\documentclass[a4paper,11pt]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsthm} \usepackage{graphicx} \usepackage{enumerate} \title{Baer elements in lattice modules} \author{C S Manjarekar,U N Kandale} \date{} \begin{document} \maketitle \begin{abstract} Let L be a compactly generated multiplicative lattice with 1 compact in which every finite product of compact elements is compact and M be a module over L. In this paper we generalize the concepts of Baer elements,${\ast}$-elements and closed elements and obtain the relation between ${\ast}$-elements and Baer elements and also closed elements and Baer elements. Some characterization are also obtain for closed elements of M and minimal prime elements of M. \end{abstract} \textbf{keywords:} Prime element,primary element,lattice modules,Baer element, ${\ast}$-element, closed element. \section{Introduction} A multiplicative lattice L is a complete lattice provided with commutative, associative and join distributive multiplication in which the largest element 1 acts as a multiplicative identity. An element $ a \in L $ is called proper if $ a < 1 $. A proper element p of L is said to be prime if $ ab \leq p $ implies $ a \leq p $ or $ b \leq p $. If $ a \in L$,$ b\in L $ , $(a:b) $ is the join of all elements c in L such that $ cb \leq a $. A proper element p of L is said to be primary if $ ab \leq p $ implies $ a \leq p $ or $ b^{n} \leq p $ for some positive integer n. If $ a \in L$ then $\sqrt{a} = \vee \{x \in L \mid x^{n} \leqslant a, n \in Z_+\} $.An element $a \in L$ is called a radical element if $a = \sqrt{a}$. An element $a \in L$ is called compact if $ a \leqslant \underset{\alpha}{\vee} b_\alpha $ implies $ a \leqslant b_{\alpha 1} \vee b_{\alpha 2} \vee ........ \vee b_{\alpha n}$ for some finite subset $ \{\alpha _1 , \alpha _2 ,.......\alpha _n\} $. Throughout this paper,L denotes a compactly generated multiplicative lattice with 1 compact and every finite product of compact elements is compact. We shall denote by $L_{\ast}$ the set compact elements of L. A nonempty subset F of $L_{\ast}$ is called a filter of $L_{\ast}$ if the following conditions are satisfied, \begin{enumerate}[(i)] \item $x,y\in F \ implies \ xy\in F$ \item $x\in F,x \leqslant y \ implies \ y\in F$. \end{enumerate} Let $F(L_{\ast})$ denote the set of all filters of L. For a nonempty subset $\{{F_{\alpha }}\}\subseteq F(L_{\ast})$,define $\Cup F_{\alpha} = \{x\in L_{\ast} \mid x\geq f_1f_2----f_n\in F_{\alpha_{i}} ,\ for \ some \ i=1,2,---,n\}$. Then it is observed that ,$F(L_{\ast}) = \langle F(L_{\ast}),\Cup,\cap\rangle $ is a complete distributive lattice with $ \Cup $ as the supremum and the set theroretic $\bigcap $ as the infimum. For $a\in L_{\ast}$ the smallest filter containing a is denoted by [a) and it is given by $[a) = \{x\in L{\ast}\mid x\geq a^{n} \ for \ some \ nonnegative \ integer \ n\}$. For a filter $F\in F(L_{\ast})$ we denote,$0_{F} = \vee \{x\in L_{\ast}\mid xs = 0,\ for \ s\in F\}$. Let M be a complete lattice and L be a multiplicative lattice. Then M is called L-module or module over L if there is a multiplication between elements of L and M written as $ aB $ where $a \in L$ and $B \in M$ which satisfies the following properties, \begin{enumerate}[(i)] \item $(\underset{\alpha}{\vee}a_\alpha)A = \underset{\alpha}{\vee}a_\alpha A \ \ \forall \ a_\alpha \in L, \ A \in M$ \item $a(\underset{\alpha}{\vee}A_\alpha) = \underset{\alpha}{\vee}aA_\alpha \ \ \forall \ a \in L, \ A_\alpha \in M$ \item $(ab)A = a(bA) \ \ \ \ \forall \ a, b \in L, \ A \in M$ \item $1 B = B$ \item $ 0 B = 0_M, \ for \ all \ a, a_\alpha\ ,b \in L \ and \ A, A_\alpha \in M $, where 1 is the supremum of L and 0 is the infimum of L.We denote by $ 0_M $ and $ I_M $ the least element and the greatest element of M. Elements of L will generally be denoted by $a,b,c,\dots$ and elements of M will generally be denoted by $A,B,C\dots $. \end{enumerate} Let M be a L-module. If $N \in M$ and $a \in L$ then $ (N:a) = \vee \{X \in M \ \mid \ aX \leqslant N \} $. If $ A,B \in M $, then $ (A:B) = \vee \{ x \in L \ \mid \ xB \leqslant A \} $. An L-module M is called a multiplication L-module if for every element $ N \in M $ there exists an element $ a \in L $ such that $ N = aI_M $ see [1]. In this paper a lattice module M will be a multiplication lattice module, which is compactly generated with the largest element $I_{M}$ compact. A proper element N of M is said to be prime if $aX \leqslant N$ implies $X \leqslant N$ or $aI_M \leqslant N$ that is $a \leqslant (N:I_M)$ for every $a \in L \ , \ X \in M$. If N is a prime element of M then $(N:I_M)$ is prime element of L [4]. An element $N < I_M$ in M is said to be primary if $aX \leqslant N$ implies $X \leqslant N$ or $a^n I_M \leqslant N$ that is $a^n \leqslant (N:I_M)$ for some integer n. An element N of M is called a radical element if $(N:I_{M}) = \sqrt{(N:I_{M})}$. If $aN = 0_{M}$ impiles $a = 0 \ or \ N = 0_{M}$ for any $a \in L \ and \ N \in M$ then M is called torsion free L-module. \section{Residuation properties} We state some elementary properties of residuation in the following theorem. \newtheorem{t2}{Theorem}[section] \begin{t2} Let L be a multiplicative lattice and M be a multiplication lattice module over L.For $x,y\in L \ and \ Z,A,B\in M$,where $(0_{M}:I_{M})$ is a radical element. We have the following identities, \begin{enumerate}[(i)] \item $x \leqslant y \ implies \ (0_{M}:y) \leqslant (0_{M}:x) \ and \ 0_{M}:(0_{M}:x) \leqslant 0_{M}:(0_{M}:y)$ \item $x \leqslant 0_{M}:(0_{M}:x)$ \item $0_{M}:[0_{M}:(0_{M}:x)] = (0_{M}:x)$ \item $(0_{M}:x) = (0_{m}:x^{n})$ for every $n\in Z_{+}$ \item $0_{M}:(0_{M}:x) \wedge 0_{M}:(0_{M}:y) = 0_{M}:(0_{M}:xy) = 0_{M}:[0_{M}:(x \wedge y)]$ \item $(0_{M}:a) = 0_{M} \ implies \ (0_{M}:a^{n}) = 0 \ for \ every \ n\in Z_{+}$ \item $x \vee y = 1 \ implies \ (0_{M}:x) \vee (0_{M}:y) = 0_{M}:(x \wedge y) = 0_{M}:xy$ \item For Z in M ,$Z \leqslant 0_{M}:(0_{M}:Z)$ \item $A \leqslant B \ implies \ (0_{M}:B) \leqslant (0_{M}:A)$ \item $0_{M}:[0_{M}:(0_{M}:A)] = 0_{M}:A$ \item $0_{M}:xI_{M} = 0_{M}:x^{n}I_{M}$ for some positive integer n. \end{enumerate} \end{t2} We define,\ \ $0_{FM} = \vee \{X \in M_{\ast} \mid sX = 0_{M} \ for \ some \ s \in F\}$, where $M_{\ast}$ is the set of comapact elements of M. The proofs of the following theorems are simple \begin{t2} Let $F \subseteq L$ be a filter of $F(L {\ast})$ and let X be a compact element of M. Then $X \leqslant 0_{FM}$ if and only if $sX = 0_{M} \ for \ some \ s \in F$. \end{t2} \begin{t2} For $F \in F(L_{\ast}),\ 0_{FM} = \vee \{(0_{M}:x) \mid x\in F\}$. \end{t2} \begin{t2} For $F_{1},F_{2} \in F(L_{\ast})$ \begin{enumerate}[(i)] \item $F_{1} \subseteq F_{2} \ implies \ 0_{F_{1}M} \leqslant 0_{F_{2}M}$. \item $0_{F_{1}M} \wedge 0_{F_{2}M} = 0_{(F_{1} \bigcap F_{2})M}$ \end{enumerate} \end{t2} \section{Baer elements} \newtheorem{d3}{Definition}[section] A study of Baer elements,${\ast}$-elements and closed elements carried out by D D Anderson, C Jayram, P A Phiri [2]. We generalize these concepts for lattice modules. \begin{d3} An element $A \in M$ is said to be Baer element if for $x \in L{\ast} , xI_{M} \leqslant A \ implies \ 0_{M}:(0_{M}:xI_{M}) \leqslant A$. \end{d3} \begin{d3} An element A of M is said to be ${\ast}$-element if $A = 0_{FM}$ for some filter $F \in F(L_{\ast})$ such that zero does not belong to F. \end{d3} \begin{d3} An element A of M is said to be closed element if $A = 0_{M}:(0_{M}:A)$. \end{d3} The next result establishes the relation between closed element and Baer element. \newtheorem{t3}{Theorem}[section] \begin{t3} Every closed element is a Baer element. \end{t3} \begin{proof} Let A be a closed element of M and x be a compact element of $L_{\ast}$ such that $xI_{M} \leqslant A$. Then $0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:(0_{M}:A) = A$ as A is a closed. This shows that A is a Baer element. \end{proof} \begin{d3} An element P of M is called a minimal prime element over $A \in M$ if $A \leqslant P$ and there is no other prime element Q of M such that $A \leqslant Q < P$. \end{d3} The following result gives the characterization of a minimal prime element over an element. \begin{t3} Let a be proper element of L and P be a prime element of M with $aI_{M} \leqslant P$. Then the following statements are equivalent, \begin{enumerate}[(i)] \item P is minimal prime element over $aI_{M}$. \item For each compact element x in L, $xI_{M} \leqslant P$, there is compact element y in L such that $yI_{M} \nleq P$ and $x^{n}yI_{M} \leqslant aI_{M} = A$ for some positive integer n. \end{enumerate} \end{t3} \begin{proof} $(i) \Rightarrow (ii)$ \newline Let P be a minimal prime over $aI_{M}$ and suppose $xI_{M} \leqslant P$. Let $S = \{x^{n}y \mid y \nleqslant (P:I_{M}) \ and \ n \ is \ a \ positive \ integer \}$. It is clear that, S is a multiplicatively closed set. Suppose $x^{n}y \nleqslant aI_{M}$ for any integer n and for any $yI_{M} \nleq P$, where y is compact in L. By separation lemma see[5], there is prime element $(Q:I_{M})$ of L such that $(P:I_{M}) \leqslant (Q:I_{M}) \ and \ t \nleqslant (Q:I_{M}) \ for \ all \ t \in S$. Then we have $(Q:I_{M}) \leqslant (P:I_{M})$ since otherwise $x^{n}(Q:I_{M}) \in S \ and \ x^{n}(Q:I_{M}) \nleqslant (Q:I_{M})$ a contradiction. Hence $(P:I_{M}) = (Q:I_{M})$. It follows that $P=Q$ see[3]. But then for $t \in S ,t \leqslant x \leqslant (P:I_{M}) = (Q:I_{M})$ a contraduction. $(ii) \Rightarrow (i)$ \newline Suppose for any x in L, $xI_{M} \leqslant P$, there is y in L such that $yI_{M} \nleqslant P \ and \ x^nyI_{M} \leqslant aI_{M}$ for some positive integer n. Also suppose that there is a prime element Q of M with $aI_{M} \leqslant Q < P$. Choose, $xI_{M} \leqslant P \ and \ xI_{M} \nleqslant Q$. By hyphothesis, there is a compact element y in L such that $yI_{M} \nleqslant P$ and integer n such that $x^nyI_{M} \leqslant aI_{M} \leqslant Q$. As $xI_{M} \nleq Q, x \nleq (Q:I_{M})$. Since Q is a prime element of M,$(Q:I_{M})$ is also prime element of L see [4]. Hence $x^{n} \nleq (Q:I_{M})$. Thus, $x^n \nleqslant (Q:I_{M}) \ and \ y \nleqslant (Q:I_{M})$ where $(Q:I_{M})$ is a prime element of L, which is a contradiction. \end{proof} In the next result, we prove the important property of a minimal prime element. \begin{t3} Let M be an lattice module.Every minimal prime element of M is a $\ast$-element where $0_{FM}$ is prime element. \end{t3} \begin{proof} Let p be a minimal prime element of M. Define the set $F = \{x \in L_{\ast} \mid xI_{M} \nleq P \}$. We first show that F is a filter of $F(L_{\ast})$. Let x and y be compact element of L such that $x,y \in F$. So $xI_{M} \nleq P \ and \ yI_{M} \nleq P$. As P is prime, $xyI_{M} \nleq P$. This shows that $xy \in F$. Now let $x \in F \ and \ x \leqslant y$. Hence $xI_{M} \nleq P \ implies \ yI_{M} \nleq P$ and $y \in F$. If $0 \in F$ then we have $0I_{M} \nleq P \ that \ is \ 0_{M} \nleq P$ a contradction. Thus $F \in F(L_{\ast}) \ and \ 0 \notin F$. Now we show that $P=0_{FM}$. Let x be a compact element of L such that $xI_{M} \leqslant P$. By theorem (3.2) it follows that there exist a compact element $y \in L$ such that $yI_{M} \nleq P \ and \ x^{n}yI_{M} = 0_{M}$ for some positive integer n. We have $y \in F$ and $x^{n}I_{M} \leqslant 0_{FM}$. As $0_{FM}$ is prime element, so $xI_{M} \leqslant 0_{FM}$ implies $P \leqslant 0_{FM}$. Now let x be a compact element of L such that $xI_{M} \leqslant 0_{FM}$. Then by lemma (2.2), $rxI_{M} = 0_{M}$ for some $r \in F$. So we have $rxI_{M} \leqslant P \ and \ rI_{M} \nleq P$. As P is prime, $xI_{M} \leqslant P$ and $0_{FM} \leqslant P$ which shows that $P=0_{FM}$. Thus every minimal prime element of M is $\ast$-element. \end{proof} The relation between ${\ast}$-element and Baer element is proved in the next result. \begin{t3} Each $\ast$-element of M is a Baer element. \end{t3} \begin{proof} Suppose an element A of M is $\ast$-element. Hence $A=0_{FM}$ for some filter $F \in F(L_{\ast})$ such that $0 \notin F$. Let $x \in L_{\ast}$ such that $xI_{M} \leqslant A$. Then we have $rxI_{M} = 0_{M} \ that \ is \ xI_{M} \leqslant (0_{M}:r)$ for some $r \in F$ by theorem (2.2). Therefore by (i) and (iii) of thereom (2.1) we get $0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:[0_{M}:(0_{M}:r)] = (0_{M}:r)$. Hence by theorem (2.3), $0_{M}:(0_{M}:xI_{M}) \leqslant \underset{s \in F}{\vee}(0_{M}:s) = 0_{FM} = A$. This shows that A is a Baer element. \end{proof} The next result we prove the existence of closed and Baer elements. \begin{t3} Let M be multiplication lattice module.For any $x \in L, (0_{M}:x)$ is both Baer and closed element. \end{t3} \begin{proof} For an element $x \in L_{\ast}$, let $xI_{M} \leqslant (0_{M}:x)$, then $0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:[0_{M}:(0_{M}:x)] = (0_{M}:x)$ by (i) and (iii) of theorem (2.1). Thus $(0_{M}:x)$ is a Bare element. Again from (iii) of theorem (2.1),$(0_{M}:x) = 0_{M}:(0_{M}:(0_{M}:x))$. This show that $(0_{M}:x)$ is a closed element. \end{proof} In the following theorem we prove the characterization of closed element in terms of Baer element. \begin{t3}% theorem(2.14) For $a \in L_{\ast}, aI_{M}$ is closed if and only if $aI_{M}$ is a Baer element. \end{t3} \begin{proof} Let $L_{\ast}$ be the set of all compact element of L and $aI_{M}$ be a Baer element of M. We show that $aI_{M}=0_{M}:(0_{M}:aI_{M})$. As $aI_{M} \leqslant aI_{M}$, we have $[0_{M}:(0_{M}:aI_{M})] \leqslant aI_{M}$. But $aI_{M}(0_{M}:aI_{M}) \leqslant 0_{M}$ implies $aI_{M} \leqslant 0_{M}:(0_{M}:aI_{M})$. Therefore $0_{M}:(0_{M}:aI_{M}) = aI_{M}$. Thus $aI_{M}$ is closed. The converse is proved in theorem (3.1). \end{proof} \begin{t3} For a nonzero compact element a in L, $0_{M}:a = 0_{[a)}$. \end{t3} \begin{proof} We note that $F = [a) = \{z \in L_{\ast} \mid z \geq a^{n} \ for \ some \ n \in Z_{+}\} \in F(L_{\ast})$ and $0_{FM} = \vee \{X \in M_{\ast} \mid sX = 0_{M} \ for \ some \ s \in F \}$. Now let z be compact element of L such that $z \in F\cap \{0\}$. Then $z \in F \ and \ z= 0$. As $z \in F,z \geq a^{n} \ for \ some \ n \in Z_{+}$. Hence $a \leqslant \sqrt{z}=0$ which shows that $a=0$. This contradiction implies that $0 \notin F$. Now we show that $0_{M}:a = 0_{FM}$. As a is a compact element in L,$a \in F$. So we have $0_{M}:a \leqslant 0_{FM} = \vee \{(0_{M}:x) \mid x \in F\}$. Let Z be a compact element in M and $Z \leqslant 0_{FM}$. Then by theorem (2.2) $sZ = 0_{M} \ for \ some \ s \in F$. So $s \geq a^{n} \ for \ some \ n \in Z_{+}$. We note that $0_{M}:a^{n} = 0_{M}:a$. Consequently,we have $a^{n}Z \leqslant sZ = 0_{M}$. This implies that $Z \leqslant (0_{M}:a^{n}) = (0_{M}:a)$ Consequently,$0_{F} \leqslant (0_{M}:a)$ and $(0_{M}:a) = 0_{F}$. \end{proof} The following theorem establishes the property of Baer,closed and ${\ast}$-element. \begin{t3} Suppose L has no divisors of zero then the element $0_{M}$ is always a Baer,closed and ${\ast}$-element whereas $1_{M}$ is Baer and closed. \end{t3} \begin{proof} Let x be a nonzero element of L. From theorem (3.5),for any $x \in L, 0_{M}:x$ is both Baer and closed and by theorem (3.7) for a nonzero compact element x of L, $0_{M}:x = 0_{[x)}$. To show that $0_{M}$ a is Baer element,take $x \in L_{\ast}$ such that $xI_{M} \leqslant 0_{M}$. We have $0_{M}:(0_{M}:xI_{M}) \leqslant O_{M}:(0_{M}:0_{M}) = 0_{M}$. Hence $0_{M}$ is a Baer element. As $0_{M} = 0_{M}:(0_{M}:0_{M})$, $0_{M}$ is closed. Every Baer element is a ${\ast}$-element. To show that $1_{M}$ is a Baer element. Take any $x \in L_{\ast}$ such that $xI_{M} \leqslant 1_{M}$. We have $0_{M}:(0_{M}:xI_{M}) = 0_{M}:[\vee \{a \in L \mid axI_{M} = 0_{M}\}] = 0_{M}:0 = 1_{M}$. So $1_{M}$ is a Baer element. Now $0_{M}:(0_{M}:1_{M}) = 0_{M}:[\vee \{a \in L \mid aI_{M} = 0_{M}\}] = 1_{M}$ and $1_{M}$ is closed. \end{proof} \newtheorem{r3}{Remark}[section] \begin{r3} For defining the ${\ast}$-element, the condition $0 \notin F$ is necessary. \newline Suppose if possible X is a ${\ast}$-element. Hence $X = 0_{FM}$, for some filter F such that $0 \notin F$. Then we have $X = \vee \{(0_{M}:r) \mid r \in F \}$. Now $0_{M}:0 = \vee \{A \in M \mid 0A = 0_{M}\} = 1_{M}$. Thus only $1_{M}$ will be a ${\ast}$-element. Hence, for defining a ${\ast}$-element we take F such that $0 \notin F$. \end{r3} \begin{t3} If $\{A_{\alpha\}_{\alpha}}$ is a family of Baer elements then $\underset{\alpha}{\wedge}A_{\alpha}$ is a Baer element. \end{t3} \begin{proof} Let $x \in L_{\ast} \ such \ that \ xI_{M} \leqslant \underset{\alpha}{\wedge}A_{\alpha}$. Then for each $\alpha, xI_{M} \leqslant A_{\alpha}$. As each $A_{\alpha}$ is a Baer element, $0_{M}:(0_{M}:xI_{M}) \leqslant A_{\alpha}$. Hence $0_{M}:(0_{M}:xI_{M}) \leqslant \underset{\alpha}{\wedge}A_{\alpha}$. Thus $\underset{\alpha}{\wedge}A_{\alpha}$ is a Baer element. \end{proof} The next result we prove the relation between minimal prime element and Baer element. \begin{t3} If A is a meet of minimal prime elements then A is a Baer element. \end{t3} \begin{proof} From theorem (3.3),every minimal prime element of M is a ${\ast}$-element and by theorem (3.4),each ${\ast}$-element of M is a Baer element. From these two results,every minimal prime element is a Baer element. So meet of all minimal prime elements is a Baer element,by theorem (3.9). \end{proof} \begin{t3} If $\{A_{\alpha}\}_{\alpha}$ is a family of closed elements then $\underset{\alpha}{\wedge}A_{\alpha}$ is a closed element. \end{t3} \begin{proof} We have $\underset{\alpha}{\wedge}A_{\alpha} \leqslant A_{\alpha}$ for each ${\alpha}$. As each $A_{\alpha}$ is closed element we have $0_{M}:[0_{M}:(\wedge A_{\alpha})] \leqslant 0_{M}:(0_{M}:A_{\alpha}) =A_{\alpha}$. This gives $0_{M}:[0_{M}:(\underset{\alpha}{\wedge}A_{\alpha})] \leqslant \underset{\alpha}{\wedge}A_{\alpha}$. Now let Z be an element of M such that $Z \leqslant \underset{\alpha}{\wedge}A_{\alpha}$. Then we have $Z \leqslant 0_{M}:(0_{M}:Z) \leqslant 0_{M}:(0_{M}:\underset{\alpha}{\wedge}A_{\alpha})$, by (ix) of theorem (2.1). This gives $\underset{\alpha}{\wedge}A_{\alpha} \leqslant 0_{M}:[0_{M}:(\underset{\alpha}{\wedge}A_{\alpha})]$. Thus we get $0_{M}:[0_{M}:(\underset{\alpha}{\wedge}A_{\alpha})] = \underset{\alpha}{\wedge}A_{\alpha}$. \end{proof} Here is an important property of largest element of M which is compact. \begin{t3} $1_{M}$ is never a ${\ast}$-element where $1_{M}$ is compact and M is torsion free L-module. \end{t3} \begin{proof} Suppose that $1_{M}$ is a ${\ast}$-element. Then there exist some filter $F \in F(L_{\ast})$ such that $1_{M} = 0_{FM}$, where $0 \notin F$. Then as $1_{M}$ is compact and $1_{M}= 0_{FM} = \vee \{(0_{M}:x) \mid x \in F\}$, $1_{M} = (0_{M}:x_{1}) \vee (0_{M}:x_{2}) \vee \dots \vee (0_{M}:x_{n})$ for some $x_{1},x_{2}, \dots ,x_{n} \in F$. Consequently,as $1_{M}$ is closed, $1_{M} = 0_{M}:(0_{M}:1_{M}) = 0_{M}:[0_{M}:((0_{M}:x_{1}) \vee (0_{M}:x_{2}) \vee \dots \vee (0_{M}:x_{n}))] = 0_{M}:[0_{M}:(0_{M}:x_{1}) \wedge 0_{M}:(0_{M}:x_{2}) \wedge \dots \wedge 0_{M}:(0_{M}:x_{n})]$. Therefore $1_{M} = 0_{M}:[0_{M}:(0_{M}:(x_{1}x_{2} \dots x_{n})] = 0_{M}:(x_{1}x_{2} \dots x_{n})$, by (iii) and (v) of theorem (2.1). This implies that $x_{1}x_{2} \dots x_{n} = 0$. Since $x_{1},x_{2},\dots ,x_{n}$ are in F. We have $0 = x_{1}x_{2} \dots x_{n} \in F$. Which is a contraduction as $0 \notin F$. \end{proof} The next result we prove the characterization of a Baer element. \begin{t3} The following statements are equivalent, \begin{enumerate}[(i)] \item An element $A \in M$ is a Baer element. \item For any element $x,y \in L$ such that x is compact $0_{M}:xI_{M} = 0_{M}:yI_{M} \ \ and \ \ xI_{M} \leqslant A \ \ implies \ \ yI_{M} \leqslant A$. \item For any element $x,y \in L_{\ast}, 0_{M}:x = 0_{M}:y \ \ and \ \ xI_{M} \leqslant A \ \ implies \ \ yI_{M} \leqslant A$. \end{enumerate} \end{t3} \begin{proof} $(i) \Rightarrow (ii)$ \newline Assume that, A is a Baer element of M.Let $x,y \in L$ be such that x is compact,$0_{M}:xI_{M} = 0_{M}:yI_{M} \ and \ xI_{M} \leqslant A$. Then by theorem (2.1),$yI_{M} \leqslant 0_{M}:(0_{M}:yI_{M}) = 0_{M}:(0_{M}:xI_{M}) \leqslant A$, since A is a Baer element. $(ii) \Rightarrow (iii)$ \newline Obvious. $(iii) \Rightarrow (i)$ \newline Assume that for any element $x,y \in L_{\ast}, 0_{M}:xI_{M} = 0_{M}:yI_{M}$ and $xI_{M} \leqslant A \ \ implies \ \ yI_{M} \leqslant A$. We show that $A \in M $ is a Baer element. Let $x \in L_{\ast}$ be such that $xI_{M} \leqslant A$. We have $0_{M}:xI_{M} = 0_{M}:[0_{M}:(0_{M}:xI_{M})]$. Hence by (iii), we have $0_{M}:(0_{M}:xI_{M}) \leqslant A$. Hence, A is a Baer element. \end{proof} In the following theorem we prove the relation between Baer element of a lattice module and radical element of a multiplicative lattice. \begin{t3} If A is Baer element of M then $A:I_{M}$ is a radical element. \end{t3} \begin{proof} Let A be Baer element of a lattice module M. We show that $(A:I_{M}) = \sqrt{(A:I_{M})}$. Assume that x is compact element such that $x^{n}I_{M} \leqslant A$ for some positive integer n. We have $0_{M}:xI_{M} = 0_{M}:x^{n}I_{M}$,by (xii)of theorem (2.1) and hence by above theorem $xI_{M} \leqslant A \ \ that \ \ is \ \ x \leqslant (A:I_{M})$. Hence $\sqrt{(A:I_{M})} \leqslant (A:I_{M})$ and we have $\sqrt{(A:I_{M})} = (A:I_{M})$ i.e.$(A:I_{M})$ is a radical element. \end{proof} \begin{t3} If A is a Baer element then every minimal prime element over A is a Baer element. \end{t3} \begin{proof} Let A be a Baer element and P be a minimal prime in M over A. Assume that $0_{M}:x = 0_{M}:z$ for some $x,z \in L$ such that x is compact and $xI_{M} \leqslant P$. There exists a compact element $y \in L$ such that $yI_{M} \nleq P$ and $x^{n}yI_{M} \leqslant A \leqslant P$ for some positive integer n, by theorem (2.10). Note that $0_{M}:yx = (0_{M}:x):y = (0_{M}:x^{n}):y = 0_{M}:x^{n}y = 0_{M}:yx^{n} = 0_{M}:yz$. As A is a Baer element. By theorem (3.13),$xyI_{M} \leqslant A$ implies $yzI_{M} \leqslant A \leqslant P$. Hence $zI_{M} \leqslant P$ as P is prime. So again by theorem (3.13), P is a Baer element. \end{proof} The characterization of minimal prime element of M is proved in the next theorem. \begin{t3} Let L be a lattice module and P be a prime element of M. Then P is a minimal prime element if and only if for $x \in L_{\ast}$, P contains precisely one of $xI_{M} \ and \ 0_{M}:x$. \end{t3} \begin{proof} If part: \newline Assume that for $x \in L_{\ast}$,P contains precisely one of $xI_{M}$ and $0_{M}:x$. First assume that P contains $xI_{M}$. But $0_{M}:x \nleq P$. Therefore there exists a compact element y in L such that $yI_{M} \leqslant 0_{M}:x$ but $yI_{M} \nleq P$. Thus $xyI_{M} \leqslant 0_{M}$. This shows that for each compact element x in L,$xI_{M} \leqslant P$,there exist a compact element y in L such that $yI_{M} \nleq P$ and $xyI_{M} \leqslant 0_{M}$. By theorem (3.2),it follows that P is a minimal prime element of M. Next assume that $0_{M}:x \leqslant P$ but $xI_{M} \nleq P$. Let z be a compact element of L such that $zI_{M} \leqslant (0_{M}:x) \leqslant P$. But $xI_{M} \nleq P$ and $xzI_{M} \leqslant 0_{M}$. Consequently, by theorem (3.2) P is a minimal prime element. Thus the condition is sufficient. \newline Only if part: \newline Assume that P is a minimal prime element of M. Let x be a compact element of L. Suppose if possible $xI_{M} \leqslant P$. Then by lemma (3.2) there exist a compact element y in L such that $yI_{M} \nleq P$ and $x^{n}yI_{M} = 0_{M}$ for some positive integer n. Consequently,$yI_{M} \leqslant 0_{M}:x^{n} = 0_{M}:x$. This implies that $0_{M}:x \nleq P$. Now suppose if possible $xI_{M} \nleq P$ and $0_{M}:x \nleq P$. Then there exist a compact element y in L such that $yI_{M} \leqslant 0_{M}:x$ but $yI_{M} \nleq P$. Hence we have $xyI_{M} \leqslant 0_{M}$ and so $xyI_{M} \leqslant P$. But $xI_{M} \nleq P$ and $yI_{M} \nleq P$ which contradicts the fact that P is prime element of M. This shows that P contains precisely one of $xI_{M}$ and $(0_{M}:x)$. \end{proof} The relation between ${\ast}$-element of M and a minimal prime element over it is established in the next theorem. \begin{t3} If A is a ${\ast}$-element of M then every minimal prime over A is a minimal prime. \end{t3} \begin{proof} Let P be a minimal prime element of M over A. We know by theorem (3.4) and theorem (3.14), a ${\ast}$-element A is a Baer element and $(A:I_{M})$ is a radical element. Let $x \in L_{\ast}$ be such that $xI_{M} \leqslant P$. But P is a minimal prime over A. Then by theorem (2.2) there exists $y \in L_{\ast}$ such that $yI_{M} \nleq P$ and $x^{n}yI_{M} \leqslant A$ i.e.$x^{n}y \leqslant A:I_{M}$. So $x^{n}y^{n} \leqslant A:I_{M}$ i.e. $xy \leqslant \sqrt{(A:I_{M})} = (A:I_{M})$. By hyphothesis, xy is compact and $xyI_{M} \leqslant A = 0_{FM}$,for some filter F of $L_{\ast}$ such that $0 \notin F$. Hence $xyI_{M}d =0_{M}$ for some $d \in F$. We show that there is no compact element x in F such that $xI_{M} \leqslant P$. Suppose there is comapct element z in L such that $zI_{M} \leqslant P$ and $z \in F$. Then by theorem (2.3),$0_{M}:z \leqslant 0_{F} = A \leqslant P$. This contradict the fact that P contains precisely one of $zI_{M}$ and $0_{M}:z$ where $z \in L_{\ast}$. Hence there is no compact element x in F such that $xI_{M} \leqslant P$. This implies that $dI_{M} \nleqslant P$. As P is prime ,$dI_{M} \nleqslant P$ and $yI_{M} \nleqslant P$ implies $ydI_{M} \nleqslant P$. Thus $xydI_{M} = 0_{M} \leqslant P \ \ and \ \ ydI_{M} \nleqslant P$. Therefore by theorem (3.2), P is minimal prime. \end{proof} \begin{r3} By theorem (3.3),we infer that every minimal prime element is a ${\ast}$-element and it is a Baer element. Therefore by theorem (3.17),if A is the meet of all minimal prime elements containing it ,A is a Baer element. \end{r3} Notation : For a family $\{A_{\alpha}\}$ of Baer elements of L we define, $\veebar A_{\alpha} = \vee \{xI_{M},x \in L_{\ast} \mid 0_{M}:(x_{1} \vee x_{2} \dots \vee x_{n})I_{M} \leqslant 0_{M}:xI_{M}$,for some compact elements $x_{j}I_{M} \leqslant {A_{\alpha}}_{j} \ and \ some \ j=1,2,\dots ,n\}$. The important property of a family of Baer elements is established in the next theorem. \begin{t3} If $\{A_{\alpha}\}$ is a family of Baer elements of L,$\veebar A_{\alpha}$ is the smallest Baer element greater than each $A_{\alpha}$. \end{t3} \begin{proof} We first show that $\veebar A_{\alpha}$ is a Baer element greater than each $A_{\alpha}$. Let x be a compact element of L such that $xI_{M} \leqslant \veebar A_{\alpha}$. Then there exist compact elements $x_{1},x_{2},\dots,x_{n}$ such that $0_{M}:(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M} \leqslant 0_{M}:xI_{M}$ and $x_{j}I_{M} \leqslant {A_{\alpha}}_{j} \ j=1,2,\dots,n$. Next we show that $0_{M}:(0_{M}:xI_{M}) \leqslant \veebar A_{\alpha}$. Let z be compact element in L such that $zI_{M} \leqslant 0_{M}:(0_{M}:xI_{M})$. Then $0_{M}:zI_{M} \geq 0_{M}:[0_{M}:(0_{M}:xI_{M})]$. That is $0_{M}:xI_{M} \leqslant 0_{M}:zI_{M}$ (by theorem (2.1),(x) and (xi)). Therefore $0_{M}:(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M} \leqslant 0_{M}:zI_{M}$. This implies that $zI_{M} \leqslant \veebar A_{\alpha}$. Thus $0_{M}:(0_{M}:xI_{M}) \leqslant \veebar A_{\alpha}$. This shows that $\veebar A_{\alpha}$ is a Baer element. Let z be a compact element in L such that $zI_{M} \leqslant A_{\alpha}$ for some $\alpha$. But $0_{M}:zI_{M} \leqslant 0_{M}:zI_{M}$. Thus $zI_{M} \leqslant \veebar A_{\alpha}$. Hence each $A_{\alpha} \leqslant \veebar A_{\alpha}$. Let B be a Baer element such that $A_{\alpha} \leqslant B$ for each $\alpha$ and let x be a compact element in L such that $0_{M}:(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M} \leqslant 0_{M}:xI_{M}$ for some compact elements $x_{j}I_{M} \leqslant {A_{\alpha}}_{j} , j = 1,2,\dots ,n$ so that $xI_{M} \leqslant \veebar A_{\alpha}$. Note that B is a Baer element and the compact element $(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M} \leqslant B$. Hence $0_{M}:[0_{M}:(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M}] \leqslant B$. Again note that $0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:[0_{M}:(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M}]$ and $xI_{M} \leqslant 0_{M}:(0_{M}:xI_{M})$. Therefore $xI_{M} \leqslant B$ and hence $\veebar A_{\alpha} \leqslant B$. Consequently $\veebar A_{\alpha}$ is the smallest Baer element greater than each $A_{\alpha}$. \end{proof} \begin{t3} For any proper element $A \in M$, $\veebar \{0_{M}:(0_{M}:xI_{M}) \mid x \in L_{\ast} \ \ and \ \ xI_{M} \leqslant A\}$ is the smallest Baer element greater than A. \end{t3} \begin{proof} First we show that $0_{M}:(0_{M}:xI_{M})$ is a Baer element i.e. we show that for any $x \in L_{\ast},xI_{M} \leqslant 0_{M}:(0_{M}:xI_{M})$ implies $0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:(0_{M}:xI_{M})$ which holdes obviously. Hence by theorem (3.18),$B = \veebar \{0_{M}:(0_{M}:xI_{M}) \mid x \in L_{\ast} \ \ and \ \ xI_{M} \leqslant A\}$ is the smallest Baer element containing each $0_{M}:(0_{M}:xI_{M}) \ \ for \ \ xI_{M} \leqslant A$. Let a compact element x in L be such that $xI_{M} \leqslant A$. Then we have $xI_{M} \leqslant 0_{M}:(0_{M}:xI_{M}) \leqslant B$. Thus $A \leqslant B$. Let $zI_{M}$ be a Baer element in M such that $A \leqslant zI_{M}$ and let y be compact element in L such that $yI_{M} \leqslant B$. Then $0_{M}:(z_{1} \vee z_{2} \vee \dots \vee z_{n})I_{M} \leqslant 0_{M}:yI_{M}$, for some compact elements $z_{i}I_{M} \leqslant 0_{M}:(0_{M}:x_{i}I_{M})$, where $i = 1,2,\dots ,n$. Thus $0_{M}:x_{i}I_{M} \leqslant 0_{M}:z_{i}I_{M}$ for each i. This gives $0_{M}:(x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M} = 0_{M}:x_{1}I_{M} \wedge 0_{M}:x_{2}I_{M} \wedge \dots 0_{M}:x_{n}I_{M} \leqslant 0_{M}:z_{1}I_{M} \wedge 0_{M}:z_{2}I_{M} \wedge \dots \wedge 0_{M}:z_{n}I_{M}$ = $0_{M}:(z_{1} \vee z_{2} \vee \dots \vee z_{n})I_{M} \leqslant 0_{M}:yI_{M}$. Thus if $x=x_{1} \vee x_{2} \vee \dots \vee x_{n}$ is compact element such that $xI_{M} = (x_{1} \vee x_{2} \vee \dots \vee x_{n})I_{M} \leqslant A \leqslant zI_{M}$, we get $0_{M}:xI_{M} \leqslant 0_{M}:yI_{M}$. As $zI_{M}$ is a Baer element we have $yI_{M} \leqslant 0_{M}:(0_{M}:yI_{M}) \leqslant 0_{M}:(0_{M}:xI_{M}) \leqslant zI_{M}$. Therefore $B \leqslant zI_{M}$. This shows that $\veebar \{0_{M}:(0_{M}:xI_{M}) \mid x \in L_{\ast} \ \ and \ \ xI_{M} \leqslant A\}$ is the smallest Baer element greater than A. \end{proof} Notation : For a family $\{A_{\alpha}\}$ of closed elements of M we define, $A \bigtriangledown B = \vee \{zI_{M},z \in L_{\ast} \mid 0_{M}:(x \vee y)I_{M} \leqslant 0_{M}:zI_{M} \ \ for \ \ some \ \ xI_{M} \leqslant A \ \ and \ \ yI_{M} \leqslant B\}$. Then we have the following important result. The property of closed elements is proved in the next theorem. \begin{t3} If A and B are closed elements of M $A \bigtriangledown B$ is the smallest closed element greater than A as well as B. \end{t3} \begin{proof} We show that $A \bigtriangledown B$ is closed greater than A as well as B. Let $C = A \bigtriangledown B$. We always have $C \leqslant 0_{M}:(0_{M}:C)$ where $C \in M$. Let x be compact element in L such that $xI_{M} \leqslant 0_{M}:(0_{M}:C)$. Then $0_{M}:C \leqslant 0_{M}:xI_{M}$. This implies that $0_{M}:(y \vee z)I_{M} \leqslant 0_{M}:C \leqslant 0_{M}:xI_{M}$ where $y,z \in L_{\ast},yI_{M} \leqslant A \ and \ zI_{M} \leqslant B$. But $yI_{M} \leqslant A \bigtriangledown B , zI_{M} \leqslant A \bigtriangledown B$. Hence $0_{M}:(r \vee s)I_{M} \leqslant 0_{M}:yI_{M} \ and \ 0_{M}:(u \vee v)I_{M} \leqslant 0_{M}:zI_{M}$ where $rI_{M},uI_{M} \leqslant A$ and $sI_{M},vI_{M} \leqslant B$. Therefore $0_{M}:(r \vee s)I_{M} \wedge 0_{M}:(u \vee v)I_{M} \leqslant 0_{M}:yI_{M} \wedge 0_{M}:zI_{M}$. Consequently $0_{M}:(r \vee s \vee u \vee v)I_{M} \leqslant 0_{M}:(y \vee z)I_{M} leqslant 0_{M}:xI_{M}$, where $(r \vee u)I_{M} \leqslant A \ and \ (s \vee v)I_{M} \leqslant B$. This implies that $xI_{M} \leqslant C$. Hence $0_{M}:(0_{M}:C) \leqslant C$. This gives $0_{M}:(0_{M}:C) = C$ and C is closed. As $0_{M};sI_{M} \leqslant 0_{M}:sI_{M}$ for any element s in L, it follows that $A,B \leqslant A \bigtriangledown B$. Suppose that W is closed element such that $A,B \leqslant W$ and let $x \in L_{\ast}$ be such that $0_{M}:(u \vee v)I_{M} \leqslant 0_{M}:xI_{M}$ for some $uI_{M} \leqslant A \ \ and \ \ vI_{M} \leqslant B$. Note that W is a closed element and $(u \vee v)I_{M} \leqslant W$. Hence we have $0_{M}:[0_{M}:(u \vee v)I_{M}] \leqslant 0_{M}:(0_{M}:W) = W$. Again note that $0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:[0_{M}:(u \vee v)I_{M}] \leqslant W$ and $xI_{M} \leqslant 0_{M}:(0_{M}:xI_{M})$. Therefore $xI_{M} \leqslant W$ and hence $A \bigtriangledown B \leqslant W$. Consequently,it proves that $A \bigtriangledown B$ is the smallest closed element greater than A as well as B. \end{proof} \begin{t3} If A and B are closed elements of M then $A \bigtriangledown B = 0_{M}:[0_{M}:(A \vee B)]$. \end{t3} \begin{proof} By theorem (3.20) we have $A \vee B \leqslant A \bigtriangledown B$. Hence $0_{M}:[0_{M}:(A \vee B)] \leqslant A \bigtriangledown B$ as $A \bigtriangledown B$ is closed element. Let $xI_{M} \leqslant A \bigtriangledown B,x \in L_{\ast}$. Then $0_{M}:(u \vee v)I_{M} \leqslant 0_{M}:xI_{M}$,for some $uI_{M} \leqslant A \ \ and \ \ vI_{M} \leqslant B$. Consequently,we have $xI_{M} \leqslant 0_{M}:(0_{M}:xI_{M}) \leqslant 0_{M}:[0_{M}:(u \vee v)I_{M}] \leqslant 0_{M}:[0_{M}:(A \vee B)]$. Hence $A \bigtriangledown B \leqslant (0_{M}:0_{M}:(A \vee B))$. Thus $A \bigtriangledown B = 0_{M}:[0_{M}:(A \vee B)]$. \end{proof} \begin{thebibliography}{9} \bibitem{Callialp} Callialp F, Tekir U, Multiplication lattice modules,Iranian journal of science and technology (2011),309-313. \bibitem{Anderson} Anderson D D, Jayaram C and Phiri P A. Baer lattices. Acta. Sci.Math.(Szeged) ,59(1994), 61-74. \bibitem{John} Johnson J A , a-adic completions of Noetherian lattice modules,1970, Fund. Math., 66, 341 – 371. \bibitem{kh} Khouja-Al Eaman A. Maximal elements and prime elements in Lattice Modules, Damascus university Journal for Basic Sciences, Vol 19, No 2, 2003. \bibitem{thare} Thakare N k, Manjarekar C S, Abstract spectral theory: Multiplicative lattices in which every character is contained in a unique maximal character, Proceedings of the international symposium on Algebra and its applications (1984),Marcel Dekker and Bessel, pp 265-276. \end{thebibliography} \vspace{0.10 in} \footnotesize \textbf{C S Manjarekar} Department of Mathematics Shivaji University Kolhapur csmanjrekar@yahoo.co.in +91 \ 8552 \ 8603 \ 17 \ \ / \ \ +91 \ 9611 \ 6879 \ 25 \vspace{0.10 in} \textbf{U N Kandale} General Engineering Department Sharad Institute of Technology - College of Engineering Yadrav (Ichalkaranji) ujwalabiraje@gmail.com +91 \ 8421 \ 360061 \ \ / \ \ +91 \ 9 \ 6655 \ 9 \ 2207 \end{document}