 The generators of $3$-class group of some fields of degree $6$ over $\mathbb{Q}$ 



**Abstract:** $\\ $ Let $\mathrm{k}=\mathbb{Q}\left(\sqrt[3]{p},\zeta_3\right)$, where $p$ is a prime number such that $p \equiv 1 \pmod  9$, and let $C_{\mathrm{k},3}$ be the $3$-component of the class group of $\mathrm{k}$. In @GERTH3, Frank Gerth III proves a conjecture made by Calegari and Emerton @Cal-Emer which gives necessary and sufficient conditions for $C_{\mathrm{k},3}$ to be of $\operatorname{rank}\,$ two. The purpose of the present work is to determine generators of $C_{\mathrm{k},3}$, whenever it is isomorphic to $\mathbb{Z}/9\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.

<span>**Keywords:**</span> <span>Pure cubic fields, $3$-class groups, generators.\
AMSC: 11R11, 11R16, 11R20, 11R27, 11R29, 11R37.</span>

Introduction {#sec1}
============

Let $\Gamma=\mathbb{Q}(\sqrt[3]{p})$ be a pure cubic field, where $p$ is a prime number such that $p \equiv 1  \pmod  9$. We denote by $\zeta_3=-1/2+i\sqrt{3}/2$ the normalized primitive third roots of unity, $\mathrm{k}=\mathbb{Q}(\sqrt[3]{p},\zeta_3)$ the normal closure of $\Gamma$ and $C_{\mathrm{k},3}$ the $3$-component of the class group of $\mathrm{k}$.\
Assuming $9$ divides exactly the $3$-class number of $\Gamma$. Then $C_{\mathrm{k},3} \simeq \mathbb{Z}/9\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ if and only if $u=1$, where $u$ is an index of units that will be defined in the notations below. In this paper, we will determine the generators of $C_{\mathrm{k},3}$ when $C_{\mathrm{k},3}$ is of type $(9,3)$ and $3$ is not a cubic residue modulo $p$. We spot that Calegari and Emerton (@Cal-Emer [Lemma 5.11]) proved that the $\operatorname{rank}\,$ of the $3$-class group of $\mathbb{Q}(\sqrt[3]{p},\zeta_3)$, with $p \equiv 1  \pmod  9$, is equal to two if $9$ divides the $3$-class number of $\mathbb{Q}(\sqrt[3]{p})$. Moreover, in his work @GERTH3 [Theorem 1, p.471], Frank Gerth III proves that the converse to Calegari-Emerton's result is also true. The present work can be viewed as a continuation of the works @Cal-Emer and @GERTH3 .\
After reviewing some basic properties of the norm residue symbols and prime factorization in the normal closure of a pure cubic field that will be needed later, we will establish in section \[Sect3\] some preliminary results of the $3$-class group $C_{\mathrm{k},3}$. Using this, we arrive to determine the generators of $3$-class groups $C_{\mathrm{k},3}$ of type $(9,3)$. All the study cases are illustrated by numerical examples and summarized in tables in section \[sect4\]. The usual notations on which the work is based is as follows:

-   $\Gamma= \mathbb{Q}(\sqrt[3]{d})$ : a pure cubic field, where $d$ is a cube-free natural number;

-   $\mathrm{k}_0= \mathbb{Q}(\zeta_3)$ : the third cyclotomic field ;

-   $\mathrm{k}=\mathbb{Q}(\sqrt[3]{d},\zeta_3)$ : the normal closure of the pure cubic field $\Gamma$;

-   $u=[E_{\mathrm{k}}: E_0]$ : the index of the sub-group $E_0$ generated by the units of intermediate fields of the extension $\mathrm{k}/\mathbb{Q}$ in $E_{k}$ the group of units of $\mathrm{k}$;

-   $\lambda= 1-\zeta_3$ prime integer of $\mathrm{k}_0$;

-   $\langle\tau \rangle=\operatorname{Gal}\left(\mathrm{k}/\Gamma\right)$, $\tau^2=id, \tau(\zeta_3)=\zeta_3^2$ and $\tau(\sqrt[3]{d})=\sqrt[3]{d}$;

-   $ \langle \sigma \rangle=\operatorname{Gal}\left(\mathrm{k}/\mathbb{Q}(\zeta_3)\right)$, $\sigma^3=id, \sigma(\zeta_3)=\zeta_3$ and $\sigma(\sqrt[3]{d})=\zeta_3\sqrt[3]{d}$;

-   For an algebraic number field $L$:

    -   $\mathcal{O}_{L}$ : the ring of integers of $L$;

    -   $E_{L}$ : the group of units of $L$;

    -   $\mathcal{D}_{L}$ : the discriminant of $L$;

    -   $h_{L}$ : the class number of $L$;

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

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

    -   $L_3^{(1)}$ : the Hilbert $3$-class field of $L$;

    -   $[\mathcal{I}]$ : the class of a fractional ideal $\mathcal{I}$ in the class group of $L$;

-   $\left(\dfrac{c}{p}\right)_3 = 1 \Leftrightarrow X^3\equiv c (mod~p)$ resolved on $\mathbb{Z}
    \Leftrightarrow c^{(p-1)/3} \equiv 1 \pmod p$, where $c \in \mathbb{Z}$ and $p$ is a prime number congruent to $1 (mod~3)$.

Norm residue symbol and ideal factorization theory {#sec2}
==================================================

The norm residue symbol
-----------------------

$\\ $ Let $L/K$ an abelian extension of number fields with conductor $f$. For each finite or infinite prime ideal $\mathcal{P}$ of $\mathrm{K}$, we note by $f_{\mathcal{P}}$ the largest power of $\mathcal{P}$ that divides $f$. Let $a \in K^{*}$, we determine an auxiliary number $a_0$ by the two conditions $a_0 \equiv a ~ (mod~ f_{\mathcal{P}})$ and $a_0 \equiv 1 ~ (mod~ \frac{f}{f_{\mathcal{P}}})$. Let $\mathcal{Q}$ an ideal co-prime with $\mathcal{P}$ such that $(a_0)=\mathcal{P}^{e}\mathcal{Q}$ ($b=0$ if $\mathcal{P}$ is infinite). We note by $$\left(\frac{a,
L}{\mathcal{P}}\right)=\left(\frac{L/K}{\mathcal{Q}}\right)$$ the Artin map in $L/K$ applied to $\mathcal{Q}$.

Let $K$ be a number field containing the $l^{th}$-roots of units, where $l\in \mathbb{N}$, then for each $a,b \in K^{*}$ and prime ideal $\mathcal{P}$ of $K$, we define the ***norm residue symbol*** by:

$$\left(\dfrac{a,
b}{\mathcal{P}}\right)_{l}=\dfrac{\left(\dfrac{a, K(\sqrt[l]{b})}{\mathcal{P}}\right) \sqrt[l]{b} }{\sqrt[l]{b}}.$$ Therefore, if the prime ideal $\mathcal{P}$ is unramified in the field $ K(\sqrt[l]{b})$, then we write $$\left(\dfrac{
b}{\mathcal{P}}\right)_{l}=\dfrac{\left(\dfrac{ K(\sqrt[l]{b})}{\mathcal{P}}\right) \sqrt[l]{b} }{\sqrt[l]{b}}.$$

Notice that $ \left(\dfrac{a,b}{\mathcal{P}}\right)_{l} $ and $\left(\dfrac{b}{\mathcal{P}}\right)_{l}$ are two $l^{th}$-roots of units.

Following @Hass2, the principal properties of the norm residue symbol are given as follows:

Properties {#properties .unnumbered}
----------

1.  The product formula:

    -   $ \left(\dfrac{a_1 a_2,b}{\mathcal{P}}\right)_{l} = \left(\dfrac{a_1,b}{\mathcal{P}}\right)_{l} \left(\dfrac{a_2,b}{\mathcal{P}}\right)_{l}$;

    -   $ \left(\dfrac{a,b_1b_2}{\mathcal{P}}\right)_{l} = \left(\dfrac{a,b_1}{\mathcal{P}}\right)_{l} \left(\dfrac{a,b_2}{\mathcal{P}}\right)_{l}$;

2.  The inverse formula: $ \left(\dfrac{a,b}{\mathcal{P}}\right)_{l}= \left(\dfrac{b,a}{\mathcal{P}}\right)_{l}^{-1};$

3.  $\left(\dfrac{a,b}{\mathcal{P}}\right)_{l}=1 \Leftrightarrow $ $a$ is norm residue of $K(\sqrt[l]{b})$ modulo $f_{b}$;

4.  $ \left(\dfrac{\sigma a,\sigma b}{\sigma\mathcal{P}}\right)_{l}=\sigma \left(\dfrac{a,b}{\mathcal{P}}\right)_{l},$ for each automorphism $\sigma$ of $K$;

5.  If $\mathcal{P} $ is not divisible by the conductor $f_{b}$ of $K(\sqrt[l]{b})$ and appears in $(a)$ with the exponent $e$, then:

    -   $ \left(\dfrac{a,b}{\mathcal{P}}\right)_{l}= \left(\dfrac{b}{\mathcal{P}}\right)_{l}^{-e};$

    -   $\mathcal{P}$ is infinite $(e=0)$ $\Rightarrow \left(\dfrac{a,b}{\mathcal{P}}\right)_{l}=1;$

6.  The classical reciprocity law: let $a,b\in K^{*}$, and the conductors $f_{a}$ and $f_{b}$ of respectively $K(\sqrt[l]{a})$ and $K(\sqrt[l]{b})$ are co-prime, then: $$\left(\dfrac{a}{(b)}\right)_{l} =\left(\dfrac{b}{(a)}\right)_{l};$$

7.  $ \prod_{\mathcal{P}} \left(\dfrac{a,b}{\mathcal{P}}\right)_{l}=1$, where the product is taken on the finite and infinite prime ideals;

8.  Let $L$ is a finite extension of $K$, $a \in L$ and $b \in K^{*}$, then: $$\prod_{\mathfrak{P}|\mathcal{P}} \left(\dfrac{a,b}{\mathfrak{P}}\right)_{l}=  \left(\dfrac{\mathcal{N}_{L/K}(a),b}{\mathcal{P}}\right)_{l}.$$

\[RqMes\] From property $(3)$, we have: $$a \text{ is a norm  in} \  K(\sqrt[l]{b}) \Rightarrow \left(\dfrac{a,b}{\mathcal{P}}\right)_{l}=1,$$ for each prime ideal $\mathcal{P}$ of $K$.

For more basic properties of the norm residue symbol in the number fields, we refer the reader to the papers @DED, @Hass1 and @Hass2. Notice that in section \[Sect3\], we will use the norm cubic residue symbols $(l=3)$. As the ring of integer $ \mathcal{O}_{\mathrm{k}_0}$ is principal, $h_{\mathrm{k}_0} = 1,$ we will write the norm cubic residue symbol as follows:

$\left(\dfrac{a, b}{(\pi)}\right)_3 = \left(\dfrac{a, b}{\pi}\right)_3 $ and $\left(\dfrac{a}{(\pi)}\right)_3 = \left(\dfrac{a}{\pi}\right)_3$

where $a, b \in \mathrm{k}_0^*$ and $\pi$ is a prime integer of $\mathcal{O}_{\mathrm{k}_0}.$

Prime factorization in a pure cubic field and in its normal closure {#subfactor}
-------------------------------------------------------------------

$\\ $ Let be $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ a pure cubic field, and $\mathcal{O}_{\Gamma}$ the ring of integers of $\Gamma$. We write the natural integer $d$ in form $d=ab^2,$ where $a$ and $b$ are cube-free and co-prime positive integers. In his paper @DED, Dedekind has defined two different types of pure cubic fields as follows:

\[3inG\] Using the same notations as above:

1.  We say that $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ is of the ***first kind*** if $3\mathcal{O}_{\Gamma}=\mathcal{P}^3$, where $\mathcal{P}$ is a prime ideal of $\mathcal{O}_{\Gamma}$, in this case, $a^2-b^2\not\equiv 0 \ (mod~9).$

2.  We say that $\Gamma=\mathbb{Q}(\sqrt[3]{d})$ is of the ***second kind*** if $3\mathcal{O}_{\Gamma}=\mathcal{P}^2\mathcal{P}_1$, where $\mathcal{P}\neq \mathcal{P}_1$ are two primes of $\mathcal{O}_{\Gamma}$, in this case, $a^2-b^2\equiv 0 \ (mod~9).$

Now, let $p$ be a prime number. In the following Proposition, we give the decomposition of the prime $p$ in the pure cubic field $\Gamma=\mathbb{Q}(\sqrt[3]{ab^2})$. We denote by $\mathcal{P}, \mathcal{P}_{i}$ prime ideals of $\Gamma$, and by $\mathcal{N}$ the absolute norm $\mathcal{N}_{\Gamma/\mathbb{Q}}$.

\[pGAMMA\] $\\ $ Let $p$ a prime number such that $p\neq 3$, then:

1.  If $p$ divides $ab$ and $p\neq 3$, then $p\mathcal{O}_{\Gamma} = \mathcal{P}^3 , ~ \mathcal{N}(\mathcal{P}) = p$.

2.  If $p \nmid 3ab$ and $p\equiv -1\ \ (mod\
    3)$, then $p\mathcal{O}_{\Gamma}=\mathcal{P}\mathcal{P}_1$, with $\mathcal{N}(\mathcal{P}) = p$ and $\mathcal{N}(\mathcal{P}_1) =
    p^2.$

3.  If $p \nmid 3ab$ and $p \equiv 1 \ \
    (mod\  3)$, then:

    1.  $p\mathcal{O}_{\Gamma} = \mathcal{P}\mathcal{P}_1 \mathcal{P}_2$ with $\mathcal{N}(\mathcal{P})=\mathcal{N}(\mathcal{P}_1)=\mathcal{N}(\mathcal{P}_2)$, if $ab^2$ is a cubic residue modulo $p$;

    2.  $p\mathcal{O}_{\Gamma} = \mathcal{P}$ with $\mathcal{N}(\mathcal{P})=p^3$, if $ab^2$ is not a cubic residue modulo $p$.

See @DED.

The ramification of the prime $3$ need a particular treatment, it is the purpose of the following Proposition:

\[3GAMMA\] $ ~~\\ $ The decomposition into prime factors of $3$ is: $$\begin{aligned}
   3 \mathcal{O}_{\Gamma}= \left
    \lbrace
      \begin{aligned}
        \mathcal{P}^3 , \ if \ \ a^2 \not\equiv b^2  \ \ (mod\  9),\\
        \mathcal{P}^2 \mathcal{P}_1, \ if \ \ a^2 \equiv b^2 \ \ (mod\  9).\\
      \end{aligned}
    \right.\end{aligned}$$

See @DED.

The ideal factorization rules for the $3$rd cyclotomic field $\mathrm{k_0}$ (see @Clas) is as follows:

-   $3\mathcal{O}_{\mathrm{k_0}}=\lambda^2=(1-\zeta_3)^2$;

-   $p\mathcal{O}_{\mathrm{k_0}}=\pi_1\pi_2$ in $\mathrm{k_0}$ if $p\equiv 1  \pmod  3$;

-   $q\mathcal{O}_{\mathrm{k_0}}=\mathsf{q}$ in $\mathrm{k_0}$ if $q\equiv -1  \pmod  3$.

Next, let $\mathrm{k}$ be the normal closure of $\Gamma$. We note by $\mathcal{O}_{\mathrm{k}}$ the ring of integers of $\mathrm{k}$, $\mathfrak{P}$ and $\mathfrak{P}_s$ are prime ideals of $\mathrm{k},$ $\mathcal{N}=\mathcal{N}_{k/\mathbb{Q}}$ the norm of $\mathrm{k}$ on $\mathbb{Q}.$ Combining the ideal factorization rules for $\Gamma$ with those of the field $\mathrm{k_0}$. The decomposition of the prime $  3 $ in $\mathrm{k}$ is the purpose of the following Theorem:

\[3K\] $\\ $ The prime $ 3$ decomposes in $ \emph{k} $ as follows: $$\begin{aligned}
   3 \mathcal{O}_{\mathrm{k}}= \left
    \lbrace
      \begin{aligned}
        \mathfrak{P}^6 , \ si \ \ a^2 \not\equiv b^2  \ \ (mod\  9),\\
        \mathfrak{P}_1^2 \mathfrak{P}_2^2 \mathfrak{P}_3^2, \ si \ \ a^2 \equiv b^2 \ \ (mod\  9).\\
      \end{aligned}
    \right.\end{aligned}$$

$\\ $ We have $3$ ramifies in the quadratic field $\mathrm{k}_0=\mathbb{Q}(\zeta_3)$.

-   Suppose that $\Gamma$ is the first kind, then by Proposition \[3GAMMA\] we have $3\mathcal{O}_{\Gamma}=\mathcal{P}^3$. Hence, $3\mathcal{O}_{k}=\mathfrak{P}^6$.

-   Conversely, suppose that $\Gamma$ is of second kind, then $3\mathcal{O}_{\Gamma}=\mathcal{P}^2\mathcal{P}_1$. It follows that $3\mathcal{O}_{k}=\mathfrak{P}_1^2\mathfrak{P}_2^2\mathfrak{P}_3^2$.

However, we have the following Proposition in which we characterize the decomposition of prime ideals of $p\neq 3$ in $\mathrm{k}$.

\[pK\] $\\ $ Let $p$ a prime number such that $p\neq 3$, then:

1.  If $p$ divides $\mathcal{D}_{\Gamma}$, then:

    1.  $p\mathcal{O}_{\mathrm{k}}=\mathfrak{P}_1^3\mathfrak{P}_2^3$, with $\mathcal{N}(\mathfrak{P}_1)=\mathcal{N}(\mathfrak{P}_2)=p$, if and only if $-3$ is a quadratic residue modulo $p$.

    2.  $p \mathcal{O}_{\mathrm{k}}=\mathfrak{P}^3$, with $\mathcal{N}(\mathfrak{P})=p^2$, if and only if $-3$ is not a quadratic residue modulo $p$.

2.  If $p$ does not divides $\mathcal{D}_{\Gamma}$ and $p \equiv 1~(mod~3),$ then:

    1.  $p$ decomposes completely in $\emph{k}$ if and only if $\mathcal{D}_{\Gamma}$ is a cubic residue modulo $p$.

    2.  $p \mathcal{O}_{\mathrm{k}}=\mathfrak{P}_1\mathfrak{P}_2$, with $\mathcal{N}(\mathfrak{P}_1)=\mathcal{N}(\mathfrak{P}_2)=p^3$, if and only if $\mathcal{D}_{\Gamma}$ is not a cubic residue modulo $p$.

3.  If $p$ does not divides $\mathcal{D}_{\Gamma}$ and $p \equiv - 1~(mod~3),$ then: $p \mathcal{O}_{\mathrm{k}}=\mathfrak{P}_1\mathfrak{P}_2\mathfrak{P}_3$, with $\mathcal{N}(\mathfrak{P}_1)=\mathcal{N}(\mathfrak{P}_2)=\mathcal{N}(\mathfrak{P}_3)=p^2$, if and only if $-3$ is not a quadratic residue modulo $p$.

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1.  We use Proposition \[pGAMMA\] and the decomposition of prime ideals in the quadratic fields $\mathrm{k}_0=\mathbb{Q}(\zeta_3)$.

2.  Suppose that $p$ does not divide $\mathcal{D}_{\Gamma}$ and $p\equiv 1~(mod~3)$, then $-3$ is a quadratic residue modulo $p$, then the multiplication formula gives $$\left( \dfrac{-3}{p}\right)=\left( \dfrac{-1}{p}\right)\left( \dfrac{3}{p}\right).$$ On the one hand, by the Euler's Theorem we have $$\left( \dfrac{-1}{p}\right)=(-1)^{(p-1)/2},$$ On the other hand, the quadratic reciprocity law gives $$\left( \dfrac{p}{3}\right)\left( \dfrac{3}{p}\right)=(-1)^{(p-1)/2},$$ since $p\equiv 1~(mod~3)$, then $p$ is a square modulo $3$, which gives $\left( \dfrac{p}{3}\right)=1$, so $$\left( \dfrac{3}{p}\right)=(-1)^{(p-1)/2}.$$ Then $$\left( \dfrac{-3}{p}\right)=((-1)^{(p-1)/2})^2=(-1)^{p-1}=1.$$ Thus, $p$ decomposes completely in $\mathrm{k}_0$.

    1.  If $\mathcal{D}_{\Gamma}$ is a cubic residue modulo $p$, then by Proposition \[pGAMMA\] we have $p$ split completely in $\Gamma$. Hence $p$ split completely in $\mathrm{k}$.

    2.  If $\mathcal{D}_{\Gamma}$ is not a cubic residue modulo $p$, we have $p$ remains prime in $\Gamma$. Hence $p\mathcal{O}_{\mathrm{k}}=\mathfrak{P}_1\mathfrak{P}_2$.

3.  We have $p\mathcal{O}_{\Gamma}=\mathcal{P}\mathcal{P}_1$, and $p$ remains inert in $\mathrm{k}_0$, hence the result.

In the preceding Proposition \[pK\], the situation $p\mathcal{O}_{\mathrm{k}}=\mathfrak{P}_1\mathfrak{P}_2$ is never happens because if $p\equiv -1~(mod~3),$ we have always $ \left( \dfrac{-3}{p}\right)=-1. $

**The generators of $C_{\mathrm{k},3}$**  {#Sect3}
=========================================

First, we let $C_{\mathrm{k},3}^{(\sigma)}=\lbrace \mathcal{A}\in C_{\mathrm{k},3} ~| ~ \mathcal{A}^{\sigma}=\mathcal{A} \rbrace$ be the group of ambiguous ideal classes of $\mathrm{k}/\mathrm{k}_0$, where $\sigma$ is a generator of $\operatorname{Gal}\left(\mathrm{k}/\mathrm{k}_0\right)$, and put $q^{*}=0$ or $1$ according to $\zeta_{3}$ is not norm or norm of an element of $\mathrm{k}\backslash \{0\}$. Let $t$ be the number of primes ramifies in $\mathrm{k}/\mathrm{k}_{0}$. Then according to @GERTH1, we have $$|C_{\mathrm{k},3}^{(\sigma)}|=3^{t-2+q^{*}}.$$

If we denote by $C_{\mathrm{k}_0,3}$ the Sylow $3$-subgroup of the ideal class group of $\mathrm{k}_0$, $C_{\mathrm{k}_0,3}=\lbrace 1 \rbrace$. Let be $ C_{\mathrm{k},3}^{(1-\sigma)}=
 \lbrace \mathcal{A}^{(1-\sigma)}~|~ \mathcal{A} \in C_{\mathrm{k},3} \rbrace $. By the exact sequence : $$1 \longrightarrow C_{\mathrm{k},3}^{(\sigma)}  \longrightarrow C_{\mathrm{k},3} \overset{1-\sigma}{\longrightarrow } C_{\mathrm{k},3} \longrightarrow C_{\mathrm{k},3}/C_{\mathrm{k},3}^{1-\sigma} \longrightarrow 1$$ we deduce that $$|C_{\mathrm{k},3}^{(\sigma)}|= | C_{\mathrm{k},3}/C_{\mathrm{k},3}^{1-\sigma} |.$$ The fact that $C_{\mathrm{k},3}^{(\sigma)}$ and $C_{\mathrm{k},3}/C_{\mathrm{k},3}^{1-\sigma}$ are elementary abelian $3$-groups imply that: $$\operatorname{rank}\, C_{\mathrm{k},3}^{(\sigma)}= \operatorname{rank}\,( C_{\mathrm{k},3}/C_{\mathrm{k},3}^{1-\sigma}).$$ Define the $3$-group $  C_{\mathrm{k},3}^{(1-\sigma)^{i}} $ for each $i\in \mathbb{N}$ by $$C_{\mathrm{k},3}^{(1-\sigma)^{i}}=
 \lbrace \mathcal{A}^{(1-\sigma)^{i}}~|~ \mathcal{A} \in C_{\mathrm{k},3} \rbrace,$$ and let $s$ be the positive integer such that $C_{\mathrm{k},3}^{(\sigma)}\subseteq C_{\mathrm{k},3}^{(1-\sigma)^{s-1}} $ and $C_{\mathrm{k},3}^{(\sigma)} \not \subseteq C_{\mathrm{k},3}^{(1-\sigma)^{s}} $.\
The following Proposition gives the structure of the $3$-class group $C_{\mathrm{k},3}$ when $27$ divides exactly the class number of $\mathrm{k}$:

\[39\] Let be $\Gamma$ a pure cubic field, $\mathrm{k}$ its normal closure and $u$ the index of units defined as above, then:

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

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

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-   Assume that $C_{\mathrm{k},3}\simeq \mathbb{Z}/9\mathbb{Z}\times
    \mathbb{Z}/3\mathbb{Z}$, then $h_{\mathrm{k},3}=27$. According to Theorem $14.1$ of @B-C, we have $27=\frac{u}{3}\cdot h_{\Gamma,3}^{2}$, then $u=1$ because otherwise $27$ will be a square in $\mathbb{N}$, which is a contradiction. Then $h_{\Gamma,3}^{2}=3^4$ and $h_{\Gamma,3}=9$.\
    On the other hand, by Lemma $2.1$ and Lemma $2.2$ of @GERTH2 we have $ C_{\mathrm{k},3}\simeq  C_{\Gamma,3}
     \times C_{\mathrm{k},3}^{-}$, then $ | C_{\mathrm{k},3}^{-}|=3$. Since $C_{\mathrm{k},3}$ is of type $\left(9,3\right)$, we deduce that $C_{\mathrm{k},3}^{-}$ is a cyclic $3$-group of order $3$ and $C_{\mathrm{k},3}^{+}$ is a cyclic $3$-group of order $9$. Therefore: $$u=1 \,\, \text{and}\,\,C_{\Gamma,3}\simeq \mathbb{Z}/9\mathbb{Z} .$$ Reciprocally, assume that $u=1$ and $C_{\Gamma,3}\simeq
      \mathbb{Z}/9\mathbb{Z}$. By Theorem $14.1$ of @B-C, we deduce that $|C_{\mathrm{k},3}|=
        \frac{1}{3}\cdot |C_{\Gamma,3}|^{2}
        $, then $ | C_{\mathrm{k},3}|=27$ and $ | C_{\mathrm{k},3}^{-}|=3$. Thus: $$C_{\mathrm{k},3}\simeq C_{\Gamma,3}
         \times C_{\mathrm{k},3}^{-} \simeq \mathbb{Z}/9\mathbb{Z}\times
    \mathbb{Z}/3\mathbb{Z}.$$

-   We have the same proof as above.

\[11\] Let $\mathrm{k}=\mathbb{Q}(\sqrt[3]{p}, \zeta_3)$, where $p$ is a prime number such that $p\equiv 1~(\bmod~3)$. Let $C_{\mathrm{k},3}^{(\sigma)}$ be the ambiguous ideal class group of $\mathrm{k}/\mathbb{Q}(\zeta_3) $, where $\sigma$ is a generator of $\operatorname{Gal}\left(\mathrm{k}/\mathbb{Q}(\zeta_3)\right)$. Then $\vert C_{\mathrm{k},3}^{(\sigma)}\vert =3$.

Since $p\equiv 1  \pmod  3$, then according to section \[subfactor\], we have $p=\pi_1\pi_2$, where $\pi_1$ and $\pi_2$ are two primes of $\mathrm{k}_{0}$ such that $\pi_2=\pi_1^{\tau}$ and $\pi_1 \equiv \pi_2 \equiv 1 ~(\bmod~3\mathcal{O}_{\mathrm{k}_0})$. We study all cases depending on the congruence class of $p$ modulo $9$, then:

-   If $p \equiv 4\ \text{or} \ 7  \pmod  9$, then according to section \[subfactor\], the prime $3$ is ramified in the field $L$, so the prime ideal $(1-\zeta_3)$ is ramified in $\mathrm{k}/\mathrm{k}_0$. Also $\pi_1$ and $\pi_2$ are totally ramified in $\mathrm{k}$. So $t=3$. In addition, the fact that $p \equiv 4\ \text{or} \ 7  \pmod  9$ imply that $\pi_{i}\not\equiv 1  \pmod { (1-\zeta_3)^3 }$ for $i= \lbrace 1, 2\rbrace$, then according to section $5$ of @GERTH1 we obtain $$\left(\frac{\zeta_3,p}{p} \right)_3 \neq 1$$ where the symbol $(\dfrac{ \ , \ }{})_3$ is the cubic Hilbert symbol. We deduce that $\zeta_3$ is not a norm in the extension $\mathrm{k} / \mathrm{k}_{0} $, so $q^*=0$. Hence $\operatorname{rank}\,   C_{\mathrm{k},3}^{(\sigma)}=1$ and then $\vert
    C_{\mathrm{k},3}^{(\sigma)}\vert= 3 $.

-   If $p \equiv 1  \pmod  9$, the prime ideals which ramified in $\mathrm{k}
    / \mathrm{k}_0$ are $\pi_{1}$ and $\pi_{2}$, so $t=2$. Moreover, $\pi_{1} \equiv \pi_2 \equiv 1  \pmod  {(1-\zeta_3)^3} $, then according to @GERTH1, the cubic Hilbert symbol: $$\left(\frac{\zeta_3,p}{ \pi_1}\right)_{3}=\left(\frac{\zeta_3,p}{ \pi_2}\right)_{3}=1,$$ We conclude that $\zeta_3$ is a norm in the extension $\mathrm{k} / \mathrm{k}_{0} $, then $q^*=1$, so $\operatorname{rank}\,   
    C_{\mathrm{k},3}^{(\sigma)}=1$ and $\vert
    C_{\mathrm{k},3}^{(\sigma)}\vert= 3 $.

The basic result for determining the generators of the $3$-class group of $\mathrm{k}=\mathbb{Q}(\sqrt[3]{p},\zeta_3)$ when the $3$-class number of $\mathrm{k}$ is divisible by $27$ exactly, where $p$ is a prime number such that $p \equiv 1 ~(mod~9)$, is summarized in the following Theorem:

\[Cara39\] Let $\Gamma= \mathbb{Q}(\sqrt[3]{p})$, where $p$ is a prime number such that $p \equiv 1 ~(mod~9)$, $\mathrm{k}=\mathbb{Q}(\sqrt[3]{p},\zeta_3)$ its normal closure and $C_{\mathrm{k},3}$ the $3$-class group of $\mathrm{k}$. Assuming $9$ divides the $3$-class number of $\Gamma$ exactly, then:\
The $3$-class group $C_{\mathrm{k},3}$ is isomorphic to $ \mathbb{Z}/9\mathbb{Z}\times
\mathbb{Z}/3\mathbb{Z}$ if and only if $u=1$.

-   By Proposition \[39\], it is clear that if $C_{\mathrm{k},3}$ is isomorphic to $ \mathbb{Z}/9\mathbb{Z}\times
    \mathbb{Z}/3\mathbb{Z}$ then $u=1$.

-   Assume that $u=1$, then according to Theorem $14.1$ of @B-C, $h_{\mathrm{k},3}=27$. Since $9$ divides the $3$-class number of $\Gamma$, then by Lemma $5.11$ of @Cal-Emer we have $\operatorname{rank}\, C_{\mathrm{k},3}=2$. Hence $C_{\mathrm{k},3} \simeq \mathbb{Z}/9\mathbb{Z}\times
    \mathbb{Z}/3\mathbb{Z}$.

\[prop:carrr39\] Let $\Gamma= \mathbb{Q}(\sqrt[3]{p})$, where $p$ is a prime number such that $p \equiv 1 ~(mod~9)$, $\mathrm{k}=\mathbb{Q}(\sqrt[3]{p},\zeta_3)$ it's normal closure, and $C_{\mathrm{k},3}$ be the $3$-class group of $\mathrm{k}$. Assume that $9$ divides the $3$-class number of $\Gamma$ exactly and $u=1$. Put $\langle \mathcal{A} \rangle=C_{\mathrm{k},3}^{+}$, where $\mathcal{A} \in C_{\mathrm{k},3}$ such that $\mathcal{A}^{9}=1$ and $\mathcal{A}^{3} \neq 1$. Let $C_{\mathrm{k},3}^{(\sigma)}$ be the $3$-group of ambiguous ideal classes of $\mathrm{k}/\mathrm{k}_0$ and $C_{\mathrm{k},3}^{1-\sigma}=\{ \mathcal{A}^{1-\sigma}\, \vert\,
\mathcal{A} \in C_{\mathrm{k},3} \}$ be the principal genus of $C_{\mathrm{k},3}$. Then:

1.  $C_{\mathrm{k},3}^{(\sigma)}$ is a subgroup of $C_{\mathrm{k},3}^{+}$, $\mathcal{A} \not\in C_{\mathrm{k},3}^{(\sigma)} $ and $C_{\mathrm{k},3}^{(\sigma)}=\langle\mathcal{A}^{3}\rangle=\langle\mathcal{B}^{1-\sigma}\rangle$, where $ \mathcal{B} \in C_{\mathrm{k},3}$ such that $C_{\mathrm{k},3}^{-}=\langle\mathcal{B}\rangle$.

2.  $C_{\mathrm{k},3}^{-}=\langle(\mathcal{A}^{2})^{\sigma-1}\rangle$, and we have $C_{\mathrm{k},3}^{1-\sigma}= C_{\mathrm{k},3}^{-}\times
    C_{\mathrm{k},3}^{(\sigma)} $ is a $3$-group of type $(3,3)$,

where $C_{\mathrm{k},3}^{+}$ and $C_{\mathrm{k},3}^{-}$ are defined in Lemma $2.1$ of @GERTH2.

1.  Since $9$ divides the $3$-class number of $\Gamma$ exactly and $u=1$, then according to Theorem \[Cara39\], $C_{\mathrm{k},3}$ is of type $(9,3)$, this implies by @GERTH3 that the integer $s$ defined above is equal $3$, and according to Case $4$ of @GERTH3, we conclude that$\vert \left(C_{\mathrm{k},3}^{(\sigma)}
      \right)^{+}\vert =3$ and $\vert \left(C_{\mathrm{k},3}^{(\sigma)}
      \right)^{-}\vert= 1$, this implies that $C_{\mathrm{k},3}^{(\sigma)}$ is a subgroup of $C_{\mathrm{k},3}^{+}$. Therefore, $\langle\mathcal{A}^{3}\rangle$ is the unique subgroup of order $3$ of $C_{\mathrm{k},3}^{+}$ and $C_{\mathrm{k},3}^{(\sigma)}$ is cyclic of order $3$, then $C_{\mathrm{k},3}^{(\sigma)}=\langle\mathcal{A}^{3}\rangle$.\
    Moreover, if $C_{\mathrm{k},3}^{-}=\langle\mathcal{B}\rangle$ where $ \mathcal{B} \in C_{\mathrm{k},3}$, then $\mathcal{B} \not\in
    C_{\mathrm{k},3}^{(\sigma)} $, so $\mathcal{B}^{\sigma} \neq \mathcal{B}$. Furthermore, $\mathcal{B}^{\sigma} \neq \mathcal{B}^2$ because otherwise we will have $\mathcal{B}^{\sigma^2}=(\mathcal{B}^2)^{\sigma}=(\mathcal{B}^{\sigma})^2=\mathcal{B}^{4}$, as $ \mathcal{B} \in C_{\mathrm{k},3}^{-}$, then $\mathcal{B}^3=1 $. Therefore, $\mathcal{B}^{\sigma^2}=\mathcal{B}$, so $\mathcal{B}^{\sigma^3}=\mathcal{B}^{\sigma}$, since $\sigma^3=1$, then $\mathcal{B}^{\sigma}=\mathcal{B}$. This is impossible because $\mathcal{B}^{\sigma}\neq \mathcal{B}$. As $\mathcal{B}^{3}=1$ and $\mathcal{B}^{1+\sigma+\sigma^2}=1$, then $\mathcal{B}^{\sigma^2}=\mathcal{B}^{2+2\sigma}$. This equality makes it possible to show that $\mathcal{B}^{1-\sigma}$ is an ambiguous class. We conclude that $C_{\mathrm{k},3}^{(\sigma)}=\langle \mathcal{B}^{1-\sigma}\rangle$.

2.  We reason as in the assertion $1$. Since $\mathcal{A}^{2}\not\in C_{\mathrm{k},3}^{-} $, we deduce that $C_{\mathrm{k},3}^{-}=\langle(\mathcal{A}^{2})^{1-\sigma}\rangle$. then $C_{\mathrm{k},3}^{-}$ and $C_{\mathrm{k},3}^{(\sigma)}$ are contained in $C_{\mathrm{k},3}^{1-\sigma}$ which is of order $9$, because $ | C_{\mathrm{k},3}|=27 $ and $ | C_{\mathrm{k},3}^{(\sigma)}|=3$. Consequently, $$C_{\mathrm{k},3}^{1-\sigma}=
    C_{\mathrm{k},3}^{(\sigma)}\times
    C_{\mathrm{k},3}^{-}=\langle\mathcal{A}^{3},\mathcal{B}\rangle.$$

Our principal result can be stated as follows:

\[thm:main\] Let $\mathrm{k}=\mathbb{Q}(\sqrt[3]{ p}, \zeta_3 )$, where $p$ is a prime number such that $p \equiv 1  \pmod  9$. The prime $ 3$ decomposes in $ \emph{k} $ as $3\mathcal{O}_{\mathrm{k}}=
\mathcal{P}^{2}\mathcal{Q}^{2}\mathcal{R}^{2}$, where $\mathcal{P},
\,\,\mathcal{Q}$ and $\mathcal{R}$ are prime ideals of $\mathrm{k}$ . Put $h=\frac{h_{\mathrm{k}}}{27}$, where $h_{\mathrm{k}}$ is the class number of $\mathrm{k}$. Assume that $9$ divides exactly the $3$-class number of $\mathbb{Q}(\sqrt[3]{ p})$ and $u=1$. If $3$ is not a cubic residue modulo $p$, then:

1.  The class $[\mathcal{R}^{h}]$ generate $C_{\mathrm{k},3}^{+}$;

2.  The $3$-class group $C_{\mathrm{k},3}$ is generated by classes $[\mathcal{R}^{h}]$ and $[\mathcal{R}^{h}][\mathcal{P}^{h}]^2$, and we have: $$C_{\mathrm{k},3}=\langle[\mathcal{R}^{h}]\rangle\times\langle[\mathcal{R}^{h}][\mathcal{P}^{h}]^2\rangle=\langle
    [\mathcal{R}^{h}],[\mathcal{P}^{h}]^2\rangle.$$

In Appendix of this paper, we illustrated this results by the numerical examples with the aid of *Pari* programming @PARI and summarized in some tables in section \[Exp\].

$\\ $ We start our proof by showing that $[\mathcal{R}^{h}]$ is of order $9$:\
Since the field $\Gamma=\mathbb{Q}(\sqrt[3]{ p} )$ with $p \equiv 1  \pmod  9$ is of second kind, then by Proposition \[3GAMMA\] we have $3\mathcal{O}_{\Gamma}=
\mathcal{H}^{2}\mathcal{S}$, where $\mathcal{H}$ and $\mathcal{S}$ are prime of $\Gamma$, since $\mathcal{H}\mathcal{O}_{k}=\mathcal{P}\mathcal{Q}$ and $\mathcal{S}\mathcal{O}_{k}=\mathcal{R}^2$, then $3\mathcal{O}_{\mathrm{k}}=
\mathcal{P}^{2}\mathcal{Q}^{2}\mathcal{R}^{2}$, where $\mathcal{P},
\,\,\mathcal{Q}$ and $\mathcal{R}$ are prime ideals of $\mathrm{k}$. Moreover, the prime ideal $\mathcal{R}$ is invariant by $\tau$, then $[\mathcal{R}]\in \{ \chi\in C_{\mathrm{k},3} \vert
\chi^{\tau}=\chi\}$.\
If $9$ divides the $3$-class number of $\mathbb{Q}(\sqrt[3]{ p})$ exactly and $u=1$, then by Theorem \[Cara39\] we have $C_{\mathrm{k},3}$ is of type $(9,3)$. According to Proposition \[39\], we have $C_{\mathrm{k},3}^{+}$ is cyclic of order $9$, thus $[\mathcal{R}^{h}]^{9}=1$. Hence the class $[\mathcal{R}^{h}]$ is of order $9$ if and only if $\mathcal{R}^{h} \, \text{and}\, \mathcal{R}^{3h}$ are not principal.\
We argue by the absurd: assume that $\mathcal{R}^{h}$ is principal, we have $$\begin{aligned}
[\mathcal{R}^{h}]=1
& \Rightarrow & \exists \alpha \in
\mathrm{k} ~\mid ~ \mathcal{R}^{h}=\alpha\mathcal{O}_{\mathrm{k}}, \\
&\Rightarrow &  \mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}(\mathcal{R}^{h})=\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}(\alpha \mathcal{O}_{\mathrm{k}} ),  \\
&\Rightarrow & \lambda^{h}\mathcal{O}_{\mathrm{k}_0}  =\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}(\alpha) \mathcal{O}_{\mathrm{k}_0}, \ where \ \lambda=1-\zeta_3, \\
&\Rightarrow & \exists \epsilon \in E_{\mathrm{k}_{0}} ~|~
\lambda^{h}=\epsilon\cdot \mathcal{N}_{\mathrm{k}\vert
\mathrm{k}_{0}}(\alpha ), \\
&\Rightarrow & \exists \beta \in \mathcal{O}_{\mathrm{k}} ~|~
\lambda^{h}=\mathcal{N}_{\mathrm{k}\vert
\mathrm{k}_{0}}(\beta ),~ because ~ E_{\mathrm{k}_{0}}\subseteq
\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}(\mathrm{k}^{*} ),\end{aligned}$$

that is to say $\lambda^{h}$ is a norm in $\mathrm{k}=\mathrm{k}_{0}(\sqrt[3]{
p})=\mathrm{k}_{0}(\sqrt[3]{
\pi_{1}\pi_{2}})$, where $\pi_1$ and $\pi_2$ are two primes of $\mathrm{k}_0$ such that $p=\pi_1\pi_2$. Hence, by property $(5)$ we have: $$(*)\,\,\,\,\,\,\,\,\,\left(\frac{\lambda^{h}, \pi_{1}\pi_{2}}{\mathcal{P}}\right)_{3}=1,$$ for all ideal $\mathcal{P}$ of $\mathrm{k}_{0}$.\
In particular, we calculate this symbol for $\mathcal{P}=\pi_1 \mathcal{O}_{\mathrm{k}_0}$ or $\mathcal{P}=\pi_2 \mathcal{O}_{\mathrm{k}_0}$.\
For $\mathcal{P}=\pi_1 \mathcal{O}_{\mathrm{k}_0}$, using the property $(1)$ of the norm residue symbol, we have: $$\left(\frac{\lambda^{h}, \pi_{1}\pi_{2}}{\mathcal{P}}\right)_{3}=
\left(\frac{\lambda^{h},
\pi_{1}\pi_{2}}{\pi_1}\right)_{3}=\left(\frac{\lambda^{h},
\pi_{1}}{\pi_1}\right)_{3}\cdot \left(\frac{\lambda^{h},
\pi_{2}}{\pi_1}\right)_{3}$$ the properties $(2)$ and $(5)$ imply that: $$\left(\frac{\lambda^{h},\pi_{2}}{\pi_1}\right)_{3}=\left(\frac{\lambda,\pi_{2}}{\pi_1}\right)_{1}^{h}=
 \left(\frac{\lambda}{\pi_1}\right)_{3}^{0\times h}=1.$$ and from the properties $(1)$ and $(6)$ we have $$\left(\frac{\lambda^{h},
\pi_{1}}{\pi_1}\right)_{3}=\left(\frac{\lambda,
\pi_{1}}{\pi_1}\right)_{3}^{h}=\left(\frac{\lambda}{\pi_1}\right)_{3}^{h}$$ consequently $$\left(\frac{\lambda^{h},
\pi_{1}\pi_2}{\pi_1}\right)_{3}=\left(\frac{\lambda}{\pi_1}\right)_{3}^{h}$$.

Since the two primes $\pi_1$ and $\pi_2$ play symmetric roles, then we obtain a similar relation when $\mathcal{P}=\pi_2$:

$$\left(\frac{\lambda^{h},
\pi_{1}\pi_{2}}{\pi_2}\right)_{3}=\left(\frac{\lambda}{\pi_2}\right)_{3}^{h}.$$\
The equation $(*)$ imply that $$\left(\frac{\lambda}{\pi_1}\right)_{3}^{h}=\left(\frac{\lambda}{\pi_2}\right)_{3}^{h}=1.$$

The fact that $3$ is not a cubic residue modulo $p$ imply that $$\left(\frac{\lambda}{\pi_1\pi_2}\right)_{3} \neq 1$$ then $$\left(\frac{\lambda}{\pi_1}\right)_{3} \neq 1  \ \text{or} \ \left(\frac{\lambda}{\pi_2}\right)_{3} \neq 1.$$ Since $3$ does not divide $h$, then $$\left(\frac{\lambda}{\pi_1}\right)_{3}^{h} \neq 1  \ \text{or} \ \left(\frac{\lambda}{\pi_2}\right)_{3}^{h} \neq 1.$$ which is a contradiction. Consequently, the ideal $\mathcal{R}^{h}$ is not principal.\
Since the class $[\mathcal{R}^{h}]$ is invariant by $\tau$, we deduce that the ideal $\mathcal{R}^{3h}$ is principal if and only if $\langle[\mathcal{R}^{h}]\rangle=C_{\mathrm{k},3}^{(\sigma)}$.\
Since $9$ divides exactly the $3$-class number of $\mathbb{Q}(\sqrt[3]{ p})$ and $u=1$, then by we get $|C_{\mathrm{k},3}|=27$, so the positive integer $s$ defined above is equal $3$, then $C_{k,3}^{(1-\sigma)^3}=1$, this implies that $C_{k,3}^{(\sigma)} = C_{k,3}^{(1-\sigma)^2}$. Suppose that $[\mathcal{R}^{h}]\in
C_{\mathrm{k},3}^{(\sigma)}$, then $[\mathcal{R}^h] = [ \mathcal{L}^{(1-\sigma)^2}]$ with $\mathcal{L}$ is prime ideal of $\mathrm{k}$, then there exist $\alpha \in \mathrm{k}^{*}$ such that $\mathcal{R}^{h}=(\alpha)\cdot\mathcal{L}^{(1-\sigma)^{2}}$, so $\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}( \mathcal{R}^h )=\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}( \alpha.\mathcal{L}^{(1-\sigma)^2})$, since $\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}(\mathcal{L}^{(1-\sigma)^2})= \mathcal{L}^{(1-\sigma)(1-\sigma^3)} =1$, then $\lambda^{h}\mathcal{O}_{\mathrm{k}}=\mathcal{N}_{\mathrm{k}\vert
\mathrm{k}_{0}}(\alpha)\mathcal{O}_{\mathrm{k}}$, where $\lambda=1-\zeta_3$, so there exist $\varepsilon \in E_{\mathrm{k}}$ such that $\lambda^{h}=\varepsilon\cdot \mathcal{N}_{\mathrm{k}\vert
\mathrm{k}_{0}}(\alpha )$, as $\lambda^{h} $ and $\mathcal{N}_{\mathrm{k}\vert
\mathrm{k}_{0}}(\alpha ) $ are in $\mathrm{k}_{0}$ then $\varepsilon \in E_{\mathrm{k}_{0}} $, since $E_{\mathrm{k}_{0}}\subseteq \mathcal{N}_{\mathrm{k}\vert
\mathrm{k}_{0}}(\mathrm{k}^{*} )$ then $\lambda^{h}=
\mathcal{N}_{\mathrm{k}\vert \mathrm{k}_{0}}(\alpha_{1} ) $ where $\alpha_1
\in \mathcal{O}_{\mathrm{k}}$, that means $\lambda^{h}$ is a norm in $\mathrm{k}=\mathrm{k}_{0}(\sqrt[3]{ p})$ which is impossible. Finally, $[\mathcal{R}^{h}]$ is of order $9$. This completes the proof of the first statement.\
The second step in the proof is showing that the class $[\mathcal{R}^{h}][\mathcal{P}^{h}]^2$ is of order $3$. We know that $(\mathcal{R}^{h})^{\tau}=\mathcal{R}^{h}$ and $(\mathcal{P}^{h})^{\tau}=\mathcal{Q}^{h}$, then: $$\begin{aligned}
\left(\mathcal{R}^{h}\cdot
(\mathcal{P}^{h})^{2}\right)^{1+\tau} &=&
\left(\mathcal{R}^{h}\right)^{1+\tau} \cdot \left((\mathcal{P}^{h})^{2}\right)^{1+\tau}\\
&=&
(\mathcal{P}^{h})^{2}\cdot(\mathcal{R}^{h})^{2}\cdot
(\mathcal{Q}^{h})^{2}\\
&=& 3^{h}\mathcal{O}_{\mathrm{k}},\end{aligned}$$ which imply that $[\mathcal{R}^{h}\cdot (\mathcal{P}^{h})^{2}]^{1+\tau}=1$. Hence $[\mathcal{R}^{h}\cdot (\mathcal{P}^{h})^{2}]\in
C_{\mathrm{k},3}^{-}$.\
On the other hand $\mathcal{R}^{h}\cdot (\mathcal{P}^{h})^{2}$ is not principal, because otherwise we have $[\mathcal{R}^{h}]=[\mathcal{P}^{h}]^{7}$, the fact that $[(\mathcal{R}^{h})^{2}\cdot(\mathcal{P}^{h})^{2}\cdot(\mathcal{Q}^{h})^{2}]=1$ imply that $[(\mathcal{Q}^{h})^{2}]=1$, which is a contradiction because the class $[\mathcal{Q}^{h}]$ is of order $9$ (reasoning as $\mathcal{R}^{h}$). Hence $[\mathcal{R}^{h}][\mathcal{P}^{h}]^{2}$ is of order $3$ and generate the group $C_{\mathrm{k},3}^{-}$.\
Since $[\mathcal{R}^{h}]$ is a generator of $C_{\mathrm{k},3}^{+}$, we deduce that $$C_{\mathrm{k},3}=\langle[\mathcal{R}^{h}],[\mathcal{R}^{h}][\mathcal{P}^{h}]^{2}\rangle.$$

Using the same notation as above, we have the following properties:

1.  $\mathcal{P}^{\sigma}=\mathcal{Q},\,\,\mathcal{Q}^{\sigma}=\mathcal{R}$;

2.  $\mathcal{R}^{\tau}=\mathcal{R}\,\, \text{and}\,\, \langle[\mathcal{R}]\rangle=
    \{ \chi \in C_{\mathrm{k},3} \vert \chi^{\tau }=\chi \}$;

3.  $\mathcal{P}^{\tau\sigma}=\mathcal{P}\,\, \text{and}\,\, \langle[\mathcal{P}]\rangle=
    \{ \chi \in C_{\mathrm{k},3} \vert \chi^{\tau \sigma}=\chi \}$;

4.  $\mathcal{Q}^{\tau\sigma^{2}}=\mathcal{Q}\,\, \text{and}\,\, \langle[\mathcal{Q}]\rangle=
    \{ \chi \in C_{\mathrm{k},3} \vert \chi^{\tau \sigma^{2}}=\chi \}$;

5.  The $3$-class group can be generated also by:\
    $$C_{\mathrm{k},3}=\langle[\mathcal{P}^{h}],[\mathcal{P}^{h}][\mathcal{Q}^{h}]^2\rangle=
    \langle[\mathcal{Q}^{h}],[\mathcal{Q}^{h}][\mathcal{R}^{h}]^2\rangle.$$

6.  The $3$-group $C_{\mathrm{k},3}^{(\sigma)}$ of ambiguous ideal classes is given by:\
    $$C_{\mathrm{k},3}^{(\sigma)}=\langle[\mathcal{R}^{3h}]\rangle=\langle[\mathcal{P}^{3h}]\rangle=\langle[\mathcal{Q}^{3h}]\rangle.$$

7.  The principal genus $C_{\mathrm{k},3}^{1-\sigma}=\{ \mathcal{A}^{1-\sigma}\, \vert\,
    \mathcal{A} \in C_{\mathrm{k},3} \}$ is of type $(3,3)$ and generated by: $$C_{\mathrm{k},3}^{1-\sigma}=\langle[\mathcal{R}^{3h}],[\mathcal{R}^{h}][\mathcal{P}^{h}]^2\rangle.$$

$\\ $ The fact that the ideals $\mathcal{P}^{h},\, \mathcal{Q}^{h}$ and $\mathcal{R}^{h}$ are not principals, we prove the assertions $(1), (2), (3) $ and $(4)$ by applying the decomposition of $3$ in the normal closure $\mathrm{k}$.\
For the assertion $(5)$, since the ideals $\mathcal{P}^{h},\,
\mathcal{Q}^{h}$ and $\mathcal{R}^{h}$ are not principal, we obtain the result by the same reasoning above.\
The assertions $(6)$ and $(7)$ follows by using Proposition \[prop:carrr39\].

 **Appendix** {#Exp}
=============

\[sect4\] Using the *Pari* programming @PARI, we illustrate the results of our main Theorem \[thm:main\] by numerical examples. We have $$C_{\mathrm{k},3}=\langle[\mathcal{R}^{h}],[\mathcal{R}^{h}][\mathcal{P}^{h}]^2\rangle$$ The following table verifies, for each prime number $p \equiv 1 \pmod  9$ such that $\left(\dfrac{3}{p}\right)_{3}\neq 1$ and $9$ divides the $3$-class number of $\mathbb{Q}(\sqrt[3]{ p})$ exactly and $u=1$, that the ideals $\mathcal{R}^{h}$ and $\mathcal{R}^{3h}$ are not principal. Therefore, the ideal $\mathcal{R}^{9h}$ is always principal. $$\text{Table 1}$$

<span> | c | c | c | c | c | </span> $p$ & Type of $C_{\mathrm{k},3}$ & Is principal $\mathcal{R}^{h}$ & Is principal $\mathcal{R}^{3h}$ & Is principal $\mathcal{R}^{9h}$\

$199$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$487$ & $[9, 3]$ & $[10, 0]~$ & $[12, 0]~$ & $[0, 0]~$\
$1297$ & $[9, 3]$ & $[16, 0]~$ & $[12, 0]~$ & $[0, 0]~$\
$1693$ & $[9, 3]$ & $[2, 2]~$ & $[6, 0]~$ & $[0, 0]~$\
$1747$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$1999$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$2017$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$2143$ & $[9, 3]$ & $[14, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$2377$ & $[9, 3]$ & $[7, 0]~$ & $[3, 0]~$ & $[0, 0]~$\
$2467$ & $[9, 3]$ & $[20, 0]~$ & $[15, 0]~$ & $[0, 0]~$\
$2593$ & $[9, 3]$ & $[4, 2]~$ & $[3, 0]~$ & $[0, 0]~$\
$2917$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$3511$ & $[9, 3]$ & $[10, 0]~$ & $[12, 0]~$ & $[0, 0]~$\
$3673$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$3727$ & $[9, 3]$ & $[5, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$4159$ & $[9, 3]$ & $[4, 2]~$ & $[12, 0]~$ & $[0, 0]~$\
$4519$ & $[9, 3]$ & $[4, 4]~$ & $[12, 0]~$ & $[0, 0]~$\
$4591$ & $[9, 3]$ & $[1, 2]~$ & $[3, 0]~$ & $[0, 0]~$\
$4789$ & $[9, 3]$ & $[25, 5]~$ & $[30, 0]~$ & $[0, 0]~$\
$5347$ & $[9, 3]$ & $[8, 0]~$ & $[6, 0]~$ & $[0, 0]~$\
$5437$ & $[9, 3]$ & $[77, 0]~$ & $[33, 0]~$ & $[0, 0]~$\
$6949$ & $[9, 3]$ & $[7, 2]~$ & $[3, 0]~$ & $[0, 0]~$\
$8209$ & $[9, 3]$ & $[2, 2]~$ & $[6, 0]~$ & $[0, 0]~$\
$8821$ & $[9, 3]$ & $[4, 0]~$ & $[3, 0]~$ & $[0, 0]~$\

\[tab:genor9\]

However, we verify in the following table that the ideal $\mathcal{R}^{h}\mathcal{P}^{2h}$ is not principal and $\mathcal{R}^{h}\mathcal{P}^{2h}$ is of order $3$.

$$\text{Table 2}$$

<span>| c | c | c | c | </span> $p$ & Type of $C_{\mathrm{k},3}$ & Is principal $\mathcal{R}^{h}\mathcal{P}^{2h}$ & Is principal $(\mathcal{R}^{h}\mathcal{P}^{2h})^3$\
$199$ & $[9, 3]$ & $[0, 1]~$ & $[0, 0]~$\
$487$ & $[9, 3]$ & $[0, 2]~$ & $[0, 0]~$\
$1297$ & $[9, 3]$ & $[6, 4]~$ & $[0, 0]~$\
$1693$ & $[9, 3]$ & $[6, 2]~$ & $[0, 0]~$\
$1747$ & $[9, 3]$ & $[0, 1]~$ & $[0, 0]~$\
$1999$ & $[9, 3]$ & $[0, 2]~$ & $[0, 0]~$\
$2017$ & $[9, 3]$ & $[0, 2]~$ & $[0, 0]~$\
$2143$ & $[9, 3]$ & $[0, 4]~$ & $[0, 0]~$\
$2377$ & $[9, 3]$ & $[3, 2]~$ & $[0, 0]~$\
$2467$ & $[9, 3]$ & $[0, 10]~$ & $[0, 0]~$\
$2593$ & $[9, 3]$ & $[0, 2]~$ & $[0, 0]~$\
$2917$ & $[9, 3]$ & $[0, 1]~$ & $[0, 0]~$\
$3511$ & $[9, 3]$ & $[0, 2]~$ & $[0, 0]~$\
$3673$ & $[9, 3]$ & $[0, 1]~$ & $[0, 0]~$\
$3727$ & $[9, 3]$ & $[3, 1]~$ & $[0, 0]~$\
$4159$ & $[9, 3]$ & $[6, 2]~$ & $[0, 0]~$\
$4519$ & $[9, 3]$ & $[24, 4]~$ & $[0, 0]~$\

\[tab:genor3\]

<span>99</span> P. Barrucand and H. Cohn, Remarks on principal factors in a relative cubic field, *J. Number Theory* **3** (1971), 226–239. F.Calegari and M. Emerton, On the ramification of Hecke algebras at Eisenstein primes, Invent. Math. 160 (2005), 97–144. R. Dedekind, Über die Anzahl der Idealklassen in reinen kubischen Zahlkörpern. *J. für reine und angewandte Mathematik,* Bd. <span>**121** </span> (1900), 40-123. F. Gerth III, On $3$-Class Groups of Cyclic Cubic Extensions of Certain Number Fields, *J. Number Theory* **8** (1976), 84–94.

F. Gerth III, On $3$-class groups of pure cubic fields, *J. Reine Angew. Math* **278/279** (1975), 52–62.

F. Gerth III, On $3$-class groups of certain pure cubic fields, *Australian Mathematical Society* Volume 72, Number 3, December 2005, pp.471-476. G. Gras, Sur les l-classes d'idéaux dans les extensions cycliques relatives de degré premier l, *Ann. Inst. Fourier,* Grenoble, tome 23, fasc <span>**3**</span> (1973).

H.Hasse, Bericht Über neuere Untersuchungen und probleme aus der Theorie der algebraischen Zahlköberrper, *I, Jber. Deutsch. Math. Verein.* **35** (1926), 1–55. H. Hasse, Newe Begründung Und Verallgemeinerung der Theorie der Normenrest Symbols, *Journal für reine und ang. Math.* <span>**162** </span> (1930), 134-143. C. S. Herz, *Construction of class fields*, Seminar on Complex Multiplication, Springer-Verlag, New York, 1966. Kenneth Ireland - Michael Rosen: *A classical Introduction to Modern Number Theory*, Second Edition, Springer 1992. The PARI Group, PARI/GP, Version 2.9.4, Bordeaux, 2017, http://pari.math.u-bordeaux.fr.

> Siham AOUISSI and Moulay Chrif ISMAILI\
> Department of Mathematics and Computer Sciences,\
> Mohammed 1st University,\
> Oujda - Morocco,\
> albert32@example.org, albert32@example.org.\
> Mohamed TALBI\
> Regional Center of Professions of Education and Training in the Oriental,\
> Oujda - Morocco,\
> albert32@example.org.\
> Abdelmalek AZIZI\
> Department of Mathematics and Computer Sciences,\
> Mohammed 1st University,\
> Oujda - Morocco,\
> albert32@example.org.

{'source': 'arXiv_out/1804'}