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SOME STRUCTURE THEOREMS FOR LATTICE- ORDERED GROUPS

BY

PAUL CONRADO)

To Reinhold Baer on his 60th birthday

1. Introduction. Throughout this paper L = L(

SOME STRUCTURE THEOREMS FOR LATTICE-ORDERED GROUPS 213

cardinal sum of the Ay, and this sublattice contains the small cardinal sum of the Ay.

In particular, £ is Archimedean and has a basis if and only if £ can be embedded in a large cardinal sum of subgroups of the real numbers (Theorem 7.3). £ is a small cardinal sum of Archimedean ordered groups if and only if £ is Archimedean and satisfies (F). We also show that if £ is complete and has a basis, then £ is /-isomorphic to the large cardinal sum of the Ay if and only if every basis of £ has an upper bound.

In §8 we explore the possibility of generalizing Theorem 6.1 in order to obtain a structure theorem for an arbitrary /-group with a basis. In §10 we obtain some results for /-groups that do not have a basis. In §11 the previous results are applied to commutative /-groups. Each abelian /-group £ has a unique divisible closure D, and if £ satisfies (F), then so does D. Applying Theorems 6.1, 7.2, and 7.3 to £ we get a complete structure theorem because a lexico-extension of a divisible abelian /-group by a divisible abelian /-group is necessarily direct (Proposition 11.2).

The theory in §9 is entirely due to A. H. Clifford. Also Clifford was ex- posed to an earlier version of the proof of Theorem 6.1 and suggested im- provements, many of which are incorporated in this proof. In particular, the original proof of Theorem 6.1 made use of the special case when £ has a finite basis [4, Theorem 1 ] and this version does not.

2. Cardinal sums, lexico-extensions and lexico-sums. A subset 5 of an /-group £ is convex if

(i) a

214 PAUL CONRAD [May

(a) LaC\(sub semigroup generated by all the Ly for y9ía) = {o} for a//a£r, and aa+aß = aß+aa for all aaELa and aßELß provided a^ß. Moreover, if x = xai + • • • +Xan, where 0?^xai£Pai and a^ay if i^j, then x = xa\J ■ • • Wxa„ and this representation is unique.

(b) [i4 ] = {a — b : a, b EA \. [A] is convex and A is the convex subsemigroup of all positive elements of [A].

(c) [A]=2Z+[LA Corollary I. If Ay(yEY) is a convex subgroup of L and if the subgroup

G of L generated by the Ay is the small direct sum of the Ay, then G= 22+^ ■,■ This theorem and corollary are proven in [4] for finite T, and the exten-

sion to infinite T is immediate. In particular, if C is a convex semigroup of positive elements of L that contains 0, then [C] — {x — y:x, yEC\, [C] is convex and C is the semigroup of all positive elements of [C]. Conversely if 5 is a convex subgroup of L, then S = {a — b: a, b £ S+}, where S+ = {sES: s = 0}. In particular, L= {a — b: a, o£P+}.

Corollary II. If L is a small direct sum of subgroups Ay for 7£T, then the following are equivalent.

(1) L=Z+Ay. (2) Each Ay is convex. (3) ayi+ • • • 4-a7n = 0, where ayiEAyi and y,9£yj for iy^j if and only if

ayi^0for i=l, ■ • • , «. For it follows from the definition of cardinal sum that (1) implies (2) and

(3), and by Corollary I, (2) implies (1). The proof that (3) implies (2) is straightforward, and we omit it.

L is a lexico-extension of an /-group S (notation L = (S)) if 5 is an /-ideal of L, L/S is an o-group, and each positive element in L\S exceeds every ele- ment in 5. Trivially, L = (L), and L— (0) if and only if L is an o-group. Let S be an /-ideal of P. In Lemma 9.1 we show that L = (S) if and only if each nonzero element in L/S consists entirely of positive elements or entirely of negative elements. If S9*0, then L= (S) if and only if each positive element in L\S exceeds every element in S [4]. Let 5 be an /-group, P an o-group, and L = S®T. Define that s4-/£L is positive if />0 or i = 0 and 5 = 0. Then L = (S) and we say that L is a direct lexico-extension of S. In §11 we prove that if L is divisible and abelian, and if L = (S), then L is a direct lexico- extension of 5. In [4] examples are given of lexico-extensions that are not direct. Also, see the example in this paper after Theorem 2.3.

Let Ai, • ■ • , An he o-groups; then by a finite alternating sequence of cardinal summations and lexico-extensions we can construct /-groups from the i4 i in which each A, is used exactly once to make a cardinal extension and the o-groups used to make the lexico-extensions are arbitrary. We call such groups lexico-sums of the Ai. For example, if « = 3, then there are two ways of constructing lexico-sums of Ai, A2, A3 in this order, namely, (i4i+(i42+-43))

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1961] SOME STRUCTURE THEOREMS FOR LATTICE-ORDERED GROUPS 215

and ((Ai+A^+Az). A subset {a7:7Gr} of £ is disjoint or the elements ay are disjoint if each ay>0 and aaC\aß = 0 for all a^ß. In particular, the null set O is disjoint. In [4] the following theorem is proven.

Theorem 2.2. Suppose that L contains n disjoint elements oi, ■ • • , a„ but does not contain m + 1 such elements. Let £,= {xG£: xOo, = 0 for all j^i\. Then the [£,•] are o-groups and L is a lexico-sum of the [£,].

In Theorem 6.1 we generalize this result to the case where each positive element of £ exceeds at most a finite number of disjoint elements, and we prove this result without using Theorem 2.2.

Corollary I. £ is a lexico-sum of n ordered subgroups if and only if L con- tains n disjoint elements but does not contain m + 1 such elements.

Corollary II. Suppose that L contains n disjoint elements but notn + l such elements. Then the following are equivalent.

(a) £ is a proper lexico-extension of an l-ideal. (b) For each proper convex subgroup C of L there exists an element a in £+

such that a>C.

Proof. Suppose that £ satisfies (b). Then since £ is a lexico-sum of a finite number of o-groups, either L = (I) for some /-ideal I^L or £ = .4+£i where A and B are nonzero /-ideals of £. In the latter case no a in £+ exceeds A. Conversely suppose that I^L is an /-ideal of £ and £ = (/). If C is a proper convex subgroup of £, then either CÇ.I or CD£ If CQI, then each aG£+V exceeds C. If CZ)I, then C/I is a proper convex subgroup of the o-group L/I. Pick an X in L/I that exceeds every element in C/I, and let cGC Then X = I+x>I+c and hence I+x—c>I. It follows that x —c>0, and hence x>C.

Note that (a) implies (b) in Corollary II with no restrictions on £, but that the converse is, in general, not true.

For each a(E:L, let La be the intersection of all convex subgroups of £ that contain a. Thus £° is the smallest convex subgroup of L that contains a.

Lemma 2.1. Let 0

216 PAUL CONRAD [May

P={x£Lv:x>0}. It suffices to show that P does not contain a pair of dis- joint elements. Consider a, bEP- By (i) aH\y>0 and bi~\y>0. Thus a(~\y and b(~\y belong to i4 which is ordered, and so aC\bi\y > 0. Therefore aC\b > 0, and hence L" is ordered. We say that L has finite rank if it contains only a finite number of convex subgroups, and that a lexico-extension L of an /-group S is of finite rank if L/S has finite rank.

Theorem 2.3. L has finite rank if and only if L is a lexico-sum of a finite number of o-groups each of which has finite rank and the lexico-extensions used in the construction of L are also of finite rank.

Proof. Suppose that L has finite rank m. Assume (by way of contradiction) that Xi, x2, ■ ■ • , xm+i are disjoint elements in L. Then by Lemma 2.1, XiELx' for i^j. Thus L*1, • • • , LXm+1 are distinct convex subgroups of L, a contradiction. It follows that there exists a positive integer « = m such that L contains w disjoint elements but not «4-1 such elements. By Theorem 2.2, L is a lexico-sum of « o-groups Ai, ■ ■ ■ , An. Each convex subgroup of each A i is also a convex subgroup of L. Therefore each At has finite rank. If B and C are convex subgroups of L and B = (C), then there exists a 1-1 correspond- ence between the convex subgroups of B/C and the convex subgroups of L that are contained in B and contain C. Therefore B/C has finite rank.

Conversely suppose that L is a lexico-sum of o-groups Ai, • • ■ , An each of which has finite rank and the lexico-extensions used in the construction of L are also of finite rank. Then L = (X-\-Y), where X is a lexico-sum of Ai, ■ • • , A, and F is a lexico-sum of A,+i, • • ■ , An for a suitable ordering of the subscripts, and L/(X-\- Y) has finite rank. Now any convex subgroup of L is contained in X+ Y or contains X+ Y. By induction X and Y have finite rank. Therefore X+ Y and hence L has finite rank.

Remarks. This includes as a special case Birkhoff's result: If the lattice of /-ideals of a commutative /-group G is finite, then either G = B + C or G contains a maximal /-ideal that contains every other proper /-ideal [2, p. 237]. If L contains only a finite number of /-ideals and L is nonabelian, then L may contain an infinite number of disjoint elements. For example, let / be the o-group of integers, and for each »£/ let P = I, and let N = ¿j.er+J». For ( • • • , ait • • ■ ) in N and j in / define that ( • • • , »,-,••• )r(j) = (■■■, ai+i, • • • )• That is, the (i+j)th component is replaced by the ¿th component. Let G = IXN and define (i, a) + (j, b) = (i+j, ar(j)+b) and (i, a) positive if i>0 or i = 0 and a is positive in N. It follows that G = (N) and that N is the only proper /-ideal of G. But clearly G contains an infinite number of disjoint