#  $3$-rank of ambiguous class groups
of cubic Kummer extensions

S. AOUISSI, D. C. MAYER, M. C. ISMAILI, M. TALBI and A. AZIZI

Dedicated to the memory of Frank Emmett Gerth III

Abstract: Let $k=k_{0}(\sqrt[3]{d})$ be a cubic Kummer extension of
$k_{0}=\mathbb{Q}(\zeta_{3})$ with $d>1$ a cube-free integer and $\zeta_{3}$ a
primitive third root of unity. Denote by $C_{k,3}^{(\sigma)}$ the $3$-group of
ambiguous classes of the extension $k/k_{0}$ with relative group
$G=\operatorname{Gal}(k/k_{0})=\langle\sigma\rangle$. The aims of this paper
are to characterize all extensions $k/k_{0}$ with cyclic $3$-group of
ambiguous classes $C_{k,3}^{(\sigma)}$ of order $3$, to investigate the
multiplicity $m(f)$ of the conductors $f$ of these abelian extensions
$k/k_{0}$, and to classify the fields $k$ according to the cohomology of their
unit groups $E_{k}$ as Galois modules over $G$. The techniques employed for
reaching these goals are relative $3$-genus fields, Hilbert norm residue
symbols, quadratic $3$-ring class groups modulo $f$, the Herbrand quotient of
$E_{k}$, and central orthogonal idempotents. All theoretical achievements are
underpinned by extensive computational results.

Keywords: Pure cubic fields, cubic Kummer extensions, $3$-group of ambiguous
ideal classes, $3$-rank, Hilbert $3$-class field, relative $3$-genus field,
multiplicity of conductors, Galois cohomology of unit groups, principal
factorization types.

Mathematics Subject Classification 2010: 11R11, 11R16, 11R20, 11R27, 11R29,
11R37.

##  1 Introduction

Let $d>1$ be a cube-free integer and $k=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ be
a cubic Kummer extension of the cyclotomic field
$k_{0}=\mathbb{Q}(\zeta_{3})$. Denote by $f$ the conductor of the abelian
extension $k/k_{0}$, by $m=m(f)$ its multiplicity, and by $C_{k,3}^{(\sigma)}$
the $3$-group of ambiguous ideal classes of $k/k_{0}$. Let
$k^{\ast}=(k/k_{0})^{\ast}$ be the maximal abelian extension of $k_{0}$
contained in the Hilbert $3$-class field $k_{1}$ of $k$, which is called the
relative $3$-genus field of $k/k_{0}$ (cf. [14, § 2, p. VII-3]).

We consider the problem of finding the radicands $d$ and conductors $f$ of all
pure cubic fields $L=\mathbb{Q}(\sqrt[3]{d})$ for which the Galois group
$\operatorname{Gal}(k^{\ast}/k)$ is non-trivial cyclic. The present work gives
the complete solution of this problem by characterizing all cubic Kummer
extensions $k/k_{0}$ with cyclic $3$-group of ambiguous ideal classes
$C_{k,3}^{(\sigma)}$ of order $3$. In fact, we prove the following Main
Theorem:

######  Theorem 1.1.

Let $k=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$, where $d>1$ is a cube-free integer,
and $C_{k,3}^{(\sigma)}$ be the $3$-group of ambiguous ideal classes of
$k/\mathbb{Q}(\zeta_{3})$. Then, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$
if and only if the integer $d$ can be written in one of the following forms:

$d=\left\\{\begin{array}[]{ll}p_{1}^{e_{1}}&\text{ with }p_{1}\equiv
1\,(\mathrm{mod}\,3),\\\ 3^{e}p_{1}^{e_{1}}&\text{ with }p_{1}\equiv 4\text{
or }7\,(\mathrm{mod}\,9),\\\ p_{1}^{e_{1}}q_{1}^{f_{1}}\equiv\pm
1\,(\mathrm{mod}\,9)&\text{ with }p_{1},-q_{1}\equiv 4\text{ or
}7\,(\mathrm{mod}\,9),\\\ 3^{e}q_{1}^{f_{1}}&\text{ with
}q_{1}\equiv-1\,(\mathrm{mod}\,9),\\\ q_{1}^{f_{1}}q_{2}^{f_{2}}&\text{ with
}q_{1}\equiv q_{2}\equiv-1\,(\mathrm{mod}\,9),\\\
q_{1}^{f_{1}}q_{2}^{f_{2}}\not\equiv\pm 1\,(\mathrm{mod}\,9)&\text{ with
}q_{i}\not\equiv-1\,(\mathrm{mod}\,9)\text{ for some }i\in\\{1,2\\},\\\
3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}&\text{ with
}q_{i}\not\equiv-1\,(\mathrm{mod}\,9)\text{ for some }i\in\\{1,2\\},\\\
q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 1\,(\mathrm{mod}\,9)&\text{
with }q_{1},q_{2}\equiv 2\text{ or }5\,(\mathrm{mod}\,9)\text{ and
}q_{3}\equiv-1\,(\mathrm{mod}\,9),\\\
q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 1\,(\mathrm{mod}\,9)&\text{
with }q_{1},q_{2},q_{3}\equiv 2\text{ or }5\,(\mathrm{mod}\,9),\\\
\end{array}\right.$ (1)

where $p_{1}\equiv 1\,(\mathrm{mod}\,3)$ and
$q_{1},q_{2},q_{3}\equiv-1\,(\mathrm{mod}\,3)$ are primes and
$e,e_{1},f_{1},f_{2}$ and $f_{3}$ are integers equal to $1$ or $2$.

We should point out that Theorem 1.1 is expressed in terms of radicands $d>1$.
However, in Theorem 2.4 of section 2.5 and Theorem 2.5 of section 2.6 we
emphasize that an ambiguous class group $C_{k,3}^{(\sigma)}$ of $3$-rank $1$
is rather a characteristic property of the class field theoretic conductor
$f=3^{e}\cdot\ell_{1}\cdots\ell_{n}$ of the cyclic extension $k/k_{0}$, i.e.
an essentially square-free positive integer, divisible by the prime factors
$\ell_{j}$ of the radicand
$d=3^{e_{0}}\cdot\ell_{1}^{e_{1}}\cdots\ell_{n}^{e_{n}}$ (and possibly
additionally by the prime $3$) but independent of the exponents
$e_{1},\ldots,e_{n}$. Theorems 2.3, 2.4 and 2.5 show that pure cubic fields
$L$ can be collected in multiplets $(L_{1},\ldots,L_{m})$ sharing a common
conductor $f$ with multiplicity $m$ and a common type of ambiguous class group
$C_{k,3}^{(\sigma)}$ of their associated Galois closures $k$.

One of the motivations of this work is an earlier article of Honda [16].
Honda’s result concerning trivial ambiguous class groups $C_{k,3}^{(\sigma)}$
of $3$-rank $0$ is restated in terms of conductors $f$ and multiplicities $m$
in section 2.4, Theorem 2.3, to enable immediate comparison with the results
of our paper, which continues Honda’s explicit description of radicands $d$
from trivial to cyclic ambiguous $3$-class groups $C_{k,3}^{(\sigma)}$ of
$3$-rank $1$. Another motivation is that our results on the factorization of
the conductor $f$ of $k$ into prime numbers provide information about the
ramification in $k$, allow the study of the class group of the relative genus
field $k^{\ast}$ of $k/k_{0}$, and can be applied to analyze the class field
tower of $k$.

A further main result of our paper is Theorem 2.1, where we use the Galois
cohomology and the Herbrand quotient of the unit group $E_{k}$ of the normal
closure $k$ as a module over the group
$G=\mathrm{Gal}(k/k_{0})=\langle\sigma\rangle$ to give a brief and elegant new
proof for the existence of exactly three principal factorization types
$\alpha,\beta,\gamma$ of pure cubic fields, unifying concepts introduced by
Barrucand and Cohn [3, Thm. 15.6] and corrected by Halter-Koch [12, Korollar,
p. 594].

In the numerical Examples 2.1, 2.2, 2.3 and 2.4, we present statistical
results of the most extensive computation of invariants of pure cubic fields
up to now. They cover all $827600$ non-isomorphic fields $L$ with normalized
radicands $1<d<10^{6}$, determined with Voronoi’s algorithm [29] and confirmed
with Magma [25], significantly extending results of H. C. Williams in [31].

The outline of this work is as follows. After having illuminated the broader
scope of our main results in section 2, we give the proof of Theorem 1.1 in
section 3. The section 4 summarizes the structures of the $3$-class group of
the pure cubic field $L$ and the full $3$-class group of the normal closure
$k$. In section 5, we conclude the paper with an important open problem to a
broader audience. The Outlook of section 6 investigates another related
scenario for the location of the relative $3$-genus field $k^{\ast}$.

Notations.  Throughout this paper, we use the following notation:

  * •

$p$ and $q$ are prime numbers such that $p\equiv 1\,(\mathrm{mod}\,3)$ and
$q\equiv-1\,(\mathrm{mod}\,3)$;

  * •

$L=\mathbb{Q}(\sqrt[3]{d})$: a pure cubic field, where $d\neq 1$ is a cube-
free positive integer;

  * •

$k_{0}=\mathbb{Q}(\zeta_{3})$, with
$\zeta_{3}=e^{2i\pi/3}=\frac{1}{2}(-1+\sqrt{-3})$ a primitive cuberoot of
unity;

  * •

$k=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$: the normal closure of $L$, a cubic
Kummer extension of $k_{0}$;

  * •

$f$: the conductor of the abelian extension $k/k_{0}$;

  * •

$m=m(f)$: the multiplicity of the conductor $f$;

  * •

$\langle\tau\rangle=\operatorname{Gal}(k/L)$ such that $\tau^{2}=id,\
\tau(\zeta_{3})=\zeta_{3}^{2}$ and $\tau(\sqrt[3]{d})=\sqrt[3]{d}$;

  * •

$\langle\sigma\rangle=\operatorname{Gal}(k/k_{0})$ such that $\sigma^{3}=id,\
\sigma(\zeta_{3})=\zeta_{3}$ and $\sigma(\sqrt[3]{d})=\zeta_{3}\sqrt[3]{d}$;

  * •

$\lambda=1-\zeta_{3}$ and $\pi$ are prime elements of $k_{0}$;

  * •

$q^{\ast}=0$ or $1$ according to whether $\zeta_{3}$ is not norm or is norm of
an element of $k^{\times}=k\setminus\\{0\\}$;

  * •

$t$: the number of prime ideals of $k_{0}$ ramified in $k$, $2s$ among them
split over $\mathbb{Q}$.

  * •

For a number field $F$, denote by $F_{1}$ the Hilbert $3$-class field of $F$
and by

    * –

$\mathcal{O}_{F}$, $\mathcal{I}_{F}$: the ring of integers, and the group of
ideals of $F$;

    * –

$E_{F}$, $\mathcal{P}_{F}$: the group of units, and the group of principal
ideals of $F$;

    * –

$C_{F}$, $h_{F}$, $C_{F,3}$: the class group, class number, and $3$-class
group of $F$.

##  2 Viewing the main result in a wider perspective

In this section, our intention is to shed light on the structure of the entire
$3$-class group $C_{k,3}$ of the normal closure $k$ of $L$ which is minorized
by the $3$-group $C_{k,3}^{(\sigma)}$ of ambiguous ideal classes of $k/k_{0}$.
The investigation of this relationship requires the distinction of principal
factorization types in the sense of Barrucand and Cohn [3]. We present a brief
and elegant new proof of this classification by means of the Galois cohomology
of the unit group $E_{k}$ as a module over
$\operatorname{Gal}(k/k_{0})=\langle\sigma\rangle$, since the original paper
[3] contained a superfluous type, which was shown to be impossible by Halter-
Koch [12], and therefore no complete proof is available in a single article of
the existing literature.

###  2.1 Galois cohomology of the unit group and principal factorization types

Let $E_{k}$ be the unit group of $k=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ and
denote by $Q=(E_{k}:E_{0})$ the index of the subgroup $E_{0}$ which is
generated by all units of proper subfields of $k$, that is, $E_{0}=\langle
E_{k_{0}},E_{L},E_{L^{\sigma}},E_{L^{\sigma^{2}}}\rangle$. Pierre Barrucand
and Harvey Cohn have based their classification of pure cubic fields
$L=\mathbb{Q}(\sqrt[3]{d})$ into principal factorization types [3, Thm. 15.6,
pp. 235–236] on the class number relation $h_{k}=\frac{Q}{3}\cdot h_{L}^{2}$
[3, Thm. 14.1, p. 232], [16, Lem. 1, p. 7], which involves the subfield unit
index $Q\in\\{1,3\\}$ [3, Thm. 12.1, p. 229].

Since a superfluous type of Barrucand and Cohn was eliminated by Franz Halter-
Koch [12, Korollar, p. 594], we present an alternative access to this
classification by means of the Galois cohomology of the unit group $E_{k}$ of
the normal closure $k$ with respect to the relative automorphism group
$G=\operatorname{Gal}(k/k_{0})=\langle\sigma\rangle$. Instead of $Q$, the
primary invariant is the cohomology group
$H^{0}(G,E_{k})=E_{k_{0}}/N_{k/k_{0}}(E_{k})$ and its order, the unit norm
index $3^{U}=(E_{k_{0}}:N_{k/k_{0}}(E_{k}))$ where $U$ can only take two
values $U\in\\{0,1\\}$, according to whether $\zeta_{3}$ is the relative norm
of a unit of $k$ or not. (Observe that $E_{k_{0}}=\langle-1,\zeta_{3}\rangle$
and $-1=(-1)^{3}=N_{k/k_{0}}(-1)$ is a norm.) The Theorem on the Herbrand
quotient of $E_{k}$ [13, Thm. 3, p. 92], which is a quantitative version of
Hilbert’s Theorem $92$ [15, § 56, p. 275], admits the calculation of the order
of the cohomology group
$H^{1}(G,E_{k})\simeq(E_{k}\cap\ker(N_{k/k_{0}}))/E_{k}^{1-\sigma}$ by the
general formula

$\\#H^{1}(G,E_{k})=\\#H^{0}(G,E_{k})\cdot[k:k_{0}],$ (2)

since no real archimedean place of $k_{0}$ becomes complex in $k$.

According to Iwasawa [21, § 2, pp. 189–190], the quotient group
$(E_{k}\cap\ker(N_{k/k_{0}}))/E_{k}^{1-\sigma}$ is isomorphic to the group of
primitive ambiguous principal ideals $\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}$
of $k$ with respect to $k_{0}$. Equation (2) implies that the order
$\\#H^{1}(G,E_{k})=3^{U}\cdot 3\in\\{3,9\\}$ can take precisely two values in
dependence on the unit norm index $3^{U}$, whereas Hilbert’s Theorem $92$ only
states that $H^{1}(G,E_{k})$ is non-trivial. Since $k_{0}$ is a principal
ideal domain, we have $\mathcal{I}_{k_{0}}=\mathcal{P}_{k_{0}}$ and
$\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}$ is a subgroup of the group
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}$ of primitive ambiguous ideals of
$k/k_{0}$. By determining the order of the group
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}$ we occasionally can estimate the
order of the subgroup $\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}$ by a better
upper bound than $3^{U+1}$. As a refinement of the classification by the
Galois cohomology, however, we first intend to split the group
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}$ into the direct product of the
absolute component $\mathcal{I}_{L}^{G}/\mathcal{I}_{\mathbb{Q}}$ and a
relative complement. The splitting can be accomplished by means of image and
kernel
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}\simeq\mathrm{img}(N_{k/L})\times\ker(N_{k/L})$
of the relative norm $N_{k/L}$ or, equivalently, by the action
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}\simeq(\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}})^{1+\tau}\times(\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}})^{1-\tau}$
of the central orthogonal idempotents $\frac{1}{2}(1+\tau)$ and
$\frac{1}{2}(1-\tau)$ in the group ring $\mathbb{F}_{3}[\tau]$ of the relative
group $\operatorname{Gal}(k/L)=\langle\tau\rangle$ over the finite field
$\mathbb{F}_{3}$. We denote by $t$ the number of all prime ideals of $k_{0}$
which ramify in $k$, and by $2s$ the number of those among them lying over
primes which split in $k_{0}/\mathbb{Q}$. Then Hilbert’s Theorem $93$ [15, §
57, p. 277] yields the orders of elementary abelian $3$-groups, respectively
dimensions of vector spaces over $\mathbb{F}_{3}$, involved:

$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}\simeq\mathbb{F}_{3}^{t}\simeq\mathcal{I}_{L}^{G}/\mathcal{I}_{\mathbb{Q}}\times\ker(N_{k/L})\simeq\mathbb{F}_{3}^{t-s}\times\mathbb{F}_{3}^{s},$
(3)

since the norm map $N_{k/L}$ induces an epimorphism from
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}$ onto
$\mathcal{I}_{L}^{G}/\mathcal{I}_{\mathbb{Q}}$. Finally, the splitting of
$\mathcal{I}_{k}^{G}/\mathcal{I}_{k_{0}}$ restricts to the subgroup
$\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}$ and we obtain the relation

$\\#(\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}})=3^{U+1}=\\#(\mathcal{P}_{L}^{G}/\mathcal{P}_{\mathbb{Q}})\cdot\\#(\ker(N_{k/L})\cap(\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}))=3^{A+R},$
(4)

where $A=\dim_{\mathbb{F}_{3}}(\mathcal{P}_{L}^{G}/\mathcal{P}_{\mathbb{Q}})$
denotes the dimension of the subspace of absolute principal factors, and
$R=\dim_{\mathbb{F}_{3}}(\ker(N_{k/L})\cap(\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}))$
denotes the dimension of the subspace of relative principal factors.

With the preceding developments we have given a concise proof of the following
Theorem.

######  Theorem 2.1.

Each pure cubic field $L=\mathbb{Q}(\sqrt[3]{d})$ belongs to precisely one of
the following three principal factorization types in dependence on the unit
norm index $3^{U}=(E_{k_{0}}:N_{k/k_{0}}(E_{k}))$ and on the pair of
invariants $(A,R)$, where
$3^{A}=\\#(\mathcal{P}_{L}^{G}/\mathcal{P}_{\mathbb{Q}})$ and
$3^{R}=\\#(\ker(N_{k/L})\cap(\mathcal{P}_{k}^{G}/\mathcal{P}_{k_{0}}))$:

Type | $U$ | $A$ | $R$
---|---|---|---
type $\alpha$ | $1$ | $1$ | $1$
type $\beta$ | $1$ | $2$ | $0$
type $\gamma$ | $0$ | $1$ | $0$

The invariants satisfy the following relations: the equation $A+R=U+1$ and the
estimates $1\leq A+R\leq t$, $1\leq A\leq t-s$, and $0\leq R\leq s$, since for
each type, the group $\mathcal{P}_{L}^{G}/\mathcal{P}_{\mathbb{Q}}$ of
absolute principal factors contains the subgroup
$\Delta=\langle\sqrt[3]{d}\mathcal{O}_{L}\rangle$ of principal ideals
generated by radicals.

######  Remark 2.1.

Our deduction of three possible principal factorization types $\alpha$,
$\beta$, and $\gamma$ has the advantage of being very brief and elegant. In
the given order, they correspond to type III, I, and IV of Barrucand and Cohn
[3, Thm. 15.6, pp. 235–236]. Type II has been proven to be impossible by
Halter-Koch [12, Korollar, p. 594]. A drawback of our access is the lack of
connections to the class number relation $h_{k}=\frac{Q}{3}\cdot h_{L}^{2}$
and the subfield unit index $Q$, which is given in the following manner by
combining [3] and [12]:

$Q=1\text{ for type }\alpha\quad\text{ and }\quad Q=3\text{ for both types
}\beta,\ \gamma.$ (5)

######  Example 2.1.

By means of our own implementation of Voronoi’s algorithm [29], we have
determined the statistical distribution of the principal factorization types
over all pure cubic fields $L=\mathbb{Q}(\sqrt[3]{d})$ with normalized
radicands $d<10^{6}$. The total number of these fields, which are pairwise
non-isomorphic by the normalization, is $827\,600$. The dominating part of
$635\,463$ fields ($76.78\%$) is of type $\beta$, $163\,527$ fields
($19.76\%$) are of type $\alpha$, and only $28\,610$ fields ($3.46\%$) are of
type $\gamma$. ($254\,254$, $382\,231$, $191\,115$ fields are of species
$1\mathrm{a}$, $1\mathrm{b}$, $2$, respectively.) Our results were confirmed
with Magma [25]. They significantly extend the computations of H. C. Williams
[31, pp. 272–273].

Before we come to the application of the principal factorization types to the
relationship between the full $3$-class group $C_{k,3}$ and its subgroup
$C_{k,3}^{(\sigma)}$ of ambiguous classes in §§ 2.4 – 2.6, we show in § 2.2
how to replace the radicands $d$ of pure cubic fields
$L=\mathbb{Q}(\sqrt[3]{d})$ by class field theoretic conductors $f$ of the
corresponding cyclic relative extensions $k/k_{0}$, which enables the
determination of the multiplicity $m(f)$ of non-isomorphic fields sharing a
common conductor.

###  2.2 Conductors and their multiplicity

The class field theoretic conductor $f$ of the Kummer extension $k/k_{0}$ is
the smallest positive integer such that $k$ is contained in the $3$-ring class
field modulo $f$ of the quadratic field
$k_{0}=\mathbb{Q}(\zeta_{3})=\mathbb{Q}(\sqrt{-3})$. If
$L=\mathbb{Q}(\sqrt[3]{d})$ is a pure cubic field with normalized cube-free
radicand $d=d_{1}d_{2}^{2}$ constituted by square-free coprime integers
$d_{1}>d_{2}\geq 1$, $\gcd(d_{1},d_{2})=1$, then $d$ is strictly smaller than
the co-radicand, $d_{1}d_{2}^{2}<d_{1}^{2}d_{2}$, and the conductor $f$ of the
Galois closure $k=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ is given by the following
formula (which also holds without normalization):

$f=\begin{cases}d_{1}d_{2}&\text{ if }d\equiv\pm 1\,(\mathrm{mod}\,9)\text{
(Dedekind's species 2)},\\\ 3d_{1}d_{2}&\text{ if }d\not\equiv\pm
1\,(\mathrm{mod}\,9)\text{ (Dedekind's species 1)}.\\\ \end{cases}$ (6)

We see that $f$ is essentially square-free with the possible exception of its
$3$-part. However, equation (6) is too coarse for determining the multiplicity
$m(f)$ of the conductor $f$, that is the number of non-isomorphic pure cubic
fields sharing the common conductor $f$. For this purpose, we need the prime
factorization of the radicand $d$. If
$d=3^{e_{0}}\cdot\ell_{1}^{e_{1}}\cdots\ell_{n}^{e_{n}}$ with $n\geq 0$,
pairwise distinct prime numbers $\ell_{j}\neq 3$, and exponents $0\leq
e_{0}\leq 2$, $1\leq e_{j}\leq 2$ for $1\leq j\leq n$, then the prime
factorization of the conductor $f$ is given by

$f=3^{e}\cdot\ell_{1}\cdots\ell_{n},\text{ where }e=\begin{cases}0&\text{ if
}e_{0}=0,\ d\equiv\pm 1\,(\mathrm{mod}\,9)\text{ (species 2)},\\\ 1&\text{ if
}e_{0}=0,\ d\not\equiv\pm 1\,(\mathrm{mod}\,9)\text{ (species 1b)},\\\
2&\text{ if }e_{0}\geq 1\text{ (species 1a)}.\end{cases}$ (7)

With equation (7) we are in the position to express the multiplicity by the
formula in [27, Thm. 2.1, p. 833].

######  Theorem 2.2.

The multiplicity $m(f)$ of the conductor $f=3^{e}\cdot\ell_{1}\cdots\ell_{n}$
of the Kummer extension $k/k_{0}$ is given in dependence on the numbers

$u:=\\#\\{1\leq j\leq n\mid\ell_{j}\equiv\pm 1\,(\mathrm{mod}\,9)\\}$ and
$v:=\\#\\{1\leq j\leq n\mid\ell_{j}\equiv\pm 2,\pm 4\,(\mathrm{mod}\,9)\\}$,

of prime divisors, respectively, on the total number $n:=u+v$ of all prime
divisors of $f$ distinct from $3$, by the formulas

$m(f)=\begin{cases}2^{n}&\text{ if }e=2\text{ (species $\mathrm{1a}$)},\\\
2^{u}\cdot X_{v}&\text{ if }e=1\text{ (species $\mathrm{1b}$)},\\\ 2^{u}\cdot
X_{v-1}&\text{ if }e=0\text{ (species $\mathrm{2}$)},\end{cases}$ (8)

where the sequence $(X_{k})_{k\geq-1}$ is defined by
$X_{k}=\frac{1}{3}(2^{k}-(-1)^{k})$.

From the broader perspective of arbitrary non-Galois cubic fields, equation
(8) can also be derived from [28], namely from Thm. 3.4, eqn. (3.4), p. 2217,
for species 1a, and, taking notice of Cor. 3.2, p. 2219, from Thm. 3.3, eqn.
(3.3), p. 2217, for species 1b and 2.

######  Definition 2.1.

When $f$ is a conductor with multiplicity $m:=m(f)$ we say that the
corresponding pairwise non-isomorphic pure cubic fields $L$ which share the
common conductor $f$ form a multiplet $(L_{1},\ldots,L_{m})$. Their normalized
companion radicands $d_{1},\ldots,d_{m}$ such that
$L_{i}=\mathbb{Q}(\sqrt[3]{d_{i}})$ can be constructed from $f$ by varying the
exponents $e_{j}$ of the prime factors $\ell_{j}$.

###  2.3 Determination of 3-ranks of class groups

An estimate for the $3$-class rank $r:=\operatorname{rank}\,(C_{L,3})$ of a
pure cubic field $L=\mathbb{Q}(\sqrt[3]{d})$ has been given by Barrucand,
Williams and Baniuk [4, (2.1), p. 313] in the following form.

Let $\tilde{t}$ be the number of all prime divisors of the conductor $f$ of
$k/k_{0}$, $\tilde{s}$ be the number of those which are congruent to $1$
modulo $3$, and $\tilde{v}$ be the number of those which are congruent to
either $\pm 2$ or $\pm 4$ modulo $9$. Put $\tilde{\varepsilon}:=0$ if
$\tilde{v}=0$, and $\tilde{\varepsilon}:=1$ if $\tilde{v}\geq 1$, that is
$\tilde{\varepsilon}=\min(1,\tilde{v})$. Then we have a lower bound and an
upper bound for $r$ in terms of $\tilde{s}$ and
$\tilde{\delta}:=\tilde{t}-1-\tilde{\varepsilon}$:

$\max(\tilde{s},\tilde{\delta})\quad\leq\quad
r\quad\leq\quad\tilde{s}+\tilde{\delta}.$ (9)

In all of our applications, the lower bound (maximum) and the upper bound
(sum) will coincide, and we shall obtain the precise $3$-class rank of $L$.

In Table 1, we start with the fewest possible prime factors of the conductor
$f$, and none of them split in $k_{0}$, that is $\tilde{s}=0$.

Table 1: Pure cubic fields $L$ with $\tilde{s}=0$ and $r=0$ Item | $f$ | $\tilde{t}$ | $\tilde{s}$ | $\tilde{v}$ | $\tilde{\varepsilon}$ | $\tilde{\delta}=\tilde{t}-1-\tilde{\varepsilon}$ | $\max(\tilde{s},\tilde{\delta})$ | $\tilde{s}+\tilde{\delta}$ | $r$
---|---|---|---|---|---|---|---|---|---
$(1)$ | $9$ | $1$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$(2)$ | $q\equiv 8\,(9)$ | $1$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$(3)$ |  $3q$, $q\equiv 2,5\,(9)$ | $2$ | $0$ | $1$ | $1$ | $0$ | $0$ | $0$ | $0$
$(4)$ |  $9q$, $q\equiv 2,5\,(9)$ | $2$ | $0$ | $1$ | $1$ | $0$ | $0$ | $0$ | $0$
$(5)$ |  $q_{1}q_{2}$, $q_{j}\equiv 2,5\,(9)$ | $2$ | $0$ | $2$ | $1$ | $0$ | $0$ | $0$ | $0$

In Table 2, we proceed with conductors $f$ having the smallest numbers of
prime divisors such that exactly one of them splits in $k_{0}$, that is
$\tilde{s}=1$.

Table 2: Pure cubic fields $L$ with $\tilde{s}=1$ and $r=1$ Item | $f$ | $\tilde{t}$ | $\tilde{s}$ | $\tilde{v}$ | $\tilde{\varepsilon}$ | $\tilde{\delta}=\tilde{t}-1-\tilde{\varepsilon}$ | $\max(\tilde{s},\tilde{\delta})$ | $\tilde{s}+\tilde{\delta}$ | $r$
---|---|---|---|---|---|---|---|---|---
$(1)$ | $p\equiv 1\,(9)$ | $1$ | $1$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$
$(2)$ |  $3p$, $p\equiv 4,7\,(9)$ | $2$ | $1$ | $1$ | $1$ | $0$ | $1$ | $1$ | $1$
$(3)$ |  $9p$, $p\equiv 4,7\,(9)$ | $2$ | $1$ | $1$ | $1$ | $0$ | $1$ | $1$ | $1$
$(4)$ |  $pq$, $p\equiv 4,7\,(9)$, $q\equiv 2,5\,(9)$ | $2$ | $1$ | $2$ | $1$ | $0$ | $1$ | $1$ | $1$

Finally, Table 3 lists further conductors $f$ with a small number of prime
divisors such that none of them splits in $k_{0}$, that is $\tilde{s}=0$.

Table 3: Pure cubic fields $L$ with $\tilde{s}=0$ and $r=1$ Item | $f$ | $\tilde{t}$ | $\tilde{s}$ | $\tilde{v}$ | $\tilde{\varepsilon}$ | $\tilde{\delta}$ | $\max(\tilde{s},\tilde{\delta})$ | $\tilde{s}+\tilde{\delta}$ | $r$
---|---|---|---|---|---|---|---|---|---
$(1)$ |  $9q$, $q\equiv 8\,(9)$ | $2$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$
$(2)$ |  $q_{1}q_{2}$, $q_{1},q_{2}\equiv 8\,(9)$ | $2$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$
$(3)$ |  $3q_{1}q_{2}$, $q_{1},q_{2}\equiv 2,5\,(9)$ | $3$ | $0$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$
$(4)$ |  $3q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$, $q_{2}\equiv 8\,(9)$ | $3$ | $0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
$(5)$ |  $9q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$, $q_{2}\equiv 2\,(3)$ | $3$ | $0$ |  $1$ or $2$ | $1$ | $1$ | $1$ | $1$ | $1$
$(6)$ |  $q_{1}q_{2}q_{3}$, $q_{1},q_{2},q_{3}\equiv 2,5\,(9)$ | $3$ | $0$ | $3$ | $1$ | $1$ | $1$ | $1$ | $1$
$(7)$ |  $q_{1}q_{2}q_{3}$, $q_{1},q_{2}\equiv 2,5\,(9)$, $q_{3}\equiv 8\,(9)$ | $3$ | $0$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$

###  2.4 Class groups with 3-rank zero

To enable comparison with our main result, we mention the related earlier
result by Taira Honda [16] on ambiguous $3$-class groups having the minimal
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=0$. We extend Honda’s theorem by
information on the multiplicity $m(f)$ of conductors $f$ and on the Galois
cohomology of the unit group $E_{k}$, expressed by the principal factorization
type (PFT), briefly called the type. (Partially, the following theorem is due
to Barrucand and Cohn, [2, Cor. 4.2.1, pp. 14–15] and [3, Cor. 14.1.1, p.
232]. It is also mentioned by Gerth [11, Cases 1–2, p. 473].) Honda’s result
can also be obtained as a consequence of [30, Thm. 3, p. 399] by H. Wada.

######  Theorem 2.3.

Let the conductor of $k/k_{0}$ be $f=3^{e}\cdot\ell_{1}\cdots\ell_{n}$ with
$0\leq e\leq 2$, pairwise distinct primes $\ell_{j}\neq 3$ for $1\leq j\leq
n$, and $n\geq 0$ if $e=2$, but $n\geq 1$ if $e\leq 1$. Denote the
multiplicity of $f$ by $m:=m(f)$. Then, $3\nmid\\#C_{k,3}^{(\sigma)}$
$\Longleftrightarrow$ $3\nmid h_{k}$ $\Longleftrightarrow$ $3\nmid h_{L}$
$\Longleftrightarrow$ $L$ belongs to one of the following multiplets, where
always $\ell_{j}=q_{j}\equiv-1\,(\mathrm{mod}\,3)$:

  1. $(1)$

singulet with $m=1$ of type $\gamma$ such that $f=3^{2}$,

  2. $(2)$

singulets with $m=1$ of type $\gamma$ such that $f=q_{1}$ with $q_{1}\equiv
8\,(\mathrm{mod}\,9)$,

  3. $(3)$

singulets with $m=1$ of type $\beta$ such that $f=3q_{1}$ with $q_{1}\equiv
2,5\,(\mathrm{mod}\,9)$,

  4. $(4)$

doublets with $m=2$ of type $(\beta,\beta)$ such that $f=3^{2}q_{1}$ with
$q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$,

  5. $(5)$

singulets with $m=1$ of type $\beta$ such that $f=q_{1}q_{2}$ with
$q_{j}\equiv 2,5\,(\mathrm{mod}\,9)$ for $1\leq j\leq 2$.

The singulet $(1)$ is unique, but there exist infinitely many multiplets of
each shape $(2)$–$(5)$.

###### Proof.

We have $3\nmid\\#C_{k,3}^{(\sigma)}$ $\Longleftrightarrow$ $3\nmid h_{k}$ by
[16, Lem. 2, p. 7], and the equivalence $3\nmid h_{k}$ $\Longleftrightarrow$
$3\nmid h_{L}$ is due to the class number formula $h_{k}=\frac{Q}{3}\cdot
h_{L}^{2}$ [16, Lem. 1, p. 7], where $Q\in\\{1,3\\}$, and here necessarily
$Q=3$. The final equivalence follows from Table 1 in § 2.3.

Items (1) and (4) correspond to case (i) and (iii) in Honda’s paper [16, Thm.,
§ 1, p. 8]. However, we must split Honda’s case (ii) into items (2) and (3),
because the PF types of the singulets are different. Our item (5) unifies
Honda’s cases (iv) and (v), since the conductor $f$ is independent of the
exponents $e_{j}$ of the prime factors $q_{j}$ occurring in the radicand $d$.

The multiplicity $m(f)$ of each conductor is calculated by means of [27, Thm.
2.1, p. 833], which is exactly our Theorem 2.2, using the sequence
$(X_{k})_{k\geq-1}=(\frac{1}{2},0,1,1,3,\ldots)$:

  1. $(1)$

For the unique conductor $f=3^{2}$ of species 1a with $n=0$, we have
$m(f)=2^{n}=1$, a singulet with prime radicand $d=3$.

  2. $(2)$

For $f=q_{1}\equiv 8\,(\mathrm{mod}\,9)$ of species 2, we must take into
consideration that $u=1$, $v=0$, and we obtain $m(f)=2^{u}\cdot
X_{v-1}=2\cdot\frac{1}{2}=1$, a singulet with prime radicand $d=q_{1}$.

  3. $(3)$

For $f=3q_{1}$ of species 1b with $q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$, we
have $u=0$, $v=1$, $m(f)=2^{u}\cdot X_{v}=1\cdot 1=1$, a singulet with prime
radicand $d=q_{1}$.

  4. $(4)$

For $f=3^{2}q_{1}$ of species 1a with $n=1$, we get $m(f)=2^{n}=2$
(independently of $u$ and $v$), a doublet with two associated composite
radicands $d=3q_{1}$ and $d=3^{2}q_{1}$.

  5. $(5)$

For $f=q_{1}q_{2}$ of species 2 with $q_{1},q_{2}\equiv
2,5\,(\mathrm{mod}\,9)$, we have $u=0$, $v=2$, $m(f)=2^{u}\cdot X_{v-1}=1\cdot
1=1$, a singulet with composite radicand either $d=q_{1}q_{2}$ or
$d=q_{1}^{2}q_{2}$.

The principal factorization type, as a refinement of the Galois cohomology of
the unit group $E_{k}$, is a consequence of the estimates in Theorem 2.1.
Since $s=0$, we have $0\leq R\leq 0$ and type $\alpha$ is generally
impossible. For items $(1)$ and $(2)$, we have $t=1$ and thus $1\leq A\leq
1-0=1$, which discourages type $\beta$. For all other cases, there exists a
prime factor $q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$, and thus type $\gamma$ is
impossible, because $\zeta_{3}$ can be norm of a unit in $k$ only if the prime
factors of $f$ are $3$ or $\ell_{j}\equiv 1,8\,(\mathrm{mod}\,9)$. This is our
new proof of [3, Thm. 15.7, p. 236].

All claims on the infinitude of the various sets of conductors $f$ are
consequences of Dirichlet’s theorem on primes $q\in r+9\mathbb{Z}$ in
arithmetic progressions, here: invertible residue classes $r$ modulo $9$ with
$\gcd(r,9)=1$. ∎

######  Example 2.2.

Among the $827\,600$ pure cubic fields $L=\mathbb{Q}(\sqrt[3]{d})$ with
normalized radicands $d<10^{6}$, there are $73\,885$, that is $8.93\%$, whose
class number is not divisible by $3$. Table 4 shows the contribution (absolute
and relative frequency) of each item in Theorem 2.3 together with all
paradigms $d<100$. Due to the cut off at $d=10^{6}$, only $3\,519$ of the
doublets in item $(4)$ are complete, the other $6\,060$ pseudo-singulets have
companion radicands outside the range of investigations. Consequently, we have
$2\cdot 3\,519+6\,060=13\,098$.

Table 4: $73\,885$ pure cubic fields $L$ with $3\nmid h_{L}$ Item | $f$ | Type | $\\#$ | $\%$ | Paradigms for $d$
---|---|---|---|---|---
$(1)$ | $9$ | $\gamma$ | $1$ | $0.00$ | $3$
$(2)$ | $q\equiv 8\,(9)$ | $\gamma$ | $13\,099$ | $17.73$ | $17,53,71,89$
$(3)$ |  $3q$, $q\equiv 2,5\,(9)$ | $\beta$ | $26\,167$ | $35.42$ | $2,5,11,23,29,41,47,59,83$
$(4)$ |  $9q$, $q\equiv 2,5\,(9)$ | $(\beta,\beta)$ | $13\,098$ | $17.73$ | $6,12,15,33,45,69,87,99$
$(5)$ |  $q_{1}q_{2}$, $q_{j}\equiv 2,5\,(9)$ | $\beta$ | $21\,520$ | $29.13$ | $10,44,46,55,82$

###  2.5 Conductors divisible by a splitting prime

Now we come to ambiguous class groups $C_{k,3}^{(\sigma)}$ of $3$-rank one,
and we first give more details concerning the leading three lines of equation
(1) in our Theorem 1.1, where $d$ is divisible by a prime $p_{1}\equiv
1\,(\mathrm{mod}\,3)$ which splits in $k_{0}$.

######  Theorem 2.4.

Let the conductor of $k/k_{0}$ be $f=3^{e}\cdot\ell_{1}\cdots\ell_{n}$ with
$0\leq e\leq 2$, $n\geq 1$, and pairwise distinct primes $\ell_{j}\neq 3$ for
$1\leq j\leq n$. Briefly denote the multiplicity of $f$ by $m:=m(f)$. Assume
that $\ell_{j}\equiv 1\,(\mathrm{mod}\,3)$ for at least one $1\leq j\leq n$.
Then, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ $\Longleftrightarrow$ $L$
belongs to one of the following multiplets, where $\ell_{1}=p_{1}\equiv
1\,(\mathrm{mod}\,3)$ and $\ell_{2}=q_{2}\equiv-1\,(\mathrm{mod}\,3)$.

  1. $(1)$

singulets with $m=1$ of type $\alpha$ or $\gamma$ such that $f=p_{1}$ with
$p_{1}\equiv 1\,(\mathrm{mod}\,9)$,

  2. $(2)$

singulets with $m=1$ of type $\alpha$ or $\beta$ such that $f=3p_{1}$ with
$p_{1}\equiv 4,7\,(\mathrm{mod}\,9)$,

  3. $(3)$

doublets with $m=2$ of type $(\alpha,\alpha)$ or $(\beta,\beta)$ such that
$f=3^{2}p_{1}$ with $p_{1}\equiv 4,7\,(\mathrm{mod}\,9)$,

  4. $(4)$

singulets with $m=1$ of type $\alpha$ or $\beta$ such that $f=p_{1}q_{2}$ with
$p_{1}\equiv 4,7\,(\mathrm{mod}\,9)$ and $q_{2}\equiv 2,5\,(\mathrm{mod}\,9)$.

There exist infinitely many multiplets with conductors of all these shapes
$(1)$–$(4)$. For each of these conductors, the ambiguous $3$-class group
$C_{k,3}^{(\sigma)}\simeq(3)$ of the normal closure $k$ of $L$ is cyclic of
order $3$.

###### Proof.

Given the assumption that $\ell_{j}\equiv 1\,(\mathrm{mod}\,3)$ for at least
one $1\leq j\leq n$, the equivalence of the rank condition
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ to the four shapes of conductors
will be proved in §§ 3.2–3.4 and in the Cases (1)–(4) of § 3.6.

The multiplicity $m(f)$ of each conductor is calculated by means of [27, Thm.
2.1, p. 833], that is our Theorem 2.2, using the sequence
$(X_{k})_{k\geq-1}=(\frac{1}{2},0,1,1,3,\ldots)$:

  1. $(1)$

For $f=p_{1}\equiv 1\,(\mathrm{mod}\,9)$ of species 2, we must take into
consideration that $u=1$, $v=0$, and we obtain $m(f)=2^{u}\cdot
X_{v-1}=2\cdot\frac{1}{2}=1$, a singulet with prime radicand $d=p_{1}$.

  2. $(2)$

For $f=3p_{1}$ of species 1b with $p_{1}\equiv 4,7\,(\mathrm{mod}\,9)$, we
have $u=0$, $v=1$, $m(f)=2^{u}\cdot X_{v}=1\cdot 1=1$, a singulet with prime
radicand $d=p_{1}$.

  3. $(3)$

For $f=3^{2}p_{1}$ of species 1a with $n=1$, we get $m(f)=2^{n}=2$
(independently of $u$ and $v$), a doublet with two associated composite
radicands $d=3p_{1}$ and $d=3^{2}p_{1}$.

  4. $(4)$

For $f=p_{1}q_{2}$ of species 2 with $p_{1},-q_{2}\equiv
4,7\,(\mathrm{mod}\,9)$, we have $u=0$, $v=2$, $m(f)=2^{u}\cdot X_{v-1}=1\cdot
1=1$, a singulet with composite radicand either $d=p_{1}q_{2}$ or
$d=p_{1}^{2}q_{2}$.

The principal factorization type is a consequence of the estimates in Theorem
2.1. Since $s=1$, we have $0\leq R\leq 1$ and type $\alpha$ may generally
occur. For item $(1)$, we have $t=2$ and thus $1\leq A\leq 2-1=1$, which
discourages type $\beta$, but type $\gamma$ may occur. For items $(2)$–$(4)$,
there exists a prime factor $p_{1}\equiv 4,7\,(\mathrm{mod}\,9)$, and type
$\gamma$ is impossible, because $\zeta_{3}$ can be norm of a unit in $k$ only
if the prime factors of $f$ are $3$ or $\ell_{j}\equiv
1,8\,(\mathrm{mod}\,9)$.

All claims on the infinitude of the various sets of conductors $f$ are a
consequence of Dirichlet’s theorem on primes in arithmetic progressions. ∎

######  Corollary 2.4.1.

For each of the pure cubic fields $L$ belonging to one of the items $(2)$,
$(3)$, $(4)$ in Theorem 2.4, the following equivalences specify the
correlation between types, subfield unit indices $Q$, $3$-class group
structures of the normal closure $k$, and cubic residue symbols.

  1. 1.

$L$ is of type $\alpha$ $\Longleftrightarrow$ $Q=1$, and $\biggl{(}Q=1,\
C_{k,3}\simeq(3)\biggr{)}$ $\Longleftrightarrow$
$\left(\frac{c}{p_{1}}\right)_{3}\neq 1$,

  2. 2.

$L$ is of type $\beta$ $\Longleftrightarrow$ $Q=3$, and $\biggl{(}Q=3,\
C_{k,3}\simeq(3,3)\biggr{)}$ $\Longrightarrow$
$\biggl{[}\stackrel{{\scriptstyle
conjecture}}{{\Longleftrightarrow}}\biggr{]}$
$\left(\frac{c}{p_{1}}\right)_{3}=1$,

where $c=3$ in items $(2)$, $(3)$, and $c=q_{2}$ in item $(4)$. For each of
these conductors, the $3$-class group $C_{L,3}\simeq(3)$ of $L$ is cyclic of
order $3$. If $C_{k,3}\simeq(3,3)$ is elementary bicyclic, then the fields
$k^{\ast}=k\cdot L_{1}=k\cdot L_{1}^{\sigma}=k\cdot L_{1}^{\sigma^{2}}=K_{4}$
coincide with one of the four unramified cyclic cubic extensions
$K_{1},\ldots,K_{4}$ of $k$ within $k_{1}$, as illustrated in Figure 1 (valid
for $L$ of type $\beta$).

Figure 1: First possible location of the relative $3$-genus field
$k^{\ast}=(k/k_{0})^{\ast}$Degree$54$Hilbert of
bicubic$18$compositum$9$Hilbert of conjugate cubics$6$bicubic$3$conjugate
cubics$2$quadratic$1$baseScenario I: $k^{\ast}=L_{1}\cdot
k$$k_{1}$$k^{\ast}=L_{1}\cdot k=L_{1}^{\sigma}\cdot k=L_{1}^{\sigma^{2}}\cdot
k$$K_{3}$$K_{2}$$K_{1}$$k$$k_{0}$$L_{1}$$L_{1}^{\sigma},L_{1}^{\sigma^{2}}$$L$$L^{\sigma},L^{\sigma^{2}}$$\mathbb{Q}$

###### Proof.

The claims for item $(2)$ have been shown partially by Gerth [11, Case 3, pp.
474–475]. For all items $(2)$, $(3)$, $(4)$, they are due to Ismaili and El
Mesaoudi [20, Thm. 3.2, p. 104]. The statement concerning the $3$-genus field
$k^{\ast}$ was proved by Ismaili [18, Thm. 3.2, pp. 29–31]. We also emphasize
the connection with Table 2 in § 2.3. ∎

######  Corollary 2.4.2.

For each of the pure cubic fields $L$ belonging to item $(1)$ in Theorem 2.4,
the following equivalences specify the correlation between types, subfield
unit indices $Q$, and $3$-class group structures of $L$ and $k$.

  1. 1.

$L$ is of type $\alpha$ $\Longleftrightarrow$ $Q=1$ $\Longleftrightarrow$
$C_{L,3}\simeq(3^{w})$ and $C_{k,3}\simeq(3^{w},3^{w-1})$ for some $w\geq 1$,

  2. 2.

$L$ is of type $\gamma$ $\Longleftrightarrow$ $Q=3$ $\Longleftrightarrow$
$C_{L,3}\simeq(3^{w})$ and $C_{k,3}\simeq(3^{w},3^{w})$ for some $w\geq 2$.

Here, the $3$-class group of $k$ is either heterocyclic if $Q=1$ (in
particular, $(3^{w},3^{w-1})$ is to be interpreted as the cyclic type $(3)$
for $w=1$) or homocyclic if $Q=3$, but never of elementary bicyclic type
$(3,3)$. Also, $Q=1$ $\Longleftrightarrow$ $\mathfrak{P}\in\mathcal{P}_{k}$,
where $p_{1}\mathcal{O}_{k}=\mathfrak{P}^{3}(\mathfrak{P}^{\tau})^{3}$
($\mathfrak{P}$ a prime ideal of $k$).

###### Proof.

These statements have been proved by Gerth in [11, Formulas p. 474, and Case
4, pp. 475–476]. ∎

Table 5: $81\,894$ pure cubic fields $L$ with $s=1$, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ Item | $f$ | Type | $\\#$ | $\%$ | Paradigms for $d$
---|---|---|---|---|---
$(1)$ | $p_{1}\equiv 1\,(9)$ |  | $13\,063$ | $15.95$ |
|  | $\alpha$ | $11\,958$ | $91.54$ | $19,37,73$
|  | $\gamma$ | $1\,105$ | $8.46$ | $541,919,1279$
$(2)$ |  $3p_{1}$, $p_{1}\equiv 4,7\,(9)$ |  | $26\,168$ | $31.95$ |
|  | $\alpha$ | $17\,485$ | $66.82$ | $7,13,31,43,79,97$
|  | $\beta$ | $8\,683$ | $33.18$ | $61,67,103,151$
$(3)$ |  $9p_{1}$, $p_{1}\equiv 4,7\,(9)$ |  | $13\,048$ | $15.93$ |
|  | $(\alpha,\alpha)$ | $8\,709$ | $66.75$ | $21,39,63,93$
|  | $(\beta,\beta)$ | $4\,339$ | $33.25$ | $183,201,309,453$
$(4)$ |  $p_{1}q_{1}$, $p_{1}\equiv 4,7\,(9)$, $q_{1}\equiv 2,5\,(9)$ |  | $29\,615$ | $36.16$ |
|  | $\alpha$ | $19\,898$ | $67.19$ | $26,28,35$
|  | $\beta$ | $9\,717$ | $32.81$ | $62,172,287$

######  Example 2.3.

Among the $827\,600$ pure cubic fields $L=\mathbb{Q}(\sqrt[3]{d})$ with
normalized radicands $d<10^{6}$, there are $81\,894$, that is $9.90\%$, with
$s=1$ and $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$. Table 5 shows the
contribution (absolute and relative frequency) of each item in Theorem 2.4
together with all paradigms $d<100$ or even bigger. Here, we have to take into
account that different PF types are possible. Due to the cut off at
$d=10^{6}$, only $3\,514$ of the doublets in item $(3)$ are complete, the
other $6\,020$ pseudo-singulets have companion radicands outside the range of
investigations. Consequently, we have $2\cdot
3\,514+6\,020=13\,048=8\,709+4\,339$. Here, $2\,348$ doublets are of type
$(\alpha,\alpha)$ and $1\,166$ of type $(\beta,\beta)$, but inhomogeneous
doublet types $(\alpha,\beta)$ do not occur, which could be explained when the
implication for cubic residue symbols in Corollary 2.4.1 were an equivalence.
Among the pseudo-singulets, $4\,013$ are of type $\alpha$ and $2\,007$ of type
$\beta$. The exponent $w$ in Corollary 2.4.2 seems to be unbounded: in
addition to Table 5, we mention that for type $\alpha$, occurrences of $w=2$
set in with $d\in\\{199,271,487\\}$, $w=3$ with $d\in\\{3061,3583,4177\\}$,
and $w=4$ with $d\in\\{6733,8263\\}$, for type $\gamma$ we have $w=3$ for
$d=8389$.

###  2.6 Conductors divisible by non-split primes only

We continue the discussion of ambiguous class groups $C_{k,3}^{(\sigma)}$ of
$3$-rank one by giving more details about the trailing six lines of equation
(1) in our Theorem 1.1, where $d$ is only divisible by primes
$q_{j}\equiv-1\,(\mathrm{mod}\,3)$ which do not split in $k_{0}$.

######  Theorem 2.5.

Let the conductor of $k/k_{0}$ be $f=3^{e}\cdot\ell_{1}\cdots\ell_{n}$ with
$0\leq e\leq 2$, $n\geq 1$, and pairwise distinct primes $\ell_{j}\neq 3$ for
$1\leq j\leq n$. Denote the multiplicity of $f$ by $m:=m(f)$. Assume that
$\ell_{j}\equiv-1\,(\mathrm{mod}\,3)$ for all $1\leq j\leq n$. Then,
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ $\Longleftrightarrow$ $L$
belongs to one of the following multiplets, where
$\ell_{j}=q_{j}\equiv-1\,(\mathrm{mod}\,3)$ for $1\leq j\leq 3$.

  1. $(1)$

doublets with $m=2$ of type $(\beta^{x},\gamma^{y})$, $x+y=2$, such that
$f=3^{2}q_{1}$ with $q_{1}\equiv 8\,(\mathrm{mod}\,9)$,

  2. $(2)$

doublets with $m=2$ of type $(\beta^{x},\gamma^{y})$, $x+y=2$, such that
$f=q_{1}q_{2}$ with $q_{1},q_{2}\equiv 8\,(\mathrm{mod}\,9)$,

  3. $(3)$

singulets with $m=1$ of type $\beta$ such that $f=3q_{1}q_{2}$ with
$q_{1},q_{2}\equiv 2,5\,(\mathrm{mod}\,9)$,

  4. $(4)$

doublets with $m=2$ of type $(\beta,\beta)$ such that $f=3q_{1}q_{2}$ with
$q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$, $q_{2}\equiv 8\,(\mathrm{mod}\,9)$,

  5. $(5)$

quartets with $m=4$ of type $(\beta,\beta,\beta,\beta)$ such that
$f=3^{2}q_{1}q_{2}$ with $q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$,
$q_{2}\equiv-1\,(\mathrm{mod}\,3)$,

  6. $(6)$

singulets with $m=1$ of type $\beta$ such that $f=q_{1}q_{2}q_{3}$ with
$q_{1},q_{2},q_{3}\equiv 2,5\,(\mathrm{mod}\,9)$.

  7. $(7)$

doublets with $m=2$ of type $(\beta,\beta)$ such that $f=q_{1}q_{2}q_{3}$ with
$q_{1},q_{2}\equiv 2,5\,(\mathrm{mod}\,9)$, $q_{3}\equiv
8\,(\mathrm{mod}\,9)$,

There exist infinitely many multiplets with conductors of all these shapes
$(1)$–$(7)$. For each of these conductors, the ambiguous $3$-class group
$C_{k,3}^{(\sigma)}\simeq(3)$ of the normal closure $k$ of $L$ is cyclic of
order $3$.

###### Proof.

Given the assumption that $\ell_{j}\equiv-1\,(\mathrm{mod}\,3)$ for all $1\leq
j\leq n$, the equivalence of the rank condition
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ to the seven shapes of
conductors will be proved in §§ 3.1–3.5 and in the Cases (5)–(10) of § 3.6.

The multiplicity $m(f)$ of each conductor is calculated by means of [27, Thm.
2.1, p. 833] using the sequence
$(X_{k})_{k\geq-1}=(\frac{1}{2},0,1,1,3,\ldots)$:

  1. $(1)$

For $f=3^{2}q_{1}$ of species 1a with $n=1$, we get $m(f)=2^{n}=2$
(independently of $u$ and $v$), a doublet with two associated companion
radicands $d=3q_{1}$ and $d=3^{2}q_{1}$.

  2. $(2)$

For $f=q_{1}q_{2}$ with $q_{1},q_{2}\equiv 8\,(\mathrm{mod}\,9)$ of species 2,
we must take into consideration that $u=2$, $v=0$, and we obtain
$m(f)=2^{u}\cdot X_{v-1}=4\cdot\frac{1}{2}=2$, a doublet with two associated
companion radicands $d=q_{1}q_{2}$ and $d=q_{1}^{2}q_{2}$.

  3. $(3)$

For $f=3q_{1}q_{2}$ of species 1b with $q_{1},q_{2}\equiv
2,5\,(\mathrm{mod}\,9)$, we have $u=0$, $v=2$, $m(f)=2^{u}\cdot X_{v}=1\cdot
1=1$, a singulet with composite radicand $d=q_{1}^{e_{1}}q_{2}^{e_{2}}$
satisfying $d\not\equiv\pm 1\,(\mathrm{mod}\,9)$.

  4. $(4)$

For $f=3q_{1}q_{2}$ of species 1b with $q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$,
$q_{2}\equiv 8\,(\mathrm{mod}\,9)$ we have $u=1$, $v=1$, $m(f)=2^{u}\cdot
X_{v}=2\cdot 1=2$, a doublet with two associated companion radicands of the
shape $d=q_{1}^{e_{1}}q_{2}^{e_{2}}$ satisfying $d\not\equiv\pm
1\,(\mathrm{mod}\,9)$.

  5. $(5)$

For $f=3^{2}q_{1}q_{2}$ of species 1a with $n=2$, we get $m(f)=2^{n}=4$
(independently of $u$ and $v$), a quartet with four associated companion
radicands $d=3q_{1}q_{2}$, $d=3^{2}q_{1}q_{2}$, $d=3q_{1}^{2}q_{2}$,
$d=3q_{1}q_{2}^{2}$.

  6. $(6)$

For $f=q_{1}q_{2}q_{3}$ of species 2 with $q_{1},q_{2},q_{3}\equiv
2,5\,(\mathrm{mod}\,9)$, we have $u=0$, $v=3$, $m(f)=2^{u}\cdot X_{v-1}=1\cdot
1=1$, a singulet with composite radicand
$d=q_{1}^{e_{1}}q_{2}^{e_{2}}q_{3}^{e_{3}}$ satisfying $d\equiv\pm
1\,(\mathrm{mod}\,9)$.

  7. $(7)$

For $f=q_{1}q_{2}q_{3}$ of species 2 with $q_{1},q_{2}\equiv
2,5\,(\mathrm{mod}\,9)$, $q_{3}\equiv 8\,(\mathrm{mod}\,9)$ we have $u=1$,
$v=2$, $m(f)=2^{u}\cdot X_{v-1}=2\cdot 1=2$, a doublet with two associated
companion radicands of the shape $d=q_{1}^{e_{1}}q_{2}^{e_{2}}q_{3}^{e_{3}}$
satisfying $d\equiv\pm 1\,(\mathrm{mod}\,9)$.

The principal factorization type is a consequence of the estimates in Theorem
2.1. Since $s=0$, we have $0\leq R\leq 0$ and type $\alpha$ is generally
forbidden. For all cases, we have $t\geq 2$ and thus $1\leq A\leq 2-0=2$,
which enables type $\beta$. For items $(3)$–$(7)$, there exists a prime factor
$q_{1}\equiv 2,5\,(\mathrm{mod}\,9)$, and type $\gamma$ is impossible, because
$\zeta_{3}$ can be norm of a unit in $k$ only if the prime factors of $f$ are
$3$ or $\ell_{j}\equiv 1,8\,(\mathrm{mod}\,9)$. The latter condition is
satisfied by items $(1)$–$(2)$, whence type $\gamma$ may occur.

All claims on the infinitude of the various sets of conductors $f$ are
consequences of Dirichlet’s theorem on prime numbers arising from invertible
residue classes. ∎

######  Corollary 2.5.1.

For each of the pure cubic fields $L$ belonging to one of the items
$(1)$–$(7)$ in Theorem 2.5, the $3$-class group $C_{L,3}\simeq(3^{w})$ of $L$
is cyclic of order $3^{w}$. If the $3$-class group $C_{k,3}\simeq(3,3)$ of $k$
is elementary bicyclic, then the four distinct fields $k^{\ast}$, $k\cdot
L_{1}$, $k\cdot L_{1}^{\sigma}$, $k\cdot L_{1}^{\sigma^{2}}$ are the four
unramified cyclic cubic extensions of $k$ within $k_{1}$, as illustrated in
Figure 2.

###### Proof.

The statement concerning the $3$-genus field $k^{\ast}$ has been proved by
Ismaili [18, Thm. 3.2, pp. 29–31]. The connection with Table 3 in § 2.3 should
be emphasized. ∎

Figure 2: Second possible location of the relative $3$-genus field $k^{\ast}=(k/k_{0})^{\ast}$Degree$54$Hilbert of bicubic$18$composita$9$Hilbert of conjugate cubics$6$bicubic$3$conjugate cubics$2$quadratic$1$baseScenario II: $k^{\ast}\neq L_{1}\cdot k$, $k^{\ast}\neq k_{1}$$k_{1}$$L_{1}\cdot k$$L_{1}^{\sigma}\cdot k$$L_{1}^{\sigma^{2}}\cdot k$$k^{\ast}$$k$$k_{0}$$L_{1}$$L_{1}^{\sigma},L_{1}^{\sigma^{2}}$$L$$L^{\sigma},L^{\sigma^{2}}$$\mathbb{Q}$ Table 6: $107\,595$ pure cubic fields $L$ with $s=0$, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ Item | $f$ | Type | $\\#$ | $\%$ | Paradigms for $d$
---|---|---|---|---|---
$(1)$ |  $9q_{1}$, $q_{1}\equiv 8\,(9)$ |  | $6\,538$ | $6.08$ |
|  | $(\beta,\ast)$ | $4\,835$ | $73.95$ | $51,159,213$
|  | $(\gamma,\ast)$ | $1\,703$ | $26.05$ | $153,321,477$
$(2)$ |  $q_{1}q_{2}$, $q_{1},q_{2}\equiv 8\,(9)$ |  | $3\,007$ | $2.79$ |
|  | $(\beta,\ast)$ | $2\,259$ | $75.12$ | $901,1819,3043$
|  | $(\gamma,\ast)$ | $748$ | $24.88$ | $1207,1513,3763$
$(3)$ |  $3q_{1}q_{2}$, $q_{1},q_{2}\equiv 2,5\,(9)$ | $\beta$ | $21\,460$ | $19.95$ | $20,22,58,92,94$
$(4)$ |  $3q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$, $q_{2}\equiv 8\,(9)$ | $(\beta,\beta)$ | $27\,510$ | $25.57$ | $34,68,85,106$
$(5)$ |  $9q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$, $q_{2}\equiv 2\,(3)$ | $(\beta,\beta,\beta,\beta)$ | $34\,170$ | $31.76$ |
|  $9q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$, $q_{2}\equiv 2,5\,(9)$ |  | $20\,999$ | $61.45$ | $30,60,66,90$
|  $9q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$, $q_{2}\equiv 8\,(9)$ |  | $13\,171$ | $38.55$ | $102,204,255,306$
$(6)$ |  $q_{1}q_{2}q_{3}$, $q_{1},q_{2},q_{3}\equiv 2,5\,(9)$ | $\beta$ | $5\,249$ | $4.88$ | $460,550,638,820$
$(7)$ |  $q_{1}q_{2}q_{3}$, $q_{1},q_{2}\equiv 2,5\,(9)$, $q_{3}\equiv 8\,(9)$ | $(\beta,\beta)$ | $9\,661$ | $8.98$ | $170,530,710,748$

######  Example 2.4.

Among the $827\,600$ pure cubic fields $L=\mathbb{Q}(\sqrt[3]{d})$ with
normalized radicands $d<10^{6}$, there are $107\,595$, that is $13.00\%$, with
$s=0$ and $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$. Table 6 shows the
contribution (absolute and relative frequency) of each item in Theorem 2.5
together with all paradigms $d<100$ or even bigger. Again, we partially have
to take into account that different PF types are possible. Due to the cut off
at $d=10^{6}$, only $1\,758$ of the doublets in item $(1)$ are complete, the
other $3\,022$ pseudo-singulets have companion radicands outside the range of
investigations. Consequently, we have $2\cdot
1\,758+3\,022=6\,538=4\,835+1\,703$. Here, $800$ doublets are of type
$(\beta,\beta)$, $17$ of type $(\gamma,\gamma)$, and also $941$ inhomogeneous
doublet types $(\beta,\gamma)$ occur. Among the pseudo-singulets, $2\,294$ are
of type $\beta$ and $728$ of type $\gamma$.

##  3 Proof of the Main Theorem 1.1

In this section, we investigate the $3$-rank of the group of ambiguous classes
of $k/k_{0}$. More results on $3$-class groups of cubic fields can be found in
Gerth [8], [9], [10], [11], and Kobayashi [22], [23], [24]. For the prime
factorization in pure cubic fields $L=\mathbb{Q}(\sqrt[3]{d})$, we refer to
the papers by Markoff [26], Dedekind [7], and Barrucand/Cohn [2], [3]. For the
prime decomposition rules in the cyclotomic field $k_{0}$, we refer to
Ireland/Rosen [17, Prop. 1.4.2, p. 13, and Prop. 9.1.1–4, pp. 109–111]. In the
following lemmas, we introduce Gerth’s convention [9, § 5, pp. 91–92] for
characterizing a unique primary prime element among its six associates in the
ring $\mathcal{O}_{k_{0}}$, and we point out that Ireland/Rosen [17, Prop.
9.3.5, pp. 113–114] use a different convention.

######  Lemma 3.1.

Let $p$ be a prime number congruent to $4$ or $7$ mod $9$, and thus
$p=\pi_{1}\pi_{2}$ with two prime elements $\pi_{1}$ and
$\pi_{2}=\pi_{1}^{\tau}$ of $k_{0}$. If $\pi_{1}\equiv\pi_{2}\equiv
1(\mathrm{mod}\,3\mathcal{O}_{k_{0}})$, then $\pi_{1}\text{ and
}\pi_{2}\not\equiv 1(\mathrm{mod}\,\lambda^{3})\text{ in
}\mathcal{O}_{k_{0}}$, where $\operatorname{Gal}(k/L)=\langle\tau\rangle$ and
$\lambda=1-\zeta_{3}$.

###### Proof.

The prime element $\lambda=1-\zeta_{3}$ divides $3$ in $k_{0}$:
$\lambda^{2}=1-2\zeta_{3}+\zeta_{3}^{2}=1-2\zeta_{3}-1-\zeta_{3}=-3\zeta_{3}$,
because $1+\zeta_{3}+\zeta_{3}^{2}=0$. Since
$3=-\zeta_{3}^{-1}\lambda^{2}=-\zeta_{3}^{2}\lambda^{2}$, we have
$9=3^{2}=(-\zeta_{3}^{2}\lambda^{2})^{2}=\zeta_{3}\lambda^{4}$. Consequently,
$p\equiv 4$ or $7\,(\mathrm{mod}\,9)$ implies $\pi_{1}\pi_{2}=p\equiv 4$ or
$7(\mathrm{mod}\,\lambda^{3})$. Then $\pi_{1}$ or $\pi_{2}\not\equiv
1(\mathrm{mod}\,\lambda^{3})$, because otherwise $\pi_{1}\equiv\pi_{2}\equiv
1(\mathrm{mod}\,\lambda^{3})$, $p=\pi_{1}\pi_{2}\equiv\pi_{2}\equiv
1(\mathrm{mod}\,\lambda^{3})$, and we obtain the contradiction $1\equiv 4$ or
$7(\mathrm{mod}\,\lambda^{3})$. ∎

######  Lemma 3.2.

Suppose that $q$ is a prime number congruent to $2$ or $5$ mod $9$, and thus
$\pi=-q$ congruent to $4$ or $7$ mod $9$ is a prime element of $k_{0}$.
If $\pi\equiv 1(\mathrm{mod}\,3\mathcal{O}_{k_{0}})$, then $\pi\not\equiv
1(\mathrm{mod}\,\lambda^{3})\text{ in }\mathcal{O}_{k_{0}}$, where
$\lambda=1-\zeta_{3}$.

###### Proof.

Similarly as in the proof of Lemma 3.1, the assumption $\pi\equiv
1(\mathrm{mod}\,\lambda^{3})$ would yield the contradiction $1\equiv 4$ or
$7(\mathrm{mod}\,\lambda^{3})$. ∎

The invariant $q^{\ast}$, defined in the summary of our notation at the end of
the Introduction, is related to the norm index $(E_{k_{0}}:(E_{k_{0}}\cap
N_{k/k_{0}}(k^{\times})))$ by the relation

$(E_{k_{0}}:(E_{k_{0}}\cap
N_{k/k_{0}}(k^{\times})))=\frac{(E_{k_{0}}:E_{k_{0}}^{3})}{((E_{k_{0}}\cap
N_{k/k_{0}}(k^{\times})):E_{k_{0}}^{3})}=3^{1-q^{\ast}}\text{ i.e.
}3^{q^{\ast}}=((E_{k_{0}}\cap N_{k/k_{0}}(k^{\times})):E_{k_{0}}^{3}).$

######  Lemma 3.3.

The invariant $q^{\ast}$ of a cubic Kummer extension
$K=k_{0}(\sqrt[3]{\vartheta})$ of $k_{0}$ with radicand
$\vartheta=\lambda^{e_{\lambda}}\zeta_{3}^{e_{\zeta}}\pi_{1}^{e_{1}}\cdots\pi_{g}^{e_{g}}$,
where $0\leq e_{\lambda},e_{\zeta}\leq 2$, $g\geq 0$, $\pi_{i}\equiv
1\,(\mathrm{mod}\,3\mathcal{O}_{k_{0}})$ are prime elements of $k_{0}$, and
$1\leq e_{i}\leq 2$, for $1\leq i\leq g$, is given by

$\displaystyle q^{\ast}=1$ $\displaystyle\Longleftrightarrow(\exists\,Z\in
K)\,N_{K/k_{0}}(Z)=\zeta_{3}\Longleftrightarrow(\forall\,1\leq i\leq
g)\,\pi_{i}\equiv 1\,(\mathrm{mod}\,\lambda^{3})\text{ in
}\mathcal{O}_{k_{0}},$ (10) $\displaystyle q^{\ast}=0$
$\displaystyle\Longleftrightarrow(\forall\,Z\in
K)\,N_{K/k_{0}}(Z)\neq\zeta_{3}\Longleftrightarrow(\exists\,1\leq i\leq
g)\,\pi_{i}\equiv 4,7\,(\mathrm{mod}\,\lambda^{3})\text{ in
}\mathcal{O}_{k_{0}}.$

###### Proof.

This lemma is due to Gerth [9, § 5, p. 92]. ∎

Now, we start our proof of Theorem 1.1: According to equation $(3.2)$ of [10,
p. 55], the radicand $d$ of $k=k_{0}(\sqrt[3]{d})$ can be written in the form

$\displaystyle d$ $\displaystyle=$ $\displaystyle
3^{e}.p_{1}^{e_{1}}......p_{v}^{e_{v}}p_{v+1}^{e_{v+1}}....p_{w}^{e_{w}}.q_{1}^{f_{1}}...q_{I}^{f_{I}}q_{I+1}^{f_{I+1}}...q_{J}^{f_{J}},$
(11)

where $p_{i}$ and $q_{i}$ are positive rational prime numbers such that

$\left\\{\begin{array}[]{l l l l l l l}p_{i}\equiv
1\,(\mathrm{mod}\,9),&\quad\text{ for }\quad 1\leq i\leq v,\\\ p_{i}\equiv
4\text{ or }7\,(\mathrm{mod}\,9),&\quad\text{ for }\quad v+1\leq i\leq w,\\\
q_{i}\equiv-1\,(\mathrm{mod}\,9),&\quad\text{ for }\quad 1\leq i\leq I,\\\
q_{i}\equiv 2\text{ or }5\,(\mathrm{mod}\,9),&\quad\text{ for }\quad I+1\leq
i\leq J,\\\ e_{i}=1\text{ or }2,&\quad\text{ for }\quad 1\leq i\leq w,\\\
f_{i}=1\text{ or }2,&\quad\text{ for }\quad 1\leq i\leq J,\\\ e=0,1\text{ or
}2.\end{array}\right.$

Let $C_{k,3}^{(\sigma)}=\\{\mathcal{A}\in
C_{k,3}\mid\mathcal{A}^{\sigma}=\mathcal{A}\\}$ be the $3$-group of ambiguous
ideal classes of $k/k_{0}$, where
$\operatorname{Gal}(k/k_{0})=\langle\sigma\rangle$ and $C_{k,3}$ is the
$3$-class group of $k$. According to [9, § 5, p. 92], the $3$-rank of
$C_{k,3}^{(\sigma)}$ is specified as follows:

$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=t-2+q^{\ast},$

where $t$ is the number of prime ideals of $k_{0}$ ramified in $k$, and
$q^{\ast}$ is the invariant in Lemma 3.3.

If we assume $\operatorname{rank}(C_{k,3}^{(\sigma)})=1$, $t+q^{\ast}=3$,
according to the main goal of this paper, then we have the following cases,
according to Lemma $3.1$ of [10, p. 55]:

  * •

Case 1: $2w+J=1$,

  * •

Case 2: $2w+J=2$,

  * •

Case 3: $2w+J=3$,

where $w$ and $J$ are the positive integers defined in equation (11). We shall
successively treat these cases, distinguishing subcases in dependence on $w$
and $J$, and necessarily also obtaining some side results concerning the
situation $\operatorname{rank}(C_{k,3}^{(\sigma)})=0$, $t+q^{\ast}=2$, studied
by Honda [16] and refined in our section 2.4, and the situation
$\operatorname{rank}(C_{k,3}^{(\sigma)})=2$, $t+q^{\ast}=4$, which will be
investigated in our forthcoming paper [1].

###  3.1 Radicands divisible by a single non-split prime

In the Case 1, $2w+J=1$, we must have $w=0$ and $J=1$. Then
$d=3^{e}q^{f_{1}}$, where $q$ is a prime number such that
$q\equiv-1(\mathrm{mod}\,3)$, $e\in\\{0,1,2\\}$ and $f_{1}\in\\{1,2\\}$.

####  3.1.1 Species 2

If $d\equiv\pm 1\,(\mathrm{mod}\,9)$, then $e=0$ and
$q\equiv-1\,(\mathrm{mod}\,9)$. Since $q\equiv-1\,(\mathrm{mod}\,3)$, $q$
remains inert in $k_{0}$. As $d\equiv\pm 1(\mathrm{mod}\,9)$, $3$ is
decomposed in $L$ and $\lambda$ is not ramified in $k/k_{0}$. We get $t=1$.
Put $x=\pi^{f_{1}}$, where $\pi=-q$ is a prime element of $k_{0}$, then
$k=k_{0}(\sqrt[3]{x})$, because $d=q^{f_{1}}$. In addition, since
$q\equiv-1\,(\mathrm{mod}\,9)$, then by Lemma 3.2, $\pi$ is congruent to
$1(\mathrm{mod}\,\lambda^{3})$, and according to Lemma 3.3, $\zeta_{3}$ is
norm of an element of $k\setminus\\{0\\}$ and $q^{\ast}=1$. Thus
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=0$, which is item $(2)$ of Theorem
2.3.

####  3.1.2 Species 1

If $d\not\equiv\pm 1\,(\mathrm{mod}\,9)$, then

  * •

either $q\equiv-1\,(\mathrm{mod}\,9)$, then $d=3^{e}q^{f_{1}}\equiv\pm 3^{e}\
(\mathrm{mod}\,9)$, so $e\neq 0$, which is item $(1)$ of Theorem 2.5,

  * •

or $q\equiv 2$ or $5\,(\mathrm{mod}\,9)$, whence $q$ remains inert in $k_{0}$.
Moreover, since $d\not\equiv\pm 1\ (\mathrm{mod}\,9)$, $3$ is totally ramified
in $L$, so $\lambda$ is ramified in $k/k_{0}$. Thus, $t=2.$ We have
$3=-\zeta_{3}^{2}\lambda^{2}$, and $k=k_{0}(\sqrt[3]{x})$ with
$x=\zeta_{3}^{2}\lambda^{2}\pi^{f_{1}}$, where $\pi=-q$ is a prime element of
$k_{0}$. Since $q\equiv 2$ or $5\,(\mathrm{mod}\,9)$, so by Lemma 3.2, the
prime $\pi$ is not congruent to $1(\mathrm{mod}\,\lambda^{3})$, and according
to Lemma 3.3, $\zeta_{3}$ is not norm of an element of $k\setminus\\{0\\}$. So
$q^{\ast}=0$, and we conclude that
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=0$, which gives items $(3),(4)$ of
Theorem 2.3.

We conclude that $\operatorname{rank}(C_{k,3}^{(\sigma)})=1$ at most for
$d=3^{e}q^{f_{1}}\not\equiv\pm 1(\mathrm{mod}\,9)$, with
$q\equiv-1\,(\mathrm{mod}\,9)$, and $e,f_{1}\in\\{1,2\\}$. This is item $(1)$
of Theorem 2.5.

In the Case 2, $2w+J=2$, we either have $w=1$, $J=0$, treated in § 3.2, or
$w=0$, $J=2$, treated in § 3.3.

###  3.2 Radicands divisible by a single split prime

If $w=1$ and $J=0$, then $d=3^{e}p^{e_{1}}$, where $p$ is a prime number such
that $p\equiv 1(\mathrm{mod}\,3)$, $e\in\\{0,1,2\\}$ and $e_{1}\in\\{1,2\\}$.

####  3.2.1 Species 2

If $d\equiv\pm 1\,(\mathrm{mod}\,9)$, then $p\equiv 1\pmod{9}$ and $e=0$. So
the radicand $d$ can be written in the form $d=p^{e_{1}}$, where $p\equiv
1\,(\mathrm{mod}\,9)$ and $e_{1}\in\\{1,2\\}$. This is item $(1)$ of Theorem
2.4.

####  3.2.2 Species 1

If $d\not\equiv\pm 1\pmod{9}$, we have two possibilities: $e=0$ and
$p\not\equiv 1\ (\mathrm{mod}\,\ 9)$, or $e=1$ or $2$. Then the possible forms
of $d$ are:

$d=\left\\{\begin{array}[]{l l l}p^{e_{1}}&\quad\text{ with }p\equiv 4\text{
or }7\,(\mathrm{mod}\,9),\\\ 3^{e}p^{e_{1}}\not\equiv\pm
1\,(\mathrm{mod}\,9)&\quad\text{ with }p\equiv 1\pmod{3},\\\
\end{array}\right.$

where $e,e_{1}\in\\{1,2\\}$. The former is item $(2)$ of Theorem 2.4, the
latter must be examined further.

Now, assume that $d=3^{e}p^{e_{1}}\not\equiv\pm 1\pmod{9}$, with $p\equiv
1\pmod{3}$. As $\mathbb{Q}(\sqrt[3]{ab^{2}})=\mathbb{Q}(\sqrt[3]{a^{2}b})$ for
square-free coprime $a,b\in\mathbb{N}$, we can choose $e_{1}=1$, i.e.
$d=3^{e}p$ with $e\in\\{1,2\\}$. The fact that $p\equiv 1\pmod{3}$ implies
that $p=\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$ are prime elements of
$k_{0}$ such that $\pi_{1}^{\tau}=\pi_{2}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\
3\mathcal{O}_{k_{0}})$, furthermore the prime $p$ is totally ramified in $L$.
Thus $\pi_{1}$ and $\pi_{2}$ are totally ramified in $k$ and we have
$\pi_{1}\mathcal{O}_{k}=\mathcal{P}_{1}^{3}$ and
$\pi_{2}\mathcal{O}_{k}=\mathcal{P}_{2}^{3}$ where
$\mathcal{P}_{1},\mathcal{P}_{2}$ are two prime ideals of $k$. Since
$d\not\equiv\pm 1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\ 9)$,
$3$ is totally ramified in $L$, so $\lambda$ is ramified in $k/k_{0}$. Hence,
the number of prime ideals of $k_{0}$ which are ramified in $k/k_{0}$ is
$t=3.$
Since $3=-\zeta_{3}^{2}\lambda^{2}$, we have $k=k_{0}(\sqrt[3]{x})$, where
$x=\zeta_{3}^{2}\lambda^{2}\pi_{1}\pi_{2}$. If $p\equiv 1\,(\mathrm{mod}\,9)$,
the primes $\pi_{1}$ and $\pi_{2}$ are congruent to
$1(\mathrm{mod}\,\lambda^{3})$, and according to Lemma 3.3, we have
$q^{\ast}=1$, i.e. $\zeta_{3}$ is the norm of an element of
$k\setminus\\{0\\}$. Thus $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=2$, which
we remember for [1]. For $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ we
necessarily have $p\equiv 4$ or $7\,(\mathrm{mod}\,9)$, which yields item
$(3)$ of Theorem 2.4.

Hence, the forms of the integer $d$ with
$\operatorname{rank}(C_{k,3}^{(\sigma)})=1$ are:

$d=\left\\{\begin{array}[]{l l l}p^{e_{1}}&\quad\text{ with }p\equiv
1\,(\mathrm{mod}\,3),\\\ 3^{e}p^{e_{1}}\not\equiv\pm 1\pmod{9}&\quad\text{
with }p\equiv 4\text{ or }7\,(\mathrm{mod}\,9).\\\ \end{array}\right.$

where $e,e_{1}\in\\{1,2\\}$. These are the items $(1),(2),(3)$ of Theorem 2.4.

###  3.3 Radicands divisible by two non-split primes

If $w=0$ and $J=2$, then $d=3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}$, with
$q_{1}\equiv q_{2}\equiv 2(\mathrm{mod}\,3)$, $e\in\\{0,1,2\\}$ and
$f_{1},f_{2}\in\\{1,2\\}$.

####  3.3.1 Species 2

If $d\equiv\pm 1\,(\mathrm{mod}\,9)$, we have
$d=3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}\equiv\pm 3^{e}\,\text{ or }\,\pm
3^{e}\times 2\,\text{ or }\,\pm 3^{e}\times 5\,(\mathrm{mod}\,9)$ whence
$e=0$. If $q_{1},q_{2}\equiv 2,5\,(\mathrm{mod}\,9)$, then $t=2$ and
$q^{\ast}=0$, which is item $(5)$ of Theorem 2.3 with
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=0$. Consequently, we must have
$q_{1}\equiv q_{2}\equiv-1\,(\mathrm{mod}\,9)$ and $q^{\ast}=1$ for rank $1$.
So the radicand $d$ has the form $d=q_{1}^{f_{1}}q_{2}^{f_{2}}$ with
$q_{1}\equiv q_{2}\equiv-1\,(\mathrm{mod}\,9)$ and $f_{1},f_{2}\in\\{1,2\\}$,
i.e. item $(2)$ of Theorem 2.5.

####  3.3.2 Species 1

If $d\not\equiv\pm 1\,(\mathrm{mod}\,9)$, we either have $e\neq 0$ or ($e=0$
and there exist $i\in\\{1,2\\}$ such that
$q_{i}\not\equiv-1\,(\mathrm{mod}\,9)$). The latter case yields items
$(3),(4)$ of Theorem 2.5.
Assume that $e\neq 0$, then $d=3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}$ with
$q_{1}\equiv q_{2}\equiv 2\ (\mathrm{mod}\,3)$, $e,f_{1},$ and
$f_{2}\in\\{1,2\\}$. For each $i\in\\{1,2\\}$, $q_{i}$ is inert in $k_{0}$,
and $q_{i}$ is ramified in $L$. As $d\not\equiv\pm 1\pmod{9}$, $3$ is ramified
in $L$, so $\lambda$ is ramified in $k/k_{0}$. Thus $t=3$. The fact that
$3=-\zeta_{3}^{2}\lambda^{2}$ implies that $k=k_{0}(\sqrt[3]{x})$ with
$x=\zeta_{3}^{2}\lambda^{2}\pi_{1}^{f_{1}}\pi_{2}^{f_{2}}$, where
$-q_{i}=\pi_{i}$ is a prime element of $k_{0}$, for $i\in\\{1,2\\}$.
If $q_{1}\equiv q_{2}\equiv-1\,(\mathrm{mod}\,9)$, then all primes $\pi_{1}$,
$\pi_{2}$ are congruent to $1\leavevmode\nobreak\
(\mathrm{mod}\,\leavevmode\nobreak\ \lambda^{3})$, and according to Lemma 3.3,
$\zeta_{3}$ is norm of an element of $k\setminus\\{0\\}$ and $q^{\ast}=1$. We
conclude that $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=2$, which we keep in
mind for [1]. Thus, for rank $1$, there exists $i\in\\{1,2\\}$ such that
$q_{i}\not\equiv-1\,(\mathrm{mod}\,9)$, which is item $(5)$ of Theorem 2.5.

Hence, the form of $d$ for $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ is
$d=3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}\not\equiv\pm 1\,(\mathrm{mod}\,9)$, with
$e\in\\{0,1,2\\}$, $f_{1},f_{2}\in\\{1,2\\}$, and there exists $i\in\\{1,2\\}$
such that $q_{i}\not\equiv-1\,(\mathrm{mod}\,9)$. These are the items
$(3),(4),(5)$ of Theorem 2.5.

In the Case 3, $2w+J=3$, we either have $w=1$, $J=1$, treated in § 3.4, or
$w=0$, $J=3$, treated in § 3.5.

###  3.4 Radicands divisible by a split prime and a non-split prime

If $w=1$ and $J=1$, then $d=3^{e}p^{e_{1}}q^{f_{1}}$, where $p$ and $q$ are
prime numbers such that $p\equiv 1(\mathrm{mod}\,3)$ and $q\equiv
2(\mathrm{mod}\,3)$, $e\in\\{0,1,2\\}$ and $e_{1},f_{1}\in\\{1,2\\}$.

####  3.4.1 Species 2

If $d\equiv\pm 1\pmod{9}$, then necessarily $e=0$, and we distinguish the
following cases:

  * 1)

If $p\equiv 4\,\ \text{or}\ \,7\pmod{9}$ and $q\equiv-1(\mathrm{mod}\,9)$:
then we have $d=p^{e_{1}}q^{f_{1}}\equiv\pm 4\,\text{ or }\,\pm
7(\mathrm{mod}\,\ 9)$ and $d\not\equiv\pm 1\pmod{9}$ cannot be of species 2.

  * 2)

If $p\equiv-q\equiv 4\,\ \text{or}\,\ 7\pmod{9}$:
we have $p^{e_{1}}\equiv\pm 4\,\text{or}\,\pm 7(\mathrm{mod}\,9)$ and
$q^{f_{1}}\equiv\pm 2\,\text{or}\,\pm 5\pmod{9}$, so
$p^{e_{1}}q^{f_{1}}\equiv\pm 1\,\text{ or }\,\pm 5\,\text{ or }\,\pm 2\
(\mathrm{mod}\,9)$, and the radicand $d=p^{e_{1}}q^{f_{1}}\equiv\pm 1\pmod{9}$
belongs to item $(4)$ of Theorem 2.4.

  * 3)

If $p\equiv-q\equiv 1\,(\mathrm{mod}\,9)$ :
then $d=p^{e_{1}}q^{f_{1}}$. Since $d\equiv\pm 1\pmod{9}$, $3$ is not ramified
in $L$, so $\lambda$ is not ramified in $k/k_{0}$. Since $p\equiv 1\pmod{3}$,
$p=\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$ are prime elements of
$k_{0}$ such that $\pi_{1}^{\tau}=\pi_{2}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\
3\mathcal{O}_{k_{0}})$. The prime $p$ is totally ramified in $L$, so $\pi_{1}$
and $\pi_{2}$ are totally ramified in $k$. Since $q\equiv-1\pmod{3}$, $q$
remains inert in $k_{0}$. Thus the prime ideals ramified in $k/k_{0}$ are
those generated by $\pi_{1},\pi_{2}$ and $q$, whence $t=3$.
The fact that $p\equiv-q\equiv 1\,(\mathrm{mod}\,9)$ implies that
$\pi_{1}\equiv\pi_{2}\equiv\pi\equiv 1\leavevmode\nobreak\
(\mathrm{mod}\,\leavevmode\nobreak\ \lambda^{3})$, and $-q=\pi$ is a prime
element of $k_{0}$. Put $x=\pi_{1}^{e_{1}}\pi_{2}^{e_{1}}\pi^{f_{1}}$, then
$k=k_{0}(\sqrt[3]{x})$. Since all primes $\pi$, $\pi_{1}$ and $\pi_{2}$ are
congruent to $1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\
\lambda^{3})$, $\zeta_{3}$ is the norm of an element of $k\setminus\\{0\\}$
and $q^{\ast}=1$, according to Lemma 3.3. We conclude that
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=2$, which we remember for [1].

  * 4)

If $p\equiv 1\pmod{9}$ and $q\equiv 2\,\ \text{or}\,5\ (\mathrm{mod}\,9)$:
then we have $d=p^{e_{1}}q^{f_{1}}\equiv\pm 2\,\text{ or }\,\pm
5(\mathrm{mod}\,9)$ and $d\not\equiv\pm 1\,(\mathrm{mod}\,9)$ cannot be of
species 2.

####  3.4.2 Species 1

Let $d\not\equiv\pm 1\pmod{9}$.
On the one hand, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=t-2+q^{*}$, and
our desired fact $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ implies $t\leq
3.$
On the other hand, we have $d=3^{e}p^{e_{1}}q^{f_{1}}$, with $p\equiv
1\pmod{3}$ and $q\equiv 2\,(\mathrm{mod}\,3)$. We calculate the number $t$ of
prime ideals which ramify in $k/k_{0}$. Since $p\equiv 1\ (\mathrm{mod}\,\
3$), $p=\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$ are two prime elements
of $k_{0}$ such that $\pi_{2}=\pi_{1}^{\tau}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\
3\mathcal{O}_{k_{0}})$. The prime $p$ is ramified in $L$, whence $\pi_{1}$ and
$\pi_{2}$ are ramified in $k$. $q$ remains inert in $k_{0}$, and $q$ is
ramified in $L$. Since $d\not\equiv\pm 1\pmod{9}$, $3$ is ramified in $L$, and
we have $3\mathcal{O}_{k_{0}}=(\lambda)^{2}$ where $\lambda=1-\zeta_{3}.$
Hence, $t=4$ and $\operatorname{rank}\,(C_{k,3}^{(\sigma)})\geq 2$, which we
note for examination in [1].

###  3.5 Radicands divisible by three non-split primes

If $w=0$ and $J=3$, then $d=3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}$,
where $q_{i}$ is a prime number such that $q_{i}\equiv 2\pmod{3}$,
$e\in\\{0,1,2\\}$ and $f_{i}\in\\{1,2\\}$ for each $i\in\\{1,2,3\\}$.

####  3.5.1 Species 2

If $d\equiv\pm 1\pmod{9}$, then generally $e=0$, and we distinguish the
following cases:

  * 1)

If $q_{1}\equiv 2\,\ \text{or}\ \,5\pmod{9}$ and $q_{2}\equiv q_{3}\equiv-1\
(\mathrm{mod}\,\ 9)$:
we get $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 2\,\text{ or }\,\pm
5(\mathrm{mod}\,\ 9)$, and $d\not\equiv\pm 1\pmod{9}$ cannot be of species 2.

  * 2)

If $q_{1}\equiv q_{2}\equiv 2\,\ \text{or}\ \,5\pmod{9}$ and
$q_{3}\equiv-1\pmod{9}$:
we have $q_{1}^{f_{1}}\equiv q_{2}^{f_{2}}\equiv\pm 2\,\text{ or }\,\pm
5(\mathrm{mod}\,9)$, so $q_{1}^{f_{1}}q_{2}^{f_{2}}\equiv\pm 1\,\text{or}\,\pm
2\,\text{ or }\,\pm 5(\mathrm{mod}\,9)$, and the possible form of the radicand
$d$ is $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 1\pmod{9}$, i.e.
item $(7)$ of Theorem 2.5.

  * 3)

If $q_{1}\equiv q_{2}\equiv q_{3}\equiv-1\pmod{9}$:
then we have $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 1\pmod{9}$.
For each $i\in\\{1,2,3\\}$, $q_{i}$ remains inert in $k_{0}$, and $q_{i}$ is
ramified in $L$. As $d\equiv\pm 1\pmod{9}$, $3$ is decomposed in $L$, so
$\lambda$ is not ramified in $k/k_{0}$. Thus $t=3$. The fact that
$q_{i}\equiv-1\,(\mathrm{mod}\,9)$ implies that $\pi_{i}\equiv
1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\
3\mathcal{O}_{k_{0}})$, where $-q_{i}=\pi_{i}$ is a prime element of $k_{0}$.
Put $x=\pi_{1}^{f_{1}}\pi_{2}^{f_{2}}\pi_{3}^{f_{3}}$, then
$k=k_{0}(\sqrt[3]{x})$. Since all primes $\pi_{1}$, $\pi_{2}$ and $\pi_{3}$
are congruent to $1\leavevmode\nobreak\ (\mathrm{mod}\,\leavevmode\nobreak\
\lambda^{3})$, then according to Lemma 3.3, $\zeta_{3}$ is a norm of an
element of $k\setminus\\{0\\}$ and $q^{\ast}=1$. We conclude that
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=2$, which we note for [1].

  * 4)

If $q_{1}\equiv q_{2}\equiv q_{3}\equiv 2\,\text{or}\,5\pmod{9}$:
then we have $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 1\,\text{ or
}\,\pm 2\,\text{or}\,\pm 5\pmod{9}$, and the possible form of the radicand $d$
is $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm 1\pmod{9}$, which is
item $(6)$ of Theorem 2.5.

####  3.5.2 Species 1

If $d\not\equiv\pm 1\pmod{9}$, we should have $t\leq 3$ for
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$, but we reason similar as in the
case $w=1$, $J=1$, and we get $t=4$,
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})\geq 2$, which we remember for
examination in [1].

Finally, all forms of the integer $d$ for which
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$ can be summarized as:

$d=\left\\{\begin{array}[]{l l l l l l l l l}p_{1}^{e_{1}}&\quad\text{where \
$p_{1}\equiv 1\,(\mathrm{mod}\,3)$,}\\\ 3^{e}p_{1}^{e_{1}}&\quad\text{where \
$p_{1}\equiv 4$ or $7\,(\mathrm{mod}\,9)$,}\\\
p_{1}^{e_{1}}q_{1}^{f_{1}}\equiv\pm 1\,(\mathrm{mod}\,9)&\quad\text{where \
$p_{1},-q_{1}\equiv 4$ or $7\,(\mathrm{mod}\,9)$,}\\\
3^{e}q_{1}^{f_{1}}&\quad\text{where \ $q_{1}\equiv-1\,(\mathrm{mod}\,9)$,}\\\
q_{1}^{f_{1}}q_{2}^{f_{2}}&\quad\text{where \ $q_{1}\equiv
q_{2}\equiv-1\,(\mathrm{mod}\,9)$,}\\\ q_{1}^{f_{1}}q_{2}^{f_{2}}\not\equiv\pm
1\,(\mathrm{mod}\,9)&\quad\text{where \ $q_{i}\not\equiv-1\,(\mathrm{mod}\,9)$
for some $i\in\\{1,2\\}$,}\\\ 3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}&\quad\text{where
\ $q_{i}\not\equiv-1\,(\mathrm{mod}\,9)$ for some $i\in\\{1,2\\}$,}\\\
q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm
1\,(\mathrm{mod}\,9)&\quad\text{where \ $q_{1},q_{2}\equiv 2$ or
$5\,(\mathrm{mod}\,9)$ and $q_{3}\equiv-1\,(\mathrm{mod}\,9)$,}\\\
q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm
1\,(\mathrm{mod}\,9)&\quad\text{where \ $q_{1},q_{2},q_{3}\equiv 2$ or
$5\,(\mathrm{mod}\,9)$,}\\\ \end{array}\right.$

and where $e,e_{1},f_{1},f_{2}$ and $f_{3}$ are positive integers equal to $1$
or $2$.

###  3.6 Proof of rank one for each case

Conversely, we prove $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$, if the
radicand $d$ takes one of the forms in Theorem 1.1.

####  3.6.1 Case where $d=p^{e_{1}}$, with $p\equiv 1\pmod{9}$

The fact that $d\equiv 1\pmod{9}$ implies that the pure cubic field $L$ is of
Dedekind’s species 2 (see [3, §13]), so $3$ decomposes in $L$, and then
$\lambda$ decomposes in $k/k_{0}$. Moreover, as $p\equiv 1\pmod{3}$, we have
$p=\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$ are primes of $k_{0}$ such
that $\pi_{1}^{\tau}=\pi_{2}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\ 3\mathcal{O}_{k_{0}})$. It
follows that $p$ is totally ramified in $L$, and therefore $\pi_{1}$ and
$\pi_{2}$ are totally ramified in $k$. Thus, $t=2$. Taking into account that
$p\equiv 1\pmod{9}$, the Hilbert cubic symbol:

$\left(\frac{\zeta_{3},p}{\pi_{1}}\right)_{3}=\left(\frac{\zeta_{3},p}{\pi_{2}}\right)_{3}=1.$

It follows by Lemma 3.3 that $\zeta_{3}$ is a norm of an element of
$k\setminus\\{0\\}$, so $q^{\ast}=1$. Hence
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1.$

####  3.6.2 Case where $d=p^{e_{1}}$, with $p\equiv 4$ or $7\pmod{9}$

Since $d\not\equiv\pm 1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\ 9)$,
$3$ is totally ramified in $L$, and then $\lambda$ is ramified in $k/k_{0}$.
In the same manner of case (1), the congruence $p\equiv 1\pmod{3}$ implies
that $\pi_{1}$ and $\pi_{2}$ are totally ramified in $k$, where $\pi_{i}$ for
$i=1$ or $2$ are defined as above. Then $t=3.$ As $p\equiv 4$ or $7\pmod{9}$,
the Hilbert cubic symbol

$\left(\frac{\zeta_{3},p}{\pi_{i}}\right)_{3}\neq 1,$

for $i=1$ or $2$. It follows by Lemma 3.3 that $\zeta_{3}$ is not a norm of an
element of $k\setminus\\{0\\}$ and $q^{\ast}=0$. Thus
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1.$

####  3.6.3 Case where $d=3^{e}p^{e_{1}}\not\equiv\pm 1\pmod{9}$, with
$p\equiv 4$ or $7\pmod{9}$

Taking into account that
$\mathbb{Q}(\sqrt[3]{ab^{2}})=\mathbb{Q}(\sqrt[3]{a^{2}b})$ for square-free
coprime $a,b\in\mathbb{N}$, we can choose $e_{1}=1$, i.e. $d=3^{e}p$ with
$e\in\\{1,2\\}$. The fact that $p\equiv 1\pmod{3}$ implies that
$p=\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$ are primes of $k_{0}$ such
that $\pi_{1}^{\tau}=\pi_{2}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\ 3\mathcal{O}_{k_{0}})$, then
the prime $p$ is totally ramified in $L$, and so the two primes $\pi_{1}$ and
$\pi_{2}$ are totally ramified in $k$. It follows by $d\not\equiv\pm
1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\ 9)$ that $3$ is totally
ramified in $L$, then $\lambda$ is ramified in $k/k_{0}$. Hence $t=3.$
Further, by $3=-\zeta_{3}^{2}\lambda^{2}$ we obtain $k=k_{0}(\sqrt[3]{x})$,
where $x=\zeta_{3}^{2}\lambda^{2}\pi_{1}\pi_{2}$. Having as a fact $p\equiv 4$
or $7\pmod{9}$ and by Lemma 3.1 entails that the primes $\pi_{1}$ and
$\pi_{2}$ are not congruent to $1\leavevmode\nobreak\
(\bmod\leavevmode\nobreak\ \lambda^{3})$, then according to Lemma 3.3, we
obtain $q^{\ast}=0$ and $\zeta_{3}$ is a not norm of an element of
$k\setminus\\{0\\}$. Therefore, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$.

####  3.6.4 Case where $d=p^{e_{1}}q^{f_{1}}\equiv\pm 1\pmod{9}$, with
$p\equiv-q\equiv 4$ or $7\pmod{9}$

It follows by $d\equiv\pm 1\pmod{9}$ that $3$ is not ramified in the pure
cubic field $L$, then $\lambda$ is also not ramified in $k/k_{0}$. The
congruence $p\equiv 1\pmod{3}$ implies that $p=\pi_{1}\pi_{2}$ with
$\pi_{1}^{\tau}=\pi_{2}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\pmod{3\mathcal{O}_{k_{0}}}$, then $\pi_{1}$ and $\pi_{2}$ are totally
ramified in $k$. As $q\equiv-1\pmod{3}$, $q$ is inert in $k_{0}$. Therefore,
the primes ramified in $k/k_{0}$ are $\pi_{1},\pi_{2}$ and $q$.
Afterwards, let be $x=\pi_{1}^{e_{1}}\pi_{2}^{e_{1}}\pi^{f_{1}}$, where
$-q=\pi$ is a prime number of $k_{0}$, then $k=k_{0}(\sqrt[3]{x})$. It is
clear that the congruence $q\equiv 2$ or $5\pmod{9}$ implies that $\pi$ is not
congruent to $1\pmod{\lambda^{3}}$, then from Lemma 3.3 we get $\zeta_{3}$ is
not a norm of an element of $k\setminus\\{0\\}$, and so $q^{\ast}=0$.
Consequently, $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$.

####  3.6.5 Case where $d=3^{e}q^{f_{1}}\not\equiv\pm 1\pmod{9}$, with
$q\equiv-1\pmod{9}$

The congruence $d\not\equiv\pm 1\pmod{9}$ entails that $3$ is totally ramified
in $L$, then $\lambda$ is ramified in $k/k_{0}$. The fact that
$q\equiv-1\pmod{3}$ implies that $q$ is inert in $k_{0}$. Then, $t=2.$
Taking into account that $3=-\zeta_{3}^{2}\lambda^{2}$, we obtain
$k=k_{0}(\sqrt[3]{x})$ with $x=\zeta_{3}^{2}\lambda^{2}\pi^{f_{1}}$, where
$-q=\pi$ is a prime number of $k_{0}$. As $q\equiv-1\pmod{9}$, the prime $\pi$
is congruent to $1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\
\lambda^{3})$, then by Lemma 3.3, $\zeta_{3}$ is a norm of an element of
$k\setminus\\{0\\}$ and $q^{\ast}=1$. Hence,
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1.$

####  3.6.6 Case where $d=q_{1}^{f_{1}}q_{2}^{f_{2}}$, with $q_{1}\equiv
q_{2}\equiv-1\pmod{9}$

For each $i\in\\{1,2\\}$, the prime $q_{i}$ is inert in $k_{0}$ because
$q_{i}\equiv 2(\bmod 3)$, however $q_{i}$ is ramified in $L$. The congruence
$d\equiv\pm 1\pmod{9}$ entails that $3$ decomposes in $L$ and then $\lambda$
is not ramified in $k/k_{0}$. So $t=2$. The fact that $q_{i}\equiv-1\pmod{9}$
implies that $\pi_{i}\equiv 1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\
\lambda^{3})$, where $-q_{i}=\pi_{i}$ is a prime number of $k_{0}$.
Afterwards, put $x=\pi_{1}^{f_{1}}\pi_{2}^{f_{2}}$, then
$k=k_{0}(\sqrt[3]{x})$, and from Lemma 3.3 we see that $\zeta_{3}$ is a norm
of an element of $k\setminus\\{0\\}$, so $q^{\ast}=1$. We conclude that
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$.

####  3.6.7 Case where $d=q_{1}^{f_{1}}q_{2}^{f_{2}}\not\equiv\pm 1\pmod{9}$,
and $\exists i\in\\{1,2\\}\ |\ q_{i}\not\equiv-1\pmod{9}$

Having as a fact $q_{i}\equiv 2\pmod{3}$, for each $i\in\\{1,2\\}$, entails
that $q_{i}$ is inert in $k_{0}$. Further, $q_{i}$ is ramified in $L$. The
congruence $d\not\equiv\pm 1\pmod{9}$ implies that $3$ is ramified in $L$, and
therefore $\lambda$ is ramified in $k/k_{0}$. Then we obtain $t=3$. From
hypothesis, there exist $i\in\\{1,2\\}$ such that $q_{i}\not\equiv-1\pmod{9}$,
then $\pi_{i}\not\equiv 1\leavevmode\nobreak\ (\bmod\leavevmode\nobreak\
\lambda^{3})$ where $-q_{i}=\pi_{i}$ is a prime number of $k_{0}$.
Afterwards, put $x=\pi_{1}^{f_{1}}\pi_{2}^{f_{2}}$, then we get
$k=k_{0}(\sqrt[3]{x})$. According to Lemma 3.3 we have $\zeta_{3}$ is not norm
of an element of $k\setminus\\{0\\}$ and $q^{\ast}=0$. Then
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$.

####  3.6.8 Case where $d=3^{e}q_{1}^{f_{1}}q_{2}^{f_{2}}\not\equiv\pm
1\pmod{9}$, $e\in\\{0,1,2\\}$ and $\exists i\in\\{1,2\\}\ |\
q_{i}\not\equiv-1\pmod{9}$

For each $i\in\\{1,2\\}$, $q_{i}$ is ramified in $L$, and $\lambda$ is
ramified in $k/k_{0}$ because $d\not\equiv\pm 1\pmod{9}$, so $t=3$. We reason
as above, we get $\zeta_{3}$ is not a norm of an element of
$k\setminus\\{0\\}$ and $q^{\ast}=0$. Thus,
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$.

####  3.6.9 Case where $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm
1\pmod{9}$, with $q_{1}\equiv q_{2}\equiv 2\,\ \text{or}\
\,5\pmod{9},q_{3}\equiv-1\pmod{9}$

As $d\equiv\pm 1\pmod{9}$, $\lambda$ is not ramified in $k/k_{0}$. In the same
manner as above, we obtain $t=3$ and $\zeta_{3}$ is not a norm of an element
of $k\setminus\\{0\\}$, therefore $q^{\ast}=0$. Thus
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$.

####  3.6.10 Case where $d=q_{1}^{f_{1}}q_{2}^{f_{2}}q_{3}^{f_{3}}\equiv\pm
1\pmod{9}$, with $q_{1}\equiv q_{2}\equiv q_{3}\equiv
2\,\text{or}\,5\pmod{9}.$

Reasoning as in the above case, we get $t=3$ and $q^{\ast}=0$, then
$\operatorname{rank}\,(C_{k,3}^{(\sigma)})=1$. This completes the proof of
Theorem 1.1.

##  4 The full $3$-class group of the Galois closure $k$ of $L$

In Table 7 we finally summarize the structures of the $3$-class group
$C_{L,3}$ of the pure cubic field $L=\mathbb{Q}(\sqrt[3]{d})$ and the full
$3$-class group $C_{k,3}$ of the normal closure
$k=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$, which occur in dependence on the
various shapes of the conductor $f$ in our Theorems 2.4–2.5, where the
$3$-class group $C_{k,3}^{(\sigma)}\simeq(3)$ of the ambiguous classes of
$k/k_{0}$ is generally cyclic of order $3$.

In contrast to the previous tables, it was not possible to compute the
structure of the $3$-class group $C_{k,3}$ of the field $k$ of absolute degree
$6$ by means of Voronoi’s algorithm which is only able to determine the
fundamental unit $\varepsilon$ in the unit group
$E_{L}=\langle-1,\varepsilon\rangle$, the principal factorization type
(briefly: type), the absolute principal factors, and the class number $h_{L}$
of the field $L$ of degree $3$. Therefore we had to employ the computational
algebra system MAGMA [5, 6, 25] for the numerical investigation of the normal
closures $k$.

We emphasize that the paradigms for the radicand $d$ in Table 7 are ordered
according to the values of the exponent $w$, as opposed to the Tables 4, 5, 6.
For instance, the smallest radicand $d=1605$ with $w=3$ occurs earlier than
the smallest radicand $d=2091$ with $w=2$, for the conductor $f=9q_{1}q_{2}$
with $q_{1}\equiv 2,5\,(9)$ and $q_{2}\equiv 8\,(9)$.

The restriction of the exponent $w$ to the smallest value $w=1$ has been
proved by Gerth [11] for conductors $f=3p_{1}$ with $p_{1}\equiv 4,7\,(9)$.
However, for conductors $f=9p_{1}$ with $p_{1}\equiv 4,7\,(9)$ or
$f=p_{1}q_{1}$ with $p_{1}\equiv 4,7\,(9)$ and $q_{1}\equiv 2,5\,(9)$ or
$f=3q_{1}q_{2}$ with $q_{1},q_{2}\equiv 2,5\,(9)$ or $f=q_{1}q_{2}q_{3}$ with
$q_{1},q_{2},q_{3}\equiv 2,5\,(9)$, this restriction seems to be an open
problem and can be stated as a conjecture with very strong computational
support by our investigations.

Table 7: Structure of $3$-class groups $C_{L,3}$ and $C_{k,3}$ | $f$ | Type | $C_{L,3}$ | $C_{k,3}$ | $w$ |  Paradigms for $d$
---|---|---|---|---|---|---
Items in Theorem 2.4
$(1)$ | $p_{1}\equiv 1\,(9)$ |  |  |  |  |
|  | $\alpha$ | $(3^{w})$ | $(3^{w},3^{w-1})$ | $\geq 1$ | $19,199,3061,6733$
|  | $\gamma$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 2$ | $541,8389$
$(2)$ |  $3p_{1}$, $p_{1}\equiv 4,7\,(9)$ |  |  |  |  |
|  | $\alpha$ | $(3)$ | $(3)$ |  | $7$
|  | $\beta$ | $(3)$ | $(3,3)$ |  | $61$
$(3)$ |  $9p_{1}$, $p_{1}\equiv 4,7\,(9)$ |  |  |  |  |
|  | $\alpha$ | $(3)$ | $(3)$ |  | $21$
|  | $\beta$ | $(3)$ | $(3,3)$ |  | $183$
$(4)$ |  $p_{1}q_{1}$, $p_{1}\equiv-q_{1}\equiv 4,7\,(9)$ |  |  |  |  |
|  | $\alpha$ | $(3)$ | $(3)$ |  | $26$
|  | $\beta$ | $(3)$ | $(3,3)$ |  | $62$
Items in Theorem 2.5
$(1)$ |  $9q_{1}$, $q_{1}\equiv 8\,(9)$ |  |  |  |  |
|  | $\beta$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $51,159,213$
|  | $\gamma$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $153,321,477$
$(2)$ |  $q_{1}q_{2}$, $q_{1},q_{2}\equiv 8\,(9)$ |  |  |  |  |
|  | $\beta$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $901,8857,61273$
|  | $\gamma$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 2$ | $1207,11917$
$(3)$ |  $3q_{1}q_{2}$, $q_{1},q_{2}\equiv 2,5\,(9)$ | $\beta$ | $(3)$ | $(3,3)$ |  | $20$
$(4)$ |  $3q_{1}q_{2}$, $q_{1}\equiv 2,5$, $q_{2}\equiv 8$ | $\beta$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $34,535,1003,5972$
$(5)$ |  $9q_{1}q_{2}$, $q_{1}\equiv 2,5\,(9)$ | $\beta$ |  |  |  |
| $q_{2}\equiv 2,5\,(9)$ |  | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $30,165,2514,7374$
| $q_{2}\equiv 8\,(9)$ |  | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $102,2091,1605,17265,13833$
$(6)$ |  $q_{1}q_{2}q_{3}$, $q_{1},q_{2},q_{3}\equiv 2,5\,(9)$ | $\beta$ | $(3)$ | $(3,3)$ |  | $460$
$(7)$ |  $q_{1}q_{2}q_{3}$, $q_{1},q_{2}\equiv 2,5$, $q_{3}\equiv 8$ | $\beta$ | $(3^{w})$ | $(3^{w},3^{w})$ | $\geq 1$ | $170,1394,4301,26452,46079$

##  5 Conclusion

In this article, we have illuminated the genus theory and Galois cohomology of
an essential part of the pure cubic fields $\mathbb{Q}(\sqrt[3]{d})$, namely
those which possess $3$-class groups with abelian type invariants $1$, $(3)$,
and $(3^{w})$ with $w\geq 2$. As a result of our systematic analysis of
ambiguous ideal class groups with $3$-rank one of the normal closures
$\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ over $\mathbb{Q}(\zeta_{3})$, it turned
out that these fields can be characterized by radicands $d$, respectively
conductors $f$, having at most three prime divisors, and at most one among
them congruent to $1$ modulo $3$. Extensive computational investigations
showed that they constitute approximately one third (precisely
$31.83\%=8.93\%+9.90\%+13.00\%$) of the pure cubic fields
$\mathbb{Q}(\sqrt[3]{d})$ with $d<10^{6}$. (The asymptotic limit among all
pure cubic fields may be different.) An important open problem is the proof of
the conjectured equivalence for cubic residue symbols in Corollary 2.4.1,
mentioned in Example 2.3.

##  6 Outlook

It should be pointed out that Ismaili [18] has investigated another related
scenario for the relative $3$-genus field $k^{\ast}$. If the conductor $f$ is
divisible by exactly one prime $p\equiv 1\,(\mathrm{mod}\,3)$, that is $s=1$,
but is not contained in Theorem 2.4, and if $C_{k,3}\simeq(3,3)$ is elementary
bicyclic, then the genus field coincides with the Hilbert $3$-class field,
$k^{\ast}=k_{1}$, and the composita $k\cdot L_{1}=k\cdot L_{1}^{\sigma}=k\cdot
L_{1}^{\sigma^{2}}=K_{4}$ coincide with one of the four unramified cyclic
cubic extensions $K_{1},\ldots,K_{4}$ of $k$ within $k_{1}$, as illustrated in
Figure 3 and studied in detail by Ismaili and El Mesaoudi [19]. However, this
situation lies outside of the present article, because the $3$-group of
ambiguous ideal classes has $\operatorname{rank}\,(C_{k,3}^{(\sigma)})=2$
under the given conditions. This will be the topic of our forthcoming paper
[1], which also sheds light on peculiar phenomena revealed by the lattice
minima of the discrete geometric Minkowski image of the maximal order
$\mathcal{O}_{L}$ of the pure cubic field $L=\mathbb{Q}(\sqrt[3]{d})$.

Figure 3: Third possible location of the relative $3$-genus field
$k^{\ast}=(k/k_{0})^{\ast}$Degree$54$Hilbert of
bicubic$18$compositum$9$Hilbert of conjugate cubics$6$bicubic$3$conjugate
cubics$2$quadratic$1$baseScenario III:
$k^{\ast}=k_{1}$$k_{1}=k^{\ast}$$L_{1}\cdot k=L_{1}^{\sigma}\cdot
k=L_{1}^{\sigma^{2}}\cdot
k$$K_{3}$$K_{2}$$K_{1}$$k$$k_{0}$$L_{1}$$L_{1}^{\sigma},L_{1}^{\sigma^{2}}$$L$$L^{\sigma},L^{\sigma^{2}}$$\mathbb{Q}$

##  7 Acknowledgements

This paper is dedicated to the memory of Frank Emmett Gerth III, born 08
October 1945, who contributed numerous great ideas to algebraic number theory,
class field theory, and in particular the theory of cubic field extensions,
starting with his Ph.D. thesis [8] under supervision of Kenkichi Iwasawa, and
then continuously until [11], shortly before his early passing away on 23 May
2006.

The second author gratefully acknowledges that his research was supported by
the Austrian Science Fund (FWF): P 26008-N25.

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