THE SET OF NONSQUARES IN A NUMBER FIELD IS DIOPHANTINE

BJORN POONEN

Abstract. Fix a number field k. We prove that k^{×} − k^{×2} is diophantine over k. This is deduced from a theorem that for a nonconstant separable polynomial P (x) ∈ k[x], there are at most finitely many a ∈ k^{×} modulo squares such that there is a Brauer-Manin obstruction to the Hasse principle for the conic bundle X given by y^{2} − az^{2} = P (x).

1. Introduction

Throughout, let k be a global field; occasionally we impose additional conditions on its characteristic. Warning: we write k^{n} = ∏_{i=1}^{n} k and k^{×n} = {a^{n} : a ∈ k^{×}}.

1.1. Diophantine sets. A subset A ⊆ k^{n} is diophantine over k if there exists a closed subscheme V ⊆ 𝔸_{k}^{n+m} such that A equals the projection of V (k) under k^{n+m} → k^{n}. The complexity of the collection of diophantine sets over a field k determines the difficulty of solving polynomial equations over k. For instance, it follows from [Mat70] that if ℤ is diophantine over ℚ, then there is no algorithm to decide whether a multivariable polynomial equation with rational coefficients has a solution in rational numbers. Moreover, diophantine sets can be built up from other diophantine sets. In particular, diophantine sets over k are closed under taking finite unions and intersections. Therefore it is of interest to gather a library of diophantine sets.

1.2. Main result. Our main theorem is the following:

Theorem 1.1. For any number field k, the set k^{×} − k^{×2} is diophantine over k.

In other words, there is an algebraic family of varieties (V_{t})_{t∈k} such that V_{t} has a k-point if and only if t is not a square. This result seems to be new even in the case k = ℚ.

Corollary 1.2. For any number field k and for any n ∈ ℤ_{≥0}, the set k^{×}−k^{×2^{n}} is diophantine over k.

Proof. Let A_{n} = k^{×} − k^{×2^{n}}. We prove by induction on n that A_{n} is diophantine over k. The base case n = 1 is Theorem 1.1. The inductive step follows from

A_{n+1} = A_{1} ∪ {t^{2} : t ∈ A_{n} and −t ∈ A_{n}}.  □

Date: October 12, 2008; minor correction March 23, 2014.

2000 Mathematics Subject Classification. Primary 14G05; Secondary 11G35, 11U99, 14G25, 14J20.

Key words and phrases. Brauer-Manin obstruction, nonsquares, diophantine set, Châtelet surface, conic bundle, Hasse principle, rational points.

This research was supported by NSF grants DMS-0301280 and DMS-0841321.

1.3. Brauer-Manin obstruction. The main ingredient of the proof of Theorem 1.1 is the fact the Brauer-Manin obstruction is the only obstruction to the Hasse principle for certain Châtelet surfaces over number fields, so let us begin to explain what this means. Let Ω_{k} be the set of nontrivial places of k. For v ∈ Ω_{k}, let k_{v} be the completion of k at v. Let A be the adèle ring of k. For a projective k-variety X, we have X(A) = ∏_{v∈Ω_{k}} X(k_{v}); one says that there is a Brauer-Manin obstruction to the Hasse principle for X if X(A) ≠ ∅ but X(A)^{Br} = ∅. See [Sko01, §5.2].

1.4. Conic bundles and Châtelet surfaces. Let E be a rank-3 vector sheaf over a base variety B. A nowhere-vanishing section s ∈ Γ(B, Sym^{2} E) defines a subscheme X of ℙE whose fibers over B are (possibly degenerate) conics. As a special case, we may take (E, s) = (L_{0} ⊕ L_{1} ⊕ L_{2}, s_{0} + s_{1} + s_{2}) where each L_{i} is a line sheaf on B, and the s_{i} ∈ Γ(B, L_{i}^{⊗2}) ⊂ Γ(B, Sym^{2} E) are sections that do not simultaneously vanish on B.

We specialize further to the case where B = ℙ^{1}, L_{0} = L_{1} = O, L_{2} = O(n), s_{0} = 1, s_{1} = −a, and s_{2} = −P̃(w, x) where a ∈ k^{×} and P̃(w, x) ∈ Γ(ℙ^{1}, O(2n)) is a separable binary form of degree 2n. Let P (x) := P̃(1, x) ∈ k[x], so P (x) is a separable polynomial of degree 2n − 1 or 2n. We then call X the conic bundle given by

y^{2} − az^{2} = P (x).

A Châtelet surface is a conic bundle of this type with n = 2, i.e., with deg P equal to 3 or 4. See also [Poo09].

The proof of Theorem 1.1 relies on the Châtelet surface case of the following result about families of more general conic bundles:

Theorem 1.3. Let k be a global field of characteristic not 2. Let P (x) ∈ k[x] be a nonconstant separable polynomial. Then there are at most finitely many classes in k^{×}/k^{×2} represented by a ∈ k^{×} such that there is a Brauer-Manin obstruction to the Hasse principle for the conic bundle X given by y^{2} − az^{2} = P (x).

Remark 1.4. Theorem 1.3 is analogous to the classical fact that for an integral indefinite ternary quadratic form q(x, y, z), the set of nonzero integers represented by q over ℤ_{p} for all p but not over ℤ fall into finitely many classes in ℚ^{×}/ℚ^{×2}. J.-L. Colliot-Thélène and F. Xu explain how to interpret and prove this fact (and its generalization to arbitrary number fields) in terms of the integral Brauer-Manin obstruction: see [CTX07, §7], especially Proposition 7.9 and the very end of §7. Our proof of Theorem 1.3 shares several ideas with the arguments there.

1.5. Definable subsets of k_{v} and their intersections with k. The proof of Theorem 1.1 requires one more ingredient, namely that certain subsets of k defined by local conditions are diophantine over k. This is the content of Theorem 1.5 below, which is proved in more generality than needed. By a k-definable subset of k_{v}^{n}, we mean the subset of k_{v}^{n} defined by some first-order formula in the language of fields involving only constants from k, even though the variables range over elements of k_{v}.

Theorem 1.5. Let k be a number field. Let k_{v} be a nonarchimedean completion of k. For any k-definable subset A of k_{v}^{n}, the intersection A ∩ k^{n} is diophantine over k.

1.6. Outline of paper. Section 2 shows that Theorem 1.5 is an easy consequence of known results, namely the description of definable subsets over k_{v}, and the diophantineness of the valuation subring O of k defined by v. Section 3 proves Theorem 1.3 by showing that for most twists of a given conic bundle, the local Brauer evaluation map at one place is enough to rule out a Brauer-Manin obstruction. Finally, Section 4 puts everything together to prove Theorem 1.1.

2. Subsets of global fields defined by local conditions

Lemma 2.1. Let m ∈ ℤ_{>0} be such that char k ∤ m. Then k_{v}^{×m} ∩ k is diophantine over k.

Proof. The valuation subring O of k defined by v is diophantine over k: see the first few paragraphs of §3 of [Rum80]. The hypothesis char k ∤ m implies the existence of c ∈ k^{×} such that 1 + cO ⊂ k_{v}^{×m}; fix such a c. The denseness of k^{×} in k_{v}^{×} implies k_{v}^{×m} ∩ k = (1 + cO)k^{×m}. The latter is diophantine over k. □

Proof of Theorem 1.5. Call a subset of k_{v}^{n} simple if it is of one of the following two types: {x⃗ ∈ k_{v}^{n} : f(x⃗) = 0} or {x⃗ ∈ k_{v}^{n} : f(x⃗) ∈ k_{v}^{×m}} for some f ∈ k[x_{1}, …, x_{n}] and m ∈ ℤ_{>0}. It follows from the proof of [Mac76, Theorem 1] (see also [Mac76, §2] and [Den84, §2]) that any k-definable subset A is a boolean combination of simple subsets. The complement of a simple set of the first type is a simple set of the second type (with m = 1). The complement of a simple set of the second type is a union of simple sets, since k_{v}^{×m} has finite index in k_{v}^{×}. Therefore any k-definable A is a finite union of finite intersections of simple sets. Diophantine sets in k are closed under taking finite unions and finite intersections, so it remains to show that for every simple subset A of k_{v}^{n}, the intersection A ∩ k is diophantine. If A is of the first type, then this is trivial. If A is of the second type, then this follows from Lemma 2.1. □

3. Family of conic bundles

Given a k-variety X and a place v of k, let Hom′(Br X, Br k_{v}) be the set of f ∈ Hom(Br X, Br k_{v}) such that the composition Br k → Br X →^{f} Br k_{v} equals the map induced by the inclusion k ↪ k_{v}. The v-adic evaluation pairing Br X × X(k_{v}) → Br k_{v} induces a map X(k_{v}) → Hom′(Br X, Br k_{v}).

Lemma 3.1. With notation as in Theorem 1.3, there exists a finite set of places S of k, depending on P (x) but not a, such that if v ∉ S and v(a) is odd, then X(k_{v}) → Hom′(Br X, Br k_{v}) is surjective.

Proof. The function field of ℙ^{1} is k(x). Let Z be the zero locus of P̃(w, x) in ℙ^{1}. Let G be the group of f ∈ k(x)^{×} having even valuation at every closed point of ℙ^{1} − Z. Choose P_{1}(x), …, P_{m}(x) ∈ G representing a 𝔽_{2}-basis for the image of G in k(x)^{×}/k(x)^{×2}k^{×}. We may assume that P_{m}(x) = P (x). Choose S so that each P_{i}(x) is a ratio of polynomials whose nonzero coefficients are S-units, and so that S contains all places above 2.

Let κ(X) be the function field of X. A well-known calculation (see [Sko01, §7.1]) shows that the class of each quaternion algebra (a, P_{i}(x)) in Br κ(X) belongs to the subgroup Br X, and that the cokernel of Br k → Br X is an 𝔽_{2}-vector space with the classes of (a, P_{i}(x)) for i ≤ m − 1 as a basis.

Suppose that v ∉ S and v(a) is odd. Let f ∈ Hom′(Br X, Br k_{v}). The homomorphism f is determined by where it sends (a, P_{i}(x)) for i ≤ m − 1. We need to find R ∈ X(k_{v}) mapping to f.

Let O_{v} be the valuation ring in k_{v}, and let 𝔽_{v} be its residue field. For i ≤ m − 1, choose c_{i} ∈ O_{v}^{×} whose image in 𝔽_{v}^{×} is a square or not, according to whether f sends (a, P_{i}(x)) to 0 or 1/2 in ℚ/ℤ ≃ Br k_{v}. Since v(a) is odd, we have (a, c_{i}) = f((a, P_{i}(x))) in Br k_{v}.

View ℙ^{1} − Z as a smooth O_{v}-scheme, and let Y be the finite étale cover of ℙ^{1} − Z whose function field is obtained by adjoining \sqrt{c_{i}P_{i}(x)} for i ≤ m − 1 and also \sqrt{P (x)}. Then the generic fiber Y_{k_{v}} := Y ×_{O_{v}} k_{v} is geometrically integral. Assuming that S was chosen to include all v with small 𝔽_{v}, we may assume that v ∉ S implies that Y has a (smooth) 𝔽_{v}- point, which by Hensel’s lemma lifts to a k_{v}-point Q. There is a morphism from Y_{k_{v}} to the smooth projective model of y^{2} = P (x) over k_{v}, which in turn embeds as a closed subscheme of X_{k_{v}}, as the locus where z = 0. Let R be the image of Q under Y (k_{v}) → X(k_{v}), and let α = x(R) ∈ k_{v}. Evaluating (a, P_{i}(x)) on R yields (a, P_{i}(α)), which is isomorphic to (a, c_{i}) since c_{i}P_{i}(α) ∈ k_{v}^{×2}. Thus R maps to f, as required. □

Lemma 3.2. Let X be a projective k-variety. If there exists a place v of k such that the map X(k_{v}) → Hom′(Br X, Br k_{v}) is surjective, then there is no Brauer-Manin obstruction to the Hasse principle for X.

Proof. If X(A) = ∅, then the Hasse principle holds. Otherwise, pick Q = (Q_{w}) ∈ X(A), where Q_{w} ∈ X(k_{w}) for each w. For A ∈ Br X, let ev_{A}: X(L) → Br L be the evaluation map for any field extension L of k. Let inv_{w}: Br k_{w} → ℚ/ℤ be the usual inclusion map. Define

η: Br X → ℚ/ℤ ≃ Br k_{v}
A ↦ − ∑_{w≠v} inv_{w} ev_{A}(Q_{w}).

By reciprocity, η ∈ Hom′(Br X, Br k_{v}). The surjectivity hypothesis yields R ∈ X(k_{v}) giving rise to η. Define Q′ = (Q′_{w}) ∈ X(A) by Q′_{w} := Q_{w} for w ≠ v and Q′_{v} := R. Then Q′ ∈ X(A)^{Br}, so there is no Brauer-Manin obstruction. □

Proof of Theorem 1.3. Let S be as in Lemma 3.1. Enlarge S to assume that Pic O_{k,S} is trivial. Then the set of a ∈ k^{×} such that v(a) is even for all v ∉ S has the same image in k^{×}/k^{×2} as the finitely generated group O_{k,S}^{×}, so the image is finite.

Suppose that a ∈ k^{×} has image in k^{×}/k^{×2} lying outside this finite set. Then we can fix v ∉ S such that v(a) is odd. Let X be the corresponding surface. Combining Lemmas 3.1 and 3.2 shows that there is no Brauer-Manin obstruction to the Hasse principle for X. □

4. The set of nonsquares is diophantine

Proof of Theorem 1.1. For each place v of k, define S_{v} := k^{×} ∩ k_{v}^{×2} and N_{v} := k^{×} − S_{v}. By Theorem 1.5, the sets S_{v} and N_{v} are diophantine over k.

By [Poo09, Proposition 4.1], there is a Châtelet surface

X_{1} : y^{2} − bz^{2} = P (x)

over k, with P (x) a product of two irreducible quadratic polynomials, such that there is a Brauer-Manin obstruction to the Hasse principle for X_{1}. For t ∈ k^{×}, let X_{t} be the (smooth projective) Châtelet surface associated to the affine surface

U_{t}: y^{2} − tbz^{2} = P (x).

We claim that the following are equivalent for t ∈ k^{×}:

(i) U_{t} has a k-point.

(ii) X_{t} has a k-point.

(iii) X_{t} has a k_{v}-point for every v and there is no Brauer-Manin obstruction to the Hasse

principle for X_{t}. The implications (i) ⟹ (ii) ⟹ (iii) are trivial. The implication (iii) ⟹ (ii) follows from [CTCS80, Theorem B]. Finally, in [CTCS80], the reduction of Theorem B to Theorem A combined with Remarque 7.4 shows that (ii) implies that X_{t} is k-unirational, which implies (i).

Let A be the (diophantine) set of t ∈ k^{×} such that (i) holds. The isomorphism type of U_{t} depends only on the image of t in k^{×}/k^{×2}, so A is a union of cosets of k^{×2} in k^{×}. We will compute A by using (iii).

The affine curve y^{2} = P (x) is geometrically integral so it has a k_{v}-point for all places v outside a finite set F. So for any t ∈ k^{×}, the variety X_{t} has a k_{v}-point for all v ∉ F. Since X_{1} has a k_{v}-point for all v and in particular for v ∈ F, if t ∈ ⋂_{v∈F} S_{v}, then X_{t} has a k_{v}-point for all v.

Let B := A ∪ ⋃_{v∈F} N_{v}. If t ∈ k^{×} − B, then X_{t} has a k_{v}-point for all v, and there is a Brauer-Manin obstruction to the Hasse principle for X_{t}. By Theorem 1.3, k^{×} − B consists of finitely many cosets of k^{×2}, one of which is k^{×2} itself. Each coset of k^{×2} is diophantine over k, so taking the union of B with all the finitely many missing cosets except k^{×2} shows that k^{×} − k^{×2} is diophantine. □

Acknowledgements

I thank Jean-Louis Colliot-Thélène and Anthony Várilly-Alvarado for a few comments, and Alexandra Shlapentokh for suggesting some references.

References

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Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139-4307, USA

E-mail address: poonen@math.mit.edu
URL: http://math.mit.edu/~poonen