An Existential Divisibility Lemma for Global Fields

Jeroen Demeyer^{∗}  Jan Van Geel^{†}

2004–06–16

1 Introduction

In [Phe00] Thanases Pheidas proves, what he calls, existential divisibility lemmas for K = 𝔽_{q}(z) where q is a prime congruent to 3 mod 4 and for K = ℚ. Let R be the set of primes p in ℤ, respectively in 𝔽_{q}[t], such that −1 is not a square in the residue field of p. The existential divisibility lemmas state the existence of an existential formula ϕ(x, y) such that: if ϕ(x, y) holds for x, y ∈ K \ {0} then for all primes p ∈ R for which v_{p}(x) is odd it follows that v_{p}(xy^{−2}) > 0.

Actually putting extra local conditions on the elements x (conditions in the point at infinity in the function field case and in the real prime and the prime 2 in case K = ℚ) the truth of ϕ(x, y) for x, y ∈ K^{∗} is equivalent to the statement: for all primes p ∈ R for which v_{p}(x) is odd it follows that v_{p}(xy^{−2}) > 0.

The formula ϕ(x, y) expresses “almost” that all poles of x in R, with odd multiplicity, are poles of y.

These existential divisibility lemmas play a role in strategies to obtain undecidability results for the existential theory of the field K. In [Phe91] Pheidas proved that the existential theory of 𝔽_{p^{n}}(t) is undecidable. His proof worked for all odd primes p. Videla ([Vid94]) extended this result to rational global function fields of characteristic 2. Shlapentokh ([Shl96]) showed that the existential theory of global function fields of characteristic not 2 is undecidable and finally in her thesis ([Eis03]) Kirsten Eisenträger completed the results by proving the same result for global function fields in characteristic 2. This is essentially the (negative) solution of Hilbert’s 10th problem for global function fields.

∗The first author is a Research Assistant of the Fund for Scientific Research - Flanders (Belgium) (F.W.O.- Vlaanderen).

†Work partially supported by the European Community’s Human Potential Programme under contract HPRN-CT-2002-00287.

One of the main open questions directly related to Hilbert’s 10th problem is whether or not the existential theory of ℚ (or more general of any number field) is decidable or undecidable. In [Phe00] a program to come to a uniform way to attack Hilbert’s 10th problem for global fields is described. This program generates a series of “possible facts” (cf. [Phe00, sections 3 and 4]), two of which are related to the existential divisibility lemma and one of them is forced into the strategy by the fact that the existential divisibility lemma as proved by Pheidas uses the set of primes p for which −1 is not a square in the residue field of p.

In view of this Thanases Pheidas and Gunther Cornelissen raised the question to what extent the existential divisibility lemma holds. For which sets of primes does it hold? Why does the condition in the prime 2 occur? Can the lemma be generalized to all global fields?

Together with Karim Zahidi the second author worked out a more general version of the existential divisibility lemma. Namely for K = ℚ and for a set of primes that are inert in a quadratic extension of ℚ. This result was presented at the Oberwolfach meeting on Hilbert’s 10th problem in January 2003 (cf. [VZ03]).

In this paper we generalize the existential divisibility lemma to all global fields K (of characteristic not 2), and for all sets of primes that are inert in a quadratic extension L of K. We first prove the direct generalisation of Pheidas’ existential divisibility lemma, with conditions in the real primes and in the primes ramifying in L. In the last section we remove all these conditions and prove:

Let K be a global field and R(L/K) be the set of primes which are inert in a quadratic extension L of K. Then there is an existential formula Ω(x, y) which is equivalent with the formula

∀𝔯 ∈ R(L/K) : (v_{𝔯}(x) odd → v_{𝔯}(xy^{−2}) > 0)

2 Preliminaries

Our discussion of the existential divisibility lemma relies on facts about norms and norm groups of quadratic extensions of local and global fields. We use the Hasse–Minkowski local– global principle and Hilbert’s reciprocity law. In this section we give a survey of these facts, for more details and proofs we refer to the literature (e.g. [O’M63]).

We start fixing terminology and notation. Throughout the paper K will be a global field of characteristic not 2, so it is either a number field or the function field of a curve over a finite field 𝔽_{q} with q odd. L will be a quadratic extension of K.

With M_{K} we denote the set of all “primes” 𝔭 of K. In the number field case the finite or nonarchimedean primes correspond (one to one) to (equivalence classes of) discrete valuations of K and the infinite or archimedean primes correspond (one to one) to the different embeddings of K in the complex numbers. In the function field case all the elements of M_{K} correspond (one to one) to (equivalence classes of) discrete valuations on K. With every element of M_{K} there corresponds a normalized absolute value | |_{𝔭} on K, we let K_{𝔭} denote the completion of K with respect to this absolute value.

If 𝔭 is a non-archimedean prime, then K_{𝔭} is the fraction field of a complete discrete valuation ring O_{𝔭} and its maximal ideal is a principal ideal. A generator for this ideal is called a uniformizing element. We can choose such a uniformizing element in the base field K and denote it with π_{𝔭}. The quotient ring O_{𝔭}/(π_{𝔭}) is a finite field 𝔽_{𝔭}, the residue field of the prime 𝔭. The discrete valuation associated with a non-archimedean prime 𝔭 will be denoted with v_{𝔭} and we normalize it by v_{𝔭}(π_{𝔭}) = 1. We use the analogous terminology and notation for primes 𝔓 ∈ M_{L}.

All absolute values | |_{𝔭} extend to the quadratic extension L/K in different ways, depending on the prime 𝔭. The possible extensions correspond to what is called the splitting behavior of the prime 𝔭 in L. For a quadratic extension we have the following possibilities (in the number field case we distinguish between non-archimedean and archimedean primes, in the function field case points 4 and 5 are empty):

1. 𝔭 is a non-archimedean prime and L ⊗_{K} K_{𝔭} ≅ K_{𝔭} × K_{𝔭}. In this case there are two primes 𝔓_{1}, 𝔓_{2} in M_{L} lying over 𝔭, L_{𝔓_{1}} ≅ L_{𝔓_{2}} ≅ K_{𝔭} and 𝔽_{𝔓_{1}} ≅ 𝔽_{𝔓_{2}} ≅ 𝔽_{𝔭}. We say that the prime 𝔭 is completely split in L.

2. 𝔯 is a non-archimedean prime and L⊗_{K} K_{𝔯} is a quadratic field extension L_{ℜ} over K_{𝔯} and [𝔽_{ℜ} : 𝔽_{𝔯}] = 2. In this case there is only one prime ℜ in L lying over 𝔯, the uniformizing element π_{𝔯} of O_{𝔯} is also a uniformizing element of O_{ℜ}. The prime 𝔯 is said to be inert in L.

3. 𝔰 is a non-archimedean prime and L ⊗_{K} K_{𝔰} is a quadratic field extension L_{𝔖} over K_{𝔰} and [𝔽_{𝔖} : 𝔽_{𝔰}] = 1. Here we also have a unique prime 𝔖 in L lying over 𝔰, but the uniformizing element π_{𝔰} of O_{𝔰} is not a uniformizing element of O_{𝔖}. We have π_{𝔖}^{2} = wπ_{𝔰} with w a unit in O_{𝔖}. The prime 𝔰 is said to ramify in L.

4. 𝔞 is an archimedean prime and L ⊗ K_{𝔞} ≅ K_{𝔞} × K_{𝔞}, the archimedean prime 𝔞 is said to split in L.

5. 𝔠 is an archimedean prime and L ⊗_{K} K_{𝔠} is quadratic extension L_{ℭ} of K_{𝔠}. Note that this case can only occur if K_{𝔠} ≅ ℝ and we have L_{ℭ} ≅ ℂ, the archimedean prime 𝔠 does not split in L.

We can now partition the set of primes M_{K} according to the splitting behavior in the quadratic extension L:

P (L/K) = {𝔭 ∈ M_{K} | 𝔭 is a non-archimedean prime, completely split in L}
R(L/K) = {𝔯 ∈ M_{K} | 𝔯 is a non-archimedean prime, inert in L}
S(L/K) = {𝔰 ∈ M_{K} | 𝔰 is a non-archimedean prime, ramified in L}
A_{s}(L/K) = {𝔞 ∈ M_{K} | 𝔞 is an archimedean prime, split in L}
A_{ns}(L/K) = {𝔠 ∈ M_{K} | 𝔠 is an archimedean prime, not split in L}

In the number field case we sometimes need to refer to all archimedean primes therefore we define A(L/K) = A_{s}(L/K)∪A_{ns}(L/K). In the function field case the sets A(L/K), A_{s}(L/K) and A_{ns}(L/K) are empty.

The sets S(L/K), A_{s}(L/K) and A_{ns}(L/K) are finite. The sets P (L/K) and R(L/K) are not empty, Chebotarev’s density theorem yields that these sets are infinite and both have density \frac{1}{2}. (An other consequence of Chebotarev’s theorem states that the set of primes P (L/K) that split completely determine uniquely the (Galois) extension L/K.)

The strong approximation theorem [Neu92, page 204] implies

Proposition 1. Let T be a finite set of primes in K such that A(L/K) ⊂ T. Let a_{𝔱} ∈ K for 𝔱 ∈ T. Choose a prime 𝔭_{0} not in T. Then for all ε > 0 there exists an element x ∈ K such that

|x − a_{𝔱}|_{𝔱} < ε for all 𝔱 ∈ T and |x|_{𝔮} ≤ 1 for all 𝔮 ∈ M_{K} \ (T ∪ {𝔭_{0}})

As an immediate consequence of this one has

Corollary 2. Let T, 𝔭_{0} ∉ T and a_{𝔱} ∈ K be as in the preceding proposition. Then there exists an element a ∈ K such that

v_{𝔱}(a) = v_{𝔱}(a_{𝔱}) for all 𝔱 ∈ T \ A(L/K)
sign(a) = sign(a_{𝔱}) for all 𝔱 with K_{𝔱} ≅ ℝ

and

v_{𝔮}(a) ≥ 0 for all 𝔮 ∈ M_{K} \ (T ∪ {𝔭_{0}})

Let σ ∈ Gal(L/K) be the non-trivial automorphism of L fixing K. The norm map N_{L/K} : L^{∗} → K^{∗}; z ↦ zσ(z) defines a quadratic form on the two dimensional vector space L over K. If L = K(α) then we can write this form as x^{2} + Tr_{L/K}(α)xy + N_{L/K}(α)y^{2} with respect to the basis {1, α} for L over K. Here Tr_{L/K}(z) = z + σ(z) stands for the trace map.

Facts. We will denote this quadratic form with N and we have the following facts (cf. [O’M63] and [Sch85]):

1. For all 𝔭 ∈ P (L/K) the quadratic form N has a non-trivial zero over K_{𝔭}, it follows from this that every element of K_{𝔭} is represented by N.

2. For all 𝔯 ∈ R(L/K) the norm form N has no non-trivial zero over K_{𝔯}. The elements of K_{𝔯} represented by N are exactly the norms of the extension L_{ℜ}/K_{𝔯}, where ℜ is the unique prime in L lying over 𝔯. The norm map N_{L_{ℜ}/K_{𝔯}} is surjective on units U_{𝔯} in O_{𝔯} and the uniformizing element π_{𝔯} is not a norm. It follows that an element x ∈ K_{𝔯} is a norm if and only if its valuation v_{𝔯}(x) is even (we recall that v_{𝔯} is the discrete valuation associated to 𝔯 and normalized by v_{𝔯}(π_{𝔯}) = 1).

3. For all 𝔰 ∈ S(L/K) the norm form N has no non-trivial zero over K_{𝔰}. Again the elements represented by N are the norms of the extension L_{𝔖}/K_{𝔰}, where 𝔖 is the unique prime in L lying over 𝔰. The image N_{L_{𝔖}/K_{𝔰}}(U_{𝔖}) of the group of units in O_{𝔖} is a subgroup of index 2 in the group U_{𝔰} of units in O_{𝔰}. It follows that N_{L_{𝔖}/K_{𝔰}}(L_{𝔖}) is a subgroup of index 2 in K_{𝔰}, since N_{L_{𝔖}/K_{𝔰}}(π_{𝔖}) = uπ_{𝔰} for some unit u in O_{𝔰}.

4. For primes 𝔞 ∈ A_{s}(L/K) we have the same as completely split primes, namely that every element of K_{𝔞} is represented by N.

5. If 𝔠 ∈ A_{ns}(L/K), the norm form N has no non-trivial zero over K_{𝔠}. The elements represented by N are the norms of the extension L_{ℭ} = ℂ over K_{𝔭} = ℝ. So these are the elements represented as sums of 2 squares in ℝ, which are exactly the positive elements in ℝ.

For inert primes 𝔯 ∈ R(L/K) we need slightly more than the fact that the norms in K_{𝔯} are the elements of even valuation.

Lemma 3. Let 𝔯 ∈ R(L/K) be an inert prime and ℜ the unique prime in L lying over 𝔯. Let α ∈ O_{ℜ} be such that O_{ℜ}/(π_{ℜ}) = 𝔽_{ℜ} = 𝔽_{𝔯}[\overline{α}] = (O_{𝔯}/(π_{𝔯}))[\overline{α}], with \overline{α} the reduction of α mod ℜ. Then

v_{𝔯}(N_{L_{ℜ}/K_{𝔯}} (x + yα)) = min(v_{𝔯}(x^{2}), v_{𝔯}(y^{2}))

for x, y ∈ K_{𝔯}.

Proof. Because 𝔯 is inert, we may replace v_{𝔯} with v_{ℜ} in the statement.

Note that σ ∈ Gal(L/K) extends to a generator of Gal(L_{ℜ}/K_{𝔯}), so

v_{ℜ}( N_{L_{ℜ}/K_{𝔯}}(x + yα)) = v_{ℜ}((x + yα)(x + yσ(α))) = 2v_{ℜ}(x + yα)

It remains to prove that v_{ℜ}(x + yα) = min(v_{ℜ}(x), v_{ℜ}(y)).

The hypotheses imply that v_{𝔯}(α) = 0, since \overline{α} is non-zero in 𝔽_{ℜ}. If v_{𝔯}(x) ≠ v_{𝔯}(y), the properties of discrete valuations immediatly give us v_{ℜ}(x + yα) = min(v_{ℜ}(x), v_{ℜ}(y)).

Now suppose v_{ℜ}(x) = v_{ℜ}(y) = m for a certain integer m. Then we have x = π_{𝔯}^{m}x_{0} and y = π_{𝔯}^{m}y_{0}, with x_{0} and y_{0} elements of O_{𝔯} having valuation 0.

Since \{1, \overline{α}\} is a basis for 𝔽_{ℜ} over 𝔽_{𝔯}, we have that \overline{x_{0} + y_{0}α} = \overline{x_{0}}+\overline{y_{0}} \overline{α} is a non-zero element of 𝔽_{ℜ}, so v_{ℜ}(x_{0}+y_{0}α) = 0. Hence v_{ℜ}(x+yα) = v_{ℜ}(π_{𝔯}^{m}(x_{0}+y_{0}α)) = m = min(v_{ℜ}(x), v_{ℜ}(y)).

Information on the global norm group N_{L/K}(L^{∗}) ⊂ K^{∗} can be obtained from Hilbert’s reciprocity law as expressed by the exact sequence (cf. [Rei75] and [O’M63])

0 → _{2}Br(K) → ⨁_{𝔭∈M_{K}} _{2}Br(K_{𝔭}) \underrightarrow{∑ inv_{𝔭}} ℤ/2ℤ → 0  (1)

Here _{2}Br(F) denotes the two component (the elements of exponent two) of the Brauer group of the field F. The elements of the Brauer group of F are classes of central simple algebras over F, the group law is induced by the tensor product of algebras. In the case of local and global fields (of characteristic not 2) the (non-trivial) elements of the two component of the Brauer group are given by quaternion division algebras, which in turn can be represented by symbols (a, b) with a, b non-zero elements of the field. The symbol (a, b) corresponds to the quaternion algebra with basis {1, i, j, k} satisfying i^{2} = a, j^{2} = b, ij = −ji = k = ab. If one is not familiar with the theory of Brauer groups, it is enough to know that the subgroups of elements of exponent 2 are generated multiplicatively by these symbols modulo the following relations (cf. [Sch85])

• (ab, c) = (a, c)(b, c) and (a, bc) = (a, b)(a, c)

• for all a ≠ 0, 1; (a, 1 − a) = 1.

Remark. This is true for the two component of the Brauer group of any field (of characteristic not equal to 2) by a theorem of Merkuriev. For local and global fields one has moreover that the product of symbols is equal to a symbol, this fact does not hold in general.

The following facts are important for us:

Proposition 4 ([Rei75, section 31]). Over a local field K_{𝔭} there exists a unique quaternion division algebra ℍ_{𝔭}, so _{2}Br(K_{𝔭}) ≅ ℤ/2ℤ.

This element ℍ_{𝔭} is split over any quadratic extension E of K_{𝔭}, i.e., ℍ_{𝔭}⊗_{K_{𝔭}} E is a full matrix algebra over K_{𝔭} and is therefore trivial in Br(K_{𝔭}).

In the exact sequence (1) presenting Hilbert’s reciprocity law the maps inv_{𝔭} : _{2}Br(K_{𝔭}) → ℤ/2ℤ are defined by inv_{𝔭}(ℍ_{𝔭}) = 1 ∈ ℤ/2ℤ, proposition 4 says that inv_{𝔭} is an isomorphism.

Proposition 5 ([Rei75, Theorem 30.4]). Let F be a field, char(F) ≠ 2. A quaternion division algebra ℍ over F is represented by a symbol of the form (a, b) with a ∈ F^{∗} \ F^{∗2} if and only if ℍ ⊗ F (\sqrt{a}) is trivial in _{2}Br(F). If a ∈ F^{∗} \ F^{∗2} then a symbol (a, b) is trivial in 2Br(F) if and only if b ∈ N_{F (\sqrt{a})/F} (F (\sqrt{a})^{∗}).

The second part of this proposition together with Hilbert’s reciprocity law yield the Hasse norm theorem for quadratic extensions L/K. This theorem states that an element in K is a norm of an element in L if and only if it is a norm locally everywhere. More generally the following holds for quadratic forms over global fields:

Theorem 6 (Hasse–Minkowski, [Sch85]). A quadratic form Q over a global field K has a non-trivial zero in K if and only if it has a non-trivial zero in K_{𝔭} for all 𝔭 ∈ M_{K}.

An element x in a global field K is represented by a quadratic form Q over K if and only if x is represented by Q over K_{𝔭} for all 𝔭 ∈ M_{K}.

Let _{2}Br(L/K) be the kernel of the natural group morphism _{2}Br(K) → _{2}Br(L). Clearly for any element ω in this kernel we have inv_{𝔭}(ω) ≡ 0 mod 2 for 𝔭 ∈ P (L/K) ∪ A_{s}(L/K). Hilbert’s reciprocity law together with the approximation theorem allows us to give different parameterizations of the finite subgroups of _{2}Br(L/K). We will use this in section 4.

Let Q be a finite set of primes satisfying

S(L/K) ∪ A_{ns}(L/K) ⊆ Q ⊆ R(L/K) ∪ S(L/K) ∪ A_{ns}(L/K)

Denote with _{2}Br^{Q}(L/K) the subgroup of _{2}Br(L/K) defined by

2Br^{Q}(L/K) = {ω ∈ _{2}Br(L/K) | inv_{𝔩}(ω) ≡ 0 mod 2 for all 𝔩 ∈ M_{K} \ Q}

Hilbert’s reciprocity law gives a one-to-one correspondence between _{2}Br^{Q}(L/K) and the set {Q_{0} ⊆ Q|#Q_{0} even}. This correspondence is given by

ω ↦ Q_{ω} = {𝔮 ∈ M_{K} | inv_{𝔮} ω ≡ 1 mod 2}

The latter is a subset of Q by the definition of _{2}Br^{Q}(L/K).

Let T be a finite subset of primes in R(L/K) \ Q, and fix a Δ ∈ K^{∗} for which L = K(\sqrt{Δ}). Take an ω ∈ _{2}Br^{Q}(L/K), then ω is of the form (Δ, λ) with λ ∈ K^{∗}. We claim that we may choose λ such that v_{𝔯}(λ) ≥ 0 for all 𝔯 ∈ R(L/K) and v_{𝔱}(λ) = 0 for all 𝔱 ∈ T. To see this, we note that for 𝔱 ∈ T we have inv_{𝔱}(ω) ≡ 0 mod 2, so λ is a norm of L_{𝔗}/K_{𝔱} and v_{𝔱}(λ) is even. Now we apply the approximation theorem (corollary 2) with 𝔭_{0} ∈ P (L/K) to find an element c ∈ K^{∗} such that

v_{𝔯}(c) ≥ −\frac{v_{𝔯}(λ)}{2}  for all 𝔯 ∈ R(L/K)
v_{𝔱}(c) = −\frac{v_{𝔱}(λ)}{2}  for all 𝔱 ∈ T

Since ω = (Δ, λ) = (Δ, c^{2}λ), we may replace λ by c^{2}λ to obtain the desired conditions.

By choosing one such λ for every ω ∈ _{2}Br^{Q}(L/K), we obtain a set Λ_{Q,T} (L/K) parametrizing the elements of _{2}Br^{Q}(L/K), we have:

Lemma 7. Let Q, T and Λ_{Q,T} = Λ_{Q,T} (L/K) be as above.

There is a one-to-one correspondence between Λ_{Q,T} and the subsets of Q with an even number of elements, given by

λ ↦ Q_{λ} = {𝔮 ∈ Q | inv_{𝔮}(Δ, λ) ≡ 1 mod 2}

Conversely for every subset Q_{0} of Q with #Q_{0} ∈ 2ℤ there is a unique λ ∈ Λ_{Q,T} determined by the fact that λ is a norm of L_{𝔔}/K_{𝔮}, where 𝔔 ∈ M_{L} and 𝔮 is the prime in K lying under 𝔔, if and only if 𝔮 ∉ Q_{0}.

Proof. Except for the last statement this follows from the contruction of the set Λ_{Q,T}. The last statement follows from the fact that inv_{𝔩}(Δ, λ) ≡ 0 mod 2 for all 𝔩 ∉ Q and that inv_{𝔮}(Δ, λ) ≡ 1 mod 2 if and only if λ is not a norm from L_{𝔔}/K_{𝔮}.

3 Existential Divisibility Lemma

As before K is a global field (of characteristic not 2) and L is a quadratic extension of K. We fix an element Δ ∈ K^{∗} such that L = K(\sqrt{Δ}).

Proposition 8. There exists an existential formula ϕ(x, y) such that:

1. Let x and y be elements of K^{∗} for which ϕ(x, y) is true. If 𝔯 is any prime in R(L/K) such that v_{𝔯}(x) is odd, then v_{𝔯}(xy^{−2}) > 0.

2. ϕ(x, y) is true for all elements x and y of K^{∗} satisfying the following conditions:

(a) There exists at least one 𝔯 ∈ R(L/K) for which v_{𝔯}(x) is odd.

(b) For every 𝔯 ∈ R(L/K) with v_{𝔯}(x) odd, we have v_{𝔯}(xy^{−2}) ≥ 0.

(c) For every 𝔠 ∈ A_{ns} the element x is positive in K_{𝔠} ≅ ℝ.

(d) For every 𝔰 ∈ S(L/K), the element x is a norm from L_{𝔖}, where 𝔖 is the unique prime lying above 𝔰.

To prove the proposition we need to use lemma 3 for all inert primes 𝔯 ∈ R(L/K). However in general it is not possible to find a primitive element α ∈ L such that 𝔽_{ℜ} = 𝔽_{𝔯}[\overline{α}] for all 𝔯 ∈ R(L/K). The following lemma will solve this difficulty:

Lemma 9. Let K and L be as above. There exist elements α_{0}, α_{1} ∈ L^{∗} such that for all 𝔯 ∈ R(L/K) either 𝔽_{ℜ} = 𝔽_{𝔯}[\overline{α_{0}}] or 𝔽_{ℜ} = 𝔽_{𝔯}[\overline{α_{1}}].

Proof. We start by setting α_{0} = \sqrt{Δ}. Consider a 𝔯 ∈ R(L/K) such that v_{𝔯}(Δ) = 0 and (in the number field case) 𝔯 is not a 2-adic prime (i.e., it is not lying above the prime 2 in ℚ). Since L_{ℜ} = K_{𝔯}(\sqrt{Δ}) is a quadratic extension of K_{𝔯}, it follows from the assumptions on 𝔯 that \overline{Δ} is not a square in 𝔽_{𝔯}. This is a consequence of Hensel’s lemma since for non-2-adic primes a square in the residue field lifts to a square in the completion. So for these primes 𝔯 we have 𝔽_{ℜ} = 𝔽_{𝔯}(\overline{\sqrt{Δ}}).

The remaining primes are exactly the inert 2-adic primes or the inert primes 𝔯 for which v_{𝔯}(Δ) ≠ 0. This set T of inert primes is finite, say T = {𝔯_{1}, …, 𝔯_{n}}. For all i = 1, …, n, we let ℜ_{i} be the unique prime in L lying over 𝔯_{i}.

Consider the semi-local ring O_{T} = ∩_{i=1}^{n}(O_{𝔯_{i}}∩K) and its integral closure Õ_{T} = ∩_{i=1}^{n}(O_{ℜ_{i}}∩L). The canonical morphisms

O_{T} ↪ O_{𝔯_{i}} and Õ_{T} ↪ O_{ℜ_{i}}

induce isomorphisms O_{T} /(π_{𝔯_{i}}) ≅ O_{𝔯_{i}}/(π_{𝔯_{i}}) = 𝔽_{𝔯_{i}} and Õ_{T} /(π_{ℜ_{i}}) ≅ O_{ℜ_{i}}/(π_{ℜ_{i}}) = 𝔽_{ℜ_{i}}. For all i = 1, …, n the residue fields 𝔽_{ℜ_{i}} are degree 2 extensions of 𝔽_{𝔯_{i}}, so 𝔽_{ℜ_{i}} = 𝔽_{𝔯_{i}}[\overline{β_{i}}]. By the Chinese remainder theorem there exists an element α_{1} ∈ Õ_{T} such that α_{1} ≡ \overline{β_{i}} mod (π_{ℜ_{i}}). This implies 𝔽_{ℜ_{i}} = 𝔽_{𝔯_{i}}[\overline{α_{1}}] for all i = 1, …, n.

We can now prove the proposition.

Proof. Take α_{0}, α_{1} as in lemma 9. We define ϕ(x, y) as

ϕ(x, y) ↔ ϕ_{0}(x, y) ∧ ϕ_{1}(x, y)

where ϕ_{i}(x, y) is the formula

ϕ_{i}(x, y) ↔ ∃a, b, c ∈ K : 1 + x N(y^{−1} + α_{i}c) = a^{2} − Δb^{2}

or, written in an other way

ϕ_{i}(x, y) ↔ ∃a, b, c ∈ K : (1 + xy^{−2}) + xy^{−1} Tr(α_{i})c + x N(α_{i})c^{2} − a^{2} + Δb^{2} = 0

We will prove part 1 of the theorem. For the sake of contradiction, assume that ϕ(x, y) holds, but v_{𝔯}(xy^{−2}) ≤ 0 for some prime 𝔯 ∈ R(L/K) for which v_{𝔯}(x) is odd. Since v_{𝔯}(xy^{−2}) is odd, necessarily v_{𝔯}(xy^{−2}) < 0.

We have 𝔽_{ℜ} = 𝔽_{𝔯}[α_{i}], with i either equal to 0 or 1. Lemma 3 implies v_{𝔯}( N(y^{−1} + α_{i}c)) = min(v_{𝔯}(y^{−2}), v_{𝔯}(c^{2})) ≤ v_{𝔯}(y^{−2}). It follows that

v_{𝔯}(x N(y^{−1} + α_{i}c)) ≤ v_{𝔯}(xy^{−2}) < 0

Thus v_{𝔯}(1 + x N(y^{−1} + α_{i}c)) = v_{𝔯}(x N(y^{−1} + α_{i}c)), which is odd. But ϕ_{i}(x, y) states that 1 + x N(y^{−1} + α_{i}c) is equal to a^{2} − Δb^{2}. The latter however has even valuation in 𝔯 since it is a norm, so we found our contradiction.

To prove part 2 of the theorem, we assume x and y satisfy all the given conditions.

We claim that ϕ_{i}(x, y) (i = 0, 1) will be true if the following quadratic form is isotropic:

Q_{i}(a, b, c, d) = (1 + xy^{−2})d^{2} + xy^{−1} Tr(α_{i})cd + x N(α_{i})c^{2} − a^{2} + Δb^{2}  (2)

Indeed, if Q_{i} is isotropic but ϕ_{i}(x, y) does not hold, then Q_{i} must have a solution with d equal to zero:

x N(α_{i})c^{2} − a^{2} + Δb^{2} = 0

Now choose a prime 𝔯 ∈ R(L/K) for which v_{𝔯}(x) is odd. Then v_{𝔯}(x N(α_{i})c^{2}) will be odd, but v_{𝔯}(a^{2} − Δb^{2}) is even, which gives a contradiction.

It remains to prove that Q_{i} (i = 0, 1) is isotropic in K, or by applying the theorem of Hasse– Minkowski (cf. theorem 6) that Q_{i} is isotropic over every completion of K. We check this by considering all possible primes in M_{K}.

Case 1: 𝔞 ∈ A(L/K)

K_{𝔞} ≅ ℝ or ℂ. Set d = 0, c = \frac{1}{\sqrt{x}} and a and b such that α_{i} = a + b\sqrt{Δ}. It follows that Q_{i}(a, b, c, d) = 0.

Case 2: 𝔭 ∈ P (L/K)

Take 𝔓 as one of the two primes in M_{L} lying over 𝔭. Since L = K(\sqrt{Δ}) and L_{𝔓} = K_{𝔭} it follows that Δ is a square in K_{𝔭}. This means that the form ⟨−1, Δ⟩, which is a subform of Q_{i}, is isotropic in K_{𝔭}. So also Q_{i} is isotropic in K_{𝔭}.

Case 3: 𝔯 ∈ R(L/K)

Suppose v_{𝔯}(x) is even. We know that v_{𝔯}( N(α_{i})) will also be even, so we can write x N(α_{i}) = π_{𝔯}^{2m}u with v_{𝔯}(u) = 0. If we set d = 0 and c = π_{𝔯}^{−m}, then the first three terms of (2) will be equal to the unit u. Since every unit in K_{𝔯} is a norm of the extension L_{ℜ}, there exist elements a and b in L_{ℜ} such that u = N(a + b\sqrt{Δ}). This proves that Q_{i} is isotropic.

If v_{𝔯}(x) is odd, it is given that v_{𝔯}(xy^{−2}) > 0. This implies that v_{𝔯}(1 + xy^{−2}) = 0, so 1 + xy^{−2} is a unit. If we set c = 0, we can conclude as above that Q_{i} is isotropic.

Case 4: 𝔰 ∈ S(L/K)

Setting d = 0 and c = 1 in Q_{i} yields the equation x N_{L_{𝔖}/K_{𝔰}}(α_{i}) = N_{L_{𝔖}/K_{𝔰}}(a + b\sqrt{Δ}). If we write this as

x = N_{L_{𝔖}/K_{𝔰}} (\frac{a + b\sqrt{Δ}}{α_{i}})

we see that this will always have a solution, because by assumption x is a norm from L_{𝔖} over K_{𝔰}, and a + b\sqrt{Δ} represents all elements of L_{𝔖} = K_{𝔰}(\sqrt{Δ}).

Remark. Proposition 8 applied to K = ℚ and L = ℚ(i) is a direct generalistation of the existenial divisibility lemma proven by Pheidas in [Phe00]. In the ramified primes, only the prime 2 in that case, Pheidas formulates alternatives for condition 2d (x being a local norm). It would be possible to formulate such alternatives in the general case. We did not work this out explicitly since in theorem 13 all conditions in the ramified primes will be removed.

We also like to point out that condition 2a, the existence of an inert prime where x has an odd value, is not necessary. Even without this condition one can show that if the quadratic form (2) is isotropic there also is a zero (a, b, c, d) with d = 1. Since removing this condition would not substantially simplify the following section, we preferred to keep the formulation of the lemma analogous to that of Pheidas.

4 Removing all extra conditions

In this section we will prove a version of the existential divisibility lemma without the local conditions (2a), (2c) and (2d). from proposition 8. This way we will find a formula which is equivalent with For all primes 𝔯 in R(L/K) for which v_{𝔯}(x) is odd, it holds that v_{𝔯}(xy^{−2}) > 0.

As an application of this, we give an existential definition of the set of elements of K having non-negative valuation in a prime.

Lemma 10. There exists a number k and elements u_{1}, …, u_{k} ∈ K with v_{𝔯}(u_{i}) ≥ 0 for all 𝔯 ∈ R(L/K), such that the following holds:

Take any finite set T ⊂ R(L/K) and consider the semi-local ring O_{T} = ∩_{𝔱∈T}(O_{𝔱}∩ K). Every x ∈ O_{T} can be written as

x = u_{1}x_{1}^{2} + u_{2}x_{2}^{2} + ⋯ + u_{k}x_{k}^{2}

with x_{i} ∈ O_{T} \ {0}.

Proof. Here we need to give a different arguments for K a global function field and for K a number field.

Let K be a global function field of characteristic not 2. Let L and T be as in the statement of the theorem and let x ∈ O_{T}. Choose y ∈ O_{T} such that y ≠ ±1 and x + y ≠ ±1 (this is possible since O_{T} is infinite). Then

x = (\frac{x + y + 1}{2})^{2} − (\frac{x + y − 1}{2})^{2} + (\frac{y − 1}{2})^{2} − (\frac{y + 1}{2})^{2}

Since \frac{1}{2} ∈ O_{T} we see that every x ∈ O_{T} is equal to an alternating sum of four non-zero squares in O_{T}. It follows that the lemma holds in this case with k = 4 and u_{1} = u_{3} = 1, u_{2} = u_{4} = −1.

To prove the lemma in the number field case we first note that every integer x ∈ ℤ is represented over ℤ by the quadratic form x_{1}^{2} − x_{2}^{2} + x_{3}^{2} − x_{4}^{2}. Namely if x is odd choose y ∈ ℤ such that x + 4y ≠ ±1 and y ≠ ±1, then

x = (\frac{x + 4y + 1}{2})^{2} − (\frac{x + 4y − 1}{2})^{2} + (y − 1)^{2} − (y + 1)^{2} .

If x is even choose y ∈ ℤ such that y is odd, x + y ≠ ±1 and y ≠ ±1, then

x = (\frac{x + y + 1}{2})^{2} − (\frac{x + y − 1}{2})^{2} + (\frac{y − 1}{2})^{2} − (\frac{y + 1}{2})^{2} .

Note that in these representations of x ∈ ℤ all the coordinates are non-zero.

We know that the ring of integers O_{K} of K is a finitely generated ℤ-module, say

O_{K} = a_{1}ℤ + a_{2}ℤ + ⋯ + a_{n}ℤ.

Set k = 4n and for the u_{i}’s we take a_{1}, −a_{1}, a_{1}, −a_{1}, a_{2}, −a_{2}, a_{2}, −a_{2} …, a_{n}, −a_{n}, a_{n}, −a_{n}. Since any integer is represented with non-zero coordinates by the form x_{1}^{2} − x_{2}^{2} + x_{3}^{2} − x_{4}^{2} it follows that every element of O_{K} is represented with non-zero coordinates by the form u_{1}x_{1}^{2} + u_{2}x_{2}^{2} + ⋯ + u_{4n}x_{4n}^{2}.

Now consider O_{T} with T a finite subset of R(L/K). Every x in O_{T} can be written as y/z with y and z in O_{K}. And since O_{T} is a principal ideal domain we may assume that v_{𝔱}(z) = 0 for all 𝔱 ∈ T. It follows from the above that yz = u_{1}x_{1}^{2} + ⋯ + u_{k}x_{k}^{2}, with x_{1}, …, x_{k} non-zero elements of O_{K}. Then

x = \frac{yz}{z^{2}} = u_{1} (\frac{x_{1}}{z} )^{2} + ⋯ + u_{k} (\frac{x_{k}}{z} )^{2}

For all i = 1, …, k and 𝔱 ∈ T we have x_{i}/z ≠ 0 and v_{𝔱}(x_{i}/z) = v_{𝔱}(x_{i}) ≥ 0, which means that x_{i}/z ∈ O_{T}.

Definition 11. Given any subset Λ ⊆ K^{∗}, we define the support of Λ to be

supp(Λ) = {𝔭 ∈ M_{K} | ∃λ ∈ Λ : |λ|_{𝔭} ≠ 1}

If Λ is finite, then supp(Λ) will also be finite.

Choose four primes 𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4} in R(L/K) and define the following finite set of primes:

Q = {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}} ∪ S(L/K) ∪ A_{ns}(L/K)

Let T be any finite subset of R(L/K) \ Q and Λ_{Q,T} the subset of K^{∗} parameterizing 2Br^{Q}(L/K) as given in lemma 7. By construction of Λ_{Q,T} it holds that supp(Λ_{Q,T}) ∩ T = ∅.

Lemma 12. Let Λ = Λ_{Q,T} be the set as defined above. There exists an existential formula ψ_{Λ}(x, y) such that:

1. Let x and y be elements of K^{∗} for which ψ_{Λ}(x, y) is true. If 𝔯 is any prime in R(L/K)\ supp(Λ) such that v_{𝔯}(x) is odd, then v_{𝔯}(xy^{−2}) > 0.

2. Suppose x and y are elements of K^{∗} such that for every 𝔯 ∈ R(L/K) with v_{𝔯}(x) odd, it holds that v_{𝔯}(xy^{−2}) > 0. Then ψ_{Λ}(x, y) is true.

Proof. To construct the formula ψ_{Λ}(x, y), we use the ϕ(x, y) obtained in lemma 8, and the u_{i} from lemma 10. We also set u_{0} = 1. We define ψ_{Λ}(x, y) as

∃y_{0}, y_{1}, …, y_{k}, z_{0}, z_{1}, …, z_{k} ∈ K : (y_{0}z_{0} = 1) ∧ ⋯ ∧ (y_{k}z_{k} = 1) ∧ ((u_{0}z_{0}^{2} + ⋯ + u_{k}z_{k}^{2})y^{2} = 1)

∧ ( ⋁_{λ∈Λ} ϕ(λx, y_{0})) ∧ ⋯ ∧ ( ⋁_{λ∈Λ} ϕ(λx, y_{k}))

Note that the first two lines are equivalent with

∃y_{0}, y_{1}, … y_{k} ∈ K^{∗} : (u_{0}y_{0}^{−2} + ⋯ + u_{k}y_{k}^{−2} = y^{−2})

To prove part 1, assume that ψ_{Λ}(x, y) holds. Take any prime 𝔯 ∈ R(L/K)\supp(Λ) for which v_{𝔯}(x) is odd. ψ_{Λ}(x, y) implies that for all i = 1, …, k, ϕ(λ_{i}x, y_{i}) is true for a certain λ_{i} ∈ Λ. Because 𝔯 is outside of the support of Λ, we know that v_{𝔯}(λ_{i}x) = v_{𝔯}(x) is odd. But now ϕ(λ_{i}x, y_{i}) implies that v_{𝔯}(λ_{i}xy_{i}^{−2}) > 0, and this valuation is equal to v_{𝔯}(xy_{i}^{−2}). Lemma 10 gave us v_{𝔯}(u_{i}) ≥ 0, so we also have v_{𝔯}(xu_{i}y_{i}^{−2}) > 0. We conclude the proof of part 1 by observing that

v_{𝔯}(xy^{−2}) = v_{𝔯} (x(u_{0}y_{0}^{−2} + ⋯ + u_{k}y_{k}^{−2})) ≥ min_{i=0}^{k} v_{𝔯}(xu_{i}y_{i}^{−2}) > 0

For the proof of part 2, suppose we have x, y ∈ K^{∗} such that for every 𝔯 ∈ R(L/K) with v_{𝔯}(x) odd, the inequality v_{𝔯}(xy^{−2}) > 0 holds.

We can use approximation to get a y_{0}^{−1} for which the following finitely many conditions are satisfied:

(I) v_{𝔯}(y_{0}^{−1}) > v_{𝔯}(y^{−1}) for all 𝔯 ∈ R(L/K) \ {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}} for which v_{𝔯}(x) is odd.

(II) v_{𝔯_{j}}(y_{0}^{−1}) > −v_{𝔯_{j}}(x)/2 for j = 1, 3.

(III) v_{𝔯_{j}}(y_{0}^{−1}−y^{−1}) = −v_{𝔯_{j}}(2y^{−1})+an even number greater than max (−v_{𝔯_{j}}(x), 2v_{𝔯_{j}}(2y^{−1})), for j = 2, 4.

We define these sets of primes:

T_{0} = {𝔯_{1}, 𝔯_{3}} ∪ {𝔯 ∈ R(L/K) \ {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}} | v_{𝔯}(x) is odd}
T_{1} = T_{2} = ⋯ = T_{k} = {𝔯_{2}, 𝔯_{4}} ∪ {𝔯 ∈ R(L/K) \ {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}} | v_{𝔯}(x) is odd}

Claim. There exist y_{0}, y_{1}, …, y_{k} ∈ K^{∗} such that u_{0}y_{0}^{−2} + ⋯ + u_{k}y_{k}^{−2} = y^{−2} and for all 0 ≤ i ≤ k we have v_{𝔯}(xy_{i}^{−2}) > 0 for all 𝔯 ∈ T_{i}.

Proof of claim. We have already constructed y_{0}. The case i = 0 follows easily from (I) and (II) above, together with the hypotheses of part 2.

For every 𝔯 ∈ T_{1} we will prove that v_{𝔯}(y^{−2} − y_{0}^{−2}) is even, and v_{𝔯}(y^{−2} − y_{0}^{−2}) ≥ −v_{𝔯}(x). Set

a = y^{−2} − y_{0}^{−2} = −(y_{0}^{−1} − y^{−1})((y_{0}^{−1} − y^{−1}) + 2y^{−1})

If 𝔯 is either 𝔯_{2} or 𝔯_{4},then from (III) it follows that v_{𝔯}(y_{0}^{−1} − y^{−1}) > v_{𝔯}(2y^{−1}), so

v_{𝔯}(a) = v_{𝔯}(y_{0}^{−1} − y^{−1}) + v_{𝔯}(2y^{−1}) = an even number greater than − v_{𝔯}(x)

If 𝔯 is not 𝔯_{2} nor 𝔯_{4}, we know that v_{𝔯}(x) is odd and the hypotheses say that v_{𝔯}(y^{−2}) > −v_{𝔯}(x). Then (I) implies that v_{𝔯}(a) = v_{𝔯}(y^{−2} − y_{0}^{−2}) = v_{𝔯}(y^{−2}) is even and v_{𝔯}(a) > −v_{𝔯}(x).

Now we use approximation to find a μ ∈ K for which

v_{𝔯}(μ) = −v_{𝔯}(a)/2 for all 𝔯 ∈ T_{1}.

This way μ^{2}a is in the ring O_{T_{1}}, and by applying lemma 10 for this ring, we write μ^{2}a as

μ^{2}a = u_{1}w_{1}^{2} + ⋯ + u_{k}w_{k}^{2}.

If we set y_{i}^{−1} = w_{i}/μ, then v_{𝔯}(y_{i}^{−2}) = 2v_{𝔯}(w_{i}) − 2v_{𝔯}(μ) ≥ 0 + v_{𝔯}(a) > −v_{𝔯}(x) for 𝔯 ∈ T_{1} = T_{i}.

This concludes the proof of the claim.

Using this, we will show that ⋁_{λ∈Λ} ϕ(λx, y_{0}) is true. The argument proving that ⋁_{λ∈Λ} ϕ(λx, y_{i}) is true for i = 1, … k is completely analogous but with the role of the pairs {𝔯_{1}, 𝔯_{3}} and {𝔯_{2}, 𝔯_{4}} exchanged.

We make a subset Q_{0} of Q of even cardinality. We start by taking the primes in {𝔯_{2}, 𝔯_{4}} ∪ S(L/K) ∪ A_{ns}(L/K) for which x is not a local norm. We add 𝔯_{1} to Q_{0} if x is a norm from L_{ℜ_{1}}. If necessary, we add 𝔯_{3} to make sure Q_{0} has an even number of elements. If we take the λ ∈ Λ such that λ is a local norm everywhere, except for the primes in Q_{0}, then λx will have the following properties:

• λx is a local norm for all primes in S(L/K) ∪ A_{ns}(L/K).

• v_{𝔯_{1}}(λx) is odd.

• v_{𝔯_{2}}(λx) is even and v_{𝔯_{4}}(λx) is even.

• v_{𝔯}(λx) ≡ v_{𝔯}(x) mod 2 for all 𝔯 ∈ R(L/K) \ {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}}.

We will now prove that ϕ(λx, y_{0}) is true for this λ ∈ Λ. The choice of λ already implies conditions 2a, 2c and 2d for part 2 of lemma 8.

In order to prove condition 2b, take any prime 𝔯 ∈ R(L/K) for which v_{𝔯}(λx) is odd. We see that 𝔯 cannot be 𝔯_{2} or 𝔯_{4}. If 𝔯 is not 𝔯_{1} nor 𝔯_{3}, the fact that v_{𝔯}(λx) is odd implies that v_{𝔯}(x) is odd. In any case we have 𝔯 ∈ T_{0}. Hence v_{𝔯}(λxy_{0}^{−2}) ≥ v_{𝔯}(xy_{0}^{−2}) > 0 by the preceding claim.

The previous lemma is a form of the existential divisibility lemma without conditions in the real primes and ramifying primes. However, in one direction, it does not work for primes in supp(Λ_{Q,T}). By applying the lemma two times for well chosen sets Λ_{Q,T} and Λ_{Q′,T′}, we can solve this problem and obtain the main theorem.

Theorem 13. There is an existential formula Ω(x, y) which is equivalent with the formula

∀𝔯 ∈ R(L/K) : (v_{𝔯}(x) odd → v_{𝔯}(xy^{−2}) > 0)  (3)

Proof. Take eight different primes {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}, 𝔯′_{1}, 𝔯′_{2}, 𝔯′_{3}, 𝔯′_{4}} in R(L/K), and define Q as {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}} ∪ S(L/K) ∪ A_{ns}(L/K). Take T = {𝔯′_{1}, 𝔯′_{2}, 𝔯′_{3}, 𝔯′_{4}} and let Λ = Λ_{Q,T} be the corresponding set parameterizing _{2}Br^{Q}(L/K).

Let Q′ = {𝔯′_{1}, 𝔯′_{2}, 𝔯′_{3}, 𝔯′_{4}} ∩ S(L/K) ∩ A_{ns}(L/K) and take T′ = {𝔯_{1}, 𝔯_{2}, 𝔯_{3}, 𝔯_{4}} ∪ (supp(Λ_{Q,T}) ∩ R(L/K)). Since T′ ∩ Q′ = ∅ by the choice of T, the parameterizing set Λ′ = Λ_{Q′,T′} is defined. Note that the choice of T′ now implies R(L/K) ∩ supp(Λ_{Q,T}) ∩ supp(Λ_{Q′,T′}) = ∅.

Now we define

Ω(x, y) ↔ ψ_{Λ}(x, y) ∧ ψ_{Λ′}(x, y)

Suppose Ω(x, y) is true. ψ_{Λ}(x, y) says that v_{𝔯}(x) odd implies v_{𝔯}(xy^{−2}) > 0 for 𝔯 ∈ R(L/K) \ supp(Λ). ψ_{Λ′}(x, y) says the same thing for 𝔯 ∈ R(L/K) \ supp(Λ′). Because R(L/K) ∩ supp(Λ) ∩ supp(Λ′) = ∅, we have it for all 𝔯 in R(L/K).

If (3) is satisfied, then we know by lemma 12 that ψ_{Λ}(x, y) and ψ_{Λ′}(x, y) are both true.

As mentioned in the introduction our motivation for theorem 13 finds its origin in Pheidas’ paper [Phe00]. The next corrolary shows that the theorem also generalises the fact that in a global field the elements integral in a prime 𝔭 form a diophantine set. This result in different cases is due to different authors as indicated in the introduction of chapter 5 in [Eis03], there one also finds a uniform proof for this fact based on Hilbert’s reciprocity law for the Brauer group of a global field.

Corollary 14. For every non-archimedean prime 𝔭 ∈ M_{K}, the set {z ∈ K | v_{𝔭}(z) ≥ 0} is diophantine.

Proof. Let 𝔭 ∈ M_{K} be a non-archimedean prime. Choose an other non-archimedean prime 𝔮 ≠ 𝔭. Choose L/K a quadratic extension such that 𝔭 ∈ R(L/K) and 𝔮 ∈ S(L/K). To see that such an extension exists, let K_{𝔭}(α_{𝔭}) be the unique quadratic unramified extension of K_{𝔭} and K_{𝔮}(β_{𝔮}) a totally ramified extension of degree two of K_{𝔮}. Let X^{2} + a_{𝔭,1}X + a_{𝔭,2} be the minimal polynomial of α_{𝔭} over K_{𝔭} and X^{2} + b_{𝔮,1}X + b_{𝔮,2} be the minimal polynomial of β_{𝔮} over K_{𝔮}. Then lemma (33.8) in [Rei75] states that if c_{1} and c_{2} are taken sufficiently close to a_{𝔭,1} and b_{𝔮,1}, respectively a_{𝔭,2} and b_{𝔮,2}, the polynomial f(X) = X^{2} + c_{1}X + c_{2} is separable and irreducible over K having a root in K_{𝔭} and in K_{𝔮}. It follows that K(γ), with f(γ) = 0, is a quadratic extension of K with the desired properties.

Let ω ∈ _{2}Br(K) such that inv_{𝔭} ω ≡ 1 mod 2, inv_{𝔮} ω ≡ 1 mod 2 and inv_{𝔯} ω ≡ 0 mod 2 for all primes 𝔯 ≠ 𝔭, 𝔮. (Such an element ω exists by Hilbert’s reciprocity law). By construction L is a splitting field of ω. So ω = (γ, x) with L = K(γ). It follows from the choice of L and ω that v_{𝔭}(x) is odd and v_{𝔯}(x) is even for all 𝔯 ∈ R(L/K) \ {𝔭}. After multiplying x with a suitable square, we may assume that v_{𝔭}(x) = 1. Now apply theorem 13 with this fixed x. Then Ω(x, y) is equivalent with “v_{𝔭}(xy^{−2}) > 0”, or “v_{𝔭}(y^{−1}) > −v_{𝔭}(x)/2”. If we set y = z^{−1} and use the fact that v_{𝔭}(x) = 1, we find that Ω(x, z^{−1}) is equivalent with “v_{𝔭}(z) ≥ 0”.

References

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[O’M63] Timothy O’Meara, Introduction to quadratic forms, Springer, 1963.

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[Rei75] Irving Reiner, Maximal orders, London Academic Press, 1975.

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