C. C. Chen and G. Grätzer, on the Construction of Complemented

On the Construction of Complemented lattices

C. C. CHEN AND G. GRATZER*

Department of Mathematics, The University of Manitoba, Winnipeg, Canada Department of Mathematics, Nanyang Umversity, Singapore Communicated by Saunders MacLane Received Dec. 3, 1967

1. INTRODUCTION

R. P. Dilworth's classical resultl that every lattice can be embedded in a uniquely complemented lattice is well known but the proof itself is in accessible to the general reader. The reason for this is a simple one: the proof is too specialized, in the sense that a difficult machinery has to be developed for which no other application has been found. We suggest that all the deeper results developed by Dilworth can be avoided. As the first step of the proof let us observe that it suffices to show that an at most uniquely complemented lattice L o can be embedded in a lattice lv[ of the same type in such a way that 0 and 1 are preserved and such that each element of Lo has a complement in M. The second step consists in applying a method of Dilworth (Theorem 2.2. of UC) for the description of the free lattice L generated by a partially ordered set obtained from Lo by adjoining an unordered set S. Finally, if S is in one-to-one correspondence with the non-complemented elements ofL o , then the method of singular elements used in the final step of Dilworth's proof can be applied directly to yield a subset M ofL, which is a lattice and which has the properties required in the first step. The simplicity of this approach makes it possible to prove some far reaching generalizations. It should be emphasized that we not only get the embedding theorem but we also get the solution to the word problem in the free uniquely comple mented lattice, as in UC.

* This research was supported by National Research Council of Canada. 1 R. P. Dilworth, Lattices with unique complements, Trans. Amer. Math. Soc. 57 (1945), 123-154; this paper will be referred to as UC. 56 57 ON THE CONSTRUCTION OF COMPLEMl>NTED LATIICES

2. RESULTS

A lattice L is called at most uniquely complemented if every element of L has at most one complement; L is uniquely complemented if every element has exactly one complement.

THEOREM 1. Every at most uniquely complemented lattice can be embedded in a uniquely complemented lattice with 0 and 1 preserved. Since every lattice can be embedded into an at most uniquely complemented lattice, Theorem 1 yields Dilworth's result. Let m be a cardinal; a latticeL is anm-eomplemented(resp., m-complemented) lattice, if every element of L, different from 0 and 1, has exactly (resp., at most) m complements.

THEOREM 2. Let L be an m-complemented lattice with incomparable complements (i.e. if b, c are complements of a, then b ;::?: c implies b c). Then L can be embedded in an m-complemented lattice with 0 and 1 preserved.

DEFINITION 1. The lattice L is the free tn-complemented lattice on n generators if it satisfies the following properties: (i) L is an m-complemented lattice; (ii) L contains a subset G of cardinality n such that the smallest sublattice of L which is closed under complementation and which contains G isL; (iii) for any m--eomplemented lattice K, and for any map rp of G (as given in (ii» into K, there exists a lattice homomorphism ip of L into K such that ip extends rp and Lip (the image ofLunder ip), as a sublattice of K, is closed under complementation in K.

THEOREM 3. For any (non-zero) cardinalsm andn, thefree m-complemented lattice on n generators exists and it is unique up to isomorphism. A special kind of (complemented) lattices, typified by the five element modular, non-distributive lattice, are the homogeneous lattices: The lattice L is called homogeneous if the relation a"" b; "a = b or a is a complement of b" is an equivalence relation. Note that every lattice can be embedded in a homogeneous lattice (by adding a new 0 and a new 1) and that any sublattice of a homogeneous lattice, containing 0 and 1, is again homogeneous.

THEOREM 4. Every m-complemented homogeneous lattice can be embedded in an m-complemented homogeneous lattice. CHEN AND GRATZER 58

By adding the adjective "homogeneous" wherever necessary in the definition of "free m-complemented lattice on tt generators," we get the definition of ''free homogeneous m-complemented lattice on tt generators."

THEOREM 5. For any (non.zero) cardinals m, tt, the free m-complemented homogeneous lattice on n generators exists and it is unique up to isomorphism. It may be of some interest to point out that the extensions of a map of the "generators" to a homomorphism is not unique (see Section 6). Thus the uniqueness statements in Theorems 3 and 5 are not as obvious as the unique. ness statement is for free algebras in the ordinary sense.

3. A VARIANT ON A THEME BY DILWORTH

Let Lo be a lattice with 0 and 1; lattice operations will be denoted by v and A. Let S be a set disjoint to Lo ; we make Q = Lo V S a partially ordered set by defining

x ::;;;: y in Q if x = y or x, y E Lo and x::;;;: y in Lo.

DEFINITION 2. Let S(Q, V, fi) be the set of all finite sequences whose entries are elements ofQ, and v, ii, (,). The set P(Q) of (lattice) polynomials over Q is the smallest subset of S(Q, V, fi) such that (i) Q k:. P(Q); (ii) if A, B EP(Q), then (A V B), (A fi B) E P(Q). The parentheses will be omitted whenever there is no danger of confusion.

DEFINITION 3. For some A E P(Q) we define the upper cover A of A as follows: (i) for A ELo , A = A; (il) if A = Bo V B1 , then A exists iff 11., and 111 exist and A = 110 V 111 ; (iii) if A = Bofi B1 , then A exists iff 110 or B1 exists; A = 110 A 111 if both exist, and A = 11i if 11$ (i =f: j) does not exist. Dually, we define 4 the lower cover of A.

Remark. Intuitively, A is the smallest element of Lo containing A in the lattice freely generated by Q. Note that A and 4 are always in Lo, and if both exist, then A :?= 4. .

DEFINITION 4. Set A C B; for A, B E P(Q) if it follows from the rules (1)-(6) below: (1) .A = B; (2) A and lJ existand A ~ lJ (in~); (3) A = All V At , Aok:. B and At C B 59 ON THE CONSTRUCTION OF COMPLEMENTED LATTICES

(4) A = Ao (\ A1 ,Ao k B or A1 k B; (5) B = Bo U B1 , A k Bo or A k B1 ; (6) B = Bo (\ B1 , A k Bo and A k B1 • Set A - B, if A k Band B k A.

THEOREM A. The relation is an equivalence relation. For A E P(Q), let (A) denote the equivalence class containing A, andputL = {(A) I A eP(Q)}. Define (A) ~ (B) iff A k B. This makes L into a lattice. Identifying A EQ with (A), we have Q CL; in fact, L is the lattice freely generated by Q. Finally, tJ (resp., A) is the largest (resp., S1nallest) element of Lo contained in (resp., containing) A, and tJ (resp., A) e.msts ifand only if0 ~ A (resp. A ~ I).

This theorem is contained in Theorem 2.2 of UC if the cardinality of S does not exceed the cardinality ofLo . Since the proof of Theorem 2.2 of UC applies without any essential change we refer the reader to UC. It follows from Theorem A that (1)-(6) apply to ~ in L. Since there is no danger of confusion we zvill write A for (A).

4. THE BAste. LEMMA

Let Lo, S, Q and L be given as in Sectt'on 3. Let C(Lo) denote the set of all two element sets {x, y} such that x is a complement ofy in Lo• Let C denote an arbitrary set of two element sets {x,y} such that x 0:/= y, x, y EQ and ifx ELo ' then x 0,1 and y ¢:Lo (i.e. either x,y E S, or x ELo , x 0,1 and YES, or the symmetric case).

DEFINITION 5. Callan element A ELjoin-singular, if 1 ~ A or x,y ~ A for some {x, y} E C; meet-singular, if A 0 or A ~ x, y for some {x, y} E C. The element A is singular if it is either join-, or meet-singular. Compare this definition with Definition 4.1 of UC.

DEFINITION 6. The only component of A E Q is A. The components of Au B (resp. A (\ B) are Au B (resp. A (\ B) and the components of A and B. Now we are ready to define a new lattice M: M is the set of all those A, A E L, for which A has no singular component, and two new elements u and z. For these new elements the order is given by u ~ X ~ z for all X EM. For the other elements X ~ Y in M if and only if X Y inL. CHEN AND GRATZER 60

THE BASIC LEMMA. M is a lattice. If we identify u with I and z with 0, then M contains Lo as a sublattice, and the 0 and 1 of M and Lo are the same. Furthermore, ifLois a lattice with incomparable complements, and the condition (*) {x,y}, {x, z} e C, y ;;:;;:: z, and y, z eLo imply y = z is satisfied, then C(M) = C(Lo) u C. Proof. Let A,BeM. We claim that AvB=AuB if AuB is non~singular, and A v B = u, otherwise. To prove this assume that A U B if: M, B =j::. u, z. Since A U B is the only component of A U B, that is not a component of A or B, we get that A U B is singular. A U B cannot be meet-singular (because then A, B if: M), thus Au B is join singular. But then every C eL with C ;;:;;:: A, C B, is also join~singular, hence such a C if: M, and so u is the least upper bound of A and B. This, and the dual argument shows that M is a lattice. Since A E Lo , A 0, 1 is non~singular, we have that Lo C M. The argument given above shows that Lo is a sublattice. Similarly, SCM. Now we come to the non-trivial part of the proof. Since Lois a sublattice of M with 0 and 1 preserved,

Furthermore, if {x, y} E C, then x v y = u and x " y = z in lvI, thus CC C(M).

So we shall conclude the proof of the Basic Lemma, if we verify that

x v Y #, X " Y = z in M imply {X, Y} e C(Lo) U C. We can assume X, Y u, z and then the statement can be rewritten as follows: If X u Yand X n Yare singular, then {X, Y} E C(Lo) U C. Since Xu Y can only be join-singular and X () Y meet~singular, we have four cases to distinguish: 1. 1 C Xu Y and X () Y k 0; 2. 1 C X u Y and X () Y C Po ,Pt , with {Po, Pt} E C;

3. to , t1 C Xu Yand X nYC 0, with {to, t1} E C;

4. to, t1 C Xu Y and X nYCPo, PI' with {to, tl }, {Po, PI} e C. Case 1. Since X, Y EM, only (1) or (2) can be applied, and we get

XuY=1 and n =0, 61 ON THE CONSTRUCTION OF COMPLEMENTED LATTICES and since X u Y = g would imply X¢: M and so on, we get

gvY=1 and X A Y=O. Since X g, Y ;;;::: y, the non-existence of incomparable complements implies X g(=X), Y Y(= Y) and so {X, Y} E C(Lo). Case 2. Again, we get g v Y = 1. Let Pl ¢Lo. Since PI ~ Po (in Q) and PI does not exist, therefore only (4) applies to X nYC;Pl' hence X C; PI or Y C; PI . Either will imply that PI exists (since g exists) a contradiction. Hence this case cannot occur. Case 3. This is the dual of Case 2.

Case 4. Let us assume that t1 , PI ¢: Lo• Then tl C; Xu Y implies tl C; X or t1 C; Y, and X nYC; PI implies X C; PI or Y C; PI . Thus we have four subcases to consider. Since the whole situation is symmetric, it suffices to consider the following two cases.

4.a. tl C; X, X ~ PI , to C; X U Y, X nYC; Po. Then tl C; PI' hence tl = PI = X. Thus to C; tl U Y, t1 nYC; Po. We can apply (2) or (5) to to C; t1 U Y, and in both cases we conclude that to C; Y. Similarly, t1 () Y C; Po leads to Y C; Po. Hence condition (*) yields Y = to = Po, {X, Y} = {Po ,PI} E C. ~ ~ 4.b. t1 C; X, Y C; PI , to C; Xu Y, X () Y Po. Since t1 to , to tJ;. X ~ (because to C; X, t1 X would imply X ¢ M), (1) does not apply to to C; Xu Y and (5) gives to ~ Y. Thus to = PI = Y and we can continue as in 4.a. This concludes the proof of the Basic Lemma.

5. ApPLICATIONS

As a first application we prove Theorem 1. Let L o be an at most uniquely complemented lattice. Let No be the set of all elements of Lo that have no complements in Lo ; let N~ = {x' Ix E No} be a set disjoint from Lo' such that x -. x' is 1 - 1 and onto between No and N~ ; set S N~ and

Co = {{n, n'} In E No}. Since Lo is a lattice with incomparable complements, and Co satisfies (*), the Basic Lemma gives a lattice L1 , containing Lo as a sublattice, with the same 0 and 1 as Lo, such that

C(L1) =: , C(Lo) U Co'

Thus, L1 is an at most uniquely complemented lattice and every x ELo has a complement in L 1 • CHEN AND GRATZER 62

By induction, we can construct Ln , n 1,2,... , and then L=U(Lnln 0,1,2,...) is obviously a uniquely complemented lattice as required in Theorem 1. The proof of Theorem 2 is quite similar to that of Theorem 1. Let Lo be an m-complemented lattice with incomparable complements. For every a ELo ' a -=1= 0,1, choose a set S(a) such that for a -=1= h, S(a) is disjoint from S(h) and from Lo , and for all a ELo , a -=1= 0, 1

I S(a) U {x I x is a complement of a in Lo}1 = nt.

This is possible, since by assumption I{x I x is a complement of a in Lo}1 « tn. Now set S = U (S(a) I a ELo ' a -=1= 0,1) and c = {{x,y} I x ELo , x 0, l,y E S(x)}.

Then, again, the Basic Lemma yields a 0, I preserving extension L 1 of Lo , such that c(L,.) = C(Lo) U C.

Therefore, for a E L o , a -=1= 0, 1

I{x I x is a complement of a in L 1}1 = nt, while every a E S is uniquely complemented in L1 ; the elements a¢LouS have no complements. Thus L 1 is an m-complemented lattice. Since any two elements of S are incomparable, L,. is an m-eomplemented lattice with incomparable complements. Now, we can proceed as in the proof of Theorem I, constructing Lo r;;;. L1 r;;;. L 2 r;;;. ... and L = U (Li Ii = 0, 1, 2,...), thus completing the proof of Theorem 2. Next we prove Theorem 4. First, we note that every homogeneous lattice is a lattice with incomparable complements. Indeed, ifL is not a lattice with incomparable complements, then L has elements a, h, c such that h, care complements of a and h > c. But then h ,....., a ,....., c while h ,....., c cannot hold, so L is not homogeneous. Let L o be a homogeneous nt-complemented lattice with incomparable complements. For every a ELo , a 0,1, we again choose the set S(a) as in the proof of Theorem 2, with the only difference that S(a) = S(h) if a""'" h inLo. Then we set

C = {{x, y} I x E Lo , x =1= 0, 1, y E S(x) or x, y E S(z), X =1= y}. The rest of the proof is identical with the proofs of Theorems 1 and 2. 63 ON THE cONsTRUcTION OF' COMPLEMENTED LATTICES

6. FREE ALGEBRAS

It is obvious that the lattice, constructed in Theorem A, is freely generated by Q, since two polynomials were made equivalent only if their equivalence followed from rules (1)-(6), and each one of these rules follows from the lattice axioms. It is almost equally obvious that the construction given in the Basic Lemma is free. We only have to observe that M is nothing but a convenient representation of Lie, i.e. the lattice L modulo the congruence relation e, where e is the smallest congruence relation under which xU y - l(e), x ()y = O(e). for all {x, y} E C. Indeed, it is easy to check that every polynomial is congruent to 0, 1 or a polynomial with no singular component and the rest follows. Using the equivalent definition of freeness we get the following statement: if rp is an arbitrary map of Q into a lattice K which preserves all existing joins and meets of pairs of elements of Q, and if {x, y} E C, then xrp, yrp are complementary in K, then rp can be extended to a homomorphism ip of M into K. This statement implies Theorems 3 and 5; we leave it to the reader to formalize the proof. To conclude the paper, we would like to direct the reader's attention to the curious nature of the free algebras discussed above. Let us take the simplest case: tn = 2, i.e. free bicomplemented lattice L, say on n = No generator. Itis easy to see thatL should be called "free".L has No "generators", say Xo , Xl'"'' Xn , ..., and if we take any map rp of these Xi into any bicom plemented lattice K, then rp can be extended to a lattice homomorphism ip. Moreover, if we can find No elementsao ,... , an ,... in K, such that K is the smallest bicomplemented lattice containing all the ai , then the map Xi -> ai can be extended to a homomorphism of L onto K. In other words, L is a bicomplemented lattice and every "small enough" bicomplemented lattice is a homomorphic image ofL; and that is exactly what "free" should mean. However, this is not how "free" usually is defined. The uniqueness of the extension of rp to a homomorphism ip is always required. In this case, ip is as far from being unique as possible. For mstance, if K is chosen to be the five element modular, non-distributive lattice, then there are 2No extensions, which is maximal since I K IILI = 2No • If tn < No , then free tn-complemented lattices are special cases of "free L'-structures" introduced in a paper by the second author (Trans. Am. Math. Soc. (1968)); free L'-structures are also unique up to isomorphism even though the extension ip may not be unique. The case tn ~ No shows that there are cases not covered by free L'-structures when ip is not unique, and the free algebra is unique up to isomorphism. It would be of considerable interest to find a concept of freeness covering all these cases.

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