A maximal Boolean sublattice that is not the range of a Banaschewski function

0.17 MB·

by  Samuel Mokriš

A MAXIMAL BOOLEAN SUBLATTICE THAT IS NOT THE
RANGE OF A BANASCHEWSKI FUNCTION
7
1
SAMUELMOKRISˇANDPAVELR˚UZˇICˇKA
0
2
Dedicated to Ja´ra Cimrman on the occasion of his 50th birthday.

n
a
J Abstract. We construct acountable bounded sublattice of the lattice of all
4 subspacesofavectorspacewithtwonon-isomorphicmaximalBooleansublat-
2 tice. WerepresentoneofthemastherangeofaBanschewskifunctionandwe
prove that this isnot the case of the other. Herebywe solve aproblem of F.
] Wehrung.
O
C
.
h
t 1. Introduction
a
m In [14] Friedrich Wehrung defined a Banaschewski function on a bounded com-
[ plementedlatticeLasanantitone(i.e., order-reversing)mapsendingeachelement
of L to its complement, being motivated by the earlier result of Bernhard Ba-
1
naschewski that such a function exists on the lattice of all subspaces of a vector
v
4 space[1]. WehrungextendedBanaschewski’sresultbyprovingthateverycountable
2 complemented modular lattice has a Banaschewski function with a Boolean range
0 andthatallthepossiblerangesofBanaschewskifunctionsonLareisomorphic[14,
7
Corollary 4.8].
0
Still in [14] Wehrung defined a ring-theoretical analogue of Banaschewski func-
.
1 tion that, for a von Neuman regular ring R, is closely connected to the lattice-
0 theoreticalBanaschewskifunctiononthelatticeL(R)ofallfinitelygeneratedright
7
ideals of R. He made use of these ideas to construct a unit-regular ring S (in fact
1
: of bounded index 3) of size ℵ1 with no Banaschewski function [15].
v
Furthermore in [14] Wehrung defined notions of a Banaschewski measure and a
i
X Banaschewski trace on sectionally complemented modular lattices and he proved
r that a sectionally complemented lattice which is either modular with a large 4-
a
frame or Arguesian with a large 3-frame is coordinatizable (i.e. isomorphic to
L(R) for a possibly non-unital von Neumann regular ring R) if and only if it has
a Banaschewski trace. Applying this results, he constructed a non-coordinatizable
sectionally complemented modular lattice, of size ℵ , with a large 4-frame [14,
1
Theorem 7.5].
The aim of our paper is to solve the second problem from [14]:
Date:January26,2017.
2010 Mathematics Subject Classification. 06A15,06C20,06D75.
Key words and phrases. lattice, complemented, modular, Boolean, Banaschewski function,
Schmidt’sconstruction, closureoperator,adjunction.
ThefirstauthorwaspartiallysupportedbytheprojectSVV-2015-260227ofCharlesUniversity
inPrague. ThesecondauthorwaspartiallysupportedbytheGrantAgencyoftheCzechRepublic
underthegrantno. GACR14-15479S.
1
2 S.MOKRISˇANDP.R˚UZˇICˇKA
Problem(Problem2from[14]). Is everymaximalBoolean sublatticeofanat most
countablecomplementedmodularlatticeLtherangeofsomeBanaschewskifunction
on L? Are any two such Boolean sublattices isomorphic?
WeconstructacountablecomplementedmodularlatticeSwithtwonon-isomorphic
maximal Boolean sublattices B and E. We represent E as the range of a Ba-
naschewskifunctiononSandweprovethatB isnottherangeofanyBanaschewski
function. FinallywerepresentthelatticeS asaboundedsublatticeofthesubspace-
lattice of a vector space.
2. Basic concepts
We startwithrecallingsamebasicnotionsaswellastheprecisedefinitionofthe
Banaschewski function adopted from [14]. Next we outline the Schmidt’s M [L]
3
construction, which we then apply to define the bounded modular lattice S con-
taining a pair of non-isomorphic maximal Boolean sublattices.
2.1. Some standard notions, notation, and the Banaschewski function.
A lattice L is bounded if it has both the least element and the greatest element,
denoted by 0L and 1L, respectively. A bounded sublattice of a bounded lattice is
itssublattice containingthe bounds. Givenelementsa,b,cofa latticeLwith zero,
we will use the notation c=a⊕b when a∧b=0L anda∨b=c. A complement of
anelementa ofa bounded lattice L is anelementa′ ofL suchthata⊕a′ =1L. A
latticeLissaidtobecomplemented providedthatitisboundedandeachelementof
Lhasa(notnecessarilyunique)complement. AlatticeLisrelativelycomplemented
ifeachofits intervalis complemented. Note thatarelativelycomplementedlattice
is not necessarily bounded.
We say that a lattice L is uniquely complemented if it is bounded and each
element of L has a unique complement. By a Boolean lattice we mean a lattice
reduct of a Boolean algebra, that is, a distributive uniquely complemented lattice.
For the clarity, let us recallthe formaldefinition of the Banaschewskifunction [14,
Definition 3.1]:
Definition 2.1. Let L be a bounded lattice. A Banaschewski function on L is a
map f: L→L such that both
(1) x≤y implies f(x)≥f(y), for all x,y ∈L, and
(2) f(x)⊕x=1L for all x∈L,
hold true.
2.2. The M [L]-construction. LetL be a lattice. We willcalla triple ha,b,ci∈
3
L3 balanced, if it satisfies
a∧b=a∧c=b∧c
and we denote by M [L] the set of all balanced triples. It is readily seen that
3
M [L] is a meet-subsemilattice of the cartesian product L3. However, it is not
3
necessarily a join-subsemilattice, for one easily observes that the join of balanced
triples may not be balanced. The M [L]-construction was introduced by E. T.
3
Schmidt [12, 13] for a bounded distributive lattices L. He proved [13, Lemma 1]
that in this case M [L] is a bounded modular lattice and that it is a congruence-
3
preservingextensionofthedistributivelatticeL. Thisresultwaslaterextendedby
Gr¨atzer and Schmidt in various directions [2, 3]. In particular, in [2] they proved
A MAX. BOOLEAN SUBLATTICE NOT THE RANGE OF A BANASCH. FUNCTION 3
that every lattice with a non-trivial distributive interval has a proper congruence-
preserving extension. This was further improved by Gr¨atzer and Wehrung in [7],
where they introduced a modification of the M [L]-construction, called M hLi-
3 3
construction. Using this new idea they provedthat everynon-triviallattice admits
a proper congruence-preserving extension.
ThelatticeconstructionsM [L]andM hLiappearedintheseriesofpapersby
3 3
Gr¨atzer and Wehrung [4, 5, 6, 7, 8, 9, 10] dealing with semilattice tensor product
and its related structures, namely the box product and the lattice tensor product
[6, Definition 2.1 and definition 3.3]. Indeed, M ⊠L ≃ M hLi for every lattice
3 3
L and M ⊗L ≃ M [L] whenever L has zero and M ⊗L is a lattice (see [10,
3 3 3
Theorem 6.5] and [5, Corollary 6.3]). In particular, the latter is satisfied when the
latticeLismodularwithzero. Notealso,thatifLisaboundeddistributivelattice
both the constructions M [L] and M hLi coincide. In our paper we get by with
3 3
this simple case.
Let L be a distributive lattice. Given a triple ha,b,ci∈L3, we define
µha,b,ci=(a∧b)∨(a∧c)∨(b∧c)
and we set
(2.1) ha,b,ci=ha∨µha,b,ci,b∨µha,b,ci,c∨µha,b,cii.
UsingthedistributivityofLoneeasilyseesthatha,b,ciistheleastbalancedtriple
≥ha,b,ci in L3 and that the map h−i: L3 →L3 determines a closure operator on
the lattice L3 (see [14, Lemma 2.3] for a refinement of this observation). It is also
clear that
a∨µha,b,ci =a∨(b∧c),
b∨µha,b,ci =b∨(a∧c),
c∨µha,b,ci =c∨(a∧b).
A triple ha,b,ci∈L3 is closed with respect to the closure operatorif and only if it
is balanced. Therefore the set of all balanced triples, denoted by M [L], forms a
3
lattice [14, Lemma 2.1], where
(2.2) ha,b,ci∨ha′,b′,c′i=ha∨a′,b∨b′,c∨c′i
and
′ ′ ′ ′ ′ ′
(2.3) ha,b,ci∧ha,b,ci=ha∧a,b∧b,c∧ci.
By [5, Lemma 2.9] the lattice M [L] is modular if and only if the lattice L is
3
distributive. The “if” part of the equivalence is included in the above mentioned
[13, Lemma 1].
3. The lattice
Fix an infinite cardinal κ. As it is customary, we identify κ with the set of all
ordinalsof cardinalityless than κ. Let us denote by P(κ) the Booleanlattice of all
subsets of κ and set
F(κ):={X ⊆κ|X is finite or κ\X is finite}.
It is well-known that F(κ) is a bounded Boolean sublattice of P(κ). Next, let us
define
T ={hA,B,Ci∈F(κ)3 |C\µhA,B,Ci is finite}.
Lemma 3.1. The set T forms a bounded join-subsemilattice of F(κ)3.
4 S.MOKRISˇANDP.R˚UZˇICˇKA
Proof. Being a lattice polynomial, the map µ: P(κ)3 → P(κ) is monotone. It
follows that for all hA,B,Ci, hA′,B′,C′i∈P(κ), the inclusion
′ ′ ′ ′ ′ ′
µhA∪A,B∪B ,C∪C i⊇µhA,B,Ci∪µhA,B ,C i
holds, whence also
′ ′ ′ ′
C∪C \µhA∪A,B∪B ,C∪C i
′ ′ ′ ′
⊆(cid:0)C∪C (cid:1)\ µhA,B,Ci∪µhA,B ,C i
′ ′ ′ ′
⊆(cid:0)C\µhA(cid:1),B(cid:0),Ci ∪ C \µhA,B ,C i(cid:1).
Thus if both C \ µhA(cid:0),B,Ci and C′(cid:1)\ µ(cid:0)hA′,B′,C′i are fi(cid:1)nite, then (C ∪ C′) \
µhA∪A′,B∪B′,C∪C′iisfiniteaswell. ItfollowsthatT isjoin-subsemilatticeof
F(κ)3. Finally, it is clear that both 0F(κ)3 = h∅,∅,∅i and 1F(κ)3 = hκ,κ,κi belong
to T. (cid:3)
Let S :=T ∩M [F(κ)] denote the set of all balanced triples from T.
3
Lemma 3.2. The join-semilattice T is closed under the h−i operation.
Proof. Let hA,B,Ci∈T. Since F(κ) is a lattice, we have that all A∪µhA,B,Ci,
B∪µhA,B,CiandC∪µhA,B,CibelongtoF(κ). Sincethemapµ: P(κ)3 →P(κ)
is monotone, the inclusion µhA,B,Ci⊆µhA,B,Ci holds. It follows that
C∪µhA,B,Ci \µhA,B,Ci⊆C\µhA,B,Ci,
which is finite due(cid:0) to hA,B,Ci be(cid:1)ing an element of T. (cid:3)
Lemma 3.3. The set S forms a bounded sublattice of the lattice M [F(κ)].
3
Proof. ApplyingLemmas3.1and3.2,wededucethatSisaboundedjoin-subsemilattice
of M [F(κ)]. Therefore, it suffices to verify that S is a meet-subsemilattice of
3
F(κ)3. It is easy to observe that if at least one of hA,B,Ci,hA′,B′,C′i∈P(κ)3 is
balanced, then
′ ′ ′ ′ ′ ′
µhA∩A,B∩B ,C∩C i=µhA,B,Ci∩µhA,B ,C i.
¿From this we we get that if hA,B,Ci,hA′,B′,C′i∈S, then
′ ′ ′ ′
C∩C \µhA∩A,B∩B ,C∩C i
′ ′ ′ ′
=(cid:0)C∩C (cid:1)\ µhA,B,Ci∩µhA,B ,C i
′ ′ ′ ′
⊆(cid:0)C\µhA(cid:1),B(cid:0),Ci ∪ C \µhA,B ,C i(cid:1),
sotheset(C∩C′)\µhA(cid:0)∩A′.B∩B′,C(cid:1)∩C(cid:0)′iisfinite. Thisco(cid:1)ncludestheproof. (cid:3)
As discussed in Section 2, since the lattice F(κ) is distributive, the lattice
M [F(κ)] is modular. Observe that the mapping A 7→ hA,A,Ai embeds F(κ)
3
into S, from which we deduce that
|F(κ)|≤|S|≤|F(κ)3|.
Since the size of both F(κ) and F(κ)3 is κ, we get that |S| = κ. Let us sum up
these observations in the following corollary to Lemma 3.3.
Corollary 3.4. For κ countable infinite, S forms a countable bounded modular
lattice.
Remark 3.5. NotethatunlikeS,thelatticeT isnotameet-subsemillaticeofF(κ)3.
Indeed, both hκ,∅,κi,h∅,κ,κi∈T while hκ,∅,κi∧h∅,κ,κi=h∅,∅,κi∈/ T.
A MAX. BOOLEAN SUBLATTICE NOT THE RANGE OF A BANASCH. FUNCTION 5
4. A Banaschewski function on S
In this section we define a Banaschewski function f: S → S and describe,
element-wise, its range.
Lemma 4.1. The map f: S →S defined by
(4.1)
fhA,B,Ci:=hκ\A,κ\(B∪C),κ\(A∪B∪C)i, for all hA,B,Ci∈S,
is a Banaschewski function on S.
Proof. First we prove that S contains the range of the map f. Observe that if
we put A′ := κ\A and B′ := κ\(B∪C), then fhA,B,Ci = hA′,B′,A′∩B′i.
Since F(κ) is a Boolean lattice, the sets A′,B′ and A′ ∩B′ all belong to F(κ).
Furthermore, we have that
′ ′ ′ ′ ′ ′
A ∩B =µhA,B ,A ∩B i=µfhA,B,Ci.
In particular, A′∩B′\µfhA,B,Ci=∅, whence fhA,B,Ci∈S.
It is clear from (4.1) that the map f is antitone. Finally, we check that
1S =hκ,κ,κi=hA,B,Ci⊕fhA,B,Ci, for all hA,B,Ci∈S.
It follows immediately from the definition of f that
hA,B,Ci∧fhA,B,Ci=h∅,∅,∅i=0S.
To prove that hA,B,Ci∨fhA,B,Ci=1S, let us verify that
(4.2) κ=µhA∪(κ\A),B∪(κ\(B∪C)),C∪(κ\(A∪B∪C))i.
Notethateachelementofκthatisnot containedinC belongstoB∪(κ\(B∪C)).
TogetherwithA∪(κ\A)=κ,wegetthat(4.2)holds,whichconcludestheproof. (cid:3)
Lemma 4.2. Let E denote the range of the Banaschewski function f: S → S.
Then
E ={hA,B,A∩Bi|A,B ∈F(κ)}
and the mapping
(4.3) hA,B,A∩Bi7→hA,Bi
determines an isomorphism from E onto the Boolean lattice F(κ)×F(κ).
Proof. While proving Lemma 4.1, we have observed that
(4.4) E ⊆{hA,B,Ci∈S |C =A∩B}={hA′,B′,A′∩B′i|A′,B′ ∈F(κ)}.
Astraightforwardcomputationgivesthatf(fhA′,B′,A′∩B′i)=hA′,B′,A′∩B′i,
so the lattice E is equal to the right-hand side of (4.4). Finally, it is readily seen
that the correspondence (4.3) determines an isomorphism E →F(κ)×F(κ). (cid:3)
It was noted in [14] that if the range of a Banaschewski function on a lattice
L is Boolean, then it is a maximal Boolean sublattice of L. Thus we derive from
Theorem 4.2 that E is a maximal Boolean sublattice of B.
6 S.MOKRISˇANDP.R˚UZˇICˇKA
5. The counter-example
In the present section, we construct another maximal Boolean sublattice B of
the lattice S. We show that the lattices B and S are not isomorphicand we prove
directly that the lattice B is not the range of any Banaschewskifunction on S.
Lemma5.1. TheassignmenthA,Ci7→ghA,Ci:=hA,A∩C,Cidefinesabounded
lattice embedding g: F(κ)×F(κ) → M [F(κ)]. In particular, the range of g is a
3
bounded Boolean sublattice of M [F(κ)] isomorphic to F(κ)×F(κ).
3
Proof. It is clear from the definition of the map g that it is injective and that its
range is included in M [F(κ)]. Further, for any A,A′,C,C′ ⊆κ, the equality
3
′ ′ ′ ′
ghA,Ci∧ghA,C i=ghA∩A,C∩C i
holds by (2.3), while
′ ′ ′ ′
(5.1) ghA,Ci∨ghA,C i=ghA∪A,C∪C i
canbe easilydeducedfrom(2.1)and(2.2). Finally,observethatghκ,κi=hκ,κ,κi
and gh∅,∅i=h∅,∅,∅i, which concludes the proof. (cid:3)
For any A,C ∈F(κ), we say that hA,Ci is finite if both A andC are finite, and
wesaythathA,Ciisco-finite ifbothκ\Aandκ\C arefinite. LetuswriteA∼C
ifhA,Ciiseitherfiniteorco-finite. NotethattherearepairsA,C ∈F(κ)suchthat
hA,Ci is neither finite nor co-finite; namely, A ∼ C if and only if the symmetric
difference (A\C)∪(C\A) is finite.
Lemma 5.2. The set
A={hA,Ci∈F(κ)|A∼C}
form a bounded Boolean sublattice of F(κ)×F(κ).
Proof. Let hA,Ci, hA′,C′i be a pair of elements from A. If at least one of them is
finite, then hA∩A′,C∩C′i is clearlyfinite as well. If bothhA,Ci andhA′,C′iare
co-finite, then so is hA∩A′,C∩C′i. In either case, hA∩A′,C∩C′i∈A.
If at least one of the pairs hA,Ci,hA′,C′i is co-finite, then hA∪A′,C∪C′i is
co-finite,whileif bothhA,CiandhA′,C′iarefinite, thensois hA∪A′,C∪C′i. In
particular, hA∪A′,C∪C′i∈A whenever hA,Ci,hA′,C′i∈A.
We have shown that A is a sublattice of F(κ)×F(κ). To complete the proof,
observe that h∅,∅i is finite and hκ,κi is co-finite and that the unique complement
in F(κ)×F(κ) of each hA,Ci∈A, namely hκ\A,κ\Ci belongs to A. (cid:3)
Lemma 5.3. The g-image B =g(A) of A is a bounded Boolean sublattice of S.
Proof. Due to Lemma 5.1 and Lemma 5.2, B is a bounded Boolean sublattice of
M [F(κ)]. Thus in view of Lemma 3.3, it suffices to verify that B ⊆ S, that is,
3
that C\(A∩C) is finite for every hA,Ci∈A. This is clear when hA,Ci is finite.
IfhA,Ciisco-finite,thenC\(A∩C)=C\A⊆κ\Aisfinite andwearedone. (cid:3)
Observe that if hA,B,Ci is a balanced triple then B ⊆ A if and only if B =
A∩B =A∩C. It follows that
(5.2) B ={hA,B,Ci∈S |A∼C and B ⊆A}.
A MAX. BOOLEAN SUBLATTICE NOT THE RANGE OF A BANASCH. FUNCTION 7
Lemma 5.4. Let hA,B,Ci ∈ S \ B and let hA′,B′,C′i be a complement of
hA,B,Ci in S. If B ⊆A, then B′ 6⊆A′.
Proof. Since hA,B,Ci 6∈ B and B ⊆ A, it follows from (5.2) that A ≁ C. Hence
exactly one of the two sets A,C is finite. From B ⊆ A and C \B being finite we
conclude that C and κ\A are finite. It follows that the set B =B∩C is finite as
well.
Suppose now that B′ ⊆ A′. Since hA,B,Ci∧hA′,B′,C′i = 0S, we have that
A∩A′ =∅, whence the set A′ ⊆κ\A is finite. A fortiori, the set B′ is also finite
duetotheassumptionthatB′ ⊆A′. AsC′\B′ =C′\(B′∩A′)=C′\µhA′,B′,C′i
is also finite, we conclude that so is C′. But then
′ ′ ′ ′ ′
µhA∪A,B∪B ,C∪C i⊆B∪B ∪C∪C
is a finite set, which contradicts the assumption that hA,B,Ci ∨hA′,B′,C′i =
hκ,κ,κi=1S. (cid:3)
Corollary 5.5. Every complemented bounded sublattice C of S such that B (C
contains an element hA,B,Ci with B 6⊆A.
Proof. LethA,B,Ci∈C\B andlethA′,B′,C′ibeitscomplementinC. Applying
Lemma 5.4, we get that either B 6⊆A or B′ 6⊆A′. (cid:3)
Proposition 5.6. The lattice B is a maximal Boolean sublattice of S.
Proof. LetC beacomplementedboundedsublatticeofS satisfyingB (C. There
is hA,B,Ci ∈ C with B 6⊆ A by Corollary 5.5. We can pick a finite nonempty
F ⊆(B\A). Since the triple hA,B,Ci is balanced,
(5.3) ∅=F ∩A=F ∩B∩A=F ∩B∩C =F ∩C.
Now observe that both ghF,∅i and gh∅,Fi are in B. Applying (5.1) and (5.3), we
get that
(5.4) hA,B,Ci∧ ghF,∅i∨gh∅,Fi =hA,B,Ci∧ghF,Fi=h∅,F,∅i,
while (cid:0) (cid:1)
(5.5) hA,B,Ci∧ghF,∅i ∨ hA,B,Ci∧gh∅,Fi =h∅,∅,∅i.
It follows fro(cid:0)m (5.4) and (5.5) tha(cid:1)t th(cid:0)e lattice C is not di(cid:1)stributive, a fortiori it is
not Boolean. (cid:3)
Proposition 5.7. The sublattice B of S is not the range of any Banaschewski
function on S.
Proof. The range of a Banaschewski function on S must contain a complement of
each element of S. We show that no complement of hκ,∅,∅i in S belongs to B.
Suppose the contrary, that is, that there is hA,B,Ci = ghA,Ci ∈ B satisfying
hκ,∅,∅i⊕hA,B,Ci = 1S. Then A = A∩κ = ∅, and by (5.2) also B = ∅. Then
from B = ∅ and hκ,∅,∅i∨hA,B,Ci = 1S, one infers that C = κ. It follows that
hA,B,Ci∈/ S;indeed,C\µhA,B,Ci=C\∅=κisnotfinite. ThushA,B,Ci6∈B,
which is a contradiction. (cid:3)
Remark 5.8. Note that for the particular case of κ=ℵ , the assertion of Proposi-
0
tion5.7followsfromProposition5.9togetherwith[14,Corollary4.8],whichstates
that the ranges of two Boolean Banaschewski functions on a countable comple-
mented modular lattice are isomorphic.
8 S.MOKRISˇANDP.R˚UZˇICˇKA
Proposition 5.9. The lattices B and E are not isomorphic.
Proof. In B, every finite element ghA,Ci is a join of a finite set of atoms, namely
ghA,Ci= gh{α},∅i ∨ gh∅,{γ}i ,
α∈A ! γ∈C 
_ _
 
and, dually, every co-finite element is a meet of a finite set of co-atoms. On the
other hand, there are elements in F(κ)×F(κ) that areneither finite joins ofatoms
nor finite meets of co-atoms. Recall that in Lemma 4.2, we have observedthat the
lattice E is isomorphic to F(κ)×F(κ). Therefore the lattices B and E are not
isomorphic. (cid:3)
6. Representing S in a subspace-lattice
Although the construction in the three previous sections was performed for an
infinite cardinal κ, the results of the present section on embedding the lattice
M [P(κ)] into Sub(V) (namely Theorem 6.4) work just as well for κ finite. In
3
particular, Proposition 6.5 (an enhancement of [5, Lemma 2.9]) holds for lattices
of any cardinality.
Let F be an arbitrary field and let V denote the vector space over the field
F presented by generators x ,y ,z , α ∈ κ, and relations x + y + z = 0.
α α α α α α
For a subset X of the vector space V we denote by Span(X) the subspace of V
generated by X. Given subspaces of V, say X and Y, we will use the notation
X +Y = Span(X ∪Y). Let Sub(V) denote the lattice of all subspaces of the
vector space V.
For all A,B,C ⊆ κ we put X = Span({x | α ∈ A}), Y = Span({y | β ∈
A α B β
B}), and Z =Span({z |γ ∈C}).
C γ
We define a map F: P(κ)3 →Sub(V) by the correspondence
(6.1) hA,B,Ci7→X +Y +Z .
A B C
Each of the sets {x | α ∈ κ}, {y | β ∈ κ}, and {z | γ ∈ κ} is clearly linearly
α β γ
independent. It follows that XA∪A′ = XA +XA′ for all A,A′ ⊆ κ and, simi-
larly, YB∪B′ = YB +YB′ and ZC∪C′ = ZC +ZC′ for all B,B′,C,C′ ⊆ κ. A
straightforwardcomputation gives the following lemma:
Lemma 6.1. The map F: P(κ)3 →Sub(V) is a bounded join-homomorphism.
Proof. Clearly F h∅,∅,∅i = 0 and F hκ,κ,κi = V. Following the definitions, we
compute F(hA,B,Ci)+F(hA′,B′,C′i)=XA+YB+ZC +XA′ +YB′ +ZC′ =
XA∪A′ +YB∪B′ +ZC∪C′ =F(hA∪A′,B∪B′,C∪C′i). (cid:3)
Let G: Sub(V)→P(κ)3 be a map defined by
W 7→h{α|x ∈W},{β |y ∈W},{γ |z ∈W}i,
α β γ
for all W ∈Sub(V).
ItisstraightforwardthatGisaboundedmeet-homomorphismandthatitisthe
rightadjoint ofF (i.e., replacingthe lattice Sub(V) with its dual, the maps F and
G form a Galois correspondence [11]). Indeed, one readily sees that
F hA,B,Ci⊆W iff hA,B,Ci≤G(W).
The maps F and G induce a closure operator GF on P(κ)3.
A MAX. BOOLEAN SUBLATTICE NOT THE RANGE OF A BANASCH. FUNCTION 9
Lemma6.2. ThecompositionGF: P(κ)3 →P(κ)3 ispreciselytheclosureoperator
h−i on P(κ)3 defined by (2.1).
Proof. We shall prove that GF hA,B,Ci = hA,B,Ci for every hA,B,Ci ∈ P(κ)3.
By symmetry, it suffices to prove that
{α∈κ|x ∈F hA,B,Ci}=A∪(B∩C).
α
Letα∈A∪(B∩C). If α∈A, then x ∈F hA,B,Ci by the definition (6.1), while
α
if α∈B∩C, then x =−y −z ∈F hA,B,Ci by (6.1) andthe defining relations
α α α
of V. It follows that A∪(B∩C)⊆{α∈κ|x ∈F hA,B,Ci}.
α
In order to prove the opposite inclusion, take any ξ ∈ κ \A satisfying x ∈
ξ
FhA,B,Ci; if there is one, there is nothing to prove. We need to show that then
ξ ∈B∩C. Certainly
(6.2) x = a x + b y + c z
ξ α α β β γ γ
α∈A β∈B γ∈C
X X X
forsuitable a , b ,andc ∈Fsuchthatallbutfinitely manyofthemarezero. We
α β γ
seta =0forα∈/ A,b =0forβ ∈/ B,andc =0forγ ∈/ C. Sincez +x +y =0
α β γ γ γ γ
for every γ ∈κ, it follows from (6.2) that
(6.3) x = a x − c x + b y − c y .
ξ α α γ γ β β γ γ
   
α∈A γ∈C β∈B γ∈C
X X X X
IteasilyfollowsfromthedefiningrelationsofV that{x ,y |α∈κ}formsabasis
α α
of V. Thus, applying (6.3) we get that
(6.4) a −c =1 and b −c =0.
ξ ξ ξ ξ
Since by our assumption ξ ∈/ A, we get from (6.2) that a = 0. Substituting to
ξ
(6.4) we get that b = c = −1, hence ξ ∈ B ∩C. This concludes the proof that
ξ ξ
A∪(B∩C)⊇{α∈κ|x ∈FhA,B,Ci}. (cid:3)
α
The next lemma shows that F ↾ M [P(κ)] preserves meets. Note that with
3
Lemma6.1, this meansthat F ↾M [P(κ)]is alattice embedding ofM [P(κ)] into
3 3
the lattice Sub(V).
Lemma 6.3. Let hA,B,Ci,hA′,B′,C′i∈M [P(κ)] be balanced triples. Then
3
′ ′ ′ ′ ′ ′
F hA,B,Ci∩F hA,B ,C i=F hA∩A,B∩B ,C∩C i.
Proof. Since, by Lemma 6.1, F is a join-homomorphism, it is monotone, whence
FhA∩A′,B∩B′,C∩C′i⊆F hA,B,Ci∩F hA′,B′,C′i. Thus itremainsto prove
the opposite inclusion.
Letv ∈F hA,B,Ci∩F hA′,B′,C′ibeanon-zerovector. Thenvcanbeexpressed
as
′ ′ ′
(6.5) v = a x + b y + c z = a x + b y + c z .
α α β β γ γ α α β β γ γ
α∈A β∈B γ∈C α∈A′ β∈B′ γ∈C′
X X X X X X
Consider such an expression of v with
(6.6) |{α|a 6=0}|+|{β |b 6=0}|+|{γ |c 6=0}|
α β γ
minimal possible. Put a = 0 for α ∈/ A, b = 0 for β ∈/ B, and c = 0 for
α β γ
γ ∈/ C. By symmetry, we can assume that a 6= 0 for some α ∈ A. Suppose for a
α
contradiction that α ∈/ A′. Since the triple hA′,B′,C′i is balanced, B′∩C′ ⊆ A′,
10 S.MOKRISˇANDP.R˚UZˇICˇKA
whence α∈/ B′∩C′. Without loss of generality we can assume that α∈/ B′. If all
a ,b ,andc werenon-zero,wecouldreplacec z with−c x −c y andreduce
α α α α α α α α α
thevalueoftheexpressionin(6.6)whichisassumedminimalpossible. Thuseither
b =0 or c =0 (recall that we assume that a 6=0). We will deal with these two
α α α
cases separately. If b =0, then the equality
α
′
(6.7) a x +c z =c z
α α α α α α
must hold true. Since x and z are linearly independent, it follows from (6.7)
α α
that a =0 which contradictsour choice of α. The remaining case is when c =0.
α α
Under this assumption we have that
′
a x +b y =c z .
α α α α α α
It follows that
′ ′ ′
(6.8) a x =c z −b y =−c x −(c +b )y .
α α α α α α α α α α α
Since x and y are linearly independent, we infer from (6.8) that a =−c′ =b .
α α α α α
Then we could reduce the value of (6.6) by replacing a x +b y with c′ z in
α α α α α α
(6.5). This contradicts the minimality of (6.6). (cid:3)
Combining Lemma 6.1, Lemma 6.2, and Lemma 6.3, we conclude:
Theorem 6.4. The restrictions F ↾ M [P(κ)]: M [P(κ)] → Sub(V) and, a for-
3 3
tiory,F ↾S: S →Sub(V)areboundedlatticeembeddings. Inparticular, thelattice
S is isomorphic to a bounded sublattice of the subspace-lattice of a vector space.
It is well-known that a distributive lattice L embeds (via a bounds-preserving
lattice embedding) into the lattice P(κ), where κ is the cardinality of the set of all
maximal ideals of L. Such embedding induces an embedding M [L]֒→M [P(κ)]
3 3
(cf. Lemma 3.3). By Theorem 6.4, the lattice M [P(κ)] embeds into the lattice
3
Sub(V)for a suitable vectorspaceV (note againthat wenow alsoadmitfinite κ).
Since the lattice Sub(V) is Arguesian, so are M [P(κ)] and M [L].
3 3
On the other hand, [5, Lemma 2.9] states that a lattice L is distributive if
and only if M [L] is modular. Hence, if M [L] is modular, it follows that L is
3 3
distributive, and, by the above argument, M [L] is even Arguesian. We have thus
3
proven the following strengthening of [5, Lemma 2.9]:
Proposition 6.5. Let L be a lattice. Then L is distributive iff the lattice M [L] is
3
modular iff M [L] is Arguesian. If this is the case, then M [L] can be embedded
3 3
into the lattice of all subspaces of a suitable vector space over any given field.
References
1. B.Banaschewski,Totalgeordnete moduln, Arch.Math.7(1957), 430–440.
2. G.Gr¨atzer andE.T.Schmidt,A lattice construction and congruence-preserving extensions,
ActaMath.Hungar.66(1995), 275–288.
3. ,On the independence theorem of related structures for modular (arguesian) lattices,
StudiaSci.Math.Hungar.40(2003), 1–12.
4. G.Gr¨atzerandF.Wehrung, Flat semilattices,Colloq.Math.79(1999), 185–191.
5. ,The M3[D]construction and n-modularity,AlgebraUniversalis41(1999), 87–114.
6. ,A new lattice construction: the box product, J.Algebra221(1999), 5893–5919.
7. , Proper congruence-preserving extension of lattices, Acta Math. Hungar. 85 (1999),
169–179.
8. ,Tensor product and transferability of semilattices, Canad. J.Math. 51(1999), 792–
815.