Abstract
Let M be a module over a commutative ring R. In this paper, we continue our study about the Zariski topologygraph \(G(\tau _T)\) which was introduced in AnsariToroghy et al. (Commun Algebra 42:3283–3296, 2014). For a nonempty subset T of \(\mathrm{Spec}(M)\), we obtain useful characterizations for those modules M for which \(G(\tau _T)\) is a bipartite graph. Also, we prove that if \(G(\tau _T)\) is a tree, then \(G(\tau _T)\) is a star graph. Moreover, we study coloring of Zariski topologygraphs and investigate the interplay between \(\chi (G(\tau _T))\) and \(\omega (G(\tau _T))\).
Similar content being viewed by others
Avoid common mistakes on your manuscript.
1 Introduction
Throughout this paper, R is a commutative ring with a nonzero identity and M is a unital Rmodule. By \(N\le M\) (resp. \(N< M\)) we mean that N is a submodule (resp. proper submodule) of M.
Define \((N{:}_{R}M)\) or simply \((N:M)=\{r\in R\) \(rM\subseteq N\}\) for any \(N\le M\). We denote ((0) : M) by \(\mathrm{Ann}_{R}(M)\) or simply \(\mathrm{Ann}(M)\). M is said to be faithful if \(\mathrm{Ann}(M)=(0)\).
Let \(N, K\le M\). Then the product of N and K, denoted by NK, is defined by (N : M)(K : M)M (see [3]).
A prime submodule of M is a submodule \(P\ne M\) such that whenever \(re\in P\) for some \(r\in R\) and \(e \in M\), we have \(r\in (P:M)\) or \(e\in P\) [10].
The prime spectrum of M is the set of all prime submodules of M and denoted by \(\mathrm{Spec}(M)\).
If N is a submodule of M, then \(V(N)=\{P \in \mathrm{Spec}(M)\) \((P:M)\supseteq (N:M) \}\) [11].
The Zariski topology on \(X=\mathrm{Spec}(M)\) is the topology \(\tau _M \) described by taking the set \(\ Z(M)= \{ V(N)\) N is a submodule of \( M \}\) as the set of closed sets of \(\mathrm{Spec}(M)\) [11].
A topological space X is irreducible if for any decomposition \(X= X_1 \cup X_2\) with closed subsets \(X_i\) of X with \(i= 1, 2\), we have \(X= X_1\) or \(X= X_2\).
There are many papers on assigning graphs to rings or modules (see, for example, [1, 5, 6, 9]). In [4], the present authors introduced and studied the graph \(G(\tau _T)\) and AG(M), called the Zariski topologygraph and the annihilatingsubmodule graph, respectively.
Let T be a nonempty subset of \(\mathrm{Spec}(M)\). The Zariski topologygraph \(G(\tau _T)\) is an undirected graph with vertices \(V(G(\tau _T))\)= \(\{N < M\) there exists \(K < M\) such that \(V(N)\cup V(K)=T\) and \(V(N),V(K)\ne T\}\) and distinct vertices N and L are adjacent if and only if \(V(N)\cup V(L)=T\) (see [4, Definition 2.3]).
AG(M) is an undirected graph with vertices V(AG(M))= \(\{N \le M \) there exists \((0)\ne K<M\) with \(NK=(0)\)}. In this graph, distinct vertices \(N,L \in V(AG(M))\) are adjacent if and only if \(NL=(0)\). Let \(AG(M)^{*}\) be the subgraph of AG(M) with vertices \(V(AG(M)^{*})=\{ N<M\) with \((N:M)\ne \mathrm{Ann}(M)\) there exists a submodule \(K<M\) with \((K:M)\ne \mathrm{Ann}(M)\) and \(NK=(0)\}\). By [4, Theorem 3.4], one conclude that \(AG(M)^{*}\) is a connected subgraph.
If \(\mathrm{Spec}(M)\ne \emptyset \), the mapping \(\psi :\mathrm{Spec}(M)\rightarrow \mathrm{Spec}(R/\mathrm{Ann}(M))\) such that \(\psi (P)=(P:M)/\mathrm{Ann}(M)\) for every \(P \in \mathrm{Spec}(M)\), is called the natural map of \(\mathrm{Spec}(M)\) [11].
The prime radical \(\sqrt{N}\) 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, \(\sqrt{N}\) is defined to be M [10].
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 with \(I^{2}K\subseteq N\) implies that \(IK\subseteq N\). Further M is called a semiprime module if \((0)\subseteq M\) is a semiprime submodule. Every intersection of prime submodules is a semiprime submodule (see [17]).
The notations Nil(R), Min(M), and Min(T) will denote the set of all nilpotent elements of R and the set of all minimal prime submodules of M, and the set of minimal members of T, respectively.
A clique of a graph is a complete subgraph and the supremum of the sizes of cliques in G, denoted by \(\omega (G)\), is called the clique number of G. Let \(\chi (G)\) 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 \(\chi (G)\ge \omega (G)\).
In this article, we continue our studying about \(G(\tau _T)\) and AG(M) and we try to relate the combinatorial properties of the above mentioned graphs to the algebraic properties of M.
In Sect. 2 of this paper, we state some properties related to the Zariski topologygraph that are basic or needed in the later sections. In Sect. 3, we study the bipartite Zariski topologygraphs of modules over commutative rings (see Proposition 3.1). Also, we prove that if \(G(\tau _T)\) is a tree, then \(G(\tau _T)\) is a star graph (see Theorem 3.5). In Sect. 4, we study coloring of the Zariski topologygraph of modules and investigate the interplay between \(\chi (G(\tau _T))\) and \(\omega (G(\tau _T))\). We show that under condition over minimal submodules of \(M/(\cap _{P\in T}P:M)M \), we have \(\omega (G(\tau _T))= \chi (G(\tau _T))\) (see Theorem 4.1). Moreover, we investigate some relations between the existence of cycles in the Zariski topologygraph of a cyclic module and the number of its minimal members of T (see Proposition 4.10).
Let us introduce some graphical notions and denotations that are used in what follows: A graph G is an ordered triple \((V(G), E(G), \psi _G )\) consisting of a nonempty set of vertices, V(G), a set E(G) of edges, and an incident function \(\psi _G\) that associates an unordered pair of distinct vertices with each edge. The edge e joins x and y if \(\psi _G(e)=\{x, y\}\), and we say x and y are adjacent. A path in a graph G is a finite sequence of vertices \(\{x_0, x_1,\ldots ,x_n\}\), where \(x_{i1}\) and \(x_i\) are adjacent for each \(1\le i\le n\) and we denote \(x_{i1}  x_i\) for existing an edge between \(x_{i1}\) and \(x_i\).
A graph H is a subgraph of G, if \(V(H)\subseteq V(G)\), \(E(H)\subseteq E(G)\), and \(\psi _H\) is the restriction of \(\psi _G\) 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 \(U\subseteq V (G)\), we denote by N(U), the set of all vertices of \(G{\setminus } U\) adjacent to at least one vertex of U. For every vertex \(v\in V(G)\), the size of N(v) is denoted by deg(v). If all the vertices of G have the same degree k, then G is called kregular, or simply regular. We denote by \(C_{n}\) a cycle of order n. Let G and \(G'\) be two graphs. A graph homomorphism from G to \(G'\) is a mapping \(\phi : V(G) \longrightarrow V(G')\) such that for every edge \(\{u, v\}\) of G, \(\{\phi (u), \phi (v)\}\) is an edge of \(G'\). A retract of G is a subgraph H of G such that there exists a homomorphism \(\phi : G \longrightarrow H\) such that \(\phi (x)= x\), for every vertex x of H. The homomorphism \(\phi \) is called the retract (graph) homomorphism (see [13]).
Throughout the rest of this paper, we denote by T a nonempty subset of \(\mathrm{Spec}(M)\), \(F:=\cap _{P\in T}P\), \(Q:= (F:M)M\), \(\bar{M}:=M/ Q\), \(\bar{N}:=N/Q\), \(\bar{m}:=m+Q\), and \(\bar{I}:=I/(Q:M)\), where N is a submodule of M containing Q, \(m\in M\), and I is an ideal of R containing (Q : M).
2 Auxiliary results
In this section, we provide some properties related to the Zariski topologygraph that are basic or needed in the sequel.
Remark 2.1
Let N be a submodule of M. Set \(V^*(N):= \{P \in \mathrm{Spec}(M) \) \(P\supseteq N \}\). By [4, Remark 2.2], for submodules N and K of M, we have
By [4, Remark 2.5], we have T is a closed subset of \(\mathrm{Spec}(M)\) if and only if \(T=V(F)\) and \(G(\tau _T)\ne \emptyset \) if and only if \(T=V(F)\) and T is not irreducible. So if N and K are adjacent in \(G(\tau _T)\), then \(V^*(NK)=V^*(Q)\) and hence \(\sqrt{NK}=F\). Therefore, \(F \subseteq \sqrt{(N:M)M}\) and \(F \subseteq \sqrt{(K:M)M}\).
Lemma 2.2
(See [2, Proposition 7.6]) Let \(R_{1}, R_{2}, \ldots , R_{n}\) be nonzero ideals of R. Then the following statements are equivalent:

(a)
\(R= R_{1} \oplus \ldots \oplus R_{n}\);

(b)
As an abelian group R is the direct sum of \( R_{1}, \ldots , R_{n}\);

(c)
There exist pairwise orthogonal idempotents \(e_{1},\ldots , e_{n}\) with \(1=e_{1}+ \ldots +e_{n}\), and \(R_{i}=Re_{i}\), \(i=1, \ldots ,n\).
Proposition 2.3
Suppose that e is an idempotent element of R. We have the following statements.

(a)
\(R=R_{1}\oplus R_{2}\), where \(R_{1}=eR\) and \(R_{2}=(1e)R\).

(b)
\(M=M_{1}\oplus M_{2}\), where \(M_{1}=eM\) and \(M_{2}=(1e)M\).

(c)
For every submodule N of M, \(N=N_{1}\oplus N_{2}\) such that \(N_{1}\) is an \(R_{1}\)submodule \(M_{1}\), \(N_{2}\) is an \(R_{2}\)submodule \(M_{2}\), and \((N:_{R}M)=(N_{1}:_{R_{1}}M_{1})\oplus (N_{2}:_{R_{2}}M_{2})\).

(d)
For submodules N and K of M, \(NK=N_{1}K_{1} \oplus N_{2}K_{2}\), \(N\cap K=N_{1}\cap K_{1} \oplus N_{2}\cap K_{2}\) such that \(N=N_{1}\oplus N_{2}\) and \(K=K_{1}\oplus K_{2}\).

(e)
Prime submodules of M are \(P\oplus M_{2}\) and \(M_{1}\oplus Q\), where P and Q are prime submodules of \(M_{1}\) and \(M_{2}\), respectively.

(f)
For submodule N of M, we have \(\sqrt{N}= \sqrt{N_{1}\oplus N_{2}}= \sqrt{N_{1}}\oplus \sqrt{N_{2}}\), where \(N=N_{1}\oplus N_{2}\).
Proof
This is clear. \(\square \)
An ideal \(I< R\) is said to be nil if I consist of nilpotent elements.
Lemma 2.4
(See [15, Theorem 21.28]) Let I be a nil ideal in R and \(u\in R\) be such that \(u+I\) is an idempotent in R/I. Then there exists an idempotent e in uR such that \(eu\in I\).
Lemma 2.5
(See [5, Lemma 2.4]) Let N be a minimal submodule of M and let \(\mathrm{Ann}(M)\) be a nil ideal. Then we have \(N^{2}=(0)\) or \(N=eM\) for some idempotent \(e\in R\).
We note that M is said to be primeful if either \(M=(0)\) or \(M\ne (0)\) and the natural map of \(\mathrm{Spec}(M)\) is surjective (see [12]).
Proposition 2.6
We have the following statements.

(a)
If N, L are adjacent in \(G(\tau _T)\), then \(\sqrt{(N:M)M}/ F\) and
\(\sqrt{(L:M)M}/ F\) are adjacent in AG(M/F).

(b)
If M is a primeful module and N, L are adjacent in \(G(\tau _T)\), then \(\sqrt{N}/ F\) and \(\sqrt{L}/ F\) are adjacent in AG(M/F).
Proof

(a)
First, we see easily that for any submodule N of M, \(V(N)=V(\sqrt{(N:M)M})\). Suppose that N and L are adjacent in \(G(\tau _T)\) so that \(V(N) \cup V(L)= T\). Then we have \(V^{*}(\sqrt{(N:M)M} \sqrt{(L:M)M})= T\). It follows that \(\sqrt{(N:M)M} \sqrt{(L:M)M}\subseteq F\) (see Remark 2.1). Also by Remark 2.1, \(F \subseteq \sqrt{(N:M)M}\) and \(F \subseteq \sqrt{(L:M)M}\). Therefore, \(\sqrt{(N:M)M}/ F\) and \(\sqrt{(L:M)M}/ F\) are adjacent in AG(M/F).

(b)
This is clear by [4, Corollary 4.5]. \(\square \)
Remark 2.7
The Proposition 2.6(a) extends [4, Theorem 4.4].
Lemma 2.8
Assume that T is a closed subset of \(\mathrm{Spec}(M)\). Then \(AG(\bar{M})^{*}\) is isomorphic with a subgraph of \(G(\tau _T)\). In particular, \(AG(M/ F)^{*}\) is isomorphic with an induced subgraph of \(G(\tau _T)\).
Proof
Let \(\bar{N} \in V(AG(\bar{M})^{*})\). Then there exists a nonzero submodule \(\bar{K}\) of \(\bar{M}\) such that it is adjacent to \(\bar{N}\) (if \(N=K\), then \((N:M)=(Q:M)\), a contradiction). So we have \(NK\subseteq Q\). Hence \(V(NK)\supseteq T\). Since \(Q\subseteq N\) and \(Q\subseteq K\), then \(V(N)\subseteq T\) and \(V(K)\subseteq T\). Therefore \(V(NK)=T\) (if \(V(N)=T\), then \((N:M)=(Q :M)\), a contradiction). Hence N is a vertex in \(G(\tau _T)\) which is adjacent to K. To see the last assertion, let N/F and K/F be two vertices of \(AG(M/ F)^{*}\). If N and K are adjacent in \(G(\tau _T)\), then by Proposition 2.6, \(\sqrt{(N:M)M}/ F\) and \(\sqrt{(K:M)M}/ F\) are adjacent in AG(M/F). So
Since
we have N/F and K/F are adjacent in \(AG(M/ F)^{*}\), as desired. \(\square \)
Lemma 2.9
If \(\bar{M}\) is a faithful module and T is a closed subset of \(\mathrm{Spec}(M)\), then \(G(\tau _{\mathrm{Spec}(M)})\) and \(AG(M)^{*}\) are the same.
Proof
\(\bar{M}\) is a faithful module and T is a closed subset of \(\mathrm{Spec}(M)\) so that \(T=\mathrm{Spec}(M)\). If \(G(\tau _{\mathrm{Spec}(M)})\ne \emptyset \), then there exist nontrivial submodules N and K of M which are adjacent in \(G(\tau _{\mathrm{Spec}(M)})\). Hence \(V(NK)=\mathrm{Spec}(M)\) which implies that \(NK=(0)\) so that \(AG(M)^{*}\ne \emptyset \). By Lemma 2.8, \(AG(M)^{*}\) is isomorphic with a subgraph of \(G(\tau _{\mathrm{Spec}(M)})\). One can see that the vertex map \(\phi : V(G(\tau _{\mathrm{Spec}(M)})) \longrightarrow V(AG(M)^{*})\), defined by \(N\longrightarrow N\) is an isomorphism. \(\square \)
Recall that \(\Delta (G(\tau _T))\) is the maximum degree of \(G(\tau _T)\) and the length of an Rmodule M, is denoted by \(l_{R}(M)\).
Lemma 2.10
Let every nontrivial submodule of M be a vertex in \(G(\tau _T)\). If \(\Delta (G(\tau _T))< \infty \), then \(l_{R}(M)\le \Delta (G(\tau _T))+1\). Also, every nontrivial submodule of M has finitely many submodules.
Proof
First, we show that the descending chain of nontrivial submodules \(K_{1}\supsetneq K_{2}\supsetneq K_{3}\supsetneq \ldots \) terminates. Since \(G(\tau _T)\) is connected, there exists a submodule N such that \(V(N)\cup V(K_{1})=T\). Hence for each i, \(i\ge 1\), \(V(N)\cup V(K_{i})=T\) and so deg\((N)=\infty \), a contradiction. Next, let \(N_{1}\subsetneq N_{2}\subsetneq N_{3}\subsetneq \ldots \) be an ascending chain of nontrivial submodules of M. Since \(G(\tau _T)\) is connected, there exists a submodule K such that \(V(K)\cup V(N_{\Delta +1})=T\), where \(\Delta = \Delta (G(\tau _T))\). Hence \(V(K)\cup V(N_{i})=T\) for each \(1\le i\le \Delta +1\). Thus deg\((K)\ge \Delta +1\), a contradiction. It follows that \(l_{R}(M)\le \Delta +1\). For the proof of the last assertion, let N be a nontrivial submodule of M. Since \(G(\tau _T)\) is connected, there exists a submodule K such that \(V(N)\cup V(K)=T\). Hence for every submodule \(N'\) of N, \(V(N')\cup V(K)=T\). As \(\Delta < \infty \), the number of submodules of N should be finite. \(\square \)
Theorem 2.11
Suppose that \(\bar{M}\) is a multiplication module and \(G(\tau _T)\ne \emptyset \). If \(G(\tau _T)\) has acc (resp. dcc) on vertices, then \(\bar{M}\) is a Noetherian (resp. an Artinian) module.
Proof
Suppose that \(G(\tau _T)\) has acc (resp. dcc) on vertices. By [4, Remark 2.6], F is not a prime submodule of M and hence there exist \(r \in R\) and \(m\in M\) such that \(rm\in F\) but \(m\notin F\) and \(r \notin (F:M)\). Now \(\overline{rM}\cong \bar{M}/(\bar{0}:_{\bar{M}}r)\). Further, \(\overline{rM}\) and \((\bar{0}:_{\bar{M}}r)\) are vertices in \(AG(\bar{M})=AG(\bar{M})^{*}\) (\(\bar{M}\) is a multiplication module) because \((\bar{0}:_{\bar{M}}r)(\overline{rM})=((\bar{0}:_{\bar{M}}r):\bar{M})(\overline{rM}:\bar{M})\bar{M}\subseteq \overline{rM}((\bar{0}:_{\bar{M}}r):\bar{M})\subseteq r(\bar{0}:_{\bar{M}}r)=\bar{0}\). Then by Lemma 2.8, \(\{N\) \(\bar{N}\le \bar{M}, \bar{N}\subseteq \overline{rM}\}\cup \) \(\{N\) \(\bar{N}\le \bar{M}, \bar{N}\subseteq (\bar{0}:_{\bar{M}}r)\} \subseteq V(G(\tau _T))\). It follows that the Rmodules \(\overline{rM}\) and \((\bar{0}:_{\bar{M}}r)\) have acc (resp. dcc) on submodules. Since \(\overline{rM}\cong \bar{M}/(\bar{0}:_{\bar{M}}r)\), \(\bar{M}\) has acc on submodules and the proof is completed. \(\square \)
3 Zariski topologygraph of modules
First, in this section we give the more notation to be used throughout the remainder of this article. Suppose that e \(( e\ne 0, 1)\) is an idempotent element of R. Let \(M_{1}:=eM, M_{2}:=(1e)M, T_{1}:=\{P_{1}\in \mathrm{Spec}(M_{1}) P_{1}\oplus M_{2}\in T\}\), \(T_{2}:=\{P_{2}\in \mathrm{Spec}(M_{2}) M_{1}\oplus P_{2}\in T\}\), \(F_{1}:=\cap _{P_{1}\in T_{1}}P_{1}\), \(F_{2}:=\cap _{P_{2}\in T_{2}}P_{2}\), \(Q_{1}:=(F_{1}:M_{1})M_{1}, Q_{2}:=(F_{2}:M_{2})M_{2}\), \(\bar{M_{1}}= \overline{eM}=eM/Q_{1}\), and \(\bar{M_{2}}= \overline{(e1)M}=(e1)M/Q_{2}\). Consequently, we have, \(Q=Q_{1}\oplus Q_{2}\), where \(Q= (\cap _{P\in T}P:M)M\) and \(\bar{M}\cong \bar{M_{1}}\oplus \bar{M_{2}}\)
We recall that a submodule N of M is a prime Rmodule if and only if it is a prime \(R/\mathrm{Ann}(M)\)module (see [16, Result 1.2]).
Proposition 3.1
Suppose that \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\) and \(\mathrm{Ann}(\bar{M})\) is a nil ideal. Then the following statements hold.

(a)
If there exists a vertex of \(G(\tau _T)\) which is adjacent to every other vertex, then \(\bar{M_{1}}\) is a simple module and \(\bar{M_{2}}\) is a prime module for some idempotent element \(e\in R\).

(b)
If \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime modules for some idempotent element \(e\in R\), then \(G(\tau _T)\) is a complete bipartite graph.
Proof

(a) Suppose that N is adjacent to every other vertex of \(G(\tau _T)\). Since \(V(N)=V((N:M)M)\), we have \(N=(N:M)M\) and hence \(V(N)=V^{*}(N)\). Thus, \(N=\sqrt{N}\) because \(V(N)=V(\sqrt{N})\). We claim that \(\bar{N}\) is a minimal submodule of \(\bar{M}\). Let \(Q \subsetneq K \subsetneq N\). If \(V(K)\ne T\), then K is adjacent to N and hence \(V(K)=T\), a contradiction. So \(\bar{N}\) is a minimal submodule of \(\bar{M}\). We have \((\bar{N})^{2}\ne (0)\) because \(V(N)\ne T\). Then Lemma 2.5, implies that \(\bar{M}\cong \overline{eM}\oplus \overline{(e1)M}\) for some idempotent element e of R. Without loss of generality we may assume that \(M_{1}\oplus Q_{2}\) is adjacent to every other vertex. Since \(V(F_{1}\oplus Q_{2})=V(Q_{1}\oplus F_{2})=T\), the assumption of theorem implies that \(F=Q\). We claim that \(\bar{M_{1}}\) is a simple module and \(\bar{M_{2}}\) is a prime module. Let \(Q_{1}\subsetneq K < M_{1}\). We have \(V(K\oplus Q_{2})\ne T\) because \(Q_{1}\oplus Q_{2}\subsetneq K\oplus Q_{2}\). Since \(V(K\oplus Q_{2})\cup V(Q_{1}\oplus M_{2})=T\), we have \(K\oplus Q_{2}\) is a vertex and hence is adjacent to \(M_{1}\oplus Q_{2}\). Therefore \(V(K\oplus Q_{2})\cup V(M_{1}\oplus Q_{2})=V(K\oplus Q_{2})=T\), a contradiction. It implies that \(\bar{M_{1}}\) is a simple module. Now, we show that \(\bar{M_{2}}\) is a prime module. It is enough to show that it is a prime \(R/(Q_{2}:M_{2})\)module. Otherwise, \(\bar{I}\bar{K}= (\bar{0})\), where \((Q_{2}:M_{2})\subsetneq I< R\) and \(Q_{2}\subsetneq K< M_2\). It follows that \(V(M_{1}\oplus K) \cup V(Q_{1}\oplus IM_{2})=V(Q_{1}\oplus K(IM_{2}))=T\) because \(K(IM_{2})\subseteq IK\subseteq Q_{2}\) (note that \((Q_{2}:M_{2})\subseteq (K:M_2)\) and \((Q_{2}:M_{2})\subseteq I\)). Therefore, \(M_{1}\oplus K\) is a vertex and hence is adjacent to \(M_{1}\oplus Q_{2}\). So \(V(M_{1}\oplus K)\cup V(M_{1}\oplus Q_{2})=T=V(M_{1}\oplus Q_{2})\), a contradiction (note that \(M_{1}\oplus K\) is properly containing \(Q_{1}\oplus Q_{2}).\)

(b) Assume that \(N_{1}\oplus N_{2}\) is adjacent to \(K_{1}\oplus K_{2}\). One can see that \(\sqrt{N_{1}K_{1}}\oplus \sqrt{N_{2}K_{2}}=\sqrt{Q_{1}}\oplus \sqrt{Q_{2}}\). It implies that \(\overline{(\sqrt{(K_{1}:M_{1})M_{1}}:M_{1})}\) \( \overline{\sqrt{( N_{1}:M_{1})M_{1}}}=(\bar{0})\) and \(\overline{(\sqrt{(K_{2}:M_{2})M_{2}}:M_{2})}\) \( \overline{\sqrt{( N_{2}:M_{2})M_{2}}}=(\bar{0})\). Since \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime modules, \((\sqrt{(K_{1}:M_{1})M_{1}}:M_{1})=(Q_{1}:M_{1})\) or \(\sqrt{(N_{1}:M_{1})M_{1}}= Q_{1}\) and \((\sqrt{(K_{2}:M_{2})M_{2}}:M_{2})=(Q_{2}:M_{2})\) or \(\sqrt{(N_{2}:M_{2})M_{2}}= Q_{2}\). Therefore \(G(\tau _T)\) is a complete bipartite graph with two parts U and V such that \(N\in U\) if and only if \(V(N)=V(M_{1}\oplus Q_{2})\) and \(K\in V\) if and only if \(V(K)=V(Q_{1}\oplus M_{2})\).
\(\square \)
Corollary 3.2
Let \(\bar{M}\) be a faithful module which does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\). Then the following statements are equivalent.

(a)
There is a vertex of \(G(\tau _{\mathrm{Spec}(M)})\) which is adjacent to every other vertex of \(G(\tau _{\mathrm{Spec}(M)})\).

(b)
\(G(\tau _{\mathrm{Spec}(M)})\) is a star graph.

(c)
\(M=F\oplus D\), where F is a simple module and D is a prime module.
Proof
\(\mathrm{(a)} \Rightarrow \mathrm{(b)}\) Let \(\bar{M}\) be a faithful module. Then \(Q=(0)\) and we have \(T=\mathrm{Spec}(M)\). By Proposition 3.1, \(M=M_{1}\oplus M_{2}\), where \(M_{1}\) is a simple module and \(M_{2}\) is a prime module. Then every nonzero submodule of M is of the form \(M_{1}\oplus N_{2}\) and \((0)\oplus N_{2}\), where \(N_{2}\) is a nonzero submodule of \(M_{2}\). We show that non of the submodules of the form \((0)\oplus N_{2}\) can be adjacent to each other. Assume that \((0)\oplus N_{2}\) and \((0)\oplus K_{2}\) are adjacent in \(G(\tau _{\mathrm{Spec}(M)})\), where \((0)\ne N_2\le M_2\) and \((0)\ne K_2\le M_2\). Since (0) is a prime submodule of \(M_2\), by Remark 2.1, we have \(N_2K_2=(0)\). Hence \(V((0)\oplus N_{2})=\mathrm{Spec}(M)\) or \(V((0)\oplus K_{2})=\mathrm{Spec}(M)\), a contradiction. Similarly, we can not have any vertex of the form \(M_{1}\oplus N_{2}\), where \(N_{2}\) is a nonzero proper submodule of \(M_{2}\). Now it is easy to see that \(M_{1}\oplus (0)\) is adjacent to every other vertex and so \(G(\tau _{\mathrm{Spec}(M)})\) is a star graph.
\(\mathrm{(b)} \Rightarrow \mathrm{(c)}\) This follows by Proposition 3.1(a).
\(\mathrm{(c)} \Rightarrow \mathrm{(a)}\) Assume that \(M=F\oplus D\), where F is a simple module and D is a prime module. Using the Proposition 3.1 (b), \(G(\tau _{\mathrm{Spec}(M)})\) is a complete bipartite graph with two parts U and V such that \(N\in U\) if and only if \(V(N)=V(F\oplus (0))\) and \(K\in V\) if and only if \(V(K)=V((0)\oplus D)\). We claim that \(U=1\). Otherwise, \(V(F \oplus (0))=V(N \oplus K)\), where \(N=(0)\) or \(N=F\) and \((0)\ne K< D\). Therefore \(V(N \oplus K) \cup V((0)\oplus D)=\mathrm{Spec}(M)\) and hence \(V((0)\oplus K)=\mathrm{Spec}(M)\) that is a contradiction with our assumption. So \(F\oplus (0)\) is adjacent to every other vertex of \(G(\tau _{\mathrm{Spec}(M)})\) \(\square \)
Lemma 3.3
Let \(e\in R\) be an idempotent element of R and suppose that \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\). If \(G(\tau _T)\) is a trianglefree graph, then both \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime Rmodules. Moreover, if \(G(\tau _T)\) has no cycle, then \(\bar{M_{1}}\) is a simple module and \(\bar{M_{2}}\) is a prime module.
Proof
First recall that if \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\), then \(F=Q\) because \(V(F_{1}\oplus Q_{2})=V(Q_{1}\oplus F_{2})=T\). Without loss of generality, we can assume that \(\bar{M_{1}}\) is not a prime module. Then \(\bar{I}\bar{K}= (\bar{0})\), where \((Q_{1}:M_{1})\subsetneq I<R\) and \(Q_{1}\subsetneq K< M_1\). It follows that \(Q_{1} \oplus M_2\), \(K \oplus Q_2\), and \(IM_1 \oplus Q_2\) form a triangle in \(G(\tau _T)\), a contradiction (note that \(V(K\oplus Q_2) \cup V(IM_1\oplus Q_2)=V(K(IM_{1})\oplus Q_2)=T\). Also \(IM_{1}\ne K\). Otherwise, \(V(K\oplus Q_2)=V(K^{2}\oplus Q_2)=V(K(IM_{1})\oplus Q_2)=T\), a contradiction). So both \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime Rmodules. Now suppose that \(G(\tau _T)\) has no cycle. If none of \(\bar{M_{1}}\) and \(\bar{M_{2}}\) is a simple module, then we choose nontrivial submodules \(N_{i}\) in \(M_{i}\) for some \(i = 1, 2\). So \(N_{1} \oplus Q_{2}\), \(Q_{1} \oplus N_{2}\), \(M_{1} \oplus Q_{2}\), and \(Q_{1} \oplus M_{2}\) form a cycle, a contradiction. \(\square \)
Corollary 3.4
Assume that M is a multiplication module or a primeful module, \(\mathrm{Ann}(\bar{M})\) is a nil ideal, and \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\). Then \(G(\tau _T)\) is a star graph if and only if \(\bar{M_{1}}\) is a simple module and \(\bar{M_{2}}\) is a prime module for some idempotent \(e\in R\).
Proof
The necessity is clear by Proposition 3.1(a). For the converse, assume that \(\bar{M}=\bar{M_{1}} \oplus \bar{M_{2}}\), where \(\bar{M_{1}}\) is a simple module and \(\bar{M_{2}}\) is a prime for some idempotent \(e\in R\). Using the Proposition 3.1(b), \(G(\tau _T)\) is a complete bipartite graph with two parts U and V such that \(N\in U\) if and only if \(V(N)=V(M_{1}\oplus Q_{2})\) and \(K\in V\) if and only if \(V(K)=V(Q_{1}\oplus M_{2})\). We claim that \(U=1\). Otherwise, \(V(M_{1} \oplus Q_{2})=V(N_{1} \oplus N_{2})\), where \( N_{1}\le M_{1}\) and \( N_{2}\le M_{2}\). If \(N_1\ne M_1\), then \(\sqrt{(N_{1}:M_{1})M_{1}}=M_{1}\), a contradiction (note that if M is a multiplication module or a primeful module, then \(\sqrt{(N:M)M}\ne M\), where \(N< M\)). If \(N_2\ne Q_2\), then \(V(Q_{1}\oplus N_{2})=T\), a contradiction. So \(G(\tau _T)\) is a star graph. \(\square \)
Theorem 3.5
If \(G(\tau _T)\) is a tree, then \(G(\tau _T)\) is a star graph.
Proof
Suppose that \(G(\tau _T)\) is not a star graph. Then \(G(\tau _T)\) has at least four vertices. Obviously, there are two adjacent vertices L and K of \(G(\tau _T)\) such that \(N(L){\setminus } \{K\}\ge 1\) and \(N(K){\setminus } \{L\}\ge 1\). Let \(N(L){\setminus } \{K\} = \{L_{i}\}_{i\in \Lambda }\) and \(N(K){\setminus } \{L\} = \{K_{j}\}_{j\in \Gamma }\). Since \(G(\tau _T)\) is a tree, we have \(N(L)\cap N(K)=\emptyset \). By [4, Theorem 2.10], \(diam(G(\tau _T))\le 3\). So every edge of \(G(\tau _T)\) is of the form \(\{L, K\}\), \(\{L, L_{i}\}\) or \(\{K, K_{j}\}\), for some \(i\in \Lambda \) and \(j\in \Gamma \). Now, Pick \(p\in \Lambda \) and \(q\in \Gamma \). Since \(G(\tau _T)\) is a tree, \(L_{p}K_{q}\) is a vertex of \(G(\tau _T)\). If \(L_{p}K_{q}=L_{u}\) for some \(u\in \Lambda \), then \(V(KL_{u})=T\), a contradiction. If \(L_{p}K_{q}=K_{v}\), for some \(v\in \Gamma \), then \(V(LK_{v})=T\), a contradiction. If \(L_{p}K_{q}=L\) or \(L_{p}K_{q}=K\), then \(V(L^{2}) =T\) or \(V(K^{2}) =T\), respectively, and hence \(V(L) =T\) or \(V(K) =T\), a contradiction. So the claim is proved. \(\square \)
Theorem 3.6
Let R be an Artinian ring, M be a multiplication or a primeful module, and suppose that \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\). If \(G(\tau _T)\) is a bipartite graph, then \(T=2\) and \(G(\tau _T)\cong K_{2}\).
Proof
At first we recall that if \(G(\tau _T)\ne \emptyset \), then \(E(G(\tau _T))\ge 1\). Assume that \(G(\tau _T)\) is a bipartite graph. Therefore \(G(\tau _T)\) is not empty. We show that R can not be a local ring. Otherwise, m is the unique maximal ideal of R and hence is the unique prime ideal. Then [14, Corollary 2.11] implies that mM is the only prime submodule of M so that \(G(\tau _T)=\emptyset \), a contradiction. Hence by [8, Theorem 8.7], \(R = R_{1}\oplus \ldots \oplus R_{n}\), where \(R_{i}\) is an Artinian local ring for \(i=1, \ldots , n\) and \(n\ge 2\). By Lemma 2.2 and Proposition 2.3, since \(G(\tau _T)\) is a bipartite graph, we have \(n=2\) and hence \(\bar{M}\cong \bar{M_{1}}\oplus \bar{M_{2}}\) for some idempotent \(e\in R\) (for example, if \(n=3\), then \(M_{1}\oplus Q_2\oplus Q_3\), \(Q_{1}\oplus M_{2}\oplus Q_3\), and \(Q_{1}\oplus Q_{2}\oplus M_3\) form a triangle that is a contradiction). By Lemma 3.3, \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime modules. Then it is easy to see that \(\bar{M_1}\) and \(\bar{M_{2}}\) are vector spaces over \(R/\mathrm{Ann}(\bar{M_{1}})\) and \(R/\mathrm{Ann}(\bar{M_{2}})\), respectively and so are semisimple Rmodules. Since \(G(\tau _T)\) is a bipartite graph, \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are simple Rmodules. A Similar argument as we did in proof of Corollary 3.4 implies that \(T=\{M_{1}\oplus Q_2, Q_1\oplus M_2 \}\) and \(G(\tau _T)\cong K_{2}\). \(\square \)
Proposition 3.7
Assume that M is a multiplication module, \(\mathrm{Ann}(\bar{M})\) is a nil ideal, and \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\).

(a)
If \(G(\tau _T)\) is a finite bipartite graph, then \(T=2\) and \(G(\tau _T)\cong K_{2}\).

(b)
If \(G(\tau _T)\) is a regular graph of finite degree, then \(T=2\) and \(G(\tau _T)\cong K_{2}\).
Proof

(a) By Theorem 2.11, \(\bar{M}\) is an Artinian and Noetherian module so that \(R/\mathrm{Ann}(\bar{M})\) is an Artinian ring. A similar arguments in Theorem 3.6 says that, \(R/\mathrm{Ann}(\bar{M})\) is a nonlocal ring. So by [8, Theorem 8.7] and Lemma 2.2, there exist pairwise orthogonal idempotents modulo \(\mathrm{Ann}(\bar{M})\). By lemma 2.4, \(\bar{M}\cong \bar{M_{1}}\oplus \bar{M_{2}}\), for some idempotent e of R. Now, the proof that \(G(\tau _T)\cong K_{2}\) is similar to the proof of Theorem 3.6.

(b) We may assume that \(G(\tau _T)\) is not empty. So F is not a prime submodule by [4, Remark 2.6] and hence there exist \(r \in R\) and \(m\in M\) such that \(rm\in F\) but \(m\notin F\) and \(r \notin (F:M)\). A similar manner in proof of Theorem 2.11, shows that if the set of Rsubmodules of \(\overline{rM}\) (resp. \((\bar{0}:_{\bar{M}}r)\) is infinite, then \((\bar{0}:_{\bar{M}}r)\) (resp. \(\overline{rM}\)) has infinite degree, a contradiction. Thus \(\overline{rM}\) and \((\bar{0}:_{\bar{M}}r)\) have finite length so that \(\bar{M}\) has a finite length. Therefore \(R/\mathrm{Ann}(\bar{M})\) is an Artinian ring. As in the proof of part (a), \(\bar{M}\cong \bar{M_{1}}\oplus \bar{M_{2}}\) for some idempotent \(e\in R\). If \(\bar{M_{1}}\) has one nontrivial submodule \(\bar{N}\), then deg\((Q_{1} \oplus M_{2}) > \mathrm{deg}(N \oplus M_{2})\) (we note that by [6, Proposition 2.5], \(\bar{N} \bar{K}=(\bar{0})\) for some \((\bar{0})\ne \bar{K}< \bar{M_{1}}\)) and this contradicts the regularity of \(G(\tau _T)\). Hence \(\bar{M_{1}}\) is a simple module. Similarly, \(\bar{M_{2}}\) is a simple module. Finally a similar argument as we have seen in Theorem 3.6 gives \(G(\tau _T)\cong K_{2}\).
\(\square \)
4 Coloring of the Zariskitopology graph of modules
The purpose of this section is to study the coloring of the Zariski topologygraph of modules and investigate the interplay between \(\chi (G(\tau _T))\) and \(\omega (G(\tau _T))\). We note that since \(E(G(\tau _T))\ge 1\) when \(G(\tau _T)\ne \emptyset \), then \(\chi (G(\tau _T)))\ge 2\).
Theorem 4.1
Let \(\bar{M}\) be an Artinian module such that for every minimal submodule \(\bar{N}\) of \(\bar{M}\), N is a vertex in \(G(\tau _T)\). Then \(\omega (G(\tau _T))= \chi (G(\tau _T))\).
Proof
\(\bar{M}\) is Artinian, so it contains a minimal submodule. Since for every minimal submodule \(\bar{N}\) of \(\bar{M}\), N is a vertex in \(G(\tau _T)\), we have \(V(N)\ne T\). Also, \(N\cap L=Q\), where \(\bar{N}\) and \(\bar{L}\) are minimal submodules of \(\bar{M}\). It follows that N and L are adjacent in \(G(\tau _T)\), where \(\bar{N}\) and \(\bar{L}\) are minimal submodules of \(\bar{M}\). First, suppose that \(\bar{M}\) has infinitely many minimal submodules. Then \(\omega (G(\tau _T))=\infty \) and there is nothing to prove. Next, assume that \(\bar{M}\) has k minimal submodules, where k is finite. We conclude that \(\chi (G(\tau _T))= k=\omega (G(\tau _T))\). Obviously, \(\omega (G(\tau _T))\ge k\). If possible, assume that \(\omega (G(\tau _T))> k\). Let \(\Sigma =\{N_{\lambda }\}_{\lambda \in I}\), where \(I=\omega (G(\tau _T))\) be a maximum clique in \(G(\tau _T)\). As for every \(N_{\lambda }\in \Sigma \), \(\overline{ \sqrt{(N_{\lambda }:M)M}}\) contains a minimal submodule, there exists a minimal submodule \(\bar{K}\) and submodules \(N_{i}\) and \(N_{j}\) in \(\Sigma \), such that \(\bar{K}\subset \overline{\sqrt{(N_{i}:M)M}}\) \(\cap \overline{ \sqrt{(N_{j}:M)M}}\), and hence \(V(K)=T\), a contradiction. Hence \(\omega (G(\tau _T))=k\). Next, we claim that \(G(\tau _T)\) is kcolorable. In order to prove, put \(A=\{\bar{K_{1}}, \ldots , \bar{K_{k}}\}\) be the set of all minimal submodules of \(\bar{M}\). Now, we define a coloring f on \(G(\tau _T)\) by setting \(f(N) = min\{i\) \(K_{i}\subseteq \sqrt{(N:M)M}\}\) for every vertex N of \(G(\tau _T)\). Let N and L be adjacent in \(G(\tau _T)\) and \(f(N) = f(L) = j\). Thus \(K_{j}\subseteq \sqrt{(N:M)M}\cap \sqrt{(L:M)M}\), a contradiction. It implies that f is a proper k coloring of \(G(\tau _T)\) and hence \(\chi (G(\tau _T))\le k=\omega (G(\tau _T))\), as desired. \(\square \)
Theorem 4.2
Assume that \(\bar{M}\) is a faithful module. Then the following statements are equivalent.

(a)
\(\chi (G(\tau _{\mathrm{Spec}(M)}))=2\).

(b)
\(G(\tau _{\mathrm{Spec}(M)})\) is a bipartite graph with two nonempty parts.

(c)
\(G(\tau _{\mathrm{Spec}(M)})\) is a complete bipartite graph with two nonempty parts.

(d)
Either R is a reduced ring with exactly two minimal prime ideals or \(G(\tau _{\mathrm{Spec}(M)})\) is a star graph with more than one vertex.
Proof
By using Lemma 2.8, \(G(\tau _{\mathrm{Spec}(M)})\) and \(AG(M)^{*}\) are the same and so [5, Theorem 3.3] completes the proof. \(\square \)
Lemma 4.3
Assume that T is a finite set. Then \(\chi (G(\tau _T)))\) is finite. In particular, \(\omega (G(\tau _T)))\) is finite.
Proof
Suppose that \(T=\{P_{1}, P_{2},\ldots , P_{k}\}\) is a finite set of distinct prime submodules of M. Define a coloring \(f(N)=min\{n\in \mathbb N\) \(P_{n}\notin V(N) \}\), where N is a vertex of \(G(\tau _T)\). We can see that \(\chi (G(\tau _T)))\le k\). \(\square \)
Theorem 4.4
For every module M, \(\omega (G(\tau _T)) = 2\) if and only if \(\chi (G(\tau _T))=2\). In particular, \(G(\tau _T)\) is bipartite if and only if \(G(\tau _T)\) is trianglefree.
Proof
Let \(\omega (G(\tau _T)) = 2\). On the contrary assume that \(G(\tau _T)\) is not bipartite. So \(G(\tau _T)\) contains an odd cycle. Suppose that \(C:= N_{1}  N_{2}  \cdots  N_{2k+1}  N_{1}\) be a shortest odd cycle in \(G(\tau _T)\) for some natural number k. Clearly, \(k\ge 2\). Since C is a shortest odd cycle in \(G(\tau _T)\), \(N_{3}N_{2k+1}\) 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 \(V(N_{4} N_{1})= T\). This implies that \( N_{1}  N_{4}  \cdots  N_{2k+1}  N_{1}\) is an odd cycle, a contradiction. Thus \(N_{1}\ne N_{3}N_{2k+1}\). 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 \(N_{2}\ne N_{3}N_{2k+1}\). It is easy to check \(N_{1}, N_{2}\), and \(N_{3}N_{2k+1}\) form a triangle in \(G(\tau _T)\), a contradiction. The converse is clear. In particular, we note that empty graphs are bipartite graphs. \(\square \)
Corollary 4.5
Assume that \(e\in R\) is an idempotent element and \(\bar{M}\) does not have a nonzero submodule \(\overline{F}\ne \bar{N}\) with \(V(N)=T\). Then \(G(\tau _T)\) is a complete bipartite graph if and only if \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime modules.
Proof
Assume that \(G(\tau _T)\) is a complete bipartite graph. Therefore Theorem 4.4 states that \(G(\tau _T)\) is a trianglefree graph. So Lemma 3.3 follows that \(\bar{M_{1}}\) and \(\bar{M_{2}}\) are prime modules. The conversely holds by Proposition 3.1(b). \(\square \)
Remark 4.6
Assume that S is a multiplicatively closed subset of R such that \(S\cap (\cup _{P\in T}(P:M))=\emptyset \). Let \(T_{S}\)=\(\{S^{1}P\) \(P\in T\}\). One can see that \(V(N)=T\) if and only if \(V(S^{1}N)=T_{S}\), where M is a finitely generated module.
Theorem 4.7
Let S be a multiplicatively closed subset of R defined as in Remark 4.6 and M is a finitely generated module. Then \(G(\tau _{T_{S}})\) is a retract of \(G(\tau _T)\) and \(\omega (G(\tau _{T_{S}}))= \omega (G(\tau _T))\).
Proof
Consider a vertex map \(\phi : V(G(\tau _T)) \longrightarrow V(G(\tau _{T_{S}})), N\longrightarrow N_{S}\). Clearly, \(N_{S}\ne K_{S}\) implies that \(N\ne K\) and \(V(N)\cup V(K)=T\) if and only if \(V(N_{S})\cup V(K_{S})= T_{S}\). Thus \(\phi \) is surjective and hence \(\omega (G(\tau _{T_{S}}))\le \omega (G(\tau _T))\). If \(N \ne K\) and \(V(N)\cup V(K)=T\), then we show that \(N_{S}\ne K_{S}\). On the contrary suppose that \(N_{S}= K_{S}\). Then \(V(N_{S}^{2}) =V(N_{S}K_{S})=V(N_{S})\cup V(K_{S})= T_{S}\) and so \(V(N^{2})=T\), a contradiction. This shows that the map \(\phi \) is a graph homomorphism. Now, for any vertex \(N_{S}\) of \(G(\tau _{T_{S}})\), we can choose a fixed vertex N of \(G(\tau _T)\). Then \(\phi \) is a retract (graph) homomorphism which clearly implies that \(\omega (G(\tau _{T_{S}}))= \omega (G(\tau _T))\) under the assumption. \(\square \)
Corollary 4.8
Let S be a multiplicatively closed subset of R defined as in Remark 4.6 and let M be a finitely generated module. Then \(\chi (AG(M_{S}))= \chi (AG(M))\).
Corollary 4.9
Assume that M is a semiprime module and \(AG(M)^{*}\) does not have an infinite clique. Then M is a faithful module and \(0=(P_{1}\cap \ldots \cap P_{k}:M)\), where \(P_{i}\) is a prime submodule of M for \(i=1, \ldots , k\).
Proof
By [5, Theorem 3.8 (b)], M is a faithful module and the last assertion follows directly from the proof of [5, Theorem 3.8 (b)]. \(\square \)
Recall that the girth of a graph G is the length of a shortest cycle in G and denoted by gr(G).
Proposition 4.10
Let R be an Artinian ring, \(\bar{M}\) be a multiplication module, and let T be a closed subset of \(\mathrm{Spec}(M)\). Then we have the following statements.

(a)
If S is a finite subset of T, then there exists a clique of size S in \(G(\tau _T)\).

(b)
We have \(\omega (G(\tau _T))\ge Min(T)\) and if \(Min(T)\ge 3\), then \(gr(G(\tau _T))=3\).

(c)
If \(\sqrt{(\bar{0})}= (\bar{0})\), then \(\chi (G(\tau _{\mathrm{Spec}(M)})) = \omega (G(\tau _{\mathrm{Spec}(M)})) = Min(T)\).
Proof

(a)
Let R be an Artinian ring and let \(\bar{M}\) be a multiplication module. Then [14, Corollary 2.9] implies that \(\bar{M}\) is a cyclic module. We show that \(T=Min(T)\). Suppose that \(P_1\subseteq P_2\), where \(P_1, P_2\in T\). Then \((P_1:M)=(P_2:M)\) because every prime ideal in R is maximal. Since \(\bar{M}\) is multiplication, we have \(P_1=P_2\) and finally the proof is straightforward by the facts that \(AG(\bar{M})=AG(\bar{M})^{*}\), [6, Theorem 3.6], and \(AG(\bar{M})\) is isomorphic with a subgraph of \(G(\tau _T)\) by Lemma 2.8.

(b)
This is clear by item (a).

(c)
If \(Min(T)=\infty \), then by part (b), there is nothing to prove. Otherwise, [6, Theorem 3.8] implies that \(AG(\bar{M})\) does not have an infinite clique. So \(\bar{M}\) is a faithful module by Corollary 4.9. Next, Lemma 2.8 says that \(G(\tau _{\mathrm{Spec}(M)})\) and \(AG(M)^{*}\) are the same. Now the result follows by [6, Theorem 3.8].
\(\square \)
Lemma 4.11
Assume that \(\bar{M}\) is a semiprime module. Then the following statements are equivalent.

(a)
\(\chi (G(\tau _{\mathrm{Spec}(M)})))\) is finite.

(b)
\(\omega (G(\tau _{\mathrm{Spec}(M)})))\) is finite.

(c)
\(G(\tau _{\mathrm{Spec}(M)}))\) does not have an infinite clique.
Proof
\((a)\Longrightarrow (b)\Longrightarrow (c)\) is clear.
\((c)\Longrightarrow (a)\) Suppose that \(G(\tau _{\mathrm{Spec}(M)}))\) does not have an infinite clique. By Lemma 2.8, \(AG(\bar{M})^{*}\) does not have an infinite clique and so by Corollary 4.9, there exists a finite number of prime submodules \(P_{1}, \ldots ,P_{k}\) of M such that \((F:M)=(P_{1}\cap \ldots \cap P_{k}:M)\). Define a coloring \(f(N)=min\{n\in \mathbb N\) \(P_{n}\notin V(N) \}\), where N is a vertex of \(G(\tau _T)\). Then we have \(\chi (G(\tau _{\mathrm{Spec}(M)})))\le k\). \(\square \)
Corollary 4.12
Assume that \(AG(M/F)^{*}\) does not have an infinite clique. Then \(G(\tau _{\mathrm{Spec}(M)})\) and \(AG(M)^{*}\) are the same. Also, \(\chi (G(\tau _{\mathrm{Spec}(M)})))\) is finite.
Proof
Since M/F is a semiprime module, by Corollary 4.9, M/F is a faithful module and there exists a finite number of prime submodules \(P_{1}, \ldots ,P_{k}\) of M such that \((F:M)=(P_{1}\cap \ldots \cap P_{k}:M)\). So the result follows by Lemma 2.8 and from the proof of \(\mathrm{(c)}\Longrightarrow \mathrm{(a)}\) of Lemma 4.11. \(\square \)
We recall that M is said to be Xinjective if either \(\mathrm{Spec}(M)=\emptyset \) or the natural map of \(X=\mathrm{Spec}(M)\) is injective (see [7]).
Proposition 4.13
Suppose that \(\sqrt{(\bar{0})}= (\bar{0})\), for every minimal member P of \(\mathrm{Spec}(M)\), (P : M) is a minimal ideal of R, and \(\bar{M}\) is an Xinjective module. Then the following statements are equivalent.

(a)
\(\chi (G(\tau _{\mathrm{Spec}(M)}))\) is finite.

(b)
\(\omega (G(\tau _{\mathrm{Spec}(M)}))\) is finite.

(c)
\(G(\tau _{\mathrm{Spec}(M)})\) does not have an infinite clique.

(d)
\(Min(\mathrm{Spec}(M))\) is a finite set.
Proof
\(\mathrm{(a)} \Longrightarrow \mathrm{(b)} \Longrightarrow \mathrm{(c)}\) is clear.
\(\mathrm{(c)}\Longrightarrow \mathrm{(d)}\) Suppose \(G(\tau _{\mathrm{Spec}(M)})\) does not have an infinite clique. By Lemma 2.8, \(AG(\bar{M})^{*}\) does not have an infinite clique and hence by Corollary 4.9, there exists a finite number of prime submodules \(P_{1}, \ldots ,P_{k}\) of M such that \((F:M)=(P_{1}\cap P_{2}\cap \cdots \cap P_{k}:M)\). By assumptions, one can see that \(Min(\mathrm{Spec}(M))\) is a finite set.
\(\mathrm{(d)}\Longrightarrow \mathrm{(a)}\) Assume that \(Min(\mathrm{Spec}(M))\) is a finite set (equivalently, \(\bar{M}\) has a finite number of minimal prime submodules) so that \((F:M)=(P_{1}\cap P_{2}\cap \cdots \cap P_{k}:M)\), where \(\mathrm{Min}(\mathrm{Spec}(M))=\{P_{1}, \ldots ,P_{k}\}\). Define a coloring \(f(N)=\mathrm{min}\{n\in N\) \(P_{n}\notin V(N) \}\), where N is a vertex of \(G(\tau _{\mathrm{Spec}(M)})\). Then we have \(\chi (G(\tau _{\mathrm{Spec}(M)}))\le k\). \(\square \)
Example 4.14
If M is a faithfully flat Rmodule (for example, free modules), then pM is a pprime submodule of M, where p is a prime ideal of R by [10, Theorem 3]. So for every minimal prime submodule P of M, (P : M) is a minimal ideal of R.
Proposition 4.15
Assume that \(\sqrt{(\bar{0})}= (\bar{0})\) and \(\bar{M}\) is a faithful module. Then the following statements are equivalent.

(a)
\(\chi (G(\tau _{\mathrm{Spec}(M)}))\) is finite.

(b)
\(\omega (G(\tau _{\mathrm{Spec}(M)}))\) is finite.

(c)
\(G(\tau _{\mathrm{Spec}(M)})\) does not have an infinite clique.

(d)
R has a finite number of minimal prime ideals.

(e)
\(\chi (G(\tau _{\mathrm{Spec}(M)}))=\omega (G(\tau _{\mathrm{Spec}(M)}))=Min(R)=k\), where k is finite.
Proof
This is clear by Lemma 2.8, [5, Proposition 3.10], and [5, Corollary 3.11]. \(\square \)
References
Anderson, D.F.; Livingston, P.S.: The zerodivisor graph of a commutative ring. J. Algebra 217, 434–447 (1999)
Anderson, W.; Fuller, K.R.: Rings and Categories of Modules. Springer, New York (1974)
AnsariToroghy, H.; Farshadifar, F.: Product and dual product of submodules. Far East J. Math. Sci 25(3), 447–455 (2008)
AnsariToroghy, H.; Habibi, S.: The Zariski topologygraph of modules over commutative rings. Commun. Algebra 42, 3283–3296 (2014)
AnsariToroghy, H.; Habibi, S.: The annihilatingsubmodule graph of modules over commutative rings. Math. Rep. 20(70), 245–262 (2018)
AnsariToroghy, H.; Habibi, S.: The annihilatingsubmodule graph of modules over commutative rings II. Arab. J. Math. 5, 187–194 (2016)
AnsariToroghy, H.; OvlyaeeSarmazdeh, R.: On the prime spectrum of Xinjective modules. Commun. Algebra 38, 2606–2621 (2010)
Atiyah, M.F.; Macdonald, I.G.: Introduction to Commutative Algebra. AddisonWesley, Boston (1969)
Beck, I.: Coloring of commutative rings. J. Algebra 116, 208–226 (1988)
ChinPi, L.: Prime submodules of modules. Comment. Math. Univ. St. Pauli 33(1), 61–69 (1984)
ChinPi, L.: The Zariski topology on the prime spectrum of a module. Houst. J. Math. 25(3), 417–432 (1999)
ChinPi, L.: A module whose prime spectrum has the surjective natural map. Houst. J. Math 33(1), 125–143 (2007)
Diestel, R.: Graph theory. In: Grad, Texts in Math. Springer (2005)
Elbast, Z.A.; Smith, P.F.: Multiplication modules. Commun. Algebra 16, 755–779 (1988)
Lam, T.Y.: A First Course in Noncommutative Rings. Springer, New York (1991)
McCasland, R.L.; Moor, M.E.: Prime submodules. Commun. Algebra 20(6), 1803–1817 (1992)
Tavallaee, H.A.; Varmazyar, R.: Semiradicals of submodules in modules. IUST Int. J. Eng. Sci. 19, 21–27 (2008)
Acknowledgements
The authors express their deep gratitude to learned referees for their meticulous reading, contributions and valuable suggestions which have definitely improved the manuscript. This research was in part supported by a grant from IPM (No. 96130028).
Author information
Authors and Affiliations
Corresponding author
Additional information
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.
About this article
Cite this article
AnsariToroghy, H., Habibi, S. The Zariski topologygraph of modules over commutative rings II. Arab. J. Math. 10, 41–50 (2021). https://doi.org/10.1007/s40065020003011
Received:
Accepted:
Published:
Issue Date:
DOI: https://doi.org/10.1007/s40065020003011