Vol. 7, No. 2, 2014, 6-17
ISSN: 2279-087X (P), 2279-0888(online) Published on 26 August 2014
www.researchmathsci.org
6
Annals of
0-Ideals of a 0-Distributive Lattice
Y. S. PawarDepartment of Mathematics, SPSPM’s SKN Sinhgad College of Engineering At Post Korti, Pandharpur-413304, India
Email: pawar_y_s@yahoo.com
Received 9 June 2014; accepted 30 June 2014
Abstract. Somepropertiesof 0-ideals in 0-distributive lattices and quasi-complemented lattices are derived. Preservation of ideals byan onto homomorphism defined on a 0-distributive lattice is discussed. It is proved that the set of all of0-ideals in a normal lattice forms a sub lattice of all of its ideals but not in general. The set of all ofideals in a 0-distributive lattice forms a 0-distributivelattice under thespecially defined operations on it.
Keywords: 0-distributive lattice; prime ideal; minimal prime ideal; prime filter; maximal filter; 0-ideal; homomorphism; normal lattice; quasi complemented lattice.
AMS Mathematics Subject Classification (2010):06B99
1. Introduction
Cornish [1] introduced the concept of 0-ideal in a distributive lattice and studied their properties with the help of congruence relations. In [3], Sambsadaiva Rao studied prime 0-ideals in distributive lattices. As a generalization of the concept of distributive lattices, 0- distributive lattices are introduced by Varlet [6] and Almost distributive lattices are introduced by Swamy and Rao [5]. Recently, a study of 0-ideals and 0- homomorphisms of an Almost distributive lattice is carried out in [4], In this paper our aim is to study some properties of0-ideals in a 0- distributive lattice and the set of all0-ideals in a 0- distributive lattice.In section 2, we list some basic information on 0- distributive lattices which is needed for the development of this topic. In section 3, we study properties of 0-ideals in 0-distributive lattices.Here we give necessary and sufficient condition for a proper 0-ideal of a 0- distributive lattice to be prime and show that every0-ideal of a bounded 0-distributive lattice is the intersection of all minimal prime ideals containing it. In section 4, we discuss various situations in which image of a 0-ideal is a 0-ideal under lattice homomorphism of 0-ditributive lattices. In section 5, wetalk about the relation between the lattice of all ideals and the lattice of all 0-ideals of a 0- distributive lattice. Some properties of 0-ideals inaquasi complemented lattice are furnished insection 6.
2. Preliminaries
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0 and 0 imply 0. For any filter of define 0 =
| 0, for some }.An ideal in is called 0-ideal if 0( ) for some filter in .For any prime ideal of , define 0 { | 0, forsome
}. Note that for a minimal prime ideal in , 0( ) . For any non empty subset of , the set | 0, forall } is called an annihilator of in . An ideal in is called an annihilator ideal if .An ideal in is called dense in if {0}.An element is said to be dense in L if, 0 .An ideal of is called an - ideal if for each .Let denote the set of all ideals of a bounded lattice . Then ( , , ) is a lattice where and
∪
for any two ideals and of . An ideal of is called a direct factor of , if there exists an ideal of such that and 0 . A 0- distributive latticeis said to be normal if 0 for , .A 0-
distributive lattice is said to be quasi-complemented if for any ,there exists y such that . Note that a 0- distributive lattice is quasi complemented, if for any , there exists y L such that 0 and where denotes the set of all dense elements in L.
3. 0-Ideals
We begin with the following lemma.
Lemma 3.1. In any lattice with 0, we have
(a) For any filter of , 0( ) is a semi ideal in and 0 ≠ 0( ).
(b) If contains a dense element, then 0 , for any filter of .
(c) For a filter of , 0 0 if and only if has a dense element.
(d) For any prime ideal of , 0 is a semi ideal in and 0 0 \ ).
(e) For a proper filter of ,0 is contained in some minimal prime ideal of .
(f) If is a minimal prime ideal of containing 0 , then for any filter of .
Proof: (a) Obviously, for any filter of , 0 is a semi ideal in . Let be a filter of
such that 0 . Select 0 . 0 0 , for some
. As and , 0 and hence0 .
(b) 0 ,obviously. Let 0 and be a dense element in .
0 0 , for some . As 0 , we get 0. Thus 0
and hence .
(c) Assume that there exists a filter in such that 0 0 . But then 0
for some . This shows that has a dense element. Conversely, assume that has a dense element. Then the set of all dense elements in is a filter with 0 0 . Hence the result.
(d) Let be a prime ideal of . Then \ is a filter of . We have
0 0 for some 0 for some \
0 \ .Therefore 0 0 \ .
(e)Let be a proper filter of . Then must be contained in some maximal filter say
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(f) Let be a minimal prime ideal of containing 0 . Assume that . Select . being a minimal prime ideal, there exists y M such that
0. As 0 and we get ; a contradiction. Hence .□
Remarks. (1) In , for any proper filter , 0 .
(2) In , a proper semi ideal 0 contains no dense element .
(3) If is a 0- distributive lattice, then for any filter of , 0 is an ideal in andforany prime
ideal of , 0 is an ideal in . Consider the bounded 0-distributive lattice
0, , , , 1 as shown by the HasseDiagram of Fig.1.The ideal is not a 0- ideal of L. Hence the set Ω of all 0- ideals of is a subset of the set of all ideals of . The ideal 0 is a 0- ideal of which is notprime. The ideals and are prime 0- ideals of .
In general about the 0-ideals of a bounded 0-distributive lattice we have
Theorem 3.2. For any bounded 0 –distributive lattice , the following statements hold.
(a) A proper 0-ideal contains no dense elements.
(b) Every prime 0-ideal in isminimal prime
(c) Every minimal prime ideal in is an 0-ideal.
(d) Every non dense prime ideal in is an 0-ideal.
(e) Every 0-ideal in is an - ideal.
(f) If is a quasi- complemented lattice, then every prime ideal not containing any dense element is a 0- ideal.
Proof. (a) Let a proper 0-ideal contain a dense element say in . As is a 0-
ideal, 0 , for some proper filter in . But then, 0 0 for some . As {0}, we get 0. As 0 , and hence 0
(by Lemma 3.1 (b)). This contradicts the fact that is proper and the result follows.
(b) Let be a prime 0-ideal in . Then 0 for some proper filter in . Select
0 . Hence 0, forsome ..If , then 0 ).Hence
0 .Then by Lemma3.1 (a), 0 ;which is not true. Hence . Therefore is minimal prime.
(c) Let be a minimal prime ideal in L. Then \ is a filter of Since is a minimal
prime ideal in , we get 0 . Hence 0 \ ( by Lemma 3.1( c )).
(d) Let be a non dense prime ideal of . As 0 , there exists 0 .
Hence . Now let . Then 0 and imply .
Thus . From both the inclusions we get . As 0([ )), we get
0 )). Therefore is a 0-ideal of .
(e) Let be a 0-ideal in . Hence there exists a filter in such that 0 .
Let 0 . Then for some .Hence 0 ). This shows that the 0-ideal in is an - ideal.
1
a
0
c b
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(f) Let be a quasi-complemented lattice, be a prime ideal of with . Let . Since is a quasi- complemented lattice, there exists such that
0 and . But then as . Thus 0 \ ) shows that
0 \ . As 0 \ always, we get 0 \ and the result follows.□ Converse of Theorem 3.2( b ) need not be true i.e. every 0-ideal need not be a minimal prime ideal in L. For this consider the 0-distributive lattice represented in Fig.1.{0} is a 0-ideal, but not a prime ideal in L and hence not a minimal prime ideal in L. Necessary and sufficient condition for a proper 0-ideal of a 0- distributive lattice to be
prime is proved in the following theorem.
Theorem 3.3. Let be a proper 0-ideal of a 0- distributive lattice . Then is a prime I
ideal if and only if it contains a prime ideal.
Proof: If is a prime ideal, then obviously it contains a minimal prime ideal.Now
assume that contains a prime ideal but is not prime. Select , such that . As and P is prime, we have , with . Thus
.As is a 0-ideal of , there exists a filter in such that 0 .
Now 0(F) = 0for some . Hence
.ByLemma3.1(a), 0 . Hence ; which is absurd. Hence is prime. being a prime 0-ideal of ,it is minimal prime, by Theorem 3.2(b).Hence the result. □
It is well known that everyideal of a bounded 0-distributive lattice can not be expressed as the intersection of all prime ideals containing it but for 0-ideals of a bounded 0-distributive lattice we have
Theorem 3.4. Every0-ideal of a bounded 0-distributive lattice is the intersection of all
minimal prime ideals containing it.
Proof: Let be a 0-ideal of . Hence there exists a filter in such that 0 .
Define =
∩
{ / is a minimal prime ideal containing }.Clearly, . Suppose ⊂ . Select such that 0 . Hence 0 for each . Fix up any. 0 , for some maximalfilter of . As \ is a minimal
prime ideal, \ \ . Again we know that 0
∪
|}.Hence 0 \ .This in turn shows that \ ; which is absurd. Hence and the result follows.□
Theorem 3.5. Every proper 0-ideal of a bounded 0-distributive lattice is contained in
aminimal prime ideal.
Proof: Let be a prime 0-ideal in . Then 0 for some proper filter in .Clearly,
0 . Let Ψ { | is a filter of such that and
}.Clearly, Ψ and Ψ satisfies the Zorn’s lemma. Let be a maximal element of Ψ. We claim that is a maximal filter in . Suppose is a proper filter of such that
. By maximality of and we get . Select x
.Asx 0 , 0 for some .But then 0 ; a contradiction. Hence is a maximal filter of . being a 0-distributive lattice, is a prime filter. Therefore is a minimal prime ideal of such that \ . □
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Corollary 3.6. A proper filter of a bounded 0-distributive lattice is maximal if and
only if \ is a 0-ideal.
Proof: Let be amaximal filterof . Then \ is aminimal prime idealof and hence a
0-ideal.Conversely, let \ be a 0-ideal. Then \ being a proper 0-ideal, it must be contained in some minimal prime ideal say of ( byTheorem 3.5). Thus \ .□ We have the following simple property of 0- ideals.
Theorem 3.7. Intersection of any two 0-ideals in a 0- distributive lattice is a 0-ideal
of .
Proof: It is enough to prove that for any two filters and of , 0 0
0 . Obviously, 0 0 0 . Let 0 0 (G). But then
0 for some and 0 for some . As L is 0- distributive,
0. As , we get 0 . Thus 0 0 G .
Combining both the inclusions we get 0 0 0 .□
Corollary 3.8. Intersection of any family of 0-ideals in a 0- distributive lattice is a
0-ideal of .
4. Epimorphism and 0-ideals
We begin this section by the following remark.
If is a ring epimorphism, then it is an isomorphism if and only if kernel of is{ 0 }.But it is not true in the case of distributive lattices and hence for 0-distributive lattices also. It may be seen from the following example.
Consider the bounded distributive lattices = {0,a,1} and = 0’, 1’ as shown by the Hasse Diagram of Fig.2. Define f as shown in the figure. Clearly, is onto and ) { x / f (x) = { 0 } , but is not one –one.Now we state a result that we need frequently in this article.
Figure 2:
Result 4.1. Let and be boundedlattices and let be a lattice
epimorphism such that 0 0 and 1 1.Then we have the following: (1) Forany filter of , -1 is a filter of .
(2) Forany filter of , is a filter of .
Notation. Let , , , and , , denote bounded 0-distributive
lattices. Let be a lattice epimorphism such that 0 0 and 1
1.Throughout this article we assume that ) 0 , where ) x / (x) = 0}.
Theorem 4.2. For any filter of , 0 .
Proof: Let 0 ]. Then for some 0 . Now, 0
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0 0. As , we get 0 ]. This shows that, .
Now letx ]. As and is surjective there exists such that
.Now 0 0 for some
0 0 0 0 0 ].
This shows that,0 0 . Combining both the inclusionswe get 0 0 . □
From the proof of Theorem 4.2, it follows that
Corollary 4.3. Forany filter of a bounded lattice , .
Corollary 4.4. For any 0- ideal of , is an 0- ideal of 2..
Proof: Let be a 0- ideal of . Hence there exists a filter in such that
0 . Then by Result 4.1, ) is a filter in .Hence 0 ] 0 ] by Theorem 4.2.This shows that is an 0- ideal in .□
Remark. The condition that 0 is indispensible in Theorem 4.2.
Theorem 4.5. For any filters and of , 0 0 if and only if 0
0 .
Proof: Let and beany filters of such that 0 0 . Then
obviously, 0 0 . By Theorem 4.2,we get 0 0 ].
Conversely,let 0 0 ]. Now, 0 0
0 t O (G)] λ( t) λ(s)= 0 for some 0
0 0 0 . Therefore 0 0 ). Similarly we can show that 0 0 . Therefore 0 0 . □
Theorem 4.6. Forany 0- ideal of L2 ., -1 is an 0-ideal of .
Proof: Let be a 0- ideal of . .Then there exists a filter in such that 0 .
Then by Result 4.1, λ-1 is a filter in . Claim that -1 0 0 -1 ]. Let
0 -1 )].Then 0 for some -1 . Now 0 0
0 and λ( s) G λ( 0( ) -1 0( ) ]. This shows that 0 -1( ) ] -1 0( ) ].
Conversely, let x -10( )]. Then [ 0( ] s 0 and
G (x s) 0 ( ) {0} 0 and -1( ) 0 -1 )].This
The condition that ) {0} is indispensible in Theorem 4.2.For this consider the bounded 0- distributive lattices
= 0, , 1 and = 0’, 1’ as shown by the Hasse Diagram of Fig.3. Define as shown in the figure. Clearly, f is onto and ) 0}.For the filter
, 1 in , we have 0 { 0} 0’ and
0 0 { 0’,1’} { 0’,1’}. Hence 0 0 .
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shows that -1 0( )] 0 -1( )]. Combining both the inclusions we get -1 -10( )] 0 -1( )]. Hence -1 J is a 0-ideal of .□
5. The set Ω of all 0-ideals
Let be a 0-distributive lattice and let Ω denote the poset (Ω, ) of all 0-ideals of .In this article we prove that the poset (Ω, ) need not be a sub lattice of the lattice , ,
of all ideals of in general.But under the condition of normality of the poset(Ω, ) will be a sub lattice of the lattice , , .
Consider the bounded 0-distributive lattice L={0,a,b,c,d,1} as shown by the Hasse Diagram of Fig.4. For the filters
and , 0 0 0, , , , isnot a
0-ideal of . As join of two 0- ideals of a 0-distributive lattice is not a 0-ideal of , the set Ω of all 0-ideals of is not a sub lattice ofthe lattice , , .
In the following theorems we prove some properties of 0- ideals in a normal lattice.
Theorem 5.1. The poset (Ω, ) is a sub lattice of the lattice ) provided is a normal
lattice.
Proof: For any two filters and , 0 0 0 0 0 (see
Theorem 3.7). Hence 0 0 Ω .Now we prove that 0 0 0 .
Obviously, 0 0 0 . Let 0 G). Then 0 for some
. Hence for some and . Hence 0
(since is normal) 0 0 (since
0 ) and 0 .Thus 0 0 0 . Combining both the inclusions we
get 0 0 0 .Hence (Ω, , ) is a sub lattice of the lattice , , .
Theorem 5.2. Let be a normal lattice. Then for any ideal which contains a 0- ideal K,
there exists thelargest 0-ideal containing K and contained in .
Proof: Define β { | is an 0-idal such that }.Clearly β. Let {Ji / i
∆ } be a chain in β. Then ∪{ i / i ∆ is a 0-idal and
∪
I .So by Zorn’s lemma β contains a maximal element, say . We now prove that is unique. Suppose there exists a maximal element 1 in β. Then we have 1 . As is a normal lattice, 1 β (see Theorem 5.1). But then 1 1 ; and hence the uniqueness. Thus in a normal lattice for any ideal which contains a 0- ideal K, there exists a largest 0-ideal containing K and contained in . □We know that {0} is a 0-ideal in a lattice with 0, if it contains a dense element. Hence by Theorem 5.2, it follows that
Corollary 5.3. In a normal lattice , containing dense elements, there exists the largest 0-ideal in .
Corollary 5.4. In a bounded normal lattice , there exists the largest 0-ideal in .
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Theorem 5.5. Let be a normal lattice .If / ∆} is a family of 0-ideals in ,
then is a 0-ideal of .
Proof. As 0 ideal of ,let 0 for some filter of , for each
∆. 0 0 for each ∆ 0 0 . Conversely let
0 . Then 0 for some t . But then 1 2 3 …
for some 1 , ). Hence, …
0 … 0 . As L is a normal
lattice … . x
1 … … where
1 .
Thus 0 , (1 n ). Hence x ) i 0. As L is 0-distributive , ) a i 0. But then ) 0 (as
) 1 2 … ( since L is a normal lattice) 0 1 0 2 … 0 as ( ] 0 ) (1 n ). This shows that 0 . Combining both the inclusions we get 0 0 ; and the result follows.□
Though the poset (Ω, ) need not be a sub lattice of , interestingly we prove that the poset (Ω, ) forms a distributive lattice under the special operations and
defined on it.
Theorem 5.6. Let be a 0-distributive lattice.The poset (Ω, ) forms a distributive lattice on its own.
Proof. Let and be any filters in .
Claim I: The poset (Ω, ) is a lattice.
(1) is an infimum of 0 and 0 in Ω . Let 0 and 0
for some filter in . Hence 0 ) is in Ω. Let 0 . Then 0 and
0 imply 0 for some and 0 for some . As is a 0 -
distributive lattice, 0. But then 0 as . Hence
0 0 ). So 0 is an infimum of 0 and 0(G) in ΩIf we denote the infimum of 0 and 0(G) in Ω by 0 0 , then we have 0 0( )
O .
(2) G) is the supermum of 0 and 0 in Ω . Let 0 and 0
0 for some filter in .Let 0( G ). Then 0 for some F and
. As 0 0 , we get 0, for some . But then
0 0 . Hence 0 for some . As , we get
0 . Therefore 0 G ) is the supermom of and 0 G in Ω . If we denote the supremumof 0 and 0 G in Ω by 0 0 G , then we have 0 0 G 0 G). From (1) and (2) we get that the poset ( Ω , ) is a lattice under the binary operations
and defined on it.
Claim II: The lattice ( Ω, , ) is a distributive lattice.
Now x 0 (0 0 ) x 0 (0 )) 0, 0 and
0 for some F, K and G . 0 and 0
(as is 0- distributive ) 0 0 ) and 0 (0 ) (as
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(as k g 0 ) O G ) )
).
) O G ) always, we get
( O G ) ( ) ( ). Hence ( Ω, , ) is a
distributive lattice.□
Corollary 5.7. If a 0-distributive lattice contains dense elements, then the lattice
(Ω, , ) is a bounded, complete distributive lattice .
Proof: The lattice ( Ω , , ) is distributive lattice (by Theorem 5.6). Clearly, {0} and are the bounds of the poset (Ω, ) Let { | ∆} be any family of filters of L. Then 0( ) = 0 i) (see Corollary 3.8). Hence the poset (Ω , ) is a complete lattice. Thus it follows that thelattice (Ω , , ) is a bounded ,complete, distributive lattice. □
Corollary 5.8. For abounded 0-distributive lattice , the lattice (Ω, , ) is a bounded,
complete distributive lattice.
Any two distinct 0-ideals I and J of a lattice are said to be co-maximal if .
Lemma 5.9. Let be a 0- distributive lattice. 0 = for ,
i.e. and are co- maximal.
Proof: As 0 and 0 ( [ ) ), we get , Ω. Now
0 0 ( [ ) ) 0 [ ) ) 0 ) 0 .□
In the following theorem we show that any two distinct prime 0-ideals of a 0-distributive lattice are co- maximal.
Theorem 5.10. Any two distinct prime 0-ideals and ofa 0- distributive lattice are
co-maximal.
Proof. P and Q are distinct prime 0-ideals of and are distinct minimal prime
ideals of .Select \ and \ . As and is minimal there exists
∉
such that 0. Similarly for , there exists∉
such that y0. Now being a prime ideal,
∉
and∉
∉
. Similarly∉
and∉
y
∉
.But then * and (y a]* .Again ( (y a) (y 0 * (y a]* , by Lemma 3.12. As *
(y a]* P Q, we get P Q .
Corollary 5.11. Any two distinct minimal prime ideals of a 0-distributive lattice are
co- maximal.
6. Quasi-complemented lattices
In this article we derive necessary and sufficient conditions for every 0- ideal to be an annihilator ideal in a quasi-complemented lattice.
Theorem 6.1. Let L be a quasi-complemented, bounded lattice. Then for the following
statements, (a) ( b) (c) ( d) ( e) and all these statements are equivalent
in provided = for any 0 - ideal of .
(a) Every 0- ideal is an annihilator ideal.
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(c) Every prime 0- ideal is of the form for some
(d) Every 0- ideal is of the form for some .
(e) Every minimal prime ideal is non dense.
Proof.
(a) ( b). Since everyminimal prime ideal isan0- ideal, the implication follows.
(b) (c).Let be a prime 0- ideal of . Then by Theorem 3.3(b), is a minimal prime ideal of . Hence by assumption, is an annihilator ideal. Thus being non dense, for some . As is quasi-complemented, there exists such
that = and the implication follows.
(c) (d). Suppose the condition (d) does not hold. Hence there exists an 0- ideal I
which cannot be expressed in the form of for some . Define K = { | is a 0- ideal which cannot be expressed in the form of for some }.Then clearly K
. Let { / ∆} be a chain of 0- ideals in K .Then is a 0- ideal of . Suppose if possible i = for some . Now =
for some ∆. Hence since every 0-ideal in is an - ideal. Further as
= , we get as i = , a contradiction . Hence i for any . Thus K. Hence by Zorn’s lemma K contains a maximal element say . Claim that is prime. Let , such that and . As is quasi-complemented, there exist , such that = and = . But then
we have = . Similarly we get
. By maximality of , = and = for some , .
Now, = ] = ] =
] = [ ] [ ] = = =
. If , then = will contradicts the fact that K. Hence . This shows that is a prime ideal. But then is a prime 0- ideal which cannot be expressed in the form of for any ; which is absurd to our assumption .Hence the implication follows.
(d) (e).Let be aminimal prime ideal in .We know that every minimal
prime ideal in L is a 0- ideal. Hence by condition (d) P = for some .Thus = = . If is dense , then must be a dense element contained in proper 0- ideal , which is absurd. Hence is non dense. Thus we have proved that
(a) ( b) (c) ( d) ( e).
Now we prove that (e ) ( a) provided for any 0 - ideal of .Let be a 0-ideal of . Then by assumption = . As 1 ,,1 = for some and . Hence = = 1 = (0] . Thus . Select
. Then as *, we get 0. Also 0 0 for some .
Now, 0 ) = shows that . As
always, we get and hence the 0- ideal is an annihilator ideal.
Corollary 6.2. Let be a 0 - distributive lattice containing dense elements.Then (e ) (
a) also holds provided * contains a dense element for any 0 - ideal of .
Proof: By assumption ( *) . Select *) .
0 . *) for some and *. Hence 0. Also
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0. Now ]* ( ]**. ( ]* 0( ) = shows that ** . As ** always, we get ** and hence the 0- ideal is an annihilator ideal.
Remarks 6.3. (1) The condition = need not hold for all 0-ideals of a
bounded0 - distributive lattice . Consider the bounded 0 - distributive lattices as shown by the Hasse Diagrams of Fig.5. and Fig.6.
(i) In a bounded 0-distributive lattice represented in Fig 5, for each 0-ideal of
, holds.
(ii) In a bounded 0-distributive lattice represented in Fig 6, for an 0-ideal {0,a},
.
Figure 5: Figure 6:
(2) Recall that an ideal I of L is adirect factor of if there exists an ideal J in L such that
and 0 . If holds for 0- ideal of , then is a direct factor of .
(3) If every 0-ideal of a bounded0 - distributive lattice is a direct factor of , then holds for all 0- ideals of .
(4) If = holds for all 0- ideals of a bounded 0-distributive lattice , then = holds for all .
(5) If = holds for all 0- ideals of a bounded 0-distributive lattice , then is a normal lattice.
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