Matrices over Zhou nil-clean rings

TL;DR

This paper investigates matrices over Zhou nil-clean rings, proving that if such rings are 2-primal with bounded index, then every square matrix can be expressed as the sum of two tripotents and a nilpotent, expanding the class of rings with this property.

Contribution

It establishes that matrices over 2-primal Zhou nil-clean rings of bounded index can be decomposed into tripotents and nilpotents, broadening understanding of matrix decompositions in ring theory.

Findings

Matrices over certain Zhou nil-clean rings can be decomposed into tripotents and nilpotents.

The class of rings where such decompositions are possible is expanded.

Provides structural insights into matrices over Zhou nil-clean rings.

Abstract

A ring R is Zhou nil-clean if every element in R is the sum of two tripotents and a nilpotent that commute. Let R be a Zhou nil-clean ring. If R is 2-primal (of bounded index), we prove that every square matrix over R is the sum of two tripotents and a nilpotent. These provides a large class of rings over which every square matrix has such decompositions by tripotent and nilpotent matrices.

Equations

$$W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E_{1}=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&1\end{array}\right),$$

$$E_{2}=\left(\begin{array}[]{cccccc}0&&&&&0\\ 0&0&&&&0\\ &0&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&0&-1\end{array}\right).$$

$$W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&1\end{array}\right).$$

$$\begin{array}[]{c}W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E_{1}=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&1\end{array}\right),\\ E_{2}=\left(\begin{array}[]{cccccc}0&&&&&0\\ 0&0&&&&0\\ &0&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&0&1\end{array}\right).\end{array}$$

$$\begin{array}[]{c}W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E_{1}=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&1\end{array}\right),\\ E_{2}=\left(\begin{array}[]{cccccc}0&&&&&0\\ 0&0&&&&0\\ &0&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&0&-1\end{array}\right).\end{array}$$

$$W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&1\end{array}\right).$$

$$W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&-1\end{array}\right).$$

$$\begin{array}[]{c}W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E_{1}=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&1\end{array}\right),\\ E_{2}=\left(\begin{array}[]{cccccc}0&&&&&0\\ 0&0&&&&0\\ &0&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&0&1\end{array}\right).\end{array}$$

$$\begin{array}[]{c}W=\left(\begin{array}[]{cccccc}0&&&&&0\\ 1&0&&&&0\\ &1&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&1&0\end{array}\right),E_{1}=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 0&0&&&&c_{1}\\ &0&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&0&-1\end{array}\right),\\ E_{2}=\left(\begin{array}[]{cccccc}0&&&&&0\\ 0&0&&&&0\\ &0&0&&&0\\ &&&\ddots&&\vdots\\ &&&\ddots&0&0\\ &&&&0&-1\end{array}\right).\end{array}$$

Full text

1. Introduction

Throughout, all rings are associative with an identity. A ring $R$ is strongly nil-clean provided that every element in $R$ is the sum of an idempotent and a nilpotent that commutate (see [5]). A ring $R$ is strongly weakly nil-clean provided that every element in $R$ is the sum or difference of a nilpotent and an idempotent that commutate (see [3]). An element $a$ in a ring $R$ is tripotent if $a=a^{3}$. A ring $R$ is strongly 2-nil-clean if every element in $R$ is the sum of a tripotent and a nilpotent that commute (see [2]). The subjects of strongly and strongly weakly nil-clean rings are interested for so many mathematicians, e.g.,  [1, 4, 8, 9] and  [11].

Very recently, Zhou investigated a class of rings in which elements are the sum of two tripotents and a nilpotent that commute (see [12]). For the conventience, we call such ring a Zhou nil-clean ring. The purpose of this paper is to explore matrices over Zhou nil-clean rings. A ring $R$ is 2-primal if its Baer-McCoy radical (i.e., it is equal to the intersection of all prime ideals in $R$) coincides with the set of nilpotents in $R$. For instances, every commutative (reduced) ring is 2-primal. A ring $R$ is of bounded index if there exists $m\in{\mathbb{N}}$ such that $x^{m}=0$ for all nilpotent $x\in R$. Let $R$ be Zhou nil-clean. If $R$ is 2-primal (of bounded index), we prove that every square matrix over $R$ is the sum of two tripotents and a nilpotent. These provides a large class of rings over which every square matrix has such decompositions by tripotent and nilpotent matrices.

We use $N(R)$ to denote the set of all nilpotent elements in $R$ and $J(R)$ the Jacobson radical of $R$. ${\mathbb{N}}$ stands for the set of all natural numbers.

2. Zhou Rings

A ring $R$ is a Zhou ring if every element in $R$ is the sum of two tripotents that commute. The structure of Zhou rings was studied in [10]. This section is devoted to preliminary observations concerning on matrices over Zhou rings which will be used in the sequel. We begin with

Lemma 2.1**.**

Every square matrix over ${\mathbb{Z}}_{3}$ is the sum of two idempotents and a nilpotent.

Proof.

As every matrix over a field has a Frobenius normal form, we may assume that A=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 1&0&&&&c_{1}\\ &1&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&1&c_{n-1}\end{array}\right).

Case I. $c_{n-1}=0$. Choose

[TABLE]

[TABLE]

Then $E_{1}^{3}=E_{1}$ and $E_{2}^{3}=E_{2}$, and so $A=E_{1}+E_{2}+W$.

Case II. $c_{n-1}=1$. Choose

[TABLE]

Then $E^{3}=E$, and so $A=E+0+W$.

Case III. $c_{n-1}=2$. Choose

[TABLE]

Then $E_{1}^{3}=E_{1}$ and $E_{2}^{3}=E_{2}$, and so $A=E_{1}+E_{2}+W$. ∎

Lemma 2.2**.**

Every square matrix over ${\mathbb{Z}}_{5}$ is the sum of two tripotents and a nilpotent.

Proof.

As every matrix over a field has a Frobenius normal form, we may assume that A=\left(\begin{array}[]{cccccc}0&&&&&c_{0}\\ 1&0&&&&c_{1}\\ &1&0&&&c_{2}\\ &&&\ddots&&\vdots\\ &&&\ddots&0&c_{n-2}\\ &&&&1&c_{n-1}\end{array}\right).

Case I. $c_{n-1}=0$. Choose

[TABLE]

Then $E_{1}^{3}=E_{1}$ and $E_{2}^{3}=E_{2}$, and so $A=E_{1}+E_{2}+W$.

Case II. $c_{n-1}=1$. Choose

[TABLE]

Then $E^{3}=E$, and so $A=E+0+W$.

Case III. $c_{n-1}=-1$. Choose

[TABLE]

Then $E^{2}=-E$, and so $E^{3}=E$ and $A=E+0+W$.

Case IV. $c_{n-1}=2$. Choose

[TABLE]

Then $E_{1}^{3}=E_{1}$ and $E_{2}^{3}=E_{2}$, and so $A=E_{1}+E_{2}+W$.

Case IV. $c_{n-1}=-2$. Choose

[TABLE]

Then $E_{1}^{3}=E_{1}$ and $E_{2}^{3}=E_{2}$, and so $A=E_{1}+E_{2}+W$.

Therefore we complete the proof.∎

Recall that a ring $R$ is a Yaqub ring if it is the subdirect product of ${\mathbb{Z}}_{3}$’s. A ring $R$ is a Bell ring if it is the subdirect product of ${\mathbb{Z}}_{5}$’s. We have

Lemma 2.3**.**

Every Zhou ring is isomorphic to a strongly nil-clean ring of bounded index, a Yaqub ring, a Bell ring or products of such rings.

Proof.

Let $R$ be a Zhou ring. In view of [10, Theorem 5.2], $R$ is a ring $R_{1}$, a Yaqub ring $R_{2}$, a Bell ring $R_{3}$ or products of such rings. Here, $R_{1}/J(R_{1})$ is Boolean and $U(R_{1})$ is a group of exponent $2$. Let $x\in J(R_{1})$. By the proof of  [10, Theorem 5.2], $x^{6}=x^{4}$, and so $x^{4}=0$. Thus, $J(R)^{4}=0$. It follows by  [9, Theorem 2.7], $R_{1}$ is a strongly nil-clean ring of bounded index$4$.∎

Theorem 2.4**.**

Let $R$ be a Zhou ring. Then every square matrix over $R$ is the sum of two tripotents and a nilpotent.

Proof.

In view of Lemma 2.3, $R$ is isomorphic to $R_{1},R_{2},R_{3}$ or the products of these rings, where $R_{1}$ is a strongly nil-clean ring of bounded index, $R_{2}$ is a Yaqub ring and $R_{3}$ is a Bell ring.

Step 1. Let $A\in M_{n}(R_{1})$. In view of [9, Corollary 6.8], there exist an idempotent $E\in M_{n}(R_{1})$ and $W\in N(M_{n}(R_{1}))$ such that $A=E+W$.

Step 2. Let $A\in M_{n}(R_{2})$, and let $S$ be the subring of $R_{2}$ generated by the entries of $A$. That is, $S$ is formed by finite sums of monomials of the form: $a_{1}a_{2}\cdots a_{m}$, where $a_{1},\cdots,a_{m}$ are entries of $A$. Since $R_{2}$ is a commutative ring in which $3=0$, $S$ is a finite ring in which $x=x^{3}$ for all $x\in S$. Thus, $S$ is isomorphic to finite direct product of ${\mathbb{Z}}_{3}$. As $A\in M_{n}(S)$, it follows by Lemma 2.1 that $A$ is the sum of two tripotents and a nilpotent matrix over $S$.

Step 3. Let $A\in M_{n}(R_{3})$, and let $S$ be the subring of $R_{3}$ generated by the entries of $A$. Analogously, $S$ is isomorphic to finite direct product of ${\mathbb{Z}}_{5}$. As $A\in M_{n}(S)$, it follows by Lemma 2.2 that $A$ is the sum of two tripotents and a nilpotent matrix over $S$.

Let $A\in M_{n}(R)$. We may write $A=(A_{1},A_{2},A_{3})$ in $M_{n}(R_{1})\times M_{n}(R_{2})\times M_{n}(R_{3})$, where $A_{1}\in M_{n}(R_{1}),A_{2}\in M_{n}(R_{2}),A_{3}\in M_{n}(R_{3})$. According to the preceding discussion, we easily complete the proof.∎

Corollary 2.5**.**

Let $R$ in which for any $x\in R$, $x^{5}=x$. Then every square matrix over $R$ is the sum of two tripotents and a nilpotent.

Proof.

In view of  [12, Theorem 2.11], every element in $R$ is the sum of two tripotents and a nilpotent. But $N(R)=0$, and so $R$ is a Zhou ring. This completes the proof, by Theorem 2.4.∎

3. Zhou nil-clean Rings

The goal of this section is to explore matrix decompositions for Zhou nil-clean rings. The following lemma is crucial.

Lemma 3.1**.**

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

1. (1)

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

$R/J(R)$* is a Zhou ring and $J(R)$ is nil.*

Proof.

$(1)\Rightarrow(2)$ In view of [12, Theorem 2.11], $J(R)$ is nil and $R/J(R)$ has the identity $x=x^{5}$. By using [12, Theorem 2.11] again, every element in $R/J(R)$ is the sum of two tripotents and a nilpotent $\overline{w}$. Clearly, $R/J(R)$ is reduced, and so $\overline{w}=\overline{0}$. Thus, $R/J(R)$ is a Zhou ring.

$(2)\Rightarrow(1)$ Let $a\in R$. As $S:=R/J(R)$ is a Zhou ring, it is Zhou nil-clean. It follows by [12, Theorem 2.11] that $\overline{a-a^{5}}\in N(R/J(R))$. As $J(R)$ is nil, we see that $a-a^{5}\in N(R)$. According to [12, Theorem 2.11], $R$ is Zhou nil-clean, as asserted.∎

We are ready to prove the following.

Theorem 3.2**.**

Let $R$ be a 2-primal Zhou nil-clean ring. Then every square matrix over $R$ is the sum of two tripotents and a nilpotent.

Proof.

In view of [12, Theorem 2.11], $R\cong S,T$ or $S\times T$, where $S$ is a 2-primal strongly nil-clean ring, and $T$ is a Zhou ring with $\frac{1}{2}\in T$.

Step 1. Let $A\in M_{n}(S)$. In view of  [9, Theorem 6.1], $A$ is the sum of an idempotent and a nilpotent.

Step 2. Let $A\in M_{n}(T)$. In view of [12, Theorem 2.11], $T/J(T)$ is commutative; and so $N(T)\subseteq J(T)$. Thus, $J(T)=N(T)$. Then $\overline{A}\in M_{n}(T/J(T))$. By virtue of Theorem 2.4, there exist tripotents $\overline{C},\overline{D}\in M_{n}(T/J(T))$ and a nilpotent $\overline{W}\in M_{n}(T/J(T))$ such that $\overline{B}=\overline{C+D+W}$. Clearly, $M_{n}(R/J(R))\cong M_{n}(R)/J(M_{n}(R))$. As $R$ is 2-primal, it follows by [7, Theorem 10.21] that $J(M_{n}(T))=M_{n}(J(T))=M_{n}(N(T))=M_{n}(N_{*}(T))=N_{*}(M_{n}(T))$ is nil, where $N_{*}(T)$ is the Baer-McCoy radical of $T$. As $\frac{1}{2}\in M_{n}(T)$, by [12, Lemma 2.6], we may assume that $B,C\in M_{n}(T)$ are tripotents. Clearly, $W\in M_{n}(T)$ is nil. Thus, we can find $V\in J(M_{n}(T))$ such that $A=B+C+(W+V)$. Write $W^{k}=0$. Then $(W+V)^{k}\in J(M_{n}(T))$, and so $(W+V)^{kl}=0$ for some $l\in{\mathbb{N}}$. Therefore $A$ is the sum of two tripotents and a nilpotent.

Let $A\in M_{n}(R)$. We may write $A=(A_{1},A_{2})$ in $M_{n}(S)\times M_{n}(T)$, where $A_{1}\in M_{n}(S),A_{2}\in M_{n}(T)$. Thus, by the preceding discussion, we obtain the result.∎

Corollary 3.3**.**

Let $R$ be a commutative Zhou nil-clean ring. Then every square matrix over $R$ is the sum of two tripotents and a nilpotent.

Proof.

This is obvious by Theorem 3.2, as every commutative ring is 2-primal.∎

Corollary 3.4**.**

Let $R$ be a Zhou nil-clean ring, and let $A\in M_{n}(R)$ with central entries. Then $A$ is the sum of two tripotents and a nilpotent.

Proof.

Let $C(R)$ be the center of $R$, and let $A\in C(R)$. In view of [12, Theorem 2.11], $a-a^{2}\in N(R)$; and so $a-a^{2}\in U(C(R))$. By using [12, Theorem 2.11] again, $C(R)$ is a commutative Zhou nil-clean. Since $A\in M_{n}(C(R))$, By Corollary 3.3, $A$ is the sum of two tripotents and a nilpotent in $M_{n}(C(R))$. This completes the proof from $M_{n}(C(R))\subseteq M_{n}(R)$.∎

Example 3.5**.**

Let $n\geq 2$ be an integer. If $n=2^{k}3^{l}5^{m}$, then every square matrix over ${\mathbb{Z}}_{n}$ is the sum of two tripotents and a nilpotent.

Proof.

As $n=2^{k}3^{l}5^{m}$. It follows 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 $R_{1}/J(R_{1})$ is a Boolean ring, $R_{2}/J(R_{2})$ is a Yaqub ring and $R_{3}/J(R_{3})$ is a Bell ring. It follows by [12, Theorem 2.11], $R$ is a commutative Zhou nil-clean ring. Therefore we complete the proof by Corollary 3.3.∎

Lemma 3.6**.**

(see [9, Lemma 6.6]). Let $R$ be of bounded index. If $J(R)$ is nil, then $J(M_{n}(R))$ is nil for all $n\in{\mathbb{N}}$.

Theorem 3.7**.**

Let $R$ be a Zhou nil-clean ring of bounded index. Then every square matrix over $R$ is the sum of two tripotents and a nilpotent.

Proof.

As in the proof of Theorem 3.3, $R=S\times T$, where $S$ is strongly nil-clean of bounded index and $T$ is a Zhou nil-clean ring with $\frac{1}{2}\in S$.

Step 1. Let $A\in M_{n}(S)$. In view of [9, Corollary 6.8], $A$ is the sum of an idempotent and a nilpotent. Hence, it is the sum of two tripotents and a nilpotent.

Step 2. Let $A\in M_{n}(T)$. Then $\overline{A}\in M_{n}(T/J(T))$. In light of Lemma 3.1, $T/J(T)$ is a Zhou ring. By virtue of Theorem 2.4, there exist tripotents $\overline{C},\overline{D}\in M_{n}(T/J(T))$ and a nilpotent $\overline{W}\in M_{n}(T/J(T))$ such that $\overline{A}=\overline{C+D+W}$. Clearly, $M_{n}(T/J(T))\cong M_{n}(T)/J(M_{n}(T))$. In light of Lemma 3.6, $J(M_{n}(T))$ is nil. Since $\frac{1}{2}\in M_{n}(T)$, it follows by [12, Lemma 2.6] that we may assume that $C,D\in M_{n}(R)$ are tripotents. Clearly, $W\in M_{n}(R)$ is nil. Thus, we can find $V\in J(M_{n}(R))$ such that $A=B+C+(W+V)$. Write $W^{k}=0$. Then $(W+V)^{kl}=0$ for some $l\in{\mathbb{N}}$. This shows that $A$ is the sum of two tripotents and a nilpotent.

Let $A\in M_{n}(R)$. Then $A=(A_{1},A_{2})$ with $A_{1}\in M_{n}(S)$ and $A_{2}\in M_{n}(T)$. By the discussion in Step 1 and Step 2, we complete the proof.∎

Example 3.8**.**

Let $n\geq 2$ be an integer, and let $S=T_{s}({\mathbb{Z}}_{n})(s\in{\mathbb{N}})$. If $n=2^{k}3^{l}5^{m}$, then every square matrix over $S$ is the sum of two tripotents and a nilpotent.

Proof.

As in the proof of Example 3.5, ${\mathbb{Z}}_{n}$ is a Zhou nil-clean ring. Let $A\in S$. Then $A-A^{2}\in N(S)$, as its diagonal entries are nilpotent. It follows by [12, Theorem 2.11], $S$ is a Zhou nil-clean ring. Clearly, $S$ is of bounded index. Therefore we complete the proof, by Theorem 3.7.∎