Fields in which varieties have rational points: A note on a problem of 𝐴𝑥

Abstract

If k is a field, let G(k) denote the Galois group over k of the algebraic closure of k. In [1 ] Ax proved the following result: Let k be a perfect field such that G(k) is abelian, and every (absolutely irreducible) variety defined over k has a k-rational point. Then G(k) is pro-cyclic. He asked [1, Problem 1] if the assumption that G(k) is abelian can be removed. In this note we construct an example to show that this assumption is necessary. Recall that if X is a variety defined over k, then k(X) is a regular extension of k [2]. Further X contains a k(X)-rational point (in fact a k(X)-rational point which is generic over k). [ADDED IN PROOF. A negative answer to Probelm 1 of [1] has also been given by Moshe Jarden (Rational points on algebraic varieties over large numberfields, Bull. Amer. Math. Soc. 75 (1969), 603-606).] If E and F are regular extensions of k, then the free composite of E and F over k is again a regular extension of k. Let { Ea } be any set of regular extensions of k. By an easy transfinite induction we can construct the free composite of { Ea } over k, and it will again be a regular extension of k. Now let { Ea } be a complete set of nonisomorphic finitely generated regular extensions of k, and let E(k) denote their free composite. Then E(k) is a regular extension of k and every variety X, defined over k, has an E(k)-rational point. Set En(k) =E(En-1(k)) and E??o(k) = Un. 01 E'n(k) . If X is a variety defined over EX(k), then X is actually defined over En(k) for some n, and hence has an En+l(k)-rational point. Since k is algebraically closed in EX(k), the map G(EX(k))-*G(k) is surjective, so G(k) nonabelian implies G(E0(k)) nonabelian. Finally if k is of characteristic zero, then EX(k) is perfect. Hence if we take, say k = Q, we get the desired example. REMARK. In an oral communication, S. Garfunkel has pointed out to Ax that Problem 5 of [1 ] has an affirmative solution, i.e. that the theory of the rings Z/m is decidable. This follows from Corollaries