On the Roman domination number of the complement of sum annihilating ideal graph of a reduced ring

The rings considered in this paper are commutative with identity that admit at least one nonzero zero-divisor. Let $R$ be a ring. Let $Z(R)$ denote the set of all zero-divisors of $R$⁠, and let us denote $Z(R)\{0}$ by $Z(R)*$⁠. Motivated by the work of Beck [1], several researchers have introduced graphs with algebraic structures and studied the interplay between the algebraic properties of the algebraic structures and the graph-theoretic properties of the graphs associated with them. The graphs considered in this article are undirected and simple. For a graph $G$⁠, we denote the vertex set of $G$ by $V(G)$ and the edge set of $G$ by $E(G)$⁠. In this paper, we allow graphs to admit an infinite number of vertices. Recall that the zero-divisor graph of $R$⁠, denoted by $Γ(R)$⁠, is an undirected graph with $V(Γ(R))=Z(R)*$ and distinct vertices $x$ and $y$ are adjacent in $Γ(R)$ if and only if $xy=0$ [2]. For an excellent and inspiring survey on the zero-divisor graphs of commutative rings, one can refer to Ref. [3].

With any commutative ring $R$⁠, Anderson and Badawi [4] have introduced and investigated an undirected graph called the total graph of $R$⁠, denoted by $T(Γ(R))$ with $V(T(Γ(R)))=R$ and distinct vertices $x$ and $y$ are adjacent in $T(Γ(R))$ if and only if $x+y∈Z(R)$⁠. For several interesting theorems which illustrate the interplay between the graph-theoretic properties of $T(Γ(R))$ and the ring-theoretic properties of $R$⁠, see Ref. [4]. For an excellent, interesting, and inspiring book on graphs associated with commutative rings, one can refer to Ref. [5]. Anderson et al. [5] have characterized commutative rings according to the properties of graphs constructed over them. They have investigated in detail several graphs such as zero-divisor graphs, generalized zero-divisor graphs, total graphs, generalized total graphs, annihilator graphs, dot product graphs, and more.

Recall that an ideal $I$ of a ring $R$ is said to be an annihilating ideal of $R$ if there exists $r∈R\{0}$ such that $Ir=(0)$ [6]. We denote the set of all annihilating ideals of $R$ by $A(R)$ and $A(R)\{(0)}$ by $A(R)*$⁠. Recall that the annihilating-ideal graph of $R$⁠, denoted by $AG(R)$⁠, is an undirected graph with $V(AG(R))=A(R)*$ and distinct vertices $I$ and $J$ are adjacent in $AG(R)$ if and only if $IJ=(0)$ [6]. For several interesting and inspiring results on $AG(R)$⁠, the reader is referred to Refs. [6, 7].

Motivated by the research work of Anderson and Badawi [4] on the total graph of a commutative ring and the research work of Behboodi and Rakeei [6, 7] on the annihilating-ideal graph of a commutative ring, with $R$⁠, we have introduced and investigated an undirected graph, denoted by $Ω(R)$ such that $V(Ω(R))=A(R)*$ and distinct vertices $I$ and $J$ are adjacent in $Ω(R)$ if and only if $I+J∈A(R)$ [8]. The graph $Ω(R)$ was also investigated in Ref. [9], and the authors of [9] called $Ω(R)$⁠, the sum annihilating ideal graph of $R$⁠. Let $G=(V,E)$ be a simple graph. Recall that the complement of $G$⁠, denoted by $Gc$⁠, is a graph with vertex set $V$ and distinct vertices $u$ and $v$ are adjacent in $Gc$ if and only if they are not adjacent in $G$ [10, Definition 1.2.13]. We [11] studied the interplay between the graph-theoretic properties of $(Ω(R))c$ and the ring-theoretic properties of $R$⁠.

Let $R$ be a ring (⁠$R$ can be an integral domain). We denote the set of all prime ideals of $R$ by $Spec(R)$⁠, the set of all maximal ideals of $R$ by $Max(R)$⁠, and the set of all minimal prime ideals of $R$ by $Min(R)$⁠. For a subset $E$ of $R$⁠, we denote the set ${p∈Spec(R)∣p⊇E}$ by $V(E)$⁠. For any subset $A$ of $R$⁠, we denote the set $A\{0}$ by $A*$⁠. For an element $x∈R$⁠, the annihilator of $x$ in $R$⁠, denoted by $AnnR(x)$ or $Ann(x)$⁠, is defined as $Ann(x)={r∈R∣rx=0}$⁠. We denote the set of all proper ideals of $R$ by $I(R)$ and $I(R)\{(0)}$ by $I(R)*$⁠. We denote the nilradical of $R$ by $Nil(R)$⁠. We say that $R$ is reduced if $Nil(R)=(0)$⁠. If $A$ is a proper subset of a set $B$⁠, then we denote it by $A⊂B$⁠. For any $n∈N\{1}$⁠, we denote the ring of integers modulo $n$ by $Zn$⁠. For definitions and standard results from commutative ring theory, one can refer to Refs. [12–14].

Let $G=(V,E)$ be a graph. Let us recall the following definitions. A set $S⊆V$ is called a dominating set of $G$ if every vertex $u∈V\S$ has a neighbor $v∈S$ [10, Definition 10.2.1]. A $γ$-set of $G$ is a minimum dominating set of $G$⁠; that is, a dominating set of $G$ whose cardinality is minimum [10, Definition 10.2.2]. The domination number of $G$ is the cardinalty of a minimum dominating set of $G$ and it is denoted by $γ(G)$ [10, Definition 10.2.3]. The domination number of graphs associated with commutative rings has been studied by several authors, for example, see Refs. [15–20].

The graph-theoretic properties of Roman domination number of a graph $G=(V,E)$ was investigated by Cockayne et al., see Ref. [21]. For a clear motivation for the concept of Roman domination number of a graph, see [21, p. 12], where Cockayne et al. mentioned that the definition of a Roman dominating function was given implicitly in Refs. [22, 23]. Recall that a Roman dominating function (RDF) on a graph $G=(V,E)$ is a function $f:V→{0,1,2}$ such that every vertex $u$ for which $f(u)=0$ is adjacent to at least one vertex $v$ for which $f(v)=2$ [21, p. 12]. Let $i∈{0,1,2}$⁠. Let us denote ${v∈V∣f(v)=i}$ by $Vi$⁠. Observe that $V=⋃i=02Vi$ with $Vi∩Vj=∅$ for all distinct $i,j∈{0,1,2}$⁠. We denote $f$ by $f=(V0,V1,V2)$⁠. The domination number of $Ω(R)$ (respectively, $(Ω(R))c$⁠) was determined in Ref. [19], where we had proved that $γ(Ω(R))<∞$⁠, but there are several reduced rings such that $(Ω(R))c$ does not admit any finite dominating set, and if $|Min(R)|<∞$⁠, then $γ((Ω(R))c)<∞$⁠. If $R$ is not reduced and $Z(R)$ is an ideal of $R$⁠, then we were able to characterize $R$ such that $γ((Ω(R))c)<∞$⁠. However, if $R$ is not reduced and $Z(R)$ is not an ideal of $R$⁠, the problem of characterizing $R$ such that $γ((Ω(R))c)<∞$ was left open, see [19, Section 3]. As $R$ can have an infinite number of nonzero annihilating ideals and the vertex set of $(Ω(R))c$ is $A(R)*$⁠, in this paper, we consider graphs $G=(V,E)$ which admit at least one Roman dominating function $f=(V0,V1,V2)$ such that both $V1$ and $V2$ are finite. Note that a graph $G=(V,E)$ admits a Roman dominating function $f=(V0,V1,V2)$ such that both $V1$ and $V2$ are finite if and only if $γ(G)<∞$⁠. Hence in this paper, we restrict ourselves to reduced rings $R$ such that $γ((Ω(R))c)<∞$ and determine $γR((Ω(R))c)$⁠. We use RDF to denote Roman dominating function. We denote the cardinality of a set $A$ by $|A|$⁠.

Let $G=(V,E)$ be a graph such that $γ(G)<∞$⁠. Let $f=(V0,V1,V2)$ be an RDF on $G$ such that both $V1$ and $V2$ are finite. Then, the weight of $f$ is the value $f(V)=∑v∈Vf(v)$⁠. The minimum weight of a Roman dominating function on $G$ is called the Roman domination number of $G$⁠, denoted by $γR(G)$⁠. We say that an RDF $f=(V0,V1,V2)$ on $G$ is a $γR$-function if $f(V)=γR(G)$⁠, see [21, p. 13].

Throughout this paper, we use $R$ to denote a ring. Unless otherwise specified, we assume that $R$ is reduced, $A(R)*≠∅$ and $γ((Ω(R))c)<∞$⁠. This paper aims to compute $γR((Ω(R))c)$⁠. This paper consists of three sections including the introduction. In Section 2, we state and prove some preliminary results that are used in proving the main results of this paper. In Section 3, we try to compute $γR((Ω(R))c)$⁠. As $|Min(R)|<∞$ is a sufficient condition for the domination number of $(Ω(R))c$ to be finite, see [19, Section 3], we assume that $|Min(R)|<∞$⁠, and try to determine $γR((Ω(R))c)$⁠. If $|Min(R)|=2$⁠, then $γR((Ω(R))c)∈{2,3,4}$ (Proposition 3.2); $γR((Ω(R))c)=2$ if and only if $R$ is ring-isomorphic to $R1×R2$⁠, where $Ri$ is an integral domain for each $i∈{1,2}$ with $Ri$ is a field for at least one $i∈{1,2}$ (Proposition 3.3); $γR((Ω(R))c)≠3$ (Proposition 3.4); $γR((Ω(R))c)=4$ if and only if $Min(R)∩Max(R)=∅$ (Remark 3.5). Let $|Min(R)|=n$ for some $n∈N$ with $n≥3$⁠. If $Min(R)∩Max(R)=∅$⁠, then $γR((Ω(R))c)=2n$⁠. If $n=3$⁠, and if $|Min(R)∩Max(R)|≥2$⁠, then $γR((Ω(R))c)=4$ and $R$ is ring-isomorphic to $D×F1×F2$⁠, where $D$ is an integral domain (⁠$D$ can be a field) and $Fi$ is a field for each $i∈{1,2}$⁠. If $n=3$⁠, and if $|Min(R)∩Max(R)|=1$⁠, then $γR((Ω(R))c)=5$⁠, and $R$ is ring-isomorphic to $T×F$⁠, where $T$ is a reduced ring with $|Min(T)|=2$⁠, $Min(T)∩Max(T)=∅$⁠, and $F$ is a field. If $n≥4$⁠, and $|Min(R)∩Max(R)|≥1$⁠, then $γR((Ω(R))c)=2n−1$⁠, and $R$ is ring-isomorphic to $T×F$⁠, where $T$ is a reduced ring with $|Min(T)|=n−1$⁠, $Min(T)∩Max(T)$ can be nonempty, and $F$ is a field (Theorem 3.22). Examples 3.7 and 3.23 illustrate the results proved in this section.

In this paper, we consider graphs $G=(V,E)$ with no restriction on $|V|$ and such that they admit at least one Roman dominating function $f=(V0,V1,V2)$ with both $V1$ and $V2$ are finite. The graph $G$ satisfies the property that it admits at least one Roman dominating function $f=(V0,V1,V2)$ with both $V1$ and $V2$ are finite if and only if $γ(G)<∞$⁠. In this section, we state and prove some preliminary results that are used for proving the main results of this paper.

Proof. Assume that $γ(G)=2$⁠. We obtain from Remark 2.1 that $2≤γR(G)$⁠. By Remark 2.2, $γR(G)≤4$⁠. Let $D={v1,v2}$ be a dominating set of $G$⁠. As $|V|≥3$ by assumption, there exists $v∈V\D$⁠. Hence, $v$ and $vi$ are adjacent in $G$ for some $i∈{1,2}$⁠. Thus, $G$ admits at least one edge. So, $γR(G)≥γ(G)+1=3$ by Remark 2.3. Therefore, $3≤γR(G)≤4$⁠. □

Proof. Assume that $|V|≥3$ and $γ(G)=2$⁠. By Lemma 2.5, $3≤γR(G)≤4$⁠. Note that $γR(G)=3$ if and only if there exists a $γR$-function $f=(V0,V1,V2)$ on $G$ such that the weight of $f$ equals 3, if and only if $|V1|+2|V2|=3$⁠, if and only if either $V2=∅=V0$⁠, and $|V1|=3$ or $|V1|=|V2|=1$⁠, and $V0=V\(V1∪V2)$⁠. □

For a graph $G$⁠, we denote the degree of a vertex $v$ in $G$ by $deg(v)$⁠.

1. $γR(G)=3$⁠.

2. There exists a $γR$-function $f=(V0,V1,V2)$ on $G$ such that $|V1|=|V2|=1$⁠, and $V0=V\(V1∪V2)$⁠.

3. There exists $x∈V$ such that $deg(x)=1$ in $Gc$⁠.

Proof. By hypothesis, $|V|≥4$ and $γ(G)=2$⁠.

$(1)⇔(2)$ If there exists a $γR$-function $f=(V0,V1,V2)$ on $G$ such that $V2=V0=∅$⁠, and $|V1|=3$⁠, then $|V|=3$⁠, since $V=⋃i=02Vi$⁠, contradicting the assumption that $|V|≥4$⁠. Therefore, we obtain from Lemma 2.6 that $γR(G)=3$ if and only if there exists a $γR$-function $f=(V0,V1,V2)$ on $G$ such that $|V1|=|V2|=1$⁠, and $V0=V\(V1∪V2)$⁠.

$(2)⇒(3)$ Assume that there exists a $γR$-function $f=(V0,V1,V2)$ on $G$ such that $|V1|=|V2|=1$⁠, and $V0=V\(V1∪V2)$⁠. Let $V2={x}$⁠, and let $V1={y}$⁠. As $y$ and $x$ are not adjacent in $G$ by [21, Proposition 3$(b)$], it follows that $y$ and $x$ are adjacent in $Gc$⁠. Since $f$ is a $γR$-function on $G$⁠, each element of $V0$ is adjacent to $x$ in $G$⁠. Therefore, $deg(x)=1$ in $Gc$⁠.

$(3)⇒(1)$ Assume that there exists $x∈V$ such that $deg(x)=1$ in $Gc$⁠. Hence, there exists $y∈V$ such that $y$ and $x$ are adjacent in $Gc$⁠, but $z$ and $x$ are adjacent in $G$ for all $z∈V\{x,y}$⁠. Let $V1={y}$⁠, $V2={x}$⁠, and let $V0=V\(V1∪V2)$⁠. Then, $f=(V0,V1,V2)$ is an RDF on $G$ and the weight of $f$ equals $|V1|+2|V2|=3$⁠. Therefore, $γR(G)≤3$⁠. As $γR(G)≥3$ by Lemma 2.5, we obtain that $γR(G)=3$⁠. □

Proof. By hypothesis, $R$ is a reduced ring. We obtain from [12, Proposition 1.8] and [14, Theorem 10] that $⋂p∈Min(R)p=(0)$⁠. Let $I∈A(R)*$⁠. Then, $Ir=(0)$ for some $r∈R\{0}$⁠. As $r≠0$⁠, we obtain that $r∉p$ for some $p∈Min(R)$⁠. From $Ir=(0)⊂p$⁠, it follows that $I⊆p$⁠. □

Proof. By hypothesis, $R$ is a reduced ring with $|Min(R)|=n$ for some $n∈N\{1}$⁠. Let $Min(R)={pi∣i∈{1,2,…,n}}$⁠. Then, $⋂i=1npi=(0)$ and $Z(R)=⋃i=1npi$⁠.

Let $i∈{1,2,…,n}$⁠. Let us denote ${1,2,…,n}\{i}$ by $Ai$⁠. Since distinct minimal prime ideals of a ring are not comparable under inclusion, it follows that $pi⊉⋂j∈Aipj$⁠. Hence, we can find $xi∈(⋂j∈Aipj)\pi$⁠. Observe that $xi≠0$ and $pixi=(0)$⁠. Thus, $pi∈A(R)$⁠. As $pi≠(0)$⁠, we get that $pi∈A(R)*$⁠. From $pixi=(0)$⁠, it follows that $pi⊆Ann(xi)$⁠. If $r∈Ann(xi)$⁠, then $rxi=0$⁠. From $xi∉pi$⁠, we obtain that that $r∈pi$⁠. Therefore, $Ann(xi)⊆pi$⁠. So, $pi=Ann(xi)$⁠. □

Proof. By hypothesis, $R$ is a reduced ring with $|Min(R)|=n$ for some $n∈N\{1}$⁠. Let $Min(R)={pi∣i∈{1,2,…,n}}$⁠. Let $W$ be a nonempty proper subset of $Min(R)$ such that $|{I∈A(R)*∣V(I)∩Min(R)=W}|<∞$⁠. Let $|W|=t$⁠. Then, $1≤t<n$⁠. Without loss of generality, we can assume that $W={pk∣k∈{1,…,t}}$⁠. Note that $Min(R)\W={pt+j∣j∈{1,…,n−t}}$⁠. Since distinct minimal prime ideals of a ring are not comparable under inclusion, it follows that $⋂k=1tpk⊈⋃j=1n−tpt+j$ by [12, Proposition 1.11$(i)$]. Let $x∈(⋂k=1tpk)\(⋃j=1n−tpt+j)$⁠. Let $j∈{1,…,n−t}$⁠. We claim that $pt+j∈Max(R)$⁠. Since $pt+j≠R$⁠, there exists $mt+j∈Max(R)$ such that $pt+j⊆mt+j$ by [12, Corollary 1.4]. Suppose that $pt+j≠mt+j$⁠. Then, $pt+j⊂mt+j$⁠. Observe that $mt+j⊈⋃i=1npi$ by [12, Proposition 1.11$(i)$]. Let $y∈mt+j\(⋃i=1npi)$⁠. From $Z(R)=⋃i=1npi$⁠, we get that $y∉Z(R)$⁠. So, $ym∉Z(R)$ for all $m∈N$⁠. By the choice of $x$ and $y$⁠, it follows that $Rxym∈A(R)*$ and $V(Rxym)∩Min(R)=W$ for all $m∈N$⁠. As $|{I∈A(R)*∣V(I)∩Min(R)=W}|<∞$ by assumption, we can find $m1,m2∈N$ with $m1<m2$ such that $Rxym1=Rxym2$⁠. So, $xym1=rxym2$ for some $r∈R$⁠. Hence, $xym1(1−rym2−m1)=0∈pt+j$⁠. As $xym1∉pt+j$⁠, we obtain that $1−rym2−m1∈pt+j⊂mt+j$⁠. Thus, $rym2−m1,1−rym2−m1∈mt+j$⁠, so $1=1−rym2−m1+rym2−m1∈mt+j$⁠, a contradiction, since $mt+j≠R$⁠. Therefore, $pt+j=mt+j∈Max(R)$ for each $j∈{1,…,n−t}$⁠. □

Proof. By hypothesis, $R$ is reduced. As $⋂p∈Min(R)p=(0)$ and $R$ is not an integral domain, we get that $|Min(R)|≥2$⁠.

Assume that $|A(R)*|=2$⁠. Then, we obtain from [6, Theorem 1.1] that $R$ is Artinian. By [12, Proposition 8.1], we get that $Spec(R)=Max(R)=Min(R)$ and $|Max(R)|<∞$ by [12, Proposition 8.3]. As each member of $Max(R)$ belongs to $A(R)*$ by Lemma 2.9 and $|A(R)*|=2$ by assumption, we obtain that $|Max(R)|=2$⁠. Let $Max(R)={mi∣i∈{1,2}}$⁠. From $m1+m2=R$ and $⋂i=12mi=(0)$⁠, it follows that the mapping $ϕ:R→Rm1×Rm2$ given by $ϕ(r)=(r+m1,r+m2)$ is an isomorphism of rings by [12, Proposition 1.10$(ii)$ and $(iii)$]. Let $i∈{1,2}$⁠. Let us denote $Rmi$ by $Fi$⁠. Then, $F1$ and $F2$ are fields and $R$ is ring-isomorphic to $F1×F2$⁠.

Conversely, assume that $R$ is ring-isomorphic to $F1×F2$⁠, where $Fi$ is a field for each $i∈{1,2}$⁠. Let us denote $F1×F2$ by $T$⁠. Since (0) and $F$ are the only ideals of any field $F$⁠, it follows that $I(T)*=A(T)*={(0)×F2,F1×(0)}$⁠. Therefore, $|A(T)*|=2$⁠, so $|A(R)*|=2$⁠.□

Proof. By hypothesis, $R$ is a reduced ring with $|Min(R)|=n$ for some $n∈N$ $(n≥2)$⁠. Assume that $|Min(R)∩Max(R)|≥1$⁠. Let $Min(R)={pi∣i∈{1,2,…,n}}$⁠. We can assume without loss of generality that $pn∈Max(R)$⁠. Note that $⋂i=1npi=(0)$⁠. Since distinct minimal prime ideals of a ring are not comparable under inclusion, it follows that $⋂k=1n−1pk⊈pn$⁠. So, $(⋂k=1n−1pk)+pn=R$⁠. Since $(⋂k=1n−1pk)∩pn=(0)$⁠, we obtain that the mapping $ϕ:R→R⋂k=1n−1pk×Rpn$ given by $ϕ(r)=(r+⋂k=1n−1pk,r+pn)$ is an isomorphism of rings by [12, Proposition 1.10$(ii)$ and $(iii)$]. Let us denote $⋂k=1n−1pk$ by $I$⁠, $RI$ by $T$⁠, and $Rpn$ by $F$⁠. Thus, $R$ is ring-isomorphic to $T×F$⁠. Since $I$ is a radical ideal of $R$⁠, we get that $T$ is a reduced ring. As $Min(T)={pkI∣k∈{1,…,n−1}}$⁠, it follows that $|Min(T)|=n−1$⁠. Observe that $F$ is a field, since $pn∈Max(R)$⁠. □