CLASSIFICATION OF RINGS WITH UNIT GRAPHS HAVING DOMINATION NUMBER LESS THAN FOUR

S. KIANI, H. R. MAIMANI, M. R. POURNAKI, AND S. YASSEMI

Dedicated with gratitude to our friend Alberto Facchini on the occasion of his 60th birthday

Abstract. Let R be a finite commutative ring with nonzero identity. The unit graph of R is the graph obtained by setting all the elements of R to be the vertices and defining distinct vertices x and y to be adjacent if and only if x + y is a unit element of R. In this paper, a classification of finite commutative rings with nonzero identity in which their unit graphs have domination number less than four is given.

1. Introduction

Throughout the paper by a graph we mean a finite undirected graph without loops or multiple edges. Also, all rings are finite commutative with nonzero identity and all fields are finite. Therefore, in this paper, we restrict ourselves to the finite case, although some of the results hold true in the infinite case. For undefined terms and concepts, the reader is referred to [12] and [2].

Let G be a graph with vertex set V. A subset D of V is called a dominating set of G if every vertex in V \ D is adjacent to at least a vertex in D. In other words, D dominates the vertices outside D. A γ-set of G is a minimum dominating set of G, that is, a dominating set of G whose cardinality is minimum. The domination number of G, denoted by γ(G), is the cardinality of a γ-set of G. The study of domination has long been a topic of interest both in graph theory and complexity theory. It was first considered by Ore who introduced the concept of minimum dominating sets of vertices in a graph. The dominating set problem concerns testing whether γ(G) ≤ k for a given graph G and integer k. The problem is a classical NP-complete decision problem in computational complexity theory (see, for example, [3]). Therefore, it is believed that there is no efficient algorithm that finds a smallest dominating set of a given graph. The first volume of the two-volume book by Haynes, Hedetniemi and Slater [5, 6] provides a comprehensive introduction to “domination in graphs”.

2000 Mathematics Subject Classification. Primary: 05C75; Secondary: 13M05.

Key words and phrases. Unit graph, Domination number, Total domination number, Finite ring.

The research of H. R. Maimani, M. R. Pournaki, and S. Yassemi was in part supported by a grant from IPM (No. 93050113, No. 93130115, and No. 93130214). The research of M. R. Pournaki was also partially supported by a grant from INSF.

In this paper, we consider graphs which are generated by rings, known as unit graphs. We then give a classification of rings in which their unit graphs have domination number less than four.

2. Preliminaries and the statement of the main result

Let n be a positive integer and ℤ_{n} be the ring of integers modulo n. Grimaldi [4] defined a graph G(ℤ_{n}) based on the elements and units of ℤ_{n}. The vertices of G(ℤ_{n}) are the elements of ℤ_{n} and distinct vertices x and y are defined to be adjacent if and only if x + y is a unit of ℤ_{n}. For a positive integer m, it follows that G(ℤ_{2m}) is a φ(2m)-regular graph, where φ is the Euler phi function. In case m ≥ 2, G(ℤ_{2m}) can be expressed as the union of φ(2m)/2 Hamiltonian cycles. The odd case is not very easy, but the structure is clear and the results are similar to the even case. We recall that a cone over a graph is obtained by taking the categorical product of the graph and a path with a loop at one end, and then identifying all the vertices whose second coordinate is the other end of the path. When p is an odd prime, G(ℤ_{p}) can be expressed as a cone over a complete partite graph with (p − 1)/2 partitions of size two. This leads to an explicit formula for the chromatic polynomial of G(ℤ_{p}). The paper [4] also concludes with some properties of the graphs G(ℤ_{p^{m}}), where p is a prime number and m ≥ 2.

Recently, Ashrafi et al. [1] generalized G(ℤ_{n}) to G(R), the unit graph of R, where R is an arbitrary ring and studied the properties of this graph. Later, more properties of the unit graph of a ring and its applications were given in [7, 8, 9]. Let us first define this notion.

Definition 2.1. Let R be a ring and U(R) be the set of unit elements of R. The unit graph of R, denoted by G(R), is the graph obtained by setting all the elements of R to be the vertices and defining distinct vertices x and y to be adjacent if and only if x + y ∈ U(R).

For a given ring R, if we omit the word “distinct” in the definition of the unit graph G(R), we obtain the closed unit graph of R which is denoted by \overline{G}(R). This graph may have loops and if 2 ∉ U(R), then \overline{G}(R) = G(R).

The graphs in Fig. 1 are the unit graphs of the rings indicated.

[figure: 0; 1; 2; G(ℤ_{3}); 0; 1; 2; 3; G(ℤ_{4}); 10; 11; 00; 01; G(ℤ_{2} × ℤ_{2}); 0; 1; 2; \overline{G}(ℤ_{3})]

Figure 1. The unit graphs of some specific rings.

It is easy to see that, for given rings R and S, if R ≅ S as rings, then G(R) ≅ G(S) as graphs. This point is illustrated in Fig. 2, for the unit graphs of two isomorphic rings ℤ_{3} × ℤ_{2} and ℤ_{6}.

[figure: 00; 11; 21; 10; 20; 01; G(ℤ_{3} × ℤ_{2}); ≅; 0; 5; 1; 2; 4; 3; G(ℤ_{6})]

Figure 2. The unit graphs of two isomorphic rings.

For a graph G, let V (G) denote the set of vertices of G. Let G_{1} and G_{2} be two vertex-disjoint graphs. The categorical product of G_{1} and G_{2} is denoted by G_{1} ˙×G_{2}. That is, V (G_{1} ˙×G_{2}) := V (G_{1}) × V (G_{2}) and two distinct vertices (x, y) and (x′, y′) are adjacent if and only if x is adjacent to x′ in G_{1} and y is adjacent to y′ in G_{2}. Clearly, for given rings R_{1} and R_{2}, two distinct vertices (x, y) and (x′, y′) of \overline{G}(R_{1}) ˙×\overline{G}(R_{2}) are adjacent if and only if x is adjacent to x′ in \overline{G}(R_{1}) and y is adjacent to y′ in \overline{G}(R_{2}). This implies that \overline{G}(R_{1}) ˙×\overline{G}(R_{2}) ≅ G(R_{1} × R_{2}).

In Fig. 3, we illustrate the above point for the direct product of ℤ_{2} and ℤ_{3}.

[figure: 0; 1; 2; \overline{G}(ℤ_{3}); ˙×; 0; 1; \overline{G}(ℤ_{2}); ≅; 00; 11; 21; 10; 20; 01; G(ℤ_{3} × ℤ_{2}); ≅; 0; 5; 1; 2; 4; 3; G(ℤ_{6})]

Figure 3. The categorical product of two closed unit graphs.

We are now in a position to state the main result of this paper.

Theorem 2.2. Let R be a ring. Then the following statements hold true for the unit graph G(R):

(1) γ(G(R)) = 1 if and only if R is a field.

(2) γ(G(R)) = 2 if and only if either R is a local ring which is not a field, R is isomorphic to the product of two fields such that only one of them has characteristic 2, or R ≅ ℤ_{2} × 𝔽, where 𝔽 is a field.

(3) γ(G(R)) = 3 if and only if R is not isomorphic to the product of two fields such that only one of them has characteristic 2, and R ≅ R_{1}×R_{2}, where for i = 1, 2, R_{i} is a local ring with maximal ideal 𝔪_{i} in such a way that R_{i}/𝔪_{i} ≇ ℤ_{2}.

The rest of the paper is organized as follows. In Section 3, we deal with unit graphs associated with products of fields. By the results of this section, we give a proof for parts (1) and (2) of Theorem 2.2. In Section 4, we deal with unit graphs associated with products of local rings. The results of this section lead to a proof for part (3) of Theorem 2.2. In both sections, we state and prove some lemmas that will be used in the proof of Theorem 2.2. Furthermore, for the convenience of the reader, we state without proof a few known results in the form of propositions which will be used in the proofs. We also recall some definitions and notations concerning graphs for later use.

3. Unit graphs associated with products of fields

We start this section by unit graphs associated with fields. The following lemma gives us the domination number of such graphs.

Lemma 3.1. Let 𝔽 be a field. Then γ(G(𝔽)) = 1.

Proof. Since 𝔽 is a field, every nonzero element of 𝔽 is a unit element of 𝔽. That is, for every x ∈ 𝔽 \ {0}, x = x + 0 ∈ U(𝔽) and so x is adjacent to 0. Hence, every nonzero element of 𝔽 is adjacent to 0 and so D = {0} is a dominating set of G(𝔽) with, of course, minimum cardinality. Therefore, we conclude that γ(G(𝔽)) = 1. □

The degrees of all vertices of a unit graph is given by the following proposition. We recall that for a graph G and for a vertex x of G, the degree of x, denoted by deg(x), is the number of edges of G incident with x.

Proposition 3.2 ([1], Proposition 2.4). Let R be a ring. Then the following statements hold true for the unit graph G(R):

(1) If 2 ∉ U(R), then deg(x) = |U(R)| for every x ∈ R.

(2) If 2 ∈ U(R), then deg(x) = |U(R)|−1 for every x ∈ U(R) and deg(x) = |U(R)| for every x ∈ R \ U(R).

We are now ready to prove part (1) of Theorem 2.2, that is:

Theorem 3.3. Let R be a ring. Then γ(G(R)) = 1 if and only if R is a field.

Proof. (⇒): Let γ(G(R)) = 1 and consider D = {x} as a γ-set of G(R). We claim that deg(x) = |U(R)|. In order to prove the claim, we first suppose that x = 0. In this case, x ∉ U(R) and so by Proposition 3.2, we have deg(x) = |U(R)|. Second, we suppose that x ≠ 0. Note that we have x = −x; otherwise, since D = {x} is a dominating set of G(R), −x is adjacent to x and so −x + x = 0 ∈ U(R), which is a contradiction. Hence, 2x = 0, which together with x ≠ 0, imply that 2 ∉ U(R). Now, by Proposition 3.2(1), we conclude that deg(x) = |U(R)|. Therefore, the claim holds true. On the other hand, since D = {x} is a dominating set of G(R), every element of R \ {x} is adjacent to x and so deg(x) = |R| − 1. Therefore, |U(R)| = |R| − 1, which implies that R is a field.

(⇐): This implication, by Lemma 3.1, is obvious. □

We now state and prove some results that will be needed in the proof of part (2) of Theorem 2.2. In Lemmas 3.4 and 3.6, we give a lower bound for the domination number of two classes of unit graphs associated with products of two fields. Lemma 3.5 is useful for this purpose.

Lemma 3.4. Let 𝔽_{1} and 𝔽_{2} be two fields in which both of them have characteristic 2 and none of them is isomorphic to ℤ_{2}. Then γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3.

Proof. Suppose, in contrary, that γ(G(𝔽_{1} × 𝔽_{2})) ≤ 2. Therefore, the unit graph G(𝔽_{1} × 𝔽_{2}) has a dominating set in the form of D = {(x_{1}, x_{2}), (x_{3}, x_{4})}, where x_{1}, x_{3} ∈ 𝔽_{1}, x_{2}, x_{4} ∈ 𝔽_{2} and (x_{1}, x_{2}) ≠ (x_{3}, x_{4}). Hence, we have either x_{1} ≠ x_{3} or x_{2} ≠ x_{4}.

Case 1: Suppose that x_{1} ≠ x_{3}. The assumption Char(𝔽_{1}) = Char(𝔽_{2}) = 2 implies that

(x_{1}, x_{4}) + (x_{1}, x_{2}) = (0, x_{4} + x_{2}) ∉ U(𝔽_{1} × 𝔽_{2}) and
(x_{1}, x_{4}) + (x_{3}, x_{4}) = (x_{1} + x_{3}, 0) ∉ U(𝔽_{1} × 𝔽_{2}).

Hence, (x_{1}, x_{4}) is not adjacent to the elements of D. Therefore, (x_{1}, x_{4}) ∈ D and so (x_{1}, x_{4}) = (x_{1}, x_{2}) or (x_{1}, x_{4}) = (x_{3}, x_{4}). Now, x_{1} ≠ x_{3} implies that the second equality does not hold and so the first one holds. Therefore, we have x_{4} = x_{2} and so D = {(x_{1}, x_{2}), (x_{3}, x_{2})}. Let a ∈ 𝔽_{1} be given. Then Char(𝔽_{2}) = 2 implies that

(a, x_{2}) + (x_{1}, x_{2}) = (a + x_{1}, 0) ∉ U(𝔽_{1} × 𝔽_{2}) and
(a, x_{2}) + (x_{3}, x_{2}) = (a + x_{3}, 0) ∉ U(𝔽_{1} × 𝔽_{2}).

Hence, (a, x_{2}) is not adjacent to the elements of D. Therefore, (a, x_{2}) ∈ D and so (a, x_{2}) = (x_{1}, x_{2}) or (a, x_{2}) = (x_{3}, x_{2}). This implies that a = x_{1} or a = x_{3}. Since a ∈ 𝔽_{1} is arbitrary, we conclude that 𝔽_{1} = {x_{1}, x_{3}} ≅ ℤ_{2}, which is a contradiction.

Case 2: Suppose that x_{2} ≠ x_{4}. By a similar argument as in Case 1, this case leads to 𝔽_{2} = {x_{2}, x_{4}} ≅ ℤ_{2}, which is again a contradiction.

Therefore, we conclude that γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3, as required. □

Lemma 3.5. Let 𝔽_{1} and 𝔽_{2} be two fields. If Char(𝔽_{1}) ≠ 2 and a ∈ 𝔽_{1}, then D_{1} = {(a, 0), (−a, 0)} is not a dominating set of the unit graph G(𝔽_{1} ×𝔽_{2}). If Char(𝔽_{2}) ≠ 2 and b ∈ 𝔽_{2}, then D_{2} = {(0, b), (0, −b)} is not a dominating set of the unit graph G(𝔽_{1} × 𝔽_{2}).

Proof. We prove the first part. The proof of the second part is similar. Suppose, in contrary, that D_{1} = {(a, 0), (−a, 0)} is a dominating set of the unit graph G(𝔽_{1} × 𝔽_{2}).

Since Char(𝔽_{1}) ≠ 2, |𝔽_{1}| ≥ 3 and so we may choose x ∈ 𝔽_{1} for which x ≠ a and x ≠ −a. Now, we have

(x, 0) + (a, 0) = (x + a, 0) ∉ U(𝔽_{1} × 𝔽_{2}) and
(x, 0) + (−a, 0) = (x − a, 0) ∉ U(𝔽_{1} × 𝔽_{2}).

Hence, (x, 0) is not adjacent to the elements of D_{1}. Therefore, (x, 0) ∈ D_{1} and so (x, 0) = (a, 0) or (x, 0) = (−a, 0). Hence, x = a or x = −a, which contradicts x ≠ a and x ≠ −a. Therefore, D_{1} is not a dominating set of the unit graph G(𝔽_{1} × 𝔽_{2}). □

Lemma 3.6. Let 𝔽_{1} and 𝔽_{2} be two fields in which none of them has characteristic 2. Then γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3.

Proof. Suppose, in contrary, that γ(G(𝔽_{1} × 𝔽_{2})) ≤ 2. Therefore, the unit graph G(𝔽_{1} ×𝔽_{2}) has a dominating set in the form of D = {(x_{1}, x_{2}), (x_{3}, x_{4})}, where x_{1}, x_{3} ∈ 𝔽_{1}, x_{2}, x_{4} ∈ 𝔽_{2} and (x_{1}, x_{2}) ≠ (x_{3}, x_{4}). Note that

(−x_{1}, −x_{4}) + (x_{1}, x_{2}) = (0, −x_{4} + x_{2}) ∉ U(𝔽_{1} × 𝔽_{2}) and
(−x_{1}, −x_{4}) + (x_{3}, x_{4}) = (−x_{1} + x_{3}, 0) ∉ U(𝔽_{1} × 𝔽_{2}).

Hence, (−x_{1}, −x_{4}) is not adjacent to the elements of D. Therefore, (−x_{1}, −x_{4}) ∈ D and so (−x_{1}, −x_{4}) = (x_{1}, x_{2}) or (−x_{1}, −x_{4}) = (x_{3}, x_{4}). Also,

(−x_{3}, −x_{2}) + (x_{1}, x_{2}) = (−x_{3} + x_{1}, 0) ∉ U(𝔽_{1} × 𝔽_{2}) and
(−x_{3}, −x_{2}) + (x_{3}, x_{4}) = (0, −x_{2} + x_{4}) ∉ U(𝔽_{1} × 𝔽_{2}).

Hence, (−x_{3}, −x_{2}) is not adjacent to the elements of D. Therefore, (−x_{3}, −x_{2}) ∈ D and so (−x_{3}, −x_{2}) = (x_{1}, x_{2}) or (−x_{3}, −x_{2}) = (x_{3}, x_{4}). Hence, one of the following four possibilities occurs:

(1) (−x_{1}, −x_{4}) = (x_{1}, x_{2}) and (−x_{3}, −x_{2}) = (x_{1}, x_{2}),

(2) (−x_{1}, −x_{4}) = (x_{3}, x_{4}) and (−x_{3}, −x_{2}) = (x_{3}, x_{4}),

(3) (−x_{1}, −x_{4}) = (x_{3}, x_{4}) and (−x_{3}, −x_{2}) = (x_{1}, x_{2}), or

(4) (−x_{1}, −x_{4}) = (x_{1}, x_{2}) and (−x_{3}, −x_{2}) = (x_{3}, x_{4}).

The first two cases lead to x_{1} = x_{2} = x_{3} = x_{4} = 0, which contradicts (x_{1}, x_{2}) ≠ (x_{3}, x_{4}). The third case leads to x_{1} = −x_{3} := a and x_{2} = x_{4} = 0. Therefore, D = {(a, 0), (−a, 0)}, which contradicts Lemma 3.5. The fourth case leads to x_{1} = x_{3} = 0 and x_{2} = −x_{4} := b. Therefore, D = {(0, b), (0, −b)}, which again contradicts Lemma 3.5. Therefore, we conclude that γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3, as required. □

In Lemma 3.12, we prove that the domination number of unit graphs associated with products of three fields is equal to four. The following results are useful for this purpose.

We recall that a bipartite graph is one whose vertex set is partitioned into two (not necessarily nonempty) disjoint subsets, called partite sets, in such a way that the two end vertices for each edge lie in distinct partite sets.

Proposition 3.7 ([1], Theorem 3.5). Let R be a ring and 𝔪 be a maximal ideal of R such that |R/𝔪| = 2. Then the unit graph G(R) is a bipartite graph with partite sets V_{1} = 𝔪 and V_{2} = R \ 𝔪.

Lemma 3.8. Let 𝔽_{1}, 𝔽_{2} and 𝔽_{3} be three fields such that one is isomorphic to ℤ_{2}, one is not isomorphic to ℤ_{2} and the latter one has characteristic 2. Then γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4.

Proof. Without loss of generality, we may assume that 𝔽_{1} ≇ ℤ_{2}, Char(𝔽_{2}) = 2 and 𝔽_{3} ≅ ℤ_{2}. Since 𝔽_{1} × 𝔽_{2} × 𝔽_{3} is a ring with maximal ideal 𝔽_{1} × 𝔽_{2} × {0} such that |(𝔽_{1} × 𝔽_{2} × 𝔽_{3})/(𝔽_{1} × 𝔽_{2} × {0})| = 2, Proposition 3.7 implies that the unit graph G(𝔽_{1}×𝔽_{2}×𝔽_{3}) is bipartite with partite sets V_{1} = 𝔽_{1}×𝔽_{2}×{0} and V_{2} = 𝔽_{1}×𝔽_{2}×{1}. This implies that |V_{1}| = |V_{2}| = |𝔽_{1} × 𝔽_{2}|.

Assume now, in contrary, that γ(G(𝔽_{1} ×𝔽_{2} ×𝔽_{3})) ≤ 3 and consider D with |D| = 3 as a dominating set of G(𝔽_{1} × 𝔽_{2} × 𝔽_{3}). Note that |𝔽_{1}| ≥ 3, therefore |𝔽_{1} × 𝔽_{2}| ≥ 6 and so |V_{1}| = |V_{2}| ≥ 6. Now, |D| = 3 implies that D cannot completely lie either in V_{1} or V_{2}. Hence, D ∩ V_{1} ≠ ∅ and D ∩ V_{2} ≠ ∅. This implies that either |D ∩ V_{1}| = 1 or |D ∩ V_{2}| = 1. Without loss of generality, suppose |D ∩ V_{1}| = 1 is the case. Since D ∩ V_{1} ⊆ V_{1} = 𝔽_{1} × 𝔽_{2} × {0}, we may consider D ∩ V_{1} = {(x, y, 0)} with x ∈ 𝔽_{1} and y ∈ 𝔽_{2}. Since |𝔽_{1}| ≥ 3, we may choose three distinct elements x_{1}, x_{2} and x_{3} of 𝔽_{1}. Therefore, (x_{1}, y, 1), (x_{2}, y, 1) and (x_{3}, y, 1) are three distinct elements of V_{2}. But Char(𝔽_{2}) = 2 implies that for i = 1, 2, 3,

(x_{i}, y, 1) + (x, y, 0) = (x_{i} + x, 0, 1) ∉ U(𝔽_{1} × 𝔽_{2} × 𝔽_{3})

and so (x_{i}, y, 1) is not adjacent to (x, y, 0). Hence, for i = 1, 2, 3, (x_{i}, y, 1) ∈ D and so |D| ≥ 4, which is a contradiction. This shows that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4. □

We need the following result from [11] for later use. Note that in [11], the author has used a different notion of product. But it causes no problem here as loops have no influence on the domination number. Here, K_{1} denotes the graph with one vertex and no edges.

Proposition 3.9 ([11], Theorem 3.1 and Proposition 4.1). Let G and H be two graphs. Then γ(G ˙×H) ≥ γ(G) + γ(H) − 1. Moreover, if the equality holds and γ(G) = 1, then G = K_{1} and H is an edgeless graph.

Lemma 3.10. Let R be a ring and 𝔽 be a field. Then γ(G(R × 𝔽)) > γ(G(R)).

Proof. By using Proposition 3.9, we conclude that

γ(\overline{G}(R) ˙×\overline{G}(𝔽)) ≥ γ(\overline{G}(R)) + γ(\overline{G}(𝔽)) − 1.

Note that, again by Proposition 3.9, if the equality holds, then since, by Lemma 3.1, γ(\overline{G}(𝔽)) = γ(G(𝔽)) = 1, we obtain that \overline{G}(𝔽) = K_{1}, which is a contradiction. Therefore, the equality does not hold, that is,

γ(\overline{G}(R) ˙×\overline{G}(𝔽)) > γ(\overline{G}(R)) + γ(\overline{G}(𝔽)) − 1 = γ(\overline{G}(R)).

Now, since for every ring S, γ(\overline{G}(S)) = γ(G(S)), we conclude that

γ(G(R × 𝔽)) = γ(\overline{G}(R) ˙×\overline{G}(𝔽)) > γ(\overline{G}(R)) = γ(G(R)),

as required. □

Lemma 3.11. Let 𝔽_{1}, 𝔽_{2} and 𝔽_{3} be three fields. If a ∈ 𝔽_{1}, b ∈ 𝔽_{2} and c ∈ 𝔽_{3} are nonzero elements, then D = {(0, 0, 0), (−a, −b, 0), (a, 0, −c), (0, b, c)} is a dominating set of the unit graph G(𝔽_{1} × 𝔽_{2} × 𝔽_{3}).

Proof. A given element in (𝔽_{1} × 𝔽_{2} × 𝔽_{3}) \ D has one of the following forms, where x ∈ 𝔽_{1}, y ∈ 𝔽_{2} and z ∈ 𝔽_{3} are nonzero elements: either (x, 0, 0), (0, y, 0), (0, 0, z), (x, y, 0) with x ≠ −a or y ≠ −b, (x, 0, z) with x ≠ a or z ≠ −c, (0, y, z) with y ≠ b or z ≠ c, or (x, y, z).

For (x, 0, 0), we have

(x, 0, 0) + (0, b, c) = (x, b, c) ∈ U(𝔽_{1} × 𝔽_{2} × 𝔽_{3})

and so (x, 0, 0) is adjacent to (0, b, c). By a similar argument, we obtain that (0, y, 0) is adjacent to (a, 0, −c) and (0, 0, z) is adjacent to (−a, −b, 0).

For (x, y, 0) with x ≠ −a or y ≠ −b, we have either

(x, y, 0) + (a, 0, −c) = (x + a, y, −c) ∈ U(𝔽_{1} × 𝔽_{2} × 𝔽_{3}) or
(x, y, 0) + (0, b, c) = (x, y + b, c) ∈ U(𝔽_{1} × 𝔽_{2} × 𝔽_{3}).

This implies that (x, y, 0) with x ≠ −a or y ≠ −b is adjacent to (a, 0, −c) or (0, b, c). By a similar argument, we obtain that (x, 0, z) with x ≠ a or z ≠ −c is adjacent to (−a, −b, 0) or (0, b, c). Also, (0, y, z) with y ≠ b or z ≠ c is adjacent to (−a, −b, 0) or (a, 0, −c).

Finally, for (x, y, z), we have

(x, y, z) + (0, 0, 0) = (x, y, z) ∈ U(𝔽_{1} × 𝔽_{2} × 𝔽_{3})

and so (x, y, z) is adjacent to (0, 0, 0).

Therefore, D is a dominating set of the unit graph G(𝔽_{1} × 𝔽_{2} × 𝔽_{3}). □

Lemma 3.12. Let 𝔽_{1}, 𝔽_{2} and 𝔽_{3} be three fields. Then γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) = 4.

Proof. We first show that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4. In order to do this, note that there are two possibilities: either at least two of 𝔽_{1}, 𝔽_{2} and 𝔽_{3}, say 𝔽_{1} and 𝔽_{2}, have characteristic 2, or at least two of 𝔽_{1}, 𝔽_{2} and 𝔽_{3}, say again 𝔽_{1} and 𝔽_{2}, do not have characteristic 2.

First, assume that Char(𝔽_{1}) = 2 and Char(𝔽_{2}) = 2. Now, one of the following cases may occur:

Case 1: Both of 𝔽_{1} and 𝔽_{2} are isomorphic to ℤ_{2}. In this case, if 𝔽_{3} ≅ ℤ_{2}, then it is easy to see that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) = 4. If 𝔽_{3} ≇ ℤ_{2}, then we have 𝔽_{1} ≅ ℤ_{2}, Char(𝔽_{2}) = 2 and 𝔽_{3} ≇ ℤ_{2}, and so by Lemma 3.8, γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4.

Case 2: One of 𝔽_{1} or 𝔽_{2} is isomorphic to ℤ_{2} and the other one is not, say 𝔽_{1} ≅ ℤ_{2} and 𝔽_{2} ≇ ℤ_{2}. In this case, if 𝔽_{3} ≅ ℤ_{2}, then we have Char(𝔽_{1}) = 2, 𝔽_{2} ≇ ℤ_{2} and 𝔽_{3} ≅ ℤ_{2}, and so by Lemma 3.8, γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4. If 𝔽_{3} ≇ ℤ_{2}, then we have 𝔽_{1} ≅ ℤ_{2}, Char(𝔽_{2}) = 2 and 𝔽_{3} ≇ ℤ_{2}, and so again by Lemma 3.8, γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4.

Case 3: None of 𝔽_{1} and 𝔽_{2} is isomorphic to ℤ_{2}. In this case, by using Lemmas 3.10 and 3.4, we obtain that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) > γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3 and so γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4.

Second, assume that Char(𝔽_{1}) ≠ 2 and Char(𝔽_{2}) ≠ 2. In this case, by using Lemmas 3.10 and 3.6, we obtain that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) > γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3 and so γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4.

All in all, we have proven that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) ≥ 4 and in view of Lemma 3.11, we conclude that γ(G(𝔽_{1} × 𝔽_{2} × 𝔽_{3})) = 4, as required. □

In Lemma 3.16, we give a lower bound for the domination number of unit graphs associated with rings which are products of more than two local rings. The following results are useful for this purpose.

Lemma 3.13. Let 𝔽_{1}, …, 𝔽_{n} be fields. If n ≥ 3, then γ(G(𝔽_{1} × ⋯ × 𝔽_{n})) ≥ n + 1.

Proof. We prove the lemma by induction on n. By Lemma 3.12, the inequality is true for n = 3 and we assume that it is true for n − 1. Now, by using Lemma 3.10 and the induction hypothesis, we obtain that γ(G(𝔽_{1} ×⋯×𝔽_{n})) > γ(G(𝔽_{1} ×⋯×𝔽_{n−1})) ≥ n and so γ(G(𝔽_{1} × ⋯ × 𝔽_{n})) ≥ n + 1. □

Lemma 3.14. Let R be a ring. Then γ(G(R/J(R))) ≤ γ(G(R)), where J(R) denotes the Jacobson radical of R.

Proof. Let γ(G(R)) = n and consider D = {x_{1}, …, x_{n}} as a γ-set of G(R). Among x_{1}+J(R), …, x_{n}+J(R), choose distinct ones and call them x_{i_{1}} +J(R), …, x_{i_{k}} +J(R). Note that {x_{i_{1}}, …, x_{i_{k}}} ⊆ {x_{1}, …, x_{n}} and, in particular, k ≤ n. We now claim that D′ = {x_{i_{1}} + J(R), …, x_{i_{k}} + J(R)} is a dominating set of the unit graph G(R/J(R)). In order to prove the claim, let y + J(R) ∈ (R/J(R)) \ D′ be given. Therefore, y ∈ R \ D and so there exists x_{ℓ} ∈ D such that y is adjacent to x_{ℓ} in G(R), that is, y + x_{ℓ} ∈ U(R). This implies that (y + J(R)) + (x_{ℓ} + J(R)) ∈ U(R/J(R)) and so y + J(R) is adjacent to x_{ℓ} + J(R) in G(R/J(R)). But, x_{ℓ} + J(R) = x_{i_{j}} + J(R) for some 1 ≤ j ≤ k. Hence, y + J(R) is adjacent to x_{i_{j}} + J(R) in G(R/J(R)) and so the claim holds true. Therefore, we have γ(G(R/J(R))) ≤ |D′| = k ≤ n = γ(G(R)). □

We note that Lemma 3.14 remains valid for every arbitrary proper ideal of R, but in this paper we shall apply it in the above form.

Remark 3.15. It is known that every finite commutative ring is isomorphic to a direct product of local rings (see [10, Page 95]). Thus for a given ring R, we may write R ≅ R_{1} × ⋯ × R_{n}, where for every 1 ≤ i ≤ n, R_{i} is a local ring with maximal ideal 𝔪_{i}. It is easy to see that

{R_{1} × ⋯ × R_{i−1} × 𝔪_{i} × R_{i+1} × ⋯ × R_{n} | 1 ≤ i ≤ n}

is the set of all maximal ideals of R_{1} × ⋯ × R_{n} and thus we conclude that the Jacobson radical of R is isomorphic to J(R_{1} × ⋯ × R_{n}) = 𝔪_{1} × ⋯ × 𝔪_{n}. Also, note that the function f : R_{1} × ⋯ × R_{n} ⟶ (R_{1}/𝔪_{1}) × ⋯ × (R_{n}/𝔪_{n}), given by f(r_{1}, …, r_{n}) = (r_{1} + 𝔪_{1}, …, r_{n} + 𝔪_{n}), is an epimorphism with ker f = 𝔪_{1} × ⋯ × 𝔪_{n} and thus we have

\frac{R}{J(R)} ≅ \frac{R_{1} × ⋯ × R_{n}}{𝔪_{1} × ⋯ × 𝔪_{n}} ≅ \frac{R_{1}}{𝔪_{1}} × ⋯ × \frac{R_{n}}{𝔪_{n}} .

Lemma 3.16. Let R be a ring and in view of Remark 3.15, write R ≅ R_{1} × ⋯ × R_{n}, where for every 1 ≤ i ≤ n, R_{i} is a local ring. If n ≥ 3, then γ(G(R)) ≥ n + 1.

Proof. For every 1 ≤ i ≤ n, let 𝔪_{i} be the maximal ideal of R_{i}. By using Remark 3.15, we conclude that R/J(R) ≅ 𝔽_{1} × ⋯ × 𝔽_{n}, where for every 1 ≤ i ≤ n, 𝔽_{i} is the field R_{i}/𝔪_{i}. Now, by using Lemmas 3.14 and 3.13, we conclude that γ(G(R)) ≥ γ(G(R/J(R))) = γ(G(𝔽_{1} × ⋯ × 𝔽_{n})) ≥ n + 1, as required. □

The following three lemmas are needed for the proof of part (2) of Theorem 2.2.

Lemma 3.17. Let R be a local ring with maximal ideal 𝔪 and let x ∈ R. If x ∉ 𝔪, then x ∈ U(R). If x ∈ 𝔪, then x + 1 ∈ U(R). In particular, D = {0, 1} is a dominating set of the unit graph G(R).

Proof. First, let x ∉ 𝔪. If x ∉ U(R), then ⟨x⟩ is a proper ideal of R and since R is local, ⟨x⟩ ⊆ 𝔪. Hence, x ∈ 𝔪, which contradicts x ∉ 𝔪. Therefore, x ∈ U(R).

Second, let x ∈ 𝔪. Then x + 1 ∉ 𝔪; otherwise, we obtain that 1 ∈ 𝔪, which is a contradiction. Now, by the previous case, x + 1 ∈ U(R).

In particular, for a given a ∈ R, if a ∉ 𝔪, then it is adjacent to 0 and if a ∈ 𝔪, then it is adjacent to 1. Therefore, D = {0, 1} is a dominating set of G(R). □

We recall that for a graph G and for a given vertex x ∈ V (G), the neighbor set of x is the set N(x) := {v ∈ V (G) | v is adjacent to x}.

Lemma 3.18. Let 𝔽_{1} and 𝔽_{2} be two fields with Char(𝔽_{1}) = 2 and Char(𝔽_{2}) ≠ 2. If b ∈ 𝔽_{2} is nonzero, then D = {(0, −b), (1, b)} is a dominating set of the unit graph G(𝔽_{1} × 𝔽_{2}).

Proof. Since Char(𝔽_{1}) = 2, we may conclude that 2(1, 1) = (0, 2) ∉ U(𝔽_{1} × 𝔽_{2}). Therefore, Proposition 3.2(1) implies that the degree of every element of G(𝔽_{1} × 𝔽_{2}) is equal to |U(𝔽_{1} × 𝔽_{2})| = |U(𝔽_{1})||U(𝔽_{2})| = (|𝔽_{1}| − 1)(|𝔽_{2}| − 1). In particular, we conclude that

|N(0, −b)| = |N(1, b)| = (|𝔽_{1}| − 1)(|𝔽_{2}| − 1).

Also, for a given element (x, y) ∈ 𝔽_{1} × 𝔽_{2}, it is adjacent to both of (0, −b) and (1, b) if and only if

(x, y) + (0, −b) = (x, y − b) ∈ U(𝔽_{1} × 𝔽_{2}) and
(x, y) + (1, b) = (x + 1, y + b) ∈ U(𝔽_{1} × 𝔽_{2}),

if and only if x ≠ 0, −1 and y ≠ b, −b. Therefore,

N(0, −b) ∩ N(1, b) = {(x, y) ∈ 𝔽_{1} × 𝔽_{2} | x ≠ 0, −1 and y ≠ b, −b}.

Since Char(𝔽_{2}) ≠ 2 and b ∈ 𝔽_{2} is nonzero, we obtain that

|N(0, −b) ∩ N(1, b)| = (|𝔽_{1}| − 2)(|𝔽_{2}| − 2).

Now, the inclusion-exclusion principle implies that

|N(0, −b) ∪ N(1, b)| = |N(0, −b)| + |N(1, b)| − |N(0, −b) ∩ N(1, b)|
= 2(|𝔽_{1}| − 1)(|𝔽_{2}| − 1) − (|𝔽_{1}| − 2)(|𝔽_{2}| − 2)
= |𝔽_{1}||𝔽_{2}| − 2
= |𝔽_{1} × 𝔽_{2}| − 2.

Since

(0, −b) + (1, b) = (1, 0) ∉ U(𝔽_{1} × 𝔽_{2}),

(0, −b) and (1, b) are not adjacent and so (0, −b), (1, b) ∉ N(0, −b) ∪ N(1, b). This implies that N(0, −b)∪N(1, b) = (𝔽_{1}×𝔽_{2})\D. Therefore, every element of (𝔽_{1}×𝔽_{2})\D is adjacent to at least an element of D, which implies that D is a dominating set of G(𝔽_{1} × 𝔽_{2}). □

Lemma 3.19. Let 𝔽 be a field. Then D = {(0, 0), (1, 0)} is a dominating set of the unit graph G(ℤ_{2} × 𝔽).

Proof. A given element in (ℤ_{2} ×𝔽)\D has one of the following forms, where y_{1}, y_{2} ∈ 𝔽 are nonzero elements: either (0, y_{1}) or (1, y_{2}). Since

(0, y_{1}) + (1, 0) = (1, y_{1}) ∈ U(ℤ_{2} × 𝔽) and
(1, y_{2}) + (0, 0) = (1, y_{2}) ∈ U(ℤ_{2} × 𝔽),

(0, y_{1}) is adjacent to (1, 0) and (1, y_{2}) is adjacent to (0, 0). Therefore, D is a dominating set of G(ℤ_{2} × 𝔽). □

We are now ready to prove part (2) of Theorem 2.2, that is:

Theorem 3.20. Let R be a ring. Then γ(G(R)) = 2 if and only if either R is a local ring which is not a field, R is isomorphic to the product of two fields such that only one of them has characteristic 2, or R ≅ ℤ_{2} × 𝔽, where 𝔽 is a field.

Proof. (⇒): Let γ(G(R)) = 2. By Remark 3.15, we may write R ≅ R_{1} × ⋯ × R_{n}, where for every 1 ≤ i ≤ n, R_{i} is a local ring. If n ≥ 3, then by using Lemma 3.16, we conclude that 2 = γ(G(R)) ≥ n + 1 ≥ 4, which is a contradiction. Therefore, n ≤ 2 and so either R ≅ R_{1} or R ≅ R_{1} × R_{2}. In the first case, R is a local ring which, by Lemma 3.1, is not a field. In the second case, by using Proposition 3.9 and the fact that for every ring S, γ(\overline{G}(S)) = γ(G(S)), we may conclude that

2 = γ(G(R))
= γ(G(R_{1} × R_{2}))
= γ(\overline{G}(R_{1}) ˙×\overline{G}(R_{2}))
≥ γ(\overline{G}(R_{1})) + γ(\overline{G}(R_{2})) − 1
= γ(G(R_{1})) + γ(G(R_{2})) − 1.

Since for i = 1, 2, we have γ(G(R_{i})) ≥ 1, we obtain that

2 ≤ γ(G(R_{1})) + γ(G(R_{2})) ≤ 3,

which implies that either γ(G(R_{1})) + γ(G(R_{2})) = 2 or γ(G(R_{1})) + γ(G(R_{2})) = 3. If γ(G(R_{1}))+γ(G(R_{2})) = 3 is the case, then one of its terms is equal to one and the other one is equal to two. For example, say γ(G(R_{1})) = 2 and γ(G(R_{2})) = 1. Therefore, by Theorem 3.3, we conclude that R_{2} := 𝔽_{2} is a field and so R ≅ R_{1} × 𝔽_{2}. Now, by Lemma 3.10, 2 = γ(G(R)) = γ(G(R_{1} ×𝔽_{2})) > γ(G(R_{1})) = 2, which is a contradiction and so this case does not hold. Therefore, we have γ(G(R_{1})) + γ(G(R_{2})) = 2. This implies that γ(G(R_{1})) = γ(G(R_{2})) = 1. Therefore, by Theorem 3.3, we conclude that R_{1} := 𝔽_{1} and R_{2} := 𝔽_{2} are fields and also R ≅ 𝔽_{1} × 𝔽_{2}. Now, in view of Lemmas 3.4 and 3.6, either R is isomorphic to the product of two fields such that only one of them has characteristic 2, or R ≅ ℤ_{2} × 𝔽, where 𝔽 is a field.

(⇐): Let R be one of the rings in the statement of the theorem. It is easy to see that R is not a field and so by Theorem 3.3, we conclude that γ(G(R)) ≠ 1. Now, Lemmas 3.17, 3.18 and 3.19 imply that R has a dominating set with two elements and so γ(G(R)) = 2. □

4. Unit graphs associated with products of local rings

We now state and prove some results that will be needed in the proof of part (3) of Theorem 2.2.

We start by introducing the notion of a total dominating set. Let G be a graph with vertex set V. A subset D of V is called a total dominating set of G if every vertex in V is adjacent to at least a vertex in D. In other words, D dominates not only vertices outside D but also vertices in D. A γ_{t}-set of G is a minimum total dominating set of G, that is, a total dominating set of G whose cardinality is minimum. The total domination number of G, denoted by γ_{t}(G), is the cardinality of a γ_{t}-set of G. It is obvious that γ(G) ≤ γ_{t}(G).

In Lemma 4.4, we give the domination number and the total domination number of a class of unit graphs associated with products of two local rings. The following lemmas are useful for this purpose.

Lemma 4.1. Let R be a local ring with maximal ideal 𝔪 such that |R/𝔪| = 2 and let S be a local ring. Then γ_{t}(G(R × S)) ≤ 4.

Proof. Since R×S is a ring with maximal ideal 𝔪×S such that |(R×S)/(𝔪×S)| = 2, Proposition 3.7 implies that the unit graph G(R × S) is bipartite with partite sets V_{1} = 𝔪 × S and V_{2} = (R \ 𝔪) × S.

We now claim that D = {(0, 0), (0, 1), (1, 0), (1, 1)} is a total dominating set of G(R × S). In order to prove the claim, note that for every (a, b) ∈ R × S, we have either (a, b) ∈ V_{1} = 𝔪 × S or (a, b) ∈ V_{2} = (R \ 𝔪) × S. We now apply Lemma 3.17 to get the following observations. Let 𝔫 be the maximal ideal of S. In the first case, that is, (a, b) ∈ 𝔪 × S, if b ∉ 𝔫, then

(a, b) + (1, 0) = (a + 1, b) ∈ U(R × S)

and so (a, b) is adjacent to (1, 0); and if b ∈ 𝔫, then

(a, b) + (1, 1) = (a + 1, b + 1) ∈ U(R × S)

and so (a, b) is adjacent to (1, 1). In the second case, that is, (a, b) ∈ (R \ 𝔪) × S, if b ∉ 𝔫, then

(a, b) + (0, 0) = (a, b) ∈ U(R × S)

and so (a, b) is adjacent to (0, 0); and if b ∈ 𝔫, then

(a, b) + (0, 1) = (a, b + 1) ∈ U(R × S)

and so (a, b) is adjacent to (0, 1). Therefore, the claim holds true and so γ_{t}(G(R×S)) ≤ 4, as required. □

Lemma 4.2. Let R be a local ring with maximal ideal 𝔪 such that |R/𝔪| = 2 and let S be a local ring. If R is not a field, then γ(G(R × S)) = γ_{t}(G(R × S)) = 4.

Proof. Since R×S is a ring with maximal ideal 𝔪×S such that |(R×S)/(𝔪×S)| = 2, Proposition 3.7 implies that the unit graph G(R × S) is bipartite with partite sets V_{1} = 𝔪 × S and V_{2} = (R \ 𝔪) × S. Note that |R/𝔪| = 2 implies that |𝔪| = |R \ 𝔪| = |R|/2 and so |V_{1}| = |V_{2}| = (|R|/2)|S|.

Assume now, in contrary, that γ(G(R × S)) ≤ 3 and consider D with |D| = 3 as a dominating set of G(R × S). Since R is not a field, |R| ≥ 4. Also, S is a ring and so |S| ≥ 2. Therefore, |V_{1}| = |V_{2}| = (|R|/2)|S| ≥ 4. Now, |D| = 3 implies that D cannot completely lie either in V_{1} or V_{2}. Hence, D ∩V_{1} ≠ ∅ and D ∩V_{2} ≠ ∅. This implies that either |D ∩ V_{1}| = 1 or |D ∩ V_{2}| = 1. Without loss of generality, suppose |D ∩ V_{1}| = 1 is the case. Since D ∩ V_{1} ⊆ V_{1} = 𝔪 × S, we may consider D ∩ V_{1} = {(x, y)} with x ∈ 𝔪 and y ∈ S. Since |R \ 𝔪| = |R|/2 ≥ 2, we may choose distinct elements x_{1} and x_{2} of R \ 𝔪. Therefore, (x_{1}, −y) and (x_{2}, −y) are distinct elements of (R \ 𝔪) × S = V_{2}. Also, (x, y) ∈ V_{1} implies that (x, y) ≠ (x_{1}, −y) and (x, y) ≠ (x_{2}, −y). Note that, if (x_{1}, −y) ∉ D, then (x_{1}, −y) is adjacent to (x, y) and so

(x_{1}, −y) + (x, y) = (x_{1} + x, 0) ∈ U(R × S),

which is impossible. Therefore, (x_{1}, −y) ∈ D. A similar argument shows that (x_{2}, −y) ∈ D. Hence, D = {(x, y), (x_{1}, −y), (x_{2}, −y)}. Since |𝔪| = |R|/2 ≥ 2, we may choose a ∈ 𝔪 \ {x}. Now, (a, y) ∈ 𝔪 × S = V_{1} and (a, y) ≠ (x, y) imply that (a, y) ∉ D. Since (a, y) ∈ V_{1}, it cannot be adjacent to (x, y) and so it is either adjacent to (x_{1}, −y) or (x_{2}, −y). Therefore, we have either

(a, y) + (x_{1}, −y) = (a + x_{1}, 0) ∈ U(R × S) or
(a, y) + (x_{2}, −y) = (a + x_{2}, 0) ∈ U(R × S),

which both of them are contradictions. This shows that γ(G(R × S)) ≥ 4.

Now, by Lemma 4.1, we have 4 ≤ γ(G(R × S)) ≤ γ_{t}(G(R × S)) ≤ 4 and so γ(G(R × S)) = γ_{t}(G(R × S)) = 4, as required. □

Lemma 4.3. Let S be a local ring which is not a field. Then γ(G(ℤ_{2} × S)) = γ_{t}(G(ℤ_{2} × S)) = 4.

Proof. Let 𝔪 be the maximal ideal of S. There are two possibilities: either |S/𝔪| = 2 or |S/𝔪| ≥ 3. If |S/𝔪| = 2 is the case, then by Lemma 4.2, we conclude that γ(G(ℤ_{2}×S)) = γ_{t}(G(ℤ_{2}×S)) = 4, as required. Therefore, we suppose that |S/𝔪| ≥ 3. Since ℤ_{2} × S is a ring with maximal ideal {0} × S such that |(ℤ_{2} × S)/({0} × S)| = 2, Proposition 3.7 implies that the unit graph G(ℤ_{2} × S) is bipartite with partite sets V_{1} = {0} × S and V_{2} = {1} × S. This implies that |V_{1}| = |V_{2}| = |S|.

We now claim that γ(G(ℤ_{2} × S)) ≥ 4. In order to prove the claim, in contrary, suppose that γ(G(ℤ_{2} × S)) ≤ 3 and consider D with |D| = 3 as a dominating set of G(ℤ_{2} × S). Since S is not a field, |𝔪| ≥ 2. Therefore, |S/𝔪| ≥ 3 implies that |S| ≥ 3|𝔪| ≥ 6 and so |V_{1}| = |V_{2}| = |S| ≥ 6. Now, |D| = 3 implies that D cannot completely lie either in V_{1} or V_{2}. Hence, D ∩V_{1} ≠ ∅ and D ∩V_{2} ≠ ∅. This implies that either |D ∩ V_{1}| = 1 or |D ∩ V_{2}| = 1. Without loss of generality, suppose |D ∩ V_{1}| = 1 is the case. Therefore, |D ∩V_{2}| = 2 and so we may consider D = {(0, y_{1}), (1, y_{2}), (1, y_{3})}, where y_{1}, y_{2}, y_{3} ∈ S with y_{2} ≠ y_{3}. Let y ∈ S \ {y_{2}, y_{3}} be given. Therefore, (1, y) ∉ D and so it is adjacent to an element of D. Since (1, y), (1, y_{2}) and (1, y_{3}) are elements of V_{2}, (1, y) is not adjacent to (1, y_{2}) and (1, y_{3}), therefore it is adjacent to (0, y_{1}). Hence, for every y ∈ S \ {y_{2}, y_{3}}, (1, y) is adjacent to (0, y_{1}), which implies that deg(0, y_{1}) ≥ |S| − 2. On the other hand, by Proposition 3.2 and Lemma 3.17,

deg(0, y_{1}) ≤ |U(ℤ_{2} × S)| = |U(ℤ_{2})||U(S)| = |U(S)| = |S| − |𝔪| ≤ |S| − 2.

Therefore,

|S| − 2 ≤ deg(0, y_{1}) ≤ |S| − |𝔪| ≤ |S| − 2,

which implies that deg(0, y_{1}) = |S| − 2 and |𝔪| = 2. Now, deg(0, y_{1}) = |S| − 2 implies that (0, y_{1}) is not adjacent to (1, y_{i}) for i = 2, 3. That is, for i = 2, 3,

(0, y_{1}) + (1, y_{i}) = (1, y_{1} + y_{i}) ∉ U(ℤ_{2} × S)

and so y_{1} + y_{i} ∉ U(S). This means that, by Lemma 3.17, for i = 2, 3, we have y_{1} + y_{i} ∈ 𝔪. Therefore, by using y_{2} ≠ y_{3} and |𝔪| = 2, we may conclude that 𝔪 = {y_{1} + y_{2}, y_{1} + y_{3}}. Without loss of generality, we may suppose that y_{1} + y_{2} = 0 and therefore D = {(0, y_{1}), (1, −y_{1}), (1, y_{3})}. Since y_{1} + y_{3} ∈ 𝔪, we conclude that −y_{3} − y_{1} ∈ 𝔪 and so −y_{3} − y_{1} ∉ U(S). Therefore,

(0, −y_{3}) + (0, y_{1}) = (0, −y_{3} + y_{1}) ∉ U(ℤ_{2} × S),
(0, −y_{3}) + (1, −y_{1}) = (1, −y_{3} − y_{1}) ∉ U(ℤ_{2} × S) and
(0, −y_{3}) + (1, y_{3}) = (1, 0) ∉ U(ℤ_{2} × S),

and so (0, −y_{3}) is not adjacent to the elements of D. This implies that (0, −y_{3}) ∈ D and therefore (0, −y_{3}) = (0, y_{1}). This means that y_{1}+y_{3} = 0, which is a contradiction. This shows that γ(G(ℤ_{2} × S)) ≥ 4.

Now, by Lemma 4.1, we have 4 ≤ γ(G(ℤ_{2} × S)) ≤ γ_{t}(G(ℤ_{2} × S)) ≤ 4 and so γ(G(ℤ_{2} × S)) = γ_{t}(G(ℤ_{2} × S)) = 4, as required. □

Lemma 4.4. Let R be a ring such that R ≅ R_{1} × R_{2}, where R_{1} and R_{2} are local rings with maximal ideals 𝔪_{1} and 𝔪_{2}, respectively. Suppose that either |R_{1}/𝔪_{1}| = 2 or |R_{2}/𝔪_{2}| = 2. If R ≇ ℤ_{2} × 𝔽, where 𝔽 is a field, then γ(G(R)) = γ_{t}(G(R)) = 4.

Proof. Without loss of generality, let |R_{1}/𝔪_{1}| = 2. If 𝔪_{1} ≠ {0}, then R_{1} is not a field. Therefore, by Lemma 4.2, γ(G(R)) = γ_{t}(G(R)) = 4. If 𝔪_{1} = {0}, then R_{1} ≅ ℤ_{2} and since R ≇ ℤ_{2} × 𝔽, where 𝔽 is a field, we conclude that R_{2} is not a field. Now, Lemma 4.3 implies that γ(G(R)) = γ_{t}(G(R)) = 4. □

We also need Lemmas 4.5, 4.6 and 4.8 for proving part (3) of Theorem 2.2 and completing the paper.

Lemma 4.5. Let 𝔽_{1} and 𝔽_{2} be two fields in which either both of them have characteristic 2 and none of them is isomorphic to ℤ_{2} or none of them has characteristic 2. Then γ(G(𝔽_{1} × 𝔽_{2})) = γ_{t}(G(𝔽_{1} × 𝔽_{2})) = 3.

Proof. The assumption implies that |𝔽_{1}| ≥ 3 and |𝔽_{2}| ≥ 3 and so we may choose nonzero elements a, c ∈ 𝔽_{1} with a ≠ c and b, d ∈ 𝔽_{2} with b ≠ d. We claim that D = {(0, 0), (a, b), (c, d)} is a total dominating set of the unit graph G(𝔽_{1} × 𝔽_{2}). In order to prove the claim, note that a given element in 𝔽_{1} × 𝔽_{2} has one of the following forms, where x ∈ 𝔽_{1} and y ∈ 𝔽_{2} are nonzero elements: either (0, 0), (x, 0), (0, y) or (x, y).

For (0, 0), we have

(0, 0) + (a, b) = (a, b) ∈ U(𝔽_{1} × 𝔽_{2})

and so (0, 0) is adjacent to (a, b).

For (x, 0), we have either x ≠ −a or x ≠ −c, and so either

(x, 0) + (a, b) = (x + a, b) ∈ U(𝔽_{1} × 𝔽_{2}) or
(x, 0) + (c, d) = (x + c, d) ∈ U(𝔽_{1} × 𝔽_{2}).

This implies that (x, 0) is adjacent to (a, b) or (c, d). By a similar argument, we obtain that (0, y) is also adjacent to (a, b) or (c, d).

For (x, y), we have

(x, y) + (0, 0) = (x, y) ∈ U(𝔽_{1} × 𝔽_{2})

and so (x, y) is adjacent to (0, 0).

Therefore, the claim holds true and so γ_{t}(G(𝔽_{1} × 𝔽_{2})) ≤ 3. On the other hand, by Lemmas 3.4 and 3.6, we conclude that γ(G(𝔽_{1} × 𝔽_{2})) ≥ 3. Therefore,

3 ≤ γ(G(𝔽_{1} × 𝔽_{2})) ≤ γ_{t}(G(𝔽_{1} × 𝔽_{2})) ≤ 3

and so γ(G(𝔽_{1} × 𝔽_{2})) = γ_{t}(G(𝔽_{1} × 𝔽_{2})) = 3, as required. □

Lemma 4.6. Let 𝔽_{1} and 𝔽_{2} be two fields in which only one of them has characteristic 2 and none of them is isomorphic to ℤ_{2}. Then γ_{t}(G(𝔽_{1} × 𝔽_{2})) ≤ 3.

Proof. Without loss of generality, we may assume that Char(𝔽_{1}) = 2 and Char(𝔽_{2}) ≠ 2. Since 𝔽_{1} ≇ ℤ_{2} and 𝔽_{2} ≇ ℤ_{2}, we conclude that |𝔽_{1}| ≥ 3 and |𝔽_{2}| ≥ 3, and so we may choose a ∈ 𝔽_{1} and b ∈ 𝔽_{2} such that a ≠ 0, 1 and b ≠ −1, 1. We claim that D = {(0, 1), (1, −1), (a, b)} is a total dominating set of the unit graph G(𝔽_{1} × 𝔽_{2}). In order to prove the claim, note that a given element in 𝔽_{1} × 𝔽_{2} has one of the following forms, where x ∈ 𝔽_{1} and y ∈ 𝔽_{2} are nonzero elements: either (0, 0), (x, 0), (0, y) or (x, y).

For (0, 0),

(0, 0) + (1, −1) = (1, −1) ∈ U(𝔽_{1} × 𝔽_{2})

implies that (0, 0) is adjacent to (1, −1).

For (x, 0),

(x, 0) + (0, 1) = (x, 1) ∈ U(𝔽_{1} × 𝔽_{2})

implies that (x, 0) is adjacent to (0, 1).

For (0, y), if y ≠ 1, then we have

(0, y) + (1, −1) = (1, y − 1) ∈ U(𝔽_{1} × 𝔽_{2})

and if y = 1, then we have

(0, y) + (a, b) = (a, 1 + b) ∈ U(𝔽_{1} × 𝔽_{2}).

Therefore, (0, y) is either adjacent to (1, −1) or (a, b).

Finally, for (x, y), if x = y = 1, then we have

(x, y) + (a, b) = (1 + a, 1 + b) ∈ U(𝔽_{1} × 𝔽_{2}),

if x ≠ 1 and y = 1, then we have

(x, y) + (0, 1) = (x, 2) ∈ U(𝔽_{1} × 𝔽_{2}),

if x = 1 and y ≠ 1, then for y ≠ −1 we have

(x, y) + (0, 1) = (1, y + 1) ∈ U(𝔽_{1} × 𝔽_{2})

and for y = −1 we have

(x, y) + (a, b) = (1 + a, −1 + b) ∈ U(𝔽_{1} × 𝔽_{2}),

and if x, y ≠ 1, then we have

(x, y) + (1, −1) = (x + 1, y − 1) ∈ U(𝔽_{1} × 𝔽_{2}).

Therefore, (x, y) is either adjacent to (0, 1), (1, −1) or (a, b).

Hence, the claim holds true and so γ_{t}(G(𝔽_{1} × 𝔽_{2})) ≤ 3, as required. □

We need the following proposition for the proof of Lemma 4.8.

Proposition 4.7 ([1], Lemma 2.7(a)). Let R be a ring, x, y ∈ R and suppose that J(R) denotes the Jacobson radical of R. If x + J(R) and y + J(R) are adjacent in the unit graph G(R/J(R)), then every element of x + J(R) is adjacent to every element of y + J(R) in the unit graph G(R).

Lemma 4.8. Let R be a ring. Then γ_{t}(G(R)) ≤ γ_{t}(G(R/J(R))), where J(R) denotes the Jacobson radical of R.

Proof. Let γ_{t}(G(R/J(R))) = n and consider D = {x_{1} + J(R), …, x_{n} + J(R)} as a γ_{t}−set of G(R/J(R)). We claim that D′ = {x_{1}, …, x_{n}} is a total dominating set of G(R). In order to prove the claim, let y ∈ R be given. Therefore, y + J(R) ∈ R/J(R) and so there exists x_{i} + J(R) ∈ D such that y + J(R) is adjacent to x_{i} + J(R) in G(R/J(R)). Now, Proposition 4.7 implies that y is adjacent to x_{i} in G(R) and so the claim holds true. This implies that γ_{t}(G(R)) ≤ |D′| = n = γ_{t}(G(R/J(R))). □

We are now ready to prove part (3) of Theorem 2.2, that is:

Theorem 4.9. Let R be a ring. Then γ(G(R)) = 3 if and only if R is not isomorphic to the product of two fields such that only one of them has characteristic 2, and R ≅ R_{1} × R_{2}, where for i = 1, 2, R_{i} is a local ring with maximal ideal 𝔪_{i} in such a way that R_{i}/𝔪_{i} ≇ ℤ_{2}.

Proof. (⇒): Let γ(G(R)) = 3. Note that, by Theorem 3.20, R is not isomorphic to the product of two fields such that only one of them has characteristic 2. Now, by Remark 3.15, we may write R ≅ R_{1} × ⋯ × R_{n}, where for every 1 ≤ i ≤ n, R_{i} is a local ring with maximal ideal 𝔪_{i}. If n ≥ 3, then by using Lemma 3.16, we conclude that 3 = γ(G(R)) ≥ n + 1 ≥ 4, which is a contradiction. Therefore, n ≤ 2. If n = 1, then R ≅ R_{1} is either a field or a local ring which is not a field. In the first case, by Lemma 3.1, we have γ(G(R)) = 1, and in the second case, by Theorem 3.20, we have γ(G(R)) = 2, which both of them are contradictions. Therefore, n = 2 and so we have R ≅ R_{1} × R_{2}. By Theorem 3.20, we conclude that R ≇ ℤ_{2} × 𝔽, where 𝔽 is a field, and so Lemma 4.4 implies that |R_{1}/𝔪_{1}| ≥ 3 and |R_{2}/𝔪_{2}| ≥ 3. Therefore, R_{1}/𝔪_{1} ≇ ℤ_{2} and R_{2}/𝔪_{2} ≇ ℤ_{2}.

(⇐): By the assumption and Remark 3.15, we have R/J(R) ≅ 𝔽_{1} × 𝔽_{2}, where 𝔽_{1} = R_{1}/𝔪_{1} and 𝔽_{2} = R_{2}/𝔪_{2} are fields. Also, 𝔽_{1} and 𝔽_{2} are not isomorphic to ℤ_{2}. Now, one of the following cases occurs:

Case 1: Either both of 𝔽_{1} and 𝔽_{2} have characteristic 2 or none of them has characteristic 2. In this case, by Lemma 4.5, we conclude that γ(G(R/J(R))) = γ_{t}(G(R/J(R))) = 3. Now, in view of Lemmas 3.14 and 4.8, we conclude that 3 ≤ γ(G(R)) ≤ γ_{t}(G(R)) ≤ 3, which implies that γ(G(R)) = 3, as required.

Case 2: Only one of 𝔽_{1} or 𝔽_{2} has characteristic 2. In this case, by Theorem 3.20, we have γ(G(R/J(R))) = 2, and also by Lemma 4.6, we have γ_{t}(G(R/J(R))) ≤ 3. Therefore, in view of Lemmas 3.14 and 4.8, we conclude that either γ(G(R)) = 2 or γ(G(R)) = 3. Since R ≅ R_{1} × R_{2}, R is not a local ring. Also, by the assumption, R is not isomorphic to the product of two fields such that only one of them has characteristic 2. Finally, in contrary, suppose that R ≅ ℤ_{2} × 𝔽, where 𝔽 is a field. Therefore, R/J(R) ≅ ℤ_{2} × 𝔽 and so ℤ_{2} × 𝔽 ≅ 𝔽_{1} × 𝔽_{2}, which implies that |ℤ_{2} × 𝔽| = |𝔽_{1} × 𝔽_{2}|. The assumption implies that Char(𝔽) = 2 and so |ℤ_{2} × 𝔽| is a power of 2, while, by the assumption in Case 2, |𝔽_{1} × 𝔽_{2}| is divisible by an odd prime. This contradiction shows that R ≇ ℤ_{2} × 𝔽, where 𝔽 is a field. Hence, by Theorem 3.20, γ(G(R)) = 2 is impossible and so γ(G(R)) = 3, as required. □

Acknowledgment

The authors would like to thank the referee for a careful reading of the paper and for valuable comments.

References

[1] N. Ashrafi, H. R. Maimani, M. R. Pournaki, S. Yassemi, Unit graphs associated with rings, Comm. Algebra 38 (2010), no. 8, 2851–2871.

[2] M. F. Atiyah, I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont. 1969.

[3] M. R. Garey, D. S. Johnson, Computers and Intractability. A Guide to the Theory of NP-Completeness, A Series of Books in the Mathematical Sciences, W. H. Freeman and Co., San Francisco, Calif., 1979.

[4] R. P. Grimaldi, Graphs from rings, Proceedings of the Twentieth Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1989), Congr. Numer. 71 (1990), 95–103.

[5] T. W. Haynes, S. T. Hedetniemi, P. J. Slater, Fundamentals of Domination in Graphs, Monographs and Textbooks in Pure and Applied Mathematics, 208, Marcel Dekker, Inc., New York, 1998.

[6] T. W. Haynes, S. T. Hedetniemi, P. J. Slater (Editors), Domination in Graphs. Advanced Topics, Monographs and Textbooks in Pure and Applied Mathematics, 209, Marcel Dekker, Inc., New York, 1998.

[7] H. R. Maimani, M. R. Pournaki, S. Yassemi, Rings which are generated by their units: a graph theoretical approach, Elem. Math. 65 (2010), no. 1, 17–25.

[8] H. R. Maimani, M. R. Pournaki, S. Yassemi, Weakly perfect graphs arising from rings, Glasg. Math. J. 52 (2010), no. 3, 417–425.

[9] H. R. Maimani, M. R. Pournaki, S. Yassemi, Necessary and sufficient conditions for unit graphs to be Hamiltonian, Pacific J. Math. 249 (2011), no. 2, 419–429.

[10] B. R. McDonald, Finite Rings with Identity, Pure and Applied Mathematics, 28, Marcel Dekker, Inc., New York, 1974.

[11] G. Mekiš, Lower bounds for the domination number and the total domination number of direct product graphs, Discrete Math. 310 (2010), no. 23, 3310–3317.

[12] D. B. West, Introduction to Graph Theory, Prentice Hall, Inc., Upper Saddle River, NJ, 1996.

S. Kiani, Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran.

E-mail address: s.kiani@srbiau.ac.ir

H. R. Maimani, Mathematics Section, Department of Basic Sciences, Shahid Rajaee Teacher Training University, P.O. Box 16785-163, Tehran, Iran, and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran.

E-mail address: maimani@ipm.ir

M. R. Pournaki, Department of Mathematical Sciences, Sharif University of Technology, P.O. Box 11155-9415, Tehran, Iran, and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran.

E-mail address: pournaki@ipm.ir
URL: http://math.ipm.ac.ir/∼pournaki/

S. Yassemi, School of Mathematics, Statistics and Computer Science, College of Science, University of Tehran, Tehran, Iran, and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran.

E-mail address: yassemi@ut.ac.ir
URL: http://math.ipm.ac.ir/∼yassemi/