Varieties of lattices

Varieties of Lattices Peter

.J

iI)SC1I

dud lleuirv Rose

Synopsis An interesting pro blem iii universal algebra is the connection bet ween the internal structure of an algebra and the identities which it satisfies. I he stud of varieties of algebras provides some insight into this problem. Here we are concerned mainly with lattice varieties, about which a wealth of information has been obtained in the last twenty years. e begin with some preliminary results from universal algebra and lattice theory. I he second chapter presents sonic properties of the lattice of all lattice suhvarieties. Here we also discuss the important notion of a splitting pair of varieties and give several characterizations of the associated splitting lattice. I he more detailed study of lattice varieties splits naturally into the study of modular lattice varieties and nonmodular lattice varieties, dealt th in the third and fourth chap ter respectively. Among the results discussed there 11reese's are theorem that the variety of all modular lattices is not generated by its finite mem hers, and several results concerning the questioii which varieties cover a given variety. I he fifth chapter contains a proof of Baker's finite basis theorem and sonic results about the join of finitely based lattice varieties. Included in the final chapter is a characterization of the amalgamation classes of certain congruence distributive varieties and the result that there are only three lattice varieties which have the amalgamation property.

Acknowledgements I he second author acknowledges the grants from the niversi ty of Cape lown Research Coninut tee and the South African Council for Scientific and Industrial Research.

e dedicate

this monograph to our supervisor Bjarni JOnsson.

Contents Introduction 1

Preliminaries 1.1 Ihe Concept 1.2 1.3 1

2

A

2.—I

2.5

Congruence Distri hutivity Congruences on Lattices

5

10

1

13 13

he Structure of the Bottom of A

17

Splitting Lattices and Bounded Honioniorphisnis Splitting lattices generate all lattices Finite lattices that satisfy ( \\ )

20 -11 -1-1

46

Introduction

-16

3.2 3.3

Pro jective Spaces and Arguesian Lattices ii— Frames and Freese's I heoreni Covering Relations bet ween Modular \ arieties

-17

Nonmodular Varieties —1.2

-1.3 -1.-I

5.2 5.3 5.—I

Introduction

11

Semidistri hutivi tv Almost Distri hutive \ arieties Further Sequences of \ arieties

78

86 10-I

115

Introduction Baker's Finite Basis I heorem joins of finitely based varieties Equational Bases for some \ arieties .

Amalgamation in Lattice Varieties 6.1 Introduction 6.2

57 69

77

Equational Bases 5.1

6

1

3.1

-1.1

5

of a \ arietv

Congruences and Free Algebras

Modular Varieties

3.—I

4

1

General Results 2.1 Ihe Lattice A 2.2 2.3

3

ix

15 16

20 26

128 128 129

Preliminaries vii

('OX TEX £5

viii 6.3 6.-I

Xmal( V) for Residually Small \ arieties Products of absolute retracts

32

6.5

Lattices and the Anialgamation Property Anialgamation iii modular varieties 1 he Day — Je2ek 1 heorem Xnial( V) for sonic lattice varieties

35 37

GM

6.7

6.8

Bibliography

.

-Ii -15

149

Introduction I he study of lattice varieties evolved out of the study of varieties iii general, which was initiated by Garrett Birkhotf in the 1930's. He derived the first significant results in this subiect, and further developments by Alfred Iarski and later, for congruence distributive varieties, by Bjarni JOnsson, laid the groundwork for nianv of the results about lattice varieties. During the same period, investigations in proJective geometry and modular lattices, by Richard Dedekind, John von Xeumann, Garrett Birkhotf, George Griitzer, Bjarni JOnsson and others, generated a wealth of information about these structures, which was used by Hr by Baker and Rudolf \\ ille to o btain some structural results about the lattice of all modular su bvarieties. Xonmodular varieties were considered by Ralph Mckenzie, and a paper of his published in 1972 stimulated a lot of research in this direction. Since then the efforts of many people have advanced the subject of lattice varieties in several directions, and many interesting results have been obtained. I he purpose of this book is to present a selection of these results in a (more or less) self-contained framework and uniforni notation. In Chapter 1 we recall sonic preliminary results from the general study of varieties of algebras, and some basic results about congruences on lattices. I his chapter also serves to introduce most of the notation which we use su bsequentlv. Chapter 2 contains some general results about the structure of the lattice A of all lattice subvarieties and about the important concept of 'splitting". \\ e present several characterizations of splitting lattices and Alan Day's result that splitting lattices generate all lattices. I hese results are applied in Chapter -1 and 6. Chapters 3 — 6 each begin with an introduction in which we mention the important results that fall under the heading of the chapter. Chapter 3 then proceeds with a review of projective spaces and the coordinatization of (complemented) modular lattices. I hese concepts are used to prove the result of Ralph 11reese, that the finite modular lattices do not generate all modular lattices. In the second part of the chapter we give sonic structural results about covering relations between modular varieties. In Chapter -1 we concentrate on nonmodular varieties .Ac haracterization of semidistn butive varieties is followed by several technical lemmas which lead up to an essentially complete description of the 'almost distributive" part of A. \\ e derive the result of Bjarni JOnsson and Ivan Rival, that the smallest nonmodular variety has exactly 16 covers, and conclude the chapter th results of Henry Rose about covering chains of join-irreduci ble semidistri butive varieties. Chapter .5 is concerned with the questioii which varieties are finitely based .A proof of kirby Baker's finite basis theorem is followed by an example of a nonfinitely based variety, ix

x

LVTLtODtCT1O.N

and a discussion about when the join of two finitely based varieties is again finitely based. In Chapter 6 we study amalgamation in lattice varieties, and the amalgamation prop-

erty. The first half of the chapter contains a characterization of the amalgamation class of certain congruence distributive varieties, and in the remaining part we prove that there are only three lattice varieties that have the amalgamation property. By no means can this monograph be regarded as a full account of the subject of lattice varieties. In particular, the concept of a congruence variety (i.e. the lattice variety generated by the congruence lattices of the members of some variety of algebras) is not included, partly to avoid making this monograph too extensive, and partly because it was felt that this notion is somewhat removed from the topic and requires a wider background

of universal algebra. For the basic concepts and tiscts from lattice theory and universal algebra we refer the reader to the books of George Grätzer [ULT], [UA] and Peter Crawley and Robert P. Dilworth [ATL]. however, we denote the join of two elements a and 6 in a lattice by a + 6 (rather than a V 6) and the meet by a •6, or simply a6 (instead of a A 6; for the meet of two congruences we use the symbol fl). When using this plus, dot notation, it is traditionally assumed that the meet operation has priority over the join operation, which reduces the apparent complexity of a lattice expression. As a final remark, when we consider results that are applicable to wider classes of algebras (not only to lattices) then we aim to state and prove them in an appropriate general form.

Chapter

1

Preliminaries The Concept of a Variety

1.1

Lattice varieties.

Let F be a set of lattice identities (equations), and denote by Mod F the class of all lattices that satisfy even identity in F. A class V of lattices is a lattice ('al/ely if

ModE

V

for sonic set of lattice identi ties F. Ihe class of all lattices, which we will denote by L, is of course a lattice variety since L Mod 0. e will also frequently encounter the following lattice varieties:

I P

Modtry

z

Mod

z

z

y

£(

all trivial lattices, all distributive lattices, all modular lattices.

)} z

)}

J

be the (countable) set of all lattice identities. For any class K of lattices, we denote by Id K the set of all identities which hold in every mem her of K. A set of identities is said to he if F IdK F

Let

J

for some class of lattices K. It is easy to see that for any lattice variety V, and for any closed set of identi ties F, V

Mod Id

V

and

Id Mod F,

F

whence there is a bijection between the collection of all lattice varieties, denoted by A, and the set of all closed su hsets of J. I hus A is a set, although its mem hers are proper classes. A is partially ordered by inclusion, and for any collection { i E I } of lattice varieties

fl

U

Id

tEl

lattice variety, which implies that A is closed under arbitrary intersections. Since A also has a largest element, namely L, we conclude that A is a complete lattice with intersection as the meet operation. For any class of lattices K,

is again a

Ky

Mod IdK

fl{V 1

E

A: K

I

Cli.tl'TEit 1. P1tELL%IL'Lti(JES

2

is the smallest variety containing K, and we call it the variety generated by K. Now the join of a collection of lattices varieties is the variety generated by their union. We discuss the lattice A in more detail in Section 2.1.

Varieties of algebras. Many of the results about lattice varieties are valid for varieties of other types of algebras, which are defined in a completely analogous way. When we consider a class K of algebras, then the members of K are all assumed to be algebras of the same type with only finitary operations. We denote by

IlK — the class of all homomorphic images of members of K SK — the class of all sa&slgebms of members of K

PK — the class of all di,rtct pivdacts of members of K P5K — the class of all subdirect pivd acts of members of K. (Recall that an algebra .1 is a subdirect product of algebras (1 I) if there is an embedding f from .1 into the direct product such that f followed by the ith projection irj maps .1 onto for each i I.) The first significant results in the general study of varieties are due to Birkhoff [35], who showed that varieties are precisely those classes of algebras that are closed under the formation of homomorphic images, subalgebras and direct products, i.e.

Visavariety ifandonlyif HV=SV=PV=V. Tarski [46] then put this result in the form

C = HSPK and later KogalovskiY [65] showed that

C = HP5K for any class of algebras K.

1.2

Congruences and Free Algebras

Cougruences of algebras. Let .1 be an algebra, and let (Jon(A) be the lattice of all congruences on A. For a,b .1 and U (Jon(A) we denote by: a/U — .1/U — 0,1 — con(a,b) —

congruence class of a modulo U quotient algebra of .1 modulo U zern and unit of (Jon(A) principal congruence generated by (a,b) (i.e. the smallest congruence that identifies a and 6). the the the the

(Jon(A) is of course an algebraic (= compactly generated) lattice with the finite joins of principal congruences as compact elements. (Recall that a lattice element c is compact if whenever c is below the join of set of elements C then c is below the join of a finite subset of C. A lattice is algebraic if it is complete and every element is a join of compact elements.)

('OXOR

1,2.

IRET

LVI)

L

1± (JKBIL lB

For later reference we recall here a description of the join operation iii Con( 1). LEMMA 1.1 Let

J

be an

u

if and univ

£(YTC)fJ

for

5011W

/iE

W,

E C

E

J and C I

Con( J).

Ihen

II.

and

E A.

I he connection between congruences and homomorphisms is exhibited by the itoino— theorem: for any homomorphism Ii A — B the na

h

i( X

K

h(oo')

K

Qf(oo'). I his establishes the

h( i(b)

)

by y

i(e))

Ii

i.

b,e E L.

for all

first observe that, by the preceding leninia, the set S

e

LTS

I hen y is clearly ioi i-preserving. for all b E L (since f/i f). Ibis we need oniy show i is orderpreservi ig, it suffices to show that

L —

Also fy(b) f/i i(b) f i(b) that y is meet—preserving. Since

if(oo')

c;LIVERjL

2.

E

X

be} is

finite.

Let!

{ i(b) i(e)flS i(b) i(e)

—

if if

0

S

0.

i( be). lo see this, let I be the set of all I hen clearly f( a) be. \\ e claim that a K K i( X such that be f(o) implies a (I. By definition X I I and I is closed under E meets. Suppose o, G' E I and be K f(o —i—f (i'). Since L satisfies ( \\ ), we have (i

)

)

f(o)

b K

—t—f(o')

or e

(o) H—f(o') or be

K

f(o) or

K

be K f(G').

Applviig i to the first two cases we obtain a K i(b) K or a K i(e) o', and in the third and fourth case we have a K 0 ± o' or a K 0 ± o' (since (1, o' E 1). 1 hus I 1(X), and it follows that a is the least element for which f( a) be, i.e. a i(be). consider h( a) Ii i(be). If 5 i(b) i(e) and so (ic) holds. 0 then i(be) (ic), fl/i(S) prove If 5 then Ii i(be) Ii i(b)/i i(e) so to it is enough to show that 0 C:.

H—

H—

h( i(b) i(e) K for all £ E 5. But any such £ satisfies f( i(b) i(e) be K ), and applviiig Q we get i(b) i(e) K Qf(J h( i(b) i(e) K hh(t and, since we K showed that h(o) K Qf(o), we have as required. )

)

)

)

A complete

)

characterization of the projective lattices in

L

)

can he found in Freese and

ation [78]. e now describe a particularity elegant algorithm, due to JOnsson, to determine whether a finitely generated lattice is bounded. Let 1J0(L) be the set of all (I E L that have no nontrivial join-cover (i.e. the set of all join-prime elements of L ). For k E w let L ) be the set of all (I E L such that if is any nontrivial join-cover of o, then there exists a join-cover I L)1JL) of (I with Xote that 1J0(L) and if we assume that 1J1JL ) for some I 1, then for any o E L ) any nontrivial join—cover of (I there exists a join—cover , whence (I E L)1JL) of (I with I So, by induction we have

I

.

I

I I

I

I

I)0(L)

I ... I

D1JL)

I

Finally, let D( L I)1JL and define the sets and L)'(L dually. JOnsson's algorithm states that a finitely generated lattice is bounded if and only if D( L L L)'( L). I his result will follow from I heorem 2.23. )

)

)

)

)

2.20 (JOnsson and Xation [75]). Suppose L is a lattice generated by ase t X I L. let be the set of all (finite) meets of elements aIX. and for k E w let be the set of all meets of joins of elements of 11k' Ihen Dk ( L) I 11k for all k E w, LEMMA

I'I-tOOF.

Suppose

I I and L Xote that the are closed under meets, L and let iii E w be the smallest number for which (I E 11,,. If .

o E L)0(

)

.

.

IT,

0,

SJ'LJJIIXG

2,3.

AXL) HOt XIJLI) BWJWJORPBTh11S

then (I is of the form (I join-cover of (I for each i exists (I

29

where each I is a finite subset of is a 1 I ii, and since (I E I)0(L), it must he trivial. Hence there

37 .

.

.

,

E

(I

1

I

proceed by induction. Suppose L)1JL ) 11k, G E and iii 1 is the with I smallest number for which (I E 11,. Again (I is of the form (I I and each I is a join—cover of (I. If I is trivial, pick (Lj E I with (I K and if I is nontrivial, pick a join—cover I of If in ;> k ± 1 then 37 and (I is the meet of these elements 37 E and the so (I E 1 '

H—

I

I

I

For the next lemma, note that \\ hitman's condition for any two finite subset I , of L, if (I fi I K 37 of (I or I is a trivial meet-cover of b.

i(X

Suppose L

LEMMA 2.21

lemma, Ihen L)1JL

)

is freely

)

b,

is equivalent to the following: then is a trivial join—cover

generated by X. and let

and therefore L)(L

)

(\\

he as in the I)re;'iotls

L.

)

I

I'I-tOOF. By the previous lemma, it is enough to show that L) for each k E w. If G E 11w, then (I for some E X , and \\ hitman's condition ( \\ ) and ( \\ 1 imply that any join—cover of (I must be trivial, hence (I E D0( L ). suppose (I E 11k. Ihen (i I)0(L), some u E w. If is a 37 1 for some finite sets I nontrivial join—cover of o, then ( \\ ) implies that for some we have 37 1 (see remark above). Since I L)0( L), is a trivial join-cover of each a E I and therefore Since I is also a join-cover of o, it follows that (I E I ( L). (I E 11 kH—L for some Proceeding h induction, suppose now that Dk ( L ) and 11k

I

I

.

I

/

Ihin

Y

(i

/

foi

11

/

of (I. Let be any nontrivial join—cover of (I. As before for some Let be the set of all a E I such that intl -,it U' 111 / U U I)(L) thiji I

I L)1(L) of join-cover

a

of (I which refines

indi

/i E (

ich

/

1

10111

\\ implies that 37 I is a nontrivial join—cover of a, itch a EU 110111 )

i

i

and is contained iii

L). Hence

L).

E

(I

A join-cover of (I in a lattice L is said to he iiiidandaril if 110 proper subset of is a join—cover of o, and minimal if for any join—cover I of o, I Observe implies I that every join—cover contains an irredundant join—su bcover (since it is a finite set) and

I

.

that the elements of an irredundant join-cover are noncomparable. Also, every nunimal join-cover is irredundant. 2.22 (JOnsson and Xation [75]). If I is a free lattice and is a nontrivial cover of some (I E I). then their exists a minimal cover o of (I with o and LEMMA

I

I'I-tOOF. First assume that I is freely generated by a finite set X Suppose is a nontrivial join-cover of (I E I), and let C he the collection of all irredundant join-covers I iifin intl Li nimi 220 (i E 0(1 of (I D1J I) is finite, hence C is finite. Xote that if I E C and I I for sonic subset of I, then for each a E I there exists a' E and a' E I such that a K a' K a', and since the elements of I are nonconiparable, we niust have a a' a' and therefore I I In particular, it follows that C is partially ordered by the relation Let o he a minimal .

)

C

)

I

I

I

I

.

CBAJ'TEJ

30

2.

with respect to nieniber of C, and suppose I is any join—cover of Because is assumed to he nontrivial, so are and I , and since (I E )

I)

I I u

I u

ifS

c;EIVERjL (I

with

I

I

I

I

a

that is a of that IJ {o} belong to the sublattice I' generated by I I I. By the first part of the proof, there I exists a set f1) I I) such that u and u is a nunimal cover of (I in P. \\ e show that u is also a nunimal cover of (I in I. Let be the lattice obtained by adjoining a smallest element 0 to I, and let Ii be the endomorphism of lb that maps each mem ber of I onto itself and all the remaining elements of X onto 0. 1 hen Ii maps every mem her of fI onto itself, and h( a) K a for all a E lb. Hence, if I is ally join-cover of (I in f then the set I ' h( I {0} is a join-cover of (I in P and I ' u, so that that )

.

give an internal characterization of lower For finitely generated lattices we can boundedness. I his result, together th its dual, implies that an upper and lower bounded fini telv generated lattice is bounded. IHEOREM 2.2:3 (JOnssoll and \ation [75]). following statements are equivalent:

kor any fini telv generated lattice L, the

(i) L is lower bounded:

(ii) D(L) (iii) Every homomorphism of a finitely generated lattice into L is lower bounded, I'I-tOOF. Suppose (i ) holds, let f be a lower bounded epimorphism that maps some free ij( (I) the smallest elenient of the set lattice I onto L, and denote by i( (I) [o) for all (I E L. \\ e show by induction that

i(o)

E

D1J1)

implies

o E

D1JL)

then L)(L) L follows froni the result L)( I) I of Lemma 2.21. Since i is a joinpreserving map, the image i( I of a join—cover I of (I is a join—cover of i(o ). If i( I is trivial, then i( (I K i( a) for some a E I , hence (I f i( (I K f i( a) a and I is also trivial. It follows that i(o) E implies o E 1J0(L). Suppose that i(o E D1Jf and that I is a nontrivial join—cover of i. i hen i( I is a nontrivial join—cover of i( (I and by Lemma 2.22 there exists a minimal join—cover of i(,) i ioin of G intl t(1 o) to I if( if( Furthermore, the set I o) is a join-cover of i(o) with I o) I o. By the minimality of hw intl 't(l 't(l o) follows froni the induction hypothesis that ,f( I o) It and therefore (I E L)1JL). \Ow assume D( L L, and consider a homomorphism f K — L where K is generated by a finite set I. Let I and for k E w let he the set of all joins of nieets of elenients ill For each k E w define maps L — K by )

)

)

)

)

)

)

)

tol

)

JJ{ti

E 11k

.f(ti)

(I].

2.3. SPLiTTING LATTICES

SRI) BOtRDED .IIOMOMO1tP.WSMS

31

(In particular 110 = El' = iK.) We claim that for every k w and a Dk(L), if the set f'[a) is nonempty, then f'[a) = [Jk(a)), and 3k(a) is therefore the smallest element of this set. Since 11(L) = L it then follows that is lower bounded. So suppose that a Dk(L) for some k, and f'[a) is nonempty. If x K and x 3k(a) then

f

,( z

>

,4 a _J'f(lK)

if

{p€JIk:fQc)a}=0 otherwise,

and in both cases f(x) a (f(lK) a since f'[a) is nonempty). Thus [3k(a)) For the reverse inclusion we have to show that for all x K (*)

f(x)

a

implies

x

ç f'[a).

Jk(a).

The set of elements x K that satisfy (it) contains 1' and is closed under meets, hence is enough to show that it is also closed under joins. For k = 0, we have a 110(L), so Ma), hence EL.' Ma). >21(L') a implies f(u) a for some u L, and by (it) u Suppose now that (it) holds for all values less than some fixed k > 0. Let x = EL.' and assume f(x) a, i.e. f(L.') is a join-cover of a. If it is trivial, then (it) is satisfied as before, so assume it is nontrivial. Then there exists a join-cover I' ç Dk_1(L) of a with I' C f(L.'), and by the inductive hypothesis x 3k_1(v) for all v 1'. Now the elements are meets of elements in and the element z = E 4..i( V) therefore belongs to Since f(z) a, it follows from the definition of 3k that z Jk(a), hence E follows x as required. (lii) implies (i) immediately from the assumption that L is ik(a)

it

finitely generated.

0

The equivalence of (i) and (ill) was originally proved by McKenzie [72]. Note that is true for any lattice L, so the above theorem implies that if L is lower bounded then 11(L) = L, and the converse holds whenever L is finitely generated. Together with Lemma 2.21 we also have that every finitely generated sublattice of a free lattice is bounded. It is fairly easy to compute 11(L) and 11'(L) for any given finite lattice L. Thus one can check that the lattices IV, L6, L1,. . . , L15 are all bounded, and since they also satisfy Whitman's condition (W), Theorem 2.19 implies that they are projective (hence sublattices of a free lattice). On the other hand L1 fails to be upper bounded (dually for L2) and A4, L3, L4 and L5 are neither upper nor lower bounded (see Figures 2.1 and 2.2). Note that if L.' is a finite subset of 11k(L) then EL.' 11k+1(L). Since every join irreducible element of a distributive lattice is join prime, and dually, it follows that every

finite distributive lattice is bounded. We now recall a construction which is usually used to prove that the variety of all lattices is generated by its finite members. In Section 1.2 it was shown that every free lattice 1"(X) can be constructed as a quotient algebra of a word algebra I1'(X), whence elements of .L"(X) are represented by lattice terms (words) of I1'(X). The length A of a lattice term is defined inductively by A(x) = 1 for each x X and A(p + q) = A(pq) = A(p)+A(q) for any terms p,q W(X). Let X be a finite set, and for each k w, construct a finite lattice P(X, k) as follows: Take II' to be the finite subset of the free lattice 1"(X) which contains all elements that can be represented by lattice terms of length at most k, and let P(X, k) be the set of all

(JiLtl'itit 2.

32

(JERE1LtL 1tESELTS

finite meets of elements from II', together with largest element lp = EX of 1"(X). Then P(X, k) is a finite subset of .L"(X), and it is a lattice under the partial order inherited from .L"(X), since it is closed under meets and has a largest element, however, P(X, k) is not a sublattice of .L"(X) because for a, 6 P(X, k), a +p 6 a 6 and equality holds if and only if a +p 6€ P(X,k). Nevertheless, P(X,k) is clearly generated by the set X. LEMMA

2.24

ifp =

(i)

q is a lattice identity that fails in some lattice, then p = q fails in a finite lattice of the form P(X, k) from some finite set X and k w. (II) if h .L"(X) —' P(X, k) is the extension of the identity map on X, then h is upper

bounded. (i) Let X be the set of variables that occur in p and q, and let k be the greater of the lengths of p and q. Since p = q fails in some lattice, p and q represent different elements of the free lattice .L"(X) and therefore also different elements of P(X, k). Thus p = q fails in P(X,k). (II) Note that for a P(X,k) ç .L"(X), h(a) = a, and in general h(6) bfor any 6 h(6)

p

p a implies 6 h(6) h(a) = a, whence a is the largest element of

1". Therefore h(6)

a, and conversely 6

f'(a] for all a

P(X, k).

a implies

0

S is a splitting lattice if and only if S is a finite subdirrxtly irreducible bounded lattice THEoREM 2.25 (McKenzie [72]).

Suppose S is a splitting lattice, and p = q is its conjugate identity. We have to show that S is a bounded epimorphic image of some free lattice .L"(X). As we noted in the beginning of Section 2.3, every splitting lattice is finite, so there exists an epimorphism h .L"(X) —, S for some finite set X. The identity p = q does not hold in 5, hence it fails in 1"(X) and also in P(X, k) for some large enough k w (by Lemma 2.24 (i)). Therefore S {P(X, k)}" and since S is subdirectly irreducible and P(X, k) is finite, it follows from Jónsson's Lemma (Corollary 1.7(i)) that S HS{P(X, k)}. So there exists a sublattice L ofP(X,k)and anepimorphismg L—w S. Sincegisonto,wecan chooseforeachz X an element L such that = h(z). Let 1: .L"(X) —' L be the extension of the map z By Lemma 2.24 (II) P(X, k) is an upper bounded image of 1"(X), hence by the equivalence of (i) and (lii) of (the dual of) Theorem 2.23 is upper bounded. Since L is finite,8 is obviously bounded, and therefore h = yf is upper bounded. A dual argument shows that h is also lower bounded, whence S is a bounded lattice. Conversely, suppose S is a finite subdirectly irreducible lattice, and let u/v be a prime critical quotient of S. If S is bounded, then there exists a bounded epimorphism h from some free lattice .L"(X) onto S. Let r be the smallest element of and let s be the largest element of h'(v]. Now r + 8/8 is a prime quotient of 1"(X), for if 8 < I F +8

f

h'[u)

By

Lemma 1.10 there exists a largest congruence (1 on .L"(X) which does not identify r +8 and 8. Since h(r +8) = u v = h(8) we have kerh ç 0, and equality follows from the fact that u/v is a critical quotient of S. Now Corollary 2.12 implies that S is a splitting 0 lattice. Referring to the remark after Theorem 2.23 we note that the lattices L.3, £1,. . . , are examples of splitting lattices. In fact McKenzie [72] shows how one can effectively

2.3. SPL1fl2LVG LATTICES

SRI) BOtRDED .IIOMOMO1tP.WSMS

33

compute a conjugate identity for any such lattice. For the details of this procedure we refer the reader to his paper and also to the more recent work of Freese and Nation [83]. From Corollary 2.12 and the proof of the above theorem we obtain the following:

Coaoisuw 2.26 (McKenzie [72]). A lattice L is a splitting lattice if and only if .L is where 1"(n) is some finitely generated free lattice and cbr. is the isomorphk to largest congruence that does not identify some covering pair r >- s of 1"(n). Canonical representations and semidistributivity. A finite set

Li of a lattice .L is said to be a join rtpiuentation of an clement a in L if a = E C. Thus a join representation is a special case of a join-cover. t is a canoniasl join representation of a if it is irredundant (i.e. no proper subset of t is a join representation of a) and refines every other join representation of a. Note that an clement can have at most one canonical join representation, since if Li and I' are both canonical join representations of a then Li C C and because the clements of an irredundant join representation are noncomparable it follows that Li ç ç Li. however canonical join representations do not cxist in general (consider for cxample the largest clement of A4). Canonical meet representations are defined dually and have the same uniqueness property. A fundamental result of Whitman's [41] paper is that every clement of a free lattice has a canonical join representation and a canonical meet representation. We briefly outline the proof of this result. Denote by p the clement of 1"(X) represented by the term p. A term p is said to be minimal if the length of p is minimal with respect to the lengths of all terms that represent p. If p is formally a join of simpler terms pi,. . . none of which is itself a join, then these terms will be called the join components of p. The meet components of p are defined dually. Note that every term is either a variable or it has join components or meet components.

THEoREM 2.27 (Whitman[41]). A term p is minimal join components pi,. . . and for each i = 1,. . . , ii

(1)

if and only if p = z

X

orp has

is minimal,

(2) (3) for any meet component

r of pj, r

or the duals of(1), (2) and (3) hold for the meet components ofp.

1 are minimal, so by duality we may assume that p has join components if(1), (2) or (3) fail, then we can easily construct a term q such that ?J = P Pi,...,P7e. All z

but A(q)A(p), which shows that p is not minimal. (If (1) üils, replace a nonmiimal by a minimal term; if (2) fails, omit the for which if (3) üils, replace by its meet component r which satisfies v p.) Conversely, suppose p satisfies (1), (2) and (3), and let q be a minimal term such that = p. We want to show that A(p) = A(q), then p is also minimal. First observe that then q must have join components, for if q X or if q has meet components Qi,. . . ?JPi+...+P7 together with (W3)or(W) and since ?J = p we must have equality throughout, which contradicts the miimality of q.

CBAJ'TEJ

ifS

c;EIVERjL

2.

K So let ,q, he the join coniponents of q. For each i E {i, ., II}, ± ... and either has meet components (whose images iii X ) are not below E A or K for some p by condition (3)), so we can use ( \\ 3) or ( \\ ) to conclude that unique , ITL}. Similarly, since q is minimal, it satisfies (1 )—(3 ), and so for E { 1, K p1. each j E {i,. ., ITI} there exists a unique 1 hus E {i,. ., II} such that K p1 K , whence (2) implies i and and j It follows that .

.

.

.

i(

.

.

.

.

the niap

j.

.

is a bijection, iii

i

.

AL')

).

ii, 1h, and since both terms are minimal by (1), A(q).

(:oRoLL;\RY 2.28 Every element of a free lattice has a canonical join representation and a canonical meet tation, I'I-tOOF. Suppose a is an element of a free lattice, and let p he a nunimal term such that a. If p has no join components, then (\\ 3) or (\\ ) imply that a is join irreducible, in p which case { a } is the canonical join representation of a. If p has join components , then condition (2) above implies that I , p,J is irredundant, and (\\3) or and condition (3) imply that I is a canonical join representation of a. I he canonical meet representation is constructed dually. .

.

.

.

.

I he existence of canonical representations is closely connected to the following weak form of distri butivity: A lattice L is said to be if it satisfies the following two implications for all a, £, y,: E L: a

)

(SD)

a

LEMM;\ 2.29 If every element

j:.

I'I-tOOF. Let a

37

,

of a lattice

he a canonical join

which implies

a

±y:

a

£(fJ± :).

L has a canonical join

and dually

representation then L

representation of a, and suppose

a

—f—

y

wehavepKj

Ihenforeach yE

each v E satisfied.

implies implies

H—

a

H—

or pK j,:. Itfollowsthat ±y:for y:. I he reverse inclusion always holds, hence ( SD+ is )

\ow Corollary 2.28 and the preceding lemma together with its dual imply that every free lattice is semidistribu tive. 1 he next lemma extends this observation to all bounded

lattices. LEMMA

2.30

(i ) Bounded epimorphisms

semidis t ri bu ti vity.

(ii) Every bounded lattice is semidistribu tive, L L' is a hounded epiniorphisni. Let i(y) i(:) ± :. Ihen i(a) L is the join-preserving map associated with ,f.

I'I-tOOF. (i) Suppose L is semidistri hutive and f a

i(

i

:

—

:

Hence a

f i(a)

f(

± i(u) i(:))

±

2.3. SPLiTTING LATTICES

SRI) BOtRDED .IIOMOMO1tP.WSMS

35

that holds in ii. (SlY) follows by duality (using aj), whence 1! is semidistributive. Now (II) follows immediately from the fact that every free lattice is 0 semidistributive. which shows

Since there are lattices in which the semidistributive laws fail (the simplest one is the diamond A4) it is now clear that these lattices cannot be bounded. however, there are also semidistributive lattices which are not bounded. An example of such a lattice is given at the end of this section (Figure 2.2). For finite lattices the converse of Lemma 2.29 also holds. To see this we need the

following equivalent form of the semidistributive laws. LEMMA 2.31 (Jónsson and Kiefer [62]). A

lattice

L

Tm

7$

u=Eaj=E6j

(it)

i=1

if and only if for

satisfS

all

7$

u=EEaj6j.

implies

i=1j=1

1=1

Assuming that L satisfies (SD+), we will prove by induction that the statement

for all

I?(rn ii)

'

implies

holds for all in, vi 1. Then (it) follows if we choose any to P(1,1) is precisely (SD+), so we assume that vi> 1, and that P(1,n') holds whenever 1 Sn' 1 and that P(in', vi) holds for

hence

u = to +

holds

+ a,,4 =

to +

E7-.i

u = to + = to + + u = to + a,,465 + u = (to + E7=ii a71465) +

E7i

=

by

to +

where

=

1

in' c in. By hypothesis

a5. Consequently

+ E7=i

+

I!(in — 1,vi)

as required. Therefore

Conversely,

if

implies by P(1, vi), and now

= (to + =

a,,465) +

implies

to +

I!(in,vi) holds for all in,vi. 6 = a + c for some u,a,6,c

(it) holds and u = a +

Coaoisuw 2.32 (Jónsson and Kiefer [62]). A finite lattice satisfies every clement has a canonical join representation.

.L, then u

= 0

if and only if

Let L be a finite lattice that satisfies (SD+), and suppose that I' and Ii' are two join representations of u L. By the preceding lemma the set {ab: a V, 6 W} is

CBAJ'TLJ

36

c;LIVERjL

2.

ifS

again a join representation of a, and it clearly refines both and Since L is finite, a has oniy finitely nianv distinct join representations. Coni hining these iii the same way we 01) a join representation I that refines even other join representation of a. Clearly a canonical join representation is gi veil by a subset of I which is an irredundant join I

.

tai

representation of

a.

I he converse follows from Lemma 2.29

I his finite semidistributive lattices have the same property as free lattices iii the sense that every element has canonical join and meet representations. Further results about semidistri butivity appear in Section —1.2.

Cycles in semidistributive lattices. \\ e shall now discuss another

way of characteriz-

ing splitting lattices, due to JOnsson and Xation [75]. Let L be a finite lattice and denote by 1(L) the set of all nonzero join-irreduci ble elements of L. Every element p E 1(L) has \\ e define two binary relations I and 11 oil a unique lower cover, which we denote by the set 1(L) as follows: for p,q E 1(L) we write JLIq

forsometEL

q °K, then we let if be a relative complement

E'BAJ'TLJ

LAB \JBJEIIES

3,

iii the quotient (it is easy to see that every complemented modular lattice is relatively conipleniented). Clearly = = ± ± y and ally = we filter that does not contain y must also exclude y', so can replace y by y'. \ow I implies [u) [u) ± I, where [y) is the principal filter generated by y. Hence by modularity, we see that

of

[OjJ =

H H H

([v) ±

(else [y), [y and Lr) generate a pelitagoll). So there is a maximal filter C K ).( [y whence it follows thaLr E (7 and [u ) ± (7± F, = [y) ± F± C. Ihis time £ I gives (7 and to avoid a pentagon, we must have [OK) = [u) Hence there is a (C ± [ii) K shows maximal filter II and that (7 E (7, y E II and = °K [ti) ((7 ± Colisequelitlv (7 and II are three distinct atoms of :TK, and since II K C they

I,

I).

I),

generate a diamond. I hus I K C ± II as required. lo see that f is one-one, suppose < y, and £/OK. If I is a maximal filter containing?, then I iLherefore f(u).

I,

E

he a relative complement of hut I f(y) since ?y =

ill

I he above result is not true if we allow K

to he an arbitrary modular lattice. Hall and Dilworth [44] construct a modular lattice that cannot he embedded in any conipleniented

modular lattice.

Coordinatization of projective spaces. Ihe

of a subspace is defined to be the cardinality of a minimal generating set. I his is equal to the height of the su bspace in the lattice of all subspaces. If it is finite, then it is one greater than the usual notion of Luclidian dimension, since a line is generated by a minimum of two points t wo— projective dimensional (sub-) space is called a iniieciice line and a three- dimensional one is called a iniieciice platte. It is easy to characterize the subspace lattices of projective lines: the are all the (modular) lattices of length 2, excluding the three element chain. Xote that except for the four element Boolean algebra, these lattices are all simple Aprojective 5pace ill which every line has at least three points is termed iwnu/egentemie. A simple geometric argument shows that the lines of a nondegenerate proJective space all have the same num her of points. \olldegenerate proJective spaces are characterized by the fact that their su bspace lattices are directly indeconiposable (not the direct product of suhspace lattices of smaller proJective spaces) and, in the light of the following theorem, they form the building blocks of all other projective spaces. .

.

IHEOREM :3.4 (Maeda [51]). Every (modular) indecomnposable (modular) hit tices,

lattice is the product of directly

A proof of this theorem caii he found in [GLT] p.180. 1 here it is also shown that a directly indecomposable modular geometric lattice is su hdirectly irreducible (by Lemma 1.13, it will be simple if it is finite dimensional). An important type of nondegenerate projective space is constructed in the following

way: Let I) he a

ring (i.e. a ring with ullit, in which every nonzero element has a miultiplicative inverse), and let he an Q-dimensional vector space over I). (For Q =

Cii .tl'TEit

52

3.

MODtLS1t t•ititJEflES

Conversely, we have the classical coordinatization theorem of projective geometry, due to Veblen and Young [10] for the finite dimensional case and Frink [46] in general.

THEoREM 3.5 Let P be a nondegenerate L)esarguesian projective space of dimension a 3. Then there exists a division ring II, unique up to isomorphism, such that £(P)

For a proof of this theorem and further details, the reader should consult [ATL] p.111 or [ULT] p.208. here we remark only that to construct the division ring II which coordinatizes P we may choose an arbitrary line I of P and define 1) on the set I — where p is any point of I. The 0 and 1 of II may also be chosen arbitrarily, and the addition and multiplication are then defined with reference to the lattice operations in £(P). This leads to the following observation: LEMMA 3.6 Let P and C) be two nondcgenerate Desazguesian projcctive spwn of dimension 3 and let Dp and Dq be the corresponding division rings which coordinate them. if £(P) can be embedded in £(Q) such that the atoms of £(P) arc mapped to atoms of £(Q), then Dp can be embedded in Dq.

It is interesting to note that projective spaces of dimension 4 or more automatically satisfy flesargues' Law ([ULT] p.207), hence any noncoordinatizable projective space is either degenerate, or a projective plane that does not satisfy flesargues' Law, or a projective line that has k + 1 points, where k is a finite number that is not a prime power.

Arguesian lattices. The lattice theoretic version of flesargues' Law can be generalized to any lattice L by considering arbitrary triples a,b V (also referred to as triangles in .L) instead of just triples of atoms. We now show that under the assumption of modularity this form of flesargues' Law is equivalent to the Aiyjueoisn identity:

+ bu)(ai + bi)(a2 + &i)

60(ai + d) +

+ d)

where d is used as an abbreviation for d = c2(cu

+ ci) =

A lattice is said to LEMMA

be

(au

+ a1)(6u + 61)((a1 + a2)(61 +

+ (au + a2)(6u +

Argueoisn if it satisfies this identity.

3.7 Letp= (ao+6u)(ai+bi)(a2+bj), then

%(a1 + d) is equivalent to the Siguesian identity, (ii) every Siguesian lattiw is modular and (i) the identity p

(iii) to chctk whether the Siguesian identity holds in a modular lattiw, it is enough to which satisfy consider triangles a' = (4,b&,4) and 6' = (it)

(1=0,1,2)

where/ is defined in the same manner as p. Since we always have ho(bi + d) 6o, the Arguesian identity dearly implies p Conversely,let L be a lattice which satisfies the identity p 60(ai+d)+60. We first show that 1, is modular. Given u,v,to €1, with u to, let = v, = u and

3.2.

PItOJECTIVE SPACES SRI) A1WEESLtR LATTICES

53

= a2 = = = to. Then p = (v+u)to and d = to, whence the identity implies (v + u)to no + u. Since u + no (u + v)to holds in any lattice, we have equality, and so .t is modular. This proves (II). To complete (i), observe that p and d are unchanged if we swop the aj's with their corresponding bj's, hence we also havep bu(bi+d)+au. Combining these two inequalities gives a1

p(au(ai+d)+6o)(6o(6i+d)+au) =%(ai+d)+bu(bu(61+d)+%) =%(ai+d)+%bu+bu(61+d) modularity. Also c2, c1 shows d and therefore a,(a1 + d). This means we can delete the term and obtain the Arguesian identity. Now let a,6 V and define = + p), M = + p). Since we are assuming that .L is modular, by

+

+ p) +

=

+ p) =

+ p) +

=

+ p)(aj +

+ p) + p)

Sowe have p = p', and condition (it) is satisfied, If the Arguesian identity holds for a',!? and we define d' in the same way as d, then dearly d' d and

0

hence the identity holds for the triangles a, 6. THEoREM 3.8 Ma modular lattice .L satisfies Desazgues' Law then versely, if .t is Aiguesian, then .t satisfies i)esaigues' Law.

.L

is Arguesian. Con-

Let %,a1,a2,6,61,64 L, p = (au+bu)(ai +6i)(a2+bj), Ck = ({i,j,kJ = {0,1,2}) and d = c2(cu + ci) as before. By part (lii) of the preceding lemma we may assume that

1=0,1,2.

(it)

Define

and

= bj + &,(a1 + 61). The following calculation shows that the triangles (%, a1, a2) are centrally perspective:

(au+bu)(ai+bi)=(p+bu)(ai+bi)

by(*) by modularity

= p + bt,(a1 + Therefore Desargues' Law implies that c2

=

(a1

+ a2)(6j + (61+ bu)(ai +

+ C1

= (ai +a2)(6i +&a)+ai(bu+61)+ci = cu+ci

+ai(bu+6i),

E'BAJ'rLJ whence c, = c1(coH-q

=

±

ci

=

± (o0 ± )(b0 ± ± ± bo)(oo ±

Ci

= =

llODL LAB \JBJEIIES

3.

(47i1 H-

p

H—

(1k)

H—

bo)(oo

H-

(Iii)

H—

(1u)(0u

H-

)

± (Ifl

by (Ic) by (Ic)

(Ri,

obtain b0 p. ± ci) ± b0 Conversely, suppose L is Arguesian (hence niodular) and (o0, ), (b0, b1, b2 are K centrally perspective, i.e.(ou b0 b2. Let = (12)(q c2) and take = b1, = , = q, = b0, U, = iii the (equivalent form of the) = Arguesian identi tv p' K ci') bb. \\ e claim that under these assignments p' = K and ci') q from which it follows that the two triangles are axiall perspective. Firstly, so

we finally

H—

)

H—

H—

H—

H—

H—

H—

)

H—

H—

H—

H—

=

H—

H—

H—

cfl(q

c2)

H—

± (ou H- (12)(bu H- o2 H- b2fl(q H- c2) ± ± ± bifl)(q c2) (oi ± ± ± (Lfl(0u ± b0)(b0 ± ± bifl(c1 c2) (oii ± (1u(bu bifl(q c2) ± H H—

H—

H—

=

H—

(10)(bu

H—

H—

H—

bj(q

H—

± (Ifl(b0 ± bfl(q 50 J)'

H—

q

H—

ci'

b0

H—

H-

bfl(q

K (b0

±

K (b0

H-

=

H-

(b0

H-

H—

c2)

c2) =

ou) = c2. Secondly, H-

H-

± )((o0 o2 b•fl(ou Hb2

H-

±

H—

± H—

)(b0

H-

H-

±

o2) b2) 02) H-b2 =

H—

H-

o0)

Lw

(

H-

H-

(Io))

± bfl) central persp.) 02) H-b2

which iniplies i'

T

cL

H

—

)

ii

cL

)

T

(±ii

ii

H-

H-b2

=

H-

H-

b2)

HH-

HH-

02))

H-

02)

H-

=

H-

I he first statement of this theorem appeared in (Jràtzer, JOnsson and Lakser [73], and the converse is due to JOnsson and Monk [69]. In [GLT] p.205 it is shown that for any Desarguesian projective plane P the atonis of L( satisfy the Arguesian identity and that this implies that L( is Arguesian. Hence it follows froni the preceding theorem that P is Desarguesian if and onl if L( satisfies (the generalized version of) Desargues' Law. Since modularity is characterized by the exclusion of the pentagon X, which is isomorphic to its dual, it follows that the class of all modular lattices .t1 is (i.e. ii E *1 implies that the dual of JJ is also in .\4 ). I he preceding theorem can be used to prove the corresponding result for the variety of all Arguesian lattices.

PItOJECTIVE SPACES AND A1WUESLt.N LATTICES

3.2.

LEMMA 3.9 (Jónsson[72]). The variety of all Aiguesian

55

lattices is self4ual.

For modular lattices the Arguesian identity is equivalent to Desargues' Law by Theorem 3.8. Let .L be an Arguesian lattice and denote its dual by I. Lemma 3.7 (II) implies that L is modular, and by the above remark, so is I. We show that the dual of flesargues' Law holds in L, i.e. for all z0, zi, z2, pt,, Ui, L (it) implies that Then

(n)

z0zi + uoui

+ p1p4)(z0z2 + ni).

I satisfies Desargues' Law and is therefore Arguesian.

Assume (it) holds, and let 6u =

b4=z1p1 and ck=

z0z•4, a1

=

({i,j,k}=

(au + bu)(ai +61) = (z0z2

PuPi, a2

=

Z0Th3, ho

= ZiZ.j,

61

=

UiPz,

{O,1,2}). Then

+ z1z2)(p0p4 + PiPi)

zai

it follows from flesargues' Law that c2 c0 + c1. But c, equals the right hand side of (n). Therefore (n) is satisfied.

by (it), so c2

(z1z2

a2

+

PoPi, c1

z0zi and

0

So far we have only considered the most basic properties of Arguesian lattices. Ex-

tensive research has been done on these lattices, and many important results have been obtained in recent years. mention some of the results now. Recall that the collection of all equivalence relations (partitions) on a fixed set form an algebraic lattice, with intersection as meet. If two equivalence relations permute with each other under the operation of composition then their join is simply the composite relation. A lattice is said to be linear if it can be embedded in a lattice of equivalence relations in such a way that any pair of elements is mapped to a pair of permuting equivalence relations. (These lattices are also referred to as lattices that have a type 1 ,rtprraenkstion, see [ULT] p.198). Sn example of a linear lattice is the lattice of all normal subgroups of a group (since groups have permutable congruences), and similar considerations apply to the "subobject" lattices associated with rings, modules and vectorspaces. Jónsson [53] showed that any linear lattice is Arguesian, and posed the problem whether the converse also holds. A recent example of kiaiman [86] shows that this is not the case, i.e. there exist Arguesian lattices which are not linear. Most of the modular lattices which have been studied are actually Arguesian. The question how a modular lattice fails to be Arguesian is investigated in Day and Jónsson [89]. Pickering [84] [a] proves that there is a non-Arguesian, modular variety of lattices, all of whose members of finite length are Arguesian. This result shows that Arguesian lattices cannot be characterized by the exclusion of a finite list of lattices or even infinitely many lattices of finite length. For reasons of space the details of these results are not included here.

The cardinality of AM. In this section

we discuss the result of Baker [69] which shows there are uncountably many modular varieties. begin with a simple observation

that about finite dimensional modular lattices. LEMMA

f

L

3.10 Let L and Jet be two modular lattices, both of dimension At is onc-one and order-preserving then f is an embedding.

vi

cw. if a map

E'BAJ'TLJ I'I-tOOF.

3,

,\1ODL LAB

\JBJEIIES

\\ e have to show that

(t)

±v)

=

f(s) ±f(y)

ajid

=

Sine t to b oitki t(t — t( — t(u) t(t u) t( )t(u) intl equality holds if is coniparable with For £, fJ nonconiparable we use induction on the length of the quotient ± v in L then a, v are successive elements iii some maximal U bserve firstly that if a chain of L, and since JJ has the same dimension as L and f is one—one it follows that j is 2, then a) v ). If the length of and cover and are both covered by —i—, so (t holds iii this case. Xow suppose the length of y is u ;> 2 and (Ic) holds for all quotients of length ii. i hen either or has length 2. By modularity and by synimetry we can assume that there exists such have I he quotients and /v = length 'y H-v. )

)

H—

)

ii, hence

f(?) It th it t( t') Similarly )

±

=

t(

)—

=

f(s)

t( u) intl

=

t(

—

(f(s)

= t( t'—u) = t( t')— t(

t(

u)

)—

t(

u)

).f( ii).

Let P he a finite partially ordered set and define N( to he the class of all lattices that do not contain a subset order—isomorphic to P. For example if 5 is the linearly ordered set {O, 1,2,3, -I} then N( 5) is the class of all lattices of length

Ct

LEMMA 3.11

icr any Phil te partially

ordered set P

ultraproducts. sublat tices and homomorphic images: (ii) any subdirectly irreducible lattice in the variety N( p)V is a member of N( If). (i

)

N(

is closed under

I'I-tOOF. (i) I he property of not containing a finite partially ordered set can be expressed as a first-order sentence and is therefore preserved under ultraproducts. If L is a lattice and a sublattice of L contains a copy of P, then of course so does L. Finally, if a homomorphic image of L contains P then for each nunimal p E P choose an inverse image p E L, and thereafter choose an inverse image of each q E P covering a nunimal element in P such that p C q, p E P}. Proceeding in this way one obtains a copy of P in L. (ii) I his is an immediate consequence of Corollary 1.5.

IHEOREM :3.12 (Baker [69]). IL eiv are uncoun tablv many

modular lattice varietIes,

I'I-tOOF. Let II be the set of all prime num bers, and for each p E II denote by the p-element (Jalois field. Let = Ii,) and observe that each is a finite suhdirectly irreduci ble lattice since it is the suhspace lattice of a finite nondegenerate projective space. e also let A he the class of all Arguesian lattices of length C -I. \ow define a map f froni the set of all subsets of II to AM by

= e claim

that

since

N(

that

f(i).

fl

fl{N(Lq)V

q

S}.

I.

I

is one-one. Suppose 5, 1 1 II and p E S I hen f(I) N( )V and and is su bdirectlv irreducible it follows from the preceding lemma

3.3. .N-1"iLtMES AND 1"itEESE'S THEOREM

57

On the other hand we must have L, f(S) since L, N(Lq) for some q S would imply that L, contains a subset order-isomorphic to Lq. By Lemma 3.10 Lq is actually a sublattice of .1,, and it follows from Lemma 3.6 that is a subfield of 1,. This however

0 By a more detailed argument one can show that the map f above is in tisct a lattice embedding, from which it follows that A4 contains a copy of 2u1 as a sublattice.

n-Frames and Freese's Theorem

3.3

Products of projective modular lattices. By a projective modular lattice we mean a lattice which is projective in the variety of all modular lattices. LEMMA 3.13 (Preese[76]). if .1 and B arc projoctive modular lattices with greatest and least clement then .1 x B is a projcctive modular lattice. Let f be a homomorphism from a free modular lattice F onto .1 x B, and choose elements u, v F such that f(u) = (lA, Oil) and 1(v) = (OA, 1k). By Lemma 2.9 it suffices to produce an embedding g : .1 x B - F such that fy is the identity on .1 x B. Clearly f followed by the projection irg onto the first coordinate maps the quotient u/tv onto 1. Assuming that .1 is projective modular, there exists an embedding gg: .1 u/tv such that rAffiA is the identity on 1. Similarly, if B is projective modular, there exists an embedding : B v/tv such that ir2fg2 = id2. Define g by g(a,b) = OA(a) +ga(b) for all (a, 6) .1 x B. Then g is join preserving, and clearly fy is the identity on .1 x B. To see that g is also meet preserving, observe that by the modularity of F

g(a,6) = OA(a) + vu + hence

g(a,6)g(c,d) = =

= (OA(a) + v)u +

= (OA(a) + v)(u +

(OA(a) + v)(u + o2(b)ROA(c) + v)(u + 92(d)) (OA(a)OA(c) + v)(u + 92(6)92(d)) = g(ac,bd),

where the middle equality follows from the tisct that in a modular lattice the map I u-' is an isomorphism from u/tv to u + v/v.

t+v 0

Von Neumann n-frames. Let {aj :1 = 1,. ..,n} and {c15 :5 = 2,. ..,n} be subsets of a modular lattice L for some finite n 2. We say that ô = (aj,c15) is an n-fmme in L with atoms a1,... ,a74, if the sublattice of .L generated by the aj is a Boolean algebra and for each 5 = 2,..., n the elements a1,c15,a5 generate a diamond in L (i.e. a1 + c15 = a5 + c15 = a1 + a5 and a1c15 = a5c15 = a1a5). The top and bottom element of the Boolean algebra are denoted by aj) respectively, but they need not (= a1a2) and (= equal the top and bottom of L (denoted by and 1jj. if they do, then ô is called a spanning n-flume. if the elements a1,... ,a74 L are the atoms of a sublattice isomorphic to ?', then they are said to be independent over 0 = a1a2. if L is modular this is equivalent to the conditions aj 0 and Ofor all i= 1,...,n(see [GLT]p.167). The index 1 in c15 indicates that an n-frame determines further elements cjj for distinct 1,5 1 as follows: let c51 = c11 and

=

+ a5)(cj1 + c15).

CiLtl'TEit 3. MODtLS1t tCt1UEflES

58

These elements fit nicely into the n-frame, as is shown by the next lemma. LEMMA 3.14 Let ô = (ai, c15) be an n-frame in a modular lattice and suppose as above. Then, for distinct i,j {1,.. . , n}

(i)

is defined

aj+cjj = aj+aj = cjJ+a5;

01) 45

a,.

=

(ill)

for any fixed index k;

generate a diamond; (v) 45 = (aj + aj)(cjk + Ckj) for any k distinct from

(iv)

i,j.

(i) Using modularity and the n-frame relations, we compute

aj + 45 = aj + (aj + a5)(cj1 + cij) = + + + =

;i

The second part follows by symmetry. (ii) We first show that 45 a,.

Esj

aj +45

aj. by modularity since i

(aj +

=

+ a5)

a,.

=

5

since the afs generate

?'.

Esj

hence if I = 1 then c15 a,. c15a1 = The genera! case will follow in the same way once we have proved (iv). (iii) We first fix i = k = 1 and show that a1 EfL2ci,. rn vi. Ci,. for 3 a1

Er=2 si,. +

ci,.

= = = =

+...+ + c12 + . . . + ci,,4_i) (c12 + . . . + ci,,4)(ai + a2 + . . . + a,. Cl2 + . . . + + c12 + . . . + c1774_1 + by part (II). (c12

Thus a1 11=2 C1,. a1 c1,. ... a1c12 = Let e = and suppose i 1. Then C1j + aje = (clj + aj)e = (clj + a1)e = so hence (ill) holds for k = 1 and any i. = + a1e = + Now (iv) follows from (i) and the ca!culation

Therefore (II) holds in genera!. Using this one can show in the same way as for k = that for k + 1 rn vi and, letting c' =

Then one shows as before that ckj+aje = Ckj, whence aje = For k = 1(v) holds by definition. Suppose i = 1 j,k. (a1

+ aJ)(clk + cka) = =

(a1 (a1

1,

Thus (ill) holds in genera!.

+ aJ)(clk + (ak + aJ)(ckl + c15)) + aJ)(cm + aj + aJ)(ckl + c15) by(ii).

3.3. .N-1"iLtMES LVI) 1"ItEESE'S THEOREM

The case 5 = 1 and note that 4k

1, k

59

is handled similarly. Finally suppose that

+

a5

i,j, k, 1

are all distinct

+

(aj + a5)(cjk +

Cks)

= (aj + aj)((aj + a5 + ak)(cjl + cm) + cks)

=(aj+aj)(;1+clk+ckj)

= (aj + a5)(cji + (ai + as)(clk + cks))

=(aj+aj)(cji+cij)=cjj

0 A concept equivalent to that of an n-frame is the following: A modular lattice L contains an n-diamond 5 = (ai,. . . ,a74,e) if the aj are independent over Oj = a1a2 and e is a relative complement of each aj in lj/Oj (lj = E7=i a5). The concept of an n-diamond is due to Jiulin [72] (he referred to it as an (n — 1)-diamond). Note that although e seems to be a special clement relative to the aj, this is not really true since any n elements of the set {ai,. . . ,a74,e} are independent, and the remaining clement is a relative complement of all the others.

Let S = (aj,e) bean n-diamond and define c15 = e(ai+a5), then Ôö = (aj,cij) is an n-frame. Conversely, if ô = (abcls) is an n-frame and e = E72cis then = is an n-diamond. 1"urthennore = 5. = ô and LEMMA 3.15

4

Since e is a relative complement for each a5e(ai + a5) and

in lj/Oj, aie(ai + a5) =

=

ai+e(ai+a5)=(ai+e)(ai+a5)=1(ai+a5)=ai+a5=a5+e(ai+a5), so

=

+ a5)) is an n-frame. Conversely, if e = c15 then aje =

by Lemma 3.14 (iii) and

aj+e=c12+...+aj+cij+...+c17;

=ai+...+a7=1.

4

= (ai, e) is an n-diamond. Also e(ai +a5) = c15 +a5) = c15 since +a5) = can be proved similar to Lemma 3.14 (ill). Finally, ifS = (aj,e) is any n-diamond, and we let

hence

0 LEMMA 3.16 (Frecse[76]). Suppose I = (ai, e) is an (n+ 1)-tuple ofelements of a modular lattice such that the form an independent set = a1a2, e = and e is incomparable with each Define (1 ranges over 1,. . . , n) e c=fl(aj+e) b=Eaje d=E(aj+b)c=b+Eajc=Eajc and

= (i) (II)

if + e = if =

1

+ b,e),

for all i then

.j* =

= ((ai + b)c,ed).

is an n-diamond in 1/b.

for all i then

is an n-diamond in d/O,is.

CiLtl'TEit 3. MODtLS1t t•It1tJET1ES

60

(ill)

ff6

(ai + 6)c for all (i) Since 6

shows that the

i

is an n-diamond in d/6.

then

e and aj

e we have aj

+6

6 for all i.

The following calculation

+6 are independent over 6: +6)= 6+ aj 6+ 03 + 1

=

6.

=6.

by

(II) Since ajc and 03 ajc a5c over 03. Also e c and aje ajc d imply and ajc + S = (ajc + e)d = c(aj + e)d = S = d. Now (iii) follows from (i) and (II).

a5

=

= 03, the ajc are independent = aje = 03 by assumption,

0

Suppose Al and .t are two modular lattices and f is a homomorphism from Al to .t. is an n-frame in Al and the elements f(ai), f(cij) are all distinct, then (f(ai),f(c15)) is an n-frame in .L (since the diamonds generated by a1,c1j,aj are simple lattices). Risking a slight abuse of notation, we will denote this n-frame by 1(ô). Of course similar considerations apply to n-diamonds. The next result shows that n-diamonds (and hence n-frames) can be "pulled back" along epimorpliisms.

If ô = (aj,cij)

Coaoisuw

3.17 (Jfuhn [72], Freese [76]). Let Al and .L be modular lattices and let 5 = is an n-diamond in L then there is an n-diamond 5= (âi,ê) in Al such that = 5.

1:

Al

—,

L be an epimorphism.

if

It follows from Lemma 3.13 that ?' is a projective modular lattice, so we can find vi,. . . Al such that = aj and the üj are independent over Choose f'{e} such that and let I = Since S is an n-diamond, each nj is incomparable with Defining in the same way as 6,c,d in the previous lemma, we see that f(74 = Oj, f(is) = lj and = aj. Therefore 1 + + 0 whence S = is the required n-diamond.

t

LEMMA

3.18 (Jferrmann and Jfukn[76]). Let ô = (aj,cij) be an n-frame in a modular u1 L satisfy u1 a1. Define nj = aj(ui+cij) fori land

latticeL and let

u=

aren-framesin

Thenôt'

respectively.

A proof of this result can be found in Freese [79]. obtained from ô by a reduction over (under) u.

think of ôt'

as being

The canonical n-frame. The following example shows that n- frames occur naturally in the study of il-modules: Let (il,+,—,•,OR,lR) be a ring with unit, and let £(.W',il) be the lattice of all (left-) submodules of the (left-) il-module i?!'. denote the canonical basis of BY' by e1,. . . with (i.e. Cj = (OR,.. .,Ojj, the 1R in the ith position), and let ..,OR)

aj =

= {rej : r

il]

i,j=l,...,n i#j.

3.3. .N-1"iLtMES LVI) 1"itEESE'S THEOREM

61

Then it is not difficult to check that it) is a modular lattice and that is a (spanning) n-frame in £(itT, it), referred to as the canonirxsi n-fmrne of

= (ai, c11)

it).

Definition of the auxiliary ring. Let L be a modular lattice containing an n-frame = for some n 3. We define an auxiliary ring 14 associated with the frame ô as follows:

i4 =

= {z

.1, :

za2 =

a1a2

and z + a2 =

a1

+ a2]

andforsomek€{3,...,n},z,p€i4 r(z)=(z+clk)(a2+ak), z z

ep = =

(a1 (a1

OR=al,

ir'(z)=

+ a2)[ak + + z)(a2 + ir'(p))] +a2)[ir(z)+r'(p)] 1RC12.

4, or .L is an Srgucsian lattice and n = 3, then (i4, with is a ring unit, and the operations are indqpendent of the choice of k. THEoREM 3.19 Mn

e,

lR)

This theorem is due to von Neumann [60] for n 4 and Day and Pickering [83] for n = 3. The presentation here is derived from kierrmann [84], where the theorem is stated without proof in a similar form. The proof is long, as many properties have to be checked, and will be omitted here as well. The theorem however is fundamental to the study of modular lattices. It is interesting to compare the definition of it with the definition 1) in the classical coordinatization theorem for projective ([ULT] p.209). The element a2 corresponds to the point at infinity, and the operations of addition and multiplication are defined in the same way. There is nothing special about the indices 1 and 2 in the definition of i4 = it12. We can replace them throughout by distinct indices i and 5 to obtain isomorphic rings For example the isomorphism between it12 and it11 (5 1,2) is induced by the projectivity it12

a1

+ a2/O

/

a1

+ a2 + c21/c21

\

a1

+ a1/O 2

it11.

(Since in a modular lattice every transposition is bijective, it only remains to show that this induced map preserves the respective operations. For readers more tsmiliar with von Neumann's 1,-numbers, we note that they are n(n— 1)- tuples of elements 3j5 itj1, which correspond to each other under the above isomorphisms.)

Coordinatization of complemented modular lattices. The auxiliary ring construction is actually part of the von Neumann coordinatization theorem, which we will not use, but mention here briefly (for more detail, the reader is referred to von Neumann [60]). THEoREM 3.20 Let L be a complementcd modular lattice containing a spanning n-frame (4 n w) and let it be the auxiliary ring. Then L is isomorphic to the lattice it)

of all finitely generatcd submodulcs of the it-module its. will be a vector space over Notice that if it happens to be a division ring II, then II, and £flDT',D) £(LF',D). hence the above theorem extends the coordinatization

CRSI'TEit

62

3.

MODtLS1t t•ititJEflES

of (finite dimensional) projective spaces to arbitrary complemented modular lattices containing a spanning n-frame (n 4). Moreover, Jónsson [59] [60'] showed that if L is a complemented Arguesian lattice, then the above theorem also holds for n = 3. Further generalizations to wider classes of modular lattices appear in Baer [52], Inaba [48], Jónsson and Monk [69], Day and Pickering [83] and kierrmann [84].

Characteristic of an n-frame. Recall that the chwuclerislic of a ring it with unit is the least number r = chant such that adding to itself r times equals If no such r exists, then chant = 0. define a related concept for n-frames as follows: Let ô = (abclj) be an n-frame in some modular lattice L (n 3), and choose k {3,. . . , n}. The projectivity a1

+ a2/0

/

a1

+ a2 + ak/ak

induces an automorpliism

\

+ a2/0

on the quotient a1

/

a1

+ a2 + ak/c2k

+ a2/0 given

= ((z + ak)(clk + a2) +

a1

+ a2/0

by

+ a2).

r be a natural number and denote by c% the automorphism say that ô is an n-frame of chwuclerislic r if c%(a1) = a1. Let

\

iterated

r

times. We

LEMMA 3.21 Suppose ô = (aj,c15) is an n-frame of characteristic r, and it is the auxiliary ring of ô. Then the characteristk of it divides r.

By definition °R =

a1

=

and

it, = z (since ,r(lR) = is independent of the choice of k. a,Iit z

From the definition of z

c12.

(c21

p we see

+ clk)(a2 + ak) = cwj. This

that for

also shows that

The condition c%(a1) = a1 therefore implies terms

0

whence the result follows.

The next result shows that the automorpliism is compatible with the operation of reducing an n-frame. For a proof the reader is referred to the original paper. LEMMA 3.22 (Freese [79]). Let ô, u, ôt' and then

(i)

be as in Lemma 3.1&

Hz

a1

+

z+u€ai+a2+u/u and

(II) zu

a1u +

and

= aaz)u.

It follows that if ô is an n-frame of characteristic r, then so are ôt' and below shows how one can obtain an n-frame

The lemma

of any given characteristic.

LEMMA 3.23 (Freese [79]). Let ô = (aj,clj) be an n-frame and r any natural number. if wedofinea = c%(a1), u2 = a2(a+ai), u1 = ai(u2+c12), uj = aj(ui+cij) and u = tij then ôt' is an n-frame of characteristic r.

IL1

3.3. JY-1"iLtMES AND 1"itEESE'S THEOREM

63

Note that u2, defined as above, agrees with the definition of u2 in Lemma 3.18 since

a2(u1 + c12) = a2(a1(u2 + c12) + c12) = a2(u1 + c12)(u2 + c12)

=

a2(u2

+ c12) =

a2c12 + u2

=

u2.

Let it be the auxilIary ring of ô. By definition the elements of it are all the relative complements of a2 in a1 + Since z it implies =z it, it follows that a relative complement of a2 in a1 + (this can also be verified easily from the definition of as). Thus a = c%(ai) it and a + a2 = a1 + a2. By the preceding lemma

4(ai+u)= c%(ai)+u

= a+u+a2(a+ai) =(a+a2)(a+ai)+u = (a1 + a2)(a + a1) + u = a + a1 + u.

Also ai+u2 = (ai+a2)(ai+a) = a1+a shows that a1+u

We can now

prove the result

a,

whence 4(ai+u)

corresponding to Theorem 3.17

for n-frames

= a1+u.D

of

a given

characteristic.

be modular lattices and let f : Al —w .L be an epimorphism. Mo = (aj, c15) is an n-frame of characteristk r in then there is an n-frame 3= (âj,815) of characteristic r in Al such that f(3) = 0. THEoREM 3.24 (Iireese [79]). Let Al and

.L

4

From Theorem 3.17 we obtain an Al such that = 0. If we let u2 and u be as in the preceding lemma, then we see that +

=

o = (aj + u,c15 + u) is an n-frame of characteristic r in Al. Since o has characteristic r by assumption,

f(u2) = from which

it easily follows that

+ a1)) = a2(c%(ai) + a1) = f(ui)

=

=

a2(a1

and

+ a1) =

f(uj) =

Therefore

0 M is not generated by its finite

and below, where Mp is the class of all finite modular

members.This is

follows immediately from the theorem

the

main result of I?reese [79],

lattices. 3.25 (ii'reese [79]). There exists a modular lattice .t such that .t

(Mje)1.

Paooi.The lattice L is constructed (using a technique due to Mall and Dilworth follows: Let

[44]) as

F and K be two countably infinite fields of characteristic p and q respectively, where p and q are distinct primes. Let 1,, = 1") be the subspace lattice of the over F and let 0= (aj,c15) be the canonical 4-frame in 4-dimensional vector space (the index i = 1,2,3,4 and 5 = 2,3,4 through out). Note that 0 is a spanning 4-frame of characteristic p. Similarly let .tq = £(K4,K) with canonical 4-frame 0' = of characteristic q. Since El = w, there are precisely w one-dimensional subspaces in the quotient ac+4/O.s, hence ac+4/O.s J4, (the countable two-dimensionallattice). The via the map z quotient oft, is isomorphic to a1 a4a1 +a2) and since

CHAPTER

6-I

LAJI

3,

\JJUEIIES

1'

1

(3, (13

±

(11

(ii)

(i U

I1igure 3.3

be

= w, we see that ith/03 ± (14 is also isoniorphic to any isoniorphism which satisfies

± ± ±

I he lattice L L

is

(13 (13 (13

± ± ±

Let

(14) (14)

and y

K

—

±

= = =

(14)

constructed by 'looselv gluing" the lattice the disjoint union of

and

a ith/o3 ± (14

Lq over

L

and

in in E

ith/03 ±

(14.

I hen it is easy to check that L is a modular lattice. ( I he conditions on a are needed to niake the two -I-frames fit together nicely.) Let I) be the finite distributive sublattice of L g ni itt U thi t i = 1 2 3 -I} H iguit 3 3 (i)) th it I) thi product of the four element Boolean algebra and the lattice in Figure 3.3 (ii). Both these lattices are finite projective modular lattices, so by Lemma 3.13 1) is a projecti ye modular lattice. Suppose now that L E (,tIF)V = HSP.\4F. Ihen L is a homomorphic iniage of sonic lattice ii E SP.tiF, and hence I is u idaulty /71/lie (i.e. a subdirect product of finite lattices). hereafter we show that any lattice which has L as a homomorphic image cannot be residually finite, and this contradiction will conclude the proof. Let f he the homomorphism from ii L. Since I) is a projective modular sublattice of L, we can find elements in ii which generate a su blattice isoniorphic to I), and = = Let us assume for the moment that (t

)

there exist further elements

and

in ii such that o =

)

is a -I-frame

3.3. JY-1"iLtMES LVI) 1"iWESP$S THEOILEM

65

is a 4-frame of characteristic q and

of characteristic p, 3' =

=

f(Q)=Q,. lattice Al and a homomorI is residually finite, then we can find a finite-rmodular -a. which phism g I Al maps the (finite) 4-frames o and o in a one to one fashion into If

—'

respectively. By Lemmas 3.19 Al, where we denote them by ô = (âj,Ôlj) and and 3.21 they give rise to two auxilIary rings it ç a1 + and it' ç âç + of characteristic p and q respectively. Since Al is a finite lattice, it and it' are finite rings, so = p" and Iit'I = qT' for some n,rn w. Also, since °R = = 1R, it has at least two elements. Now in Al the elements generate a sublattice 1) II, hence &i + â2/03 aç + âYO3,, + o3, = âç and Ô2 + 03, = (see Figure 3.3 (i)). It follows that the two quotients are isomorphic, and checking the definition of i4 above Theorem 3.19, we see that this isomorphism restricts to an isomorphism between it and it'. Thus liti = Iit'I, which is a contradiction, as p and q are distinct primes and liti 2. Consequently is not residually finite, which implies that L is not a member of (Mp)t'. We now complete the proof with a justification of (it). This is done by adjusting the elements in several steps, thereby constructing the required 4-frames. Since we will be working primarily with elements of I, we first of all change the notation, denoting the 4-frames ô, ô' in .L by 3 = (aj,815), 3' = and the in by Also the condition that the elements generate a sublattice isomorphic to 1) (Figure 3.3 (i)) will be abbreviated by To check that holds, one has to verify that the aj are independent over a1a2, the are independent over = 0', 1/a3 + a4 4+4/0', = a1 +0' and = a2 + 0'. Actually, once the transposition has been established, it is enough to show that a1 and a2 since then 0' aç(a2 + 0') = 0' implies

/

I

I

aç =

/

44

4

4

44

4

4

(1+ 0')aç =

+ a2 + 0')aç =

(a1

a1

+ (a2 + 0')aç =

a1

+ 0',

4

and = a2 +0' follows similarly. SUp 1: Let 1' = andê' = eç2+eç3+eç4. diamond in .L. Since ê' 13,/03,, we can choose e' 1'/O' such that f(d) = ê'. Clearly e' is incomparable with each Defining 6' = c' = + e'), d' = + 6')c' (1 = 1,2,3,4) corresponding to 6,c,din Lemma 3.16 it is easy to check that 6' d' c', f(6') = f(c') = =f(d) and = so combining = part (iii) of that lemma with Lemma 3.15 shows that

4

o =

=

is a 4-frame in d/6' and f(o) = 9. Now let 1) = (a1 +a2)6' and consider the elements = holds, +0. Since whence 1(0) = Oj and a1 + a2 = In particular it follows that (ij 4). We show that the are independent over 0. Observe that a3 + a4 6' d implies 0' '7

=

= a4+0. Sincea1

4,a2

4,d

= (a1d'+O)(a2d'+as +a4 +0)

(4c'+6')(4c'+6')=

6'.

CiLtl'TEit 3. MODtLS1t t•ititJ.ET1ES

66

Since Similarly

+ a2, the left hand side is W4

=

Ei5n

t

0', and the opposite inequality is obvious.

Also

= (as+U)(aid'+a2d'+a4+U)

=O+as(aid+a2d+a4+U)=O because aid+a2d+a4 a1 +a2+a4, and likewise = = 6'. We proceed to show that Let I = and observe that Note firstly that = since = = aft + Now

U.

holds.

=

+ 0'.

T+V=aid+a2d+as+a4+U+6'

and a1d +6'

a1d + U =

shows that T +6' = aç +

+ 0')d' = ace together with a similar computation for a2 Furthermore

(a1

TU=(aid+a2d'+as+a4+U)6'

=a=3+a=4+U+(aid'+a2d')6'

Since = aid' + (a1 + a2)6' + 6' = we have Stqp 2: Using Lemmas 3.18 and 3.23 we now construct a new 4-frame

+ üç). By Lemma 3.23 is a 4-frame of characteristic q. Since 3' in .L has characteristic q, it follows that f(u) = 03, and f(ô') = 3' (see proof of Theorem 3.24). Moreover, if we define 0 = and aj = +0 then + = Ôj and calculations similar to the in Step 1 show that the generate a copy of II. where u is derived from u2 =

derived from the Stqp 3: In this step we first construct a new 4-frame = we elements of Step 2 such that Then adjust accordingly to obtain ô' = 3. satisfying Since and â2+â3+â4,it is possible to chooser Isuch that = a1a2 a2+a3+a4. Let c12 = 42(ai+a2) and observe that f(c12) = 42(âi+â2) = by the choice of a in the construction of L. Let e = c12 + and define 6, c, d as in Lemma 3.16. Since f(e) = ê12 + = ê is a relative complement of each âb + considerations similar to the in Step 1 show that

= (ajc + 6,(a1c + a5c + 6)ed)

= is a 4-frame in d/6 and Let and

Then

0s

'4

t4

= 3.

t4=aic+0*s, t4=aie+0*s, a1

v'=EL1t4.

and two applications of Lemma 3.18 show that

=(44,)=

3.3. JY-1"iLtMES

SRI) 1"itEESE'S THEOREM

67

is a 4-frame in il/V. Also since ô' has characteristic q, Lemma 3.22 implies that has characteristic q. Furthermore, it is easy to check that f(u') = and f(v') = 03,, whence

f(o)=cJ. We now show

that

that the elements

generate a copy of II. From

we deduce

+ 0.s = 42(ai + a2)+ 0s = 42(ai + a2 + 0s = 42(4 + 4) = a1 + c12 = a1 + cç2(a1 + a2) = (ai + cç2)(a1 + a2)

c12

=(ai+0*s+42)(ai +a2)= (aç+4)(ai+a2) =ai+a2.

Similarly a2 + c12 = a1 + a2 and a1c12 = a1a2 = c12

Moreover

+ a1e =

a2c12.

Since c12

(c12 + ai)e = (c12 + a2)e c12 + a1c = c12 + a2c.

=

c12

e

+ a2e

= 4(aic+0*s+42) = 4(aic+0*s+c12) = 4(a2c + 0.s + c12) = 4(a2c + cc2) = a2c + 442 = a2c +

4+4+4 implies

A similarca!culationyieldst4 = a2e+0*s. Now a2c+4+t4

act' = ac(aic+a2c+0*s+t4+tA)= aic+0*s, and similarly

4il = a2c + Op. dv' =

Together with a3c + a4c

d(aç

+

+ a2e + Op +

V we compute

+

= = d(aie+a2e+0p) = (aic+ a2c+a3c+ a4c)0p +aie+ a2e = asc+a4c+(aic+ a2c)0p +aie+ a2e = Since

aic+b+v'= aic+0*s+v' = aiu'+V=nç

and similarly n4+V

we have

Lastly wewanttoshow and e

= = = = =

implies V the calculation,

(aic + a2c + b)ed = (aic + a2c)e + 6 (aic + a2c)(a1 + a2)(c12 + + 6 (aic + a2c)(c12 + (ai + +6 (aic + a2c)(c12 + +6 c12(a1c

+ a2c) +

+6

=9, and since we already how that dflrs-Hri

/

CiLtl'TEit 3. MODtLS1t tCt1UEflES

68

completes this step. SUp 4: As in Step 2 we use Lemma 3.18 to construct a new 4-frame

derivedfrom u2 = and u1 = a 4-frame of characteristic p and as before 1(ô) =3. where u is

of characteristic q, so is ô' (Lemma 3.22). It remains to show that holds. Note that

Since

3'

By Lemma3.23

ôis

was

= ¼ = d and Os =

w'

tVj

Therefore

Also u2 =

+

wj(ui

+

(see proof of Lemma 3.23) and u1 = +9 from Step 3 we have =

+

imply u2 +

=

Together with

= u2+9

= A last calculation shows that

= = U

+

+

w' it follows that a1

u = holds. Denoting ô,ô' by

and

3,3'

=

by

ô,ô'

we see

4

and a2

4.

hence

that condition (it) is now satisfied. 0

Note that the lattice L in the preceding theorem has finite length. Let Mn be the class of all modular lattices of finite length, and denote by Mcj the

collection of all subspaces of vector spaces over the rational numbers. By a result of henmann and huhn [75] Mq ç (Mp)t'. Furthermore herxmann [84] shows that any modular variety that contains Mcj cannot be both finitely based and generated by its members of finite length. From these results and Freese's Theorem one can obtain the following conclusions.

Coaoisuw 3.26 (i) Both (M,e)" and

arc not finitely based. (II) C (My,)" C M and all three varieties are distinct. (ill) The variety of Srguesian lattices is not generated by its members of finite length.

3.4.

CV \ ±uxc; BLLAIHOXS

ELIV

1/01) LAB L

69

\

Covering Relations between Modular Varieties

3.4

The structure of the bottom of

In Section 2.1 we saw that the distributive variety P is covered by exactly two varieties, .t13 and I he latter is nonniodular, and its covers will be studied iii the next chapter. \\ hich varieties cover .t13Y (Jriitzer [66] showed that if a finitely generated modular variety V properly contains .t13, then J14 E V or J132 E V or both these lattices are in V (see Figure 2.1 and 3.6). Ihe restriction that V should be finitely generated was removed by JOnsson [68]. In fact JOnsson showed that for any' modular variety' V the condition J132 V is equivalent to V beiig generated by K its mem hers of length 2. 1 he next lemmas lead up to the proof of this result. Recall from Section 1 .-I that principal congruences in a modular lattice can he descri bed b sequences of transpositions, which are all bijective. I wo nontrivial quotients iii a modular lattice are said to be projective to each other if the are connected by some (alternating) sequence of (hijective) transpositions. For example the sequence (10/b0 projective to each other in u steps. I his makes (10/b0 and o,, sequence is said to he rionual if for even k and k for odd k ± (k 1, ., ii 1). It is rionual if iii addition for even we have ± K and for odd k

/

/

.

.

LEMMA :3.27 (Gràtzer [66]). In a modular lattice any alternating tions can he replaced by a normal seq uence of the same th,

sequence of transposi-

Pick any three consecutive quotients froni the sequence. By' dualit we may assunie that (I/b c/il. If this part of the sequence is not normal (i.e. (I ± c then we replace by (I ± c/y(o ± c). lo see that o/b (I ± c/y(o ± c) we only have tj )(o to observe that c) (It) b and, by' modularity', (I (o c) v(° c) c/il. Xotice also that the normality of adjacent ± c) ± c. Sinularly (I ± c/y(o ± c) parts of the sequence is not disturbed by this procedure, for suppose u/c' is the quotieiit that precedes (I/b in the sequence, then b implies 'v( c) b(i c) b. I hus we can replace quotieiits as necessary, until the sequence is normal. I'I-tOOF.

/ \

/

\

H—

H—

H—

H—

H—

—

H—

/ \

(Jrätzer also observed that the six elements of a normal sequence (I/b c/il generated by' o, and c. Hence, in a modular lattice, they generate a homomorphic image of the lattice in Figure 3.-I (i) (this is the honioniorphic iniage of the free modular lattice iw(o, y, c) subject to the relations (I ± y (I ± c y ± c). If the sequence is also strongly normal, then y y ± oc, and so o, y and c generate a homomorphic image of the lattice in Figure 3.—I (ii). Of course the dual lattices are generated by a (strongly) normal sequence (I/b c/il. Figure 3.-I (ii) also shows that strongly' normal sequences cannot occur in a distributive lattice, unless all the quotients are trivial. However JOnsson [68] proved the following: are

\ /

Suppose L is a modular lattice and p/q and r/h are 11011 trivial quotients of L that are I)rojecti;'e in ii steps. if no nontrivial suhquotients of p/q and r/h are projecti;'e in frwer than n steps. then either n K 2 or else p/q and r/h are connected by a strongly normal sequence. also in ii steps. LEMMA :3.28

PROOF. Let p/q r/h (sonic ii 3) he the sequence that connects the two quotieiits. By Lemma 3.27 we can assunie that it is normal. If it is not strongly normal, then for sonic k with U ii we have .

.

.

/

\

CiLtl'TEit 3. MODtLS1t tCt1UEflES

70

z

a

6+

a

(i)

(ii) Figure 3.4

but or dually. Let Ck_1 = 6k—1 + ak_lak+1, and for i vi, i k — 1 we define to be the element of that corresponds to Ck_1 under the given (bijective) transpositions. With reference to Figure 3.4 (i) it is straightforward to verify that Ck_1/bk_1 Since

0
b> c > d. Since .L is subdirectly irreducible con(a, b) and con(b, c) cannot have trivial intersection, and therefore some nontrivial subquotients a'/b' of a/b and p/q of b/c are projective to each other. By Lemma 3.30 we can assume that they are projective in three steps. Similarly some nontrivial subquotients p'/qf of p/q and d/d' of c/d are projective to each other, again in three steps. Since all transpositions are bijective, p'/qf is also projective to a subquotient of

3.4.

('0 \ nuxc;

J1LLAIHOXS

KLIV

L

LAB

73

\

(I

Figure 3.6

(I'/b' iii three steps. From Lenima 3.31 we infer that p'/q' transposes up onto a lower edge of a dianiond and down onto an tipper edge of a diamond. It follows that the two dianionds generate a sublattice of L which has J132 as homomorphic image. I his contradicts (i thus (i ) implies (ii). Every variety is generated by its stihdiirectly irredtici hle mciii hers, so to prove that (ii) (id implies (iii), we only have to observe that the inclusion (I(b ± cci)(c ± ci) K b holds iii every lattice of dimension 2. Indeed, in such a lattice we always have c K ci or ci K c or cci ci) (i(b c)ci K (Ui K b ± (IC ± oil, in the 0. In the first case (i(b ± cci)(c second (i(b cci)(c ci) o(b ci)c K oc K b oc oci and iii the third (i(b cci)(c ci) (ib(c H- ci) K b K b H- (IC H- (Ici. Finally, Figure 3.6 shows that the inclusion fails in J132 , and therefore (iii) implies H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

(ii). For any cardinal Q 3 there exists up to isomorphism exactly one lattice iL with dimension 2 and Q atoms (see Figure 2.2). For Q n E w each generates a variety while for Q w all the lattices ii, generate the sanie variety since they all have the same finitely generated sublattices. Clearly .\4 I and by JOnsson's Lemma covers .\4 for 3 K ii E w. I he above theoreni implies that i132 for all conversely, variety if V is a of modular lattices that satisfies i132 V, then V 3, and is either I, P, or *1,, for sonic u E w, ii 3. Ihus we obtain: COROLLARY 3.33 (JOnsson [68]). In the

by exactly three varieties: *1

*1

lattice A. the variety .t132 and *1

(3

K

ii

E

w) is covered

H-

PROOF. *1 fl *132 P is covered by By *13 is covered by *132, and *1 n cover .ti Suppose a variety V properly the distrihutivity of A, .ti fl .t132 and .ti n includes .\4 If V contains a nonmodular lattice, then X E V, hence .\4 HV. If V contains only modular lattices, then either i132 E V or i132 V. In the first case we have .t1 for some V, while in the latter case I heorem 3.32 implies that V Hk E w. Hence .t1 V, and the proof is complete.

I

I

I

I he proof in fact shows that, for ii 3, C( *1 {*1 *1 H- *132, *1 Hstrongly covers .t1 (see Section 2.1). But this is to be expected in view of I heorem 2.2 and the result that every finitely generated lattice variety is finitely based (Section .5.1).

(7-IAPI LII

74

LAJI

3,

\

hUE I JES

A3

N li1iguni' 3.7

ii E

two join ini'tiuci bli'

hat ,t13 oniy one.

U

t

w)

Further results on modular varieties. of Hong [72]

t

I HEOREM 3.34

(

.t14 anti .t132

(

t

hi' lat

)

in

3.7.

•tl

(4

1 hi'

main

hi' following:

L 1)0

a

,wbclirehlv iITeclucibIe modular Ia -1L

J2, J3, i13,

Lice

and

HS{L}.

I lion lie clnnen,qon of L I hi' proof of

an'

t

hi'ori'ni a normal

list gi'ni'nati'ti by t hi' lat

t

of this

i'

1/32,

(oRoLLAI-ty 3.35 (Hong [70]). —

Li't

tI

.t14,

icr 2

on a tii't aili'ti of quot 'nt fit Li't 13

anti

ii

e

—

t

ogi't hi'n.

A2,

hi' thi'

anti

varieft

w —

of how t hi' tiianiontk that

—

covered by A3,

—

P?,

—

li'ngt h modular Ia t not i'xci'i'ti width not i'xci'i'ti ii (I ITt, it ). Xoti' that I hi'oni'ni 3.32 i'vi'ny lat tici' of li'ngth at anti 3 can hi' i'm hi'titii'ti in thi' — lat tici' of a pnoji'cti vi' plani' ( [GLT] t hi' vanii'tv gi'ni'nati'ti by all such lat IT?

anti

hi' t hi' vanii'ty gi'ni'na t i'ti by all

3.-i.

CO \ ±uxc; J1LLAIHOXS

KLIV

Ihe variety

COROLLARY :3.36 (Hong [72]).

±

±

L

is

± A2,

LAB

\

7.5

strongly covered by the collection

t

± A3,

From I heorem 2.2 one niav now dednce that is finitely based. we first of all note that since i13 has width 3, Considering the varieties and are both equal to the distributive variety P. 1 he two modular varieties which cover .t1 are generated by modular lattices of width -I, hence I he variety is investigated in Freese [77]. It is not finitely generated since it contains simple lattices of ar hi trarv length (Figure 3.8 (i ). Freese o htains the following result: )

IHEOREM

;3

.37

Ihe variety

is s

varieties: I

1

r

,

.

covered by the following collection of ten

it

n

j.

He also gives a complete list of the suhdirectly irreducible members of this variety, and shows that it has uncountablv many su bvarieties. Further remarks about the varieties appear at the end of Chapter .5.

76

3,

15

14

(i

3.8

\1OL)L

\JJUL11LS

Chapter

4

Nonmodular Varieties 4.1

Introduction

I he first significant results specifically about nonniodular varieties appear iii a by Mckenzie [72], although earlier studies by JOnsson concerning su blat tices of free lattices contributed to some of the results in this paper (see also kostinskv [72], J Onsson and Xation [75]). Splitting lattices are characterized as su bdirectlv irreducible bounded homomorphic images of finitely generated free lattices, and an effective procedure for deciding if a lattice is splitting, and to find its conjugate equation (see Section 2.3) is given .Xlso included in Mckenzie's paper are several problenis which stimulated a lot of research in this direction. One of these pro blems was solved when Day [77] showed that the class of all splitting lattices generates the variety of all lattices (Section 2.3). Mckenzie [72] also lists fifteen subdirectly irreducible lattices L2,. , (see Figure 2.2) each of which generates a join irreducible variety that covers the smallest Davey, Poguntke and Rival [75] proved that a variety, generated nonmodular variety by a lattice which satisfies the double chain condition, is semidistri butive if and only if it does not contain one of the lattices J13, , L5. JOnsson proved the same result without the double chain condition restriction, and in JOnsson and Rival [79] this is used to show that Mckenzie's list of join irreducible covers of is complete. Further results in this direction by Rose [84] prove' that there are eight chains of semidistri hutive varieties, each generated by a finite subdirectly irreducible lattice }V is the Lq0, Lq3, Lq4, Lq5 (ii a> 0, see Figure 2.2), such that = and only join irreducible cover of for i = 6, 7,8,9, 10, 13, U, 1.5. Extending sonic results of Rose, Lee [85] gives a fairly complete description of all the varieties which do not contain any of i13, L2, L3, , In particular, these varieties turn out to be locally finite. Ruckelshausen [78] obtained some partial results about the covers of .t13 A, and \ation [85] [86] has developed another approach to finding the covers of finitel generated }V has tell join irreducible covers, and that above varieties, which he uses to show that there are exactly two join irreducible covering chains of varieties. I hese results { are intentioned again ill more detail at the end of Section -1.-I. I he notions of splitting lattices and bounded homomorphic images have been discussed ill Section 2.3, so this chapter covers the results of JOnsson and Rival [79], Rose [84] alId .

.

.

.

.

.

.

.

—t—

Lee [85].

f

S'JLT'JHTIT\

JUT

x

'.r 'fi (z

)

fi = fi + 2 = .r + 2 = zfl = zr

t7 '.r)

'fi (z

(p?np .r)fi + (z = zr .r) + z(fi = fir .r) + .r)(fl 4- (z =

+

fir

'.r

'fi (z

¶27) $1 (JT?np i

.r

.r

= fi + 2 = fir zr 4- zfl = .r 4- fi

+

+ z = 4-fi 2 .r) + z(fi = fir

z

.1

zfl=z(fl+a) Vt

P!UTBS JJT?33fl

I1TO.TJ

&T!A!

110ii.)3S

+as)

141 iT

= .r fi = = fir = zr ——

n

(

n

y

ft

11? OJ1T%LFT

014$ $i 1TIJi

1Ji OA1illq puT? 3)11314 A314i TI1TI1TOJ

$OiJdliTi

J/qwmiwo:infou 1?

119

$PIOI1TOJO

71

7 (

Ofl?

7i11

$OiJdIiTi

1)

$OiJdIiTi

1)

$)Mi1TJ

1?

7

3M

1

71

p111?

7

(

$1

3141

'.r)?7

=

fROpi pill? jo:jjq

71

1)111?

71 71

fir

'fi (z 01 'J—

))iJJPJ

110991191) Pill?

7-/f-f

9

J-

Jo 3$Jll0.)

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3.ThJfRJ 141 iT

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393141 $111311T313

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$1141

(yj—

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pccopJo

AJOAipOd$OJ [.r) pill? iii i: 7

f)

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3fn3ipIIi

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'Wfl0iw!i?f0J cri JOJ 33$)

pill?

O$fl? \)1TJJ IIOA1[)

$IIOUJTJOJ

$1 11T3 poppoq 111

71

tJ

JO AID? iTi$1P1I1T)$ OA1illq

JJi; 31LTM

'0 '.r 'fi

pill? AJfnhlp

——

'Z.7

))iiiTIqn$

AID '1)

= .r zfl = fl).r 4-

.j

(

Yr AJfnhlp) ( $TYjq) ( Jo 7 O.)IIOff 14loq

twdIiTo. $111311T313 Thiioq )14J fRdL)IILTd .)1UTOJT 31) 111 AlIT? lilOilOflI) /i) Oi 11Th) 9 3J3141 qiqix) 71 D '1) D '7

=

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2 D

$11

o$p? 1TIJi .r

2

1P141 )14J

3$JOAOJ 110i$llpIIi

Ajfnhlp)

3141

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——

J Ji JOJ

91

lpIq 3)11 JO 7 .)1IJdJO11TO$i 01 '?7 MI

)141

.r

7

JOJ 3111111 $3.ThJfRJ 314J 3$JOAIIO.)

141 iT

POIJ1JOA MI

3

3.)iifRJ

1?

'J 1)

.r D

0 14)119 JT?14J

D

J

71

J

+

1)

Pill?

XIIT? IIIT?14.)

:

13

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Vi

[GLI •( .11

t?

7 3110

110.)

u1tq

Jo

ituip .iutpru 'Z.J

Jo 3$0dd115

fP14(J

7

$1 1011

iTJQIfui1iTOQ oAqllq

Aff \!lifpflp 3M

ATTIT OI1T1T$$T 1!p14!1

)j)141

4.2. SEMJ.D1STLtIBETIV1fl'

79

U

x p

I,

xv'

x

(i)

(ii)

(iii)

Figure 4.2

exist

u,x,p,x€

L such that

u=x+p=x+x

(it) As a first observation we have

that x, p and

x

but

x+px ej. It follows that (lb = and (I ± b = c ± b, whence X(c/o,b). .

.

Figure duality)

.

.

.

-LU

= c such that X ( a/ c') implies eb

shows that the above result does not hold if L includes LH or (by I he next result shows just how useful the preceding lemmas are.

-L7 (ii)

IHEOREM 4.8 (Rose [84]). Let L be asubdirectlv irreducible seinidistributive lattice that excludes LH and 1lien L has a unique critical quotient.

PROOF. Suppose to the contrary that c/o and p/q are two distinct critical quotieiits of L. I hen (p, q) E con( o, c), hence by Leninia 1.11 there exists p' E p/q such that p' > q

CBAJ'TLJ

86

-1,

XOXUODL LAB \JBJEIIES

and c/o prolects weakly onto p'/q in k steps. \\ e niav assunie that c/o (7 p/q (else p/c or (I/q is critical and can replace p/q), and p/q (7 c/i. Consequently k 1, p'/q (7 c/o and therefore we can find a nontrivial quotient u/v J7 c/o such that c/o u/v. By duality, suppose that c/o u/c', and put o' = cc'. Since c/o' is also critical, we get (o', c) E con( o', c') Again by Lemma 1.11 there exists c' E c/o, c' ;> o' such that v/o' .

projects weakly onto c'/o' (see Figure

(iii)). Consider a shortest sequence

-1.7

v/o' = Lu/go

.

.

= c'/o'.

.

\

siice c' IT v. U bserve also that if n = 2, then we cannot have '/i' K v. c'/o', since that would imply c' = o' ± First suppose that v/o' I hen only (i) or (ii) of Lemma -1.2 can

Clearly

n

a>

2

apply, since the sequence cannot be shortened if n 3, and for u = 2 this follows from the observation above. If (i) holds, then there exist o", b, c" E L such that X(c"/o", b) and b/bc" v/o'. Since c'/o' is critical, (o', c') E con(o", c"), whence by Corollary t7 we have X ( c'/o', b). If (ii) holds, then there exist o", b, c", / E L such that X ( c"/o", b), I/o" v/o' and I/o' o" ± b/b Again we get X ( c'/o', b) from Corollary -1.7. But in both cases we also have b K o', so this is a contradiction. suppose that v/o' .As we already noted, this implies ii a> 3, so we may only apply the dual parts of (i or (ii) of Leninia —L2. I hat is, there exist o", b, c", E L with X ( c"/o", b) and either o" ± b/bI v/o' or v/I 1 c'/o', v/I b/bc". Again Corollary -L7 gives X(c'/o', b). In the first case this contradicts b o', and in the second, since b/bc" v/I, we have c' = b ± I b ± o' c', and this contradicts o' =

I

I

.

)

I

/

Xotice that if a lattice has a unique critical quotieiit c/o, then this quotieiit is prime (i.e. c covers o), c is join irreducible, o is nieet irreducible, and con(o, c) identifies no two distinct elements except c and o. lo get a for the above theoreni, the reader should check that the lattices X, L6, L7, and each have a unique critical quotieiit, where as LH and each have two.

4.3

Almost Distributive Varieties

Recall the definition of the identities and and (5Dj) only hold in the trivial variety, while

in I heoreni -1.2. Of course (5Db)

holds in the distributive variety, hut fails in J13 and X (Figure —1.8). 1 hus (and by duality (SDjF)) is equivalent to the distributive identity. I he first identities that are of interest are therefore )

± z) =

(S]L)

(Slit)

—I—Liz

=

± L(z ± Ly)) ± z(L ± tifl.

—t—U(L

Neardistributive lattices. A lattice, or a lattice variety, is said to be if it satisfies the identities (SJL and (Slit). I his definition appears in )

lice Lee [85]. By

I heorem -1.2 every neardistributive lattice is semidistributive, and it is not difficult (though somewhat tedious) to check that X, L6,. , and are all neardistri hutive. .

.

DJSJJUBL LiVE

4.3,

\JIUEIILS

87

-V

z

1/

1/

(i)

(ii)

(Hi)

li1igure -1.8

the other hand Figure fails in

-1.8

(iii) shows that (SD;) fails iii LH, and by duality (SDt)

IHEOREM 4.9 (Lee [85]). lattice vanety seinidistribu tive and contains neither LH or

V

is

neardistribu tn'e ii and unIv if

V

is

I'I-tOOF. I he forward iniplication follows inimediately from the remarks above. Conversel, suppose that LH, V and V is semidistri butive hut not neardistributive. \\ e show that this leads to a contradiction. By duality we may assume that (SD;) does not hold in V, sofor somelattice L E V, £,y,Z,G,CE L = (I - 6 such that 6= (a' + 6)z. Consider a shortest sequence

z/6 =

's's,

z1/p1 's's,

... 's's,

z74/p74

= c/a.

Clearly vi 2. Thecasevi = 2canalso beruledout,since z/bisatransposeofz1/p1, while Theorem 4.8 implies that c/a is a subquotient of z74_1/p74_1, hence z1/p1 = wouldimply z = zi+6 a'+bor 6 = p1z ab, bothof which areimpossible. Thus vi 3. If z/6 z1/p1 z2/p4, then z1/p1 is prime, since c 1 - c. On theother hand a'+b a'+ho implies .N(a'+b/a',zi), whence a'z1 = (a'+b)zi =

/ \ /,,

\ /

E'JTIAPTER

108

(I'H-b

LAB \JBJEIIES

-I.

z

(i)

(ii)

Figure

a contradiction. I herefore Since L excludes and

-1.21

('

and G' b0 are nonconiparable and b )( G' b0 = c'. it follows as iii the proof of Lenima —1.3.5 that the o', c', b and b0 generate L7. I his G' b b b0, and as G' o'b, we can only have (I'b0 By Leninia —1.6 X(o' bu/o', and G' b0/i' bu/i'bu imply X(bu/o'b, ). Hence K z this implies b0 K = and together with (I'b0 and = K z. It follows that G' b b z, which is a contradiction. \ow suppose that X(c'/o'bu, Since we are also assunung that L excludes we can dualize the above argument to again o btaiii a contradiction. I hus = e complete the proof by showing that o', c', b and generate (Figure —L21 (ii)). we have Clearly G' implies In fact must , since = ;> = would imply (I'b = = b by the dual of Lemma -L3-1 (i), a contradiction. = Also (I'(iYb , since o' o'b, and now the dual of Leninia —L3—1 (i (I'b implies G'b b. Hence = b = z. Finally G' b/c' b/o'b, X ( b/o'b, whence and Lemma -LU imply X(o' H- b/c', = (o' HG'

H—

b

H—

H—

H—

)

H—

H—

H—

)

\

H—

H—

H—

H—

H—

H—

H—

.

H—

\

H—

)

H—

H—

H—

H—

The sequence

I he next theorem is in preparation to proviig the result due to L7j Rose [84] that is the only join irreducible cover of L7j. A quotieiit c/o of a lattice is an if for sonic b, b0, , b,, E L the set {o, c, b, b0, , } generates a su blattice , with c/i as critical quotient (Figure 2.2). In this case we shall of L isomorphic to write b, b0, , bJ. .

.

.

.

.

.

.

.

.

IHEOREM 4.37 (Rose [84]). Let L be a subdirectlv irreducible lattice, and assume that the variety {L}v contains none of the lattices i13. , L5. L7, , Suppose further that. fbr some k E w. c/o is an oIL, Ihen .

(i)

.

.

.

if L does not have any -quotients. then c/o L/ con( c) has no -quotients. L3. then c/o is an

(ii) if L is finite and L

.

.

is a critical quotient

of

and

L

-quotient.

ti

{L}v is semidistri hutive, and by I heorem I'FtOOF.(i) By I heorem critical quotieiit, which we denote by Choose b, b0,. , so that . .

tS

L has a unique b, b0,. . , b1J .

4.4. FU1LL'JiEit SEQUERCES 01" tCt1UETIES

109

Figure 4.22

holds. We will prove several statements, the last of which shows

that z/p = c/a. The first

three are self-evident. (A) Any nontrivial subquotient d/a' of c/a is an i.4J-quotient. (B) Suppose that for some a',c', z L we have JV(c'/a', z) with a a'z c a' + z c. Then .Lr'(c'/a', ;b,bcj,. . . ,b,j holds (see Figure 4.22). .L we have .N(a + (1 (C) Suppose that for some z {0,. ..,k}). Then 147'(c/a,b,bu,. . . ,6b z) holds, and similarly if f'I(a + b/ab, z) then we have 14j(c/a,b, z). (0) l'br any quotients u/v and p/q in ifu/v p/q c/a, then u/v tic/va c/a, and all four transpositions are bijeetive. By Lemma 4.5 the lattice generated by q, c, 6 is a homomorphic image of the lattice in Figure 4.23 (i). The pentagon .N(r/d, 6) is contained in a + b/ab, whence it follows that i.4J(r/d,b,b,,,. . . ,6,). From this we infer that r/d is distributive, for otherwise r/d would contain a pentagon .N(d/a',b') (by semidistributivity L excludes A4), and we would have

/ \

4

\ /

,6j).

\

\

hence the transposition r/s e/d is bijective. By Lemma 4.6, the transpositions p/q r/s and e/d c/a are also bijective, and we consequently have p/q c/a. Again by Lemma 4.5, the lattice generated by q, u,b is a homomorphic image of the lattice in Figure 4.23 (II). Note that ab bq c 6 a + 6, whence .N(6/bq,bo). Since v + 6/I 6/bq, it follows by still another application of Lemma 4.5 that the lattice generated by d',b and is a homomorphic image of the lattice in Figure 4.23 (ill) and by Lemma 4.12 the transposition v + 6/d' r"/# is bijective. Put I = r'(b + 6,) to

\

\

\

obtain .N(t/d',b) and therefore . . ,6j). This implies that t/s" is distributive, and so is r'/d', since the two quotients are isomorphic. The transposition e//I r'/d is therefore bijective, and the bijectivity of u/v e'/d' and r'/d p/q follows from Lemma 4.12. Consequently u/v p/q. Now semidistributivity (Lemma 4.4) implies u/v tic/va c/a. By duality, these two transpositions must also be bijective. (E) if c/a projects weakly onto a quotient u/v, then u/v wi/va' c'/a' for some subquotient c'/a' of c/a Assume that c/a = zo/pu 'S.,,, ",,, ... .'v,,, z74/p74 = u/v, where the transpositions alternate up and down. We use induction on vi. The vi = 0,1 are trivial, so by duality we may assume that c/cyi zi/pi 2 pi + z2/pi Since c/cyi is also

\

/

/

\

/

/

/

\

/

CiLtl'TEit 4.

110

tCt1UEflES

v+6 d'

(ii)

(i)

'1

I

bob

(iii)

Figure 4.23

(0) to conclude that the first transpose must be bijective. hence p' + z2/p1 transposes bijectively onto a subquotient c'/a' of c/a (a' = cpi). A second application of (0) gives c'/a' dz2/a'pj z2/p4, proving the case vi = 2, while a .14-quotient we can apply

for

vi

>

2 the sequence

/

\

can now be shortened by one step. The result follows by induction.

(F)z/p=c/a. z/p is critical and prime, c/a projects weakly onto z/p. By (E) z/p projects onto a subquotient c'/a' of c/a and since z/p is the only critical quotient of L, we must have z/p = c'/a'. if z c c, then the hypothesis of part (i) (of the theorem) is satisfied with a replaced by z, and we infer that z/p is a subinterval of c/z, which is impossible. hence z = c, and similarly p = a, which also shows that c/a is the only of L. Since

To complete the proof of part (i), suppose

I = L/con(a,c) contains an La-quotient,

i.e. for some u,v,d,4. . .,4 L . .,4) in I. if c = u in L, then u = c > a > v and . .,4), which would contradict the fact that c/a is the only 14.quotient of L. Thus c u and, similarly, a u and c,a v. if a = then we must have .N(u/v,a) in 1,. But by Corollary 4.7 this would imply .N(c/a,a), which is impossible. So a d and, more generally, c, a {d, . . , 4}. Since con(a, c) identifies only a and c, we infer that L with u/v c/a, and this contradiction concludes part (i). For the proof of part (ii), we will use the concept of an isolated quotient and all its implications (Lemmas 4.33 — 4.36). Let c'/a' be an isolated quotient of L such that we

have

4.

c/a ç c'/a'.

(C) Suppose that for some 6

a'b-(b-(a'+b.

L we have .N(c'/a',b) with a'b

-


0. Hence by

H—

Although P (u) characterizes all those pairs of finitely based congruence distributive sti bvarieties whose join is finitely based, it is not a property that is easily verified. Fortunately, for lattice varieties, k— boundedness can be expressed ill terms of weak projec— tivities. More precisely, if we exclude the use of the polynomials in the definition of a k-translation then a k-translation from ouie quotient of a lattice to another is nothing else hut a sequence of k weak transpositions. I wo quotients (I/b and (I'/b' are then said to

5,3,JOJXS OF FLVJ JELl BASH) \JIUEIIES

123

he k- bounded if they both proiect weakly onto sonic nontrivial quotient c/il in less than or equal to k steps. Furthermore, if p = q is a lattice identity then we can assume that the inclusion p K q holds iii any lattice (if not, replace p = q by' the equivalent identi tv pq = p q) and the sentence Uk can be rephrased as: H—

Lt

IL fbi all a, v E the quotients q( a )/p( a) and weakly trivial on to a common non quotient in k (or less) q'( v)/p'( v) do not both project styps.

L E

Uk II.

Ihe following

and only

is a slightly sharpened version (for

lattices) of Lemma 5.8.

5.15 Let I he a homomorphic image of L and let,r /y he a prime quotient in L. quotient (I/b of L. if?i/i; projects weakly on to 1/7 in nsteps. then o/b projects weakly onto in ii ± 1 steps if ii ;> 0. and in two steps if ii = 0. LEMMA for ally'

I'I-tOOF. Suppose ?I/7

projects onto

in

0

steps, i.e. it =

and

/

=

y

suppose that it/i; y y ±y (LV H—gil ,t. proJects weakly onto in n ;> 0 steps. Since the other cases can be treated similarly, we may assume that it/h for sonic E L, K = 1,. ., ii 1. In this case implies that there exists b K Letting = = and (I/b Xext, there exists (If', E L ± (I we have we have such that and (If, K (Ii. Let ting U, = and = = iitepeatiiig this process we get .

.

.

.

(I/b

where T' =

and

(Iç/bç ' =

.

projects weakly onto £/y hence the result follows.

Ii

(Ic/bc

£'/y',

...

the first argument t J' ± 2 steps. One of the steps ±

50 (I/b ± can still be eliminated,

Given a variety V, we denote by' ( the variety that is defined by' the identities which have u or less variables for some positive integer u. Clearly VI ( V and iii) = (iii) for any iii n. Another nice consequence of this definition is the following lemma, which appears iii JOnsson [74]. of

V

)

LEMM;\ 5.16 If (u) and is finitely based. V ± V' I'I-tOOF. In general, if

hi, (u) are finite

hi( u)

is finite, then

subdirect product of the two finite lattices result follows.

fbi some lattice varietIes

(

V

and V'. then

\ow u) is a u) and Iìp( u), hence finite, and so the is finitely based.

e can now give sufficient conditions for the join of two finitely based lattice varieties to he finitely based. I his result appeared in Lee [85'] and is a generalization of a result of JOnsson [74].

IHEOI-tEM

R( fi

= i

5.17 If V and V' are finitely based lattice varietIes with md if I v(' — 3) md I v'(' — 3) mu finift

V fl V'

=

it' h

and ms

d

(JiLtl'TEit 5. EQtAflONtL BASES

124

6k—r—2

Figure 5.2

We can assume that V and V' are defined by the identities p = q and/ = respectively, relative to the variety of all lattices, and that the inequalities p q and q' hold in any lattice. Let K and K' be the classes of (r + 3)-generated subdirectly irreducible lattices in V and V' respectively, and define h = max(i1(K),i1(K)). We only consider h > 0 since if h = 0, then V, V' ç 2', the variety of distributive lattices, in which case the theorem holds trivially. Let V, = (V + then V, is finitely based by Lemma 5.16. If we can show that the condition P(n) in Lemma 5.14 holds for some vi, then V + V' will be finitely based relative to V, and hence relative to the variety of all lattices. So let L V, and suppose that for some u, u' M the quotients q(u)/p(u) and q'(u')/p'(u') are bounded, that is they project weakly onto a common quotient c/d of L in in and in' steps respectively. Property P(n) demands that in, in' vi for some fixed integer vi. Taken = max(2h+ 5,h+ 5) and assume that u,u',c,d have been chosen so as to minimize the number in + in'. We will show that if in> vi then there is another choice for u, u', c, d such that the corresponding combined number of steps in the weak projectivitics is strictly less than in + in'. This contradiction, together with the same argument for in', proves the theorem. By assumption q(u)/p(u) = au/bc., 's',,, ai/bi 's',,, a,,4/b,,4 = c/d for some quotients in L which transpose weakly alternatingly up and down onto (1 = 0,1,...,in— 1). Since in> max(2h+5,h+r+5), we can alwaysfind an integer ksuch that max(h+2,r+2)< k< in—h—2. (Jonsiderther+3quotientsup to and including ak/bk in the above sequence. Since the other cases can be treated similarly, we may assume that

i/

r+

ak_r_2/bk_r_2

ak_r_1/bk_r_1

Let L0 be the sublattice of L generated by the

...

/,

ak/bk.

r +3 elements

ak_r_2, bk_r_1, ak_r,.

. .

Notice that L0 V+V', and L0 is a finite lattice because a = subdirect product of J'Wr+3) and Eys(r+3), and is therefore finite. ak/bk (= ak_1+bk/bk) can be divided into (finitely many) prime quotients in L0 and at least one of these prime quotients, say z/p, must project weakly onto a nontrivial subquotient of c/d. Let be the unique subdirectly irreducible quotient lattice of A3 in which r/y is a critical quotient.

5,3,JOJXS OF FLVJ TELl BASLI) \JIUEIHES

12.5

I hen V', and hence I heoreni 2.3 (i implies E V E V IJ V'. \\ e examine each of the three cases that arise: Case I V'. Since V',there exists v such that p'(7) E V and q'(71). L0 E V implies 11(L0) Ii .Also v) and v). So by Lemma .5.1.5 q'(7) p'( = q'( = in Ii ± 1 steps. \ow q( a )/p( a) projects weakly onto q'( v)/p'( v) prolects weakly onto c/il in k steps, hence onto in k ± 1 steps. Ii 1 ± ± 1 iii K iii ± iii', so this contradicts the minimalitv of iii Case 2: V and V, p( v) .As above, q( v) for some v E E V'. Since since in Ii 1 steps E V', 11(L0) K Ii, and hence q( v)/p( v) proJects weakly' onto and froni there onto a nontrivial subquotient c'/il' of c/il in iii k steps. By the choice of we have Ii 1 iii 1 .Also q'( a' )/p'( a') projects weakly' onto c/il in iii' 1 steps so again we get a contradiction. Case .3: P. 11( = r implies E Vfl V' = fl'. First suppose that r ;> 0, hence fl' projects weakly onto in r steps, so by Lemma .5.1.5 r ± 1 steps. \ow either H—

)

:

H—

H—

H—

H—

H—

H—

(1

for sonic quotieiits

.

.

.

.

.

.

.

.

.

,r/j

or

in L. Since

,

and

that q( a )/p( a) projects weakly onto in k 2 steps and hence onto a nontrivial suhquotient c'/il' of c/il in iii 2 steps. As before q( a' )/p( a') projects weakly onto c'/il' in iii' 1 steps which again contradicts the minimality of iii \ow suppose that r = 1, which iniplies it' = P and = 2. Hence iii L0 we have we have H—

H—

/

H-

H-

\

\ /

and

It follows that we can shorten the sequence of weak pro jectivi ties froni q( a /p( a) onto a nontrivial suhquotient c'/il' of c/il to iii 2 steps Again q'( a' )/p'( a') projects weakly onto c'/il' in iii' H- 1 steps, giving rise to another contradiction. I his concludes the proof. E )

.

th a theorem that summarizes sonic conditions under which the two varieties join of finitely based is known to he finitely based. Parts (i) and (ii) are froni Lee [85'], and they follows easily froni the preceding theorem. Part (iii) is clue to JOnsson and the remaining results are froni hang [87]. e end this section

IHEOREM 5.18 Let

lattice varieties,

and V' he two finitely conditions holds then V H- V' is finitely based: V

modular and

V

is

(ii)

V

and V' are locally finite and Ji( V

(iii)

V

and V' are modular and V' is generated

(iv)

V

is

modular and

V' is

the following

generated by a finite lattice that excludes i13.

(i)

V' is

ii one of

generated

fl V') is

by'

finite.

by'

a lattice of finite lens th.

a lattice with finite projective radius.

(JiLtl'TEit 5. EQtAflONtL BASES

126

(v) V fl V = 2), the distributive variety. Lee [851 also showed that any almost distributive (see Section 4.3) subdirectly irreducible lattice has a projective radius of at most 3. Since any almost distributive variety is locally finite, it follows from Theorem 5.17 that the join of two finitely based almost distributive varieties is again finitely based.

5.4

Equational Bases for some Varieties

A variety V is usually specified in one of two ways: either by a set S of identities that determine V (i.e. V = ModE) or by a class K of algebras that generate V (i.e. V = K"). In the first case S is of course an equational basis for V, so here we are interested in the second case. A lattice inclusion or inequality of the form p q will also be referred to as a lattice identity, since it is equivalent to the identity p = pq. Theorem 3.32 shows that the variety generated by all lattices of length 2, has an equational basis consisting of one identity e: z0(zi + z2z3)(z2 + z3) z1 + z0z2 + z0z3. Jónsson [68] observed that if one adds to this the identity

e,4:

z0

II (zj+zj)

1 p and £ a> Suppose L is a finite lattice. As iii Section 2.2, we let = Y:{L E L p4, where p is ally join irreducible of L and is its unique dual cover. Dually we define < iii and £ K A( in) = fl{L E L for any meet irreducible iii E L.

CoI-toLLAI-n 6.24 Let L be a finite seinidistribu tive lattice, and let I = u/v be a quotient where X denotes the in L with 0L ' IL. Ihen L[I] is asuhlattice of L x

v en tag on, v is join irreducible and a is meet irreducible. By semidistri butivity, L is the disjoint union of the quotients u/c', un( v)/OL, 1L/ v-i- A( a) and n( v)/A( a), where the last quotient might be empty if n( v) > A ( a ). I his defines an equivalence relation 0 L with the quotients as 0—classes. 0 is a congruence relation since L is semidistri huti ye, and L/0 is isomorphic to a sublattice of 2 x 2. Letting I = (a/0)/( v/U) = { a/0} we have U I = a/c'. I hus L/0[.J] is isomorphic to a su blattice of X, and the result follows from the preceding lemnia.

I'I-tOOF. Clearly

I he following crucial lemma forces larger and larger bounded lattices into any modular variety that has the amalgamation property. LEMMA 6.2.5 Let V be a variety L a EL. thuji

that has the = (L

property and contains

11011—

V. 11.

(/=0,1).

It follows froni Section 2.2 that all lattices in B are semidistributive, 2 E Bp, BF is closed under the formation of finite products and L E BF implies L[I] E BF for any quotient I of L, so L E BF implies L1 E BF (1 = 0. 1). \\ proceed by induction on Assume I = 1. If a = 'L then L1 is a sublattice of L x 3 E V, hence E V. If a then there is a co-atom a' such that a K a' 1, and L = (a'] IJ [A( a')). I herefore L x 2 call he pictured as ill Figure 6.7 (i). Let I = (a'. 1 )/(0. 1), then (L x 2)[I] is a sublattice of (L x 2) x X Corollary 6.21), hence a lattice in V. I he congruence classes modulo the induced homomorphism Ii (L x 2 )[I] — X produce the diagram in Figure 6.7 (ii). Since is one of these congruence classes, we can double it, again using Day's construction, to obtain a lattice L' ill I'I-tOOF.

h

Figure 6.8.

CBJPrLR

LU

LV

6,

(i ) (0.0)

6.

•T

6.8

LJ11JCL \JIULitILS

iilAL(V) 1011

6.S.

LATIICE \JRJEIIES

LL5

Clearly L' = (L x 2 E V by Lemma 6.23. LetJ be the quotient u/v considered as lying in the congruence class labeled 11 in Figure 6.8, and consider the lattice L'[J] = = If we define and = 110 IJB[.]], then we have L'[J] = C1). = (L x 2 )[J] E V ajid = J(110 IJ 'J 11[J]), hence C E V if and only if these two lattices belong to V. 11 = w/0, so and IJ 11[J] = (11 x 2)[( a. i)/( v. 1)]. By induction then C E V and this gives L'[J] E V. Since is isomorphic to I 'J 11[J] iJ C iJ I) which is a su blattice of L'[J], is also in V. I he proof for i = 0 follows b synimetry.

4/],

IHEOREM 6.26 and Je2ek [84]). ILe only lattice varieties that have the ination property are the variety of all trivial lattices. the variety P of all distributive lattices. and the variety L of all lattices,

I

I'I-tOOF. If V is a nondistri hutive lattice variety that has the amalgamation property, then E V by I heorem 6.20. 1 he preceding lemma implies that for every L E BF and any a E L, if L E V then L[a/ v] E V, since L[a/ v] is a sublattice of (L x 2 )[( a. i)/( v. 1)]. It follows I heorem 2.38 that BF V, and since the finite bounded lattices generate all lattices ( Iheorem 2.36), we have V = C.

I

Amal(V) for some finitely generated lattice varieties

6.8

he a variety which fails to satisfy the amalgamation property. In this case Anial( V) sti bclass of V, and it is interesting to find out whether or not Anial( V) is an elementary class. In this section we outline the proofs of two results in this direction.

Let is a

V

I hey concern the amalgamation classes of finitely generated lattice varieties, and they are surprisiiglv contrasting th each other: If V is a finitely generated nondistri butive modular lattice variety then Amal( V is not elementary; on the other hand if V is a variety generated by a pentagon then Amal( V) is an elementary class determined by Horn sent! ences. )

Finitely generated modular varieties. \\e

begin with the following:

L)EFINiIIoN 6.27 (Al bert and Burns [88]), he a variety, and suppose that the diagram ( I, f, 11,y, C) cannot he amal— in V. ohs truction is any su C of such that the diagram ft) ( fi) and f' and y' where I', 11,y', C cannot he amalgamated in V. I' ft ( are the restrictions off and y to I'.

Let

V

(ii) Let

V

(i

)

he a locally finite variety,

pivperiy with respect to folIo wins

if C V.

V

if for

bnal(V) is said to have the boa ru/ed every k

E w

there exists an

u E w

such

that the

Ii olds:

bnal(V). k and the diagram ( I, f, 11,y, C) cannot he amalgamated in then there is an ohs truction C C such that C E

IHEOREM 6.28 (Albert and Burns [88]). Let V he a finitely generated variety of finite type. Ihen Amal(V) is elementar if and onl if it has the hounded obstruction property.

CiLtl'TEit 6. AMAL(LtMATIO.N VI LATTICE tCt1tJET1ES

146

Using the preceding theorem, Bergman [89] proved the following result.

Let V be a finitely generated nondistribu live modular variety. Then Amal(V) is not elementary. THEoREM 6.29

OUTLINE OF PROOF. Let V be as in the theorem and let 1, be a finite modular nondistributive lattice which generates V. If S is a subdirectly irreducible member of V then ILl, since S is an image of a sublattice of L. Thus S is simple, and since S has a diamond as a sublattice, we have 5. Pick S with largest possible cardinality. Let z

S such that a covers z. DefineB = 5x2 andlet f:2 '-' B beanembedding with 1(2) = {(z,O),(a,1)J. Then there is (1 Amal(V) and embeddings :2 (1 such that for each vi w, the diagram cannot be amalgamated in V, and every obstruction has cardinality at least vi. (For the details the reader is referred to Bergman [89].) Thus by Theorem 6.28, 0 Ainal(V) is not elementary. be the bottom element of S and let a

The variety generated by the pentagon. As before, let .V

be the variety generated by the pentagon. The following result appears in Bruyns, Natuxman and Rose [a].

it is dosed under reduced products and therefore is determined by Horn sentences. Furthermore, if B is an image of .1 Amal(X) and B is a subdirect power of the pentagon then B Amal(X). THEoREM 6.30 Amal(V) is an elementary class.

The full proof of the above theorem is too long to give here. It requires several definitions and intermediate results. We will list some of them first, and then outline the proof of the theorem. For more details the reader is referred to Bruyns, Natuxman and Rose [a]. DIWINI'rIoN 6.31

(i) Let

1)

be a congruence on a lattice

shall say that

1.

1)

is a 2-congruence

be a subdirrxt product of lattices : I I) and let B = rqpresentation .1 B is said to be irtgukwif for any kernels and projections we have t1ilA

(ii) Let

.1.

THEoREM 6.32 Let A be

(iii)

A subdirect of two distinct

a nontrivial member of.V. The following are equivalent:

AmaI(X). (ii) iLL B .V, then every 2-congruence on B, and3isnotanimageofd. (i)

Oj

if

.1

.1

is asubdirect power of

and any homomorphism

g=

.1

can be extended to a 2-congruence on

a, and for any regular subdirect representation 1: .1 —'

g:

.1

IV

there is a homomorphism h

: 1Y1

-. IV such that

hi.

(iv) A is a subdirect power of IV, 3 is not an image of..L, and if..L IV1 is a regular subdirect representation, then every 2-congruence on A can be extended to a 2congruence of a1.

FOR SOME LATTICE VARIETiES

6.&

147

PItoPosn'IoN 6.33 (i) Let B .V, and assume that for any distinct 2-congruences is .1 Amal(X) and embcddings fo,fi : .1 - B such that two distinct congruences on A. Then B AmaL(X).

(A3

and on B there and 1111(A) are

(ii) Let B be an image of .1 Amal(X) and assume that B is a subdirrxt power of dY. JIB C R1 is a regular subdirrxt representation, then every 2-congruence on B can be extendcd to a 2-congruence on OUTLINE OF THE PROOF OF THEoREM 6.30. We first consider the last statement of the theorem. Let B be an image of .1 Amal(X). Since 3is not an image of .1 we have that 3 is not an image of B. Thus B Amal(X) by Proposition 6.33 (ii) and Theorem 6.32 (iv). Next we consider direct products. Let .1 = be a direct product of members of Smal(X). Without loss of generality we may assume that each is nontrivial. We use Proposition 6.33 (i) to prove that .1 Smal(X). Thus we have to show that for any distinct on .1 there is a 7 a and embeddings fo,fi : i_v -t .1 such that 0o110(A,,) and 1111(A.,) are two distinct congruences on A. Now if and are distinct 2-congruences on 1, then A/(00 fl is isomorphic to either 3 or 2 x 2. In either case we have u, v, z .1 with u> v> z such that

(v,z)'$Oo and (u,v)'$Oi, By Jónsson's Lemma there are congruences ô0, ô1 on Defining

.1

(v,z)€Oi.

induced by ultrafilters V0 and

Dionasuchthat*%,co0andoic01.

R={3€a:u,ig>v,ig}

5={i€a:v,,>z,is}

Thereare three possible cases:

a the set b] belongs to both V0 and a the set ft] belongs to neither V0 nor Vi. (iii) There exists a 7 a such that ft] belongs to one ultrafilter and not the other. If (i) holds then we can choose u,v,zso that B = S = {7},forsome7 a. For 3 a with 3 7 let be an arbitrary but fixed element of A3. In this case we can have (i) For some 7 (ii) For each 7

f' so that for i

1] the embedding A, A is defined as follows: For z i_v the 7th coordinate of A is z, and if 3 a with 3 a then the ith coordinate of Mx) is Suppose now that (ii) holds. Pick any 7 a. We have (R—ft}) V0 and (S—ft}) V1. First observe that since A, is nontrivial it is a subdirect power of IV (Theorem 6.32 (iii)). Thus there are at least two distinct epimorphisms r,s: A —' 2 = {O,1}. The embedding : A, A is defined as follows: For x A, the 7th coordinate of fo(x) A is x. If 3 (B — {'y}) then the ith coordinate of fo(x) is if r(x) = 1 and if r(x) = 0. For 3€ a with 3 (BUft}) the ith coordinate is an arbitrary but fixed element of A,3. The embedding f' : A, A is defined as follows: 10

=

{O,

148

CiLtl'TEit 6. AMAL(LtMATIO.N VI LATTICE tCt1tJET1ES

For z i_v the 7th coordinate of fi(z) .1 is z. If I (5 — {'y}) then the Ith coordinate of fi(z) is if s(z) = 1 and if s(z) = 0. For I (S U {'y}) the Ith coordinate is an arbitrary but fixed element of The case (lii) is a combination of (i) and (II). For instance if {'y} V0 and {'y} then (5— {'y}) hence 10 is defined as in case (i) and f' is as in case (II). Thus we have shown that every direct product of members of Ama!(X) belongs to Sma!(X). Now if B is a reduced product of members of Ainal(X) then it must be a subdirect power of IV (see Bruyns, Natuxman and Rose [a] Lemma 0.1.9). On the other hand B is an image of a product .1 of members of AmaI(X). Since .1 AmaI(X) it follows that B tinal(X). In particular every ultraproduct of members of AmaI(X) belongs to Ainal(X), so that AmaI(X) is elementary (see Yasuhara [74]). It is determined by horn 0 sentences since it is dosed under reduced products (Chang and Keisler [73]).

B1BL1OG1LtPiI 1'

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GLI, x, 10, 11, 24,48—50, 52, 54, 55, 57, 61, 74, 94, 121, 128, 140, 151 U., ix, x, 46, 54, 69, 128, 129, 131, 137, 151

151

dimension of a lattice, 11 of a subspace, 50 distributive lattice, 1 division ring, 50

haiman, Id. 0., 55,

doubled quotient, 26

151 Flail, Id., 50, 63, 139, 151 height of a variety, 16 .llerxmann, (2., 46, 60-62, 68, 115, 151

elementary class, 116 strictly, 116 equational basis, 115 essential embedding, 130 extension, 130 excluding a lattice, 18

Mule,

D(L),28

homomorphism bounded, 26 theorem, 3 hong, 0. X., 46, 74, 75, 151 .llowie, J. Id., 128, 152 liSP Theorem, 2 .lluhn, A., 59, 60, 68, 151, 152 .11.,

128, 152

1RDLI

159

ideal, 5 completion, 136 lattice,

Jordan-kiölder Chain condition, 11 I'iang, 1. 1., 116, 125, 153 k-bounded pair, 118

proper, 5

identities, dosed set of,

1

Ident.52 conjugate, 22 XL,

Inaba, E., 62, 152 including a lattice, 18 independent elements, 57 interval, 10 inverse limit, 43 irreducible finitely subdirectly, 3 maximal, 131 subdirectly, 3 weakly maximal, 131 irredundant join-cover, 29 subset, 24 isolated quotient, 104

Keisler, IL. J., 116, 148, 149 K-freely generated, 4 luefer, J., 35, 153 Kimura, N., 128, 153 Kiss, E. w., 153 Kochen, S. B., 116, 153 Kcgalovskii, S. It., 2, 153 Kostinsky, A., 27, 43, 77, 153 Kruse, It. L., 115, 153 k-translation, 118 2

(L1,L2)-filter, 137 (L1, L2)-ideal, 136 Lakser, ii, 54, 128, 129, 131, 137, 151

lattice

Je2ek, J., 128, 141, 145, 150 Jipsen, P., 129, 132, 133, 135, 152

algebraic, 2 almost distributive, 88 Arguesian, 52 bounded, 27 complemented distributive, 11 completion of two, 136 cycle in, 37 dimension of, 11 distributive, 1 excluding, 18

J(L),36

filter,5

join

geometric, 48 ideal, 5 including, 18 length of, 11 linear, 55 lower bounded, 27 modular, 1 neardistributive, 86 nonmodular, 18 powerset, 6 semidistributive, 78 semimodular, 48 simple, 12 splitting, 22 trivial, 1 upper bounded, 27

Iw(L), 42

components, 33 isolated, 104 representation, 33 join-cover, 24 irredundant, 29 minimal, 29 nontrivial, 24 Jónsson polynomials, 9 Jónsson's Lemma, 6 Jónsson, B., v, ix, 6—8,

24, 25, 28—30, 35, 36, 38, 39, 45, 46, 49, 54, 55, 62, 69, 70, 72, 73, 77, 78, 81—83, 89,94, 102, 115—118, 120, 122, 123, 125, 126, 128, 129, 132— 135, 137, 150—153 14—16, 18,

1RDLI

160

upper continuous, 78 upper-semimodular, 48 variety, 1 width of, 47 Lee, J. U., 16, 20, 46, 77, 86—88, 94, 97, 101—103, 116, 123, 125, 153

length of a chain, 11 of a lattice, 11 of a lattice term, 31 tEl], 25 linear lattice, 55 lines in a projective space, 47 locally finite variety, 16 lower bounded homomorphism, 26

lattice, 27

£(P), 48 Lvov,

V., 115, 153 Lyndon, it, 115, 153 1.

Al3, 17 116

MacLane, S., 49 Macda, K, 50, 154 Makical, M., 115, 154 Mal'cev, A. L, 118, 119, 154 Márki, L., 129, 153 maximal irreducibles, 131 McKenzie, It., ix, 15, 16, 18, 22, 23, 27, 31, 32, 43, 46, 77, 115, 116, 126, 154

meet components, 33 isolated, 104 meet-cover, 24 minimal join-cover, 29 term, 33 S

18

modular lattice, 1 Monk, U. S., 54, 62, 153 Morel, A. C., 7, 116, 150 Murskii, V. L., 115, 154

JV, 18

Nakayama, I., 8, 151 Nation, J. B., 20, 28—30, 33, 36, 38, 39, 45, 77, 112, 151, 153, 154

natural partitions, 95 Natuxman, C., 129, 146, 148, 149 neardistributive lattice, 86 Nelson, 0. I., 127, 154 Neumann, B. 11., 14, 128, 154 Neumann, 11., 128, 154 n-frame, 57 canonical, 61 characteristic of, 62 spanning, 57 nondegenerate projective space, 50 nonmodular lattice, 18 nontrivial join-cover, 24 quotient, 10 normal sequence, 69 strong, 69

Oates, S., 115, 154 pentagon IV, 18 perspective axially, 51 centrally, 51 Pickering, 0., 55, 61, 62, 150, 154 Pierce, It. 5., 128, 135, 154 .L'(Al), 48 Poguntke, w., 77, 149 points in a projective space, 47 Powell, M. B., 115, 154 powerset lattice, 6 Pröhle, P., 129, 153 prime filter, 5 ideal, 5

quotient, 10 principal congruence, 2 filter, 5 ideal, 5 projective geometry, 47

inaclass,22

1RDLI

161

line, 50 plane, 50 quotients, 11 radius, 118 space, 47 flesarguesian, 51 nondegenerate, 50 subspace of, 48 projects weakly onto, 10 proper filter, 5 ideal, S property

Rival, 1., ix, 18, 78, 81—83, 89, 94, 149, 153, 155 Rose, IL, ix, 10, 20,23, 77, 83, 85,87, 92, 99, 104, 108, 129, 132, 133, 135, 146, 148, 149, 152, 155 Ruckeishausen, 'V., 20, 77, 112, 155

Pu,6

Sands, B., 45, 150 Sankappanavar, .11. P., 115, 116, 149 Schiitzenberger, Id., 46, 155 Schreier, 0., 128, 155 Scott, 0. 5., 7, 116, 150 SlY, 34 second isomorpliism theorem, 3 self-dual variety, 54 semidistri butive lattice, 34, 78 variety, 81 semimodular lattice, 48

PudlIk, P., 15, 155

Si, 3

(P),

134 132

(Q), amalgamation, 130 bounded obstruction, 145 strong amalgamation, 130

quotient, 10 algebra, 2 critical, 10 doubled, 26 isolated, 104 nontrivial, 10 prime, 10

projective,

11

relinesc,24 regular subdirect representation, 146 relatively free algebras, 4 representation canonical join, 33 join, 33 regular subdirect, 146 residually finite, 64 small, 131

retract, 22 absolute, 130 retraction, 22 ring auxilIary, 61

characteristic of, 62 division, 50

simple, 12 spanning n-frame, 57

splitting algebra, 22 lattice, 22 pair, 20 strictly elementary class, 116 strong amalgamation, 129 property, 130 normal sequence, 69 strongly covers, 13 subdirect product, 2 subdirectly irreducible, 3 subspace dimension of, 50 of a projective space, 48

2,11 Tarski, A., ix, 2, 155 Taylor, W., 115, 131, 134, 155 term length of, 31 minimal, 33 Tholen, W., 153 translation, 118

1RDLI

162

transposition bijective, 11 down onto, 10 up onto, 10 weakly down onto, 10 weakly into, 10 weakly up onto, 10 triangle, 47

trivial lattice, 1 Tuma, J., 15, 155 type 1 representation, 55 UA, x, 4, 118, 151 ultrafilter, 5 ultraproduct, 6 upper bounded homomorphism, 26 lattice, 27 upper continuous lattice, 78 upper-semimodular lattice, 48

variety almost distributive, 20 congruence distributive, S conjugate, 22 finitely generated, 4 generated, 2 height of, 16

lattice,

1

self-dual, 54 semidistributive, 81 Veblen, 0., 52, 155 V151, 3

ViMn, V. V., 115, 155 MMI, 131

i';;', 127 von Neumann, J., ix, 46, 57, 61, 155 V51, 3

VWMI, 131

weak transpositions, 10 weakly

atomic, 14 maximal irreducibles, 131 (W )-failure, 26 Whitman's condition (W), 24, 25, 27, 29,

31,45

Whitman, P.,

15, 22, 24, 27, 33, 43, 87,

155

width of a lattice, 47 Wille, IL, ix, 16, 17, 23, 24, 46, 156 word algebra, 4 I1'(X), 4 Yasuhara, Id., 129, 130, 148, 156 Young, W. 11., 52, 155