Nonessential sum graph of an Artinian ring Open Access

The growth of interdisciplinary study of graph and algebra took place after the introduction of zero-divisor graph by Istvan Back [1]. Some of the interesting graphs are comaximal graph of commutative ring [2], intersection graph of ideals of rings [3], total graph of commutative ring [4], etc. In [5], Atani et al. introduced a graph associated to proper nonsmall ideals of a commutative ring, namely, small intersection graph. The small intersection graph of a ring R, denoted by G(R), is an undirected graph with vertex set is the collection of all nonsmall proper ideals of R and any two distinct vertices are adjacent if and only if their intersection is not small in R. Taking this insight of small intersection graph of a ring, we, in this paper, define nonessential sum graph of an Artinian ring.

To continue this sequel, we are going to remember some definitions and notations from ring and graph. Let R be a commutative ring with unity. An ideal I of R is said to be essential in R if $I∩J≠0$⁠, whenever J is a nonzero ideal of R. The sum of all minimal ideals of R is known as socle of R, denoted by $soc(R)$⁠. We use $min(R)$ to denote the collection of all minimal ideals of R. The ring R is said to be an Artinian ring if every descending chain of R terminates. In an Artinian ring, every ideal contains a minimal ideal.

Let G be an undirected simple graph with vertex set $V(G)$ and edge set $E(G)$⁠. G is said to be a null graph if $V(G)=φ$ and that G is said to be empty if $E(G)=φ$⁠. We denote degree of $v∈V(G)$ by $deg(v)$⁠. If $deg(v)=1$⁠, then v is called an end vertex. G is complete if any two vertices are adjacent. G is said to be r-regular if degree of each vertex of G is r. A walk in G is an alternating sequence of vertices and edges, $v0x1v1…xnvn$ in which each edge $xi$ is $vi−1vi$⁠. A closed walk has the same first and last vertices. A path is a walk in which all vertices are distinct; a circuit is a closed walk which all its vertices are distinct (except the first and last). The length of a circuit is the number of edges in the circuit. The length of the smallest circuit of G is called the girth of G, denoted by $girth(G)$⁠. G is connected if there is a path between every two distinct vertices. G is disconnected if it is is not connected. A vertex of the connected graph G is said to be a cut vertex if removal of it makes G disconnected. If x and y are two distinct vertices of G, then $d(x,y)$ is the length of the shortest path from x to y and if there is no such path then $d(x,y)=∞$⁠. The diameter of G is the maximum distance among distances between all pair of vertices of G, denoted by $diam(G)$⁠. G is said to be a bipartite graph if the vertex set of G can be partitioned into two disjoint subsets $V1$ and $V2$ such that every edge of G joins $V1$ and $V2$⁠. If $|V1|=m$⁠, $|V2|=n$ and if every vertex of $V1$ (or $V2$⁠) is adjacent to all vertices of $V2$⁠, then the bipartite graph is said to be complete and is denoted by $Km,n$⁠. If either m or n is equal to 1, then $Km,n$ is said to be a star. An r-partite graph is a graph whose vertex set is partitioned into r subsets with no edge has both ends in any one subset. If each vertex of a partite subset is joined to every vertex that is not in that partite subset, then the r-partite graph is said to be complete. A complete subgraph of G is called a clique. The number of vertices in the largest clique of G is called the clique number of G, denoted by $ω(G)$⁠. The neighborhood $N(v)$ of a vertex v in G is the set of vertices which are adjacent to v. For each $S⊆V(G)$⁠, $N(S)=∪v∈SN(v)$ and $N[S]=N(S)∪S$⁠. A set of vertices S in G is a dominating set, if $N[S]=V$⁠. The domination number, $γ(G)$⁠, of G is the minimum cardinality of a dominating set of G. An independent set of G is a set of vertices of G such that no two vertices are adjacent in that vertex set. The independence number of G is the number of vertices in the largest independence set in G, denoted by $α(G)$⁠.

In this paper, we introduce nonessential sum graph of commutative ring with unity. Let R be a commutative ring with unity. The nonessential sum graph of R, denoted by $NES(R)$⁠, is an undirected graph with vertex set as the collection of all nonessential ideals of R and any two vertices A and B are adjacent if and only if $A∩B$ is also a nonessential ideal of R. In this article, we are mainly interested in nonessential sum graph of Artinian ring.

Any undefined terminology can be obtained in [7–8, 15–20].

In this section, we obtain some results related to connectedness, diameter, girth, completeness, cut vertex, partiteness and regular character. We start with a remark.

Proof. If possible suppose $K=∑μmi$ is an essential ideal of R. Since each $mj≠(0)$⁠, so $K∩mj≠(0)$ for $j∉μ$⁠, which implies that $mj⊆K$⁠. But it is a contradiction by Remark 2.1. Hence the lemma. □

From this onwards, R is an Artinian ring.

Proof. First consider that $NES(R)$ is a null graph. On the contrary, assume that $m1$ and $m2$ are two distinct minimal ideals of R. So $m1∩m2=0$ and this provides that both $m1$ and $m2$ are nonessential ideals of R, a contradiction. Conversely, suppose that R has exactly one minimal ideal m, say. If m is the only nontrivial proper ideal of R, then obviously $NES(R)$ is a null graph. If A is a nontrivial proper ideal of R with $A≠m$⁠, then it is easy to observe that A is essential in R. The proof is complete.$□$

Proof. Let $NES(R)$ be an empty graph. Then by Theorem 2.3 $|min(R|≠1$min (R|≠1. If $|min(R)|≥3$ and $m1,m2,m3∈min(R)$⁠, then $m1$ and $m2$ are adjacent by Lemma 2.2. Therefore, $|min(R)|=2$ and so we take $|min(R)|={m1,m2}$ with $m1≠m2$⁠. Clearly $m1$ and $m2$ are nonessential. If I is any other nonessential ideal which is different from $m1$ and $m2$⁠, then $mi⊂I$ for $i=1,2$⁠. This gives that I and $mi$ are adjacent, a contradiction. Thus $m1$ and $m2$ are the only nonessential ideals of R. For the other direction, we consider R has exactly two minimal ideals, which are the only nonessential ideals of R. Then $m1+m2=soc(R)$ is essential. So, $NES(R)$ is an empty graph. This completes the proof.$□$

Proof.$(i)⇒(ii)$ Suppose that $NES(R)$ is disconnected. We consider $G1$ and $G2$ are two components of $NES(R)$ and I, J be two ideals such that $I∈V(G1)$ and $J∈V(G2)$⁠. Take the minimal ideals $m1$ and $m2$ with $m1⊆I$ and $m2⊆J$⁠. If $m1=m2$⁠, then $I−m1−J$ is a path, a contradiction. This asserts that $m1≠m2$⁠. Again, if $|min(R)|≥3$⁠, then $m1+m2$ is nonessential in R. From this we get $I−m1−m2−J$ is a path, a contradiction. Therefore $|min(R)|=2$⁠.

$(ii)⇒(iii)$ Assume that $|min(R)|=2$⁠. Then we obtain $soc(R)=m1+m2$⁠, where $m1$ and $m2$ are the minimal ideals of R. Let $Gi={I⊆R:mi⊆IandIisnonessentialinR}$⁠. Let I and J be two nonadjacent vertices in $G1$⁠, then $I+J$ is essential in R, which implies $soc(R)⊆I+J$⁠. Hence $m2⊆I$ or $m2⊆J$⁠, a contradiction because in that case either I is essential or J is essential. So, $G1$ is complete subgraph of $NES(R)$⁠. In the same way, $G2$ is also a complete subgraph of $NES(R)$⁠. Suppose K and L are two adjacent vertices where $K∈V(G1)$ and $L∈V(G2)$⁠. Since $soc(R)=m1+m2⊆K+L$⁠, so $K+L$ is essential, a contradiction. Thus $NES(R)=G1∪G2$⁠, where $G1$ and $G2$ are two disjoint complete subgraphs of $NES(R)$⁠.

$(iii)⇒(i)$ The proof is obvious.

Proof. If $NES(R)$ is disconnected then $diam(NES(R))=∞$⁠. Suppose that $NES(R)$ is connected. If I and J are two nonadjacent vertices of $NES(R)$ then $I+J$ is essential in R. Consider the minimal ideals $m1$ and $m2$ with $m1⊆I$ and $m2⊆J$⁠. If $m1+J$ is nonessential, then $I−m1−J$ is a path, which gives $d(I,J)=2$⁠. Similarly, if $m2+I$ is nonessential in R, then $d(I,J)=2$⁠. Suppose that $m1+J$ and $m2+I$ are both essential in R. Since $NES(R)$ is connected, so $|min(R)|≥3$⁠. Let $m3∈min(R)$⁠. Since $I+J$ is essential in R, therefore $m3⊆I+J$⁠. This implies $m3⊆I$ or $m3⊆J$⁠. If we take $m3⊆I$ then obviously $m3+I$ is nonessential in R. We assert that $m3+J$ is nonessential. If possible, $m3+J$ is essential in R, then $m1⊆soc(R)⊆m3+J$⁠, which gives $m1⊆J$⁠. Hence $m1+J=J$ is nonessential, a contradiction. Therefore $I−m3−J$ is a path. Thus $diam(I,J)=2$⁠. $□$

Proof. First if we consider $|min(R)|=2$⁠, then by Theorem 2.5 $NES(R)=G1∪G2$⁠, where $G1$ and $G2$ are two disjoint complete subgraphs of $NES(R)$⁠. Therefore in this case, $girth(NES(R))=3$⁠, whenever $NES(R)$ contains a cycle. Next, when $|min(R)|≥3$⁠, $m1+m2,m2+m3,m3+m1$ are nonessential in R where $mi∈min(R),i=1,2,3$⁠. Thus $m1−m2−m3−m1$ is a cycle. Hence $girth(NES)(R)=3$⁠.$□$

1. There exists no vertex in $NES(R)$ which is adjacent to every other vertex.

2. $NES(R)$ is not a complete graph.

Proof. To prove (i), let $min(R)={m1,m2,…,mt}$⁠. Assume that there exists a vertex I in $NES(R)$ such that I is adjacent to every other vertex. Let $mi⊆I$ for some i. Let $K=∑j≠imj$⁠, which is nonessential in R. Thus K is a vertex in $NES(R)$⁠. Now, $K+I⊇∑j≠i+mi=soc(R)$⁠. Hence $K+I$ is essential, a contradiction to the fact that I is adjacent to every other vertex. Hence the result.

(ii) Clearly $NES(R)$ is not complete by (i). $□$

Proof. On the contrary assume that I is a cut vertex of $NES(R)$⁠. Then $NES(R)∖{I}$ is disconnected. Thus, there are vertices J and K with I lies in every path joining K to J. By Theorem 2.6, $d(K,J)=2$ and therefore $J−I−K$ is a path. We claim that I is a minimal ideal of R. If not, there exists an ideal L of R such that $L⊆I$⁠. As I is nonessential in R, therefore L is also nonessential in R. Since $J+L⊆J+I$ and $J+I$ is nonessential in R, so $J+L$ is nonessential in R. In the same direction, $K+L$ is also nonessential in R. So, $J−L−K$ is a path in $NES(R)∖{I}$⁠, which is a contradiction. Thus, I is a minimal ideal of R. Now, we assert that there exist a minimal ideal $mi≠I$ of R such that $mi⊈J$⁠. If not then $mi⊆J$ for each $I(≠mi)∈min(R)$ and so $∑mi≠Imi⊆J$⁠. This gives that $soc(R)=I+∑mi≠Imi⊆I+J$⁠, a contradiction to the fact that $I+J$ is nonessential. Similarly, there exists $mj(≠I)$ such that $mj⊈K$⁠. Now we see that for each $mt∈min(R)$ either $mt⊆J$ or $mt⊆K$⁠. Since $J+K$ is essential, $mt⊂soc(R)⊆J+K$⁠, which implies $mt⊆J$ or $mt⊆K$⁠. Let $I≠mi,mj∈min(R)$ such that $mi⊈J$ and $mj⊈K$⁠. Therefore, $mi⊆K$ and $mj⊆J$⁠. So, $K−mi−mj−J$ is a path in $NES(R)∖{I}$⁠, a contradiction. Therefore, $NES(R)$ has no cut vertex.$□$

Proof. If possible assume that $NES(R)$ is a complete r-partite graph with r parts $V1,V2,…,Vr$⁠. Since two minimal ideals are always adjacent, by Remark 2.1, so each $Vi$ contains at most one minimal ideal. Thus we get $|min(R)|≤r$⁠. Our claim is $|min(R)|=r$⁠. Suppose $min(R)={m1,m2,…,mt}$ and $t<r$⁠. Without loss of generality we can take $mi∈Vi$ for $1≤i≤t$⁠. So, $Vt+1$ contains no minimal ideal. Since $min(R)$ is finite, so $∑j≠imj$ is nonessential in R. Now, $∑j≠imj+mi=soc(R)$⁠, so $∑j≠imj$ and $mi$ are not adjacent. Thus $∑j≠imj∈Vi$ as $mi∈Vi$⁠. Let $I∈Vt+1$ and $mk⊆I$ for some $mk∈min(R)$⁠. So, I is adjacent to $mk$⁠. Since $NES(R)$ is assumed to be complete r-partite and $mk∈Vk$⁠, so I is adjacent to every element of $Vk$⁠, which implies I is adjacent to $∑i≠kmi$⁠, a contradiction. Therefore, $|min(R)|=r$⁠. Now, consider $J=∑i=3rmi$⁠. Clearly J is nonessential in R by Remark 2.1. As J is adjacent to $m1$ and $m2$⁠, so $J∉V1,V2$⁠. Moreover, $J+mi=J$ for $3≤i≤r$⁠. So, J is adjacent to all minimal ideals of R. We get that $J∉Vi$ for each i, a contradiction. Hence the theorem.$□$

Proof. (i) Let I be an end vertex of $NES(R)$⁠. So, $deg(I)=1$⁠. Suppose $|min(R)|≥3$⁠. For each $mi∈min(R)$⁠, $mi$ is adjacent to every other minimal ideal of R, so $deg(mi)≥2$⁠. Hence I is not a minimal ideal. We can assume $m1⊆I$⁠. Hence I and $m1$ are adjacent. Since $deg(I)=1$⁠, so the only vertex adjacent to I is $m1$ and $mj⊈I,j≠1$⁠. Again I and $m2$ are not adjacent, so $I+m2$ is essential. So we get, $mj⊆soc(R)⊆I+m2$ for $j≠1,2$⁠, which implies $mj⊆I$ for $j≠1$⁠, a contradiction. So, $|min(R)|=2$⁠. By Theorem 2.5, $NES(R)=G1∪G2$⁠, where $G1$ and $G2$ are two disjoint complete subgraphs of $NES(R)$⁠. Let $I∈V(Gi)$⁠. Since $Gi$ is a complete subgraph and $deg(I)=1$⁠, so $|V(Gi)|=2$⁠. The converse part is clear.

(ii) Suppose that $NES(R)$ is a star graph. So, $NES(R)$ contains an end vertex. By the previous part $|min(R)|=2$ and then by Theorem 2.5, the graph is disconnected. Hence, $NES(R)$ is not a star graph. $□$

Proof. (i) Suppose I and J are two vertices of $NES(R)$ such that $I⊆J$⁠. Let K be a vertex adjacent to J. So, $J+K$ is nonessential in R. As $I+K⊆J+K$ , so $I+K$ is nonessential in R. Thus, each vertex adjacent to J is also adjacent to I. Hence $deg(I)≥deg(J)$⁠.

(ii) Let $NES(R)$ be an r-regular graph. So, for each $mi∈min(R)$⁠, $deg(mi)=r$⁠. Since $mi$ is adjacent to each minimal ideal, by Remark 2.1, so $min(R)$ is finite. Suppose, $|min(R)|≥3$⁠, so $deg(m1+m2)≤deg(m1)$ by (i). Also, $deg(m1+m2)≠deg(m1)$⁠, since $∑j≠2mj$ is adjacent to $m1$ but not to $m1+m2$⁠. Thus, $deg(m1+m2)<deg(m1)$⁠, a contradiction. So, $|min(R)|≤2$⁠. If $|min(R)|=1$ then $NES(R)$ is null. Therefore, $|min(R)|=2$⁠. By Theorem 2.5, $NES(R)=G1∪G2$⁠, where $G1$ and $G2$ are two disjoint complete subgraphs of $NES(R)$⁠. Let $min(R)={m1,m2}$ and $m1∈G1$⁠. Since $deg(m1)=r$⁠, so $|G1|=r+1$⁠. In the same direction, $|G2|=r+1$⁠. Hence, $|V(NES(R))|=2r+2$⁠. $□$

In this section, we will find clique number, independence number, domination number of $NES(R)$⁠.

Proof. (i) Since any two minimal ideals of R are adjacent, by Lemma 2.2, the subgraph with vertex set ${mi}mi∈min(R)$ of $NES(R)$ is complete. So, $ω(NES(R))≥|min(R)|$⁠.

(ii) If $ω(NES(R))<∞$⁠, then by (i) the number of minimal ideals of R is finite.

(iii) It is clear from Theorem 2.4.

(iv) Let $min(R)={m1,m2,…,mt}$ and for each $1≤i≤t$⁠, take $Ai={m1,m2,…,mi−1,mi+1,…,mt}$⁠. Let $P(Ai)$ be the power set of $Ai$⁠. For each $X(≠{})∈P(Ai)$⁠, consider $RX=∑T∈XT$⁠. Clearly $TX$ is nonessential. Also, subgraph with vertex set ${RX}X∈P(Ai)$ is a complete subgraph which is clear by Lemma 2.2. Now, $|P(Ai)∖{}|=2|min(R)|−1−1$⁠. Therefore $|{RX}X∈P(Ai)|=2|min(R)|−1−1$⁠. Hence, $ω(NES(R))≥2|min(R)|−1−1$⁠.$□$

Proof. (i) Since $NES(R)$ is not a null graph, $|min(R)|≥2$⁠. Consider $T={m1,m2}$⁠, where $m1,m2∈min(R)$⁠. Take a vertex I in $NES(R)$⁠. If $m1⊆I$ or $m2⊆I$⁠, then $m1+I$ or $m2+I$ is non-essential in R. Then I is adjacent to $m1$ or $m2$⁠. Suppose that $m⊈I$ and $m⊈J$⁠. If I is not adjacent to $m1$⁠, then $m1+I$ is essential in R. So, $m2⊆soc(R)⊆m1+I$⁠, which implies $m2⊆I$⁠, a contradiction. Therefore I is adjacent to $m1$⁠. In the same way, I is adjacent to $m2$⁠. Thus $γ(NES(R))≤2$⁠.

(ii) If $min(R)$ is finite, then by Theorem 2.8, there exists no vertex which is adjacent to every other vertex. So, $γ(NES(R))≠1$⁠. Therefore, $γ(NES(R))=2$ by part (i). In the opposite direction, let $γ(NES(R))=1$⁠. So, the graph has a vertex which is adjacent to every other vertex. So the graph does not contain finite minimal ideals. Hence the result. $□$

Proof. Let $min(R)$ be finite and $min(R)={m1,m2,…,mt}$⁠. Since ${∑j=1,j≠1tmj}i=1t$ is an independent set in $NES(R)$⁠, therefore $t≤α(NES(R))$⁠. Assume that $α(NES(R))$ is equal to p and $S={I1,I2,…,Ip}$ is the maximal independent set. For each $I∈S$⁠, I is nonessential in R. So, there exists a minimal ideal m such that $m⊈I$⁠. If $p>t$⁠, then there exists $1≤i,j≤p$ and $m∈min(R)$ such that $m⊈Ii$ and $m⊈Ij$⁠. Thus $m⊈Ii+Ij$⁠. Otherwise, $m⊆Ii+Ij$ which leads to a contradiction. As S is independent, so $Ii+Ij$ is essential, which implies $m⊆soc(R)⊆Ii+Ij$⁠, a contradiction. Therefore, $α(NES(R))=|min(R)|$⁠. If we take $α(NES(R))=∞$⁠, then by similar argument we get a contradiction. Hence, $α(NES(R))=|min(R)|$⁠. $□$