Formal triangular matrix ring with nil clean index $4$

TL;DR

This paper characterizes formal triangular matrix rings with a nil clean index of 4, providing insights into their algebraic structure and properties related to idempotents and nilpotent elements.

Contribution

It offers a complete characterization of formal triangular matrix rings with nil clean index 4, advancing understanding of their algebraic structure.

Findings

Characterization of formal triangular matrix rings with nil clean index 4

Analysis of idempotent elements in these rings

Insights into the structure of nilpotent elements in the rings

Abstract

For an element $a \in R$, let $\eta(a)=\{e\in R\mid e^2=e\mbox{ and }a-e\in \mbox{nil}(R)\}$. The nil clean index of $R$, denoted by NinA$(R)$, is defined as Nin$(R)=\sup \{\mid \eta(a)\mid: a\in R\}$. In this article we have characterized formal triangular matrix ring $\begin{pmatrix}A & M0 & B\end{pmatrix}$ with nil clean index $4$.

Equations

$$\eta(\alpha)=\left\{\left(\begin{array}[]{cc}e&w\\ 0&f\\ \end{array}\right)\in R:~{}e\in\eta(a),~{}f\in\eta(b),~{}w=ew+fw\right\}.$$

$$\eta(\alpha)=\left\{\left(\begin{array}[]{cc}e&z\\ 0&f_{0}\\ \end{array}\right)\in R:e\in\eta(a),z=ez+zf_{0}\right\}$$

$$T_{i}=\left\{\left(\begin{array}[]{cc}e_{i}&z\\ 0&f_{0}\\ \end{array}\right)\in R:(1_{A}-e_{i})z=zf_{0}\right\}\hskip 28.45274pt(i=1,~{}2).$$

$$\beta =$$

$$=$$

$$\alpha =$$

$$=$$

$$\eta(\gamma)\supseteq\left\{\left(\begin{array}[]{cc}1_{A}-e&0\\ 0&f\\ \end{array}\right),\left(\begin{array}[]{cc}1_{A}-(e+j)&0\\ 0&f\\ \end{array}\right),\left(\begin{array}[]{cc}1_{A}-e&w\\ 0&f\\ \end{array}\right):w\in M\right\}$$

Full text

**Formal triangular matrix ring with nil clean index $4$

**

Dhiren Kumar Basnet

*Department of Mathematical Sciences, Tezpur University,

Napaam, Tezpur-784028, Assam, India.

Email: dbasnet@tezu.ernet.in*

Jayanta Bhattacharyya

*Department of Mathematical Sciences, Tezpur University,

Napaam, Tezpur-784028, Assam, India.

Email: jbhatta@tezu.ernet.in*

**Abstract: ** For an element $a\in R$, let $\eta(a)=\{e\in R\mid e^{2}=e\mbox{ and }a-e\in\mbox{nil}(R)\}$. The nil clean index of $R$, denoted by $\operatorname{Nin}(R)$, is defined as $\operatorname{Nin}(R)=\sup\{\mid\eta(a)\mid:a\in R\}$. In this article we have characterized formal triangular matrix ring $\left(\begin{smallmatrix}A&M\\ 0&B\end{smallmatrix}\right)$ with nil clean index $4$.

Key words: Nil clean ring, nil clean index.

$2010$** Mathematics Subject Classification:** 16U99

1 Introduction

Throughout this article $R$ denotes a associative ring with unity. The set of nilpotents and set of idempotents are denoted by $\operatorname{nil}(R)$ and $\operatorname{idem}(R)$ respectively. The cyclic group of order $n$ is denoted by $C_{n}$ and $|S|$ denotes the cardinality of the set $S$. For an element $a\in R$, if $a-e\in\operatorname{nil}(R)$ for some $e\in\operatorname{idem}(R)$, then $a=e+(a-e)$ is said to be a nil clean expression of $a$ in $R$ and $a$ is called a nil clean element[3, 2]. The ring R is called nil clean if each of its elements is nil clean.

For an element $a\in R$, let $\eta(a)=\{e\in R\mid e^{2}=e\mbox{ and }a-e\in\mbox{nil}(R)\}$. The nil clean index of $R$, denoted by $\operatorname{Nin}(R)$, is defined as $\operatorname{Nin}(R)=\sup\{\mid\eta(a)\mid:a\in R\}$[1]. Characterization of arbitrary ring with nil clean indices $1$, $2$ and few sufficient condition for a ring to be of nil clean index $3$ is given in [1]. In this article we have characterized formal triangular matrix ring $\left(\begin{smallmatrix}A&M\\ 0&B\end{smallmatrix}\right)$ with nil clean index $4$, where $A$ and $B$ are rings and $M$ is a $A-B-$bimodule. Following results about nil clean index will be used in this article.

Lemma 1.1**.**

([$1$, Lemma $2.5$]) Let R=\left(\begin{array}[]{cc}A&M\\ 0&B\\ \end{array}\right), where $A$ and $B$ are rings, ${}_{A}M_{B}$ is a bimodule. Let $\mbox{Nin}(A)=n$ and $\mbox{Nin}(B)=m$. Then

1. (i)

$\mbox{Nin}(R)\geq|M|.$** 2. (ii)

If $(M,+)\cong C_{p^{k}}$, where $p$ is a prime and $k\geq 1$, then $\mbox{Nin}(R)\geq n+[\frac{n}{2})(|M|-1)$, where $[\frac{n}{2})$ denotes the least integer greater than or equal to $\frac{n}{2}$. 3. (iii)

Either $\mbox{Nin}(R)\geq nm+|M|-1$ or $\mbox{Nin}(R)\geq 2nm$.

Lemma 1.2**.**

([$1$, Lemma $2.6$]) Let R=\left(\begin{array}[]{cc}A&M\\ 0&B\\ \end{array}\right), where $A$ and $B$ are rings, ${}_{A}M_{B}$ is a bimodule with $(M,+)\cong C_{2^{r}}$. Then $\mbox{Nin}(R)=2^{r}\mbox{Nin}(A)\mbox{Nin}(B)$.

Theorem 1.3**.**

([$1$, Theorem $4.1$]) $\mbox{Nin}(R)=2$ if and only if R=\left(\begin{array}[]{cc}A&M\\ 0&B\\ \end{array}\right), where $\mbox{Nin}(A)=\mbox{Nin}(B)=1$ and ${}_{A}M_{B}$ is a bimodule with $|M|=2.$

Theorem 1.4**.**

([$1$, Proposition $4.2$]) If R=\left(\begin{array}[]{cc}A&M\\ 0&B\\ \end{array}\right), where $\mbox{Nin}(A)=\mbox{Nin}(B)=1$ and ${}_{A}M_{B}$ is a bimodule with $|M|=3$ then $\mbox{Nin}(R)=3.$

2 Main result

Theorem 2.1**.**

Let R=\left(\begin{array}[]{cc}A&M\\ 0&B\\ \end{array}\right), where $A$ and $B$ are rings, ${}_{A}M_{B}$ is a non trivial bimodule. Then $\operatorname{Nin}(R)=4$ if and only if one of the following holds:

1. $(1)$

$(M,+)\cong C_{2}$* and $\operatorname{Nin}(A)\operatorname{Nin}(B)=2$.* 2. $(2)$

$(M,+)\cong C_{4}$* and $\operatorname{Nin}(A)=\operatorname{Nin}(B)=1$.* 3. $(3)$

$(M,+)\cong C_{2}\oplus C_{2}$* plus one of the following*

1. $(a)$

$\operatorname{Nin}(A)=\operatorname{Nin}(B)=1$. 2. $(b)$

\operatorname{Nin}(A)=1,~{}~{}B=\left(\begin{array}[]{cc}S&W\\ 0&T\\ \end{array}\right), where $\operatorname{Nin}(S)=\operatorname{Nin}(T)=1$ and $|W|=2$, and $eM(1_{B}-f)+(1_{A}-e)Mf\neq 0$ for all $e^{2}=e\in A$ and $f\in\eta(b)$, where $b\in B$ with $|\eta(b)|=2$. 3. $(c)$

\operatorname{Nin}(B)=1,~{}~{}A=\left(\begin{array}[]{cc}S&W\\ 0&T\\ \end{array}\right), where $\operatorname{Nin}(S)=\operatorname{Nin}(T)=1$ and $|W|=2$, and $eM(1_{B}-f)+(1_{A}-e)Mf\neq 0$ for all $e^{2}=e\in B$ and $f\in\eta(a)$, where $a\in A$ with $|\eta(a)|=2$.

$Proof:~{}~{}(\Leftarrow)$ If $(1)$ holds then by Lemma 1.2, we get $\operatorname{Nin}(R)=4$.

If $(2)$ holds then $\operatorname{Nin}(R)\geq|M|=4$. Now, for any \alpha=\left(\begin{array}[]{cc}a&x\\ 0&b\\ \end{array}\right)\in R,

[TABLE]

Because $|M|=4,~{}|\eta(a)|\leq 1$ and $|\eta(b)|\leq 1$, it follows that $|\eta(\alpha)|\leq 4$. Hence $\operatorname{Nin}(R)=4$.

Let $(3)~{}(a)$ holds, then $\operatorname{Nin}(R)\geq|M|=4$. Now, for any \alpha=\left(\begin{array}[]{cc}a&x\\ 0&b\\ \end{array}\right)\in R,

[TABLE]

Because $|M|=4,~{}|\eta(a)|\leq 1$ and $|\eta(b)|\leq 1$, it follows that $|\eta(\alpha)|\leq 4$. Hence $\operatorname{Nin}(R)=4$.

Suppose $(3)(c)$ holds, then clearly $\operatorname{Nin}(R)\geq|M|=4.$ Let \alpha=\left(\begin{array}[]{cc}a&w\\ 0&b\\ \end{array}\right)\in R. We show that $|\eta(\alpha)|\leq 4$ and hence $\operatorname{Nin}(R)=4$ holds. Since $\operatorname{Nin}(B)=1$, we can assume that $\eta(b)=\{f_{0}\}$. Then as above we have

[TABLE]

If $|\eta(a)|\leq 1$, then $|\eta(\alpha)|\leq|\eta(a)|.|M|\leq 4$. So we can assume that $|\eta(a)|=2$. Write $\eta(a)=\{e_{1},~{}e_{2}\}$. Thus $\eta(\alpha)=T_{1}\bigcup T_{2}$, where

[TABLE]

Since $\eta(1_{A}-a)=\{1_{A}-e_{1},~{}1_{A}-e_{2}\}$, the assumption $(3)(c)$ shows that $\{z\in M:(1_{A}-e_{i})z=zf_{0}\}$ is a proper subgroup of $(M,+)$; so $|T_{i}|\leq 2$ for $i=1,~{}2$. Hence $|\eta(\alpha)|\leq|T_{1}|+|T_{2}|\leq 4$

$(\Rightarrow)$ Suppose $\operatorname{Nin}(R)=4$. Then $2\leq|M|\leq\operatorname{Nin}(R)=4$. If $|M|=2$ then $\operatorname{Nin}(A)\operatorname{Nin}(B)=2$ by Lemma 1.2, so $(1)$ holds.

Suppose $|M|=3$, then we have by Lemma 1.1, $\operatorname{Nin}(A)+|M|\leq\operatorname{Nin}(R)$, showing $\operatorname{Nin}(A)\leq 2$, similarly $\operatorname{Nin}(B)\leq 2$. But $\operatorname{Nin}(A)=2=\operatorname{Nin}(B)$ will give $\operatorname{Nin}(R)\geq 6$ by Lemma 1.1 and $\operatorname{Nin}(A)=\operatorname{Nin}(B)=1$ will give $\operatorname{Nin}(R)=3$ by Theorem 1.4. Hence the only possibility is $\operatorname{Nin}(A)\operatorname{Nin}(B)=2$, so without loss of generality we assume that $\operatorname{Nin}(A)=2$ and $\operatorname{Nin}(B)=1$. Write $M=\{0,x,2x\}$. Now by Theorem 1.3, we have A=\left(\begin{array}[]{cc}T&N\\ 0&S\\ \end{array}\right), where $T$ & $S$ are rings, ${}_{T}N_{S}$ is bimodule with $\operatorname{Nin}(T)=\operatorname{Nin}(S)=1$ and $|N|=2$. Note that for $e\in\operatorname{idem}(A),~{}ex\in\{0,~{}x\}$, for if $ex=2x$, we have $2x=ex=e(ex)=e(2x)=e(x+x)=ex+ex=2x+2x=4x=x$ which is not true.

Now Let a=\left(\begin{array}[]{cc}1_{T}&0\\ 0&0\\ \end{array}\right)\in A such that

a=\left(\begin{array}[]{cc}1_{T}&0\\ 0&0\\ \end{array}\right)+\left(\begin{array}[]{cc}0&0\\ 0&0\\ \end{array}\right)=\left(\begin{array}[]{cc}1_{T}&y\\ 0&0\\ \end{array}\right)+\left(\begin{array}[]{cc}0&-y\\ 0&0\\ \end{array}\right).

Let us denote, e_{1}=\left(\begin{array}[]{cc}1_{T}&0\\ 0&0\\ \end{array}\right),~{}e_{2}=\left(\begin{array}[]{cc}1_{T}&y\\ 0&0\\ \end{array}\right),~{}n_{1}=\left(\begin{array}[]{cc}0&0\\ 0&0\\ \end{array}\right) and

n_{2}=\left(\begin{array}[]{cc}0&-y\\ 0&0\\ \end{array}\right), where $e_{1},~{}e_{2}\in\operatorname{idem}(A)~{}\&~{}n_{1},~{}n_{2}\in\operatorname{nil}(R)$. Now we have following cases:

Case I: Let $e_{1}x=e_{2}x=0$, then we have an element

\beta=\left(\begin{array}[]{cc}(1_{A}-a)&0\\ 0&0\\ \end{array}\right)\in R such that

[TABLE]

are six nil clean expressions for $\beta,$ which implies $|\eta(\beta)|\geq 6$, that is $\operatorname{Nin}(R)\geq 6$, which is not possible.

Case II: Let $e_{1}x=e_{2}x=x$, then we have an element

\alpha=\left(\begin{array}[]{cc}a&0\\ 0&0\\ \end{array}\right)\in R such that

[TABLE]

are six nil clean expressions for $\alpha,$ which implies $|\eta(\alpha)|\geq 6$, that is $\operatorname{Nin}(R)\geq 6$, which is also not possible.

Case III: Let $e_{1}x=x$ and $e_{2}x=0$, then we have $(e_{1}-e_{2})x=x$.

Let $j=e_{1}-e_{2}$, then clearly $j\in\operatorname{nil}(A)$ and we have

$jx=x~{}~{}\Rightarrow~{}~{}(1_{A}-j)x=0~{}~{}\Rightarrow~{}~{}x=0,~{}~{}($ as $(1-j)\in\operatorname{U}(A))$.

Which is not possible.

Case IV: Let $e_{1}x=0$ and $e_{2}x=x$, as in case III, we get a contradiction.

Hence if $M\cong C_{3},~{}~{}\operatorname{Nin}(R)$ is never 3.

Suppose $|M|=4$. If $(M,+)\cong C_{4}$, then $\operatorname{Nin}(A)\operatorname{Nin}(B)=1$ by

Lemma 1.2, So $(2)$ holds. Let $(M,+)\cong C_{2}\oplus C_{2}$. Since $\operatorname{Nin}(R)=4$, by Lemma 1.1, we have $\operatorname{Nin}(A)\operatorname{Nin}(B)\leq 2$. If $\operatorname{Nin}(A)\operatorname{Nin}(B)=1$ then $(3)(a)$ holds. If $\operatorname{Nin}(A)\operatorname{Nin}(B)=2$, without loss of generality we can assume $\operatorname{Nin}(A)=2$ and $\operatorname{Nin}(B)=1$. So by Theorem 1.3, we have A=\left(\begin{array}[]{cc}S&W\\ 0&T\\ \end{array}\right) where $\operatorname{Nin}(S)=\operatorname{Nin}(T)=1$, and $|W|=2$. To complete the proof suppose in contrary that $eM(1_{B}-f)+(1_{A}-e)Mf=0$ for some $f^{2}=f\in B$ and $e\in\eta(a)$, where $a\in A$ with $|\eta(a)|=2.$ Then $ew=wf$ for all $w\in M$. It is easy to check that $\eta(a)=\{e,~{}e+j\}$ where j=\left(\begin{array}[]{cc}0&w_{0}\\ 0&0\\ \end{array}\right)\in A with $0\neq w_{0}\in W$. Thus, for \gamma:=\left(\begin{array}[]{cc}1_{A}-e&0\\ 0&f\\ \end{array}\right),

[TABLE]

So $|\eta(\gamma)|\geq 5$, a contradiction. Hence $(3)(c)$ holds, similarly $(3)(b)$ can be proved$\hfill\Box$