Integral points on generalised affine Châtelet surfaces
Vladimir Mitankin
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(∖ 
Max Planck Institut für Mathematik 

	Vivatsgasse 7 

	53111 Bonn 

	Germany
	vmitankin@mpim-bonn.mpg.de2010  Mathematics Subject Classification 14G05 (primary), 11D57, 11G35, 11R29 (secondary).
	We show, conditionally on Schinzel's hypothesis, that the only obstruction to the integral Hasse principle for generalised affine Châtelet surfaces is the Brauer–Manin one.
Integral points on generalised affine Châtelet surfaces
    Vladimir Mitankin
    December 30, 2023
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§ INTRODUCTION

This article is dedicated to the study of the set of integral points on an integral model of the affine surfaces X over  of the shape

    X :   _(√(a))(x, y) = P(t).

In the current set-up a ∈ is non-zero squarefree, P(t) ∈[t] is separable as a polynomial over  and _(√(a))(x, y) = _(√(a))/(x + ω y) is the norm form of (√(a)) for the basis 1, ω over  with ω = -(1 + √(a))/2 when a ≡ 1  4 and ω = √(a) when a ≡ 2, 3  4. We shall refer to X as a generalised affine Châtelet surface. Let  be the integral model of X over  given by the same equation, () its set of integral points and (_) = X() ×∏_p () its set of adeles. One can embed () diagonally in (_) and hence a necessary condition for the existence of integral points on  is (_) ≠∅. We say that Xsatisfies the integral Hasse principle if this is also a sufficient condition.

Colliot-Thélène and Xu <cit.> found that the Brauer–Manin obstruction <cit.> plays an important role when it comes to the existence of integral points on . More precisely, given a -variety X there is a pairing between the Brauer group X of X and the set of -adeles X(_) of X. This pairing allowed Manin to construct a sequence of inclusions X() ⊆ X(_)^ X⊆ X(_), where the intermediate set is known as the Brauer–Manin set. A Brauer–Manin obstruction to the existence of rational points on X is then present if X(_) ≠∅ but X(_)^ X = ∅. It is said to be  the only obstruction to the Hasse principle if X(_)^ X≠∅ implies X() ≠∅. A similar sequence of inclusions () ⊆(_)^ X⊆(_) is available. It induces a Brauer–Manin obstruction to the integral Hasse principle, an overview of which is included in section <ref>.

The surface X given in (<ref>) is closely related to a Châtelet surface. The latter is a smooth proper model of X when P(t) = 3 or 4. The arithmetic of a Châtelet surface is very well understood by the work of Colliot-Thélène, Sansuc and Swinnerton-Dyer <cit.>, <cit.>. They have shown that the only obstruction to the Hasse principle and weak approximation for a Châtelet surface is the Brauer–Manin one. Under the following hypothesis this is still true for a smooth proper model of X when the restriction on the degree of P(t) is dropped, as the results of Colliot-Thélène and Sansuc <cit.>, Swinnerton-Dyer <cit.>, Serre <cit.>, and Colliot-Thélène and Swinnerton-Dyer <cit.> show. 

[Schinzel's hypothesis] 
	Let P_1(t), …, P_n(t) ∈[t] be irreducible polynomials over  with positive leading coefficients and such that the set of values ∏_i = 1^n P_i(m) for m running in  has no fixed prime divisor. Then there exist infinitely many natural numbers m ∈ such that each of P_1(m), …, P_n(m) is a prime.


Unconditional results when P(t) is of arbitrary degree were attained by Browning, Matthiesen and Skorobogatov <cit.> if P(t) ∈[t] is a product of linear factors over . In this setting a variant of Hypothesis [hyp:H]H is available via recent developments in additive combinatorics.

Let X be as in (<ref>) with P(t) = ∏_i = 1^n P_i(t), where P_i(t) ∈[t] are irreducible over  and pairwise coprime polynomials. As explained in section <ref> for each i ∈{1, …, n} the quaternion algebra (a, P_i(t)) belongs to X and thus the set (_)^𝒜 of adeles orthogonal to 𝒜 contains (). We are now ready to state our main result.

 
	Let a ∈ be non-zero squarefree and let P_1(t), …, P_n(t) ∈[t] be irreducible such that P(t) = ∏_i = 1^n P_i(t) is separable as a polynomial over . Assume that P_1(t) is linear or that the discriminant of the splitting field of P_1(t) is coprime to the discriminant of (√(a)). Define X by (<ref>) and let 𝒜 = ⟨ (a, P_1(t)), …, (a, P_n(t)) ⟩ be the subgroup of X generated by the quaternion algebras (a, P_i(t)). Assume that the image of the natural projection (_)^𝒜→ X() is not bounded. Then, conditionally on Hypothesis [hyp:H]H, we have 
 	
    (_)^𝒜≠∅() ≠∅.

 	Moreover, let S be a finite set of finite places of . Then for each > 0 and for each adelic point (x_p, y_p, t_p) ∈ (_)^𝒜 with (x_∞, y_∞, t_∞) in an unbounded component of X() there exists a point (x, y, t) ∈() such that |t - t_p|_p < for any p ∈ S. 


In the case a = -1 and P(t) = - t^k + m the surfaces X were already studied by Gundlach <cit.>, again conditionally on Hypothesis [hyp:H]H. However, our methods and Gundlach's ideas are different in nature. We continue with a few remarks regarding the conditions assumed in Theorem <ref>.

There are only finitely many points with integral coordinates inside the image of the projection (_)^𝒜→ X() when it is bounded. If none of them satisfies (<ref>) the inclusion () ↪ X() yields () = ∅ and then we say that there is an obstruction to the existence of integral points coming from the archimedean place. Berg <cit.> points out that if one takes a = -1 and P(t) = -t^4 + m with m = 33, 43, 67, … then there is no Brauer–Manin obstruction, however, () = ∅ because of the obstruction at ∞. One can produce many examples in a similar manner where the lack of integral points on  is explained via the obstruction at the real place. Thus a certain unboundedness condition is absolutely necessary. Such a condition features in Theorem <ref>.

If one drops the requirement that P_1(t) is linear or the discriminant of its splitting field is coprime to the discriminant of (√(a)), then there may be extra elements of X that need to be taken into account. Such an example is a = -5 and  P(t) = t^5 + 20t + 32, where X/≅/5 is generated by a non-cyclic algebra of order 5<cit.>. However, Remark <ref> allows us to forget this assumption when the narrow class number of (√(a)) is 1 or 2.

By (<ref>) we see that Theorem <ref> implies that the Brauer–Manin obstruction is the only obstruction to the integral Hasse principle for X, conditionally on Hypothesis [hyp:H]H. As explained in section <ref>, when P(t) is irreducible the quaternion algebra (a, P(t)) is a trivial element of X and thus (_)^𝒜 =  (_). Then the statement of Theorem <ref> becomes equivalent to X satisfying the integral Hasse principle. 

In general, it is seldom an easy task to give an explicit description of the elements of X which is, usually, the first non-trivial step in the analysis of (_)^. Progress in this direction for generalised affine Châtelet surfaces was made by Berg, who describes the quotient X / in the case when P(t) is irreducible over  with additional assumptions on its splitting field <cit.>. Our method has the advantage that it requires only basic knowledge of X. We work with the quaternion algebras (a, P_1(t)), …, (a, P_n(t)) ∈ X. This allows us to study infinitely many surfaces which are not covered in Berg's work, for example those with (√(a)) not contained in the splitting field of P(t). Other difficulties were pointed out by Colliot-Thélène and Harari <cit.>. The projection to the t-coordinate endows X with a conic bundle structure over the affine line. The two main problems with the currently known techniques for studying integral points according to them are that for any smooth fibre X_t the quotient X_t / is infinite and the fibres corresponding to the zeros of P(t) are not split.

Our strategy exploits the structure of the narrow ideal class group of (√(a)) to adapt a method of Colliot-Thélène and Sansuc <cit.> to this setting. More precisely, let X be as in (<ref>). Under the assumption of Hypothesis [hyp:H]H we reduce the study of the existence of points on  to understanding if a suitable fibre X_t over an integral t with X_t() ≠∅ has an integral point. This is answered positively with the help of the next proposition.


	Let a be a non-zero squarefree integer and let K = (√(a)). Let S be a finite set of finite primes of  which are not inert in O_K. Let _1, …, _h-1 be prime ideals of O_K above split primes p_1, …, p_h-1∉S, respectively, such that the non-trivial elements of the narrow class group of K are precisely the distinct classes [_1], …, [_h-1]. Then if
	
    _K(x, y)
    		= p_1 … p_h-1∏_p ∈ S p

	has a solution with x, y ∈, it also has a solution with x, y ∈. 


One can view Proposition <ref> as a separate result, a corollary of which is that certain affine conics satisfy the integral Hasse principle. 



 §.§ Examples

Keep notation as in Theorem <ref>. We conclude the introduction with two numerical examples which demonstrate how Theorem <ref> with some of its assumptions relaxed still yields the existence of integral points on . 


	Let X ⊂Å^3 be the affine surface defined over  by
	
    X:    x^2 - 2y^2 = t^5 + 6t^2 + 3t + 3.

	Eisenstein's criterion verifies that P(t) = t^5 + 6t^2 + 3t + 3 is irreducible and thus (_)^𝒜 = (_) as explained in section <ref>. Hence it suffices to verify that (_) ≠∅. It is clear that X() ≠∅. Hensel's lemma and the existence of non-trivial solution to x^2 - 2y^2 ≡ 3  p imply that () ≠∅ for each p ≠ 2, 3. When p = 2 or 3 the solutions (1, 1, 4)  8 and (1, 0, 1)  3 to x^2 - 2y^2 ≡ P(t)  8 and mod 3, respectively, and Hensel's lemma show the existence of a point in (). Thus (_) ≠∅. The narrow class group of (√(2)) is trivial. Thus by Remark <ref> we are in position to apply Theorem <ref> which states, conditionally on Hypothesis [hyp:H]H, that  has an integral point. In fact, (59, 47, -4) ∈().


In the second example Theorem <ref> is applied without the assumption of Hypothesis [hyp:H]H.


	Let a ≡ 1  8 be a positive squarefree integer such that the narrow class number of (√(a)) is 1 or 2, e.g. a  = 17, 33, 41, 57, 73, 89, 97, … Let X ⊂Å^3 be the affine surface over  given by 
	
    X:    x^2 - xy + 1 - a/4y^2 = ∏_i = 1^n (t^2k_i - a^2m_i - 1),

	where n, k_i and m_i are positive integers such that t^2k_i - a^2m_i + 1 are irreducible over  and pairwise coprime. We claim that (_)^𝒜≠∅. Indeed, since a > 0 we have X() ≠∅ and (a, t_∞^2k_i - a^2m_i - 1)_∞ = 1 for any real point (x_∞, y_∞, t_∞). The later implies the unboundedness condition in Theorem <ref>. One easily sees that () ≠∅ for any p ∤ a odd by a similar argument as in Example <ref> via Remark <ref>. Moreover, for each such p either a ∈^* 2 or (t_p^2k_i - a^2m_i - 1) = 0 for any (x_p, y_p, t_p) ∈(). Thus (a, t_p^2k_i - a^2m_i - 1)_p = 1. If p | a is odd, then (1, 0, 1)  p lifts to a -point by Hensel's lemma. For this point the corresponding Hilbert symbols for the quaternion algebras (a, t^2k_i - a^2m_i - 1) are all equal to 1. Finally, a ≡ 1  8 implies that (1 - a)/4 is even and thus depending on the parity of n one of (1, 0, 0) or (1, 2, 0)  8 lifts to a point in (_2) by Hensel's lemma. Furthermore, a ≡ 1  8 implies that the corresponding Hilbert symbols are again all 1 at p = 2. This gives a point (x_p, y_p, t_p) ∈(_)^𝒜. Let a finite set S of finite primes be given which we extend as in the proof of Theorem <ref>. Instead of Hypothesis [hyp:H]H we can use here the fact that  satisfies strong approximation away from ∞. Thus there is λ∈ arbitrarily close to t_p when p ∈ S and such that
	
    ∏_i = 1^n (λ^2k_i - a^2m_i - 1)
    		= ∏_p ∈ S' p^n_p,

	where S' is some finite set containing S. All we need to guarantee now is that no prime p of  which is inert in (√(a)) divides ∏_i = 1^n (λ^2k_i - a^2m_i - 1). However, this is a straightforward corollary of the shape of each of the polynomials t^2k_i - a^2m_i - 1, the fact that for odd such primes a is not a quadratic residue mod p and 2 is split in (√(a)). Following the same circle of ideas as in the proof of Theorem <ref> puts us in position where Proposition <ref> is applicable. Hence () ≠∅.


 §.§ Structure

This article is organised as follows. In section 2 we give an overview of the Brauer–Manin obstruction for integral points. Section 3 is dedicated to the proof of Proposition <ref>. In section 4 we state and prove Proposition <ref> and its Corollary <ref> which are needed in the proof of Therorem <ref> when the narrow class number of (√(a)) is greater than 2. In the last section we prove Theorem <ref>.



 §.§ Notation

Throughout the rest of this article the following convention will be in force. For any number field k denote its ring of integers by O_k and its discriminant by d_k. We preserve a for a fixed non-zero squarefree integer. Let K = (√(a)) and let  be the narrow class group of K. The symbol [·] denotes an element  and we shall write h = # for the narrow class number of K. For any b ∈ K^× let (b) denote the principal fractional ideal b O_K. The subscript i will usually be preserved for an irreducible factor of P(t) while the subscript j will usually be used for one of the primes p_1, …, p_h-1 appearing in Proposition <ref>
	An earlier version of this work is part of the author's PhD thesis. The author is very grateful to his supervisor Tim Browning for his constant support and guidance. The author thanks Roger Heath-Brown for suggesting the use of Proposition <ref> and Martin Bright, Jean-Louis Colliot-Thélène, Yonatan Harpaz, Daniel Loughran, Carlo Pagano, Arne Smeets, Efthymios Sofos, Jesse Thorner and Ulrich Derenthal for many useful discussions and comments. While working on this paper the author was supported by ERC grant . Part of this work was done while in residence at MSRI, the author would like to express his gratitude for their hospitality. The author is also grateful to Max Planck Institute for Mathematics in Bonn for its hospitality and financial support.


§ THE BRAUER–MANIN OBSTRUCTION

This section contains a brief overview of the Brauer–Manin obstruction for integral points, as introduced by Colliot-Thélène and Xu <cit.>. Let  be a separated scheme of finite type over . Let X = ×_ and denote by X = H_ét^2(X, 𝔾_m) the Brauer group of X. We use standard notation for the completions of  and  with the convention that _∞ =.

Denote by X(_) the set of adeles of X. Its elements are points (_p) ∈∏_p ≤∞ X() such that, for all but finitely many primes _p belongs to (). Thus the set of rational points X() on X can be embedded diagonally in X(_) and we identify X() with its image under this embedding.

For each α∈ X and for each prime p ≤∞ class field theory yields evaluation maps _α : X() →.  Moreover, there is a group homomorphism _p : → / and thus a map

    X() [r, "_α"]  
    		   [r, "_p"] 
    		    / .

It gives rise to a natural pairing, the Brauer–Manin pairing, between X (_) and X defined via

    X(_) × X 
    		   ⟶ / , 
    		
    (_p), α   ⟼∑_p ≤∞_p _α_p.

The key point is that we have a commutative diagram 

    [column sep=large]
    		   X() [hookrightarrow]r[d, swap, "_α"]
    		   X(_) [d, "_α"] 
       [r]
    		   p ≤∞⊕[r, "∑_p _p"]
    		    /

with the bottom line fitting into a short exact sequence by class field theory. Thus any rational point on X lies is in the left kernel of the pairing (<ref>).

The Brauer–Manin obstruction for integral points can be described in the following way. Recall that (_) = X() ×∏_p (). By separateness of  one can view () and (_) as subsets of X() and X(_), respectively. Therefore, there is an induced pairing

    (_) × X
    	→ / ,

which vanishes on the image of () in (_). For any α∈ X let (_)^α denote the set of points in (_) orthogonal to α under (<ref>). Define the Brauer-Manin set (_)^ X as the left kernel of (<ref>), i.e.

    (_)^ X
    	=	⋂_α∈ X(_)^α.
 

It follows that we have inclusions

    ()
    	⊆(_)^ X⊆(_).

A Brauer–Manin obstruction to the integral Hasse principle for X is present if the set of adeles (_) is non-empty but the Brauer–Manin set (_)^ X is empty. Alternatively, the Brauer–Manin obstruction is said to be the only obstruction to the integral Hasse principle if (_)^ X≠∅ is sufficient to imply the existence of an integral point on .

Let 𝒜 be a subgroup of X and let (_)^𝒜 be the set of adeles orthogonal to the elements of 𝒜 under (<ref>). Thus by (<ref>) the Brauer–Manin set is a subset of (_)^𝒜 and we have a further chain of inclusions 

    ()
    	⊆(_)^ X⊆(_)^𝒜⊆(_).


In this article we are interested in X defined by (<ref>) with P(t) = ∏_i = 1^n P_i(t) ∈[t] the factorisation of P(t) into irreducible polynomials of [t]. The quaternion algebras (a, P_i(t)) for i = 1, …, n belong to X and we take 𝒜 to be the subgroup of X generated by them.  Indeed, let us show that (a, P_i(t)) ∈ X. Since _K(x, y) = _K/Q(x + ω y), then (a, _K(x, y)) is clearly trivial in ((X)) and thus we have

    (a, P_i(t))
    	= ∏_j ≠ i (a, P_j(t))

in the Brauer group of the function field of X. On the other hand, (a, P_i(t)) is an Azumaya algebra on the open subvariety U given by P_i(t) ≠ 0 while ∏_j ≠ i (a, P_j(t)) is Azumaya on V given by ∏_j ≠ i P_j(t) ≠ 0. Moreover, the above verifies that these two quaternion algebras are compatible on U ∩ V. The condition P(t) separable implies that X = U ∪ V is smooth and thus X injects into ((X)) and (a, P_i(t)) lies in the image of this injection.




§ INTEGRAL POINTS ON AFFINE CONICS

In this section we prove Proposition <ref>. With notation as in its statement assume that (<ref>) has a solution (x, y) = m/k, ℓ/k over . Therefore

    _K(m, ℓ) 
    	= k^2 p_1 … p_h-1∏_p ∈ S p.

We can further assume that (m, ℓ) = 1. Indeed, if a prime p divides (m, ℓ), then p^2 divides the right hand side and since {p_1, …, p_h-1}∩ S = ∅, then p | k. Thus we can divide m, l and k through the common factor p. 

Recall that we took a basis 1, ω of K over  with ω  = -(1 +√(a))/2 when a ≡ 1  4 and ω  = √(a) when a ≡ 2,3  4. Factoring (<ref>) into ideals of O_K yields

    '
    	= (k)^2 _1 _1' …_h-1_h-1' ∏',

where ' denotes the conjugate of an ideal in O_K, = (m + ωℓ) and the product is taken over the corresponding prime ideals above the primes in S.

If a prime ' from the product ∏' divides , then we swap the labels of  and ' so that it is  which divides . Let 𝒮 denote the set of primes  appearing in the product ∏' which divide . Thus there is a prime ideal ∈𝒮 above p for each prime p ∈ S and the ideals in 𝒮 are pairwise coprime and non-conjugate. Since _1, …, _h-1 are pairwise coprime and coprime to the ideals in 𝒮, then

    = _1 …_h-1∏_∈𝒮,

where _j is either _j or _j' for all j = 1, …, h-1 and '= (k)^2.

We claim that [] is a square in . To show this we analyse the common divisors of  and '. If  is a prime factor of  and ', then m + ωℓ and m + ωℓ are elements of , where ω denotes the conjugate of ω. Their sum and their difference are in  as well, whence a(2m - ℓ), 2aℓ∈ when a ≡ 1  4 and 2am, 2aℓ∈ when a ≡ 2,3  4. It follows that | (2a) since (m,ℓ) = 1. 

If a ≡ 2,3  4 the discriminant d_K of K equals 4a. Thus  lies over a prime which divides d_K but all such primes ramify in O_K. If a ramified prime divides k the ideal above it must appear to an exponent which is a multiple of 4 in the ideal prime factorisation of (k)^2. Since we took ' = (k)^2, then a standard application of the unique prime ideal factorization theorem shows that = ^2 and '= '^2 for some ideal  of O_K.

If a ≡ 1  4, then d_K = a. In this case  either lies over an odd prime dividing a and hence is ramified in O_K or  divides 2. In the former case or if a ≡ 1  8 where 2 is split the same argument as before shows that = ^2. If, however, a ≡ 5  8 and | (2), then = (2) since 2 is inert in O_K. Thus  is a principal ideal of O_K generated by an element of positive norm which implies that its class in  is trivial. Hence [] = []^2. 

We have shown that the following equality in the narrow class group  holds:

    []
    	= [^2 _1 …_h-1∏_∈𝒮].

The class of  is trivial since _K/(m + ωℓ) > 0 by (<ref>). We deduce that 

    [_1 …_h-1∏_∈𝒮]
    	= []^-1^2.

Swapping _j and _j' for some j ∈{1, …, h-1} above will preserve the fact that the class on the left hand side is a square in . Indeed [_j _j'] = [O_K] and so

    [_1 …_j - 1_j' _j + 1…_h-1∏_∈𝒮]
    	= []^-1[_j']^2.

Therefore 

    [_1 …_h-1∏_∈𝒮]
    	= []^2

for some ideal  of O_K. Since [_1], …, [_h-1] are all of the non-trivial elements of , then [] = [_j] for some j ∈{1, …, h-1} or [] is trivial. In the former case we can multiply both sides of the last equation above by []^-2. As above this is equivalent to substituting [_j] with [_j'] in the left hand side. In this way we get a trivial class

    [_1 …_j - 1_j' _j + 1…_h-1∏_∈𝒮]
    	= [][]^-1^2
    	= [O_K].

When [] is trivial the same conclusion is still valid if instead of ' one has  above. 

In both cases the assumptions of Proposition <ref> imply that all prime ideals on the left hand side lie above primes of  which are not inert in O_K. Therefore the norm of the ideal on the left hand side is

    ( _1 …_j - 1_j' _j + 1…_h-1∏_∈𝒮 )
    	= p_1 … p_h-1∏_p ∈ S p.


We have shown that there is a principal fractional ideal  of O_K whose class in  is trivial and whose norm is p_1 … p_h-1∏_p ∈ S p. The triviality of [] implies that  has a generator m_* + ωℓ_* with positive norm. But then () = _K/(m_* + ωℓ_*). In other words, there are integers m_∗, ℓ_∗ such that

    N_K(m_*, ℓ_*)
    	= p_1 … p_h-1∏_p ∈ S p.

This concludes the proof of Proposition <ref>.

	If h = 1, then the set {p_1, …, p_h-1} is clearly empty. If h = 2, then the class of ^2 is trivial in  and hence already (<ref>) implies that [_1 ∏_∈𝒮] is trivial as well. In this case our argument proves the existence of a principal fractional ideal of O_K of norm the right hand side of (<ref>) and hence the existence of an integral solution to (<ref>). What is important is that we need no assumption on any prime appearing on the right hand side of (<ref>) except that it is not inert in O_K. In conclusion, when h = 1, 2 our argument works without the requirement that “specials” primes p_1, …, p_h-1 appear in (<ref>).  

	When a > 1 and the leading coefficient of P(t) is negative we will need to show that
	
    _K(x, y) 
    	= - p_1 … p_h-1∏_p ∈ S p
 
	has an integral solution provided that it has a rational one. A minor modification of the proof of Proposition <ref> suffices to see this. If the ideal class group of K coincides with , then there is an element of O_K of norm -1 and all principal ideals of O_K lie in the same class inside . Since [(-1)] is clearly the trivial element of , then the same argument as above will give an integral solution satisfying (<ref>). Multiplying both sides of (<ref>) by -1 and exploiting the fact that -1 is a norm of an element of O_K gives the desired integral solution. If, however, the ideal class group of K is a subgroup of  of order 2, then the ideal  in the proof of Proposition <ref> has a generator of negative norm and hence its class in  is no longer the trivial one. Applying the same steps as in the proof of Proposition <ref> will now produce a principal fractional ideal  of O_K of norm p_1 … p_h-1∏_p ∈ S p and such that [] = []. Thus  has a generator of negative norm, m_* + ωℓ_* say. This implies that there are m_*, ℓ_* ∈ satisfying _K/(m_* + ωℓ_*) = - () which shows the existence of the desired integral solution.


§ A CHEBOTAREV DENSITY RESULT

In this section we prove a preparatory result and its corollary needed to implement Proposition <ref> in the proof of Theorem <ref>. Throughout this section we assume that h > 2. The notation fixed in the introduction is still in power. 


	Let Q(t) be an irreducible polynomial of [t]. Assume that Q(t) is linear or that the discriminant of the splitting field of Q(t) is coprime to d_K. Then there exist infinitely many (h-1)-tuples (p_1, …, p_h-1) of distinct primes of  such that
	
  * Q(t)  p_j is a product of non-repeated linear factors  for j = 1, …, h-1,
		
  *  each (p_j) = _j _j' is split in O_K,
		
  *  the non-trivial elements of  are precisely the classes of _1, …, _h-1.
	
	If Q(t) is linear, then (i) is clearly satisfied for any prime not dividing the leading coefficient of Q(t). Thus in this case we only need to check that there are infinitely many (h-1)-tuples (p_1, …, p_h-1) satisfying the remaining conditions (ii) and (iii). A minor modification of our method does this. 

	When Q(t) in not linear we need a finer set-up.	Let c be the leading coefficient of Q(t) and let α be a root of Q(t) defined over a fixed algebraic closure  of . Let L = (α) and let F ⊆ be the splitting filed of Q(t). By construction F/ is Galois. 

	Denote by [1/c] the localisation of  with respect to the multiplicative system generated by powers of c. By construction it is a Dedekind domain. Let O be the integral closure of [1/c] in L. By <cit.> the ring O is also a Dedekind domain. In the case when Q(t) is monic O is simply the ring of integers of L. 

	Let p ∈ be a prime not dividing c, the conductor of [1/c][α] ⊆ O and the discriminant of L. The set of primes which fail these assumptions is finite <cit.>. We can now begin translating the conditions (i), (ii), (iii) into conditions on certain associated to p elements of the Galois group of an appropriate number field. 

	Clearly, for any prime p ∤ c of  the polynomial Q(t) is a product of linear factors mod p if and only if Q_1(t) = 1/cQ(t) is a product of linear factors mod p. Moreover, by definition α∈ O is a root of Q_1(t) and Q_1(t) is monic irreducible polynomial with coefficients in [1/c]. Then L = (α) is a separable extension of  given by a primitive element α∈ O such that Q_1(t) is its minimal polynomial with coefficients in [1/c]. By <cit.> the polynomial Q_1(t) is a product of distinct linear factors mod p if and only if p splits completely in O. Moreover, since p ∤ c the prime p is completely split in O if and only if it is completely split in O_L by the commutative diagram
	
    O[hook]rd   O_Kd
       [1/c][hook]r   
    		.

	If a prime splits completely in any field extension of L, then it must split completely in L since the inertia degree and the ramification index are multiplicative in towers of field extensions. Therefore, in order to show that Q_1(t)  p is a product of non-repeated linear factors it suffices to check that p is completely split in F. 
	
	The prime p does not divide the discriminant of L and thus it must be coprime to the discriminant of F<cit.>. The latter implies that p is unramified in F and hence the Artin symbol is well defined on prime ideals of O_F above p. We can understand how p decomposes in O_F by looking at these Artin symbols. Let | p be a prime ideal of O_F. The Artin map associates an element of the Galois group (F / ) to  to it. The prime p splits completely in F if and only if this element is the identity <cit.>, i.e.
	
    F / / 
    		= 1.

	In conclusion, Q(t) is a product of distinct linear factors mod p if and only if for any prime  of O_F above p the associated Artin symbol in the Galois group (F / ) is the identity.
	
	Once again, by <cit.> the condition that (p) = ' splits in O_K is equivalent to  
	
    K//
    		= 1.

	
	Let H denote the narrow Hilbert class field of K. By definition it is an Abelian extension of K and hence for any prime  of O_H lying over a prime  of K the Artin symbol associated to  only depends on . Let  be a fixed fractional ideal of O_K for which the Artin symbol is well-defined. There is an isomorphism (H/K) ≅ given by the Artin map <cit.>. It implies that [] and [] are the same element of  if and only if the corresponding Artin symbols satisfy  
	
    H/K/
    		= H/K/.


	We need to combine all the conditions listed above. To do so, we first examine the structure of the Galois group of a certain compositum. 
	We know that K/ is cyclic, H/K is Abelian and H/ is Galois. Moreover every prime of  that ramifies in K is unramified in H by construction. Thus  
	
    (H/) 
    			   = (H/K) ⋊(K/) 
       = ⋊ /2,
	
	by <cit.>.
	
	Let M = FH. Since (d_K, d_F) = 1 and H is unramified over finite primes of K, then no finite primes simultaneously ramify in F ∩ H. Thus F ∩ H =. By <cit.> the extension M / is Galois and its Galois group (M/ ) is the direct product (F/) ×(H/). This together with (<ref>) implies that
	
    (M / ) 
    			   = (F / ) ×(H / ) 
       = (F / ) ×(H / K) ⋊(K / ).

	
	In view of (<ref>), (<ref>), (<ref>) and (<ref>) what is left is to take the element 
	
    σ 
    		= 1,  H/K/, 1∈(M/).

	We apply Chebotarev's density theorem <cit.> to the conjugacy class < σ> in (M / ). By doing so we conclude that there are infinitely many primes p of  which split completely in F. Moreover, they split in K as (p) = ' and the classes [] = [] coincide in .
	
	Now let _1, …, _h-1 be fractional ideals of O_K for which the Artin symbol is well defined and such that each of them represents a different non-trivial element of . We apply the above argument with each of the ideals _1, …, _h-1. As a result we get infinitely many (h-1)-tuples (p_1, …, p_h-1) of distinct primes of  satisfying the assumptions of Proposition <ref>. This concludes its proof.



The following corollary of Proposition <ref> will be needed in the proof of Theorem <ref>.



	Let P(t) ∈[t] be separable as a polynomial of [t] with factorisation P(t) = ∏_i = 1^n P_i(t) into irreducible polynomials P_i(t) ∈[t].	Assume that P_1(t) is linear or that the discriminant of the splitting field of P_1(t) is coprime to d_K. Then there exist infinitely many (h-1))-tuples (p_1, …, p_h-1) of distinct primes of  satisfying (ii) and (iii) of Proposition <ref> and such that for each j = 1, …, h-1 there exists a λ_p_j∈_p_j with p_j  P(λ_p_j).

	We apply Proposition <ref> with Q(t) = P_1(t) and thus (ii) and (iii) of Proposition <ref> are satisfied. For the last part it is enough to show the claim for one of the primes p_1, …, p_h-1. Let p = p_1, say. Assume that p is big enough so that it is unramified in the splitting field of P(t) and does not divide the leading coefficient c of P_1(t). Then P(t)  p is separable. We will first show that P_1(t)  p has a simple root which is equivalent to P(t)  p having a simple root since P(t)  p is separable. This is clearly the case if P_1(t) is linear. Assume from now on that P_1(t) is not linear.

	Let d =  P_1(t). By (i) of Proposition<ref> we have
	
    P_1(t)
    		≡ c(t - α_1) … (t - α_d)  p.

	We can assume that α_1, …, α_d are the reductions mod p of some elements lying in the set {0, …, p - 1}. The same symbols will be used for the preimages of α_1, …, α_d in {0, …, p - 1} where it will be clear from the context which one we mean. 
	
	The factorisation of P_1(t)  p implies that there exists a polynomial R(t) ∈[t] such that in [t] we have the identity
	
    P(t)
    		= (c(t - α_1) … (t - α_d) + pR(t)) P_2(t) … P_n(t).

	Note that none of the linear polynomials t - α_i ∈[t], i = 1, …, d divides R(t). Moreover, none of t - α_i is a factor of P_i(t)  p for i = 2, …, n. Indeed, the later follows from the separability of P(t) mod p while if t - α_i was a factor of R(t), then t - α_i would have been a factor of P_1(t) as a polynomial in [t]. This contradicts the assumption that P_1(t) is irreducible in [t]. 
	
	The above implies that we can choose λ_p ∈ in the following way,
	
    λ_p
    		=
    		α_1
    			    R(α_1) ≢0  p, 
    α_1 + p 
    			   .

	This concludes the proof of Corollary <ref>.


§ PROOF OF THEOREM <REF>

Recall the statement of Hypothesis [hyp:H_1]H_1.


	Let P_1(t), …, P_n(t) ∈[t] be irreducible polynomials. Let S be a finite set of primes of  containing ∞, all finite primes p where one of the polynomials P_i does not have all of its coefficients in  or has all its coefficients divisible by p, and all finite primes less or equal to ∏_i = 1^n P_i(t). Given elements λ_p ∈ at finite primes p ∈ S, one may find λ∈, integral away from S, arbitrarily close to each λ_p for the p-adic topology for each finite p ∈ S and arbitrarily big in . Moreover, each P_i(λ) ∈ is a unit in  for all primes p ∉S except perhaps for one prime q_i, where it is a uniformising parameter.


The above conjecture is usually formulated over an arbitrary number field but our application in the proof of Theorem <ref> only requires its version over . It was noticed by Serre <cit.> that Hypothesis [hyp:H]H implies Hypothesis [hyp:H_1]H_1. A proof of this fact is available in <cit.>.

We can now begin the proof of Theorem <ref>. Notation is as set in the introduction. 


	We have _K(x, y) = x^2 - ay^2 when a ≡ 2,3  4 while when a ≡ 1  4 these two quadratic forms represent 1 and are equivalent over , over  for all primes p ≤∞ and over  when p ≠ 2. Hence in either case they represent the same elements of the relevant domain.


Let an adelic point

    (x_p, y_p, t_p) ∈ (_)^𝒜

with (x_∞, y_∞, t_∞) in unbounded component of X() be given. The map described in (<ref>) is constant on each connected component of X()<cit.>. Thus without loss of generality we can assume that (x_∞, y_∞, t_∞) is such that for each i = 1, …, n and t ∈ with |t| ≥ t_∞ the sign of P_i(t) does not chance. It is convenient to write M_p = (x_p, y_p, t_p) for the p-adic component of the adelic point above. Let > 0 be given and let S be a finite set of non-archimedean primes p of  for which we would like to approximate the t-coordinate of M_p with an integral point on  of a distance at most .  

Our argument works in the following way. In view of the inclusions () ⊆ X() and (_) ⊆ X(_), as sets, we use Hypothesis [hyp:H_1]H_1 to find a point (x, y, t) ∈×× lying on X with |t - t_p|_p < for p ∈ S. Fixing t now defines an affine conic with a rational point. This puts us in the set-up of Proposition <ref> which, when applied, gives an integral point on  with the same t coordinate.

We first need to adjust the adele (M_p) and the set S. Let S ⊂ S_1 be the finite extension of S obtained by adding 2, all primes dividing a, all primes from Hypothesis [hyp:H_1]H_1 and h-1 primes p_1, …, p_h-1 not from the set S which satisfy the assumptions of Proposition <ref>.

It is clear that ∏_i = 1^n P_i(t) ≠ 0 defines a dense Zariski open subvariety U of X. For each prime p the set () is open in X() with respect to the p-adic topology. Thus there exists a point M_p' = (x_p', y_p', t_p') ∈() ∩ U() with |t_p - t_p'|_p <. For our purposes it suffices to assume that each M_p lies in () ∩ U(). Indeed, for any element α∈ X the map (<ref>) is a locally constant map <cit.>. Thus _p(_α(M_p)) = _p(_α(M_p')) and hence this small shift is compatible with (<ref>). 

We claim that for any of p_1, …, p_h-1 there exists a point M_p_j' = (x_p_j', y_p_j', t_p_j') ∈(_p_j) with p_j ∏_i = 1^n P_i(t_p_j'). By Corollary <ref> there is an element t_p_j' ∈_p_j such that p_j ∏_i = 1^n P_i(t_p_j'). Each of the primes p_j is split in O_K and thus (a/p_j) = 1. This implies that the congruence x^2 - ay^2 ≡∏_i = 1^n P_i(t_p_j') ≡ 0  p_j has a non-trivial solution (x, y) over _p_j. Since Hensel's lemma lifts this to a _p_j-solution Remark <ref> shows that such M_p_j' exists. Replace M_p with M_p' for each of the primes p_1, …, p_h-1. Since we chose p_1, …, p_h-1∉ S we are not replacing any of the components of (M_p) that we want to approximate. Moreover, for any j = 1, …, h-1 the equality (a/p_j) = 1 implies that the map in (<ref>) is zero on (_p_j) for each of the quaternion algebras (a, P_i(t)). Therefore (<ref>) is preserved.

The assumption (<ref>) means, in particular, that for each i = 1, … n the Hilbert symbols (a, P_i(t_p))_p associated to the quaternion algebras (a, P_i(t)) ∈ X satisfy (a, P_i(t_p))_p = 1 for almost all p and ∏_p ≤∞ (a, P_i(t_p))_p = 1. Thus by <cit.> for each i = 1, …, n there exists c_i ∈^* such that (a, P_i(t_p))_p = (a, c_i)_p. Therefore for each i = 1, …, n and each prime p we have (a, c_iP_i(t_p))_p = 1 and hence c_iP_i(t_p) is represented by x^2 - ay^2 over . 

We have shown that the affine variety Y ⊂Å_^2n + 3 given by

    [scale = 1.5]	
    		 (Y)    at (-1.7, 0.3) Y:;
    		 (W.eq) at (0.6, 0.6)    x_i^2 - ay_i^2 = c_iP_i(t),    i = 1, …, n,;
    		 (X.eq) at (0.03, 0)   x^2 - ay^2 = ∏_i = 1^n P_i(t);

has a local point with t = t_p for each prime p including ∞. 


	There is a point in Y() with t ∈ satisfying |t - t_p|_p < when p ∈ S_1 ∖{∞}.

	Let W ⊂Å^2n + 1 be the variety defined over  by
	
    W:    x_i^2 - ay_i^2 = c_iP_i(t),    i = 1, …, n.

	Then Y is the fibre product Y = W ×_Å^1_ X. The change of variables 
	
    u + √(a)v
    		= (x + √(a)y)∏_i = 1^n1/x_i + √(a)y_i

	defines a birational map over  between Y and W × C, where C ⊂Å^2_ is the affine conic
	
    C:    u^2 - av^2 = 1/c_1 … c_n.

	Moreover, this birational map sends any point on Y with ∏_i = 1^n P_i(t) ≠ 0 to a point on W × C with the same t. For each prime p we now map the local point on Y that we showed to exist above to a local point on W × C. 

	We will show that there exists a rational point on W × C which can be mapped back to a rational point on Y with t coordinate less than  away from t_p for each p ∈ S_1 finite. Before doing so let S_1 ⊆ S_2 be the finite extension obtained by adding all primes p ∉ S_1 for which there is a c_i with non-zero p-adic valuation. Applying Hypothesis [hyp:H_1]H_1 now gives λ∈ with |λ| > t_∞ which satisfies
	
  * |λ - t_p|_p < for each p ∈ S_2 ∖{∞},
		
  * |λ|_p ≤ 1 for each p ∉S_2,
	
	and such that
	
    c_iP_i(λ) 
    		=  δ_i q_i ∏_p ∈ S_2 ∖{∞} p^l_p, i,    i = 1, …, n,

	with primes q_1, …, q_n ∉S_2, exponents l_p, i∈ and δ_i ∈{± 1}. 

	The above conditions imply that, in fact, λ∈. Indeed, without loss of generality we can assume that ≤ 1. Then
		
    | λ|_p 
    			≤max{| λ - t_p |_p, | t_p |_p}≤ 1      p ∈ S_2 ∖{∞}.

	Thus λ∈ for each finite prime p and hence it must be an integer. 

	It is now easy to show that for each p ≤∞ and each i = 1, …, n the rational number c_i P_i(λ) is represented by x^2 - a y^2 over . For each finite prime p ∈ S_2 we have |λ - t_p|_p < and hence |c_i P_i(λ) - c_i P_i(t_p)|_p <. If  is small enough, which we can freely assume, then this together with the fact that c_iP_i(t_p) is represented by x^2 - a y^2 over  is enough to claim that c_iP_i(λ) is represented as well. On the other hand, when p ∉S_2 and p ≠ q_i, then a and c_iP_i(λ) are both units in . Thus the Hilbert symbol (a, c_iP_i(λ))_p = 1 which is equivalent to c_iP_i(λ) being represented by x^2 - ay^2 over . If a > 0, then over  the form x^2 - ay^2 is indefinite and represents every real number, in particular, it represents c_iP_i(λ). On the other hand, we already saw that (a, c_iP_i(t_∞))_∞ = 1 which is only possible if c_iP_i(t_∞) > 0 when a < 0. Recall that we chose t_∞ so that the polynomials P_i(t) do not chance their sign when |t| ≥ t_∞. Since λ is in this range we have c_iP_i(λ) > 0 and therefore it is represented by the positive definite form x^2 - ay^2 over . 

	We have shown that for each i = 1, …, n the product of Hilbert symbols satisfies ∏_p ≠ q_i (a, c_i P_i(λ))_p = 1. By the product formula we conclude that (a, c_i P_i(λ))_q_i = 1 and therefore c_i P_i(λ) is represented by x^2 - a y^2 over each completion of . The Hasse–Minkowski theorem implies now that it must be represented over . In other words, we have shown that for each i = 1, …, n there exist x_i, y_i ∈ such that 
	
    x_i^2 - a y_i^2 
    		= c_i P_i(λ) 
    		= δ_i q_i ∏_p ∈ S_2 ∖{∞} p^l_p, i≠ 0.

	Thus there is a point in W() with t-coordinate being λ. On the other hand, C has points everywhere locally and since affine conics satisfy the Hasse principle we must have C() ≠∅. Therefore W × C has a rational point.

	Under the birational map defined by (<ref>) we map the rational point we showed to exist above on W × C to a -point on Y with the same t coordinate λ. Since S_1 ⊆ S_2 and |λ - t_p|_p < for each p ∈ S_2 ∖{∞} this concludes the proof.


Lemma <ref> implies that there is a rational point (x, y, t) on the affine variety defined by x^2 - ay^2 = P(t) over . Moreover, t ∈ satisfies |t - t_p|_p < when p ∈ S_2, where S_2 is as in the proof of Lemma <ref>.
In view of Remark <ref> we get (x_*, y_*, t) ∈ X() with the same t. On the other hand, Hypothesis [hyp:H_1]H_1 allows us to freely choose the sign of t so that under the assumptions in Theorem <ref> we can take t with P(t) > 0. Indeed, this is true except when a > 0 and the leading coefficient of P(t) is negative in which case there is a negative sign on the far right hand side bellow. Using (<ref>) we then get that 

    _K(x_*, y_*) 
    	= ∏_i = 1^n P_i(t)
    	= ∏_i = 1^nq_i ∏_p ∈ S_2 ∖{∞} p^l_p, i'.

The new exponents l_p, i' differ to l_p, i only when p divides any of c_1, …, c_n. 

Recall that 2 ∈ S_2. If a = 1, then by Lemma <ref> we have |t - t_2|_2 < which implies that P(t)  2  4. In this case there is an elementary argument verifying that P(t) is a difference of two integral squares. Indeed, if P(t) is odd, then P(t) = (P(t) + 1/2)^2 - (P(t) - 1/2)^2. On the other hand, if P(t) is even, then it must be divisible by 4 and similarly we get P(t) = (P(t) + 4/4)^2 - (P(t) - 4/4)^2.

Assume now that a ≠ 1 and we are not in the case when a > 0 and the leading coefficient of P(t) is negative. Recall that we chose p_1, …, p_h-1 in S_2 covering all of the non-trivial elements of  in the sense of Proposition <ref>. Moreover, we have (P(t)) =  (P(t_p)) = 1 since |t - t_p|_p < for each p ∈{p_1. …, p_h-1}. Let S_3 be the subset of S_2 ∖{p_1, …, p_h-1, ∞} consisting of all primes dividing P(t) to odd multiplicity and let S_* = S_3 ∪{ q_1, …, q_n }. Writing x_* = m/k and y_* = ℓ/k and redefining k if needed gives

    _K(m,ℓ) 
    	= k^2 p_1 … p_h-1∏_p ∈ S_* p.


For any prime p ∈ S_* the p-adic valuation of _K(m,ℓ) is clearly odd. Thus if p ∈ S_* is odd, then either p | a or a is a quadratic residue modulo p because N_K(x, y) is properly equivalent to x^2 - ay^2 over . Hence p is not inert in O_K. On the other hand, if 2 ∈ S_* then 2 is ramified if a ≡ 2,3  4 and 2 is split when a ≡ 1  8. If, however, a ≡ 5  8 then it is easy to see that the 2-adic valuation of the left hand side is zero and hence 2 ∉S_*. Thus the primes on the right hand side above are not inert in O_K. This puts us in position to apply Proposition <ref> which ensures that there exist x, y ∈ such that

    _K(x,y)
    	= p_1 … p_h-1∏_p ∈ S_* p	.

Multiplying both sides above by a suitable square and redefining x and y if needed gives

    _K(x,y)  
    	= q_1 … q_n ∏_p ∈ S_2 ∖{∞} p^∑_i = 1^n l_p, i'
    	= ∏_i = 1^n P_i(t) = P(t).

Thus () ≠∅. What is left is to recall that |t - t_p|_p < for each p ∈ S_2 ∖{∞} by Hypothesis [hyp:H_1]H_1. Since S ⊂ S_2, then the point (x, y, t) ∈() fulfils the desired approximation property. 

If a > 1 and the leading coefficient of P(t) is negative a small modification is needed. Following the same steps as above will lead us to the situation where instead of (<ref>) we need to show that there are x, y ∈ such that 

    _K(x,y)
    	= - p_1 … p_h-1∏_p ∈ S_* p,

provided that there is a rational solution to the above equation. We do so by applying Proposition <ref> with Remark <ref> taken into account. This concludes the proof Theorem <ref>.


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