Varieties of lattices

Varieties of Lattices Peter .J iI)SC1I dud lleuirv Rose Synopsis An interesting pro blem iii universal algebra is t...

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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
1/kerh. where

keh

{(o, o')

E A2

:

h(o)

h(o')}

E

Eon(J).

theoii iii which states that for (fixed) e will also make use of the Con( 1) and aiiy o E Con( 1) containing 0 there exists a congruence (o/0) E Con( defined by o E

(1/0 (o/0) b/0

if and only if

oob,

0/0 defines an such that ( 1/0 )/( o/0) is isomorphic to I/o. Furthermore the map C) K isomorphism from the principal filter [0) {o E Con( 1): 0 o} to Con( 1/0). I he homomorphism theorem implies that an algebra I is a su hdirect product of quotient aige bras 0 E Con( I)) if and only if the meet (intersection) of the iriidacibie if and only if A satisfies anyone of the is following equivalent conditions: I

(i

)

whenever

I

is a subdirect product of algebras

I

I

the factors (ii) the 0 of Con( I) is completely meet irreducible; (iii) there exist o,

b E

A such

that con(o, b) is the smallest

11011-0

element of Con( A).

iriidacibie if whenever I is a su bdirect product of finitely many algebras then A is isomorphic to one of the I K u), or equivalently if the 0 of Con( I) is meet irreducible. For any variety V of algebras we denote by I

is said to be firti/ely

the class of all su bdirectly irreducible mem hers of V the class of all finitely su bdirectly irreducible mem hers of

V.

state Birkhotf's [44] su bdirect representation theorem: Every a su hdirect product of its su hdirecti irreducible homomorphic images. e can

I his can be deduced from the following result concerning decompositions in algebraic lattices: IHEOREM 1 .2 ([ATL] p .-13). in an meet irred ii cihie elements.

lattice every element is the meet of completely

E'BAJ'TLJ

-I

1.

PJIL±L\JJXAIUES

be a class of algebras, and let I be an algebra generated is K-freely gerien led by the set X if ally map f ff0111 the set of generators A to any I E K extends to a honioniorphisni 7 I — I. If, in addition, I E K, then ± is called the K- fiie algebra Q A or the (i-generated fiie algebra ocer K, and is denoted by ±k (A ) or (Q). I he extension J is unique (since A generates 1±), and it follows that ) is uniquely deternuned (up to isoniorphisni ) for

Relatively free algebras. Let K by a set A

I.

I

\\ e say that

:

each cardinal a. However, (a) need not exist for every' class K of algebras. Birkhotf [35] found a simple method of constructing K—freely' generated algebras ill general, and from this he could deduce that fbr any' non trivial variety' V and any' cardinal a (3. the V-free a generators exists. \\ e briefly' outline his method below (further details can be found ill [UA]). Let (X be the word algebra over the set X , i.e. (X is the set of all terms polynomials or words) ill the language of the algebras ill K, with variables taken froni the set X, and with the operations defined 011 (X ill the obvious way (e.g. for lattices the join of two terms p, q E (X) is the term (p ± q) E (X )). It is easy to check that, for any class K of aige bras, (X is K-freely generated by the set X Other K-freely generated algebras call he constructed as quotient algebras of (X ill the following way: I

I

)

I

)

)

I

I

I

.

)

I

)

Let!

fl{kerh h: that F

(X)

I

J

—

is a homomorphism, p q E IdK}).

I

J E K}

(X )/Ok is K-freely generated by the set E A }. Indeed, given a niap f — J E K, define f' A — J by' extends to a honiomorphism F — J. If contains a nontrivial algebra, then K f' 7' (if not, then 0k identifies all of (X ), hence F is trivial), and by construction F is a subdirect product of the algebras (X )/ker Ii, which are all members of SK. if K is closed under the formation of subdirect products, then F E K, whence Birkhotf's result follows. If an identity' holds in every' member of a variety' V, then it must hold in u) for each u E w, and if all identity fails ill sonic nieni her of V, then it niust fail ill sonic finitely generated algebra (since an identity' has only' finitely' niany' variables), and hence it fails in u) for sonic u E w. Ihus e claini

I

:

I

:

:

:

I

I

V

E

follows froni Birkhotf's subdirect representation theorem that every' variety' and it is generated by' its finitely' generated subdirectly' irreducible members. In fact, a similar argument to the one above shows that V

whence every' variety' is generated by' a single (countably' generated) algebra. lo obtain all interesting notion of a finitely generated variety, we define this to he a variety that is generated by finitely finite algebras, or equivalently, by a finite algebra (the product of the former algebras).

1.3. (JORGIt t

1.3

flUE DiSTItIB t TI Vifl'

5

Congruence Distributivity

An algebra .1 is said to be congruence distributive if the lattice Con(J.) is distributive. A variety V of algebras is congruence distributive if every member of V is congruence distributive.

Congruence distributive algebras have factorable congruences. What is interesting about algebras in a congruence distributive variety is that they satisfy certain conditions which do not hold in general. The most important ones are described by Jónsson's Lemma (1.4) and its corollaries, but we first point out a result which follows directly from the definition of congruence distributivity. the product of two algebras A1, and Uon(A) is distributive. Then Uon(A) is isomorphk to Uon(Ai) x Uon(A2) (to. congruenws on the product of two algebras can be factored into two congruences on the algebras). LEMMA 1.3 Suppose .1 is

Let L = Con(Li) x Con(12), and for .9=

L define a relation # on

.1

by

(ai,a2)#(bi,bj) if and only if aiOibi and One easily checks that 9

Con(A), and that if

çb

= (bi,

L, then

ifandonlyif Thus the map # from L to Con(A) is one-one and order preserving, so it remains to show that it is also onto. Let ô Con(L), and for i = 1,2 define = where irj is the projection from .1 onto Clearly = 0 in Con(A), hence ô = (ô+pi)fl(ô+pj) by the distributivity of (Jon(Jj. Observe that for i = 1,2 ç Ô+pi and so from the second isomorphism theorem we obtain such that for a = (ai,a2),b = (bi,&z) A

(i€{1,2}).

ifandonlyif

Therefore aôb if!

weseethatô=t

and a(ô-J-p'j)b

if!

a10161

and a29464. Letting U =

L,

0

A short review of filters, ideals and ultraproducts. Let L be a lattice and F a filter in L (i.e. F is a sublattice of L, and for all p L, if p z F then p 1"). A filter F is proper if F L, it is principal if F = [a) = {z L a z} for some a L, and F is An ultniJilterisamaximal proper filter (maximal with respect to inclusion). In a distributive lattice the notions of a proper prime filter and an ultrafilter coincide. The notion of a (proper / principal / prime) ideal is defined dually, and principal ideals are denoted by (a]. Let IL be the collection of all ideals in L. Then IL is dosed under arbitrary intersections and has a largest element, hence it is a lattice, partially ordered by inclusion. Dually, the collection of all filters of L, denoted by FL, is also a lattice. The order on FL is itoerse inclusion i.e. F U if! F 2 U. With these definitions it is not difficult to see that the map z [z) (z (z]) is an embedding of L into FL (IL), and that FL and IL satisfy the same identities as L. It follows that whenever L is a member of some lattice variety then so are FL and IL.

E'BAJ'TLJ

6

1.

PJIL±L\JJXAIUES

For an arbitrary set I we say that :T I if :T is a filter in the lattice the collection of all subsets of 1 ordered by inclusion). Since P1 is distributive, P1 ( a filter [/ is an ultrafilter if and only if [/ is a prime filter, and this is equivalent to the condition that whenever I is partitioned into finitely nianv disjoint blocks, then [/ contains exactly one of these blocks. Let I he a direct product of a fanuly of algebras { I E 1}. If is a filter over I (i.e. a filter in the powerset lattice P1), then we can define a congruence Cy on J by

if and only if

(lOyb

{i E

I

:

bJ

E :T

is the Ith coordinate of (I. If I is a direct product of algebras (I E 1) and U is an ultrafilter over 1, then the quotient algebra 1/oil is called an ultit product. For any given class of algebras K, we

where

denote

b

the class of all ultraproducts of mem hers of K.

Jónsson's Lemma. \\ e are

now ready to state and prove' this remarkable result.

LEMM;\ 1.4 (JOnsson [67]). Suppose B is a distribu tn'e via direct and 0 E Eon( B) is meet irreducible, Ihen there exists an ultralilter

product 1 U over I such that

I

0.

\\ e will denote by [I the principal filter generated by a subset .J of I, and to Ci>[J) II. Clearly (I siiiplifv the notation we set if and only if I I {i E I bJ, for o, b E B 1 1. If 0 1 E Con( B) then any ultrafilter over I will do. So assume 0 1 1 0. \\ e claim that C has and let C be the collection of all subsets I of I such that the following properties: I'I-tOOF.

)

:

(i) I E C, 0

K 9 1

(H)

I 'J

(iii) (i

)

K

E C E C

C;

implies K implies

I

E C;

E C

or K

E C.

are the zero and unit of Con( B), and K 9 1 implies lo prove (iii), observe that! IJ K E C implies

and (ii) hold because

0

0±

,

0

(0±

n

n(0±

the distri hutivity of Con( B), and since 0 is assumed to he meet irreducible, it follows whence I E C or K E C. that 0 0 ± or 0 0 ± C itself need not he a filter, hut using Zorn's Lemma we can choose a filter U over I, maximal with respect to the property U I C. It is easy to see that by

ozdB

U

JEll

it remains to show that U is an ultrafilter. Suppose the contrary. I hen there exists a set II I I such that neither II nor I II belong to U. If II fl I E C for all I E U, then by (i) and (ii) U IJ {ll } would generate a filter contained iii C, contradicting the maximality of so

II\11Y

COXGRL LIVCE DISTIUBL

1,3.

II. Hence

U, and similarly for I But I fl K E U, and

InK which contradicts (Hi ).

we can find

(II n

I, K

such that II fl

E U

I, (I II

)

fl K

C.

I n K) IJ ((I II) ninE)

I his conipletes the

CoI-tOLLAI-n 1.5 (JOnsson [67]). ence c/is till) II ti ye, Ihen

(i)

7

proof.

Let K be

a

class of

such that

Ky

V

Vcq

V

wn

I ilg bit in PROOF (i) \\ HSPk isomorphic to a quotieiit algebra 11/0, where 11 is a suhalgebra of a direct product A 1, and { I E I} I K. If 11/0 is finitely su bdirectlv irreducible, then 0 is meet irreducible, hence by the preceding lemma there exists an ultrafilter Ii over I such that C) 11 I 0. 1 hus 11/0 is a homomorphic image of 11/c, which is isomorphic to a su balgebra of the ultraproduct 1/cu. (ii) follows from (i and Birkhotf's su bdirect representation theoreni. )

lo exhibit the full strength of JOnsson's Lemma for finitely generated varieties,

we

need the following result:

and Scott [62]). JfK is a finite set of finite aLgebras. then every ultraproc/uct of members of K is isomorphic to a member of K. LEMMA 1.6

(

li1ravne, Morel

product of mem hers of K, and define an equivalence i if and only if 1, partitions I into 11. Since K is finite, finitely niany blocks , If// is an ultrafilter over I, then U contains exactly one of these blocks, say I Let Ti {( fl I I E U} he the ultrafilter over I induced by U, and let )CEJAJ. \\ e claini that I'I-tOOF. Let A relation on I by

j

he a direct .

.

.

:

I

A/cu where

11

11

(j

is an algebra isomorphic to each of the

E

I),

and hence to a member of K

as required.

Consider the epimorphism restriction of (I to I. \\ e have

Ii

:

given by h(i

I —

(Ikerhb itt itt

)

where

is

the

itt {i E

I

:

bJ

itt

E U

ocub

Ii cu, whence the first isoniorphisni follows. lo establish the second isoniorphisni, observe that over the and therefore is isoniorphic to an ultrapower given by b finite algebra 11. In this case the canonical eni bedding 11 we (b, b,. .) E (b ), is always onto (hence an isomorphism) because for each c E can partition I into finitely many blocks .Jch b} (one for each b 11), and E {j E I then one of these blocks, say niust he in Ti, hence b' c/ca.

so ker

J

'

.

:

'

CiLtl'TEit 1. P1tELL%IL'LtitJES

8

Coaoisuw 1.7 (Jónsson [67]). Let K be a finite set of finite algebras such that V = is congruence distributive. Then

K"

(i)

has up to isomorphism only finitely many subdirectly irreducible members, each one finite; (ill) V has only finitely many subvarieties; V

(iv)

if A,B in

.1

V51 are nonisomorphk and but not in B.

IBI, then there is an identity that holds

(i) follows immediately from 1.5 and 1.6, and (II) follows from (i), since HSK has only finitely many non isomorphic members. (ill) holds because every subvariety is determined by its subdirectly irreducible members, and (iv) follows from the observation that if both .1 and B are finite, nonisomorphic and IBI, then B HS{A}, which 0 implies B {A}" by part (i).

Lattices are congruence distributive algebras. This is one of the most important results about lattices, since it means that we can apply Jónsson's Lemma to lattices. We first give a direct proof of this result. THEoREM 1.8 (ii'unayama and Nakayama [42]).

l'br any lattice .L, (Jon(L) is distributive.

Ccn(L) and observe that the inclusion (On cb)+(Oflô) c On(çb+ô) holds in any lattice. So suppose for some z, p .L the congruence 0 n (b + ô) identifies z have to show that (On çb) + (On ô) identifies z and p. By assumption z0p and and p. z( cb+ô)p, hence by Lemma 1.1 Let 0, cb,ô

z = Mbziôz2cbzsôz4

=

p

for some L. Ifwecanreplacetheelements; by same 0-class as z and p, then

4,

which all belong to the

= 4(On çb)4(Onç4)4(On çb)4(Onc4)4...4 =

z(0 n çb) + (On ô)p follows. One way of making this replacement is by taking = zzj + p; + zp (the median polynomial), then any congruence which identifies; with 4 also identifies with and (since Lisa lattice) zp zt; z/0 z+p, whence

whence

foralli=0,1,...,n.

0

Consequently every lattice variety is congruence distributive. Notice that the proof appeals to the lattice properties of L only in the last few lines.

Jónsson polynomiak.The next theorem is a generalization of the above result. THEoREM 1.9 (Jónsson [67]). S variety V of algebras is congruence distributive if and only if for some positive integer vi there exist ternary polynomials such that . . for i = 0,1,... ,n, the following identities hold in V: 10(z,p,z) = z,

174(z,p,z) = z,

forieven

(it) z,

z, z)

for i odd.

=

COXGRL LIVCE DISTIUBL

1,3.

II\11Y

9

Suppose V is congruence distributive and consider the algebra ( {o, b, c} ). Define 0 con(i,c), o con(i,b) and L' con(b.c). Ihen (o,c) E On (o ± (0 n o) ± (0 n &'). By Leninia 1.1 there exist d0, £'L. , il,, E such that I'I-tOOF.

.

od3On

d0On Each

d1

is of the forni

J

.

.d,,

Lj(o, b, c) for some ternary polynomial and it remains to (Ic) are satisfied for o, j b and z c in ( {o, b, c}), since

d1

show that the identities then they must hold iii every meni ber of V. I he first two identities follow froni the fact that d0 (I and il,, c. For the third identity let Ii' {o, b, c} ) — {o, b} ) be the homomorphism induced by the niap o, c o, b b. I hen h(i ) h( c) implies 0 1 ker Ii and since each b, c) is 0-related to (I we have '

I

h(o)

(I

\ow suppose (I,b o, c

i

I

and consider Ii Ihen 01 kerh, and since b, c)

b,o).

{o, b, c]) —

is even

c.

I

b,c))

{o, c]) induced by the map

(o, b, c)

C)

it follows that

I he proof for odd

i is sinular. \ow assunie the identities (Ic) hold in V for some ternary polynomials , let prove distributive show I E V and 0, o, Con( I). lo that V is congruence it suffices to E I (On o) — (On Ihin (i E 0 intl thit On (o— Ut (i E On (o— Lemma 1.1 there exist b0, ,b, E J such that .

)

)

)

.

.

i

ü,i,..., ii

we

c.

have

o

together with (o, c)

y,

)

)

•b,,,

o

Ihe identity

.

.

L'&20b3 So for each

.

E 0

implies that the elements

(Jo, b1, c)

E

(I/O whence

(Jo,b0,c) on

o

on

lJi,&2,c) on o

It follows that (Jo,o,c) (On o) ± (On H /Jo,c,c) holds for each remaining identities now give (o, c) E (0 n o) ± (0 n b').

i

0,1,...,n and the

I he polynomials ., / are known as JOnsson polynomials, and will he of use in Chapter .5. Here we just note that for lattices we can deduce I heorem 1.8 froni I heorem 1.9 if we take ii 2, (the median and y, z) y, z) ± zy ± y, :) .

.

('lIFTER

10

1.

FJILTLL\JJXIIUES

Congruences on Lattices

1.4

Prime quotients and unique maximal congruences.

Let L be a lattice and a, v E L quotient a/v (alternatively intercal a]) we niean the su blattice K \\ a}. e say that a/v is nontricial if a ;> v, and prime if a covers v E L (i.e. a/v a, v}, in hols: a v). If L is su hdirectlv irreducible and con( a, v) is the { smallest non-0 congruence of L, then a/v is said to he a critical quotient.

with

.10 11 a/v is a pri lie quotient of L. then there exists that does not identify a and v,

LEMM;\ O

By a

a.

K

v

1

Let

C

lake

0

I'I-tOOF.

and

v.

I

E C

unique maximal

the set of all congruences of L that do not identify a 37 C, and suppose 0 identifies a and v. By Lemma 1.1 we can find and EL such that Con( L)

be

v.

a

lteplaciiig

a

by

a

v

H—

v

for all i 1, , ii. Since a/v is assumed to he prime, we niust have a which or v for all i, implies a v for sonic i, a contradiction. I hus 0 E C, and it is clearly the largest element of C. a

.

.

.

Weak transpositions. Given two quotients r/h and a/v in L, we say that r/h weakly up onto a/v (in synihols r/h a/ v) if r ± v a and v. Dually, we say that r/h weakly down onto a/v (in synibols r/h a/ v) if v and r a. e write r/h a/v if r/h transposes weakly up or down onto a/c'. I he quotieiit r/h weakly onto a/v in ii

if there exists a sequence of quotieiits .

.

in L such that

a/v.

.

Xote that the sym bols and define nonsymmetric binary relations on the set of quotients of a lattice. Some authors (in particular [GLT], [ATL] and Rose [84]) define weak transpositions iii terms of the of the above relations, but denote these inverse relations by the same sym hols. sually the phrase 'transposes weakly into" (rather than 'onto" is used to distinguish the two definitions. I he usefulness of weak transpositions lies in the fact that they can he used to characterize principal congruences in arbitrary lattices. )

(Dilworth [50]). Let r/h and a/v he quotients in a lattice L. Ihen con( r, identifies a and v if and only if for some finite chain a v. the ;> > ;> /, quotient r/h projects weakly onto (all i 0, 1, , in 1). LEMMA 1.11

.

.

.

Xotice that if a/v is a prime critical quotient of a subdirectlv irreducible lattice L, then by the above lemma every nontrivial quotieiit of L projects weakly onto a/v.

Bijective transpositions and modularity. \\e say that r/h up onto a/v (in sym hols r/h a/ v) or equivalently a/v down onto r/h (in syni hols a/v r/h) \\e write r/h a/v if either r/h if r ± v a and rv a/v or a/v. Xote that

/

is a symmetric relation, and

that

a/v and a/v

/

r/h if and only if r/h

\

a/v.

\

1.4. CORG1tEERCES OR

Suppose

r/s

/

LSfl1ICES

11

u/v and, in addition, for every

t=(t+v)F

and

r/s

I

u/v

and I'

we have

t'=t'F+v. from r/s to u/v,

I + v is an isomorpliism and in this case we say transposes bijectively up onto u/v (in symbols r/s u/v), or equivalently u/v tmnsposes bijectively down onto F/s (in symbols u/v F/s). In a modular lattice every transpose is bijective, since I F and modularity imply

Then the map I

that

/,,

r/s

(I+v)F= I+vF

I+s =1

and similarly I' = I'F+v. It follows that for any sequence of weak transpositions z0/p0 z74/p74 we can find subquotients a4/zd of zj/pj (1 = 0,1,...,n —1) such that I

Zoflhcj

•-'

II)

II)

...

I

Z74fl174.

In this

case we say that the two quotients z&/ut and z74/p74 are pivjectioe to each other, and by Lemma 1.11 this concept is clearly sufficient for describing principal congruences

in modular lattices. Congruence lattices of modular lattices. The symbol 2 denotes a two element lattice, and a complemented distributive lattice will be referred to as a Boolean algebra (although we do not include complementation, zero and unit as basic operations). We need the following elementary result about distributive lattices: LEMMA 1.12 Lot II boa finite distributive atoms, then 1) is a Boolcan algebra.

It

lattiw.

if the largest element of II

suffices to show that 1) is complemented. Let a join of all atoms that are not below a. By assumption a + = Ojj, whence is the complement of a.

is ajoin of

1) and define to be the = 'D and by distributivity

0

A chain C? is a linearly ordered subset of a lattice, and if ICI is finite then the length of C? is defined to be ICI — 1. A lattice L is said to be of length n if there is a chain in L that has length vi and all chains in L are of length vi. J.tecall the Jordan-kiölder Chain condition ([ULT] p.172): if Al is a (semi-) modular lattice of finite length then any two maximal chains in Al have the same length. In such lattices the length is also referred to as the dimension of the lattice. LEMMA 1.13 Let Al be a

modular lattice.

(i) Mu/v is a prime quotient of Al, then con(u,v) is an atom of (Jon(M). (ii) MM has finite length in, then (Jon(M) is isomorphk to a Boolean algebra 2w', where n in.

(i) If con(u,v) 2 con(F,s) for some F 5 Al, then u/v and a prime subquotient of F+s/Fs are projective to each other, which implies that con(u,v) = con(F,s). It follows that con(u,v) is an atom. (ii) Let z, -< z1 -< ... -< be a maximal chain in Al. Then the principal congruences con(zj, (1 = 0,1,.. .,rn— 1) are atoms (not necessarily distinct) of Con(M), and since

12

E'BAJ'TLR

1.

J'RLLL\JJXAIULS

their join collapses the whole of JJ , the result follows from the ctistri butivit of Con( JJ and the preceding Ieiiiiiia. is

As a corollary we have that every su hdirectly irreducible modular lattice of finite length (i.e. Con(JJ) 2).

Chapter

2

General Results 2.1

The Lattice

A

is a dually algebraic distributive lattice. In Section 1.1 it was shown that the collection A of all lattice subvarieties of L is a coniplete lattice, with intersection as meet. A completely analogous argument shows that this result is true iii general for the collection of all suhvarieties of an arbitrary variety V of algebras. \\ e denote by A

Xv

the lattice of all su bvarieties of the variety

V.

then we usually drop the subscript V. iilaiice to V if it can be defined by sonic finite set Call a variety V' Xv flit//ely of identities together with the set Id V. If V is finitely based relative to the variety Mod 0 the class of all alge bras of the sanie type as V) then we flay onut the phrase 'relati ye (If

V

L

to V".

br

IHEOREM 2.1 lattice, and the any variety V of algebras. Xv is a dually varieties which are finitely relative to V are the dually compact elements,

Let V', (1 E I) be subvarieties of V, and suppose that V' I' RE] Mod( If V' is finitely based relative to V, then V' V fl ModE for some finite ). set F IdV'. It follows that F and since F is finite, it will be included iii the union of fini telv many Clearly the finite intersection of the corresponding suhvarieties is included iii V', whence V' is dually conipact. Conversely, suppose V' is dually conipact. \\ e always have I'I-tOOF.

(t)

V

Mod Id V

V fl

fl

Mod{s},

I

so by dual conipactness V' V fl Mod{sJ for sonic finite set F ldV'. , Hence V' is fini telv based relative to V. Finally (Ic) implies that every elenient of Xv is a nieet of dually conipact elenients, and so Xv is dually algebraic. .

.

Let Cv ( V') denote the collection of all varieties iii Xv that cover V'. \\ e say that V' if ally variety that properly contains V', contains sonic nieni her Cv( V') of Cv(V'). 13

CBAJ'TLJ

LI

2.

US

c;LIVERjL

Recall that a lattice L is weakly atomic if every nontrivial quotient of L contains a prime subquotient. An algebraic (or dually algebraic) lattice L is always weakly atomic, since for ally nontrivial quotient u/v ill L we can find a compact element c K a, c P and K using Zorn's Lenima we can choose a maximal element il of the set c ± v}, E L which then satisfies v K il c± In particular, if a is compact, then there exists il E L such

that

v K

IHEOREM 2.2 Let

il

12'

a.

be

a

su hvariety of a variety V.

(i) 1112' is finitely based relative to

(ii)

If Cv( 12')

V

then Cv( F) strongly covers

is finite and strongly covers

12'

then

12'

12'.

is finitely based relative to V.

I'I-tOOF. (i) 12' is dually conipact, so by the remark above, ally variety which contains 12', contains a variety that covers 12'. (ii) Suppose Cv( 12') ., } for sonic u E w. I hen for each i 1, ., ii there { exists all identity fails in V 1, ., II}. E Id 12' such that e claim that 12' 12". Since each holds in 12', we certainly have 12' 1 12". If 12' 12" then the assumption that Cv (12') strongly covers 12' iniplies that 12" for sonic i E {i, , n}. But this is a contradiction since fails in .

.

.

1

I

.

.

.

.

.

.

focus our attention 011 congruence distributive varieties, since we can then apply J Onsson's Lemma to o btain further results. e

IHEOREM 2.3 (JO1155011 [67]). Let let 12', 12" E Xv, Ihen

(i) (V'-F V")sj

(i)

and

IJ

(ii) Xv is a distributive lattice: (iii) if 12' is finitely generated. then I'I-tOOF.

dis t ri hti ti ye variety of

he a

V

\\e always have

12'

u

± V'7V" is a finite quotient in Xv.

I

(12'

J

if E (12' 12"). It is iiot difficult to see that V", and since I is su bdirectlv irreducible, Coiiversely,

H—

H—

then JOnsson's Lemma implies that J E 12' IJ 12") HSPti 12' IJ HSPti 12" 12' IJ we niust have J E or J E vçy (ii) If , V2, V3 E Xv, then (i implies that every su bdirectlv irreducible mciii ber of I whence or V2±V3) belongs to either I he reverse iiiclusioii is always satisfied. (iii) By Corollary 1 .7(iii), the quotieiit V'/V fl 12" is fiuiite, aiid it traiisposes biiectivelv up oiito 12' ± V"/ 12" silice Xv is distributive. 12' IJ

)

Ihe fact that, for any congruence distributive variety

the lattice Xv is dually algebraic and distributive can also be derived froni the following more general result, due to B. H.

12,

\euniann [62]:

IHEOREM 2.4 fully' in variant

for any variety on

V

of algebras. Xv is dually isomorphic to the lattice of all (w).

(A congruence 0 E Con( I) is fully iii ('unaPt if (lOb implies f(o )Of( b) for all endonior— phisms f I — I). However, we will not make use of this result. :

2± lEE LATIICE Some

A

15

properties of the variety

C. For ally class K of algebras, denote by

the class of all finite mem hers of

I he variety (P 1)

V

V

K.

of all lattice varieties has the following interesting properties:

L

is generated by

(P2) Every member of

its finite (suhdirectlv irreducible) mem hers (i.e. V

can he embedded in a nieni her of

(P3) Every finite nieni her of

V

(i.e.

V

V

(VF )V).

I

call he eni bedded in a finite nieni her of

I hat L satisfies (P1) was proved by' Dean [56], who showed that if an identity fails in sonic lattice, then it fails in sonic finite lattice (see Lemma 2.23). In Section 2.3 we prove an even stronger result, nanielv that L is generated by the class of all splitting lattices (which are all finite). (P2) follows from the result of \\ hitnian [46] that every lattice can he em bedded in a partition lattice, which is siniple (hence su hdirectly irreducible, see also JO1155011 [53]). (P3) follows from the analogous result for finite lattices and finite partition lattices, due to Pudlák and luma [80]. IHEOREM 2.5 (Mchellzie [72]). Let V be a variety a sti hvariety V' of V:

(i) V' is coiiipletelv' join prime in Xv (i.e. V'

implies V'

K

K

some

i E

I):

(ii) V' can be generated by a finite subdirectlv irreducible member: (iii) V' is completely join irreducible in Xv:

(iv) V' can be generated

by'

a finitely generated subdirectly irreducible member:

(v) V' can be genera ted

by'

a (single) su bdirectly irreducible member:

(vi) V' is join irreducible in Xv:

lhn

ush i;

)

11

(P1) holds

md if

(ii)

is

distributive then (iik-(Hi) and

in

tilt K gint itt U iut ik finiti gin' itt U duci ble mem hers, so if V' is completely join irreducible, then it must he generated by one of them. is obvious. Suppose that V (VF )v (i.e. (P1) holds). I hen V is the join of all its finitely generated suhvarieties. If V' V is join prime, then it is included in one of these, and therefore V' is itself finitely generated. I his means that V' call he generated by finitely itiany finite suhdirectly irreducible algebras, and since it is also join irreducible, it must he generated by just one of them, i.e. (ii) holds. If V is congruence distributive and (ii holds, then I heorem 2.3(i) implies that V' is filliti iii iiit hi hi nct it intl h 10111 iiit dun hIt vi) also follows from I heorem 2.3(i). is completely join irreducible. (v I'I-tOOF

(iii

)

1

1

I

)

Ihus for

we have (i

vi). Mckenzie also gives examples of lattice varieties which show that, in general, none of the reverse implications hold. If V' is assumed to he finitely generated, then of course (ii )—( vi) are equivalent. V

L

v

16

CiLtl'TEit 2. (JERE1LtL 1tESELLS

THE0ItEM 2.6 (Jónsson [67]). Let V be a congruenw distributive variety

of algebras.

Then (i) (P1) implies that every proper subvariety of V has a cover in Ay; (ii) (P2) implies that V is join irreducible in Ay; (ill) (P1) and (P2) imply that V has no dual cover.

(i) If V' is a proper subvariety of V = (Vp)1', then there exists a finite algebra .1 V such that .1 V'. By Theorem 2.3(111) the quotient {A}1' + V'/V' is finite and therefore contains a variety that covers V'. (ii) If V' and V" are proper subvarieties of V, then there exist algebras 1' and 1" in V such that 1' V' and 1" V". Assuming that V ç SV51, we can find a subdirectly irreducible algebra .1 V which has 1' x 1" as subalgebra. Then A V' and .1 V", so Theorem 2.3(i) implies that .1 V' + V", whence V' + V" V. (lii) Again we let V' be a proper subvariety of V. By (P1) there exists a finite algebra .1 V such that .1 V'. Now (P2) implies that {j}1' V, whence by (ii) V properly contains V' + {L}" which in turn properly contains V'. Consequently V' is not a dual 0 cover of V.

The cardinality of A. Let if be the (countable) set of all lattice identities. Since every variety in A is defined by some subset of 3, we must have Al 2". The same argument shows that if V is any variety of algebras (of finite or countable type), then IAyI 2". Whether this upper bound on the cardinality is actually attained depends on the variety V. For V = 4 the answer is affirmative, as was proved independently by McKenzie [70] and Baker [69] (see also Wille [72] and Lee [85]). Baker in fact shows that the lattice A4 of all modular subvarieties contains the Boolean algebra 2" as a sublattice. We postpone the proof of this result until we have covered some theory of projective spaces in the next chapter. In Section 4.3 we give another result, from Lee [85], which shows that there are 2'' almost distributive lattice varieties (to be defined). In contrast, Jónsson's Lemma implies that any finitely generated congruence distributive variety V has only finitely many subvarieties and therefore Ay is finite. An as yet unsolved problem about lattice varieties is whether the converse of the above observation is true, i.e. if a lattice variety has only finitely many subvarieties, is it finitely generated? This problem can also be approached from below: if a lattice variety V is finitely generated, is every cover of V finitely generated? Sometimes these problems are phrased in terms of the height of a variety V in A (= length of the ideal (V]). Since A is distributive, to be of finite height is of course equivalent to having only finitely many subvarieties. Call a variety V lousily finite if every finitely generated member of V is finite. For locally finite congruence distributive varieties, the above problem is easily solved. THEoREM 2.7 Every finitely generated variety of algebras is locally finite. Conversely, V

if

is a locally finite congruence distributive variety, then

(i) every join irreducible subvariety of V that has finite height in Ay is generated by a finite subdirectly irreducible member, and (ii) every variety of finite height in Ay is finitely generated.

2.2.

THE STLttCTEitE 01" THE BOTTOM 01" A

a

a3

17

a4

Iv

Figure 2.1

if V is finitely generated, then V = for some finite algebra .1 V. For w the elements of are represented by vi-ary polynomial functions from A7' to 1, of which we can have at most JIVI". hence is finite for each vi w, and this is equivalent to V being locally finite. vi

Conversely, assume that V is a locally finite congruence distributive variety. (i) if a subvariety V1 of V is join irreducible and has finite height in Ay, then it isin ftct completely join irreducible, whence Theorem 2.5 implies that V' is generated by a finitely generated subdirectly irreducible algebra, which must be finite. (ii) follows from (i) and the fact that 0 a variety of finite height is the join of finitely many join irreducible varieties.

Nonfinitely based and nonfinitely generated varieties. It is easy to see that a variety can have at most countably many finitely based or finitely generated subvarieties, hence McKenzie's and Baker's result (IAI = IA.wI = 2u1) imply that there are both nonfinitely based and nonfinitely generated (modular) lattice varieties. An example of a modular variety that is not finitely based is given in Section 5.3, and £ and M are examples of varieties that are not finitely generated. In tisct Freese [79] showed that, unlike £, M is not even generated by its finite members (see Section 3.3). Other such varieties were previously discovered by Baker [69] and Wille [69].

2.2

The Structure of the Bottom of A

Covering relations between modular varieties. The class of all trivial (one-element) lattices, denoted by T = Mod{z = p}, is the smallest lattice variety and hence the least element of A. if a lattice variety V properly contains T, then V must contain a lattice which has the two-element chain 2 as sublattice, hence 2 V. It is well known that, up to isomorphism, 2 is the only subdirectly irreducible distributive lattice, and therefore generates the variety of all distributive lattices, V = {2}" = Mod{a4p + z) = zy + zz}. It follows that V is the unique cover of T in the lattice A. The next important identity is the modular identity zp+zz = a4p+zz), which defines the variety M of all modular lattices. Of course every distributive lattice is modular, and a lattice L satisfies the modular identity if and only if, for all u, v, to L with u to we have u + no = (u + v)to (for arbitrary lattices we only have instead of equality). The diamond M3 (see Figure 2.1) is the smallest example of a nondistributive modular lattice. A well known result due to Birkhoff [35] states that M3 is in fact a sublattice of every

CBAJ'TLJ

18

nondistri hutive niodular lattice. \\ e give a sketch of the proof. a (J H—yz and v z)(fj—t— z). Ihen clearly a H—

H—

(I

a

b

a

c

a

(a

H—

(u

K v,

ifS

L and define and the elements

-f-mr)v H—

(a ± z)v

±

generate a diamond with least element a and greatest element a v for all choices of y, z E L, then the identity H-

c;LIVERjL

2.

H-LIZ

H-

z)(y

H-

v.

On the other hand, if

z)

holds iii L, and it is not difficult to see that this identity is equivalent to the distributive

identity. It follows that every nondistri hutive modular lattice contains a su blat tice isomorphic to J13, and consequently the variety .t13 {113} V covers P. More generally, since the lattices Jl,, (see Figure 2.1) are simple (hence su hdirectly irreducible) modular lattices for each u is a su blat tice of , it follows from Corollary 1.7(i) that, 3, and since {lJ,JV. Hence the IJ K II}, where up to isomorphism, {2} 3 K varieties .\4 form a countable chain of join irreducible modular subvarieties of .ti, with covering .\4 for u -I, .\4 is in fact the 3. JOnsson [68] proved that for u only join irreducible cover of .\4 and that .t13 has exactly two join irreducible covers, .t132 and .\44. I his result is presented in Section 3.-I. Further remarks about the covers of .t132 and various other modular varieties appear at the end of Chapter 3. :

Covering relations between nonmodular varieties. A lattice variety is said to be .\4 ). If L E V is rwr,modaiai if it contains at least one nonmodular lattice L (i.e. L nonmodular, then we can the existence of three elements a, v, a' E L such that a K a' and a a v a'. In that case the elements (I a ('U', b v and c v a' ( ( a generate a sublattice of L that is isomorphic to the ierdauori X with critical quotieiit c/o (see Figure 2.1). Since the pentagon is nonmodular, one obtains the well known result of Dedekind [00]: Every nonmodular lattice con tains a sub] at tice isomorphic to V Many of the later results will he of a sinular flavor, in the sense that a certain property is shown to fail precisely because of the presence of sonic particular su blat tices. If L and K are lattices, we say that L K if L does not have a sublattice isomorphic to K. Otherwise we say that L K. In this terminology, modulari tv is said to be characterized by the exclusion of the pentagon. An immediate consequence is that the variety generated by the pentagon (denoted by is the smallest nonmodular variety .Ag am, JOnsson's Lemma enables us to conipute {2, X} and hence is a join irreducible cover of the distributive variety P. Since every lattice is either modular or nonmodular, we conclude that .t13 and are the only covers of P. In a paper of Mckenzie [72] there is a list of 1.5 subdirectlv irreducible lattices {LJV (1 L2,. , (see Figure 2.2) with the following property: If we let L1 1, ., 1.5) then each of them satisfy (L1 {2, X, U. Hence each L1 is a join irreduci ble cover of the variety It is a nontrivial result, due to JOnsson and Rival [79], that Mckenzie's list is complete. A proof of this result appears in Chapter -1. H—

H—

.

.

.

.

)

H—

)

2,2.

IL/IL

SILT/U

('ft RE

01 IL/IL

B0fift\1

01 A

11igurc 2.2

19

CiLtl'TtJit 2. (JERE1LtL

20

1LESELTS

'43

Figure 2.3

4, 4, 4, 4,

Rose [84] proves that above each of the varieties £13, £14 and there is a chain of varieties £? (vi w), each one generated by a finite subdirectly irreducible lattice .14' (14 = such that is the unique join irreducible cover of £}'

= 6,7,8,9,10,13,14,15). Lattice varieties which do not include any of the lattices A4, L1,. . . , L12 are called almost distributive by Lee [85]. They are all locally finite, and Lee shows that their finite subdirectly irreducible members can be characterized in a certain way which, in principle, enables us to determine the position of any finitely generated almost distributive variety in the lattice A. Ruckelshausen [78] investigates the covers of M3 + .V, and further results by Nation [85] [86] include a complete list of the covers of the varieties and £12. Nation also shows that above and £12 there are exactly two covering chains of join irreducible varieties. These results are discussed in more detail at the end of Chapter 4. A diagram of A is shown in Figure 2.4. (1

2.3

Splitting Lattices and Bounded Homomorphisms

The concept of splitting. A pair of elements (z,p) in a lattice L is said to be a splitting pair of .L if L is the disjoint union of the principal ideal (z] and the principal filter [p) (or equivalently, if for any z L we have z z if and only an immediate consequence of this definition.

if p

z). The following lemma is

2,3.

SJ'LJJILTLVG

LAILIICLS AXL) HOt XDLI) BWJOiJOIUU-JlSiJS

21

"V

A

r2 r2 r2 r2 r2 r2

A1A A P

l"l

C

T

11igurc 2.-I

C

r2 ('2 r2 r2 r2 12 11 1,

C' C'

L1

C

G2

C

C

C

C)CIIJCII

C, C, C,1C,

CiLtl'TEit 2. (JERE1LtL 1tESELLS

22

LEMMA 2.8

in a compl etc lattiw L the following conditions arc equivalent:

(i) (z, p) is a splitting pair of L; (ii) z is completely meet prime in (ill) p is completely join prime in

.L

z;

and p

= and z =

.L

z.

The notion of "splitting" a lattice into a (disjoint) ideal and filter was originally investigated by Whitman [43]. In McKenzie [72] this concept is applied to the lattice A, as a generalization of the familiar division of A into a modular and a nonmodular part. What is notewortby about McKenzie's and subsequent investigations by others is that they yield greater insight, not only into the structure of A, but also into the structure of free lattices. In this section we first present some basic tiscts about splitting pairs of varieties in general and then discuss some related concepts and their implications for lattices. Let V be a variety of algebras and suppose (V0,V1) is a splitting pair in A. By Lemma 2.8 V0 is completely meet prime, hence completely meet irreducible, and since V0

= Mod ldV0 =

fl

Mod{e}

it follows that

V0 can be defined by a single identity Dually, since every variety is generated by its finitely generated subdirectly irreducible members, V1 = {A}" for some finitely generated subdirectly irreducible algebra 1. We shall refer to such an algebra .1 (which generates a completely join prime subvariety of V) as a splitting algebra in V, and to the variety V0 as its conjugate variety, defined by the conjugate identity Note that if V is generated by its finite members, then we can assume, by Theorem 2.5, that .1 is a finite algebra. If, in addition, V is congruence distributive, then Corollary 1.7(iv) implies that .1 is unique. In particular, if(Vo, V1) is a splitting pair in A, then V1 = {L}" where L is a finite subdirectly irreducible lattice, and we refer to such a lattice as a splitting lattice. The two standard examples of splitting pairs in A are (T,D) and (M,X).

Projective Algebras. An algebra P in a class K of algebras said to

be pivjective in K

if for any homomorphism h P—' B and epimorphism g .1 —w B with A,B K, there exists a homomorphism f P —, .1 such that h = yf (Figure 2.5(i)). An algebra B is a irttract of an algebra .1 if there exist homomorphisms 1: B —' .1 such that yf is the identity on B. Clearly f must be an embedding, and g is called a ,rttmction of f. 2.9 Let K be a class of algebras in which K-free algebras exist. Then, for any K, the following conditions arc equivalent:

LEMMA

P

(i)

P is projective in K;

(ii) l'br any algebra .1 K and any epimorphism g P ; .1 such that yf is the identity on P;

.1 —w

P there exists an embedding

f

(ill)

P is a retract of a K-free algebra.

f

(ii) is a special case of (i), with B = P and h the identity on P. Clearly must be an embedding in this case. Suppose (ii) holds, and let 1 be a set with X = P1. Then

2.3. SPLiTTING LATTICES SRI) BOtRDED 1IOMOMO1tP.WSMS

fi

.1'

.1'

'>

1')c(_V)

(i)

23

'

(11)

Figure 2.5

there exists an epimorphism g : l')ç(X) —' P. Since l')ç(X) K, it follows from (11) that P is a retract of l')ç(X). Lastly, suppose P is a retract of some K-free algebra l')ç(X),

andleth:P—tBandg:A-*B begiven(A,B€K). Thenthereexistf':P—'l')ç(X) and g': 1')ç(X) —, P such that g'f' is the identity on P (Figure 2.5(11)). Since g is onto, we can define a map k : X —' .1 satisfying gk(z) = hg'(z) for all z X. Let ii be the 0 extension of k to l')ç(X), then hf' is the required homomorphism from P to 1. Thus, for any variety V, every V-free algebra is projective in V. LEMMA 2.10 (Rose [84]). Let K be a class of algebras and suppose that P K" is subdirrxtly irreducible and projective in K". Then P is isomorphk to a subalgebra of some member of K.

P

K" = HSPK, Lemma 2.9 (11)

implies that P SPK. hence we can where assume that P is a subalgebra of a direct product K for i in some we see that index set I. Denoting the projection map from the product to each by P is a subdirect product of the family of algebras : I I]. But P is assumed to be subdirectly irreducible, so there exists 5 such that is isomorphic to P. Therefore Since

I

0

ir5:P—'L5isanembedding.

Recall from Section 1.1 that can be constructed as a quotient algebra of the word algebra I1'(X), whence every element of can be represented by a term of W(X). Also if .1. is an algebra and p,q are terms in W(X), then the identity p = q holds in .1 if and only if h(p) = h(q) for every homomorphism h: I1'(X) —' A. Notice that if V is a variety containing 1, then any such h can be factored through The following theorem was proved by McKenzie [72] for £-free lattices, and then generalized to projective lattices by Wille [72] and to projective algebras by Day [75].

a variety of algebras, suppose P V is projective in V, and for P there is a largest congruence çb (Jon(P) which does not identify a and 6. Then P/çb is a splitting algebra in V, and if P J'WX) is an embedding and —w P is a retraction off (i.e. yf = idp) then for any terms p, q which represent g : f(a),f(6) respectively, the identity p = q is a conjugate identity of P/çb. Paootlt is enough to show that p = q fails in P/r/' but holds in any subvariety of V that does not contain P/çb. Let 7 : P -. P/çb be the canonical epimorphism. Then does THEoREM 2.11 Let V be

some a,b

f:

CBAJ'TLJ

2.

US

c;LIVERjL

not identify f(o) and f(b), hence by the remark above, p Suppose now q fails in 17 that V' is aiiy su bvariety of V not containing 17 and let Ii X X ) be the ) ( extension of the identity map 011 the generating set X. Clearly the identity p q will hold ill V' if and only if hf( (I) hf( b). Suppose to the contrary that (I and b are not identified by ker hf. Since is assumed to be the largest such congruence, we have ker hf and it follows from the second homomorphism theorem that P 17 factors through X ), hence 17 E V', a contradiction. I herefore p q holds ill V'. :

I

:

In particular, any pro jective subdirectlv irreducible algebra is splitting. Combining Lemma 1.10 with the above theorem we obtain the result of \\ ille [72]: (:oRoLL;\RY 2.12 Let V I)e a lattice variety. u/v a pri lie quotient in some lattice P E V which projecti ye in V. and suppose 0 is the t on P that does not collapse a / v. Ihen 170 is a split tins lattice in V.

If we take V to he the variety L of all lattices, then the con verse of the above corollary is also true. In fact it follows froni a result of Mckenzie ( Corollary 2.26) that every splitting lattice in L is isomorphic to a quotient lattice of some finitely generated free lattice 1(u) modulo a congruence 0 which is maximal with respect to not collapsing sonic prime quotient of 1( ii). Before we can prove' this result, however, we need some more information about free lattices, which is due to \\ hi titian [41] [42] and can also be found ill [ATL] and [GLT]. \\ hi titian showed that a lattice L is (L- ) freely generated by a set X L if and only if for all y E A and o, b, c, il E L the following four conditions are

I

satisfied:

\\

1)

(\\2) (\\3)

K

implies

,bK j implies tKc—frd implies

(\\)

implies

i

(i.e. generators are incomparable) Kg or bK v

tKc or (,Kc±d

or

bKc±d

or

(Ib<.c or

I hese conditions are

also known as \\ hi tman's solution to the word pro blem for lattices since they provide an algori thin for testing when two lattice terms represent the same element of a free lattice.) In fact JOnsson [70] showed that if V is a nontrivial lattice variety then ( \\ 1), ( \\ 2) and (\\ 3) hold in ally V-freely generated lattice. \\ e give a proof of this result. A subset X of a lattice L is said to he iiiidaru/ard if for any distinct E A

(\\ 2') (W3')

and

>

...

also require the important notion of a join-cover. Let I and he two nonemptv finite subsets of a lattice L. \\ e say that I (in symbols I ) if for every a E I there exists v E such that a K v. is a join—conic of (I E L if (I K , and a join—cover of (I is rwn,icu'uzi if (I v for all v E U bserve that aiiy join—cover of o, which refines a nontrivial join-cover of o, is itself nontrivial. I he lIOtiOll of a ;ueei-cocec is defined dually. e

.

2.3. SI'L1fl1LNG LATTICES LEMMA 2.13

(i)

SRI) BOtRDED .IIOMOMO1tP.WSMS

Let

.L

be a lattice gvnerated by a set X

if every join-cover Li ç X of an element a

25

ç .t.

L is trivial, then every join-cover of a

is trivial. (II) X is irredundant (W3) hold.

if and only if for

all

z,p

X and a,b,c,d

.L

(Wi), (W2) and

of L for which every join-cover Li ç of a is trivial, and let 5(1') and P(l') denote the sets of all elements of 1, that are (finite) joins and meets of elements of I' respectively. Since every join-cover I' ç 5(1') of a is refined by a join-cover t.' ç 1', I' is trivial. We claim that the same holds for every join-cover I' ç P(1') of a. I', a v. Each Suppose I' is a finite nonempty subset of P(1') such that for all v element v is the meet of a nonempty set ti,, ç 1'. Since a v, there exists an element such that a u,. Each it.,, belongs to 1', so the set H' = : v cannot be a join-cover of a, i.e. a EW. But v = fiti,, tie,, and hence E E which means that I' is not a join-cover of a. This contradiction proves the

(i) Let 1'

be a subset

t;

claim. Now let lb = P(X) and = P5(1;4) for vi w. If every join-cover Li ç X of a is trivial, then by the above this is also true for .1 replaced by 1. Since .1 generates I.., we have L = so if I' is any join-cover of a, then I' ç I;,, for some vi w, and therefore I' is trivial. (II) If X is irredundant, then clearly the elements of X must be incomparable, so (Wi) is satisfied. It also follows that any join-cover Li ç X of a generator z X is trivial, every join-cover trivial. hence by part (i) of z is This implies (W3), and (W2) follows by duality. Conversely, if z z1 + z2 + ... + z74 with all z,z1,. . .,z74 1 distinct, then repeated application of (W3) yields z z = for some i = 1,.. .,vi. 0 This contradiction, and its dual argument, shows that 1 is irredundant.

1,

a class of lattices that contains at least one is a lattice that is K-freely generated by a set X, then X is

THEoREM 2.14 (Jónsson[7O]). Let K be

nontrivial lattice. irredundant.

if 1"

Suppose z, z1,. . . , z74 X are distinct, and let L K be a lattice with more than one element. Then there exists a,b L such that a 6. Choose a map : 1 —' L such that f(z) = a and = 6 (all 1). Then the extension of to a homomorphism By 0 duality X is irredundant.

f

f

The last condition (W) is usually referred to as Whitman's condition, and it may be considered as a condition applicable to lattices in general, since it makes no reference to the generators of L. Clearly, if a lattice L satisfies (W), then every sublattice of L again satisfies (W). In particular Whitman's result and Lemma 2.9 show that every projective lattice satisfies (W). Day [70] found a very simple proof of this fact based on a construction which we will use several times in this section and in Chapters 4 and 6. Given a lattice L and a quotient = u/v in L, we construct a new lattice

I

with the ordering z

p

in

L[I]=(L—I)u(1x2) L if and only if one of the following conditions holds:

CBAJ'TLJ

26

I and (Hi)

I, I,

EL

(H) £

gEL

(iv) £

y

£

1),

£ K

c;LIVERjL

2.

in L;

(b,j) and £
<jin

2.

L[I] is refrrred to as L wit/i doubled quotient u/c', and it is easy to check that L[I] ill fact a lattice (2 1). Also there is a natural {O, 1 } is the two-element chain with U epimorphism 7 L[ u/c'] L defined by if

Lv e

(l,b <

say

that

(o, b, c, ci)

a

P

E

EL u/c' (o,i) some

£ £

if

—

a

I E

2.

(\\ )-kzilaii of L if (lb

K c

±

ci

hut (lb < c,

ci

and

(—fr ci.

LEMM;\ 2.15 (Day [70]). Let (o, b, c, ci) be a

Ihen there does not exist an f on L (i,e, there exists no caret raction 017). I'I-tOOF. Suppose the contrary.

(

\\ )-fttilure oiL, and let u/c' c ± ci/ob. L[u/ c'] such that 7f is the identity map

L —

I hell

From the equivalence of (i) and (ii) of Lemma 2.9 we CoI-tOLLAI-n

f( u).

f(

But f(v) f(ob) f(o)f(b) (lb (v.1) since o,b < u, and dually is a contradiction since by definition (c', 1) < (u, U).

f(u)

(u,U). Ihis

0 htain:

2.16

(i) Every lattice which is prajective in L satisfies (H) Evury fixi]

(\\

).

lattice satisfies (\\ 1), (\\2), (\\3) ajid (\\).

As mentioned before, the converse of (ii) is also true. A partial converse of (i) is given I heorem 2.19.

by

Bounded homomorphisms. A lattice homomorphism f L upper bounded if for every b E L' the set (b] or has a greatest element, denoted by Qj ( b);

tower boa nc/ed if for every b E L' the set or has a least elenient, denoted by ij( b);

if f

boa nc/ed

)

(ii)

ilL is 11

E

L

K b} is

if a

b} is either empty

f(

gi veil one of the above three properties holds.

L — L' and y

a finite lattice. then f

t md y ii

uppii

i)

L'

—

L" he two lattice homomorphisms.

hounded, haunch

€1

is

uppii

be

either empty

following lemma lists sonic easy consequences of the above defilli tiOlls.

LEMM;\ 2.17 Let f (i

L

L' is said to

is both tipper and lower bounded;

(apper/iowei) bounded

I he

[b)

E

—

i)

haunch

€1

2.3. SPLiTTING LATTICES

SRI) BOtRDED .IIOMOMO1tP.WSMS

27

(ill)

is (upper/lower) bounded and bounded.

is an qpimorphism, then g is (upper/lower)

(iv)

if yf

(v)

ii then aj(6) is the greatest member L is meet preserving. Dually, 1ff is a lower bounded qpimorphism then ij(b) is the least member off—' {6}, and ij : ii .L is join preserving.

is (upper/lower) bounded and g is an embedding, then bounded.

f is

(upper/lower)

1ff is an upper bounded epimorphism and 6 of f'{b}, and the map

aj

:

1!

A lattice is said to be upper bounded if it is an upper bounded epimorphic image of some free lattice, and lower bounded if the dual condition holds. A lattice which is a bounded epimorphic image of some free lattice is said to be boundS (not to be confused with lattices that have a largest and a smallest element: such lattices will be referred to as 0,1 - lattices). Of course every bounded lattice is both upper and lower bounded. We shall see later (Theorem 2.23) that, for finitely generated lattices, the converse also holds. The notion of a bounded homomorphism was introduced by McKenzie [72], and he used it to characterize splitting lattices as subdirectly irreducible finite bounded lattices (Theorem 2.25). We first prove a result of Kostinsky [72] which shows that every bounded lattice that satisfies Whitman's condition (W) is projective. For finitely generated lattices this result was already proved in McKenzie [72].

2.18 Suppose 1"(X) is a free lattice generated by the set be a lower bounded qpimorphism. Then for each 6 .L the set {z LEMMA

1 and let f : 1"(X) -. L X : f(z) 6} is finite.

f

Since is lower bounded and onto, ij(b) exists. Suppose it is represented by a term t(z,,. ..,z74) .L"(X)for some . .,z74 then clearly

{z€X:f(z)6}={z€X:zt(z,,...,z7j}c{z,,...,z7j. 0 In particular note that if a finite lattice is a (lower) bounded homomorphic image of a free lattice F, then F must necessarily be finitely generated. 2.19 (Kostinsky [72]). Every bounded lattiw that satisfies (W) is projcctive (in £).

f be a

bounded homomorphism from some free lattice F(X) onto a lattice that .L satisfies (W). We show that .L is a retract of F(X), and then apply Lemma 2.9. To simplify the notation we will denote the maps aj and ij simply by a and Let

.L, and suppose

3.

Let h : .L"(X) —, .L"(X) be the endomorphism that extends the map z af(z), z X. that for each a .L"(X), 3f(a) h(a) af(a), from which it follows that fh(a) = f(a). This is clearly true for z X. Suppose it holds for a,a' 1"(X). Since I is join-preserving We claim

if(a + a') = 3f(a) + if(a') = h(a + a')

h(a) + h(a') af(a) + af(a') af(a + a')

CBAJ'TLJ

28

and similarly since claini.

Q

is nieet-preserving

the map y

e now define

(t)

b Ii

i(be)>

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'
implies implies

byi?(1,n—1).

u=to+aib;+Erjaibj=to+E7iaibj P(1,n)

for all vi. Now assume that in> 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
if

pllq if

third relation a is defined by paq if JLIq or pllq. Xote that if JLIq then y and So, depending on whether or not the last inequality is p q strict, the elements p, q, generate a su blattice of L that is isomorphic to either or (Figure 2.6) .Also if pllq, then p q 12 generate a su blattice of p q q, p q L isomorphic to A

H—

H—

H—

H—

if if if

112 113

If we assume that L is semidistri huti ye, then the last case is excluded since 113 fails (SD+) q ± ± ± pq). Observe also that in general the element in the definition of I is not unique, hut in the presence of (SD) we can always take n(q), where

n(q)

E L

since L is finite and by (SD

)

:

n( q)

Kr

and

q

E L

:

q4,

itself satisfies qn(q)

q

I he following lemma froni JOnsson and Xation [75] motivates the above definitions. Xote that if L)( L) L for sonic finite lattice L, then sonic join-irreduci ble element of L is not in I) ( L), since for any nonemptv subset I of L), 37 is an element of ( L). 1

2.33 if L is a finite seinidistribu tive lattice and p 1(L) IJ(L) with paq.

LEMMA q E

E

1(L)

IJ( L), there exists a nontrivial join-cover < L)(L) of p refines Since p K 37 , we have 37 Choose ti E vu minimal with respect to the 1(L) and p y since is a nontrivial join—cover. Xote

I'I-tOOF. Since p

I

.

.

y E

— p yields the desired conclusion.

p

—

u

D( L) then there exists

of p such that no join-cover n(p), whence v0 < for property < Clearly that < n(p) if and only if intl if j L)( L) thin q

2,3. SJ'LJJILTLVG LAILIICLS AXL) BOL XIJLi)

BWJWJORJ'BJSUS

37

(J(

A1

p

113

11igure 2.6

K Otherwise E D( L ), and we can choose an element :. e claim that IJ( L). Assume the contrary. Since , is a p y p K join-cover of :, which is either trivial or nontrivial. If it is trivial, we let I {y, :}, and I D( L of which refines , and we if it is nontrivial, then there exists a join—cover ij {j}. In both cases / is a su hset of L)( L) and a join-cover of p which refines let / (since y K v0), contradicting the assumption p L)( L). Corollary 2.32 every element of L has a canonical join representation, so there exists a finite set / I L such that D(L), there exists q E / u such 37 / u. Since we see that v—fr: >1), that q D(L). Lettingq' Y:(/u—{q} andy Kr and q p (since : K : Furthermore, and therefore p by the minimali tv of :. Consequently p lq. H—

I

)

I

—I—

By a repeated application of the preceding lemma we obtain elements E 1(L) with for i E w. Since L is finite, this sequence must repeat itself eventually, so we can assume that up,, up0. Such a sequence is called a cycle, and it follows that the nonexistence of such cycles iii a finite semidistri butive lattice L implies that D( L ) L. I he next result, also from JOnsson and Xation [75], shows that the converse is true is an

arbitrary lattice. IHEOREM 2.34

IlL

is any

lattice which contains a cycle, then L)(L)

I'I-tOOF. Suppose

each Pu E

has a nontrivial cover, so L)1JL). If then Pu

for sonic

E

L.

1(L). From Figure 2.6 we can see that

I)0(L). Suppose no belongs to L), but say Kr ± £, and for sonic E L. Th

CiLtl'TEit 2. (JERE1LtL

38 Since

{pi,z}

ç Dk_1(L) of p1, with Li C fri,z}. For z. Dk_1(L)), but if u < p', then u z, a contradiction. covers p1,, and therefore there exists a cover u ç Dk_1(L) of For each u L.', either u pu* or u
covers ptj, there exists a cover Li, either u z or u
each u

hence u If p0Bpi, then with L.' C Therefore p1,

1tESELTS

L!

E L•' foralli,andsoll(Lh&L.

0

(Joaoisuw 2.35 l'bra finite semidistributive lattiw L, 11(L) = L ifand only ifL contains no cycles. An example of a finite semidistributive lattice which contains a cycle is give in Figure 2.7. at the end of this section.

Day's characterization of finite bounded lattices. The results of this section are essentially due to Alan Day, but the presentation here is taken from Jónsson and Nation [75]. A more general treatment can be found in Day [79]. We investigate the relationship between J(L) and J(Con(L)). By transitivity, a congruence relation on a finite lattice is determined uniquely by the prime quotients which it collapses. The next lemma shows that we need only consider prime quotients of the form p/ps, where p J(L). LEMMA 2.36 Let L be

(i) (ii) (iii)

a finite lattice, and suppose 0

(Jon(L). Then

0€ J(Con(L)) if and only if 0 = con(u,v) for some prime quotient u/v of L; if u/v is a prime quotient of L then there exists p J(L) such that p/ps if L is semidistributive, then the element p in (ii) is unique.

/

u/v;

(i) In a finite lattice

0= E {con( u,v) : u/v is prime and

u$v}

so if 0 is join-irreducible then 0 = con(u, v) for some prime quotient u/v. Conversely, where suppose ô (Jon(L) is strictly below 0. Then ô ç is the unique largest congruence that does not identify u and v. hence fl con(u, v) is the unique dual cover of con(u, v), and it follows that con(u, v) J(con(L)). To prove (ii) we simply choose p minimal with respect to the condition u = v+p. Then p J(L) and p/ps u/v. (lii) Supposeforsomeq J(L),q p, u/v. Thenu = v+p = v+q,

/

and bysemidistributivity u= v+pq. But thenpq v,henceu= v+pq= v, whichisacontradiction. pq

/

0

From (i) and (ii) we conclude that for any finite lattice L the map p from J(L) to J(Ccn(L)) is onto. Day [79] shows that the map is one-one if and only if L is lower bounded. Let us say that a set C) of prime quotients in L corresponds to a congruence relation 0 on L if 0 collapses precisely those prime quotients in L that belong

toQ.

2.3. SPLiTTING LATTICES

SRI) BOtRDED .IIOMOMO1tP.WSMS

2.37 A set of prime quotients Con(L) if and only if LEMMA

C)

in a finite lattice

(it) l'br any two prime quotients r/s and u/v in F + v or su v < u F, then u/v C).

r/s

\

4 if r/s

.L

corresponds to some

C)

and

ifeithers

r/s

/

The two conditions of (it) can be rewritten as either

u/st 2 u/v, hence if F/s is collapsed

39

by some congruence, so is

1)

v< u

F + v/v 2 u/v or u/v. Therefore (it)

is dearly necessary. (J}. If z/p is a prime quotient Suppose (it) holds and let 1) = E{con(u,v) u/v that is collapsed by 0, then con(z,p) ç (1. But con(z,p) is join irreducible and Con(L) is distributive, so con(z,p) is in fact join prime, whence con(z,p) ç con(u,v) for some u/v C). By Lemma 1.11 u/v projects weakly onto z/p, and since (it) forces each quotient 0 in the sequence of transposes to be in C), it follows that z/p C). THEoREM 2.38 (Jónsson and Nation [75]). if .L is a finite semidistributive lattice and S ç J(L), then the following conditions are equivalent:

(i) There exists (ii)

1)

J(L), pOpe if and only if p S.

Con(L) such that for all p

l'braIlp,q€J(L),puqandq€Simplyp€S. Assume (i) and let p,q J(L) with puq and q S. Then Figure 2.6 shows that projects weakly onto p/ps, so if 1) collapses it also collapses p/ps, which implies

p€S.

/

Conversely, suppose (ii) holds, and let C) be the set of all prime quotients u/v in .L, such that the unique member p J(L) with p/ps u/v belongs to S. Then p/ps C) if and only if p 5, so by the proceeding lemma it suffices to show that C) satisfies (it). Consider two prime quotients F/s and u/v in .L and let p and q be the corresponding members of J(L), so that F/s and p/ps u/v. If F/s C), then by uniqueness by definition. Assuming thatp9&q, wearegoing q€S. to show that

/

(1)

/

ifs v< u F+v thenpBq, and

(2) Statement (ii) then implies p

5, whence u/v

C)

as required. Under the hypothesis of

(1)weneedtoshow wemust havep v. For thesamereasonp q, andp q by assumption. would imply which together with pv = gives p(v+q) = p* Finally, p = by semidistributivity. This, however, is impossible since v+q = s+v+q = F+v u p. This proves (1). Now suppose that the hypothesis of (2) is satisfied. Clearly q

provepLqitsufflcestoshow

s and

s, so to

thatqcpandpcq+s. Wecertainlyhaveq+s= Fp

the hypothesis, and this indusion must be strict, since p = F > s would imply p = q by the join irreducibility of p. Observe that p s because ps su v and p v. Since by

pF,thisimpliesF=s+pwhich,togetherwithF=s+qyieldsF=s+pq. s, whichisimpossible becauses+pq= F >8. required.

qas 0

CiLtl'TtJit 2. (JERE1LtL

40

1LESELTS

Figure 2.7

THEoREM 2.39

l'br any finite semidistribu live lattice L, the following conditions are

cquivalent:

(i) IJ(L)I = IJ(Con(L))I (II) 11(L)=L

(ill) 11'(L)=L (iv) L is boundcd.

It follows from

the preceding theorem that, for two distinct elements p and q of if and only if there exists a cycle containing both p and q. Consequently the map p from J(L) to J(Con(L)) is one-one if and only if L contains no cycles if and only if 11(L) = L by Corollary 2.35. Therefore (i) is equivalent

J(L),

=

to(ii). Lemma 2.36(h) and its dual imply that the number of meet irreducible elements of L is equal to the number of join irreducible elements (to every prime quotient nC/rn where rn is meet irreducible and rrC is its dual cover, corresponds a unique p J(L) such that nC/rn, and vice versa). Therefore the condition IJ(L)I = IJ(Con(L))I is equivalent p/ps to its own dual, and hence to the dual of 11(L) = L, namely 11'(L) = L. Lastly (II) and (lii) together are equivalent to (iv) by Theorem 2.23 and its dual. 0

/

It is interesting to examine how the above conditions fail in the semidistributive lattice in Figure 2.7, which contains the cyclep., dp4 Bp4 B p1j. If we add an element a on the edge c,p4, then we obtain an example of a subdirectly irreducible semidistributive lattice which is not a splitting lattice. It is not difficult to prove that every critical quotient of a splitting lattice must be prime, but the example we just mentioned shows that the converse does not hold even for semidistributive lattices.

2,4.

2.4

SJ'Lirnxc;

c;EXERAJI ALL

Splitting lattices generate all lattices

e will now prove a few lenimas which lead up to the result of Day [77], that the variety L of all lattices is generated by the class of all splitting lattices. I his result and sonic of the characterizations of splitting lattices will he used at the end of Chapter 6.

Let B be the class of all bounded lattices and let BF he the class of finite meni hers of is the class of all splitting lattices, and it is clearly sufficient B. By I heorem 2.2.5 (Bp (BF)V. to show that L

and direct products

LEMMA 2.40 Bp is closed under sti blat tices. homomorphic

with finitely many factors, PROOF. If L is a sublattice of a lattice B E BF, then by Lemma 2.17 (iv), any f 1(X) — — L, where X ) is a finitely generated free lattice, is hounded. If L is a homomorphic image of B, say Ii B L, then there exists y X ) — B such that f fig, and by the equivalence of (i ) and (iii) of I heorem 2.23 g is bounded. Ii is bounded since B is finite, hence f is also bounded. Lastly, if x B2 is , B2 E Bp and f ) an epimorphism, then is bounded, where B1 is the proJection map x B2 1,2). For b1 E B1, let i1(b1 ) be the least preimage of b1 under the map ilIf, and denote the zero of B1 by Since (bk, 02) is the least elenuent of is also the {bd, i2( least elenuent of and sinularly b2) is the least elenuent of b2 )}. 02 )}, {( {( i2( It follows that b2) is the least preimage of (bk, b2) (bk, b2 under (Ok, 02) H— ,f. ) ± ) Hence f is lower bounded, and a dual argument shows that f is also upper bounded. E :

i(

:

:

i(

:

i(X

:

I he two element chain is a splitting lattice, so the above lemma implies that every finite distributive lattice is bounded. Recall the construction of the lattice L[u/ c'] from a lattice L and a quotient u/c' of L (see above Lemma 2.1.5). if L

LEMMA 2.41 (Day [77]).

E

Bp

and I

u/v is a quotient oiL. then L[l]

E

Bp.

assumption L is a finite lattice, hence L[l] is also finite. Let X be a finite with f IJX L[l] a lattice epimorphism and let L[l] L be the natural epimorphism. Since L E BF, Ii i( X L is bounded, so for each b E L there we have exists a least mem her b) of {b}. By definition of ( I'I-tOOF. By

set

:

:

)

:

)

if b

El.

hence 7(b) i5 the least nuember of {b} for each b E (L I) 'J I x {O}. Xote that for any (1k, is the least member of (1 1,2) then ± (Li is the 112 E L if least member of ± 114. Since for any E u/c', (1, 1) ± (c', 1) it is enough to show that {( c', 1 )} has a least nuember. f is surjective, so there exists a a' E with f( a') ( v, 1). Define 1

i(X

E

X

1), and

if

S

{p

a'

Clearly f(m)

(

c',

:

E

(v,

1) K

1(X):

fl{ (c', 1) K

b E

f(p) implies

I

L

K

ajid

v

b}.

CBAJ'TLJ

2.

c;LIVERjL

I

S S is closed under meets. \\ e need to show that S is also closed tinder joins. Let 1o, q E 5 and suppose that (c', 1) K f(p q) f(p) (q). Xote that by construction of L[l], r ± E I x {1} implies r E I x {1} or E I x {1}, so if f(p ± q) E I x {1}, K q, which certainly implies then ( v, 1) K f(p) or ( v, 1) K f(q ), whence m K p or K p ± q. On the other hand, if f(p ± q) E L I, then 7f(p ± q) f(p ± q), and so 1(X), where ± q) Kp ± q. Iherefore f is lower bounded by ij L[l] —t—

if ih7(b)

—

A dual

Let

x {1}.

argument shows that f is also upper hounded, hence L[I] I

E

BF.

(L) he the set of all (\\ )-fttilures of the lattice L (see Lemma 2.1.5) and define

{c± il/ob: (o,b,c,d)

Iu'(L)

a

bE(L—l)IJlx{O}

if bE]

E

U(L)}.

2.42 (Day [77]). if L is a lattice that fiuils (\\ ). then there exists a lattice ii and bounded cpniioiphisiii p L L satisfying: icr any ( , (14) E and any ( L (1 E ± £4 :

I

)

I'I-tOOF. For each I E L) we construct the lattice L[I] and denote by 7] the natural epimorphism from L[I] onto L. Xote that is hounded with the upper and lower hounds

given by

of

if bEL—I l(b,0) if bE]

if bEL—I

lb

if

lb

bE]

respectively. Let L' he the product of all the L[I] as I ranges through I th pro jection map. Recall that for f, y — L we can define a suhlattice of L' by :

Lq(f,y)

fl{Lq(71 71], 7j71j) 1,1 E I — L to he the restriction of 7jQj( y), hence the I-tuple (Qj( y)) is ii. Xow, for every u E L, 7jQj( y) y an element of I, and clearly (Gi ( y)) is the greatest element of y) { y}. Sinularly i1 verify and therefore is a hounded epimorphism. lo the last part of the p u) v)), ( ( limmi it to thit foi ill (i b il) E U (L) i,(i) i,(b) o,(c) — I his is indeed the case, since c ± ci/ob I E L) implies Let

I

L': f(s)

:

7)71] to

ij(o) ij(b)

Qj(c) ± Qj(ci).

E BF, then ii is a su hlattice of a finite product of lattices L[I], hence and 2.-Il, E BF.

Xote that if L by Lemmas 2.-lU

(ob, 1) < (c ± ci,O)

I

IHEOREM 2.4:3 (Day [77]). bounded epinorphisin f L :

icr any L.

lattice L. there is a lattice

I

(\\ and )

a

2,4.

SJ'LJrnxc;

c;EXERAJI ALL L

and, for each

sublattice of the product E

I

I,, and

w, let be the

u E

L

limit of the L,, ,

defined by if and only if

for all

I

I

< ilk,

< c1,

is order- preserving, we have (If

\ow if (If bf < previous

c

H—

Let f define the maps

<

, b1

Jf, then

C

—f—

(lb < c

<

il1

H—

±

L — L, let L —

il

q

H—il1

that for ally and

(If bf

(lb < c, il.

and

b,,, <

max{j, k, 1,

i

< c,

,

±

il,.

ITI},

il1.

and we are done. If (If bf K then by the ± and again ob < c ± il. Hence L satisfies (\\ ). he the identity map 011 L0 L, and for u 1

by

Q,7,

I hell it

0/ <

u E w

Lf. \\ e claim that Ihen there exist

where under the projection 7Ff E Lf is the image of (\\ with o, b < c ± il and satisfies ). Suppose o, b, c, il E indices j, k, 1, in such that

Since each

i.e. L is the

—

and

i,.,

—

is easy to check that for y E L the sequences (h,, ( y)) and ( y)) are the greatest {y} respectively, hence is a hounded epimorphism.

and least elements of

IHEOREM 2.44 (Day [77]). L is generated by the class of all split tin,g I'I-tOOF. Let L IHp(), the free distributive lattice 011 three generators, say y, :, and consider the lattice L constructed ill the preceding theorem. L is a finite distributive )V• Choose elements lattice, hence L E BF, and it follows that L E E L which flap to £, y,: under L. Since the set X L y, :} satisfies (\\2') and (\\3'), so does the set X In addition L satisfies (\\ ), hence the sublattice of L generated by X is isomorphic to ic( 3). By a well known result of \\ hitman [42], the free lattice 011 countably many generators is a su blat tice of I (3), and therefore

I he two statements of the following corollary were proven equivalent to the above theoreni by A. hostinsky (see Mchellzie [72]). CoI-tOLLAI-n

(i)

2.45

ii) is weakly atomic fbr each ii E w,

(ii) kor any proper suhvariety

V

of L. there is a split tins

pair (Vi, V2) of

L

such that

V

I'I-tOOF. (i ) By the above theorem ic( ii) is a subdirect product of splitting lattices (1 E I). Let f ii) — he the subdirect representation, and suppose r/h is a S intli i E I iF t( i ) iF t( s) finiti il quotit nt of I (ii) I h n foi we call choose a prinie quotient p/q r)/iFff(h). By I heorem 2.2.5 ii) I*( f( f 1

I

CiLtl'TEit 2. (JERE1LtL 1tESELLS

44

is a bounded epimorphism, so there exists a greatest preimage v of q and a least preimage u of p, and it is easy to check that u + v/v is a prime subquotient of v/s (see proof of Theorem 2.25). (II) This is an immediate consequence of the preceding theorem, since £ = (lip)1' if 0 and only if every proper subvariety of £ does not contain all splitting lattices.

Note that if is weakly atomic for vi w, then by Corollary 2.26 is a subdirect product of splitting lattices, hence (lip)1' for each vi w. This dearly implies £ = (lip)1'. Using some of the results of this section, we prove one last characterization of finite bounded lattices.

of lattices ui/rn

2.46 (Day [79]). S finite lattiw .L is bounded if and only if there is a sequenw with . . , 4k-i-i = .L and a sequenw of quotients no/vu,. . . , (1 =

1 = L1, £1,. such that

The reverse implication follows from Lemma 2.41, since the trivial lattice 1 is obviously bounded. To prove the forward implication, let U be an atom in Con(L). We need only show that .L can be obtained from L,4 = .L/U by finding a suitable quotient u,4/v74 in L,4 such that .L. Since L,4 is again a finite bounded lattice, we can then repeat this to obtain L,4_1, .L74_2,..., 4, = 1. By Theorem 2.39 the map p is a bijection from J(L) to J(Con(14), and since U J(Con(14), there exists a unique p J(L) with U = con(p,p4. .L is semidistributive, so by the dual of Lemma 2.36 (iii) we can find a unique meet irreducible in .L such that rn*/rn p/ps, where is the unique cover of in. We daim that

\

(1) rn/ps transposes bijectively up onto rn*/p, and (2) ziJy

if and only if z = p or {z,y} = {z,p+ z} for some z

rn/ps.

Letting t¼ = rn/U and = p/U, we then have L74[u74/v74] L. Toprove(1),supposez rn/ps butz c (p+z)rn. Then wecanflndq€ J(14 such that q (p+ z)rn and q z. Now pJip implies z U (p+z)rn, which in turn implies Since the map p u—' is one-one, this forces p = q in, a contradiction. Dually rn*/p, one proves that for z z = ins + p. Since L is a finite lattice we need only check the forward implication of (2) for pairs (z,y) U of the form z -< p. Clearly con(z,p) con(p,p4, and since is an atom of Con(L), equality holds. This means that p is the unique join irreducible for which and therefore {z, p} = {z,p + z}. The reverse implication follows from the p/ps observation that if z = z, say, and z rn/ps, then z p andy = p+z. This 0 implies that p/ps p/z, whence zUp.

/

2.5

/

Finite lattices that satisfy (W)

We condude this chapter with a result about finite lattices that satisfy Whitman's condition (W), and some remarks about finite sublattices of a free lattice.

111115111511(U)

2,5. FL\1ILTE

IHEOREM 2.47 ( Davev and Sands[77] ). Suppose f is an epnnorphisin K onto a lattice L. ilL satisfies ( \\ ). then there exists an that y is the identity map on L.

a

from

y

:

finite lattice K such

L —

I'I-tOOF. Let f he the epiiiiorphisiii from K onto L. Since K is finite, j is hounded, so we 01) tam the meet preserving map Qj L — K and the join preserving map L — K (see Lemma 2.17(v)). Let he the collection of all join preserving maps 7 L — K which are K Qf(b) for all b E L ). JJ is not empty since ij E JJ pointwise below Qj (i.e. define a map y L — K by y(b) Y:{ 7(b): 7 E }. y is clearly join preserving and is in :

:

ii

:

.

ii

:

fact the largest element of JJ (in the pointwise order) .Also, since if( b) for all b E L, we have b f if(b) Kfy(b) K .tQf(b) b

K

y(b)

K

b)

which implies that fy is the identity map on L. It remains to show that y is meet preserviig, then y is the desired embedding of L into K. Suppose y((lb) )y(b) for some (I, b E L. Since y is order preserving, we actually have y((lb) )y(b). Define Ii: L — JJ

r) —

H—

y((l)y(b)

if

(lb < £

if

(lb

below Of since for (lb K I hen Ii ii because h( (lb) y((l )y( b) ;> y( (lb), hut Ii is we have h(s) )y(b) K It follows ± (Yf((l )Qf(b) ± (Yf((lb) that Ii is not join preserving, so there exist c, ci E L such that il) h(c) ± h(ci). From the definition of Ii we see that this is only possi ble if (lb K c ci, (lb < c and (lb < ci. I hus \\ implies that (I K c ci or b K c ci. However, either one of these conditions leads to a contradiction, since then )

)

)

—f—

H—

)

h(c ±

ci)

y(c±

H—

ci)

± y((l)y(b)

y(c ±

ci)

y(c) ± y(ci)

h(c) ± h(ci).

Actually the result proved in Davev and Sands [77] is somewhat more general, since it suffices to require that every chain of elements in K is finite.

Finite sublattices of a free lattice. Another result worth mentioning is that any finite semidistri butive lattice which satisfies \\ hitman's condition ( \\ can be em bedded in a free lattice. I his longstanding conjecture of JOnsson was finally proved by Xation [83]. Following an approach originally suggested b JOnsson, \ation that a finite semidis— tn hutive lattice L which satisfies (\\ cannot contain a cycle. By Corollary 2.3.5 and I heorem 2.39 L is hounded, and it follows from I heorem 2.19 that L can he embedded in a free lattice. (Xote that (\\ fails in the lattice of Figure 2.7) Of course finite su blattice of a free lattice is semidistri hutive and satisfies ( \\ Corollary 2.16, Lemma 2.30). So in particular Xation's result shows that the finite sub— lattices of free lattices can he characterized by the first-order conditions (SD+), (SD and )

)

)

)

Chapter

3

Modular Varieties 3.1

Introduction

Modular lattices were studied iii general by Dedekind around 1900, and for quite sonie time they were referred to as Dedekind lattices. I he importance of modular lattices stems from the fact that nian alge braic structures give rise to such lattices. For example the lattice of normal subgroups of a group and the lattice of su bspaces of a vector space and a proJecti ye space (pro jecti ye geometry) are modular. I he Jordan Holder I heorem of group theory depends only on the (senu-) modularity of normal subgroup lattices and the theorem of and Ore holds in any modular lattice. Pro jective spaces play an important role in the study of modular varieties because their su bspace lattices provide us th infinitely many su bdirectlv irreducible (complemented) modular lattices of arbitrary dimension. I hey also add a geometric flavor to the study of modular lattices. I he Arguesian identi tv was introduced by JOnsson [53] (see also Schiitzenberger [45]). It implies modularity and is a lattice equivalent of Desargues' Law for projective spaces. Some of the results about Arguesian lattices are discussed in Section 3.2, hut to keep the length of this presentation thin reasonable hounds, niost proofs have been omitted. As we have mentioned before McKenzie [70] and Baker [69] (see also \\ ille [72] and Lee [85]) showed independently that the lattice A of all lattice su hvarieties has mem hers. Moreover, Baker's proof shows that the lattice Ati of all modular lattice subvarieties contains the Boolean algebra as a sublattice. Continuous geometries, as introduced by von Xeumann [60], are complemented modular lattices and von Xeumann's coordinatization of these structures demonstrates an important connection between rings and modular lattices. sing the notion of an n-frame and its associated coordinate ring (clue to von Xeumann), Freese [79] shows that the variety .t1 of all modular lattices is not generated by its finite nienibers. Herrmann [84] extends this result by showing that .t1 is not even generated by its nienibers of finite leng t It I he structure of the hot toni end of AM is investigated in Grätzer [66] and JOnsson [68], where it is shown that the variety .t13 generated by the diamond J13 is covered by exactly two varieties, .t14 and .t132. Furthermore, JOnsson [68] proved that above .t14 we have a chain of varieties .t1 each generated by a finite modular lattice of length 2, such that is the only join irreduci ble cover of .\4 Hong [72] adds further detail to this picture -16

3,2.

PROJECII\E SJ'AEIS AXI)

LA1IICLS

ARC;L

-17

(J

-V

ii

p

li1igure 3.1

showing, among other things, that .t132 has exactly five join irreducible covers. I he methods developed by JOnsson and Hong have proved to very useful for the investigation of modular varieties generated by lattices of finite length and / or finite width number of pairwise incomparable elements). Freese [72] extends these methods and gives a complete description of the variety generated by all modular lattices of width -1. by

3.2

Projective Spaces and Arguesian Lattices

discussion of projective spaces, since many of the results about modular varieties make use of some of the properties of these structures. A some of the results reviewed here will also be used iii Chapter 6. e begin with a

Definition of a projective space. In this section we will he concerned with pairs of sets (If, L), where P is a set of and a collection L of su hsets of P, called If a pout p E P is an element of a line I E L, then we say that p I, and I i/ui ag/u p. A set of points is coilineai if all the points lie on the same line .A triangle is an ordered triple of noncollinear (hence distinct) points (p, q, r). (If, L) is said to he a piijeciice (sometimes also called pro jective geometry) if it satisfies:

(P 1) each line contains at least two

(P2) any two distinct poi its p and

poits;

are contained in enicity one line (denoted by {p, q}); (P3) for any triangle (p, q, r), if a line intersects two of the lilies {p, q}, {p, r} or {q, r} ill distinct points, then it meets the third side (i.e. coplanar lines intersect, see Figure 3.1). q

I he two simplest pro jective spaces, which have 110 lilies at all, are (0, 0) and ({p}, 0), while ({p, q}, {{p, q}}) and ({p, q, r}, {{p, q, r}}) have one line each, and ({p, q, r}, {{p, q}, {p, r}, {q, r}})

has three lines.

E'BAJ'TLJ

-18

3,

,\1ODL LAB

\JBJEIIES

I he last two examples show that we can have different pro iecti ye spaces defined on the same set of points. However we will usually be dealing only with one space (If, L at a time which we then simply denote by the letter P. A 5 of a proJective space P is a subset 5 of P such that every line which passes through any two distinct points of 5 is included iii 5 (i.e. 1o, q E 5 implies {p, q} 1 5, where we define {p, jJ {jJ). I he collection of all su hspaces of P is denoted by L( F). )

Projective spaces and modular geometric lattices.

A lattice L is said to be upper.Eenhinloc/ular or simply if (I -< b in L implies (I c -< b c or (I c b c for all c E L. Clearly every modular lattice is semimodular. A geometric lattice is a semimodular algebraic lattice in which the compact elements H—

H—

H—

H—

are exactly the finite joins of atoms. I he next theorem summarizes the connection bet ween projective spaces and (modular) geometric lattices. IHEOREM :3.1 Let

1) is

(i)

P a

be an arbitrary projecti ;'e space.

Ihen

complete modular lattice:

ii

(ii) associated with every modular lattice is the set vial] atoms of and a line of atoms below p ± q (i.e. {p, q} {r E

ii.

is a projec ti ye space

ii ).

where il) is the set q

two distinct atoms p and P(JJ): r

F:

(iii)

ii'

(iv) fbr any modular lattice ii. if is the subl at tice of ii generated by the atoms of ajid hi]' is the ideal lattice of ii'. thai hi]' L( i]fl: (v) C(

is a

(vi) L(

ii))

I'I-tOOF.

(i

ordered

by

modular

i]

lattice:

fbr any modular geometric lattice

i]

is closed under arbitrary intersections and P E hence L( inclusion, is a complete lattice. For 5, 1 E the join can he described by )

qEIJ (here we use (P3), see [GLT] p.203). Suppose 11 E L( and 11 1' 1. lo prove L( modular, we need only show that 11(5 1) 1 1(5 I. Let r E 11(5 I). I hen r E 11 and r E 5 ± I, which implies r E {p, q} for sonic p E 5, q E 1 1 11. If r q then r E I, and if r q then p E {r,q} 1 11 (P2)), hence p E 1(5. In either case r E 1(5 H- I as required. (ii) \\ e have to show that J] satisfies (P1 ), (P2) and (P3). (P1 holds by construction, and (P2) follows from the fact that, by modularity, the join of two atoms covers both atoms. lo prove (P3), suppose (p, q, r) form a triangle, and are two distinct K p H- q and y K q H- r (see Figure 3.1). It suffices to show that poits (atoms) such y} fl {p, r} 0. Since J), q, r are noncollinear, p H- q H- r covers p H- r by (upper senu-) modularity. Also H—/IC PH- q H- r, hence H—V H—y)(pH— q H- r), which covers or equals H—K PH- r, H—y)(pH— r) by (lower semi—) modularity. lU H-V H—y)(pH— r), then and since H-V and p H- r are elements of height 2, we must have H—V r. In this case p H{p, r} and there intersection is certainly nonemptv. If H-V H-V )(p H- r) v} then must he an atom, and is in fact the point of intersection of v} and {p, r}. H—

)

H—

H—

)

3.2. l'itOJECTIVE SPACES AND A1WUESLt.N LATTiCES

49

(iii) A point p P corresponds to the one clement subspace (atom) {p} £(P), and it is easy to check that this map extends to a correspondence between the lines of P and (iv) Each idea! of Al' is generated by the set of atoms it contains, every subspace of P(AI) is the set of atoms of some idea! of Al', and the (infinite) meet operations (intersection) of both lattices are preserved by this correspondence. hence the result follows. (v) By (iii) P P(Al) for some modular lattice Al, hence (iv) implies that £(P) LW'. It is easy to check that LW' is a geometric lattice, and modularity follows from (i). Now (vi) follows from (iv) and the observation that if Al is a geometric lattice, then

0 LEMMA 3.2 (Birkhoff [35']). Every geometric lattice is complementcd.

Let .L be a geometric lattice and consider any nonzero clement z .L. By Zorn's Lemma there exists an clement in L that is maxima! with respect to the property Every element of L is the join of all the atoms xiii = We want to show x + in = below it, so if x + in < 1L, then there is an atom p x + in, and by semimodularity in -< in + p. We show that x(in + p) = °L, which then contradicts the maximality of in, and we are done. Suppose x(in + p) > Then there is an atom q x(in + p),

0 + in, a contradiction. MacLane [38] showed that every geometric lattice is relatively complemented in + q

x

In tisct (see [ULT] p.179). The next theorem is a significant result that is due to Frink [46], a!though Jónsson [54] made the observation that the lattice L is in the same variety as K.

THEoREM 3.3 Let V be a variety of lattices. Then every complementcd modular lattice

K

V

can be 0,1-embedded in some modular geometric lattice

.L

V.

Let Al = 7K be the filter lattice of K, ordered by reverse inclusion. Then Al satisfies all the identities which hold in K, hence Ar! is modular and Ar! V. For .L we projective with take the subspace lattice of the space P(Al) associated Al. Note that the points of P(Al) are the maxima! (proper) filters of K. By Theorem 3.1 (v), Lisa modular geometric lattice, and by (iv) L LW', which implies that L is also in V. Define a map

f:K—'Lby

f(x) = {.L"

P(Al) x

I',

K. It is easy to check that f(x)isin fact a subspace of P(M), that f(OK) =0, f(lK) = P(Al) and that f is meet preserving, hence isotone. To conclude that f is a!so join preserving, it is therefore sufficient to show that f(x + p) ç f(x) + f(p). This is for each x

Thenx+p€1",

and we have to show that there exist two maxima! filters U f(x), B such that F (3+11 (i.e. F2 (3+11). If x €1", then we simply take F = U, and B as any maxima! filter containing p, and similarly for p I" (here, and subsequently, we use Zorn's Lemma to extend any filter to a maxima! filter). Thus we may assume that x, p I". Further we may assume that zy = °K, since if zy > °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<

k<

vi

and

vi

\ ak_lak+1/bk_lbk+1 / Ck+1/bk+1.

3 we have

k>

1

or

k

c

vi — 1

(or

both).

In the first

Ck_2/bk_2

\ ak_lak+1/bk_lbk+1 / Ck+1/bk+1

Ck_1/bk_1

\ ak_lak+1/bk_lbk+1 / Ck+2/bk+2.

case

and in the second

Either way it follows that the nontrivial subintervals q,/bu of p/q and of v/s are 0 projective in vi — 1 steps. This however contradicts the assumption of the lemma. LEMMA 3.29 (Jónsson

Let

.1,

be a modular lattice such that J4.z is not a homomor-

phk image of a sublattice of .L. if(v c z, y, z c u) and (9< a?, jf,9 c u') arc diamonds in L such that V = yu' and z = 9+ v, then u/v u'/V (refer to Figure 3.5(i)).

\

p+

\

Observe firstly that the conditions imply u/p 9/9, since p + 9 = (p + v) + 9 = = uandp9 = p(u'9) = = 9. Let to = v+u'. Then to v+ 9 = z and p', 9 imply u u', hence to u/z. We show that to = u and dually vu' = 9, which

z

u gives the desired conclusion. Note that we cannot have to = z since then the two diamonds would generate a sublattice that has as homomorphic image. So suppose z c to c u. Because all the edges of a diamond are projective to another, the six elements to, v, z, p, z, u generate the lattice in Figure 3.5 (ii). Under the transposition u/p 9/9 the element zto + p

\

3.4.

COVE1tLVG ItELATIORS BET WEER MODtLS1t tCt1UETIES

71

'a

'a'

z 'a

z

(ii)

(iii)

(i) Figure 3.5

is sent to to' = z'(no + p), and together with 9, a?, if, u' these elements generate the lattice in Figure 3.5 (iii). It is easy to check that (no + pto < z + pto,no + p,to < u) and ((z'+to')(p'+to') < p'+tt#,z'+z'(p'+u#),z'+tt# < u') are diamonds (they appear in Figure 3.5 (II) and (iii)). We claim that to/no + pto i/yf + to'. Indeed, since we have 'a'

\

no + yx + 'a' =

'a' +

no + 'a' + yto =

('a' + z)to + ('a'

zf+to'= V+z'(no+v)=Qf+z')(no+v)

+ p)to =

to

= 'a'(no + p) = 'a'(no + p)to = 'a'(no + pto). But this means that the two diamonds form a sublattice of L that is isomorphic to the lattice in Figure 3.5 (iv), and therefore has as a homomorphic image. This contradiction 0 shows that we must have 'a = to. LEMMA 3.30 (Jónsson [68]). if .L isa mod ular Iattiw such that is not a homomorphic image of a sublattice of then any two quotients in L that arc projcctive to each other have nontrivial subquotients that arc projcctive to each other in three stqps or less.

4

It is enough to prove the theorem for two quotients a/b and c/d that are projective in four steps, since longer sequences can be handled by repeated application of this case. We assume that no nontrivial subquotients of a/b and c/d are projective in less than four steps and derive a contradiction. By Lemma 3.28 there exists a strongly normal sequence

of transpositions a/b =

/ \ / \ ai/bi

a2/bj

as/h

a4/b4

= c/d

or dually. Associated with this sequence are three diamonds (hij+62
ai), (b4 < bias,a2,aih < aia3) and (b4+b4
= = bs(aias),

+ a2 = a2 + (ho + a2+b4 =a2+(h.j+b4) 6o

CRSI'TEit

72

whence we conclude

t•It1UETJES

ai/btj+bj\ aia2/bj/as/bj+b4.

to show that a/b and c/d are projective in 2 steps. In tisct

(n) as

MODtLS1t

that (s)

Ibis enables us

3.

au/bij

/

a2+btj+b4/ai +a3

\

a4/b4

is shown by the following calculations

au+(a2+bu+b4)=ai+b4=ai+aias+6j+b4 (ai aias+6j) =ai+a3 (aias+(6j+b4)=asby(s)) (bymodularity)

%(a2+b.J+b4)=b.J+%(a2+b4)

=

(ai%)

bu+auai(a2+b4) bu + %(a2 + aib4)

(by modularity)

where the inequality holds since a1b4 = a1a3b4 aia3(6j + b4) = by (it). The second part of (n) follows by symmetry. Since a/b and c/d were assumed to be projective in not 0 less than 4 steps, this contradiction completes the proot LEMMA 3.31 Suppose L is a modular lattice with b c a d c c in L. Ma/b and c/d arc projcctive in three stcps, then a/b transposes up onto a lower edge of a diamond and c/d transposes down onto an upper edge of a diamond. Since a d, no nontrivial subintervals of a/b and c/d are projective to each other in less than three steps. hence by Lemma 3.29 a/b and c/d are connected by a strongly

normal sequence of length 3, say

a/b =

au/bu

/ ai/bi

\ / as/h a2/bj

= c/d

(the dual case cannot apply). Then a/b transposes up onto + b2/bu + 62 of the diamond + 62 < % + b1,60 + a2 cai) (see Figure 3.4(11)) and c/d transposes down onto 0 c b1a3, a2, c aia3) as required. aias/bibs of THEoREM 3.32 (Jónsson [68]). ditions are

(i) A4z (11)

l'br any variety

V

of modular lattices the following con-

cquivalent: V;

every subdirectly irreducible member of V has dimension two or less;

(ill) the inclusion a(b + cd)(c + d)

b + ac

+ ad holds in

V.

Suppose V but some subdirectly irreducible lattice L in V has dimension greater than two. Then .L contains a four clement chain a > 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.)

d

1c11n

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3.ThJfRJ 141 iT

3141 iX)11

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p111?

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393141 $111311T313

(i

$1141

(yj—

)

pccopJo

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

f)

71

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))ii1TJ

f) 7

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JO

2 D

JFPJ 01

3fn3ipIIi

T1A 3141 (Tl?I1T .r

'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

=

it

'.r 'fi

2 D

$11

o$p? 1TIJi .r

2

1P141 )14J

3$JOAOJ 110i$llpIIi

Ajfnhlp)

3141

1011111?.)

——

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

'71

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
must be noncomparable. By the weak -< u, whence it follows that

9 IL such that x + px u=x'+p=x'+x,

atomicity of IL, we can find

=9+

In ThL we can then find minimal elements if, 9 subject to the conditions u V= 9+9, V p and 9 x. Now 9V < V since equality would imply 9= u. Furthermore, if 9V < to then 9< 9+to (equality would imply to 9p') and 9+to 9+V = u. hence 9+ to = u and by the minimality of V covers 9V, and similarly 9 covers 99. So, dropping the primes, we have found u, x, p, x ThL satisfying

(*)and

(n)

xp-
Since xp -< p ,we have either p(xp+x) = xp or p xp+x, and similarly x(xx+p) = xx or x xx + p. We will show that in each of the four that arise, the lattice ThL or IEEL must contain M3 or one of the L1 (1 = 1,.. .,5) as a sublattice. Case 1: p xp + x and x xx + p. Since x, p and x are noncomparable, so are xp and x p and xx -< x. The following calculations show that we in fact have L4to,p,x): top+tox = xp+xx = to; to = px; p+xx p+x and equality follows from the assumption that p + xx x; similarly x + xp = p + x (see Figure 4.1 (ii), 4.2 (i) and (ii)).

Case2:

pxp+xandx(xx+p)=xx. Lets=x(p+x)andt=xx+p. Themost

general relationship between x,s,t and xis pictured in Figure 4.2 (iii). We will show that either L5(s,t,x) or Ls(x,s+ t,x) (see Figure 4.1). Clearly sx = x(p+ x)x = xx = xt by assumption. Furthermore I and x are noncomparable (x I since Ix = xx c x, p I £ x

sincep

x but p

Lx.

Suppose now that

s+I = p+ x.

Then

I(else

p+ x = s+I = I

x,

CBAJ'TLJ

80

LAB \JBJEIIES

-I,

-V

:7,÷j

[/2

x

ill' LI

(iii)

(ii

(i

li1igure -1.3

contradicting thus z =

/, z are nonconiparable .Also = since / K ± z (by assumption), whence z = z. Ihis shows z implies that z, /, are iioiiconiparable, and y ± ±

z) and therefore

/

z

z

a>

a>

L5(h, /, z). On the other hand ± / follows from the calculations: so L3( z, ± /, = Z ±£ z(hH-/) = /) = zH(zH- ± = (u H-

h

(I

±/±

(since (since

£

a)

zH- (hH- /)

=

±

=

and z Kj z y. I his case is symmetric to the preceding case. \\e claim that for u = 0,1,... one can z) = and H—y) = find increasing chains of elements and (tic) y E y Hhold th and z replaced by y,, and Indeed, let flu = fj, z0 = z and Case 3: ti( Case -I:

z) =

H—

H—

H—

.

I hen

=

±£

=

±

and if we suppose

that

z then clearly ic (tic) \ow suppose that and hold with = ( and z replaced by y,, and for some u 0. \\ e show that the same is true for and Firstly = H-y,, = = a. Further we may assunie that = and = for otherwise Hprevious would \ow one of the three cases apply. z and z so and hence If / then put = (Figure -1.3 (i )). \\ e show Hthat either L3( 1, or L4(h, 1, or L3( froni which it follows that we may assunie £ a = since a, and = / since -<( H- £ = / Hshows Also /) / that / = /) and /, are = ( flu

Li

H—

H—

z

£

and z0 K K z and

H—

z

K

H—

.

H—

y,,

,

z.

K

H—

)

H—

-C.

H—

noncompara ble. \ow either have L4(h, 1, (if L5(

H—

H—

/

= =

which implies L3(

or L3(

gu) (Figure

t3 (ii)).

,

1,

(if

that

), or ;>

/ iii which case we ;> Similarly we may assume

H—

).

we

obtain

SL\JJDJSILJUBL

-1,2.

II\J1Y

Si

In I:TIL we now form the join

Furthermore,

IT

of all the

K if then there exists iii (Figure -1.3 (iii)). \\ e compute

=

=

E

of all the

E

H>Iu

IT

=

and =

/

= H>Iu

=

y,

K.r

Clearly

y,,

where the second and last equality make use of the tipper continuit

Lastly,

.

for all u. I herefore w such that for all u a> / hence

is compact and

since

and the join

of I:T1L.

Kr,

Ew m

I his hence

implies =

= and sinularly

Consequently = = Dropping the subscripts we now have a, £, = E I:TIL satisfying ( ), ( ) and we Let / = £( ± z). 1ff = uz then L4( z, £) holds, and if / > uz then consider = the four cases depending on whether or not the equations z = / and ti z = / hold. If both hold, then we get J13( a', y, z), if both fail then we let = (ti w)( a' z) j, z) (here we use to obtaiii z, see Figure —L3 (iv)), and if just one equation holds, say u / z / = z, then L4( y, /, z) follows. I his conipletes the p roof. H—

H—

H—

H—

H—

H—

H—

H—

H—

Semidistributive varieties. If L is a finite lattice, then I:TIL :Ti:TL L, so L is semidistri huti ye if and only if L excludes J13, L2, L3, L4 and L5. \\ e say that a cariely V of if every member of V is semidistrihutive. I he next theorem characterizes all the semidistri huti ye varieties. IHEOREM 4.2 (JOnsson and Rival [79]).

br a

yen

lattice variety

V.

the following

statements are equivalent: (i

)

V

is semidistrihutive,

(ii)

V.

(iii) Both the Jilter and ideal lattice of

(iv) Let flu = 1' z0 = z and. for u E w let some natural number in the identit

(SD;) and its dual

(SDt) hold

are semidistribu tive,

=

H—

and

z

H—

,

Ihen for

± z) =

in V.

lattices iii (ii) fail to be semidistri butive, (i ) implies (ii), and (ii) implies (i ) follows froni Lemma —Li and the fact that L E V iniplies I:TIL, :TIZ?TL E V. Also (i) implies (iii) since E V. (iii) implies (iv): By duality it suffices to show that, for sonic iii, £( y ± z) = Ly, in the free lattice of V generated by v' z. By induction one easily sees that I'I-tOOF. Since each of the

CBAJ'TLJ

82

and all the

-1,

XOXUODL LAB \JBJETJES

and as the join of hence by the upper continuity of = = v, say. Also g ± z for each u implies g ± z. By semidistri butivity we therefore have v = z). Hence v = is g a compact element of so for sonic iii E z) = (V ± (iv) implies (i ): If L E V is not senudistri butive, then there are elements g, z in L such that g = z z) or dually. I hen, for all u, g g z). Consequently the identity fails for each u. g ± K

.

In 11A (3) we define

\ow

as the join of all the

and

H—

H—

I he fourth statement shows that senudistributivity cannot be characterized by a set of identities, and so the class of all semidistri hutive lattices does not forni a variety.

Semidistributivity and weak transpositions.

For the notions of weak pro jectivitv I he next result concerns the possi hili tv of shortening a sequence of weak transpositions. Suppose in sonic lattice L a quotient L u/go projects weakly onto another quotient in ii ;> 2 steps, say we refer the reader to Section

1.—I.

L2/u2

If there exists

/gH.

a quotieiit u/v such that

Lu/ Vu

u/v

/

then we can shorten the sequence of weak transpositions by replaci ig the quotients by the single quotient u/v. In a distributive lattice this can always be done, since we may take u/v = I he nonexistence of such a quotient u/v is therefore connected with the presence of a diamond or a pentagon as a su blattice of L. If L is semidistri hutive, then this sublattice niust of course he a pentagon. I he aini of Lemma -1.3 is to describe the location of the pentagon relative to the quotients Li/gi. e introduce the following and notation: A quotient c/o in a lattice L is

and

said to he an X-quotient if there exists b E L such that (I ± b = ± c and (lb = (IC. In this case (I, b, c E L generate a su blattice isoniorphic to the pentagon X, a condition which we abbreviate by writing X ( c/o, b).

asemidis tn

LEMMA 4.3 (JOnsson and Rival [79]). Let L be

1)11

ti ye

lattice and suppose

L2/u2 in L. Then either

Lu/go

(i) there exist o, b, c

E

(ii) there exist (I,b,c,1

L with V(c/( b).

and b/bc is asuhquotient otto/go oi

EL with X(c/o,b).gu

1

and //go

(iii) there exists a subquotient p/q of Lu/go such that p/q quotient u/c',

o

± b/b oi

\ / u/v

Ion some

then the elements (I = Lb ± ± j2). (i) If ± ± £2 give X(c/o, b) and b/bc = Lu/Lb Lu/go (Figure -1.-I (i )). (ii) Suppose = By the semidistri butivit of L, = = L0L2 hence (L0L2 -i-U2) ± = £2 ±gp If L0L2 j-ff2 then the elements £2, = L0L2 b = , c = satisfy ii( c/i, b), and ± b/b transposes down onto the su hquotient L of /gu (Figure —IA (ii)). I'I-tOOF. b = Lu and c =

=

.

SflhiDiSiHUJJL ll\iiLY

-1,2.

83

Lu .1

£2

+112

V2 V2

Vu

Vu

Lu

V2

(ii) 11igure

-1.-I

= £2, then

(iii) If LuL2

Lu/Vu 9

p/q

LEMM;\ 4.4 (Rose [84]).

IlL

in L. then Lu/vu N LOL2/VuIh

/

is

asejnidistrihutive lattice and L2/u2

I'I-tOOF. Since we are dealing with transpositions, Vu = LOVL V2 = and ± = \ow bijectivitv By selludistri butivit L0L2. the of the transpositions = implies Vu Lwt2 = (Lb .Xlso Lwt2 = = Lu, and sinularly V2 L0L2 = and 112(L0L2) LIuLh. H—

H—

H—

H—

H—

H—

3-generated semidistributive lattices.

Let be the lattices in Figure It is easy to check that each of these lattices is freely generated by the elements subject to the defining relations listed below. LEMMA

-1.5. z

4.5 (JOnsson and Rival [79], Rose [84]). Let L be asejnidistnhutive lattice L, g, z with L L ± z and LZ g.

generated by the three (i

)

ilL

excludes

then L

(ii) 1! L

E

L

and excludes

then L

E

L E HF4.

I'I-tOOF. (i £ Lf) ± z is equivalent to £11 z = z, so it suffices to show that under the above assumptions )z = I he free lattice determined by the elements £, z, £V, Vz and the defining relations £ L V ± z and £Z pictured in Figure -1.6(i). we must Let Vu t and a' = LV tVZ. lo avoid = £ j—(ft)Z, II? = LIZ have = H—

)

H—

H—

CBJJ'IIJIR

LAJI

-I,

LI

I

(t' -t.

[I

±

(t'

±

Z

-t.

[I

(t'

±y): ±

±

Z

-t.

±

Z

[I

±y): ± (t'

(t'

±

Z

±

± z)

[I

LI

I

I

±y):

-t.

[I

±y): ± (t'

±

Z

± ?JZ ± z) ± z)

?JZ

±

±[/z -t.

-t. 1,/i -t.

LI

(i)

\JJULiJLS

(ii)

(iii)

-1,2.

SflhLL)JSIIJUBL

II\J1Y

85

C,

q

P C

z=b

(I' =

c'

(I

(I, (I

(i)

li1igure

Since a' ± vu z = Further we compute

(iii)

(H)

= z

=

a' ± z

,

= u'z =

—1.7

selludistri biitivitv implies = (1'. /12 Again by selludistri butivity LIZ = (vu ± y )z =

= —f—y)z, as required. H—iiz v)z = (ii) Let = ± y)( ± z) and / = ± ± z), then the sublattice of z and / is isoniorphic to L7 (Figure tO (iii)) with critical quotient h//. (iii) is dual to (ii), and (iv ) follows from (ii) and (iii). .

—t—

LEMMA

ij generated

b

4.6 (Rose [84]). Let L be asejnidistributive lattice that excludes LH and L with X( a/v,b). (i c and a/v projects weakly onto c/o. then X(c/o,b).

PROOF. \\ e show that a/v c/o implies X(c/o, b), then the result follows by repeated application of this result and its dual. Let-v = a, = (I and z = b (Figure —1.7 (i ). I hen hence by Lemma -1.5 z generate a homomorphic image ± z and z K of Lj (Figure —1.5). Coniputiiig in this lattice, we have be = = H—v z ± = z ± (J H—v) = b H—c ,which implies X(c/G,b). )

)

(:0R0LL;\RY 4.7 Suppose a/v and c/o are nontrivial quotients in a seinidistrihu tive lattice L that excludes LH and L E con( a, v). then X ( a/c', b) implies

X(c/o, b). PROOF. By Lemma 1.11 there is a sequence (I = a/v projects weakly onto for each i = ., ii. By Lemma b), hence and ± b ;> 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
For finite lattices we can get an even stronger result.

finite lattice L is tn/MI tive and excludes LH and IHEOREM 4.10 (Lee [85]).

tive if and unIv

ilL

is seinidis—

I'I-tOOF. I he forward direction follows immediately from I heorem -1.9. Conversely, supis finite, semidistri butive and excludes LH and hut is not neardistribu— tive. I hen by I heorem -L9, {L}v contains a lattice K, where K is one of the lattices J13, , , L5, LH, Since Ii is subdirectlv irreducible, JOnsson's Lemma implies K E HS{L}. It is also easy to check that every choice of K satisfies \\ hitman's condition (\\ ), hence I heorem 2.-IT implies that K is isomorphic to a sublattice of L. I his however contradicts the assumption that L is semidistributive and excludes LH, .

.

.

It is not known whether the above theorem also holds for infinite lattices. Xote that I heorem —L8 implies that any finite su bdirectlv irreducible neardistri butive lattice has a unique critical quotient. Rose [84] observed that any semidistri butive lattice which contains a cycle must include either LH or (refer to Section 2.3 for the definition of a cycle). Ihis follows easily from Corollary —1.7 and the fact that if mama ... ap,aJ'0 is a cycle then

E'BAJ'TLR

88

LAB

-I,

\JBJEIIES

V L—J.

fl

fl

I

j

Figure

-1.9

con(pj, Pj* (see Figure 2.6), whence all the quotients generate the same congruence. In particular, I heorem —1.10 and Corollary 2.3.5 therefore imply that every finite neardistributive lattice is bounded. )

)

Almost distributive lattices. Ihe next definition

is also from Lee [85]. A lattice or if it is neardistributive and satisfies the

a lattice variety is said to be

inclusion

(AD)

v(u±c)
and it dual

Every distributive lattice is almost distributive, since the distributive identity implies (u ± c) v. On the other hand LH and fail to he ± c( v i1) = ( a c)( a ± ± G) almost distributive, since they are not neardistributive. Further investigation shows that (AD) fails in L6, and (see Figure -1.9), and by duality (AD+) does not hold iii L7 while the next lemma shows that X, and and are almost distributive. Recall from Section 2.3 Day's construction of 'dou bling" a quotient a / v in a lattice L to obtaiii a new lattice L[a/ r]. Here we only need the case where L is a distributive lattice I), and a = = il E I). In this case we denote the new lattice by L)[d]. Xote that X, and can he obtained from a distributive lattice in this way. a

H—

LEMM;\ 4.11 (Lee [85]).

almost

icr

any distributive lattice I) and il

E

I). the lattice

L)[d] is

c/is till) II ti ye,

PROOF. By duality it suffices to show that D[d] satisfies (SD;) and (AD). If (SD;) fails, then we can find £,/J,Z,G,C E L)[d] such = c = ± £(z ± ± z). Let I) (i.e. = a for -u be the image of a E D[d] tinder the natural epimorphism L)[d] whence all a il, and ((1,0) = (il, 1) = il). Since I) is distributive, we must have ii = = (il, 0) and c = (il, 1). Clearly (I is meet irreducible by the construction of D[d], and this leads to a contradiction as in I heorem -L9. lo show that (AD ) holds in L)[d], let us now denote by a, v, z, o, c the elements of L)[d] corresponding to an assignment of the (same) variables of (AD ). If (I = c, then (AD) obviously holds. If (I il (i.e. G then the distributivity of I) again implies that ii = hence (I = (d,0) and c = (il, 1). \Ow

DJSJJIJBL LiVE \JRIELILS

4.3,

89

C,

(I,

(i

C

C

b

b (JI!

(ii)

(Hi)

Figure

(i implies of (I imply v cases.

c(

a

'

(i

)

=

a

c, whence

H—

(I

a

H—

v( (i

C'

c(

v

H—

)

=

-1.10

c ), a

H—

while c

' IT

v( a

H—

(I

and the meet irreducibility c). I his (AD ) holds iii all

Of course not every almost distributive lattice is of the form L)[d] (take for example or ally one element lattice), but we shall see shortly that all su bdirectly irreducible almost distributive lattices can indeed be characterized ill this way. 2 x 2,

4.12 (JOnsson and Rival [79]). Let L be aseinidistribu tive lattice which excludes and suppose that (I, b, c, o', b' E L with X ( c/o, b) and c/o ë/o'. Set r = (I'b ± c and

/

LEMMA

=

(b—i—

c)Y, Ihen

r/o'r or L

(i) c/o

L c

±

or

or L includes

in Figure -1.10 (i is isomorphic to ill Figure -1.5. Assume L excludes and (as well as = c, j = o', z = b in ), and take Lemma -L5 (iii). I hell L is a homomorphic image of and since we have X(c'/o', b) and (i )

)

E'BAJ'TLJ

90

(1

±

)

LAB \JBJEIIES

-I.

r

1'

l'r ± (I'!, 1'

(ii

(i

Figure -1.11

= (I'b ± (I in the sanie is true iii L. For E c/o o'r, 1, (1 ± (I'b)c generate a sublattice isoniorphic to

we must have

(1 ± (I'b)C = 1, else (see Figure -1.10 (ii)). Sinularlv, we must have ci' ± (I'b = 1' to avoid for 1' E 1', c, b) (see Figure -1.10 (iii)). I his shows that c/o transposes up onto r/o'r and also proves that this transposition is biiective. (ii) is dual to (i ). Lastly (iii) hold because for I E r/i'r and 1' E c b/b, we must have (1 = I and 1',' = 1', to avoid L(3((1 h)r/1,h,b) and L(3(l'r h/1',r,b) (see I

b,

—f—

H—

H—

—i—

H—

Figure -1.11).

Characterizing almost distributive varieties. I he next theorem is implicit iii JOnsson and lii val [79] and appeared in the present form in Rose [84]. IHEOREM 4.1:3 Let L be

following conditions are

asuhdirectlv irreducible semidistrihu tive lattice,

Ihen the

uivalen t:

(i) L excludes L6, L7, L L

D[d] fbr some distributive lattice I) and some il

E

I).

I'I-tOOF. Assume that (i) holds, and consider an X-quotient u/v in L. By I heorem -1.8, L has a unique critical quotient which we denote b c/i. It follows that c/i is prime, and Lemma 1.11 implies that u/v projects weakly onto c/o, say

u/v =

.

.

.

= c/o.

Of course this implies that , ii, and since u/v E con( u, v for each i = 0, 1, is assumed to he an X-quotient, we have X( u/c', b) for some b E L. I hus Corollary -Li iniplies b) for each i. In particular, it follows that c/o is an X-quotient. \\ e show that it is the only one. Xote that iniplies ± whence by Lemma —1.12 (i (ii) and (iii this transposition is hijecti ye. Similarly the dual of Lemma t12 shows that iniplies Hence c/o is pro jective to a su hquotient of u/ v (see Figure -1. 12). By I heoreni -L8 this su hquotient must equal c/o, otherwise L would have two critical quotieiits. Furthermore u v implies )

.

.

.

/

)

4.3. ALMOST D1STIUBUT1VE tCt1UETIES

91

U

6

V

Figure 4.12

that u/c is also an R-quotient, and for the same reason as above, c/a would have to be a subquotient of u/c. But this is clearly impossible, hence u = c, and similarly v = a. If (ii) holds then L excludes L11 and L12, so again Theorem 4.8 implies that c/a is a prime quotient and con(a, c) identifies no two distinct elements of L except a and c. Let 1) = L/con(a,c). Clearly 1) cannot include A4, otherwise the same result would be true for L, contradicting semidistributivity. 1) also excludes the pentagon IV, since con(a,c) collapses the only R-quotient of L. hence 1) is distributive. Let d = a,c II, then it is easy tocheckthat themap z {z}(z c,a),c (d,1)anda (d,O)isanisomorphism from L to DV]. To prove that (lii) implies (i), we first note that since II is distributive, the natural homomorphism from D[d] to II must collapse any R-quotient in D[d]. hence (d, 1)/(d,O) is the only dY-quotient, and as each of the lattices L.3, L1,. . . , has at least 0 two dY-quotients, (i) must hold. The following corollary summarizes the results that have been obtained about almost

distributive lattices and varieties.

Coaoisuw 4.14 (i) A subdirectly irreducible lattice L is almost distributive if and only if L some distributive lattice II and d €11.

DV] for

(ii) A lattice variety is almost distributive if and only ifit is semidistributive and contains

none of the lattices

L.3, L1,. . . ,

(iii) Every finitely generated subdirectly irreducible almost distributive lattice is finite.

(iv) Every almost distributive variety that has finitely many subvarieties is generated by a finite lattice.

(v) Every join irreducible almost distributive variety of finite height is generated by a finite subdirectly irreducible lattice. (vi) Every finite almost distributive lattice is bounded (in the sense of Section 2.3). (i) The forward implication follows from Theorem 4.13, since .L is certainly semidistributive and, as L6, L1,. . . , L12 all fail to be almost distributive, L must exclude these lattices. Theorem 4.11 provides the reverse implication.

CBAJ'TLJ

92

-1,

XOXUODL LAB

\JBJEIIES

(ii) I he forward direction is trivial. Conversely, if V is senudistributive and contains none of L6, L7, , then I heorem -1.11 and I heorem -1.13 imply that every subdirectlv irreducible mem her, and hence every mem her of V is almost distributive. (iii) If the lattice L IJ[d] in (i) is finitely generated, so is the distributive lattice I). It follows that I) is finite, and since L is also finite. Xow (iv) and (v) follow from Lemma 2.7, and (vi) is a consequence of Lemmas 2.-lU and 2.-Il. .

.

.

In particular the last result shows that an finite subdirectlv irreducible almost dis— tn hutive lattice is a splitting lattice. Part (iii) above says that almost distributive varieties are locally finite, and this is the reason why they are much easier to describe than ar hi trarv varieties. More generally we have the following result:

4.15 (Rose [84]). Let L be a finitely generated subdireetly irreducible lattice all of whose critical quotients are prime. if c/o is a critical quotient of L. and if L/con(o, c) belongs to a variety that is generated by a finite lattice. then L is finite, LEMMA

I'I-tOOF. Since every critical quotieiit of L is prime, each congruence class of con(o, c) has at most two elements. I hus the assumption that L is infinite implies that L/con(i, c) is infinite as well. However, this leads to a contradiction, since the lattice L/con(o, c) is finitely generated and belongs to a variety generated by a finite lattice, hence L/con(o, c) must be finite.

Subdirectly irreducible lattices of the form I)[d]. \\e continue our investigation

of

which will then lead to a characterizasemidistri hutive lattices that exclude LH and tion of all the finite su bdirectlv irreducible almost distributive lattices.

asubdirectlv irreducible semidistribu tive lattice that and suppose c/o is the (unique) critical quotient of L. Ihen

LEMM;\ 4.16 (Rose [84]). Let L be

excludes LH and

(i) the su blat tices [o) and (c] of L are distributive;

(ii) fbr any non trivial quotient u/v of (c] there exist b, v'

X(c/o,b).b< (iii)

if a

v

=

c

E L

with

v K

v'

a

such that

v'aiid

in (ii). then we also have

a

=

±

b.

I'I-tOOF. (i semidistri butivity, L excludes J13. Suppose that for some a, v, b E L we have X( a/v, b). Ihen Corollary —1.7 implies X(c/o, b), whence b (c]. It k') and b follows that [o) and (c] also exclude the pentagon, and are therefore distributive. (ii) Choose a shortest possi ble sequence )

a/v =

.

.

.

= c/o.

Since v c, we must have ii r2/y2. By the nuni3. Suppose that a/v mality of ii, only part (i) or (ii) of Lemma t3 can apply. I hat is, there exist o', b, c', a' E L with X (c'/o', b) and v a' a such that b/bc' I a/v or a'/ v i' ± b/b. But Corollary -1.7 implies X ( c/o, b, which is inipossi ble since in both cases b I hus we must have a/v By the dual of Lemma t3 (i) and (ii), there exist o', b, c', v' E L £2/u2. with X(c'/o', b) and v K v' a such that either i' ± b/b 1 a/v or a/v' b/bc'. UnIv the latter is possi ble, since we again have X (c/o, b) by Corollary t7. o, b a implies

/

DJS1JIJBL LiVE

4.3,

\JRIELILS

H-

93

b

C'' (I H-

b

C,

C

b (I

Figure

-1.13

\

a, and (I v' implies v' b = v' (o b), hence v' b/ v' b/(o b) r'. Also by' = be' b implies b IT v, and the hiiectivitv of the transposition follows from the distii hutivitv of k') (part (i )). (iii) is a special case if (ii). (I

H—

b K

H—

H—

H—

H—

H—

THEOREM 4.17 (Rose [84]). Let I) be a finite distributive lattice and d E I). is sti bdirectly irreducible if and unIv if oil of the following conditions hold:

(i) every cover

old

H—

Ihun D[d]

is join reducible.

(ii) every dual cover

old

is meet reducible. and

(iii) eveiy priiiie quotient in I) is projective to a pri lie quotient p/q with p = d or

q

=

irreducible. Let (I = (d, 0) and e = (d, 1). Xotice are satisfied vacuously. If a E I) covers d, then a covers e in L, whence by Lemma —1.16 (iii there exists b E L nonconiparable with e, and a = ± b. I hus a is join reduci ble in L and also in I). Dually, every element that is covered by d is meet irreducible. lo prove (iii), consider a prime quotieiit a/c' e/o in L, and choose a PROOF. Suppose L = L)[d] is su hdirectlv that I) = d implies that (i )' (ii) and (iii)

)

a/v =

.

.

.

I'

with u as small as possi ble. For i n none of the quotieiits £j/!Jj contains e/i, and is therefore isomorphic to in I) ( where denotes the image of under the natural epitiorphism D[d] — I)). Since I) is distributive and 71/7 is prime, each is prime, whence It follows that = d or = d, so (iii) holds with p/q = Conversely, suppose (i), (ii) and (iii) hold. Since I) and hence L are finite, it suffices to show that every prime quotieiit of L projects weakly onto e/o. \\ e begin by showing that every prime quotieiit a/c' e/o is pro jective to a prime quotient /y with = (I or = Since e is the only cover of (I (and dually), we cannot have c' = (I or a = e. Also if a =

CBAJ'TLJ

9-1

or

-1.

XOXUODL LAB

\JBJEIIES

= c, then we take = u/c'. Otherwise, using (iii), we niav assunie by duality, that is pro iective to a prime quotient f/il with c. Since I) is distributive, this means

v

\

c/IIJIdTI/u/

that

=

=

and further more il implies a( v c) = and v ± c) = c (since (I c). I his a/v a ± v ± £/c. \ow we apply (i) to obtaiii b E L with b and = = c b = b (since (I is meet b. It follows that irreducible), while the join-irreduci hility of c implies eb = ob. I hus we have X ( c/o, b), and since con( a, v) identifies and c, it also identifies c and (I. Consequently L is su hdirectly H—

=

a

H—

c

a

H—

=

H—

v,

/

\

H—

H—

H—

—

irreducible.

Varieties covering that smallest nonmodular variety. From the results obtaiiied

so

far one can now prove the following: IHEOREM 4.18 and

Ihe variety

is

covered by precisely three almost distributive varieties.

\\ ith the help of JOnsson's Lemma it is not difficult to check that each of the varieties L1( = {L1}v) cover (1 = 1,. ., 1.5). So let V he an almost distributive variety \\ that properly includes e have to show that V includes at least one of or Every variety is determined by its finitely generated su hdirectly irreducible mem hers, hence we can find such a lattice L E V not isomorphic to X or 2. By Corollary —1.1—1 (i and (iii), L L)[il] for some finite distributive lattice I), il E I). \\ e show that L)[il] contains one of or as a su blattice. I) is nontrivial since L)[il] 2. Let t3D and 1L) be the smallest and largest element of I) respectively. I heorem —1.17 (i and (ii) would imply L)[il] imply that il t3D, Also, t3D (1 X, so by duality we can find a, v E I) such that v a il. By I heorem —1.17 (iii) a/v is projective to a prime (luotielit p/q such that p = il or q = il. Since I) is distributive and a il, the case q = il is excluded, hence a/v is projective to il/q Again, by the distributivity of I), a q and therefore il = a q (see Figure -1.1-1 (i )). I heorem -1.17 (ii) a and q are meet reducible, so there exist! g E I) such that a = q = gil and since I) is finite we may assume that a and g v. I he su blat tice Ii' of I) generated by il, g is a homomorphic image of the lattice in Figure t1 (ii) (the distributive lattice with generators £, il, g and defining relation il = a and u v, Ii' must in fact he isomorphic to the lattice = ud). Since iii Figure ti-I (iii) or (iv). Consequently D'[il], as a su blat tice of I) [il], is isomorphic to or A sublattice isoniorphic to is obtained from the dual case when il v a. I'I-tOOF.

.

)

.

H—

I he above theorem and Corollary

-1. 1-1

(ii) now imply:

IHEOREM 4.19 (JOnsson and Rival [79]). In the lattice A of all variety is covered by exactly 16 varieties: *13 ± L2,. .

lattice .

su bvarieties.

the

,

Representing finite almost distributive lattices. Building on Iheorem -1.17, Lee [85] gives another criterion for the su hdirect irreducibility of a lattice I) [il], where I) is distri hutive, and he also sets up a correspondence between finite subdirectlv irreducible almost distributive lattices and certain matrices of zeros and ones. Before discussing his results, we recall sonic facts about distributive lattices which can he found iii [GLT].

4.3. ALMOST D1STIUBUTIVE

(i)

tCtitJEflES

95

"U

(iv)

(u)

Figure 4.14

LEMMA 4.20 Let 1) and II' be finite distributive lattices and denote by J(D) the poset of all nonzero join irreducible elements of .0. Then

(i) any poset isomorphism from 1)

J(D) to J(D') can be extended to an isomorphism from

toll'.

(ii) every maximal chain of II has length IJ(D)l.

II, let B =

Given a finite distributive lattice II and d meet reducible dual covers of d, C? = {ci,. . . , and consider the set

. .,

6,4

be

the set of all

the set of all join reducible covers of d,

Ad(D)={z€D:zd-
B1={z €Xd(D):zd=6j},

{C?1,. .

C?5

.,C?4 of Ad(D), referred to as the

={z€Xd(D):z+d=cJ}.

(We assume here, and subsequently, that the index i ranges from 1 to rn and 5 ranges from 1 to vi.) By the distributivity of II, two blocks B1 and C?5 have at most one element in common, so we can define an vii x vi matrix A(D[d]) of 0's and l's by aq = lB fl (If any, and hence all, of the sets B, C? or X411) is empty, then J(D[d]) = (), the 0 x 0 matrix with no entries.) A(D[d]) is called the matrix associated with DV], but notice that because the elements of B and C? were labeled arbitrarily, A(D[d]) is determined only up to the interchanging of any rows or any columns. Observe also that A(D[d]) does not have any rows or columns with just zeros, since {B1} and {Cj} are partitions of the same set As examples we note

that J(2) = (), J(JV) = (1), J(L13) = (1,1), J(L14)

or

(see

=

0)

and

= @0

Figure 4.15).

will also be concerned with the sublattice D of II generated by the set Li = will be atoms A4D) U {d}. Let 1 = EL', 0 = fit.', then clearly the elements c1,. . . in the quotient 1/d and = 1, so by Lemma 1.12, 1/d is isomorphic to the Boolean algebra 2", and the elements c1,. . . are the only covers of d in D. Dually we have

LAR \JJU.L11.LS

-I,

(d :1)

(do)

J(X)

(ii)

(1)

(d J)

C,

4.15

DJSIIUBL

4.3.

II\E

97

that d/O* is isomorphic to the Boolean algebra 2' and that bk,. ,b, are the only dual covers of din L)*. It follows that L)* has length iii n and therefore, by Lemma -1.20 (ii), e can in fact describe the elements of .J( L)*): .J( L)* = iii ± ii. .

.

H—

4.21 (Lee [85])..J(L)* = d/0*. in and = fi Cj. 111 elements )

only if

fl

C1

.

.

the complement of K are incomparable except: if and

,

.

off

.

(

.

.

,

where

is

= 0.

that the

distinct atoms of L)* and therefore pairwise incomparable we first note that in a distributive lattice every and join irreducible in L)*. As for the join irreducible element is a nieet of generators. I hus if c is join irreducible and c then I'I-tOOF. It is clear

are

j,

j,

IT d and therefore But cannot be a join of join irreduci bles strictly less that itself. It follows that is join irreducible. Also , K K are incomparable, since for sonic j k iniplies = for any K d as above, which contradicts IT d. Clearly also IT for any i and E Ck, and j, since K d. I herefore it reniains to show that K if and only if fl C1 = 0. If = 0*j = 0* and hence K d and £ E so = = fl C1, then IT Conversely and since is a nieet of generators, fl C1 = 0 iniplies Cj I

c K

d.

K

.

j

. .

j

j.

given an iii x n matrix I = of U's and l's, we define a finite distributive and an elenuent il,1 as follows: , Suppose he the Boolean algebra generated by the iii ± ii atonis So,

)

lattice

.

.

.

Put and

=

and let X,1 = XA u

:

(Iij = 1}.

\ow we let

he the

=

H-

sublattice of

generated

by

{d1}.

LEMMA 4.22 (Lee [85]).

for

no proper subset

I

of Xd( I)) does

I

IJ

{d} generate J)*,

\\ e may assume that Xd( I)) is Suppose to the contrary that I = I)), for sonic and I u {d} generates L)*. I hell E Xd( E C1 for sonic block of the natural partition of Xd(D). Let , E D*. By = IJ Lemma —1.21 is join irreducible, and since I {d} is a generating set, is the meet I'I-tOOF.

Xd( I))

.

.

j

of a subset of I IJ {d}. Xotice d = for each E C1, so by the dual of Lemma 1.12 C generates a Boolean alge bra th least element Hence is not a , we also cannot have proper subset of C and, as Choose = C C E Ihen ± d ± d = c1, so .

However this contradicts

IT

d.

IHEOI-tEM 4.2:3 (Lee [85]). Let I) be a finite folIo wins are eq iii valeii t:

(i) D[d]

[il1] where I =

1(

D[d]) and d

distributive lattice and d E

I) corresponds to

il,1 E

E

I). Ihun the

CBAJ'TLJ

98

(ii) the set Xd(D)

IJ

{d} generates I) (i.e. D*

XOXUODL LAB \JBJEIIES

-1,

I));

(iii) I) [il] is sti hdirectlv irreducible.

and A4 he defined as above. Clearly IJ{d} Sine '0 1 intl Ltiiiiui 22 = (ii) implies (i ): Again suppose is the Boolean algebra generated b the atoms \\ e claim that the elements bk,. , b,,, and let = d4 pt,. , p,, , are all in I his follows because q1d21 = b1 and JjJ ± = (b1 q1 )d,1 = = rows for all i, and I = I( L)[d]) has no or columns of zeros, j, ± ± il,1 = il,1 ± = hence for any given i (or j) there exists j (respectively 1) such jj E X1. Clearly the are covers of il1, and they are the only ones, since by Lemma 1.12 37 = XA = , Dually the are all the dual covers of il1. Let , B, and {( } } he the J(D*1) liltul il piitition' of (4(1) i) = th Li nimi —121 = where and itt (Ijj = 1 in J(IJ[d]) itt in = flC1. C1 0 = s— c5 s— from .J( D*) to .J( is a n C1 0. Consequently the flap Th itt poset isomorphism which extends to an isomorphism L)* by Lemma —1.20 (i ). J)* is the sublattice of 1) generated by Ad(L)) IJ {d}, so by assumption L)* = 1), and we always have = Clearly also d = 37 is mapped to = 37 by the isomorphism. (ii) i tiplies (iii ): \\ e verify that the conditions (i )' (ii) and (Hi of I heorem —1.17 hold. By Lemma 1.12, the join reduci ble covers of din L)* = 1) are in fact all the covers of d, and dually, which implies that the first two conditions hold. Also, if a v in 1), then the length of D/con( a, v) is less that the length of 1). It follows that con( a, v) identifies d with one of its covers or dual covers, hence condition (iii) of I heorem -1.17 holds. (iii) implies (ii): Suppose D[d] is subdirectly irreducible, hut D* is a proper sublattice of 1). Let 0* be the smallest and i*the largest element of J)*, and choose an element PROOF. (i) implies (ii): Let il1, foi {d} thit =

illi j

.

.

.

.

.

—I—

.

. .

—t—

.

.

)

)

z E 1)

or z 0*. 1 hell one of the quotients z ± i*/i* or 0*/zO* is nontrivial. a K a', then the quotients a/v and U hserve that in any distributive lattice, if v a'/ v' cannot proJect onto each other. Hence prime quotients in Z i*/ 1* or 0*/ z0* project onto aiiy prime quotient p/q with p = d or q = d, since J), q E L)* by condiition (i) and (ii) of I heoreni -Lii. I his however contradicts condition (iii) of the sanie theorem. Case 2: 0* K z K 1*. Choose z such that the height of z is as large as possi ble, and let he a cover of z. I hell is the only cover of z, else z would he the nieet E L)*, and J)* of two elements froni and would also belong to J)*• By I heorem 17 (iii), the prime quotient z*/ z proJects onto a prime quotient p/q with p = d or q = £1, alId since 1) is distributive, this implies z*/z a/v p/q for sonic quotient a/v. Since is the unique = z*/ z*/ cover of z, we must have a/v z z d/ zd, and p/q. Suppose p = d. I hell the two quotients are distinct, otherwise I heorem -1.17 (ii) iniplies z E L)*. Consequently z*/d z/zd. As before Lemma 1.12 implies that 1*/il is a Boolean algebra, hence z*/d is a Boolean alge bra, and so is z/ zd ( via the hijecti ye transposition). I herefore z is the join of the atoms of z/ zd, which are in fact elements of Xd( 1)). 1 his implies z E D* a contradiction. \ext suppose q = d. Since, by Lemma 1.12, 1* 1 J)* we would again have = 1). z E J)*, a contradiction. I hus we conclude that Case 1: z

iT

1*

—f—

f

f

f

—1.

/ \\

Given aiiy matrix

I

of U's and l's with

f

\

rows or columns of zeros, the equivalence of

DJSIIUBL

4,3.

II\E

99

(ii) and (iii) tells us that [il1] is su bdirectlv irreducible. Conversely, for ally su bdirectlv irreduci ble lattice I) [il], the matrix 1(1) [il] ) has 110 rows or columns of zeros, and it is not difficult to see that, up to the interchanging of some rows or columns, the matrices I and I( [il1]) are the same. Furthermore, given ally lattice L)[d] and A' Xd( I)), the sublattice I)' generated by A' IJ {d} is subdirectlv irreducible, and by Lemma —1.22 Ad( I)') = A'. Rephrased ill terms of the matrices that represent the lattices I) and Ii' we have the following:

I

C0R0LL;\RY 4.24 Let I = I( L)[d] ) fbr some finite distnbu tive lattice I). il E I). and which suppose Ii' is the sublat tice generated by some A' Aa( Ii), Ihen the matrix represents D'[d] is 0/) tamed from I by each I to an element of Aa( I)) X' to 0 and deleting any rows or columns of zeros that may have arisen, Eon versel any matrix 0/) tamed from I in this way ts a (su hdirec tlv irreducible) 5 ii blat tice off) [il].

Covering chains of almost distributive varieties. Ihe next lemma, which by Rose [84]

directly from I heorem -1.17, can

LEMMA 4.25 Let L he a finite

(i)

if

{L}v

he

was proved derived from the above corollary.

suhdirectly irreducible almost distributive lattice. L

2, X.

L L

if

fbr some

L

k E w

(see

2.2).

PROOF. (i ) Corollary —1.1—1 (i ) L IJ[d] for some finite distributive lattice I) and E I). Let I = 1(1) [il] ) be the matrix representing I) [il] and suppose I has more than one row. If I has 110 column with two l's ill it, then it has at least two columns (since it has at least two rows, and 110 rows of U's), and we can therefore find two entries equal to 1 ill two different columns and rows. Deleting all other rows and columns, it follows from Corollary -1.2-1 that is a sublattice of L. Hence if then 1 has only , where k ± 2 is one row th all entries equal to 1. 1 his is the matrix representing prove the number of columns of 1 (see Figure -L16 (i )). Similar arguments (ii) and (iii ).E

il

(Rose [84]). only join irreducible cover of THEOREM 4.26

for each

i E

{13, i-I, 15} and

u E w

the variety

is

the

= PROOF. Let i = 13 and suppose V is a join irreducible variety that covers V must he almost distributive, otherwise, by Corollary -Li-I (ii), V contains one, say L, of the lattices i13, L2,. , ill which case V hence either V is ± {L}v ;> not a cover of or V is join reducible. V is of finite height, thus by Corollary -Li-I (i (iii) and Lemma 2.7 V is generated by a finite subdirectlv irreducible lattice L = where I) is distributive. Since V is join irreducible, we must have k = L V covers i = U and 15 is completely analogous. .

I he or

.

smallest subdirectlv irreducible almost distributive lattice that is not of the form 2,

for i = 13, U, 15, ii

E w

is represented by the

matrix

(see Figure

t

16

(iii)).

cI-IJ1'rliJi

100

/ (d..

LAJI

-I,

N

fl

J(Lq3)

(11.. .1)

U

—

b•,

J(L) 4.16

—

U

\JIULi1LS

\JIUEIIES

DJSIIUBL LiVE

4.3.

101

Further results on almost distributive varieties. IHEOREM 4.27 (Lee [85]). Every almost distributive only' finite] many covers,

lattice variety of finite height

I'I-tOOF. Let V be an almost distributive variety' of finite height. I hen V is generated by' finitely' nianv 511 bdirectlv irreducible almost distributive lattices, and by' Corollary -1.1-1 (i) these lattices are of the form , [d,j for sonic finite distributive lattices Di,. , Let k = max{ Xd? (Di) I = 1,. ., n}. By Corollary t1 (iv), each join irreducible cover of V is generated by a finite su hdirectly irreducible lattice L)[d], and clearly' we must have Xd( I)) = 1. By' I heorem -1.23 1) [il] can be represented by' a matrix of U's and l's with at most k ± 1 rows and k ± 1 columns, hence V has only finitely many join irreducible covers. On the other hand, each join reducible cover of V is a join of V and a join irreducible cover of a su bvariety' of V. I herefore V also has only' fini telv many join reducible covers. .

. .

.

.

H—

LEMMA 4.28 (Lee [85]). Let I) be asublat tice of a finite distribu tive lattice I). and il E if 1J[d] is su hdirectlv irreducible. then 1J[d] L)'[d]. where Ii' is generated by il and a subset of Xd( I)),

natural partitions of Xd(1J), and let ,r E c1 = ,r ± il with ,r E Choose E I) such that K c1. For each ,r E Xd(D) we have ,r E fl C1 for unique i,j, iii which case we define ,t' = By' distri hutivity' ,t'd = )d = ,td ± and and ,t ± = c1 implies ,t' ± = it ± = ± d)j, hence ,td = b1 implies ,t'd = I)), follows that the set X' = {,t' and since the elements E Xd(D)} is a subset of Xd( and all have to be distinct, the map £' is bijective. Let Ii' be the sublattice \\ e show that of I) generated by' X' IJ {d}. By' Lemma -1.22 Xd( Ii') = and Ii' have the same matrix representation, then it follows from I heorem —1.23 that I) Ii'. By Lemma 1.12 the elements , are all the (join reducible) covers of il, and dually for Let = and = = he E X':,r'd = E the blocks of the natural partitions of X'. Clearly' ,t E implies ,t' E for all I, and the converse must also hold, since the map ,t £' is bijecti ye and the blocks are which implies finite. Similarly £ E if and only if E Hence = fl fl J(D[d]) = J(d'[d]). I'I-tOOF. Let = ,rd with

.

.

,

B,,, } and

.

.

he the

,

.

.

H—

.

H—

j.

'

.

.

.

X':,r'±d

'

4.29 (Lee [85]). Let I) he a finite distributive lattice. il E I). and let D* he the su blat tice off) generated by Xd( I)) 'J {d}. lThen D*[d] is a retract of D[d]. in particular, is the smallest homomorphic image of D[d] separating (il, 0) and (il, 1). LEMMA

I'I-tOOF. Since (il,

1), by' Lemma 1.11 there is a unique subdirectly' irreducible homomorphic image L)[d] of L)[d] such that (il, 1 )/( il, 0) is a critical quotient of L)[d]. By' I heorem —1.23 D*[d] is also subdirectly' irreducible with critical quotient (il, 1 )/(d,0),hence IJ*[d] is isomorphic to its image D*[d] L)[d]. \\ e have to show that D*[d] = L)[d]. I he ii, where II is obtained from epimorphism L)[d] L)[d] induces an epimorphism I) ii, and it suffices to show L)[d] by' collapsing the quotient (il, 1 )/( il, 0). I hen L)* that must be noncomparable with il, = 77. Consider ,r E I) such that E X—1(L)) KJ ±d. Let ,t0 = ,tc±b, then one easily' so we can find b,cE I) with ,td Kb U

I

.t

I

E'BAJ'TLJ

102

LAB

-1,

\JBJEIIES

1

01

1000...01

1

Figure

checks that J0

E

-1.17

Xd( D* ) and b, c E L)*. Xotice that d

for otherwise the epimorphism

L)[d] would collapse the quotient c/( il, 1) and, as (il, 0) and (il, 1) generate a whence 1 pentagon, it would also identify (il, 0) and (il, 1). Similarly 1 in 7 L)*, we iii fact Because it follows that Td = b, T ± d = and since b, E I I I I) Ihiojini -123 \-1(D) IJ {d} hw 7 E Ihu-, L)*[d] = L)[d]. generating set for ii, hence = II, and therefore L)*[d] L)[d]

LEMMA 4.30 (Lee [85]). Let I) be a finite distrihu tn'e lattice and il E I). Ihun every su hdirectly irreducible member of {D[d]}" is isomorphic to L)'[d], where Ii' is a sti blat tice oil) generated by il and a sti bset of Xd( Ii). I'I-tOOF. Let L he a suhdirectlv irreducible member of By JOnsson's Lemma L E HS{l) [il] }, so there is a su blat tice L0 of I) [il] and an epimorphism f L. If (il, 1 )/(d, 0) L0, then L0 is distributive and hence L 2. If (il, 1 )/(d, 0) 1 L0 then = l)0[d] for a sublattice D0 of I). But if (il, 1 )/(d, 0) is collapsed by f, then again L 2. Suppose (il, 1) /( d,0) is not collapsed by It Since L is su hdirectly irreducible, and il, f( 1 )/f( d,0) is critical, L is a smallest homomorphic image of D0[d] separatiig (il, 0) Also and (il, 1). By Lemma -1.29, the same holds for [il]. Hence L is a su blat tice of I), and is su bdirectlv irreduci ble, therefore Lemma —L28 implies that L is isomorphic to L)'[d], where Ii' is a sublattice of I) generated by il and a

subset Xd(lfl. Xotice that there are at least Xd(D) ± 1 nonisomorphic suhdirectly irreducible memhers in since if I , are two subset of different cardinality, then I IJ {d} and IJ {d} generate two nonisomorphic su blat tices. e now consider an interesting sequence of almost distributive lattices which is given iii Lee [85], and was originally suggested by J Onsson. Let Kj be the finite subdirectlv irreducible almost distributive lattice represented by the (1 ± 1) x (1 ± 1) matrix J(Kj) in Figure -1.17, and let V0 = K2, A3,. .

ri .

4.31 K1

V1

PROOF. By Corollary is a splitting

and

fbr

I

E

.,i

i

{1, 2,3.

.

.jiV

iOl i

1

/

.

-1. 1-1 (vi) K1 = for sonic finite distributive lattice d1 E lattice, so it generates a completely join prime variety for each

i

DJSIIUBL LiVE

4.3.

103

(Lemnia 2.8). Since it suffices to show that Kj for ally i = j. By the preceding lemma an subdirectlv irreducible lattice in {K1} is isomorphic to a sublattice of A1 = D1 [d1] generated by a subset of (D1). If j i then ( D1) = which certainly implies Kj \ow suppose j > i and let Ad? ( Xa, ( with corresponding natural partitions {

p2,.

{

C2,...,

(D1) =

and

.

£4},.

= =

. .

,

. .

., 1121+4 with natural partitiolls

= = {{u2,

,

f is an enibedding of Kj into Kj, then we can assume without loss of generality that .Xs an em bedding ,t must map B- blocks onto B'- blocks and C- blocks onto = C'-hlocks, hence = i,,. = 112i+2. But = )} = fj,. which is a contradiction. I herefore Kj is hot isoniorphic to {( } } , a su blat tice of Kj, and consequently Kj {K1 f(

.

.

.

.

IHEOREM 4.32 (Lee [85]). Let A be the variety Vj be defined as above,

(i)

=

(ii)

Ihere

(iii)

V0

vial] almost distrihu tive lat tices and let

2w,

chain of almost distributive varieties,

is an infinite

has infinitely many dual covers,

(iv) Ihere is an almost distributive variety with infinitely in any covers in I'I-tOOF. (i) By the preceding lemma,

distinct su hsets of

K2, A3,. .} generate distinct .

su b varieties of .}V for each i E w. Ihen V0 = (ii) Let = {Kj, V follows again by Lemma —1.31. (iii) \\ e claim that Kj is the only fini telv generated (hence finite) su bdirectlv irreducible mem her of that is not ill Vj, froni which it then follows that Vj for each i E w. By Lemma —1.31 Kj V2. Every finite su hdirectlv irreducible mem ber L E V0 is a splitting lattice, so L E for sonic j. If I and j then L E V1, and if L E L is not isomorphic to Kj then, by looking at the matrix that represents L, we see that L E {K1}v for any j ;> i, so we also have L E I his proves the claim. (iv) Let he the conjugate variety of Kj relative to AA (I E w), and let V = for each i. By I heorem 2.3 (i) every suhdirectly irreducible V V± nieniher of V ± belongs to V or Let L he a suhdirectly irreducible lattice ill Lemma -1.30 implies that L is a sublattice of so Kj for any j I. It follows that L E V or L Kj, hence Kj is the only su bdirectlv irreducible lattice ill which is hot ill V. V± .

E'BAPILLB

10-1

LAB \JIUEiJES

-I.

Further Sequences of Varieties

4.4

In Section -1.3 we saw that above each of the varieties and there is exactly one covering sequence of join irreducible varieties ( I heorem -1.26). 1 hese results are due to Rose [84], and he also proved the corresponding results for L6, Since these , varieties are not almost distributive, the proofs are more involved. Here we only consider above L6. the sequence .

Some technical results. Let

.

.

lattice and A a subset of L. An element z E L is j and z implies I he notion of an E X A —meet element is defined dually. A quotient u/v of L is said to be if every element of u/v is u/ ('-join isolated and u/ v-nieet isolated. I he next four lemmas (-1.32---1.35) appear in Rose [84], where they are used to prove that the variety is the only join irreducible cover of for i E {6, 7,8,9, 10} (see Figure 2.2). I hese lemmas only apply to lattices certain conditions summarized said to be A

—join

if

z

L be a

=

.

here as

subdirectly irreducible neardistributive lattice with critical (which quotient c/o is unique by I heorem -L8). Furthermore c'/o' is a quotient of L such CONDITION (Ic). L is a finite

that (i) G'

K

K

(ii) ally

z E

(iii) ally

z E

c'/o' c'/o'

{o'} is c'/o'-joill isolated; {c'} is c'/o'-nieet isolated.

that if b c'/i' and b is noncoitiparable with some z E c'/i', then b is = noncoitiparable with all the elements of c'/ii'. Moreover, G' b = z b = b and zb = c'b, which implies X (c'/o', b). Hence, for any b c'/o', the conditions X (c'/o', b) and X ( c/o, b) are equivalent. U bserve

H—

LEMM;\ 4.33 Suppose L satisfies condition

(i) (ii)

H—

H—

(ic ).

if a c' in L. then there exists b E L such that X(c'/o', b) and if L excludes L7. then we also have (I'

a

=

±

b

b

PROOF. (i) By Lemma -1.16 (iii) there exists b E L such that X(c/o, b), b K a, b and a/c' 0 ± b/(o ± b)c'. b is noncoitiparable with c', so X(c', o', b) follows from the remark above, and we cannot have (o ± b)c' c', since (o ± b)c' is not c'/o'-nieet isolated. So (o b )c' = c' and therefore a = b = b. Since L is finite we can choose / such that b K / o' ± b. / is also noncomparable with c', so we get X(c'/o', /), and of course a = o' ± /. Hence we may assume that a b. Also b o'b, since (I'b b would imply \:(b// o'), hence X(b//, (I), and by Corollary X(c/o, o), which is impossible. (ii) Suppose to the contrary, that / o' for some / E L. By the dual of part (i ) there exists b0 E L with X(c'/o', b0) and / = (I'b0 o' b0 (Figure -LiS (i )). Since o' b b, we have I b = o' b and so X(o'//, b). o'// o' b0/b0 and Corollary —Li imply X ( o' b0/b0, b). I hus o' b0 o' b, which clearly implies that b0 (I'b and b0 >'- y, we must and b0 are llollcolilparaI)le with o' b. Since o' b >'- c, b have (o' b )( o' b0 ) = c', (o' b0 )b = and (o' b)b0 = I. Hence the elements 6!', b and b0 generate L7 (Figure -LiS (i and this contradiction completes the proof —

—

—

ti

H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

H—

/

H—

-1.4.

IL

OF

111111KB SKQL

10.5

(I'

(i)

V

(ii)

(J'b

li1igure -1.18

now add the following condition. (:oNDUIoN (tic ). b is an element of L such b, (I'b} ij c'/o'. e

LEMMA

(ii)

G'

4.34

if L

=

±

±

b

that X(c'/o', b) and

satisfies conditions (ic), (tic) and excludes

b

>t

iinplie £

G'

H—

b/i'b =

then fbr

u

E

{b,

G'

H—

L.

<. C'.

I'I-tOOF. (i If tj IT b, then is nonconiparable with b and with i'. \\ e claim that can he chosen so that o', C', b, y generate (see Figure -1.18 (ii)). \\ e may assume that we would have X( y/I, o'), hence X(y/I, o), and by (I' If o'y then y y, ± y. Corollary -1.7 X(c/o, o), which is impossi hle. I herefore y o'y. By semidistri hutivity )

±

b

=

G'

±

y

=

b

±

= (I'b ± G'y ± by.

From this it follows that the elements (I'b = c'b, o'y = c'y and by are noncomparahle, and

therefore (I'

=

o'b±o'y,

b= (I'b±by and

y

= (I'y±by.

I his shows that o', b and y generate an eight element Boolean algebra. Since X (c'/o', b) and X ( c'/o', y) hold, L includes (ii )If IT c, then o' ± = o' ± b, and since we cannot have K b, part (i) implies that L includes L satisfies

conditions (Ic), (tic) and excludes L7.

and

then c' is

meet ilTedu cible, I'I-tOOF. Suppose c' is meet reduci hle. I hen there exists an elenient t covering c' such that c' = £(o' b). By Leninia —1.33 (i ) there exists b0 E L with X(c'/o', b0) and = (I'b0. I he elements G' b0 b, G' ± b0 and b ± b0 generate a lattice K that H- b0 H—

H—

is a homomorphic image of the lattice in Figure —1.19 (i ). If K is isomorphic to that lattice, then bb0 K o', since bb0 IT o' would imply bb0 ± (I' E c'/o' {o'}, contradicting the

CiLtl'TEit 4.

106

tCt1UEflES

+60

=a'+60

66u

(i)

(ii)

Figure 4.19

(i)

(ii)

Figure 4.20

assumption that every clement of c'/a' — {a'} is c'/a'-join isolated. In ftct we must have 660 c a', since a' is c'/a'-meet isolated. Thus K U {a'} is a sublattice of L isomorphic to contrary to the hypothesis. We infer that K is a proper homomorphic image of the lattice in Figure 4.19 (i), and since a' +6, a' and c' are distinct, it follows that c' c 6+60. Now Figure 4.19 (II) shows that if a'b and a'b, are noncomparable, then L while a'b c a'b.j or a'b.j c a'b imply that L includes Lr. Finally, we cannot includes have a'b = a'bt,,, since then L includes .L1, which contradicts the semidistributivity of .L.D

4.36 if .1, satisfies conditions is meet irreducible. LEMMA

(*),(n) and

excludes £9,

and

then 6

To avoid repetition, we first establish two technical results: (A) .N(u/v, z) for some u/v c'/a' and z c'/a', then there exists p L with R(c'/a',p) such that JV(a' + p/a',z), R(p/a'p,z) and a' + p >- c' (Figure 4.20 (i)) or

if

dual1. Consider a sequence u/v = z1/p1 ... .'v,,, z74/p74 = c/a. Since c/a is a subquotient of c'/a' and u/v is not, there is an index i > 0 such that c'/a' and

4.4. FU1LL'iiEit SEQUERCES 01" tCti(Jfl1ES

107

\,

ç c'/a'. By duality, suppose that Since is c'/a'-meet isolated, c'/a', and since c'/a', we must have = c c' would also imply Now Lemma 4.6, and the fact that u/v projects weakly onto and therefore > c' and c/a imply R(z1/p1, z) and R(c/a, z). Since z c'/a' we must have R(c'/a', z). Choose z .L such that c' -< z then dearly .N(z/a', z) holds (Figure 4.20 (ii)). By Lemma 4.33 (i) there exists p L with .N(c'/a',p) and z = a' + p. Since z/a' p/a'p, Lemma 4.6 again implies JV(p/a'p, z). This proves (A). (B) if for some u, v, z .1, with z 6 we have .N(u/v, z), then u/v ç c'/a'. Suppose u/v c'/a'. Since dearly z c'/a', (A) implies that there exists ho L such that n(c'/a',ho) and either

\

(1) .N(a' + ho/a', z), .N(ho/a'ho, z) and c' -< a' + ho or (2) JY(c'/a'btj, z), .N(a' + ho, z) and a'ho -< a'. We will show that, contrary to the hypothesis of the lemma, the elements a', c', 6 and ho generate L15. Since we already know that .N(c'/a',b) and JV(c'/a',btj), it suffices to check that a'b + a'ho = a' and (a' + 6)(a' + ho) = c'. Either of (1) or (2) imply that z is noncomparable with a'ho and a' + ho. Since a'b c 6 z we must have a'ho a'b. Strict

inclusion a'b c a'ho is also not possible, because a'b -< a' and a'ho c a'. Thus a'b and a'ho are noncomparable, and since a'b -< a', it follows that a'b + a'ho = a'. Next note that a'z = a'b, because a'b -< a' and a'b a'z c a'. hence a'z = a'b > a'bho = a'hoz, so we cannot have .N(c'/a'ho, z) in (2). Therefore (1) must hold, and in particular .N(a'+ho/a', z), whence it follows that a'+ho 6. Thusc' (a'+b)(a'+ho) c a'+b, and since c' -< a'+b, c = (a' + 6)(a' + ho). This proves (B). Proceeding now with the proof of the lemma, suppose 6 is meet reducible. Then we can find z >- 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', or dually R(c'/a'ho,zi) and a'ho -< a'. First suppose that .N(a' + ho/a', z1). We cannot have a' + ho c a' +6 since a' +6 >- 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

-<

a', c'

-<

a' +6 and

Then

a'+b/a'b= c'/a'U{a'b,b,a'+b}; (2) a' + b/a'b is an isolated quotient of I... (1)

Assume (1) fails. Then there exists z 1, such that z a' + b/a'b but z c'/a' U {a'b,b,a'+b}. Since a'b c z c a'+b and a'b -< 6 -< a'+b, it follows that band z are noncomparable and zb = a'b, z +6 = a' + 6. Furthermore, as c'/a' is isolated, z is noncomparable with a' and c' , whence a' + z = a' +6. This, however, contradicts the semidistributivity of L, since a' +6 a' + z6 = a'. Therefore (1) holds.

FUitTliElt SEQUENCES 01" tCt1tIETJES

4.4.

To prove (2), (3)

111

it suffices to show that

a' is join irreducible and c' is meet irreducible;

(4) 6 is both join and meet irreducible;

6= z + 6> z imply z il/a'; p€Landa'+b=a'+p>pimply p=6; L and a' +

(5) z (6)

(7) z

L and a'b = z6 c z imply z

(8) p

L and a'b = a'p

c'/a';

c p imply p =

6.

(3) and (4) follow from Lemmas 4.35 and 4.36 and their duals respectively. Suppose 6,wehave z. Then d by Lemma4.34 (ii) and, z c'b = a'b. Now z a', because a'b is the only dual cover of a'. Since d/a' is isolated, this implies z d/a', whence(S) holds. If a'+b = a'+p, then Lemma 4.34 (i)implies that 6, and from the join irreducibility of 6 we infer p = 6, thereby proving (6). Finally, p (7) and (8) are the duals of (5) and (6). (ii) then the elements .L can be chosen such that

a'+b= z+6>

if

a+6/a6 = c/au {a6,6,a+6], a+bu/abij a+ = a+

u {abj, 61, a +

for

i

{1, 2,..., k],

and all these quotients arc isolated. By Lemma 4.35 and its dual, the quotient c/a is isolated. Choose z L with c -< z a +6. Since L excludes L1, Lemma 4.33 (i), (II) and (C) above imply the existence of 6' L with .N(c/a,6') and a + 6' = z, such that this sublattice is an interval in L and is isolated. Since a is join irreducible and a -< a6', we infer that a6' a6. Thus a6 alt -<6' -< z a+6, whence it follows that 14ftc/a,6',bu,...,b,j. So we may replace 6 by 6', and continuing in this way we prove (IL). Since c/a is a prime of .L, (II) implies that we can find . . ,6k in such that the sublattice generated by c/a and these b's is an interval of L. Since L there exists u 1, with u >- a + 6k or u -< a6 = k and from Lemma 4.33 (i) or its dual we obtain 6' 1, such that JV(a + bk/abk,U), which implies Lr'(c/a,b,bo,. . as 0 required. After much technical detail we can finally prove: THEoREM 4.38 (Rose [84]).

is the only join irreducible cover of £4.

Suppose to the contrary that for some natural number vi, the variety = has a join irreducible cover V Choose vi as small as possible. Since V has finite height in A, it is completely join irreducible, so it follows from Theorem 2.5 that V = {L}" for some finitely generated subdirectly irreducible lattice .L. Note that 14 {L}" Using the results of Section 2.3 one can check that 14 is a splitting lattice, and since it also satisfies Whitman's condition (W), Theorem 2.19 implies that is projective in . .L we have £. By Lemma 2.10 14 is a sublattice of .L, so for some i41(c/a,6,bu,. .,67j. By Theorem 4.37 (i) c/a is critical, and .L/con(a,c) has no L!jquotients. Again, since 14 is subdirectly irreducible and projective, Lemma 2.10 implies .

(iii .tl'TEit 4.

112

.NO.NMODULS1t tCtitJ.ET1ES

that 14 is not

a member of the variety generated by .L/con(a,c). This, together with the implies that, for vi = o, L/con(a,c) is a member of .V and, for vi > o, .L/con(a, c) is in By Lemma 4.15 .L is finite and, since .L 14, it follows from 0 Theorem 4.38 (II) that L includes 14". This contradiction completes the proot

minimality of

vi

By a similar approach Rose [84] proves that is the only join irreducible cover i = 7 and 9 (the cases i = 8 and 10 follow by duality). A slight complication arises due to the fact that 14 and 14 are not projective for vi 1, since the presence of doubly reducible elements implies that (W) fails in these lattices. As a result the final step requires an inductive argument. For the details we refer the reader to the original paper of Rose [84].

of £}' for

Further results about nonmodular varieties. The variety M3+X is the only join reducible cover of .V (and M3), and its covers have been investigated by Ruckelshausen [78]. 143 in FigHis results show that the varieties V1,. . ., V8 generated by the lattices ure 2.4 are the only join irreducible covers of M3 + .V that are generated by a planar lattice of finite length. The techniques used in the preceding investigations make extensive use of Theorem 4.8, and are therefore unsuitable for the study of varieties above or £12. Rose [84] showed that £12 has at least two join irreducible covers, generated by the two subdirectly irreducible lattices and U respectively, (see Figure 4.24, dual considerations apply to Using methods developed by Freese and Nation [83] for the study of covers in free lattices, Nation [85] proves that these are the only join irreducible covers of £12, and that above each of these is exactly one covering sequence of join irreducible varieties £1k4 and a result of Rose [84], any semidistributive lattice which fails to be bounded contains a sublattice isomorphic to L11 or L12 (see remark after Theorem 4.10). Thus it is interesting to note that the lattices 14j and U7' are again splitting lattices. In Nation [86] similar techniques are used to find a complete list of covering varieties of (and £2 by duality). The ten join irreducible covers are generated by the subdirectly irreducible lattices Liu,. . . , L25 in Figure 4.25.

4

-1.-I.

IL

11111-1

VI

KB SLQL

113

ill '-'112

7 '2

11igurc -1.2-I

il-I

E'JTIAPTER

-I.

XOX4JOL)L LAB \ABJL1JLS

L

P L 2:3

11igurc -1.2.5

Chapter

5

Equational Bases 5.1

Introduction

for a variety V of algebras is a collection F of identities such that V = ModE. An interesting pro blem iii the study of varieties is that of finding equational bases. Of course the set IdV of all identities satisfied by nienibers of V is aiwa s a basis, hut this set is generally highly redundant, so we are interested iii finding proper (possi bly minimal) equational basis for V. In particular we would like to know under what conditions V has a /711/fe equational basis. It might seem reasonable to conjecture that every finitely generated variety is finitely based, hut this is not the case in general. Lyndon [54] constructed a seven—element algebra with one binary operation which generates a nonfinitel based variety, and later a four—element and three—element example were found by \ [63] and Murskir [65] respec— ti vely. On the other hand Lyndon [511 proved that any two element alge bra with finitely many operations does generate a finitely based variety. I he same is true for finite groups Oates and Powell [64]), finite lattices (even th finitely nianv additional operations, Mckenzie [70]), finite rings (kruse [77], Lvov [77]) and various other finite algebras. Shortly after Mckenzie's result, Baker discovered that any finitely generated congruence distributive variety is finitely based .Ac tuallv his result is somewhat more general and moreover, the proof is constructive, meaning that for a particular finitely generated congruence distributive variety one can follow the proof to obtain a finite basis. However the proof, which only appeared in its final version in Baker [77], is fairly complicated and several nonconstructive shortcuts have been published (see Herrmann [73], Makkai [73], lavlor [78] and also Burns and Sankappanavar [81]). 1 he proof that is presented in this chapter is clue to JOnsson [79] and is a further generalization of Baker's theorem. In contrast to these results on finitely based lattice varieties, Mckenzie [70] gives an example of a lattice variety that is not finitely based. Another example by Baker [69], constructed from lattices corresponding to pro jective planes, shows that there is a nonfinitely based niodular variety. Clearly an equational basis for the meet (intersection) of two varieties is given by the union of equational bases for the two varieties, which implies that the meet of two finitely based varieties is always finitely based. An interesting question is whether the same is true for the join of two finitely based varieties. I his is not the case, as was independently discovered by JOnsson [74] and Baker. I he example given in Baker [77'] is included in Au

equational

115

CBAJ'TLJ

116

5,

EQL

AIIOXAL

this chapter and actually shows that even with the requirement of niodularitv the above question has a negative answer. In JOnsson's paper, however, we find sufficient conditions for a positive answer and these ideas are generalized further by Lee [85']. One consequence is that the join of the variety .\4 of all modular lattices and the smallest nonmodular variety is finitely based. Ihis variety, denoted by .tft( = is a cover of .t1, and an ± equational basis for consisting of just eight identities is presented iii JOnsson [77]. Recently J Onsson showed that the join of two finitely based modular varieties is fini telv based whenever one of them is generated by a lattice of finite length .Ageneralization of this result and further extensions to the case where one of the varieties is nonmodular appear in hang [87]. Although Baker's theoreni allows one to construct, in principle, finite equational bases for any finitely generated lattice variety, the resulting basis is usually too large to be of any practical use. In Section 5.—I we give some examples of finitely based varieties for which reasonably small equational bases have been found. I hese include the varieties .\4 (u E w, froni JOnsson [68]), (McKenzie [72]) and the variety .tft refrrred to above.

5.2

Baker's Finite Basis Theorem

Some results from model theory.

A class K of algebras is an eiemerdaiy if it is the class of all algebras which satisflv sonic set S of first-order sentences (i.e. K = Mod 5), and K is said to be eiemerdanj if S may be taken to be finite or, equivalently, if K is determined by a single first—order sentence (the conjunction of the finitely many

sentences in 5). I hese concepts from model theory are applicable to ally class of models of sonic given first—order language. Here we assume this to be the language of the algebras in k. For a general treatment consult (hang and Keisler [73] or Burns and Sankappanavar [81].) I he problem of finding a finite equational basis is a particular case of the following more general question: \\ hen is all elenuentary class strictly elenuentaryY Recall the definition of an ultraproduct from Section 1.3. 1 he nonconstructive shortcuts to Baker's finite basis theorem make use of the following well—known result about tilt rap roduc t s:

J

IHEOREM 5.1 (Los[55] ). Let 1, and suppose is the congruence induced = by some ultra Jilter U over the index set I, Ihen. for any first—order sentence a. the ultraproduct satisfies a if and only if the set {i E I satisfies a} is in U.

In particular this theorem shows that elementary classes are closed under ultraprod— ucts. But it also has many other consequences. For example we can deduce the following wo important results: IHEOREM 5.2 (Fravne, Morel and Scott [62], Kochen [61]). Anelemen tarv class K of is strictly elemen tar if and only if the complement of K is closed under ultraproducts. Ihe complement can be taken relative to any strictly elementary class con tainin,g K. I'I-tOOF. Suppose B is an elementary class that contains K. If K is strictly elementary, then mem hership in K call he descri bed by a first-order sentence. By the preceding theorem the

5,2. BAKER'S FLVJTE BASiS

iflEOfifli

117

negation of this sentence is preserved by' ultraproducts, so any ultraproduct of meni hers in B K must again he in B K. Conversely, suppose K is elementary and is contained in a strictly elementary class B. Assuming that B K is closed under ultraproducts, let S he the set of all sentences that hold iii every member of K, and let I he the collection of all finite subsets of S. Since B is strictly elementary, B = Mod for sonic E I. If K is not strictly elementary then, for each i E I, there must exist an algebra not IJ every in K such that satisfies sentence in the finite set i Xote that this iniplies K. \\ e construct an ultraproduct E B E K as follows: Let I = I, and, for each i E I define .J, = {j E I j I' i}. I hen .J, 0 and for all i, k E I, whence :T = {.J I fl 1k = I for sonic i} is a proper filter over I, and by Zorn's Lemma can he extended to an ultrafilter U. \\ e claim that satisfies every sentence in S. I his follows from I heorem .5.1 and the o hservation that for each a E 5, E U. {j E I J1 satisfies a} I' :

:

Since K is an elementary class, we have so this contradicts the assumption that B

are all mem hers of B K, K. But the K is closed under ultraproducts. I herefore K E

must he strictly elementary. IHEOREM 5.3 Let K be an elementary class, and suppose 11. K is strictly elementary, then K =

that K = iiod 5, sentences s.

is some set of sentences such

iiod

fbr some finite set

that for every finite subset of 5, Mod properly contains K. As in the proof of the previous theorem we can then construct an ultraproduct not in K. Ihis, however, contradicts the result that the E K of algebras complenuent of K is closed under ultraproducts. I'I-tOOF. Suppose to the contrary

Every identity is a first-order sentence and every variety is an elementary class, so the second result tells us that if a variety' is definable by a finite set of first—order sentences, then it is fini telv based. I he following theorem, from JOnsson [79], uses I heorem .5.2 to give another sufficient condition for a variety' to he finitely based. he a variety of contained in some strictly elementar class is contained in C and V fl C is strictly B. if there exists an elementary class C such that elementary, then V is finitely based, IHEOREM 5.4 Let

V

I hen I heorem .5.2 implies that B V is not closed under ultraproducts. Hence, for some index set I, there exist I, E B V and an ultrafilter U over I such that the ultraproduct J= Each has at least one su hdirectly irreducible image not in V. On the other hand, if we let I' = then l'/ou E V since it is a homomorphic image of are not necessarily' in B need not he closed under homomorphic images, so the B, hut I'/ou E B and B strictly elementary imply that { I E I E B} is in U. I herefore, restricting the ultraproduct to this set, we can assume that every' I C and, because C is an elementary' class, it follows that J'/o71 E V fl C. I his contradicts I heorem .5.2 since V fl C is strictly elementary (by assumption), and is an ultraproduct of algebras not in V fl C. I'I-tOOF. Suppose V is not finitely' based.

(JiiAl'TEit 5. EQEAflOPLtL BASES

118

Finitely based congruence distributive varieties. Let V

be a congruence distributive variety of algebras (with finitely many operations). By Theorem 1.9 this is equivalent to the existence of vi + 1 ternary polynomials . ., such that V satisfies the following identities: 1o(z,p,z) = z, 174(z,p,z) = z, = z

z,

z, z)

for i odd.

In the remainder of this section we let Vg be the finitely based congruence distributive variety that satisfies these identities. Clearly V ç

Translations, boundedness and projective radius. The notion of weak projectivity in lattices and its application to principal congruences can be generalized for an arbitrary algebra A by considering translations of A (i.e. polynomial functions on A with all but one variable fixed in A). A 0-Imnolalion is any map 1: A —, A that is either constant or the identity map. A 1-Imnolulion is a map f : A —, A that is obtained from one of the basic operations of A by fixing all but one variable in A. For our puxposcs it is convenient to also allow maps that are obtained from one of the polynomials above. Equivalently we could assume that the are among the basic operations of the variety. A k-translation is any composition of k 1-translations and a translation is a map that is a k-translation for some k w. For a,b A define the relation fMa,b) on A by (c,d)

J.'k(a,b)

if

{c,dJ = {f(a),f(b)J

for some k-translation f of A. Let I'(a,b) = UkEWJ.'Ma,b). This relation can be used to characterize the principal congruences of A (implicit in Mal'cev [54], see [UA] p.54) as follows: A we have (c,d) con(a,b) if and only if there exists a sequenw c = = dmA such that (cj,cj+l) J.'k(a,b) fori cm. Two pairs (a,b),(a',b') A x A are said to be k-bounded if I'k(a,b) fl J.'k(a',b') 0 and they are boundS if f(a,b) fl f(a',b') 0. Observe that if A has only finitely many operations, then k-boundedness can be expressed by a first order formula. The pivjectioc radius (2-radius in Baker [77]) of an algebra A, written .11(A), is the smallest number k > 0 such that for all a,b,a',b' A

l'iw a,b

con(a,b) fl con(a',b')

0 implies I'k(a,b) fl J.'k(a',I/)

0

(if it exists, else = co). For a class K of algebras, we let .11(K) = sup{i1(A): A K]. few The next lemmas show that under certain conditions a class of finitely subdirectly irreducible algebras (see Section 1.2) is elementary if and only if it has finite projective radius. These results first appeared in a more general form in Baker [77] (using n-radii) but we follow a later presentation due to Jónsson [79]. LEMMA 5.5 if_I c0,c1,. . .,cm A and c0 such that and are 1-boundcd.

f74(c5)

=

c,,4,

then there exists a number p c in

Consider the 1-translations fj(z) = i for all i in, hence there exists a smallest index q

c,,4

vi. vi

Then fo(c5) = e0 and such that the elements

5,2. BAKER'S FLVJTE BASiS

iflEOfifli

fq(ej) are not all equal to in such that c fq(ep) the 1-translation ,f( = /q(

q

p

,fq(e,u)

)

again see that (c, ci) E (c,ci) E In either case (c, ci) E are 1- hounded. we

is odd, then fq(eu) so we can choose = ci. It follows that (c, il) E c2+fl and shows that (c, ci) E even (e0, e, ). For q we have Choosing p in such that c = fq(ep) e0 ci, now gives tjfl and the 1-translation (e0, eJ), y( e2+fl, ( = tq

If

e2+fl, which implies that (e0, e,

(e0, e,,,

icr all J E

5.6

and o, b, o', b'

ccn(o, b)

119

fl ccn( o', b')

0

E

)

and (e2, e2+1)

A. 11(o, b) fl 11(o!, b')

implies

0.

I'I-tOOF. Suppose (c, ci) E con( b) fl con( o', b') for some c, ci E 1, c ci. Since (c, ci) E con(o, b), there exists (by Mal'cev) a sequence c = , e,,, = ci in I such that (Lj, i iii. As before let iii, = /Je0, e, ), and choose p ii such that c' = ,fq ( e2 = ci'. By composition of translations (c', ci') q E .fq ( 11( (I, b). with Also (c', ci') E con( o', b'), since con( o', b') identifies e, and hence all elements of the form ej, e, with ej, e0) = Again there exists a sequence = , = ci' with iii'. From Lemma 5.5 we obtain an E 11(o', b') for i index p iii' such that (c', ci') and (e, are 1- hounded and it follows via a composition of translations that (o, b) and (o', b') are bounded. .

)

)

.

)

.

.

. .

.

Recall from Section 1.2 that an alge bra I is finitely su bdirectlv irreducible if the Con( I) is not the meet of fini telv many non—0 congruences, and that denotes the class of all fini telv su bdirectlv irreducible mem bers of V.

o

E

LEMM;\ 5.7 Let C be an elementary subclass is elementary,

Ihen

of

)

is finite

if and unIv if

have onl finitely nianv basic operations, so there I'I-tOOF. assumption algebras iii exists a first order formula I satisfies (o, b, o', b') y, y') such that for all I E if and only if (o, b) and (o', b') are k- bounded. Suppose = I hen an algebra I E C is finitely su bdirectlv irreduci ble itt it satisfies the sentence for all fj, £', f" or £' or Hence is elementary. Conversely, suppose is y y' y, y'). = = an elementary class. Lemma .5.6 implies that I E itt I satisfies the negation of for each k E w. So is also elementary and hence (by I heorem .5.2) strictly ttt tint U finiti m nt in i ' it iii in fact have A E itt A satisfies for just one particular k (the largest). It follows that all algebras in satisfy Uk, whence = )

5.8 I'I-tOOF. Let 11(

=

and b0, c, ci E I, image I' of I with c' I'). By assumption fl

a, v) E a, p

then 11(V)

=

=

v.

K

2.

and suppose (c, ci) E con(o0, b0 for sonic I E V ci. I hen there exists a su hdirectly irreducible epimorphic and hence and (primes denote images in and are k— bounded, i.e. for some distinct a, v E I', )

c

ci'

)

= 0,1 choose E A such that ( Such elements exist since if f' is a k-translation in I' with f'( of.) = For

i

a

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120

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5.

AIIOXAL

= v, then we can construct a correspondiiig k—translation f in I b replacing each fixed element of I' by one of its preimages in I and we let = and = ,f( bfl. = v*. choose j ii such that = a0, UL U0, I his is possi ble since in I', a, a, v) and a, v, v) must he distinct for sonic j ii, else and f'(

)

(I

a, a, v) = H( a, a, v) = H( a, v, v) = /2( a, v, v) =

.

.

.

=

a, v, v) =

I he 1-translations

now show that (a*, and a0, E /j ( a0, ( a0, (a*, v*) and a0), a0, a0, v0). LemniaS.5 applied to the sequence a0, E ( ( v*) and (ao, v*) are 1implies that either (a*, p*) and ( a0) are 1-hounded or ( bounded. In either case ( a0, vo and ( are 2—bounded and therefore (o0, b0) and bfl are (k ± 2)-hounded. )

)

\\ ith the hell) of these four lemmas and I heorem 5.-I, we can now prove the following result: IHEOREM 5.9 (JOnsson [79]).

is a is finitely based,

11 V

V

c/is tn

1)11

ti ye

variety of

and

I'I-tOOF. By Lemma 5.7 R( by the above Lemma R( V) = ) = k for some k E w, and K with R(1) k—t-2. Since the condition R(1) K k±2. Let B he the class of all E can be expressed by a first—order formula, and since is strictly elementary, so is B. K Clearly 2, hence Lemma implies that is elementary. By assumption 5.7 ) flip = ill ni' nt in ondutti Ih m 5 -I ith C = lip that V is finitely based.

J

Assuming that V is a finitely generated congruence distributive variety, Corollary 1.7 implies that up to isomorphism is a finite set of finite algebras. Since such a collection is always strictly elementary, one o htains Baker's result from the preceding theorem: IHEOREM 5.10 (Baker [77]). 1! V is a finitely generated congruence distributive variety of algebras then V is finitely based,

5.3

Joins of finitely I)ased varieties

In this section we first give an example which shows that the join of two finitely based modular varieties need not be finitely based.

(Baker [77']). IL eiv exist finitely based modular varieties that the complement of V ± V' is not closed under ultraproducts. LEMMA 5.11

V

and V' such

JJ be the modular lattice of Figure 5.1 (i and let N( JJ be the class of all lattices that do not contain a subset order—isomorphic to JJ regarded as a partially ordered set). By Lemma 3.10 (H) JJ N(JJ ), so there exists an identity s E Id N(JJ that does not hold in JJ. Let V he the variety of modular lattices that satisfy s (i.e. V = Mod{s, s,4, where s, is the modular identity) and let V' he the variety of all the dual lattices of mem hers in V. Since the modular variety is self-dual, V' is defined by s, and the dual identity s' of s. Hence V and V' are both fini telv based. Let K,, he the lattice of Figure 5.1 (ii). I'I-tOOF. Let

)

)

5,3,JOLVS OF FLVJ TELl BASLI) \JIUEIIES

(i)

121

(ii

Figure

.5.1

is subdirectlv irreducible ± V. \ote that (in fact siniple), and since K,, contains a copy of JJ and its dual as su blat tices, both s and g' fail in Hence K,, V IJ V', and the claim follows from I heoreni 2.3 (i). \ow let K = K,, and choose any nonprincipal ultrafilter U over w. e show that the ultraproduct K = K/c71 is in V V. Xotice that an order—isomorphic copy of JJ is situated only at the bottom of each I his fact can be expressed as a first— order sentence and, by I heorem .5.1, also holds in K. Similarly, the dual of JJ can only be situated at the top of K. I he local structure of the middle portion of K,, can also be descri bed by a first order sentence, whence K looks like an infinite version of Interpreting Figure .5.1 (ii) as a diagram of K, we see that JJ is not order-isomorphic to ally subset of A7/con(c, il) since con(c, ci) collapses the only copy of JJ in K. Consequently K/con(c, ci) E V and by a dual argument K/con(i, b) E V. U bserve that K is not a simple lattice since principal congruences can only identify quotients reachable by finite sequences of transpositions ( I heorem 1.11). In fact con( o, b) fl con( c, ci) = 0, and hence K call he embedded in K/con( b) x K/con( c, ci). I herefore K E V ± V'. e claim

that, for each

u E

w,

V

—t—

.

logether with I heorem

.5.2

IHEOREM 5.12 (Baker [77']). he finitely based,

and I heorem .5.3, the above lemma implies:

Ihe join of two finitely based (modular) varieties need not

In view of this theorem it is natural to look for sufficient conditions under which the join of two finitely based varieties is finitely based. In what follows, we shall assume that is a congruence distributive variety, and that V and V' are two subvarieties defined relative to by the identities p = q and p' = q' respectively. By all elenuentary result of lattice theory, any finite set of lattice identities is equivalent to a single identity (relative to the class of all lattices, [GLT] p.28). Moreover Baker [74]

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5.

AIIOXAL

that this result extends to congruence distributive varieties iii general. Conse— quentlv, the above condition on the varieties V and V' is equivalent to them being finitely based relative to Vt.. I he next two lemmas are due to JOnsson [74], though the second one has been generalized to congruence distributive varieties. If p is an n—arv polynomial function ( = word or term with at most u variables) on an algebra I and , ., by p( a), a E E I then we will abbreviate p( thereby assunung that only the first u components of a are used to evaluate p. showed

.

LEMMA

.

.

.

5.13 An algebra

1

(t)

V

E

.

± V' if and only if for all a,

v

E

1W

con(p( a), q( a)) fl con(p'( v), q'( v)) = 0.

Y:{conu( a), q( a)) a E IW} and 0' = Y:{con(p'( a), q'( a)) a E IW}. By the (infinite) distributivity of Con(J) we have that (t) holds if and only if 0 fl 0' = 0. I his in turn is equivalent to I being a su bdirect product of 1/0 and 1/0'. Since 1/0 E V and I/O' E V', it follows that I E V V'. On the other hand JOnsson's Lemma implies that aiiy I E V ± V' can he written as a su bdirect product of two algebras I/o and I/o'. Xotice that 0 and 0' above are the smallest congruences On I for which I/O E V alId I/O' E V', hence 0 0 and 0' 1 o'. Since 0 fl 0' = 0 we conclude that 0 n 0' = 0. I'I-tOOF. Let 0 =

H—

Recall the notion of k- houndedness defined ill the previous section. It is au elementary property, so we call construct a first-order sentence such that all algebra I E satisfies if and only if for a, v E (p( a), q( a)) and (p'( v), q'( v)) are not k- hounded. LEMM;\ 5.14 V

H—

V' is finitely based

relative to

holds fbr some positive in u: P( u): kor any 1 E if 1 satisfies

if and only ii

then 1 satisfies

the following

fbr all

k

>

property

1.

I'I-tOOF. Firstly, we claim that, relative to Vt., the variety V V' is defined by the set of sentences S = a2, a3, .}. Indeed, by Lemma .5.6 we have that H—

.

.

con(p'( v), q'( v)) = 0 if and only if a)) fl v), q'( v)) = 0

con(p( a), q( a))

l1Jp( a), q(

fl

Leniniaö. 13 an alge bra I E belongs to V ± V' if and only if I for all k > 0. \\ e can make use of I heorem .5.3 to conclude that V V' will have a finite basis relative to if and only if it is defined, relative to V:, by a finite subset of 5, or equivalently by a single sentence since implies a, for all iii k. If P( u) holds, then clearly V ± V' is defined relative to by the sentence 011 the other hand, if P( u) fails, then there must exist an algebra 1 E such that 1 satisfies a, for sonic iii ;> ii. If this is true for any positive integer ii, then V ± V' callulot he finitely based relative to for all k satisfies

;>

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<$J<7$

then one obtains an equational basis for M7 = (3 vi w). To see this, note that e74 holds in a lattice of length 2 whenever two of the variables z0, z1,. . . , z74 are assigned to the same element or one of them is assigned to 0 or 1, but fails when they are assigned to vi + 1 distinct atoms. Therefore e74 holds in M7 and fails in M74+i. For M3 this basis may be simplified even further by observing that implies e, hence

= Mod{zo(zi +z2)(z2+z3)(z3+zi)

z0z1

+zoz•i+zozs.}

An equational basis for .V was found by McKenzie [72]. It is given by the identities 111:

z(p+u)(p+v)z(p+uv)+zu+zv z(p+u(z+v))=z(p+uz)+z(zp+uv)

McKenzie shows that 'A and hold in any lattice of width 2, whence .V ç Mod{qi, 'pj}, and then proves by direct computation that any identity which holds in .V is implied by 'A and Ipj. In view of Theorem 4.19 the second part may now also be verified by checking that either qi or üil in each of the lattices A4, L1, L2,. . . , L15 (see Figure 2.2). Theorem 5.17 implies that the variety M +X is finitely based (M is the variety of all modular lattices). Note that since .V is the only nonmodular variety that covers the distributive variety, M+ is the unique cover of M. Jónsson [77] derives the following equational basis for consisting of 8 identities: (i)

(ii) (iii)

((z+c)p+z)(z+z+a)=(z+a)p+z (z+c)pz+(p+a)c ((t+z)p+a)c=((ct+z)p+a)c+((bt+z)p+a)c

5.4.

(iv)

EQLtT1OPLtL BASES FOR SOME tCtitJEflES

127

((ct+z)p+a)c=(((ct+z)c+a+zp)p+a)c

(v) ((61 + z)p + a)c = ((61 + a)c+ zp)((6+ z)p + a)c and the duals of (ill), (iv) and (v), where a = pq + p?,b = q and c = p(q + rq). (Note that (ii) is the identity (MY) which forms part of the equational basis for the variety of all almost distributive lattices in Section 4.3.)

Varieties generated by lattices of bounded length or width. Let be the lattice variety generated by all lattices of length at most in and width at most vi (1 in, vi oo) which are defined similarly for modular and recall from Section 3.4 the varieties lattices. For in, vi c co all these varieties are finitely generated, hence finitely based (Theorem 5.10), and it would be interesting to find a finite equational basis for each of them. Apart from several trivial cases, and the case = M74, not much is known about these varieties. Nelson [68] showed that Vj° = .V (= for vi 3). With the help of Theorem 4.19, this follows from the observation that each of the lattices A4, L1, L2,. . . , L15 has width

>3. Baker [77] proves that Vt and are not finitely based, and the same holds for vi 5. The proofs are similar to the proof of Lemma 5.11. As mentioned at the end of Chapter 3 = M3, and by a result of Freese [77] is finitely based. Whether is finitely based is apparently still an unresolved question.

Chapter

6

Amalgamation in Lattice Varieties 6.1

Introduction

I he word amalgamation generally refers to a process of com hining or merging certail structures which have something in common, to form a larger or more complicated structure which incorporates all the individual features of its substructures. In the study of varieties, amalgamation, of course, has a very specific meaning, which is defined in the following section. 1 his leads to the formulation of the amalgamation property, which has been of interest for quite some time in several related areas of mathematics such as the theory of field extensions, universal algebra, model theory and theory. Amalgamations of groups were considered by Schreier [27] in the form of free products with amalgamated subgroup. Implicit in his work, and in the su bsequent investigations of B. H. Xeumann [54] and H. Xeumann [67], is the result that the variety of all groups has the amalgamation property. I he first definition of this property in a universal algebra setting can be found in 11raissé [54]. 1 he strong amalgamation property appears in JOnsson [56] and [60] among a list of properties used for the construction of universal (and homogeneous) models of various first order theories, including lattice theory. One of the results in the [56] is that the variety L of all lattices has the strong amalgamation property. Interesting applications of the amalgamation property to free products of algebras can be found in JOnsson [61], and Lakser [71] and [GLT]. I he property also plays a role in the theory of algebraic field extensions (JOnsson [62]) and can be related to the solvability of algebraic equations (Utile [76] ,[78] ,[79] ). However, it sooii became clear that not many of the better known varieties of algebras satisfy the amalgamation property. Counterexamples showing that it fails in the variety of all semigroups are given in Kimura [57] and Howie [62], and these can be used to construct counterexamples for the variety of all rings .As far as lattice varieties are concerned, it follows from Pierce [68] that the variety of all distributive lattices does have the ainalgaitiation property, hut JOnsson and Lakser [73] showed that this was not true for any nondistri butive modular su bvariety. Finally Day and Je2ek [84] completed the picture for lattice varieties, by showing that the amalgamation fails in every nondistri hutive proper su hvariety of L. A comprehensive survey of the amalgamation 128

6,2.

J'RELE\JJX hUES

129

I ff1

J,B,C,1

E

K

1/'

Figure 6.1

property and related concepts for a wide range of algebras can be found iii Kiss, Márki, Pröhle and I holen [83]. Because of all the negative results, investigations in this field are now focusing on the amalgamation class Amal(K of all amalgamation bases of K, which was first defined in (Jràtzer and Lakser [71]. A syntactic characterization, and sonic general facts about the structure of Amal(K ), K an elementary class, appear in Yasuhara [74]. Bergman [85] gives sufficient conditions for a mem ber of a residually small variety V of alge bras to be an amalgamation base of V, and JOnsson [90] showed that for finitely generated lattice varieties these conditions are also necessary and, nioreover, that it is effectively decidable whether or not a finite lattice is a mem her of the amalgamation class of such a variety. In Section 6.3 we present sonic of Bergman's results, and a generalization of J Onsson's results to residually small congruence distributive varieties whose mciii bers have one—element subalgebras (due to Jipsen and Rose [89]). In C JOnsson and Lakser [73] it is shown that the two—element chain does not belong to the amalgamation class of any finitely generated nondistri butive lattice variety, and that the amalgamation class of the variety of all modular lattices does not contain ally nontrivial distributive lattice. On the other hand Bernian [81] constructed a nonniodular variety V such that the two-elenuent chain is a nieni her of Amal( V). Lastly, whenever the anialganiation property fails in 5011K' variety V, then Amal( V) is a proper subclass of V, and it would he of interest to know what kind of class we are dealing with. In particular, is Amal( V an elenientarv class (i.e. can nienibership be characterized by some collection of first order sentences)? sing results of Albert and Burns [88], Bergman [89] showed that the anialganiation class of an fini telv generated nondistri butive modular variety is not elementary. In contrast Bruvns, \aturnian alId Rose [a] show that for the variety generated by the pentagon, the anialganiation class is elenientary, and is ill fact determined by Horn sentences. )

)

6.2

Preliminaries

The amalgamation class of a variety.

By a duigram ill a class K of algebras we mean a quintuple (J, f, B, g, C) with J, B, C E K and f I B, g C embeddings. An I), amaigamaliori in K of such a diagram is a triple (f', g', I)) with I) E V and B I) em beddings such that f'f = y'y (see Figure 6.1). C A amaigamaliori is an amalgamation th the additional property that

(B)

fl

g'(C) =

J

f(J).

r

CBAJ'TEJ

130

6.

UJALC;U11110X lx LXLIICE \JIUEIIES

J E K is called an amalgamation for K if every diagram (,1, f, B,y, C) can amalgamated iii K. I he class of all amalgamation bases for K is called the amalgamation of K, and is denoted by' JmaI(k ). K is said to have the amalgamation piiperty if every diagram can he (strongly) amalgamated in K. \\ e are interested mainly An algebra be

iii the case where K is a variety.

Some general results about Amal(V). \\ e summarize below some results, the first of which is due to C ràtzer, J Onsson and Lakser [73] and the others are from Yasuhara [74]. IHEOREM 6.1 Let V be a variety of

(i)

1ff

:

V). and fbr every y

1' E

1

J E Amal( V). Every J' E V can he urn bedded

:

1

C.f

in

and y can be

V. then

(ii) (iii)

is a

in some

J

E

Amal( V). with

K

±

propel class,

Ihe complement of Amal(V) is closed under reduced (v) if xJ'e V), and if I' has a one element su

(iv)

powers. then I E

1

),

In general we know very' little about the mem bers of Amal( V). lake for example = , the variety' of all modular lattices: as vet nobody has been able to construct a nontrivial amalgamation base for .\4 In fact, we do not even know whether Anial(.t1 has ally finite members except the trivial lattices. As we shall see below, the situation is somewhat better if we restrict ourselves to residually small varieties (defined below). V

.

Essential extensions and absolute retracts.

All extellsioll 11 of all algebra 1 is said to he if every llolltrivial collgruellce 011 11 restricts to a llolltri vial cOllgruellce 011 1. All em beddillg f I — 11 is all em bedding if 11 is all esselltial extellsioll of f( 1). Xotice that if I is (o, b)-irreduci ble (i.e. coll( a,h) is the smallest llolltrivial collgruellce 011 1) alld f I 11 is all esselltial embeddillg, thell 11 is (f(o ), f(b) )-irreduci ble. :

:

LEMM;\ 6.2

If/i

fbllowed

that

h

of

into 11/0.

1

I'I-tOOF. By

:

J

—

by' the

11 is any embedding. then there exists a 0 canonical epimorphism from 11 onto 11/0 is an essential

Zorn's Lemma we call choose

0

11

such

to he maximal with respect to llot

ally two mem hers of h( I). All alge bra

I

ill a variety'

V

is

—

retract of V if, for every' em beddillg f epimorphism (retractioll) y 11 1 such that the

1 11 with 11 E V, there exists composite yt is the idelltity map 011 A.

:

IHEOREM 6.3 (Bergmall [85]). Every absolute ret ract of a variety V is an amalgamation

of V.

retract of V alld let (A, 1, 11,y, C) he a diagram ill V. I hell there exist epimorphisms Ii alld k such that f/i = id,1 = lo amalgamate the I) by f'(b) = (b,yh(b)) and y' C I) diagram, we take I) = 11 x C and define f': 11 by y'(c) = (fk(c),c), then f'f(o) = (f(o),y(o)) = y'y(o) for all (I E A. I'I-tOOF. Suppose A is all absolute

:

6.2.

l'itELL%ILVti(JES

131

Recall that Vgj denotes the class of all subdirectly irreducible algebras of V and consider the following two subclasses (referred to as the class of all maximal inreducibles and all weakly maximal inr,Aucibles respectively): VMI VWMI

Ill

I

LEMMA

VMI

—

6.4 Al

= {Al = {Al

V51 : V51 :

Al has no proper extension in V51} Al has no proper essential extension in V51}.

VWMI. VWMI

if and only if Al

V51

and Al is an absolute retract in V.

f f':

Let Al VWMI and suppose : Al B V is an embedding. By Lemma 6.2, induces an essential embedding Al -. B/U for some U Ccn(B). Since Al has no proper essential extensions, must be an isomorphism, so the canonical epimorphism from B to B/U followed by the inverse of is the required retraction. The converse follows from the observation that an absolute retract of V cannot have a proper essential 0 extension in V.

f

f'

f'

Theorem 6.3 together with Lemma 6.4 implies that VWMI ç Ainal(V). Observe also that if V is a finitely generated congruence distributive variety, then Vgj is a finite set of finite algebras (Corollary 1.7), and so we can determine the members of VMI by inspection.

Residually varieties. A variety V is said to be residually small if the subdirectly irreducible members of V form, up to isomorpliism, a set, or equivalently, if there exists an upper bound on the cardinality of the subdirectly irreducible members of V. For example, any finitely generated congruence distributive variety is residually small. THEoREM 6.5 (Taylor [72]). iSV is

has an essential extension in

a residually small variety, then every member of Vgj

VWMI.

The union of a chain of essential extensions is again an essential extension,

so

we can apply Zorn's Lemma to the set Vgj (ordered by essential inclusion) to obtain maximal elements. Clearly these are all the elements of VWMI.

its 0

In fact Taylor [72] also proved the converse of the above theorem, but we won't make use of this result. Note that if V is a finitely generated congruence distributive variety then every member of V51 has an essential extension in VMI.

Igamations constructed from factors. The following lemma is valid in any class of algebras that is closed under products, and makes the problem of amalgamating a diagram somewhat more accessible. LEMMA 6.6 (Crätzer and Lakser [71]). S diagram (A,f,B,g,C) in a variety V can be v B there exists a 1) V and homomorphisms amalgamated if and only if for all u : B —' 1) and? : C? —' 1) such that and f'(u) f'(v), and the same holds = g'cj

f'

forU

ff

The condition is dearly necessary. To see that it is sufficient, we need only observe that the diagram can be amalgamated by the product of these generated as

uandvrunthroughalldistinctpairsofBandC?.

0

CiLtl'TEit 6. AMALGAMtTiO.N VI LATTICE tCt1UETIES

132

I

'B€V

1

Al

VWMI

Figure 6.2

6.3

Amal(V) for Residually Small Varieties

Property (Cfl.

An algebra

.1

in a variety

V

is said to have pivperly (C)) if for any

homomorphism h = yf. This property was used in (.irätzer and Lakser [71], Bergman [85] and Jónsson [90] to characterize amalgamation classes of certain varieties. THEoREM 6.7 (Bergman [85]). Let V be a residually small variety. iLL

(Q), then

.1

V

has property

Amal(V).

We use Lemma 6.6. Let (A, f, B,g, C?) be any diagram in V and let u v B. Choose a maximal congruence (1 on B such that .9 does not identify u and v. Then B/U V51 and hence by Theorem 6.5 B/U has an essential extension Al VWMI. Let —w be the canonical homomorphism B B/U, but considered as a map into Al. Clearly 7 f'(u) 7(v), and since .1 has property (Q), the homomorphism .1 —' Al can be C? —' Al such that extended to a = g'g. We argue similarly for 0 u v C?, hence Lemma 6.6 implies .1 Ainal(V).

ff

f'f:

We show that, for certain congruence distributive varieties of algebras (including all residually small lattice varieties), the converse of the above theorem also holds. We first prove two simple results.

and B be algvbras in a congruence distributive Let ha: B '-' AxB bean embedding such that ha(6) = (a, 6) for all 6 B. Then the projection irjg : 1 x B —' B is the only retraction of ha onto B. LEMMA 6.8 (Jipsen and Rose [89]). Let A

varietyV,a

Aandsuppose{a}isasubalgebraofi..

Suppose g

: .1

Lemma 1.3 there exist

—w B is a retraction, that is fl/ia is an identity map on B. By (Jon(A) and ô Con(B) such that for (z,p),(z',p') .1 x B

xB U

(z, p) ker g (z',p')

if and only if

xlix' and trW.

Since fl/ia is an identity map on B, ô must be a trivial congruence on B. To prove that it suffices to show that for any a' .1 we have aUa'. Suppose the contrary. First o = observe that for 6,6' B with 6 6' we always have

g(a,6) =

6

6' = g(a,6').

6.3. AMtL(V) 1"Oit 1LESII)LtLLY SMALL tCt1UEflES Now

if (a,a')

1)

for some a'

.1,

then there exist 6,6'

133

B such that

Thus g(a,6') = g(a',b) and so Lemma 1.3 implies aDa' and bob', a contradiction.

0

C0UNTERExiMPLE. The assumption that h = ha is a one-element subalgebra embedding cannot be dropped. Indeed, let 2 = {0, 1} be the two-element chain and consider a lattice embedding h :2 2 x 2 given by h(0) = (0,0) and h(1) = (1,1). Then both projections on 2 x 2 are retractions onto 2.

Coaoisuw

6.9 Let V be congruena distributive, A, B V, and suppose .1 has a oneC? clement subalgebra {a} and B is an absolute retract in V. ilk : .1 x B V is an embedding, then the projection irjg : .1 x B —' B can be extended to an epimorphism of C?

ontoB.

If ha : B -. .1 x B is an embedding as in Lemma 6.8, then k/ia is an embedding of B into C?. Since B is an absolute retract in V there is a retraction p of C? onto B. It 0 follows from Lemma 6.8 that pIAxB = irjg.

The characterization theorem. The following theorem is a generalization of a result of Jónsson [90]. There the result was proved for finitely generated lattice varieties. THEoREM 6.10 (Jipsen and Rose [89]). Let V be a residually small congruence distributive variety, .1 V and suppose .1 has a one-element subalgebra. Then the following conditions

are equivalent: (i) (II)

.1

satisfies property (Q);

A€Amal(V);

(ill) Let h: .1 —' Ar! VWMI be a homomorphism and k : .1 -. .1 x Al be an embedding given byk(a)= (a,h(a)) for aIla Ed. 1ff: d'-. B V is an essential embedding then the diagram (A, f, B, k,d x Al) can be amalgamated in V.

(i) implies (ii) by Theorem 6.7, and trivially (ii) implies (iii). Suppose (iii) holds. It follows from Lemma 6.2, that in order to prove (i), we may assume that the embedding f : .1 B is essential. Let h: .1 —' Al VWMI be any homomorphism, and define an embedding k : A A x Al by k(a) = (a,h(a)) for all a A. Notice that h = irMk where lrf4 is the projection from .1 x Al onto Al. By (iii) the diagram (1, 1' B, k, A x Al) has an amalgamation (C?, f', k') in V. It follows from Corollary 6.9 that there is a retraction p: C? —, Al such that h = p1/k = pf'f (see Figure 6.3). Letting g = pf' we have h = gf.0

a finitely generated congruence distributive variety, we have that each VMI (= the set of all maximal extensions in the finite set VWMI). Since members of VWMI are absolute retracts, we only have to test property (Q) for all homomorphisms h : A —' Al' VMI (this is how property (Q) is defined in Jónsson [90]). If A V is a finite algebra, then A has only finitely many nonisomorphic essential extensions B V and there are only finitely many possibilities for the homomorphisms h : A —' Al' VMI. In each case one can effectively determine if there exists a homomorphism g : B —' Al' such that vIA = h. Thus we obtain: In case

Al

VWMI

V is

is embedded in some Al'

CBAJ'TLJ

13-I

lx LA1IIE'E \JIUEIIES

6.

I >11

Pit

P

xi]

Figure 6.3

(:oRoLL;\RY 6.11 (JOnsson [90]). Let V be a finitely generated with a one—el ement variety, 11 1 E V is a finite

decidable whether or not

1

distrihu tn'e then it is etkicti ;'ely

V).

E

Property (P). \\ e

conclude this section by' stating without proof further interesting results from Bergman [85] and J Onsson [90]. For an algebra J in a variety V, we let be the direct product of all algebras J/O with 0 E Con( 1) and 1/0 E flAmal( V), and we let he the canonical homomorphism of J into An algebra 1 in a variety V is said to have Invperiy (F) if is an em bedding of 1 into and for every homomorphism y I — 11 E Vuj there exists a homomorphism with h1i,1 = y. Ii: —

ii

IHEOREM 6.12 (Bergman [85]). For any finitely genera ted variety of modular lattices and 1 E V. we have ext ensile and has property' E Amal( V) if and only if 1 is

J

(I'). Bergman also showed that the above theorem holds for V = the smallest non— modular variety' (see JOnsson [90]), and that the reverse implication holds for all finitely' generated lattice varieties, hut it is not known whether the same is true for the forward i iii ph

ca ti on.

.1

IHEOREM 6.13 (JOnsson [90]) finite lattice L E .V to Amal(.Vj if and only if L is a subdirect power of X and L does not have the three element chain as a homomorphic

image. It is not known whether this theorem is also true for infinite mem hers of

6.4

Products of aI)solute retracts

It is shown in lavlor [73] that, in general, the product of absolute retracts is not an absolute retract even if V is a congruence distributive variety. I heorem 6.1-I however shows that absolute retracts are preserved under arbitrary products in a congruence distributive varieties, provided that every mem her of this variety has a one-element su halgebra. I he

6.5. LATTiCES

SRI) TIlE SMtL(LtMtTiOR I'itOI'Eitfl'

135

theorem is a generalization of a result of Jónsson [90], which states that if V is a finitely generated lattice variety then any product of members of VMI is an absolute retract in V.

THEoREM 6.14 (Jipsen and Rose [89]). Let V be a congruence distributive variety and

assume that every member of V has a onc-element subalgebra. Then every direct product of absolute retracts in V is an absolute retract in V. Suppose

f

.1

=

is a direct product of absolute retracts in V, and consider I, let : .1 —' be a projection. By Corollary 6.9 such that = !zjf. Consider a homomorphism

an embedding : .1 B V. For i there is a homomorphism hi : B —, h : B —, .1 given by the identity map.

i I.

Then irihf =

hif =

and so

hi: .1 —'

.1

is

0

Lattices and the Amalgamation Property

6.5

Given the characterization of Amal(V) (Theorem 6.10), some well known results about the amalgamation classes of finitely generated lattice varieties can be derived easily.

6.15 (Pierce [68]). The variety V of all distributive lattices has the amalgamation property. THEoREM

We show that property (Q) is satisfied for any A V. V51 = VWMI = DM1 = {2}, so let A, B V with embedding : A B and homomorphism h : A —' 2. If h(L) = {0} or h(L) = {1} then trivially there exists g : B —' 2 such that h = yf. On the other hand, if h is onto, then kerh splits A into an ideal and a filter F, say. Extend f(I) to the ideal = {b B : 6 a 1(1)] and f(.L") to the filter .1" = {b B : 6 a 1(1")]. Clearly PilE' = 0, hence by Zorn's Lemma and the distributivity of B, can be enlarged

f

I

I'

I'

toamaximalidealP,whichisalsodisjointfromP. Defineg:B—i2 byg(b)=Oif o

P, and g(O) =

1

otherwise. Then one easily checks that g is a homomorphism and

thath=gf.

0

In fact, one can show more generally that if V is any congruence distributive variety which is generated by a finite simple algebra that has no nontrivial subalgebra, then V has the amalgamation property. This result is essentially contained in Day [72].

l'br any nondistributive finitely generatcd lattice variety V we have 2 Amal(V) and consequently the amalgamation property hills in V. THEoREM 6.16

f

.i4 or R is a member of V. Let be a map that into a prime critical quotient of L = or L = IV, depending on which lattice is

Since V is nondistributive,

embeds 2

mV. and h(1) = such that h =

Nowitiseasy toseethat theredocs notexist ahomomorphismg : L

yf (see Figure 6.4).

—,

Al 0

Berman [81] showed that there exists a lattice variety V such that 2 Amal( V). In fact Berman considers the variety V = : vi w}t' (see Figure 2.2) and proves that all of its finitely generated subdirectly irreducible members are amalgamation bases.

£ has the strong amalgamation property. Next we would like to prove the result of Jónsson [56], that the variety £ of all lattices has the strong amalgamation property. Since £ is not residually small, we cannot make use of the results in the previous section.

CiLtl'TEit 6. AMALGAMtTiO.N Vi LATTICE tCt1UETLES

136

I

21

21

JV

h

h

Al

Al

Figure 6.4

We first consider a notion weaker than that of an amalgamation: Let L1 and L2 be two lattices with L' = Li fl L2 a sublattice of both Li and L2. A completion of L1 and L2 is a triple (f1,f4,L3) such that L3 is any lattice and L3 are embeddings (1 = 1,2) with filL' = filL'. Suppose L1 and L2 are members of some lattice variety V, and let us denote the inclusion map L' by Ji (1 = 1,2). Then dearly (f1,f4,L3) is an amalgamation of the diagram (L',J1, L1,J2, L2) in V if and only if L3 V. how to construct a completion of L1 and L2? This can be done in various ways, of which we consider two, namely the ideal completion and the filter completion. We begin by setting P = L1 U L2 and defining a partial order on P as follows: on L1 and L2, respectively), and if agrees with the existing order (which we denote by and z p L5 ({i,J} = {1,2}) then

p

p

ifandonlyif

z
(equivalently indeed

U

U

z€1

z<;p

and

—

z€L1flL2

It is straightforward to

U

a partial order on P. Define a subset (1) (2)

— forsome

1

zpz

and implies

verify that

is

of P to be a (Li, L2)-ideal of P if 1 satisfies imply

z€1

and

(1=1,2).

Let I(L1, i.'2) be the collection of all (Li, L2)-ideals of P together with the empty set. I(L1, L2) is dosed under arbitrary intersections, so it forms a complete lattice with 1. J = I fl J and 1+ J equal to the (L1,L2)-ideal generated by I U J for any I,J I(L1,L2). For each z

L1 U L2

the principal ideal (z] is in I(L1,L2), by

LEMMA

so we

can define the maps

(1= 1,2).

6.17 (1" f4,I(L1, L2)) is a completion (the ideal completion) of L1 and L2. Let

Z,7lpZ+jll.

z,p

(1 = 1 or 2). Then p) = (z On theotherhand wehavez,p€(z]+(p] so by

ii]

2 (z] +

(p] since

6,6.

U11L;U111J0x lx

,\1ODL LAB \JB1EIL1LS

137

C

Figure 6.5

hence

y) =

£

(

y). Similarly

£)

(

y) =

y) and clearly

tj

and

12

are

one-one, with fjjL =

I he notion of a (Li, L2 )-tiller and the tiller completion dually. As an easy consequence we now obtain:

112,

:T(

L2)) are defined

IHEOREM 6.18 (JOnsson [56]). L has the strong amalgamation property. I'I-tOOF. Let (J, f, B,y, C) he a diagrani in L. Since L is closed under taking isomorphic copies, we may assume that I is a su blat tice of B and C, and that I = B fl C with ,f and y as the corresponding inclusion maps. 1(11, C) E L, so the ideal completion 1(11, is in fact an amalgamation of the diagram. lo see that this is a strong C)) (fL 12, amalgamation we need only observe that if (B) fl C) then E 1. E U bserve that we could not have used the above approach to prove the amalgamation property for the variety P, since the ideal completion of two distributive lattices need not be distributive. Indeed, let B = J12( o, b) ( = 2 x 2 generated by o, b) and C = J12 (o, c) with G b = H- c and (lb = (IC (see Figure 6.5). I hen B U C = i13(o, b, c) is already a lattice, and therefore a su blat tice of I(B, C). However B U C is nondistri hutive, hence I(B, C) P. 1 he same holds for ally other lattice I) in which we nught try to amalgamate B and C, so it follows that P does not have the strong amalgamation property. H—

6.6

Amalganiat ion in nio dular varieties

No nondistributive modular variety has the amalgamation property. In our presentation of this result we follow C JOnsson and Lakser [73]. \\ e begin th a technical lemnia. LEMMA

6.19 Let

J,B,C,f)

V

be

a

variety of

that has the

property. and let

E

(i) 1ff) is an extension of C and f is any embedding of C into I). then there exists an extension F E V off) and an embedding y I) F such that = f. :

(ii) if B is an extension of 1. and Q is an au tomorphism of 1. then there exis ts an extension B E V of B and an au tomorphism a of B such that aLl =

CBAJ'TEJ

138

lx

6.

LAILIICE

=

B

8 =

B

\JIUEIIES

Figure 6.6

j

I'I-tOOF. (i) Let he the inclusion niap C — I) and consider the diagrani (C, 1, D,j, I)) which by assuniption has an anialganiation (11,9, li). I he result follows if we identify I)

with its isomorphic image f'( I)) in F. lo prove (ii), let = 1, = B, and consider with C = h0( = Q as an emheddiig of into e now apply part (i and — = to ohtaii an extension of and an embedding satisfying B? 1 we get a sequence of extensions h0( = Repeating this process for u = 2,3, — of I = and a sequence of em beddings Ii such that = Figure 6.6). .

)

.

.

.

.

.

.

e can now define

and

B =

then clearly a is an embedding of B into B. E B, then there is an u E w such that j E is an extension of = y, since

=

lo

see that

a

onto, choose any Putt = = tfl, then Hence ö is an automorphism of B, is also

)

and by construction ö I = IHEOREM 6.20 ( Griitzer, JOnsson and Lakser [73]). the variety .\4 of all modular hit tices does not have the

nondistrihu tive suhvaiiety of

property.

contrary that there exists a variety V such that P C and V has the amalgamation property. nder these assuniptions we will prove a nuni her of statenients about V that will eventually lead to a contradiction. I he proof does require the coordinatization theoreni of projective spaces (see Section 3.2). As a simple observation we have that i13 E V. Statement I: Every member of V can he embedded in a simple complemented lattice that also to V. Clearly ally lattice in V can he em bedded into sonic lattice L E V, which has a least and a greatest element, denoted by 0L and 'L respectively (e.g. we can take L to be the ideal we can enihed the three elenuent chain 3 = {O lattice). Given E L, 1 2} into L and into i13(o, b, c) such that I'I-tOOF. Let us assunie to the

1

2'

in i13(o,b,c).

By the amalgamation property, there exists a lattice L as a 0,1—sublattice (i.e. °L = (3L1 and 'L = 'L1 ) such that {OL, is contained ill a diamond sublattice of

lx

6,6.

LAB

\JBIEIIES

139

we obtaii a lattice Iterating this for each E L, £ E V that 0L' contains L as a sublattice, and for each E L £ 0L' 'L there exists y, z E such that we z, forni a diamond su blat tice. Repeating this process, o btain an infinite { sequence

of lattices in V, each with the sanie 0 and 1 as L and satisfving: for all u E w, all E if belongs to a dianiond {OL, y, z, IL} in Lettiiig 0L, 'L then = we have that is complemented. In a complemented lattice, a E V, and clearly congruence 0 is determined by the ideal OL/O, hence if 0 is a nontrivial congruence on then £OOL for sonic £ E Lx,, £ 0 collapses all of and implies belongs to a diamond {OL, u, z, 1JJ in so again 0 collapses all of since the diamond J13 is simple. Hence is a simple complemented lattice in V containing L as a su blat tice. Xote that since Hall and Dilworth [46] constructed a modular lattice that cannot be embedded iii any complemented modular lattice, this statement suffices to conclude that .\4 does not have the amalgamation property. Statement 2: icr every L E V there exists an infinite dimensional prajecti ;'e space P such that L can be embedded in L( and L( E V e may assume that L has a greatest and least element, and that it contains an infinite chain, for if it does not, then we adjoin an infinite chain above the greatest element of L and the resulting lattice is still a member of V. By Statement 1, L can be embedded in a simple complemented lattice C E V, and by I heorem 3•3, C can be 0,1—em bedded in sonic modular geometric lattice JJ E V. JJ can he represented as a product of modular geometric lattices (1 E I) which correspond to nondegenerate projective spaces in the sense that Let ñ denote the embedding of C into ii followed by the ith L( projection. Since preserves 0 and 1, it cannot map C onto a single element, hence by the simplicity of C, must he an embedding of C into is infinite L( ) .Also dimensional since C contains an infinite chain. Stat ement 3: 1 here exists a projecti;'e plane Q such that L( 0) E V and Q has at least six points on each line. By Statement 2, there exists an infinite dimensional nondegenerate projective space P such that L( belongs to V. If every line of P has at least six points, then we can take Q to he any projective plane in P, and L( Q) E V since L( Q) is a sublattice of L( If). If the lilies of P have less than six points, then by I heorem 3.5, L( is isomorphic to L( , I), where is an infinite dimensional vector space and I is a field with 2, 3 or elements. Let K be a finite field extension of I with K be a three dimensional 5, and let vector space over Ii. As ill Section 3.2, L( , K) deternunes a projective plane Q, such that L( , K) L( Q). Since I is em bedded in K, L( , K) is a sublattice of L( , I), and since is infilute dimensional, L( , I) is isomorphic to a sublattice of L( , I). It follows that L( Q) can he em bedded in L( If), and is therefore a mem her of V. By construction, every line of Q —I

I

I

I

I

I

I

has

at least K

H—

1

6

points.

\\ i th the help of these three statements and Lemma 6.19 we can produce the desired contradiction. Let Q he the projective plane ill the last statement. By Statement 2, there exists an infilute dimiensional nondegenerate projective space P such that L( E

Cii.tl'TEit 6. AMAL(LtMATIO.N VI LATTICE tCtitJ.ET1ES

140

V and £(Q) is isomorphic to a sublattice of £(P). By the coordinatization theorem (see Section 3.2), there exists a vector space V over a division ring 1) such that £(P) is isomorphic to £(V,D). So £(Q) is embedded in £(%;D) ç 8(V) (= the lattice of subgroups of the abelian group V), which implies that the arguesian identity holds in £(Q). This means C) can be coordinatized in the standard way by choosing any line I in C) and two distinct points a,, aa, on I. The division ring structure is then defined on the set K = I — {a4. here we require only the definition of the addition operation ([ULT] p.208): Choose two distinct points p and q of C) that are collinear with &, but are not on

I.

(iivenz,p€K,let (1) (2)

is independent of the choice of p and q, and is an abelian group. Any permutation of the points of I induces an automorphism of the quotient 1/0 ç (&j. £(Q). Since I has at least six points, we can find z,p K —{au} such that Let a be an automorphism of I/O that keeps z, p fixed and maps z p to a point z By Lemma 6.19 (ii) there exists an extension L of £(Q) such that a extends to an automorphism 3 of L. By Statement 2 and the same argument as above, there exists an abelian group .1 and an embedding f : .L 8(A). We claim that f(z p) is the subgroup of .1 which satisfies: The operation

a,

(3)

a€f(zi?

forsome 6€ f(z),

)

c€f(p)

wehave

and eb is the additive inverse of 6). Indeed, let u,v

a

be

as in (1),

f(u+v)= f(u)+f(v) {nøs : r f(u), So there exists .1 satisfying 8€ d€ 1(v)]). (= d€f(u) and aed€f(v). Since u z +p and v p+q, follows that there exist 6,c€ A such that and assumea€

it

and 6€ f(z), deb€f(p) c€f(p), aedec€f(q). a e 6 belongs to f(z p) + f(z) and f(v)+ 1(p), and since

that Similarly Conversely, assume the right hand side of (3) holds. Since such that

we have

d€J

d€f(p)

Now (1) implies

6+c f(u) and

and

aebec

Since f and

f'

agree on

and

f

is

p + q, there exists

aebeced€f(q). 1(v), and from (2) we get a

Also, in the above argument we can replace

f'(t)=f(i(t))

a.,

f by f' which we define by

forall t€A.

p, it follows

from

an embedding and z

(3)

z

that

p.

0

IL/IL DÀY

6,7.

The Day

6.7

Lii

JEZEK ILLILOJILU

Je2ek

Theorem

In this section we will prove the result of Day and Je2ek [84]: if V is a lattice variety that has the amalgamation property and con t ai is the tagon X then V in us t be the variety P and L are the L of all lattices. logether with the preceding result, this implies that only varieties that have the amalgamation property'. I he proof makes use of the result that L is generated by the class BF of all finite hounded lattices (see Section 2.2). Partial results iii this direction had previously been obtaiied by' Berman [81], who showed that if a variety has the amalgamation property and includes X, then it must also include L3, L6, L7, LH and

I,

The notion of J-decomposability of a finite lattice. Ihis concept was introduced by Slavik [83]. Let L he a finite lattice th and L2 proper su blat tices of L and L = (written L = L is said to he by (hUt Lv)) if whenever of (R. L3) is a completion of and L2, then f = U is an em bedding of L into L3. So in a sense 1( L2) is the smallest completion of and L2. In particular, if we let L2 is by' definition fl L2 into L1 (1 = 1. 2), then 1( j1 denote the inclusion map of embeddable into any amalgamation of the diagram L2 ). Hence if V fl is a variety' having the amalgamation property, and L2 E V, then 1( L2) E V. I his, together with the fact that J-decomposabilitv can be characterized by three easily verifiable conditions on and L2, makes it a very useful concept. For ally element z E L we define C f( z) to he the set of all covers of z, and C z) the set of all dual covers of z. )

LEMM;\ 6.21 (Day and Je2ek [84]). Let L = U L2 be a finite sublat tices of L. Ihen L is 1-decomposable by' means of and L2 also satisfy:

(1) £

L1 fl L2

(3)

z E

h fl L2

aiid,r Kg imply,r

I implies C(z) I implies

L =

H( z)

K z

or

L2) and let

1(

Kg for somezE

H(z)

I

(ft. 12' I( Lp Lv))

lattice with and L2

if

and L2 and only'

if

h

he the ideal completion of

and L2 (see Lemma 6.17). By' definition the map

I

=

.R

'J

h

:

L —

L2)

given by

f(s)

= (H

is a lattice embedding. Let t E = {i. 2}) and ,t K y. Ihen = g E L1 fl L2 such that (g] = f(g) ,hence £ Kp g, which implies that there exists a z E K z K g. I herefore (1) holds. Suppose to the contrary that (2) fails. I hen there with covers exists z E L2 dual ,t L2 and L2 Clearly z = fl g E g E so f( z) = ( z] = But IJ ( g] is already a (Li. L2 )-ideal, so we should have ( H. IJ (g] = ± (g]. I his is a contradiction, since (z] (H 'J (g]. Dually, the existence of the filter completion implies that (3) holds. Conversely', suppose that (1), (2) and (3) hold, and let ( f2. L3 ) be any completion J ,h is all em heddlillg of L into L3. Firstly, ,r of and L2. \\ e must show that f = y implies this follows from the fact that ñ is all ciii heddlillg, f(g), since if £, g E and if ,r E L1, g E L1 then by (1) there exists a z E fl L2 such that,r K z K g.

I

h

h

H—

—f—

Cii.tl'TEit 6. AMAL(LtMATIO.N VI LATTICE

142

tCti(JET1ES

z,p This shows fl L2, we must have z c z c p, giving f(z) c f(z) c that f is one-one and order preserving. To see that f is in fact a homomorphism, requires a bit more work. Since L2, p L2 — = is a homomorphism, we only have to consider z and show that f(z + p) = f(z) + 1(P) (f(zp) = f(z)f(p) follows by duality). We define two maps pj : .1, —, (1 = 1,2) by v u}. Note that the join is taken = E{v orderpreserving, in 1,, so Also, clearly is and is the identity map on pj = Define two increasing sequences of elements L1, L2 by z0 = z, y1, = p and Because

Z74+1

=

z74

+

P74+1

By induction one can easily see that

z

=

+p

P74

+ ,i2(z7j.

and f(z74)

+

= f(z) + 1(P)

foralln€w. Weshowthatforsomen=kwehavezh=Z+PorPh=Z+P,then f(z +

+ f(Ph) = f(z) + f(p) as required.

p) = f(Zh)

c z + p for all vi. Since 1, is finite, this implies that there exists a k such that Zh+1 = Zh and Ph+1 = Ph, so by definition 1'1(Ph) Zh and 142(Zh) Ph. We always have 1'1(Ph) and ZJ, hence ZJipJi. If ZJpJ 142(Zh) 142(Zh),141(Ph) Ph then ZhPh 1'1(Ph), so we have 1'1(Ph) = ZhPh Since Ph L2, condition (1) implies that there exists z such that 1'1(Ph) z fl Ph. But then z 1'1(Ph), so we L2. Similarly, if L2 then also get L1 n fl = ZhPh Zhph ZhPh = i'i(Pii) hence we actually have Suppose

z74, p74

142(Zh)

fl L2

=

ZhPh

=

1'1(Ph)

L1

fl

gives Ph-u = Ph + Zh, contrary to the initial Similarly L1 fl so there exist covers u,v Pu Ph ofzhph such that u Zh and v Ph. But 142(Zh) -< u Z, implies u L1 —L2 (else u and 1'1(Ph) -< v Ph implies v L1 — L2. Since this contradicts condition 0 (3), it follows that Zn = Z + p or Pu = z + p for some vi. however Zh

L1

assumption that

Two easy

else 142(Zh)

=

Zh which

of the above characterization are:

Coaoisuw 6.22 (i)

if L = A(L1,L2) and, for i = 1,2, sublattiw of L

(II)

is a

sublattiw of 14, which in turn is a proper

then L =

ifL=[a)U(b]forsomea,b€L

thenL=J([a),(b]).

£ is the only nonmodular variety that has the amalgamation property.We also need the following lemma about the lattice LU] constructed by Day [70] (see Section 2.2). LEMMA 6.23 Let I = u/v be a quotient in a Iattiw L, U Gon(L) and put J = (u/U)/(v/U). if I = U J, then L[I] is a sublattiw of the direct product of L and L/U[J].

Recall that if çb, *5 are two congruences on an algebra .1 such that çb fl *5 is the zero of (Jon(A), then .1 is a subdirect product of A/çb and i/o. So we need only define b and *5 on L[I] in such a way that L[I]/çb is a sublattice of L, L[I]/O is a sublattice of

U3

IL/IL 1)11— JEZEK ILLILORLU where L/0[.J], and fl c = 0. Let = ker L[l] then L[l]/ is of course isoniorphic to L. Define c by

if and oniy if

7(L)O7(g)

\\ i th this definition c is a congruence, since Moreover, it follows that I = U I i) E

I

X

{i}

implies implies

and Ii

L is the natural epitiorphisni,

or

t

{/}

(1

= 1,2).

is a homomorphism and 0 E Con( L ).

/c = /0 and 1)/c (L. = (L/O. 1)

(1

= 0,1)

L[l]/c

is a subset of L/0[.J]. By examining several cases of meets and joins iii L[l]/c, one sees that it is in fact a sublattice of L/0[.J]. c)tj. Ihen h(L ) = h(g) and j, g E L I or Suppose now that L, u E L[l] and follows we have L'flc = 0 as thatL = x {i} (1 = 0 or 1). In all cases it

whence

El

required. > 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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over 1, 6 prime, 5 principal, 5 proper, 5 finitely based, 13 relative to a variety, 13 finitely generated variety, 4 finitely subdirectly irreducible, 3 l')c(_V), 4

FL,5

Davey, B. A., 45, 77, 149, 150 Day, A., ix, 23, 25, 26, 38, 41, 42, 44, 55, 61, 77,88, 128, 135, 141, 142, 145, 150

Dean, It. A., 15, 150 Dedekind, It., ix, 18,46, 150 Desargues' Law, 51,53 Dcsargucsian projective space, 51 diagram, 129 diamond Al3, 17 Dilworth, It. P., x, 10, 50, 63, 139,

Frthsé, It., 128, 150 frame, 57 Frayne, I., 7, 116, 150 free algebra, 4 freely generated, 4 Ifrecse, It., v, ix, 17, 28, 33, 46, 47, 57, 59, 62, 63, 75, 112, 127, 150, 151 I!'rink, 0., 49, 52, 151 151, 3 fully invariant congruence, 14 I"unayama, N., 8, 151 Galois field, 56

generated compactly, 2 freely, 4 geometric lattice, 48 149—

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