### Article

Kyungpook Mathematical Journal 2016; 56(4): 1103-1113

**Published online** December 23, 2016

Copyright © Kyungpook Mathematical Journal.

### On the Diameter, Girth and Coloring of the Strong Zero-Divisor Graph of Near-rings

Prohelika Das

Department of Mathematics, Cotton College State University, Guwahati 781001, Assam, India

**Received**: May 8, 2014; **Accepted**: January 13, 2015

In this paper, we study a directed simple graph Γ_{s}^{*}(^{*}(_{s}_{s}

**Keywords**: Near-ring, N-subsets, diameter, girth, essential ideal, chromatic number, left annihilator.

### 1. Introduction

In this paper by a near-ring _{l}_{s}^{*}(^{*}(_{s}

The concept of zero-divisor graph of a commutative ring was first introduced by Beck in [5]. Beck [5] was mainly interested in the coloring of the ring. This notion was redefined in [3] and they proved that such a graph is always connected and its diameter is always less than or equal to 3. Anderson and Mulay in [4] studied diameter and girth of zero-divisor graph of a commutative ring. The notion of zero-divisor graph was extended to a non-commutative ring [1] and various properties of diameter and girth were established. In [10], Redmond has generalised the notion of zero-divisor graph. For an ideal _{I}

In this paper, we study the graph theoretic aspect of a near-ring _{1} + _{2}) = _{1} + _{2} for any _{1}, _{2} ∈ _{d}_{d}_{d}

For basic definitions and results related to near-ring, we would like to mention Pilz [9].

Recall that a graph

A left ^{n}^{n}^{−1} ≠ 0. The near-ring _{s}_{s}_{s}

Moreover, in this paper, we deal with coloring of Γ_{s}_{4}[[^{2} − 2, ^{2} − 2, ^{2}, 2_{s}_{s}_{s}_{s}_{1} and _{2} (say) provided _{1}) and _{2}) are essential. In addition to it we show that if Γ_{s}_{s}

The following are some examples of strong zero-divisor graphs.

### Example 1.1

Γ

(_{s}Z _{4}) ≅ Γ(Z _{4})Γ

(_{s}Z _{2}× Z _{2}) ≅ Γ(Z _{2}× Z _{2}) ≅ Γ (_{s}Z _{6}) (Z _{2}× Z _{2}≇Z _{6})${\mathrm{\Gamma}}_{s}({\scriptstyle \frac{{Z}_{3}[x]}{\langle {x}^{2}\rangle}})\cong {\mathrm{\Gamma}}_{s}({\scriptstyle \frac{{Z}_{2}[x,y]}{\langle {x}^{2},{y}^{2},xy\rangle}})$ but${\scriptstyle \frac{{Z}_{3}[x]}{{x}^{2}}}\ncong {\scriptstyle \frac{{Z}_{2[x,y]}}{\langle {x}^{2},{y}^{2},xy\rangle}}$ ${\mathrm{\Gamma}}_{s}({\scriptstyle \frac{{Z}_{4}[x]}{\langle {x}^{2}\rangle}})\ncong {\mathrm{\Gamma}}_{s}({\scriptstyle \frac{{Z}_{2}[x,y]}{\langle {x}^{2},xy,{y}^{2}\rangle}})\hspace{0.17em}({\scriptstyle \frac{{Z}_{4}[x]}{\langle {x}^{2}\rangle}}\cong {\scriptstyle \frac{{Z}_{2}[x,y]}{\langle {x}^{2},xy,{y}^{2}\rangle}})$

### 2. Main Results

In this section, we present results regarding diameter and girths of Γ_{s}

A vertex _{s}^{2} =

### Lemma 2.1

For a subset

### Lemma 2.2

**Proof**

Since ^{m}^{m}^{−1} ≠ 0. Again ^{m}^{−1}^{m}^{m}^{−1}^{m}^{−2}^{m}^{−1}(^{m}^{−1} ⊆

Thus in this lemma, we see that the nilpotency of _{s}

### Theorem 2.3

_{s}

**Proof**

We give the proof in two steps such as

(i) Step1:

Suppose

d (I ,J ) = 2. LetM ∈V ^{*}(N ) be such that,I →M →J is a directed path. ThenIM = 0 andMJ = 0 which gives thatM ∈r (I ) =l (I ). Now,M (I +J ) = 0 gives thatM (≠ 0) ⊆l (I +J ). Thusl (I +J ) ≠ 0, a contradiction.(i) Step2:

CaseI: If

IJ ≠ 0, considerM =l (I ),N =l (J ). Claim:I →M =l (I ) →N =l (J ) →J is a directed path. It is enough to show thatl (I )l (J ) = 0. Suppose there exists anx ∈l (I ),y ∈l (J ) such thatxy ≠ 0. Nowx ∈l (I ) =r (I ) givesIxy = 0. Thusxy ∈r (I ) =l (I ) givesxyI = 0. Againy ∈l (J ) givesxyJ = 0. Now we getxy (I +J ) = 0 which gives (0 ≠)xy ∈l (I +J ), a contradiction.CaseII: If

IJ = 0, then (I +J )^{2}⊆I ^{2}+J ^{2}. Andl (I +J )^{2}= 0, asx ∈l (I +J )^{2}givesx (I +J ) ⊆l (I +J ) giving therebyx ∈l (I +J ) = 0. SinceJ is nilpotent,qJ is also so whereq ∈l (I ) withqJ ^{2}≠ 0. NowqJ ⊆l (I ) givesl (I +qJ ) ≠ 0 [Lemma 2.2]. AgainI +qJ =J , otherwiseI ⊆J impliesl (I +J ) =l (J ) ≠ 0, a contradiction. HenceI +qJ ,J are distinct andI +J =I +qJ +J which givesl (I +qJ +J ) = 0 and (I +qJ )J ≠ 0. Henced (I +qJ ,J ) = 3[caseI].

### Theorem 2.4

_{1}_{1}) _{1}) ∩

**Proof**

Let _{1} + _{i}_{1} + _{i}_{i}_{1} + _{i}_{1} + _{i}_{1} = 0 for each _{1} ∈ _{1}. Thus _{i}_{1}_{1} = 0 as _{1}) giving thereby _{i}_{1}) = _{1}_{1}), since _{1}) is maximal. Now we get _{i}_{i}_{1})∩_{1} +

### Theorem 2.5

_{1} = _{1}) _{2} = _{2}) _{1} ∩ _{2} = 0_{1}_{2}_{1}_{2} = (0) = _{2}_{1}

**Proof**

For _{1}_{2} ≠ 0, we get _{1} ⊈ _{2}) = _{2} and _{2} ⊈ _{1}) = _{1}) = _{1}. Now _{1}_{1} ⊆ _{2} gives _{1} ⊆ _{2} as _{1} ⊈ _{2} = _{2}) giving thereby _{1} ∩ _{2} = _{1} ≠ 0, a contradiction. Similarly, _{2}_{1} = 0

### Definition 2.6

Invariant associated of a near-ring _{i}_{i}_{i}_{i}_{j}

### Corollary 2.7

_{s}

**Proof**

Let _{1}, _{2}, _{3}}, then _{1} = _{1}), _{2} = _{2}) and _{3} = _{3}) for some invariant subsets _{1}, _{2} and _{3} respectively. Then _{1}_{2} = 0, _{2}_{3} = 0 and _{3}_{1} = 0 [theorem 2.5]. Hence _{1} → _{2} → _{3} → _{1} is a cycle of length 3. Thus _{s}

### Theorem 2.8

_{s}

**Proof**

Let _{1}, _{2}, _{3}, _{4}, _{5}} where _{i}_{i}_{i}_{j}_{s}_{s}

Next we determine the girth of the graph Γ_{s}

### Theorem 2.9

_{s}_{s}

**Proof**

Assume _{s}_{1} → _{2} → _{3}_{n}_{1}...(i) be a cycle with minimal length _{i}_{s}_{1}, _{2}, ...., _{i}_{i}_{+2}. If _{i}_{i}_{+2} ≠ _{k}_{i}_{−1} → _{i}_{i}_{+1} → _{i}_{i}_{+2} → _{i}_{−1}, (_{i}_{i}_{+2} = _{k}

(i)

I _{i}I _{i}_{+2}≠I _{i}_{+1}. IfI _{i}I _{i}_{+2}=I _{i}_{+1}, then (I _{i}I _{i}_{+2})I _{i}_{−1}=I _{i}_{+1}I _{i}_{−1}. NowI _{i}_{+1}I _{i}_{−1}= (I _{i}I _{i}_{+2})I _{i}_{−1}⊆I _{i}I _{i}_{−1}= 0, which givesI _{i}_{+1}I _{i}_{−1}= 0. ThusI _{i}_{−1}→I →_{i}I _{i}_{+1}→I _{i}_{−1}is a cyclic, a contradiction to (i).(ii)

I _{i}I _{i}_{+2}≠I _{i}_{−1}. For otherwise,I _{i}_{+1}(I _{i}I _{i}_{+2}) =I _{i}_{+1}I _{i}_{−1}which givesI _{i}_{+1}I _{i}_{−1}= (I _{i}_{+1}I )_{i}I _{i}_{+2}= 0. ThusI _{i}_{−1}→I →_{i}I _{i}_{+1}→I _{i}_{−1}is a cycle, a contradiction to (i).I _{i}I _{i}_{+2}≠I _{i}_{+3}. IfI _{i}I _{i}_{+2}=I _{i}_{+3}, then we getI _{i}_{+3}I _{i}_{+1}= (I _{i}I _{i}_{+2})I _{i}_{+1}⊆I _{i}I _{i}_{−1}= 0, gives the cycleI _{i}_{+1}→I _{i}_{+2}→I _{i}_{+3}→I _{i}_{+1}, a contradiction.

Now _{i}_{i}_{+2} is adjacent to three distinct vertices _{i}_{−1}, _{i}_{+1} and _{i}_{+3}. Thus there exists an extra edge in

Now we present coloring of the strong zero-divisor graph Γ_{s}

### Theorem 2.10

_{i}_{i}_{i}_{j}_{s}

**Proof**

First we give _{i}_{i}_{j}_{i}_{k}_{k}_{k}_{k}_{k}_{k}^{*}(^{*}(^{*}(_{n}_{n}_{n}^{2} = 0, a contradiction. Suppose ^{/} has the color of _{k}^{/} ∈ ^{*}(^{/}^{/}) are given the color of _{k}_{k}^{/}^{/})) ⊈ _{k}^{/}^{/})) = 0, then ^{/}^{/})) ⊈ _{k}^{/}^{/}) ⊆ _{k}^{/}^{/})) ⊈ _{k}^{/} are not adjacent. If ^{/} = 0, then ^{/}^{/}) = 0 gives (^{/}^{/}))^{/}^{/}))(_{2} = (^{/}^{/}))^{2} = 0, a contradiction.

### Example 2.11

Consider _{6} = {0, 1, 2, 3, 4, 5} which is a near-ring with respect to the tables given below. The only left _{1} = {0, 3}, _{2} = {0, 2, 4} and _{3} = {0, 2, 3.4} which are invariant also and _{1}) = _{2} and _{2}) = _{1} are two maximal ideals of the annihilator ideal form. Here the chromatic number _{s}_{6})) is 2 + 1 = 3, i.e., _{s}_{6})) is equal to

In the results below, we deal with the essentiality of annihilator ideals in a near-ring _{s}

### Theorem 2.12

(i) If for a left N-subset I of N, l (I )is essential, then I = 0.(ii) N is strongly semi-prime .

**Proof**

(

i ) ⇒ (ii ) SupposeJ is an invariantN -subset ofN such thatJ ^{2}= 0. LetA be a nonzero ideal ofN . IfAJ = 0 thenA =A ∩l (J ) ≠ 0. IfAJ ≠ 0, thenAJ (≠ 0) ⊆A ∩l (J ). Thus in either casesl (J ) is essential. HenceJ = 0.(

ii ) ⇒ (i ) LetI be a leftN -subset such thatl (I ) is essential. LetJ =l (I )∩IN . NowJ ^{2}⊆l (I )IN = 0. ThusJ = 0, i.e.,l (I )∩IN = 0 which givesIN = 0 asl (I ) is essential. HenceI = 0.

### Example 2.13

Consider the ring _{6} = {0, 1, 2, 3, 4, 5} which is strongly semi-prime with unity. Here _{1} = _{2}) = {0, 3} and _{2} = _{1}) = {0, 2, 4} are the only nonzero ideals and _{6} =

### Example 2.14

_{4} = {0, 1, 2, 3} is a ring with unity. Here _{4} is not strongly semi-prime as for _{2} = 0 and _{4}

### Theorem 2.15

Let

### Theorem 2.16

**Proof**

Let _{1} < _{2} < _{3} < ......be an ascending chain for left _{i}_{i}_{i}_{+1}. Now _{i}_{i}_{i}_{+1} ∩ _{i}_{i}_{i}_{i}_{+1} ∩ _{i}_{i}_{i}_{i}_{+1} ∩ _{i}_{i}_{i}^{2} = (_{i}_{+1} ∩ _{i}_{i}_{i}_{i}_{i}_{n}_{n}_{n}_{−1}) such that _{n}_{n}_{−1} ∩_{n}_{−1}). Here for _{i}_{j}_{i}_{i}_{−1}))(_{j}_{j}_{−1})) ⊆ _{i}_{−1})_{j}

### Theorem 2.17

_{s}

**Proof**

Let _{1} < _{2} < _{3}.... be an ascending chain of invariant subsets with essential left annihilators. Suppose _{i}_{i}_{+1}. Let _{i}_{+1}(≠ 0) ∈ _{i}_{+1}_{i}_{i}_{+1} = _{i}_{+1}) ∩ 〈_{i}_{+1}〉 ≠ 0, where 〈_{i}_{+1}〉 is the ideal generated by _{i}_{+1}. Here _{i}_{j}

### Theorem 2.18

_{1})_{2})_{n}_{i}_{s}

**Proof**

We give _{i}_{i}_{i}_{+1} = 0 since for otherwise (_{i}_{i}_{i}_{+1}) ≠ 0. Now(_{i}_{i}_{i}_{+1})^{2} ⊆ _{i}_{i}_{i}_{+1} = 0 which gives _{i}_{i}_{i}_{+1} = 0, a contradiction. Now let

(i) CaseI: If

I ⊆_{i}I for somei , then give the color ofI to_{k}I ifk is themax {i |I ⊂_{i}I }. HereI andI are not adjacent since for otherwise_{k}I ⊆l (I ) together with_{k}I ⊆_{k}I gives that (I )_{k}^{2}= 0, a contradiction.(ii) CaseII: If

I ⊈_{i}I for anyi , then there exists anx ∈I such that_{i}x ∉I . Now consider the ideal generated byx denoted <x > which is clearly non zero. Thusl (I )∩ <_{i}x >≠ 0. But (l (I )∩ <_{i}x >)^{2}⊆l (I )_{i}I = 0, a contradiction._{i}

Suppose two distinct vertices _{k}_{k}_{k}

Now we mention the following notes:

(i) Note 1: In a near-ring

N ,χ (Γ (_{s}N )) = 2 if and only if for any two nonzeroI ,J ∈V ^{*}(N ),IJ ≠ 0 wheneverI ≠ 0,J ≠ 0. For, suppose there existsI ≠ 0 andJ ≠ 0 such thatIJ = 0. Then {0,I ,J } is a clique. Thus clique(Γ (_{s}n )) >χ (Γ (_{s}N )), a contradiction.(ii) Note 2: In a strongly semi-prime near-ring without unity, every essential ideal of the form

l (I ) with invariant_{i}I is maximal. For suppose_{i}l (I ) is not maximal, there exists a proper ideal_{i}K ofN such thatl (I ) ⊂_{i}K ⊂N . Now consider the idealJ generated byI _{i}x (≠ 0) for somex (≠ 0) ∈K . Herel (I ) ∩_{i}J = 0 but (l (I ) ∩_{i}J )^{2}= 0, a contradiction.

### Example 2.19

Consider the set _{(p∞)} of all rational numbers of the form ^{∞}) is a ring with respect to addition modulo 1 and multiplication defined as ^{∞}). It is to be noted that each subgroup of ^{∞}) is an ideal of it and the only proper ideals of ^{∞}) are of the form _{1} < _{2} < ..... and each _{i}_{p∞} is a reduced ring without unity. But here _{k}_{−1}) = 0 for all _{i}_{j}_{s}_{p∞})) = 2.

### Example 2.20

Consider the set _{2}) with unity which is not strongly semi-prime as _{i}

is essential and _{s}

### Theorem 2.21

_{s}_{1}_{2}

(i)

If N is strongly semi-prime without unity, then N has exactly two invariant N subsets I _{1}and I _{2}, where l (I _{1})and l (I _{2})are essential .(ii)

If N is not strongly semi-prime, then Γ (_{s}N )is a star graph with more than one vertices .

**Proof**

(i) Let

I _{1},I _{2}andI _{3}be three distinct invariantN -subsets ofN such thatl (I )’s are essential. Now_{i}J _{1}=l (I _{1})∩I _{2}≠ 0,J _{2}=l (I _{3})∩I _{2}≠ 0 andJ _{3}=l (I _{2})∩I _{3}≠ 0. HereJ _{3}≠J _{1}for otherwise (l (I _{2}) ∩I _{3})_{2}= (l (I _{2}) ∩I _{3})(l (I _{1}) ∩I _{2}) = 0, a contradiction. Thus (l (I _{2}) ∩I _{3})(l (I _{1}) ∩I _{2}) = 0. Similarly (l (I _{1}) ∩I _{2})(l (I _{3}) ∩I _{1}) = 0 and (l (I _{3}) ∩I _{1})(l (I _{2}) ∩I _{3}) = 0. ThusJ _{1}→J _{2}→J _{3}→J _{1}is a cycle, a contradiction.(ii) Suppose

N is not strongly semi-prime and letI ≠ 0 be an invariantN subset such thatI _{2}= 0. Assume thatI ∈V _{1}. We show thatV _{1}= {I }. Here eitherl (I ) is not essential orI is minimal. Supposel (I ) is essential and there exists anI _{1}(≠ 0) such thatI _{1}⊊I . Nowl (I ) ∩I _{1}≠ 0 and (l (I ) ∩I _{1})I = 0 givesl (I )∩I _{1}∈V _{2}andII _{1}= 0 givesI _{1}∈V _{2}. But (l (I )∩I _{1})I = 0, a contradiction. Now we consider the following cases.(a) CaseI: If

l (I ) is essential. Suppose there exists aP ∈V _{1}{I }. IfIP = 0, thenP ∈V _{2}, a contradiction. HenceIP ≠ 0. Since Γ (_{s}N ) is connected, there exists aK ∈V _{2}such thatPK = 0. Nowl (I )∩IP ≠ 0 andI (l (I ) ∩IP ) = 0 gives thatl (I ) ∩IP ∈V _{2}. But (l (I ) ∩IP )K = 0, a contradiction.(b) CaseII: If

l (I ) is not essential. Now supposeI is minimal. ThenI ∩P =P which gives that (I ∩P )I =I _{2}= 0. ThusI ∩P ∈V _{2}. But (I ∩P )K = 0, a contradiction. IfI is not minimal, thenIP ⊂I which gives (IP ∩I )K = (IP )K = 0. ThusIP ∩I ∈V _{2}, a contradiction to (IP ∩I )K = 0.

### Theorem 2.22

([7])

In the example 2.11, we see that every essential left ideal is essential as left

### Theorem 2.23

[7])

### Theorem 2.24

_{s}

**Proof**

Here _{1}(≠ 0) ⊆ _{1} is non nilpotent and _{1}) is as large as possible. If _{1}) = 0, we stop. If not, there exists a left _{1} = 0. But _{1}(≠ 0) ∈ _{1}) is as large as possible. Now _{1} ⊆ _{1}) = 0, we stop. Suppose _{1}) ≠ 0. Now _{1}_{1} ⊆ _{1} + _{1} ⊆ _{1} + _{1} + _{1}) ∩ _{1} + _{1}) = _{1}) ∩ _{1})[theorem 2.4] gives _{1} + _{1}) ∩ _{1}) ∩ _{1}) ∩ _{2}(≠ 0) ∈ _{1}) ∩ _{1}) ∩ _{2} non nilpotent and _{2}) is as large as possible. If _{2} + _{1} + _{1}) = 0, we stop. If not, proceeding in the same way, we get _{1}) ⊇ _{1}) ∩ _{1}) ⊇ _{1}) ∩ _{1}) ∩ _{2}) ⊇ ....... which is stationary. Hence we get a positive integer _{1}) ∩ _{1}) ∩ .... ∩ _{t}_{1}) ∩ _{1}) ∩ .... ∩ _{t}_{+1}). Now _{1})+_{1})+....+_{t}_{1})+_{1})+....+_{t}_{+1})=_{1}+_{1}+....+_{t}_{t}_{+1}) ⊆ _{t}_{+1}). Now _{t}_{+1} ⊆ _{1}+_{1}+....+_{t}_{t}_{+1} ⊆ _{t}_{+1}) which gives (_{t}_{+1})_{2} = 0, a contradiction. Thus _{1} +_{1} +....+_{t}_{1} + _{1} + .... + _{t}

### Acknowledgement

I would like to acknowledge the referees for their valuable suggestions.

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