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\begin{document}
\thispagestyle{empty}
\begin{center}
\null\vspace{-1cm}
{\footnotesize Available at:
{\tt http://publications.ictp.it}}\hfill IC/2010/074\\
\vspace{1cm}
United Nations Educational, Scientific and Cultural Organization\\
and\\
International Atomic Energy Agency\\
\medskip
THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS\\
\vspace{2.5cm}
{\bf SOME RESULTS ON THE INTERSECTION GRAPHS\\
OF IDEALS OF RINGS}\\
\vspace{2cm}
S. Akbari\footnote{s$_-$akbari@sharif.edu}\\
{\it Department of Mathematical Sciences, Sharif University of Technology,\\
Tehran, Iran\\
and\\
The Abdus Salam International Centre for Theoretical Physics,
Trieste, Italy,}\\[1.5em]
R. Nikandish\footnote{r$_-$nikandish@sina.kntu.ac.ir} and
M.J. Nikmehr\footnote{nikmehr@kntu.ac.ir}\\
{\it Department of Mathematics, K.N. Toosi University of Technology, Tehran, Iran.}
\end{center}
\vspace{0.5cm}
\centerline{\bf Abstract}
\baselineskip=18pt
\bigskip
Let $R$ be a ring with unity and $I(R)^*$ be
the set of all non-trivial left ideals of $R$. The intersection
graph of ideals of $R$, denoted by $G(R)$, is a graph with the
vertex set $I(R)^*$ and two distinct vertices $I$ and $J$ are
adjacent if and only if $I\cap J\neq 0$. In this paper, we study
some connections between the graph-theoretic properties of this
graph and some algebraic properties of rings. We characterize all
rings whose intersection graphs of ideals are not connected. Also we
determine all rings whose clique number of the intersection graphs of
ideals are finite. Among other results, it is shown
that for every ring, if the clique number of $G(R)$ is finite,
then the chromatic number is finite too and if $R$ is a reduced ring both are equal.
\vfill
\begin{center}
MIRAMARE -- TRIESTE\\
August 2010\\
\end{center}
\vfill
%{\it Key Words}: Intersection graph of ideals of a ring, Clique
%number, Chromatic number.}
%2010{ \it Mathematics Subject Classification}: 05C15, 05C69, 16A40, 16A66.
\newpage
\noindent{\bf\large 1. Introduction}\\
{\noindent The interplay between ring-theoretic and
graph-theoretic properties was first studied in \cite{beck}, and
this approach has since became
increasingly popular. Many researchers have obtained ring-theoretic
properties in terms of graph-theoretic properties by suitable defining
graph structure on some elements of a ring, for example, the zero-divisor
graph and the total graph \cite{akbari}, \cite{akb}, \cite{ak}, \cite{anderson-badawi}, \cite{andersson}
and \cite{wu}.
\noindent The field of graph theory and ring theory both benefit
from the study of algebraic concepts using graph theoretic
concepts. For instance, knowledge of algebraic structures of rings
can innovate new ideas for studying the graphs. Usually after
translating of algebraic properties of rings into graph-theoretic
language, some problems in ring theory might be more easily
solved. When one assigns a graph to
an algebraic structure numerous interesting algebraic problems
arise from the translation of some graph-theoretic parameters
such as clique number, chromatic number, independence number and
so on. The main purpose of this paper is the study of the
intersection of ideals in a ring using graph-theoretic concepts.
\noindent Throughout this paper all rings have unity. Let $R$ be a ring. By
$I(R)$ and $I(R)^*$ we mean the set of all left ideals of $R$ and
the set of all non-trivial left ideals of $R$, respectively. A ring $R$ is said to
be \textit{local} if it has a unique maximal left ideal.
The ring of $n\times n$ matrices over $R$ is denoted by $M_{n}(R)$.
The ring $R$ is said to be \textit{reduced} if it has no non-zero
nilpotent element. The \textit{socle} of ring $R$, denoted by
$soc(R)$, is the sum of all minimal left ideals of $R$. If there are no
minimal ideals, this sum is defined to be zero. A prime ideal
$\mathfrak{p}$ is said to be an \textit{associated prime ideal}
of a commutative Noetherian ring $R$, if there exists a non-zero
element $x$ in $R$ such that $\mathfrak{p}=Ann(x)$. By ${\rm
Ass}(R)$ and ${\rm Min}(R)$ we denote the set of all associated
prime and minimal prime ideals of $R$, respectively. The set of
nilpotent elements of $R$ is denoted by ${\rm Nil}(R)$. The
intersection of all maximal left ideas of $R$ is called the
\textit{Jacobson radical} of $R$ and is denoted by $J(R)$.
A ring $R$ is said to be \textit{semisimple}, if $J(R)=0$. Let $M$
be a left $R$-module. A chain of left submodules of {\it length $n$} is a sequence
$M_i$ $(0\leq i\leq n)$ of left submodules of $M$ such that
$0=M_0\subset M_1\subset \cdots\subset M_n=M$. A \textit{composition series} of $M$ is a
maximal chain, that is one in which no extra left submodules can
be inserted. It is known that every pair of composition series
for $M$ are equivalent. The length of composition series of $M$
is denoted by $l(M)$. An $R$-module $M$ is said to be
\textit{finite length} if $l(M)<\infty$.
\noindent Let $G$ be a graph with the vertex set $V(G)$. The
\textit{Complement graph} of $G$, denoted by $\overline{G}$, is a
graph with the same vertices such that two vertices of
$\overline{G}$ are adjacent if and only if they are not adjacent
in $G$. The \textit{degree} of a vertex $v$ in a graph $G$ is the
number of edges incident with $v$. The degree of a vertex $v$ is
denoted by $deg(v)$. Let $r$ be a non-negative integer. The
graph $G$ is said to be \textit{$r$-regular}, if the degree of
each vertex is $r$. If $u$ and $v$ are two adjacent vertices of
$G$, then we write $u-\hspace{-.2cm}-v$. The \textit{complete
graph} of order $n$, denoted by $K_n$, is a graph with $n$
vertices in which any two distinct vertices are adjacent. A {\it star graph} is a graph with a vertex adjacent to all other
vertices and has no other edges. Recall that a graph $G$ is
\textit{connected} if there is a path between every two distinct
vertices. For distinct vertices $x$ and $y$ of $G$, let $d(x,y)$
be the length of the shortest path from $x$ to $y$ and if there
is no such a path we define $d(x,y)=\infty$. The
\textit{diameter} of $G$, $diam(G)$, is the supremum of the set $\{d(x,y)
: x~{\rm and}~y~{\rm are~distinct~vertices~of}~G\}$. A \textit{clique} of $G$ is a
complete subgraph of $G$ and the number of vertices in the
largest clique of $G$, denoted by $\omega(G)$, is called the
\textit{clique number} of $G$. For a graph $G$, let $\chi(G)$
denote the \textit{chromatic number} of $G$, i.e., the minimum
number of colors which can be assigned to the vertices of $G$ in
such a way that every two adjacent vertices have different
colors. An \textit{independent set} of $G$ is a subset of the
vertices of $G$ such that no two vertices in the subset represent
an edge of $G$. \textit{The independence number} of $G$, denoted
by $\alpha(G)$, is the cardinality of the largest independent set.
\noindent The \textit{intersection graph of ideals} of a ring $R$,
denoted by $G(R)$, is a graph with the vertex set $I(R)^*$ and two
distinct vertices $I$ and $J$ are adjacent if and only if $I\cap
J \neq 0$. This graph was first defined in \cite{chakrabarty} and
the intersection graph of ideals of $\mathbb{Z}_n$ was studied.
They determined the values of $n$ for which the graph of
$\mathbb{Z}_n$ is complete, Eulerian or Hamiltonian. Since the
most properties of a ring are closely tied to the behavior of its
ideals, one may expect that the intersection graph of ideals
reflect many properties of a ring.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{9mm} \noindent{\bf\large 2. Diameter and Some Finiteness Conditions }\vspace{5mm}
\noindent
\noindent In this section, all rings whose the intersection
graphs are not connected will be characterized. We prove that if
$G(R)$ is a connected graph, then its diameter is at most $2$.
Next, some conditions under which the intersection graph of
ideals of a ring is finite are given. Furthermore, the regularity of the
intersection graph of ideals of a ring is studied.
First we need the following results.
\begin{thm}\label{1111} {\rm \cite [Corollary 2.5]{chakrabarty}}
For any graph $G=G(R)$ of a ring $R$, whenever $G$ is not
connected, it is a null graph (i.e., it has no edge).
\end{thm}
\begin{lem}\label{ja} {\rm \cite[p.232]{jac}}
Let $D$ be a division ring and $n$ be a natural number. If $H_r$, $0\leq r\leq n$, denotes the left ideal of $M_n(D)$ containing all matrices whose
$d_{ij}=0$, for every $r1$, then $M_{n_1}(D_1)$ has at
least a non-zero left ideal, say $I$. Then $(0,M_{n_2}(D_2))$ and
$(I,M_{n_2}(D_2))$ are adjacent. Now, Theorem \ref{1111}, implies that $G(R)$ is a
connected graph, a contradiction. Thus $n_1=1$. Similarly, $n_2=1$. Now, assume
that $R\cong M_{n}(D)$, where $D$ is a division ring. By Lemma \ref{ja}, the dimension of every
maximal left ideal of $M_n(D)$ over $D$ is $n^2-n$. So if $n>2$,
then every two maximal left ideals of $M_n(D)$ has non-zero
intersection, a contradiction. Thus $R\cong M_2(D)$, where $D$ is
a division ring. Note that
if $R\cong D_{1}\times D_{2}$, where $D_{1}$ and $D_{2}$ are
two division rings, then it is clear that $G(R)$ is not connected.
Finally, assume that $R\cong M_2(D)$.
By Lemma \ref{ja}, the dimension of every non-trivial left ideal
of $M_2(D)$ over $D$ is $2$. So the intersection of every two
distinct non-trivial left ideals of $M_2(D)$ is zero. Thus $G(R)$
is not a connected graph.}
\end{proof}
\begin{thm}\label{1} Let $R$ be a ring and $G(R)$ be a
connected graph. Then diam$ (G(R))\leq 2$.
\end{thm}
\begin{proof}{ Let $I$ and $J$ be two left ideals of $R$. If $I\cap J$ is non-zero, then $I$ and $J$ are adjacent.
Thus assume that $I\cap J=0$. If $I+J\neq R$, then consider the
path $I-\hspace{-.2cm}-(I+J)-\hspace{-.2cm}-J$. Thus suppose that
$I+J=R$. If there exists a left ideal $0\neq L\subset I$, and
$L+J\neq R$, then consider the path
$I-\hspace{-.2cm}-(L+J)-\hspace{-.2cm}-J$. Thus assume that
$L+J=R$. Let $x\in I$. Thus there exists $a\in L$ and $b\in J$
such that $x=a+b$. We have $x-a=b\in I\cap J$. Thus $x=a\in L$.
This implies that $L=I$. Thus $I$ is a minimal left ideal. Now, since $G(R)$ is connected, there
exists a non-trivial left ideal $I_1$ such that $I_1\neq I$ and
$I_1\cap I\neq 0$. If $I_1\cap J$ is non-zero, then consider the
path $I-\hspace{-.2cm}-I_1-\hspace{-.2cm}-J$. Hence assume that
$I_1\cap J=0$. By a similar argument one can see that $I_1+J=R$
and $I_1$ is a minimal left ideal of $R$. This implies that
$I\cap I_1=0$, a contradiction. The proof is complete.}
\end{proof}
\begin{lem}\label{art} Let $R$ be a ring and $I$ be a left ideal of $R$. If $deg(I)$ is
finite, then $R$ is a left Artinian ring.
\end{lem}
\begin{proof}{ Since $deg(I)$ is finite, so $I$ and $R/I$ are left Artinian $R$-modules.
Thus by \cite[Proposition 4.5]{good}, $R$ is a left Artinian
$R$-module and the proof is complete.}
\end{proof}
\begin{thm} Let $R$ be a commutative ring and $\mathfrak{m}$ be a maximal ideal of $R$. If $deg(\mathfrak{m})$ is finite, then $G(R)$ is finite.
\end{thm}
\begin{proof}{ By Lemma \ref{art}, $R$ is an Artinian ring. So by \cite[Theorem 8.7]{ati}, $R\cong R_1\times \cdots \times R_n$, where
$(R_i, \mathfrak{m}_i)$ is a local Artinian ring, for $i=1, \ldots , n$. Since $\mathfrak{m}$ is a maximal ideal of $R$, there exists $j$, $1\leq j\leq n$, such that
$\mathfrak{m}=R_1\times \cdots \times R_{j-1}\times \mathfrak{m}_{j}\times R_{j+1}\times \cdots \times R_n$. Now, if one of the
$R_i$ has an infinite number of ideals, then $deg(\mathfrak{m})$ is infinite, a contradiction. Therefore every $R_i$, $1\leq i\leq n$, has finitely many ideals
and the proof is complete.
}
\end{proof}
\noindent Now, we wish to investigate the properties of a ring
(not necessarily commutative) with at least a maximal left ideal
of finite degree. Before stating our results we need the following
lemma.
\begin{lem}\label{clique}
Let $D$ be an infinite division ring and $n\geq 2$. Then $M_n(D)$ has an infinite number of maximal left ideals. Moreover, if $n\geq 3$, then
every two distinct maximal left ideals are adjacent in $G(M_n(D))$ and $\omega(G(M_n(D))$ is infinite.
\end{lem}
\begin{proof}{ We show that $M_n(D)$ has infinitely many maximal left ideals. For every $x\in D$,
let $A_x=I_n+ xE_{1n}$, where $E_{1n}$ is an $n$ by $n$ matrix whose $(1,n)$-th entry is 1 and
other entries are zero. Clearly, $A_x H_{n-1}(A_{x})^{-1}$ is a maximal left ideal of $M_n(D)$ (see Lemma 2).
By an easy calculation one can see that if $a_{21}\neq 0$, and $$A=\left [ \begin{array}{cccc}
a_{12} & \ldots & a_{1,n-1} & 0 \\
a_{21} & \ldots & a_{2,n-1} & 0 \\
\vdots & \vdots & \ldots & \vdots \\
a_{n1} & \ldots & a_{n,n-1} & 0
\end{array}\right ],$$
\noindent{then the inverse of $(2,n)$th entry times $(2,1)$th of
the matrix $A_x A(A_{x})^{-1}$ is $-x$. So by Lemma 2, $M_n(D)$
has an infinite number of maximal left ideals. Now, since every
two maximal left ideals have a non-zero intersection, we conclude
that $\omega(G(M_n(D))$ is infinite.~~}}
\end{proof}
\begin{thm}\label{33}
Let $R$ be a ring and $\mathfrak{m}$ be a maximal left ideal of
$R$ of finite degree. If $G(R)$ is infinite and it is not null, then the following hold:
\noindent {\rm (i)} The number of maximal left ideals of $R$ is
finite.
\noindent {\rm (ii)} $\chi(G(R))<\infty$.
\noindent {\rm (iii)} There exists a two sided ideal of infinite
degree.
\end{thm}
\begin{proof}{ (i) Clearly, $R$ is a left Artinian ring. Toward a contradiction, let $\mathfrak{m}, \mathfrak{m}_1, \mathfrak{m}_2, \ldots $
be an infinite number of maximal left ideals of $R$. Since
$deg(\mathfrak{m})<\infty$, there exists some $j$ such that
$\mathfrak{m}\cap \mathfrak{m}_j=0$. This implies that $J(R)=0$
and so $R$ is a semisimple left Artinian ring. Thus by
Wedderburn-Artin Theorem (see \cite [3.5]{lam}), $R\cong
M_{n_1}(D_1)\times \cdots \times M_{n_k}(D_k)$, where $D_i$ is a
division ring for every $i$, $1\leq i\leq k$. Since $G(R)$ is
infinite, at least one of the $D_i$, say $D_1$ is infinite and
$n_1\geq 2$. Hence by Lemma \ref{clique}, $M_{n_{1}}(D_1)$ has
infinitely many left ideals. If $k\geq 2$, then it is not hard to
see that the degree of every maximal left ideal of $R$ is
infinite, a contradiction. So $R\cong M_{n_1}(D_1)$. If $n_1=2$,
then $G(R)$ is a null graph, a contradiction. If $n_1>2$, then by
Lemma \ref{clique}, the degree of every maximal left ideal of $R$
is infinite, a contradiction. Thus the number of maximal left
ideals of $R$ is finite.
\noindent (ii) and (iii) Since $deg(\mathfrak{m})$ is finite and
$G(R)$ is infinite, there are non-trivial left ideals
$\{I_i\}_{i=1}^{\infty}$ such that for every $i$, $i\geq 1$,
$I_i\cap \mathfrak{m}=0$. Thus $I_i+ \mathfrak{m}=R$. We prove
that each $I_i$ is a minimal left ideal. Let $I'_i\subset I_i$ be
a non-zero left ideal. Clearly, $I'_i+\mathfrak{m}=R$. Suppose
that $x\in I_i$. Thus there are two elements $a\in I'_i$ and
$b\in \mathfrak{m}$ such that $a+b=x$. So we have $b=x-a\in
I_i\cap \mathfrak{m}=0$. This implies that $I_i=I'_i$. Therefore
there are infinitely many minimal left ideals
$\{I_i\}_{i=1}^{\infty}$ such that $I_i+\mathfrak{m}=R$ and
$I_i\cap \mathfrak{m}=0$, for every $i$, $i\geq 1$. The argument
shows that if $I$ is a non-trivial left ideal and $I\cap
\mathfrak{m}=0$, then $I$ is a minimal left ideal of $R$. This
implies that the number of left ideals of $R$ which are not
minimal is finite. Now, since the intersection of every two
distinct minimal left ideals of $R$ is zero, one can color all
minimal left ideals of $R$ by a color and color each other vertex
with a new color to obtain a proper vertex coloring of $G(R)$.
This completes the proof of Part (ii).
\noindent To prove the Part (iii),
consider $soc(R)$. By \cite [Exercise 19, p.69]{lam}, $soc(R)$ is a two sided ideal containing all minimal
left ideals of $R$. If $soc(R)=R$, then since every minimal left ideal is a simple left module, we conclude that
$R$ is a semisimple left Artinian ring. Since $deg(\mathfrak{m})<\infty$, by Wedderburn-Artin Theorem and Lemma \ref{clique}, we conclude that
$R\cong D_1\times \cdots \times D_k$, where $D_i$ is a
division ring for every $i$, $1\leq i\leq k$. This yields that $G(R)$ is a finite graph, a contradiction. Therefore $soc(R)\neq R$.
Now, since the number of minimal left ideals of $R$ is infinite, we conclude that $deg(soc(R))$ is infinite and Part (iii) is proved.
The proof is complete.}
\end{proof}
\begin{remark}\label{ramark2} Let $R$ be a ring. If for every maximal left ideal
$\mathfrak{m}$ of $R$, $deg(\mathfrak{m})<\infty $, then $G(R)$ is null
or a finite graph. To see this as the previous proof shows, if
$G(R)$ is infinite, then the number of minimal left ideals of $R$
is infinite. Now, by Part (i) of Theorem \ref{33}, there exists
a maximal left ideal $\mathfrak{m}_j$ which contains infinitely
many minimal left ideals. So $deg(\mathfrak{m}_j)=\infty $, a
contradiction.
\end{remark}
\noindent Now, we propose a question: If $R$ is a ring and
$deg(\mathfrak{m})<\infty $, for a maximal left ideal $\mathfrak{m}$ of $R$, then is it true that $G(R)$ is null or
finite?
\noindent In the next theorem we will show that every
intersection graph of ideals which is regular, is a complete graph or a null graphs.
\begin{thm}\label{9}
Suppose that $R$ is a ring and $G(R)$ is an $r$-regular graph , for some
non-negative integer $r$. Then either $G(R)$ is a complete graph
or a null graph.
\end{thm}
\begin{proof}
{Suppose that $G(R)$ is not null. By Lemma \ref{art}, $R$ is a left Artinian ring.
Toward a contradiction suppose that $G(R)$ is not a complete graph. It follows
from \cite[Theorem 2.11]{chakrabarty} that $R$ has at least two
minimal left ideals, say $I_1$ and $I_2$. Since $I_1$ and $I_2$ are not
adjacent and $diam(G(R))\leq 2$, there exists an ideal $J$ of $R$
such that $I_1-\hspace{-.2cm}-J-\hspace{-.2cm}-I_2$. So both
$I_1$ and $I_2$ are contained in $J$. Thus each vertex adjacent
to $I_1$ is adjacent to $J$ too. This argument shows that
$deg(J)>deg(I_1)$, a contradiction. Therefore, $G(R)$ is
complete.}
\end{proof}
In the sequel of this section we study some rings whose
intersection graph of ideals are complete, i.e., $diam(G(R))=1$.
\begin{thm}\label{53}
Let $R$ be a commutative ring. Then $R$ is an integral domain if and only if $R$ is a
reduced ring and $G(R)$ is a complete graph.
\end{thm}
\begin{proof}
{One side is clear. For the other side suppose that $R$ is a
reduced ring and $G(R)$ is a complete graph. We claim that if
$I$, $J$ $\in I(R)^*$ and $IJ=0$, then $I\cap J=0$. By contrary,
suppose that $I\cap J\neq 0$. Then there exists a non-zero
element $x$ in $I\cap J$. So $(x)\subseteq I$ and $(x)\subseteq
J$ and hence $(x)^2\subseteq IJ=0$. Therefore, $x=0$, a
contradiction. Now, if $R$ is not a domain, then there exist
non-zero elements $x,y \in R$ such that $xy=0$. If $(x)=(y)$,
then $x^2=0$, a contradiction. Thus $(x)\neq (y)$, and by the
claim $(x)(y)=(x)\cap (y)=0$, a contradiction. The proof is
complete.~}
\end{proof}
\noindent The condition of $R$ to be a reduced ring in the previous
theorem is necessary. To see this let $R=\mathbb{Z}_{p^3}$, where
$p$ is a prime number. Then $R$ is not a reduced ring, $G(R)$ is
a complete graph but $R$ is not an integral
domain.
\begin{thm}\label{3}
Suppose that $R$ is a commutative Noetherian ring and $G(R)$ is a complete
graph. Then ${\rm Ass}(R)={\rm Min}(R)$ has just one element.
\end{thm}
\begin{proof}
{Suppose that $0=\cap_{i=1}^n Q_{i}$ is a minimal primary
decomposition of the ideal $0$, see \cite [Corollary 4.35]{sharp}.
Since $G(R)$ is complete and $\cap_{i=1}^n Q_{i}$ is a minimal
primary decomposition, the ideal $0$ is primary and so by \cite [Remarks 9.33]{sharp}, ${\rm
Ass}(R)=\{\sqrt{0}\}$. Therefore by \cite [Corollary
9.36]{sharp}, $Z(R)={\rm
Ass}(R)=\{\sqrt{0}\}={\rm Nil}(R)$. Since $R$ is a
Noetherian ring and $0$ is a primary ideal, $\sqrt{0}=\mathfrak{p}$ is a prime ideal
and so by \cite[Proposition 1.8]{ati}, ${\rm
Min}(R)=\{\mathfrak{p}\}$.~}
\end{proof}
\noindent The following example shows that the converse of the
previous result is not true.
\begin{example}\label{77777}
Let $R=\frac{k[x,y]}{(x,y)^2}$, where $k$ is an infinite field
and $x$, $y$ are indeterminates. Clearly, $R$ is an Artinian local
ring with the unique maximal ideal $\mathfrak{m}= \overline{(x,y)}$. Since
$\mathfrak{m}^2=0$, by \cite [Theorem 4.9]{sharp}, $0$ is a
primary ideal of $R$. Thus ${\rm Ass}(R)={\rm
Min}(R)=\{\sqrt{0}\}$, but $\overline{(x)}\cap \overline{(y)}=0$.
\end{example}
\noindent We close this section with the following theorem.
\begin{thm}\label{7}
Let $(R, \mathfrak{m})$ be a commutative Artinian local ring.
Then the
following statements are equivalent.\\
{\rm(i)} $R$ is a Gorenstein ring.\\
{\rm(ii)} $I=AnnAnn I$ for all ideals $I$ of $R$.\\
{\rm(iii)} $G(R)$ is a complete graph.\\
{\rm(iv)} $R$ has a unique minimal ideal.
\end{thm}
\begin{proof}
{ By \cite[Exercise 3.2.15]{herzog} and \cite [Theorem
2.11]{chakrabarty} the proof is complete.}
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{9mm} \noindent{\bf\large 3. The Clique Number and the
Chromatic Number of the Intersection Graph of Ideals}\vspace{5mm}
\noindent In this section we characterize all rings whose
intersection graph of ideals are triangle-free. There are many graphs whose clique numbers are finite
whereas chromatic numbers are infinite. We show that if the clique number of an intersection graph of ideals is finite, then its chromatic number
is also finite and they are equal when $R$ is a reduced ring. Finally,
it is proved that if $R$ is a commutative reduced ring with
$|{\rm Min}(R)|<\infty$, then
$\alpha(G(R))=|{\rm Min}(R)|$.
\begin{lem}\label{88}
Let $R$ be a ring such that $\omega(G(R))<\infty$. Then $R$ is a
left Artinian ring.
\end{lem}
\begin{proof}
{Suppose to the contrary that $R$ is not a left Artinian ring.
Then there exists a descending chain $I_1\supset I_2\supset \cdots
\supset I_n \supset \cdots$ of left ideals of $R$. Hence
$\{I_t\}_{t=1}^ \infty$ is an infinite clique of $G(R)$, a
contradiction.}
\end{proof}
Note that the converse of the above lemma is not true. In fact,
the following example is an Artinian ring such that
$\omega(G(R))=\infty$.
\begin{example}\label{888}
Let $R= \frac{k[[x,y]]}{(x^2,y^2)}$, where $k$ is a field and
$x$, $y$ are indeterminates. It is a routine exercise in
commutative algebra that $R$ is an Artinian Gorenestien ring with
infinitely many non-trivial ideals. Therefore, by Theorem
\ref{7}, $G(R)$ is a complete graph and so $\omega(G(R))=\infty$.
\end{example}
\begin{thm}\label{local} Let $R$ be a ring and $G(R)$ be a triangle-free graph which is not null.
Then $R$ is a local ring and one of the following holds:
\noindent {\rm(i)} The maximal left ideal of $R$ is principle and moreover $G(R)=K_1$ or $G(R)=K_2$.
\noindent {\rm(ii)} The minimal generating set of $\mathfrak{m}$ has size $2$, $\mathfrak{m}^2=0$ and $G(R)$ is a star graph.
\end{thm}
\begin{proof}{Toward a contradiction suppose that $R$ has two maximal left ideals $\mathfrak{m}_1$ and $\mathfrak{m}_2$.
If $\mathfrak{m}_1\cap \mathfrak{m}_2$ is non-zero, then
$\mathfrak{m}_1$, $\mathfrak{m}_2$, $\mathfrak{m}_1\cap
\mathfrak{m}_2$ form a triangle, a contradiction. Thus assume
that $\mathfrak{m}_1\cap \mathfrak{m}_2=0$. By Theorem \ref{1},
there exists an ideal $J$ which is adjacent to both
$\mathfrak{m}_1$ and $\mathfrak{m}_2$. If $J\cap
\mathfrak{m}_1=J$, then $J\subset \mathfrak{m}_1$ and so
$\mathfrak{m}_1\cap\mathfrak{m}_2$ is non-zero, a contradiction.
Thus $J$, $J\cap \mathfrak{m}_1$, $J\cap \mathfrak{m}_2$, form a
triangle, a contradiction. Therefore $R$ is a local ring. Let
$\mathfrak{m}$ be the unique maximal left ideal of $R$. Since $R$
is triangle-free, by Lemma \ref{88}, $R$ is a left Artinian ring. Thus by Hopkins
Theorem (see \cite[Theorem 4.15]{good}), $R$ is a left Noetherian
ring. So $\mathfrak{m}$ is a finitely generated left ideal. If a
minimal generating set of $\mathfrak{m}$ has at least size $3$ and
contains, $x,y,z, \ldots$, then $\mathfrak{m}, (x,y), (y,z)$ form
a triangle, a contradiction. Thus $\mathfrak{m}$ is generated
with at most two elements. First assume that $\mathfrak{m}=(x)$
is a principle left ideal. Since $R$ is a left Artinian ring,
$J(R)=\mathfrak{m}$ is nilpotent. If $\mathfrak{m}^3\neq 0$, then
by Nakayama's Lemma (see \cite[4.22]{lam}), $\mathfrak{m}$,
$\mathfrak{m}^2$, $\mathfrak{m}^3$ are distinct and so form a
triangle, a contradiction. Thus $\mathfrak{m}^3=0$. Let $I$ be a
non-trivial left ideal of $R$. We show that $I=\mathfrak{m}$ or
$I=(x^2)$. Let $a\in I$ be a non-zero element. Thus $a=bx$, for
some $b\in R$. If $b\not \in \mathfrak{m}$, then $b$ is a unit
and so $I=\mathfrak{m}$. Otherwise $b$ is contained in
$\mathfrak{m}$ and so $a=cx^2$, for some $c\in R$. Again if $c$
is not in $\mathfrak{m}$, then $I=(x^2)$, otherwise $I=0$, a
contradiction. If $\mathfrak{m}^2$ is non-zero, then we find
$G(R)=K_2$. If $\mathfrak{m}^2=0$, then we have $G(R)=K_1$.
\noindent Now, suppose that $\mathfrak{m}$ is not principle and
$\mathfrak{m}=(x,y)$. We show that $\mathfrak{m}^2=0$. Clearly,
if $I, J\neq \mathfrak{m}$ be two non-trivial left ideals of $R$,
then they are minimal left ideals and so $IJ=0$ or $IJ=J$. If
$IJ=J$, then since $I\subseteq J(R)$, by Nakayama's Lemma we
conclude that $J=0$, a contradiction. Therefore $IJ=0$ and this
implies that $\mathfrak{m}^2=0$. The proof is complete.}
\end{proof}
\begin{thm} Let $R$ be a ring and $\omega(G(R))<\infty$. Then $\chi(G(R))<\infty$.
\end{thm}
\begin{proof}{ By Lemma \ref{88}, $R$ is a left Artinian ring.
Thus the length of every left ideal of $R$ is finite. Note that
$l(R)\leq \omega(G(R))+1$. If $G(R)$ is finite, then we are done.
Thus assume that $G(R)$ is infinite. Since $R$ is a left Artianian ring,
every left ideal of $R$ contains a minimal left ideal. So if $G(R)$ is
infinite and the number of minimal left ideals of $R$ is finite,
then $G(R)$ has an infinite clique, a contradiction. Thus $R$ has
an infinite number of minimal left ideals.
For every $r$, $1\leq r\leq \omega(G(R)) +1$, let
$$\mathcal{S}_r=\{\, I\in I(R)^* \,|\, l(I)=r\, \}.$$ Let $t$ be
the maximum natural number such that $\mathcal{S}_t$ is infinite.
Note that since $\mathcal{S}_1$ is infinite, $t$ exists and
$t\geq 1$. By the definition of $t$, the number of left ideals of
$R$ of length $t+1$ is finite. By Schreier Refinement Theorem
(see \cite[Theorem 4.10]{good}) every left ideal of length $t$ is
contained in at least one left ideal of length $t+1$. So there is
a left ideal $J$ of length $t+1$ such that $J$ contains
infinitely many left ideals of length $t$. Since
$\omega(G(R))<\infty$, there exist left ideals $L, K\in
\mathcal{S}_t$ such that $L,K\subseteq J$ and $L\cap K= 0$.
Now, by \cite[Proposition 4.12]{good}, we have
$$t+1=l(J)\geq l(L+K)=l(L\oplus K)=l(L)+l(K)=2t.$$ Therefore
$t=1$. This implies that the set of all left ideals of $R$ which
are not minimal is finite. Since the intersection of two distinct
minimal left ideals of $R$ is zero, so one can color all minimal
left ideals of $R$ with the same color and color every non-minimal left ideal
with a new color. Thus $\chi(G(R))$ is finite and the proof is
complete.}
\end{proof}
\begin{thm} Let $R$ be a ring and $\omega(G(R))<\infty $. Then the following hold:
\noindent {\rm(i)} If $R$ is not local and it is commutative,
then $G(R)$ is finite.
\noindent {\rm(ii)} If $(R, \mathfrak{m})$ is a local ring, then
either $G(R)$ is finite or the size of every minimal generating
set of $\mathfrak{m}$ is $2$.
\end{thm}
\begin{proof}{ (i) Since $\omega(G(R))$ is finite, by Lemma \ref{88}, $R$ is an Artinian ring and so by
\cite[Theorem 8.7]{ati}, $R\cong R_1\times \cdots \times R_n$, where
$R_i$ is a local Artinian ring for $i=1,\ldots ,n$. If $n>1$ and at least one of the $R_i$ has infinite
number of ideals, then
clearly, $\omega(G(R))$ is infinite, a contradiction. So if $R$ is not a local ring, then $G(R)$ is finite.
\noindent (ii) If $(R, \mathfrak{m})$ is a local ring, then $R/\mathfrak{m}$ is a division ring
(see \cite[Proposition
15.15]{anderson}) and $\mathfrak{m}/\mathfrak{m}^2$ is a left ($R/\mathfrak{m}$)-module. If its dimension
is more than $2$ and $R/\mathfrak{m}$ is
infinite, then every $1$-dimensional subspace is contained in
infinitely many $2$-dimensional subspaces and so $\mathfrak{m}$
has infinitely many left ideals whose intersection is non-zero. Thus
$\omega(G(R))$ is infinite, a contradiction. Hence suppose that
$R/\mathfrak{m}$ is infinite and the dimension of $\mathfrak{m}/\mathfrak{m}^2$ over $R/\mathfrak{m}$ is at most $2$.
\noindent First assume that the dimension of
$\mathfrak{m}/\mathfrak{m}^2$ over $R/\mathfrak{m}$ is $2$. We
show that every minimal generating set of $\mathfrak{m}$ has two
elements. Suppose that
$\{{a}+\mathfrak{m}^2,{b}+\mathfrak{m}^2\}$ is a basis for
$\mathfrak{m}/\mathfrak{m}^2$ over $R/\mathfrak{m}$. Then
$\mathfrak{m}=(a, b)+\mathfrak{m}^2$. By Nakayama's Lemma we have
$\mathfrak{m}=(a, b)$. Now, if $\{{a},{b}\}$ is not a minimal
generating set for $\mathfrak{m}$, then $\mathfrak{m}$ is
principle and so the dimension of $\mathfrak{m}/\mathfrak{m}^2$
over $R/\mathfrak{m}$ is $1$, a contradiction.
\noindent Now, assume that the dimension of $\mathfrak{m}/\mathfrak{m}^2$
over $R/\mathfrak{m}$ is $1$. Thus $\mathfrak{m}=(x)$ is principle. We
show that every proper left ideal of $R$ is a power of $\mathfrak{m}$.
Let $I$ be a non-zero left ideal of $R$. Since $\mathfrak{m}$ is
nilpotent, there exists a natural number $i$ such that $I\subseteq
\mathfrak{m}^i$ but $I\not \subseteq \mathfrak{m}^{i+1}$. Let
$a\in I\setminus \mathfrak{m}^{i+1}$. We have $a=bx^i$, for some
$b\in R$. If $b\in \mathfrak{m}$, then $a\in \mathfrak{m}^{i+1}$,
a contradiction. Thus $b$ is unit. Hence $x^i\in I$. This implies
that $I=\mathfrak{m}^i$. By Hopkins Theorem (see \cite[Theorem
4.15]{good}), $\mathfrak{m}$ is a nilpotent ideal and so $G(R)$ is a finite graph.
If $\mathfrak{m}=0$, then $R$ is a division ring and we are done.
\noindent Now, we prove that
if $(R,\mathfrak{m})$ is a local left Artinian ring, and
$R/\mathfrak{m}$ is finite, then $R$ is a finite ring. By
Wedderburn's ``Little" Theorem (see \cite[13.1]{lam}),
$F:=R/\mathfrak{m}$ is a field. Since $R$ is a left Artinian ring by Hopkins Theorem (see \cite[Theorem
4.15]{good}), $\mathfrak{m}$ is a nilpotent left ideal and $R$ is a
left Noetherian ring. Suppose that $\mathfrak{m}^r=0$. Consider
the $F$-vector space $\mathfrak{m}^{i-1}/\mathfrak{m}^i$, $2\leq
i\leq r$. Since $R$ is a left Noetherian ring, for each $i$,
$\mathfrak{m}^i$ is a finitely generated $R$-module. This implies
that for every $i$, $2\leq i\leq r$,
$\mathfrak{m}^{i-1}/\mathfrak{m}^i$ is a finite dimensional
$F$-vector space. Since $F$ is finite, we conclude that $R$
is a finite ring. The proof is complete.~ }
\end{proof}
\begin{cor}\label{8}
Let $R$ be a Gorenstein ring. Then $\omega(G(R))=n$ if and only if
$R$ has $n$ non-trivial ideals.
\end{cor}
\begin{proof}
{Let $R$ be a Gorenstein ring and $\omega(G(R))=n$. By Lemma
\ref{88}, $R$ is an Artinian ring. It turns out by Theorem
\ref{7}, $G(R)$ is a complete graph, and so $R$ has exactly
$n$ non-trivial ideals.\\
Conversely, suppose that $R$ has $n$ non-trivial ideals. Then $R$
is an Artinian ring. Again Theorem \ref{7} implies that
$G(R)$ is a complete graph, and hence $\omega(G(R))=n$. }
\end{proof}
\begin{thm}\label{12}
Let $R$ be a reduced ring with $\omega(G(R))<\infty$. Then
$\omega(G(R))=\chi(G(R))$.
\end{thm}
\begin{proof}
{ By Lemma \ref{88}, $R$ is a left Artinian ring. By
\cite[Theorem 4.15]{good}, since $J(R)$ is nilpotent, we conclude
that $J(R)=0$ and so $R$ is a semisimple ring. Since $R$ is
reduced by Wedderburn-Artin Theorem, $R\cong D_1\times \cdots
\times D_n$, where $D_i$ is a division ring, for each $i$, $1\leq
i\leq n$. Therefore, $R$ has $2^n-2$ non-trivial left ideals.
Now, consider the ideal $I=D_1\times 0\times \cdots \times 0$.
Clearly, every ideal of the form $D_1\times I_2\times \cdots
\times I_n$ contains $I$, where $I_i$ is a left ideal of $D_i$,
for $i=2,\ldots ,n$. But the set of these left ideals forms a
clique $\mathcal{C}$ of $G(R)$ and so $\omega(G(R))\geq
2^{n-1}-1$. Now, we show that $\chi(G(R))=2^{n-1}-1$. First we
color the vertices of the clique $\mathcal{C}$ by $2^{n-1}-1$
different colors. Let $J=(0=J_1)\times J_2\times \cdots \times
J_n$ be a vertex not contained in $\mathcal{C}$. We color $J$
with the color of the vertex $L=D_1\times L_2\times \cdots \times
L_n$ of $\mathcal{C}$ in which $L_i=0$ if $J_i=D_i$ and $L_i=D_i$
if $J_i=0$. Thus we obtain a proper vertex coloring for $G(R)$
using $2^{n-1}-1$ colors, as desired.}
\end{proof}
\noindent It follows from Theorem \ref{12}, if $R$ is a reduced
ring and $\omega(G(R))<\infty$, then $R$ has finitely many
ideals. This is not true necessarily in the case non-reduced. See
the following example.
\begin{example}\label{1333}
Let $R=\frac{k[x,y]}{(x,y)^2}$, where $k$ is an infinite field and
$x$, $y$ are indeterminates. Then $\overline{(x,y)}, \overline{(x)}, \overline{(y)}$
and $\{\overline{(x+ay)}|~a \in
k\}$ are all non-trivial ideals of $R$. Therefore, $G(R)$ is an
infinite star graph.
\end{example}
\begin{thm}
Let $R$ be a commutative reduced ring and $|{\rm
Min}(R)|<\infty$. Then $\alpha(G(R))=|{\rm Min}(R)|$.
\end{thm}
\begin{proof}
{Suppose that ${\rm Min}(R)=
\{\mathfrak{p}_1,\ldots,{\mathfrak{p}_n}\}$. Let
$\widehat{\mathfrak{p}}_j=\bigcap_{i=1,i\neq j}^n
\mathfrak{p_i}$, for $j=1,\ldots ,n$. Since $R$ is reduced, the
intersection of all minimal prime ideals is zero and so it is
easily checked that $
\{\widehat{\mathfrak{p}}_1,\ldots,\widehat{\mathfrak{p}}_n\}$ is
an independent set of $G(R)$. Hence $|{\rm Min}(R)| \leq
\alpha(G(R))$. Obviously,
$\alpha(G(R))=\omega(\overline{G(R)})$. Thus it suffices to show
that $\chi(\overline{G(R)}) \leq \alpha(G(R))$. Define a coloring
$f$ on $V(\overline{G(R)})$ by $f(I)={\rm min}\{i, I\nsubseteq
\mathfrak{p}_i \}$. We show that $f$ is a proper vertex coloring
of $\overline{G(R)}$. First note that, since $R$ is a reduced
ring, for each non-zero ideal $I$, there exists a minimal prime
ideal $\mathfrak{p}$ such that $I \nsubseteq \mathfrak{p}$. Now,
suppose that $I$ and $J$ are adjacent vertices in
$\overline{G(R)}$ and $f(I)=f(J)= i$, $1\leq i \leq n$. Since $R$
is a commutative ring, we deduce that $IJ=0$. Therefore $IJ
\subseteq \mathfrak{p}_i$ and so either $I\subseteq
\mathfrak{p}_i$ or $J\subseteq \mathfrak{p}_i$, a contradiction.
Then $f$ is a proper vertex coloring of $\overline{G(R)}$ and
hence $\chi(\overline{G(R)}) \leq n$. Therefore,
$\alpha(G(R))=|{\rm Min}(R)|$.}
\end{proof}
\noindent We end this paper with with following result about the
clique number and the chromatic number of the intersection graph
of a direct product of rings.
\noindent Obviously, if $R$ is a ring and $R\cong R_1\times \cdots\times R_k$, where
$R_i$ is a ring, then $\omega({G}(R))<\infty$ if and only if
$\omega({G}(R_i))<\infty$, for every $i$, $1\leq i\leq k$. Now,
we prove this fact for the chromatic number instead of the clique
number.
\begin{thm}\label{10}
Let $R_1,\ldots,R_k$ be rings and $R\cong R_1\times \cdots\times R_k$. Then
$\chi({G}(R))<\infty$ if and only if
$\chi({G}(R_{i}))<\infty$, for every $i$, $1\leq i\leq k$.\\
\end{thm}
\begin{proof}
{First suppose that $R\cong R_1\times R_2$. \noindent Let $c_i:
V(G(R_i))\longrightarrow \{1,\ldots,\chi(G(R_i))\}$ be a proper
vertex coloring of $G(R_i)$, $i=1,2$. We extend the map $c_i$ to
${I}(R_i)$ by defining $c_i(R_i)=-1$ and $c_i(0)=0$, for $i=1,2$.
Define a map $c$ on ${I}(R)^*$ by
$$c(I\times J)= (c_1(I),c_2(J)),~~{\rm for~every}~~I\in I(R_1)~~{\rm and}~~J\in I(R_2).$$It is not hard to check that $c$ is a proper coloring of
${G}(R_1\times
R_2)$. Conversely, let $c$ be a proper vertex coloring of ${G}(R)$, then
the restriction of $c$ to ${I}(R_1\times 0)^{*}$ and ${I}(0\times
R_2)^{*}$
provides a proper coloring of ${G}(R_1\times
0)$ and ${G}(0\times
R_2)$, respectively.
\noindent If $k>2$, then the assertion follows from the case $k=2$
and induction.}
\end{proof}
\noindent {\bf Acknowledgments.} {\rm This work was completed while the first
author was visiting the Abdus Salam International Centre
for Theoretical Physics (ICTP) under the auspices of Combinatorics Program 2010.
He would like to express his gratitude for the support.}
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\end{document}