# Absolute keV X-ray yield and conversion efficiency in over dense Si petawatt
laser plasma

Sergey N. Ryazantsev1,2  Artem S. Martynenko 2,3  Maksim V. Sedov2  Igor Yu.
Skobelev1,2  Mikhail D. Mishchenko2,4  Yaroslav S. Lavrinenko1,5  Christopher
D. Baird6  Nicola Booth7  Phil Durey6  Leonard N. K. Döhl6  Damon Farley6
Kathryn L. Lancaster6  Paul McKenna8  Christopher D. Murphy6  Tatiana. A.
Pikuz2,9  Christopher Spindloe7  Nigel Woolsey6 and Sergey A. Pikuz1,2
1National Research Nuclear University MEPhI (Moscow Engineering Physics
Institute), 115409 Moscow, Russia
2Joint Institute for High Temperatures of the Russian Academy of Sciences,
Moscow 125412, Russia
3Plasma Physics Department, GSI Helmholtzzentrum fur Schwerionenforschung,
64291 Darmstadt, Germany
4European XFEL, 22869 Schenefeld, Germany
5Moscow Institute of Physics and Technology, State University, 141701 Moscow,
Russia
6York Plasma Institute, Department of Physics, University of York, York YO10
5DD, United Kingdom
7Central Laser Facility, STFC Rutherford Appleton Laboratory, Didcot OX11 0QX,
United Kingdom
8Department of Physics, SUPA, University of Strathclyde, Glasgow G4 0NG,
United Kingdom
9Open and Transdisciplinary Research Initiative, Osaka University, Osaka
565-0871, Japan<EMAIL_ADDRESS>
###### Abstract

Laser-produced plasmas are bright, short sources of X-rays often used for
time-resolved imaging and spectroscopy. Absolute measurement requires accurate
knowledge of laser-to-x-ray conversion efficiencies, spectrum, photon yield
and angular distribution. Here we report on soft X-ray emission from a thin Si
foil irradiated by a sub-PW picosecond laser pulse. These absolute
measurements cover a continuous and broad spectral range that extends from
4.75 to 7.5 Å(1.7–2.6 keV). The X-ray spectrum consists of spectral line
transitions from highly charged ions and broadband emission with contributions
from recombination, and free-free processes that occur as electrons decelerate
in plasma electromagnetic fields. These quantitative measurements are compared
to particle-in-cell simulations allowing us to distinguish bremsstrahlung and
synchrotron contributions to the free-free emission. We found that experiment
and simulation estimations of laser-to-bremsstrahlung conversion efficiency
are in a good agreement. This agreement illustrates the accuracy of experiment
and physical interpretation of the measurements.

††journal: osajournal††articletype: Research Article

##  1 Introduction

Laser produced plasmas (LPP) are widely used as an X-rays source for both
fundamental and applied research. This is due to its relatively short duration
of the emission and possibility to precisely synchronise a measurement with
phase of evolution of the probed object or experiment. It is possible to
implement measurement techniques involving X-rays to investigate objects on
temporal scales as short as femtoseconds. Broadband emission sources are often
required, especially for bioimaging and absorption spectroscopy. LPP at the
table-top scale are used as X-ray sources in commercial applications [1] and
at the very large scale in inertial confinement fusion (ICF) research as very
bright X-ray backlighters for capsule explosions. For this purposes petawatt-
class (PW) lasers facilities (such as ARC [2] and PETAL [3] or any other
mentioned in TABLE E.1 in [4]) able to generate picosecond duration pulses of
about kJ are used. Even for ICF experiments, there is a need for bright short-
duration sources of soft X-rays. For example, the backlighting of direct-drive
cryogenic DT implosions [5] uses low energy photons of <2 keV due to low
opacity of the plastic shell and deuterium-tritium fuel [6]. In addition, the
low energy part of PW short-duration plasma sources spectrum is used to study
warm-dense-matter via absorption spectroscopy [7].

These two applications, imaging of an ICF experiment and diagnosing warm-
dense-matter, requires different spectral composition of the probing
radiation. The source with narrow emission band, ideally monochromatic, is
required for recording high-quality backlit images. In turn, absorption
spectroscopy is most effective when using a source of radiation with
continuous spectrum absent of spectral line features or sharp intensity drops
[8]. Both types of the probe radiation can be obtained via LPP based due
different processes illustrated in Fig. 1.

Plasma ions produce characteristic spectral lines during transitions of
electrons between bound energy levels. For the case of highly charged ions of
even relatively low-Z (around 10) the lines lie in keV range of photon energy.
Typically, an element is chosen and stripped to the K-shell so that intense
emission, usually transitions from the excited state with principal quantum
number, n = 2, to the ground state in hydrogen-like (Lyα) and helium-like
transition (Heα) is at the appropriate wavelength for high-contrast quasi-
monochromatic backlighting imaging [6, 9]. The plasma spectrum also contains
lower intensity continuous emission from free-bound (photorecombination) and
free-free (bremsstrahlung and synchrotron emission). The contributions of each
process depend on the element and irradiation conditions. As a result, a
significant amount of experimental work is needed to characterize a of LPP
source, including those produced by picosecond PW pulses. For example, Li et
al., [10] estimated photon yield and conversion efficiency from laser- gas jet
interaction up to 3$\times$1019 W/cm2.

In this paper we investigate the soft X-ray emission properties of near solid
density plasma created in a Si foil by a sub-PW laser pulse. The goal of the
work is not only to present exact values for the X-ray source characteristics
of photon yield in absolute units and angular distribution of the radiation,
but also to describe the physical processes contributing to the emission.

##  2 Experimental setup

Experimental investigation of emissivity in soft X-ray range as a
relativistic-intensity, picosecond duration laser pulse strikes a Si foil was
carried out on Vulcan PW laser facility [11]. A schematic of the experimental
setup is shown in Fig. 1. The plasma was created by irradiation of 2 $\mu$m
thick Si foil by laser pulses with intensity of 3$\times$1020 W/cm2 ($\lambda$
= 1054 nm, $\tau\approx$ 1 ps, on target energy $\approx$200 J, P = 200 TW)
focused on foil surface. The laser beam was focused by an off-axis parabolic
(OAP) mirror to a spot with diameter of $\approx$7 $\mu$m with an angle of
incidence of 45∘. The laser contrast was enhanced by using a plasma mirror.

Figure 1: Experimental setup and scheme of processes forming the shape of the
plasma soft X-ray radiation spectrum registered in the experiments.

X-ray emission from a front side of the target was recorded using three
focusing spectrometers with spatial resolution (FSSR) [12] at angles close to
the target surface normal. A magnet was installed between the target and FSSR
crystals to help prevent fast electrons produced during irradiation reaching
the crystals and causing fluorescence. Fujifilm BAS-TR image plates and an
Andor CCD DX-434 (for only one of the spectrometers) were used as X-ray
detectors. Their sensitive layers were protected and shielded from visible
light by filter foils made of Be, Al and polypropylene of different
thicknesses. Observation ranges of the spectrometers overlap to provide cross-
calibration between the separate FSSRs enabled measurement of continuous and
high-resolution spectra across photon wavelengths from 4.7 to 7.3 Å(1.7 to 2.6
keV). Raw spectra registered by the individual FSSR spectrometers are shown in
Fig. 2(a).

All the data gathered with the FSSRs were corrected for distance from the LPP,
filtering, crystal reflectivity and detector response functions. Numerical
modelling [13], via ray tracing through spectrometer, used the actual
diffraction profiles or crystallographic rocking curves of the spherically
bent crystals installed in the FSSRs. They were calculated with XOP [14]
software. Rocking curves for the crystals used in experiments and image plates
sensitivity functions are presented in the Supplemental Document uploaded with
this Pre-print. Transmission functions calculated using the Henke tables [15]
for filter foils was also taken into account. A summary of the response
functions for each spectrometer is shown in 2(b). After convolution of the
registered spectra with them an initial plasma radiation spectrum was restored
(Fig.2(c)). The most significant corrections were needed around the Si K-edge
at 6.7135 Å[16]. This causes a decrease in reflectivity of the dispersive
$\alpha$-quartz (SiO2) crystals in the wavelength region close to the Heα
line. The detailed Heα spectral line structure and the overlap between the two
FSSRs in this region enables accurate calibration of the spectrometers.

(a)

(b)

(c)

Figure 2: (a)—normalized experimental emission spectra registered by the FSSRs
from the front side of Si foil before convolution with the spectrometer routes
response functions $\chi$($\lambda$) shown in (b). $\chi$($\lambda$) is a
ratio of the number of registered photons of $\lambda$ wavelength to the one
emitted by a source. (c)—combined experimental emission spectrum after
convolution by $\chi$($\lambda$). The blue and purple polygons qualitatively
demonstrate contributions of bremsstrahlung, synchrotron emission and
recombination continuum. Shape of the latter one was calculated for solid-
state Si plasma (Ni = 6$\times$1022 cm-3).

##  3 Discussion

###  3.1 Registered spectrum description

The spectrum shown in 2(c) contains both characteristic and continuous
emission parts. Its shape is formed by all types of electron transitions:
free-free, free-bound, bound-bound (1). Characteristic emission of the plasma
ions is produced by transitions between energy states of H- and He-like Si
ions. All of them are broadened due to strong Stark effect. The observed
widths correspond to the electron density close to Si solid state value
(N${}_{SS}\approx$6$\times$1023 cm-3). It was achieved both by extremely high
value of the laser pulse temporal contrast of 10-10, this contrast was
improved by a plasma mirror installed in a laser beam path. The most intense
lines are the Lyα (2p$\rightarrow$1s transition in an H-like ion). About
2$\times$1014 of photons ($\approx$20% of the total number) were emitted in
the narrow wavelength range from 6.05 to 6.35 Å (1.95–2.05 keV), which
contains the line itself and its dielectronic satellites. This part of the
plasma spectrum is the most suitable for implementation of quasi-monochromatic
backlights schemes.

In the Heα region, from 6.53 to 6.89 Å (1.8–1.9 keV), the number of photons is
smaller by approximately two: $\approx$1014. This spectral region contains a
broad peak formed by overlapped Heα (1s2p 1P${}_{1}\rightarrow$1s2 1S0 in He-
like ion), intercombination line (1s2p 3P${}_{2,1}\rightarrow$1s2 1S0) and
dielectronic satellites produced by Li-like ions. In Lyβ (3p$\rightarrow$1s),
and Heβ (1s3p 1P${}_{1}\rightarrow$1s2 1S0) lines contain roughly
0.5$\times$1014 and 0.7$\times$1014 photons respectively. In comparison the Kα
line at 7.2 Å is weak as most of the Si is highly ionised. The mentioned
numbers of registered photons correspond to following values of conversion
efficiency (CE) of laser energy to X-rays for Heα, Lyα, Lyβ, Heβ lines as
$\approx$1.5$\times$10-4, $\approx$3.2$\times$10-4, $\approx$0.94$\times$10-4,
$\approx$1.2$\times$10-4 respectively.

In the short wavelength region of the spectrum there is a pedestal in
intensity, this results from the recombination continuum produced by free-
bound transitions of plasma electrons. This spectral region contains
unresolved spectral lines corresponding to 4p$\rightarrow$1s transitions in H-
and He-like ions. This is due to ionization potential depression [17] in the
dense plasma. The shape of the recombination continuum calculated using the
Hummer-Mihalas [18] model for He- and H-like ions in a solid-density Si plasma
is shown in Fig. 2(c) by the magenta area. Approximately 3$\times$1014
recombination continuum photons were registered between 4.75 and 5.4 Å.
Simulations indicate that the recombination continuum extends to 2 Å. The
conversion efficiency to this part of the spectrum is considerable and
suggests that the recombination continuum produced by PW solid-density Si
plasma is suitable for X-ray absorption spectroscopy in the range $\leq$5 Å,
where the spectral lines are absent.

Other continuous emission results from free-free processes, this contributes
to the spectrum through a monotonic growth in intensity in the long wavelength
region of the spectrum. This is approximated by an exponential function shown
by the blue line in Fig.2(c).

###  3.2 Modelling of the continuous emission.

There is no an obvious way to distinguish contributions of bremsstrahlung and
synchrotron emission experimentally. It can be done analytically on the base
of particle-in-cell (PIC) simulation results. There are several PIC codes that
can self-consistently describe both synchrotron radiation and bremsstrahlung -
EPOCH [19], OSIRIS [20], CALDER [21], PICLS [22] and some others [23, 24].
These codes were used in a variety of works [25, 26, 27, 28, 29] for
theoretical investigation of high intensity (from 1019 to 1024 W/cm2) laser
pulse energy conversion into continuous radiation. It was found that
synchrotron emission dominates roughly at intensities exceeding $\sim$1022
W/cm2, but all these investigations are focused on a short laser pulse (30–120
fs).

We have performed two-dimensional (2D) PIC simulation via the EPOCH code for
long 700 fs pulse. The code simulates bremsstrahlung radiation using a Monte
Carlo method through the elastic and inelastic cross section for the electrons
[28]. Synchrotron emission and radiation reaction are calculated in accordance
with principles described in [30, 31]. All the modelling parameters were
chosen close to experimental conditions: laser wavelength $\lambda_{laser}$ =
0.8 $\mu$m, focal spot radius 4 $\mu$mm, angle if incidence $\alpha$=45∘,
intensity Ilas=3$\times$1020 W/cm2. The laser pulse had Gaussian spatial and
3rd order super-Gauss temporal profile. A 2 $\mu$m thick layer of fully
ionized Si ions with solid-state density (5$\times$1022 ions/cm-3) was used as
a target. A long laser pulse requires a large simulation box to accurately
describe the laser-target interaction and to accommodate the expanding plasma.
In our simulations the box size was 120×120 $\mu$m with 10$\times$10 nm grid.
The simulation included a current smoothing algorithm and third order particle
weighting to limit noise and numerical heating. All boundary conditions were
absorbing for radiation and thermalizing for particles. The target was divided
into 3 zones: Zone 1 with 50 ion macro-particles per cell (ppci) and
50$\times$14 electron-macro-particles per cell (ppcel), zone 2 with 25 ppci
25$\times$14 ppcel, and zone 3 with 10 ppci , 10$\times$14 ppcel. This allows
efficient use of computational time without loss in accuracy. As a result, a
temporally integrated angular distribution of bremsstrahlung and synchrotron
X-rays in the wavelength range observed in the experiment was obtained, this
is shown in Fig. 3.

Figure 3: Theoretically calculated angular distribution of bremsstrahlung
(magenta curve) and synchrotron radiation (blue) emitted in the 4.8–7.3 Å
(1.7–2.6 keV) wavelength range by a 2 $\mu$m thick layer of fully ionized Si
ions with solid-state density irradiated by a subpicosecond laser pulse with
intensity 3$\times$1020 W/cm2. The green curve is the sum of two components.

In summary the distribution is slightly elongated along the laser beam
propagation axis. The ratio of maximum values for back (180∘) and transverse
(90∘) directions is $\approx$0.8. For all directions the synchrotron radiation
is significantly less intense than the bremsstrahlung. Thus 95% of
$\approx$1.5$\times$1014 photons associated with free-free emission
corresponds to bremsstrahlung, i.e. integral over the blue curve in Fig.2(c).
It contains about 63 mJ. Therefore, conversion efficiency of laser energy to
the bremsstrahlung is $\approx$3.2$\times$10-4, which is close to the value
$\approx$5$\times$10-4 predicted by the PIC simulation.

##  4 Conclusions

The laser plasma produced by sub-PW (1 ps, 200 J, 200 TW, with focal spot
diameter 7 $\mu$m, and on target intensity 3$\times$1020 W/cm2) laser pulse in
a 2 $\mu$m thick Si foil is a very bright source of soft X-rays. We recorded
about 1015 photons emitted by plasma in the wavelength range of 4.75–7.3 Å
(1.7–2.6 keV) with total energy $\approx$0.33 J and a CE of about
1.5$\times$10-3. About 40% of this energy were emitted in Si XIV (Si13+) Lyα,
Lyβ and Si XIII (Si12+) Heα and Heβ resonance spectral lines and associated
satellites. Lyα line is the most intense spectral line containing about a half
of all emitted photons. This makes this line the best choice for quasi-
monochromatic X-ray backlighter imaging and is sufficiently bright for use in
appoint-project Bragg crystal imaging system.

About 170 mJ (CE$\approx$8.4$\times$10-4) of incident laser energy was re-
emitted in continuous spectrum, of this approximately 64 mJ (CE
$\approx$3.2$\times$10-4) of it corresponds to bremsstrahlung radiation. The
rest is associated with recombination continuum emission. The experimentally
obtained laser-to-bremsstrahlung CE is close to 5$\times$10-4 predicted by the
EPOCH PIC code. The simulations are in qualitative and quantitative agreement
with experimental results. The recombination continuum is sufficiently bright
and featureless for use as an absorption spectroscopy source.

Acknowledgments The authors would like to thank the Central Laser Facility
staff, whose dedication and expertise were essential to the success of their
experiment. Calculations were carried out on the computational resources of
the JSCC RAS. The reported study was funded by RFBR, project number
19-32-60050. This study was done in the frame of the State Assignment to JIHT
RAS (topic #075-00892-20-00). The work of UK team received financial support
from UK EPSRC grants EP/P026796/1, EP/L01663X/1 and EP/H012605/1.

Disclosures The authors declare no conflicts of interest.

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# A Note on Finite Number Rings ††thanks:  Mathematics Subject Classification:
11T06, 11T30, 11Z05, 97H40

Suk-Geun Hwang , Woo Jeon , and Ki-Bong Nam , Tung T. Nguyen  Department of
Mathematics Education, Kyungpook University, Taegu 702-701, Korea and
Department of Mathematics, University of Wisconsin-Whitewater, Whitewater, WI
53190. USA, Email<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Mathematics
& Statistics, University of Wisconsin-Green Bay, Green Bay, WI 54311, USA,
Email<EMAIL_ADDRESS>Dept. of Math., Univ. of Wisconsin-Whitewater,
Whitewater, WI 53190, USA, Email<EMAIL_ADDRESS>of Kinesiology, Western
University, Ontario, Canada, Email<EMAIL_ADDRESS>
###### Abstract

We define the finite number ring ${\mathbb{Z}}_{n}[\sqrt[m]{r}]$ where $m,n$
are positive integers and $r$ in an integer akin to the definition of the
Gaussian integer ${\mathbb{Z}}[i]$. This idea is also introduced briefly in
[9]. By definition, this finite number ring ${\mathbb{Z}}_{n}[\sqrt[m]{r}]$ is
naturally isomorphic to the ring ${\mathbb{Z}}_{n}[x]/{\langle
x^{m}-r\rangle}$. From an educational standpoint, this description offers a
straightforward and elementary presentation of this finite ring, making it
suitable for readers who do not have extensive exposure to abstract algebra.
We discuss various arithmetical properties of this ring. In particular, when
$n=p$ is a prime number and $\mathbb{Z}_{p}$ contains a primitive $m$-root of
unity, we describe the structure of $\mathbb{Z}_{n}[\sqrt[m]{r}]$ explicitly.

Keywords: Radical Extension, Unital Determinant, Finite Fields, Polynomial
Factorization

##  1 Introduction

For each positive integer $n$, we will denote by $\mathbb{Z}_{n}$ the ring of
integers modulo $n$. We recommend [6, 7] for some good textbooks on the ring
$\mathbb{Z}_{n}$. Throughout the paper, $p$ is a prime number.

Recall that the Gaussian integers ${\mathbb{Z}}[i]$ is isomorphic to
${\mathbb{Z}}[x]/\langle x^{2}+1\rangle$ where $\langle x^{2}+1\rangle$ is the
ideal of ${\mathbb{Z}}[x]$ which is generated by $x^{2}+1$ (see [5, 10]). Note
also that $\mathbb{Z}[i]$ can also be considered as the set of all elements of
the form $a+bi$ where $a,b\in\mathbb{Z}$ and $i^{2}=-1.$ Addition and
multiplication on $\mathbb{Z}[i]$ can be defined in a natural way. For example

$(a+bi)(c+di)=(ac-bd)+(ad+bc)i.$

The advantage of this presentation of $\mathbb{Z}[i]$ is that it is rather
straightforward and can be understood by readers who do not have serious
exposure to the theory of quotient rings.

Motivated by this observation, we can define an extension ring
${\mathbb{Z}}_{n}[i]$ of ${\mathbb{Z}}_{n}$ as the set of all formal elements
of the form $a+bi$ where $a,b\in\mathbb{Z}_{n}$. Multiplication and addition
are also defined in a natural way. More specifically, if $r\in\mathbb{Z}$ (or
$\mathbb{Z}_{n}$) and $m$ is a positive integer, we can define

$\displaystyle{\mathbb{Z}}_{n}[\sqrt[m]{r}]=\\{x_{0}+x_{1}\sqrt[m]{r}+\cdots+x_{m-1}(\sqrt[m]{r})^{m-1}|x_{0},\cdots,x_{m-1}\in{\mathbb{Z}}_{n}\\}$

The addition on $\mathbb{Z}_{n}[\sqrt[m]{r}]$ is given naturally as

$\sum_{i=0}^{m-1}x_{i}(\sqrt[m]{r})^{i}+\sum_{i=0}^{m-1}x_{i}(\sqrt[m]{r})^{i}=\sum_{i=0}^{m-1}(x_{i}+y_{i})(\sqrt[m]{r})^{i}$

Similarly, the multiplication on $\mathbb{Z}_{n}[\sqrt[m]{r}]$ can defined by
the following rule (and extended by linearity)

$(\sqrt[m]{r})^{i}(\sqrt[m]{r})^{j}=\begin{cases}(\sqrt[m]{r})^{i+j}&\text{if
$i+j<m$}\\\ {r}(\sqrt[m]{r})^{k}&\text{if $i+j>m$ and $k\equiv i+j$ mod
$m$}\\\ {r}&\text{if $i+j=m$}\end{cases}$ (1)

We can see that $\mathbb{Z}_{n}[\sqrt[m]{r}]$ is a commutative ring which is
naturally isomorphic to the quotient ring $\mathbb{Z}_{n}[x]/\langle
x^{m}-r\rangle$. The goal of this paper is to prove some fundamental
properties of these finite rings using the explicit description mentioned
above.

##  2 Ring theoretic properties of $\mathbb{Z}_{n}[\sqrt[m]{r}]$

We begin with a lemma about finite commutative rings.

######  Lemma 1.

Let $R$ be a finite commutative ring. Then $R$ is an integral domain if and
only if $R$ is a field.

###### Proof.

If $R$ is a field then clearly $R$ is an integral domain. Conversely, suppose
that $R$ is an integral domain. We claim that each element $a\in R$ such that
$a\neq 0$ has an inverse. In fact, let $m_{a}:R\to R$ be the multiplication
map

$m_{a}(x)=ax.$

Because $R$ is an integral domain, the map $m_{a}$ is injective. Since $R$ is
finite, $m_{a}$ must be surjective as well. Therefore, we can find $b\in R$
such that $ab=1.$ This shows that $a$ is invertible in $R.$ Since this is true
for all $a\in R$ such that $a\neq 0$, we conclude that $R$ is a field. ∎

Suppose that $n=p_{1}^{\alpha_{1}}\ldots p_{s}^{\alpha_{s}}$ be the prime
factorization of $n.$ Then, by the Chinese remainder theorem

$\mathbb{Z}_{n}\cong\prod_{i=1}^{s}\mathbb{Z}_{p_{i}^{\alpha_{i}}}.$

Consequently

$\mathbb{Z}_{n}[\sqrt[m]{r}]\cong\prod_{i=1}^{s}\mathbb{Z}_{p_{i}^{\alpha_{i}}}[\sqrt[m]{r}].$

A direct consequence of this observation is the following

######  Proposition 1.

If $n$ is not a prime integer, then the ring ${\mathbb{Z}}_{n}[\sqrt[m]{r}]$
is not an integral domain.

###### Proof.

If $n$ is not prime, then ${\mathbb{Z}}_{n}$ is not an integral domain. Since
the ring ${\mathbb{Z}}_{n}[\sqrt{r}]$ contains ${\mathbb{Z}}_{n}$, we conclude
that is also not an integral domain. ∎

######  Proposition 2.

If $x^{m}-r$ is reducible over $\mathbb{Z}_{n}$; namely there exists
$f,g\in\mathbb{Z}_{n}[x]$ such that $\deg(f)>1$ and $\deg(g)>1$ and

$x^{m}-r=f(x)g(x),$

then $\mathbb{Z}_{n}[\sqrt[m]{r}]$ is not an integral domain.

###### Proof.

By our assumption

$f(\sqrt[m]{r})g(\sqrt[m]{r})=(\sqrt[m]{r})^{m}-r=0.$

Since $\deg(f),\deg(g)<m$, we conclude that $f(\sqrt[m]{r})$ must be a non-
zero divisor in $\mathbb{Z}_{n}[\sqrt[m]{r}].$ Therefore,
$\mathbb{Z}_{n}[\sqrt[m]{r}]$ is not an integral domain. ∎

We have the following condition, which provides a necessary and sufficient
condition for $\mathbb{Z}_{n}[\sqrt[m]{r}]$ to be a field.

######  Theorem 1.

The finite number ring $\mathbb{Z}_{n}[\sqrt[m]{r}]$ is a field if and only
the following two conditions are satisfied

  1. 1.

$n$ is a prime number.

  2. 2.

$x^{m}-r$ is irreducible in $\mathbb{Z}_{n}[x].$

###### Proof.

The necessary part of this Theorem follows from Lemma 1, Proposition 1, and
Proposition 2. Conversely, suppose that $n$ is a prime number and $x^{m}-r$ is
irreducible over $\mathbb{Z}_{n}[r]$, we claim that
$\mathbb{Z}_{n}[\sqrt[m]{r}]$ is a field. Let
$a=a_{0}+a_{1}\sqrt[m]{r}+a_{2}(\sqrt[m]{r})^{2}+\cdots+a_{m-1}(\sqrt[m]{r})^{m-1}$
be a non-zero element in $\mathbb{Z}_{n}[\sqrt[m]{r}]$. Let

$h(x)=a_{0}+a_{x}+\cdots+a_{m-1}x^{m-1}\in\mathbb{Z}_{p}[x].$

Since $p$ is prime, $\mathbb{Z}_{p}$ is a field, and $\mathbb{Z}_{p}[x]$ is a
Euclidean domain. By the Euclidean algorithm and the fact that $x^{m}-r$ is
irreducible of degree $m$, we can find $f_{1},f_{2}\in\mathbb{Z}[x]$ such that

$h(x)f_{1}(x)+(x^{m}-r)f_{2}(x)=1.$

We then see that

$af_{1}(\sqrt[m]{r})=h(\sqrt[m]{r})f_{1}(\sqrt[m]{r})=1.$

This shows that $a$ is invertible as required. ∎

###  2.1 Units in $\mathbb{Z}_{n}[\sqrt[m]{r}]$

In this section, we study units in the ring $\mathbb{Z}_{n}[\sqrt[m]{r}]$ by
embedding it in a matrix ring. We start with the following lemma.

######  Lemma 2.

A non-zero element
$a=a_{0}+a_{1}\sqrt[m]{r}+a_{2}(\sqrt[m]{r})^{2}+\cdots+a_{m-1}(\sqrt[m]{r})^{m-1}$
of the ring ${\mathbb{Z}}_{n}[\sqrt[m]{r}]$ is a unit, if and only if $A$ is
invertible where

$A=\begin{bmatrix}a_{0}&a_{m-1}r&a_{m-2}r&a_{m-3}r&\cdots&a_{1}r\\\
a_{1}&a_{0}&a_{m-1}r&a_{m-2}r&\cdots&a_{2}r\\\
a_{2}&a_{1}&a_{0}&a_{m-1}r&\cdots&a_{3}r\\\ &&&\cdots&\\\
a_{m-1}&a_{m-2}&a_{m-3}&a_{m-4}&\cdots&a_{0}\end{bmatrix}.$

Equivalently,
$a_{0}+a_{1}\sqrt[m]{r}+a_{2}(\sqrt[m]{r})^{2}+\cdots+a_{m-1}(\sqrt[m]{r})^{m-1}$
is a unit if and only if $\Delta=\det(A)\in\mathbb{Z}_{n}^{\times}.$

###### Proof.

Suppose that $A$ is invertible. We want to find a non-zero element

$b_{0}+b_{1}\sqrt[m]{r}+b_{2}(\sqrt[m]{r})^{2}+\cdots+b_{m-1}(\sqrt[m]{r})^{m-1},$

of the ring $\mathbb{Z}_{n}[\sqrt[m]{r}]$ such that

$\displaystyle(a_{0}+a_{1}\sqrt[m]{r}+\cdots+a_{m-1}(\sqrt[m]{r})^{m-1})(b_{0}+b_{1}\sqrt[m]{r}+\cdots+b_{m-1}(\sqrt[m]{r})^{m-1})=1.$

This is equivalent to

$\displaystyle
a_{0}b_{0}+a_{m-1}rb_{1}+a_{m-2}rb_{2}+a_{m-3}rb_{3}+\cdots+a_{1}rb_{m-1}=1$
$\displaystyle
a_{1}b_{0}+a_{0}b_{1}+a_{m-1}rb_{2}+a_{m-2}rb_{3}+\cdots+a_{2}rb_{m-1}=0$
$\displaystyle
a_{2}b_{0}+a_{1}b_{1}+a_{0}b_{2}+a_{m-1}rb_{3}+\cdots+a_{3}rb_{m-1}=0$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\cdots$ $\displaystyle
a_{m-1}b_{0}+a_{m-2}b_{1}+a_{m-3}b_{2}+a_{m-4}b_{3}+\cdots+a_{0}b_{m-1}=0.$
(2)

Since $A$ is invertible, the system (2.1) has a unique solution. This implies
that the system of linear equations (2.1) has a unique non-zero solution. We
conclude that $a$ is invertible. Conversely, suppose that $a$ is invertible.
Then the multiplication
$m_{a}:\mathbb{Z}_{n}[\sqrt[m]{r}]\to\mathbb{Z}_{n}[\sqrt[m]{r}]$ is
invertible. In particular, for each $0\leq i\leq m-1$, we can find

$v_{i}=b_{0i}+b_{1i}\sqrt[m]{r}+\ldots+b_{(m-1)i}(\sqrt[m]{r})^{m-1},$

such that $av_{i}=(\sqrt[m]{r})^{i}.$ In other words, we have

$\displaystyle A\begin{bmatrix}b_{0i}\\\ b_{1i}\\\ \vdots\\\ b_{ii}\\\
\vdots\\\ b_{(m-1)i}\end{bmatrix}=e_{i}=\begin{bmatrix}0\\\ 0\\\ \vdots\\\
1\\\ \vdots\\\ 0\end{bmatrix}.$

Therefore, if we let $B=(b_{ij})_{0\leq i,j\leq m-1}$ then $AB=I_{m}$. This
shows that $A$ is invertible.

∎

######  Remark 2.1.

When $r=1$, the matrix $A$ above is often called a circulant matrix (see [2,
3]). Such matrices have diverse applications in many scientific fields
including engineering, physics, network theory, as well as pure and applied
mathematics.

Let us call the determinant $\Delta$ of (2) the unital determinant of the
element
$a_{0}+a_{1}\sqrt[m]{r}+a_{2}(\sqrt[m]{r})^{2}+\cdots+a_{m-1}(\sqrt[m]{r})^{m-1}$
in the ring ${\mathbb{Z}}_{p}[\sqrt[m]{r}]$. From Lemma 2, we have the
following

######  Proposition 3.

Let $p$ be a prime number. If the unital determinant of any non-zero element
of the ring ${\mathbb{Z}}_{p}[\sqrt[m]{r}]$ is not zero, then
${\mathbb{Z}}_{p}[\sqrt[m]{r}]$ is a field.

######  Corollary 2.

An element $a+b\sqrt{r}$ of the ring $\mathbb{Z}_{n}[\sqrt{r}]$ is a unit if
and only if $a^{2}-rb^{2}\in\mathbb{Z}_{n}^{\times}.$

###### Proof.

This follows form the fact that the unital determinant of $a+b\sqrt{r}$ is
$a^{2}-rb^{2}.$ ∎

######  Corollary 3.

${\mathbb{Z}}_{p}[\sqrt{r}]$ is an integral domain (equivalently a field by
Lemma 1) if and only if there does not exist
$(a,b)\neq(0,0)\in\mathbb{Z}_{p}^{2}$ such that

$a^{2}-rb^{2}=0.$

In other words, $\mathbb{Z}_{p}[\sqrt{r}]$ is an integral domain if and only
$c$ is not a square in $\mathbb{Z}_{p}.$

######  Remark 2.2.

This corollary also follows from Theorem 1. In fact,
$\mathbb{Z}_{p}[\sqrt{r}]$ is not a field if and only if $x^{2}-r$ is
reducible. In other words, it must have a linear factor; i.e. a root over
$\mathbb{Z}_{p}$. This shows that $c$ must be a square.

The case $m=3$ is also quite interesting. In this case the unital determinant
of the element $a_{0}+a_{1}\sqrt[3]{r}+a_{2}(\sqrt[3]{r})^{2}$ is given by

$\det\begin{bmatrix}a_{0}&ra_{2}&ra_{1}\\\ a_{1}&a_{0}&ra_{2}\\\
a_{2}&a_{1}&a_{0}\end{bmatrix}=a_{0}^{3}+ra_{1}^{3}+r^{2}a_{2}^{3}-3ra_{0}a_{1}a_{2}.$
(3)

By Lemma 2, we conclude that

######  Corollary 4.

An element $a_{0}+a_{1}\sqrt[3]{r}+a_{2}(\sqrt[3]{r})^{2}$ is a unit in in
$\mathbb{Z}_{n}[\sqrt[3]{r}]$ if and only if

$a_{0}^{3}+ra_{1}^{3}+r^{2}a_{2}^{3}-3ra_{0}a_{1}a_{2}\in\mathbb{Z}_{n}^{\times}.$

###  2.2 $m$-power map

We discuss a simple case where $x^{m}-r$ is reducible over ${\mathbb{Z}}_{p}$.

######  Proposition 4.

$r$ belongs to the image of the $m$-power map
$f^{[m]}:{\mathbb{Z}}_{p}\to{\mathbb{Z}}_{p}$ if and only if $x^{m}-r$ has a
linear factor.

###### Proof.

Suppose, there is an element $a\in{\mathbb{Z}}_{p}$ such that $a^{m}=r.$ This
implies that $x-a$ a factor of $x^{m}-r.$ Conversely, if $x^{m}-r$ has a
linear factor, say $x-a$, we must have $a^{m}=r.$ ∎

######  Corollary 5.

If the $m$-power map $f^{[m]}$ is onto then $x^{m}-r$ is reducible.

######  Corollary 6.

Suppose that $m\leq 3.$ Then $x^{m}-r$ is reducible if and only if $r$ belongs
to the image of the $m$-power map.

###### Proof.

If $r$ belongs to the image of the $m$-power map then by Proposition 4,
$x^{m}-r$ is reducible. Conversely, suppose that $x^{m}-r$ is reducible. Since
the degree of $x^{m}-r$ is at most $3$, at least one of its factors must be
linear. By Proposition 4, we conclude that $r$ belongs to the image of the
$m$-power map. ∎

We have a quite interesting consequence of this corollary.

######  Proposition 5.

Suppose that $n=p$ is a prime number. Then $r$ belongs to the image of the
$3$-power map if and only there exists
$(a_{0},a_{1},a_{2})\in\mathbb{Z}_{n}^{3}$ and
$(a_{0},a_{1},a_{2})\neq(0,0,0)$ such that
$a_{0}^{3}+ra_{1}^{3}+r^{2}a_{2}^{3}-3ra_{0}a_{1}a_{2}=0$.

###### Proof.

Suppose that $n$ belongs to the image of the $3$-power map, say $r=b^{3}$. We
see that

$a_{0}^{3}+ra_{1}^{3}+r^{2}a_{2}^{3}-3ra_{0}a_{1}a_{2}=(a_{0}+ba_{1}+b^{2}a_{2})(a_{0}^{2}+b^{2}a_{1}^{2}+b^{4}a_{2}^{4}-ba_{0}a_{1}-b^{2}a_{0}a_{2}-b^{3}a_{1}a_{2}).$

Therefore, if we take $a_{0}=-b,a_{1}=1,a_{2}=0$ then

$a_{0}^{3}+ra_{1}^{3}+r^{2}a_{2}^{3}-3ra_{0}a_{1}a_{2}=0.$

Conversely, suppose that there exists a non-trivial triple
$(a_{0},a_{1},a_{2})$ such that
$a_{0}^{3}+ra_{1}^{3}+r^{2}a_{2}^{3}-3ra_{0}a_{1}a_{2}=0$. By Corollary 4, we
know that $a_{0}+a_{1}\sqrt[3]{r}+a_{2}(\sqrt[3]{r})^{2}$ is not a unit in
$\mathbb{Z}_{n}[\sqrt[3]{r}]$. In particular, this implies that
$\mathbb{Z}_{n}[\sqrt[3]{r}]$ is not a field. Since $n$ is a prime number, by
Theorem 1, this also implies that $x^{m}-r$ is reducible over
$\mathbb{Z}_{n}[x].$ By corollary 6, $r$ must belongs to the image of the
$3$-power map. ∎

######  Lemma 3.

The $m$-power map is onto on $\mathbb{Z}_{p}$ if and only $\gcd(p-1,m)=1.$

###### Proof.

This follows from the fact that the unit group $\mathbb{Z}_{p}^{\times}$ of
$\mathbb{Z}_{p}$ is cyclic of order $p-1.$ ∎

######  Proposition 6.

For any prime $p>2,$ there is an element $a\in{\mathbb{Z}}_{p}$ such that
$x^{2}-a$ is irreducible over ${\mathbb{Z}}_{p}.$

###### Proof.

By Lemma 3, the $2$-power map is not onto. Let $a$ be an element in
$\mathbb{Z}_{p}$ which does not belong to the image of $f^{[2]}.$ Then by
Corollary 6, $x^{2}-a$ is an irreducible polynomial over ${\mathbb{Z}}_{p}.$ ∎

######  Remark 2.3.

There are exactly $\frac{p-1}{2}$ elements $a\in\mathbb{Z}_{p}$ such that
$x^{2}-a$ is irreducible.

###  2.3 Gaussian Integers over ${\mathbb{Z}}_{p}$

A prime $p$ is called Pythagorean if there are integers $a$ and $b$ such that
$a^{2}+b^{2}=p$. It is well-known that this condition is equivalent to the
condition that either $p=2$ or $p$ is a prime of the form $4k+1$ (see [5]).

######  Proposition 7.

${\mathbb{Z}}_{p}[i]$ is a field if and only if $p$ is not an Pythagorean
prime.

Proof. If $p$ is Pythagorean then there are integers $a,b$ such that
$a^{2}+b^{2}=p^{2}$. This implies that $(a+bi)((a-bi)=a^{2}+b^{2}\equiv p$
(mod $p$) i.e., $a^{2}+b^{2}\equiv 0$ mod $p$. This implies that $a+bi$ has a
zero divisor. Thus ${\mathbb{Z}}_{p}[i]$ is not an integral domain, i.e., it
is not a field.

Conversely, suppose that $\mathbb{Z}_{p}[i]$ is not a field. By Corollary 3,
there exists $a,b\in\mathbb{Z}_{p}$ such that $a^{2}+b^{2}=0.$ This is
equivalent to the condition that $p=2$ or $p\equiv 1\pmod{4}.$ In other words,
$p$ is a Pythagorean prime. $\Box$

######  Example 2.1.

Since $(3+4i)(3-4i)=25\equiv 0$ mod $5,$ the ring ${\mathbb{Z}}_{5}[{i}]$ is
not a field. We can also see that $x^{2}+1$ is reducible over
${\mathbb{Z}}_{5}$. In fact $x^{2}+1=(x+2)(x+3)$ in $\mathbb{Z}_{5}[x].$
$\Box$

##  3 Finite rings over finite fields

In this section, we study the finite ring $\mathbb{F}_{q}[\sqrt[m]{r}]$ where
$\mathbb{F}_{q}$ is the finite field with $q$ elements. This section is a bit
less elementary than the previous section in the sense that we need Galois
theory for some of our arguments. In fact, we have the following statement
about the Galois group of a finite field.

######  Proposition 8.

Let $\mathbb{F}_{q^{n}}/\mathbb{F}_{q}$ be a finite extension. Then the Galois
group of $\mathbb{F}_{q^{m}}/\mathbb{F}_{q}$ is a cyclic group of order $n$
generated by the Frobenius automorphism $\text{Frob}_{q}$ defined by
$\text{Frob}(x)=x^{q}.$ Note that $\text{Frob}_{q}$ is exactly the $q$-power
map $f^{[q]}.$

######  Definition 1.

Let $a,b$ be two relatively prime integers. The order of $a$ modulo $b$,
denoted by $\text{ord}_{b}(a)$ is the smallest number $h$ such that
$a^{h}\equiv 1\pmod{b}$.

We also need the following definition.

######  Definition 2.

Let $\mathbb{F}_{p}$ be the prime finite field of characteristics $p$ and
$r\in\overline{\mathbb{F}_{p}}^{\times}.$ The order of $r$, denoted by
$\text{ord}(r)$ is defined to be the smallest positive integer $h$ such that
$r^{h}=1.$

We are now ready to discuss the factorization of $x^{m}-r$ over a finite field
$\mathbb{F}_{q}$ under the assumption that $\mathbb{F}_{q}$ contains a
primitive $m$-roots of unity. We first observe the following phenomenon.

######  Theorem 7.

Suppose that $\mathbb{F}_{q}$ contains a primitive $m$-root of unity. Then
$x^{m}-r$ can be factorized into a product of $\frac{m}{t}$ irreducible
polynomials of degree $t$ where $t=\text{ord}_{\text{ord}(r)m}(q)$; namely,
$t$ is the smallest number such that $\text{ord}(r)m|q^{t}-1.$

###### Proof.

Since $\gcd(q,m)=1$, the polynomial $x^{m}-r$ is separable over
$\mathbb{F}_{q}.$ Let $\alpha$ be any root of $x^{m}-r$ over
$\overline{\mathbb{F}_{q}}.$ Then, by definition $\alpha^{m}=r.$ The finite
extension $\mathbb{F}_{q}(\alpha)/\mathbb{F}_{q}$ is a Galois extension whose
Galois group is generated by the Frobenius map $a\mapsto a^{q}$. Consequently,
the Galois conjugates of $\alpha$ are
$\\{\alpha,\alpha^{q},\ldots,\alpha^{q^{t-1}}\\}$ where $t$ is the smallest
positive integer such that $\alpha^{q^{t}}=\alpha.$ Equivalently,
$\alpha^{q^{t}-1}=1.$ We remark that since the Galois orbit of $\alpha$ over
$\mathbb{F}_{q}$ has $t$ elements, the minimal polynomial of $\alpha$ over
$\mathbb{F}_{q}$ must have degree $t.$ We claim that
$t=\text{ord}_{\text{ord}(r)m}(q)$, independent of $\alpha$. If this is true,
our theorem is proved. In fact, since $\mathbb{F}_{q}$ contains a primitive
$m$-root of unity, we know that $m|q-1.$ We then see that

$1=\alpha^{q^{t}-1}=(\alpha^{m})^{\frac{q^{t}-1}{m}}=r^{\frac{q^{t}-1}{m}}.$

The above equality implies that $\text{ord}(r)|\frac{q^{t}-1}{m}$ or
equivalently $\text{ord}(r)m|q^{t}-1.$ Since $t$ is the smallest positive
integer with this property, we conclude that
$t=\text{ord}_{\text{ord}(r)m}(q).$ ∎

######  Corollary 8.

Let $\mathbb{F}_{q},r,m,t$ be as in Theorem 7. Then
$\mathbb{F}_{q}[\sqrt[m]{r}]$ is isomorphic to
$\mathbb{F}_{q^{t}}^{\frac{m}{t}}.$

We discuss a consequence of Theorem 7. First, we recall the $k$-power map
$f^{[k]}:\mathbb{F}_{q}\to\mathbb{F}_{q}$ defined by $f^{[k]}(a)=a^{k}.$

######  Theorem 9.

Suppose that $\mathbb{F}_{q}$ contains a primitive $m$-root of unity. Then
$x^{m}-r$ is irreducible over $\mathbb{F}_{q}[x]$ if and only if $r$ does not
belong to the image for each $k|m$ and $k>1.$

###### Proof.

If $r$ belongs to the image of $f^{[k]}$ for some $k|m$ and $k>1$, then we can
write $x=b^{k}$ for some $k>1$ and $k|m$ then we can see that $x^{m}-r$ is
reducible since $x^{m/k}-b$ is a factor of $x^{m}-r.$ Conversely, suppose that
$r$ does not belong to the image of $f^{[k]}$ for all $k|m$ and $k>1.$ We
claim that $x^{m}-r$ is irreducible. In fact, let $t$ be the degree of an
irreducible factor of $x^{m}-r$ over $\mathbb{F}_{q}[x]$ (by Theorem 7, $t$ is
independent of the choice of the irreducible factor). We claim that $t=m.$ We
know that

$t=\text{ord}_{\text{ord}(r)m}(q)=\text{ord}_{(q-1)m}(q).$

By definition, $t$ is the smallest positive integer such that
$\text{ord}(r)m|q^{t}-1.$ Put it equivalently

$m|\frac{q^{t}-1}{\text{ord}(r)}=\frac{q-1}{\text{ord}(r)}(1+q+\ldots+q^{t-1}).$

Since $\mathbb{F}_{q}$ contains a primitive $m$-root of unity, we know that
$m|q-1.$ Consequently

$0\equiv\frac{q-1}{\text{ord}(r)}(1+q+\ldots+q^{t-1})\equiv\frac{q-1}{\text{ord}(r)}t\pmod{m}.$
(4)

Let $d=\gcd(m,\frac{q-1}{\text{ord}(r)})=1.$ We claim that $d=1$. Indeed,
since $\mathbb{F}_{q}^{\times}$ is a cyclotomic group of order $q-1$, there
exists a primitive root in $g\in\mathbb{F}_{q}^{\times}$ such that
$r=g^{\frac{q-1}{\text{ord}(r)}}.$ Since $d|\frac{p-1}{\text{ord}(r)}$, $r$
belongs to the image of $f^{[d]}.$ By our assumption on $r$, we must have
$d=1.$ From this fact and Equation 4, we conclude that $t\equiv 0\pmod{m}$.
Since $t|m$, we must have $t=m$. This shows that $x^{m}-r$ is irreducible. ∎

We discuss two corollaries Theorem 9.

######  Corollary 10.

If $r$ is a primitive root in $\mathbb{F}_{q}^{\times}$ and $\mathbb{F}_{q}$
contains a primitve $m$-root of unity, then $x^{m}-r$ is irreducible over
$\mathbb{F}_{q}[x]$.

###### Proof.

Since $r$ is a primitive root and $m|q-1$, $r$ can not belong to the image of
$f^{[k]}$ for all $k|m$ and $k>1.$ By Theorem 9, $x^{m}-r$ must be
irreducible. ∎

The next corollary is a direct consequence of Theorem 9.

######  Corollary 11.

Suppose that $m$ is a prime number and $\mathbb{F}_{q}$ contains a primitive
$m$-root of unity. Then $x^{m}-r$ is irreducible if and only $r$ is not an
$m$-power.

######  Corollary 12.

Suppose that $\mathbb{F}_{q}$ contains a primitive $m$-root of unity. Let $M$
be the positive integer such that

  1. 1.

$m|M$ and $\text{rad}(M)=\text{rad}(m)$. Here, for a positive integer $a$,
$\text{rad}(a)$ is the product of all distinct prime factors of $a.$

  2. 2.

$\gcd(M,\frac{q-1}{M})=1.$

Then, the number of $r\in\mathbb{F}_{q}$ such that $x^{m}-r$ is irreducible is
$\varphi(M)\dfrac{q-1}{M}.$

###### Proof.

By Theorem 9, we know that $x^{m}-r$ is irreducible if and only if $r$ does
not belong to the image of $f^{[k]}$ for each $k|m$ and $k>1$. By the proof of
this theorem, this is equivalent to saying that
$\gcd(m,\frac{q-1}{\text{ord}(r)})=1.$ Because $m|q-1$, this is also
equivalent to $M|\text{ord}(r).$ Let us write $\text{ord}(r)=Mh$ where
$h|\frac{q-1}{M}.$ Since the group $\mathbb{F}_{q}^{\times}$ is cyclic, the
number of elements of order $Mh$ is exactly
$\varphi(Mh)=\varphi(M)\varphi(h).$ We conclude that the number of irreducible
polynomials of the form $x^{m}-r$ is exactly

$\sum_{h|\frac{q-1}{M}}\varphi(M)\varphi(h)=\varphi(M)\sum_{h|\frac{q-1}{M}}\varphi(h)=\varphi(M)\frac{q-1}{M}.$

The last equality follows from the fact that for any positive integer $n$

$\sum_{d|n}\varphi(d)=n.$

∎

######  Remark 3.1.

If we assume that $m$ is a squarefree number then we can relax the condition
that $\mathbb{F}_{q}$ contains a primitive $m$-root of unity in Corollary 12.
In fact, suppose that $\mathbb{F}_{q}$ does not contain a primitive $m$-root
of unity. This implies that $m\nmid q-1.$ Because $m$ is a squarefree number,
there exists a prime divisor $d$ of $m$ such that $d\nmid q-1.$ This implies
that the $f^{[d]}$ power map is surjective. In other words, we can write
$r=b^{d}$ for some $b\in\mathbb{F}_{q}.$ This implies that $x^{m/d}-b$ is a
factor of $x^{m}-r.$ As a corollary, $x^{m}-r$ is always reducible.

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