Nil clean graph of rings

TL;DR

This paper introduces the nil clean graph of a ring, exploring its properties such as girth, dominating set, and diameter specifically for finite commutative rings, linking algebraic and graph-theoretic concepts.

Contribution

It defines the nil clean graph of a ring and investigates its fundamental graph properties within the context of finite commutative rings.

Findings

Analyzed girth, diameter, and dominating sets of nil clean graphs.

Established relationships between ring properties and graph structure.

Provided new insights into the interplay between ring theory and graph theory.

Abstract

In this article, we have defined nil clean graph of a ring $R$. The vertex set is the ring $R$, two ring elements $a$ and $b$ are adjacent if and only if $a + b$ is nil clean in $R$. Graph theoretic properties like girth, dominating set, diameter etc. of nil clean graph have been studied for finite commutative rings.

Equations

$$GF(25)$$

$$=\{ax+b+⟨x^2+x+1⟩:a,b\in {\mathbb{Z}}_5\}$$

$$(x+\mathrm{Nil}(R))+(y+\mathrm{Nil}(R))=e+\mathrm{Nil}(R),$$

Full text

**NIL CLEAN GRAPH OF RINGS

**

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: ** In this article, we have defined nil clean graph of a ring $R$. The vertex set is the ring $R$, two ring elements $a$ and $b$ are adjacent if and only if $a+b$ is nil clean in $R$. Graph theoretic properties like girth, dominating set, diameter etc. of nil clean graph have been studied for finite commutative rings.

1 Introduction

In past, lot of mathematicians have studied graph theoretic properties of graph associated with rings. By graph, in this article we mean simple undirected graph. If $G$ denotes a graph, let $V(G)$ denotes the set of vertices and $E(G)$ denotes the set of edges. In $1988$, Istavan Beck [3], studied colouring of graph of a finite commutative ring $R$, where vertices are elements of $R$ and $xy$ is an edge if and only if $xy=0$. For a positive integer $n$, let $\mathbb{Z}_{n}$ be the ring of integers modulo $n$. P. Grimaldi[9] defined and studied various properties of the unit graph $G(\mathbb{Z}_{n})$, with vertex set $\mathbb{Z}_{n}$ and two distinct vertices $x$ and $y$ are adjacent if $x+y$ is a unit. Further in [1], authors generalized $G(\mathbb{Z}_{n})$ to unit graph $G(R)$, where $R$ is an arbitrary associative ring with non zero identity. They have studied properties like diameter, girth chromatic index etc.

In this paper we have introduced nil clean graph $G_{N}(R)$ associated with a finite commutative ring $R$. The properties like girth, diameter, dominating sets etc. of $G_{N}(R)$ have been studied.

Ring $R$ in this article, is finite commutative ring with non zero identity. An element $r$ of ring $R$ is said to be nil clean element[7, 6] if for an idempotent $e$ in $R$ and a nilpotent $n$ in $R$, $r=n+e$. The ring $R$ is said to be nil clean if each element of $R$ is nil clean. The set of nil clean elements of $R$ is denoted by $NC(R)$. The sets of idempotents and nilpotents of $R$ are denoted by $\operatorname{Idem}(R)$ and $\operatorname{Nil}(R)$ respectively. A ring $R$ is said to be a weak nil clean ring[5], if for each $r\in R$, $r=n+e$ or $r=n-e$, for some $n\in\operatorname{Nil}(R)$ and $e\in\operatorname{Idem}(R)$.

2 Basic properties

In this section we will define nil clean graph of a finite commutative ring and discuss its basic properties.

Definition 2.1**.**

The nil clean graph of a ring $R$, denoted by $G_{N}(R)$, is defined by setting $R$ as vertex set and defining two distinct verities $x$ and $y$ to be adjacent if and only if $x+y$ is a nil clean element in $R$. Here we are not considering loops at a point (vertex) in the graph.

For illustration following is the nil clean graph of $GF(25)$, where $GF(25)$ is the finite field with $25$ elements.

[TABLE]

let us define $\alpha:=x+\langle x^{2}+x+1\rangle$, then we have $GF(25)=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\alpha,2\alpha,3\alpha,4\alpha,1+\alpha,1+2\alpha,1+3\alpha,1+4\alpha,2+\alpha,2+2\alpha,2+3\alpha,2+4\alpha,3+\alpha,3+2\alpha,3+3\alpha,3+4\alpha,4+\alpha,4+2\alpha,4+3\alpha,4+4\alpha\}$. Observe that $NC(GF(25))=\{\overline{0},\overline{1}\}$.

In graph theory, a complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. So by this definition the following theorem follows.

Theorem 2.2**.**

The nil clean graph $N_{G}(R)$ is a complete graph if and only if $R$ is a nil clean ring.

$Proof.$ Let $N_{G}(R)$ be complete nil clean graph of a ring $R$. For $r\in R$, $r$ is adjacent to [math] implies $r=r+0$ is nil clean, hence $R$ is nil clean. Converse is clear from definition of nil clean graph.$\hfill\square$

Two graphs $G_{1}$ and $G_{2}$ are said to be isomorphic if there exists an isomorphism from $G_{1}$ to $G_{2}$ we mean a bijective mapping $f:V(G_{1})\rightarrow V(G_{2})$ from $V(G_{1})$ onto $V(G_{2})$ such that two vertices $u_{1}$ and $v_{1}$ are adjacent in $G_{1}$ if and only if the vertices φ$f(u_{1})$ and φ$f(v_{1})$ are adjacent in $G_{2}$ [4]. For rings $R$ and $S$ if $R\cong S$ it is easy to see that $G_{N}(R)\cong G_{N}(S)$.

Lemma 2.3**.**

Let $R$ be a commutative ring and idempotents lift modulo $\operatorname{Nil}(R)$. If $x+\operatorname{Nil}(R)$ and $y+\operatorname{Nil}(R)$ are adjacent in $G_{N}(R/\operatorname{Nil}(R))$ then every element of $x+\operatorname{Nil}(R)$ is adjacent to every element of $y+\operatorname{Nil}(R)$ in the nil clean graph $G_{N}(R)$.

$Proof.$ Let $x+\operatorname{Nil}(R)$ and $y+\operatorname{Nil}(R)$ be adjacent in $G_{N}(R/\operatorname{Nil}(R))$, then

[TABLE]

where $e$ is an idempotent in $R$, as idempotents lift modulo $\operatorname{Nil}(R)$. Thus from above we have $x+y=e+n$, for some $n\in\operatorname{Nil}(R)$ and hence $x$ and $y$ are adjacent in $G_{N}(R)$. Now for $a\in x+\operatorname{Nil}(R)$ and $b\in y+\operatorname{Nil}(R)$, we have $a=x+n_{1}$ and $b=y+n_{2}$, for some $n_{1}$, $n_{2}\in\operatorname{Nil}(R)$. Therefore $a+b=e+(n-n_{1}-n_{2})$. Hence, $a$ and $b$ are adjacent in $G_{N}(R)$.$\hfill\square$

Let $G$ be a graph, for $x\in V(G)$, the degree of $x$ denoted by $deg(x)$, is defined to be number of edges of $G$ whose one of the end point is $x$. The neighbour set of $x\in V(G)$, is denoted and defined as by $N_{G}(x):=\{y\in V(G)|y\mbox{ is adjacent to }x\}$ and the set $N_{G}[x]=N_{G}(x)\cup\{x\}$.

Lemma 2.4**.**

Let $G_{N}(R)$ be the nil clean graph of a ring $R$. For $x\in R$ we have the following:

1. (i)

If $2x$ is nil clean, then $deg(x)=|NC(R)|-1$. 2. (ii)

If $2x$ is not nil clean, then $deg(x)=|NC(R)|$.

$Proof.$ Let $x\in R$, observe that $x+R=R$, so for every $y\in NC(R)$, there exists a unique element $x_{y}\in R$, such that $x+x_{y}=y$. Thus we have $deg(x)\leq|NC(R)|$. Now if $2x\in NC(R)$, define $f:NC(R)\rightarrow N_{G_{N}(R)}[x]$ given by $f(y)=x_{y}$. It is easy to see that $f$ is a bijection and therefore $deg(x)=|N_{G_{N}(R)}(x)|=|N_{G_{N}(R)}[x]|-1=|NC(R)|-1$. If $2x\notin NC(R)$, define $f:NC(R)\rightarrow N_{G_{N}(R)}(x)$ given by $f(y)=x_{y}$. Then $f$ is a bijection and therefore $deg(x)=|N_{G_{N}(R)}(x)|=|NC(R)|$.$\hfill\square$

A graph $G$ is said to be connected if for any two distinct vertices of $G$, there is a path in $G$ connecting them.

Theorem 2.5**.**

For a ring $R$. The following hold :

1. (i)

$G_{N}(R)$* need not be connected.* 2. (ii)

Let $R=\mathbb{Z}_{n}$. For $\overline{a}\in\mathbb{Z}_{n}$ there is a path from $\overline{a}$ to $\overline{0}$. 3. (iii)

$G_{N}(\mathbb{Z}_{n})$* is connected.* 4. (iv)

Let $R=\mathbb{Z}_{n}$. For $A\in M_{n}(\mathbb{Z}_{n})$ there is a path from $A$ to [math], where [math] is the zero matrix of $M_{n}(\mathbb{Z}_{n})$. 5. (v)

$G_{N}(M_{n}(\mathbb{Z}_{n}))$* is connected.*

$Proof.$ $(i)$ Clear by the graph $G_{N}(GF(25))$, figure 1. For $(ii)$ and $(iii)$ replacing $p$ by $n$ in the figure 4, we get a Hamiltonian path in $G_{N}(\mathbb{Z}_{n})$. Now for the proof of $(iv)$, let $A=[a_{ij}]\in M_{n}(Z_{n})$, now we define $A_{1}=[a1_{ij}]$ where $a1_{ij}=-a_{ij}$ for $i\geq j$, otherwise $a1_{ij}=0$, observe that $A_{1}+A$ is nilpotent hence nil clean. Thus there exists an edge between $A$ and $A_{1}$. Again define $A_{2}=[a2_{ij}]$ where $a2_{ij}=a_{ij}$ for $i=j$, otherwise $a1_{ij}=0$, then clearly we have an edge between $A_{1}$ and $A_{2}$ in $G_{N}(M_{n}(\mathbb{Z}_{n}))$. For each element $a_{ii}$ of $A_{2}$, by $(ii)$ we have a path $\{a_{ii},b_{i1},b_{i2},b_{i3},\dots,b_{ik_{i}}=\overline{0}\}$ of length $k_{i}\in\mathbb{N}$ to $\overline{0}$. Now let $K=\max\{k_{i}:1\leq i\}$, we will construct a path of length $K$ from $A_{2}$ to [math], as follows. Define $B_{i}=[b1_{jl}]$, where $b1_{jj}=b_{ji}$, if $b1_{ij}$ appears in some above paths, otherwise $b1_{ij}=0$, $1\leq i\leq K$. Thus $\{A,A_{1},A_{2},B_{1},B_{2},\dots,B_{K}=0\}$ is a path from $A$ to [math] in $G_{N}(M_{n}(\mathbb{Z}_{n}))$. Lastly $(v)$ follows from $(iv)$.$\hfill\square$

Following result related to finite field will be used in upcoming results, we state it here to maintain the self containment of the article.

Lemma 2.6**.**

A ring $R$ is a finite commutative reduced ring with no non trivial idempotents if and only if $R$ is a finite field.

$Proof.(\Rightarrow)$ Let $0\neq x\in R$. Observe the set $A=\{x^{k}:k\in\mathbb{N}\}$ is a finite set. Therefore there exist $m>l$ such that $x^{l}=x^{m}$. Note that $x^{l}=x^{m}=x^{m-l+l}=x^{m-l}.x^{l}=x^{m-l}.x^{m}=x^{2m-l+l-l}=x^{2(m-l)+l}=\dots=x^{k(m-l)+l}$, where $k$ is a natural number. Now we have $[x^{l(m-l)}]^{2}=x^{l(m-l)}.x^{l(m-l)}=x^{l(m-l)+l(m-l)+l-l}\ =x^{l(m-l)+l}.x^{l(m-l)-l}=x^{l}.x^{l(m-l)-l}=x^{l(m-l)}$ that is $x^{l(m-l)}$ is an idempotent. Thus $x^{l(m-l)}=1$ which gives that $x$ is a unit, therefore $R$ is a finite field. $(\Leftarrow)$ Obvious.$\hfill\square$

3 Invariants of nil clean graph

In this section, we prove some results related to invariants of graph theory. Following subsection is for girth of $G_{N}(R)$.

3.1 Girth of $G_{N}(R)$

For a graph $G$, the girth of $G$ is the length of the shortest cycle in $G$. We have following results on girth of nil clean graph.

Theorem 3.1**.**

The following hold for nil clean graph $G_{N}(R)$ of $R$:

1. (i)

If $R$ is not a field, then girth of $G_{N}(R)$ is equal to 3. 2. (ii)

If $R$ is a field

1. (a)

girth is $2p$ if $R\cong GF(p^{k})$ (field of order $p^{k}$), where $p$ is a odd prime and $k>1$; 2. (b)

girth is infinite, in fact $G_{N}(R)$ is a path, otherwise.

$Proof.$ (i) Let $R$ has at least one non-trivial idempotent or non trivial nilpotent. If $e\in R$ be an nontrivial idempotent, then we have

so the girth of $G_{N}(R)$ is 3. Again if $R$ contains a nontrivial nilpotent $n\in R$, then we have the cycle

so the girth is $3$. By Lemma 2.6 rings without non trivial idempotents and nilpotents are field. This proves (i).

(ii) As set of nil clean elements of a finite field is $\{0,1\}$, so the nil clean graph of $\mathbb{F}_{p}$, where $p$ is a prime, is

From the graph, it is clear that the grith of $G_{N}(\mathbb{F}_{p})$ is infinite, which proves $(b)$. It is easy to observe from characterization of finite field that the nil clean graph of $GF(p^{k})$ for $p>2$, is a disconnected graph consisting of a path of length $p$ and $(\frac{p^{k-1}-1}{2})$ number of $2p$ cycles. For the proof let $GF(p^{k})=Z_{p}[X]/\langle f(x)\rangle$, where $f(x)$ is a irreducible polynomial of degree $k$ over $\mathbb{Z}_{p}$. Let $A\subseteq GF(p^{k})$, such that $A$ consists of all linear combinations of $x,x^{2},\dots,X^{k-1}$ with coefficients from $\mathbb{Z}_{p}$ such that if $g(x)+\langle f(x)\rangle\in A$ then $-g(x)+\langle f(x)\rangle\notin A$. Clearly, $A$ can be written as A=\{g_{i}(x)+\langle f(x)\rangle~{}\big{|}~{}1\leq i\leq(\frac{p^{k-1}-1}{2})\}, let us denote $\overline{g_{i}(x)}=g_{i}(x)+\langle f(x)\rangle$ for $1\leq i\leq(\frac{p^{k-1}-1}{2})$. So $(a)$ follows from Figure 5.

Here, $A=-\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{1}$, $B=-\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{p-1}$, $C=\overline{-g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{2}$,

$D=\overline{-g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{\frac{p-1}{2}}$, $E=-\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{\frac{p+1}{2}}$,

$A^{\prime}=\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\bar{1}$, $B^{\prime}=-\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{p-1}$, $C^{\prime}=\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{2}$,

$D^{\prime}=\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{\frac{p-1}{2}}$, $E^{\prime}=-\overline{g_{\frac{p^{k-1}-1}{2}}(x)}+\overline{\frac{p+1}{2}}$

Corollary 3.2**.**

$G_{N}(R)$* is not cyclic.*

A graph $G$ is said to be bipartite if its vertex set can be partitioned into two disjoint subsets $V_{1}$ and $V_{2}$, such that $V(G)=V_{1}\cup V_{2}$ and every edge in $G$ has the form $e=(x,y)\in E(G)$, where $x\in V_{1}$ and $y\in V_{2}$. Note that no two vertices both in $V_{1}$ or both in $V_{2}$ are adjacent.

Theorem 3.3**.**

$G_{N}(R)$* is bipartite if and only if $R$ is a field.*

$Proof.$ Let $G_{N}(R)$ be bipartite, so girth is of $G_{N}(R)$ is not a odd number, hence by Theorem 3.1 $R$ can not be a non field ring. Now if $R$ is a field, it is clear from the nil clean graph of $R$ that $G_{N}(R)$ is bipartite.$\hfill\square$

3.2 Dominating set

Let G be a graph, a subset $S\subseteq V(G)$ is said to be dominating set for $G$ if for all $x\in V(G)$, $x\in S$ or there exists $y\in S$ such that $x$ is adjacent to $y$. Following theorem shows that for a finite commutative weak clean ring dominating number is $2$, where dominating number is the carnality of smallest dominating set.

Theorem 3.4**.**

Let $R$ be a weak nil clean ring such that $R$ has no non trivial idempotents, then $\{1,2\}$ is a dominating set for $G_{N}(R)$.

$Proof.$ Let $a\in R$, then $a=n,n+1$ or $n-1$ for some $n\in\operatorname{Nil}(R)$. If $a=n,$ then $n+1\in$ NC($R$) implies $a$ is adjacent to $1$. If $a=n-1,$ then $n-1+1=n\in$ NC($R$) implies $a$ is adjacent to $1$. If $a=n+1$ and $2=n_{1}$ for some nilpotent $n_{1}\in R$, then $a$ is adjacent to $2$. If $a=n+1$ and $2=n_{1}-1$ for some nilpotent $n_{1}\in R$, then $a$ is adjacent to $2$. If $a=n+1$ and $2=n_{1}+1$ for some nilpotent $n_{1}\in R$. Now $a+2=(n+1)+(n_{1}+1)=(n+n_{1})+2=n+n_{1}+(n_{1}+1)=(n+2n_{1})+1$ is nil clean. Hence $a$ is adjacent to $2$.$\hfill\square$

Theorem 3.5**.**

Let $R=A\times B$, such that $A$ is nil clean and $B$ weak nil clean with no non trivial idempotents. Then $\{(1_{A},1_{B}),(2_{A},2_{B})\}$ is a dominating set for $G_{N}(R)$.

$Proof.$ Let $(a,b)\in R$, where $a\in A$ and $b\in B$. For $n_{1}\in\operatorname{Nil}(A)$, $n_{2}\in\operatorname{Nil}(B)$ and $0\neq e\in\operatorname{Idem}(A)$, $(a,b)$ has one of the following form $(a,b)=(n_{1},n_{2})+(e,1_{B}),(n_{1},n_{2})+(e,0),(n_{1},n_{2})-(e,1_{B})$ or $(n_{1},n_{2})+(0,0)$ If $(a,b)=(n_{1},n_{2})+(e,1_{B})$, we have $(a,b)+(2_{A},2_{B})=(n_{1}+e+2_{A},n_{2}+1_{B}+2_{B})$. Since $A$ is nil clean $n_{1}+e+2_{A}=n^{\prime}_{1}+f$ for some $n_{1}^{\prime}\in\operatorname{Nil}(A)$ and $f\in\operatorname{Idem}(A)$. Since $B$ is weak nil clean $2_{B}=n_{2}^{\prime},$ or $n_{2}^{\prime}-1_{B}$, for some $n_{2}^{\prime}\in\operatorname{Nil}(B)$. If $2_{B}=n_{2}^{\prime}$, we have $(a,b)+(2_{A},2_{B})=(n^{\prime}_{1},n_{2}+n_{2}^{\prime})+(f,1_{B})$ which is a nil clean expression, hence $(a,b)$ is adjacent to $(2_{A},2_{B})$. If $2_{B}=n_{2}^{\prime}-1_{B}$, we have $(a,b)+(2_{A},2_{B})=(n^{\prime}_{1},n_{2}+n^{\prime}_{2})+(f,0)$, thus $(a,b)$ is adjacent to $(2_{A},2_{B})$. In other three cases it is easy to see that $(a,b)+(1_{A},1_{B})$ is nil clean hence $(a,b)$ is adjacent to $(1_{A},1_{B})$. Therefore $\{(1_{A},1_{B}),(2_{A},2_{B})\}$ is a dominating set for $R$.$\hfill\square$

Theorem 3.6**.**

Let $R$ be a weak nil clean ring. $\{1,2\}$ is a dominating set in $G_{N}(R)$.

$Proof$ If $R$ has no non trivial idempotents, then by Theorem 3.4 we are done. If $R$ has a non trivial idempotent say $e$, then by Peirce decomposition $R\cong eR\oplus(1-e)R$, now by Theorem 2.3 of [2] we have one of $eR$ or $(1-e)R$ must be a nil clean ring. Without lost of generality suppose that $eR$ is a nil clean ring and $(1-e)R$ be a weak nil clean ring. Now if $(1-e)R$ has no non trivial idempotents, then we have the result by Theorem 3.5. If $f\in\operatorname{Idem}((1-e)R)$, repeating as above we get a direct sum decomposition of $R$ where only one summand is weak nil clean. As ring $R$ is a finite ring, so after repeating above to the weak nil clean summand of $R$, we will have a direct sum decomposition of $R$, where idempotents of weak nil clean summand of $R$ is trivial, then again by Theorem 3.5 we have the result.$\hfill\square$

3.3 Chromatic index

An edge colouring of a graph $G$ is a map $C:E(G)\rightarrow S$, where $S$ is a set of colours such that for all $e,f\in E(G)$, if $e$ and $f$ are adjacent, then $C(e)\neq C(f)$. The chromatic index of a graph denoted by $\chi^{\prime}(G)$ and is defined as the minimum number of colours needed for a proper colouring of $G$. Let $\vartriangle$ be the maximum vertex degree of $G$, then Vizing’s theorem [8] gives $\vartriangle\leq\chi^{\prime}(G)\leq\vartriangle+1$. Vizing’s theorem divides the graphs into two classes according to their chromatic index; graphs satisfying $\chi^{\prime}(G)=\vartriangle$ are called graphs of class $1$, those with $\chi^{\prime}(G)=\vartriangle+1$ are graphs of class $2$.

Following theorem shows that $G_{N}(R)$ is of class $1$.

Theorem 3.7**.**

Let $R$ be a finite commutative ring then the nil clean graph of $R$ is of class $1$.

$Proof.$ We colour the edge $ab$ by the colour $a+b$. By this colouring, every two distinct edges $ab$ and $ac$ has a different colour and $C=\{a+b|ab\mbox{ is an edge in }G_{N}(R)\}$ is the set of colours. Therefore nil clean graph has a $|C|$-edge colouring and so $\chi^{\prime}(G_{N}(R))\leq|C|$. But $C\subset NC(R)$ and $\chi^{\prime}(G_{N}(R))\leq|C|\leq|NC(R)|$. By Lemma 2.4, $\vartriangle\leq|NC(R)|,$ and so by Vizing’s theorem, we have $\chi^{\prime}(G_{N}(R))\geq\vartriangle=|NC(R|$. Therefore $\chi^{\prime}(G_{N}(R))=|NC(R|=\vartriangle,$ i.e., $G_{N}(R)$ is class of $1$.$\hfill\square$

3.4 Diameter

For a graph $G$, number of edges on the shortest path between vertices $x$ and $y$ is called the distance between $x$ and $y$ and is denoted by $d(x,y)$. If there is no path between $x$ and $y$ then we say $d(x,y)=\infty$. The diameter of a graph $diam(G)$ is the maximum of distances of each pair of distinct vertices in $G$.

The following are some results related to diameter of nil clean graph of a ring.

Lemma 3.8**.**

$R$* is nil clean ring if and only if $diam(G_{N}(R))=1$.*

Theorem 3.9**.**

Let $R$ be a non nil clean, weak nil clean ring with no non trivial idempotents then $diam(G_{N}(R))=2$.

$Proof.$ Let $a,b\in R$, then for some $n_{1},n_{2}\in\operatorname{Nil}(R)$, we have $a=n_{1}+1,n_{1}-1$ or $n_{1}$ and $b=n_{2}+1,n_{2}-1$ or $n_{2}$. If $a=n_{1}+1$ and $b=n_{2}-1$ or $n_{2}$, clearly $a+b$ is nil clean, thus $ab$ is an edge in $G_{N}(R)$, therefore $d(a,b)=1$. If $a=n_{1}+1$ and $b=n_{2}+1$, we have the path $a\rule[2.15277pt]{14.22636pt}{0.4pt}(-1)\rule[2.15277pt]{14.22636pt}{0.4pt}b$ in $G_{N}(R)$ thus $d(a,b)\leq 2$. If $a=n_{1}-1$ and $b=n_{2}-1$ or $n_{2}$ similarly as above a path of length $2$ from $a$ to $b$ through $1$, thus in this case also $d(a,b)\leq 2$. Finally if $a=n_{1}$ and $b=n_{2}$, then $d(a,b)=2$. Thus from above we conclude that $diam(G_{N}(R))\leq 2$. Now as $R$ is a non nil clean, weak nil clean ring thus we have at least one $x\in R$, such that $x=n-1$ but $x\neq n+1$, i.e., $x$ is not nil clean. Then we have $d(0,x)=2$, therefore $diam(G_{N}(R))\geq 2$, and hence the result follows.$\hfill\square$

Theorem 3.10**.**

Let $R=A\times B$, such that $A$ is nil clean and $B$ weak nil clean with no non trivial idempotents, then $diam(G_{N}(R))=2$.

$Proof.$ We have $\operatorname{Idem}(R)=\{(e,0_{B}),(e,1_{B})|e\in\operatorname{Idem}(A)\}$. Now let $(a_{1},b_{1}),(a_{2},b_{2})\in R$, in case $(a_{1},b_{1})+(a_{2},b_{2})$ is nil clean then $d((a_{1},b_{1}),(a_{2},b_{2}))=1$ in $G_{N}(R)$. If $(a_{1},b_{1})+(a_{2},b_{2})$ is not nil clean, thus $b_{1}+b_{2}$ is not nil clean. So we have followings cases:

CASE I: If $b_{1}=n_{1}+1$ and $b_{2}=n_{2}+1$, we have the path $(a_{1},b_{1})\rule[2.15277pt]{14.22636pt}{0.4pt}(0,-1)\rule[2.15277pt]{14.22636pt}{0.4pt}(a_{2},b_{2})$ in $G_{N}(R)$, thus $d((a_{1},b_{1}),(a_{2},b_{2}))\leq 2$.

CASE II: If $b_{1}=n_{1}-1$ and $b_{2}=n_{2}-1$, we have the path $(a_{1},b_{1})\rule[2.15277pt]{14.22636pt}{0.4pt}(0,1)\rule[2.15277pt]{14.22636pt}{0.4pt}(a_{2},b_{2})$ in $G_{N}(R)$, thus $d((a_{1},b_{1}),(a_{2},b_{2}))\leq 2$.

CASE III: If $b_{1}=n_{1}-1$ and $b_{2}=n_{2}$, we have the path $(a_{1},b_{1})\rule[2.15277pt]{14.22636pt}{0.4pt}(0,1)\rule[2.15277pt]{14.22636pt}{0.4pt}(a_{2},b_{2})$ in $G_{N}(R)$, thus $d((a_{1},b_{1}),(a_{2},b_{2}))\leq 2$.

CASE IV: If $b_{1}=n_{1}$ and $b_{2}=n_{2}-1$, by CASE III and symmetry we have $d((a_{1},b_{1}),(a_{2},b_{2}))\leq 2$.

Therefore $diam(R)\leq 2$, $R$ is not nil clean implies $diam(R)\geq 2$, Thus $diam(R)=2$.$\hfill\square$

Theorem 3.11**.**

If $R$ is weak nil clean ring but not nil clean then $diam(G_{N}(R))=2$.

$Proof.$ If $R$ has no non trivial idempotents, then by Theorem 3.10 we are done. If $R$ has a non trivial idempotent say $e$, then by Peirce decomposition $R\cong eR\oplus(1-e)R$, so by Theorem 2.3 of [2] we have one of $eR$ or $(1-e)R$ must be a nil clean ring. Without lost of generality suppose that $eR$ is a nil clean ring and $(1-e)R$ be a weak nil clean ring. Now if $(1-e)R$ has no non trivial idempotents, then we have the result by Theorem 3.10. If $f\in\operatorname{Idem}((1-e)R)$, repeating as above we get a direct sum decomposition of $R$ where only one summand is weak nil clean. As $R$ is a finite ring, so after repeating above to the weak nil clean summand of $R$ we will have a direct sum decomposition of $R$, where idempotents of weak nil clean summand of $R$ are trivial, then again by Theorem 3.10 we have the result.$\hfill\square$

Theorem 3.12**.**

Let $n$ be a positive integer, then the following hold for $\mathbb{Z}_{n}$.

1. (i)

If $n=2^{k}$, for some integer $k\geq 1$, then $diam(G_{N}(\mathbb{Z}_{n}))=1$. 2. (ii)

If $n=2^{k}3^{l}$, for some integer $k\geq 0$ and $l\geq 1$, then $diam(G_{N}(\mathbb{Z}_{n}))=2$. 3. (iii)

For a prime $p$, $diam(G_{N}(\mathbb{Z}_{p}))=p-1$. 4. (iv)

If $n=2p$, where $p$ is a odd prime, then $diam(G_{N}(\mathbb{Z}_{2p}))=p-1$. 5. (v)

If $n=3p$, where $p$ is a odd prime, then $diam(G_{N}(\mathbb{Z}_{3p}))=p-1$.

$Proof.$ $(i)$ and $(ii)$ follow from Lemma 3.8 and Theorem 3.11 respectively. $(iii)$ follows from $G_{N}(\mathbb{Z}_{p})$ in Figure 4, $(iv)$ follows from graph $G_{N}(\mathbb{Z}_{2p})$ in Figure 6 and $(v)$ follows from graphs in Figure 8 and Figure 7.