# ON GRADED U-NIL CLEAN RINGS

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###### Abstract

In this paper we introduce and study the class of graded U-nil clean rings, as
a generalization of graded nil-good class defined in [3]. We also investigate
the transfer of the graded U-nil cleaness to matrix rings, and to graded group
rings as well. The question of when trivial ring extensions are graded U-nil
clean is also investigated.

keywords: Graded rings and modules, U-nil clean ring, group ring, matrix ring,
idealization of a module.

Mathematics Subject Classification. Primary: 16W50. Secondary: 16U99, 16S34,
16S50.

##  1 Introduction

The year 1977 was the beginning of a new era of the introduction of new
classes of rings, which generalize the concept of clean rings introduced by
Nicholson [17] in the same year. Indeed, Nicholson defined a clean ring as a
ring in which every element is a sum of an idempotent and a unit. Ever since,
and particularly over the last decade, many authors have studied numerous
classes of rings as a generalization or a variation of clean rings; for
instance, in 2016 Danchev introduced in [6], an interesting class, called nil-
good rings, in which every element is either nilpotent or a sum of a nilpotent
and a unit. Danchev has also given significant results related to extensions
of such rings, such as group rings and matrix rings as well. In his turn,
Khashan took the initiative in [14], to enrich the literature related to
classes of rings, by introducing the class of U-nil clean rings, whose
definition is given in the next section. The aim of this paper is studying the
U-nil clean property from graded ring theory point of view. Indeed, the idea
of extending the properties of rings to graded rings has been interesting many
authors recently; for more mdetails, the reader is referred to [3, 4, 10, 11,
12]. In particular, in [3] the authors examined the nil-good property in
graded rings; indeed in this paper, we introduce graded U-nil clean rings as a
graded version of the concept given by Khashan in [14] and as a generalization
of the graded nil-good notion defined in [3]. We begin by giving basic
properties and examples of graded U-nil clean rings. In addition, we establish
a sufficient conditions for a graded ring to be graded U-nil clean, if the
component which corresponds to the identity element of the grading group is a
U-nil clean ring. Then, we emphasis on the extensions of such rings, in
particular the matrix rings, since it is the main goal of this paper. We also
study the question of when group rings and trivial ring extensions are graded
U-nil clean.

##  2 Preliminaries

Throughout this paper, all rings are assumed to be unital and associative. If
$R$ is a ring then $J(R)$ denotes the Jacobson radical of $R$, $U(R)$ is the
multiplicative group of units of $R$, $N(R)$ denotes the set of nilpotent
elements of $R$, $Z(R)$ is the center of the ring $R$ and $Idem(R)$ stands for
the set of idempotents of $R$.

Let $R$ be a ring, $G$ a group with identity element $e$, and let
$\\{R_{g}\\}_{g\in G}$ be a family of additive subgroups of $R$. the ring $R$
is said to be $G$-graded (or simply graded) if $R=\bigoplus_{g\in G}R_{g}$ and
$R_{g}R_{h}\subseteq R_{gh}$ for every $g,h\in G$. Let $h(R)=\bigcup_{g\in
G}R_{g}$; elements of $h(R)$ are said homogeneous, and for every $g\in G$, the
subgroup $R_{g}$ is called the $g$-th component of $R$. If $a\in R_{g}$, then
we say that $a$ is of degree $g$. In addition, the degree of every homogeneous
idempotent element is $e$. Furthermore, if $t\in Idem(R_{e})$, then $tR$ is a
graded ring with unity $t$ and the $g$-component $(tR)_{g}=tR_{g}$ for each
$g\in G$. The set $sup(R)=\\{g\in G,R_{g}\neq 0\\}$ is called the support of
the graded ring $R$.

If $A=\oplus_{g\in G}A_{g}$ and $B=\oplus_{g\in G}B_{g}$ are two $G$-graded
rings; a ring homomorphism $\epsilon:A\longrightarrow B$ is said to be graded-
homomorphism if $\epsilon(h(A))\subseteq h(B)$. The ring homomorphism
$\epsilon$ is called degree-preserving homomorphism if
$\epsilon(A_{g})\subseteq B_{g}$ for every $g\in G$.

Given $G$-graded rings $R_{1},R_{2},\dots,R_{k}$. The direct product
$R=\prod_{i=1}^{k}R_{i}$ is a $G$-graded ring, where its $g$-th component is
$R_{g}=\prod_{i=1}^{k}R_{i}^{g}$, with $R_{i}^{g}$ is the $g$-th component of
the ring $R_{i}$.

A proper right (left or two-sided) ideal $I$ of a $G$-graded ring
$R=\oplus_{g\in G}R_{g}$ is called homogeneous (or graded), if $I=\oplus_{g\in
G}I\cap R_{g}$. If $I$ is a two-sided homogeneous ideal, then $R/I$ is a
$G$-graded ring where its $g$-th component is $(R/I)_{g}=R_{g}/I\cap R_{g}$.
In addition, a homogeneous ideal is called graded-nil, if every homogeneous
element of this ideal is nilpotent. A homogeneous ideal $I$ of a graded ring
$R$ is called graded-maximal, if there is no other proper homogeneous ideal of
$R$ which contains $I$; equivalently, according to page 46 in [16], $I$ is
graded-maximal if $I+Rx=R$ for any $x\in h(R)\backslash I$. A graded ring is
said to be graded-local, if it has a unique graded-maximal right ideal.
Moreover, the intersection of all graded-maximal right ideals of a graded ring
$R$ defines the graded Jacobson radical of $R$, denoted $J^{g}(R)$. Further,
according to Proposition 2.9.1 in [16] we have that, $J^{g}(R)$ is a
homogeneous two-sided ideal.

If $R=\oplus_{g\in G}R_{g}$ is a $G$-graded ring, then a $G$-graded $R$-module
is an $R$-module $M$ such that $M=\oplus_{g\in G}M_{g}$, where
$\\{M_{g}\\}_{g\in G}$ are additive subgroups of $M$, and $R_{h}M_{g}\subseteq
M_{hg}$ for all $g,h\in G$.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded ring. According to [15], we have
that the group ring $R[G]$ is $G$-graded where its $g$-th component is
$(R[G])_{g}=\oplus_{h\in G}R_{gh^{-1}}h$ and with the multiplication defined
by
$(r_{g}g^{\prime})(r_{h}h^{\prime})=r_{g}r_{h}(h^{-1}g^{\prime}hh^{\prime})$,
where $g,g^{\prime},h$ and $h^{\prime}\in G$, $r_{g}\in R_{g}$ and $r_{h}\in
R_{h}$. Now, if $H$ is a normal subgroup of $G$, then according to [16] $R[H]$
can be viewed as a $G$-graded ring, where its $g$-th component is
$(R[H])_{g}=\oplus_{h\in H}R_{gh^{-1}}h$. In addition, the ring $R$ can be
seen as a $G/H$-graded ring (see page 178 in [16]) as follow
$R=\bigoplus_{C\in G/H}R_{C}$ where $R_{C}=\oplus_{x\in C}R_{x}$. Clearly
$R[H]$ can also be observed as a $G/H$-graded ring.

Let $A$ be a ring and $E$ an $(A,A)$-bimodule. We denote by $R:=A\propto E$
the set of pairs $(a,e)$ with pairwise addition and multiplication given by
$(a,e)(b,f)=(ab,af+eb)$; the ring $R$ is called the idealization of $E$ (over
$A$). Considerable background concerning the idealization of modules over
commutative rings can be found in [2, 7, 9]. If $A$ is a $G$-graded ring and
$E$ a $G$-graded $(A,A)$-bimodule, then according to [2] $R=A\propto E$ is
$G$-graded with the $g$-th component $R_{g}=A_{g}\oplus E_{g}$.

If $R=\oplus_{g\in G}R_{g}$ is a $G$-graded ring, $n$ an arbitrary natural
number and $\sigma=(g_{1},\dots,g_{n})\in G^{n}$. For every $\lambda\in G$, we
denote by $M_{n}(R)_{\lambda}(\sigma)$ the additive subgroup of $M_{n}(R)$
formed by matrices $(a_{ij})_{n\times n}$, such that $a_{ij}\in
R_{g_{i}\lambda g_{j}^{-1}}$ for $i,j\in\\{1,\dots,n\\}$. We have
$M_{n}(R)=\bigoplus_{\lambda\in G}M_{n}(R)_{\lambda}(\sigma)$ which makes the
matrix ring $M_{n}(R)$ a $G$-graded ring with respect to usual matrix addition
and multiplication. Throughout this paper, if the matrix ring $M_{n}(R)$ is
observed as a graded ring in the previous way then we denote it by
$M_{n}(R)(\sigma)$.

An element $r$ of a ring $R$ is called unit regular if $r=eu$ for some $e\in
Idem(R)$ and $u\in U(R)$. Khashan [14] defined a ring to be U-nil clean, if
every element is a sum of a nilpotent and a unit regular. Obviously,
idempotents and units are unit regular elements. Let $R=\oplus_{g\in G}R_{g}$
be a $G$-graded ring; we define a homogeneous element $r\in R_{g}$ (where
$g\in G$) to be graded unit regular if $r=eu$ for some $e\in Idem(R_{e})$ and
$u\in U(R)\cap R_{g}$. The symbol $gur(R)$ will stand for the set of graded
unit regular elements of $R$; while $ur(R)$ is the set of unit regular
elements of $R$.

##  3 Graded U-nil clean rings

Let $G$ be a group with identity $e$.

######  Definition 3.1.

A homogeneous element $x$ of a $G$-graded ring $R$ is called graded U-nil
clean if $x=r+n$ where $r\in gur(R)$ and $n\in N(R)\cap h(R)$. The ring $R$ is
called graded U-nil clean if every element of $h(R)$ is graded U-nil clean.

######  Remark 3.1.

If $x=r+n$ is a graded U-nil clean decomposition of a homogeneous element $x$
of the ring $R$. Then, we can easily show that $x$, $r$ and $n$ are all of the
same degree. In fact, let $g$ (resp. $h$) be the degree of $x$ (resp. $r$). If
$g\neq h$, then we will have either $x=n$ or $x=0$. Both cases yield to the
contradiction $g=h$. Therefore, $g=h$ and necessarily the degree of $n$ is
also $g$.

######  Example 3.1.

Let $A$ be any U-nil clean ring (for instance $\mathbb{Z}_{3}$ since it is a
field), and let $G=\\{e,g,h\\}$ be a cyclic group of order three. The ring
$R=M_{2}(A)$ of $2\times 2$ matrices with coefficients in $A$ is clearly
$G$-graded, with the grading: $R_{e}=\begin{pmatrix}A&0\\\ 0&A\end{pmatrix}$,
$R_{g}=\begin{pmatrix}0&A\\\ 0&0\end{pmatrix}$ and
$R_{h}=\begin{pmatrix}0&0\\\ A&0\end{pmatrix}$. It is obvious that $R_{e}$ is
U-nil clean. Since every element of $R_{g}$ and $R_{h}$ is nilpotent, then $R$
is a graded U-nil clean ring.

The following example shows that the class of graded nil-good rings is a
proper subclass of graded U-nil clean rings class. According to [3], a graded
ring is said to be graded nil-good if every homogeneous element is either
nilpotent, or a sum of a homogeneous unit and a homogeneous nilpotent.

######  Example 3.2.

Consider $R=M_{2}(\mathbb{Z}_{2})$ and $G$ a cyclic group of order two. Then,
$R=\begin{pmatrix}\mathbb{Z}_{2}&0\\\
0&\mathbb{Z}_{2}\end{pmatrix}\bigoplus\begin{pmatrix}0&\mathbb{Z}_{2}\\\
\mathbb{Z}_{2}&0\end{pmatrix}$ is a $G$-graded ring which is not graded nil-
good by [3, Example 2.6]. On the other hand, the homogeneous elements
$\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\\
0&1\end{pmatrix}$ are idempotents. Moreover, $\begin{pmatrix}0&1\\\
0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\\ 1&0\end{pmatrix}$ are both
nilpotents. Further, $A=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ is a unit
since $A^{2}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$. Hence, every homogeneous
element of $R$ is graded U-nil clean; thus $R$ is graded U-nil clean.

######  Proposition 3.1.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded U-nil clean ring. Then,

  1. 1.

$R_{e}$ is a U-nil clean ring.

  2. 2.

Every graded-homomorphic image of $R$ is $G$-graded U-nil clean.

###### Proof.

  1. 1.

Let $a\in R_{e}$. Since $R$ is graded U-nil clean, then $a=r+n$ where $r\in
gur(R)$ and $n\in N(R)\cap h(R)$. According to Remark 3.1 we have $r$ and $n$
are both elements of $R_{e}$. Therefore, the ring $R_{e}$ is U-nil clean.

  2. 2.

Consider $S$ a G-graded ring and $\varphi:R\longrightarrow S$ a graded-
homomorphism. Suppose that $R$ is graded U-nil clean and let $a\in h(R)$; our
goal is to show that $\varphi(a)$ is graded U-nil clean of the ring
$\varphi(R)$. By assumption $a$ can be written $a=fu+n$ where $f\in
Idem(R_{e})$, $u\in U(R)\cap h(R)$ and $n\in N(R)\cap h(R)$. Hence,
$\varphi(a)=\varphi(f)\varphi(u)+\varphi(n)$. Since $\varphi$ is a graded-
homomorphism, then $\varphi(a)$ has been written as a sum of a graded unit
regular element and a homogeneous nilpotent element of the ring $\varphi(R)$.
Therefore, $\varphi(R)$ is graded U-nil clean.

∎

The first point of Proposition 3.1, makes us wondering if the U-nil clean
property of $R_{e}$ implies that $R=\oplus_{g\in G}R_{g}$ is graded U-nil
clean. Well, the next example proves that is false in general.

######  Example 3.3.

Let $R:=\mathbb{Z}_{2}[X]$. We have that $R=\oplus_{n\in\mathbb{Z}}R_{n}$
where $R_{n}=\mathbb{Z}_{2}X^{n}$ for $n\geq 0$ and $R_{n}=0$ for $n<0$.
Hence, $R$ is a $\mathbb{Z}$-graded ring. Obviously, $R_{0}=\mathbb{Z}_{2}$ is
a U-nil clean ring since it is a field. On the other hand, it is clearly that
the homogeneous element $X\in R_{1}$ is not graded U-nil clean. Hence, $R$ is
not graded U-nil clean.

######  Proposition 3.2.

Let $R_{1},R_{2},\dots,R_{k}$ be $G$-graded rings. If $R_{i}$ is graded U-nil
clean for every $i\in\\{1,2,\dots,k\\}$, then $R=\prod_{i=1}^{k}R_{i}$ is a
$G$-graded U-nil clean ring.

###### Proof.

Let $(r_{1},r_{2},\dots,r_{k})$ be a homogeneous element of $R$ of degree $g$
where $g\in G$. For each $i$ we choose $a_{i}\in gur(R_{i})$ and $n_{i}\in
N(R_{i})\cap h(R_{i})$ such that $r_{i}=a_{i}+n_{i}$. It is clear that
$a=(a_{1},a_{2},\dots,a_{k})\in ur(R)\cap h(R)$ and
$n=(n_{1},n_{2},\dots,n_{k})\in N(R)\cap h(R)$; it follows that $r=a+n$ and
therefore $R$ is $G$-graded U-nil clean. ∎

It has been proved in [14, Proposition 2.1] by Khashan that any U-nil clean
ring is a $G$-clean ring ($G$ refers to the word "generalization" and not to
the group mentioned in the head of this section). Let us recall first the
definition of a $G$-clean ring introduced and studied by Hongbo and Weting in
[8]. A ring is defined to be $G$-clean if every element is a sum of a unit and
a unit regular. The class of $G$-clean rings is a generalization of the class
of clean rings introduced by Nicholson. In graded ring theory, if $H$ is a
group; we define naturally in this paper, a $H$-graded ring to be graded
$G$-clean if every homogeneous element is a sum of a homogeneous unit and a
graded unit regular element. The next result gives a graded version of
Proposition 2.1 in [14].

######  Proposition 3.3.

Let $H$ be a group with identity $e^{\prime}$ and $R=\oplus_{g\in H}R_{g}$ a
$H$-graded commutative ring. If $R$ is graded U-nil clean, then it is a graded
$G$-clean ring.

###### Proof.

Suppose that $R$ is graded U-nil clean, then by Proposition 3.1
$R_{e^{\prime}}$ is a U-nil clean ring. Thus, according to [14, Proposition
2.1] the ring $R_{e^{\prime}}$ is a $G$-clean ring. Now, let $a\in R_{g}$
(where $g\in H)$ be a homogeneous element of $R$. Since $R$ is graded U-nil
clean, then there exist $r\in gur(R)\cap R_{g}$ and $n\in N(R)\cap R_{g}$ such
that $a=r+n$. We know that there exist $x\in U(R)\cap R_{g}$ and $f\in
Idem(R_{e^{\prime}})$ such that $r=fx$. Now, since $R_{e^{\prime}}$ is
$G$-clean, then $f=s+u$, where $s\in ur(R_{e^{\prime}})$ and $u\in
U(R_{e^{\prime}})$. Thus, $a$ is written $a=sx+(ux+n)$. Since $R$ is
commutative, then $ux+n\in U(R)\cap R_{g}$. Moreover, $s=fv$ for some $f\in
Idem(R_{e^{\prime}})$ and $v\in U(R_{e^{\prime}})$. Hence, $sx=f(vx)$ is then
graded unit regular of degree $g$. Finally, $a$ has been written as a sum of
graded unit regular and a homogeneous unit, which means that $R$ is $H$-graded
$G$-clean. ∎

######  Remark 3.2.

It is well known that, if $I$ is a nil ideal of a ring $R$ then, the ideal $I$
is contained in the classical Jacobson radical of $R$ (i.e. $I\subseteq
J(R))$. Next, we will give a graded approach of this result.

######  Lemma 3.1.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded ring and $I$ a homogeneous ideal
of $R$. If $I$ is graded-nil then $I\subseteq J^{g}(R)$.

###### Proof.

Suppose that $I$ is a graded-nil ideal. Let $a$ be a homogeneous element of
$I$; since $I$ is a proper ideal then $Ra\neq R$. Assume that $a\notin
J^{g}(R)$, hence there exists a graded-maximal ideal $M$ of $R$ such that
$a\notin M$. Therefore, as mentioned in the preliminaries of this paper, we
have that $M+Ra=R$. Thus, $m+ra=1$ for some $m\in M$ and $r\in R$. Hence,
$\overline{r}\overline{a}=\overline{1}$ in $R/M$ which is a contradiction
since $\overline{a}$ is nilpotent in $R/M$. Therefore $I\subseteq J^{g}(R)$. ∎

######  Theorem 3.1.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded ring and $I$ a graded-nil ideal
of $R$. Then $R$ is a graded U-nil clean ring if and only if so is $R/I$.

###### Proof.

”$\Longrightarrow$” Suppose $R$ is graded U-nil clean. The canonical map
$\pi:R\longrightarrow R/I$ is clearly a graded-homomorphism. Now, according to
proposition 3.1 $R/I$ is graded U-nil clean as a the graded-homomorphic image
of $R$ by $\pi$.

”$\Longleftarrow$” Let us suppose that $\overline{R}=R/I$ is graded U-nil
clean and let $a\in R_{g}$ where $g\in G$. There exist then $\overline{f}\in
Idem(\overline{R}_{e})$, $\overline{u}\in U(\overline{R})\cap\overline{R}_{g}$
and $\overline{n}\in N(\overline{R})\cap\overline{R}_{g}$ such that
$\overline{a}=\overline{f}\overline{u}+\overline{n}$. On the other hand,
$I_{e}=I\cap R_{e}$ is clearly a nil ideal of the ring $R_{e}$, hence by
applying [1, Proposition 27.1] to $R_{e}$, we have that $\overline{f}$ can be
lifted to an idempotent element of $R_{e}$, thus we can suppose that $f\in
Idem(R_{e})$. Now, since $I$ is graded-nil, then according to Lemma 3.1
$I\subseteq J^{g}(R)$; therefore, $\overline{u}\in U(R/J^{g}(R))$. According
to [16, Proposition 2.9.1], we have that $u\in U(R)$. We know that
$\overline{a-fu}$ is a homogeneous nilpotent element of $R/I$. Thus, since $I$
is graded-nil, it follows that $a-fu$ is a homogeneous nilpotent element of
$R$. Finally, $a$ is graded U-nil clean, and the proof is achieved. ∎

Concerning the condition ”$I$ is graded-nil” in Theorem 3.1, it can’t be
neglected. Indeed, let $S=\mathbb{Z}_{3}[X]$; we have that
$S=\oplus_{n\in\mathbb{Z}}S_{n}$ where $S_{n}=\mathbb{Z}_{3}X^{n}$ for $n\geq
0$ and $S_{n}=0$ for $n<0$, which means that $S$ is $\mathbb{Z}$-graded. It is
clear that $S$ is not graded U-nil clean (the homogeneous element $X$ is not
graded U-nil clean). Now, the homogeneous ideal $I=(X)$ of $S$ is clearly not
graded-nil. Yet, the ring $S/I$ is graded U-nil clean as an isomorphic image
of the field $\mathbb{Z}_{3}$.

######  Proposition 3.4.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded U-nil clean ring. If
$Idem(R_{e})\subseteq Z(R)$, then $J^{g}(R)$ is graded-nil.

###### Proof.

Let $j$ be a homogeneous element of $J^{g}(R)$ of degree $g\in G$. Our goal is
to prove that $j\in N(R)$. Indeed, according to Proposition 3.1 $R_{e}$ is
U-nil clean; further, since $Idem(R_{e})\subseteq Z(R)$ then $R_{e}$ is an
abelian ring (i.e. $Idem(R_{e})\subseteq Z(R_{e})$). Hence, by [14,
Porposition 2.3] the Jacobson radical of $R_{e}$, $J(R_{e})$ is nil. If $g=e$,
then $j\in J^{g}(R)\cap R_{e}$. In addition, according to [16, Corollary
2.9.3] $J^{g}(R)\cap R_{e}=J(R_{e})$; therefore $j$ is nilpotent. Now, suppose
that $g\neq e$. Since $R$ is graded U-nil clean, then there exist $f\in
Idem(R_{e})$, $u\in U(R)\cap R_{g}$ and $n\in N(R)\cap R_{g}$ such that
$j=fu+n$. Let $k\in\mathbb{N}$ such that $n^{k}=0$. Using the binomial
expansion we obtain $(fu-j)^{k}=(fu)^{k}+jx+yj=0$ for some $x,y\in R$. The
condition $Idem(R_{e})\subseteq Z(R)$ allows us to write
$(fu)^{k}=f^{k}u^{k}$; moreover $f^{k}=f$ since $f$ is an idempotent element.
Hence, since $J^{g}(R)$ is a two-sided ideal, we have $fu^{k}=-jx-yj\in
J^{g}(R)$. Thus, $f=fu^{k}u^{-k}\in J^{g}(R)\cap R_{e}=J(R_{e})$. But
$J(R_{e})$ is a nil ideal; therefore the idempotent element $f$ is nlipotent.
It follows $f=0$ and then $j=n\in N(R)$. Therefore, $J^{g}(R)$ is graded-nil.
∎

By combining Theorem 3.1 and Proposition 3.4 we obtain the following
corollary.

######  Corollary 3.1.

Let $R=\oplus_{g\in G}R_{g}$ be $G$-graded ring such that
$Idem(R_{e})\subseteq Z(R)$. Then, $R$ is graded U-nil clean if and only if so
is $R/J^{g}(R)$.

Let’s go back to Example 3.3, which proves that the following implication:

$R_{e}\leavevmode\nobreak\ \text{is U-nil clean}\Longrightarrow R=\oplus_{g\in
G}R_{g}\leavevmode\nobreak\ \text{is graded U-nil clean}.$

is false in general. Indeed, the above implication holds true under some
sufficient conditions given below. Let us recall first the definition of a
$PI$-ring. A ring $R$ is called $PI$-ring if there exist a natural integer
$n$, and an element $P$ of $\mathbb{Z}[X_{1},X_{2},\dots,X_{n}]$ such that for
every $(r_{1},r_{2},\dots,r_{n})\in R^{n}$, we have
$P(r_{1},r_{2},\dots,r_{n})=0$.

######  Theorem 3.2.

Let $R=\oplus_{g\in G}R_{g}$ be a graded-local $PI$-ring of finite support. If
$R_{e}$ is U-nil clean, then $R$ is graded U-nil clean.

###### Proof.

Suppose that $R_{e}$ is U-nil clean. According to Proposition 2.3 in [14], the
Jacobson radical $J(R_{e})$ of $R_{e}$ is nil. Since $R$ is a $PI$-ring then
according to [13, Theorem 3] the Jacobson radical $J(R)$ of $R$ is also nil.
Now, let $a$ be a homogeneous element of $R$. We have indeed two cases:
\- Case 1: $a\notin J^{g}(R)$. Hence, the homogeneous ideals $aR$ and $Ra$
cannot be contained in $J^{g}(R)$. Since $R$ is graded-local, then $aR=Ra=R$.
Thus, $a$ is a unit and in particular $a$ is graded U-nil clean.
\- Case 2: $a\in J^{g}(R)$. Since the support of $R$ is finite, according to
Corollary 2.9.4 in [16] we have that $J^{g}(R)\subseteq J(R)$, and hence
$J^{g}(R)$ is a nil ideal; therefore, $a$ is nilpotent.
Finally, every homogeneous element is graded U-nil clean; thus $R$ is graded
U-nil clean. ∎

##  4 Extensions of graded U-nil clean rings

In this section, we’ll discuss the ”graded U-nil clean” property of graded
group rings as well as of trivial ring extensions. The question of when the
matrix ring is graded U-nil clean is also investigated.

###  4.1 Trivial ring extensions

In this subsection, we investigate the behaviour of the graded U-nil clean
property in trivial ring extensions of graded U-nil clean rings.

######  Theorem 4.1.

Let $A$ be a $G$-graded ring, $E$ a $G$-graded $(A,A)$-bimodule and
$R=A\propto E$ the idealization of $E$. Then, $A$ is graded U-nil clean if and
only if $R$ is graded U-nil clean.

###### Proof.

It is easy to check that $0\propto E$ is a homogeneous ideal of $R$; moreover,
it is a graded-nil ideal since $(0\propto E)^{2}=0$. On the other hand, the
epimorphism $\epsilon:R\longrightarrow A$ given by $\epsilon((a,e))=a$, is a
graded-homomorphism. It is clear that the kernel of $\epsilon$ is exactly
$0\propto E$; hence $R/(0\propto E)$ and $A$ are graded-isomorphic. Now, the
Theorem 3.1 completes the proof. ∎

The previous theorem allows us give more examples in addition the one given in
Example 3.2, of graded U-nil clean rings which are not graded nil-good. Let’s
consider the graded ring $R$ taken in Example 3.2; and let $E$ be any
$G$-graded $(R,R)$-bimodule (where $G$ is the cyclic group of order two).
Hence, the graded ring $R\propto E$ is graded U-nil clean (by Theorem 4.1),
which is not graded nil-good according to theorem 3.1 in [3].

###  4.2 Group rings

We continue our investigation of extensions of graded U-nil clean rings; this
subsection will be dedicated to graded group rings.

######  Lemma 4.1.

Let $A=\oplus_{g\in G}A_{g}$ and $B=\oplus_{g\in G}B_{g}$ be two G-graded
rings. If $\varphi:A\longrightarrow B$ is a degree-preserving homomorphism,
then its kernel $\ker(\varphi)$ is a homogeneous ideal of the ring $A$.

###### Proof.

It is well known that $\ker(\varphi)$ is an ideal of $A$. Hence, it remains to
show that $\ker(\varphi)$ is a homogeneous ideal. Let $\sum_{g\in
G}a_{g}\in\ker(\varphi)$; therefore $0=\varphi(\sum_{g\in G}a_{g})=\sum_{g\in
G}\varphi(a_{g})$. Since $\forall g\in G$: $\varphi(a_{g})\in B_{h}$ for some
$h\in G$, then $\varphi(a_{g})=0$ for every $g\in G$; it follows that
$a_{g}\in\ker(\varphi)$ for each $g\in G$. Hence, $\ker(\varphi)$ is a
homogeneous ideal of $A$. ∎

######  Theorem 4.2.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded ring. Then, if $R[G]$ is graded
U-nil clean then $R$ is a U-nil clean ring.

###### Proof.

Suppose that $R[G]$ is graded U-nil clean. According to Proposition 3.1, we
have that $(R[G])_{e}$ is a U-nil clean ring. Furthermore, Proposition 2.1 in
[15] proves that the two rings $(R[G])_{e}$ and $R$ are isomorphic. Hence, by
Proposition 2.2 in [14] $R$ is a U-nil clean ring. ∎

######  Theorem 4.3.

Let $R$ be a $G$-graded ring, and $H$ a normal subgroup of $G$. Suppose that
$G$ is locally finite $p$-group where $p$ is a nilpotent element of $R$. If
$R$ is a $G/H$-graded U-nil clean ring then $R[H]$ is also $G/H$-graded U-nil
clean.

###### Proof.

Suppose that $R$ is $G/H$-graded U-nil clean. As has been shown in the proof
of Theorem 2.3 in [18] , we can admit that $H$ is a finite $p$-group, hence
$H=\\{h_{1},\dots,h_{n}\\}$ where $h_{i}\in H$ for each $1\leq i\leq n$. Now,
Let’s consider the epimorphism $\epsilon:R[H]\longrightarrow R$ given by
$\epsilon(\sum_{i=1}^{n}r_{i}h_{i})=\sum_{i=1}^{n}r_{i}$, where $r_{i}\in R$
for every $1\leq i\leq n$. It has been proved in page 180 of [16] that the
epimorphism $\epsilon$ is a degree-preserving homomorphism of $G/H$-graded
rings. Therefore, by Lemma 4.1 the kernel of $\epsilon$ is a graded ideal of
the ring $R[H]$. It follows that $R[H]/\ker(\epsilon)$ and $R$ are
$G/H$-graded-isomorphic rings; therefore, $R[H]/\ker(\epsilon)$ is
$G/H$-graded U-nil clean. Moreover, since $p\in N(R)$ then according to [5,
Theorem 9] we have $\ker(\epsilon)$ is a nilpotent ideal of $R[H]$; hence
$\ker(\epsilon)$ is a graded-nil ideal. Thus, by Theorem 3.1 $R[H]$ is
$G/H$-graded U-nil clean. ∎

###  4.3 Matrix rings

As mentioned in the introduction; the main result of this paper, which is the
U-nil clean property of graded matrix ring will be discussed in this
subsection

######  Theorem 4.4.

Assume that $G$ is an Abelian group. Let $R$ be a $G$-graded ring and $n$ a
natural number. Then, $R$ is graded U-nil clean if and only if
$T_{n}(R)(\sigma)$ is graded U-nil clean, for any $\sigma\in G^{n}$.

###### Proof.

Let $n$ be a natural number and suppose that $T_{n}(R)(\sigma)$ is graded
U-nil clean for any $\sigma\in G^{n}$. Let’s choose $\sigma=(e,e,\dots,e)$ and
consider the epimorphism of rings $\varphi:T_{n}(R)(\sigma)\longrightarrow R$
defined by $\varphi((a_{ij})_{n\times n})=a_{11}$. It is clear that the image
of $T_{n}(R)_{g}(\sigma)$ by $\varphi$ is $R_{g}$ for any $g\in G$; hence
$\varphi$ is a degree-preserving homomorphism. According to Lemma 4.1,
$\ker(\varphi)$ is a graded ideal of $T_{n}(R)(\sigma)$. It follows, the rings
$R$ and $T_{n}(R)(\sigma)/\ker(\varphi)$ are graded-isomorphic. Furthermore,
according to Theorem 3.1, $T_{n}(R)(\sigma)/\ker(\varphi)$ is graded U-nil
clean, and thus $R$ is graded U-nil clean as well.

Conversely, suppose that $R$ is graded U-nil clean and let $\sigma\in G^{n}$.
Let’s consider the ideal $I$ of the ring $T=T_{n}(R)(\sigma)$ formed by
matrices of $T$ with zeros in the main diagonal. Let’s prove that $I$ is a
graded ideal of $T$. Let $\sum_{g\in G}A_{g}\in I$, with $A_{g}\in T_{g}$
($T_{g}$ is the $g$-th component of the $G$-graded ring $T$) for every $g\in
G$; our aim is to show that $A_{g}\in I$ for any $g\in G$. Indeed, since the
sum $\sum_{g\in G}R_{g}$ is direct, then every element of the main diagonal of
$A_{g}$ is nil for each $g\in G$. Consequently, $A_{g}\in I$ for each $g\in
G$; thus $I$ is a graded ideal. On the other hand, $I$ is a nilpotent ideal
and in particular a graded-nil ideal of $T$. Now, let’s consider the ring
epimorphism $\epsilon:T\longrightarrow\prod_{1=1}^{n}R$ given by
$\epsilon((a_{ij})_{n\times n})=(a_{11},a_{22},\dots,a_{nn})$. Let $A\in
T_{g}$ where $g\in G$. Since $G$ is an abelian group, then elements of the
main diagonal of $A_{g}$ are all from $R_{g}$. Hence, $\epsilon$ is a graded-
homomorphism. Further, the kernel of $\epsilon$ is exactly the ideal $I$;
therefore, the two rings $T/I$ and $\prod_{1=1}^{n}R$ are graded-isomorphic.
According to Proposition, 3.2 $\prod_{1=1}^{n}R$ is graded U-nil clean; it
follows that $T/I$ is a graded U-nil clean ring. Now, since $I$ a graded-nil
ideal, then by Theorem 3.1 we conclude that $T$ is graded U-nil clean. ∎

######  Lemma 4.2.

Let $R=\oplus_{g\in G}R_{g}$ be G-graded ring and $t\in Idem(R_{e})\cap Z(R)$.
If the two rings $tR$ and $(1-t)R$ are $G$-graded U-nil clean, then $R$ is
graded U-nil clean.

###### Proof.

Assume that $tR$ and $(1-t)R$ are graded U-nil clean. Since $t\in Z(R)$ then
$1-t$ is a central element of $R$ as well; therefore we can apply peirce
decomposition as follow: $R=tR\oplus(1-t)R$. On the other hand, the ring
homomorphism $\epsilon:tR\times(1-t)R\longrightarrow R$ defined by
$\epsilon((tx,(1-t)y)=tx+(1-t)y$ is a graded-isomorphism. According to
Proposition 3.2, the ring $tR\times(1-t)R$ is graded U-nil clean. Therefore,
$R$ is graded U-nil clean.

∎

We recall that idempotent elements $f_{1},f_{2},\dots,f_{n}$ of a ring are
called orthogonal if $f_{i}f_{j}=0$ whenever $i\neq j$. Next, we give a
generalization of Lemma 4.2 using again the peirce decomposition.

######  Lemma 4.3.

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded ring, and suppose that
$1=f_{1}+\dots+f_{n}$ in $R$ where $f_{i}\in Idem(R_{e})\cap Z(R)$ are
orthogonal for $1\leq i\leq n$. If $f_{i}R$ is graded U-nil clean for each
$i\in\\{1,\dots,n\\}$, then $R$ is graded U-nil clean.

######  Theorem 4.5 (Main result).

Let $R=\oplus_{g\in G}R_{g}$ be a $G$-graded ring. If $R$ is graded U-nil
clean, then $M_{n}(R)(\sigma)$ is graded U-nil clean for every natural number
$n$ and every $\sigma\in G^{n}$.

###### Proof.

Let $n$ be an arbitrary natural number and $\sigma\in G^{n}$. It is clear that
$e_{11},\dots,e_{nn}$ are orthogonal idempotents of $M_{n}(R)_{e}(\sigma)$,
where $e_{ii}$ is the matrix having 1 on the $(i,i)$-position and 0 elsewhere.
Moreover, $e_{ii}$ is in the center of $M_{n}(R)(\sigma)$ for each $1\leq
i\leq n$, and we have that $e_{11}+\dots+e_{nn}=1$. On the other hand, the
ring homomorphism $\phi_{i}:e_{ii}M_{n}(R)(\sigma)\longrightarrow R$ given by
$\phi(e_{ii}(a_{kj})_{n\times n})=a_{ii}$ is a graded-isomorphism for each
$i\in\\{1,\dots,n\\}$. Hence, $e_{ii}M_{n}(R)(\sigma)$ is a graded U-nil clean
ring for every $i\in\\{1,\dots,n\\}$. Therefore, according to Lemma 4.3
$M_{n}(R)(\sigma)$ is $G$-graded U-nil clean. ∎

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