On the Annihilator Submodules and the Annihilator Essential Graph

Abstract

Let R be a commutative ring and let M be an R-module. For a ∈ R, Ann_M(a) = {m ∈ M : am = 0} is said to be an annihilator submodule of M. In this paper, we study the property of being prime or essential for annihilator submodules of M. Also, we introduce the annihilator essential graph of equivalence classes of zero divisors of M, AE_R(M), which is constructed from classes of zero divisors, determined by annihilator submodules of M and distinct vertices [a] and [b] are adjacent whenever Ann_M(a) + Ann_M(b) is an essential submodule of M. Among other things, we determine when AE_R(M) is a connected graph, a star graph, or a complete graph. We compare the clique number of AE_R(M) and the cardinal of m −Ass_R(M).

1 Introduction

Throughout this paper, R is a commutative ring with non-zero identity and all modules are unitary. Let M be an R-module. A proper submodule P of M is said to be prime if rm ∈ P for r ∈ R and m ∈ M, implies that m ∈ P or $r\in \text {Ann}_{R}(M/P)=\{ r\in R : rM\subseteq P \}$. Let Spec_R(M) denote the set of prime submodules of M. For a ∈ R we call Ann_M(a) = {m ∈ M : am = 0} the annihilator submodule of a in M. Let m −Ass_R(M) = {P ∈Spec_R(M) : P = Ann_M(a), for some 0≠a ∈ R}. The properties of prime submodules and m −Ass_R(M) are studied in [8, 9] and [4]. By [8, Proposition 3.2], any maximal element of {Ann_M(a) : a∉Ann_R(M)} is a prime submodule of M. Thus, m −Ass_R(M) is a non-empty set, when M is a Noetherian R-module. In Section 2, we study some properties of the elements of m −Ass_R(M). In particular, we show that Ann_M(a) = {m ∈ M|rm ∈Ann_R(aM)M for some r ∉Ann_R(aM)} whenever Ann_M(a) is a prime submodule of M and a∉r(Ann_R(M)). Also, we compare m −Ass_R(M) and the set of associated prime ideals of R, Ass_R(R), and we show that:

$$\mathrm{m}-\text{Ass}_{R}(M)=\left\{ \text{Ann}_{M}(a) | \text{Ann}_{R}(a) \in \text{Ass}_{R}(R) \right\}, $$

where M is either a free or a faithful multiplication R-module.

There are many studies of various graphs associated to rings or modules (see for instance [3, 5, 6, 10]). A submodule N of M is called an essential submodule if it has a non-zero intersection with any other non-zero submodule of M. In the third section, we investigate the property of being essential for an annihilator submodule, Ann_M(a), in two cases, a ∈ r(Ann_R(M)) = {r ∈ R : r^tM = 0 for some positive integer t} or a∉r(Ann_R(M)). We prove that, if Ann_M(a), Ann_M(b) ∈m −Ass_R(M), then Ann_M(a) + Ann_M(b) is an essential submodule of M. By relying on this fact, we introduce the annihilator essential graph of equivalence classes of zero divisors of M, AE_R(M), which is constructed from classes of zero divisors, determined by annihilator submodules and distinct vertices [a] and [b] are adjacent whenever Ann_M(a) + Ann_M(b) is an essential submodule of M. Among other things, we determine when AE_R(M) is a connected graph, a star graph, or a complete graph. An aspect of AE_R(M) is the connection to elements of m −Ass_R(M). We compare the clique number of AE_R(M) and the cardinal number of m −Ass_R(M) under the additional assumption r(Ann_R(M)) = 0 or r(Ann_R(M)) = Ann_R(M)≠ 0.

The zero-divisor graph determined by equivalence classes, Γ_E(R), was introduced in [10], and further studied in [2, 7, 11]. We shall compare Γ_E(R) and AE_R(R) to determine some properties of the ring R.

Let Γ be a (undirected) graph. We say that Γ is connected if there is a path between any two distinct vertices. For vertices x and y of Γ, we define d(x, y) to be the length of a shortest path between x and y, if there is no path, then $\mathrm {d}(x,y) = \infty$. The diameter of Γ is

$$\text{diam}({\Gamma}) =\sup\left\{\mathrm{d}(x,y) | x \text{ and } y \text{ are vertices of } {\Gamma}\right\}. $$

The girth of Γ, denoted by gr(Γ), is the length of a shortest cycle in Γ ($\text {gr}({\Gamma }) = \infty$ if Γ contains no cycle). A graph Γ is complete if any two distinct vertices are adjacent. The complete graph with n vertices is denoted by K_n (we allow n to be an infinite cardinal). The clique number, ω(Γ), is the greatest integer n > 1 such that $K_{n}\subseteq {\Gamma }$, and $\omega ({\Gamma })=\infty$ if $K_{n}\subseteq {\Gamma }$ for all integers n ≥ 1.

2 Annihilators Which Are Prime Submodules

Let R be a commutative ring and M be an R-module. In this section, we investigate the primeness of annihilator submodules of M.

Theorem 1

LetM be a NoetherianR-module withr(Ann_R(M))≠Ann_R(M).Then, there existsa ∈ r(Ann_R(M)) suchthat Ann_M(a) isa prime submodule of M.

Proof

Assume that a ∈ r(Ann_R(M)) ∖Ann_R(M). If Ann_M(a) is a maximal element of X = {Ann_M(b) : b∉Ann_R(M)}, then [8, Proposition 3.2] shows that Ann_M(a) is a prime submodule of M and we are done. Otherwise, there exists b ∈ R such that Ann_M(b) is a maximal element of X and $\text {Ann}_{M}(a)\subseteq \text {Ann}_{M}(b)$. We show that b ∈ r(Ann_R(M)). By assumption a ∈ r(Ann_R(M)), so that there is an integer t such that $0=a^{t}M\subseteq \text {Ann}_{M}(b)$. Thus, abM = 0, since Ann_M(b) is a prime submodule. Hence, $b\in \text {Ann}_{R}(aM)\subseteq \text {Ann}_{R}(bM)$ and so b²M = 0 which implies that b ∈ r(Ann_R(M)). □

The following example shows that Ann_M(a) can be a prime submodule of M but a∉r(Ann_R(M)).

Example 1

Let $\Bbb Z_{p^{2}q}$ be the ring of integers modulo p²q for some prime integers p, q. Then, $\text {Ann}_{\Bbb Z}(\Bbb Z_{p^{2}q})=p^{2}q\Bbb Z$, $r(\text {Ann}_{\Bbb Z}(\Bbb Z_{p^{2}q}))=pq\Bbb Z$ and $\text {Ann}_{\Bbb Z_{p^{2}q}}(p^{2})=q\Bbb Z_{p^{2}q}$ is a prime submodule of $\Bbb Z_{p^{2}q}$ while $\Bbb Zp^{2}\not \in r(\text {Ann}_{\Bbb Z}(\Bbb Z_{p^{2}q}))$.

Lemma 1

Ifa ∈ R,then$\text {Ann}_{R}(M/ \text {Ann}_{M}(a)) =\text {Ann}_{R}(aM)$.

Proof

If $a\in \text {Ann}_{R}(M)$, there is nothing to prove. Thus, assume that $a\not \in \text {Ann}_{R}(M)$ and $r\in \text {Ann}_{R}(M/ \text {Ann}_{M}(a))$. Then, $rM\subseteq \text {Ann}_{M}(a)$ and so arM = 0. Hence, r ∈Ann_R(aM). The converse is similar. □

Theorem 2

Let Ann_M(a) bea prime submodule ofM anda∉r(Ann_R(M)).Then,$\text {Ann}_{M}(a)$ = {m ∈ M : rm ∈Ann_R(aM)M for some r ∈ R ∖Ann_R(aM)} andit is a minimal prime submodule of M.

Proof

By assumption and Lemma 1, $\text {Ann}_{R}(M/ \text {Ann}_{M}(a))=\text {Ann}_{R}(aM)=\frak p$ is a prime ideal of R. Let $H:=\{m\in M : rm\in {\frak p}M \text {for some } r\not \in {\frak p} \}$ and m ∈ H. Then, there exists $s\in R\setminus {\frak p}$ such that $sm\in {\frak p}M$. This implies that $sm={\sum }_{i = 1}^{k}s_{i}m_{i}$, where $s_{i}\in \frak p$. Thus, $sam={\sum }_{i = 1}^{k}s_{i}am_{i}= 0$ and so $sm\in \text {Ann}_{M}(a)$. Hence, by assumption and $s\notin \frak p$, it follows that m ∈Ann_M(a). Therefore, $H\subseteq \text {Ann}_{M}(a)$. Let m ∈Ann_M(a). Then, $am= 0\in \frak pM$. If $a\notin \frak p$, we are done. Otherwise, a²M = 0 and so a ∈ r(Ann_R(M)), contrary to assumption. Thus, m ∈ H.

Assume that P is a prime submodule of M and $P\subseteq \text {Ann}_{M}(a)$. Let m ∈Ann_M(a). Then, am = 0 ∈ P which implies that a ∈Ann_R(M/P) or m ∈ P. If $aM\subseteq P\subseteq \text {Ann}_{M}(a)$, then a²M = 0 and so a ∈ r(Ann_R(M)); it is a contradiction. Hence, m ∈ P and P = Ann_M(a) which implies that Ann_M(a) is a minimal prime submodule of M. □

Corollary 1

Let$\text {Ann}_{R}(M)=\frak p$bea prime ideal of R and$a\not \in \frak p$.Then, the following statements are true:

• (i) Ann_R(aM) = Ann_R(M).

• (ii) If Ann_M(a) isa prime submodule of M, then Ann_M(a) = {m ∈ M : rm = 0 for $\text {some } r\in R\setminus \frak p \}=\cup _{b\not \in \frak p}\text {Ann}_{M}(b)$.

• (iii) |m −Ass_R(M)|≤ 1.

Proof

(i) It is clear that $\text {Ann}_{R}(M) \subseteq \text {Ann}_{R}(aM)$. To establish the reverse inclusion, let r ∈Ann_R(aM). Then, arM = 0 and so ar ∈Ann_R(M). By assumption, Ann_R(M) is a prime ideal of R and a∉Ann_R(M), thus r ∈Ann_R(M). Hence, $\text {Ann}_{R}(aM) \subseteq \text {Ann}_{R}(M)$.

(ii) It follows by (i) and Theorem 2.

(iii) It follows by (ii).

□

The following lemma shows that there is a natural injective map from

$$\text{Spec}_{R}(M)\cap \{\text{Ann}_{M}(a) : a\not\in r(\text{Ann}_{R}(M))\} \longrightarrow \text{Spec}(R) \cap \{\text{Ann}_{R}(aM) : a\in R\} $$

given by $\text {Ann}_{M}(a)\rightarrow \text {Ann}_{R}(aM)$.

Lemma 2

Let Ann_M(a) and Ann_M(b) beprime submodules ofM anda, b∉r(Ann_R(M)).Then, Ann_M(a) = Ann_M(b) ifand only if Ann_R(aM) = Ann_R(bM).

Proof

In view of Lemma 1, if Ann_M(a) = Ann_M(b), then Ann_R(aM) = Ann_R(bM). For the converse, assume that m ∈Ann_M(a). Thus, am = 0 and am ∈Ann_M(b). If m ∈Ann_M(b), we are done. Otherwise, a ∈Ann_R(bM) = Ann_R(aM) which implies that a ∈ r(Ann_R(M)) contrary to assumption. Thus, m ∈Ann_M(b). □

The following result shows that the above injective map from Spec_R(M) ∩{Ann_M(a) : a∉r(Ann_R(M))} to Spec(R) ∩{Ann_R(aM) : a ∈ R}, could be a bijection.

An R-module M is called a multiplication module if for each submodule N of M, N = IM for some ideal I of R. Multiplication module has been studied in [1].

Theorem 3

LetM be either a free or a faithful multiplication module anda ∈ R. Then,Ann_M(a) isa prime submodule ofM if and only ifAnn_R(a) is a prime ideal ofR. Inparticular,

$$\mathrm{m}-\text{Ass}_{R}(M)=\{\text{Ann}_{M}(a) : \text{Ann}_{R}(a) \in \text{Ass}_{R}(R)\}. $$

Proof

Assume that M is a free R-module, thus M≅ ⊕_i∈IR_i (R_i = R), where I is an index set. Let a ∈ R and Ann_R(a) be a prime ideal of R. It is easy to see that Ann_M(a)≅ ⊕_i∈IAnn_R(a). Let rm ∈Ann_M(a) and r∉Ann_R(aM) = Ann_R(a) for some r ∈ R, m = (m_i)_i∈I ∈ M. Thus, rm_i ∈Ann_R(a) and so m_i ∈Ann_R(a), for all i ∈ I. Hence, m ∈Ann_M(a) and Ann_M(a) is a prime submodule of M. By the same argument, the converse follows.

By [1, Corollary 2.11], Ann_M(a) is a prime submodule of M if and only if Ann_R(aM) is prime ideal of R. On the other hand, Ann_R(aM) = Ann_R(a) since M is faithful. Thus, Ann_M(a) is a prime submodule of M if and only if Ann_R(a) is a prime ideal of R. □

Theorem 4

LetM be a projective module anda ∈ R.If Ann_R(a) isa prime ideal ofR, then Ann_M(a) isa prime submodule of M. Furthermore, |Ass_R(R)|≤∣m −Ass_R(M)∣,wheneverM is a faithful projective module.

Proof

By assumption, there exists a free R-module F and an R-module A such that F ≅M ⊕ A. By assumption and Theorem 3, Ann_F(a) is a prime submodule of F. Let x ∈ M, r ∈ R, and rx ∈Ann_M(a). Then, arx = 0 and so ar(x,0) = 0. Thus, r(x,0) ∈Ann_F(a). Hence, r ∈Ann_R(aF) or (x,0) ∈Ann_F(a). Therefore, $r\in \text {Ann}_{R}(a(M\oplus A))\subseteq \text {Ann}_{R}(aM)$ or x ∈Ann_M(a). □

3 The Annihilator Essential Graph of Zero Divisors

Recall that R is a commutative ring and M is an R-module. A submodule N of M is called an essential submodule if it has a non-zero intersection with any other non-zero submodule of M. In this section, we investigate the essentialness of the annihilator submodules of M and we introduce the annihilator essential graph of equivalence classes of zero divisors of M, AE_R(M), which is constructed from classes of zero divisors, determined by annihilator submodules of M.

Theorem 5

LetM be anR-module. Then, the following statements are true: (i) For alla ∈ R,aM + Ann_M(a) isan essential submodule of M. (ii) Ifa ∈ r(Ann_R(M)),then Ann_M(a) isan essential submodule of M. (iii) Ifa∉r(Ann_R(M)) and Ann_M(a) isa prime submodule of M, then Ann_M(a) isnot an essential submodule of M.

Proof

(i) Let a ∈ R. We have to show that aM + Ann_M(a) is an essential submodule of M. Let N be a submodule of M. Then, $aN\subseteq aM\cap N\subseteq (aM+ \text {Ann}_{M}(a))\cap N$. If (aM + Ann_M(a)) ∩ N = 0, then aN = 0 which implies that $N\subseteq \text {Ann}_{M}(a)$. Hence, $N\subseteq (aM+ \text {Ann}_{M}(a))\cap N$ and so N = 0. Therefore, aM + Ann_M(a) is an essential submodule of M.

(ii) By assumption, there exists an integer t, such that a^tM = 0. Thus, $aM\subseteq \text {Ann}_{M}(a^{t-1})$ and so $aM+ \text {Ann}_{M}(a)\subseteq \text {Ann}_{M}(a^{t-1})$. Hence, Ann_M(a^t− 1) is an essential submodule of M by (i). Suppose that N is a non-zero submodule of M. Then, Ann_M(a^t− 1) ∩ N≠ 0 and so there is 0≠x ∈ N such that a^t− 1x = 0. Thus, $0\neq a^{i}x\in N \cap \text {Ann}_{M}(a)$, for some i with 0 ≤ i ≤ t − 2. Hence, Ann_M(a) is an essential submodule of M.

(iii) By assumption, a∉r(Ann_R(M)) so aM≠ 0. Let am ∈ aM ∩Ann_M(a). Then, by hypotheses, m ∈Ann_M(a) which shows that am = 0 and so aM ∩Ann_M(a) = 0. Thus, Ann_M(a) is not an essential submodule of M.

□

Theorem 6

LetM be anR-module anda, b ∈ R.Then, the following statements are true: (i) If eitherabM = 0 orAnn_M(a),Ann_M(b) areprime submodules of M, then Ann_M(a) + Ann_M(b) isan essential submodule of M. (ii) If Ann_M(a),Ann_M(b) areprime submodules ofM anda, b∉r(Ann_R(M)),thenabM = 0.

Proof

(i) Let a, b ∈ R and abM = 0. Then, $bM\subseteq \text {Ann}_{M}(a)$ and so $bM+\text {Ann}_{M}(b)\subseteq \text {Ann}_{M}(a)$ + Ann_M(b). Thus, by using Theorem 5 (i), the assertion follows.

Assume that a, b ∈ R and Ann_M(a),Ann_M(b) are prime submodules of M. If either a ∈ r(Ann_R(M)) or b ∈ r(Ann_R(M)), then by Theorem 5 (ii), either Ann_M(a) or Ann_M(b) is an essential submodule of M and the assertion follows. Thus, assume that a, b∉r(Ann_R(M)). Without loss of generality, we can assume that $\text {Ann}_{M}(a)\not \subseteq \text {Ann}_{M}(b)$; see Theorem 2. Thus, there is m ∈ M such that am = 0 and bm≠ 0. By am ∈Ann_M(b), it follows that a ∈Ann_R(bM). Hence, abM = 0 and the result follows by previous paragraph.

(ii) We suppose that abM≠ 0 and look for a contradiction. By (i), Ann_M(a) + Ann_M(b) is an essential submodule of M. Thus, (Ann_M(a) + Ann_M(b)) ∩ abM≠ 0. Hence, there are m ∈ M, $m^{\prime }\in \text {Ann}_{M}(a)$ and $m^{\prime \prime }\in \text {Ann}_{M}(b)$ such that abm≠ 0 and $abm=m^{\prime }+m^{\prime \prime }$. Thus, a²b²m = 0. By assumption, Ann_M(a),Ann_M(b) are prime submodules of M, thus Ann_M(a) = Ann_M(a²) and Ann_M(b) = Ann_M(b²). Therefore, abm = 0 that is a contradiction. □

Corollary 2

Leta, b∉r(Ann_R(M)) and Ann_M(a) bea prime submodule of M. Then,Ann_M(a) + Ann_M(b) isan essential submodule ofM if and onlyif$\text {Ann}_{M}(b) \not \subseteq \text {Ann}_{M}(a)$.

Proof

If $\text {Ann}_{M}(b) \not \subseteq \text {Ann}_{M}(a)$, then by a similar argument to that of Theorem 6 (ii), one can show that abM = 0 and so Ann_M(a) + Ann_M(b) is an essential submodule of M. Conversely, assume that $\text {Ann}_{M}(b) \subseteq \text {Ann}_{M}(a)$. Thus, Ann_M(a) is an essential submodule of M and so Ann_M(a) ∩ aM≠ 0. Now, by a similar argument to that of Theorem 6 (ii), we achieve a contradiction. □

Assume Z(M) denotes the set of zero divisors of M and Z(M)^∗ = Z(M) ∖{0}. For a, b ∈ R, we say that $a\sim b$ if and only if Ann_M(a) = Ann_M(b). As noted in [10], $\sim$ is an equivalence relation. If [a] denotes the class of a, then [0] = Ann_R(M) and [a] = R ∖ Z(M) for all a ∈ R ∖ Z(M); the other equivalence classes form a partition of Z(M).

Definition 1

The annihilator essential graph of equivalence classes of zero divisors of M, denoted AE_R(M), is a graph associated to M whose vertices are the classes of elements of Z(M)^∗, and each pair of distinct classes [a] and [b] are adjacent if and only if Ann_M(a) + Ann_M(b) is an essential submodule of M.

The following remark which we include for the reader’s convenience is based on the Theorems 5 and 6 and Corollary 2.

Remark 1

Let M be an R-module and let [a],[b] be two distinct vertices of AE_R(M). Then, the following statements hold: (i) If a ∈ r(Ann_R(M)), then [a] is a universal vertex of AE_R(M). (ii) If abM = 0, then [a],[b] are adjacent in AE_R(M). (iii) If Ann_M(a),Ann_M(b) are prime submodules of M, then [a],[b] are adjacent in AE_R(M). (iv) Let a, b∉r(Ann_R(M)) and Ann_M(a) be a prime submodule of M. Then, [a],[b] are adjacent in AE_R(M) if and only if $\text {Ann}_{M}(b) \not \subseteq \text {Ann}_{M}(a)$.

The following example shows that AE_R(M) can be an empty graph.

Example 2

Consider M = ℤ × ℤ₆ as a ℤ-module. Thus, Ann_M(2) = 0 × 3ℤ₆, Ann_M(3) = 0 × 2ℤ₆ and Ann_M(6) = 0 × ℤ₆. Hence, AE_ℤ(M) is an empty graph with vertices [2, 3] and [6].

Theorem 7

LetM be a NoetherianR-module. Then,AE_R(M) isnot a connected graph if and only if$r({\text {Ann}_{R}(M)})= 0$andthere exists an elementa ∈ Z(M)^∗suchthat$\text {Ann}_{M}(a) \subseteq \cap _{P\in \mathrm {m}-\text {Ass}_{R}(M)} P$.

Proof

(⇒) Assume that AE_R(M) is not connected. Thus, r(Ann_R(M)) = 0, otherwise, by Remark 1 (i), AE_R(M) has a universal vertex which is a contradiction. Also, Remark 1 (iii) implies that the induced subgraph of elements of m −Ass_R(M) is complete; hence, by hypothesis, there exists a ∈ Z(M)^∗ such that Ann_M(a) is not prime and [a] is not joint with any element of m −Ass_R(M). Therefore, $\text {Ann}_{M}(a) \subseteq \cap _{P\in \mathrm {m}-\text {Ass}_{R}(M)} P$ by Remark 1 (iv).

(⇐) Let r(Ann_R(M)) = 0 and let there be an element a in Z(M)^∗ such that $\text {Ann}_{M}(a) \subseteq \cap _{P\in \mathrm {m}-\text {Ass}_{R}(M)} P$. We have to show that [a] is an isolated vertex in AE_R(M). By hypothesis and Remark 1 (iv), [a] is not adjacent with any element of $\mathrm {m}-\text {Ass}_{R}(M)$. On the other hand, if $\text {Ann}_{M}(b)\not \in \mathrm {m}-\text {Ass}_{R}(M)$, then $\text {Ann}_{M}(b)\subseteq \text {Ann}_{M}(c)$, for some c ∈ R with $\text {Ann}_{M}(c)\in \mathrm {m}-\text {Ass}_{R}(M)$. Thus, $\text {Ann}_{M}(a)+ \text {Ann}_{M}(c)$ is not an essential submodule of M. Hence, Ann_M(a) + Ann_M(b) is not an essential submodule of M, and so [a] and [b] are not adjacent. Therefore, [a] is an isolated vertex and AE_R(M) is not connected. □

Corollary 3

LetM be a NoetherianR-module. Then,AE_R(M) isa connected graph if and only if eitherr(Ann_R(M))≠ 0 orfor eacha ∈ Z(M)^∗thereexistsP ∈m −Ass_R(M) suchthat$\text {Ann}_{M}(a) \not \subseteq P$.

Theorem 8

LetM be a NoetherianR-module. IfAE_R(M) isa connected graph, thendiamAE_R(M) ≤ 3.

Proof

If r(Ann_R(M))≠ 0, then AE_R(M) has a universal vertex and therefore we have diamAE_R(M) ≤ 2. Now, let r(Ann_R(M)) = 0 and [a],[b] are two distinct vertices of AE_R(M). By assumption, AE_R(M) is connected, so there exist $a^{\prime },b^{\prime }\in Z(M)^{*}$ such that $\text {Ann}_{M}(a^{\prime }),$$\text {Ann}_{M}(b^{\prime })$ are prime submodules of M and $\text {Ann}_{M}(a)\not \subseteq \text {Ann}_{M}(a^{\prime })$ and $\text {Ann}_{M}(b)\not \subseteq \text {Ann}_{M}(b^{\prime })$. Thus, by Remark 1 (iv), if $[a^{\prime }]=[b^{\prime }]$, then $[a]-[a^{\prime }]-[b]$ is a path, and if $[a^{\prime }]\neq [b^{\prime }]$, then $[a]-[a^{\prime }]-[b^{\prime }]-[b]$ is a path and so diamAE_R(M) ≤ 3. □

To see that the bound is sharp, notice that equality holds for ℤ₃₀. It is easy to see that $\mathrm {diam } AE_{\Bbb Z_{30}}(\Bbb Z_{30})= 3$.

Corollary 4

If R is Noetherian, thenAE_R(R) isconnected and diamAE_R(R) ≤ 3.

Proof

It is an immediate consequence of Theorems 7 and 8. □

Theorem 9

LetM be a Noetherian module and letAE_R(M) bea connected graph. IfAE_R(M) hasa cycle, then gr(AE_R(M)) ≤ 4.

Proof

If r(Ann_R(M))≠ 0 or ∣m −Ass_R(M)∣ ≥ 3, then the result follows by Remark 1 (i) and (iii). Assume that r(Ann_R(M)) = 0 and ∣m −Ass_R(M)∣ ≤ 2. Thus, we have ∣m −Ass_R(M)∣ = 2 since AE_R(M) is connected. Suppose that Ann_R(a) and Ann_R(b) are two prime submodules of M and [c] is an arbitrary vertex of AE_R(M). Then, [c] is adjacent to [a] or [b], by Corollary 3 and Remark 1 (iv). Thus, we have a cycle of length at most four and therefore gr(AE_R(M)) ≤ 4. □

Theorem 10

LetM be a NoetherianR-module. Then, the following statements are true: (i) Ifr(Ann_R(M)) = 0,thenω(AE_R(M)) = ∣m −Ass_R(M)∣. (ii) Ifr(Ann_R(M)) = Ann_R(M)≠ 0,thenω(AE_R(M)) = ∣m −Ass_R(M)∣ + 1.

Proof

(i) Let n := ω(AE_R(M)) and k := ∣m −Ass_R(M)∣. Then, by Remark 1 (iii), k ≤ n. If Ann_M(a) is not a prime submodule, then there is an annihilator prime submodule Ann_M(b) such that [a] and [b] are not adjacent by Remark 1 (iv), where a, b ∈ Z(M)^∗. Assume that H is a maximal clique and k < n. So, there is a vertex of m −Ass_R(M) which is not a vertex of H. Assume that H contains k − t vertices of m −Ass_R(M), where t > 0. Thus, at least t + 1 vertices of H are adjacent to k − t vertices of m −Ass_R(M) and could not be adjacent to all other t vertices of m −Ass_R(M). Hence, there exist at least two vertices [c],[d] of H that are not adjacent to a vertex [e] of m −Ass_R(M). Then, $\text {Ann}_{M}(c)+ \text {Ann}_{M}(d) \subseteq \text {Ann}_{M}(e)$ and Ann_M(e) is not essential, by Theorem 5 (iii). Thus, [c],[d] are not adjacent and it is a contradiction. Hence, k = n.

(ii) Assume that a ∈Ann_R(M). Thus, [a] is a universal vertex of AE_R(M). Now, by a similar argument to that of (i), the result follows. □

Theorem 11

LetM be a NoetherianR-module andr(Ann_R(M)) = 0.Then, the following statements are equivalent: (i) AE_R(M) isa complete graph; (ii) Ann_M(a) isa prime submodule of M, for alla ∈ Z(M)^∗; (iii) AE_R(M) = K₁orK₂.

Proof

(i) ⇒ (ii) Assume that a ∈ Z(M)^∗. If Ann_M(a) is a maximal element of X = {Ann_M(b) : b∉Ann_R(M)}, then it is a prime submodule of M and the result follows. Otherwise, there is b ∈ Z(M)^∗ such that Ann_M(b) is a maximal element of X and $\text {Ann}_{M}(a) \subseteq \text {Ann}_{M}(b)$. Thus, [a] is not adjacent to [b] by Remark 1 (iv) contrary to assumption. Hence, Ann_M(a) is a prime submodule of M, for all a ∈ Z(M)^∗.

(ii) ⇒ (i) It is obvious by Remark 1 (iii).

(i) ⇒ (iii) Assume that a, b, c ∈ Z(M)^∗ and assume [a],[b],[c] are all distinct vertices of AE_R(M). Thus, in view of (i) ⇔ (ii) and Theorem 6 (ii), we have ab = ac = bc = 0. Hence, a(b + c) = 0 and so b + c ∈ Z(M)^∗. Therefore, [b + c] is a vertex of AE_R(M). If [b + c] and [b] are two distinct vertices of AE_R(M), then by assumption [b + c] and [b] are adjacent and so as above 0 = b(b + c) = b² which implies that b ∈ r(Ann_R(M)) = 0; this is a contradiction. Now, assume that [b + c] = [b]. Thus, Ann_M(b + c) = Ann_M(b). So that, $cM\subseteq \text {Ann}_{M}(b)=\text {Ann}_{M}(b+c)$. Thus, (b + c)cM = 0 and so c² = 0. It leads to a similar contradiction. Hence, AE_R(M) has at most two vertices.

(iii) ⇒ (i) It is obvious. □

Theorem 12

LetM be anR-module and Ann_R(M)≠ 0.Then, the following statements are true: (i) IfAE_R(M) isa star graph with more than two vertices, thenAnn_R(M) isa prime ideal ofR. (ii) IfM is Noetherian and Ann_R(M) isa prime ideal ofR, thenAE_R(M) isa star graph.

Proof

(i) Assume that AE_R(M) is a star graph with center vertex [c], where c ∈Ann_R(M). We have to show that Ann_R(M) is a prime ideal of R. Assume that a, b ∈ R and abM = 0. If either aM = 0 or bM = 0, we are done. Otherwise, with condition Ann_M(a)≠Ann_M(b), [a] and [b] are adjacent in AE_R(M), by Remark 1 (ii). So that we can assume that [a] = [c] and then Ann_M(a) = Ann_M(c) which implies that a ∈Ann_R(M), a contradiction. Hence, Ann_M(a) = Ann_M(b). Thus, a²M = 0 and so a ∈ r(Ann_R(M)) and [a] is the center vertex of AE_R(M) by Remark 1 (i) is again a contradiction. Therefore, either aM = 0 or bM = 0 and Ann_R(M) is a prime ideal of R.

(ii) In view of Corollary 1, ∣m −Ass_R(M)∣ = 1. Let Ann_M(a) be the unique element of m −Ass_R(M) and so the unique maximal element of X = {Ann_M(b) : b∉Ann_R(M)}. Then, Ann_M(a) is not an essential submodule of M, by Theorem 5. Assume that b, c ∈ Z(M) ∖Ann_R(M). Thus, $\text {Ann}_{M}(b)+\text {Ann}_{M}(c)\subseteq \text {Ann}_{M}(a)$ and so Ann_M(b) + Ann_M(c) is not essential submodule of M. Hence, [b] and [c] are not adjacent in AE_R(M). Therefore, all vertices of AE_R(M) are adjacent to [d] where d ∈Ann_R(M) that implies AE_R(M) is a star graph. □

The annihilator essential graph AE_R(R) and the graph of equivalence classes of zero divisors Γ_E(R) have the same vertex sets (see [10]). Moreover, if [a],[b] are adjacent vertices of Γ_E(R), then ab = 0, so by Remark 1 (ii), [a],[b] are adjacent in AE_R(R). Therefore, Γ_E(R) is a subgraph of AE_R(R). Now, the question arises as when these graphs are the same?

Theorem 13

If R is a reduced ring, thenAE_R(R) = Γ_E(R).

Proof

Assume that [a],[b] are adjacent in AE_R(R). Then, Ann_R(a) + Ann_R(b) is an essential ideal of R. If ab≠ 0, then (Ann_R(a) + Ann_R(b)) ∩ abR≠ 0. Thus, there exist r ∈ R, s ∈Ann_R(a) and t ∈Ann_R(b) such that 0≠abr = s + t. Hence, a²b²r = 0, and so abr ∈Nil(R) = 0 is a contradiction. Thus, ab = 0. The converse is true by the previous paragraph. □

Let p be a prime number and $R= \Bbb Z_{p^{3}}$. Then, $AE_{R}(R)={\Gamma }_{E}(R)= K_{2}$, but $\text {Nil}(R)\neq 0$. So that the converse of Theorem 13 is not true necessarily.

Theorem 14

Let R be a Noetherian non-reduced ring andAE_R(R) = Γ_E(R).Then, the following statements are true: (i) Z(R) = Ann_R(a),for somea ∈Nil(R). (ii) If Ann_R(a) isan essential ideal ofR, thena ∈Nil(R). (iii) Nil(R) = r(Ann_R(Z(R))).

Proof

(i) Assume that 0≠x ∈Nil(R) thus Ann_R(x) is an essential ideal of R, see Theorem 5 (ii), and so x is a universal vertex of AE_R(R). Hence, by assumption, either $\text {Ann}_{R}(x)=\text {Ann}_{R}(y)$ or xy = 0, for all y ∈ Z(R). We first show that Z(R) is an ideal of R. Let a, b ∈ Z(R). It is enough to show that a + b ∈ Z(R). If $\text {Ann}_{R}(b)=\text {Ann}_{R}(a)$, then we are done. Thus, assume that $\text {Ann}_{R}(b)\neq \text {Ann}_{R}(a)$. If $\text {Ann}_{R}(x)= \text {Ann}_{R}(a)$, then there is an integer n such that $x^{n}a = 0$ and xⁿ≠ 0. Moreover xb = 0 since $\text {Ann}_{R}(x)\neq \text {Ann}_{R}(b)$. Hence, xⁿ(a + b) = 0 and so a + b ∈ Z(R). If $\text {Ann}_{R}(x)\neq \text {Ann}_{R}(a)$ and Ann_R(x)≠Ann_R(b), then xa = xb = 0 which implies that x(a + b) = 0 and a + b ∈ Z(R). Thus, Z(R) is an ideal of R. We have $Z(R)= \cup _{\text {Ann}_{R}(x)\in \text {Ass}_{R}(R)}\text {Ann}_{R}(x)$. Thus, by Prime Avoidance Theorem $Z(R)=\text {Ann}_{R}(a)$, for some a ∈ Z(R). Furthermore, by a ∈ Z(R) = Ann_R(a), it follows that a² = 0 and a ∈Nil(R).

(ii) Let a ∈ R and assume that Ann_R(a) is an essential ideal of R. If Ann_R(a) = Ann_R(a²), then Ann_R(a) ∩ Ra = 0. Thus, by hypotheses Ra = 0 and so a = 0 ∈Nil(R). Now, assume that Ann_R(a)≠Ann_R(a²). Thus, [a],[a²] are two distinct adjacent vertices of AE(R). Hence, by assumption, a³ = 0 and therefore a ∈Nil(R).

(iii) Let a ∈ r(Ann_R(Z(R))). Then, aⁿZ(R) = 0, for some integer n, and so a^n+ 1 = 0 since a ∈ Z(R). Hence, a ∈Nil(R) and therefore $r(\text {Ann}_{R}(Z(R))) \subseteq \text {Nil}(R)$. Assume that Z(R) is generated by $a_{1}, a_{2}, \dots , a_{n}$ and a ∈Nil(R). Then, by hypotheses and Remark 1 (i), for all $i = 1,2,\dots , n$, either aa_i = 0 or Ann_R(a_i) = Ann_R(a). If aa_i = 0 for all $i = 1,2,\dots , n$, then a ∈Ann_R(Z(R)). Now, assume that there is j with 1 ≤ j ≤ n such that Ann_R(a_j) = Ann_R(a). Thus, a^ka_j = 0 for some integer k. Thus, a^ka_i = 0, for all $i = 1,2,\dots , n$ that implies a ∈ r(Ann_R(Z(R))). So, $\text {Nil}(R)\subseteq r(\text {Ann}_{R}(Z(R)))$. □

Theorem 15

Let R be a Noetherian ring and letAE_R(R) bea graph with at least four vertices. Then,AE_R(R) isnot a cycle graph.

Proof

Let AE_R(R) be a cycle graph with vertices [a₁],⋯ ,[a_n]. Then, [11, Proposition 1.8] implies that AE_R(R)≠Γ_E(R) so we can assume that [a₁],[a_n] are not adjacent in Γ_E(R). Thus, in view of [11, Corollary 3.3], it follows that Ann_R(a₂),Ann_R(a_n− 1) are prime ideal of R. Hence, [a₂],[a_n− 1] are adjacent in AE_R(R), by Remark 1 (iii). So that n = 4 and moreover Γ_E(R) is the path [a₁] − [a₂] − [a₃] − [a₄]. Now, a₁a₄ ∈ Z(R)^∗ and Ann_R(a₁a₄)≠Ann_R(a_i), for all i with 1 ≤ i ≤ 4 since a₁a₃≠ 0, a₂a₄≠ 0 and ${a_{2}^{2}}\neq 0$. Therefore, AE_R(R) is not a cycle graph. □

Theorem 16

IfM is a faithful multiplicationR-module, thenAE_R(R) andAE_R(M) areisomorphic.

Proof

Let a, b ∈ Z(M); we show that Ann_R(a) = Ann_R(b) if and only if Ann_M(a) = Ann_M(b). Let Ann_R(a) = Ann_R(b). Then, Ann_R(aM) = Ann_R(bM) since M is faithful. Thus, Ann_M(a) = Ann_R(aM)M = Ann_R(bM)M = Ann_M(b) since M is multiplication. Conversely, if Ann_M(a) = Ann_M(b), then Ann_R(aM) = Ann_R(bM) by Lemma 1. So that Ann_R(a) = Ann_R(b). Now, in view of [1, Theorem 2.13], Ann_R(a) + Ann_R(b) is an essential ideal of R if and only if Ann_M(a) + Ann_M(b) is an essential submodule of M. Therefore, AE_R(R) and AE_R(M) are isomorphic. □