**2. Definition and Basic Structure of** *TI***(Γ(***R***))**

In this section, we define an ideal-based dot total graph, denoted by *TI*(Γ(R)), and show that this graph is always connected and has a small diameter of at most two. By dividing the element of R into three disjoint subsets, we study the basic results on the structure of this graph and the relationship between *TI*(Γ(R)) and *TZ*(R)(Γ(R)), *TZ*(R/*I*)(Γ(R/*I*)), Γ*I*(R), or Γ(R/*I*) with some examples clarification. Moreover, we find the degree of each vertex of *TI*(Γ(R)) that depends on the three sets *I*, *X*, and *Y*. Furthermore, we determine the case when *TI*(Γ(R)) is a complete graph or a regular graph.

**Definition 1.** *Let* R *be a commutative ring with* 1 = 0 *and ideal I. Then, a simple graph that is not directed is defined as TI*(Γ(R))*, possessing vertices of* R*. In this case, the graph has vertices x and y that are both distinct and adjacent if and only if xy* ∈ *I.*

**Proposition 1.** (*a*) *If I* = (0)*, then TI*(Γ(R)) = Γ0(R)*.*

(*b*) *If Z*(R) *is an ideal and I* = *Z*(R)*, then TI*(Γ(R)) = *TZ*(R)(Γ(R))*.*

(*c*) *If I* = R*, then TI*(Γ(R)) = *Kn, where n* = |R|*.*

(*d*) *Let* R *have a proper nonzero ideal I. Consequently, this means that TZ*(R/*I*)(Γ(R/*I))= K*1,*n. The value of n is given by n* = |R/*I*| − 1 *if and only if the prime ideal of* R *is I.*

**Proof.** The proofs of (*a*) and (*b*) follow by the definition of the zero-divisor graph of R, which appeared in Beck [4], and definition of the dot total graph of R, which appeared recently in Ashraf et al. [1], respectively.

(*c*) Let *x* and *y* be distinct vertices of R. Then, *xy* ∈ R = *I*. Thus, *x* is adjacent to *y* for all *x*, *y* ∈ R. Hence, *TI*(Γ(R)) = *Kn*, where *n* = |*I*| = |R|. This completes the proof.

(*d*) Let *I* be a prime ideal of R. Then, R/*I* is an integral domain. Thus, *TZ*(R/*I*)(Γ(R/*I*)) is a star graph. Hence, *TZ*(R/*I*)(Γ(R/*I*)) = *K*1,*n*, where *n* = |R/*I*| − 1.

Conversely, let *TZ*(R/*I*)(Γ(R/*I*)) = *K*1,*n*, where *n* = |R/*I*| −1. Then, *TZ*(R/*I*)(Γ(R/*I*)) is a star graph, and we have the following two cases:

Case(*i*) If *Z*(R/*I*) = 1, then R/*I* is an integral domain.

Case(*ii*) If *Z*(R/*I*) > 1, then there exists at least two vertices in *Z*(R/*I*). Therefore, *TZ*(R/*I*)(Γ(R/*I*)) is not a star graph, which is a contradiction.

Thus, we obtain R/*I* as an integral domain. Hence, *I* is the prime ideal of R. This completes the proof.

In view of the following examples, we shall find *TI*(Γ(R)) and *TZ*(R/*I*)(Γ(R/*I*)) with several ideals *I* of the same ring R.

**Example 1.** *Let* R = Z8*. Then,* (0),(2),(4)*, and* Z<sup>8</sup> *are ideals of* R*:* (*i*) *Let I* = (0)*, then* R/*I* = Z<sup>8</sup> *(see Figure 1).*

**Figure 1.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>8</sup> and *I* = (0).

(*ii*) *Let I* = (2)*, then* R/*I* = Z<sup>2</sup> *(see Figure 2).*

**Figure 2.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>8</sup> and *I* = (2).

(*iii*) *Let I* = (4)*, then* R/*I* = Z<sup>4</sup> *(see Figure 3).*

**Figure 3.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>8</sup> and *I* = (4).

(*iv*) *Let I* = Z8*, then* R/*I* = (0) *(see Figure 4).*

**Figure 4.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>8</sup> and *I* = Z8.

**Example 2.** *Let* R = Z6*. Then* (0),(2),(3)*, and* Z<sup>6</sup> *are ideals of* R*:* (*i*) *Let I* = (0)*, then* R/*I* = Z<sup>6</sup> *(see Figure 5).*

**Figure 5.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>6</sup> and *I* = (0).

(*ii*) *Let I* = (2)*, then* R/*I* = Z<sup>2</sup> *(see Figure 6).*

**Figure 6.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>6</sup> and *I* = (2).

(*iii*) *Let I* = (3)*, then* R/*I* = Z<sup>3</sup> *(see Figure 7).*

**Figure 7.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>6</sup> and *I* = (3).

(*iv*) *Let I* = Z6*, then* R/*I* = (0) *(see Figure 8).*

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**Figure 8.** (**left**) *TZ*(R/*I*)(Γ(R/*I*)) and (**right**) *TI*(Γ(R)), when R = Z<sup>6</sup> and *I* = Z6.

**Theorem 1.** *TI*(Γ(R)) *is connected and diam*(*TI*(Γ(R))) ≤ 2*. Therefore, where TI*(Γ(R)) *has a cycle, it implies gr*(*TI*(Γ(R))) ≤ 5*.*

**Proof.** Assume that *x* and *y* are distinct vertices of *TI*(Γ(R)). In such a scenario, various cases will be true as shown:

Case(*i*): If *x* ∈ *I* and *y* ∈ *I*, then *xy* ∈ *I*. Thus, *x* − *y* is a path of length one in *TI*(Γ(R)).

Case(*ii*): If *x* ∈ *I* and *y* ∈/ *I*, then *xy* ∈ *I*. Thus, *x* − *y* is a path of length one in *TI*(Γ(R)).

Case(*iii*): If *x* ∈/ *I* and *y* ∈ *I*, then this will result in a path of length one as in the previous case.

Case(*iv*): If *x* ∈/ *I* and *y* ∈/ *I*, then we will consider the following subcases:

Subcase(*a*): If *xy* ∈ *I*, then *x* − *y* is a path of length one in *TI*(Γ(R)).

Subcase(*b*): If *xy* ∈/ *I*, then there is some *z* ∈ *I* such that *xz* ∈ *I* and *zy* ∈ *I*. Thus, *x* − *z* − *y* is a path of length two in *TI*(Γ(R)). Hence, *TI*(Γ(R)) is connected and *diam*(*TI*(Γ(R))) ≤ 2. Since for any undirected graph, *H* contains a cycle, *gr*(*H*) ≤ 2 *diam*(*H*) + 1 (for reference see [8]). Thus, *gr*(*TI*(Γ(R))) ≤ 5.

Suppose that R is a commutative ring, and *I* is an ideal of R. We construct a graph *TI*(Γ(R)) with the following method:

First, the set of vertices of *TI*(Γ(R)) can be classified into three disjoint subsets of R:

(*i*) *I* = (*a*) is the subset of R such that *I* is the ideal generated by the element *a*.

(*ii*) *X* = {*x* ∈R\ *I* : *xb* ∈/ *I*, for all *b* ∈R\ *I*}.

(*iii*) *Y* = {*y* ∈R\ *I* : *yb* ∈ *I*, for some *b* ∈R\ *I*}.

Second, we will connect the edges between the vertices defined in the three previous sets as follows:

We define a complete graph (*Kn*, where *n* = |*I*|) by using the first set *I* = (*a*) as its vertex set. Thus, we have an edge between each vertex of *I* (i.e., *ab* ∈ *I*). Then, join each vertex of the second set *X* to all vertices of the complete graph *Kn*, and similarly, join each vertex of the third set *Y* to all vertices of the complete graph *Kn*. Thus, we have an edge between each vertex of the sets *X* and *Y* with all vertex of *I* (i.e., *xa* ∈ *I* and *ya* ∈ *I*). Finally, in this part of the edges the relationship between *TI*(Γ(R)) and Γ*I*(R) is identical. Thus, for distinct *y*1, *y*<sup>2</sup> ∈ *Y*, *y*<sup>1</sup> is adjacent to *y*<sup>2</sup> in *TI*(Γ(R)) if and only if *y*<sup>1</sup> is adjacent to *y*<sup>2</sup> in Γ*I*(R) (i.e., *y*1*y*<sup>2</sup> ∈ *I*).

Henceforth, we shall rely on the three sets *I*, *X*, and *Y* defined above in this paper.

**Theorem 2.** *Let y*<sup>1</sup> *and y*<sup>2</sup> *be any two distinct vertices of the third set Y and y*<sup>1</sup> *is adjacent to y*<sup>2</sup> *in TI*(Γ(R))*. If y*<sup>1</sup> + *I* = *y*<sup>2</sup> + *I, then y*<sup>1</sup> + *I is adjacent to y*<sup>2</sup> + *I in* Γ(R/*I*) *and if y*<sup>1</sup> + *I* = *y*<sup>2</sup> + *I, then y*<sup>2</sup> <sup>1</sup>, *<sup>y</sup>*<sup>2</sup> <sup>2</sup> ∈ *I.*

**Corollary 1.** *Let TI*(Γ(R)) *have two vertices (u and v) that are both distinct and adjacent. This implies that the elements, u* + *I and v* + *I, are adjacent in TI*(Γ(R))*. Assuming that u*<sup>2</sup> ∈ *I, this implies that TI*(Γ(R)) *has all the distinct elements of u* + *I adjacent to it.*

**Corollary 2.** *Let I be an ideal of* R*. Then,* Γ(R/*I*) *and* Γ*I*(R) *are subgraphs of TI*(Γ(R))*.*

**Corollary 3.** *Assume that a nonzero ideal of* R *is I. Then, if and only if the prime ideal of* R *is I, TZ*(R/*I*)(Γ(R/*I*)) *is subgraph of TI*(Γ(R))*.*

**Corollary 4.** *TI*(Γ(R)) *contains* |*I*| *disjoint subgraphs isomorphic to* Γ(R/*I*)*.*

**Proof.** Since Γ*I*(R) is subgraph of *TI*(Γ(R)) (see Corollary 2) and Γ*I*(R) contains |*I*| disjoint subgraphs isomorphic to Γ(R/*I*), *TI*(Γ(R)) contains |*I*| disjoint subgraphs isomorphic to Γ(R/*I*).

**Remark 1.** *Let I*, *X, and Y be three disjoint sets defined above, and u*, *v* ∈ R*. Then, we have the following results:*


**Theorem 3.** *Let I be a prime ideal of* R*. Then, TI*(Γ(R)) *contains a subgraph isomorphic to Kn, where n* = |*I*| + 1*.*

**Proof.** Since *I* is a prime ideal, *Y* will vanish, and there is at least one element *v* in *X*. Moreover, we have a complete subgraph *Kn*, where *n* = |*I*| and the vertex *v* is adjacent to each vertex of *Kn*. Thus, we have a complete subgraph of order *n* = |*I*| + 1.

**Theorem 4.** *Let I be a non-prime ideal of* R*, and there exists u* ∈ *Y such that u*<sup>2</sup> ∈ *I. Then, TI*(Γ(R)) *contains a subgraph isomorphic to Km, where m is at least* 2|*I*|*.*

**Proof.** Since there exists *u* ∈ *Y* such that *u*<sup>2</sup> ∈ *I*, all the distinct elements of *u* + *I* are adjacent in *TI*(Γ(R)) (see Corollary 1). Thus, we have a complete subgraph of order *n* = |*I*|. Moreover, by element of the set *I*, we have a complete subgraph *Kn*, where *n* = |*I*| and all the distinct elements of *u* + *I* are adjacent to each vertex of *Kn*. Thus, we have a complete subgraph of order *m* = *n* + *n* = |*I*| + |*I*| = 2|*I*|. Therefore, if *Y* consists of the elements *u* + *I* only, then *m* = 2|*I*|. If there exists *v* ∈ *Y* other than the elements *u* + *I*, then *v* is adjacent to all the elements *u* + *I* or there exists an element *w* ∈ *Y* such that *v* is adjacent to *w*, and *w* is adjacent to all the elements *u* + *I*. Hence, in both cases we have a complete subgraph of order |*I*| + 1, and all elements of this subgraph are adjacent to each vertex of *Kn*. Thus, we have a complete subgraph of order *m* = |*I*| + 1 + |*I*| = 2|*I*| + 1.

**Theorem 5.** *Let I be an ideal of* R *that is not prime, and u*<sup>2</sup> ∈/ *I for all u* ∈ *Y. Then, TI*(Γ(R)) *contains a subgraph isomorphic to Km, where m is at least* |*I*| + 1*.*

**Proof.** Assume that *u*<sup>2</sup> ∈/ *I* for all *u* ∈ *Y*. By the same arguments as used in Theorems 3 and 4, we obtain the result.

**Corollary 5.** *TI*(Γ(R)) *is a complete graph if and only if either I* = R *or* R ∼= Z2*.*

**Theorem 6.** *Let I be a prime ideal of* R*. Then, the degree of each vertex of TI*(Γ(R)) *is either* |R| − 1 *or* |*I*|*.*

**Proof.** Since *I* is a prime ideal of R, the set *Y* will vanish. Thus, R consists of two disjoint subsets *I* and *X*. Then, we have the following two cases:

Case (*i*): If *u* ∈ *I*, then *u* is adjacent to each vertex in *TI*(Γ(R)) except *u*; that is, *u* is adjacent to (|R| − 1) vertices, and hence the degree of *u* is |R| − 1.

Case (*ii*): If *u* ∈ *X*, then *u* is adjacent to the vertices, which belongs to *I*; that is, *u* is adjacent to |*I*| vertices and, hence, *deg*(*u*) = |*I*|.

**Corollary 6.** *Let I be an ideal of* R *and u* ∈ *Y. Then, the number of elements of TI*(Γ(R)) *adjacent to u is either* |*I*| *or at least* |*I*| + 1*.*

**Proof.** Let *u* ∈ *Y*. Then, we have two types of adjacencies. First, *u* is adjacent to each element of *I*, i.e., *u* is adjacent to |*I*| vertices. Second, if |*Y*| ≥ 2, then *u* is adjacent to some elements of *Y*, i.e., *u* is adjacent to at least one vertex of *Y*. Thus, *u* is adjacent to at least |*I*| + 1 vertices. If |*Y*| = 1, then *u* is adjacent to |*I*| elements only. Hence, *deg*(*u*) ≥ |*I*| + 1 or *deg*(*u*) = |*I*|.

**Corollary 7.** *Let I be an ideal of* R *and u* ∈ *TI*(Γ(R))*. Then, the degree of u depends on the three sets I*, *X, and Y defined earlier as follows.*

$$\deg(u) = \begin{cases} |\mathcal{R}| - 1 & \text{if } u \in I \\ |I| & \text{if } u \in X \text{ or } u \in Y \text{ and } |Y| = 1 \\ \text{at } I \text{ast } |I| + 1 & \text{if } u \in Y \text{ and } |Y| \ge 2. \end{cases}$$

**Corollary 8.** *TI*(Γ(R)) *is regular graph if and only if either <sup>I</sup>* = R *or* R ∼= Z<sup>2</sup> *(i.e., TI*(Γ(R)) *is a complete graph).*

**Remark 2.** *Minimum degree of TI*(Γ(R)) *is δ*(*TI*(Γ(R))) = |*I*|*, and maximum degree of TI*(Γ(R)) *is* Δ(*TI*(Γ(R))) = |R| − 1*.*
