CORRECTION TO ‘ON THE COMPLEMENT OF THE ZERO-DIVISOR GRAPH OF A PARTIALLY ORDERED SET’

The purpose of this note is to communicate an error in [2] and correct the results that are affected. The error lies in the proof of [2, Lemma 3.8], where it was not taken into account that edges can appear between vertices of $G^c(A)$ when passing to $G^c(A\cup \{a\})$ . Consequently, it can happen that $F(x)=F(y)$ for $x,y\in V(G^c(A))$ with $\{x,y\}^\wedge \neq \{0\}$ , and there are three results, that is, [2, Theorem 1.1, Lemma 3.8 and Corollary 4.3], that require modification. In this note, we correct these results by introducing a condition on the set $\mathscr {A}$ of atoms of P (although, Lemma 3.8 is not addressed since it can be omitted under the new assumptions). By using the correction to [2, Theorem 1.1], we confirm that [2, Corollaries 4.1 and 4.2] do not require revision.

The following definitions will be used and all other notation is the same as in [2]. Let P be a partially ordered set with zero and $Z(P)\neq \{0\}$ . For $x\in P$ , we define $\textbf {a}(x)=\{a\in \mathscr {A} \mid a\leq x\}$ and let $\mathcal {C}_x=\{y\in P \mid \textbf {a}(y)=\mathscr {A}\setminus \textbf {a}(x)\}$ . Let $[x]=\{y\in P \mid \textbf {a}(x)=\textbf {a}(y)\}$ . If $\{\textbf {a}(x) \mid x\in P\}\cup \{\mathscr {A}\}$ is assumed to be a Boolean algebra (under inclusion), then $\mathcal {C}_x\neq \emptyset $ for every $x\in V(G^c(P))$ and the sets $[x]$ ( $x\in V(G^c(P))$ ) and $\mathscr {A}$ are finite if $\omega (G^c(P))<\infty $ . (For example, if $|\mathscr {A}|\geq 3$ , then $\bigcup _{a\in \mathscr {A}}\mathcal {C}_a$ induces a clique.) These facts will be freely applied throughout.

The following result serves as a correction to the statement of [2, Theorem 1.1].

Theorem 1. Let P be a partially ordered set with $0$ such that $Z(P)\neq \{0\}$ and $\omega (G^c(P))<\infty $ . Then $G(P)$ is weakly perfect. Furthermore, if $\{\mathbf{a}(x) | x\in P\}\cup \{\mathscr {A}\}$ is a Boolean algebra such that, for all $x,y\in V(G^c(P))$ , the inequality $|[x]|\leq |[y]|$ holds whenever $\mathbf{a}(x)\subseteq \mathbf{a}(y)$ , then $G^c(P)$ is also weakly perfect.

Proof. The ‘ $G(P)$ is weakly perfect’ part of [2, Theorem 1.1] does not require revision, so it is sufficient to prove the last statement. Let K be a clique of maximum cardinality in $G^c(P)$ . If $x\in V(G^c(P))\setminus V(K)$ , then there exists ${k\in V(K)}$ with ${\{x,k\}^\wedge =\{0\}}$ , which implies $\textbf {a}(k)\subseteq \textbf {a}(p)$ for every $p\in \mathcal {C}_x$ . Thus, if $p\in \mathcal {C}_x$ , then ${\{0\}\neq \textbf {a}(k)\cap \textbf {a}(k')\subseteq \{p,k'\}^\wedge }$ for every $k'\in V(K)$ . This shows that if ${x\in V(G^c(P))\setminus V(K)}$ , then $\mathcal {C}_x\subseteq V(K)$ .

Continue to assume $x\in V(G^c(P))\setminus V(K)$ and suppose that $\textbf {a}(x)$ is maximal in ${\{\textbf {a}(y) \mid y\in V(G^c(P))\setminus V(K)\}}$ . Then $|[x]|\leq |\mathcal {C}_x|$ since, otherwise, $(V(K)\setminus \mathcal {C}_x)\cup [x]$ induces a ‘larger’ clique. Indeed, if ${|[x]|>|\mathcal {C}_x|}$ , then $|(V(K)\setminus \mathcal {C}_x)\cup [x]|=|V(K)|-|\mathcal {C}_x|+|[x]|>|V(K)|$ . Also, it is a clique since if $k\in V(K)$ and $\alpha \in [x]$ with $\{k,\alpha \}^\wedge =\{0\}$ , then $k\in \mathcal {C}_x$ (clearly $\textbf {a}(k)\subseteq \mathscr {A}\setminus \textbf {a}(x)$ , and if the inclusion is proper, then $\textbf {a}(x)\subsetneq \textbf {a}(y)$ whenever $\textbf {a}(y)=\mathscr {A}\setminus \textbf {a}(k)$ , contradicting the maximality of $\textbf {a}(x)$ ). It will be shown that ${|[x]|\leq |\mathcal {C}_x|}$ for every $x\in V(G^c(P))\setminus V(K)$ , and hence a colouring ${F:V(G^c(P))\rightarrow V(K)}$ is obtained by setting $F|_{V(K)}=\iota $ and, for every $x\in V(G^c(P))\setminus V(K)$ , letting $F|_{[x]}:[x]\rightarrow \mathcal {C}_x$ be any injection.

Let $x\in V(G^c(P))\setminus V(K)$ . Since $|\mathscr {A}|<\infty $ , there exists $z\in V(G^c(P))\setminus V(K)$ such that $\textbf {a}(z)$ is maximal in $\{\textbf {a}(y) \mid y\in V(G^c(P))\setminus V(K)\}$ and $\textbf {a}(x)\subseteq \textbf {a}(z)$ . Let $x'\in \mathcal {C}_x$ and $z'\in \mathcal {C}_z$ . Thus, $\textbf {a}(z')\subseteq \textbf {a}(x')$ . The inequalities $|[x]|\leq |[z]|$ and $|[z']|\leq |[x']|$ hold by hypothesis, and $|[z]|\leq |\mathcal {C}_z|=|[z']|$ by the choice of z. Hence, $|[x]|\leq |[x']|=|\mathcal {C}_x|$ .

The following arguments show that [2, Corollaries 4.1 and 4.2] remain valid and that [2, Corollary 4.3] holds for ‘distributive modules’.

Let R be a reduced commutative ring. Consider the partial order on R such that $r\leq s$ if and only if either $\text {ann}(s)\subsetneq \text {ann}(r)$ , or r is less than or equal to s in some fixed well-order on $\text {ann}(s)=\text {ann}(r)$ . It can be shown that $\Gamma (R)$ is equal to the zero-divisor graph $G(R)$ of $(R,\leq )$ [3, Remark 3.4]. One can check that if $r\in R$ and a is a (poset-theoretic) atom of R, then $a\leq r$ if and only if $ar\neq 0$ . It follows that if $a_1,\ldots ,a_k$ are atoms, then $\textbf {a}(a_1+\cdots +a_k)=\{a_1,\ldots ,a_k\}$ (for example, if $a_1\not \leq a_1+\cdots +a_k$ , then $a_1^2=a_1(a_1+\cdots +a_k)=0$ , which is a contradiction). Hence, if $\omega (G^c(R))<\infty $ , then $\{\textbf {a}(r) \mid r\in R\}$ is the Boolean algebra of all subsets of $\mathscr {A}$ . Similarly, in the lattice of ideals of R, if $\omega (\mathbb {AG}^c(R))<\infty $ , then $\{\textbf {a}(I) \mid I$ is an ideal of $R\}$ is a Boolean algebra since if $I_1,\ldots ,I_k$ are minimal ideals, then $\textbf {a}(I_1+\cdots +I_k)=\{I_1,\ldots ,I_k\}$ . To see this, if $I,I_1,\ldots ,I_k$ are minimal ideals and $I\not \in \{I_1,\ldots ,I_k\}$ , then $I\cap I_j=\{0\}$ implies $II_j=\{0\}$ , so $I(I_1+\cdots +I_k)=II_1+\cdots +II_k=\{0\}$ , and thus $I\cap (I_1+\cdots +I_k)=\{0\}$ (alternatively, the claim is easily checked by noting that the condition ‘ $\omega (\mathbb {AG}^c(R))<\infty $ ’ implies that R is a reduced Artinian ring [1, Theorem 1.1], that is, R is a finite direct product of fields).

Now, [2, Corollary 4.2] remains valid since, as noted above, the condition ‘ $\omega (\mathbb {AG}^c(R))<\infty $ ’ guarantees that R is reduced and Artinian, which implies every ideal is a sum of minimal ideals, and hence $|[I]|=1$ for every ideal I of R. To see that [2, Corollary 4.1] remains valid, note that if $\textbf {a}(x)=\{a_1,\ldots ,a_k\}$ and $\textbf {a}(y)=\{a_1,\ldots ,a_k,a_{k+1}\}$ , then $[x]\rightarrow [y]$ by $\alpha \mapsto \alpha +a_{k+1}$ is injective. (The property ‘ $a\leq r$ if and only if $ar\neq 0$ ’ mentioned above can be used to verify $\alpha +a_{k+1}\in [y]$ .)

However, [2, Corollary 4.3] needs revision. For this, observe that the argument in [2] given prior to Corollary 4.3 shows that if $(F\cup \{\emptyset \},\cap )$ is a meet-semilattice with $\omega (\mathbb {I}(F))<\infty $ , then $\mathbb {I}(F)$ is weakly perfect if $G^c(F\cup \{\emptyset \})$ is weakly perfect. Hence, to show that the intersection graph of an R-module M is weakly perfect, it is enough to show that the subgraph of $IG(M)$ induced by the nonessential submodules of M is weakly perfect.

An R-module M is called distributive if $N_1\cap (N_2+N_3)=(N_1\cap N_2)+(N_1\cap N_3)$ for all submodules $N_1,N_2,N_3\leq M$ (for example, this is the case for every cyclic group). If M is distributive with $\omega (IG(M))<\infty $ , then clearly $\{\textbf {a}(N) \mid N$ is a submodule of $M\}$ is the Boolean algebra of all subsets of minimal submodules (for example, similar to the argument given above for ideals, if $N,N_1,\ldots ,N_k$ are minimal submodules and $N\not \in \{N_1,\ldots ,N_k\}$ , then $N\cap (N_1+\cdots +N_k)=N\cap N_1+\cdots +N\cap N_k=\{0\}$ ). Also, if S and T are nonessential submodules of M with $\textbf {a}(S)=\{N_1,\ldots ,N_k\}$ and ${\textbf {a}(T)=\{N_1,\ldots ,N_k,N_{k+1}\}}$ , then $|[S]|\leq |[T]|$ since $[S]\rightarrow [T]$ by $N\mapsto N+N_{k+1}$ is injective; for example, if $N+N_{k+1}=N'+N_{k+1}$ , then

$$ \begin{align*}N=N\cap(N'+N_{k+1})=N\cap N'+N\cap N_{k+1}=N\cap N',\end{align*} $$

that is, $N\subseteq N'$ , and the reverse inclusion holds symmetrically.

By Theorem 1, this verifies the following correction to [2, Corollary 4.3].

Corollary 2. Let F be a collection of nonempty subsets of a set S such that $F\cup \{\emptyset \}$ is closed under intersection. If $\omega (\mathbb {I}(F))<\infty $ , then $\mathbb {I}^c(F)$ is weakly perfect. If, furthermore, $\{\mathbf{a}(x) \mid x\in F\}\cup \{\emptyset ,\mathscr {A}\}$ is a Boolean algebra such that for all $x,y\in V(G^c(F\cup \{\emptyset \}))$ , the inequality $|[x]|\leq |[y]|$ holds whenever $\mathbf{a}(x)\subseteq \mathbf{a}(y)$ , then $\mathbb {I}(F)$ is also weakly perfect. In particular, if R is a (not necessarily commutative) ring and M is an R-module such that $\omega (IG(M))<\infty $ , then $IG^c(M)$ is weakly perfect. In this case, if M is a distributive module, then $IG(M)$ is also weakly perfect.