# Rings whose clean elements are
uniquely strongly clean

Peter Danchev  Institute of Mathematics and Informatics, Bulgarian Academy of
Sciences, 1113 Sofia, Bulgaria<EMAIL_ADDRESS><EMAIL_ADDRESS>,
Omid Hasanzadeh  Department of Mathematics, Tarbiat Modares University,
14115-111 Tehran Jalal AleAhmad Nasr, Iran<EMAIL_ADDRESS>and
Ahmad Moussavi  Department of Mathematics, Tarbiat Modares University,
14115-111 Tehran Jalal AleAhmad Nasr, Iran<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We define the class of CUSC rings, that are those rings whose clean elements
are uniquely strongly clean. These rings are a common generalization of the
so-called USC rings, introduced by Chen-Wang-Zhou in J. Pure & Applied Algebra
(2009), which are rings whose elements are uniquely strongly clean. These
rings also generalize the so-called CUC rings, defined by Calugareanu-Zhou in
Mediterranean J. Math. (2023), which are rings whose clean elements are
uniquely clean. We establish that a ring is USC if, and only if, it is
simultaneously CUSC and potent. Some other interesting relationships with CUC
rings are obtained as well.

###### Key words and phrases:

clean element (ring), uniquely clean element (ring), strongly clean element
(ring), uniquely strongly clean element (ring)

###### 2010 Mathematics Subject Classification:

16S34, 16U60

##  1\. Introduction and Basic Notion

Everywhere in the current paper, let $R$ be an associative but not necessarily
commutative ring having identity element, usually stated as $1$. Standardly,
for such a ring $R$, the letters $U(R)$, $\rm{Nil}(R)$ and ${\rm Id}(R)$ are
designed for the set of invertible elements (also termed as the unit group of
$R$), the set of nilpotent elements and the set of idempotent elements in $R$,
respectively. Likewise, $J(R)$ denotes the Jacobson radical of $R$. For all
other unexplained explicitly notions and notations, we refer to the classical
source [6].

In order to present our achievements here, we now need the necessary
background material as follows: Mimicking [5], an element $a$ from a ring $R$
is called clean if there exists $e\in{\rm Id}(R)$ such that $a-e\in U(R)$.
Then, $R$ is said to be clean if each element of $R$ is clean. In addition,
$a$ is called strongly clean provided $ae=ea$ and, if each element of $R$ are
strongly clean, then $R$ is said to strongly clean too. On the other hand,
imitating [7], $a\in R$ is called uniquely clean if there exists a unique
$e\in{\rm Id}(R)$ such that $a-e\in U(R)$. In particular, a ring $R$ is said
to be uniquely clean (or just UC for short) if every element in $R$ is
uniquely clean.

Generalizing these notions, in [3] was defined an element $a\in R$ to be
uniquely strongly clean if there exists a unique $e\in{\rm Id}(R)$ such that
$a-e\in U(R)$ and $ae=ea$. In particular, a ring $R$ is uniquely strongly
clean (or just USC for short) if each element in $R$ is uniquely strongly
clean.

In a similar vein, expanding the first part of the above concepts, in [1] a
ring is called CUC if any clean element is uniquely clean, and a ring is
called UUC if any unit is uniquely clean.

Our work targets to extend these two last definitions by defining and
exploring the following key instruments: A ring is called CUSC ring if every
clean element is uniquely strongly clean, and a ring is called UUSC ring if
every unit is uniquely strongly clean.

The relations between all of these notions are showed by the next diagrams:

UCUSCCUSCCleanUUSCUUCCUCStrongly Clean

Our plan to do that is structured thus: In the next section, we give some
critical examples and properties of both CUSC and UUSC rings and their
extensions (see, for instance, Propositions 2.9 and 2.11). The subsequent
third section is devoted to the exhibition of some other remarkable
characterization behaviors of these two classes of rings under some standard
extending procedures like matrix rings and group rings (see Theorems 3.1, 3.6,
3.7, 3.11, 3.15). We also pose some challenging properties which, hopefully,
will stimulate a further research on the subject.

##  2\. Examples and Major Properties

We start our consideration with the following.

######  Example 2.1.

The following are fulfilled:

  1. (i)

$\mathbb{Z}_{2}[x]$ is CUSC, but is not USC.

  2. (ii)

$\mathbb{Z}[x]$ is CUC, but is not UC.

  3. (iii)

$R=\begin{pmatrix}\mathbb{Z}_{2}&\mathbb{Z}_{2}\\\
0&\mathbb{Z}_{2}\end{pmatrix}$ is UUC, but is not CUC.

  4. (iv)

${\rm T}_{2}(\mathbb{Z}_{2})$ is CUSC, but is not CUC.

###### Proof.

  1. (i)

By [1, Example 2.2 (1)], $\mathbb{Z}_{2}[x]$ is CUC, and hence so is CUSC.
However, it is not USC; otherwise, if $\mathbb{Z}_{2}[x]$ is USC, then it has
to be clean, but this contradicts [5, Example 2].

  2. (ii)

$\mathbb{Z}[x]$ is CUC by [1, Example 2.2(1)], but is not UC; otherwise, if
$\mathbb{Z}[x]$ is UC, then it should be clean, which is a contradiction with
[5, Example 2].

  3. (iii)

$R$ is UUC by [1, Corollary 2.13(5)], but is not CUC, because CUC rings are
always abelian in view of [1, Proposition 2.1]. If $R$ is CUC, then
$\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}$ $\in{\rm Id}(R)$ is central. But
$\begin{pmatrix}0&1\\\ 0&0\end{pmatrix}\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}$
$=0$ and $\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}\begin{pmatrix}0&1\\\
0&0\end{pmatrix}=\begin{pmatrix}0&1\\\ 0&0\end{pmatrix},$ a contradiction.

  4. (iv)

${\rm T}_{2}(\mathbb{Z}_{2})$ is CUSC in virtue of [3, Theorem 10], but is not
CUC, because CUC rings are always abelian.

∎

We now show the validity of a series of technical claims.

######  Proposition 2.2.

Let $R$ be an abelian ring. Then, the following are equivalent:

  1. (i)

$(U(R)+U(R))\cap{\rm Id}(R)=\\{0\\}$.

  2. (ii)

$R$ is UUSC.

  3. (iii)

$R$ is UUC.

  4. (iv)

$R$ is CUC.

  5. (v)

$R$ is CUSC.

###### Proof.

(i) $\Rightarrow$ (ii): Let $u=e+v$ where $ev=ve$, $u,v\in U(R)$ and $e\in{\rm
Id}(R)$. Then we have $e=u+(-v)$ and by (i), $e=0$.
(ii) $\Rightarrow$ (iii): Let $u=e+v$ where $u,v\in U(R)$ and $e\in{\rm
Id}(R)$. By hypothesis we have $ev=ve$ and hence by (ii) we get $e=0$.
(iii) $\Rightarrow$ (iv): Follows from [1, Proposition 2.1].
(iv) $\Rightarrow$ (v): It is clear.
(v) $\Rightarrow$ (i): Let $e=u_{1}+u_{2}$ where $u_{1},u_{2}\in U(R)$ and
$e\in{\rm Id}(R)$, whence $u_{1}=e+(-u_{2})=0+u_{1}$, so we have two strongly
clean decompositions for $u_{1}$. Then $e=0$ by (v). ∎

An element in a ring is called 2-good if it is the sum of two units.

######  Example 2.3.

  1. (i)

Let ${\rm Id}(R)=\\{0,1\\}$. Then, $R$ is CUSC (resp., UUSC) if, and only if,
$1$ is not 2-good.

  2. (ii)

Let $R$ be a commutative ring. If $R$ is USC (or, resp., UC), then ${\rm
T}_{n}(R)$ is USC for any $n\geq 2$ using [3, Theorem]. So, ${\rm T}_{n}(R)$
is CUSC for any $n\geq 2$. But, ${\rm T}_{n}(R)$ is not CUC for any ring $R$
and any $n\geq 2$, because CUC rings are always abelian.

A subring of a USC ring need not be USC; for example, $\mathbb{Z}_{2}[[x]]$ is
USC, but $\mathbb{Z}_{2}[x]$ is not USC. Nevertheless, we obtain the
following.

######  Proposition 2.4.

Let $S$ be a subring of a ring $R$. If $R$ is CUSC (resp., UUSC), then so is
$S$.

###### Proof.

In view of Proposition 2.2, it suffices to show the UUSC case. To that aim,
write $u=v+e$, where $u,v\in U(S)$ and $e\in Id(S)$ with $ev=ve$. Then,

$u-(1_{R}-1_{S})=[v-(1_{R}-1_{S})]+e,$

where both $u-(1_{R}-1_{S}),v-(1_{R}-1_{S})\in U(R)$ and $e\in{\rm
Id}(S)\subseteq{\rm Id}(R)$ with

$e[v-(1_{R}-1_{S})]=ev-e(1_{R}-1_{S})=ve-(1_{R}-1_{S})e=[v-(1_{R}-1_{S})]e.$

Since $R$ is UUSC, it follows at once that $e=0$. So, $S$ is UUSC, as
promised. ∎

######  Proposition 2.5.

Let $R=\prod_{i\in I}R_{i}$ be a direct product of rings $\\{R_{i}\\}$ for all
$i\in I$. Then, $R$ is CUSC (resp., UUSC) if, and only if, so are $R_{i}$ for
all indices $i$.

###### Proof.

The necessity follows quickly from Proposition 2.4.

For the sufficiency, let $(u_{i})=(v_{i})+(e_{i})$, where $(e_{i})\in{\rm
Id}(R)$ and $(u_{i}),(v_{i})\in U(R)$ with $(v_{i})(e_{i})=(e_{i})(v_{i})$.
Then, for each index $i$, we write $u_{i}=v_{i}+e_{i}$, where $e_{i}\in{\rm
Id}(R_{i})$ and $u_{i},v_{i}\in U(R_{i})$ with $v_{i}e_{i}=e_{i}v_{i}$. Since
$R_{i}$ is UUSC, it follows at once that $e_{i}=0$. So, $(e_{i})=0$. Hence,
$R$ is UUSC, as expected. ∎

Two immediate consequences are the following.

######  Corollary 2.6.

Any subdirect product of CUSC (resp., UUSC) rings is CUSC (resp., UUSC).

######  Corollary 2.7.

Let $R$ be a ring and $e$ a central idempotent of $R$. Then, the following are
equivalent:

  1. (i)

$R$ is CUSC (resp., UUSC)

  2. (ii)

$eR$ and $(1-e)R$ are CUSC (resp., UUSC).

###### Proof.

It follows directly from Propositions 2.4 and 2.5. ∎

######  Lemma 2.8.

Let $I\subseteq J(R)$ be an ideal of $R$. The following hold:

  1. (i)

If $R/I$ is UUSC, then $R$ is UUSC. The converse holds, provided $R$ is an
abelian ring and idempotents lift modulo $I$.

  2. (ii)

If $R/I$ is CUSC and $R$ is abelian, then $R$ is CUSC. The converse holds,
provided idempotents lift modulo $I$.

###### Proof.

  1. (i)

Write $u=e+v$, where $u,v\in U(R)$ and $e\in{\rm Id}(R)$ with $ev=ve$, so it
is immediate that $\bar{u}=\bar{e}+\bar{v}$, where $\bar{u},\bar{v}\in
U(\bar{R}=\dfrac{R}{I})$ and $\bar{e}\in{\rm Id}(\bar{R})$ with
$\bar{e}\bar{v}=\bar{v}\bar{e}$; here $\bar{R}:=R/I$. Consequently, we have
$\bar{e}=\bar{0}$, so that $e\in I\subseteq J(R)$ and hence $e=0$.

Conversely, let $\bar{u}=\bar{e}+\bar{v}$, where $\bar{u},\bar{v}\in
U(\bar{R}=\dfrac{R}{I})$ and $\bar{e}\in{\rm Id}(\bar{R})$. It suffices to
show that $\bar{e}=\bar{0}$. By hypothesis, we can assume $e\in{\rm Id}(R)$
and $u,v\in U(R)$. Then, we have $u-(e+v)\in I\subseteq J(R)$, so
$u=e+v+j=e+(v+j)$ for $j\in J(R)$. As $R$ is abelian, we obtain
$e(v+j)=(v+j)e$, where $v+j\in U(R)$. Then, $e=0$ and hence $\bar{e}=\bar{0}$.

  2. (ii)

The proof is similar to (i).

∎

A ring $R$ is called local if $R/J(R)$ is a division ring.

######  Proposition 2.9.

Let $R$ be a ring. Then, the following are equivalent:

  1. (i)

$R$ is CUSC and local.

  2. (ii)

$\dfrac{R}{J(R)}\cong\mathbb{Z}_{2}$.

  3. (iii)

$R$ is UC and local.

  4. (iv)

$R$ is UC and ${\rm Id}(R)=\\{0,1\\}$.

  5. (v)

$R$ is USC and local.

  6. (vi)

$R$ is USC and ${\rm Id}(R)=\\{0,1\\}$.

  7. (vii)

$R$ is CUC and local.

###### Proof.

(i) $\Rightarrow$ (iii): Since $R$ is local, it must be that ${\rm
Id}(R)=\\{0,1\\}$. Thus, $R$ is abelian and also is CUSC, so $R$ is both CUC
and clean (because every local ring is known to be clean). Therefore, $R$ is
uniquely clean.
(iii) $\Rightarrow$ (i): It follows directly from Proposition 2.2 .
(ii) $\Leftrightarrow$ (iii) $\Leftrightarrow$ (iv): The equivalences are
proved in [7, Theorem 15].
(iii) $\Leftrightarrow$ (v): It is clear owing to [3, Example 4].
(iv) $\Leftrightarrow$ (vi): It is clear.
(i) $\Leftrightarrow$ (vii): It is clear by Proposition 2.2. ∎

For a subring $C$ of a ring $D$, the set

$R[D,C]:=\\{(d_{1},\ldots,d_{n},c,c,\ldots)|d_{i}\in D,c\in C,n\geq 1\\}$

with the addition and the multiplication defined component-wise is called the
tail ring extension and is denoted by $R[D,C]$.

######  Proposition 2.10.

The ring $R[D,C]$ is CUSC (resp., UUSC) if, and only if, so is $D$.

###### Proof.

Straightforward. ∎

If $R$ is a ring and $\alpha:R\rightarrow R$ is a ring homomorphism, let
$R[[x,\alpha]]$ denote the ring of skew formal power series over $R$; that is,
all formal power series in $x$ with coefficients from $R$ with multiplication
defined by $xr=\alpha(r)x$ for all $r\in R$.

We now offer the following:

######  Proposition 2.11.

$R[[x,\alpha]]$ is UUSC if, and only if, $R$ is UUSC.

###### Proof.

Assuming that $R[[x,\alpha]]$ is a UUSC ring, as we know that

$U(R[[x,\alpha]])=\\{\sum_{i=0}^{\infty}a_{i}x^{i}\mid a_{0}\in U(R)\\},$

we can readily conclude that $R$ is a UUSC ring.

Reciprocally, let us now assume that $R$ is a UUSC ring and

$f=\sum_{i=0}^{\infty}f_{i}x^{i}\in U(R[[x,\alpha]]).$

Apparently, $f_{0}\in U(R)$. Now, if we have $f=e+g$, where $e\in{\rm
Id}(R[[x,\alpha]])$, $g\in U(R[[x,\alpha]])$ and $eg=ge$, then $g_{0}$ is a
unit and $e_{0}$ is an idempotent in $R$ such that $f_{0}=e_{0}+g_{0}$ and
$e_{0}g_{0}=g_{0}e_{0}$. Since $R$ is a UUSC ring, we infer $e_{0}=0$. From
$e^{2}=e$ and $e_{0}=0$, we can conclude with the aid of a simple calculation
that $e_{i}=0$, which implies that $e=0$, as required. ∎

As an interesting consequence, we deduce:

######  Corollary 2.12.

Let $R$ be a ring and $\alpha:R\rightarrow R$ is a ring homomorphism. Then,
the following hold:

  1. (i)

For $n\geq 1,\frac{R[[x,\alpha]]}{\langle x^{n}\rangle}$ is UUSC if, and only
if, $R$ is UUSC.

  2. (ii)

For $n\geq 1,\frac{R[x,\alpha]}{\langle x^{n}\rangle}$ is UUSC if, and only
if, $R$ is UUSC.

We are now ready to attack the following.

######  Proposition 2.13.

Suppose that $R=S+I$, where $S$ is a subring of $R$ with $1_{R}\in S$, and $I$
is an ideal of $R$ such that $S\cap I=\\{0\\}$. If $S$ is UUSC and
$(U(R)+U(R))\cap{\rm Id}(I)=\\{0\\}$, then $R$ is UUSC.

###### Proof.

Write $u=e+v$, where $ev=ve$, $u,v\in U(R)$ and $e\in Id(R)$. Write
$u=u_{s}+u_{i}$, $e=e_{s}+e_{i}$, $v=v_{s}+v_{i}$, where $u_{s},e_{s},v_{s}\in
S$ and $u_{i},e_{i},v_{i}\in I$. So, employing [1, Proposition 2.12], we
deduce $u_{s}=e_{s}+v_{s}$, where $u_{s},v_{s}\in U(S)$, $e_{s}\in{\rm
Id}(S)$. Also, one may write

$ev=(e_{s}+e_{i})(v_{s}+v_{i})=e_{s}v_{s}+e_{s}v_{i}+e_{i}v_{s}+e_{i}v_{i}=$
$=v_{s}e_{s}+v_{s}e_{i}+v_{i}e_{s}+v_{i}e_{i}=(v_{s}+v_{i})(e_{s}+e_{i})=ve.$

If, however, $e_{s}v_{s}=v_{s}e_{i}$ or $v_{i}e_{s}=v_{i}e_{i}$, then one sees
that $e_{s}v_{s}=0$, because $S\cap I=\\{0\\}$. Therefore, $e_{s}=0$, and thus
$e=e_{i}\in I$. From the equality $(U(R)+U(R))\cap Id(I)=\\{0\\}$, it follows
automatically that $e=0$. But, if $e_{s}v_{s}=v_{s}e_{s}$, we derive
$e_{s}=0$, because $S$ is UUSC. So, by what we have established above, we
receive $e=0$. Processing analogously, we will have a similar result for
double, triple and quadruple states. Therefore, in all cases, we obtain
$e_{s}=0$ and hence $e=0$, so that $R$ is UUSC, as claimed. ∎

Let $A$ be a ring and $V$ a bi-module over $A$. The trivial extension of $A$
and $V$ is ${\rm T}(A,V)=\\{(a,x)|a\in A,x\in V\\}$ with the addition defined
component-wise and the multiplication defined by $(a,x)(b,y)=(ab,ay+xb)$.

######  Corollary 2.14.

Let $A,B$ be rings, $V$ a bi-module over $A,M$ an $(A,B)$-bi-module and $n\geq
1$

  1. (i)

$A[[x]]$ is UUSC if, and only if, $A$ is UUSC.

  2. (ii)

$\dfrac{A[[x]]}{\langle x^{n}\rangle}$ is UUSC if, and only if, $A$ is UUSC.

  3. (iii)

$\frac{A[x]}{\langle x^{n}\rangle}$ is UUSC if, and only if, $A$ is UUSC.

  4. (iv)

${\rm T}(A,V)$ is UUSC if, and only if, $A$ is UUSC.

  5. (v)

The formal triangular matrix ring $\begin{pmatrix}A&M\\\ 0&B\end{pmatrix}$ is
UUSC if, and only if, both $A,B$ are UUSC.

  6. (vi)

The upper triangular matrix ring ${\rm T}_{n}(A)$ is UUSC if, and only if, $A$
is UUSC.

###### Proof.

  1. (i)

Let $R=A[[x]],S=A$ and $I=xR$. Since ${\rm Id}(I)=\\{0\\}$, point (i) follows
from Propositions 2.13 and 2.4.

  2. (ii)

Let $R=\frac{A[[x]]}{\langle x^{n}\rangle}$, $S=A$ and $I=xR$. Since ${\rm
Id}(I)=\\{0\\}$, point (ii) follows from Propositions 2.13 and 2.4.

  3. (iii)

As $\frac{A[[x]]}{\langle x^{n}\rangle}\cong\frac{A[x]}{\langle
x^{n}\rangle}$, point (iii) follows automatically from (ii).

  4. (iv)

For the sufficiency, let $R={\rm T}(A,V)$, $S={\rm T}(A,0)$ and $I={\rm
T}(0,V)$. Since ${\rm Id}(I)=\\{0\\}$, $R$ is UUSC utilizing Proposition 2.13.

For the necessity, let $R$ be UUSC and $u=e+v$, where $u,v\in U(A)$ and
$e\in{\rm Id}(A)$ with $ev=ve$. Then, we have $(u,0)=(e,0)+(v,0)$, where
$(u,0)(v,0)\in U(R)$ and $(e,0)\in{\rm Id}(R)$ with $(e,0)(v,0)=(v,0)(e,0)$.
So, $(e,0)=(0,0)$ and hence $e=0$. Thus, $A$ is UUSC.

  5. (v)

For the sufficiency, let $R=\begin{pmatrix}A&M\\\ 0&B\end{pmatrix}$,
$S=\begin{pmatrix}A&0\\\ 0&B\end{pmatrix}$ and $I=\begin{pmatrix}0&M\\\
0&0\end{pmatrix}$. Since ${\rm Id}(I)=\\{0\\}$, $R$ is UUSC consulting with
Proposition 2.13.

For the necessity, let $R$ be UUSC. Write $u=e+v$ and
$u^{\prime}=e^{\prime}+v^{\prime}$, where $u,v\in
U(A),u^{\prime},v^{\prime}\in U(B),e\in{\rm Id}(A)$ and $e^{\prime}\in{\rm
Id}(B)$ with $ev=ve$ and $e^{\prime}v^{\prime}=v^{\prime}e^{\prime}$.
Therefore, one infers that

$\begin{pmatrix}u&0\\\ 0&u^{\prime}\end{pmatrix}=\begin{pmatrix}e&0\\\
0&e^{\prime}\end{pmatrix}+\begin{pmatrix}v&0\\\ 0&v^{\prime}\end{pmatrix},$

where $\begin{pmatrix}u&0\\\ 0&u^{\prime}\end{pmatrix},\begin{pmatrix}v&0\\\
0&v^{\prime}\end{pmatrix}\in U(R)$ and $\begin{pmatrix}e&0\\\
0&e^{\prime}\end{pmatrix}\in{\rm Id}(R)$ with

$\begin{pmatrix}e&0\\\ 0&e^{\prime}\end{pmatrix}\begin{pmatrix}v&0\\\
0&v^{\prime}\end{pmatrix}=\begin{pmatrix}v&0\\\
0&v^{\prime}\end{pmatrix}\begin{pmatrix}e&0\\\ 0&e^{\prime}\end{pmatrix}.$

So, by hypothesis,

$\begin{pmatrix}e&0\\\ 0&e^{\prime}\end{pmatrix}=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}.$

Consequently, $e=e^{\prime}=0$ whence $A,B$ are both UUSC.

  6. (vi)

It follows from (v).

∎

As a consequence, we derive:

######  Corollary 2.15.

Let $R$ be a ring and let $T$ be a subring of $R[[x]]$ with $R\subseteq
T\subseteq R[[x]]$. Then, $R$ is UUSC if, and only if, so is $T$.

###### Proof.

It follows from a combination of Proposition 2.4 and Corollary 2.14. ∎

A Morita context is a $4-tuple\begin{pmatrix}A&M\\\ N&B\end{pmatrix}$, where
$A$ and $B$ are rings, ${}_{A}M_{B}$ and ${}_{B}N_{A}$ are bi-modules and
there exist two context products $M\times N\rightarrow A$ and $N\times
M\rightarrow B$, written multiplicatively as $(x,y)=xy$ and $(y,x)=yx$, such
that $\begin{pmatrix}A&M\\\ N&B\end{pmatrix}$ is an associative ring with the
obvious matrix operations. A Morita context $\begin{pmatrix}A&M\\\
N&B\end{pmatrix}$ is called trivial, provided the context products are
trivially, i.e., $MN=0$ and $NM=0$. We know

$\begin{pmatrix}A&M\\\ N&B\end{pmatrix}\cong{\rm T}(A\times B,M\oplus N),$

where $\begin{pmatrix}A&M\\\ N&B\end{pmatrix}$ is a trivial Morita context.

We, thereby, extract the following.

######  Corollary 2.16.

The trivial Morita context $\begin{pmatrix}A&M\\\ N&B\end{pmatrix}$ is UUSC
if, and only if, $A,B$ are both UUSC.

###### Proof.

It is easy to see that the next relations are valid:

$\begin{pmatrix}A&M\\\ N&B\end{pmatrix}\cong{\rm T}(A\times B,M\oplus
N)\cong\begin{pmatrix}A\times B&M\oplus N\\\ 0&A\times B\end{pmatrix}.$

Thus, the rest of the proof follows applying Corollary 2.14 and Proposition
2.5. ∎

######  Proposition 2.17.

If ${\rm T}(A,V)$ is CUSC, then $A$ is CUSC. The converse holds, provided that
$ex=xe$ for all $x\in V$ and $e\in{\rm Id}(A)$.

###### Proof.

Let $T(A,V)$ be CUSC. It is easy to see that $A$ remains CUSC too.

For the converse, suppose A is CUSC.

Claim. If $(e,x)^{2}=(e,x)\in{\rm T}(A,V)$, then $e^{2}=e$ and $x=0$.

In fact, $(e,x)^{2}=(e,x)$ gives $e^{2}=e$ and $ex+xe=x$. So, by hypothesis,
we have $2ex=x$. Then, multiplying by $e$ gives $2ex=ex$ and hence $ex=0$, so
that $x=0$, proving the claim.

Let us write

$(e,0)+(u,x)=(f,0)+(v,x^{\prime}),$

where $(u,x),(v,x^{\prime})\in U(T(A,V))$ and $(e,0),(f,0)\in{\rm Id}(T(A,V))$
with $(e,0)(u,x)=(u,x)(e,0)$ and $(f,0)(v,x^{\prime})=(v,x^{\prime})(f,0)$.
Therefore, we write $e+u=f+v$, where $u,v\in U(A)$ and $e,f\in{\rm Id}(A)$
with $eu=ue$ and $fv=vf$. Thus, by hypothesis, $e=f$ forcing that
$(e,0)=(f,0)$ and hence ${\rm T}(A,V)$ is CUSC, as claimed. ∎

Let $R$ be a ring and let $M$ be an $(R,R)$-bi-module which is a general ring
(possibly with no unity) in which the equalities $(mn)r=m(nr)=(mr)n$ and
$(rm)n=r(mn)$ hold for all $m,n\in M$ and $r\in R$. Then, the ideal-extension
${\rm I}(R,M)$ of $R$ by $M$ is defined to be the additive abelian group ${\rm
I}(R,M)=R\oplus M$ with multiplication $(r,m)(s,n)=(rs,rn+ms+mn)$. Note that,
if $R^{\prime}$ is a ring and $R^{\prime}=R\oplus K$, where $R$ is a subring
and $K$ is an ideal of $R^{\prime}$, then $R^{\prime}\cong{\rm I}(R,K)$ holds.

######  Proposition 2.18.

If ${\rm I}(R,M)$ is UUSC, then $R$ is UUSC.

######  Proposition 2.19.

An ideal-extension ${\rm I}(R,M)$ is UUSC if the following conditions are
satisfied:

  1. (a)

$R$ is UUSC.

  2. (b)

If $e\in{\rm Id}(R)$, then $em=me$ for all $m\in M$.

  3. (c)

If $m\in M$, then $m+n+mn=0$ for some $n\in M$.

###### Proof.

In view of points (b) and (c), every unit of ${\rm I}(R,M)$ is of the form
$(u,m)$, where $u\in U(R)$ and $m\in M$ as well as every idempotent of ${\rm
I}(R,M)$ is of the form $(e,0)$, where $e\in{\rm Id}(R)$ (see, for instance,
[7, Proposition 7]). Now, let $(u,m)=(e,0)+(v,m^{\prime})$, where
$(u,m),(v,m^{\prime})$ are units of ${\rm I}(R,M)$, and $(e,0)$ is an
idempotent of ${\rm I}(R,M)$. Moreover,
$(e,0)(v,m^{\prime})=(v,m^{\prime})(e,0)$, so $u=e+v$ such that $ev=ve$.
Therefore, in virtue of (a), we have $e=0$, and thus $(e,0)=(0,0)$, as
pursued. ∎

######  Remark 2.20.

It is easy to see that Propositions 2.18 and 2.19 hold also for CUSC rings.

##  3\. Main Theorems

A ring $R$ is called semi-potent if every one-sided ideal not contained in
$J(R)$ contains a non-zero idempotent. Moreover, a semi-potent ring $R$ is
called potent if idempotents lift modulo $J(R)$. Hereafter, the center of a
ring $R$ is denoted by ${\rm Z}(R)$.

######  Theorem 3.1.

Let $R$ be a semi-potent ring. Then, the following are equivalent:

  1. (i)

$\dfrac{R}{J(R)}$ is UUSC.

  2. (ii)

$\dfrac{R}{J(R)}$ is Boolean.

  3. (iii)

$U(R)=1+J(R)$.

  4. (iv)

$U(R)\subseteq ucn_{0}(R)=\left\\{e+j|e^{2}=e\in{\rm Z}(R),j\in
J(R)\right\\}$.

  5. (v)

For each $a\in U(R)$, there exists a unique $e\in{\rm Id}(R)$ such that
$a-e\in J(R)$.

  6. (vi)

For each $a\in U(R)$, there exists $e\in{\rm Id}(R)$ such that $a-e\in J(R)$.

###### Proof.

(i) $\Rightarrow$ (ii): Since $R$ is semi-potent, $\bar{R}=\frac{R}{J(R)}$ is
semi-potent. We now show that $\bar{R}$ is a reduced ring. Assume $a^{2}=0$
for some $0\not=a\in\bar{R}$. Then, by [10, Theorem 2.1], there exists
$0\not=e^{2}=e\in\bar{R}$ such that $e\bar{R}e\cong{\rm M}_{2}(S)$ for a non-
trivial ring $S$. As $\bar{R}$ is UUSC, so $e\bar{R}e$ is UUSC by Proposition
2.4. It is, however, not 2-good, so $1$ is not 2-good in $e\bar{R}e$ whence
$1$ is not 2-good in ${\rm M}_{2}(S)$ too, and this is a contradiction,
because in any $2\times 2$ matrix ring it is always true that

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}=\begin{pmatrix}1&1\\\
1&0\end{pmatrix}+\begin{pmatrix}0&-1\\\ -1&1\end{pmatrix}.$

Hence, $\bar{R}$ is reduced and thus abelian. We next show that $\bar{R}$ is a
Boolean ring. Assume on the contrary that $a^{2}\not=a$ for some
$a\in\bar{R}$. As $\bar{R}$ is semi-potent with $J(\bar{R})=\\{0\\}$,
$(a-a^{2})\bar{R}$ contains a non-zero idempotent, say $e$. Write
$e(a-a^{2})=b$ with $b\in\bar{R}$. Then, it must be that

$e=e(a-a^{2})b=ea\cdot e(1-a)b=e(1-a)\cdot eab$

. As $e\bar{R}e$ is reduced, both $ea$ and $e(1-a)$ are units of $e\bar{R}e$.
Since $ea+e(1-a)=e$, it manifestly follows from Proposition 2.2 that
$e\bar{R}\bar{e}$ is not UUSC. This, however, contradicts Proposition 2.4.
Hence, $\bar{R}$ is Boolean, as wanted.

(ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv) $\Rightarrow$ (v) $\Rightarrow$
(vi): These implications follow immediately from [1, Theorem 3.1].

(vi) $\Rightarrow$ (i): Let $\bar{u}=\bar{e}+\bar{v}$, where $\bar{e}\in{\rm
Id}(\bar{R})$ and $\bar{u},\bar{v}\in U(\bar{R})$ with
$\bar{e}\bar{v}=\bar{v}\bar{e}$. As units lift modulo $J(R)$, we can assume
$u,v\in U(R)$. By (vi), $u=f+j$ and $v=g+j^{\prime}$, where $f,g\in{\rm
Id}(R)$ and $j,j^{\prime}\in J(R)$. Then, $f=u-j\in U(R)$, so $f=1$.
Similarly, $g=1$. Therefore, $\bar{1}=\bar{e}+\bar{1}$, so that
$\bar{e}=\bar{0}$ and hence $\bar{R}$ is UUSC, as desired. ∎

A ring $R$ is said to be left quasi-duo (resp., right quasi-duo) if every
maximal left ideal (resp., maximal right ideal) of $R$ is an ideal.

We now extract a series of assertions as follows.

######  Corollary 3.2.

Every potent UUSC ring is left and right quasi-duo.

###### Proof.

Let $M$ be a maximal left ideal in a ring $R$. Since $R$ is potent and UUSC, a
consultation with Theorem 3.1 assures that the quotient-ring
$\bar{R}=\frac{R}{J(R)}$ is Boolean, so we have
$\frac{\bar{R}}{\bar{M}}\cong\mathbb{Z}_{2}$. But then

$\frac{R}{M}\cong\frac{\bar{R}}{\bar{M}}\cong\mathbb{Z}_{2}$

has two elements, so that $\frac{R}{M}=\\{M,1+M\\}$, and thus we conclude
$R=M\cup(1+M)$.

Now, let $x\in M$ and $r\in R$; we must show that $xr\in M$. This is, however,
clear if $r\in M$; for otherwise $r=1+y,y\in M$. Then, $xr=x+xy\in M$, as
required. ∎

######  Corollary 3.3.

Let $R$ be a regular ring. Then, $R$ is UUSC if, and only if, $R$ is Boolean.

###### Proof.

We know that if $R$ is regular, then $J(R)=\\{0\\}$ and $R$ is semi-potent, as
required. ∎

######  Proposition 3.4.

A ring $R$ is USC if, and only if, $R$ is simultaneously clean and CUSC.

###### Proof.

It is self-evident. ∎

######  Theorem 3.5.

Let $R$ be a ring which is simultaneously CUSC (resp., UUSC) and semi-potent.
Then, $2\in J(R)$.

###### Proof.

Let us assume in a way of contradiction that $2\notin J(R)$. Since $R$ is
semi-potent, there exists $0\neq e^{2}=e\in 2R$. Write $e=2a$ with $a\in R$.
Therefore, $ea=ae$ and so $(1-3e)(1-3ae)=1$. Similarly, $(1-3ae)(1-3e)=1$ and
thus $1-3e\in U(R)$. We, however, know that $1-2e\in U(R)$, whence

$1-2e=0+(1-2e)=e+(1-3e).$

We, therefore, have two strongly clean decompositions for the element $1-2e$
while $R$ is CUSC (resp., UUSC). Hence, $e=0$, which is the desired
contradiction. ∎

The following statement is pivotal.

######  Theorem 3.6.

Let $R$ be a CUSC (resp., UUSC) and let $\bar{R}=\dfrac{R}{J(R)}$. Then, the
following hold:

  1. (i)

$1$ is not 2-good in $R$.

  2. (ii)

For any $0\neq e^{2}=e\in R$ and any $u_{1},u_{2}\in U(eRe)$, the inequality
$u_{1}+u_{2}\neq e$ is true.

  3. (iii)

For any $n>1$, there does not exist $0\neq e^{2}=e\in R$ such that
$eRe\cong{\rm M}_{n}(S)$ for some ring $S$.

  4. (iv)

$\bar{1}$ is not 2-good in $\bar{R}$.

  5. (v)

If $R$ is potent, then for any $\bar{0}\neq\bar{e}^{2}=\bar{e}\in\bar{R}$ and
any $\bar{u}_{1},\bar{u}_{2}\in U(\bar{e}\bar{R}\bar{e})$, the inequality
$\bar{u}_{1}+\bar{u}_{2}\neq\bar{e}$ is true.

  6. (vi)

If $R$ is potent, then, for any $n>1$, there does not exist
$\bar{0}\neq\bar{e}^{2}=\bar{e}\in\bar{R}$ such that
$\bar{e}\bar{R}\bar{e}\cong{\rm M}_{n}(S)$ for some ring $S$.

###### Proof.

  1. (i)

Write $1=u_{1}+u_{2}$, where $u_{1},u_{2}\in U(R)$. Then,
$u_{1}=1+(-u_{2})=0+u_{1}$. As $R$ is UUSC, we get $1=0$ that is an absurd.
For the CUSC case the arguments are analogous, because every unit is obviously
a clean element.

  2. (ii)

We know that $eRe$ is a subring of $R$ and $1_{eRe}=e$. Therefore, the result
follows from (i).

  3. (iii)

Since $n>1$, it is well known that ${\rm M}_{n}(S)$ contains a corner ring
isomorphic to the $2\times 2$ matrix ring. So, $eRe$ contains a corner ring
isomorphic to the $2\times 2$ matrix ring. Thus, without loss of generality,
we can assume that $n=2$. Furthermore, in ${\rm M}_{2}(S)$, we have

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}=\begin{pmatrix}1&1\\\
1&0\end{pmatrix}+\begin{pmatrix}0&-1\\\ -1&1\end{pmatrix}.$

That is why, in the corner ring $eRe$, we have $u_{1}+u_{2}=e$. This is,
however, against (ii), as wanted.

  4. (iv)

And because units lift modulo $J(R)$, point (iv) follows immediately from (i).

  5. (v)

Given $\bar{e},\bar{u}_{1},\bar{u}_{2}$ as in (v), we can assume $e^{2}=e$
because idempotents lift modulo $J(R)$. Then, one has that
$\bar{e}\bar{R}\bar{e}\cong\dfrac{eRe}{J(eRe)}$ exploiting [6, Theorem 21.10]
since $eRe$ is CUSC (resp., UUSC) using Proposition 2.4.

  6. (vi)

Follows from (v).

∎

We now arrive at our chief necessary and sufficient condition that is the
transversal between USC and CUSC rings.

######  Theorem 3.7.

A ring $R$ is USC if, and only if, $R$ is CUSC and potent.

###### Proof.

Suppose that $R$ is CUSC and potent. So, $R$ is semi-potent, and thus
$\bar{R}=\dfrac{R}{J(R)}$ is semi-potent. We show that $\bar{R}$ is reduced.
To that goal, assume that $\bar{t}^{2}=\overline{0}$, where
$\overline{0}\neq\bar{t}\in\bar{R}$. Then, utilizing [3, Lemma 14], there
exists $\overline{0}\neq\bar{e}^{2}=\bar{e}\in\bar{R}$ such that
$\bar{e}\bar{R}\bar{e}\cong{\rm M}_{2}(S)$ for some ring $S$. However, this
contradicts Theorem 3.6. Hence, $\bar{R}$ is reduced, so it has to be abelian.

Now, we prove that $\bar{R}$ is USC. To that aim, suppose

$\bar{a}=\bar{e}+\bar{u}=\bar{f}+\bar{v}$

are two strongly clean decompositions for $\bar{a}$ in $\bar{R}$.
Consequently, $\bar{g}:=\bar{e}-\bar{f}$ is a central idempotent of $\bar{R}$,
because $\bar{R}$ is abelian and because $\overline{2}=\overline{0}$ taking
into account Theorem 3.5. Thus, $\bar{g}=\bar{v}+(-\bar{u})$, so that
$\bar{g}=\bar{g}\bar{v}\bar{g}+\bar{g}(-\bar{u})\bar{g}$, where both
$\bar{g}\bar{v}\bar{g}$ and $\bar{g}(-\bar{u})\bar{g}$ are units of
$\bar{g}\bar{R}\bar{g}$. Bearing in mind Theorem 3.6, it must be that
$\bar{g}=\overline{0}$, whence $\bar{e}=\bar{f}$. Hence, $\bar{R}$ is USC, as
promised, and, moreover, it is also abelian. Therefore, $\bar{R}$ is UC and
thus it is clean. As all idempotents lift modulo $J(R)$, we conclude that $R$
is clean referring to [5, Proposition 6]. Then, $R$ is CUSC and clean, so $R$
is USC by virtue of Proposition 3.4. The other implication is elementary.
∎

As three challenging consequences, we yield:

######  Corollary 3.8.

Let $R$ be a CUSC ring. Then, the following are equivalent:

  1. (i)

$R$ is clean.

  2. (ii)

$R$ is exchange.

  3. (iii)

$R$ is potent.

  4. (iv)

$R$ is USC.

  5. (v)

$R$ is strongly clean.

###### Proof.

It is straightforward, so we omit the full details leaving them to the
interested reader for an inspection. ∎

We shall say that a ring $R$ is semi-boolean, provided the factor-ring
$\frac{R}{J(R)}$ is Boolean and idempotents lif modulo $J(R)$ or,
equivalently, provided every element of $R$ is a sum of an idempotent and an
element from $J(R)$.

######  Corollary 3.9.

A ring $R$ is semi-boolean if, and only if, $R$ is potent and
$\dfrac{R}{J(R)}$ is UUSC.

###### Proof.

Supposing that $R$ is semi-boolean, we get $R$ is clean, and thus it is potent
and hence semi-potent. Besides, $\dfrac{R}{J(R)}$ is Boolean, so it is UUSC by
Theorem 3.1.

Conversely, assuming $R$ is potent and $\dfrac{R}{J(R)}$ is UUSC, we get $R$
is semi-potent. Thus, $\dfrac{R}{J(R)}$ is Boolean by Theorem 3.1 and,
moreover, idempotents lift modulo $J(R)$, so that $R$ is semi-boolean. ∎

######  Corollary 3.10.

If $R=ucn_{0}(R)=\\{e+j|e^{2}=e\in{\rm Z}(R),j\in J(R)\\}$, then $R$ is USC.
The converse is not true.

###### Proof.

If $R=ucn_{0}(R)$, then $R$ is uniquely clean by [1, Corollary 3.4]. Thus, $R$
is USC.

Next, put $R={\rm T}_{2}(\mathbb{Z}_{2})$. So, $R$ is USC, but it is
manifestly not UC. Also, $R\neq ucn_{0}(R)$, because

$\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}+\begin{pmatrix}0&1\\\ 0&0\end{pmatrix},$

where $\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}$ is obviously not a central
element. ∎

We are now prepared to establish the following crucial result, which gives a
satisfactory necessary and sufficient condition for the triangular matrix ring
to be CUSC, and also expands the corresponding one from [3].

######  Theorem 3.11.

Let $R$ be a commutative semi-potent ring. Then, the following statements are
equivalent:

  1. (i)

$R$ is CUSC.

  2. (ii)

$R$ is CUC.

  3. (iii)

${\rm T}_{n}(R)$ is CUSC for all $n\geq 1$.

  4. (iv)

${\rm T}_{n}(R)$ is CUSC for some $n\geq 1$.

###### Proof.

(i) $\Leftrightarrow$ (ii): This implication follows from Proposition 2.2.
(iii) $\Rightarrow$ (iv): This implication is obvious.
(iv) $\Rightarrow$ (i): Set $e=\mathrm{diag}(1,0,\ldots,0)\in{\rm T}_{n}(R)$.
Then, one inspects that $R\cong e{\rm T}_{n}(R)e$. Therefore, $R$ is CUSC by
Proposition 2.4.
(i) $\Rightarrow$ (iii): Clearly, the result is true for $n=1$. Assume now
that the result holds for $n\geq 2$. Let $A=\begin{pmatrix}a_{11}&x\\\
0&A_{1}\end{pmatrix}\in{\rm T}_{n+1}(R)$ is a clean element, where $a_{11}\in
R$, $x\in M_{1\times n}(R)$ and $A_{1}\in{\rm T}_{n}(R)$. Then, both $A_{1}$
and $a_{11}$ are clean elements. By (i) and by hypothesis, $A_{1}$ and
$a_{11}$ have unique strongly clean expressions in ${\rm T}_{n}(R)$ and in
$R$, respectively, like this: $A_{1}=E_{1}+U_{1}$, $a_{11}=e_{11}+u_{11}$.

Note that, if

$A=\begin{pmatrix}e_{22}&y\\\ 0&E_{2}\end{pmatrix}+\begin{pmatrix}u_{22}&z\\\
0&U_{2}\end{pmatrix}$

is a strongly clean expression in ${\rm T}_{n+1}(R)$, then $A_{1}=E_{2}+U_{2}$
and $a_{11}=e_{22}+u_{22}$ are strongly clean expressions in ${\rm T}_{n}(R)$
and in $R$, respectively (notice that $A$ is clean, so $A_{1}$ and $a_{11}$
are both clean). Hence, $E_{1}=E_{2}$, $U_{1}=U_{2}$, $e_{11}=e_{22}$ and
$u_{11}=u_{22}$. Now, it suffices to prove that there exists a unique
$x_{1}\in{\rm M}_{1\times n}(R)$ such that

$A=\begin{pmatrix}a_{11}&x\\\
0&A_{1}\end{pmatrix}=\begin{pmatrix}e_{11}&x_{1}\\\
0&E_{1}\end{pmatrix}+\begin{pmatrix}u_{11}&x-x_{1}\\\ 0&U_{1}\end{pmatrix}$

is a strongly clean decomposition in ${\rm T}_{n+1}(R)$. Set
$E:=\begin{pmatrix}e_{11}&x_{1}\\\ 0&E_{1}\end{pmatrix}$ and
$U:=\begin{pmatrix}u_{11}&x-x_{1}\\\ 0&U_{1}\end{pmatrix}$. Apparently, $U$ is
invertible. It is also easy to see that

$\displaystyle E^{2}=E$ $\displaystyle\Leftrightarrow
e_{11}x_{1}+x_{1}E_{1}=x_{1}$ (1)
$\displaystyle\Leftrightarrow(e_{11}I+E_{1})x_{1}=x_{1}$ $\displaystyle EU=UE$
$\displaystyle\Leftrightarrow
e_{11}(x-x_{1})+x_{1}U_{1}=u_{11}x_{1}+(x-x_{1})E_{1}$
$\displaystyle\Leftrightarrow(U_{1}-u_{11}I-2e_{11}I)x_{1}+(E_{1}+e_{11}I)x_{1}$
(2) $\displaystyle=(E_{1}-e_{11}I)x$

Combining (1) with (2), we deduce

$\displaystyle(U_{1}-u_{11}I-2e_{11}I)x_{1}+x_{1}=(E_{1}-e_{11}I)x,\text{so
that}$ $\displaystyle[U_{1}+(1-2e_{11}-u_{11})I]x_{1}=(E_{1}-e_{11}I)x.$

Since $R$ is commutative semi-potent and CUSC, one verifies that $R$ is semi-
potent CUC and $\dfrac{R}{J(R)}$ is Boolean (see, e.g., [1, Theorem 3.2]).
Thus, $2\in J(R)$ and $1-u^{\prime}\in J(R)$ for all $u^{\prime}\in U(R)$,
because $U(R)=1+J(R)$. In particular, $1-u_{11}\in J(R)$ and so
$1-2e_{11}-u_{11}\in J(R)$. Therefore, $U_{1}+(1-2e_{11}-u_{11})I$ is
invertible and hence we can write

$x_{1}=[U_{1}+(1-2e_{11}-u_{11})I]^{-1}(E_{1}-e_{11}I)x.$

Then, $x_{1}$ has to be unique.

Next, we show that $x_{1}$ satisfies (1) and (2). As $E_{1}U_{1}=U_{1}E_{1}$,
the elements $E_{1}+e_{11}I,U_{1}+(1-2e_{11}-u_{1})I$ and
$[U_{1}+(1-2e_{11}-u_{11})I]^{-1}$ all commute. Consequently,

$\displaystyle(E_{1}+e_{11}I)x_{1}=[U_{1}+(1-2e_{11}-u_{11})I]^{-1}(E_{1}+e_{11}I)$
$\displaystyle(E_{1}-e_{11}I)x=[U_{1}+(1-2e_{11}-u_{11})I]^{-1}(E_{1}-e_{11}I)x$
$\displaystyle=x_{1}.$

So, $x_{1}$ satisfies (1). Similarly, $x_{1}$ satisfies (2), as pursued. ∎

A question which naturally arises is what happens in the case of UUSC rings?

We are now concerned with group rings.

We denote by $RG$ the group ring of $G$ over $R$. The augmentation mapping
$\varepsilon:RG\rightarrow R$ is given by $\varepsilon(\sum a_{g}g)=\sum
a_{g}$ and its kernel, denoted by $\triangle(RG)$, is an ideal generated by
$\\{1-g|g\in G\\}$. A group $G$ is called a $p$-group if every element of $G$
is power of $p$, where $p$ is a prime. Also, a group $G$ is called locally
finite if every finitely generated subgroup is finite.

######  Lemma 3.12.

Suppose that every idempotent of $RG$ is contained in $R$. Then, $RG$ is CUSC
(resp., UUSC) if, and only if, $R$ is CUSC (resp., UUSC).

###### Proof.

In view of Proposition 2.2, it suffices to show the UUSC case. The necessity
is clear by Proposition 2.4. For the sufficiency, let $u=e+v$ where $u,v\in
U(RG)$ and $e\in Id(RG)=Id(R)$ with $ev=ve$. Then
$\varepsilon(u)=\varepsilon(e+v)=\varepsilon(e)+\varepsilon(v)=e+\varepsilon(v)$,
where $\varepsilon(u),\varepsilon(v)\in U(R)$ and $e\in Id(R)$ with
$e\varepsilon(v)=\varepsilon(v)e$. Since $R$ is UUSC, it follows that $e=0$.
Then $RG$ is UUSC. ∎

As an immediate consequence, we extract:

######  Corollary 3.13.

The integral group ring $\mathbb{Z}G$ of an arbitrary group $G$ is CUSC.

The following two assertions are worthwhile.

######  Proposition 3.14.

Let $G$ be a locally finite $2$-group, and let $R$ be UUSC and semi-potent.
Then, $RG$ is UUSC.

###### Proof.

Let $u=e+v$ with $ev=ve$, where $u,v\in U(RG)$ and $e\in{\rm Id}(RG)$.
Consequently,

$\varepsilon(u)=\varepsilon(e)+\varepsilon(v),\varepsilon(e)\varepsilon(v)=\varepsilon(ev)=\varepsilon(ve)=\varepsilon(v)\varepsilon(e),$

where $\varepsilon(u),\varepsilon(v)\in U(R)$ and $\varepsilon(e)\in{\rm
Id}(R)$. So, $\varepsilon(e)=0$, and thus
$e\in\operatorname{ker}\varepsilon=\Delta(RG)$. We also know by Theorem 3.5
that if $R$ is UUSC and semi-potent, then $2\in J(R)$. Consequently, using [9,
Lemma 2], we have $\Delta(RG)\subseteq J(RG)$, so that $e\in J(RG)$ which
assures that $e=0$. ∎

We finish off our work with the following criterion.

######  Theorem 3.15.

Let $R$ be a potent ring and let $G$ be a locally finite group. Then, $RG$ is
CUSC (resp., UUSC) if, and only if, $R$ is CUSC (resp., UUSC) and $G$ is a
$2$-group.

###### Proof.

Firstly, we prove the CUSC case. To that purpose, assume that $R$ is CUSC and
$G$ is a $2$-group. Since $R$ is a potent ring, one observes that $R$ is USC
in virtue of Theorem 3.7. Therefore, $RG$ is USC owing to [8, Theorem 3.1].
Secondly, suppose that $RG$ is CUSC, so $R$ is CUSC by virtue of Proposition
LABEL:prop2.4. Assume, in a way of contradiction, that $G$ is not a $2$-group.
Then, $G$ contains an element $g$ of prime order $p>2$. As $R\langle g\rangle$
is a subring of $RG$, so $R\langle g\rangle$ is UUSC by Proposition 2.4. Since
$R$ is potent and UUSC, the factor $\dfrac{R}{J(R)}$ is Boolean employing
Theorem 3.1, so $p\in U(R)$. And since

$x^{p-1}+x^{p-2}+\cdots+x+1=(x-1)[x^{p-2}+2x^{p-3}+\cdots+(p-2)x+(p-1)]+p,$

we derive $(x-1)(x^{p-1}+x^{p-2}+\cdots+x+1)=R[x]$. Hence, we obtain

$R\langle g\rangle\cong\dfrac{R[x]}{\langle
x^{p}-1\rangle}\cong\dfrac{R[x]}{\langle x-1\rangle}\oplus\dfrac{R[x]}{\langle
x^{p-1}+\cdots+x+1\rangle}.$

Thus, $S:=\dfrac{R[x]}{\langle x^{p-1}+\cdots+x+1\rangle}$ is UUSC by
Proposition 2.5. However, in $S$, we have that

$(x^{2}+x)[x^{p-3}+x^{p-5}+\cdots+x^{2}+1]=x^{p-1}+\cdots+x=-1\in U(S),$

whence $x,x+1\in U(S)$. But, we also have $x+1=0+(x+1)=1+x$ that are two
strongly clean decompositions for $1+x\in U(R)$, contradicting our initial
assumption. Finally, $G$ is a $2$-group, as asserted.

Next, we show the UUSC case. Let $R$ be UUSC and $G$ a $2$-group. Then, $RG$
is UUSC exploiting Proposition 3.14. Conversely, letting $RG$ be UUSC, then
$R$ is UUSC consulting with Proposition 2.4. Additionally, $G$ has to be a
$2$-group using the same arguments as in the proof of the previous case. ∎

Funding: The first-named author, P.V. Danchev, of this research paper was
partially supported by the Junta de Andalucía under Grant FQM 264, and by the
BIDEB 2221 of TÜBÍTAK.

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