Dependent Fields have no Artin-Schreier Extensions

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have no Artin-Schreier Extensions La Roche-en-Ardenne Itay Kaplan Joint work with Thomas Scanlon Hebrew University of Jerusalem and Université Claude Bernard Lyon 1 21 April 2008

In [5], Macintyre showed that an innite ω-stable eld is algebraically closed; this was subsequently generalized by Cherlin and Shelah to the superstable case, and commutativity need not be assumed but follows [3]. It is known [6] that separably closed elds are stable; the converse has been conjectured early on, but little progress has been made. In 1992, Thomas Scanlon noted that an innite stable eld of characteristic p at least has no Artin-Schreier extensions, and hence has no nite Galois extension of degree divisible by p. Here we show the same for dependent (NIP) innite elds.

Notation Notation is standard.

theories Denition A theory T has the independence property if there is a formula φ( x, ȳ) and tuples (ā i : i ω) and ( b A : A ω) such that = φ(ā i, b A ) if and only if i A. Denition A theory T is dependent if it does not have the independence property (also known as NIP). Example Every stable theory is NIP as well as any o-minimal (and many others).

groups Denition A group G is dependent if Th(G) is dependent. We use only the following theorem about dependent groups: Theorem (Baldwin-Saxl) [1] Let G be a dependent group. Given a family of uniformly dened subgroups, there is a number n < ω such that any nite intersection of groups from this family is already an intersection of n of them.

Artin-Schreier theory Let k be a eld of characteristic p. Let ρ be the polynomial X p X. Theorem (Artin-Schreier) 1 Given a k, either the polynomial ρ a has a root in k, in which case all its root are in k, or it is irreducible. In the latter case, if α is a root then k(α) is cyclic of degree p over k. 2 Conversely, let K be a cyclic extension of k of degree p. Then there exists α K such that K = k(α) and for some a k, ρ(α) = a. Such extensions are called Artin-Schreier extensions.

Main Theorem Theorem Let K be an innite dependent eld of characteristic p > 0. Then K is Artin-Schreier closed - i.e. ρ is onto. Corollary If L/K is a nite separable extension, then p does not divide [L : K]. Corollary K contains F alg p.

Beginning of proof First we work in a more general setting. Let F be an algebraically closed eld For a number n < ω and b F n+1, we dene the algebraic group G b = {(t, x 1,..., x n ) t = b i (x p i x i ) for 1 i n}. G b is an algebraic subgroup of (K, +) n+1. If b K F, then by Baldwin-Saxl, for some number n 0, for every n 0 + 1 tuple b, there is a sub n 0 tuple b such that the projection π : G b(k) G b (K) is onto. (Consider the family of subgroups of the form {t x[t = a(x p x)]}).

First claim Claim Let F be an algebraically closed eld. Suppose b F is algebraically independent, then G b is a connected group. Proof. (sketch) By induction on n := length(ā). For n = 1 it's easy. Assume the claim for n, and let length(ā) = n + 1, ā = ā n. π : G b is onto. Let H = G G b 0 b (the connected component of 0). : π(h)] [G b [G b : H] <, so π(h) = by G b induction. Assume that H G b. Then for every (t, x) G b there is exactly one x n+1 such that (t, x, x n+1 ) H.

First claim (cont) Proof. So H is a graph of a denable function over b. Now choose x i such that b i (x p x i i ) = 1 for i = 1,..., n. Then we can nd a solution to b n+1 (x p x) = 1 in dcl(b n+1, x 1,..., x n ) = F p (b n+1, x 1,..., x n ) ins. (for a eld L, L ins = n<ω Lp n ). Denote L = F p (b n+1, x 1,..., x n ). As x is separable over L, it follows that x L. By our assumption, b n+1 is transcendental over x 1,..., x n. From this one can easily derive a contradiction.

Second claim Claim Let k be a eld of characteristic p, and let f : k k be an additive polynomial (i.e. f (x + y) = f (x) + f (y)). Then f is of the form a i x pi. Moreover, if k is algebraically closed and ker(f ) = F p then f = (a (x p x)) pn for some n < ω, a k. Proof. (sketch) By induction on deg(f ). First show that f (x) is constant (a 0 ), and prove the rst part. The induction base for the moreover: if a 0 0, then (f, f ) = 1, hence f has no multiple factors, hence deg(f ) = p, hence f = a 0 x + a 1 x p, but letting x = 1 we get a 0 = a 1.

Important fact We use the following: Fact Let k be a perfect eld, and G a closed 1-dimensional connected algebraic subgroup of some n < ω. Then G is isomorphic over k to ( k alg, + ) n dened ( over k, for k alg, + ). The proof of this fact uses Hilbert 90, and the following, from [4] Fact A closed connected subgroup of a isomorphic to ( k alg, + ) dim(g). ( k alg, + ) n group is

Proof continues Let us use the fact and the 2 claims to nish the proof. We may assume that K is ℵ 0 saturated. Let k = n K pn. k is an innite perfect eld. choose b k n 0+1 that are algebraically independent. By Baldwin-Saxl, there is some sub sub n 0 tuple b such that the projection π : G b(k) G b (K) is onto. By the rst claim, both G b and G b are connected. Of course, their dimension is 1.

Proof continues By the fact, we know that both these groups are isomorphic over k to (K alg, +). So we have some ν k[x ] such that G b(k alg ) (K alg, +) π G b (K alg ) ν (K alg, +) commutes. As the sides are isomorphisms, dened over k K, we can restrict them to K. As π G b(k) is onto G b (K), then so is ν K. ker(ν) = p = ker(π) (even when restricted to K ).

Proof ends Suppose that 0 c ker(ν) k. let ν = ν m c (m c (x) = c x). ν is an additive polynomials over K whose kernel is F p. So WLOG ker(ν) = F p. By the second claim, ν is of the form a (x p x) pn for a K. But ν is onto, hence so is ρ and we are done (given y K, there is some x K such that a (x p x) pn = a y pn )

On Simple elds It is known that bounded (i.e. with only nitely many extensions of each degree) PAC (pseudo-algebraically closed: every absolutely irreducible variety has a rational point) elds are simple [2] and again the converse is conjectured. In 2006, Frank Wagner adapted Scanlon's argument to the simple case and showed that simple elds have only nitely many Artin-Schreier extensions. He even showed this is true when the eld is type denable in a simple theory.

type denable elds As of yet, I do not know the situation with type denable elds in a dependent theory. What is known is that For a eld type denable in a dependent theory there are either innitely many Artin-Schreier extensions, or none.

References J.T. Baldwin and J. Saxl. Logical stability in group theory. Journal of the Australian Mathematical Society, 21:267276, 1976. Z. Charzidakis and A. Pillay. Generic structures and simple theories. Annals of Pure and Applied Logic, 95:7192, 1998. G. Cherlin and S. Shelah. Superstable elds and groups. Annals of Mathematical Logic, 18:227270, 1980. James E. Humphreys. Linear Algebraic Groups. Springer-Verlag, second edition, 1998. A. Macintyre. ω 1 -categorical elds. Fundamenta Mathematicae, 70(3):253270, 1971. C. Wood. Notes on stability of separably closed elds. Journal of Symbolic Logic, 44(3):337352, 1979.