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A Generalized Study of Zero Divisor Graphs of Boolean Rings 

ℤ 2n = ℤ2 × ℤ 2 × · · · × ℤ 2 
 

S. G. Jakkewad1, R. G. Metkar2, G. A. Dhanorkar3, P. N.Tekalkar4 

1Rayat Shikshan Sanstha’s K. B. P. College Vashi, Navi Mumbai 

2Indira Gandhi (Sr) College Cidco, Nanded 

3Rayat Shikshan Sanstha’s K. B. P. College Vashi, Navi Mumbai 

4Rayat Shikshan Sanstha’s K. B. P. College Vashi, Navi Mumbai 

S. G. Jakkewad:  shrikantjakkewad171991@gmail.com , 

R. G. Metkar:  rammetkarmath@gmail.com . 

 

Article History: 

Received: 29-10-2024 

Revised: 15-11-2024 

Accepted: 19-12-2024 

Abstract:  

The Zero-divisor graph of a commutative ring R is defined as a graph in which the vertices 

represent the non-Zero Zero divisors of R, and two vertices x and y are connected if and only 

if x× y = 0. In this study, we focus on examining the Zero divisor graphs of Boolean rings 

and deriving insights from their graphical representations. 

Keywords: Zero-divisor graph, Commutative Ring, Boolean Ring ℤ 2n = ℤ 2 × ℤ 2 × ℤ 2 × 

... × ℤ 2, Girth, Diameter. 

 

1 Introduction 

The roots of graph theory can be traced back to 1735 when the Swiss mathematician Leonhard Euler 

provided a groundbreaking solution to the Konigsberg bridge problem,1 introducing a novel 

conceptual framework. Euler’s subsequent theorem marked a seminal moment in the field, paving the 

way for the development of Eulerian graphs. 

The exploration of cycles on polyhedra by the Revd. Thomas Penyngton Kirkman (1806-95) and Sir 

William Rowan Hamilton (1805-65) led to the conception of Hamiltonian graphs. The foundational 

notion of a tree, defined as a connected graph devoid of cycles, first emerged implic- itly in the work 

of Gustav Kirchho (1824-87), who applied graph-theoretical principles to analyℤe currents in 

electrical networks. Later, Arthur Cayley (1821-95), James Joseph Sylvester (1806- 97), Georg Polya 

(1887-1985), and others introduced trees in connection with the enumeration of specific chemical 

structures. 

I. Beck introduced the concept of a Zero-divisor graph in 1988.2 Denoted by Γ(R), the Zero divisor 

graph of a ring R is a simple graph whose vertices represent elements of R, with two vertices x and y 

being adjacent if and only if their product equals Zero. Beck’s research focused on colorings of R.2 

This perspective originated in a paper by D.F. Anderson and P.S. Livingston4 and has since seen 

further development.6, 12, 13 For instance, it is established in15 that all Zero-divisor graphs are 

connected, meaning a path exists between any pair of vertices. 



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A ring R is classified as Boolean if every element r in R satisfies the condition r2 = r. In 

particular, a Boolean ring is invariably commutative with a characteristic of 2. A graph is deemed 

Boolean if it exhibits isomorphism to the Zero-divisor graph of a Boolean ring.10–12, 16, 17. 

2 Preliminaries 

Graph theory is a branch of mathematics concerned with the study of graphs, which consist of nodes 

and edges, often used to represent mathematical concepts visually. The field explores the connections 

between the vertices and edges within these structures. Below, we will delve into fundamental 

definitions to facilitate comprehension of key terminologies. 

A Zero-divisor graph is an undirected graph representing the Zero-divisors of a commutative ring. Its 

vertices correspond to elements of the ring, and its edges connect pairs of elements whose product 

equals Zero. 

In this context, two vertices u and v within the graph G are considered connected if there exists a path 

from u to v. This notion of connection forms an equivalence relation on the vertex set V. 

Consequently, the vertex set V can be partitioned into nonempty subsets V1, V2, ..., Vw, where two 

vertices u and v are connected if and only if they both belong to the same subset Vi. The subgraphs 

G[V1], G[V2], ..., G[Vw] are referred to as the components of G. If G contains only one component, 

it is termed connected; otherwise, it is classified as disconnected. 

Additionally, a graph is considered simple when it does not contain loops, meaning there are no edges 

connecting a vertex to itself, and when each pair of vertices is connected by at most one edge, ensuring 

that no two edges join the same pair of vertices. (A self-loop is an edge that joins a single endpoint to 

itself) [Refer Figure1] 

 

Fig 1: SIMPLE GRAPH 

Graph theory describes a walk-in graph G as a finite, non-null sequence denoted by W = 

v0e1v1e2v2...ekvk. This sequence alternates between vertices and edges, where for 1 ≤ i ≤ k, the ends 

of ei are vi−1 and vi. This sequence is called a walk from v0 to vk, or a (v0, vk) walk. The vertices v0 

and vk are respectively termed the origin and terminus of the walk, while v1, v2,. . . , vk−1 are its 

internal vertices. The integer k is the length of W. Example:uavfyfvgyhwbv [Refer Figure 2] 



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Fig 2: WALK 

If the edges e1, e2, ..., ek of a walk W are distinct, W is called a trail. 

Example: wcxdyhwbvgy 

 

Fig 3: TRAIL and PATH 

A path is a trail in which no vertices (except possibly the end vertices) are repeated i.e. the vertices 

v0, v1, ..., vk are distinct, W is called a path. 

Example: xcwhyeuav A circuit is a closed trail (that is end vertices are the same) with at least one 

edge known as Circuit. 

 

Fig 4: CYCLE 

A walk is closed if it has a positive length and its origin and terminus are the same. A closed trail 

whose origin and internal vertices are distinct is a cycle. The graph’s diameter is the maximum distance 

between the pair of vertices. It can also be defined as the maximal distance between the pair of 

vertices. 

Example: BC → CF → FG A radius of the graph exists only if it has a diameter. The minimum 

among all the maximum distances between vertexes to all other vertices is considered as the radius of 

Graph G. It is denoted as r(G). 



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Example: BC → CA The girth of a graph is the length of the shortest cycle contained in the graph. If 

the graph does not contain any cycles (i.e., it’s an acyclic graph), its girth is defined as infinity. 

• a 4-cycle (square) has girth 4. A grid also has a girth 4, and a triangular mesh has a girth 3. 

A graph with a girth of four or more is triangle-free. 

3 Main results 

3.1 Theorem 

Statement: Let ℤ 2
n = ℤ 2 × ℤ 2 × ℤ 2 ×... × ℤ 2 be a Boolean ring and ℤ 2

n has (2n − 2) Zero divisors 

i.e | ℤ 2
n| = 2n and | ℤ (ℤ 2

n)| = 2n − 2, then n vertices in Zero divisor graph Γ(ℤ n) have 2n−1 –   

degree. 

Proof Let ℤ 2
n = ℤ 2 × ℤ 2 × ℤ 2 × ... × ℤ 2, be a Boolean ring and ℤ 2

n has (2n – 2) Zero divisors i.e.    

Zero divisors and ℤ 2
n is given by, 

 ℤ 2
n = {(b1, b2, b3, ..., bn) | bi′s ∈ {0, 1}, i = 1, 2, ..., n} 

Since, each bis has two choices 0 and 1, so there are 2n elements in ℤ 2
n   

| ℤ 2
n | = 2n 

The set of Zero divisors of ℤ 2
n is denoted by ℤ (ℤ 2

n) and is given by, 

ℤ (ℤ 2
n)  = {(b1, b2, b3, ..., bn) | bis ∈ {0, 1}, all bi ≠ 0, and all bi≠1, i = 1, 2, ..., n} 

∴ |ℤ (ℤ 2
n) | = 2n − 2 

The vertices (1, 0, 0, ..., 0), (0, 1, 0, ..., 0), (0, 0, 1, 0,…, 0), ..., (0, 0, 0, .., 1) are adjacent to 

(0, b2, b3, ..., bn), (b1, 0, b3,..., bn), ..., (b1, b2, b3, ..., 0) respectively. 

The vertex (1, 0, 0, .., 0) is adjacent to (0, b2, b3, .., bn) and ℤ 2
n has two choices 0 or 1 but 

all bi ≠0, ∀i = 1, 2, ..., n. 

i.e the vertex (1, 0, 0, ..., 0) is adjacent to (0, b2, b3, ..., bn) except (0, 0, 0, ..., 0). i.e the degree of 

(1, 0, 0, ..., 0) is 2n−1 − 1 since (1, 0, 0, ..., 0) is adjacent with 2n−1 − 1 number of vertices. Similarly, 

the degree of 

(0, 1, 0, ..., 0), (0, 0, 1, 0, ..., 0), (0, 0, 0, 1, 0, ..., 0), ....., (0, 0, 0, ..., 1) is 2n−1 − 1. 

∴ These n elements have 2n−1 − 1 degree. 

∴ The n vertices in the Zero divisor graph of ℤ 2
n, Γ(ℤ 2

n) have 2n−1 − 1 degree.  

3.1.1 Example 

B3 = ℤ2
3 = ℤ2 × ℤ2 × ℤ2 



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Fig 5: Γ(ℤ2 × ℤ2 × ℤ2) 

 ℤ 2
n = ℤ2 ×ℤ2 ×ℤ2 = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0), (1, 1, 1)} 

ℤ (ℤ2 × ℤ2 × ℤ2) = {(0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 is given in the figure above. 

Here, n = 3. Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2) = 23 − 2 = 6 

Degree of vertex (0, 0, 1) = 23−1 − 1 = 22 − 1 = 3 

Degree of vertex (0, 1, 0) = 23−1 − 1 = 22 − 1 = 3 

Degree of vertex (0, 0, 1) = 23−1 − 1 = 22 − 1 = 3 

∴ 3 vertices have degree 3. 

∴ The theorem holds. 

3.1.2 Example 

ℤ2
4 = ℤ2 × ℤ2 × ℤ2 × ℤ2 

 

Fig 6: Γ(ℤ2 × ℤ2 × ℤ2 × ℤ2) 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 is shown in the figure above. Here, n = 4. The number of 

vertices in the graph of Zero divisors of ℤ2 × ℤ2 × ℤ2 × ℤ2 is: 24 − 2 = 14. 

The degree of the vertices is computed as follows: 

Degree of vertex (0, 0, 0, 1) = 24−1 − 1 = 22 − 1 = 7 

Degree of vertex (0, 0, 1, 0) = 24−1 − 1 = 22 − 1 = 7 



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2 

Degree of vertex (0, 1, 0, 0) = 24−1 − 1 = 22 − 1 = 7 

Degree of vertex (1, 0, 0, 0) = 24−1 − 1 = 22 − 1 = 7 

∴ 4 vertices have degree 7, and the theorem holds. 

3.1.3 Example 

  Let ℤ5 = ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2. 

ℤ2 ×ℤ2 ×ℤ2 ×ℤ2 ×ℤ2 = {(0, 0, 0, 0, 0), (0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 0, 1, 1), (0, 0, 1, 0, 0), 

(0, 0, 1, 0, 1), (0, 0, 1, 1, 0), (0, 0, 1, 1, 1), (0, 1, 0, 0, 0), (0, 1, 0, 0, 1), (0, 1, 0, 1, 0), (0, 1, 0, 1, 1), (0, 1, 1, 

0, 0), 

(0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 1, 1), (1, 0, 0, 0, 0), (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 0, 0, 1, 1), (1, 0, 1, 

0, 0), 

(1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 1, 1), (1, 1, 1, 

0, 0), 

(1, 1, 1, 0, 1), (1, 1, 1, 1, 0), (1, 1, 1, 1, 1)} 

ℤ(ℤ2×ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 0, 1, 1), (0, 0, 1, 0, 0), (0, 0, 1, 0, 1), (0, 0, 1, 

1, 0), 

(0, 0, 1, 1, 1), (0, 1, 0, 0, 0), (0, 1, 0, 0, 1), (0, 1, 0, 1, 0), (0, 1, 0, 1, 1), (0, 1, 1, 0, 0), (0, 1, 1, 0, 1), (0, 1, 1, 

1, 0), 

(0, 1, 1, 1, 1), (1, 0, 0, 0, 0), (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 0, 0, 1, 1), (1, 0, 1, 0, 0), (1, 0, 1, 0, 1), (1, 0, 1, 

1, 0), 

(1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 1, 1), (1, 1, 1, 0, 0), 

 (1, 1, 1, 0, 1), (1, 1, 1, 1, 0)} The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure below, 

 

Fig 7: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2) 

Here, n = 5 Number of vertices in ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2) = 25 − 2 = 30 



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Degree of vertex (0, 0, 0, 0, 1) = 25−1 − 1 = 24 − 1 = 15 

Degree of vertex (0, 0, 0, 1, 0) = 25−1 − 1 = 24 − 1 = 15 

Degree of vertex (0, 0, 1, 0, 0) = 25−1 − 1 = 24 − 1 = 15 

Degree of vertex (0, 1, 0, 0, 0) = 25−1 − 1 = 24 − 1 = 15 

Degree of vertex (1, 0, 0, 0, 0) = 25−1 − 1 = 24 − 1 = 15 

∴ 5 vertices have degree 15. 

∴ The theorem holds true. 

3.2 Theorem 

Statement: Let ℤ2
n = ℤ2 × ℤ2 × ℤ2 × ℤ2 × ... × ℤ2 be a Boolean ring then n vertices of Zero divisor 

graph Γ(ℤ2
n ) have degree 1. 

Proof: Let ℤ2
n = ℤ2 × ℤ2 × ℤ2 × ℤ2...ℤ2, be a Boolean ring ℤ2

n has (2n − 2) Zero divisors and 

ℤ2
n is given by, 

ℤ2
n = {(b1, b2, b3, ..., bn) | bis ∈ {0, 1}, ∀i = 1, 2, ..., n} 

Since, each bis has two choices 0 and 1, so there are 2n elements in ℤ2
n 

|ℤ2
n| = 2n 

The set of Zero divisors of ℤ2
n is denoted by ℤ(ℤ2

n) and is given by, 

ℤ(ℤ2
n) = {(b1, b2, b3, ..., bn) | bis ∈ {0, 1}, all bi≠  0, all bi ≠1, i = 1, 2, ..., n} 

|ℤ(ℤ2
n) | = 2n − 2 

The vertices (0, 1, 1, ..., 1), (1, 0, 1, 1, ..., 1), (1, 1, 0, 1, ..., 1), ..., (1, 1, 1, ..., 0) are adjacent to (1, b2, b3, ..., 

bn), 

(b1, 1, b3, ..., bn), (b1, b2, 1, b4, ..., bn), ..., (b1, b2, b3, ..., 1) respectively. 

Clearly the vertex (0, 1, 1, ..., 1) is adjacent to (1, b2, b3, ..., bn) and b2, b3, ..., bn have only one choice 

which is 0. 

∴ The vertex (0, 1, 1, ..., 1) is adjacent to (1, 0, 0, ..., 0) only.  

∴ The degree of (0, 1, 1, ..., 1) is 1. 

Similarly, the degree of (1, 0, 1, 1, ..., 1), (1, 1, 0, 1, ..., 1), ..., (1, 1, 1, ..., 0) is 1. 

∴ These n elements have degree 1. 

∴ The n vertices of the Zero divisor graph of ℤ2
n have degree 1 

3.2.1 Example 

ℤ2
2 = ℤ2 × ℤ2 = {(0, 0), (0, 1), (1, 0), (1, 1)} 

ℤ (ℤ2 × ℤ2) = {(0, 1), (1, 0)} 



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Number of vertices in ℤ (ℤ2 × ℤ2) = 22 − 2 = 2 

Degree of vertex (0,1) = 1 

Degree of vertex (1,0) = 1 

∴ 2 vertices have degree 1. 

3.2.2 Example 

 

Fig 8: Γ(ℤ2 × ℤ2 × ℤ2) 

ℤ2
3 = ℤ2 × ℤ2 × ℤ2 

ℤ2 × ℤ2 × ℤ2 = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0), (1, 1, 1)} 

ℤ (ℤ2 × ℤ2 × ℤ2) = (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0) 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 is given in the figure above. Number of vertices in the graph 

of ℤ (ℤ2 × ℤ2 × ℤ2) = 23 − 2 = 6  

Degree of vertex (0, 1, 1) = 1 

Degree of vertex (1, 0, 1) = 1 

Degree of vertex (1, 1, 0) = 1 

∴ 3 vertices have degree 1. 

3.2.3 Example 

ℤ2
4 = ℤ2×ℤ2×ℤ2×ℤ2 = {(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), ((0, 1, 1, 0), 

(0, 1, 1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1)} 

ℤ(ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure below. 



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Fig 9: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2) 

Here, n = 4. Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2) = 24 − 2 = 14 

Degree of vertex (0, 1, 1, 1) = 1 

Degree of vertex (1, 0, 1, 1) = 1 

Degree of vertex (1, 1, 0, 1) = 1 

Degree of vertex (1, 1, 1, 0) = 1 

∴ 4 vertices have degree 1 and the theorem holds true. 

3.2.4 Example 

ℤ2
5 = ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 

ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 = {(0, 0, 0, 0, 0), (0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 0, 1, 1), (0, 0, 1, 0, 0), 

(0, 0, 1, 0, 1), (0, 0, 1, 1, 0), (0, 0, 1, 1, 1), (0, 1, 0, 0, 0), (0, 1, 0, 0, 1), (0, 1, 0, 1, 0), (0, 1, 0, 1, 1), (0, 1, 1, 

0, 0), 

(0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 1, 1), (1, 0, 0, 0, 0), (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 0, 0, 1, 1), (1, 0, 1, 

0, 0), 

(1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 1, 1), (1, 1, 1, 

0, 0), 

(1, 1, 1, 0, 1), (1, 1, 1, 1, 0), (1, 1, 1, 1, 1)} 

ℤ(ℤ2×ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 0, 1, 1) (0, 0, 1, 0, 0), (0, 0, 1, 0, 1), (0, 0, 1, 1, 

0), 

(0, 0, 1, 1, 1), (0, 1, 0, 0, 0), (0, 1, 0, 0, 1), (0, 1, 0, 1, 0), (0, 1, 0, 1, 1), (0, 1, 1, 0, 0), (0, 1, 1, 0, 1), (0, 1, 1, 

1, 0), 

(0, 1, 1, 1, 1), (1, 0, 0, 0, 0), (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 0, 0, 1, 1), (1, 0, 1, 0, 0), (1, 0, 1, 0, 1), (1, 0, 1, 

1, 0), 

(1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 1, 1), (1, 1, 1, 0, 0), 

(1, 1, 1, 0, 1), (1, 1, 1, 1, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure below, 



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Fig 10: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2) 

 

Here, n = 5 

Number of vertices in ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2) = 25 − 2 = 30 

Degree of vertex (0, 1, 1, 1, 1) = 1 

Degree of vertex (1, 0, 1, 1, 1) = 1 

Degree of vertex (1, 1, 0, 1, 1) = 1 

Degree of vertex (1, 1, 1, 0, 1) = 1 

Degree of vertex (1, 1, 1, 1, 0) = 1 

∴ 5 vertices have degree 1. 

∴ The theorem holds. 

3.3 Corollary: 

All vertices of the Zero divisor graph Γ(ℤ2
n) have odd degrees. 

3.3.1 Example 

 

Fig 11: Γ (ℤ2 × ℤ2 × ℤ2) 

ℤ2
3 = ℤ2 × ℤ2 × ℤ2  

ℤ2 × ℤ2 × ℤ2 = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0), (1, 1, 1)} 



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ℤ (ℤ2 × ℤ2 × ℤ2) = (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0) 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 is given in the figure above. Number of vertices in the graph 

of ℤ (ℤ2 × ℤ2 × ℤ2) = 23 − 2 = 6  

The degree of (0,1,1), (1,0,1), (1,1,0) is 1. 

The degree of (0,0,1), (0,1,0), (1,0,0) is 3. 

∴ All vertices of ℤ2 × ℤ2 × ℤ2 have odd degrees. 

3.3.2 Example 

ℤ2
4 = ℤ2 × ℤ2 × ℤ2 × ℤ2 

ℤ2×ℤ2×ℤ2×ℤ2 = {(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), ((0, 1, 1, 0), (0, 1, 

1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1)} 

ℤ(ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure below. 

 

Fig 12: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2) 

Here, n = 4. 

Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2) = 24 − 2 = 14 

Degree of vertices (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (1, 0, 0, 0) is 7 

Degree of vertices (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0) is 3. 

Degree of vertices (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 1), (1, 1, 1, 0) is 1. 

∴ All vertices of Zero divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 have odd degree. 

3.4 Theorem 

Statement: Let ℤ2
n = ℤ2 × ℤ2 × ℤ2 × ℤ2 × ... × ℤ2 be any Boolean ring then the Zero divisor graph 

of the Boolean ring Γ(ℤ2
n) has a cycle of length n. 

Proof: Let ℤ2
n = ℤ2 × ℤ2 × ℤ2 × ℤ2 × ... × ℤ2 be a Boolean ring ℤ2

n has (2n − 2) Zero divisors 



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and ℤ2
n is given by, 

Bn = {(b1, b2, b3, ..., bn) | bi′s ∈ {0, 1}, i = 1, 2, ..., n} 

Since each bis has two choices 0 and 1, there are 2n elements in ℤ2
n. 

|ℤ2
n| = 2n 

The set of Zero divisors of ℤ2
n is denoted by ℤ(ℤ2

n) and is given by, 

ℤ(ℤ2
n ) = {(b1, b2, b3, ..., bn) | bis ∈ {0, 1},all bi ≠0, and all bi ≠ 1, i = 1, 2, ..., n} 

∴ |ℤ(ℤ2
n) | = 2n − 2 

The n vertices (1, 0, 0, ..., 0), (0, 1, 0, ..., 0), (0, 0, 1, 0, ..., 0), ..., (0, 0, 0, ..., 1) are adjacent to each other. 

But we consider the case where these vertices are only adjacent to its consecutive elements 

i.e. (1, 0, 0, ..., 0) is adjacent to (0, 1, 0, ..., 0); (0, 1, 0, ..., 0) is adjacent to (0, 0, 1, 0, ..., 0); and so 

on till (0, 0, 0, ..., 1, 0) is adjacent to (0, 0, 0, ..., 1) and (0, 0, 0, ..., 1) is adjacent to (1, 0, 0, ..., 0). So, we 

get these n vertices forming edges between them. 

These n vertices form “n” edges between them. Hence, we get a closed loop formed by these n edges. 

∴ The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 has a cycle of length n. 

3.4.1 Example 

ℤ2
3 = ℤ2 × ℤ2 × ℤ2 

ℤ2 × ℤ2 × ℤ2 = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0), (1, 1, 1)} 

ℤ (ℤ2 × ℤ2 × ℤ2) = {(0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0)} 

 

Fig 13: Γ(ℤ2 × ℤ2 × ℤ2) 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 is given in the figure 10 above. 

Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2) = 23 − 2 = 6 

The vertices (0, 0, 1), (0, 1, 0) and (1, 0, 0) form a cycle. 

∴ ℤ2
3 has a cycle of length 3. 

∴ The theorem holds. 

3.4.2 Example 



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ℤ2
4 = ℤ2 × ℤ2 × ℤ2 × ℤ2 

ℤ2×ℤ2×ℤ2×ℤ2 = {(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), ((0, 1, 1, 0), (0, 1, 

1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1)} 

Γ(ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure below. 

 

 

Fig 14: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2) 

Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2) = 24 − 2 = 14 

The vertices (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (1, 0, 0, 0) form a cycle of length 4. 

The vertices (0, 1, 1, 0), (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 0, 1) form a cycle of length 4. 

The vertices (1, 0, 0, 0), (0, 1, 0, 1), (1, 0, 1, 0), (0, 1, 0, 0) form a cycle of length 4 and etc. 

∴ ℤ2
n has a cycle of length 4 and the theorem holds. 

3.4.3 Example: 

B5 = ℤ2
5 = ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 

ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 = {(0, 0, 0, 0, 0), (0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 0, 1, 1), (0, 0, 1, 0, 0), 

(0, 0, 1, 0, 1), (0, 0, 1, 1, 0), (0, 0, 1, 1, 1), (0, 1, 0, 0, 0), (0, 1, 0, 0, 1), (0, 1, 0, 1, 0), (0, 1, 0, 1, 1), (0, 1, 1, 

0, 0), 

(0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 1, 1), (1, 0, 0, 0, 0), (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 0, 0, 1, 1), (1, 0, 1, 

0, 0), 

(1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 1, 1), (1, 1, 1, 

0, 0), 

(1, 1, 1, 0, 1), (1, 1, 1, 1, 0), (1, 1, 1, 1, 1)} 

ℤ(ℤ2×ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 0, 1, 1) (0, 0, 1, 0, 0), (0, 0, 1, 0, 1), (0, 0, 1, 1, 

0), 



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(0, 0, 1, 1, 1), (0, 1, 0, 0, 0), (0, 1, 0, 0, 1), (0, 1, 0, 1, 0), (0, 1, 0, 1, 1), (0, 1, 1, 0, 0), (0, 1, 1, 0, 1), (0, 1, 1, 

1, 0), 

(0, 1, 1, 1, 1), (1, 0, 0, 0, 0), (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 0, 0, 1, 1), (1, 0, 1, 0, 0), (1, 0, 1, 0, 1), (1, 0, 1, 

1, 0), 

(1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 1, 1), (1, 1, 1, 0, 0), 

(1, 1, 1, 0, 1), (1, 1, 1, 1, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure below, 

 

Fig 15: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2) 

Here, n = 5 

Number of vertices in ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2 × ℤ2) = 25 − 2 = 30 

The vertices (0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0), (0, 0, 1, 0, 0) form a cycle of 

length 5. 

The vertices (1, 0, 1, 1, 0, 0), (0, 0, 1, 0, 0), (0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (0, 0, 0, 1, 0) form a cycle of 

length 5, etc. 

∴ ℤ2
5 has a cycle of length 5. 

∴ The theorem holds. 

3.5 Theorem 

Statement: Let ℤ2
n = ℤ2 × ℤ2 × ℤ2 × ℤ2...ℤ2 be a Boolean ring. For n ≥ 3, the girth of the Zero-

divisor graph Γ(ℤ2
n) is 3. 

Proof: Let ℤ2
n = ℤ2 × ℤ2 × ℤ2 × ℤ2...ℤ2 be a Boolean ring then ℤ2

n has (2n − 2) Zero divisors 

and ℤ2
n is given by, 

ℤ2
n = {(b1, b2, b3, ..., bn) | bi′s ∈ {0, 1}, i = 1, 2, ..., n} 

Since each bis has two choices 0 and 1, so there are 2n elements in ℤ2
n. 



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|ℤ2
n| = 2n 

The set of Zero divisors of ℤ2
n is denoted by ℤ(ℤ2

n) and is given by, 

ℤ (ℤ2
n) = {(b1, b2, b3, ..., bn) | bis ∈ {0, 1}, all bi≠  0, and all bi ≠ 1, i = 1, 2, ..., n} 

∴ |ℤ(ℤ2
n) | = 2n − 2 

Consider the vertices (1, 0, 0, ..., 0), (0, 1, 0, ..., 0), (0, 0, 1, ..., 0), ..., (0, 0, 0, ..., 1), they all form their 

edges between them. By Theorem (2.1), the above elements have degree 2n−1 − 1 

For n ≥ 3, the degree of the above elements would be greater than or equal to 3. 

Any combination of 3 elements between these vertices will form a cycle, and the length of that cycle 

will be 3. 

The cycle formed will be the shortest cycle of the graph. 

We know, the girth of the graph is the length of the shortest cycle in the graph. 

∴ The girth of the Zero-divisor graph Γ(ℤ2
n) is 3. (For n≥3). 

3.5.1 Example 

ℤ2
3 = ℤ2 × ℤ2 × ℤ2 

ℤ2 × ℤ2 × ℤ2 = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0), (1, 1, 1)} 

ℤ (ℤ2 × ℤ2 × ℤ2) = (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 0) 

 

Fig 16: Γ(ℤ2 × ℤ2 × ℤ2) 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 is given in the figure above. 

Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2) = 23 − 2 = 6 

The vertices (0,0,1), (0,1,0) and (1,0,0) form a cycle. 

∴ ℤ2
3 has a cycle of length 3. 

∴The girth of ℤ2
3 is 3. 

∴ The theorem holds. 

3.5.2 Example 

ℤ2
4 = ℤ2 × ℤ2 × ℤ2 × ℤ2 



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ℤ2×ℤ2×ℤ2×ℤ2 = {(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), ((0, 1, 1, 0), (0, 1, 

1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1)} 

ℤ(ℤ2×ℤ2×ℤ2×ℤ2) = {(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), 

(1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0)} 

The Zero-divisor graph of ℤ2 × ℤ2 × ℤ2 × ℤ2 is given in the figure. 

 

Fig 17: Γ (ℤ2 × ℤ2 × ℤ2 × ℤ2) 

Number of vertices in the graph of ℤ (ℤ2 × ℤ2 × ℤ2 × ℤ2) = 24 − 2 = 14 

The vertices (0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0) form a cycle of length 3. 

The vertices (0, 1, 1, 0), (1, 0, 0, 0), (0, 0, 0, 1) form a cycle of length 3. 

The vertices (1, 0, 0, 0), (0, 1, 0, 1), (0, 0, 1, 0) form a cycle of length 3 and etc. 

∴ ℤ2
4 has the shortest cycle of order 3. 

∴The girth of ℤ2
3 is 3 and the theorem holds. 

4 Conclusion 

An attempt has been made to generalize the Zero divisor graphs of Boolean rings. In this project, we 

have tried to prove the results regarding the degree of vertices, length of cycles, and girth of the graph 

for the Zero divisor graphs of these Boolean rings. By studying the graphs for different values of n, we 

could develop these results and hence form the proofs of the theorems stated in the project before. The 

ultimate goal of this project is to understand the relationship between different aspects of the Zero-

divisor graphs of Boolean rings and their various properties. 

5 Sage 

Sage is free, open-source mathematics software that can be used alternatively to Mathematica or 

Matlab. Sage Math (previously Sage or SAGE,” System for Algebra and Geometry Experimentation”) 

is a computer algebra system (CAS) with features covering many aspects of mathematics, including 

algebra, combinatorics, graph theory, numerical analysis, number theory, calculus, and statistics. We 

have been using this tool to form graphs. 

Commands that that been used to form a zero-divisor graph of ℤ2
3 are: 



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We have defined the function G1=Graph() 

G1.add_vertex(′(0, 0, 1)′) 

G1.add_edge(′(0, 1, 0)′,′ (1, 0, 0)′) 

G1.add_vertices([′(1, 1, 0)′,′ (1, 0, 1)′,′ (0, 1, 1)′]) 

G1.add_ edges([(′(0, 1, 0)′,′ (0, 0, 1)′), (′(0, 0, 1)′,′ (1, 0, 0)′), (′(1, 1, 0)′,′ (0, 0, 1)′), 

(′(1, 0, 1)′,′ (0, 1, 0)′), (′(0, 1, 1)′,′ (1, 0, 0)′)]) 

G1. Show () 

Similarly, we have used the commands for generating graphs of ℤ2
4 and ℤ2

5. 

6 Acknowledgement 

The authors thank K.B.P. College Vashi, Navi Mumbai for providing financial support for the MRP 

under institutional seed money Ref.No.:1643/2024-2025/sr. 

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