# On the capitulation problem of some pure metacyclic fields
of degree 20 II

###### Abstract

Let $n$ be a $5^{th}$ power-free naturel number and
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ be the cyclotomic field generated by a
primitive $5^{th}$ root of unity $\zeta_{5}$. Then
$k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ is a pure metacyclic field of
absolute degree $20$. In the case that $k$ possesses a $5$-class group
$C_{k,5}$ of type $(5,5)$ and all the classes are ambiguous under the action
of $Gal(k/k_{0})$, the capitulation of $5$-ideal classes of $k$ in its
unramified cyclic quintic extensions is determined.

Fouad ELMOUHIB (the corresponding author)
Department of Mathematics and Computer Sciences,
Mohammed First University, Oujda, Morocco,
Correspondence<EMAIL_ADDRESS>
Mohamed TALBI
Regional Center of Professions of Education and Training,
<EMAIL_ADDRESS>

Abdelmalek AZIZI
Department of Mathematics and Computer Sciences,
Mohammed First University, Oujda, Morocco,
<EMAIL_ADDRESS>
Key words: pure metacyclic fields, 5-class groups, Hilbert 5-class field,
Capitulation.
AMS Mathematics Subject Classification: 11R04, 11R18, 11R29, 11R37

##  1 Introduction

Let $k$ be a number field, and $L$ be an unramified abelian extension of $k$.
We say that an ideal $\mathcal{I}$ of $k$ or its class capitulates in $L$ if
$\mathcal{I}$ becomes principal in $L$.
Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field, where $n$ is
a $5^{th}$ power free naturel number and $k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ be
the cyclotomic field generated by a primitive $5^{th}$ root of unity
$\zeta_{5}$. Then $k\,=\,\Gamma(\zeta_{5})$ is the normal closure of $\Gamma$.
Let $k_{5}^{(1)}$ be the Hilbert $5$-class field of $k$, $C_{k,5}$ be the
$5$-ideal classes group of $k$, and $C_{k,5}^{(\sigma)}$ be the group of
ambiguous ideal classes under the action of
$Gal(k/k_{0})\,=\,\langle\sigma\rangle$. In the case that $C_{k,5}$ is of type
$(5,5)$ and rank $C_{k,5}^{(\sigma)}\,=\,1$, the capitulation of the $5$-ideal
classes of $k$ in the six intermediate extensions of $k_{5}^{(1)}/k$ is
determined in [References].
In this paper, we investigate the capitulation of $5$-ideal classes of $k$ in
the unramified cyclic quintic extensions of $k_{5}^{(1)}/k$, whenever
$C_{k,5}$ is of type $(5,5)$ and rank $C_{k,5}^{(\sigma)}\,=\,2$, which mean
that all classes are ambiguous. Let $p$ and $q$ primes such that
$p\,\equiv\,1\,(\mathrm{mod}\,5)$, $q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$.
According to [References, theorem 1.1], if $C_{k,5}$ is of type $(5,5)$ and
rank $C_{k,5}^{(\sigma)}\,=\,2$, we have three forms of the radicand $n$ as
follows:

  * -

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

  * -

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

  * -

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

We will study the capitulation of $C_{k,5}$ in the six intermediate extensions
of $k_{5}^{(1)}/k$ in these cases. The theoretical results are underpinned by
numerical examples obtained with the computational number theory system
PARI/GP [References].

Notations.

Throughout this paper, we use the following notations:

  * •

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

  * •

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

  * •

$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$: the cyclotomic field, where
$\zeta_{5}\,=\,e^{2i\pi/5}$ is 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 $\tau$ is
identity on $\Gamma$, and sends $\zeta_{5}$ to its square. Hence $\tau$ has
order 4.

  * •

$\langle\sigma\rangle\,=\,\operatorname{Gal}(k/k_{0})$ such that $\sigma$ is
identity on $k_{0}$, and sends $\sqrt[5]{n}$ to $\zeta_{5}\sqrt[5]{n}$. Hence
$\sigma$ has order 5.

  * •

For a number field $L$, denote by:

    * –

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

    * –

$C_{L}$, $h_{L}$, $C_{L,5}$: the class group, class number, and $5$-class
group of $L$.

    * –

$L_{5}^{(1)},L^{*}$: the Hilbert $5$-class field of $L$, and the absolute
genus field of $L$.

    * –

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

  * •

$(\frac{a}{b})_{5}\,=\,1$ $\Leftrightarrow$ $X^{5}\equiv a\,(\mathrm{mod}\,b)$
resolved on $\mathcal{O}_{k_{0}}$, where $a,b$ are primes in
$\mathcal{O}_{k_{0}}$.

$k_{5}^{(1)}$$K_{4}$$K_{5}$$K_{6}$$K_{3}$$K_{2}$$K_{1}$$k$

Figure 1: The unramified quintic sub-extensions of $k_{5}^{(1)}/k$

##  2 Preliminaries

###  2.1 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 these extensions.

######  Proposition 2.1.

.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.
(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]. ∎

###  2.2 Relative genus field $(k/k_{0})^{*}$ of $k$ over $k_{0}$

Let $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$ be a pure quintic field,
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ the $5^{th}$-cyclotomic field and
$k\,=\,\Gamma(\zeta_{5})$ be the normal closure of $\Gamma$. The relative
genus field $(k/k_{0})^{*}$ of $k$ over $k_{0}$ is the maximal abelian
extension of $k_{0}$ which is contained in the Hilbert $5$-class field
$k_{5}^{(1)}$ of $k$. Let $q^{*}\in\\{0,1,2\\}$ such that
$q^{*}$ = $\begin{cases}2&\text{if }\,\zeta_{5},\zeta_{5}+1\,\text{are norm of
element in}\,k-\\{0\\}.\\\
1&\text{if}\,\,\zeta_{5}^{i}(\zeta_{5}+1)^{j}\,\text{is the norm of an element
in}\,k-\\{0\\}\,\text{for some exponents i and j}.\\\ 0&\text{if for no
exponents i, j the element}\,\,\zeta_{5}^{i}(\zeta_{5}+1)^{j}\,\text{is a norm
from}\,\,k-\\{0\\}.\\\ \end{cases}$

######  Proposition 2.2.

. Let $k\,=\,k_{0}(\sqrt[5]{n})$ such that $n$ =
$\mu\lambda^{e_{\lambda}}\pi_{1}^{e_{1}}....\pi_{f}^{e_{f}}\pi_{f+1}^{e_{f+1}}.....\pi_{g}^{e_{g}}$
in $k_{0}$, where $\mu$ is unity of $\mathcal{O}_{k_{0}}$,
$\lambda\,=\,1-\zeta_{5}$ the unique prime above 5 in $k_{0}$ and each prime
$\pi_{i}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$ for $1\leq i\leq
f$ and $\pi_{j}\,\not\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$ for
$f+1\leq j\leq g$. Then we have:

  * $(i)$

there exists $h_{i}\in\\{1,..,4\\}$ such that
$\pi_{f+1}\pi_{i}^{h_{i}}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$,
for $f+2\leq i\leq g$.

  * $(ii)$

if $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,\lambda^{5})$ and $q^{*}\,=\,1$,
then the genus field $(k/k_{0})^{*}$ is given as:

$(k/k_{0})^{*}$ =
$k(\sqrt[5]{\pi_{1}},....\sqrt[5]{\pi_{f}},\sqrt[5]{\pi_{f+1}\pi_{f+2}^{h_{f+2}}},....\sqrt[5]{\pi_{f+1}\pi_{g}^{h_{g}}})$

where $h_{i}$ is chosen as in (i).

  * $(iii)$

in the other cases of $q^{*}$ and the congruence of $n$, the genus field
$(k/k_{0})^{*}$ is given by deleting an appropriate number of $5^{th}$ root
from the right side of (ii).

###### Proof.

see [References, proposition 5.8]. ∎

##  3 Study of capitulation

This being the case, let $\Gamma$, $k_{0}$ and $k$ as above. If $C_{k,5}$ is
of type $(5,5)$ and the group of ambigous classes $C_{k,5}^{(\sigma)}$ under
the action of $Gal(k/k_{0})\,=\,\langle\sigma\rangle$ has rank 2, we have
$C_{k,5}\,=\,C_{k,5}^{(\sigma)}$. By class field theory $C_{k,5}^{1-\sigma}$
correspond to $(k/k_{0})^{*}$, and since $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$ we
get that $C_{k,5}^{1-\sigma}\,=\,\\{1\\}$, hence
$(k/k_{0})^{*}\,=\,k_{5}^{(1)}$ is the Hilbert $5$-class field of $k$.
When $C_{k,5}$ is of type $(5,5)$, it has $6$ subgroups of order $5$, denoted
$H_{i}$, $1\leq i\leq 6$. Let $K_{i}$ be the intermediate extension of
$k_{5}^{(1)}/k$, corresponding by class field theory to $H_{i}$. Its easy to
see that $C_{k,5}\,\cong\,C_{k,5}^{+}\times C_{k,5}^{-}$ such that
$C_{k,5}^{+}\,=\,\\{\mathcal{A}\in
C_{k,5}\,|\,\mathcal{A}^{\tau^{2}}=\mathcal{A}\\}$ and
$C_{k,5}^{-}\,=\,\\{\mathcal{X}\in
C_{k,5}\,|\,\mathcal{X}^{\tau^{2}}=\mathcal{X}^{-1}\\}$ with
$Gal(k/\Gamma)\,=\,\langle\tau\rangle$. As each $K_{i}$ is cyclic of order $5$
over $k$, there is at least one subgroup of order $5$ of $C_{k,5}$, i.e at
least one $H_{l}$ for some $l\in\\{1,2,3,4,5,6\\}$, which capitulates in
$K_{i}$ (by Hilbert’s theorem $94$).

######  Definition 3.1.

Let $\mathcal{S}_{j}$ be a generator of $H_{j}$ $(1\leq j\leq 6)$
corresponding to $K_{j}$. For $1\leq j\leq 6$, let
$i_{j}\in\\{0,1,2,3,4,5,6\\}$. We say that the capitulation is of type
$(i_{1},i_{2},i_{3},i_{4},i_{5},i_{6})$ to mean the following:

when $i_{j}\in\\{1,2,3,4,5,6\\}$, then only the class $\mathcal{S}_{i_{j}}$
and its powers capitulate in $K_{j}$;

when $i_{j}\,=\,0$, then all the $5$-classes capitulate in $K_{j}$.

Throught the paper we order the subgroups $H_{i}$ of $C_{k,5}$ as follows:
$H_{1}\,=\,C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$,
$H_{6}\,=\,C_{k,5}^{-}\,=\,\langle\mathcal{X}\rangle$,
$H_{2}\,=\,\langle\mathcal{AX}\rangle$,
$H_{3}\,=\,\langle\mathcal{AX}^{2}\rangle$,
$H_{4}\,=\,\langle\mathcal{AX}^{3}\rangle$ and
$H_{5}\,=\,\langle\mathcal{AX}^{4}\rangle$. By class field theory we have
$H_{6}$ correspond to $K_{6}\,=\,k\Gamma_{5}^{(1)}$, with $\Gamma_{5}^{(1)}$
is the Hilbert $5$-class field of $\Gamma$. By the action of
$Gal(k/\mathbb{Q})$ on $C_{k,5}$, we can give the following:

######  Proposition 3.1.

For all continuations of the automorphisms $\sigma$ and $\tau$ we have:

$K_{i}^{\sigma}\,=\,K_{i}$ $(i\,=\,1,2,3,4,5,6)$ i.e $\sigma$ sets all $K_{i}$

$K_{1}^{\tau^{2}}\,=\,K_{1}$, $K_{6}^{\tau^{2}}\,=\,K_{6}$,
$K_{2}^{\tau^{2}}\,=\,K_{5}$ and $K_{3}^{\tau^{2}}\,=\,K_{4}$. i.e $\tau^{2}$
sets $K_{1}$, $K_{6}$ and permutes $K_{2}$ with $K_{5}$ and $K_{3}$ with
$K_{4}$.

###### Proof.

We will agree that for all $1\leq i\leq 6$, and for all $w\in
Gal(k/\mathbb{Q})$ we have
$H_{i}^{w}\,=\,\\{\mathcal{C}^{w}\,|\,\mathcal{C}\in H_{i}\\}$.

Since all classes are ambigous because $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$, then
$\sigma$ sets all $H_{i}$.

We have $H_{1}\,=\,C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$ and
$H_{6}\,=\,C_{k,5}^{-}\,=\,\langle\mathcal{X}\rangle$, then
$H_{1}^{\tau^{2}}\,=\,H_{1}$ and $H_{6}^{\tau^{2}}\,=\,H_{6}$.
\- Since
$(\mathcal{AX})^{\tau^{2}}\,=\,\mathcal{A}^{\tau^{2}}\mathcal{X}^{\tau^{2}}\,=\,\mathcal{AX}^{-1}\,=\,\mathcal{AX}^{4}\in
H_{5}$ then $H_{2}^{\tau^{2}}\,=\,H_{5}$.
\- Since
$(\mathcal{A}\mathcal{X}^{2})^{\tau^{2}}\,=\,\mathcal{A}^{\tau^{2}}(\mathcal{X}^{2})^{\tau^{2}}\,=\,\mathcal{AX}^{-2}\,=\,\mathcal{AX}^{3}\in
H_{4}$ then $H_{3}^{\tau^{2}}\,=\,H_{4}$.
\- Since $\tau^{4}\,=\,1$ we get that $H_{5}^{\tau^{2}}\,=\,H_{2}$ and
$H_{4}^{\tau^{2}}\,=\,H_{3}$.
The relations between the fields $K_{i}$ in $(1)$ and $(2)$ are nothing else
than the translations of the corresponding relations for the subgroups $H_{i}$
via class field theory. ∎

To study the capitulation problem of $k$ whenever $C_{k,5}$ is of type $(5,5)$
and $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$, we will investigate the three forms of
the radicand $n$ proved in [References, theorem 1.1] and mentioned above.

###  3.1 The case $n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, where
$p\,\equiv\,1\,(\mathrm{mod}\,25)$

Let $k\,=\,\Gamma(\zeta_{5})$ be the normal closure of
$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, where $n\,=\,p^{e}$ such that
$p\,\equiv\,1\,(\mathrm{mod}\,25)$ and $e\in\\{1,2,3,4\\}$. Since
$p\,\equiv\,1\,(\mathrm{mod}\,5)$ we have that $p$ splits completely in
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ as $p\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$,
with $\pi_{i}$ are primes in $k_{0}$ such that
$\pi_{i}\,\equiv\,1\,(\mathrm{mod}\,5\mathcal{O}_{k_{0}})$, then the primes of
$k_{0}$ ramified in $k$ are $\pi_{i}$.
If $\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{P}_{3}$ and $\mathcal{P}_{4}$ are
respectivly the prime ideals of $k$ above $\pi_{1},\pi_{2},\pi_{3}$ and
$\pi_{4}$, then $\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k}\,(i=1,2,3,4)$
and since $\tau$ acte transitively on $\pi_{i}$, we have that $\tau^{2}$
permutes $\pi_{1}$ with $\pi_{3}$, hence $\tau^{2}$ permutes $\mathcal{P}_{1}$
with $\mathcal{P}_{3}$. Since $\pi_{i}^{\sigma}\,=\,\pi_{i}$, we have
$\mathcal{P}_{i}^{\sigma}\,=\,\mathcal{P}_{i}$. In fact
$[\mathcal{P}_{i}]\,(i=1,2,3,4)$ generate the group of strong ambigous ideal
classes denoted $C_{k,s}^{(\sigma)}$. The next theorem allow us to determine
explicitly the intermediate extensions of $k_{5}^{(1)}/k$.

######  Theorem 3.1.

Let $k$ and $n$ as above. Let $\pi_{1},\pi_{2},\pi_{3}$ and $\pi_{4}$ a primes
of $k_{0}$ congrus to $1$ modulo $\lambda^{5}$ such that
$p\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$, then:

$k_{5}^{(1)}\,=\,k(\sqrt[5]{\pi_{1}},\sqrt[5]{\pi_{3}})$.

The six intermediate extensions of $k_{5}^{(1)}/k$ are:
$k(\sqrt[5]{\pi_{1}})$, $k(\sqrt[5]{\pi_{3}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}})$,
$k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$ and
$k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$. Furthermore $\tau^{2}$ permutes
$k(\sqrt[5]{\pi_{1}})$ with $k(\sqrt[5]{\pi_{3}})$ and
$k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$ with $k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$, and
sets $k(\sqrt[5]{\pi_{1}\pi_{3}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$.

###### Proof.

We have that $k_{5}^{(1)}\,=\,(k/k_{0})^{*}$. Since $k\,=\,k_{0}(\sqrt[5]{n})$
with $n\,=\,p\,=\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ in $k_{0}$ and
$\pi_{i}\,\equiv\,1\,(\mathrm{mod}\,\lambda^{5})\,(i\,=\,1,2,3,4)$, then by
proposition 2.2 we have
$(k/k_{0})^{*}\,=\,k(\sqrt[5]{\pi_{1}},\sqrt[5]{\pi_{3}})$.

If $k_{5}^{(1)}\,=\,k(\sqrt[5]{\pi_{1}},\sqrt[5]{\pi_{3}})$, then the six
intermediate extensions are: $k(\sqrt[5]{\pi_{1}})$, $k(\sqrt[5]{\pi_{3}})$,
$k(\sqrt[5]{\pi_{1}\pi_{3}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$,
$k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$ and $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$. We
have $\tau^{2}(\pi_{1})\,=\,\pi_{3}$ then its easy to see that $\tau^{2}$ sets
the fields $k(\sqrt[5]{\pi_{1}\pi_{3}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$.
Since
$\tau^{2}(\pi_{1})\,=\,\tau^{2}(\sqrt[5]{\pi_{1}^{5}})\,=\,(\tau^{2}(\sqrt[5]{\pi_{1}}))^{5}\,=\,\pi_{3}$,
then $\tau^{2}(\sqrt[5]{\pi_{1}})$ is $5^{th}$ root of $\pi_{3}$. Hence
$k(\sqrt[5]{\pi_{3}})\,=\,k(\tau^{2}(\sqrt[5]{\pi_{1}}))$ i.e
$k(\sqrt[5]{\pi_{3}})\,=\,k(\sqrt[5]{\pi_{1}})^{\tau^{2}}$. By the same
resoning we prove that
$k(\sqrt[5]{\pi_{1}})\,=\,k(\sqrt[5]{\pi_{3}})^{\tau^{2}}$. Hence $\tau^{2}$
permutes $k(\sqrt[5]{\pi_{1}})$ with $k(\sqrt[5]{\pi_{3}})$.
We have $\tau^{2}(\pi_{1}\pi_{3}^{2})\,=\,\pi_{1}^{2}\pi_{3}$ then
$\tau^{2}(\pi_{1}\pi_{3}^{2})\,=\,\tau^{2}(\sqrt[5]{(\pi_{1}\pi_{3}^{2})^{5}})\,=\,(\tau^{2}(\sqrt[5]{\pi_{1}\pi_{3}^{2}}))^{5}\,=\,\pi_{1}^{2}\pi_{3}$,
hence $\tau^{2}(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$ is $5^{th}$ root of
$\pi_{1}^{2}\pi_{3}$. Then
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}})\,=\,k(\tau^{2}(\sqrt[5]{\pi_{1}\pi_{3}^{2}}))$
i.e
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}})\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})^{\tau^{2}}$.
By the same resoning we prove that
$k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})^{\tau^{2}}$.
Hence $\tau^{2}$ permutes $k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$ with
$k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$. ∎

The generators of $C_{k,5}$ when its of type $(5,5)$ and the radicand $n$ is
as above are determined as follows:

######  Theorem 3.2.

Let $k$ and $n$ as above. Let $\pi_{1},\pi_{2},\pi_{3}$ and $\pi_{4}$ a primes
of $k_{0}$ congrus to 1 $(\mathrm{mod}\,\lambda^{5})$ such that
$n\,=\,p\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$. Let
$\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{P}_{3}$ and $\mathcal{P}_{4}$ prime
ideals of $k$ such that
$\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k_{0}}\,(i=1,2,3,4)$. Then:

$C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$

###### Proof.

According to [References, theorem 1.1], for that case of the radicand $n$, we
have that $\zeta_{5}^{i}(1+\zeta_{5})^{j}$ is norm of element in $k-\\{0\\}$.
By [References, section 5.3], if $\zeta_{5}$ is not norm of unit of $k$ we
have $C_{k,5}\,=\,C_{k,5}^{(\sigma)}\,\neq\,C_{k,s}^{(\sigma)}$, so
$C_{k,s}^{(\sigma)}$ contained in $C_{k,5}^{(\sigma)}$. Hence we discuss two
cases:

$1^{th}$ case: $C_{k,5}\,=\,C_{k,5}^{(\sigma)}\,\neq\,C_{k,s}^{(\sigma)}$: In
this case, $C_{k,s}^{(\sigma)}$ is contained in
$C_{k,5}\,=\,C_{k,5}^{(\sigma)}$, and by [References, section 5.3] we have
$C_{k,5}^{(\sigma)}/C_{k,s}^{(\sigma)}\,=\,C_{k,5}/C_{k,s}^{(\sigma)}$ is
cyclic group of order $5$. Since $C_{k,5}$ has order $25$ then
$C_{k,s}^{(\sigma)}$ is cyclic of order $5$. We have that
$C_{k,s}^{(\sigma)}\,=\,\langle[\mathcal{P}_{1}],[\mathcal{P}_{2}],[\mathcal{P}_{3}],[\mathcal{P}_{4}]\rangle$,
$\mathcal{P}_{1}^{\tau^{2}}\,=\,\mathcal{P}_{3}$ and
$\mathcal{P}_{2}^{\tau^{2}}\,=\,\mathcal{P}_{4}$, so $\mathcal{P}_{1}$ and
$\mathcal{P}_{2}$ can not be both principals in $k$, otherwise
$\mathcal{P}_{3}\,=\,\mathcal{P}_{1}^{\tau^{2}}$ and
$\mathcal{P}_{4}\,=\,\mathcal{P}_{2}^{\tau^{2}}$ will be principals too, hence
$C_{k,s}^{(\sigma)}\,=\,\\{1\\}$, which is impossible. by the same reasoning
we have that $\mathcal{P}_{3}$ and $\mathcal{P}_{4}$ can not be both
principals in $k$. Since $C_{k,s}^{(\sigma)}$ is cyclic of order $5$ and
without loosing generality we get that
$C_{k,s}^{(\sigma)}\,=\,\langle[\mathcal{P}_{1}]\rangle$, so $\mathcal{P}_{1}$
and $\mathcal{P}_{3}\,=\,\mathcal{P}_{1}^{\tau^{2}}$ are not principals. Since
$C_{k,5}\cong C_{k,5}^{+}\times C_{k,5}^{-}$ its sufficient to find generators
of $C_{k,5}^{+}$ and $C_{k,5}^{-}$. As
$[\mathcal{P}_{1}\mathcal{P}_{3}]^{\tau^{2}}\,=\,[(\mathcal{P}_{1}\mathcal{P}_{3})^{\tau^{2}}]\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}]$
then $C_{k,5}^{+}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}]\rangle$, and
$[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]^{\tau^{2}}\,=\,[(\mathcal{P}_{1}\mathcal{P}_{3}^{4})^{\tau^{2}}]\,=\,[\mathcal{P}_{1}^{4}\mathcal{P}_{3}]\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]^{-1}$
then $C_{k,5}^{-}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$.
Hence
$C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$.

$2^{th}$ case: $C_{k,5}\,=\,C_{k,5}^{(\sigma)}\,=\,C_{k,s}^{(\sigma)}$: We
admit the same reasoning of $1^{th}$ case because none of
$\mathcal{P}_{i}\,(i=1,2,3,4)$ is principal, otherwise
$C_{k,5}\,=\,C_{k,s}^{(\sigma)}\,=\,\\{1\\}$, which is impossible. Hence
$C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$.
∎

Now we are able to stat the main theorem of capitulation in this case.

######  Theorem 3.3.

We keep the same assumptions as theorem 3.2 Then:

If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,=\,1$ then
$K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}})$ or $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$,
$K_{2}\,=\,k(\sqrt[5]{\pi_{3}})$, $K_{3}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$
or $k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$,
$K_{4}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$ or
$k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$, $K_{5}\,=\,k(\sqrt[5]{\pi_{1}})$ and
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ or $k(\sqrt[5]{\pi_{1}\pi_{3}})$.
Otherwise we just permute $K_{2}$ and $K_{5}$.

$[\mathcal{P}_{1}\mathcal{P}_{3}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}})$, $[\mathcal{P}_{i}]$ capitulates in
$k(\sqrt[5]{\pi_{i}})\,(i=1,3)$, $[\mathcal{P}_{1}\mathcal{P}_{3}^{2}]$
capitulates in $k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$,
$[\mathcal{P}_{1}\mathcal{P}_{3}^{3}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$ and $[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$
capitulates in $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$.

$(i)$ If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,=\,1$ and
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ then the possible types of
capitulation are: $(0,0,0,0,0,0)$, $(1,0,0,0,0,0)$, $(0,2,0,0,5,0)$,
$(1,2,0,0,5,0)$, $\\{(0,0,3,4,0,0)$ or $(0,0,4,3,0,0)\\}$, $\\{(1,0,3,4,0,0)$
or $(1,0,4,3,0,0)\\}$, $\\{(0,2,3,4,5,0)$ or $(0,2,4,3,5,0)\\}$,
$\\{(1,2,3,4,5,0)$ or $(1,2,4,3,5,0)\\}$.

$(ii)$ If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,=\,1$ and
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}})$ then the same possible types of
capitulation accur as in $(i)$ with $i_{6}=0$ or $1$ and $i_{1}=0$ or $6$

\- $(iii)$ If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,\neq\,1$ then the same possible
types of capitulation accur as $(i)$ and $(ii)$ by permuting 2 and 5.

###### Proof.

According to theorem 3.1, we have that $\tau^{2}$ permutes
$k(\sqrt[5]{\pi_{1}})$ with $k(\sqrt[5]{\pi_{3}})$ and
$k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$ with $k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$, and
sets $k(\sqrt[5]{\pi_{1}\pi_{3}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$. By
class field theory $K_{i}$ correspond to $H_{i}\,(i=1,2,3,4,5,6)$, for that we
determine explicitly the six subgroups $H_{i}$ of $C_{k,5}$ as follows:
We have that $C_{k,5}\,=\,\langle\mathcal{A},\mathcal{X}\rangle$, where
$H_{1}\,=\,C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$ and
$H_{6}\,=\,C_{k,5}^{-}\,=\,\langle\mathcal{X}\rangle$. By theorem 3.2 we have
$\mathcal{A}\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}]$ and
$\mathcal{X}\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$, then
$\mathcal{A}\mathcal{X}\,=\,[\mathcal{P}_{1}]^{2}$,
$\mathcal{A}\mathcal{X}^{2}\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}^{3}]^{3}$,
$\mathcal{A}\mathcal{X}^{3}\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}^{2}]^{4}$ and
$\mathcal{A}\mathcal{X}^{4}\,=\,[\mathcal{P}_{3}]^{4}$. Hence
$H_{2}\,=\,\langle[\mathcal{P}_{1}]\rangle$,
$H_{3}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}^{3}]\rangle$,
$H_{4}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}^{2}]\rangle$ and
$H_{5}\,=\,\langle[\mathcal{P}_{3}]\rangle$. Since $\tau^{2}$ sets
$k(\sqrt[5]{\pi_{1}\pi_{3}})$ and $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$, if
$K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}})$ then
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ and vise versa. If
$(\frac{\pi_{1}}{\pi_{3}})_{5}\,=\,1$ then
$X^{5}\equiv\pi_{1}\,(\mathrm{mod}\,\pi_{3})$ resolved on
$\mathcal{O}_{k_{0}}$ and by proposition 2.1 we have that $\pi_{1}$ splits
completly in $k_{0}(\sqrt[5]{\pi_{3}})$, which equivalent to say that
$\mathcal{P}_{1}$ splits completly in $k(\sqrt[5]{\pi_{3}})$, hence
$K_{2}\,=\,k(\sqrt[5]{\pi_{3}})$ and we get that
$K_{5}\,=\,k(\sqrt[5]{\pi_{1}})$ and if
$K_{3}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$ then
$K_{4}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$ and vise versa. Since $\pi_{1}$
and $\pi_{3}$ divide $\pi_{1}\pi_{3}$, $\pi_{1}\pi_{3}^{2}$,
$\pi_{1}\pi_{3}^{3}$ and $\pi_{1}\pi_{3}^{4}$, if
$(\frac{\pi_{1}}{\pi_{3}})_{5}\,\neq\,1$ then $K_{2}\,=\,k(\sqrt[5]{\pi_{1}})$
and $K_{5}\,=\,k(\sqrt[5]{\pi_{3}})$.

\- Since $\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k}$ $i=1,3$ we have
$(\mathcal{P}_{1}\mathcal{P}_{3})^{5}\,=\,\pi_{1}\pi_{3}\mathcal{O}_{k}$, then
$(\mathcal{P}_{1}\mathcal{P}_{3})^{5}\,=\,\pi_{1}\pi_{3}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}})}$
in $k(\sqrt[5]{\pi_{1}\pi_{3}})$ and
$\pi_{1}\pi_{3}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}})}\,=\,(\sqrt[5]{\pi_{1}\pi_{3}}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}})})^{5}$,
hence
$\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}})}\,=\,\sqrt[5]{\pi_{1}\pi_{3}}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}})}$.
Thus $\mathcal{P}_{1}\mathcal{P}_{3}$ seen in
$\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}})}$ becomes principal, i.e
$[\mathcal{P}_{1}\mathcal{P}_{3}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}})$.
\- Since
$(\mathcal{P}_{1}\mathcal{P}_{3}^{2})^{5}\,=\,\pi_{1}\pi_{3}^{2}\mathcal{O}_{k}$,
we have
$(\mathcal{P}_{1}\mathcal{P}_{3}^{2})^{5}\,=\,\pi_{1}\pi_{3}^{2}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})}$
in $k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$ and
$\pi_{1}\pi_{3}^{2}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})}\,=\,(\sqrt[5]{\pi_{1}\pi_{3}^{2}}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})})^{5}$,
hence
$\mathcal{P}_{1}\mathcal{P}_{3}^{2}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})}\,=\,\sqrt[5]{\pi_{1}\pi_{3}^{2}}\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})}$.
Thus $\mathcal{P}_{1}\mathcal{P}_{3}^{2}$ seen in
$\mathcal{O}_{k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})}$ becomes principal, i.e
$[\mathcal{P}_{1}\mathcal{P}_{3}^{2}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}^{2}})$. By the same reasoning we have
$[\mathcal{P}_{1}\mathcal{P}_{3}^{3}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}^{3}})$ and $[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$
capitulates in $k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ .
\- We have $\mathcal{P}_{1}^{5}\,=\,\pi_{1}\mathcal{O}_{k}$, then
$\mathcal{P}_{1}\mathcal{O}_{k(\sqrt[5]{\pi_{1}})}\,=\,\sqrt[5]{\pi_{1}}\mathcal{O}_{k(\sqrt[5]{\pi_{1}})}$.
Hence $[\mathcal{P}_{1}]$ capitulates in $k(\sqrt[5]{\pi_{1}})$. By the same
reasoning we have $[\mathcal{P}_{3}]$ capitulates in $k(\sqrt[5]{\pi_{3}})$.

$(i)$ If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,=\,1$ and
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ we have
$[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$ capitulates in $K_{6}$. According to
[References, Lemma 6.2] we have that $C_{k,5}^{+}\cong C_{\Gamma,5}$. We
denote by $j_{k/\Gamma}:C_{\Gamma,5}\longleftarrow C_{k,5}$ the homomorphism
induced by extension of ideals of $\Gamma$ in $k$. By class field theory we
have that $C_{k,5}^{-}$ correspond to $K_{6}\,=\,k\Gamma_{5}^{(1)}$, and since
$C_{k,5}^{+}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}]\rangle$ and
$\mathcal{P}_{1}\mathcal{P}_{3}\,=\,j_{k/\Gamma}(\mathcal{J})$ such that
$C_{\Gamma,5}\,=\,\langle\mathcal{J}\rangle$, then
$[\mathcal{P}_{1}\mathcal{P}_{3}]$ capitulates in $K_{6}$, and as
$C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$,
then all classes capitulate in $K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$. We
determine possible types of capitulation
$(i_{1},i_{2},i_{3},i_{4},i_{5},i_{6})$. We have that $i_{6}\,=\,0$,
$K_{2}\,=\,K_{5}^{\tau^{2}}$, $K_{3}\,=\,K_{4}^{\tau^{2}}$ and
$C_{k,5}\,=\,C_{k,5}^{\tau^{2}}$. If $i_{1}\neq 0$ we have $i_{1}=1$, if
$i_{2}\neq 0$ we have $i_{2}=2$ and if $i_{5}\neq 0$ we have $i_{5}=5$.
$i_{3}$ and $i_{4}$ are both nulls or non nulls, so if $i_{3}$ and $i_{4}$
$\neq 0$, then $(i_{3},i_{4})\,=\,(3,4)$ or $(4,3)$. Thus the possible types
of capitulation are $(0,0,0,0,0,0)$, $(1,0,0,0,0,0)$, $(0,2,0,0,5,0)$,
$(1,2,0,0,5,0)$, $\\{(0,0,3,4,0,0)$ or $(0,0,4,3,0,0)\\}$, $\\{(1,0,3,4,0,0)$
or $(1,0,4,3,0,0)\\}$, $\\{(0,2,3,4,5,0)$ or $(0,2,4,3,5,0)\\}$,
$\\{(1,2,3,4,5,0)$ or $(1,2,4,3,5,0)\\}$.

$(ii)$ If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,=\,1$ and
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}})$ we have
$[\mathcal{P}_{1}\mathcal{P}_{3}]$ capitulates in $K_{6}$ then if $i_{6}\neq
0$ we have $i_{6}=1$. $[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$ capitulates in
$K_{1}$ then if $i_{1}\neq 0$ we have $i_{1}=6$, so the same possible types of
capitulation accur as in $(i)$ with $i_{6}=0$ or $1$ and $i_{1}=0$ or $6$.

$(iii)$ If $(\frac{\pi_{1}}{\pi_{3}})_{5}\,\neq\,1$, by $(1)$ we have
$K_{2}\,=\,k(\sqrt[5]{\pi_{3}})$ and $K_{5}\,=\,k(\sqrt[5]{\pi_{1}})$ then the
same possible types of capitulation accur as $(i)$ and $(ii)$ by permuting 2
and 5. ∎

###  3.2 The case $n\,=\,p^{e}q\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$,
where $p\,\not\equiv\,1\,(\mathrm{mod}\,25)$, $q\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$

Let $k\,=\,\Gamma(\zeta_{5})$ be the normal closure of
$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, where $n\,=\,p^{e}q\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ such that $p\,\not\equiv\,1,(\mathrm{mod}\,25)$,
$q\,\not\equiv\,\pm 7\,(\mathrm{mod}\,25)$ and $e\in\\{1,2,3,4\\}$. Since
$q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ we have that $q$ is inert in
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$, so we can take in the following
$q\,=\,\pi_{5}$ a prime in $k_{0}$. As before, by
$\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{P}_{3},\mathcal{P}_{4}$ and
$\mathcal{P}_{5}$ we denote respectivly the prime ideals of $k$ above
$\pi_{1},\pi_{2},\pi_{3},\pi_{4}$ and $\pi_{5}$ in $k_{0}$, such that
$\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k}\,(i=1,2,3,4,5)$. We have that
$\tau^{2}$ permutes $\pi_{1}$ with $\pi_{3}$, then $\tau^{2}$ permutes
$\mathcal{P}_{1}$ with $\mathcal{P}_{3}$, but $\tau^{2}$ sets $q\,=\,\pi_{5}$
and also $\mathcal{P}_{5}$.
The six intermediate extensions of $k_{5}^{(1)}/k$ are determined as follows:

######  Theorem 3.4.

Let $k,n,\pi_{1},\pi_{2},\pi_{3},\pi_{4}$ and $\pi_{5}$ as above. Put
$x_{1}\,=\,\pi_{1}\pi_{5}^{h_{1}}$ and $x_{2}\,=\,\pi_{1}\pi_{3}^{4}$ are
choosen such that $x_{1}\,\equiv\,x_{2}\,\equiv
1\,(\mathrm{mod}\,\lambda^{5})$, where $h_{1}\in\\{1,2,3,4\\}$. Then:

$k_{5}^{(1)}\,=\,k(\sqrt[5]{x_{1}},\sqrt[5]{x_{2}})$.

The six intermediate extensions of $k_{5}^{(1)}/k$ are: $k(\sqrt[5]{x_{1}})$,
$k(\sqrt[5]{x_{2}})$, $k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$,
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$,
$k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ and
$k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$. Furthermore $\tau^{2}$ permutes
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$ with
$k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ and $k(\sqrt[5]{x_{1}})$
with $k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$, and sets $k(\sqrt[5]{x_{2}})$,
$k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$.

###### Proof.

Since $k\,=\,k_{0}(\sqrt[5]{n})$ we can write $n$ in $k_{0}$ as
$n\,=\,\pi_{1}^{e}\pi_{2}^{e}\pi_{3}^{e}\pi_{4}^{e}\pi_{5}$ with
$\pi_{i}\,\not\equiv\,1\,(\mathrm{mod}\,\lambda^{5})$ because
$p\,\not\equiv\,1\,(\mathrm{mod}\,25)$ and
$q\,\not\equiv\,1\,(\mathrm{mod}\,25)$. By proposition 2.2 there exists
$h_{1},h_{2}\in\\{1,..,4\\}$ such that $\pi_{1}\pi_{5}^{h_{1}}\,\equiv\,\pm
1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$ and $\pi_{1}\pi_{3}^{h_{2}}\,\equiv\,\pm
1,\pm 7\,(\mathrm{mod}\,\lambda^{5})$. to investigate the correspondence
between the six intermediate extension of $k_{5}^{(1)}/k$ and the six
subgroups of $C_{k,5}$, we assume that $h_{2}\,=\,4$. Put
$x_{1}\,=\,\pi_{1}\pi_{5}^{h_{1}}$ and $x_{2}\,=\,\pi_{1}\pi_{3}^{4}$.

The fact that $k_{5}^{(1)}\,=\,k(\sqrt[5]{x_{1}},\sqrt[5]{x_{2}})$ follows
from proposition 2.2.

The six intermediate extensions are: $k(\sqrt[5]{x_{1}})$,
$k(\sqrt[5]{x_{2}})$, $k(\sqrt[5]{x_{1}x_{2}})$,
$k(\sqrt[5]{x_{1}x_{2}^{2}})$, $k(\sqrt[5]{x_{1}x_{2}^{3}})$ and
$k(\sqrt[5]{x_{1}x_{2}^{4}})$. Since $x_{1}\,=\,\pi_{1}\pi_{5}^{h_{1}}$ and
$x_{2}\,=\,\pi_{1}\pi_{5}^{4}$, we have
$k(\sqrt[5]{x_{1}x_{2}})\,=\,k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$,
$k(\sqrt[5]{x_{1}x_{2}^{2}})\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$,
$k(\sqrt[5]{x_{1}x_{2}^{3}})\,=\,k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$
and $k(\sqrt[5]{x_{1}x_{2}^{4}})\,=\,k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$.
Since $\pi_{1}^{\tau^{2}}\,=\,\pi_{3}$, $\pi_{3}^{\tau^{2}}\,=\,\pi_{1}$ and
$\pi_{5}^{\tau^{2}}\,=\,\pi_{5}$, and by the same reasoning as theorem 3.1 we
prove that $\tau^{2}$ permutes
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$ with
$k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ and $k(\sqrt[5]{x_{1}})$
with $k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$, and sets $k(\sqrt[5]{x_{2}})$,
$k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$. ∎

The generators of $C_{k,5}$ in this case are determined as follows:

######  Theorem 3.5.

Let $k,n,\pi_{1},\pi_{2},\pi_{3},\pi_{4},\pi_{5}$ and $h_{1}$ as above. Let
$\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{P}_{3},\mathcal{P}_{4}$ and
$\mathcal{P}_{5}$ prime ideals of $k$ such that
$\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k_{0}}\,(i=1,2,3,4,5)$. Then:

$C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$

###### Proof.

According to [References, theorem 1.1], for this case of the radicand $n$, we
have that $\zeta_{5}^{i}(1+\zeta_{5})^{j}$ is not norm of element in
$k-\\{0\\}$ for any exponents $i$ and $j$, then by [References, section 5.3],
we have
$C_{k,5}\,=\,C_{k,5}^{(\sigma)}\,=\,C_{k,s}^{(\sigma)}\,=\,\langle[\mathcal{P}_{1}],[\mathcal{P}_{2}],[\mathcal{P}_{3}],[\mathcal{P}_{4}],[\mathcal{P}_{5}]\rangle$.
Since $\mathcal{P}_{1}^{\tau^{2}}\,=\,\mathcal{P}_{3}$,
$\mathcal{P}_{2}^{\tau^{2}}\,=\,\mathcal{P}_{4}$ and
$\mathcal{P}_{5}^{\tau^{2}}\,=\,\mathcal{P}_{5}$, as the proof of theorem 3.2
we have that $\mathcal{P}_{1},\mathcal{P}_{3}$ and $\mathcal{P}_{5}$ are non
principals. As
$[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]^{\tau^{2}}\,=\,[(\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}})^{\tau^{2}}]\,=\,[\mathcal{P}_{3}\mathcal{P}_{1}\mathcal{P}_{5}^{2h_{1}}]\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]$
then
$C_{k,5}^{+}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]\rangle$,
and we have that
$C_{k,5}^{-}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$. Hence
$C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$.
∎

The main theorem of capitulation in this case is as follows:

######  Theorem 3.6.

We keep the same assumptions as theorem 3.5 Then:

$K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ or
$k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$,
$K_{2}\,=\,k(\sqrt[5]{\pi_{1}\pi_{5}^{h_{1}}})$ or
$k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$,
$K_{3}\,=\,k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$ or
$k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ ,
$K_{4}\,=\,k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ or
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$,
$K_{5}\,=\,k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$ or
$k(\sqrt[5]{\pi_{1}\pi_{5}^{h_{1}}})$ and
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$ or
$k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$.

$[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$,
$[\mathcal{P}_{1}\mathcal{P}_{5}^{h_{1}}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{5}^{h_{1}}})$,
$[\mathcal{P}_{1}^{2}\mathcal{P}_{3}^{4}\mathcal{P}_{5}^{h_{1}}]$ capitulates
in $k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$,
$[\mathcal{P}_{1}^{4}\mathcal{P}_{3}^{2}\mathcal{P}_{5}^{h_{1}}]$ capitulates
in $k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$,
$[\mathcal{P}_{3}\mathcal{P}_{5}^{h_{1}}]$ capitulates in
$k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$ and
$[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$ capitulates in
$k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$.

\- If $K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$, then the
possible types of capitulation are: $(0,0,0,0,0,0)$, $(1,0,0,0,0,0)$,
$\\{(0,5,0,0,2,0)$ or $(0,2,0,0,5,0)\\}$, $\\{(1,5,0,0,2,0)$ or
$(1,2,0,0,5,0)\\}$, $\\{(0,5,4,3,2,0)$ or $(0,2,4,3,5,0)\\}$,
$\\{(1,5,4,3,2,0)$ or $(1,2,4,3,5,0)\\}$, $\\{(0,5,3,4,2,0)$ or
$(0,2,3,4,5,0)\\}$, $\\{(1,5,3,4,2,0)$ or $(1,2,3,4,5,0)\\}$,
$\\{(0,0,3,4,0,0)$ or $(0,0,4,3,0,0)\\}$, $\\{(1,0,3,4,0,0)$ or
$(1,0,4,3,0,0)\\}$.
\- If $K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$, then the same possible
types occur, with $i_{6}$ takes value 0 or 1.

###### Proof.

According to theorem 3.4, we have that $\tau^{2}$ permutes
$k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$ with
$k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ and $k(\sqrt[5]{x_{1}})$
with $k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$, and sets $k(\sqrt[5]{x_{2}})$,
$k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$. We determine first the six
subgroups $H_{i}$ of $C_{k,5}$. We have that
$C_{k,5}\,=\,\langle\mathcal{A},\mathcal{X}\rangle$, where
$H_{1}\,=\,C_{k,5}^{+}\,=\,\langle\mathcal{A}\rangle$ and
$H_{6}\,=\,C_{k,5}^{-}\,=\,\langle\mathcal{X}\rangle$. By theorem 3.5 we have
$\mathcal{A}\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]$ and
$\mathcal{X}\,=\,[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$, then
$\mathcal{A}\mathcal{X}\,=\,[\mathcal{P}_{1}\mathcal{P}_{5}^{h_{1}}]^{2}$,
$\mathcal{A}\mathcal{X}^{2}\,=\,[\mathcal{P}_{1}^{2}\mathcal{P}_{3}^{4}\mathcal{P}_{5}^{h_{1}}]^{4}$,
$\mathcal{A}\mathcal{X}^{3}\,=\,[\mathcal{P}_{1}^{4}\mathcal{P}_{3}^{2}\mathcal{P}_{5}^{h_{1}}]$
and
$\mathcal{A}\mathcal{X}^{4}\,=\,[\mathcal{P}_{3}\mathcal{P}_{5}^{h_{1}}]^{3}$.
Hence $H_{2}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{5}^{h_{1}}]\rangle$,
$H_{3}\,=\,\langle[\mathcal{P}_{1}^{2}\mathcal{P}_{3}^{4}\mathcal{P}_{5}^{h_{1}}]\rangle$,
$H_{4}\,=\,\langle[\mathcal{P}_{1}^{4}\mathcal{P}_{3}^{2}\mathcal{P}_{5}^{h_{1}}]\rangle$
and $H_{5}\,=\,\langle[\mathcal{P}_{3}\mathcal{P}_{5}^{h_{1}}]\rangle$. Since
$\tau^{2}$ sets $k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$ and
$k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$, so if
$K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$ then
$K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ and inversly. By class field
theory, the fact that $H_{i}\,(i=2,5)$ correspond to $K_{i}\,(i=2,5)$ mean
that $\mathcal{P}_{1}\mathcal{P}_{5}^{h_{1}}$ splits completly in $K_{2}$ and
$\mathcal{P}_{3}\mathcal{P}_{5}^{h_{1}}$ splits completly in $K_{5}$. As
$\pi_{1}\pi_{5}^{h_{1}}$ divides $\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}$ and
$\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}$, by proposition 2.1,
$\pi_{1}\pi_{5}^{h_{1}}$ can not split in
$k_{0}{\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}}}$ and
$k_{0}{\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}}}$, this equivalent to
say that $\mathcal{P}_{1}\mathcal{P}_{5}^{h_{1}}$ can not split completly in
$k{\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}}}$ and
$k{\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}}}$. By the same reasoning we
have that $\mathcal{P}_{3}\mathcal{P}_{5}^{h_{1}}$ can not split completly in
$k{\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}}}$ and
$k{\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}}}$. Hence if
$K_{2}\,=\,k(\sqrt[5]{\pi_{1}\pi_{5}^{h_{1}}})$ then
$K_{5}\,=\,k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$ and inversly, which allow us
to deduce that if
$K_{3}\,=\,k(\sqrt[5]{\pi_{1}^{2}\pi_{3}^{4}\pi_{5}^{h_{1}}})$ then
$K_{5}\,=\,k(\sqrt[5]{\pi_{1}^{4}\pi_{3}^{2}\pi_{5}^{h_{1}}})$ and inversly.

We keep the same proof as $(2)$ theorem 3.3.

-If $K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$, then $K_{6}\,=\,k\Gamma_{5}^{(1)}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ and we have that $[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]$ capitulates in $K_{6}$, moreover since $C_{k,5}^{+}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]\rangle\cong C_{\Gamma,5}$ then $\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}\,=\,j_{k/\Gamma}(\mathcal{J})$ such that $C_{\Gamma,5}\,=\,\langle\mathcal{J}\rangle$, then $[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]$ capitulates in $K_{6}$. As $C_{k,5}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}],[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$, then all classes capitulate in $K_{6}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$. We determine the possible types of capitulation $(i_{1},i_{2},i_{3},i_{4},i_{5},i_{6})$. We have that $i_{6}\,=\,0$, $K_{2}\,=\,K_{5}^{\tau^{2}}$, $K_{3}\,=\,K_{4}^{\tau^{2}}$ and $C_{k,5}\,=\,C_{k,5}^{\tau^{2}}$. If $i_{1}\neq 0$ we have $i_{1}=1$. $i_{2}$ and $i_{5}$ are both nulls or non nulls, so if $i_{2}$ and $i_{5}$ $\neq 0$, then $(i_{2},i_{5})\,=\,(2,5)$ or $(5,2)$ depending on $\mathcal{P}_{1}\mathcal{P}_{5}^{h_{1}}$ splits completely in $k(\sqrt[5]{\pi_{1}\pi_{5}^{h_{1}}})$ or in $k(\sqrt[5]{\pi_{3}\pi_{5}^{h_{1}}})$. Similarly if $i_{3}$ and $i_{4}$ $\neq 0$, then $(i_{3},i_{4})\,=\,(3,4)$ or $(4,3)$. Hence the possible types given are proved.
-If $K_{1}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}^{4}})$ then $K_{6}\,=\,k\Gamma_{5}^{(1)}\,=\,k(\sqrt[5]{\pi_{1}\pi_{3}\pi_{5}^{2h_{1}}})$ and we have $C_{k,5}^{+}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]\rangle$ capitulates in $K_{6}$, the possible values of $i_{2},i_{3},i_{4},i_{5}$ are as above, $(i_{2},i_{5})\,=\,(2,5)$ or $(5,2)$ if they are non nulls, $(i_{3},i_{4})\,=\,(3,4)$ or $(4,3)$ if they are non nulls. If $i_{1}\neq 0$ then $i_{1}=6$ because $H_{6}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}^{4}]\rangle$, and if $i_{6}\neq 0$ then $i_{1}=1$ because $H_{1}\,=\,\langle[\mathcal{P}_{1}\mathcal{P}_{3}\mathcal{P}_{5}^{2h_{1}}]\rangle$. Hence the possible types given are proved. ∎

###  3.3 The case $n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$,
where $p\,\not\equiv\,1\,(\mathrm{mod}\,25)$

Let $k\,=\,\Gamma(\zeta_{5})$ be the normal closure of
$\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, where $n\,=\,5^{e}p$ such that
$p\,\not\equiv\,1,(\mathrm{mod}\,25)$ and $e\in\\{1,2,3,4\\}$. Since
$n\,=\,5^{e}p\,\not\equiv\,\pm 1\pm 7,(\mathrm{mod}\,25)$ then
$\lambda\,=\,1-\zeta_{5}$ is ramified in $k/k_{0}$. Let
$\pi_{1},\pi_{2},\pi_{3}$ and $\pi_{4}$ primes of $k_{0}$ such that
$p\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$. Let
$\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{P}_{3},\mathcal{P}_{4}$ and
$\mathcal{I}$ prime ideals of $k$ above $\pi_{1},\pi_{2},\pi_{3},\pi_{4}$ and
$\lambda$, we have $\mathcal{P}_{i}^{5}\,=\,\pi_{i}\mathcal{O}_{k}$ and
$\mathcal{I}^{5}\,=\,\lambda\mathcal{O}_{k}$. According to [References,
theorem 1.1], for this case of the radicand $n$, we have that
$\zeta_{5}^{i}(1+\zeta_{5})^{j}$ is not norm of element in $k-\\{0\\}$ for any
exponents $i$ and $j$, then we have
$C_{k,5}\,=\,C_{k,5}^{(\sigma)}\,=\,C_{k,s}^{(\sigma)}$. Hence the results
about the six intermediate extensions of $k_{5}^{(1)}/k$, the generators of
$C_{k,5}$ and the capitulation problem in this case are the same as case $2$
by substituting $q$ by $5$, $\pi_{5}$ by $\lambda\,=\,1-\zeta_{5}$ and
$\mathcal{P}_{5}$ by $\mathcal{I}$.

##  4 Numerical examples

The task to determine the capitulation in a cyclic quintic extension of a base
field of degree $20$, that is, in a field of absolute degree $100$, is
definitely far beyond the reach of computational algebra systems like MAGMA
and Pari/GP. For this reason we give exemples of a pure metacyclic fields
$k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ such that $C_{k,5}$ is of type
$(5,5)$ and $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$.

Table 1: $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$ with $C_{k,5}$ is of type
$(5,5)$ and $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$.

$n$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$ |  | $n$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$ |
---|---|---|---|---|---|---|---|---|---
$55$ | 25 | $(5,5)$ | 2 |  | $1457$ | 25 | $(5,5)$ | 2 |
$655$ | 25 | $(5,5)$ | 2 |  | $6943$ | 25 | $(5,5)$ | 2 |
$1775$ | 25 | $(5,5)$ | 2 |  | $8507$ | 25 | $(5,5)$ | 2 |
$1555$ | 25 | $(5,5)$ | 2 |  | $12707$ | 25 | $(5,5)$ | 2 |
$2155$ | 25 | $(5,5)$ | 2 |  | $151$ | 25 | $(5,5)$ | 2 |
$5125$ | 25 | $(5,5)$ | 2 |  | $1301$ | 25 | $(5,5)$ | 2 |
$8275$ | 25 | $(5,5)$ | 2 |  | $2111$ | 25 | $(5,5)$ | 2 |
$30125$ | 25 | $(5,5)$ | 2 |  | $251^{2}$ | 25 | $(5,5)$ | 2 |
$38125$ | 25 | $(5,5)$ | 2 |  | $601^{3}$ | 25 | $(5,5)$ | 2 |
$113125$ | 25 | $(5,5)$ | 2 |  | $2131^{2}$ | 25 | $(5,5)$ | 2 |
$93$ | 25 | $(5,5)$ | 2 |  | $1901^{4}$ | 25 | $(5,5)$ | 2 |
$382$ | 25 | $(5,5)$ | 2 |  | $1051^{4}$ | 25 | $(5,5)$ | 2 |
$943$ | 25 | $(5,5)$ | 2 |  | $1801^{3}$ | 25 | $(5,5)$ | 2 |

## References

  * [1] F.Elmouhib, M. Talbi, and A. Azizi, _5-rank of ambiguous class groups of quintic Kummer extensions_ , Accepted for publication in Proceedings-Mathematical Sciences.
  * [2] F.Elmouhib, M. Talbi, and A. Azizi, _On the capitulation problem of some pure metacyclic fields of degree 20._ , https://arxiv.org/abs/2010.15935 V2.
  * [3] Ph. Furtwängler, _Beweis des Hauptidealsatzes für Klassenkörper algebraischer Zahlkörper._ , Abh Math. Sem. Univ. Hamburg 7 (1930), 14-36.
  * [4] G.Gras, _Sur les $l$-classes d’idéaux dans les extensions cycliques relatives de degré premier impaire $l$_. _Annales de l’institut Fourier_ , (1973).
  * [5] E.Hecke, _Algebraic Number Theory_ , GTM 77, Springer-Verlag 1981.
  * [6] E.Hecke, _Lectures on the Theory of Algebraic Numbers_ , GTM , Vol. 77, Springer-Verlag 1981.
  * [7] D.Hilbert, _Über die Theorie der relativ-Abelschen Zahlköper_ , Acta Math. 26 (1902), 99-132.
  * [8] M.Ishida, _The genus Fields of Algebraic Number Fields_. Lecture notes in Mathematics Vol 555, Springer-Verlag (1976).
  * [9] 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.
  * [10] K. Iimura, _A criterion for the class number of a pure quintic field to be divisible by 5_ , J. Reine Angew. Math. 292 (1977) 201-210,
  * [11] D.C. Mayer, _Transfers of metabelian p-groups_ , Monatsh. Math. 166 (2012), no. 3-4, 467-495, DOI 10.1007/s00605-010-0277-x.
  * [12] K. Miyake, _Algebraic Investigations of Hilbert’s Theorem 94, the Principal Ideal Theorem and Capitulation Problem_ , Expos. Math. 7 (1989), 289-346.
  * [13] H. Suzuki, _A Generalisation of Hilbert’s Theorem 94_ , Nagoya Math. J., vol 121 (1991).
  * [14] T. Tannaka, _A generalized principal ideal theorem and a proof of a conjecture of Deuring_ , Ann. Math. 67 (1958).
  * [15] O. Taussky, _A remark concerning Hilbert’s theorem 94_ , J. reine angew. Math. 239/240 (1970), 435-438.
  * [16] The PARI Group, PARI/GP, Version 2.4.9, Bordeaux, 2017, http://pari.math.u-bordeaux.fr.


# some ordered matrix semigroups

P. G. Romeo and Sathi P H  Dept. of Mathematics, CUSAT, Kochi, Kerala INDIA.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

A semigroup together with compatible partial order is called an odered
semigroup. In this paper we discuss the ordered matrix semigroups.

##  1\. Intoduction and preliminaries

A semigroup $(S,\cdot)$ is a non empty set together with an assosiativebinary
operation $\cdot$. Obviously every group is a semigroup, however there are
semigroups which fails to be a group such as integers under usual
multiplicaton, natural numbers under adition and the like. More familiar
examples of semigroups are

  1. (1)

Full transformation semigroup $\mathcal{T}(X)$ of all transformations on a set
$X$ with composition of transformations as binary operation.

  2. (2)

All square matrices over a field $F$ [ring $R$] $M_{n}(F)\,[M_{n}(R)]$ with
the usual multiplication of matrices as binary operation.

As a mathematical structure semigroups attracted much attention and has grown
to an extend that it pocesses a vivd and deep structure theory. Regular
semigroups and inverse semigroups include the class of semigroups which are
well developed. An element $a$ in a semigroup $S$ is said to be von Neumann
regular if there exists an alement $x\in S$ such that $axa=a$. A semigroup $S$
is said to be regular if every element $a\in S$ are regular. A regular
semigroup in which the idempotent elements ( those elements $e$ in a semigroup
$S$ such that $e\cdot e=e)$ commutes are called inverse semigroups.

######  Definition 1.

Let $S$ be a semigroup and $a\in S$.

  1. (1)

A particular solution to $axa=a$ is called the inner inverse of $a$ and is
denoted by $a^{-}$.

  2. (2)

A solution of the equation $xax=x$ is called the outer inverse of $a$ and is
denoted by $a^{=}$

  3. (3)

An inner inverse of $a$ that is also an outer inverse is called a reflexive
inverse and is denoted by $a^{+}$.

The set of all inner (resp. outer, resp. reflexive) inverses of $a$ is denoted
by $a\\{1\\}(resp.a\\{2\\},resp.a\\{1,2\\})$

######  Definition 2.

A semigroup $S$ is said to be weakly separative if, for any $a,b\in S$,

$asa=asb=bsa=bsb\Rightarrow a=b.$

######  Lemma 1.

Every regular semigroup (in the sense of von Neumann) is weakly separative.

###### Proof.

Let $S$ be regular. For $a,b\in S,$ we have $axa=a$ and $byb=b$ for some
$x,y\in S$. Assume that $asa=asb=bsa=bsb$ for all $s\in S$. In particular,
$a=axa=ax$ and $ayb=byb=b$ so that $a=axb=ax(byb)=(axb)yb=ayb=b$. ∎

######  Example 1.

(1) The set of all transformations on a set $X$ with composition of maps is a
regular semigroup called the full transformation semigroup.

(2) All $n\times n$ matrices $M_{n}$ over a field $F$ or a ring $R$ are
regualr semigroups with respect to usual mutliplication of matrices.

However, here it should be noted that the usual matrix product is not the only
matrix product with respect to which $M_{n}$ is a semigroup. In fact there are
other interseting matrix products such as

  1. (1)

Kronecker product (Tensor product)

  2. (2)

Hadamard product (Schur product)

with respect to which $M_{n}$ is a semigroup.

Consider $A=[a_{ij}],B=[b_{ij}]\in M_{n}$, then their Kronecker product
$A\otimes B$ is defined to be the $n^{2}\times n^{2}$ matrix partitioned into
$n^{2}$ blocks with the $(i,j)^{th}$ block as the $n\times n$ matrix
$a_{ij}B$, i.e.,

$\mathbf{{A}\otimes{B}}\;=\;\begin{pmatrix}a_{11}B&a_{12}B&\cdots&a_{1n}B\\\
a_{21}B&a_{22}B&\cdots&a_{2n}B\\\ \cdot&\cdot&\cdots&\cdot\\\
a_{n1}B&a_{n2}B&\cdots&a_{nn}B\end{pmatrix}.$

Clearly $(A\otimes B)\otimes C=A\otimes(B\otimes C)$ for all $A,B,C\in M_{n}$,
thus $(M,\otimes)$ is a semigroup.

The Hadamard product or Schur product of $A,B\in M_{n}$ is defined by the
entrywise product $A\circ B=(a_{ij}b_{ij})$ that is,

$\mathbf{{A}\circ{B}}\;=\;\begin{pmatrix}a_{11}b_{11}&a_{12}b_{12}&\cdots&a_{1n}b_{1n}\\\
a_{21}b_{21}&a_{22}b_{22}&\cdots&a_{2n}b_{2n}\\\ \cdot&\cdot&\cdots&\cdot\\\
a_{n1}b_{n1}&a_{n2}b_{n2}&\cdots&a_{nn}b_{nn}\end{pmatrix}$

Since $A\circ(B\circ C)=(A\circ B)\circ C$ for all $A,B,C\in M_{n}(F)$, we
have $(M_{n},\circ)$ is a semigroup with respect to the multiplication
$\circ$.

A Partial order on a semigroup $S$ is a binary relation $\leq$ on $S$ which is
reflexive, antisymmetric and transitive.

######  Definition 3.

An ordered semigroup $(S,\cdot,\leq)$ is a semigroup $(S,\cdot)$ together with
a compatible partial order $"\leq"$ on $S$ such that for any $x,y,z\in S$,

$x\leq y\Rightarrow\,xz\leq xz\quad\text{and}\quad zx\leq zy.$

######  Example 2.

The set of all relations on a set $X$ denoted by $B(X)$ is an ordered
semigroup with the composition of relations as the binary operation and
inclusion as the compatible partial order on $B(X)$.

Generally, any semigroup$(S,\cdot)$ can be considered as an ordered semigroup
with respect to the order $\leq$ as the identity relation on $S$, that is

$x\leq y\Longleftrightarrow x=y.$

If $P$ is any partial order on a semigroup $S$ then the relation

$P_{1}=\\{(x,y)\,|\,(axb,ayb)\in P\,\text{for all}\,(a,b)\in S\\}$

is the largest compatible partial order contained in $P$. Thus, every
semigroup can be endowed with a partial order so that $S$ becomes an ordered
semigroup.

######  Lemma 2.

Let $(S,\cdot,\leq)$ be an ordered semigroup. For every $s\in S$ define a
binary composition $\rho$ on $S$ by

$x\rho y=xsy\quad\text{for all}\quad x,y\in S$

Then, $(S,\rho)$ is a semigroup such that $(S,\rho,\leq)$ is an ordered
semigroup and is called the $s$-invariant of $S$.

######  Definition 4.

Let $(S,\cdot,\leq)$ be an ordered semigroup. A non empty subset $A\in S$ is a
subsemigroup of $S$ if for any $a,b\in A$ implies $ab\in A$.

##  2\. Ordered Matrix Semigroups

In the following we proceed to describe ordered semigroup $n\times n$
matrices. At the outset it is seen that the semigroup $M_{n}(\mathbb{R})$
square matrices over real nuumbers with order defined by, for $A,B\in
M_{n}(\mathbb{R})$ with $A=[a_{ij}],B=[b_{ij}],\,1\leq i\leq n,\,1\leq j\leq
n$, then

$A\leq B\Rightarrow a_{ij}\leq b_{ij}\,\text{for all}\quad
i\quad\text{and}\,j$

fails to be an ordered semigroup. However for $M_{n}(\mathbb{Z}_{+})$ of all
$n\times n$ matrices with non negative integer entries we have the following
lemma.

######  Lemma 3.

Consider the semigroup $M_{n}(\mathbb{(}\mathbb{Z}_{+}))$ of all $n\times n$
matrices with non negative integer entries. For $A,B\in M_{n}(\mathbb{Z}_{+})$
with $A=[a_{ij}],B=[b_{ij}],\,1\leq i\leq n,\,1\leq j\leq n$, define the
relation $\leq$ by

$A\leq B\Rightarrow a_{ij}\leq b_{ij}\,\text{for all}\quad
i\quad\text{and}\,j$

is a partial order with respect to which $M_{n}(\mathbb{Z}_{+})$ is an ordered
semigroup.

###### Proof.

Clearly the order $\leq$ is reflexive as, $a_{ij}\leq a_{ij}$ for all $i$ and
$j$. Suppose $A\leq B$ and $B\leq A$, ie., $a_{ij}\leq b_{ij}$ and $b_{ij}\leq
a_{ij}$ for all $i$ and $j$.This gives $a_{ij}=b_{ij}$ and hence anti
symmetric. If $A\leq B$ and$B\leq C$, then $a_{ij}\leq b_{ij}$ and $b_{ij}\leq
c_{ij}$. This implies $a_{ij}\leq c_{ij}$,ie, $A\leq C$, hence transitive.Thus
$\leq$ is a partial order.

To prove the compatibility, suppose $A\leq B$. ie,$a_{ij}\leq b_{ij}$. For
$C=[c_{ij}]\in M$,

$(CA)_{ij}=\sum c_{ik}a_{kj}\leq\sum c_{ik}b_{kj}=(CB)_{ij}\,\text{for}\,1\leq
k\leq n$

thus $CA\leq CB$. Similarly it is seen that $AC\leq BC$, ie., $\leq$ is
compatible. ∎

.

###  2.1. Conrad order on Regular matrix semigroups

In (cf. [2]) Abian described a partial order on a semiprime ring $R$ as
follows

$a\leq b\Longleftrightarrow arb=ara\quad\text{for all}\quad r\in R.$

In the light of this partial order, Conrad defined a relation $\rho$ on a
semigroup $S$ by

$a\rho b\Leftrightarrow asa=asb=bsa\quad\text{for all}\quad s\in S$

which turned out to be a partial order for ’weakly separative’ semigroups.

Ler $R$ a ring. Consider matrix semigroup $M_{n}(R)$. Now we proceed to
describe the relation $\rho$ on $M_{n}(R)$ as follows, for $A,B\in M_{n}(R)$.,

$A\rho B\Leftrightarrow ASA=ASB=BSA\quad\forall\,\,S\in\,\,M_{n}(R)$

and this relation $\rho$ is called the Conrad relation on $M_{n}(R)$

######  Lemma 4.

The Conrad relation on $M_{n}(R)$ is a partial order with respect to which
$M_{n}(R)$ is an ordered semigroup.

###### Proof.

It is easy to see that the relation $\rho$ on $M_{n}(R)$ is reflexive. For
transitivity, suppose $A\rho B$ and $B\rho C$, then $ASA=ASB=BSA$ and
$BSB=BSC=CSB$ for all $S\in M_{n}(F)$. For $P\in M_{n}(R),$ we obtain
$(ASC)P(ASC)=(ASC)P(ASB)=(ASC)P(BSA)=(ASC)P(CSA)\Rightarrow ASC=CSA$.
Similarly, $(ASC)P(ASC)=(ASC)P(ASB)=(ASC)P(BSA)=(ASC)P(CSA)\Rightarrow
ASC=CSA$. This gives $ASC=CSA=ASA$, ie, $A\rho C$.

Burgess and Raphael proved that $\rho$ is a partial order on a semigroup $S$
if and only if it is weakly separative. Since $M_{n}(R)$ is a regular
semigroup, obviously $M_{n}(R)$ is weakly seperative and hence $\rho$ is a
partial order. Suppose $A\rho B$ and $C\rho D$. Then, $ASB=BSA=BSB$ and
$CSC=CSD=DSC\quad\forall\,S\in M_{n}(F)$.
Then,$(AC)S(BD)=(AC)S(AC)=(BD)S(AC)\Rightarrow AC\rho BD$, hence the
compatibility and $(M_{n}(R),\rho)$ is an ordered matrix semigroup. ∎

###  2.2. More partial orders on $M_{n}(R)$

Let $R$ be any ring. Jerzy K Baksalary and Sujit Kumar Mitra introduced the
following two orderings on $M_{n}(R)$

For $A,B\in M_{n}(R)$, two orders defined by

$A\star\leq B\Leftrightarrow
A^{\star}A=A^{\star}B\,\,\text{and}\,\,R(A)\subseteq R(B)$

where $A^{*}$ stands for the conjugate transpose of $A$ and $R(A)$ is the
range space of $A$. In a similar way another order is given by

$A\leq\star B\Leftrightarrow
AA^{\star}=BA^{\star}\,\text{and}\,R(A^{*})\subseteq R(B^{\star}).$

These orders are called the left star order and the right star order
respectively.

######  Lemma 5.

The left star order and the right star order are partial orders on $M_{n}(R)$

###### Proof.

Clearly $A\star\leq A$. Suppose $A\star\leq B$ and $B\star\leq A$. Then
$A^{\star}A=A^{\star}B,\,\,R(A)\subseteq R(B),\,\,B^{\star}B=B^{\star}A$ and
$R(B)\subseteq R(A)$. This gives $R(A)=R(B)$. Let $A^{\dagger}$ be the Moore-
Penrose inverse of $A$ ( ie., the unique matrix satisfying $AA^{\dagger}A=A$,
and $A^{\dagger}AA^{\dagger}=A^{\dagger}$ ). Then
$AA^{\dagger}=(AA^{\dagger})^{\star}$ and
$A^{\dagger}A=(A^{\dagger}A)^{\star}$ and

$A=AA^{\dagger}A=(AA^{\dagger})^{\star}A=(A^{\dagger})^{\star}(A^{\star}A=(A^{\dagger})^{\star}A^{\star}B=(AA^{\dagger})^{\star}B=AA^{\dagger}B=B$

hence the relation ${}^{\prime}\star\leq^{\prime}$ is antisymmetric.

For transitivity, let $A\star\leq B$ and $B\star\leq C$. Then,
$A^{\star}A=A^{\star}B,\,\,R(A)\subseteq R(B),B^{\star}B=B^{\star}C$ and
$R(B)\subseteq R(C)$, from this we get $R(A)\subseteq R(C)$. Now,
$A^{\star}A=A^{\star}B=A^{\star}(BB^{\dagger}B)=A^{\star}(BB^{\dagger})^{\star}B=(A^{\star}(B^{\dagger})^{\star}B^{\star})B=A^{\star}(B^{\dagger})^{\star}(B^{\star}B)=A^{\star}(B^{\dagger})^{\star}(B^{\star}C)=(BB^{\dagger}A)^{\star}C=A^{\star}C$.
Thus $A^{\star}A=A^{\star}C$ with $R(A)\subseteq R(C)$ implying that
$A\star\leq C*$.

Similarly it can be seen that the right star order $"\leq\star"$ is a partial
order on $M_{n}(R)$. ∎

######  Lemma 6.

$M_{n}(R)$ is an ordered semigroup under the left star ordering.

###### Proof.

It is already seen that the left star order is a partial order on $M_{n}(R)$.
Consider $A,B,C,D\in M_{n}(F)$ such that $A\star\leq B$ and $C\star\leq D$,
then $A^{\star}A=A^{\star}B,\,\,R(A)\subseteq R(B),\,\,C^{\star}C=C^{\star}D$
and $R(C)\subseteq R(D)$.
Further

$\displaystyle(AC)^{\star}AC=C^{\star}(A^{\star}A)C=C^{\star}(A^{\star}B)C$
$\displaystyle=(AC)^{\star}B(CC^{\dagger}C)=(AC)^{\star}B(CC^{\dagger})^{\star}C$
$\displaystyle=(AC)^{\star}B(C^{\dagger})^{\star}(^{\star}C)$
$\displaystyle=(AC)^{\star}B(C^{\dagger})^{\star}(C^{\star}D)$
$\displaystyle=(AC)^{\star}B(CC^{\dagger})^{\star}=(AC)^{\star}B(CC^{\dagger}D)$
$\displaystyle=(AC)^{\star}BD.$

and since $R(C)\subseteq R(D)$, we have $R(AC)\subseteq R(BD)$. Thus
$AC\star\leq BD$ and hence $(M_{n}(R),\cdot,\star\leq)$ is an ordered matrix
semigroup.
Similarly we can prove that $(M_{n}(R),\cdot,\leq\star)$ is an ordered
semigroup. ∎

###  2.3.  Positive semidefinite matrices (PSD)

From here onwards we restrict to the matrices $M_{n}(\mathbb{C})$. A matrix
$A\in M_{n}(\mathbb{C})$ is said to be positive semi definite (positive
definite) if $v^{\star}Av\geq 0\,(v^{\star}Av>0)$ for all
$v\in\mathbb{C}_{n}$. We write $A\geq 0\,(A>0)$ to mean that $A$ is positive
semi definite (positive definite).

In the following we list some properties of PSD matrice

  1. (1)

$A$ is PSD if and only if it is Hermitian, ie., $A=A^{\star}$

  2. (2)

If $A$ is PSD, then all principal submatrices of $A$, $Trace\,(A),\,\mid
A\mid$ and all principal minors of $A$ are PSD.

  3. (3)

$A$ is PSD if and only if $A=M^{\star}M$ for some matrix $M$.

  4. (4)

$A$ is PSD if and only if $A=P^{\star}P$ for some upper triangular matrix $P$
with non negative diagonal entries only(Cholesky decomposition of $A$

  5. (5)

$A$ and $B$ be congruent matrices then $A$ is PSD $\Leftrightarrow B$ is PSD.

  6. (6)

$A$ is PSD if and only if $A=B^{2}$ for some PSD matrix $B$. Then the unique
$B=A^{1/2}$ is called the positive square root of $A$.

  7. (7)

Schur product theorem: Let $A$ and $B$ be PSD matrices of size n. Then the
Schur product $A\circ B$ is also PSD.(but the conventional product neednot be
PSD)

  8. (8)

Let $A$ be Hermitian and PSD. Then there exists a sequence of PSD matrices
$A_{1},A_{2}....$ such that $A_{k}\rightarrow A$ as $k\rightarrow\infty$. We
can define $A_{k}=A+k^{-1}I$

It is easy to observe that the set of all $n\times n$ Hermitian matrices
denoted by $M_{n}^{H}\subset M_{n}(\mathbb{C})$. For $A,B\in
M_{n}^{H},\,A\circ B$ is always Hermitian, ie., $M_{n}^{H}$ is a semigroup and
is a subsemigroup of $M_{n}(\mathbb{C})$.

###  2.4.  Loewner partial order on $M_{n}^{H}$

Every partial order $\preceq$ on a real linear space $S$ can be defined by
$A\preceq B$ for all $A,B\in S$ if their difference lies in a special closed
convex cone. For the Loewner order the elements of the real linear space are
the $n\times n$ Hermitian matrices and elements of the closed convex cone are
the positive semi-definite matrices (PSD).

Also, the collection of all PSD matrices of order $n$ is a semigroup under
’$\circ$’ and is denoted by $M_{n}^{\geq}$. Then

$M_{n}^{\geq}=\\{K\in M_{n}\,\mid\,K=LL^{\ast}\,\text{ for some}\,L\in
M_{n}\\}.$

For $A,B\in M_{n}^{\geq}$, $A\circ B\in M_{n}^{\geq}$ and $A\circ(B\circ
C)=(A\circ B)\circ C$ for all $A,B,C\in M_{n}^{\geq}$ and hence $M_{n}^{\geq}$
is a semigroup. Now we define the Loewner order $\preceq$ on $M_{n}^{H}$ by
$A\preceq B$ if and only if f $B-A$ is Hermitian and positive semidefinite.

######  Lemma 7.

The Loewner partial order $\preceq$ is a compatible partial order on
$M_{n}^{H}$.

###### Proof.

Clearly $0\preceq A$ means that $A$ is PSD. For $A\in M_{n}^{H}$,
$A-A=\textbf{0}\in M_{n}^{\geq}$ implying that $A\preceq A$. Suppose $A\preceq
B$ and $B\preceq A$. Since $A\preceq B\Rightarrow B-A\in M_{n}^{\geq}$, that
is., $B-A=KK^{\ast}$ for some $K\in M_{n}$. Similarly, $B\preceq A\Rightarrow
A-B=LL^{\star}$ for some $L\in M_{n}$. Since, $A-B=-(B-A)$ we have
$LL^{\star}+KK^{\star}=0$ which implies $LL^{\star}=KK^{\star}=0$, ie.,
$A-B=0=B-A\Rightarrow A=B$ proving the antisymmetry.

For transitivity of $\preceq$, consider $A,B\in M_{n}^{\geq}$ such that
$A\preceq B$ and $B\preceq C$, then, $B-A=KK^{\star}$ and $C-B=PP^{\star}$ for
some $L,P\in M_{n}$. Now, $C-A=(C-B)+(B-A)\in M_{n}^{\geq}$, being the sum of
two PSD matrices. Thus $\preceq$ is a partial order.

Let $A\preceq B$ and $C\preceq D$. Then $B-A=KK^{\star}$ and $D-C=PP^{\star}$
for some $L,P\in M_{n}$. Now, $B\circ D-A\circ C=B\circ(D-C)+(B-A)\circ
C=B\circ(LL^{\star})+(KK^{\star})\circ C\in M_{n}\geq$ as the Hadamard product
and sum of two PSD matrices are PSD. This implies that $A\circ C,\preceq
B\circ D$, that is., $\preceq$ is compatible under $\circ$. Thus
$(M_{n}^{H},\circ,\preceq)$ is an ordered matrix semigroup. ∎

Ordered semigroups have many applications in the theory of computer
arithmetics, formal languages, error-correcting codes and the like. This class
of semigroups is currently a hot topic of research and have been studied by
several authors.

## References

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  * [3] Burgess; Raphael R, On Conrad’s relation on semiprime rings and semigroups, Semigroup Forum(1978) 133-140
  * [4] Michael P Drazin, A partial order in completely regular semigroups, Journal of Algebra, 98, 362-374(1986).
  * [5] Jerzy K Baksalary; Sujith Kumar Mitra, Left star and right star partial orderings, Linear Algebra and its applications 149: 73-89, Elsevier Science Publishing Co,Inc, 1991 (1991).
  * [6] Roger A Horn; Chatles R Johnson, Matrix Analysis, Second edition, Cambridge University Press, 1994.