The Zariski topology-graph of modules over commutative rings II

Abstract

Let M be a module over a commutative ring R. In this paper, we continue our study about the Zariski topology-graph $G(\tau _T)$ which was introduced in Ansari-Toroghy et al. (Commun Algebra 42:3283–3296, 2014). For a non-empty 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 topology-graphs and investigate the interplay between $\chi (G(\tau _T))$ and $\omega (G(\tau _T))$.

1 Introduction

Throughout this paper, R is a commutative ring with a non-zero identity and M is a unital R-module. 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 topology-graph and the annihilating-submodule graph, respectively.

Let T be a non-empty subset of $\mathrm{Spec}(M)$. The Zariski topology-graph $G(\tau _T)$ 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 topology-graph that are basic or needed in the later sections. In Sect. 3, we study the bipartite Zariski topology-graphs 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 topology-graph 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 topology-graph 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 $x_{i-1}$ and $x_i$ are adjacent for each $1\le i\le n$ and we denote $x_{i-1} - x_i$ for existing an edge between $x_{i-1}$ 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 k-regular, 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 non-empty 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 topology-graph 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

$$\begin{aligned}V(N)\cup V(K)=V(N\cap K)=V(NK)=V^*(NK).\end{aligned}$$

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 non-zero 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 $R_{2}=(1-e)R$.

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

• (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 $e-u\in I$.

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

$$\begin{aligned} \sqrt{(N:M)M} \sqrt{(K:M)M} \subseteq F. \end{aligned}$$

Since

$$\begin{aligned}NK=((N:M)M:M)((K:M)M:M)M\subseteq \sqrt{(N:M)M} \sqrt{(K:M)M},\end{aligned}$$

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 non-trivial 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 R-module 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 non-trivial submodule of M has finitely many submodules.

Proof

First, we show that the descending chain of non-trivial 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 non-trivial 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 non-trivial 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 R-modules $\overline{rM}$ 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 topology-graph 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 $M_{1}:=eM, M_{2}:=(1-e)M, T_{1}:=\{P_{1}\in \mathrm{Spec}(M_{1})| P_{1}\oplus M_{2}\in T\}$, $T_{2}:=\{P_{2}\in \mathrm{Spec}(M_{2})| M_{1}\oplus P_{2}\in T\}$, $F_{1}:=\cap _{P_{1}\in T_{1}}P_{1}$, $F_{2}:=\cap _{P_{2}\in T_{2}}P_{2}$, $Q_{1}:=(F_{1}:M_{1})M_{1}, Q_{2}:=(F_{2}:M_{2})M_{2}$, $\bar{M_{1}}= \overline{eM}=eM/Q_{1}$, and $\bar{M_{2}}= \overline{(e-1)M}=(e-1)M/Q_{2}$. 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 R-module 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 non-zero 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 $\bar{M}\cong \overline{eM}\oplus \overline{(e-1)M}$ 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 non-zero 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 non-zero submodule of M is of the form $M_{1}\oplus N_{2}$ and $(0)\oplus N_{2}$, where $N_{2}$ is a non-zero 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 non-zero 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 non-zero submodule $\overline{F}\ne \bar{N}$ with $V(N)=T$. If $G(\tau _T)$ is a triangle-free graph, then both $\bar{M_{1}}$ and $\bar{M_{2}}$ are prime R-modules. 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 non-zero 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 R-modules. 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 non-trivial 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 non-zero 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 non-zero 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 R-modules. Since $G(\tau _T)$ is a bipartite graph, $\bar{M_{1}}$ and $\bar{M_{2}}$ are simple R-modules. 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 non-zero 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 non-local 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 R-submodules 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 non-trivial 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 Zariski-topology graph of modules

The purpose of this section is to study the coloring of the Zariski topology-graph 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 k-colorable. 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 non-empty parts.

• (c) $G(\tau _{\mathrm{Spec}(M)})$ is a complete bipartite graph with two non-empty 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 triangle-free.

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 non-zero 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 triangle-free 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 X-injective 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 X-injective 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 R-module (for example, free modules), then pM is a p-prime 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]. $\square$