CUBO, A Mathematical Journal

Vol. 24, no. 02, pp. 333–341, August 2022

DOI: 10.56754/0719-0646.2402.0333

Ideal based graph structures for commutative
rings

M. I. Jinnah
1

Shine C. Mathew
2, B

1Formerly of Department of

Mathematics, University of Kerala,

Kariavattom, Thiruvananthapuram,

Kerala, India.

jinnahmi@yahoo.co.in

2Department of Mathematics, St.

Berchmans College, Changanacherry,

Kottayam, Kerala, India.

shinecmathew@gmail.com B

ABSTRACT

We introduce a graph structure Γ∗2(R) for commutative rings

with unity. We study some of the properties of the graph

Γ∗2(R). Also we study some parameters of Γ
∗

2(R) and find

rings for which Γ∗2(R) is split.

RESUMEN

Introducimos una estructura de grafo Γ∗2(R) para ani-

llos conmutativos con unidad. Estudiamos algunas de las

propiedades del grafo Γ∗2(R). También estudiamos algunos

parámetros de Γ∗2(R) y encontramos anillos para los cuales

Γ∗2(R) se escinde.

Keywords and Phrases: Maximal ideal, idempotent, clique number, domination number, split graph.

2020 AMS Mathematics Subject Classification: 13A15, 05C75.

Accepted: 20 July, 2022

Received: 17 November, 2021

c©2022 M. I. Jinnah et al. This open access article is licensed under a Creative Commons

Attribution-NonCommercial 4.0 International License.

http://cubo.ufro.cl/
http://dx.doi.org/10.56754/0719-0646.2402.0333
https://orcid.org/0000-0002-1234-8907
https://orcid.org/0000-0002-8999-2645
mailto:jinnahmi@yahoo.co.in
mailto:shinecmathew@gmail.com


334 M. I. Jinnah & S. C. Mathew CUBO
24, 2 (2022)

1 Introduction

The idea of relating a commutative ring to a graph was introduced by Istvan Beck [3]. He introduced

a graph, Γ(R), whose vertices are the elements of R and two distinct vertices x and y are adjacent

if and only if xy = 0. In [1], Anderson and Livingston modified the definition of Beck to introduce

the zero-divisor graph, Γ∗(R), and investigated many of its properties. Γ∗(R) is the subgraph of

Γ(R) induced by the set of non-zero zero-divisors of R. Cherian Thomas introduced many graph

structures for R in [10] and obtained many interesting results.

Throughout the paper, the word ‘ring’ shall mean a commutative ring with 1 6= 0 which is not a

field. We denote the Jacobson radical of a ring R by J(R) and the set of all maximal ideals by

max R.

For the basic concepts from graph theory refer [4, 9]; for commutative ring theory, see [2].

We give two ideal based graphs, Γ1(R) and Γ2(R), introduced in [10].

The graph Γ1(R) has all ideals of R as vertices and two distinct vertices a and b are adjacent if

and only if ab = 0.

The graph Γ2(R) has the same vertex set as that of Γ1(R) and two distinct vertices a and b are

adjacent if and only if a + b = R.

In [5], the authors have studied the subgraph Γ∗1(R) of Γ1(R) induced by all the non-zero proper

ideals of R.

We state the following result:

Theorem 1.1 ([5]). Let R be an Artin ring. Γ∗1(R) is complete if and only if one of the following

holds:

(i) R ∼= F1 ⊕ F2 where F1 and F2 are fields.

(ii) R is local with maximal ideal m having index of nilpotency 2.

(iii) R is local with principal maximal ideal m having index of nilpotency 3.

In [8], S. C. Mathew has introduced and studied some basic properties of Γ∗2(R) which is the

subgraph of Γ2(R) induced by the set of all non-zero proper ideals of R. In this paper we include

those results, for the sake of completeness. We compare the graphs Γ∗1(R) and Γ
∗
2(R) and find the

clique number and domination number of Γ∗
2
(R). Also we investigate the properties of rings for

which Γ∗
2
(R) is split.



CUBO
24, 2 (2022)

Ideal based graph structures for commutative rings 335

2 The graph Γ∗2(R) and its properties

In this section we define the graph Γ∗2(R) and investigate some properties of the graph.

Definition 2.1. Let R be a ring. We associate a graph Γ∗
2
(R) to R whose vertex set is the set

of all non-zero proper ideals of R and for distinct ideals a and b, the corresponding vertices are

adjacent if and only if a + b = R.

Remark 2.2. Γ∗
2
(R) is totally disconnected if and only if R is local.

Remark 2.3. Γ∗2(R) = K1 if and only if (R, m) is local with m principal and m
2 = 0.

Theorem 2.4. Let R be a non-local ring. Then Γ∗
2
(R) is connected if and only if J(R) = 0.

Proof. (⇒): Assume Γ∗
2
(R) is connected. If J(R) 6= 0, then J(R) is an isolated vertex in Γ∗

2
(R).

(⇐): Assume that J(R) = 0. Now, max R induces a complete subgraph in Γ∗
2
(R). Let a be any

proper non-zero non maximal ideal. Since J(R) = 0, there exists a maximal ideal m such

that a * m. Thus a is adjacent to m and hence Γ∗
2
(R) is connected.

Corollary 2.5. If Γ∗
2
(R) is connected, diam Γ∗

2
(R) ≤ 3.

Remark 2.6. a is an isolated vertex of Γ∗
2
(R) if and only if a ⊆ J(R).

Next result follows from the proof of Theorem 2.4 and Remark 2.6.

Theorem 2.7. Γ∗
2
(R) is connected except for isolated vertices. That is, Γ∗

2
(R) has at most one

component different from K1.

Theorem 2.8. Γ∗
2
(R) ∼= K2 if and only if R is a direct sum of two fields.

Proof. (⇒): Let R ∼= F1 ⊕ F2 where F1 and F2 are fields. Then the ideals of R are F1 ⊕ 0, 0 ⊕ F2,

0 ⊕ 0 and F1 ⊕ F2. Then, Γ
∗
2
(R) ∼= K2.

(⇐): Suppose Γ∗2(R)
∼= K2. Then R is non-local. Also, R cannot have more than two maximal

ideals. Therefore R has exactly two maximal ideals, say m1 and m2 with m1 ∩ m2 = 0. This

implies R ∼=
R

m1
⊕

R

m2
, a direct sum of two fields.

Theorem 2.9. The only triangle free connected graphs that can be realized as Γ∗2(R) are K1

and K2.

Proof. Let G be a triangle free connected graph. Since G is triangle free R can have at most two

maximal ideals. Also since G is connected the result follows.

Theorem 2.10. Γ∗
2
(R) is complete if and only if either R is a direct sum of two fields or R is

local with principal maximal ideal having index of nilpotency 2.



336 M. I. Jinnah & S. C. Mathew CUBO
24, 2 (2022)

Proof. (⇒): If Γ∗2(R) is complete, R can have at most two maximal ideals. For, assume R has 3

maximal ideals say, m1, m2 and m3. Then m1m2 = 0; otherwise m1m2 is a vertex of Γ
∗
2
(R) and

will not be adjacent to m1 and m2. For the same reason, m1m3 = 0. Then m1(m2 + m3) = 0.

This implies m1 = 0 which is not possible. Now assume that R has exactly 2 maximal ideals

say, m1 and m2. Then J(R) = m1m2 = {0}. Thus R is a direct sum of 2 fields. Now, if R

is local with maximal ideal m, since Γ∗
2
(R) is complete, m must be principal with index of

nilpotency 2.

(⇐): If R is a direct sum of two fields, Γ∗2(R)
∼= K2 and if R is local with principal maximal ideal

having index of nilpotency 2, Γ∗
2
(R) ∼= K1.

The following corollary is immediate.

Corollary 2.11. The only complete graphs that can be realized as Γ∗2(R) are K1 and K2.

3 Comparison between Γ∗1(R) and Γ
∗
2(R)

Theorem 3.1. Assume diam Γ∗
2
(R) = 2. Then any two vertices in Γ∗

2
(R) which are not adjacent

are also not adjacent in Γ∗1(R). That is, Γ
∗
1(R) is a subgraph of Γ

∗
2(R).

Proof. Let diam Γ∗
2
(R) = 2. Suppose a and b are not adjacent in Γ∗

2
(R). Then, there exists a

maximal ideal m such that a + m = R = b + m. Therefore, (a + m)(b + m) = R. That is,

ab + am + bm + m2 = R.

But, ab + am + bm + m2 ⊆ ab + m. Therefore, ab + m = R. This implies, in particular, ab 6= 0.

Thus, a and b are not adjacent in Γ∗
1
(R).

Remark 3.2. Suppose a and b are adjacent in Γ∗
2
(R). Then, a + b = R. This implies ab = a ∩ b.

Hence a is adjacent to b in Γ∗1(R) if and only if a ∩ b = 0. This must hold for every pair of

comaximal ideals a and b.

Theorem 3.3. Let R be a non-local ring. Then, Γ∗
2
(R) is a subgraph of Γ∗

1
(R) if and only if R is

a direct sum of two fields; and hence Γ∗
1
(R) = Γ∗

2
(R) only when R is a direct sum of two fields.

Proof. (⇒): Γ∗
2
(R) is a subgraph of Γ∗

1
(R) if and only if for any pair of comaximal ideals a and b

of R, ab = 0. So, if Γ∗2(R) is a subgraph of Γ
∗
1(R), in particular, m1m2 = 0 where m1 and m2

are two maximal ideals of R. Hence, R ∼= R
m1

⊕ R
m2

.

(⇐): If R is a direct sum of two fields, Γ∗
1
(R) = Γ∗

2
(R) = K2.

Theorem 3.4. If R is a finite direct sum of fields, Γ∗1(R)
∼= Γ∗2(R).



CUBO
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Ideal based graph structures for commutative rings 337

Proof. Let R = F1 ⊕ F2 ⊕ · · · ⊕ Fn where Fi’s are fields. Thus, an ideal a of R is of the form,

a1 ⊕ a2 ⊕ · · · ⊕ an where, ai = 0 or Fi.

Define ϕ : V (Γ∗
1
(R)) → V (Γ∗

2
(R)) by ϕ(a1 ⊕ a2 ⊕ · · · ⊕ an) = b1 ⊕ b2 ⊕ · · · ⊕ bn where

bi =











Fi, if ai = (0)

0, if ai = Fi.

Clearly, ϕ is a bijection.

Suppose a and b are adjacent in Γ∗
1
(R). Thus b must contain 0 at the positions in which a contains

Fi’s. Therefore, ϕ(b) contains Fi’s at the positions where ϕ(a) contains 0. Then, ϕ(a) is adjacent

to ϕ(b).

Similarly, if ϕ(a) and ϕ(b) are adjacent in Γ∗
2
(R) then, a and b are adjacent in Γ∗

1
(R). Thus, ϕ is

a graph isomorphism. That is, Γ∗1(R)
∼= Γ∗2(R).

Remark 3.5. In the context of Theorem 3.4, we can explicitly determine Γ∗1(R) and Γ
∗
2(R) by

identifying the vertex set with the power set P(X) \ {X, ∅} where X = {1, 2, . . . , n} and A ⊂ X

with
⊕

i∈A Fi. Then A and B are adjacent in Γ
∗
1(R) if and only if A ∩ B = ∅ and A and B are

adjacent in Γ∗
2
(R) if and only if A ∪ B = X.

Theorem 3.6. Γ∗
1
(R) and Γ∗

2
(R) are edge disjoint if and only if R has no non-trivial idempotents.

Proof. (⇒): Suppose that R contains a non-trivial idempotent e. Then, R = Re ⊕ R(1 − e). This

implies, Re + R(1 − e) = R and Re ∩ R(1 − e) = ReR(1 − e) = 0. That is, Γ∗
1
(R) and Γ∗

2
(R)

are not edge disjoint.

(⇐): Assume that Γ∗
1
(R) and Γ∗

2
(R) are not edge disjoint and then there exist two ideals a and b

such that a+b = R and a∩b = ab = 0. Then, R = a⊕b and hence, a = Re and b = R(1−e)

for some idempotent e. Since a and b are non-zero proper ideals, e must be non-trivial.

Theorem 3.7. Let R be a non-local ring. If Γ∗
1
(R) = Γ∗

2
(R), R is not semi-local.

Proof. Assume that R is semi-local with maximal ideals m1, m2, . . . , mn. Then, there are the

following possibilities.

Case (I): Γ∗2(R) is connected.

This assumption implies m1m2 · · · mn = 0, by Theorem 2.4. Therefore, (m1 · · · mn−1), mn are

adjacent in Γ∗1(R) as well as in Γ
∗
2(R), which means Γ

∗
1(R) 6= Γ

∗
2
(R).

Case (II): Γ∗
2
(R) is disconnected.

This implies m1m2 · · · mn = J(R) 6= 0. We subdivide this case into two.



338 M. I. Jinnah & S. C. Mathew CUBO
24, 2 (2022)

Case (II)(a): J(R) is nilpotent.

Then there exist least positive integers k1, k2, . . . , kn such that m
k1
1
m

k2
2

· · · mknn = 0 with

at least one kj > 1 for 1 ≤ j ≤ n, say kn > 1.

If kn > 2, we have (m1 · · · mn) + mn 6= R and (m1 · · · mn)mn 6= 0. That is, Γ
∗
1
(R) 6=

Γ∗
2
(R).

Now consider the case when kn = 2. If ki > 1 for some i 6= n, (m1 · · · mn)+ mn 6= R and

(m1 · · · mn)mn 6= 0. If ki = 1 ∀i 6= n, (m1 · · · ml−1ml+1 · · · m
2
n) + ml = R where l 6= n.

But, (m1 · · · ml−1ml+1 · · · m
2
n)ml = 0. So, Γ

∗
1
(R) 6= Γ∗

2
(R).

Case (II)(b): J(R) is not nilpotent.

In this case we have (m1 · · · mn) + m1 6= R and (m1 · · · mn)m1 6= 0.

Thus, if Γ∗
1
(R) = Γ∗

2
(R), R cannot be semi-local.

Theorem 3.8. Let (R, m) be an Artin local ring. Then, Γ∗1(R) = Γ
∗
2
(R) if and only if either m

has index of nilpotency 2 or m is principal with index of nilpotency 3.

Proof. Follows from Remark 2.2 and Theorem 1.1.

4 Some parameters of Γ∗2(R)

In this section we find the clique number and the domination number of Γ∗
2
(R).

Theorem 4.1. cl(Γ∗2(R)) = | max R|.

Proof. Clearly max R induces a complete subgraph. Let a be any non-zero non-maximal proper

ideal of R. Then a is contained in a maximal ideal. That is, there exists a maximal ideal m such

that a is not adjacent to m. Thus, max R induces a maximal complete subgraph.

Now suppose S = {ai : i ∈ Λ}, where Λ is an index set, induces a complete subgraph in Γ
∗
2
(R).

Then one maximal ideal can contain at most one ai ∈ S. That is, there exists an injective map

from S to max R. This implies, |S| ≤ | max R|. Thus, cl(Γ∗
2
(R)) = | max R|.

Theorem 4.2. Let R be a semi local ring with | max R| = n > 2. Then, γ(Γ∗
2
(R)) = | max R| +

Number of isolated vertices in Γ∗2(R).

Proof. Let Γ∗∗
2
(R) be the connected component of Γ∗

2
(R) induced by the non-isolated vertices of

Γ∗2(R). Now, by Theorem 2.7, it is enough to show that γ(Γ
∗∗
2 (R)) = | max R|.

Let max R = {m1, m2, . . . , mn}. Clearly max R is a dominating set for Γ
∗∗
2 (R). Now consider,

S = {m2 · · · mn, m1m3 · · · mn, . . . , m1m2 · · · mn−1}, which is an independent set in Γ
∗∗
2
(R). Note

that any ideal a /∈ S can be adjacent only to at most one element of S. So every dominating set



CUBO
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Ideal based graph structures for commutative rings 339

in Γ∗∗2 (R) must contain at least n elements. Thus, γ(Γ
∗∗
2 (R)) = n = | max R|. Hence the result

follows.

Remark 4.3. If R is a semi-local ring with | max R| = 2 then, the above result is not true. For

example, if R is a direct sum of two fields, γ(Γ∗
2
(R)) = γ(K2) = 1 but | max R| = 2.

5 Splitness

A graph (V, E) is said to be a split graph if V is the disjoint union of two sets K and S where

K induces a complete subgraph and S is an independent set. Then, we can assume either K is

a clique or S is a maximal independent set. In [6] & [7], the authors have carried out a detailed

study on splitness of some graphs associated with a ring. In this section we continue the study in

the case of Γ∗2(R).

Lemma 5.1. Let R = R1 × R2 × R3 be a ring. If Γ
∗
2(R) is split, each Ri must be a field.

Proof. Suppose R1 is not a field. Then there exists a proper non-zero ideal I of R1. Then,

{I × R2 × R3, R1 × R2 × 0, 0 × R2 × R3, R1 × 0 × 0} induces a C4 in Γ
∗
2
(R), a contradiction.

Lemma 5.2. If Fi (1 ≤ i ≤ 3) are fields and R = F1 × F2 × F3 then Γ
∗
2
(R) is split.

Proof. V (Γ∗
2
(R)) can be partitioned into K = {F1 × F2 × 0, F1 × 0 × F3, 0 × F2 × F3} and S =

{F1 × 0 × 0, 0 × F2 × 0, 0 × 0 × F3} where K induces a complete subgraph and S is an independent

set.

Lemma 5.3. Let F be a field and R1 a local ring. Let R = R1 × F. Then Γ
∗
2
(R) is split.

Proof. Let {Ij : j ∈ J} be the collection of non-zero proper ideals of R1. Then {Ij × F : j ∈

J} ∪ {Ij × 0 : j ∈ J} is an independent set and {0 × F, R1 × 0} is a K2. This forms a partition of

V (Γ∗2(R)). Thus, Γ
∗
2(R) is split.

Lemma 5.4. Suppose R has exactly n maximal ideals mi (1 ≤ i ≤ n) with each mi being generated

by an idempotent ei. Then R ∼=

n
∏

i=1

Fi where each Fi ∼= R/mi, a field.

Proof. Let e =
∏n

i=1
ei. Then e ∈ J(R). Therefore, 1 − e is a unit (and an idempotent). So,

1 − e = 1 ⇒ e = 0. Then by the Chinese Remainder Theorem,

R ∼=
R

∏n
i=1

Rei
∼=

R
⋂n

i=1
Rei

∼=

n
∏

i=1

R

Rei
.



340 M. I. Jinnah & S. C. Mathew CUBO
24, 2 (2022)

Theorem 5.5. Let R be a ring. Γ∗2(R) is a split graph if and only if one of the following conditions

holds:

(i) R is local.

(ii) R ∼= R1 × F where R1 is a local ring and F is a field.

(iii) R ∼= F1 × F2 × F3 where Fi’s are fields.

Proof. First we note that Γ∗2(R) is split if and only if Γ2(R) is split. Also, if R is local, Γ
∗
2(R)

is split. Sufficiency of other conditions follows from the lemmas. To prove the necessity of the

conditions, we assume that R is not local and V (Γ2(R)) is the disjoint union of two sets K and S

where K induces a complete subgraph and S is an independent set. We assume that K and S are

non-empty. Also, S can contain at most one maximal ideal.

Case (I): S contains a maximal ideal, say m1.

In this case, R can have only one maximal ideal other than m1. For, if m2 and m3 are distinct

maximal ideals other than m1, then m2 and m3 are in K. Then, m2m3 ∈ S, m1 ∈ S. Clearly,

m1 + m2m3 = R, a contradiction. Thus, R contains only one maximal ideal other than m1,

say m2 which belongs to K. Let xi ∈ mi (i = 1, 2) with x1+x2 = 1. As m
2
2
+m1 = R, m

2
2
∈ K

which implies m22 = m2. Similarly, as Rx2 + m1 = R, Rx2 ∈ K which implies m2 = Rx2.

Then, m2 is a finitely generated maximal ideal which is idempotent. Hence, m2 is generated

by an idempotent. So, R ∼= R1 × F where F is a field and m2 is isomorphic to the ideal

R1 × {0}. Further, R1 must be local.

Case (II): S contains no maximal ideal.

In this case, R can have at most three maximal ideals, for, if m1, m2, m3 and m4 are distinct

maximal ideals, m1m2 and m3m4 are in S which leads to a contradiction. If R has only two

maximal ideals, say, m1 and m2, then m1, m2 ∈ K. Since, m
2
i + mi 6= R (i = 1, 2), we have

m2
1
, m2

2
∈ S. But m2

1
+ m2

2
= R. So, to avoid a contradiction we have to assume m2

1
= m1 or

m22 = m2. That is, R
∼= R1 × F where F is a field and R1 is a local ring. So, let us assume

R has exactly 3 maximal ideals m1, m2 and m3. Note that mi ∈ K (i = 1, 2, 3). Then, as

m1 + m2m3 = R, there exists x1 ∈ m1 such that Rx1 + m2m3 = R which implies Rx1 ∈ K

and hence, Rx1 = m1. Similarly arguing with m
2
1
+ m2m3 = R, we get m1 = m

2
1
. Then m1

is generated by an idempotent. Similarly each mj (j = 2, 3) is generated by an idempotent.

Then by the Lemma 5.4, R ∼= F1 × F2 × F3 where Fi (1 ≤ i ≤ 3) are fields.



CUBO
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Ideal based graph structures for commutative rings 341

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[3] I. Beck, “Coloring of commutative rings”, J. Algebra, vol. 116, no. 1, pp. 208–226, 1988.

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[6] M. I. Jinnah and S. C. Mathew, “When is the comaximal graph split?”, Comm. Algebra, vol.

40, no. 7, pp. 2400–2404, 2012.

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	Introduction
	The graph 2*(R) and its properties
	Comparison between 1*(R) and 2*(R)
	Some parameters of 2*(R)
	Splitness