text:
\section{Introduction}\label{section-introduction}
One of the most important problems in number theory is the determination of the structure of class group of a number field, particularly its rank. The case of quadratic fields, Gauss's genus theory, determines the rank of $2$-class group. In a series of papers ( see [\ref{GER1}], [\ref{GER2}], [\ref{GER3}]), Frank Gerth III proved several results on the $3$-class groups of pure cubic extension of $\mathbb{Q}$ and cyclic cubic extension of $\mathbb{Q}$. Recently, S.Aouissi, M.C.Ismaili, M.Talbi, A.Azizi in [\ref{SIH}] had classified all fields $\mathbb{Q}(\sqrt[3]{n},\zeta_3)$ whose $3-$class group is of type $(9,3)$. \\
Let $k \,=\, \mathbb{Q}(\sqrt[5]{n},\zeta_5)$, a number of researchers have studied the $5$-class group $C_{k,5}$. M.Kulkarni, D.Majumdar and B.Sury in [\ref{Mani}] proved some results that are seen as generalisation of Gerth's work to the case of any odd prime, and they give more details in case of $5$-class group of $k$. In [\ref{Pa}], C.Parry gives a formula between class numbers of pure quintic field $\mathbb{Q}(\sqrt[5]{n})$ and its normal closure $k$. In \ref{Kobaya} H.Kobayashi proved that if the radicand $n$ has a prime factor $p$ congruent to $-1$ modulo $5$, then the class number of the pure quintic field $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ is a multiple of $5$, and the class number of $k$ is multiple of $25$. \\
Let $n>1$ be a $5^{th}$ power-free integer and
$k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_5)$ be a quintic Kummer extension of the cyclotomic field $k_0\,=\,\mathbb{Q}(\zeta_5)$. By $C_{k,5}^{(\sigma)}$ we denote the $5$-group of ambiguous ideal classes under the action of $Gal(k/k_0)$ = $\langle\sigma\rangle$, i.e $C_{k,5}^{(\sigma)}\,=\, \{\mathcal{A}\,\in\,C_{k,5}|\, \mathcal{A}^\sigma\,=\,\mathcal{A}\}$. Let $k^{\ast}\,=\,(k/k_0)^{\ast}$ be the maximal abelian extension of $k_0$ contained in the Hilbert $5$-class field $k_5^{(1)}$ of $k$,
which is called the relative $5$-genus field of $k/k_0$.

We consider the problem of finding the radicands \(n\)
of all pure quintic fields $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, for which the Galois group $\operatorname{Gal}(k^{\ast}/k)$ is non-trivial. The present work gives the complete solution of this problem
by characterizing all quintic Kummer extensions $k/k_0$ with  $5$-group of ambiguous ideal classes $C_{k,5}^{(\sigma)}$ of order $5$ or $25$. This paper can be viewed as the continuation of the work of M.Kulkarni, D.Majumdar and B.Sury in [\ref{Mani}]. \\
In fact, we shall prove the following Main Theorem:
\begin{theorem}
\label{thm:Rank1}
Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n>1$ is a $5^{th}$ power-free integer, and $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_5)$ be its normal closure. We assume that the $5-$class group $C_{k,5}$ is of type $(5,5)$.\\
(1) If rank $(C_{k,5}^{(\sigma)})\,=\,1$, then the integer $n$ can be written in one of the following forms:
\begin{equation}
\label{eqn:Rank1}
 n\,=\,\left\lbrace
   \begin{array}{ll}

  5^{e}q_1^{2}q_2 \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25) & \text{ with } \quad q_1 \,\text{ or }\, q_2$  $\not\equiv \,\pm7\, (\mathrm{mod}\, 25)\\

   5^ep \not\equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \text{ with } \quad p\, \not\equiv \, -1\,(\mathrm{mod}\,25)\\

   5^{e}q_1 \not\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)& \text{ with }\quad q_1\,\equiv \,\pm7\, (\mathrm{mod}\, 25)\\

   p^{e}q_1 \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \text{ with } \quad p\, \not\equiv \, -1\,(\mathrm{mod}\,25),\, q_1\, \not\equiv \, \pm7\,(\mathrm{mod}\,25)\\

   p^{e} \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) &  \text{ with } \quad p\, \equiv \, -1\,(\mathrm{mod}\,25)\\

   q_1^{e_1}q_2\,\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25) & \text{ with } \quad q_i\,\equiv \,\pm7\, (\mathrm{mod}\, 25)

   \end{array}
   \right.
\end{equation}
where $p\,\equiv \, -1\, (\mathrm{mod}\, 5)$ and $q_1,q_2\,\equiv \, \pm2\, (\mathrm{mod}\, 5)$ are primes and
$e, e_1$ are integers in $\{1,2,3,4\}$.\\\\
(2) If rank $(C_{k,5}^{(\sigma)})\,=\,2$, then the integer $n$ can be written in one of the following forms:
\begin{equation}
\label{eqn:Rank1}
 n\,=\,\left\lbrace
   \begin{array}{ll}
   5^el \not \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \quad \text{ with } \quad l\, \not\equiv \, 1\,(\mathrm{mod}\,25),\\

   l^{e_1}q_1 \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \quad \text{ with } \quad l\, \equiv \, 1\,(\mathrm{mod}\,5),\, q_1\, \equiv \, \pm2,\pm7,\pm3\,(\mathrm{mod}\,25)\\

   l^{e_1} \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \quad \text{ with } \quad l\, \equiv \, 1\,(\mathrm{mod}\,25),\\

   \end{array}
   \right.
\end{equation}
where $l\,\equiv \, 1\, (\mathrm{mod}\, 5)$ and  $q_1\,\equiv \, \pm2\, (\mathrm{mod}\, 5)$ are primes and
$e, e_1$ are integers in $\{1,2,3,4\}$.
\end{theorem}
This result will be underpinned by numerical examples obtained with the computational number theory system \textbf{PARI/GP} [\ref{PRI}] in section 3.

\begin{center}

$\textbf{Notations. \ }$
\end{center}
Throughout this paper, we use the following notations:
\begin{itemize}

\item The lower case letter $p$,$q$ and $l$ will denote a prime numbers such that, $p\,\equiv\,-1\, (\mathrm{mod}\, 5)$, $q\,\equiv\,\pm2\, (\mathrm{mod}\, 5)$ and $l\,\equiv\,1\, (\mathrm{mod}\, 5)$.

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

 \item $k_0\,=\,\mathbb{Q}(\zeta_5)$, the cyclotomic field, where $\zeta_5\,=\,e^{\frac{2i\pi}{5}}$ a primitive $5^{th}$ root of unity.

 \item $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_5)$: the normal closure of $\Gamma$, a quintic Kummer extension of $k_0$.

\item $\Gamma^{'},\,\Gamma^{''},\,\Gamma^{'''},\,\Gamma^{''''},\,$ the four conjugates quintic fields of $\Gamma$, contained in $k$.

 \item $\langle\tau\rangle\,=\,\operatorname{Gal}(k/\Gamma)$ 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 $\tau(\sqrt[5]{n})\,=\,\sqrt[5]{n}$.

 \item $\langle\sigma\rangle\,=\,\operatorname{Gal}(k/k_0)$ 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} $.

 \item $\lambda\,=\,1-\zeta_5$ is prime element above $5$ of $k_0$.

 \item $q^{\ast}\,=\,0,\,1$ or $2$ according to whether $\zeta_{5}$ and $1+\zeta_5$ is not norm or is norm of an element of\\\\ $k^*\,=\,k\setminus\lbrace 0\rbrace$.

 \item $d$: the number of prime ideals of $k_{0}$ ramified in $k$.

 \item For a number field $L$, denote by:

  \begin{itemize}
   \item $\mathcal{O}_{L}$: the ring of integers of $L$;
   \item $E_{L}$: the group of units of $L$;
   \item $C_{L}$, $h_{L}$, $C_{L,5}$: the class group, class number, and $5$-class group of $L$.
   \item $L_5^{(1)}, L^{\ast}$: the Hilbert $5$-class field of $L$, and the absolute genus field of $L$.
  \end{itemize}

\end{itemize}
\begin{center}
\begin{tikzpicture}
\begin{scope}[xscale=2,yscale=2]

  \node (A) at (0,3) {$\mathbf{k}$};
  \node (B) at (1,2){$\mathbf{\Gamma}$ };
  \node (C) at (1.5,2){$\mathbf{\Gamma'}$ };
  \node (D) at (2,2){$\mathbf{\Gamma''}$ };
  \node (E) at (2.5,2){$\mathbf{\Gamma'''}$ };
  \node (F) at (3,2){$\mathbf{\Gamma''''}$ };
  \node (G) at (-1,1) {$\mathbf{k_0}$};
  \node (H)   at (0,0) {$\mathbb{Q}$ };
  \node (I)   at (0,-0.5) {\underline{Figure 1}};

  \draw [-,>=latex] (G) -- (A)   node[midway,above,rotate=40] {5};
  \draw [-,>=latex] (B) -- (A)   node[midway,above,rotate=-40] {4};
  \draw [-,>=latex] (C) -- (A)   node[midway,below left] {};
  \draw [-,>=latex] (D) -- (A)   node[midway,below left] {};
  \draw [-,>=latex] (E) -- (A)   node[midway,below left] {};
  \draw [-,>=latex] (F) -- (A)   node[midway,below left] {};
  \draw [-,>=latex] (H) -- (G)   node[midway,below right] {4};
  \draw [-,>=latex] (H) -- (B)   node[midway,below right] {5};
  \draw [-,>=latex] (H) -- (C)   node[midway,below right] {};
  \draw [-,>=latex] (H) -- (D)   node[midway,below right] {};
  \draw [-,>=latex] (H) -- (E)   node[midway,below right] {};
  \draw [-,>=latex] (H) -- (F)   node[midway,below right] {};

\end{scope}

\end{tikzpicture}

\end{center}

\section{\Large Proof of Main Theorem}
\begin{theorem}{(Decompositon in cyclotomic fields)}\\
Let $m$ a positive integer and $p$ a prime number. Suppose $p$ do not divides $m$, and let $f$ be the smallest positive integer such that $p^f \, \equiv \, 1\, (\mathrm{mod}\, m)$. Then $p$ splits into $\phi(m)/f$ distinct primes in $\mathbb{Q}(\zeta_m)$ each of which has a residue class degree $f$. In particular, p splits completely if and only if $p \, \equiv \, 1\, (\mathrm{mod}\, m)$
\end{theorem}
\begin{proof}
see [\ref{cyc}] page 14.
\end{proof}
\begin{corollary}
\label{deco}
Let $p$ a prime integer, we have :
\item[-] If $p\,=\,5$, then $\lambda\,=\,1-\zeta_5$ is the unique prime over 5 in $\mathbb{Q}(\zeta_5)$.
\item[-] If $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$, then $l$ splits completely in $\mathbb{Q}(\zeta_5)$ as $l\,=\,\pi_1\pi_2\pi_3\pi_4$, with $\pi_i$ are primes in $\mathbb{Q}(\zeta_5)$
\item[-] If $q \, \equiv \, \pm2\,(\mathrm{mod}\, 5)$, then $q$ is inert in $\mathbb{Q}(\zeta_5)$.
\item[-] If $p \, \equiv \, -1\, (\mathrm{mod}\, 5)$, then $p$ splits in $\mathbb{Q}(\zeta_5)$ as $p\,=\,\pi_1\pi_2$, with $\pi_i$ are primes in $\mathbb{Q}(\zeta_5)$.
\end{corollary}
Before proving the main theorem, we give a proof of the existance of a unique prime $l \, \equiv \, 1\,(\mathrm{mod}\,5)$ divides the radicand $n$ in the case of rank $(C_{k,5}^{(\sigma)})\,=\,2$
\begin{theorem}
\label{ran2}
If rank $C_{k,5}^{(\sigma)} = 2$, so  $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$, and there is unique prime $l \, \equiv \, 1\,(\mathrm{mod}\,5)$ that divides the radicand  $n$. Furthermore, we have $(k/k_0)^*\,=\, k_5^{(1)}$.
\end{theorem}
\begin{proof}
If rank $(C_{k,5}^{(\sigma)}) = 2$ so the order of $C_{k,5}^{(\sigma)}$ is at least $25$, since $C_{k,5}^{(\sigma)} \subseteq C_{k,5}$ and $|C_{k,5}|\,=\,25$ because $C_{k,5}$ is of type $(5,5)$ we have $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$ it means that all ideal classes are ambiguous.\\
By class fields theory $C_{k,5}^{1-\sigma}$ correspond to $(k/k_0)^*$ and $Gal(k_5^{(1)}/k)\cong C_{k,5}$. Since $C_{k,5}^{(\sigma)}\,=\,C_{k,5}$, we get $C_{k,5}^{1-\sigma}\,=\,\{1\}$, hence $(k/k_0)^*\,=\, k_5^{(1)}$, and by [\ref{Mani}, proposition 5.8] we know explicitly in this case the Hilbert $5-$class field of $k$.\\
We assume now that there is no prime $l \, \equiv \, 1\,(\mathrm{mod}\,5)$ divides $n$.\\ We can write $n$ as $n\,=\,5^eq_1^{f_1}...q_r^{f_r}p_1^{g_1}....p_s^{g_s}$ with $q_i\, \equiv \, \pm2 \, (mod\, 5)$ and $p_j\, \equiv \, -1\, (mod\, 5)$,\\ $f_i\,=\,1,2,3,4$ for $1\leq i\leq r$ and $g_j\,=\,1,2,3,4$ for $1\leq j\leq s$, and $e\,=\,0,1,2,3,4$, by Corollary \ref{deco} each $q_i$ is inert in $k_0$, and $q_i$ is ramified in $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, also by Corollary \ref{deco} $p_j$ splits in $k_0$ as $p_j\,=\,\pi_1\pi_2$, where $\pi_1,\pi_2$ are primes in $k_0$, and $p_j$ is ramified in $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, so the prime ideals ramified in $k/k_0$ are those above $q_i$ and $\pi_j$ and the ideal above $\lambda$ with $\lambda\,=\,1-\zeta_5$ if 5 is ramified in $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$.\\
If $\lambda$ is ramified in $k/k_0$, we note $\mathcal{I}$ the prime ideal in $k$ above $\lambda$, and for $1\leq i\leq r$ ,$\mathcal{Q}_i$ the prime ideal above $q_i$ in $k$, and for $1\leq j\leq s$ ,$\mathcal{P}_j$ the prime ideal above $\pi_j$ in $k$, with $\pi_j$ is prime of $k_0$ above $p_j$. We have evidently $\mathcal{I}^5\,=\,(\lambda)$, $\mathcal{Q}_i^5\,=\,(q_i)$, $\mathcal{P}_j^5\,=\,(\pi_j)$ in $k$.\\
we note by $C_{k,s}^{(\sigma)}$ the group of strong ambiguous adeal classes. We have to treat two cases:\\

(i) $C_{k,s}^{(\sigma)}\,\neq\, C_{k,5}^{(\sigma)}\,=\,C_{k,5}$:\\

Let $C_{k,5}^+\,=\,\{\mathcal{A}\in\, C_{k,5}| \mathcal{A}^{\tau^2}\,=\,\mathcal{A}\}$ and $C_{k,5}^-\,=\,\{\mathcal{A}\in\, C_{k,5}| \mathcal{A}^{\tau^2}\,=\,\mathcal{A}^{-1}\}$ be a nontrivial subgoups of $C_{k,5}$. We have $(C_{k,5}^{(\sigma)})^+\,=\, C_{k,5}^+$, by [\ref{Mani}, Lemma 6.2] $C_{k,5}^+\, \simeq\, C_{\Gamma,5}$ i.e $C_{k,5}^+$ can be generated by $5-$class comes from $\Gamma$ ($|C_{k,5}^+|\,=\,5$). The strong ambiguous classes are those of primes ramified in $k/k_0$, namly $[\mathcal{Q}_i]$ for $1\leq i\leq r$, $[\mathcal{P}_j]$ for $1\leq j\leq s$ and $[\mathcal{I}]$ if $\lambda$ is ramified in $k/k_0$. Its easy to see that:  $[\mathcal{Q}_i^{\tau^2}]\,=\,[\mathcal{Q}_i]^{\tau^2}\,=\,[\mathcal{Q}_i]$ and $[\mathcal{P}_j^{\tau^2}]\,=\,[\mathcal{P}_j]^{\tau^2}\,=\,[\mathcal{P}_j]$ and $[\mathcal{I}^{\tau^2}]\,=\,[\mathcal{I}]^{\tau^2}\,=\,[\mathcal{I}]$, we now that $C_{k,5}/C_{k,s}^{(\sigma)}$ is generated by image in $C_{k,5}/C_{k,s}^{(\sigma)}$  of element in $C_{k,5}^+$. Since $C_{k,s}^{(\sigma)}$ is generated by $[\mathcal{Q}_i],[\mathcal{P}_j]$ and $[\mathcal{I}]$ if $\lambda$ is ramified in $k/k_0$, all elements of $C_{k,5}$ will be fixed by $\tau^2$, in particular whose of $C_{k,5}^-$, therefore $\forall \mathcal{A} \in C_{k,5}^-$, $\mathcal{A}^{\tau^2}\,=\,\mathcal{A}^{-1}\,=\,\mathcal{A}$ i.e $\mathcal{A}^2\,=\,1$ i.e $\mathcal{A}^4\,=\,1$, hence $\mathcal{A}\,=\,1$ because $\mathcal{A}$ is $5-$class, so we get $C_{k,5}^-\,=\,{1}$, and this contradict the fact that $|C_{k,5}^{-}|=5$.\\

(ii) $C_{k,s}^{(\sigma)}\,=\, C_{k,5}^{(\sigma)}\,=\,C_{k,5}$:\\

In this case $C_{k,5}$ will be generated by $[\mathcal{Q}_i]$, $[\mathcal{P}_j]$ and $[\mathcal{I}]$ if $\lambda$ is ramifed in $k/k_0$, and as in (i) all the classes are fixed by $\tau^2$, which give the same contradiction. Thus we proved the existence of a prime $l$ divides $n$ such that $l\,\equiv\, 1\, (\mathrm{mod}\,5)$.\\
According to [\ref{Mani},section 5.1], we have rank $C_{k,5}^{(\sigma)}\,=\, d-3+q^*$ where $d$ is the number of prime ramified in $k/k_0$ and $q^*$ is an index has as value 0,1 or 2. Assuming that there is two prime $l_1$ and $l_2$ divides $n$ such that $l_i\,\equiv\, 1\, (\mathrm{mod}\,5)$, $(i = 1,2)$, then $d \geq 8$ and rank $C_{k,5}^{(\sigma)}$ is at least $5$, that is impossible. Thus if rank $C_{k,5}^{(\sigma)}\,=\,2$ its exsite unique prime $l\,\equiv\, 1\, (\mathrm{mod}\,5)$ divides $n$.
\end{proof}
\subsection{Proof of Theoreme \ref{thm:Rank1}}

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n\geq 2$ is a $5^{th}$ power-free integer, $k\,=\,\Gamma(\zeta_5)$ be its normal closure, and $C_{k,5}$ be the $5$-class group of $k$. Let $C_{k,5}^{(\sigma)}$ be the ambigous ideal classes group under the action of $Gal(k/k_0)\,=\,\langle\sigma\rangle$. Since $k_0$ has class number $1$, $C_{k,5}^{(\sigma)}$ is un elementary abelian $5$-group, so rank $(C_{k,5}^{(\sigma)})\,=\,1 \, \mathrm{or}\,2$.\\
According to [\ref{Mani},section 5.1], the rank of $C_{k,5}^{(\sigma)}$ is given as follows:
\begin{center}
rank $(C_{k,5}^{(\sigma)})\,=\, d-3+q^*$
\end{center}
where $d$ is the number of prime ideals of $k_{0}$ ramified in $k$, and $q^{\ast}\,=\,0,\,1$ or $2$ according to whether $\zeta_{5}$ and $1+\zeta_5$ is not norm or is norm of an element of $k^*\,=\,k\setminus\lbrace 0\rbrace$ as follows:
\begin{center}
$q^*$ = $ \begin{cases}
2 & \text{if }\, \zeta, 1+\zeta \in N_{k/k_0}(k^*),\\
1 & \text{if }\, \zeta^i(1+\zeta)^j \in N_{k/k_0}(k^*)\,\text{ for some i and j },\\
0 & \text{if }\, \zeta^i(1+\zeta)^j \notin N_{k/k_0}(k^*)\, \text{for}\hspace{2mm} 0\leq i,j\leq 4 \text{ and}\hspace{2mm} i+j\neq 0.\\
\end{cases}$
\end{center}
We can writ $n$ as $n\,=\,\mu\lambda^{e}\pi_1^{e_1}....\pi_g^{e_g}$, where $\mu$ is a unit in $\mathcal{O}_{k_0}$, $\lambda = 1-\zeta_5$, $\pi_1,,,,\pi_g$ are primes in $k_0$ and $e \in \{0,1,2,3,4\}$, $e_i \in \{1,2,3,4\}$ for $1\leq i\leq g$. According to [\ref{Mani}, Lemma 5.1] we have, $\zeta_5\, \in N_{k/k_0}(k^*)\, \Longleftrightarrow\, N_{k_0/\mathbb{Q}}((\pi_i))\, \, \equiv \, 1 \,(\mathrm{mod}\,25)$ for all i, and from [\ref{Jan}, proposition 8.2] if $\pi$ is a prime of $\mathcal{O}_{ k_0}$ over a prime $p\,\in\, \mathbb{Z}$ we have $N_{k_0/\mathbb{Q}}((\pi))\, =\, p^f$, with $f$ is the least positif integer such that $p^f\,\, \equiv \,\,1\, (\mathrm{mod},\,5)$, so we can get that $\zeta_5$ is norm of an element of $k^*\,=\,k\setminus\lbrace 0\rbrace$ if and only if $p^f\,\, \equiv \,\,1\, (\mathrm{mod},\,25)$ for all prime $p\,\neq\,5$ dividing $n$:
\begin{enumerate}
\item[$-$] If $l\,\, \equiv \,\,1\, (\mathrm{mod},\,5)$, by corollary \ref{deco} $l\,=\,\pi_1\pi_2\pi_3\pi_4$, we have $N_{k_0/\mathbb{Q}}((\pi_i))\, =\, l$, so to have $N_{k_0/\mathbb{Q}}((\pi_i))\, \, \equiv \,\, 1\, (\mathrm{mod},\,25)$ the prime $l$ must verify $l\,\, \equiv \,\,1\, (\mathrm{mod}\,25)$.

\item[$-$] If $q\,\, \equiv \,\,\pm2\, (\mathrm{mod},\,5)$ we have $q$ is inert in $k_0$, so $N_{k_0/\mathbb{Q}}((q))\, =\, q^4$, so to have $N_{k_0/\mathbb{Q}}((q))\, \, \equiv \,\, 1\, (\mathrm{mod},\,25)$ the prime $q$ must verify $q\, \equiv \,\pm7\, (\mathrm{mod}\,25)$.

\item[$-$] If $p\,\, \equiv \,\,-1\, (\mathrm{mod}\,5)$,by corollary \ref{deco} $p\,=\,\pi_1\pi_2$ we have $N_{k_0/\mathbb{Q}}((\pi))\, =\, p^2$, so to have $N_{k_0/\mathbb{Q}}((\pi))\, \, \equiv \,\, 1\, (\mathrm{mod}\,25)$ the prime $p$ must verify $p\,\, \equiv \,\,-1\, (\mathrm{mod}\,25)$.
\end{enumerate}
(1) If rank $(C_{k,5}^{(\sigma)})\,=\,1$, we get that $d+q^*\,=\,4$, so there are three possible cases as follows:
\begin{itemize}
    \item Case 1: \(q^*=0\,\, \mathrm{and}\,\, d=4\),
    \item Case 2: \(q^*=1\,\, \mathrm{and}\,\, d=3\),
    \item Case 3: \(q^*=2\,\, \mathrm{and}\,\, d=2\), \end{itemize}
We will successively treat the three cases to prove the first point of the main theorem.

\begin{itemize}
 \item Case 1: we have $q^* = 0$ and $d=4$, so the number of prime ideals which are ramified in $k/k_0$ should be $4$. According to the proof of theorem \ref{ran2}, if $n$ is divisible by a prime $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ then the prime $l$ is unique.\\\\
 - If $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ divides $n$, then by Corollary \ref{deco}, $l\,=\,\pi_1\pi_2\pi_3\pi_4$ where $\pi_i$ are primes of $k_0$. The prime $l$ is ramified in $\Gamma$, because disk$(\Gamma/\mathbb{Q})\,=\, 5^5n^4$ and $l$ divides this discriminent, then $\pi_1,\pi_2,\pi_3$ and $\pi_4$ are ramified in $k$. Hence we have $d=4$, so $l$ is the unique prime divides $n$, because if $n$ is dividing by other prime we obtain $d>4$, which is impossible in the three cases. So $n\,=\,l^{e_1}$ with  $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ and $e_1\in\{1,2,3,4\}$. According to [\ref{Mani}, Lemma 5.1] we have $(\lambda)$ ramifies in $k/k_0\, \Longleftrightarrow \, n\not \equiv \, \pm1\pm7\, (\mathrm{mod}\,25)$, with $\lambda\,=\,1-\zeta_5$, so in the case $n\,=\,l^{e_1}$
where  $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ we should have $n \, \equiv \, \pm1\pm7\, (\mathrm{mod}\,25)$, so the only $l$ verifiy $n\,=\, l^{e_1} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\,25)$ is $l \, \equiv \, 1\, (\mathrm{mod}\, 25)$, and for this prime we have $q^*\,\geq\,1$, because $\zeta_5$ is norme, which is impossible in this case.\\
We note that if $n\,=\,l^{e_1}$ with  $l \, \equiv \, 1\, (\mathrm{mod}\, 25)$ and $q*\,=\,2$, there is no fields $k$ of type $(5,5)$, because we have rank$(C_{k,5}^{(\sigma)})\,=\,3$, and if $q^*\,=\,1$ we have rank$(C_{k,5}^{(\sigma)})\,=\,2$, which will treat in the second point of the proof.\\\\
-If no prime $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ divides $n$, we have two forms of $n$:

\begin{enumerate}
\item[$(i)$] $n\,=\, 5^{e}p^{e_1}q_1^{e_2} \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, \pm2\, (\mathrm{mod}\, 5)$ and $e,e_1,e_2 \in \{1,2,3,4\}$. By Corollary \ref{deco}, $p\,=\,\pi_1\pi_2$, where $\pi_1,\,\pi_2$ are primes in $k_0$ and $q_1$ is inert in $k_0$, the prime $p$  is ramified in $\Gamma$, then $\pi_1,\,\pi_2$ are ramified in $k$. We have $q_1$ is ramified in $\Gamma$. Since $n\,\not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$, then $\lambda\,=\,1-\zeta_5$ is ramified in $k$, so we get $d=4$. To verifiy that $n\,=\, 5^{e}p^{e_1}q_1^{e_2} \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ we can choose $e_1\,=\,2$ and $e_2\,=\,1$ because $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$, so $n\,=\, 5^{e}p^{2}q_1$ with $e \in \{1,2,3,4\}$. In one hand if $e\,=\,2,3,4$ we have $n\, \equiv \, 0\, (\mathrm{mod}\, 25)$, in the other hand if $n\,=\, 5p^{2}q_1$ we have $p^2q_1\,=\,5\alpha+2$ or $p^2q_1\,=\,5\alpha'+3$ with $\alpha,\alpha' \in \mathbb{Z}$, so $5p^{2}q_1 \, \equiv \, 10\, (\mathrm{mod}\, 25)$ or $5p^{2}q_1\, \equiv \, 15\, (\mathrm{mod}\, 25)$, therefore we conclude that $n\, \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$. According to [\ref{Mani}, Theorem 5.18], if $p\,\equiv\,-1\,(\mathrm{mod}\, 25)$ and $q_1\,\equiv\,\pm7\,(\mathrm{mod}\, 25)$, rank $(C_{k,5})$ is at least $3$ which contradict the fact that $C_{k,5}$ is of type $(5,5)$, and by the proof of [\ref{Mani}, Theorem 5.13, Theorem 5.15], for the other congruence cases of $p$ and $q_1$ we have $q^*\,=\,1$ which is impossible in this case.\\

\item[$(ii)$] $n\,=\, p^{e}q_1^{e_1}q_2^{e_2} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, 2\, (\mathrm{mod}\, 5),\,\, \\q_2 \, \equiv \, 3\, (\mathrm{mod}\, 5)$ and $e,e_1,e_2 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_1\,=\,2, e_2\,=\,1$ i.e $n\,=\, p^{e}q_1^{2}q_2 $ with $e \in \{1,2,3,4\}$. By Corollary \ref{deco} $p\,=\,\pi_1\pi_2$ where $\pi_1,\,\pi_2$ are primes in $k_0$ and $q_1,\, q_2$ are inert in $k_0$. We have $p,\,q_1$ and $q_2$ are ramified in $\Gamma$,so $\pi_1,\,\pi_2,\,q_1$ and $q_2$ are ramified in $k$ then $d=4$. The condition $n\,=\, p^{e}q_1^{2}q_2 \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ is not verified for all $p \, \equiv \, -1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, 2\, (\mathrm{mod}\, 5)$ and $q_2 \, \equiv \, 3\, (\mathrm{mod}\, 5)$, so we combine all the cases of congruence and we obtain that $p \, \equiv \, -1\, (\mathrm{mod}\, 25),\,\, q_1 \, \equiv \, 12\, (\mathrm{mod}\, 25),\,\, q_2 \, \equiv \, 3\, (\mathrm{mod}\, 25)$. By [\ref{Mani}, Lemma 5.1], since $N_{k_0/\mathbb{Q}}(\pi_i)\,\equiv\,1\,(\mathrm{mod}\, 25),\, N_{k_0/\mathbb{Q}}(q_1)\,\not\equiv\,1\,(\mathrm{mod}\, 25), N_{k_0/\mathbb{Q}}(q_2)\,\not\equiv\,1\,(\mathrm{mod}\, 25)$, we have $q^*\,=\,1$ which is impossible in this case.

\end{enumerate}

We deduce that in the case 1, there is no radicand $n$ who verifiy rank$(C_{k,5}^{(\sigma)})\,=\,1$.

\item Case 2: we have $q^* = 1$ and $d=3$, so the number of prime ideals which are ramified in $k/k_0$ should be $3$. According to case 1, n is not divisible by any prime $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ in this case.\\
We can writ $n$ as $n\,=\,\mu\lambda^{e}\pi_1^{e_1}....\pi_g^{e_g}$, where $\mu$ is a unit in $\mathcal{O}_{k_0}$, $\lambda = 1-\zeta_5$, $\pi_1,,,,\pi_g$ are primes in $k_0$ and $e \in \{0,1,2,3,4\}$, $e_i \in \{1,2,3,4\}$ for $1\leq i\leq g$.\\
By [\ref{Mani}, proposition 5.2] $d\,=\,g$ or $g+1$ according to whether $n \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ or $n \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$, to obtain $d=3$, $n$ must be written in $\mathcal{O}_{k_0}$ as: $n\,=\, \pi_1^{e_1}\pi_2^{e_2}\pi_3^{e_3}$ or $n\,=\, \lambda^{e}\pi_1^{e_1}\pi_2^{e_2}$, therefore we have three forms of $n$:
\begin{enumerate}
\item[$(i)$] $n\,=\, 5^ep^{e_1} \not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 5)$ and $e,e_1 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_1\,=\,1$ i.e $n\,=\, 5^ep $ with $e \in \{1,2,3,4\}$. By Corollary \ref{deco} $p\,=\, \pi_1\pi_2$ with $\pi_1,\,\pi_2$ are primes in $k_0$, we have $p$ is ramified in $\Gamma$, so $\pi_1,\,\pi_2$ are ramified in $k$, and since $n\,\not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$, according to [\ref{Mani}, Lemma 5.1], $\lambda \,=\, 1-\zeta_5$ is ramified in $k$ so we obtain $d=3$. The condition $n\,\not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$ is verified for all $p \, \equiv \, -1\, (\mathrm{mod}\, 5)$ because, if $e\,=\,2,3,4$ we have $n\,=\,5^ep\, \equiv \, 0\, (\mathrm{mod}\, 25)$, if $e=1$ i.e $n\,=\, 5p$ we have $p\,=\,5\alpha+4$ that implie $5p\,=\,25\alpha+20$ with $\alpha \in \mathbb{Z}$, so $n\,=\,5p \, \equiv \, 20\, (\mathrm{mod}\, 25)$. According to the proof of [\ref{Mani}, theorem 5.15] if $p \, \equiv \, -1 (\mathrm{mod}\, 25)$ we have $q^*\,=\,2$, and   if $p \, \not\equiv \, -1 (\mathrm{mod}\, 25)$ we have $q^*\,=\,1$, we conclude that $n\,=\,5^ep \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \not\equiv \, -1 (\mathrm{mod}\, 25)$.\\
We note that the computational number theory system \textbf{PARI} [\ref{PRI}], show that if $n\,=\,5^ep \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \not\equiv \, -1 (\mathrm{mod}\, 25)$, the field $k$ is not always of type $(5,5)$.

\item[$(ii)$] $n\,=\, 5^{e}q_1^{e_1}q_2^{e_2} \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1 \, \equiv \, 2\, (\mathrm{mod}\, 5),\,\, q_2 \, \equiv \, 3\, (\mathrm{mod}\, 5)$ and $e,e_1,e_2 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$, we can choose $e_1\,=\,2$ and $e_2\,=\,1$ i.e $n\,=\, 5^{e}q_1^{2}q_2$ with $e \in \{1,2,3,4\}$. By Corollary \ref{deco}, $q_1$ and $q_2$ are inert in $k_0$ , and $q_1$, $q_2$  are ramified in $\Gamma$, then $q_1,\,q_2$ are ramified in $k$. Since $n\,\not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$, then $\lambda\,=\,1-\zeta_5$ is ramified in $k$, so we get $d=3$.The condition  $n\, \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ is verified for all $q_1 \, \equiv \, 2\, (\mathrm{mod}\, 5),\,\, q_2 \, \equiv \, 3\, (\mathrm{mod}\, 5)$, if $e\,=\,2,3,4$ we have $n\,= \,5^{e}q_1^{2}q_2 \,\, \equiv \, 0\, (\mathrm{mod}\, 25)$, if $n\,=\, 5q_1^{2}q_2$ we have $q_1^2q_2\,=\,5\alpha+2$  with $\alpha, \in \mathbb{Z}$, so $5q_1^{2}q_2 \, \equiv \, 10\, (\mathrm{mod}\, 25)$. If $q_1 \, \equiv \, 7\, (\mathrm{mod}\, 25)$ and $q_2 \, \equiv \, -7\, (\mathrm{mod}\, 25)$ we have $q^*\,=\,2$, and if $q_1 \, \not\equiv \, 7\, (\mathrm{mod}\, 25)$ or $q_2 \, \not\equiv \, -7\, (\mathrm{mod}\, 25)$, according to the proof of [\ref{Mani}, theorem 5.13] we have $q^*\,=\,1$, but for this form of the radicand $n$ the computational number theory system \textbf{PARI} [\ref{PRI}] show that $C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}$.\\

\item[$(iii)$] $n\,=\, p^{e}q_1^{e_1} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, \pm2\, (\mathrm{mod}\, 5)$ and $e,e_1 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_1\,=\,1$ i.e $n\,=\, p^{e}q_1$ with $e \in \{1,2,3,4\}$. By Corollary \ref{deco} $p\,=\,\pi_1\pi_2$ where $\pi_1,\,\pi_2$ are primes in $k_0$ and $q_2$ is inert in $k_0$. We have $p$ is ramified in $\Gamma$,so $\pi_1,\,\pi_2$ are ramified in $k$. $q_2$ is ramified in $\Gamma$ too, we obtain $d=3$. The condition $n\,=\, p^{e}q_1 \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ is not verified for all $p \, \equiv \, -1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, \pm2\, (\mathrm{mod}\, 5)$, so we combine all the cases of congruence and we obtain that $p \, \not\equiv \,-1\, (\mathrm{mod}\, 25)$ and $ q_1 \, \not\equiv \,\pm7\, (\mathrm{mod}\, 25)$. According to [\ref{Mani}, theorem 5.18] If $p \, \equiv \, -1\, (\mathrm{mod}\, 25)$ and $q_1 \, \equiv \, \pm7\, (\mathrm{mod}\, 25)$ we have rank $C_{k,5} \geq 3$ which is impossible in our cases. If $p \, \not\equiv \,-1\, (\mathrm{mod}\, 25)$ and $q_1 \, \not\equiv \, \pm7\, (\mathrm{mod}\, 25)$, according to [\ref{Mani}, theorem 5.13] we have $q^*\,=\,1$. Using the computational number theory system \textbf{PARI/GP} [\ref{PRI}], if $n\,=\,p^{e}q_1$ with $p \, \not\equiv \, -1\, (\mathrm{mod}\, 25)$ and $ q_1 \, \not\equiv \, \pm7\, (\mathrm{mod}\, 25)$, the field $k$ is not always of type $(5,5)$.
\end{enumerate}

We summarize all forms of integer $n$ in the case 2, for which $k$ is of type $(5,5)$ and rank $(C_{k,5}^{(\sigma)})\,=\,1$ as follows:
\begin{equation}
\label{}
 n=\left\lbrace
   \begin{array}{ll}

   5^ep \not \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \text{ with } p\, \not\equiv \, -1\,(\mathrm{mod}\,25),\\

   p^{e}q_1 \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \text{ with } p\, \not\equiv \, -1\,(\mathrm{mod}\,25) \text{ and } q_1\, \not\equiv \, \pm7\,(\mathrm{mod}\,25)\\

   \end{array}
   \right.
\end{equation}
\item Case 3: we have $q^* = 2$ and $d=2$, so the number of prime ideals which are ramified in $k/k_0$ should be $2$. Let $n\,=\,\mu\lambda^{e}\pi_1^{e_1}....\pi_g^{e_g}$, where $\mu$ is a unit in $\mathcal{O}_{k_0}$, $\lambda = 1-\zeta_5$, $\pi_1,,,,\pi_g$ are primes in $k_0$ and $e \in \{0,1,2,3,4\}$, $e_i \in \{1,2,3,4\}$ for $1\leq i\leq g$.\\
$d\,=\,g$ or $g+1$ according to whether $n \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ or $n \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$, to obtain $d=2$, $n$ must be written in $\mathcal{O}_{k_0}$ as: $n\,=\, \pi_1^{e_1}\pi_2^{e_2}$ or $n\,=\, \lambda^{e}\pi_1^{e_1}$, therefore we have three forms of $n$:
\begin{enumerate}
\item[$(i)$] $n\,=\, 5^eq_1^{e_1} \not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1 \, \equiv \, \pm2\, (\mathrm{mod}\, 5)$ and $e,e_1 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_1\,=\,1$ i.e $n\,=\, 5^eq_1 $ with $e \in \{1,2,3,4\}$. Since $q^*=2$ so we have $q_1\,\, \equiv \,\,\pm7\, (\mathrm{mod}\, 25)$. By Corollary \ref{deco} $q_1$ is inert in $k_0$, we have $q_1$ is ramified in $\Gamma$, so $q_1$ is ramified too in $k$, and since $n\,\not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$, according to [\ref{Mani}, Lemma 5.1], $\lambda \,=\, 1-\zeta_5$ is ramified in $k$ so we obtain $d=2$. The condition $n\,\not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$ is verified for all $q_1 \,\, \equiv \,\, \pm7\, (\mathrm{mod}\, 5)$, because if $e\,=\,2,3,4$ we have $n\,=\,5^eq_1\, \equiv \, 0\, (\mathrm{mod}\, 25)$, if $e=1$ i.e $n\,=\, 5q_1$ we have $n\,=\,5q_1 \, \equiv \, \pm10\, (\mathrm{mod}\, 25)$, we conclude that $n\,=\,5^eq_1 \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1 \, \equiv \, \pm7\, (\mathrm{mod}\, 25)$, but for this form of the radicand $n$ the computational number theory system \textbf{PARI/GP} [\ref{PRI}] show that $C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}$.

\item[$(ii)$] $n\,=\, q_1^{e_1}q_2^{e_2} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1 \, \equiv \, 3\, (\mathrm{mod}\, 5),\,\, q_2 \, \equiv \, 2\, (\mathrm{mod}\, 5)$ and $e_1,e_2 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_2\,=\,1$ i.e $n\,=\, q_1^{e_1}q_2$ with $e_1 \in \{1,2,3,4\}$. Since $q^*=2$, we have $\zeta_5\,\in\,N_{k/k_0}(k^*)$, we get that $q_1\,\, \equiv \,\,-7\, (\mathrm{mod}\, 25)$ and $q_2\,\, \equiv \,\,7\, (\mathrm{mod}\, 25)$.  By Corollary \ref{deco} $q_1$ and $q_2$ are inert in $k_0$, and we have $q_1,\,q_2$ are ramified in $\Gamma$,so $q_1,\,q_2$ are ramified in $k$, so we obtain $d=2$. The condition $n\,=\, q_1^{e_1}q_2 \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ is verified, because we have $n\,=\,q_1^{e_1}q_2 \, \equiv \, \pm7\, (\mathrm{mod}\, 25)$ for all $e_1$, so we conclude that $n\,=\, q_1^{e_1}q_2 \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1 \, \equiv \, -7\, (\mathrm{mod}\, 25),\,\, q_2 \, \equiv \, 7\, (\mathrm{mod}\, 25)$, but for this form of the radicand $n$ the computational number theory system \textbf{PARI/GP} [\ref{PRI}] show that $C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}$

\item[$(iii)$] $n\,=\, p^{e} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 5)$ and $e  \in \{1,2,3,4\}$. Since $q^*=2$, we have $\zeta_5\,\in\,N_{k/k_0}(k^*)$, we get that $p\,\, \equiv \,\,-1\, (\mathrm{mod}\, 25)$. By Corollary \ref{deco}, $p\,=\,\pi_1\pi_2$ where $\pi_1,\,\pi_2$ are primes of $k_0$. The prime $p$ is ramified in $\Gamma$, then $\pi_1,\pi_2$ are ramified in $k$. hence we have $d=2$. The condition $n\,=\, p^{e} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ is verified for all $p\,\, \equiv \,\,-1\, (\mathrm{mod}\, 25)$, we conclude that $n\,=\, p^{e} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 25)$. Using the computational number theory system \textbf{PARI/GP} [\ref{PRI}], if $n\,=\, p^{e} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \equiv \, -1\, (\mathrm{mod}\, 25)$, the field $k$ is not always of type $(5,5)$.

\end{enumerate}
We deduce that in the case 3, there is one form of $n$ for which the fields $k$ is of type $(5,5)$ and rank$(C_{k,5}^{(\sigma)})\,=\,1$ as follows:
\begin{center}
 $n\,=\,p^{e} \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25)  \text{ with } p\, \equiv \, -1\,(\mathrm{mod}\,25)$\\

\end{center}
\end{itemize}

(2) If rank $(C_{k,5}^{(\sigma)})\,=\,2$, so  $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$. According to [\ref{Mani},section 5.1], the rank of $C_{k,5}^{(\sigma)}$ is given as follows:
\begin{center}
rank $(C_{k,5}^{(\sigma)})\,=\, d-3+q^*$
\end{center}
where $d$ et $q^*$ are defined previously. Since rank $(C_{k,5}^{(\sigma)})\, =\, 2$ and $q^*\,=\,0, 1\, \mathrm{or}\, 2$, there are three possible cases as follows:\\
\begin{itemize}
    \item Case 1: \(q^*=0\,\, \mathrm{and}\,\, d=5\),
    \item Case 2: \(q^*=1\,\, \mathrm{and}\,\, d=4\),
    \item Case 3: \(q^*=2\,\, \mathrm{and}\,\, d=3\), \end{itemize}
We will treat the three cases to prove the forms of the radicand $n$. By theorem \ref{ran2}, $n$ must be divisible by one prime $l \, \equiv \, 1\,(\mathrm{mod}\,5)$ in all cases, and since rank $(C_{k,5}^{(\sigma)})\, =\, 2$, the invariant $q^*$ should be $0$ or $1$, because if $q^*\,=\,2$ and $l\,\equiv\,1\,(\mathrm{mod}\,5)$ divides $n$, we get that the invariant $d$ is at least $4$, so we obtain that rank $(C_{k,5}^{(\sigma)})$ is at least $3$.
\begin{itemize}
\item Case 1: we have $q^* = 0$ and $d=5$, so the number of prime ideals which are ramified in $k/k_0$ should be $5$. The radicand $n$ must be divisible by one prime $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$.\\
We can writ $n\,\in\,\mathcal{O}_{k_0}$ as $n\,=\,\mu\lambda^{e}\pi_1^{e_1}....\pi_g^{e_g}$, where $\mu$ is a unit in $\mathcal{O}_{k_0}$, $\lambda = 1-\zeta_5$, $\pi_1,,,,\pi_g$ are primes in $k_0$ and $e \in \{0,1,2,3,4\}$, $e_i \in \{1,2,3,4\}$ for $1\leq i\leq g$.\\
$d\,=\,g$ or $g+1$ according to whether $n \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ or $n \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$. To obtain $d=5$, $n$ must be written in $\mathcal{O}_{k_0}$ as: $n\,=\, \pi_1^{e_1}\pi_2^{e_2}\pi_3^{e_3}\pi_4^{e_4}\pi_5^{e_5}$ or $n\,=\, \lambda^{e}\pi_1^{e_1}\pi_2^{e_2}\pi_3^{e_3}\pi_4^{e_4}$, therefore we have two forms of $n$:
\begin{enumerate}
\item[$(i)$] $n\,=\, 5^el^{e_1} \not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ and $e,e_1 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_1\,=\,1$ i.e $n\,=\, 5^el $ with $e \in \{1,2,3,4\}$. By Corollary \ref{deco} $l\,=\, \pi_1\pi_2\pi_3\pi_4$ with $\pi_i$ are primes in $k_0$, we have $l$ is ramified in $\Gamma$, so $\pi_1,\,\pi_2,\pi_3$ and $\pi_4$ are ramified in $k$, and since $n\,\not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$, according to [\ref{Mani}, Lemma 5.1], $\lambda \,=\, 1-\zeta_5$ is ramified in $k$ so we obtain $d=5$. The condition $n\,\not \equiv \,\pm1\pm7\, (\mathrm{mod}\, 25)$ is verified for all $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$, because if $e\,=\,2,3,4$ we have $n\,=\,5^el\, \equiv \, 0\, (\mathrm{mod}\, 25)$, if $e=1$ i.e $n\,=\, 5l$ we have $l\,=\,5\alpha+1$ that implie $5l\,=\,25\alpha+5$ with $\alpha \in \mathbb{Z}$, so $n\,=\,5l \, \equiv \, 5\, (\mathrm{mod}\, 25)$. If $l \, \equiv \, 1 (\mathrm{mod}\, 25)$ we have $\zeta_5\,\in\,N_{k/k_0}(k^*)$, so $q^*\,\geq\,1$ which in impossible in this case. We conclude that $n\,=\,5^el \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \not\equiv \, 1 (\mathrm{mod}\, 25)$. Using the computational number theory system \textbf{PARI/GP} [\ref{PRI}], if $n\,=\,5^el \not\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \not\equiv \, 1 (\mathrm{mod}\, 25)$, the field $k$ is not always of type $(5,5)$.\\

\item[$(ii)$] $n\,=\, l^{e}q_1^{e_1} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \equiv \, 1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, \pm2\, (\mathrm{mod}\, 5)$ and $e, e_1 \in \{1,2,3,4\}$. As $\mathbb{Q}(\sqrt[5]{ab^2c^3d^4})\,=\,\mathbb{Q}(\sqrt[5]{a^2b^4cd^3})\,=\,\mathbb{Q}(\sqrt[5]{a^3bc^4d^2})\,=\,\mathbb{Q}(\sqrt[5]{a^4b^3c^2d})$ we can choose  $e_1\,=\,1$ i.e $n\,=\, l^{e}q_1$ with $e \in \{1,2,3,4\}$. By Corollary \ref{deco} $l\,=\,\pi_1\pi_2\pi_3\pi_4$ where $\pi_i$ are primes in $k_0$ and $q_1$ is inert in $k_0$. We know that $l$ is ramified in $\Gamma$,so $\pi_1,\,\pi_2,\pi_3$ and $\pi_4$ are ramified in $k$. $q_1$ is ramified in $\Gamma$ too, we obtain $d=5$. The condition $n\,=\, l^{e}q_1 \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ is not verified for all $l \, \equiv \, 1\, (\mathrm{mod}\, 5),\,\, q_1 \, \equiv \, \pm2\, (\mathrm{mod}\, 5)$, so we combine all the cases of congruence and we obtain that $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ and $ q_1 \, \equiv \, \pm2,\pm3\,\pm7 \,(\mathrm{mod}\, 25)$. Using the computational number theory system \textbf{PARI/GP} [\ref{PRI}], if $n\,=\, l^{e}q_1 \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$, with $l \, \equiv \,1\, (\mathrm{mod}\, 5)$, and $ q_1 \, \equiv \, \pm2,\pm3\,\pm7\, (\mathrm{mod}\, 25)$, the field $k$ is not always of type $(5,5)$
\end{enumerate}

We summarize all forms of integer $n$ in this case  as follows:
\begin{equation}
\label{}
 n=\left\lbrace
   \begin{array}{ll}

   5^el \not \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \text{ with } l\, \not\equiv \, 1\,(\mathrm{mod}\,25),\\

   l^{e}q_1 \, \equiv \, \pm1,\pm7\, (\mathrm{mod}\,25) & \text{ with } l \, \equiv \, 1\, (\mathrm{mod}\, 5)\, q_1 \, \equiv \, \pm2,\pm3\,\pm7\, (\mathrm{mod}\, 25)\\

   \end{array}
   \right.
\end{equation}

\item Case 2: We have $q^*\,=\,1$ and $d=4$, so the number of prime ideals which are ramified in $k/k_0$ should be $4$. The radicand $n$ must be divisible by one prime $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$, and according to Corollary \ref{deco} $l$ splits in $k_0$ as $l\,=\, \pi_1\pi_2\pi_3\pi_4$ with $\pi_i$ are primes in $k_0$. Since $l$ is ramified in $\Gamma$, so $\pi_1,\,\pi_2,\pi_3$ and $\pi_4$ are ramified in $k$, hence if $n$ is devisible by another prime than $l$ the number of primes which are ramified in $k/k_0$ surpass $4$, therefore we have unique form of $n$ in this case, its $n\,=\, l^{e} \, \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$ and $e\in \{1,2,3,4\}$. The condition $n\, \equiv \,\, \pm1\pm7\, (\mathrm{mod}\, 25)$ is verified only for $l\,  \, \equiv \, 1, (\mathrm{mod}\, 25)$, and we have $q^*\,=\,1$. In conclusion we get $n\,=\, l^{e}$ with $l\, \, \equiv \, 1, (\mathrm{mod}\, 25)$. Using the computational number theory system \textbf{PARI/GP} [\ref{PRI}], if $n\,=\,l^{e} \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \equiv \, 1 (\mathrm{mod}\, 25)$, the field $k$ is not always of type $(5,5)$

\item case 3: We have $q^*\,=\,2$ and $d=3$, so the number of prime ideals which are ramified in $k/k_0$ should be $3$. The radicand $n$ must be divisible by one prime $l \, \equiv \, 1\, (\mathrm{mod}\, 5)$, and according to Corollary \ref{deco} $l\,=\, \pi_1\pi_2\pi_3\pi_4$ with $\pi_i$ are primes in $k_0$. Since $p$ is ramified in $\Gamma$, so $\pi_1,\,\pi_2,\pi_3$ and $\pi_4$ are ramified in $k$, so we deduce that the number of primes ramified in $k/k_0$ is at least $4$, so the case 3 does not exist.
\end{itemize}

\section{Numerical examples}
Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n$ is a positive integer, $5^{th}$ power-free, and let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_5)$ its normal closure. We assume that $C_{k,5}$ is of type $(5,5)$. Using the system \textbf{PARI/GP} [\ref{PRI}], we illustrate our main result Theorem \ref{thm:Rank1}.
\subsection{rank $(C_{k,5}^{(\sigma)})\, =\, 1$}
\begin{center}
Table 1: $n\,=\, p^{e}\,\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p\,\equiv \,-1\, (\mathrm{mod}\, 25)$\\
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$p$ &  $n\,=\,p^{e}$ & $p\,(\mathrm{mod}\,5)$ & $p\,(\mathrm{mod}\,25)$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$ \\
\hline
149 & 22201 = $149^2$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
199 & 7880599 = $199^3$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
349 & 42508549 = $349^3$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
449 &  $449$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
559 &  $559$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
1249 &  $1249$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
1499 &  $1499$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
1949 &  $1949$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
1999 &  $1999$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
2099 &  $449$ & -1 & -1 & 25 & $(5,5)$ & 1 \\
\hline
\end{tabular}
\end{center}
\newpage
\hspace{4cm} Table 2: $n\,=\, q_1^{e_1}q_2\,\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_i\,\equiv \,\pm7\, (\mathrm{mod}\, 25)$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
$q_1$ & \small$q_1\,(\mathrm{mod}\,5)$ & \small$q_1\,(\mathrm{mod}\,25)$& $q_2$ & \small$q_2\,(\mathrm{mod}\,5)$ & \small$q_2\,(\mathrm{mod}\,25)$& $n\,=\,q_1^{e_1}q_2$ & $h_{k,5}$ & $C_{k,5}$ & \small rank $(C_{k,5}^{(\sigma)})$\\
\hline
7 & 2 & 7 & 43 & 3 & -7 & 2107 = $7^2\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
7 & 2 & 7 & 193 & 3 & -7 & 1351 = $7\times193$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
7 & 2 & 7 & 293 & 3 & -7 & 2051 = $7\times293$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
107 & 2 & 7 & 43 & 3 & -7 & 492307 = $107^2\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
107 & 2 & 7 & 193 & 3 & -7 & 20651 = $107\times193$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
107 & 2 & 7 & 293 & 3 & -7 & 31351 = $107\times293$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
107 & 2 & 7 & 443 & 3 & -7 & 47401 = $107\times443$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
157 & 2 & 7 & 43 & 3 & -7 & 6751 = $157\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
157 & 2 & 7 & 193 & 3 & -7 & 30301 = $157\times193$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
157 & 2 & 7 & 443 & 3 & -7 & 69551 = $157\times443$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
257 & 2 & 7 & 193 & 3 & -7 & 49601 = $257\times293$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
257 & 2 & 7 & 293 & 3 & -7 & 75301 = $257\times293$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
307 & 2 & 7 & 193 & 3 & -7 & 59251 = $307\times193$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
457 & 2 & 7 & 43 & 3 & -7 & 19651 = $457\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
457 & 2 & 7 & 443 & 3 & -7 & 202451 = $457\times443$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
557 & 2 & 7 & 43 & 3 & -7 & 23251 = $557\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
607 & 2 & 7 & 43 & 3 & -7 & 26101 = $607\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
\hline
\end{tabular}

\begin{center}
 Table 3 : $n\,=\, 5^{e}q_1 \not\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1\,\equiv \,\pm7\, (\mathrm{mod}\, 25)$\\
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$q_1$ &  $n\,=\,5^eq_1$ & $q_1\,(\mathrm{mod}\,5)$ & $q_1\,(\mathrm{mod}\,25)$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$ \\
\hline
7 & 175 = $5^2\times7$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
107 & 535 = $5\times107$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
157 & 19625 = $5^3\times157$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
257 & 6425 = $5^2\times257$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
307 & 38375 = $5^3\times307$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
457 & 2285 = $5\times457$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
557 & 2785 = $5\times557$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
607 & 3053 = $5\times607$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
757 & 3785 = $5\times457$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
857 & 4285 = $5\times457$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
907 & 4535 = $5\times907$ & 2 & 7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
43 & 1075 = $5^2\times43$ &3 & -7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
193 & 120625 = $5^4\times193$ & 3 & -7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
293 & 120625 = $5^4\times193$ & 3 & -7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
443 & 11075 = $5^2\times443$ & 3 & -7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
643 & 3215 = $5\times643$ & 3 & -7 & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
\hline
\end{tabular}
\end{center}
\newpage
\begin{center}
Table 4: $n\,=\, 5^{e}q_1^{2}q_2 \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $q_1$ or $q_2$  $\not\equiv \,\pm7\, (\mathrm{mod}\, 25)$\\
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
$q_1$ & \small$q_1\,(\mathrm{mod}\,5)$ & \small$q_1\,(\mathrm{mod}\,25)$& $q_2$ & \small$q_2\,(\mathrm{mod}\,5)$ & \small$q_2\,(\mathrm{mod}\,25)$& $n\,=\,5^eq_1^2q_2$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$\\
\hline
2 & 2 & 2 & 3 & 3 & 3 & 60 = $5\times2^2\times3$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
2 & 2 & 2 & 13 & 3 & 13 & 260 = $5\times2^2\times13$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
2 & 2 & 2 & 53 & 3 & 3 & 1060 = $5\times2^2\times53$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
2 & 2 & 2 & 23 & 3 & -2 & 460 = $5\times2^2\times23$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
7 & 2 & 7 & 3 & 3 & 3 & 735 = $5\times7^2\times3$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
17 & 2 & 17 & 3 & 3 & 3 & 108375 = $5^3\times17^2\times3$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
17 & 2 & 17 & 23 & 3 & -2 & 33235 = $5\times17^2\times23$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1\\
37 & 2 & 17 & 3 & 3 & 3 & 20535 = $5\times37^2\times13$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
37 & 2 & 17 & 13 & 3 & 13 & 88985 = $5\times37^2\times13$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
47 & 2 & -3 & 3 & 3 & 3 & 33135 = $5\times47^2\times3$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
47 & 2 & -3 & 13 & 3 & 13 & 143585 = $5\times47^2\times13$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
47 & 2 & -3 & 23 & 3 & -2 & 254035 = $5\times47^2\times23$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
47 & 2 & -3 & 43 & 3 & -7 & 474935 = $5\times47^2\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
107 & 2 & 7 & 23 & 3 & -2 & 1316635 = $5\times2^2\times3$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
67 & 2 & 17 & 3 & 3 & 3 & 67335 = $5\times67^2\times3$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
67 & 2 & 17 & 53 & 3 & 3 & 1189585 = $5\times67^2\times53$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
97 & 2 & -3 & 43 & 3 & -7 & 2022935 = $5\times97^2\times43$ & 5 & $\mathbb{Z}/5\mathbb{Z}$ & 1 \\
\hline
\end{tabular}
\end{center}
\vspace{1cm}
\begin{center}
Table 5: $n\,=\, p^{e}q_1 \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p\,\not\equiv \,-1\, (\mathrm{mod}\, 25)$, $q_1\,\not\equiv \,\pm7\, (\mathrm{mod}\, 25)$
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
$p$ & \small$p\,(\mathrm{mod}\,5)$ & \small$p\,(\mathrm{mod}\,25)$& $q_1$ & \small$q_1\,(\mathrm{mod}\,5)$ & \small$q_1\,(\mathrm{mod}\,25)$& $n\,=\,p^{e}q_1$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$\\
\hline
59 & -1 & 9 & 2 & 2 & 2 & 118 = $59\times2$ & 25 & $(5,5)$ & 1 \\
19 & -1 & 19 & 3 & 3 & 3 & 57 = $19\times3$ & 25 & $(5,5)$ & 1 \\
59 & -1 & 9 & 23 & 3 & -2 & 1357 = $59\times23$ & 25 & $(5,5)$ & 1 \\
359 & -1 & 9 & 2 & 2 & 2 & 718 = $359\times2$ & 25 & $(5,5)$ & 1 \\
409 & -1 & 9 & 2 & 2 & 2 & 816 = $409\times2$ & 25 & $(5,5)$ & 1 \\
59 & -1 & 9 & 127 & 2 & 2 &  7493 = $59\times127$ & 25 & $(5,5)$ & 1 \\
109 & -1 & 9 & 23 & 3 & -2 & 2507 = $109\times23$ & 25 & $(5,5)$ & 1 \\
509 & -1 & 9 & 2 & 2 & 2 & 1018 = $509\times2$ & 25 & $(5,5)$ & 1 \\
709 & -1 & 9 & 2 & 2 & 2 & 1418 = $709\times2$ & 25 & $(5,5)$ & 1 \\
19 & -1 & 19 & 53 & 3 & 3 & 1007 = $19\times53$ & 25 & $(5,5)$ & 1 \\
\hline
\end{tabular}
\end{center}
\newpage

\begin{center}
Table 6: $n\,=\,5^ep \not\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $p \, \not\equiv \, -1 (\mathrm{mod}\, 25)$\\
\hspace{1,5cm}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$p$ &  $n\,=\,5^ep$ & $p\,(\mathrm{mod}\,5)$ & $p\,(\mathrm{mod}\,25)$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$ \\
\hline
19 & 475 = $5^2\times19$ & -1 & 19 & 25 & (5,5) & 1 \\
29 & 145 = $5\times29$ & -1 & 4 & 25 & (5,5) & 1 \\
59 & 7375 = $5^3\times59$ & -1 & 9 & 25 & (5,5) & 1 \\
89 & 55625 = $5^4\times89$ & -1 & 14 & 25 & (5,5) & 1 \\
109 & 2725 = $5^2\times109$ & -1 & 9 & 25 & (5,5) & 1 \\
229 & 28625 = $5^3\times229$ & -1 & 4 & 25 & (5,5) & 1 \\
239 & 1195 = $5\times239$ & -1 & 14 & 25 & (5,5) & 1 \\
269 & 6725 = $5^2\times19$ & -1 & 19 & 25 & (5,5) & 1 \\
379 & 168125 = $5^4\times379$ & -1 & 4 & 25 & (5,5) & 1 \\
389 & 1945 = $5^2\times389$ & -1 & 14 & 25 & (5,5) & 1 \\
\hline
\end{tabular}
\end{center}

\subsection{rank $(C_{k,5}^{(\sigma)})\, =\, 2$}
\begin{center}
Table 1: $n\,=\,5^el \not\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \not\equiv \, 1 (\mathrm{mod}\, 25)$\\
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$l$ &  $n\,=\,5^el$ & $l\,(\mathrm{mod}\,5)$ & $l\,(\mathrm{mod}\,25)$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$ \\
\hline
11 & 55 = $5\times11$ & 1 & 11 & 25 & (5,5) & 2 \\
41 &   5125 = $5^3\times41$ & 1 & -9 & 25 & (5,5) & 2 \\
61 & 5125 = $5^4\times61$ & 1 & 11 & 25 & (5,5) & 2 \\
71 & 1775 = $5^2\times71$ & 1 & -4 & 25 & (5,5) & 2 \\
131 & 655 = $5\times131$ & 1 & 6 & 25 & (5,5) & 2 \\
181 & 113125 = $5^4\times181$ & 1 & 6 & 25 & (5,5) & 2 \\
241 & 30125 = $5^3\times241$ & 1 & -9 & 25 & (5,5) & 2 \\
311 & 1555 = $5\times311$ & 1 & 11 & 25 & (5,5) & 2 \\
331 & 8275 = $5^2\times331$ & 1 & 6 & 25 & (5,5) & 2 \\
431 & 2155 = $5\times431$ & 1 & 6 & 25 & (5,5) & 2 \\
\hline
\end{tabular}
\end{center}

\begin{center}
Table 2: $n\,=\, l^{e}q_1\,\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \equiv \,1\, (\mathrm{mod}\, 5)$ and $ q_1 \, \equiv \, \pm2,\pm3\,\pm7\, (\mathrm{mod}\, 25)$
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
$l$ & \small$l\,(\mathrm{mod}\,5)$ & \small$l\,(\mathrm{mod}\,25)$& $q_1$ & \small$q_1\,(\mathrm{mod}\,5)$ & \small$q_1\,(\mathrm{mod}\,25)$& $n\,=\,l^{e}q_1$ & $h_{k,5}$ & $C_{k,5}$ & \small rank $(C_{k,5}^{(\sigma)})$\\
\hline
31 & 1 & 6 & 2 & 2 & 2 &  $31\times2$ & 25 & $(5,5)$ & 2 \\
131 & 1 & 6 & 23 & 3 & -2 &  $131^3\times23$ & 25 & $(5,5)$ & 2 \\
181 & 1 & 6 & 47 & 2 & -3 &  $181\times47$ & 25 & $(5,5)$ & 2 \\
11 & 1 & 11 & 3 & 3 & 3 &  $11\times3$ & 25 & $(5,5)$ & 2 \\
41 & 1 & 16 & 23 & 3 & -2 &  $41\times23$ & 25 & $(5,5)$ & 2 \\
191 & 1 & 16 & 2 & 2 & 2 &  $191\times2$ & 25 & $(5,5)$ & 2 \\
41 & 1 & 16 & 47 & 2 & -3 &  $41^2\times47$ & 25 & $(5,5)$ & 2 \\
311 & 1 & 11 & 2 & 2 & 2 &  $311^4\times2$ & 25 & $(5,5)$ & 2 \\
\hline
\end{tabular}
\end{center}
\newpage
\begin{center}
Table 3: $n\,=\,l^{e} \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)$ with $l \, \equiv \, 1 (\mathrm{mod}\, 25)$
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$l$ &  $n\,=\,l^{e}$ & $l\,(\mathrm{mod}\,5)$ & $l\,(\mathrm{mod}\,25)$ & $h_{k,5}$ & $C_{k,5}$ & rank $(C_{k,5}^{(\sigma)})$ \\
\hline
151 & $151$  & 1 & 1 & 25 & (5,5) & 2 \\
251 &   $251^2$  & 1 & 1 & 25 & (5,5) & 2 \\
601 & $601^3$ & 1 & 1 & 25 & (5,5) & 2 \\
1051 & $1051^4$ & 1 & 1 & 25 & (5,5) & 2 \\
1301 & $1301$ & 1 & 1 & 25 & (5,5) & 2 \\
1451 & $1451^2$ & 1 & 1 & 25 & (5,5) & 2 \\
1801 & $1801^3$ & 1 & 1 & 25 & (5,5) & 2 \\
1901 & $1901^4$ & 1 & 1 & 25 & (5,5) & 2 \\
2111 & 2111 & 1 & 1 & 25 & (5,5) & 2 \\
2131 & $2131^2$ & 1 & 1 & 25 & (5,5) & 2 \\
\hline
\end{tabular}
\end{center}

\section{Conjecture}
In this article, we have classified some pure quintic fields $\mathbb{Q}(\sqrt[5]{n})$, more precisely, we focused on the ones whose normal closures $\mathbb{Q}(\sqrt[5]{n},\zeta_5)$ possesses a $5$-class groups of type $(5,5)$, by treating the rank of the ambiguous classes, that can be characterized by the radicand $n$.\\
As to provide numerical examples, we use the system \textbf{PARI/GP} \ref{PRI}. Thus, we have noticed that the done calculations for some $n$ forms, show that $5$-class groupe $C_{k,5}$ of the field $k$, is isomorphic to $\mathbb{Z}/5\mathbb{Z}$, which allows us to give this conjecture as follows:\\
\begin{conjecture}
Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n$ is a positive integer, $5^{th}$ power-free. Let $k\,= \Gamma(\zeta_5)$ be the normal closure of $\Gamma$. Denote by $C_{k,5}$ the 5-class group of $k$, $q_1,q_2\,\equiv \, \pm2\, (\mathrm{mod}\, 5)$ are primes and $e \in \{1,2,3,4\}$.\\
 If the radicand $n$ take one form as follows:
\begin{equation}
 n\,=\,
   \begin{cases}

  q_1^{e_1}q_2\,\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25) & \text{ with } \quad q_i\,\equiv \,\pm7\, (\mathrm{mod}\, 25)\\

   5^{e}q_1 \not\equiv \, \pm1\pm7\, (\mathrm{mod}\, 25)& \text{ with } \quad  q_1\,\equiv \,\pm7\, (\mathrm{mod}\, 25)\\

  5^{e}q_1^{2}q_2 \not \equiv \, \pm1\pm7\, (\mathrm{mod}\, 25) & \text{ with } \quad q_1 \,\text{ or }\, q_2$  $\not\equiv \,\pm7\, (\mathrm{mod}\, 25)\\

   \end{cases}
\end{equation}
Then $C_{k,5}$ is a cyclic groupe of order $5$.
\end{conjecture}

redpajama_set_name: RedPajamaArXiv


text:
TechEngage News Internet
Amazon had another record-breaking holiday season
Amazon had a record-breaking holiday season. The online retail giant announced this achievement via a press release. The company says it added "tens of millions" of Prime members, including both trial-based members and paid users.
Amazon has also received more worldwide orders than ever before. These orders included more Amazon devices as well. The company claims it received orders of "millions more Amazon devices" compared with this time last year in 2017.
Amazon's best-selling devices included Amazon Echo speakers and Fire TV. These two products were best sellers during their Black Friday sale as well. People ordered more Echo Dots, Fire TV Sticks, and Echos than ever this holiday season.
Echo Dot has been an extremely popular choice because of its low price. The device was also deeply discounted in November and December. Even people who did not normally buy smart home devices purchased the item.
This helped Echo Dot become the most popular device during the Black Friday sale. Amazon faced stiff competition from the like of Google, Apple, and Samsung, but even beat Chinese companies such as Xiaomi.
Fire TV dongles sold in the millions, and Fire tablets and Kindle devices were hot sellers in 2018.
Smart home devices were a popular choice. Other top selling items included L.O.L. Surprise! Glam Glitter Series Doll, fashion items from Carhartt, and Bose QuietComfort Wireless Headphones.
According to Amazon, a lot of customers utilized Prime Free Shipping. These include millions of orders in the U.S that were shipped with Prime Free Same-Day, Prime Free One-Day, or Prime Now's free two-hour delivery.
Prime members also received exclusive discounts, which attracted even more people to sign up. Free Prime delivery options offered customers the opportunity to avoid the hassle that comes with holiday shopping in retail stores.
Another huge factor in Prime's popularity is the fact that it is now available in so many countries. Prime is now available in Luxembourg, Australia, Netherlands, and Singapore.
Jeff Wilke, Amazon's CEO Worldwide Consumer, said:
"This season was our best yet, and we look forward to continuing to bring our customers what they want, in ways most convenient for them in 2019. We are thrilled that in the U.S. alone, more than one billion items shipped for free this holiday with Prime."
Wilke further added:
"Thank you to our employees all around the world who are committed to bringing our customers the widest selection of products with low prices and fast and free delivery options throughout the holidays and all year long."
The press release gave an in-depth list of best-selling products by category. Some of the hottest electronics included Bose QuietComfort 35 (Series II) Wireless Headphones, Samsung's Flat 65 inch 4K UHD 8 Series Smart LED TV, and Apple iPads (Wi-Fi, 32GB) in Space Gray.
These devices all have hundreds or thousands of 4-star reviews.
Amazon hasn't always released Prime Membership subscriber figures, but in April it finally disclosed that it had hit more than 100 million members.
This number is going to grow due to the convenience and exclusive discounts it offers. Prime comes with other services too, such as Prime Video, Prime Music, free e-books – and more services are being added. Market experts predict that the number of Prime users will grow by 35 to 40 percent annually.
This is good news for Amazon, which continues to dominate over its competitors.
Amazon,
RecordBreakingSales
Russia to briefly "turn off" country's Internet access. But why?
If I asked you to picture the Internet for a moment – the actual, physical Internet – what would you imagine? Many of us have no idea how this thing…
Elon Musk faces a lawsuit because of his insensitive tweets
Tesla's CEO Elon Musk is under serious political stress these days. Musk is facing a defamation lawsuit filed against him by Vernon Unsworth, the British cave rescuer who accused Musk…

redpajama_set_name: RedPajamaCommonCrawl


Each of our editors has spent the last few weeks philosophizing about the topic "escape." This theme is quite general, as the variety of responses will show, but we've nailed down particular questions. We've all condensed our thoughts down to a couple pages of writing and will publish one article every day for the next couple weeks. This document will house individual links to each contribution as well as an overview of our project. We hope you enjoy.
At a young age, many hold the superstition that if monsters are under their bed, they'll be protected if they hide beneath the covers. The rationale is something like: no one can hurt me if I'm under here — this place is safe! Why does that make any sense? If there really were monsters under there, why should they care if I'm under the covers? There's something similar going on when a self-help book tells you the best way to avoid stress is to go to your happy place: nothing can stress you when you're there — that place is happy! Why does that make any sense? Going to my happy place doesn't change what I'm stressed about, unless of course I'm stressed that I'm not there. But these ideas have a real power. One might argue the whole program of stoicism is something along the lines of a self-help book pleading you to escape pain by looking inward — ignore what's "wrong" and imagine yourself happy!
Escape, as discussed above, seems puzzling because it doesn't resolve the situation that we find problematic. Escaping underneath the covers doesn't magically make the monsters that I believe are under my bed disappear. And, going to my happy place doesn't change the stressful situation. Is this a genuine cause for worry: are most of our attempts at escape moot and ultimately unhelpful? There do seem to be situations in which this is so; situations from which we can never escape, no matter how much we try – an issue that our editors will investigate further in this collection. But of course, as any college student knows, the occasional Netflix binge-imbued weekend provides much needed solace from the stress and worry caused by an overwhelming number of pages to be read and essays to be written. This escape from work might even be energizing, providing the necessary stamina needed to get through the work at some later point. So, while escaping to my happy place might not resolve the stressful situation, it makes the situation less stressful, and maybe even bearable, for the time being. The monsters underneath my bed don't disappear when I hide beneath my covers, but when I do, I don't feel like the monsters are a problem worth losing sleep over anymore. Escape, then, seems to be a way to deal with problematic situations, not by resolving them, but by taking away their problematic nature (at least temporarily).
Escape is an important way in which we face problems in our daily lives. But as such a rich topic, it has received surprisingly little philosophical attention, though reflection promptly invites some provocative questions. What should we make of our use of spatial metaphors when we talk of escape (we "escape from" distress and "escape to" our comfort zone)? Or, how do we escape, and when? And, where are we heading? These are some of the questions guiding our editors, who have interpreted our topic in multiple ways.
To start, we have Daniel's article tackling escapism in video games. Why do we choose to escape reality when we act through our avatars? And when we do, who do we choose to be?
Perhaps there is a more general point to take from Daniel's treatment of escape and video games. Here is where Anna decides to treat personal identity by means of escape. When we run away to far off lands in hopes of discovering ourselves, what do we mean?
Besides escaping to far off lands, David posits there is not just a common escape across time and distance, but across species, which he understands through transhumanism. How have capitalism, technology, and modernity altered our biological self-understanding?
And yet, these threats to our biology are supposed to simplify our lives. Varun analyzes several of these "convenient" technologies and how they force us to escape. But what are we escaping from? And what should we do about it?
But why should we care? Is escape always deficient? Steven argues for a Platonic account of escape that suggests we might call running away courageous, as we might for the soldier who knows when to retreat.
Perhaps though, not all escapists are soldiers. Perhaps some are showmen moguls, like Elon Musk, whose entire brand is built upon building offering escapes from — rather than solutions to — our problems. Felix guides us through questions concerning the viability of the Hyperloop to improve public transit or SpaceX and Tesla as legitimate forms of solving climate crisis.
Mere escapes like the "solutions" Musk offers have a particular intellectual history that Wen Li traces back through English theories of Utopia as far back as the 17th century. When imagining Utopia, why do so many escape into the abstract?
But many escape in more ways than this and those above. Perhaps beyond escape-from and escape-to, there is a distinctive escape-into as Andy argues. And what is the relation of this radical third to its counterparts?
More pragmatically, is there a sense in which studying the various forms of escape might have therapeutic value? Meera directs us to one sense in which depression might be understood in contrast to sadness and its parallel escapes by a feeling of valuelessness.
And if we take this further, we might ask how we can escape the end of this life. Henry discusses escapes from life as philosophical problems.
And how have philosophers made use of any of the above escapes? Ellie argues that externalists merely escape from skepticism while Lucy argues in our final article that Dewey derides abstract theoretical thinking as an escape from action-centric considerations.
UCPR is proud to present these twelve pieces on escape and we hope that you find them as interesting as we found them engaging to write.


text:
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Despite repeated declarations of "Never Again", from Solferino to Auschwitz, from Rwanda to South Sudan, history is filled with a litany of incomprehensible atrocities perpetrated by people against other people. The Marikana shootings of August 2012, in which 41 striking miners were killed by police, and described as the single most lethal use of force in post-apartheid South Africa, now form part of this growing sore on the human race's conscience. When such atrocities are committed, we try to expose the truth and bring those responsible to account through storytelling. Partly due to the complex relationship between human rights and power, the authenticity of such stories is often contested, with far reaching implications for both victims and perpetrators. The purpose of this paper is to add to the body of knowledge of the emerging human rights storytelling genre. It will achieve this by critically reflecting on the stories that emerged from the Marikana massacre in order to gain a better understanding of what they tell us about the protection and promotion of human rights in South Africa. The paper seeks to answer three critical questions: 1) How authentic are the stories told by different parties after an atrocity? 2) What can we learn about the relationship between human rights and power from stories of atrocity? 3) Does storytelling make a difference in our understanding, protection and promotion of human rights? The paper concludes that despite, one may also argue because of, the contestations around the stories emerging out of the Marikana massacre, it has become an important site shaping the human rights discourse in post-apartheid South Africa. Storytelling ultimately improves our understanding, protection and promotion of human rights.


text:
Tech stocks plunge as investors brace for 'major white-knuckle period'
Backlash against Facebook contributed to a sharp slide in all major market indices Tuesday.
Facebook stock shares at the New York Stock Exchange on Nov. 19, 2018.Drew Angerer / Getty Images
Nov. 20, 2018, 3:45 PM UTC
By Alyssa Newcomb
Technology stocks fell sharply on Tuesday, with analysts saying the market is in for a "major white knuckle period" for the rest of the year.
The Dow Jones Industrial Average plunged by more than 500 points mid-morning, dragged down partly by Boeing, which unsettled investors after ditching a planned conference call to discuss its 737 MAX airplane, the model that crashed in Indonesia last month, killing all 189 people on board.
The S&P 500, a broader measure of the market, and the blue-chip Dow have now erased all gains for 2018.
Apple and Facebook pulled down the tech-heavy Nasdaq to its lowest level in seven months, continuing what has been a six-week slide for the technology sector.
Each of the FAANG stocks, the acronym used for Facebook, Amazon, Apple, Netflix and Google, also slipped into a bear market on Monday, which Wall Street defines as a loss of more than 20 percent compared to a stock's 52-week high.
"Right now, there is a buyer's strike on tech stocks and we believe many of these names are oversold," said Daniel Ives, managing director of equity research at Wedbush Securities.
Apple and Facebook especially "are facing a slew of worries around fundamentals, valuations, and overall growth concerns heading into 2019," he said.
On Monday, Apple shares fell 4 percent after the Wall Street Journal reported the iPhone maker had slashed orders from key suppliers, signaling that their new devices may not be the blockbuster hits Apple has experienced in previous years. Several analysts have also downgraded or cut their price targets on Apple over future growth concerns.
"The straw that broke the camel's back for Apple has been the very soft iPhone supplier checks, which has caused an avalanche of selling," Ives told NBC News.
Last week, Apple lost its status as the world's only publicly traded, trillion-dollar company after Lumentum, a company that makes facial recognition and augmented reality components, said it received a request from one of its largest suppliers to reduce shipments. Apple is Lumentum's biggest customer, though the company did not say who made the request.
Facebook shares hit their lowest level since February 2017. The embattled social media company is on track to complete its third straight month of losses, making it the worst losing streak in Facebook's history as a publicly traded company.
The downward slide comes as Facebook grapples with the continued fallout from its handling of user privacy and a recent report from the New York Times that detailed leadership failures and dirty tactics Facebook allegedly used to smear competitors in an attempt to deflect attention.
Alyssa Newcomb
Alyssa Newcomb is an NBC News contributor who writes about business and technology.


text:
Destinations, Asia
Sea of green algae off the Chinese coast isn't as bad for tourism as you'd think
Julie Makinen, Los Angeles Times
Although disruptive to the usual tourism season, the photos of swimmers in Qingdao's waters are so unusual that they've spread around the world and brought the city a new wave of attention.
Like a bright green shag carpet mutating uncontrollably, an algae bloom twice the area of Los Angeles County has engulfed the Chinese coast around the city of Qingdao and buried beaches knee-high in grass-like material.
Bulldozers and brigades of workers with pitchforks have been deployed to clean up the algae, called Enteromorpha prolifera, which began to appear a month ago. Though crews are hauling away hundreds of tons of the material per day, it's unclear whether they're even keeping pace with the growth.
The algae, known as hutai in Mandarin, is now covering an area of some 11,500 square miles. Scientists say the growth is spurred by an abundance of phosphorus and other nutrients in the water, probably from agricultural runoff, industrial pollution or even human waste.
A woman who answered the Qingdao Tourism Hotline on Friday said officials have delayed the traditional July 1 opening of the city's designated beach swimming zone due to the algae bloom, which is 16 inches deep in some areas, though the plant poses no danger to humans.
Intrepid beachgoers have been frolicking in the green seas, lounging on plushy-looking green beds of the material on shore, even burying themselves in big green haystacks of the stuff.
This is the seventh year running that Qingdao has lived up to its name, which means "Green Island." In 2008, when Qingdao was to host the Olympic sailing competition, the government dispatched 1,000 soldiers to the area to scoop up the algae using boats, trucks, bulldozers and nets.
Conditions for the algae growth are now ideal, with water temperatures about 68 degrees. Once temperatures rise to about 73, the algae will begin to die off naturally, experts said.
In the meantime, Chinese netizens are actively discussing how to put the mass of algae to good use. One Weibo user posted a recipe for a hutai dish that resembled guacamole.
Hutai is enjoyed as a food in southern China; dried it sells for about $6 a pound. It can be fried with peanuts or shrimp, mixed into spring rolls, and even made into pancakes with tofu, egg and flour. Low in calories and high in fiber, hutai is also said to lower cholesterol.
Some local seafood farms have been harvesting the algae, planning to dry it and use it as feed in the winter months.
Local authorities have tried to keep the algae away from the beach by installing more than seven miles of netting along the shore, but this year's bloom has proved overwhelming.
At the Qingdao Holiday Inn, a concierge who wouldn't give her name said Friday that some visitors were canceling their reservations due to the algae and that the water in the area was "not very clean."
"Some people who love swimming very much are still going in," she said. "But I wouldn't recommend it."
Tommy Yang and Nicole Liu in Beijing bureau contributed to this report.
(c)2013 Los Angeles Times. Distributed by MCT Information Services.
Stay up-to-date on the business of travel in Asia with our weekly newsletter straight from Skift's Singapore bureau.
Tags: china, nature
Photo Credit: A boy swims in the algae-filled coastline of Qingdao, Shandong province in this July 15, 2011 file photo. China Daily / Reuters
How a Drop in Chinese Tourism Will Hurt Luxury Retail in the U.S.
Lonely Planet to Revamp Its Guidebooks, Starting With China
Rentals & Shares
Dennis Schaal and Deanna Ting, Skift
Skift Forum Asia: China Is on Track to Be Airbnb's Top Origin Market by 2020


text:
\section{Introduction}
\label{section:introduction}
Let $K_0 \cup K_1$ be a $2$-component split link in $S^3$,
and $b: I\times I\to S^3$ an embedding satisfying
$b(I\times I) \cap K_i = b(\{i\} \times I )$ for $i=0,1$.
The image $b(I\times I)$ is called a band connecting $K_0$ and $K_1$;
by abuse of notation, $b$ also denotes $b(I\times I)$.
Then the knot $K_0\cup K_1 \cup \partial b -\mathrm{int}( (K_0 \cup K_1) \cap b)$ is called the \textit{band-connected sum} of $K_0$ and $K_1$ along $b$, and
denoted by $K_0 \#_b K_1$,
or $K_b$ for short.

A core of a band $b$ is $b(I\times \{*\})$, and a cocore of $b$
is $b(\{*\}\times I)$, where $*\in \mathrm{int}{I}$.
A band $b$ is \textit{trivial} if there is a splitting sphere for $K_0\cup K_1$ meeting $b$ in a cocore of $b$.
Obviously, if $b$ is a trivial band, then $K_0 \#_b K_1 \cong K_0\#K_1$.
We prove  the converse.

\begin{theorem}
\label{th}
If a band-connected sum $K_0 \#_b K_1$ is equal to the connected sum $K_0 \# K_1$, then $b$ is a trivial band.
\end{theorem}

Regarding the problem when a band-connected sum is composite,
Eudave-Mu\~noz proved the following.

\begin{theorem}[\cite{EM}]
\label{EMth}
If $K_b$ is a composite knot,
then there is a decomposing sphere for $K_b$ which is either
disjoint from $b$ or intersects $b$ in a core of $b$.
\end{theorem}

To prove Theorem~\ref{th}
we use Theorem~\ref{EMth} and the fact that
if $K_b \cong K_0 \# K_1$, $K_b$ bounds a Seifert surface
of type~1a (Lemma~\ref{lemma}), which is  defined below.
Any band-connected sum $K_b$ bounds a compact connected surface $S$ containing $b$.
Such a surface $S$ is called \textit{type~1}
if $S-b$ is disconnected,
and \textit{type~2} otherwise.
In other words,
a type~1 surface for $K_0\#_b K_1$ is a compact connected,
possibly nonorientable, surface that is the union of $b$ and
two connected surfaces bounded by $K_i$ $(i=0,1)$.
A surface $S$ of type~1 is called \textit{type~1a} if there is
a splitting sphere for $K_0\cup K_1$ that is disjoint from $S -b$,
and\textit{ type~1b} otherwise.

For a composite band-connected sum bounding a surface of type~1a,
we study the configuration of a splitting sphere, a decomposing sphere,
and a type~1a surface,
and obtain Theorem~\ref{mainth} below.
Theorem~\ref{th} follows from Theorem~\ref{mainth}.

\begin{theorem}
\label{mainth}
If $K_b$ is a composite knot and bounds a surface of type~$1$a,
then there is a decomposing sphere disjoint from $b$.
\end{theorem}

As another corollary to Theorem~\ref{mainth},
we give sufficient conditions for band-connected sums to be prime.

\begin{proposition}
\label{prop}
Assume that $K_b = K_0 \#_b K_1$ bounds
a surface $S_0 \cup b \cup S_1$ of type~$1$a, where
$S_i$ is a surface bounded by $K_i$ $(i=0,1)$.
Then $K_b$ is prime if $b$ is a nontrivial band
 and either $(1)$ or $(2)$ below holds.
\begin{enumerate}[$(1)$]
\item $K_i$ is a trivial knot for $i=0,1$.
\item $K_i$ is prime and $S_i$ is an incompressible Seifert surface for
$i=0,1$.
\end{enumerate}
\end{proposition}

\section{Proofs}
\label{section:proofs}
\begin{lemma}
\label{lemma}
If $K_b \cong K_0\# K_1$, then $K_b$ bounds a Seifert surface of type~$1$a.
\end{lemma}

\noindent
\textit{Proof of Lemma}~\ref{lemma}.
The assumption implies $g(K_b) = g(K_0) +g(K_1)$,
where $g(\cdot)$ denotes the genus of a knot.
Then, by \cite{Ga, Sch}
$K_b$ bounds a Seifert surface $S$ of type~1 that is the union of
$b$ and minimal genus Seifert surfaces $S_i$ for $K_i$ $(i=0,1)$.
Let $P$ be a splitting sphere for $K_0 \cup K_1$ such that
$|P \cap (S -b)|$ is minimal among all splitting spheres.
Suppose  $P \cap (S-b) \ne \emptyset$ for a contradiction;
then $P\cap (S-b)$ consists of simple closed curves.
Let $\Delta_1$ be an innermost one
among disks in $P$ bounded by components of $P\cap(S-b)$.
Without loss of generality $\partial \Delta_1 \subset S_0$.
If $\partial \Delta_1$ is essential in $S_0$, then
surgery of $S_0$ along $\Delta_1$ yields a Seifert surface for $K_0$
with fewer genus than $S_0$ and possibly a closed surface.
Since $S_0$ has minimal genus, we see $\partial \Delta_1$ bounds a disk
$\Delta_2$ in $S_0$.

Each component of $\Delta_2 \cap P$ bounds a disk in $\Delta_2$,
and let $\Delta_3 (\subset S_0)$ be an innermost one among all
such disks in $\Delta_2$;
note $\Delta_3 \cap P = \partial \Delta_3$.
Let $B$ be the $3$-ball satisfying $\partial B = P$ and $\Delta_3\subset B$,
and $B'$ the closure of a component of $B -\Delta_3$ containing $K_0$
or $K_1$.
Shrinking $B'$ slightly, we obtain a splitting sphere $\partial B'$ with
$|\partial B' \cap (S-b)| < |P \cap (S-b)|$.
This contradicts the minimality of $|P\cap (S-b)|$.
It follows $P\cap (S-b) =\emptyset$, so that $S$ is type~1a
\hfill $\square$

\medskip
\noindent
\textit{Proof of Theorem}~\ref{mainth}.
By Theorem~\ref{EMth} we may assume that there is a decomposing
sphere $Q$ for $K_b= K_0\#_b K_1$ intersecting $b$ in a core of $b$.
Take a splitting sphere $P$ for $K_0 \cup K_1$,
a decomposing sphere $Q$ for $K_b$, and a type~1a surface $S$
bounded by $K_b$ that satisfy conditions~(1)---(4) below.
\begin{enumerate}[(1)]

\item $P, Q, S$ are in general position, i.e. any two of these
are in general position, and the intersection of any two of these
and the rest are in general position.

\item $P \cap (S -b) = \emptyset$.

\item $Q \cap b$ is a core of $b$, and $P\cap b$ consists of
cocores of $b$.

\item $|Q \cap S| + |P \cap Q| + |P \cap Q \cap S|$ is minimal
among all $P, Q, S$ satisfying (1)---(3) above.
\end{enumerate}

We first study the configuration of $P \cap Q$ and $Q \cap S$ on $Q$.
Note that $Q \cap S$ consists of an arc connecting the two points $K_b \cap Q$ and possibly simple closed curves.
Let $\alpha$ be the arc component of $Q \cap S$, and
set $c = Q\cap b$, a core of $b$.
Since $c$ connects  $K_0$ and $K_1$, and $S -b$ is disconnected,
$c$ is contained in $\alpha$.
Note also that $P\cap Q$, consisting of simple closed curves,
is disjoint from any circle component of $Q\cap S$
because $P\cap (S-b) =\emptyset$.
For the same reason $P\cap Q$ intersects $\alpha$ only in $c$.
See Figure~1.

\begin{figure}[h!]
\includegraphics[clip]{b_c_s_rev_fig1.eps}
\caption{$P\cap Q$ consists of simple closed curves.
\newline
$Q\cap S$ consists of the arc $\alpha$ and simple closed curves.}
\end{figure}

Assume for a contradiction that
either $Q \cap S$ or $P\cap Q$ has
a circle component disjoint from $\alpha$.
Each such component bounds a disk in $Q$ disjoint from $\alpha$.
Let $\Delta$ be an innermost one among all these disks.
If $\partial \Delta$ is a component of $Q \cap S$,
surger $S$ along $\Delta$ to yield a bounded surface $S'$ and
possibly a closed surface.
Since $\Delta \cap P = \emptyset$,
we see that $P \cap (S' -b) = \emptyset$ and
$|Q \cap S'| < |Q \cap S|$.
This contradicts the minimality assumption~(4).

If $\partial \Delta$ is a component of $P \cap Q$,
surger $P$ along $\Delta$ to yield two spheres,
one of which is  a splitting sphere $P'$.
Since $\Delta \cap S =\emptyset$,
we see that $P' \cap (S -b) = \emptyset$ and
$|P' \cap Q| < |P\cap Q|$.
This contradicts the minimality assumption.
We thus obtain the following.

\begin{claim}
\label{no disjoint circles}
$Q\cap S = \alpha$, and each component of $P\cap Q$ intersects
$\alpha$. \hfill $\square$
\end{claim}

Assume for a contradiction that
there is a component of $P\cap Q$ intersecting $\alpha$
more than once.
Each such component bounds a disk in $Q$ containing
at most one endpoint of $\alpha$.
Let $\Delta_1$ be an innermost one among all such disks.
Then, $\Delta_1 \cap \alpha$ consists
of at most one non-properly embedded arc and at least one
properly embedded arc (Figure~2(1)).
Each properly embedded arc of $\Delta_1 \cap \alpha$
and a subarc of $\partial \Delta_1$ cobound a disk in $\Delta_1$
containing no endpoint of $\alpha$.
Let $\Delta_2$ be an innermost one among all such disks;
note $\Delta_2 \cap (S-b) =\emptyset$.
We can isotop $b$ along $\Delta_2$ to reduce $|P\cap Q\cap S|$
(Figure~2(2)), a contradiction to the minimality.
Thus Claim~\ref{intersectonce} below is proved.

\begin{figure}[h!]
\includegraphics[clip]{b_c_s_rev_fig2.eps}
\caption{Circles in (1) are components of $P\cap Q$.}
\end{figure}

\begin{claim}
\label{intersectonce}
Each component of $P\cap Q$ intersects $\alpha$ in one point of the core $c$.
\hfill $\square$
\end{claim}

\begin{figure}[h!]
\includegraphics[clip]{b_c_s_rev_fig3.eps}
\caption{Circles are the components of $P\cap Q$.
Arcs are the components of $P\cap b$, i.e. cocores of the band.}
\end{figure}

Let us take a look at $P\cap Q$ on the splitting sphere $P$.
Let $\Delta_3$ be an innermost one among all disks in $P$
bounded by components of $P\cap Q$; see Figure~3.
Surger the decomposing sphere $Q$ along $\Delta_3$ to obtain
two spheres, at least one of which is a decomposing sphere for $K_b$.
Let $Q'$ be a decomposing sphere obtained from $Q$ after the surgery.
Without loss of generality $Q'$ is disjoint from $K_0$
and intersects $K_1$ in two points.
Let $\Delta_4$ be the closure of the component of $b - (b\cap Q')$
disjoint from $K_0$ (Figure~4).
Sliding $Q'$ along the disk $\Delta_4$,
we obtain a decomposing sphere for $K_b$ disjoint from $b$ as claimed.
This completes the proof of Theorem~\ref{mainth}.
\hfill $\square$

\begin{figure}[h!]
\includegraphics[clip]{b_c_s_rev_fig4.eps}
\caption{}
\end{figure}

\medskip\noindent
\textit{Proof of Theorem} \ref{th}.
By Lemma \ref{lemma}
$K_b$ bounds a Seifert surface that is the union of minimal genus
Seifert surfaces $S_i$ for $K_i$ $(i=0, 1)$ and $b$.
We proceed by induction on the product of genera $g(K_0)g(K_1)$.

\textit{Case $1$. $g(K_0)g(K_1) =0$.}
\\
Since $K_0$ or $K_1$ is a trivial knot,
$S_0$ or $S_1$ is a disk.
The boundary of a regular neighborhood of the disk
 is a splitting sphere for $K_0\cup K_1$
intersecting $b$ in a single cocore of $b$.
It follows that  $b$ is a trivial band.

\textit{Case $2$. $g(K_0)g(K_1)\ge 1$.}
\\
By Theorem \ref{mainth} there is a decomposing sphere $Q$ for $K_b$
disjoint from $b$.
Without loss of generality $Q$ intersects $K_1$ in two points.
Let $B$ be the $3$-ball which is bounded by $Q$ and
disjoint from $K_0$.
Then,  $B\cap K_1$ is an arc knotted in $B$.
Let $K'_1$ be the knot obtained from $K_1$ by replacing
$(B, B\cap K_1)$ with $(D^2\times I, \{0\}\times I)$.
We can regard $b$ as a band connecting $K_0$ and $K'_1$.
If $b$ is a trivial band for $K_0\#_b K'_1$,
then it is trivial for $K_0\#_b K_1$.
Since $g(K_0)g(K'_1) < g(K_0)g(K_1)$,
Theorem~\ref{th} is proved inductively.
\hfill $\square$

\medskip\noindent
\textit{Proof of Proposition} \ref{prop}.
Assume for a contradiction that $K_b$ is not prime,
but satisfies either condition~(1) or (2) in Proposition~\ref{prop}.
If $K_b$ is a trivial knot, then $K_0$ and $K_1$ are trivial knots
and especially $b$ is a trivial band (\cite{Sch2}),
a contradiction to the assumption of Proposition~\ref{prop}.
It follows that $K_b$ is a composite knot.

Then, following the arguments in Case~2 in the proof of
Theorem~\ref{th},
without loss of generality we can take a $3$-ball $B$ such that
$B\cap (K_0 \cup b) =\emptyset$ but
 $B\cap K_1$ is an arc knotted in $B$.
This implies that $K_1$ is not a trivial knot,
which violates condition~(1).
It follows that $K_b$ satisfies condition~(2).

Let $K'_1$ be $K_1$ with $(B, B\cap K_1)$ replaced by
$(D^2\times I, \{0\}\times I)$.
Since $K_1$ is prime and $B\cap K_1$ is a knotted arc in $B$,
we see $K'_1$ is a trivial knot.
The Seifert surface $S_1$ for $K_1$ intersects $\partial B$ in
an arc connecting the two points $\partial B \cap K_1$ and
possibly simple closed curves.
However, using the incompressibility of $S_1$, we can isotop $S_1$
without meeting $S_0\cup b$ so that $S_1\cap \partial B$ is just one arc.
The arc $S_1\cap \partial B$ in $\partial B=
\partial (D^2\times I)$ and $\{0\}\times I$ cobound a disk
in $D^2\times I$.
Then, the union $S'$ of the disk and $S_1-\mathrm{int}B$ is
an incompressible Seifert surface for the trivial knot $K'_1$.
It follows that $S'$ and thus $S_1-\mathrm{int}B$ are disks.
Isotopy of $\partial B$ along the disk $S_1 -\mathrm{int}B$
yields a splitting sphere for $K_0\cup K_1$ intersecting $b$
in a cocore of $b$.
This contradicts the assumption that $b$ is a nontrivial band.
\hfill $\square$


Understand exactly what you need to do.
Although there are some general guidelines to writing an MBA dissertation proposal, different schools and professors might have their own sets of requirements. Therefore, the first thing you need to do is to determine the exact directions you need to follow when writing this paper. Your dissertation advisor should be able to provide you with this information.
Introduce your hypothesis in the introductory paragraph.
The purpose of a proposal is to get people interested in your dissertation. Every project like this must be built around a hypothesis you will need to prove. You have to introduce this hypothesis in the very beginning of the paper in order to grab the readers' attention right away.
You should not only present your main idea in this paragraph but explain it as well. Add a few of your best arguments to make the introduction stronger.
A dissertation is a formal and serious academic paper, and this is exactly how your proposal should look like. You have to use formal professional language when writing this paper. Don't overuse terms, though, as this will make the text difficult to read.
Your grammar, spelling, and syntax must be impeccable. If you aren't 100% sure of your English skills, hire a professional to edit the paper for you.
Support your arguments, but not extensively.
You need to offer some evidence to support the arguments you introduce in this piece of work. However, you shouldn't go in too much detail. Include only a couple of the strongest arguments and pieces of proof you have. Your goal is to get people interested in reading more, so you shouldn't uncover all your cards in one go.
A proposal usually has a 1000-2000 word count limit. It will be difficult to squeeze every important thing in such a small paper, so avoid including any information that is not absolutely necessary. Your writing must be clear, brief and to the point. Don't forget that some of your readers may not be experts in the subject, so try to make the paper easy to understand.