# The generators of $5$-class group of some
fields of degree 20 over $\mathbb{Q}$

Fouad ELMOUHIB  Mohamed TALBI  Abdelmalek AZIZI

###### Abstract

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n$ is
a positive integer, $5^{th}$ power-free. Let $k_{0}\,=\,\mathbb{Q}(\zeta_{5})$
be the cyclotomic field containing a primitive $5^{th}$ root of unity
$\zeta_{5}$, and $k\,=\,\Gamma(\zeta_{5})$ be the normal closure of $\Gamma$.
Let $C_{k,5}$ be the $5$-component of the class group of k. The purpose of
this paper is to determine generators of $C_{k,5}$, whenever it is of type
$(5,5)$ and the rank of the group of ambiguous classes under the action of
$Gal(k/k_{0})\,=\,\langle\sigma\rangle$ is $1$.

##  1 Introduction

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n$ is
a positive integer, $5^{th}$ power-free, We denote by
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ the cyclotomic field containing a primitve
$5^{th}$ root of unity, $k\,=\,\Gamma(\zeta_{5})$ the normal closure of
$\Gamma$, and $C_{k,5}$ the $5$-class group of $k$. In References we gave all
forms of the radicand $n$ whenever $C_{k,5}$ is of type $(5,5)$, using the
rank of ambiguous ideal classes $C_{k,5}^{(\sigma)}\,=\,\\{\mathcal{A}\in
C_{k,5}|\mathcal{A}^{\sigma}\,=\,\mathcal{A}\\}$, which takes $1$ or $2$ as
value. According to References, if $C_{k,5}$ is of type $(5,5)$ and rank
$C_{k,5}^{(\sigma)}\,=\,1$, we have three forms of $n$ as follows:

  * (1)

$n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$e\in\\{1,2,3,4\\}$ and p prime such that
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$.

  * (2)

$n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$e\in\\{1,2,3,4\\}$, p and q are primes such that
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$, $q\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$.

  * (3)

$n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with $e\in\\{1,2,3,4\\}$
and p prime such that $p\,\equiv\,-1\,(\mathrm{mod}\,25)$.

In this paper, we will prove that $C_{k,5}$ is of type $(5,5)$ if and only if
$5$ divides exactly the $5$-class number of $\Gamma$ and $u\,=\,5^{3}$, where
$u$ is the index of subgroup $E_{0}$ generated by the units of intermediate
fields of the extension $k/\mathbb{Q}$ in the unit group of $k$. Next we
determine generators of $C_{k,5}$, when $C_{k,5}$ is isomorphic to
$\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$ and rank
$C_{k,5}^{(\sigma)}\,=\,1$, for each of the three forms of $n$.
Since $k$ is a Kummer extension of $k_{0}$, we begin with the decomposition
laws in Kummer extensions, which is useful to give the prime factorization in
the normal closure $k$. As the proof of our main results is etablished by the
norm residue symbol, we recall the definition and the most important
properties of this concept. Making use this, we can determine generators of
$5$-class group $C_{k,5}$ of type $(5,5)$ if rank $C_{k,5}^{(\sigma)}\,=\,1$.
Our main result will be underpinned by numerical examples, using the PARI/GP
[References] in section 7. In fact, we shall prove the following Main Theorem:

######  Theorem 1.1.

Let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$, where $n$ is a positive integer
$5^{th}$ power-free, be the normal closure of the pure quintic field
$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$. Let
$Gal(k/\Gamma)\,=\,\langle\tau\rangle$. Let $p$,$q$ and $l$ primes such that,
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$, $q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ and
$l\,\neq\,p$, $l\,\neq\,q$ . Assume that $C_{k,5}$ is of type $(5,5)$ and
rank$(C_{k,5}^{(\sigma)})\,=\,1$, then we have:

  * (1)

If $n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, with
$e\in\\{1,2,3,4\\}$ and $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$. The prime $p$
decompose in $k$ as
$p\mathcal{O}_{k}\,=\,\mathcal{P}^{5}_{1}\mathcal{P}^{5}_{2}$, where
$\mathcal{P}_{1}$ and $\mathcal{P}_{2}$ are prime ideals of $k$. Let
$\mathcal{L}$ a prime ideal of $k$ above $l$. If $5$ and $l$ are not a quintic
residue modulo $p$, then the $5$-class group $C_{k,5}$ is generated by classes
$[\mathcal{P}_{1}]$ and $[\mathcal{L}]^{1-\tau^{2}}$ and we have:

$C_{k,5}$ =
$\langle[\mathcal{P}_{1}]\rangle\times\langle[\mathcal{L}]^{1-\tau^{2}}\rangle$
= $\langle[\mathcal{P}_{1}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

  * (2)

If $n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, with
$e\in\\{1,2,3,4\\}$ and $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$,
$q\,\not\equiv\,\pm 7\,(\mathrm{mod}\,25)$. The prime $p$ decompose in $k$ as
$p\mathcal{O}_{k}\,=\,\mathcal{P}^{5}_{1}\mathcal{P}^{5}_{2}$, where
$\mathcal{P}_{1}$ and $\mathcal{P}_{2}$ are prime ideals of $k$. Let
$\mathcal{L}$ a prime ideal of $k$ above $l$. If $q$ and $l$ are not a quintic
residue modulo $p$, then the $5$-class group $C_{k,5}$ is generated by classes
$[\mathcal{P}_{1}]$ and $[\mathcal{L}]^{1-\tau^{2}}$ and we have:

$C_{k,5}$ =
$\langle[\mathcal{P}_{1}]\rangle\times\langle[\mathcal{L}]^{1-\tau^{2}}\rangle$
= $\langle[\mathcal{P}_{1}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

  * (3)

If $n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, with
$e\in\\{1,2,3,4\\}$ and $p\,\equiv\,-1\,(\mathrm{mod}\,25)$. $5$ decompose in
$k$ as
$5\mathcal{O}_{k}\,=\,\mathcal{B}^{4}_{1}\mathcal{B}^{4}_{2}\mathcal{B}^{4}_{3}\mathcal{B}^{4}_{4}\mathcal{B}^{4}_{5}$,
where $\mathcal{B}_{i}$ are prime ideals of $k$. If $5$ is not a quintic
residue modulo $p$, then the $5$-class group $C_{k,5}$ is generated by classes
$[\mathcal{B}_{i}]$ and $[\mathcal{B}_{j}]$, $i\,\neq\,j\in\\{1,2,3,4,5\\}$
and we have:

$C_{k,5}$ =
$\langle[\mathcal{B}_{i}]\rangle\times\langle[\mathcal{B}_{j}]\rangle$ =
$\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle$

Notations.

Throughout this paper, we use the following notations:

  * •

The lower case letter $p$,$q$ and $l$ will denote a prime numbers such that,
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$, $q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ and
$l\,\neq\,p$, $l\,\neq\,q$ .

  * •

$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$: a pure quintic field, where $n\neq 1$ is
a $5^{th}$ power-free positive integer.

  * •

$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$: the cyclotomic field, where
$\zeta_{5}\,=\,e^{\frac{2i\pi}{5}}$ a primitive $5^{th}$ root of unity.

  * •

$k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$: the normal closure of $\Gamma$, a
quintic Kummer extension of $k_{0}$.

  * •

$\langle\tau\rangle\,=\,\operatorname{Gal}(k/\Gamma)$ such that
$\tau^{4}\,=\,id,\,\tau^{3}(\zeta_{5})\,=\,\zeta_{5}^{3},\,\tau^{2}(\zeta_{5})\,=\,\zeta_{5}^{4},\,\tau(\zeta_{5})\,=\,\zeta_{5}^{2}$
and $\tau(\sqrt[5]{n})\,=\,\sqrt[5]{n}$.

  * •

$\langle\sigma\rangle\,=\,\operatorname{Gal}(k/k_{0})$ such that
$\sigma^{5}\,=\,id,\ \sigma(\zeta_{5})\,=\,\zeta_{5}$ and
$\sigma(\sqrt[5]{n})\,=\,\zeta_{5}\sqrt[5]{n},\,\sigma^{2}(\sqrt[5]{n})\,=\,\zeta_{5}^{2}\sqrt[5]{n},\,\\\
\\\
\sigma^{3}(\sqrt[5]{n})\,=\,\zeta_{5}^{3}\sqrt[5]{n},\,\sigma^{4}(\sqrt[5]{n})\,=\,\zeta_{5}^{4}\sqrt[5]{n}$.

  * •

$u$: the index of subgroup $E_{0}$ generated by the units of intermediate
fields of the extension

$k/\mathbb{Q}$ in the unit group of $k$.

  * •

$\lambda\,=\,1-\zeta_{5}$ is prime element above $5$ in $k_{0}$.

  * •

$q^{\ast}\,=\,0,\,1$ or $2$ according to whether $\zeta_{5}$ and $1+\zeta_{5}$
is not norm or is norm of an element of

$k^{*}\,=\,k\setminus\\{0\\}$.

  * •

For a number field $L$, denote by:

    * –

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

    * –

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

    * –

$C_{L,5}$: the $5$-class group of $L$;

    * –

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

##  2 Decomposition laws in Kummer extension

Since the extensions of $k$ and $k_{0}$ are all Kummer’s extensions, we recall
the decomposition laws of ideals in this extensions. Let $L$ a number field
contains the $l^{th}$ root of unity, where $l$ is prime, and $\theta$ element
of $L$, such that $\theta\,\neq\,\mu^{l}$, for all $\mu\in L$, therefore
$L(\sqrt[l]{\theta})$ is cyclic extension of degree $l$ over $L$. We note by
$\zeta$ a $l^{th}$ primitive root of unity.

######  Proposition 2.1.

.
(1) We assume that a prime $\mathcal{P}$ of $L$, divides exactly $\theta$ to
the power $\mathcal{P}^{a}$.

  * -

If $a=0$ and $\mathcal{P}$ don’t divides $l$, then $\mathcal{P}$ split
completly in $L(\sqrt[l]{\theta})$ when the congruence
$\theta\,\equiv\,X^{l}\,(\mathrm{mod}\,\mathcal{P})$ has solution in $L$.

  * -

If $a=0$ and $\mathcal{P}$ don’t divides $l$, then $\mathcal{P}$ is inert in
$L(\sqrt[l]{\theta})$ when the congruence
$\theta\,\equiv\,X^{l}\,(\mathrm{mod}\,\mathcal{P})$ has no solution in $L$.

  * -

If $l\nmid a$, then $\mathcal{P}$ is totaly ramified in $L(\sqrt[l]{\theta})$.

(2) Let $\mathcal{B}$ a prime factor of $1-\zeta$ that divides $1-\zeta$
exactly to the ath power. Suppose that $\mathcal{B}\nmid\theta$, then
$\mathcal{B}$ split completly in $L(\sqrt[l]{\theta})$ if the congruence

$\theta\,\equiv\,X^{l}\,(\mathrm{mod}\,\mathcal{B}^{al+1})$ $(*)$

has solution in $L$. the ideal $\mathcal{B}$ is inert in $L(\sqrt[l]{\theta})$
if the congruence

$\theta\,\equiv\,X^{l}\,(\mathrm{mod}\,\mathcal{B}^{al})$ $(**)$

has solution in $L$, without $(*)$ has. The ideal $\mathcal{B}$ is totaly
ramified in $L$ if the congruence $(**)$ has no solution.

###### Proof.

see References. ∎

##  3 Prime factorization in pure quintic field and in its normal closure

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ a pure quintic field, and
$k\,=\,\Gamma(\zeta_{5})$ its normal closure, and
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ the $5^{th}$ cyclotomic field. We begin with
the factorization of primes in $\Gamma$, next we give the decomposition of
primes in $k_{0}$, knowing that $5$ is the unique prime ramified in $k_{0}$.
Finally the results of proposition 2.1 allow us to state the prime
factorization in $k$. For more details on decomposition laws, we refer the
reader to References, References, References.
Let $p\,\neq\,5$ a prime number, $\mathcal{P}$ a prime ideal of $\Gamma$,
$\pi$ a prime of $k_{0}$ and $\mathcal{L}$ a prime ideal of $k$. We denote by
$\mathcal{N}$ the absolute norm.

######  Proposition 3.1.

Using the same notations as above, we have:

If $p$ divides the radicand $n$ then
$p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}^{5}$ and
$\mathcal{N}(\mathcal{P})\,=\,p$.

If $p\nmid 5n$ and $p\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ then
$p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}\mathcal{P}_{2}$ and
$\mathcal{N}(\mathcal{P}_{1})\,=\,p$,
$\mathcal{N}(\mathcal{P}_{2})\,=\,p^{4}$.

If $p\nmid 5n$ and $p\,\equiv\,-1\,(\mathrm{mod}\,5)$ then
$p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}\mathcal{P}_{2}\mathcal{P}_{3}$ and
$\mathcal{N}(\mathcal{P}_{1})\,=\,p$,
$\mathcal{N}(\mathcal{P}_{2})\,=\,\mathcal{N}(\mathcal{P}_{3})\,=\,p^{2}$.

If $p\nmid 5n$ and $p\,\equiv\,1\,(\mathrm{mod}\,5)$ then:

  * -

$p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}\mathcal{P}_{2}\mathcal{P}_{3}\mathcal{P}_{4}\mathcal{P}_{5}$
and $\mathcal{N}(\mathcal{P}_{i})\,=\,p$, if $n$ is quintic residu modulo $p$.

  * -

$p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}$ and
$\mathcal{N}(\mathcal{P}_{i})\,=\,p^{5}$, if $n$ is not quintic residu modulo
$p$.

###### Proof.

It follow from [References,theorem 7.4]. ∎

the ramification of the prime $5$ need a particular treatment.

######  Theorem 3.1.

Using the same notations as above. Let $f$ be the conductor of $k/k_{0}$, and
$R\,=\,q_{1}...q_{s}$ denotes the square free product of all prime divisors of
the radicand $n$ of $\Gamma$, then $f$ satisfies the relation:

$f^{4}=\left\\{\begin{array}[]{rl}5^{2}R^{4}&\text{if
}n^{4}\,\not\equiv\,1\,(\text{mod}\,25)\,\,\text{(field of the first
kind)},\\\ R^{4}&\text{if }n^{4}\,\equiv\,1\,(\text{mod}\,25)\,\,\text{(field
of the second kind)}.\end{array}\right.$

###### Proof.

see References. ∎

######  Proposition 3.2.

Using the same notations as above. The decomposition into prime factors of $5$
is:

  * -

If $\Gamma$ is of first kind then: $5\mathcal{O}_{\Gamma}\,=\,\mathcal{P}^{5}$
and $\mathcal{N}(\mathcal{P})\,=\,5$.

  * -

If $\Gamma$ is of second kind then:
$5\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}\mathcal{P}_{2}^{4}$ and
$\mathcal{N}(\mathcal{P}_{i})\,=\,5$.

The decomposition Laws in the $5^{th}$ cyclotomic field $k_{0}$ is as follows:

######  Proposition 3.3.

Using the same notations as above, we have :

$5\mathcal{O}_{k_{0}}\,=\,\lambda^{4}\,=\,(1-\zeta_{5})^{4}$.

$p\mathcal{O}_{k_{0}}\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ in $k_{0}$, if
$p\,\equiv\,1\,(\mathrm{mod}\,5)$.

$p\mathcal{O}_{k_{0}}\,=\,\pi_{1}\pi_{2}$ in $k_{0}$, if
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$.

$p\mathcal{O}_{k_{0}}\,=\,p$ in $k_{0}$ (inert), if $p\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$.

Next, let $k$ the normal closure of $\Gamma$. The decomposition of the prime
$5$ in $k$ is the purpose of the following proposition.

######  Proposition 3.4.

The prime $5$ decompose in $k$ as follows:

If $\Gamma$ is of first kind then: $5\mathcal{O}_{k}\,=\,\mathcal{L}^{20}$.

If $\Gamma$ is of second kind then:
$5\mathcal{O}_{k}\,=\,\mathcal{L}_{1}^{4}\mathcal{L}_{2}^{4}\mathcal{L}_{3}^{4}\mathcal{L}_{4}^{4}\mathcal{L}_{5}^{4}$.

###### Proof.

We have $5$ ramifies in $k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ then:

Suppose that $\Gamma$ is of first kind, by proposition 3.2 we have
$5\mathcal{O}_{\Gamma}\,=\,\mathcal{P}^{5}$. Hence
$5\mathcal{O}_{k}\,=\,\mathcal{L}^{20}$.

Suppose that $\Gamma$ is of second kind, by proposition 3.2 we have
$5\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}^{4}\mathcal{P}_{2}$. It follows
that,
$5\mathcal{O}_{k}\,=\,\mathcal{L}_{1}^{4}\mathcal{L}_{2}^{4}\mathcal{L}_{3}^{4}\mathcal{L}_{4}^{4}\mathcal{L}_{5}^{4}$.
∎

However, we have the following Proposition in which we characterize the
decomposition of prime ideals of $p\,\neq\,5$ in $k$.

######  Proposition 3.5.

Using the same notations as above, we have :

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

  * (a)

If $p\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$, then
$p\mathcal{O}_{k}\,=\,\mathcal{L}^{5}$.

  * (b)

If $p\,\equiv\,-1\,(\mathrm{mod}\,5)$, then
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}^{5}\mathcal{L}_{2}^{5}$.

  * (c)

If $p\,\equiv\,1\,(\mathrm{mod}\,5)$, then
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}^{5}\mathcal{L}_{2}^{5}\mathcal{L}_{3}^{5}\mathcal{L}_{4}^{5}$.

If $p$ does not divides $\mathcal{D}_{\Gamma}$ and
$p\,\equiv\,1\,(\mathrm{mod}\,5)$ then:

  * (a)

$p$ decompose completly in $k$ if and only if $\mathcal{D}_{\Gamma}$ is a
quintic residue modulo $p$.

  * (b)

$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}\mathcal{L}_{3}\mathcal{L}_{4}$
if and only if $\mathcal{D}_{\Gamma}$ is not a quintic residue modulo $p$.

If $p$ does not divides $\mathcal{D}_{\Gamma}$ and
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$ then:
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}\mathcal{L}_{3}\mathcal{L}_{4}\mathcal{L}_{5}\mathcal{L}_{6}$.

If $p$ does not divides $\mathcal{D}_{\Gamma}$ and $p\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$ then:
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}$.

###### Proof.

We use 1 of Proposition 3.1 and the decomposition of prime ideals in the
cyclotomic fields $k_{0}\,=\,\mathbb{Q}(\zeta_{5})$.

Suppose that $p$ does not divide $\mathcal{D}_{\Gamma}$ and
$p\,\equiv\,1\,(\mathrm{mod}\,5)$.

  * (a)

If $\mathcal{D}_{\Gamma}$ is a quintic residue modulo $p$, then by 4 of
proposition 3.1 we have $p$ split completely in $\Gamma$. Hence $p$ split
completely in $k$.

  * (b)

If $\mathcal{D}_{\Gamma}$ is not a quintic residue modulo $p$, we have $p$
remains inert in $\Gamma$. Hence
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}\mathcal{L}_{3}\mathcal{L}_{4}$.

If $p\,\nmid\,\mathcal{D}_{\Gamma}$ and $p\,\equiv\,-1\,(\mathrm{mod}\,5)$,
then $p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}\mathcal{P}_{2}\mathcal{P}_{3}$
and $p\mathcal{O}_{k_{0}}\,=\,\pi_{1}\pi_{2}$. Hence
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}\mathcal{L}_{3}\mathcal{L}_{4}\mathcal{L}_{5}\mathcal{L}_{6}$.

If $p\,\nmid\,\mathcal{D}_{\Gamma}$ and $p\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$,
then $p\mathcal{O}_{\Gamma}\,=\,\mathcal{P}_{1}\mathcal{P}_{2}$ and $p$
remains inert in $k_{0}$, hence
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}$. ∎

######  Remark 3.1.

Let $p\,\neq\,5$, according to proposition 3.5, if
$p\,\nmid\,\mathcal{D}_{\Gamma}$, then $p$ is unramified in $k$ and not inert
in $k$. Let
$p\mathcal{O}_{k}\,=\,\mathcal{L}_{1}\mathcal{L}_{2}....\mathcal{L}_{g}$,
since the elements of $Gal(k/\mathbb{Q})$ permut the set
$\\{\mathcal{L}_{1}....\mathcal{L}_{g}\\}$ then
$\mathcal{L}_{i}^{\sigma}\,\neq\,\mathcal{L}_{i}$ for $1\leq i\leq g$. This
result has great importance in the proof of our main results.

##  4 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 $K$, we note by
$f_{\mathcal{P}}$ the largest power of $\mathcal{P}$ that divides $f$. Let
$\beta\in K^{*}$, we determine an auxiliary number $\beta_{0}$ by the two
conditions $\beta_{0}\,\equiv\,\beta\,(\mathrm{mod}\,f_{\mathcal{P}})$ and
$\beta_{0}\,\equiv\,1\,(\mathrm{mod}\,\frac{f}{f_{\mathcal{P}}})$. Let
$\mathcal{Q}$ an ideal co-prime with $\mathcal{P}$ such that
$(\beta_{0})\,=\,\mathcal{P}^{a}\mathcal{Q}$ ($b=0$ if $\mathcal{P}$
infinite). We note by

$\left(\frac{\beta,L}{\mathcal{P}}\right)\,=\,\left(\frac{L/K}{\mathcal{Q}}\right)$

the Artin map in $L/K$ applied to $\mathcal{Q}$.
Let $K$ be a number field containing the $m^{th}$-roots of units, where
$m\in\mathbb{N}$, then for each $\alpha,\beta\in K^{*}$ and prime ideal
$\mathcal{P}$ of $K$, we define the norm residue symbol by:

$\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}\,=\,\frac{\left(\frac{\beta,K(\sqrt[m]{\alpha})}{\mathcal{P}}\right)\sqrt[m]{\alpha}}{\sqrt[m]{\alpha}}$

Therefore, if the prime ideal $\mathcal{P}$ is unramified in the field
$K(\sqrt[m]{\alpha})$, then we write:

$\left(\frac{\alpha}{\mathcal{P}}\right)_{m}\,=\,\frac{\left(\frac{K(\sqrt[m]{\alpha})}{\mathcal{P}}\right)\sqrt[m]{\alpha}}{\sqrt[m]{\alpha}}$

######  Remark 4.1.

Notice that $\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}$ and
$\left(\frac{\alpha}{\mathcal{P}}\right)_{m}$ are two $m^{th}$-roots of unity.

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

######  Properties 4.1.

.

$\left(\frac{\beta_{1}\beta_{2},\alpha}{\mathcal{P}}\right)_{m}\,=\,\left(\frac{\beta_{1},\alpha}{\mathcal{P}}\right)_{m}\left(\frac{\beta_{2},\alpha}{\mathcal{P}}\right)_{m}$;

$\left(\frac{\beta,\alpha_{1}\alpha_{2}}{\mathcal{P}}\right)_{m}\,=\,\left(\frac{\beta,\alpha_{1}}{\mathcal{P}}\right)_{m}\left(\frac{\beta,\alpha_{2}}{\mathcal{P}}\right)_{m}$;

$\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}\,=\,\left(\frac{\alpha,\beta}{\mathcal{P}}\right)_{m}^{-1}$;

If $\mathcal{P}$ is not divisible by the conductor $f(\sqrt[m]{\alpha})$ of
$K(\sqrt[m]{\alpha})$ and appears in $(\beta)$ with the exponent b, then:
$\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}\,=\,\left(\frac{\alpha}{\mathcal{P}}\right)_{m}^{-b}$
;

$\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}\,=\,1$ if and only if
$\beta$ is norm residue of $K(\sqrt[m]{\alpha})$ modulo $f(\sqrt[m]{\alpha})$
;

$\left(\frac{\tau\beta,\tau\alpha}{\tau\mathcal{P}}\right)_{m}\,=\,\tau\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}$
for each automorphism $\tau$ of $K$ ;

${\displaystyle\prod_{\mathcal{P}}\left(\frac{\beta,\alpha}{\mathcal{P}}\right)_{m}}\,=\,1$
for all finite or infinite prime ideals;

If $K^{\prime}$ is a finite extension of $K$, $\alpha\in
K^{*},\beta^{\prime}\in K^{\prime}$ then:

${\displaystyle\prod_{\mathcal{P^{\prime}}|\mathcal{P}}\left(\frac{\beta^{\prime},\alpha}{\mathcal{P^{\prime}}}\right)_{m}}\,=\,\left(\frac{\mathcal{N}_{K^{\prime}/K}(\beta^{\prime}),\alpha}{\mathcal{P}}\right)_{m}$

Let $\alpha,\beta\in K^{*}$ and the conductors $f(\sqrt[m]{\alpha})$,
$f(\sqrt[m]{\beta})$ of respectively $K(\sqrt[m]{\alpha})$,
$K(\sqrt[m]{\beta})$ are co-prime then, the classical reciprocity law:

$\left(\frac{\beta}{(\alpha)}\right)_{m}\,=\,\left(\frac{\alpha}{(\beta)}\right)_{m}$

For more basic properties of the norm residue symbol in the number fields, we
refer the reader to References.
Notice that in the rest of the article, we will use the norm quintic residue
symbols $(m\,=\,5)$. As the ring of integer $\mathcal{O}_{k_{0}}$ is
principal, we will write the norm quintic residue symbol as follows:

$\left(\frac{\beta,\alpha}{(\pi)}\right)_{5}\,=\,\left(\frac{\beta,\alpha}{\pi}\right)_{5}$
and
$\left(\frac{\alpha}{(\pi)}\right)_{5}\,=\,\left(\frac{\alpha}{\pi}\right)_{5}$

Where $\alpha,\beta\in k_{0}^{*}$ and $\pi$ a prime integer of
$\mathcal{O}_{k_{0}}$.

##  5 Fields $\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ whose $5$-class group is of
type $(5,5)$

In this section we use class number formula of [References] to give necessary
and sufficient condition such that the $5$-class group of the fields
$\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ is of type $(5,5)$.

######  Lemma 5.1.

Let $Gal(k/\Gamma)\,=\,\langle\tau\rangle\,\cong\,\mathbb{Z}/4\mathbb{Z}$, and
$C$ be a $\mathbb{Z}_{5}[\langle\tau\rangle]$ module. Let
$C^{+}\,=\,\\{\mathcal{A}\in C\,|\,\mathcal{A}^{\tau^{2}}\,=\,\mathcal{A}\\}$
and $C^{-}\,=\,\\{\mathcal{A}\in
C\,|\,\mathcal{A}^{\tau^{2}}\,=\,\mathcal{A}^{-1}\\}$. Then

$C\,\cong\,C^{+}\times C^{-}$.

###### Proof.

Let $\mathcal{A}\in C$. Write
$\mathcal{A}=\mathcal{A}^{\frac{1+\tau^{2}}{2}}.\mathcal{A}^{\frac{1-\tau^{2}}{2}}$.
Then $\mathcal{A}^{\frac{1+\tau^{2}}{2}}\in C^{+}$ and
$\mathcal{A}^{\frac{1-\tau^{2}}{2}}\in C^{-}$. Let $\mathcal{A}\in C^{+}\cap
C^{-}$, then $\mathcal{A}=\mathcal{A}^{\tau^{2}}=\mathcal{A}^{-1}$, that is
$\mathcal{A}^{2}=1$. Thus $\mathcal{A}=1$, since $C$ is
$\mathbb{Z}_{5}[<\tau>]$-module. Hence $C$ $\cong$ $C^{+}\times C^{-}$. ∎

######  Lemma 5.2.

$C_{\Gamma,5}\,\cong\,C_{k,5}^{+}$.

###### Proof.

Since $C_{k,5}$ is $\mathbb{Z}_{5}[<\tau>]$-module, then
$C_{k,5}\,\cong\,C_{k,5}^{+}\times C_{k,5}^{-}$. We have a naturel inclusion
$C_{\Gamma,5}\,\hookrightarrow\,C_{k,5}$ as $5$ relativly prime to
$[k:\Gamma]\,=\,4$. Moreover, $C_{\Gamma,5}\,\hookrightarrow\,C_{,5}^{+}$ as
$\mathcal{A}^{\tau^{2}}\,=\,\mathcal{A}$ for all $\mathcal{A}\in
C_{\Gamma,5}$. Let $\mathcal{A}\in C_{k,5}^{+}$, then
$\mathcal{A}\,=\,4(\frac{1}{4}\mathcal{A})\,=\,(1+\tau+\tau^{2}+\tau^{3})(\frac{1}{4}\mathcal{A})\,=\,\mathcal{N}(\frac{1}{4}\mathcal{A}))$.
Thus, $\mathcal{A}\in C_{\Gamma,5}$. So $C_{\Gamma,5}\,\cong\,C_{k,5}^{+}$. ∎

Now let $u$ the index of subgroup $E_{0}$ generated by the units of
intermediate fields of the extension $k/\mathbb{Q}$ in the units group of $k$.
In [[Pa]], C.Parry proved that $u$ is a divisor of $5^{6}$, and he presented
the relation formula between the class numbers of $k$ and $\Gamma$:
$h_{k}\,=\,(\frac{u}{5})(\frac{h_{\Gamma}}{5})^{4}$. The structure of the
$5$-class group $C_{k,5}$ is given by the following proposition:

######  Proposition 5.1.

Let $\Gamma$ be a pure quintic field, $k$ its normal closure, and $u$ be the
index of units defined above, then

$C_{k,5}\simeq\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}\,\Longleftrightarrow\,h_{\Gamma}$
is exactly divisible by $5$ and $u\,=\,5^{3}$.

###### Proof.

Let $C_{k,5}\simeq\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$ then
$|C_{k,5}|\,=\,25$. Acoording to [References]
$|C_{k,5}|\,=\,(\frac{u}{5})\frac{|C_{\Gamma,5}|^{4}}{5^{4}}\,=\,25$, namly
$uh_{\Gamma}^{4}\,=\,5^{7}$. Let $n\,=\,v_{5}(u)$, and
$n^{\prime}\,=\,v_{5}(h_{\Gamma})$, so we get that $n+4n^{\prime}\,=\,7$, the
unique natural values of $n$ and $n^{\prime}$ verify this equality is
$n\,=\,3$ and $n^{\prime}\,=\,1$. Therfore we have $u\,=\,5^{3}$ and
$h_{\Gamma}$ is exactly divisible by $5$.
Conversely according to lemma 5.2, $C_{k,5}\,\cong\,C_{k,5}^{+}\times
C_{k,5}^{-}$ and $C_{k,5}^{+}\,\cong\,C_{\Gamma,5}$, then $C_{k,5}^{+}$ is
cyclic of order $5$ because $(|C_{k,5}^{+}|\,=\,|C_{\Gamma,5}|\,=\,5)$. Since
$|C_{k,5}|\,=\,|C_{k,5}^{+}|.|C_{k,5}^{-}|\,=\,25$ by the formula
$|C_{k,5}|\,=\,(\frac{u}{5})\frac{|C_{\Gamma,5}|^{4}}{5^{4}}$, we get that
$|C_{k,5}^{-}|\,=\,5$ and $C_{k,5}^{-}$ is a cyclic group of order $5$, then
$C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$. ∎

##  6 Proof of Main Theorem

We start the proof by some results as follows.

######  Proposition 6.1.

Let $p$ a prime number such that $p\,\equiv\,-1\,(\mathrm{mod}\,5)$, and let
$Gal(k/\Gamma)\,=\,\langle\tau\rangle$, then we have:

$p\,=\pi_{1}\pi_{2}$, where $\pi_{i}$ are primes of $k_{0}$, and
$\pi_{1}\,\equiv\,-\pi_{2}\,\equiv 1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$.

$\pi_{1}^{\tau}\,=\,-\pi_{2}$ and $\pi_{2}^{\tau}\,=\,-\pi_{1}$

###### Proof.

It follow from $p\,\equiv\,-1\,(\mathrm{mod}\,5)$ and proposition 3.3. Since
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$ and $p\,=\,\pi_{1}\pi_{2}$, then
$\pi_{1}\pi_{2}\,\equiv\,-1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$ hence,
$\pi_{1}\,\equiv\,-\pi_{2}\equiv 1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$,
because the other cases of congruence show that
$\pi_{1}\pi_{2}\,\not\equiv\,-1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$

It is known that any prime of the form $p\,\equiv\,-1\,(\mathrm{mod}\,5)$, can
be written as: $p\,=\,a^{2}+ab-b^{2}$, with $a,b\in\mathbb{Z}$ co-prime. We
define:

$\pi_{1}\,=\,a\zeta_{5}^{3}+a\zeta_{5}^{2}+b$ and
$\pi_{2}\,=\,a\zeta_{5}^{3}+a\zeta_{5}^{2}+a-b$,

then $p\,=\,\pi_{1}\pi_{2}$. Let $Gal(k/\Gamma)\,=\,\langle\tau\rangle$ with
$\tau:\,\zeta_{5}\longrightarrow\zeta_{5}^{2}$, so we get that:
$\pi_{1}^{\tau}\,=\,(a\zeta_{5}^{3}+a\zeta_{5}^{2}+b)^{\tau}\,=\,a\zeta_{5}+a\zeta_{5}^{4}+b\,=\,a(\zeta_{5}+\zeta_{5}^{4})+b\,=\,a(-1-\zeta_{5}^{2}-\zeta_{5}^{3})+b\,=\,-\pi_{2}$.
$\pi_{2}^{\tau}\,=\,(a\zeta_{5}^{3}+a\zeta_{5}^{2}+a-b)^{\tau}\,=\,a\zeta_{5}+a\zeta_{5}^{4}+a-b\,=\,a(\zeta_{5}+\zeta_{5}^{4})+a-b\,=\,a(-1-\zeta_{5}^{2}-\zeta_{5}^{3})+a-b\,=\,-\pi_{2}$.
∎

######  Proposition 6.2.

Let $p$ a prime such that $p\,\equiv\,-1\,(\mathrm{mod}\,5)$, and
$p\,=\,\pi_{1}\pi_{2}$ in $k_{0}$.

  * (1)

$\left(\frac{c}{\pi_{1}}\right)_{5}\,=\,\left(\frac{c}{\pi_{2}}\right)_{5}^{2}$
for all $c\in\mathbb{Z}$ such that $p\nmid c$.

  * (2)

$\left(\frac{c}{\pi_{1}}\right)_{5}\,=\,\left(\frac{c}{\pi_{2}}\right)_{5}\,=\,1$
if and only if $c$ is rest quintic modulo $p$.

  * (3)

$\left(\frac{\pi_{2}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi_{1}}{\pi_{2}}\right)_{5}\,=\,1$.

###### Proof.

  * (1)

$\left(\frac{c}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi_{1},c}{\pi_{1}}\right)_{5}^{-1}\,=\,\left(\frac{c,\pi_{1}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{c,-\pi_{2}^{\tau}}{-\pi_{2}^{\tau}}\right)_{5}\,=\,\tau\left(\frac{c,\pi_{2}}{\pi_{2}}\right)_{5}\,=\,\left(\frac{c,\pi_{2}}{\pi_{2}}\right)_{5}^{2}\,=\,\left(\frac{\pi_{2},c}{\pi_{2}}\right)_{5}^{-2}\,=\,\left(\frac{c}{\pi_{2}}\right)_{5}^{2}$,
because $\pi_{1}$, $\pi_{2}$ are unramified in $k_{0}(\sqrt[5]{c})$
(proposition 2), and we use properties $(3),(4),(6)$ of 4.1.

  * (2)

Let $\omega\in\mathcal{O}_{k_{0}}$ and $\pi$ a prime element of
$\mathcal{O}_{k_{0}}$, such that $\pi\nmid\omega$, then
$X^{5}\,\equiv\,\omega\,(\mathrm{mod\,\pi})$ is resolved on
$\mathcal{O}_{k_{0}}$ if and only if
$\omega^{m}\,\equiv\,1\,(\mathrm{mod\,\pi})$, when
$m\,=\,\frac{\mathcal{N}(\pi)-1}{5}$. If
$\left(\frac{c}{\pi_{1}}\right)_{5}\,=\,\left(\frac{c}{\pi_{2}}\right)_{5}\,=\,1$,
we get that $\pi_{1}$ and $\pi_{2}$ split completly in $k_{0}(\sqrt[5]{c})$,
so the equations $X^{5}\,\equiv\,c\,(\mathrm{mod\,\pi_{1}})$ and
$X^{5}\,\equiv\,c\,(\mathrm{mod\,\pi_{2}})$ are resolved on
$\mathcal{O}_{k_{0}}$, therefore
$c^{\frac{p^{2}-1}{5}}\,\equiv\,1\,(\mathrm{mod\,\pi_{1}})$ and
$c^{\frac{p^{2}-1}{5}}\,\equiv\,1\,(\mathrm{mod\,\pi_{2}})$, thus
$c^{\frac{p^{2}-1}{5}}\,\equiv\,1\,(\mathrm{mod\,p})$, and by Euler’s
citernion, we have $c$ is rest quintic modulo $p$.
If $c$ is rest quintic modulo $p$ we have
$c^{\frac{p^{2}-1}{5}}\,\equiv\,1\,(\mathrm{mod\,p})$, so
$c^{\frac{p^{2}-1}{5}}\,\equiv\,1\,(\mathrm{mod\,\pi_{1}})$ and
$c^{\frac{p^{2}-1}{5}}\,\equiv\,1\,(\mathrm{mod\,\pi_{2}})$, therefore the
equations $X^{5}\,\equiv\,c\,(\mathrm{mod\,\pi_{1}})$ and
$X^{5}\,\equiv\,c\,(\mathrm{mod\,\pi_{2}})$ are resolved on
$\mathcal{O}_{k_{0}}$, namly
$\left(\frac{c}{\pi_{1}}\right)_{5}\,=\,\left(\frac{c}{\pi_{2}}\right)_{5}\,=\,1$.

  * (3)

Accoding to [References, theorem 5.15] we have
$\left(\frac{\pi_{1}^{\alpha_{1}}\pi_{2}^{\alpha_{2}},\pi_{1},\pi_{2}}{\pi_{1}}\right)_{5}\,=\,1$,
so using properties 4.1 we get that
$\left(\frac{\pi_{2}}{\pi_{1}}\right)_{5}\,=\,1$ and since $\pi_{1}$ and
$\pi_{2}$ play symmetric roles we have also
$\left(\frac{\pi_{2}}{\pi_{1}}\right)_{5}\,=\,1$. Thus we deduce that
$\left(\frac{\pi_{2}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi_{1}}{\pi_{2}}\right)_{5}\,=\,1$.

∎

###  6.1 Case 1: $n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
with $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{5^{e}p})$ be a pure quintic field, where
$e,\in\\{1,2,3,4\\}$, and $p$ a prime number such that
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$, and $k\,=\,\Gamma(\zeta_{5})$ its
normal closure. We have $p\,=\,\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$
are primes of $k_{0}$, such that
$\pi\,\equiv\,-\pi_{2}\,\equiv\,1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$ and
$\pi_{1}^{\tau}\,=\,-\pi_{2}$. By proposition 3.1, $p$ is totaly ramified in
$\Gamma$, therefore $\pi_{1},\,\pi_{2}$ are ramified in $k$. According to
[References, Lemma 5.1] we have $\lambda\,=\,1-\zeta_{5}$ is ramified in
$k/k_{0}$ because $5^{e}p\,\not\equiv\pm 1\pm 7\,(\mathrm{mod}\,\lambda^{5})$
$(e\neq 0)$.
If we note by $\mathcal{P}_{1},\mathcal{P}_{2}$ and $\mathcal{I}$
respectively, the prime ideals of $k$ above $\pi_{1},\pi_{2}$ and $\lambda$
respectively, we get that $\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k}$
$(i=1,2)$, $\mathcal{I}^{5}\,=\,\lambda\mathcal{O}_{k}$,
$\mathcal{P}_{i}^{\sigma}\,=\,\mathcal{P}_{i}$,
$\mathcal{P}_{1}^{\tau}\,=\,\mathcal{P}_{2}$,
$\mathcal{I}^{\sigma}\,=\,\mathcal{I}^{\tau}\,=\,\mathcal{I}$. .
According to proposition 5.1, we have $C_{k,5}\,\cong\,C_{k,5}^{+}\times
C_{k,5}^{-}$, such that $C_{k,5}^{+}$ and $C_{k,5}^{-}$ are cyclic subgoups of
order $5$. Since $\mathcal{P}_{i}^{\sigma}\,=\,\mathcal{P}_{i}$ $(i\,=\,1,2)$,
$\mathcal{P}_{1}^{\tau}\,=\,\mathcal{P}_{2}$,
$\mathcal{P}_{2}^{\tau}\,=\,\mathcal{P}_{1}$ and
$\mathcal{I}^{\sigma}\,=\,\mathcal{I}^{\tau}\,=\,\mathcal{I}$, we get that
$[\mathcal{P}_{1}]$,$[\mathcal{P}_{2}]$ and $[\mathcal{I}]$ are ambiguous
classes. As $C_{k,5}^{(\sigma)}$ is elementary group of rank $1$, we can
deduce that $C_{k,5}^{(\sigma)}\,=\,\langle[\mathcal{P}_{1}]\rangle$. Also as
$[\mathcal{P}_{1}]^{\tau^{2}}\,=\,[\mathcal{P}_{1}]$, we have
$C_{k,5}^{+}\,=\,\langle[\mathcal{P}_{1}]\rangle$, therefore
$C_{k,5}^{+}\,=\,C_{k,5}^{(\sigma)}\,=\,\langle[\mathcal{P}_{1}]\rangle$ if
and only if $\mathcal{P}_{1}$ is not principal.
we argue by absurd: Assume that $\mathcal{P}_{1}$ is principal, we have:

$[\mathcal{P}_{1}]\,=\,1\Longrightarrow\,\exists\beta\in\mathcal{O}_{k}\,|\,\mathcal{P}_{1}\,=\,\beta\mathcal{O}_{k}$

$\Longrightarrow\,\mathcal{N}_{k/k_{0}}(\mathcal{P}_{1})\,=\,\mathcal{N}_{k/k_{0}}(\beta\mathcal{O}_{k})$

$\Longrightarrow\,\pi_{1}\mathcal{O}_{k_{0}}\,=\,\mathcal{N}_{k/k_{0}}(\beta)\mathcal{O}_{k_{0}}$

$\Longrightarrow\,\exists\epsilon\in
E_{k_{0}}\,|\,\pi_{1}\,=\,\epsilon\mathcal{N}_{k/k_{0}}(\beta)$

According to [References, theorem 5.15]
$E_{k_{0}}\subset\mathcal{N}_{k/k_{0}}(k^{*})$ because $q^{*}=1$

$\Longrightarrow\,\exists\alpha\in\mathcal{O}_{k}\,|\,\pi_{1}\,=\,\mathcal{N}_{k/k_{0}}(\alpha)$

that is to say $\pi_{1}$ is norm in
$k\,=\,k_{0}(\sqrt[5]{5^{e}\pi_{1}\pi_{2}})$, where $\pi_{1},\pi_{2}$ are two
primes of $k_{0}$ such that $p\,=\,\pi_{1}\pi_{2}$. Hence we have:

$\left(\frac{\pi_{1},5^{e}\pi_{1}\pi_{2}}{\mathcal{P}}\right)_{5}\,=\,1$

for all ideal $\mathcal{P}$ of $k_{0}$.
In particular, we calculate this symbol for
$\mathcal{P}\,=\,\pi_{1}\mathcal{O}_{k_{0}}$ or
$\mathcal{P}\,=\,\pi_{2}\mathcal{O}_{k_{0}}$. In one hand,

$A$ =
$\left(\frac{\pi_{1},5^{e}\pi_{1}\pi_{2}}{\pi_{1}\mathcal{O}_{k_{0}}}\right)_{5}\,=\left(\frac{\pi_{1},5^{e}}{\pi_{1}}\right)_{5}\,.\left(\frac{\pi_{1},\pi_{1}}{\pi_{1}}\right)_{5}.\left(\frac{\pi_{1},\pi_{2}}{\pi_{1}}\right)_{5}$

In the other hand, we have:

  * -

$\left(\frac{\pi_{1},\pi_{1}}{\pi_{1}}\right)_{5}\,=\,1$, because $\pi_{1}$ is
norm in $k_{0}(\sqrt[5]{\pi_{1}})/k_{0}$.

  * -

$\left(\frac{\pi_{1},\pi_{2}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi_{2}}{\pi_{1}}\right)_{5}^{-1}\,=\,1$,
by (4) of properties 4.1, and (3) of 6.2.

  * -

$\left(\frac{\pi_{1},5^{e}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{5}{\pi_{1}}\right)_{5}^{-e}$,
by (4) of properties 4.1.

we get that, $A\,=\,\left(\frac{5}{\pi_{1}}\right)_{5}^{-e}=\,1$. Since
$\pi_{1}$ and $\pi_{2}$ play symmetric roles, then $B$ =
$\left(\frac{\pi_{1},5^{e}\pi_{1}\pi_{2}}{\pi_{2}\mathcal{O}_{k_{0}}}\right)_{5}\,=\,\left(\frac{5}{\pi_{2}}\right)_{5}^{-e}$,
and since
$\left(\frac{\pi_{1},5^{e}\pi_{1}\pi_{2}}{\mathcal{P}}\right)_{5}\,=\,1$, for
all prime ideals of $k_{0}$, then $A\,=\,B\,=\,1$, namly

$\left(\frac{5}{\pi_{1}}\right)_{5}^{-e}\,=\,\left(\frac{5}{\pi_{2}}\right)_{5}^{-e}\,=\,1$.

In fact that $5$ is not a quintic residue modulo $p$, imply that

$\left(\frac{5}{\pi_{1}\pi_{2}}\right)_{5}\,=\,\left(\frac{5}{\pi_{1}}\right)_{5}\left(\frac{5}{\pi_{2}}\right)_{5}\,\neq\,1$.

then

$\left(\frac{5}{\pi_{1}}\right)_{5}\,\neq\,1\,\mathrm{or}\,\left(\frac{5}{\pi_{2}}\right)_{5}\,\neq\,1$.

Since $5$ does not divide $e$, then

$\left(\frac{5}{\pi_{1}}\right)_{5}^{e}\,\neq\,1\,\mathrm{or}\,\left(\frac{5}{\pi_{2}}\right)_{5}^{e}\,\neq\,1$.

Which is contradiction. Consequently, the ideal $\mathcal{P}_{1}$ is not
principal.
The second step in the proof of the case 1, is to find a no ambiguous class,
which generates the group $C_{k,5}^{-}$. By class fields theory, the classes
arise from unramified primes of $k_{0}$, that are not ambigous. Let $l$ prime
integer such that $l\,\neq\,p$, then $l$ is unramified in $\Gamma$. Let
$\pi^{\prime}$ a prime of $k_{0}$ above $l$, then $\pi^{\prime}$ is unramified
in $k$. Let $\mathcal{L}$ a prime ideal of $k$ above $\pi^{\prime}$, then
$C_{k,5}^{-}\,=\,\langle[\mathcal{L}]^{1-\tau^{2}}\rangle$ if and only if
$\mathcal{L}$ is not principal.
we argue by absurd: Assume that $\mathcal{L}$ is principal, we have:

$[\mathcal{L}]\,=\,1\Longrightarrow\,\exists\beta\in\mathcal{O}_{k}\,|\,\mathcal{L}\,=\,\beta\mathcal{O}_{k}$

$\Longrightarrow\,\mathcal{N}_{k/k_{0}}(\mathcal{L})\,=\,\mathcal{N}_{k/k_{0}}(\beta\mathcal{O}_{k})$

$\Longrightarrow\,\pi^{\prime}\mathcal{O}_{k_{0}}\,=\,\mathcal{N}_{k/k_{0}}(\beta)\mathcal{O}_{k_{0}}$

$\Longrightarrow\,\exists\epsilon\in
E_{k_{0}}\,|\,\pi^{\prime}\,=\,\epsilon\mathcal{N}_{k/k_{0}}(\beta)$

According to [References, theorem 5.15]
$E_{k_{0}}\subset\mathcal{N}_{k/k_{0}}(k^{*})$ because $q^{*}=1$

$\Longrightarrow\,\exists\alpha\in\mathcal{O}_{k}\,|\,\pi^{\prime}\,=\,\mathcal{N}_{k/k_{0}}(\alpha)$

that is to say $\pi^{\prime}$ is norm in
$k\,=\,k_{0}(\sqrt[5]{5^{e}\pi_{1}\pi_{2}})$, where $\pi_{1},\pi_{2}$ are two
primes of $k_{0}$ such that $p\,=\,\pi_{1}\pi_{2}$. Hence we have:

$\left(\frac{\pi^{\prime},5^{e}\pi_{1}\pi_{2}}{\mathcal{P}}\right)_{5}\,=\,1$

for all ideal $\mathcal{P}$ of $k_{0}$.
In particular, we calculate this symbol for
$\mathcal{P}\,=\,\pi_{1}\mathcal{O}_{k_{0}}$ or
$\mathcal{P}\,=\,\pi_{2}\mathcal{O}_{k_{0}}$. In one hand,

$A$ =
$\left(\frac{\pi^{\prime},5^{e}\pi_{1}\pi_{2}}{\pi_{1}\mathcal{O}_{k_{0}}}\right)_{5}\,=\left(\frac{\pi^{\prime},5^{e}}{\pi_{1}}\right)_{5}\,.\left(\frac{\pi^{\prime},\pi_{1}}{\pi_{1}}\right)_{5}.\left(\frac{\pi^{\prime},\pi_{2}}{\pi_{1}}\right)_{5}$

In the other hand, we have:

  * -

$\left(\frac{\pi^{\prime},5^{e}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi^{\prime},5}{\pi_{1}}\right)_{5}^{e}\,=\,\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}^{0\times
e}\,=\,1$, by (1), (4) of properties 4.1.

  * -

$\left(\frac{\pi^{\prime},\pi_{2}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}^{0}\,=\,1$,
by (4) of properties 4.1.

  * -

$\left(\frac{\pi^{\prime},\pi_{1}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}$,
by (4) of properties 4.1.

we get that, $A\,=\,\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}=\,1$. Since
$\pi_{1}$ and $\pi_{2}$ play symmetric roles, then $B$ =
$\left(\frac{\pi^{\prime},5^{e}\pi_{1}\pi_{2}}{\pi_{2}\mathcal{O}_{k_{0}}}\right)_{5}\,=\,\left(\frac{\pi^{\prime}}{\pi_{2}}\right)_{5}$,
and since
$\left(\frac{\pi^{\prime},5^{e}\pi_{1}\pi_{2}}{\mathcal{P}}\right)_{5}\,=\,1$,
for all prime ideals of $k_{0}$, then $A\,=\,B\,=\,1$, namly

$\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\pi^{\prime}}{\pi_{2}}\right)_{5}\,=\,1$.

In fact that $l$ is not a quintic residue modulo $p$, imply that

$\left(\frac{\pi^{\prime}}{\pi_{1}\pi_{2}}\right)_{5}\,=\,\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}\left(\frac{\pi^{\prime}}{\pi_{2}}\right)_{5}\,\neq\,1$.

then

$\left(\frac{\pi^{\prime}}{\pi_{1}}\right)_{5}\,\neq\,1\,\mathrm{or}\,\left(\frac{\pi^{\prime}}{\pi_{2}}\right)_{5}\,\neq\,1$.

Which is contradiction. Consequently, the ideal $\mathcal{L}$ is not
principal.
Its easy to see that $[\mathcal{L}]^{1-\tau^{2}}\,\in\,C_{k,5}^{-}$, and
$[\mathcal{L}]^{1-\tau^{2}}\,\neq\,1$, otherwise
$[\mathcal{L}]^{1-\tau^{2}}\,=\,1$ then
$[\mathcal{L}]\,=\,[\mathcal{L}]^{\tau^{2}}$, therefore
$[\mathcal{L}]\,\in\,C_{k,5}^{+}\,=\,C_{k,5}^{(\sigma)}$, which contredict the
fact that the class $[\mathcal{L}]$ is not ambigous. Finaly we deduce that

$C_{k,5}\,\cong\,\langle[\mathcal{P}_{1}]\rangle\,\times\langle[\mathcal{L}]^{1-\tau^{2}}\rangle\,\cong\,\langle[\mathcal{P}_{1}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

######  Remark 6.1.

Since $[\mathcal{P}_{2}]$ and $[\mathcal{I}]$ are also ambiguous classes, we
can prove by the same reasoning that:

$C_{k,5}\,\cong\,\langle[\mathcal{P}_{2}],[\mathcal{L}]^{1-\tau^{2}}\rangle\,\cong\,\langle[\mathcal{I}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

###  6.2 Case 2: $n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $q\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{p^{e}q})$ be a pure quintic field, where
$e\in\\{1,2,3,4\\}$, and $p$, $q$ primes such that
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $q\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$, and $k\,=\,\Gamma(\zeta_{5})$ the normal closure. We
have $p\,=\,\pi_{1}\pi_{2}$, where $\pi_{1}$ and $\pi_{2}$ are primes of
$k_{0}$, such that
$\pi_{1}\,\equiv\,-\pi_{2}\,\equiv\,1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$
and $\pi_{1}^{\tau}\,=\,-\pi_{2}$. $q$ is inert in $k_{0}$ from 3.3. By
proposition 3.1, $p$ and $q$ are totaly ramified in $\Gamma$, therefore
$\pi_{1},\,\pi_{2}$ and $q$ are ramified in $k$. According to
[References,section 5, Lemma 5.1] we have $\lambda\,=\,1-\zeta_{5}$ is not
ramified in $k/k_{0}$ because $n\,\equiv\pm 1\pm
7\,(\mathrm{mod}\,\lambda^{5})$.
If we note by $\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{Q}$ respectively, the
prime ideals of $k$ above $\pi_{1},\pi_{2},q$ respectively, we get that
$\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k}$ $(i=1,2)$,
$\mathcal{Q}^{5}\,=\,q\mathcal{O}_{k}$. Let $l$ be a prime integer such that
$l\,\neq\,p$ and $l\,\neq\,q$, then $l$ is unramified in $\Gamma$. Let
$\mathcal{L}$ a prime ideal of $k$ above $l$.
We prove the result of the second point of the main theorem 1.1, by the same
reasoning of case $1$ 6.1, it sufficient to replace $5$ by $q$. Then we have,

$C_{k,5}\,\cong\,\langle[\mathcal{P}_{1}]\rangle\,\times\langle[\mathcal{L}]^{1-\tau^{2}}\rangle\,\cong\,\langle[\mathcal{P}_{1}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

######  Remark 6.2.

Since $[\mathcal{P}_{2}]$ and $[\mathcal{Q}]$ are also ambiguous classes, we
can prove by the same reasoning that:

$C_{k,5}\,\cong\,\langle[\mathcal{P}_{2}],[\mathcal{L}]^{1-\tau^{2}}\rangle\,\cong\,\langle[\mathcal{Q}],[\mathcal{L}]^{1-\tau^{2}}\rangle$

###  6.3 Case 3: $n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\equiv\,-1\,(\mathrm{mod}\,25)$

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{p^{e}})$ be a pure quintic field, where
$e\in\\{1,2,3,4\\}$, and $p$ a prime such that,
$p\,\equiv\,-1\,(\mathrm{mod}\,25)$, and $k\,=\,\Gamma(\zeta_{5})$ its normal
closure. Without loosing generality, we can choose $e\,=\,1$. Since the field
$\Gamma$ is of second kind, then by proposition 3.5, $5$ decompose in $k$ as
$5\mathcal{O}_{k}\,=\,\mathcal{B}^{4}_{1}\mathcal{B}^{4}_{2}\mathcal{B}^{4}_{3}\mathcal{B}^{4}_{4}\mathcal{B}^{4}_{5}$,
where $\mathcal{B}_{i}$ are prime ideals of $k$. Since $[\mathcal{B}_{i}]$ and
$[\mathcal{B}_{j}]$ with $i\,\neq\,j\in\\{1,2,3,4\\}$ are in $C_{k,5}$ we have
$\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle\,\subset\,C_{k,5}$. To
prove that $\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle\,=\,C_{k,5}$, we
should prove that $\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle$ has
order $25$. Its sufficient to prove that
$[\mathcal{B}_{i}^{a_{1}}\mathcal{B}_{j}^{a_{2}}]\,=\,1$ for
$a_{1},a_{2}\in\\{0,1,2,3,4,\\}$ if and only if $a_{1}\,=\,a_{2}\,=\,0$.

$[\mathcal{B}_{i}^{a_{1}}\mathcal{B}_{j}^{a_{2}}]\,=\,1\Longrightarrow\,\exists\,\beta\in\mathcal{O}_{k}\,|\,\mathcal{B}_{i}^{a_{1}}\mathcal{B}_{j}^{a_{2}}\,=\,\beta\mathcal{O}_{k}$

$\Longrightarrow\,\mathcal{N}_{k/k_{0}}(\mathcal{B}_{i}^{a_{1}}\mathcal{B}_{j}^{a_{2}})\,=\,\mathcal{N}_{k/k_{0}}(\beta\mathcal{O}_{k})$

$\Longrightarrow\,\lambda^{a_{1}+a_{2}}\mathcal{O}_{k_{0}}\,=\,\mathcal{N}_{k/k_{0}}(\beta)\mathcal{O}_{k_{0}}$
with $\lambda\,=\,1-\zeta_{5}$

$\Longrightarrow\,\exists\,\epsilon\in
E_{k_{0}}\,|\,\lambda^{a_{1}+a_{2}}\,=\,\epsilon\mathcal{N}_{k/k_{0}}(\beta)$

According to [References, theorem 5.15]
$E_{k_{0}}\subset\mathcal{N}_{k/k_{0}}(k^{*})$ because $q^{*}=2$

$\Longrightarrow\,\exists\alpha\in\mathcal{O}_{k}\,|\,\lambda^{a_{1}+a_{2}}\,=\,\mathcal{N}_{k/k_{0}}(\alpha)$

that is to say $\mathcal{B}_{i}^{a_{1}}\mathcal{B}_{j}^{a_{2}}$ is norm in
$k\,=\,k_{0}(\sqrt[5]{p})$, where $p\,=\,\pi_{1}\pi_{2}$ in $k_{0}$. Hence we
have:

$\left(\frac{\lambda^{a_{1}+a_{2}},\pi_{1}\pi_{2}}{\mathcal{P}}\right)_{5}\,=\,1$

for all ideal $\mathcal{P}$ of $k_{0}$.
In particular, we calculate this symbol for
$\mathcal{P}\,=\,\pi_{1}\mathcal{O}_{k_{0}}$ or
$\mathcal{P}\,=\,\pi_{2}\mathcal{O}_{k_{0}}$. Using (1) of properties 4.1 We
have,

$A\,=\,$$\left(\frac{\lambda^{a_{1}+a_{2}},\pi_{1}\pi_{2}}{\pi_{1}\mathcal{O}_{k_{0}}}\right)_{5}\,=\,\left(\frac{\lambda^{a_{1}},\pi_{1}}{\pi_{1}}\right)_{5}.\left(\frac{\lambda^{a_{1}},\pi_{2}}{\pi_{1}}\right)_{5}.\left(\frac{\lambda^{a_{2}},\pi_{1}}{\pi_{1}}\right)_{5}.\left(\frac{\lambda^{a_{2}},\pi_{2}}{\pi_{1}}\right)_{5}$

then, we get:

  * -

$\left(\frac{\lambda^{a_{1}},\pi_{1}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\lambda,\pi_{1}}{\pi_{1}}\right)_{5}^{a_{1}}\,=\,\left(\frac{\lambda}{\pi_{1}}\right)_{5}^{a_{1}}$,
by (1) and (9) of properties 4.1.

  * -

$\left(\frac{\lambda^{a_{1}},\pi_{2}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\lambda,\pi_{2}}{\pi_{1}}\right)_{5}^{a_{1}}\,=\,\left(\frac{\lambda}{\pi_{1}}\right)_{5}^{0\times
a_{1}}\,=\,1$, by (3) and (4) of properties 4.1.

  * -

$\left(\frac{\lambda^{a_{2}},\pi_{1}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\lambda,\pi_{1}}{\pi_{1}}\right)_{5}^{a_{2}}\,=\,\left(\frac{\lambda}{\pi_{1}}\right)_{5}^{a_{2}}$,
by (1) and (9) of properties 4.1.

  * -

$\left(\frac{\lambda^{a_{2}},\pi_{2}}{\pi_{1}}\right)_{5}\,=\,\left(\frac{\lambda,\pi_{2}}{\pi_{1}}\right)_{5}^{a_{2}}\,=\,\left(\frac{\lambda}{\pi_{1}}\right)_{5}^{0\times
a_{2}}\,=\,1$, by (3) and (4) of properties 4.1.

we get that,
$A\,=\,\left(\frac{\lambda}{\pi_{1}}\right)_{5}^{a_{1}+a_{2}}=\,1$. Since
$\pi_{1}$ and $\pi_{2}$ play symmteric roles, then $B$ =
$\left(\frac{\lambda^{a_{1}+a_{2}},\pi_{1}\pi_{2}}{\pi_{2}\mathcal{O}_{k_{0}}}\right)_{5}\,=\,\left(\frac{\lambda}{\pi_{2}}\right)_{5}^{a_{1}+a_{2}}$,
and since
$\left(\frac{\lambda^{a_{1}+a_{2}},\pi_{1}\pi_{2}}{\mathcal{P}}\right)_{5}\,=\,1$,
for all prime ideals of $k_{0}$, then $A\,=\,B\,=\,1$,
namly
$\left(\frac{\lambda}{\pi_{1}}\right)_{5}^{a_{1}+a_{2}}\,=\,\left(\frac{\lambda}{\pi_{2}}\right)_{5}^{a_{1}+a_{2}}\,=\,1$.
Since $5$ is not a quintic residue modulo $p$, then $a_{1}\,=\,a_{2}\,=\,0$
because $\left(\frac{\lambda}{\pi_{1}}\right)_{5}$ is a $5^{th}$ root of
unity, therfore $\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle$ has order
$25$, and since $\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle\,\subset
C_{k,5}$ we deduce that
$C_{k,5}\,=\,\langle[\mathcal{B}_{i}],[\mathcal{B}_{j}]\rangle$.

######  Corollary 6.1.

Using the same notation as above, we have:

$C_{k,5}^{(\sigma)}\,=\,C_{k,5}^{+}\,=\,\langle[\mathcal{B}_{1}\mathcal{B}_{2}\mathcal{B}_{3}\mathcal{B}_{4}\mathcal{B}_{5}]\rangle$.

$C_{k,5}^{-}\,=\,\langle[\mathcal{B}_{i}]^{1-\tau^{2}}\rangle$ for
$(i=1,2,3,4,5)$.

The $5$-class group can be generated also by:

$C_{k,5}\,=\,\langle[\mathcal{B}_{1}\mathcal{B}_{2}\mathcal{B}_{3}\mathcal{B}_{4}\mathcal{B}_{5}],[\mathcal{B}_{i}]^{1-\tau^{2}}\rangle$
for $(i=1,2,3,4,5)$.

###### Proof.

The fact that the ideals $\mathcal{B}_{i}$ are not principals by the same
reasoning above, we prove that
$[\mathcal{B}_{1}\mathcal{B}_{2}\mathcal{B}_{3}\mathcal{B}_{4}\mathcal{B}_{5}]^{\sigma}\,=\,[\mathcal{B}_{1}\mathcal{B}_{2}\mathcal{B}_{3}\mathcal{B}_{4}\mathcal{B}_{5}]$
and
$[\mathcal{B}_{1}\mathcal{B}_{2}\mathcal{B}_{3}\mathcal{B}_{4}\mathcal{B}_{5}]^{\tau^{2}}\,=\,[\mathcal{B}_{1}\mathcal{B}_{2}\mathcal{B}_{3}\mathcal{B}_{4}\mathcal{B}_{5}]$
by applying the decomposition of $5$ in the normal closure $k$.

Since $\mathcal{B}_{i}$ are not principals then
$[\mathcal{B}_{i}]^{1-\tau^{2}}\in C_{k,5}^{-}$

That comes from $C_{k,5}\cong C_{k,5}^{+}\times C_{k,5}^{-}$ ∎

##  7 Numerical examples

Using the system PARI/GP [References], we illustrates our main results Theorem
1.1 by numerical examples.

###  7.1 Case 1: $n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
with $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$

In this case we have:
$C_{k,5}\,=\,\langle[\mathcal{P}],[\mathcal{L}]^{1-\tau^{2}}\rangle$, where
$\mathcal{P}$ is prime ideals of $k$ above $p$, and $\mathcal{L}$ is prime
ideals of $k$ above $l$ such that $l\,\neq\,p$. The following table verifies,
for each primes $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $l\,\neq\,p$, such
that $5$ and $l$ are not a quintic residue modulo $p$ and $C_{k,5}$ is of type
$(5,5)$, that the ideals $\mathcal{P}$ and $\mathcal{L}$ are not principals,
and are of order $5$

Table 1:

$p$ | $l$ |  Type of $C_{k,5}$ |  Is principal $\mathcal{P}$ |  Is principal $\mathcal{L}$ |  Is principal $\mathcal{P}^{5}$ |  Is principal $\mathcal{L}^{5}$
---|---|---|---|---|---|---
19 | 2 | (5,5) | [4,0] | [1,0] | [0,0] | [0,0]
19 | 113 | (5,5) | [4,0] | [4,0] | [0,0] | [0,0]
29 | 43 | (5,5) | [6,0,0,0] | [6,0,1,0] | [0,0,0,0] | [0,0,0,0]
29 | 149 | (5,5) | [6,0,0,0] | [2,0,0,0] | [0,0,0,0] | [0,0,0,0]
59 | 67 | (5,5) | [4,0,0,0] | [6,0,0,0] | [0,0,0,0] | [0,0,0,0]
79 | 97 | (5,5) | [21,21,0,0] | [28,28,0,0] | [0,0,0,0] | [0,0,0,0]
79 | 307 | (5,5) | [21,21,0,0] | [7,7,0,0] | [0,0,0,0] | [0,0,0,0]
89 | 19 | (5,5) | [1,0] | [4,0] | [0,0] | [0,0]
89 | 101 | (5,5) | [1,0] | [4,4] | [0,0] | [0,0]
109 | 13 | (5,5) | [2,0] | [1,0] | [0,0] | [0,0]
109 | 103 | (5,5) | [2,0] | [1,0] | [0,0] | [0,0]
179 | 157 | (5,5) | [3,0] | [2,0] | [0,0] | [0,0]
299 | 83 | (5,5) | [8,12,0,0] | [12,8,0,0] | [0,0] | [0,0]
239 | 3 | (5,5) | [3,0] | [2,0] | [0,0] | [0,0]
269 | 157 | (5,5) | [1,0] | [3,0] | [0,0] | [0,0]

###  7.2 Case 2: $n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $q\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$

In this case we have:
$C_{k,5}\,=\,\langle[\mathcal{P}],[\mathcal{L}]^{1-\tau^{2}}\rangle$, where
$\mathcal{P}$ is prime ideal of $k$ above $p$, and $\mathcal{L}$ is prime
ideals of $k$ above $l$ such that $l\,\neq\,p$ and $l\,\neq\,q$. The following
table verifies, for each primes $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$,
$q\,\not\equiv\,\pm 7\,(\mathrm{mod}\,25)$, $l\,\neq\,p$ and $l\,\neq\,q$ such
that $q$ and $l$ are not a quintic residue modulo $p$ and $C_{k,5}$ is of type
$(5,5)$, that the ideals $\mathcal{P}$ and $\mathcal{L}$ are not principals,
and are of order $5$.
. Table 2
$p$ $q$ $l$ Type of $C_{k,5}$ Is principal $\mathcal{P}_{1}$ Is principal
$\mathcal{P}_{2}$ Is principal $\mathcal{P}_{1}^{5}$ Is principal
$\mathcal{P}_{2}^{5}$ 19 3 53 (5,5) [1,0] [3,0] [0,0] [0,0] 19 3 67 (5,5)
[1,0] [4,0] [0,0] [0,0] 19 53 7 (5,5) [3,0] [1,0] [0,0] [0,0] 19 53 23 (5,5)
[3,0] [1,0] [0,0] [0,0] 29 17 157 (5,5) [28,14,0,0] [14,7,0,0] [0,0,0,0]
[0,0,0,0] 59 2 13 (5,5) [0,2] [0,2] [0,0] [0,0] 59 2 47 (5,5) [0,2] [0,3]
[0,0] [0,0] 59 23 127 (5,5) [2,0,0,0] [6,0,0,0] [0,0,0,0] [0,0,0,0] 89 37 61
(5,5) [3,0] [3,0] [0,0] [0,0] 89 37 73 (5,5) [3,0] [2,0] [0,0] [0,0] 109 23 47
(5,5) [28,0,0,0] [14,0,0,0] [0,0,0,0] [0,0,0,0]

###  7.3 Case 3: $n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\equiv\,-1\,(\mathrm{mod}\,25)$

In this case we have:
$C_{k,5}\,=\,\langle[\mathcal{B}_{1}],[\mathcal{B}_{2}]\rangle$, where
$\mathcal{B}_{i}$ are prime ideals of $k$ above $5$. The following table
verifies, for each prime $p\,\equiv\,-1\,(\mathrm{mod}\,5)$ and
$p\,\equiv\,-1\,(\mathrm{mod}\,25)$, such that,
$\left(\frac{5}{p}\right)_{5}\,\neq\,1$ and $C_{k,5}$ is of type $(5,5)$, that
the ideals $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are not principals, and are
of order $5$

Table 3:

$p$ |  Type of $C_{k,5}$ |  Is principal $\mathcal{B}_{1}$ |  Is principal $\mathcal{B}_{2}$ |  Is principal $\mathcal{B}_{1}^{5}$ |  Is principal $\mathcal{B}_{2}^{5}$
---|---|---|---|---|---
149 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
199 | (5,5) | [6,0,0,0] | [6,0,0,0] | [0,0,0,0] | [0,0,0,0]
499 | (5,5) | [1,4] | [1,4] | [0,0] | [0,0]
599 | (5,5) | [3,0] | [3,0] | [0,0] | [0,0]
2099 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
2549 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
2699 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
2749 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
3299 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
4049 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
4099 | (5,5) | [8,0,0,0] | [8,0,0,0] | [0,0,0,0] | [0,0,0,0]
4349 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
4549 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
4999 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
5099 | (5,5) | [8,0,0,0] | [8,0,0,0] | [0,0,0,0] | [0,0,0,0]
5399 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
5749 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
6199 | (5,5) | [0,4,0,0,0,0,0,0] | [0,4,0,0,0,0,0,0] | [0,0,0,0,0,0,0,0] | [0,0,0,0,0,0,0,0]
6299 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]
6599 | (5,5) | [1,0] | [1,0] | [0,0] | [0,0]

## References

  * [1] W.E.H. Berwich, _Integral bases_ , Cambridge Tracts in Math. and Math.Phys, Vol 22, 1927.
  * [2] R.Dedekind. _Ueber die Anzahl der Idealklassen in reinen kubischen Zahlkorpern_ , J fur reine und angewandte mathematik, Bd 121 (1900), 40-123.
  * [3] F.Elmouhib, M. Talbi, and A. Azizi, 5-rank of ambiguous class groups of quintic Kummer extensions, arXiv:2003.00761v2.
  * [4] F. Gerth III, _On $3$-class groups of cyclic cubic extensions of certain number fields_, J. Number Theory 8 (1976), No. 1, 84–98.
  * [5] F. Gerth III, _On $3$-class groups of pure cubic fields,_ J. Reine Angew. Math. 278/279 (1975), 52–62.
  * [6] F. Gerth III, _On $3$-class groups of certain pure cubic fields_,Bull. Austral. Math. Soc. 72 (2005), 471–476.
  * [7] D. Grant. _A proof of quintic reciprocity using the arithmetic of $y^{2}=x^{5}+\frac{1}{4}$_. ACTA ARITHMETICA (1996).
  * [8] G.Gras, _Sur les $l$-classes d’idéux dans les extensions cycliques relatives de degré premier impaire $l$_. _Annales de l’institut Fourier_ , (1973).
  * [9] H. Hasse, Newe Begrundung Und Verallgemeinerung der Theorie der Normenrest Symbols, Journal fur reine und ang. Math. 162 (1930), 134-143.
  * [10] E.Hecke, _Lectures on the Theory of Algebraic Numbers_ , GTM , Vol. 77, Springer-Verlag 1981.
  * [11] K.Ireland and M.Rosen, _A Classical Introduction to modern Number Theory_. Graduate Texts in Mathematics 84, Springer-Verlag (1982).
  * [12] G.J Janus, _Algebraic Number Fields_. Academic Press, New York-London (1973).
  * [13] M.Kulkarni,D. Majumdar, B.Sury _$l$ -class groups of cyclic extension of prime degree $l$_, J. Ramanujan Math. Soc. 30, No.4 (2015), 413-454.
  * [14] H. Kobayashi, _Class numbers of pure quintic fields_ , Journal of Number Theory 160 (2016) 463-477.
  * [15] C. Parry, _Class number relations in pure quintic felds_ , Symposia Mathematica. 15 (1975), 475-485.
  * [16] Pierre Samuel. _Théorie algébrique des nombres_. Hermann, editeurs des sciences et des arts (1971).
  * [17] Lawrence C. Washington, _Introduction to Cyclotomic Fields_ , Springer-Verlag New York Inc (1982).
  * [18] The PARI Group, PARI/GP, Version 2.4.9, Bordeaux, 2017, http://pari.math.u-bordeaux.fr.

Fouad ELMOUHIB
Department of Mathematics and Computer Sciences,
Mohammed 1st University,
Oujda - Morocco,
<EMAIL_ADDRESS>

Mohamed TALBI
Regional Center of Professions of Education and Training in the Oriental,
Oujda - Morocco,
<EMAIL_ADDRESS>

Abdelmalek AZIZI
Department of Mathematics and Computer Sciences,
Mohammed 1st University,
Oujda - Morocco,
<EMAIL_ADDRESS>