1. Introduction
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].
2. Connectedness of nonessential sum graph
In this section, we obtain some results related to connectedness, diameter, girth, completeness, cut vertex, partiteness and regular character. We start with a remark.
Remark 2.1. An ideal A is nonessential ideal in R if and only if A⊆soc(R). If B is a nonessential ideal of R then every ideal which is contained in B is also a nonessential ideal of R. If m is a minimal ideal of R and if A and B are two ideals such that m⊆A+B, then m⊆A or m⊆B.
Lemma 2.2. If min(R)={mi}i∈λ, where λ is an index set and μ is a finite subset of λ, then ∑μmi is a nonessential ideal of R.
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.
Theorem 2.3. NES(R) is a null graph if and only if R contains exactly one minimal ideal.
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.□
Theorem 2.4. NES(R) is an empty graph if and only if R has exactly two minimal ideals, which are the only nonessential ideals of R.
Proof. Let NES(R) be an empty graph. Then by Theorem 2.3 |min(R|≠1min (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.□
Theorem 2.5. The following statements are equivalent:
NES(R)=G1∪G2, where G1 and G2 are two disjoint complete subgraphs of NES(R).
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⊆I and I is nonessential in R}. 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.
Theorem 2.6. The diameter of NES(R) is 1,2 or ∞.
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. □
Theorem 2.7. If NES(R) contains a cycle, then girth(NES(R))=3.
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.□
Theorem 2.8. Let R contain finitely many minimal ideals, then the following holds:
There exists no vertex in NES(R) which is adjacent to every other vertex.
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). □
Theorem 2.9. If NES(R) is connected, then NES(R) has no cut vertex.
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.□
Theorem 2.10. NES(R) is not a complete r-partite graph.
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.□
Theorem 2.11. The following statements holds:
NES(R) contains an end vertex if and only if NES(R)=G1∪G2, where G1 and G2 are two disjoint complete subgraphs of NES(R) and |V(Gi)|=2 for some i=1,2.
NES(R) is not a star graph.
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. □
Theorem 2.12. The following statements holds:
If I and J are two vertices of NES(R) such that I⊆J, then deg(I)≥deg(J).
If NES(R) is an r-regular graph then |V(NES(R))|=2r+2.
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. □
3. Clique number, independence number, domination number of nonessential sum graph
In this section, we will find clique number, independence number, domination number of NES(R).
Theorem 3.1. The following holds:
If ω(NES(R))<∞, then number of minimal ideals of R is finite.
ω(NES(R))=1 if and only if min(R)={m1, m2} and these two are the only nonessential ideals in R.
If the number of minimal ideals of R is finite, then ω(NES(R))≥2|min(R)|−1−1.
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.□
Theorem 3.2. The following holds:
min(R) is finite if and only if γ(NES(R))=2 and min(R) is infinite if and only if γ(NES(R))=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. □
Theorem 3.3. Let R contain finite number of minimal ideals. Then α(NES(R))=|min(R)|.
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)|. □