Lattices with doubly replaceable decompositions1. Abstract. We study meet decompositions of an element of a complete lattice. We give

Series I: COMMENTATIONES MATHEMATICAE XXX (1991) ROCZN1KI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXX (1991)

An d r z e j Wa l e n d z i a k

(Warszawa)

Lattices with doubly replaceable decompositions

1. Abstract. We study meet decompositions of an element of a complete lattice. We give a generalization of some results of [3].

2. Basic notions. Let L be a complete lattice. Lattice join, meet, inclusion and proper inclusion are denoted respectively by the symbols v , л , ^ and <.

Let 0 be the least element and 1 the greatest element of L.

An element m e L is called irreducible iff, for all x, y e L , m = x л у implies m = x or m = у. M(L) is defined to be the set of all irreducible elements of L.

If aeL, then a representation a = f \ T with T Ç M(L) is called a (meet) decomposition of a. A decomposition a = Д T is irredundant if Д (T— {t}) Ф a for all t e T.

If a = Д T is a decomposition of a, we define t i , • • •, tn. = Д T—({tl , , tn}), for each subset {tl5 ..., tn} of T.

We say that a e L has doubly replaceable irredundant decompositions iff it has at least one irredundant decomposition and whenever a = Д T = Д R are two irredundant decompositions of a, for each t0 e T there exists r0 e R such that

a = t0 A r 0 = t0 A r 0 (1).

The lattice L has doubly replaceable irredundant decompositions iff each a e L has doubly replaceable irredundant decompositions.

A complete lattice L is called upper continuous iff, for every a e L and for every chain C ^ L, a

C

(a

c: ceC). It is well known that every algebraic lattice is upper continuous.

For two elements a, b e L (a ^ b) we define b/a : = [xeL : a ^ x ^ b}.

If b/a = {a, b}, then we say that b covers a, in notation: a -<b. A lattice L is said to be strongly atomic if, for any a, b e L with a < b, there exists peb/a such that a-<p.

O Cf. [1], p. 252.

In a complete strongly atomic lattice L, for each aeL , let Pa denote the set of all elements covering a, and set ua: = \ f P0.

If, for every a eL , the sublattice u ja is modular, then we say that L is locally modular.

3. Some lemmas. Throughout this section L will denote an upper continuous, strongly atomic, locally modular lattice. In the proofs of Theorems 3.7 and 7.3 of [2] it was not used that L is an algebraic lattice but only that L is upper continuous. Therefore, from 3.7 of [2] it follows that the lattice L is semimodular, and by 7.3 of [2] we have

Le m m a

1. I f a , b , p 1,p 2eL, and if pt , p2eP a, b л {px v p2) = a and Pi v b = p2 v b, then = p2.

Now we shall prove

Le m m a 2 .

For each a e L the sublattice u ja is a point lattice, i.e., for every pair of elements x > y in ua/a, x is the join of elements covering y.

P ro o f. Let x, y e u ja with y < x. By the proof of Lemma 4.1 of [3], every element of u ja is the join of elements of Pa. Then

x = \ / ( p v y: a ^ p ^ x , p £ y ).

By semimodularity, y -< pv y whenever a-<p y. Therefore, x is the join of elements covering y and hence ua/a is a point lattice.

Le m m a

3. Let a be an element of L, and let x e l / a and y e u ja . I f x —<x v y, then x

у —су.

P ro o f. Since ua/a is a point lattice, we need only show that b = x л у is irreducible in the sublattice y/b. Suppose on the contrary that y/b contains two distinct atoms p 1 and p2. Obviously we have b ^ x л (pt v p2) ^ x л y = b, i.e.,

(1) x A ( p j v p 2) = b.

By semimodularity, x —c x v p ^ < x v j and x - < x v p 2 ^ x v y . Since x —<x v y we have

(2) x v p j = x v p 2 = x v y .

From (1) and (2) we conclude, by Lemma 1, that pA = p2. This contradiction shows that b is irreducible in y/b. Thus the proof is complete.

Le m m a

4. Let a e L and let p t , p2 be two distinct elements covering a. I f m eM (L), m ^ a and m ^ pt , p2, then a —cm л (p2 v p2).

P ro o f. Apply Lemma 3 and the proof of Lemma 4.2 of [3].

De f in it io n

(cf. [3], p. 10). If a = Д T is a decomposition of a, let

H T: = {peP a: p < t for some te T } and hT: = \J H T.

Lemma 5.

Let a = Д

be an irredundant decomposition of a. Then (i) For each p e H T there exists a unique t(p)eT such that p < t(p).

(ii) For each t e T there is a unique t e P a with t ^ t.

(iii) H T is an independent subset of Pa.

P ro o f, (i) Let p e H T and suppose that there are two distinct tt , t2e T such * that p < Fx and p ^ F2. Since ф t2, t2 ^ t1 and therefore p ^

= a, contrary to the assumption that p e H T. Thus t t = t2 and (i) holds.

(ii) Let te T . Since the decomposition a = Д T is irredundant and the lattice L is strongly atomic there is p e P a with p ^ t. Suppose that there are two distinct such elements p1, p2. Hence p l5 p2 ^ t, and therefore, by Lemma 4, a -<t л (pl v p2). Since this contradicts t л (pt v p2) ^ t

t = a we must have

and (ü) follows.

(iii) Now, let p e H T and q e H T — {p}. Then p ^ t(p) and q < t(q) < tip).

Therefore, \ / (H T — {p}) ^ t(p). Since p ^ t(p) it follows that p ^ \ J (HT — {p}).

Thus H T is independent and (iii) holds.

The next lemma is a generalization of Lemma 4.4 of [3].

Lemma 6.

Let a = Д

be an irredundant decomposition of aeL. I f T = (tl5 ..., tn} is a finite subset of T, then p < Д (Г — T') implies p ^ \ / ( t : te T ') for each p e P e.

P ro o f. We use induction cn n. If n = 1, the lemma follows from Lemma

(ii). Assume the lemma holds for n = к — 1 and.let a —<p ^ tx, ..., tk. We shall consider two cases.

C ase 1. Let p < tk. Then p ^ t lt ..., tk- t and hence by the induction hypothesis

p ^ v ... v tk- ! ^ t± v ... v tk.

C ase 2. Let p £ t k, and set tk : = tk л (p v tk). Obviously, tk ^ tk and hence from Lemma 4 it follows that tke P a. Since p ^ ^ , . . . , ^ and tk ^ t k

^ tl t ..., tk we have

Ik

^

> • • • »

Л

• • • »

Now by the induction hypothesis we obtain

(3) t'k < v ... v fk_!.

Observe that

(4) P v tk = £* v tk.

Indeed, £k v tk = [£k

(p v tk)] v < p v tk and since tk ~<tk v tk and tk —< p v tk we conclude that (4) holds. From (3) and (4) we obtain

p ^ p v t k = t'k v t k ^ t l v . . . v tk.

Thus the lemma follows by induction.

L^emma 7.

I f a = f \ T is an irredundant decomposition of a eL , then hT = \J(t: te T )

ua.

P ro o f. Apply Lemma 6 and the proof of Theorem 4.4 of [3].

Lemma 8.

I f A is an independent set of atoms of the lattice L, and St (i e I) is a collection of subsets of A such that (r')f6jS. =

then / \ ( \ / Sp. ie l) = 0.

P ro o f. As the proof is similar to that of Theorem 6.6 in [2] it will be omitted.

Lemma 9.

Let a = f \ T be an irredundant decomposition of a sL . I f b, c e L are such that b v c = ua and b л c = a, then b v (tip)

c ) ^ u a for each p e H T.

P ro o f. Let p e H T. Since p ^ t{p) we conclude that p £ t(p). Hence, by semimodularity of the lattice L, t(p) -^ p v tip). Since ua/a is a point lattice (cf.

Lemma 2) and c e u ja there exists a subset Q ^ P a such that c -= \J Q. For each qsQ , if q ^ t(p), then t(p) - ^ q v t(p), by semimodularity. Since t(p)eM(L) and hence is covered by a unique element, p v tip) = q v tip) for each q e Q such that q £ t(p ). Then

c v

= V ô v

>-

By Lemma 3, we obtain t{p)

c -<c. Since ua/a is modular,

C

(b V (tip)

с)) = (Ь Л с) V (tip)

c) = tip)

C.

Now b v (tip)

с) ф ua, since otherwise с = ua

c = tip)

, contrary to tip)

—(c. Thus

ua — b y c >— b V (tip)

c).

Now we prove the following

Lemma

10. I f a, b e L and a ^ b , then there exists meM(L) such that m ^ a and m^-b.

P ro o f. Since L is strongly atomic and a < a v b, there exists p e L such that a ^.p < a v b. Let

S : = {x e L : = x ^ a , x ^ p } .

S is nonempty, since a e S. Let C be a chain in S. Then upper continuity yields p

\J C = \J ip

c: ceC ) — a.

Thus \J C e S, and S contains a maximal element m, by Zorn’s lemma. Clearly, meM(L), m ^ a and m ^ b .

The following lemma is a generalization of Theorem 4.3 of [3].

Lemma 11.

Let a — f \ T be an irredundant decomposition of aeL. Then if J is an independent subset of Pa such that J 3 H T, there exists an irredundant decomposition a = / \ R such that HR = J.

P ro o f. Since the lattice L is upper continuous and J is an independent subset of Pa, J is contained in a maximal independent subset M of Pa. Let := J — H T and M x := M — J t , and let b : = \ f and c : = \ J M x. By 6.5 of [2], b v c = \J M = ua and by Lemma 8, b л c = a. Finally, we define sp: = b v (t(p)

c) for each p e H T and sp: = c v V (Л — {p}) f°r each р е Д . From Lemma 8 we conclude that

(5) Д (\f(Ji~{p}y- p e J i) = a.

Since Mfl/a is modular, by the Isomorphism Theorem (cf. [4]) the lattices b/a = b/b

c and ua/c = b v c/c are isomorphic. Therefore, (5) implies that / \ ( s p: p e J 1) = c. Hence

/ \ ( s p: p e J) = /\{ s p: р е Д ) л Д (sp: p e H T) = с л Д (sp: p e H T)

= A (c

(b v (t(p) л с)): p e H T) = Д ({b л c) v (t{p) л c): p e H T)

= Д (t(p) л c: p e H T) = с л f\(t(p): p e H T) = с л Д T= a.

Thus we have

(6) a = / \ ( s p: peJ).

Observe that

(7)

P < / \ ( Sq: Qe J ~{p})»

for each p e J. Indeed, if p e H T, then

sp = b v (t{p) л c) = V Д v (t(p) л c) ^ \ / J i v V (я г “ {р}) = V ( J ~ {P})»

and if р е Д , then

Sp = c V \ / ( Д - {р} )

> \ / H T

Therefore, for each p e J , sp ^ \ / { J — {p})- Then for each p, q s J with p Ф q we have p ^ \ / ( J — {<?}) ^ sq> and hence we obtain (7).

By Lemma 9, sp-<ua for each p e H T. Moreover, if р е Д , then from the semimodularity of L and the maximality of M, sp -<ua. Thus sp-^ua for each p e J. From Lemma 10 it follows that for every p e J there exists rpe M (L ) such that rp ^ sp and rp > ua, i.e., ua

rp = sp.

Let Я = {rp: p e J } and suppose that a < Д (rp: peJ). Since L is strongly atomic, there is q e P a such that q ^ f \ ( r p: peJ). Therefore,

a < q = q A u a ^ Д (rp

ua\ peJ) = f\{ s p. peJ),

contrary to (6). Therefore,

(8) a = f\{ r p: peJ).

Now we shall prove that the decomposition (8) of a is irredundant. Suppose on the contrary that there is p e J such that a = Д (rq: q e J — {p}). Hence using (7) we have

a = ua

a — / \ ( r q

ua: q e J - { p } ) = / \ ( s q: q e J - { p } ) ^ p .

This contradiction shows that the decomposition a = Д R of a is irredundant.

Moreover, HR = J, and the proof is complete.

4. Main results.

Theorem

1. I f a is an element of an upper continuous, strongly atomic, locally modular lattice L and a = f \ T = Д R are two irredundant decom­

positions of a with hT ^ hR, then there exists a subset S Я R and a one-to-one mapping f of T onto S such that, for each te T , a = t

f (t).

P ro o f. For each t e T we define Rt = {reR: a = г л Г}. Let {tl f ..., tn} be a finite subset of T. Suppose that jRtl vj ... и Rtn contains m < n elements, say rt , . . . , rm. Then, for each r e R — {r1, ..., rm}, r

a (i= 1, . . . , и). Hence, since tt contains only one element of Pa, tt < r for each i = 1, . . . , n, r e R —{/*!,..., rm}. Thus

h v ... v t n ^ r i, . . . , r m.

From Lemma 6 we conclude that t x v ... v tn v ... v fm. But this is impossible, since the dimension m of the sublattice f x v ... v r j a is less than the dimension n of the sublattice tx v ... v tn/a. Therefore,

(9) |Rt l u . . . u R J ? n ( 2).

Let te T . Since hT < hR, t ^ \ / H R, and hence there exists a finite subset {Pi> •••, Pk} ^ HR such that t ^ px v ... v pk. By Lemma 5 (i), pt < r(pt) for each i = l , . . . , k . Then t ^ г ( р х) , ..., r(pk), and therefore JRt s {r(px) , ..., r{pff}. Hence for each te T , Rt is finite. From this and (9) we obtain the theorem, by Theorem 1 of [5] on representatives of subsets.

Our second theorem is a generalization of Theorem 4.2 of [3].

Theorem 2.

I f an element a of an upper continuous, strongly atomic, locally modular lattice L has two decompositions a = Д T = Д R, then for each t0e T

there exists r0e R such that a = r0

t0. Moreover, the resulting decomposition is irredundant if so is a = Д T.

P ro o f. Let t0eT. Set S := { s: s = T0

r, reR }. Now we prove that a = t0

t0 is irreducible in t0/a. Suppose on the contrary that Tja contains two distinct atoms px and p2. By semimodularity, t0 -<t0 v px and t0 -<t0 v p2.

(2) For every set X , |X| denotes the cardinality of X.

Hence, since t0 is irreducible we have t0 v p l = t0 v p 2. Moreover,

*o л (Pi v Pi) = a- Then, from Lemma 1 it follows that pt = p2. This contradic­

tion shows that a is irreducible in t ja . Therefore, since L is strongly atomic, a is completely irreducible in t j a , i.e., for every subset U £ t j a , a = / \ U implies aeU. Consequently, from a = / \ R = ï0 л / \ R = f \ S and S Ç t j a we get a e S, i.e., a = T0

r0 with r0 e R.

By the second part of the proof of Theorem 4.2 of [3] we conclude that if the decomposition a = Д T is irredundant, then so is a = r0 л F0. Therefore the proof is complete.

R em ark 1. Theorem 2 does not yield that |T| = \R\ if a = Д T = Д R are two irredundant decompositions of a. In [3] (p.15) an example of an algebraic modular strongly atomic lattice is constructed in which there is an element with two irredundant decompositions of different cardinalities.

As an immediate consequence of Theorem 2 we get

Theorem 3.

Let L be an upper continuous, strongly atomic, locally modular lattice, and let a be an element of L. I f a = / \ Т = Д R are two irredundant decompositions of a such that hT = hR, then \T\ — |R| and for each t0e T there exists r0ER such that a = r0 л t0 is an irredundant decomposition of a.

R e m ark 2. Since every algebraic lattice is upper continuous, this theorem implies the Corollary of [3] (p. 15).

The next theorem is a generalization of Theorem 4.6 of [3].

Theorem 4.

Let L be an upper continuous, strongly atomic, locally modular lattice, and let a = \ J T — f \ R be two irredundant decompositions of a e L with hT ^ h R. Then for each t0e T there exists r0e R such that

(10) a = r0 A t 0 = t0 A r0.

P ro o f. The proof is similar to that of Theorem 4.6 of [3]. Here we apply Lemmas 5 and 6.

The main result of our paper is

Theorem 5.

Let L be an upper continuous, strongly atomic, locally modular lattice. An element a e L has doubly replaceable irredundant decompositions iff for every irredundant decomposition a = / \ T of a, ua ^ \ J ( t : teT).

P ro o f. Sufficiency. We know that each element of an upper continuous strongly atomic lattice has an irredundant decomposition (cf. [6], Theorem 10).

Let a = Д T = Д R be two such decompositions. By assumption,

ua ^ \J (T : t e T ) and ua ^ \ J (f: reR). Therefore, from Lemma 7 it follows that

ua = hT = hR. Hence, by Theorem 4, for every t0e T there exists r0 such that

the decompositions (10) hold. Thus a has doubly replaceable irredundant

decompositions.

Necessity. Let a = f \ T be an irredundant decomposition of a. Suppose that uaf , \ J (t : te T ). Hence, from Lemma 7 we conclude that hT < ua. Then, by Lemma 6.5 of [2], H T is not a maximal independent subset of Pa, and hence there exists an independent subset J ç Pa such that J 3 H T. By Lemma 11, there exists an irredundant decomposition a = f \ R such that HR = J. Now observe that r(p) ^ hT for each p e H R — H T. Indeed, if q e H T, then q e H R and q ф p. Hence q ^ r(q) ^ r(p), since r(p) Ф r(q). Therefore, hT = \ f H T ^ r(p).

Suppose that there exists t0 e T such that r(p) л F0 = a for some p e H R — H T. Then a ^ hT л t0 ^ hT л t0 = t0, a contradiction. Consequently, r(p) л t > a for all te T, and hence a does not have doubly replaceable irredundant decompositions. Thus ua ^ \/ (T : teT ).

Co r o l l a r y 1.

Let L be an upper continuous, strongly atomic, locally modular lattice. L has doubly replaceable irredundant decompositions iff for each a e L and for every irredundant decomposition a = f \ T of a, ua ^ \ J (t: teT ).

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