# Quasi-Regular Graphs Associated with Commutative Rings

Authors: Nasr Zeyada; Najat Muthana; Sultanah Al-Rashidi
Journal: Journal of Mathematics
 (2022)

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## Abstract

One of the most important branches of mathematics is algebraic graph theory, which solves graph problems with algebraic methods. In graph theory, several algebraic properties of a ring can be represented. In this paper, we define an innovative graph on rings, explore its characteristics, and examine how it relates to other notions in the field. LetS be a ring; the quasi-regular graph of S is a graph with a vertex set of S−0 and any two different vertices w and z are adjacent if 1−wz is a unit in S. We study this graph by providing different examples and proving some crucial characteristics. This study provides important results and paves the way for a lot of different inquiries and studies utilizing this novel approach.


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## Body

## 1. Introduction

The relationship between graph theory and rings has been extensively studied and has produced amazing results in algebraic graph theory. The investigation of the theoretical aspects of a graph over a ring graph was first thought of by Beck in 1988 [1]. He proposed the concept of a zero-divisor graph ΓS for a commutative ring S. He defined ΓS as an undirected graph with components at its vertices ZS∗=ZS−0, the set of nonzero zero-divisors of S, where any two vertices w and z are adjacent if wz=0 [1]. The first main rearrangements of Beck’s zero-divisor graph were presented by Anderson and Livingston [2].Grimaldi was the first to study the unit graph forℤn [3]. The unit graph of S, denoted by GS, is the graph produced by making all elements of S vertices and defining different vertices w and z to be adjacent if and only if w+z∈US where US is the set of unit elements of S.One sort of graph related to rings, the unit graph was first introduced in 2010 by Ashrafi et al. [4]. Such graph extended the unit graph Gℤn to GS and derived numerous characterization findings for finite commutative rings in terms of connectedness, chromatic index, diameter, girth, and planarity of GS. If we remove the phrase “distinct” from the definition, we get the closed unit graph G¯S, which may contain loops. It is worth noting that if 2∉US, then G¯S=GS.A graph’s diameter is an important invariant. Many articles in this field are devoted to the diameter of the generated graph; see, for example, [5, 6]. In 2014, Su and Zhou [7] proved that, for any arbitrary ring S, the girth of GS is 3,4,6, or ∞ and recently explored the diameter of GS with S/JS self-injective ring S and provided a thorough characterization of the diameter of GS=1,2,3 or ∞, respectively [8].In this paper, we introduce a new concept that we call a quasiregular graph and symbolise it withQS, where S is a commutative ring with a nonzero identity. Furthermore, we determine when a quasiregular graph is isomorphic to several well-known graphs. We also provide some number theory approaches to prove some of the properties of a quasiregular graph over S.We refer to [9–12] for the undefined notions in this paper.


## 2. Quasiregular Graph

Definition 1.
A quasiregularQS is a graph with vertex set VQS=S/0 in which any two distinct vertices w and z are adjacent if and only if 1−wz is a unit in S. Therefore, the edge set of QS is EQS=wz:1−wz is a unit in S for w,z∈S.Remark 1.
Using the notions of Jacobson radical of ringJS=∩m:m is a maximal ideal of S, wz∈EQS if and only if w∈JS for any z∈S.
Hence, the elements of Jacobson Radical of a ringS are adjacent to each vertex in the quasiregular graph [13].
The following examples show the difference between the notion of a unit graph and the notion of a quasiregular graph over the same ringℤ6.Example 1.
Figure1, instant unit graph of the ring ℤ6.Figure 1
Unit graph ofℤ6.Example 2.
The quasiregular graph of the ringℤ6 is shown in Figure 2.
We now illustrate some examples of quasiregular graphs over an infinite ring.Figure 2
Quasiregular graph ofℤ6.Example 3.
For the ring of integersℤ. We find that the edge set of the quasiregular graph of ℤ contains edges 12, and −1−2, and otherwise, all elements in ℤ/1,−1,2,−2 of the graph will be isolated points.Example 4.
The quasiregular graph of rational numbersℚ. For all a∈ℚ∗, a is adjacent to b for all b≠1/a in ℚ∗.Example 5.
LetF4 be the extension field over ℤ2x by the irreducible polynomial Px=x2+x+1 is over ℤ2. Hence, F4=0,1,u,1+u. The quasiregular graph of F4 is a path shown in Figure 3.Figure 3
Quasiregular graphs of extension fieldF4.Example 6.
LetF8 be an extension field over ℤ2x by the irreducible polynomial Px=x3+x+1 over ℤ2. Hence, F8=0,1,u,u2,1+u,1+u2,u+u2,1+u+u2. The quasi-regular graph of F8 is shown in Figure 4.Figure 4
Quasiregular graphs of extension fieldF8.Proposition 1.
IfS≅R as rings, then QS≅QR as graphs.Proof.
Assume thatf:S⟶R is an isomorphism of rings. This isomorphism induces an isomorphism h:QS⟶QR where ha=fa for all a∈S∗. Indeed, if a,b∈S are adjacent vertices in QS, then 1S−ab is an invertible element in S. Hence, h1S−ab=1R−hahb is an invertible element in R and QS≅QR as graphs.
In the following example, we construct the quasiregular graphs of two isomorphic ringsℤ2×ℤ3≅ℤ6.Example 7.
By Proposition1, since ℤ2×ℤ3≅ℤ6 as rings Qℤ2×ℤ3≅Qℤ6 as graphs as shown in Figure 5.Figure 5
Quasiregular graphs of two isomorphic rings.Example 8.
In Figure6, the quasiregular graph over ℤ3 is complete.
The concept in the following theorem is based on the work of Ashrafi et al. [4].Figure 6
Complete quasiregular graphQℤ3.Theorem 1.
LetS be a ring. If S is a division ring with Char S = 2, then the unit graph GS is complete.
We prove the following result for quasiregular graphs over a ringS.Proposition 2.
For a commutative ringS with identity, the quasiregular graph QS is a complete graph if and only if either S≅ℤ2 or ℤ3.Proof.
Assume the commutative ringS, which has the identity 1, and suppose that QS is a complete graph. Such that S>3; and a∈S/0,1,−1 so 1 is adjacent to 1+a. Thus, 1−1+a=−a is a unit. Therefore, S is a field.
Ifa−1≠a, a−1 is not adjacent to a. Hence, a2=1 for all a∈S∗ and 22=1 whenever 2≠0. In this case, char S=3. Now, suppose a∈S/0,1,−1; 1=a+12=a2+2a+1 and 1+2a=0, so 1=−2a=a which is a contradiction. Therefore, S≅ℤ3.
Now, let charS=2; a∈S/0,1 implies that a+1≠0 and a+12=1 so a2+1=1 and a2=0 which is a contradiction. Hence, S≅ℤ2. The converse is clear.
Note that, as a result of Theorem1 and Proposition 2, the completeness property indeed gives an important difference between the graphs G(S) and Q(S).Example 9.
In Figure7, the quasiregular graph over ℤ4 is the path graph.Figure 7
Quasiregular graphs ofℤ4 and ℤ2.Proposition 3.
For a commutative ringS with 1, QS is a path graph if and only if S≅ℤ4.Proof.
Assume thatQS is a path a1,a2,…. So 1−a1a2 and 1−a2a3 are units; 1−a1a21−a2a3=1−a1a2−a2a3+a1a22a3=1−a2a1+a3−a1a2a3 is a unit. Hence, a1+a3−a1a2a3=a1 or a3 or a2, the first and second cases give a contradiction. Thus, a1+a3−a1a2a3=a2.
Now, if 1 has two adjacent elementsar,ar+2, then 1−ar and 1−ar+2 are units and ar+ar+2−arar+2=1. Since 1−2=−1 is a unit; So, 1 is adjacent to 2 and we may assume that ar=2. Thus, 1−2=ar+21−2 and ar+2=1 which is a contradiction. Hence, 1 is adjacent to only one element and the graph must be started by 1, 1−2−a3−⋯. The element 1−2a3 is a unit. Thus, we get 1−2a3=2−a3=1−a3−1 is a unit.
It follows thata3=3 and 1+3−6=2. Thus, −2=2 and 4=0, which means that char S=4. Assume that a∈S/0,1,2,3 with a is adjacent to 3, so 1−3a is a unit. Thus, 1−3a2=1−6a+9a2=1−2a+a2=1+a2 is a unit, and 1 is adjacent to a, which is a contradiction. Therefore, S≅ℤ4. The converse is clear from the preceding example.Example 10.
The quasiregular graph overℤ2, is cyclic as seen in Figure 7.Proposition 4.
The quasiregular graphQS over a ring S is cyclic if and only if S≅ℤ2.Proof.
LetQS be a cyclic graph. Then, 1 is adjacent to two elements a1,a2. Then, 1−a1 and 1−a2 are units and 1−a11−a2=1−a1−a2+a1a2 is a unit. It follows that a1+a2−a1a2=a1, a1+a2−a1a2=a2 or a1+a2−a1a2=1, which implies that a21−a1=0 or a11−a2=0 or a21−a1=1−a1. Since 1−a1 and 1−a2 are units; So, a2=1 the other possibility gives a contradiction. Thus, 1 is adjacent to only one element if S has more than two elements, which is a contradiction. Hence, S≅ℤ2.Definition 2.
A star graphSr is a bipartite graph that is complete K1,r with r+1 vertices [10], Figure 8.Figure 8
Star graphSr.Proposition 5.
For a commutative ringS with 1, the quasiregular graph QS is a star graph over the ring S if and only if either S≅F4 or S≅ℤ4.Proof.
Assume thatQS is a star graph; there are two cases;Case I: 1 is adjacent to everya in S/0,1, so 1−a is a unit for all a in S/0,1. If a∈S/0,1, then 1−a∉S/0,1 and 1−1−a=a is a unit. Hence S is a field. Now, if a≠b∈S/0,1, 1−ab is not a unit and 1=ab. Suppose that S has three different elements a1,a2,a3 in S/0,1, we have 1=a1a2, 1=a1a3 and 1=a2a3, which is in contradiction with the uniqueness of the inverse. Therefore, S is a field of four elements 0,1,a,b with ab=1 and S≅F4, where K is an expression filed of ℤ2 of degree 2 over the minimal polynomial x2+x+1.Case II: there ise∈S/0,1 in which e is adjacent to each element in S/0,e. In this case, 1−ea is a unit for every a in S/0,e. Assume that char S=2, so 1−ea is a unit for every a∈S∗ and 1−e2=1+e2 is a unit. Thus, JS=0,e and e2=e. It follows that e1−e=e−e2=0, but 1−e is a unit; So e=0 which is a contradiction. Hence; char S≠2. Now, we have two subcases:Case II.I:e=−1, which means that 1+a is a unit for all a∈S/0,−1. Thus 2 is a unit; Also, 1+2=3 is a unit and by induction 1,2,3,…⊆US. Thus 1+2=3 is a unit, which implies that 1 is adjacent to −2 and −2=−1. Thus 1=0 which is a contradiction.Case II.II:e∈S/0,1,−1, we know that −1 is a unit and 1−2=−1, So 1 is adjacent to 2 and 2=e. Thus 1−2−1=3 is a unit and 1 is adjacent to −2, which implies that −2=2 and 4=0. Hence char S=4. Now, assume that a∈S/0,1,2,3, so 1−2a is a unit. Thus 2a=2 for every a∈S/0,2. Since a≠1 and a≠3, So 2a+1=2 and 2a+2=2. It follows that 2=0 which is a contradiction. Therefore, S≅ℤ4. The converse is clear from Examples 5 and 9.
The diameter of the quasiregular graph for a commutative ring S is discussed in both cases: ifJS≠0 and if JS=0.Proposition 6.
For finite commutative ringS with 1, diamQS≤3.Proof.
For the Jacobson radicalJS, there are two cases;Case I:JS≠0; Let w be a nonzero element in JS. Then, wz−1 is a unit and w is adjacent to all z∈S. If a and b are nonadjacent elements, then we have a path a−w−b of length 2 where w is a nonzero element in JR. Therefore, dimR=2.Case II:JS=0; If S>6, then a direct product of finite fields is isomorphic to S. It is clear to see that diamQS=2 when S is a field. Since 1 is adjacent to all elements in S/0,1. We can assume, without losing generality, that S=F1×F2, F1 and F2 are finite fields, respectively. Thus, 1,1 is not adjacent to all elements 1,b and a,1, where a∈F1/1 and b∈F2/1. The elements of the form a,1 are adjacent to c,d where c≠1≠d and c≠a−1. The elements of the form 1,b are adjacent to 0,d where bd≠1. Note that, 1,1 is adjacent to all elements of the form c,d where c≠1≠d. (that case of S=6) Therefore; the diameter of QS≤3.
From the previous proposition, it was found that there are four possibilities for the value of the diameter of a quasiregular graph, if the diameter equals 0 then the example of that from the quasiregular graph over the ringℤ2 is as drawn in Figure 7. In Figure 6, The quasiregular graph of ring ℤ3 is the case if diam=1 . If the quasiregular graph is complete, for example, Qℤ4 then diam=2 as drawn in Figure 7. Finally; diam=3 as shown in the graph of ring ℤ6 is drawn Qℤ6 in Figure 2.Remark 2.
For a given prime numberp, Let S≅ℤp; then we have the following formula to calculate the number of edges in the Quasiregular graph if Qℤp.(1)EQℤp=p2−4p+52Example 11.
EQℤ7=13 and EQℤ11=41, See Figure 9,Figure 9
Number of edges ofQℤ11 and Qℤ11.Proposition 7.
For a finite commutative ringS with identity, if QS is not a star graph, then grQS=3.Proof.
There are three possibilities forJS;Case I:JS≥3, there are nonzero elements w,z∈JS in this case, so w and z are adjacent. Also, both elements w and z are adjacent to 1. Thus, QS has a 3-cycle and grQS=3.Case II:JS=2, So JS has only one nonzero element a. Thus, a is adjacent to all elements in S/0,a. By an assumption, QS is not a star graph. So, there exists b≠c∈S/0,a with b and c adjacent. Thus, QS has a 3-cycle and grQS=3.Case III:JS=1, if S is a finite field with S>4 (otherwise QS is a star graph) For a,b∈S∗ with ab≠1, then a and b are adjacent. Hence, we get a 3-cycle 1−a−b−1 and grQS=3. Now, suppose that S>4, and S≅F1×F2, such that F1 and F2 are finite fields, respectively, with F2>2; So, QS has a 3-cycle 0,1−0,a−1,0−0,1 where a≠1; Thus grQS=3. Given the fact that any finite commutative ring with zero radical is isomorphic to a direct sum of finite fields; our case is handled.



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