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The M_3[D] construction and n-modularity

George Grätzer, Friedrich Wehrung

To cite this version:

George Grätzer, Friedrich Wehrung. The M_3[D] construction and n-modularity. Algebra Universalis,

Springer Verlag, 1999, 41, no. 2, pp.87-114. �10.1007/s000120050102�. �hal-00004046�

ccsd-00004046, version 1 - 24 Jan 2005

G. GR ¨ATZER AND F. WEHRUNG

Abstract. In 1968, E. T. Schmidt introduced the M3[D] construction, an extension of the five-element nondistributive lattice M3 by a bounded dis- tributive latticeD, defined as the lattice of all tripleshx, y, zi ∈D³ satisfying x∧y=x∧z=y∧z. The latticeM3[D] is a modular congruence-preserving extension ofD.

In this paper, we investigate this construction for an arbitrary latticeL.

For everyn >0, we exhibit an identity µn such thatµ1 is modularity and µn+1is properly weaker thanµn. LetM_ndenote the variety defined byµn, the variety ofn-modular lattices. IfLisn-modular, thenM3[L] is a lattice, in fact, a congruence-preserving extension ofL; we also prove that, in this case, IdM3[L]∼=M3[IdL].

We provide an example of a latticeLsuch thatM3[L] is not a lattice. This example also provides a negative solution to a problem of R. W. Quackenbush:

Is the tensor productA⊗Bof two latticesAandBwith zero always a lattice.

We complement this result by generalizing theM3[L] construction to anM4[L]

construction. This yields, in particular, a bounded modular latticeLsuch that M4⊗Lis not a lattice, thus providing a negative solution to Quackenbush’s problem in the varietyMof modular lattices.

Finally, we sharpen a result of R. P. Dilworth: Every finite distributive lattice can be represented as the congruence lattice of a finite 3-modularlattice.

We do this by verifying that a construction of G. Gr¨atzer, H. Lakser, and E. T.

Schmidt yields a 3-modular lattice.

1. Introduction

E. T. Schmidt [11] and [12] introduced the following construction. LetM3be the five-element, modular, nondistributive lattice and letD be a bounded distributive lattice. The latticeM3extended byD, denoted byM3[D], is the lattice of all triples hx, y, zi ∈D³ satisfyingx∧y=x∧z=y∧z; we call such triplesbalanced. Then M3[D] is a (modular) lattice andM3[D] andDhave isomorphic congruence lattices.

Meet inM3[D] is performed componentwise, while the join is the smallest balanced triple inM3[D] containing the triple formed by componentwise joins.

Note that the elementshx,0,0i, x∈D, form a sublattice ofM3[D] isomorphic toD. We identifyhx,0,0i ∈M3[D] withx∈D, makingM3[D] an extension ofD.

LetL be a lattice. A lattice K is acongruence-preserving extension ofL, ifK is an extension of L and every congruence of L has exactly one extension to K.

Of course, then the congruence lattice ofL is isomorphic to the congruence lattice ofK. E. T. Schmidt proved thatM3[D]is a congruence-preserving extension ofD.

Date: Sept. 17, 1998.

1991Mathematics Subject Classification. Primary: 06B05, Secondary: 06C05.

Key words and phrases. Lattice, modular, congruence-preserving extension.

The research of the first author was supported by the NSERC of Canada.

1

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.

This paper started with a problem proposed in G. Gr¨atzer and E. T. Schmidt [4]:

Does every lattice have a proper congruence-preserving extension? (We solved this problem in G. Gr¨atzer and F. Wehrung [6].) Of course, if the lattice is a bounded distributive lattice D, then M3[D] is such an extension. So two problems were raised:

1. For what classes of latticesC, isM3[L] a lattice forL∈C?

2. When isM3[L] a congruence-preserving extension ofL?

Surprisingly, in addition to Schmidt’s result (M3[D] is a (modular) lattice pro- vided thatDis a bounded distributive lattice), we could only find one other relevant result in the literature, see R. W. Quackenbush [10]: ifL is modular, thenM3[L]

is a lattice.

In Section 2, we define a lattice identity µ_n, for every n > 0, such that µ₁ is equivalent to the modular identity andµ_n+1is weaker thanµ_n. LetM_ndenote the variety defined byµ_n; we callM_n the variety ofn-modular lattices. We prove that ifLis an-modular lattice, thenM3[L] is a lattice and it is a congruence-preserving extension ofL. In Section 3, we verify thatM_n ⊂M_n+1 andM_n gets very large asngets large: S

(M_n|n < ω) generates the varietyL.

We show, in Section 4, that M3[L] is, in general, not a lattice. In Section 5, 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 M3[D] in the literature can be reconciled using tensor products, and we obtain the isomorphismM3[L]∼=M3⊗L, for any latticeLwith zero which satisfiesµ_nfor some n. It follows, then, that the result of Section 4 can be reinterpreted: there is a lattice Lwith zero such thatM3⊗Lis not a lattice. This solves, in the negative, a problem proposed in R. W. Quackenbush [10]: Is the tensor product of two lattices with zero always a lattice? In fact, our counterexample consists of two planar lattices. In Section 7, we show the there is a counterexample consisting of two modular lattices, M4 and the subspace lattice of any infinite dimensional vector space. There is another result onn-modular lattices in Section 6: IdM3[L]∼=M3[IdL].

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 representD 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

LetLbe a lattice. The triplehx, y, zi ∈L³ isbalanced, if x∧y=x∧z=y∧z.

We denote byM3[L] the set of all balanced triples. We regardM3[L] as a subposet ofL³, in fact, a meet-subsemilattice ofL³.

Lemma 2.1. Let L be a lattice. Then M3[L] is a lattice iff M3[L] is a closure system inL³.

Proof. IfM3[L] is a closure system and hx0, y0, z0i, hx1, y1, z1i ∈M3[L], then the closure ofhx0∨x1, y0∨y1, z0∨z1iinM3[L] is the join ofhx0, y0, z0iandhx1, y1, z1i.

Conversely, ifM3[L] has joins, then the closure ofhx, y, zi ∈M3[L] is hx, o, oi ∨ ho, y, oi ∨ ho, o, zi,

whereo is any element ofLcontained inx,y, andz.

Let us define the lattice polynomialspn, qn, andrn, for n < ω, in the variables x,y, andz:

p0=x, q0=y, r0=z,

p1=x∨(y∧z), q1=y∨(x∧z), r1=z∨(x∧y), . . .

pn+1=pn∨(qn∧rn), qn+1=qn∨(pn∧rn), rn+1=rn∨(pn∧qn).

Lethx, y, zi ∈L³. Define, forn >0,

hx, y, zi⁽ⁿ⁾=hpn(x, y, z), qn(x, y, z), rn(x, y, z)i Note that

(1) hx, y, zi ≤ hx, y, zi⁽¹⁾≤ · · · ≤ hx, y, zi⁽ⁿ⁾≤ · · ·

Definition 2.2. For n >0, define the identity µ_n as pn =pn+1. LetM_n be the lattice variety defined by µ_n. The lattices in M_n are called n-modular; lattices in M_n−M_n−1 are called exactly n-modular or of modularity rank n. A lattice L /∈M_n, for alln < ω, is ofof modularity rank ∞.

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.

Corollary 2.4. The following inclusions hold:

M₁⊆M₂⊆ · · · ⊆M_n⊆ · · ·

Corollary 2.5. For every finite latticeL, there is an integern >0 such that Lis n-modular.

Proof. Indeed, ifLis finite, then the increasing sequenceha, b, ci⁽ⁿ⁾must terminate

inL³, so Lemma 2.3 yields this result.

Corollary 2.6. S

(M_i|i < ω)generatesL, the variety of all lattices.

Proof. Indeed, by Corollary 2.5,S

(M_i|i < ω) contains all finite lattices and it is well-known that all finite lattices generate the varietyL.

Lemma 2.7. M₁ is the variety of modular lattices.

Proof. IfLis a modular lattice, then computing in F_M(3):

p2= (x∨(y∧z))∨(y∨(x∧z))∧(z∨(x∧y)) =x∨(y∧z) =p1.

Conversely, if L is nonmodular, then it contains a pentagon N5 ={o, a, b, c, i}

(with zeroo, uniti, and withb < a) as a sublattice and p1(b, a, c) =b∨(a∧c) =b,

p2(b, a, c) = (b∨(a∧c))∨((a∨(b∧c))∧(c∨(a∧b)) =a,

soµ₁: p1=p2 fails withx=b,y=a, andz=c.

On the other hand,p2=p3 holds inN5, so we obtain Lemma 2.8. The variety N₅ generated byN5 is2-modular.

E. T. Schmidt [11] and [12] proved that M3[L] is a modular lattice, if L is distributive. We now prove the converse.

Lemma 2.9. L be a lattice. ThenM3[L]is a modular lattice iff Lis distributive.

Proof. So letM3[L] be modular. Since M3[L] is an extension ofL, it follows that L is modular. IfL is not distributive, thenLcontainsM3={o, a, b, c, i}, the five element modular nondistributive lattice, as a sublattice. Then ho, o, oi, ha, a, ai, ha, a, ii, hb, c, oi, hi, i, ii ∈M3[L], ha, a, ii ∧ hb, c, oi=ho, o, oi, ha, a, ai ∨ hb, c, oi= hi, i, ii, which easily imply that

N5={ho, o, oi,ha, a, ai,ha, a, ii,hb, c, oi,hi, i, ii}

is the five-element nonmodular lattice, a sublattice ofM3[L], a contradiction.

The following lemma is due to E. T. Schmidt [11], for n = 1, and to R. W.

Quackenbush [10], forn= 2:

Lemma 2.10. Let n > 0 and let L∈ M_n be a lattice. Then M3[L] is a lattice.

Furthermore, ifL is bounded, then M3[L] has a spanningM3.

Proof. By Lemma 2.1, we have to prove that M3[L] is a closure system. For hx, y, zi ∈M3[L], define

(2) hx, y, zi=hpn(x, y, z), qn(x, y, z), rn(x, y, z)i.

By (1),hx, y, zi ≤ hx, y, zi. Since a polynomial is isotone, hx, y, zi ≤ hx^′, y^′, z^′iim- plies thathx, y, zi ≤ hx^′, y^′, z^′i. Finally, lethx, y, zi=hx^∗, y^∗, z^∗i; thenpn(x, y, z) = pn+1(x, y, z) =p1(x^∗, y^∗, z^∗), sohx, y, ziis closed.

The spanningM3 is{h0,0,0i,h1,0,0i,h0,1,0i,h0,0,1i,h1,1,1i}.

Theorem 1. Let n >0 and letL be a bounded n-modular lattice. Then M3[L]is a lattice with a spanningM3. The map

ε:x7→ hx,0,0i

embeds L into M3[L]. If we identify x∈ L with xε =hx,0,0i ∈ M3[L], then the latticeM3[L]is a congruence-preserving extension ofL.

Proof. By Lemma 2.10,M3[L] is a lattice. Furthermore,εis, obviously, an embed- ding; we identifyx∈Lwithxε=hx,0,0i. Now let Θ be a congruence ofL. Form Θ³, a congruence ofL³, and letM3[Θ] be the restriction of Θ³ toM3[L]. We claim that M3[Θ] is the unique extension of the congruence Θ to M3[L]. Since M3[Θ]

restricted toLequals Θ, it is sufficient to prove the following two statements:

(i) M3[Θ] is a congruence ofM3[L].

(ii) Every congruence Φ of M3[L] is of the formM3[Θ], for some congruence Θ ofL.

Re: (i). M3[Θ] is obviously a meet-congruence on M3[L]. It remains to prove the join substitution property. So let hx0, y0, z0i ≡ hx1, y1, z1i (M3[Θ]) and let hu, v, wi ∈ M3[L]. Then x0 ≡ x1 (Θ) and so x0∨u ≡ x1∨u (Θ). Similarly,

y0∨v≡y1∨v (Θ) andz0∨w≡z1∨w (Θ). Since a polynomial has the substitution property, we conclude that

pn(x0∨u, y0∨v, z0∨w)≡pn(x1∨u, y1∨v, z1∨w) (Θ), and similarly forqn andrn. Thus

hpn(x0∨u, y0∨v, z0∨w), qn(x0∨u, y0∨v, z0∨w), rn(x0∨u, y0∨v, z0∨w)i

≡ hpn(x1∨u, y1∨v, z1∨w), qn(x1∨u, y1∨v, z1∨w), rn(x0∨u, y0∨v, z0∨w)i

modulo Θ³, and therefore, moduloM3[Θ]. By (2), this last congruence is the same as

hx0, y0, z0i ∨ hu, v, wi ≡ hx1, y1, z1i ∨ hu, v, wi (M3[Θ]), which was to be proved.

Re: (ii). Let Φ be a congruence ofM3[L] and let Θ be the restriction of Φ toL.

We want to show thatM3[Θ] = Φ. Lethx0, y0, z0i,hx1, y1, z1i ∈M3[L].

If hx0, y0, z0i ≡ hx1, y1, z1i (M3[Θ]), thenx0 ≡x1 (Θ) in L and so hx0,0,0i ≡ hx1,0,0i (M3[Θ]). SinceM3[Θ] and Φ agree onL, we conclude that

(3) hx0,0,0i ≡ hx1,0,0i (Φ).

Similarly,hy0,0,0i ≡ hy1,0,0i (Φ); therefore,

h0, y0,0i= (hy0,0,0i ∨ h0,0,1i)∧ h0,1,0i

≡(hy1,0,0i ∨ h0,0,1i)∧ h0,1,0i=h0, y1,0i (Φ) that is,

(4) h0, y0,0i ≡ h0, y1,0i (Φ).

Similarly,

(5) h0,0, z0i ≡ h0,0, z1i (Φ).

Joining the three congruences (3)–(5), we obtain thathx0, y0, z0i ≡ hx1, y1, z1i (Φ).

Conversely, let hx0, y0, z0i ≡ hx1, y1, z1i (Φ). Meeting with hx0 ∨x1,0,0i ∈ M3[L], we derive that hx0,0,0i ≡ hx1,0,0i (Φ) and so x0 ≡ x1 (Θ). Similarly, h0, y0,0i ≡ h0, y1,0i (Φ). Therefore,

hy0, y0,1i=h0, y0,0i ∨ h0,0,1i ≡ h0, y1,0i ∨ h0,0,1i=hy1, y1,1i (Φ) and meeting withh1,0,0i ∈M3[L], we conclude thathy0,0,0i ≡ hy1,0,0i (Φ), that is,y0≡y1 (Θ). Similarly,z0≡z1 (Θ) and sohx0, y0, z0i ≡ hx1, y1, z1i (Θ³), from which it follows thathx0, y0, z0i ≡ hx1, y1, z1i (M3[Θ]).

3. The varietyM_n

In this section, we prove that M_n is properly contained in M_n+1, for every n >0. It is obvious that M_n ⊆M_n+1. To show that the equality fails, we have to construct, for eachn >0, an exactly (n+ 1)-modular latticeLn. The latticeL3

is shown in Figure 1. The definition ofLn follows the pattern of L3, except that there aren+ 1x-s: x0,x1, . . . ,xn; there aren+ 1y-s: y0,y1, . . . ,yn; and so there aren+ 1 sublattices of the formC₂² in the middle.

SinceLn is planar and bounded, it is a lattice.

x0∨(y0∧z0) =x1

y0

z1=z0∨(x0∧y0) x1∨(y0∧z1) =x2 z2=z1∨(x1∧y0) z3=z2∨(x2∧y0) x2∨(y0∧z2) =x3

1

x0 z0

Figure 1 0

L

Theorem 2. Ln is an exactly (n+ 1)-modular lattice.

Proof. It is easy to check that the free lattice onC2+C1 (see Figure VI.1.1 in [2]) is 2-modular. This shows that µ₂ holds in any lattice at any triplehx, y, zisuch that two of the variables x, y and z are comparable. So to check thatµ₂ holds in L1 at hx, y, zi, we can assume thatx, y, and z form an antichain; since there are very few antichains of three elements inL1, it is very easy to compute that µ₂ holds. SoL1 is (exactly) 2-modular.

Now we induct onn. The interval [x1∧z1,1] ofLn is isomorphic toLn−1. So if x,y, z ∈[x1∧z1,1], thenµ_n holds athx, y, zi, therefore, µ_n+1 holds at hx, y, zi.

If two of x, y, z are in [x1∧z1,1], say, y, z ∈ [x1∧z1,1], then replacing x by x=x∨(y∧z), and, similarly, fory andz, we have all three elements in [x1∧z1,1]

andµ_nholds forx,y,z, so µ_n+1holds forx,y,z. Since there is no three element antichain outside of [x1∧z1,1], we are left with the case that two ofx,y,zare not in [x1∧z1,1], say,xandz. We cannot then havex≤x1∧z1, because there is no such antichain. Similarly, z x1∧z1. Theorefore, by symmetry, we can assume that x=x0 andz =z0. It follows thaty ∈[x1∧z1, y0]. So we havep1=x1 and q1=y,r1 =z1, all in [x1∧z1,1]. By induction,µ_n holds forx1,y, andz1and so µ_n+1 holds forx, y,z.

It is clear that µ_n fails inLn with the substitutionx=x0, y=y0, andz=z0

becausepn(x0, y0, z0) =xn <1 =pn+1(x0, y0, z0).

Corollary 3.1. The following proper inclusions hold:

M₁⊂M₂⊂ · · · ⊂M_n⊂ · · · 4. M3[L] is not always a lattice

In this section, we construct a bounded latticeLsuch thatM3[L] is not a lattice.

Theorem 3. For the latticeLof Figure 2,M3[L]is not a lattice. Moreover,Lhas modularity rank∞.

Proof. The reader can easily verify that Lis a lattice by exhibiting the join- and meet-tables; for instance,xi∧zj=cmin(i,j),xi∨zj= 1, and so on. By Lemma 2.1, to show thatM3[L] is not a lattice, we have to verify thatM3[L] is not a closure system.

We claim that hx0, y0, z0i has no closure. So let us assume to the contrary that

hx, y, zi is the closure ofhx0, y0, z0i. Since p1(x0, y0, z0) = x1, q1(x0, y0, z0) = y0, r1(x0, y0, z0) =z1, so by induction, hx, y, zi must contain allhxn, y0, zni, that is, hxn, y0, zni ≤ hx, y, zi, for alln ≥0. On the other hand, hun, y0, vni is balanced, so hx, y, zi ≤ hun, y0, vni, for all n > 0. But there is no hx, y, zi ∈ L satisfying hxn, y0, zni ≤ hx, y, zi ≤ hun, y0, vni, for alln >0, sohx, y, zidoes not exist.

x0 z0

Figure 2 y0

x1 z1

u0

u1

v0

v1

c0= 0 c1

c2

L

IfLwasn-modular, for somen < ω, then hx0, y0, z0i⁽ⁿ⁾=hxn, y0, zniwould be

closed, but it is not.

We shall see in Section 6 thatL provides a negative solution to Quackenbush’s problem, namely,M3⊗L is not a lattice.

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 M3[L] has a spanning M3. If we do not assume that L has a unit, then the appropriate statement is that inM3[L], for everya∈M3[L], there is a i∈M3[L] such that (i]

has a spanningM3.

If we do not assume thatL has a zero, the definition of the embeddingε: x7→

hx,0,0iin Theorem 1 does not make sense, affecting the crucial part about congru- ence-preserving extensions. So we need to reformulate Theorem 1:

Theorem 4. Letn >0and letLbe ann-modular lattice. ThenM3[L]is a lattice.

The map

ψ:x7→ hx, x, xi

embeds L into M3[L]. If we identify x∈L with xψ =hx, x, xi ∈M3[L], then the latticeM3[L]is a congruence-preserving extension ofL.

Proof. The first part of the proof requires little change.

Let Φ be a congruence of M3[L] and let Θ be the restriction of Φ to L. We want to show that M3[Θ] = Φ. Let hx0, y0, z0i, hx1, y1, z1i ∈ M3[L], and put o=V

(xi∧yi |i <3 ).

Ifhx0, y0, z0i ≡ hx1, y1, z1i (M3[Θ]), thenx0≡x1 (Θ) inLand sohx0, x0, x0i ≡ hx1, x1, x1i (Φ). Therefore,

hx0, o, oi=hx0, x0, x0i ∧ hx0∨y0, o, oi

≡ hy0, y0, y0i ∧ hx0∨y0, o, oi=hy0, o, oi (Θ), that is,

hx0, o, oi ≡ hy0, o, oi (Θ).

Similarly, ho, y0, oi ≡ ho, y1, oi (Φ) and ho, o, z0i ≡ ho, o, z1i (Φ). Joining the three congruences, we obtainhx0, y0, z0i ≡ hx1, y1, z1i (Φ).

The proof of the converse is similar to the original proof withoplaying the role of 0 andi=W

(xi∨yi|i <3 ) playing the role of 1.

6. Two views ofM3[D]

For a finite distributive lattice D, in the literature, M3[D] is presented either as the lattice of balanced triples hx, y, zi ∈ D³ (as we presented it in Section 1) or as the lattice M₃^P, the lattice of isotone maps from P = J(D) (the poset of join-irreducible elements ofD) toM3. Either approach is convenient; both present a modular lattice with a spanningM3 withDembedded as the ideal generated by an atom ofM3and the lattice is generated byM3andD. The second approach has the advantage that it yields with no computation thatM3[D] is a modular lattice.

The first approach, however, better lends itself to generalization, as we did it in this paper.

It follows from A. Mitchke and R. Wille [9] that the two constructions yield iso- morphic lattices; indeed, both constructions yield a modular lattice with a spanning M3 withD embedded as the ideal generated by an atom of M3 and the lattice is generated byM3 andD and, up to isomorphism, there is only one such lattice.

In this section we shall give a more direct explanation why the two constructions yield isomorphic lattices. To this end, we introduce the concept of a capped tensor product from G. Gr¨atzer and F. Wehrung [7].

Definition 6.1. LetAandBbe{∨,0}-semilattices. Abi-ideal ofA×Bis a subset I ofA×B satisfying the following conditions:

(i) I is hereditary;

(ii) I contains∇A,B= (A× {0})∪({0} ×B);

(iii) ifha0, bi,ha1, bi ∈I, thenha0∨a1, bi ∈I;

(iv) ifha, b0i,ha, b1i ∈I, thenha, b0∨b1i ∈I.

Fora∈Aandb∈B, we define the bi-ideal

a⊗b=∇A,B∪ { hx, yi ∈A×B | hx, yi ≤ ha, bi }.

The bi-ideal lattice ofA×B is an algebraic lattice. Thetensor product A⊗B is the{∨,0}-subsemilattice of compact elements of the bi-ideal lattice ofA×B.

A bi-ideal I is capped, if there is a finite subsetC of A×B such thatI is the hereditary subset of A×B generated by C along with ∇A,B. A tensor product

A⊗B iscapped, if all bi-ideals ofA×B are capped. A capped tensor product is a lattice.

For a latticeLwith zero, letL⁻ denote the join-subsemilatticeL− {0}.

Let A⊗B be a capped tensor product and let I ∈ A⊗B. We define a map ϕI:A⁻→B:

Forx∈A,x >0, letϕI(x) be the largest elementy inB such thathx, yi ∈I.

Lemma 6.2. ϕI maps A⁻ intoB and

ϕI(x0)∧ϕI(x1) =ϕI(x0∨x1), for x0,x1∈A⁻.

Proof. First we show that ϕI(x) is defined, for all x∈A⁻. SinceI is capped, we can writeIin the formI=S

(ai⊗bi|i < n)∪ ∇A,B, wherenis a natural number, ai∈A, bi∈B, fori < n. Now define

yi=

(bi, ifx≤ai; 0, otherwise, for i < n and let y = W

(yi | i < n). By definition, hx, yii ∈ I, so by 6.1(iv), hx, yi ∈I. Now lethx, zi ∈I, for somez∈B. Thenhx, zi ∈ai⊗bi, for somei < n, and so z≤bi ≤y. This proves thaty satisfies the requirements in the definition ofϕI(x).

Now ϕI(x0∨x1) ≤ϕI(x0) is obvious, hence, ϕI(x0∨x1) ≤ϕI(x0)∧ϕI(x1).

Conversely, ϕI(x0)∧ϕI(x1) ≤ ϕI(x0), so hx0, ϕI(x0)∧ϕI(x1)i ∈ I; similarly, hx1, ϕI(x0)∧ϕI(x1)i ∈ I, therefore, by 6.1(iii), hx0∨x1, ϕI(x0)∧ϕI(x1)i ∈ I, implying thatϕI(x0)∧ϕI(x1)≤ϕI(x0∨x1).

LetB^ddenote the dual lattice ofB and let Hom^∨(A⁻, B^d) denote the lattice of join-homomorphisms fromA⁻ to B^d, ordered componentwise, as a subset ofB^A− (not (B^d)^A−).

Theorem 5. Let A and B be lattices with zero and let A⊗B be capped. Then ε:I7→ϕI defines an isomorphism betweenA⊗B andHom^∨(A⁻, B^d).

Proof. Lemma 6.2 states that the map is well-defined. Sincehx, yi ∈Iiffy≤ϕI(x), it follows thatϕI determinesI and soεis one-to-one. To show that εis onto, let ϕ ∈ Hom^∨(A⁻, B^d) and define I = { hx, yi ∈ A×B | y ≤ ϕ(x)}. Since ϕ is a join-homomorphism, it follows that Iis a bi-ideal andϕ=ϕI. Corollary 6.3. Let L be a lattice with zero. Then the following conditions are equivalent:

(i) M3⊗Lis a lattice.

(ii) For allhx, y, zi ∈L³, there existsn >0such thathx, y, zi⁽ⁿ⁾=hx, y, zi⁽ⁿ⁺¹⁾. Furthermore, if (i)is satisfied, thenM3⊗L∼=M3[L].

In particular, if L is n-modular, for some n, then M3 ⊗L is a lattice, and M3⊗L∼=M3[L].

Proof. SinceM3is finite,M3⊗Lis a lattice iffM3⊗Lis a capped tensor product, see Theorem 3 of [8]. Furthermore, in the same theorem, it is stated that this is equivalent to saying that, for every antitone mapξ: J(M3)→L, the adjustment sequence of ξ terminates after a finite number of steps. Here, J(M3) = {a, b, c}, and the ordering on J(M3) is trivial, thus every map from J(M3) toLis antitone.

Identifyξ: J(M3)→Lwith the triplehξ(a), ξ(b), ξ(c)i. With this identification, the adjustment sequence ofξis easily seen to be the sequence of allhξ(a), ξ(b), ξ(c)i⁽ⁿ⁾, n >0. The equivalence between (i) and (ii) follows.

Now assume thatM3⊗Lis a lattice. Again,M3⊗Lis a capped tensor product, thus, by Theorem 5, M3⊗L ∼= Hom^∨(M₃⁻, L^d). For all ξ ∈ Hom^∨(M₃⁻, L^d), we can identify ξwith the triplehξ(a), ξ(b), ξ(c)i, which, by Lemma 6.2, is a balanced

triple. The isomorphismM3⊗L∼=M3[L] follows.

Now the solution of Quackenbush’s problem (discussed in the Introduction) easily follows:

Corollary 6.4. Let Lbe the lattice of Theorem 3. ThenM3⊗Lis not a lattice.

Proof. This is obvious by Theorem 3 and Corollary 6.3.

Corollary 6.5. Let Lbe a lattice with zero. If Lisn-modular, for some n, then

(i) M3⊗L∼=M3[L].

If L is a finite distributive lattice and P = J(L), the poset of join-irreducible ele- ments ofD, then

(ii) M3⊗L∼=M₃^P.

Proof. Part (i) follows immediately from Corollary 6.3.

If L is a finite distributive lattice, then, by Theorem 5, M3⊗L is isomorphic to Hom^∨(L⁻, M3), and anyϕ∈Hom^∨(L⁻, M3) can be identified with an isotone

map fromP intoM3.

Combining (i) and (ii) of Corollary 6.5, we obtain the desired isomorphism:

Corollary 6.6. Let D be a finite distributive lattice. Then M₃^P ∼=M3[D],

whereP = J(D), the poset of join-irreducible elements ofD.

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) M3⊗L^ω is a lattice.

(ii) L isn-modular, for some n >0.

Proof.

(ii)implies(i). IfLisn-modular, thenL^ω isn-modular, thus, by Corollary 6.3, M3⊗L^ωis a lattice.

(i)implies(ii). Let us assume that the modularity rank ofLis∞. For alln >0, there exists, by definition, a triplehxn, yn, zni ∈L³ such that

hxn, yn, zni⁽ⁿ⁾<hxn, yn, zni⁽ⁿ⁺¹⁾. Define the following elements ofL^ω:

x=hxn|n >0i, y=hyn|n >0i, z=hzn |n >0i.

Then,hx, y, zi⁽ⁿ⁾<hx, y, zi⁽ⁿ⁺¹⁾ holds, for alln >0. By Corollary 6.3,M3⊗L^ωis

not a lattice.

Another algebraic consequence ofn-modularity is the following:

Proposition 6.8. Let Lbe an n-modular lattice, for somen >0. Then M3[IdL]∼= IdM3[L].

Proof. LetU ∈M3[IdL], that is,U =hI, J, Ki ∈M3[IdL]. We define Ub ={ ha, b, ci ∈M3[L]|a∈I, b∈J, c∈K},

that is,Ub = (I×J×K)∩M3[L]. ThenUb is obviously a hereditary subset ofM3[L].

We claim that it is join closed. Indeed, letha0, b0, c0i,ha1, b1, c1i ∈U, setb ha0∨a1, b0∨b1, c0∨c1i=hx, y, zi ∈L³,

and define hx, y, zi⁽ⁿ⁾ as in Section 2. We have x∈ I, y ∈ J, z ∈ K, soy∧z ∈ J∧K⊆I andx∨(y∧z)∈I; similarly,y∨(x∧z)∈J,z∨(x∧y)∈K. Thus

hx, y, zi⁽¹⁾=hx∨(y∧z), y∨(x∧z), z∨(x∧y)i ∈I×J×K.

By induction,hx, y, zi⁽ⁿ⁾∈I×J×K. SinceLsatisfiesµ_n,ha0, b0, c0i∨ha1, b1, c1i= hx, y, zi⁽ⁿ⁾, so ha0, b0, c0i ∨ ha1, b1, c1i ∈ I ×J ×K, and therefore, ha0, b0, c0i ∨ ha1, b1, c1i ∈Ub, proving thatUb is an ideal ofM3[L].

Since everya∈I can be augmented to an ha, b, ci ∈M3[L] with suitableb∈J, c ∈ K (and similarly for b ∈ J and for c ∈ K), it follows that Ub determines U. Therefore,ϕ:U 7→Ub is a one-to-one map fromM3[IdL] into IdM3[L].

To complete the proof, we have to prove that ϕis onto. So let X ∈IdM3[L].

Define

I={a∈L| ha, b, ci ∈X, for someb, c∈M3[L]}.

Since X is hereditary, so is I. Now let a0, a1 ∈ I. Then ha0, b0, c0i ∈ X and ha1, b1, c1i ∈X, for someb0,b1, c0,c1∈M3[L]. Let

ha0, b0, c0i ∨ ha1, b1, c1i=hx, y, zi ∈M3[L].

Thenx∈Ianda0∨a1≤x, hence,a0∨a1∈I, proving thatIis an ideal. Similarly, one can define the idealsJ andKofL, by permuting the three coordinates inL³.

Now we prove that

hI, J, Ki ∈M3[IdL].

Indeed, let w ∈ J ∧K. Then w ∈ J, so there exist i ∈ I and k ∈ K such that hi, w, ki ∈ X. Similarly, since w ∈ K, there exist i^′ ∈ I and j ∈ J such that hi^′, j, wi ∈X. Puto=i∧i^′∧j∧k∧w. SinceX is an ideal, we haveho, w, oi ∈X. Similarly,ho, o, wi ∈X. SinceX is join closed,

ho, w, oi ∨ ho, o, wi=ho, w, wi=hw, w, wi ∈X,

therefore, we conclude that w ∈ I, so J ∧K ⊆ I. Similarly, I∧K ⊆ J and I∧J ⊆K, so

I∧K=I∧J =J∧K, that is, hI, J, Ki ∈M3[IdL].

Finally, we prove that, for U = hI, J, Ki, we have Ub = X. Indeed, by the definitions ofI, J, and K, it is obvious thatX ⊆Ub. So let hx, y, zi ∈Ub, that is, hx, y, zi ∈ M3[L] and x ∈ I, y ∈ J, z ∈ K. Since x ∈ I, there exist j ∈ J and

k ∈ K such that hx, j, ki ∈ X. Similarly, there are i ∈ I and k^′ ∈ K such that hi, y, k^′i ∈X, and there are i^′∈I andj^′∈J such thathi^′, j^′, zi ∈X. Therefore,

hx, y, zi ≤ hx, j, ki ∨ hi, y, k^′i ∨ hi^′, j^′, zi ∈X,

proving thatUb⊆X.

Remark 6.9. E. T. Schmidt suggested that we consider, for an arbitrary latticeL with zero, the least join-congruence Θ ofL³identifying all triplesh0, x, xi,hx,0, xi, andhx, x,0i, for allx∈L, and relate the quotient latticeL/Θ toM3[L].

An alternative description of Θ is the following. For allha, b, ci ∈L³, denote by Clha, b, ci the least idealI of L³ containing ha, b, ciand such that if hx, y, zi ∈ I, then hx, y, zi⁽¹⁾ ∈I. In particular, if L isn-modular, for some n, then Clhx, y, zi is just the ideal ofL³ generated by the closurehx, y, ziofhx, y, zi. Then it is not hard to verify that

hx0, y0, z0i ≡ hx1, y1, z1i (Θ) iff Clhx0, y0, z0i= Clhx1, y1, z1i, for all xi, yi, zi in L and for i < 2. Hence, if L is n-modular, for somen, then L³/Θ is isomorphic to M3[L]. Furthermore, Clhx0, y0, z0i = Clhx1, y1, z1i is also equivalent to

(a⊗x0)∨(b⊗y0)∨(c⊗z0) = (a⊗x1)∨(b⊗y1)∨(c⊗z1), thusL³/Θ is isomorphic toM3⊗L, in general.

7. The M4[L] construction

LetM4 denote the lattice of height 2 with four atoms,a,b,c, andd.

In this section, we prove that M4⊗L is not a lattice, for a suitable modular latticeLwith zero, thereby showing that R. W. Quackenbush’s problem (discussed in the Introduction) has a negative solution also for modular lattices. We also find new examples of nonmodular tensor products that are not lattices, for instance, (M3⊗M3)⊗L.

The M3[L] construction has a natural extension to M4. For every lattice L, define

M4[L] ={ hx0, x1, x2, x3i ∈L⁴|xi∧xj =x0∧x1, fori6=j}.

As in Lemma 2.1, it is easy to prove thatM4[L] is a meet-subsemilattice ofL⁴, and that it is a lattice if and only if it is a closure system inL⁴.

Just as for triples, define the lattice polynomialsq0,q1,q2, and q3 in four vari- ables,x0,x1,x2, andx3, as follows:

qi0=xi, fori <4,

q0,n+1=q0n∨(q1n∧q2n)∨(q1n∧q3n)∨(q2n∧q3n), and, cyclically, defineq1,n+1, q2,n+1, andq3,n+1.

Lethx0, x1, x2, x3i ∈L⁴. Define, forn < ω, hx0, x1, x2, x3i⁽ⁿ⁾

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=hq0n(x0, x1, x2, x3), . . . , q3n(x0, x1, x2, x3)i.

The proof of the following result is very similar to the proof of Corollary 6.3, thus we will omit it.

Proposition 7.1. Let Lbe a lattice with zero. Then the following are equivalent:

(i) M4⊗Lis a lattice.

(ii) For all hx0, x1, x2, x3i ∈L⁴, there existsn >0 such that hx0, x1, x2, x3i⁽ⁿ⁾=hx0, x1, x2, x3i⁽ⁿ⁺¹⁾. Furthermore, if (i) is satisfied, thenM4⊗L∼=M4[L].

Again, it is not difficult to verify that if L is distributive, then it satisfies the identity hx0, x1, x2, x3i⁽²⁾ =hx0, x1, x2, x3i⁽¹⁾. However, this is no longer true for modular lattices, as witnessed by the main result of this section:

Theorem 6. LetV be an infinite dimensional vector space over a fieldK. Denote by L(V)the lattice of all subspaces of V. ThenM4⊗L(V)is not a lattice.

Note the contrast with theM3case: L(V) is a modular lattice, soM3⊗L(V) is a lattice.

Proof. We shall work with the lattice F_M(J₁⁴) introduced in A. Day, C. Herrmann, and R. Wille [1] as follows. By definition, F_M(J₁⁴) is the modular lattice generated by the elementsa, b,c, andd, subject to the relations

a∧b=a∧c=a∧d=b∧c=b∧d=c∧d= 0, a∨b=a∨c=a∨d=b∨d=c∨d= 1.

It is proved in [1] that F_M(J₁⁴) is isomorphic to a sublattice of the lattice of all subgroups of a free abelian group of infinite rank. Replacing the free Abelian group on a countably infinite number of generators by a vector space U of countably infinite dimension over a fieldK, it is easy to see that the same construction shows that F_M(J₁⁴) is isomorphic to a sublattice of the subspace latticeL(U).

The classical dualization map

X7→X^⊥={ξ∈U^∗|ξ[X] ={0} },

where U^∗ denotes the dual of U, is a dual embedding from L(U) into L(U^∗).

Therefore, the dual lattice L= F_M(J₁⁴)^d of F_M(J₁⁴) embeds into L(U^∗). Now let V0be a vector subspace of U^∗ satisfying the following properties:

(i) V0 has countably infinite dimension.

(ii) (X+Y)∩V0= (X∩V0) + (Y ∩V0), for allX,Y ∈L.

(iii) (X−Y)∩V06=∅, for allX,Y inLsuch thatX *Y.

ThenX 7→X∩V0is a lattice embedding fromLintoL(V0). Hence we have reached the following conclusion:

L embeds into L(V0), where V0 is a countably infinite dimensional vector space overK.

L(V0) embeds intoL(V), thusLembeds intoL(V).

Now suppose thatM4⊗L(V) is a lattice. By Proposition 7.1, the adjustment se- quence (6) based on any quadruple of elements ofL(V) terminates. Thus,a fortiori, the same holds for quadruples of elements ofL.

However, we shall now prove that there exists a quadruple of elements ofLwhose adjustment sequence does not terminate, thus completing the proof. For this, we need the following description of L, obtained by dualizing the one given in [1] for F_M(J₁⁴).

Put N =ω∪ {∞}. For hi, jiand hk, li in N×N, let us write hi, ji ∼ hk, li, if x≡y (mod 2), for allx,y∈ {i, j, k, l} − {∞}. Then, we have

L={ hi, ji ∈N×N| hi, ji ∼ h∞,∞i }.

The least element ofLish0,0i, the largest element ofLish∞,∞i.

Denote by ∧ and ∨ the infimum and the supremum on N. The meet and the join of Lare given as follows:

hi, ji ∧ hk, li=





















hi∧k, j∧li,

ifhi, ji ∼ hk, li;

h(i−1)∧(j−1)∧k,(i−1)∧(j−1)∧li, ifhi, ji 6∼ hk, liandi∧j≥k∧l;

hi∧(k−1)∧(l−1), j∧(k−1)∧(l−1)i, ifhi, ji 6∼ hk, liandi∧j < k∧l;

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hi, ji ∨ hk, li=





















hi∨k, j∨li,

ifhi, ji ∼ hk, li;

h(i+ 1)∨(j+ 1)∨k,(i+ 1)∨(j+ 1)∨li, ifhi, ji 6∼ hk, liandi∨j≤k∨l;

hi∨(k+ 1)∨(l+ 1), j∨(k+ 1)∨(l+ 1)i, ifhi, ji 6∼ hk, liandi∨j > k∨l.

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The base quadruplehx, y, z, tiof elements ofLis given by

x=h0,∞i; y=h1,∞i; z=h∞,1i; t=h∞,0i.

For alln >0, put

hx, y, z, ti⁽ⁿ⁾=hx⁽ⁿ⁾, y⁽ⁿ⁾, z⁽ⁿ⁾, t⁽ⁿ⁾i.

Then an easy (though somehow tedious) induction proof, based on the formulas (7) and (8), gives that for alln >0, we have

x⁽²ⁿ⁺¹⁾=x⁽²ⁿ⁺²⁾=h2n+ 2,∞i, t⁽²ⁿ⁺¹⁾=t⁽²ⁿ⁺²⁾ =h∞,2n+ 2i, y⁽²ⁿ⁾=y⁽²ⁿ⁺¹⁾ =h2n+ 1,∞i, z⁽²ⁿ⁾=z⁽²ⁿ⁺¹⁾ =h∞,2n+ 1i.

In particular, the sequencehx, y, z, ti⁽ⁿ⁾,n >0, is not eventually constant.

Corollary 7.2. LetV be an infinite dimensional vector space over a fieldK. Then M3⊗M3[L(V)] is not a lattice.

Proof. Define K =M3[L(V)] andL =L(V). Since L is modular, it follows from Corollary 6.3 thatM3[L]∼=M3⊗L. By Proposition 2.9 of [7], the tensor product of semilattices with zero is associative, thus we have

M3⊗(M3⊗L)∼= (M3⊗M3)⊗L.

Thus, in order to prove that M3⊗K is not a lattice, it suffices to prove that (M3⊗M3)⊗Lis not a lattice.

Now, we note that the following four elements

t=h1,0,0i, u=h0, a, bi, v=h0, b, ci, w=h0, c, ai

ofM3[M3] have pairwise meeth0,0,0iand pairwise joinh1,1,1i, thus they generate a 0-sublattice isomorphic toM4. Hence, there exists a zero preserving embedding f ofM4 intoM3⊗M3∼=M3[M3].

Now we need a very special case of Corollary 3.8 of [7]:

LetA,A^′,B be lattices with zero such thatAis a{0}-sublattice ofA^′. IfA^′⊗B is a lattice, thenA⊗B is a lattice.

Apply this withA=M4,A^′ =M3[M3], andB=L. By Theorem 6,M4⊗Lis not a lattice. Therefore, by the above statement, (M3⊗M3)⊗L is not a lattice

either.

Corollary 7.3. There is a modular lattice M such that M3[M] is of modularity rank ∞.

Proof. We can takeM =L(V), for any infinite dimensional vector spaceV. Indeed, ifM3[M] isn-modular, for somen >0, then, by Theorem 1, M3⊗M3[M] would

be a lattice, contradicting Corollary 7.2.

8. Congruence lattices

In this section, we prove that every finite distributive lattice can be represented as the congruence lattice of a finite 3-modular latticeL.

In G. Gr¨atzer, H. Lakser, and E. T. Schmidt [3], it is proved that every finite distributive latticeDcan be represented as the congruence lattice of a finite planar latticeL. This latticeL has the following properties:

(C1) L has a {0,1}-sublattice,G, thegrid, which is of the formC×D, where C andD are finite chains.

(C2) Every element ofH=L−Gis doubly irreducible inL.

It follows from (C1) that for every elementx∈L, there is a largest grid element xwithx≤x; dually, there is a smallest grid elementx∈Gwith x≤x.

(C3) Forx,y∈H, ifx=y, thenx=y; and dually.

(C4) For everyx∈H, either the interval [x, x]G is a prime interval and [x, x]L

is the three-element chain, or [x, x]G is a prime square and [x, x]L is an M3.

Recall that aprime squareis an interval of length two isomorphic toC₂². For a grid elementx, we shall use the notationhxC, xDi, wherexC∈CandxD∈D.

By (C4), if [a, b] is a prime interval inG, then [a, b]L−[a, b]G is either empty or it is a singleton; in the latter case, we denote the new element byn(a, b). Similarly, if [a, b] is a prime square inG, then the set [a, b]L−[a, b]G is either empty or it is a singleton; in the latter case, we denote the new element bym(a, b).

For a grid elementx, theC-line throughxis defined as {a∈G|aD=xD};

symmetrically, we define theD-line throughx.

Note the following immediate consequences of (C1)–(C4):

Ifx,y ∈Landxky, then (C5) x∧y,x∨y∈G.

(C6) x∧y=x∧y andx∨y=x∨y.

(C7) xis on theC-line or on theD-line throughx∧y; and symmetrically and dually.

The goal of this section is to prove the following result:

Theorem 7. Let L be a finite lattice satisfying conditions (C1)–(C4). Then L is 3-modular.

Proof. In this proof, we shall use the notation

hx, y, zi⁽ⁿ⁾=hx⁽ⁿ⁾, y⁽ⁿ⁾, z⁽ⁿ⁾i,

for a triple hx, y, zi in L and n > 0, and with this notation we can restate the theorem:

Let L be a finite lattice satisfying conditions (C1)–(C4). Then, for any triple hx, y, ziin L, the triplehx⁽³⁾, y⁽³⁾, z⁽³⁾iis balanced.

Ifhx, y, ziinLis not an antichain, thenhx⁽²⁾, y⁽²⁾, z⁽²⁾iis balanced by Lemma 2.8.

So from now on, we assume that

(A1) hx, y, ziis an antichain.

If xk y∧z, y k x∧z, z k x∧y, then by (C5), x⁽¹⁾, y⁽¹⁾, z⁽¹⁾ ∈ G, hence, by Lemma 2.7, the triplehx⁽²⁾, y⁽²⁾, z⁽²⁾iis balanced. So by symmetry, we can assume:

(A2) y∧z≤x.

Equivalently,x⁽¹⁾=x.

Ifx∧z ≤y and x∧y ≤z, thenhx, y, ziis balanced. So we have two cases to consider: x∧z≤y,x∧ykz(see Figure 3) andx∧zky,x∧ykz (see Figure 6).

Figure 3 x∧y

u=x∧z=y∧z z

x y

Case 1. x∧z≤y andx∧ykz; see Figure 3.

Note that the assumptions for Case 1 are symmetric inxandy.

Let u = y ∧z. By (A2), y ∧z ≤ x∧z and by the assumption for Case 1, x∧z≤y∧z, so

(9) u=x∧z=y∧z∈G.

Hence,u≤z. We distinguish two subcases: u=zand u < z.

Case 1a. u=z. Obviously,z /∈ G, because z ∈G would imply thatz =z = u≤x, contradicting the assumption (A1). So either z =m(u, z) or z =n(u, z).

In either case,zx∧y, see Figures 4.1 and 4.2. LetLC (resp,L^′_C) be theC-line throughu(resp., throughz). By symmetry and (C7), we can assume thatx∧y is onLC.

Since x∧y =x∧y, it follows that eitherx or y is on the lineLC. Since the assumptions are symmetric inxandy, we can assume thatxis on the lineLC; but L^′_C “covers” LC and y cannot contain an element on L^′_C (that would contradict (A1)), therefore,yalso is on the lineLC. We conclude thatxandyare comparable, so we can assume that x≤y (and so x= x∧y). In this case,x= xleads to a contradiction with (A1), therefore, x < x. Also, x < y because x = y would

contradict either (A1) or (C3). Ifx=n(x, x) withx,xonLC, then we cannot find room fory by (A1).

Figure 4.1

u=z u=z

x∧y x∧y

z z

Figure 4.2 z

z

So there are two possibilities forx:

(i) x=n(x, x) withxonLC andxonL^′_C; (ii) x=m(x, x) withxonLC andxonL^′_C. If (i) holds, then

hx⁽¹⁾, y⁽¹⁾, z⁽¹⁾i=hx, y, xi and x < x, sohx⁽³⁾, y⁽³⁾, z⁽³⁾iis balanced by Lemma 2.8. If (ii) holds, then

hx⁽¹⁾, y⁽¹⁾, z⁽¹⁾i=hx, y,(x∧y)∨zi is balanced.

Figure 5 x∧y z

x∧y

x∧y z

x∧y z

z z⁽¹⁾

z⁽¹⁾ z⁽¹⁾

z⁽¹⁾

Case 1b. u < z. Then x⁽¹⁾ = x, y⁽¹⁾ = y, z⁽¹⁾ = (x∧y)∨z, see Figure 5.

We cannot have x=y (= x∧y); indeed, then x=x or y = y would contradict (A1);x < xandy < y would contradict (C3). It follows that by symmetry we can assume without loss of generality that either x < y or x∧y < x, x∧y < y. We consider these cases separately.

Subcase I. x < y. Obviously, x /∈ G because x ∈ G would imply that x < y, contradicting (A1). We cannot have both x_C < y_C and x_D < y_D because this would again imply thatx < y, contradicting (A1).

y andz⁽¹⁾ cannot be on the same line throughx. Indeed, ify andz⁽¹⁾ are, say, on theD-line throughx, thenx < y < z⁽¹⁾(since z⁽¹⁾≤y would imply thatz < y, contradicting (A1)). Soz⁽¹⁾=x∨z=y∨z, which implies thatx∧z < y∧z(since Gis distributive). Sou < y∧z, contradictingu=y∧z.

Therefore, by symmetry, we can assume thatxandy are on theC-line through x=x∧y andz⁽¹⁾ is on theD-line throughx. Thenxis not on theC-line through x(this would contradict (A1)), so eitherx=n(x, x) withx_C =xC, in which case hx⁽³⁾, y⁽³⁾, z⁽³⁾i is balanced (since x =x⁽¹⁾ ≤z⁽¹⁾) or x=m(x, x), in which case hx⁽¹⁾, y⁽¹⁾, z⁽¹⁾i=hx, y, z⁽¹⁾iis balanced.

Subcase II.x∧y < x andx∧y < y. LetLC be theC-line throughx∧y and let LD be theD-line throughx∧y. Sincez⁽¹⁾= (x∧y)∨z, it follows thatz⁽¹⁾ is on LC or onLD, say, onLD. Then by (A1),y < z⁽¹⁾, which contradicts (9) as above, so this subcase cannot occur.

This completes Case 1b and, therefore, Case 1.

Figure 6

y z

x

y∧z y⁽¹⁾ z⁽¹⁾

u x∧z x∧y

Case 2. x∧z k y and x∧y k z. This case is illustrated in Figure 6; the grey filled elements are inG. Define u= (x∧y)∨(x∧z). The elementsy∧z, x∧y, x∧z, anduare distinct; indeed, any equality would contradict withx∧z ky or withx∧ykz. Souis at least a “prime square” abovey∧z.

We start with two observations:

(10) x=u and y⁽¹⁾∧z⁽¹⁾=u.

Without loss of generality, we can assume thatx∧y is on the C-line through y∧zand thatx∧zis on theD-line throughy∧z.

Ifu < x, then (x∧y)C < x_C or (x∧z)D< x_D, say, (x∧y)C < x_C. It follows thaty_C< x_C and soy_C≤x_C. Therefore,y≤x, a contradiction.

y∧z=y∧z≤uimplies thaty∧z≤usincey∧zis at most one “prime square”

abovey∧z. Thus (y∨u)∧(z∨u) =u, as claimed.

Now this case is easy. Ifx=uorx=m(x, x), thenhx⁽¹⁾, y⁽¹⁾, z⁽¹⁾iis balanced since x = x⁽¹⁾. If x = n(x, x), then by symmetry we can assume that x≤ z⁽¹⁾ (since u =x, x≺x, and u < z⁽¹⁾) and then x⁽²⁾ =x, y⁽²⁾ =y∨x, z⁽²⁾ =z⁽¹⁾; hencehx⁽²⁾, y⁽²⁾, z⁽²⁾iis balanced, which completes the proof.

Note that in Theorem 7, “3-modular” cannot be changed to “2-modular”. In- deed, let A0 = {0,1,2} with 0 < 1 < 2 and A1 = {0,1} with 0 < 1. We take a = h1,0i, b = h1,1i, and L = G∪ {n(a, b)}. Then L satisfies (C1)–(C4). Set x=n(a, b), y=h2,0i, andz=h0,1i. Thenx=x⁽¹⁾ =x⁽²⁾ < x⁽³⁾=h1,1i. SoL is not 2-modular.

Corollary 8.1. Every finite distributive latticeD can be represented as the con- gruence lattice of a finite planar 3-modular latticeL.

Proof. This is immediate by combining the representation theorem in G. Gr¨atzer,

H. Lakser, and E. T. Schmidt [3] with Theorem 7.

9. Discussion

Problem 1. Which lattice varieties are closed under tensor product?

There are several ways to define what it means for a lattice variety V to be

“closed under tensor product”:

(i) IfAand B are finite lattices,A,B∈V, then A⊗B∈V.

(ii) If A andB ∈V are lattices with zero, A, B ∈V, andA⊗B is capped, thenA⊗B∈V.

(iii) IfA andB ∈V are lattices with zero,A,B ∈V, andA⊗B is a lattice, thenA⊗B∈V.

The trivial variety,T, the variety of all distributive lattices,D, and the variety of all lattices, L, are closed under tensor product under any of the three inter- pretations. The problem, whether there are any more, is open in all of its three variants.

It is easy to verify that T and D are the only two finitely generated varieties that are closed under tensor product (under any one of the interpretations).

Problem 2. Compute the modularity rank ofA[B], for small latticesAandB?

The following examples were computed by B. Wolk:

Example 9.1. The modularity rank of M3[Mk]is 3, for allk >2.

LetM4={0, a, b, c, d,1}. Then the computations can be arranged in an array, as follows:

n pn qn rn

0 hb, c, ai hb, a, di ha,0, ci 1 hb, c, ai hb, a, di h1, c, ci 2 hb, c, ai h1,1,1i h1, c, ci 3 h1,1,1i h1,1,1i h1,1,1i

Among the 89,217 three-element antichains inM3[M4], 936 do not satisfyµ₂but they all satisfy µ₃. More generally, this is true in all the lattices M3[Mk], k >2, that is, they are all exactly 3-modular.

Example 9.2. Let F7 denote the lattice of subspaces of the Fano plane. Then M3[F7] is not3-modular.

Notation: the points are 1, 2, 3, 4, 5, 6, 7; the lines are 124, 235, 346, 457, 561, 672, 713, and the plane is P L. The following example shows that M3[F7] is not 3-modular.

n pn qn rn

0 h3,6,4i h3,457,2i h7,2,561i 1 h3,6,4i h3,457,2i h713,124,561i 2 h346,346,346i h3,457,2i h713,124,561i 3 h346,346,346i h713,457,672i h713,124,561i 4 hP L,346,346i h713,457,672i h713,124,561i

M3[F7] is too large for a complete search of the type B. Wolk was conducting;

it has 1,090 elements, and around 190 million three-element antichains.

Corollary 7.3 show that the modularity rank of M3[L] may be ∞ even if L is modular.

Problem 3. LetK be a field, and letV be ad-dimensional vector space overK.

What is the modularity rank ofM3[L(V)].

Problem 4. Is it possible to represent every finite distributive lattice as the con- gruence lattice of a finite (planar) 2-modular lattice?

For n > 1, define s(n) as the smallest integer so that there is an exactly n- modular lattice of size s(n). Obviously, s(2) = 5, as realized byN5. The lattice presented after the proof of Theorem 7 shows thats(3) = 7.

Problem 5. Determine the function s(n). Computes(n) for small values ofn.

Problem 6. Describe the free lattice with three generators overM₂. Is it finite?

What about F_Mn(3)?

References

[1] A. Day, C. Herrmann, and R. Wille, On modular lattices with four generators, Algebra Universalis2(1972), 317–323.

[2] G. Gr¨atzer,General Lattice Theory, Pure and Applied Mathematics75, Academic Press, Inc.

(Harcourt Brace Jovanovich, Publishers), New York-London; Lehrb¨ucher und Monographien aus dem Gebiete der Exakten Wissenschaften, Mathematische Reihe, Band 52. Birkh¨auser Verlag, Basel-Stuttgart; Akademie Verlag, Berlin, 1978. xiii+381 pp.

[3] G. Gr¨atzer, H. Lakser, and E. T. Schmidt,Congruence lattices of small planar lattices, Proc.

Amer. Math. Soc.123(1995), 2619–2623.

[4] G. Gr¨atzer and E. T. Schmidt,A lattice construction and congruence-preserving extensions, Acta Math. Hungar.66(1995), 275–288.

[5] ,On the Independence Theorem of related structures for modular (arguesian) lattices, manuscript. Submitted for publication in Studia Sci. Math. Hungar., March 1997.

[6] G. Gr¨atzer and F. Wehrung,Proper congruence-preserving extensions of lattices, AMS Ab- stract 97T-06-189. Acta Math. Hungar., to appear.

[7] ,Tensor products of semilattices with zero, revisited, J. Pure Appl. Algebra, to appear.

[8] ,Tensor products and transferability of semilattices, AMS Abstract 97T-06-190.

[9] A. Mitchke and R. Wille,Freie modulare Verb¨andeF M(DM3).Proceedings of the University of Houston Lattice Theory Conference (Houston, Tex., 1973), pp. 383–396. Dept. Math., Univ.

Houston, Houston, Tex., 1973.

[10] R. W. Quackenbush, Nonmodular varieties of semimodular lattices with a spanning M3. Special volume on ordered sets and their applications (L’Arbresle, 1982). Discrete Math.53 (1985), 193–205.

[11] E. T. Schmidt,Zur Charakterisierung der Kongruenzverb¨ande der Verb¨ande, Mat. ˇCasopis Sloven. Akad. Vied.18(1968), 3–20.

[12] ,Every finite distributive lattice is the congruence lattice of a modular lattice, Algebra Universalis4(1974), 49–57.

Department of Mathematics, University of Manitoba, Winnipeg MN, R3T 2N2, Canada E-mail address: gratzer@cc.umanitoba.ca

URL:http://www.maths.umanitoba.ca/homepages/gratzer.html/

C.N.R.S., D´epartement de Math´ematiques, Universit´e de Caen, 14032 Caen Cedex, France

E-mail address: wehrung@math.unicaen.fr URL:http://www.math.unicaen.fr/~wehrung