Some results on the comaximal ideal graph of a commutative ring

The rings considered in this article are commutative with identity which admit at least two maximal ideals. Let R be a ring such that R admits at least two maximal ideals. Recall from Ye and Wu (J. Algebra Appl. 11(6): 1250114, 2012) that the comaximal ideal graph of R, denoted by C (R) is an undirected simple graph whose vertex set is the set of all proper ideals I of R such that I 6⊆ J(R), where J(R) is the Jacobson radical of R and distinct vertices I1, I2 are joined by an edge in C (R) if and only if I1 + I2 = R. In Section 2 of this article, we classify rings R such that C (R) is planar. In Section 3 of this article, we classify rings R such that C (R) is a split graph. In Section 4 of this article, we classify rings R such that C (R) is complemented and moreover, we determine the S-vertices of C (R). 2010 MSC: 13A15, 05C25


Introduction
The rings considered in this article are commutative with identity which admit at least two maximal ideals.Let R be a ring.We denote the set of all maximal ideals of R by M ax(R).We denote the Jacobson radical of R by J(R).We denote the cardinality of a set A using the notation |A|.Motivated by the research work done on comaximal graphs of rings in [9,12,13,15,16] and on the annihilating-ideal graphs of rings in [5,6], M. Ye and T. Wu in [18] introduced a graph structure on a ring R, whose vertex set is the set of all proper ideals I of R such that I ⊆ J(R) and distinct vertices I 1 and I 2 are joined by an edge if and only if I 1 + I 2 = R. M. Ye and T. Wu called the graph introduced by them in [18] as the comaximal ideal graph of R and denoted it using the notation C (R) and investigated the influence of certain graph parameters of C (R) on the ring structure of R. Let R be a ring such that |M ax(R)| ≥ 2.
The aim of this article is to classify rings R such that C (R) is planar; C (R) is a split graph; and C (R) is complemented.
It is useful to recall the following definitions from commutative ring theory.A ring R is said to be quasilocal (respectively, semiquasilocal) if R has a unique maximal ideal (respectively, R has only a finite number of maximal ideals).A Noetherian quasilocal (respectively, semiquasilocal) ring is referred to as a local ring (respectively, a semilocal ring).Recall that a principal ideal ring is said to be a special principal ideal ring (SPIR) if R has a unique prime ideal m.It is clear that m is nilpotent.If R is a SPIR with m as its only prime ideal, then we denote it by saying that (R, m) is a SPIR.Suppose that a ring T is quasilocal with m as its unique maximal ideal such that m = (0) and nilpotent.Let n ≥ 2 be least with the property that m n = (0).If m is principal, then it follows from (iii) ⇒ (i) of [3,Proposition 8.8] that {m i |i ∈ {1, . . ., n − 1}} is the set of all nonzero proper ideals of T .Hence, (T, m) is a SPIR.
We next recall the following definitions and results from graph theory.The graphs considered in this article are undirected and simple.Let G = (V, E) be a graph.Recall from [4, Definition 1.2.2] that a clique of G is a complete subgraph of G.The clique number of G, denoted by ω(G) is defined as the largest integer n ≥ 1 such that G contains a clique on n vertices [4, Definition, page 185].We set ω(G) = ∞ if G contains a clique on n vertices for all n ≥ 1.
Let n ∈ N. A complete graph on n vertices is denoted by K n .A graph G = (V, E) is said to be bipartite if V can be partitioned into two nonempty subsets V 1 and V 2 such that each edge of G has one end in V 1 and the other in V 2 .A bipartite graph with vertex partition V 1 and V 2 is said to be complete if each element of V 1 is adjacent to every element of V 2 .Let m, n ∈ N. Let G = (V, E) be a complete bipartite graph with vertex partition V 1 and V 2 .If |V 1 | = m and |V 2 | = n, then G is denoted by K m,n [4,Definition 1.1.12].
Let G = (V, E) be a graph.Recall from [4,Definition 8.1.1]that G is said to be planar if G can be drawn in a plane in such a way that no two edges of G intersect in a point other than a vertex of G. Recall that two adjacent edges are said to be in series if their common end vertex is of degree two [7, page 9].Two graphs are said to be homeomorphic if one graph can be obtained from the other by insertion of vertices of degree two or by the merger of edges in series [7, page 100].It is useful to note from [7, page 93] that the graph K 5 is referred to as Kuratowski's first graph and the graph K 3,3 is referred to as Kuratowski's second graph.The celebrated theorem of Kuratowski states that a graph G is planar if and only if G does not contain either of Kuratowski's two graphs or any graph homeomorphic to either of them [7,Theorem 5.9].
Let G = (V, E) be a graph.It is convenient to name the following conditions satisfied by G so that it can be used throughout Section 2 of this article.
(i) We say that G satisfies (Ku 1 ) if G does not contain K 5 as a subgraph (that is, equivalently, if ω(G) ≤ 4).
(ii) We say that G satisfies (Ku * 1 ) if G satisfies (Ku 1 ) and moreover, G does not contain any subgraph homeomorphic to K 5 .
(iii) We say that G satisfies (Ku 2 ) if G does not contain K 3,3 as a subgraph.
(iv) We say that G satisfies (Ku * 2 ) if G satisfies (Ku 2 ) and moreover, G does not contain any subgraph homeomorphic to K 3,3 .
Suppose that a graph G = (V, E) is planar.It follows from Kuratowski's theorem [7, Theorem 5.9] that G satisfies both (Ku * 1 ) and (Ku * 2 ).Hence, G satisfies both (Ku 1 ) and (Ku 2 ).It is interesting to note that a graph G can be nonplanar even if it satisfies both (Ku 1 ) and (Ku 2 ).For an example of this type, refer [7, Figure 5.9(a), page 101] and the graph G given in this example does not satisfy (Ku * 2 ).It is not hard to construct an example of a graph G such that G satisfies (Ku 1 ) but G does not satisfy (Ku * 1 ).
Let R be a ring such that |M ax(R)| ≥ 2. In Section 2 of this article, we try to classify rings R such that C (R) is planar.It is proved in [18,Theorem 3.1] In Section 2 of this article, we first focus on classifying rings R such that is the disjoint union of two nonempty subsets K and S such that the subgraph of G induced on K is complete and S is an independent set of G. Let R be a commutative ring with identity.In [16] P. K. Sharma and S.M. Bhatwadekar introduced and investigated a graph associated with R, whose vertex set is the set of all elements of R and distinct vertices x, y are joined by an edge if and only if Rx + Ry = R.The graph studied in [16] is named as the comaximal graph of R in [12].In [9], M.I.Jinnah and S.C. Mathew classified rings R such that the comaximal graph of R is a split graph.Let R be a ring such that |M ax(R)| ≥ 2. In Section 3 of this article, we try to classify rings R such that rings, where F is a field and S is a quasilocal ring.
Let G = (V, E) be a graph.Recall from [2,11] that two distinct vertices u, v of G are said to be orthogonal, written u ⊥ v if u and v are adjacent in G and there is no vertex of G which is adjacent to both u and v in G; that is, the edge u − v is not an edge of any triangle in . Moreover, we recall from [2] that G is complemented if each vertex of G admits a complement in G. Furthermore, G is said to be uniquely complemented if G is complemented and whenever the vertices u, v, w of G such that u ⊥ v and u ⊥ w, then a vertex x of G is adjacent to v in G if and only if x is adjacent to w in G. Let R be a ring such that R is not an integral domain.Recall from [1] that the zero-divisor graph of R denoted by Γ(R) is an undirected graph whose vertex set is Z(R)\{0} (here Z(R) denotes the set of all zero-divisors of R) and distinct vertices x, y are joined by an edge if and only if xy = 0.The authors of [2] determined in Section 3 of [2] rings R such that Γ(R) is complemented or uniquely complemented.For a ring R, we denote the set of all units of R by U (R) and we denote the set of all nonunits of R by N U (R).The Krull dimension of a ring R is simply denoted by dimR.In [15,Proposition 3.11] it is proved that the subgraph of the comaximal graph of R induced on N U (R)\J(R) is complemented if and only if dim( R J(R) ) = 0. Section 4 of this article is devoted to find a classification of rings R such that Moreover, in Section 4, a discussion on the S-vertices of C (R) is included.Let G = (V, E) be a graph.Recall from [14, Definition 2.9] a vertex a of G is said to be a Smarandache vertex or simply a S-vertex if there exist distinct vertices x, y, and b of G such that a − x, a − b, and b − y are edges of G but there is no edge joining x and y in G.In [14], A.M. Rahimi investigated the S-vertices of the zero-divisor graph of a commutative ring and the zero-divisor graph of a ring with respect to an ideal.For a ring R with Let A, B be sets.If A is a subset of B and A = B, then we denote it symbolically using the notation A ⊂ B. Let G be a graph.We denote the vertex set of G by V (G).Let R be a ring.For a proper ideal I of R, as in [15], we denote {m ∈ M ax(R)|m ⊇ I} by M (I).

Some preliminary results and on the planarity of C (R)
As is already mentioned in the introduction, the rings considered in this article are commutative with identity which admit at least two maximal ideals.
Proof.Assume that C (R) satisfies (Ku 2 ).We assert that R admits a nontrivial idempotent.Suppose that R does not have any nontrivial idempotent.By hypothesis, Since we are assuming that R has no nontrivial idempotent, we obtain that Ra i = Ra j and Rb i = Rb j for all distinct i, j ∈ N.
This is a contradiction.Therefore, R admits at least one nontrivial idempotent.Let e be a nontrivial idempotent of R. Observe that the mapping quasilocal ring for each i ∈ {1, 2} and try to classify such rings R in order that C (R) satisfies (Ku 2 ).Lemma 2.4.Let R 1 , R 2 be rings and let R = R 1 × R 2 .Suppose that R i has at least two nonzero proper ideals for each i ∈ {1, 2}.Then C (R) does not satisfy (Ku 2 ).
Proof.We are assuming that R i has at least two nonzero proper ideals for each i ∈ {1, 2}.Let I 1 , I 2 be distinct nonzero proper ideals of R 1 and let Let I be an ideal of a ring R. Then the annihilator of I in R, denoted by Ann R I is defined as Ann R I = {r ∈ R|Ir = (0)}.Lemma 2.5.Let (R, m) be a local ring which is not a field.The following statements are equivalent: We are assuming that R has only one nonzero proper ideal.Hence, m is the only nonzero proper ideal of R.
The following statements are equivalent: . We know from Lemma 2.4 that for at least one i ∈ {1, 2}, R i has at most one nonzero proper ideal.Hence, for that i, either m i = (0) or in the case Since there are at most two proper ideals of R 1 , we obtain that Proposition 2.7.Let R be a ring such that |M ax(R)| = 2.The following statements are equivalent: Proof.(i) ⇒ (ii) This follows from Kuratowski's theorem [7, Theorem 5.9].
(iv) ⇒ (i) Let us denote the ring R 1 × R 2 by T .Without loss of generality, we can assume that (R 1 , m 1 ) is a SPIR with m 2 1 = (0).Let V 1 , V 2 be as in the proof of (ii) ⇒ (i) of Lemma 2.6 and it is already noted there that |V 1 | ≤ 2 and C (T ) is a complete bipartite graph with vertex partition V 1 and V 2 .It is now clear that C (T ) is planar.Since R ∼ = T as rings, we get that C (R) is planar.
Let R be a ring such that |M ax(R)| = 3.We next try to classify such rings R in order that C (R) Proof.We are assuming that C (R 1 ) does not satisfy (Ku 2 ).Then there exist subsets Proof.Assume that C (R) satisfies (Ku 2 ).As |M ax(R)| = 3, it follows from Lemma 2.2 that there exist nonzero rings T 1 and T 2 such that R ∼ = T 1 × T 2 as rings.Since R has exactly three maximal ideals, it follows that either T 1 or T 2 is not quasilocal.Without loss of generality, we can assume that T 1 is not quasilocal.Hence, the number of maximal ideals of T 1 is exactly two.Let us denote the ring T 1 × T 2 by T .Since R ∼ = T as rings, we obtain that C (T ) satisfies (Ku 2 ).Now, it follows from Lemma 2.8 that C (T 1 ) satisfies (Ku 2 ).Hence, we obtain from Lemma 2.2 that there exist nonzero rings T 11 and T 12 such that T 1 ∼ = T 11 × T 12 as rings.Therefore, R ∼ = T 11 × T 12 × T 2 as rings.Hence, on renaming the rings T 11 , T 12 , and T 2 , we obtain that there exist rings R 1 , R 2 , and Proof.Assume that C (R) satisfies (Ku 2 ).We know from Lemma 2.9 that Proof.Suppose that C (R) does not satisfy (Ku 2 ).Then there exist subsets After renaming the maximal ideals of R (if necessary), we can assume without loss of generality that I 1 ⊆ m 1 .Since I 1 + J k = R for each k ∈ {1, 2, 3}, it follows that J k ⊆ m 1 for each k ∈ {1, 2, 3}.By hypothesis, |W 2 | ≤ 2 and |W 3 | ≤ 2. Therefore, we obtain that W 2 ∩V 2 = ∅ and W 3 ∩V 2 = ∅.This implies that J 1 J 2 J 3 ⊆ m 2 m 3 .It follows from I i + J k = R for all i, k ∈ {1, 2, 3} that I i + J 1 J 2 J 3 = R for each i ∈ {1, 2, 3} and so, I i + m 2 m 3 = R.Hence, we get that I i ∈ W 1 for each i ∈ {1, 2, 3}.This implies that |W 1 | ≥ 3.This is in contradiction to the assumption that |W 1 | ≤ 2. Therefore, C (R) satisfies (Ku 2 ).
Let R be a ring such that |M ax(R)| = 3.Let i ∈ {1, 2, 3} and let m i , W i be as in the statement of Lemma 2.11.In Proposition 2.12, we classify such rings R in order that |W i | ≤ 2 for each i ∈ {1, 2, 3}.Proposition 2.12.Let R be a ring such that |M ax(R)| = 3.Let {m i |i ∈ {1, 2, 3}} denote the set of all maximal ideals of R. Let i ∈ {1, 2, 3}.Let us denote the set of all proper ideals I of R such that M (I) = {m i } by W i .Then the following statements are equivalent: by assumption, it follows from [10, Theorem 81] that there exists a 2 ∈ m 1 \(Ra 1 ∪ m 2 ∪ m 3 ).Note that m 1 = Ra 2 and it is clear from the choice of the elements a 1 , a 2 that Ra 1 = Ra 2 and {Ra 1 , Ra 2 , m 1 } ⊆ W 1 .This is in contradiction to the assumption that |W 1 | ≤ 2. Therefore, m 1 is principal.Similarly, it can be shown that m 2 and m 3 are principal.Let i ∈ {1, 2, 3}.Observe that i=1 m i is principal, there exists a ∈ J(R) such that J(R) = Ra.From (J(R)) 2 = (J(R)) 3 , we obtain that Ra 2 = Ra 3 .Hence, a 2 = ra 3 for some r ∈ R. Since 1 − ra is a unit in R, we obtain that a 2 = 0 and so, (J(R)) 2 = (0).Since , we obtain from the Chinese remainder theorem [3, Proposition 1.10(ii) and (iii)] that the mapping f : R → R given by as its set of all maximal ideals.Let us denote the set of all proper ideals A of T such that M (A) = {N i } by U i for each i ∈ {1, 2, 3}.Since R i has at most one nonzero proper ideal for each i ∈ {1, 2, 3}, it follows that |U i | ≤ 2. From R ∼ = T as rings, we obtain that |W i | ≤ 2 for each i ∈ {1, 2, 3}.Proposition 2.13.Let R be a ring such that |M ax(R)| = 3.The following statements are equivalent: 3}} denote the set of all maximal ideals of R. Let i ∈ {1, 2, 3} and let us denote the set of all proper ideals I of R such that M (I) = {M i } by W i .We know from (ii) ⇒ (i) of Proposition 2.12 that |W i | ≤ 2. Hence, we obtain from Lemma 2.11 that C (R) satisfies (Ku 2 ).
Let R be a ring such that |M ax(R)| = 3.We try to classify such rings R in order that C (R) is planar.If C (R) is planar, then we know from Kuratowski's theorem [7, Theorem 5.9] that C (R) satisfies (Ku 2 ).Hence, we obtain from (i) ⇒ (ii) of Proposition 2.13 that R ∼ = R 1 × R 2 × R 3 as rings, where (R i , m i ) is a SPIR with m 2 i = (0) for each i ∈ {1, 2, 3}.
Proof.The proof of this lemma closely follows the proof given in [17,Lemma 3.13].Note that |M ax(R)| = 3 and 3 are in series and moreover, in H, v i is adjacent to v 4 and v 5 for each i ∈ {1, 2, 3}.Furthermore in H, v 1 and v 2 are adjacent to v 6 .Therefore, on merging the edges v 3 − v 7 and v 7 − v 6 , we obtain a graph H 1 which contains K 3,3 as a subgraph.Hence, H contains a subgraph which is homeomorphic to K 3,3 .This shows that C (R) does not satisfy Proof.The proof of this lemma closely follows the proof given in [17,Lemma 3.14].Observe that Note that in H, the edges e 1 : v 1 − v 2 , e 2 : v 2 − v 3 are edges in series and the edges e 3 : v 4 − v 5 , e 4 : v 5 − v 6 are edges in series.Observe that in H, v 1 is adjacent to all the elements of W except v 3 and v 5 ; v 3 is adjacent to all the elements of W except v 1 and v 5 ; v 4 is adjacent to all the elements of W except v 2 and v 6 ; v 6 is adjacent to all the elements of W except v 2 and v 4 ; v 7 is adjacent to all the elements of W except v 2 and v 5 .Let H 1 be the graph obtained from H on merging the edges e 1 and e 2 and on merging the edges e 3 and e 4 .It is clear that H 1 is a complete graph on five vertices.This proves that C (R) contains a subgraph H such that H is homeomorphic to K 5 .Therefore, C (R) does not satisfy (Ku * 1 ).Lemma 2.16.Let R = F 1 × F 2 × R 3 , where F 1 and F 2 are fields and (R 3 , m 3 ) is a SPIR with m 3 = (0) but m 2 3 = (0).Then C (R) is planar.
and the pendant edges e 4 : v 1 − v 5 , e 5 : v 1 − v 9 , e 6 : v 2 − v 6 , and e 7 : v 2 − v 8 .Note that Γ can be represented by means of a rectangle.The edges e 1 , e 2 are edges in series and their common end vertex v 4 can be plotted inside the rectangle representing Γ and the edges e 1 , e 2 can be drawn inside this rectangle.The edges e i (i ∈ {3, 4, 5, 6, 7}) are such that one of their end vertices ∈ {v 1 , v 2 } and they can be drawn outside the rectangle representing Γ in such a way that there are no crossing over of the edges.This proves that C (R) is planar.
and the pendant edges e 1 : v 1 − v 5 , e 2 : v 2 − v 6 , and e 3 : v 3 − v 4 .The cycle Γ can be represented by means of a triangle and the one of the end vertex of e i is v i for each i ∈ {1, 2, 3} and the edges e 1 , e 2 , e 3 can be drawn outside the triangle representing Γ in such a way that there are no crossing over of the edges.This proves that C (R) is planar.
(ii) ⇒ (iii) Since C (R) satisfies (Ku * 2 ), we get that C (R) satisfies (Ku 2 ).Therefore, we obtain from Proposition 2.13 that R ∼ = R 1 × R 2 × R 3 as rings, where (R i , m i ) is a SPIR with m 2 i = (0) for each i ∈ {1, 2, 3}.Let us denote the ring R 1 × R 2 × R 3 by T .Since R ∼ = T as rings, we obtain that C (T ) satisfies (Ku * 1 ) and (Ku * 2 ).As C (T ) satisfies (Ku * 2 ), it follows from Lemma 2.14 that R i is a field for at least one value of i ∈ {1, 2, 3}.Suppose that R i is a field for exactly one value of i ∈ {1, 2, 3}.Without loss of generality, we can assume that R 1 is a field and R 2 , R 3 are not fields.In such a case, we obtain from Lemma 2.15 that C (T ) does not satisfy (Ku * 1 ).This is a contradiction.Therefore, R i is a field for at least two values of i ∈ {1, 2, 3}.This proves that 3}, in which case, we obtain from Lemma 2.17 that C (T ) is planar or there are exactly two values of i ∈ {1, 2, 3} such that R i is a field and in such a case, we obtain from Lemma 2.16 that C (T ) is planar.Since R ∼ = T as rings, we get that C (R) is planar.

When is C (R) a split graph?
Let R be a ring such that |M ax(R)| ≥ 2. The aim of this section is to classify rings R such that C (R) is a split graph.Throughout this section, we assume that K is a nonempty subset of V (C (R)) such that the subgraph of C (R) induced on K is complete and S is a nonempty subset of V (C (R)) such that S is an independent set of C (R).
Proof.First, we claim that M ax(R)) ⊆ K. Suppose that M ax(R) ⊆ K. Then there exists m ∈ M ax(R) such that m / ∈ K. Hence, m ∈ S. We are assuming that |M ax(R)| ≥ 3. Therefore, there exist distinct maximal ideals m , m of R such that m = m and m = m.Since m+m = m+m = m+m m = R, we get that m , m , and m m are adjacent to m in C (R).As m ∈ S, we obtain that m , m , m m ∈ K. Hence, m and m m must be adjacent in C (R).This is impossible, since m + m m = m = R.Therefore, M ax(R) ⊆ K.We next verify that K ⊆ M ax(R).Let I ∈ K. Then I is a proper ideal of R and so, there exists a maximal ideal m of R such that I ⊆ m.We assert that I = m.Suppose that I = m.Then I and m are adjacent in C (R).This is impossible, since I + m = m = R.Hence, I = m and this proves that K ⊆ M ax(R) and so, K = M ax(R).
Let R be a ring such that |M ax(R)| = 3.In Theorem 3.3, we classify such rings R in order that C (R) is a split graph.Theorem 3.3.Let R be a ring such that |M ax(R)| = 3.The following statements are equivalent: As |M ax(R)| = 3, we obtain from Lemma 3.1 that M ax(R) = K.Let {m 1 , m 2 , m 3 } denote the set of all maximal ideals of R. Let i ∈ {1, 2, 3} and let us denote the set of all proper ideals I of R such that As K = M ax(R), it follows that I must be in S. Note that m 2 m 3 ∈ S. It is clear that I = m 2 m 3 .Now, I, m 2 m 3 ∈ S and S is an independent set of C (R), we get that I and m 2 m 3 cannot be adjacent in C (R).However, I + m 2 m 3 = R.This is a contradiction and so, I = m 1 .This shows that W 1 = {m 1 }.Similarly, it can be shown that W 2 = {m 2 } and W 3 = {m 3 }.We next show that m i is principal for each i ∈ {1, 2, 3}.Note that m 1 ⊆ m 2 ∪m 3 .Hence, there exists x 1 ∈ m 1 \(m 2 ∪ m 3 ).Observe that Rx 1 ∈ W 1 = {m 1 } and so, m 1 = Rx 1 .Similarly, using the facts that W 2 = {m 2 } and W 3 = {m 3 }, it can be proved that m 2 = Rx 2 for any x 2 ∈ m 2 \(m 1 ∪ m 3 ) and m 3 = Rx 3 for any x 2 3 for some r ∈ R. As x 1 x 2 x 3 ∈ J(R), we obtain that 1 − rx 1 x 2 x 3 is a unit in R and so, x 1 x 2 x 3 = 0. Since m i + m j = R for all distinct i, j ∈ {1, 2, 3} and J(R) = (0), we obtain from the Chinese remainder theorem [3, Proposition 1.10(ii) and (iii)] that the mapping f : R → R m1 × R m2 × R m3 given by f (r) = (r + m 1 , r + m 2 , r + m 3 ) is an isomorphism of rings.Let us denote the field R mi by F i for each i ∈ {1, 2, 3}.Therefore, R ∼ = F 1 × F 2 × F 3 as rings, where F i is a field for each i ∈ {1, 2, 3}.
(ii) ⇒ (i) We are assuming that R ∼ = F 1 × F 2 × F 3 as rings, where F i is a field for each i ∈ {1, 2, 3}.Let us denote the ring induced on K is complete and S is an independent set of C (T ).Therefore, C (T ) is a split graph.As R ∼ = T as rings, we obtain that C (R) is a split graph.
Let R be a ring such that |M ax(R)| = 2.We next try to classify such rings in order that C (R) is a split graph.Lemma 3.4 is well-known.We include a proof of Lemma 3.4 for the sake of completeness.Lemma 3.4.Let G = (V, E) be a complete bipartite graph.The following statements are equivalent: (ii) G is a star graph.
Proof.(i) ⇒ (ii) Assume that G is a split graph.Hence, there exist nonempty subsets K, S of V such that V = K ∪ S, K ∩ S = ∅, the subgraph of G induced on K is complete, and S is an independent set of G.By hypothesis, G is a complete bipartite graph.Let G be complete bipartite with vertex partition V 1 and V 2 .We claim that S ∩ V i = ∅ for some i ∈ {1, 2}.Suppose that S ∩ V i = ∅ for each i ∈ {1, 2}.Let s i ∈ S ∩ V i for each i ∈ {1, 2}.Then s 1 and s 2 are adjacent in G.This is impossible, since S is an independent set of G. Therefore, either Since no two distinct elements of V 2 are adjacent in G, whereas any two distinct vertices of K are adjacent in G, it follows that |V 2 | = 1.This shows that G is a star graph.
(ii) ⇒ (i) Suppose that G is a star graph.Hence, G is a complete bipartite graph with vertex partition V 1 and V 2 such that |V i | = 1 for at least one i ∈ {1, 2}.Without loss of generality, we can assume that the subgraph of G induced on K is complete, and S is an independent of G. Therefore, G is a split graph.Theorem 3.5.Let R be a ring such that |M ax(R)| = 2.The following statements are equivalent: (ii) R ∼ = F × S as rings, where F is a field and S is a quasilocal ring.
Proof.Let {m 1 , m 2 } denote the set of all maximal ideals of R.
(i) ⇒ (ii) Assume that C (R) is a split graph.As |M ax(R)| = 2, we know from (3) ⇒ (1) of [18, Theorem 4.5] that C (R) is a complete bipartite graph with vertex partition V 1 and V 2 , where for each i ∈ {1, 2}, V i is the set of all proper ideals I of R such that M (I) = {m i }.As we are assuming that C (R) is a split graph, we obtain from Lemma 3.4 that C (R) is a star graph.Hence, there exists a vertex I of C (R) such that I is adjacent to each vertex J of C (R) with J = I.We can assume without loss of generality that I ∈ V 1 .In such a case, we obtain that (ii) ⇒ (i) Assume that R ∼ = F × S as rings, where F is a field and S is a quasilocal ring.Let us denote the ring F × S by T .Let V 1 = {(0) × S} and V 2 = {F × I|I is a proper ideal of S}.Note that C (T ) is a complete bipartite graph with vertex partition V 1 and V 2 and as |V 1 | = 1, it follows that C (T ) is a star graph.Hence, we obtain from (ii) ⇒ (i) of Lemma 3.4 that C (T ) is a split graph.Since R ∼ = T as rings, we get that C (R) is a split graph.

Some more results on C (R)
Let R be a ring such that |M ax(R)| ≥ 2. The aim of this section is to classify rings R such that C (R) is complemented and to determine the S-vertices of C (R).
Let R be a ring.Let X be the set of all prime ideals of R. Recall from [3, Exercise 15, page 12] that for a subset E of R, the set of all prime ideals p of R such that p ⊇ E is denoted by V (E).We know from [3, Exercise 15, page 12] that the collection {V (E)|E ⊆ R} satisfies the axioms for closed sets in a topological space.The resulting topology is called the Zariski topology.The topological space X is called the prime spectrum of R and is denoted by Spec(R).Let E ⊆ R. We know from [3, Exercise 15(i), page 12] that V (E) = V (I), where I is the ideal of R generated by E. The subspace of Spec(R) consisting of all the maximal ideals of R with the induced topology is called the maximal spectrum of R and is denoted by M ax(R).The collection {V (I) ∩ M ax(R)|I varies over all ideals of R} is the collection of all closed sets of M ax(R).As in [15], we denote V (I) ∩ M ax(R) by M (I).Thus M (I), as mentioned in the introduction, is the set of all maximal ideals m of R such that m ⊇ I.As in [15], for an element a ∈ R, we denote M (Ra) simply by M (a) and M ax(R)\M (a) by D(a).Let G = (V, E) be a simple graph.Let v ∈ V .Then the set of all u ∈ V such that v is adjacent to u in G is called the set of neighbours of v in G and we use the notation N G (v) to denote the set of all neighbours of v in G. (i) Let I, J ∈ V (C (R)) be such that I ⊥ J in C (R).Then IJ ⊆ J(R).
(ii) If C (R) is complemented, then it is uniquely complemented.
Proof.(i) Suppose that IJ ⊆ J(R).Then there exists a maximal ideal m of R such that IJ ⊆ m.Hence, I ⊆ m and J ⊆ m and so, I + m = J + m = R. Now, m ∈ V (C (R)) is such that m is adjacent to both I and J in C (R).This is impossible, since I ⊥ J in C (R).Therefore, IJ ⊆ J(R).
(ii) Let I ∈ V (C (R)).We are assuming that C (R) is complemented.Hence, there exists at least one J ∈ V (C (R)) such that I ⊥ J in C (R).Let J 1 , J 2 ∈ V (C (R)) be such that I ⊥ J 1 and I ⊥ J 2 in C (R).We know from (i) that IJ i ⊆ J(R) for each i ∈ {1, 2}.Let A ∈ V (C (R)) be such that J 1 is adjacent to A in C (R).Hence, J 1 + A = R.We claim that J 2 + A = R. Suppose that J 2 + A = R. Then there exists m ∈ M ax(R) such that J 2 + A ⊆ m.It follows from I + J 2 = R that I ⊆ m.As IJ 1 ⊆ J(R) ⊆ m, we get that J 1 ⊆ m.Therefore, J 1 + A ⊆ m.This is impossible, since J 1 + A = R.Hence, J 2 + A = R.This shows that N C (R) (J 1 ) ⊆ N C (R) (J 2 ).Similarly, it can be shown that N C (R) (J 2 ) ⊆ N C (R) (J 1 ).Therefore, N C (R) (J 1 ) = N C (R) (J 2 ).This proves that C (R) is uniquely complemented.

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and so, Ra = Ra 2 = m 1 .Now, there exists r ∈ R such that a = ra 2 .This implies that e = ra is a nontrivial idempotent element of R and moreover, m 1 = Ra = Re.Note that the mapping f : R → R(1 − e) × Re defined by f (x) = (x(1 − e), xe) is an isomorphism of rings.Hence, f (m 1 ) = (0) × Re is a maximal ideal of R(1 − e) × Re.Therefore, R(1 − e) is a field.Since |M ax(R)| = 2, it follows that the ring Re is quasilocal.Thus with F = R(1 − e) and S = Re, we obtain that F is a field and S is a quasilocal ring and R ∼ = F × S as rings.

Remark 4 . 1 .
Let R be a ring such that |M ax(R)| ≥ 2. The following statements hold.

Lemma 4 . 2 .
Let R be a ring such that |M ax(R)| ≥ 2. The following statements are equivalent:(i) C (R) is complemented.

(
ii) M (I) is a closed and open subset of M ax(R) for each I ∈ V (C (R)).Proof.We adapt an argument found in the proof of[15, Proposition 3.10].
(i) ⇒ (ii) Assume that C (R) is complemented.Let I ∈ V (C (R)).It is clear that M (I) is a closed subset of M ax(R).Since C (R) is complemented, there exists J ∈ V (C (R)) such that I ⊥ J in C (R).Hence, I and J are adjacent in C (R) and there is no A ∈ V (C (R)) such that A is adjacent to both I and J in C (R).That is, I + J = R and there is no proper ideal A of R with A + I = A + J = R.Note that there exist a ∈ I and b ∈ J such that a + b = 1.We claim that M (I) = D(b).Let m ∈ M (I).As m ⊇ I, a ∈ I , and a + b = 1, it follows that b / ∈ m.Hence, m ∈ D(b).This shows that M (I) ⊆ D(b).Let m ∈ D(b).

Remark 4 . 8 . 3 . 4 . 9 .
Let R be a ring such that |M ax(R)| = 2. Let {m 1 , m 2 } denote the set of all maximal ideals of R. We know from (3) ⇒ (1) of[18, Theorem 4.5] that C (R) is a complete bipartite graph with vertex partition V 1 and V 2 , whereV i = {I|M (I) = {m i }} for each i ∈ {1, 2}.Hence, it is clear that no vertex of C (R) is a S-vertex of C (R).Therefore, for a ring R with |M ax(R)| ≥ 2, in determining the S-vertices of C (R), we assume that |M ax(R)| ≥ Proposition Let R be a ring such that |M ax(R)| ≥ 3. Let I ∈ V (C (R)).Then the following statements are equivalent:(i) I is a S-vertex of C (R). (ii) |M ax(R)\M (I)| ≥ 2. Proof.(i) ⇒ (ii) Now, I ∈ V (C (R))and we are assuming that I is a S-vertex of C (R).Hence, there exist distinct I 1 , I 2 , I 3 ∈ V (C (R)) such that I − I 1 , I − I 2 , I 2 − I 3 are edges of C (R), but there is no edge joining I 1 and I 3 in C (R).Since I 1 and I 3 are not adjacent in C (R), there exists m ∈ M ax(R) such that I 1 + I 3 ⊆ m.It follows from I + I 1 = I 2 + I 3 = R that I ⊆ m and I 2 ⊆ m.As I 2 is proper ideal of R, there exists m ∈ M ax(R) such that I 2 ⊆ m .It is clear that m = m and it follows from I + I 2 = R that I ⊆ m .Hence, {m, m } ⊆ M ax(R)\M (I) and so, |M ax(R)\M (I)| ≥ 2. (ii) ⇒ (i) Let I ∈ V (C (R)) be such that |M ax(R)\M (I)| ≥ 2. Let {m 1 , m 2 } ⊂ M ax(R) be such that I ⊆ m i for each i ∈ {1, 2}.By hypothesis, |M ax(R)| ≥ 3. Hence, there exists m 3 ∈ M ax(R)\{m 1 , m 2 }.Observe that I − m 1 , I − m 2 , m 2 − m 1 ∩ m 3 are edges of C (R), but there is no edge of C (R) joining m 1 and m 1 ∩ m 3 .This proves that I s a S-vertex of C (R).
3 as rings, where R i is a quasilocal ring for each i ∈ {1, 2, 3}.Let m i denote the unique maximal ideal of R i for each i ∈ {1, 2, 3}.Let us denote the ring R 1 × R 2 × R 3 by T .Since R ∼ = T as rings, we obtain that C (T ) satisfies (Ku 2 ).Let i ∈ {1, 2, 3}.It follows from Lemma 2.4 that R i has at most one nonzero proper ideal.Hence, either m i = (0) in which case, R i is a field or m i = (0) is the only nonzero proper ideal of R i in which case, we obtain from Lemma 2.5 that is an isomorphism of rings.Let i ∈ {1, 2, 3}.Let us denote the ring R i by n i .Since m i is a principal ideal of R, we obtain that n i is a principal ideal of R i and it is clear that n