Varieties Generated by Modular Lattices of Width Four Citation Freese, Ralph Stanley (1972) Varieties Generated by Modular Lattices of Width Four. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/W67C-JR90. https://resolver.caltech.edu/CaltechTHESIS:04082016-123408947 Abstract A variety (equational class) of lattices is said to be finitely based if there exists a finite set of identities defining the variety. Let M ∞ n denote the lattice variety generated by all modular lattices of width not exceeding n. M ∞ 1 and M ∞ 2 are both the class of all distributive lattices and consequently finitely based. B. Jónsson has shown that M ∞ 3 is also finitely based. On the other hand, K. Baker has shown that M ∞ n is not finitely based for 5 ≤ n ˂ ω. This thesis settles the finite basis problem for M ∞ 4 . M ∞ 4 is shown to be finitely based by proving the stronger result that there exist ten varieties which properly contain M ∞ 4 and such that any variety which properly contains M ∞ 4 contains one of these ten varieties. The methods developed also yield a characterization of sub-directly irreducible width four modular lattices. From this characterization further results are derived. It is shown that the free M ∞ 4 lattice with n generators is finite. A variety with exactly k covers is exhibited for all k ≥ 15. It is further shown that there are 2 Ӄo sub- varieties of M ∞ 4 . Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Dilworth, Robert P. Thesis Committee: Unknown, Unknown Defense Date: 13 December 1971 Funders: Funding Agency Grant Number NSF UNSPECIFIED Caltech UNSPECIFIED Ford Foundation UNSPECIFIED Record Number: CaltechTHESIS:04082016-123408947 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:04082016-123408947 DOI: 10.7907/W67C-JR90 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 9661 Collection: CaltechTHESIS Deposited By: INVALID USER Deposited On: 08 Apr 2016 20:23 Last Modified: 01 Jul 2024 17:11 Thesis Files Preview PDF - Final Version See Usage Policy. 29MB Repository Staff Only: item control page

VARIETIES GENERATED BY MODULAR LATTICES OF WIDTH FOUR Thesis by Ralph Stanley Freese In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1972 (Submitted December 13, 1971) • ii ACKNOWLEDGMENTS I am particularly indebted to my adviser, R. P. Dilworth, for his advice and encouragement. I also wish to thank Richard Dean, Kirby Baker, and Dang Hong for their helpful discussions with me. I am grateful to the National Science Foundation for its support throughout my graduate career, and to the California Institute of Technology for providing me with a teaching assistantship for the past three years. I am also indebted to the Ford Foundation for its support. Finally, I wish to thank my wife Anne for proofreading this thesis. iii ABSTRACT A variety (equational class) of lattices is said to be finitely based if there exists a finite set of identities defining the variety. '!Ti. 00 n Let denote the lattice variety generated by all modular lattices of width not exceeding n. m1 and '!Tl. 2 are both the class of all distributive 00 00 lattices and consequently finitely based. m3 is also finitely based. 00 On the other hand, K. Baker has shown that mn is not finitely based for 5 ~ n < w. 00 00 basis problem for '!Tl. • 4 '!Tl. B. Jonsson has shown that 00 4 This thesis settles the finite • is shown to be finitely based by proving the stronger result that there exist ten varieties which properly contain m: and such that any variety which properly contains m: contains one of these ten varieties. The methods developed also yield a characterization of subdirectly irreducible width four modular lattices. ization further results are derived. with n generators is finite. exhibited for all k :2: 15. varieties of m4 . 00 From this character- It is shown that the free '!Tl. 00 4 lattice A variety with exactly k covers is It is further shown that there are 2 ~ ° sub- iv TABLE OF CONTENTS Page ACKNOWLEDGMENTS • ii ABSTRACT • • • iii INTRODUCTION . 1 CHAPTER I Hong's Theorem • • • • • • • • • 6 II Some Useful Modular Lattices with Four Generators 30 III .............. The Fundamental Theorem on Weak Atomicity • . . • • • • • .... 45 . . . . . . . . 80 108 Applications . . . . . . . . . . . . . . v 127 REFERENCES . . . . . . . . . . . . . . . . . IV The Main Structure Theorem 1 IN TR OD UC TION A variety of lattices is a class of lattices which is closed with respect to the formation of sublattices, homomorphic images and A fundamental theorem of Birkhoff [4 J states that direct products. varieties of lattices are exactly those lattices defined by their identities. That is, a class C, of lattices is a variety if the class of lattices which satisfy all the identities satisfied by all the members of If C, is any class of lattice then the class of all sub- C, is the class C,. direct products of homomorphic images of sublattices of ultraproducts of members of C, is the smallest variety containing C, and is called the variety generated by C-. This theorem, which is due to Bjarni Jonsson [15], has made possible many advances in the theory of lattice varieties. Let '71m be the variety generated by all modular lattices n whose width does not exceed n and whose length does not exceed m, where n and m are cardinals. It follows from the finite nature of identities that the variety generated by the finitely generated members of a class C, is the same as the variety generated by C-. It follows from this that if n 1 and n 2 are infinite cardinals and m is any cardinal then m n1 nz = '71 nz and '71 = ?JI • m m infinite cardinal. The symbol o:i is used to replace any For example, the variety generated by all modular CXl lattices of width no t exceeding n, 1 s: n < w, is denoted by ?JI • n This CXl thesis makes a careful study of '71 . 4 A variety is finitely based if i t is defined by a finite set of identities, A basic problem in the theor y of modular varieties is to 2 determine the values of m and n for which 'mm is finitely based (Wille n [22 ]). R. McKenzie has shown that the variety generated by a finite lattice is finitely based D.8 ]. From this it follows that 'fl? m is finitely - based if both m and n are finite. 'm finite! y based for all n [2, 3 ]. K. Baker has shown that '111. n00 is 00 1 and 'm 00 2 are both equal to the variety of all distributive lattices and thus are finitely based. shown that '111. ~ is finitely based [ 16]. has shown that '111. CXl n B. Jc5nsson has On the other hand K. Baker [2 J is not finitely based for 5 ~ n < co variety for which the above problem is not solved. pletes the solution by showing that '111. by showing that '111. co 4 n co 4 '111. CIO • is the only 4 This thesis com- is finitely based. This is done is covered by ten varieties and that any variety properly containing '111. ~ contains one of these ten varieties. It follows from this result that an independent set of identities which defines 00 '111. 4 has ten or less elements and there exist sets of independent identities defining 'Ill_ 00 4 with n elements, n =1, 2, ••. , 1 O. A problem closely related to Wille 1 s problem but which appears to be more difficult is to determine which of the varieties '111. m n have the property that 'mm is covered by a finite set of varieties such that any n variety properly containing 'mm contains one of these covering n varieties. It is a classical theorem that the variety of all distributive lattices, which is equal to '111.~. 'fl?~ and '111.:, has this property. mentioned above this thesis shows that '111. CXl 4 has this property. '111.: have this property as was shown by B. Jo'nsson [16 ]. has shown that '111.:has this property [14]. Of course, variety of all modular lattices, has this property, and m3 and 00 D. X. Hong m:. '111. As 00 n , the 5 ~ n < oo 3 must fail to have this property. '17( m n' At the present time the question for 5 ::; n ::; cc and 4 ::; m < cc, remains unsettled. The techniques us ed to show that m4cc is finitely based are also used to characterize the subdirectly irre ducible members of 771 : . results of interest follow from this characterization. 2 ~0 cc subvarieties of 711. . 4 First, there are Sinc e there are countably many finite sets of identities the above implies that there exists a subvariety of ?71 which is not finitely based. 711.: are locally finite. Two cc 4 Secondly, it is shown that all members of This fact has the corollary that the free 711.: lattice on a finite number of generators is finite (compare with Birkhoff's Problem 46 [4 ]) . lary that 711. cc 4 This local finiteness also has the coral- is generated by its finite members. This fact is known to be true for the variety of all lattice s (R. Dean [7 ]) , false for the variety of Desargian projective planes (K. Baker [l ]) , and unsolved for the variety of all modular lattices. The proofs of the above results depend heavily on the development of a detailed structure theory for modular lattices. techniques are employed. Two basic First, the classical result that a modular lattice which is generated by three elements is finite is applied several times in order to obtain some of the local structure of modular lattices. In order to piece these bits of local structure together to obtain an overall picture of the lattice a second technique , the theory of projectivities, is employed. In [8 J and [9 J Dilworth showed that there is a strong connection between the structure of a lattice and the notion of projectivity. 4 R. Thrall [21 J showed that two projective quotients in a modular lattice could be connected by a sequence of transposes of a standard form (for definitions see Chapter I). G. Grfitzer called such a sequence normal and applied it to the study of lattice varieties [13 J. B. Jonsson defined the concept of a strongly normal sequence and showed that in most cases projective quotients in a modular lattice have subquotients connected by a strongly normal sequence. this concept to solve the finite basis problem for 'm~ and He employs m:. The lattice generated by the six endpoints of three consecutive quotients in a sequence of transposes is in fact generated by three elements and thus a homomorphic image of the free modular lattice on three generators which has 28 elements. For a normal sequence the lattice generated by the endpoints of three consecutive quotients is a homomorphic image of a lattice with 15 elements. normal sequence this number is reduced to 10. For a strongly D. X. Hong further develops the theory of projectivity by showing how these various lattices generated by three consecutive quotients can fit together. Chapter I of this thesis proves a slight extension of Hong's theorem. Chapter II studies the structure of a modular lattice gen- erated by four elements satisfying certain relations. It is shown that any such lattice contains as a sublattice one of three speeific lattices. Chapter III applies the result of Chapters I and II to prove that a modular subdirectly irreducible lattice is weakly atomic if it does not have any of the lattices A , A , ... , A diagramed in Chapter III as a 2 3 10 homomorphic imag e of a sublattice. Chapter IV applies the first three 5 chapters to prove that a modular subdirectly irreducible lattice, which does not have any of A , •.• , A as a homomorphic image of a sub2 10 lattice, has width not exceeding four. Chapter V applies this result to derive the applications mentioned above. General references to lattice theory are [2] and [6 ], to universal algebra [5 ], [12 ], and [19 ], and to the theory of varieties [20 ]. 6 CHAPTER I HONG'S THEOREM We begin with several definitions. Let L be a modular lattice. An ordered pair (a, b) in L X L with. b ~ a will be called a quotient of L. Instead of (a, b) we shall write b/a for this quotient. We shall use the term quotient and the symbol b/a to denote the sublattice of L consisting of the elements in the set (x E LI a s: x ~ b }. sometimes be referred to as a quotient sublattice. is called a nontrivial quotient if b >a. a ~ e ~ f s: b. This will The quotient b/a f/e is a subquotient of b/a if If b/a and d/c are quotients in L we write b/a /d/c and we say that b/a transposes up to d/c if a;: b /\ c and d =b V c. In this situation we also say that d/c transposes down to b/a, written d/c ""-.,.b/a. We also say that b/a and d/c are transposes. The quotient b/a is said to be projective to d/c in n steps if there exists a sequence of quotients b/a = b /a , b /a 1 , ••. , bn/an = d/c 0 0 1 such that bk/ak and bk+l /ak+l are transposes, k =0, 1, ••. , n-1. Much of the following notation is taken from [14 J and [16 ]. The projective distance between b/a and d/c, written p. d. (b/a, d/c), is the smallest integer n such that there are nontrivial subquotients b /a 1 1 of b/a and a /c of d/c which are projective in n steps. 1 1 integer exists then we write p, d, (b/a, d/c) = If no such CQ. A sequenc e of transposes b /a , b /a ,.,. , bn/an is called 0 0 1 1 normal if the transposes alternate up and down and 7 bk-l /ak-l/ bk/ak ~bk+l /ak+l implies bk= bk-l V bk+l and bk-l /ak-l ~bk/ak/bk+l /ak+l implies ak = ak-l /\ ak+l' The sequence is called strongly normal if it is normal and bk-l /ak-l/ bk/ak ~bk+l /ak+l implies bk-l /\ bk+l ~ ak and bk-1 /ak-1 ~bk/ak./bk+l /ak+l implies ak-1 v ak+l ~bk. Suppose we have a sequence of transposes (1) in a modular lattice. b 3 Since a 0 =b 0 /\ a , b = b /\ a , a = b /\ a , 1 2 1 1 1 0 2 = b /\ a , •.• , the lattice L generated by a , b , a , bl, ... is gen1 3 0 0 1 2 erated by b , a , b , a ,. • . . 0 1 2 3 Thus L 1 is a homomorphic image of the free modular lattice on n generators, FM(n). little information concerning the structure of L This fact furnishes 1 when n > 3, since in this case, FM(n) is infinite. However, FM(3) is finite and has only a few homomorphic images. Hence useful information on the structure of L 1 can be obtained by consider ing consecutive sets of three quotients and determining the various ways in which the corresponding images of FM(3) can fit together. FM(3) and some of its homomorphic images are exhibited below. It follows immediately from the definition of normal sequence that the endpoints of consecutive quotients generate a lattice which is a homomorph i c imag e of G 2 (Fig. 1. 2) or its dual. If the sequence is strongly normal then the endpoints of three consecutive quotie nts ge ne rate a homomorphic image of G 3 or its dual. More spe cific a lly 8 Figure 1. 1 9 x G 2 3 = FM(x , x , x ) / (x /\ x = x /\ x = x /\ x ) 1 3 3 2 2 1 1 2 3 Figure 1. 2 G 3 = FM(xl 'x2' X3) I (xl /\ Xz = xl /\ X3 = Xz /\ X3; xl v X3 :i? Xz) Figure 1. 3 10 if the sequence bk-l /ak-l ~ bk/ak / bk+l /ak+l is strongly normal then the lattice it generates is a homomorphic image of ak Figure 1. 4 We denote the five element modular non-distributive lattice by M ; M with an addition atom is called M , etc. 4 3 3 z x v Figure 1. 5 We call nn ordered five-tuple (v, x, y, z, u) of elements frorn a modular lattice a diamond if these elements form a copy of M 3 with v 11 and u as the bottom and top elements, respectively. Any nonidentity permutation of x, y and z yields a diamond, which by definition is distinct from the original diamond, even though they represent the same sublattice of L. We see from Fig. 1. 4 that if bk-l /ak-l is a nontrivial quotient then the figure contains a nontrivial diamond. More specifically, if bk-l /ak_ 1":. bk/ak/bk+l /ak+l is part of a strongly normal sequence we let Dk = (vk, xk' yk, zk' ~) = (ak' ak-1 /\ bk+l' bk, ak+l /\ bk-1, bk-1 /\ bk+l) and if bk-l/ak-1/bk/ak~bk+l/ak+l' Dk= (vk,xk,yk' zk' ~) = (ak-l V ak+l' bk-l V ak+l' ak, bk+l V ak-l' bk). In this way a strongly normal sequence b /a , b /a , ••• , bn/an of n+ 1 quotients 0 0 1 1 generates a sequence of n - 1 diamonds D , D , .•• , D n-l which is 2 1 called the associated sequence of diamonds. The remainder of this chapter will be devoted to the proof of a theorem which extends slightly a result of D. X. Hong on the structure of the lattice generated by two consecutive diamonds in an associated sequence. In order to state the theorem concisely the fol- lowing notation will be used. The diamond D said to translate up to the diamond D 2 1 = (v , x , y , z , u ) is 1 1 1 1 1 = (v , x , y , z , u ) if one of the 2 2 2 2 2 quotients u /x , u /y , u /z transposes up to one of the quotients 1 1 1 1 1 1 x /v , y /v , z /v • 2 2 2 2 2 2 The notation D/D 1 <2) 2 is used when u /z transposes up to x /v and 1 1 2 2 Dl ~ Dz (2) 12 is used when z /v transposes down to u /x . 1 1 2 2 ~to D 2 if u /v -........,.u /v 1 1 2 2 and if x 2 D 1 is said to transpose = u 2 /\ x 1 , y 2 = u 2 /\ y 1 and The notation means that D 1 transposes down to D • 2 If D = (v, x, y, z, u) is a diamond then D* is defined to be the diamond (v, z, x, y, u). x 1 /\ u So D 'JlD[ means u /v -.......u /v and 1 1 2 2 1 = z , y A u = x and z /\ u = y • 2 1 2 2 2 1 2 2 The theorem mentioned above can then be formulated as follows. Theorem 1. 1 Let b/a and d/c be nontrivial quotients in a modular lattice L such that p. d. (b/a, d/c) = n, 2 < n <co, Then some nontrivial subquotients 'b/a and a/c of b/a and d/c can be connected by a strongly normal sequence of transposes b'/i' . . , b /a n n =b 0 /a 0 , b 1 /a 1 , .. = a/c such that the associated diamonds D , •.. , D satisfy: n- 1 1 * or Dk / '2> Dk+l (i) Dk / <l> Dk+l Dk~Dk:l or Dk~Dk+l if bk/ak ,/' bk+l /ak+l if and bk/ak~ bk+l /ak+l k= 1,2, ... ,n-2 (ii) / * If Dk <l>Dk+l or Dk~D:+l ~~ Dk= Dk+l then k = 2, .•. ,n-2. The proof of this theorem is a slight modification of Hong's proof. First we need Lemma 1. 2 (B. Jbnsson [16]). Let b/a and d/c be nontrivial quotients of a modular lattice L such that p. d. (b/a, d/c) = n, 2 < n < co , 13 Then (i) Any normal sequence of n transposes from b/a to d/c is also strongly normal. There exist nontrivial subquotients 'b/a and d./c of b /a and (ii) d/c which can be connected by a strongly normal sequence of transposes. We give a sketch of the proof. A detailed proof appears in [16 ]. Suppose b /a = b 0 /a , b /a 1 , .•• , bn/an = d/c is a normal 0 1 sequence. Then, as mentioned above, the lattice generated by ak-l' bk-l' ak' bk' ak+l' bk+l is a homomorphic image of G 2 or its dual (Fig. 1. 2). former. Since n > 2 either k- 1 > 1 or k+ 1 < n. Assume the Then L contains the configuration pictured below: f --... bk-2 ck-2 ak-2 - ak+l F i g ur e 1. 6 14 It is easily checked that bk-l /ak-l '-.,. bk/ak/ bk+l /ak+l is strongly normal if and only if ck-l = bk_ 1 . be the image of ck-l in b/ai. But if ck-l :F. bk-l' let ci Then since bk_ 2 /ck_ /f/e""'-:. 2 bk+l /ck+l' we have p. d. (b/a, d/c) ~ n - 1, contrary to assumption. To prove (ii) we take a sequence of n transposes connecting subquotients of b/a and d/c, which we know exists by definition of projective distance. It is an easy matter to replace this sequence by a normal sequence (see [13 J or [21 ]), which, by (i), must be strongly normal. The following lemma characterizing direct product sublattices will be needed in the proof of Theorem 1. 1. If L The Direct Product Lemma. modular lattice L with greatest elements u element v such that u L 2 1 /\ u 2 and L 1 1 and u are sublattices of a 2 and a common least =v, then the lattice generated by L 1 and is isomorphic to the direct product of L Proof. 2 1 and L . 2 First we show that if a., b. EL., i = 1, 2, 1 1 1 For (a 1 V a ) /\ (b 2 1 Vb ) 2 = (a 1 V a 2 ) /\ (b 1 V b 2) /\ (b 1 V u 2) = ( [a 1 /\ (b 1 V u )JVa 2 2 ) /\ (b 1 Vb ) 2 = ( [a /\ u /\ (b V u ) J V a ) /\ (b V b ) 1 2 2 1 1 1 2 :: ( [a /\ (b 1 1 V (u 1 /\ u )) J V a ) /\ (b V b ) 2 2 2 1 = (( a 1 /\ b 1) v a 2 ) /\ (b v b ) 1 2 15 = (a 1 /\ b 1 ) V (a 2 /\ (b 1 vb 2 )) = (a 1 /\ b 1 ) V (a 2 /\ u 2 /\ (b 1 V b 2)) = (a 1 /\ b 1 ) V (a 2 /\ b 2) With the aid of this it is easy to show that (x , x ) .... x V x is 2 1 1 2 an isomorphism of L 1 X L 2 onto the sublattice generated by L For example, to show the map is one-to-one, let a Then u 1 /\ (a 1 1 V a 2 V a ) = u /\ (b V b ) which by (1) gives a 1 1 2 2 1 1 and L • 2 =b 1 V b 2 • = b 1 • Simi- larly az = b2. The proof of Theorem 1. 1 will be preceded by some lemmas which are more easily stated w i th the following notation. Let D = (v, x, y, z, u) be a diamond. We call u/x, u/y and u/z upper quotients of D and x/v, y Iv and z /v lower quotients of D. Suppose b/a is a subquotient of an upper or lower quotient of D, say z s: a s: b ~ _u. 1£ we assu1ne that z <a < b < u then the lattice gen- erated by a, b and D is isomorphic to the lattice diagramed below (see [l 6 J). This lattice has three new diamonds as sublattices. the upper-most diamond by Du/b' lowest diamond by D D u/b (1) I . a z We denote the middle one by Db/a and the More formally we have = ((x /\ b) V (y /\ b), x V (y /\ b), y V (x /\ b), b, u) Db /a = ((x /\ a) V (y /\ a), (x /\ b ) V (y /\ a) , (x /\ a) V (y /\ b), a/\ [(x /\ b) V (y /\ b) ], (x /\ b) V (y /\ b~ z /\ (x V (y /\ a), (x /\ a) V (y /\ a)) 16 u x v Figure 1. 7 With these equations the definitions of Du/b' Db/a and Da/z can be extended to include the possibilities u = b, b = a, or a = z. If u =b then the elements of Du/bare all the same; that is, Du/bis a single element. In this case Da/b is called a degenerate diamond. It should also be noted that this is the only way in which Du/b can be degenerate; that is, if u-:!- b the five elements of Du/bare distinct. Similar remarks apply for Db/a and Da/z" Similarly thr e e d i amonds (some of which may be possibly de generate) are obtained if b/a is a subquotient of any upper or lower 17 quotient of D. As an illustration note that if b /a is a subquotient of u/z then x /\b/x/\ a is a subquotient of x/v and z /\ (x V (y /\ b)) /z /\ (x V (y /\a)) is a subquotient z/v. It is easily checked that the diamonds Db/a' Dx /\ b/x/\ a' and Dz /\ (x V (y /\ b)) I z /\ (x V (y /\ a)) are the same. The next few lemmas are due to Hong [14 J. = (v,x,y,z,u) andD = (v',x',y',z',u') are diamonds in L with u = u', x ~ x', y ~ y', z ~ z' then D' = D I = u x Lemma 1.3. 1 IfD 1 D/ u y 1 =D/'" u z Proof: Du I z, Taking b = (( x /\ z 1 ) = u and z = z' in (1) gives V ( y /\ z 1) , x V ( y /\ z 1), y v (x /\ z '), z' , u) Now (x/\ z') V{y/\ z') = ((x/\ z') v0 /\ z' = ((x /\ x' /\ z') VYJ /\ z' = ((x /\ v') V 0/\ z' = ((x /\ x' /\ y') =((x /\ y') V VYJ /\ z 1 0/\ z' = (x V y) /\ y' /\ z' = u /\ v' x V (y /\ z') = x V (x /\ z') V (y /\ z') = x V v' = x V (x' /\ y') =v' 18 =x 1 /\ (x V y') = x' /\ u = x' Similarly y V (x /\ z') = y'. So D' = D u I z ,. The other statements in the lemma follow by symmetry. Corollary 1. 4. Let (1) be a strongly normal sequence with associated diamond D. Let c. Eb. /a., i = k-1, k, k+l, be images of one another under the given 1 1 1 transpositions. Let bk-l /ck-l and bk+l /ck+l be quotients such that and (3) is strongly normal. Then the diamond associated with (3) is Db = D u I ck . I k ck Proof: It is easily checked that the diamonds associated with (1) and (3) satisfy the hypothesis of Lemma 1. 3. follows. Corollary 1. 5. (1) Let The corollary readily 19 be a strongly normal sequence in L with associated diamond D. Let c. Eb. /a., i = k-1, k, k+l, be images of one another under the 1 1 1 given transpositions. Then is strongly normal with assoCiated diamond Db Proof: = D I Y ck . The strong normality of ( 1) easily implies the strong normality of (2). D I k ck The diamonds associated with (1) and (2) are = (ak, ak-1 /\ bk+l' bk, bk-1 /\ ak+l' bk-1 /\ bk+l) D' = (ck-1 /\ ck+l' ck-1 /\ bk+l' bk V (ck-1 /\ ck+l)' bk-1 /\ ck+l' bk-1 /\ bk+lJ These satisfy the hypothesis of Lemma 1. 3, and thus D' = I , But bk/\ (bk-1 /\ ck+l) =bk .". ck+l = ck. bk-1 /\ bk+l bk-1 /\ ck+l. Thus by the remark preceding Lemma 1. 3, D' =Db /c D k k The following lemma of Hong is the key to the proof of Theorem 1. 1. Lemma 1. 6. Suppose is a strongly normal sequence such that 20 Then the associated diamonds, satisfy D1 ~* (1) 2 or else one of the following holds: There exists c , a s c < b such that if ci E bi /ai is the 0 0 0 0 (i) image of c 0 under the given transpositions, i = 1, 2, 3 then is a strongly normal sequence with associated diamonds (1D ) I 1 ul c1 = =(Dz)bz/cz with (Dl )bl /c1~Dz)bz/cz. (ii) There exists co, ao <co s:: bo, Ci i = 1, 2, 3, the images of (Dl )bl /q and (D2)yz/cz c 0 in b. /c. under the given transpositions such that 1 1 is a strongly normal sequence with associated diamonds (D ) I 1 cl al (D ) and (D ) = (D ) I with (D ) ~(D ) 2 cz / az 1 cl / Yl 2 cz vz 1 c1 / a1 <2> 2 cz / az • Proof: Note the following relations hold. (1) Ht>ncc either v v 2 1 v u 2 < u 1 or v 2 < u 2 /\ v 1 or else v 1 V u 2 = u 1 and = u 2 /\ v 1 . So we have three cases. = 21 (2) From (1) we see that this transposition maps y x 2 onto y 1 • If this transposition sends z then n ~D~, as asserted. 1 Note that y 1 So let xi 2 2 onto z 1 and onto x , i.e., if z vv = x 1 1 2 1 = z 2 V v 1 and suppose xl -:/:. x 1 • is a relative complement to both x 1 and xi in u /v • 1 1 Thus x 1 and xl are incomparable. Note that and u /\ b = u . 2 3 1 It follows easily from the Direct Product Lemma that the lattice generated by u , xl, u , z , b , a is an eight-element 2 2 3 1 3 Boolean algebra. Consequently Now it is easily checked that b 0 /b (5) 0 /\ xl / x 1 /x 1 /\ xl / x 1 V xl V a Since x 3 x 1 v xl /xl / /xi V a 3 "--.. b 3 /\ (x 1 V xl V a 3 ) /a 3 and xl are incomparable these quotients must be nontrivial. 1 Thus we have p. d. (b /a , b /a ) s 2, a contradiction. 0 0 3 3 Case 2: v 1 V u 2 < u 1 • Let w = v 1 V u 2 and c 1 = y 1 V{x 1 /\ w) and let c. E b./a., i = 0, 2, 3 be the images of c l positions. l l Consider 1 under the given trans- 22 This is clearly a normal sequence and so by Lemma 1. 2 it is a strongly normal sequence. By Corollary 1. 5 the associated diamonds are x 1 V (y 1 /\ w), c 1 , w, u ) 1 (D 2 )y /c = (<c 1 /\ c 3 , c 2 vx 2 , h 2 V(c 1 /\ c), c 2 vz 2 , u 2 ) 2 2 Now u2 s: w s: (xl v u2) /\ (yl v uz) and Xz s: Y1 so that (x1 /\ w) V (y 1 /\ w) V u 2 = Gx 1 V u 2 ) /\ "1 V (cy1 V u 2 ) /\ w) =w [<xi Aw) v (y 1Aw)J AUz = [(<x1 Aw) v y 1) AwJ AUz = (<x 1 /\ w) V y 1) /\ u 2 = (cxl /\ w) VY1) /\ (xz VYz) = x 2 v (<x1 /\ w) v y 1) /\ Yz = x 2 V ( c 1 /\ b 2 ) Thus w/(x /\ w) V (y /\ w) .........,.u /c V x and thus (i) holds. 2 1 1 2 2 Case 3: v 2 <v 1 /\ u • 2 If we reverse the order of the reference of b. /a . and apply the dual of Case 2, we get the third alternative of the l 1 l e mma. Lemma 1. 7. Suppose D . = (v , x . , y., z., u .), i l 1 l l l l diamonds in L such that either = 1, 2 are two 23 (1) D /n*2 1 (l) or (2) Let c E y /v and let c = c V Yz be its image in u /y • 1 1 1 2 2 2 1 (i) 1 1 (D 1 )y /c ~(D~)uz/cz Then if (1) holds (D 1 )y /c ,...(;(D 2 )uz/cz if (2) holds. 1 1 Furthermore, if n 1 = D~ then (D 1 )y /cl= (D~)u /c 1 (ii) 2 2• Proof: Let us suppose that (2) holds. Hence Now y 1 /\ (u /\ c ) = c /\ y 1 2 2 1 = c 1 and thus by the remark preceding Lemma 1. 3 (D ) I /\ = (D 1 ) I . 1 u1 u1 cz Yl cl Let us suppose (1) holds. x 1 Vv 2 = z . 2 Then y This gives conclusion (i). 1 Vv 2 = x 2 • z 1 V v 2 = Yz and Hence ( 3) Recall that (Dz)u2/c2 = (<xz /\ c2) v(z2 /\ c2), xz v(z2 /\ c2). c2, Zz V (x2 /\ c 2 ), u 2) 24 Now by (3) u 1 A [(x 2 Ac ) v 2 (z 2 A c )] 2 = u 1 A c 2 A (z 2 V (x 2 A c 2)) = u 1 A c 2 A (zz V (yl A c 2)) = u 1 A c z A ( x 1 V yl ) A ( z z V ( yl A c z~ = ul A Cz A (x1 v Y1 (z2 v (yl A Cz)~ = u 1 A c 2 A (x 1 v (y1 A c 2 ~ Also u 1 A [z 2 V (x 2 A c ) J = (x 2 = xl 1 Vy ) A 1 (z 2 V (y A 1 c 2~ v (Y1 A (zz v Y1 A c2)~ =x 1 V (y l A c 2) Similarly u 1 A Gz V 2 c 2 ~ = y1 V (x1 c 2 ). (z A A These calculations show that But we have already seen that (D ) I 1 u 1 u1 A cz holds. = (D 1 ) Yl I c . Hence (ii) 1 The last statement of the Lemma is obvious. One more lemma is needed. be a strongly normal sequence with associated diamonds D , D , and 1 2 D 3 and let p. d. (b /a , b /a ) 0 0 4 4 = 4. Then at least one of the relations 25 D .........._ D* 1 (l~ 2 (1) D /In* 2-{i, 3 fails to hold. Proof: Suppose both relations hold. Since u 1 /I. u 3 = b 1 /I. b 3 = u , it follows from the Direct Product Lemma that z , u , y , u , x , u 2 1 1 2 2 3 3 generates an eight element Boolean algebra. Whence But this clearly contradicts p.d. (b /a ,b /a ) 0 0 4 4 Proof of Theorem 1. 1: = 4. It will be convenient to make an induc- If n =3 property (ii) holds automatically and (i) follows from tion on n. Thus we may suppose that 3 < n = p. d. (b./a, d/c) and Lemma 1. 6. that the theorem holds for pairs of quotients of projective distance less than four. Since p. d. (b/a, d/c) = n, subquotients b(/a 0 of b/a and b' /a' of d/c exist which can be connected by a sequence of n transn n poses. 0 0 transposes to a quotient bl /al which can be Thus b /a connected by a sequence of n - 1 transposes (n - 1 arrows) to b' /a'. n n 0 0/ duality it will suffice to consider the case where b /a bl /al. By By the induction hypothesis there exist subquotients bl' /al' of bl /al and b /a n n of b' /a' which can be connected by a strongly normal sequence n n ( 1) which satisfies all the conditions of Theorem 1, 1. b /a n n (Note that 26 bl' /al•/ b /a would imply p. d. (b/a, d/c) ~ n - 1, a contradiction.) 2 2 Let bo = bo /\bi' and ao = bo /\ a1' and bl = bo v b2 and a 1 = ai' /\ b 1 • Then is normal and hence by Lemma 1. 2 strongly normal. Let D. = 1 (v.,x.,y.,z.,u.), i = 1,2, .•. ,n-1, be the diamonds associated with (2). 1 l l l l Then by Corollary 1. 4 D ,D , ••• ,Dn-l are the diamonds associated 2 3 with (1). Now we can apply Lemma 1. 6 to If D ~D~ then property (i) of Theorem 1. 1 holds. 1 By Lemma 1. 8 D 2 ~D~ cannot hold. So by our induction hypothesis D 2 ~n 3 and * * ./l'>l< if Dk"'G> Dk+l or Dk..........~Dk+l then Dk= Dk+l' k = 3, •.• , n-1. Thus (ii ) holds in this case. So we may now assume that either condition (i) or (ii) of Lemma 1. 6 applies. If condition (i) holds, then we get a sequence which can be normalized to (4) by l e tting cl< = ck-l /\ ck+l and bk = bk V cl< for the even and 0 < k < n and ck= ck and bk= bk otherwise. By Lemma 1. 2 the sequence (4) is 27 By Corollary 1. 4 and Corollary 1. 5 the diamonds strongly normal. associated with (4) are (D )b I , (D 2 )b I , ... , (D _ 1 )b I · 1 1 cl 2 c2 n n-1 cn-1 By Lemma 1. 6, (i) (D )b I ~(D )b I 1 1 Cl(2) 2 2 Cz . Applying Lemma 1. 7 to D , D , ••• , Dn-l we see that the rest of the diamonds associated with 2 3 (4) satisfy (i) of Theorem 1. 1. We may suppose that D otherwise (ii) holds by the induction assumptions. may be described as follows: 1 f 0 /e 0 / f 1 /e ~f 2 _.fun; since Thus the situation there is a strongly normal sequence 2 /e 2 / ••• fn/en, where fi =bi and ei = ci, i = 0, 1, •.. , n, and p. d. (f /e , f /e ) = n. . 0 0 n n Furthermore the associ- ated diamonds, which we again denote D. = (v.,x.,y.,z . ,u.), i = 1, .• 1 1 1 1 1 1 .. , n-1, satisfy property (i) of the theorem, D ~D , n ~n; and 1 2 2 property (ii) holds for i ~ 3. ul /\ v 3 = f 1 /\ f 3 /\ v 3 = u2 /\ v 3 = v 2 Thus by the Direct Product Lemma the lattice generated by the sublattices u /v and v /v is isomorphic to their direct product. 1 2 3 2 we obtain two new diamonds Dl = (vl, xl, yl, zl, up = (v 1 Vv 3 Hence , x 1 vv 3 , y 1 vv 3 , z 1 vv , u 1 vv 3 ) and DZ= (vz,x2.'Yz•zz,uz) = 3 ( v 2 V v 3' x2 V v 3' y 2 V v 3' z2 V v 3' u2 V v 3 ) = ( v 3' z3' x3' y 3' u3 ) = n*3 . See Fig. 1. 8. Cons i d e r the sequ enc e The following calculations show that (5) is a normal sequence, 28 v =ui £1 V 3 Figure 1. 8 and hence, by Lemma 1. 2, a strongly normal sequence. ( 6) = u'1 Since 29 (7) y'1 Clearly x 3 V£ 4 = £ . 3 The rest of the sequence is normal because the sequence (4) is normal. It is easily checked that the diamonds associated with (5) are Dl,DZ,D ,D 4 , .•. ,Dn-l" 3 DZ = D~ are satisfied. Furthermore, the relations Dl~DZ and Thus the sequence (5) satisfies properties (i) and (ii) of the theorem. A similar argument applies i£ (ii) of Lemma 1. 6 holds. the proof of the theorem is complete. Thus 30 CHAPTER II SOME USEFUL MODULAR LATTICES WITH FOUR GENERATORS In this chapter a theorem on modular lattices with four generators satisfying certain specific relations between the generators is proved. In addition, several corollaries are observed, which will be useful in Chapter III. Let M 4 and A 4 be the lattices diagramed in Figure 2. 1. Figure 2. 1 Theorem 2. 1. Let L be a modular lattice with four distinct generators a, b, c, d which satisfy ( 1) aVb=aVc=aVd=bVd=cVd=aVbVcVd ( 2) aAb=aAc=aAd=bAd=cAd=aAbAcAd Then either A 4 is isomorphic to a homomorphic image of a sublattice of Lor L has a aublattice L' which is isom orphic to M 4 and if u is the 31 greatest element of L' then, for one of the atoms x of L', u/x transposes up to a s ubquotient of a V d/ d. The hypotheses of the theorem just says that any pair of generators except possibly b and c join to the top element of Land any two except possibly b and c intersect to the bottom element of L. We say that an ordered four-tuple (x, y, z, w) satisfies Proof: property (P) if x, y, z and w satisfy (1) with x =a, y = b, z = c and w = d. The dual property, which is given by (2), is denoted (Pd). Let a d 1 = d /\ (b 0 0 = a, b v c ). 0 0 = b, c 0 0 = c, d 0 = d, a 1 = a 0 /\ (b Then (b , a , d , c ) satisfies (P). 0 1 1 0 Vc ) and 0 0 For example, bo val = bo v (ao /\ {bo v co)) = (bo v ao) /\ (bo v co) = bo v co = b c 0 0 v a v d 1 /\ (a 1 1 v c . 0 Now if we set b 1 = b 0 /\ (a 1 v d ) and c = 1 1 V d ) then as above (a , b , c , d ) satisfies (P). 1 1 1 1 1 Inductively we define d.+l = d.1 /\ (b.1 V c.) 1 1 a.+l = a.1 /\ {b.1 V c.) 1 1 {3) Thus we obtain four descending chains a c 0 ~c 0 ~ a 1 ~ a 2 ~ ••• , b 0 ~ b 1 ~ ... , and(b.,a.+ ,d. ,c.) 1 ~•.• , d 0 ~d 1 ~.•• , suchthat(a.,b.,c.,d.) 1 1 1 1 1 1 1 i+ 1 1 satisfy (P). Let e. = b. V c. and£. =a. V d.. 1 1 1 1 1 1 Then the lattice generated by ei' di V ai+l and di+l V ai is a (possibly degenerate) diamond with greatest element fi and least element fi+l • Indeed, since (a., b., c., d.) 1 1 1 1 32 has {P) we have a. V d. = c. V d . = f.. l l l l e.l V d.l V a.+l l Hence l = b. V c. V d. V a.+l =f. l l l 1 l and From (3) we have e. = b. V c. ~ d.+l" l l l l Hence and (ai+l V di) /\ (ai V di+l) =ai+l V (di /\ (ai V di+l ~ = ai+l V di+l V {di/\ ai) =£.+l V {d. /\a.) l But ai /\ di ~ a 0 /\ d 0 l 1 which is the least element of L by hypothesis. The remaining two calculations are similar. The lattice generated by fi+l' bi+l V ci, bi V ci+l is a homomorphic image of the lattice diagramed in Fig. 2. 2. The proof is exactly the same as in the previous case except that b. /\ c. is not nec1 l essarily the least element of L. Let us suppose that f 2 < e 1 < f 1 < f 1 V {b /\ c ) < e 0 0 0 <f . 0 Then the above agruments show that {f , a V d , a V d , e , f ) = D is 0 1 1 1 0 0 0 0 a nondegenerate diamond. As w a s seen in Chapter I the fact that 33 e. i+ 1 V(b.t\ c . ) 1 1 Figure 2. 2 f 1 <f 1 V {b 0 t\ c ) < e 0 0 implies that D 0 and £ V(b t\ c ) generate the 1 0 0 lattice diagramed in Fig. 2. 3. Figure 2 . 3 34 As remarked above, the elements -f , b V c , and b V c 0 1 1 1 0 generate a sublattice which is a. homomorphic image of the one diagramed in Fig. 2. 2. Furthermore, since e homomorphism must be an isomorphism. 1 < £ < £ V (b 1 1 0 /\ c ) this 0 Hence the sublattice gen- erated by f , b V c , and b V c is isomorphic to the lattice diagramed 1 1 0 1 0 in Fig. 2. 4. Figure 2. 4 As above, the sublattice generated by e , a V d , and a V d 2 1 2 1 1 is diagramed in Fig. 2. 5. f Figure 2. 5 35 With these facts it is easy to see that the sublattice L generated 1 by a 0 v d 1 , a 1 v d 0 , b 0 v c 1 , b 1 v c 0 , a 1 v dz, and az v d 1 is iso- morphic to the lattice diagramed in Fig. Z. 6. f Figure Z. 6 Now a = dz v (a 0 since a /\ d a 0 0 0 V dz V f 1 = a 0 V dz V a 1 V d 1 = a 0 V d 1 , and f 1 /\ (a 0 /\dz) /\ f ) = dz v (a /\ (a v d )) 1 1 0 1 1 = dz v (a 1 v (a 0 /\ d 1)) = dz v a 1 , is the least element of L. v d 1 /£ 1 • Similarly az V d /a 2 V a / 1 0 Hence a a 1 0 V dz /a 1 V dz / v a0 /f 1 • With these facts it is easy to show that the lattice Lz generated by L , a 1 0 V dz and 36 a 2 Vd 0 is isomorphic to the lattice diagramed in Fig. 2. 7. Figure 2. 7 Now if f 0 > e 0 > f 1 > e 1 > f 2 but e 1 = e 1 V (b 0 A c ) then Fig. 2. 7 0 sugget5ts, and arguments similar to those above, prove that the lattice L 3 sub~ generated by L , a V d and a V d is isomorphic to the 0 2 0 2 1 lattice diagramed by Fig. 2. 8. In Fig. 2. 8 note that L is isomorphic to A . 4 Hence A lattice of Lin thes e c ases. 3 4 is a homomorphic image of L 2 and L is a homomorphic image of a sub- 3 37 f L 3 -- Figure 2. 8 For the remaining cases we have f 0 ::?: e 0 ~ f 1 ~e 1 ::?: f 2 and we know that there is at least one equality. It follows immediately from the definitions of these elements that any equality implies e 1 =f 2 • It has already been shown that (f , a V d , a V d , e , f ) 2 2 1 2 1 1 1 forms a diamond, and since e a 1 Vd 1 = b 1 Vc . 1 1 = f , it follows that f = e . 2 1 1 But then This, together with the fact that (a , b , c , d ) 1 1 1 1 satisfies (P), shows that any two elements of (a , b , c , d } join to f • 1 1 1 1 1 We must show that a , b , e , d are distinct. 1 1 1 1 If 0 is the bottom element of L, we note that Now if any two of [a , b , c , d } ar e even comparable, say a ~ b , 1 1 1 1 1 1 38 (5) v ho = al v bl v bo = al v bo fl = ( ao /\ {bo v co)) v bo = (ao v bo) /\ (bo v co) = eo It follows that f /b / e 0 e /b . 0 0 1 1 Since f 1 =bl, e 0 = b . 0 Similarly = c , a contradiction to a , b , c , d being distinct. 0 0 0 0 0 We conclude that a , b , c , d are distinct. 1 1 1 1 As we have pointed out a , b , c , d satisfy (P) and hence 1 1 1 1 equation (1). d 1 Since a also satisfy (2). ~ a 1 0 , b 1 ~ b 0 , c 1 ~ c 0 and d 1 ~ d 0 , a , b , c , 1 1 1 So the same procedure can be applied to the dual of the lattice generated by a , b , e , d • 1 1 1 1 As above either A homomorphic image of a sublattice of L or there exists al bl ~ b /\ dl. 1 , ci ~ c 1 and di But since al wise join to f . 1 ~ a ~ d 1 ~ a is a 4 1 , which pairwise intersect to al/\ hi/\ cl 1 etc., we also have that al, hi, ci, di pair- Hence the lattice generated by al, bi, ci and dl is isomorphic to M 4 . Moreover, fl /al~ fl /a 1 / fl Vd 0 /d 0 ~ a 0 V d 0 /d 0 and so the last statement of the theorem is also true. Let A 2. 10. A 9 7 and A 9 be the lattices diagramed in Fig. 2. 9 and Fig. is the lattice of subspaces of projective plane of order two. 39 Figure 2. 9 A9 Figure 2. 10 Theorem 2. 2. Let the modular lattice L have diamonds D= (v,x,y,z,u) and(v,z,c 1 ,v 1 ,z 1 ) suchthatu/\z 1 = z. Then either A , A or A is a homomorphic image of a sublattice of L. 4 7 9 The situation described in the hypotheses of the theorem is pictured in Fig. 2. 11. 40 u z' x v Figure 2. 11 Proof: Since u /\ v' = u /\ z · /\ v' = z /\ v' = v the Direct Product Lemma shows that D and v' generate the lattice M 3 In particular, there is another diamond D' below. y Vv', z Vv 1 , u Vv') = (v ,x ,y ,z ,u 1 1 1 1 X 2, = (v V v', x V v', 1). u' z' x v Figure z. 12 Note tha t diagramed 41 a' I v / u' ly~x' I v / u' lz~ y' I v / u' Ix ( 6) Let b Eu' ly, a' Ex' Iv, c Eu' lz, b' E y' Iv and a Ea' Ix be the images of c' under the sequence of transposes (6). Since c' is a relative complement of both z and v' in x' Iv, b is a relative complement of y' and u in u' ly. Similar statements hold for a, a', b' and c. Now let us suppose that one of the following statements fails a' Vy = c' Vy =b a' Vz = b' Vz = c b' Vx = c' Vx =a ( 7) a /\ y' = c /\ y' = b' a /\ z' =b /\ z' = c' b /\ x' = c /\ x' = a' Say, for example, c' V x :I a. Then, since c' V xis the image of c' under the transposition z' I v / u' Ix, we conclude that c' V x is a relative complement of u and x' in u' Ix. Since a is also a relative complement of both u and x' in u' Ix, the elements u, a, c' V x, x' satisfy the hypotheses of Theorem 2. 1. Since all of the quotients of (6) are isomorphic it follows that there exists elements r, s E z' Iv such that z, r, s, v' satisfy the hypotheses of Theorem 2. 1. either A 4 Hence is a homomorphic image of a sublattice of L or there exists a sublattice L' isomorphic to M 4 and such that if u is the greatest element of L' there is an atom of L' w such that ulw/f/e c z' Iv'. In this case L' and (D')f le together form a sublattice with A homomorphic image (see Fig. 2. 13). 7 as a 42 u' y' v' Figure 2. 13 We conclude from this that the equations in (7) must all hold. In this case we claim that the sixteen element set S U D U D' form a lattice isomorphic to A • 9 closed under joins. First we show that S is If g, h ED U D 1 then clearly g Vh ED U D' !:= S. Suppose g E (a, b, c, a', b', c'} and h ED U D'. g V h ES. = {a, b, c, a', b', c'} We wish to show that The equations of (7) show that for several choices of g and h, g V h ES. Examples of cases not covered by (7) are 43 a Vy =a v xv y =a Vu= u' ES a Vy' = a V xv y' = u' ES a V x' = u' a Vx = a All other cases are similar to one of the above. Now if both g and h E [a,b,c,a',b',c'} then by (7) c' = b /\ z', a' ~band hence a' V c' = a' V (b/\ z') = b/\ (a' V z') = b/\ u' = b ES Also c' Va = a as a ~ c' and c' V c = c 1 V z V c = z' V c = u'. The remaining cases are similar to these. Similarly S is closed under meets. Now since we have virtually calculated all meets and joins, it can be verified directly that S is isomorphic A . 9 Alternatively, it is known that a modular, simple, length three lattice, with sixteen elements whose top element is a join of its atoms is isomorphic to the projective plane of order 2, that is, A . 9 It is easy to check that S has these properties. Corollary 2. 3. Let n 1 = (v , x , y , z , u ) and n = (v , x , Yz· 1 1 1 1 1 2 2 2 z , u ) be diamond sublattice s of L, a modular lattice. 2 2 z A 1 9 /v 1~ b / a / x 2 /v 2 and that u 1 /\ u 2 =b. Suppose Then either A , A or 4 7 is a homomorphic image of a sublattice of L. 44 Proof: From the hypotheses we have From the Direct Product Lemma we obtain a diamond Dl = n ( v 1 1 Vv 2 = Vv , x vv , y vv , z Vv , u Vv ) = ( v I ,x I ,y I ,z I ,u I ) • 2 1 1 1 1 1 2 1 2 2 1 1 1 2 Similarly we obtain a diamond D~ = D 2 Vv 1 = (v 2 Vv , x Vv , 2 1 1 Furthermore, z'1 = Also, u~ /\ u~ = (u 1 V v ) /\ (u V v ) 2 2 1 = ((u1 V v 2) /\ u 2) V v 1 =v 2 V (u /\ u ) V v 2 1 1 =v 2 Vb V u 1 Thus D~ and D~ satisfy the hypothesis of Theorem 2. 2. Since the con- clusions of Theorem 2. 2 are the same as Corollary 2. 3, the proof is complete. 45 CHAPTER III THE FUNDAMENTAL THEOREM ON WEAK ATOMICITY Let A 1 through A 10 be the lattices diagramed below. 47 Before stating the main result of this chapter we make some standard definitions. Let L be an arbitrary lattice. H(L} is the class of all lattices isomorphic to a homomorphic image of L. we identify isomorphic lattices. Within H(L} Similarly, S(L) is the class of lattices isomorphic to a sublattice of L. If a ~ b are elements of L and a < x ~ b implies x = b, then b covers a, written b >a. if b >a. The quotient b /a is called a prime quotient L is called atomic if L has a least element 0 and if x > 0 there is a y E L such that x ~ y > O. Lis weakly atomic if x > y irnplies there exists b and a such that x :l'! b > a ~ y. A sublattice L' of L is called an isometric sublattice if 48 (x E L'!a <x Sb}= (b} implies (x E Lia <x Sb}= (b}for a, bin L'. This means that a prime quotient in L' is a prime quotient in L. We mention that in a modular, subdirectly irreducible lattice weak atomicity is equivalent to the existence of elements a and b such that b > a. The goal of this chapter is to prove Theorem 3. 1. If L is a modular, subdirectly irreducible lattice such that none of A , .•. , A is a homomorphic image of a sublattice 10 2 of L, then L is weakly atomic. As we shall see in the next chapter, the weak atomicity of L is a powerful tool for analyzing the structure of L. In proving Theorem 3. 1 we shall use techniques similar to those explained by Hong [14 ]. Lemma 3. 2 (cf. [14 ]). A i. S(L). 4 b / a / u/x. or D I u x1 Let L be a modular lattice such that Let D = (v, x, y, z, u) be a diamond in L. Suppose that Then either (i) a Vv = x (ii) there exists x' and b', x s x' < u and b s b 1 < u such that has u = x' Vb' as its greatest element and b' /I. x' as its smallest element. Proof: It may be assumed that ( 1) v<aVv<x for a V v ~ x and if a V v == x them (i) holds. If v =a V v then (ii) holds 49 with x 1 = x and b = b + v. 1 Let u 1 be the greatest element of Da V v/v' which is, -:>£ cours·e , the least element of Dx/a V v· (a V y) /\(a V z). That is, u 1 = (a V v Vy) /\(a V v V z' = By {l) both these diamonds are nondegenerate. by the definition of u A l so, 1 (2) z x v Figure 3. 1 Let b' = b V v and t =b' /\ (u1 V x). Now, since u/x~b/a, we have X /\ t : X /\ b I /\ ( X =x /\ (b V v) =(x /\ b) V v V Ul ) 50 =a V v x V t = x V [(b V v) /\ (u = (u = (u 1 1 1 V x)] V x) /\ (x Vb V v) V x) /\ u =u 1 V x It follows that t Ia V v/ ( 3) u 1 V x Ix Consider the sublattice generated by x, u x V u 1 = x Vt and x /\ u 1 1 and t. By (2) and (3) = a V v = x /\ t The free modular lattice with three generators subject to the above restrictions is L', which is diagramed in Fig. 3. 2. x Vu 1 x a Vv L' Figure 3. 2 Sl That is, the sublattice generated by x, t and u image of L'. t = u • 1 (u 1 1 is a homomorphic Notice that if the diamond in L' is collapsed then In this case (ii) holds with x' = u 1 V x, since x' /\ b' = V x) /\ b' = t = u • 1 Let Dl =Dav v/v' D2 = Dx/a V v' D3 = (Dl)u /t /\ u ' 1 1 D 4 = (D 2 )t v ul /ul and let DS = {vs, xS' t, u 1 , us) be the nondegenerate diamond of L'. Then we have n 1 = (v, a Vv, z /\(y Va Vv), u 1 , y /\(z Va Vv~ D 2 = ( u , u V x, y V a V v, z V a v v, u) 1 1 D3 = (vs /\ (yl v (zl /\vs)), vs, Y1 v (zl /\ vs), zl v (yl /\ vs), u1) n 4 =(u 1 , us, y 2 /\{z 2 Vus), z 2 /\(y 2 Vus), us V(y 2 /\(z 2 Vus))) Note that u /v S is an upper quotient of n and a lower quo3 1 tient of DS and uS/u of n • 4 1 is a lower quotient of DS and an upper quotient Hence D3' DS and D Now let v 3 = vS /\ (y 1 4 together form a lattice isomorphic to V {z 1 /\ vS)) be the least element of n U4 = us v (y2 /\ (z2 v us)) be the greatest element of D 4 and let y' = (y vv ) /\ u = {y /\ u ) Vv and let z' = (z vv ) /\ u = 4 3 4 3 3 4 {z /\ u ) V v • 4 3 Since 3 , S2 y 4 = Yz A(z 2 V uS) = {y Va V v) A (z 2 V uS) ~ {y V u ~ (y V u 1 1 ) A (z 2 V uS) ) it follows that y' V u 1 = (y A u 4 ) V v 3 Vu 1 = (y A U4) v ul = U4 A (ul v y) = (US V y 4 ) A ( u 1 V y) y4 = V ( uS A ( u 1 A y)) = y 4 v ( u 1 v (us A y)) Now since us A y ~ x 2 A y 2 = v 2 = u 1 we have Similar calculations show that y' A u z' A u 1 = z • 3 1 = y , z' V u 3 1 = z 4 and With these facts it follows easily that n , DS' D , 4 3 y' and z' form a lattice which is isomorphic to A • 4 diction proves the theorem. This contra- 53 u z x v Figure 3. 3 Corollary 3. 3. A , A {/. HS(L). 8 9 Let L be a modular lattice such that A , A , 4 7 LetD = (v,x,y,z,u) andD' = (v 1 ,x 1 ,y 1 ,z 1 ,u 1 ) b e diamond sublattices of L such that u Proof: = u' and x = x'. Let us suppose that v :f. v'. may assume that v' /; v. Then v = v'. Then, by symmetry, w e Apply Lemma 3. 2 with b = z' and a= v'. 54 The sublattice generated by D and v V v' is denoted L' (see Fig. 3. 4). u z x v L' Figure 3. 4 As before we let u 1 denote the top element of Dv V v' /v• By Lemma 3. 2 there is an element b', z' = b ~ b' < u such that b' /\ (u and b 1 V u 1 V x = u. (u 1 Now V x) /\ z 1 = (u 1 V x) /\ b 1 /\ z 1 = u 1 Hence x V (u 1 /\ z 1 ) = x V ( ( u = (u 1 1 V x) /\ z ~ V x) /\ (x V z') = (u1 V x) /\ (x' V z') = (u 1 V x) /\ u =(u1 V x) I\ u =u 1 v x 1 /\ z 1 1 V x) =u1 55 Also x /\ u 1 /\ z' = v' Hence (1) u /\ z'/v•/u Vx/x 1 1 Since u 1 V x > x we have u /\ z 1 > v 1 • 1 Note that, since u 1 is the top element of Dv Vv' /v' u 1 depends only on D and v' and not on z'. Hence, if we now let b = y' and a= v', the above argument yields that (2) Recall that (x V u ) /\ b' 1 b' =b V v = z' V v. (3) = u 1 so that b' :<?: u 1 • Also recall that Hence (v V v') V (u 1 /\ z 1) = v V (u1 /\ z') = u 1 /\ (v V z') = u 1 /\ b' Similarly (4) (v V v') V (u /\ y') = u 1 1 Now consider the sublattice L" generated by v V v', u u 1 /\ z'. 1 /\ y' and Since they are less than x, y' and z', respective! y, any two of them intersect to the bottom element of L", v'. Using this and (3) and (4) we see that L" is a homomorphic image of the lattice diagramed 56 in Fig. 3. 5. vVv' u 1 /\ z' . v' F igure 3. 5 Since u 1 /\ z ' > u' w e know the diamond in L" is nondegenerate. Now the diamond Dv V v ' Iv has u 1 as i ts top e lement and v V v' as one of its atoms and v as its bottom ele m e n t . A , Corollary 2. 3 either 8 A 4 or A 9 Hence by the dual of E HS ( L), a contradiction. This completes the proof. Let L be a modular lattice such that A , A ti. 2 3 Lemma 3. 4. HS(L}. Suppose a strongly normal sequence satisfies the conditions of Theorem 1. 1. Then the associated diamonds must alternately trans- pose and translate. That is, the numbers below the arrows between the associated diamonds must alternate. Proof: We have already seen that Dk-1''~ D~, Dk ~Dk+l is impossible. Suppose 57 Then it is easy to verify that Dk-l U Dk U Dk+l U [~-l /\ ~+l' ~-l /\ vk+l' vk-l /\ ~+l' vk-l /\ vk+l} forms a sublattice with A 3 as a homomorphic image. As an illustration of the last lemma suppose b /a / 0 0 b /a / 2 2 •.. ~b b /a 1 1 10 /a 10 /b 11 /a 11 is a strongly normal sequence satisfying all the conditions of Theorem 1. 1 in a modular lattice L such that A , A ti. HS(L). 2 3 Suppose D 1 ~ D~. Let D , ... , D be the associated diamonds. 10 1 Then we must have D D , D 3 -- D~:c4' 2 3 4) D4~> Ds = D~, D6~>D7 = D~, D8~>D9 = D1co· Notice that D , ••• , D form a sublattice which is a homomorphic image of the 10 2 sublattice pictured in Fig. 3. 6. Notice that a 11 ;::: y 2 = b • 2 Now we are ready to begin the proof of Theorem 3. 1. Since L is subdirectly irreducible and modular we need only show that there e x ist elements a and b in L such that b covers a. By the results of Jonsson [16 J we may assume that L has a sublattice L the lattice diagramed in Fig. 3. 7. 1 isomorphic to A direct proof of this assumption will be indicated below. If x > v we are done. Thus let x > x~~ > v. Now x~:c and L 1 g enerate the sublattice diagramed in Fig. 3. 8. We conclude from these observations that L has a sublattice L 2 which is isomorphic to the lattice diagramed in Fig. 3. 9. There exist subquotients b/a of u' /x' and d/c of e/u' which are connected by a sequence of transposes satisfying the conditions of Theorem 1. 1. If b / a / b /a then it is clear that the sublattice 1 1 58 v2 Figure 3. 6 59 u x v Figure 3. 7 Figure 3. 8 60 x' x v Figure 3. 9 generated by D 1 , Db/a and Du/\ b/u /\ a has A 5 as a homomorphic image. Here D 1 is the first diamond associated with the sequence from b/a to d/c. Hence it may be assumed that (1) Furthermore, by applying Theorem 1. 1 to the sequence ( 2) w e may assume that(l) is strongly normal satisfies condition (i) of Theorem 1. 1, and 61 ( 3) or k imply = 1, 2 ••.. ,n-3. Here D , ... , Dn-l are the diamonds associated with (1). 1 Note that b 1 ~ b 0 = b ~ c. It is well known ~nd easy to see that Hence n ~ 4 and so n - 3 ~ 1. this implies p. d. (b /a , d/c) <!: 3. 1 1 n 1,-(i'> D~ implies n 1 = D~. But if n 1 Thus = D~ we may apply Lemma 3. 4 with the aid of (3) to the sequence (2) and, as the example after that lemma illustrates, bn_ 2 ~ b 0 ~ c. But by (1) p. d. (bn_ /an_ , d/c) 2 2 As pointed out above these two statements are contradictory. = 2. It fol- lows that (4) The next part of the argument again uses techniques developed in [14]. LetDl = (v~,x~,y~,zi,u~} = (D')b/a and D~ = (v~,x~,y~, z~, u~) = (D)b /\ u/a /\a· b'/a 1/ b / a . Let b' = u 1 V ui and a' = b' /\ a. Then If b'/a 1 /x*/v* where x* and v* are elements of a diamond D):c = (v):C,x;'<,y*,z):C,u):C) thenD~, D~ and D):c form a sublattice with A 5 as a homomorphic image. From this and the fact that b/a~b'/a' we conclude p.d. (b'/a', d/c) = p.d. (b/a, d/c) = n. Now it is easy to check that the sequence ( 5) has D , ... , Dn-l as its associated diamonds and satisfies all the con1 ditions of Theorem 1. 1. The situation is diagramed in Fig. 3 . 1 0. 62 Figure 3. 10 Consider the sublattice generated by y~, a' and u . 1 The fact that u~ /x~/ b/a and the definition of b' and a' imply that u~ /x~/ b' /a'. Hence it follows that Yi Va' = b'. Also a' V u 1 = b'. The free lattice subject to these restrictions is given in Fig. 3. 11. Suppose the diamond in Fig. 3. 11, which we denote by Then let (D 1 ) =D1 =(v1,X:1 =u1/\vo,y1,z1,\ii =u1). By (5) b' I\ Uz = ul. I\ v0 Let(Dz)x2/v2vz1 =Dz. Hence uo I\ Uz = ul = ul. ul rxl/ zo Iv 0 and ul .rzl/xl /v2. I ul ul Also As we noted in the proof of 63 b' a' Figure 3. 11 Lemma 3. 4, D , D , D generate a lattice with A as a homomorphic 0 1 2 2 image. We conclude from this that the diamond D be degenerate. 1 in Fig. 3. 11 must That is, that sublattice generated by a', y~ and u must be distributive. and u 0 1 Similarly the sublattice generated by a', z~ is distributive. A similar argument shows that if the sublattice generated by ui, a' and y 1 or the sublattice lattice generated by ui, a' and z 1 is not distributive then there exist s /r and s /r subquotients of 1 1 2 2 ui/x~ and of u~/z~, respectively, such that the diamond in (ui,a',y ) 1 64 and (D ') I and (D 2') I form a sublattice with A 5 as a homomor1 sl rl Sz rz phic image. We conclude that (u~,a 1 ,y } and {u~,a 1 ,z } generate distributive sublattices. (6) 1 1 vi Vu 1 Thus =(a'/\ yi) Vu 1 V(a' /\ zi) Vu 1 = [(a' V u ) /\ (yi V u ) JV [a' V u ) /\ (zi V u )] 1 1 1 1 = [b 1 /\ (Yi V u ) J V [b 1 /\ ( z~ V u ) J 1 1 =y1 V u1 V z 1 V u1 I I Similarly ( 7) v Vu 1 I 1 = b' By the Direct Lemma Product there exist diamonds D; = D~ /\ u (v ( I 1 I I I I I = 1 I /\ u , x /\ u , y /\ u , z /\ u , u /\ u ) andD = D /\ u = 1 4 1 1 1 1 1 1 1 1 1 1 I I I I I ) v 1 /\ ul' xl /\ ul' y 1 /\ ul' zl /\ ul' ul /\ ul • Since (8) I I we have ~ = x . 4 I I I I Since '3 = u , Corollary 3. 3 implies that v = v . 4 4 3 By the construction of D3 and D4, we know that (9) and Now z , c_'I' u v = u 'I' v and u 'I' v transposes up to u 'I' v . 4 3 3 4 4 1 1 3 3 This transposition is of course an isomorphism; let z~ be the image of z~ Then, as z~ < ui s; b', we have 65 z4 /\. u2 =z4/\. b { 10) 1 /\. u2 =z'4 "u1 = z4 /\. ui /\. ul =z4 /\. u4 -- z'4 Hence the Direct Prpduct Lemma may be applied to the sublat- z4/z4 and u 2 /z4, to obtain a diamond DS =D 2 vz4. Since u4/z4 x I v 2 , u 4• v' z 4 I.z 4 v-• z 4 = u 1• 1-,/ z4 x 2 v-, z 4 I v 2 v-, z 4 • (S ee F.ig. 3 • 12 . ) 2 tices / 1 z'5 -- z 2 v z'4 V I -yl 3- 4 Fig ure 3, 12 66 Since x3 = x4 and since z4 is a relative complement of x4 in u4Jv4 = u3 lv3, and :z4 is the image of z4 under the isomorphism u3 / v 3 / ul /vl, it follows that z4 is a relative complement of xl in I ulI I v 1. Hence the sublattice generated by zl, xl and z4 is a homo- morphic image of the lattice diagramed in Fig. 3. 13. z'4 x' 1 v' 1 Figure 3. 13 Let D 0 denote the diamond in this sublattice lattice. If this diamond is nondegenerate then D 0 , (Dz>u' /u' /\ v and (Ds' } v 'I ' UO V5 V 5 2 2 form a sublattice which has A as a homomorphic image. Hence D 5 0 is degenerate, which implies z4 = zl. In this case Dz, Dl and DS form one of the lattices pictured in Fig. 3. 14. 67 Figure 3. 14 Remarks, The above arguments show that if L has two diamond sublattices D = (v, x, y, z, u) and D 1 = (v', x', y', z 1 , u 1 ) such that u/ z / z' /v' and u' is not the greatest element of L then one of the lattices of Fig. 3. 14 is a sublattice of L. Furthermore, the two lower diamonds of Fig. 3. 14 are (D)b/a and (D')b Vv' /a Vv' for some a, b such that z~a<b~u. The same arguments can also be used to show that if D = (v, x, y, z, u) is a sublattice of L such that u is not the greatest element of L 68 then L has one of the following sublattices. Figure 3. 15 Furthermore, the lower diamond of these lattices is Db/a for some z s: a < b s: u. Before continuing the proof of Theorem 3. 1 three additional lemmas will be needed. Lemma 3. 5. f HS(L). Let L be a modular lattice such that A , .•. , A 10 2 Let d/c = b 0 /a 0/ b 1 /a ~b 1 2 /a 2/ ••• ~bn/an = f/e be a strongly normal sequence from d/c to f/e. Let us also assume that the associated diamonds satisfy Dl ~D~. D2~>D3 =n:. D4~ Ds =ni .... ,Dn-2~ Dn-1 69 Then f ;. c. Proof: Since n 1 ~D~ u 2 /\ x 1 = z 2 . We also know that b 1 /\ b 3 Hence U3 /\ xl = u3 /\ ul /\ xl = Uz /\ Xl = z2 Applying the Direct Product Lemma we obtain a diamond n3 = (v 3 Vx , x Vx , y Vx , z Vx , u Vx ) = (v3,x3,y3,z3,u3) 1 3 1 3 1 1 3 1 3 such that u /x 1/ 1 x3/v3 and n 3 ~n3 (see Fig. 3. 16). Figure 3. 16 70 We also setD2 = (v2,x2•Y2•z2,u2) = (v 2 Vv 1 , x 2 Vv 1 , y 2 Vv 1 , z 2 vv , u vv ) = (v ,y ,z ,x ,u ) =(Dr)*. 1 1 1 1 1 1 1 2 Let r = z4 /\ v 5 ands = r V u . 3 is a normal sequence of transposes. y 5 /\ s = r = s /\ z4. Also setn4 = (D3)>:<. Then it follows that From this it follows that Hence the lattice generated by y , s and z4 is a 5 homomorphic image of the lattice given in Fig. 3. 1 7. z' 4 s r Figure 3. 1 7 71 H the diamond of this lattice, which we denote D , is non- 0 degenerate then, since v 0 V v4 = v Corollary 2. 3 on the diamonds D contradiction. 0 V z 0 4 V v4 = z4, we can invoke and (D 1 ) 1 4 Z4 V I 1 UO Z4 to arrive at a Hence the sublattice generated by z4, y tributive. Similarly, the sublattice generated by z4, z tributive. Hence u 5 /\ z4 5 5 and s is disand s is dis- = (s V y 5 )" /\ z4 /\ (s V z 5) /\ z4 = [(sf\ z4) V (y 5 /\ z4) J /\ ((s /\ z4) V (z 5 /\ z4) J = [r V (y 5 /\ z4) J /\ [r V (z 5 /\ z4) J = YS /\ z4 /\ zS /\ z4 = v 5 /\ z4 =r The Direct Product Lemma yields a diamond DS = D 5 V z4 = (vs V z4, x 5 V z4. y 5 v z4, zs v z4, us v z4> such that D 5,..(i'> n5 and u4lz4,~r~/v5. diamonds Dl vk ::!: c. Let D(, = (D5)*. Continuing in this way we obtain = D 1 , Dl' D3• ... , D~-l such that Dk~ Dk and such that From the definition of the associated diamonds we know f l e / zn-l Iv n-l • We also know zn-l lvn_ 1/ f l e / z~-l lv~-l. But, since v~-l Lemma 3. 6. lattice. Set w 0 Let D = v, w 4 C!: c, l l I w.1 w.1- 1 i = l, 2, 3, 4. Hence this clearly implies f 1 c. = (v, x, y, z, u) be a diamond in a modular =x and let w 0 ~ w 1 !!:: w 2 ~ w 3 !!:: w 4 • Then there exist elements x = t 0 ' t1 ' t2 ~ t 3 ' t4 z., u.) = D z~-l lv~_ 1 • such that w. lw. l / l- 1 =u and diamonds Di= (vi.xi' yi, x. Iv. and u. I x . / t. It. 1 l 1 l 1 1- 1 , 72 Lemma 3. 7. Assume the hypothesis of the previous lemma. Suppose also that there is another diamond D' = (v', x', y', z', u') such that u I z / z' Iv • . Letw!1 =w.1 Vv', i=0 ., l,2,3,4andletD!1 =(v!,x!, 1 1 y!, z!, u!) be the diamonds obtained by applying Lemma 3. 6 to D' and 1 1 1 . the r o 1e o f x ) . Th en w I , w I , w I , w I , w I ( w1. th z I p 1a y1ng 4 2 3 0 1 w./w. /x./v./u./z./w.Vz/w. lV /w!/w! /z!/v! 1 1- 1 1 1 1 1 1 1z 1 1- 1 1 1 Furthermore w. V z = z! /\ u, i = 0,1,2,3,4 (see Fig. 3.18). 1 1 u' x' x Figure 3. 18 73 Proofs: 1 1- 1 1 z. = u. /\ (z V w. 1 1 l 1- 1 1 l l l 1- 1 l ) and Straightforward calculations show that v., x., y., ). l z. and w. form a diamond and that w./w. 1 = v, 1 , i = 2,3,4, x. = v. Vw., i = 1,2,3,4, y. = u. /\{y Vw. v. = u. 1 Let u. = (w. Vy) /\ (w. V z), i = 1, 2, 3, 4, v 1 l 1- 1 / x . / v .• 1 1 1 This is the con- l clusion of Lemma 3. 6. The proof of Lemma 3. 7 will also be complete if we show u. I z. / l w. V z /w. 1 1- 1 l V z/ ~· w ! /w ! 1- 1 1 , and w. V z = z ! /\ u. l l ui V w i-l V z = w i-l V z V ((w i V y) /\ (w i V z~ = wi-l V ((wi V z) /\ (wi Vy V z)) Also u. /\ (w. 1- 1 l V z) = z. by definition. 1 Hence u . / z . / w. V z/w. l 1 1- 1 1 Vz Now, as z ~ v' (w. V z) V(w. 1- 1 1 and as w. /\ v' ~ x /\ v' = v l Vv') =w. Vv' l ~ z (wi V z) /\ (wi-l V v') = wi-l V ((wi V z) /\ v') = wi-l V (z V (wi /\ v')J = wi-l V z To see that w. V z = z! /\ u, first note u! = (w! Vy')/\ {w! V x') 1 l l 1 1 1 = {w. V v' Vy') /\(w. V v' V x ) = {w. Vy')/\ (w. V x') and z! = v! V w! 1 1 1 = v ! V w. w he re v ! = u ! l 1 l 1- 1 , i = 2, 3 • 4 and v 1 1 1 l 1 l Al s o, as u ~ z' , 1 = v • u/\{w. Vy') =w. Y.(u/\ y 1) =w. v(y/\ z' /\ y')=w.V{u/\v 1) =w. V z. l 1 1 l l 74 Similarly u /\ (w. V x') = w. V z. 1 Hence 1 u /\ u! = u /\ (w. V y') /\ (w. V x') /\ u = w. V z 1 1 1 1 Thus if i is 2, 3, or 4 u /\ z! = u /\ (v! V w .) = u /\ (u! 1 1 = w. V ( u /\ u! 1- 1 1 V w .) 1- 1 1 1 ) = w. V w. V z = w. V z 1 1 1 If i = 1 then u /\ zl = u /\ (vl V w ) = u /\ (v' V w ) 1 1 =w 1 V(u/\v 1 ) =w 1 Vz This completes the proof. Now we return to the proof of Theorem 3. 1. Recall that we have shown that L has three diamond sublattice s D = (v, x, y, z, u), D' = (v 1 ,x 1 ,y',z',u') and D" = (v",x",y",z",u") such that u / z / z 1 /v 1 (11) u' /z•/ z" /v" and The diamonds D, D', D" form one of the sublattices of Fig. 3. 14. If these diamonds are isometric diamonds the theorem is true. Hence there exists w 1 E L such that v < w 1 < x. Applying the previous two lemmas to the diamonds D and D' and also D' and D" we obtain up diamonds DZ= (vz, xz, Yz• zz, uz) = Dw /v' D l = (vl, xl, yl, zl, = 1 r x' y' 1 r) - D r r r r 1) (D ') w1 Vv'/v'' D'4-- ( V4, 4' 4•Z4,U4 - x/w1' D'3 -_ ( V3,X3,Y3•Z3,U3 = (D') x Vv /w 1 Vv = (D') z /w 1 Vv DS = (vS' x5, Ys• z5, uS) = (D")u' Vv" /v' Vv" a nd D6 = (v(>• xb, Y6• zb' u6) = (D")u' Vv" /v' Vv" such 1 1 that 1 1 1 1' 3 3 75 xlw 1/ x'4Iv' / u'4lz' / 4 4 u'1 lz'1 / z'5 Iv'5 z' Iv' Vw 1/ and z'3 Iv'3 u'3 lz'3 /z•6 Iv'6 This is represented in Fig. 3. 19. u' 6 x' Figure 3. 19 76 Since L is subdirectly irreducible and xl ~ v3 there exist sub- quotients d/c of x:/v3 and f/e of xl /vl which are connected by a strongly normal sequence of transposes. If d/c ~bl /a / 1 .•. f/e, then the first associated diamond, D , together with (D3)d/c and 1 (D_5)v(, Vd/v Ve form a sublattice which has A 5 as a homomorphic 6 , (D 1 )f/ and image. Similarly, if b /a / f /e then D n- 1 n- 1 n- 1 e 1 (D_5)vS Vf/v/:; Ve form a sublattice with A 5 as a homomorphic image. Hence it may be assumed that the sequence connecting d/c to f /e has the form: Furthermore, we may assume this sequence satisfies the conditions of Theorem 1. 1. With the aid of Lemma 3. 5, Lemma 3. 4 and Theor e m 1.1, we can conclude that the diamonds associated with (12) satisfy It follows from (12) and (13) that (14) k = 1 , .•• , n-1 Applying Lemma 3. 6 and the dual of Lemma 3. 7 to the elements v' < xl ~ c /\ x' ~ d /\ x' ~ x', diamonds n7 = (v7, x7, y7, z7, u7) and I D 10 -- ( 15) a nd ( I I • d s uc h th a t v 10' xlI 0' y 10 • z 1I 0 • ulI 0 ) are o bta1ne d / c / x 1 /v' 7 7 77 u' /x'~z' /v'~ z 1 /\z 1 /z 1 /\ v'~ U/\Z 1 /u/\V 1 ~u' /z' 7 7 7 7 7 7 7 7 10 10 ( 16) and v'10 V v'1 -- z' /\ v'7 (1 7) Since u' /z' / z 1 /\z 1 /z 1/\v 1 by(l6) z • ~z' Av 1 • 10 10 7 7 , 10 7 Hence by(l7) ( 18) Now let DS = (D(,)v(, Vd/v(, Ve' and let c' = v 1 /\ v7 and d' = d V c'. The situation is represented in Fig. 3. 20. u' 8 CI = 7 V 1 /\ V 1 Figure 3. 20 Notice that this situation is the dual of the situation represented in Fig. 3. 10. By using the dual arguments used in that case, we can conclude that there exists a diamond n9 =D 2 /\ z7 such that 78 1 ' I V7' ' ~Uzi\ Z7' I Xz/\ Z7 -- U9' /x'9· Z7 Let s = u9 V ul 0 and r = s /\ v7. This situation is represented in Fig. 3. 21. x' 7 Figure 3. 21 Let L' be the sublattice generated by u9, r and xl . 0 L' is a homomorphic image of the lattice given in Fig. 3. 11 with a' = r, b' = s, u 1 = u9, x 1 = x9 and yl = xlO" If the diamond D nondegenerate then as before D , (D 1 ) 0 7 V71 V / uo V7 0 V in this sublattice is XO and (D 1 ) 1 I 1 form a sublattice with A as a homomorphic image 5 9 U9 /\ ZQ U9/\ VQ (see Fig. 3, 11). by fu9, r, Similar arguments show that the sublattices generated yl 0 }, (ulO' r, z9} and (ulO' r, y9} are distributive . As before this implies that 79 v 1 V u' 10 9 (19) =v'9 V u 10 = s 1 By the Direct Product Lemma this yields two new diamonds DI 1 = 1 1 D 91 /\ ul1 0 a n d D 12 -- D'10 /\ u9. Thus 3. 3. (20) V By definition v9 :<: vl. 1 1 1 1 1 9 /\ u 10 -- v 11 -- v'12 -- v 10 /\ u 9 =v 2 /\ z7. By (14) vl ~ v • 2 Moreover, z7 :<: c Hence Now by (20) and (21) we have (22) Also v7 :<: v3 :<:vi by their definitions and v7 ~ zl I v7 ~vlO Vvl. 0 ~ vl Hence, by(22), (16) and(l8) I (23) vi 0 = vl 0 V (vi /\ ul 0) = ul 0 /\ (vl 0 V vp -- ulI 0 /\ v 7I /\ ( v 1I 0 V v 11) = zlO /\ (vlO v vp = zio This last contradiction proves the theorem. 0 by ( 1 6). Thus 80 CHAPTER IV THE MA.IN S TR UC TURE THE OREM Let tJ be the variety (equatorial class) of all distributive lattices and 'f!I. ~be the variety generated by all modular width four lattices. is well-known that if L l tJ then either M 3 E S(L) or N 5 It E S(L). Figure 4. 1 In this chapter we prove an analogous result for either Ak E HS(L) for some k, 2 ~ k ~ 10 or N 5 m4 : If L ~ '!11. 4 then ro ro E S(L). We begin with Lemma 4. 1. (i.e., x > v). Let D = {v, x, y, z, u) be an isometric diamond in L Let us suppose that A , A , A (/. HS(L) and that there is 4 7 9 another diamond D = (v , x , y , z , u ) such that 1 1 1 1 1 1 Then either (2) 81 or ( 3) Nate that z s: u Proof: 1 /\ u. Equality cannot hold, for otherwise Corollary 2. 3 would give a contradiction. u s: u • 1 Suppose u S: x dict (1), u V v 1 1 Since u > z this means Then, since z s: u S:v as well. = x , again since v -< x . 1 1 1 1 would contra- Thus we see that (3) holds in this case. Now suppose that x 1 'I:. u; then u /\ x 1 = z and u V x 1 = u 1 . Thus, by (1). ( 4) u Vv 1 =uVzVv 1 = u V x 1 -- u 1 (5) u /\ v 1 = U/\ xl /\ v 1 = z /\ vl = v Hence (2) holds in this case. Theorem 4. 2. Let L be a modular, subdirectly irreducible lat- tice such that A , ... ,A l:. HS(L). 10 2 Then M M Figure 4. 2 3 3 x 2 t/. S(L). x2 82 Proof: If the conclusion of this theorem fails, then there exist diamonds D = (v, x, y, z, u} and D' = (v', x', y', z', u'} such that D~D' (l} By Theorem 3. 1, L is weakly atomic. such that v ~a< b ~ x. Consequently there exist a, b E L Let a' =a V x' and b' = b V x'. Then Db/a~ D'b' /a', and so Db /a and D'b• /a' form a lattice isomorphic to M 3 x 2. Hence we may assume v < x. v ~ e < f ~ v 1• There also must exist e and f such that Now the diamonds ( v V e, x V e, y V e, z V e, u V e} and (v V f, x V f, y V f, z V f, u V f} together form an isometric sublattice isomorphic to M 3 x 2. Hence we assume v -< v', i.e., D and D' together form an isometric sublattice. Recall that a sublattice L' of Lis called isometric if a covers b in L' implies that a covers b in L. Since L is subdirectly irreducible there is a strong! y normal sequence of transposes (2} which satisfies the conditions of Theorem 1. 1. Furthermore it may be assumed that p.d. (v'/v,z'/v'} ~min [p.d. (v'/v,x'/v'}, p.d. (v'/v,y'/v'}} (3} Suppose v' / v / b /a ~ b /a / b /a = z' /v' and D ~ D~. 1 1 3 3 2 2 1 It follows immediately from the definitions of the associated diamonds that z x1 2 =v' /\ (z 2 V v'} = v' and Thus x = z 2 so that D = D~. 1 1 = v' /\ x1, and x = z 2 V v' = v'. 1 = z 2 V v'. Thus z 2 83 z' v' Y2 Figure 4. 3 3 The sequence v' /v ~bl /a 1/ b /a ~b /a 2 2 3 sible because b 1 = z' /v' is impos- ~ v' and p. d. (b /a , z' /v') = 2 are contradictory. 1 1 Recall that if the number under the arrow between Dk and D~+l is one, we say Dk transposes to D~+l; if it is a two, Dk translates to If Dn_ 2 transposes to D~_ 1 , then Dn_ 2 = D~_ 1 , provided n > 3, since the sequence (2) satisfies the conditions of Theorem 1. 1. above argument shows that this is the case even if n Let us suppose that ( 4) Also suppose that b n- 1 /an- 1 ~ b n /an = z' /v' = 3. The 84 D (5) n-2 = n~:< n-1 Lemma 3. 4 together with (4) and (5) imply ( 6) In fact, since the sequence (2) satisfies the conditions of Theorem 1. 1, we have either or (8) In either case v depending on whether n is odd or even. =v'. Thus a 2 :<?: v 2 ~ v'. 2 ~ v n- 1 But this contradicts p. d. (b /a , v' /v) 2 2 ~ a n = 2. We conclude that (5) cannot hold and hence (9) D ~D n-1 n-2 <2 > Applying Lemma 4. 1 to the diamonds D' and D n- 1 w e conclude that either u'/v 1 / u ( 1 O) n-1 /v n-1 or (11) Suppose ( 1 O) holds. Consider the set [x x' V vn-l' y' V vn-l }. n- 1 = v n- 1 V z', yn- 1 , z n- 1 , By (10) these are all atoms in un-l /vn-l" If 85 there are four distinct elements in this set then u sublattice isomorphic to M lattice which has A y' V v 4 /v n- 1 which, together with Dn_ as a homomorphic image. 8 n- 1 2 contains a form a sub- Thus we may assume = y n- 1 and x' V v n- 1 = z n- 1 • n- 1 An argument dual to one used above shows that (4) implies that n ~ 4. Thus Since v n- 1 = a n V a n-2 = v' V v n- 2 we have that v (13) V x1 = v n-2 n-2 V v' V x' = v n-1 = z n-1 V x' Thus we may apply the Direct Product Lemma to the sublattices z x n- 1 /x' and z x' n-2 /\ u'n- y ' n- 1 n-2 /v n- 2 /\ x' ' z to obtain a new diamond D' n- 2 A n-2 " x' u ' A n-2 " = (v n- 2 /\ x', x') - (v' x' y' z' n-2' n-2' n-2' n-2' 2>· Now it is easy to check that ( 14) Consequent! y, p. d. (x' /v 1 , v 1 /v) ~ n - 1 < n = p. d. ( z 1 /v', v' /v), contradieting (3). Hence we conclude (1 O) cannot hold and so ( 11) must hold. As before, if u' /\ v n-l ~ (x', y'} then u' /v' contains M which together with Dn-l form a sublattice with A ima ge . hf1Vl' z Thus we may assume v n- 1. /v n- 1 ~u n-".>/xn- 2 . the proof of Ll' rnma 3. -!, D', D n- 1 /\ u' = x'. Moreover v n- 1 , D n- 2 7 as a sublattice as a homomorphic Since D n- 1 4 n- 2 ~ D n- 1 we = v n- 2 Vv'. Now :1s in generate a sublattice with A 3 86 u v'n-2 a n-1 =b n-1 n-4 Figure 4. 4 as a homomorphic image. hold. This contradiction shows that ( 11) cannot It follows that assumption (4) cannot hold. Hence, it may be assumed that ( 1 5) b n- 1 /a n- 1/ b n /a n = z' /v' This leads to the following four cases 87 (l 6a) V' /v - 0 b /a O~l ' b /a 1 //J ... b 1 /a 1/ nn- b /a n n = z' /v' with or with or with (l 8b) Dl ~D>2!<' D2~<2>D3 = n>4!<' ••• ' D n- 3/':(2)D n- 2 = n>l<n- 1 or with Let us suppose that the situation of equations ( l 8a) and (l 8b) holds. If w is any e lement of L, let D~ V w denote (v Zz v w, x2 v w, Uz v w). 2 V w, Yz V w, The n, sinc e D l (~ n>~· Dl = D~ v vl. 88 Furthermore, as everything in D~ is greater than or equal to v , 2 D~ Vv = D~ vv 2 Vv = D~ Vv 1 = n 1 , since v 2 Vv = a 2 va 0 = v 1 • Now as in the example after Lemma 3. 4 (l 8b) implies u . since D 2 ~ z'. Hence, *2 V v = D 1 , u 1 = u 2 V v ~ z'. Since v 1 ;;::: v, we have D 1 ~ z' I v. But the dimension of z' Iv is two; thus u 1 = z' and v 1 = v. Now the set (z, x , y , z } has at least three elements, so we may assume that z, 1 1 1 x , y are distinct. 1 1 Then the diamonds (v, z, x , y , z') and D = (v, x, y, 1 1 z, u) satisfy the hypotheses of Theorem 2. 2, which gives a contradiction. Now we suppose (1 7a) and (l 7b) hold. u (20) 2 As before ~ z1 From the definition of the associated diamonds (21) and (22) Now if u 2 ~v', then it would follow from (21) and (22) that u /x /v•/v. 2 But v' > v and x 1 < u 1 <u 2 by (1 7b). (23) Since v u 2 = v' V v 2 ~ z ;;? 1 v'. and v V z 1 Hence we have 2 f;.v' =v' and v 2 ~ u 2 ~ z', z' :<!: v V v 2 =v V z 1 V v 2 Thus, since v' -< z', either v V v 2 In either cas e (24) 1 z v v 2 = z V (v V v ) = z' 2 = v' or v v v 2 = z '. 89 Thus we may apply the Direct Product Lemma to the sublattices z' /z and z' /v z x 2 1 2 /\ z, u /\ (v 2 to obtain a new diamond D 2 /\ z). /\ z) = v Now x 1 1 V (v /\ z = v • 1 2 2 /\ z = (v /\ z) = z /\(x 1 2 /\ z, x V v ) 2 2 /\ z, y 2 /\ z, = z /\ x 2 and Hence Moreover, (26) u 1 /\ ( u 2 /\ z) = u 1 /\ z = u 1 /\ v' /\ z By (25) and (26) we may apply Corollary 2. 3 to the diamonds D and D 1 2 /\ z to arrive at a contradiction. In both of the two remaining cases we have the following situa- ti on: = bb I a n ~ u n- 1 I z n- 1 (2 7) z 1 Iv' (28) u n-1 /xn-1/ z n-2 /v n-2 x v' Figure 4. 5 n-2 90 We would like to show that D', D with A 2 as a homomorphic image. do this we must show that u' A u z' A u y' A x • n- 2 n- 2 n- 2 As pointed out before, in order to n- 2 = u n- 1 • , x' A x n- 2 n- zA u'. By its definition u = v n- 1 n- 1 n- 1 = z' A x = n- 2 , • All three pairs of these generators intersect to . Also x For example, x' A x n- 1 V (x' A x n- 2 ) = x u', the greatest element of L'. Similarly x x generate a sublattice n- 2 Consider the sublattice L' generated by x the least element of the L'. v' A x and D n- 1 n- 2 n- 1 n- 2 A (x A y' A x n- 1 V (x' Ax A x') = x n- 2 = n- 2 ) = x n- 2 n-2 A Au' It follows that L' is a homomorphic image of the lattice dia- gramed in Fig. 4. 6. u 1 A x n-2 x n-1 x 1 A x n-2 L' Figure 4. 6 Since x either w = x n- 1 or w = v n- 1 . If w = v n- 1 then x n- 1 n- 1 = u' A x > v n- 1 n- 2 , which impliesu'Aun- z=U 1 f\(U n- 1 Vxn- )=un- 1 V(u 1 f\X n- z)=u n- 1 Vx n- l 2 the desired conclusion. which is nontrivial as x n- 1 If w ::: x > v n- 1 . n- 1 then L' is a diamond, Moreover, u n- 1 A (u' A x n- 2 ) 91 = u n- 1 /\ x n- 2 =x monds L' and D n- 1 n- 1 . Hence we can apply Theorem 2. 2 to the dia- , arriving at a contradiction. This final contra- diction proves the theorem. Remark. Let L be a modular subdirectly irreducible lattice such that A , ... , A f. HS(L). 2 10 The dual to the last part of the above proof shows that the following situation cannot occur: L has three iso- metric diamonds D. = (v.,x.,y.,z.,u.), i = 1,2,3 such that 1 1 1 1 1 1 and (1) and (2) We improve upon this in the next lemma. Lemma 4. 3. Let L be a modular subdirectly irreducible lattice such that A , ... , A l. HS(L). 10 2 Then L cannot have three isometric diamonds D., i = 1, 2, 3, which satisfy (1 ). 1 Proof: As remarked we need only show that (2) holds. By (1) ( 3) The Direct Product Lemma, applied to u /u V u and D , now yields 3 2 1 2 M 3 X 2 as a sublattice unless u have u 2 =u1 V u3. Hence 2 =u 1 V u3" Thus by Theorem 4. 2 we 92 (4) z Vx 3 = z 3 V (v 2 /\ u 1 ) = v 2 /\ (z 3 V u 1 ) 1 =v /\ [(z 2 2 /\ u ) Vu ] 1 3 = v 2 /\ z 2 /\ (u 1 V u ) 3 = v 2 /\ z 2 /\ u2 = v 2 Clearly v 3 V x Let w = u s: v • 2 1 3 /\ (v 3 V x 1 ). Now v 3 The second inequality shows that w.;. u , w -:f. x and w -:f. y • 3 3 3 then by the Direct Product Lemma v lattice M u 1 v 3 s: v 3 3 X 2 unless v then u 1 :S: v s: v 3 by semimodularity. again violating (1 ). then u 1 D 1 3 3 V x /v and D generate the sub3 1 3 2 which violates ( 1). If u 1 Vv 3 s: u 3 3 then u -< u V v /v and n generate M X 2. 3 3 3 3 If w # z If w = v =v 3 V x 1 • Thus we must have x 1 s: v 3 . If Hence, since v follows that w.;. v • 3 u /v3" 3 3 3 s: w s: v • 2 1 Since x 1 :S: u 1 1 Vv -< u , v 1 3 :S: u 3 V v , u /\ (u V v ) 3 3 3 1 3 -< u :S: x 2 1 V , =v 3 . But From this contradiction it Hence w is an atom in the two-dimensional lattice then u /v contains a copy of M which together with 4 3 3 forms a sublattice with A 7 as a homomorphic image. Thus z 3 =w = u 3 /\ (v 3 Vx 1 ). Hence z 3 s:v 3 Vx 1 , which implies v 3 Vx 1 = v 3 vz 3 V x = z V x = v by (4). Thus (2) holds and the proof is complete. 1 3 1 2 Theorem 4. 4. such that A 2 , ..• , A 10 If L is a subdirectly irreducible modular lattice ~ HS(L) then M;, 3 is not a sublattice of L, where 93 u' z' z v Figure 4. 7 Proof: As seen above Theorem 3. 1 implies that the existence of a sublattice isomorphic to M~ , 3 such that both diamonds are iso- metric sublattices. Since Lis subdirectly irreducible there is a sequence of transposes x' /v' = b /a , b /a , ••• , bn/an 5= v' /x which satisfy the con0 0 1 1 ditions of Theorem 1. 1. Let us suppose that b 0 /a 0/ b 1 /a 1 • Then By Lemma 4. 1 either (1) u1 /v•/ u1 /v 1 or Suppose that (2) holds. Since x' / v • / x /v , x' ~ v , and so u' /\ v f. x'. 1 1 1 1 94 Trivially [y',z'} - [u' /\ v } #.¢,let us say that y' 1. u' /\ v • 1 1 x 1 > v , u' >U 1 /\v by(2). 1 1 Thusv'-< U 1 /\v -<u'. 1 Since Hence (v',x',y', u' /\ v , u') is a diamond, which together with D = (v,x,y,z,u) and D 1 1 form a sublattice with A 5 as a homomorphic image. Thus (2) cannot hold. Now suppose that (1) holds. u1 = u 1 and v' = v • 1 By Theorem 4. 2 we must have Thus, since x' /vV x /v , x' = x . 1 1 1 Furthermore, if [y', z'} 1. [y , z } then u' Iv' has M as a sublattice which together 1 1 4 with D would form a sublattice with A we may assume y' = y 1 as a homomorphic image. and z' = z ; that is, D' = D . 1 1 Lemma 4. 3 it cannot happen that D n 1 ~ D~. 8 1 ~D 2 • Theorem 4. 2 implies that D D 2 ~ D 3 = D:, D 4 ~>Ds = D~, •••• implies that an-2 <!: v' = ao· 1 Thus Consequently, by Thus we may assume that = D~. By Lemma 3. 4 As pointed out before, this But this contradicts p. d~ (b .., /a , v' /z) n-c. n- 2 = p. d. (b n- 2 /a n- 2 , b n /a n ) = 2. The remaining possibility is that x' Iv' = b /a ~ b /a • 1 1 0 0 this case x 1 /v' ~u /x • 1 1 Lets= u V u 1 and r = s /\ v'. In Now we have the situation already encountered in Theorem 3. 1 (see Fig. 3. 21). Exactly as in the proof of Theorem 3. 1 we conclude that ( 3} v Vu 1 =v 1 V u = s But now the Direct Product Lemma yields M x 2 as a sub- /\ v 1 = x . 1 Also v = v by Theorem 2. 2. Moreover we may assume that y = y and z = z , 1 for otherwise A E HS(L) as seen several times before. lattice unless u 1 7 = u = s. 3 Then x = u I\ v' = u 1 1 1 Thus D = n . 1 95 1 Now either D ~ D 2 or n1 ~ n2: contradiction as above when D 1 Both of these lead to the same equaled D'. The proof is complete. We now introduce the following class of lattices: B 3 =A 1 B4 = = Fig u re 4 . 8 96 ' X =u4 5 vl Figure 4. 8 (Continued) In general Bn consists of n diamonds D , D , ••• , Dn such that for 1 2 i=2, ••• ,n-l = zi = v i+l (1) ui-1 (2) zl = v2 (3) z = u n-1 n B 00 consists of the diamonds n ,D , ••• , D n'... which satisfy (1) and 1 2 (2). 00 Bd is the dual of B 00 and B 00 consists of diamonds [D.1 [ i E Z} 00 97 satisfying (1). d Note that the dimension of B is n+ 1 and that m4 • co co B n , B co , B co , B co E Theorem 4. 5. Let L be a modular subdirectly irreducible lat- tice such that Az, .•. , A 1 s: n < co, then B n n 10 i:. HS(L). If the dimension of L is n + 1, ES( L); if L is infinite dimensional then either B co or Bd is a sublattice of L. co Proof: Since L is subdirectly irreducible and of dimension at least two, L is nondistributive, hence B =(v 1 , x 1 , y 1 , z 1 , u 1 ) is a sub- 1 lattice of L, which by Theorem 3. 1 we may take to be an isometric sublattice. If the dimension of L is two we are done. exists s E L such that either s > u former. 1 or s < v • 1 Otherwise there Let us assume the Now with the aid of Theorem 4. 4 and the second remark preceding Lemma 3. 5 there is a diamond sublattice D and B such that DZ If the dimension of L is three we are done. form BZ. 1 2 If not we may assume by duality that there exists s E L such that s > u . 2 By the first remark preceding Lemma 3. 5 and by Theorem 4. 4 there is a diamond D 3 such that Bz and D in L such that s > u 3 formed by Dz and D D3' D 4 3 form B • 3 If there still exists an s then we apply the same procedure to the lattice of B • 3 This yields a diamond D form a sublattice isomorphic to B • 3 with D 1 form B 4 • 4 such that Dz, This sublattice together If Lis finite dimensional this argument can be repeated to obtain B of L. 3 n as a sublattice of L with u n the greatest element By a dual argument and a possible renumbering, it may also be assumed that v 1 is the least element of L. Since B n is an isometric 98 sublattice of dimension n+ 1 L must have dimension n+ 1. If L is infinite dimensional, then as before, B sublattice of L. Either there are elements sk ~ u dimension of sk/u \. s: v k ~ O. 1 1 1 1 is an isometric in L such that the is greater than for all k > 0, or there are elements such that the dimension of v /\_ is greater than k for all 1 If the former is the case then the process above yields B 00 as If the latter holds B d<X> is a s ubla ttice of L. a sublattice of L. Remark. The above arguments also show that if B 00 is a sub- lattice of L then we may assume that either v 1 is the least element of <X> L or that B <X> is a sublattice of L. In summary, if L satisfies the conditions of Theorem 4. 5 then exactly one of the following four situations occur: (i) for some n, Bn is a sublattice of L with v 1 and un the least and greatest elements of L, respectively; (ii) B 00 is a sublattice of Land v (iii) the dual situation to (ii); (iv) B <X> is a sublattice of L. 1 is the least element of L; <X> We define a~ of L, denoted core (L), to be if (i) holds B <X> if (ii) holds Bd if (iii) holds if (iv) holds B core (L) = n <X> co B <X> The core of L is to be a specific sublattice of L whose elements are numbered in accordance with equations (1). (2) and (3) preceding 99 Theorem 4. 5. There may be more than one core of L, however, it is easy to see that they are all isomorphic. specific core of L. Core (L) stands for some Actually we will see below that the only lattice satisfying the conditions of Theorem 4. 5 with more than one core is M . 4 Consequently we will often refer to the core of L. Lemma 4. 6. Let B , n ;;: 4 be a sublattice of L, where L is a n modular subdirectly irreducible lattice such that A , ••• ,A ~ HS(L). 10 2 Then, ifs E un/v 1 either s ;;: v or s ~ un-l" Let us suppose thats "/! v Proof: s /\ u • 1 2 Since u 1 > z 1 = v 2 ands {:. un-l" Consider f. s , s /\ u 1 < u 1 • If s /\ u 1 = v 1 then 2 Theorem 4. 2 implies thats = v 1 ~ un-l' is an atom of u /v ands /\ u -:# z = v • 1 2 1 1 1 a contradiction. Ifs /\ u 1 Hence s /\ u 1 # x 1 or y 1 then u /v would contain M as a sublattice which with n would form Ar 2 4 1 1 Thus we may assume s /\ u 1 = x • 1 Dually we may assume s V v It will now be shown that s V v n = s Vv n- 1 = x • n = x • n First note that n (1) =z Since s {:. v n- 1 , s Vv n- 1 >v Since v = v ; thus v 1 V s n n- ~ v n . n- 1 n- 1 . n /\x n =v n Hence, by Theorem 4. 2, u < v n (1) now yields u Hence s V v n- 1 = s Vv n n- 1 Vv /\ (v n- 1 n- 1 n- 1 /\ Vs) = xn' as desired. Now s/s /\ vn/xn/vn. Ifs/\ vn ~ vn-l then s/s /\ vn/xn/vn-l' which is impossible because s / s /\ v n has dimension one (sinc e 100 s/s /\ v / x /v ) and x /v has dimension two. n n n n n- 1 un_ 2 and s /\ v n {;. v n-l = x • 1 Vv x . 1 n = 4 Repeating the above argument we s: v 4 = u 2 and s Av 4 ~ u 1 and u 1 A (s A v ) = x • 4 V(s/\ v ) = x = u since u -< u . 4 2 4 2 1 2 s: v This reduction shows that we may assume n is S or 4. obtain s /\ v 1 n = un_ 3 • Since v n ~ x 1 , we have that u 1 A (s Av n) Let us suppose that n = 4. u Thus s /\ v = (s /\ v ) u = u • 4 1 2 1 Thus Hence (s/\ v ) Vv = (s Av ) Vx 2 4 1 4 Furthermore, s /\ v 4 Hence, by the Direct Product Lemma, M .;. u 3 2 since u /\ s /\ v = 1 4 x 2 is a sublattice of L, contradicting Theorem 4. 2. Thus u Let n = S. As before we have that s /\ v S ~ u is A vs· Since s "i! v , it follows thats "/!.vs; thus s /\ vs< 2 1 vs= u • 3 Hence u 1 3 and s /\ v S -J. u . 2 =v 3 V (s /\vs) = u 3 , which is again a V (s /\vs) contradiction by Theorem 4. 2. before we may assume that u 1 By the argument used several times V (s Av S) = x • 3 Since u /\ (s Av S) = x , 1 1 we have that (2) and Since s /\ v S s: x 3 ~ v 5 , s Ax 3 = s /\ v • 5 Thus (3) As s V x since v 5 3 ~ s f;. v = u implies that (4) 3 5 , ( s V x ) V v = s V v = xS and ( s V x ) A v = x 3 5 5 3 5 3 > x . 3 This together with the fir s t transposition of ( 3) 101 (5) With the aid of (2), (3), (4) and (5) it is easy to verify that B together with S, s V x 3 and s /\ v 5 form the sublattice A 10 • 5 This con- tradiction proves the lemma. Lemma 4. 7. Let L be a modular, nondistributive subdirectly irreducible lattice such that A , ••• ,A ~ HS(L). 2 10 Lets EL, then one of the following holds (i) For some vk, up, E core (L) with 0 s: J, - ks: 2, vk s: s s: u,f 00 (ii) The core (L) is B 00 or B 00 ands :a: uk for all k. (iii) The core (L) is Bd or B oo ands :$; u.K for all k. co Proof: Lemma 4. 6. 00 If core ( L) = Bn then v 1 S: s s: un by the remark preceding A straightforward application of Lemma 4. 6 gives vk' u J, E core (L) with 0 s: J, - k ~ 2 and vk s: s S: uJ,. d 00 Hence we may assume the core (L) is B 00 , B 00 or B 00 . Suppose also that for some n s s: u (1) n and If s <?: vk for some k then the proof may be completed as above. Thus, in particular, we may assumes~ v Since s ~ v n- 4 , t > s. By Lemma 4. 6, t <?: v Direct Product Lemma applied to t/v isomorphic to M 3 X 2, n- 4 n- 4 • n- 2 Lett= s V v = u n- 4 • n- 4 • Now the and t/ s yields a sublattice which is impossible by Theorem 4. 2. 102 t s /\ v n- 4 Figure 4. 9 We conclude that if ( 1) holds the lemma is true. s ~ v n and s "/!. v n+ 1 for some n the lemma is true. then either this last statement holds or s ~ v the lemma is true. If core (L) = B for all n. c:o In either case Similarly the lemma is true if core (L) = B~. Hence it may be assumed that core (L) s ~ v n for all n. n Dually if =B . c:o c:o Ifs s: u n for all n then Hence s ~ un for some n . If s s: un then n > n 0 1 0 0 1 . and by choosing the smallest such n, we have s s: u ands ~ u nl n1- 1 This is the cas e considered above. By this and the dual argument we may assume (2) s .,. u n and s -;:. u n for all n 103 Suppose ( 3) and Let n > m. s Vu n Then s /\ u = s Vu m n s: u m for all n and k implies that s /\ u n = s /\ u m • Similarly • Then s, s /\ u , s V u , u , u form a sublattice is on n n m morphic to N , contradicting modularity. 5 Hence by duality we may assume that for some n and k s /\ un f;. uk. Since s /\ un s: un, be chosen such that s /\ un f;. uk and s /\ un s: uk+l. implies that uk_ 3 = vk-l s: s /\ un s: uk+l · (2). k may But then Lemma 4. 6 Then s ~ ~- 3 , contradicting This proves the lemma. Lemma 4. 8. Let L be a modular subdirectly irreducible lattice such that A , ... ,A ~ SH(L). 2 10 Let C = core (L) and suppose that the dimension of C is greater than two. Let s E L such that vk s: s s: uk+l for some vk' ~+l EC thens EC. If vk s: s s: uk+ 2 , vk, ~+ 2 EC, then either s EC or s V uk E [xk+ 2 , yk+ 2 } and s /\ uk E [xk, yk} (see Fig. 4. 10). Proof: If s E u /v n n for n equal k, k + 1 or k + 2 then s E [v. , x , y , z , u } for otherwise u /v had M as a sublattice and 4 n n n n n n n since core (L) has dimension greater than two, A 7 or A 8 If E HS(L). vk s: s s: uk+l and s ~C then uk /\ s cannot be uk = zk+l or zk = vk+l. For then we would have s E uk+l /vk+l' contradicting the above. s /\ uk = vk then s = vk EC by Theorem 4. 3. If Thus s /\ uk is an atom of u /vk which must be either xk or yk' for otherwise A E HS(L). 1 7 Say thats /\ uk = xk. Thus s ~ xk, and therefore s V vk+l = 104 "k+l Figure 4. 10 s V xk V vk+l = s V ~· s V vk+l Hence ~ S: s V vk+l s: uk+l" = ~ or s V vk+l = ~+l' If s V vk+l which is the case considered above, Theorem 4. 2, s Thus either = ~ then s E uk/vk' If s V vk+l = ~+l then by = uk+l' contradicting s f. C. Now suppose vk S: s s: ~+ 2• If vk+l S: s, then the above applies. Similarly, it may be assumed that s '/;. ~-l' that above shows that if s An argument similar to l C then s V ~ E [xk+ 2 , Yk+ 2 } and s /\ ~ = [xk, yk }. Now we are ready to prove the main theorem of this thesis. Theorem 4. 9. l HS(L). Let L be a modular lattice such that A , •.• , A 2 10 CX) Then L E 711 . 4 Proof: Assume L l 711~. Lis a subdir e ct product of subdirectly 105 irreducible lattices. If all these subdirectly irreducible lattices lie in m4co then L E ?71 4co• Hence it may be assumed that Lis subdirectly irreducible. Since L s , s , s in L. 3 4 5 then s s 1 2 ~ uk for ~ ?71 co 4 there exist five noncomparable elements s , s , 1 2 It follows from Lemma 4. 7 that if s all k (~ E core (L) = C). 1 ~~for all k Then the nontrivial quotient I s 1 /\ s 2 lies entirely above uk for all k. Since L is subdirectly irreducible there exists a sequence of transposes x /v = b /a , 1 1 0 0 b /a , •.• , bn/an ,:= s /s /\ s • 1 1 1 1 2 It will be shown that this is impossible by showing that for some j. and £,., i 1 1 = 1, .•. , n v. sa. Su. 2 J·1 1 J.+ 1 (1) i=O, .•• ,n Indeed, bn s u p,n+ 2 contradicts bn ~ s /\ s 2 ~ uk, for all k. 1 (1) by induction. For i = 0, (1) holds with j 0 = t 0 = 1. We prove Let us suppose that (1) holds for i = k and suppose that bk/ak/bk+l /ak+l' Since bk s: u ~+ 2 and ak ~ v jk this transposition implies v jk s: ak+l i u tk+ 2 . It follows from Lemma 4. 7 that v. s ak+l s u. + for some jk+l • lk+l lk+l 2 . By semimodularity bk+l Vu. + is either u. or covers u. lk+l 2 lk+l+ 2 Jk+l+ 2 In either case bk+l 1:. ujk+l + 4 . Since vjk+l s ak+l s bk+l Lemma 4. 7 again implies that (1) holds. It follows fr om thi s that ( 2) 0 s: r. s: 2' 1 vk. S:s.i s:~ .+r. 1 1 1 Clearly the k.'s may be picke d so that 1 ( 3) s. 1.vk. + l 1 1 i = 1,2,3,4,5 106 Since the s.'s are incomparable, k. - 3 ~ k. ~ k. + 3, 1 ~ i, j, 1 1 J 1 k 0 =min (k , k , k , k , k }. 1 2 3 4 5 Then k 0 Hence two of the ki 1 s are equal, say k s 1 f. C. ~ kj ~ k 1 = k • 2 0 ~5. Let + 3, j = 1, 2, 3,4, 5. Let us suppose that Thus, by Lemma 4. 8 it may be assumed that (4) and E (~ + ' yk + 2 } and s /\ uk = 2 1 2 1 {~ , yk }. Suppose s 2 /\ ~l = ~l and s 2 V ~l = yk + 2 • Since s 1 and 1 1 s are incomparable s /\ s = ~ . Since s > xk , it follows that 2 2 1 2 1 1 s V s > s . By a dimension argument s V s < uk + . But 2 2 1 2 1 1 1 Now suppose s 2 f. C. Then s 2 V uk 1 1 1 sl V s2 V vk1+2 = sl V s2 V ~l = xk1+2 V yk1+2 = ~1+2' which is = yk and s 2 V uk = 1 1 If s 2 /\ ~l = xk and s 2 V ~l = xk +z then it 1 1 impossible by Theorem 4. 2. ~l + 2 cannot both hold. Similarly s 2 /\ uk 1 is easy to see that uk + /vk contains A as a sublattice. If s /\ uk = 2 2 1 1 1 2 yk and s 2 V ~l = yk + 2 then uk +z /vk contains A 4 as a sublattice. 1 1 1 1 We conclude that one of s , s is an element of the core C.. By (3) we 1 2 may assume we have the following situation: ( 5) and Here either s 1 f. C or s 1 = ~ . Let us suppose that k 1 = k as well. E {xk ' y k } • 1 1 Since yk s we must have s /\ uk xk . If either s xk or 2 3 1 1 1 1 1 s xk then s and s are comparable. Thus s 1- xk 1- s . By (3) 3 1 3 3 1 1 1 = 3 1 = Then s 3 /\ Uk 1 = = s , s f. C. 1 3 diction. But it has already been shown that this leads to a contra- 107 Suppose we have another pair of equal ki's, say k as before we may assume s parable we must have k assume that k = k 3 1 3 4 = k 1 = yk 3 ± 1. = k . 4 3 Then = yk . Since s and s are incom2 4 4 The situation is symmetric so we - 1; that is, (6) Also as before ( 7) Since the lattice generated by s , s , s , s and C has width 2 4 1 3 four, s S i. C. As pointed out above ks :i!: k - 3 and ks s: k 1 3 + 3 = k 1 + 2. If ks = kl - 3, then by Lemma 4. 8 sS V uk _ 3 E [xk -l' Ykl _ 1 }. Since 1 1 sS S: sS V ~l _ 3 and xk _ 1 s: s 3 and yk _ 1 = s 4 , it follows that sS is 1 1 comparable with s or s , a contradiction. Similarly ks = k - 2, 4 1 3 implies that sS is comparable with s 1 If ks ~ k or s . 2 1 + 1 then 1 <?vk +l = ~ _ 1 ~ yk _ 1 = s • If ks= k 1 or ks= k 1 - 1 then we 4 1 1 have three equal k. 's, a situation already shown to be impossible. s 5 1 For the remaining case we have k distinct. Recall k 0 1 = k kl - 2. So s k 0 3 :<!: k 0 s: k 1 - 2. By Lemma 4. 8 s 3 S: s 3 V ~ - 1. Also k 1 :<!: k 0 :<!: k 1 Thus - 3. Then one of k , k , ks must be k - 2, say k = 4 1 3 3 is comparable to s 1 and k , k , k , ks are 4 1 3 = min [k , k , k , k , ks} and k s: k ~ k + 3. 0 0 1 1 2 3 4 [k , k , k ,ks} = [k , k + 1, k + 2, k + 3 }. 1 3 4 0 0 0 0 Suppose k 2 1 3 E [~ 3 + 2 ' yk3+ 2 } = fxk 1 • Yk1 }. or s , contrary to our assumption. 2 Hence Then one of k , k , ks must be k + 2, say k = k + 2. 3 4 1 1 3 = vk +2 = ~ :<!: Yk = s 2 . 1 3 1 1 proves the theorem. But then s 3 :<!: vk This final contradiction 108 CHAPTER V APPLICATIONS In this chapter we present some applications of Theorem 4. 9. We begin with the characterization of the subdirectly irreducible width four modular lattices announced in [11 ]. Let L be such a lattice. Clearly A , .•. ,A ~ HS(L) so that the previous theorems apply. 10 2 particular L has a core. d In Recall that the core is one of the sublattices CXl B n , B co , B co , B co and, in some sense, it is the largest such sublattice that will fit in L (see the definition following Theorem 4. 5). Recall that B co is a sequence of diamonds D. = (v., x., y., z., u.) i E Z such that 1 CXl 1 1 1 1 1 (1) d and B , B , B co have similar definitions which are given befor e n co Theorem 4. 5. We would like to find the elements of L which are not in cor e (L). With regard to Theorem 4. 7, suppose s E L - core L such that s ~ uk for all k. If t E L - core (L), t ~~for all. k and t -/. s then, as in the proof of Theorem 4. 9, Lis not subdirectly contrary to assumption. of L. irreduc ible, It follows that s must be the greatest e lement A simila r argument shows that if t ~ uk for all k then t i s t h e least element of L. It is cle a r that the only subdirectly i rreducible w i d t h four modular lattic e of d i mension two is M , and that there i. s none of 4 109 dimension one. Hence assume that the dimension of L is greater than two. Let 0 and 1 denote the least and greatest elements of L, if they exist. Now by Lemma 4. 7 and Lemma 4. 8 it follows that if s E L - (core L U (0, 1 }) then s f. ~+l and (2) for some k Lemma 4. 8 also tells us that (3) Thus, for each s E L - (core L U (0, 1 }), there corresponds a k = k(s) such that (2) and (3) hold. It was shown in the proof of Theorem 4. 9 that if s, t E L (core L U (O, l }) and k(s) = k(t) then either A k(s) 2 or A 4 is in HS(L). Thus = k(t) implies s = t. Theorem 5. 1. tice of width four. Let L be a modular subdirectly irreducible lat- Then either (i) L = M . 4 (ii) L has dimension n + 1 > 2, L has B n as a sublattice and for each k, 2 :S: k :S: n - 1 there is at most one element wk E L - Bn dimension k. (iii) Also wk V zk E (xk+l' Yk+l} and wk/\ zk = (xk-l, Yk- l }. L has Bc:o as a sublattice with v equal to the least element of L. 1 (the least element of B,) For each k ~ 2 there is at most one element wk EL - Bc:o of dimension k, and wk V zk E (xk+l' yk+l} a nd 110 wk t\ zk E [xk- l, yk- l }. L may also have a greatest element. (iv) Lis the dual of one of the lattices of (ii). (v) L has B oo as a sublattice. 00 For all k there is at most one ele- ment wk E L - B: which is incomparable with zk and zk V wk E (xk+l, Yk+l} and Zkt\ wk E (xk-l' yk-l }. L may also have either a top ele- ment, a bottom element or both. Furthermore, all the lattices described in (i)-(v) are subdirectly irreducible modular lattices of width four. complete list of such lattices. Hence this is a All the lattices of (i) and (ii) are simple; all those of (iii) without a greatest element and all those of (iv) without a least element and all those of (v) without a least or a greatest element are simple. Now we turn to the subject of lattice varieties. If ;;l., is a class of lattices, we let V(..f:J denote the variety (equational class) generated by .e. Also we let P (zl) denote all ultraproducts of elements of ~. u The next theorem, which is basic to the study of lattice varieties, is due to B. Jonsson. Theorem 5. 2. Let ~ be a class of lattices. Then every sub- directly irreducible member of V( il) is a member of HSP (;!). u More- over, if gf, has only finitely many members each of which is finite then every subdirectly irreducible member of V(;;l) is a member of HS(:l). Furthermore, if V and W are lattice varieties then every subdirectly irreducible member of V V W, the variety generated by V and W, is a member of either V or W. 111 A proof of this theorem appears in [15 ]. If -:!, is a class of modular lattices, each of which has width at most four, then P (;l) is a class of modular lattices, each of width at u most four. Consequently the subdirectly irreducible members of 'mz, the variety generated by width four modular lattices, are just the subdirectly irreducible lattices of width four or less. The subdirectly irreducible modular lattices of width exactly four are given by Theorem 5. 1. M 3 three. is the only subdirectly irreducible modular lattice of width This follows from Theorem 4. 5 and is also in [16 ]. The remaining subdirectly irreducible modular lattices of width less than three are 2 and l, the lattices with two and one elements, respectively. Now we answer the problem suggested in the introduction. Vi =7Jl 00 4 VV(Ai), i = 2, ... ,10 and v 1 = 00 ?r'4 VV(N 5). Let Let 'mbe the variety of all modular lattices and A the variety of all lattices. The quotient sublattice A I'm Theorem 5. 3. varieties is atomic with atoms V l' ... , V 10 . 00 4 of the lattice of all Consequently ?II: is finitely based. Proof: Let W be a variety of modular lattices such that W # 7J!~. Since every lattice is a subdirect product of subdirectly irreducible lattices, there exists a subdirectly irreducible lattice Lin W - "'frl. Hence L has width greater than four. for some i, 2 s: i s: 10. 00 show that vi>- 'JJ/ 4' i = z. 3, •.. , 10. 2 s: i, j s: 10. Then A. E V. l By Theorem 4. 9, A. E HS( L} 1 But then W :::>V(L) ::iv .. - J =. - It only remains to 1 Suppose V. c V. for some i ~ j 1 J =V(A J.) V ?JI:. Ai l. ?JI: and the last pa rt of Theorem 5. 2 imply A. E V(A .) , but this contradicts the second part of 1 J 112 that same theorem. Hence the varieties V , ••• , V are incomparable. 2 10 Now suppose that for some variety V and some j, 2 ~ j ~ 10, V . :::::i V :::::i J:f. 711 4°'. Then, by the first part of the proof V :::::i V. for some i = 2, ... , 1 O. i By the above i = j. Hence V = V.J and V.J > 711 4°', j = 2, ... , 10. If Wis a variety which contains 711 ~ and which is not contained in'!//., then N 5 E W, thus W ~ V . 1 As above it is easy to see that V parable with V 2 , ... , V 10 and that V 1 > 711 =· 1 is incom- Since varieties are determined by the identities all of their members satisfy, Ai ~ 711 :, i = 2, 3, •.. , 10, implies there exist identities E: , E: , ••• , E:lO' such that €i holds in all members of 711~ but fails in 2 3 A., i i = 2, ••• , 10. It follows easily from the first part of the theorem that the modular law together with e , ••• , E:lO' determine the variety 2 7" ' 'I CX) 4 • ,,,, That is, all identities of "' CX) 4 are derivable from the modular identity, x /\ (y V (x /\ y)) = (x /\ y) V (x /\ z), and € , ... , E:l o· 2 This completes the proof. In [2 J K. Baker gives an infinite set of identities ak' k = 0, 1, 2, •.. , which define '!!1 °'. Let r .. ands .. , i ~ i, j ~ 5, i-:/. j be the lat4 iJ iJ . 1 . 1 . th . bl . by t ice po ynomia s in e varia e xi' xj' z iJ , z iJ , .•. , z ij given 2 6 1 (1) Then a ( 2) 6 is the identity 113 The identity holds in all members of 1'1~. a lattice of width four. Hence, if x , ... , xS are substituted into L, 1 x. s: x. for some i :# j, 1 S: i, j s: S. 1 J from this that a 6 To see this let L be But then r .. = s ..• lJ lJ It follows easily Since each member of 711: is the sub- holds in L. direct product of width four lattices a holds in?JI ~· 6 It can be checked that a fails inA , ... ,A ([2] gives an easy method for this; see 10 2 6 also[3]). co Hence m is defined by a and the modular law. 4 6 a has 127 variables. 6 One might ask what is the least number n such that there exists an identity which together with the modular '77(:. defines The following five variable identity was used by Jons son in [16 J as an example of an identity which holds in M (3) a/\ 4 but fails in M : 5 /\ (x. V x.) S: V (a/\ x.)1 1 lS:iS:jS:4 J lS:iS:4 One can show that this identity holds in '771 ~ (use the modular law). This identity fails in A , A , AS, A , A , A but holds in A 4 6 2 3 8 9 and A 10 . J. B. Nation points out that no five variable identity can hold in 'm: and fail in A 10 . Indeed, A 10 has eight elements which are both join and meet irreducible. Now A Fig. S. 1.) A • 4 4 Thus n ~ 8. is generated by four elements a , a , a , a . 4 2 1 3 Let e : f(z , ... , zk) = g(z , •.. , zk) hold in 4 1 1 (See m: but fail in Then for some substitution bi E A , i = 1, ... , k, f(b , ... , bk) 4 1 :# g(b 1 , ... , bk). Each bi= wi(a , a , a , a ). 1 2 3 4 Hence the four-variable identity e~: f(w (x , ... , x ), ... , wk(x , ... , x )) = g(w (x , ... , x ), .. 4 1 1 4 4 1 1 1 114 Figure 5. 1 •. , wk(x , ••• , x )) does not hold in A • 4 1 4 from e4 , e4 holds in '!Ti.=. Moreover, since e4 is derived Similarly there is an eight-variable identity el O which holds in '!Ti. 4 and fails in Al 0 • Since for any two lattice CXl identities in r and s variables, respectively, there is a lattice identity in r + s variable equivalent to the conjunction of the first two, we conclude using (3), e4, es that n ~ 1 7. In [l 7 J McKenzie raises the following question: For which integers k is there a variety which possesses an independent basis with k elements but not one with k + 1? He shows that such varieties exist for any k ~ 12. Let K n be the lattice B adjoined such that w 2 V z 2 = x 3 , w 2 I\ z 2 n with w 2 and w n- 1 = x 1 , wn-l V zn-l = xn and KS be B 5 with w 2 and w 4 adjoined such that w V z 2 = x , w I\ z = x , w V z = y and w I\ z = y • L et KS be 2 3 2 4 2 1 4 4 4 4 3 wn-l A zn-l = xn-z· Let 115 B n 5 with w ~ 5, m adjoined so that w 3 V z 3 = x 4 and w 3 A z 3 = x 2 . n n = (M 4 , Bn+l' KS, KS' A 2 , A 3 , A 4 , A 6 , A 7 , A 8 , A 9 , N 5 } U I 4 s: m < n }. Furthermore if V is any variety properly containing V(K ) then V contains V(K ) V V(L) for some L in S. n n L 0 Then, if V(K ) is covered by V(K ) V V(L) where Lis any member of the set S (K 3 To see this let be a subdirectly irreducible lattice in V but not in V(Kn). If L 0 has width greater than four then one of A , ..• ,A , N is in HS(L ). 10 2 5 0 If A5 E HS(Lo) then M4 E HS(Lo); if AlO E HS( Lo) then K5 E HS(Lo). L 0 has width less than four and is modular then it is M element chain, contrary to L 0 not in V(Kn). If L 0 3 or a two- is modular and has width four then it is one of the lattices described in Theorem 5. 1. it is easily checked that L 0 If Now not in V(Kn) implies that one of M , Bn+l' 4 Ks, KS' K , K 5 , .•. , Kn_ 2 , or Kn-l is a sublattice of L • 4 0 clusion, it has been shown that if n ~ 5 In con- V(K ) is covered by exactly n n + 8 varieties and that any variety properly containing V(K ) contains n one of these n + 8 covering varieties. Now we apply to above result to show that V(K ) has an inden pendent basis with n + 8 equation but no independent basis with more equations. The second part of this statement follows immediately from the fact that all varieties properly containing V(K ) contain one of n + 8 n covering carieties. V( (S - L) V Kn). Let L ES, then by Theorem 5. 2 Lis not in Consequently there is an equation e:L which holds in V( (S - L) V Kn) but fails in L. Now it is easy to verify that ( e:L IL E S} is an independent basis with n + 8 elements. A lattice is called locally finite if its finitely generated sublattice s are finite. A variety is locally finite if all its members are 116 locally finite. CX) Theorem 5. 4. 4 is locally finite. We must show that finitely generated members of 711 Proof: are finite. ?'11 CX) 4 If L is a finitely generated subdirectly irreducible member CX) of 771 4 then it follows from Theorem 5. 1 that L is finite. suppose that L = (G) where !G ! = n. Since Lis finite, it is finite dimensional; say the dimension of L is m + 1. core of L is B m Furthermore By Theorem 5. 1 the , see Fig. 5. 2. The only other possible elements of L - B m are the elements wk such that wk V zk E [xk+l' Yk+ l} and wk/\ zk E [~ _ 1 , yk-l }, k= 2, ... ,m-1. .• , r. Letk , ... ,kr b e those k's such that wk· EL, i= 1, .. 1 l Since the wki is a join and meet irreducible wk , ... , wk E G. 1 1 Let U1' ... 'j m-r- 2} be such that [ kl' ..• ' k r } n U1' •.• ' j m-r- 2} = ¢ and [k1·· .• 'kr } u U1·· .. ,j m-r- 21 = [2, ... ,m-1 }. Note that if wk rf:. L then either ~+l or yk+l is both meet and join irreducible; say Yk+l is join and meet irreducible. Then Yk+l E G, k = j 1 , .•. , jm-r- 2 . Thus there must be at least r plus m - r - 2 elements in G. Therefore r+m-r-2!5:n Thus dim( L) = m + 1 ~ n + 3 We conclude that if Lis a subdirectly irreducible member of 771~ whiCh is generated by n elements then the dimension of Lis less than or equal to n+ 3. Since L has width four or less it follows immediately 117 u x x m m m-2 Figure 5. 2 118 from this that "m. ~ has only finitely many subdirectly irreducible lattices with n generators for any fixed n. 00 Now let L be any member of 111 ments. 4 which is generated by n ele- Then L is a sublattice of L 1 which is the direct product of sub- directly irreducible lattices L., i E I each of which is a homomorphic 1 image of L. TI That is, L' = i EI L.. 1 Since Lis generated by n elements each L. is generated by n elements. Thus, by the above, each L . is 1 1 finite and there are only finitely many distinct members of the set [L. Ii E I}. 1 In order to complete the proof it is sufficient to show that L 1 is locally finite. Lemma 5. 5. Let L' = TI L.1 where each L.1 is finite and there i EI are only finitely many distinct L. 's. 1 Proof: Let f , ••• , fn E 1 erated by f , ••. , fn. 1 Then L 1 is locally finite. TI L. and let L be the sublattice geni EI 1 Since each Li is finite and there are only finitely many different Li 1 s, the set on the n-tuples [(£ (i), ••. , fn (i)) Ii E I} is 1 finite. Pick i , •.• , it such that [f (i), .•• , fn(i) Ii E I} 1 1 . • , fn(ik) I k = 1, ..• , t}. = [£ 1 (ik), .. Let cp be the projection homomorphism from L' to t TI L. 1 k=l k that is, cp(f} = (f(i 1), ••• ,f(ip,)). To prove the lemma we ne e d to show that <D restricted to L is an isomorphism. finite and so that L' is locally finite. It then follows that L is Pick i EI. Then for some k, 119 Since f and g are words in f , ••• , fn' f(i} = 1 f, g E L = (fl , • • • , f n ) • Consequently, if cp(f} = cp(g), i.e., if f(i.} = g(i.} J j = 1, ••• , f, then f(i} = g(i} for all i. J Thus f = g and so cp restricted to L is one-to-one. Corollary 5. 6. co If V is a subvariety of '1t by its finite members. 4 , then V is determined That is, the variety generated by the finite members of V is V. Proof: hers. Any variety is determined by its finitely generated mem- Since the finitely generated members of V are finite the . corollary follows. We now turn to the problem of showing that there are 2~ 0 distine t s ubvar ie ties of 'm ~. Recall that B 00 consists of diamonds D. = (v.,x.,y.,z.,u.}, i = 1,2, ••. 1 1 1 1 1 i = 2, 3, ••• and z 1 1 = v • 2 such that u. 1- 1 = z 1. = v .+l' 1 (See Fig. 5. 3.} Let C co be the lattice B 00 together with elements wk, k = 2, 3, ..• such that wk V zk = xk+l and wk/\ zk = ~-l · Let?( be the class of all sublattices of C co obtained by deleting some of the wk's from C co . Let LE 1' We associate with Lan infinite sequence (a , a , a , ... } of 1 2 3 zeros and ones as follows: wk ~ L. if wk EL then ak-l = 1 and ak-l = 0 if This is clearly a one-to-one and onto correspondence. N I'< I = 2 °. It will be shown that I [V(L} IL E '<}I Hence ~ = 2 °. With each finite sequence ci. zeros and ones (a , a , ••• , an} associate the lattice L 1 2 obtained by appending wk to Bn+Z if ak-l = 1 in such a way that wk V zk =~+land Wk/\ zk = xk_ 1 • 120 Figure 5. 3 Lemma 5. 7. Suppose Land L' are the lattices associated with (a , a , ••. ,) and (b , ••• , bn), respectively. 1 2 1 Then L' E HSP ( L) if and only if for some k (b 1 ,b 2 , ••• ,bn) ~ (ak+l'ak+z•• •• ,ak+n>· u Here the less than or equal to sign means that ai ~ bk+i' i = 1, .•• , n. Proof: Suppose L' E HSP (L). u of L 1 Let L where L 2 1 ESP u(L). Then L' is a homomorphic image Choose an inverse image of each element of L'. be the sublattice of L 1 generated by these inverse images. If 121 we restrict the homomorphism cp which maps L 1 onto L' to L 2 we I 2 from L 2 into L'. But since L 2 has an inverse image of each element of L', cp I L maps L onto L'. Since 2 2 obtain a homomorphism cp L L 2 E SSP u(L) = SP u(L) _:: "'1 Theorem 5. 4. Hence L 2 Q:) 4 and is finitely generated, L The fact that L 2 is finite and L 2 may be regarded as a sublattice of L. 2 is finite by ESP u(L) imply L E S(L). 2 In order to avoid con- fusion we label the elements of L' with primes: D! = (v!,x!,y!,z!,u!), 1 i = 1, 2, ••• , nt2, and w~ (if a. 1 l- 1 = 1). Since L L2 onto L', there is a smallest element b E L the greatest element of L'. 2 2 1 1 1 1 is finite and cp maps such that cp(b) = u~+ 2 ' It is easy to verify that cp restricted to the quotient sublattice of elements of L by replacing L 2 1 2 lying below b is onto L'. Hence, with this quotient sublattice we may assume that u~+z has exactly one inverse image in L • 2 Now by the dual of this argument we may also assume that vi, the least element of L 1 , has exactly one inverse image. Let~ be the class of lattices associated with all the (0, 1)- sequences, (c , c ,, •• , en)' for all n < w together with the lattices M 3 1 2 and M 3 3 ' (Fig. 5. 4). M .)., , .3 Fig ur e 5. 4 122 Lemma 5. 8. Let M be a finite sublattice of the lattice L (of Lemma 5. 7), let N E 'K and let "¥ be a homomorphism of M onto N. Let N have dimension n + 2, n ~ 0 so that vi and u~+l are least and greatest elements of N. Suppose vi and u~+l have unique inverse images under Then for some k, and r such that k - r = n, cp cp -1 I (v ) = v r· 1 -1 I (un+l) = ~ and Consequently "¥ is an isomorphism and thus N ::' M. Furthermore, Mis an isometric sublattice of L. Proof: Let dim X denote the dimension of any finite modular lattice X and let dL be the dimension function on the elements of L. The first conclusion of the lemma implies that Since N is a homomorphic image of M we must have dim M = dim N Also, the fact that dim M = and therefore "¥ must be one-to-one. dL(~) - dL(ur) implies that M is an isometric sublattice of L. Hence it only remains to prove the first conclusions of the lemma. We do this by induction on n. If n= 0 th en N = M 3 -_ n'1 = (v 1 ,x 1 ,y1 ,z 1 ,u1 • I I I I I) -I , -xI , -yI , Letv 1 1 1 -z ' , -• · · images o f v 1 , x 1 , y' , z' , u' , respec t•ive 1y. It f o 1 u b e inverse 1 1 1 1 1 1 1 (-, · a lows f rom the uniqueness o f _, v an d -• u t h at n' v ,-, x , -, y , -, z , -, u ) is 1 1 1 1 1 1 1 1 = diamond sublattice of L. Hence Di = Dk for some k, which proves the lemma in this case. Now suppose dim N =n+ Z, n > O. unique inverse images of u~+l and vi. Let u~+l and ;i denote the Let~· denote the smallest n inverse image of u~. Applying the induction hypothesis to u~/;i, u~/vi 123 and wlu~/vl it follows that u~ =um and v1 = vr, where m - r = n - 1. Now let x~+l' y~+l and u~ denote the largest inverse images of x~+l' Y~+l and u~ = z~+l. Then x~+l, Y~+l and u~ are incomparable and are covered by u~+l • The only way this can happen in L is u~+l = uk, for some k, and fx~+l' y~+l' u~} = fxk, yk' zk = ~-l }. Since x~+l' Y~+l I • bl e, :;f= • an d zn+l = unI are incompara xn+l' ~ yn+l an d~ un =um are incompar- able. Thus um is incomparable with xk' yk. It follows that k= m+ 1 so that u~+l = ~' vl = v r and k - r = m+ 1 - r = n, proving the lemma. Now we return to the proof of Lemma 5.7. By the remarks pre1 ceding Lemma 5. 8 we may apply that lemma with M = L , N = L and 2 w = CD· We conclude that L ~ L and L is an isometric sublattice of L. 2 2 1 Moreover, L 1 2 is simple, since L "'!!: L' and L is simple. 2 some k, r, k - r = n + 1, L 2 Also, for is a sublattice of uk/v r" But the only simple sublattices of uk/v r with greatest element uk and least element v r are those obtained by possibly deleting some of the w 1 L ~ L , (b , .•. , bn) describes L as well as L • 2 1 2 1 m 's from u. /v . Since k r Consequently (b , b , .•• , bn) ~ (ar+l' .•. , ak-l), the desired conclusion. 1 2 The con- version of the lemma is obvious. ~ Now we return to the problem of showing that there are 2 ° varieties generated by single members of'<· Recall that'< consists of all sublattices of CCX> obtained by deleting some of the wk's and associated with each member of'<, a sequence of zeros and ones (a , a , ... ) such 1 2 that wk is in the lattice if and only if ak-l = 1. By a finite block subsequence of (a , a , a , ... ) we mean a sub1 2 3 sequence of the form (ak, ak+l' ... , ak+r). Suppose there exists a set of g sequences such that if a = (a , a , ... ) and b = (b , b , ... ) are in 8 1 2 1 2 124 then either (i) there exists a finite block subsequence (ak, ak+l, ... , ak+r) of (a , a , .•. ) such that 1 2 (ak, ak+l' • • • 'ak+r) 1. (bm' bm+l' · · • 'bm+r) for all choices of m or (ii) there exists a finite block sequence (bk' bk+ 1 , ••. , bk+r) of (b , b , .•• ) such that 1 2 for all choices of m. Let La and ~ be the members of 'C' associated with a and b, respectively. Then the above conditions imply that L a and Lb generate distinct varieties, since, by Lemma 5. 7 and Theorem 5.2, the lattice associated with (ak, ak+l' •.• , ak+r) cannot be in V(~) if the first condition holds and the lattice associated with (bk' bk+l' ••. , bk+r) is not in V(L ) if the second condition holds. ~ a Thus to show the existence ° varieties it is sufficient to construct a set g which satisfies (i) ~ and (ii) such that Igl = 2 ° . Let of 2 s 1 -1001 s2 = 1 0 0 0 0 0 0 1 (n+l)! zeroes s ~ n =10 01 Let N be the set of positive integers, and let T = (i 1 , i 2 , i3' ... } and U = [j , j , j , ••. } be distinct infinite subsets of N. Assume also that 1 2 3 i 1 <i 2 <i 3 <· • · and j 1 <j 2 <j 3 < •.•. Associate the sequence s. s. s. • • · with T and the sequence s. s. s .••. with U. Here 11 12 13 Ji J2 J3 s. s. s. • • • denote the concatenation of the sequences s. s. s . • . . • 11 12 13 11 12 13 We may assume that T ¢. U. Letn ET, n ~ U. Thens n is a finite block 125 subsequence of the sequence associated with T. Suppose it is less than or equal to a finite block subsequence (am, am+l, •.. , am+(n+l) ! + 2> of the sequence associated with U. Then am= am+(n+l) ! + 2 = 1, so that am and am+(n+l) ! + 2 must be either the beginning or end of one of the s j's. It follows that for some jr, jr+l, •.. , jk' (am' am+l' •.• ,am+(n+l) ! + 2 ) has one of the following four for ms. (1) s. s. Jr Jr+l s. Jk 1 s. s. Jr Jr+l s. Jk s. s. Jr Jr+l s. 1 Jk 1 s. s. Jr Jr+l s. 1 Jk Clearly jt < n, t = r, r+l, ••• , k. However, each of these four sequences has length less than or equal to k 2(k-r+l) + ~ (jt+l) ! + 2 t=r Now if n = 1 then the condition jt ~ n = 1 shows that there can be no such jt's and, in fact, it is clear that s s. s. s . . . . in this case. Ji Jz J3 1 is not a block subsequence of H n ~ 2 then since jt < n k 2(k-r+l)+ ~ (jt+l)! +2 t=r n-1 ~ 2(n-1) + ~ t=l <(n+1)!+2 (t+l)! + 2 126 The first inequality expresses the fact that the length of the sequences in (1) is not greater than the length of the sequence 1 s s s ••. sn-l 1. 1 2 3 The second inequality is proved easily by induction. is the length of s n we see that s n Since (n+l) ! + 2 is not less than or equal to a finite block subsequence of the sequence s. s. s .••. associated with U. Thus J1 Jz J3 for g we take the sequences associated with the infinite subsets of N. We have proved the following theorem. Theorem 5, 9. N There exist 2 ° distinct varieties contained in Since there are only countably many varieties defined by a finite set of equation, Theorem 5. 9 has the following corollary, which contrasts Theorem 5. 3, Corollary 5. 10. N There exist 2 ° distinct varieties contained in m4 which are not defined by any finite set of identities, CX) 127 REFERENCES [l J Baker, Kirby A., Equational Classes of Modular Lattices, Pacific J. Math. 28 (1969), 9-15. [2 J Baker, Kirby A., Equational Axioms for Classes of Lattices, Bulletin Amer. Math. Soc. 77 (1971),97 - 102. [3 J Baker, Kirby A., Notes on Finite Basis Theorems for Lattice Theorie s. [4 J Birkhoff, G., Lattice Theory, 3rd edition, Amer. Math. Soc . Colloq. Publ., Vol. 25, Amer. Math. Soc . , Providence, R.I., 1967. [5 J Cohn, P. M., Universal Alge bra, Harper and Row, New York, 1965. [6 J Crawley, P. and Dilworth , R. P. , The Algeb raic Theory of Lattices, Pren ti c e - Hall, t o appear. [7 J Dean, R., Component Sub se ts of the Free Lattice on n Genera to rs, Proceeding of Amer. Math. Soc. 7 (1956), 220-226. [8 J Dilworth, R. P., The Structure of Relatively Complemented Lattices, Annals of Math. 51 {1950), 348-359. [9 J Dilworth, R. P., Struc ture and Decomposition Theory of Lattices, Proc. of Symp. in Pure Math. , II, Providence ( 1961), 3 -1 6. [l 0 J Freese, R., Varieties G enerate d by Modular Lattices of Width Four, to appear. [11 J Freese, R., Sub directly Irreducible Modula r Lattices of Width Four, Not. Amer. Math. Soc. 17 (1970), 680-Al. [12 J Gra'.tzer, G. , Universal Alg ebra , Van Nostrand, Princeton, N.J.,1967. [13 J Gra'.tzer, G. Equational Classes of Lattices, Duke Math. J. (1966), 613-622. [14 J Hong, D. X., Covering Relations among Lattice Varieties, Ph.D. Thesis, Vander bilt University {1970}. 33 128 [15 J Jc{nsson, B., Algebras whose Congruence Lattices are Distributive, Math. Sc and. 21 ( 19 67}, 110-121. [16 J Jcinsson, B., Equational Classes of Lattices, Math. Scand. 22 (1968), 187-196. [l 7 J McKenzie, R., Equational Bases and Non-Modular Lattice Varieties, to appear. [18 J McKenzie, R. , Equational Bases for Lattice Theories, Math. Sc and. 2 7 ( 1970), 24-38. [19 J Pierc e, R. S., Introduction to the Theory of Abstract Algebras, Holt, Rinehart and Winston, New York, 1968. [20 J Tarski, A., Equational Log ic and Equational Theories of Algebras, Proceedings of the 1966 Hannover Logic Colloquium, North Holland, Amsterdam, 1968. [21 J Thrall, R., On the Projective Structure of a Modular Lattice, Proceedings of Amer. Math. Soc. 2 (1951), 146-152~ [22 J Wille, R., Primitive L~n Verb~nden, Math. z.