Discrete 224 (2000) 1–14 www.elsevier.com/locate/disc View metadata, citation and similar papers at core.ac.uk brought to you by CORE Free Q-distributive lattices from meet provided by Elsevier - Publisher Connector Manuel Abad ∗, J. Patricio Diaz Varela Departamento de Matematica, Universidad Nacional del Sur, 8000 Bahia Blanca, Argentina Received 18 November 1997; revised 3 September 1999; accepted 21 February 2000

Abstract

In this paper we provide an explicit construction of the free Q-distributive generated by a ÿnite . The starting point is the construction of the free meet over a poset given by Horn and Kimura ( Universalis 1 (1971) 26–38). c 2000 Elsevier Science B.V. All rights reserved.

Keywords: Distributive lattices; Meet semilattices; Free ; Quantiÿers

1. Introduction

Quantiÿers on distributive lattices were considered for the ÿrst time in [16], but it was Cignoli [4] who studied them as algebras, which he named Q-distributive lat- tices. Cignoli’s work was the source of various further investigations (see [2,5,14,15]). We reproduce here the deÿnition given in [4]. A Q-distributive lattice is an algebra (L; ∧; ∨; 0; 1; ∇) of type (2; 2; 0; 0; 1) such that the following conditions are satisÿed, for any x; y ∈ L: (Q1) (L; ∧; ∨; 0; 1) is a distributive lattice with 0; 1, (Q2) ∇0=0, (Q3) x6∇x, (Q4) ∇(x ∧∇y)=∇x ∧∇y, (Q5) ∇(x ∨ y)=∇x ∨∇y. The variety of Q-distributive lattices will be denoted by Q.AQ-distributive lattice will be simply denoted by L. In [5], Cignoli gave a construction of the free Q-distributive lattice over a set X . This construction generalizes that given by Halmos [8,9,10] for the free monadic , and is similar in spirit to the construction of the free distributive p-algebra generated by a distributive lattice given by Davey and Goldberg in [6]. Also Priestley,

∗ Corresponding author. E-mail addresses: [email protected] (M. Abad), [email protected] (J.P. Diaz Varela).

0012-365X/00/$ - see front matter c 2000 Elsevier Science B.V. All rights reserved. PII: S0012-365X(00)00106-0 2 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 in [15], gave a method to obtain, via natural dualities, the ÿnitely generated free alge- bras in all the proper subvarieties of Q-distributive lattices. In this paper we present an approach to free Q-distributive lattices that is completely di erent from those of Halmos–Cignoli and Priestley. Our starting point is the construction of the free meet semilattice over a poset due to Horn and Kimura [11]. After some technical diculties we end up with a concise and clear description of the free Q-distributive lattice over a poset. Our method can be applied in similar situations, such as De Morgan algebras, pseudocomplemented distributive lattices and Ockham algebras (see [1,13]).

2. Preliminaries

In this section we summarize the deÿnitions, properties and notations we need. Let X be a poset (partially ordered set). A Q-distributive lattice L containing X such that X generates L is said to be free on the poset X if every order-preserving function f : X → L0;L0 ∈ Q, can be extended to a homomorphism g : L → L0. A meet semilattice is an algebra (A; ∧), where ∧ is a binary operation satisfying a ∧ (b ∧ c)=(a ∧ b) ∧ c; a ∧ b = b ∧ a and a ∧ a = a, for any a; b; c ∈ A. The variety of meet semilattices will be denoted by S. Any poset in which any two elements x; y have a greatest lower bound x ∧ y is a meet semilattice. If A is a meet semilattice, then A has a canonical ordering deÿned by x6y if and only if x ∧ y = x. In this ordering, any two elements have a greatest lower bound which is equal to x ∧ y. A partially ordered set is called an if x6y only when x = y. The deÿnition of the free meet semilattice over a poset X is similar to that of free Q-distributive lattice over X . Horn and Kimura in [11] gave the following construction of the free meet semilattice over a poset X .   Let X be the set of all non-empty ÿnite of X .IfS1;S2 ∈ X , deÿne S16S2 if for every y ∈ S2 there exists x ∈ S1 such that x6y. This is a partial ordering and  S1 ∧ S2 is the set of minimal elements of S1 ∪ S2. Hence (X;∧) is a meet semilattice. The mapping h : X → X deÿned by h(x)={x} is an order embedding. We have that the subalgebra generated by h(X )isX since any element S ∈ X can be written V S = {x}.Iff : X → B is an order preserving function, where B ∈ S, then the x ∈ S V function f : X → B deÿned by f(S)= (f(S)) is a homomorphism extending f. X is then the free meet semilattice over the poset X .

We collect here the notation we are going to use throughout the paper. FV (I)is the free algebra in the variety V over the poset I. For A ∈ V and X ⊆ A; [X ]V is the (V)-subalgebra of A generated by X . D0;1; Q and S are the varieties of all distributive lattices with 0; 1;Q-distributive lattices and meet semilattices, respectively. If L is a distributive lattice, J(L) denotes the set of join irreducible elements of L. Let A; B be posets, then A[B] is deÿned to be the set of all increasing maps from B to A. A[B] is a poset with the order inherited from the direct power A|B|. 2[B] can be thought of as the set of all increasing subsets of B. M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 3

If L is a meet semilattice, let L ⊕ 1 denote the result of adding to L an element 1 such that x ∧ 1=1∧ x = x, for all x ∈ L. Similarly, we let L 1 denote the result of removing 1 from L if L hasa1. Let X ∗ denote the poset (X; 6∗), where 6∗ is the dual ordering of 6.

 Theorem 2.1. For ÿnite X; FS (X )=X is isomorphic with the meet semilattice of all ∗ ∼ [X ∗] ∗ proper increasing subsets of X ; that is; FS (X ) = 2 \ X .

Proof. If S ∈ X, the mapping ’(S)=C(S], where C denotes the set-theoretical com- plementation and (S] is considered in X ∗, is the required .

For ÿnite X , the set of meet irreducible elements of FS (X )isX and every element x ∈ F (X ) has a unique representation as a meet of an antichain in X . In fact, if S V Ix ={p ∈ X : p¿x}, and Min Ix is the set of minimal elements in Ix, then x = Min Ix. In addition, Ix = Iy if and only if x = y. Let Bn denote the Boolean algebra with n atoms.

Corollary 2.2. If X is a ÿnite antichain with |X | = n; then FS (X )=Bn 1.

We say that the equivalence classes or blocks of an equivalence relation  deÿned on a meet semilattice R are meet quadrilateral-closed if whenever x; y ∈ A for some block A, with x6y, and c ∈ B for some block B, with c6y, then x ∧ c = d ∈ B. If  is an equivalence relation on a meet semilattice R, then  is a congruence if and only if

xÂy ⇔ (x ∧ c)Â(y ∧ c) for every c ∈ R:

The following characterization is an adaptation of a similar result for lattices [7, p. 120].

Theorem 2.3. Let R be a meet semilattice and let  be an equivalence relation on R. Then  is a congruence if and only if (i) Each block of  is convex. (ii) Each block of  is a meet subsemilattice. (iii) The blocks of  are meet quadrilateral-closed.

3. Construction of the free Q-distributive lattice over a ÿnite poset I

In this section we focus on the construction of the free Q-distributive lattice over a ÿnite poset I. We summarize, for further reference, some easy properties of the operator ∇. 4 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14

Lemma 3.1 (Cignoli [4]). In a Q-distributive lattice L; the following conditions hold: (1) ∇(L) is a (0; 1)-sublattice of L. (2) ∇(∇x)=∇x. (3) ∇1=1. (4) x ∈∇(L) if and only if ∇x = x. (5) For each x ∈ L; ∇x is the smallest element in [x) ∩∇(L). (6) If x6y; then ∇x6∇y.

Let G be a ÿnite poset, G = {g1;g2;:::;gn}, and consider the following subsets of FQ(G):

G1 = {∇g1; ∇g2;:::;∇gn};

G2 = {∇(g1 ∧ g2); ∇(g1 ∧ g3);:::;∇(gn−1 ∧ gn)}

= {∇(gi ∧ gj);i¡j}; . . n ^  o Gk = ∇ X ;X⊆ G; |X | = k :

Let [n J = Gk ∪ G: k=1

Theorem 3.2. [J] = F (G). D0; 1 Q

Proof. Since G ⊆ [J] ⊆ F (G), it is sucient to prove that [J ] is closed D0; 1 Q D0; 1 under ∇. Let x ∈ [J] . Then there exist non-empty antichains H in J ∪{0; 1}, such that D0; 1 i _m ^  x = Hi ;m¿1: i=1 W V Since ∇0=0 and ∇1 = 1 we can assume that H ⊆ J . From (Q5), ∇x = m ∇( H ). S i i=1 i 1 2 n Let Hi = Hi ∩ G and Hi = Hi ∩ k=1 Gk . Then _m ^ ^  1 2 ∇x = ∇ Hi ∧ Hi : i=1 S But H 2 ⊆ n G ⊆∇(F (G)), which is a (0; 1)-sublattice of F (G). So, by Lemma Vi k=1V k Q W V V Q W V 3.1, H 2=∇( H 2). Then, by (Q4), ∇x= m ∇( H 1∧∇( H 2))= m ∇( H 1)∧ V i W i V V i=1 i i V i=1 i ∇( H 2)= m ∇( H 1) ∧ H 2. Since H 2 ⊆ J , it follows that H 2 ∈ [J ] , and i i=1 V i i i i D0; 1 since H 1 ⊆ G, ∇( H 1) ∈ G ⊆ J for some j. Therefore ∇x ∈ [J ] . i i j D0; 1 S Corollary 3.3. ∇(F (G))=[ n G ] . Q k=1 k D0; 1 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 5

Fig. 1.

Let I be a ÿnite poset isomorphic with G and let I 0 be a disjoint isomorphic copy 0 0 0 of I; f0 : I → I the isomorphism. Let FS (I ) be the free meet semilattice over I , and 0 0 denote by ∧ the meet operation in FS (I ). Let G=I∪˙FS (I ) and deÿne the following ordering on G. For x; y ∈ G;x6y, if and only if one of the following hold: (i) x; y ∈ I and x6y in I. 0 0 (ii) x; y ∈ FS (I ) and x6y in FS (I ). 0 0 (iii) x ∈ I; y ∈ FS (I ) and f0(x)6y in FS (I ).

Example. Let I be a three-element antichain, I = {x; y; z};I0 = {x0;y0;z0} a disjoint 0 0 0 isomorphic copy of I and f0(x)=x , f0(y)=y ;f0(z)=z . Then G is the poset depicted in Fig. 1. It is clear that if f; g : I → I 0 are two order , then G, with the ordering associated to f is isomorphic to G with the ordering associated to g.

Let G1 = FS (G) ⊕ 1. ∼ [G∗] By Theorem 2.1, G1 = 2 . In addition, G is the set of meet irreducible elements of G1 [12], and every element x ∈ G1;x6= 1, has a unique representation as the meet of a non-empty antichain Hx of G;Hx = Min {p ∈ G: x6p}. We agree that ∅ is an antichain corresponding to 1 and H 6∅ for every antichain H . x V V x V 1 2 0 1 2 If Hx = Hx ∩ I and Hx = Hx ∩ FS (I ), then x = Hx = Hx ∧ Hx . Observe that 1 2 1 2 f0(Hx ) ∩ Hx = ∅, since Hx is an antichain, and f0(Hx ) ∪ Hx can be not an antichain. The following lemma will play an important role.

1 1 Lemma 3.4. For every x; y ∈ G1 = FS (G) ⊕ 1;x6y if and only if Hx 6Hy and 1 2 2 Min{f0(Hx ) ∪ Hx }6Hy .

Proof. x6y if and only if Hx6Hy, that is, for every p ∈ Hy there exists q ∈ Hx such 1 1 1 1 that q6p.Ifp ∈ Hy = Hy ∩ I, then q ∈ I, and so q ∈ Hx ∩ I = Hx . Thus Hx 6Hy . 2 0 1 Suppose p ∈ Hy = Hy ∩ FS (I ). If q ∈ I, then q ∈ Hx ∩ I = Hx , and f0(q)6p. Thus 1 2 2 0 0 2 1 Min {f0(Hx )∪Hx }6Hy .Ifq ∈ FS (I ), then q ∈ Hx ∩FS (I )=Hx .SoMin{f0(Hx )∪ 2 2 Hx }6Hy . 6 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 S ˙ 0 Let x ∈ G1 and let Hx be the non-empty antichain of G = I FS (I ) corresponding 1 0 V 1 0 to x. Observe that f0(Hx ) ⊆ I , and so, f0(Hx ) is an element of FS (I ). Then V 1 2 0 { f0(Hx )}∪Hx ⊆ FS (I ). We will deÿne an operator m : G → G in the following way: m(1) = 1 and if V V 1 1 V 1 2 V 1 2 x ∈ G1;x6=1;x= Hx ∧ Hx , then m(x)= f0(Hx ) ∧ Hx . It is clear that m is well deÿned being that the representation of x is unique. The following lemma proves that the operator m has some of the properties of an existential quantiÿer.

Lemma 3.5. The following conditions are satisÿed: (i) m(x)6m(y) whenever x6y. (ii) m(x ∧ m(y)) = m(x) ∧ m(y).

Proof. (i) Assume that x6y. Then from Lemma 3.4, H 16H 1 and Min {f (H 1) ∪ V V V x y V V 0 x H 2}6H 2. Since X 6 X , it follows that m(x)= f (H 1) ∧ H 2 = f (H 1) ∧ Vx y V V V 0 x x 0 x f (H 1) ∧ H 26 f (H 1) ∧ H 2 = m(y). 0 x V x V 0 y yV V V V (ii) Let x = H 1 ∧ H 2, and y = H 1 ∧ H 2. Then m(y)= f (H 1)∧ H 2, and Vx Vx V y V y V V0 y V y m(x ∧ m(y)) = m( H 1 ∧ H 2 ∧ f (H 1) ∧ H 2)= f (H 1) ∧ H 2 ∧ f (H 1) ∧ V x x 0 y y 0 x x 0 y 2 Hy = m(x) ∧ m(y).

Observe that the condition x6m(x) does not necessarily hold, as it can be seen in the example of Fig. 3, where m(x ∧ y)=x0∧y0 and x ∧ y m(x ∧ y).

In order to obtain this property we will make some identiÿcations on G1 (for instance, in the above-mentioned example, we want to identify the elements x ∧ y and x ∧ y ∧ x0∧y0).

To this end, we distinguish now two types of elements in G1. V 1 Let x ∈ G1 and consider the element f0(Hx ). We say that x is of type I if V 1 V 1 { f0(Hx )}∪Hx is an antichain. We say that x is of type II if Hx6{ f0(Hx )} V 1 2 and f0(Hx ) 6∈ Hx . ∗ V 1 For x of type I, let x be the element associated to Hx ∪{ f0(Hx )}, that is, ^   ∗ V 1 V 1 x = Hx ∪ f0(Hx ) = x ∧ f0(Hx ):

1 1 2 2 V 1 Observe that Hx∗ = Hx and Hx∗ = Hx ∪{ f0(Hx )}. Consider now the interval =[x∗;x]. For x of type II, we deÿne x∗ = x and I = {x∗} = {x}. IHx Hx have the properties of The next remarks and Lemma 3.6 prove that the intervals IHx a partition, and so they will provide us with the desired identiÿcation. , then x cannot be of type II. Indeed, if Observe that if y is of type I and x ∈ IHy 1 1 1 1 V 1 V 1 x ∈ IHy , then Hx = Hy , that is, f0(Hx )=f0(Hy ), and f0(Hx )= f0(Hy ). From ∗ 2 V 1 2 V 1 2 y 6x, Min {Hy ∪{ f0(Hx )}} = Hy ∪{ f0(Hx )}6Hx .Ifx were of type II, there exist two possibilities:

2 V 1 V 1 2 2 V 1 (i) Hx 6{ f0(Hx )} and f0(Hx ) 6∈ Hx . Then there exists a ∈ Hx with a¡ f0(Hx ). 2 V 1 Thus there exists b ∈ Hy such that b6a¡ f0(Hy ), a contradiction. M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 7

1 V 1 1 1 1 1 (ii) Hx 6{ f0(Hx )}. Then |Hx | = 1, that is, Hx = {x1}. Since Hy = Hx = {x1} and V 1 x16f0(x1), it follows that f0(Hy ) ∪ Hy = f0(x1) ∪ Hy is not an antichain, a contradiction.

Lemma 3.6. Let Hx;Hy be non-empty antichains in G associated to x and y; respectively; and suppose that x and y are of type I. If IHx 6= IHy ; then IHx ∩ IHy = ∅.

∗ ∗ Proof. Suppose that there exists z ∈ IHx ∩ IHy . Then x 6z6x and y 6z6y. From 1 1 1 1 1 1 1 1 1 1 Lemma 3.4, Hx∗ 6Hz 6Hx and Hy∗ 6Hz 6Hy . Since Hx = Hx∗ and Hy = Hy∗ we 1 1 1 have that Hx = Hy = Hz . In a similar way, from x∗6z,    1 2 V 1 2 V 1 2 Min f0(Hx ) ∪ Hx ∪ f0(Hx ) = Hx ∪ f0(Hx ) 6Hz and from z6x,

1 2 2 Min{f0(Hz ) ∪ Hz }6Hx : 1 1 But Hz = Hx ,so 1 2 2 Min{f0(Hx ) ∪ Hz }6Hx : Then  2 V 1 1 2 2 Hx ∪ f0(Hx ) 6Min{f0(Hx ) ∪ Hz }6Hx : Similarly,  2 V 1 1 2 2 Hy ∪ f0(Hx ) 6Min{f0(Hx ) ∪ Hz }6Hy : 2 V 1 2 2 V 1 2 2 V 1 So Hx ∪{ f0(Hx )}6Hy and Hy ∪{ f0(Hx )}6Hx . Since Hx ∪{ f0(Hx )} and 2 V 1 2 2 Hy ∪{ f0(Hx )} are antichains, it follows that Hx = Hy .

If x and y are of type II, it is trivial that if I 6= I , then I ∩ I = ∅. Hx Hy Hx Hy

= F (G) ⊕ 1, then z ∈ I for some r either of type I or of Remark 3.7. If z ∈ G1 S Hr , and 1 is of type II. Suppose that z 6=1.Ifz type II. Indeed, for z =1; 1 ∈ IH1 is of type II, we take r = z.Ifz is not of type II, let H 1 = H 1 and H 2 = H 2 \ V r z r z {y ∈ H 2: y is comparable to f (H 1)}. It is clear that H = H 1 ∪ H 2 is a non-empty z V V 0 z r r r 1 2 antichain and r = Hr ∧ Hr is of type I. In addition, z ∈ IHr .

Hence the intervals IH constitute a partition of G1, and therefore they determine an equivalence relation  on G1.

Theorem 3.8. Â is a congruence of meet semilattices on G1.

Proof. It is clear that IH is convex and a meet subsemilattice. All we need to prove is that the blocks IH are meet quadrilateral-closed, that is, whenever x6y; c6y and for some block I , then c and d = x ∧ c ∈ I for some block I . x; y ∈ IHz Hz Hr Hr 8 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14

Suppose x; y ∈ I ;x6y. Let c6y and d = x ∧ c. Assume that x 6= y, otherwise the Hz result is trivial. From c6y6z and Lemma 3.4 1 1 1 1 2 2 Hc 6Hy 6Hz and Min{f0(Hc ) ∪ Hc }6Hz : (1) V 1 V 1 2 (1) If c is of type II, Hc6{ f0(Hc )} and f0(Hc ) 6∈ Hc . 2 V 1 1 2 2 2 (a) If Hc 6{ f0(Hc )} then, from (1), Min{f0(Hc ) ∪ Hc } = Hc 6Hz . From 1 1 V 1 V 1 Hc 6Hz , it follows that f0(Hc )6 f0(Hz ). Thus ^ ^ ^ ^ ^ ^ ∗ 1 2 V 1 1 2 V 1 1 2 z = Hz ∧ Hz ∧ f0(Hz )¿ Hc ∧ Hc ∧ f0(Hc )= Hc ∧ Hc = c: Hence c6x and, consequently, d = x ∧ c = c. 1 V 1 1 1 2 (b) If Hc 6{ f0(Hc )} then Hc ={c1} and Min {f0(Hc )∪Hc }=Min {{f0(c1)}∪ 2 2 Hc }6Hz .So ^ ^ ^ ^ ∗ 1 2 V 1 2 2 z = Hz ∧ Hz ∧ f0(Hz )¿c1 ∧ Hz ∧ f0(c1)¿c1 ∧ Hc ∧ f0(c1) ^ 2 = c1 ∧ Hc = c: Thus, c6x, and, consequently, d = x ∧ c = c. (2) If c is not of type II, by Remark 3.7, c ∈ I , where r is deÿned by Hr 1 1 2 2 2 V 1 Hr = Hc ;Hr = Hc \{s ∈ Hc : s is comparable to f0(Hc )}: ∗ 1 1 2 2 V 1 The element r is associated to Hr∗ = Hr ; Hr∗ = Hr ∪{ f0(Hr )}. Let us see that r∗6z∗ ∧ c. 1 1 1 1 1 1 1 1 Hr∗ = Hr = Hc = Min (Hz ∪ Hc ) = Min (Hz∗ ∪ Hc )=Hz∗∧c: ∗ 1 2 2 Since r 6c, Min {f0(Hr∗ ) ∪ Hr∗ }6Hc . From this and c6z, we have 1 2 1 2 V 1 Min{f0(H ∗ ) ∪ H ∗ } = Min{f0(H ) ∪ H ∪{ f0(H )}} r r  c r  r 1 2 2 = Min f0(H ) ∪ H \ s ∈ H : s c c c V 1 V 1 is comparable to f0(Hc ) ∪{ f0(Hc )} 1 2 V 1 6 Min{f0(Hc ) ∪ Hc ∪{ f0(Hc )}} 2 2 V 1 6 Min{Hz ∪ Hc ∪{ f0(Hc )}} 2 2 V 1 6 Min{Hz ∪ Hc ∪{ f0(Hz )}} 2 2 2 = Min(Hz∗ ∪ Hc )=Hz∗∧c: Hence, by Lemma 3.4, r∗6z∗ ∧ c. Then r∗6x ∧ c = d, and consequently, d and c belong to the block IHr .

Let M = G1=Â and letm  : M → M be deÿned by

m(|x|Â)=|m(x)|Â: m is well deÿned. Indeed, suppose that |x| = |y| . Let r ∈ G such that x; y ∈ I . Then   V Hr r∗6x; y6r. Since m is increasing, m(r∗)6m(x);m(y)6m(r). But m(r)= f (H 1) ∧ V 0 r 2 V 1 ∗ Hr = m(r ∧ f0(Hr )) = m(r ). So m(x)=m(y), and then |m(x)| = |m(y)|Â, that is, m(|x|Â)=m (|y|Â). M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 9

Fig. 2.

Fig. 3.

Lemma 3.9. m(|x|Â ∧ m(|y|Â)) =m (|x|Â) ∧ m(|y|Â).

Proof. By Lemma 3.5 and deÿnitions.

Lemma 3.10. m(|x|Â) ∧|x| = |x|Â. V V V Proof. Let x = H 1 ∧ H 2. Thenm (|x| ) ∧|x| = |m(x)| ∧|x| = | f (H 1) ∧ V V Vx x V V   V   0 x H 2| ∧| H 1 ∧ H 2| = | H 2| ∧| f (H 1) ∧ H 1| . But from the deÿnition x  x V x  V x  0 x x  V V V 1 1 1 1 2 of Â, | f0(Hx ) ∧ Hx | = | Hx |Â. Hencem (|x|Â) ∧|x| = | Hx ∧ Hx | = |x|Â. S ˙ 0 Example. Let I = {x; y} be a two-element antichain. Then G = I FS (I ) is the poset of Fig. 2.

G1 = FS (G) ⊕ 1 is depicted in Fig. 3. The congruence  identiÿes the elements x ∧ y ∧ x0∧y0 and x ∧ y, the other blocks being singletons. Fig. 4 shows the quotient

M = G1=Â, in which we have ignored the classes and referred directly to their representatives. Now we consider the distributive lattice L whose poset of join-irreducible elements is M, that is, let L ∼= 2[M ∗]. The set J(L) of join irreducible elements of L is isomorphic with M, so identifying isomorphic algebras, we may (and will) assume that J(L)=M. It is known that each element x 6=0inL has a unique representation as a join of a non-empty set of mutually incomparable elements of J(L). 10 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14

Fig. 4.

Let ∇ : L → L be deÿned as follows: (1) ∇0=0, ∇1=1. (2) If x ∈ M, then ∇x =m(x). W (3) If x = S, S ⊆ M is the unique representation of x as a join of mutually incom- W parable elements of M, then ∇x = m(S).

It is clear that ∇ is well deÿned, being thatm  is well deÿned and the representation of x is unique.

Theorem 3.11. ∇ is a quantiÿer on L.

Proof. (Q2) is obviously satisÿed. W W Let x = S1, y = S2, where S1 and S2 are non-empty antichains in M. Taking the set S to be the set of maximal elements of S ∪ S , then, sincem  is increasing, W 3 W W 1W 2 W W m(S3)= m(S1 ∪S2). Then ∇(x∨y)= m(S3)= m(S1 ∪S2)= m(S1)∨ m(S2)= ∇x ∨∇y. So (Q5) holds. W W W W From (Q5) and Lemma 3.9, ∇(x ∧∇y)=∇( S ∧∇( S ))=∇( S ∧ m(S ))= W W 1 2 1 2 ∇( (S ∧ m(S ))) = ∇(S ∧ m(S )). But x ∧ m(y) ∈ M, for x ∈ S and y ∈ S ,so 1 2 1 2W W 1 W 2 ∇(x ∧ m(y)) =m (x ∧ m(y)). Hence ∇(S ∧ m(S )) = m(S ∧ m(S ))) = m(S ) ∧ W W 1 2 1 2 1 m(S2)= m(S1) ∧ m(S2)=∇∨S1 ∧∇∨S2 = ∇x ∧∇y, and (Q4) is proved. Property (Q3) is an immediate consequence of Lemma 3.10.

We summarize here the essential steps in the given construction. We started with a ÿnite poset I (of generators), and having in mind the properties of S 0 ˙ 0 the operator ∇, we considered an isomorphic copy I of I and the poset G=I FS (I ). Then we constructed G1=FS (G)⊕1 and we deÿned an operator m on G1. By means of some identiÿcations on G1 (the most dicult part) we considered the quotient M =G1=Â M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 11 and the operatorm  on M having the properties of Lemmas 3.9 and 3.10. Finally, we considered the distributive (0; 1)-lattice L having M as its set of join-irreducible elements and we extendedm  to a quantiÿer ∇ on L. Now we will prove that (L; ∇) is the free Q-distributive lattice over the poset I.

Since G generates G1 =FS (G)⊕1 as a meet 1-semilattice (meet semilattice with 1), then |G|Â = {|x|Â: x ∈ G} generates M = G1=Â. Since M generates L, then [|G|Â]D = L. Sn 0; 1 Now we return to the set J of Theorem 3.2: J = k=1 Gk ∪G, and [J ]D =FQ(G). ∼ 0; 1 We want to prove that FQ(G) = L. Let f : G → I be an . Then if |I|Â = {|x|Â: x ∈ I}, f induces an ∗ ∗ ∗ order mapping f : G →|I|Â ⊆ M deÿned by f (g)=|f(g)|Â, and then f can be extended to a Q-homomorphism

∗ f : FQ(G) → L:

∗ Lemma 3.12. |G|Â ⊆ f (J).

0 Proof. Let z ∈|G|Â = |I∪˙FS (I )|Â.Ifz = |x|Â, with x ∈ I then there exists y ∈ G ⊆ J such that f(y)=x. Then f∗(y)=z, that is, z ∈ f∗(J ). If z = |x| , with x ∈ F (I 0), V Â V S then there exists an antichain X of I 0 such that x = X . Then x = f (X )=m(x ), V 0 1 1 for an antichain X of I and x = X . Hence 1 1 1     ^ ^ z = |x|Â = |m(x1)|Â =m(|x1|Â)=∇(|x1|Â)=∇ X1 = ∇ |X1|Â : V Â Let H = {y ∈ G: f(y) ∈ X } and let y = ∇( H). 1 V V V It is clear that y ∈ J, and f∗(y)=f∗(∇( H)) = ∇f∗( H)=∇( f∗(H)) = V ∗ ∇( |X1|Â)=z. Therefore, z ∈ f (J ).

Corollary 3.13. f∗ is onto.

Proof. From Theorem 3.2 and the previous lemma, f∗(F (G)) = f∗([J] )=[f∗(J )] ⊇ [|G| ] = L: Q D0; 1 D0; 1 Â D0; 1

Corollary 3.14. There exists an order embedding g from M into the set J(FQ(G)) of join irreducible elements of FQ(G).

Proof. Immediate from Birkho ’s .

Our next objective is to prove that g is an order isomorphism, that is, that ∼ M = J(FQ(G)).

Lemma 3.15. J(FQ(G)) ⊆ [J]S ∪{1}.

Proof. Let x ∈ J(FQ(G)), x 6= 1. Then, from Theorem 3.2, there exist antichains Wm V V Hi ⊆ J such that x = i=1( Hi). Since x is join-irreducible, it follows that x = H for an antichain H. Thus x ∈ [J]S . 12 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14

0 Let t : G = FS (I )∪˙I → J be such that t I : I → G is an order isomorphism and 0 t : FS (I ) → J deÿned by V  ^  t f0(X ) = ∇ t(X ) ; X an antichain in I.

Lemma 3.16. t is an onto order-preserving mapping.

Proof. For x; y ∈ G, with x6y, we have the following cases: (1) x, y ∈ I. Then by deÿnition t(x)6t(y). (2) x, y ∈ F (I 0). Then x, y are associated to antichains X , Y , respectively, of I S V such that X 6Y and, consequently, f (X )6f (Y ). Hence, t(x)=t( f (X )) = V V V 0 0 0 ∇( t(X ))6∇( t(Y )) = t( f0(Y )) = t(y). 0 0 (3) x ∈ I, y ∈ FS (I ). Then f0(x)6y. Since f0(x), y ∈ FS (I ), we are in case 2, that is, t(f0(x))6t(y). On the other hand, t(x)6∇t(x)=t(f0(x)). Thus t(x)6t(y).

Let us see that t is onto. Indeed, if y ∈ G, since t  is an order isomorphism, I V there exists x ∈ I such that t(x)=y. Suppose that y ∈ J \ G. Then y = ∇( Y ), where V V Y is an antichain in G. Let H = {s ∈ I: t(s) ∈ Y }. Then t( f (H)) = ∇( t(H)) = V 0 ∇( Y )=y.

From the previous lemma, t can be extended to an onto homomorphism of meet 1-semilattices  t : G1 → [J]S ⊕ 1: Suppose that z is a type I element. Then ^ ^  ∗ 1 2 V 1 t(z )=t Hz ∧ Hz ∧ f0(Hz ) ^  ^   1 2 V 1 = t Hz ∧ t Hz ∧ t f0(Hz ) ^ ^  ^  1 2 1 = t(Hz ) ∧ t Hz ∧∇ t(Hz ) ^  ^  1 2 = t Hz ∧ t Hz = t(z): Hence, if xÂy, t(x)=t(y), that is, Â ⊆ Ker t.

∼ Theorem 3.17. FQ(G) = L.

 Proof. From  ⊆ Ker t, there exists an onto homomorphism k : M = G1= → [J ]S ⊕ 1. Then by Lemma 3.15 and Corollary 3.14,

|M|¿|[J]S ⊕ 1|¿|J(FQ(G))|¿|M|: So the embedding g of Corollary 3.14 is an order isomorphism. Since J(L)=M,it ∼ follows that FQ(G) = L. M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14 13

Fig. 5.

Let L(P) be the free (0; 1)-distributive lattice over the poset P [3]. Then,

∼ 0 Corollary 3.18. ∇(FQ(G)) = L(FS (I )).

0 Proof. Â is trivial on FS (I ).

Let m denote an m-element chain. Then,

∼ Corollary 3.19. FQ(m) = L(2 × m).

· 0 ∼ ∼ [2×m] Proof. It is clear that I ∪ FS (I ) = 2 × m, and then G1 = 2 . Since  is the trivial congruence, it follows that ∼ [2[2×m]] ∼ FQ(m) = 2 = L(2 × m):

The following example was also worked out by Priestley [15].

Example. The poset of join-irreducible elements of the free Q-distributive lattice over a two-element antichain is shown in Fig. 4, where ∇ =m. The corresponding free Q-distributive lattice is depicted in Fig. 5 where the element p denotes ∇x ∧∇y. 14 M. Abad, J.P. Diaz Varela / Discrete Mathematics 224 (2000) 1–14

Acknowledgements

We would like to thank the referees for their valuable suggestions.

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