# Photonic quadrupole topological insulator using orbital-induced synthetic
flux

Julian Schulz111These authors contributed equally Physics Department and
Research Center OPTIMAS, TU Kaiserslautern, 67663 Kaiserslautern, Germany
Jiho Noh11footnotemark: 1 Department of Mechanical Science and Engineering,
University of Illinois at Urbana–Champaign, Urbana, IL 61801 USA  Wladimir A.
Benalcazar  Department of Physics, Princeton University, Princeton, New Jersey
08542, USA  Gaurav Bahl  Department of Mechanical Science and Engineering,
University of Illinois at Urbana–Champaign, Urbana, IL 61801 USA  Georg von
Freymann  Physics Department and Research Center OPTIMAS, TU Kaiserslautern,
67663 Kaiserslautern, Germany  Fraunhofer Institute for Industrial Mathematics
ITWM, 67663 Kaiserslautern, Germany

Abstract

The rich physical properties of multiatomic molecules and crystalline
structures are determined, to a significant extent, by the underlying geometry
and connectivity of atomic orbitals. This orbital degree of freedom has also
been used effectively to introduce structural diversity in a few synthetic
materials including polariton lattices nonlinear photonic lattices and
ultracold atoms in optical lattices. In particular, the mixing of orbitals
with distinct parity representations, such as $s$ and $p$ orbitals, has been
shown to be especially useful for generating systems that require alternating
phase patterns, as with the sign of couplings within a lattice. Here we show
that by further breaking the symmetries of such mixed-orbital lattices, it is
possible to generate synthetic magnetic flux threading the lattice. This
capability allows the generation of multipole higher-order topological phases
in synthetic bosonic platforms, in which $\pi$ flux threading each plaquette
of the lattice is required, and which to date have only been implemented using
tailored connectivity patterns. We use this insight to experimentally
demonstrate a quadrupole photonic topological insulator in a two-dimensional
lattice of waveguides that leverage modes with both $s$ and $p$ orbital-type
representations. We confirm the nontrivial quadrupole topology of the system
by observing the presence of protected zero-dimensional states, which are
spatially confined to the corners, and by confirming that these states sit at
the band gap. Our approach is also applicable to a broader range of time-
reversal-invariant synthetic materials that do not allow for tailored
connectivity, e.g. with nanoscale geometries, and in which synthetic fluxes
are essential.

When studying materials and their physical properties, much emphasis is put on
how atoms are combined to form molecules and crystalline structures through
different orbital connections. In particular, the richness of macroscopic
properties in multiatomic molecules and crystalline structures is closely
related to the way in which orbitals connect. For example, a water molecule
has an angled shape due to the hybridization of $s$ and $p$ orbitals, which in
turn, causes the six-fold rotational symmetry of snowflakes. Similarly, the
electronic properties of monolayer 2D transition metal dichalcogenides can be
deliberately tuned over a wide range, in part, due to the interplay between
$d$ orbitals on metal atoms and $p_{z}$ orbitals on chalcogen atoms [1, 2, 3].
In this context, the theory of “topological quantum chemistry” [4] has greatly
advanced the understanding of the intimate relationship between electronic
orbitals and topological phases in crystalline structures and has led to the
realization that topologically nontrivial materials are much more common than
previously thought.

The ability of synthetic systems to replicate, and in many cases extend, the
properties of chemical compounds and crystalline structures has recently
become of great interest. As in real materials, the orbital degree of freedom
can also be incorporated into synthetic systems by using analogous
wavefunctions with distinct nodal structures. Such synthetic multi-orbital
systems have been demonstrated in polariton lattices [5, 6], photonic lattices
[7, 8, 9, 10], and ultracold atoms in optical lattices [11, 12, 13, 14, 15,
16], with the possibility of negative [16, 8] and even complex-valued [17, 10]
coupling coefficients. These features have made synthetic platforms well-
suited to explore novel physics that typically are difficult to study in
solid-state systems.

Due to their inherent robustness and potential for disorder-resilient
technologies, topological phases in synthetic periodic platforms are a very
active area of study. Notably, the initial demonstrations of higher-order
topological phases [18] were produced using synthetic materials [19, 20, 21,
22, 23]. A quadrupole topological insulator (QTI) is the first member of the
multipole higher-order topological insulators, but is not straightforward to
implement as it requires a $\pi$ flux of synthetic magnetic field threading
each plaquette in the lattice [24, 18]. In previous experimental realizations
of QTIs [19, 20, 21, 22, 23], the $\pi$ flux was achieved by tailored
connectivities within the system [19, 20, 21, 22] or couplings with arbitrary
phases by exploiting additional coupling links [23]. However, these approaches
are not always practical, especially in nanoscale geometries, and a good
solution is needed to enable synthetic fluxes in a broader range of
experimental platforms.

Figure 1: Synthetic $\pi$ flux threading a unit cell plaquette induced by $s$
and $p$ orbitals. (a)-(c) Schematic and corresponding eigenvalues and
eigenmodes of a unit cell plaquette (a) composed of only $s$ orbitals without
synthetic $\pi$ flux, (b) with synthetic $\pi$ flux, and (c) composed of both
$s$ and $p$ orbitals, respectively. The solid and dashed lines in the
schematic indicate the positive and negative couplings, respectively. The
areas and colors of the circles indicate amplitudes and phases of the
corresponding eigenmodes, respectively. As can be seen in (a) and (b), the
artificial introduction of a $\pi$ flux leads to a significant change in the
eigenspectrum and eigenmodes. When a wavefunction in (c) crosses the $p$
orbital at site 3 a $\pi$-phase is accumulated, thereby inducing a flux in the
plaquette. This effect is confirmed by the fact that the eigenvalues and
eigenmodes from the full continuum calculations (c) line up with the states in
(b).

Here, we present a QTI in a photonic system that uses the symmetry
representations of on-site orbitals to generate the necessary synthetic $\pi$
fluxes. We consider $s$ and $p$ orbitals, which have inherent even and odd
parities, respectively. We exploit the property that, as the wavefunctions of
the collective system traverse a $p$ orbital, they accumulate a phase of
$\pi$. The combination and judicious control of the orbitals in a four-site
unit cell [Fig. 1(c)], breaking the symmetries of mixed-orbital lattices,
creates a synthetic $\pi$ flux that opens a gap at “half-filling” which, along
with the modulation in the hopping amplitudes, results in the gapped system
having a QTI phase. We experimentally demonstrate the photonic QTIs in a
waveguide lattice fabricated using direct laser writing by showing the
existence of mid-gap modes, which are localized at the corners of the lattice.
The fabrication of the waveguides by direct laser writing allows for
unprecedented control over the waveguide parameters including both the cross-
sections and the trajectories of the waveguides [25, 10], which was not
straightforward in the conventional femtosecond direct laser writing technique
[26].

We start by presenting the implementation of our QTI by using $s$ and $p$
orbitals to induce $\pi$ flux threading a unit cell, taking advantage of the
$\pi$-phase accumulated by the wavefunction as it crosses the $p$ orbital. A
square unit cell is composed of four sites with different orbitals: two $p$
orbitals, which have the major axes tilted from the $y$-axis by $\pm
45^{\circ}$, respectively, and two $s$ orbitals as shown in Fig. 1(c). We
first validate that the $\pi$ flux threads through this square unit cell by
comparing three unit cell plaquettes as shown in Fig. 1: those composed of
only $s$ orbitals without and with $\pi$ flux, and a unit cell plaquette
composed of both $s$ and $p$ orbitals.

To compare the three unit cell plaquettes, we computed the eigenvalues and
eigenmodes of the first two cases, having only $s$ orbitals, using the tight-
binding Hamiltonians, and the other case, having both $s$ and $p$ orbitals,
using the full-continuum wave equation. For a unit cell plaquette without a
$\pi$ flux [Fig. 1(a)], the eigenvalues are $-2\gamma,\,0,\,2\gamma$, where
only the zero-energy states are two-fold degenerate and $\gamma$ is the
coupling rate between the nearest-neighbor sites. Whereas for a unit cell
plaquette, in that a $\pi$ flux threading the plaquette is induced by
introducing a negative coupling [Fig. 1(b)], the eigenvalues are
$\pm\sqrt{2}\gamma$, each of which is two-fold degenerate. On the other hand,
the eigenvalues and eigenmodes of the unit cell plaquette with both $s$ and
$p$ orbitals calculated using the full-continuum wave equation show a great
resemblance with the unit cell with the synthetic $\pi$ flux [Fig. 1(c)],
where the bulk gap is opened. The similarity between these two unit cell
plaquettes validates that the specific arrangement of having two $p$ orbitals
in the same unit cell aligned at different angles induces the effective
magnetic flux of $\pi$ per plaquette.

Figure 2: Quadrupole topological insulator using orbital-induced synthetic
flux. (a) Schematic of the quadrupole topological insulator with orbital-
induced synthetic flux. For the tight-binding model, $\gamma$ and $\lambda$
are the nearest-neighbor coupling terms within (black) and across (red) unit
cells, respectively. Dashed lines represent coupling terms with negative signs
due to the overlap with the “negative part” of the $p$ orbital. Numbers
indicate the basis of the Hamiltonian. (b) A bulk band structure where
$|\gamma/\lambda|=1/2$. The band structure consists of two two-fold degenerate
bands. (c) The numerically calculated density of states in the non-trivial
phase ($|\gamma/\lambda|$=1/2) and (d) trivial phase ($|\gamma/\lambda|$=2),
respectively, where the system has 10$\times$10 unit cells. (e) Combined
eigenmode local density of states of the four topologically protected corner
modes in the non-trivial phase ($|\gamma/\lambda|$=1/2). Here, the system has
5$\times$5 unit cells as in the experiment.

For the tight-binding description of the model, we choose a base where all $s$
orbitals have a phase of zero and one fixed site of the $p$ orbital has a
phase of zero while the opposite site has a phase of $\pi$. In Fig. 1(c) and
Fig. 2(a), zero and $\pi$ phases of the basis states are colored in blue and
red, respectively. As the hopping is determined by the overlap of the base
states, the “negative part” of the $p$ orbital results in some negative
hoppings. Then, the tight-binding bulk Hamiltonian of this system becomes:

$\displaystyle h^{oq}(\textbf{k},\delta)=$
$\displaystyle\left[\gamma-\lambda\cos(k_{x}a)\right]\Gamma_{4}+\lambda\sin(k_{x}a)\Gamma_{3}$
$\displaystyle+\left[\gamma-\lambda\cos(k_{y}a)\right]\Gamma_{2}+\lambda\sin(k_{y}a)\Gamma_{1}+\delta\Gamma_{0},$
(1)

where $a$ is the lattice constant, $\gamma$ and $\lambda$ are the nearest-
neighbor coupling terms within and across unit cells, respectively, the
$\Gamma$-matrices [18] are $\Gamma_{j}=-\tau_{2}\bigotimes\sigma_{k}$ for
$j\in\\{1,2,3\\}$, $\Gamma_{0}=\tau_{3}\bigotimes\sigma_{0}$,
$\Gamma_{4}=\tau_{1}\bigotimes\sigma_{0}$ where $\tau$ and $\sigma$ are Pauli
matrices for the degrees of freedom within a unit cell. When the on-site
energy of the $s$ and $p$ orbitals are the same, $\delta=0.$ This Hamiltonian
closely resembles those Hamiltonians studied in the previous realizations of
the quadrupole insulators [18, 19, 20], where the only difference is the sign
of $\lambda\cos(k_{x,y}a)$ in the first and third terms on the right-hand
side. The consequent difference is that at the phase transition point in the
original model the bandgap closes at the M point, while in our model it occurs
at the $\boldsymbol{\Gamma}$ point. In addition, this Hamiltonian has two
mirror symmetries $M_{x}$ and $M_{y}$, which do not commute with each other,
and also has $C_{4}$ symmetry. Fig. 2(b) shows the 2D bulk band structure of
our model in the topologically non-trivial phase, where the bandgap has opened
at the $\boldsymbol{\Gamma}$ point due to the synthetic $\pi$ flux in the
system. For the system with open boundary conditions, as shown in Fig. 2(a),
the non-trivial quadrupole phase leads to gapped edge states and in-gap corner
states, as shown in Fig. 2(c-e) (see Supplementary Section I for a more
detailed discussion of the symmetries and the topological quadrupole phases of
the Hamiltonian).

We experimentally verify the quadrupole topology of the system by considering
a two-dimensional lattice of evanescently-coupled waveguides. A square unit
cell is composed of four waveguides: two elliptical waveguides, which have the
major axes tilted from the $y$-axis by
$\pm$45\text{\,}\mathrm{\SIUnitSymbolDegree}$$, respectively, and two circular
waveguides as shown in Fig. 2(a). We control the radii of waveguides such that
the lowest-energy mode ($s$ orbital) of the circular waveguides and the
second-lowest-energy mode ($p$ orbital) of the elliptical waveguides have the
same propagation constant, enabling them to couple to one another. The lowest-
energy mode of the elliptical waveguides has a propagation constant far
detuned from the other modes, such that they are well separated from our main
system and can be neglected (see Supplementary Section II for the Eigenmode
calculations with the full-continuum Hamiltonian).

The radii of the major and minor axes of the elliptical waveguides in the
experiment are $0.6\text{\,}\mathrm{\SIUnitSymbolMicro m}$ and
$1.3\text{\,}\mathrm{\SIUnitSymbolMicro m}$, and the radius of the circular
waveguides is $0.5\text{\,}\mathrm{\SIUnitSymbolMicro m}$. The center to
center distances between the waveguides determine the coupling strength, which
in our structure are dimerized to be $1.6\text{\,}\mathrm{\SIUnitSymbolMicro
m}$ and $2.1\text{\,}\mathrm{\SIUnitSymbolMicro m}$ for strong and weak
couplings, respectively. The core of the waveguide is made out of the resin
SU8-2 (Microchem) with a refractive index of $n_{\text{core}}=1.59$ and is
surrounded by IP-Dip (Nanoscribe), which has a refractive index of
$n_{\text{clad}}=1.54$. The sample was fabricated using a Nanoscribe Photonic
Professional GT [25, 27, 28, 29] (see Supplementary Section III for Details
about the Fabrication). For the measurements, light with a wavelength of
$760\text{\,}\mathrm{nm}$ from a white light laser (NKT photonics and a VARIA
filter box) is injected with a 20$\times$ objective (NA=0.4) to a selected
waveguide at the input facet of the waveguide array. After the propagation
through the $1\text{\,}\mathrm{mm}$ long structure, the diffracted light at
the output facet is imaged with another 20$\times$ objective onto a CMOS
camera (Thorlabs DDC1545M) [the measurement setup is sketched out in Fig.
S3(a)]. The equation governing the diffraction of light through the waveguide
array is:

$i\partial_{z}\psi(\textbf{r},z)=\hat{H}\psi(\textbf{r},z),$ (2)

where $\psi(\textbf{r},z)$ is the transverse electric field amplitudes at
propagation distance $z$. $\hat{H}$ is the wavelength ($\lambda$) dependent
continuum Hamiltonian for the wave propagation in the waveguide array. Since
we only consider a single bound mode for each waveguide and it evanescently
couples to the neighboring waveguides, we can approximate the diffraction of
light in our waveguide array using a tight-binding model,
$i\partial_{z}\psi_{i}(z)=-\sum_{j}c_{ij}(\lambda)\psi_{j}(z)$, where
$\psi_{i}$ is the amplitude in the $i$-th waveguide, and $c_{ij}(\lambda)$ is
the coupling constant between waveguides $i$ and $j$ at wavelength $\lambda.$

To experimentally observe the corner-localized topological modes, the light
was injected into a waveguide at one of the corners of the waveguide array. In
Fig. 3(b-c), the diffracted light observed from the output facet for two
different topological phases is shown. When the waveguides located at the
corners of the waveguide array are excited in a trivial phase, the injected
light spreads significantly into the bulk [Fig. 3(b)], which indicates that
there is no corner-localized eigenmode. On the other hand, for the non-trivial
phase, the light does not diffract into the bulk and is tightly confined close
to the corner where the light was initially injected [Fig. 3(c)]. This
confinement of the light at the injected corner is an indication of the
presence of the corner modes and their localization is due to the non-trivial
topology of the system. To prove further that the spatial localization of the
corner modes in the non-trivial phase is not due to the weak coupling between
the waveguides, light is injected at a waveguide in the center of the
waveguide array. The injected light diffracts significantly into the bulk of
the structure as shown in Fig. 3(d), which supports further that the corner-
localized mode shown in Fig. 3(c) emerges due to the non-trivial topology of
the model.

Figure 3: Experimentally measured diffracted light at the output facet. (a)
Cross-sectional image of the output facet of the waveguide array in the non-
trivial phase with a broad illumination of the input facet (see Supplementary
Fig. S3). (b-d) Measured intensity profiles at the output facet of the
waveguide structures. Waveguides, where light is injected at the input facet,
are indicated with yellow dashed circles. The intensity profiles are
normalized to their respective maximum value to increase visibility. (b) Light
is injected into the waveguide at the left corner of the waveguide array in
the trivial phase and (c) non-trivial phase, respectively. (d) When light is
injected into a waveguide in the center of the waveguide array in the non-
trivial phase, it spreads into the bulk of the structure. In (a) and (b-d),
black and white lines are overlapped to indicate the positions of the
waveguides, respectively.

However, the localization of the corner modes alone does not prove the
quadrupole properties, as this behavior was also observed in a system similar
to ours, just without a $\pi$ flux. In this system, however, the corner state
lies not in a bandgap but is a bound state in the continuum [30]. To
demonstrate that the corner localized modes in our system are indeed in a
bandgap and topologically non-trivial modes due to the quadrupole topology, we
introduce auxiliary waveguides as shown in Fig. 4(a). The auxiliary waveguides
are weakly coupled to the lattice such that they can be used as an external
drive injecting light into the lattice at the energy of their bound modes
without significantly altering the intrinsic modes of the lattice [31]. In the
experiment, the center to center distance from the auxiliary waveguide to the
waveguide at the corner is $2.1\text{\,}\mathrm{\SIUnitSymbolMicro m}$. The
auxiliary waveguide is identical to the circular waveguides in the lattice,
therefore the energy of the light injected from the auxiliary waveguide into
the lattice is at zero energy. In our system, the light initially injected at
the auxiliary waveguide couples only to the corner state in the non-trivial
phase [Fig. 4(b)] but does not couple into the system in the trivial phase
[Fig. 4(c)]. This proves experimentally that in our system with $s$ and $p$
orbitals, the $\pi$ flux is induced in the unit cell, which opens the bandgap,
and that the corner localized modes in the non-trivial phase are pinned at
midgap due to the quadrupole topology of the system.

Figure 4: Direct excitation of the corner mode using an auxiliary waveguide
weakly coupled to the system. (a) Cross-sectional image of the output facet of
the waveguide array in the non-trivial phase with an auxiliary waveguide with
a broad illumination of the input facet (see Supplementary Fig. S3). (b)
Diffracted light measured at the output facet when light is injected into the
auxiliary waveguide directly at the left corner of the waveguide array in the
non-trivial phase and (c) trivial phase, respectively. Waveguides, where light
is injected at the input facet, are indicated with yellow dashed circles. In
(a) and (b-c), black and white lines are overlapped to indicate the positions
of the waveguides, respectively.

In this work, we have demonstrated synthetic crystalline structures composed
of multiple orbitals. We realized the quadrupole topological insulator in
which synthetic $\pi$ flux threading each plaquette is induced due to the
different symmetry representations of the orbitals. To prove the non-trivial
quadrupole topology of the system, we have experimentally verified that our
realization of the quadrupole topological insulator has zero-dimensional
corner-localized modes in the middle of the band gap. Previously, the orbital
degree of freedom has been nearly exclusively utilized in the ultracold atoms
in optical lattices among diverse metamaterial systems. Different experimental
platforms have their unique features, such as abilities to more precisely or
actively control the gain/loss, on-site energies, or coupling strengths, which
can be associated with the orbital degree of freedom to open the possibility
to study richer physics. Furthermore, the realization of such photonic
quadrupole topological insulators in the time-reversal symmetric system can
provide a more straightforward route to utilize the quadrupole topology in
practical applications, since the geometry we present is easier to implement
than the one proposed previously.

## Data availability

The data that support the findings of this study are available from the
corresponding author on reasonable request.

## Acknowledgements

We thank Christina Jörg for useful discussions. G.v.F. and J.S. acknowledge
funding by the Deutsche Forschungsgemeinschaft through CRC/Transregio 185
OSCAR (project No. 277625399). G.B. and J.N. acknowledge the support of the US
Office of Naval Research (ONR) Multidisciplinary University Research
Initiative (MURI) grant N00014-20-1-2325 on Robust Photonic Materials with
High-Order Topological Protection. W.A.B. is thankful for the support of the
Moore Postdoctoral Fellowship at Princeton University.

## Author contributions

J.S., J.N. and W.A.B. conceived the project. J.S. and J.N. performed numerical
simulations, designed the sample, and performed data analysis. J.S. fabricated
and characterized the sample, and performed the optical probing experiment.
G.B. and G.v.F. supervised all aspects of the project. All authors contributed
in writing the manuscript.

## Competing interests

The authors declare no competing interests.

## Additional Information

Supplemental Materials: Photonic quadrupole topological insulator using
orbital-induced synthetic flux

##  I Quadrupole phases

We discuss the bulk properties of the model by imposing the periodic boundary
condition on a unit cell. The symmetry group, that protects the quantization
of both components of the polarization $p_{x}$ and $p_{y}$ and the quadrupole
moment $q_{xy}$, includes two mirror symmetries
$M_{x}=\tau_{3}\bigotimes\sigma_{1}$ and $M_{y}=\tau_{1}\bigotimes\sigma_{1}$
that do not commute with each other [18]. When $\delta=0$, the Hamiltonian
(Eq. 1) has the required mirror symmetries,
$M_{x}h^{oq}(k_{x},k_{y})M_{x}^{\dagger}=h^{oq}(-k_{x},k_{y})$ and
$M_{y}h^{oq}(k_{x},k_{y})M_{y}^{\dagger}=h^{oq}(k_{x},-k_{y})$.

These two mirror symmetries satisfy the condition for the quantization of the
quadrupole moment such that they do not commute with each other and satisfy
$\\{M_{x},M_{y}\\}=0$. In addition, the Hamiltonian (Eq. 1) has $C_{4}$ and
chiral symmetries,
$C_{4}h^{oq}(k_{x},k_{y})C_{4}^{\dagger}=h^{oq}(k_{y},-k_{x})$ and
$\mathcal{C}h^{oq}(k_{x},k_{y})\mathcal{C}^{\dagger}=-h^{oq}(k_{x},k_{y})$,
where
$C_{4}=\left\\{\left(\tau_{1}+i\tau_{2}\right)\bigotimes\sigma_{0}-\left(\tau_{1}-i\tau_{2}\right)\bigotimes
i\sigma_{2}\right\\}/2$ and $\mathcal{C}=\tau_{3}\bigotimes\sigma_{0}$. The
bulk Hamiltonian is gapped for $|\gamma/\lambda|\neq 1$ but closes at
$\boldsymbol{\Gamma}$ point when $|\gamma/\lambda|=1$, where the topological
transition occurs. Since the inversion symmetry $\mathcal{I}$ is related to
the mirror symmetries as $\mathcal{I}=M_{y}M_{x}$, the Hamiltonian is also
inversion symmetrical.

We further study the quadrupole topology from the Bloch Hamiltonian by
considering the Wannier bands and their polarization through the nested Wilson
loop formulation [18]. The Wannier centers $\nu^{j}_{x}(k_{y})$ are
proportional to the phases of the eigenvalues of the Wilson loop operator,
$\mathcal{W}_{x}$ [Fig. S1(a)], and the polarization $p_{x}$ can be obtained
by taking the integral of the Wannier bands over the whole Brillouin zone in
the $y$-direction. The polarization $p_{y}$ can be determined similarly by
considering $\mathcal{W}_{y}$ and the corresponding $\nu^{j}_{y}(k_{x})$ [Fig.
S1(b)]. Note that for the Bloch Hamiltonian in our model, the Wannier bands
are gapped and $p_{x}=p_{y}=0$, which indicates that the bulk dipole moments
are zero. Here, we denote the upper and lower Wannier bands as $\nu_{x}^{\pm}$
($\nu_{y}^{\pm}$), respectively. Using the nested Wilson loop method, we can
also compute the polarizations of each Wannier band. The aforementioned
existence of non-commuting mirror symmetries $M_{x,y}$ quantizes the Wannier-
sector polarizations $p_{x}^{\nu_{\pm}}$ and $p_{y}^{\nu_{\pm}}$ to be either
0 or 1/2, which subsequently quantizes the bulk quadrupole
$q_{xy}=2p_{x}^{\nu_{\pm}}p_{y}^{\nu_{\pm}}$ to be either 0 or 1/2. From the
calculation we find that $p_{x,y}=0$ for $|\gamma/\lambda|\neq 1$. On the
other hand, we find that $q_{xy}=0$ for $|\gamma/\lambda|>1$ but $q_{xy}=1/2$
for $|\gamma/\lambda|<1$, which proves the quadrupole topology of the model.

Figure S1: (a) Wannier bands $\nu_{x}(k_{y})$ and (b) $\nu_{y}(k_{x})$
computed for the bottom two bands of the bulk band structure where
$\gamma/\lambda=1/2$.

##  II Eigenmode calculations using the full-continuum Hamiltonian

While studying the evanescently-coupled waveguide system for realizing the
photonic quadrupole topological insulators, we only considered the case where
the lowest-energy mode ($s$ orbital) of the circular waveguides and the
second-lowest-energy mode ($p$ orbital) of the elliptical waveguides have the
same energy. The lowest-energy mode of the elliptical waveguide is ignored in
the model since due to the judiciously controlled waveguide radii, the energy
of this mode is well separated from the other modes of interest. To confirm
this, we calculated the eigenmodes using the full-continuum calculation by
diagonalizing a continuum Hamiltonian for the propagation of the wave in the
photonic lattice (Eq. 2). In Fig. S2, we show the eigenmodes of the 5$\times$5
unit cell quadrupole topological insulator with orbital-induced synthetic
flux, identical to the experiment. As shown in Fig. S2, the energy of the bulk
band consisting of the lowest-energy-mode of the elliptical waveguide is
spectrally well separated from the eigenmodes consisting of both $s$ orbital
and $p$ orbital.

Figure S2: (a) Eigenenergies of the 5$\times$5 unit cell waveguide array in
the non-trivial phase and (b) trivial phase, respectively. Eigenmodes are
calculated using the full-continuum calculation by diagonalizing a continuum
Hamiltonian for the propagation of the wave in the photonic lattice. Black
dots indicate the bulk eigenmodes composed of the lowest-energy-mode of the
elliptical waveguide. Blue(red) dots indicate the bulk(corner) eigenmodes
composed of the lowest-energy mode of the circular waveguide ($s$ orbital) and
the second-lowest-energy mode ($p$ orbital) of the elliptical waveguides.

##  III Details on the sample fabrication

The sample was fabricated using the Nanoscribe Photonic Professional GT
similar to the way laid out in [29]. First, the structure is 3D-printed by
two-photon lithography in a negative-tone photoresist (IP-Dip, Nanoscribe).
After the development ($1\text{\,}\mathrm{h}$ in Propylene glycol methyl ether
acetate and $1\text{\,}\mathrm{h}$ in Isopropanol) one is left with the
inverse waveguide structure like the one shown in Fig. S3(d). A surrounding
support grid is rotationally symmetrically designed and written with higher
laser power to minimize distortion of the waveguide structure due to shrinkage
of the photoresist during development. The next step is to dip the structure
in gamma-butyrolactone. After one hour, where the structure is left to soak in
the solvent, most of the gamma-butyrolactone is removed so that only the
structure is wetted. A drop of SU8-2 (MicroChem) is then placed on the
structure so that the SU8 can diffuse into the channels filled with gamma-
butyrolactone. Finally, the sample is slowly heated up
($10\text{\,}\mathrm{K}\text{\,}{\mathrm{min}}^{-1}$) with a hotplate at
$150\text{\,}\mathrm{\SIUnitSymbolCelsius}$ for $5\text{\,}\mathrm{min}$ to
solidify the SU8. The pre-infiltration with gamma-butyrolactone helps to
increase the likelihood that the thin channels will be fully infiltrated, as
the SU8 enters the channels by diffusion. Otherwise, the infiltration process
would only rely on capillary forces, which can be hindered, e.g., by thin
polymer threads. However, the SU8 should be diluted as little as possible with
gamma-butyrolactone because it decreases in volume when baked out, allowing it
to retract into the channels. After the baking process, the channels are
filled with a material with a higher refractive index as IP-Dip and become
waveguides and look brighter than the surrounding material [see Fig. S3(b,c)].

$50\text{\,}\mathrm{\SIUnitSymbolMicro m}$
---
$10\text{\,}\mathrm{\SIUnitSymbolMicro m}$
---
$5\text{\,}\mathrm{\SIUnitSymbolMicro m}$
---
(d)
---
(b)
---
(c)
---
expanded
---
laser beam
$y$
---
$x$
---
$z$
---
input facet
---
camera
output facet
---
camera

$\times 20$
---

input facet
---
illumination
output facet
---
illumination
substrate fixed
---
on a $xy$-stage
input facet of the
---
waveguidestructure
(a)
---

$\times 20$
---

Figure S3: (a) Schematic of the measurement setup. To image the input or
output facet on the cameras a light-emitting diode is put into the beam path
to serve as transmission illumination. (b) Microscope image of the input facet
in reflection illumination of a final structure. Only after the baking
process, the channels become waveguides and look brighter than the surrounding
material. (c) Microscope image of the input facet in transmission illumination
of the waveguide array marked in (b) with a white box. (d) SEM Image of an
inverse waveguide structure after development before the infiltration.

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# Duo property for rings by the quasinilpotent perspective

Abdullah Harmanci  Abdullah Harmanci, Department of Mathematics, Hacettepe
University, Ankara, Turkey<EMAIL_ADDRESS>,  Yosum Kurtulmaz
Yosum Kurtulmaz, Department of Mathematics, Bilkent University, Ankara, Turkey
<EMAIL_ADDRESS>and  Burcu Ungor  Burcu Ungor, Department of
Mathematics, Ankara University, Ankara, Turkey<EMAIL_ADDRESS>
###### Abstract.

In this paper, we focus on the duo ring property via quasinilpotent elements
which gives a new kind of generalizations of commutativity. We call this kind
of ring qnil-duo. Firstly, some properties of quasinilpotents in a ring are
provided. Then the set of quasinilpotents is applied to the duo property of
rings, in this perspective, we introduce and study right (resp., left) qnil-
duo rings. We show that this concept is not left-right symmetric. Among others
it is proved that if the Hurwitz series ring $H(R;\alpha)$ is right qnil-duo,
then $R$ is right qnil-duo. Every right qnil-duo ring is abelian. A right
qnil-duo exchange ring has stable range 1.

2010 MSC: 16U80, 16U60, 16N40

Keywords: Quasinilpotent element, duo ring, qnil-duo ring

##  1\. Introduction

Throughout this paper, all rings are associative with identity. Let $N(R)$,
$J(R)$, $U(R)$, $C(R)$ and Id$(R)$ denote the set of all nilpotent elements,
the Jacobson radical, the set of all invertible elements, the center and the
set of all idempotents of a ring $R$, respectively. We denote the $n\times n$
full (resp., upper triangular) matrix ring over $R$ by $M_{n}(R)$ (resp.,
$U_{n}(R))$, and $D_{n}(R)$ stands for the subring of $U_{n}(R)$ consisting of
all matrices which have equal diagonal entries and $V_{n}(R)=\\{(a_{ij})\in
D_{n}(R)\mid a_{ij}=a_{(i+1)(j+1)}$ for $i=1,\dots,n-2$ and $j=2,\dots,n-1\\}$
is a subring of $D_{n}(R)$. Let $\mathbb{Z}$ and $\mathbb{Z}_{n}$ denote the
ring of integers and the ring of integers modulo $n$ where $n\geq 2$.

In [4], Feller introduced the notion of duo rings, that is, a ring is called
right (resp., left) duo if every right (resp., left) ideal is an ideal, in
other words, $Ra\subseteq aR$ (resp., $aR\subseteq Ra$) for every $a\in R$,
and a ring is said to be duo if it is both right and left duo. The duo ring
property was studied in different aspects. For example, in [5], the concept of
right unit-duo ring was introduced, namely, a ring $R$ is called right unit-
duo if for every $a\in R$, $U(R)a\subseteq aU(R)$. Left unit-duo rings are
defined similarly. In [9], the normal property of elements on Jacobson and nil
radicals were concerned. A ring $R$ is called right normal on Jacobson radical
if $J(R)a\subseteq aJ(R)$ for all $a\in R$. Left normal on Jacobson radical
rings can be defined analogously. Also in [9], on the one hand, a ring $R$ is
said to satisfy the right normal on upper nilradical if $N^{*}(R)a\subseteq
aN^{*}(R)$ for all $a\in R$ where $N^{*}(R)$ is the upper nilradical of $R$.
Similarly, left normal on upper nilradical rings are defined similarly. On the
other hand, a ring $R$ is said to satisfy the right normal on lower nilradical
if $N_{*}(R)a\subseteq aN_{*}(R)$ for all $a\in R$ where $N_{*}(R)$ is the
lower nilradical of $R$. Similarly, left normal on lower nilradical rings are
defined similarly. Also, a ring $R$ is called right nilpotent-duo if
$N(R)a\subseteq aN(R)$ for every $a\in R$. Left nilpotent duo rings are
defined similarly (see [7]).

Motivated by the works on duo property for rings, the goal of this paper is to
approach the notion of duo rings by the way of quasinilpotent elements, in
this regard, we introduce the notion of qnil-duo rings. Firstly, we
investigate some properties of quasinilpotent elements which we need for the
investigation of qnil-duo property. Then we study some properties of this
class of rings and observe that being a qnil-duo ring need not be left-right
symmetric. It is proved that any right (resp., left) qnil-duo ring is abelian,
and any exchange right (resp., left) qnil-duo ring has stable range 1. It is
observed that regularity and strongly regularity coincide for right (resp.,
left) qnil-duo rings. We also study on some extensions of rings such as Dorroh
extensions, Hurwitz series rings and some subrings of matrix rings in terms of
qnil-duo property.

##  2\. Some properties of quasinilpotents

Let $R$ be a ring and $a\in R$. The commutant and double commutant of $a$ in
$R$ are defined by comm$(a)=\\{b\in R\mid ab=ba\\}$ and comm${}^{2}(a)=\\{b\in
R\mid bc=cb$ for all $c\in~{}\mbox{comm}(a)\\}$, respectively, and
$R^{qnil}=\\{a\in R\mid 1+ax$ is invertible in $R$ for every $x\in$
comm$(a)\\}$. Elements of the set $R^{qnil}$ are called quasinilpotent (see
[6]). Note that $J(R)=\\{a\in R\mid 1+ax$ is invertible for $x\in R\\}$. If
$a\in N(R)$ and $x\in$ comm$(a)$, then $ax\in N(R)$ and $1+ax\in U(R)$. So
$J(R)\subseteq R^{qnil}$, $N(R)\subseteq R^{qnil}$ and $R^{qnil}$ does not
contain invertible elements, $0\in R^{qnil}$ but the identity is not in
$R^{qnil}$. In this section, we start to expose some properties of $R^{nil}$
and continue to study some other properties of quasinilpotent elements in
rings.

######  Example 2.1.

There are rings $R$ such that $J(R)$ is strictly contained in $R^{qnil}$.

###### Proof.

Let $F$ be a field and $R=M_{n}(F)$ for some positive integer $n$. Then
$J(R)=0$ and the matrix unit $E_{1n}$ belongs to $R^{qnil}$ but not $J(R)$. ∎

We now mention some of the known facts about quasinilpotents for an easy
reference.

######  Proposition 2.2.

  1. (1)

Let $R$ be a ring, $a\in R$ and $n$ a positive integer. If $a^{n}\in
R^{qnil}$, then $a\in R^{qnil}$, in particular every nilpotent element is in
$R^{qnil}$ ([3, Proposition 2.7]).

  2. (2)

If $R$ is a local ring, then $U(R)\cap R^{qnil}=\emptyset$ and $R=U(R)\cup
R^{qnil}$ ([3, Theorem 3.2]).

  3. (3)

Let $R$ be a ring, $a$, $b\in R$. Then $ab\in R^{qnil}$ if and only if $ba\in
R^{qnil}$ ([11, Lemma 2.2]).

  4. (4)

Let $a\in R^{qnil}$ and $r\in U(R)$. Then $r^{-1}ar\in R^{qnil}$ ([3, Lemma
2.3]).

  5. (5)

Let $e^{2}=e\in(R)$. Then $(eRe)^{qnil}=(eRe)\cap R^{qnil}$ ([16, Lemma 3.5]).

In the following, we determine quasinilpotent elements in some classes of
rings.

######  Lemma 2.3.

Let $R$ be a ring. Then the following hold.

  1. (1)

$\left\\{\begin{bmatrix}a&b\\\ 0&c\end{bmatrix}\mid a,c\in R^{qnil},b\in
R\right\\}\subseteq U_{2}(R)^{qnil}$.

  2. (2)

$\left\\{\begin{bmatrix}a&b\\\ 0&a\end{bmatrix}\mid a\in R^{qnil},b\in
R\right\\}\subseteq D_{2}(R)^{qnil}$.

  3. (3)

Let $A=\begin{bmatrix}a&b\\\ 0&a\end{bmatrix}\in D_{2}(R)^{qnil}$ with $b\in$
comm${}^{2}(a)$. Then $a\in R^{qnil}$.

###### Proof.

(1) Let $a$, $c\in R^{qnil}$, $b\in R$, $A=\begin{bmatrix}a&b\\\
0&c\end{bmatrix}\in U_{2}(R)$ and $B=\begin{bmatrix}x&y\\\
0&z\end{bmatrix}\in$ comm$(A)$. Then $AB=BA$ implies $1-ax$ and $1-cz$ are
invertible. Hence $I-AB$ is invertible. So $A\in R^{qnil}$.
(2) Let $A=\begin{bmatrix}a&b\\\ 0&a\end{bmatrix}\in D_{2}(R)$ with $a\in
R^{qnil}$, $b\in R$ and $B=\begin{bmatrix}x&y\\\ 0&x\end{bmatrix}\in$
comm$(A)$. Then $AB=BA$ implies $x\in$ comm$(a)$. Then $1-ax$ is invertible.
Hence $I_{2}-AB$ is invertible. So $A\in D_{2}(R)^{qnil}$.
(3) Clear. ∎

######  Proposition 2.4.

Let $(R_{i})_{i\in I}$ be a family of rings for some index set $I$ and let
$R=\prod_{i\in I}R_{i}$. Then $R^{qnil}=\prod_{i\in I}R^{qnil}_{i}$.

###### Proof.

Let $(a_{i})$, $(x_{i})\in R$. Then $(x_{i})\in$ comm$(a_{i})$ if and only if
$x_{i}\in$ comm$(a_{i})$ for all $i\in I$. Hence $1+(a_{i})(x_{i})$ is
invertible in $R$ if and only if $1+a_{i}x_{i}$ is invertible in $R_{i}$ for
every $i\in I$. So the result follows. ∎

Let $R$ be an algebra over a commutative ring $S$. The Dorroh extension (or
ideal extension) of $R$ by $S$ denoted by $I(R,S)$ is the direct product
$R\times S$ with usual addition and multiplication defined by
$(a_{1},b_{1})(a_{2},b_{2})=(a_{1}a_{2}+b_{1}a_{2}+b_{2}a_{1},b_{1}b_{2})$ for
$a_{1}$, $a_{2}\in R$ and $b_{1}$, $b_{2}\in S$.

######  Lemma 2.5.

Let $I(R,S)$ be an ideal extension. Then the following hold.

  1. (1)

For $(a,b)\in I(R,S)$, $(c,d)\in$ comm$(a,b)$ if and only if $c\in$ comm$(a)$.

  2. (2)

$(a,b)$ has an inverse $(c,d)$ in $I(R,S)$ if and only if
$(a+b)(c+d)=1=(c+d)(a+b)$ and $bd=db=1$.

###### Proof.

(1) $(c,d)\in$ comm$(a,b)$ if and only if $(a,b)(c,d)=(c,d)(a,b)$ if and only
if $ac+da+bc=ca+da+bc$ and $bd=db$ if and only if $ac=ca$ and $bd=db$ if and
only if $c\in$ comm$(a)$.
(2) $(a,b)(c,d)=(0,1)=(c,d)(a,b)$ if and only if $ac+da+bc=ca+da+bc=0$ and
$bd=db=1$ if and only if $ac+da+bc+bd+(-bd)=(a+b)(c+d)-1=0$ and $bd=db=1$. ∎

######  Proposition 2.6.

Let $I(R,S)$ be an ideal extension of $R$ by $S$. Then

  1. (1)

$(R,0)^{qnil}=(R,0)\cap I(R,S)^{qnil}$.

  2. (2)

$(0,S)\cap I(R,S)^{qnil}\subseteq(0,S)^{qnil}$.

###### Proof.

(1) Let $(x,0)\in(R,0)^{qnil}$ and $(a,b)\in I(R,S)$ with $(a,b)\in$
comm$(x,0)$. Then $a\in$ comm$(x)$ and so $1+xa$ is invertible in $R$. We
prove $(0,1)+(x,0)(a,b)$ is invertible. Since $S$ lies in the center of $R$,
$a+b\in$ comm$(x)$. Hence $1+x(a+b)$ is invertible, say
$(1+x(a+b))(u+1)=(u+1)(1+x(a+b))=1$. This implies that $u(x(a+b))+x(a+b)+u=0$.
Hence $((0,1)+(x,0)(a,b))(u,1)=(u,1)((0,1)+(x,0)(a,b))=(0,1)$ for all
$(a,b)\in$ comm$(x,0)$. So $(x,0)\in I(R,S)^{qnil}$.

Conversely, let $(x,0)\in(R,0)\cap I(R,S)^{qnil}$ and $(r,0)\in$ comm$(x,0)$.
Hence $(0,1)+(x,0)(r,0)=(rx,1)$ is invertible. Let $(a,b)$ be the inverse of
$(rx,1)$. Then $(rx,1)(a,b)=(0,1)$ implies $b=1$ and $rxa+rx+a=0$.
$(a,1)(rx,1)=(0,1)$ implies $arx+a+rx=0$. Hence $(1+a)(1+rx)=1$ and
$(1+rx)(1+a)=1$. Hence $(1,0)+(r,0)(x,0)$ is invertible in $(R,0)$ for all
$(r,0)\in$ comm$(x,0)$. Thus $(x,0)\in(R,0)^{qnil}$ or $(R,0)\cap
I(R,S)^{qnil}\subseteq(R,0)^{qnil}$.
(2) Let $(0,s)\in(0,S)\cap I(R,S)^{qnil}$. Let $(0,b)\in(0,S)$ with $(0,b)\in$
comm$(0,s)$. Then $(0,1)+(0,s)(0,b)=(0,1+sb)$ is invertible in $I(R,S)$. There
exists $(u,v)\in I(R,S)$ such that
$(0,1+sb)(u,v)=((1+sb)u,(1+sb)v)=(0,1)=(u,v)(0,1+sb)=((1+sb)u,v(1+sb))$. Hence
$(1+sb)v=v(1+sb)=1$ and $(1+sb)u=0$. Hence $u=0$. Thus
$(0,1)+(0,s)(0,b)=(0,1+sb)$ is invertible in $(0,S)$ with inverse
$(0,v)\in(0,S)$. It follows that $(0,s)\in(0,S)^{qnil}$ and so $(0,S)\cap
I(R,S)^{qnil}\subseteq(0,S)^{qnil}$. ∎

The following gives us necessary and sufficient conditions for $(0,S)^{qnil}$
to be contained in $I(R,S)^{qnil}$.

######  Theorem 2.7.

Let $I(R,S)$ be the ideal extension of an algebra $R$ by a commutative ring
$S$. Let $(0,i)\in(0,S)^{qnil}$. Then $(0,i)\in I(R,S)^{qnil}$ if and only if
for every $(a,b)\in$ comm$(0,i)$ there exists $(u,v)\in I(R,S)$ such that
$(i(a+b)+1)(u+v)=(1+ib)v=1$.

###### Proof.

Assume that $(0,i)\in I(R,S)^{qnil}$. Let $(a,b)\in$ comm$(0,i)$ in $I(R,S)$.
Then $(0,1)+(0,i)(a,b)$ must be invertible. There exists $(u,v)\in I(R,S)$
such that $(0,1)=((0,1)+(0,i)(a,b))(u,v)$. It follows that
$(0,1)=((0,1)+(0,i)(a,b))(u,v)=(ia,1+ib)(u,v)=(iau+(1+ib)u+v(ia),(1+ib)v)$.
Then $iau+(1+ib)u+iav=0$ and $(1+ib)v=1$. They lead us
$((i(a+b)+1)(u+v)=(1+ib)v$. Hence $(i(a+b)+1)(u+v)$ is invertible. Conversely,
note that $iau+(1+ib)u+iav=0$ if and only if $((i(a+b)+1)(u+v)=(1+ib)v$.
Assume that for $(0,i)\in(0,S)$ there exists $(u,v)\in I(R,S)$ such that
$(i(a+b)+1)(u+v)=(1+ib)v=1$. Then by concealing paranthesis we may reach that
$(0,1)=((0,1)+(0,i)(a,b))(u,v)$ for $(a,b)\in$ comm$(0,i)$. Hence $(0,i)\in
I(R,S)^{qnil}$. ∎

Let $R$ be a ring and $S$ a subring of $R$ with the same identity as that of
$R$ and

$T[R,S]=\\{(r_{1},r_{2},\cdots,r_{n},s,s,\cdots):r_{i}\in R,s\in S,n\geq
1,1\leq i\leq n\\}.$

Then $T[R,S]$ is a ring under the componentwise addition and multiplication.
Note that N$(T[R,S])=T[N(R),N(S)]$ and $C([T,S])=T[C(R),C(R)\cap C(S)]$.

######  Proposition 2.8.

Let $R$ be a ring and $S$ a subring of $R$ with the same identity as that of
$R$.

  1. (1)

If $A=(a_{1},a_{2},a_{3},\dots,a_{n},s,s,s,\dots)\in T[R,S]^{qnil}$, then
$a_{i}\in R^{qnil}$ for $i=1,2,3,\dots,n$ and $s\in S^{qnil}$.

  2. (2)

If $a\in R^{qnil}$ and $s\in S^{qnil}$, then $A=(a,s,s,s,\dots,..)\in
T[R,S]^{qnil}$.

###### Proof.

(1) Let $A=(a_{1},a_{2},a_{3},\dots,a_{n},s,s,s,\dots)\in T[R,S]^{qnil}$ and
$b_{i}\in$ comm$(a_{i})$ and $t\in$ comm$(s)$. Then
$B=(b_{1},b_{2},b_{3},\dots,b_{n},t,t,t,\dots)\in$ comm$(A)$. Let
$1=(1,1,1,\dots,1,\dots)$ denote the identity of $T[R,S]$. So $1+AB$ is
invertible. Therefore $1+a_{i}b_{i}$ is invertible for $i=1,2,3,\dots,n$ and
$1+st$ is invertible in $S$. Hence $a_{i}\in R^{qnil}$ for $i=1,2,3,\dots,n$
and $s\in S^{qnil}$.
(2) Let $a\in R^{qnil}$, $s\in S^{qnil}$, $A=(a,s,s,\dots)$. If
$B=(b_{1},b_{2},\dots,b_{m},t,t,t,\dots.)\in T[R,S]$ lies in comm$(A)$, then
$b_{1}\in$ comm$(a)$, $b_{i}\in$ comm$(s)$ for $i=2,3,\dots,m$ and $t\in$
comm$(s)$. Hence $1+ab_{1}$ and $1+sb_{i}$ are invertible in $R$ where
$i=2,3,\dots,m$ and $1+st$ is invertible in $S$. Hence $1+AB$ is invertible in
$T[R,S]$. So $A=(a,s,s,\dots)\in T[R,S]^{qnil}$. ∎

Let $R$ be a ring with an endomorphism $\alpha$ and let $H(R;\alpha)$ be the
set of formal expressions of the type $f(x)=\sum^{\infty}_{n=0}a_{n}x^{n}$
where $a_{n}\in R$ for all $n\geq 0$. Define addition as componentwise and
$*$-product on $H(R;\alpha)$ as follows: for
$f(x)=\sum^{\infty}_{n=0}a_{n}x^{n}$ and $g(x)=\sum^{\infty}_{n=0}b_{n}x^{n}$,
$f*g=\sum^{\infty}_{n=0}c_{n}x^{n}$ where
$c_{n}=\sum^{n}_{n=0}\binom{n}{i}a_{i}b_{n-i}$. Then $H(R;\alpha)$ becomes a
ring with identity containing $R$ under these two operations. The ring
$H(R;\alpha)$ is called the Hurwitz series ring over $R$. The Hurwitz
polynomial ring $h(R;\alpha)$ is the subring of $H(R;\alpha)$ consisting of
formal expressions of the form $\sum^{n}_{i=0}\binom{n}{i}a_{i}x^{i}$. Let
$\epsilon:H(R;\alpha)\rightarrow R$ defined by $\epsilon(f(x))=a_{0}$. Then
$\epsilon$ is a homomorphism with ker$(\epsilon)=xH(R;\alpha)$ and
$H(R;\alpha)/$ker$(\epsilon)\cong R$. There exist one-to-one correspondences
between $H(R;\alpha)$ and $R$ relating to invertible elements, commutants and
ideals. Let $R[[x;\alpha]]$ be the skew formal power series ring over $R$. The
sum is the same but multiplication in $H(R,\alpha)$ is similar to the usual
multiplication of $R[[x;\alpha]]$, except that binomial coefficients appear in
each term in the multiplication defined in $H(R,\alpha)$. Also, there is a
ring homomorphism $\epsilon$ between $R[[x;\alpha]]$ and $R$, defined by
$\epsilon(f(x))=a_{0}$ where $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots\in
R[[x;\alpha]]$. Clearly, $\epsilon$ is an onto map and
$R[[x;\alpha]]/$ker$(\epsilon)\cong R$.

######  Lemma 2.9.

Let $R$ be a ring and $\alpha$ a ring endomorphism of $R$. Then

  1. (1)

$U(H(R;\alpha))=\epsilon^{-1}U(R)$,

  2. (2)

$U(R[[x;\alpha]])=\epsilon^{-1}U(R)$.

###### Proof.

It is routine. ∎

In the next result, we determine the quasinilpotent elements of $H(R;\alpha)$
and $R[[x;\alpha]]$.

######  Proposition 2.10.

  1. (1)

Let $H(R;\alpha)$ be a Skew Hurwitz series ring over $R$. Then
$H(R;\alpha)^{qnil}=\epsilon^{-1}R^{qnil}$.

  2. (2)

Let $R[[x;\alpha]]$ be a Skew formal power series ring over $R$. Then
$R([[x;\alpha]])^{qnil}=\epsilon^{-1}R^{qnil}$.

###### Proof.

(1) Let $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots\in H(R;\alpha)^{qnil}$ and $r\in
R$ with $r\in$ comm${}_{R}(a_{0})$. Then $r\in$ comm${}_{H(R;\alpha)}(a)$.
Then $1+f(x)r\in U(H(R;\alpha))$. Hence $1+a_{0}r\in U(R)$. So $a_{0}\in
R^{qnil}$. Since $a_{0}=\epsilon(f(x))$, $f(x)\in\epsilon^{-1}(R^{qnil})$.
Conversely, let
$g(x)=b_{0}+b_{1}x+b_{2}x^{2}+\dots\in\epsilon^{-1}(R^{qnil})$. Then
$\epsilon(g(x))=b_{0}\in R^{qnil}$. Let
$h(x)=c_{0}+c_{1}x+c_{2}x^{2}+\dots\in$ comm${}_{H(R,\alpha)}(g(x))$. Then
$c_{0}\in$ comm${}_{R}(b_{0})$ and so $1+b_{0}c_{0}\in U(R)$. So
$1+g(x)h(x)\in U(H(R,\alpha))$. Hence $g(x)\in H(R,\alpha)^{qnil}$. This
completes the proof.
(2) Similar that of (1). ∎

##  3\. Qnil-duo Rings

In this section, we deal with the right duo property on the set of
quasinilpotent elements. By this means, we give a generalization of
commutativity from the perspective of quasinilpotents.

######  Definition 3.1.

A ring $R$ is called right qnil-duo if $R^{qnil}a\subseteq aR^{qnil}$ for
every $a\in R$. Similarly, $R$ is called left qnil-duo if $aR^{qnil}\subseteq
R^{qnil}a$ for every $a\in R$. If $R$ is both right and left qnil-duo, then it
is called qnil-duo, i.e. $R^{qnil}a=aR^{qnil}$ for every $a\in R$.

The qnil-duo property of rings is not left-right symmetric as the following
example shows.

######  Example 3.2.

Let $S=F(t)$ denote the quotient field of the polynomial ring $F[t]$ over a
field $F$ and $\alpha\colon S\rightarrow S$ defined by
$\alpha(f(t)/g(t))=f(t^{2})/g(t^{2})$. Let $R=S[[x;\alpha]]$ denote the skew
power series ring with $xa=\alpha(a)x$ for $a\in S$. Every element of $R$ is
of the form $a=\sum^{\infty}_{i=0}a_{i}x^{i}$. For any
$r=a_{0}+\sum^{\infty}_{i=1}a_{i}x^{i}$ with $a_{0}\neq 0$ is invertible.
Hence $R^{qnil}=xR$. This ring is considered in [2, Lemma 1.3(3)], [8, Example
1] and in [7, Example 1.5]. As in the proof of [8, Example 1], for $tx^{m}\in
tR^{qnil}$, there is no $g(x)\in R^{qnil}$ such that $tx^{m}=g(x)t$. Hence $R$
is not left qnil-duo. We show that $R$ is right qnil-duo. Let $f(x)\in
R^{qnil}$, $g(x)\in R$. We show that there exists $f_{1}(x)\in R^{qnil}$ such
that $f(x)g(x)=g(x)f_{1}(x)$. Assume that $g(x)$ is invertible. Then
$f(x)g(x)=g(x)(g(x)^{-1}f(x)g(x)\in g(x)R^{qnil}$, otherwise, let
$g(x)=h(x)x^{m}$ where $h(x)=a_{0}+a_{1}x^{+}a_{2}x^{2}+\cdots$ is invertible.
Then
$f(x)g(x)=f(x)h(x)x^{m}=f(x)x^{m}h_{1}(x)=x^{m}f_{1}(x)h_{1}(x)=x^{m}h_{1}(x)(h_{1}(x)^{-1}f_{1}(x)h_{1}(x))=g(x)(h_{1}(x)^{-1}f_{1}(x)h_{1}(x))\in
g(x)R^{qnil}$ since $f(x)$ is not invertible and $f_{1}(x)$ is an application
of $x^{m}$ to $f(x)$ from the right, therefore $f_{1}(x)=x^{k}f_{2}(x)\in
R^{qnil}$ for some $k\geq 1$, by Proposition 2.2(4),
$h_{1}(x)^{-1}f_{1}(x)h_{1}(x)\in R^{qnil}$. Thus $R$ is right qnil-duo.

######  Examples 3.3.

(1) All commutative rings, all division rings are qnil-duo.
(2) There are local rings that are not right qnil-duo.

###### Proof.

(1) When $R$ is a commutative ring, it is both right and left qnil-duo. If $R$
is a division ring, then $R^{qnil}=\\{0\\}$, therefore $R$ is both right and
left qnil-duo.
(2) Let $A=\mathbb{Z}_{4}[x,y]$ be the polynomial ring with non-commuting
indeterminates $x$ and $y$ and $I$ be the ideal generated by the set
$\\{x^{3},y^{2},yx,x^{2}-xy,x^{2}-2,2x,2y\\}$. Consider the ring $R=A/I$. By
[15, Example 7], $R$ is a local ring. It is easily checked that
$R^{qnil}=\\{0,2,x,y,2+x,2+y,2+x+y,x+y\\}$ and $(R^{qnil})^{2}\neq 0$. $2+x$
belongs to $R^{qnil}$ since it is nilpotent. For $x\in R^{qnil}$ and $y\in R$,
$xy\in R^{qnil}y$. It is easily checked that there is no $t\in R^{qnil}$ such
that $xy=yt\in yR^{qnil}$. Hence $R$ is not right qnil-duo. ∎

######  Lemma 3.4.

Let $R$ be a ring with $R^{qnil}$ central in $R$. Then $R$ is qnil-duo.

###### Proof.

Assume that $R^{qnil}$ is central in $R$. Let $a\in R$ and $b\in R^{qnil}$.
Then $b$ being central implies $ab=ba\in aR^{qnil}$. ∎

######  Theorem 3.5.

Let $\\{R_{i}\\}_{i\in I}$ be a family of rings for some index set $I$ and
$R=\prod_{i\in I}R_{i}$. Then $R_{i}$ is right (resp., left) qnil-duo for each
$i\in I$ if and only if $R$ is right (resp., left) qnil-duo.

###### Proof.

Assume that $R_{i}$ is right (resp., left) qnil-duo for each $i\in I$. Let
$a=(a_{i})\in R$, $b=(b_{i})\in R^{qnil}$. By Proposition 2.4, $b_{i}\in
R_{i}^{qnil}$ for each $i\in I$. By assumption there exists $c_{i}\in
R_{i}^{qnil}$ such that $b_{i}a_{i}=a_{i}c_{i}$ for each $i\in I$. Set
$c=(c_{i})$. Then $ba=ac\in aR^{qnil}$. Hence $R^{qnil}a\subseteq aR^{qnil}$.
Conversely, suppose that $R$ is right qnil-duo. Let $a_{i}\in R_{i}$ and
$b_{i}\in R_{i}^{qnil}$ where $i\in I$. Let $a=(a_{i})$, $b=(b_{i})\in R$
where $i^{th}$-entry of $a$ is $a_{i}$ and the other entries are 0 and
$i^{th}$-entry of $b$ is $b_{i}$ and the other entries are 0, respectively.
Then $a=(a_{i})\in R$ and by Proposition 2.4, $b\in R^{qnil}$. Supposition
implies there exists $c=(c_{i})\in R^{qnil}$ such that $ba=ac$. Comparings
entries of both sides we have $b_{i}a_{i}=a_{i}c_{i}$. By Proposition 2.4,
$c_{i}\in R_{i}^{qnil}$. Thus for each $i\in I$, $R_{i}$ is right qnil-duo.
Similarly, it is proven that for each $i\in I$, $R_{i}$ is left qnil-duo. ∎

Recall that a ring $R$ is called abelian if every idempotent in $R$ is
central.

######  Theorem 3.6.

Let $R$ be a ring. Then the following hold.

  1. (1)

$ex-exe$ and $xe-exe\in R^{qnil}$ for every $x$ and $e^{2}=e\in R$.

  2. (2)

Right (resp., left) qnil-duo rings are abelian.

  3. (3)

Let R be a ring and $e\in$ Id$(R)$. If $R$ is right (resp., left) qnil-duo
ring, then $eR$ and $(1-e)R$ are right (resp., left) qnil-duo rings. The
converse holds if $e$ is central.

###### Proof.

(1) Let $t\in$ comm$(ex-exe)$. Then $t(ex-exe)=(ex-exe)t$. So we have $(t(xe-
exe))^{2}=0$. Hence $1-(ex-exe)t$ is invertible and so $ex-exe\in R^{qnil}$.
Similarly, $xe-exe\in R^{qnil}$.
(2) Let $e^{2}=e\in R$. By hypothesis, $R^{qnil}e\subseteq eR^{qnil}$. By (1),
$xe-exe\in R^{qnil}$ for all $x\in R$. It implies for any $x\in R$, there
exists $t\in R^{qnil}$ such that $(xe-exe)e=et$. Multiplying the latter
equality by $e$ from the left we have $et=0$. So $xe=exe$. Similarly, $ex=exe$
since $ex-exe\in R^{qnil}$ by (1). Hence $R$ is abelian.
(3) It is clear by Theorem 3.5. ∎

######  Corollary 3.7.

Let $R$ be a right (resp., left) qnil-duo ring and $e\in$ Id$(R)$. Then the
corner ring $eRe$ is a right (resp., left) qnil-duo ring.

###### Proof.

The ring $R$ being right (resp., left) qnil-duo, $e$ is central in $R$ by
Theorem 3.6(2). Hence Theorem 3.6(3) completes the proof. ∎

######  Theorem 3.8.

Every right (resp., left) qnil-duo ring is directly finite.

###### Proof.

Let $R$ be a right qnil-duo ring and $a$, $b\in R$ with $ab=1$. Set $e=1-ba$.
Then $e$ is an idempotent. By Theorem 3.6, $e$ is central. So $0=ae=ea$. Hence
$0=a-ba^{2}$. Multiplying the latter by $b$ from the right, we get $1=ba$. ∎

There is a directly finite ring that is neither right nor left qnil-duo.

######  Example 3.9.

Consider the ring $R=M_{2}(\mathbb{Z}_{2})$. Then $R$ is a directly finite
ring but not abelian. Hence it is neither right nor left qnil-duo.

We apply Theorem 3.6 to show that full matrix rings and upper triangular
matrix rings need not be right (resp., left) qnil-duo. But there are some
subrings of full matrix rings that are qnil-duo.

######  Examples 3.10.

  1. (1)

For any ring $R$ and any positive integer $n$, $M_{n}(R)$ and $U_{n}(R)$ are
neither right nor left qnil-duo.

  2. (2)

If $R$ is commutative, then $V_{n}(R)$ is qnil-duo.

  3. (3)

$V_{n}(R[[x;\sigma]])$ is neither right nor left qnil-duo.

###### Proof.

(1) The rings $M_{n}(R)$ and $U_{n}(R)$ are not abelian. By Theorem 3.6(2),
they are neither right nor left qnil-duo.
(2) If $R$ is a commutative ring, $V_{n}(R)$ is also commutative, therefore it
is right and left qnil-duo.
(3) Let $R$ be a ring with an endomorphism $\sigma$. Assume that there exists
$a_{1}\in R$ such that $\sigma(a_{1})\notin a_{1}R$. Let
$A=\begin{bmatrix}x&x&x\\\ 0&x&x\\\ 0&0&x\end{bmatrix}\in
V_{3}(R[[x;\sigma]])^{qnil}$, $B=\begin{bmatrix}a_{1}&a_{2}&a_{3}\\\
0&a_{1}&a_{2}\\\ 0&0&a_{1}\end{bmatrix}\in V_{3}(R[[x;\sigma]])$. Assume that
there exists $D\in V_{3}(R[[x;\sigma]])^{qnil}$ such that $AB=BD$. Then
$(1,1)$ entry of $AB$ is $\sigma(a_{1})x$ and that of $BD$ is $a_{1}xf(x)$ for
some $f(x)\in R[[x;\sigma]]$. This contradicts with the choice of $\sigma$ and
$a_{1}$. Therefore $V_{3}(R[[x;\sigma]])$ is not right qnil-duo. Similarly, it
can be shown that $V_{3}(R[[x;\sigma]])$ is not left qnil-duo. ∎

######  Theorem 3.11.

Let $R$ be a local ring with $(R^{qnil})^{2}=0$. Then $R$ is right (resp.,
left) qnil-duo.

###### Proof.

By Proposition 2.2(2), we have $R=U(R)\cup R^{qnil}$. We prove
$R^{qnil}a\subseteq aR^{qnil}$. Let $a\in R$, $b\in R^{qnil}$. If $ba=0$, then
we are done since $ba=0=a0\in aR^{qnil}$. Otherwise, i.e., if $ba\neq 0$, then
we divide the proof in some cases.
Case I. Let $a\notin R^{qnil}$. Then $a\in U(R)$. By Proposition 2.2(4),
$a^{-1}ba\in R^{qnil}$ since $b\in R^{qnil}$. Then $ba=a(a^{-1}ba)\in
aR^{qnil}$.
Case II. Let $a\in R^{qnil}$. By hypothesis, $ba=0$, this contradicts with
$ba\neq 0$.
Therefore $R$ is right qnil-duo ring. Similarly, we may prove
$aR^{qnil}\subseteq R^{qnil}a$ for each $a\in R$. ∎

As an illustration of Theorem 3.11, we give the following examples. Also, the
condition $(R^{qnil})^{2}=0$ in Theorem 3.11 is not superfluous.

######  Example 3.12.

(1) Consider the ring $R=\left\\{\begin{bmatrix}a&b&c\\\ 0&a&0\\\
0&0&a\end{bmatrix}\in D_{3}(\mathbb{Z}_{4})\right\\}$. Then
$R^{qnil}=\left\\{\begin{bmatrix}a&b&c\\\ 0&a&0\\\ 0&0&a\end{bmatrix}\in R\mid
a\in 2\mathbb{Z}_{4},b,c\in\mathbb{Z}_{4}\right\\}$. So $(R^{qnil})^{2}=0$. By
Theorem 3.11, $R$ is qnil-duo.
(2) Let $R$ denote the ring in Examples 3.3. Then
$R^{qnil}=\\{0,2,x,y,2+x,2+y,2+x+y,x+y\\}$ and $(R^{qnil})^{2}\neq 0$ and
$2+x$ belongs to $R^{qnil}$ since it is nilpotent and $(2+x)^{2}\neq 0$. Since
$R$ is local and $R^{qnil}$ does not contain invertible elements,
$R^{qnil}=J(R)$. To complete the proof we may assume that $x$, $y\in
R^{qnil}$. Then $xy=2$ and $xy\in R^{qnil}y$. It is easily checked that there
is no $t\in R^{qnil}$ such that $xy=yt\in yR^{qnil}$. Hence $R$ is not right
qnil-duo. Compare to Theorem 3.11.

Note that by Theorem 3.11, if $R$ is a division ring, $D_{2}(R)$ is a qnil-duo
ring. One may ask whether $D_{2}(R)$ is qnil-duo over a domain $R$. The
following example answers negatively.

######  Example 3.13.

Consider the ring $D_{2}(R[[x]])$ in [9, Example 1.4(1)]. It is proved that
$D_{2}(R[[x]])$ is neither right nor left normal property of elements on
Jacobson radical. Since $J(D_{2}(R[[x]]))=D_{2}(R[[x]])^{qnil}$,
$D_{2}(R[[x]])$ is neither right nor left qnil-duo.

######  Theorem 3.14.

Let $R$ be a domain. If $D_{2}(R)$ is right (resp., left) qnil-duo, then $R$
is right (resp., left) qnil-duo.

###### Proof.

Assume that $D_{2}(R)$ is right qnil-duo. Let $a\in R$ and $b\in R^{qnil}$.
Consider $A=\begin{bmatrix}a&0\\\ 0&a\end{bmatrix}\neq 0$,
$B=\begin{bmatrix}b&0\\\ 0&b\end{bmatrix}\neq 0$. Let $X=\begin{bmatrix}x&y\\\
0&x\end{bmatrix}\in$ comm$(B)$. Then $I_{2}-BX$ is invertible since $1-bx$ is
invertible in $R$. Hence $BA\in D_{2}(R)^{qnil}A$. There exists
$C=\begin{bmatrix}c&d\\\ 0&c\end{bmatrix}\in D_{2}(R)^{qnil}$ such that
$BA=AC$. Then $ba=ac$ and $ad=0$. By hypothesis $d=0$. By Lemma 2.3(3), $c\in
R^{qnil}$. It follows that $ba=ac\in aR^{qnil}$. Hence $R^{qnil}a\subseteq
aR^{qnil}$. ∎

Recall that a ring $R$ is said to have stable range 1 if for any $a$, $b\in R$
satisfying $aR+bR=R$, there exists $y\in R$ such that $a+by$ is right
invertible (cf. [14]). In [12], a ring $R$ is called exchange if for any $x\in
R$, there exists $e\in$ Id$(R)$ such that $e\in Rx$ and $1-e\in R(1-x)$, and
it is proved that for an abelian ring $R$, $R$ is exchange if and only if it
is clean, and $R$ is exchange if and only if idempotents lift modulo every
left (or right) ideal.

######  Theorem 3.15.

The following hold.

  1. (1)

Right (resp., left) qnil-duo exchange rings have stable range 1.

  2. (2)

Right (resp., left) qnil-duo regular rings (in the sense of von Neumann) are
strongly regular.

###### Proof.

(1) Let $R$ be a right qnil-duo exchange ring. By Theorem 3.6, $R$ is abelian.
Hence [17, Theorem 6] implies $R$ has stable range 1.
(2) Let $R$ be a qnil-duo regular ring and $a\in R$. There exists $b\in R$
such that $a=aba$. Then $ab=(ab)^{2}$, $ba=(ba)^{2}\in$ Id$(R)$. By Theorem
3.6, $ab$ is central. So $a=aba=a^{2}b$. Hence $R$ is strongly regular. ∎

Let $R$ be a ring. The Jacobson radical of the polynomial ring $R[x]$ is
$J(R[x])=N[x]$ where $N=J(R[x])\cap R$ is a nil ideal of $R$. Then $N\subseteq
R^{qnil}$ and $J(R[[x]])=xR[[x]]$. Therefore, ${R[[x]]}^{qnil}=xR[[x]]$. One
may wonder whether or not $R[x]$ and $R[[x]]$ are qnil-duo. The following
example shows that $R[x]$ and $R[[x]]$ need not be right qnil-duo.

######  Example 3.16.

(1) Let $F$ be a field, $R=M_{n}(F)$ and consider the ring $R[x]$. Observe
that $M_{n}(F[x])$ is not right (or left) qnil-duo for any positive integer
$n\geq 2$ by Examples 3.10. It follows that $R[x]$ is not right (or left)
qnil-duo since $M_{n}(F)[x]\cong M_{n}(F[x])$.
(2) Let $R=A/(ab-ba-1)$ denote the Weyl algebra discussed in [9, Example
1.2(2)]. Let $S=R[[x]]$. Then $S^{qnil}=xR[[x]]=J(R[[x]])$, $R$ is a domain
and $R[[x]]$ is abelian. It is proved that $R[[x]]$ is neither right normal
nor left normal on $J(R)$. Therefore, $R[[x]]$ is neither right qnil-duo nor
left qnil-duo.

######  Theorem 3.17.

Let $R$ be an algebra over a commutative ring $S$. Consider the Dorroh
extension (or ideal extension) $I(R,S)$ of $R$ by $S$. If $I(R,S)$ is right
qnil-duo, then so is $R$.

###### Proof.

Assume that $I(R,S)$ is right qnil-duo. Let $a\in R$, $b\in R^{qnil}$. Then
$(a,0)\in I(R,S)$ and $(b,0)\in I(R,S)^{qnil}$. Indeed, let $(x,y)\in$
comm$(b,0)$. By Lemma 2.5, $x\in$ comm$(b)$. Since $R$ is an algebra over $S$,
we have $x+y\in$ comm$(b)$. Then $1+b(x+y)$ is invertible in $R$ with inverse
$t$. Again by Lemma 2.5, $(0,1)+(b,0)(x,y)$ is invertible in $I(R,S)$ with the
inverse $(t-1,1)$. Then $(b,0)(a,0)\in I(R,S)^{qnil}(a,0)$. There exists
$(c,s)\in I(R,S)^{qnil}$ such that $(b,0)(a,0)=(a,0)(c,s)$. So $ba=a(c+s)$. To
complete the proof we show $c+s\in R^{qnil}$. Let $x\in$ comm$(c+s)$. Then
$cx+sx=sc+xs$. Since $R$ is an algebra over $S$, $sx=xs$, this implies
$cx=xc$, and so $x\in$ comm$(c)$. Hence $(x,0)\in$ comm$(c,s)$. Since
$(c,s)\in I(R,S)^{qnil}$, $(0,1)+(c,s)(x,0)$ is invertible in $I(R,S)$. Thus
$1+(c+s)x$ is invertible in $R$ by Lemma 2.5(2). So $c+s\in R^{qnil}$.
Therefore $R$ is right qnil-duo. ∎

######  Proposition 3.18.

Let $R$ be a ring and $S$ a subring of $R$. If $T[R,S]$ is right qnil-duo,
then so are $R$ and $S$. The converse holds if $S^{qnil}\subseteq R^{qnil}$.

###### Proof.

Assume that $T[R,S]$ is right qnil-duo. Let $a\in R$, $b\in R^{qnil}$. Let
$A=(a,0,0,0,\dots)$, $B=(b,0,0,0,\dots)\in T[R,S]$. By Proposition 2.8, $B\in
T[R,S]^{qnil}$. By supposition there exists
$C=(c_{1},c_{2},\dotsb,c_{m},t,t,\dots.)\in T[R,S]^{qnil}$ such that $BA=AC$.
Hence $ba=ac_{1}$. By Proposition 2.8, $c_{1}\in R^{qnil}$. Similarly, let
$s\in S$, $t\in S^{qnil}$ and $C=(0,s,s,s,s,\dots,s,\dots)$,
$D=(0,t,t,t,\dots.)\in T[R,S]$. By Proposition 2.8, $D\in T[R,S]^{qnil}$.
There exists $D^{\prime}=(d_{1},d_{2},d_{3},\dots.,d_{l},u,u,u,\dots)\in
T[R,S]^{qnil}$ such that $DC=CD^{\prime}$. By Proposition 2.8, $u\in S^{qnil}$
and $ts=su\in sS^{qnil}$.

Suppose that $R$ and $S$ are right qnil-duo and $S^{qnil}\subseteq R^{qnil}$.
Let $A\in T[R,S]$, $B\in T[R,S]^{qnil}$ where
$A=(a_{1},a_{2},\dotsb,a_{n},s,s,\dotsb)$,
$B=(b_{1},b_{2},\dots,b_{m},t,t,\dots)$, we prove $BA=AC$ for some $C\in
T[R,S]^{qnil}$. By Proposition 2.8, $b_{i}\in R^{qnil}$ for $i=1,2,\dots,m$
and $t\in S^{qnil}$. By supposition $b_{i}\in R^{qnil}$ implies
$b_{i}a_{i}=a_{i}c_{i}$ for some $c_{i}\in R^{qnil}$. We divide the proof in
some cases:
Case I. $n\leq m$. Then $b_{i}a_{i}\in R^{qnil}a_{i}$. Since $R$ is right
qnil-duo, there exist $c_{i}\in R^{qnil}$ such that $b_{i}a_{i}=a_{i}c_{i}$
for each $1\leq i\leq n$. For $n+1\leq i\leq m$, $b_{i}s\in R^{qnil}s$. There
exist $c_{i}\in R^{qnil}$ such that $b_{i}s=sc_{i}$. For $ts\in S^{qnil}s$,
there exists $l\in S^{qnil}$ such that $ts=sl\in sS^{qnil}$. Let
$C=(c_{1},c_{2},c_{3},\dots,c_{m},l,l,l,\dots)$. By Proposition 2.8(2), $C\in
T[R,S]^{qnil}$. Then $BA=AC\in AT[R,S]^{qnil}$.
Case II. $n>m$. Let $1\leq i\leq m$. Then $b_{i}a_{i}\in R^{qnil}a_{i}$ and
since $R$ is right qnil-duo, there exist $c_{i}\in R^{qnil}$ such that
$b_{i}a_{i}=a_{i}c_{i}$. For $m+1\leq i\leq n$, $ta_{i}\in S^{qnil}a_{i}$. By
$S^{qnil}\subseteq R^{qnil}$, we have $ta_{i}=a_{i}c_{i}\in a_{i}R^{qnil}$ for
some $c_{i}\in R^{qnil}$. For $ts\in S^{qnil}s$, by supposition there exists
$l\in S^{qnil}$ such that $ts=sl\in sS^{qnil}$. Let
$C=(c_{1},c_{2},c_{3},\dots,c_{n},l,l,l,\dots)$. By Proposition 2.8(2), $C\in
T[R,S]^{qnil}$. Then $BA=AC$. Hence $T[R,S]^{qnil}A\subseteq AT[R,S]^{qnil}$.
It completes the proof. ∎

######  Theorem 3.19.

  1. (1)

Let $H(R;\alpha)$ be a Skew Hurwitz series ring over a ring $R$. If
$H(R;\alpha)$ is right qnil-duo, then $R$ is right qnil-duo.

  2. (2)

Let $R[[x;\alpha]]$ be a skew formal power series ring over a ring $R$. If
$R[[x;\alpha]]$ is right qnil-duo, then $R$ is right qnil-duo.

###### Proof.

Suppose that $H(R;\alpha)$ is a right qnil-duo ring. Let $a\in R^{qnil}$ and
$b\in R$. By the definition of $\epsilon$ and Proposition 2.10, there exist
$f(x)$, $g(x)\in H(R;\alpha)$ with $f(x)\in H(R;\alpha)^{qnil}$ and
$\epsilon(f(x))=a$, $\epsilon(g(x))=b$. There exists
$h(x)=c_{0}+c_{1}x+c_{2}x^{2}+\dots\in H(R;\alpha)^{qnil}$ such that
$f(x)g(x)=g(x)h(x)$. Hence $\epsilon(f(x)g(x))=\epsilon(g(x)h(x))$ implies
$ab=bc_{0}\in bR^{qnil}$. Thus $R^{qnil}b\subseteq bR^{qnil}$.
(2) Similar to that of (1). ∎

##  4\. Some subrings of matrix rings

Besides, for any ring $R$ and any positive integer $n\geq 2$, $M_{n}(R)$ is
not right (or left) qnil-duo, in this section, quasinilpotent elements of some
subrings of full matrix rings are determined for the purpose of the use
whether or not their subrings to be right (or left) qnil-duo.

The rings $L_{(s,t)}(R)$: Let $R$ be a ring and $s$, $t\in C(R)$.
Let $L_{(s,t)}(R)=\left\\{\begin{bmatrix}a&0&0\\\ sc&d&te\\\
0&0&f\end{bmatrix}\in M_{3}(R)\mid a,c,d,e,f\in R\right\\}$, where the
operations are defined as those in $M_{3}(R)$. Then $L_{(s,t)}(R)$ is a
subring of $M_{3}(R)$.

######  Lemma 4.1.

Let $A=\begin{bmatrix}a&0&0\\\ sc&d&te\\\ 0&0&f\end{bmatrix}\in L_{(s,t)}(R)$.
Then the following hold.

  1. (1)

$A$ is invertible in $L_{(s,t)}(R)$ if and only if $a$, $d$ and $f$ are
invertible in $R$.

  2. (2)

If $a$, $d$, $f\in R^{qnil}$, then $A\in L_{(s,t)}(R)^{qnil}$.

###### Proof.

(1) One way is clear. Let $A=\begin{bmatrix}a&0&0\\\ sc&d&te\\\
0&0&f\end{bmatrix}\in L_{(s,t)}(R)$. Assume that $a$, $d$ and $f$ are
invertible with $ax=xa=1$, $dz=zd=1$ and $fv=vf=1$ where $x,z,v\in R$.
Consider $B=\begin{bmatrix}x&0&0\\\ sy&z&tu\\\ 0&0&v\end{bmatrix}\in
L_{(s,t)}(R)$ where $y=-zcx$ and $u=-zev$. Then $AB=BA=I_{3}$.
(2) Assume that $a$, $d$, $f\in R^{qnil}$. We prove that $A\in
L_{(s,t)}(R)^{qnil}$. Let $B=\begin{bmatrix}x&0&0\\\ sy&z&tu\\\
0&0&v\end{bmatrix}\in L_{(s,t)}(R)$ with $B\in$ comm$(A)$. It is easily
checked that $x\in$ comm$(a)$, $z\in$ comm$(d)$, $v\in$ comm$(f)$. Then
$1+ax$, $1+dz$, $1+fv$ are invertible in $R$. By (1),
$I_{3}+AB=\begin{bmatrix}1+ax&0&0\\\ scx+sdy&1+dz&tdu+tev\\\
0&0&1+fv\end{bmatrix}$ is invertible. So $A\in L_{(s,t)}(R)^{qnil}$. ∎

######  Lemma 4.2.

Let $A=\begin{bmatrix}a&0&0\\\ sc&d&te\\\ 0&0&f\end{bmatrix}\in L_{(s,t)}(R)$.
Then the following hold.

  1. (1)

If $A\in L_{(0,t)}(R)^{qnil}$, then $a\in R^{qnil}$.

  2. (2)

If $A\in L_{(s,0)}(R)^{qnil}$, then $f\in R^{qnil}$.

  3. (3)

$A\in L_{(0,0)}(R)^{qnil}$ if and only if $a,d,f\in R^{qnil}$.

###### Proof.

(1) Let $A=\begin{bmatrix}a&0&0\\\ 0&d&te\\\ 0&0&f\end{bmatrix}\in
L_{(0,t)}(R)^{qnil}$ and $x\in$ comm$(a)$. Consider $B=\begin{bmatrix}x&0&0\\\
0&0&0\\\ 0&0&0\end{bmatrix}\in L_{(0,t)}(R)$. Then $B\in$ comm$(A)$. Since
$A\in L_{(0,t)}(R)^{qnil}$, $I+AB$ is invertible in $L_{(0,t)}(R)$. By Lemma
4.1(1), $1+ax\in U(R)$. Therefore $a\in R^{qnil}$.
(2) Similar to the proof of (1).
(3) The sufficiency follows from Lemma 4.1(2). For the necessity, $a,f\in
R^{qnil}$ by (1) and (2), respectively. Also, by the similar discussion in
(1), we obtain $d\in R^{qnil}$. ∎

######  Theorem 4.3.

Let $R$ be a ring. If $L_{(0,t)}(R)$ is right qnil-duo, then $R$ is a right
qnil-duo ring.

###### Proof.

Assume that $L_{(0,t)}(R)$ is right qnil-duo and let $a\in R$ and $b\in
R^{qnil}$. Consider $A=\begin{bmatrix}a&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}$,
$B=\begin{bmatrix}b&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\in L_{(0,t)}(R)$. By
Lemma 4.1, $B\in L_{(0,t)}(R)^{qnil}$. By supposition there exists
$B^{\prime}=\begin{bmatrix}x&0&0\\\ 0&z&tu\\\ 0&0&v\end{bmatrix}\in
L_{(0,t)}(R)^{qnil}$ such that $BA=AB^{\prime}$. It implies $ba=ax$. By Lemma
4.2(1), $x\in R^{qnil}$. Hence $ba=ax\in aR^{qnil}$. Thus $R^{qnil}a\subseteq
aR^{qnil}$. ∎

There are right qnil-duo rings $R$ such that the rings $L_{(s,t)}(R)$ need not
be right qnil-duo as shown below.

######  Example 4.4.

The ring $L_{(1,1)}(\mathbb{Z}_{4})$ is not right qnil-duo.

###### Proof.

Let $A=\begin{bmatrix}0&0&0\\\ 1&2&1\\\ 0&0&3\end{bmatrix}\in
L_{(1,1)}(\mathbb{Z}_{4})$ and $B=\begin{bmatrix}2&0&0\\\ 1&2&3\\\
0&0&2\end{bmatrix}\in L_{(1,1)}(\mathbb{Z}_{4})^{qnil}$. Assume that there
exists $C=\begin{bmatrix}x&0&0\\\ y&z&v\\\ 0&0&u\end{bmatrix}\in
L_{(1,1)}(\mathbb{Z}_{4})^{qnil}$ such that $BA=AC$. Then
$BA=\begin{bmatrix}0&0&0\\\ 2&0&3\\\ 0&0&2\end{bmatrix}$ and
$AC=\begin{bmatrix}0&0&0\\\ x+2y&2z&2v+u\\\ 0&0&3u\end{bmatrix}$. $BA=AC$
implies $3=2v+u$ and $2=3u$. These equations lead us a contradiction. Hence
$L_{(1,1)}(\mathbb{Z}_{4})$ is not right qnil-duo. ∎

The rings $H_{(s,t)}(R)$: Let $R$ be a ring and $s$, $t\in C(R)$ be invertible
in $R$. Let

$H_{(s,t)}(R)=\left\\{\begin{bmatrix}a&0&0\\\ c&d&e\\\ 0&0&f\end{bmatrix}\in
M_{3}(R)\mid a,c,d,e,f\in R,a-d=sc,d-f=te\right\\}$.

Then $H_{(s,t)}(R)$ is a subring of $M_{3}(R)$.

######  Lemma 4.5.

Let $A=\begin{bmatrix}a&0&0\\\ c&d&e\\\ 0&0&f\end{bmatrix}$,
$B=\begin{bmatrix}x&0&0\\\ y&z&u\\\ 0&0&v\end{bmatrix}\in H_{(s,t)}(R)$. Then

  1. (1)

$AB=BA$ if and only if $ax=xa$, $dz=zd$, $fv=vf$.

  2. (2)

$A$ is invertible with inverse $B$ if and only if $ax=xa=1$, $dz=zd=1$,
$fv=vf=1$.

  3. (3)

$A\in H_{(s,t)}(R)^{qnil}$ if and only if $a$, $d$, $f\in R^{qnil}$.

###### Proof.

(1) The necessity is clear. For the sufficiency, suppose that $ax=xa$,
$dz=zd$, $fv=vf$. The matrix $AB$ has $cx+dy$ as $(2,1)$ entry, $du+ev$ as
$(2,3)$ entry and $BA$ has $ya+zc$ as $(2,1)$ entry, $ze+uf$ as $(2,3)$ entry.
To show $AB=BA$ it is enough to get $cx+dy=ya+zc$ and $du+ev=ze+uf$. Now
$scx+sdy=ax+d(sy-x)=ax-dz=xa-za+za-dz=sya+szc$. So $cx+dy=ya+zc$ since $s$ is
invertible. Similarly, we get $du+ev=ze+uf$.
(2) One way is clear. Assume that $ax=xa=1$, $dz=zd=1$, $fv=vf=1$. Let $B\in
H_{(s,t)}(R)$ with $y=-zcx$ and $u=-zev$. Then $AB=BA=I_{3}$.
(3) Assume that $A\in H_{(s,t)}(R)^{qnil}$. Let $x\in$ comm$(a)$, $y\in$
comm$(f)$. Let $D=\begin{bmatrix}x&0&0\\\ s^{-1}x&0&-t^{-1}y\\\
0&0&y\end{bmatrix}$. Then $D\in$ comm$(A)$. In fact, $scx=(a-d)x$ and
$tey=(d-f)y$. Hence $I_{3}+AD$ is invertible in $H_{(s,t)}(R)$. It follows
that $1+ax$, $1+fy\in U(R)$. So $a,f\in R^{qnil}$. As for $d\in R^{qnil}$, let
$r\in$ comm$(d)$ and $D=\begin{bmatrix}0&0&0\\\ -s^{-1}r&r&t^{-1}r\\\
0&0&0\end{bmatrix}$. Then $D\in$ comm$(A)$. By assumption $I_{3}+AD\in
U(H_{(s,t)}(R))$. Hence $1+dr\in U(R)$. Hence $d\in R^{qnil}$. Conversely,
suppose that $a$, $d$, $f\in R^{qnil}$. Let $B\in$ comm$(A)$. Then $x\in$
comm$(a)$, $z\in$ comm$(d)$ and $v\in$ comm$(f)$. By supposition, $1+ax$,
$1+dy$ and $1+fv$ are invertible. By part (2), $I_{3}+AB\in U(H_{(s,t)}(R))$.
Hence $A\in H_{(s,t)}(R)^{qnil}$. This completes the proof. ∎

######  Theorem 4.6.

Let $R$ be a ring. Then $R$ is right qnil-duo if and only if $H_{(s,t)}(R)$ is
right qnil-duo.

###### Proof.

Assume that $R$ is a right qnil-duo ring. Let $A=\begin{bmatrix}a&0&0\\\
c&d&e\\\ 0&0&f\end{bmatrix}\in H_{(s,t)}(R)$ and $B=\begin{bmatrix}x&0&0\\\
y&z&u\\\ 0&0&v\end{bmatrix}\in H_{(s,t)}(R)^{qnil}$. By Lemma 4.5, $x$, $z$,
$v\in R^{qnil}$. There exist $x^{\prime}$, $z^{\prime}$, $v^{\prime}\in
R^{qnil}$ such that $xa=ax^{\prime}$, $zd=dz^{\prime}$, $vf=fv^{\prime}$. Let
$y^{\prime}=s^{-1}(x^{\prime}-z^{\prime})$ and
$u^{\prime}=t^{-1}(z^{\prime}-v^{\prime})$ and
$B^{\prime}=\begin{bmatrix}x^{\prime}&0&0\\\
y^{\prime}&z^{\prime}&u^{\prime}\\\ 0&0&v^{\prime}\end{bmatrix}$. Then
$B^{\prime}\in H_{(s,t)}(R)^{qnil}$. We next show that $BA=AB^{\prime}$. It is
enough to see $ya+zc=cx^{\prime}+dy^{\prime}$ and
$ze+uf=du^{\prime}+ev^{\prime}$. We start with,
$cx^{\prime}+dy^{\prime}=cx^{\prime}+ds^{-1}x^{\prime}-ds^{-1}z^{\prime}$.
Multiplying the latter from the left by $s$ and using the fact that $s$ is
central, we have
$s(cx^{\prime}+dy^{\prime})=scx^{\prime}+dx^{\prime}-dz^{\prime}=(sc+d)x^{\prime}-zd=ax^{\prime}-zd=xa-
zd=(xa-za)+(za-zd)=sya+szc=s(ya+zc)$. Since $s$ is invertible,
$ya+zc=cx^{\prime}+dy^{\prime}$. Similarly,
$du^{\prime}+ev^{\prime}=dt^{-1}z^{\prime}-dt^{-1}v^{\prime}+ev^{\prime}$.
Multiplying the latter from the left by $t$ and using the fact that $t$ is
central, we have
$t(du^{\prime}+ev^{\prime})=dz^{\prime}-dv^{\prime}+tev^{\prime}=zd+(te-d)v^{\prime}=zd-
fv^{\prime}=zd-vf=zd-zf+zf-vf=z(d-f)+(z-v)f=t(ze+uf)$. By using invertibility
of $t$, we get $du^{\prime}+ev^{\prime}=ze+uf$. Conversely, suppose that
$H_{(s,t)}(R)$ is a right qnil-duo ring. Let $a\in R$ and $b\in R^{qnil}$.
Consider $A=aI_{3}$, $B=bI_{3}\in H_{(s,t)}(R)$. By Lemma 4.5, $B\in
H_{(s,t)}(R)^{qnil}$. By supposition there exists
$B^{\prime}=\begin{bmatrix}x&0&0\\\ y&z&u\\\ 0&0&v\end{bmatrix}\in
H_{(s,t)}(R)^{qnil}$ such that $BA=AB^{\prime}$. It implies $ba=ax$. Again by
Lemma 4.5, $x\in R^{qnil}$. Hence $ba=ax\in aR^{qnil}$. Thus
$R^{qnil}a\subseteq aR^{qnil}$. ∎

Generalized matrix rings: Let $R$ be a ring and $s\in U(R)$. Then
$\begin{bmatrix}R&R\\\ R&R\end{bmatrix}$ becomes a ring denoted by $K_{s}(R)$
with addition defined componentwise and multiplication defined in [10] by

$\begin{bmatrix}a_{1}&x_{1}\\\
y_{1}&b_{1}\end{bmatrix}\begin{bmatrix}a_{2}&x_{2}\\\
y_{2}&b_{2}\end{bmatrix}=\begin{bmatrix}a_{1}a_{2}+sx_{1}y_{2}&a_{1}x_{2}+x_{1}b_{2}\\\
y_{1}a_{2}+b_{1}y_{2}&sy_{1}x_{2}+b_{1}b_{2}\end{bmatrix}.$

In [10], $K_{s}(R)$ is called a generalized matrix ring over $R$.

######  Lemma 4.7.

Let $R$ be a ring. Then the following hold.

  1. (1)

$U(K_{0}(R))=\left\\{\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in K_{0}(R)\mid
a,d\in U(R)\right\\}$.

  2. (2)

$C(K_{0}(R))=\left\\{\begin{bmatrix}a&0\\\ 0&a\end{bmatrix}\in K_{0}(R)\mid
a\in C(R)\right\\}$.

###### Proof.

(1) Let $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in U(K_{0}(R))$. There
exists $B=\begin{bmatrix}x&y\\\ z&t\end{bmatrix}\in K_{0}(R)$ such that
$AB=BA=I$, where $I$ is the identity matrix. Then we have $ax=xa=1$ and
$dt=td=1.$ So $a$ and $d$ are invertible. Conversely, let
$A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in K_{0}(R)$ with $a$, $d\in U(R)$.
Let $x=a^{-1}$, $t=d^{-1}$, $k=-a^{-1}bd^{-1}$ and $l=-d^{-1}ca^{-1}$. Then
$B=\begin{bmatrix}x&k\\\ l&t\end{bmatrix}$ is the inverse of $A$ in
$K_{0}(F)$.
(2) Let $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in C(K_{0}(R))$. By
commuting $A$ in turn with the matrices $\begin{bmatrix}1&0\\\
0&0\end{bmatrix}$ and $\begin{bmatrix}0&1\\\ 0&0\end{bmatrix}$ in $K_{0}(R)$
we reach at $A=\begin{bmatrix}a&0\\\ 0&a\end{bmatrix}$. For the converse, let
$A=\begin{bmatrix}a&0\\\ 0&a\end{bmatrix}\in K_{0}(R)$ where $a\in C(R)$. Then
clearly, $A$ commutes with every element of $K_{0}(R)$. So $A\in C(K_{0}(R))$.
∎

######  Proposition 4.8.

Let $R$ be a ring and $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in K_{0}(R)$.
If $a$, $d\in R^{qnil}$, then $A\in K_{0}(R)^{qnil}$.

###### Proof.

Suppose that $a$, $d\in R^{qnil}$. Let $B=\begin{bmatrix}x&y\\\
z&t\end{bmatrix}\in K_{0}(R)$ with $B\in$ comm$(A)$. Then $x\in$ comm$(a)$,
$t\in$ comm$(d)$. Let $r=1+ax$, $v=1+dt$, $s=ay+bt$ and $u=cx+dz$. By
assumption, $r=1+ax$ and $v=1+dt$ are invertible in $R$. Let
$k=-r^{-1}sv^{-1}$ and $l=-v^{-1}ur^{-1}$. Then
$I_{2}+AB=\begin{bmatrix}r&s\\\ u&v\end{bmatrix}$ is invertible with the
inverse $C=\begin{bmatrix}r^{-1}&k\\\ l&v^{-1}\end{bmatrix}$. ∎

Let $R$ be a ring, $a$, $b\in R$. Define $l_{a}-r_{b}\colon R\rightarrow R$ by
$(l_{a}-r_{b})(r)=ar-rb$ and $l_{b}-r_{a}\colon R\rightarrow R$ by
$(l_{b}-r_{a})(r)=br-ra$. In [1], a local ring $R$ is called bleached if for
any $a\in J(R)$ and any $b\in U(R)$, the abelian group endomorphisms
$l_{b}-r_{a}$ and $l_{a}-r_{b}$ of $R$ are surjective. Such rings is called
uniquely bleached if the appropriate maps are injective as well as surjective.
In [13], $R$ is a weakly bleached ring provided that for any $a\in J(R)$,
$b\in 1+J(R)$, $l_{a}-l_{b}$ and $l_{b}-l_{a}$ are surjective and it is proved
that matrices over 2-projective free rings are strongly J-clean. It is proved
that all upper triangular matrices over bleached local rings are strongly
clean. In [12, Example 2] and [1, Theorem 18] it is proved that a local ring
$R$ is weakly bleached if and only if the $2\times 2$ upper triangular matrix
ring $U_{2}(R)$ is strongly clean. In the preceding, maps of the form
$l_{a}-r_{b}$ play a central role. In this vein, we make use of the abelian
group endomorphisms $l_{a}-r_{b}$ to get the following result as partly the
converse of Proposition 4.8.

######  Theorem 4.9.

Let $R$ be a ring and $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in
K_{0}(R)^{qnil}$. If for any $x\in$ comm$(a)$ and $y\in$ comm$(d)$ and for the
abelian group endomorphisms $l_{y}-r_{x}$ and $l_{x}-r_{y}$, $b\in$
Ker$(l_{x}-r_{y})$ and $c\in$ Ker$(l_{y}-r_{x})$, then $a$, $d\in R^{qnil}$.

###### Proof.

Assume that $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in K_{0}(R)^{qnil}$,
$x\in$ comm$(a)$ and $y\in$ comm$(d)$, for $l_{y}-r_{x}$ and $l_{x}-r_{y}$,
$b\in$ Ker$(l_{x}-r_{y})$ and $c\in$ Ker$(l_{y}-r_{x})$. Then $b\in$
Ker$(l_{x}-r_{y})$ implies $(l_{x}-r_{y})(b)=0$. So $xb=by$. $c\in$
Ker$(l_{y}-r_{x})$ implies $(l_{y}-r_{x})(c)=0$. So $yc=cx$. Let
$B=\begin{bmatrix}x&0\\\ 0&y\end{bmatrix}\in K_{0}(R)$. Then $xb=by$ and
$yc=xc$ give rise to $B\in$ comm$(A)$. By hypothesis, $I_{2}+AB$ is
invertible. Then Lemma 4.7 implies $1-ax$ and $1-dy$ are invertible. Hence
$a$, $d\in R^{qnil}$. ∎

We may determine the set $K_{0}(R)^{qnil}$ for some rings $R$.

######  Proposition 4.10.

  1. (1)

If $R$ is a local ring, then $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}\in
K_{0}(R)^{qnil}$ if and only if $a$, $d\in R^{qnil}$.

  2. (2)

Let $R$ be a ring. Then $A=\begin{bmatrix}a&0\\\ 0&d\end{bmatrix}\in
K_{0}(R)^{qnil}$ if and only if $a$, $d\in R^{qnil}$.

###### Proof.

(1) Assume that $R$ is a local ring and $A=\begin{bmatrix}a&b\\\
c&d\end{bmatrix}\in K_{0}(R)^{qnil}$, and $d\notin R^{qnil}$. By Proposition
2.2, $d\in U(R)$. In this case, $1+d$ can not belong to $U(R)$. By Lemma 4.7,
$I+A$ can not belong to $U(K_{0}(R))$. This contradicts $A\in
K_{0}(R)^{qnil}$. It follows that $d\in R^{qnil}$. Similarly, we obtain $a\in
R^{qnil}$. The converse is clear by Proposition 4.8.
(2) Clear. ∎

There are some classes of rings $R$ in which $K_{0}(R)$ being a right qnil-duo
ring implies $R$ being a right qnil-duo ring.

######  Theorem 4.11.

Let $R$ be a ring. Then $K_{0}(R)$ being a right qnil-duo ring implies $R$
being a right qnil-duo ring if $R$ is one of the following rings.

  1. (1)

$R$ is local.

  2. (2)

$R$ has no nonzero zero divisors.

###### Proof.

(1) Let $R$ be a local ring. Assume that $K_{0}(R)$ is a right qnil-duo ring.
Let $a\in R$, $b\in R^{qnil}$. Consider $A=\begin{bmatrix}a&0\\\
0&a\end{bmatrix}$, $X=\begin{bmatrix}b&0\\\ 0&b\end{bmatrix}\in K_{0}(R)$. By
Proposition 4.8, $X\in K_{0}(R)^{qnil}$. There exists
$X^{\prime}=\begin{bmatrix}x^{\prime}&y^{\prime}\\\
z^{\prime}&t^{\prime}\end{bmatrix}\in K_{0}(R)^{qnil}$ such that
$XA=AX^{\prime}$. Hence $ba=ax^{\prime}$. By Proposition 4.10, $x^{\prime}\in
R^{qnil}$. So $ba=ax^{\prime}\in aR^{qnil}$.
(2) Let $R$ be a ring having no nonzero zero divisors. Assume that $K_{0}(R)$
is a right qnil-duo ring. Let $a\in R$, $b\in R^{qnil}$. If $a=0$ or $b=0$,
there is nothing to do. Let $a\neq 0$ and $b\neq 0$ and consider
$A=\begin{bmatrix}a&0\\\ 0&a\end{bmatrix}$, $B=\begin{bmatrix}b&0\\\
0&b\end{bmatrix}\in K_{0}(R)$. By Proposition 4.8, $B\in K_{0}(R)^{qnil}$.
There exists $B^{\prime}=\begin{bmatrix}x^{\prime}&y^{\prime}\\\
z^{\prime}&t^{\prime}\end{bmatrix}\in K_{0}(R)^{qnil}$ such that
$BA=AB^{\prime}$. It implies $ba=ax^{\prime}=at^{\prime}$, $ay^{\prime}=0$ and
$az^{\prime}=0$. Hence $x^{\prime}=t^{\prime}$ and $y^{\prime}=z^{\prime}=0$.
Hence $x^{\prime}\in R^{qnil}$ by Proposition 4.10. ∎

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