The M\_{3}[D] construction and n-modularity

0.28 MB·English

by  George Grätzer

THE M [D] CONSTRUCTION AND n-MODULARITY
3
5
0
0 G.GRA¨TZERANDF.WEHRUNG
2

n Abstract. In 1968, E. T. Schmidt introduced the M3[D] construction, an
a extension of the five-element nondistributive lattice M3 by a bounded dis-
J tributivelatticeD,definedasthelatticeofalltripleshx,y,zi∈D3 satisfying

5 x∧y=x∧z=y∧z. Thelattice M3[D]isamodularcongruence-preserving
2 extension ofD.
In this paper, we investigate this construction for an arbitrary lattice L.

] For every n > 0, we exhibit an identity µn such that µ1 is modularity and
M µn+1 isproperlyweaker thanµn. LetMn denotethevarietydefinedbyµn,
thevarietyofn-modular lattices. IfLisn-modular,thenM3[L]isalattice,in
G
fact, acongruence-preserving extension of L;wealsoprove that, inthiscase,
h. IdM3[L]∼=M3[IdL].
t WeprovideanexampleofalatticeLsuchthatM3[L]isnotalattice. This
a examplealsoprovidesanegativesolutiontoaproblemofR.W.Quackenbush:
m IsthetensorproductA⊗BoftwolatticesAandBwithzeroalwaysalattice.
[ WecomplementthisresultbygeneralizingtheM3[L]constructiontoanM4[L]
construction. Thisyields,inparticular,aboundedmodularlatticeLsuchthat

1 M4⊗Lis not a lattice, thus providinga negative solution to Quackenbush’s
v probleminthevarietyMofmodularlattices.
0 Finally, we sharpen a result of R. P. Dilworth: Every finite distributive
3 latticecanberepresentedasthecongruencelatticeofafinite3-modular lattice.
4 WedothisbyverifyingthataconstructionofG.Gra¨tzer,H.Lakser,andE.T.
1 Schmidtyieldsa3-modularlattice.
0
5
0
/
h
1. Introduction
t
a
m E.T.Schmidt[11]and[12]introducedthefollowingconstruction. LetM3 bethe
five-element, modular, nondistributive lattice and let D be a bounded distributive
:
v lattice. ThelatticeM extendedby D,denotedbyM [D],isthelatticeofalltriples
3 3
Xi hx,y,zi∈D3 satisfying x∧y = x∧z = y∧z; we call such triples balanced. Then
M [D]isa(modular)latticeandM [D]andDhaveisomorphiccongruencelattices.
r 3 3
a MeetinM [D]isperformedcomponentwise,whilethejoinisthesmallestbalanced
3
triple in M [D] containing the triple formed by componentwise joins.
3
Note that the elements hx,0,0i, x ∈ D, form a sublattice of M [D] isomorphic
3
to D. We identify hx,0,0i∈M [D] with x∈D, making M [D] an extension of D.
3 3
Let L be a lattice. A lattice K is a congruence-preserving extension of L, if K
is an extension of L and every congruence of L has exactly one extension to K.
Of course, then the congruence lattice of L is isomorphic to the congruence lattice
ofK. E.T.SchmidtprovedthatM [D] is a congruence-preserving extension of D.
3
Date:Sept. 17,1998.
1991 Mathematics Subject Classification. Primary: 06B05,Secondary: 06C05.
Key words and phrases. Lattice,modular,congruence-preserving extension.
TheresearchofthefirstauthorwassupportedbytheNSERCofCanada.
1
2 G.GRA¨TZERANDF.WEHRUNG
This construction plays a central role in a number of papers dealing with con-
gruences of modular lattices, see G. Gr¨atzer and E. T. Schmidt [4] and [5], as two
recent references.
ThispaperstartedwithaproblemproposedinG.Gr¨atzerandE.T.Schmidt[4]:
Does everylattice have a propercongruence-preservingextension? (We solvedthis
problem in G. Gr¨atzer and F. Wehrung [6].) Of course, if the lattice is a bounded
distributive lattice D, then M [D] is such an extension. So two problems were
3
raised:
1. For what classes of lattices C, is M [L] a lattice for L∈C?
3
2. When is M [L] a congruence-preservingextension of L?
3
Surprisingly, in addition to Schmidt’s result (M [D] is a (modular) lattice pro-
3
videdthatDisaboundeddistributivelattice),wecouldonlyfindoneotherrelevant
result in the literature, see R. W. Quackenbush [10]: if L is modular, then M [L]
3
is a lattice.
In Section 2, we define a lattice identity µ , for every n > 0, such that µ is
n 1
equivalenttothemodularidentityandµ isweakerthanµ . LetM denotethe
n+1 n n
varietydefinedbyµ ; wecallM the varietyofn-modular lattices. We provethat
n n
ifLisan-modularlattice,thenM [L]isalattice anditisacongruence-preserving
3
extension of L. In Section 3, we verify that M ⊂ M and M gets very large
n n+1 n
as n gets large: (M |n<ω) generates the variety L.
n
We show, in Section 4, that M [L] is, in general, not a lattice. In Section 5,
3
S
we show how we can remove the condition that L be bounded in the results of
the previous sections. In Section 6, we explain how the two different definitions of
M [D] in the literature canbe reconciledusing tensor products, andwe obtainthe
3
isomorphismM [L]∼=M ⊗L,foranylatticeLwithzerowhichsatisfiesµ forsome
3 3 n
n. Itfollows,then,thattheresultofSection4canbereinterpreted: thereisalattice
LwithzerosuchthatM ⊗Lisnotalattice. Thissolves,inthenegative,aproblem
3
proposedinR.W.Quackenbush[10]: Isthetensorproductoftwolatticeswithzero
always a lattice? In fact, our counterexample consists of two planar lattices. In
Section7,weshowthethereisacounterexampleconsistingoftwomodularlattices,
M and the subspace lattice of any infinite dimensional vector space. There is
4
another result on n-modular lattices in Section 6: IdM [L]∼=M [IdL].
3 3
In Section 8, we prove that every finite distributive lattice can be represented
as the congruence lattice of a finite 3-modular lattice L. Without 3-modularity,
this is a result of R. P. Dilworth. We prove this by verifying that the lattice L
constructed by G. Gr¨atzer, H. Lakser, and E. T. Schmidt [3] to represent D is, in
fact, 3-modular.
The paper concludes with a discussion of some additional results and a list of
open problems in Section 9.
2. The identities
Let L be a lattice. The triple hx,y,zi∈L3 is balanced, if
x∧y =x∧z =y∧z.
We denote byM [L]the setofallbalancedtriples. We regardM [L]asa subposet
3 3
of L3, in fact, a meet-subsemilattice of L3.
Lemma 2.1. Let L be a lattice. Then M [L] is a lattice iff M [L] is a closure
3 3
system in L3.
THE M3[D] CONSTRUCTION AND n-MODULARITY 3
Proof. If M [L] is a closure system and hx ,y ,z i, hx ,y ,z i ∈ M [L], then the
3 0 0 0 1 1 1 3
closureofhx ∨x ,y ∨y ,z ∨z iinM [L]isthejoinofhx ,y ,z iandhx ,y ,z i.
0 1 0 1 0 1 3 0 0 0 1 1 1
Conversely, if M [L] has joins, then the closure of hx,y,zi∈M [L] is
3 3
hx,o,oi∨ho,y,oi∨ho,o,zi,
where o is any element of L contained in x, y, and z. (cid:3)
Let us define the lattice polynomials p , q , and r , for n< ω, in the variables
n n n
x, y, and z:
p =x, q =y, r =z,
0 0 0
p =x∨(y∧z), q =y∨(x∧z), r =z∨(x∧y),
1 1 1
...
p =p ∨(q ∧r ), q =q ∨(p ∧r ), r =r ∨(p ∧q ).
n+1 n n n n+1 n n n n+1 n n n
Let hx,y,zi∈L3. Define, for n>0,
hx,y,zi(n) =hp (x,y,z),q (x,y,z),r (x,y,z)i
n n n
Note that
(1) hx,y,zi≤hx,y,zi(1) ≤···≤hx,y,zi(n) ≤···
Definition 2.2. For n > 0, define the identity µ as p = p . Let M be the
n n n+1 n
lattice variety defined by µ . The lattices in M are called n-modular; lattices
n n
in Mn −Mn−1 are called exactly n-modular or of modularity rank n. A lattice
L∈/ M , for all n<ω, is of of modularity rank ∞.
n
Lemma 2.3. L is an n-modular lattice iff, for all a, b, c ∈ L, ha,b,ci(n) is the
closure of ha,b,ci.
Proof. This statement immediately follows from the definitions. (cid:3)
Corollary 2.4. The following inclusions hold:
M ⊆M ⊆···⊆M ⊆···
1 2 n
Corollary 2.5. For every finite lattice L, there is an integer n>0 such that L is
n-modular.
Proof. Indeed,ifLisfinite,thentheincreasingsequenceha,b,ci(n) mustterminate
in L3, so Lemma 2.3 yields this result. (cid:3)
Corollary 2.6. (M |i<ω) generates L, the variety of all lattices.
i
Proof. Indeed, bySCorollary2.5, (Mi |i<ω) contains all finite lattices and it is
well-known that all finite lattices generate the variety L. (cid:3)
S
Lemma 2.7. M is the variety of modular lattices.
1
Proof. If L is a modular lattice, then computing in FM(3):
p =(x∨(y∧z))∨(y∨(x∧z))∧(z∨(x∧y))=x∨(y∧z)=p .
2 1
Conversely, if L is nonmodular, then it contains a pentagon N = {o,a,b,c,i}
5
(with zero o, unit i, and with b<a) as a sublattice and
p (b,a,c)=b∨(a∧c)=b,
1
p (b,a,c)=(b∨(a∧c))∨((a∨(b∧c))∧(c∨(a∧b))=a,
2
4 G.GRA¨TZERANDF.WEHRUNG
so µ : p =p fails with x=b, y =a, and z =c. (cid:3)
1 1 2
On the other hand, p =p holds in N , so we obtain
2 3 5
Lemma 2.8. The variety N generated by N is 2-modular.
5 5
E. T. Schmidt [11] and [12] proved that M [L] is a modular lattice, if L is
3
distributive. We now prove the converse.
Lemma 2.9. L be a lattice. Then M [L] is a modular lattice iff L is distributive.
3
Proof. So let M [L] be modular. Since M [L] is an extension of L, it follows that
3 3
L is modular. If L is not distributive, then L contains M = {o,a,b,c,i}, the five
3
element modular nondistributive lattice, as a sublattice. Then ho,o,oi, ha,a,ai,
ha,a,ii, hb,c,oi, hi,i,ii ∈ M [L], ha,a,ii∧hb,c,oi = ho,o,oi, ha,a,ai∨hb,c,oi =
3
hi,i,ii, which easily imply that
N ={ho,o,oi,ha,a,ai,ha,a,ii,hb,c,oi,hi,i,ii}
5
is the five-element nonmodular lattice, a sublattice of M [L], a contradiction. (cid:3)
3
The following lemma is due to E. T. Schmidt [11], for n = 1, and to R. W.
Quackenbush[10], for n=2:
Lemma 2.10. Let n > 0 and let L ∈ M be a lattice. Then M [L] is a lattice.
n 3
Furthermore, if L is bounded, then M [L] has a spanning M .
3 3
Proof. By Lemma 2.1, we have to prove that M [L] is a closure system. For
3
hx,y,zi∈M [L], define
3
(2) hx,y,zi=hp (x,y,z),q (x,y,z),r (x,y,z)i.
n n n
′ ′ ′
By (1), hx,y,zi≤hx,y,zi. Since a polynomial is isotone, hx,y,zi≤hx,y ,z i im-
pliesthathx,y,zi≤hx′,y′,z′i. Finally,lethx,y,zi=hx∗,y∗,z∗i;thenp (x,y,z)=
n
p (x,y,z)=p (x∗,y∗,z∗), so hx,y,zi is closed.
n+1 1
The spanning M is {h0,0,0i,h1,0,0i,h0,1,0i,h0,0,1i,h1,1,1i}. (cid:3)
3
Theorem 1. Let n > 0 and let L be a bounded n-modular lattice. Then M [L] is
3
a lattice with a spanning M . The map
3
ε: x7→hx,0,0i
embeds L into M [L]. If we identify x ∈ L with xε = hx,0,0i ∈ M [L], then the
3 3
lattice M [L] is a congruence-preserving extension of L.
3
Proof. By Lemma 2.10, M [L]is a lattice. Furthermore,ε is, obviously,anembed-
3
ding; we identify x∈L with xε=hx,0,0i. Now let Θ be a congruence of L. Form
Θ3, a congruenceofL3, andlet M [Θ]be the restrictionofΘ3 to M [L]. We claim
3 3
that M [Θ] is the unique extension of the congruence Θ to M [L]. Since M [Θ]
3 3 3
restricted to L equals Θ, it is sufficient to prove the following two statements:
(i) M [Θ] is a congruence of M [L].
3 3
(ii) Every congruence Φ of M [L] is of the form M [Θ], for some congruence
3 3
Θ of L.
Re: (i). M [Θ] is obviously a meet-congruence on M [L]. It remains to prove
3 3
the join substitution property. So let hx ,y ,z i ≡ hx ,y ,z i (M [Θ]) and let
0 0 0 1 1 1 3
hu,v,wi ∈ M [L]. Then x ≡ x (Θ) and so x ∨ u ≡ x ∨u (Θ). Similarly,
3 0 1 0 1
THE M3[D] CONSTRUCTION AND n-MODULARITY 5
y ∨v ≡y ∨v (Θ)andz ∨w ≡z ∨w (Θ). Sinceapolynomialhasthesubstitution
0 1 0 1
property, we conclude that
p (x ∨u,y ∨v,z ∨w)≡p (x ∨u,y ∨v,z ∨w) (Θ),
n 0 0 0 n 1 1 1
and similarly for q and r . Thus
n n
hp (x ∨u,y ∨v,z ∨w),q (x ∨u,y ∨v,z ∨w),
n 0 0 0 n 0 0 0
r (x ∨u,y ∨v,z ∨w)i
n 0 0 0
≡hp (x ∨u,y ∨v,z ∨w),q (x ∨u,y ∨v,z ∨w),
n 1 1 1 n 1 1 1
r (x ∨u,y ∨v,z ∨w)i
n 0 0 0
modulo Θ3, andtherefore, modulo M [Θ]. By (2), this lastcongruence is the same
3
as
hx ,y ,z i∨hu,v,wi≡hx ,y ,z i∨hu,v,wi (M [Θ]),
0 0 0 1 1 1 3
which was to be proved.
Re: (ii). Let Φ be a congruence of M [L] and let Θ be the restrictionof Φ to L.
3
We want to show that M [Θ]=Φ. Let hx ,y ,z i, hx ,y ,z i∈M [L].
3 0 0 0 1 1 1 3
If hx ,y ,z i ≡ hx ,y ,z i (M [Θ]), then x ≡ x (Θ) in L and so hx ,0,0i ≡
0 0 0 1 1 1 3 0 1 0
hx ,0,0i (M [Θ]). Since M [Θ] and Φ agree on L, we conclude that
1 3 3
(3) hx ,0,0i≡hx ,0,0i (Φ).
0 1
Similarly, hy ,0,0i≡hy ,0,0i (Φ); therefore,
0 1
h0,y ,0i=(hy ,0,0i∨h0,0,1i)∧h0,1,0i
0 0
≡(hy ,0,0i∨h0,0,1i)∧h0,1,0i=h0,y ,0i (Φ)
1 1
that is,
(4) h0,y ,0i≡h0,y ,0i (Φ).
0 1
Similarly,
(5) h0,0,z i≡h0,0,z i (Φ).
0 1
Joiningthethreecongruences(3)–(5),weobtainthathx ,y ,z i≡hx ,y ,z i (Φ).
0 0 0 1 1 1
Conversely, let hx ,y ,z i ≡ hx ,y ,z i (Φ). Meeting with hx ∨ x ,0,0i ∈
0 0 0 1 1 1 0 1
M [L], we derive that hx ,0,0i ≡ hx ,0,0i (Φ) and so x ≡ x (Θ). Similarly,
3 0 1 0 1
h0,y ,0i≡h0,y ,0i (Φ). Therefore,
0 1
hy ,y ,1i=h0,y ,0i∨h0,0,1i≡h0,y ,0i∨h0,0,1i=hy ,y ,1i (Φ)
0 0 0 1 1 1
andmeetingwithh1,0,0i∈M [L],weconcludethathy ,0,0i≡hy ,0,0i (Φ),that
3 0 1
is, y ≡y (Θ). Similarly, z ≡z (Θ) and so hx ,y ,z i≡hx ,y ,z i (Θ3), from
0 1 0 1 0 0 0 1 1 1
which it follows that hx ,y ,z i≡hx ,y ,z i (M [Θ]). (cid:3)
0 0 0 1 1 1 3
3. The variety M
n
In this section, we prove that M is properly contained in M , for every
n n+1
n > 0. It is obvious that M ⊆ M . To show that the equality fails, we have
n n+1
to construct, for each n>0, an exactly (n+1)-modular lattice L . The lattice L
n 3
is shown in Figure 1. The definition of L follows the pattern of L , except that
n 3
therearen+1x-s: x , x ,...,x ; therearen+1y-s: y , y ,...,y ; andsothere
0 1 n 0 1 n
are n+1 sublattices of the form C2 in the middle.
2
Since L is planar and bounded, it is a lattice.
n
6 G.GRA¨TZERANDF.WEHRUNG
1
x ∨(y ∧z )=x y z =z ∨(x ∧y )
2 0 2 3 0 3 2 2 0
x ∨(y ∧z )=x z =z ∨(x ∧y )
1 0 1 2 2 1 1 0
x ∨(y ∧z )=x z =z ∨(x ∧y )
0 0 0 1 1 0 0 0
L x z
3 0 0
0
Figure 1
Theorem 2. L is an exactly (n+1)-modular lattice.
n
Proof. It is easy to check that the free lattice on C +C (see Figure VI.1.1 in [2])
2 1
is 2-modular. This shows that µ holds in any lattice at any triple hx,y,zi such
2
that two of the variables x, y and z are comparable. So to check that µ holds
2
in L at hx,y,zi, we can assume that x, y, and z form an antichain; since there
1
are very few antichains of three elements in L , it is very easy to compute that µ
1 2
holds. So L is (exactly) 2-modular.
1
Now we induct on n. The interval [x1∧z1,1] of Ln is isomorphic to Ln−1. So if
x, y, z ∈ [x ∧z ,1], then µ holds at hx,y,zi, therefore, µ holds at hx,y,zi.
1 1 n n+1
If two of x, y, z are in [x ∧z ,1], say, y, z ∈ [x ∧z ,1], then replacing x by
1 1 1 1
x=x∨(y∧z),and, similarly,for y andz,we have allthree elements in[x ∧z ,1]
1 1
and µ holds for x, y, z, so µ holds for x, y, z. Since there is no three element
n n+1
antichainoutside of [x ∧z ,1], we areleft with the casethat twoof x, y,z arenot
1 1
in [x ∧z ,1], say, x and z. We cannot then have x≤x ∧z , because there is no
1 1 1 1
such antichain. Similarly, z (cid:2) x ∧z . Theorefore, by symmetry, we can assume
1 1
that x = x and z = z . It follows that y ∈ [x ∧z ,y ]. So we have p = x and
0 0 1 1 0 1 1
q =y, r =z , all in [x ∧z ,1]. By induction, µ holds for x , y, and z and so
1 1 1 1 1 n 1 1
µ holds for x, y, z.
n+1
It is clear that µ fails in L with the substitution x = x , y = y , and z = z
n n 0 0 0
because p (x ,y ,z )=x <1=p (x ,y ,z ). (cid:3)
n 0 0 0 n n+1 0 0 0
Corollary 3.1. The following proper inclusions hold:
M ⊂M ⊂···⊂M ⊂···
1 2 n
4. M [L] is not always a lattice
3
Inthissection,weconstructaboundedlatticeLsuchthatM [L]isnotalattice.
3
Theorem 3. For the lattice L of Figure 2, M [L] is not a lattice. Moreover, L has
3
modularity rank ∞.
Proof. The reader can easily verify that L is a lattice by exhibiting the join- and
meet-tables;forinstance,x ∧z =c ,x ∨z =1,andsoon. ByLemma2.1,to
i j min(i,j) i j
showthatM [L]isnotalattice,wehavetoverifythatM [L]isnotaclosuresystem.
3 3
We claim that hx ,y ,z i has no closure. So let us assume to the contrary that
0 0 0
THE M3[D] CONSTRUCTION AND n-MODULARITY 7
hx,y,zi is the closure of hx ,y ,z i. Since p (x ,y ,z ) = x , q (x ,y ,z ) = y ,
0 0 0 1 0 0 0 1 1 0 0 0 0
r (x ,y ,z ) = z , so by induction, hx,y,zi must contain all hx ,y ,z i, that is,
1 0 0 0 1 n 0 n
hx ,y ,z i ≤ hx,y,zi, for all n ≥ 0. On the other hand, hu ,y ,v i is balanced,
n 0 n n 0 n
so hx,y,zi ≤ hu ,y ,v i, for all n > 0. But there is no hx,y,zi ∈ L satisfying
n 0 n
hx ,y ,z i≤hx,y,zi≤hu ,y ,v i, for all n>0, so hx,y,zi does not exist.
n 0 n n 0 n
u y v
0 0 0
u v
1 1
L x c z
1 2 1
x c z
0 1 0
c =0
0
Figure 2
If L was n-modular, for some n<ω, then hx ,y ,z i(n) =hx ,y ,z i would be
0 0 0 n 0 n
closed, but it is not. (cid:3)
We shall see in Section 6 that L provides a negative solution to Quackenbush’s
problem, namely, M ⊗L is not a lattice.
3
5. Removing the bounds
Most results of Sections 2–4 remain valid without assuming that the lattice
L has a unit. The only exception is, of course, the statement that M [L] has
3
a spanning M . If we do not assume that L has a unit, then the appropriate
3
statement is that in M [L], for every a∈M [L], there is a i∈M [L] such that (i]
3 3 3
has a spanning M .
3
If we do not assume that L has a zero, the definition of the embedding ε: x 7→
hx,0,0iinTheorem1doesnotmakesense,affectingthe crucialpartaboutcongru-
ence-preserving extensions. So we need to reformulate Theorem 1:
Theorem 4. Let n>0 and let L be an n-modular lattice. Then M [L] is a lattice.
3
The map
ψ: x7→hx,x,xi
embeds L into M [L]. If we identify x ∈ L with xψ = hx,x,xi ∈ M [L], then the
3 3
lattice M [L] is a congruence-preserving extension of L.
3
8 G.GRA¨TZERANDF.WEHRUNG
Proof. The first part of the proof requires little change.
Let Φ be a congruence of M [L] and let Θ be the restriction of Φ to L. We
3
want to show that M [Θ] = Φ. Let hx ,y ,z i, hx ,y ,z i ∈ M [L], and put
3 0 0 0 1 1 1 3
o= (x ∧y |i<3).
i i
Ifhx ,y ,z i≡hx ,y ,z i (M [Θ]),thenx ≡x (Θ)inLandsohx ,x ,x i≡
0 0 0 1 1 1 3 0 1 0 0 0
V
hx ,x ,x i (Φ). Therefore,
1 1 1
hx ,o,oi=hx ,x ,x i∧hx ∨y ,o,oi
0 0 0 0 0 0
≡hy ,y ,y i∧hx ∨y ,o,oi=hy ,o,oi (Θ),
0 0 0 0 0 0
that is,
hx ,o,oi≡hy ,o,oi (Θ).
0 0
Similarly, ho,y ,oi ≡ ho,y ,oi (Φ) and ho,o,z i ≡ ho,o,z i (Φ). Joining the
0 1 0 1
three congruences, we obtain hx ,y ,z i≡hx ,y ,z i (Φ).
0 0 0 1 1 1
The proof of the converse is similar to the original proof with o playing the role
of 0 and i= (x ∨y |i<3) playing the role of 1. (cid:3)
i i
W
6. Two views of M [D]
3
For a finite distributive lattice D, in the literature, M [D] is presented either
3
as the lattice of balanced triples hx,y,zi ∈ D3 (as we presented it in Section 1)
or as the lattice MP, the lattice of isotone maps from P = J(D) (the poset of
3
join-irreducible elements of D) to M . Either approachis convenient; both present
3
a modular lattice with a spanning M with D embedded as the ideal generated by
3
anatomofM andthelatticeisgeneratedbyM andD. Thesecondapproachhas
3 3
the advantage that it yields with no computation that M [D] is a modular lattice.
3
The first approach, however, better lends itself to generalization, as we did it in
this paper.
It follows fromA. Mitchke and R. Wille [9] that the two constructions yield iso-
morphiclattices;indeed,bothconstructionsyieldamodularlatticewithaspanning
M with D embedded as the ideal generated by an atom of M and the lattice is
3 3
generated by M and D and, up to isomorphism, there is only one such lattice.
3
Inthissectionweshallgiveamoredirectexplanationwhythetwoconstructions
yieldisomorphic lattices. To this end, we introduce the concept ofa cappedtensor
product from G. Gr¨atzer and F. Wehrung [7].
Definition6.1. LetAandB be{∨,0}-semilattices. Abi-ideal ofA×B isasubset
I of A×B satisfying the following conditions:
(i) I is hereditary;
(ii) I contains ∇ =(A×{0})∪({0}×B);
A,B
(iii) if ha ,bi, ha ,bi∈I, then ha ∨a ,bi∈I;
0 1 0 1
(iv) if ha,b i, ha,b i∈I, then ha,b ∨b i∈I.
0 1 0 1
For a∈A and b∈B, we define the bi-ideal
a⊗b=∇ ∪{hx,yi∈A×B |hx,yi≤ha,bi}.
A,B
The bi-ideal lattice of A×B is an algebraic lattice. The tensor product A⊗B
is the {∨,0}-subsemilattice of compact elements of the bi-ideal lattice of A×B.
A bi-ideal I is capped, if there is a finite subset C of A×B such that I is the
hereditary subset of A×B generated by C along with ∇ . A tensor product
A,B
THE M3[D] CONSTRUCTION AND n-MODULARITY 9
A⊗B is capped, if all bi-ideals of A×B are capped. A capped tensor product is a
lattice.
For a lattice L with zero, let L− denote the join-subsemilattice L−{0}.
Let A⊗B be a capped tensor product and let I ∈ A⊗B. We define a map
−
ϕ : A →B:
I
For x∈A, x>0, let ϕ (x) be the largest element y in B such that hx,yi∈I.
I
Lemma 6.2. ϕ maps A− into B and
I
ϕ (x )∧ϕ (x )=ϕ (x ∨x ),
I 0 I 1 I 0 1
for x , x ∈A−.
0 1
Proof. First we show that ϕ (x) is defined, for all x ∈ A−. Since I is capped, we
I
canwriteI intheformI = (a ⊗b |i<n)∪∇ ,wherenisanaturalnumber,
i i A,B
a ∈A, b ∈B, for i<n. Now define
i i
S
b , if x≤a ;
i i
y =
i
(0, otherwise,
for i < n and let y = (y | i < n). By definition, hx,y i ∈ I, so by 6.1(iv),
i i
hx,yi∈I. Nowlethx,zi∈I,forsomez ∈B. Thenhx,zi∈a ⊗b ,forsomei<n,
i i
W
and so z ≤ b ≤ y. This proves that y satisfies the requirements in the definition
i
of ϕ (x).
I
Now ϕ (x ∨x ) ≤ ϕ (x ) is obvious, hence, ϕ (x ∨x ) ≤ ϕ (x )∧ϕ (x ).
I 0 1 I 0 I 0 1 I 0 I 1
Conversely, ϕ (x ) ∧ ϕ (x ) ≤ ϕ (x ), so hx ,ϕ (x ) ∧ ϕ (x )i ∈ I; similarly,
I 0 I 1 I 0 0 I 0 I 1
hx ,ϕ (x )∧ϕ (x )i ∈ I, therefore, by 6.1(iii), hx ∨ x ,ϕ (x )∧ϕ (x )i ∈ I,
1 I 0 I 1 0 1 I 0 I 1
implying that ϕ (x )∧ϕ (x )≤ϕ (x ∨x ). (cid:3)
I 0 I 1 I 0 1
Let Bd denote the duallattice of B andlet Hom∨(A−,Bd) denote the lattice of
join-homomorphisms from A− to Bd, ordered componentwise, as a subset of BA−
(not (Bd)A−).
Theorem 5. Let A and B be lattices with zero and let A⊗B be capped. Then
ε: I 7→ϕI defines an isomorphism between A⊗B and Hom∨(A−,Bd).
Proof. Lemma6.2statesthatthemapiswell-defined. Sincehx,yi∈I iffy ≤ϕ (x),
I
it follows that ϕ determines I and so ε is one-to-one. To show that ε is onto, let
I
ϕ ∈ Hom∨(A−,Bd) and define I = {hx,yi ∈ A×B | y ≤ ϕ(x)}. Since ϕ is a
join-homomorphism, it follows that I is a bi-ideal and ϕ=ϕ . (cid:3)
I
Corollary 6.3. Let L be a lattice with zero. Then the following conditions are
equivalent:
(i) M ⊗L is a lattice.
3
(ii) Forallhx,y,zi∈L3,thereexistsn>0suchthathx,y,zi(n) =hx,y,zi(n+1).
Furthermore, if (i) is satisfied, then M ⊗L∼=M [L].
3 3
In particular, if L is n-modular, for some n, then M ⊗ L is a lattice, and
3
M ⊗L∼=M [L].
3 3
Proof. Since M is finite, M ⊗Lis alattice iff M ⊗Lis a cappedtensorproduct,
3 3 3
see Theorem 3 of [8]. Furthermore, in the same theorem, it is stated that this is
equivalent to saying that, for every antitone map ξ: J(M ) → L, the adjustment
3
sequence of ξ terminates after a finite number of steps. Here, J(M ) = {a,b,c},
3
and the ordering on J(M ) is trivial, thus every map from J(M ) to L is antitone.
3 3
10 G.GRA¨TZERANDF.WEHRUNG
Identifyξ: J(M )→Lwiththetriplehξ(a),ξ(b),ξ(c)i. Withthisidentification,the
3
adjustmentsequenceofξ iseasilyseentobe thesequenceofallhξ(a),ξ(b),ξ(c)i(n),
n>0. The equivalence between (i) and (ii) follows.
NowassumethatM ⊗Lisalattice. Again,M ⊗Lisacappedtensorproduct,
3 3
thus, by Theorem 5, M3 ⊗L ∼= Hom∨(M3−,Ld). For all ξ ∈ Hom∨(M3−,Ld), we
can identify ξ with the triple hξ(a),ξ(b),ξ(c)i, which, by Lemma 6.2, is a balanced
triple. The isomorphism M ⊗L∼=M [L] follows. (cid:3)
3 3
NowthesolutionofQuackenbush’sproblem(discussedintheIntroduction)easily
follows:
Corollary 6.4. Let L be the lattice of Theorem 3. Then M ⊗L is not a lattice.
3
Proof. This is obvious by Theorem 3 and Corollary 6.3. (cid:3)
Corollary 6.5. Let L be a lattice with zero. If L is n-modular, for some n, then
(i) M ⊗L∼=M [L].
3 3
If L is a finite distributive lattice and P = J(L), the poset of join-irreducible ele-
ments of D, then
(ii) M ⊗L∼=MP.
3 3
Proof. Part (i) follows immediately from Corollary 6.3.
If L is a finite distributive lattice, then, by Theorem 5, M ⊗L is isomorphic
3
to Hom∨(L−,M3), and any ϕ ∈ Hom∨(L−,M3) can be identified with an isotone
map from P into M . (cid:3)
3
Combining (i) and (ii) of Corollary 6.5, we obtain the desired isomorphism:
Corollary 6.6. Let D be a finite distributive lattice. Then
MP ∼=M [D],
3 3
where P =J(D), the poset of join-irreducible elements of D.
For a given lattice, n-modularity has the following algebraic meaning:
Proposition 6.7. Let L be a lattice with zero. Then the following conditions are
equivalent:
(i) M ⊗Lω is a lattice.
3
(ii) L is n-modular, for some n>0.
Proof.
(ii) implies (i). If L is n-modular,then Lω is n-modular,thus, by Corollary6.3,
M ⊗Lω is a lattice.
3
(i)implies (ii). LetusassumethatthemodularityrankofLis∞. Foralln>0,
there exists, by definition, a triple hx ,y ,z i∈L3 such that
n n n
hx ,y ,z i(n) <hx ,y ,z i(n+1).
n n n n n n
Define the following elements of Lω:
x=hx |n>0i,
n
y =hy |n>0i,
n
z =hz |n>0i.
n