A Study on the Structure of Ideal-Based Non-Zero Divisor Graphs Associated with Z_n

Definition 2.1:

Let $R$ be a ring and $I$ be a proper ideal for $R$ . The ideal-based non-zero divisor graph, denoted by ${\varnothing }_I$ (R), has vertex set V( ${\varnothing }_I$ (R)) = $R$ \ { $I,1,-1$ }. Moreover, two different vertices δ and γ in V( ${\varnothing }_I$ (R)) are adjacent if and only if either δγ $\notin$ I or γδ $\notin$ I.

Example 2.2:

Let R = $Z_6$ and let $I_0=\left\{0\right\},I_1=\{0,2,4\},I_2=\{0,3\}$ . The following Figure 1 are used to illustrate the graph.

Lemma 2.3:

If there is an invertible element α in $V\left({\varnothing }_I\left(R\right)\right)$ , then for any proper ideal $I$ of a ring $R$ , the graph ${\varnothing }_I\left(R\right)$ is connected.

Proof:

Because $\alpha$ is invertible then there is $\beta \in R$ , such that $\mathit{\alpha \beta }=\mathit{\beta \alpha }=1$ . Let $\gamma \in V(\left({\varnothing }_I\left(R\right)\right)=R\backslash \{I,1,-1\}$ , if $\mathit{\alpha \gamma }\in I\mathrm{\text{and}}\mathit{\gamma \alpha }\in I$ , then we have $\mathit{\beta \alpha \gamma }\in I\mathrm{\text{and}}\mathit{\gamma \alpha \beta }\in I$ , which is a contradiction. Thus ${\varnothing }_I\left(R\right)$ is connected.

Definition 2.4:

¹³ $\varphi \left(n\right)$ denotes the number of positive integers that are co-prime to $n$ , where $n\ge 1.$

Theorem 2.5:

¹³If the integer $n>$ 1 has prime factorization $n={P_1}^{S_1}{P_2}^{S_2}\dots {P_r}^{S_r}$ , then $\varphi \left(n\right)={\prod }_{i=1}^r({P_i}^{S_i}-{P_i}^{S_{i-1}})$ . $\varphi \left(n\right)$ is the number of invertible elements in the integer ring of module $n$ , which is noteworthy.

Lemma 2.6:

For n $>$ 6, $\varphi \left(n\right)>$ 3.

Proof:

We shall partition the proof into two distinct cases.

Theorem 2.7:

⁹ The non-zero divisor graph ${\varnothing }_0$ ( ${\mathit{Z}}_{\mathit{n}}$ ) is connected if and only if n $\notin$ {1, 2, 3, 6}.

Lemma 3.1:

For any prime number $n\notin \{2,3\}$ , ${\varnothing }_I$ ( $Z_n$ ) represents a completely connected graph.

Proof:

Given that $n$ is a prime number, the only proper ideal for Z _n is the zero ideal. According to Theorem 2.7, ${\varnothing }_I$ ( $Z_n$ ) is connected. Assume that $\gamma ,\beta \in V\left({\varnothing }_I\left(Z_n\right)\right)$ with $\mathit{\gamma \beta }=0$ ; therefore, either $\gamma =0\text{or}\beta =0$ , leading to a contradiction. This indicated that the graph was complete.

Lemma 3.2:

For any prime number $p\ge 2$ , the graph ${\varnothing }_I$ ( $Z_{P^2}$ ) is connected.

Proof:

The proper ideal of the ring $Z_{P^2}$ are $I=\left\{0\right\}$ and $J=⟨p⟩$ which is maximal ideal of $Z_{P^2}$ .

For the integer module $n,{\mathit{Z}}_{\mathit{n}}=\{1,2,\dots ,n-1\}$ and the ideal $I=\mathit{d}{\mathit{Z}}_{\mathit{n}}=\{0,d,2d,\dots ,((n/d)-1)d\}$ for some integer $d\backslash n$ . Since $V\left({\varnothing }_I\left(\mathit{Zn}\right)\right)={\mathit{Z}}_{\mathit{n}}\backslash \{I,1,-1\}$ , we can now develop a method to calculate the order of the graph ${\varnothing }_I$ (Z_n) as following: $\left|V\left({\varnothing }_I\left({\mathit{Z}}_{\mathit{n}}\right)\right)\right|=n-\frac{n}{d}-2$ , where $\left|I\right|=\frac{n}{d}$ .

Lemma 3.3:

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 $p\backslash n$ , where $p=\sqrt{r^2+n}-r$ , for some integer r such that $\sqrt{r^2+n}\in Z$ .

Proof:

If $n=2k+1$ 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 $\left|V\left({\varnothing }_I\left(\mathit{Zn}\right)\right)\right|=n-\frac{n}{p}-2$ , where $\left|I\right|=\frac{n}{p}$ . Therefor if $p\backslash (n=2k+1)$ , then $k=\frac{p-1}{2},\left(\frac{p-1}{2}\right)p,\left(\frac{p-1}{2}\right)p+p,\left(\frac{p-1}{2}\right)p+2p,\left(\frac{p-1}{2}\right)p+3p,\dots$ . Then, the general formula for $p$ is $p=\sqrt{r^2+n}-r$ for some integer $r$ that makes $\sqrt{r^2+n}\in Z$ .

Example 3.4:

If $n=15$ take $r=1$ then $p=3$ is the greatest order of ideal $I$ such that $\left|V\left({\varnothing }_I\left(\mathit{Zn}\right)\right)\right|$ = $15-\frac{15}{3}-2=8$ .

Theorem 3.5:

For any integer $n\ge 10$ , the graph ${\varnothing }_I\left(Z_n\right)$ is connected for every non-zero proper ideal $I$ of $Z_n$ .

Proof:

The proof will be divided into two cases.

Case 1: I is a prime ideal; let $\gamma ,\delta \in$ V ( ${\varnothing }_I$ ( $Z_n$ )), with $\mathit{\gamma \delta }\in$ I, then either $\gamma \in$ I or $\delta \in$ I, contradiction. Thus ${\varnothing }_I$ ( $Z_n$ ) connected (complete).

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

Corollary 3.5:

For any integer n ≥ 10, then $\gamma \left({\varnothing }_I\left(Z_n\right)\right)$ = 1 and $\mathit{gr}\left({\varnothing }_I\left(Z_n\right)\right)=3.$

Proof:

For any integer $n\ge 10$ , from the previous proof of the theorem 3.3 then $\left|V\left({\varnothing }_I\left(Z_n\right)\right)\right|\ge 3$ for any ideal $I$ , and by Lemma 2.6 we have $\gamma \left({\varnothing }_I\right(Z_n$ )) = 1 and gr ( ${\varnothing }_I\left(Z_n\right)$ ) $=3$ .

Definition 4.1:

¹⁴A cut-set is defined as a set of vertices $\{\beta ,\gamma ,\delta ,\dots \}$ in a connected graph G, where G can be represented as the union of two subgraphs $X$ and $Y$ , such that

Here, an example that illustrates the definition of the cut-set in graph

Example 4.1:

In the graph ${\varnothing }_0\left(Z_8\right)$ of ring $Z_8$ , the set $C=\{3,5\}$ represents the cut set. As shown Figure 2. $V\left({\varnothing }_0\left(Z_8\right)\right)=\{2,3,4,5,6\}$ .

Proposition 4.2:

For every prime number $p$ greater than 2, the cut-set of the non-zero divisor graph ${\varnothing }_0\left(Z_{P^2}\right)$ is

Proof:

$\mathit{ann}\left(P\right)=\{x\in Z_{P^2}|\mathit{Px}=0\}$ = $\{p,2p,\dots ,(p-1)p\}$ , for $x$ and $y$ in $\mathit{ann}\left(p\right)$ , then $x=\mathit{kp}$ and $y=\mathit{hp}$ , for some $k,h\in Z_P$ and $\mathit{xy}=\mathit{kh}{P}^2=0$ , it follows that $x$ is not connected with $y$ . The graph that excludes the vertices of set $\mathit{ann}\left(p\right)$ 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.

Example 4.4:

The cut set in the graph ${\varnothing }_0\left(Z_{3^2}\right)$ of the ring $Z_{3^2}$ is the set C = {2,4,5,7}, as illustrated in Figure 3.

Theorem 4.5:

Let $=x.y$ , where $x\ne y$ and $x>y$ . Then, the Cut-set of $Z_n$ , for $n>4$ , corresponds the set

Proof:

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 $\gamma \in X\mathrm{\text{and}}\delta \in Y$ then, $\delta =0$ , Consequently, subgraphs $X$ and Y have no edges between them. Now, if $C^{~}$ = C $\backslash$ {a}, then either $a\in P_n$ , which means that $a$ is invertible, and by Lemma 2.3, a connect all vertices of ${\varnothing }_0\left(Z_n\right)$ contradiction, or a $\in D_n$ , then there exists path from $x$ to $y$ by $a$ , since if $\mathit{ax}\equiv 0(\mathit{mod}n)$ then $\mathit{ax}=\mathit{kn},k\in Z$ , thus $\mathit{kny}=\mathit{axy}$ , therefore $a=\mathit{xy}\in ⟨y⟩$ , a contradiction.

Example 4.3:

For the ring $Z_{12}$ , $n=3.4$ , then the cut set of $\mathit{Zn}$ , for $n>4$ , corresponds to the set C = $P_n\cup D_n\backslash (⟨x⟩\bigcup ⟨y⟩)$ = {3,7,210}, as illustrated in the Figure 4, where $P_n=\{5,7\}\mathrm{\text{and}}{D}_n=\{2,10\}$ .

Definition 5.1:

Let $x$ be a vertex of any graph; the degree of $x$ is represented as $\mathit{deg}\left(x\right)$ , and the degree of $x$ is defined as follows: $\mathit{deg}\left(x\right)=\left|\right\{y\in V\backslash \left\{x\right\}:d(x,y)=1\left\}\right|$ , $(d(x,y)$ length of the short path).

From the above definition, it is clear that $\mathit{deg}\left(x\right)$ = $\mathit{deg}(-x) in {\varnothing }_I(Z_n\mathit{)}$

Theorem 5.2:

Let $R=Z_n$ be the ring of integers module, and let ${\varnothing }_I$ ( $Z_n$ ) denoted the ideal based non-zero divisor graph of $R$ . Suppose $I=⟨d⟩$ is ideal of, where $d\backslash n$ . for any vertex $x\in V$ ( ${\varnothing }_I$ (R)) and $g=\mathit{gcd}(x,d)$ . Then, the degree of $x$ is given by

Proof:

Case 1: if $g=\mathit{\text{gcdgcd}}(x,d)=1$ , assume that for all y ∈ V( ${\varnothing }_I$ (R)), $\mathit{xy}\in I=⟨d⟩$ , thus $d\backslash \mathit{xy}$ , but $\mathit{gcd}(x,d)=1$ , therefore, $d\backslash y$ , that means $y\in I$ , a contradiction. Then, by the definition of an ideal-based non-zero divisor, y must be a universal vertex, so $\mathit{deg}\left(x\right)=\left|V\left({\varnothing }_I\left(R\right)\right)\right|-1=n-2-\left|I\right|-1=n-3-\frac{n}{d}.$

Case 2: if $g=\mathit{gcd}(x,d)>1$ , let $x=\mathit{gl}$ and $d=\mathit{g\tau }$ , so we need to exclude the ideal generated by $\tau$ , when $I$ want an account for deg ( x). But $\mathit{deg}\left(x\right)=n-2-|⟨\tau ⟩|=n-2-\frac{n}{g}$ .

Example 5.3:

Applying Theorem 5.2, we will present a Table 2, illustrating examples of calculating the degree of a vertex in certain ${\varnothing }_I$ ( $Z_n)$ .

Corollary 5.4:

Let $=Z_{P^2}$ , $P>2$ , prime number, for any vertex $x$ ∈ V ( ${\varnothing }_{⟨P⟩}$ (R)). The degree of $x$ is then given by $deg\left(x\right)=P^2-P-3$ .

Corollary 5.5:

Let $=Z_{2P}$ , $P>2$ , prime number, for any vertex x ∈ V ( ${\varnothing }_I$ (R)), For any non-zero ideal of R, then

Theorem 5.6:

Let R = $Z_n$ be the ring of integers module $n$ , and let ${\varnothing }_0$ ( $Z_n$ ) be a non-zero divisor graph of R. Then, for any vertex $x\in V\left({\varnothing }_0\left(R\right)\right),$ set $g$ = $\mathit{deg}$ ( $x,n$ ). Thus, the degree of $x$ is given by:

Proof:

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

Case 1: if $g=\mathit{gcd}(x,n)=1$ , assume that for all y ∈ $V\left({\varnothing }_0\left(R\right)\right)$ , $\mathit{xy}=0$ , but $x$ is invertible thus, $y=0$ , which is a contradiction. Therefore $\mathit{deg}\left(x\right)=\left|V\left({\varnothing }_0\left(R\right)\right)\right|-1=n-3-1$ .

Case 2: Let $S=\{y\in V\left({\varnothing }_0\left(R\right)\right):\mathit{xy}=0\}$ , representing the non-neighbors of $x$ , then $S=\mathit{ann}\left(x\right)\cap V\left({\varnothing }_0\left(R\right)\right)$ ,which contains exactly $g$ elements if $x^2\ne 0$ , and $g+1$ elements when $x^2=0$

Example 5.7:

Let $R=Z_{12}$ then the following Table 3, representing the degree of every vertex $x$ of the graph ${\varnothing }_0\left(R\right)$ .

Example 5.8:

Let $R=Z_{30}$ then the following Table 4, representing the degree of every vertex $x$ of the graph ${\varnothing }_0\left(R\right)$ .