Coloring of the annihilator graph of a commutative ring

2nd Reading August 27, 2015 15:54 WSPC/S0219-4988

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Journal of Algebra and Its Applications Vol. 15, No. 6 (2016) 1650124 (13 pages) c World Scientific Publishing Company DOI: 10.1142/S0219498816501243

J. Algebra Appl. Downloaded from www.worldscientific.com by SWISS FEDERAL INSTITUTE OF TECHNOLOGY ZURICH (ETH) on 09/01/15. For personal use only.

Coloring of the annihilator graph of a commutative ring

R. Nikandish Department of Mathematics Jundi-Shapur University of Technology P. O. Box 64615–334, Dezful, Iran [email protected] M. J. Nikmehr∗ and M. Bakhtyiari† Faculty of Mathematics K. N. Toosi University of Technology P. O. Box 16315–1618, Tehran, Iran ∗[email protected] †[email protected] Received 23 January 2015 Accepted 15 July 2015 Published 31 August 2015 Communicated by D. Passman Let R be a commutative ring with identity, and let Z(R) be the set of zero-divisors of R. The annihilator graph of R is defined as the graph AG(R) with the vertex set Z(R)∗ = Z(R)\{0}, and two distinct vertices x and y are adjacent if and only if annR (xy) = annR (x) ∪ annR (y). In this paper, we study annihilator graphs of rings with equal clique number and chromatic number. For some classes of rings, we give an explicit formula for the clique number of annihilator graphs. Among other results, bipartite annihilator graphs of rings are characterized. Furthermore, some results on annihilator graphs with finite clique number are given. Keywords: Annihilator graph; chromatic number; clique number; bipartite graph. Mathematics Subject Classification: 05C25, 05C15, 05C69

1. Introduction In recent years, the field of graphs associated with rings has been received much attention since its appearance, see for example [2, 3, 8, 10]. On the other hand, computing the clique number and the chromatic number of a given graph is an NPcomplete problem, see [6]. Therefore, it is natural to consider the problem of finding the clique number and chromatic number of graphs associated with rings. Let us

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first recall some standard notations and definitions that will be used throughout the paper. Throughout this paper, all rings are assumed to be commutative with identity different from zero that are not field. We denote by Max(R), Min(R), Nil(R) and U (R), the set of all maximal ideals of R, the set of all minimal prime ideals of R, the set of all nilpotent elements of R and the set of all invertible elements of R, respectively. Also, the set of all zero-divisors of an R-module M is denoted by Z(M ). We say that depth(R) = 0 and dim(R) = 0 whenever every non-unit element of R is a zero-divisor and for every two distinct prime ideals p, q we have p q and p q, respectively. The ring R is said to be reduced if it has no nonzero nilpotent element. Two ideals I and J of a ring R are called coprime if I + J = R. For every ideal I of R, we denote the annihilator of I by annR (I). A prime ideal p of R is called an associated prime ideal, if annR (x) = p, for some nonzero element x ∈ R. The set of all associated prime ideals of R is denoted by Ass(R). For a subset A of a ring R we let A∗ = A\{0}. For any undefined notation or terminology in ring theory, we refer the reader to [1, 4]. Let G = (V, E) be a graph, where V = V (G) is the set of vertices and E = E(G) is the set of edges. By G, we mean the complement graph of G. The girth of a graph G is denoted by gr(G). Also G is called a null graph if it has no edge. We write u − v, to denote an edge with ends u, v. A graph H = (V0 , E0 ) is called a subgraph of G if V0 ⊆ V and E0 ⊆ E. Moreover, H is called an induced subgraph by V0 , denoted by G[V0 ], if V0 ⊆ V and E0 = {{u, v} ∈ E | u, v ∈ V0 }. Let G1 and G2 be two disjoint graphs. The join of G1 and G2 , denoted by G1 ∨ G2 , is a graph with the vertex set V (G1 ∨ G2 ) = V (G1 ) ∪ V (G2 ) and edge set E(G1 ∨ G2 ) = E(G1 ) ∪ E(G2 ) ∪ {uv | u ∈ V (G1 ), v ∈ V (G2 )}. A complete bipartite graph of part sizes m, n is denoted by Km,n . Also, a complete graph of n vertices is denoted by Kn . A clique of G is a complete subgraph of G and the number of vertices in a largest clique of G, denoted by ω(G), is called the clique number of G. For a graph G, let χ(G) denote the vertex chromatic number of G, i.e. the minimal number of colors which can be assigned to the vertices of G in such a way that every two adjacent vertices have different colors. Clearly, for every graph G, ω(G) ≤ χ(G). A graph G is said to be weakly perfect if ω(G) = χ(G). For any undefined notation or terminology in graph theory, we refer the reader to [12]. The annihilator graph of a ring R is defined as the graph AG(R) with the vertex set Z(R)∗ = Z(R)\{0}, and two distinct vertices x and y are adjacent if and only if annR (xy) = annR (x) ∪ annR (y). This graph was first introduced and investigated in [2] and many of interesting properties of annihilator graph were studied. This paper is devoted to the study of coloring of annihilator graphs associated with some classes of rings. Our main aim in this paper is to explore some weakly perfect annihilator graphs. For example, we show that the annihilator graph of direct n product . Furof n integral domains is weakly perfect with ω(AG(R)) = χ(AG(R)) = [n/2] thermore, under some assumptions, we show that the annihilator graph associated with an Artinian ring is weakly perfect. Moreover, in this case, we give an explicit 1650124-2


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formula for the vertex chromatic number of AG(R). Finally, some results on the annihilator graphs with finite clique numbers are given.


2. Some Weakly Perfect Annihilator Graphs of Rings In this section, we show that if R is a finite direct product of integral domains, then AG(R) is weakly perfect. Also, for some Artinian rings we find the chromatic number (clique number) of annihilator graphs. The following two lemmas will be used frequently in this paper. Lemma 2.1. Let R be a ring and x, y be distinct elements of Z(R)∗ . Then the following statements are equivalent : (1) x − y is an edge of AG(R). (2) Rx ∩ annR (y) = (0) and Ry ∩ annR (x) = (0). (3) x ∈ Z(Ry) and y ∈ Z(Rx). Proof. (2) ⇔ (3) is clear. (1) ⇒ (2) Assume that x − y is an edge of AG(R). So there exists an element r ∈ R such that rxy = 0, rx = 0 and ry = 0. Thus Rx ∩ annR (y) = (0) and Ry ∩ annR (x) = (0). (2) ⇒ (1) Since Rx ∩ annR (y) = (0) and Ry ∩ annR (x) = (0), there exist r1 , r2 ∈ R such that r1 xy = r2 yx = 0, r1 x = 0 and r2 y = 0. If r1 = r2 or r1 y = 0 or r2 x = 0, then there is nothing to prove. So let r1 = r2 , r1 y = 0 and r2 x = 0. Therefore, (r1 − r2 )xy = 0, (r1 − r2 )x = 0 and (r1 − r2 )y = 0. Hence x − y is an edge of AG(R). Lemma 2.2. Let R be a ring. (1) Let x, y be elements of Z(R)∗ . If annR (x) annR (y) and annR (y) annR (x), then x − y is an edge of AG(R). Moreover, if R is a reduced ring, then the converse is also true. (2) Let R ∼ = R1 × · · · × Rn , x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ), where n is a positive integer, every Ri is a ring and xi , yi ∈ Ri , for every 1 ≤ i ≤ n. If Ri xi ∩ annRi (yi ) = (0) and Rj yj ∩ annRj (xj ) = (0), for some 1 ≤ i, j ≤ n, then x − y is an edge of AG(R). In particular, if xi − yi is an edge of AG(Ri ) or xi = yi ∈ Nil(Ri )∗ , for some 1 ≤ i ≤ n, then x − y is an edge of AG(R). Proof. (1) If annR (x) annR (y) and annR (y) annR (x), then by part (4) of [2, Lemma 2.1], x − y is an edge of AG(R). To prove the converse, assume to the contrary, annR (y) ⊆ annR (x). Since R is reduced, we deduce that Rx ∩ annR (x) = (0) and so Rx ∩ annR (y) = (0), a contradiction, by Lemma 2.1. Thus annR (y) annR (x). Similarly, annR (x) annR (y). (2) Since Ri xi ∩ annRi (yi ) = (0), there exists an element 0 = ai ∈ Ri xi such that ai yi = 0 and thus (0, . . . , 0, ai , 0, . . . , 0) ∈ Rx ∩ annR (y). Similarly, Rj yj ∩ 1650124-3


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annRj (xj ) = (0) implies that Ry ∩ annR (x) = (0). By Lemma 2.1, x − y is an edge of AG(R). The “in particular” statement is now clear. Theorem 2.1. Suppose that R ∼ = D1 × · · · × Dn , where 2 ≤ n < ∞ and Di is an integral domain for every 1 ≤ i ≤ n. Then the following statements hold : n -partite graph, where [n/2] denotes the largest integer less (1) AG(R) is an [n/2] than or equal to n/2. n − 1 -partite graph. (2) AG(R) is not an [n/2] Proof. (1) Assume that X = (x1 , . . . , xn ) and Y = (y1 , . . . , yn ) are vertices of AG(R), where xi , yi ∈ Di , for every 1 ≤ i ≤ n. Define the relation ∼ on V (AG(R)) as follows: X ∼ Y , whenever “xi = 0 if and only if yi = 0”, for every 1 ≤ i ≤ n. It is easily seen that ∼ is an equivalence relation on V (AG(R)). By [X], we mean the equivalence class of X. For every equivalence class [X], let [X]t be the number of nonzero components in every element of [X]. Obviously, 1 ≤ t ≤ n − 1. Now, we put Aj = {[X] | [X]t = j} and Aj = [X]∈Aj [X], for every 1 ≤ j ≤ n − 1. We claim that AG(R)[Aj ] is a complete nj -partite subgraph of AG(R), for every 1 ≤ j ≤ n − 1. To see this, suppose that [X] and [Y ] are two distinct arbitrary equivalence classes of Aj . We show that there is no adjacency between elements of [X] whereas each element of [X] is adjacent to each element of [Y ] in AG(R)[Aj ]. Let X1 and X2 be two elements of [X] and Y1 be an element of [Y ]. There exist elements xi , yi and zi of Di , such that X1 = (x1 , . . . , xn ), X2 = (y1 , . . . , yn ) and Y1 = (z1 , . . . , zn ), for every 1 ≤ i ≤ n. Since X1 ∼ X2 , xi = 0 if and only if yi = 0, for every 1 ≤ i ≤ n. This implies that annR (X1 ) = annR (X2 ), and so by part (1) of Lemma 2.2, X1 and X2 are not adjacent. Also, X1 Y1 and [X]t = [Y ]t imply that xi = 0, zi = 0, xj = 0, zj = 0, for some i = j, 1 ≤ i, j ≤ n and hence annR (X1 ) annR (Y1 ) and annR (X1 ) ann R (Y1 ). By part (1) of Lemma 2.2, X1 is adjacent to Y1 . The equality |Aj | = nj implies that AG(R)[Aj ] is a complete nj -partite subgraph of AG(R), for every 1 ≤ j ≤ n − 1 and so the claim is proved. To complete the proof, let j = [n/2]. Clearly, |Aj | ≥ |Ai |, for every 1 ≤ i ≤ n − 1. In what follows, we start from AG(R)[Aj ] to get AG(R). If n = 2, then AG(R) = AG(R)[A1 ] and so AG(R) is a bipartite graph. So let n ≥ 3, [Xj ] ∈ Aj and Xj = (x1 , . . . , xn ) ∈ [Xj ]. By replacing one of the nonzero components of Xj by 0 (if it is possible), and one of the zero components of Xj by a nonzero element (if it is possible), we get Xj−1 ∈ [Xj−1 ] ∈ Aj−1 , Xj+1 ∈ [Xj+1 ] ∈ Aj+1 . Thus annR (Xj+1 ) ⊂ annR (Xj ) ⊂ annR (Xj−1 ). Similarly, replace one of the nonzero components of Xj−1 by 0, and one of the zero components of Xj+1 by a nonzero element, we obtain Xj−2 ∈ [Xj−2 ] ∈ Aj−2 , Xj+2 ∈ [Xj+2 ] ∈ Aj+2 , annR (Xj−1 ) ⊂ annR (Xj−2 ) and annR (Xj+2 ) ⊂ annR (Xj+1 ). Continue this procedure up to Max{s, t}, where j + t = n − 1 and j − s = 1. Let B1 = n−1 i=1 [Xi ]. Then by part (1) of Lemma 2.2, there is no adjacency between every pair of vertices contained in B1 in AG(R). Indeed, we obtained B1 corresponding to the [Xj ] ∈ Aj 1650124-4


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of AG(R)[Aj ]. Consider [Xj ] = [Yj ] ∈ Aj , and apply this method on V (AG(R))\B1 . m We have B2 = i=1 [Yi ] where 1 ≤ m ≤ n − 1 (n − 1 − m is the number of cases that [Xi ] = [Yi ] for some 1 ≤ i ≤ n−1) and so B2 is a subset of V (AG(R)) corresponding to the part [Yj ] ∈ Aj in AG(R)[Aj ]. Similarly, if we let [Zj ] ∈ Aj \{[Xj ], [Yj ]}, k and apply our method on the V (AG(R))\B1 ∪ B2 , we get B3 = i=1 [Zi ] where 1 ≤ k ≤ n − 1. We continue this procedure for |Aj | times, and we obtain all Bi ’s, for [n/2] every 1 ≤ i ≤ |Aj |. It is not hard to see that V (AG(R)) = i=1 Bi and AG(R)[Bi ] is a null graph, for every 1 ≤ i ≤ |Aj |. This, together n with AG(R) is connected (see -partite graph. [2, Theorem 2.2]), implies that AG(R) is an [n/2] n (2) By part (1), AG(R)[Aj ] is a complete j -partite subgraph of AG(R), for every n 1 ≤ j ≤ n − 1. Let j = [n/2]. Then it is easily seen that AG(R) cannot be an − 1)-partite graph. ( [n/2] In view of Theorem 2.1, we have the following corollary. Corollary 2.1. Let R be a reduced ring with |Min(R)| < ∞ and suppose that p, q are coprime, for every two distinct p, q ∈ Min(R). Then the following statements are equivalent: (1) |Min(R)| = n. n -partite graph. (2) AG(R) is an [n/2] Proof. The proof follows from Theorem 2.1 and [9, Theorem 1.4]. ∼ D1 × · · · × Dn , where Di is an Theorem 2.2. Let R be a ring such that R = integral domain, for every 1 ≤ i ≤ n < ∞. Then n ω(AG(R)) = χ(AG(R)) = . [n/2] Proof. The proof is obtained by Theorem 2.1. If R is a reduced Artinian ring, then by [1, Theorem 8.7], R is a direct product of finitely many fields. Thus by Theorem 2.2, we have the following immediate corollary. Corollary 2.2. Let R be a reduced Artinian ring. Then |Max(R)| ω(AG(R)) = χ(AG(R)) = . [|Max(R)|/2] It is known that if R is a ring such that depth(R) = 0, then R is infinite. Also, Ganesan [5] proved that if R is infinite and Z(R) = (0), then Z(R) must be infinite (in fact |R| ≤ |Z(R)|2 , when 2 ≤ |Z(R)| < ∞). By using these facts and following lemma, we prove Theorem 2.3. Lemma 2.3. Let R be a non-reduced ring such that R ∼ = R1 × F, where F is a field and Z(R1 ) = Nil(R1 ). Then ω(AG(R)) = χ(AG(R)) = |Nil(R1 )∗ ||F |. 1650124-5


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Proof. Let V1 := {(x, y) ∈ R | x ∈ Z(R1 )∗ , y ∈ F ∗ }, V2 := {(x, 0) ∈ R | x ∈ Z(R1 )∗ }, / Z(R1 )}, V3 := {(x, 0) ∈ R | x ∈ J. Algebra Appl. Downloaded from www.worldscientific.com by SWISS FEDERAL INSTITUTE OF TECHNOLOGY ZURICH (ETH) on 09/01/15. For personal use only.

V4 := {(0, y) ∈ R | y ∈ F ∗ }. Then {V1 , V2 , V3 , V4 } is a partition of V (AG(R)). By Lemma 2.2, one may check that AG(R)[V1 ∪ V2 ] = K|Nil(R1 )∗ ||F | and AG(R)[V3 ∪ V4 ] = K|V3 |,|V4 | , also there is no adjacency between two vertices of V1 (respectively, V2 ) and V4 (respectively, V3 ). Hence ω(AG(R)) = χ(AG(R)) = |Nil(R1 )∗ ||F |. Theorem 2.3. Let R be a non-reduced ring. Then the following statements hold : (1) If |Z(R)| < ∞, then the following statements are equivalent : (i) ω(AG(R)) = |Nil(R)|. (ii) χ(AG(R)) = |Nil(R)|. (iii) AG(R) = K2,3 . (2) If |Z(R)| = ∞, ω(AG(R)) < ∞ and Z(R) is an ideal of R, then the following statements are equivalent : (i) (ii) (iii) (iv)

ω(AG(R)) = |Nil(R)|. χ(AG(R)) = |Nil(R)|. AG(R) = K|Nil(R)∗ | ∨ K ∞ . x − y is not an edge of AG(R), for every x, y ∈ Z(R)\Nil(R).

Proof. (1) (i) ⇒ (iii) Let ω(AG(R)) = |Nil(R)|. Since |Z(R)| < ∞, R is an Artinian ring (indeed R is finite). By [1, Theorem 8.7], there exists a positive integer n such that R = R1 ×· · ·×Rn , where each Ri , 1 ≤ i ≤ n, is an Artinian local ring. If n = 1, then by [2, Theorem 3.10], ω(AG(R)) = |Nil(R)| − 1, a contradiction. So we may assume that R ∼ = R1 × R2 , where R1 is an Artinian local ring with Nil(R1 ) = (0). We show that R2 ∼ = Z2 . Assume to the contrary, x ∈ R2 \{0, 1}. By part (2) of Lemma 2.2, one may check that {(a, x), (a, 1) ∈ R | a ∈ Nil(R1 )∗ } ∪ {(a, 0) ∈ R | a ∈ Nil(R1 )∗ } is a clique of AG(R), a contradiction. Thus R ∼ = R1 × Z2 . By Lemma 2.3, we have ω(AG(R)) = 2|Nil(R1 )∗ |. Also by hypothesis, ω(AG(R)) = |Nil(R)|. Therefore, |Nil(R)| = 2. This, together with R1 is an Artinian local ring, implies that R1 ∼ = Z4 or R1 ∼ = Z2 [X]/(X 2). Hence AG(R) = K2,3 . (iii) ⇒ (i) Since |Z(R)| < ∞, R is finite. If R is local, then by [2, Theorem 3.10], AG(R) = K|Nil(R)∗ | , a contradiction. Thus we may assume that R ∼ = R1 × R2 , where R1 is a local ring with Nil(R1 ) = (0). This implies that |U (R1 )| ≥ 2. Since AG(R) = K2,3 , |Z(R)∗ | = 5. Hence V (AG(R)) = {(u, 0), (v, 0), (a, 0), (a, 1), (0, 1)}, where u, v ∈ U (R1 ) and a ∈ Nil(R1 )∗ . Therefore, ω(AG(R)) = |Nil(R)|.

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(ii) ⇒ (iii) It is proved by a similar argument to the proof of (i) ⇒ (iii). Moreover, (iii) ⇒ (ii) is obvious. (2) (iii) ⇒ (i),(ii),(iv) are clear. (ii),(iv) ⇒ (iii) are obtained by (i) ⇒ (iii) and its proof. Hence, we have only to prove (i) ⇒ (iii). (i) ⇒ (iii) Let ω(AG(R)) = |Nil(R)|. Since ω(AG(R)) < ∞, |Nil(R)| < ∞. Let a ∈ Nil(R)∗ and x ∈ Z(R)\Nil(R). We claim that a − x is an edge of AG(R). If ax = 0, then there is nothing to prove. If ax = 0, then we show that axl = 0, for some positive integer l. Assume to the contrary, axl = 0, for every positive integer l. Since |Nil(R)| < ∞, there exist positive integers n, m such that axn = axm . With no loss of generality, assume that n < m. Hence axn (1 − xm−n ) = 0 and thus Z(R) is not an ideal of R, a contradiction. Let n be the least positive integer such that axn = 0. Then we have axn−1 ∈ Ra ∩ annR (x). As a ∈ Nil(R)∗ , there exists a least positive integer s such that xas = 0. Thus xas−1 ∈ Rx ∩ annR (a). It follows from Lemma 2.1, x − a is an edge of AG(R) and so the claim is proved. Hence every vertex contained in Nil(R)∗ is adjacent to the other vertices. This, together with ω(AG(R)) = |Nil(R)| and [2, Theorem 3.10], implies that AG(R) = K|Nil(R)∗ | ∨K ∞ .

In the following theorem we classify all bipartite annihilator graphs of rings. Theorem 2.4. Let R be a non-reduced ring. Then the following statements are equivalent : (1) ω(AG(R)) = 2. (2) χ(AG(R)) = 2. (3) AG(R) = K2,3 or AG(R) = K2 or AG(R) = K1 ∨ K ∞ . Proof. (3) ⇒ (1), (2) are clear. If we prove (1) ⇒ (3), then (2) ⇒ (3) is easily obtained. Thus (1) ⇒ (3) is the only thing to prove. (1) ⇒ (3) Suppose that ω(AG(R)) = 2. If Z(R) = Nil(R), then by [2, Theorem 3.10], we have |Nil(R)| = 3 and so AG(R) = K2 . Now, let Z(R) = Nil(R). Since ω(AG(R)) = 2, gr(AG(R)) ∈ {4, ∞} (note that gr(AG(R)) ∈ {3, 4, ∞}). By [2, Theorems 3.6 and 3.18], we conclude that AG(R) = K2,3 or AG(R) = K1 ∨ K ∞ .

To state our main result in this section, we need to fix some notations. Notation 1. Let R be a ring such that R = R1 × · · · × Rn , where Ri is a local ring, for every 1 ≤ i ≤ n. We define an n × n matrix M (AG(R)) whose entries aij are given by  ∗ if i < j,  |Rj | ∗ aij = |Nil(Rj ) | if i = j,   if i > j. |U (Rj )|

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In other notation, we have   |Nil(R1 )∗ | |R2∗ | ··· |Rn∗ |    |U (R1 )| |Nil(R2 )∗ | · · · |Rn∗ |    M (AG(R)) =  . .. .. .. ..   . . . .   |U (R2 )| · · · |Nil(Rn )∗ | |U (R1 )| Let Vi = (ai1 , . . . , ain ) be the ith row of M (AG(R)). Put Λi = Λ(Vi ) = ai1 · · · ain n and Λ(AG(R)) = i=1 Λi . Notation 2. Let R be a ring and A, B, . . . , be nonempty subsets of R. Then (A, B, . . .) = {(a, b, . . .) | a ∈ A, b ∈ B, . . .} and if A = {a} (or B = {b}, . . .) then we put (A, B, . . .) = (a, B, . . .) (or (A, B, . . .) = (A, b, . . .)). Now, we are ready to present the following theorem. Theorem 2.5. Let R = R1 × · · · × Rn , where Ri is an Artinian local ring with |Nil(Ri )∗ | ≥ 2n−1 − 1, for every 1 ≤ i ≤ n. Then ω(AG(R)) = χ(AG(R)) = Λ(AG(R)). Proof. Let An := {(x1 , . . . , xn ) ∈ V (AG(R)) | xi ∈ Nil(Ri )∗ , for all 1 ≤ i ≤ n} and B := V (AG(R))\An . By (x1 , . . . , xn )t , we mean the number of zero components of (x1 , . . . , xn ) ∈ B. Define the relation ∼ on B as follows: If (x1 , . . . , xn ), (y1 , . . . , yn ) ∈ B, then we write (x1 , . . . , xn ) ∼ (y1 , . . . , yn ) if and only if “(x1 , . . . , xn )t = (y1 , . . . , yn )t ”. It is not hard to check that ∼ is an equivalence relation on B. We denote by [(x1 , . . . , xn )] the equivalence class of (x1 , . . . , xn ). Moreover, for every 0 ≤ i ≤ n − 1, if (x1 , . . . , xn )t = i, then we put Ai = [(x1 , . . . , xn )]. Clearly, V (AG(R)) = ∪ni=0 Ai and Ai ∩ Aj = ∅, for every 0 ≤ i, j ≤ n and i = j. So {A0 , . . . , An } is a partition of V (AG(R)). We show that ω(AG(R)) = χ(AG(R)) = |A0 |. Indeed, we have the following claims: Claim 1. AG(R)[Ai ] is a complete subgraph of AG(R), for every 0 ≤ i ≤ n − 1. Let (x1 , . . . , xn ), (y1 , . . . , yn ) ∈ [Ai ] and (x1 , . . . , xn ) = (y1 , . . . , yn ). Since (x1 , . . . , xn ) ∼ (y1 , . . . , yn ), (x1 , . . . , xn )t = (y1 , . . . , yn )t . Two following cases may occur. Case 1. xi = 0 if and only if yi = 0, for every 1 ≤ i ≤ n. In this case, xi ∈ Nil(Ri )∗ and yi = 0 for some 1 ≤ i ≤ n. Also, yj ∈ Nil(Rj )∗ and xj = 0 for some 1 ≤ j ≤ n. By part (2) of Lemma 2.2, we conclude that (x1 , . . . , xn ) is adjacent to (y1 , . . . , yn ). Case 2. xi = 0, yi = 0, xj = 0, yj = 0 for some i = j, 1 ≤ i, j ≤ n. It is easily seen that annR ((x1 , . . . , xn )) annR ((y1 , . . . , yn )) and annR ((x1 , . . . , xn )) annR ((y1 , . . . , yn )). Now, part (1) of Lemma 2.2 implies that (x1 , . . . , xn ) is adjacent to (y1 , . . . , yn ). Claim 2. If i < j, then ω(AG(R))[Ai ∪ Aj ] = χ(AG(R))[Ai ∪ Aj ] = ω(AG(R)) × [Ai ] = χ(AG(R))[Ai ] = |Ai |. 1650124-8


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Define the equivalence relation ∼j on Aj as follows: If (x1 , . . . , xn ), (y1 , . . . , yn ) ∈ Aj , then we write (x1 , . . . , xn ) ∼j (y1 , . . . , yn ), whenever “xi = 0 if and only if yi = 0”, for every 1 ≤ i ≤ n. Using an appropriate labeling, [(x1 , . . . , xn )]t1 ,t2 ,...,tj means the equivalence class of (x1 , . . . , xn ) ∈ Aj such that xt1 = · · · = xtj = {0} where {t1 , . . . , tj } ⊂ {1, . . . , n}. We complete the proof of this claim in two steps. Step 1. Let [(x1 , . . . , xn )]t1 ,...,tj be an equivalence class in Aj . We are going to find a subset Ω of Ai , such that there is no adjacency between two vertices contained in [(x1 , . . . , xn )]t1 ,...,tj and Ω. Fix t1 , . . . , ti ∈ {t1 , . . . , tj } (Note that i < j) and let [(y1 , . . . , yn )]t1 ,...,ti be the equivalence class of (y1 , . . . , yn ) ∈ Ai under the equivalence relation ∼i on Ai , where ∼i is defined as the same as ∼j . We put Ω = {(y1 , . . . , yn ) ∈ [(y1 , . . . , yn )]t1 ,t2 ,...,ti | ytl ∈ Nil(Rl )∗ for every l ∈ {t1 , . . . , tj }\ {t1 , . . . , ti }, and ytk ∈ U (Rk ) for every k ∈ {1, . . . , n}\{t1 , . . . , tj }}. If (x1 , . . . , xn ) ∈ [(x1 , . . . , xn )]t1 ,...,tj , (y1 , . . . , yn ) ∈ Ω, then R(x1 , . . . , xn ) ∩ annR ((y1 , . . . , yn )) = (0) and thus by Lemma 2.1, (x1 , . . . , xn ) is not adjacent to (y1 , . . . , yn ). Step 2. |Ai | > |Aj |. Let [(x1 , . . . , xn )]t1 ,...,tj be the equivalence class of Aj considered in Step 1 and Ω = (Ω1 , . . . , Ωn ) ⊆ [(x1 , . . . , xn )]t1 ,...,tj . Obviously, Ωt1 = · · · = Ωtj = {0} and either Ωtk = U (Rk ) or Ωtk = Nil(Rk )∗ , for every k ∈ {1, . . . , n}\ {t1 , . . . , tj }. This implies that every nonzero component of Ω has two choices so the number of all this selections for Ω is 2n−j . Since 2n−1 − 1 ≤ |Nil(Ri )∗ | < |U (Ri )|, for every 1 ≤ i ≤ n, it is not hard to see that |Ω| > |[(x1 , . . . , xn )]t1 ,...,tj | (since for every choice of Ω , |Ω | ≤ |Ω| and for some |Ω | < |Ω|). By applying this argument to every equivalence class of Aj , we have |Ai | > |Aj |. Therefore, by Steps 1 and 2, we deduce that ω(AG(R))[Ai ∪ Aj ] = χ(AG(R)) [Ai ∪ Aj ] = ω(AG(R))[Ai ] = χ(AG(R))[Ai ] = |Ai |. n Claim 3. AG(R)[An ] is a complete [n/2] -partite subgraph of AG(R). Let (x1 , . . . , xn ) be a vertex of AG(R)[An ]. Then xi ∈ {0} ∪ U (Ri ), for every 1 ≤ i ≤ n to that of proof of Theorem 2.1 and (x1 , . . . , xn ) is not unit. By a similar nargument parts and show that AG(R)[An ] is an one may partition V (AG(R)[An ]) into [n/2] n [n/2] -partite subgraph of AG(R). Moreover, by the proof of Theorem 2.1, it is not hard to see that there are enough colors to color AG(R)[An ]. Claim 4. ω(AG(R)) = χ(AG(R)) = |A0 |. This is easily obtained by using induction on i < j < k < · · · < n in Claim 2. To complete the proof, we show that Λ(AG(R)) = |A0 |. For this, for every 1 ≤ i ≤ n, let Wi = (Ω1 , . . . , Ωn ), where Ωi = Nil(Ri )∗ , Ωj = U (Rj ), for j < i and Ωk = Rk∗ , for i < k. Clearly, A0 = ni=1 Wi and Wi ∩ Wj = ∅, for every 1 ≤ i, j ≤ n and i = j. It is not hard to check that Λi = |Wi |, for every 1 ≤ i ≤ n. Thus n n |A0 | = i=1 |Wi | = i=1 Λi = Λ(AG(R)). The following example shows that hypothesis “|Nil(Ri )∗ | ≥ 2n−1 − 1, for every 1 ≤ i ≤ n” is needed in Theorem 2.5. 1650124-9


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Example 2.1. Let R ∼ = Z8 × Z4 × Z9 . We have A0 = (Nil(Z8 )∗ , 2, Nil(Z9 )∗ ) ∪ (Nil(Z8 )∗ , U (Z4 ), U (Z9 )) ∪ (Nil(Z8 )∗ , U (Z4 ), Nil(Z9 )∗ ) ∪ (U (Z8 ), 2, Nil(Z9 )∗ )


∪ (U (Z8 ), 2, U (Z9 )) ∪ ((U (Z8 ), U (Z4 )), Nil(Z9 )∗ ) ∪ (Nil(Z8 )∗ , 2, U (Z9 )), A1 = (0, 2, Nil(Z9 )∗ ) ∪ (0, 2, U (Z9 )) ∪ (0, U (Z4 ), Nil(Z9 )∗ ) ∪ (Nil(Z8 )∗ , 0, Nil(Z9 )∗ ) ∪ (U (Z8 ), 0, Nil(Z9 )∗ ) ∪ (Nil(Z8 )∗ , 0, U (Z9 )) ∪ (U (Z8 ), Nil(Z4 )∗ , 0) ∪ (Nil(Z8 )∗ , U (Z4 ), 0) ∪ (Nil(Z8 )∗ , Nil(Z4 )∗ , 0), A2 = (0, 0, Nil(Z9 )∗ ) ∪ (0, 2, 0) ∪ (Nil(Z8 )∗ , 0, 0), [(2, 0, 2)]t1 = (Nil(Z8 )∗ , 0, Nil(Z9 )∗ ) ∪ (U (Z8 ), 0, Nil(Z9 )∗ ) ∪ (Nil(Z8 )∗ , 0, U (Z9 )). Now, by part (2) of Lemma 2.2, every element of [(2, 0, 2)]t1 is adjacent to all of the elements of A0 \(U (Z8 ), 2, U (Z9 )), but there is no adjacency between two vertices of [(2, 0, 2)]t1 and (U (Z8 ), 2, U (Z9 )). This implies that [(2, 0, 2)]t1 ∪ A0 \(U (Z8 ), 2, U (Z9 )) is a clique of AG(R). Now, we show that |A0 | < |[(2, 0, 2)]t1 ∪ A0 \(U (Z8 ), 2, U (Z9 ))|. For this, it is enough to show that |(U (Z8 ), 2, U (Z9 ))| < |[(2, 0, 2)]t1 |. To this end, we have |(U (Z8 ), 2, U (Z9 ))| = 4 × 6 = 24, but |[(2, 0, 2)]t1 | = 3 × 2 + 3 × 6 + 4 × 2 = 32 and thus A0 is not a largest clique of AG(R). We close this section with the following result. Theorem 2.6. Let R ∼ = S × F1 × · · · × Fm , where Fi is a field for every 1 ≤ i ≤ m and S ∼ = R1 ×· · ·×Rn , where Ri is an Artinian local ring with |Nil(Ri )∗ | ≥ 2n−1 −1, for every 1 ≤ i ≤ n. Then ω(AG(R)) = χ(AG(R)) = Λ(AG(S))|F1 | · · · |Fm |. Proof. We put: A0 := {(x1 , . . . , xn ) ∈ V (AG(S)) | xi = 0,

for all 1 ≤ i ≤ n}.

By proof of Theorem 2.5, we have A0 is the largest clique of AG(S). Let A0 = {(x1 , . . . , xn , y1 , . . . , ym ) | (x1 , . . . , xn ) ∈ A0 , yi ∈ Fi }. Since for every (x1 , . . . , xn ) ∈ A0 , xi ∈ Nil(Ri )∗ for some 1 ≤ i ≤ n, by part (2) of Lemma 2.2, A0 is the largest clique of AG(R). Also, by a similar argument to that of proof of Theorem 2.5, we have ω(AG(R)) = χ(AG(R)) = Λ(AG(S))|F1 | · · · |Fm |. 3. Annihilator Graphs with Finite Clique Numbers In this section, we study annihilator graphs of rings with finite clique numbers. We begin with recalling the following lemma from [3]. 1650124-10


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Lemma 3.1 ([3, Lemma 3.6]). Let p1 = Ann(x1 ) and p2 = Ann(x2 ) be two distinct elements of Ass(R). Then x1 x2 = 0.


Theorem 3.1. Let R be a non-reduced ring such that ω(AG(R)) < ∞. Then: (1) Max(R) ∩ Ass(R) = ∅. (2) |Ass(R)| ≤ ω(AG(R)). (3) If dim(R) = 0 and 1 < |Ass(R)| < ∞, then AG(R)[Z(R)\Nil(R)] contains a complete |Ass(R)|-partite subgraph. Proof. (1) Since ω(AG(R)) < ∞, |Nil(R)| < ∞. Thus there exists an element a ∈ Nil(R)∗ such that Ra is a minimal ideal of R. Thus annR (a) is maximal and so Max(R) ∩ Ass(R) = ∅. (2) It is clear by Lemma 3.1. (3) Let Ass(R) = {p1 , . . . , pn }, where n ≥ 2 is a positive integer. Since dim × n n (R) = 0, pi j=1,j=i pj , for every 1 ≤ i ≤ n. Let Vi = pi \ j=1,j=i pj . We have only to prove that every vertex of Vi is adjacent to each vertex of Vj , for every 1 ≤ i, j ≤ n and i = j. Let a ∈ Vi , b ∈ Vj and i = j. Thus a ∈ pi \pj and b ∈ pj \pi . Assume that pi = annR (x) and pj = annR (y), for some x, y ∈ Z(R)∗ . Hence R/pi ∼ = Rx and R/pj ∼ = Ry (as R-modules) and so Z(Rx) = pi , Z(Ry) = pj . Therefore, x ∈ annR (a)\annR (b) and y ∈ annR (b)\annR (a) and so by part (1) of Lemma 2.2, a is adjacent to b, as desired. Hence AG(R)[Z(R)\Nil(R)] contains a complete |Ass(R)|-partite subgraph. In the following theorem, it is shown that any Artinian ring with ω(AG(R)) < ∞ is either finite or direct product of finitely many fields. Theorem 3.2. Let R be an Artinian ring. Then ω(AG(R)) < ∞ if and only if one of the following statements holds: (1) R is a reduced ring. (2) R is a finite ring. Proof. One side is clear. To prove the converse, suppose that ω(AG(R)) < ∞ and R is not a reduced ring. We show that R is a finite ring. By [1, Theorem 8.7], R∼ = R1 ×· · ·×Rn , where Ri is an Artinian local ring for every 1 ≤ i ≤ n. Therefore, it suffices to show that |Ri | < ∞, for every 1 ≤ i ≤ n. Since Nil(Ri ) = (0), for some 1 ≤ i ≤ n, without loss of generality, we may assume that Nil(R1 ) = (0). This implies that Z(R1 ) = Nil(R1 ) < ∞ (since Nil(R)∗ is a clique), and so |R1 | < ∞. Also, if |Rj | = ∞, for some 1 < j ≤ n, then by part (2) of Lemma 2.2, the set (Nil(R1 )∗ , 0, . . . , 0, Rj , 0, . . . , 0) is an infinite clique, a contradiction. So we deduce that |Ri | < ∞, for every 1 ≤ i ≤ n. Now, we state the following corollary. 1650124-11


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Corollary 3.1. Let R be a non-reduced infinite ring such that ω(AG(R)) < ∞. Then:


(1) For every a ∈ Nil(R)∗ , |annR (a)| = ∞. (2) For every a ∈ Nil(R)∗ , annR (a) Nil(R). Proof. (1) Let |annR (a)| < ∞, for some a ∈ Nil(R)∗ . Since ω(AG(R)) < ∞, |Nil(R)| < ∞ and so |Ra| < ∞. It follows that Ra and annR (a) are Artinian R-modules. Since Ra ∼ = R/annR (a) (as R-modules), by [11, Proposition 7.17], R is an Artinian ring. Now, by Theorem 3.2, R is a finite ring, a contradiction. (2) It is obvious. Theorem 3.3. Let R be a reduced ring. If ω(AG(R)) < ∞, then |Min(R)| ≤ ω(AG(R)). Proof. If |Min(R)| = 2, then since AG(R) is connected, we have ω(AG(R)) ≥ 2 and thus in this case the proof is complete. Now, let |Min(R)| ≥ 3 and ω(AG(R)) = n, for some positive integer n. Suppose to the contrary, |Min(R)| > ω(AG(R)). n+1 Let p1 , . . . , pn+1 be distinct elements of Min(R). Since pi j=1,j=i pj , for every 1 ≤ i ≤ n, one may take xi ∈ pi \ n+1 j=1,j=i pj . We show that {x1 , . . . , xn+1 } is a n+1 n+1 clique in AG(R). Since xi ∈ pi \ j=1,j=i pj , annR (xi ) ⊆ j=1,j=i pj . So xi xj = 0. Obviously, annR (xi ) ⊆ pj and annR (xj ) ⊆ pi . It follows from [7, Corollary 2.2], / pi , b ∈ / pj . This there exist elements a ∈ annR (xi ) and b ∈ annR (xj ) such that a ∈ implies that annR (xi ) annR (xj ) and annR (xj ) annR (xi ), and so by part (1) of Lemma 2.2, xi − xj is an edge of AG(R) for 1 ≤ i, j ≤ n + 1 and i = j. Thus that {x1 , . . . , xn+1 } is a clique in AG(R). This contradicts ω(AG(R)) = n. We close this paper with the following result. Proposition 3.1. Let R be a non-reduced ring. If AG(R) is a complete graph, then the following statements hold : (1) If Ass(R) = ∅, then |Ass(R)| = 1. (2) If R is a Noetherian ring with depth(R) = 0, then R is an Artinian local ring. Proof. (1) First, we claim that R is an indecomposable ring. If not, since R is non-reduced, one may suppose that R ∼ = R1 × R2 and Nil(R1 ) = (0), where R1 , R2 are two rings. Thus U (R1 ) > 1. If u ∈ U (R1 ), then (1, 0) is not adjacent to (u, 0), a contradiction and so the claim is proved. If p ∈ Ass(R), then it is proved that p = Z(R). Let x ∈ Z(R)\p. Since p = annR (y) for some y ∈ Z(R)∗ , we conclude that R/p ∼ / Z(Ry). Now, by = Ry (as an R-module) and hence Z(Ry) = p. Thus x ∈ Lemma 2.1, x is not adjacent to y, which is impossible, unless x = y. By a similar / p). Hence x = x2 and so R is argument, x2 = y (note that since p is prime, x2 ∈ decomposable, again a contradiction. This implies that Ass(R) = {Z(R)}. (2) It is clear by part (1). 1650124-12


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References [1] M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra (AddisonWesley, 1969). [2] A. Badawi, On the annihilator graph of a commutative ring, Comm. Algebra 42 (2014) 108–121. [3] I. Beck, Coloring of commutative rings, J. Algebra 116 (1988) 208–226. [4] W. Bruns and J. Herzog, Cohen–Macaulay Rings (Cambridge University Press, 1997). [5] N. Ganesan, Properties of rings with a finite number of zero-divisors, Math. Ann. 157 (1964) 215–218. [6] M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness (W. H. Freeman and Company, New York, 1979). [7] J. A. Huckaba, Commutative Rings with Zero-Divisors (Marcel Dekker, New York, 1988). [8] S. Kiani, H. R. Maimani and R. Nikandish, Some results on the domination number of a zero-divisor graph, Canad. Math. Bull. 57(3) (2014) 573–578. [9] H. Matsumura, Commutative Ring Theory (Cambridge University Press, 1986). [10] M. J. Nikmehr and F. Heydari, The M-principal graph of a commutative ring, Period. Math. Hungar. 68 (2014) 185–192. [11] R. Y. Sharp, Steps in Commutative Algebra (Cambridge University Press, 1990). [12] D. B. West, Introduction to Graph Theory, 2nd edn. (Prentice Hall, Upper Saddle River, 2001).

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