The annihilating-submodule graph of modules over commutative rings II

Let M be a module over a commutative ring R. In this paper, we continue our study of annihilating-submodule graph AG(M) which was introduced in (The Zariski topology-graph of modules over commutative rings, Comm. Algebra., 42 (2014), 3283{3296). AG(M) is a (undirected) graph in which a nonzero submodule N of M is a vertex if and only if there exists a nonzero proper submodule K of M such that NK = (0), where NK, the product of N and K, is defined by (N : M)(K : M)M and two distinct vertices N and K are adjacent if and only if NK = (0). We prove that if AG(M) is a tree, then either AG(M) is a star graph or a path of order 4 and in the latter case M\cong F\times ?S, where F is a simple module and S is a module with a unique non-trivial submodule. Moreover, we prove that if M is a cyclic module with at least three minimal prime submodules, then gr(AG(M)) = 3 and for every cyclic module M, cl(AG(M))\geq |Min(M)|.


Introduction
Throughout this paper, R is a commutative ring with a non-zero identity and M is a unital R-module. By N ≤ M (resp., N < M) we mean that N is a submodule (resp., proper submodule) of M.
There are many papers on assigning graphs to rings or modules (see, for example, [4,7,10,11]). The annihilating-ideal graph AG(R) was introduced and studied in [11]. AG(R) is a graph whose vertices are ideals of R with non-zero annihilators and in which two vertices I and J are adjacent if and only if I J = (0). Later, it was modified and further studied by many authors (see [1][2][3]).
In [7,8], we generalized the above idea to submodules of M and defined the (undirected) graph AG(M), called the annihilating-submodule graph, with vertices V ( In this work, we continue our study in [7,8] and we generalize some results related to annihilating-ideal graph obtained in [1][2][3] for annihilating-submodule graph. A prime submodule of M is a submodule P = M, such that whenever re ∈ P for some r ∈ R and e ∈ M, we have r ∈ (P : M) or e ∈ P [14].
The prime radical rad M (N ) or simply rad(N ) is defined to be the intersection of all prime submodules of M containing N , and in case N is not contained in any prime submodule, rad M (N ) is defined to be M [14].
The notations Z (R), Nil(R), and Min(M) will denote the set of all zero-divisors, the set of all nilpotent elements of R, and the set of all minimal prime submodules of M, respectively. In addition, Z R (M) or simply Z (M), the set of zero divisors on M, is the set {r ∈ R| rm = 0 for some 0 = m ∈ M}.
A clique of a graph is a complete subgraph and the supremum of the sizes of cliques in G, denoted by cl(G), is called the clique number of G. Let χ(G) denote the chromatic number of the graph G, that is, the minimal number of colors needed to color the vertices of G, so that no two adjacent vertices have the same color. Obviously χ(G) ≥ cl(G).
In Sect. 2, we prove that if AG(M) is a tree, then either AG(M) is a star graph or is the path P 4 and in this case, M ∼ = F × S, where F is a simple module and S is a module with a unique non-trivial submodule (see Theorem 2.7). Next, we study the bipartite annihilating-submodule graphs of modules over Artinian rings (see Theorem 2.8). In Sect. 3, we study coloring of the annihilating-submodule graph and investigate the interplay between χ(AG(M)), cl(AG(M)), and Min(M) (see Theorems 3.5 and 3.8). In Corollary 3.7, we prove that if M is a cyclic module with at least three minimal prime submodules, then gr(AG(M)) = 3 and for every cyclic module M, cl(AG(M)) ≥ |Min(M)|.
Let us introduce some graphical notions and denotations that are used in what follows: a graph G is an ordered triple (V (G), E(G), ψ G ) consisting of a non-empty set of vertices, V (G), a set E(G) of edges, and an incident function ψ G that associates an unordered pair of distinct vertices with each edge. The edge e joins x and y if ψ G (e) = {x, y}, and we say x and y are adjacent. A path in graph G is a finite sequence of vertices {x 0 , x 1 , . . . , x n }, where x i−1 and x i are adjacent for each 1 ≤ i ≤ n and we denote x i−1 − x i for existing an edge between x i−1 and , and ψ H is the restriction of ψ G to E(H ). A bipartite graph is a graph whose vertices can be divided into two disjoint sets U and V , such that every edge connects a vertex in U to one in V ; that is, U and V are each independent sets and complete bipartite graph on n and m vertices, denoted by K n,m , where V and U are of size n and m, respectively, and E(G) connects every vertex in V with all vertices in U . Note that a graph K 1,m is called a star graph and the vertex in the singleton partition is called the center of the graph. For some U ⊆ V (G), we denote by V (U ), the set of all vertices of G\U adjacent to at least one vertex of U .
If all the vertices of G have the same degree k, then G is called k-regular, or simply regular. An independent set is a subset of the vertices of a graph, such that no vertices are adjacent. We denote by P n and C n , a path and a cycle of order n, respectively. Let G and G be two graphs. A graph homomorphism from G to G is a mapping φ : The homomorphism φ is called the retract (graph) homomorphism (see [12]).

Cycles in the annihilating-submodule graphs
An ideal I ≤ R is said to be nil if I consist of nilpotent elements.

Proposition 2.5 Let M be a finitely generated R-module such that R/Ann(M) is Artinian. Then, every non-zero proper submodule N of M is a vertex in AG(M).
Proof Let N be a non-zero submodule of M. Therefore, there exists a maximal submodule K of M, such that N ⊆ K . Hence, we have Proof If none of M 1 and M 2 is a prime module, then there exist r ∈ R i (R 1 = Re and and R 1 m 1 × R 2 m 2 form a triangle in AG(M), a contradiction. Thus, without loss of generality, one can assume that M 1 is a prime module. We prove that AG(M 2 ) has at most one vertex. On the contrary suppose that {N , K } is an edge of AG(M 2 ). Therefore, M 1 × (0), (0) × N , and (0) × K form a triangle, a contradiction. If AG(M 2 ) has no vertex, then M 2 is a prime module and so part (a) occurs. If AG(M 2 ) has exactly one vertex, then by [7, Theorem 3.6] and Proposition 2.5, we obtain part (b). Now, suppose that AG(M) has no cycle. If none of M 1 and M 2 is a simple module, then choose non-trivial submodules N i in M i for some i = 1, 2. Therefore, for some i ∈ and j ∈ . Now, consider the following claims: Claim 2 Our claim is to show that N is a minimal submodule of M and K 2 = (0). To see that, first, we show that for every 0 = m ∈ N , Rm = N . Assume that 0 = m ∈ N and Rm = N . If Rm = K , then K ⊆ N , a contradiction. Thus Rm = K , and the induced subgraph of AG(M) on N , K , and Rm is K 3 , a contradiction. Therefore, Rm = N . This implies that N is a minimal submodule of M. Now, if K 2 = (0), then we obtain the induced subgraph on N , K , and (N : Let S be a multiplicatively closed subset of R. A non-empty subset S * of M is said to be S-closed if se ∈ S * for every s ∈ S and e ∈ S * . An S-closed subset S * is said to be saturated if the following condition is satisfied: whenever ae ∈ S * for a ∈ R and e ∈ M, then a ∈ S and e ∈ S * .
We need the following result due to Chin-Pi Lu.  [18,Theorem 4.4]). In addition, we know that if M is a finitely generated module, then for every prime ideal p of R with p ⊇ Ann(M), there exists a prime submodule P of M, such that (P : M) = p (see [15,Theorem 2]). We prove that AG(M) is a bipartite graph with parts V 1 and V 2 . We may assume that V 1 is an independent set because AG(M) is triangle-free. We claim that one end of every edge of AG(M) is adjacent to Rm and another end contains Rm. To prove this, suppose that {N , K } is an edge of AG(M) and Rm = N , Rm = K . Since N (Rm) ⊆ Rm, by the minimality of Rm, either N (Rm) = (0) or Rm ⊆ N . The latter case follows that K (Rm) = (0). If N (Rm) = (0), then K (Rm) = (0) and hence Rm ⊆ K . So, our plain is proved. This gives that V 2 is an independent set and V (C) ⊆ V 1 . Since every vertex of A contains Rm and AG(M) is triangle-free, all vertices in A are just adjacent to Rm and so by [7,Theorem 3.4], V (C) ⊆ B. Since one end of every edge is adjacent to Rm and another end contains Rm, we also deduce that every vertex of C contains Rm and so every vertex of A ∪ V 2 contains Rm. Note that if Rm = P, then one end of each edge of AG(M) is contained in Rm, and since Rm is a minimal submodule of M, AG(M) is a star graph with center Rm = P. Now, suppose that P = Rm. We claim that P ∈ A. Since Rm ⊆ P, it suffices to show that (Rm)P = (0). To see this, let r ∈ (P : M). We prove that rm = 0. Clearly, (

On the coloring of the annihilating-submodule graphs
We recall that N < M is said to be a semiprime submodule of M if for every ideal I of R and every submodule K of M, I 2 K ⊆ N implies that I K ⊆ N . Furthermore, M is called a semiprime module if (0) ⊆ M is a semiprime submodule. Every intersection of prime submodules is a semiprime submodule (see [20]).  Since the chromatic number χ(G) of a graph G is the least positive integer r , such that there exists a retract homomorphism ψ : G −→ K r , the following corollaries follow directly from the proof of Theorem 3.1.   { p 1 , . . . , p n } is a finite set of distinct minimal prime ideals of R and S = R\ ∪ n i=1 p i , then R p 1 × · · · × R p n ∼ = R S . In [19], this result was generalized to finitely generated multiplication modules. In Theorem 3.6, we use this generalization for a cyclic module.  = (0, . . . , 0, m/1, . . . , 0, . . . , 0) and φ(e i ) = n i /t i , where m ∈ M, 1 ≤ i ≤ n, and m/1 is in the ith position of e i . Consider the principal submodules N i = (n i /t i ) = (n i /1) in the module M S . By Lemma 2.2 and Proposition 2.1, the product of submodules (0) × · · · × (0) × (m/1)R p i × (0) × · · · × (0) and (0) × · · · × (0) × (m/1)R p j × (0) × · · · × (0) are zero, i = j. Since φ is an isomorphism, there exists t i j ∈ S, such that t i j r i n j = 0, for every i, j, 1 ≤ i < j ≤ n, where n i = r i m for some r i ∈ R. Let t = 1≤i< j≤n t i j . We show that {(tn 1 ), . . . , (tn n )} is a clique of size n in AG(M). For every i, j, 1 ≤ i < j ≤ n,   Proof For the first assertion, we use the same technique in [3,Theorem 13]. Let cl(AG(M)) = 2. On the contrary assume that AG(M) is not bipartite. Therefore, AG(M) contains an odd cycle. Suppose that C := N 1 − N 2 −· · ·− N 2k+1 − N 1 be a shortest odd cycle in AG(M) for some natural number k. Clearly, k ≥ 2. Since C is a shortest odd cycle in AG(M), N 3 N 2k+1 is a vertex. Now, consider the vertices N 1 , N 2 , and N 3 N 2k+1 . If N 1 = N 3 N 2k+1 , then N 4 N 1 = (0). This implies that N 1 − N 4 − · · · − N 2k+1 − N 1 is an odd cycle, a contradiction. Thus, N 1 = N 3 N 2k+1 . If N 2 = N 3 N 2k+1 , then we have C 3 = N 2 − N 3 − N 4 − N 2 , again a contradiction. Hence, N 2 = N 3 N 2k+1 . It is easy to check N 1 , N 2 , and N 3 N 2k+1 form a triangle in AG(M), a contradiction. The converse is clear. In particular, we note that empty graphs and the isolated vertex graphs are bipartite graphs.