On the Maximal Spectrum of a Module and Zariski Topology

Abstract

For any module M over a commutative ring R, $Spec_{R}(M)$ (resp. $Max_{R}(M)$) of M is the collection of all prime (resp. maximal) submodules. In this article, we investigate the interplay between the topological properties of $Max_{R}(M)$ and module theoretic properties of M. Also, for various types of modules M, we obtain some conditions under which $Max_{R}(M)$ is homeomorphic with the maximal ideal space of some ring.

1 Introduction

Throughout this article, R is a commutative ring with non zero identity and M is a unitary R-module. For any ideal I of R containing $Ann_{R}(M)$, $\bar{R}$ and $\bar{I}$ denote $R/Ann(M)$ and $I/Ann(M)$, respectively. Also $\mathbb N$, $\mathbb Z$, and $\mathbb Q$ denote the set of positive integers, the ring of integers, and the field of rational numbers, respectively. Moreover, the notation “$\subset$” will denote the strict inclusion.

For $M$ as an $R$-module and N a submodule, we recall the colon ideal of M into N, $(N:M)=\{r\in R| rM\subseteq N\}$.

A submodule P of M is said to be a prime submodule if $P\ne M$ and whenever $r\in R$ and $e \in M$ satisfy $re\in P$, then $r\in P$ or $e\in (P:M)$. If P is a prime submodule, then $(P:M)$ is a prime ideal of R. Moreover, if Q is a maximal submodule of M, then Q is a prime submodule and $(Q:M)$ is a maximal ideal of $R$ [8] and [9].

If $Spec_{R}(M)\ne \emptyset$ (resp. $Max_{R}(M)\ne \emptyset$), the mapping $\psi :Spec_{R}(M)\rightarrow Spec(\bar{R})$ (resp. $\phi :Max_{R}(M)\rightarrow Max(\bar{R})$) such that $\psi (P)=\overline{(P:M)}$ (resp. $\phi (Q)=\overline{(Q:M)}$) for every $P \in Spec_{R}(M)$ (resp. $Q \in Max_{R}(M)$), is called the natural map of $Spec_{R}(M)$ (resp. $Max_{R}(M)$) [10] (resp. [1]).

M is said to be primeful (resp. $Max$ -surjective) if either $M=(0)$ or $M\ne (0)$ and the natural map of $Spec_{R}(M)$ (resp. $Max_{R}(M)$) is surjective [11] (resp. [1]).

M is said to be $X$ -injective if either $Spec_{R}(M)=\emptyset$ or $Spec_{R}(M)\ne \emptyset$ and the natural map of $Spec_{R}(M)$ is injective [2].

The Zariski topology on $X=Spec_{R}(M)$ is the topology $\tau _M$ described by taking the set $\ Z(M) =\{ V_{M}(N)| N$ is a submodule of $M \}$ as the set of closed sets of $X$, where

$$\begin{aligned} V_{M}(N)=\{P \in X|(P:M)\supseteq (N:M) \}. \end{aligned}$$

When $M=R$, $\tau _M=\tau _R$ is the well-known Zariski topology on $Spec(R)$ [10].

There exists a topology on $Max_{R}(M)$ having $\ Z^{m}(M) =\{ V_{M}^{m}(N)| N$ is a submodule of $M \}$ as the set of closed sets of $Max_{R}(M)$, where

$$\begin{aligned} V_{M}^{m}(N)=\{Q \in Max_{R}(M)|(Q:M)\supseteq (N:M) \}. \end{aligned}$$

We denote this topology by $\tau _M^{m}$. In fact $\tau _M^{m}$ is the same as the subspace topology induced by $\tau _M$ on $Max_{R}(M)$. When $M=R$, this topology is denoted by $\tau _R^{m}$ and for every ideal $I$ of $R$, we have

$$\begin{aligned}V_{R}^{m}(I)=\{q \in Max (R)|q\supseteq I \}.\end{aligned}$$

In the rest of this article $Spec_{R}(M)$ (resp. $Max_{R}(M)$) is always equipped with the Zariski topology $\tau _M$ (resp. $\tau _M^{m})$ and $Max_{R}(M)$ is assumed to be a non-empty subset of $Spec_{R}(M)$.

The present authors introduced the concept of $Max$-injective modules and investigated some important properties of this family of modules. An $R$-modules $M$ is called $Max$-injective if the natural map of $Max_{R}(M)$ is injective [4]. Clearly, every X-injective module is $Max$-injective.

A topological space W is said to be $Max$-spectral if it is homeomorphic with the maximal ideal space of some ring (see Definition 3.17). $Max$-spectral spaces have been characterized by Hochster in [12, p. 57, Proposition 11].

In this article, we investigate the interplay between the topological properties of $Max_{R}(M)$ and module theoretic properties of M (see Proposition 3.2, Theorem 3.6, Theorem 3.13, Corollary 3.15, Proposition 3.19, and Theorem 3.24). Theorem 3.14 provides useful information about the relationship between topological properties of $Max_{R}(M)$ and $Max(\bar{R})$. Also we consider the conditions under which $Max_{R}(M)$ is a Noetherian topological space (see Proposition 3.2, Theorem 3.6, Theorem 3.14, and Corollary 3.15). Moreover, we study the topological space $Max_{R}(M)$ from the point of view of $Max$-spectral spaces (see Theorem 3.24). It is shown that if M is a $Max$-injective module over a PID, then $Max_{R}(M)$ is a $Max$-spectral topological space (see Theorem 3.24 (g)). These results enable us to provide a large family of modules such that their maximal submodules are $Max$-spectral.

2 Preliminaries

In this section, we review some preliminary results which will be needed in next section.

Definition 2.1

For a topological space $X$, we recall

• (a) X is quasi compact if it satisfies one of the following two equivalent conditions.

  • (1) Every collection of open subsets whose union is $X$ contains a finite subcollection whose union is X.

  • (2) Every collection of closed subsets whose intersection is empty set contains a finite subcollection whose intersection is empty set (see [14, Definition 2.135]).

• (b) X is said to be Noetherian if the open subsets of X satisfy the ascending chain condition (or maximal condition). (see [6], Chap. 6, Example 5]).

• (c) X is said to be connected if it is not the union $X = X_{0} \cup X_{1}$ of two disjoint closed non-empty subsets $X_{0}$ and $X_{1}$ (see [14, Definition 2.105]).

• (d) X is said to be irreducible if X is not the union of two proper closed subsets. For $X'\subseteq X$, $X'$ is irreducible if it is irreducible as a space with the relative topology. This is equivalent to say that, if $F$, $G$ are closed subsets of $X$ such that $X'\subseteq F\cup G$, then $X'\subseteq F$ or $X'\subseteq G$ (see [7], Ch. II, p. 119]).

• (e) A maximal irreducible subset of $X$ is called an irreducible component of X. It is well known that every irreducible component of X is closed in $X$ (see [7], Ch. II]).

Remark 2.2

Let $X$ and $Y$ be two topological spaces.

• (a) Let $f$ be a continuous mapping from $X$ to $Y$.

  • (1) If $X$ is a connected (resp. quasi compact) topological space, then $f(X)$ is a connected (resp. quasi compact) topological space (see [14, Theorem 2.107 and Theorem 2.138]).

  • (2) For every irreducible subset $E$ of $X$, $f(E)$ is an irreducible subset of $Y$ (see [7], Ch. II]).

• (b) If $X$ is a Noetherian topological space, then every subspace of $X$ is a Noetherian topological space, and $X$ is a quasi compact topological space (see [6], Chap. 6, Exc. 5]).

• (c) Every Noetherian topological space has only finitely many irreducible components (see [7, p. 124, Proposition 10]).

• (d) Closed subspaces of quasi compact topological spaces are quasi compact (see [14, Theorem 2.137]).

• (e) Every finite topological space is quasi compact (see [14, p. 51]).

• (f) Closure of any connected (resp. irreducible) subspace is connected (resp. irreducible) (see [14, Corollary 2.112] and [7], Ch. II]).

• (g) Let $A$ and $B$ be subsets of $X$ such that $A\subseteq B\subseteq X$, where $B$ is closed in $X$ and equipped with the relative topology. Then $A$ is an irreducible closed subset of $B$ if and only if $A$ is an irreducible closed subset of $X$ (see Definition 2.1 (d)).

3 Main Results

As it was mentioned before, $Spec_{R}(M)$ (resp. $Max_{R}(M)$) is always equipped with Zariski topology $\tau _M$ (resp. $\tau _M^{m})$.

Lemma 3.1

Let $M$ be an $R$-module and let $\phi :Max_{R}(M)\rightarrow Max(\bar{R})$ be the natural map of $Max_{R}(M)$. Then the following hold.

• (a) $\phi$ is a continuous map.

• (b) If $M$ is $Max$-surjective, then $\phi$ is closed and open mapping.

Proof

• (a) This follows from the fact that $\phi ^{-1}(V_{\bar{R}}^{m}(\bar{I}))= V_{M}^{m}(IM)$ for every ideal $I$ of $R$ containing $Ann(M)$.

• (b) Let $N$ be a submodule of $M$ and let $V_{M}^{m}(N)$ be a closed subset of $Max_{R}(M)$. Then as in the proof part (a), we have

  $$\begin{aligned} \phi ^{-1}(V_{\bar{R}}^{m}(\overline{(N:M)}))= V_{M}^{m}((N:M)M)=V_{M}^{m}(N). \end{aligned}$$

  Hence $\phi (V_{M}^{m}(N))=V_{\bar{R}}^{m}(\overline{(N:M)})$ because $\phi$ is surjective. Also $\phi$ is open by similar arguments and the proof is completed.$\square$

A topological space $W$ is a cofinite topological space when its open sets are empty and $W$ and all subsets with a finite complement. This topology is denoted by $\tau ^{fc}$.

Proposition 3.2

Let $R$ be a ring such that the intersection of every infinite collection of maximal ideals of $R$ is zero (for example, when $R$ is PID or one dimensional Noetherian domain) and let $M$ be an $R$-module. Then $Max_{R}(M)$ is a Noetherian topological space.

Proof

Let $V_{M}^{m}(N)$ be a closed subset of $Max_{R}(M)$ for some submodule $N$ of $M$. If $V_{M}^{m}(N)$ is infinite, then $(N:M)$ is contained in an infinite number of maximal ideals of $R$. Since the intersection of every infinite collection of maximal ideals of $R$ is zero, $(N:M)=(0)$ so that $V_{M}^{m}(N)=Max_{R}(M)$. It follows that $\tau _M^{m}\subseteq \tau ^{fc}$ and hence $Max_{R}(M)$ is a Noetherian topological space because every cofinite topological space is Noetherian. $\square$

Notation 3.3

Let $M$ be an $R$-module and W be a subset of $Max_{R}(M)$. We will denote the intersection of all elements in $W$ by $\mathfrak {I}(W)$ and the closure of $W$ in $Max_{R}(M)$ (resp. $Spec_{R}(M)$) by $Cl^{m}(W)$ (resp. $Cl(W)$).

Lemma 3.4

Let M be an R-module and $W$ be a subset of $Max_{R}(M)$. Then $Cl^{m}(W)=V_{M}^{m}(\mathfrak {I}(W))$. Hence, W is closed if and only if $V_{M}^{m}(\mathfrak {I}(W))= W$.

Proof

Let $W$ be a subset of $Max_{R}(M)$. It is well known that

$$\begin{aligned} Cl^{m}(W)=Cl(W)\cap Max_{R}(M). \end{aligned}$$

But $Cl(W)=V(\mathfrak {I}(W))$ by [10, Proposition 5.1]. It follows that $Cl^{m}(W)=V_{M}^{m}(\mathfrak {I}(W))$. $\square$

For a proper ideal $I$ of $R$, we recall that the J-radical $I$, denoted by $J_{R}^{m}(I)$, is the intersection of all maximal ideals containing $I$. An ideal $I$ of $R$ is a J-radical ideal if $I=J_{R}^{m}(I)$.

Definition 3.5

Let $M$ be an $R$-module. The J-radical of a submodule $N$ of $M$, denoted by $J_{M}^{m}(N)$, is the intersection of all members of $V_{M}^{m}(N)$. In case that $V_{M}^{m}(N)=\emptyset$, we define $J_{M}^{m}(N)=M$. A submodule $N$ of $M$ is said to be a J-radical submodule if $N=J_{M}^{m}(N)$.

Theorem 3.6

Let $M$ be an $R$-module. Then the following are equivalent.

• (a) $Max_{R}(M)$ is a Noetherian topological space.

• (b) The ascending chain condition for J-radical submodules of $M$ holds.

Proof

(a)$\Rightarrow$(b) Straightforward.

(b)$\Rightarrow$(a) Let

$$\begin{aligned} V_{M}^{m}(N_{1})\supseteq V_{M}^{m}(N_{2})\supseteq \cdots \supseteq V_{M}^{m}(N_{i})\supseteq \cdots \end{aligned}$$

be a descending chain of closed sets $V_{M}^{m}(N_{i})$ of $Max_{R}(M)$, where $N_{i}$ is a submodule of $M$. Hence

$$\begin{aligned} J_{M}^{m}(N_{1})\subseteq J_{M}^{m}(N_{2})\subseteq \cdots \subseteq J_{M}^{m}(N_{i})\subseteq \cdots \end{aligned}$$

is an ascending chain of J-radical submodules of $M$. So by hypothesis, there exists a $k\in \mathbb N$ such that for all $n>k$, we have $J_{M}^{m}(N_{k+n})= J_{M}^{m}(N_{k}).$ Now using Lemma 3.4, for all $n>k$, $V_{M}^{m}(N_{k+n})= V_{M}^{m}(N_{k})$ and the proof is completed. $\square$

Corollary 3.7

Let $M$ be a Noetherian $R$-module. Then $Max_{R}(M)$ is a Noetherian topological space.

We recall that if $I$ is an ideal of $R$, then the J-components of $I$ are the minimal members of the family of J-radical prime ideals containing $I$ (see [15, p. 631]).

Definition 3.8

Let $M$ be an $R$-module and $L$ a submodule of $M$. A submodule $P$ of $M$ is a J-component of $L$, if $(P:M)$ is a J-component of $(L:M)$. Clearly, this definition is the generalization of J-component of an ideal in rings.

Definition 3.9

A module $M$ is said to have property (JFC) if every closed subset of $Max_{R}(M)$ has a finite number of irreducible components.

Example 3.10

Let $M$ be an $R$-module. Then $M$ has property (JFC) in each of the following cases:

• (a) $Max_{R}(M)$ is a Noetherian topological space (see parts (b) and (c) of Remark 2.2);

• (b) $R$ is PID (see Proposition 3.2 and part (a));

• (c) $M$ is Noetherian (see Corollary 3.7 and part (a));

• (d) $M$ is semi local (see Remark 2.2 (e) and part (a)).

When $M$ is the $R$-module $R$, then $R$ has property (JFC) if and only if every ideal of $R$ has a finite number of J-components (see [15, p. 632]). Theorem 3.13 (d) extends this property for modules.

The proof of the following lemma is easy and is omitted.

Lemma 3.11

Let $M$ be a $Max$-surjective $R$-module. Then the following hold.

• (a) If $N$ is a submodule of $M$, then

  $$\begin{aligned} J_{R}^{m}((N:M))=(J_{M}^{m}(N):M). \end{aligned}$$

• (b) If $q$ is a J-radical ideal of $R$ containing $Ann_{R}(M)$, then there exists a submodule $Q$ of $M$ such that $(Q:M)=q$.

Remark 3.12

If $S$ is a commutative ring with non zero identity, then there exists a one-to-one correspondence between the J-radical prime ideals of ring $S$ and irreducible closed subsets of $Max(S)$ (see [15, p. 631]).

Theorem 3.13

Let $M$ be a $Max$-surjective $R$-module. Then the following hold.

• (a) If $Y\subseteq Max_{R}(M)$, then $Y$ is an irreducible closed subset of $Max_{R}(M)$ if and only if $Y=V_{M}^{m}(N)$ for some submodule $N$ of $M$ such that $(N:M)$ is a J-radical prime ideal of $R$.

• (b) If $W\subseteq Max_{R}(M)$ and $L$ is submodule of $M$, then $W$ is an irreducible component of $V_{M}^{m}(L)$ if and only if $W=V_{M}^{m}(N')$ for some J-component $N'$ of $L$.

• (c) If $Z\subseteq Max_{R}(M)$, then $Z$ is an irreducible component of $Max_{R}(M)$ if and only if $Z=V_{M}^{m}(pM)$ for some J-component ideal $p$ of $Ann_{R}(M)$.

• (d) $M$ has property (JFC) if and only if every submodule of $M$ has a finite number of J-components.

Proof

(a) $(\Rightarrow )$ Let $Y$ be an irreducible closed subset of $Max_{R}(M)$. Since $Y$ is closed, $Y=V_{M}^{m}(N)$ for some submodule $N$ of $M$. It turns out that $\phi (V_{M}^{m}(N))=V_{\bar{R}}^{m}(\overline{(N:M)})$ is an irreducible closed subset of $Max(\bar{R})$ by Lemma 3.1 and Remark 2.2 (a). Now by Remark 3.12, $\overline{(N:M)}$ is a J-radical prime ideal of $\bar{R}$ so that $(N:M)$ is a J-radical prime ideal of $R$. Conversely, let $V_{M}^{m}(K)$ be a closed subset of $Max_{R}(M)$, where $K$ is a submodule of $M$ such that $(K:M)$ is a J-radical prime ideal of $R$. We show that $V_{M}^{m}(K)$ is irreducible. To see this, let $E$ and $E'$ be submodules of $M$ with

$$\begin{aligned} V_{M}^{m}(K)\subseteq V_{M}^{m}(E) \cup V_{M}^{m}(E'). \end{aligned}$$

Hence as in the proof of Lemma 3.1 (b), we have

$$\begin{aligned} V_{\bar{R}}^{m}(\overline{(K:M)}) \subseteq V_{\bar{R}}^{m}(\overline{(E:M)}) \cup V_{\bar{R}}^{m}(\overline{(E':M)}). \end{aligned}$$

Since $(K:M)$ is a J-radical prime ideal of $R$, it is easy to check that $\overline{(K:M)}$ is a J-radical prime ideal of $\bar{R}$. Therefore, $V_{\bar{R}}^{m}(\overline{(K:M)})$ is an irreducible closed subset of $Max(\bar{R})$ by Remark 3.12. Hence by Definition 2.1 (d),

$$\begin{aligned} V_{\bar{R}}^{m}(\overline{(K:M)}) \subseteq V_{\bar{R}}^{m}(\overline{(E:M)}) \,\, or \,\, V_{\bar{R}}^{m}(\overline{(K:M)}) \subseteq V_{\bar{R}}^{m}(\overline{(E':M)}). \end{aligned}$$

Suppose that $V_{\bar{R}}^{m}(\overline{(K:M)}) \subseteq V_{\bar{R}}^{m}(\overline{(E:M)})$. This implies that $V_{M}^{m}(K)\subseteq V_{M}^{m}(E)$. By similar arguments, $V_{M}^{m}(K)\subseteq V_{M}^{m}(E')$ when $V_{\bar{R}}^{m}(\overline{(K:M)}) \subseteq V_{\bar{R}}^{m}(\overline{(E':M)})$.

(b) $(\Rightarrow )$ Let $W$ be an irreducible component of $V_{M}^{m}(L)$. By Definition 2.1 (e) and Remark 2.2 (g), $W$ is an irreducible closed subset of $Max_{R}(M)$. So by part (a), $W=V_{M}^{m}(N_{1}')$ for some submodule $N_{1}'$ of $M$ such that $(N_{1}':M)$ is a J-radical prime ideal of $R$. We claim that $N_{1}'$ is a J-component of $L$ or equivalently, $(N_{1}':M)$ is a J-component of $(L:M)$. Clearly $(N_{1}':M)\supseteq (L:M)$ using Lemma 3.11 (a). So by the above arguments, it is enough to show that $(N_{1}':M)$ is a minimal member of the family of J-radical prime ideals containing $(L:M)$. To see this, let $q$ be a J-radical prime ideal of $R$ with

$$\begin{aligned} (L:M)\subseteq q \subseteq (N_{1}':M). \end{aligned}$$

Since $M$ is $Max$-surjective, there exists a submodule $Q$ of $M$ such that $q=(Q:M)$ by Lemma 3.11 (b). Hence

$$\begin{aligned} V_{M}^{m}(L) \supseteq V_{M}^{m}(Q) \supseteq V_{M}^{m}(N_{1}'). \end{aligned}$$

Also $V_{M}^{m}(Q)$ is an irreducible closed subset of $V_{M}^{m}(L)$ by part (a), and Remark 2.2 (g). Since $W=V_{M}^{m}(N_{1}')$ is an irreducible component of $V_{M}^{m}(L)$, by the above arguments, we have $V_{M}^{m}(Q)=V_{M}^{m}(N_{1}')$. Now using Lemma 3.11 (a), $q = (N_{1}':M)$ as desired.

$(\Leftarrow )$ Let $N_{2}''$ be a J-component of $L$. Then $V_{M}^{m}(N_{2}'')$ is an irreducible closed subset of $V_{M}^{m}(L)$ by part (a) and Remark 2.2 (g). Let $L'$ be a submodule of $M$ such that $(L':M)$ is a J-radical prime ideal of $R$ and

$$\begin{aligned} V_{M}^{m}(N_{2}'') \subseteq V_{M}^{m}(L') \subseteq V_{M}^{m}(L). \end{aligned}$$

Since $N_{2}''$ be a J-component of $L$, using Lemma 3.11 (a), we have $V_{M}^{m}(N_{2}'') = V_{M}^{m}(L')$ as required.

(c) This follows from part (b) and Lemma 3.11 (b) and the fact that if $N$ is a submodule of $M$, then

$$\begin{aligned} V_{M}^{m}((N:M)M)=V_{M}^{m}(N). \end{aligned}$$

(d) Follows from part (b). $\square$

Let $X$ be a topological space. We consider strictly decreasing chain $Z_{0}$, $Z_{1}$,..., $Z_{r}$ of length $r$ of irreducible closed subsets $Z_{i}$ of $X$. The supremum of the lengths, taken over all such chains, is called the combinatorial dimension of $X$ and denoted by $dim(X)$. For the empty set, $\emptyset$, the combinatorial dimension of $\emptyset$ is defined to be $-1$.

Theorem 3.14

Let $M$ be a $Max$-surjective $R$-module. Then the following hold.

• (a) $Max_{R}(M)$ is a Noetherian topological space if and only if $Max(\bar{R})$ is a Noetherian topological space.

• (b) $Max_{R}(M)$ is a connected topological space if and only if $Max(\bar{R})$ is a connected topological space.

• (c) $Max_{R}(M)$ is an irreducible topological space if and only if $Max(\bar{R})$ is an irreducible topological space.

• (d) $Max_{R}(M)$ is a quasi compact topological space if and only if $Max(\bar{R})$ is a quasi compact topological space.

• (e) $dim(Max_{R}(M))= dim(Max(\bar{R}))$.

Proof

Let $\phi :Max_{R}(M)\rightarrow Max(\bar{R})$ be the natural map of $Max_{R}(M)$.

(a) $(\Rightarrow )$ Let $V_{\bar{R}}^{m}(\bar{I}_{1}) \supseteq V_{\bar{R}}^{m}(\bar{I}_{2}) \supseteq \cdots \supseteq V_{\bar{R}}^{m}(\bar{I}_{i}) \supseteq \cdots$ be a descending chain of closed sets in $Max(\bar{R})$, where each $\bar{I}_{i}$ is an ideal of $\bar{R}$. Since $\phi$ is continuous by Lemma 3.1 (a),

$$\begin{aligned} \phi ^{-1}(V_{\bar{R}}^{m}(\bar{I}_{1})) \supseteq \phi ^{-1}(V_{\bar{R}}^{m}(\bar{I}_{2})) \supseteq \cdots \supseteq \phi ^{-1}(V_{\bar{R}}^{m}(\bar{I}_{i})) \supseteq \cdots \end{aligned}$$

is a descending chain of closed sets in $Max_{R}(M)$. By hypothesis, there exists a $t\in \mathbb N$ such that for all $n>t$, $\phi ^{-1}(V_{\bar{R}}^{m}(\bar{I}_{t+n}))=\phi ^{-1}(V_{\bar{R}}^{m}(\bar{I}_{t}))$. Hence for all $n>t$, we have $V_{\bar{R}}^{m}(\bar{I}_{t+n})=V_{\bar{R}}^{m}(\bar{I}_{t})$ because $\phi$ is surjective. Therefore, $Max(\bar{R})$ is a Noetherian topological space. To show the converse, by Theorem 3.6, it is enough to show that the ascending chain condition for J-radical submodules of $M$ holds. To see this, let

$$\begin{aligned} N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{i}\subseteq \cdots \end{aligned}$$

be an ascending chain of J-radical submodules of $M$. Then by Lemma 3.11 (a), one can see that

$$\begin{aligned} \overline{(N_{1}:M)}\subseteq \overline{(N_{2}:M)}\subseteq \cdots \subseteq \overline{(N_{i}:M)}\subseteq \cdots \end{aligned}$$

is an ascending chain of J-radical ideals of $\bar{R}$. So by Theorem 3.6, there exists a $k\in \mathbb N$ such that for all $n>k$, $\overline{(N_{k+n}:M)}= \overline{(N_{k}:M)}.$ Hence for all $n>k$,

$$\begin{aligned} V_{M}^{m}(N_{k+n})=V_{M}^{m}((N_{k+n}:M)M)= V_{M}^{m}((N_{k}:M)M)=V_{M}^{m}(N_{k}). \end{aligned}$$

So for all $n>k$, we have

$$\begin{aligned} N_{k+n}=J_{M}^{m}(N_{k+n})= J_{M}^{m}(N_{k})= N_{k}, \end{aligned}$$

as desired.

(b) First assume that $Max_{R}(M)$ is a connected topological space. Then $Max(\bar{R})=\phi (Max_{R}(M))$ is connected by Lemma 3.1 and Remark 2.2 (a). To see the reverse implication, we assume that $Max(\bar{R})$ is a connected topological space. If $Max_{R}(M)$ is a disconnected topological space, then there exist submodules $N$ and $K$ of $M$ such that

$$\begin{aligned} Max_{R}(M)=V_{M}^{m}(N) \cup V_{M}^{m}(K) \end{aligned}$$

and

$$\begin{aligned} V_{M}^{m}(N) \cap V_{M}^{m}(K)=\emptyset , \end{aligned}$$

where $V_{M}^{m}(N)\ne \emptyset$, and $V_{M}^{m}(K)\ne \emptyset$. Hence as in the proof of Lemma 3.1 (b), we have

$$\begin{aligned} Max(\bar{R})=V_{\bar{R}}^{m}(\overline{(N:M)}) \cup V_{\bar{R}}^{m}(\overline{(K:M)}). \end{aligned}$$

It is easy to cheek that

$$\begin{aligned} V_{\bar{R}}^{m}(\overline{(N:M)}) \cap V_{\bar{R}}^{m}(\overline{(K:M)})=\emptyset , V_{\bar{R}}^{m}(\overline{(N:M)})\ne \emptyset , \,\, and \,\, V_{\bar{R}}^{m}(\overline{(K:M)})\ne \emptyset . \end{aligned}$$

Therefore, $Max(\bar{R})$ is a disconnected topological space, a contradiction. Hence $Max_{R}(M)$ is a connected topological space.

(c) We have similar argument as in part (b).

(d) $(\Rightarrow )$ This follows from Lemma 3.1 (a) and Remark 2.2 (a). To show the converse, let $\{V_{M}^{m}(N_{\alpha }) :\alpha \in \Lambda \}$ be a family of closed subset of $Max_{R}(M)$ such that $\cap _{\alpha \in \Lambda } V_{M}^{m}(N_{\alpha })=\emptyset$, where $N_{\alpha }$ is a submodule of $M$ for every $\alpha \in \Lambda$. Then $\{\phi (V_{M}^{m}(N_{\alpha })) :\alpha \in \Lambda \}$ is a family of closed subset of $Max(\bar{R})$ because $\phi$ is closed by Lemma 3.1 (b). Since $\phi$ is surjective, it is easy to see that $\cap _{\alpha \in \Lambda } \phi (V_{M}^{m}(N_{\alpha }))=\emptyset$. As $Max(\bar{R})$ is quasi compact, there exists a finite subset $\Gamma$ of $\Lambda$ such that $\cap _{\alpha \in \Gamma } \phi (V_{M}^{m}(N_{\alpha }))=\emptyset$. This implies that $\cap _{\alpha \in \Gamma } V_{M}^{m}(N_{\alpha })=\emptyset$ and hence $Max_{R}(M)$ is quasi compact.

(e) Let $Z_0 \supset Z_1 \supset \cdots \supset Z_n$ be a descending chain of irreducible closed subset of $Max_{R}(M)$. Then by Theorem 3.13 (a), for $i$ $(1\le i \le n)$, there exists submodule $L_{i}$ of $M$ such that $(L_{i}:M)$ is a J-radical prime ideal of $R$ and $Z_{i}=V_{M}^{m}(L_{i})$. It follows that

$$\begin{aligned} V_{\bar{R}}^{m}(\overline{(L_{0}:M)}) \supset V_{\bar{R}}^{m}(\overline{(L_{1}:M)}) \cdots \supset V_{\bar{R}}^{m}(\overline{(L_{n}:M)}) \end{aligned}$$

is a descending chain of irreducible closed subset of $Max(\bar{R})$ by Remark 3.12. Hence

$$\begin{aligned} dim(Max_{R}(M))\le dim(Max(\bar{R})). \end{aligned}$$

Now let

$$\begin{aligned} A_0 \supset A_1 \supset \cdots \supset A_t \end{aligned}$$

be a descending chain of irreducible closed subset of $Max(\bar{R})$. By Remark 3.12, for each $i$ $(1\le i \le t)$, there exists a J-radical prime ideal $\bar{p}_{i}$ of $\bar{R}$ such that $A_{i}=V_{\bar{R}}^{m}(\bar{p}_{i})$. This yields that

$$\begin{aligned} p_{0}\subset p_{1}\subset \cdots \subset p_{t} \end{aligned}$$

is an ascending chain of J-radical prime ideal of $R$. Since $M$ is $Max$-surjective, by Lemma 3.11 (b), for every $p_{i}$ $(1\le i \le t)$, there exists a submodule $Q_{i}$ of $M$ such that $p_{i}=(Q_{i}:M)$. Hence by Theorem 3.13 (a),

$$\begin{aligned} V_{M}^{m}(Q_{0}) \supset V_{M}^{m}(Q_{1}) \supset \cdots \supset V_{M}^{m}(Q_{t}) \end{aligned}$$

is a descending chain of irreducible closed subset of $Max_{R}(M)$. It follows that $dim(Max_{R}(M))\ge dim(Max(\bar{R}))$ and the proof is completed. $\square$

Corollary 3.15

Let $M$ be a $Max$-surjective $R$-module. Then the following hold.

• (a) If $R$ is Noetherian, then $Max_{R}(M)$ is a Noetherian topological space.

• (b) If $\Psi$ is the family of all J-radical prime ideal of $R$, then we have

  $$\begin{aligned}dim(Max_{R}(M)) = sup\{n| p_0 \subset p_1 \subset \cdots \subset p_n \,\, is \,\, an \,\, ascending \,\, chain \,\, of \Psi \}. \end{aligned}$$

Proof

• (a) Follows from Theorem 3.14 (a).

• (b) Apply the technique of Theorem 3.14 (e). $\square$

Remark 3.16

We recall that an $R$-module $M$ is a Hilbert module if every prime submodule in $M$ is the intersection of all the maximal submodules containing it. For example, every finitely generated divisible module over an integral domain is a Hilbert module (see [16, p. 2]). Let $M$ be a Hilbert $R$-module. If $Max_{R}(M)$ is connected (resp. irreducible) topological space, then $Spec_{R}(M)$ is connected (resp. irreducible) topological space. Since if $M$ is Hilbert, by [10, Proposition 5.1] it is easy to see that $Cl(Max_{R}(M))=Spec_{R}(M)$. Now the result follows from the Remark 2.2 (f).

Definition 3.17

We say that a topological space $W$ is a $Max$ -spectral space if $W$ is homeomorphic with the maximal ideal space of some ring $S$ (with the topology inherited from $Spec(S)$).

Remark 3.18

$Max$-spectral spaces have been characterized by Hochster [12, p. 57, Proposition 11] as the topological spaces W which satisfy the following conditions:

• (a) W is a $T_1$ space;

• (b) W is quasi compact.

Proposition 3.19

Let $M$ be an $R$-module. Then the following are equivalent.

• (a) $M$ is $Max$-injective.

• (b) $Max_{R}(M)$ is a $T_0$ space.

• (c) $Max_{R}(M)$ is a $T_1$ space.

• (d) $Max_{R}(M)$ is a $T_2$ space.

Proof

Straightforward. $\square$

Corollary 3.20

Let $M$ be an $R$-module.

• (a) If $Max_{R}(M)$ is a $Max$-spectral topological space, then $M$ is $Max$-injective.

• (b) If $M$ is primeful and $Spec_{R}(M)$ is a $Max$-spectral topological space, then $Spec_{R}(M)=Max_{R}(M)$.

Proof

This follows from Remark 3.18, Proposition 3.19, and [2, Theorem 4.3]. $\square$

We recall that a topological space $X$ is spectral if it is homeomorphic to $Spec(S)$ with the Zariski topology for some ring $S$ (see [12]).

Remark 3.21

Let $M=\mathbb Z(p^{\infty })\oplus \mathbb Z$. Then $M$ is a primeful $\mathbb Z$-module and $Spec_{R}(M)$ is a spectral topological space but $Spec_{R}(M)\ne Max_{R}(M)$ by [3], Table of examples 3.1 and 3.2]. This shows that part (b) in Corollary 3.20 is not valid in general if the word “$Max$-spectral” is replaced with “spectral”.

Example 3.22

• (a) $Max_{\mathbb Z}(\mathbb Z_{2} \oplus \mathbb Z_{3})$ is a $Max$-spectral topological space by [3], Table of examples 3.1 and 3.2] and Remark 3.18.

• (b) $Max_{\mathbb Q}(\mathbb Q \oplus \mathbb Q)$ is not a $Max$-spectral topological space because $0 \oplus \mathbb Q$ and $\mathbb Q\oplus 0$ are maximal submodules of the $\mathbb Q$-module $\mathbb Q\oplus \mathbb Q$ with $(0 \oplus \mathbb Q:\mathbb Q\oplus \mathbb Q)=(\mathbb Q \oplus 0:\mathbb Q\oplus \mathbb Q)$, while $0 \oplus \mathbb Q\ne \mathbb Q \oplus 0$. Thus $Max_{\mathbb Q}(\mathbb Q \oplus \mathbb Q)$ is not $Max$-spectral by Corollary 3.20 (a).

Let $M$ be an $R$-module such that $Max_{R}(M)$ is a $Max$-spectral topological space. For a submodule $N$ of $M$, it is natural to ask the following question: Is $Max_{R}(M/N)$ a $Max$-spectral topological space?

In Proposition 3.23 (c), we give a positive answer to this question under some additional conditions.

Proposition 3.23

Let $M$ be an $R$ module and let $N$ be a submodule of $M$. Then the following hold.

• (a) If $Max_{R}(M)$ is a $T_1$ topological space, then so is $Max_{R}(M/N)$.

• (b) If $Max_{R}(M)$ is a Noetherian topological space, then so is $Max_{R}(M/N)$.

• (c) Let $Max_{R}(M)$ be a $Max$-spectral space. Then $Max_{R}(M/N)$ is a $Max$-spectral space in the following cases

  • (i) The subspace $H:=\{Q\in Max_{R}(M)| Q\supseteq N\}$ of $Max_{R}(M)$ is closed;

  • (ii) $R$ is a ring such that the intersection of every infinite collection of maximal ideals of $R$ is zero (for example, when $R$ is PID or one dimensional Noetherian domain).

Proof

(a) Follows from Proposition 3.19 and the fact that if $N$ is a submodule of $M$, then

$$\begin{aligned} Max_{R}(M/N)=\{Q/N| Q\in Max_{R}(M), Q\supseteq N\}. \end{aligned}$$

(b) We define the map $f:Max_{R}(M/N)\rightarrow H$, where $H:=\{Q\in Max_{R}(M)| Q\supseteq N\}$ and $f(Q/N)=Q$ for every $Q/N\in Max_{R}(M/N)$. Clearly $f$ is a bijection map. Now let $V_{M}^{m}(E)\cap H$ be a closed set of $H$, where $E$ is a submodule of $M$. Then

$$\begin{aligned} f^{-1}(V_{M}^{m}(E)\cap H)&= f^{-1}(V_{M}^{m}(E))\cap f^{-1}(H)=f^{-1}(V_{M}^{m}(E))\cap Max_{R}(M/N)\\&= f^{-1}(V_{M}^{m}(E))=V_{M}^{m}(K/N), \end{aligned}$$

where $K=(E:M)M+N$. So $f:Max_{R}(M/N)\rightarrow H$ is a continuous map. It is easy to check that

$$\begin{aligned} f(V_{M}^{m}(L/N))=V_{M}^{m}(L) \cap H \end{aligned}$$

for every submodule $L$ of $M$ containing $N$. Hence $f:Max_{R}(M/N)\rightarrow H$ is a closed map so that $Max_{R}(M/N)$ is homeomorphic with $H$. Now since $Max_{R}(M)$ is Noetherian, $H$ is Noetherian by Remak 2.2 (b). Hence $Max_{R}(M/N)$ is a Noetherian space as desired.

(c)(i) As in the proof part (b), we see that $Max_{R}(M/N)$ is homeomorphic with $H$. Now the result follows by part (a), Remark 3.18, and Remark 2.2 (d).

(c)(ii) This follows from Proposition 3.2, Remark 3.18, Remark 2.2 (b), and part (a). $\square$

The next theorem is an important result about an $R$-module $M$ for which $Max_{R}(M)$ is $Max$-spectral. This result is obtained by combining Lemma 3.1, Proposition 3.2, Theorem 3.6, Proposition 3.19, Remark 2.2 (e), and Remark 3.18.

Theorem 3.24

Let $M$ be a $Max$-injective $R$-module. Then $Max_{R}(M)$ is a $Max$-spectral topological space in each of the following cases:

• (a) $M$ is $Max$-surjective;

• (b) $Im(\phi )$ is quasi compact, where $\phi :Max_{R}(M)\rightarrow Max(\bar{R})$ is the natural map of $Max_{R}(M)$;

• (c) $Ann_{R}(M)$ is a maximal ideal of $R$;

• (d) $Max_{R}(M)$ is a finite set;

• (e) $Max(R)$ is a finite set;

• (f) $Max(\bar{R})$ is Noetherian, in particular when $R$ is Noetherian;

• (g) The intersection of every infinite of maximal ideals of $R$ is zero, in particular when $R$ is PID or one dimensional Noetherian domain;

• (h) The ascending chain condition for J-radical submodules of $M$ holds.

Remark 3.25

Let $M= \oplus _{i \in \mathbb N } \mathbb Z/p_i \mathbb Z$, where $p_i$ is a prime integer for each $i \in \mathbb N$. Then $M$ is an $X$-injective module over $\mathbb Z$ but $Spec_{\mathbb Z}(M)$ is not a spectral topological space by [3], Table of examples 3.1 and 3.2]. This shows that the words “$Max$-injective,” “$Max_{R}(M)$,” and “$Max$-spectral” in part (g) of Theorem 3.24, can not be replaced with “$X$-injective,” “$Spec_{R}(M)$,” and “spectral,” respectively.

An $R$-module $M$ is multiplication if for every submodule $N$ of $M$, there exits an ideal $I$ of $R$ such that $N=IM$ (see [13]).

Corollary 3.26

Let $M$ be an $R$-module. Then $Max_{R}(M)$ is a $Max$-spectral topological space in each of the following cases

• (a) $M$ is finitely generated and multiplication;

• (b) $M$ is primeful and top; (We refer the reader to [13] and [5] for the concept and properties of top modules.

• (c) $M$ is primeful and $X$-injective;

• (d) $M$ is $X$-injective and $R$ is PID.

Proof

This follows from parts (a) and (g) of Theorem 3.24 and taking into account the following facts from [13, Theorem 3.5], [2, Proposition 3.3], [11, Theorem 2.2 3.3], and [1, Proposition 3.3 (c)],

Fact 1. Let denote the class of multiplication, top, $X$-injective, and $Max$-injective modules, respectively by $\Gamma _{1}$, $\Gamma _{2}$, $\Gamma _{3}$, and $\Gamma _{4}$, then

$$\begin{aligned} \Gamma _{1} \subseteq \Gamma _{2} \subseteq \Gamma _{3} \subseteq \Gamma _{4}. \end{aligned}$$

Fact 2. If we denote the class of finitely generated, primeful, and $Max$-surjective modules, respectively by $\Omega _{1}$, $\Omega _{2}$, and $\Omega _{3}$, then

$$\begin{aligned} \Omega _{1} \subseteq \Omega _{2} \subseteq \Omega _{3}. \end{aligned}$$

$\square$

In below, using Corollary 3.26 (d) and [3], Table of examples 3.1], we provide further examples about of modules such that their maximal submodules are $Max$-spectral.

Example 3.27

Let $p$ (resp. $p_i$, $i \in \mathbb N$) be a prime number. Then for each of the following cases, $Max_{\mathbb Z}(M)$ is a $Max$-spectral topological space.

• (a) $M=\mathbb Z_{( p ) }=S^{-1}\mathbb Z$, where $S=\mathbb Z {\setminus } (p)$.

• (b) $M=\mathbb Z(p^{\infty })\oplus \mathbb Z$.

• (c) $M=\oplus _ {i \in \mathbb N} \mathbb Z/p_i \mathbb Z$.

• (d) $M=\prod _ {i \in \mathbb N} \mathbb Z/p_i \mathbb Z$.

• (e) $M=\mathbb Q \oplus (\oplus _{i \in \mathbb N } \mathbb Z/p_i \mathbb Z)$.

Unfortunately, we have not been able to find a $Max$-injective $R$-module $M$ such that $Max_{R}(M)$ is not $Max$-spectral. This motivates the following question.

Question 3.28

Let $M$ be a $Max$-injective $R$-module. Is $Max_{R}(M)$ a $Max$-spectral topological space?