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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240308171106883-0082:S0004972723000849:S0004972723000849_eqnu1.png?pub-status=live)
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.