#  Fields $\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ whose $3$-class group is of type
$(9,3)$

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

Abstract: Let $\mathrm{k}=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$, with $d$ a cube-
free positive integer. Let $C_{\mathrm{k},3}$ be the $3$-class group of
$\mathrm{k}$. With the aid of genus theory, arithmetic properties of the pure
cubic field $\mathbb{Q}(\sqrt[3]{d})$ and some results on the $3$-class group
$C_{\mathrm{k},3}$, we determine all integers $d$ such that
$C_{\mathrm{k},3}\simeq\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$.

Keywords: Pure cubic fields; $3$-class groups; structure of groups.
Mathematics Subject Classification 2010: 11R11, 11R16, 11R20, 11R27, 11R29,
11R37.

##  1 Introduction

####

Let $d$ be a cube-free positive integer,
$\mathrm{k}=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$, and $C_{\mathrm{k},3}$ be the
$3$-class group of $\mathrm{k}$. A number of researchers have studied the
$3$-class group $C_{\mathrm{k},3}$ and the calculation of its rank. Calegari
and Emerton [7, Lemma 5.11] proved that the rank of the $3$-class group of
$\mathbb{Q}(\sqrt[3]{p},\zeta_{3})$, with a prime $p\equiv 1\pmod{9}$, is
equal to two if $9$ divides the class number of $\mathbb{Q}(\sqrt[3]{p})$. The
converse of the Calegari-Emerton result was proved by Frank Gerth III in [11,
Theorem 1, p. 471].

####

The purpose of this paper is to classify all integers $d$ for which
$C_{\mathrm{k},3}$ is of type $(9,3)$, i.e.
$C_{\mathrm{k},3}\simeq\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$.
This investigation can be viewed as a continuation of the previous more
general works [7, Lemma 5.11] and [11, Theorem 1, p. 471]. Effectively, we
shall prove the following main theorem:

######  Theorem 1.1.

Let $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ be a pure cubic field, where $d\geq 2$ is
a cube-free integer, and let $\mathrm{k}=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ be
its normal closure. Denote by $u$ the index of the subgroup generated by the
units of intermediate fields of the extension $\mathrm{k}/\mathbb{Q}$ in the
group of units of $\mathrm{k}$.

  * 1)

If the field $\mathrm{k}$ has a $3$-class group of type $(9,3)$, then
$d=p^{e}$, where $p$ is a prime number congruent to $1\pmod{9}$ and $e=1$ or
$2$.

  * 2)

If $p$ is a prime number congruent to $1\pmod{9}$, $9$ divides the class
number of $\Gamma$ exactly, and $u=1$, then the $3$-class group of
$\mathrm{k}$ is of type $(9,3).$

This result will be underpinned by numerical examples obtained with the
computational number theory system PARI [17] in § 3. In section 2, where
Theorem 1.1 is proved, we only state results that will be needed in this
paper. More information on $3$-class groups can be found in [10] and [9]. For
the prime ideal factorization in the pure cubic field
$\mathbb{Q}(\sqrt[3]{d})$, we refer the reader to the papers [8], [5], [4] and
[16]. For the prime factorization rules of the third cyclotomic field
$\mathbb{Q}(\zeta_{3})$, we refer the reader to [13, Chap. 9, Sec. 1,
Propositions 9.1.1–4, pp. 109-111].

Notations:

$\bullet$ The lower case letter $p$, respectively $q$, will denote a prime
number congruent to $1$, respectively $-1$, modulo $3$;
$\bullet$ $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ : a pure cubic field, where $d\geq
2$ is a cube-free integer;
$\bullet$ $\mathrm{k}_{0}=\mathbb{Q}(\zeta_{3})$ : the cyclotomic field, where
$\zeta_{3}=e^{2i\pi/3}$;
$\bullet$ $\mathrm{k}=\Gamma(\zeta_{3})$ : the normal closure of $\Gamma$;
$\bullet$ $\Gamma^{\prime}$ and $\Gamma^{\prime\prime}$ : the two conjugate
cubic fields of $\Gamma$, contained in $\mathrm{k}$;
$\bullet$ $u$ : the index of the subgroup $E_{0}$ generated by the units of
intermediate fields of the extension $\mathrm{k}/\mathbb{Q}$ in the group of
units of $\mathrm{k}$;
$\bullet$
$\langle\tau\rangle=\operatorname{Gal}\left(\mathrm{k}/\Gamma\right)$, such
that $\tau^{2}=id$, $\tau(\zeta_{3})=\zeta_{3}^{2}$, and
$\tau(\sqrt[3]{d})=\sqrt[3]{d}$;
$\bullet$
$\langle\sigma\rangle=\operatorname{Gal}\left(\mathrm{k}/\mathrm{k}_{0}\right)$,
such that $\sigma^{3}=id$, $\sigma(\zeta_{3})=\zeta_{3}$, and
$\sigma(\sqrt[3]{d})=\zeta_{3}\sqrt[3]{d}$;
$\bullet$ $\lambda=1-\zeta_{3}$ is a prime element above $3$ of
$\mathrm{k}_{0}$;
$\bullet$ $q^{*}=1$ or $0$, according to whether $\zeta_{3}$ is norm of an
element of $\mathrm{k}$ or not;
$\bullet$ $t$ : the number of prime ideals ramified in
$\mathrm{k}/\mathrm{k}_{0}$;
$\bullet$ For an algebraic number field $L$:

  * $-$

$\mathcal{O}_{L}$, $E_{L}$ : the ring of integers of $L$, and the group of
units of $L$;

  * $-$

$C_{L,3}$, $h_{L}$ : the $3$-class group of $L$, and the class number of $L$;

  * $-$

$L_{3}^{(1)}$, $L^{*}$ : the Hilbert $3$-class field of $L$, and the absolute
genus field of $L$.

##  2 Fields $\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ whose $3$-Class Group is of
Type $(9,3)$

###  2.1 Preliminary results

In [15, Chap. 7, pp. 87–96], Ishida has explicitly given the genus field of
any pure field. For the pure cubic field $\Gamma=\mathbb{Q}(\sqrt[3]{d})$,
where $d$ is a cube-free natural number, we have the following theorem.

######  Theorem 2.1.

Let $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ be a pure cubic field, where $d\geq 2$ is
a cube-free integer, and let $p_{1},\ldots,p_{r}$ be all prime divisors of $d$
such that $p_{i}$ is congruent to $1\pmod{3}$ for each
$i\in\\{1,2,\ldots,r\\}$. Let $\Gamma^{\ast}$ be the absolute genus field of
$\Gamma$, then

$\Gamma^{\ast}=\prod_{i=1}^{r}M(p_{i})\cdot\Gamma,$

where $M(p_{i})$ denotes the unique subfield of degree $3$ of the cyclotomic
field $\mathbb{Q}(\zeta_{p_{i}})$. The genus number of $\Gamma$ is given by
$g_{\Gamma}=3^{r}$.

######  Remark 2.1.

(1) If no prime $p\equiv 1\pmod{3}$ divides $d$, i.e. $r=0$, then
$\Gamma^{\ast}=\Gamma$.
(2) For any value $r\geq 0$, $\Gamma^{\ast}$ is contained in the Hilbert
$3$-class field $\Gamma_{3}^{(1)}$ of $\Gamma$.
(3) The cubic field $M(p)$ is determined explicitly in [12, § 4, Proposition
1, p. 11].

Assuming that $h_{\Gamma}$ is divisible exactly by $9$, we can explicitly
construct the absolute genus field $\Gamma^{*}$ as follows:

######  Lemma 2.1.

Let $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ be a pure cubic field, where $d\geq 2$ is
a cube-free integer. If $h_{\Gamma}$ is exactly divisible by $9$, then there
are at most two primes congruent to $1\pmod{3}$ dividing $d$.

###### Proof.

If $p_{1},\ldots,p_{r}$ are all prime numbers congruent to $1\pmod{3}$
dividing $d$, then $3^{r}|h_{\Gamma}$. Therefore, if $h_{\Gamma}$ is exactly
divisible by $9$, then $r\leq 2$. So there are two primes $p_{1}$ and $p_{2}$
dividing $d$ such that $p_{i}\equiv 1\pmod{3}$ for $i\in\\{1,2\\}$, or there
is only one prime $p\equiv 1\pmod{3}$ with $p|d$, or there is no prime
$p\equiv 1\pmod{3}$ such that $p|d$. ∎

######  Lemma 2.2.

Let $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ be a pure cubic field, where $d\geq 2$ is
a cube-free integer. If $h_{\Gamma}$ is exactly divisible by $9$ and if the
integer $d$ is divisible by two primes $p_{1}$ and $p_{2}$ such that
$p_{i}\equiv 1\pmod{3}$ for $i\in\\{1,2\\}$, then
$\Gamma^{\ast}=\Gamma_{3}^{(1)}$,
$(\Gamma^{\prime})^{\ast}={\Gamma^{\prime}}_{3}^{(1)}$ and
$(\Gamma^{\prime\prime})^{\ast}={\Gamma^{\prime\prime}}_{3}^{(1)}$.
Furthermore,
$\mathrm{k}\cdot\Gamma_{3}^{(1)}=\mathrm{k}\cdot{\Gamma^{\prime}}_{3}^{(1)}=\mathrm{k}\cdot{\Gamma^{\prime\prime}}_{3}^{(1)}$.

###### Proof.

If $h_{\Gamma}$ is exactly divisible by $9$ and $d$ is divisible by two prime
numbers $p_{1}$ and $p_{2}$ which are congruent to $1\pmod{3}$, then
$g_{\Gamma}=9$ so $\Gamma^{\ast}=\Gamma_{3}^{(1)}=\Gamma\cdot M(p_{1})\cdot
M(p_{2})$, where $M(p_{1})$ (respectively $M(p_{2})$) is the unique cubic
subfield of $\mathbb{Q}(\zeta_{p_{1}})$ (respectively
$\mathbb{Q}(\zeta_{p_{2}})$). The equations

$\begin{cases}(\Gamma^{\prime})^{\ast}=\Gamma^{\prime}\cdot M(p_{1})\cdot
M(p_{2})={\Gamma^{\prime}}_{3}^{(1)},\\\
(\Gamma^{\prime\prime})^{\ast}=\Gamma^{\prime\prime}\cdot M(p_{1})\cdot
M(p_{2})={\Gamma^{\prime\prime}}_{3}^{(1)}\end{cases}$

can be deduced by the fact that in the general case we have

$\begin{cases}(\Gamma^{\prime})^{\ast}=\Gamma^{\prime}\prod\limits_{i=1}^{r}M(p_{i}),\\\
(\Gamma^{\prime\prime})^{\ast}=\Gamma^{\prime\prime}\prod\limits_{i=1}^{r}M(p_{i}),\end{cases}$

where $p_{i}$, for each $1\leq i\leq r$, is a prime divisor of $d$ such that
$p_{i}\equiv 1\pmod{3}$. From the fact that $h_{\Gamma}$ is exactly divisible
by $9$, we conclude that $h_{\Gamma^{\prime}}$ is exactly divisible by $9$ and
$h_{\Gamma^{\prime\prime}}$ is exactly divisible by $9$, because $\Gamma$,
$\Gamma^{\prime}$ and $\Gamma^{\prime\prime}$ are isomorphic.
Moreover,

$\begin{cases}\mathrm{k}\cdot\Gamma_{3}^{(1)}=\mathrm{k}\cdot\Gamma\cdot
M(p_{1})\cdot M(p_{2})=\mathrm{k}\cdot M(p_{1})\cdot M(p_{2}),\\\
\mathrm{k}\cdot{\Gamma^{\prime}}_{3}^{(1)}=\mathrm{k}\cdot\Gamma^{\prime}\cdot
M(p_{1})\cdot M(p_{2})=\mathrm{k}\cdot M(p_{1})\cdot M(p_{2}),\\\
\mathrm{k}\cdot{\Gamma^{\prime\prime}}_{3}^{(1)}=\mathrm{k}\cdot\Gamma^{\prime\prime}\cdot
M(p_{1})\cdot M(p_{2})=\mathrm{k}\cdot M(p_{1})\cdot M(p_{2}).\end{cases}$

Hence,
$\mathrm{k}\cdot\Gamma_{3}^{(1)}=\mathrm{k}\cdot{\Gamma^{\prime}}_{3}^{(1)}=\mathrm{k}\cdot{\Gamma^{\prime\prime}}_{3}^{(1)}$.
∎

Now, let $u$ be the index of units defined in the above notations. We assume
that $h_{\Gamma}$ is exactly divisible by $9$ and $u=1$. From [5, § 14,
Theorem 14.1, p. 232], we have $h_{\mathrm{k}}=\dfrac{u}{3}\cdot
h_{\Gamma}^{2}$, whence $h_{\mathrm{k}}$ is exactly divisible by $27$. The
structure of the $3$-class group $C_{\mathrm{k},3}$ is described by the
following Lemma:

######  Lemma 2.3.

Let $\Gamma$ be a pure cubic field, $\mathrm{k}$ its normal closure, and $u$
be the index of units defined in the notations above, then

$C_{\mathrm{k},3}\simeq\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\quad\Leftrightarrow\quad[C_{\Gamma,3}\simeq\mathbb{Z}/9\mathbb{Z}\quad\text{and}\quad
u=1].$

Lemma 2.3 will be underpinned in section 3 by numerical examples obtained with
the computational number theory system PARI [17].

###### Proof.

Assume that
$C_{\mathrm{k},3}\simeq\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$.
Let $h_{\Gamma,3}$(respectively, $h_{\mathrm{k},3}$) be the $3$-class number
of $\Gamma$ (respectively, $\mathrm{k}$), then $h_{\mathrm{k},3}=27$.
According to [5, § 14, Theorem 14.1, p. 232], we get $27=\frac{u}{3}\cdot
h_{\Gamma,3}^{2}$ with $u\in\\{1,3\\}$, and thus $u=1$, because otherwise $27$
would be a square in $\mathbb{N}$, which is a contradiction. Thus
$h_{\Gamma,3}^{2}=81$ and $h_{\Gamma,3}=9$.
Let $C_{\mathrm{k},3}^{+}=\\{\mathcal{A}\in
C_{\mathrm{k},3}\mid\mathcal{A}^{\tau}=\mathcal{A}\\}$ and
$C_{\mathrm{k},3}^{-}=\\{\mathcal{A}\in
C_{\mathrm{k},3}\mid\mathcal{A}^{\tau}=\mathcal{A}^{-1}\\}$. According to [9,
§ 2, Lemmas 2.1 and 2.2, p. 53], we have $C_{\mathrm{k},3}\simeq
C_{\mathrm{k},3}^{+}\times C_{\mathrm{k},3}^{-}$ and
$C_{\mathrm{k},3}^{+}\simeq C_{\Gamma,3}$, hence $\lvert
C_{\mathrm{k},3}^{-}\rvert=3$. Since $C_{\mathrm{k},3}$ is of type
$\left(9,3\right)$, we deduce that $C_{\mathrm{k},3}^{-}$ is a cyclic group of
order $3$ and $C_{\mathrm{k},3}^{+}$ is a cyclic group of order $9$.
Therefore, we have

$u=1\quad\text{and}\quad C_{\Gamma,3}\simeq\mathbb{Z}/9\mathbb{Z}.$

Conversely, assume that $u=1$ and $C_{\Gamma,3}\simeq\mathbb{Z}/9\mathbb{Z}$.
By [5, § 14, Theorem 14.1, p. 232], we deduce that
$|C_{\mathrm{k},3}|=\frac{1}{3}\cdot|C_{\Gamma,3}|^{2}$, and so
$|C_{\mathrm{k},3}|=27$. Furthermore, $|C_{\mathrm{k},3}^{-}|=3$ and

$C_{\mathrm{k},3}\simeq C_{\Gamma,3}\times
C_{\mathrm{k},3}^{-}\simeq\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}.$

∎

###  2.2 Proof of Theorem 1.1

Let $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ be a pure cubic field, where $d\geq 2$ is
a cube-free integer, $\mathrm{k}=\mathbb{Q}(\sqrt[3]{d},\zeta_{3})$ be its
normal closure, and $C_{\mathrm{k},3}$ be the $3$-class group of $\mathrm{k}$.
(1) Assume that the $3$\- class group $C_{\mathrm{k},3}$ is of type $(9,3)$.
We first write the integer $d$ in the form given by equation (3.2) of [9, p.
55]:

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

where $p_{i}$ and $q_{i}$ are positive rational primes such that:

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

Let $C_{\mathrm{k},3}^{(\sigma)}$ be the ambiguous ideal class group of
$\mathrm{k}/\mathrm{k}_{0}$, where $\sigma$ is a generator of
$\operatorname{Gal}\left(\mathrm{k}/\mathrm{k}_{0}\right)$. It is known that
$C_{\mathrm{k},3}^{(\sigma)}$ is an elementary abelian 3-group, because an
ambiguous class $\mathcal{C}$ satisfies $\sigma(\mathcal{C})=\mathcal{C}$, by
definition, and therefore
$\mathcal{C}^{3}=\mathcal{C}\cdot\sigma(\mathcal{C})\cdot\sigma^{2}(\mathcal{C})=\mathcal{N}_{k/k_{0}}(\mathcal{C})=1$,
since $\mathrm{k}_{0}$ has class number $1$.

The fact that the $3$-class group $C_{\mathrm{k},3}$ is of type $(9,3)$
implies that $\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=1$. In fact, it
is clear that if $C_{\mathrm{k},3}$ is of type $(9,3)$, then
$\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=1$ or $2$.

Let us assume that $\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=2$. From
[10, § 5, Theorem 5.3, pp. 97–98], we have

$\operatorname{rank}\,C_{\mathrm{k},3}=2t-s,$

where the integers $t$ and $s$ are defined in [10, § 5, Theorem 5.3, pp.
97–98] as follows:

$\displaystyle t$ $\displaystyle:=$
$\displaystyle\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)},$
$\displaystyle s$ $\displaystyle:$ $\displaystyle\text{the rank of the matrix
}(\beta_{i,j})\text{ defined in \cite[cite]{[\@@bibref{}{GERTH1}{}{}, \S\ 5,
Theorem. 5.3, pp. 97--98]}}.$

Since $C_{\mathrm{k},3}$ is of type (9,3), then
$\operatorname{rank}\,C_{\mathrm{k},3}=2,$ and according to our hypothesis we
have $t=\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=2$. So we get $s=2$.

By [10, § 5, Theorem 5.3, pp. 97–98], the $3$-class group $C_{\mathrm{k},3}$
is isomorphic to the direct product of an abelian $3$-group of rank $2(t-s)$
and an elementary abelian $3$-group of rank $s$. Here $t=s=2$. Thus
$C_{\mathrm{k},3}$ is isomorphic to the direct product of an abelian $3$-group
of rank $0$ and an elementary abelian $3$-group of rank $2$, we get

$C_{\mathrm{k},3}\simeq\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z},$

which contradicts the fact that $C_{\mathrm{k},3}$ is of type $(9,3)$. We
conclude $\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=1$.

####

On the one hand, suppose that $d$ is not divisible by any rational prime $p$
such that $p\equiv 1\pmod{3}$, i.e. $w=0$ in Eq. (1) above. According to [9, §
5, Theorem 5.1, p. 61], this implies that $C_{\mathrm{k},3}\simeq
C_{\Gamma,3}\times C_{\Gamma,3}$. Since the $3$-group $C_{\mathrm{k},3}$ is of
type $(9,3)$, then $|C_{\mathrm{k},3}|=3^{3}$, and by Lemma 2.3 we have
$C_{\Gamma,3}\simeq\mathbb{Z}/9\mathbb{Z}$, so we get $|C_{\Gamma,3}\times
C_{\Gamma,3}|=3^{4}$, which is a contradiction. Hence, the integer $d$ is
divisible by at least one rational prime $p$ such that $p\equiv 1\pmod{3}$.

On the other hand, the fact that the $3$-class group $C_{\mathrm{k},3}$ is of
type $(9,3)$ implies that $C_{\Gamma,3}\simeq\mathbb{Z}/9\mathbb{Z}$ according
to Lemma 2.3. Since $h_{\Gamma}$ is exactly divisible by $9$, then according
to Lemma 2.1, there are at most two primes congruent to $1\pmod{3}$ dividing
$d$.

Now, assume that there are exactly two different primes $p_{1}$ and $p_{2}$
dividing $d$ such that $p_{1}\equiv p_{2}\equiv 1\pmod{3}$, then, according to
Lemma 2.2, we get:

$\Gamma^{\ast}=\Gamma_{3}^{(1)}=\Gamma\cdot M(p_{1})\cdot M(p_{2}),$

where $M(p_{1})$ (respectively, $M(p_{2})$) is the unique subfield of degree
$3$ of $\mathbb{Q}(\zeta_{p_{1}})$ (respectively,
$\mathbb{Q}(\zeta_{p_{2}})$). If $M(p_{1})\neq M(p_{2})$, then $\Gamma\cdot
M(p_{1})$ and $\Gamma\cdot M(p_{2})$ are two different subfields of
$\Gamma_{3}^{(1)}$ over $\Gamma$ of degree $3$. According to class field
theory, we have

$\operatorname{Gal}(\Gamma_{3}^{(1)}/\Gamma)\cong
C_{\Gamma,3}\simeq\mathbb{Z}/9\mathbb{Z}.$

Since $\operatorname{Gal}(\Gamma_{3}^{(1)}/\Gamma)$ is a cyclic $3$-group,
there exist only one sub-group of
$\operatorname{Gal}(\Gamma_{3}^{(1)}/\Gamma)$ of order $3$. By the Galois
correspondence, there exist a unique sub-field of $\Gamma_{3}^{(1)}$ over
$\Gamma$ of degree $3$. We conclude that $\Gamma\cdot M(p_{1})=\Gamma\cdot
M(p_{2})$, which is a contradiction.

Thus, $M(p_{1})=M(p_{2})$, and then $p_{1}=p_{2}$, which contradicts the fact
that $p_{1}$ and $p_{2}$ are two different primes. Hence, there is exactly one
prime congruent to $1\pmod{3}$ which divides $d$. Thus the integer $d$ can be
written in the following form:

$d=3^{e}p_{1}^{e_{1}}q_{1}^{f_{1}}\ldots q_{I}^{f_{I}}q_{I+1}^{f_{I+1}}\ldots
q_{J}^{f_{J}},$

with $p_{1}\equiv-q_{i}\equiv 1\pmod{3}$, where $p_{1}$, $e$, $e_{1}$, $q_{i}$
and $f_{i}$ (for $1\leq i\leq J$) are defined in Eq. (1).

Next, since $\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=1$, then
according to [9, § 3, Lemma 3.1, p. 55], there are three possible cases as
follows:
$\bullet$ Case 1 : $2w+J=1$,
$\bullet$ Case 2 : $2w+J=2$,
$\bullet$ Case 3 : $2w+J=3$,
where $w$ and $J$ are the integers defined in Eq. (1) above. We will
successively treat these cases as follows:
$\bullet$ Case 1: We have $2w+J=1$, then $w=0$ and $J=1$. This case is
impossible, because we have shown above that the integer $d$ is divisible by
exactly one prime number congruent to $1\pmod{3}$ and thus $w=1$.
$\bullet$ Case 2: We have $2w+J=2$, and as in Case $1$, we necessarily have
$w=1$ and $J=0$, which implies that $d=3^{e}p_{1}^{e_{1}}$, where $p_{1}$ is a
prime number such that $p_{1}\equiv 1\pmod{3}$, $e\in\\{0,1,2\\}$ and
$e_{1}\in\\{1,2\\}$. Then,

  * $-$

If $d\equiv\pm 1\pmod{9}$, then we necessarily have $e=0$.
Assume that $p_{1}\equiv 4$ or $7\pmod{9}$, then $d\not\equiv\pm 1\pmod{9}$
which is an absurd. So we necessarily have $p_{1}\equiv 1\pmod{9}$. Thus the
integer $d$ will be written in the form $d=p_{1}^{e_{1}}$, where $p_{1}\equiv
1\pmod{9}$ and $e_{1}\in\\{1,2\\}$.

  * $-$

If $d\not\equiv\pm 1\pmod{9}$:

    * $*$

If $e\neq 0$, the integer $d$ is written as $d=3^{e}p_{1}^{e_{1}}$, where
$p_{1}\equiv 1\pmod{3}$ and $e,e_{1}\in\\{1,2\\}$.

    * $*$

If $e=0$, then $d$ is written as $d=p_{1}^{e_{1}}$ with $p_{1}\equiv 4$ or
$7\pmod{9}$ and $e_{1}\in\\{1,2\\}$.

$\bullet$ Case 3: We have $2w+J=3$, then we necessarily get $w=1$ and $J=1$,
because $w\neq 0$. Thus $d=3^{e}p_{1}^{e_{1}}q_{1}^{f_{1}}$, where
$p_{1}\equiv 1\pmod{3}$, $q_{1}\equiv-1\pmod{3}$, $e\in\\{0,1,2\\}$ and
$e_{1},f_{1}\in\\{1,2\\}$. Then:

  * $-$

If $d\equiv\pm 1\pmod{9}$, we necessarily have $e=0$. If $p_{1}$ or
$-q_{1}\not\equiv 1\pmod{9}$, then $d\not\equiv\pm 1\pmod{9}$ which is an
absurd. It remain only the case where $p_{1}\equiv-q_{1}\equiv 1\pmod{9}$.
Then the integer $d$ will be written in the form
$d=p_{1}^{e_{1}}q_{1}^{f_{1}}$, where $p_{1}\equiv-q_{1}\equiv 1\pmod{9}$ and
$e_{1},f_{1}\in\\{1,2\\}$.

  * $-$

If $d\not\equiv\pm 1\pmod{9}$:
According to [10, § 5, p. 92], the rank of $C_{\mathrm{k},3}^{(\sigma)}$ is
specified as follows:

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

where $t$ and $q^{*}$ are defined in the notations.
On the one hand, the fact that
$\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=1$ imply that $t=2$ or $3$
according to whether $\zeta_{3}$ is norm of an element of $\mathrm{k}$ or not.
On the other hand, we have $d=3^{e}p_{1}^{e_{1}}q_{1}^{f_{1}}$ with
$p_{1}\equiv 1\pmod{3}$ and $q_{1}\equiv-1\pmod{3}$. By [13, Chap. 9, Sec. 1,
Proposition 9.1.4, p.110] we have $q_{1}$ is inert in $\mathrm{k}_{0}$, and by
[8, Sec 4, pp. 51-54] we have $q_{1}$ is ramifed in
$\Gamma=\mathbf{Q}(\sqrt[3]{d})$. Since $d\not\equiv\pm 1\pmod{9}$, then $3$
is ramifed in $\Gamma$ by [8, Sec 4, pp. 51-54], and
$3\mathcal{O}_{\mathrm{k}_{0}}=(\lambda)^{2}$ where $\lambda=1-\zeta_{3}.$
Since $p_{1}\equiv 1\pmod{3}$, then by [13, Chap. 9, Sec. 1, Proposition
9.1.4, p.110] we have $p_{1}=\pi_{1}\pi_{2}$ where $\pi_{1}$ and $\pi_{2}$ are
two primes of $\mathrm{k}_{0}$ such that $\pi_{2}=\pi_{1}^{\tau}$, the prime
$p_{1}$ is ramifed in $\Gamma$, then $\pi_{1}$ and $\pi_{2}$ are ramified in
$\mathrm{k}$. Hence, the number of prime ideals which are ramified in
$\mathrm{k}/\mathrm{k}_{0}$ is $t=4$, which contradicts the fact that $t=2$ or
$3.$

We summarize all forms of integer $d$ in the three cases $1,2$ and $3$ as
follows:

$d=\left\\{\begin{array}[]{l l l}p_{1}^{e_{1}}&\quad\text{with \ $p_{1}\equiv
1\pmod{9},$ }\\\ p_{1}^{e_{1}}&\quad\text{with \ $p_{1}\equiv 4\ \text{or}\
7\pmod{9},$ }\\\ 3^{e}p_{1}^{e_{1}}\not\equiv\pm 1\pmod{9}&\quad\text{with \
$p_{1}\equiv 1\pmod{9},$ }\\\ 3^{e}p_{1}^{e_{1}}\not\equiv\pm
1\pmod{9}&\quad\text{with \ $p_{1}\equiv 4\ \text{or}\ 7\pmod{9},$ }\\\
p_{1}^{e_{1}}q_{1}^{f_{1}}\equiv\pm 1\pmod{9}&\quad\text{with \ $p_{1}\equiv-
q_{1}\equiv 1\pmod{9},$ }\\\ \end{array}\right.$

where $e,e_{1},f_{1}\in\\{1,2\\}$.

Our next goal is to show that the only possible form of the integer $d$ is the
first form $d=p_{1}^{e_{1}}$, where $p_{1}\equiv 1\pmod{9}$ and $e_{1}=1$ or
$2$.
$\bullet$ Case where $d=p_{1}^{e_{1}}$, with $p_{1}\equiv 4\ \text{or}\
7\pmod{9}$:

  * $-$

If $\left(\frac{3}{p}\right)_{3}\neq 1$, then according to [3, § 1, Conjecture
1.1, p. 1], we have $C_{\mathrm{k},3}\simeq\mathbb{Z}/3\mathbb{Z}$, which
contradict the fact that $C_{\mathrm{k},3}$ is of type $(9,3)$. We note that
in this case, the fields $\Gamma$ and $\mathrm{k}$ are of Type III,
respectively, $\alpha$, in the terminology of [5, § 15, Theorem 15.6, pp.
235-236], respectively, [1, § 2.1, Theorem 2.1, p. 4].

  * $-$

If $\left(\frac{3}{p}\right)_{3}=1$, then according to [3, § 1, Conjecture
1.1, p. 1], we have
$C_{\mathrm{k},3}\simeq\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$,
which is impossible. We note that in this case, the fields $\Gamma$ and
$\mathrm{k}$ are of Type I, respectively, $\beta$, in the terminology of [5, §
15, Theorem 15.6, pp. 235-236], respectively, [1, § 2.1, Theorem 2.1, p. 4].
For this case, we see that in [5, § 17, Numerical Data, p. 238], and also in
the tables of [6] which give the class number of a pure cubic field, the prime
numbers $p=61,67,103,$ and $151$, which are all congruous to $4$ or $7\
\pmod{9}$, verify the following properties:

    * (i)

$3$ is a residue cubic modulo $p$;

    * (ii)

$3$ divide exactly the class number of $\Gamma$;

    * (iii)

$u=3$;

    * (iv)

$C_{\Gamma,3}\simeq\mathbb{Z}/3\mathbb{Z}$, and
$C_{\mathrm{k},3}\simeq\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}.$

$\bullet$ Case where $d=3^{e}p_{1}^{e_{1}}\not\equiv\pm 1\pmod{9}$, with
$p_{1}\equiv 1\pmod{9}$: Here $e,e_{1}\in\\{1,2\\}$. As
$\mathbb{Q}(\sqrt[3]{ab^{2}})=\mathbb{Q}(\sqrt[3]{a^{2}b})$, we can choose
$e_{1}=1$, i.e. $d=3^{e}p_{1}$ with $e\in\\{1,2\\}$. On the one hand, the fact
that $p_{1}\equiv 1\pmod{3}$ implies by [13, Chap. 9, Sec. 1, Proposition
9.1.4, p.110] that $p_{1}=\pi_{1}\pi_{2}$ with $\pi_{1}^{\tau}=\pi_{2}$ and
$\pi_{1}\equiv\pi_{2}\equiv 1\pmod{3\mathcal{O}_{\mathrm{k}_{0}}}$, the prime
$p_{1}$ is totally ramified in $\Gamma$, then $\pi_{1}$ and $\pi_{2}$ are
totally ramified in $\mathrm{k}$ and we have
$\pi_{1}\mathcal{O}_{\mathrm{k}}=\mathcal{P}_{1}^{3}$ and
$\pi_{2}\mathcal{O}_{\mathrm{k}}=\mathcal{P}_{2}^{3}$, where
$\mathcal{P}_{1},\mathcal{P}_{2}$ are two prime ideals of $\mathrm{k}$.

We know that
$3\mathcal{O}_{\mathrm{k}_{0}}=\lambda^{2}\mathcal{O}_{\mathrm{k}_{0}}$, with
$\lambda=1-\zeta_{3}$ a prime element of $\mathrm{k}_{0}$. Since
$d\not\equiv\pm 1\pmod{9}$, then $3$ is totally ramified in $\Gamma$, and then
$\lambda$ is ramified in $\mathrm{k}/\mathrm{k}_{0}$. Hence, the number of
ideals which are ramified in $\mathrm{k}/\mathrm{k}_{0}$ is $t=3.$

On the other hand, from [13, Chap. 9, Sec. 1, Proposition 9.1.4, p.110] we
have $3=-\zeta_{3}^{2}\lambda^{2}$, then
$\mathrm{k}=\mathrm{k}_{0}(\sqrt[3]{x})$ with
$x=\zeta_{3}^{2}\lambda^{2}\pi_{1}\pi_{2}$. The primes $\pi_{1}$ and $\pi_{2}$
are congruent to $1\pmod{\lambda^{3}}$ because $p_{1}\equiv 1\pmod{9}$, then
according to [1, § 3, Lemma 3.3, p. 17] we have $\zeta_{3}$ is a norm of an
element of $\mathrm{k}\setminus\\{0\\}$, so $q^{*}=1$. We conclude according
to [10, § 5, p. 92] that $\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=2$
which is impossible.
$\bullet$ Case where $d=3^{e}p_{1}^{e_{1}}\not\equiv\pm 1\pmod{9}$, with
$p_{1}\equiv 4\ \text{ or }\ 7\pmod{9}$:
Here $e,e_{1}\in\\{1,2\\}$. So we can choose $e=1$, then $d=3p_{1}^{e_{1}}$
with $e_{1}\in\\{1,2\\}$. As above we get $\pi_{1}$ and $\pi_{2}$, and
$\lambda$ are ramified in $\mathrm{k}/\mathrm{k}_{0}$.
Put $p\mathcal{O}_{\Gamma}=\mathcal{P}^{3}$,
$\pi_{1}\mathcal{O}_{\mathrm{k}}=\mathcal{P}_{1}^{3}$,
$\pi_{2}\mathcal{O}_{\mathrm{k}}=\mathcal{P}_{2}^{3}$ and
$\lambda\mathcal{O}_{\mathrm{k}}=I^{3}$, where $\mathcal{P}$ is a prime ideal
of $\Gamma$, and $\mathcal{P}_{1},\mathcal{P}_{2}$ and $I$ are prime ideals of
$\mathrm{k}$. According to [14, § 3.2, Theorem 3.5, pp 36-39] we get $C_{k,3}$
is cyclic of order $3$ which contradict the fact that $C_{k,3}$ is of type
$(9,3)$.
$\bullet$ Case where $d=p_{1}^{e_{1}}q_{1}^{f_{1}}\equiv\pm 1\pmod{9}$, with
$p_{1}\equiv-q_{1}\equiv 1\pmod{9}$ : Since $d\equiv\pm 1\pmod{9}$, then
according to [8, Sec. 4, pp. 51-54] we have $3$ is not ramified in the field
$\Gamma$, so $\lambda=1-\zeta_{3}$ is not ramified in
$\mathrm{k}/\mathrm{k}_{0}$. As $p_{1}\equiv 1\pmod{3}$, then by [13, Chap. 9,
Sec. 1, Proposition 9.1.4, p.110] $p_{1}=\pi_{1}\pi_{2}$ with
$\pi_{1}^{\tau}=\pi_{2}$ and $\pi_{1}\equiv\pi_{2}\equiv
1\pmod{3\mathcal{O}_{\mathrm{k}_{0}}}$, so $\pi_{1}$ and $\pi_{2}$ are totally
ramified in $\mathrm{k}$. Since $q_{1}\equiv-1\pmod{3}$, then $q_{1}$ is inert
in $\mathrm{k}_{0}$. Hence, the primes ramifies in $\mathrm{k}/\mathrm{k}_{0}$
are $\pi_{1},\pi_{2}$ and $q_{1}$. Put
$x=\pi_{1}^{e_{1}}\pi_{2}^{e_{1}}\pi^{f_{1}}$, where $-q_{1}=\pi$ is a prime
number of $\mathrm{k}_{0}$, then we have
$\mathrm{k}=\mathrm{k}_{0}(\sqrt[3]{x})$. The fact that $p_{1}\equiv-
q_{1}\equiv 1\pmod{9}$ imply that $\pi_{1}\equiv\pi_{2}\equiv\pi\equiv
1\pmod{\lambda^{3}}$, then by [1, § 3, Lemma 3.3, p. 17], $\zeta_{3}$ is a
norm of an element of $\mathrm{k}\setminus\\{0\\}$, so $q^{*}=1$. Then by [10,
§ 5, p. 92] we get
$\operatorname{rank}\,C_{\mathrm{k},3}^{(\sigma)}=t-2+q^{*}=2$ which is
impossible.

Finally, we have shown that if the $3$-class group $C_{\mathrm{k},3}$ is of
type $(9,3)$, then $d=p^{e}$, where $p$ is a prime number such that $p\equiv
1\pmod{9}$ and $e=1$ or $2$. We can see that this result is compatible with
the first form of the integer $d$ in [1, § 1, Theorem 1.1, p. 2].
(2) Suppose that $d=p^{e}$, with $p$ is a prime number congruent to
$1\pmod{9}$. Here $e\in\\{1,2\\}$, since
$\mathbb{Q}(\sqrt[3]{p})=\mathbb{Q}(\sqrt[3]{p^{2}})$ we can choose $e=1$.
From [5, § 14, Theorem $14.1$, p. 232], we have
$h_{\mathrm{k}}=\frac{u}{3}\cdot h_{\Gamma}^{2}$, the fact that $u=1$ and that
$h_{\Gamma}$ is exactly divisible by $9$ implies that $h_{\mathrm{k}}$ is
exactly divisible by $27$.
Since $h_{\Gamma}$ is exactly divisible by $9$, then by [7, Lemma $5.11$] we
have $\operatorname{rank}\,C_{\mathrm{k},3}=2$. We conclude that
$C_{\mathrm{k},3}\simeq\mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$.

##  3 Numerical Examples

Let $\Gamma=\mathbb{Q}(\sqrt[3]{ab^{2}})$ be a pure cubic field, where $a$ and
$b$ are coprime square-free integers. We point out that
$\mathbb{Q}(\sqrt[3]{ab^{2}})=\mathbb{Q}(\sqrt[3]{a^{2}b})$. Assume that
$\mathrm{k}=\mathbb{Q}(\sqrt[3]{ab^{2}},\zeta_{3})$ and $C_{\mathrm{k},3}$ is
of type $(9,3)$. Using the system Pari [17], we compute class groups for $b=1$
and a prime $a=p\equiv 1\pmod{9}$. The following table illustrates our main
result Theorem 1.1 and Lemma 2.3. Here we denote by:

$\displaystyle C_{\Gamma,3}\ (\text{respectively,}\ C_{\mathrm{k},3})$
$\displaystyle:$ $\displaystyle\text{the 3-class group of}\ \Gamma\
(\text{respectively,}\ \mathrm{k});$ $\displaystyle h_{\Gamma,3}\
(\text{respectively,}\ h_{\mathrm{k},3})$ $\displaystyle:$
$\displaystyle\text{the 3-class number of}\ \Gamma\ (\text{respectively,}\
\mathrm{k}).$

Table : Some fields $\mathbb{Q}(\sqrt[3]{p},\zeta_{3})$ whose $3$-class group
is of type $(9,3)$.

$p$ | $p^{2}$ | $p\pmod{9}$ | $h_{\Gamma,3}$ | $h_{\mathrm{k},3}$ | $u$ | $C_{\Gamma,3}$ | $C_{\mathrm{k},3}$
---|---|---|---|---|---|---|---
$199$ | $39601$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$271$ | $73441$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$487$ | $237169$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$523$ | $273529$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$1297$ | $1682209$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$1621$ | $2627641$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$1693$ | $2866249$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$1747$ | $3052009$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$1999$ | $3996001$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$2017$ | $4068289$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$2143$ | $4592449$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$2377$ | $5650129$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$2467$ | $6086089$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$2593$ | $6723649$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$2917$ | $8508889$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$3511$ | $12327121$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$3673$ | $13490929$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$3727$ | $13890529$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$3907$ | $15264649$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$4159$ | $17297281$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$4519$ | $20421361$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$4591$ | $21077281$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$4789$ | $22934521$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$4933$ | $24334489$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$4951$ | $24512401$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$5059$ | $25593481$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$5077$ | $25775929$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$
$5347$ | $28590409$ | $1$ | $9$ | $27$ | $1$ | $[9]$ | $[9,3]$

######  Remark 3.1.

Let $p$ is a prime number such that $p\equiv 1\pmod{9}$. Let
$\Gamma=\mathbb{Q}(\sqrt[3]{p})$,
$\mathrm{k}=\mathbb{Q}(\sqrt[3]{p},\zeta_{3})$ be the normal closure of the
pure cubic field $\Gamma$ and $C_{\mathrm{k},3}$ be the $3$-part of the class
group of $\mathrm{k}$. If $9$ divide exactly the class number of
$\mathbb{Q}(\sqrt[3]{p})$ and $u=1$, then according to Theorem 1.1, the
$3$-class group of $\mathbb{Q}(\sqrt[3]{p},\zeta_{3})$ is of type $(9,3).$
Furthermore, if $3$ is not residue cubic modulo $p$, then a generators of
$3$-class group of $\mathbb{Q}(\sqrt[3]{p},\zeta_{3})$ can be deduced by [2, §
3, Theorem 3.2, p. 10].

##  4 Acknowledgements

The authors would like to thank Professors Daniel C. Mayer and Mohammed Talbi
who were of a great help concerning correcting the spelling mistakes that gave
more value to the work.

## References

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> Abdelmalek AZIZI, Moulay Chrif ISMAILI and Siham AOUISSI
> Department of Mathematics and Computer Sciences,
> Mohammed first University,
> 60000 Oujda - Morocco,
><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
>
>
> Mohamed TALBI
> Regional Center of Professions of Education and Training,
> 60000 Oujda - Morocco,
><EMAIL_ADDRESS>
>