Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case
The rings considered in this article are nonzero commutative with identity which are not fields. Let R be a ring. We denote the collection of all proper ideals of R by I(R) and the collection I(R)\{(0)} by I(R)*. Recall that the intersection graph of ideals of R, denoted by G(R), is an undirected g...
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| Zitieren: | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case / P. Vadhel, S. Visweswaran // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 130–143. — Бібліогр.: 19 назв. — англ. |
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| citation_txt | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case / P. Vadhel, S. Visweswaran // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 130–143. — Бібліогр.: 19 назв. — англ. |
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| description | The rings considered in this article are nonzero commutative with identity which are not fields. Let R be a ring. We denote the collection of all proper ideals of R by I(R) and the collection I(R)\{(0)} by I(R)*. Recall that the intersection graph of ideals of R, denoted by G(R), is an undirected graph whose vertex set is I(R)* and distinct vertices I, J are adjacent if and only if I ∩ J ≠ (0). In this article, we consider a subgraph of G(R), denoted by H(R), whose vertex set is I(R)* and distinct vertices I, J are adjacent in H(R) if and only if IJ ≠ (0). The purpose of this article is to characterize rings R with at least two maximal ideals such that H(R) is planar.
|
| first_indexed | 2025-12-07T17:34:08Z |
| format | Article |
| fulltext |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 26 (2018). Number 1, pp. 130–143
c© Journal “Algebra and Discrete Mathematics”
Planarity of a spanning subgraph
of the intersection graph of ideals
of a commutative ring I, nonquasilocal case
P. Vadhel and S. Visweswaran
Communicated by D. Simson
Abstract. The rings considered in this article are nonzero
commutative with identity which are not fields. Let R be a ring.
We denote the collection of all proper ideals of R by I(R) and the
collection I(R)\{(0)} by I(R)∗. Recall that the intersection graph of
ideals of R, denoted by G(R), is an undirected graph whose vertex
set is I(R)∗ and distinct vertices I, J are adjacent if and only if
I ∩J 6= (0). In this article, we consider a subgraph of G(R), denoted
by H(R), whose vertex set is I(R)∗ and distinct vertices I, J are
adjacent in H(R) if and only if IJ 6= (0). The purpose of this article
is to characterize rings R with at least two maximal ideals such that
H(R) is planar.
1. Introduction
The rings considered in this article are commutative with identity
1 6= 0. Let R be a ring. As in [10], we denote the collection of all proper
ideals of R by I(R) and I(R) \ {(0)} by I(R)∗. The rings R considered in
this article are such that I(R)∗ 6= ∅. The idea of associating a ring with a
graph and studying the interplay between ring-theoretic properties of the
ring and the graph-theoretic properties of the graph associated with it was
2010 MSC: 13A15, 05C25.
Key words and phrases: quasilocal ring, special principal ideal ring, clique
number of a graph, planar graph.
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P. Vadhel, S. Visweswaran 131
initiated by I. Beck in [9] and subsequently, a lot of research activity has
been carried out by several researchers in this area (see for example, [2, 3,
4, 7, 11, 15]). The study of the intersection graph of ideals of a ring has
begun with the work of Chakrabarthy, Ghosh, Mukherjee and Sen [12]. Let
R be a ring with identity which is not necessarily commutative. We denote
the collection of all proper left ideals of R by LI(R) and the collection
LI(R) \ {(0)} by LI(R)∗. Recall from [12] that the intersection graph of
ideals of R, denoted by G(R), is an undirected graph whose vertex set is
LI(R)∗ and distinct vertices I, J are adjacent if and only if I ∩ J 6= (0).
For any n > 2, we denote the ring of integers modulo n by Zn. In [12],
among other results, the planarity of intersection graph of ideals of Zn
was discussed. Inspired by their work, in [14], S.H. Jaffari and N. Jaffari
Rad characterized commutative ring R with identity such that G(R) is
planar. An improvement of the results presented in [14] regarding the
planarity of G(R) was given in [16]. The intersection graph of ideals of
a ring has also been studied by other researchers (see for example, [1,
5, 17]). Inspired by the work done on G(R), with any ring R such that
|I(R)∗| > 1, in [19], we introduced and investigated an undirected graph,
denoted by H(R), whose vertex set is I(R)∗ and distinct vertices I, J are
adjacent in H(R) if and only if IJ 6= (0). Note that for any ideals I, J of
a ring R, IJ ⊆ I ∩ J . Hence, if IJ 6= (0), then I ∩ J 6= (0). Thus distinct
I, J ∈ I(R)∗ are adjacent in H(R), then I and J are adjacent in G(R). For
a graph G, we denote the vertex set of G by V (G) and the edge set of G
by E(G). As V (H(R)) = V (G(R)) = I(R)∗, it follows from the arguments
given above that H(R) is a spanning subgraph of G(R). For any set A,
we denote the cardinality of A by |A|. For any ring R, we denote the set
of all maximal ideals of R by Max(R). Motivated by the work done on
the planarity of G(R) in [14, 16], in this article, we focus our study on
characterizing rings R with |Max(R)| > 2 such that H(R) is planar.
It is useful to recall the following results from graph theory. The graphs
considered in this article are undirected and simple. Let G = (V,E) be a
graph. Recall from [8, 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. A graph G = (V,E) is said to be complete
if any two distinct vertices of G are adjacent in G and for any n ∈ N, a
complete graph on n vertices is denoted by Kn. G is said to be bipartite
if the vertex set V of G is partitioned into two nonempty subsets V1 and
V2 such that each edge of G has one end in V1 and the other end in V2. A
bipartite graph with vertex partition V1 and V2 is said to be complete if
each element of V1 is adjacent to all the vertices of V2. Let m,n ∈ N. Let
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132 Planarity of a spanning subgraph, I
G = (V,E) be a complete bipartite graph with vertex partition V1 and
V2. If |V1| = m and |V2| = n, then G is denoted by Km,n [8, Definition
1.1.12].
Recall from [13, page 9] that two adjacent edges of a graph G are
said to be in series if their common end vertex is of degree two. Two
graphs are said to be homeomorphic if one graph can be obtained from the
other by the insertion of vertices of degree two or by the merger of edges
in series [13, page 100]. We note from [13, page 93] that K5 is referred
to as Kuratowski’s first graph and K3,3 is referred to as Kuratowski’s
second graph. The celebrated theorem of Kuratowski states that a finite
graph G = (V,E) is planar if and only if G does not contain either of
Kuratowski’s two graphs or any graph homeomorphic to either of them
[13, Theorem 5.9].
Let G = (V,E) be a graph. A clique of G is a complete subgraph of G
[8, Definition 1.2.2]. Suppose that there exists k ∈ N such that any clique
of G contains at most k vertices. Then the clique number of G, denoted
by ω(G), is defined as the largest positive integer n such that G contains
a clique on n vertices [8, page 185]. If G contains a clique on n vertices
for all n > 1, then we set ω(G) = ∞.
It is convenient to name the conditions satisfied by a graph G = (V,E):
(C1) G does not contain K5 as a subgraph (that is, equivalently,
if ω(G)64);
(C2) G does not contain K3,3 as a subgraph;
(C∗
1 ) G satisfies (C1) and moreover, G does not contain any subgraph
homeomorphic to K5;
(C∗
2 ) G satisfies (C2) and moreover, G does not contain any subgraph
homeomorphic to K3,3.
If a graph G is planar, then it follows from Kuratowski’s theorem
[13, Theorem 5.9] that G satisfies both (C∗
1) and (C∗
2) and hence, G
satisfies both (C1) and (C2). It is interesting to note that a graph G can
be nonplanar, even if it satisfies both (C1) and (C2). For an example of
this type, refer [13, Figure 5.9(a), page 101] and the graph given in this
example does not satisfy (C∗
2 ). It is not hard to construct an example of
a graph G such that G satisfies (C1) but G does not satisfy (C∗
1 ).
As is already mentioned in the beginning, the rings considered in
this article are commutative with identity 1 6= 0. A ring R is said to be
quasilocal (semiquasilocal) if R has only one maximal ideal (respectively,
R has only a finite number of maximal ideals). A Noetherian quasilocal
(semiquasilocal) ring R is referred to as a local (respectively, a semilocal)
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P. Vadhel, S. Visweswaran 133
ring. Whenever a set A is a subset of a set B and A 6= B, we denote it
symbolically by A ⊂ B.
The aim of this article is to characterize rings R with |Max(R)| > 2
(that is, to characterize nonquasilocal rings R) such that H(R) is planar.
Moreover, our aim is to investigate whether the algebraic structure of
R plays a role to arrive at the conclusion that H(R) is planar if H(R)
satisfies at least one between (C1) and (C2).
This article consists of four sections. In Section 2, we state and prove
several preliminary results that are needed for proving the main results
of this article. Let R be a ring. We denote the nilradical of R by nil(R)
and the Jacobson radical of R by J(R). In Section 2, we assume that R
is a ring such that I(R)∗ 6= ∅ and we do not put any restriction on the
number of maximal ideals of R. It is proved in Corollary 2.3 that if H(R)
satisfies either (C1) or (C2), then |Max(R)| 6 3. If H(R) satisfies either
(C1) or (C2), then it is verified in Corollary 2.11 that J(R) is nilpotent.
In Section 3, we consider rings R such that |Max(R)| = 3. It is proved
in Theorem 3.2 that H(R) satisfies (C1), if and only if H(R) satisfies
(C2), if and only if R ∼= F1 × F2 × F3 as rings, where Fi is a field for each
i ∈ {1, 2, 3}, if and only if H(R) is planar.
In Section 4, we consider rings R such that |Max(R)| = 2. The main
result proved in Section 4 is Theorem 4.8. It is shown in Theorem 4.8
that H(R) satisfies (C1), if and only if H(R) satisfies (C2), if and only
if H(R) is planar and moreover, in Theorem 4.8, we characterize up to
isomorphism of rings, rings R such that H(R) is planar.
2. Some preliminary results
The aim of this section is to state and prove some preliminary results
that are needed for proving the main results of this article. The rings R
considered in this section are such that I(R)∗ 6= ∅.
Lemma 2.1. Let R be a ring such that H(R) does not contain any infinite
clique. Then R is semiquasilocal.
Proof. Assume that H(R) does not contain any infinite clique. Suppose
that Max(R) is infinite. Then it is possible to find a subset {mi|i ∈ N} of
Max(R). It is clear that for any distinct i, j ∈ N, mimj 6= (0) and so, mi
and mj are adjacent in H(R). Therefore, the subgraph of H(R) induced
by {mi|i ∈ N} is an infinite clique. This is a contradiction. Therefore, R
is semiquasilocal.
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134 Planarity of a spanning subgraph, I
Lemma 2.2. Let R be a ring. Let n > 4. If ω(H(R)) 6 n + 1, then
|Max(R)| 6 n− 1.
Proof. Assume that ω(H(R)) 6 n + 1, where n > 4. Suppose that
|Max(R)| > n. Let {mi|i ∈ {1, 2, . . . , n}} ⊆ Max(R). As |Max(R)| > 4,
it follows that for any three distinct m, n, p ∈ Max(R) and for any nonneg-
ative integers i, j, k, minjpk 6= (0). Hence, the subgraph of H(R) induced
by {mi|i ∈ {1, 2, . . . , n}} ∪ {m1m2,m1m3} is a clique on n + 2 vertices.
This is a contradiction and so, we obtain that |Max(R)| 6 n− 1.
Corollary 2.3. Let R be a ring. If H(R) satisfies either (C1) or (C2),
then |Max(R)| 6 3.
Proof. Observe that if ω(G) > 6 for a graph G, then G neither satisfies
(C1) nor satisfies (C2). Thus if H(R) satisfies either (C1) or (C2), then
ω(H(R)) 6 5. Hence, on applying Lemma 2.2 with n = 4, it follows that
|Max(R)| 6 3.
It is well-known that for any ring R, nil(R) ⊆ J(R).
Lemma 2.4. Let R be a ring. If a ∈ J(R) \ nil(R), then Ran 6= Ram for
all distinct n,m ∈ N.
Proof. Suppose that Ran = Ram for some distinct n,m ∈ N. We can
assume without loss of generality that n < m. Now, an = ram for some
r ∈ R. This implies that an(1− ram−n) = 0. Since a ∈ J(R), 1− ram−n
is a unit in R, and so, we obtain that an = 0. This is in contradiction to
the assumption that a /∈ nil(R). Therefore, Ran 6= Ram for all distinct
n,m ∈ N.
Lemma 2.5. Let R be a ring such that H(R) does not contain any infinite
clique. Then nil(R) = J(R).
Proof. Assume that H(R) does not contain any infinite clique. Suppose
that nil(R) 6= J(R). Then there exists a ∈ J(R) \ nil(R). Then ak 6= 0
for all k ∈ N. We know from Lemma 2.4 that Ran 6= Ram for all distinct
n,m ∈ N. Note that the subgraph of H(R) induced by {Ran|n ∈ N} is
an infinite clique. This is a contradiction and so, nil(R) = J(R).
Lemma 2.6. Let R be a ring such that H(R) does not contain any infinite
clique. Then for any m ∈ Max(R), nil(Rm) = mRm.
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P. Vadhel, S. Visweswaran 135
Proof. Assume that H(R) does not contain any infinite clique. We know
from Lemma 2.1 that R is semiquasilocal. Let {mi|i ∈ {1, . . . , n}} denote
the set of all maximal ideals of R. Hence, J(R) = ∩n
i=1mi. We know
from Lemma 2.5 that nil(R) = J(R). Let i ∈ {1, . . . , n}. It follows
from [6, Corollary 3.12] that nil(Rmi
) = (nil(R))R\mi
= (∩n
k=1
mk)R\mi
=
∩n
k=1
mkRmi
. Since mkRmi
= Rmi
for all k ∈ {1, . . . , n} \ {i}, it follows
that nil(Rmi
) = miRmi
.
Lemma 2.7. Let R be a ring such that H(R) does not contain any infinite
clique. Then for any m ∈ Max(R), Rm satisfies descending chain condition
(d.c.c.) on principal ideals.
Proof. Assume that H(R) does not contain any infinite clique. Let m ∈
Max(R). Suppose that Rm does not satisfy d.c.c. on principal ideals. Then
for each i ∈ N, there exists xi ∈ Rm such that Rmx1 ⊃ Rmx2 ⊃ Rmx3 ⊃ · · ·
is a strictly descending sequence of principal ideals of Rm. It is clear that
for each i ∈ N, xi 6=
0
1
and xi+1 = yixi for some yi ∈ mRm. Hence, for each
i ∈ N, xi+1 = (
∏i
j=1
yj)xi. Therefore,
∏i
j=1
yj 6= 0
1
for each i ∈ N. We
know from Lemma 2.6 that each element of mRm is nilpotent. As yi ∈ mRm
for each i ∈ N, it follows that there exist integers 1 6 i1 < i2 < i3 < · · ·
such that Rmyij 6= Rmyik for all distinct j, k ∈ N. Note that for each i ∈ N,
there exist ri ∈ m and si ∈ R \m such that yi =
ri
si
. As Rm
ri
1
= Rmyi for
each i ∈ N, it follows from the above discussion that the subgraph of H(R)
induced by {Rrij |j ∈ N} is an infinite clique. This is in contradiction to
the assumption that H(R) does not contain any infinite clique. Therefore,
Rm satisfies d.c.c. on principal ideals for each m ∈ Max(R).
Lemma 2.8. Let R be a ring such that H(R) does not contain any infinite
clique. Then R satisfies d.c.c. on principal ideals.
Proof. Assume that H(R) does not contain any infinite clique. We know
from Lemma 2.1 that R is semiquasilocal. It is shown in Lemma 2.7 that
Rm satisfies d.c.c. on principal ideals for each m ∈ Max(R). Therefore, we
obtain that R satisfies d.c.c. on principal ideals.
Lemma 2.9. Let I be an ideal of a ring R such that I ⊆ nil(R) and
I = I2. If R satisfies d.c.c. on principal ideals, then I = (0).
Proof. This is [18, Lemma 2.8].
Lemma 2.10. Let R be a ring such that H(R) does not contain any
infinite clique. Then nil(R) is nilpotent.
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136 Planarity of a spanning subgraph, I
Proof. Assume that H(R) does not contain any infinite clique We know
from Lemma 2.8 that R satisfies d.c.c. on principal ideals. Suppose that
nil(R) is not nilpotent. If (nil(R))i 6= (nil(R))j for all distinct i, j ∈ N,
then it follows that the subgraph of H(R) induced by {(nil(R))i|i ∈ N} is
an infinite clique. This is impossible and so, there exist i, j ∈ N with i < j
such that (nil(R))i = (nil(R))j . Let us denote (nil(R))i by I. Note that
I ⊆ nil(R) and I = I2 and so, it follows from Lemma 2.9 that I = (0).
This proves that nil(R) is nilpotent.
Corollary 2.11. Let R be a ring. If H(R) satisfies either (C1) or (C2),
then J(R) is nilpotent.
Proof. Assume that H(R) satisfies either (C1) or (C2). Then ω(H(R)) 6 5.
We know from Lemma 2.5 that J(R) = nil(R) and so, we obtain from
Lemma 2.10 that J(R) is nilpotent.
3. The case when R has exactly three maximal ideals
The aim of this section is to characterize rings R with |Max(R)| = 3 such
that H(R) is planar.
Lemma 3.1. Let R be a ring such that |Max(R)| = 3. If H(R) satisfies
either (C1) or (C2), then J(R) = (0).
Proof. Let {mi|i ∈ {1, 2, 3}} denote the set of all maximal ideals of R.
Then J(R) = ∩3
i=1mi. As distinct maximal ideals of a ring are comaximal,
it follows from [6, Proposition 1.10(i)] that J(R) =
∏
3
i=1
mi.
Assume that H(R) satisfies (C1). Suppose that J(R) 6= (0). We
consider two cases.
Case 1: m1 = m2
1. Note that the subgraph of H(R) induced by {m1,m2,m3,
m1m2,m1m3} is a clique on five vertices. This is impossible.
Case 2: m1 6= m2
1. Observe that the subgraph of H(R) induced by
{m1,m2,m3,m1m2,m
2
1} is a clique on five vertices. This is impossible.
Thus, if H(R) satisfies (C1), then J(R) = (0).
Assume that H(R) satisfies (C2). Suppose that J(R) 6= (0). Let A =
{m1,m2,m3} and let B = {m1m2,m1m3,m2m3}. Note that A∩B = ∅ and
the subgraph of H(R) induced by A∪B contains K3,3 as a subgraph. This
is in contradiction to the assumption that H(R) satisfies (C2). Therefore,
J(R) = (0).
This shows that if H(R) satisfies either (C1) or (C2), then J(R) = (0).
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P. Vadhel, S. Visweswaran 137
Theorem 3.2. Let R be a ring such that |Max(R)| = 3. Then the follo-
wing statements are equivalent:
(i) H(R) satisfies (C1).
(ii) R ∼= F1 × F2 × F3 as rings, where Fi is a field for each i ∈ {1, 2, 3}.
(iii) H(R) is planar.
(iv) H(R) satisfies (C2).
(v) H(R) satisfies both (C∗
1 ) and (C∗
2 ).
Proof. Let {m1,m2,m3} denote the set of all maximal ideals of R.
(i) ⇒ (ii) and (iv) ⇒ (ii). We know from Lemma 3.1 that ∩3
i=1mi = (0).
Hence, we obtain from [6, Proposition 1.10 (ii) and (iii)] that the mapping
f : R → R
m1
× R
m2
× R
m3
defined by f(r) = (r + m1, r + m2, r + m3) is an
isomorphism of rings. Let i ∈ {1, 2, 3} and let us denote the field R
mi
by Fi.
Observe that R ∼= F1 × F2 × F3 as rings.
(ii) ⇒ (iii). Assume that R ∼= F1×F2×F3 as rings, where Fi is a field
for each i ∈ {1, 2, 3}. Let us denote the ring F1 ×F2 ×F3 by T . Note that
Max(T ) = {M1 = (0)×F2×F3,M2 = F1× (0)×F3,M3 = F1×F2× (0)}
and V (H(T )) = {v1 = M1, v2 = M1 ∩M2, v3 = M2, v4 = M2 ∩M3, v5 =
M3, v6 = M1 ∩ M3}. Observe that H(T ) is the union of the cycles
Γ1 : v1 − v2 − v3 − v4 − v5 − v6 − v1 and Γ2 : v1 − v3 − v5 − v1. The cycle
Γ1 can be represented by means of a hexagon. The edges of Γ2 are three
chords of this hexagon, two of them pass through v1 and the third joins
v3 with v5. It is clear that Γ2 can be represented by means of a triangle
and it can be drawn inside the hexagon representing Γ1 in such a way
that there are crossing over of the edges. This proves that H(T ) is planar.
As R ∼= T as rings, it follows that H(R) is planar.
(iii) ⇒ (v). This follows from Kuratowski’s theorem [13, Theorem 5.9].
The statements (v) ⇒ (i) and (v) ⇒ (iv) are clear.
4. The case when R has exactly two maximal ideals
Our aim in this section is to characterize rings R with |Max(R)| = 2
such that H(R) is planar.
Lemma 4.1. Let (R1,m1), (R2,m2) be quasilocal rings such that mi is
nilpotent for each i ∈ {1, 2}. Let R = R1 × R2. If H(R) satisfies either
(C1) or (C2), then m4
i = (0) for each i ∈ {1, 2}.
Proof. Let i ∈ {1, 2} and let ni > 1 be least with the property that
m
ni
i = (0). Assume that H(R) satisfies either (C1) or (C2). First, we show
that m4
1 = (0). Suppose that m4
1 6= (0). Then n1 > 5 and mi
1 6= m
j
1
for
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138 Planarity of a spanning subgraph, I
all distinct i, j ∈ {1, 2, . . . , n1}. Let A = {m1 × R2,m
2
1 × R2,m
3
1 × R2}
and let B = {m4
1 × R2, (0) × R2, R1 × (0)}. It is clear that A ∪ B ⊆
V (H(R)) and A ∩ B = ∅. Note that the subgraph of H(R) induced by
A ∪ {m4
1 × R2, (0) × R2} is a clique on five vertices and the subgraph
of H(R) induced by A ∪ B contains K3,3 as a subgraph. Thus if H(R)
satisfies either (C1) or (C2), then m4
1 = (0). Similarly, it can be shown
that m4
2 = (0). Thus, if H(R) satisfies either (C1) or (C2), then m4
i = (0)
for each i ∈ {1, 2}.
Lemma 4.2. Let R = R1 × R2, where (R1,m1) and (R2,m2) are as in
the statement of Lemma 4.1. If H(R) satisfies either (C1) or (C2), then
mi is principal for each i ∈ {1, 2}.
Proof. Assume that H(R) satisfies either (C1) or (C2). Then we know
from Lemma 4.1 that m4
i = (0) for each i ∈ {1, 2}. We first prove that m1 is
principal. Suppose that m1 is not principal. Since m1 is nilpotent, it follows
that dimR1
m1
(m1
m2
1
) > 2. Let x, y ∈ m1 be such that x+m2
1, y+m2
1 are linearly
independent over R1
m1
. Let A = {R1x × R2, R1y × R2, R1(x + y) × R2}
and let B = {m1 ×R2, (0)×R2, R1 × (0)}. Note that A ∪B ⊆ V (H(R)),
A ∩ B = ∅, the subgraph of H(R) induced by A ∪ {m1 × R2, (0)× R2}
is a clique on five vertices, and the subgraph of H(R) induced by A ∪B
contains K3,3 as a subgraph. Thus, if H(R) satisfies either (C1) or (C2),
then m1 is principal. Similarly, it can be shown that m2 is principal. This
proves that if H(R) satisfies either (C1) or (C2), then mi is principal for
each i ∈ {1, 2}.
Lemma 4.3. Let R = R1 × R2, where (R1,m1) and (R2,m2) are as in
the statement of Lemma 4.1. Suppose that mi 6= (0) for each i ∈ {1, 2}. If
H(R) satisfies either (C1) or (C2), then m2
i = (0) for each i ∈ {1, 2}.
Proof. Assume that H(R) satisfies either (C1) or (C2). Let i ∈ {1, 2}. As
mi 6= (0) and mi is nilpotent, it follows that mi 6= m2
i . We first verify that
m2
1 = (0). Suppose that m2
1 6= (0). Let A = {m1 ×R2,m
2
1 ×R2,m1 ×m2}
and let B = {R1 ×m2, (0)×R2, R1 × (0)}. Note that A ∪B ⊆ V (H(R)),
A ∩ B = ∅, the subgraph of H(R) induced by A ∪ {R1 ×m2, (0)× R2}
is a clique on five vertices, and the subgraph of H(R) induced by A ∪B
contains K3,3 as a subgraph. Thus, if H(R) satisfies either (C1) or (C2),
then m2
1 = (0). Similarly, it can be shown that m2
2 = (0). This shows that
if H(R) satisfies either (C1) or (C2), then m2
i = (0) for each i ∈ {1, 2}.
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P. Vadhel, S. Visweswaran 139
Remark 4.4. Recall that a principal ideal ring R is said to be a special
principal ideal ring (SPIR) if R has a unique prime ideal. If m is the unique
prime ideal of a SPIR R, then m is necessarily nilpotent. If R is a SPIR
with m as its unique prime ideal, then we denote it by mentioning that
(R,m) is a SPIR. Let (R,m) be a quasilocal ring such that m is principal
and nilpotent. Let n > 2 be least with the property that mn = (0).
Then it follows from the proof of (iii) ⇒ (i) of [6, Proposition 8.8] that
{mi|i ∈ {1, . . . , n− 1}} is the set of all nonzero proper ideals of R and so,
(R,m) is a SPIR.
Lemma 4.5. Let T = F1 × F2, where F1 and F2 are fields. Then H(T )
is planar.
Proof. It is clear that V (H(T )) = {(0)× F2, F1 × (0)} and has no edges.
Therefore, H(T ) is planar.
Lemma 4.6. Let T = T1 × F2, where (T1, n1) is a SPIR with n1 6= (0)
and F2 is a field. If n41 = (0), then H(T ) is planar.
Proof. We consider the following cases.
Case 1: n21 = (0). It is clear from Remark 4.4 that I(T1)
∗ = {n1}. Hence,
V (H(T )) = {v1 = (0) × F2, v2 = n1 × F2, v3 = T1 × (0), v4 = n1 × (0)}.
Observe that H(T ) is the path v1−v2−v3−v4. Therefore, H(T ) is planar.
Case 2: n21 6= (0) but n31 = (0). Note that it follows from Remark 4.4
that I(T ∗
1 ) = {n1, n
2
1}. In this case, V (H(T )) = {v1 = (0) × F2, v2 =
n1 × F2, v3 = n1 × (0), v4 = T1 × (0), v5 = n21 × F2, v6 = n21 × (0)}.
Observe that H(T ) is the union of the cycle Γ of length five given by
Γ : v1 − v2 − v3 − v4 − v5 − v1 and the three edges e1 : v2 − v4, e2 : v2 − v5,
and e3 : v4 − v6. Note that Γ can be represented by means of a pentagon.
The edges e1, e2 are two chords of the pentagon representing Γ through v2
and they can be drawn inside the pentagon and the edge e3 which joins
the vertex v4 of the pentagon with the pendant vertex v6 can be drawn
outside this pentagon so that there are no crossing over of the edges. This
shows that H(T ) is planar.
Case 3: n31 6= (0) but n41 = (0). It follows from Remark 4.4 that I(T1)
∗ =
{n1, n
2
1, n
3
1}. In this case,V (H(T )) = {v1 = (0)×F2, v2 = n1×F2, v3 = n21×
(0), v4 = T1×(0), v5 = n31×F2, v6 = n1×(0), v7 = n21×F2, v8 = n31×(0)}. It
is easy to verify thatH(T ) is the union of cycles Γ1 : v1−v2−v3−v4−v5−v1,
Γ2 : v5 − v4 − v3 − v6 − v7 − v5, and the edges e1 : v2 − v4, e2 : v2 − v5, e3 :
v4 − v6, e4 : v4 − v7, e5 : v7 − v1, e6 : v7 − v2, e7 : v2 − v6, and e8 : v4 − v8.
Note that the cycles Γ1 and Γ2 have exactly two edges in common and
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140 Planarity of a spanning subgraph, I
they can be represented by means of two pentagons and they can be
drawn side by side without any crossing over of the edges. The edges
e1, e2 are chords of the pentagon representing Γ1 and they pass through
the vertex v2. The edges e3, e4 are chords of the pentagon representing
Γ2 and they pass through the vertex v4. The edges e1, e2 can be drawn
inside the pentagon representing Γ1 and the edges e3, e4 can be drawn
inside the pentagon representing Γ2 without any crossing over of the edges.
The edges e5, e6, and e7 can be drawn outside the pentagons representing
Γ1 and Γ2 and finally the vertex v8 can be plotted inside the pentagon
representing Γ1 and the edge e8 which joins v4 with the pendant vertex v8
can be drawn inside the pentagon representing Γ1 in such a way that there
are no crossing over of the edges. This proves that H(T ) is planar.
Lemma 4.7. Let T = T1×T2, where (Ti, ni) is a SPIR for each i ∈ {1, 2}.
If ni 6= (0) but n2i = (0) for each i ∈ {1, 2}, then H(T ) is planar.
Proof. It follows from Remark 4.4 that I(Ti)
∗ = {ni} for each i ∈ {1, 2}.
Hence, V (H(T )) = {v1 = (0) × T2, v2 = n1 × n2, v3 = T1 × n2, v4 =
n1 × (0), v5 = T1 × (0), v6 = n1 × T2, v7 = (0) × n2}. Note that H(T ) is
the union of the cycle Γ : v1 − v2 − v3 − v4 − v5 − v6 − v1 and the edges
e1 : v3 − v1, e2 : v3 − v5, e3 : v3 − v6, e4 : v2 − v5, e5 : v2 − v6, e6 : v7 − v1,
and e7 : v7 − v6. Observe that the cycle Γ can be represented by means
of a hexagon. The edges e1, e2, and e3 are the chords of the hexagon
representing Γ through the vertex v3 and they can be drawn inside the
hexagon representing Γ without any crossing over of the edges. The edges
e4 and e5 can be drawn outside the hexagon representing Γ in such a way
that there are no crossing over of the edges. The vertex v7 can be plotted
inside the hexagon representing Γ and the edges e6 and e7 can be drawn
inside the hexagon representing Γ in such a way that there are no crossing
over of the edges. This shows that H(T ) is planar.
Theorem 4.8. Let R be a ring such that |Max(R)| = 2. The following
statements are equivalent:
(i) H(R) satisfies (C1).
(ii) R is isomorphic to one of the rings of the following type:
(a) F1 × F2, where Fi is a field for each i ∈ {1, 2}.
(b) T1 × F2, where (T1, n1) is a SPIR with n1 6= (0) but n41 = (0)
and F2 is a field.
(c) T1×T2, where (Ti, ni) is a SPIR with ni 6= (0) but n2i = (0) for
each i ∈ {1, 2}.
(iii) H(R) is planar.
“adm-n3” — 2018/10/20 — 9:02 — page 141 — #147
P. Vadhel, S. Visweswaran 141
(iv) H(R) satisfies (C2).
(v) H(R) satisfies both (C∗
1 ) and (C∗
2 ).
Proof. Let {m1,m2} denote the set of all maximal ideals of R.
(i) ⇒ (ii) and (iv) ⇒ (ii). If H(R) satisfies either (C1) or (C2), then
we know from Corollary 2.11 that J(R) is nilpotent. Let n > 1 be such
that (J(R))n = (0). Hence, mn
1m
n
2 = (0). As mn
1 and mn
2 are comaximal,
it follows from [6, Proposition 1.10 (i)] that mn
1 ∩ mn
2 = mn
1m
n
2 = (0).
Hence, we obtain from [6, Proposition 1.10 (ii) and (iii)] that the mapping
f : R → R
mn
1
× R
mn
2
defined by f(r) = (r +mn
1 , r +mn
2 ) is an isomorphism
of rings. Let us denote the ring R
mn
i
by Ti and mi
mn
i
by ni for each i ∈ {1, 2}.
Note that (Ti, ni) is a quasilocal ring and ni is nilpotent for each i ∈ {1, 2}.
Let us denote the ring T1 × T2 by T . Since R ∼= T as rings, it follows that
H(T ) satisfies either (C1) or (C2). Hence, we obtain from Lemma 4.1 that
n4i = (0 + mn
i ) for each i ∈ {1, 2}. Moreover, we know from Lemma 4.2
and Remark 4.4 that (Ti, ni) is a SPIR for each i ∈ {1, 2}.
If ni = (0 +mn
i ) for each i ∈ {1, 2}, then we get that Ti is a field for
each i ∈ {1, 2} and we obtain that R is isomorphic to a ring of the type
mentioned in (ii) (a).
If n1 6= (0 + mn
1 ) but n2 = (0 + mn
2 ). Then (T1, n1) is a SPIR with
n1 6= (0 + mn
1 ) but n41 = (0 + mn
1 ) and T2 is a field. In this case, R is
isomorphic to a ring of the type mentioned in (ii) (b).
If ni 6= (0 + mn
i ) for each i ∈ {1, 2}, then we know from Lemma 4.3
that n2i = (0 +mn
i ) for each i ∈ {1, 2}. Thus in this case, we obtain that
(Ti, ni) is a SPIR with ni 6= (0+mn
i ) but n2i = (0+mn
i ) for each i ∈ {1, 2}
and R is isomorphic to a ring of the type mentioned in (ii)(c).
(ii) ⇒ (iii). Let T be a ring. If T is a ring of the form mentioned
in (ii) (a), then we know from Lemma 4.5 that H(T ) is planar. If T is a
ring of the form mentioned in (ii) (b), then we obtain from Lemma 4.6
that H(T ) is planar. If T is a ring of the form mentioned in (ii) (c), then
from Lemma 4.7, we get that H(T ) is planar. As R is isomorphic to one
of the rings of the type mentioned in (ii) (a), (ii) (b) or (ii) (c), it follows
that H(R) is planar.
(iii) ⇒ (v). This follows from Kuratowski’s theorem [13, Theorem 5.9].
The statements (v) ⇒ (i) and (v) ⇒ (iv) are clear.
Acknowledgement
We are very much thankful to the referee and all the members of the
Editorial Board of Algebra and Discrete Mathematics for their suggestions
“adm-n3” — 2018/10/20 — 9:02 — page 142 — #148
142 Planarity of a spanning subgraph, I
and support. As per the suggestion of the members of the Editorial Board,
we shall present the results that we obtained on the planarity of H(R),
where R is quasilocal in the article — Planarity of a spanning subgraph of
the intersection graph of ideals of a commutative ring II, Quasilocal case.
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Contact information
P. Vadhel,
S. Visweswaran
Department of Mathematics, Saurashtra
University, Rajkot, 360 005 India
E-Mail(s): pravin_2727@yahoo.com,
s_visweswaran2006@yahoo.co.in
Received by the editors: 22.09.2015
and in final form 24.08.2018.
|
| id | nasplib_isofts_kiev_ua-123456789-188380 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T17:34:08Z |
| publishDate | 2018 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Vadhel, P. Visweswaran, S. 2023-02-26T12:36:01Z 2023-02-26T12:36:01Z 2018 Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case / P. Vadhel, S. Visweswaran // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 130–143. — Бібліогр.: 19 назв. — англ. 1726-3255 2010 MSC: 13A15, 05C25. https://nasplib.isofts.kiev.ua/handle/123456789/188380 The rings considered in this article are nonzero commutative with identity which are not fields. Let R be a ring. We denote the collection of all proper ideals of R by I(R) and the collection I(R)\{(0)} by I(R)*. Recall that the intersection graph of ideals of R, denoted by G(R), is an undirected graph whose vertex set is I(R)* and distinct vertices I, J are adjacent if and only if I ∩ J ≠ (0). In this article, we consider a subgraph of G(R), denoted by H(R), whose vertex set is I(R)* and distinct vertices I, J are adjacent in H(R) if and only if IJ ≠ (0). The purpose of this article is to characterize rings R with at least two maximal ideals such that H(R) is planar. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case Article published earlier |
| spellingShingle | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case Vadhel, P. Visweswaran, S. |
| title | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case |
| title_full | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case |
| title_fullStr | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case |
| title_full_unstemmed | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case |
| title_short | Planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring I, nonquasilocal case |
| title_sort | planarity of a spanning subgraph of the intersection graph of ideals of a commutative ring i, nonquasilocal case |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/188380 |
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