% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{e-jc} \specs{P3.8}{25(3)}{2018} % Please remove all commands that change parameters such as % margins or page sizes. % Packages amssymb and amsthm are already loaded. % We recommend these AMS packages: \usepackage{amsmath,amssymb} % We recommend the graphicx package for importing figures: \usepackage{graphicx} % Use this command to create hyperlinks (optional and recommended): \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb} \usepackage{enumerate} \usepackage{yfonts} %\mathgoth %\usepackage{mathscr} %\mathscr \usepackage{euscript, mathrsfs} \def\widthe{{\rm width}} \def\hte{{\rm ht}} \def\depth{{\rm depth}} \def\max{{\rm max}} \def\supp{{\rm supp}} \def\hom{{\rm hom}} \def\deh{{\rm deh}} \def\ass{{\rm ass}} \def\Hom{{\rm Hom}} \def\Pos{{\rm Pos}} \def\Lat{{\rm Lat}} % Theorem-like environments that are declared in the style file are: % theorem, lemma, corollary, proposition, fact, observation, claim, % definition, example, conjecture, open, problem, question, remark, note %%%%%%%%%%%%%METADATA%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Give the submission and acceptance dates in the format shown. % The editors will insert the publication date in the third argument. \dateline{Feb 27, 2018}{Jun 22, 2018}{Jul 13, 2018} % Give one or more subject codes separated by commas. % Codes are available from http://www.ams.org/mathscinet/freeTools.html \MSC{13C05, 05E40, 13P25.} % Uncomment exactly one of the following copyright statements. 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Released under the CC BY-ND license (International 4.0).} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If needed, include a line break (\\) at an appropriate place in the title. \title{Primary decomposition of ideals \\ of lattice homomorphisms} % Input author, affiliation, address and support information as follows; % The address should include the country, but does not have to include % the street address. Give at least one email address. \author{Leila Sharifan\thanks{The research of the first author was in part supported by a grant from IPM (No. 96130021).}\\ \small Department of Mathematics and Computer Sciences\\[-0.8ex] \small Hakim Sabzevari University\\[-0.8ex] \small Sabzevar, Iran\\ \small School of Mathematics\\[-0.8ex] \small Institute for research in Fundamental Sciences (IPM)\\[-0.8ex] \small P. O. Box: 19395-5746, Tehran, Iran.\\ \small\tt leila-sharifan@aut.ac.ir\\ \and Ali Akbar Estaji \qquad Ghazaleh Malekbala\\ \small Department of Mathematics and Computer Sciences\\[-0.8ex] \small Hakim Sabzevari University\\[-0.8ex] \small Sabzevar, Iran\\ \small\tt aaestaji@hsu.ac.ir\\ \small \tt gmalekbala@gmail.com} \begin{document} \maketitle % E-JC papers must include an abstract. The abstract should consist of a % succinct statement of background followed by a listing of the % principal new results that are to be found in the paper. The abstract % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract may be distributed without the rest of the % paper so it must be entirely self-contained. Try to include all words % and phrases that someone might search for when looking for your paper. \begin{abstract} For two given finite lattices $L$ and $M$, we introduce the ideal of lattice homomorphism $J(L,M)$, whose minimal monomial generators correspond to lattice homomorphisms $\phi : L\to M$. We show that $L$ is a distributive lattice if and only if the equidimensinal part of $J(L,M)$ is the same as the equidimensional part of the ideal of poset homomorphisms $I(L,M)$. Next, we study the minimal primary decomposition of $J(L,M)$ when $L$ is a distributive lattice and $M=[2]$. We present some methods to check if a monomial prime ideal belongs to $\ass(J(L,[2]))$, and we give an upper bound in terms of combinatorial properties of $L$ for the height of the minimal primes. We also show that if each minimal prime ideal of $J(L,[2])$ has height at most three, then $L$ is a planar lattice and $\widthe(L)\leq 2$. Finally, we compute the minimal primary decomposition when $L=[m]\times [n]$ and $M=[2]$. \end{abstract} \section{Introduction} The study of monomial ideals and interaction between their algebraic and combinatorial properties is an important topic in combinatorial and computational commutative algebra. Such ideals serve as a useful tool for studying polynomial ideals and also have grown into an active research area. Recently, some researchers have focused on some classes of monomial ideals and algebras associated to the ordered algebraic structures. Among them, we can point out to {\it Hibi rings} which is defined by Hibi in 1987 \cite{Hi}. Corresponding to a distributive lattice $L$ the {\it joint-meet ideal $I_L$} is defined in \cite{EH} as a monomial ideal in a specific polynomial ring which is closely related to Hibi rings. In 2005, Herzog and Hibi \cite{HH2} associated to a poset $P$ its so-called {\it Hibi ideal} which is again a monomial ideal. In 2011, Ene, Herzog and Mohammadi \cite{EHM} considered {\it generalized Hibi ideals} and studied some of their algebraic properties. In 2014 Fl{\o}ystad et al. \cite{FGH} introduced a further generalization of such ideals corresponding to isotone maps between two posets. These ideals are called {\it ideals of poset homomorphisms} and further studied in \cite{HQSh} and \cite{JKS}. Often, in the above researches, algebraic properties of mentioned monomial ideals are studied in terms of combinatorial properties of $L$ or even of the underlying poset $P$. In this paper, we consider lattice homomorphisms instead of poset homomorphisms and introduce the {\it { ideals of lattice homomorphisms}}. Our main goal is to study minimal primary decomposition of such ideals and carefully relate it to the combinatorial properties of the corresponding lattices. Given two finite posets $P$ and $Q$, a map $\phi:P \to Q$ is called {\it isotone} (or, {\it poset homomorphism}) if it is order preserving. In other words, $\phi : P \to Q $ is isotone if and only if $\phi(p_1)\leq \phi(p_2)$ for all $p_1,p_2 \in P$ with $p_1\leq p_2$. The set of isotone maps $P \to Q $ is denoted by $\Hom_{\Pos}(P,Q)$. Given two finite lattices $L$ and $M$, a map $\phi: L\to M$ is called a {\it lattice homomorphism} if for any $l_1,l_2\in L$, $\phi(l_1\vee l_2)=\phi(l_1)\vee \phi(l_2)$ and $\phi(l_1\wedge l_2)=\phi(l_1)\wedge \phi(l_2)$. We denote by $\Hom_{\Lat}(L,M)$ the set of lattice homomorphism $L\to M$. It is clear that $\Hom_{\Lat}(L,M)\subseteq \Hom_{\Pos}(L,M)$. Now, let $S$ be the polynomial ring over a field ${\bf{k}}$ with variables $x_{l,m}$ where $l\in L$ and $m\in M$, i.e., $S={\bf{k}}[x_{l,m}; l\in L, m\in M]$. As in \cite{FGH} we associate to any $\phi \in \Hom_{\Pos}(L,M)$ the monomial $$u_{\phi}=\prod_{l\in L} x_{l,\phi(l)}.$$ The ideal of poset homomorphisms associated to $L$ and $M$ (as defined in \cite{FGH} for any posets $P$ and $Q$) is the ideal of $S$ whose minimal monomial generators correspond to poset homomorphisms, i.e., $$I(L,M)=(u_{\phi}\ ; \ \phi\in \Hom_{\Pos}(L,M))\subset S.$$ We define the ideal of lattice homomorphisms, in a similar way, as $$J(L,M)=(u_{\phi}\ ; \ \phi\in \Hom_{\Lat}(P,Q))\subset S.$$ It is clear that $J(L,M)\subseteq I(L,M)$. Both ideals $J(L,M)$ and $ I(L,M)$ are square-free monomial ideals and so are radical ideals. Thus $\ass(J(L,M))=\min(J(L,M))$ and $\ass(I(L,M))=\min(I(L,M))$, where by $\ass(J)$ we mean the set of associated prime ideals of the ideal $J$ and by $\min(J)$ we mean the set of minimal prime ideals of it. As we have pointed out before, we are going to study the minimal primary decomposition of $J(L,M)$. Note that since $J(L,M)$ is radical, we have $$J(L,M)=\bigcap_{\textswab{p}\in \ass(J(L,M))} \textswab{p}.$$ By \cite[Proposition 1.5]{FGH}, the height of each $\textswab{p}\in \ass(I(L,M))$ is at least $|M|$ and the associated primes of $I(L,M)$ of the minimum height is of the form \begin{equation}\label{a} \textswab{p}=\textswab{p}_\psi:=(x_{\psi(m),m};\ m\in M), \ \ {\text{where}}\ \psi\in \Hom_\Pos(M,L).\end{equation} Proposition \ref{basic} shows that each $\textswab{p}_\psi$ described above belongs to $\ass(J(L,M))$ and Theorem \ref{distributive} says that any associated prime of $J(L,M)$ of the minimum height is as (\ref{a}) if and only if $L$ is a distributive lattice. An immediate consequence of Theorem \ref{distributive} is that the equidimensional part of $J(L,M)$ coincides with the equidimensional part of $I(L,M)$ if and only if $L$ is a distributive lattice (Corollary \ref{equidimensional part}). Moreover, by Theorem \ref{distributive} and the main result of \cite{HQSh}, we conclude that if $L$ is a distributive lattice, then $J(L,M)$ is an unmixed ideal if and only if $L$ is a chain and this is the case if and only if $I(L,M)=J(L,M)$ (Corollary \ref{unmixed}). The rest of the paper is devoted to the study of the minimal primes of $J(L,[2])$ when $L$ is a distributive lattice. We should point out that in general, finding exactly all associated prime ideals of $J(L,M)$ seems to be pretty hard even if we restrict ourselves to the case that $M=[2]$. Given two nonempty subsets $A$,$B$ of $L$, in Lemma \ref{E15}, we observe that if $\bigwedge A\leq \bigvee B$ then \begin{equation}\label{key property}J(L,[2])\subset \textswab{p}:=\textswab{p}_{A,B}=(x_{a,1};\ a\in A )+(x_{b,2}; \ b\in B).\end{equation} Replacing the chain $[2]$ with an arbitrary chain $[n]$ with $n>2$, we can not generalize Lemma \ref{E15} (see Remark \ref{chain[n]}). So, characterizing prime ideals that containing $J(L,M)$ is more complicated in the general situation. This is the reason that we assume that $M=[2]$. The next easy fact is that any associated prime ideal of $J(L,[2])$ has the form $\textswab{p}_{A,B}$ for some non-empty subsets $A$ and $B$ of $L$ (see Lemma \ref{E20}). Let $\varnothing\neq A\subset L$ and $\varnothing\neq B\subset L$, in Theorem \ref{E25}, we prove that $\textswab{p}_{A,B}$ is an associated prime of $J(L,[2])$ if and only if the following statements hold: (i) $\bigwedge A\leq \bigvee B.$ (ii) $\forall \varnothing\neq A_1\subset A$ and $ \forall \varnothing\neq B_1\subset B, \bigwedge A_1 \nleq\bigvee B$ and $ \bigwedge A\nleq \bigvee B_1.$ \medskip \noindent In the sequel, we give some interesting corollaries of Theorem \ref{E25}. It is clear that by Theorem \ref{distributive}, for any distributive lattice $L$ and any $\textswab{p}\in ass(J(L,[2]))$, $\hte(\textswab{p})\geq 2$. It would be nice to find an upper bound for the height of such prime ideals. In Theorem \ref{upper bound} we prove that if $\textswab{p}$ is an associated prime ideal of $J(L,[2])$ then $\hte(\textswab{p})\leq m(L)+M(L)$ (for the definition of $m(L)$ and $M(L)$ see the paragraph just before Theorem \ref{upper bound}). While Theorem \ref{distributive} shows that any minimal prime ideal of $J(L,[2])$ of height $2$ is of the form $\textswab{p}_\psi$ for some $\psi\in \Hom_\Pos([2],L)$, in Corollary \ref{antichain in minimal prime} we give another method to check when a prime ideal of height bigger than two belongs to $\ass(J(L,[2]))$. Indeed, we prove that if $A$ and $B$ are two non-empty subsets of $L$, and $|A|>1$ or $|B|>1$, then $\textswab{p}_{A,B}\in ass(J(L,[2]))$ if and only if the following statements hold: \begin{enumerate} \item[{\rm (i) }] $A$ and $B$ are antichains. \item[{\rm (ii) }] $\forall a\in A$ and $\forall b\in B$, $a\nleq b$. \item[{\rm (iii)}] If $a,a'$ are two arbitrary distinct elements of $A$ and $A'=(A\setminus \{a,a'\})\cup \{a\wedge a'\}$, then $\textswab{p}_{A',B}\in ass(J(L,[2]))$. \item[{\rm (iv)}] If $b,b'$ are two arbitrary distinct elements of $B$ and $B'=(B\setminus \{b,b'\})\cup \{b\vee b'\}$, then $\textswab{p}_{A,B'}\in ass(J(L,[2]))$. \end{enumerate} \noindent Next, we try to describe distributive lattices $L$ in which any minimal prime of $J(L,[2])$ has height at most $3$. As we see in Corollary \ref{ht3}, if any associated prime of $J(L,[2])$ has height at most $3$ then $L$ should be a planar lattice with $\widthe(L)\leq 2$, where by width we mean the maximum number of elements in an antichain contained in $L$. Given the positive integers $11$ and $|M|>1$. We are going to study $\ass(J(L,M))$ and relate it to $\ass(I(L,M))$. We remark that all pairs of posets $P,Q$ for which $I(P,Q)$ is an unmixed ideal characterized in \cite{HQSh} (recall that an ideal is called unmixed if all of its associated prime ideals have the same height). For each $\psi\in \Hom_{\Pos}(M,L)$, we consider the prime ideal $$\textswab{p}_\psi=(x_{\psi(m),m}; \ m\in M)\subset S,$$ we also define monomial $u_\psi^\tau$ as $$u_\psi^\tau:=\prod_{m\in M} x_{\psi(m),m}.$$ By using \cite[Proposition 1.5]{FGH}, we immediately get the first result. %-------- \begin{proposition}\label{basic} Let $L$ and $M$ be two lattices. Then the following statements hold. \begin{itemize} \item[{\rm (1)}] If $\textswab{p}\in \ass(J(L,M))$ then $\hte (\textswab{p})\geq |M|$. \item[{\rm (2)}] For each $\psi\in \Hom_{\Pos}(M,L)$, $\textswab{p}_\psi\in \ass(J(L,M))$. \item[{\rm (3)}] $I(M,L)^\tau\subseteq J(L,M)^\vee,$ where $I(M,L)^\tau:= (u_\psi^\tau; \ \psi\in \Hom_{\Pos} (M,L))$. \item[{\rm (4)}] $\hte (J(L,M))=|M|$. \end{itemize} \end{proposition} \begin{proof} (1). Let $\textswab{p}=(x_{l_1,m_1},\ldots,x_{l_n,m_n})\in \ass(J(L,M))$. For any $m\in M$, let $\phi_m:L\to M$ be the constant lattice homomorphism with $\phi_m(l)=m$ for each $l\in L$. It is clear that $u_{\phi_m}=\prod_{l\in L} x_{l,m}\in \textswab{p}$. So, for some $1\leq i\leq n$, we should have $m_i=m$ which shows that $n\geq |M|.$ (2) and (3). Let $\psi\in \Hom_{\Pos}(M,L)$. By \cite[Proposition 1.5]{FGH}, $J(L,M)\subseteq I(L,M)\subseteq \textswab{p}_\psi$. Now part (1), shows that $\textswab{p}_\psi\in \ass(J(L,M))$ and the conclusion follows. (4). It is followed by parts (1) and (2). \end{proof} In the next theorem, we detect when each associated prime ideal $\textswab{p}$ of $J(L,M)$ with $\hte(\textswab{p})=|M|$ is of the form $\textswab{p}_\psi$ for some $\psi\in \Hom_\Pos(L,M)$. \begin{theorem}\label{distributive} Let $L$ and $M$ be two lattices. Then $$\{\textswab{p}; \ \textswab{p}\in \ass(J(L,M)), \hte (\textswab{p})\leq |M|\}=\{\textswab{p}_\psi; \ \psi\in \Hom_{\Pos}(M,L)\}$$ if and only if $L$ is a distributive lattice. \end{theorem} \begin{proof} First, assume that $L$ is a distributive lattice. Let $\textswab{p}$ be any minimal prime ideal of $J(L,M)$ with $\hte (\textswab{p})=|M|$. Since $J(L,M) \subset\textswab{p}$, it follows that for each $m \in M$ there exists an element $\psi(m) \in L $ such that $x_{\psi(m),m} \in \textswab{p}$. Then $\textswab{p} = (x_{\psi(m),m}; \ m\in M)$. It remains to be shown that $\psi: M \to L$ is isotone. Suppose this is not the case. Then there exist $m,m' \in M$ such that $m< m'$ and $\psi(m)\nleq \psi(m')$. So, $\psi(m')< \psi(m)\vee \psi(m')$. Thus, there exists $\phi\in \Hom_{\Lat}(L,\{m2$. We start with the following easy lemma. \begin{lemma} \label{E15} Let $L$ be a lattice. If $A$ and $B$ are nonempty subsets of $L$ such that $\bigwedge A\leq \bigvee B$, then \[ J(L,[2])\subseteq \textswab{p}_{A,B}=(x_{a,1};\ a\in A)+(x_{b,2};\ b\in B). \] \end{lemma} %--------- \begin{proof} Consider $\phi \in \Hom_{\Lat}(L, [2])$. If $\phi(\bigvee B)=2$, then there exists $b\in B$ such that $\phi(b)=2$, which implies that $$ x_{b,2}= x_{b,\phi(b)}|u_{\phi}.$$ Hence, $u_{\phi}\in \textswab{p}_{A,B}$. If $\phi(\bigvee B)=1$, then $\phi(\bigwedge A)=1$, which follows that there exists $a\in A $ such that $\phi(a)=1$. Thus, $$ x_{a,1}= x_{a,\phi(a)}|u_{\phi},$$ it implies that $u_{\phi}\in \textswab{p}_{A,B}$. Therefore, $J(L,[2])\subseteq \textswab{p}_{A,B}$. \end{proof} %--------- %--------- \begin{lemma} \label{E20} Let $L$ be a lattice, and $A$, $B$ be subsets of $L$. If $\textswab{p}_{A,B}\in \ass(J(L,[2]))$, then $A\neq \varnothing$ and $B\neq \varnothing$. \end{lemma} %--------- \begin{proof} It is enough to follow the proof of Proposition \ref{basic}. \end{proof} %--------- %------- Now, let $L$ be an arbitrary distributive lattice. It is clear that each minimal prime ideal of $J(L,[2])$ is of the form $\textswab{p}_{A,B}$ for some $A\subseteq L$ and $B\subseteq L$. Which ones of the prime ideals presented in this form are the minimal prime ideals of $J(L,M)$? The following results help to find them. \begin{theorem} \label{E25} Let $L$ be a distributive lattice, and $A$, $B$ be nonempty subsets of $L$. Then $\textswab{p}_{A,B}\in \ass(J(L,[2]))$ if and only if the following statements hold: \begin{enumerate} \item[{\rm (1)}] $\bigwedge A\leq \bigvee B$. \item[{\rm (2)}] For every $ \varnothing \not=A_1\subset A$ and for every $ \varnothing \not=B_1\subset B$, $\bigwedge A_1 \not\leq \bigvee B$ and $\bigwedge A \not\leq \bigvee B_1$. \end{enumerate} \end{theorem} %--------- \begin{proof} First assume that (1) and (2) hold. By Lemma \ref{E15}, $J(L,[2])\subseteq \textswab{p}_{A,B}$. So, there exists a prime ideal $\textswab{p}_1\in \ass(J(L,[2]))$ such that $J(L,[2])\subseteq \textswab{p}_1\subseteq \textswab{p}_{A,B}$. Hence, by Lemma \ref{E20}, we can assume that there exist $\varnothing \neq A_1\subseteq A$ and $\varnothing \neq B_1\subseteq B$ such that $\textswab{p}_1=\textswab{p}_{A_1,B_1}$. If $A_1=A$ and $B_1=B$, then $\textswab{p}_{A,B}=\textswab{p}_{A_1,B_1}\in \ass(J(L,[2]))$ and the proof is now complete. Now assume that $A_1\subset A$ or $B_1\subset B$. By (2), $\bigwedge A_1\nleq \bigvee B_1$, which follows that $ \bigvee B_1< \bigwedge A_1\vee\bigvee B_1$. Thus, there exists a lattice homomorphism $\phi: L\rightarrow [2]$ such that $\phi(\bigvee B_1)=1$ and $\phi(\bigwedge A_1\vee \bigvee B_1)=2$. It implies that $$\forall a\in A_1, \phi(a)= 2\ {\text{and}} \ \forall b\in B_1, \phi(b)=1, $$ which follows that $x_{a,1}\nmid u_{\phi}$ for every $a\in A_1$ and $x_{b,2}\nmid u_{\phi}$ for every $b\in B_1$. Therefore, $u_{\phi}\not\in \textswab{p}_{A_1,B_1}$ and this is a contradiction, which proves $A=A_1$ and $B=B_1$. Conversely, suppose that $\textswab{p}_{A,B}\in \ass(J(L,[2]))$. If $\bigwedge A\not\leq \bigvee B$, then $ \bigvee B< \bigwedge A\vee\bigvee B$. Thus, there exists a lattice homomorphism $\phi: L\rightarrow [2]$ such that $\phi(\bigvee B)=1$ and $\phi(\bigwedge A\vee \bigvee A)=2$. It implies that $$\forall a\in A, \phi(a)= 2\ {\text{and}} \ \forall b\in B, \phi(b)=1, $$ which follows that $x_{a,1}\nmid u_{\phi}$ for every $a\in A$ and $x_{b,2}\nmid u_{\phi}$ for every $b\in B$. Therefore, $u_{\phi}\not\in \textswab{p}_{A,B}$ and again we get a contradiction. Thus, the statement (1) holds. Now, suppose that $\varnothing \neq A_1\subseteq A$ and $\bigwedge A_1\leq \bigvee B$. By Lemma \ref{E15}, $$ J(L,[2])\subseteq \textswab{p}_{A_1,B}\subseteq \textswab{p}_{A,B}\in \ass(J(L,[2]))=\min(J(L,[2])),$$ and so, $A=A_1$. By a similar argument, we can show that if $\varnothing \neq B_1\subseteq B$ and $\bigwedge A\leq \bigvee B_1$ then $B=B_1$. Thus, the statement (2) holds. \end{proof} \begin{remark}\label{chain[n]}\rm Let $L$ be a distributive lattice, $n>2$ and $\textswab{p}\subset {\bf{k}}[x_{l,m}; \ l\in L, m\in [n]] $ be a monomial prime ideal. If $J(L,[n])\subset \textswab{p}$ then, by the proof of Proposition \ref{basic}, there exist nonempty subsets $A_1,\ldots, A_n$ of $L$ such that $$\textswab{p}=\textswab{p}_{A_1,\ldots,A_n}:=\sum_{i=1}^n(x_{a,i};\ a\in A_i).$$ By a similar argument as the proof of Theorem \ref{E25}, one can see that if $J(L,[n])\subset \textswab{p}_{A_1,\ldots,A_n}$ for some nonempty subsets $A_1,\ldots A_n$ of $L$ then \begin{equation}\label{eqchain[n]}\forall 1\leq i1$, this bound is not sharp. In this case, the minimal prime ideals of $J(L,[2])$ are of height $2$ or $3$ while $m(L)+M(L)=4$. But, if we consider the lattices described in Lemma \ref{ht maximal product of chains} the given bound is sharp. \\ The next corollary is followed by Theorem \ref{E25}. It suggests a recursive method to check if a monomial prime ideal belongs to $\ass(J(L,[2])) $ or not. \begin{corollary}\label{antichain in minimal prime} Let $L$ be a distributive lattice and $A$, $B$ be two nonempty subsets of $L$ where $|A|>1$ or $|B|>1$. Then $\textswab{p}_{A,B}\in \ass(J(L,[2]))$ if and only if the following statements hold. \begin{itemize} \item[{\rm (1)}] Both $A$ and $B$ are antichains in $L$. \item[{\rm (2)}] for every $a\in A$ and every $b\in B$, $a\not\leq b$. \item[{\rm (3)}] If $a,a'$ are two arbitrary distinct elements of $A$ and $A'=(A\setminus \{a,a'\})\cup \{a\wedge a'\}$, then $\textswab{p}_{A',B}\in ass(J(L,[2]))$. \item[{\rm (4)}] If $b,b'$ are two arbitrary distinct elements of $B$ and $B'=(B\setminus \{b,b'\})\cup \{b\vee b'\}$, then $\textswab{p}_{A,B'}\in ass(J(L,[2]))$. \end{itemize} \end{corollary} \begin{proof} If $\textswab{p}_{A,B}\in \ass(J(L,[2]))$, it is easy to see that (1),(2),(3) and (4) hold. Conversely, assume that the monomial prime ideal $\textswab{p}_{A,B}$ satisfies the given conditions. We show that the necessary conditions of Theorem \ref{E25} hold for $A$ and $B$. We suppose that $|A|>1$ (the case $|B|>1$ can be discussed similarly). Choose two distinct elements $a,a'\in A$ and let $A'=A\setminus \{a,a'\}$. By the statement (3), $ \textswab{p}_{A',B}\in\ass(J(L,[2])). $ So, by applying Theorem \ref{E25} for $\textswab{p}_{A',B}$, $$\bigwedge A=\bigwedge A' \leq \bigvee B,$$ and for every $\varnothing\neq B_1\subset B$ we have $\bigwedge A\not\leq \bigvee B_1$. We prove that for every $\varnothing\neq A_1\subset A$, $\bigwedge A_1\not\leq \bigvee B$. By contrary assume that there exists $A_1\subset A$ with $|A_1|=|A|-1$ such that $\bigwedge A_1\leq \bigvee B$. We conclude from the statement (2) that $|A|>2$ or $|B|>1$. Assume that $|A|>2$. Choose two distinct elements $a_1,a_2\in A_1$. Let $A'=(A\setminus \{a_1,a_2\})\cup \{a_1\wedge a_2\}$ and $A'_1=(A_1\setminus\{a_1,a_2\})\cup \{a_1\wedge a_2\}$. It is clear that $\bigwedge A_1=\bigwedge A'_1\leq \bigvee B $. Now by Lemma \ref{E15} and condition (3) we have: $$J(L,[2])\subseteq \textswab{p}_{A'_1,B}\subseteq \textswab{p}_{A',B}\in \ass(J(L,[2]))=\min(J(L,[2])$$ which shows that $\textswab{p}_{A'_1,B}= \textswab{p}_{A',B}$ and so $A'=A'_1$. But, by statement (1), $A'_1\subset A'$ so we get a contradiction. Assume that $|B|>1$. Choose two distinct elements $b,b'\in B$ and let $B'=(B\setminus \{ b,b'\})\cup \{b\vee b'\}$. By the statement (4), $\textswab{p}_{A,B'}\in \ass(J(L,[2]))$. Since $\bigwedge A_1\leq \bigvee B=\bigvee B'$, by applying Theorem \ref{E25} for $\textswab{p}_{A,B'}$, we get a contradiction. \end{proof} The next result is an immediate consequence of Corollary \ref{antichain in minimal prime} and shows that the Alexander dual of $J(L,[2])$ is generated in successive degrees. \begin{corollary}\label{successive degree} Let $L$ be a distributive lattice and $$s=\max\{\deg(u);\ u\in G(J(L,[2]))^{\vee} \}.$$ Then for each $2\leq i\leq s$, $G(J(L,[2]))^{\vee}$ has an element of degree $i$. \end{corollary} In the following we are trying to detect when each element of $\ass(J(L,[2]))$ has height at most $3$. First, note that by Theorem \ref{E25}, we have: \begin{corollary}\label{ht 3 prime} Let $L$ be a distributive lattice. A monomial prime ideal $\textswab{p}\subset S$ of height $3$ is an associated prime of $J(L,[2])$ if and only if $\textswab{p}$ has one of the following shapes: \[\textswab{p}=(x_{a,1}, x_{b_1,2}, x_{b_2,2}), \ {\text{where}} \ a\not\leq b_1, a\not\leq b_2, a\leq b_1\vee b_2\ {\text{and} } \ \{b_1,b_2\}\ {\text{is an antichain}}, \] or \[\textswab{p}=(x_{a_1,1}, x_{a_2,1}, x_{b,2}), \ {\text{where}} \ a_1\not\leq b, a_2\not\leq b, a_1\wedge a_2\leq b\ {\text{and} } \ \{a_1,a_2\}\ {\text{is an antichain}}. \] \end{corollary} To proceed our goal we see that if $L$ has an antichain with $3$ elements, then $J(L,[2])$ has an associated prime of height $4$. For this, we need the following lemmas. \begin{lemma}\label{1} Assume that $\{a,b,c\}$ is an antichain contained in a distributive lattice $L$. Then the following statements hold. \begin{enumerate} \item[{\rm (1)}] If $a\wedge b=a\wedge c $ then $\{a\vee b, b\vee c, a\vee c\}$ is an antichain. \item[{\rm (2)}] If $a\vee b=a\vee c $ then $\{a\wedge b, b\wedge c, a\wedge c\}$ is an antichain. \end{enumerate} \end{lemma} \begin{proof} (1). From $a\wedge b=a\wedge c $, we infer that $$ b =b\vee(a\wedge b) =b\vee(a\wedge c) =(b\vee a)\wedge (b\vee c) $$ %$$a\wedge b=a\wedge c\Longrightarrow b\vee (a\wedge b)=b\vee(a\wedge c)\Longrightarrow b=(b\vee a)\wedge (b\vee c)$$ Now, if $b\vee a$ and $ b\vee c$ are comparable then $b= b\vee a$ or $b=b\vee c$, which shows that $b,c$ are comparable or $b,a$ are comparable which is a contradiction. So $b\vee a$ and $ b\vee c$ are incomparable. Similarly, we can see that $a\vee c$ and $b\vee c$ are incomparable. Now, if $a\vee b\leq a\vee c$ then $a\vee b\vee c=a\vee c$, which shows that $b\vee c\leq a\vee c$ which is again a contradiction. So $\{a\vee b, b\vee c, a\vee b\}$ is an antichain. The proof of (2) is similar. \end{proof} %By a similar argument we can show that %\begin{lemma}\label{2} %Assume that $\{a,b,c\}$ is an antichain contained in a distributive lattice $L$. If $a\vee b=a\vee c $ then $\{a\wedge b, b\wedge c, a\wedge c\}$ is an antichain. %\end{lemma} \begin{lemma}\label{3} Assume that $\{a, b, c\}$ is an antichain contained in a distributive lattice $L$. If $a\wedge b$, $ a\wedge c$ and $ b\wedge c $ are three different elements of $L$ and $a\wedge b < a\wedge c$ then the following statements hold. \begin{enumerate} \item[{\rm (1)}] $a\wedge c$ and $b\wedge c$ are incomparable and $a\wedge b< b\wedge c$. \item[{\rm (2)}] $\{a\vee b, a\vee c,b\vee c\}$ is an antichain or, $a\vee c$ and $b\vee c$ are incomparable, $a\vee b> a\vee c$ and $a\vee b > b\vee c$. \end{enumerate} %If $a\wedge b\neq a\wedge c\neq b\wedge c $ and $a\wedge b < a\wedge c$ then $a\wedge b< b\wedge c$ and, $a\wedge c$ and $b\wedge c$ are incomparable. \end{lemma} %------------ \begin{proof} (1). From $a\wedge b < a\wedge c$, we conclude that $a\wedge b\wedge c=a\wedge b$, which implies that $a\wedge b< b\wedge c$. %$$a\wedge b < a\wedge c\Longrightarrow a\wedge b\wedge c=a\wedge b\Longrightarrow a\wedge b< b\wedge c$$ On the other hand, if for example we have $b\wedge c < a\wedge c$ then $a\wedge b\wedge c= b\wedge c$, which follows that $a\wedge b=b\wedge c$ and this is a contradiction. (2). Note that, by Lemma \ref{1}(2), $a\vee c$, $ b\vee c$ and $ a\vee b$ are three different elements of $L$. Now, if $\{a\vee b, a\vee c,b\vee c\}$ is not an antichain, then similar to the proof of (1), one can see that one of the $a\vee b, a\vee c,b\vee c$ is bigger than the others and the other two elements are incomparable. We show that the case that $a\vee c> b\vee c$, $a\vee c> a\vee b$ and, $b\vee c, a\vee b$ are incomparable does not happen (by a similar argument one can show that the case that $b\vee c> a\vee c, b\vee c > a\vee b $ and, $a\vee c $ and $a\vee b $ are incomparable does not happen). Assume that $a\vee c> b\vee c$ and $a\vee c> a\vee b$. Since $$b\geq (a\vee c)\wedge b\geq (b\vee c)\wedge b=b,$$ we conclude from (1) that $$b=(a\wedge b)\vee (c\wedge b)=c\wedge b\leq c,$$ which is a contradiction. \end{proof} Using the above lemmas, next we show that if $\widthe(L)>2$ then $\ass(J(L,[2]))$ has an element of height $4$. \begin{theorem}\label{antichain of size 3} Assume that $\{a,b,c\}$ is an antichain contained in a distributive lattice $L$. Then $J(L,[2])$ has a minimal prime of height $4$. \end{theorem} \begin{proof} %The proof is completed by the following cases. Case 1: Assume that $\{a\wedge b, a\wedge c, b\wedge c\}$ is an antichain. If we let $A=\{a,b,c\}$ and $B=\{a\wedge b\wedge c\}$, then by Theorem \ref{E25}, $\textswab{p}_{A,B}\in \ass(J(L,[2]))$. Case 2: Assume that $a\wedge b=a\wedge c$. Then, by Lemma \ref{1}(1), $\{a\vee b, a\vee c, b\vee c\}$ is an antichain. If we let $A=\{a\vee b\vee c\}$ and $B=\{a,b,c\}$, then by Theorem \ref{E25}, $\textswab{p}_{A,B}\in \ass(J(L,[2])).$ Case 3: Assume that $a\wedge b$, $ a\wedge c$ and $ b\wedge c $ are three different elements of $L$ and $a\wedge b < a\wedge c$. If $\{a\vee b, a\vee c, b\vee c\}$ is an antichain, then, if we define $A$, $B$ as the case 2, we get $\textswab{p}_{A,B}\in \ass(J(L,[2]))$. If $\{a\vee b, a\vee c, b\vee c\}$ is not an antichain, then, by Lemma \ref{3}(1), $a\wedge c$ and $ b\wedge c$ are incomparable and $a\wedge ba\vee c$, $a\vee b>b\vee c$ and $a\vee c, b\vee c$ are incomparable. So we have $$ (a\wedge c)\vee(b\wedge c) =(a\vee b)\wedge c =(a\vee b\vee c)\wedge c =c $$ and $$ (a\vee c)\wedge(b\vee c) =(a\wedge b)\vee c =(a\wedge b\wedge c)\vee c =c. $$ Thus, if we put $A=\{a\vee c, b\vee c\}$ and $B=\{a\wedge c, b\wedge c\}$, then by Theorem \ref{E25}, $\textswab{p}_{A,B}\in \ass(J(L,[2])).$ \end{proof} We recall that a distributive lattice is {\it planar} if and only if it is a sublattice of a direct product of two chains if and only if no element covers more than two elements (See \cite[page 3]{Q}). The next corollary is an immediate consequence of Theorem \ref{antichain of size 3} and shows that if $J(L,[2])^\vee$ does not have any generator of degree $4$, then $L$ must be a planar lattice of width at most $2$. \begin{corollary}\label{ht3} If $L$ is a distributive lattice and any associated prime of $J(L,[2])$ is of height at most $3$, then $L$ is a planar lattice and $\widthe(L)\leq 2$. \end{corollary} \begin{proof} First note that if $L$ has an antichain with $3$ elements then, by Theorem \ref{antichain of size 3}, $\ass(J(L,[2]))$ has an element of the height $4$. So $\widthe(L)\leq 2$. It is clear that in this case, no element of $L$ covers more than two other elements. So $L$ should be a planar lattice. \end{proof} Note that the converse of Corollary \ref{ht3} does not hold. For example if $L$ is the lattice of Remark \ref{chain[n]}, then $\widthe(L)=2$ and $(x_{a,1},x_{b,1}, x_{c,2},x_{d,2})\in \ass(J(L,[2]))$. \section{The case $J([m]\times [n],[2])$} In this section, we are going to study $J([m]\times [n],[2])$ more carefully. We assume that $m\leq n$ and describe $\ass(J([m]\times [n],[2]))$. We first need the following lemma. \begin{lemma}\label{ht maximal product of chains} Let $L=[m_1]\times \cdots \times [m_\ell]$, where $\ell\geq 2$ and each $m_i$ is at least $3$. Then $$\max\{\deg(u); \ u\in G(J(L,[2])^\vee)\}=2\ell.$$ \end{lemma} \begin{proof} For each $1\leq i\leq \ell$, let ${\bf{a_i}}=(a_{i1},\cdots,a_{i\ell})$, where $a_{ij}=\left\{ \begin{array}{ll} 3, & \hbox{if $j\neq i$} \\ 2, & \hbox{if $j= i$} \end{array} \right.$ and let ${\bf{b_i}}=(b_{i1},\cdots,b_{i\ell})$, where $b_{ij}=\left\{ \begin{array}{ll} 1, & \hbox{if $j\neq i$} \\ 2, & \hbox{if $j= i$} \end{array} \right.$. Using Theorem \ref{E25}, one can easily see that $\textswab{p}_{A,B}\in \ass(J(L,[2]))$ where $A=\{{\bf{a_1}},\ldots,{\bf{a_\ell}}\}$ and $B=\{{\bf{b_1}},\ldots,{\bf{b_\ell}}\}$. Note that $\hte(\textswab{p}_{A,B})=2\ell.$ On the other hand, if ${\bf{a}}=(a_1,\ldots,a_\ell), {\bf{b}}=(b_1,\ldots, b_\ell)\in L$, then ${\bf{a}}$ covers ${\bf{b}}$ if and only if there exists $1\leq i\leq \ell$ such that $a_i=b_i+1$ and for each $j\neq i$, $a_j=b_j$. This shows that $m(L)+M(L)=2\ell$ and the conclusion follows by Theorem \ref{upper bound}. \end{proof} Note that by Lemma \ref{ht maximal product of chains}, each minimal prime of $J([m]\times [n],[2])$ has height at most $4$. \begin{theorem}\label{primary section 4} Let $L=[m]\times [n]$ and $ \textswab{p}$ is a monomial prime ideal of $S$. Then $ \textswab{p}\in \ass(J(L,[2]))$ if and only if one of the following conditions hold: \begin{enumerate} \item[{\rm (1)}] $\textswab{p}=\textswab{p}_\psi$ for some $\psi\in \Hom_\Pos([2],[m]\times[n])$. \item[{\rm (2)}] $\textswab{p}=(x_{(i_1,j_1),1},x_{(i_2,j_2),2}, x_{(i_3,j_3),2})$, where \begin{center} $1\leq i_2