# Hofstadter butterflies in magnetically modulated graphene bilayer: an
algebraic approach

Manisha Arora1, Rashi Sachdeva2 and Sankalpa Ghosh1 1 Department of Physics,
Indian Institute of Technology Delhi, New Delhi-110016, India  2 Mathematical
Physics and NanoLund, Lund University, Box 118, 22100 Lund, Sweden

###### Abstract

It has been shown that Bernal stacked bilayer graphene (BLG) in a uniform
magnetic field demonstrates integer quantum Hall effect with a zero Landau-
level anomaly Geimbilayer . In this article we consider such system in a two
dimensional periodic magnetic modulation with square lattice symmetry. It is
shown algebraically that the resulting Hofstadter spectrum can be expressed in
terms of the corresponding spectrum of monolayer graphene in a similar
magnetic modulation. In the weak-field limit, using the tight-binding model,
we also derive the Harper-Hofstadter equation for such BLG system in a
periodic magnetic modulation. We further demonstrate the topological
quantisation of Hall conductivity in such system and point out that the
quantised Hall plateaus are equally spaced for all quantum numbers for the
quantised Hall conductivity.

##  I Introduction

A two-dimensional electron gas in a transverse magnetic field and a periodic
potential, shows fractal energy spectrum of the shape of a butterfly, first
shown in the seminal work by D. Hofstadter Hofstadter , thereby acquiring its
name as the Hofstadter butterfly. Within linear response theory, the Hall
conductivity of this system can be identified with a topological invariant
called Chern number defined over the topological space of the Brillouin zone
TKNN ; Kohmoto  corresponding to the periodic potential. This explains the
perfect quantisation of Hall conductivity leading to the integer quantum Hall
effect Klitzing . These two remarkable facets of the same system, namely
topologically quantised Hall conductance and the fractal like energy spectrum,
both stem from the magnetic translation symmetry Zak ; Fischbeck ; Florek ;
Tapash , that generalises the lattice translational symmetry of Bloch
electrons in the presence of time-reversal symmetry breaking magnetic field.
To explore the domain of application of such magnetic translation symmetry, in
our recent work Manisha  we studied the charge carriers in monolayer graphene
(MLG) Novoselov1 ; Gusynin  that are massless Dirac fermions in a periodically
modulated magnetic field. It was pointed out that the magnetic translation
symmetry of a system in a uniform magnetic field and a periodic potential is
retained when the magnetic field is periodically modulated. We showed that in
such a system the energy spectrum is again given by Hofstadter butterfly like
fractal structure and the Hall conductivity is topologically quantised.

A natural yet non-trivial question can be asked by seeking the consequences of
application of such two dimensional magnetic modulation to the quasiparticles
of Bernal stacked bilayer graphene (BLG), implying that two hexagonal lattices
of MLG with inequivalent sites $A,B$ and $\tilde{A},\tilde{B}$ are placed on
top of each other such that $\tilde{A}$ site of the top layer is directly
above the $B$ sites of the bottom layer and the two layers are coupled through
interlayer hopping Bernal . Such Bernal stacked BLG can be thought of as an
intermediate step between monolayer graphene and different types of Moiré
superlattices andp  that are realised in van der Waals heterostructures such
as graphene-hexagonal boron-nitride (hBN) heterostructure Hofstadter1 ;
Hofstadter2 ; Hofstadter3  or twisted bilayer graphene TBLG , where in the
first case the Hofstadter butterfly structures have already been demonstrated
experimentally. This itself provides a sufficient motivation to study the
effect of two-dimensional periodic magnetic modulation on the quasiparticles
of BLG. However more intriguing aspect in the present context is that the
Bernal stacked BLG shows a different chirality exhibiting a $2\pi$ Berry phase
and a dominantly parabolic behaviour with a finite mass, as opposed to a $\pi$
Berry phase and linear dispersion for the massless chiral quasiparticles in
MLG McCann . Consequently in a transverse magnetic field, double degenerate
Landau levels are formed incorporating two different orbital states with same
energy Falkobilayer  and giving rise to a new kind of integer quantum Hall
effect in bilayer graphene with a zero level anomaly in the Landau level
structure Geimbilayer . The different nature of chirality of the
quasiparticles also manifests itself, not only when the graphene bilayer is
exposed to a uniform magnetic field, but also when we consider the
transmission of such massive chiral fermions through scalar potential
barriersKSN3 ; kleinexp ; Tan  or through vector potential barriers created by
localised magnetic field dubbed as magnetic barriers Masir ; levitov2 ; Neetu
.

In this work we start by showing that the low energy hamiltonian for the BLG
in a periodic magnetic modulation commutes with the magnetic translation
operator for a uniform field like in the corresponding problem for the MLG. As
a result it is possible to obtain algebraically the Hofstadter like spectrum
for the BLG in such periodically modulated magnetic field. The existence of
two layers is reflected with the doubling of the number of Hofstadter
butterflies in the spectrum when compared to the corresponding cases of MLG
under similar approximation. In the present case, the separation between such
doubled Hofstadter butterflies is controlled by the strength of the interlayer
hopping. The Hall conductivity in this case also turns out to be topologically
quantised. However unlike the experimental discovery of quantum Hall effect in
bilayer graphene in uniform transverse magnetic field Geimbilayer , that shows
a topologically quantised Hall conductivity with somewhat similar expression
but with a zero Landau level (LL) anomaly, here the quantised Hall plateaus
are equally spaced. We explain various aspects of this feature. Accordingly
the rest of the paper is organised in the following way. After introducing the
periodic magnetic modulation in section II.1, in section II.2 we write the
effective four band low energy hamiltonian of BLG in the presence of such
magnetic modulation and algebraically solve the problem using suitable
commutator relations between different components of the canonical momentum
operator. This is followed by a calculation involving the use of tight binding
approximation to write the lattice hamiltonian in such periodic modulation and
to show that it again comes in the form of standard Harper-Hofstadter
equation. We plot the spectrum as a function of strength of the magnetic
modulation. In the next section II.3, we work out algebraically the same
energy spectrum, though this time, by projecting the hamiltonian to the lowest
two bands. Thereafter it’s comparative analysis with the results from the
four-band case is done. In section III using results obtained from the four-
band model, within linear response theory we show that the Hall conductance is
topologically quantised and there exists a $0$ value in the quantum number. We
finally conclude our work.

##  II Bilayer graphene in the presence of two dimensional (2D) periodic
magnetic modulation

###  II.1 The model

In effective mass approximation, the single particle hamiltonian for Bernal
stacked BLG in an arbitrary magnetic modulation pointed along the transverse
direction ($z$-axis) can be written as Zarenia1

$H=\begin{bmatrix}0&\Pi&t&0\\\ \Pi^{\dagger}&0&0&0\\\ t&0&0&\Pi^{\dagger}\\\
0&0&\Pi&0\end{bmatrix}$ (1)

giving the following eigenvalue equation

$v_{F}H\Psi=E\Psi$ (2)

where
$\Psi=\begin{bmatrix}\psi_{a}&\psi_{b}&\psi_{c}&\psi_{d}\end{bmatrix}^{T}$.
Here $t\approx 400~{}\text{meV}$ is the interlayer coupling, $v_{F}$ is the
Fermi velocity, and

$\Pi=\Pi_{x}-i\Pi_{y},~{}\Pi^{\dagger}=\Pi_{x}+i\Pi_{y},$

where the canonical momentum operator
$\Pi_{x,y}=(p_{x,y}+\frac{eA_{x,y}}{c})$. The vector potential
$\boldsymbol{A}(x,y)$ satisfies

$\boldsymbol{\nabla}\times\boldsymbol{A}(x,y)=\boldsymbol{B}(x,y)\hat{z}.$ (3)

For a periodic magnetic field profile $\boldsymbol{B}(x,y)$ in the $x-y$
plane,

$\boldsymbol{B}=\sum_{\boldsymbol{G}}\boldsymbol{B}_{\boldsymbol{G}}e^{i\boldsymbol{G}.\boldsymbol{r}}=\boldsymbol{B}_{u}+\boldsymbol{B}_{p}$
(4)

Here $\boldsymbol{G}$ is the reciprocal lattice vector of the periodic
lattice, $\boldsymbol{B}_{u}$ is the uniform part of the magnetic field given
by the spatial average of the field given in Eq. $(\ref{B})$. The residual
periodic magnetic field is defined as
$\boldsymbol{B}_{p}=\sum_{\boldsymbol{G}\neq
0}\boldsymbol{B}_{\boldsymbol{G}}e^{i\boldsymbol{G}.\boldsymbol{r}}$ that
satisfies $\boldsymbol{\nabla}\cdot\boldsymbol{B}_{p}=0$. It can be easily
checked that the net flux through the unit cell due to $\boldsymbol{B}_{p}$ is
zero and hence comes entirely from $\boldsymbol{B}_{u}$. The results that are
reported in this paper are generally true for any arbitrary periodic magnetic
modulation in a two dimensional plane. However for the sake of computation we
choose a specific form of two dimensional periodic magnetic modulation.

Figure 1: (Color online): Schematic plot of the magnetic field profile given
in Eq. (5). $B_{1}$ and $B_{2}$ are the values of the field in the colored and
uncolored regions in an arbitrary unit. The height of the coloured area is
indicative of the fact that $B_{1}\neq B_{2}$.

Our chosen two dimensional periodic magnetic field modulation is given by

$\boldsymbol{B}=\Big{[}B_{2}+(B_{1}-B_{2})\sum_{m,n}\Theta(R_{s}-r_{mn})\Big{]}\boldsymbol{\hat{z}}$
(5)

Here $R^{s}$ is radius of the circular region within each unit cell and
$\boldsymbol{r}_{mn}=\boldsymbol{r}-\boldsymbol{R}_{mn}$ where
$\boldsymbol{r}=x\boldsymbol{\hat{x}}+y\boldsymbol{\hat{y}}$. Apart from this,
the lattice vector
$\boldsymbol{R}_{mn}=ma\boldsymbol{\hat{x}}+na\boldsymbol{\hat{y}}$ and

$\displaystyle\Theta(R_{s}-r_{mn})$ $\displaystyle=$
$\displaystyle\begin{cases}1,~{}~{}\text{if}~{}\ r_{mn}\leq R_{s},\\\
0,~{}~{}\text{if}~{}\ r_{mn}>R_{s}\end{cases}$ (6)

where $m,n\in I(-N_{max},+N_{max})$ and $a$ denotes the lattice constant. We
assume that the lattice constant $a$ is much larger than the carbon-carbon
(c-c) bond length in bilayer graphene which justifies the consideration of
continuum model for bilayer graphene. A schematic presentation of the magnetic
field profile is given in Fig. 1. In general for any periodic magnetic field
profile, because of the decomposition given in Eq. (4), the vector potential
can be decomposed as

$\boldsymbol{A}=\boldsymbol{A}_{u}+\boldsymbol{A}_{p}$

with $A_{x}=A_{ux}+A_{px},~{}~{}A_{y}=A_{uy}+A_{py}$, where in a suitable
gauge $\boldsymbol{A}_{p}$ can be written as a periodic function. Accordingly,
for the current magnetic field profile given in Eq. (5) we have chosen a
vector potential that has two parts, namely

$\displaystyle\boldsymbol{A}_{u}(\boldsymbol{r})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\boldsymbol{B}_{u}\times\boldsymbol{r},~{}\text{Symmetric
Gauge}$ $\displaystyle\boldsymbol{A}_{p}(\boldsymbol{r})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{m,n}\Big{[}(B_{p}^{mn}\boldsymbol{\hat{z}})\times(r_{mn}\boldsymbol{\hat{r}}_{mn})\Big{]}$
(7)

where

$B_{p}^{mn}=\begin{cases}\Big{[}(B_{1}-B_{2})-\frac{(B_{1}-B_{2})\pi(R_{s})^{2}}{a^{2}}\Big{]}\Theta(R_{s}-r_{mn}),\\\
\text{if}\ r_{mn}\leq R_{s}\\\ \\\
(B_{2}-B_{1})\frac{\pi(R_{s})^{2}}{Na^{2}}\Theta(r_{mn}-R_{s}),~{}\text{otherwise}\end{cases}$
(8)

Because of the periodic structure of the magnetic modulation given in Eq.(5),
the resulting quasiparticle spectrum of Bernal stacked BLG subjected to such
periodic magnetic modulation will be fundamentally different from the
corresponding four band dispersion of the BLG without such magnetic modulation
McCann . The uniform field $\boldsymbol{B}_{u}$ in the expression (4) will
lead to Landau levels for such BLG Zarenia1 , whereas the additional
$\boldsymbol{B}_{p}$ field which forms a lattice of the magnetic modulation
will convert these Landau levels into Hofstadter bands Hofstadter . It may be
recalled that unlike in Moiré pattern in TBLG where a Hofstadter pattern
arises MoireA ; MoireB  due to the presence of the Moiré periodicity that
results from the relative twist between the two layers each of which is
individually treated in continuum, in the present problem the periodic
structure is provided by the periodic modulation of the magnetic field itself
given in Eq. (5).

###  II.2 Energy spectrum using the four-band model in the presence of 2D
periodic magnetic modulation

The eigenvalue equation for the hamiltonian given in (1) leads to the
following set of coupled equations

$\displaystyle v_{F}\Pi\psi_{b}$ $\displaystyle=$ $\displaystyle
E\psi_{a}-v_{F}t\psi_{c}$ (9) $\displaystyle v_{F}\Pi^{\dagger}\psi_{a}$
$\displaystyle=$ $\displaystyle E\psi_{b}$ (10) $\displaystyle
v_{F}\Pi^{\dagger}\psi_{d}$ $\displaystyle=$ $\displaystyle
E\psi_{c}-v_{F}t\psi_{a}$ (11) $\displaystyle v_{F}\Pi\psi_{c}$
$\displaystyle=$ $\displaystyle E\psi_{d}$ (12)

Figure 2: (Color online): Energy eigenvalues $\epsilon$ vs flux $\alpha$ plot
for $B_{1}=0$. $\epsilon=\frac{aE}{\hbar v_{F}}$. In this figure as well as in
subsequent figures $\alpha=\frac{p}{q}$ represents the magnetic flux through
the unit cell and $(p,q)$ are co-prime. For a fixed $B_{1}$, $B_{2}$ changes
with change of $\alpha$ according to Eq. (17).

We define $H_{G}=v_{F}^{2}\Pi\Pi^{\dagger}$. Since $H_{G}^{\dagger}=H_{G}$
coupling Eqs. (9) - (10), and Eqs. (11)-(12) we get

$\displaystyle(H_{G}-E^{2})\psi_{a}$ $\displaystyle=$ $\displaystyle-
v_{F}t\psi_{c}$ (13) $\displaystyle(H_{G}-E^{2})\psi_{c}$ $\displaystyle=$
$\displaystyle-v_{F}tE\psi_{a}$ (14)

The operator $H_{G}$ determines the energy spectrum for the monolayer graphene
in the same periodic magnetic modulation. This tells us that $\psi_{a,c}$ will
be obtained by solving the same eigenvalue equation upto an arbitrary phase
factor like the MLG in a periodic magnetic modulation and $\psi_{b,d}$ can be
obtained from Eqs. (10) and (12) subsequently for a given eigenvalue $E$. This
is one of the major findings of this work, and has consequences like the
topological quantisation of the Hall conductivity in magnetically modulated
BLG which is studied in details in subsequent sections.

To get $\psi_{a,c}$ we combine Eqs. (13) and (14), giving

$(H_{G}-E^{2})^{2}\psi_{a,c}=v_{F}^{2}t^{2}E^{2}\psi_{a,c}$ (15)

Square-rooting of the above equation directly gives us the the eigenvalue
spectrum $E$ of the BLG in the magnetic modulation defined in (5) as

$\epsilon=\frac{1}{2}\bigg{[}\pm\frac{at}{\hbar}\pm\sqrt{(}\frac{a^{2}t^{2}}{\hbar^{2}}+4\epsilon_{M})\bigg{]}$
(16)

where $\epsilon=\frac{aE}{\hbar v_{F}}$ and
$\epsilon_{M}=\frac{a^{2}E_{M}^{2}}{\hbar^{2}v_{F}^{2}}$ are dimensionless
energies in BLG and MLG respectively with $E_{M}$ representing the energy of
the MLG modulated by the same magnetic modulation.It can be readily checked
that for $t=0$, the expression (16) reduces to the corresponding spectrum of
MLG in a way similar to the corresponding problem in uniform magnetic field
case Zarenia1 . The plots for the spectrum obtained using the above set of
equations is shown in Fig. (2)- Fig. (6). In these plots the $x$-axis
corresponds to $\alpha=\frac{p}{q}$ which is the total number of flux quanta
through a unit cell and is given in terms of the co-prime $(p,q)$. $B_{1,2}$
defined in Eq. (5) is related to $\alpha$ through

$\alpha\phi_{0}=B_{1}\pi R_{s}^{2}+B_{2}(a^{2}-\pi R_{s}^{2}).$ (17)

And the $y$-axis corresponds to the dimensionless energy eigenvalue.

For a detailed analysis of the plots, we rewrite the LHS of Eq. (16) as
$\epsilon_{+-,+-}$ where the first and second sign show the respective signs
in the RHS of the spectrum. This implies that, $\epsilon_{+,+}$ and
$\epsilon_{-,-}$ correspond to the two outer butterflies (blue and yellow in
Fig 2) and $\epsilon_{+,-}$ and $\epsilon_{-,+}$ correspond to the two inner
butterflies (green and red in Fig. 2)). For the case, when $\epsilon_{M}=0$,
implying $\epsilon_{+,-}=\epsilon_{-,+}=0$, the two inner butterflies touch
each other. This is the result obtained by setting interlayer coupling
strength $t=400$ meV in Eq. (16).

Figure 3: (Color online): Energy spectrum $\epsilon$ defined in Eq. (16) as a
function of magnetic flux $\alpha$ for $B_{1}=1$ for three different values of
interlayer coupling strengths $t$. The energy is plotted in the dimensionless
unit $\epsilon=\frac{aE}{\hbar v_{F}}$.

Figure 4: (Color online): The corresponding energy spectrum for monolayer
graphene (MLG) as a function of magnetic flux $\alpha$ for the same value of
$B_{1}$ as in Fig. 4. Here $y$-axis corresponds to $\epsilon=\epsilon_{M}$
which is the dimensionless energy of MLG, and is defined in the text after Eq.
(16). Figure 5: (Color online):Energy spectrum $\epsilon$ vs flux $\alpha$
plot for a fixed value of $B_{2}=0$, but varying $B_{1}$. As explained in the
text, $\epsilon$ is the dimensionless energy. The relation between $B_{1}$ and
$B_{2}$ is given in Eq. (17) that gives how $B_{1}$ varies with $\alpha$.
Figure 6: (Color online):$\epsilon$ vs. $\alpha$ plot for $B_{2}=1$. Notations
are same as in Fig. 5.

As the value of $B_{1}$ is increased, the energy eigenvalue $\epsilon_{M}$ of
the MLG in a similar magnetic modulation increases (see Fig. (4)). This causes
increase in the the gap between the outermost butterflies $\epsilon_{+,+}$ and
$\epsilon_{-,-}$. Similarly the gap between the inner butterflies
$\epsilon_{+,-}$ and $\epsilon_{-,+}$ also increases with increase in $B_{1}$.
This can be seen in the plots as we move from Fig (2) $(B_{1}=0)$ as well as
in the leftmost figure of Fig. (4) $(B_{1}=1)$.

We now study the effect of fixing the value of magnetic field strength $B_{1}$
and varying the interlayer coupling strength $t$. The effect is demonstrated
in Fig. (4). As $B_{1}$ is fixed, $\epsilon_{M}$ attains a fixed set of
values. Going from left to right in Fig. (4), we can clearly observe that as t
increases, the gap between the outer butterflies increases whereas the gap
between the two inner butterflies decreases. This can be understood from Eq.
(16), where as we increase t, the separation between $\epsilon_{+,+}$ and
$\epsilon_{-,-}$ increases whereas that between $\epsilon_{+,-}$ and
$\epsilon_{-,+}$ decreases. The spectrum is also plotted as a function of
fixed value of $B_{2}$, but varying values of $B_{1}$ in Fig. (5) and Fig.
(6).

Written in a slightly modified form, the eigenvalue equation for the BLG in a
periodic magnetic modulation can be written in terms of the corresponding
hamiltonian operator for the MLG as

$H_{G}\psi_{a,c}=\big{[}(E^{\pm})^{2}-\frac{1}{4}v_{f}^{2}t^{2}\big{]}\psi_{a,c}$
(18)

where $E^{\pm}=E\pm\frac{1}{2}v_{f}t$. It is known that $H_{G}$ commutes with
the magnetic translation operator Zak  corresponding to the uniform field part
of the magnetic modulation, namely $B_{u}$. Hence the eigenfunctions of the
$H_{G}$, which are also the magnetic Bloch functions, can be expanded in terms
of localised Wannier functions in the presence of uniform transverse magnetic
field $\boldsymbol{B}_{u}$Luttinger ; Wannier .

$\psi_{\boldsymbol{k}}^{a,c}=\sum_{i}g(\boldsymbol{R}_{i})\exp\big{(}-i\frac{e\boldsymbol{A}_{u}\cdot\boldsymbol{R}_{i}}{\hbar
c}\big{)}w_{0}(\boldsymbol{r}-\boldsymbol{R}_{i})$ (19)

To simplify further calculation here we set the condition
$|\frac{ea^{2}}{\hbar c}(B_{u}+B_{p})|<<1$. For $B_{p}=0$, this implies
$|\frac{ea^{2}}{\hbar c}(B_{u})|\ll 1$ which leads to $a\ll l_{B_{u}}$. Here
$l_{B_{u}}$ is the magnetic length which is defined in terms of the uniform
part of the magnetic modulation. Thus this condition corresponds to a weak and
slowly varying magnetic field Luttinger ; exact . This is a reasonable
approximation as long as the modulation part $B_{p}$ ( represented by either
the r.m.s value, or the peak value) is smaller than $B_{u}$, and hence, can be
treated as a perturbative effect. This type of condition is generally used in
lattice gauge theory Governale . Under these approximations we can write the
eigenvalue equation ($\ref{MLGBLG})$ as a discrete Schrödinger equation which
takes the form of Hofstadter-Harper equation

$\displaystyle\epsilon g(m,n)$ $\displaystyle=$ $\displaystyle
e^{\frac{iea}{\hbar c}(A_{ux}+A_{px})}g(m+1,n)+e^{-\frac{iea}{\hbar
c}(A_{ux}+A_{px})}g(m-1,n)+e^{\frac{iea}{\hbar c}(A_{uy}+A_{py})}g(m,n+1)$
(20) $\displaystyle+$ $\displaystyle e^{-\frac{iea}{\hbar
c}(A_{uy}+A_{py})}g(m,n-1)-[\frac{ea^{2}}{\hbar c}(B_{u}+B_{p})+4]g(m,n)$

with

$\epsilon=-\frac{[(E^{\pm})^{2}-\frac{1}{4}v_{f}^{2}t^{2}]a^{2}}{v_{F}^{2}\hbar^{2}}$
(21)

and $g(m,n)=g(\boldsymbol{R_{mn}})$.

Figure 7: (Color online):Bilayer graphene energy spectrum for $B_{1}=1$. The
$y$-axis plots the dimensionless energy $\epsilon$ which is in the unit of
$\frac{\hbar^{2}}{2ma^{2}}$. The $x$-axis plots the magnetic flux
$\alpha=\frac{p}{q}$ like the previous figures. The expression of the energy
is given in Eq. (33).

Once $\psi_{a,c}$ is calculated from Eqs. (19) and (20), using Eq. (10) we can
calculate ( again under the weak magnetic field limit)

$\displaystyle\psi_{\boldsymbol{k}}^{b}$ $\displaystyle=$
$\displaystyle\frac{v_{F}}{E}\bigg{[}-i\hbar(\frac{\psi_{\boldsymbol{k}}^{a}(\boldsymbol{r}+a\boldsymbol{\hat{x}})-\psi_{\boldsymbol{k}}^{a}(\boldsymbol{r}-a\boldsymbol{\hat{x}})}{2a})$
(22) $\displaystyle-$ $\displaystyle
i\hbar(\frac{\psi_{\boldsymbol{k}}^{a}(\boldsymbol{r}+a\boldsymbol{\hat{y}})-\psi_{\boldsymbol{k}}^{a}(\boldsymbol{r}-a\boldsymbol{\hat{y}})}{2a})$
$\displaystyle+$
$\displaystyle\frac{e}{c}(A_{u}^{x}+iA_{u}^{y}+A_{p}^{x}+iA_{p}^{y})\psi_{k}^{a}(\boldsymbol{r})\bigg{]}$

In a similar manner $\psi_{\boldsymbol{k}}^{d}$ also can be determined from
$\psi_{\boldsymbol{k}}^{c}$. Thus all the eigenfunctions can be determined.

###  II.3 Energy spectrum- Effective two-band model in 2D periodic magnetic
modulation

In the previous section we studied the BLG spectrum in a periodic magnetic
modulation using the four-band model. In several BLG problems, an effective
two-band model is used Geimbilayer ; Falkobilayer  to analyse the behaviour of
the charge carriers in BLG in an external field. In such two-band model, the
low energy hamiltonian in effective-mass approximation is further projected on
the lowest two energy bands McCann . In this section we shall study the
magnetically modulated BLG spectrum using this effective two-band model within
the framework of our algebraic approach.

Using the form of the effective $2\times 2$ hamiltonian at low energies for
BLG charge carriers in the presence of 2d periodic magnetic modulation, we can
write the eigenvalue equation as

$\displaystyle\frac{1}{2m}\left[{\begin{array}[]{cc}0&(\Pi^{\dagger})^{2}\\\
(\Pi)^{2}&0\\\ \end{array}}\right]\begin{bmatrix}\psi_{1}\\\
\psi_{2}\end{bmatrix}=E\begin{bmatrix}\psi_{1}\\\ \psi_{2}\end{bmatrix}$ (25)

Using the commutator

$[\Pi^{\dagger},\Pi]=\frac{2\hbar e}{c}(B_{u}+B_{p})$ (26)

a straightforward decoupling yields (we retain the same notation for the
hamiltonian and the energy eigenvalues)

$H\psi_{1}=E^{2}\psi_{1}$ (27)

with

$\displaystyle H$ $\displaystyle=$
$\displaystyle\frac{1}{4m^{2}}(\Pi^{\dagger})^{2}(\Pi)^{2}=\frac{1}{4m^{2}}(\Pi^{\dagger})\Big{(}[\Pi^{\dagger},\Pi]+\Pi\Pi^{\dagger}\Big{)}(\Pi)$
(28) $\displaystyle=$
$\displaystyle\frac{1}{4m^{2}}\Big{[}\frac{H_{G}^{2}}{v_{F}^{4}}+\Big{(}\frac{2\hbar
e}{c}(B_{u}+B_{p})\Big{)}\frac{H_{G}}{v_{F}^{2}}+\frac{2\hbar
e}{c}\Big{(}\Pi^{\dagger}(B_{u}+B_{p})\Big{)}\Pi\Big{]}$

Under the approximation that the magnetic field is weak and slowly varying, we
can ignore the last term appearing in the above hamiltonian which invokes the
following argument. In the current problem $B_{u}+B_{p}=B(\vec{r})$ is
spatially constant both inside and outside circular regions, whereas at the
boundary of the two regions it is a step function and hence strictly speaking
it’s derivative will give discontinuity in the form of a delta function. But
in reality there will be an intermediate region where the magnetic field will
smoothly go from $B_{1}$ to $B_{2}$ over a finite region that is sufficiently
large as compared to the bond-length of graphene, but sufficiently small
compared to the relevant length scales that appear in the current problem.
Therefore we can set

$\Big{(}p_{x}-ip_{y}\Big{)}\Big{(}B_{u}+B_{p}\Big{)}\approx 0$ (29)

throughout the considered region. Moreover, we also impose the condition
$|\frac{ea^{2}}{\hbar c}B(\vec{r})|\ll 1$ and $|\frac{eaA_{x,y}}{\hbar c}|\ll
1$. Given this

$\frac{2a^{4}e}{\hbar^{3}c}\Big{(}\Pi^{\dagger}(B_{u}+B_{p})\Big{)}\Pi\\\
=\frac{2a^{4}e}{\hbar^{3}c}\Big{[}\Big{(}p_{x}-ip_{y}\Big{)}\Big{(}B_{u}+B_{p}\Big{)}+\Big{(}\frac{e}{c}(A_{x}-iA_{y})\Big{)}\Big{(}B_{u}+B_{p}\Big{)}\Big{]}\Pi\approx=0$
(30)

The hamiltonian $H$ can therefore be approximated as

$H\approx\Big{[}\frac{H_{G}^{2}}{v_{F}^{4}}+\Big{(}\frac{2\hbar
e}{c}(B_{u}+B_{p})\Big{)}\frac{H_{G}}{v_{F}^{2}}\Big{]}=H^{\prime}$ (31)

and the eigenvalue equation to be solved is

$H^{\prime}\psi_{1}=4m^{2}E^{2}\psi_{1}$

Since $[H^{\prime},H_{G}]=0$, $\psi_{1}$ can be taken same as the
eigenfunction of $H_{G}$. The other component of the spinorial wavefunction
$\psi_{2}$ can be written as

$\psi_{2}=\frac{1}{2mE}\hat{\Pi}^{2}\psi_{1}$ (32)

$E$ can now be determined as

$E=\pm\frac{\hbar^{2}}{2ma^{2}}\sqrt{\epsilon_{M}^{2}+4\pi\frac{(B_{u}+B_{p})a^{2}}{\phi_{0}}\epsilon_{M}}$
(33)

$\epsilon_{M}$ is the dimensionless energy of the MLG in the same magnetic
modulation. It is interesting to note that similar to the case of dispersion
obtained from the effective two-band low energy description of BLG, here also
the energy scales in the unit of $\frac{\hbar^{2}}{2ma^{2}}$. We can therefore
set the dimensionless energy here as
$\epsilon=\frac{E}{\frac{\hbar^{2}}{2ma^{2}}}$. This points out to the fact
that the quasiparticles are massive, unlike the case of magnetically modulated
MLG where the quasiparticles are massless. The energy spectrum is also plotted
in the same unit in Fig. 7. In this case we get two copies of the Hofstadter
butterfly spectrum which is consistent with the two-band approximation. The
analytical form of the energy spectrum in a magnetically modulated Bernal
stacked BLG, derived respectively under four band and the two band
approximation and given in Eq. (16) and Eq. (33), forms one of the major
result in this work.

###  II.4 Approximation involved in obtaining the Hofstadter spectrum for
magnetically modulated graphene layers

The spectrum presented in Figs. 2, 4, 4, 5, 6, 7 are all enumerated with the
help of Eq. (20) which is a discrete eigenvalue equation. This is derived from
the eigenvalue Eq. 18 by first expanding the solutions in terms of gauged
Wannier functions defined in Eq. (19) and then using the nearest neighbour
tight-binding approximation. As a result, for the MLG spectrum we obtained
only two copies of the Hofstatdter butterflies both of which are bounded, and
the BLG spectrum was obtained using the expression from the MLG spectrum by
use of relation (16) and Eq. (33). It may be recalled BookChap  that the
eigenvalue equation (20) is similar in form to the celebrated Harper equation
Harper  that was obtained in the original work of Hofstadter Hofstadter , by
considering the single lowest band of Bloch equation in the presence of a
uniform magnetic field. In absence of the magnetic field, the band dispersion
for such a single band in a square lattice periodic potential in the tight
binding approximation is given by

$E(k_{x},k_{y})=2t(\cos k_{x}a_{sl}+\cos k_{y}a_{sl})$ (34)

where $k_{x},k_{y}$ are Bloch momentums along $x,y$ directions respectively,
$t$ is the energy scale set by hopping and $a_{sl}$ is the lattice constant of
the square lattice. As can be seen from this expression, the width of such
single band is finite, resulting in the finiteness of the energy spectrum
given by the Hofstadter butterfly when the effect of a uniform magnetic field
is added through Peierls’ substitution. This situation may be contrasted
carefully with the approach in the current problem where we start from the
low-energy dispersion of Bernal-stacked bi-layer graphene or monolayer
graphene in the $\boldsymbol{k}\cdot\boldsymbol{p}$ approximation that are
respectively quadratic McCann  and linear Novoselov1  in quasi
(Bloch)-momentum and hence unbounded. When a periodic magnetic modulation
given by Eq. 4 is added to such problem, it is expected that the resulting
spectrum should show higher energy copies of the Hofstadter butterflies
corresponding to the higher-energy part of the spectrum of the unperturbed
problem. However, as can be seen, the spectrum presented in Figs. 2, 4, 4, 5,
6, 7 are all bounded.

The absence of these higher energy copies in our calculation can be explained
by noting our restriction to the nearest neighbour tight-binding which is
clearly visible in the form of Eq. (20). Here we first provide a brief
justification for ignoring the higher order terms in the tight-binding
approximation. Then we shall write some comments about how such restrictions
can be relaxed and more accurate spectrum can be obtained. As already
mentioned that in the current problem we use the weak field limit
$a\ll\ell_{B_{u}}$, where $a$ accounts for the periodicity of the periodic
magnetic modulation studied here. Even though the effective bilayer graphene
hamiltonian is continuous and hence unbounded, the lattice present due to the
periodic modulation justifies the discretisation and the nearest neighbour
truncation of the resultant equation under this weak field limit. A single
dispersion like Eq. (34) for the current problem by considering only the
$\boldsymbol{B}_{p}$ in Eq. 4 is also a non-trivial problem, because the
lattice parameter $a$ here is not the carbon-carbon bond length, and is purely
due to the magnetic modulation. To extend the current calculation in the
strong field limit (where the magnetic length is comparable or may be even
lower than the lattice-spacing) one can follow the methods described in ref.
Janecek  where the Scrhödinger equation in a periodic potential and magnetic
field is numerically diagonalised exploiting the magnetic translational
symmetry of the problem. Alternatively one can use the method adopted in ref.
Gumbs  by first solving the corresponding problem in an uniform magnetic field
Zarenia1  and then applying the zero-flux periodic magnetic modulation
represented by the fiield $B_{p}\hat{z}$ as a perturbation. Direct relaxation
of our scheme can also be used, however a comparison between these various
schemes is not clear at this stage. All these methods are numerically
demanding and we hope to address some of them in future work.

##  III Topological quantization of Hall conductance for the modulated BLG-
four band model

The Landau level spectrum in a transverse uniform magnetic field from the four
band hamiltonian (1) is given by Zarenia1

$E_{LL}^{\text{4band}}=s_{1}[\frac{t^{2}}{2}+2N+1+s_{2}\sqrt{\frac{t^{4}}{4}+(2N+1)t^{2}+1}]^{\frac{1}{2}}$
(35)

This expression may be compared with the expression (16) that gives the
corresponding spectrum under periodic magnetic modulation considered in the
current work. Here $s_{1}=\pm{1}$ and $s_{2}=\pm{1}$ and $N$ is the Landau
level index. The experimental results of quantised Hall conductivity in Bernal
stacked bilayer graphene reported in the pioneering work Geimbilayer  was
however theoretically analysed Falkobilayer ; McCann  by the low energy two
band hamiltonian and its Landau level spectrum for the AB (Bernal) stacked BLG
in a transverse uniform magnetic field. The corresponding spectrum appears
very different from the one given in (35) and is given by

$E=\pm\hbar\omega_{c}\sqrt{N(N-1)},~{}~{}N\geq 2$ (36)

where $N$ is the Landau level index and $E_{0}=E_{1}=0$. Particularly in the
expression (36) for zero energy eigenvalues thus there are two levels
respectively for $N=0,1$. This causes the double degeneracy. The valley
structure and electronic spin degrees of freedom, each contributes a double
degeneracy to the Landau level spectrum in (36). Thus each level in bilayer
graphene is four-fold degenerate except the zero energy level that shows an
eight-fold degeneracy. Accordingly the quantized Hall conductivity in bilayer
graphene Geimbilayer ; Falkobilayer ; Novoselov1  occurs at

$\sigma_{xy}=4N\frac{e^{2}}{h}$ (37)

where $N\in Z$ but $N\neq 0$. As has been pointed out McCann  that this
quantization rule resembles that for 2DEG in a conventional semiconductor with
degeneracy $4$ coming from the spin and valley degrees of freedom. But, with
the exception that here $N=0$ plateau is missing and there is a step of height
$\frac{8e^{2}}{\hbar}$ across the zero-electron density (the Hall conductivity
being plotted at a fixed magnetic field). Additionally a crucial assumption in
these works is that the degeneracy of the Landau level is preserved and any
splitting of the Landau level ( such as Zeeman splitting) is negligible as
compared to the level broadening ( due to disorder etc.) and temperature
effect in the experiment.

A direct comparison between these mentioned results and that of ours, namely
the fate of quantised Hall conductivity in the presence of periodic magnetic
modulation is somewhat out of scope of the current work. This is due to the
fact that the model hamiltonian (1) is based on a single valley and single
spin. We assumed that the magnetic modulation given in Eq. (5) is sufficiently
strong to freeze the spin-degrees of freedom and there is no term that allows
scattering between the two valleys. Given this form if we extend the treatment
of the TKNN model TKNN ; Kohmoto  to the current problem described by the
hamiltonian (1) we can demonstrate the topological quantisation of the Hall
conductivity for the same without any anomalous behaviour at the zero density
as expected from the spectrum given in (16). Accordingly in this section, we
shall derive the topological quantization of Hall conductivity for the BLG
under two dimensional periodic magnetic field, using the four band model.
Extension of this treatment using an effective two-band model described in
section II.3 and including spin and valley degeneracy is not considered in the
current work.

For the magnetic modulation defined in Eq. (5), the eigenfunctions of the
hamiltonian (1) are simultaneous eigenfunctions for the magnetic translation
group defined for the uniform magnetic field part $B_{u}$ of the magnetic
modulation, and hence can be written as

$\displaystyle\begin{bmatrix}\psi^{a1}_{k_{x},k_{y}}\\\
\psi^{b2}_{k_{x},k_{y}}\\\ \psi^{a2}_{k_{x},k_{y}}\\\
\psi^{b1}_{k_{x},k_{y}}\end{bmatrix}=e^{i\boldsymbol{k}.\boldsymbol{r}}\begin{bmatrix}u^{a1}_{k_{x},k_{y}}\\\
u^{b2}_{k_{x},k_{y}}\\\ u^{a2}_{k_{x},k_{y}}\\\
u^{b1}_{k_{x},k_{y}}\end{bmatrix}$ (38)

Since $B_{u}\neq 0$, the commutation relation between the magnetic translation
operator ensures that each magnetic unit cell contains zero’s of such wave-
function and thus Kohmoto

$u^{a,b}_{k_{x},k_{y}}(x,y)=|u^{a,b}_{k_{x},k_{y}}|e^{i\theta^{a,b}_{k_{x},k_{y}}}$
(39)

where the functions $u^{a,b}_{k_{x},k_{y}}(x,y)$ can be found out by the
solving the Schrödinger equation and the full four-component wave-function is
normalised to $1$.

$H(k_{x},k_{y})u^{n_{b}}_{k_{x},k_{y}}=E^{n_{b}}u^{n_{b}}_{k_{x},k_{y}}$ (40)

with $n_{b}$ being the band-index. For the current problem we solve for a
single band and hence the band index is omitted. Rest of the derivation that
will lead to the expression of Hall conductivity is very similar to the one
carried out for the non-relativistic two dimensional gas in the TKNN work
Kohmoto  or the one done for MLG in such magnetic modulation Manisha , but
with an important difference coming from the 4-component pseudo-spinorial
nature of the magnetic Bloch function. It may be noted that this four
component nature of the magnetic Bloch function is due to the existence of the
two layers (each having two sublattices conventionally known as $A$ and $B$)
in the bilayer graphene McCann  and not related with spin or valley
degeneracy. This directly follows from the hamiltonian (1) which is written
for single valley and single spin. We therefore provide here some essential
steps of the derivation for the Hall conductivity involving this four
component pseudo-spinor for the magnetically modulated BLG

To this purpose we first define
$u^{\prime}_{k_{x},k_{y}}=\begin{bmatrix}u^{a1}_{k_{x},k_{y}}&u^{b2}_{k_{x},k_{y}}&u^{a2}_{k_{x},k_{y}}&u^{b1}_{k_{x},k_{y}}\end{bmatrix}^{T}$.
Within the linear response regime, the Hall conductance given by Kubo formula
for a filled single band can be written as

$\sigma_{xy}=\frac{e^{2}}{h}\frac{1}{2\pi i}\int
d^{2}k[\boldsymbol{\nabla_{k}}\times\boldsymbol{\hat{A}}(k_{x},k_{y})]_{3}$
(41)

where the integration is over the magnetic Brillouin zone (MBZ) and

$\boldsymbol{\hat{A}}(k_{x},k_{y})=\int
d^{2}ru^{\prime*}_{k_{x},k_{y}}\boldsymbol{\nabla_{k}}u^{\prime}_{k_{x},k_{y}}$
(42)

Using Stoke’s law, Eq. $(\ref{5014})$ yields

$\sigma_{xy}=\frac{e^{2}}{h}\frac{1}{2\pi i}\int_{\partial
MBZ}\boldsymbol{dk}.\boldsymbol{\hat{A}}(k_{x},k_{y})$ (43)

To understand the non trivial topology of $\boldsymbol{\hat{A}}(k_{x},k_{y})$,
we consider the gauge transformation

$\displaystyle\begin{bmatrix}u^{a1a}_{k_{x},k_{y}}\\\ u^{b2b}_{k_{x},k_{y}}\\\
u^{a2a}_{k_{x},k_{y}}\\\
u^{b1b}_{k_{x},k_{y}}\end{bmatrix}=e^{if(k_{x},k_{y})}\begin{bmatrix}u^{a1}_{k_{x},k_{y}}\\\
u^{b2}_{k_{x},k_{y}}\\\ u^{a2}_{k_{x},k_{y}}\\\
u^{b1}_{k_{x},k_{y}}\end{bmatrix}$ (44)

where $f(k_{x},k_{y})$ is some arbitrary smooth function of only $k_{x}$ and
$k_{y}$. We divide the MBZ into two regions, one contains the zero of the
wavefunction and other does not. For region I, we can have

$\begin{bmatrix}u^{a11}_{k_{x},k_{y}}\\\ u^{b21}_{k_{x},k_{y}}\\\
u^{a21}_{k_{x},k_{y}}\\\
u^{b11}_{k_{x},k_{y}}\end{bmatrix}=e^{ig(k_{x},k_{y})}\begin{bmatrix}u^{a1}_{k_{x},k_{y}}\\\
u^{b2}_{k_{x},k_{y}}\\\ u^{a2}_{k_{x},k_{y}}\\\
u^{b1}_{k_{x},k_{y}}\end{bmatrix}$

and for region II

$\begin{bmatrix}u^{a12}_{k_{x},k_{y}}\\\ u^{b22}_{k_{x},k_{y}}\\\
u^{a22}_{k_{x},k_{y}}\\\
u^{b12}_{k_{x},k_{y}}\end{bmatrix}=e^{ih(k_{x},k_{y})}\begin{bmatrix}u^{a1}_{k_{x},k_{y}}\\\
u^{b2}_{k_{x},k_{y}}\\\ u^{a2}_{k_{x},k_{y}}\\\
u^{b1}_{k_{x},k_{y}}\end{bmatrix}$

Hence Eq. $(\ref{5014})$ becomes

$\displaystyle\sigma_{xy}$ $\displaystyle=$
$\displaystyle\frac{e^{2}}{h}\frac{1}{2\pi
i}\Big{[}\int_{H_{1}}d^{2}k[\boldsymbol{\nabla}\times\boldsymbol{\hat{A}}^{1}(k_{x},k_{y})]_{3}$
(45) $\displaystyle+$
$\displaystyle\int_{H_{2}}d^{2}k[\boldsymbol{\nabla}\times\boldsymbol{\hat{A}}^{2}(k_{x},k_{y})]_{3}\Big{]}$

where the vectors $\boldsymbol{\hat{A}}^{1}(k_{x},k_{y})$ and
$\boldsymbol{\hat{A}}^{2}(k_{x},k_{y})$ are defined for regions I and II
respectively, and are given as

$\displaystyle\boldsymbol{\hat{A}}^{1}(k_{x},k_{y})$ $\displaystyle=$
$\displaystyle\int d^{2}r[\boldsymbol{\hat{k_{x}}}(i\frac{\partial g}{\partial
k_{x}}u^{a1*}_{k_{x},k_{y}}u^{a1}_{k_{x},k_{y}}+i\frac{\partial g}{\partial
k_{x}}u^{b2*}_{k_{x},k_{y}}u^{b2}_{k_{x},k_{y}}+u^{a1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{a1}_{k_{x},k_{y}}$ (46) $\displaystyle+$ $\displaystyle
u^{b2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{b2}_{k_{x},k_{y}}+i\frac{\partial g}{\partial
k_{x}}u^{a2*}_{k_{x},k_{y}}u^{a2}_{k_{x},k_{y}}+i\frac{\partial g}{\partial
k_{x}}u^{b1*}_{k_{x},k_{y}}u^{b1}_{k_{x},k_{y}}+u^{a2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{a2}_{k_{x},k_{y}}$ $\displaystyle+$ $\displaystyle
u^{b1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{b1}_{k_{x},k_{y}})+\boldsymbol{\hat{k_{y}}}(i\frac{\partial
g}{\partial k_{y}}u^{a1*}_{k_{x},k_{y}}u^{a1}_{k_{x},k_{y}}+i\frac{\partial
g}{\partial k_{y}}u^{b2*}_{k_{x},k_{y}}u^{b2}_{k_{x},k_{y}}$ $\displaystyle+$
$\displaystyle u^{a1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{a1}_{k_{x},k_{y}}+u^{b2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{b2}_{k_{x},k_{y}}+i\frac{\partial g}{\partial
k_{y}}u^{a2*}_{k_{x},k_{y}}u^{a2}_{k_{x},k_{y}}+i\frac{\partial g}{\partial
k_{y}}u^{b1*}_{k_{x},k_{y}}u^{b1}_{k_{x},k_{y}}$ $\displaystyle+$
$\displaystyle u^{a2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{a2}_{k_{x},k_{y}}+u^{b1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{b1}_{k_{x},k_{y}})]$

and

$\displaystyle\boldsymbol{\hat{A}}^{2}(k_{x},k_{y})$ $\displaystyle=$
$\displaystyle\int d^{2}r[\boldsymbol{\hat{k_{x}}}(i\frac{\partial h}{\partial
k_{x}}u^{a1*}_{k_{x},k_{y}}u^{a1}_{k_{x},k_{y}}+i\frac{\partial h}{\partial
k_{x}}u^{b2*}_{k_{x},k_{y}}u^{b2}_{k_{x},k_{y}}+u^{a1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{a1}_{k_{x},k_{y}}$ (47) $\displaystyle+$ $\displaystyle
u^{b2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{b2}_{k_{x},k_{y}}+i\frac{\partial h}{\partial
k_{x}}u^{a2*}_{k_{x},k_{y}}u^{a2}_{k_{x},k_{y}}+i\frac{\partial h}{\partial
k_{x}}u^{b1*}_{k_{x},k_{y}}u^{b1}_{k_{x},k_{y}}+u^{a2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{a2}_{k_{x},k_{y}}$ $\displaystyle+$ $\displaystyle
u^{b1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{x}}u^{b1}_{k_{x},k_{y}})+\boldsymbol{\hat{k_{y}}}(i\frac{\partial
h}{\partial k_{y}}u^{a1*}_{k_{x},k_{y}}u^{a1}_{k_{x},k_{y}}+i\frac{\partial
h}{\partial k_{y}}u^{b2*}_{k_{x},k_{y}}u^{b2}_{k_{x},k_{y}}$ $\displaystyle+$
$\displaystyle u^{a1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{a1}_{k_{x},k_{y}}+u^{b2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{b2}_{k_{x},k_{y}}+i\frac{\partial h}{\partial
k_{y}}u^{a2*}_{k_{x},k_{y}}u^{a2}_{k_{x},k_{y}}+i\frac{\partial h}{\partial
k_{y}}u^{b1*}_{k_{x},k_{y}}u^{b1}_{k_{x},k_{y}}$ $\displaystyle+$
$\displaystyle u^{a2*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{a2}_{k_{x},k_{y}}+u^{b1*}_{k_{x},k_{y}}\frac{\partial}{\partial
k_{y}}u^{b1}_{k_{x},k_{y}})]$

Since the full four-component spinorial wavefunction in Eq. (38) is
normalised, a straightforward algebra now gives

$\boldsymbol{\hat{A}}^{1}(k_{x},k_{y})-\boldsymbol{\hat{A}}^{2}(k_{x},k_{y})=i\boldsymbol{\nabla_{k}}t(\boldsymbol{k})$
(48)

where $t(\boldsymbol{k})=g(\boldsymbol{k})-h(\boldsymbol{k})$.

We now write, using Stokes theorem and the fact that the two regions have
opposite directions for circulation, the Eq. $(\ref{5024})$ as

$\sigma_{xy}=\frac{e^{2}}{h}\frac{1}{2\pi
i}\int_{C}\boldsymbol{dk}.(\boldsymbol{\hat{A}}^{1}(k_{x},k_{y})-\boldsymbol{\hat{A}}^{2}(k_{x},k_{y}))$

which on using Eq. $(\ref{5044})$ leads to

$\sigma_{xy}=\frac{e^{2}}{h}\frac{1}{2\pi}\int_{C}\boldsymbol{dk}.\boldsymbol{\nabla_{k}}t(\boldsymbol{k})$
(49)

where $C$ denotes closed boundary between the two regions. At each point on
the closed loop C, the wavefunction has to be single valued therefore after
traversing the complete loop C the two wavefunctions still should have the
same phase relationship. This is possible only if

$\begin{bmatrix}u^{a12}_{k_{x},k_{y}}\\\ u^{b22}_{k_{x},k_{y}}\\\
u^{a22}_{k_{x},k_{y}}\\\
u^{b12}_{k_{x},k_{y}}\end{bmatrix}=e^{i(t(k_{x},k_{y})+2\pi\sigma_{H})}.\begin{bmatrix}u^{a11}_{k_{x},k_{y}}\\\
u^{b21}_{k_{x},k_{y}}\\\ u^{a21}_{k_{x},k_{y}}\\\
u^{b11}_{k_{x},k_{y}}\end{bmatrix}$

Thus expression $(\ref{sigma42})$ finally comes out to be

$\sigma_{xy}=\frac{e^{2}}{h}\sigma_{H}$ (50)

Figure 8: Quantum Hall plateaus in bilayer graphene in periodic magnetic
modulation at zero temperature: The y-axis plots the quantised Hall
resistivity and the $x$-axis plots the magnetic field strength $B_{u}$ at a
fixed electron density $n$. All other scaling factors are shown in the axis
label and have their usual meaning.

where $\sigma_{H}\in Z$. Some remarks on this finding of our work are in
order. As explained earlier, this four component nature of the wave function
stems from the existence of two layers each having two sub-lattices. We have
not considered the effect of the valley and spin in our model. Unlike in the
case of the Landau level spectrum given by the two band hamiltonian (36) in
the presence of a uniform transverse magnetic field considered in Falkobilayer
; McCann2 , the relevant energy spectrum for the four band model given in (16)
does not distinguish between the $0,1$ LL index with the rest. As a result the
quantised Hall plateaus in our case are always equally spaced. In this
connection it may be pointed out that the lifting of the degeneracy of $0,1$
LL index was earlier predicted with the application of an out of plane
electric field to the bilayer graphene in addition to the transverse magnetic
field and was observed experimentally Geim3 . The current situation is
different from that. Thus in a way quite similar to the case of two-
dimensional electron system in a uniform magnetic field and a periodic lattice
potential TKNN ; Kohmoto , if the Fermi energy lies in one of the gaps of the
spectrum, the Hall conductivity is quantised in terms of an integer. Our
treatment of evaluating quantised Hall conductivity is also different from the
method adopted in ref. Falkobilayer , where they have used the two-band model
to evaluate the the quantised Hall effect. However we expect that even in that
calculation, the periodic magnetic modulation should lift the zero level
anomaly. This expectation is based on the expression of the energy spectrum
given in (33) within the two-band model in the presence of the periodic
modulation that does not show any $N=0$ degeneracy that appears when the field
is uniform. We have not considered this problem any further here. Such
topological quantisation is naturally robust against effect of disorder or any
other perturbation as long as it does not close the gap. As a notional
demonstration we have plotted such quantised Hall conductivity plateaus given
in Eq. (50) at absolute zero temperature in Fig. (8). Please note the plotted
quantities are not directly calculated from Kubo formula, within linear
response regime, but the plot of $\sigma_{H}$ given in Eq. (50) at the
corresponding values of $B_{u}$. The construction of the plateaus are done
just by connecting the intermediate regions with lines parallel to $x$-axis.
Under ideal situation this represents the zero temperature quantised Hall
conductivity plot and hence the phrase ”notional demonstration” .

To summarise, we demonstrated rigorously that the quantised Hall conductance
can survive in Bernal stacked bilayer graphene when the magnetic field is
periodically modulated. We pointed out that this is a result of the
preservation of magnetic translational symmetry in the resulting system, and
provides some algebraic results for the spectrum. We have pointed out that in
the framework of considered theoretical model in this work, the resulting Hall
conductivity in our system is quantised in the unit of $\frac{e^{2}}{h}$ for
all quantum numbers for the quantised Hall conductivity, unlike the anomalous
behaviour of the quantised Hall conductivity in bilayer graphene Geimbilayer
in uniform magnetic field due to a degeneracy between $0,1$ LL index. The type
of magnetic modulation that has been considered in this work can be created in
a number of ways. Such techniques span from the use of nano-lithography Bader
to more recent days structure such as two dimensional magnetic van der Waals
materials Vanm  or by arranging ad-atoms of a ferromagnetic material in
periodic arrays on a suitable surface adatom , to name a few. The presence of
periodic pseudo-magnetic field (superlattice) was experimentally confirmed in
buckled monolayer graphene Mao . The order of lattice constant in such
superlattices is of the order of $40$ nm and the field strength is $\sim
100~{}\text{Tesla}$. Several other recent experiments also realised graphene
in periodic magnetic modulation by subjecting graphene to strain induced
superlattice strain , or by exposing epitaxial graphene to an abrikosov vortex
lattice created on a type II superconductor sc . In the work strain , it was
demonstrated that periodic modulation of strain applied to graphene modulates
the $C-C$ bond and gives rise to intense pseudo-magnetic field superlattice of
lattice constant $1-10$ nm and field strength $\sim 100~{}\text{Tesla}$. Where
in some such cases, the weak field limit that we adopt in this work is
approximately respected, in other cases the strength of the modulation is
stronger. As discussied in sec. II.4, the nearest neighbour tight-binding
approximation that is used in the current work need to be relaxed in the
stronger field cases, which will show higher energy copies of the Hofstatder
butterflies. We can therefore hope that our work may stimulate new
experimental works and further theoretical analysis in future in similar
systems.

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# Characterization and examples of commutative isoartinian rings

Asghar Daneshvar and Kamran Divaani-Aazar  A. Daneshvar, Department of
Mathematics, Faculty of Mathematical Sciences, Alzahra University, Tehran,
Iran<EMAIL_ADDRESS><EMAIL_ADDRESS>K. Divaani-Aazar,
Department of Mathematics, Faculty of Mathematical Sciences, Alzahra
University, Tehran, Iran<EMAIL_ADDRESS>
###### Abstract.

Noetherian rings have played a fundamental role in commutative algebra,
algebraic number theory, and algebraic geometry. Along with their dual,
Artinian rings, they have many generalizations, including the notions of
isonoetherian and isoartinian rings. In this paper, we prove that the Krull
dimension of every isoartinian ring is at most one. We then use this result to
provide a characterization of isoartinian rings. Specifically, we prove that a
ring $R$ is isoartinian if and only if $R$ is uniquely isomorphic to the
direct product of a finite number of rings of the following types: (i)
Artinian local rings; (ii) non-Noetherian isoartinian local rings with a
nilpotent maximal ideal; (iii) non-field principal ideal domains; (iv)
Noetherian isoartinian rings $A$ with $\operatorname{Min}A$ being a singleton
and $\operatorname{Min}A\subsetneq\operatorname{Ass}A$; (v) non-Noetherian
isoartinian rings $A$ with $\operatorname{Min}A$ being a singleton and
$\operatorname{Min}A\subsetneq\operatorname{Ass}A$; (vi) non-Noetherian
isoartinian rings $A$ with a unique element in $\operatorname{Min}A$ that is
not maximal, and $\operatorname{Min}A=\operatorname{Ass}A$. Several examples
of these types of rings are also provided.

###### Key words and phrases:

Isoartinian ring; Marot ring; perfect ring; principal ideal domain; Prüfer
ring; subperfect ring.

###### 2020 Mathematics Subject Classification:

13C05; 16P20; 13E10; 16P60; 13E05.

##  1\. Introduction

Throughout this article, the term “ring” refers to commutative rings with
nonzero identity. In 1890, David Hilbert proved his celebrated basis theorem,
which states that every ideal of a polynomial ring with coefficients in a
field or in the ring of integers is finitely generated. Then in 1921, Emmy
Noether showed out that in a ring $R$, every ideal of $R$ is finitely
generated if and only if $R$ satisfies ACC on its ideals, meaning that every
ascending chain of ideals of $R$ eventually terminates. Noether’s subsequent
deep research had a significant impact on module theory and representation
theory. Assuming ACC on a ring leads to many beautiful properties. For
instance, Emanuel Lasker and Noether demonstrated that every ideal of a ring
satisfying ACC has a primary decomposition, which is a generalization of the
fundamental theorem of arithmetic.

In 1927, Emil Artin began studying rings with both ACC and DCC properties,
which led to several interesting applications of ACC property. It was clear
from simple examples that ACC does not imply DCC, so Artin assumed both ACC
and DCC. At that time, no example of a ring satisfying DCC but not ACC was
known. However, in 1935, Yasuo Akizuki proved that for a ring $R$, DCC implies
ACC. In honor of Noether and Artin, the rings satisfying ACC (respectively,
DCC) are now referred to as Noetherian (respectively, Artinian) rings.

The method of decomposition is a common approach in mathematical research,
which enables the transfer of properties from basic building blocks to more
complex objects and vice versa. This technique has been used for a long time,
with the decomposition of natural numbers into prime factors being one of the
earliest examples. Algebra is no exception, and the fundamental theorem of
finitely generated Abelian groups is just one of many instances of this
approach. Another famous example is the theorem attributed to Akizuki-Cohen,
which states that an Artinian ring is uniquely isomorphic to the direct
product of a finite number of Artinian local rings.

Due to the importance and wide range of applications of Noetherian and
Artinian rings, many authors have proposed and examined some generalizations
of these concepts. One of the most natural generalizations was introduced by
Alberto Facchini and Zahra Nazemian in 2016 [8]. They define a ring $R$ to be
isonoetherian (respectively, isoartinian) if every ascending (respectively,
descending) chain of ideals of $R$ terminates up to isomorphism. Several
interesting results and examples pertaining to these types of rings can be
found in [8]. To explore the topic of isonoetherian and isoartinian rings in
more depth, we recommend consulting the following references: [6, 7, 8, 2, 3].

Numerous natural questions arise regarding isonoetherian and isoartinian
rings. For instance, it is unknown whether an isoartinian ring is necessarily
isonoetherian. In [8, Corollary 4.8], Facchini and Nazemian showed that a
reduced isoartinian ring is isomorphic to a finite direct product of principal
ideal domains. In particular, every reduced isoartinian ring is Noetherian.
The aim of this paper is to establish a decomposition result for general
isoartinian rings. We prove that:

######  Theorem 1.1.

A ring $R$ is isoartinian if and only if it is uniquely isomorphic to the
direct product of a finite number of the following types of rings with various
choices:

  1. (i)

Artinian local rings.

  2. (ii)

non-Noetherian isoartinian local rings with nilpotent maximal ideal.

  3. (iii)

non-field principal ideal domains.

  4. (iv)

Noetherian isoartinian rings $A$ with a unique minimal prime ideal
$\mathfrak{p}$ such that $A/\mathfrak{p}$ is a principal ideal domain and
$\operatorname{Min}A\varsubsetneq\operatorname{Ass}A$.

  5. (v)

non-Noetherian isoartinian rings $A$ with a unique minimal prime ideal
$\mathfrak{p}$ such that $A/\mathfrak{p}$ is a principal ideal domain and
$\operatorname{Min}A\varsubsetneq\operatorname{Ass}A$.

  6. (vi)

non-Noetherian isoartinian rings $A$ with a unique minimal non-maximal prime
ideal $\mathfrak{p}$ such that $A/\mathfrak{p}$ is a principal ideal domain
and $\operatorname{Min}A=\operatorname{Ass}A$.

In the proof of this theorem, a key ingredient is to demonstrate that for an
isoartinian ring $R$, the set of associated primes of $R$ is finite, and every
minimal prime ideal of $R$ is also an associated prime of $R$. This implies,
in particular, that the Krull dimension of every isoartinian ring is at most
one.

While there are many examples of rings falling under types (i) and (iii) in
the above theorem, the other four types of isoartinian rings deserve more
attention. We provide some examples of these types. One of these examples
demonstrates that isoartinian rings may not be isonoetherian. Furthermore, we
provide some examples to show that the condition of being “isoartinian” cannot
be relaxed for types (ii), (v) and (vi).

##  2\. Main results

In the proof of our main result, we will utilize the fact that the Krull
dimension of every isoartinian ring is at most one, as stated in Proposition
2.3. To demonstrate this, we will rely on the following two lemmas.

We denote the set of associated primes of $R$ by $\operatorname{Ass}R$, and we
use $Z(R)$ to represent the set of zero-divisors of $R$. An ideal $I$ of a
ring $R$ is said to be regular if it contains a regular element of $R$. A ring
$R$ is called Marot if every regular ideal of $R$ can be generated by regular
elements. It is well-known that if $Z(R)$ is a finite union of prime ideals,
then $R$ is a Marot ring.

######  Lemma 2.1.

Let $R$ be an isoartinian ring. Then $\operatorname{Ass}R$ is nonempty and
finite. In particular, $R$ is a Marot ring.

###### Proof.

Since $R$ is isoartinian, it satisfies both ACC and DCC on annihilator ideals,
as shown in [8, Lemma 4.10(1)]. Let $\Sigma:=\\{(0:_{R}x)\mid 0\neq x\in
R\\}$, which is nonempty since $R$ is nonzero. As $R$ has ACC on annihilator
ideals, $\Sigma$ has a maximal element $(0:_{R}z)$ with respect to inclusion.
One can easily observe that the ideal $(0:_{R}z)$ is prime, and hence
$(0:_{R}z)$ belongs to $\operatorname{Ass}R$.

Now, we show that $\operatorname{Ass}R$ is finite. Suppose to the contrary
that $\operatorname{Ass}R$ is infinite. Then there are prime ideals
$\mathfrak{p}_{1},\mathfrak{p}_{2},\ldots,\mathfrak{p}_{n},\ldots\in\operatorname{Ass}R$
such that $\mathfrak{p}_{1}$ is a maximal element of $\operatorname{Ass}R$ and
$\mathfrak{p}_{n+1}$ is a maximal element of
$\operatorname{Ass}R\setminus\\{\mathfrak{p}_{1},\mathfrak{p}_{2},\ldots,\mathfrak{p}_{n}\\}$
for all $n\geq 1$. Consider the descending chain of ideals

$\mathfrak{p}_{1}\supseteq\mathfrak{p}_{1}\cap\mathfrak{p}_{2}\supseteq\mathfrak{p}_{1}\cap\mathfrak{p}_{2}\cap\ldots\cap\mathfrak{p}_{n}\supseteq\cdots.$

Since $R$ has DCC on annihilator ideals, this chain stabilizes. Thus, there is
a natural number $\ell$ such that
$\mathfrak{p}_{1}\cap\mathfrak{p}_{2}\cap\ldots\cap\mathfrak{p}_{\ell}=\mathfrak{p}_{1}\cap\mathfrak{p}_{2}\cap\ldots\cap\mathfrak{p}_{\ell}\cap\mathfrak{p}_{\ell+1}$.
Thus, $\mathfrak{p}_{i}\subseteq\mathfrak{p}_{\ell+1}$ for some $1\leq
i\leq\ell$. However, $\mathfrak{p}_{i}$ is a maximal element of
$\operatorname{Ass}R\setminus\\{\mathfrak{p}_{1},\mathfrak{p}_{2},\ldots,\mathfrak{p}_{i-1}\\}$,
which implies that $\mathfrak{p}_{i}=\mathfrak{p}_{\ell+1}$. This is a
contradiction, since
$\mathfrak{p}_{\ell+1}\in\operatorname{Ass}R\setminus\\{\mathfrak{p}_{1},\mathfrak{p}_{2},\ldots,\mathfrak{p}_{i},\ldots,\mathfrak{p}_{\ell}\\}$.

Finally, note that $Z(R)$ is the union of members of $\operatorname{Ass}R$,
and that if $Z(R)$ is a finite union of prime ideals, then $R$ is a Marot
ring. Therefore, $R$ is a Marot ring. ∎

We use $\operatorname{Spec}R$ to denote the set of all prime ideals of $R$.
Also, let $\operatorname{Max}R$ (respectively $\operatorname{Min}R$) denote
the set of all maximal (respectively minimal) prime ideals of $R$.

######  Lemma 2.2.

Let $R$ be an isoartinian ring. Then $\operatorname{Min}R$ is a nonempty
subset of $\operatorname{Ass}R$. As a consequence, $\operatorname{Min}R$ is
finite.

###### Proof.

First, note that since $R$ is nonzero, $\operatorname{Min}R$ is nonempty. It
follows from the definition that the localization of an isoartinian ring at
any multiplicative set is also isoartinian, as shown in [4, Lemma 3.2]. Let
$\mathfrak{p}\in\operatorname{Min}R$. Then $\mathfrak{p}R_{\mathfrak{p}}$ is
the unique prime ideal of the isoartinian ring $R_{\mathfrak{p}}$. According
to Lemma 2.1, we have
$\operatorname{Ass}R_{\mathfrak{p}}=\\{\mathfrak{p}R_{\mathfrak{p}}\\}$.
Hence, there exists $x/t\in R_{\mathfrak{p}}$ such that
$\mathfrak{p}R_{\mathfrak{p}}=(0:_{R_{\mathfrak{p}}}x/t)$. Clearly,
$(0:_{R_{\mathfrak{p}}}x/t)=(0:_{R_{\mathfrak{p}}}x/1)$.

Let $\Sigma$ denote the nonempty set of all annihilator ideals of the form
$(0:_{R}I)$, where $I$ is a finitely generated ideal of $R$ and
$I\subseteq\mathfrak{p}$. Since $R$ has DCC on annihilator ideals, $\Sigma$
admits a minimal element $(0:_{R}J)$. We claim that
$(0:_{R}J)=(0:_{R}\mathfrak{p})$. Suppose the contrary holds. Then as
$(0:_{R}\mathfrak{p})\subseteq(0:_{R}J)$, there exists an element
$y\in(0:_{R}J)\setminus(0:_{R}\mathfrak{p})$. Since
$y\notin(0:_{R}\mathfrak{p})$, there exists $a\in\mathfrak{p}$ such that
$ay\neq 0$. As $(0:_{R}J+Ra)\subseteq(0:_{R}J)$, by the minimality of
$(0:_{R}J)$ in $\Sigma$, we deduce that $(0:_{R}J+Ra)=(0:_{R}J)$. Thus,
$y\in(0:_{R}J+Ra)$, which implies $ay=0$, a contradiction.

As
$x/1\in(0:_{R_{\mathfrak{p}}}\mathfrak{p}R_{\mathfrak{p}})\subseteq(0:_{R_{\mathfrak{p}}}JR_{\mathfrak{p}})$
and $J$ is finitely generated, there exists $s\in R\setminus\mathfrak{p}$ such
that $J(sx)=0$. Consequently, $sx\in(0:_{R}J)=(0:_{R}\mathfrak{p})$, and hence
$\mathfrak{p}\subseteq(0:_{R}sx)$. On the other hand, we have

$(0:_{R}x)\subseteq(0:_{R_{\mathfrak{p}}}x/1)\cap
R=\mathfrak{p}R_{\mathfrak{p}}\cap R=\mathfrak{p},$

so $(0:_{R}sx)\subseteq\mathfrak{p}$. Therefore, $\mathfrak{p}=(0:_{R}sx)$,
and so $\mathfrak{p}\in\operatorname{Ass}R$. ∎

We can now prove the claim we made at the beginning of this section:

######  Proposition 2.3.

Let $R$ be an isoartinian ring. Then $R/\mathfrak{p}$ is a principal ideal
domain for every $\mathfrak{p}\in\operatorname{Ass}R$. Consequently $\dim
R\leq 1$, and so
$\operatorname{Spec}R=\operatorname{Min}R\cup\operatorname{Max}R$.

###### Proof.

Let $\mathfrak{p}\in\operatorname{Ass}R$. Then, $\mathfrak{p}=(0:_{R}x)$ for
some nonzero element $x$ of $R$. As $Rx$ is a submodule of $R$ and
$R/\mathfrak{p}\cong Rx$, the ring $R/\mathfrak{p}$ is isoartinian. By [8,
Corollary 4.8], we deduce that $R/\mathfrak{p}$ is a principal ideal domain.

Clearly, $\dim R=\dim R/\mathfrak{q}$ for some
$\mathfrak{q}\in\operatorname{Min}R$. By Lemma 2.2, we have
$\mathfrak{q}\in\operatorname{Ass}R$. As $R/\mathfrak{q}$ is a principal ideal
domain, it follows that $\dim R=\dim R/\mathfrak{q}\leq 1$, as desired. ∎

To prove the main result, we need the following six additional lemmas. We
begin by showing that isoartinian rings are a subclass of a well-known class
of rings called Prüfer rings.

Recall that the localization of a ring $R$ at the set of all regular elements
is called the total ring of fractions of $R$ and is denoted by $T(R)$. An
ideal $I$ of a ring $R$ is said to be invertible if $I^{-1}I=R$, where

$I^{-1}=\\{r\in T(R)\mid rI\subseteq R\\}.$

A ring $R$ for which every finitely generated regular ideal is invertible is
called a Prüfer ring.

######  Lemma 2.4.

Let $R$ be an isoartinian ring. Every regular ideal of $R$ is principal and
generated by a regular element. Consequently, $R$ is a Prüfer ring.

###### Proof.

Let $I$ be a regular ideal of $R$. Then, $I$ contains an element $z$ such that
$(0:_{R}z)=0$. Thus, $I\cong Iz^{k}$ for every natural number $k$. Consider
the following descending chain of ideals of $R$

$R\supseteq I\supseteq Rz\supseteq Iz\supseteq Rz^{2}\supseteq Iz^{2}\supseteq
Rz^{3}\supseteq\cdots.$

As $R$ is isoartinian, there is a natural number $k$ such that $I\cong
Iz^{k}\cong Rz^{k+1}$. It follows that $I$ is principal. Clearly, $z^{k+1}$ is
regular. Let $\varphi:Rz^{k+1}\longrightarrow I$ be the mentioned isomorphism.
Then, $I=R\varphi(z^{k+1})$ and $\varphi(z^{k+1})$ is regular.

For the last statement, let $J$ be a finitely generated regular ideal. By the
first statement, $J=Rz_{0}$ for some regular element $z_{0}$. As
$1=(\frac{1}{z_{0}})z_{0}$, it follows that $R\subseteq J^{-1}J$. Therefore,
$J^{-1}J=R$, and so $R$ is a Prüfer ring. Note that $J^{-1}J\subseteq R$ is
evident. ∎

A discrete valuation ring (DVR) is a principal ideal domain with a unique
nonzero maximal ideal. An Artinian local principal ideal ring is called
special, and it has only finitely many ideals, each of which is a power of the
maximal ideal.

######  Lemma 2.5.

Assume that $R$ is a local ring with a principal maximal ideal
${\mathfrak{m}}$. Then the following are equivalent:

  1. (i)

$R$ is Noetherian.

  2. (ii)

$\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}=0$.

  3. (iii)

$\\{0\\}\cup\\{\mathfrak{m}^{i}\mid i\in\mathbb{N}_{0}\\}$ is the set of all
ideals of $R$, and so $R$ is either a DVR or a special ring.

###### Proof.

(i)$\Rightarrow$(ii) is immediate by the Krull intersection theorem.

(ii)$\Rightarrow$ (iii) Let ${\mathfrak{m}}=Rz$ and $I$ be a nonzero ideal of
$R$. Then there exists a natural number $\ell$ such that
$I\subseteq\mathfrak{m}^{\ell}$ and $I\nsubseteq\mathfrak{m}^{\ell+1}$. Choose
$x\in I\setminus\mathfrak{m}^{\ell+1}$. Since
$x\in\mathfrak{m}^{\ell}\setminus\mathfrak{m}^{\ell+1}$, we have $x=rz^{\ell}$
for some $r\in R\setminus\mathfrak{m}$. Since $r$ is a unit, we have
$z^{\ell}\in I$, and so $\mathfrak{m}^{\ell}=Rz^{\ell}=I$. Thus,
$\\{0\\}\cup\\{\mathfrak{m}^{i}\mid i\in\mathbb{N}_{0}\\}$ is the set of all
ideals of $R$. In particular, $R$ is a principal ideal ring. If $\mathfrak{m}$
is nilpotent, then $R$ is Artinian, and so $R$ is a special ring.

Now, assume that $\mathfrak{m}$ is not nilpotent, so $z^{k}\neq 0$ for all
$k\in\mathbb{N}$. We will show that $R$ is a DVR. Since $R$ is a local
principal ideal ring, and $\mathfrak{m}\neq 0$, it suffices to show that $R$
is a domain. Let $x$ be a nonzero element of $R$. Since
$\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}=0$, there exists a natural number
$t$ such that $x\in{\mathfrak{m}}^{t}\setminus{\mathfrak{m}}^{t+1}$. Hence,
$x=rz^{t}$ for some unit $r$. Thus, every nonzero element of $R$ can be
written as $uz^{n}$, where $u$ is a unit of $R$ and $n$ is a natural number.
Now, a direct examination shows that $R$ is a domain. Therefore, $R$ is a DVR.

(iii)$\Rightarrow$(i) is evident. ∎

######  Lemma 2.6.

Let $R$ be an isoartinian ring. Then $R$ is isomorphic to the direct product
of finitely many indecomposable isoartinian rings. Moreover, if

$R\cong R_{1}\times\cdots\times R_{l}\cong S_{1}\times\cdots\times S_{m}$

where $R_{i}$’s and $S_{j}$’s are indecomposable isoartinian rings, then $l=m$
and $R_{i}\cong S_{\sigma(i)}$ for some permutation $\sigma$ on the set
$\\{1,\dots,l\\}$.

###### Proof.

First, note that the (minimal) prime ideals of $\prod_{i=1}^{l}R_{i}$ are of
the form $\prod_{i=1}^{l}{\mathfrak{p}}_{i}$, where for some $j$,
$\mathfrak{p}_{j}$ is a (minimal) prime ideal of the ring $R_{j}$ and
${\mathfrak{p}}_{i}=R_{i}$ for all $i\neq j$.

By Lemma 2.2, $\operatorname{Min}R$ is finite. Let $n$ be the number of
elements of $\operatorname{Min}R$. If $R$ is indecomposable, then we are done.
Otherwise, suppose $R\cong R_{1}\times R_{2}$, where $R_{1}$ and $R_{2}$ are
two rings. If $R_{1}$ and $R_{2}$ are indecomposable, then we are done.
Without loss of generality, assume that $R_{1}$ is not indecomposable, so
$R_{1}\cong A_{1}\times A_{2}$ for some rings $A_{1}$ and $A_{2}$. Thus, we
have $R\cong A_{1}\times A_{2}\times R_{2}$. If $A_{1}$, $A_{2}$ and $R_{2}$
are all indecomposable, then we are done. Otherwise, we repeat the process.
This process must stop after at most $n$ steps. All of the finite
decomposition components are isoartinian by [8, Lemma 4.1].

For the second assertion, let $\phi:R_{1}\times\cdots\times
R_{l}\longrightarrow R$ and $\psi:S_{1}\times\cdots\times S_{m}\longrightarrow
R$ be two ring isomorphisms, in which $R_{i}$’s and $S_{j}$’s are
indecomposable isoartinian rings and $l,m\in\mathbb{N}$. For each $1\leq i\leq
l$ and $1\leq j\leq m$, set $A_{i}=\phi(0\times\cdots\times 0\times
R_{i}\times 0\times\cdots\times 0)$ and $B_{j}=\psi(0\times\cdots\times
0\times S_{j}\times 0\times\cdots\times 0)$. Then

$R=A_{1}\oplus\cdots\oplus A_{l}=B_{1}\oplus\cdots\oplus B_{m}$

and each $A_{i}$ as well as each $B_{j}$ is indecomposable as an ideal,
because $R_{i}$’s and $S_{j}$’s are indecomposable as rings. Now, by [11,
Lemma 3.8], $l=m$ and $A_{i}=B_{\sigma(i)}$ for some permutation $\sigma$ of
the set $\\{1,\dots,l\\}$. It follows that $R_{i}\cong S_{\sigma(i)}$ for
every $i=1,\dots,l$. ∎

Let $\operatorname{Nil}R$ denote the nilradical of $R$.

######  Lemma 2.7.

Let $R$ be an isoartinian ring. Then the following statements hold.

  1. (i)

$\operatorname{Nil}R$ is nilpotent.

  2. (ii)

Every two distinct minimal prime ideals of $R$ are coprime.

###### Proof.

(i) follows by [4, Proposition 2.3].

(ii) Let $\mathfrak{p}$ and $\mathfrak{q}$ be two distinct minimal prime
ideals of $R$, and let $A:=R/\mathfrak{p}\times R/\mathfrak{q}$. By Lemma 2.2,
$\mathfrak{p}$ and $\mathfrak{q}$ belong to $\operatorname{Ass}R$, and so
$R/\mathfrak{p}$ and $R/\mathfrak{q}$ are principal ideal domains by
Proposition 2.3. Since $A$ is the product of two principal ideal domains, we
can conclude that every ideal of $A$ is isomorphic to a direct summand of $A$.
Thus, by [2, Theorem 2.3], there exist prime ideals $P_{1},\dots,P_{k}$ of $A$
such that $A\cong A/P_{1}\times\dots\times A/P_{k}$, and $P_{i}$’s are
comparable or coprime. By applying Lemma 2.6 and [8, Lemma 4.1], we can
conclude that $k=2$, $A/\mathfrak{p}\cong A/P_{\sigma(1)}$, and
$A/\mathfrak{q}\cong A/P_{\sigma(2)}$ for some permutation $\sigma$ of the set
$\\{1,2\\}$. As every ring homomorphism between two quotient rings of $A$ is
an $A$-homomorphism, we conclude that

$\mathfrak{p}=\operatorname{Ann}_{A}(A/\mathfrak{p})=\operatorname{Ann}_{A}(A/P_{\sigma(1)})=P_{\sigma(1)}.$

Similarly, we have $\mathfrak{q}=P_{\sigma(2)}$. Since $\mathfrak{p}$ and
$\mathfrak{q}$ are two distinct minimal primes, they are not comparable and
therefore coprime. ∎

The next result provides a simple criterion for an isoartinian ring to be
indecomposable.

######  Lemma 2.8.

An isoartinian ring $R$ is indecomposable if and only if $\operatorname{Min}R$
is a singleton.

###### Proof.

First, assume that $R$ is decomposable. Then, there is a natural number $t\geq
2$ and rings $R_{1},\dots,R_{t}$ such that $R\cong R_{1}\times\dots\times
R_{t}$. Keeping in mind the form of prime ideals of the direct products of
rings, this implies that $\operatorname{Min}R$ has at least $t$ elements.

Next, assume that
$\operatorname{Min}R=\\{\mathfrak{p}_{1},\dots,\mathfrak{p}_{t}\\}$, with
$t\geq 2$. Note that by Lemma 2.2, $\operatorname{Min}R$ is finite. By Lemma
2.7(ii), every two distinct elements $\mathfrak{p}$ and $\mathfrak{q}$ of
$\operatorname{Min}R$ are coprime, and so
$\mathfrak{p}^{k}+\mathfrak{q}^{k}=R$ for every natural number $k$. By Lemma
2.7(i), $\operatorname{Nil}A$ is nilpotent, and hence
$\mathfrak{p}_{1}^{k}\cdots\mathfrak{p}_{t}^{k}=0$ for some natural number
$k$. Now, the Chinese remainder theorem [1, Proposition 1.10] yields

$\begin{array}[]{ll}R&\cong R/0\\\ &\cong
R/(\mathfrak{p}_{1}^{k}\cdots\mathfrak{p}_{t}^{k})\\\ &\cong
R/(\mathfrak{p}_{1}^{k}\cap\cdots\mathfrak{\cap}p_{t}^{k})\\\
&\cong(R/\mathfrak{p}_{1}^{k})\times\cdots\times(R/\mathfrak{p}_{t}^{k}),\end{array}$

and so $R$ is decomposable. ∎

Although any local ring is indecomposable, the indecomposable isoartinian ring
$\mathbb{Z}$ shows that $\operatorname{Max}R$ cannot be replaced by
$\operatorname{Min}R$ in the above lemma.

Using Proposition 2.3 and Lemma 2.8, we can completely determine the spectrum
of an isoartinian local ring.

######  Corollary 2.9.

Let $(R,\mathfrak{m})$ be an isoartinian local ring. Then
$\operatorname{Spec}R=\\{\mathfrak{p},\mathfrak{m}\\}$, where $\mathfrak{p}$
is the unique minimal prime ideal of $R$. Moreover if $\mathfrak{m}$ contains
a regular element, then
$\mathfrak{p}=\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}$ and
$\mathfrak{p}\neq\mathfrak{m}$.

###### Proof.

Since $R$ is local, it is indecomposable. Hence, Lemma 2.8 yields that $R$ has
a unique minimal prime ideal $\mathfrak{p}$. Proposition 2.3 implies that
$\operatorname{Spec}R=\operatorname{Min}R\cup\operatorname{Max}R=\\{\mathfrak{p},\mathfrak{m}\\}$.

Now, assume that $\mathfrak{m}$ contains a regular element. By Lemma 2.4,
$\mathfrak{m}$ is principal and is generated by a regular element $z$. For
every natural number $i$, we show that there is no ideal between
${\mathfrak{m}}^{i}$ and ${\mathfrak{m}}^{i-1}$. Clearly,
${\mathfrak{m}}/{\mathfrak{m}}^{i}$ is the unique prime ideal of the ring
$R/{\mathfrak{m}}^{i}$, and ${\mathfrak{m}}/{\mathfrak{m}}^{i}$ is cyclic.
Hence, by Cohen’s theorem, $R/{\mathfrak{m}}^{i}$ is a Noetherian ring. As
$\dim R/{\mathfrak{m}}^{i}=0$, by [1, Theorem 8.5], we deduce that
$R/{\mathfrak{m}}^{i}$ is Artinian. Hence, by [1, Proposition 8.8],
$R/{\mathfrak{m}}^{i}$ is a special ring, and so
$R/\mathfrak{m}^{i},\mathfrak{m}/\mathfrak{m}^{i},\dots,\mathfrak{m}^{i-1}/\mathfrak{m}^{i},0$
are the only ideals of $R/\mathfrak{m}^{i}$. Thus, there is no ideal strictly
between ${\mathfrak{m}}^{i}$ and ${\mathfrak{m}}^{i-1}$.

Set ${\mathfrak{r}}=\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}$. We show that
${\mathfrak{r}}$ is prime. Let $a$ and $b$ be two elements of
$R\setminus\mathfrak{r}$. We show that $ab\not\in\mathfrak{r}$. There are
natural numbers $l$ and $t$ such that
$a\in{\mathfrak{m}}^{l}\setminus{\mathfrak{m}}^{l+1}$ and
$b\in{\mathfrak{m}}^{t}\setminus{\mathfrak{m}}^{t+1}$. Thus, by the above
argument, ${\mathfrak{m}}^{l}=Ra+{\mathfrak{m}}^{l+1}$ and
${\mathfrak{m}}^{t}=Rb+{\mathfrak{m}}^{t+1}$. So,

${\mathfrak{m}}^{l+t}={\mathfrak{m}}^{l}{\mathfrak{m}}^{t}=(Ra+{\mathfrak{m}}^{l+1})(Rb+{\mathfrak{m}}^{t+1})\subseteq
Rab+{\mathfrak{m}}^{l+t+1}.$

Suppose that $ab\in{\mathfrak{m}}^{l+t+1}$. Then
${\mathfrak{m}}^{l+t}={\mathfrak{m}}^{l+t+1}$, and so $z^{l+t}=rz^{l+t+1}$ for
some $r\in R$. Since $z$ is regular, we get $rz=1$, which is a contradiction.
Consequently $ab\notin{\mathfrak{m}}^{l+t+1}$, and so $ab\not\in\mathfrak{r}$.
If $\mathfrak{r}={\mathfrak{m}}$, then $z\in\mathfrak{m}^{2}$, and so
$z=rz^{2}$ for some $r\in R$. As $z$ is regular, we deduce that $z$ is a unit,
a contradiction. Therefore $\mathfrak{r}=\mathfrak{p}$. ∎

Next, we state the following easy observation.

######  Lemma 2.10.

Let $\mathfrak{p}$ and ${\mathfrak{q}}=Rz$ be two prime ideals of a ring $R$
with ${\mathfrak{p}}\subsetneq\mathfrak{q}$. Then
$\mathfrak{p}\mathfrak{q}={\mathfrak{p}}z=\mathfrak{p}$.

###### Proof.

Clearly, $\mathfrak{p}\mathfrak{q}={\mathfrak{p}}z\subseteq\mathfrak{p}$. If
$x\in{\mathfrak{p}}$, then $x=rz$ for some $r\in R$. Since
$z\not\in{\mathfrak{p}}$, we have $r\in{\mathfrak{p}}$. Hence
$x=rz\in{\mathfrak{p}}z$, and the proof is complete. ∎

Finally, we are ready to prove Theorem 1.1. This theorem is analogous to the
classical decomposition theorem for Artinian rings, which was proved by
Akizuki and Cohen.

Proof of Theorem 1.1: Based on Lemma 2.6 and [8, Lemma 4.1], we can conclude
that $R$ is isoartinian if and only if it is uniquely isomorphic to the direct
product of finitely many indecomposable isoartinian rings. Hence, to complete
the proof, it remains to establish that every indecomposable isoartinian ring
$A$ takes on one of the six given forms.

Let $A$ be an indecomposable isoartinian ring. We note that $\dim A\leq 1$ by
Proposition 2.3, and as $A$ is indecomposable, by Lemma 2.8,
$\operatorname{Min}A$ has a unique element $\mathfrak{p}$. Hence
$\operatorname{Nil}A=\mathfrak{p}$, and so $\mathfrak{p}$ is nilpotent by
Lemma 2.7(i). The proof is broken into two cases.

Case 1. $\dim A=0$.

Then $\operatorname{Spec}A=\\{\mathfrak{p}\\}$. If $A$ is Noetherian, then $A$
is Artinian (type (i)). If $A$ is not Noetherian, then type (ii) occurs.

Case 2. $\dim A=1$.

Lemma 2.2 implies that $\mathfrak{p}\in\operatorname{Ass}A$, and so
$A/\mathfrak{p}$ is a principal ideal domain by Proposition 2.3. As $\dim
A=1$, it follows that $\mathfrak{p}$ is not maximal, and so $A/\mathfrak{p}$
is not a field. If $\mathfrak{p}=0$, then $A$ is a non-field principal ideal
domain (type (iii)).

In the rest of the proof, we may and do assume that $\mathfrak{p}\neq 0$.
Assume that $A$ is Noetherian. We claim that
$\operatorname{Min}A\varsubsetneq\operatorname{Ass}A$. Suppose on the contrary
$\operatorname{Min}A=\operatorname{Ass}A$. Let $\mathfrak{m}$ be a maximal
ideal of $A$. Then $\mathfrak{p}\subsetneq\mathfrak{m}$, and so $\mathfrak{m}$
contains a regular element. Now, Lemma 2.4 implies that $\mathfrak{m}=Az$ for
some regular element $z$. By Lemma 2.10, it follows that
$\mathfrak{p}=\mathfrak{m}\mathfrak{p}$. By localizing at $\mathfrak{m}$ and
applying Nakayama’s lemma, it follows that $\mathfrak{p}A_{\mathfrak{m}}=0$.
Thus $\mathfrak{p}A_{\mathfrak{n}}=0$ for every maximal ideal $\mathfrak{n}$
of $A$, and so $\mathfrak{p}=0$. We have arrived at a contradiction, and so
type (iv) occurs.

Next, suppose that $A$ is not Noetherian. Then either type (v) or type (vi)
occurs. $\square$

##  3\. Examples

In this section, we present examples that demonstrate the occurrence of types
(ii), (iv), (v), and (vi) stated in Theorem 1.1. These examples are provided
in Examples 3.1, 3.3, 3.4, and 3.7. Furthermore, we illustrate with some
examples that the condition “isoartinian” in types (ii), (v) and (vi) of
Theorem 1.1 cannot be relaxed; see Examples 3.2, 3.5 and 3.8.

Specifically, our first example demonstrates the occurrence of type (ii) in
Theorem 1.1 and provides an instance of an isoartinian ring that is not
isonoetherian.

######  Example 3.1.

Let $F$ be a field and $R=F[x_{1},x_{2},x_{3},\dots]$, where $x_{i}x_{j}=0$
for each $i,j\in\mathbb{N}$. Clearly, $R$ is a local ring with the unique
prime ideal ${\mathfrak{m}}=\langle x_{1},x_{2},\dots\rangle$. Since
$\mathfrak{m}^{2}=0$, we can deduce that $R$ can be expressed as
$R=F\oplus(\oplus_{i\in\mathbb{N}}Fx_{i})$. Thus, $R$ is an $F$-vector space
with a countable basis and any ideal $I$ of $R$ can be written as
$F^{(\Gamma)}$, where $\Gamma$ is a countable set.

Now, consider a descending chain of ideals of $R$:

$I_{1}\supseteq I_{2}\supseteq\cdots\supseteq I_{n}\supseteq\cdots.$

If there exists an $I_{j}$ isomorphic to $F^{(\Gamma)}$ with $\Gamma$ finite,
then the chain is stationary. If there is no such ideal $I_{j}$, then we must
have $I_{1}\cong I_{2}\cong\cdots$. This shows that $R$ is an isoartinian
ring.

However, $R$ is not isonoetherian. To see this, for each natural number $k$,
let $J_{k}:=\langle x_{1},x_{2},\dots,x_{k}\rangle$. Then $J_{k}$ is an
$F$-vector space of dimension $k$, and the chain

$J_{1}\subseteq J_{2}\subseteq\cdots\subseteq J_{k}\subseteq\cdots$

does not stabilize up to isomorphism. Hence, $R$ is not isonoetherian and,
consequently, not Noetherian.

We now present an example of a non-isoartinian local ring with a nilpotent
maximal ideal.

######  Example 3.2.

Let $F$ be a field and $R=F[x_{1},x_{2},x_{3},\dots]$, where $x_{i}x_{j}=0$
and $x_{i}^{3}=0$ for all natural numbers $i<j$. Clearly, $R$ is a non-
Noetherian local ring with the unique prime ideal ${\mathfrak{m}}=\langle
x_{1},x_{2},\dots\rangle$ and $\mathfrak{m}^{3}=0$. For each natural number
$k$, set $J_{k}:=\langle x_{k},x_{k+1},\dots\rangle$. These ideals form the
following descending chain:

$J_{1}\supsetneq J_{2}\supsetneq\cdots\supsetneq J_{k}\supsetneq\cdots.$

It is easy to see that $x_{k}\in(0:_{R}J_{k+1})\setminus(0:_{R}J_{k})$. Since
isomorphic modules have the same annihilator, this chain does not stabilize up
to isomorphism. Thus, $R$ satisfies all conditions in type (ii) of Theorem 1.1
except the isoartinian condition.

Next, we provide an example of rings of type (iv) in Theorem 1.1. It is well-
known that every Artinian ring is perfect. This raises the question of whether
every isoartinian ring is also perfect. However, the following example shows
that an isoartinian ring may not even be subperfect. Furthermore, as every
Artinian ring is Cohen-Macaulay, one may conjecture that any Noetherian
isoartinian ring is also Cohen-Macaulay. Nevertheless, the next example
demonstrates that this is not true.

Let $J(R)$ denote the Jacobson radical of $R$. Recall that a ring $R$ is said
to be perfect if $R/J(R)$ is a semisimple ring and $J(R)$ is $t$-nilpotent. A
ring $R$ is called subperfect if its total ring of fractions is a perfect
ring. An $R$-module $M$ is called isoartinian if every descending chain of
submodules of $M$ terminates up to isomorphism.

######  Example 3.3.

Let $F$ be a field and $R=F[[X,Y]]/\langle XY,Y^{2}\rangle$. Let $x$ and $y$
denote the residue classes of $X$ and $Y$ in $R$, and set
$\mathfrak{m}=\langle x,y\rangle$ and $\mathfrak{p}=Ry$. As $0=\langle
x,y^{2}\rangle\cap\mathfrak{p}$ is a minimal primary decomposition of the zero
ideal of $R$, it follows that
$\operatorname{Ass}R=\\{\mathfrak{p},\mathfrak{m}\\}$, and so
$\operatorname{Min}R\subsetneq\operatorname{Ass}R$. It is easy to verify that
$(0:_{R}y)=\mathfrak{m}$, and so $\mathfrak{p}$ is a simple $R$-module. We
prove the following claim:

Claim. Let $J$ be an ideal of $R$ such that $J\cap\mathfrak{p}=0$. Then $J$ is
an isoartinian $R$-module.

Proof. As $R/\mathfrak{p}$ is a principal ideal domain and
$(J+\mathfrak{p})/\mathfrak{p}$ is an ideal of $R/\mathfrak{p}$, it follows
that $(J+\mathfrak{p})/\mathfrak{p}$ is an isoartinian
$R/\mathfrak{p}$-module. Thus, $(J+\mathfrak{p})/\mathfrak{p}$ is also
isoartinian as an $R$-module. Note that the set of $R$-submodules of
$(J+\mathfrak{p})/\mathfrak{p}$ coincides with the set of its
$R/\mathfrak{p}$-submodules, and every $R/\mathfrak{p}$-homomorphism between
two ideals of $R/\mathfrak{p}$ is also an $R$-homomorphism. Since $J\cong
J/J\cap\mathfrak{p}\cong(J+\mathfrak{p})/\mathfrak{p}$, we deduce that $J$ is
an isoartinian $R$-module.

Next, we show that $R$ is isoartinian. Let

$I_{1}\supseteq I_{2}\supseteq\cdots\supseteq I_{n}\supseteq\cdots$

be a descending chain of ideals of $R$. Assume that
$I_{\ell}\cap\mathfrak{p}=0$ for some natural number $\ell$. Then $I_{\ell}$
is an isoartinian $R$-module by the above claim. This implies that the chain
must be stationary up to isomorphism. Now, assume that
$I_{\ell}\cap\mathfrak{p}=\mathfrak{p}$ for all natural numbers $\ell$. Since
$\mathfrak{m}=\mathfrak{p}\oplus Rx$, we can easily see that
$I_{\ell}=\mathfrak{p}\oplus(I_{\ell}\cap Rx)$. By the above claim,
$I_{\ell}\cap Rx$ is an isoartinian $R$-module for all $\ell\in\mathbb{N}$. As
$I_{1}\cap Rx$ is an isoartinian $R$-module, there is a natural number $k$ and
$R$-isomorphisms $\phi_{i}:I_{k}\cap Rx\longrightarrow I_{k+i}\cap Rx$ for all
$i\geq 1$. In view of the expression for $I_{\ell}$, $\phi_{i}$ can be
extended to an $R$-isomorphism $\psi_{i}:I_{k}\longrightarrow I_{k+i}$ for all
$i\geq 1$. Hence, the chain must be stationary up to isomorphism, which
implies that $R$ is isoartinian.

Since $\operatorname{Min}R\subsetneq\operatorname{Ass}R$, [9, Lemma 2.2]
implies that $R$ is not subperfect. Additionally, since
$\operatorname{Min}R\subsetneq\operatorname{Ass}R$, we deduce that $R$ is not
Cohen-Macaulay.

We follow by giving an example of rings of type (v) in Theorem 1.1.

######  Example 3.4.

Let $F$ be a field and $R$ be the quotient of the algebra
$F[[x_{1},x_{2},\dots]]$ modulo the relations:

$\begin{cases}x_{1}x_{i}=0\ \ \ \ \text{for all}\ i\geq 2\\\ x_{i}x_{j}=0\ \ \
\ \text{for all}\ i\geq 2\ \text{and}\ j\geq 2.\end{cases}$

Set $\mathfrak{p}=\langle x_{2},x_{3},\dots\rangle$ and $\mathfrak{m}=\langle
x_{1},x_{2},\dots\rangle$. Clearly, $\mathfrak{p}^{2}=0$. It follows that
$\mathfrak{p}$ is the unique minimal prime ideal of $R$, and $\mathfrak{m}$ is
the unique maximal ideal of $R$. It is easy to check that
$\mathfrak{m}=Rx_{1}\oplus\mathfrak{p}$, $(0:_{R}x_{1})=\mathfrak{p}$,
$(0:_{R}x_{2})=\mathfrak{m}$, and $\text{Soc}(R)=\mathfrak{p}$. Hence,
$\operatorname{Min}R\subsetneq\operatorname{Ass}R$. We have
$\mathfrak{m}=Rx_{1}\oplus(\oplus_{i>1}Fx_{i})$, and $Rx_{1}\cong
R/\mathfrak{p}$ is a principal ideal domain. Thus, by the equivalence of the
conditions 3(ii) and 4(iii) of [2, Theorem 4.5], there exists $0\neq x\in R$
such that every proper ideal of $R$ is semisimple or is isomorphic to
$Rx\oplus J$, where $J\subseteq\text{Soc}(R)$. Every semisimple ideal of $R$
is contained in $\text{Soc}(R)$, which is an $F$-vector space with a countable
basis. Thus, by an argument similar to that given in Example 3.1, we can show
that $R$ is isoartinian. Clearly, $R$ is not Noetherian.

In the following example, we demonstrate that the condition of being
“isoartinian” in type (v) of Theorem 1.1 cannot be relaxed.

######  Example 3.5.

Let $F$ be a field and $R$ be the quotient of the algebra
$F[[x,y_{1},y_{2},\dots]]$ modulo the relations:

$\begin{cases}xy_{i}=0\ \ \ \ \text{for all}\ i\geq 1\\\ y_{i}y_{j}=0\ \ \
\text{for all}\ j>i\\\ y_{i}^{3}=0\ \ \ \ \ \text{for all}\ i\geq 2\\\
y_{1}^{2}=0.\end{cases}$

Set $\mathfrak{p}=\langle y_{1},y_{2},\dots\rangle$ and $\mathfrak{m}=\langle
x,y_{1},y_{2},\dots\rangle$. As $\mathfrak{p}^{3}=0$, it follows that
$\mathfrak{p}$ is the unique minimal prime ideal of $R$. Since
$R/\mathfrak{p}\cong\mathbb{F}[[x]]$ is a DVR, we deduce that $\dim R=1$. It
is easy to check that $\mathfrak{p}=(0:_{R}x)$ and
$\mathfrak{m}=(0:_{R}y_{1})$, and so
$\operatorname{Min}R\subsetneq\operatorname{Ass}R$. Clearly, $R$ is not
Noetherian.

We will now show that $R$ is not isoartinian. To do this, assume that
$J_{k}=\langle y_{\scalebox{0.5}{2k}},y_{\scalebox{0.5}{2k+1}},\dots\rangle$,
and consider the following descending chain of ideals in $R$:

$J_{1}\supsetneq J_{2}\supsetneq J_{3}\supsetneq\cdots.$

Suppose $J_{k}\cong J_{k+1}$ for some $k\in\mathbb{N}$. It is clear that
$y_{\scalebox{0.5}{2k+1}}\in(0:_{R}J_{k+1})$, but
$y_{\scalebox{0.5}{2k+1}}\not\in(0:_{R}J_{k})$. This contradicts the fact that
isomorphic modules have the same annihilator. Thus, $R$ satisfies all
conditions in type (v) of Theorem 1.1 except for being isoartinian.

Our next example shows that type (vi) in Theorem 1.1 can occur. To present it,
we need the following lemma.

We recall that a ring $R$ is a chain ring if the set of ideals of $R$ is
totally ordered with respect to inclusion (this is equivalent to the condition
that the set of principal ideals of $R$ is totally ordered with respect to
inclusion). It is well-known that a chain ring is local. A chain domain is
called a valuation domain.

######  Lemma 3.6.

Let $A$ be a DVR with the maximal ideal $Az$. Let $K$ be the field of
fractions of $A$, and set $R=A+xK[[x]]$, $\mathfrak{p}=xK[[x]]$, and
$\mathfrak{m}=Az+\mathfrak{p}$. Then

  1. (i)

$R$ is a valuation domain.

  2. (ii)

$R$ is a non-Noetherian domain, $\mathfrak{m}=Rz$, and $\dim R=2$.

  3. (iii)

$\operatorname{Spec}R=\\{0,\mathfrak{p},\mathfrak{m}\\}$.

  4. (iv)

$\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}=\mathfrak{p}$.

###### Proof.

(i) Since $R$ is a subring of $K[[x]]$, it follows that $R$ is a domain. Let
$f$ and $g$ be two nonzero elements of $R$. As $K[[x]]$ is a DVR, without loss
of generality, we may assume that $g|f$ in $K[[x]]$, that is $f=hg$ for some
$h\in K[[x]]$. We want to show that $f|g$ or $g|f$ in $R$. Let
$h=h_{0}+h_{1}x+h_{2}x^{2}+\cdots$. If $h_{0}=0$, then

$h=x\left(h_{1}+h_{2}x+\cdots\right)\in xK[[x]]\subseteq R,$

and so $g|f$ in $R$, and we are done. Next, assume that $h_{0}\neq 0$. Since
$A$ is a valuation domain, we have either $h_{0}\in A$ or $h^{-1}_{0}\in A$.
We break the proof into two cases:

Case 1. $h_{0}\in A$. Then $h=h_{0}+x\left(h_{1}+h_{2}x+\cdots\right)\in R$,
and so $g|f$ in $A$, and we are done again.

Case 2. $h^{-1}_{0}\in A$. As $h_{0}$ is nonzero, $h$ is an invertible element
in $K[[x]]$. Clearly, $h^{-1}$ has the form

$h^{-1}=h^{-1}_{0}+l_{1}x+l_{2}x^{2}+\dots=h^{-1}_{0}+x(l_{1}+l_{2}x+\cdots)\in
R.$

So, from $h^{-1}f=g$, we deduce that $f|g$ in $R$. This means that $R$ is a
valuation domain.

(ii) By [5, Corollary 14(a)], we conclude that $\dim R=2$. By [5, Proposition
6], $\mathfrak{m}$ is the unique maximal ideal of $R$. Since $R$ is a
valuation domain and $z\notin\mathfrak{p}$, it follows that
$\mathfrak{p}\subseteq Rz$. Hence, $\mathfrak{m}=Az+\mathfrak{p}\subseteq Rz$,
and so $\mathfrak{m}=Rz$. If $R$ were Noetherian, then Krull’s principal ideal
theorem would imply that $\dim R=\operatorname{ht}\mathfrak{m}\leq 1$. So, $R$
is non-Noetherian.

(iii) Obviously, $\mathfrak{p}$ is a prime ideal of $R$. Suppose that $R$
possesses a prime ideal $\mathfrak{q}$ other than the prime ideals $0$,
$\mathfrak{p}$, and $\mathfrak{m}$. As $R$ is a valuation domain, either
$\mathfrak{p}\subsetneq\mathfrak{q}$ or $\mathfrak{q}\subsetneq\mathfrak{p}$.
Thus, we can conclude that either the chain
$0\subsetneq\mathfrak{p}\subsetneq\mathfrak{q}\subsetneq\mathfrak{m}$ or the
chain $0\subsetneq\mathfrak{q}\subsetneq\mathfrak{p}\subsetneq\mathfrak{m}$
exists, implying that $\dim R\geq 3$. Since, by (ii), $\dim R=2$, we deduce
that $\operatorname{Spec}R=\\{0,\mathfrak{p},\mathfrak{m}\\}$.

(iv) For a proper ideal $I$ of a chain ring $T$, by [10, Lemma 1.3, Chapter
II], either some power of $I$ is zero or $\bigcap_{i=1}^{\infty}I^{i}$ is a
prime ideal of $T$. Since $R$ is a domain and $\mathfrak{m}\neq 0$, no power
of $\mathfrak{m}$ is zero, and so $\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}$
is a prime ideal of $R$. As $R$ is non-Noetherian, Lemma 2.5 asserts that
$\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}\neq 0$. If
$\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}=\mathfrak{m}$, then
$\mathfrak{m}=\mathfrak{m}^{2}$, which implies that $\mathfrak{m}=0$ by
Nakayama’s lemma. Thus,
$\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}=\mathfrak{p}$. ∎

######  Example 3.7.

Let $A$, $K$, $R$, $\mathfrak{p}$, and $\mathfrak{m}=Rz$ be as in the lemma
above. By Lemma 3.6, $R$ is a non-Noetherian valuation domain of dimension
two, and $\bigcap_{i=1}^{\infty}{\mathfrak{m}}^{i}=\mathfrak{p}$. Set
$I=Rx^{2}$, $D=R/I$ and $\bar{\mathfrak{p}}=\mathfrak{p}/I$. Since every
factor of a chain ring is again a chain ring, the ring $D$ is a chain ring as
well. As $\operatorname{Spec}R=\\{0,\mathfrak{p},\mathfrak{m}\\}$, it follows
that $\operatorname{Spec}D=\\{\bar{\mathfrak{p}},\mathfrak{m}/I\\}$. Lemma
2.10 yields that $\mathfrak{m}\mathfrak{p}=\mathfrak{p}$. Since $\mathfrak{p}$
is nonzero, by Nakayama’s lemma, it follows that $\mathfrak{p}$ is not
finitely generated. This implies that the ideal $\bar{\mathfrak{p}}$ is also
not finitely generated.

We claim that $D$ is an isoartinian ring. Since $z+I$ is a regular element of
$D$, we have that $\mathfrak{m}/I$ is not an associated prime ideal of $D$. On
the other hand, we can verify that $\bar{\mathfrak{p}}=(0:_{D}(x+I))$, and so
$\bar{\mathfrak{p}}\in\operatorname{Ass}D$. Hence,
$\operatorname{Min}D=\operatorname{Ass}D$. Suppose that we have a decreasing
chain of ideals

$I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq\cdots$

in $D$. Assume that $\bar{\mathfrak{p}}\subsetneq I_{j}$ for every
$j\in\mathbb{N}$. Since $R/\mathfrak{p}\cong A$ is a DVR, it follows that
$\\{0\\}\cup\\{\mathfrak{m}^{i}/\mathfrak{p}\mid i\in\mathbb{N}_{0}\\}$ is the
set of all ideals of $R/\mathfrak{p}$. It can be easily checked that each
$I_{j}$ is a power of the maximal ideal $\mathfrak{m}/I$, and hence they are
principal and generated by a power of the regular element $z+I$. Therefore,
they are isomorphic to $D$, and the chain stops up to isomorphism.

Now, assume that there is a natural number $j_{1}$ such that
$I_{j_{1}}\subseteq\bar{\mathfrak{p}}$. If $I_{j}=\bar{\mathfrak{p}}$ for all
$j\geq j_{1}$, then we are done. So, without loss of generality, we may assume
that $I_{j_{1}}\subsetneq\bar{\mathfrak{p}}$. Let
$\bar{y}\in\bar{\mathfrak{p}}\setminus I_{j_{1}}$. Then $I_{j_{1}}\subsetneq
D\bar{y}\subsetneq\bar{\mathfrak{p}}$, because $D$ is a chain ring. As
$\bar{\mathfrak{p}}\bar{y}=0$, we see that $D\bar{y}$ is a well-defined
$D/\bar{\mathfrak{p}}$-module. Note that $D/\bar{\mathfrak{p}}\cong
R/\mathfrak{p}$ is a DVR. Thus, by the structure theorem for finitely
generated modules over a principal ideal domain, every decreasing chain of
submodules of the $D/\bar{\mathfrak{p}}$-module $D\bar{y}$ must stop up to
isomorphism after a finite number of steps. This forces the chain to stop up
to isomorphism. Note that $D$-submodules of $D\bar{y}$ coincide with its
$D/\bar{\mathfrak{p}}$-submodules.

Our last example below shows that the condition of being “isoartinian” in type
(vi) of Theorem 1.1 cannot be relaxed.

######  Example 3.8.

Let $R$ be the quotient of the algebra $\mathbb{Z}[x_{1},x_{2},\dots]$ modulo
the relations:

$\begin{cases}x_{i}x_{j}=0\ \ \ \text{for all}\ j>i\\\ x_{i}^{3}=0\ \ \ \ \ \
\text{for all}\ i\geq 1.\end{cases}$

Set $\mathfrak{p}=\langle x_{1},x_{2},\dots\rangle$. As $\mathfrak{p}^{3}=0$,
it follows that $\mathfrak{p}$ is the unique minimal prime ideal of $R$. It is
easy to see that $\mathfrak{p}=(0:_{R}x_{1}^{2})$, so
$\mathfrak{p}\in\operatorname{Ass}R$. On the other hand, it is straightforward
to check that the zero ideal of $R$ is $\mathfrak{p}$-primary. Thus,
$Z(R)=\mathfrak{p}$, which implies that
$\operatorname{Ass}R=\\{\mathfrak{p}\\}=\operatorname{Min}R.$ Clearly, $R$ is
not Noetherian. Let $J_{k}=\langle
x_{\scalebox{0.5}{2k}},x_{\scalebox{0.5}{2k+1}},\dots\rangle$ for all
$k\in\mathbb{N}$. With the same argument as in Example 3.5, we see that the
following descending chain of ideals of $R$ does not terminate up to
isomorphism:

$J_{1}\supsetneq J_{2}\supsetneq J_{3}\supsetneq\cdots.$

Hence, $R$ satisfies all conditions in type (vi) of Theorem 1.1 except for
being isoartinian.

We have not provided an example to show that the condition of being
“isoartinian” in type (iv) of Theorem 1.1 cannot be relaxed. In fact, we
conclude the paper by proposing the following question:

######  Question 3.9.

Let $R$ be a (one-dimensional) Noetherian ring with a unique minimal prime
ideal $\mathfrak{p}$ such that $R/\mathfrak{p}$ is a principal ideal domain
and $\operatorname{Min}R\subsetneq\operatorname{Ass}R$. Is $R$ necessarily
isoartinian?

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