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.
Chapter 3. Integral Basis for Ki, K and L 6
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}
Chapter 3. Integral Basis for Ki, K and L 7
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
Chapter 3. Integral Basis for Ki, K and L 8
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.
Chapter 3. Integral Basis for Ki, K and L 9
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
Chapter 3. Integral Basis for Ki, K and L 10
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)
Chapter 3. Integral Basis for Ki, K and L 11
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),
Chapter 3. Integral Basis for Ki, K and L 12
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
b1b2d1d2d3) if m2 ≡ 6 (mod 9) and m1 ≡ 2 (mod 9)
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
b1b2d1d2d3) if m2 ≡ 3 (mod 9) and m1 ≡ 5 (mod 9)
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
b1b2d1d2d3) if m2 ≡ 6 (mod 9) and m1 ≡ 5 (mod 9)
.
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}
Chapter 3. Integral Basis for Ki, K and L 13
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
b1b2d1d2d3) if m2 ≡ 3 (mod 9) and m1 ≡ 7 (mod 9)
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
b1b2d1d2d3) if m2 ≡ 6 (mod 9) and m1 ≡ 7 (mod 9)
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
b1b2d1d2d3) if m2 ≡ 6 (mod 9) and m1 ≡ 4 (mod 9)
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
b1b2d1d2d3) if m2 ≡ 3 (mod 9) and m1 ≡ 4 (mod 9)
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.
Chapter 3. Integral Basis for Ki, K and L 14
We can factor p in each of the subfields as follows:
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.
We can factor p in each of the subfields as follows:
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:
Chapter 3. Integral Basis for Ki, K and L 15
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).
We can factor (3) in each of the subfields as follows:
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).
We can factor (3) in each of the subfields as follows:
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 .
Chapter 3. Integral Basis for Ki, K and L 16
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
Chapter 5. Unit group of K 21
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 >.
Example 2 K = Q( 3√
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
Chapter 5. Unit group of K 22
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
Chapter 5. Unit group of K 23
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.
Chapter 5. Unit group of K 24
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)
Chapter 5. Unit group of K 25
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.
Chapter 5. Unit group of K 26
5.4 Example Type IV units in K
Example 3 K = Q( 3√
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 >.
Example 4 K = Q( 3√
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
Chapter 5. Unit group of K 27
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.
Chapter 5. Unit group of K 28
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
Chapter 5. Unit group of K 29
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.
Chapter 5. Unit group of K 30
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
Example 5 K = Q( 3√
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) =
Chapter 5. Unit group of K 31
(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).
Chapter 6. Unit Group of L 34
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
Chapter 6. Unit Group of L 35
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.
Chapter 6. Unit Group of L 36
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.
Chapter 6. Unit Group of L 37
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:
Chapter 6. Unit Group of L 38
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.
Suppose a55 = 3. Then
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
Chapter 6. Unit Group of L 39
(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.
Suppose a66 = 3. Then
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.
Chapter 6. Unit Group of L 40
Suppose a77 = 3, then
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 >
Chapter 6. Unit Group of L 41
We know that a11 = a22 = a33 = a44 = 1 by previous work and a55 = 1 by a similar proof to Case 2.
Suppose a66 = 3, then
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.
Suppose a77 = 3. Then
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.
Suppose a88 = 3. Then
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 =
Chapter 6. Unit Group of L 42
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:
Chapter 6. Unit Group of L 43
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 = 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
Suppose a66 = 3. Then
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.
Suppose a77 = 3. Then
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.
Chapter 6. Unit Group of L 44
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
Chapter 6. Unit Group of L 45
which has no solution because all exponents must be divisible by 3 so a55 = 1.
Suppose that a66 = 3. Then E36 = εa1
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.
Suppose that a77 = 3. Then E37 = εa1
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.
Suppose that a88 = 3. Then E38 = εa1
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
Chapter 6. Unit Group of L 46
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.
Suppose a77 = 3. Then E37 = ea1
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
Chapter 6. Unit Group of L 47
= 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 .
Chapter 6. Unit Group of L 48
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
√ε1ε2ε3ε′1ε
′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
√ε1ε2ε3ε′1ε
′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
Suppose a77 = 3. Then E37 = ea1
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
Chapter 6. Unit Group of L 49
=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
so det(A) = 3b∗ where b∗ ≤ 2.
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
so det(A) = 3b∗ where b∗ ≤ 1.
(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.
Chapter 6. Unit Group of L 50
(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
Suppose a88 = 3. Then E38 = ea1
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.
Suppose N(B1) � N(B2): E0 =< ε1, ε′1, ε2, ε
′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
Suppose a88 = 3. Then E38 = ea1
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)
Chapter 6. Unit Group of L 51
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 >
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 = e−1
3 , e5e5 = e25, e6e6 = e−15 , e7e7 = e27, e8e8 = e7
Suppose a88 = 3. Then E38 = ea1
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
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.
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.
Chapter 6. Unit Group of L 52
e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e1e3, e5e5 = e25, e6e6 = e5, e7e7 = e27, e8e8 = e7
Suppose a88 = 3. Then E38 = ea1
1 ea22 ea3
3 ea44 ea5
5 ea66 ea7
7 e8 and multiplying by the complex conjugate gives:
(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
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 = e1e5, e7e7 = e27, e8e8 = e7
Suppose a88 = 3, then E38 = ea1
1 ea22 ea3
3 ea44 ea5
5 ea66 ea7
7 e8 and multiplying by the complex conjugate gives:
(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
Chapter 6. Unit Group of L 53
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 >
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, ε4, u4 >. Then E0 ⊂ E1 and we will show that
[E : E1
]= 1, thus[
E : E0
]=[E : E1
] [E1 : E0
]= 31
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 = e−1
3 , e5e5 = e25, e6e6 = e1e3e5, e7e7 = e27, e8e8 = e7
Suppose a88 = 3. Then E38 = ea1
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.
We will again take conjugates with σ1.
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 >
The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1.
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
Chapter 6. Unit Group of L 54
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.
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 = e−1
5 , e7e7 = e27, e8e8 = e−17
Suppose a88 = 3. Then E38 = ea1
1 ea22 ea3
3 ea44 ea5
5 ea66 ea7
7 e8 and multiplying by the complex conjugate gives:
(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
Chapter 6. Unit Group of L 55
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
Suppose a88 = 3. Then E38 = ea1
1 ea22 ea3
3 ea44 ea5
5 ea66 ea7
7 e8 and multiplying by the complex conjugate gives:
(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.
We will again take conjugates with σ1.
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.
e1e1 = e21, e2e2 = e−11 , e3e3 = e23, e4e4 = e1e3, e5e5 = e25, e6e6 = e−1
5 , e7e7 = e27, e8e8 = e5e7
Chapter 6. Unit Group of L 56
Suppose a88 = 3. Then E38 = ea1
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.
We will again take conjugates with σ1.
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 >
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, ε4, ε
′4 >. Then E0 ⊂ E1 and we will show that
[E : E1
]= 1, thus
[E : E0
]=[
E : E1
] [E1 : E0
]= 31.
The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1.
Suppose a88 = 3 then E38 = εa1
1 ε′1a2εa3
2 ε′2a4εa5
33
√ε1ε2ε3ε′1ε
′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
√ε1ε2ε3ε′1ε
′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.
Chapter 6. Unit Group of L 57
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 >
We know from [9] Theorem X that e3a = ζa ε1ε2ε3ε′1ε
′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
√ε1ε2ε3ε′1ε
′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.
The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1.
Suppose a88 = 3 then E38 = εa1
1 ε′1a2εa3
2 ε′2a4εa5
33
√ε1ε2ε3ε′1ε
′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
√ε1ε2ε3ε′1ε
′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.
The same reasoning as in the previous case gives us that a11 = a22 = a33 = a44 = a55 = a66 = a77 = 1.
Suppose a88 = 3 then E38 = εa1
1 ε′1a2εa3
2 ε′2a4εa5
3 ε′3a6εa7
43
√ε1ε2ε3ε4ε′1ε
′′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
√ε1ε2ε3ε4ε′1ε
′′2ε
′3ε
′4
.
Chapter 6. Unit Group of L 58
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:
Chapter 6. Unit Group of L 59
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
.
Chapter 6. Unit Group of L 60
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
Since E1+τL ∈ K then all the exponents must be divisible by 3 which gives the following relations:
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
Chapter 6. Unit Group of L 61
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
.
Without loss of generality suppose there are two independent solutions to (6.15). 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 = (ea513 e
a714 ε−t1
1 )d1(ea523 e
a724 ε−t2
1 )d2
= ea51d1+a52d23 e
a71d1+a72d24 ε−t1d1−t2d2
1
and we can rename the exponents to get ed11 ed2
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
Chapter 6. Unit Group of L 62
Then we can square both sides, let(
Bd11 B
d22
αj
)2
= Ej and reduce modulo cubes to get
(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 .
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), 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)
Chapter 6. Unit Group of L 63
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
.
Without loss of generality suppose there are two independent solutions to (6.17). 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 = (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
and we can rename the exponents to get ed11 ed2
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
Then we can square both sides, let(
Bd11 B
d22
αj
)2
= Ej and reduce modulo cubes to get
(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
Chapter 6. Unit Group of L 64
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.
Chapter 6. Unit Group of L 65
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 .
Chapter 6. Unit Group of L 66
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.
Chapter 6. Unit Group of L 67
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.
Chapter 6. Unit Group of L 68
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
where v = 0, 1 or 2.
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
where v = 0, 1 or 2.
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
Chapter 6. Unit Group of L 69
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
Chapter 6. Unit Group of L 70
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.
Chapter 6. Unit Group of L 71
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
Chapter 6. Unit Group of L 72
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
Chapter 6. Unit Group of L 73
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
e38 = εa1ε′12aεb3ε
′32bu4, where 0 ≤ a, b ≤ 2 and a + b > 0 if [E : E0] = 9
e8 = u4 if [E : E0] = 3
N(B1)N(B2) ∼ ζaN(B3) : E =< ε1, ε′1, ε2, ε
′2, ε3, 3
√ε1ε2ε3ε′1ε
′2ε
′′3, 3√ε4, e8 > where
e38 = εa1ε′12aεb2ε
′22bu4, where 0 ≤ a, b ≤ 2 and a + b > 0 if [E : E0] = 9
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 >
Chapter 6. Unit Group of L 74
(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
√ε1ε2ε3ε′1ε
′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
√ε1ε2ε3ε′1ε
′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
√ε1ε2ε3ε4ε′1ε
′′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.
Chapter 6. Unit Group of L 75
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
.
Chapter 7. Rank of the Class Group of K and L 78
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).
Chapter 7. Rank of the Class Group of K and L 79
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)
Chapter 7. Rank of the Class Group of K and L 80
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)
Chapter 7. Rank of the Class Group of K and L 81
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.
Chapter 7. Rank of the Class Group of K and L 82
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,
Chapter 7. Rank of the Class Group of K and L 83
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
Chapter 7. Rank of the Class Group of K and L 84
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
Chapter 7. Rank of the Class Group of K and L 85
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
Appendix A. Units of Cubic Fields and their Normal Closures 89T
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
Appendix A. Units of Cubic Fields and their Normal Closures 90T
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
Appendix A. Units of Cubic Fields and their Normal Closures 91T
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
Appendix A. Units of Cubic Fields and their Normal Closures 92T
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
Appendix A. Units of Cubic Fields and their Normal Closures 93T
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
Appendix A. Units of Cubic Fields and their Normal Closures 94T
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
3I
315
-106
9891
1995
8174
227
0829
21
-525
6904
126
0619
981
0912
677
2609
2-1
6888
4-3
5754
041
I
316
-135
-113
391
-135
00
-113
-39
-39
1II
I
325
263
14-3
83
-55
153
308
-16
-22
3I
340
1-4
914
134
3-3
40-9
8-4
914
283
I
342
7-1
01
70
0-1
00
1II
I
348
4885
9201
-511
5420
-520
590
123
9266
25-2
5942
24-3
7704
60-3
4016
4010
4870
1072
090
1I
350
121
-15
199
49-7
-14
-55
1I
356
2478
2940
1-7
4788
635
1123
7145
110
8324
125
3022
2842
627
3592
45-1
5284
170
-120
3368
5-7
7205
951
I
364
17
-21
-49
5214
7-2
-43
I
369
9854
9569
2093
0412
-145
0113
01
-548
1277
788
4733
2719
9769
2476
4198
8-5
1592
04-8
3555
381
I
380
-170
931
2-2
11
1311
970
-47
-181
-37
131
I
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 95T
able
A.1
:co
ntin
ued
Ma
bc
d1
de
fg
hi
d2
Typ
e
388
3791
37-2
9430
-618
41
3791
370
0-2
9430
6184
6184
1II
I
396
199
-81
1-2
4224
366
33-2
7-5
41
I
412
4441
361
-112
9537
1431
691
9484
034
-483
1727
-192
3898
-127
4561
1745
2951
7106
3I
414
-260
81-7
233
4320
127
829
4682
725
49-3
734
-252
9-1
026
1I
420
1-9
024
1-4
49-2
2530
608
-81
I
423
115
0-6
01
-375
376
100
50-2
0-4
01
I
425
154
-36
127
113
5-1
8-3
6-1
212
1I
436
1765
918
00-1
089
1-5
778
1198
923
4376
2-4
17-6
181
I
460
3272
671
-112
368
-807
271
-172
661
1602
549
2299
6622
367
-537
86-5
9581
1I
468
-95
20-6
1-9
50
020
66
1II
I
476
3681
6481
9-3
9992
352
-183
4125
118
2236
923
-291
0998
3-2
7068
254
-233
3999
395
4994
6933
533
1I
477
-799
7626
597
-628
21
1599
632
2155
3918
4-2
047
-158
31-1
5045
3IV
490
11
-11
-710
21
-1-2
3I
492
-143
6885
2526
24-1
7889
111
4894
283
6645
-395
59-1
4553
8-2
6849
1002
21
I
495
-166
319
1724
414
341
-120
56-9
4071
-103
6815
2445
1039
321
I
Appendix A. Units of Cubic Fields and their Normal Closures 96T
able
A.2
:U
nits
ofki
andK
ino
ton
Tab
leA
.1
Ma
bc
d1
Typ
ed
ef
gh
id2
29-3
2246
1439
1038
1946
237
0284
1
I-1
0939
9864
-602
0785
73-1
6036
0646
3560
8262
6378
5384
5219
5348
3
41-2
1199
1370
839
3054
7847
5184
-707
6118
3382
1
I-1
0695
9528
7171
8564
4846
2821
5585
6218
3216
3101
8790
6564
-720
2976
4756
-161
9857
8693
83
5946
3342
2739
3-4
2285
5550
0478
0468
4934
1
I16
9839
7537
49-1
2297
6557
075
-752
1617
7860
-436
2699
9772
8114
3609
4819
3208
9573
83
6913
7536
1147
5894
0080
5940
1-5
6306
6830
8465
4381
2072
055
5253
6974
5961
5284
770
1
I41
2341
1326
1689
4647
1551
1311
6586
4084
4962
9284
375
2192
6199
0089
0712
1383
20
-100
5326
1112
5628
8955
940
-779
6890
4241
8751
8716
60-5
3458
1166
9178
3412
2790
1
7113
8622
6224
-309
6725
29-6
0612
661
IV-5
2345
1755
2009
7418
2461
1761
669
1264
1265
2-1
1721
0910
-147
7393
373
8311
4607
2952
4019
1345
4318
1491
8875
2-1
6438
3888
3638
741
I-3
3048
2048
5678
901
-543
9894
1306
3660
0-4
8946
3331
9284
60
7576
2653
4412
996
2858
9400
7747
822
1122
0894
1302
002
3
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 97T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
8544
5707
361
-213
0919
7425
4096
281
I-5
3378
2711
-343
9542
1643
1739
48
1214
0172
617
7918
60-9
8193
361
87-2
0254
8707
4437
1534
9522
9558
2499
9034
0774
4-4
1490
6405
3997
7134
21
I-3
4417
1471
1709
4883
6027
6295
6548
4732
7625
714
0027
1852
1475
1006
8
7767
2708
7034
5547
84-1
4072
1747
6201
7658
6-3
1601
3722
1933
7566
81
97-1
2891
2513
6879
8783
3890
-112
7854
887
1
III
-128
9125
1368
00
7987
8338
9011
2785
4887
1127
8548
871
99-5
0708
0573
9910
0254
5608
260
6946
152
1
I-4
8132
0327
3-3
0205
7587
37-5
4890
1829
4
1040
4530
1242
3435
7684
3559
6246
161
101
-759
1749
839
4748
2842
28-6
6955
1396
1
I38
6657
5960
323
1964
3174
4-3
3217
2992
8
-830
2701
452
-106
9564
388
7132
7531
63
103
-239
9726
95-2
9351
1054
7353
6248
1
I16
2892
8520
3017
2053
7129
6162
686
-347
5011
82-1
3731
3548
-631
8072
43
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 98T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
107
6387
4463
3822
-285
7693
6259
-223
2090
3094
3
IV38
7694
4300
4317
1097
2800
53-4
5623
8025
03
-816
6355
8863
-759
1377
479
9610
1511
723
111
4352
5986
334
-126
3423
7836
7443
8736
11
IV45
2785
0888
5-6
2670
2663
16-2
2461
7398
45
-942
1450
917
2713
3941
7146
7378
8618
3
113
2462
9299
21-1
4773
8928
-748
1516
81
I-3
4771
4563
3615
6528
6181
9794
224
7192
2272
-154
6919
68-1
6956
8576
3
119
1712
9469
29-2
4488
4818
-210
1542
01
I-2
7483
3597
1-2
5144
7453
947
5455
22
5587
5424
410
3931
996
-966
6308
3
123
-589
5405
1929
598
0156
6106
2186
5679
148
1
I29
5378
8110
8735
1721
6559
7511
3293
1715
4
-593
9423
6092
-142
2096
3932
-227
8078
564
1
146
1392
6589
61-4
9984
6141
189
9032
809
1
I26
3455
2976
1-2
4952
4142
65-9
7420
1978
9
-500
3293
546
8999
3401
718
5011
2151
3
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 99T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
159
-172
9417
0949
567
-294
7566
3573
2411
3333
5052
934
1
I-1
8891
2950
4929
5-6
4313
3179
8152
2299
9464
2672
4
3487
0862
5130
421
9130
5016
34-4
2454
0061
570
1
178
1558
6562
89-4
1371
0656
2428
8330
1
I38
4361
184
1330
2511
3716
8151
448
-683
2812
8-4
2039
122
-298
9238
81
188
-218
4978
1589
5956
3189
211
2364
5827
1
I17
0539
1523
-108
0859
1927
-218
4454
282
-297
6949
6165
8709
434
7626
4133
31
190
1179
1115
3-8
9330
395
1197
1022
3
I13
5091
922
3097
7339
730
3852
15
-234
9875
5-9
3729
23-5
2853
993
194
2807
9544
4218
1-4
0640
2766
104
-135
8630
0107
1
I41
6358
3020
909
-645
6134
7988
7-8
3075
3052
656
-719
2281
0890
219
2651
0747
814
3506
4328
073
195
-507
2427
359
2018
9521
68-1
9731
8508
1
I67
0446
7809
4223
1936
64-4
2788
7776
-115
6163
992
-125
5889
0073
7878
761
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 100T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
202
3790
8124
3208
9244
251
-524
0909
3286
3673
845
-207
8963
4078
2524
001
I-6
4025
3479
6362
7966
25-5
6020
3157
6267
7396
2213
6430
8631
4870
6970
1091
1927
9807
9789
905
1627
2138
7764
9479
50-2
3252
0993
6980
7575
3
226
-564
9001
1642
0417
7254
8314
2879
0030
2245
7802
169
-823
1322
6665
0561
4370
3
I-1
4689
0335
3210
5693
5593
-496
2765
7829
8684
1869
39-5
7359
1870
7694
0357
717
2411
5174
5176
5989
3922
1337
5750
3124
6497
2611
9416
7312
2011
8818
184
3
246
-130
0032
9324
4806
7653
7814
3099
912
0087
8654
5890
0330
8316
5272
513
9474
0378
5035
9541
8414
5185
1
I-2
2790
7362
2792
9793
8032
3924
9-6
6379
8133
3356
7766
6780
0217
50-1
0230
2129
0763
2653
5659
2150
50
3637
3072
2898
7051
7558
4237
516
9074
9581
9986
1731
9558
8395
1632
6996
6318
1917
3907
1497
301
255
6917
7846
5273
8931
2196
41-1
1543
8486
7582
0778
4511
410
0104
8594
0314
0782
001
I-4
4910
7311
5396
4201
0241
5-3
5897
3430
5195
6772
6202
414
2137
2836
0228
1136
228
7082
2306
3133
7294
0280
289
2693
1147
0743
9270
48-2
2414
4430
0543
9607
132
1
258
1352
2534
3601
1625
4206
1713
-124
1570
9061
0555
8329
0698
-138
6371
5000
5588
6366
024
1
I25
4288
1466
2802
0769
1425
6916
9516
1055
7203
7451
408
1046
5853
2766
2042
8660
068
-399
4412
2151
7607
5435
02-1
7067
4353
6431
3419
1458
4-1
6439
9846
0987
1148
3126
01
261
-356
5763
3489
6601
0007
141
2742
0997
9535
7365
868
1717
7788
8615
0760
1
I-1
9202
9048
1447
8094
32-1
9384
5390
3614
5942
503
-273
2769
9152
1779
5694
3004
8312
9276
5791
8214
2391
1708
4153
7947
612
8285
4750
8579
4129
21
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 101T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
279
-146
3404
502
2322
7059
1-3
8172
721
III
-146
3404
502
00
2322
7059
138
1727
238
1727
21
282
2606
8371
3193
7809
-532
8750
9891
6735
2063
9996
0845
771
I-2
0231
2204
2970
869
-149
0409
2222
4814
781
2350
4890
1381
3085
1222
5914
202
3465
8209
5842
11-1
2387
7874
0308
31
318
-851
9943
0998
501
-222
8136
3882
8235
3447
2441
4307
321
I50
7458
2914
7381
7-5
3586
7461
1618
21-1
5285
4262
4954
21
-743
4606
3724
074
1150
1980
8460
8122
3941
8082
4154
1
330
-234
1850
9399
1175
0712
930
-121
0033
749
1
I48
0453
7924
927
9728
1932
4-2
9046
7881
0
-695
2595
760
-585
7697
5942
0332
983
1
332
-298
2473
8933
6396
1613
2633
90-1
5334
0198
943
3
I-2
4159
7945
0702
1389
4036
3521
954
9564
9444
93
3489
1036
7956
-579
5617
3519
-158
7338
8901
83
333
7309
3831
90-2
8853
2187
-331
5459
601
IV55
0608
9603
-752
7428
976
-188
0385
661
-794
3804
1847
0043
912
8138
6700
33
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 102T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
372
-142
0410
7237
1122
4075
916
0436
8313
3242
2224
110
3067
2313
5502
0337
1
I-1
0036
2635
7733
5269
79-8
0014
5235
6436
2968
77-9
7300
5943
0137
5754
7
1395
4825
5532
6467
6730
9388
0434
6127
8722
2705
8133
9208
9471
011
387
2599
0955
62-2
5338
7618
189
6298
960
1
III
2599
0955
620
0
-253
3876
181
-896
2989
60-8
9629
8960
1
404
3280
3442
3892
6020
701
-598
3356
9270
3660
915
4182
5587
9032
4206
01
I67
7422
1125
1048
1651
2558
5707
8272
2640
375
2544
6574
7129
1855
65
-916
3588
5409
5086
30-9
3635
3536
5512
2730
-688
4391
1687
9706
201
428
2408
8550
6522
2742
0377
3-1
0169
0663
2962
2336
4813
1850
4621
6787
0022
4252
1
I18
4294
4455
2515
7668
292
4500
5770
6605
0537
4551
735
2652
5440
4009
5711
87
-244
5481
1700
7667
3971
5-1
5849
0177
7357
4567
490
-935
8992
3838
6936
4598
1
444
5630
9509
1251
1549
1781
300
-234
1174
1025
1
I-9
5375
7612
4930
6866
2812
5052
7261
9982
5
1250
1928
975
-105
4525
7425
-138
2278
2575
1
452
7661
1901
591
-368
1300
0572
6992
1158
571
I-5
0415
2787
170
-290
7848
4856
927
8024
4667
7
6569
2535
974
9874
3718
87-7
2454
7508
23
cont
inue
don
next
page
Appendix A. Units of Cubic Fields and their Normal Closures 103T
able
A.2
:co
ntin
ued
Ma
bc
d1
Typ
ed
ef
gh
id2
475
3087
9248
41-7
8686
8522
2506
2963
21
I36
2565
4648
1047
0332
315
8772
4681
0
-464
6813
38-8
5993
9828
-562
1608
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
Bas
is,
Cla
ssN
umbe
rsan
dR
ank
ofth
e3-
Cla
ss
Gro
upfo
rK
andL
{m1,m
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 106T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 107T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 108T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 109T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 110T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 111T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 112T
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
Appendix B. Some invariants of K and L where mi ≤ 500 for all i 113T
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..