Weakly $r$-clean rings and weakly $\star$-clean rings

TL;DR

This paper introduces and explores the properties of weakly $r$-clean, weakly $g(x)$-$r$-clean, and weakly $igstar$-clean rings, generalizing the concept of weakly clean rings in ring theory.

Contribution

It defines new classes of rings—weakly $r$-clean, weakly $g(x)$-$r$-clean, and weakly $igstar$-clean—and investigates their properties and relationships.

Findings

Defined weakly $r$-clean rings and their properties.

Generalized to weakly $g(x)$-$r$-clean rings with polynomial functions.

Introduced weakly $igstar$-clean and $igstar$-$r$-clean rings and discussed their properties.

Abstract

Motivated by the concept of weakly clean rings, we introduce the concept of weakly $r$-clean rings. We define an element $x$ of a ring $R$ as weakly $r$-clean if it can be expressed as $x=r+e$ or $x=r-e$ where $e$ is an idempotent and $r$ is a regular element of $R$. If all the elements of $R$ are weakly $r$-clean then $R$ is called a weakly $r$-clean ring. We discuss some of its properties in this article. Also we generalise this concept of weakly $r$-clean ring to weakly $g(x)$-$r$-clean ring, where $g(x) \in C(R)[x]$ and $C(R)$ is the centre of the ring $R$. Finally we introduce the concept of weakly $\star$-clean and $\star$-$r$-clean ring and discuss some of their properties.

Equations

$$x_{\alpha }$$

$$=0if\alpha \notin \{{\alpha }_1,{\alpha }_2\}$$

$$a-e$$

$$=(a-1)e+a(1-e)$$

$$=-(1-a)e+a(1-e)$$

$$=-(1-e^{\prime }-u)e+(-u^{\prime }-e^{\prime \prime })(1-e)$$

$$=-\{(1-e^{\prime })e+e^{\prime \prime }(1-e)\}+\{ue-u^{\prime }(1-e)\}$$

$$u(x+f)$$

$$=(u-e{)}^2+(u-e)$$

$$=x^2+x$$

$${\theta }^{\prime }(g(\theta (s_0)))$$

$$=\theta (a_0+a_1{s}_0+\cdot \cdot \cdot +a_n{s}_0^n)$$

$$=\theta (0)$$

$$=0$$

Full text

**Weakly $r$-clean rings and weakly $\star$-clean rings

**

Ajay Sharma

*Department of Mathematical Sciences, Tezpur University,

Napaam, Tezpur-784028, Assam, India.

Email: ajay123@tezu.ernet.in*

Dhiren Kumar Basnet

*Department of Mathematical Sciences, Tezpur University,

Napaam, Tezpur-784028, Assam, India.

Email: dbasnet@tezu.ernet.in*

**Abstract: ** Motivated by the concept of weakly clean rings, we introduce the concept of weakly $r$-clean rings. We define an element $x$ of a ring $R$ as weakly $r$-clean if it can be expressed as $x=r+e$ or $x=r-e$ where $e$ is an idempotent and $r$ is a regular element of $R$. If all the elements of $R$ are weakly $r$-clean then $R$ is called a weakly $r$-clean ring. We discuss some of its properties in this article. Also we generalise this concept of weakly $r$-clean ring to weakly $g(x)$-$r$-clean ring, where $g(x)\in C(R)[x]$ and $C(R)$ is the centre of the ring $R$. Finally we introduce the concept of weakly $\star$-clean and $\star$-$r$-clean ring and discuss some of their properties.

Key words: $r$-clean ring, Weakly $r$-clean ring, $\star$-clean ring, Weakly $\star$-clean ring, $\star$-$r$-clean ring.

$2010$** Mathematics Subject Classification:** 16N40, 16U99.

1 INTRODUCTION

Here rings $R$ are associative with unity unless otherwise indicated. The Jacobson radical, set of units, set of idempotents, set of regular elements, set of nilpotent elements and centre of a ring $R$ are denoted by $J(R)$, $U(R)$, $Idem(R)$, $Reg(R)$, $Nil(R)$ and $C(R)$ respectively. Nicholson[8] called an element $x$ of $R$, a clean element, if $x=e+u$ for some $e\in Idem(R)$, $u\in U(R)$ and called the ring $R$ as clean ring if all its elements are clean. Weakening the condition of clean element, M.S. Ahn and D.D. Anderson[1] defined an element $x$ as weakly clean if $x$ can be expressed as $x=u+e$ or $x=u-e$, where $u\in U(R)$, $e\in Idem(R)$. Generalising the idea of clean element, N. Ashrafi and E. Nasibi [2, 3] introduced the concept of $r$-clean element as follows: an element $x$ of a ring $R$ is said to be $r$-clean if it can be written as a sum of an idempotent and a regular element. Further if all the elements of $R$ are $r$-clean then the ring $R$ is called $r$-clean ring. Motivated by these ideas we define an element of a ring $R$ as weakly $r$-clean if $x$ can be expressed as $x=r+e$ or $x=r-e$, where $r\in Reg(R)$, $e\in Idem(R)$ and call the ring $R$ to be weakly $r$-clean if all its elements are so. We show that in case of an abelian ring the concept of weakly $r$-clean rings coincides with that of weakly clean rings and hence with that of weakly exchange rings. We also show that the center of weakly $r$-clean rings are not so. However if $R$ has no non-trivial idempotents then the center of a weakly $r$-clean ring is weakly $r$-clean. Further we discuss some interesting properties of weakly $g(x)$-$r$- clean ring, where $g(x)\in C(R)[x]$. Finally we define the concept of weakly $\star$-clean ring and $\star$-$r$-clean ring. A ring $R$ is a $\star$-ring (or ring with involution) if there exists an operation $\star\,\,:\,\,R\rightarrow R$ such that for all $x,y\in R$, $(x+y)^{\star}=x^{\star}+y^{\star}$, $(xy)^{\star}=y^{\star}x^{\star}$ and $(x^{\star})^{\star}=x$. An element $p$ of a $\star$-ring is a projection if $p^{2}=p=p^{\star}$. Obviously, [math] and $1$ are projections of any $\star$-ring. Henceforth $P(R)$ will denote the set of all projections in a $\star$-ring. We define an element $x$ in a $\star$-ring $R$ is weakly $\star$-clean element if $x=u+p$ or $x=u-p$, where $u\in U(R)$ and $p\in P(R)$. If all the elements of $R$ are weakly $\star$-clean then the $\star$-ring $R$ is called weakly $\star$-clean. We show that for an abelian $\star$-clean ring $R$, if every idempotent of the form $e=ry$ or $e=yr$ is projection, where $r\in Reg(R)$, then $R$ is $\star$-$r$-clean if and only if $R$ is $\star$-clean.

2 Weakly $r$-clean ring

Definition 2.1**.**

An element $x$ in a ring $R$ is called weakly $r$-clean if it can be written as $x=r+e$ or $x=r-e$, for some $r\in Reg(R)$ and $e\in Idem(R)$. If all the elements of $R$ are weakly $r$-clean, then the ring $R$ is called a weakly $r$-clean ring.

For example all clean rings, weakly clean rings and $r$-clean rings are weakly $r$-clean ring. The following theorem is obvious.

Theorem 2.2**.**

Homomorphic image of weakly $r$-clean ring is weakly $r$-clean.

However the converse is not true as $\mathbb{Z}_{6}\cong\mathbb{Z}/\langle 6\rangle$ is a weakly $r$-clean ring, but $\mathbb{Z}$ is not weakly $r$-clean ring.

Theorem 2.3**.**

Let $\{R_{\alpha}\}$ be a collection of rings. Then the direct product $R=\underset{\alpha}{\prod}R_{\alpha}$ is weakly $r$-clean if and only if each $R_{\alpha}$ is weakly $r$-clean ring and at most one $R_{\alpha}$ is not $r$-clean.

Proof.

Let $R$ be weakly $r$-clean ring. Then being homomorphic image of $R$ each $R_{\alpha}$ is weakly $r$-clean. Suppose $R_{\alpha_{1}}$ and $R_{\alpha_{2}}$ are not $r$-clean, where ${\alpha}_{1}\neq{\alpha}_{2}$. Since $R_{\alpha_{1}}$ is not $r$-clean so not all elements $x\in R_{\alpha_{1}}$ of the form $x=r-e$ where $r\in Reg(R_{\alpha_{1}})$ and $e\in Idem(R_{\alpha_{1}})$. As $R_{\alpha_{1}}$ is weakly $r$-clean so there exists $x_{\alpha_{1}}\in R_{\alpha_{1}}$ with $x_{\alpha_{1}}=r_{\alpha_{1}}+e_{\alpha_{1}}$, where $r_{\alpha_{1}}\in Reg(R_{\alpha_{1}})$ and $e_{\alpha_{1}}\in Idem(R_{\alpha_{1}})$ but $x_{\alpha_{1}}\neq r-e$ for any $r\in Reg(R_{\alpha_{1}})$ and $e\in Idem(R_{\alpha_{1}})$. Similarly there exists $x_{\alpha_{2}}\in R_{\alpha_{2}}$ with $x_{\alpha_{2}}=r_{\alpha_{2}}-e_{\alpha_{2}}$, where $r_{\alpha_{2}}\in Reg(R_{\alpha_{2}})$ and $e_{\alpha_{2}}\in Idem(R_{\alpha_{2}})$, but $x_{\alpha_{2}}\neq r+e$ for any $r\in Reg(R_{\alpha_{2}})$ and $e\in Idem(R_{\alpha_{2}})$. Define $x=(x_{\alpha})\in R$ by

[TABLE]

Then clearly $x\neq r\pm e$ for any $r\in Reg(R)$ and $e\in Idem(R)$. Hence at most one $R_{\alpha}$ is not $r$-clean.

$(\Leftarrow)$ If each $R_{\alpha}$ is $r$-clean then $R=\prod R_{\alpha}$ is $r$-clean and hence weakly $r$-clean. Assume $R_{\alpha_{0}}$ is weakly $r$-clean but not $r$-clean and that all other $R_{\alpha}$’s are $r$-clean. If $x=(x_{\alpha})\in R$ then in $R_{\alpha_{0}}$ we can write $x_{\alpha_{0}}=r_{\alpha_{0}}+e_{\alpha_{0}}$ or $x_{\alpha_{0}}=r_{\alpha_{0}}-e_{\alpha_{0}}$, where $r_{\alpha_{0}}\in Reg(R_{\alpha_{0}})$ and $e_{\alpha_{0}}\in Idem(R_{\alpha_{0}})$. If $x_{\alpha_{0}}=r_{\alpha_{0}}+e_{\alpha_{0}}$ then for $\alpha\neq\alpha_{0}$ let, $x_{\alpha}=r_{\alpha}+e_{\alpha}$ and if $x_{\alpha_{0}}=r_{\alpha_{0}}-e_{\alpha_{0}}$ then for $\alpha\neq\alpha_{0}$ let, $x_{\alpha}=r_{\alpha}-e_{\alpha}$ then $r=(r_{\alpha})\in Reg(R)$ and $e=(e_{\alpha})\in Idem(R)$ such that $x=r+e$ or $x=r-e$ and consequently $R$ is weakly $r$-clean ring. ∎

Lemma 2.4**.**

Let $R$ be a ring with no zero divisor. Then $R$ is weakly clean if and only if $R$ is weakly $r$-clean.

Proof.

Let $r(\neq 0)\in Reg(R)$, then $r=ryr$ for some $y\in R$, i.e., $r(1-yr)=0$, which implies that $r$ is a unit. Now the result follows immediately. ∎

Lemma 2.5**.**

Let $R$ be a ring and $e^{2}=e\in R$, $a=u-f\in eRe$, where $u\in U(eRe)$, $f\in Idem(eRe)$. Then there exist elements $v\in U(R)$ and $\overline{e}\in Idem(R)$ such that $a=v-\overline{e}$.

Proof.

Since $u\in U(eRe)$, so there exists $w\in U(eRe)$ such that $uw=e=wu$. Assume $v=u+(1-e)$, then $v(w+(1-e))=1=(w+(1-e))v$, because $v(1-e)=0=(1-e)w$. Clearly $a-v=a-u-(1-e)=-(f+(1-e))$. It is easy to see that $\overline{e}=f+(1-e)$ is idempotent and hence the result follows. ∎

For the sake of completeness of the article we prove a result on weakly clean element.

Proposition 2.6**.**

Let $R$ be an abelian ring and $a\in R$ is weakly clean element then $ae$ is weakly clean for any idempotent $e\in R$.

Proof.

Let $a\in R$ be a weakly clean element. So $a=u+e^{\prime}$ or $a=u-e^{\prime}$, where $u\in U(R)$ and $e^{\prime}\in Idem(R)$

Case $1$: If $a=u+e^{\prime}$, then $ae$ is clean for any idempotent $e$ by Lemma 2.1 [2].

Case $2$: If $a=u-e^{\prime}$, then $ae=ue-e^{\prime}e$. But $ue=ue^{2}=eue\in U(eRe)$ and $e^{\prime}e\in Idem(eRe)$, as $R$ is abelian. So by Lemma 2.5, $ae=v-\overline{e}$, for some $v\in U(R)$ and $\overline{e}\in Idem(R)$. Hence $ae$ is weakly clean. ∎

Lemma 2.7**.**

Let $R$ be an abelian ring such that $a$ and $-a$ both are clean. Then the followings hold:

1. (i)

$a+e$* is clean for any idempotent $e$ of $R$.* 2. (ii)

$a-e$* is weakly clean for any idempotent $e$ of $R$.*

Proof.

$(i)$ It follows from Lemma 2.1 [2] .

$(ii)$ Consider $a=u+e^{\prime}$ and $-a=u^{\prime}+e^{\prime\prime}$ where, $u,u^{\prime}\in U(R)$ and $e,e^{\prime}\in Idem(R)$. Now

[TABLE]

Since $\{(1-e^{\prime})e+e^{\prime\prime}(1-e)\}\in Idem(R)$ and $\{ue-u^{\prime}(1-e)\}\in U(R)$, so $a-e$ is weakly clean. ∎

Here we define a weakly $r$-clean (respectively weakly clean) element $x\in R$ is of type $1$ form if $x=r+e$, where $r\in Reg(R)$ (respectively $r\in U(R)$) and $e\in Idem(R)$ and of type $2$ form if $x=r-e$, where $r\in Reg(R)$ (respectively $r\in U(R)$) and $e\in Idem(R)$.

Theorem 2.8**.**

Let $R$ be an abelian ring. Then $R$ is weakly clean if and only if $R$ is weakly $r$-clean.

Proof.

$(\Rightarrow)$ Obviously weakly clean rings are weakly $r$-clean.

$(\Leftarrow)$ Let $R$ be weakly $r$-clean ring and $x\in R$. So $x=r\pm e^{\prime}$, where $r=ryr$ for some $y\in R$ and $e^{\prime}\in Idem(R)$.

First let $x=r+e^{\prime}$. Assume $yr=e$, then clearly $e$ is an idempotent and therefore

$(ye+(1-e))(re+(1-e))=1$. Let $u=re+(1-e)$, then $u\in U(R)$ and $ue=r$, so $ue+f\in U(R)$, where $(1-e)=f$. Thus $-r=-(ue+f)+f$ is clean. Hence by lemma 2.7, $r+e^{\prime}$ is clean.

If $x=r-e^{\prime}$, then clearly $r$ and $-r$ both are clean and so by lemma 2.7, $r-e^{\prime}$ is weakly clean. Hence $R$ is weakly clean. ∎

Consider the ring $\mathbb{Z}_{(6)}$ of all rational numbers $\frac{a}{b}$ of lowest form, where $b$ is relatively prime to $6$. Since $\frac{21}{11}+1=\frac{32}{11}$ and $\frac{21}{11}-1=\frac{10}{11}$ both are not units and 0, 1 are the only idempotents of $\mathbb{Z}_{(6)}$ so $\mathbb{Z}_{(6)}$ is not weakly clean and hence not weakly $r$-clean. The following is an example of weakly $r$-clean but not $r$-clean ring.

Example 2.9**.**

$\mathbb{Z}_{(3)}\cap\mathbb{Z}_{(5)}$ is weakly $r$-clean but not $r$-clean.

Since $\mathbb{Z}_{(3)}\cap\mathbb{Z}_{(5)}$ is weakly clean but not clean [1] and if $R$ is an abelian ring then $R$ is clean if and only if $R$ is $r$-clean by Theorem 2.2 [2]. So by Theorem 2.8, $\mathbb{Z}_{(3)}\cap\mathbb{Z}_{(5)}$ is weakly $r$-clean but not $r$-clean.

Definition 2.10**.**

A ring $R$ is called weakly exchange ring if for any $x\in R$, there exists an idempotent $e\in xR$ such that $1-e\in(1-x)R$ or $1-e\in(1+x)R$.

The relation between weakly $r$-clean ring and weakly exchange ring is given below.

Theorem 2.11**.**

If $R$ is abelian ring then $R$ is weakly $r$-clean ring if and only if $R$ is weakly exchange ring.

Proof.

$(\Rightarrow)$ It is equivalent to show $R$ is weakly clean ring if and only if $R$ is weakly exchange ring.

Let $x\in R$, if $x=u+e$ where $u\in U(R)$ and $e\in Id(R)$ then clearly it satisfies the exchange property [8].

Suppose, $x=u-e$ where $u\in U(R)$ and $e\in Idem(R)$. Let $f=u^{-1}(1-e)u$ then $f^{2}=f$. Now

[TABLE]

$\therefore\,\,\,x+f\in R(x^{2}+x)$. So $x$ satisfies weakly exchange property by Lemma 2.1 [5]. Hence $R$ is weakly exchange ring. $(\Leftarrow)$ Suppose $x\in R$ then there exists an idempotent $e\in Rx$ such that $1-e\in R(1-x)$ or $1-e\in R(1+x)$.

Case $1$: Suppose $1-e\in R(1+x)$ and $e=ax$, $1-e=b(1+x)$, for some $a,b\in R$. Assume that $ea=a$ and $(1-e)b=b$ so that $axa=ea=a$ and $b(1+x)b=b$. Here $ax,\,xa,\,b(1+x),\,(1+x)b$ all are central idempotent and $ax=(ax)(ax)=(ax)(xa)=x(ax)a=xa$ similarly $(1+x)b=b(1+x)$. Now $(a+b)(x+(1-e))=ax+bx+a(1-e)+b(1-e)=1$ so $x+(1-e)$ is unit. Hence $x$ is weakly clean element.

Case $2$: If $1-e\in R(1-x)$ then clearly $x$ is weakly clean by Proposition 1.8 [8]. ∎

A polynomial ring over a commutative ring $R$ can never be weakly $r$-clean as $x$ can not be represented in weakly clean decomposition. N. Ashrafi and E. Nasibi [2], showed that $R$ is $r$-clean if and only if $R[[x;\alpha]]$ is $r$-clean, where $R[[x;\alpha]]$ is the ring of skew formal power series over $R$ i.e. all formal power series in $x$ with coefficients from $R$ and $\alpha$ is an ring endomorphism. Multiplication is defined by $xr=\alpha(r)x$, for all $r\in R$. In particular $R[[x]]=R[[x;I_{R}]]$ is the ring of formal power series over $R$.

Proposition 2.12**.**

Let $R$ be an abelian ring and $\alpha$ be an endomorphism of $R$. Then the following are equivalent.

1. (i)

$R$* is weakly $r$-clean ring.* 2. (ii)

The formal power series ring $R[[x]]$ over $R$ is weakly $r$-clean. 3. (iii)

The skew power series ring $R[[x;\alpha]]$ over $R$ is weakly $r$-clean.

Proof.

Follows from Theorem 2.8, and similar results of weakly clean rings. ∎

Consider a ring $R[D,C]$ constructed in [10], for any subring $C$ of a ring $D$. The following result gives the relation between $R[D,C]$ with $D$ and $C$, about weakly $r$-clean ness.

Proposition 2.13**.**

Let $C$ be a subring of a ring $D$ then $R[D,C]$ is weakly $r$-clean ring if and only if $D$ is $r$-clean and $C$ is weakly $r$-clean ring.

Proof.

$(\Rightarrow)$ As $R[D,C]=D\otimes D$ so by Theorem $2.3$ $D$ is $r$-clean. Clearly $C$ is weakly $r$-clean ring by Theorem $2.2$.

$(\Leftarrow)$ Let $D$ be $r$-clean and $C$ is weakly $r$-clean ring. Let $x=(a_{1},a_{2},\cdot\cdot\cdot,a_{n},c,c,c,\cdot\cdot\cdot)\in R[D,C]$. Since $C$ is weakly $r$-clean ring, so $c=r\pm e$, where $r\in Reg(R)$ and $e\in Idem(R)$. If $c=r+e$, write $a_{i}=r_{i}+e_{i}$ then $x=\overline{r}+\overline{e}$, where $\overline{r}=(r_{1},r_{2},\cdot\cdot\cdot,r_{n},r,r,\cdot\cdot\cdot)\in Reg(R[D,C])$ and $\overline{e}=(e_{1},e_{2},\cdot\cdot\cdot,e_{n},e,e,\cdot\cdot\cdot)\in Idem(R[D,C])$. If $c=r-e$, write $a_{i}=r_{i}-e_{i}$ then $x=\overline{r}-\overline{e}$, where $\overline{r}=(r_{1},r_{2},\cdot\cdot\cdot,r_{n},r,r,\cdot\cdot\cdot)\in Reg(R[D,C])$ and $\overline{e}=(e_{1},e_{2},\cdot\cdot\cdot,e_{n},e,e,\cdot\cdot\cdot)\in Idem(R[D,C])$. ∎

Clearly every abelian semi regular ring is weakly clean and hence weakly $r$-clean. However the converse is not true as shown by the following example.

Example 2.14**.**

Let $\mathbb{Q}$ be a field of rational numbers and $L$ the ring of all rational numbers with odd denominators. Then by example 2.7 [2] and Theorem $2.8$, $R[\mathbb{Q},L]$ is commutative exchange ring and hence $R[\mathbb{Q},L]$ is weakly $r$-clean ring but not semi regular.

If $R$ is weakly $r$-clean ring then $R/I$ is weakly $r$-clean being homomorphic image of weakly $r$-clean ring. The theorem is a partial converse of this statement.

Theorem 2.15**.**

Let $I$ be a regular ideal of a ring $R$, where idempotents can be lifted modulo $I$. Then $R$ is weakly $r$-clean if and only if $R/I$ is weakly $r$-clean.

Proof.

Following theorem 2.8 [2], it is enough to show that for any $a\in R$ if $a+I$ has type $2$ weakly $r$-clean decomposition in $R/I$ then $a$ is weakly $r$-clean in $R$. Let $a+I\in R/I$ has type $2$ weakly $r$-clean decompositions in $R/I$ then there exists idempotent $e+I\in R/I$ such that $(a+I)+(e+I)\in Reg(R/I)$ i.e. $(a+e)+I\in Reg(R/I)$. Therefore $((a+e)+I)(x+I)((a+e)+I)=(a+e)+I$, for some $x\in R$, so $(a+e)x(a+e)-(a+e)\in I$ but $I$ is regular ideal, Hence $(a+e)$ is regular by Lemma 1, [4] . Since idempotents can be lifted modulo $I$, we may assume that $e$ is an idempotent of $R$. ∎

Proposition 2.16**.**

Let $M$ be a $A-B$ bi-module. If T=\left(\begin{array}[]{cc}A&0\\ M&B\\ \end{array}\right), a formal triangular matrix ring is weakly $r$-clean then $A$ and $B$ are weakly $r$-clean ring.

Proof.

Let $a\in A$ and $b\in B$ and $m\in M$, consider t=\left(\begin{array}[]{cc}a&0\\ m&b\\ \end{array}\right)\in T.

Case $I$: If $t$ is $r$-clean then clearly $a$ and $b$ both are $r$-clean in their respective rings $A$ and $B$ by Theorem 16, [3] .

Case $II$: If t=-\left(\begin{array}[]{cc}f_{1}&0\\ f_{2}&f_{3}\\ \end{array}\right)+\left(\begin{array}[]{cc}r_{1}&0\\ r_{2}&r_{3}\\ \end{array}\right) where \left(\begin{array}[]{cc}f_{1}&0\\ f_{2}&f_{3}\\ \end{array}\right)^{2}=\left(\begin{array}[]{cc}f_{1}&0\\ f_{2}&f_{3}\\ \end{array}\right) and \left(\begin{array}[]{cc}r_{1}&0\\ r_{2}&r_{3}\\ \end{array}\right)\in Reg(T). It is simple calculation to show that $f_{1}$ and $f_{3}$ are idempotents in $A$ and $B$ respectively and $r_{1}$ and $r_{3}$ are regular elements in $A$ and $B$ respectively. Finally $a=-f_{1}+r_{1}$ and $b=-f_{3}+r_{3}$. Hence $A$ and $B$ both are weakly $r$-clean rings. ∎

In [10], T. Kosan, S. Sahinkaya and Y. Zhou showed that the center of weakly clean rings need not be weakly clean. In that example $End(_{R}E(R))$ is weakly clean whereas the center is not weakly clean as $R$ is not weakly clean. But since $R$ is abelian in that example so by Theorem 2.8, the center of $End(_{R}E(R))$ is not weakly $r$-clean. Hence the center of weakly $r$-clean ring need not be weakly $r$-clean.

Theorem 2.17**.**

Let $R$ be a weakly $r$-clean ring with no nontrivial idempotents. Then the center of $R$ is also a weakly $r$-clean ring.

Proof.

Let $R$ be weakly $r$-clean ring and $Z(R)$ be the center of $R$. Let $x\in Z(R)$ then there exists regular element $r\in R$ such that either $x=r$ or $x=r-1$ or $x=r+1$. If $x=r$ then clearly $x$ is weakly $r$-clean in $Z(R)$. If $x=r\pm 1$ then $r\in Z(R)$ as $x\pm 1\in Z(R)$. Hence $Z(R)$ is weakly $r$-clean ring. ∎

A ring $R$ is called graded ring(or more precisely, $\mathbb{Z}$-graded) if there exists a family of subgroups $\{R_{n}\}_{n\in Z}$ of $R$ such that $R=\underset{n}{\oplus}R_{n}$ (as abelian groups) and $R_{n}R_{m}\subseteq R_{n+m}$, for all $n,m\in\mathbb{Z}$. The relation of graded ring and weakly $r$-clean ring is given below.

Proposition 2.18**.**

Let $R=R_{0}\oplus R_{1}\oplus R_{2}\oplus\cdot\cdot\cdot$ be a graded ring then the following hold:

1. (i)

If $R$ is weakly $r$-clean ring then $R_{0}$ is weakly $r$-clean ring. 2. (ii)

If $R_{0}$ is weakly $r$-clean ring and each $R_{n}$ is a torsion-less $R_{0}$ module, i.e. $r\notin Z(R_{0})\Rightarrow r\notin Z(R_{n})$, then $R$ is weakly $r$-clean ring.

Proof.

1. (i)

Let $r_{0}\in R_{0}$, so $r_{0}=r+e$ or $r_{0}=r-e$, where $r\in Reg(R)$ and $e\in Idem(R)$. Since $Idem(R)=Idem(R_{0})$, so $e\in R_{0}$ and $r\in R_{0}\cap Reg(R)\subseteq Reg(R_{0})$. Hence $R_{0}$ is weakly $r$-clean ring. 2. (ii)

Let $x=x_{0}+x_{1}+x_{2}+\cdot\cdot\cdot+x_{n}\in R$, where $x_{i}\in R_{i}$. Write $x_{0}=r_{0}\pm e_{0}$, where $r_{0}\in Reg(R_{0})$ and $e_{0}\in Idem(R_{0})=Idem(R)$. Put $x^{\prime}=r_{0}+x_{1}+x_{2}+\cdot\cdot\cdot+x_{n}$, so $x=x^{\prime}\pm e_{0}$ where $e_{0}\in Idem(R_{)}$. If $x^{\prime}\in Z(R)$ then there exists a non zero homogeneous $t\in R_{n}$ with $tx^{\prime}=0$. But then $r_{0}t=0$, a contradiction.

∎

3 Weakly $g(x)$-$r$-clean rings

Definition 3.1**.**

Let $g(x)$ be a fixed polynomial in $C(R)[x]$. An element $x\in R$ is called weakly $g(x)$-$r$-clean if $x=r\pm e$, where $r\in Reg(R)$ and $g(e)=0$. We say that $R$ is weakly $g(x)$-$r$-clean if every element is weakly $g(x)$-$r$-clean.

If $g(x)=x^{2}-x$, then weakly $g(x)$-$r$-clean ring is similar to weakly $r$-clean ring. An element $s$ of $R$ is called root of the polynomial $g(x)\in C(R)[x]$ if $g(s)=0$. For $g(x)=x^{2}-x$, the ring $\mathbb{Z}_{(3)}\cap\mathbb{Z}_{(5)}$ is weakly $g(x)$-$r$-clean ring but not $g(x)$-$r$-clean ring.

Let $R$ and $S$ be rings and $\theta:C(R)\rightarrow C(S)$ be a ring homomorphism with $\theta(1)=1$. Then $\theta$ induces a map $\theta^{\prime}$ from $C(R)[x]$ to $C(S)[x]$ such that for $g(x)=\sum_{i=0}^{n}a_{i}x^{i}\in C(R)[x]$, $\theta^{\prime}(g(x)):=\sum_{i=0}^{n}\theta(a_{i})x^{i}\in C(S)[x]$.

Proposition 3.2**.**

Let $\theta:R\rightarrow S$ be a ring epimorphism. If $R$ is weakly $g(x)$-$r$-clean ring then $S$ is weakly $\theta^{\prime}(g(x))$-$r$-clean ring.

Proof.

Let $g(x)=a_{0}+a_{1}x+\cdot\cdot\cdot+a_{n}x^{n}\in C(R)[x]$, then $\theta^{\prime}(g(x))=\theta(a_{0})+\theta(a_{1})x+\cdot\cdot\cdot+\theta(a_{n})x^{n}\in C(S)[x]$. Since $\theta$ is ring epimorphism so for any $s\in S$, there exists $x\in R$ such that $\theta(x)=s$. Let $x=r\pm s_{0}$ where $r\in Reg(R)$ and $g(s_{0})=0$, as $R$ is weakly $g(x)$-$r$-clean ring. Now $s=\theta(x)=\theta(r\pm s_{0})=\theta(r)\pm\theta(s_{0})$. Clearly $\theta(r)\in Reg(S)$ and

[TABLE]

Hence $S$ is $\theta^{\prime}(g(x))$-$r$-clean ring. ∎

Next we extend the Theorem 2.3 to $g(x)$-$r$-clean ring.

Theorem 3.3**.**

Let $g(x)\in\mathbb{Z}[x]$ and $\{R_{i}\}_{i\in I}$ be a family of rings. Then $\underset{i\in I}{\prod}R_{i}$ is $g(x)$-$r$-clean ring if and only if $R_{i}$’s are weakly $g(x)$-$r$-clean ring and at most one $R_{\alpha}$ is not $g(x)$-$r$-clean ring.

Proof.

$(\Rightarrow)$ Similar to Theorem 2.3.

$(\Leftarrow)$ If each $R_{\alpha}$ is $g(x)$-$r$-clean ring, then $R=\prod R_{\alpha}$ is $g(x)$-$r$-clean. Assume $R_{\alpha_{0}}$ is weakly $g(x)$-$r$-clean but not $g(x)$-$r$-clean and that all other $R_{\alpha}$’s are $g(x)$-$r$-clean. Let $x=x_{\alpha}\in R$ so in $R_{\alpha_{0}}$ we can write $x_{\alpha_{0}}=r_{\alpha_{0}}+s_{\alpha_{0}}$ or $x_{\alpha_{0}}=r_{\alpha_{0}}-s_{\alpha_{0}}$, where $r_{\alpha_{0}}\in Reg(R_{\alpha_{0}})$ and $g(s_{\alpha_{0}})=0$ in $R_{\alpha_{0}}$ . If $x_{\alpha_{0}}=r_{\alpha_{0}}+s_{\alpha_{0}}$ then for $\alpha\neq\alpha_{0}$ assume, $x_{\alpha}=r_{\alpha}+s_{\alpha}$ and if $x_{\alpha_{0}}=r_{\alpha_{0}}-s_{\alpha_{0}}$ then for $\alpha\neq\alpha_{0}$ assume, $x_{\alpha}=r_{\alpha}-s_{\alpha}$, where $r_{\alpha}\in R_{\alpha}$ and $g(s_{\alpha})=0$ in $R_{\alpha}$. Then $r=(r_{\alpha})\in Reg(R)$ and $g(s=\{s_{\alpha}\})=a_{0}\{1_{R_{i}}\}+a_{1}\{s_{i}\}+\cdot\cdot\cdot+a_{n}\{s_{i}^{n}\}=\{a_{0}+a_{1}s_{i}+\cdot\cdot\cdot+a_{n}s_{i}^{n}\}$=0. ∎

Theorem 3.4**.**

Let $R$ be a ring and $a,b\in R$ then $R$ is weakly $(ax^{2n}-bx)$-$r$-clean ring if and only if $R$ is weakly $(ax^{2n}+bx)$-$r$-clean ring, for $n\in\mathbb{N}$.

Proof.

Suppose $R$ is weakly $(ax^{2n}-bx)$-$r$-clean ring. Since $(as^{2n}-bs)=0\Rightarrow(a(-s)^{2n}+b(-s))=0$. So for $a\in R$, $-a=r\pm s$, where $(as^{2n}-bs)=0$ and $r\in R$. Hence $a=(-r)\pm(-s)$ and $(a(-s)^{2n}+b(-s))=0$. Similarly the converse is also true. ∎

Proposition 3.5**.**

Let $2\leq n\in\mathbb{N}$. If for every $a\in R$, $a=r\pm t$, where $r\in Reg(R)$ and $t^{n-1}=1$ then $R$ is weakly $(x^{n}-x)$-$r$-clean ring.

Proof.

The proof follows from Theorem 3.4. ∎

Proposition 3.6**.**

Let $g(x)=(x^{n}-x)\in C(R)[x]$ and $g(e)=0$. Let $a\in e^{n-1}Re^{n-1}$ be strongly $g(x)$-clean in $e^{n-1}Re^{n-1}$ then $a$ is strongly $g(x)$-clean in $R$.

Proof.

Let $a=u+f$, where $u\in U(e^{n-1}Re^{n-1})$ and $g(f)=0$ in $e^{n-1}Re^{n-1}$. Then $uw=e^{n-1}=wu$, for some $w\in U(e^{n-1}Re^{n-1})$. Consider $v=u-(1-e^{n-1})$, then $\{u-(1-e^{n-1})\}\{w-(1-e^{n-1})\}=1$, so $v=u-(1-e^{n-1})$ is a unit in $R$. Now $a-v=f+(1-e^{n-1})$ and $\{f+(1-e^{n-1})\}^{n}=f^{n}+(1-e^{n-1})^{n}=f+(1-e^{n-1})$, so $a$ is $g(x)$-clean ring in $R$. ∎

Proposition 3.7**.**

Let $R$ be an abelian ring and $g(x)\in C(R)[x]$ then if $a$ is $g(x)$-clean element in $R$ then $ae^{n-1}$ is also $g(x)$-clean element in $R$ for any root $e$ of g(x).

Proof.

Let $a=u+f$, where $u\in U(R)$ and $g(f)=0$ in $R$. Suppose $e$ be any root of $g(x)$ in $R$. Then $ae^{n-1}=ue^{n-1}+fe^{n-1}$ but $ue^{n-1}\in U(e^{n-1}Re^{n-1})$ and $(fe^{n-1})^{n}=f^{n}e^{n-1}=fe^{n-1}$. So $ae^{n-1}$ is $g(x)$-clean in $R$. ∎

4 Weakly $\star$-clean rings and $\star$-$r$-clean ring

A ring $R$ is a $\star$-ring (or ring with involution) if there exists an operation $\star\,\,:\,\,R\rightarrow R$ such that for all $x,y\in R$, $(x+y)^{\star}=x^{\star}+y^{\star}$, $(xy)^{\star}=y^{\star}x^{\star}$ and $(x^{\star})^{\star}=x$. An element $p$ of a $\star$-ring is a projection if $p^{2}=p=p^{\star}$. Obviously, [math] and $1$ are projections of any $\star$-ring. Henceforth $P(R)$ will denote the set of all projections in a $\star$-ring.

Here we define the concept of weakly $\star$-clean ring and discuss some properties of weakly $\star$-clean ring.

Definition 4.1**.**

An element $x$ of a $\star$-ring $R$ is said to be weakly $\star$-clean if $x=u+p$ or $x=u-p$, where $u\in U(R)$ and $p\in P(R)$. A $\star$-ring $R$ is said to be weakly $\star$-clean ring if every elements are weakly $\star$-clean.

Example 4.2**.**

1. (i)

Units, elements in $J(R)$ and nilpotents of a $\star$-ring $R$ are weakly $\star$-clean. 2. (ii)

Idempotents of a $\star$-regular rings are weakly $\star$-clean.

Lemma 4.3**.**

Let $R$ be a boolean $\star$-ring. Then $R$ is weakly $\star$-clean if and only if $\star=1_{R}$ is the identity map of $R$.

Proof.

It is clear that boolean rings are clean. Suppose that $R$ is weakly $\star$-clean. Given any $a\in R$, we have $-a=u\pm p=\pm p+1=\pm p-1$, for some $p\in P(R)$. So we have $a=1\pm p\in P(R)$. Hence $a^{\star}=a$, which implies $\star=1_{R}$. Conversely if $\star=1_{R}$ then every idempotent of $R$ is a projection. Thus, $R$ is a weakly $\star$-clean. ∎

Example 4.4**.**

$R=\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ with involution $\star$ defined by $(a,b)^{\star}=(b,a)$ is weakly clean but not weakly $\star$-clean.

Lemma 4.5**.**

Let $R$ be a $\star$-ring. If $2\in U(R)$, then for any $u^{2}=1,\,u^{\star}=u\in R$ if and only if every idempotent of $R$ is a projection.

Proof.

Proof is given in Lemma 2.3 [11]. ∎

Corollary 4.6**.**

Let $R$ be a $\star$-ring with $2\in U(R)$. The following are equivalent:

1. (i)

$R$* is weakly clean and every unit of $R$ is self-adjoint.(i.e., $u^{\star}=u$ for every unit $u$ ).* 2. (ii)

$R$* is weakly $\star$-clean and $\star=1_{R}$.*

Proof.

$(1)\Rightarrow(2)$. Let $a\in R$. Then $a=u\pm p$, for some $p\in P(R)$ and $u\in U(R)$. Note that $(1-2p)\in U(R)$, so $(1-2p)^{\star}=(1-2p)\Rightarrow p^{\star}=p$ by Lemma 4.5. Also $a^{\star}=(u\pm p)^{\star}=u^{\star}\pm p^{\star}=u\pm p=a$.

$(2)\Rightarrow(1).$ is trivial. ∎

For a $\star$-ring $R$, an element $x\in R$ is called self adjoint square root of 1 if $x^{2}=1$ and $x^{\star}=x$.

Theorem 4.7**.**

*Let $R$ be a $\star$-ring, the following are equivalent:

(1) $R$ is weakly $\star$-clean and $2\in U(R)$.

(2) Every element of $R$ is a sum of unit and a self adjoint square root of 1 or an element of the form $2p+1$, where $p^{2}=p=p^{\star}$.*

Proof.

$(1)\Rightarrow(2)$. Consider $a\in R$, then $\frac{a-1}{2}=u\pm p$, where $u\in U(R)$ and $p\in P(R)$. If $\frac{a-1}{2}=u-p$, then $a=2u+(1-2p)$, where $2u\in U(R)$ and $1-2p$ is a self adjoint square root of 1. If $\frac{a-1}{2}=u+p$, then $a=2u+(1+2p)$, where $p^{2}=p=p^{\star}$.

$(2)\Rightarrow(1)$ First we show that $2\in(R)$. By assumption, $1=u+x$ or $1=u+(2p+1)$, where $u\in U(R)$, $x$ is self adjoint square root of $1$ and $p\in P(R)$. If $1=u+x$, then clearly $2\in U(R)$ by Theorem 2.5 [11]. If $1=u+(2p+1)$, then $u=-2p$, so $2\in U(R)$. Next for showing $R$ is weakly $\star$-clean ring, let $a\in R$, so $2a+1=u+x$ or $2a+1=u+(2p+1)$, where $u\in U(R)$, $x$ is self adjoint square root of $1$ and $p\in P(R)$.

Case I: If $2a+1=u+x$, then $a=\frac{u}{2}-\frac{1-x}{2}$. Since $(\frac{1-x}{2})^{2}=\frac{1-x}{2}=(\frac{1-x}{2})^{\star}$ and $2\in(R)$, so $a$ is weakly $\star$-clean element.

Case II: If $2a+1=u+(2p+1)$, then $a=\frac{u}{2}+p$, a $\star$-clean element. ∎

Lemma 4.8**.**

Let $R$ be weakly $\star$-clean ring. If $I$ is a $\star$ invariant ideal of $R$, then $R/I$ is weakly $\star$-clean. In particular, $R/J(R)$ is weakly $\star$-clean ring.

Proof.

The result follows from Lemma 2.7 [11]. ∎

Let $R$ be a $\star$-ring. Then $\star$ induces an involution in the power series ring $R[[x]]$, defined by $(\sum_{i=0}^{\infty}a_{i}x^{i})^{\star}=\sum_{i=0}^{\infty}a_{i}^{\star}x^{i}$.

Proposition 4.9**.**

Let $R$ be a $\star$-ring. Then $R[[x]]$ is weakly $\star$-clean if and only if $R$ is weakly $\star$-clean.

Proof.

Let $R[[x]]$ be weakly $\star$-clean. Since $R\cong R[[x]]/<x>$ and $<x>$ is $\star$-invariant ideal of $R[[x]]$. So $R$ is weakly $\star$-clean ring. Conversely, suppose that $R$ is weakly $\star$-clean ring. Let $g(x)=\sum_{i=0}^{\infty}a_{i}x^{i}\in R[[x]]$. Write $a_{0}=u\pm p$ with $p\in P(R)$ and $u\in U(R)$. Then $g(x)=\pm p+(u+\sum_{i=1}^{\infty}a_{i}x^{i})$, where $p\in P(R)\subseteq P(R[[x]])$ and $u+\sum_{i=1}^{\infty}a_{i}x^{i}\in U(R[[x]])$. Hence $g(x)$ is weakly $\star$-clean in $R[[x]]$. ∎

Theorem 4.10**.**

Homomorphic image of weakly $\star$-clean ring is weakly $\star$-clean.

Similarly we extend the Theorem 2.3 to $\star$-clean ring given below.

Theorem 4.11**.**

Let $\{R_{\alpha}\}$ be a collection of $\star$-rings. Then the direct product $R=\underset{\alpha}{\prod}R_{\alpha}$ is weakly $\star$-clean if and only if each $R_{\alpha}$ is weakly $\star$-clean ring and at most one $R_{\alpha}$ is not $\star$-clean.

Proof.

Similar to the proof of Theorem 2.3. ∎

Definition 4.12**.**

An element $x$ in a $\star$-ring $R$ is said to be $\star$-$r$-clean if $x=r+p$ where $r\in Reg(R)$ and $p\in P(R)$. A $\star$-ring $R$ is said to be $\star$-$r$-clean ring if every element of $R$ is $\star$-$r$-clean.

Proposition 4.13**.**

Let $R$ be a $\star$-ring and $e\in P(R)$. If $a\in eRe$ is strongly $\star$-clean in $eRe$ then $a$ is strongly $\star$-clean in $R$.

Proof.

Let $a=f+v$, where $f\in P(eRe)$ and $v\in U(eRe)$, so there exists $w\in U(eRe)$ such that $vw=e=wv$. Clearly $(v-(1-e))(w-(1-e))=1$ implies $v-(1-e)\in U(R)$ and $a-(v-(1-e))=f+(1-e)\in P(R)$. Hence $a$ is strongly $\star$ clean in $R$. ∎

Proposition 4.14**.**

Let $R$ be an abelian $\star$-ring. If $a$ is $\star$-clean element in $R$ then $ae$ is $\star$-clean for any $e\in P(R)$.

Proof.

Let $a=u+p_{1}$, where $u\in U(R)$ and $p_{1}\in P(R)$. Now $ae=ue+p_{1}e$. Clearly $ue\in U(eRe)$ and $p_{1}e\in P(eRe)$ imply $ae$ is strongly $\star$-clean in $eRe$, so $ae$ is strongly $\star$-clean in $R$. ∎

Proposition 4.15**.**

Let $R$ be an abelian $\star$-ring and $a$ be a $\star$-clean element in $R$ and $e\in P(R)$. If $-a$ is $\star$-clean then $a+e$ is also $\star$-clean.

Proof.

Clearly $a$ and $a+1$ are $\star$-clean, as $a$ is $\star$-clean implies $1-a$ is so. Let $a=u+f$ and $1+a=v+g$, where $f,g\in P(R)$ and $u,v\in U(R)$. $a+e=(1+a)e+a(1-e)=(ve+u(1-e))+(ge+f(1-e))$. But $(ve+u(1-e))\in U(R)$ and $(ge+f(1-e))\in P(R)$. So $a+e$ is $\star$-clean. ∎

Lemma 4.16**.**

Let $R$ be an abelian $\star$-ring where every idempotent of the form $ry$ or $yr$ is projection, for any regular element $r$, then $r$ is $\star$-clean.

Proof.

Let $r\in Reg(R)$ then $r=ryr$, for some $y\in R$. So $f=yr\in P(R)$. Let $e=f+(1-f)rf=(y+(1-f)ry)r=ar$, where $a=y+(1-f)ry$. Here $e^{2}=e\in P(R)$ and $(1-e)=(1-f)(1-r)=b(1-r)\in R(1-r)$, where $b=(1-f)$. Assume $ea=a$ so that $ara=a$. Since idempotents are central so $ra=r(ar)a=ra(ra)=(ra)ra=a(ra)r=ar$. Similarly $(1-r)b=b(1-r)$, where it is assumed that $(1-e)b=b$. By Proposition 1.8 [8], $a-b$ is the inverse of $r-(1-e)$ and so $r$ is $\star$-clean. ∎

Theorem 4.17**.**

Let $R$ be an abelian $\star$-clean, where any idempotent of the form $e=ry$ or $yr$ is projection, for any $r\in Reg(R)$. Then $R$ is $\star$-$r$-clean if and only if $R$ is $\star$-clean.

Proof.

$(\Leftarrow)$ Obviously $R$ is $\star$-clean $\Rightarrow$ $R$ is $\star$-$r$-clean.

$(\Rightarrow)$ Let $R$ be a $\star$-$r$-clean ring and $x\in R$, then $x=r+e^{\prime}$, where $e^{\prime}\in P(R)$ and $r\in Reg(R)$. Therefore $r=ryr$, for some $y\in R$. Taking $e=ry$ we see that $(re+(1-e))(ye+(1-e))=1$ and $e\in P(R)$. Hence $u=re+(1-e)$ is a unit and $r=eu$. Set $f=1-e$ then $-(eu+f)$ is a unit and $f\in P(R)$. So, $-r=-(ue+f)+f$ is $\star$-clean. Also by Lemma 4.16, $r$ is $\star$-clean. Hence by Proposition 4.15, $x=r+e^{\prime}$ is $\star$-clean. ∎