# Zero Divisor Graph of a Lattice and Its Unique Ideal

Authors: Rajendra Deore; Pramod Tayade
Journal: Journal of Mathematics
 (2019)

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## Abstract

LetL be a lattice with the least element 0. Let AL be the finite set of atoms with AL>1 and ΓL be the zero divisor graph of a lattice L. In this paper, we introduce the smallest finite, distributive, and uniquely complemented ideal B of a lattice L having the same number of atoms as that of L and study the properties of ΓL and ΓB.


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## Body

## 1. Introduction

LetL be a lattice with the least element 0. An element a∈L is said to be a zero divisor if there exists a nonzero element b∈L such that a∧b=0. The set of zero divisors in L is denoted by ZL. We associate a simple graph ΓL to L with the vertex set ZL∗=ZL−0; the set of nonzero zero divisors of L and distinct x,y∈ZL∗ are adjacent if and only if x∧y=0.In [1], the authors have introduced the notion of coloring in graphs derived from lattices. In [2], the authors associated to any finite lattice L a simple graph GL whose vertex set is ZL∗, and two vertices x and y are adjacent if and only if x∧y=0.In [3], the authors associated a simple, undirected graph ΓL with a lattice L with the vertex set ZL∗=ZL−0; the set of nonzero zero divisors of L and distinct x,y∈ZL∗ are adjacent if and only if x∧y=0. They studied the structure of ΓL and some basic properties of the zero divisor graph of a lattice. The zero divisor graph of various algebraic structures has been studied by several authors [4-6].We now give here some preliminaries.LetL,∧,∨ be a lattice with the least element 0. An element a∈L is called as an atom if there is no b∈L such that 0<b<a. The set of all atoms is denoted by AL. In this paper, we consider lattices with at least two atoms. The lattice L is called atomic if for any a∈L, there exists an element b∈AL such that b≤a. A nonempty subset I of L is called an ideal if a,b∈I implies that a∨b∈I and for l≤a implies l∈I. A proper ideal I of L is said to be a prime ideal if a,b∈L and a∧b∈I implies a∈I or b∈I.The undefined terms and notations are as in [2, 3]. Let G=V,E be a graph. A graph G is said to be connected if for each pair of distinct vertices x and y, there is a finite sequence of distinct vertices x=v1,v2,v3,…,vn=y such that each pair of vertices vk,vk+1;1≤k≤n−1 is an edge, and such a sequence of vertices is called as a path. The number of edges in a path is the length of a path. For distinct vertices x and y of G, let dx,y be the length of the shortest path from x to y. dx,y=∞ if there is no path from x to y and dx,x=0. The diameter of a graph G is diamG=supdx,y:wherexandyare distinct  vertices ofG. The length of a smallest cycle in a graph G is called as girth, and it is denoted by grG and grG=∞ if G contains no cycle. The degree of a vertex v in G is the number of edges incident on v which is denoted by degv. The maximum degree vertex of a graph G, ΔG=maxdegv:vis a vertex ofG. A graph G is said to be bipartite if the vertex set of G is partitioned into two disjoint subsets P and Q such that no edge has both end points in any one of P or Q. A complete bipartite graph is a graph whose vertices can be partitioned into two subsets P and Q such that no edge has both endpoints in the same subset, and every possible edge that could connect vertices in different subsets is part of the graph. However, in a complete bipartite graph G if P=m, Q=n, then it is denoted by km,n. A star graph is a complete bipartite graph k1,n. A graph in which each pair of distinct vertices is joined by an edge is a complete graph, and it is denoted by kn. A subset C=c1,c2,c3,… of L is called a clique in ΓL if ci,cj are adjacent for all i,j;i≠j. The clique number clΓL is the cardinality of the maximum possible complete subgraph of ΓL. If the size of a clique is not bounded, then we say clΓL=∞. A (vertex) coloring of G is assigning of colors to the vertices so that adjacent vertices have different colors. The minimum number of colors required to color G is the chromatic number of G and is denoted by χG.


## 2. Diameter and Line Graph on Four Vertices

Lemma 1.
LetL be a lattice with the least element 0 and AL, the set of atoms in L. If AL≥2, then AL⊆ZL∗.Proof.
LetAL=a,b. Then, 0<a,b and a∧b=0.Proposition 1.
A connected graph on four vertices with vertex setVL=a1,a2,a3,a4 and edges a1−a2,a2−a3,a3−a4 cannot be realised as ΓL.Proof.
On the contrary, suppose thatL=0,a1,a2,a3,a4,1 is the required lattice, it must be closed under meet. In particular, a1∧a4∈L. We have the following cases:Case 1:a1∧a4=a1 implies that a1∧a3=a1∧a4∧a3=0Case 2:a1∧a4=a2 implies that a1∧a4=a1∧a4∧a1=a2∧a1=0Case 3:a1∧a4=a3 implies that a1∧a4=a1∧a4∧a1=a3∧a1=0Case 4:a1∧a4=a4 implies that a2∧a4=a2∧a1∧a4=0
In each of the above cases, we get a contradiction. Thus, we havea1∧a4 is not in L. Hence, L is not a lattice.Proposition 2 ([3], Proposition 1.4).
IfL is a lattice, then diamΓL≤3.Theorem 1.
Let L be a lattice with the least element 0. IfΓL contains a cycle, then grΓL≤4.Proof.
Assume thatΓL contains the smallest cycle Cn:a1−a2−a3−⋯−an−a1 of length n. If n≤4, then we are done. Suppose that n≥5. Therefore, a1∧a4≠0. We have the following cases for consideration:Case 1: ifa1∧a4=a1, then a1−a2−a3−a1 is a cycle of length three smaller than n≥5. Similarly, a1∧a4=a4.Case 2: ifa1∧a4=a2, then an−a2−⋯−an−1−an is a cycle of length n−1, smaller than n≥5.Case 3: ifa1∧a4=a3, then it is similar to Case 2.
From Cases 1, 2, and 3, we conclude thata1∧a4≠a1,a2,a3, and a4. However, a2−a1∧a4−a3−a2 is a cycle of length 3, smaller than n≥5, a contradiction to our assumption. Hence, grΓL≤4.
LetL,∧,∨ be a finite bounded lattice and a∈L. Then, we define a↓=x∈L:x≤a, a↑=y∈L:a≤y as given in Definition 3.6 of [2]. The base of k∈L is denoted by Bk and is defined to be the set of all atoms a of L with a≤k and Bkc=AL\Bk.Lemma 2 ([2], Lemma 5.7).
LetL be a finite bounded lattice and a,b∈L such that Ba=Bb. Then, dega=degb.
Following example shows that the converse of the above lemma is not true.Example 1.
For latticeL=0,a1,a2,b,1, a1=Ba1≠Ba2=a2 and dega1=dega2 in Γa1∨a2.


## 3. Unique Ideal of Lattices

Lemma 3.
LetL be a lattice with a nonempty subset S=s1,s2,…,sn. Then, I=S=l∈L:l≤s1∨s2∨,…,∨sn is the smallest ideal of L containing S.
An Ideal of a latticeL is a sublattice but not conversely.Theorem 2.
LetL be a nondiamond lattice with the least element 0 and having a finite set of atoms. Then there exists a finite, distributive, and uniquely complemented ideal B of L such that AL=AB.Proof.
LetAL=a1,a2,a3,…,an be the set of atoms of L. Let B=AL=1J↓, where 1J≠a1∨a2∨⋯∨ak−1∨ak+1⋯∨an,1J=a1∨a2∨⋯∨an is the largest element of B which is a locally maximal element of a lattice L and 0 is the least element. By Lemma 3, B is finite and is the smallest ideal of L containing all atoms of L.Corollary 1.
LetL be a lattice with the least element 0 and having a finite set of atoms. Then, B is a finite, distributive, and uniquely complemented sublattice of L.Corollary 2.
LetL be a lattice with the least element 0 and has a finite set of atoms. Then,(a)
ZB∗=B−0,1J(b)
ZB⊆ZLProof.
(a) Forx∈ZB∗,x=a1∨a2∨…∨ak−1∨ak+1…∨an,x∧ak=0. (b) For x∈ZB,x=a1∨a2∨⋯∨ak−1∨ak+1⋯∨an,x∧ak=0, x∈ZL. If x=0, then x∈ZL.Corollary 3.
LetL be a lattice with least element 0, AL=a1,a2,a3,…,an be the set of all atoms of L, and a unique ideal B=1J↓ of L. Let 1J and 1 be the largest elements of B and L, respectively. Then,(a)
If1J≤x<1, then x has no complement in L(b)
Ifb<1J, then b has a complement in B but has no complement in LCorollary 4.
([2], Corollary 5.11). Let a be an element of L such that Ba=AL and a be comparable with all elements of L. Then, GL=Ga↓.Theorem 3.
LetL be a lattice with the least element 0 and having a finite set of atoms. If B is the unique ideal, then ΓB is the induced subgraph of ΓL.Proof.
Proof follows by Corollary 2 and Corollary 4.Theorem 4.
LetL be a lattice with the least element 0 and having a finite set of atoms and B be its unique ideal. Then, ΓB is the n-partite induced subgraph of ΓL.Proof.
LetAL=a1,a2,…,an,n≥2. Then, AB=a1,a2,…,an is the set of atoms in a finite sublattice B. Therefore, B−0=a1↑∪a2↑∪⋯∪an↑. There is no adjacency between any two arbitrary elements in ai↑, for all 1≤i≤n. If x∈ai↑∪ak↑ for some 1≤i,k≤n, then there is no adjacency between x and both elements ai and ak. Thus, there exist disjoint subsets Bi of B−0, B−0=B1∪B2⋯∪Bn. This shows that ΓB is n-partite.Theorem 5.
LetL be a finite lattice with AL=a,b. a=x0<x1<x2<x3<⋯<xn,B=b∨a=B0 and Bi=b∨xi. Then,(1)
AB=ABi for all 1≤i≤n(2)
B⊂B1⊂B2⊂B3⊂⋯⊂Bn(3)
ΓB⊂ΓB1⊂ΓB2⊂ΓB3⊂⋯⊂ΓBn(4)
ΓL=K1,n(5)
grΓL=grΓB=∞Theorem 6.
LetL be a finite lattice with AL=a,b and a=x0<x1<x2<x3<⋯<xn and b=y0<y1<y2<y3<⋯<yp. Then,(1)
B=a∨b⊆x1∨y1(2)
grΓL=grΓx1∨y1↓=4(3)
grΓB=∞Example 2.
LetL=1,2,4,8,3,0 be a lattice under the divisibility relation and 1 is the least and 0 is the largest element of L. Here, B=1,2,3,0 is the unique ideal, and B1=1,2,4,3,0 is not since 3 is the complement of 2 and 4.Example 3.
LetA=2k:gcd2k,3=1 for positive integer k and let L=A∪0,1,3. Then, L is a lattice under divisibility relation and 1 is the least and 0 is the largest element of L. Here, B=1,2,3,0 is the unique ideal and B1=1,2,4,3,0 and B2=1,2,4,8,3,0 …. are not since 3 is the complement of more than two elements.Proposition 3.
LetL be a lattice with the least element 0. Let AL=a,b with a unique ideal B=1J↓. Then, ΓB=K1,1.Proof.
By Lemma 5.6 of [2], ΓB=Kr,s,. where r is the number of elements in a set a↑\b↑ and s is the number of elements in a set b↑\a↑. Now a↑=a,a∨b and b↑=b,a∨b, a↑\b↑=a. Similarly, b↑\a↑=b.Proposition 4.
LetL be a lattice with the least element 0 and let AL=a1,a2,a3,…,an. Then, a unique sublattice B=1J↓ is isomorphic to a Boolean ring A.Proof.
LetB be the unique ideal of L. Then, by Definition 4.1 and Proposition 4.2 of [2], ΓB=GB, where B is a Boolean ring under the operations p+q=p∧q′∨p′∧q and p⋅q=p∧q (p′ is a complement of p).Proposition 5.
LetLi be the lattices with the least element 0i:1≤i≤k. Then, L1⊗L2⊗L3⊗⋯⊗Lk is a lattice.Proposition 6.
LetLi be the lattices with the least element 0i:1≤i≤k;ni=ALi. Then, AL1⊗L2⊗L3⊗⋯⊗Lk=n1+n2+n3+⋯+nk.Proof.
An orderedk-tuple b1,b2,b3,…,bk−1,bk is an atom if at most one component is an atom in ALi and the rest of the components are corresponding zero’s of Li. Hence, the result.Proposition 7.
LetL be a lattice with the least element 0 and B be its unique ideal. Let AL=a1,a2,…,an be the set of atoms in L. Then, clΓB=n.Proof.
Suppose thatC=a1,a2,…,an,b is a clique. Then, by definition of a unique ideal B, b=ai∨aj for some i≠j, and b∧ai≠0, shows that C is not a clique.Theorem 7.
LetL be lattice with the element 0 and having a finite set of atoms. Let B be its unique ideal. Then, AL=n if and only if clΓB=χΓB=n.Proof.
Case 1: clearlyclΓB≤χΓB. Let AL=a1,a2,…,an be the set of atoms in L. We decompose ZB∗ as P1=B1, Pk=x∈Bk:x∉∪j<kBj for k=2,3,4,…,n and Bj=x∈ZB∗:x≥aj. Define a map g on ZB∗ by gx=k for x∈Pk. If x,y are adjacent, then gx≠gy, so g is a coloring on ΓB; hence, χΓB≤n by Proposition 7. Conversely, assume that AB is the set of atoms of B is a clique in B. Then, clΓB=n=AL.
Case 2: ifB has an infinite number of atoms, then clΓB=∞; hence, χΓB=∞. Suppose clΓB=∞=clΓL. Let c1,c2,… be an infinite clique in ΓB. Since B is atomic, for each i, there exists an atom ai such that ai≤ci. Again ci,cj are adjacent for i≠j, and it shows that ai≰cj. Hence, B has an infinite number of atoms.Proposition 8.
LetL be a lattice with the least element 0 and B be its unique ideal. Then, χΓB=χΓL.Proof.
By Theorem 7,XTB=n, applying Theorem 7, for atomic lattice with n number of atoms, XTL=n, similar to Theorem 2.1 in [1]. Following example shows that for nonatomic lattice L with the least element 0 and at least two atoms, χΓB=χΓL=2.Example 4.
By Example 2 of [1], we know that the set A=1/n:nis a positive integer∪0 is a bounded distributive lattice having no atom. Let D2=1,2,5,10 be a lattice of divisors of 10 under the divisibility relation. Then, A⊗D2 is nonatomic lattice with two atoms. The unique ideal B=0,2∨0,5. Here, χΓB=χΓL=2.Lemma 4.
LetL be a lattice with the least element 0 and B be its unique ideal. If B contains an infinite chain, then clΓL=∞.Proof.
Letb1<b2<⋯ be an increasing chain in B. Set xi=bi+1∧bi′. If xi=0, then by distributivity in B, we have bi=bi∨bi+1∧bi=bi+1, which contradicts the hypothesis. Suppose that xi=xj:i<j and i+1≤j. Therefore, bi+1≤bj and xj=bj+1∧bj′=bi+1∧bi′∧bj′=0, a contradiction. This shows that xi are distinct. Again bi+1∧bj′=0 implies that xi∧xj=0:i≠j. Hence, xi:i=1,2,3,… is an infinite clique in L.


## 4. Diamond Lattices

An elementa∈L is called a coatom if a<1. A lattice L is said to be a diamond lattice if every element a∈L,a∉0,1 is an atom and coatom. A diamond with n atoms is denoted by Dn.Theorem 8.
LetD2=0,a1,a2,1 be a diamond. Then, B=0,a1,a2,1 is a distributive and uniquely complemented ideal of D2. Moreover, ΓB=ΓD2=K1,1.Theorem 9.
LetDn=0,a1,a2,…,an,1, be a diamond lattice. Then,(1)
Forn=2,B=0,a1,a2,1 is a distributive and complemented ideal of Dn(2)
Forn≠1,Kn=ΓDn(3)
Forn≥3,grΓDn=3Theorem 10.
LetL=D2⊗Dn, be a lattice. Then, B=D2⊗D2 is a distributive and complemented ideal of L, which can be generalized.


## 5. Some Combinatorial Results

Theorem 11.
IfL is a lattice which is nondiamond or nondecomposable into diamonds with the least element 0 and AL=a1,a2,a3,…,an, then dega1=n−1C1+n−1C2+n−1C3+⋯+n−1Cn−1 in ΓB.Proof.
LetAL=a1,a2,a3,…,ann≥2. ΓB=Γ1J↓. Then, in a finite subgraph ΓB, a1 is adjacent to n−1 vertices; hence, contribution of dega1 is n−1C1. Again a1 is adjacent to ai∨aj,i,j≠1, and contribution of dega1 for this pair is n−1C2. Continuing in this way, contribution of dega1 for a2∨⋯∨ak−1∨ak∨⋯∨an is n−1Cn−1. Total contribution is dega1=n−1C1+n−1C2+n−1C3+n−1+⋯+n−1Cn−1.Corollary 5.
LetL be a lattice with the least element 0 and B be its unique ideal. Then, for the induced subgraph ΓB,(1)
ΔΓB=dega1:a1∈AB(2)
ΔΓB≤ΔΓLProposition 9.
LetL be a finite lattice with the least element 0 with AL=a,b and a=x0<x1<x2<⋯<xk. Then, ΔΓL=k+1.Proof.
Here,Bxi=a; hence, xi∧b=0 for 1≤i≤k.Proposition 10.
LetL be a lattice with the least element 0, AL=a1,a2,a3,…,an set of atoms, and B be the unique sublattice of L. Then, ZB∗=2n−2,n≥2.Proof.
LetAL=a1,a2,a3,…,an=AB. For ai,1≤i≤n, zero divisors, their contribution is nC1. Also ai∨aj,i≠j are zero divisors, and their contribution is nC2. Continuing is this way, contribution of a1∨a2∨⋯∨ak−1∨ak+1⋯∨an is a set of zero divisors and is n=nCn−1.Corollary 6.
Let L be a finite lattice with the least element 0 andB be its unique ideal. Then, ZL∗≥2n−2.Proof.
Proof follows by Corollary 2 and Proposition 10.



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