# The Graph of Equivalence Classes of Zero Divisors

Authors: Vinayak Joshi; B. N. Waphare; H. Y. Pourali
Journal: ISRN Discrete Mathematics
 (2014)

---

## Abstract

We introduce a graphG
E
(
L
) of equivalence classes of zero divisors of a meet semilattice L with 0. The set of vertices of G
E
(
L
) are the equivalence classes of nonzero zero divisors of L and two vertices [
x
] and [
y
] are adjacent if and only if [
x
]
∧
[
y
]
=
[
0
]. It is proved that G
E
(
L
) is connected and either it contains a cycle of length 3 or G
E
(
L
)
≅
K
2. It is known that two Boolean lattices L
1 and L
2 have isomorphic zero divisor graphs if and only if L
1
≅
L
2. This result is extended to the class of SSC meet semilattices. Finally, we show that Beck's Conjecture is true for G
E
(
L
)
.


---

## Body

## 1. Introduction

The idea of a zero divisor graph was introduced by Beck in [1] to investigate the interplay between ring theoretic properties and graph theoretic properties. The concept of zero divisor graph is also well studied in ordered structures; see Alizadeh et al. [2, 3], Halaš and Jukl [4], Halaš and Länger [5], Joshi [6], Joshi et al. [7–11], and Lu and Wu [12].The graph of equivalence classes of zero divisors of commutative rings is well studied in Allen et al. [13], Anderson and LaGrange [14], and Spiroff and Wickham [15].In this paper, we extend this concept to a meet semilatticeL with 0. The main aspect of the graph G
E
(
L
) of equivalence classes of zero divisors of L is the connection to the associated primes of L. In general, every vertex of G
E
(
L
) either corresponds to an associated prime or is adjacent to an associated prime of L. Further, we consider the equivalence relation ~ on a meet semilattice L with 0, x
~
y if and only if ann
(
x
)
=
ann
(
y
) for x
,
y
∈
L and construct the meet semilattice L
′
=
{
[
x
]
∣
x
∈
Z
{
0
}
(
L
)
*
}. It is proved that L
′ is an SSC meet semilattice with the property that G
E
(
L
)
≅
G
{
0
}
(
L
′
), the zero divisor graph of L
′. Hence it is clear that the study of G
E
(
L
) is nothing, but the study of zero divisor graph of SSC meet semilattices. This observation helps us to prove that for bounded SSC meet semilattices L
1 and L
2, G
{
0
}
(
L
1
)
≅
G
{
0
}
(
L
2
) if and only if L
1
≅
L
2. This extends the result of LaGrange [16] for Boolean algebra. It is proved that for an SSC meet semilattice L, G
{
0
}
(
L
) is connected and either it contains a cycle of length 3 or G
{
0
}
(
L
)
≅
K
2. In the class of a finite semicomplemented meet semilattice L, G
E
(
L
) is isomorphic to the zero divisor graph G
{
0
}
(
L
) if and only if L is SSC.


## 2.  Properties of the GraphG
E
(
L
)

We begin with the necessary definitions and terminology.A nonempty subsetI of a meet semilattice L is said to besemi-ideal, if y
⩽
x
∈
I implies that y
∈
I. A semi-ideal I of a meet semi-lattice L is said to be anideal, if a
,
b
∈
I implies that a
∨
b
∈
I. Dually we have the concept of a filter. A proper semi-ideal (ideal) is said to beprime, if a
∧
b
∈
I implies either a
∈
I or b
∈
I.A prime semi-ideal (ideal)P of a meet semilattice L is said to beminimal prime semi-ideal (ideal), if there is no prime ideal Q such that Q
⫋
P.The concept ofSSC lattices is introduced by Janowitz [17]. We write this definition for a meet semilattice with 0.A meet semilatticeL with 0 is said to besemicomplemented if, for a
≠
1, there exists b
≠
0 such that a
∧
b
=
0 and L is said to besection semicomplemented  (in brief SSC) if for a
,
b
∈
L with a
<
b, there exists nonzero c such that 0
<
c
≤
b and a
∧
c
=
0. It is clear that every SSC lattice is a semicomplemented lattice, but not conversely. More details about SSC poset can be found in Thakare et al. [18, 19], Joshi [20].LetL be a meet semilattice with 0. The set of allzero divisors of L is denoted by Z
{
0
}
(
L
)
=
{
x
∈
L
∣
x
∧
y
=
0
for
some
y
∈
L
∖
{
0
}
} and the set of all nonzero zero divisors is denoted by Z
{
0
}
(
L
)
*. Clearly, Z
{
0
}
(
L
)
*
∪
{
0
}
=
Z
{
0
}
(
L
).LetG be a graph, and let x
,
y be distinct vertices in G. We denote by d
(
x
,
y
) the length of a shortest path from x to y if it exists and put d
(
x
,
y
)
=
∞ if no such path exists. Thediameter of G, denoted by diam
⁡
(
G
), is zero if G is the graph on one vertex and is diam
⁡
(
G
)
=
sup
⁡
{
d
(
x
,
y
)
where
x
and
y
are
distinct
vertices
of
G
} otherwise. Acycle in a graph G is a path that begins and ends at the same vertex. Thegirth of G, written girth
(
G
), is the length of the shortest cycle in G (and girth
(
G
)
=
∞ if G has no cycles). The chromatic number of G is denoted by χ
(
G
). Thus, χ
(
G
) is the minimum number of colors which can be assigned to the elements of L such that adjacent elements receive different colors. If this number is not finite, write χ
(
G
)
=
∞. A subset C of G is a clique if any two distinct vertices of C are adjacent. If G contains a clique with n elements and every clique has at most n elements then the clique number of G is Clique
(
G
)
=
n. If the sizes of the cliques are not bounded, then Clique
(
G
)
=
∞. We always have χ
(
G
)
⩾
Clique
(
G
).Joshi [6] introduced the concept of the zero divisor graph of a poset P having the smallest element 0 with respect to an ideal I of P. We mention below this definition when the corresponding poset is a meet semilattice and I
=
{
0
}. Note that this definition coincides with the definition given by Lu and Wu [12] when I
=
{
0
}.Definition 1 (Joshi [6], Lu and Wu [12]).
LetL be a semilattice with 0. We associate a simple undirected graph with L, called thezero divisor graph of L
with respect to
{
0
}, denoted by G
{
0
}
(
L
), with the set of vertices is V
(
G
{
0
}
(
L
)
) = Z
{
0
}
(
L
)
*
=
{
x
∈
L
∖
{
0
}
∣
x
∧
y
=
0
for
some
y
∈
L
∖
{
0
}
} and two distinct vertices x
,
y are adjacent if and only if x
∧
y
=
0.
For anyx
,
y
∈
L, we say that x
~
y if and only if ann
(
x
)
=
ann
(
y
), where ann
(
a
)
=
{
b
∣
a
∧
b
=
0
}. Note that ~ is an equivalence relation on L. Furthermore, if x
1
~
x
2 and x
1
∧
y
=
0, then y
∈
ann
(
x
1
)
=
ann
(
x
2
) and hence x
2
∧
y
=
0. We define [
x
], the equivalence class of x, as follows: [
x
]
=
{
z
∈
L
∣
ann
(
z
)
=
ann
(
x
)
}.Lemma 2.
LetL be a meet semilattice with 0 and L
′
=
{
[
x
]
∣
x
∈
L
}, where [
x
]
=
{
z
∈
L
∣
a
n
n
(
z
)
=
a
n
n
(
x
)
}. Then L
′ is a meet semilattice under the partial order [
x
]
⩽
[
y
] if and only if a
n
n
(
y
)
⊆
a
n
n
(
x
).Proof.
Sinceann
(
x
)
,
ann
(
y
)
⊆
ann
(
x
∧
y
), we have [
x
∧
y
]
⩽
[
x
]
,
[
y
]. We claim that [
x
]
∧
[
y
]
=
[
x
∧
y
]. Let [
t
] be another lower bound of [
x
] and [
y
]. Then ann
(
x
)
,
ann
(
y
)
⊆
ann
(
t
). Now, we claim that ann
(
x
∧
y
)
⊆
ann
(
t
). Let z
∈
ann
(
x
∧
y
). Then z
∧
x
∈
ann
(
y
)
⊆
ann
(
t
). This gives x
∧
z
∧
t
=
0; that is, z
∧
t
∈
ann
(
x
)
⊆
ann
(
t
). Hence z
∧
t
=
0; that is, z
∈
ann
(
t
). This proves that ann
(
x
∧
y
)
⊆
ann
(
t
). Thus [
t
]
⩽
[
x
∧
y
] and this yields [
x
]
∧
[
y
]
=
[
x
∧
y
]. Clearly, if L is a finite lattice, then L
′ is also a lattice.Lemma 3.
L
′ is an S
S
C meet semilattice.Proof.
Let0
<
[
x
]
≨
[
y
]. Then ann
(
y
)
⫋
ann
(
x
). Hence there exists z
∈
ann
(
x
) such that z
∉
ann
(
y
). Thus 0
≠
z
∧
y
=
t. But then [
0
]
<
[
t
]
⩽
[
y
] and [
t
]
∧
[
x
]
=
[
0
]. Thus L
′ is an SSC meet semilattice.Now, we introduce the graphG
E
(
L
) of equivalence classes of zero divisors of a meet semilattice L with 0, denoted by G
E
(
L
).Definition 4.
LetL be a meet semilattice with 0. We associate a simple undirected graph with L whose vertices are the equivalence classes of elements in Z
{
0
}
(
L
)
* and with each pair of distinct classes [
x
] and [
y
] joined by an edge if and only if [
x
]
∧
[
y
]
=
{
0
}
=
[
0
].We illustrate the concept of the graph of equivalence classes of zero divisor with an example, see Figure1.Figure 1
The graphG
E
(
L
) of equivalence classes of zero divisors.Consider the following two infinite latticesL and L
1.We can see that the zero divisor graph of latticesL, G
{
0
}
(
L
) is infinite, but the graph of equivalence classes of zero divisors of the lattice L, namely, G
E
(
L
), is finite. Also for the lattices L
1, G
{
0
}
(
L
1
), and G
E
(
L
1
) both are infinite, see Figure 2.Figure 2
The graphG
E
(
L
) is finite, whereas G
E
(
L
1
) is infinite.Corollary 5.
LetL be a meet semilattice with 0. Then there exists an S
S
C meet semilattice L
′ (as constructed in Lemma 2) such that G
E
(
L
)
=
G
{
0
}
(
L
′
).Proof.
From the construction ofL
′, it is clear that G
E
(
L
)
=
G
{
0
}
(
L
′
).Hence from Corollary5, it is clear that the study of the graph of equivalence classes of zero divisors of L is nothing, but the study of zero divisor graphs of SSC meet semilattices.Remark 6.
Recall that a prime semi-idealP of a meet semilattice L is an associated prime if it is the annihilator of some elements of L; that is, P
=
ann
(
x
) for some nonzero x
∈
L. The set of associated primes is denoted by Ass
(
L
). If Clique
(
G
{
0
}
(
L
)
)
<
∞, then Ass
(
L
) is nonempty and finite. Further, any maximal element of the family of annihilator semi-ideals is prime; see Halaš and Jukl [4, Lemma 2.4, 2.6]. Note that every zero divisor is contained in an annihilator ideal and maximal annihilators are associated primes; we have Z
{
0
}
(
L
)
⊆
⋃
P
∈
Ass
(
L
)
‍
P. Clearly, the reverse inclusion is trivial. Hence Z
{
0
}
(
L
)
=
⋃
P
∈
Ass
(
L
)
‍
P; that is, the set of zero-divisors of L equals the union of all associated primes of L. It is obvious to prove that a map ϕ
:
Ass
(
L
)
→
V
(
G
E
(
L
)
) defined by ϕ
(
P
)
=
[
x
], where P
=
ann
(
x
) is an injective map. We adopt the conventions of Spiroff and Wickham [15], and by a slight abuse of terminology we will refer to the vertex [
y
] as an associated prime if ann
(
y
)
∈
Ass
(
L
). It will be clear from context whether [
y
] refers to an equivalence class, a vertex, or a specific annihilator.Theorem 7.
LetL be a meet semilattice with 0, and let C
l
i
q
u
e
(
G
{
0
}
(
L
)
)
<
∞. If a
n
n
(
x
) and a
n
n
(
y
) are distinct prime semi-ideals of L, then [
x
] and [
y
] are adjacent in G
E
(
L
). Furthermore, every vertex of G
E
(
L
) is either an associated prime semi-ideal or adjacent to a maximal semi-ideal in ℒ
=
{
a
n
n
(
x
)
∣
(
0
≠
)
x
∈
L
}.Proof .
Sinceann
(
x
) and ann
(
y
) are distinct, we have x
∧
y
=
0. This further gives that [
x
]
∧
[
y
]
=
[
x
∧
y
]
=
[
0
]. Thus [
x
] and [
y
] are adjacent. Furthermore, suppose that [
x
]
∈
V
(
G
E
(
L
)
) is not an associated prime semi-ideal. Then there exists [
z
]
∈
V
(
G
E
(
L
)
) such that [
x
]
∧
[
z
]
=
[
0
]
=
{
0
}. Hence x
∈
ann
(
z
). We claim that ann
(
z
) is maximal. Otherwise, since Clique
(
G
{
0
}
(
L
)
)
<
∞, ℒ has a maximal element, say ann
(
z
′
)
(
z
′
≠
0
) containing ann
(
z
), then x
∈
ann
(
z
)
⊆
ann
(
z
′
). It is clear that every maximal element of ℒ is prime. Therefore x
∧
z
′
=
0. Thus there is an edge between [
x
] and [
z
′
]. This completes the proof.Remark 8.
For a Boolean lattice,[
a
]
=
{
a
} for every a
∈
Z
{
0
}
(
L
), and hence in this case, G
E
(
L
) and G
{
0
}
(
L
) are isomorphic. This fact is illustrated in Figure 3.Figure 3
The isomorphism ofG
E
(
L
) and G
{
0
}
(
L
).In view of Remark8, we raise the following problem.Question 1.
Find a class of meet semilatticesℒ such that G
{
0
}
(
L
)
≅
G
E
(
L
) for L
∈
ℒ.Before we answer this question, we need the following result which follows from the definition ofSSC meet semilattice.Lemma 9.
LetL be a meet semilattice with 0. Then L is S
S
C if and only if for a
,
b
∈
L, a
n
n
(
a
)
=
a
n
n
(
b
) implies a
=
b.Proof.
Suppose thatL is SSC. Further, assume that there exist a
,
b
∈
L such that ann
(
a
)
=
ann
(
b
) and a
≠
b. Since L is SSC, there exist c
∈
L such that 0
<
c
⩽
a and c
∧
b
=
0. This gives c
∈
ann
(
b
)
=
ann
(
a
), a contradiction to the fact that c
≠
0. Hence a
=
b.Theorem 10.
LetL be a finite semicomplemented meet semilattice. Then ϕ
:
G
{
0
}
(
L
)
→
G
E
(
L
) is a graph isomorphism if and only if L is S
S
C.Proof .
LetL be a SSC meet semilattice. Then G
E
(
L
)
≅
G
{
0
}
(
L
) follows from Lemma 9.
Conversely, assume thatG
E
(
L
)
≅
G
{
0
}
(
L
). First, we prove that [
a
]
=
{
a
}, ∀
a
∈
Z
{
0
}
(
L
)
*. Let b
∈
[
a
] for a
≠
b
∈
Z
{
0
}
(
L
)
*. Then we have [
a
]
=
[
b
] for a
≠
b, a contradiction to |
V
(
G
E
(
L
)
)
|
=
|
V
(
G
{
0
}
(
L
)
)
|. Now, let a
<
b
(
a
≠
0
). Since L is semicomplemented, a
,
b
∈
Z
{
0
}
(
L
)
*. This gives ann
(
b
)
⫋
ann
(
a
). Otherwise, if ann
(
a
)
=
ann
(
b
), then b
∈
[
a
]
=
{
a
}, a contradiction. From the proof of Lemma 9, it is clear that L is SSC.If we consider that theϕ
:
G
{
0
}
(
L
)
→
G
E
(
L
) is a graph isomorphism defined by ϕ
(
a
)
=
[
a
], then we do not need even the finiteness of L.Theorem 11.
LetL be a semicomplemented meet semilattice. Then ϕ
:
G
{
0
}
(
L
)
→
G
E
(
L
) is a graph isomorphism defined by ϕ
(
a
)
=
[
a
] if and only if L is S
S
C.Proof.
LetL be a SSC meet semilattice. Then G
E
(
L
)
≅
G
{
0
}
(
L
) follows from Lemma 9.
Conversely, assume thatϕ
:
G
{
0
}
(
L
)
→
G
E
(
L
) is a graph isomorphism defined by ϕ
(
a
)
=
[
a
]. This gives [
a
]
=
{
a
}, for all a
∈
Z
{
0
}
(
L
)
*. Let a
<
b
(
a
≠
0
). Since L is semicomplemented, a
,
b
∈
Z
{
0
}
(
L
)
*. This gives ann
(
b
)
⫋
ann
(
a
). From the proof of Lemma 9, it is clear that L is SSC.From Figure4, it is clear that for the posets P
1
,
P
2, G
{
0
}
(
P
1
)
≅
G
{
0
}
(
P
2
)
⇏
P
1
≅
P
2. Hence, it is worth to study the following problem.Figure 4
Nonisomorphic lattices with isomorphic zero divisor graphs.Problem 12.
Find a class of posets𝒫 such that G
{
0
}
(
P
1
)
≅
G
{
0
}
(
P
2
) if and only if P
1
≅
P
2 for the posets P
1
,
P
2
∈
𝒫.In [8], Joshi and Khiste answered this problem in the case of Boolean posets. More details about Boolean posets can be found in Waphare and Joshi [21].Theorem 13 (Joshi and Khiste [8, Theorem 2.11]).
LetP
1 and P
2 be Boolean posets. Then G
{
0
}
(
P
1
)
≅
G
{
0
}
(
P
2
) if and only if P
1
≅
P
2.It is well known that there is a1
-
1 correspondence between Boolean algebras and Boolean rings.A consequence of the above theorem is the following result.Corollary 14 (LaGrange [18,  Theorem 4.1], Mohammadian [22,Theorem 7]).
LetP
1 and P
2 be Boolean algebras. Then G
{
0
}
(
P
1
)
≅
G
{
0
}
(
P
2
) if and only if P
1
≅
P
2.Now, we are ready to extend this result to the class ofSSC meet semilattices. Note that every Boolean lattice is SSC but not conversely. Consider the lattice L
1 depicted in Figure 2 is SSC but not a Boolean lattice.Theorem 15.
LetL
1 and L
2 be bounded S
S
C meet semilattices. Then the following statements are equivalent: (1)
G
E
(
L
1
)
≅
G
E
(
L
2
);
(2)
G
{
0
}
(
L
1
)
≅
G
{
0
}
(
L
2
);
(3)
L
1
≅
L
2.Proof.
The equivalence of statements(
1
) and (
2
) follows from Lemma 9.
(
2
)
⇒
(
3
)
: Since L
1 and L
2 are semicomplemented, V
(
G
{
0
}
(
L
i
)
)
=
L
i
∖
{
0
L
i
,
1
L
i
} for i
=
1,2. Let ϕ
:
V
(
G
{
0
}
(
L
1
)
)
→
V
(
G
{
0
}
(
L
2
)
) be a graph isomorphism. Define a map f
:
L
1
→
L
2 such that f
(
a
)
=
ϕ
(
a
), for all a
∈
V
(
G
{
0
}
(
L
1
)
) along with f
(
0
L
1
)
=
0
L
2 and f
(
1
L
1
)
=
1
L
2, where 0
L
i
(
1
L
i
) is the smallest (largest) element of L
i. Since ϕ is bijective, f is bijective. We claim that f is a biorder preserving map.
Leta
⩽
b. First, we show that f
(
a
)
⩽
f
(
b
). If any of a
,
b is either 0
L
1 or 1
L
1, then we are through. Hence assume that a
,
b
∉
{
0
L
1
,
1
L
1
}. Suppose that f
(
a
)

f
(
b
); that is, ϕ
(
a
)

ϕ
(
b
). Since L
2 is SSC, there exists ϕ
(
c
)
∈
L
2 such that 0
L
2
<
ϕ
(
c
)
⩽
ϕ
(
a
) and ϕ
(
c
)
∧
ϕ
(
b
)
=
0
L
2. Since b
≠
0
L
1, we have ϕ
(
b
)
≠
0
L
2. This gives that ϕ
(
c
) and ϕ
(
b
) are adjacent in G
{
0
}
(
L
2
). Since ϕ is a graph isomorphism, we have that c and b are adjacent which further imply that c
∧
b
=
0
L
1. Hence, c
∧
a
=
0
L
1 in L
1. Thus a and c are adjacent in G
{
0
}
(
L
1
). Hence, we have ϕ
(
c
)
∧
ϕ
(
a
)
=
0
L
2. But ϕ
(
c
)
⩽
ϕ
(
a
) gives ϕ
(
c
)
=
0
L
2, a contradiction. Hence f
(
a
)
⩽
f
(
b
) in L
2.
Conversely, assume thatf
(
a
)
⩽
f
(
b
) in L
2. We claim that a
⩽
b. If any of f
(
a
) or f
(
b
) is 0
L
2 or 1
L
2, then we are through. Let us assume that f
(
a
)
,
f
(
b
)
∈
L
2
∖
{
0
L
2
,
1
L
2
}. This yields a
,
b
∉
{
0
L
1
,
1
L
1
}. Now, suppose on the contrary that a

b for a
,
b
∉
{
0
L
1
,
1
L
1
}. Since L
1 is SSC, there exists c
∈
L
1 such that 0
L
1
<
c
⩽
a and c
∧
b
=
0
L
1. Thus c and b are adjacent in G
{
0
}
(
L
1
). Then ϕ
(
c
) and ϕ
(
b
) are adjacent in G
{
0
}
(
L
2
). Then ϕ
(
c
)
∧
ϕ
(
b
)
=
0
L
2 in L
2. Replacing the role of a
,
b in the above proof by c
,
a, respectively, we have ϕ
(
c
)
⩽
ϕ
(
a
). This implies that ϕ
(
c
)
⩽
ϕ
(
a
)
⩽
ϕ
(
b
). But this gives ϕ
(
c
)
=
ϕ
(
c
)
∧
ϕ
(
b
)
=
0
L
2, a contradiction. Hence a
⩽
b. Thus L
1
≅
L
2.
(
3
)
⇒
(
2
)
: Obvious.Lemma 16.
LetL
1 be a bounded S
S
C meet semilattice, and L
2 be a bounded semicomplemented meet semilattice such that G
{
0
}
(
L
1
)
≅
G
{
0
}
(
L
2
). Then L
2 is also S
S
C.Proof.
SinceL
1 and L
2 are semicomplemented, V
(
G
{
0
}
(
L
i
)
)
=
L
i
∖
{
0
L
i
,
1
L
i
} for i
=
1,2. Let ϕ
:
V
(
G
{
0
}
(
L
1
)
)
→
V
(
G
{
0
}
(
L
2
)
) be a graph isomorphism. Define a map f
:
L
1
→
L
2 such that f
(
a
)
=
ϕ
(
a
), for all a
∈
V
(
G
{
0
}
(
L
1
)
) along with f
(
0
L
1
)
=
0
L
2 and f
(
1
L
1
)
=
1
L
2, where 0
L
i(1
L
i) is the smallest (largest) element of L
i. Since ϕ is a bijective map, f is bijective. Let f
(
a
)
<
f
(
b
) in L
2. Since L
2 is semicomplemented, and if f
(
a
)
=
0
L
2 or f
(
b
)
=
1
L
2, then we are through. Hence assume that f
(
a
)
≠
0
L
2 and f
(
b
)
≠
1
L
2. Then clearly, a
≠
b and b
≠
1
L
1. Since L
1 is SSC and a
∧
b
<
b, there exists 0
L
1
<
c
⩽
b and c
∧
a
=
0
L
1. Since c and a are adjacent in G
{
0
}
(
L
1
), we have ϕ
(
c
) and ϕ
(
a
) are adjacent in G
{
0
}
(
L
2
); that is, ϕ
(
c
)
∧
ϕ
(
a
)
=
0
L
2. Further b and c are nonadjacent in G
{
0
}
(
L
1
). Then ϕ
(
b
) and ϕ
(
c
) are nonadjacent in G
{
0
}
(
L
2
). Consider 0
L
2
≠
ϕ
(
b
)
∧
ϕ
(
c
)
=
f
(
b
)
∧
f
(
c
)
=
f
(
d
) say. Clearly, 0
L
2
<
f
(
d
)
⩽
f
(
b
) and f
(
d
)
∧
f
(
a
)
=
0
L
2 (as ϕ
(
c
)
∧
ϕ
(
a
)
=
0
L
2 implies f
(
c
)
∧
f
(
a
)
=
0
L
2). Thus L
2 is SSC.Theorem 17.
LetL
1 and L
2 be bounded semicomplemented meet semilattices. If G
E
(
L
1
)
≅
G
{
0
}
(
L
2
), then L
2 is S
S
C.Proof.
For the given latticeL
1, we can construct an SSC meet semilattice L
1
′ (as given in Lemma 2), such that G
E
(
L
1
)
≅
G
{
0
}
(
L
1
′
). This gives G
{
0
}
(
L
1
′
)
≅
G
{
0
}
(
L
2
). By Lemma 16, L
2 is SSC.The above result is analogues to the following result of Anderson and LaGrange [14].Theorem 18 (Anderson and LaGrange [14, Theorem 2.6]).
LetR and S be commutative reduced rings with 1
≠
0 and Z
(
S
)
≠
{
0
}. If Γ
E
(
R
)
≅
Γ
(
S
), then S is a Boolean ring.We quote the result of Joshi [6, Theorem 2.4] when the corresponding poset is a meet semilattice.Theorem 19 (Joshi [6, Theorem 2.4]).
LetL be a meet semilattice with 0. Then the graph G
{
0
}
(
L
) is connected and d
i
a
m
(
G
{
0
}
(
L
)
)
⩽
3.Theorem 20.
LetL be a meet semilattice with 0. Then the graph G
E
(
L
) is connected and d
i
a
m
(
G
E
(
L
)
)
⩽
d
i
a
m
(
G
{
0
}
(
L
)
)
⩽
3.Proof.
Follows from Corollary5, Theorem 19, and the fact that adjacency in G
E
(
L
) implies adjacency in G
{
0
}
(
L
). Hence diam
⁡
(
G
E
(
L
)
)
⩽
diam
⁡
(
G
{
0
}
(
L
)
)
⩽
3.Theorem 21.
LetL be an S
S
C lattice. Then G
{
0
}
(
L
) is either K
2 or contains a cycle of length 3. Hence g
i
r
t
h
(
G
{
0
}
(
L
)
)
=
3.Proof.
IfV
(
G
{
0
}
(
L
)
)
=
2, then G
{
0
}
(
L
)
≅
K
2, by Theorem 19. Now assume that |
V
(
G
{
0
}
(
L
)
)
|
⩾
3. Choose a
,
b
,
c
∈
V
(
G
{
0
}
(
L
)
) such that a
-
b
-
c is a path with a
≠
c and a
∧
c
≠
0. Since L is SSC and without loss of generality assume that a
∧
c
<
a, there exists x such that 0
<
x
⩽
a and c
∧
x
=
0. Clearly, x
∈
V
(
G
{
0
}
(
L
)
). Since a
∧
b
=
0, we have x
∧
b
=
0. Thus we have a cycle b
-
x
-
c
-
b of length 3.An immediate consequence of the above theorem is the following corollary.Corollary 22.
LetL be a meet semilattice with 0; then G
E
(
L
) is either K
2 or contains a cycle of length 3. Hence, if G
E
(
L
)
≇
K
2, then g
i
r
t
h
(
G
E
(
L
)
)
=
3
=
g
i
r
t
h
(
G
{
0
}
(
L
)
).Proof .
It follows from Corollary5 and Theorem 21.Remark 23.
In the above result, the condition of existence of a cycle inG
E
(
L
) is necessary. In Figure 1, we can observe that G
{
0
}
(
L
) has a cycle but G
E
(
L
) has no cycle.


## 3.  Correlation between Diameter ofG
{
0
}
(
L
) and G
E
(
L
)

The following result gives the correlation between diameter ofG
{
0
}
(
L
) and G
E
(
L
).Theorem 24.
The following statements are true for a meet semilatticeL with 0.(a)
Ifd
i
a
m
(
G
{
0
}
(
L
)
)
=
1, then d
i
a
m
(
G
E
(
L
)
)
=
1.
(b)
Ifd
i
a
m
(
G
{
0
}
(
L
)
)
=
2, then d
i
a
m
(
G
E
(
L
)
)
=
1 or 2.
(c)
Ifd
i
a
m
(
G
{
0
}
(
L
)
)
=
3, if and only if d
i
a
m
(
G
E
(
L
)
)
=
3.
(d)
Ifd
i
a
m
(
G
E
(
L
)
)
=
1, then d
i
a
m
(
G
{
0
}
(
L
)
)
=
1 or 2.
(e)
Ifd
i
a
m
(
G
E
(
L
)
)
=
2, then d
i
a
m
(
G
{
0
}
(
L
)
)
=
2.Proof.
(
a
) It is clear that diam
⁡
(
G
E
(
L
)
)
≠
0. Hence the result follows from Theorem 20.
(
b
) Suppose that diam
⁡
(
G
{
0
}
(
L
)
)
=
2. So we have a path a
-
b
-
c in G
{
0
}
(
L
). We have two possibilities.
Case  1. If [
a
]
,
[
b
]
,
and
[
c
] are distinct vertices of G
E
(
L
), we get the path [
a
]
-
[
b
]
-
[
c
]. Hence by Theorem 20, diam
⁡
(
G
E
(
L
)
)
=
2.
Case  2. The only possibility in this case is [
a
]
=
[
c
], and hence, diam
⁡
(
G
E
(
L
)
)
=
1.
(
c
) Let diam
⁡
(
G
{
0
}
(
L
)
)
=
3. Then there exists a path a
-
c
-
d
-
b. Since c
∈
ann
(
a
) gives [
a
]
≠
[
b
]; otherwise d
(
a
,
b
)
=
2, a contradiction. Now assume that d
(
[
a
]
,
[
b
]
)
=
2 in G
E
(
L
). Then there exists [
x
]
∈
V
(
G
E
(
L
)
) such that [
a
]
-
[
x
]
-
[
b
]. This yields that d
(
a
,
b
)
=
2, again a contradiction. Thus by Theorem 20, diam
⁡
(
G
E
(
L
)
)
=
3. The converse follows from Theorem 20.
(
d
) Suppose that diam
⁡
(
G
E
(
L
)
)
=
1. Then diam
⁡
(
G
{
0
}
(
L
)
)
≠
3, by (
c
). Hence the result.
(
e
) Suppose that diam
⁡
(
G
E
(
L
)
)
=
2; then there exists a path [
a
]
-
[
c
]
-
[
b
] in G
E
(
L
). This yields a path a
-
c
-
b in G
{
0
}
(
L
). Thus diam
⁡
(
G
{
0
}
(
L
)
)
≠
1. The result follows from (
c
) and diam
⁡
(
G
{
0
}
(
L
)
)
≠
3.Remark 25.
The assertions(
a
) and (
d
) of Theorem 24 are justified in Figures 1 and 5, whereas the assertions (
b
) and (
e
) are justified in Figures 5 and 6. Figure 3 verifies the assertion (
c
).Figure 5
diam
⁡
(
G
E
(
L
)
)
=
diam
⁡
(
G
{
0
}
(
L
)
)
=
1.Figure 6
diam
⁡
(
G
E
(
L
)
)
=
diam
⁡
(
G
{
0
}
(
L
)
)
=
2.Corollary 26.
LetL be a meet semilattice with 0. Then d
i
a
m
(
G
E
(
L
)
)
⩽
2 if and only if d
i
a
m
(
G
{
0
}
(
L
)
)
⩽
2.


## 4. Cut Vertices inG
E
(
L
)

In this section, we examine the properties of cut vertices ofG
E
(
L
).Definition 27.
A latticeL with 0 is said to be 0-distributive, if a
∧
b
=
a
∧
c
=
0 implies that a
∧
(
b
∨
c
)
=
0; Varlet [23]. More details about 0-distributive posets can be found in Joshi and Waphare [24]; see also Joshi and Mundlik [25].Theorem 28.
LetL be a 0-distributive lattice. If [
a
] is a cut vertex of G
E
(
L
), then [
a
]
∪
{
0
} forms an ideal of L.Proof.
Let[
a
] be a cut vertex of G
E
(
L
). Let a
1 and a
2 be nonzero elements of [
a
]. Since a
1
,
a
2
∈
[
a
], ann
(
a
1
)
=
ann
(
a
)
=
ann
(
a
2
). Now, it is enough to show that ann
(
a
1
∨
a
2
)
=
ann
(
a
). Clearly, ann
(
a
1
∨
a
2
)
⊆
ann
(
a
).
Lett
∈
ann
(
a
). This gives t
∧
a
1
=
t
∧
a
2
=
t
∧
a
=
0. Since L is 0-distributive, we have t
∧
(
a
1
∨
a
2
)
=
0; that is, t
∈
ann
(
a
1
∨
a
2
). Thus ann
(
a
)
⊆
ann
(
a
1
∨
a
2
). This implies that a
1
∨
a
2
∈
[
a
].
Now suppose thatx
∈
[
a
]
∪
{
0
} and y
⩽
x, to show that y
∈
[
a
]
∪
{
0
}. If x
=
0 or y
=
0, then y
∈
[
a
]
∪
{
0
}. Let y
≠
0 and y
⩽
x. Then ann
(
x
)
⊆
ann
(
y
). Since [
a
] is a cut vertex of G
E
(
L
), therefore for any two arbitrary vertices [
b
]
,
[
c
]
∈
V
(
G
E
(
L
)
), we have a path [
b
]
-
[
a
]
-
[
c
]. Therefore b
∈
ann
(
a
)
=
ann
(
x
); that is, b
∧
x
=
0, which further yields b
∧
y
=
0. Similarly, c
∧
y
=
0. Therefore, we get a path [
b
]
-
[
y
]
-
[
c
]. Since [
a
] is cut vertex, [
a
]
=
[
y
]. Thus [
a
]
∪
{
0
} is an ideal.Remark 29.
It is clear from the proof of the above theorem that in a general lattice, if[
a
] is a cut vertex of G
E
(
L
), then [
a
]
∪
{
0
} is a semi-ideal.Lemma 30.
LetL be a meet semilattice with 0. If [
a
] is a cut vertex of G
E
(
L
), then a
n
n
(
a
) is maximal in ℒ
=
{
a
n
n
(
x
)
∣
0
≠
x
∈
L
}. Hence a
n
n
(
a
) is a prime semi-ideal.Proof.
Let[
a
] be a cut vertex of G
E
(
L
), and let G
1 and G
2 be mutually separated subgraphs of G
E
(
L
) with V
(
G
1
∪
G
2
)
=
V
(
G
E
(
L
)
)
∖
[
a
]. Let [
b
]
∈
G
1 and [
c
]
∈
G
2, therefore we have the path [
b
]
-
[
a
]
-
[
c
]. Suppose that ann
(
a
)
⊆
ann
(
x
) for some ann
(
x
)
∈
ℒ. Since [
b
]
-
[
a
] is an edge, b
∈
ann
(
a
)
⊆
ann
(
x
). Then [
b
]
∧
[
x
]
=
{
0
}. Similarly, [
c
]
∧
[
x
]
=
{
0
}. Thus we have another path [
b
]
-
[
x
]
-
[
c
] passing through of [
x
]. Since [
a
] is a cut vertex of G
E
(
L
), then [
x
]
=
[
a
]. Thus ann
(
x
)
=
ann
(
a
). It is easy to prove that ann
(
a
) is a prime semi-ideal.Lemma 31.
If[
a
] is a cut vertex of G
E
(
L
), then all other associated primes of G
E
(
L
) are contained in only one component of G
E
(
L
)
∖
[
a
].Proof.
Suppose thatG
1 and G
2 are two mutually separated connected components of G
E
(
L
)
∖
[
a
], and each contains an associated prime. It is easy to observe that the associated primes are adjacent, and hence G
1, G
2 are connected, a contradiction.Alizadeh et al. [2] raised the following problem.Problem 32.
Characterize those posetsP for which (
diam
⁡
(
G
{
0
}
(
P
)
)
,
girth
(
G
{
0
}
(
P
)
)
)
=
(
2,3
)
or
(
3,3
).In view of Corollary5 and the following result, it is clear that for an SSC meet semilattice L, (
diam
⁡
(
G
{
0
}
(
L
)
)
,
girth
(
G
{
0
}
(
L
)
)
)
=
(
3,3
).Theorem 33.
LetL be a meet semilattice with 0. If |
V
(
G
E
(
L
)
)
|
⩾
4 and G
E
(
L
) have at least 2 cut vertices, then (
d
i
a
m
(
G
E
(
L
)
)
,
g
i
r
t
h
(
G
E
(
L
)
)
=
(
3,3
).Proof.
Let[
a
] and [
b
] be cut vertices of G
E
(
L
). Since [
a
] is a cut vertex of G
E
(
L
), there is some [
a
1
] such that any path connecting [
a
1
] and [
b
] must include [
a
]. Similarly, since [
b
] is a cut vertex, there is some [
b
1
] such that any path connecting [
b
1
] and [
a
] must include [
b
]. Clearly, [
a
1
]
≠
[
b
1
] and any path from [
a
1
] to [
b
1
] must include [
a
] and [
b
] and so diam
⁡
(
[
a
]
,
[
b
]
)
⩾
3. By Theorem 20, diam
⁡
(
G
E
(
L
)
)
=
3. Thus (
diam
⁡
(
G
E
(
L
)
)
,
girth
(
G
E
(
L
)
)
=
(
3,3
), by Corollary 22.It is easy to see that Figure3 fulfills the conditions of Theorem 33.


## 5. Beck’s Conjecture forG
E
(
L
)

Joshi [6] (see also Halaš and Jukl [4], Nimbhorkar et al. [26], and Lu and Wu [12]) proved that Beck’s Conjecture is true for the zero divisor graphs of posets. We quote this result when the poset is a meet semilattice and ideal is a zero ideal.Theorem 34 (Joshi [6, Theorem 2.9]).
LetL be a meet semilattice with 0. If C
l
i
q
u
e
(
G
{
0
}
(
L
)
)
<
∞, then C
l
i
q
u
e
(
G
{
0
}
(
L
)
)
=
χ
(
G
{
0
}
(
L
)
)
=
n, where n is the number of minimal prime semi-ideals of L.
From Theorem34 and Corollary 5, it is clear that Beck’s Conjecture is true for the graph G
E
(
L
). In the sequel, we calculate the chromatic number of G
E
(
L
). As a preparation, we need the following easy lemma.Lemma 35.
LetP be a minimal prime semi-ideal of a meet semilattice L with 0. Then for any x
∈
P, there exists y
∉
P such that x
∧
y
=
0.Proof.
It follows from the fact thatL
∖
P is a maximal filter of L and x
∉
L
∖
P.Lemma 36.
LetL be a meet semilattice with 0, and let L
′ be a meet semilattice with 0 (as constructed in Lemma 2). If P is a minimal prime semi-ideal of L, then P
′
=
{
[
x
]
∣
x
∈
P
} is a minimal prime semi-ideal of L
′.Proof.
First, we prove thatP
′ is a semi-ideal. Let [
y
]
<
[
x
]
∈
P
′. On the contrary, assume that [
y
]
∉
P
′. Hence ann
(
y
)
⊆
P. Further, [
y
]
<
[
x
] gives ann
(
x
)
⫋
ann
(
y
) which yields ann
(
x
)
⊆
P. Since x
∈
P, by Lemma 35, there exists y
∉
P such that x
∧
y
=
0; that is, y
∈
ann
(
x
). But then y
∈
ann
(
x
)
⊆
P, a contradiction. Thus P
′ is a semi-ideal.
Let[
x
]
∧
[
y
]
∈
P
′. Then x
∧
y
∈
P. By primeness of P, either x
∈
P or y
∈
P. This gives either [
x
]
∈
P
′ or [
y
]
∈
P
′. Thus P
′ is prime. Let Q
′ be a prime semi-ideal of L
′ such that Q
′
⫋
P
′. Hence there exists [
x
]
∈
P
′ such that [
x
]
∉
Q
′. Again by Lemma 35, there exists y
∉
P such that x
∧
y
=
0. This gives [
x
]
∧
[
y
]
=
[
0
]
∈
Q
′. But then [
y
]
∈
Q
′
⊆
P
′, a contradiction. Thus P
′ is a minimal prime semi-ideal of L
′.On similar lines, we can prove the following result.Lemma 37.
LetL be a meet semilattice with 0, and let L
′ be a meet semilattice with 0 (as constructed in Lemma 2). If P
′is a minimal prime semi-ideal of L
′, then P
=
{
x
∣
[
x
]
∈
P
′
} is a minimal prime semi-ideal of L.Let us denote the set of all minimal prime semi-ideals ofL by Mi
n
s
(
L
).Theorem 38.
LetL be a meet semilattice with 0, and let L
′ be a meet semilattice with 0 (as constructed in Lemma 2). Let ϕ
:
Min
s
(
L
)
→
Min
s
(
L
′
) be a map defined by ϕ
(
P
)
=
P
′
=
{
[
x
]
∣
x
∈
P
}. Then ϕ is bijective.Proof.
Letϕ
:
Min
s
(
L
)
→
Min
s
(
L
′
) be a map defined by ϕ
(
P
)
=
P
′
=
{
[
x
]
∣
x
∈
P
}. By Lemma 36, P
′ is a minimal prime semi-ideal. First, we show that ϕ is one-to-one. Let ϕ
(
P
)
=
ϕ
(
Q
). Then for x
∈
P, we have [
x
]
∈
P
′
=
ϕ
(
P
)
=
ϕ
(
Q
)
=
Q
′. This gives x
∈
Q. Hence P
⊆
Q. Similarly, Q
⊆
P. Thus ϕ is one-to-one. Let Q
′ be any minimal prime semi-ideal in L
′. Then Q
=
{
x
∣
[
x
]
∈
Q
′
} is a minimal prime semi-ideal of L, by Lemma 37. It is clear that ϕ
(
Q
)
=
Q
′. Thus ϕ is onto.With this preparation, we now prove Beck’s Conjecture forG
E
(
L
).Theorem 39.
LetL be a meet semilattice with 0. If C
l
i
q
u
e
(
G
E
(
L
)
)
<
∞, then χ
(
G
E
(
L
)
)
=
χ
(
G
{
0
}
(
L
)
)
=
C
l
i
q
u
e
(
G
E
(
L
)
)
=
C
l
i
q
u
e
(
G
{
0
}
(
L
)
)
=
n, where n is the number of all minimal prime semi-ideals of L.Proof.
LetClique
(
G
E
(
L
)
)
<
∞. Then Clique
(
G
{
0
}
(
L
′
)
)
<
∞, by Corollary 5, where L
′ is the semilattice constructed as in Lemma 2. By Theorem 34, Clique
(
G
{
0
}
(
L
′
)
)
=
χ
(
G
{
0
}
(
L
′
)
)
=
n, where n is the number of minimal prime semi-ideals of L
′. Now, by Theorem 38, the number of minimal prime semi-ideals of L is also n. Thus we have Clique
(
G
E
(
L
)
)
=
χ
(
G
E
(
L
)
)
=
n.



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*Source: 896270-2014-05-08.xml*