**5.** *TI***(Γ(***R***)) Is Eulerian and Hamiltonian**

In this section, we determine when *TI*(Γ(R)) is Eulerian, Hamiltonian, and *TI*(Γ(R)) contains a Eulerian trail.

**Theorem 14.** *Let I be an ideal of a ring* R *such that* |R| = *n* ≥ 3*. If I* = R *and* |R| *are odd, then TI*(Γ(R)) *is a Eulerian.*

**Proof.** Suppose that *I* = R and |R| is odd. Then, by Corollary 5, *TI*(Γ(R)) is a complete graph *Kn* of odd vertices. Thus, the degree of each vertex of *TI*(Γ(R)) is even. Hence, *TI*(Γ(R)) is Eulerian.

**Theorem 15.** *Let I be a zero ideal of a ring* R *such that* |*X*| = 1 *and* |*Y*| *are even. If each vertex of* Γ(R) = Γ0(R) *has odd degree, then TI*(Γ(R)) *contains a Eulerian trail.*

**Proof.** Suppose that each vertex of Γ(R) = Γ0(R) has odd degree. Then, each vertex of *Y* in *TI*(Γ(R)) has even degree. Since |*X*| = 1 and |*Y*| are even, the vertex 0 of *I* has odd degree. Moreover, the degree of the vertex of *X* has degree one. Thus, each vertex of *TI*(Γ(R)) has an even degree except for two vertices that have odd degrees. Hence, *TI*(Γ(R)) contains a Eulerian trail.

**Remark 8.** *In view of Theorem 15, the element of X is the unity of the ring* R*. Moreover, the Eulerian trail of TI*(Γ(R)) *begins at unity and ends at zero of* R *or begins at zero and ends at unity of* R *(for example,* R = Z<sup>2</sup> × Z<sup>2</sup> *and I* = (0)*).*

**Theorem 16.** *Assume that a prime ideal of Ring* R *is I. Thus, if and only if* |*I*| *is even and* |*X*| *is odd, then TI*(Γ(R)) *is Eulerian.* |R| *also is odd.*

**Proof.** Suppose that *TI*(Γ(R)) is Eulerian. Then, every vertex of *TI*(Γ(R)) has an even degree. Since *I* is a prime ideal of R, the degree of each vertex of *TI*(Γ(R)) either (|R| − 1) or |*I*| (Theorem 6). Therefore, we have the following cases.

Case(*i*): If *u* ∈ *X*, then *deg*(*u*) = |*I*|, which is even. Thus, |*I*| is even.

Case(*ii*): If *u* ∈ *I*, then *deg*(*u*) = |R| − 1, which is even, and we obtain |R| as odd.

Now, we have |R| as odd and |*I*| as even. Therefore, |*X*| is odd.

Conversely, suppose that |*I*| is even and |*X*| is odd. Then, |R| is odd. Thus, |R| − 1 is even, and |*I*| is also even. Since *I* is a prime ideal of R, the degree of each vertex of *TI*(Γ(R)) is either |R| − 1 or |*I*|. Thus, the degree of each vertex of *TI*(Γ(R)) is even. Hence, *TI*(Γ(R)) is Eulerian.

**Theorem 17.** *Let I be a prime ideal of a ring* R*. Then, TI*(Γ(R)) *contains a Eulerian trail if and only if either* |*I*| = 2 *or* |*X*| *is even or* |*X*| = 2 *and* |*I*| *are odd.*

**Proof.** Suppose that *TI*(Γ(R)) contains a Eulerian trail. Then, exactly two vertices of *TI*(Γ(R)) have odd degree. Since *I* is a prime ideal of R, the vertex set of *TI*(Γ(R)) consists of *I* and *X* only, and *Y* will vanish. Let *u* and *v* be the two vertices of odd degree and let *w*1, *w*2, ..., *wn* be the vertices of even degree. Then, we have the following cases.

Case(*i*): If *u*, *v* ∈ *I* and *wi* ∈ *X* for all 1 ≤ *i* ≤ *n*, then *deg*(*u*) = *deg*(*v*) is odd and *deg*(*wi*) for all 1 ≤ *i* ≤ *n* is even. Therefore, |R| − 1 is odd, and |*I*| = 2 is even; thus, |R| is even and |*I*| = 2. Hence, |*I*| = 2 and |*X*| is even. Moreover, |R| is even.

Case(*ii*): If *u*, *v* ∈ *I* and there exists at least one *wj* ∈ *I*, then *deg*(*u*) = *deg*(*v*) = *deg*(*wj*) is odd. Hence, there are more than two odd vertices in *TI*(Γ(R)), and we obtain a contradiction.

Case(*iii*): If *u*, *v* ∈ *X* and *wi* ∈ *I* for all 1 ≤ *i* ≤ *n*, then *deg*(*u*) = *deg*(*v*) is odd and *deg*(*wi*) for all 1 ≤ *i* ≤ *n* is even. Note that |*I*| is odd, and |R| − 1 is even. We obtain |*I*| as odd and |R| as odd. Since *u*, *v* ∈ *X* only, we have |*X*| = 2. Hence, |*X*| = 2 and |*I*| are odd. Moreover, |R| is odd.

Case(*iv*): If *u*, *v* ∈ *X* and there exists at least one *wj* ∈ *X*, then *deg*(*u*) = *deg*(*v*) = *deg*(*wj*) is odd. Thus, there are more than two odd vertices in *TI*(Γ(R)), we obtain a contradiction.

Case(*v*): If *u* ∈ *I* and *v* ∈ *X*, then *deg*(*u*) = *deg*(*v*) = *deg*(*wi*) for all 1 ≤ *i* ≤ *n* is odd. Thus, all the vertices of *TI*(Γ(R)) have odd degree, and we obtain a contradiction. Therefore in all the cases, we obtain either |*I*| = 2 and |*X*| as even or |*X*| = 2 and |*I*| as odd.

Conversely, suppose that either |*I*| = 2 and |*X*| are even or |*X*| = 2 and |*I*| are odd. We first assume that |*I*| = 2 and |*X*| are even. Thus, |R| is even, and let *u* be any vertex of *TI*(Γ(R)). Therefore, we have the following cases.

Case(*i*): If *u* ∈ *I*, then *deg*(*u*) = |R| − 1, which is odd. Since |*I*| = 2 and |*X*| are even, and there are only two vertices in *I* posessing an odd degree, and each vertices in *X* has an even degree. Hence, *TI*(Γ(R)) contains a Eulerian trail.

Case(*ii*): If *u* ∈ *X*, then *deg*(*u*) = |*I*| = 2, which is even by the same argument; there are only two vertices *w*1, *w*<sup>2</sup> ∈ *I* such that *w*<sup>1</sup> and *w*<sup>2</sup> are adjacent to each vertices in *X* and *w*<sup>1</sup> adjacent to *w*<sup>2</sup> and *deg*(*w*1) = *deg*(*w*2) = |*X*| + 1, which is odd. Therefore, there are only two vertices in *I* that have an odd degree, and each other vertices in *X* has even degree. Hence, *TI*(Γ(R)) contains a Eulerian trail.

Now, we assume that |*X*| = 2 and |*I*| are odd. Thus, |R| is odd, and let *u* be any vertex of *TI*(Γ(R)). Then, we have the following cases.

Case(*i*): If *u* ∈ *I*, then *deg*(*u*) = |R| − 1, which is even. Since |*X*| = 2 and |*I*| are odd, there are only two vertices in *X* that have an odd degree, and each other vertices in *I* has an even degree. Hence, *TI*(Γ(R)) contains a Eulerian trail.

Case(*ii*): If *u* ∈ *X*, then *deg*(*u*) = |*I*|, which is odd; thus, |R| is odd. By the same argument, there are only two vertices in *X* possessing an odd degree and each other vertices in *I* has degree |R| − 1, which is even. Hence, *TI*(Γ(R)) contains a Eulerian trail.

From all the above cases, we conclude that *TI*(Γ(R)) contains a Eulerian trail. Hence, if either |*I*| = 2 and |*X*| are even or |*X*| = 2 and |*I*| are odd, then *TI*(Γ(R)) contains a Eulerian trail.

**Remark 9.** *In view of Theorem 17, if* |*I*| = 2*, then Eulerian trail of TI*(Γ(R)) *begins at one of these two elements of I and ends at other(for example* R = Z<sup>6</sup> *and I* = (3) *see Example 2(iii) Figure 7). Moreover, if* |*X*| = 2*, then the Eulerian trail of TI*(Γ(R)) *begins at one of these two elements of X and ends at the other (for example* R = Z<sup>3</sup> *and I* = (0)*).*

**Theorem 18.** *Let I be an ideal of a ring* R *that is not prime such that* |*I*| *is even,* |*X*| *is odd, and* |*Y*| *is even. Then, we have the following case:*


**Theorem 19.** *Let I be an ideal of a ring* R *that is not prime such that* |*I*| *is even,* |*X*| *is even, and* |*Y*| *is odd. Then, we have the following case.*


**Remark 10.** *In view of Theorems 16, 18, and 19, since* |*I*| *is even and* |R| *is odd, there is no graph on n vertices that can be realized as TI*(Γ(R)) *for some ring* R *and an ideal I of* R*.*

**Theorem 20.** *Let I be a non-prime ideal of a ring* R *such that* |*I*| = 2*,* |*X*| *is even, and* |*Y*| *is even. Then, TI*(Γ(R)) *contains a Eulerian trail if and only if* Γ*I*(R) *is Eulerian.*

**Proof.** Suppose that *TI*(Γ(R)) contains a Eulerian trail. Then, each vertex of *TI*(Γ(R)) has an even degree except for two vertices that have odd degrees. Since *I* is not a prime ideal of R, the vertex set of *TI*(Γ(R)) consists of *I*, *X*, and *Y*. Let *u* and *v* be the two vertices of odd degree and let *w*1, *w*2, ..., *wn* be the vertices of even degree. Since |*X*| and |*Y*| are even and |*I*| = 2, *u*, *v* ∈ *I*. If we assume that at least one of *u*, *v* ∈ *X*, then |*I*| is odd, which is a contradiction, and if we assume that at least one of *u*, *v* ∈ *Y*, then we have more than two odd vertices in *TI*(Γ(R)). Therefore, *TI*(Γ(R)) has no Eulerian trail, which is a contradiction. Thus, the two elements of *I* are the only ones that are odd and all other elements of *X* and *Y* are even. Now, we know that Γ*I*(R) is a subgraph of *TI*(Γ(R)), i.e., the set *Y* is the set of vertex of Γ*I*(R), but to obtain the exact number of edges that incident on each vertex of Γ*I*(R), we remove the edges join all vertex of *I* with each vertex of *Y*. Thus, the degree of each vertex of Γ*I*(R) is same as the degree of each vertex of *Y* in *TI*(Γ(R)) subtract |*I*|. Therefore, the degree of each vertex of Γ*I*(R) is even. Hence, Γ*I*(R) is Eulerian.

Conversely, suppose that Γ*I*(R) is Eulerian. Then, each vertex of Γ*I*(R) has an even degree. Thus, each vertex of *Y* in Γ*I*(R) has an even degree. Now, we will construct *TI*(Γ(R)). First, we have two vertices of the set *I* adjacent to all vertex of R, i.e., adjacent to (|*I*| − 1 + |*X*| + |*Y*|) vertices. Since |*I*| = 2, |*X*| and |*Y*| are even, and the degree of each vertex of *I* is odd, i.e., we have two vertices of *I* being odd. Now, if we have an even vertex of set *X*, which is adjacent to all vertices of *I* only. Thus, the degree of each vertex of *X* is even. Finally, we have an even vertex of set *Y* adjacent to all vertex of *I* and at least one vertex of *Y*. Since |*I*| = 2 is even and each vertex of *Y* in Γ*I*(R) has an even degree, the degree of each vertex of *Y* in *TI*(Γ(R)) is even. Thus, each vertex of *TI*(Γ(R)) has an even degree except for two vertices, which have odd degrees. Hence, *TI*(Γ(R)) contains a Eulerian trail.

**Theorem 21.** *Let I be an ideal of a ring* R *that is not prime such that* |*I*| = 2*,* |*X*| *is odd, and* |*Y*| *is odd. Then, TI*(Γ(R)) *contains a Eulerian trail if and only if* Γ*I*(R) *is Eulerian.*

**Proof.** By the same arguments used in the above Theorem 18, the proof is clear.

**Example 3.** *Let* R = Z<sup>12</sup> *and I* = (6) = {0, 6}*. Then, Y* = {2, 3, 4, 8, 9, 10} *and X* = {1, 5, 7, 11}*. We observe that* Γ*I*(R) *is Eulerian and TI*(Γ(R)) *contains a Eulerian trail (see Figure 9).*

**Figure 9.** (**left**)*TI*(Γ(R)) and (**right**) Γ*I*(R), when R = Z<sup>12</sup> and *I* = (6).

**Theorem 22.** *Let I be an ideal of a ring* R *such that* |R| = *n* ≥ 3*. If I* = R*, then TI*(Γ(R)) *is Hamiltonian.*

**Proof.** Suppose that *I* = R. Then, by Corollary 5, *TI*(Γ(R)) is a complete graph. Hence, *TI*(Γ(R)) is Hamiltonian.

**Theorem 23.** *Let I be a zero ideal of a ring* R*. Then, TI*(Γ(R)) *cannot be Hamiltonian.*

**Proof.** Suppose that *I* = (0). Then, by Corollary 9, *TI*(Γ(R)) has a cut-vertex. Hence *TI*(Γ(R)) cannot be Hamiltonian.

**Theorem 24.** *Let I be a nonzero ideal of a ring* R *such that* |R| = *n* ≥ 3*. If* |*I*| ≥ *<sup>n</sup>* <sup>2</sup> *, then TI*(Γ(R)) *is Hamiltonian.*

**Proof.** Suppose that *u* and *v* are any two vertices of *TI*(Γ(R)). Then, we have the following cases.

Case(*i*): If *u* and *v* are adjacent for all *u*, *v* ∈ *TI*(Γ(R)), then *TI*(Γ(R)) is complete. Therefore, *TI*(Γ(R)) is Hamiltonian.

Case(*ii*): If *u* and *v* are nonadjacent for some *u*, *v* ∈ *TI*(Γ(R)), then by Remarks 1 (*viii*),(*ix*), and (*x*), we have the following subcases:

Subcase(*a*): If *u*, *v* ∈ *X*, then by Corollary 7, *deg*(*u*) = *deg*(*v*) = |*I*|. Thus, *deg*(*u*) + *deg*(*v*) = |*I*| + |*I*| ≥ *<sup>n</sup>* <sup>2</sup> <sup>+</sup> *<sup>n</sup>* <sup>2</sup> = *n*. Hence, *TI*(Γ(R)) is Hamiltonian.

Subcase(*b*): If *u*, *v* ∈ *Y*, then by Corollary 7, *deg*(*u*) = *deg*(*v*) ≥ |*I*|. Thus, *deg*(*u*) + *deg*(*v*) ≥ |*I*| + |*I*| ≥ *<sup>n</sup>* <sup>2</sup> <sup>+</sup> *<sup>n</sup>* <sup>2</sup> = *n*. Hence, *TI*(Γ(R)) is Hamiltonian.

Subcase(*c*): If *u* ∈ *X* and *v* ∈ *Y*, then by Corollary 7, *deg*(*u*) = |*I*| and *deg*(*v*) ≥ |*I*|. Thus, *deg*(*u*) + *deg*(*v*) ≥ |*I*| + |*I*| ≥ *<sup>n</sup>* <sup>2</sup> <sup>+</sup> *<sup>n</sup>* <sup>2</sup> = *n*. Hence, *TI*(Γ(R)) is Hamiltonian.

**Corollary 13.** *Let I be a nonzero ideal of a ring* R *such that* |R| = *n* ≥ 3*. If* |*I*| ≥ *<sup>n</sup>* <sup>2</sup> *for each pair u*, *v of X, then TI*(Γ(R)) + *uv is Hamiltonian if and only if TI*(Γ(R)) *is Hamiltonian.*

**Corollary 14.** *Let I be a nonzero ideal of a ring* R*. Then, I is a maximal ideal of a ring* R *if and only if TI*(Γ(R)) *is Hamiltonian.*

#### **6. Conclusions**

We considered a generalization of dot total graph of R as well as an ideal-based zero-divisor graph. We showed that *TI*(Γ(R)) is connected and has a small diameter of at most two. Furthermore, we studied the connectivity, clique number and the girth of *TI*(Γ(R)). In addition, the cases when *TI*(Γ(R)) is Eulerian, Hamiltonian, and *TI*(Γ(R)) contains a Eulerian trail. For future work, the application of this graph to the study on Laplacian eigenvalues of an ideal-based dot total graph, which is closely related to the work in the paper [6], can be investigated. Additionally, the energy of an ideal-based dot total graph, which is related to the recent work in [5,7], requires more consideration. The purpose of studying this type of graph is beneficial from its application point of view in practical life, such as networks, especially communication networks, which will be studied in an independent manuscript.

**Author Contributions:** Investigation, M.A., J.H.A., A.M.A. and A.A.; Writing—original draft, M.A., J.H.A., A.M.A. and A.A.; Writing—review and editing, M.A., J.H.A., A.M.A. and A.A. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Acknowledgments:** The authors are grateful to the referees for their valuable comments and suggestions.

**Conflicts of Interest:** The authors declare no conflict of interest.
