+a4

3 e2a5+a65 e2a7+1

7 .

This gives the following relations on the exponents: a2 ≡ 2a1 + a4 + a6 (mod 3), a4 ≡ a3 (mod 3), a6 ≡a5 (mod 3), and a7 ≡ 2 (mod 3) and the equation can be rewritten: E3

8 = ea11 e2a1+a3+a5

2 ea33 ea3

4 ea55 ea5

6 e27e8.


e1−σ13 = e1e

−12 e33e

−34 , e1−σ1

4 = e−11 e2e

−13 e34, e1−σ1

5 = e−11 e2e

36, e1−σ1

6 = e−11 e2e

−15 e36, e1−σ1

7 = e7e−18 ,

e1−σ18 = e7e

28

Then (E1−σ18 )3 = (e1e−1

2 e33e−34 )a3(e−1

1 e2e−13 e34)2a3+a5(e−1

1 e2e36)a5(e−1

1 e2e−15 e36)a5(e7e−1

8 )2e27e28.

Combining like terms, moving the cubes to the left and renaming (E1−σ18 )3 = (E∗

8 )3

(E∗8 )3 = e−a5

1 ea52 e−a3

3 =ε′1

a5

εa51 εa3

2

.

Then a5 = a3 = 0 so E38 = ea1

1 e2a12 e27e8 = εa1

1 ε′12a1ε24ε

′4 This can have no solution because N(B1) � N(B4)

hence a88 �= 3 =⇒ a88 = 1 and b∗ = 2.

Suppose N(B1) ∼ N(B2) � N(B3) ∼ N(B4): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >

We know from [9] Theorem X that e3 = ζa(ε1ε2/ε′1ε′2) and e3 = ζc(ε3ε4/ε′3ε

′4) both have solutions in L

so b∗ is at least 2. Let E1 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, ε

′3, ε4, 3

√ε3ε4ε′3ε

′4>. Then E0 ⊂ E1 and we will show that[

E : E1

]= 1, thus

[E : E0

]=[E : E1

] [E1 : E0

]= 32

This case is similar to Case 3 with three Type III fields so we know that a11 = a22 = a33 = a44 =

a55 = a66 = a77 = 1. We will again number the basis elements e1 to e8 and find some relations between

the conjugates.


5 , e7e7 = e27, e8e8 = e5e7



1 ea22 ea3

3 ea44 ea5

5 ea66 ea7

7 e8 and the product with the complex conjugate is

(E8E8)3 = e2a1−a2+a41 e2a3+a4

3 e2a5−a6+15 e2a7+1

7

We get the following relations on the exponents: a2 ≡ 2a1 + a4 (mod 3), a4 ≡ a3 (mod 3), a6 ≡ 2a5 +

1 (mod 3), and a7 ≡ 1 (mod 3) and the equation can be rewritten:

E38 = ea1

1 e2a1+a42 ea3

3 ea34 ea5

5 e2a5+16 e7e8.


e1−σ13 = e−1

1 e2e34, e1−σ1

4 = e−11 e2e

−13 e34, e1−σ1

5 = e5e−16 , e1−σ1

6 = e5e26, e1−σ1

7 = e−15 e6e

38,

e1−σ18 = e−1

5 e−17 e38

Then (E1−σ18 )3 = (e−1

1 e2e34)a3(e−1

1 e2e−13 e34)a3(e5e−1

6 )a5(e5e26)2a5+1(e−15 e6e

38)(e−1

5 e−17 e38)

Combining like terms, moving the cubes to the left and renaming (E1−σ18 )3 we get

(E∗8 )3 = e−2a3

1 e22a3e−a3

3 e−15 e−1

7

This can only have no solutions since a11 = . . . = a77 = 1 hence a88 �= 3 =⇒ a88 = 1 and b∗ = 2.

Suppose N(B1)N(B2) ∼ ζaN(B3) and N(B4) � N(Bi) for 1 ≤ i ≤ 3:

E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >


′2ε

′′3

has a solution ea in L so b∗ is at least 1. Let E1 =<

ε1, ε′1, ε2, ε

′2, ε3, 3


′2ε

′′3, ε4, ε

′4 >. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1, thus

[E : E0

]=[

E : E1

] [E1 : E0

]= 31.


Suppose a88 = 3 then E38 = εa1

1 ε′1a2εa3

2 ε′2a4εa5

33


′2ε

′′3

a6εa74 ε′4. Using the usual technique we multiply by

the complex conjugate and simplify to get

E38 = ε2a1−a2+a6

1 ε2a3−a4+a62 ε2a5+a6

3 ε2a7−14 .

The following relations on the exponents are clear: a2 ≡ 2a1 + a5 (mod 3), a4 ≡ 2a3 + a5 (mod 3),

a6 ≡ a5 (mod 3), and a7 ≡ 2 (mod 3) and the equation can be rewritten

E38 = εa1

1 ε′12a1+a5εa3

2 ε′22a3+a5εa5

33


′2ε

′′3

a5

ε24ε′4.

We can conjugate with 1 − σ1 and simplify to get

(E1−σ18 )3 = ε1

a5ε′1−a5ε2a5

2 εa53 ε34

so a5 = 0 and E38 = εa1

1 ε′12a1εa3

2 ε′22a3ε24ε

′4. We know this can have no solution because the norms are not

similar so b∗ = 1.


Suppose N(B1)N(B2) ∼ ζaN(B3) and N(B4) ∼ N(B1): E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >


′2ε

′′3

and e3b = ζb ε1ε4ε′1ε

′4

have a solutions in L so b∗ is at least

2. Let E1 =< ε1, ε′1, ε2, ε

′2, ε3, 3


′2ε

′′3, ε4, 3

√ε1ε4ε′1ε

′4>. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1,

thus[E : E0

]=[E : E1

] [E1 : E0

]= 32.



1 ε′1a2εa3

2 ε′2a4εa5

33


′2ε

′′3

a6εa74

3

√ε1ε4ε′1ε

′4. Using the usual technique we multiply

by the complex conjugate and simplify to get

E38 = ε2a1−a2+a6+1

1 ε2a3−a4+a62 ε2a5+a6

3 ε2a7+14 .

The following relations on the exponents are clear: a2 ≡ 2a1 + a5 + 1 (mod 3), a4 ≡ 2a3 + a5 (mod 3),

a6 ≡ a5 (mod 3), and a7 ≡ 1 (mod 3) and the equation can be rewritten

E38 = εa1

1 ε′12a1+a5+1

εa32 ε′2

2a3+a5εa53

3


′2ε

′′3

a5

ε4 3

√ε1ε4ε′1ε

′4

.

Now we conjugate with σ1 and take the quotient to get

(E1−σ18 )3 ≈ ε1

a5−2ε′1−a5+2

εa53 ε−1

4

which has no solution because the exponents are not all divisible by 3. Thus a88 = 1 and b∗ = 2.

Suppose N(B1)N(B2)N(B3) ∼ ζaN(B4) or N(B1)N(B2) ∼ ζaN(B3)N(B4):

E0 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >

We know that e3a = ζa ε1ε2ε3ε4ε′1ε

′′2 ε′3ε

′4

has a solutions in L so b∗ is at least 1. Let

E1 =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, 3

√ε1ε2ε3ε4ε′1ε

′′2 ε′3ε

′4>. Then E0 ⊂ E1 and we will show that

[E : E1

]= 1, thus[

E : E0

]=[E : E1

] [E1 : E0

]= 31.



1 ε′1a2εa3

2 ε′2a4εa5

3 ε′3a6εa7

43


′′2 ε′3ε

′4. Using the usual technique we multiply by

the complex conjugate and simplify to get

E38 = ε2a1−a2+1

1 ε2a3−a4+12 ε2a5−a6+1

3 ε2a7+14 .

The following relations on the exponents are clear: a2 ≡ 2a1 + 1 (mod 3), a4 ≡ 2a3 + 1 (mod 3), a6 ≡2a5 + 1 (mod 3), and a7 ≡ 1 (mod 3) and the equation can be rewritten

E38 = εa1

1 ε′12a1+1

εa32 ε′2

2a3+1εa53 ε′3

2a5+1ε4 3


′′2ε

′3ε

′4

.


Now we conjugate with σ1 and take the quotient and simplify to get

(E∗8 )3 = ε1

−1ε′1ε′2ε

−14

which has no solution because the exponents are not all divisible by 3. Thus a88 = 1 and b∗ = 1 and the

theorem is proved.

We can see from Theorem 6.9 that to find units in L that are not contained in any of the subfields we

need to have at least two of the cubic subfields to be Type III or for K to be of Kind 1 or 2. We will now

describe a procedure for calculating the units in L as well as the basis E given that we have a basis e for

K. The units from the Type III fields will be discussed near the end of this section. Theorem 6.10 will

show that we can use Hilbert’s Theorem 90 to find units in L when K is of Kind 1 or 2.

Theorem 6.10 Let EL ∈ E and EL /∈ εεi for any i and K be of Kind 1 or 2. If {e1, e2, e3, e4} is a basis

for e then EL

EσL

= ec11 ec22 ec33 ec44 for some σ where 0 ≤ ci ≤ 2 for i = 1, 2, 3, 4.

Proof: For K being of Kind 1 or 2 the basis elements for the units of K can be chosen with elements

of the form e = εk or e3 = εbi

i εbj

j εk where 0 ≤ bi, bj ≤ 2. So for each unit in K there is some base field kl

such that the norm NK/kl(e) = 1 and by renumbering our fields we can always choose kl = k1. We will

then write σ = σ1 and N(α) = NK/k1(α) for α ∈ K.

We would like to show that the units of L, which are not contained in any of the subfields, can be

constructed from units of K using Hilbert’s Theorem 90. We will then have to consider the different Kinds

of K as separate cases since they have distinct bases.

Case 1: Suppose K is of Kind 1. Then a basis for e can be chosen as e1 = ε1, e32 = εa1ε2, e33 = εb1ε3,

e34 = εc1ε4 and ε1 =< ε1, u1, e2, e′2, e3, e

′3, e4, e

′4, >.

In this case we have that εε1 = ε1 because ε ⊆ ε1 since ui = 3

√εi

ε′i

for i = 2, 3, 4 and clearly ε1 ⊆ εε1.

Since E3 ⊆ ε1 then we can write

E3L = εa1

1 ua21 ea3

2 e′2a4ea5

3 e′3a6ea7

4 e′4a8 . (6.11)

Note: e2e′2e′′2 = 3

√ε3a1 ε2ε′2ε

′′2 = εa1 and similarly e3e

′3e

′′3 = εb1 and e4e

′4e

′′4 = εc1

Then if we multiply E3L by its complex conjugate we get

(E1+τL )3 = ε2a1+a2

1 e2a32 e′2

a4e′′2a4e2a5

3 e′3a6e′′3

a6e2a74 e′4

a8e′′4a8

= ε2a1+a2+a·a4+b·a6+c·a81 e2a3−a4

2 e2a5−a63 e2a7−a8

4

Since E1+τL ∈ K then all the exponents must be divisible by 3 which gives the following relations:


a4 ≡ 2a3 (mod 3)

a6 ≡ 2a5 (mod 3)

a8 ≡ 2a7 (mod 3)

a2 ≡ a1 + a · a3 + b · a5 + c · a7 (mod 3)

And equation (6.11) can be written as

E3L = εa1

1 ua21 ea3

2 e′22a3ea5

3 e′32a5ea7

4 e′42a7 (6.12)

Note: e′iστ = ei

σ2τ = (eτi )σ = eσi = e′i and eiστ = (eτi )σ

2= eσ

2

i = e′′i for i = 2, 3, 4.

Then

(E1+στL )3 = ε2a1+a2

1 (e2e′24e′′2)a3(e3e′3

4e′′3)a5(e4e′4

4e′′4)a7

= ε2a1+a2+a·a3+b·a5+c·a71 (e′2)3a3(e′3)3a5(e′4)3a7

So we now have two relationships for the exponent of ε1

2a1 + a2 + a · a3 + b · a5 + c · a7 ≡ 0 (mod 3)

2a1 + a2 + 2(a · a3 + b · a5 + c · a7) ≡ 0 (mod 3),

which we can add together to get

4a1 + 2a2 ≡ 0 (mod 3)

a2 ≡ a1 (mod 3).

So equation (6.12) becomes

E3L = εa1

1 ua11 ea3

2 e′22a3ea5

3 e′32a5ea7

4 e′42a7

and

a · a3 + b · a5 + c · a7 ≡ 0 (mod 3) (6.13)

To use Hilbert’s Theorem 90 we need to find e ∈ K such that N(e) = 1. We have that a·a3+b·a5+c·a7 =

3t so there are 2 possible cases for the solutions to the equation which are based on whether a, b, or c are

zero or not. If {a, b, c} = {0, 0, 0} then there are three independent solutions for {a3, a5, a7}, otherwise

there are two independent solutions for {a3, a5, a7}.

For each case let ei = ea3i

2 ea5i

3 ea7i

4 ε−ti1 where {a3i , a5i , a7i} is a solution to a · a3i + b · a5i + c · a7i = 3ti

and ti = (a · a3i + b · a5i + c · a7i)/3 where i = 1, 2 in the case of two solutions and i = 1, 2, 3 if there are

three. Then N(ei) = εa·a3i

+b·a5i+c·a7i

1 ε−3ti1 = 1 and for each i we can write B1−σ

i = ei and B3i

N(Bi)= ei

eσ2i

.


Without loss of generality suppose there are two independent solutions to (6.13). If N(Bd11 Bd2

2 ) = α3j

for some αj ∈ K1 has solution(s) {d1, d2}, not both zero, then there are units in L that are not products

of the units of the subfields, otherwise there are none. Let {d1, d2} be one of those solutions, then

ed11 ed2

2 = (ea312 e

a513 e

a714 ε−t1

1 )d1(ea322 e

a523 e

a724 ε−t2

1 )d2

= ea31d1+a32d2

2 ea51d1+a52d2

3 ea71d1+a72d2

4 ε−t1d1−t2d21

and we can rename the exponents to get ed11 ed2

2 = ea32 ea5

3 ea74 ε−t

1 . Now,

(Bd1

1 Bd22

αj

)3

=ed11 ed2

2

(ed11 ed2

2 )σ2=

ea32 ea5

3 ea74 ε−t

1

e′′2a3e′′3

a5e′′4a7ε−t

1

=e2a32 e′2

a3e2a53 e′3

a5e2a74 e′4

a7

N(ea32 ea5

3 ea74 )

= e2a32 e′2

a3e2a53 e′3

a5e2a74 e′4

a7ε3t1

Then we can square both sides, let(

Bd11 B

d22

αj

)2

= Ej and reduce modulo cubes to get

(Ej)3 = ea32 e′2

2a3ea53 e′3

2a5ea74 e′4

2a7

so Ej = EL.

If there are three solutions to (6.13) then there may be three solution for N(Bd11 Bd2

2 Bd33 ) = α3

j for some

αj ∈ K1 but Theorem 6.9 shows there will be at most two independent solutions for EL.

Case 2: Suppose K is of Kind 2. Then a basis for e can be chosen in two ways depending on whether

or not one of the subfields is Type III. If all the subfields are Type I or IV then the basis can be chosen as

e1 = ε1, e2 = ε2, e33 = εa1εb2ε3, e34 = εc1ε

d2ε4 and εε1 =< ε1, u1, ε2, u2, e3, e

′3, e4, e

′4, >. Since E3 ⊆ εε1 then we

can write

E3L = εa1

1 ua21 εa3

2 ua42 ea5

3 e′3a6ea7

4 e′4a8 (6.14)

Note: e3e′3e′′3 = 3

√ε3a1 (ε2ε′2ε′′2)bε3ε′3ε′′3 = εa1 and similarly e4e

′4e

′′4 = εc1

Then if we multiply E3L by its complex conjugate we get

(E1+τL )3 = ε2a1+a2

1 ε2a3+a42 e2a5

3 e′3a6e′′3

a6e2a74 e′4

a8e′′4a8

= ε2a1+a2+a·a6+c·a81 ε2a3+a4

2 e2a5−a63 e2a7−a8

4


a6 ≡ 2a5 (mod 3), a8 ≡ 2a7 (mod 3), a4 ≡ a3 (mod 3), 0 ≡ 2a1 + a2 + a · 2a5 + c · 2a7 (mod 3).

Then we have that

E3L = εa1

1 ua21 εa3

2 u2a3ea5

3 e′32a5ea7

4 e′42a7


Note: e′iστ = e′i and eστ

i = e′′i for i = 3, 4 and u1+στ2 = ε1+στ

2 = ε′2−1.

Then

(E1+στL )3 = ε2a1+a2

1 ε′2−2a3(e3e′3

4e′′3)a5(e4e′4

4e′′4)a7

= ε2a1+a2+a·a5+c·a71 ε′2

−2a3e′33a5e′4

3a7 .

This shows that a3 = 0 and provides a second congruence for the exponent on ε1.

2a1 + a2 + a · 2a5 + c · 2a7 ≡ 0 (mod 3)

2a1 + a2 + a · a5 + c · a7 ≡ 0 (mod 3).

These two together show that

a2 ≡ a1 (mod 3)

0 ≡ a · a5 + c · a7 (mod 3) (6.15)

and equation (6.14) becomes

E3L = εa1

1 ua11 ea5

3 e′32a5ea7

4 e′42a7 .

To use Hilbert’s Theorem 90 we need to find e ∈ K such that N(e) = 1. We have that a ·a5 + c ·a7 = 3t

so there are 2 possible cases for the solutions to the equation which are based on whether a, or c are zero

or not. If {a, c} = {0, 0} then there are two independent solutions for {a5, a7}, otherwise there is one

solutions for {a5, a7}.

For each case let ei = ea5i3 e

a7i4 ε−ti

1 where {a5i , a7i} is a solution to a ·a5i +c ·a7i = 3ti and ti = (a ·a5i +c ·a7i)/3 where i = 1 in the case of one solution and i = 1, 2 if there are two. Then N(ei) = ε

a·a5i+c·a7i

1 ε−3ti1 = 1

and for each i we can write B1−σi = ei and B3

i

N(Bi)= ei

eσ2i

.


2 ) = α3j



ed11 ed2

2 = (ea513 e

a714 ε−t1

1 )d1(ea523 e

a724 ε−t2

1 )d2

= ea51d1+a52d23 e

a71d1+a72d24 ε−t1d1−t2d2

1


2 = ea53 ea7

4 ε−t1 . Now,

(Bd1

1 Bd22

αj

)3

=ed11 ed2

2

(ed11 ed2

2 )σ2=

ea53 ea7

4 ε−t1

e′′3a5e′′4

a7ε−t1

=e2a53 e′3

a5e2a74 e′4

a7

N(ea53 ea7

4 )= e2a5

3 e′3a5e2a7

4 e′4a7ε3t1



Bd11 B

d22

αj

)2


(Ej)3 = ea53 e′3

2a5ea74 e′4

2a7

so Ej = EL.

If there are two solutions to (6.15) then there may be two solution for N(Bd11 Bd2

2 ) = α3j for some

αj ∈ K1 but Theorem 6.9 shows there will be at most one independent solution for EL.

If one of the subfields is Type III then let k2 be that field and the basis can be chosen as e1 = ε1,

e2 = ε2, e33 = εa1ε3, e34 = εb1ε4 and εε1 =< ε1, u1, ε2, ε′2, e3, e

′3, e4, e

′4, >. Since E3 ⊆ εε1 then we can write

E3L = εa1

1 ua21 εa3

2 ε′2a4ea5

3 e′3a6ea7

4 e′4a8 . (6.16)

Note: e3e′3e′′3 = εa1 and similarly e4e

′4e

′′4 = εb1

Proceeding the same way as before:

(E1+τL )3 = ε2a1+a2

1 ε2a32 ε′2

a4ε′′2a4e2a5

3 e′3a6e′′3

a6e2a74 e′4

a8e′′4a8

= ε2a1+a2+a·a6+b·a81 ε2a3−a4

2 e2a5−a63 e2a7−a8

4 .


a4 ≡ 2a3 (mod 3), a6 ≡ 2a5 (mod 3), a8 ≡ 2a7 (mod 3), 0 ≡ 2a1 + a2 + a · 2a5 + b · 2a7 (mod 3).

And we have that

E3L = εa1

1 ua21 εa3

2 ε′22a3ea5

3 e′32a5ea7

4 e′42a7 .

Note: e′iστ = e′i and eστ

i = e′′i for i = 3, 4 and εστ2 = ε′′2 .

Then

(E1+στL )3 = ε2a1+a2

1 (ε2ε′24ε2)a3(e3e′3

4e′′3)a5(e4e′4

4e′′4)a7

= ε2a1+a2+a·a5+b·a71 ε′2

3a3e′33a5e′4

3a7 .

This provides a second equation for the exponent on ε1.

2a1 + a2 + a · 2a5 + b · 2a7 ≡ 0 (mod 3)

2a1 + a2 + a · a5 + b · a7 ≡ 0 (mod 3).

These two together show that

a2 ≡ a1 (mod 3)

0 ≡ a · a5 + b · a7 (mod 3). (6.17)


Then equation (6.16) becomes

E3L = εa1

1 ua11 εa3

2 ε′22a3ea5

3 e′32a5ea7

4 e′42a7 .

To use Hilbert’s Theorem 90 we need to find e ∈ K such that N(e) = 1. We have that a ·a5 + b ·a7 = 3t

so there are 2 possible cases for the solutions to the equation which are based on whether a, or b are zero

or not. If {a, b} = {0, 0} then there are two independent solutions for {a5, a7}, otherwise there is one

solutions for {a5, a7}. The value of a3 is determined by the calculation below.

For each case let ei = εa3i

2 ea5i

3 ea7i

4 ε−ti1 where 0 ≤ a3i ≤ 2 and {a5i , a7i} is a solution to a·a5i +b·a7i = 3ti

and i = 1 in the case of one solutions and i = 1, 2 if there are two. Then N(ei) = εa·a5i

+b·a7i1 ε−3t

1 = 1. For

each choice of a3i and {a5i , a7i} we can write B1−σi = ei and B3

i

N(Bi)= ei

eσ2i

.


2 ) = α3j



ed11 ed2

2 = (εa312 e

a513 e

a714 ε−t1

1 )d1(εa322 e

a523 e

a724 ε−t2

1 )d2

= εa31d1+a32d2

2 ea51d1+a52d2

3 ea71d1+a72d2

4 ε−t1d1−t2d21


2 = εa32 ea5

3 ea74 ε−t

1 . Now,

(Bd1

1 Bd22

αj

)3

=ed11 ed2

2

(ed11 ed2

2 )σ2=

εa32 ea5

3 ea74 ε−t

1

ε′′2a3e′′3

a5e′′4a7ε−t

1

=ε2a32 ε′2

a3e2a53 e′3

a5e2a74 e′4

a7

N(εa32 ea5

3 ea74 )

= ε2a32 ε′2

a3e2a53 e′3

a5e2a74 e′4

a7ε3t1


Bd11 B

d22

αj

)2


(Ej)3 = εa32 ε′2

2a3ea53 e′3

2a5ea74 e′4

2a7

so Ej = EL.

Corollary 6.10.1 will show that our algorithm which will be presented in the next section will produce

the new units in L. Corollary 6.10.2 will show that in the case where three of the cubic subfields are of

Type IV there can be no new unit in L.

Corollary 6.10.1 Let K and L be as in Theorem 6.10 and {e1, e2, e3, e4} be a basis for e.

(1) Suppose K is of Kind 1,

E3 = εa11 ua1

1 ea32 e′2

2a3ea53 e′3

2a5ea74 e′4

2a7


has a solution in L and a · a3 + b · a5 + c · a7 = 3t. If B1−σ = ea32 ea5

3 ea74 ε−t

1 then εd1ud1N(B) = mj

2α3 for

some α ∈ K1 and 0 ≤ j, d ≤ 2.

(2) Suppose K is of Kind 2, all subfields are Type I or IV,

E3 = εa11 ua1

1 ea53 e′3

2a5ea74 e′4

2a7

has a solution in L and a · a5 + c · a7 = 3t. If B1−σ = ea53 ea7

4 ε−t1 then εd1u

d1N(B) = mj

2α3 for some α ∈ K1

and 0 ≤ j, d ≤ 2.

(3) Suppose K is of Kind 2, k2 is of Type III,

E3 = εa11 ua1

1 εa32 ε′2

2a3ea53 e′3

2a5ea74 e′4

2a7

has a solution in L and a · a5 + c · a7 = 3t. If B1−σ = εa32 ea5

3 ea74 ε−t

1 then εd1ud1N(B) = mj

2α3 for some

α ∈ K1 and 0 ≤ j, d ≤ 2.

Proof: Since

E3 = εa11 ua1

1 ea32 e′2

2a3ea53 e′3

2a5ea74 e′4

2a7

then

E3ε−a11 u−a1

1 = ea32 e′2

2a3ea53 e′3

2a5ea74 e′4

2a7 .

As in the proof of Theorem 6.10 we have that

(B)3

N(B)= γ3ea3

2 e′22a3ea5

3 e′32a5ea7

4 e′42a7

and by combining these last two equations we get

(B)3

N(B)= γ3E3ε−a1

1 u−a11 .

Moving all the cubed terms to one side of the equation shows that(

BγE

)3

= N(B)ε−a11 u−a1

1 where

N(B)ε−a11 u−a1

1 ∈ K1. Since N(B)ε−a11 u−a1

1 is also a cube in K1,

K1

(3√N(B)εd1ud

1

)⊆ L = K1 ( 3

√m2)

and hence Kummer Theory says that N(B)εd1ud1 = mj

2α3 where α ∈ K1 and j = 0, 1, or 2.

The proofs of the other two cases are the same.

Corollary 6.10.2 Let K be of Kind 2 with at least three of the subfields Type IV then E = εεi for any i.


Proof: If EL ∈ E then we know from Theorem 6.10 that

E3L = εa1

1 ua11 ea5

3 e′32a5ea7

4 e′42a7 (6.18)

with e33 = εa1εb2ε3, e34 = εc1ε

d2ε4, e′3

3 = εa1ε′2bε′3, e′4

3 = εc1ε′2dε′4, and a · a5 + c · a7 ≡ 0 (mod 3). There are two

cases to consider, either one of the subfields is Type I or not. In the first case we will assume without loss

of generality that k3 is Type I and e33 = ε3. Then a = 0 so c · a7 ≡ 0(mod 3) =⇒ a7 ≡ 0(mod 3) since

c �= 0 by [9] Corollary I to Theorem X so then (6.18) becomes

E3L = εa1

1 ua11 ea5

3 e′32a5

and we know that has no solution from the proof of Theorem 6.9 Case 2(A).

Suppose all four subfields are Type IV then abcd �= 0 so a · a5 + c · a7 ≡ 0 (mod 3) has a solution where

a5 and a7 are not divisible by 3.

Note: (ε1)σ2τ = u31ε

−21 , (u1)σ2τ = u2

1ε−11 , (e3)σ2τ = 3

√ε′′1

aεb2ε′′3 = (e3e′3)−1ua

1εb2u

−b2 , (e′3)σ2τ = 3

√ε′′1

aε′′2bε′3 =

e′3ua1ε

−a1 u2b

2 ε−b2 , (e4)σ2τ = 3

√ε′′1

cεd2ε′4 = e′4ε

−c1 uc

1ud2, and (e′4)σ2τ = 3

√ε′′1

cε′′2dε4 = e4ε

−c1 uc

1ε−d2 ud

2

Then NL/Kσ2 (E3L) = (E3

L)1+σ2τ ,

(E3L)1+σ2τ = ε−2a1−2a·a5−3c·a7

1 u6a1+3a·a5+3c·a71 ε−b·a5−2d·a7

2 u3b·a5+3d·a72 e′3

3a5e3a74 e′4

3a7

and some simplification gives

(E3L)1+σ2τ = ε′1

−(2a1+a·a5+c·a7)ε′3a5ε4

a7ε′4a7 .

Now (EL)1+σ2τ is in Kσ2 = k′1k2, a field of index 2 which is isomorphic to K, so we can write (EL)1+σ2τ

in terms of the basis for Kσ2 , which is < ε′1, ε2, e∗3, e

∗4 > where (e∗3)3 = ε′1

aεb2ε

′3 and (e∗4)3 = ε′1

cεd2ε

′′4 . So

(E3L)1+σ2τ = ε′1

−2a1−2a·a5ε−b·a5+d·a72 (e∗3)3a5(e∗4)−3a7

and since [E : Kσ2 ] = 2 then all the exponents must be divisible by 3 so we get the following congruences:

2a1 + 2a · a5 ≡ 0 (mod 3)

−b · a5 + d · a7 ≡ 0 (mod 3).

So we know that a5 ≡ a · a1 (mod 3) and a7 ≡ abd · a1 (mod 3) so our initial condition a · a5 + c · a7 ≡0 (mod 3) becomes (1 + abcd)a1 ≡ 0 (mod 3) and

E3L = εa1

1 ua11 ea·a1

3 e′32a·a1ea·b·d·a1

4 e′42a·b·d·a1 .

To find another relation on the exponents we norm to Kσ4 .


Note: (ε1)σ4τ = u31ε

−21 , (u1)σ4τ = u2

1ε−11 , (e3)σ4τ = 3

√ε′′1

aε′′2bε′3 = (e′3)ua

1ε−a1 ε−b

2 u2b2 , (e′3)σ4τ = 3

√ε′′1

aε′2bε3 =

e3ua1ε

−a1 u−b

2 , (e4)σ4τ = 3√ε′′1

cε′′2dε4 = e4ε

−c1 uc

1ud2ε

−d2 , and (e′4)σ4τ = 3

√ε′′1

cε′2dε′′4 = (e4e′4)−1uc

1εd2u

−2d2 .

Then NL/Kσ4 (E3L) = (E3

L)1+σ4τ ,

(E3L)1+σ4τ = ε−5a1−abcd·a1

1 u9a1+3abcd·a11 u−3ab·a1

2 e33a·a1e′3

3a·a1

which can be simplified to

(E3L)1+σ4τ = ε′1

−(3a1+abcd·a1)ε′22ab·a1(ε3ε′3)a·a1 .

It is again necessary to write this in terms of the basis for Kσ4τ , which is < ε′1, ε′2, e∗∗3 , e∗∗4 > where

(e∗∗3 )3 = ε′1aε′2

bε′′3 and (e∗∗4 )3 = ε′1

cε′2

dε4. So we have

(E3L)1+σ4τ = ε′1

−2a1−abcd·a1ε3ab·a12 (e∗∗3 )−3a·a1

and that gives another relation on the exponents:

−2a1 − abcd · a1 ≡ 0 (mod 3)

Now we have two congruences, one from each norm,

(1 + abcd)a1 ≡ 0 (mod 3)

(2 + abcd)a1 ≡ 0 (mod 3).

Since a, b, c, d �= 0 then adding these two together gives that a1 ≡ 0 (mod 3) =⇒ a5 ≡ a7 ≡ 0 (mod 3)

so E3L = εa1

1 ua11 ea5

3 e′32a5ea7

4 e′42a7 has no nontrivial solution in L so EL = εε1. It is easy to see that εεi = εε1

for i = 2, 3, 4.

Now we know very specifically the cases where a new unit can be found in L and we have shown that

we can find those units using Hilbert’s Theorem 90 when K is of Kind 1 or 2. The next sections will outline

the procedure for calculating the units in L which are not contained in any of the subfields.

6.2 Units in L from Type I Subfields

Let {e1, e2, e3, e4} be a basis for e where [e : e0] ≥ 9. Suppose E0 is a unit in L such that E0 /∈ E0 where

E0 = ε4∏

i=1

εi, then we know from Theorem 6.10 that

E0

Eσ0

= eb11 eb22 eb33 eb44

so we can apply Hilbert’s Theorem 90 to the basis elements of e to find E0.


Let K be of Kind 1 and e =< ε1, e2, e3, e4 > where e3i = εai1 εi with 0 ≤ ai ≤ 2 for i = 2, 3, 4. Consider

the set E = {ec22 , ec33 , ec44 }, where ci = 0 or 1 for each i and as many ci’s as possible are 1, subject to the

condition NL/Kl(eci

i ) = 1 for some l = 1, 2, 3, or 4. Let σ = σl and N = NL/Klthen for each eci

i ∈ E

there exists Bi ∈ L such that eci

i = Bi

Bσi

and(

ei

eσ2i

)ci

= B3i

N(Bi)where N(Bi) = βi ∈ Kl. We want to find a

product Bα = Bb22 Bb3

3 Bb44 , with 0 ≤ b2, b3, b4 ≤ 2 and Bα �= 1, such that N(Bα) = βb2

2 βb33 βb4

4 = (α)3 for

some α ∈ L and we will choose bi = 0 when ci = 0. If such a product exists then we let Es = Bα

α so that

E3s =

(Bb2

2 Bb33 Bb4

4

α

)3

=eb22 eb33 eb44

(eb22 eb33 eb44 )σζaεblu

cl . (6.19)

To find this product we need to be able to solve the equation N(Bα) = βb22 βb3

3 βb44 = (α)3 for the

exponents bi for i = 2, 3, and 4. Since α is unique up to multiplication by units in the base field we will

look for elements of the form

β = ζaεbluclβ

b22 βb3

3 βb44 = α3 (6.20)

where {εl, ul} is a fundamental system of units of Kl.

Let pr ∈ Z be any prime such that pr is not ramified in Kl and pr � NKl/Q(Bα). If pr splits as Pr1 · · ·Prt

in Kl then for P = Prj with 1 ≤ rj ≤ rt we define the map φP : K∗l → Z3 (where K∗

l = Kl − {0}) by

φP (z) = v

where ( zP

)3

= ζv

where(

zP

)3

is the cubic power residue symbol over Kl.

It is well known that(

zP

)3

= z(N(P )−1)/3 ≡ a (mod P ). Let z ∈ Kl then z = a1 + a2ζ + a33√ml +

a4ζ 3√ml + a5

3√m2

l + a6ζ3√m2

l (mod P ). The value of N(P ) depends on the congruence of pr modulo 3.

For pr ≡ 2 (mod 3) we know that pr factors in Kl as Pr1Pr2Pr3 with N(Prj ) = p2r for j = 1, 2, 3. If

we solve the equation x3 − ml ≡ 0 (mod pr) for x ≡ m(2pr−1)/3l ≡ b (mod pr) then we know, for some

ordering of the Prj ’s, that 3√ml ≡ ζj−1b (mod Prj ). So for P = Prj( z

P

)3≡ (a1 + a2ζ + a3ζ

j−1b + a4ζ(ζj−1b) + a5(ζj−1b)2 + a6ζ(ζj−1b)2)(p2r−1)/3 (mod P )

and then ( zP

)3≡ ζv (mod P )( zP

)3

= ζv

where v = 0, 1 or 2.


For pr ≡ 1 (mod 3) and x3 −ml ≡ 0 (mod pr) solvable we know that pr factors in Kl as Pr1 · · ·Pr6

with N(Prj ) = pr for all j so we need to solve two equations

x3 −ml ≡ 0 (mod pr)

and

w2 + w + 1 ≡ 0 (mod pr).

The first equation factors as (x− d1)(x− d2)(x− d3) ≡ 0 (mod pr) and the second as (w − f1)(w − f2) ≡0 (mod pr). Each P is generated by a distinct pair ( 3

√ml − dq, ζ − fs) for q = 1, 2, or 3 and s = 1 or 2.

So for each P we can calculate

( zP

)3≡ (a1 + a2fs + a3dq + a4fsdq + a5d

2q + a6fsd

2q)(pr−1)/3 (mod P )

( zP

)3≡ fv

s (mod P )

and then ( zP

)3

= ζv


For pr ≡ 1 (mod 3) and x3 −ml ≡ 0 (mod pr) not solvable we know that pr factors in Kl as Pr1Pr2

with N(Prj ) = p3r for all j = 1, or 2 so we need to solve

w2 + w + 1 ≡ 0 (mod pr).

This equation factors as (w − f1)(w − f2) ≡ 0 (mod pr) so each P is generated by (ζ − fs) for s = 1 or 2.

Then for each P we can calculate

( zP

)3≡ (a1 + a2fs + a3

3√ml + a4fs 3

√ml + a5

3√ml

2 + a6fs 3√ml

2)(p3r−1)/3 (mod P )

( zP

)3≡ fv

s (mod P )

and then ( zP

)3

= ζv


For each P in a set of n prime ideals, where n is yet to be specified, construct the matrix with rows

W = [φP (ζ), φP (εl), φP (ul), φP (β2), φP (β3), φP (β4)] (6.21)

and solve for vector w = [a, b, c, b2, b3, b4] such that W · w = [0]. This gives a solution to (6.20) that is

a cube modulo every prime in our set. We need to choose n sufficiently large so that we find at most 2


nontrivial solutions for w. By nontrivial we need that at least one of b2, b3 or b4 is non zero modulo 3 for

the entries where the corresponding cj �= 0.

Once we have a possible solution w it is necessary to see if α can be calculated by using the numerical

cube root function in Kl. If there is no solution to 3√β then that choice of w does not produce a new unit

in L. If there is a solution then we have solved for 3√β = α and (6.19) has a solution in L.

This method is probabilistic since it is possible for β to be a cube modulo a large number of consecutive

primes and not be cube of an element of Kl. However, if β is not the cube of an element of Kl then the

equation x3 − β ≡ 0 (mod P ) has a solution for only 1/3 of the primes P ∈ Kl so the probability of being

a cube modulo n primes and not have a solution in Kl is approximately 1/3n.

For the case where K is of Kind 2 then the basis for K can be chosen as e =< ε1, ε2, e3, e4 > where

e33 = εa11 εa2

2 ε3 and e34 = εa31 εa4

2 ε4 where 0 ≤ ai ≤ 2 for 1 ≤ i ≤ 4. Then to find the units in L we follow the

same procedure as for Kind 1 but in this case it may be that e3 and e4 do not both have norm 1 to the

same base field. In that case we simply follow our procedure for each of the units individually. In the case

of Kind 2 we will solve W · w = [0] for only one non trivial solution because there can be only one unit in

L by Theorem 6.9.

If for a particular set E there is more than one choice of l such that NL/Kl(eci

i ) = 1 for each ei ∈ E

then it is necessary to run the algorithm for all choices for l.

6.3 Example Type I units in L

Example 6

The fields k1 = Q( 3√

2), k2 = Q( 3√

5), k3 = Q( 3√

10), and k4 = Q( 3√

20) are all Type I and in Example 1

we found that [e : e0] = 33 and e =< ε1, e2, e3, e4 > where e2 = 3√ε3, e3 = 3

√ε1ε2 and e4 = 3

√ε1ε4. Let

E = {e2, e3, 1} and σ = σ4 then using Hilbert’s Theorem 90 we can find Bi ∈ L such that ei = Bi

Bσi

and

βi = NL/K4(Bi) for i = 2 and 3

B2 = 13 (3 − 3

√2 − ζ 3

√2 + 2 3

√4 + 3

√5 + 3

√20 − ζ 3

√25 + 3

√50)

β2 = 11 + 3ζ + 3 3√

20 + 3 3√

50

B3 = 13 (9 + 6 3

√2 + 2ζ 3

√2 + 2ζ 3

√4 + 3 3

√5 + ζ 3

√5 + 2 3

√10 + ζ 3

√10 − 3

√20 + 3

√25 + 3

√50)

β3 = 18 − 7ζ − 7 3√

20 − 4ζ 3√

20 + 3 3√

50 − 2ζ 3√

50

Using the primes p = 7, 11, 17, 19, 23, 29, 41, 47, 53, 59 and 61, the row-reduced matrix for W is


W =

1 0 0 0 1

0 1 0 0 2

0 0 1 0 2

0 0 0 1 1

.

Then w = {2, 1, 1, 2, 1} so we solve for α3 = ζ2ε4u4β22β3 where

ζ2ε4u4β22β3 = 1620881 + 1667320ζ + 597121 3

√20 + 614210ζ 3

√20 + 439989 3

√50 + 452568ζ 3

√50.

Using the cube root function we find that

α =13

(200 + 73ζ + 70 3√

20 + 26ζ 3√

20 + 52 3√

50 + 17ζ 3√

50)

where α3 = ζ2ε4u4β22β3. Now we can solve for the unit in L

E1 =B2

2B3

α=

19

(1 − ζ + 8 3

√2 − 2ζ 3

√2 − 2 3

√4 − ζ

3√

4 − 7 3√

5 − 5ζ 3√

5 + 3√

10 + 5ζ 3√

10+

2 3√

20 − 2ζ 3√

20 + 3√

52 − 4ζ 3

√52 − 3

√50 + ζ

3√

50 + 3√

100 + 2ζ 3√

100)

and

E1 = 3

√u1u2ε23u2

3u4ε4.

6.4 Units in L from Type III Subfields

For a Type III field, Ki we can write εi = Bi

Bσi

where NKi/k(Bi) = πaπb with 1 ≤ a, b ≤ 2 and a + b = 3.

Suppose without loss of generality that Ki for i =1, 2 are Type III fields and the other cubic fields can be

of any Type. To find a unit in L we can use the criteria outlined in [9]:

Theorem 6.11 Let k1 be a Type III field. Then:

(a) e3 = ζaε1/ε′i has no solution e in L.

(b) e3 = ζa ε1ε2ε′1ε

′′2

has a solution e in L if and only if k2 is a Type III and ζaN(B1) = N(B2).

(c) e3 = ζa ε1ε2ε′1ε

′2

has a solution e in L if and only if k2 is a Type III and ζaN(B1) = N(B2).

(d) e3 = ζa ε1ε2ε3ε′1ε

′2ε

′′3

has a solution e in L if and only if

α3 =ζaN(B1)N(B2)

N(B3)

has a solution α ∈ k. Thus either k2 or k3 is of Type III.


Note: For α3 = ζaN(B1)N(B2)N(B3) if B3 is not Type III then N(B3) = 1, ζ, ζ2 so case (d) reduces to case (b)

or (c). For units of the form in (d) we will only consider the case where k3 is Type III.

To find units in L of the form of (b) or (c) in Theorem 6.11 it is necessary that ζaN(B1) ∼ N(B2) . If

ζaN(B1) = N(B2) then the quotient of B1 and B2 can be expressed as

e3 =(B1

B2

)3

=ζaB3

1N(B2)N(B1)B3

2

=ζaε1ε

−12

εσ2

1 ε−σ2

2

.

If ζaN(B1) = N(B2) = N(B2)τ then since στ = τσ2 we get

e3 =(B1

Bτ2

)3

=ζaε1ε

−12

εσ2

1 ε−σ2

.

Similarly we can extend this idea to products of three Bi’s where ζaN(B1)N(B2) ∼ N(B3) or

ζaN(B1)N(B2) ∼ N(B3) and we generate units in L of the form of (d) in Theorem 6.11 where

e3 =(B1B2

B3

)3

=ζaε1ε2ε

−13

εσ2

1 εσ2

2 ε−σ2

3

or

e3 =(B1B

τ2

B3

)3

=ζaε1ε2ε

−13

εσ2

1 εσ2 ε−σ2

3

.

Using these quotients we can find new units in L which are not in any of the subfields.

6.5 Example Type III units in L

Example 7

Let L = Q(ζ, 3√

7, 3√

19). In Example 5 we saw that N(B2) ∼ N(B3) ∼ N(B4) so we can find two new

units in L

E1 =(

B2B3

)= 2033 + 1957ζ/3− 2819 3

√7/3− 2773ζ 3

√7/3 + 31 3

√49 + 315ζ 3

√49 + 1169 3

√19/3− 127ζ 3

√19/3−

812 3√

133/3− 1195ζ 3√

133/3 + 3 3√

931 + 184ζ 3√

931 + 745 3√

361/3− 4ζ 3√

361− 254 3√

2527/3− 229/3ζ 3√

2527−25 3

√17689 + 53ζ 3

√17689

and

E2 =(

B2B4

)= −95 − 38ζ/3 − 9 3

√7 − 32ζ 3

√7 + 79 3

√49/3 + 62ζ 3

√49/3 − 49 3

√19/3 + 20ζ 3

√19/3 − 4 3

√133 −

19ζ 3√

133+25 3√

931/3+29/3ζ 3√

931−32 3√

361/3+3ζ 3√

361+ 3√

2527−6ζ 3√

2527+7 3√

17689/3+5/3ζ 3√

17689

where

E31 = ε−2

2 ε−σ2 ε23ε

σ3


and

E32 = ε−2

2 ε−σ2 ε24ε

σ4

We are now in a position to describe the basis for the units of L

6.6 Basis for the Unit Group of L

Theorem 6.12 The basis for E can be chosen in one of 41 possible ways. If e1, e2, · · · , e8 is a basis for

E and εi, ui, where ui = 3√ε2i ε

′i for Ki Type I or IV and ui = ε′i for ki Type III, is a basis for Ui then the

basis will depend on the Kind of K and the Types of the subfields.

Case 1: If K is of Kind 1 and

(A) [E : E0] = 32 then

e1 = ε1, e32 = εa1ε2, e

33 = εb1ε3, e

34 = εc1ε4, e5 = u1, e6 = u2,

e37 = εa11 ua1

1 ua22 u3, where 0 ≤ a1, a2 ≤ 2 and a1 + a2 > 0,

e38 = εb11 ub11 ub2

2 u4, where 0 ≤ b1, b2 ≤ 2 and b1 + b2 > 0.

(B) [E : E0] ≤ 3 then

e1 = ε1, e32 = εa1ε2, e

33 = εb1ε3, e

34 = εc1ε4, e5 = u1, e6 = u2, e7 = u3,

e38 = εa11 ua1

1 ua22 ua3

3 u4, where 0 ≤ a1, a2, a3 ≤ 2 and a1 + a2 + a3 > 0 if [E : E0] = 3

e8 = u4 if [E : E0] = 1Case 2: If K is of Kind 2 and

(A) K1 and K2 are Type I or IV and K3 and K4 are Type I:

e1 = ε1, e2 = ε2, e33 = εa1ε

b2ε3, e

34 = εc1ε

d2ε4, e5 = u1, e6 = u2, e7 = u3

e38 = εa11 ua1

1 ua23 u4, where 0 ≤ a1, a2 ≤ 2 and a1 + a2 > 0 if [E : E0] = 3

e8 = u4 if [E : E0] = 1(B) K1 is Type I or IV, K2 is Type III and K3 and K4 are Type I:

e1 = ε1, e2 = ε2, e33 = εa1ε3, e

34 = εb1ε4, e5 = u1, e6 = ε′2, e7 = u3,

e38 = εa11 ua1

1 εa22 ε′2

2a2ua33 u4, where 0 ≤ a1, a2, a3 ≤ 2 and a1 + a3 > 0 if [E : E0] = 3

e8 = u4 if [E : E0] = 1(C) Three of the subfields are Type IV and K4 is Type I or IV:

Note: If k4 is Type I then c = d = 0.

e1 = ε1, e2 = ε2, e33 = εa1ε

b2ε3, e

34 = εc1ε

d2ε4, e5 = u1, e6 = u2, e7 = u3, e8 = u4

(D) K1 and K2 are Type III and two Type I:

E =< ε1, ε′1, ε2, ε

′2, 3√ε3, u3, 3

√ε4, u4 >

N(B1) � N(B2) : E =< ε1, ε′1, ε2, ε

′2, 3√ε3, u3, 3

√ε4, e8 > where


e38 = εa1ε

′12aεb2ε

′22buc

3u4, where 0 ≤ a, b, c ≤ 2 and a + b + c > 0 if [E : E0] = 3

e8 = u4 if [E : E0] = 1Case 3: If K is of Kind 3 and

(A) All fields are Type I or IV:

E =< ε1, u1, ε2, u2, ε3, u3,3√εc11 εc22 εc33 ε4,

3

√εc11 ε

′c22 ε

′c33 ε′4 >

(B) K1 Type III, K2 and K3 are Type I or IV and K4 is Type I:

E =< ε1, ε′1, ε2, u2, ε3, u3,

3√εc22 εc33 ε4, e8 > where

e38 = εa1ε′12au4, where c2 = c3 = 0 and a > 0 if [E : E0] = 3

e8 = 3√ε′2

c2ε′3c3ε′4 if [E : E0] = 1

(B) K1 and K2 are Type III, K3 is Type I or IV and K4 is Type I:

N(B1) � N(B2) : E =< ε1, ε′1, ε2, ε

′2, ε3, u3, 3

√ε4, e8 > where

e38 = εa1ε′12aεb2ε

′22bu4, where 0 ≤ a, b ≤ 2 and a + b > 0 if [E : E0] = 3

e8 = u4 if [E : E0] = 1

N(B1) ∼ N(B2) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′′2, ε3, u3, 3

√ε4, e8 > where

e38 = εa1ε′12au4, where a > 0 if [E : E0] = 9

e8 = u4 if [E : E0] = 3(C) Three Type III and K4 Type I or IV and

N(B1) ∼ N(B2) ∼ N(B3) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε3, 3

√ε1ε2ε′1ε

′′2, 3

√ε1ε3ε′1ε

′3, ε4, u4 >

N(B1) ∼ N(B2) � N(B3) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′′2, ε3, ε

′3, 3√ε4, e8 > where



e8 = u4 if [E : E0] = 3

N(B1)N(B2) ∼ ζaN(B3) : E =< ε1, ε′1, ε2, ε

′2, ε3, 3


′2ε

′′3, 3√ε4, e8 > where



e8 = u4 if [E : E0] = 3

N(B1) � N(B2) � N(B3) � N(B1) and N(B1)N(B2) � N(B3) : E =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, 3√ε4, e8 >

where e38 = εa1ε

′12aεb2ε

′22bεc3ε

′32cu4, where 0 ≤ a, b, c ≤ 2 and a + b + c > 0 if [E : E0] = 3

e8 = u4 if [E : E0] = 1(D) Four Type III :

N(B1) ∼ N(B2) ∼ N(B3) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε3, 3

√ε1ε2ε′1ε

′′2, 3

√ε1ε3ε′1ε

′3, ε4, ε

′4 >

Case 4: If K is of Kind 4 and

(A) All fields are Type I or IV:

E =< ε1, u1, ε2, u2, ε3, u3, ε4, u4 >


(B) K1 Type III:

E =< ε1, ε′1, ε2, u2, ε3, u3, ε4, u4 >

(C) K1, K2 Type III and two Type I or IV and

N(B1) ∼ N(B2) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, u3, ε4, u4 >

N(B1) � N(B2) : E =< ε1, ε′1, ε2, ε

′2, ε3, u3, ε4, u4 >

(D) Three Type III and K4 Type I or IV and

N(B1) � N(B2) � N(B3) � N(B1) and N(B1)N(B2) � N(B3) : E =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, u4 >

N(B1) ∼ N(B2) � N(B3) : E =< ε1, ε′1, ε2,

3

√εa1ε

b2

ε′a1 ε

′b2, ε3, ε

′3, ε4, u4 >

N(B1) ∼ N(B2) ∼ N(B3) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, 3

√ε1ε3ε′1ε

′3, ε4, u4 >

N(B1)N(B2) ∼ N(B3) : E =< ε1, ε′1, ε2, ε

′2, ε3,

3

√εa1ε

b2ε3

ε′a1 ε

′b2 ε′′3

, ε4, ε′4 >

Note: For the next case with 4 Type III fields only the norms that are similar will be mentioned. Any

other norms or products of norms are assumed to be dissimilar.

(E) Four Type III and

No norms are similar: E =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, ε

′4 >

N(B1) ∼ ζaN(B2) : E0 =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, ε

′3, ε4, ε

′4 >

N(B1) ∼ N(B2) ∼ N(B3) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, 3

√ε1ε3ε′1ε

′3, ε4, ε

′4 >

N(B1) ∼ N(B2) � N(B3) ∼ N(B4) : E =< ε1, ε′1, ε2, 3

√ε1ε2ε′1ε

′2, ε3, ε

′3, ε4, 3

√ε3ε4ε′3ε

′4>

N(B1)N(B2) ∼ N(B3) : E =< ε1, ε′1, ε2, ε

′2, ε3, 3


′2ε

′′3, ε4, ε

′4 >

N(B1)N(B2) ∼ ζaN(B3) and N(B4) ∼ N(B1): E =< ε1, ε′1, ε2, ε

′2, ε3, 3


′2ε

′′3, ε4, 3

√ε1ε4ε′1ε

′4>

N(B1)N(B2)N(B3) ∼ N(B4) or N(B1)N(B2) ∼ N(B3)N(B4) :

E =< ε1, ε′1, ε2, ε

′2, ε3, ε

′3, ε4, 3


′′2 ε′3ε

′4>

Proof: Let E1 be the group generated by < e1, e2, · · · , e8 >. We will show that E1 = E. Clearly

E1 ⊆ E and we know that the rank of E = 8. The rank of E1 ≤ 8 since it is generated by 8 elements. To

show it has rank 8 we will show that it contains the subgroup of rank 8 generated by < εi, ui > where i =

1, 2, 3 and 4.

First we consider the case when K is of Kind 1 then clearly ε1 and u1 ∈ E1. e2e6

= 3

√εa1ε2

εa1ε

′2

= 3√ε22ε

′′2 = u2

and e32ε−a1 = ε2 so ε2 and u2 ∈ E1. Similarly we find that ε3 and ε4 ∈ E1.

If[E : E0

]= 32 then e37ε

−a11 u−a1

1 u−a22 = u3 and e38ε

−b11 u−b1

1 u−b22 = u4 so u3 and u4 ∈ E1 and the rank

of E1 = 8. It is known from Theorem 6.10 that e7 and e8 are not in E0 but are in E1 so[E1 : E0

]= 32.

Thus since E1 ⊆ E then it must be that E1 = E.


If[E : E0

]= 31 then e3

e7= u3 and e38ε

−a11 u−a1

1 u−a22 u−a3

3 = u4 so u3 and u4 ∈ E1 and the rank of

E1 = 8.

If K is of Kind 2 then again clearly εi and ui ∈ E1 for i = 1 and 2. If the subfields are all Type I or

IV then e33ε−a1 ε−b

2 = ε3, e34ε−c1 ε−d

2 = ε4 and e3e7u−b

2 = u3 so ε3, ε4 and u3 ∈ E1. e38ε−a11 u−a1

1 u−a23 = u4 so

u4 ∈ E1 and the rank of E1 = 8.

If K2 is Type III then replace e2 with ε′2 above and e33ε−a1 = ε3, e34ε

−b1 = ε4 and e3

e7= u3 so ε3, ε4 and

u3 ∈ E1. e38ε−a11 u−a1

1 ε−a22 ε′2

−2a2u−a33 = u4 so u4 ∈ E1 and the rank of E1 = 8. Thus the rank E1 = 8 for

all the cases.

For the last three cases it is known from Theorem 6.10 that e7 is not in E0 but is in E1 so[E1 : E0

]= 31.

Thus since E1 ⊆ E then it must be that E1 = E.

For the cases when K is of Kind 3 or 4 if we choose E1 = e1, e2, · · · , e8 it was shown in the proof of

Theorem 6.9 that[E : E1

]= 1 so E = E1.

Chapter 7

Rank of the Class Group of K and L

7.1 Class numbers of L and all its subfields

The class number formula for an algebraic number field, F , is well known and is

h =2√| disc(F ) |

2r+sπsreg(F )

∏p

1 − 1/p∏P |p

(1 −N(P ))

where r is the number of real embeddings and s is half the number of non-real embeddings of F into C.

The regulator of F , reg(F ), is an (r + s − 1) × (r + s − 1) determinant which depends on fundamental

units of F . Then for a pure cubic field ki, reg(ki) = log(εi) where εi > 1 and the product is taken over

all rational primes p. Since the class number is an integer it is sufficient to take the product over a large

enough number of primes so that the value of the right side remains close to an integer. We can calculate

εi using Vornoi’s algorithm so implementing the formula is straight forward. The class number h of K

satisfies the relation from Theorem XIV in [9]

33h = (e : e0)h1h2h3h4

and (e : e0) = 3r where r is the number of new units in K calculated in chapter 5. The class number

relations Hi and H for Ki and L respectively are given in Theorem I of [9] and are summarized here.

Theorem 7.1 The following class numbers relations hold

(1) 35H = (E : ε)H1H2H3H4

(2) 3Hi = (Ui : ui)h2i

Here (Ui : ui) = 3r and r = 0 or 1 as ki is type III or not and (E : ε) is the group index.

76

Chapter 7. Rank of the Class Group of K and L 77

Using the notation from Gerth [5] let G be an abelian 3-group. Then G may be viewed as a module

over Z3 and we define G+ = {a ∈ G | aτ = a} and G− = {a ∈ G | aτ = a−1} so that G = G+ ×G−. Let

M be any finite algebraic extension field of Q, CM denote the ideal class group of M , SM be the Sylow

3-subgroups of CM . The rank of SM is the number of cyclic factors in the decomposition of SM .

We say that an ideal class a ∈ CL is an ambiguous ideal class of the extension L/K1 if aσ = a, where

< σ >= Gal(L/K1). Let C(σ)L = {a ∈ CL | aσ = a} and S

(σ)L = {a ∈ SL | aσ = a}. Then S

(σ)L is called the

group of ambiguous ideals of SL. Let C1−σ ={a1−σ | a ∈ C

}for any abelian group C on which σ acts.

In [3] Gerth talks about strong ambiguous classes. If a ∈ C(σ)L then there exists a ∈ a such that

a1−σ = (x) for some x ∈ L∗ = L− 0. We say that a ∈ CL is a strong ambiguous ideal class if there exists a

representative a ∈ a such that a1−σ = (1). If an ambiguous ideal class is not strongly ambiguous it is said

to be a weak ambiguous ideal class.

We want to calculate the rank of SK and SL. From [4] Theorem 3.1 we know that if SKi = {1} for

some 1 ≤ i ≤ 4 then the rank SL = 2t− s where t is the rank of the group of ambiguous ideal classes S(σ)L

in SL and

s = rank(S(σ)L · S1−σ

L )/S1−σL . (7.1)

The calculations for t and s will be shown in sections 7.3 and 7.4 respectively. In [4] it is also shown that

the rank of SK = t− s1 where

s1 = rank((S(σ)L · S1−σ

L )/S1−σL )−. (7.2)

The calculation of s1 will be shown in 7.4.

The calculations for t and s depend on the cubic Hilbert symbol so we will begin by discussing how to

calculate the symbol.

7.2 Calculation of the cubic Hilbert symbol for divisors of 3

Let Q = {π1, . . . , πw, . . . , πd} be the primes that ramify from K1 to L such that π1, . . . , πw are divisors of

3. For primes πj ∈ K1 that are not divisors of 3 and πj � β the cubic Hilbert symbol(

β,m2πj

)reduces to the

power symbol(

βπj

)l3

where πlj || m2 and the calculation proceeds as it did in section 6.2. If π1 is the only

divisor of 3 then we can calculate(

β,m2π1

)using the fact that the cubic Hilbert symbol is multiplicative.

Since 1 =∏πj

(β,m2πj

)then (

β,m2

π1

)=∏πj

j>1

(β,m2

πj

)−1

.


When we have three divisors of 3 we have to take into account two different cases, either 3 | m2 or not.

From Hasse [6] we have a formula for calculating the symbol for L ∈ K1 a divisor of 3 in our basefield.

We define e and e∗ to be Le || 3 and Le∗ || 1 − ζ respectively. For this case e = 2 and e∗ = e/2 = 1(β, α

L

)= ζ2( α−1

3β−11−ζ ),

where α ≡ 1 (mod Le) and β ≡ 1 (mod Le∗).

We will begin with the case were m1 ≡ ±1 (mod 9) and 3 � m2 and we let L = πj for 1 ≤ j ≤ 3. Since

m1 ≡ ±1 (mod 9) then ( 3√m1∓1)2

3 is an integer in k1. The factorization of 3 is as follows: (3) = p1p22 in k1,

p1 = π21 and p2 = π2π3 in K1 so p1p2 | ( 3

√m1 ∓ 1) and π2

1π2π3 | ( 3√m1 ∓ 1) =⇒ 3

√m1 ≡ ±1 (mod 1 − ζ)

and

3√m1 ≡ ±1 (mod L) (7.3)

Similarly (1 − ζ) = π1π2π3 =⇒ζ ≡ 1 (mod L) (7.4)

for all L.

For β ∈ K1 where L � β we will calculate(

β,m∗2

L

)where m∗

2 = ±m2 such that m∗2 ≡ 1 (mod 3) and

(m∗2 − 1)/3 ∈ Z. We need to have β ≡ 1 (mod L). We know that β ≡ b (mod L) where b = 1 or 2. To

calculate the value of b we need to find a representative for β (mod L) which has no denominators of 3

since 3 ≡ 0 (mod L), otherwise we will be dividing by zero.

Lemma 7.2 Let mi ≡ ±1 (mod 9), mi = ac2 where c ≡ 1 (mod 3), β ∈ OKi , (3) = (π1π2π3)2 ∈ Ki.

Then for j = 1, 2, 3, there exists γj ∈ OKi such that β− γjπj ≡ b1 + b2ζ + b33√mi + b4ζ 3

√mi + b5

3√m2

i +

b6ζ3√m2

i (mod πj) where bl ∈ Z for all l.

Proof:

Since β ∈ OKi then β can be expressed as a Z linear combination of the basis for OKi defined in

Theorem 3.1, A3 = {1, ζ, 3√mi,

1± 3√mi+ 3√mi2/c

3 , 13 (2 + ζ ± 3

√mi ∓ ζ 3

√mi), ζ(1± 3√mi+ 3√mi

2/c3 )} where ± or ∓

correspond as mi ≡ ±1 (mod 9). For some ordering of the πi’s we can choose ρ1 = ( 3√mi ∓ 1)(1 − ζ)/3,

ρ2 = ρσ1 and ρ3 = ρσ2 where ρi is in πi but not in π2i . For each πj it is sufficient to show that we can find

γ’s such that each basis element can be expressed with no 3 in the denominator. Let ρ = ρ1, then

1 ± 3√mi + 3

√mi

2/c

3+ ζ2ρ2 ≡

(3√mi

2

c

)(1 − c

3

)± 3

√mi (mod πj)

13

(2 + ζ ± 3√mi ∓ ζ 3

√mi) ± 2ρ ≡ ζ + 3

√mi − ζ 3

√mi (mod πj).


So the lemma is proved.

Using Lemma 7.2 we can find γ ∈ K1 such that (β) − γρ = (b1 + b2ζ + b33√m1 + b4ζ 3

√m1 + b5

3√m2

1 +

b6ζ3√m2

1) where bj ∈ Z for all j. To find the numerical value for b substitute ±1 for 3√m1 and 1 for ζ in

β− γρ to get β− γρ ≡ b (mod L). If b = 2 then let β = −β. Now β ≡ 1 (mod L) so (β− 1)(1− ζ2)/3 ≡ b1

(mod L) where b1 = 0, 1, or 2 then, as before, find γ1 such that c = (β − 1)(1 − ζ2)/3 − γ1ρ has no

denominator of 3. We know that c = a1 + a2ζ + a33√m1 + a4ζ 3

√m1 + a5

3√m2

1 + a6ζ3√m2

1 where aj ∈ Z for

all j so we can substitute ±1 for 3√m1 and 1 for ζ in c to get

c ≡ b1 (mod L).

Now we can calculate (β,m∗

2

L

)= ζ

2b1

„m∗

2−13

«= ζαi .

For the case where m2 = 3an2 with a = 1 or 2 and 3 � n2 we will never have that πi is relatively prime

to m2 so since (3) = (π1π2π3)2 we can use the product formula. Without loss of generality let L = π1,

then for β �= πj for j = 1, 2, 3

(β, 3an2

L

)=(β, n2(π2π3)2a

π1

)(β, π1

π1

)2a

.

For the second term we use the product formula again

1 =(β, π1

π1

)(β, π1

π2

)(β, π1

π3

) t∏j=1

(β, π1

pj

)where pj | β

so (β, π1

π1

)2

=(β, π1

π2

)(β, π1

π3

) t∏j=1

(β, π1

pj

).

We need to calculate each piece separately.

To find(

β, απ1

), when α �≡ 0 (mod π1), we need β ≡ 1 (mod π1) and α ≡ 1 (mod π2

1). For the first

condition note that β∗ = ±β ≡ 1 (mod π1). For the second we know that ±α ≡ 1 (mod π1) so ±α ≡ ζ2j

(mod π21) for j = 0, 1 or 2 and ±ζjα ≡ 1 (mod π2

1). Then(β, α

π1

)=(β∗,±ζjα

π1

)(ζj , β∗

π1

)

where(

β∗,±ζjαπ1

)= ζc with c ≡ (β∗−1)

1−ζ(±ζjα−1)

3 (mod π1). To calculate the value of c we need to have that

c is an integer in K1 and has no 3’s in the denominator. Find a linear combination of π1 and π2π3 such

that r1(π1) + r2(π2π3) = 1 then r2(π2π3) ≡ 1 (mod π1). Now rationalizing the denominator of c gives

c ≡ (β∗ − 1)(±ζjα− 1)(1 − ζ2)9

(mod π1)


then

c ≡ (β∗ − 1)(±ζjα− 1)(1 − ζ2)(r42π42π

43)

(π1π2π3)4(mod π1) (7.5)

and we know that π21 | (±ζjα− 1), π1 | (±β − 1) and π1 | (1 − ζ2) so the numerator of (7.5) is divisible by

(π1π2π3)4 so c is an integer in K1. As in the previous case we can eliminate the denominators which are 3

by finding γ such that c′ = c + γρ ≡ b (mod π1) and then substitute ±1 for 3√m1 and 1 for ζ in c′ to get

c ≡ c′ ≡ b (mod π1).

We perform a similar calculation to find(

ζj ,βπ1

)=(

ζj ,±ζiβπ1

)(ζi,ζ2j

π1

)where

(ζi,ζ2j

π1

)= 1.

If β | 3 then the calculation proceeds as above except for the case of(

β,πj

π1

). Without loss of generality

let β = π1 and we will show how to calculate the symbol(

π1,π2π1

). We would like to be able to use formula

(4) from [6]: (β, α

p

)=(α

p

)−l

,if p is unramified in the extension K1( 3

√α)

and divides β exactly to the l power.(7.6)

and the product formula. Hence we need to find α such that π1 is unramified in the extension K1( 3√α)

and π2 | α so that (π1, α

π1

)=(π1, α/π2

π1

)(π1, π2

π1

). (7.7)

To do this we use Theorem 119 from Hecke [7] which states that if π1 | (1 − ζ) and π1 � α then π1 is

unramified in K1( 3√α) if the congruence

α ≡ ±1 (mod π31)

can be solved. We know that π2 ≡ ±1 (mod π1) and π2 ≡ ±ζj (mod π21). There are 18 reduced residues

modulo π31 and those can all be represented by ±ζjA where A = 1, 4, 7 and j = 0, 1, 2. To see that these

residues are distinct we suppose that

±ζiA ≡ ±ζjB (mod π31)

then, since A ≡ B ≡ 1 (mod π21),

±ζi ≡ ±ζj (mod π21)

so i = j and the sign is the same. Then A ≡ B (mod π31) =⇒ A = B and the 18 residues are distinct.

Thus we can choose α = ±ζj4lπ2 ≡ 1 (mod π31) where 0 ≤ j, l ≤ 2 and

(π1, α

π1

)=(α

π1

)2

≡ (α)4 ≡ 1 (mod π21). (7.8)


We can use the product rule to find

(π1, α/π2

π1

)=(π1,±ζj4l

π1

)=(π1,±ζj4l

π2

)2(π1,±ζj4l

π3

)2∏

p|2

(π1,±ζj4l

p

)2

and each Hilbert symbol on the right can be calculated by one of the methods above. Then combining

equations (7.7) and (7.8) gives the desired result

(π1, π2

π1

)=(π1,±ζj4l

π1

)2

.

7.3 Calculation of NB

By [3] the rank of S(σ)L is

t = rank S(σ)L = d + q∗ − (r + 1 + o)

where

d = number of ramified primes in L/K1

r = 2 (the rank of the free abelian part of the group of units U1 of K1).

o = 1 since K1 contains the cube root of unity.

q∗ is defined by [V ∗K1

: U31 ] = 3q∗ , where V ∗

K1={x ∈ U1 | x = NL/K1(y), y ∈ L∗

}. Here U3

1 ={x3 | x ∈ U1

}, and L∗ = L− {0} .

To calculate d we need to know some information about the primes ramified from K1 to L. Let λ = 1−ζ

be a prime element in k dividing 3 and let P3 be a prime element in K1 dividing 3. Using a different ordering

than in the previous section, suppose that {π1, . . . , πg} are the primes, different from divisors of 3, that

ramify from K1 to L. The total number of ramified primes, d, depends on the ramification of the divisors

of 3 from K1 to L which will depend on the subfields and the base field. If m1m2 �≡ 0 (mod 3) then there

are two cases. If at least one of mi ≡ ±1 (mod 9) for i = 2, 3 or 4 then 3 is unramified from K1 to L

so d = g. If m1 ≡ ±1 (mod 9) and m2 �≡ ±1 (mod 9) then 3 has three divisors in K1 and each of those

divisors ramifies from K1 to L so d = g + 3. If m1 ≡ 0 (mod 3) then the divisor of 3 in K1 is unramified

if mi ≡ ±1 (mod 9) for i =2, 3 or 4 so d = g and it is ramified otherwise so d = g + 1. If m2 ≡ 0 (mod 3)

and m1 ≡ ±1 (mod 9) then there are three divisors of 3 in K1 which are ramified so d = g + 3 and if

m1 �≡ ±1 (mod 9) there is only one divisor of 3 to ramify so d = g + 1. These results are summarized in

the following table.


d =

g, if mi ≡ ±1 (mod 9) for some i = 2, 3, 4

g + 1, if mi �≡ ±1 (mod 9) for any i

g + 3, if m1 ≡ ±1 (mod 9) and mi �≡ ±1 (mod 9) for any i ≥ 2

To find q∗ we will use the cubic Hilbert symbol. For K1, L = K1( 3√m2) and u ∈ U1 it is known that

u ∈ NL/K1(L∗) ⇔ (u,m2P

)= 1 for all prime ideals P ∈ K1. For any prime ideal P which is unramified from

K1 to L we know that(u,m2P

)= 1 so we need only consider the ramified primes. Let Q = {π1, . . . , πd} be

the primes that ramify from K1 to L. We will calculate the matrix, NB with rows

(αi) , 1 ≤ i ≤ d

where αi ∈ Z3 is defined by (u,m2

πi

)= ζαi

for each u ∈ {ζ, ε1, u1} and πi ∈ Q where(

,m2πi

)is the cubic Hilbert symbol which can be calculated as

in Section 7.2. Since UK1 =< ζ, ε1, u1 > then q∗ = 3 − rank NB.

7.4 Calculation of ND

Let K1 = Q( 3√m1, ζ) be a subfield of L such that H1 (The class number of K1) is relatively prime to 3.

Theorem 2.7 in [5] defines ND and ND− and we restate it here.

Let π1, . . . , πd be the prime ideals of k which ramify in L. Let X = Z3 × · · · × Z3 (a product of d − 1

copies of Z3). For 1 ≤ i ≤ d− 1, we define a map ψi : K∗1 → Z3 (where K∗

1 = K1 − 0) by

ψi(z) = vi (7.9)

where

(z, L/πi) = ζvi

and ( , L/πi) =(

,m2πi

)is the cubic Hilbert symbol described above. Then we define ψ : K∗

1 → X by

ψ(z) = (ψ1(z), . . . , ψd−1(z)). (7.10)

Theorem 7.3 Let σ = σ1 be the generator of the cyclic group Gal(L/K1), and let τ be the generator

of Gal(L/K). Let SL (resp. SK , resp. SK1) be the 3-class group of L (resp. K, resp. K1). Assume

SK1 = {1}. Now let t denote the number of ambiguous ideals in L/K1. If t =0 or 1 then s1 = 0. If t ≥ 2,


let a1, . . . , at be norms of ideals chosen from a basis for the ambiguous ideal classes for L/K1. Let U1 be

the group of units of K1, and ψ be the map defined by equations (7.9) and (7.10). If

s = rank {[ψi(aj)] (mod ψ(U1))},

where [ψi(aj)] is the t × d − 1 matrix (over Z3) whose ij − th element is defined by equation (7.9), then

rank SL = 2t− s. Let A1, . . . ,Au be ideals of L whose ideal classes generate S(σ)L

−and NL/K1(Aj) = (yj),

then if

s1 = rank {[ψi(yj)] (mod ψ(U1))},

then rank SK = t− s1.

The matrices for the calculation of s and s1 can be constructed as follows. Let π1, . . . , πt be the prime

ideals in K1 that ramify in L then we can construct the matrix[(

πj , m2πi

)]for 1 ≤ i, j ≤ t. The ideals

A1, . . . ,Au can be constructed from taking appropriate products of the πj ’s. A rational prime p factors

in Kl as p1 · · · pr where r = 3 if p ≡ 2 (mod 3), r = 6 if p ≡ 1 (mod 3) and x3 − ml ≡ 0 (mod p) is

solvable and r = 2 otherwise. In L each factor is totally ramified so (p) = (p1 · · · pr)3 = P1, · · ·Pr so if

a ∈ SσL− then a1+τ =

∏1≤i≤s

Pi =∏

1≤i≤r

pi3. If p = p1p2p3 the subscripts can be choses so that p2

τ = p3

then (p2p32)1+τ = (p2

3p33) and

Aj = p2p32 ∈ Sσ

L−.

If p = p1 · · · p6 then the subscripts can be chosen so that piτ = pi+3 for i = 1, 2, 3 then (pipi+3

2)1+τ =

(pipi2τ )1+τ = (pipi

τ )3 and

Aj = pipi+32 ∈ Sσ

L−

for i =1, 2, 3. If p = p1p2 then (p1p22)1+τ = (p1p2)3 so

Aj = p1p22 ∈ Sσ

L−.

Note: In the calculation for ND it is necessary for the class number to be relatively prime to 3 so that a

representatives for the ideal classes can be found. If there are weak ambiguous ideal classes we have no way

of finding a representative and the rank of ND may be too large. It is possible to find the number of weak

ambiguous classes if they exist. If we let S(σ)L,s be the subgroup of S(σ)

L containing the strong ambiguous

classes, then [3] shows that rank of S(σ)L,s is

rank S(σ)L,s = d + q − 4

where

q is defined by [VK1 : U31 ] = 3q, where VK1 =

{x ∈ U1 | x = NL/K1(y), y ∈ E

}. So to find q it is sufficient


to count the number of units of K1 that are norms of integers of L. If q < q∗ then there are weak ambiguous

ideal classes and the calculation of the rank can be smaller by the the value of q∗ − q.

7.5 Rank of the 3-Class Group Example

Example 8

Let L = Q(ζ, 3√

2, 3√

31) then {m1,m2,m3,m4} = {2, 31, 62, 124} and we know the following information

about the cubic subfields ki = Q( 3√mi) and their normal closures Ki = Q( 3

√mi, ζ)

Subfield k1 k2 k3 k4

Type I III I III

hi 1 3 3 9

Hi 1 3 9 27

Using the techniques in Chapters 5 and 6 we can easily find e = {ε1, ε2, ε4, 3√ε3} and E ={

ε1, u1, ε2, ε′2, 3√ε3, u3, ε4,

3

√ε22ε

′2

ε24ε′4

}so [e : e0] = 3 and

[E : ε

]= 32. Then using the class number relations

we see that h = h1h2h3h4 [e : e0] /33 = 9 and H = H1H2H3H4

[E : ε

]/35 = 27.

Since gcd(h1, 3) = 1 we can use K1 as the base field for the calculation of the cubic Hilbert symbols and

the rank. From K1 to L only prime divisors of p = 31 are ramified and (31) = P1P2P3P4P5P6 where Pi ∈ K1

and P1 = 13 (−1 + ζ + 4 3

√2 + 2ζ 3

√2 + 2 3

√4 + ζ 3

√4), P2 = P σ

1 , P3 = P σ2

1 , P4 = P τ1 , P5 = P τ

2 and P6 = P τ3 so

so d = 6. To find q∗ we need to calculate the matrix NB = [αji] where(

µj ,31Pi

)= ζαij , 1 ≤ j ≤ 3, 1 ≤ i ≤ 6

and µj ∈ {ζ, ε1, u1}. We will show the calculation of(

u1,31P1

)where u1 = 1

3 (1+2ζ+2 3√

2+ζ 3√

2+ 3√

4+2ζ 3√

4)

and the rest of the entries for the NB matrix are similar.

To calculate the symbol we need to know which solutions to X3 − 2 ≡ (X − 4)(X − 7)(X − 20) ≡0 (mod 31) and ζ2 + ζ + 1 ≡ (ζ − 25)(ζ − 5) ≡ 0 (mod 31) generate the ideal P1. It is easy to see that

P1 | ( 3√

2 − 20) and P1 | (ζ − 25) so 3√

2 ≡ 20 (mod P1) and ζ ≡ 25 (mod P1) so(u1, 31P1

)≡ u

(31−1)/31 (mod P1)

≡ 33 13

(1 + 2(25) + 2(20) + (25)(20) + 202 + 2(25)202)10 (mod P1)

≡ 25 ≡ ζ1 (mod P1).

Then

NB = [αji] =

1 1 1 1 1 1

2 1 2 1 2 1

1 2 2 1 0 0


and rank(NB) = 3 and q∗ = 3 − 3 = 0 so there are no weak ambiguous ideal classes in this example. Now

we have that t = d + q∗ − 4 = 2.

We also want to calculate [ψi(Pj)] =N D. We will show the calculations for(

P2,31P1

)and

(P1,31P1

)and

the rest of the entries will be similar. We begin with(

P2,31P1

)since it is the same as the calculations for

NB. (P2, 31P1

)≡ P

(31−1)/32 ≡ 25 ≡ ζ1 (mod P1).

The diagonal element(

P1,31P1

)is calculated by the product rule. Since

(31, 31P1

)=(P1, 31P1

)(P2 · · ·P6, 31

P1

)= 1

then we can calculate (P1, 31P1

)=(P2 · · ·P6, 31

P1

)2

where P2 · · ·P6 = 13 (−26 − ζ + 8 3

√2 − 14ζ 3

√2 + 19 3

√4 − 10ζ 3

√4). So

(P2 · · ·P6, 31

P1

)≡ (P2 · · ·P6)(31−1)/3 ≡ 25 ≡ ζ1 (mod P1).

We can perform similar calculations on the rest of the entries to get

ND =

1 0 1 2 1 1

1 1 1 0 1 2

1 2 1 1 1 0

0 2 1 2 2 2

2 2 0 2 1 2

1 2 2 2 0 2

and by taking sums of the rows we can find the entries of ND−. An entry in the ND

− matrix comes from

the symbol(

PjP2j+3,31

Pi

)=(

Pj ,31Pi

)(Pj+3,31

Pi

)2

for 1 ≤ j ≤ 3, 1 ≤ i ≤ 6 which can be calculated by taking

the sum of the jth row and twice the {j + 3}th row which gives

ND− =

1 1 0 0 2 2

2 2 1 1 0 0

0 0 2 2 1 1

.

Then s = rank[ND (mod NB)] = 1 and s1 = rank[ND− (mod NB)] = 0 so rankSL = 2t − s = 3 and

rankSK = t− s1 = 2. Then SL∼= Z3 × Z3 × Z3 and SK

∼= Z3 × Z3.

Bibliography

[1] P. Barrucand and H. Cohn, A rational genus, class number divisibility, and unit theory for pure cubic

fields, J. Number Theory, vol. 2 (1970) pp. 7-21.

[2] P. Barrucand and H. Cohn, Remarks on principal factors in a relative cubic field, J. Number Theory,

vol. 3 (1971) pp. 226-239.

[3] F. Gerth, On 3-class groups of pure cubic fields, Journ. Reine Angew. Math. (1973) pp. 52-62.

[4] F. Gerth, On 3-class groups of cyclic cubic extensions of certain number fields, J. Number Theory,

vol. 8 (1976) pp. 84-98.

[5] F. Gerth, On 3-class groups of non-Galois cubic fields, Acta Arithmetica (1976) pp. 307-321.

[6] H. Hasse, Bericht Uber Neuer Untersuchungen und Probleme aus der Theorie der Algebraischen

Zahlkorpers, Physica-Verlag, Wurzburg/Wien, 1970 pp. 54-84

[7] E. Hecke, Lectures on the Theory of Algebraic Numbers, Springer-Verlag, New York 1981

[8] D. Marcus, Number Fields, Springer-Verlag, New York 1977

[9] C. J. Parry, Class number formulae for bicubic fields, Illinois Journal of Mathematics, Vol 21 (1977)

pp. 148-163

[10] P. Ribenboim, Classical Theory of Algebraic Numbers, Springer-Verlag, New York 2001

[11] T. Takagi, Collected Papers, Springer-Verlag, Tokyo with arrangement with Iwanami Shoten Publish-

ers, Tokyo, 1990 pp. 239-242

[12] H. C. Williams, G. Cormack and E. Seah, Calculation of the regulator of a pure cubic field, Mathematics

of Computation, Volume 34, Number 150 (1980) pp. 567-611

86

Appendix A

Units of Cubic Fields and their

Normal Closures

Let M = jk2 and N = j2k where (j, k) = 1 then the following table provides the fundamental units in

ki = Q( 3√M) and Ki = Q( 3

√M, ζ). Let

ε =1d1

(a + bM + cN)

be the fundamental unit of ki and

u =1d2

(d + eζ + fM + gζM + hN + iζN)

the fundamental unit of Ki where uuτ = ε when ki is Type I or IV and u = εσ if ki is Type III. Tables

A.1 and A.2 contain both ε and u as well as the Type of the cubic subfileds where M ≤ 495. Table A.2

contains those fields where the units are too large to fit in the format of Table A.1.

87

Appendix A. Units of Cubic Fields and their Normal Closures 88T

able

A.1

:U

nits

ofki

=Q

(3√ M

)an

dK

i=ki(ζ)

whe

reM

<49

5

Ma

bc

d1

de

fg

hi

d2

Typ

e

2-1

10

1-1

-2

-2

-1

21

3I

3-2

01

1-3

03

3-1

-23

IV

51

-4

21

-13

-84

82

-23

I

61

-63

1-7

-42

41

-11

I

72

-10

12

00

-10

01

III

10-7

-12

33

51

-2-1

03

I

111

4-2

1-1

1-1

9-4

44

23

I

121

3-3

15

2-1

-2-1

11

I

13-4

-32

1-4

00

-3-2

-21

III

141

2-1

111

4-2

-4-1

13

I

151

-30

121

25-2

4-2

0-1

04

81

I

1718

-70

1-7

2813

2-4

-53

IV

192

2-1

32

00

21

13

III

201

1-1

1-2

52

1-1

-23

I

21-4

76

41

-47

00

6-4

-41

III

2223

3-4

1-1

441

195

-5-7

3I

23-4

1399

-316

062

301

2127

2-6

7853

-313

40-7

480

8390

1102

03

I

263

-10

13

00

-10

01

III

28-1

-11

3-1

00

-1-1

-13

III

301

9-3

1-9

106

3-1

-21

I

cont

inue

don

next

page


able

A.1

:co

ntin

ued

Ma

bc

d1

de

fg

hi

d2

Typ

e

31-3

6754

201

-367

00

54-2

0-2

01

III

3337

4220

197

392

-394

098

1-5

1151

219

5648

176

9436

1594

72-1

9016

6-2

3988

41

I

3461

3-2

4-5

11

-52

305

110

16-2

9-3

41

I

35-2

210

-13

-22

00

101

13

III

3710

-30

110

00

-30

01

III

38-1

5155

-31

326

247

-23

-97

-22

73

I

39-2

30

21

-23

00

0-2

-21

III

421

-42

121

4897

14-1

4-8

-41

I

43-7

20

1-7

00

20

01

III

4411

3-2

-17

369

11-1

6-1

9-2

93

I

4510

8166

-312

1-5

91-7

76-5

216

618

444

1I

46-4

139

4830

91

2092

2529

122

-584

-197

-34

1I

47-5

9219

9-6

9704

6478

61

4042

03-1

0494

16-4

0279

6-1

1200

480

578

1116

143

I

51-1

1015

2592

102

164

0854

73-2

52-1

728

-398

681

I

521

-42

113

294

-4-4

-23

I

53-3

4434

045

1723

202

1-3

4462

7-6

7228

3-8

7229

9174

747

647

2322

23

IV

5565

9736

1-1

0122

54-1

8999

61

3250

985

-122

584

-887

092

-854

858

8476

2332

641

I

5710

8457

-88

113

17-1

098

-627

-342

7416

33

IV

581

-82

1-6

1-3

28

162

-23

I

601

-12

61

16-1

5-8

-42

41

I

cont

inue

don

next

page


able

A.1

:co

ntin

ued

Ma

bc

d1

de

fg

hi

d2

Typ

e

611

-16

41

64-6

1-3

2-1

64

83

I

621

-24

61

32-3

1-1

6-8

24

1I

634

-10

14

00

-10

01

III

65-4

10

1-4

00

10

01

III

661

24-6

1-3

233

168

-2-4

1I

671

16-4

1-6

7-1

31-1

616

84

3I

681

12-6

133

16-4

-8-2

21

I

69T

able

A.2

Tab

leA

.2I

701

8-2

1-3

235

168

-2-4

3I

7315

4-8

712

115

40

0-8

7-1

2-1

21

III

74-9

61-2

360

115

1419

2197

-361

-109

-23

3I

761

4-2

1-1

619

84

-2-4

3I

77-4

0232

807

-711

3592

3894

986

1-1

2002

6200

-387

5950

119

1021

5628

2127

7621

4149

8-4

4900

603

I

78-2

1340

7984

1944

-801

541

-643

839

-204

2912

-327

448

1506

8811

1906

7663

81

I

7929

295

-38

129

20

095

3838

1II

I

8242

9065

320

1605

-273

730

3-2

3335

53-3

0538

03-1

6578

553

7130

1617

9538

160

3I

8437

912

-45

1-5

720

259

13-2

1-2

71

I

85T

able

A.2

Tab

leA

.2I

86-7

6-1

1-7

00

61

11

III

8962

570

-233

0-2

617

337

019

2832

6-1

947

-829

1-1

421

436

3IV

cont

inue

don

next

page


able

A.1

:co

ntin

ued

Ma

bc

d1

de

fg

hi

d2

Typ

e

901

-54

361

-161

-81

1836

12-1

21

I

919

-20

19

00

-20

01

III

92-8

279

-737

1139

154

94-1

4539

-443

8-1

217

1427

1966

3I

93-1

6022

-644

2815

001

1-1

6022

00

-644

28-1

5001

-150

011

III

94-1

1287

5110

7457

3096

51

1672

357

1722

926

1112

2-3

6780

7-8

3339

-244

63

I

9518

6732

1-4

1948

822

461

6833

5735

6331

763

1168

-149

764

-171

148

-138

326

3I

102

-123

929

7188

3-9

708

1-1

1838

776

211

4164

925

338

-349

1-8

914

1I

105

3024

1-5

844

-120

115

135

-284

9-3

812

-320

812

880

81

I

106

-858

531

77-2

881

8533

5778

-582

-180

3-2

5812

31

I

109

1094

5-1

890

-84

110

945

00

-189

084

841

III

110

1-5

11

2247

5-5

-2-1

3I

114

6156

1-1

3758

219

1-4

0604

-321

8717

3683

7413

69-3

581

I

115

-198

2599

914

3798

054

2670

1-3

1095

997

-195

5000

5992

460

6394

480

8267

0-1

2322

703

I

116

7516

953

85-8

529

158

666

1554

3-8

842

-120

29-1

307

3626

1I

117

412

-50

-21

141

20

0-5

021

211

III

118

4664

7527

-363

175

-186

4939

3-2

3600

059

-285

6213

3-1

0116

6348

1155

811

8723

320

6257

3I

122

1-2

55

112

224

725

-25

-10

-53

I

124

5-1

01

50

0-1

00

1II

I

126

-51

01

-50

01

00

1II

I

129

2597

810

401

-307

61

2597

80

010

401

3076

3076

1II

I

cont

inue

don

next

page


able

A.1

:co

ntin

ued

Ma

bc

d1

de

fg

hi

d2

Typ

e

130

115

-31

5125

-5-1

0-1

11

I

132

-224

531

-229

5926

340

157

113

-137

764

-382

74-1

1217

1062

815

034

1I

134

-270

703

9771

7-8

758

3-2

7070

30

097

717

8758

8758

3II

I

138

2542

9951

1377

6309

-361

8144

1-2

4530

283

3309

0283

1115

0247

4746

894

-123

9123

-215

7702

1I

140

15

-21

-28

-53

-55

42

3I

141

2538

0115

288

-123

061

-143

633

-193

641

-960

827

596

7148

1846

1I

142

1405

9-5

199

480

110

153

-525

3-2

953

-194

619

356

61

I

148

125

9-9

81

2741

1369

-259

-518

-98

983

I

150

13

-31

-6-1

1-1

12

11

I

153

-50

43

184

57-5

-16

-63

3IV

154

-916

2710

973

1142

3-2

901

-478

06-8

378

541

1664

1563

3I

156

-421

19-5

0130

931

2189

327

443

1031

-406

7-1

894

-383

1I

158

1001

0881

-162

5176

-419

141

-175

2054

4-1

4935

901

4780

9632

4087

251

1046

-884

363

I

164

329

22-3

01

779

208

-104

-142

-14

383

I

165

1-6

612

112

024

122

-22

-8-4

1I

166

1-2

4244

1-2

659

-133

124

248

444

-44

3I

170

4544

11-1

5021

-120

961

-223

991

-256

581

-588

340

434

8361

1062

1I

171

58-1

63

193

156

11-1

7-1

5-6

3IV

172

-427

741

3-4

6-2

61-3

98

1714

3I

174

1566

1-2

688

-21

179

17-1

652

-171

4-1

418

5330

71

I

cont

inue

don

next

page


able

A.1

:co

ntin

ued

Ma

bc

d1

de

fg

hi

d2

Typ

e

175

281

-48

-21

8851

576

-16

-82

-68

3I

182

-17

30

1-1

70

03

00

1II

I

186

-368

27-2

676

1599

121

668

2994

514

50-3

796

-919

-254

1I

198

1-6

31

12-1

1-4

-21

21

I

204

1-9

31

18-1

7-6

-31

21

I

206

-240

67-2

0033

4082

3-2

4067

00

-200

33-4

082

-408

23

III

207

1-1

26

123

474

-4-4

-21

I

210

1-1

83

135

716

-6-2

-11

I

212

1-2

79

1-1

07-5

49

183

-31

I

214

1-5

49

1-2

15-1

0818

363

-31

I

218

154

-91

217

108

-18

-36

-33

1I

220

127

-91

109

54-9

-18

-33

1I

222

118

-31

7336

-6-1

2-1

11

I

228

19

-31

-18

196

3-1

-21

I

230

-140

759

-395

443

951

7571

698

165

3664

-123

58-2

615

-598

1I

234

16

-31

-12

134

2-1

-21

I

236

1889

-695

126

1-7

372

-466

143

911

9324

4-1

423

I

238

3095

-549

81

-618

5-4

879

211

998

127

-34

3I

244

2575

5-1

344

-889

125

755

00

-134

488

988

91

III

252

13

-31

-7-1

3-1

12

11

I

cont

inue

don

next

page


able

A.1

:co

ntin

ued

Ma

bc

d1

de

fg

hi

d2

Typ

e

260

7722

1-3

291

-276

01

-378

04-4

1755

-619

5923

2050

194

1I

268

2681

-793

117

1-8

710

-575

945

813

5127

7-1

423

I

275

133

8-2

601

-220

0-4

397

-338

338

520

260

3I

276

-119

231

-190

9211

490

189

953

1385

7674

68-1

3816

-653

8-2

294

1I

284

-472

8599

-775

803

4549

381

-559

6363

-197

9836

5501

9885

1399

9164

6-1

6740

81

I

285

6156

1-9

552

301

-390

64-3

1615

1132

5936

730

-172

1I

292

-187

1062

9-1

6380

1213

4401

91

-505

1863

1-1

5237

877

5317

942

7614

779

6924

14-1

6031

693

I

306

-881

2730

504

-775

81

-437

75-1

1779

2-1

0984

6496

7782

4890

1I

308

-769

124

-31

-319

-152

6-1

7947

6753

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523

Appendix B

Some invariants of K and L where

mi ≤ 500 for all i

Let K = Q(m1,m2), L = Q(m1,m2, ζ) then m3 = m1m2 and m4 = m21m2 where m3 and m4 are cube

free. Then we can find the fundamental units for ki and Ki using Tables A.1 and A.2. Table B.1 provides

a basis for the the fundamental units of K, e and L, E where mi ≤ 500 for 1 ≤ i ≤ 4. The basis for the

units of K has 4 elements and all of them are specified in the table. The basis for the units of L has 8

elements so the only elements listed in the table are those which are not products of units in the subfields.

The remaining basis elements are specified in Theorem 6.12. Table B.1 also lists the class numbers, hi, of

the cubic subfields ki, 1 ≤ i ≤ 4 as well as the class numbers of K and L (h and H respectively). The last

column of Table B.1 gives the rank of the 3-class group of {h,H}. The entries marked with (v) indicate

that the rank may be smaller than indicated by v because of the presence of weak ambiguous classes. The

value of v is the difference between the number of units that are norms and the number of units of Ki,

where i is the base field used for the calculation, that have Hilbert symbol 1. Since there are three units

in Ki then v ≤ 3.

104

Appendix B. Some invariants of K and L where mi ≤ 500 for all i 105T

able

B.1

:U

nit

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dR

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2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{3,2

,6,

12}

{1,1

,1,

1}IV

II

I{ε

1,

3√ ε2,

3√ ε3,

3√ ε4}

{ 3√ u2 2ε 4

u4,

3√ u2 2u

3

}{1

,1}

{0,0

}

{2,5

,10,

20}

{1,1

,1,

3}

II

II

{ε1,

3√ ε1ε 2},

3√ ε3,

3√ ε1ε 4}

{ 3√ u1u3ε 4

u2 2ε2 2

u4

}{3

,3}

{1,1

}

{2,7

,14,

28}

{1,3

,3,

3}I

III

III

I{ε

1,ε

2,ε

4,

3√ ε1ε 3}

{ 3√ ε2 2ε′ 2

ε2 4ε′ 4

}{3

,3}

{1,1

}{2

,11,

22,4

4}{1

,2,3

,1}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε4}

{ 3√ u1u2 3ε 4

u2u4ε 2

}{6

,12}

{1,1

}

{2,1

3,26

,52}

{1,3

,3,

3}I

III

III

I{ε

1,ε

2,ε

3,

3√ ε1ε 4}

{ 3√ ε2 2ε′ 2

ε′ 3ε 3

}{3

,3}

{1,1

}

{2,1

5,30

,60}

{1,2

,3,

3}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε1ε 3,

3√ ε4}

{ 3√ u1u3

u2 2ε2 2

}{1

8,10

8}{2

,3}

{2,1

7,34

,68}

{1,1

,3,

3}I

IVI

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u2 3u

4

}{3

,9}

{1,2

}{2

,19,

38,7

6}{1

,3,3

,6}

III

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√ u2 1u2 3ε 4

u4

}{1

8,10

8}{1

,2}

{2,2

1,42

,84}

{1,3

,3,

3}I

III

II

{ε1,ε

2,√ ε

2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

ε′ 2u4ε2 2

}{9,2

7}{2

,3(1

)}

{2,2

3,46

,92}

{1,1

,1,

3}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε3,

3√ ε1ε 4}

{ 3√ u1u2 3ε 4

u2 2u4ε2 2

}{3

,3}

{1,1

}{2

,29,

58,1

16}

{1,1

,6,

1}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε1ε 3,

3√ ε4}

{ 3√ u1u3ε 4

u2 2u4ε2 2

}{6

,12}

{1,1

}

{2,3

1,62

,124

}{1

,3,3

,9}

III

II

III

{ε1,ε

2,ε

4,

3√ ε3}

{ 3√ ε2 2ε′ 2

ε2 4ε′ 4

}{9

,27}

{2,2

}{2

,33,

66,1

32}

{1,1

,6,

3}I

II

I{ε

13√ ε2 1

ε 2,

3√ ε3,

3√ ε2 1ε 4}

{ 3√u2 1ε 4

u2 2u4ε2 2

}{1

8,10

8}{2

,3}

{2,3

5,70

,140

}{1

,3,9

,9}

III

II

I{ε

1,ε

2,ε

4,

3√ ε1ε 3}

{}{2

7,24

3}{3

,5(1

)}

{2,3

7,74

,148

}{1

,3,3

,6}

III

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√ u2 1u2 3ε 4

u4

}{1

8,10

8}{1

,2}

{2,3

9,78

,156

}{1

,6,3

,3}

III

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

ε′ 2ε2 2

u4

}{1

8,10

8}{2

,3(1

)}

cont

inue

don

next

page


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{2,4

1,82

,164

}{1

,1,1

,6}

II

II

{ε1,

3√ ε1ε 2,

3√ ε3,

3√ ε2 1ε 4}

{ 3√ u2 1u2 3ε 4

u2ε 2

u4

}{6

,12}

{1,1

}

{2,4

3,86

,172

}{1

,12,

9,3}

III

III

II

{ε1,ε

2,ε

3,

3√ ε4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3

}{3

6,43

2}{2

,3}

{2,4

5,90

,150

}{1

,1,3

,3}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u2 1u3

u2 2ε2 2

}{9

,27}

{2,3

}{2

,47,

94,1

88}

{1,2

,3,

1}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε4}

{ 3√ u2 1u3

u2 2ε2 2

}{6

,12}

{1,1

}{2

,51,

102,

204}

{1,3

,3,

12}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√u3ε 4

u2 2ε2 2

u4

}{1

08,3

888}

{2,3

(1)}

{2,5

3,10

6,21

2}{1

,1,6

,6}

IIV

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u2 3u

4

}{1

2,14

4}{1

,2}

{2,5

5,11

0,22

0}{1

,1,9

,9}

II

II

{ε1,ε

2,

3√ ε 1ε2 2ε 3,

3√ ε4}

{}{2

7,24

3}{3

,4}

{2,5

7,11

4,22

8}{1

,6,3

,9}

IIV

II

{ε1,ε

2,

3√ ε1ε 3,

3√ ε1ε 4}

{}{5

4,97

2}{2

,3(1

)}

{2,5

9,11

8,23

6}{1

,1,2

,6}

II

II

{ε1,

3√ ε1ε 2,

3√ ε3,

3√ ε2 1ε 4}

{ 3√ u2 1u2 3ε 4

u2ε 2

u4

}{1

2,48

}{1

,1}

{2,6

1,12

2,24

4}{1

,6,1

2,3}

II

III

I{ε

1,ε

4,

3√ ε1ε 2,

3√ ε1ε 3}

{ 3√ u2 1u2ε 3

u3

}{7

2,17

28}

{1,2

}

{2,6

3,12

6,25

2}{1

,6,9

,6}

III

III

II

{ε1,ε

2,ε

3,

3√ ε2 1ε 4}

{ 3√ ε2 2ε′ 2

ε′ 3ε 3

}{3

6,43

2}{2

,3(1

)}

{2,6

5,13

0,26

0}{1

,18,

9,3}

III

II

I{ε

1,ε

2,ε

3,

3√ ε3ε 4}

{}{5

4,97

2}{3

,5(2

)}

{2,6

7,13

4,26

8}{1

,6,3

,6}

II

III

I{ε

1,ε

3,

3√ ε1ε 2,

3√ ε2 1ε 4}

{ 3√ u1ε2 2

u4

u2 2

}{3

6,14

4}{1

,2}

{2,6

9,13

8,27

6}{1

,1,3

,3}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u2 1u3

u2 2ε2 2

}{9

,27}

{2,3

}{2

,71,

142,

284}

{1,1

,6,

3}I

IVI

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3u

4

}{6

,36}

{1,2

}{2

,73,

146,

292}

{1,3

,3,

3}I

III

II

{ε1,ε

2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

u4

}{9,2

7}{1

,2}

{2,7

7,15

4,30

8}{1

,3,3

,9}

II

II

{ε1,ε

2,

3√ ε1ε 2ε 3,

3√ ε2ε 4}

{}{2

7,24

3}{3

,5(1

)}

{2,7

9,15

8,31

6}{1

,6,3

,6}

III

II

III

{ε1,ε

2,ε

4,

3√ ε1ε 3}

{ 3√ ε2 2ε′ 2

ε′ 4ε 4

}{1

2,48

}{1

,1}

{2,8

3,16

6,33

2}{1

,2,6

,1}

II

II

{ε1,

3√ ε1ε 2,

3√ ε1ε 3,

3√ ε4}

{ 3√ u1u3

u2ε 2

}{1

2,48

}{1

,1}

cont

inue

don

next

page


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{2,8

5,17

0,34

0}{1

,3,3

,9}

II

II

{ε1,ε

2,

3√ ε2ε 3,

3√ ε1ε 4}

{}{2

7,24

3}{3

,4}

{2,8

7,17

4,34

8}{1

,1,1

2,3}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε3,

3√ ε2 1ε 4}

{ 3√u2 1ε 4

u2 2ε2 2

u4

}{3

6,43

2}{2

,3}

{2,8

9,17

8,35

6}{1

,2,3

,6}

IIV

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3u

4

}{1

2,14

4}{1

,2}

{2,9

1,18

2,36

4}{1

,9,2

7,18

}I

III

III

I{ε

1,ε

2,ε

3,ε

4}

{}{1

62,8

748}

{3,6

(2)}

{2,9

3,18

6,37

2}{1

,3,6

,3}

III

II

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√u3ε 4

ε′ 22ε 2

u4

}{1

8,10

8}{2

,3}

{2,9

5,19

0,38

0}{1

,3,3

,9}

II

II

{ε1,ε

2,

3√ ε2 1ε2 2ε 3,

3√ ε2 1ε2 2ε 4}

{}{2

7,24

3}{3

,5(1

)}

{2,9

7,19

4,38

8}{1

,3,3

,3}

III

II

III

{ε1,ε

2,ε

4,

3√ ε1ε 3}

{ 3√ ε2 2ε′ 2

ε′ 4ε 4

}{3

,3}

{1,1

}

{2,9

9,19

8,39

6}{1

,1,6

,3}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√u1ε 4

u2ε 2

u4

}{1

8,10

8}{2

,3}

{2,1

01,2

02,4

04}

{1,2

,3,

1}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε4}

{ 3√ u2 1u3ε 4

u2 2ε2 2

u4

}{6

,12}

{1,1

}{2

,103

,206

,412

}{1

,3,3

,6}

II

III

I{ε

1,ε

3,

3√ ε2 1ε 2,

3√ ε2 1ε 4}

{ 3√u2 1ε 4

u2 2ε2 2

u4

}{1

8,10

8}{1

,2}

{2,1

05,2

10,4

20}

{1,6

,18,

9}I

II

I{ε

1,ε

2,

3√ ε1ε 3,

3√ ε2 2ε 4}

{}{3

24,3

4992

}{4

,6(1

)}

{2,1

07,2

14,4

28}

{1,1

,12,

3}I

IVI

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u2 3u

4

}{1

2,14

4}{1

,2}

{2,1

09,2

18,4

36}

{1,3

,18,

9}I

III

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u2 3u

4

}{1

62,8

748}

{2,5

}{2

,111

,222

,444

}{1

,3,1

8,3}

IIV

II

{ε1,ε

2,

3√ ε1ε 3,

3√ ε1ε 4}

{}{5

4,97

2}{2

,3(1

)}

{2,1

13,2

26,4

52}

{1,4

,1,

3}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε3,

3√ ε1ε 4}

{ 3√u1ε 4

u2 2ε2 2

u4

}{1

2,48

}{1

,1}

{2,1

15,2

30,4

60}

{1,3

,9,

3}I

II

I{ε

1,ε

2,

3√ ε2 1ε 2,

3√ ε3ε 4}

{}{2

7,24

3}{3

,4}

{2,1

17,2

34,4

68}

{1,3

,6,

9}I

III

III

I{ε

1,ε

2,ε

4,

3√ ε2 1ε 3}

{ 3√ ε2 2ε′ 2

ε2 4ε′ 4

}{1

8,10

8}{2

,3(1

)}

{2,1

19,2

38,4

76}

{1,3

,9,

3}I

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε 1ε2 2ε 4}

{}{2

7,24

3}{3

,5(1

)}

{2,1

23,2

46,4

92}

{1,2

,3,

6}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε1ε 3,

3√ ε4}

{ 3√ u1u3

u2 2ε2 2

}{3

6,43

2}{2

,3}

cont

inue

don

next

page


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{2,1

75,3

50,4

90}

{1,3

,3,

9}I

II

I{ε

1,ε

2,

3√ ε2 1ε 2ε 3,

3√ ε2 1ε2 2ε 4}

{ 3√u2 1ε2 2

ε 4u2 2u3ε 3

u4

}{2

7,72

9}{3

,5}

{3,5

,15,

45}

{1,1

,2,

1}IV

II

I{ε

1,

3√ ε2,

3√ ε3,

3√ ε4}

{ 3√ u2u

3,

3√ u2 2u

4

}{2

,4}

{0,0

}{3

,7,2

1,63

}{1

,3,3

,6}

IVII

III

III

I{ε

1,ε

2,ε

4,

3√ ε 1ε 2ε 3ε2 4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3,

3√ ε2 2ε′ 2

ε′ 4ε 4

}{6

,12}

{1,1

}

{3,1

0,30

,90}

{1,1

,3,

3}IV

II

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{3√u

3u

4}

{3,9

}{1

,2}

{3,1

1,33

,99}

{1,2

,1,

1}IV

II

I{ε

1,

3√ ε2,

3√ ε3,

3√ ε4}

{3√u

2u

4,

3√ u2u

3}

{2,4

}{0

,0}

{3,1

3,39

,117

}{1

,3,6

,3}

IVII

III

III

I{ε

1,ε

2,ε

3,

3√ ε 1ε 2ε2 3ε 4}

{ 3√ ε2 2ε′ 2

ε′ 3ε 3

,3√ ε2 2

ε′ 2ε2 4

ε′ 4

}{6

,12}

{1,1

}

{3,1

4,42

,126

}{1

,3,3

,9}

IVI

III

I{ε

1,ε

2,ε

4,

3√ ε2ε 3}

{}{9

,27}

{2,3

}{3

,17,

51,1

53}

{1,1

,3,

9}IV

IVI

IV{ε

1,ε

2,

3√ ε3

3√ ε1ε 2ε 4}

{}{9

,27}

{1,2

}{3

,19,

57,1

71}

{1,3

,6,

6}IV

III

IVIV

{ε1,ε

2,ε

3,

3√ ε2 1ε 3ε 4}

{}{1

2,48

}{1

,1}

{3,2

0,60

,150

}{1

,3,3

,3}

IVI

II

{ε1,ε

2,

3√ ε2 2ε 3,

3√ ε2 2ε 4}

{ 3√ u2u3ε 4

u4

}{9

,81}

{2,4

}{3

,22,

66,1

98}

{1,3

,6,

6}IV

II

I{ε

1,ε

2,

3√ ε2ε 3,

3√ ε2ε 4}

{ 3√ u2 2u

3u

4

}{3

6,12

96}

{2,4

}{3

,23,

69,2

07}

{1,1

,1,

8}IV

II

I{ε

1,

3√ ε2,

3√ ε3,

3√ ε4}

{ 3√ u2 2u

3,

3√ u2u

4

}{8

,64}

{0,0

}{3

,26,

78,2

34}

{1,3

,3,

6}IV

III

II

{ε1,ε

2,ε

3,

3√ ε2 3ε 4}

{}{6

,12}

{1,1

}{3

,28,

84,2

52}

{1,3

,3,

6}IV

III

II

{ε1,ε

2,ε

3,

3√ ε2 3ε 4}

{}{6

,12}

{1,1

}{3

,29,

87,2

61}

{1,1

,1,

1}IV

II

I{ε

1,

3√ ε2,

3√ ε3,

3√ ε4}

{ 3√ u2 2u

3,

3√ u2 2u

4

}{1

,1}

{0,0

}{3

,31,

93,2

79}

{1,3

,3,

3}IV

III

III

III

{ε1,ε

2,ε

3,

3√ ε1ε 2ε 3ε 4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3

3√ ε2 2ε′ 2

ε2 4ε′ 4

}{3

,3}

{1,1

}{3

,34,

102,

306}

{1,3

,3,

3}IV

II

I{ε

1,ε

2,

3√ ε2 2ε 3,

3√ ε2 2ε 4}

{ 3√ u2u

3u

4

}{9

,81}

{2,4

}{3

,35,

105,

315}

{1,3

,6,

3}IV

III

II

{ε1,ε

2,ε

3,

3√ ε2 3ε 4}

{}{6

,12}

{1,1

}{3

,37,

111,

333}

{1,3

,3,

3},

IVII

IIV

IV{ε

1,ε

2,ε

3,

3√ ε1ε 3ε 4}

{}{3

,3}

{1,1

}{3

,38,

114,

342}

{1,3

,3,

27}

IVI

III

I{ε

1,ε

2,ε

4,

3√ ε2ε 3}

{}{2

7,24

3}{2

,3}

cont

inue

don

next

page


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{3,4

1,12

3,36

9}{1

,1,2

,4}

IVI

II

{ε1,

3√ ε2,

3√ ε3,

3√ ε4}

{ 3√ u2u

3,

3√ u2 2u

4

}{8

,64}

{0,0

}{3

,43,

129,

387}

{1,1

2,6,

3}IV

III

III

III

{ε1,ε

2,ε

3,

3√ ε 1ε 2ε2 3ε 4}

{ 3√ ε2 2ε′ 2

ε′ 3ε 3

3√ ε2 2ε′ 2

ε2 4ε′ 4

}{2

4,19

2}{1

,1}

{3,4

4,13

2,39

6}{1

,1,3

,3}

IVI

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3u

4

}{3

,9}

{1,2

}{3

,46,

138,

414}

{1,1

,3,

6}IV

II

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u2 3u

4

}{6

,36}

{1,2

}{3

,47,

141,

423}

{1,2

,8,

7}IV

II

I{ε

1,

3√ ε2,

3√ ε3,

3√ ε4}

{ 3√ u2 2u

3,

3√ u2u

4

}{1

12,1

2544

}{0

,0}

{3,5

2,15

6,46

8}{1

,3,3

,9}

IVI1

III

I{ε

1,ε

2,ε

4,

3√ ε2ε 3}

{}{9

,27}

{2,3

}{3

,53,

159,

477}

{1,1

,3,

9}IV

IVI

IV{ε

1,ε

4,

3√ ε3,

3√ ε1ε 2ε 4}

{}{9

,27}

{1,2

}{3

,55,

165,

495}

{1,1

,9,

9}IV

II

I{ε

1,ε

2,

3√ ε2ε 3,

3√ ε2 2ε 4}

{}{2

7,24

3}{2

,4}

{5,6

,30,

150}

{1,1

,3,

3}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε4}

{ 3√ u2 1u3

u2 2ε2 2

}{9

,27}

{2,3

}{5

,7,3

5,17

5}{1

,3,3

,3}

III

III

II

{ε1,ε

2,ε

3,

3√ ε1ε 4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3

}{3

,3}

{1,1

}{5

,11,

55,2

75}

{1,2

,1,

3}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε3,

3√ ε2 1ε 4}

{ 3√u2 1ε 4

u2 2ε2 2

u4

}{6

,12}

{1,1

}{5

,12,

60,9

0}{1

,1,3

,3}

II

II

{ε1,

3√ ε1ε 2,

3√ ε1ε 3,

3√ ε1ε 4}

{ 3√ u1u4

u2 2ε2 2

}{9

,27}

{2,3

}

{5,1

3,65

,325

}{1

,3,1

8,3}

III

III

II

{ε1,ε

2,ε

3,

3√ ε4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3

}{1

8,10

8}{2

,3}

{5,1

4,70

,350

}{1

,3,9

,3}

II

II

{ε1,ε

2,

3√ ε2 2ε 3,

3√ ε2 1ε 2ε 4}

{}{2

7,24

3}{3

,5(1

)}

{5,1

7,85

,425

}{1

,1,3

,6}

IIV

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3u

4

}{6

,36}

{1,2

}{5

,19,

95,4

75}

{1,3

,3,

3}I

III

II

{ε1,ε

2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

u4

}{9

,27}

{1,2

}{5

,28,

140,

490}

{1,3

,9,

9}I

III

II

{ε1,ε

2,ε

4,

3√ ε2 1ε 3}

{}{2

7,24

3}{3

,5(1

)}

{6,7

,42,

252}

{1,3

,3,

6}I

III

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3ε 4

ε′ 2ε2 2

u4

}{1

8,10

8}{2

,3}

{6,1

0,60

,45}

{1,1

,3,

1}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε1ε 3,

3√ ε1ε 4}

{ 3√ u1u3

u2ε 2

}{3

,3}

{1,1

}co

ntin

ued

onne

xtpa

ge


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{6,1

1,66

,396

}{1

,2,6

,3}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√u1ε 4

u2 2ε2 2

u4

}{3

6,43

2}{2

,3}

{6,1

3,78

,468

}{1

,3,3

,9}

III

II

III

{ε1,ε

2,ε

4,

3√ ε1ε 3}

{ 3√ ε2 2ε′ 2

ε2 4ε′ 4

}{9

,27}

{2,3

(1)}

{6,1

4,84

,63}

{1,3

,3,

6}I

II

III

{ε1,ε

4,

3√ ε2,

3√ ε3}

{ 3√ u2 2ε 3

u3

}{1

8,36

}{2

,3}

{6,1

5,90

,20}

{1,2

,3,

3}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}{ 3√ u

2 1u3

u2ε 2

}{1

8,10

8}{2

,3}

{6,2

1,12

6,28

}{1

,3,9

,3}

III

III

III

I{ε

1,ε

2,ε

3,

3√ ε2ε 3ε 4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3,

3√ ε2 2ε′ 2

ε2 4ε′ 4

}{9

,27}

{2,3

}{6

,22,

132,

99}

{1,3

,3,

1}I

II

I{ε

1,

3√ ε2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

u4

}{9

,27}

{2,3

}{6

,26,

156,

117}

{1,3

,3,

3}I

III

III

I{ε

1,ε

2,ε

4,

3√ ε1ε 3}

{ 3√ ε2 2ε′ 2

ε′ 4ε 4

}{3

,3}

{1,1

}

{6,3

3,19

8,44

}{1

,1,6

,1}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√u3ε 4

u2ε 2

u4

}{6

,12}

{1,1

}{6

,34,

204,

153}

{1,3

,12,

9}I

II

IV{ε

1,ε

2,

3√ ε1ε 2,

3√ ε2 1ε 3}

{}{1

08,3

888}

{2,3

}{6

,38,

228,

171}

{1,3

,9,

6}I

II

IV{ε

1,ε

2,

3√ ε1ε 2,

3√ ε2 1ε 3}

{}{5

4,97

2}{2

,3(1

)}

{6,3

9,23

4,52

}{1

,6,6

,3}

III

II

I{ε

1,ε

2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u2 3ε 4

ε′ 2ε2 2

u4

}{3

6,43

2}{2

,3(1

)}

{6,4

6,27

6,20

7}{1

,1,3

,8}

II

II

{ε1,

3√ ε1ε 2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√u3ε 4

u2 2ε2 2

u4

}{2

4,19

2}{1

,1}

{6,5

1,30

6,68

}{1

,3,3

,3}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√u3ε 4

u2 2ε2 2

u4

}{2

7,24

3}{2

,3(1

)}

{6,5

7,34

2,76

}{1

,6,2

7,6}

IIV

III

I{ε

1,ε

2,ε

3,

3√ ε2 1ε 4}

{}{1

08,3

888}

{2,3

(1)}

{6,5

8,34

8,26

1}{1

,6,3

,1}

II

II

{ε1,

3√ ε2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

u4

}{1

8,10

8}{2

,3}

{6,6

2,37

2,27

9}{1

,3,3

,3}

II

III

I{ε

1,ε

4,

3√ ε2 1ε 1,

3√ ε1ε 3}

{ 3√ u2 1u2 2ε 3

u3

}{9

,27}

{1,2

}

{6,6

9,41

4,92

}{1

,1,6

,3}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε4}

{ 3√ u2 1u3

u2 2ε2 2

}{1

8,10

8}{2

,3}

{6,7

0,42

0,31

5}{1

,9,9

,3}

II

II

{ε1,ε

3,

3√ ε2,

3√ ε2 3ε 4}

{}{8

1,21

87}

{4,7

}co

ntin

ued

onne

xtpa

ge


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{6,7

4,44

4,33

3}{1

,3,3

,3}

II

IIV

{ε1,ε

4,

3√ ε2,

3√ ε3}

{ 3√ u2u

3

}{9

,81}

{2,4}

{6,8

2,49

2,36

9}{1

,1,6

,4}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√ u2 1u3

u2 2ε2 2

}{2

4,19

2}{1

,1}

{7,1

0,70

,490

}{3

,1,9

,9}

III

II

I{ε

1,ε

2,ε

3,

3√ ε2 2ε2 3ε 4}

{}{2

7,24

3}{3

,5(1

)}

{7,1

2,84

,126

}{3

,1,3

,9}

III

II

III

{ε1,ε

2,ε

4,

3√ ε2ε 3}

{ 3√ ε2 1ε′ 1

ε′ 4ε 4

}{9

,27}

{2,3

(1)}

{7,2

0,14

0,35

0}{3

,3,9

,3}

III

II

I{ε

1,ε

2,ε

3,

3√ ε4}

{}{2

7,24

3}{}

{10,

12,1

5,15

0}{1

,1,2

,3}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√ u2 1u2 3ε 4

u4

}{6

,12}

{1,1

}

{10,

14,1

40,1

75}

{1,3

,9,

3}I

II

I{ε

1,ε

2,

3√ ε 1ε2 2ε 3,

3√ ε2 1ε 2ε 4}

{}{2

7,24

3}{3

,5(1

)}

{10,

22,2

20,2

75}

{1,3

,9,

3}I

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε2 1ε 2ε 4}

{}{2

7,24

3}{3

,4}

{10,

26,2

60,3

25}

{1,3

,3,

3}I

III

II

{ε1,ε

2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√ u2 1u4

u2 3ε2 3

}{9

,27}

{1,2

}{1

0,28

,35,

350}

{1,3

,3,

3}I

III

III

I{ε

1,ε

2,ε

3,

3√ ε1ε 4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3

}{3

,3}

{1,2

}{1

0,34

,340

,425

}{1

,3,9

,6}

II

II

{ε1,ε

3,

3√ ε2 1ε 2,

3√ ε1ε 4}

{}{5

4,97

2}{3

,4}

{10,

38,3

80,4

75}

{1,3

,9,

3}I

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε2 1ε 2ε 4}

{}{2

7,24

3}{2

,4(1

)}

{11,

12,1

32,1

98}

{2,1

,3,

6}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε1ε 3,

3√ ε4}

{ 3√ u1u3

u2ε 2

}{3

6,43

2}{2

,3}

{12,

13,1

56,2

34}

{1,3

,3,

6}I

III

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3ε 4

ε2 2ε′ 2

u4

}{1

8,10

8}{2

,3}

{12,

14,2

1,25

2}{1

,3,3

,6}

II

III

I{ε

1,ε

3,

3√ ε2 1ε 2,

3√ ε1ε 4}

{ 3√ u2 1ε2 2

u4

u2 2

}{1

8,36

}{2

,3(1

)}

{12,

17,2

04,3

06}

{1,1

,12,

3}I

IVI

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3u

4

}{1

2,14

4}{1

,2}

{12,

19,2

28,3

42}

{1,3

,9,

27}

III

II

III

{ε1,ε

2,ε

4,

3√ ε3}

{ 3√ ε2 2ε′ 2

ε2 4ε′ 4

}{8

1,21

87}

{2,5

}{1

2,20

,30,

45}

{1,3

,3,

1}I

II

I{ε

1,

3√ ε2,

3√ ε1ε 3,

3√ ε1ε 4}

{ 3√ u2 1u3ε 4

u4

}{9

,27}

{2,3

}

{12,

22,3

3,39

6}{1

,3,1

,3}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε1ε 4}

{ 3√ u1u4

u2 3ε2 3

}{9

,27}

{2,3

}

cont

inue

don

next

page


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{12,

23,2

76,4

14}

{1,1

,3,

6}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε3,

3√ ε2 1ε 4}

{ 3√u2 1ε 4

u2u4ε 2

}{1

8,10

8}{2

,3}

{12,

26,3

9,46

8}{1

,3,6

,9}

III

III

III

I{ε

1,ε

2,ε

4,

3√ ε 2ε 3ε2 4}

{ 3√ ε2 2ε′ 2

ε2 3ε′ 3,

3√ ε2 2ε′ 2

ε′ 4ε 4

}{1

8,10

8}{2

,3}

{12,

28,4

2,63

}{1

,3,3

,6}

III

II

III

{ε1,ε

2,ε

4,

3√ ε1ε 3}

{ 3√ ε2 2ε′ 2

ε′ 4ε 4

}{6

,12}

{1,2

}

{12,

44,6

6,99

}{1

,1,6

,1}

II

II

{ε1,

3√ ε1ε 2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3

u2ε 2

}{6

,12}

{1,1

}{1

2,52

,78,

117}

{1,3

,3,

3}I

II

III

{ε1,ε

4,

3√ ε2,

3√ ε3}

{ 3√ u2 2ε 3

u3

}{9

,27}

{2,3

}

{12,

68,1

02,1

53}

{1,3

,3,

9}I

II

IV{ε

1,ε

4,

3√ ε2 1ε 2,

3√ ε1ε 3}

{}{2

7,24

3}{2

,3}

{12,

76,1

14,1

71}

{1,6

,3,

6}I

II

IV{ε

1,ε

4,

3√ ε2,

3√ ε3}

{ 3√ u2 2u3

ε2 2

}{3

6,12

96}

{2,4

}{1

2,92

,138

,207

}{1

,3,3

,8}

II

II

{ε1,

3√ ε2 1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}{ 3√ u

1u3ε 4

u4

}{7

2,17

28}

{2,3

}{1

2,11

6,17

4,26

1}{1

,1,1

2,1}

II

II

{ε1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u2 1u3

u2 2ε2 2

}{1

2,48

}{1

,1}

{12,

124,

186,

279}

{1,9

,6,

3}I

III

III

I{ε

1,ε

2,ε

4,

3√ ε2 1ε 3}

{ 3√ ε2 2ε′ 2

ε2 4ε′ 4

}{1

8,10

8}{2

,3(1

)}

{12,

140,

210,

315}

{1,9

,18,

3}I

II

I{ε

1,ε

2,

3√ ε2 1ε 3,

3√ ε2 2ε 4}

{}{1

62,8

748}

{4,6

(1)}

{12,

148,

222,

333}

{1,6

,18,

3}I

II

IV{ε

1,ε

4,

3√ ε2 1ε 2,

3√ ε2 1ε 3}

{}{1

08,3

888}

{2,3

(1)}

{12,

164,

246,

369}

{1,6

,3,

4}I

II

I{ε

1,

3√ ε2,

3√ ε1ε 3,

3√ ε1ε 4}

{ 3√ u2 1u3ε 4

u4

}{7

2,17

28}

{2,3

}

{12,

172,

258,

387}

{1,3

,3,

3}I

II

III

{ε1,ε

4,

3√ ε1ε 2,

3√ ε1ε 3}

{ 3√ u2 1u2ε 3

u3

}{9

,27}

{1,2

}

{12,

188,

282,

423}

{1,1

,3,

7}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

u4

}{2

1,14

7}{1

,1}

{12,

212,

318,

477}

{1,6

,3,

9}I

II

IV{ε

1,ε

4,

3√ ε2 1ε 2,

3√ ε2 1ε 3}

{}{5

4,97

2}{2

,3}

{12,

220,

330,

495}

{1,9

,9,

9}I

II

I{ε

1,ε

4,

3√ ε2 1ε 2,

3√ ε2 1ε 3}

{}{2

43,1

9683

}{4

,6}

{14,

20,3

5,49

0}{3

,3,3

,9}

II

III

I{ε

1,ε

2,ε

3,

3√ ε2 1ε 4}

{}{2

43,1

9683

}{}

{15,

21,3

15,1

75}

{2,3

,3,

3}I

III

II

{ε1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u2 3ε 4

ε2 2ε′ 2

u4

}{1

8,10

8}{2

,3}

cont

inue

don

next

page


able

B.1

:co

ntin

ued

{m1,m

2,m

3,m

4}

{h1,h

2,h

3,h

4}

Typ

ee

E{h,H

}ra

nks

{15,

33,4

95,2

75}

{2,1

,9,

3}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε3,

3√ ε1ε 4}

{ 3√u1ε 4

u2ε 2

u4

}{5

4,97

2}{2

,3}

{20,

28,7

0,17

5}{3

,3,9

,3}

III

II

I{ε

1,ε

2,ε

33√ ε

3ε 4}

{}{3

6,12

96}

{}{2

0,44

,110

,275

}{3

,1,9

,3}

II

II

{ε1,ε

2,

3√ ε2 1ε 2ε 3,

3√ ε2 1ε2 2ε 4}

{}{2

7,24

3}{3

,5}

{20,

52,1

30,3

25}

{3,3

,9,

3}I

II

I{ε

1,ε

3,

3√ ε2 1ε 2,

3√ ε2 3ε 4}

{}{2

7,24

3}{}

{20,

68,1

70,4

25}

{3,3

,3,

6}I

II

I{ε

1,ε

2,

3√ ε2ε 3,

3√ ε2 2ε 4}

{ 3√ u2 2u4

ε2 2ε 4

}{2

7,24

3}{}

{20,

76,1

90,4

75}

{3,6

,3,

3}I

II

I{ε

1,

3√ ε2 1ε 2,

3√ ε3,

3√ ε2 1ε 4}

{}{2

7,24

3}{}

{30,

84,3

15,3

50}

{3,3

,3,

3}I

II

I{ε

1,ε

2,

3√ ε2 1ε2 2ε 3,

3√ ε2 1ε 2ε 4}

{}{2

7,24

3}{}

{42,

60,3

15,4

90}

{3,3

,3,

9}I

II

I{ε

1,ε

2,

3√ ε2ε 3,

3√ ε2 1ε2 2ε 4}

{}{2

7,24

3}{}

{45,

63,1

05,1

75}

{1,6

,6,

3}I

III

II

{ε1,ε

2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u2 3ε 4

u2 2ε 2

u4

}{3

6,43

2}{2

,3(1

)}

{45,

99,1

65,2

75}

{1,1

,9,

3}I

II

I{ε

1,

3√ ε1ε 2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√u2 1ε 4

u2ε 2

u4

}{2

7,24

3}{2

,3}

{45,

117,

195,

325}

{1,3

,6,

3}I

III

II

{ε1,ε

2,

3√ ε2 1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u3ε 4

ε2 2ε′ 2

u4

}{1

8,10

8}{1

,1}

{45,

126,

210,

350}

{1,9

,18,

3}I

III

II

{ε1,ε

2,ε

3,

3√ ε1ε 3ε 4}

{}{5

4,97

2}{3

,5(1

)}

{45,

153,

255,

425}

{1,9

,3,

6}I

IVI

I{ε

1,ε

2,

3√ ε1ε 3,

3√ ε1ε 4}

{}{5

4,97

2}{2

,3}

{45,

171,

285,

475}

{1,6

,15,

3}I

IVI

I{ε

1,ε

2,

3√ ε3,

3√ ε4}

{ 3√ u3u

4

}{9

0,81

00}

{2,4

}{6

3,90

,210

,490

}{6

,3,1

8,9}

III

II

I{ε

1,ε

2,ε

3,

3√ ε4}

{}{9

0,81

00}

{}{6

3,15

0,35

0,42

0}{6

,3,3

,9}

III

II

I{ε

1,ε

2,ε

3,

3√ ε 2ε2 3ε 4}

{}{9

0,81

00}

{}{9

0,10

5,35

0,25

2}{3

,6,3

,6}

II

II

{ε1,ε

2,

3√ ε2ε 3,

3√ ε2 2ε 4}

{}{9

0,81

00}

{}{9

0,12

6,42

0,17

5}{3

,9,9

,3}

III

II

I{ε

1,ε

2,ε

3,

3√ ε3ε 4}

{}{9

0,81

00}

{}{1

05,1

26,4

90,1

50}

{6,9

,9,

3}I

III

II

{ε1,ε

2,ε

3,

3√ ε2 1ε2 3ε 4}

{}{9

0,81

00}

{}{1

50,1

75,2

10,2

52}

{3,3

,18,

6}I

II

I{ε

1,ε

2,

3√ ε1ε 3,

3√ ε2 1ε 4}

{ 3√ u1u2 3ε 4

u4

}{9

0,81

00}

{}

Vita

Alberto Pablo Chalmeta was born in Lausanne, Switzerland in 1970. He graduated from Virgina Tech

with a Bachelors degree in Mechanical Engineering in 1994. He entered the Masters program at Virginia

Tech in 1996 and after earning his degree he entered the Ph.D. progam. In 2002 he went to work full time

at New River Community College as a mathematics teacher while completing his Ph.D..

Documents

vtechworks.lib.vt.edu€¦ · On the Units and the Structure of the 3-Sylow Subgroups of the Ideal Class Groups of Pure Bicubic Fields and their Normal Closures Alberto Pablo Chalmeta