On zero-divisor graphs of infinite posets

Abstract

It is known that the so-called Beck’s conjecture, i.e. the equality of the finite clique and chromatic numbers of a zero-divisor graph, holds for partially ordered sets Halaš and Jukl (Discrete Math 309(13):4584–4589, 2009). In this paper we present a simple direct proof of this fact. Also, the case when the finiteness assumption of the clique number is omitted is investigated. We have shown that the conjecture fails in general and a bunch of counterexamples is presented.

1 Introduction

Beck (1988) introduced the zero-divisor graph of a commutative ring R, where two elements x, y of R are adjacent if $x\cdot y=0$. He studied the colourings of such defined graphs and conjectured that the chromatic number and the clique number are equal provided both are finite. This conjecture was later on shown to be false by Anderson and Naseer (1993). The concept of zero-divisor graph was further generalized and studied in ordered structures, e.g., in Joshi (2012); Khandekar and Joshi (2020) or Patil et al. (2017). Among others, it was proved that the Beck’s conjecture holds in meet-semilattices with 0 Nimbhorkar et al. (2007) as well as for partially ordered sets Halaš and Jukl (2009) or quasiordered sets Halaš and La̋nger (2010).

Let us remark that graphs connected with posets have been intensively studied in extremal graph theory, see e.g. van Lint and Wilson (2001). For example, the so-called Dilworth’s theorem states that the Beck’s equality holds for incomparability graphs of finite posets. This statement has many important consequences in the area of extremal combinatorics.

The main aim of the paper is to show that the Beck’s conjecture fails for infinite posets in general. We present a class of (separative) posets which contain countable cliques, while their chromatic number can be arbitrarily large. As separative partial orders naturally give rise to complete Boolean algebras, our results also provide a negative answer to Beck’s conjecture for the corresponding Boolean rings.

Let $(Q,\le )$ be a quasiordered set (qoset, for short), i.e., $\le$ is a reflexive and transitive relation on Q. For any qoset there is a natural equivalence relation $\sim$ on Q defined by

$$\begin{aligned} x\sim y \quad \text{ iff } \quad x\le y \; \text{ and } \; y\le x. \end{aligned}$$

Denoting $Q_\sim =\{[x]_\sim ;\, x\in Q\}$ its blocks, then $(Q_\sim ,\le )$ defined by $[x]_\sim \le [y]_\sim$ iff $x\le y$ becomes a partially ordered set associated with $(Q,\le )$.

To any qoset $(Q,\le )$ we assign a graph $\Gamma (Q)=(Q,E)$ as follows: its vertices are elements of Q, and $p,q\in Q$ are connected by an edge iff $L(p,q)=L(Q)$, i.e., $pq\in E$ provided the sets of lower bounds of p, q and Q are the same. These pairs of elements are called disjoint or orthogonal (which is usually denoted by $p\perp q$). Hence, the graph $\Gamma ((Q,\le ))$ is nothing else but the graph $(Q,\perp )$ of the relation $\perp$. We call the graph $\Gamma (Q)$ the zero-divisor graph of $(Q,\le )$.

A set $A\subseteq Q$ is said to be linked if any two distinct elements of A have a common lower bound, i.e., if they form an independent subset of the zero-divisor graph $\Gamma (Q)$. A linked partition of Q is a partition into linked sets. In other words, a linked partition is an admissible vertex colouring of the zero-divisor graph $\Gamma (Q)$, i.e. any two elements connected by an edge have different colours. The elements of the same class are coloured by the same colour, and distinct classes are coloured by different colours. Observe that the problem of finding an admissible colouring of a graph G is equivalent with finding a graph homomorphism from G into the complete graph K. Recall that a mapping $h:V(G)\rightarrow V(H)$ is a graph homomorphism from G to H if it preserves the edges, i.e., $uv\in E(G)$ implies $h(u)h(v)\in E(H)$. Given a (possibly infinite) graph, we write $\chi (G)$ for the chromatic number of G, i.e. the least (cardinal) number needed for an admissible vertex colouring of G, and $\omega (G)$ for its clique number, i.e., the supremum of all cardinalities of cliques of G.

Clearly, if 0 is the least element of $(Q,\le )$, then all the vertices from $[0]_\sim$ are adjacent to all other vertices in $\Gamma (Q)$. Denoting $Q^{+}=Q\setminus [0]_\sim$, the zero-divisor graph $\Gamma (Q^{+})$ is the induced subgraph of the graph $\Gamma (Q)$. If $Q^{\star }\subseteq Q$ is the set of representatives of $Q_\sim$, then $Q^{\star }$ is naturally isomorphic to $Q_\sim$. Hence, orthogonal subsets of both posets correspond to each other and consequently we have $\omega (\Gamma (Q))=\omega (\Gamma (Q^{\star }))$ and $\chi (\Gamma (Q))=\chi (\Gamma (Q^{\star }))$. In conclusion, dealing with posets in fact gives the same results as for qosets in general. Consequently, without lost of generality we may restrict ourselves in what follows to posets without a least element.

Recall that a graph G is compact Lu and Wu (2010) if G contains no isolated vertices and for each pair x, y of non-adjacent vertices of G, there is a vertex z with

$$\begin{aligned} N(x)\cup N(y)\subseteq N(z), \end{aligned}$$

(1)

where N(v) denotes the neighborhood of v, i.e. the set of all vertices adjacent to v. It can be easily seen that the zero-divisor graph $\Gamma (Q)$ of a poset $(Q,\le )$ is compact. Indeed, if x and y are non-adjacent vertices, then there is $z\in Q$ with $z\le x$ and $z\le y$. If $u\in N(x)$, then u and x are disjoint in Q and the same holds for u and z. Thus $u\in N(z)$, which yields $N(x)\subseteq N(z)$. Analogously, $N(y)\subseteq N(z)$, which verifies (1).

It is proved in Lu and Wu (2010) that also the converse is true, i.e. that the compactness fully characterizes the zero-divisor graphs of posets:

Proposition 1

A simple graph G is the zero-divisor graph of a poset iff G is compact.

One of the main results of Halaš and Jukl (2009) states that the zero-divisor graphs of qosets have the same chromatic and clique numbers under the assumption that they are finite. The proof is based on dealing with certain ideals in the respective qoset (poset). In what follows, we will present a short direct proof of this statement.

1.1 Finite case

Let $G=(V,E)$ be a graph with $\omega (G)$ finite. Denote $\mathcal {K}_G$ the system of all cliques of G with the largest number of vertices. Consider the following condition for a graph G:

$$\begin{aligned} \forall x,y\in V;\quad xy\in E \quad \Longrightarrow \quad \bigcup \mathcal {K}_G\subseteq N(x) \cup N(y). \end{aligned}$$

(2)

First we show that for graphs satisfying (2) the chromatic number is the same as the clique number.

Theorem 2

Let $G=(V,E)$ be a graph with $\omega (G)$ finite, which satisfies the condition (2). Then $\chi (G)=\omega (G)$.

Proof

Let $K\subseteq V$ be a clique in G of cardinality $\omega (G)$. We show that there is a graph homomorphism $h:G\rightarrow K$, i.e., that G can be coloured by $\omega (G)$ colours. For $v\in K$ we put $h(v)=v$. Further, if $v\notin K$, then v is not adjacent to some vertex $x\in K$, otherwise $K\cup \{v\}$ would be a larger clique of G. In this case we put $h(v)=x$.

To show that such a defined mapping h is a graph homomorphism, assume that $uv\in E$ for some $u,v\in V$. Since K is a complete subgraph of G, it suffices to show that $h(u)\ne h(v)$. This is obvious provided both vertices u, v belong to K. If $u\in K$ and $v\notin K$, then $h(v)\in K$ is non-adjacent to v, thus $u=h(u)\ne h(v)$. Finally, if $u\notin K$ and $v\notin K$, then h(u) is non-adjacent to u and h(v) is non-adjacent to v. In this case $x=h(u)=h(v)$ is not possible, since $x\notin N(u)\cup N(v)$ violates the condition (2) assumed to be valid for the graph G. $\square$

Theorem 3

Let $(P,\le )$ be a partially ordered set such that $\omega (\Gamma (P))$ is finite. Then the zero-divisor graph $\Gamma (P)$ fulfils the condition (2).

Proof

Let $x,y\in P$ be two elements such that xy forms an edge in $\Gamma (P)$, and let $v\in P$ be an element with $v\in \bigcup \mathcal {K}_G$, i.e., $v\in K$, where K is a clique in $\Gamma (P)$ of cardinality $\omega (\Gamma (P))$.

We show that v is either a neighbour of x or a neighbour of y. Assume that this is not the case, i.e., both xv and yv are not edges. Then x and v, as well as y and v are compatible in P, hence there are $r_x$, $r_y\in P$ such that $r_x\le x$, $r_x\le v$ and $r_y\le y$, $r_y\le v$. Obviously, as x and y are disjoint, $r_x$ and $r_y$ are disjoint, too. Moreover, since K is a clique, the elements $r_x$ and $r_y$ are disjoint with all the members of $K\smallsetminus \{v\}$. Consequently, the set $(K\smallsetminus \{v\})\cup \{r_x,r_y\}$ forms a clique in $\omega (\Gamma (P))$ with a larger number of elements than the clique K, which is a contradiction. $\square$

In conclusion, the last two theorems yield the following statement:

Theorem 4

Let $(P,\le )$ be a partially ordered set such that $\omega (\Gamma (P))$ is finite. Then $\chi (\Gamma (P))=\omega (\Gamma (P))$.

2 Infinite posets

With respect to the previous result, a natural question arises: is the statement true when we omit the finiteness assumption of $\omega (\Gamma (P))$? The aim of this section is to show that the assertion of Theorem 4 is not anymore valid for all infinite posets P when $\omega (\Gamma (P))$ is infinite. The problem we are dealing with can be reformulated as follows:

$$\begin{aligned} \text { Given a poset }P, \text {does }\omega (\Gamma (P))\!=\!\aleph _0 \text { imply} \chi (\Gamma (P))\!=\!\aleph _0? \end{aligned}$$

Note that $\omega (\Gamma (P))=\aleph _0$ means that all subsets of pairwise disjoint elements of P are finite or countable. Such orders are called ccc or satisfying the countable chain condition. Analogously, $\chi (\Gamma (P))=\aleph _0$ means that P can be partitioned into countably many linked sets. Such orders are called $\sigma$-linked. So the above question can be rephrased as:

$$\begin{aligned} \text { Is every }ccc\text { partial order }\sigma \text {-linked?} \end{aligned}$$

To answer this question negatively, first we recall some basic definitions of combinatorial set theory. For details, we refer the reader to standard monographs Erdős et al. (1984) and Jech (2003).

Let S be a non-empty set and $0\ne r\in \mathbb N$ be a positive integer. Denote $[S]^{r}=\{X\subseteq S: |X|=r\}$ the collection of all r-element subsets of S. Let $\{A_i: i\in I\}$ be a partition of $[S]^{r}$ into $\left| I\right|$ classes ($I\ne \emptyset$), i.e. $[S]^{r}=\bigcup _{i\in I} A_i$ and $A_i\cap A_j=\emptyset$ for all $i,j\in I$ such that $i\ne j$. We call a set $H\subseteq S$ homogeneous for this partition of $[S]^{r}$ provided $[H]^{r}\subseteq A_i$ for some $i\in I$, i.e., if all r-element subsets of H belong to the same class $A_i$ of the partition.

We can think of the elements of the different classes as being coloured by different colours. Thus, we have $\left| I\right|$ colours and each r-element subset is coloured by one of them. Homogeneous sets are monochromatic in the sense that all their r-element subsets are coloured by the same colour.

Given infinite cardinal numbers $\kappa$ and $\lambda$, r a positive integer and s (finite or infinite) cardinal, we write $\kappa \rightarrow (\lambda )^{r}_s$ as the shorthand for the statement: for every set with $|S|=\kappa$ and every partition of $[S]^{r}$ into s classes there exists a homogeneous set $H\subseteq S$ with $|H|\ge \lambda$.

The fundamental result about partitions of finite sets is the finite Ramsey’s theorem. There is also an infinite version of Ramsey’s theorem for infinite cardinals, called Erdős–Rado theorem:

Theorem 5

(Erdős–Rado theorem)

If $r\ge 0$ and $\kappa$ is an infinite cardinal, then

$$\begin{aligned} exp_{r}(\kappa )^{+}\,\rightarrow (\kappa ^{+})^{r+1}_\kappa , \end{aligned}$$

where $exp_{0}(\kappa )=\kappa$ and inductively $exp_{r+1}(\kappa )=2^{exp_{r}(\kappa )}$.

Again, Erdős–Rado theorem can be rephrased as follows: if f is a colouring of the $(r+1)$-element subsets of a set of cardinality $exp_{r}(\kappa )^{+}$ by $\kappa$ many colours, then there is a homogeneous set H of cardinality $\kappa ^{+}$, i.e. $(r+1)$-element subsets of H have the same f-value.

In particular, for $r=1$, we obtain the following:

$$\begin{aligned} (2^{\kappa })^{+}\,\rightarrow (\kappa ^{+})^{2}_\kappa . \end{aligned}$$

(3)

Clearly, for any infinite cardinal $\kappa$ and $\lambda >2^{\kappa }$ we have $\lambda \ge (2^{\kappa })^{+}$, and consequently

$$\begin{aligned} \lambda \,\rightarrow (\kappa ^{+})^{2}_\kappa . \end{aligned}$$

In a weaker form, we have $\lambda \,\rightarrow (3)^{2}_\kappa$ as well.

A partially ordered set P is called separative if for all $p,q\in P$ with $p\nleq q$ there is some $r\le p$ disjoint with q.

If P is separative and $p,q\in P$ are distinct elements, then the relation $p\nleq q$ or $q\nleq p$ is valid. Consequently, there is $r\in P$ such that $r\le p$ and r, q are disjoint, or $r\le q$ and r, p are disjoint. In this case we briefly say that r separates the elements p and q.

Lemma 6

Let $p,q,s\in P$ be three distinct elements of a separative poset P, and let $r_{x,y}\in P$ be an element separating x and y, where $\{x,y\}\subseteq \{p,q,s\}$. Then at least two elements $r_{1}, r_{2}\in \{r_{p,q}, r_{p,s}, r_{q,s}\}$ are disjoint.

Proof

Assume that $r_{p,q}\le p$ and $r_{p,q}$ is disjoint with q. If $r_{q,s}$ is not disjoint with $r_{p,q}$ then necessarily $r_{q,s}\le s$. Now $r_{p,s}\le p$ implies that $r_{p,s}$ is disjoint with $r_{q,s}$, since $r_{p,s}$ is disjoint with s and $r_{q,s}\le s$. Similarly, $r_{p,s}\le s$ yields that $r_{p,s}$ and $r_{p,q}$ are disjoint. $\square$

Theorem 7

Let $\kappa$ and $\lambda > 2^{\kappa }$ be infinite cardinals. If P is a separative poset of cardinality $\lambda$, then P cannot be covered by the union of $\kappa$ linked sets, i.e., $\chi (\Gamma (P))>\kappa$.

Proof

Let $\lambda > 2^{\kappa }$ be a cardinal and denote $[\lambda ]^2$ the set of all 2-element subsets of $\lambda$. Applying the weak version of Erdős–Rado theorem, we have $\lambda \,\rightarrow (3)^{2}_\kappa$. Hence for every function $f:[\lambda ]^2\rightarrow \kappa$ there is a 3-element subset on which f is constant, i.e., there are $i,j,k<\lambda$ with $f(\{i,j\})=f(\{j,k\})=f(\{i,k\})$.

Let $P=\{p_i: i<\lambda \}$ be an enumeration of elements of P by ordinals. For a two-element subset $\{i,j\}$, $i,j<\lambda$ distinct ordinals, denote by $r_{i,j}$ an element of P which separates the elements $p_i$ and $p_j$.

Further, suppose that P can be covered by $\kappa$ linked subsets, i.e., $P=\bigcup _{r<\kappa }A_r$, with $A_r\cap A_s=\emptyset$ for $r\ne s$. For $i,j\in \lambda$, $i\ne j$, we define the function $f:[\lambda ]^2 \rightarrow \kappa$ as follows:

$$\begin{aligned} f(\{i,j\})=r \quad \text{ iff } \quad r_{i,j} \in A_r. \end{aligned}$$

Applying Erdős–Rado theorem, we can find some $r<\kappa$ and three distinct i, j, k in $\lambda$ such that all three elements $r_{i,j}$, $r_{j,k}$ and $r_{i,k}$ belong to $A_r$. Due to Lemma 6 we conclude that two of these elements are disjoint in P. This shows that $A_r$ is not linked, which is a contradiction.

$\square$

Corollary 8

Let P be a separative poset satisfying the ccc condition, and let $\left| P\right| >2^{\kappa }$ for some cardinal $\kappa \ge \aleph _0$. Then $\chi (\Gamma (P))>\kappa$ and $\omega (\Gamma (P))\le \aleph _0$.

Proof

Since P satisfies the ccc condition, each clique in $\Gamma (P)$ is at most countable. Hence $\omega (\Gamma (P))\le \aleph _0$, whereas the inequality $\chi (\Gamma (P))>\kappa$ follows from the previous theorem. $\square$

In the sequel we provide examples of separative posets satisfying the ccc condition. Given a set A, we denote $[A]^{<\omega }$ the set of all finite subsets of A. A collection $\mathcal {A}\subseteq [A]^{<\omega }$ is called a $\Delta$-system if there is a finite set $R\in [A]^{<\omega }$ such that $B\cap C=R$ for all distinct $B,C\in \mathcal {A}$. The set R is called the root of the $\Delta$-system. In what follows, we need the following lemma Jech (2003):

Lemma 9

($\Delta$-system Lemma)

Let A be an uncountable set and let $\mathcal {A}\subseteq [A]^{<\omega }$ be an uncountable collection of finite subsets of A. Then there is an uncountable subsystem $\mathcal {B}\subseteq \mathcal A$ that is a $\Delta$-system.

For arbitrary sets I, J, let $P(I,J)=\{p\subseteq I\times J: \left| p\right| <\aleph _0, p \text{ is } \text{ a } \text{ function }\}$ be the set of all finite partial functions from I to J. Further, define a partial order $\le$ on P(I, J) by $p\le q$ iff $q\subseteq p$, i.e., p is smaller than q if p extends q. In this case $\emptyset$ is the greatest element, and two partial functions p, q are compatible if and only if $p\upharpoonright S= q\upharpoonright S$ where $S=\textrm{dom}(p)\cap \textrm{dom}(q)$, i.e., if the functions p and q agree on the intersection of their domains S. In this case $p\cup q$ is a function belonging to P(I, J) and representing the infimum of p and q.

In what follows we recall the well-known fact, together with its proof, that P(I, J) is a separative poset, which fulfils the ccc condition, provided J is countable, cf. Koppelberg (1989) or Kunen (2011).

Lemma 10

If J is a countable set with at least two elements, then P(I, J) is separative and it satisfies the ccc condition.

Proof

First, assume that $p\nleq q$, i.e., the partial function p does not extend q. If $p(i)\ne q(i)$ for some $i\in \textrm{dom}(p)$, then $r=p$ is disjoint with q. Otherwise, there is $i\in \textrm{dom}(q)\smallsetminus \textrm{dom}(p)$, and $r= p\cup \{(i,j)\}$ is disjoint with q, where $j\ne q(i)$ is an element from J. Hence, P(I, J) is separative.

Further, let $\mathcal {F}\subseteq P(I,J)$ be an uncountable family. Put $\mathcal {S}=\{\textrm{dom}(p): p\in \mathcal {F}\}$. Since $\mathcal {F}$ is uncountable, the same holds for $\mathcal {S}$ (otherwise $\mathcal {F}$ would be a countable union of countable sets). According to Lemma 9, there is an uncountable $\Delta$-system $\mathcal {B}\subseteq \mathcal {S}$ with a finite root R, i.e., $R=A\cap B$ for all $A,B\in \mathcal {B}$. Let $\mathcal {G}=\{p\in \mathcal {F}: \textrm{dom}(p)\in \mathcal {B}\}$ be the subfamily of $\mathcal {F}$. Again, as $J^R$ is countable, there is an uncountable subfamily of $p\in \mathcal {G}$ with common $p\upharpoonright R$. If p, q are two such functions, i.e., $\textrm{dom}(p)\cap \textrm{dom}(q)=R$ and $p\upharpoonright R=q\upharpoonright R$, then they are compatible, hence $\mathcal {F}$ is not pairwise disjoint. $\square$

Obviously, if J contains at least two elements and I is infinite, then P(I, J) contains a countable pairwise disjoint subset. For example we may assume that $\omega \subseteq I$ and $0,1\in J$. In this case the family $\{g_n: n>0, n\in \omega \}$ is pairwise disjoint, where $g_n:n\rightarrow J$, $n>0$, is given by $g_n(n-1)=1$ and $g_n(m)=0$ for all $m<n-1$.

Hence, applying the previous results, we obtain that $\chi (\Gamma (P))>\kappa$ and $\omega (\Gamma (P))=\aleph _0$ for $P=P(\lambda ,J)$, where $\lambda >2^{\kappa }$ is an infinite cardinal and $1<\left| J\right| \le \aleph _0$.

3 Infinite Boolean algebras and rings

Let $B=(B,\vee ,\wedge ,{}^{\prime },0,1)$ be a Boolean algebra. Recall that a subset S of $B^{+}=B\smallsetminus \{0\}$ is dense in B, if for all $x\in B^+$ there is some $y\in S$ such that $y\le x$.

It can be easily seen that if S is a dense subset of a Boolean algebra B, then S is separative. Indeed, if $p\nleq q$ then $q^{\prime }\wedge p>0$ and there is $r\in S$ such that $r\le q^{\prime }\wedge p$. Obviously, r and q are disjoint and $r\le p$ holds.

On the other hand, every separative partially ordered set P determines a complete Boolean algebra B(P), cf. Jech (2003) or Koppelberg (1989). The method of completion uses the notion of a regular cut. For $p\in P$ denote by $U_p=\{x\in P: x\le p\}$ the set of all lower bounds of the element p.

A subset $U\subseteq P$ is a regular cut provided:

• 1) $p\le u$ and $u\in U$ implies $p\in U$ (U is hereditary)

• 2) whenever $p\notin U$, then there is $q\le p$ with $U_q \cap U=\emptyset$.

The family $B(P)=\{U\subseteq P: U \text{ regular } \text{ cut }\}$ ordered by the set inclusion forms a complete Boolean algebra. As P is separative, it follows that each principal ideal $U_p$ for all $p\in P$ is a regular cut. Hence, the correspondence $p\mapsto U_p$ is an order embedding of P into B(P). Moreover, with respect to this embedding, P is dense in B(P) and the algebra B(P) is unique up to isomorphism with this property.

When P is equipped with the partial order topology (a subset is open if it is hereditary), then B(P) corresponds to the regular open algebra

$$\begin{aligned} RO(P)=\{X\subseteq P: X=\textrm{int}\;\textrm{cl}(X)\} \end{aligned}$$

consisting of all regular open subsets of P.

If $S\subseteq B$ is a dense subset of a Boolean algebra B, then the condition ccc is valid in B if and only if it is valid in S. If $X\subseteq B$ is a pairwise disjoint family, then for each $x\in X$ there is $y_x\in S$ such that $y_x\le x$. Obviously, $x\ne z$ implies $y_x\ne y_z$ since x and z are disjoint, thus X and $\{y_x: x\in X\}$ form pairwise disjoint families of the same cardinality.

It is well-known that there is a 1-1 correspondence between Boolean algebras and commutative idempotent rings with a unit, the so-called Boolean rings. The correspondence is as follows: If $B=(B,\vee ,\wedge ,{}^{\prime },0,1)$ is a Boolean algebra, then $\textrm{R}(B)=(B,\Delta , \wedge , 0,1)$ is a Boolean ring, where the operation $\Delta$ of symmetric difference is given by $x \Delta y = (x\wedge y^{\prime }) \vee (x^{\prime } \wedge y)$. Conversely, if $R=(B,+, \cdot , 0,1)$ is a Boolean ring, then $\textrm{B}(R)=(B, \hat{\vee }, \hat{\wedge }, {}^{\prime }, 0,1)$ is a Boolean algebra, where $x\hat{\vee } y= x+y + (x\cdot y)$, $x\hat{\wedge } y= x\cdot y$ and $x^{\prime }=1 + x$. The operations of meet and ring product are the same.

Hence, a Boolean algebra violating the equality of the chromatic and clique numbers also violates this equality in the ring-theoretic version of the zero-divisor graphs. Summarizing the considerations in this section, we obtain the following proposition.

Theorem 11

Let $\kappa \ge \aleph _0$ be a cardinal and B be a Boolean algebra satisfying the ccc condition, where $\left| B\right| >2^{\kappa }$. Then $\chi (\Gamma (B))>\kappa$ and $\omega (\Gamma (B))= \aleph _0$.

Remark 1

Let $\kappa \ge \aleph _0$ and $\lambda >2^{\kappa }$ be cardinal numbers. Then the completion B(P) of the poset $P=P(\lambda , I)$, I countable, is an example of a Boolean algebra satisfying the assumptions of the previous theorem.

Another example represents the free Boolean algebra $F_{\textbf{BA}}(\lambda )$ with $\lambda$ generators. Obviously, the cardinality of this algebra equals to $\lambda$. It is the well-known fact that every free Boolean algebra satisfies the ccc condition. This follows from the fact that if $\{b_i:i<\lambda \}$ is a set of its free generators, then the set of elements of the form $\bigwedge _{i\in I} b_i \wedge \bigwedge _{j\in J} b_j^{\prime }$, where I and J are disjoint finite subsets of $\lambda$, is dense in $F_{\textbf{BA}}(\lambda )$. This dense subset is isomorphic to $P=P(\lambda ,2)$, thus $F_{\textbf{BA}}(\lambda )$ satisfies the ccc condition. Note, that while B(P) is complete, $F_{\textbf{BA}}(\lambda )$ is not complete.

4 Conclusion

It was known that the so-called Beck’s conjecture, i.e. the equality of the finite clique and chromatic numbers of a zero-divisor graph, holds for partially ordered sets. In this paper, we have presented a simple direct proof of this fact. Also, we have dealt with the case when the finiteness assumption of the clique number is omitted. It has been shown that the conjecture fails in general. A bunch of counterexamples is presented by sufficiently large separable partial orders satisfying the so-called countable chain condition (ccc). We have shown that in this case all cliques are countable, whereas the chromatic number is larger than an arbitrary given cardinal. Since separative partial orders naturally give rise to complete Boolean algebras, we have obtained the counterexamples to the ring-theoretic version of this problem as well.