Perfectness of G-generalized Join of Graphs

Abstract

In this paper, we prove that the G-generalized join of complete or totally disconnected graphs is perfect if and only if G is perfect. As a result, we deduce some results proved in (Saeid et al. Rocky Mountain J. Math. 48(3) (2018), 729–751) and (Nilesh et al. arXiv (2022), arXiv:2205.04916). We also characterize rings, posets and reduced semigroups whose zero-divisor graphs and ideal based zero-divisor graphs are perfect. As a consequence, we characterize distributive lattices with 0, reduced semirings and boolean rings whose zero divisor graphs are perfect, which are proved in (Patil et al. in Discrete Math. 340: 740–745, 2017).

1 Introduction

All the graphs considered in the paper are finite, simple and undirected. Let $G=(V(G),E(G))$ be a graph. For $v\in V(G)$ and $S\subseteq V(G)$, let $N_G(v)$ denote the open neighborhood of v in G and $\left\langle S \right\rangle$ denote the subgraph induced by S. Let $\overline{G}$ denote the complement of a graph G. A proper k-coloring of a graph G is a function from V(G) into a set of k colors such that no two adjacent vertices receive the same color. The chromatic number of a graph G, denoted by $\chi (G)$, is the least positive integer k such that there exists a proper k-coloring of G. A clique in a graph G is a complete subgraph of G. The clique number of G is the largest size of a clique in G and it is denoted by $\omega (G)$. Let G be a graph with $V(G)=\{u_1,u_2,\ldots , u_n\}$ and $H_1,H_2,\ldots , H_n$ be pairwise disjoint graphs. The G-generalized join graph, denoted by $G[H_1,H_2,\ldots , H_n]$, of $H_1,H_2,\ldots , H_n$ is the graph obtained by replacing each vertex $u_i$ of G by $H_i$ and joining each vertex of $H_i$ to each vertex of $H_j$ by an edge if $u_i$ is adjacent to $u_j$ in G. If $H_i\cong H$, for $1\le i\le n$, then $G[H_1,H_2,\ldots , H_n]$ becomes the standard lexicographic product G[H].

For a graph G, we define a relation $\mathtt {\sim }_G$ on V(G) as follows: For any $x, y \in V(G)$, define $x \mathtt {\sim }_G\ y$ if and only if $N_G(x) = N_G(y)$. Clearly, $\mathtt {\sim }_G$ is an equivalence relation on V(G). Let [x] be the equivalence class which contains x and S be the set of all equivalence classes of this relation $\mathtt {\sim }_G$. Based on this equivalence classes we define the reduced graph $G_r$ of a graph G as follows. The reduced graph $G_r$ of G (defined in [13]) is the graph with vertex set $V(G_r)=S$ and two distinct vertices [x] and [y] are adjacent in $G_r$ if and only if x and y are adjacent in G.

Note that if $V(G_r)=\big \{[x_1],[x_2],\ldots , [x_k]\big \}$, then G is the $G_r$-generalized join of $\left\langle [x_1]\right\rangle ,\left\langle [x_2]\right\rangle ,\ldots , \left\langle [x_k]\right\rangle$, that is, $G=G_r\big [\left\langle [x_1]\right\rangle ,\left\langle [x_2]\right\rangle ,\ldots , \left\langle [x_k]\right\rangle \big ]$ and each $[x_i]$ is an independent subset of G (that is, $\langle [x_i]\rangle$ has no edge). Clearly, $G_r$ is isomorphic to an induced subgraph of G. It is easy to observe the following observation.

Observation 1. If $G_r$ is the reduced graph of G with $\omega (G_r)=\chi (G_r)$, then $\chi (G)=\omega (G_r)$.

Let G be a graph with $\omega (G)=k$, and let $\varDelta _k(G)$ be the set of all the vertices of a graph G which lie in some clique of size k of G. A connected graph G is called a generalized complete k-partite graph (see [13]) if the vertex set V(G) of G is a disjoint union of A and H satisfying the following conditions:

• (1) $A = \varDelta _k(G)$ and the subgraph induced by A is a complete k-partite graph with parts, say, $A_i, i = 1, 2,\ldots , k$.

• (2) For any $h\in H$ and $i\in \{1,2,\ldots , k\}$, h is adjacent to some vertex of $A_i$ if and only if h is adjacent to any vertex of $A_i$. Set $W(h) = \{1\le i \le k\ | N(h)\cap A_i\ne \emptyset \}$ for any $h\in H$.

• (3) For any $h_1, h_2 \in H, h_1$ is adjacent to $h_2$ if and only if $W(h_1)\cup W(h_2) = \{1,2,\ldots , k\}.$

A graph G is called a compact graph (see [13]) if G contains no isolated vertices and for each pair x, y of non-adjacent vertices of G, there is a vertex z in G with $N(x)\cup N(y)\subseteq N(z)$. A graph G is said to be k-compact if it is compact and $\omega (G)=k$.

Throughout this paper, rings are finite non-zero commutative rings with unity. Let R be a ring. A non-zero element x of R is said to be a zero-divisor if there exists a non-zero element y of R such that $xy=0$. A non-zero element u of R is unit in R if there exists v in R such that $uv=1$. For $x\in R$, the annihilator of x is the set $Ann(x)=\{y\in R\mid xy=0\}$. A ring R is said to be local if it has unique maximal ideal M. The nilradical of a ring R is the set $J=\{x\in R : x^t=0,$ for some positive integer $t\}$. The index of nilpotency of J is the least positive integer m for which $J^m=\{0\}$, where $J^m=JJ\ldots J$ (m-times). A ring R is said to be reduced if $J=\{0\}$. A ring is said to be indecomposable if it can not be written as a direct product of two rings. Let $\mathbb {Z}_n$ be the ring of integer modulo n.

For any ring R, in [6], Beck associated a simple graph with R whose vertices are the elements of R and any two distinct vertices x and y are adjacent if and only if $xy=0$ in R. Beck conjectured that (see [6]) the chromatic number and clique number of this graph are the same and this was disproved by Anderson and Naseer in [2] (also, see [10]). It can be observed that for the graph associated with the ring, the vertex 0 is adjacent to every other vertex. Anderson and Livingston in [5] slightly modified the definition of the graph associated with a ring by considering the zero-divisors as the vertices and any two distinct vertices x and y are adjacent if and only if $xy=0$ in R. They called this zero-divisor graph of the ring R and it is denoted by $\varGamma (R)$. Zero-divisor graphs have been extensively studied in the past. This can be seen in [1, 3, 4, 11, 20].

The following definitions and results can be found in [4, 20]. For $x,y\in R$, define $x\mathtt {\sim }_R\ y$ if and only if $Ann(x)=Ann(y)$. It is proved in [4] that the relation $\mathtt {\sim }_R$ is an equivalence relation on R. For $x\in R$, let $D_x=\{r\in R\ |\ x\mathtt {\sim }_R\ r \}$ be the equivalence class of x. Let $R_E=\{D_{x_1},D_{x_2},\ldots , D_{x_k}\}$ be the set of all equivalence classes of the relation $\mathtt {\sim }_R$. The compressed zero-divisor graph $\varGamma _E(R)$ of R (defined in [20]) is a simple graph with vertex set $R_E\backslash \{D_0,D_1\}$ and two distinct vertices $D_x$ and $D_y$ are adjacent if and only if $xy=0$. The following result can be found in [18].

Theorem 1

[18]. If R is a ring, then

• (i) $\varGamma (R)\cong \varGamma _E(R)[\langle D_{x_1}\rangle ,\langle D_{x_2}\rangle ,\ldots , \langle D_{x_{k-2}}\rangle ]$, where $D_{x_i}\ne D_0, D_1$, for $1\le i\le k-2$,

• (ii) $\left\langle D_{x_i}\right\rangle$ is complete if and only if $x_i^2=0$, and

• (iii) $\left\langle D_{x_i}\right\rangle$ is totally disconnected (that is, $\left\langle D_{x_i}\right\rangle$ has no edge) if and only if $x_i^2\ne 0$.

The following result is proved in [3].

Theorem 2

[3]. If R is a non-zero reduced ring, then there exists a positive integer k such that the compressed zero-divisor graph $\varGamma _E(R) \cong \varGamma (\mathbb {Z}_2^k)$, where $\mathbb {Z}_2^k=\mathbb {Z}_2\times \mathbb {Z}_2 \times \ldots \times \mathbb {Z}_2$ (k-times).

In [9], Hala and Jukl introduced the concept of the zero-divisor graph of a poset. Let $(P,\le )$ be a finite poset with the least element 0. For any $a, b\in P$, denote $L(a,b) =\{c\in P\ |\ c \le a$ and $c\le b\}$. A non-zero element $a\in P$ is said to be a zero-divisor if $L(a,b) = \{0\}$ for some $0\ne b\in P$. We say a non-zero element $a\in P$ is an atom (primitive) if for any $0\ne b\in P,\ b \le a$ implies $a=b$. The zero-divisor graph $\varGamma (P)$ of a poset P is a graph whose vertex set $V(\varGamma (P))$ consists of the zero-divisors of P, in which a is adjacent to b if and only if $L(a,b) =\{0\}$. It is shown in [9] that for any poset P, the clique number and the chromatic number of $\varGamma (P)$ are the same.

By a semigroup, we mean a finite commutative semigroup with the zero element 0. A semigroup S is said to be reduced if for any $a\in S$ and any positive integer n, $a^n = 0$ implies $a = 0$. A semigroup S is said to be idempotent (it is a so-called semilattice, see [13]) if for each $a\in S$, $a^2=a$.

We define a zero-divisor graph of a semigroup in a similar manner in the definition of zero-divisor graph of a ring.

Let $R = \mathbb {Z}_2^k$. Clearly, it is a Boolean ring and it becomes a poset by defining $a\le b$ iff $ab=a$ for any $a,b \in R$. Note that, the zero-divisor graphs of R as a ring (or a semigroup) and as a poset coincide. Let H be a subgraph of $\varGamma (\mathbb {Z}_2^k)$. We say that H is minimal (see [13]) if H is an induced subgraph of $\varGamma (\mathbb {Z}_2^k)$ which contains all the atoms of the poset $\mathbb {Z}_2^k$, and we say H is minimal closed (see [13]) if H is minimal and $V(H)\cup \{0\}$ is a sub-semigroup of $\mathbb {Z}_2^k$. The following results can be found in [13].

Theorem 3

[13]. Let G be a simple graph with $\omega (G)=k$. Then the following statements are equivalent:

• (i) G is the zero-divisor graph of a poset.

• (ii) G is a k-compact graph.

• (iii) G is a generalized complete k-partite graph.

• (iv) The reduced graph $G_r$ of G is isomorphic to a minimal subgraph of $\varGamma (\mathbb {Z}_2^k)$.

Theorem 4

[13]. Let G be a simple graph with $\omega (G)=k$. Then the following statements are equivalent:

• (i) G is the zero-divisor graph of a reduced semigroup with 0.

• (ii) G is a generalized complete k-partite graph such that for any non-adjacent vertices $a, b\in V(G)$, there is a vertex $c\in V(G)$ with $W(c)= W(a)\cup W(b)$.

• (iii) The reduced graph $G_r$ of G is isomorphic to a minimal closed subgraph of $\varGamma (\mathbb {Z}_2^k)$.

• (iv) G is the zero-divisor graph of a semilattice (or equivalently, idempotent semigroup) with 0.

A graph G is perfect if $\omega (H)=\chi (H)$ for every induced subgraph H of G.

The following result was proved by Lovasz, see [12].

Theorem 5

[12]. The complement of every perfect graph is perfect.

In [7], Berge conjectured the following and it was proved by Chudnovsky et al., see [8].

Theorem 6

(Strong Perfect Graph Theorem [8]). A graph G is perfect if and only if it does not contain an induced subgraph which is either an odd cycle of length at least 5 or the complement of such a cycle.

The paper mainly deals with the results on perfect graph using the Strong Perfect Graph Theorem. As a result, we deduced many known results in the literature. This is precisely as follows.

In Sect. 2, we prove that the G-generalized join of complete graphs and totally disconnected graphs is perfect if and only if G is perfect. As a consequence, we deduce the results proved in [14] and [17] and prove that the lexicographic product of a perfect graph and a complete graph and the lexicographic product of a perfect graph and a complement of a complete graph are perfect.

In Sect. 3, we characterize rings, posets and reduced semigroups whose zero-divisor graphs and ideal based zero-divisors are perfect. As a result, we characterize distributive lattices with 0, reduced semirings and boolean rings whose zero divisor graphs are perfect, which are proved in [15]. Further, we completely characterize rings the ideal based zero-divisor graph of the ring $\mathbb {Z}_n$ is perfect.

2 When a G-generalized Join of Complete and Totally Disconnected Graphs is Perfect

In this section, we prove the following result on perfect graphs.

Theorem 7

If G is a graph with vertex set $V(G)=\{v_1,v_2,\ldots , v_n\}$ and $H_1, H_2,$

$\ldots , H_n$ are graphs such that each $H_i$ is either complete or a totally disconnected graph, then G is perfect if and only if $G[H_1,H_2,\ldots , H_n]$ is perfect.

Proof

Let $G'=G[H_1,H_2,\ldots , H_n]$. It is enough to prove if G is perfect, then $G'$ is perfect. Suppose $G'$ is not perfect, then by Theorem 6, $G'$ contains either an odd cycle of length at least 5 as an induced subgraph or the complement of an odd cycle of length at least 5 as an induced subgraph.

Case 1. $G'$ contains an odd cycle $C_{2k+1}$ as an induced subgraph, where $k\ge 2$.

Let $V(C_{2k+1})=\{x_0,x_1,\ldots , x_{2k}\}$ such that $x_i$ is adjacent to $x_{i+1}$ (where the addition in subscript is taken modulo $2k+1$) and $x_i$ is not adjacent to $x_j$, where $j\ne i-1,i+1$. Suppose there exists $1\le t\le n$ such that $|V(C_{2k+1})\cap V(H_t)|\ge 2$.

First, if there exists $0\le i\le 2k$ such that $x_i,x_{i+1}\in V(H_t)$. Then $H_t$ is complete and hence $x_{i-1}\notin V(H_t)$ (otherwise, $C_{2k+1}$ would not be induced in $G'$). Thus there exists $1\le s\le n$ with $s\ne t$ such that $x_{i-1}\in V(H_s)$ and hence $x_{i-1}$ is adjacent to $x_{i+1}$, which is a contradiction.

Next, if there exist $0\le i,j\le 2n$ such that $j\ne i-1,i, i+1$ and $x_i,x_j\in V(H_t)$. Then $H_t$ has no edge in $G'$ and $x_{i+1},x_{i-1}\notin V(H_t)$. Suppose if $j\ne i+2$, then there exists $1\le s\le n$ such that $s\ne t$ and $x_{i+1}\in V(H_s)$ and hence $x_j$ is adjacent to $x_{i+1}$, (because of $x_ix_{i+1}\in E(C_{2k+1})$) which is a contradiction. Therefore, if $j=i+2$, then there exists $1\le s\le n$ such that $s\ne t$ and $x_{i-1}\in V(H_s)$ and therefore $x_{i-1}$ is adjacent to $x_j$, which is again a contradiction.

Hence $|V(C_{2k+1})\cap V(H_i)|=1$, for $0\le i\le 2k$ which implies that G contains an odd cycle of length at least 5 as an induced subgraph, which is a contradiction.

Case 2. $G'$ contains a complement of an odd cycle of length at least 5 as an induced subgraph.

Let $\overline{C_{2k+1}}$ be the complement of the odd cycle $C_{2k+1}$ as an induced subgraph of $G'$, where $k\ge 2$ with $V(\overline{C_{2k+1}})=\{x_0,x_1,\ldots , x_{2k}\}$ such that $x_i$ is not adjacent to $x_j$ for $j=i-1,i+1$ and $x_i$ is adjacent to $x_j$, for $j\ne i-1,i,i+1$ (where the addition in subscripts is taken modulo $2k+1$). Suppose there exists $1\le t\le n$ such that $|V(C_{2k+1})\cap V(H_t)|\ge 2$.

First, if there exists $0\le i\le 2k$ such that $x_i,x_{i+1}\in V(H_t)$. Then $H_t$ has no edge and $x_{i-1}\notin H_t$ and hence there exists $1\le s\le n$ with $s\ne t$ such that $x_{i-1}\in V(H_s)$. But $x_{i+1}$ is adjacent to $x_{i-1}$ and hence $x_i$ is adjacent to $x_{i-1}$, which is a contradiction.

Next, if there exist $0\le i,j\le 2n$ such that $j\ne i-1,i, i+1$ and $x_i,x_j\in V(H_t)$. Then $H_t$ is complete and $x_{i-1},x_{i+1}\notin V(H_t)$. Suppose if $j\ne i+2$, then there exists $1\le s\le n$ such that $s\ne t$ and $x_{i+1}\in V(H_s)$. But $x_j$ is adjacent to $x_{i+1}$ and therefore $x_i$ is adjacent to $x_{i+1}$, which is impossible. Hence, if $j=i+2$, then there exists $1\le s\le n$ such that $s\ne t$ and $x_{i-1}\in V(H_s)$ and therefore $x_{i-1}$ is adjacent $x_i$, which is again a contradiction.

Thus $|V(\overline{C_{2k+1}})\cap V(H_i)|=1$, for $0\le i\le 2k$, which implies that G contains a complement of an odd cycle of length at least 5 as an induced, which is a contradiction.

The following corollary is an immediate consequence of Theorem 7.

Corollary 1

If G is perfect and n is a positive integer, then $G[K_n]$ and $G[K_n^c]$ are perfect.

Proof

As $G[K_n]\cong G[K_n,K_n,\ldots , K_n]$ and $G[K_n^c]\cong G[K_n^c,K_n^c,\ldots , K_n^c]$, the result follows from Theorem 7.

The following result proved in [17] is deduced from Theorem 7.

Corollary 2

(Corollary 3.2, [17]). A graph G is perfect if and only if it’s reduced graph $G_r$ is perfect.

The following relation is defined on a graph G in [14]. For $x,y\in V(G)$, define $x\approx y$ if and only if either $x=y$ or $xy\in E(G)$ and $N(x)\backslash \{y\}=N(y)\backslash \{x\}$. Clearly, it is an equivalence relation. Let [x] be the equivalence class of x, and $S=\{[x_1],[x_2],\ldots , [x_r]\}$ be the set of all equivalence classes of the relation $\approx$. Based on these equivalence classes of the relation $\approx$, we defined (This can be seen in [14]) the graph $G_{red}$ with vertex set $V(G_{red})=S$ and two distinct vertices [x] and [y] are adjacent in $G_{red}$ if and only if x and y are adjacent in G. Clearly, for any graph G, $G=G_{red}[\langle [x_1]\rangle ,\langle [x_2]\rangle ,\ldots , \langle [x_r]\rangle ]$ and $\langle [x_i]\rangle$ is complete, for $1\le i\le r$.

By Theorem 7, we deduce the following result proved in [14].

Corollary 3

(Theorem 4.4, [14]). A graph is perfect if and only if $G_{red}$ is perfect.

3 Perfect Zero-Divisor Graph of a Ring

In this section, we ask the following interesting question. When does the zero-divisor graph of a ring R perfect? To answer this question, we provide a necessary and sufficient condition for which the zero-divisor graph of a ring is perfect.

Theorem 8

If R is a ring, then $\varGamma (R)$ is perfect if and only if its compressed zero-divisor graph $\varGamma _E(R)$ of R is perfect.

Proof

The result follows from Theorems 1 and 7.

Let $R_1,R_2,\ldots , R_k$ be rings. For $x_j\in R_1\times R_2\times \ldots \times R_k$, there exists a unique $x_j(i)\in R_i$, for $1\le i\le k$, such that $x_j=(x_j(1),x_j(2),\ldots , x_j(k))$.

Note that there are several rings satisfying Beck’s conjecture; see [2, 4, 6, 9, 10, 20]. One of them is a finite reduced ring. Using Observation 1, we give a shorter proof of this result as follows.

Corollary 4

[6, 20]. If R is a non-zero reduced ring, then $\chi (\varGamma (R))=\omega (\varGamma (R))$.

Proof

By Observation 1 and Theorem 2, it is enough to prove $\omega (\varGamma (\mathbb {Z}_2^k))=\chi (\varGamma (\mathbb {Z}_2^k))$. Clearly $\{e_i\ |\ 1\le i\le k\}$, where $e_i=(0,\ldots ,0,1,0,\ldots , 0),$ induces a clique. Color first $e_i$ by i, for $1\le i\le k$.

For any $x=(x(1),x(2),\ldots , x(k))\in V(\varGamma (\mathbb {Z}_2^k))\backslash \{e_i\ |\ 1\le i\le k\},$ there exists a least j with $1\le j\le k$, such that $x(i)=0$ for $1\le i\le j-1$ and $x(j)=1$. Color x by j, then the resulting coloring is a proper k-coloring of $\varGamma (\mathbb {Z}_2^k)$.

The following result gives a necessary condition for a product of rings whose zero-divisor graphs are perfect.

Theorem 9

Let $R=R_1\times R_2\times \ldots \times R_k$, where $R_i$’s are indecomposable rings. If $\varGamma (R)$ is perfect, then $k\le 4$.

Proof

Suppose $k\ge 5$. Then the set of vertices $\{(1,1,0,0,0,0,\ldots ,0), (0,0,1,1,0,0, \ldots ,0),$ $(1,0,0,0,1,0,\ldots ,0),(0,1,0,1,0,0,\ldots ,0),(0,0,1,0,1,0,\ldots ,0)\}$ forms an induced cycle of length 5. By Theorem 6, we get a contradiction.

Next, let us prove the following result.

Theorem 10

If $R=\mathbb {Z}_2^4\ (=\mathbb {Z}_2\times \mathbb {Z}_2\times \mathbb {Z}_2\times \mathbb {Z}_2)$, then $\varGamma (R)$ is perfect.

Proof

Suppose $\varGamma (R)$ is not perfect. Then, by Theorem 6, we consider the following cases.

Case 1. $\varGamma (R)$ contains an odd cycle of length at least 5 as an induced subgraph.

Let $C_{2r+1}$ be an induced cycle in $\varGamma (R)$ of length $2r+1$ with the vertex set $\{x_0,x_1,\ldots , x_{2r}\}$, where $r\ge 2$. If exactly one co-ordinate of $x_i$ is non-zero, for $0\le i\le 2r$, then $2r+1\le 4$, a contradiction. Therefore there exists an $x_i$ containing at least two non-zero co-ordinates. WLOG, $x_i=(1,1,x_i(3),x_i(4))$, for some $i,\ 0\le i\le 2r$. Then the $1^{st}$ two coordinates of $x_{i-1}, x_{i+1}$ are zeros, that is, $x_{i-1}(1)=x_{i-1}(2)=x_{i+1}(1)=x_{i+1}(2)=0$. Since $x_{i-1}$ and $x_{i+1}$ are not adjacent, either the third coordinate or forth coordinate of $x_{i-1}$ and $x_{i+1}$ are non-zero. WLOG, $x_{i-1}(3)=x_{i+1}(3)=1$. If $x_{i-1}(4)=1$, then $x_{i+1}(4)=0$, as $x_{i-1}\ne x_{i+1}$ and hence $x_{i-1}=(0,0,1,1)$ and $x_{i+1}=(0,0,1,0)$. Since $x_{i-2}$ is adjacent to $x_{i-1}$, we have $x_{i-2}=(x_{i-2}(1),x_{i-2}(2), 0,0)$. Thus $x_{i-2}$ is adjacent to $x_{i+1}$, which is a contradiction. Hence $x_{i-1}(4)=0$, which implies that $x_{i+1}(4)=1$ and thus $x_{i+1}=(0,0,1,1)$ and $x_{i-1}=(0,0,1,0)$. Since $x_{i+2}$ is adjacent $x_{i+1}$, we have $x_{i+2}=(x_{i+2}(1),x_{i+2}(2),0,0)$ and hence $x_{i+2}$ is adjacent to $x_{i-1}$, which is a contradiction.

Case 2. $\varGamma (R)$ contains the complement of an odd cycle of length at least 5 as an induced subgraph.

Let $\overline{C_{2r+1}}$ be an induced subgraph of $\varGamma (R)$ with vertex set $\{x_0,x_1,\ldots , x_{2r}\}$, where $r\ge 2$. If no $x_i$ contains exactly two coordinates that are non-zeros, then there exists $j,\ 1\le j\le k$ such that $x_j$ contains exactly three that coordinates that are non-zero (otherwise $2r+1\le 4$), which is impossible. Thus there exists $i,\ 1\le i\le k$ such that $x_i$ contains exactly two coordinates that are non-zeros. WLOG, $x_i=(1,1,x_i(3),x_i(4))$. Since $x_i$ is adjacent to $2r-2$ vertices in $\overline{C_{2r+1}}$, namely $x_{i+2},x_{i+3},\ldots , x_{i+2r-1}$ (where the addition in subscripts taken modulo $2r+1$), we have the 1$^{st}$ two coordinates of $x_{i+2},x_{i+3},\ldots , x_{i+2r-1}$ are zero’s and hence $x_{i+2},x_{i+3},\ldots , x_{i+2r-1}\in \{(0,0,1,1),(0,0,1,0),(0,0,0,1)\}$. Thus $2r-2\le 3$, which implies $2r+1\le 6$. As it is an odd number and $r\ge 2$, we have $2r+1=5$. Therefore $\overline{C_5}\cong C_5$. By Case 1, which is impossible.

The following result in [14] is a consequence of Theorems 9 and 10.

Corollary 5

[14]. If $R=\mathbb {Z}_2^k$, then $\varGamma (R)$ is perfect if and only if $k\le 4$.

Proof

By Theorems 9 and 10, it is enough to prove that $\varGamma (R)$ is perfect if $k\le 3$. In this case we have $|V(\varGamma (R))|\le 6$, and hence $\varGamma (R)$ does not contain a cycle of length 5 as an induced subgraph of $\varGamma (R)$ and, thus the result follows.

It is well-known that any finite non-zero reduced commutative ring R is isomorphic to a finite direct product of finite fields, say $\mathbb {F}_{p_1^{\alpha _1}},\mathbb {F}_{p_2^{\alpha _2}}, \ldots , \mathbb {F}_{p_{\ell }^{\alpha _\ell }}$, where $p_i$’s are prime numbers and $\alpha _i$’s are positive integers, that is $R\cong \mathbb {F}_{p_1^{\alpha _1}}\times \mathbb {F}_{p_2^{\alpha _2}}\times \ldots \times \mathbb {F}_{p_{\ell }^{\alpha _\ell }}$.

By Theorem 2, the compressed zero-divisor graph of a reduced ring R is isomorphic to the zero-divisor graph of $\mathbb {Z}_2^k$, for some $k\ge 1$, that is $\varGamma _E(R)\cong \varGamma (\mathbb {Z}_2^k)$. So, the following result is a consequence of Theorem 9 and Corollary 5.

Theorem 11

If $R\cong \mathbb {F}_{p_1^{\alpha _1}}\times \mathbb {F}_{p_2^{\alpha _2}}\times \ldots \times \mathbb {F}_{p_{\ell }^{\alpha _\ell }}$ is a non-zero reduced ring, where $\mathbb {F}_{p_i^{\alpha _i}}$’s are finite fields, then $\varGamma (R)$ is perfect if and only if $\ell \le 4$.

Proof

The first part is clear from Theorem 9. For the second part, let us assume that $\ell \le 4$. Then $\omega (\varGamma (R))\le 4$. By the above discussion, $\varGamma _E(R)\cong \varGamma (\mathbb {Z}_2^k)$ for some $k\ge 1$. Suppose $k\ge 5$, then $\varGamma (\mathbb {Z}_2^k)$ contains a clique $\langle \{e_i: 1\le i\le k\}\rangle$ of size at least 5 (where $e_i$’s are defined in Corollary 4) and hence $\omega (\varGamma (R))\ge 5$, which is impossible. Thus $k\le 4$ and therefore, by Corollary 5 $\varGamma (\mathbb {Z}_2^k)$ is perfect, and hence $\varGamma (R)$ is perfect by Theorem 8.

The following result in [15] is an immediate consequence of Corollary 5, because every finite Boolean ring R is isomorphic to $\mathbb {Z}_2^k$, for some $k\ge 1$.

Corollary 6

[15]. Let R be a finite Boolean ring. Then the following are equivalent,

• (1) $\varGamma (R)$ is perfect.

• (2) $\varGamma (R)$ does not contain $K_5$ as a subgraph.

• (3) $|R|\le 2^4$.

3.1 Perfect Ideal Based Zero-Divisor Graph of Rings

In this subsection, we characterize rings whose ideal based zero-divisor graphs are perfect. In particular, under what values of n, the ideal based zero divisor graph of the ring $\mathbb {Z}_n$ of integers modulo n is perfect.

The following observation is observed in [16] and [21].

• (i) If I is an ideal of R and $x_1+I, x_2+I,\ldots , x_k+I$ are the distinct co-sets of I, which are zero-divisors of the quotient ring $\frac{R}{I}$, then $\varGamma _I(R)$ is a $\varGamma (\frac{R}{I})$-generalized join of $\langle x_1+I \rangle , \langle x_2+I \rangle , \ldots , \langle x_k+I \rangle$, that is,

  $$\varGamma _I(R)=\varGamma \Big (\frac{R}{I}\Big )\big [\langle x_1+I \rangle , \langle x_2+I \rangle , \ldots , \langle x_k+I \rangle \big ],$$

• (ii) $\langle x_i+I\rangle$ is a complete subgraph of $\varGamma _I(R)$ if and only if $x_i^2\in I$,

• (iii) $\langle x_i+I\rangle$ is a totally disconnected subgraph of $\varGamma _I(R)$ if and only if $x_i^2\notin I$.

Hence, by Theorems 7 and 8, we have

Theorem 12

Let I be an ideal of R, then the following are equivalent,

• (i) $\varGamma _I(R)$ is perfect;

• (ii) $\varGamma (\frac{R}{I})$ is perfect;

• (iii) $\varGamma _E(\frac{R}{I})$ is perfect.

We recall the following result proved in [19].

Theorem 13

[19]. The zero divisor graph $\varGamma (\mathbb {Z}_n)$ of a ring $\mathbb {Z}_n$ is perfect if and only if $n=p^a , p^aq^b, p^aqr$, or pqrs, where p, q, r and s are distinct primes and a and b are positive integers.

It is well known that if I is an ideal of $\mathbb {Z}_n$ generated by m, then $\frac{\mathbb {Z}_n}{I}\cong \mathbb {Z}_m$. So, we have

Corollary 7

If I is an ideal of $\mathbb {Z}_n$ generated by m, then $\varGamma _I(\mathbb {Z}_n)$ is perfect if and only if $m=p^a , p^aq^b, p^aqr$, or pqrs, where p, q, r and s are distinct primes and a and b are positive integers.

Proof

By Theorems 12 and 13, $\varGamma _I(\mathbb {Z}_n)$ is perfect if and only if $\varGamma (\mathbb {Z}_m)$ is perfect if and only if $m=p^a , p^aq^b, p^aqr$, or pqrs.

3.2 Zero-Divisor Graph of Rings, Reduced Semigroups and Posets

In [13], it is shown that the chromatic number is equal to the clique number of zero-divisor graphs of poset, reduced semiring with 0 and reduced semigroup with 0. So it is interesting to consider the following problem.

Problem. Characterize the posets, reduced rings and reduced semigroups whose zero-divisor graphs are perfect.

Now we characterize posets whose zero-divisor graphs are perfect using Theorem 3.

Theorem 14

Let G be a zero-divisor graph of a poset with 0 and $\omega (G)=k$. Then the following are equivalent,

• (i) G is perfect.

• (ii) The reduced graph $G_r$ of G is perfect.

• (iii) The reduced graph $H_r$ of H (where H is in the Definition of generalized complete k-partite graph) is perfect.

Proof

• (i) $\Leftrightarrow \ (ii)$ It follows from Corollary 2.

• (ii) $\Rightarrow \ (iii)$ It follows from the definition of perfect.

• (iii) $\Rightarrow \ (ii)$ Suppose $G_r$ is not perfect graph, then by the Theorem 6, $G_r$ contains an odd cycle of length at least 5 as an induced subgraph or the complement of an odd cycle of length at least 5 as an induced subgraph.

Let $e_1,e_2,\ldots , e_k$ be the atoms of G.

Case 1. $G_r$ contains an odd cycle of length at least 5 as an induced subgraph.

Let $C_{2s+1}$ be an odd cycle of $G_r$ as an induced subgraph with vertex set $V(C_{2s+1})=\{a_0,a_1\ldots , a_{2s}\}$, where $s\ge 2$. Then $V(C_{2s+1})$ is not a subset of $V(H_r)$. As the atoms forms a clique, we have $|V(C_{2s+1})\cap \{e_1,e_2,\ldots , e_k\}|\le 2$. First if $|V(C_{2s+1})\cap \{e_1,e_2,\ldots , e_k\}|= 2$, then there exist $i,j\in \{1,2,\ldots , k\}$ and $\ell \in \{0,1,2,\ldots , 2s\}$ such that $a_{\ell }=e_i$ and $a_{\ell +1}=e_j$. Since $a_{\ell +2}$ and $a_{\ell +3}$ are not adjacent to $a_{\ell }=e_i$, we have $i\notin W(a_{\ell +2})\cup W(a_{\ell +3})$ and hence $W(a_{\ell +2})\cup W(a_{\ell +3})\ne \{1,2,\ldots , k\}$, which is a contradiction to the definition of generalized complete k-partite graph. Next if $|V(C_{2s+1})\cap \{e_1,e_2,\ldots , e_k\}|=1$, then we get a contradiction in a similar way as above. Thus $V(C_{2s+1})\cap \{e_1,e_2,\ldots , e_k\}=\emptyset$ and hence $C_{2s+1}$ is an induced odd cycle of $H_r$, which is a contradiction.

Case 2. $G_r$ contains the complement of an odd cycle of length at least 5 as an induced subgraph.

Let $\overline{C_{2s+1}}$ be the complement of the odd cycle $C_{2s+1}$ in $G_r$ with vertex set $V(\overline{C_{2s+1}})=\{a_0,a_1,\ldots , a_{2s}\}$, where $s\ge 2$. If $V(\overline{C_{2s+1}})\cap \{e_1,e_2,\ldots , e_k\}\ne \emptyset$, then there exists $i\in \{1,2,\ldots , k\}$ such that $e_i=a_{\ell }$, for some $\ell \in \{0,1,2,\ldots , 2s\}$. Then $a_{\ell -1},a_{\ell +1}\notin \{e_1,e_2,\ldots , e_k\}$ and they are not adjacent to $e_i$ and hence $i\notin W(a_{\ell -1})\cup W(a_{\ell +1})$, which is impossible. Thus, $V(\overline{C_{2s+1}})\cap \{e_1,e_2,\ldots , e_k\}= \emptyset$ and therefore $\overline{C_{2s+1}}$ lies in $H_r$, which is a contradiction.

Next, we present equivalent conditions for a zero-divisor graph of a reduced semigroup to be perfect using Theorem 4.

Theorem 15

Let G be a zero-divisor graph of a reduced semigroup with $\omega (G)=k$. Then the following are equivalent,

• (i) G is perfect.

• (ii) The reduced graph $G_r$ of G is perfect.

• (iii) The reduced graph $H_r$ of H (where H is given in the definition of generalized complete k-partite graph) is perfect.

Proof

The proof is similar to that of Theorem 14.

A lattice $L=(L,\wedge , \vee )$ with 0 is distributive if for $x,y,z\in L$, $x\wedge (y\vee z)=(x\wedge y) \vee (x\wedge z)$. As every lattice is a poset, we have the following result proved in [15].

Corollary 8

[15]. Let L be a distributive lattice with 0. Then the following are equivalent,

• (i) $\varGamma (L)$ is perfect.

• (ii) $\varGamma (L)$ contains no induced cycle of length 5.

• (iii) $\omega (\varGamma (L))\le 4$, (equivalently, the number of atoms of $\varGamma (L)$ is at most 4).

Proof

(i) $\Rightarrow$ (ii) It is trivial from the definition of perfect graph.

(ii) $\Rightarrow$ (iii) If $\left\langle \{a_1,a_2,\ldots , a_s\}\right\rangle$ is a clique in $\varGamma (L)$, where $s\ge 5$, then the subgraph induced by $\{a_1\vee a_2, a_3\vee a_4, a_1\vee a_5, a_2\vee a_3,a_4\vee a_5\}$ is an induced cycle of length 5 (as L is distributive) which is a contradiction.

(iii) $\Rightarrow$ (i) Suppose $\varGamma (L)$ is not perfect. Then by Theorem 14, the reduced subgraph $H_r$ of H, defined in Theorem 14, is not perfect. By Theorem 6, $H_r$ contains an odd cycle of length at least 5 as an induced subgraph or its complement of an odd cycle of length at least 5 as an induced subgraph. If $H_r$ contains an induced odd cycle $C_{2s+1}$ with vertex set $V(C_{2s+1})=\{a_1,a_2,\ldots , a_{2s+1}\}$, where $s\ge 2$. Then $a_i\wedge a_{i+1}=0$, for $1\le i\le 2s$, $a_{2s+1}\wedge a_1=0$ and $a_i\wedge a_j\ne 0$, for $j\ne i-1,i,i+1$ and hence the subgraph induced by $\{a_1\wedge a_3, a_1\wedge a_4, a_2\wedge a_4, a_2\wedge a_5, a_3\wedge a_{2s+1}\}$ is a clique in $\varGamma (L)$ of size 5, which is a contradiction. Similarly if $H_r$ contains the complement $\overline{C_{2s+1}}$ of an induced odd cycle $C_{2s+1}$ with vertex set $V(\overline{C_{2s+1}})=\{a_1,a_2,\ldots , a_{2s+1}\}$, where $s\ge 2$, then the subgraph induced by $\{a_1\wedge a_2, a_2\wedge a_3, a_3\wedge a_4, a_4\wedge a_5, a_5\wedge a_1\}$ is a clique in $\varGamma (L)$ of size 5, which is again a contradiction.

As every semiring is a semigroup and by Theorem 15, we have the following result proved in [15].

Corollary 9

[15]. Let R be a reduced semiring with 0. Then the following are equivalent,

• (i) $\varGamma (R)$ is perfect.

• (ii) $\varGamma (R)$ contains no induced cycle of length 5.

• (iii) $\omega (\varGamma (R))\le 4$, (equivalently, the number of atoms of $\varGamma (R)$ is at most 4).

Proof

The proof is similar to that of Corollary 8 by replacing $\vee$ and $\wedge$ by addition and multiplication, respectively.