On Zhou nil-clean rings

TL;DR

This paper characterizes Zhou nil-clean rings, a class where elements are sums of tripotents and nilpotents, by exploring their relations with polynomials, idempotents, and 2-idempotents, and establishing conditions for their structure.

Contribution

It provides new characterizations of Zhou nil-clean rings through polynomial relations and idempotent decompositions, linking them to Kosan exchange rings.

Findings

Existence of polynomial-based idempotent decompositions in Zhou nil-clean rings.

Characterization of these rings via sums of 2-idempotents and nilpotents.

Equivalence of certain polynomial conditions with the ring being a Kosan exchange ring.

Abstract

A ring R is a Zhou nil-clean ring if every element in R is the sum of two tripotents and a nilpotent that commute. In this paper, Zhou nil-clean rings are further discussed with an emphasis on their relations with polynomials, idempotents and 2- idempotents. any a \in R, there exists e 2 Z[a] such that a-e \in R is nilpotent and e^5 = 5e^3 -4e, if and only if for any a 2 R, there exist idempotents e; f; g; h 2 Z[a] and a nilpotent w such that a = e+f +g+h+w, if and only if for any a 2 R, there exist 2-idempotents e; f 2 Z[a] and a nilpotent w 2 R such that a = e + f + w, if and only if for any a 2 R, there exists a 2-idempotent e 2 Z[a] and a nilpotent w 2 R such that a^2 = e+w, if and only if R is a Kosan exchange ring.

Equations

$$b=\frac{x^2+x}{2},c=\frac{x^2-x}{2};p=\frac{y^2+y}{2},q=\frac{y^2-y}{2}.$$

$$b-e,c-f;p-g,q-h\in N(R).$$

$$a=(x+y)-5a=(e-f)+(g-h)+w\text{mbox}forsomew\in N(R).$$

$$\begin{array}[]{c}x^{2}+x\equiv-a+2a^{2}+a^{3}-2a^{4},y^{2}+y\equiv 2a-a^{3}+a^{4};\\ x^{2}-x\equiv-2a+a^{3}+a^{4},y^{2}-y\equiv a+2a^{2}-a^{3}-2a^{4}(\mbox{mod}N(R)).\end{array}$$

$$\begin{array}[]{c}bp\equiv 2a+a^{2}-2a^{3}-a^{4},bq\equiv 0;\\ cp\equiv 0,cq\equiv-2a+a^{2}+2a^{3}-a^{4}(\mbox{mod}N(R)).\end{array}$$

$$\begin{array}[]{c}(e-f+g-h)^{5}=(e+f+g+h+2eg+2fh)^{2}(e-f+g-h)\\ =(e+f+g+h+14eg+14fh)(e-f+g-h)\\ =e-f+g-h+30(eg-fh).\end{array}$$

$$\begin{array}[]{c}(e-f+g-h)^{3}=(e+f+g+h+2eg+2fh)(e-f+g-h)\\ =e-f+g-h+6(eg-fh).\end{array}$$

$$2^3\times 3\times (3-2e)=(3+e)(24-27e+9e^2)\in N(R).$$

$$2^9\times 3^4(2^2-3^2)\in N(R),$$

$$8(e+f)=24=3^3-3=(e+f{)}^3-(e+f)=3e^{2}f+3e{f}^2+r+s.$$

$$8e^{2}f+8e{f}^2=3e{f}^2+3e^{2}f+3r{f}^2+3s{e}^2.$$

$$2^3\times 3\times 5=5\times (3e^{2}f+3e{f}^2+r+s)=9(s{e}^2+r{f}^2)+5(r+s).$$

$$3^2=e^2+2ef+f^2\text{mbox}and{3}^4=e^4+4e^{3}f+6e^2{f}^2+4e{f}^3+f^4.$$

$$2^2\times 3\times 5\times (3^4-3^2)=(2e^2+3ef+2f^2-1)\times 2^3\times 3\times 5ef=0.$$

$$A:=\left(\begin{array}[]{cc}-2&-1\\ 1&-1\end{array}\right)=\left(\begin{array}[]{cc}1&1\\ -1&0\end{array}\right)^{4}-I_{4}\in N(M_{2}(R)).$$

$$\left(\begin{array}[]{cc}-5&0\\ 0&-5\end{array}\right)=\left(\begin{array}[]{cc}1&1\\ -1&1\end{array}\right)^{4}-I_{4}\in GL_{2}(R).$$

Full text

1. Introduction

Throughout, all rings are associative with an identity. An element $a$ in a ring $R$ is (strongly) clean provided that it is the sum of an idempotent and a unit (that commute). A ring $R$ is (strongly) clean in case every element in $R$ is (strongly) clean. A ring $R$ is an exchange ring provided that for any $a\in R$, there exists an idempotent $e\in R$ such that $e\in aR$ and $1-e\in(1-a)R$. Every (strongly) clean ring is an exchange ring, but the converse is not true (see [9, Proposition 1.8]). As is well known, a ring $R$ is an exchange ring if and only if every idempotent lifts modulo every right ideal (see [1]). An element $e$ in a ring is tripotent if $e^{3}=e$. A ring $R$ is a Zhou nil-clean if every element in $R$ is the sum of two tripotents and a nilpotent that commute. Several elementary properties of such rings are established in  [12].

In this paper, we further discuss Zhou nil-clean rings with an emphasis on their relations with polynomials, idempotents and 2-idempotents. In Section 2 we establish new structure of Zhou nil-clean rings. We prove that a ring $R$ is a Zhou nil-clean ring if and only if for any $a\in R$, there exists $e\in{\mathbb{Z}}[a]$ such that $a-e\in R$ is nilpotent and $e^{5}=5e^{3}-4e$, if and only if for any $a\in R$, there exist idempotents $e,f,g,h\in{\mathbb{Z}}[a]$ and a nilpotent $w$ such that $a=e+f+g+h+w$.

An element $e\in R$ is a 2-idempotent if $e^{2}$ is an idempotent, i.e., $e^{2}=e^{4}$. Every tripotent in a ring is a 2-idempotent, but the converse is not true. For instance, every element in ${\mathbb{Z}}_{4}$ is a 2-idempotent, while $2\in{\mathbb{Z}}_{4}$ is not a tripotent. In Section 3, we characterize Zhou nil-clean rings by means of 2-idempotents. We prove that a ring $R$ is Zhou nil-clean if and only if for any $a\in R$, there exist 2-idempotents $e,f\in{\mathbb{Z}}[a]$ and a nilpotent $w\in R$ such that $a=e+f+w$, if and only if for any $a\in R$, there exists a 2-idempotent $e\in{\mathbb{Z}}[a]$ and a nilpotent $w\in R$ such that $a^{2}=e+w$. Let $R$ be a ring in which every element in $R$ is the sum of two commuting 2-idempotents. As an application, we prove that every $n\times n$ matrix over $R$ is the sum of two tripotent matrices and a nilpotent.

An element $u$ in a ring is unipotent if $u-1$ is nilpotent. A ring $R$ is called a Kosan ring if for any $u\in U(R)$, $u^{4}$ is unipotent. Finally, in Section 4, we prove that a ring $R$ is a Zhou nil-clean ring if and only if $R$ is an exchange Kosan ring.

We use $N(R)$ to denote the set of all nilpotents in $R$ and $J(R)$ the Jacobson radical of $R$. ${\mathbb{N}}$ stands for the set of all natural numbers. ${\mathbb{Z}}[u]=\{f(u)~{}|~{}f(t)$ is a polynomial with integral coefficients $\}$.

2. Structure Theorems

The purpose of this section is to explore structure of of Zhou nil-clean rings. We begin with

Lemma 2.1**.**

Let $R$ be a Zhou nil-clean ring with $5\in N(R)$. Then for any $a\in R$ there exists $e\in{\mathbb{Z}}[a]$ such that $a-e\in R$ is nilpotent and $e^{5}=5e^{3}-4e$.

Proof.

$(1)\Rightarrow(2)$ Let $a\in R$. then $a-a^{5}\in N(R)$. Set $x=3a+a^{2}+a^{4}~{}\mbox{and}~{}y=3a-a^{2}-a^{4}$. Then $x-x^{3},y-y^{3}\in N(R)$. Set

[TABLE]

Then $b-b^{2},c-c^{2};p-p^{2},q-q^{2}\in N(R)$. Thus, we can find idempotents $e,f;g,h\in{\mathbb{Z}}[a]$ such that

[TABLE]

Since $x=b-c$ and $y=p-q$, we see that $x-(e-f);y-(g-h)\in N(R)$. Therefore

[TABLE]

As $bc=\frac{x^{4}-x^{2}}{4}\in N(R)$, we see that $ef\in N(R)$; and so $ef=0$. As $pq\in N(R)$, likewise, $gh=0$.

Since $5\in R$ is nilpotent, we directly verify that

[TABLE]

Moreover, we have

[TABLE]

Thus, we see that $eh,fg\in N(R)$; hence, $eh=fg=0$. Accordingly, we see that $ef=gh=eh=fg=0$, and then we check that

[TABLE]

Moreover, we have

[TABLE]

Let $\alpha=e-f+g-h$. Then $\alpha^{5}=\alpha+5(\alpha^{3}-\alpha)$. Hence, $\alpha^{5}=5\alpha^{3}-4\alpha$, as required.∎

We now ready to present a new characterization of a Zhou nil-clean ring.

Theorem 2.2**.**

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

1. (1)

$R$* is Zhou nil-clean.* 2. (2)

$a^{5}-5a^{3}+4a\in R$* is nilpotent for all $a\in R$.* 3. (3)

For any $a\in R$, there exists $e\in{\mathbb{Z}}[a]$ such that $a-e\in R$ is nilpotent and $e^{5}=5e^{3}-4e$.

Proof.

$(1)\Rightarrow(3)$ In view of [12, Theorem 2.11], $R$ is isomorphic to $R_{1},R_{2},R_{3}$ or the product of these rings, where $R_{1}$ is strongly nil-clean and $2\in N(R_{1})$; $R_{2}$ is strongly 2-nil-clean and $3\in N(R_{2})$; $R_{3}$ is Zhou nil-clean and $5\in N(R_{3})$.

Step 1. Let $a\in R_{1}$. In view of [12, Lemma 2.4], there exists an idempotent $e\in{\mathbb{Z}}[a]$ such that $a-e\in N(R_{1})$. We easily check that $e^{5}=5e^{3}-4e$.

Step 2. Let $a\in R_{2}$. By virtue of [12, Lemma 2.6], there exists a tripotent $e\in{\mathbb{Z}}[a]$ such that $a-e\in N(R_{1})$. We easily check that $e^{5}=5e^{3}-4e$.

Step 3. Let $a\in R_{1}$. According to Lemma 2.1, there exists an idempotent $e\in{\mathbb{Z}}[a]$ such that $a-e\in N(R_{1})$. We easily check that $e^{5}=5e^{3}-4e$.

Let $a\in R$. Combining the preceding steps, we can find $e\in{\mathbb{Z}}[a]$ such that $a-e\in R$ is nilpotent and $e^{5}=5e^{3}-4e$.

$(3)\Rightarrow(2)$ Let $a\in R$. Then there exists $e\in{\mathbb{Z}}[a]$ such that $w:=a-e\in R$ is nilpotent and $e^{5}=5e^{3}-4e$. Hence, $(a-w)^{5}=5(a-w)^{3}-4(a-w)$. Thus, $a^{5}-5a^{3}+4a\in N(R)$, as required.

$(2)\Rightarrow(1)$ By hypothesis, $2^{3}\times 3\times 5=3^{5}-5\times 3^{3}+4\times 3\in N(R)$; hence, $2\times 3\times 5\in N(R)$. Write $2^{n}\times 3^{n}\times 5^{n}=0(n\in{\mathbb{N}})$. Thus, $R\cong R_{1}\times R_{2}\times R_{3}$, where $R_{1}=R/2^{n}R,R_{2}=R/3^{n}R$ and $R_{3}=R/3^{n}R$.

Step 1. Let $x\in R_{1}$. As $2\in N(R_{1})$, we see that $x^{4}(x-1)=x^{5}-x^{4}\in N(R_{1})$, and so $x-x^{2}\in N(R_{1})$. This shows that $R_{1}$ is strongly nil-clean. Hence, $R_{1}$ is Zhou nil-clean, by [12, Proposition 2.5].

Step 2. Let $x\in R_{2}$. As $3\in N(R_{2})$, we see that $x(x^{2}-1)^{2}=x^{5}-2x^{3}+x\in N(R_{2})$. This shows that $x-x^{3}\in N(R_{2})$. Hence, $R_{2}$ is strongly 2-nil-clean. In view of [12, Proposition 2.5], $R_{2}$ is Zhou nil-clean.

Step 3. Let $x\in R_{3}$. As $5\in N(R_{3})$, we have $x-x^{5}\in N(R_{3})$. In light of [12, Proposition 2.10], $R_{3}$ is Zhou nil-clean.

Therefore $R$ is Zhou nil-clean.∎

Corollary 2.3**.**

A ring $R$ has the identity $x=x^{5}$ if and only if

1. (1)

$R$* is reduced;* 2. (2)

$R$* has the identity $x^{5}=5x^{3}-4x$.*

Proof.

$\Longrightarrow$ Clearly, $R$ is reduced. In view of [12, Theorem 2.11], $R$ is Zhou nil-clean. Let $x\in R$. By virtue of Theorem 2.2, $x^{5}-5x^{3}+4x\in N(R)=0$, as required.

$\Longleftarrow$ In view of Theorem 2.2, $R$ is Zhou nil-clean. In light of [12, Proposition 2.8], $J(R)$ is nil, and so $J(R)=0$. This completes the proof by  [12, Theorem 2.11].∎

Example 2.4**.**

Let $R={\mathbb{Z}}_{25}$. Then $R$ is Zhou nil-clean. But $2\in R$ can not be written as the sum of $f^{5}=f$ and a nilpotent.

Proof.

Clearly, $J(R)=5R$ is nil and $R/J(R)\cong{\mathbb{Z}}_{5}$. In light of [12, Theorem 2.11], $R$ is Zhou nil-clean. We check that $N(R)=\{0,5,10,15,20\}$, and then $(2-x)^{5}\neq 2-x$ for all $x\in N(R)$, and we are through.∎

We come now to characterize Zhou nil-clean rings by means of their idempotents. We

Theorem 2.5**.**

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

1. (1)

$R$* is Zhou nil-clean.* 2. (2)

For any $a\in R$, there exist idempotents $e,f,g,h\in{\mathbb{Z}}[a]$ and a nilpotent $w$ such that $a=e+f+g+h+w$. 3. (3)

For any $a\in R$, there exist idempotents $e,f,g,h$ and a nilpotent $w$ that commute such that $a=e+f+g+h+w$.

Proof.

$\Longrightarrow$ Since $R$ is Zhou nil-clean, it follows by [12, Theorem 2.11] that $2^{5}-2\in N(R)$; hence, $30=2\times 3\times 5\in N(R)$. Write $2^{n}\times 3^{n}\times 5^{n}=0$. Then $R\cong R_{1}\times R_{2}\times R_{3}$, where $R_{1}\cong R/2^{n}R,R_{2}\cong R/3^{n}R$ and $R_{3}=R/5^{n}R$. Clearly, each $R_{i}$ is Zhou nil-clean. As $2\in N(R_{1})$, it follows by [12, Proposition 2.5] that $R_{1}$ is strongly nil-clean, and so every element in $R_{1}$ is the sum of an idempotent and a nilpotent that commute. As $3\in N(R_{2})$, it follows by [12, Proposition 2.8] that $R_{2}$ is strongly 2-nil-clean, and so every element in $R_{2}$ is the sum of two idempotents and a nilpotent that commute.

Let $c\in R_{3}$, and let $a=2-c$. Clearly, $R_{3}$ is a Zhou nil-clean ring with $5\in N(R_{3})$. Construct $e,f,g,h$ as in the proof of Lemma 2.1, we see that $a=(e-f)+(g-h)+w$, where $e,f,g,h\in R$ are idempotents and $w\in R$ is a nilpotent that commute. Hence, $c=(1-e)+f+(1-g)+h-w$. Thus, $c$ is the sum of four idempotents and a nilpotent that commute in $R_{3}$. Therefore every element in $R$ can be written in this form.

$\Longleftarrow$ Let $a\in R$. Then there exist idempotents $e,f,g,h$ and a nilpotent $w$ that commute such that $2-a=e+f+g+h+w$. Hence, $a=(1-e)-f+(1-h)-g-w$. Set $s=(1-e)-f$ and $t=(1-h)-g$. We easily check that $s^{3}=s,t^{3}=t$ and $st=ts$. Thus, $a$ is the sum of two tripotents and a nilpotent that commute. Therefore $R$ is Zhou nil-clean.∎

Corollary 2.6**.**

A ring $R$ has the identity $x=x^{5}$ if and only if

1. (1)

$R$* is reduced;* 2. (2)

Every element is the sum of four commuting idempotents.

Proof.

$\Longrightarrow$ This is obvious by Corollary 2.3 and Theorem 2.5.

$\Longleftarrow$ In view of Theorem 2.5, $R$ is Zhou nil-clean. Thus, $R/J(R)$ has the identity $x^{5}=x$ and $J(R)$ is nil by [12, Theorem 2.11]. Thus, $J(R)=0$, and then the result follows.∎

We note that ”four idempotents” in the proceeding corollary can not be replaced by ”three idempotents”. For instance, ${\mathbb{Z}}_{5}$. We do have

Theorem 2.7**.**

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

1. (1)

$R$* is strongly 2-nil-clean.* 2. (2)

For any $a\in R$, there exist idempotents $e,f$ and a nilpotent $w$ that commute such that $a=e+f+w$. 3. (3)

For any $a\in R$, there exist idempotents $e,f,g$ and a nilpotent $w$ that commute such that $a=e+f+g+w$.

Proof.

$(1)\Rightarrow(2)$ This is proved in [2, Lemma 2.2].

$(2)\Rightarrow(3)$ This is trivial.

$(3)\Rightarrow(1)$ Let $a\in R$. Then there exist idempotents $e,f,g$ and a nilpotent $w$ that commute such that $2-a=e+f+g+w$. Hence, $a=(1-e)-f+(1-g)-w$. Clearly, $((1-e)-f)^{3}=(1-e)-f$. Thus, $a$ is the sum of a tripotent, an idempotent and a nilpotent that commute. In light of [2, Theorem 2.3], $R$ is strongly 2-nil-clean.∎

3. Decomposition of 2-Idempotents and nilpotents

The aim of this section is to characterize Zhou nil-clean rings by their 2-idempotents and nilpotents. The next lemma will enable us to take full advantage of decompositions of rings.

Lemma 3.1**.**

Let $R$ be a ring in which every element is the sum of two 2-idempotents and a nilpotent that commute. Then $30\in R$ is nilpotent.

Proof.

Write $3=g+h+w$, where $g^{2}=g^{4},h^{2}=h^{4},w\in N(R)$ and $e,f,w$ commute with another. Let $e=g^{3}$ and $f=h^{3}$. Then $3=e+f+b$, where $e^{3}=e,f^{3}=f,b=w+(e-e^{3})+(f-f^{3})$. As $(e-e^{3})^{2}=(f-f^{3})^{2}=0$, we see that $b\in N(R)$. Hence, $3-e=f+b$, and so $(3-e)^{3}=f^{3}+3bf(b+f)$. This implies that $(3-e)^{3}-(3-e)\in N(R)$, i.e., $24-9e(3-e)=24-27e+9e^{2}\in N(R)$. We infers that

[TABLE]

Hence, we can find $w\in N(R)$ such that $2^{3}\times 3^{2}=2^{4}\times 3e+w$, and so $2^{9}\times 3^{6}=2^{12}\times 3^{3}e^{3}+w^{\prime}$ for $w^{\prime}\in N(R)$. It follows that

[TABLE]

i.e., $2^{9}\times 3^{4}\times 5\in N(R)$. Therefore $30\in N(R)$, as asserted.∎

Theorem 3.2**.**

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

1. (1)

$R$* is Zhou nil-clean.* 2. (2)

For any $a\in R$, there exist 2-idempotents $e,f\in{\mathbb{Z}}[a]$ and a nilpotent $w\in R$ such that $a=e+f+w$. 3. (3)

Every element in $R$ is the sum of two 2-idempotents and a nilpotent that commute.

Proof.

$(1)\Rightarrow(2)$ This is obvious by [2, Theorem 2.11], as every tripotent in $R$ is a 2-idempotent.

$(2)\Rightarrow(3)$ This is trivial.

$(3)\Rightarrow(1)$ In view of Lemma 3.1, $30\in N(R)$. Write $2^{n}\times 3^{n}\times 5^{n}=0(n\in{\mathbb{N}})$. Then $R\cong R_{1}\times R_{2}\times R_{3}$, where $R_{1}=R/2^{n}R,R_{2}=R/3^{n}R$ and $R_{3}=R/5^{n}R$. Set $S=R_{2}\times R_{3}$. Then $R\cong R_{1}\times S$. Step 1. Let $a^{2}=a^{4}$ in $R_{1}$. Then $a^{2}(1-a)(1+a)=0$. As $2\in N(R_{1})$, we see that $(a-a^{2})^{2}\in N(R)$; hence, $a-a^{2}\in N(R_{1})$. In light of [12, Proposition 2.5], there exists an idempotent $e\in{\mathbb{Z}}[a]$ such that $a-e\in N(R_{1})$. Thus, every element in $R_{1}$ is the sum of two idempotents and a nilpotent that commute. This shows that $R_{1}$ is strongly 2-nil-clean; hence, it is Zhou nil-clean.

Step 2. Let $a^{2}=a^{4}\in S$. Then $a(a-a^{3})=0$, and so $(a-a^{3})^{2}=a(a-a^{3})(1-a^{2})\in N(R)$. Hence, $a-a^{3}\in N(S)$. As $2\in U(S)$, it follows by [12, Lemma 2.6] that there exists $e^{3}=e\in S$ such that $a-e\in N(S)$.

Let $c\in S$. Then we can find 2-idempotents $b,c\in S$ such that $a-b-c\in N(S)$ and $a,b,c\in S$ commute. By the preceding discussion, we have tripotents $g\in{\mathbb{Z}}[b]$ and $h\in{\mathbb{Z}}[c]$ such that $b-g,c-h\in N(S)$. Therefore $a-g-h=(a-b-c)+(b-g)+(c-h)\in N(S)$ where $a,g,h$ commute. Therefore every element in $S$ is the sum of two tripotents and a nilpotent in $S$. That is, $S$ is Zhou nil-clean.

Therefore $R$ is Zhou nil-clean, as asserted.∎

Corollary 3.3**.**

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

1. (1)

$R$* is strongly 2-nil-clean.* 2. (2)

Every element in $R$ is the sum of a 2-idempotent and a nilpotent that commute.

Proof.

$(1)\Rightarrow(2)$ This is obvious, by [2, Theorem 2.8].

$(2)\Rightarrow(1)$ In view of Theorem 3.2, $R$ is Zhou nil-clean. Write $2=e+w$ where $e^{3}=e\in R,w\in N(R)$. Then $2^{3}-2\in N(R)$, and so $6\in N(R)$. In light of [11, Lemma 3.5], $R$ is strongly nil-clean.∎

Theorem 3.4**.**

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

1. (1)

$R$* is Zhou nil-clean.* 2. (2)

For any $a\in R$, there exists a 2-idempotent $e\in{\mathbb{Z}}[a]$ and a nilpotent $w\in R$ such that $a^{2}=e+w$. 3. (3)

For any $a\in R$, $a^{2}$ is the sum of a 2-idempotent and a nilpotent that commute.

Proof.

$(1)\Rightarrow(2)$ As in the proof of Theorem 3.2, $R\cong R_{1}\times S$ with $2\in N(R_{1})$ and $2\in U(S)$. Then $R$ is strongly nil-clean by [12, Proposition 2.5]. Let $a\in R$. Write $a=(a_{1},a_{2})$. Then $a_{1}^{2}$ is the sum of an idempotent and a nilpotent in $R_{1}$. As $R$ is Zhou nil-clean, so is $R_{2}$. Hence, $a_{2}^{5}-a_{2}\in N(S)$. This implies that $(a_{2}^{2})^{3}-a_{2}^{2}=a_{2}(a_{2}^{5}-a_{2})\in N(S)$. As $2\in U(S)$, it follows by [12, Lemma 2.6] that there exists $f^{3}=f\in{\mathbb{Z}}[a_{2}]$ such that $a_{2}^{2}-f\in N(S)$. Thus, we can find a tripotent $e\in{\mathbb{Z}}[a]$ such that $a^{2}-e\in N(R)$. Thus proving $(2)$ as every tripotent is a 2-idempotent.

$(2)\Rightarrow(3)$ This is obvious, as every tripotent is a 2-idempotent.

$(3)\Rightarrow(1)$ Let $a\in R$. Then there exists a 2-idempotent $e\in R$ such that $a-e\in N(R)$ and $ae=ea$. Hence, $(a^{2})^{4}-(a^{2})^{2}\in N(R)$; hence, $a^{8}-a^{4}\in N(R)$. Thus, $(a^{5}-a)^{4}=a^{3}(a^{5}-a)(a^{4}-1)^{3}=(a^{8}-a^{4})(a^{4}-1)\in N(R)$, i.e., $a^{5}-a\in N(R)$. In light of [12, Theorem 2.11], $R$ is Zhou nil-clean.∎

Corollary 3.5**.**

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

1. (1)

$R$* is Zhou nil-clean.* 2. (2)

For any $a\in R$, $a^{4}$ is the sum of an idempotent and a nilpotent that commute.

Proof.

$(1)\Rightarrow(2)$ Let $a\in R$. In view of [12, Theorem 2.11], $a^{5}-a\in N(R)$; hence, $(a^{4})^{2}-a^{4}=a^{3}(a^{5}-a)\in N(R)$. In light of [11, Lemma 3.5], $a^{4}\in R$ is strongly nil-clean, as required.

$(2)\Rightarrow(1)$ By hypothesis, $a^{4}$ is strongly nil-clean, and so $(a^{4})^{2}-a^{4}\in N(R)$. Hence, $a^{3}(a^{5}-a)\in N(R)$. Thus, $(a^{5}-a)^{4}=(a^{4}-1)^{3}a^{3}(a^{5}-a)\in N(R)$. We infer that $a^{5}-a\in N(R)$. Therefore $R$ is Zhou nil-clean, by [12, Theorem 2.11].∎

Recall that a ring $R$ is of bounded index if there exists $n\in{\mathbb{N}}$ such that $x^{n}=0$ for all $x\in N(R)$.

Lemma 3.6**.**

Let $R$ be a ring in which every element in $R$ is the sum of two commuting 2-idempotents. Then $R$ is of bounded index.

Proof.

In view of Theorem 3.4, $R$ is Zhou nil-clean. By using [12, Theoerm 2.11], $R/J(R)$ has the identity $x^{5}=x$. By using Jacobson Theorem, $R/J(R)$ is commutative. Hence, $N(R)\subseteq J(R)$.

Write $3=e+f$ where $e,f$ are commuting 2-idempotents. Set $r=e^{3}-e$ and $s=f^{3}-f$. Then $re=sf=0$. We check that

[TABLE]

Multiplying by $ef$ from two sides, we get

[TABLE]

It follows that $5(e^{2}f+ef^{2})=3(se^{2}+rf^{2}).$ Thus,

[TABLE]

This implies that $2^{3}\times 3\times 5ef=0$. Clearly,

[TABLE]

We obtain

[TABLE]

That is, $2^{5}\times 3^{3}\times 5=0$. Write $R_{1}=R/2^{5}R,R_{2}=R/3^{3}R$ and $R_{3}=R/5R$. Then $R\cong R_{1}\times R_{2}\times R_{3}$.

Step 1. Let $b\in J(R_{1})$. Write $b=e+f$ where $e,f\in R_{1}$ are 2-idempotents. Then $b^{2}-b^{4}=2ef-4e^{3}f-6e^{2}f^{2}-4ef^{3}$. As $2^{5}=0$ in $R_{3}$, we see that $(b^{2}-b^{4})^{5}=0$, and so $b^{10}=0$. Thus, $R_{1}$ is of bounded index $10$.

Step 2. Let $b\in J(R_{2})$. Write $b=e+f$ where $e,f\in R_{2}$ are 2-idempotents. Then $b^{3}=e^{3}+f^{3}+3ef(e+f)$. Hence, $b-b^{3}=(e-e^{3})+(f-f^{3})-3ef(e+f)$. It follows that $b^{2}(b-b^{3})=f^{2}(e-e^{3})+e^{2}(f-f^{3})-3efb^{3}$ and $b^{4}(b-b^{3})=f^{4}(e-e^{3})+e^{4}(f-f^{3})-3efb^{5}$. Thus, \big{(}b^{4}(b-b^{3})-b^{2}(b-b^{3})\big{)}^{3}=0, as $3^{3}=0$ in $R_{2}$. This implies that $b^{9}=0$, and so $R_{2}$ is of bounded index $9$.

Step 3. Let $b\in J(R_{3})$. Write $b=e+f$ where $e,f\in R_{3}$ are commuting 2-idempotents. Then $b^{5}=e^{5}+5e^{4}f+10e^{3}f^{2}+5e^{2}f^{3}+10ef^{4}+f^{5}=e^{5}+f^{5}=e^{3}+f^{3}$ as $5=0$. Hence, $b-b^{5}=(e-e^{3})+(f-f^{3})$. It follows that $b^{2}(b-b^{5})=f^{2}(e-e^{3})+e^{2}(f-f^{3})$ and $b^{4}(b-b^{5})=f^{4}(e-e^{3})+e^{4}(f-f^{3})$. As $e$ and $f$ are 2-idempotents, we see that $b^{2}(b-b^{5})=b^{4}(b-b^{5})$; hence, $b^{3}=0$. That is, $R_{3}$ is of bounded index $3$.

Therefore $R$ is of bounded index.∎

With this information we can now express every matrix over such rings by means of tripotent and nilpotent matrices.

Theorem 3.7**.**

Let $R$ be a ring in which every element in $R$ is the sum of two commuting 2-idempotents. Then every $n\times n$ matrix over $R$ is the sum of two tripotent matrices and a nilpotent.

Proof.

In view of Theorem 3.2, $R$ is Zhou nil-clean. By virtue of Lemma 3.6, $R$ is of bounded index. Therefore we complete the proof, by  [8, Theorem 3.7].∎

4. Kosan rings

This section is devoted to a collection of elementary properties of Kosan rings which will be used in the sequel. We start by

Proposition 4.1**.**

1. (1)

Every finite direct product of Kosan rings is a Kosan ring. 2. (2)

Every subring of a Kosan ring is a Kosan ring. 3. (3)

If $R$ is a Kosan ring, then $eRe$ is a Kosan ring for all idempotents $e\in R$.

Proof.

$(1)$ Let $S=\prod\limits_{i=1}^{n}R_{i}$, where each $R_{i}$ is a Kosan ring, if $u=(u_{1},u_{2},\cdots,u_{n})\in U(S)$, then each $u_{i}\in U(R_{i})$ and by hypothesis $u_{i}^{4}=1+w_{i}$ for some nilpotent $w_{i}\in R_{i}$. So we have $u^{4}=(1,1,\cdots,1)+(w_{1},w_{2},\cdots,w_{n})$ is a unipotent.

$(2)$ Let $S$ be a subring of a Kosan ring $R$ and $u\in U(S)$, then $u\in U(R)$ and so $u^{4}=1_{R}+w$ for some $w\in N(R)$. We infer that $w=u^{4}-1_{S}\in N(S)$. Hence, $S$ is Kosan.

$(3)$ Let $u\in U(eRe)$ with $e=e^{2}\in R$. Then $u+1-e\in U(R)$. By hypothesis, $(u+1-e)^{4}=1+w$ for some $w\in N(R)$. Hence, $u^{4}=e+w$, and so $w=u^{4}-e\in eRe$. Hence, $w\in N(eRe)$, as desired.∎

We note that the infinite products of Kosan rings may be not a Kosan ring. For instance, let $R=\prod\limits_{n=1}^{\infty|}{\mathbb{Z}}_{5^{n}}$. Then each ${\mathbb{Z}}_{5^{n}}$ is Zhou nil-clean, but $R$ is not Zhou nil-clean. Clearly, $a=(2,2,\cdots,2,\cdots)\in R$, while $a-a^{5}\in N(R)$, and we are done by [12, Theorem 2.11].

Lemma 4.2**.**

Let $I$ be a nil ideal of a ring $R$. Then $R$ is a Kosan ring if and only if so is $R/I$.

Proof.

$\Longrightarrow$ This is obvious as every unit lifts modulo every nil ideal.

$\Longleftarrow$ Let $u\in U(R)$, then $\bar{u}^{4}=\bar{1}+\bar{w}$ for $\bar{w}\in N(R/I)$. Hence, $u^{4}=1+w+r$ for some $r\in I$. Here $w+r\in N(R)$.∎

We use $T_{n}(R)$ to denote the ring of all $n\times n$ upper triangular matrices over a ring $R$. We have

Theorem 4.3**.**

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

1. (1)

$R$* is a Kosan ring.* 2. (2)

$T_{n}(R)$* is a Kosan ring for all $n\in{\mathbb{N}}$.* 3. (3)

$T_{n}(R)$* is a Kosan ring for some $n\in{\mathbb{N}}$.*

Proof.

$(1)\Rightarrow(2)$ Choose I=\{\left(\begin{array}[]{cccc}0&a_{12}&\cdots&a_{1n}\\ &0&\cdots&a_{2n}\\ &&\ddots&\vdots\\ &&&0\end{array}\right)\in T_{n}(R)| each $a_{ij}\in R\}$ Then $I$ is a nil ideal of $R$. As $T_{n}(R)/I\cong\prod\limits_{i=1}^{n}R_{i}$ be the direct product of rings $R_{i}\cong R$, it follows by Proposition 4.1, that $\prod\limits_{i=1}^{n}R_{i}$ is a Kosan ring ring. In light of Proposition 4.2, $T_{n}(R)$ is a Kosan ring, as required.

$(2)\Rightarrow(3)$ This is trivial.

$(3)\Rightarrow(1)$ Let $e=diag(1,0,\cdots,0)$. Then $R\cong eT_{n}(R)e$. We complete the proof by Proposition 3.1. ∎

Example 4.4**.**

Let $R$ be a ring. Then $M_{2}(R)$ is not a Kosan ring.

Proof.

Assume that $M_{2}(R)$ is a Kosan ring. Since \left(\begin{array}[]{cc}1&1\\ -1&0\end{array}\right)=\left(\begin{array}[]{cc}0&-1\\ 1&1\end{array}\right)^{-1}\in GL_{2}(R),

[TABLE]

Let $S=\{m\cdot 1_{R}~{}|~{}m\in{\mathbb{Z}}\}$. Then $S$ is a commutative subring of $R$. As $A\in N(M_{2}(S))$, we see that $det(A)=3\in N(S)$, and so $3\in N(R)$.

Since $2=3-1\in U(R)$, \left(\begin{array}[]{cc}1&1\\ -1&1\end{array}\right)=\left(\begin{array}[]{cc}2^{-1}&-2^{-1}\\ 2^{-1}&2^{-1}\end{array}\right)^{-1}\in GL_{2}(R), and then

[TABLE]

This implies that $5\in N(R)$, and so $1_{R}=(3\cdot 2-5)\cdot 1_{R}\in N(R)$, a contradiction. This completes the proof.∎

Recall that a ring $R$ is local if $R$ has only one maximal right ideal, i.e., $R/J(R)$ is a division ring.

Example 4.5**.**

A local ring $R$ is a Kosan ring if and only if

1. (1)

$J(R)$* is nil;* 2. (2)

$R/J(R)\cong{\mathbb{Z}}_{2},{\mathbb{Z}}_{3}$* or ${\mathbb{Z}}_{5}$.*

Proof.

$\Longrightarrow$ As $(2,3)=1$, we see that $2\in U(R)$ or $3\in U(R)$. If $2\in U(R)$, then $2^{4}-1\in N(R)$, and so $3\times 5\in N(R)$. If $3\in U(R)$, then $3^{4}-1\in N(R)$, and so $2\times 5\in N(R)$. As $(2,3,5)=1$, we see that $p\in N(R)$ where $p=2,3$ or $5$. Let $x\in J(R)$. Then $1+x^{p}\equiv(1+x)^{p}\equiv 1(mod~{}N(R))$, and so $x^{p}\equiv 0(mod~{}N(R))$. That is, $x\in N(R)$. Thus, $J(R)$ is nil.

Since $R$ is local, $R/J(R)$ is a Kosan division ring. Let $0\neq x\in R/J(R)$. Then $x^{4}=1$, and so $(x^{2}+1)(x-1)(x+1)=0$. Hence, $x=-1,1$ or $x^{2}=-1$. Suppose that $y\not\in\{0,1,-1,2,-2\}$. Then $y^{2}=-1$, and so $y+1\neq 0,-1,1$. Hence, $(y+1)^{2}=-1$, i.e., $2y=-1$. Thus, $2y=y^{2}$, and then $y=2$, a contradiction. Therefore $|R/J(R)|\leq 5$, and so $R/J(R)\cong{\mathbb{Z}}_{2},{\mathbb{Z}}_{3}$ or ${\mathbb{Z}}_{5}$.

$\Longleftarrow$ Since ${\mathbb{Z}}_{2},{\mathbb{Z}}_{3}$ and ${\mathbb{Z}}_{5}$ are all Kosan rings, we complete the proof by Lemma 4.2.∎

Example 4.6**.**

Let $R={\mathbb{Z}_{n}}(n\geq 2)$. Then $R$ is a Kosan ring if and only if $n=2^{k}3^{l}5^{s}(k,l,s~{}\mbox{are nonnegitive integers})$.

Proof.

$\Longrightarrow$ Write $n=p_{1}^{r_{1}}\cdots p_{s}^{r_{s}}$ where $p_{1},\cdots,p_{s}$ are primes and $r_{1},\cdots,r_{s}\in{\mathbb{N}}$. Then $R\cong\bigoplus\limits_{i=1}^{s}{\mathbb{Z}}_{p_{i}^{r_{i}}}$. As every homomorphic image of Kosan rings is a Kosan ring, each ${\mathbb{Z}}_{p_{i}^{r_{i}}}$ is a local Kosan ring. In light of Example 4.5, ${\mathbb{Z}}_{p_{i}^{r_{i}}}/J({\mathbb{Z}}_{p_{i}^{r_{i}}})\cong{\mathbb{Z}}_{p_{i}}$ is ${\mathbb{Z}}_{2},{\mathbb{Z}}_{3}$ or ${\mathbb{Z}}_{5}$. Thus $p_{i}=2,3$ or $5$, as desired.

$\Longleftarrow$ As $n=2^{k}3^{l}5^{m}$, we see that ${\mathbb{Z}}_{n}\cong R_{1}\times R_{2}\times R_{3}$, where $R_{1}={\mathbb{Z}}_{2^{k}},R_{2}={\mathbb{Z}}_{3^{l}}$ and $R_{3}={\mathbb{Z}}_{5^{m}}$. It is obvious that each $J(R_{i})$ is nil and $R_{1}/J(R_{1})\cong{\mathbb{Z}}_{2}$, $R_{2}/J(R_{2})\cong{\mathbb{Z}}_{3}$, and $R_{3}/J(R_{3})\cong{\mathbb{Z}}_{5}$. Then by Example 4.5, $R$ is a Kosan ring.∎

5. Exchange Properties

The class of exchange rings is very large. It includes all regular rings, all $\pi$-regular rings, all strongly $\pi$-regular rings, all semiperfect rings, all left or right continuous rings, all clean rings, all unit $C^{*}$-algebras of real rank zero and all right semi-artinian rings (see [1]). We now characterize Zhou nil-clean rings in terms of their exchange properties. We need an elementary lemma.

Lemma 5.1**.**

Let $R$ be an exchange Kosan ring. Then $30\in R$ is nilpotent.

Proof.

Since $R$ is an exchange ring, there exists an idempotent $f\in 3R$ such that $1-f\in(1-3)R$. Write $f=3a$ for some $a\in R$. We may assume that $a=fa$, then $3a^{2}=3a$. Then $1-f=(1-3)b$ for some $b\in R$ with $b(1-f)=b$, therefore $(-2)b^{2}=b$. Now we have $3=3-(1-f)+(1-f)$. It is obvious that $3-(1-f)$ is a unit with inverse $a-b$.

Set $e=1-f$ and $u=3-(1-f)$. Then $3=e+u$. By hypothesis, $u^{4}=1+w$, so $(3-e)^{4}=1+w$, then $81-65e=1+w$ which implies that $80=65e+w$. Thus, $80e=65e+we$, hence, $15e=we$. Also $80=65e+w=60e+5e+w=5e+(4e+1)w$. Then $80^{2}=5^{2}e+10e(4e+1)w+(4e+1)^{2}w^{2}$, and $5\times 80=5^{2}e+5(4e+1)w$; whence, $80^{2}-5\times 80\in N(R)$. This implies $2^{4}\times 3\times 5^{3}\in N(R)$ and so $2\times 3\times 5\in N(R)$.∎

Lemma 5.2**.**

Let $R$ be an exchange Kosan ring, then $J(R)$ is nil.

Proof.

By virtue of Lemma 5.1, $30\in R$. Write $30^{n}=0$ for some $n\in{\mathbb{N}}$. Then $R\cong R_{1}\times R_{2}\times R_{3}$ where $R_{1}=R/2^{n}R,R_{2}=R/3^{n}R$ and $R_{3}=R/5^{n}R$. Let $x\in J(R_{1})$ so $1-x\in U(R_{1})$. Let $u=1-x$, as $R_{1}$ is a Kosan ring, $u^{4}+1$ is in $N(R_{1})$. We have $(1-u)^{4}=1+u^{4}+2(3u^{2}-2u-2u^{3})\in N(R_{1})$, as $2\in N(R_{1})$, then $(1-u)^{4}\in N(R_{1})$ which implies that $1-u\in N(R_{1})$ and so $x\in N(R_{1})$. Now let $x\in N(R_{2})$ then, $(1+x)^{4}=1+w$ for some $w\in N(R_{2})$, so $x^{4}+2(x^{3}+3x^{2}+2x)=x(x^{3}+2(2x^{2}+3x+2))$, since $3\in N(R_{2}),2\in U(R_{2})$, and we have $2x^{2}+3x+2\in U(R_{2})$, so $x^{3}+2(2x^{2}+3x+2)\in U(R_{2})$, which implies that $x\in N(R_{2})$. For $R_{3}$, as $5\in N(R_{3})$ we deduce that $2\in U(R_{3})$, so in the similar way, we can prove that $J(R_{3})$ is nil. Therefore $J(R)$ is nil.∎

Lemma 5.3**.**

Let $R$ be an exchange Kosan ring. If $J(R)=0$, then $R$ is reduced.

Proof.

We claim that $N(R)=0$, if not there exists some $a\in R$ such that $a^{2}=0$. Since $R$ is an exchange ring and $J(R)=0$, by[7, Theorem 2.1], $eRe\cong M_{2}(T)$ for some idempotent $e\in R$ and some ring $T$. As $R$ is a Kosan ring, then $eRe$ is also Kosan which implies that $M_{2}(T)$ is a Kosan ring which is a contradiction with Example 4.4. This shows that $N(R)=0$ and so R is reduced.∎

We have accumulated all the information necessary to prove the following.

Theorem 5.4**.**

A ring $R$ is a Zhou nil-clean if and only if

1. (1)

$R$* is an exchange ring;* 2. (2)

$R$* is a Kosan ring.*

Proof.

$\Longrightarrow$ It is clear that every Zhou nil-clean ring is periodic and so it is strongly clean. Now let $u\in U(R)$, then $u^{5}-u\in N(R)$, $u(u^{4}-1)\in N(R)$, since $u$ is a unit then $u^{4}-1\in N(R)$. Thus, $R$ is a Kosan ring.

$\Longleftarrow$ By virtue of Lemma 5.2, $J(R)$ is nil. Set $S=R/J(R)$. Then $S$ is an exchange Kosan ring with $J(S)=0$. In view of Lemma 5.3, $S$ is reduced. Thus, $S$ is isomorphic to a subdirect product of some domains $S_{i}$. We see that each $S_{i}$ is a homomorphic image of $S$; hence, $S_{i}$ is an exchange Kosan ring with trivial idempotents. In light of [1, Lemma 17.2.1] and Lemma 5.2, $S_{i}$ is local and $J(S_{i})$ is nil. By virtue of Example 4.5, $S_{i}/J(S_{i})\cong{\mathbb{Z}}_{2},{\mathbb{Z}}_{3}$ or ${\mathbb{Z}}_{5}$. For any $x\in S_{i}$, $x-x^{5}\in J(S_{i})\subseteq N(S_{i})$. Let $\overline{a}\in S$. Then $a-a^{5}\in N(S)=0$, and so $R/J(R)$ has the identity $x^{5}=x$. Therefore $R$ is Zhou nil-clean, by [12, Theoerm 2.11].∎

Corollary 5.5**.**

A ring $R$ is Zhou nil-clean if and only if

1. (1)

$R$* is clean;* 2. (2)

$R$* is a Kosan ring.*

Proof.

$\Longrightarrow$ This is a direct result of Theorem 5.4.

$\Longleftarrow$ Since every clean ring is an exchange ring, we are through by Theorem 5.4.∎

Corollary 5.6**.**

A ring $R$ is strongly 2-nil-clean if and only if

1. (1)

$R$* is an exchange ring;* 2. (2)

The square of every unit in $R$ is a unipotent.

Proof.

$\Longrightarrow$ In light of [9, Proposition 1.8], $R$ is an exchange ring and $J(R)$ is nil. Also by [2, Theorem 3.3], $R/J(R)$ is tripotent. Now let $u\in U(R)$, then $u-u^{3}\in J(R)\subseteq N(R)$. Hence, $u^{2}-1\in N(R)$ which implies that $u^{2}$ is unipotent.

$\Longleftarrow$ Let $u\in U(R)$. Then $u^{2}\in U(R)$; hence, it is a unipotent. Thus, $R$ is a Kosan ring. In light of Theorem 5.4, $R$ is Zhou nil-clean. It follows by Lemma 5.1, $30\in N(R)$, and so $5\times 6=0$. Thus, $R\cong R_{1}\times R_{2}$, where $R_{1}=R/6R$ and $R_{2}=R/5R$. As $R_{1}$ is a Zhou nil-clean ring with $6\in N(R_{1})$, it follows by [12, Theorem 2.12] that $R_{1}$ is strongly 2-nil-clean. On the other hand, $5\in N(R_{2})$, and so $4=5-1\in U(R_{2})$. Hence, $2\in U(R_{2})$, and then $2^{2}-1\in N(R_{2})$. This implies that $6=2\times 3\in N(R_{2})$, whence $1=6-5\in N(R_{2})$, an absurd. We infer that $R_{2}=0$. Therefore $R\cong R_{1}$ is strongly 2-nil-clean, as asserted.∎