A Note on Existentially Closed Difference Fields with Algebraically Closed Fixed FieldAuthor(s): Anand PillaySource: The Journal of Symbolic Logic, Vol. 66, No. 2 (Jun., 2001), pp. 719-721Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2695040 .

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THE JOURNAL OF SYMBOLIC LOGIC

Volume 66, Number 2, June 2001

A NOTE ON EXISTENTIALLY CLOSED DIFFERENCE FIELDS WITH ALGEBRAICALLY CLOSED FIXED FIELD

ANAND PILLAY

Abstract. We point out that the theory of difference fields with algebraically closed fixed field has no

model companion.

By a difference field we mean a field K equipped with an automorphism c. It is well-known ([1]) that the class of existentially closed difference fields is an elementary class (ACFA), and moreover all completions are unstable. The fixed field (set of a E K such that a (a) = a) is responsible for the instability as it is pseudofinite. So in connection with an attempt to find stable nontrivial difference fields, one might consider the class of difference fields with algebraically closed fixed field and look for a model companion. Here we point out that this is impossible. We prove:

THEOREM 1. Let T be the theory of algebraically closed difference fields (K, +,, -, 0, 1, v) with algebraically closedfixedfield. Then T has no model com- panion. Equivalently the class of existentially closed models of T is not elementary.

Thanks to Zoe Chatzidakis for some helpful comments, and to Dugald Macpherson and Kanat Kudaibergenov for some preliminary discussions.

Let T be as in the assumptions of the theorem.

LEMMA 2. Let (K, C) be an existentially closed model of T, and let a E K be nonzero. Then there is n > 1 and nonzero x E K such that a(x)/x - a.

PROOF. One can easily find a difference field (L, z) extending (K, C) (so a = z I K) and b E L \ K such that z (b)/b = a. We may assume the field L to be algebraically closed. Let K(b)al9 be the algebraic closure of K(b) in L. Then z restricts to an automorphism of K(b)alg, which we still call z.

Case 1. K (b)alg contains no elements of Fix(-c) which are not already in K. Then (K (b)alg, -c) is also a model of T, so as (K, v) is existentially closed (for models of T) we can find a solution of u(x)/x = a in K.

Case 2. Otherwise. So let c E K(b)alg \ K with z(c) = c. Note that z(K(b)) = K(b). So all coefficients of the minimal polynomial of c over K(b) are fixed by z. Thus without loss of

Received October 20,1999 Partially supported by an NSF grant: DMS-9696268.

? 2001, Association for Symbolic Logic

0022-4812/01/6602-0017/$1.30

719

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720 ANAND PILLAY

generality c E K(b) \ K. So we may write

(i) c = d(f (b)/g(b)),

where d E K is nonzero and f [X], g [X] are monic polynomials over K with no common factors, and not both f [X], g[X] are constant. Applying z to both sides in (i) and using the fact that cr(c) = c, and z(b) = a.b we obtain

(ii) d (f (b)/g(b)) = -(d (f 'f (a b)/g(a b)),

where f (X) is the polynomial over K obtained from f by applying a to its coefficients (and similarly for g7 (X)).

Subcase 2 (a). f (X) /g (X) = xm for some (necessarily nonzero) integer m. Then from (ii) we see that d/u(d) = ambm/bm = am. So replacing d by d-1 if necessary we have found nonzero d1 E K such that a (dj )/dj is equal to some positive power of a, as required.

Subcase 2(b). Otherwise. This implies that at least one of f (X), g (X) has some nonzero root (in K of course). As b is transcendental over K, (ii) above yields:

(iii) d (f (X)Ig (X)) = u (d) (f '7(a X) /g (a X))

as rational functions over K. In particular they have the same poles and zeros. Let us assume that some root of f (X) is nonzero (a symmetric argument will work for g), and let al, .., a, (all in K) be the distinct nonzero roots of f (X), thus also the distinct nonzero zeros of d (f (X)/g(X)). Then u (ai), .., u (a,) are the distinct nonzero roots of f7(X), and thus u(aj)/a,...,u(a,)/a are the distinct nonzero roots of f '7 (aX), and so the distinct nonzero zeros of the right hand side of (iii). Thus

(iv) {a, .., an} {= a(aj)1/a,.**, (an)/a}.

Let c be the product of a , .., an. Then c e K is nonzero and by (iv) one sees that c = u(c)/an and so u(c)/c = an. This completes the proof of Lemma 2. -1

PROOF OF THEOREM 1. Suppose by way of contradiction that T had a model companion T'. By the lemma and compactness, there is N > 1 such that in any model (K, u) of T' for any a e K, a(x)/x = an has a solution in K for some n =1, .., N. So for suitable N, u(x)/x = aN has a solution in K for all a e K. As K is algebraically closed, u (x)/x = a has a solution for all a e K. The rest of the proof follows as in [2]:

Let (K, u) be a model of T' of characteristic 0. So there is x e K such that u (x)/x = -1. But then 2 (x) = x, so x e Fix(U2) \ Fix(a), contradicting the fact that Fix(a) is algebraically closed.

This contradiction shows that T' does not exist. 1

The same proof shows that even if we fix the characteristic in T, then there is no model companion T': in the characteristic 2 case, choose x such that U(x)/x = co where co 7& 1 is a cube root of 1. Then x e Fix(c 6) \ Fix(a), again contradicting the fixed field being algebraically closed.

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A NOTE ON EXISTENTIALLY CLOSED DIFFERENCE FIELDS 721

QUESTION. Is there any model complete theory T of difference fields (in the language (+, -, 0, 1, a)) such that for all models (K, u) of T, K is algebraically closed and Fix(a) is a proper algebraically closed subfield?

REFERENCES

[1] Z. CHATZIDAKIs and E. HRuSHOVSKI, The model theory ofdifferencefields, Transactions ofAmerican Mathematical Society, vol. 351 (1999), pp. 2997-3071.

[2] E. HRUSHOVSKI, On superstable fields with automorphisms, Model theory of groups (A. Nesin and A. Pillay, editors), Notre Dame Press, 1989.

DEPARTMENT OF MATHEMATICS

UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN

1409 W GREEN STREET (MC-382)

URBANA, ILLINOIS 61801-2975

E-mail: pillay~math.uiuc.edu

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