Let R = Z 6 and let I 0 = { 0 } , I 1 = { 0 , 2 , 4 } , I 2 = { 0 , 3 } . The following Figure 1 are used to illustrate the graph.

If there is an invertible element α in V ( ∅ I ( R ) ) , then for any proper ideal I of a ring R , the graph ∅ I ( R ) is connected.

Because α is invertible then there is β ∈ R , such that αβ = βα = 1 . Let γ ∈ V ( ( ∅ I ( R ) ) = R \ { I , 1 , − 1 } , if αγ ∈ I and γα ∈ I , then we have βαγ ∈ I and γαβ ∈ I , which is a contradiction. Thus ∅ I ( R ) is connected.

¹³ ϕ ( n ) denotes the number of positive integers that are co-prime to n , where n ≥ 1 .

¹³ If the integer n > 1 has prime factorization n = P 1 S 1 P 2 S 2 … P r S r , then ϕ ( n ) = ∏ i = 1 r ( P i S i − P i S i − 1 ) . ϕ ( n ) is the number of invertible elements in the integer ring of module n , which is noteworthy.

For n > 6, ϕ ( n ) > 3.

We shall partition the proof into two distinct cases. -

Case 1: n is a prime number then ϕ ( n ) = n-1 > 3.

⁹ The non-zero divisor graph ∅ 0 ( Z n ) is connected if and only if n ∉ {1, 2, 3, 6}.

Given that n is a prime number, the only proper ideal for Z _n is the zero ideal. According to Theorem 2.7, ∅ I ( Z n ) is connected. Assume that γ , β ∈ V ( ∅ I ( Z n ) ) with γβ = 0 ; therefore, either γ = 0 or β = 0 , leading to a contradiction. This indicated that the graph was complete.

For any prime number p ≥ 2 , the graph ∅ I ( Z P 2 ) is connected.

The proper ideal of the ring Z P 2 are I = { 0 } and J = ⟨ p ⟩ which is maximal ideal of Z P 2 . •

Case 1: ∅ I ( Z P 2 ) is connected by Theorem 2.7

For the integer module n , Z n = { 1 , 2 , … , n − 1 } and the ideal I = d Z n = { 0 , d , 2 d , … , ( ( n / d ) − 1 ) d } for some integer d \ n . Since V ( ∅ I ( Zn ) ) = Z n \ { I , 1 , − 1 } , we can now develop a method to calculate the order of the graph ∅ I ( Z _n ) as following: | V ( ∅ I ( Z n ) ) | = n − n d − 2 , where | I | = n d .

Let n be an odd, non-prime number then, the non-zero ideal of Z _n with the greatest order is the ideal generated by the prime number p \ n , where p = r 2 + n − r , for some integer r such that r 2 + n ∈ Z .

If n = 2 k + 1 is odd number (not prime), then the ideal has the largest of order is the ideal generated by prime number that divides n, we will provide several cases and present a general formula to determine the largest prime number p that divides n. we know that | V ( ∅ I ( Zn ) ) | = n − n p − 2 , where | I | = n p . Therefor if p \ ( n = 2 k + 1 ) , then k = p − 1 2 , ( p − 1 2 ) p , ( p − 1 2 ) p + p , ( p − 1 2 ) p + 2 p , ( p − 1 2 ) p + 3 p , … . Then, the general formula for p is p = r 2 + n − r for some integer r that makes r 2 + n ∈ Z .

If n = 15 take r = 1 then p = 3 is the greatest order of ideal I such that | V ( ∅ I ( Zn ) ) | = 15 − 15 3 − 2 = 8 .

For any integer n ≥ 10 , the graph ∅ I ( Z n ) is connected for every non-zero proper ideal I of Z n .

The proof will be divided into two cases.

Case 1: I is a prime ideal; let γ , δ ∈ V ( ∅ I ( Z n )), with γδ ∈ I, then either γ ∈ I or δ ∈ I, contradiction. Thus ∅ I ( Z n ) connected (complete).

Case 2: I not prime ideal. Therefore, we have the following subcases. •

If n is even number, then the ideal has the largest of order is the ideal generated by 2, | V ( ∅ I ( Z n ) ) | = n − n 2 − 2 = n 2 − 2 , that means when n ≥ 10 , then | V ( ∅ I ( Zn ) ) | ≥ 3. By Lemma 2.6, then ϕ ( n ) > 3, and our result follows immediately by Lemma 2.3.

For any integer n ≥ 10, then γ ( ∅ I ( Z n ) ) = 1 and gr ( ∅ I ( Z n ) ) = 3 .

For any integer n ≥ 10 , from the previous proof of the theorem 3.3 then | V ( ∅ I ( Z n ) ) | ≥ 3 for any ideal I , and by Lemma 2.6 we have γ ( ∅ I ( Z n )) = 1 and gr ( ∅ I ( Z n ) ) = 3 .

In the graph ∅ 0 ( Z 8 ) of ring Z 8 , the set C = { 3 , 5 } represents the cut set. As shown Figure 2. V ( ∅ 0 ( Z 8 ) ) = { 2 , 3 , 4 , 5 , 6 } .

For every prime number p greater than 2, the cut-set of the non-zero divisor graph ∅ 0 ( Z P 2 ) is { α ∈ V ( ∅ 0 ( Z P 2 ) ) \ ann ( p ) } .

ann ( P ) = { x ∈ Z P 2 | Px = 0 } = { p , 2 p , … , ( p − 1 ) p } , for x and y in ann ( p ) , then x = kp and y = hp , for some k , h ∈ Z P and xy = kh P 2 = 0 , it follows that x is not connected with y . The graph that excludes the vertices of set ann ( p ) is unconnected. This is minimal because any proper subset of C. C preserves the pathways between zero-divisors via units by leaving some units intact.

The cut set in the graph ∅ 0 ( Z 3 2 ) of the ring Z 3 2 is the set C = {2,4,5,7}, as illustrated in Figure 3.

Let = x . y , where x ≠ y and x > y . Then, the Cut-set of Z n , for n > 4 , corresponds the set C = P n ∪ D n \ ( ⟨ x ⟩ ⋃ ⟨ y ⟩ ) , where P n = { p ∈ v ( ( Zn ) ) | p is prime residues of n } D n = { d ∈ v ( ( Zn ) ) | d is zero divisor of Zn }

Consider the subgraphs X and Y , which are generated by the ideal ⟨ x ⟩ and ⟨ y ⟩ , respectively. Consequently, subgraphs X and Y have no edges between them if γ ∈ X and δ ∈ Y then, δ = 0 , Consequently, subgraphs X and Y have no edges between them. Now, if C ~ = C \ { a}, then either a ∈ P n , which means that a is invertible, and by Lemma 2.3, a connect all vertices of ∅ 0 ( Z n ) contradiction, or a ∈ D n , then there exists path from x to y by a , since if ax ≡ 0 ( mod n ) then ax = kn , k ∈ Z , thus kny = axy , therefore a = xy ∈ ⟨ y ⟩ , a contradiction.

For the ring Z 12 , n = 3.4 , then the cut set of Zn , for n > 4 , corresponds to the set C = P n ∪ D n \ ( ⟨ x ⟩ ⋃ ⟨ y ⟩ ) = {3,7,210}, as illustrated in the Figure 4, where P n = { 5 , 7 } and D n = { 2 , 10 } .

Let R = Z n be the ring of integers module, and let ∅ I ( Z n ) denoted the ideal based non-zero divisor graph of R . Suppose I = ⟨ d ⟩ is ideal of, where d \ n . for any vertex x ∈ V ( ∅ I ( R)) and g = gcd ( x , d ) . Then, the degree of x is given by deg ( x ) = { n − n d − 3 if g = 1 n − n g − 2 if g > 1

Case 1: if g = gcdgcd ( x , d ) = 1 , assume that for all y ∈ V( ∅ I ( R)), xy ∈ I = ⟨ d ⟩ , thus d \ xy , but gcd ( x , d ) = 1 , therefore, d \ y , that means y ∈ I , a contradiction. Then, by the definition of an ideal-based non-zero divisor, y must be a universal vertex, so deg ( x ) = | V ( ∅ I ( R ) ) | − 1 = n − 2 − | I | − 1 = n − 3 − n d .

Case 2: if g = gcd ( x , d ) > 1 , let x = gl and d = gτ , so we need to exclude the ideal generated by τ , when I want an account for deg ( x). But deg ( x ) = n − 2 − | ⟨ τ ⟩ | = n − 2 − n g .

Applying Theorem 5.2, we will present a Table 2, illustrating examples of calculating the degree of a vertex in certain ∅ I ( Z n ) .

Let = Z 2 P , P > 2 , prime number, for any vertex x ∈ V ( ∅ I ( R)), For any non-zero ideal of R, then 1.

deg ( x ) = P − 3 if x ∈ V ( ∅ ⟨ 2 ⟩ ( R))

Let R = Z n be the ring of integers module n , and let ∅ 0 ( Z n ) be a non-zero divisor graph of R. Then, for any vertex x ∈ V ( ∅ 0 ( R ) ) , set g = deg ( x , n ). Thus, the degree of x is given by: deg ( x ) = { n − 4 if g = 1 n − g − 3 if x 2 ≠ 0 n − g − 2 if x 2 = 0

To prove the result, we should consider the following subcases :

Case 1: if g = gcd ( x , n ) = 1 , assume that for all y ∈ V ( ∅ 0 ( R ) ) , xy = 0 , but x is invertible thus, y = 0 , which is a contradiction. Therefore deg ( x ) = | V ( ∅ 0 ( R ) ) | − 1 = n − 3 − 1 .

Case 2: Let S = { y ∈ V ( ∅ 0 ( R ) ) : xy = 0 } , representing the non-neighbors of x , then S = ann ( x ) ∩ V ( ∅ 0 ( R ) ) ,which contains exactly g elements if x 2 ≠ 0 , and g + 1 elements when x 2 = 0

Let R = Z 12 then the following Table 3, representing the degree of every vertex x of the graph ∅ 0 ( R ) .

Let R = Z 30 then the following Table 4, representing the degree of every vertex x of the graph ∅ 0 ( R ) .