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

Published online by Cambridge University Press:  18 September 2023

SARIKA DEVHARE
Affiliation:
Department of Mathematics, Savitribai Phule Pune University, Pune 411007, Maharashtra, India e-mail: [email protected]
VINAYAK JOSHI*
Affiliation:
Department of Mathematics, Savitribai Phule Pune University, Pune 411007, Maharashtra, India e-mail: [email protected]
JOHN LAGRANGE
Affiliation:
School of Mathematics and Sciences, Lindsey Wilson College, Columbia, KY 42728, USA e-mail: [email protected]
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Abstract

Type
Correction
Copyright
© The Author(s), 2023. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

The purpose of this note is to communicate an error in [Reference Devhare, Joshi and LaGrange2] and correct the results that are affected. The error lies in the proof of [Reference Devhare, Joshi and LaGrange2, 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, [Reference Devhare, Joshi and LaGrange2, 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 [Reference Devhare, Joshi and LaGrange2, Theorem 1.1], we confirm that [Reference Devhare, Joshi and LaGrange2, 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 [Reference Devhare, Joshi and LaGrange2]. 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 [Reference Devhare, Joshi and LaGrange2, 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 [Reference Devhare, Joshi and LaGrange2, 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 [Reference Devhare, Joshi and LaGrange2, Corollaries 4.1 and 4.2] remain valid and that [Reference Devhare, Joshi and LaGrange2, 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 $(R,\leq )$ [Reference LaGrange and Roy3, 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 [Reference Behboodi and Rakeei1, Theorem 1.1], that is, R is a finite direct product of fields).

Now, [Reference Devhare, Joshi and LaGrange2, 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 [Reference Devhare, Joshi and LaGrange2, 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, [Reference Devhare, Joshi and LaGrange2, Corollary 4.3] needs revision. For this, observe that the argument in [Reference Devhare, Joshi and LaGrange2] 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 [Reference Devhare, Joshi and LaGrange2, 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.

Acknowledgement

The authors wish to express their gratitude to Professor Peter Cameron for bringing the error in [Reference Devhare, Joshi and LaGrange2] to their attention.

References

Behboodi, M. and Rakeei, Z., ‘The annihilating-ideal graph of commutative rings I’, J. Algebra Appl. 10(4) (2011), 727739.CrossRefGoogle Scholar
Devhare, S., Joshi, V. and LaGrange, J., ‘On the complement of the zero-divisor graph of a partially ordered set’, Bull. Aust. Math. Soc. 97(2) (2018), 185193.CrossRefGoogle Scholar
LaGrange, J. D. and Roy, K. A., ‘Poset graphs and the lattice of graph annihilators’, Discrete Math. 313(10) (2013), 10531062.CrossRefGoogle Scholar