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Model Companions of TAut for stable T

John T. Baldwin

Department of Mathematics, Statistics and Computer ScienceUniversity of Illinois at Chicago

Saharon Shelah

Department of MathematicsHebrew University of Jerusalem

Rutgers University

August 17, 2011

Let T be a complete first order theory in a countable relational languageL. We assume relation symbols have been added to make each formulaequivalent to a predicate. Adjoin a new unary function symbol to obtain

the language L; TAut is obtained by adding axioms asserting that is anL-automorphism.

The modern study of the model companion of theories with an auto-morphism has two aspects. One line, stemming from Lascar [7], deals withgeneric automorphisms of arbitrary structures. A second, beginning withChatzidakis and Hrushovski [2] and questions of Macintyre about the Frobe-nius automorphism is more concerned with specific algebraic theories. Thispaper is more in the first tradition: we find general necessary and sufficientconditions for a stable first order theory with automorphism to have a modelcompanion.

Kikyo investigates the existence of model companions of TAut when Tis unstable in [4]. He also includes an argument of Kudaibergenov showing

Partially supported by NSF grant DMS 9803496 and DMS-0100594.This is paper 759 in Shelahs bibliography. The authors thank Mittag-Leffler Institute,

Rutgers University, and the The Israel Science Foundation for partial support of this

research.

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that if T is stable with the finite cover property then TAut has no model

companion. This argument was implicit in [3] and is a rediscovery of atheorem of Winkler [11] in the 70s. We provide necessary and sufficientconditions for TAut to have a model companion when T is stable. Namely,we introduce a new condition: T admits obstructions, and show that TAuthas a model companion if and only if T does not admit obstructions. Thiscondition is a weakening of the finite cover property: if a stable theory T hasthe finite cover property then T admits obstructions.

Kikyo also proved that if T is an unstable theory without the indepen-dence property, TAut does not have a model-companion. Kikyo and Shelah[6] have improved this by weakening the hypothesis to: T has the strict order

property.For p a type over A and an automorphism, (p) denotes {(x, (a)) :(x, a) p}. References of the form II.4.13. are to [9]. We thank AssafHasan for his comments on the manuscript. We also thank the referee forhis insightful remarks and in particular forcing us to make explicit and proveLemma 2.9, which may have independent interest.

Further related work is contained in [10] which investigates when TAuthas a stable model completion.

1 Example

In the following example we examine exactly why a particular TAut does nothave a model companion. Eventually, we will show that the obstructionillustrated here represents the reason TAut (for stable T) can fail to havea model companion. Let L contain two binary relation symbols E and Rand unary predicates Pi for i < . The theory T asserts that E and R areequivalence relations and that E has infinitely many infinite classes, whichare refined by R into two-element classes. Moreover, each Pi holds only ofelements from one E-class and contains exactly one element from each Rclass of that E-class. Thus, x = y Pi(x) Pj(y) implies R(x, y) if i = j.

Now, TAut does not have a model companion. To see this, let (x,y,z)be the formula: E(x, z) E(y, z) R(x, y) x = y. Let be the L-type inthe variables {z} {xiyi : i < } which asserts (xi, yi, z) holds for each i,the sequence xiyi : i < is L-indiscernible, the xi are distinct and the yi

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are distinct, and for every (x, w) L(T):

(w)

{

iU

(xi, w) (yi, (w)) : U lg(w) + 3, |U| > (lg(w) + 3)/2}.

Thus if bici : i < a realize in a model M,

(avg(bi : i < /M)) = avg(ci : i < /M).

For any finite L(T), let ,k(x, y, z) be the conjunction of the -formulas satisfied by bici : i < ka where bici : i < ka are an initialsegment of a realization of . Let ,k be the sentence

(x0, . . . xk1, y0, . . . yk1, z),k(x0, . . . xk1, y0, . . . yk1, z)

(x0, y0, x1, y1)[(x0, y0, z) (x1, y1, z) (x1) = y1].

We claim that if TAut has a model companion T

Aut, then for some k and,

TAut ,k.

For this, let M |= TAut such that bici : i < ka satisfy in M. SupposeM L N and N is an |M|+-saturated model of T. In N we can find b, crealizing the average of bi : i < and ci : i < over M respectively.

Then (avg(bi : i < /M)) = avg(ci : i < /M)

and so there is an automorphism of N extending and taking b to c.Since (M, ) is existentially closed (TAut is model complete), we can pull b, cdown to M. By compactness, some finite subset 0 of suffices and letting be the formulas mentioned in 0 and k the number of xi, yi appearing in0 we have the claim.

But now we show that if (M, ) is any model of TAut, then for any finite and any k, (M, ) |= ,k. For this, choose bi, ci for i < k which areE-equivalent to each other and to an element a in a class Pj where Pj does

not occur in and with R(bi, ci) and bi = ci. Then b, c, a satisfy ,k butthere are no bk, ck and automorphism which makes ,k true. As, for eachj,

T (x,y,z)((x,y,z) Pj(z) [Pj(x) Pj(y)]).

To put this situation in a more general framework, recall some notationfrom [9]. will denote a finite set of formulas: {i(x, yi) : lg(x) = m,i