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\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}
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\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}