Hirotaka Kikyo and Saharon Shelah- The Strict Order Property and Generic Automorphisms

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  • 8/3/2019 Hirotaka Kikyo and Saharon Shelah- The Strict Order Property and Generic Automorphisms

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    748

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    vision:2001-01-04

    modified:2001-01-09

    THE STRICT ORDER PROPERTY AND GENERIC

    AUTOMORPHISMS

    HIROTAKA KIKYO AND SAHARON SHELAH

    Abstract. If T is a model complete theory with the strict orderproperty, then the theory of the models ofT with an automorphismhas no model companion.

    1. Introduction

    Given a model complete theory T in a language L, we consider the(incomplete) theory T = T { is an L-automorphism} in the lan-guage L = L {}. For M a model ofT, and AutL(M) we call a generic automorphism ofM if (M, ) is an existentially closed modelof T. A general problem is to find necessary and sufficient conditionson T for the class of existentially models ofT to be elementary, namelyto be the class of models of some first order theory in L. This firstorder theory, if it exists, is denoted T A, and it is the model companionof T. This problem seems to be a difficult problem even if we assume

    T to be stable [1], [5], [8]. Generic automorphisms in the sense of thispaper were first studied by Lascar [7]. The work of Chatzidakis andHrushovski [2] on the case where T is the theory ACF of algebraicallyclosed fields renewed interest in the topic and Chatzidakis and Pillaystudied general properties of T A for stable T [3].

    Kudaibergenov proved that ifT A exists then T eliminates the quan-tifier there exists infinitely many. Therefore, if T is stable and T Aexists then T does not have the fcp. Pillay conjectured that if T hasthe fcp then T A does not exist after the first author observed thatthe theory of random graphs does not have T A. The first author then

    proved that if T A exists and T does not have the independence prop-erty then T is stable, and if T A exists and T has the amalgamationproperty then T is stable [4]. The latter fact covers the case of therandom graphs. The present paper extends the former case.

    So the theorem here shows that model complete theories with Thaving a model companion are low in the hierarchy of classification

    The second author is supported by the United States-Israel Binational ScienceFoundation. Publication 748.

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    2 HIROTAKA KIKYO AND SAHARON SHELAH

    theory; previous results have shown it cannot be in some intermediatepositions.

    For other examples, Hrushovski observed that there are no T A forACFA and for the theory of pseudo-finite fields P sf (unpublished).His argument depends heavily on field theory. ACFA does not havethe amalgamation property but it is not known if P sf has the amal-gamation property.

    In the rest of the paper, small letters a, b, c, etc. denote finite tuplesand x, y, etc. denote finite tuples of variables. Ifa is a tuple of elementsand A a set of elements, a A means that each element of a belongsto A.

    2. Main Theorem

    Theorem 1. LetT be a model complete theory in a language L and a new unary function symbol. If T has a model whose theory has thestrict order property then T has no model companion.

    Proof. Let M0 be a model of T with the strict order property. So,there are L-definable partial order < on k-tuples in M0 for some kand a sequence ai : i < of k-tuples in M0 such that ai < aj fori < j < . By Ramseys Theorem, we can assume that ai : i < isan L-indiscernible sequence in M0. Also, we can assume that there isan L-automorphism 0 of M0 such that 0(ai) = ai+1. So, (M0, 0) is

    a model of T.Now by way of contradiction, suppose that T has a model com-

    panion, say T A. Extend (M0, 0) to a model (N, ) of T A. N is anL-elementary extension of M0 since T is model complete. We can as-sume that (N, ) is sufficiently saturated. In the rest of the proof, wework in (N, ).

    Consider the partial type p(x) = {ai < x : i < } and let (x) y(a0 < (y) (y) < y y < x).

    Claim. In (N, ),

    (1) p(x) (x), and

    (2) ifq(x) is a finite subset of p(x) then q(x) (x).

    If this claim holds, then it contradicts the saturation of (N, ).We first show (2). Let n be such that q(x) {ai < x : i < n

    }.Then an satisfies q(x). Suppose an satisfies (x). Let b N be suchthat a0 < (b) < b < an. By a0 < (b), we have an =

    n(a0)