Reducts of Stable, CM-trivial Theories

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1 Reducts of Stable, CM-trivial Theories Herwig Nübling 27th July 2004 Abstract We show that every reduct of a stable, CM-trivial theory of finite Lascar rank is CM-trivial. AMS classification: 03C45 Keywords: Reducts; CM-triviality Introduction It is shown in [3] that every reduct of a stable, 1-based theory of finite Lascar rank is 1-based. The proof makes use of the fact that a stable theory is 1-based if and only if it does not contain a point-line system, known as a pseudoplane. It is not difficult to generalise the proof to the supersimple finite rank case. We have nevertheless found another proof for the supersimple, finite rank case which does not explicitly use the notion of a pseudoplane. On the other hand it is not true that every reduct of a stable, 1-based theory of possibly infinite rank is 1-based: the free pseudoplane from [3] is not 1 -based and has infinite Lascar rank. But it is a reduct of a 1 -based theory of infinite rank and therefore provides a counterexample. It is still an open question whether every finite Lascar rank reduct of a stable, 1 -based theory is 1 -based. When Hrushovski constructed in [4] his famous examples of non 1-based, strongly minimal sets that do not interpret an infinite group, he observed that these examples have a weaker property than 1-basedness, which he called CMtriviality. The motivation of this paper is to show analogous results to the 1-based case, this time for reducts of CM-trivial theories. 1

2 We will show (Theorem 4.2) that if T is a stable, CM-trivial theory of finite Lascar rank then every reduct of T is CM-trivial. We tried to prove this result for the wider class of simple, CM-trivial theories of finite Lascar rank but we have not fully succeeded: there is still one step in the proof for which we need stationarity. Nevertheless, we will offer partial results for the simple case whenever it is possible in this paper. Although the notion of CM-triviality was first only defined for stable theories, we can use it for simple theories with the elimination of hyperimaginaries in the same way. Pillay defined in [7] for a tuple of elements properties which he called 1 -ampleness and 2 -ampleness. He showed that a stable theory is 1 -based iff it has no 1 -ample tuple and that it is CM-trivial iff it has no 2 -ample tuple. We will recall these results in Section 1. We will then show how we can minimize the rank of 1 -ample and 2 -ample tuples in simple theories of finite Lascar rank. In Section 2 we will give the framework for our proof by introducing reducts and showing that every reduct of a supersimple theory of finite rank has finite rank. This result is probably well known but we could not find a reference for it. We show in Section 3 that every reduct of a simple, 1 -based theory of finite Lascar rank is 1 -based. The new proof of this result can be seen as an easier version of the proof for the CM-trivial case. We will finally prove in Section 4 our main theorem. It is still an open question whether a stable theory is CM-trivial iff it does not contain a point-line-plane system, which one might call a pseudospace. We had problems in finding the correct definition of a pseudospace. We assume familiarity with the basic concepts of stable and simple theories (independence, canonical bases, Lascar rank...) as found in the opening chapters of [6] for stable and in [9] for simple theories. This is part of the author s PhD Thesis at the University of East Anglia, Norwich. The author would like to thank his supervisor David Evans for all his support. 1 Ampleness Let T be a simple theory with elimination of hyperimaginaries. We are working in C eq where C is a large saturated model of T. We will firstly give the definition of 1 -basedness. Definition 1.1 A simple theory with elimination of hyperimaginaries is 1 -based 2

3 if we have for all A, B A B. acl(a) acl(b) Remark 1.2 Being 1 -based for a theory T is preserved under naming and forgetting parameters. A proof can for example be found in [9]. The next definition is from [7]: Definition 1.3 We call a, b 1 -ample over A if we have 1. acl(a, A) acl(b, A) = acl(a), 2. a A b. We call a, b 1 -ample if A =. A simple theory T with elimination of hyperimaginaries is 1 -ample if there is an A and a, b such that a, b is 1 - ample over A. Fact 1.4 T is 1 -ample if and only if T is not 1 -based. This is Remark 3.4 from [7], where it is proved for stable theories. This proof also works for simple theories with elimination of hyperimaginaries. We recall from [4] the definition of CM-triviality which also works in our context: Definition 1.5 A simple theory with elimination of hyperimaginaries is CMtrivial if for all A, B, C with A B we have C A B. acl(a,b) acl(c) Remark 1.6 A simple theory with elimination of hyperimaginaries is CM-trivial if and only if for all A B and c with acl(c, A) acl(b) = acl(a) we have Cb(c/A) Cb(c/B). The proof from [4] for the stable case also works in our case. 3

4 Remark 1.7 If T is 1 -based the T is also CM-trivial. This is Remark 12 from [4]. The proof can be generalised to our case. Fact 1.8 Whether T is CM-trivial or not is preserved under naming parameters. This is Remark 2.6 from [5] and works in our case, too. Pillay found in [7] also an equivalent notion for not being CM-trivial: Definition 1.9 We call a, b, c 2 -ample over A if we have 1. acl(a, A) acl(b, A) = acl(a), 2. acl(a, b, A) acl(a, c, A) = acl(a, A), 3. a c, b,a 4. a c. A We call a, b, c 2 -ample if A =. A simple theory T with elimination of hyperimaginaries is 2 -ample if there is an A and a, b, c such that a, b, c is 2 -ample over A. Fact 1.10 T is 2 -ample if and only if T is not CM-trivial. This is Lemma 2.3 from [7] and works also in our case. Fact 1.11 If a, b, c is 2 -ample and D is such that a, b, c D then a, b, c is 2 -ample over D. Proof: 1. and 2. of the definition follow from Lemma 2.9 of [7]. Since a D and a b,c c we have a b c, D by transitivity and hence b a c. b,d If a D c then a c, D since a D. This is a contradiction to a c. 4

5 Remark 1.12 If a, b is 1 -ample and a, b D then a, b is 1 -ample over D. The proof is similar to the proof of Fact 1.11 but easier. Remark 1.13 If a, b, c is 2 -ample then a / acl(b, c), b / acl(a, c), c / acl(a, b). Proof: The proof is straightforward, but we will give it here for the sake of completeness. If a acl(b, c) then a acl(b) since a c. But then a acl( ) since b acl(a) acl(b) = acl( ). Hence a c, which is a contradiction. If b acl(a, c) then b acl(a) as we have acl(a, b) acl(a, c) = acl(a). But then b acl( ) and a c is a contradiction. If c acl(a, b) then also c acl(a) and c acl(b). Hence c acl( ) implies a c which is a contradiction. For the rest of this section, let T be a supersimple theory of finite Lascar rank. Then T has elimination of hyperimaginaries ([1]). We will show now that we can then assume that there are rank 1 elements in a 2 -ample ( 1 -ample respectively) tuple. Lemma 1.14 ( T supersimple of finite Lascar rank) For every a such that a / acl( ) there are D and d such that a D, SU(d/D) = 1, d acl(a, D). Proof: We give the proof here, although it is essentially the proof of Lemma from [6]. Let SU(a) = n > 0. If n = 1 then take D = and d = a. If n > 1 then let C be such that SU(a/C) = n 1. Since T is supersimple 5

6 there is d C eq such that acl(cb(a/c)) = acl(d). Since a C we have d / acl( ) and so SU(d) 1. Let D be such that SU(d/D) = 1. By taking a nonforking extension of tp(d/d) to a, d we may assume that D a. By symmetry we have a d D and so d Cb(a/d, D) = Cb(a/d). Then acl(cb(a/d, D)) = acl(cb(a/d)) = acl(cb(a/c)) = acl(d). Since d / acl(d) we have a D d. Hence n = SU(a) SU(a/D) > SU(a/d, D) = SU(a/d) = n 1. Hence SU(a/D) = n and a D. Since d a we have SU(d/a, D) < SU(d/D) = 1 and so d acl(a, D). D Lemma 1.15 If a, b, c is 2 -ample then by adding a set of parameters we may also assume that there are a 1 acl(a), b 1 acl(b), c 1 acl(c) such that SU(a 1 ) = SU(b 1 ) = SU(c 1 ) = 1. We can take the parameters independent from the a, b, c so in particular adding these parameters does not change the Lascar rank of elements of acl(a, b, c). Proof: If acl(a) does not already contain a Lascar rank 1 element then by Lemma 1.14 (note that a / acl( ) by Lemma 1.13) there are d and D such that a D, SU(d/D) = 1, d acl(a, D). Let ˆd, ˆD = tp(d, D/a) be such that ˆd, ˆD a b, c. Then by transitivity and symmetry we have a, b, c ˆD. 6

7 So by Fact 1.11 a, b, c are 2 -ample after adding acl( ˆD) as a set of parameters. We have ˆd acl(a, ˆD) and SU( ˆd/ ˆD) = 1. Note further that for x acl(a, b, c) we have x ˆD and so SU(x/ ˆD) = SU(x). Continuing this way for b and c if necessary proves the lemma. Remark 1.16 If a, b is 1 -ample then we may also assume by adding a set of parameters that there are a 1 acl(a) and b 1 acl(b) such that SU(a 1 ) = SU(b 1 ) = 1. The proof is similar but easier to the proof of Lemma 1.15 and uses Lemma 1.14, Remark 1.12 and the fact that a / acl( ) since a b. Now we will show, how we can minimize the rank between the elements of an 1 -ample tuple. Lemma 1.17 ( T supersimple of finite Lascar rank) If T is 1-ample then there are a and b which are 1-ample such that SU(a/b) = SU(b/a) = 1 after possibly adding parameters. Proof: Let a be such that there is a b such that a, b is 1-ample and SU(a) is minimal after possibly adding parameters. Let then b be such that a, b is 1-ample and SU(b) is minimal after possibly adding more parameters. Firstly note that SU(a/b) 1 since a b. Assume now for a contradiction that SU(a/b) > 1. We may assume that there is a 1 acl(a) such that SU(a 1 ) = 1 by Remark Let â acl(a) be such that acl(â) = acl(a) acl(a 1, b). Note that a / acl(â, b) acl(a 1, b) since SU(a, b) > SU(b) + 1 = SU(b) + SU(a 1 ) SU(a 1, b). If a b then a, b is 1-ample over acl(â) and SU(a/â) SU(a/a â 1) = SU(a) SU(a 1 ) < SU(a) contradicts the minimality of the Lascar rank of a. Hence a b and so â â b since otherwise transitivity would give us the contradiction a, â b. We have acl(â) acl(b) acl(a) acl(b) = acl( ) and so â, b is 1-ample. But since a / acl(â) we have SU(â) = SU(a) SU(a/â) < SU(a), 7

8 which contradicts again the minimality of SU(a). Hence SU(a/b) = 1. Similarly we can show that SU(b/a) = 1. Note that it is enough to assume that for a fixed a (rather than under every possible a ) we have b such that a, b is 1-ample and SU(b) is minimal (after possibly adding parameters). The next two lemmas show how we can find a minimized 2 -ample tuple. Lemma 1.18 Let T be supersimple of finite Lascar rank. If a, b, c is 2 - ample then there are â and ĉ such that (possibly after adding parameters) â, b, ĉ is 2 -ample with SU(â/b) = SU(ĉ/b) = 1. Proof: Let a, b and c be as in the hypothesis. Fixing b and c we may assume that SU(a) is minimal amongst all elements a such that a, b, c is 2 -ample after adding parameters (using that T has finite Lascar rank). Claim: SU(a/b) = 1. Proof: We have SU(a/b) 1 by Lemma Assume now for a contradiction that SU(a/b) > 1. By Lemma 1.15 we may assume that there is a 1 acl(a) with SU(a 1 ) = 1. Let a be such that acl(a ) = acl(a) acl(a 1, b). Note that a / acl(a ) because a / acl(a 1, b) since SU(a/b) > SU(a 1 /b). If a c then a, b, c is 2 -ample over a as a acl(a) and acl(a) a acl(a, b) = acl(a ) by choice of a. But SU(a/a ) = SU(a) SU(a ) < SU(a) contradicts the minimality of a. Hence we have a c and so a a c since otherwise forking symmetry and transitivity would give us the contradiction a c. We have acl(a, b) acl(a, c) acl(a, b) acl(a, c) acl(a 1, b) = acl(a) acl(a 1, b) = acl(a ) and so acl(a, b) acl(a, c) = acl(a ). We have acl(a ) acl(b) = acl( ) and a c since b a acl(a). So a, b, c is 2 -ample and SU(a ) = SU(a) SU(a/a ) < SU(a) contradicting again the minimality of SU(a). So we have SU(a/b) = 1. Claim Let now c be such that acl(c ) = acl(a, c) acl(c, b). Then acl(a, c ) acl(b, c ) = acl(c ) and acl(c ) acl(b) = acl( ), 8

9 since acl(c ) acl(b) acl(a, c) acl(a, b) acl(b) = acl(a) acl(b) = acl( ). Since c acl(b, c) we have a b c. We also have a c as c acl(c ). So c, b, a is 2 -ample. We may further assume that SU(c ) is minimal after adding parameters. Then by the claim we have SU(c /b) = 1. Note that we still have SU(a/b) = 1. Replacing c by â and a by ĉ proves the lemma. Lemma 1.19 Let T be of finite Lascar rank. If T is 2 -ample then (possibly after adding parameters) there is a, b, c which is 2 -ample with SU(a/b) = SU(c/b) = SU(b/a, c) = 1. Proof: By Lemma 1.18 there is a 2 -ample tuple a, b, c with SU(a/b) = SU(c/b) = 1. Let now a, b, c and the set of added parameters be such that SU(b) is minimized amongst all 2 -ample tuples with SU(a/b) = SU(c/b) = 1. Claim: SU(b/a, c) = 1. Proof: We have SU(b/a, c) 1 by Lemma Assume SU(b/a, c) > 1 for a contradiction. Using Lemma 1.15 we may assume that there is b 1 acl(b) with SU(b 1 ) = 1. Note that b / acl(a, b 1, c) since SU(a, b, c) > SU(a, c) + 1 SU(a, b 1, c). Let â be such that Let ˆb be such that acl(â) = acl(a, b) acl(a, b 1, c). acl(ˆb) = acl(â) acl(b). So we have acl(â, ˆb, b) acl(â, ˆb, c) = acl(â) and acl(â) acl(b) = acl(ˆb) by construction. We also have â c and SU(â/b) = 1 since â acl(a, b) and b â / acl(b). If â c then â, b, c is 2 -ample over ˆb and ˆb SU(â/b) = SU(â/b, ˆb) = SU(c/b, ˆb) = SU(c/b) = 1. But SU(b/ˆb) SU(b/b 1 ) < SU(b) contradicts the minimality of SU(b). Hence â ˆb c and so a ˆb c since a acl(â). Since ˆb acl(b) we have acl(a) acl(ˆb) = acl( ) acl(a, ˆb) acl(a, c) = acl(a) 9

10 and a, ˆb, c is 2 -ample. By Lemma 1.18 there is a, c and a set of parameters A such that a, ˆb, c is 2 -ample over A and SU(a /ˆb, A) = SU(c /ˆb, A) = 1. We have SU(ˆb/A) SU(ˆb) = SU(b) SU(b/ˆb) < SU(b) since b / acl(ˆb) acl(â) acl(a, c, b 1 ). This contradicts again the minimality of SU(b). Hence SU(b/a, c) = 1. Claim This proves the lemma. 2 Reducts Recall that an L -theory T is Morleyized if for every L -formula φ( x) there is a relation symbol R φ in L such that T = x(r φ ( x) φ( x)). We may assume that T is Morleyized without changing any other properties like simplicity, stability or CM-triviality. So let from now on T be Morleyized. We call T a reduct of T if there is L L such that T is the set of all L -sentences in T. If M is a model of T then we can turn M into a model M for T in a canonical way. Remark 2.1 It follows from the definition that every reduct of a simple theory is simple and every reduct of a stable theory is stable. One can also show that every reduct of a supersimple theory is supersimple using the partial type counting argument from [2]. Let T be a simple theory and T be a reduct of T. Let C be a monster model for T. Then C is a monster model for T. If we are working in T eq and C eq = T then C eq contains also all imaginary elements for (T ) eq (from now on we will write T eq for (T ) eq ). Hence we can turn C eq into a model C eq for T eq (note that we forget about all imaginary elements of C eq which are not definable in T eq ). Denote for X C eq by ACL(X) the algebraic closure of X in the sense of T and by acl(x) the algebraic closure of X C eq in the sense of T. For a C eq and X C eq denote by TP(a/X) the type of a over X in the sense of T and by tp(a/x) the type of a over X C eq in the sense of T. For X, Y, Z we write X Z if X is independent from Z over Y in the Y sense of T and X Z if X is independent from Z over Y in the sense of Y 10

11 T. Let SU (or su respectively) be the Lascar rank in T ( T respectively). We first prove two general, probably widely known theorems about the reducts of simple theories. Theorem 2.2 ( T stable or supersimple with e.h.) Let T be a reduct of T, A, C C eq be small sets and B = ACL(B) C eq. If A B C then A B C eq C. Proof: The proof for the stable case is not difficult and probably well known, but we will give it anyway since we could not find a reference for it. We may assume that A is a finite tuple a. We will first show that the proposition is true if B is a model M of T (note that we are working in T eq ). So we have a C and TP(a/M, C) is finitely satisfiable in M. Hence M tp(a/m, C) is finitely satisfiable in M since tp(a/m, C) TP(a/M, C). Then a C. M Let now B not necessarily be a model. We can find a model ˆM of T which contains B and C. Let M = TP( ˆM/B, C) such that M a. Then B,C B, C M and a M by symmetry. Hence a B,C M by transitivity. B Let N = TP(M/B) be such that N a, M. Then N M = B since B B N and N M. Furthermore we have a B N. Then by transitivity M a N, M and so a B M. N In particular we have a N and a M M. Hence a N N and M a M from the previous paragraph. Then we have in T N Cb(a/M, N ) M N = B C eq. Hence a B C eq M, N. and in particular a B C eq C which proves the proposition for the stable case. The only proof we could find in the supersimple case is Theorem in [8]. Theorem 2.3 If T is supersimple and has finite Lascar rank then also T finite Lascar rank. has Proof: Note that T is at least supersimple since T is (Remark 2.1). 11

12 Claim: Let a C eq, X = ACL(X) C eq and ˆX = X C eq. If su(a/ ˆX) = n < ω then there is a 1 = tp(a/ ˆX) such that SU(a 1 /X) su(a 1 / ˆX). Proof: We are using induction on n. If n = 0 then SU(a/X) 0 = su(a/x). Let su(a/ ˆX) = n + 1. By the definition of Lascar rank there is Y ˆX such that su(a/y ) = n. Let Z = ACL(Y ) and Ẑ = Z Ceq. Note that Y Ẑ. By the existence of a non-forking extension there is a 1 = tp(a/y ) such that a 1 Y Ẑ. Hence su(a 1 /Ẑ) = su(a 1/Y ). By the induction hypothesis there is a 2 = tp(a 1 /Ẑ) such that SU(a 2 /Z) su(a 2 /Ẑ) = su(a 1 /Ẑ) = su(a 1 /Y ) = su(a/y ) = n. Further note that a 2 = tp(a/ ˆX) and so a 2 ˆXẐ. Hence a 2 Z since otherwise we have a X 2 Ẑ and so by Proposition 2.2 X a 2 ˆXẐ would be a contradiction. But then we have This proves the Claim. Claim SU(a 2 /X) SU(a 2 /Z) + 1 n + 1 = su(a 2 / ˆX). But if T has infinite Lascar rank then there is for every n < ω an element a C and X C eq such that su(a/x) n. If we take a 1 as a nonforking extension of tp(a/x) to ACL(X) C eq then there is by the Claim a 2 = tp(a 1 /ACL(X) C eq ) such that SU(a 2 /ACL(X)) su(a 2 /ACL(X) C eq ) = su(a/x) n. Hence T also has infinite Lascar rank. 12

13 3 Reducts of 1-based theories Theorem 3.1 Let T be a supersimple theory of finite Lascar rank. If T is 1-based then every reduct of T is 1-based. Proof: Assume for a contradiction that there is a reduct T which is not 1-based. Note that we are allowed to add parameters to the language of T by Remark 1.2. By Fact 1.4 T is 1-ample and by Lemma 1.17 there are (after possibly adding parameters) a, b in C eq such that a, b is 1-ample and su(a/b) = su(b/a) = 1. Let α, β = tp(a, b) such that SU(α, β) is maximal in T (using that T has finite Lascar rank). Then α / ACL(β) because otherwise we can find ˆα = tp(α/β) such that ˆα / ACL(β) since α / acl(β). But then SU(ˆα, β) > SU(β) = SU(α, β) contradicts the maximal choice of α, β. It follows by the same argument that β / ACL(α). Let X = ACL(α) ACL(β). Claim: α X β. Proof: Assume α X β. Let ˆX = X C eq. Then by Theorem 2.2 we have α ˆXβ. (1) We have α / acl(β, ˆX) ACL(β, X) = ACL(β) and so 1 su(α/β, ˆX) su(α/β) = 1, and equality holds. Hence using symmetry we have Using the same argument we also get Hence cb( ˆX/α, β) acl(α) acl(β) = acl( ) and so ˆX β α. (2) ˆX α β. (3) ˆX α, β. (4) But then α ˆX, (5) 13

14 which, using transitivity along with equation 1, gives us α ˆX, β. (6) But this is a contradiction to α β and so α X β. Claim But then α, β shows that T is not 1-based, which is a contradiction to the hypothesis. 4 Reducts of CM-trivial theories Lemma 4.1 Let T be stable with finite Lascar rank. Let C eq be a monster model for T. Denote by U the Lascar rank in C eq. Let T be a reduct of T and denote by u the Lascar rank in C eq. Let a, b, c be such that a / acl(b, c), b / acl(a, c), c / acl(a, b), u(a/b) = 1. Then there are α, β, γ = tp(a, b, c) in C eq = T such that α / ACL(β, γ), β / ACL(α, γ), γ / ACL(α, β), α γ. β Proof: Since T has finite rank, we can take α, β, γ = tp(a, b, c) such that U(α, β, γ) is maximal amongst all elements satisfying the partial type tp(a, b, c) in T. Claim 1: α / ACL(β, γ). Proof: If α ACL(β, γ) then we can find ˆα = tp(α/β, γ) such that ˆα / ACL(β, γ), as α / acl(β, γ). But then ˆα, β, γ = tp(a, b, c) and U(ˆα, β, γ) > U(β, γ) = U(α, β, γ), 14

15 which is a contradiction to the maximality of U(α, β, γ). Claim 1 Similarly to Claim 1 we can show that β / ACL(α, γ) and γ / ACL(α, β). Claim 2: α β γ. Proof: Assume for a contradiction that α γ and so α β γ. ACL(β) By the existence of non-forking extension we can find ˆα = TP(α/ACL(β)) such that ˆα γ. Let B = ACL(β) ACL(β) Ceq. Then 1 u(ˆα/b, γ) u(ˆα/acl(β)) = 1, since ˆα / acl(b, γ) ACL(B, γ) by Claim 1. Hence ˆα reverse transitivity ˆα γ. acl(β) B, γ and so by acl(β) But since tp(ˆα/acl(β)) = tp(α/acl(β)) and α acl(β) γ, we have by uniqueness of nonforking extension tp(ˆα/acl(β), γ) = tp(α/acl(β), γ) and so ˆα, β, γ = tp(a, b, c). But a rank calculation in T shows U(ˆα, β, γ) = U(ACL(ˆα, β, γ)) = U(ˆα, ACL(β), γ) = U(ˆα/ACL(β), γ) + U(ACL(β), γ) = U(ˆα/ACL(β)) + U(ACL(β), γ) = U(α/ACL(β)) + U(ACL(β), γ) > U(α/ACL(β), γ) + U(ACL(β), γ) = U(α, ACL(β), γ) = U(α, β, γ), which contradicts the maximality of U(α, β, γ). Hence α β γ. Claim 2 This proves the lemma. Theorem 4.2 Let T be a stable theory of finite Lascar rank which is CM-trivial. Then every reduct of T is CM-trivial. Proof: Note that we are allowed to add parameters to C eq by Fact 1.8. Let T be a reduct of T and assume for a contradiction that T is not CMtrivial. Then there are a, b and c from C eq which are 2 -ample by Theorem Note that T has finite Lascar rank by Theorem

16 We may assume that u(a/b) = u(c/b) = u(b/a, c) = 1, where u is the Lascar rank in T, by Lemma We know that a / acl(b, c), b / acl(a, c) and c / acl(a, b) by Lemma Hence by Lemma 4.1 we can find α, β and γ such that α, β, γ = tp(a, b, c), α / ACL(β, γ), β / ACL(α, γ), γ / ACL(α, β), α γ. β Claim: α ACL(β) ACL(α,γ) γ. Proof: Set X = ACL(β) ACL(α, γ) (= ACL(X)) and ˆX = X C eq. If we assume for a contradiction that α γ then X α ˆXγ (1) by Theorem 2.2 since X is algebraically closed in T by Theorem 2.2. We have α / ACL(X, β) = ACL(β) and hence α / acl( ˆX, β) ACL(X, β) which implies 1 = u(α/β) u(α/β, ˆX) 1 and equality holds. Hence α β ˆX (2) We also have γ / ACL(X, α, β) and so γ / acl( ˆX, α, β) which gives us 1 = u(γ/α, β) u(γ/α, β, ˆX) 1 and equality holds. Hence γ α,β ˆX (3) Using symmetry and transitivity equations (2) and (3) together gives ˆX α, γ. (4) β On the other hand β / acl(α, γ, ˆX) since β / ACL(α, γ, X) = ACL(α, γ). Hence 1 = u(β/α, γ) u(β/α, γ, ˆX) 1. 16

17 Therefore u(β/α, γ) = u(β/α, γ, ˆX) and using symmetry we get ˆX α,γ β. (5) For the canonical base of tp( ˆX/α, β, γ) using equations (4) and (5) and the definition of 2 -ampleness we get Cb( ˆX/α, β, γ) acl(β) acl(α, γ) Hence ˆX α, β, γ which gives us acl(β) acl(α, β) acl(α, γ) = acl(β) acl(α) = acl( ). α ˆX. (6) Then it follows with transitivity from equation (1) and (6) that α γ, ˆX, which contradicts α γ in the definition of 2 -ampleness. Claim Hence we have that α γ but α β γ. Thus T is not CMtrivial, which is a contradiction to the assumption. ACL(β) ACL(α,γ) Remark 4.3 Claim 2 of Lemma 4.1 was the only time we used uniqueness of non-forking extension in the proof of Theorem 4.2. References [1] Steve Buechler, Anand Pillay, Frank Wagner, Supersimple theories, J. Amer. Math. Soc 14 (2001), [2] E. Casanovas, The number of types in simple theories, Annals of Pure and Applied Logic 98 (1999), [3] D. Evans, A. Pillay, B. Poizat, Le groupe dans le groupe, Algebra i Logika 29, [4] Ehud Hrushovski, A new strongly minimal set, Annals of Pure and Applied Logic 62 (1993),

18 [5] Anand Pillay, The geometry of forking and groups of finite Morley rank, Journal of Symbolic Logic 60 (1995), [6] Anand Pillay, Geometric Stability Theory, Clarendon Press, Oxford (1996). [7] Anand Pillay, A note on CM-triviality and the geometry of forking, Journal of Symbolic Logic 65 (2000), [8] Hans Scheuermann, Unabhängigkeitsrelationen, Diplomathesis, Freiburg (1996). [9] Frank O. Wagner, Simple Theories, Mathematics and its Applications 503, Kluwer Academic Publishers, Dordrecht (2000). Institut für Mathematik und Informatik Universität Greifswald Jahnstrasse 15 a Greifswald Germany address: herwig.nuebling@uni-greifswald.de 18