1-based theories - the main gap for a-models

Arch. Math. l,ogic (1995) 34:285-300

Archive for

Mathematical

logic

9 Springer-Verlag 1995

1-based theories - the m a i n gap for a - m o d e l s B. Hart, A. Pillay, S. Starchenko* Department of Mathematics, Notre Dame University, Notre Dame, IN 46556, USA Received October 19, 1992

Abstract. We prove the "Main Gap" for the class of a-models (sufficiently saturated models) of an arbitrary stable 1-based theory T. We (i) prove a strong structure theorem for a-models, assuming NDOP, and (ii) roughly compute the number of a-models of T in any given cardinality. The analysis uses heavily group existence theorems in 1-based theories.

1 Introduction A 1-based theory is a stable theory T such that if M is a model of T and A, B are algebraically closed sets in M eq then A and B are independent over A A B; namely forking has an extremely simple form. Shelah's work on classification theory ([Sh]) left open, among other things, the problem of finding a "structure/nonstructure" theorem for the class of sufficiently saturated models of a stable, not necessarily superstable theory. We solve this problem here for 1-based theories. In fact we show that if T is 1-based with NDOP then any a-~r(T)-saturated model of T (henceforth called an a-model) is prime (in the category of a-models) over a nonforking tree of elements, where the tree has height at most ITI +. If in addition T is shallow (which we shall see has meaning here), then the number of a-models of T of cardinality Ra is less than ~,~(]w + c~]) where ~ = (21rl) +. Otherwise T has 2 ~ a-models of cardinality A, for most A. A key element of the machinery developed in this paper is that (for 1-based theories) one can deduce "finiteness" theorems from NDOP alone, which can be used to replace arguments which use superstability. Specifically it is shown that any element is interalgebraic with a set B U C where B is a finite set of independent regular trivial elements and C = {Cl,... cn } where t p ( c i / B c t , . . . , ci-1) * Authors 'partially supported by the NSERC, and by NSF grants DMS90-06628 and DMS9203399 Correspondence to: B. Hart

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is "essentially" the generic type of a group, for each i. This was already shown in [HPS] for 1-based trivial theories (in fact for the broader class of perfectly trivial theories), where the group elements disappear. The more general result in this paper uses the analysis from [HPS] together with a group existence theorem for 1-based theories due to Bouscaren and Hrushovski ([BH]), the fact that an cx~-definable group in a stable theory is enveloped in a definable group ([H]), and the understanding of the structure of 1-based groups given by [HP]. The present paper is self-contained, and can be read independently of [HPS]. Henceforth T will denote a stable theory (not necessarily countable), and we work always in N ~q where N is a sufficiently large and saturated model N of T. a, b, c . . . denote elements (which include finite tuples as we work in Neq), and A, B, C . . . d e n o t e small sets. We assume knowledge of general stability theory and stable groups, for which the reader is referred to [Sh], [M] and [Po]. For q a strong type over some set, Cb(q) denotes the canonical base of q. t~r(T) denotes the first regular cardinal > n(T). By an a-model we mean a F~,-saturated model of T, where n = ~r(T), and a-prime means prime in the category of a-models. Remember from [Sh. Ch.IV] that t~r(T) is the first regular cardinal _> t~(T). The reader can, if he or she wishes, read "lTl+-saturated model" for a-model. We begin by recalling some definitions and facts. Definition 1.1. T is said to be 1-based if for all a and A, Cb(stp(a/A)) C acl(a). Definition 1.2. A stationary type (or strong type) q over A is said to be trivial if whenever B D A, and aa,a2,a3 are realisations of qlB, then {a,a2,a3} is A-independent if it is pairwise A-independent.

Remarks 1.3. i) Triviality is a parallellism invariant. ii) If stp(a/A) is trivial and b E acl(aA) - acl(A) then stp(b/A) is trivial. iii) Clearly if q E S(A) is the generic type of a connected group G (where G is possibly infinitely definable over A) then q is not trivial. Thus, if p is a strong type over A, and for some B D A, there is q E S(B), the generic type of a connected group, such that there are a realising plB and b realising q with b E acl(aB), then p is nontrivial. Fact 1.4. ([HI) Let G be a group, c~z-definable over some set A. (This means G is the set of realisations of a partial type over A and the group operation of G is definable over A). Then G is the intersection of (at most ITl-many) d{finable (over A) groups. Fact 1.5. ([HP]) Let T be 1-based. Suppose G is a group definable over A. Let A C B and let b E G. Then stp(b/B) is the generic type o f a coset of an ~ acl(A )-definable subgroup of G. Generally when we talk of a connected group G over a set A, we will mean that G is c~-definable over A.

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2 Trivial types and nontrivial types in 1-based theories In this section T is assumed to be 1-based.

Definition 2.1. i) If {B1, B2, B3} is pairwise independent over A, but dependent over A, then we say that {B1, Bz, B3} forms a triangle over A. ii) {B1, B2, B3} is said to form an algebraic triangle over A, if it forms a triangle over A and moreover each Bi is algebraic over the other two together with A. iii) B is said to be part of an (algebraic) triangle over A, if there exist B2, B3 such that {B, B2, B3} forms an (algebraic) triangle over A. L e m m a 2.2. Let q = stp(a /A). The following are equivalent. i) q is nontrivial ii) a is part of a triangle over A iii) there is a t E acl(aA) - acl(A) such that a t is part of an algebraic triangle over A. Proof i) ~ ii) (this does not require the assumption of 1-based). As q is nontrivial we can find B 3 A such that a realises qlB, and b , c realising qlB such that {a, b, c} forms a triangle over B. Then clearly {a, b, cB} forms a triangle over A. ii) --~ iii). Suppose { a , B , C } forms a triangle over A. Let a t E C b ( s t p ( B C / aA)) be such that B forks with C over atA. Then a t C acl(aA) - acl(A), and by 1-basedness a t C acl(BC). {a t, B, C ) remains a triangle over A. Doing the same thing for B and then C in place of a, yields b E acl(AB) and c E acl(AC) such that {a t, b, c} forms an algebraic triangle over A. iii) ~ i). Let a t c acl(aA) be such that {a t, B, C } is an algebraic triangle over A. Let B t, C ' be such that stp(Bt/A) = stp(B/A), stp(Ct/A) = s t p ( C / A ) and atBC is independent from B t C t over A. Let al,az be such that s t p ( a l , B , C t / A ) = stp(a2,Bt, C /A) = stp(at,B, C /A). Then clearly a t , a l , a 2 are realisations of s t p ( a t / A ) l ( W C ' ) and { a t , a l , a 2 } forms an (algebraic) triangle over B t C t. Thus stp(at/A) is nontrivial. By Remark 1.3 ii), q is nontrivial. [] L e m m a 2.3. Let q = stp(a /A ). The following are equivalent: i) a is part of an algebraic triangle over A, ii) there are B ~ A, a connected group G over B, some generic element b of G over B, and some realisation a t of q In such that a t and b are interalgebraic over

B, iii) there are B ~ A, a B-definable group G, some b C G-acl(B) and some realisation a I of qlB such that a t is interalgebraic with b over B. Proof i) --~ ii) is given by an important result in [BH] (Propositions 2 and 3) which states, among other things, the following: (*) Suppose a is part of an algebraic triangle over A. Let c be a realisation of q independent with a over A. Let C = acl(A, c). Then there are a t interalgebraic

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with a over C, some connected (abelian) group G over C, some oo-definable set P over C and a C-definable regular action of G on P, such that stp(a~/C) is a generic type of P over C for this G-action. We see from (*) that if d is any generic element of P independent from a over C, then P has an induced C, d-definable group structure and a t is generic over C, d. As a is independent with C, d over A and a is interalgebraic with a ~ over C, d, we conclude ii). ii) --+ iii) is immediate, using Fact 1.4, namely that the oo-definable group G given by ii) is a subgroup of some B-definable group G1 say. iii) ~ i). Let B , G , d , b be as in iii). By Fact 1.5 stp(b/B) is the generic type of a coset X of a connected subgroup H of G (where X and H are over acl(B)). As above, let c realise stp(b/B)l(B,b). Then b-c realises the generic type of H over (B, b) and is interalgebraic with a suitable realisation of q fBb. As in the proof of Lemma 2.2 it is now easy to see that a is part of an algebraic triangle over A. [] We remark that Lemmas 2.2 and 2.3 hold also for infinitary types q (namely where q is the strong type of some infinite tuple over A). The definition of triviality for such types is like the one for finitary types. However the groups arising in Lemma 2.3 will now be so-called *-groups. Definition 2.4. If q = stp(a/A) satisfies the equivalent conditions of Lemma 2.3 we call q almost affine or AA (where (*) above justifies the terminology). It is clear from Lemma 2.3 that being an AA type is invariant under parallellism.

Remark 2.5. Following the terminology of [H], let us call the strong type q over A, group-internal, if for some B 3 A there is a t realising qlB, and elements b l , . . . ,bn, each in some B-definable group, such that a r C d c l ( b l , . . . ,bn,B). (So definable closure rather than algebraic closure is the main point.) Then i) I f q is group internal then there is B 3 A, a r realising qIB, and 9, some generic of a connected group over B such that a ~ E dcl(9,B) and 9 C acl(a~,B). (So in particular q is AA.) ii) If stp(a/A) is AA then there is b such that b E dcl(a,A), a C acl(b,A) and stp(b/A) is group-internal. The proofs of i) and ii) are not difficult, using [EPP] and modified canonical bases, and are not given here. The main point we want to make is that we could replace the use of AA types by group-internal types (and also replace AA-analysable types, which we define later, by group-analysable types). L e m m a 2.6. /). stp(a/A) is nontrivial iff there exist B independent with a over A, and C independent with a over A such that a forks with B U C over A. ii) stp(a /A) is AA iff there exist B independent with a over A and C independent with a over A such that a E acl(A U B U C) ",, acl(A).

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Proof i). Left to right is by Lemma 2.2. Now suppose there exist such B, C. We may suppose B and C are each algebraically closed and contain A. Then B is independent with C over D = B N C (by 1-basedness). But clearly a is independent with D over A, and also a is independent with each of B, C over D. Thus {a,B, C} forms a triangle over D. Thus stp(a/D) is nontrivial, by 2.2, whereby also stp(a/A) is nontrivial, by Remark 1.3 i). ii). Suppose again that there exist such B, C, which we again suppose to be algebraically closed and contain A. As in the proof of part i), we see that {a, B, C } forms a triangle over D, where D = B N C (and stp(a/D) does not fork over A). Replacing B by B t = Cb(stp(a, C/B)), and then C by C ' = Cb(stp(a,Bt/C)) we obtain an algebraic triangle {a, B ~, C r} over D. Thus stp(a/D) is AA whereby so is stp(a/A). [] L e m m a 2.7. Suppose stp(a/A) is AA. Then there is finite Ao C A such that stp(a/Ao) is AA.

Proof We may assume A to be algebraically closed. By 2.3 iii) and automorphisms there is B D A with B independent of a over A, and there is a B-definable group G and b C G such that a is interalgebraic with b over B. Let d c B be such that G and its group operation are d-definable, and a is interalgebraic with b over d. Let ~(y, d) define G and let X(x, y, d) witness the interalgebraicity of a and b over d (where ~(x,z) and X(x,y,z) have no parameters.) Let ~5(x,z) be the formula (3y)(~(y, z)&x(x,y, z)). Let r(z) = tp(d/A). By definability of the type r, there is a formula ~(x) over some finite subset A0 of A such that for any a t, ~(a ~) holds iff for d t realising rl(A,at), ~(at,dt). Note that cp(a) holds. We claim that stp(a/Ao) is AA. For choose a ~ realising stp(a/Ao)l A. As ~(a t) holds, there is d t realising rl(A , a ~) such that ~(a t, dr). In particular a t is independent with d t over A0, and x(a ~, bt,d t) for some b t realising ~ ( y , d ) . As the latter formula defines a group, we see that stp(d/Ao) = stp(a/Ao) is AA, as required. []

Note 2.8. One can also prove the following, although we do not need it: Let stp(a/A) be AA. Then there is a formula ~(x,y) without parameters such that for some c E A, ~(a, c) holds and moreover whenever ~p(a~, c t) then a ~ C acl(c t) or stp(at/c t) is AA. L e m m a 2.9. i) if stp(a /A) is AA then for any B D A, stp(a/B) is AA or algebraic,

ii) If stp(a/A) is AA and a ~ E acl(a,A)-acl(A) then stp(a~/A) is AA. iii) If stp(a/A) and stp(b/A) are both AA then so is stp(a, bA). Proof i) Let { a , C , D } form an algebraic triangle over A. Assume stp(a/B) is not algebraic. We may assume B = Cb(stp(a/B))U A. Thus in particular B C acl(a,A). So aB is independent with each of C,D over A. Thus a is independent with each of C , D over B, and a E acl(B U C U D ) . By Lemma 2.6 ii), stp(a/B) is AA.

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ii) clearly follows from Lemma 2.6 ii). iii) We can choose algebraic triangles {a, Bl, C1 } and {b, B2, C2} over A in such a way that first B1 is independent with {a, b} over A, and secondly such that B2 is independent with {a, b} U B1 over A. It is then easy to check that {a,b} is independent with each of B1 U B2 and CI U C2 over A. By 2.6 ii), stp(a, b/A) is AA. [] (Again 2.9 holds for strong types of infinite tuples.)

Corollary 2.10. Let p, q be strong types such that p is AA and q is trivial Then p is orthogonal to q. Proof We may assume p -- stp(a/A) and q = stp(b/A). If a forks with b over A then by 1-basedness there is c E (acl(aA) N acl(bA)) - acl(A). By Remark 1.3 ii), stp(c/A) is trivial, but by 2.9 ii) stp(c/A) is AA (so nontrivial), a contradiction. []

Definition 2.11. i) stp(a/A) is said to be AA-analysable (in A many steps (A an ordinal)) if there are a,~ E acl(aA) for c~ < A, with a = aa, such that

stp(a,~/A U {a n :/3 < c~} is AA or algebraic for all c~ < A. ii) stp(a/A) is said to be hereditarily nontrivial if stp(a/B) is nontrivial or algebraic for all B D A. iii) stp(a/A) is said to be absolutely nontrivial if stp(a'/A) is nontrivial for all a t E acl(aA) - acl(A).

Proposition 2.12. /) if stp(a/A) is AA-analysable, then so is stp(a/B) for any BDA. ii) stp(a/A) is AA-analysable iff it is hereditarily nontrivial. (Thus being AAanalysable is a parallellism invariant.) iii) if stp(a/A) is AA-analysable, then there is finite Ao C A such that stp(a/Ao) is AA-analysable in finitely many steps. iv) if stp(a/A) is AA-anaIysable and a I E acl(aA), then stp(al/A) is AAanalysable. v) if stp(a/A) and stp(b /A) are both AA-analysable, then so is stp(a, b/A). vi) If stp(a/bA) is AA-analysabte, and stp(b /A) is AA-aualysabIe, then so is stp(a/A). Proof i) follows from L e m m a 2.9 i). ii) Suppose stp(a/A) is AA-analysable and not algebraic. Then by definition there is a ' E acl(aA) such that stp(a'/A) is AA. By L e m m a 2.2 stp(aP/A) is nontrivial, so by Remark 1.3 ii), so is stp(a/A). It follows from this and part i) that stp(a/A) is hereditarily nontrivial. Conversely, suppose stp(a/A) is hereditarily nontrivial and not algebraic. By L e m m a 2.2, there is a0 E acl(aA) - acl(A) such that stp(ao/A) is AA. Note that

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a forks with a0 over A, and stp(a/A, ao) is hereditarily nontrivial. Continuing thus we find a s E acl(aA), for c~ _< A (some A < IT] +) such that stp(a~/A U {a n " 13 < c~)) is AA for all c~ < A, and a E acl(A U {as : c~ _< A)). Then stp(a/A U {as " c~ < A)) is clearly AA. Thus stp(a/A) is AA-analysable (in A many steps). iii) This clearly follows from Lemma 2.7 and the fact that there is no infinite descending chain of ordinals. iv) By induction on the (finite) length of an AA-analysis of stp(a/A). If stp(a/A) is AA-analysable in n steps then there is b E acl(aA) such that stp(b/A) is AA-analysable in n - 1 steps and stp(a/bA) is AA. By Lemma 2.9 and induction hypothesis, we may assume stp(d/bA) is AA. Let (by Lemma 2.7), c E Cb(stp(a'/bA) be such that stp(a'/cA) is AA. Then c E acl(a'), and also c E acl(bA). By induction hypothesis stp(c/A) is AA-analysable in at most n - 1 steps. So stp(d/A) is AA-analysable in at most n steps. v) and vi) are proved in a similar fashion. Details are left to the reader. [] By virtue of part ii) of the above Proposition we shall use "hereditarily nontrivial" and "AA-analysable" interchangeably. L e m m a 2.13. Let p and q be strong types such that p is AA-analysable and q is

trivial. Then p is orthogonal to q. Proof By Corollary 2.10. We finish this section with a couple of observations on trivial types, which repeat material from [HPS], but which we include here for completeness' sake. L e m m a 2.14. Let q = stp(a /A ) be non-algebraic. The following are equivalent:

i) q has weight 1. ii) there do not exist b, c E acl(aA) - acl(A) such that b is independent with c over A. If moreover q is also trivial, then another equivalent condition is iii) q is regular. Proof Clearly i) --+ ii). Conversely suppose q has weight > 1. So there is B D A with a independent from B over A, and there are c,d such that c is independent with d over B but each of c, d forks with a over B. Let c ~ E Cb(stp(a/Bc)) - acl(B) and d ~ E Cb(stp(a/Bd))-acl(B). Then c t is independent with d ~ over B, and {c ~, d ~} is independent with B over A. Thus (as c ~, d ~ E acl(a)), c ~ is independent with d ~ over A. Now suppose q is trivial but not regular. Then there is B D A, al realising q]B and a2 realising some forking extension of q over B such that al forks with a2 over B. By Lemma 2.6 i), al forks with a2 over A. But then q = stp(a2/A) has weight > 2. []

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L e m m a 2.15. Let stp(a/B) be trivial and nonorthogonal to A, where A C B. Then there is b E acl(aB) - acl(B) such that b is independent from B over A. Proof. Let C D B and c be such that a is independent with C over B, c is independent with C over A, and a forks with c over C. By L e m m a 2.6 i) (or the easier L e m m a 2.2) a forks with c over B. Let b E Cb(stp(c/aB)) - acl(B). Then b E acl(c), so is independent with B over A. [] L e m m a 2.16. Suppose stp(B/A) is trivial, and stp(b/BA) is triv&L Then

stp(Bb /A) is trivial. Proof Assume the hypotheses. We must show that Bb is not part of a triangle over A. Let C and D be such that {Bb, C , D } is pairwise A-independent. By triviality of stp(B/A), { B , C , D } is A-independent. So clearly {b, C , D } is pairwise B-independent. By triviality of stp(b/B) we conclude that {b, C, D} is B-independent, and thus {bB, C, D} is A-independent. []

3 N D O P and its local consequences Recall that a stable theory T has the DOP (dimensional order property) if there are a-models Mo,M~,M2,M', where Mo < M1, 3//0 < M2, M7 is independent with M2 over Mo, and M ' is a-prime over M1 UM2, and there is a non-algebraic p E S(M') which is orthogonal to both Ml and M2. In [Sh] it is shown that if T has DOP then for any cardinal A > tTI + such that T is A-stable, T has 2 x a-models of cardinality A. (In fact for many other A, T has 2 ;~ such models.) So as to avoid continually having to pass to a-models, it is convenient to present an equivalent definition of DOP. Definition 3.1. By a V of sets, we mean a triple B = (Bo, B1,B2) such that Bo C Bi for i = 1,2 and Ba is independent from B 2 o v e r Bo. Proposition 3.2. The following are equivalent:

i) T has DOP ii) There are B, a V of sets, a set D D UBi, and a non-algebraic strong type p over D such that BI U B2 dominates D over each Of Bl,B2, and p is orthogonal to each of B1, B2. Proof Standard. i) implies ii) is immediate. To prove ii) ~ i), let B i , D , p be as in ii). We may assume these sets all have cardinality < nr(T). Let M0 be an a-model containing Bo and independent with D over Bo. Then clearly B1 U B2 dominates D over M0 U Bi for i = 1,2, and also over M0. Moreover p is orthogonal to M0 U Bi for i = 1,2. Let Mi be a-prime over M0 U B i for i = 1,2. Then M1 is independent with M2 over Mo. Let M ' be a-prime over M1 UM2. Then stp(D/Mo U BI UB2) is realised in M ' , without loss by D itself.

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The domination conditions imply that D is independent with Mi over Mo tO Bi for i --- 1,2. Thus p is orthogonal to M1 and to 342. This gives DOP. []

Lemma 3.3. /) Let {B1,B2,B3} be an algebraic triangle over A, where A C Bi for all i. Suppose B3 C B and B3 dominates B over A. Let C = B1 to B2 tOB. Then B1 tOB2 dominates C over Bi, for i = 1, 2. ii) Assume T has NDOP. Suppose B3 is part of an algebraic triangle over A, B3 C B, B3 dominates B over A and p E S (B ). Then p is nonorthogonal to A. Proof ii) Assume A = ~ for 0 for notational convenience. We show B~ tO Be dominates C over B1. Let X 3 B1 be independent with B1 tO B2 over B1. We must show that X is independent with C over B1. Now as B3 C acl(B1 tO B2), X is independent with B3 over B1. As B3 is independent with B1 over 0, X is thus independent with B3 over ~. As B3 dominates B, X is independent with B over ~, and thus also over B1. As B2 C acl(Bl UB), easily X is independent with B t2 B2 over B1, namely X is independent with C over B1. ii) Let {B1,B2,B3} be an algebraic triangle over A (where we may assume A C Bi). By NDOP (namely Proposition 3.2) and part i), p is nonorthogonal to B1 or to B2. But each of B1,B2 is independent with B3 and thus also with B, over A. Thus p is nonorthogonal to A. [] We now again assume T to be 1-based.

Lemma 3.4. Assume T has NDOP. Suppose stp(B /A) is absolutely nontrivial (namely stp(b/A) is nontrivial for all b E acl(A tO B) \ (A). Let p E S(B) be non-algebraic. Then p is nonorthogonal to A. Proof Assume B contains A and both are algebraically closed. Let B1 3 A be a maximal algebraically closed subset of B which is part of an algebraic triangle over A. Then B1 dominates B over A. For otherwise, some C is independent with B1 over A, but forks with B over A. By looking at Cb(stp(C/B)) and using 1-basedness, we find c C B - A such that c is independent with B1 over A. By 2.2, there is c ~ C acl(cA) - A such that stp(c'/A) is AA. Then clearly B1 to {c ~) is part of an algebraic triangle over A, contradicting maximality of B1. So B1 dominates B over A as required. By L e m m a 3.3 ii), p is nonorthogonal to A. [] Corollary 3.5. Suppose T has NDOP. Let q = stp(a/A). Then q is absolutely nontriviaI iff q is AA-analysable (iff q is hereditarily nontrivial, by 2.12. ii)).

Proof Suppose first q to be absolutely nontrivial. Assume A = 0 for convenience. Let C = acl(a). Let B be a maximal subset of C which is part of an algebraic triangle over 0. As in the proof of L e m m a 3.4, B dominates C over 0. [] Claim. stp( a / B ) is hereditarily nontrivial.

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Proof of claim. Suppose by way of contradiction that for some B ~ D B, stp(a /B') is trivial and not algebraic. Replacing B ~ by Cb(stp(a/B~)), we may assume that B C B t C C. But then clearly i) B dominates B' over 0, and ii) B ~ dominates B~a over (~. By i) and Lemma 3.4, stp(a/B') is nonorthogonal to l~. By 2.15, stp(a/B ~) is not almost orthogonal to ~b. But this contradicts ii). So the claim is proved. By the claim and Proposition 2.12, stp(a/b) is AA-analysable, for some b c acl(B). But then stp(b) is AA, and thus by 2.12 again, stp(a) is AA-analysable. The converse follows from Proposition 2.12 iv) and Lemma 2.13. [] L e m m a 3.6. Suppose T has NDOP. Let A C B be algebrically closed Let B1, B2

be algebraically closed subsets of B containing A such that B1 is independent with B2 over A, and B1, B2 are maximal such. Then for any a c B - acl(Bl U Bz), stp(a /B~ U B2) is absolutely nontrivial (and thus hereditarily nontrivial by 3.5). Proof First note that 1-basedness implies that B1 U B2 dominates B over each B1 for i = 1,2: suppose for example that X is independent with B1 UB2 over B1 but forks with B over B1. By looking at Cb(stp(X/B), we find c E B - B1 such that c is independent with B2 over B~. But then B1 c is independent with B2 over A, contradicting maximality of B1. Suppose by way of contradiction that a E B and stp(a/Bt U B2) is trivial. NDOP implies that stp(a/B1 t_J B2) is nonorthogonal to say BI, and thus by triviality and 2.15 non almost orthogonal to B1. This contradicts what we said in the first paragraph. [] The next proposition is crucial, and uses the ideas in the proof of Proposition 2.4 of [HPS]. Proposition 3.7. Suppose T has NDOP. Then for any a and A (with a ~ acl(A)) there is a finite (possibly empty) set B in acl(aA) such that stp(b /A) is regular and

trivial for each b E B, B is A-independent, and stp(a /A t3 B) is AA-analysable. Proof We may assume A = (b (i.e. work over A). Let B be a maximal algebraically closed subset of acl(a) such that for all b in B, stp(b) is trivial or algebraic (or equivalently such that stp(B) is algebraic or trivial). Claim I. stp(a /B) is AA-analysable.

Proof Let c C acl(a)-acl(B). If stp(c/B) is trivial, then by Lemma 2.16, stp(cB) is trivial, contradicting maximality of B. Thus stp(a/B) is absolutely nontrivial, thus AA-analysable by Corollary 3.5.

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Claim II. B does not contain an infinite independent (over ~) set.

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Proof As in the proof of Proposition 2.4 of [HPS], a downward LowenheimSkolem argument allows us to assume that T is countable. Suppose, by way of contradiction that {ci : i < ~} is a maximal independent subset of B. For each X C w, let Ax, Bx be algebraically closed subsets of B such that {ci : i E X } C Ax, {Ci " i q{ X } C Bx, Ax is independent from Bx (over 0), and Ax, Bx are maximal such. Claim IIa). For each X C aJ there are ax E Ax, bx E Bx such that stp(a/ax, bx)

is AA-analysable. Proof Let, by Claim I, and 2.12 iii), dx E B be such that stp(a/dx) is AAanalysable. By Lemma 3.6, stp(dx/Ax U Bx) is either algebraic or absolutely nontrivial (so AA-analysable by 3.5). Thus, by 2.12 iii) again there are ax E Ax, bx E Bx such that stp(dx/ax,bx) is AA-analysable. Thus, by 2.12 vi), stp(a/ax, bx) is AA- analysable. [] Claim lib). I f X r Y, then (ax,bx) ~ (ar,br).

Proof We show that if Y - X r 0, then ax ~ a r . Let n E Y - X . Then cn E Bx is independent with ax and also cn E Ay is independent with by. Thus by triviality of stp(c,) and Lemma 2.6 i), cn is independent with axbr (over 0). (The appeal at this point to triviality is not essential.) Thus stp(cn/axbr) is trivial. If ax = a t , then by Claim IIa), stp(a/axbr) is AA-analysable, and so by Lemma 2.13, orthogonal to stp(cn/axbr). This contradicts the fact that c, C acl(a) - acl(axby). So a x r ay. By symmetry this is enough to prove Claim IIb).

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As ax, bx E acl(a) for all X C w, Claim IIb) implies that acl(a) contains an uncountable set of elements. This contradicts countability of T (even though we work in Neq), and proves Claim II. [] Claim III. For each b E B - acl(0) there is b' E acl(b) such that stp(bt/O) is

regular. Proof If stp(b) itself is not regular, then by Lemma 2.14, there are c,d E acl(b) - acl(0) such that c is independent from d. If neither of stp(c) or stp(d) is regular, we can repeat with each of c, d replacing b. So if the claim fails we easily construct a binary branching tree in B, an infinite antichain of which gives an infinite independent set in B, contradicting Claim II. [] So if B is not contained in acl(0) (which we may clearly assume), there is b E B with stp(b) regular. Let C be a maximal independent subset of B such that for all c E C stp(c) is regular. By Claim I, C is finite. Let C = { c t , . . . , cn}. Let Bl,. 9 Bn be algebraically closed subsets of B such that ci E Bi, {B1, 9 9 Bn } is 0-independent, and the Bi are maximal such.

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Claim IV. weight (Bi) = 1 for i = 1 , . . . , n. Proof Fix i. Note that if c C Bi and c is independent with ci then by Claim II, there is regular c' in B i which is independent with ci. This clearly contradicts maximality of { q , . . . , c, }. So every c in Bi forks with ci. As stp(c) is regular (so of weight 1), Bi does not contain two elements independent over 0. By L e m m a 2.14, weight(Bi) = 1 (namely weight(stp(c)) = 1 for all c in Bi). [] Claim V. For all c E B, stp(c/B1 U ... U Bn) is AA-analysable (or algebraic). Proof By induction on n. If n = 1, then B1 = B. Suppose n > 1. Let B' be a maximal algebraically closed subset of B which contains B2 U ... U B,, and is independent from B1 over 0. Clearly c 2 , . . . , c, is a maximal set of independent regular elements in B', and the maximality conditions for B2,. 9 9 Bn are satisfied inside B'. So by induction hypothesis, for all d E B', stp(d/B2 U . . . U B,) is algebraic or AA-analysable. On the other hand, clearly B1, B ' are algebraically closed, independent over 0, and maximal such. So by L e m m a 3.6, if c E B acl(Bl U B'), then stp(c/Ba U B') is absolutely trivial, hence AA-analysable. It easilly follows from the induction hypothesis mentioned above and 2.12 that stp(C /Bl U . . . O B~) is AA-analysable. [] Now Claims I and V yield (as in the proof of Claim IIa)) that there a r e bi E Bi for i = 1 , . . . , n such that s t p ( a / b l , . . . , bn) is AA-analysable. By Claim IV and L e m m a 2.14, stp(bi) is regular (and trivial, by choice of B) for all i. By choice of t h e Bi, {bl~ 9 9 bn } is 0-independent. This completes the proof of the proposition. []

4 Tree decompositions, and the number of a-models The main point of this section is a structure theorem for the a-models of a 1based theory with NDOP. We could replace a-models by some analogue of "algebraically compact" or "pure-injective" models, or even by algebraically closed sets which satisfy a pure-injectivity requirement for AA-types. T again denotes a 1-based theory. Definition 4.1. Let M be a model of T. By a decomposition tree inside M we mean a pair S consisting of a subtree I of IM I
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Remark. Clause iii) above in the definition of a decomposition tree follows from i) and ii) if we assume NDOP. T h e o r e m 4.2. Suppose T has NDOP. Let M be an a-model of T. Then any

maximal decomposition tree inside M is complete for M. Proof This is like the proof of Theorem 5.2 of [HPS] except that we systematically replace "algebraic" by "algebraic or hereditarily nontrivial". Several lemmas and claims are involved. Let S = (I, {a n : r/E I, l(r/) successor}) be a maximal decomposition tree inside M. Let J be the following subtree of I : J = {r/E I : r / = ( ) , l(r/) is limit, or 7] is a successor (of u say) and stp(an/A~, ) is trivial}. By iii)in the definition of a decomposition tree we see that J is a nonempty subtree of I. The main point is: L e m m a 4.3. Let c E M - acl(As). Then stp(c / A j ) is hereditarily nontrivial, and

is moreover nonorthogonal to A n for some rI E J. Proof We will define a subtree J ( 0 ) of J , and for each r/ E J(0) an element c n E M, such that the following are satisfied: i) c 0 = c. ii) if 7? E J(O) and stp(cv/A n) is algebraic or hereditarily nontrivial, then ~ is an endnode of J(O). iii) If ~ E J(O) and stp(cn/An) is not as in ii) then for some 0 < n < w, the set of successors of T] in J(O) is {u(i) : i < n} and moreover c~(i) E act(cvAn), stp(cn/A n U {c,(i) : i < n}) is hereditarily nontrivial, cu(i) forks with au(i) over A~, and either n = 1 in which case c~,(o) = cv, or n > 1 in which case stp(c~(i)/An) is trivial and regular for all i < n. Note that even when n = 1, c,(0) cannot fork with each of two independent trivial elements over A n . iv) if ~7 E J and l(~) is a limit ordinal, then n E J(O) iff for all 7- < r~, 7- E J(O) and {c,- : 7- < r~} is eventually constant. In this case c,~ is the eventual value of the c~- (7- < rl). The definition of J(0) and the elements cn, proceeds by induction (on l(r/)), and is like the proof of 5.2 in [HPS]. So we are brief (although there are some differences). Suppose n E J(0) and cn have been defined for all r/with l(r/) < c~, so as to satisfy i)-iv) above. Note that v) for all 7- < u with l(u) < c~, cu E acl(crAu). Suppose first c~ is a successor ordinal, with predecessor/3. Let r/E J(0), with l(r/) =/3. We want to define its successors in J(0) (and also the corresponding c's). If stp(cn/Av) is algebraic or hereditarily nontrivial, there is nothing to do, and we obtain ii). Otherwise, by Proposition 3.7, there are c o , . . . , c,-x in acl(cnAn) such that {ci : i < n} is An-independent, stp(ci/Ao) is regular and trivial for all i < n, and stp(cn/A n U {ci : i < n}) is hereditarily nontrivial. We first note that each stp(ci/An) is orthogonal to A~- for all 7- < r/. For otherwise 1-basedness gives us some d E acl(ciA n) - acl(A n) and 7- < r/ such that d

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is independent with A n over A~-. But then let u be the unique successor of 7" which is < r/. By induction hypothesis and v), c~, forks with a~ over Ar, and d E acl(c~An). Note that tp(d/A.~) is trivial, and d is independent with a~ over A~-. So by the last part of iii) above, c~, is independent with d over A~-. But then by Lemma 2.6 i), d is independent with {c~,An} over A~-, a contradiction. Thus stp(ci/An) is orthogonal to A~- for all 7" < r/. If say ci were independent with every an^(k ) over A n, for all ~ A (k) E J , then by triviality ci would be independent with {an^ (k) : ~ A (k) E J } over A n, contradicting maximality of the tree S. Thus each ci forks with some unique anA(k ) over A n. So let the successors of r/in J(0) be the set of these r/A (k), and let CnA(k) be the corresponding ci if n > 1, and c n if n = 1. So iii) is satisfied. If c~ is a limit ordinal and ~ E J , the condition iv) defines when r/ E J (0), and also tells in this case what should be c n. Let us remark that v) then holds for c~ + 1 in place of a. Thus the construction of J (0) can be carried out so as to satisfy i)-iv) and also v) for all a. Note also vi) if l(rl) is limit and ~ E J (0), then there is 7" E J(0) with 7" < r/ such that for all/3 with l(~-) < fl < 1(~7) there is unique u E J(0) such that l(u) = f3 and 7- < u (so 7" < u < ~7), and moreover for such u, c~ = c~-. Claim I. There are r/(1),..., r/(n) E J (O), such that for all ~ E J (O), ~ is comparable to some ~(i), and such that for each i = 1 , . . . , n and/3 >_ l(rl(i)), there is

at most one 7- E J (O) such that I(7-) =/3 and rl(i) < 7-. Proof If not it is easy to see, from the properties of J (0), that J (0) contains an infinite antichain X of successor nodes, each of whose predecessors has at least two successors in J(0). (Thus for all 7" E X, stp(c.~/A~- is regular and trivial by iii), where 7"- denotes the immediate predessor of 7".) Let Y = {u E J(0): for some 7" E X, u < ~-}. The orthogonality conditions on the tree S ensure that for each 7" E X, stp(a.~/Ay) does not fork over A~--, so is regular and trivial, and that Ax is Ay-independent. As Cr forks with a~- over A~-- for each 7" E X, regularity (of the strong types stp(c~/A~-)) implies that stp (c~ l A y ) does not fork over A~-for all 7" E X and that {Cr : ~- E X} is At-independent. On the other hand, by v), c~- E acl(cAy) for all 7- E X. But this contradicts the fact (Proposition 3.7 or even Claim II therein) that acl(aAr) cannot contain an infinite At-independent set of elements realising trivial types over A t ) . This contradiction proves the claim. [] Note that Claim I implies that {c~- : r E J (0)) is finite. Claim II. The height of J(O) is < IT] +, namely for some/3 < tT] +, l(~l)
Proof If not, then by Claim I, clearly J(0) contains some chain J(1) of length ]TI § such that for some c I, for eventually all T E J(1), c~ = c'. But then, by the properties of J(0), for eventually all successor 7- E J(1), tp(c~/A'r) forks over A~-. This contradicts the fact that t~(t) < ITI +. []

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By Claims I and II and the properties of J (0), J (0) has finitely many maximal elements, say ~ ( 1 ) , . . . , r/(n). As each ~(i) is maximal, stp(cn(i)/A,(i) ) is hereditarily nontrivial or algebraic. Moreover, an easy inductive argument shows that stp(c/{c,(e) : i = 1 , . . . , n} U U{A,(e) : i = 1 , . . . , n}) is hereditarily nontrivial or algebraici Thus we have C l a i m III.

stp(c/U {A~(i) : i = 1 , . . . , n }) is hereditarily nontrivial, or algebraic.

Independence of the tree A j, the fact thatstp(an/An- ) is trivial for ~7 c J , and L e m m a 2.13, imply that stp(c/Aj) does not fork over An(1) U . . . U An(n ). This gives the first part of the Lemma. The independence of the tree S, 1-basedness and NDOP then easily implies that stp(c/Aj) is nonorthogonal t o An(i) for some i = 1 , . . . , n. This completes the proof of the Lemma. [] We now complete the proof of Theorem 4.2, by showing that M is a-minimal (and thus a-prime) o v e r AI. Let M ~ < M be any a-model containing A1. Suppose by way of contradiction that M t ~ M. Let c c M - M t. Now L e m m a 4.3 and Corollary 3.5 yields that stp(M~/Aj) is absolutely nontrivial. By Remark 3.4, stp(c/M') is nonorthogonal to As. 1-basedness, and the fact that M ~ is an a-model implies that there is d E acl(cM ~) - M t such that stp(d/M t) does not fork over Aj. By L e m m a 4.3 again, stp(d/M ~) is hereditarily nontrivial and nonorthogonal to A n for some ~/ c J . 1-basedness again yields d ~ E M - M ~, such that stp(&/M ~) does not fork over A n and is hereditarily nontrivial. Now choose such d t E M - M ~, and ~7 E J , such that I07) is least possible (and d ~ is independent with M ~ over A n, and stp(d~/An) is hereditarily nontrivial). By L e m m a 2.2, there is d " E acl(&A n) such that stp(d"/An) is AA. Clearly d " is independent with AI over A n, and stp(d"/An) is orthogonal to A~ for all r < T/. The existence of d " contradicts the maximality of S. This contradiction proves the theorem. [] Finally we are able to more or less compute the number of a-models of T. Suppose T has NDOP (and is 1-based). Let A(0) = 21rl.

deep if there are ai for i < to such that for each i < to, stp(ai/{ao,..., ai-1))is regular, trivial, and orthogonal to { a 0 , . . . , ai-2} (where this makes sense). If T is not deep we call T shallow.

Definition 4.4. We say that T is

Remark 4.5. i) If T is shallow, M is a model of T of cardinality ~ and S = <w (I, {a n : r / E I ) ) is a decomposition tree inside M, then I is a subtree of ~q~ , and the depth of I is an ordinal less than A(0) +. ii) If T is shallow then the number of isomorphism types of decomposition trees S = (I, {a n : r/C I } for which I is a subtree of R~<w , is at most 3(;~(o)+)(]to +c~I).

Proof i) follows from Definition 4.1, and ii) is easy.

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Proposition

4.6. (T has NDOP). i) If T is deep, then for any A > A(0) such that A ITI = A, T has 2 ~ a-models ofcardinality A up to isomorphism,

ii) If T is shallow, then for any ~, the number of a-models o f T of cardinality Rc,, up to isomorphism, is less than ~(;~(o)+)(l~v + c~l). Proof ii) is an i m m e d i a t e c o n s e q u e n c e o f 4.5 and T h e o r e m 4.2 (and uniqueness of a - p r i m e models). i) is p r o v e d in an identical fashion to T h e o r e m 5.2 in [HM], using T h e o r e m 3.4' therein. []

References [BH] [EPP] [HI [HM]

Bouscaren, E., Hrushovski, E.: Interpreting groups in stable 1-based theories. Preprint (1992) Evans, D., Pillay, A., Poizat, B.: A group in a group. Algebra i Logika, vol. 3 (1990) Hrushovski, E.: Ph.D. thesis, Berkeley (1986) Harrington, L., Makkai, M.: An exposition of Shelah's 'Main Gap'. Notre Dame J. Formal Logic 26, 139-177 (1985) [HP] Hrushovski, E., PiUay, A.: Weakly normal groups. Proc. of Logic Colloquium '85. Amsterdam: North Holland 1987 [HPS] Hart, B., Pillay, A., Starchenko, S.: Triviality. NDOP and stable varieties. Preprint (1992) [M] Makkai, M.: A survey of basic stability theory. Israel J. Math. 49, 181-238 (1984) [Po] Poizat, B.: Groupes stables. Nnr al-Mantiq wal-Ma'rifah (1987) [Sh] Shelah, S.: Classification theory, 2nd edn. Amsterdam: North-Holland 1991