# On Graphs Related to Comaximal Ideals of a Commutative Ring

Authors: Tongsuo Wu; Meng Ye; Dancheng Lu; Houyi Yu
Journal: ISRN Combinatorics
 (2013)

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## Abstract

We study the co maximal graphΩ(R), the induced subgraph Γ(R) of Ω(R) whose vertex set is R∖(U(R)∪J(R)), and a retract Γr(R) of Γ(R), where R is a commutative ring. For a graph Γ(R) which contains a cycle, we show that the core of Γ(R) is a union of triangles and rectangles, while a vertex in Γ(R) is either an end vertex or a vertex in the core. For a nonlocal ring R, we prove that both the chromatic number and clique number of Γ(R) are identical with the number of maximal ideals of R. A graph Γr(R) is also introduced on the vertex set {Rx∣x∈R∖(U(R)∪J(R))}, and graph properties of Γr(R) are studied.


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## Body

## 1. Introduction

In 1988, Beck [1] introduced the concept of zero-divisor graph for a commutative ring. Since then a lot of work was done in this area of research. Several other graph structures were also defined on rings and semigroups. In 1995, Sharma and Bhatwadekar [2] introduced a graph Ω(R) on a commutative ring R, whose vertices are elements of R where two distinct vertices x and y are adjacent if and only if Rx+Ry=R. Recently, Maimani et al. in [3] named this graph Ω(R) as the comaximal graph of R and they noted that the subgraph Γ(R) induced on the subset R∖(U(R)∪J(R)) is the key to the co-maximal graph. Many interesting results about the subgraph were obtained in Maimani et al. [3] and Wang [4], and their works show that the properties of the graph Γ(R) are quite similar to that of the modified zero-divisor graph by Anderson and Livingston [5]. For example, both graphs are simple, connected, and with diameter less than or equal to three, and each has girth less than or equal to four if they contain a cycle. Because of this reason, in this paper we use Γ(R) to denote the graph Γ2(R)∖J(R) of [3]. In Section 2, we discover more properties shared by both zero-divisor graph and the subgraph Γ(R) of Ω(R). In particular, It is shown that the core of Γ(R) is a union of triangles and rectangles, while a vertex in Γ(R) is either an end vertex or a vertex in the core. For any nonlocal ring R, it is shown that the chromatic number of the graph Γ(R) is identical with the number of maximal ideals of R. In Section 3, we introduce a new graph Γr(R) on the vertex set
(1){Rx∣x∈R∖(U(R)∪J(R))}.This graph is in fact a retract of the graphΓ(R) and is thus simpler than the graph Γ(R) in general, but we will show that they share many common properties and invariants.Jinnah and Mathew in [6] studied the problem of when a co-maximal graph Ω(R) is a split graph, and they determined all rings R with the property. In Section 4, we give an alternative proof to their main [6, Theorem 2.3]. In the co-maximal graph Ω(R), each unit u of R is adjacent to all vertices of the graph while an element of J(R) only connects to units of R. Temporally, we say u is in the center of the graph Γ(R). Related to the co-maximal relation, there is the concept of rings with stable range one. Recall that a ring R (which needs not be commutative) has one in its stable range, if for any x,y with Rx+Ry=R, there is an element t such that x+ty is invertible. For example, the following classes of rings have one in their stable range: zero-dimensional commutative rings, von Neumann unit-regular rings, and semilocal rings. The concept comaximal graph gives an interesting graph interpretation of such rings. In fact, a commutative ringRhas one in its stable range if and only if for any pair of adjacent verticesx,yin the co-maximal graphΩ(R), the additive cosetx+Ry(andy+Rx) has at least one element in the center of the graphΩ(R).Throughout this paper, all rings are assumed to be commutative with identity. For a ringR, let U(R) be the set of invertible elements of R and J(R) the Jacobson radical of R. Recall that a graph is called complete (resp.,discrete) if every pair of vertices is adjacent (resp., no pair of vertices is adjacent). We denote a complete graph by K and a complete (resp., discrete) graph with n vertices by Kn (resp., Dn). A subset K of the vertex set of G is called a clique if any two distinct vertices of K are adjacent; the clique numberω(G) of G is the least upper bound of the size of the cliques. Similarly, we denote by Km,n the complete bipartite graph with two partitions of sizes m,n, respectively. Recall that H is called a spanning subgraphof a graph G if V(G)=V(H) and a~b in H implies a~b in G for all distinct vertices of G, where a~b means that a≠b and a is adjacent to b. Recall that a cycle in a graph is a path v1~v2~⋯~vn together with an additional edge vn~v1 (n≥3). For a simple graph G and a nonempty subset S of V(G), there is the subgraph induced onS: the vertex set is S and the edge set is
(2){x~y∣x≠y∈S,andthereisanedgex~yinthegraphG}.A discrete induced subgraph of a graphG is also called an independent subset of G. For other notations not mentioned in the paper, one can refer to [7, 8].


## 2. The SubgraphΓ(R)

As noted by Maimani et al. in [3], the main part of the co-maximal graph Ω(R) is the subgraph Γ(R) induced on the vertex subset R∖(U(R)∪J(R)). In fact, we have the following facts.(O1) A vertex inU(R) is adjacent to every vertex of Ω(R), while an element of J(R) only connects to units of R. In fact, there is a sequential sum decomposition
(3)Ω(R)=J(R)+U(R)+Γ(R).(O2)Γ(R) is empty if and only if R is a local ring, that is, a commutative ring with a unique maximal ideal.The authors of [3, 4] studied this subgraph and obtained many interesting results. In particular, it is proved that the graph is connected with diameter less than or equal to three [3, Theorem 3.1], that the girth of the graph is less than or equal to four [4, Corollary 3.8]. We include a detailed proof for the following fundamental property of graphs related to algebras.Theorem 1 (see [3, Theorem 3.1]).
The graphΓ(R) is connected with diameter less than or equal to three.Proof.
For anya∈R, set Sa={𝔪∈Max(R)∣a∈𝔪}. Then for each a∉J(R), Sa⊂Max(R). For distinct a,b∈R, we claim(1)
ab∈J(R) if and only if Max(R)=Sa∪Sb,(2)
Ra+Rb=R if and only if Sa∩Sb=∅.
Now for distincta,b∈R∖(U(R)∪J(R)), if ab∉J(R), then there exists x∈R∖(U(R)∪J(R)) such that Rab+Rx=R. Then clearly there is a path a~x~b in Γ(R) and hence d(a,b)≤2. If ab∈J(R), then take any y∈R∖(U(R)∪J(R)) such that Ra+Ry=R. We claim that by∉J(R) and it will follow that d(a,b)≤d(a,y)+d(y,b)≤3. In fact, assume to the contrary that by∈J(R). Then we have
(4)Sb∪(Sy∖Sb)=Sb∪Sy=Max(R)=Sb∪(Sa∖Sb).
It follows that(Sy∖Sb)=Sa∖Sb=∅ since Sa∩Sy=∅. Then b∈J(R), a contradiction. This completes the proof.By [4, Theorem 3.9(1)], the clique number ω(Γ(R)) is infinite whenever J(R)=0 and the ring R is indecomposable. It could be used to sharpen [3, Theorem 2.2], as the following theorem shows.Theorem 2.
For any non-local ringR, let G=Γ(R). Then the following statements are equivalent.(1)
G is a bipartite graph.(2)
G is a complete bipartite graph.(3)
R has exactly two maximal ideals.(4)
R/J(R)≅𝕂1×𝕂2, where each 𝕂i is a field.Proof.
(
3
)
⇒
(
2
): Assume that 𝔪1,𝔪2 are the maximal ideals of R. Then Γ(R) is a complete bipartite graph with two partitions 𝔪1∖𝔪2 and 𝔪2∖𝔪1.
(
2
)
⇒
(
3
): Assume that G is a complete bipartite graph with vertex partition V(G)=V1∪V2. Then for any maximal ideal 𝔪, 𝔪∖J(R) is entirely contained in a single partition. Assume 𝔪i∖J(R)⊆Vi. If R has a third maximal ideal 𝔫, then 𝔫∖J(R)⊆Vi for some i. This is impossible since 𝔫+𝔪i=R.
(
2
)
⇒
(
1
) and (4)⇒(3): Clear.
(
1
)
⇒
(
3
): Clearly, there is no loss to assume J(R)=0.
IfR has only a finite number of maximal ideals, say 𝔪i, 1≤i≤r, set 𝔫i=𝔪i∖(∪j≠i𝔪j). Then 𝔫i≠∅, and each vertex in 𝔫i is adjacent to all verices in 𝔫j. This implies ω(G)≥|Max(R)| and hence ω(G)=|Max(R)|, when R has only a finite number of maximal ideals. If further G is a bipartite graph, then r=2; that is, R has exactly two maximal ideals.
In the following, assume thatR has infinitely many maximal ideals, and we proceed to prove ω(G)=∞. Assume that R has a nontrivial idempotent e. Then R=eR×(1-e)R. If eR has no nontrivial idempotent element, then ω(Γ(eR))=∞ and hence ω(G)=∞. Then assume that both e and 1-e are nonprimitive idempotents. By induction, for any integer s≥1, there exist nontrivial corner rings Rs,i of R such that R=∏j=1s‍Rs,j. Set
(5)f1=(0,1,…,1),…,f2=(1,0,1,…,1),…,fs=(1,…,1,0).
Clearly{fj}1≤j≤s is a clique in Γ(R), and thus ω(G)=∞.
(
3
)
⇒
(
4
): This follows from the Chinese remainder theorem.Note that the proof together with [4, Theorem 3.9(1)] actually gives an alternative proof to the fact that ω(Γ(R))=|Max(R)| whenever R is not a local ring.Recall that a ringR is called an exchange ring if the left module RR has the exchange property, see [9] and the included references for details. Recall that idempotents can be lifted modulo every ideal of an exchange ring R. The class of exchange rings include artinian rings, semiperfect rings, and clean rings, the rings in which each element is a sum of an idempotent and a unit. Recall that for a ring R with all idempotents central in R, R is clean if and only if R is an exchange ring. For commutative clean rings R, we have the following.Corollary 3.
For any commutative non-local exchange ringR, let G=Γ(R). Then G is a bipartite graph if and only if G is a complete bipartite graph, if and only if R≅R1×R2, where each Ri is a local ring.A graphG is called totally disconnected if the edge set E(G) is empty. By Theorem 1, we have the following observation and hence Theorem 4.(O3)Γ(R) is totally disconnected if and only if it is an empty graph, which is true if and only if R is a local ring.Theorem 4 (see [3, 4]).
For a ringR, let G=Γ(R). Then the following statements are equivalent.(1)
A star graph is a spanning subgraph ofG, that is; G has at least two vertices, and there exists a vertex in G which is adjacent to every other vertex.(2)
G is a tree; that is, G is nonempty, connected and contains no cycles.(3)
G is a star graph.(4)
R is isomorphic to ℤ2×𝔽 for some field 𝔽.Proof.
(
1
)
⇔
(
3
)
⇔
(
4
): This is contained in [3, Corollary 2.4(2)].
(
2
)
⇔
(
4
): This is contained in [4, Theorem 3.5, Corollary 3.6].Corollary 5.
For a finite simple graphG with |G|≥2, assume G=Γ(R) for some ring R. Then the following are equivalent:(1)
A star graph is a spanning subgraph ofG.(2)
G is a tree; that is, G contains no cycle.(3)
G=K1,pn-1 for some prime number p and some positive integer n.The works of [3, 4] show that the graph Γ(R) has many properties which the zero-divisor graph of a ring (or a semigroup) already has. Recall from [10, 11] that the core of a zero-divisor graphGis always a union of triangles and squares, and a vertex inGis either an end vertex or a vertex of the core. Recall that the core of a graph G is by definition the subgraph induced on all vertices of cycles of G. In the final part of this section, we will show that the graph Γ(R) has the same property.Lemma 6.
For any patha~x~b in the graph Γ(R), if ab+x is not a unit of R, then the path is contained in a subgraph isomorphic to K1+K2+K1.Proof.
The given condition impliesRx+Rab=R. Hence ab∉J(R) and in particular, ab+x≠x. Furthermore, we have
(6)R(ab+x)+Ra=R=R(ab+x)+Rb.
Now assumeab+x∉U(R). Then ab+x∉U(R)∪J(R). If ab+x∉{a,b}, then there is a subgraph K1+K2+K1 in Γ(R) which contains the path a~x~b, see Figure 1. Since R(ab+x)+Ra=R=R(ab+x)+Rb, it follows that ab+x∉{a,b}, and this completes the proof.Figure 1
Lemma6.Lemma 7.
If a vertexx of   Γ(R) is in a cycle of five vertices, then x is in either a triangle or a rectangle.Proof.
Assumea~x~b~c~d~a is a cycle in Γ(R), and x is not in any triangle, see Figure 2. We proceed to verify that x is in a square.
In fact, ifxv≠x for some v∈U(R), then there is a square a~x~b~vx~a in Γ(R). Thus in the following, we assume uy=y,forallu∈U(R) where y∈{a,x,b}. Then by Lemma 6, we can assume further
(*)ab+x=1,xc+b=1,xd+a=1.
Note thatxd is adjacent to a, and xc is adjacent to b in Γ(R). If xd=x, then by (*) we have a=ab. Then there is a rectangle a~x~b~d~a in Γ(R). Therefore we may assume xd∉{x,d},xc∉{x,c}.
Sinced(xc,xd)≤3, we can assume there is a path xc~e~f~xd. If e≠b, then there is a rectangle x~e~xc~b~x. If e=b, then f∉{x,a} and hence by (*) there is a cycle a~x~b~f~xd~a. Then there is a rectangle a~x~f~xd~a and a triangle b~x~f~b. This completes the verification.Figure 2
Lemma7.Lemma 8.
LetG=Γ(R) and assume that G contains a cycle. Then for any path a~x~b in the core of G, x is in a cycle Cn with 3≤n≤5.Proof.
Assume(7)⋯~f~a~x~b~g~⋯
is a cycle in Γ(R). We proceed to verify that x is in a cycle Cn with 3≤n≤5. Since Rab+Rfb=Rb, we have
(8)Rb=Rab+(Ra+Rx)fb=Rab+Rxfb.
ThenRb+R(x-xfb)=Rab+Rx=R. It follows that x-xfb∉U(R)∪J(R). Consider xfb. If xfb=0, then it follows that Rb=Rab+Rxfb=Rab. In this case, there is a rectangle a~x~b~g~a. In the following, assume xfb≠0, that is, x-xfb≠x, and let y=x-xfb. Without loss of generality, assume d(y,f)=3 and y~d~e~f is a path from y to f. If further d≠b, then there is a rectangle b~x~d~y~b. If d=b, then there is a cycle C5:f~a~x~b~e~f. This completes the proof.Theorem 9.
For a ringR, let G=Γ(R) and assume that G contains a cycle. Then the following hold.(1)
The core ofG is a union of triangles and rectangles, and the girth of G is less than or equal to four.(2)
Every vertex ofG is either an end vertex or a vertex of the core.Proof.
(1) The first statement follows from Lemmas7 and 8.
(2) For the second statement, leta~x~b be a path and assume to the contrary that x is not in the core of Γ(R). Since G contains a cycle, by Theorem 1, we can assume that a~x~b~c is a path, where both b and c are in the core. Clearly, 1-ac∈U(R). By Lemma 6, both ab+x and xc+b are units of R. By assumption on x, we can assume further
(9)ab+x=1,xc+b=1.
Ifxc=x, then we have cx+ab=x+b=1, a contradiction. Then xc≠x and b(1-ac)=1-c, where 1-ac∈U(R). Thus we can replace b by 1-c. In a similar way, replace x by 1-a. Now in Γ(R) there is a path a~(1-a)~(1-c)~c, where both c and 1-c are in the core of G. Certainly, there is a vertex d∉{c,1-c} in the core such that 1-c is adjacent to d. Then a similar argument shows 1-c=v(1-d) for some v∈U(R). Since 1-d∉{1-c,c}, and the distance from a to 1-d is at most three, we must have a triangle (1-c)~d~(1-d)~(1-c) in the graph G. Then 1-c=v(1-d) implies (1-d)∈U(R), a contradiction. This completes the proof.


## 3. A RetractΓr(R) of Γ(R)

For simple graphsG and H, recall that G→φH if there exists a map φ:G→H such that for distinct u,v∈V(G), u~v in G implies φ(u)≠φ(v) and φ(u)~φ(v) in H. Such a map φ is called a graph homomorphism from G to H. If H is a subgraph of G, and there is a graph homomorphism G→φH, where the restriction of φ on H is an identity, then H is called a retractof G, see Figure 3 for an example. If G has no proper retract, then G is called a core graph (e.g., Kn is a core graph.). Note that the core of a graph needs not to be a core graph. Recall that the chromatic numberχ(G) of G is the least positive integer r such that there exists a graph homomorphism G→φKr. It is the least number of colors needed for coloring the vertices of G in such a way that no two adjacent vertices have a same color. The girth of a graph G, denoted by g(G), is the length of the minimal cycle in G.Figure 3
Retract for ringℤ12.In this section, we introduce a new graph which is a retract ofΓ(R) and we study this new graph.Definition 10.
For a ringR and any x∈R, let x¯=Rx. Construct a simple graph in the following and denote it as Γr(R)=G:
(10)V(G)={x¯∣x∈R∖(U(R)∪J(R))},E(G)={{x¯,y¯}∣x¯≠y¯∈V(G),andRx+Ry=R}.Clearly, this graph has less vertices and less edges than that of the graphΓ(R) in general, see Figure 3. More precisely, we have the following.Proposition 11.
(1) For a ringR, the graph Γr(R) is a retract of Γ(R).
(2) For any idealI of R contained in J(R), Γ(R/I) is a retract of Γ(R).
Furthermore, we have the following.
(3) If the girth ofΓ(R) is three, then so is the girth of Γr(R).
(4)Γr(R) is connected with diameter less than or equal to three.
(5)ω(Γr(R))=ω(Γ(R)) and χ(Γr(R))=χ(Γ(R)).
(6)Γr(R) and Γ(R) are homomorphically equivalent to the same unique core graph.Proof.
(
1
) Clearly, φ:x↦x¯ is a surjective graph homomorphism. For any x¯∈V(Γr(R)), fix a vertex yx in Rx to obtain a subgraph of Γ(R). Then the graph Γr(R) is isomorphic to the subgraph of Γ(R). Thus Γr(R) is a retract of the graph Γ(R). In a similar way, one checks (2). All the remaining results follow from (1). We check (3) and (5) in the following.
(
3
) The result is clear by the definition of Γr(R). In fact, if G→H and G has odd girth, then it is known that g(G)=g(H), since any cycle with odd length is a core graph.
(5) Since a composition of graph homomorphisms is still a graph homomorphism, clearlyχ(Γr(R))≥χ(Γ(R)). On the other hand, since Γr(R) is isomorphic to a subgraph of Γ(R), for any graph homomorphism ψ:Γ(R)→Ks, the restriction ψ|:Γr(R)→Ks is also a graph homomorphism. This shows the second assertion of (5). The first equality holds by the definition of a retract of a graph. Furthermore, it can be proved that the equalities hold for any graph retract and any graph blowup. See [12] for details.By Theorem2 and Proposition 11, we have the following.Corollary 12.
(1) For any non-local ringR, χ(Γr(R))≥|Max(R)|.
(2)Γr(R) is a complete bipartite graph if and only if R has exactly two maximal ideals, which is true if and only if Γr(R) is a bipartite graph.By [4, Corollary 3.8(1)], if H=Γ(R) contains a cycle, then the girth g(H)≤4. Compared with Proposition 11(3), we have the following example for the g(H)=4 case (see Figure 3).Example 13.
Consider the ringR=ℤ12. For this ring, g(Γ(R))=4, while g(Γr(R))=∞. We draw the two graphs in Figure 3.
Recall a convenient construction from graph theory,the sequential sum(11)G=G1+G2+⋯+Gr
of a sequence of graphs G1,G2,…,Gr with V(Gi)∩V(Gj)=∅ for all i≠j. The vertex set of G is ∪i=1rV(Gi) and the edge set of G is E∪(∪i=1rE(Gi)), where
(12)E={vi~vi+1∣vi∈V(Gi),vi+1∈V(Gi+1),1≤i≤r-1}.
We illustrate the construction in Figure4 for the sequence of graphs K1,K1,H, where H is a triangle K3 together with three end vertices adjacent to distinct vertices.Figure 4
Sequential sum.Theorem 14.
LetR be any commutative ring that is not a local ring, and let G=Γ(R). Then the following numbers are identical:(1)
the chromatic numberχ(G),(2)
the clique numberω(G),(3)
the cardinal number ofMax(R),(4)
χ(Γr(R)),(5)
ω(Γr(R)).Proof.
First, recall from [2, Theorem 2.3] that
(13)χ(Ω(R))=|Max(R)|+|U(R)|.
Then consider the following decomposition of the co-maximal graph into sequential sums of three subgraphs:(14)Ω(R)=J(R)+U(R)+Γ(R),
where J(R) is a discrete subgraph while U(R) is a complete subgraph. Since |J(R)|≤|U(R)|, we have
(15)χ(Ω(R))=χ(Γ(R))+|U(R)|.
By [4, Theorem 3.9(2)], ω(Γ(R))=|Max(R)|. Then
(**)χ(Γ(R))=|Max(R)|=ω(Γ(R)).
By  Proposition11,  we have χ(G)=χ(Γr(R))≥ω(Γr(R))=ω(G), and the result follows from (**).Now we study the interplay between the graph structure ofΓr(R) and the algebraic property of R. First we have the followingProposition 15.
For a ringR, let G=Γr(R). Then the following are equivalent:(1)
A star graph is a spanning subgraph ofG.(2)
G is a star graph.(3)
R≅𝔽×T, where 𝔽 is a field and T is a local ring.Proof.
We only need to prove(1)⇒(3). Assume x¯ is adjacent to every other vertex in Γr(R). Then x¯=x2¯. Assume x=rx2. Then rx=(rx)2 and x¯=rx¯. So we can assume at the start that x is a nontrivial idempotent. Let Rx=T,𝔽=R(1-x). Then T is a maximal ideal of R and R=𝔽×T. Thus 𝔽 is a field. Now J(R)=J(T), and for any y∈T∖J(T), we have T=Ty. Thus T is a local ring with the maximal ideal J(R).Note that for a finite local ring(T,𝔪), if 𝔪=Tx1∪⋯∪Txr in which Txi≠Txj, then Γr(R)=K1,r.Corollary 16.
For any commutative ringR, diam(Γr(R))=1 if and only if R≅𝔽1×𝔽2, where each 𝔽i is a field.Proof.
IfR≅𝔽1×𝔽2, then clearly diam(Γr(R))=1. Conversely, assume diam(Γr(R))=1. Then Γr(R) is a complete graph. By Proposition 15, we have R≅𝔽×T, where 𝔽 is a field and T is a local ring. If T is not a field, then take any nonzero x∈J(T). Then (1,x)∈R∖U(R)∖J(R), and (1,0)¯ is not adjacent to (1,x)¯ in Γr(R), a contradiction. This completes the proof.The following result follows from Corollary12, Proposition 15, and the proof of [3, Lemma 3.2, Proposition 3.3],Proposition 17.
For any commutative non-local ringR, diam(Γr(R))=2 if and only if one of the following conditions holds. (1)
J(R) is a prime ideal of R.(2)
R has exactly two maximal ideals, and R≇𝔽1×𝔽2 for any fields 𝔽i.Corollary 18.
For any commutative non-local ringR, the graphs Γ(R) and Γr(R) have the same diameter if and only if R≇𝔽1×𝔽2 for any fields 𝔽i but 𝔽1=𝔽2=ℤ2.By [4, Theorem 3.9(2)], Theorem 14, and Corollary 16, if diam(Γr(R))=2 (resp., diam(Γ(R))=2), then either Γr(R) (resp., Γ(R)) is a complete bipartite graph or its clique number is infinite.At the end of this section, we pose the following problem.Question 1.
Which ringsR have the property that Γ(R) is a generalized split graph? Which rings R have the property that Γr(R) is a (generalized) split graph?Recall from [13] that a simple graph G is a generalized split graph if
(16)V(G)=K∪D,K∩D=∅,
where the induced subgraph on K (resp., on D) is a core graph (resp., a discrete graph). Note that Γ(ℤ12) is not a generalized split graph, while Γr(ℤ12) is a split graph.


## 4. A RingR Whose Comaximal Graph Ω(R) Is a Split Graph

Throughout this section, assume thatG is a split graph, that is, G is simple and connected with V(G)=K∪D, where K∩D=∅ and the induced subgraph on K (resp., on D) is a complete (resp., discrete) graph, see [13] and the included references there. For the split graph G, we always assume that D is a maximal such independent subset. Under this assumption, a complete graph Kn is a split graph with |K|=n-1,|D|=1.Lemma 19.
For a commutative ringR, let G be the co-maximal graph Ω(R). If G is a split graph with V(G)=K∪D, then we have the following.(1)
For any proper idealI of R, |I∩K|≤1.(2)
𝔪∩𝔫∩K={0} holds for distinct maximal ideals 𝔪,𝔫 of R.(3)
IfR is isomorphic to neither ℤ2 nor ℤ2×ℤ2, then |K|≥|Max(R)|+1. Also, |K|=|Max(R)| if and only if R=ℤ2 or R≅ℤ2×ℤ2.Proof.
(1) Clear.
(2) Assume to the contrary that there exists some nonzerou∈𝔪∩𝔫∩K. Since 𝔪+𝔫=R, we have x∈𝔪,y∈𝔫 such that x+y=1. Clearly, x≠y and hence we can assume x∈K. Then it follows from Lemma 19(1) that x=u∈𝔫, a contradiction.
(3) IfR is a local ring, then either R≅ℤ2 or R has at least two units by assumption. The result holds in this case. In the following, we assume that R is non-local. If there is a maximal ideal 𝔪 with 𝔪⊆D, then by the proof of [6, Theorem 2.1], R≅ℤ2×𝔽 for some field 𝔽. In this case, |K|≥|Max(R)| and, |K|=|Max(R)| if and only if 𝔽=ℤ2. If no maximal ideal is contained in K, then each maximal ideal has exactly one vertex in K by (1). By (2), we have |K|≥|Max(R)|+1.We remark that Lemma19(3) is the best possible. For instance, |K|=|Max(R)|+1 holds for the local ring ℤ4 and Z2×Z2×Z2. For R=Z2×Z2×Z2, we draw its co-maximal graph Ω(R) in Figure 4, in which Ω(R)=K1+K1+H.The following is a result of [6].Lemma 20 (see [6, Theorem 2.1]).
For a commutative ringR which is non-local, if the co-maximal graph Ω(R) is a split graph and R≇ℤ2×𝔽 for any field 𝔽, then the characteristic of R is two, R has exactly three maximal ideal 𝔪i, and K∩𝔪i={xi} for all i, where each xi is idempotent and for every invertible element u of R, uxi=xi.We now give an alternative proof to the main result of [6].Theorem 21 (see [6, Theorem 2.3]).
For a commutative ringR, the co-maximal graph Ω(R) is a split graph if and only if R is one of the following: a local ring, ℤ2×ℤ2×ℤ2, ℤ2×𝔽 for some field 𝔽.Proof.
We only need to prove the necessary part. Assume thatR is not a local ring and R≇ℤ2×𝔽 for any field 𝔽. Let G=Ω(R). Assume further that the co-maximal graph G is a split graph with V(G)=K∪D. By Lemma 20, R has exactly three maximal ideals 𝔪i (1≤i≤3) such that 𝔪i∩K={xi}, where xi2=xi and uxi=xi hold for every u∈U(R). Also, R has characteristic 2. For any 1≤i≠j≤3, assume 1=rxi+sxj with rxi∈K. By Lemma 19, we have xi=rxi and hence 1-xi=sxj. It follows that (1-xi)(1-xj)=0. Note that 1+x1x2x3∈U(R) since x1x2x3∈J(R), and it follows that x1x2x3=0. Then (1-xi)(1-xj)=0 implies xk+xixk+xjxk=0 for any rearrangement i,j,k of 1,2,3. Therefore x1+x2+x3=2·f(x1,x2,x3)=0. This shows that 1-x1,1-x2,1-x3 is a complete set of orthogonal idempotent elements of R.
LetRi=R(1-xi). Then R=∑i=13‍Ri≅R1×R2×R3, where each Ri is a local ring with a unique maximal ideal 𝔫i, since R has exactly three maximal ideals. For any r(1-x1)∈𝔫1, we have
(17)1+r(1-x1)=[(1-x1)+r(1-x1)]+(1-x2)+(1-x3)∈U(R).
Sinceux2=x2 for all u∈U(R), it follows that r(1-x1)·x2=0. Now apply the fact (1-x1)(1-x2)=0; one derives r(1-x1)=0. Hence 𝔫1=0 and each Ri is a field. Thus R is a direct product of three fields.
Sincex1=x2+x3, it follows that
(18)𝔪1=R2+R3,𝔪2=R1+R3,𝔪3=R1+R2.
Then in the decompositionR=K∪D, we deduce from Lemma 19(1) that
(19)K=(U(R1)+U(R2)+U(R3))∪{x1,x2,x3},D=R∖K.
Now we claim that eachRi is isomorphic to ℤ2 and hence R≅ℤ2×ℤ2×ℤ2. In fact, assume R=R1×R2×R3. Then
(20)K=(U(R1)×U(R2)×U(R3))∪{(0,1,1),(1,0,1),(1,1,0)}.
If|R1|>2, then there exists a nonzero element 1≠v1∈R1. Then both z=(v1,0,1) and e=(0,1,0) are in the independent subset D, contradicting Rz+Re=R. This completes the proof.Corollary 22.
A finite split graph is a co-maximal graph of a non-local commutative ring if and only ifG is one of the following: (1) the sequential sum K1+Kpn-1+K1,pn-1 for some prime number p and (2) the sequential sum K1+K1+H in Figure 4, where H is a triangle K3 together with three end vertices adjacent to distinct vertices.The co-maximal graph of a field is certainly a complete graph. For a finite local ring(R,𝔪) which is not a field, assume |𝔪|=pm. Then the co-maximal graph of R is Kpn-pm+Dpm, where pn≥2pm and Dpm is a discrete graph with pm vertices.By Theorem21 and the results of [3, 4], it is natural to ask the following question: for what rings R is Γ(R) a split graph, where the vertex set of Γ(R) is R∖(U(R)∪J(R))? Note that for distinct maximal ideals 𝔪,𝔫 (if exist) and x∈𝔪,y∈𝔫, Rx+Ry=R implies x,y∈R∖(U(R)∪J(R)). Then a careful check to the proofs of Lemma 19, Lemma 20, and Theorem 21 shows the following.Theorem 23.
For any ringR, the following statements are equivalent.(1)
The co-maximal graphΩ(R) of R is a split graph.(2)
The subgraphΓ(R) is either empty or a split graph.(3)
R is one of the following: a local ring, ℤ2×ℤ2×ℤ2, ℤ2×𝔽 for some field 𝔽.Note that a split graphG is isomorphic to Γ(R) for some finite ring R if and only if either G is a star graph K1,pn-1 for some prime number p or G is the triangle together with three end vertices adjacent to distinct vertices (see the graph H in Figure 4).



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*Source: 354696-2013-02-19.xml*