# Portable solvers for batches of small systems applied to the Landau
collision operator

1st Mark F. Adams  Lawrence Berkeley National Laboratory
<EMAIL_ADDRESS>2th Peng Wang  NVIDIA Corporation
<EMAIL_ADDRESS>
###### Abstract

Many small independent sparse linear system solves occur in many applications,
such as the Landau collision operator in plasma physics and astrophysics
simulations, chemistry in combustion applications, and subdomain solves in
domain decomposition solvers. One can simply stack these systems into a global
linear system and use existing general-purpose sparse solvers. However, this
“ensemble” approach does not exploit the independent structure of these
systems and the theoretical optimality of a Krylov solver is lost.

The many independent processing elements (PEs) found in contemporary (GPU)
accelerators are well suited to solving each of these systems independently.
This “batch” approach maintains the Krylov subspace optimality, significantly
reduces the number of kernel launches, and elides (unnecessary) global
communication. This study develops portable solvers that run an entire linear
system solve on a PE in a single kernel launch within the PETSc (Portable
Extensible Toolkit for Scientific Computing) numerical library.

###### Index Terms:

Batch solvers, Landau collision operator, Kokkos, GPU

## Dedicated to the memory of Ravindra Samtaney

##  I Introduction

A solve phase with many independent sparse linear systems arises in several
applications. Single level domain decomposition solvers or multigrid smoothers
[1, 2, 3], and multiphysics models with a tensor product-like structure of a
PDE on a spatial grid with, for example, a chemistry PDE in combustion [4], or
with a velocity space Landau collision operator in plasma physics and
astrophysics [5, 6, 7, 8, 9, 10, 11], all generate many independent system
solves. Sensitivity analyses run independent simulations with solves for
implicit time integrators [12]. In addition to these PDE based solves, batched
solves appear in a pseudo-inverse at each grid point used in conservative
mapping between particle and finite element bases representations in particle-
in-cell methods [13, 14]. The related problem of batched singular value
decompositions has also been developed [15].

Modern accelerator hardware is well suited to these many small systems because
GPUs are composed of many small independent processing elements (PEs) that are
equipped with fast synchronization primitives, by definition. While one can
simply combine these linear systems into a single large linear systems and
solve with existing sparse solvers, this ensemble approach is not optimal in
several respects. For Krylov solvers, the optimality of the Krylov method is
lost for each system and direct solvers need to exploit this independent
structure. This suggests the use of batch solvers that place the entire
solution process for each system on a PE in a single kernel. Batch Krylov
solvers allow for the correct algorithm to be used for each system, with the
proper scaling of the Krylov vectors, and independent convergence checking for
each system. In addition to supporting the correct algorithm, batch solvers
drastically reduce the number of kernel launches in the solver from a dozen or
more per iteration to a single kernel launch. Batch solvers also avoid
unnecessary global communication in dot products and norms and use only the
fast local synchronization primitives on devices. The Ginkgo project [16], and
Kokkos Kernels project [12], are developing similar methods.

Performance portability is currently a significant challenge in high-
performance computing with accelerator devices (GPUs). One viable approach is
to abandon a single source model and implement a version of your solver for
each device of interest. PETSc uses this approach for portable linear algebra
(this report compares the batch solvers developed herein with PETSc ensembles
solvers). Alternatively, one can use a single source model with a portable
language like Kokkos, Raja or SYCL [17, 18, 19]. This work uses Kokkos to
write batch Jacobi preconditioned Krylov solvers [17].

This report begins with a introduction to the Landau collision operator in
§II, and builds on previos work [6], with the following new material:

  * •

a multiple grid capability with batching of multiple problems within each MPI
process for the Landau operator in §III,

  * •

a portable, batch TFQMR111transpose-free quasi-minimum residuals, a Krylov
method for asymmetric matrices solver in §IV,

  * •

and new $2V$ and $3V$ performance data for the Landau operator in §V,

and §VI concludes the report.

##  II Landau collisions for magnetized plasmas

The Vlasov-Maxwell-Landau system of equations is the fundamental model of
magnetized plasmas [20, 21]. It evolves a distribution function for each
species (one electron and potentially many ions species) in phase space with
up to three configuration space dimensions plus three velocity space
dimensions (6D). The Landau operator conserves density, momentum and energy
and admits unstructured finite element discretizations that strictly conserve
these quantities [21, 22].

The evolution of the phase space distribution or density function
$f\left(\vec{x},\vec{v},t\right)$ of a plasma in an electromagnetic field is
effectively modeled with a Vlasov-Maxwell-Landau system of the form

$\begin{split}\frac{df}{dt}&\equiv\frac{\partial f}{\partial
t}+\frac{\partial\vec{x}}{\partial
t}\cdot\nabla_{x}f+\frac{\partial\vec{v}}{\partial t}\cdot\nabla_{v}f\\\
&=\frac{\partial f}{\partial
t}+{\vec{v}}\cdot\nabla_{x}f+\frac{e}{m}\left({\vec{E}}+{\vec{v}}\times{\vec{B}}\right)\cdot\nabla_{v}f=C\end{split}$

with charge $e$, mass $m$, electric field ${\vec{E}}$, magnetic field
${\vec{B}}$, spatial coordinate ${\vec{x}}$ , velocity coordinate $\vec{v}$
and a collision term $C$ . This equation is composed of the symplectic Vlasov-
Maxwell system $\frac{df}{dt}=0$ and a metric, or diffusive, collision
operator $C$, within a metriplectic formalism [23].

Landau collisions between species $\alpha$ and $\beta$, are given by

$C_{\alpha\beta}=\nu_{\alpha\beta}\frac{m_{0}}{m_{\alpha}}\nabla\cdot\int\limits_{\bar{\Omega}}d{\bar{v}}\;\mathbf{U}(\vec{v},{\bar{v}})\cdot\left(\frac{m_{0}}{m_{\alpha}}\bar{f}_{\beta}\nabla
f_{\alpha}-\frac{m_{0}}{m_{\beta}}f_{\alpha}\bar{\nabla}\bar{f}_{\beta}\right)$
(1)

with a collision frequency
$\nu_{\alpha\beta}=e_{\alpha}^{2}e_{\beta}^{2}\ln\Lambda_{\alpha\beta}/8\pi
m_{0}^{2}\varepsilon_{0}^{2}$, the Coulomb logarithm
$\ln\Lambda_{\alpha\beta}$ (=10 herein), an arbitrary reference mass $m_{0}$ ,
the vacuum permittivity $\varepsilon_{0}$ and the effective charges $e$ of
each species. $\mathbf{U}(\vec{v},{\bar{v}})$ is the Landau tensor. Overbar
terms are evaluated on the grid for the domain $\bar{\Omega}$ of species
$\beta$ and $\bar{v}\equiv\vec{\bar{v}}$ for clarity. And in the evolution of
$f_{\alpha}$, $C_{\alpha}=\sum_{\beta}C_{\alpha\beta}$. See — for the further
details, the weak form and a kernel algorithm [24].

The Landau integral is inherently three dimensional, but in a strong magnetic
guide field, a gyrokinetic approximation allows for the use of cylindrical
coordinates, $\vec{v}=\left(r,z\right)$, to reduce the computation to a $2V$
grid [5]. Both the $2D$ and full $3V$ models are investigated in this report.
The $3V$ model is required for extension to relativistic regimes [25, 26, 27],
which is the subject of future work.

The salient feature of (1) is the inner integral over the domain
${\bar{\Omega}}$ for each species $\beta$, which results in an
$\mathcal{O}(N^{2})$ work complexity algorithm, where $N$ is the number of
integration points for a finite element formulation (§III) [24]. The two terms
in (1), a divergence and Laplacian, have rank one vector and rank two tensor,
respectively, “material” coefficients. These coefficients are computed in the
inner integral.

##  III Multiple Grids and Batching

A critical observation in (1) is that the inner integral over species $\beta$
does not include $f_{\alpha}(\bar{v})$, which naturally allows for a separate
grid for each species as is done in the XGC code [5]. A single grid with
adaptive mesh refinement (AMR) can be used [6], but separate grid simplify
meshing because only a single Maxwellian need be resolved well, for the near-
Maxwellian distributions in common plasmas. Because Maxwellian are smooth,
high-order methods are very effective. Additionally, species with similar
thermal velocities can share a grid as is done with all species in [6]. This
is common with many ionization states of impurities in some plasma
simulations. The use of multiple grids with multiple species per grid reduced
the cost of an impurity simulation from quadratic in the number of species to
linear. Given that the Jacobian constriction is an $\mathcal{O}(N^{2})$ where
$N$ is the sum of all the integration points on all grids, this capability is
critical.

Figure 1 shows the three grids used for the experiments in §V, with Maxwellian
distributions in with axisymmetric coordinates, where the third grid has eight
species of heavy ions.

Figure 1: Grids used for this study with Maxwellian distributions and
different scaling for each species group (visualization artifacts from linear
interpolation in Visit)

###  III-A Batching of spatial points

Kinetic applications commonly use operator split time integrators, where the
simplectic Vlasov system and the metric collisions are alternately advanced.
Each configuration space point advances the collision operator – independently
– which provides significant task parallelism. An application would run
thousands or more of these vertex solves in a collision advance step on each
GPU. Additionally, while the exact Jacobian of the Landau integral is dense a
simple approximate Jacobian is not only sparse but the species decouple
resulting in block diagonal system for each problem [6]. Thus, the linear
solves are composed of many independent solves from both problem and species
batching that can be exploited for increased parallelism.

##  IV Batched linear solvers

Batch linear solvers, as with linear solvers in general, can be usefully
categorized as direct and iterative. Direct methods for the most part use some
type of sparse factorization and iterative methods combine the vectors in a
Krylov subspace to generate an approximate solution that usually possesses
some optimality condition such as minimizing the residual [28]. Direct solvers
are attractive because they are robust and their poor asymptotic complexity is
not an issue with the small systems in batch solves. However, direct solvers
are inhibited by data dependencies in both matrix factorizations and solves.
Experience with a single kernel launch CUDA batch band LU solver for the $2V$
Landau examples in §III has been disappointing (§III.G and the data archive in
[29]). However, this effort used a band solver. The use of sparse solvers with
more parallelism, such as nested dissection, should perform significantly
better. Additionally, direct solvers could be critical if iterative solvers
fail. The kernels of iterative methods, with simple Jacobi preconditioning,
have minimal data dependencies and Krylov methods converge well for the Landau
collision operator problems considered here and for combustion problems in
Pele [16].

An approach to solving these systems that uses existing portable solvers in,
for example, PETSc or Trilinos Kokkos Kernels [30], is to create an ensemble
matrix, where each linear system is “stacked” to create a large block diagonal
matrix, and use a Jacobi preconditiond Krylov solver [12, 6]. However, the
ensemble approach has several limitations. Traditional implementations of
solvers abstract linear algebra operations for, among other things,
portability. On a device, this results in a kernel launch for each vector
operation, matrix-vector product, panel updates, preconditioner, etc.,
amounting to hundreds or thousands of such kernel launches. Amortizing these
kernel launch costs requires a large degree of parallelism from batching,
which may not be available from the application. Importantly, the ensemble
approach is not consistent with Krylov iterative solvers because the scaling
of each vector in the Krylov subspace is derived from the (artificial) global
operator. Direct solvers do not suffer from inconsistency, but similarly
require special techniques for batching [sherry].

The cost of a Landau collision time advance is dominated by the construction
of the Jacobian matrix and the linear solve for the implicit time integrator.
The matrix constriction is described with CPU linear solvers in [6] and §V
discusses new optimizations to the meshing batching of multiple spatial
problems in each kernel. This report introduces new linear GPU linear solvers
that also batch multiple spatial problems and place each solve on a processing
element using the Kokkos portable language.

##  V Performance experiments

The important figure of merit to understand performance of the Landau
collision operator time advance is the throughput of Newton iterations per
second. This metric factors out the specifics of the time integrator and non-
linear solver tolerance, etc., which is application dependent. Throughput is
defined as the total number of, for example, Newton iterations times the batch
size and number of GPUs, divided by the simulation time. A “warm-up” time step
is used to setup the solver and is not timed because setup costs are amortized
in a production setting.

The model problem is a deuterium plasma with two single species grids and
eight species of tungsten that share one grid. The initial electron
temperature is twice that of the ions and the model is run toward equilibrium
for 10 time steps. One level of AMR refinement from a $4\times 2$ and a
$4\times 4\times 4$ grid is used, in $2$ and $3V$ respectively, resulting in
14 (Q3 elements) and 120 (Q2 elements) in $2V$ and $3V$, respectively. We have
observed that these grids are sufficient to measure a plasma resistivity
within about $1\%$ of the fully converged resistivity [6]. The test harness
(ex2.c Landau example in PETSc, Appendix A)) replicates the model problem to
create a batch of problems to mimic an application’s use of this solver. Each
of these problems requires a linear solver solve per species, resulting in a
composition of this batch of problems with a batch of species solves per
problem. To mimic variability in a real application, the density and hence
collision frequency of each successive problem in a batch are varied within a
range of about $10\%$. Precise build parameters and instructions for
reproducing the data herein are publicly available (see Appendix A).

Two linear solvers are considered: a batched TFQMR solver in PETSc, written in
Kokkos, and an ensemble solver that uses Kokkos Kernels linear algebra
primitives within the PETSc framework. Jacobi preconditioning use throughout.

###  V-A NVIDIA A100 tests

Figure 2: Nsight Systems view of a typical Newton iteration with: CUDA device
“97.0% Kernels” (dark blue, middle row) with three large kernels (left to
right): the Jacobian construction, mass matrix construction and linear solve
(“batch-kokkos-solve”). The Jacobian is proceeded by a kernel that builds the
function values and derivatives at the integration points and each matrix
method is followed by COO matrix assembly kernel.

One node with four NVIDIA A100 Tensor Core GPUs based on the NVIDIA Ampere GPU
architecture and 256GB of memory (Perlmutter) is used to investigate
performance. Table 3 shows the $2V$ Newton iteration throughput as a function
of batch size and solver.

Figure 3: Newton iterations / sec as function of batch size for each solver:
$2V$ (left), $3V$ (right)

This data shows that the batched TFQMR solver is the fastest option with
60,000 and 550 Newton iterations per second in $2V$ and $3V$, respectively.
Note that the GPU is pretty well saturated with a batch size of 256 in $2V$
and 32 in $3V$, and this corresponds to about 57,00 and 104,000 integration
points per GPU, and thus $2V$ and $3V$ saturate at roughly the same number of
integration points total as is expected.

Tables I and II show the total component times in $2V$ and $3V$, respectively,
including mass matrix creation (“Mass”), Landau Jacobian (“Jacobian”), linear
solver (“Solve”), the total time and the total number of linear solver
iterations.

TABLE I: 2V Component times (batch size = 256), NVIDIA-A100 Component | Jacobian | Mass | Solve | Total | Krylov its
---|---|---|---|---|---
Batch TFQMR | 1.57 | 0.22 | 0.58 | 2.44 | 3,648
Ensemble TFQMR | 1.57 | 0.22 | 1.76 | 3.69 | 4,015
TABLE II: 3V Component times (batch size = 32), NVIDIA-A100 Component | Jacobian | Mass | Solve | Total | Krylov its
---|---|---|---|---|---
Batch TFQMR | 29.69 | 3.00 | 2.33 | 35.08 | 2,785
Ensemble TFQMR | 29.67 | 3.00 | 1.51 | 34.31 | 2,326

In $2V$ the solves, are subdominant and in $3V$ the time is completely
dominated by the Jacobian creation, which is expected because this the
$\mathcal{O}(N^{2})$ work complexity algorithm. Note, the mass matrix creation
is essentially only finite element assembly and sparse matrix assembly and
thus the Jacobian time minus the Mass time indicates the cost of the Landau
kernel.

####  V-A1 NVIDIA hardware utilization

To understand the hardware utilization in this data, begin with a high level
view from NVIDIA’s Nsight Systems. Figure 2 shows a plot of a time slice with
a Newton iteration with the kernels (brown, bottom row) and instrumented
sections in grey for the Jacobian and Mass matrix creation, the “kokkos-batch”
solvers and the sparse matrix assembly. This shows qualitatively that most of
the time is spent in the GPU and a quantitative analysis of hardware
utilization shows that $97\%$ of the time is in fact on the GPU. All raw data
and instruction for reproducibility can be found in Appendix A.

The analysis of the hardware utilization in the GPU kernel is divided into the
analysis of the Jacobian matrix and the mass matrix construction, and the
batch solver. The NVIDIA Nsight Compute tool is used to gather several
hardware metrics from the largest batch size in Tables I and II. Table III
presents some of the raw Nsight Compute data.

TABLE III: Nsight Compute data: Jacobian (Jac), Mass (M), Solver (Sol) Data | Jac-2V | M-2V | Sol-2V | Jac-3V | M-3V | Sol-3V
---|---|---|---|---|---|---
DRAM (GB/s) | 75.80 | 1230 | 28.18 | 38.33 | 946 | 538
L1 (TB/s) | 1.92 | 3.58 | 1.43 | 1.92 | 2.39 | 1.59
L2 (GB/s) | 747 | 4010 | 881 | 266 | 2810 | 1870
dadd/cycle | 163 | 155 | 156 | 76.20 | 91.80 | 35.12
dfma/cycle | 1155 | 0 | 168 | 546 | 0 | 36.11
dmul/cycle | 526 | 329 | 64.50 | 305 | 198 | 3.14
TFlop/sec | 4.23 | 0.68 | 0.89 | 2.06 | 0.41 | 0.16
AI-L1 | 2.20 | 0.19 | 0.50 | 1.07 | 0.17 | 0.10
Roofline-L1 % | 43.60 | 18.27 | 9.18 | 21.27 | 12.19 | 8.11
AI-L2 | 5.66 | 0.17 | 0.72 | 7.75 | 0.15 | 0.08
Roofline-L2 % | 43.60 | 54.20 | 16.70 | 21.27 | 38 | 25.30
AI-DRAM | 55.80 | 0.56 | 23.60 | 53.80 | 0.43 | 0.29
R.F.-DRAM % | 43.60 | 63.60 | 9.18 | 21.30 | 48.90 | 27.80

Before a high level analysis of this data there are a few instructive points
to be seen in this data.

  * •

The Jacobian kernel, with a high arithmetic intensity (AI) of $55.8$ with
respect to DRAM memory movement in $2V$, is not a simple loop of fused
multiply add (FMA) instructions as can be seen from lines 4-6 with about
$62\%$ of the flops in FMA instructions. This limits the achievable percent of
theoretical peak for this algorithm,

  * •

The flop rate (line 7) is about $2x$ higher in $2V$ than $3V$. This is at
least partially due to the Landau kernel $\mathbf{U}$ in (1) being more
complex with a higher AI in $2V$, but this requires futher investigation.

  * •

There are few flops and no FMAs in the mass matrix as this is essentially all
assembly.

  * •

The solver AI-DRAM is very high in $2V$ (23.6) and low in $3V$ (0.29). The
theoretical AI of the solver (no cache) is about $\frac{1}{6}$. This data
indicates that the solves are fitting in cache well in $2V$ but not at all in
$3V$.

Tables IV and V tabulate conclusions and notes from the Nsight Compute data.

TABLE IV: Nsight Compute Bottlenecks Jacobian-2V | Mass-2V | Solve-2V | Jacobian-3V | Mass-3V | Solve-3V
---|---|---|---|---|---
FP64 pipe (57%) | L2 (70%), | L1 and instruction latency bound: | FP64 pipe (31%), | L2 (50%), | L2 (28%),
| DRAM (64%) | L1 (43%) instruction issue (39%) | L1 (24%) | DRAM (49%) | DRAM (23%)
TABLE V: Nsight Compute Notes Jacobian-2V | Mass-2V | Jacobian-3V | Mass-3V | Solve-3V
---|---|---|---|---
Roofline lower than | low roofline peak b/c 1) low pipe | Low pipe utilization | Utilization not higher | Memory
FP64 pipe utilization | utilization due to being L1 latency | due to L1 latency | partly due to load imbalance: | latency
b/c DFMA instruction is | bound. 2) instruction dominated by | bound | Theoretical occupancy 44%, | bound
62% of all FP64 instructions | branch and integers. FP64 instructions |  | achieved occupancy 34% |
|  $\approx 10$% of total instructions |  |  |

###  V-B AMD MI250X tests

One node with four AMD MI250X, each with 2 Graphics Compute Dies (GCDs) for a
total of 8 GCDs per node (Crusher) is used to investigate performance. Table 4
shows the $2V$ Newton iteration throughput as a function of batch size and
solver.

Figure 4: Newton iterations / sec as function of batch size for each solver:
$2V$ (left), $3V$ (right)

This data shows that the batched TFQMR solver is the fastest option with
24,000 and 397 Newton iterations per second in $2V$ and $3V$, respectively.
Note that the GPU is pretty well saturated with a batch size of 128 in $2V$
and 16 in $3V$, and this corresponds to about 29,00 and 52,000 integration
points per grid, and thus $2V$ and $3V$ saturate at about the same number of
integration points total as is expected.

Tables VI and VII show the total component times in $2V$ and $3V$,
respectively, including mass matrix creation (“Mass”), Landau Jacobian
(“Jacobian”), linear solver (“Solve”), the total time and the total number of
linear solver iterations. Note TFQMR has two matrix-vector products per
iteration and so its work complexity is twice that of GMRES with respect to
the number of iterations. Again, batch TFQMR is the fastest.

TABLE VI: 2V Component times (batch size = 512), AMD-MI250X-GCD Component | Jacobian | Mass | Solve | Total | Krylov its
---|---|---|---|---|---
Batch TFQMR | 17.04 | 1.04 | 1.76 | 19.88 | 3,673
Ensemble TFQMR | 17.07 | 1.04 | 36.24 | 54.47 | 4,004
TABLE VII: 3V Component times (batch size = 64), AMD-MI250X-GCD Component | Jacobian | Mass | Solve | Total | Krylov its
---|---|---|---|---|---
Batch TFQMR | 168.16 | 18.07 | 11.28 | 196.81 | 2,796
Ensemble TFQMR | 168.62 | 18.06 | 39.51 | 210.84 | 2,326

##  VI Conclusion

This report demonstrates that with the effective utilization of GPUs the gold
standard for collisions in plasma simulations, the Landau collision operator,
can be used effectively with an axisymmetric ($2V$) approximation and that
full $3V$ may be feasible in the future. This Landau solver supports multiple
independent grids to efficiently resolve the domain of each species group,
with multiple species per grid for species with like velocity profiles to
reduce cost, high-order accurate finite element discretizations with adaptive
mesh refinement, and runs fully and effectively on GPUs with a portable Kokkos
implementation in PETSc. A new PETSc batch solver has been introduced and
experiments have been conducted on an NVIDIA A100 node and AMD MI250X node.
Artifacts and reproducibility instructions are publicly available (see
Appendix A).

Future work includes the development of the solver in full $3V$. This includes
the development of a single $3V$ finite element with quadrature that is
optimized to represent a Gaussian, the equilibrium Maxwellian distribution of
a plasma. Beentjes shows that, for instance, 320 integration points can
represent a Gaussian to an accuracy that we have observed is sufficient in our
verification tests [31]. For near-Maxwellian plasmas this would bring the cost
of a full $3V$ Landau time advance to less than a factor of $5x$ more
expensive than the current $2V$ solve.

## Acknowledgments

This material is based upon work supported by the U.S. Department of Energy,
Office of Science, Office of Advanced Scientific Computing Research and Office
of Fusion Energy Sciences, Scientific Discovery through Advanced Computing
(SciDAC) program.

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##  Appendix A Artifact Description and reproducibility

PETSc output files with performance data and provenance information, build
instructions for each platform and reproducibility instructions and
verification data can be found with git clone
gitlab.com/markadams4/batch_paper_data.


#  $5$-rank of ambiguous class groups of
quintic Kummer extensions

Fouad ELMOUHIB  Mohamed TALBI  Abdelmalek AZIZI

###### Abstract

Let $k\,=\,\mathbb{Q}(\sqrt[5]{n},\zeta_{5})$, where $n$ is a positive integer
$5^{th}$ power-free, whose $5-$class group denoted $C_{k,5}$ is isomorphic to
$\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$. Let
$k_{0}\,=\,\mathbb{Q}(\zeta_{5})$ be the cyclotomic field containing a
primitive $5^{th}$ root of unity $\zeta_{5}$. Let $C_{k,5}^{(\sigma)}$ the
group of the ambiguous classes under the action of $Gal(k/k_{0})$ =
$\langle\sigma\rangle$. The aim of this paper is to determine all integers $n$
such that the group of ambiguous classes $C_{k,5}^{(\sigma)}$ has rank $1$ or
$2$.

##  1 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 [References], [References],
[References]), 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
[References] 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
[References] 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 [References], C.Parry gives a formula between class
numbers of pure quintic field $\mathbb{Q}(\sqrt[5]{n})$ and its normal closure
$k$. In References 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 [References].
In fact, we shall prove the following Main Theorem:

######  Theorem 1.1.

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:

$n\,=\,\left\\{\begin{array}[]{ll}5^{e}q_{1}^{2}q_{2}\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad q_{1}\,\text{ or }\,q_{2}$
$\not\equiv\,\pm 7\,(\mathrm{mod}\,25)\\\ 5^{e}p\not\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad
p\,\not\equiv\,-1\,(\mathrm{mod}\,25)\\\ 5^{e}q_{1}\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad q_{1}\,\equiv\,\pm
7\,(\mathrm{mod}\,25)\\\ p^{e}q_{1}\,\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad
p\,\not\equiv\,-1\,(\mathrm{mod}\,25),\,q_{1}\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)\\\ p^{e}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,25)&\text{
with }\quad p\,\equiv\,-1\,(\mathrm{mod}\,25)\\\
q_{1}^{e_{1}}q_{2}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)&\text{ with }\quad
q_{i}\,\equiv\,\pm 7\,(\mathrm{mod}\,25)\par\end{array}\right.$ (1)

where $p\,\equiv\,-1\,(\mathrm{mod}\,5)$ and $q_{1},q_{2}\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$ 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:

$n\,=\,\left\\{\begin{array}[]{ll}5^{e}l\not\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\quad\text{ with }\quad
l\,\not\equiv\,1\,(\mathrm{mod}\,25),\\\ l^{e_{1}}q_{1}\,\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\quad\text{ with }\quad
l\,\equiv\,1\,(\mathrm{mod}\,5),\,q_{1}\,\equiv\,\pm 2,\pm 7,\pm
3\,(\mathrm{mod}\,25)\\\ l^{e_{1}}\,\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\quad\text{ with }\quad
l\,\equiv\,1\,(\mathrm{mod}\,25),\\\ \end{array}\right.$ (2)

where $l\,\equiv\,1\,(\mathrm{mod}\,5)$ and $q_{1}\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$ are primes and $e,e_{1}$ are integers in
$\\{1,2,3,4\\}$.

This result will be underpinned by numerical examples obtained with the
computational number theory system PARI/GP [References] in section 3.

Notations.

Throughout this paper, we use the following notations:

  * •

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

  * •

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

  * •

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

  * •

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

  * •

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

  * •

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

  * •

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

  * •

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

  * •

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

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

  * •

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

  * •

For a number field $L$, denote by:

    * –

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

    * –

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

    * –

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

    * –

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

$\mathbf{k}$$\mathbf{\Gamma}$ $\mathbf{\Gamma^{\prime}}$
$\mathbf{\Gamma^{\prime\prime}}$ $\mathbf{\Gamma^{\prime\prime\prime}}$
$\mathbf{\Gamma^{\prime\prime\prime\prime}}$ $\mathbf{k_{0}}$$\mathbb{Q}$
Figure 15445

##  2 Proof of Main Theorem

######  Theorem 2.1.

(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
$p\,\equiv\,1\,(\mathrm{mod}\,m)$

###### Proof.

see [References] page 14. ∎

######  Corollary 2.1.

Let $p$ a prime integer, we have :

If $p\,=\,5$, then $\lambda\,=\,1-\zeta_{5}$ is the unique prime over 5 in
$\mathbb{Q}(\zeta_{5})$.

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 $\mathbb{Q}(\zeta_{5})$

If $q\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$, then $q$ is inert in
$\mathbb{Q}(\zeta_{5})$.

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 $\mathbb{Q}(\zeta_{5})$.

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 $(C_{k,5}^{(\sigma)})\,=\,2$

######  Theorem 2.2.

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 $(k/k_{0})^{*}\,=\,k_{5}^{(1)}$.

###### 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 [References, 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
$n\,=\,5^{e}q_{1}^{f_{1}}...q_{r}^{f_{r}}p_{1}^{g_{1}}....p_{s}^{g_{s}}$ with
$q_{i}\,\equiv\,\pm 2\,(mod\,5)$ 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 2.1 each $q_{i}$ is inert in
$k_{0}$, and $q_{i}$ is ramified in $\Gamma\,=\,\mathbb{Q}(\sqrt[5]{n})$, also
by Corollary 2.1 $p_{j}$ 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 [References, 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 [References,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 $n$. ∎

###  2.1 Proof of Theoreme 1.1

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 [References,section 5.1], the rank of $C_{k,5}^{(\sigma)}$ is
given as follows:

rank $(C_{k,5}^{(\sigma)})\,=\,d-3+q^{*}$

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 $k^{*}\,=\,k\setminus\\{0\\}$ as
follows:

$q^{*}$ = $\begin{cases}2&\text{if }\,\zeta,1+\zeta\in N_{k/k_{0}}(k^{*}),\\\
1&\text{if }\,\zeta^{i}(1+\zeta)^{j}\in N_{k/k_{0}}(k^{*})\,\text{ for some i
and j },\\\ 0&\text{if }\,\zeta^{i}(1+\zeta)^{j}\notin
N_{k/k_{0}}(k^{*})\,\text{for}\hskip 5.69054pt0\leq i,j\leq 4\text{ and}\hskip
5.69054pti+j\neq 0.\\\ \end{cases}$

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 [References, 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 [References, 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 $k^{*}\,=\,k\setminus\\{0\\}$ if and only
if $p^{f}\,\,\equiv\,\,1\,(\mathrm{mod},\,25)$ for all prime $p\,\neq\,5$
dividing $n$:

  1. $-$

If $l\,\,\equiv\,\,1\,(\mathrm{mod},\,5)$, by corollary 2.1
$l\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$, 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 $l\,\,\equiv\,\,1\,(\mathrm{mod}\,25)$.

  2. $-$

If $q\,\,\equiv\,\,\pm 2\,(\mathrm{mod},\,5)$ 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 $q\,\equiv\,\pm 7\,(\mathrm{mod}\,25)$.

  3. $-$

If $p\,\,\equiv\,\,-1\,(\mathrm{mod}\,5)$,by corollary 2.1
$p\,=\,\pi_{1}\pi_{2}$ 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 $p\,\,\equiv\,\,-1\,(\mathrm{mod}\,25)$.

(1) If rank $(C_{k,5}^{(\sigma)})\,=\,1$, we get that $d+q^{*}\,=\,4$, so
there are three possible cases as follows:

  * •

Case 1: $q^{*}=0\,\,\mathrm{and}\,\,d=4$,

  * •

Case 2: $q^{*}=1\,\,\mathrm{and}\,\,d=3$,

  * •

Case 3: $q^{*}=2\,\,\mathrm{and}\,\,d=2$,

We will successively treat the three cases to prove the first point of the
main theorem.

  * •

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 2.2, 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 2.1,
$l\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ where $\pi_{i}$ are primes of $k_{0}$.
The prime $l$ is ramified in $\Gamma$, because
disk$(\Gamma/\mathbb{Q})\,=\,5^{5}n^{4}$ 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 [References, Lemma 5.1] we have
$(\lambda)$ ramifies in $k/k_{0}\,\Longleftrightarrow\,n\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$, with $\lambda\,=\,1-\zeta_{5}$, so in the case
$n\,=\,l^{e_{1}}$ where $l\,\equiv\,1\,(\mathrm{mod}\,5)$ we should have
$n\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, so the only $l$ verifiy
$n\,=\,l^{e_{1}}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ 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 $n$:

    1. $(i)$

$n\,=\,5^{e}p^{e_{1}}q_{1}^{e_{2}}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
with $p\,\equiv\,-1\,(\mathrm{mod}\,5),\,\,q_{1}\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$ and $e,e_{1},e_{2}\in\\{1,2,3,4\\}$. By Corollary 2.1,
$p\,=\,\pi_{1}\pi_{2}$, 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 $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, then
$\lambda\,=\,1-\zeta_{5}$ is ramified in $k$, so we get $d=4$. To verifiy that
$n\,=\,5^{e}p^{e_{1}}q_{1}^{e_{2}}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
we can choose $e_{1}\,=\,2$ and $e_{2}\,=\,1$ because
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$,
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 $p^{2}q_{1}\,=\,5\alpha+2$ or
$p^{2}q_{1}\,=\,5\alpha^{\prime}+3$ with
$\alpha,\alpha^{\prime}\in\mathbb{Z}$, so
$5p^{2}q_{1}\,\equiv\,10\,(\mathrm{mod}\,25)$ or
$5p^{2}q_{1}\,\equiv\,15\,(\mathrm{mod}\,25)$, therefore we conclude that
$n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$. According to [References,
Theorem 5.18], if $p\,\equiv\,-1\,(\mathrm{mod}\,25)$ and $q_{1}\,\equiv\,\pm
7\,(\mathrm{mod}\,25)$, 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 [References,
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.

    2. $(ii)$

$n\,=\,p^{e}q_{1}^{e_{1}}q_{2}^{e_{2}}\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ 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
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
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 2.1 $p\,=\,\pi_{1}\pi_{2}$ 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 $n\,=\,p^{e}q_{1}^{2}q_{2}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
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 [References, 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.

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

  * •

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 [References, proposition 5.2] $d\,=\,g$ or $g+1$ according to whether
$n\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ or $n\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$, 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 $n$:

    1. $(i)$

$n\,=\,5^{e}p^{e_{1}}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\equiv\,-1\,(\mathrm{mod}\,5)$ and $e,e_{1}\in\\{1,2,3,4\\}$. As
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
we can choose $e_{1}\,=\,1$ i.e $n\,=\,5^{e}p$ with $e\in\\{1,2,3,4\\}$. By
Corollary 2.1 $p\,=\,\pi_{1}\pi_{2}$ 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 $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$,
according to [References, Lemma 5.1], $\lambda\,=\,1-\zeta_{5}$ is ramified in
$k$ so we obtain $d=3$. The condition $n\,\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ is verified for all $p\,\equiv\,-1\,(\mathrm{mod}\,5)$
because, if $e\,=\,2,3,4$ we have
$n\,=\,5^{e}p\,\equiv\,0\,(\mathrm{mod}\,25)$, 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
[References, 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 $n\,=\,5^{e}p\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ with $p\,\not\equiv\,-1(\mathrm{mod}\,25)$.
We note that the computational number theory system PARI [References], show
that if $n\,=\,5^{e}p\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\not\equiv\,-1(\mathrm{mod}\,25)$, the field $k$ is not always of type
$(5,5)$.

    2. $(ii)$

$n\,=\,5^{e}q_{1}^{e_{1}}q_{2}^{e_{2}}\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ 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
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$,
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 2.1, $q_{1}$ 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 $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$,
then $\lambda\,=\,1-\zeta_{5}$ is ramified in $k$, so we get $d=3$.The
condition $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ 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 $q_{1}^{2}q_{2}\,=\,5\alpha+2$ 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
[References, theorem 5.13] we have $q^{*}\,=\,1$, but for this form of the
radicand $n$ the computational number theory system PARI [References] show
that $C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}$.

    3. $(iii)$

$n\,=\,p^{e}q_{1}^{e_{1}}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\equiv\,-1\,(\mathrm{mod}\,5),\,\,q_{1}\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$
and $e,e_{1}\in\\{1,2,3,4\\}$. As
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
we can choose $e_{1}\,=\,1$ i.e $n\,=\,p^{e}q_{1}$ with $e\in\\{1,2,3,4\\}$.
By Corollary 2.1 $p\,=\,\pi_{1}\pi_{2}$ 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 $n\,=\,p^{e}q_{1}\,\equiv\,\pm
1\pm 7\,(\mathrm{mod}\,25)$ is not verified for all
$p\,\equiv\,-1\,(\mathrm{mod}\,5),\,\,q_{1}\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$, so we combine all the cases of congruence and we obtain
that $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $q_{1}\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$. According to [References, theorem 5.18] If
$p\,\equiv\,-1\,(\mathrm{mod}\,25)$ and $q_{1}\,\equiv\,\pm
7\,(\mathrm{mod}\,25)$ we have rank $C_{k,5}\geq 3$ which is impossible in our
cases. If $p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $q_{1}\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$, according to [References, theorem 5.13] we have
$q^{*}\,=\,1$. Using the computational number theory system PARI/GP
[References], if $n\,=\,p^{e}q_{1}$ with
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$ and $q_{1}\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$, the field $k$ is not always of type $(5,5)$.

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:

$n=\left\\{\begin{array}[]{ll}5^{e}p\not\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\text{ with }p\,\not\equiv\,-1\,(\mathrm{mod}\,25),\\\
p^{e}q_{1}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,25)&\text{ with
}p\,\not\equiv\,-1\,(\mathrm{mod}\,25)\text{ and }q_{1}\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)\\\ \end{array}\right.$ (3)

  * •

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 $n\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ or $n\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, 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 $n$:

    1. $(i)$

$n\,=\,5^{e}q_{1}^{e_{1}}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$q_{1}\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ and $e,e_{1}\in\\{1,2,3,4\\}$. As
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
we can choose $e_{1}\,=\,1$ i.e $n\,=\,5^{e}q_{1}$ with $e\in\\{1,2,3,4\\}$.
Since $q^{*}=2$ so we have $q_{1}\,\,\equiv\,\,\pm 7\,(\mathrm{mod}\,25)$. By
Corollary 2.1 $q_{1}$ is inert in $k_{0}$, we have $q_{1}$ is ramified in
$\Gamma$, so $q_{1}$ is ramified too in $k$, and since $n\,\not\equiv\,\pm
1\pm 7\,(\mathrm{mod}\,25)$, according to [References, Lemma 5.1],
$\lambda\,=\,1-\zeta_{5}$ is ramified in $k$ so we obtain $d=2$. The condition
$n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ is verified for all
$q_{1}\,\,\equiv\,\,\pm 7\,(\mathrm{mod}\,5)$, because if $e\,=\,2,3,4$ we
have $n\,=\,5^{e}q_{1}\,\equiv\,0\,(\mathrm{mod}\,25)$, if $e=1$ i.e
$n\,=\,5q_{1}$ we have $n\,=\,5q_{1}\,\equiv\,\pm 10\,(\mathrm{mod}\,25)$, we
conclude that $n\,=\,5^{e}q_{1}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
with $q_{1}\,\equiv\,\pm 7\,(\mathrm{mod}\,25)$, but for this form of the
radicand $n$ the computational number theory system PARI/GP [References] show
that $C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}$.

    2. $(ii)$

$n\,=\,q_{1}^{e_{1}}q_{2}^{e_{2}}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
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
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
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 2.1 $q_{1}$ 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 $n\,=\,q_{1}^{e_{1}}q_{2}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
is verified, because we have $n\,=\,q_{1}^{e_{1}}q_{2}\,\equiv\,\pm
7\,(\mathrm{mod}\,25)$ for all $e_{1}$, so we conclude that
$n\,=\,q_{1}^{e_{1}}q_{2}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ 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
PARI/GP [References] show that $C_{k,5}\,\simeq\,\mathbb{Z}/5\mathbb{Z}$

    3. $(iii)$

$n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ 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 2.1,
$p\,=\,\pi_{1}\pi_{2}$ 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 $n\,=\,p^{e}\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ is verified for all
$p\,\,\equiv\,\,-1\,(\mathrm{mod}\,25)$, we conclude that
$n\,=\,p^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\equiv\,-1\,(\mathrm{mod}\,25)$. Using the computational number theory
system PARI/GP [References], if $n\,=\,p^{e}\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ with $p\,\equiv\,-1\,(\mathrm{mod}\,25)$, the field $k$
is not always of type $(5,5)$.

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:

$n\,=\,p^{e}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,25)\text{ with
}p\,\equiv\,-1\,(\mathrm{mod}\,25)$

(2) If rank $(C_{k,5}^{(\sigma)})\,=\,2$, so $C_{k,5}\,=\,C_{k,5}^{(\sigma)}$.
According to [References,section 5.1], the rank of $C_{k,5}^{(\sigma)}$ is
given as follows:

rank $(C_{k,5}^{(\sigma)})\,=\,d-3+q^{*}$

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:

  * •

Case 1: $q^{*}=0\,\,\mathrm{and}\,\,d=5$,

  * •

Case 2: $q^{*}=1\,\,\mathrm{and}\,\,d=4$,

  * •

Case 3: $q^{*}=2\,\,\mathrm{and}\,\,d=3$,

We will treat the three cases to prove the forms of the radicand $n$. By
theorem 2.2, $n$ 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 $3$.

  * •

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 $n\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ or $n\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$. 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 $n$:

    1. $(i)$

$n\,=\,5^{e}l^{e_{1}}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\equiv\,1\,(\mathrm{mod}\,5)$ and $e,e_{1}\in\\{1,2,3,4\\}$. As
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
we can choose $e_{1}\,=\,1$ i.e $n\,=\,5^{e}l$ with $e\in\\{1,2,3,4\\}$. By
Corollary 2.1 $l\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ 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
$n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$, according to [References,
Lemma 5.1], $\lambda\,=\,1-\zeta_{5}$ is ramified in $k$ so we obtain $d=5$.
The condition $n\,\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ is verified for
all $l\,\equiv\,1\,(\mathrm{mod}\,5)$, because if $e\,=\,2,3,4$ we have
$n\,=\,5^{e}l\,\equiv\,0\,(\mathrm{mod}\,25)$, 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
$n\,=\,5^{e}l\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\not\equiv\,1(\mathrm{mod}\,25)$. Using the computational number theory
system PARI/GP [References], if $n\,=\,5^{e}l\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)$ with $l\,\not\equiv\,1(\mathrm{mod}\,25)$, the field
$k$ is not always of type $(5,5)$.

    2. $(ii)$

$n\,=\,l^{e}q_{1}^{e_{1}}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\equiv\,1\,(\mathrm{mod}\,5),\,\,q_{1}\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$
and $e,e_{1}\in\\{1,2,3,4\\}$. As
$\mathbb{Q}(\sqrt[5]{ab^{2}c^{3}d^{4}})\,=\,\mathbb{Q}(\sqrt[5]{a^{2}b^{4}cd^{3}})\,=\,\mathbb{Q}(\sqrt[5]{a^{3}bc^{4}d^{2}})\,=\,\mathbb{Q}(\sqrt[5]{a^{4}b^{3}c^{2}d})$
we can choose $e_{1}\,=\,1$ i.e $n\,=\,l^{e}q_{1}$ with $e\in\\{1,2,3,4\\}$.
By Corollary 2.1 $l\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ 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
$n\,=\,l^{e}q_{1}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ is not verified for
all $l\,\equiv\,1\,(\mathrm{mod}\,5),\,\,q_{1}\,\equiv\,\pm
2\,(\mathrm{mod}\,5)$, so we combine all the cases of congruence and we obtain
that $l\,\equiv\,1\,(\mathrm{mod}\,5)$ and $q_{1}\,\equiv\,\pm 2,\pm 3\,\pm
7\,(\mathrm{mod}\,25)$. Using the computational number theory system PARI/GP
[References], if $n\,=\,l^{e}q_{1}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$,
with $l\,\equiv\,1\,(\mathrm{mod}\,5)$, and $q_{1}\,\equiv\,\pm 2,\pm 3\,\pm
7\,(\mathrm{mod}\,25)$, the field $k$ is not always of type $(5,5)$

We summarize all forms of integer $n$ in this case as follows:

$n=\left\\{\begin{array}[]{ll}5^{e}l\not\equiv\,\pm 1,\pm
7\,(\mathrm{mod}\,25)&\text{ with }l\,\not\equiv\,1\,(\mathrm{mod}\,25),\\\
l^{e}q_{1}\,\equiv\,\pm 1,\pm 7\,(\mathrm{mod}\,25)&\text{ with
}l\,\equiv\,1\,(\mathrm{mod}\,5)\,q_{1}\,\equiv\,\pm 2,\pm 3\,\pm
7\,(\mathrm{mod}\,25)\\\ \end{array}\right.$ (4)

  * •

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 2.1
$l$ 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
$n\,=\,l^{e}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\equiv\,1\,(\mathrm{mod}\,5)$ and $e\in\\{1,2,3,4\\}$. The condition
$n\,\equiv\,\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ 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 PARI/GP [References], if
$n\,=\,l^{e}\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\equiv\,1(\mathrm{mod}\,25)$, the field $k$ is not always of type $(5,5)$

  * •

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 2.1
$l\,=\,\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ 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.

##  3 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 PARI/GP [References], we
illustrate our main result Theorem 1.1.

###  3.1 rank $(C_{k,5}^{(\sigma)})\,=\,1$

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

$p$ | $n\,=\,p^{e}$ | $p\,(\mathrm{mod}\,5)$ | $p\,(\mathrm{mod}\,25)$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$
---|---|---|---|---|---|---
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

Table 2: $n\,=\,q_{1}^{e_{1}}q_{2}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
with $q_{i}\,\equiv\,\pm 7\,(\mathrm{mod}\,25)$
$q_{1}$ $q_{1}\,(\mathrm{mod}\,5)$ $q_{1}\,(\mathrm{mod}\,25)$ $q_{2}$
$q_{2}\,(\mathrm{mod}\,5)$ $q_{2}\,(\mathrm{mod}\,25)$
$n\,=\,q_{1}^{e_{1}}q_{2}$ $h_{k,5}$ $C_{k,5}$ rank $(C_{k,5}^{(\sigma)})$ 7 2
7 43 3 -7 2107 = $7^{2}\times 43$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 7 2 7 193 3 -7
1351 = $7\times 193$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 7 2 7 293 3 -7 2051 =
$7\times 293$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 107 2 7 43 3 -7 492307 =
$107^{2}\times 43$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 107 2 7 193 3 -7 20651 =
$107\times 193$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 107 2 7 293 3 -7 31351 =
$107\times 293$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 107 2 7 443 3 -7 47401 =
$107\times 443$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 157 2 7 43 3 -7 6751 = $157\times
43$ 5 $\mathbb{Z}/5\mathbb{Z}$ 1 157 2 7 193 3 -7 30301 = $157\times 193$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 157 2 7 443 3 -7 69551 = $157\times 443$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 257 2 7 193 3 -7 49601 = $257\times 293$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 257 2 7 293 3 -7 75301 = $257\times 293$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 307 2 7 193 3 -7 59251 = $307\times 193$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 457 2 7 43 3 -7 19651 = $457\times 43$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 457 2 7 443 3 -7 202451 = $457\times 443$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 557 2 7 43 3 -7 23251 = $557\times 43$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1 607 2 7 43 3 -7 26101 = $607\times 43$ 5
$\mathbb{Z}/5\mathbb{Z}$ 1

Table 3 : $n\,=\,5^{e}q_{1}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$q_{1}\,\equiv\,\pm 7\,(\mathrm{mod}\,25)$

$q_{1}$ | $n\,=\,5^{e}q_{1}$ | $q_{1}\,(\mathrm{mod}\,5)$ | $q_{1}\,(\mathrm{mod}\,25)$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$
---|---|---|---|---|---|---
7 | 175 = $5^{2}\times 7$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
107 | 535 = $5\times 107$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
157 | 19625 = $5^{3}\times 157$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
257 | 6425 = $5^{2}\times 257$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
307 | 38375 = $5^{3}\times 307$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
457 | 2285 = $5\times 457$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
557 | 2785 = $5\times 557$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
607 | 3053 = $5\times 607$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
757 | 3785 = $5\times 457$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
857 | 4285 = $5\times 457$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
907 | 4535 = $5\times 907$ | 2 | 7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
43 | 1075 = $5^{2}\times 43$ | 3 | -7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
193 | 120625 = $5^{4}\times 193$ | 3 | -7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
293 | 120625 = $5^{4}\times 193$ | 3 | -7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
443 | 11075 = $5^{2}\times 443$ | 3 | -7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
643 | 3215 = $5\times 643$ | 3 | -7 | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1

Table 4: $n\,=\,5^{e}q_{1}^{2}q_{2}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$
with $q_{1}$ or $q_{2}$ $\not\equiv\,\pm 7\,(\mathrm{mod}\,25)$

$q_{1}$ | $q_{1}\,(\mathrm{mod}\,5)$ | $q_{1}\,(\mathrm{mod}\,25)$ | $q_{2}$ | $q_{2}\,(\mathrm{mod}\,5)$ | $q_{2}\,(\mathrm{mod}\,25)$ | $n\,=\,5^{e}q_{1}^{2}q_{2}$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$
---|---|---|---|---|---|---|---|---|---
2 | 2 | 2 | 3 | 3 | 3 | 60 = $5\times 2^{2}\times 3$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
2 | 2 | 2 | 13 | 3 | 13 | 260 = $5\times 2^{2}\times 13$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
2 | 2 | 2 | 53 | 3 | 3 | 1060 = $5\times 2^{2}\times 53$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
2 | 2 | 2 | 23 | 3 | -2 | 460 = $5\times 2^{2}\times 23$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
7 | 2 | 7 | 3 | 3 | 3 | 735 = $5\times 7^{2}\times 3$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
17 | 2 | 17 | 3 | 3 | 3 | 108375 = $5^{3}\times 17^{2}\times 3$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
17 | 2 | 17 | 23 | 3 | -2 | 33235 = $5\times 17^{2}\times 23$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
37 | 2 | 17 | 3 | 3 | 3 | 20535 = $5\times 37^{2}\times 13$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
37 | 2 | 17 | 13 | 3 | 13 | 88985 = $5\times 37^{2}\times 13$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
47 | 2 | -3 | 3 | 3 | 3 | 33135 = $5\times 47^{2}\times 3$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
47 | 2 | -3 | 13 | 3 | 13 | 143585 = $5\times 47^{2}\times 13$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
47 | 2 | -3 | 23 | 3 | -2 | 254035 = $5\times 47^{2}\times 23$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
47 | 2 | -3 | 43 | 3 | -7 | 474935 = $5\times 47^{2}\times 43$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
107 | 2 | 7 | 23 | 3 | -2 | 1316635 = $5\times 2^{2}\times 3$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
67 | 2 | 17 | 3 | 3 | 3 | 67335 = $5\times 67^{2}\times 3$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
67 | 2 | 17 | 53 | 3 | 3 | 1189585 = $5\times 67^{2}\times 53$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1
97 | 2 | -3 | 43 | 3 | -7 | 2022935 = $5\times 97^{2}\times 43$ | 5 | $\mathbb{Z}/5\mathbb{Z}$ | 1

Table 5: $n\,=\,p^{e}q_{1}\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\not\equiv\,-1\,(\mathrm{mod}\,25)$, $q_{1}\,\not\equiv\,\pm
7\,(\mathrm{mod}\,25)$ $p$ $p\,(\mathrm{mod}\,5)$ $p\,(\mathrm{mod}\,25)$
$q_{1}$ $q_{1}\,(\mathrm{mod}\,5)$ $q_{1}\,(\mathrm{mod}\,25)$
$n\,=\,p^{e}q_{1}$ $h_{k,5}$ $C_{k,5}$ rank $(C_{k,5}^{(\sigma)})$ 59 -1 9 2 2
2 118 = $59\times 2$ 25 $(5,5)$ 1 19 -1 19 3 3 3 57 = $19\times 3$ 25 $(5,5)$
1 59 -1 9 23 3 -2 1357 = $59\times 23$ 25 $(5,5)$ 1 359 -1 9 2 2 2 718 =
$359\times 2$ 25 $(5,5)$ 1 409 -1 9 2 2 2 816 = $409\times 2$ 25 $(5,5)$ 1 59
-1 9 127 2 2 7493 = $59\times 127$ 25 $(5,5)$ 1 109 -1 9 23 3 -2 2507 =
$109\times 23$ 25 $(5,5)$ 1 509 -1 9 2 2 2 1018 = $509\times 2$ 25 $(5,5)$ 1
709 -1 9 2 2 2 1418 = $709\times 2$ 25 $(5,5)$ 1 19 -1 19 53 3 3 1007 =
$19\times 53$ 25 $(5,5)$ 1

Table 6: $n\,=\,5^{e}p\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$p\,\not\equiv\,-1(\mathrm{mod}\,25)$

$p$ | $n\,=\,5^{e}p$ | $p\,(\mathrm{mod}\,5)$ | $p\,(\mathrm{mod}\,25)$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$
---|---|---|---|---|---|---
19 | 475 = $5^{2}\times 19$ | -1 | 19 | 25 | (5,5) | 1
29 | 145 = $5\times 29$ | -1 | 4 | 25 | (5,5) | 1
59 | 7375 = $5^{3}\times 59$ | -1 | 9 | 25 | (5,5) | 1
89 | 55625 = $5^{4}\times 89$ | -1 | 14 | 25 | (5,5) | 1
109 | 2725 = $5^{2}\times 109$ | -1 | 9 | 25 | (5,5) | 1
229 | 28625 = $5^{3}\times 229$ | -1 | 4 | 25 | (5,5) | 1
239 | 1195 = $5\times 239$ | -1 | 14 | 25 | (5,5) | 1
269 | 6725 = $5^{2}\times 19$ | -1 | 19 | 25 | (5,5) | 1
379 | 168125 = $5^{4}\times 379$ | -1 | 4 | 25 | (5,5) | 1
389 | 1945 = $5^{2}\times 389$ | -1 | 14 | 25 | (5,5) | 1

###  3.2 rank $(C_{k,5}^{(\sigma)})\,=\,2$

Table 1: $n\,=\,5^{e}l\not\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\not\equiv\,1(\mathrm{mod}\,25)$

$l$ | $n\,=\,5^{e}l$ | $l\,(\mathrm{mod}\,5)$ | $l\,(\mathrm{mod}\,25)$ | $h_{k,5}$ | $C_{k,5}$ | rank $(C_{k,5}^{(\sigma)})$
---|---|---|---|---|---|---
11 | 55 = $5\times 11$ | 1 | 11 | 25 | (5,5) | 2
41 | 5125 = $5^{3}\times 41$ | 1 | -9 | 25 | (5,5) | 2
61 | 5125 = $5^{4}\times 61$ | 1 | 11 | 25 | (5,5) | 2
71 | 1775 = $5^{2}\times 71$ | 1 | -4 | 25 | (5,5) | 2
131 | 655 = $5\times 131$ | 1 | 6 | 25 | (5,5) | 2
181 | 113125 = $5^{4}\times 181$ | 1 | 6 | 25 | (5,5) | 2
241 | 30125 = $5^{3}\times 241$ | 1 | -9 | 25 | (5,5) | 2
311 | 1555 = $5\times 311$ | 1 | 11 | 25 | (5,5) | 2
331 | 8275 = $5^{2}\times 331$ | 1 | 6 | 25 | (5,5) | 2
431 | 2155 = $5\times 431$ | 1 | 6 | 25 | (5,5) | 2

Table 2: $n\,=\,l^{e}q_{1}\,\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\equiv\,1\,(\mathrm{mod}\,5)$ and $q_{1}\,\equiv\,\pm 2,\pm 3\,\pm
7\,(\mathrm{mod}\,25)$ $l$ $l\,(\mathrm{mod}\,5)$ $l\,(\mathrm{mod}\,25)$
$q_{1}$ $q_{1}\,(\mathrm{mod}\,5)$ $q_{1}\,(\mathrm{mod}\,25)$
$n\,=\,l^{e}q_{1}$ $h_{k,5}$ $C_{k,5}$ rank $(C_{k,5}^{(\sigma)})$ 31 1 6 2 2
2 $31\times 2$ 25 $(5,5)$ 2 131 1 6 23 3 -2 $131^{3}\times 23$ 25 $(5,5)$ 2
181 1 6 47 2 -3 $181\times 47$ 25 $(5,5)$ 2 11 1 11 3 3 3 $11\times 3$ 25
$(5,5)$ 2 41 1 16 23 3 -2 $41\times 23$ 25 $(5,5)$ 2 191 1 16 2 2 2 $191\times
2$ 25 $(5,5)$ 2 41 1 16 47 2 -3 $41^{2}\times 47$ 25 $(5,5)$ 2 311 1 11 2 2 2
$311^{4}\times 2$ 25 $(5,5)$ 2

Table 3: $n\,=\,l^{e}\equiv\,\pm 1\pm 7\,(\mathrm{mod}\,25)$ with
$l\,\equiv\,1(\mathrm{mod}\,25)$ $l$ $n\,=\,l^{e}$ $l\,(\mathrm{mod}\,5)$
$l\,(\mathrm{mod}\,25)$ $h_{k,5}$ $C_{k,5}$ rank $(C_{k,5}^{(\sigma)})$ 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

##  4 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 PARI/GP References. 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:

######  Conjecture 4.1.

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 $k$,
$q_{1},q_{2}\,\equiv\,\pm 2\,(\mathrm{mod}\,5)$ are primes and
$e\in\\{1,2,3,4\\}$.
If the radicand $n$ take one form as follows:

$n\,=\,\begin{cases}q_{1}^{e_{1}}q_{2}\,\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad q_{i}\,\equiv\,\pm
7\,(\mathrm{mod}\,25)\\\ 5^{e}q_{1}\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad q_{1}\,\equiv\,\pm
7\,(\mathrm{mod}\,25)\\\ 5^{e}q_{1}^{2}q_{2}\not\equiv\,\pm 1\pm
7\,(\mathrm{mod}\,25)&\text{ with }\quad q_{1}\,\text{ or }\,q_{2}$
$\not\equiv\,\pm 7\,(\mathrm{mod}\,25)\\\ \end{cases}$ (5)

Then $C_{k,5}$ is a cyclic groupe of order $5$.

## References

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

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

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