properties and two variants of Whitman's condition. 1. Introduction. It was shown by G. Gratzer ([6], [7]) that every transferable lattice L satisfies the condition.
PACIFIC JOURNAL OF MATHEMATICS Vol. 85, No. I, 1979
EMBEDDING LATTICES INTO LATTICES OF IDEALS G. GRATZER,
C. R.
PLATT, AND B. SANDS
A lattice L is transferable iff, whenever L can be embedded in the ideal lattice of a lattice M, then L can be embedded in M. This concept was introduced by the first author in 1965 who also proved in 1966 that in a transferable lattice there are no doubly reducible elements. In fact, he proved that every lattice can be embedded in the ideal lattice of a lattice containing no doubly reducible elements. In a recent paper of the first two authors, the idea emerged that one should study transferability via classes K of lattices with the property that every lattice is embeddable in the ideal lattice of a lattice in K. This approach was used to establish that transferable lattices are semi-distributive. This investigation is carried further in this paper. Our main result shows that every lattice can be embedded in the ideal lattice of a lattice satisfying the two semi-distributive properties and two variants of Whitman's condition.
It was shown by G. Gratzer ([6], [7]) that every transferable lattice L satisfies the condition 1.
(X)
Introduction.
L has no doubly reducible element.
In fact, he proved a stronger result, namely, that every lattice can be embedded in the ideal lattice of a lattice satisfying (X). In general, if (P) is a lattice-theoretic property which is preserved by sublattices and which satisfies the assertion '6"(P):
every lattice can be embedded in the ideal lattice of a lattice satisfying (P),
then (P) is a property of all transferable lattices. In addition to (X), properties of a lattice L for which this assertion is known to hold include (SF)
(SD(J
L is sectionally finite (that is, all principal ideals are finite);
for a, b, eEL, a 1\ b = a 1\ c implies that a 1\ b = a 1\ (b V c);
(SD v )
for a, b, eEL, a Vb a V b = a V (b 1\ c).
=a
V c implies that
That g'(SF) holds is a consequence of P. M. Whitman's embedding 65
G. GRATZER, C. R. PLATT, AND B. SANDS
66
theorem [10] and the observation that the partition lattice on a set S is isomorphic to the ideal lattice of the lattice of all finite partitions of S; that 'iff(SD,..) and 'iff(SD y ) hold is the content of a recent paper of G. Gratzer and C. R. Platt [8]. Consider the properties (W)
for a, b, c, d 6 L, a /\ b [a /\ b,
(WI)
V d]
n {a, b, c, d}
for a, b, c, d 6 L,
C ~
C
V d implies that
'* 0;
a /\ b
C
'* 0; for a, b, d 6L, a /\ b Vd [a /\ b, V d] n {a, b, c, d} '* 0. [a /\ b,
(W..)
C
C
V d]
n {a,
V d implies that
b, c, d}
C,
C
a implies that
C
K. Baker and A. W. Hales [2] proved that if a lattice satisfies (W), then so does its ideal lattice. Hence 'iff(W) fails; however, in this paper, we will show that 'iff( WI) and 'iff( W,,) hold. In fact, we will prove that every lattice can be embedded in the ideal lattice of a lattice satisfying the four properties (SD y ), (SDI\) , (WI)' and (WJ simultaneously. More succinctly, our main result is
It follows from the theorem and the preceding remarks that
every transferable lattice is sectionally finite and satisfies (SD y ), (SDI\), (WI)' and (Wu )' By a result of R. Antonius and 1. Rival [1], we conclude: COROLLARY.
Every transferable lattice satisfies (W).
The proof of the theorem is contained in § 2. In § 3 we shall settle the truth or falsity of 'iff(P) for most remaining combinations (P) of the above properties. In particular, it will be shown that 'iff «SDv ) & (SF) & (X» holds and that 'iff«SF) & (SDI\» and 'iff«SF) & (W..» fail. With these results, we can determine the status of 'iff(P) for all but two combinations (P) of the properties (X), (SF), (SD y ), (SDI\), (WI)' and (W,.). These two will be given at the end of the paper. 2.
Proof of the theorem.
DEFINITION 1. Let L be a lattice and let which contradicts Lemma 3. Thus there can be no failures in L",,; that is, L"" satisfies (SD v ), (SDA ), (WI)' and (W..). Next we prove that the homomorphism (j)o of L"" onto L satisfies (*). Let x e L"" and 'Uo, Voe L = L o be such that tpo(x) ~ 'Uo V Vo' Then tpo(x) = tp't(tpI(X» 'Uo V v o, and since (j)'t satisfies (*) there exist 'UlJ VI eLl such that tp't('UI ) = 'Uo, (j)'t(v l ) = VO, and PI(X) ~ 'UI V VI' Proceeding by induction, assume that we have 'U", V" e L" such that tp,,(x) ~ 'U" V V". Then tp,,(x) = tp;(tp,,+l(x» ~ u" V v'" and since tp; satisfies (*) there exist U,,+l, V,,+1 e L"+1 such that tp;(U,,+I) = U", p;(V"+l) = v", and tp"+1(x) ~ U,,+l V V"+I' Now let u = (u" In < w) and v = (v" I n < w). It follows that u, v e Leo, tpo(u) = U, Po(v) = v, and x ~ u V v, whence Po satisfies (*). From Lemma 4, J(L) is embedded in J(L",,). Since L is embedded in J(L), the theorem is proved. As mentioned earlier, we have the following corollary. COROLLARY
11. Every transferable lattice satisfies (W).
The use of homomorphisms, pullbacks, and inverse limits to repair failures stems from a proof in a recent paper of A. Day, namely, the proof (see Theorem 3.2 in [4]) that every lattice is a bounded homomorphic image of a lattice satisfying (W). REMARK.
3. Additional results. In this section we investigate the status of g'(P) for most other combinations (P) of the properties defined
74
G. GRATZER, C. R. t»LATT, AND B. SANDS
in the introduction. First, we shall indicate how certain techniques in a paper of G. Gratzer and C. R. Platt [8] can be modified so as to prove that ~«SDy) & (SF) & (X» holds. Let L be a lattice. It has already been observed that there is a lattice K satisfying (SF) such that L is embeddable in ...r(K). Hence we need only show that for every lattice K satisfying (SF) there is a lattice M satisfying (SD y ), (SF), and (X) such that ...r(K) is embeddable in ...r(M). Let K be a lattice satisfying (SF). In [8J, Gratzer and Platt construct a lattice L(Kr) satisfying (SD y) such that K can be embedded in J(L(Kr From Lemma 3 and their proof it is clear that they in fact embed J(K) in J(L(Kr The lattice L(Kr) consists of certain subsets (called closed subsets) of K x Z, ordered by inclusion. Now we replace Z by (J), and consider the set Lj(Kr) of all finitely generated closed subsets of K x (J), that is, aU closed subsets which are closures of finite subsets of K x (J). Since K satisfies (SF), each element of LAKr ) is finite. Hence Lj(Kr ), ordered by inclusion, is a lattice; in fact Lj(Kr) is embeddable in L(Kr) and· therefore satisfies (SD y). Furthermore, LAKr ) is sectionally finite. Next, it can be proved as in [8J that J(K) is embeddable in J(Lj(Kr and moreover the image under this embedding of each ideal in K is a nonprincipal ideal of Lj(Kr). Therefore (G. Gratzer [7J) the elements of Lj(Kr) may all be "split" to yield a lattice M satisfying (X) such that J(K) is embeddable in ...r(M). It is easy to see that M will still satisfy (SDy) and (SF). Thus we have:
».
».
»,
THEOREM
12.
'if'«SDy) & (SF) & (X» holds.
In contrast to the above, we now establish two negative results. LEMMA 13. If a lattice L satisfies (SF) and (SD A ), then ...r(L) satisfies (SDJ.
Proof. Let L satisfy (SF) and (SDA ), and let A, B, CeJ(L) satisfy A n B = A n C. Let peA n (B V C). There exist b e Band c e C such that p ~ b V c. By (SF), there exist largest elements boe B and Co e C such that bo, Co ~ b V c. Since peA, p /\ boe An B and p /\ coeA n C = An B. Thus the element q = (p /\ bo) V (p /\ co) e A n B, and by the choice of bo, p /\ Co ~ q ~ boo Hence p /\ Co ~ P /\ bo; by symmetry we have that p /\ bo = p /\ Co' Since L satisfies (SD A ), p /\ bo = p /\ (bo V co) = p /\ (b V c) = p. We conclude that p € A n B, and so A n (B V C) = A n B, showing that J(L) satisfies (SD A ).
EMBEDDING LATTICES INTO LATTICES OF IDEALS COROLLARY 14. LEMMA 15. satisfies (W,,).
75
g'«SF) & (SD1.) fails.
If a lattice L satisfies (SF) and (W,,), then J(L)
Proof. Let L satisfy (SF) and (W,,), and suppose that is a (W,,)-failure in J(L). Then there exists an element x e An B such that x It e and x It D, and an element b e B such that b ~ x and bite V D. Since L satisfies (SF), there exists a largest element Xo e An B such that xo ;;;;; b; note that xo.~ x and so XoIt e, XoIt D. Since X o e A n Bee V D, there exist c e e, de D such that X o c Vd. However, bite V D, so b 1: c V d. Finally, since A:::> e V D, we may choose a e A such that a> c V d. But now a 1\ b (0 V d) 1\ b ~ xo, and by the maximality of Xo we obtain that 0,1\ b = Xo' Hence the quadruple
