ELIMINATION OF QUANTIFIERS OVER p-ADIC FIELDS

after Alexander Prestel and Peter Roquette and

THE RATIONALITY OF THE POINCARÉ SERIES OF A VARIETY OVER ℚ_{p} after Jan Denef

by

Moshe Jarden*

Tel Aviv University

* Notes of a series of talks given in a seminar at the Hebrew University during December 1990 – April 1991 Introduction The goal of this note is to present Denef’s theorem about the rationality of the Poincaré series associated to the p-adic points on a variety.

Consider a formula φ(X_{1}, …, X_{m}) in the language of rings with constant symbols for the elements of ℤ_{p}. Let

Truth(φ) = {x ∈ ℤ_{p}^{m}| φ(x) is true in ℤ_{p}}.

For each positive integer n let ν(φ, t) be the number of elements in Truth(φ) modulo p^{n}.

Theorem A (Denef): The Poincaré series

P (φ, t) = ∑_{n=1}^{∞} ν(φ, n)t^{n}

associated with φ belongs to ℚ_{p}(t).

In particular if f_{1}, …, f_{r} are polynomials in m variables with coefficients in ℤ_{p}, and ν_{n} is the number of elements of V (f_{1}, …, f_{r})(ℤ_{p}) modulo p^{n}, then ∑_{n=1}^{∞} ν_{n}t^{n} is a rational function of t.

Our proof follows Section 7 of [D]. The idea of the proof is to replace φ by a simpler formula φ′. Then to express the Poincaré series associated to φ′ by Haar integrals on partial domains of ℚ_{p}^{m}, which can be directly computed.

The formula φ′ is quantifier free in an extended language. In this language there is an unary predicate P_{n} associated with positive integer n. The interpretation of P_{n} in ℚ_{p} is the set {a^{n}| a ∈ ℚ_{p}}.

The possibility to replace each φ by such φ′ is due to Macintyre:

Theorem B: The theory of p-adically closed fields admits an elimination of quantifiers in the extended language.

The proof of Theorem B is based on a model theoretic criterion for elimination of quantifiers:

Proposition C: A theory T of a first order language L admits elimination of quantifiers if and only if T is substructure complete, i.e., if two models B and B′ of T have a common substructure A of L, then they are equivalent over A.

This criterion is proved in Section 1. To deduce Theorem B from C we prove in Section 5 the elementary equivalence theorem for p-adically closed fields:

Theorem D: Let (E, v) and (F, v) be p-adically closed field extensions of a valued field (K, v). If E^{n} ∩ K = F^{n} ∩ K for each n ∈ ℕ, then (E, v) and (F, v) are elementarily equivalent in L′(valued field)(K).

This theorem follows from the embedding theorem for p-adically closed fields by standard model theoretic arguments:

Theorem E: Let (K, v) be a p-adically valued field. Let (L, v) be a countable p-adically closed extension of (K, v). Let (F, v) be a p-adically closed extension of (K, v) which is ℵ_{1} saturated. Suppose that L^{n} ∩ K ⊆ F^{n} ∩ K for each n ∈ ℕ. Then there is a K-embedding of (L, v) in (F, v).

The proof of Theorem E has two parts, algebraic and regular:

Theorem F: Let E and F be p-adic closures of a p-adically valued field (K, v). Then E ≅_{K} F if and only if K ∩ E^{n} = K ∩ F^{n} for each positive integer n.

Theorem G: Let (K, v) be a p-adically valued field. Let (L, v) be a countable p-adically closed extension of (K, v) such that K is algebraically closed in L. Let (F, v) be a padically closed extension of (K, v) which is ℵ_{1} saturated. Then there is a K-embedding of (L, v) in (F, v).

Section 3 is dedicated to the proof of Theorem F while Theorem G is proved in Section 4. Both proof follow Roquette and Prestel’s exposition [PR].

1. Elimination of quantifiers. Let L be a first order language. A theory T of L is just a set of sentences in L. The theory T is consistent if it has a model A. In this case we write A ⊧ T.

A proof from T is a sequence of formulas φ_{1}, …, φ_{n} in L such that each element of this sequence is a logical axiom, or belongs to T, or is a consequence of preceding formulas by modus ponens or by the universal quantification rule. We say that a formula φ can be proved from T if φ is the last formula of a proof from T. In this case we write T ⊢ φ.

The deduction theorem says that if T ∪ {φ} ⊢ ψ, then T ⊢ φ → ψ. The theorem of constants says that if φ(X) is a formula with a free variable X and c is a constant symbol that does not appear in T, then T ⊢ φ(c) implies T ⊢ φ(X). Gödels completeness theorem says that a theory T is inconsistent if and only if T ⊢ FALSE, where FALSE is a logically false sentence.

We say that T admits elimination of quantifiers if every formula φ(X_{1}, …, X_{n}) is equivalent modulo T to a formula φ′(X_{1}, …, X_{n}) without quantifiers.

We denote the domain of a structure A of L by A. and extend L to a first order language L(A) be adding a new constant symbol for each element of A. If we assign the element a ∈ A to the constant symbol to which it corresponds we extend A to a structure ⟨A, a⟩_{a∈A}. The diagram of A is the theory Diag(A) in L(A) which consists of all atomic sentences and negation of atomic sentences in L(A) which are true in ⟨A, a⟩_{a∈A}. Note that each element of Diag(A) has the form φ(a_{1}, …, a_{n}), where φ(X_{1}, …, X_{n}) is a formula in L (with the free variables among X_{1}, …, X_{n}) and a_{1}, …, a_{n} ∈ A. A structure B of the language L contains A if and only if it can be extended to a structure of L(A) in which Diag(A_{0}) is true.

Lemma 1.1: A theory T of a first order language L admits elimination of quantifiers if and only if T is substructure complete, i.e., if two models B and B′ of T have a common substructure A of L, then they are equivalent over A.

Proof: We prove only that if T is substructure complete, then it admits elimination of quantifiers.

It suffices to prove that each formula φ(X) in one free variable is equivalent modulo T to a quantifier free formula. To this end extend L to a language L(c) be adding a new constant symbol c. Let

S = T ∪ {φ(c)} ∪ {¬κ(c)| κ(X) is a quantifier free formula and T ⊢ κ(X) → φ(X)}.

We claim that S is inconsistent. Otherwise it would have a model A. In particular A ⊧ φ(c). Let a be the distinguished element of A which corresponds to c. Then A ⊧ φ(a). Denote the smallest L(c)-substructure of A which contains a by A_{0}. If B is a model of T which extends A_{0}, then, by assumption B is equivalent to A in the language L(A_{0}). In particular B ⊧ φ(c). By Gödel’s completeness theorem, this means that T ∪ Diag(A_{0}) ⊢ φ(c). Each member of A_{0} is named by a constant term built up from c, the individual constants of L and the functions of L. So, each element of Diag(A_{0}) has the form κ(c), where κ(X) is a quantifier free formula of L. In the proof of φ(c) from T ∪ Diag(A_{0}) there appear only finitely many elements of Diag(A_{0}). Thus there exist quantifier free formulas κ_{1}(X), …, κ_{n}(X) of L such that

T ∪ {κ_{1}(c), …, κ_{n}(c)} ⊢ φ(c).

Let κ(X) = κ_{1}(X) ∧ ⋯ ∧ κ_{n}(X). By the deduction theorem, T ⊢ κ(c) → φ(c). Since c does not appear in T, the theorem of constants implies that T ⊢ κ(X) → φ(X). Hence ¬κ(c) belongs to S and is therefore, A ⊧ ¬κ(c). It follows that A ⊧ ¬κ(a) which is a contradiction to κ(a) ∈ Diag(A_{0}).

It follows that it is possible to prove a contradiction from S. These means that there exists quantifier free formulas λ_{i}(X) in L such that T ⊢ λ_{i}(X) → φ(X), for i = 1, …, m and

{T, φ(c), ¬λ_{1}(c), …, ¬λ_{m}(c)} ⊢ FALSE.

Hence, there exists i between 1 and m such that

{T, φ(c)} ⊢ λ_{i}(c).

It follows that T ⊢ φ(c) → λ_{i}(c). By the theorem of constants T ⊢ φ(X) → λ_{i}(X) and therefore T ⊢ φ(X) ↔ λ_{i}(X).

A structure A of a language L is ℵ_{1}-saturated if it satisfies the following condition: Let F by countable set of formulas of L. Suppose that for each finite subset φ_{1}(X_{1}, …, X_{n}), …, φ_{m}(X_{1}, …, X_{n}) of F there exist a_{1}, …, a_{n} ∈ A such that φ_{(}a_{1}, …, a_{n}) is true in A for i = 1, …, m. Then there exist a_{1}, a_{2}, a_{3}, … ∈ A which satisfy each formula in F. If A is an arbitrary structure, then each nonprincipal ultrapower A^{ℕ}/D is ℵ_{1}-saturated [FJ, Lemma 6.14]

2. p-adically closed fields. An ordered group is an abelian (additive) group Γ together with a total ordering < such that α < β implies α + γ < β + γ for all α, β, γ ∈ Γ.

A subgroup Δ of Γ is convex if 0 < δ ∈ Δ and 0 < γ < δ implies γ ∈ Δ. In this case Γ/Δ becomes an ordered group under the definition:

γ + Δ > 0 if and only if γ > 0 and γ ∉ Δ.

If γ > 0 and n ∈ ℕ, then nγ = γ + γ + ⋯ + γ > 0. Hence Γ is torsion free. Its divisible hull is ℚ ⊗ Γ. Each element of the latter group can be represented as a quotient γ/n, with γ ∈ Γ and n ∈ ℕ, with the usual rule of identification of quotients. The group ℚ ⊗ Γ can be ordered by the rule

\frac{γ}{m} < \frac{δ}{n}  if and only if  nγ < mδ.

We say that Γ is a ℤ-group, if Γ has a minimal positive element 1, such that for each γ ∈ Γ and every positive integer n there exists an r between 0 and n − 1 and an element α ∈ Γ such that γ = nα + r.

The first condition means that ℤ is a convex subgroup of Γ. The second one then means that Γ/ℤ is a divisible group.

We define absolute value and the sign function on Γ in the usual way: |γ| = max{γ, −γ}, and sgn(γ) = 1 if γ > 0, σ(0) = 0, and sgn(γ) = −1 if γ < 0.

Consider a field K and let K^{×} be the multiplicative group of nonzero elements of K. A valuation of K is a map v of K^{×} into an ordered group Γ such that (1a) v(ab) = v(a) + v(b), (1b) v(a + b) ≥ min{v(a), v(b)}, and (1c) v is nontrivial, i.e., there exists a ∈ K^{×} such that v(a) ≠ 0. Add the symbol ∞ to Γ with the following convention: (2a) ∞ + ∞ = α + ∞ = ∞ + α = ∞, and (2b) α < ∞ for each α ∈ Γ. Extend v to K by defining v(0) = ∞ and observe that v still satisfies (1). We refer to the pair (K, v) as a valued field.

Next deduce the following properties of v from (1) (3a) v(1) = 0 and v(a) = v(−a) for each a ∈ K. (3b) If v(a) < v(b), then v(a + b) = v(a). Otherwise v(a + b) > v(a) and therefore v(a) ≥ min{v(a + b), v(−b)} > v(a), a contradiction. (3c) v(∑_{i=1}^{m} a_{i}) ≥ min_{1≤i≤m}{v(a_{i})}. If v(a_{i}) ≠ v(a_{j}) for all i ≠ j, then we have an

equality. (3d) If ∑_{i=1}^{m} a_{i} = 0 and m > 1, then there exist i ≠ j such that v(a_{i}) = v(a_{j}).

The value group of v is v(K^{×}) = Γ_{v}. The valuation ring of v is O_{v} = {x ∈ K| v(x) ≥ 0}. It has a unique maximal ideal M_{v} = {x ∈ K| v(x) > 0}. The quotient \overline{K}_{v} = O_{v}/M_{v} is the residue field of v. It gives a short exact sequence:

0 ⟶ M_{v} ⟶ O_{v} ⟶ \overline{K}_{v} ⟶ 0.

The group of units of O_{v} is U_{v} = {x ∈ K| v(x) = 0}. It gives another short exact sequence:

1 ⟶ U_{v} ⟶ K^{×} ⟶^{v} Γ_{v} ⟶ 0.

The quotient field of O_{v} is K and for each x ∈ K^{×} we have x ∈ O_{v} or x^{−1} ∈ O_{v}.

A theorem of Chevalley asserts that v extends to every field extension L of K. In general the extension is not unique. However, whenever there is no confusion, we will use v to also denote the extended valuation and refer to (L, v)/(K, v) as an extension of valued fields. In this case there is a natural embedding of \overline{K}_{v} into \overline{L}_{v} and of vK into vL.

If (L, v)/(K, v) is a finite algebraic extension of valued fields, then (v(L^{×}) : v(K^{×})) < [L : K]. Hence, if (L, v)/(K, v) is an algebraic extension of valued fields, then the v(L^{×})/v(K^{×}) is a torsion group.

Kurt Hensel defined the field of p-adic numbers ℚ_{p} as the completion of ℚ with respect to the p-adic valuation v_{p}. He observed that v_{p} uniquely extends to a valuation of ℚ_{p} with the same residue field, 𝔽_{p}, and valuation group, ℤ, as ℚ. He proved what we now call Hensel’s lemma, which relates the solutions of a polynomial equation over ℚ_{p} to the solutions of the same equation over 𝔽_{p}. It was later proved that v_{p} has a unique extension to each algebraic extension L of ℚ_{p}.

These properties of ℚ_{p} motivate the definition of a “Henselian field”.

Lemma – Definition 2.1: The following statements on a valued field (K, v) are equivalent. If they are satisfied, we say that K is a Henselian field:

(a) (Hensel’s Lemma) For each polynomial f ∈ O_{v}[X] and for each a ∈ O_{v} such that f̄(ā) = 0 and \overline{f′}(ā) ≠ 0 there exists x ∈ O_{v} such that f(x) = 0 and x̄ = ā.

(b) The valuation v extends uniquely to each algebraic extension of K.

(c) (Newton’s Lemma) For all monic polynomial f ∈ O_{v}[X], a ∈ O_{v}, and 0 < γ ∈ Γ such that v(f(a)) > 2v(f′(a)) + γ there exists a unique x ∈ O_{v} such that f(x) = 0 and v(x − a) > v(f′(a)) + γ.

Proof: See [J, Section 11].

Corollary 2.2: Let (K, v) be a Henselian field, a, b elements of K, and n ∈ ℕ.

(a) If v(1 − a) > 2v(n), then there exists a unique element x ∈ K such that x^{n} = a and v(1 − x) > 0.

(b) If a ≠ 0, v(b − a) > v(a) + 2v(n), and there exists x ∈ K such that a = x^{n}, then there exists y ∈ K such that b = y^{n}.

Proof:

Proof of (a): Apply Newton’s Lemma on the polynomial X^{n} − a.

Proof of (b): Apply (a) to the condition v(ba^{−1} − 1) > 2v(n).

Corollary 2.3: Every algebraic extension of a Henselian field is Henselian.

Proof: Apply Lemma 2.1(b). Corollary 2.4: If K is algebraically closed in a Henselian field L, then K is Henselian. Proof: Apply Hensel’s lemma. Proposition 2.5: Each valued field (K, v) has a separable algebraic extension (K_{v}^{h}, v^{h}) with the following properties:

(a) (K_{v}^{h}, v^{h}) is Henselian,

(b) If (L, w) is a Henselian extension of (K, v) then (K_{v}^{h}, v^{h}) can be embedded in (L, w) over K.

The valued field (K_{v}^{h}, v^{h}) is unique up to K-isomorphism. It is the Henselization of (K, v).

An extension (L, v)/(K, v) of valued fields is immediate if both fields have the same residue fields, \overline{L}_{v} = \overline{K}_{v}, and the same and value groups, v(L^{×}) = v(K^{×}). Proposition 2.6: The Henselization of a valued field is an immediate extension. Proof: See [J, Cor. 14.2].

A valuation v of a field K is p-adic if \overline{K}_{v} = 𝔽_{p} and v(p) = 1 is the smallest positive element of Γ_{v}. In this case we say that (K, v) is a p-adically valued field. Examples for such fields are ℚ and ℚ_{p}. Note that a field may admit several p-adic valuations.

If v has no extension to a p-adic valuation of a proper algebraic extension L of K, then K is p-adically closed. Proposition 2.7: Let v be a p-adic valuation of a field K. Then K is p-adically closed if and only if K is Henselian and Γ_{v} is a ℤ-group. Proof: See [PR, Thm. 3.1].

It follows from Zorn’s lemma that every p-adically valued field (K, v) has a padic closure (E, w). That is, (E, w) is p-adically closed and E is algebraic over K. In general, (E, w) is not unique, not even up to a K-isomorphism (as in the case with Henselization). A necessary and sufficient condition for (K, v) to admit a unique p-adic closure up to a K-isomorphism is that Γ_{v} is a ℤ-group [PR, Thm. 3.2].

Proposition 2.8: Let (K, v) a p-adically valued Henselian field and let L be a finite extension of K. Then

(4)  [L : K] = [\overline{L}_{v} : \overline{K}_{v}](v(L^{×}) : v(K^{×})).

Proof: See [PR, Lemma 2.9].

Note that if K is only Henselian, then the right hand side should be multiplied by p^{d}, where p = char(\overline{K}_{v}) (if char\overline{K}_{v}) = 0, then this factor is 1) and d is a nonnegative integer which is called the defect of the extension L/K. Proposition 2.8 means therefore that if (K, v) is a p-adically valued Henselian field, then every finite extension of K is defectless.

Remark 2.9: p-valued fields. Prestel and Roquette [PR] develop a more general theory than that of p-adically valued fields. They define a valued field (K, v) to be p-valued if char(K) = p and dim_{𝔽_{p}} O_{v}/pO_{v} = d is finite. They call d the p-rank of v. What we call “p-adic valuation” is in the terminology of Prestel and Roquette a p-valuation of rank 1. Everything we prove here about the latter valuation holds for an arbitrary p-valuation.

3. The isomorphism theorem. The goal of this section is to prove Macintyre’s theorem: Theorem 3.1: Let (E, v) and (F, v) be p-adic closures of a p-adically valued field (K, v). Then (E, v) ≅_{K} (F, v) if and only if K ∩ E^{n} = K ∩ F^{n} for each positive integer n.

Our proof is an elaboration of the proof of [PR, Cor. 3.11]. Lemma 3.2: Let (L, v) be a Henselian p-adically valued field and let K be a subfield. Let x be an element in L^{×} such that v(x) ∈ v(K^{×}). Then for every positive integer n there exists a ∈ K and z, u ∈ L such that v(1 − z) > 2v(n), v(u) = 0 and x = az = au^{n}. Proof: Denote the reduction modulo M_{v} = pO_{v} by a bar. Let n be a positive integer and r = 2v(n). By assumption there exists b ∈ K such that v(x) = v(b). Let y = xb^{−1}. Then v(y) = 0. As \overline{K}_{v} = \overline{L}_{v} = 𝔽_{p}, there is a_{0} ∈ K such that ȳ = ā_{0}, i.e., v(y − a_{0}) > 0. Since v(p) = 1 is the smallest positive element of v(L^{×}), we have v(y − a_{0}) ≥ v(p). As before there exists a_{1} ∈ K such that v(\frac{y−a_{0}}{p} − a_{1}) ≥ v(p). This means that v(y − a_{0} − a_{1}p) ≥ 2. Continue by induction to find a_{0}, a_{1}, …, a_{r} ∈ K such that v(y − c) > r, where c = a_{0} + a_{1}p + ⋯ + a_{r}p^{r} ∈ K and v(c) = 0.

It follows for z = x/bc that v(1−z) > 2v(n). By Corollary 2.2, there exists u ∈ L^{n} such that u^{n} = z. Obviously v(u) = 0 and x = bcu^{n}. Lemma 3.3: Let (L, v) be a p-adically valued field and let K be a subfield.

(a) If K is algebraically closed in L, then the group v(L^{×})/v(K^{×}) is torsion free.

(b) If v(K) = v(L), then, L = KL^{n} for each positive integer n.

Proof of (a): Suppose that α is an element of v(Ł^{×}) such that nα ∈ v(K^{×}). Take x ∈ L such that v(x) = α. Then v(x^{n}) = nα ∈ v(K^{×}). Hence, by Lemma 3.2, there exist a ∈ K and u ∈ L with v(u) = 0 such that x^{n} = au^{n}. Since K is algebraically closed in L the element x/u belongs to K. Hence, α = v(x/u) ∈ v(K^{×}). Conclude that v(L^{×}/K^{×}) is torsion free. Proof of (b): Apply Lemma 3.2 to each element of L. Proposition 3.4 (Radical structure theorem): Let (L, v)/(K, v) be a finite extension of p-adically valued Henselian fields. Then there exists a factorization [L : K] = n_{1} ⋯ n_{r} such that L = K(t_{1}, …, t_{r}), where t_{i}^{n_{i}} ∈ K and [K(t_{i}) : K] = n_{i}, for i = 1, …, r. In this case K(t_{1}), …, K(t_{r}) are linearly disjoint over K.

Proof: As \overline{L}_{v} = 𝔽_{p} = \overline{K}_{v}, Proposition 2.8 implies that

(1)  n = [L : K] = (v(L^{×}) : v(K^{×})).

The set of all elements t ∈ L^{×} for which there exist n such that t^{n} ∈ K forms a subgroup T of Ł^{×} which contains K^{×}. We claim that v(T) = v(L^{×}).

Indeed, each α ∈ v(L^{×}) satisfies nα ∈ v(K^{×}). Let x be an element of L such that v(x) = α. Then v(x^{n}) ∈ v(K^{×}). By Lemma 3.2, there exist a ∈ v(K) and u ∈ L such that x^{n} = au^{n} and v(u) = 0. Then t = x/u satisfies t^{n} = a and therefore t ∈ T, and v(t) = v(x) = α, as claimed. Conclude from Proposition 2.8 that K(T) = L.

We also obtain an epimorphism

(2)  v: T/K^{×} → v(L^{×})/v(K^{×}).

We claim that it is an isomorphism.

Indeed, let t be an element of T such that v(t) ∈ v(K). There exists m ∈ ℕ such that t^{m} ∈ K. By Lemma 3.2, t = az with a ∈ K and v(1 − z) > 2v(m). Thus z = 1 + p^{2v(m)+1}z_{1} where z_{1} ∈ L. It follows that b = z^{m} = 1 + p^{2v(m)+1}b_{1} with b_{1} ∈ K. By Corollary 2.2, there exists an element c ∈ K such that c^{m} = b. Apply the uniqueness part of that corollary to L to the equality (t/a)^{m} = c^{m} to conclude that t = ac ∈ K, as contended.

It follows that T/K^{×} is an abelian group of order n. Hence, T is generated modulo K^{×} by elements t_{i} of order n_{i}, i = 1, …, r such that n_{1} ⋯ n_{r} = n. In particular L = K(t_{1}, …, t_{r}). Also, t_{i}^{n_{i}} ∈ K, and therefore [K(t_{i}) : K] ≤ n_{i}. But n = [L : K] ≤ ∏_{i=1}^{r}[K(t_{i}) : K] ≤ ∏_{i=1}^{r} n_{i}. Hence [K(t_{i}) : K] = n_{i} for i = 1, …, r and K_{1}, …, K_{r} are linearly disjoint over K.

Proposition 3.5 (Algebraic embedding Theorem): Let (L, v)/(K, v) be an algebraic extension of p-adically valued fields. Suppose that L is Henselian. Let (L′, v) be an arbitrary Henselian extension of K. Then (L, v) is K-embeddable in (L′, v) if and only if K ∩ L^{n} ⊆ (L′)^{n}, for each n ∈ ℕ. Proof: The condition is obviously necessary. To prove that it is sufficient we first make two reduction steps.

Embed first the Henselian closure (K^{h}, v) in both (L, v) and (L′, v) (Proposition 2.5). Next note that v(K^{h}) = v(K) (Proposition 2.6). Hence, by Lemma 3.3(b), K^{h} = K(K^{h})^{n}. It follows that K^{h} ∩ L^{n} = K(K^{h})^{n} ∩ L^{n} = (K ∩ L^{n})(K^{h})^{n} ⊆ (L′)^{n}(K^{h})^{n} ⊆ (L′)^{n}. Thus, we may replace K by K^{h} if necessary to assume that K is Henselian.

Using compactness arguments we may in addition suppose that L/K is a finite extension.

So, by the radical structure theorem (Lemma 3.4), there exists a factorization [L : K] = n_{1} ⋯ n_{r} such that L = K(t_{1}, …, t_{r}), where t_{i}^{n_{i}} = a_{i} ∈ K, [K(t_{i}) : K] = n_{i}, for i = 1, …, r, and K(t_{1}), …, K(t_{r}) are linearly disjoint over K. For each i there exists t′_{i} ∈ L′ such that (t′_{i})^{n_{i}} = a_{i}. Since X_{i}^{n_{i}} − a_{i} is irreducible over K, the map t_{i} ↦ t′_{i} extends to a K-isomorphism σ_{i}: K(t_{i}) → K(t′_{i}). The linear disjointness of the K(t_{i})’s over K means that L = K(t_{1}) ⊗_{K} ⋯ ⊗_{K} K(t_{r}). So, the σ_{i} can be combined to a K-homomorphism σ of L onto K(t′_{1}, …, t′_{n}). But a homomorphism of fields is necessarily an isomorphism. Hence σ is the desired K-embedding of L into L′.

Finally, recall that as (K, v) is Henselian, v has a unique extension to L (Proposition 2.1(b)). Since v ∘ σ is a valuation of L which coincides with v on K, it must coincide with v on L. Thus σ is an embedding of valued fields.

By Proposition 2.7, each p-adically closed field is Henselian. Hence, Theorem 3.1 is a special case, of the following theorem: Proposition 3.6 (Algebraic isomorphism theorem): Let (K, v) be a p-valued field and (L, v), (L′, v) two p-adically valued Henselian algebraic extensions of (K, v). Then a necessary and sufficient condition that (L, v) and (L′, v) are K-isomorphic is that

K ∩ L^{n} = K ∩ L′^{n} for each n ∈ ℕ.

Indeed, by Proposition 3.5, each of the valued fields (L, v) and (L′, v) is Kisomorphic to a subfield of the other. So they are K-isomorphic, as follows from the following lemma: Lemma 3.7: Let L and L′ be algebraic extensions of a field K. Suppose that each of these fields is K-isomorphic to a subfield of the other. Then L ≅_{K} L′. Proof: Let σ: L → L′ and σ′: L′ → L be K-embeddings. Then σ′∘σ is a K-isomorphism of L with a subfield L_{0} which contains K. It suffices to prove that L ⊆ L_{0}.

Indeed, consider x ∈ L and let f = irr(x, K). Denote the set of all roots of f which are contained in L (resp., L_{0}) by S (resp., S_{0}). Then σ maps S bijectively onto S_{0}. But S_{0} ⊆ S. Hence S_{0} = S and therefore x ∈ S_{0} ⊆ L, as desired.

4. Embedding theorem for regular extensions. Each extension of fields of characteristic 0 can be decomposed into an algebraic one and a regular one. In section 3 we have proved an embedding theorem for algebraic extensions of p-adically closed fields. In this section we prove the corresponding theorem for regular extensions: Theorem 4.1: Let (K, v) be a p-adically valued field. Let (L, v) be a countable padically closed extension of (K, v) such that K is algebraically closed in L. Let (F, v) be a p-adically closed extension of (K, v) which is ℵ_{1}-saturated. Then there is a Kembedding of (L, v) in (F, v).

The proof of this theorem follows [PR, Sec. 4]. As in Section 3, it is more convenient to prove a more general theorem, in which the condition on (L, v) to be p-adically closed is relaxed. We start with the case where L is a rational field. Lemma 4.2: Let (K, v) a countable p-adically valued Henselian field. Let (L, v) be a p-adically valued field extension of (K, v) such that L = K(x) with x transcendental over K and v(K) = v(L). Let (F, v) be any valued field extension of (K, v) which is ℵ_{1}-saturated. Then there is a K-embedding of (L, v) into (F, v). Proof: We prove that for each a ∈ K there exists y ∈ K such that v(x − y) > v(x − a).

Indeed, since v(K) = v(L) there exists b ∈ K such that v(x − a) = v(b). As \overline{K} = 𝔽_{p} = \overline{L}, there exists c ∈ K such that v(\frac{x−a}{b} − c) > 0. Hence v(x − a − bc) > v(b) = v(x − a) and therefore y = a + bc satisfies the requirements.

If a_{1}, …, a_{n} are n elements of K, we may assume that v(x−a_{n}) = max_{1≤i≤n} v(x− a_{i}), apply the above statement on a_{n} and find y ∈ K such that v(x − y) > v(x − a_{i}) for i = 1, …, n. It follows that v(x − a_{i}) = v(y − a_{i}) for i = 1, …, n.

Since (F, v) is ℵ_{1}-saturated, and K is countable, there exists y ∈ F such that v(x − a) = v(y − a) for each a ∈ K.

Let f ∈ K[X] be a nonzero polynomial. By induction on deg(f) we prove that

(1)  v(f(x)) = v(f(y)).

If this is the case, then f(y) ≠ 0 and therefore y is transcendental over K. Hence, the map x ↦ y extends to a K-isomorphism of (L, v) into (F, v).

Indeed, we know that (1) holds for polynomials of degree 1. So, suppose that n > 1, (2) is true for all polynomials of degree less than n, and that deg(f) = n. Then f is irreducible over K. Assume that v(f(x)) ≠ v(f(y)).

Choose a root z of f(X) in K̃ and extend v to a valuation of K(z) in the following way. Every nonzero element of K(z) has a unique representation in the form g(z), where g ∈ K[X] is a polynomial of degree less than n. Define v(g(z)) = v(g(x)). Obviously, v satisfies the rule (1b) of Section 1 for addition. We prove that it also satisfies the rule (1a) of Section 1 for multiplication.

Otherwise there exist polynomials g_{1}, g_{2} ∈ K[X] of degree less than n such that

(2)  v(g_{1}(z)g_{2}(z)) ≠ v(g_{1}(z)) + v(g_{2}(z)).

Find polynomials q, r ∈ K[x] such that

(3)  g_{1}(x)g_{2}(x) = q(x)f(x) + r(x)  and deg(r) < n.

Note that deg(q) + n < 2n. Hence deg(q) < n. By definition, v(g_{1}(z)g_{2}(z)) = v(r(x)). Hence, by (2)

(4)  v(r(x)) ≠ v(g_{1}(x)) + v(g_{2}(x)).

It follows from (3) and (4) that

(5)  v(q(x)) + v(f(x)) = min{v(g_{1}(x)) + v(g_{2}(x)), v(r(x))}.

Substitute y for x in (3) to get

(6)  q(y)f(y) = −g_{1}(y)g_{2}(y) − r(y).

By induction, v(g_{1}(x)) = v(g_{1}(y)), v(g_{2}(x)) = v(g_{2}(y)), v(r(x)) = v(r(y)), and v(q(x)) = v(q(y)). Hence, (4) holds for y instead of for x, and therefore, by (6)

(7)  v(q(y)) + v(f(y)) = min{v(g_{1}(y)) + v(g_{2}(y)), v(r(y))}.

It follows from (5) and (7) that v(f(x)) = v(f(y)), a contradiction.

So, v satisfies also the multiplication law for valuation on K(z). It is therefore a valuation of K(z). The value group of K(z) lies between those of K and L. Since they coincide, we have v(K(z)) = v(K).

Also, let g ∈ K[x] be a polynomial of degree less than n. Then there exists a ∈ K such that v(g(x) − a) > 0. By definition, v(g(z) − a) = v(g(x) − a). Hence, v(g(z) − a)) > 0. It follows that K(z) has the same residue field as K. Since (K, v) is a p-adically valued Henselian field, Proposition 2.8 implies that [K(z) : K] = 1. This contradicts [K(z) : K] = deg(f) = n > 1, and concludes the proof.

Lemma 4.3: Let L = K(x) be a rational function field in one variable. Suppose that v is a valuation of L such that v(x) has infinite order modulo v(K^{×}). Then, for each polynomial f(x) = ∑ a_{i}x^{i} with coefficients a_{i} ∈ K we have

(8)  v(f(x)) = min_{i}{v(a_{i}) + iv(x)}.

So, we have a direct factorization for the value group of L:

v(L^{×}) = v(K^{×}) ⊕ ℤv(x).

Proof: By (3c) of Section 2, it suffices to observe that if i < j and a_{i}, a_{j} ≠ 0, then v(a_{i}) + iv(x) < v(a_{j}) + jv(x).

Lemma 4.4: Let (L, v)/(K, v) be an extension of countable p-adically valued fields such that L = K(x) with x transcendental over K. Suppose that ξ = v(x) has an infinite order modulo v(K^{×}). Let (F, v) be a p-adically closed extension of (K, v) which is ℵ_{1}-saturated. Then there is a K-embedding of (L, v) into (F, v).

Proof: By assumption, ξ does not belong to the divisible hull ℚ ⊗ v(K^{×}).

Claim: For each α ∈ ℚ ⊗ v(K^{×}) such that α < ξ there exists η ∈ v(F^{×}) ∩ (ℚ ⊗ v(K^{×})) such that α < η < ξ.

Indeed, let α ∈ ℚ ⊗ v(K^{×}). Then α ≠ ξ. Suppose without loss that α < ξ. Write α = α_{0}/n with α_{0} ∈ v(K^{×}) and n ∈ ℕ. By Proposition 2.7, v(F^{×}) is a ℤgroup. Hence, α_{0} = m + nη_{0} with η_{0} ∈ v(F^{×}) and 0 ≤ m < n, an integer. Let η = η_{0} + 1 = α + \frac{n−m}{n} ∈ v(F^{×}) ∩ (ℚ ⊗ v(K^{×})). Then α < η. We claim that η < ξ. Otherwise, 0 < ηξ − α_{0} ≤ n − m. Since ℤ is a convex subgroup of v(F^{×}) there would exist k ∈ ℤ such that nξ = k + α_{0} ∈ v(K^{×}), a contradiction. Conclude that η satisfies the claim.

Similarly, if α > ξ, there exists η ∈ v(F^{×}) ∩ (ℚ ⊗ v(K^{×})) such that ξ < η < α. If α_{1}, …, α_{n} are several elements of ℚ ⊗ v(K^{×}) we can assume that they are

ordered in an increasing order. Then either ξ < α_{1}, or there exists i such that α_{i} < ξ < α_{i+1}, or α_{n} < ξ. In each case we can use the claim and prove the existence of η ∈ v(F^{×}) ∩ (ℚ ⊗ v(K^{×})) such that sgn(ξ − α_{i}) = sgn(η − α_{i}) for i = 1, …, n.

Since (F, v) is ℵ_{1}-saturated and K is countable, there exists η ∈ v(F^{×}) such that sgn(ξ − α) = sgn(η − α) for each α ∈ ℚ ⊗ v(K^{×}). In particular we find that η ∉ ℚ ⊗ v(K^{×}). Thus η has infinite order modulo v(K^{×}).

Let y ∈ F be an element such that v(y) = η. Then y is transcendental over K. Moreover, by Lemma 4.3, each polynomial f(y) = ∑ a_{i}y^{i} with coefficients in K has the value v(f(y)) = min{v(a_{i}) + iv(y)}. Similarly v(f(x)) = min{v(a_{i}) + iv(x)}. By the choice of y, v(a_{i})+iv(y) ≤ v(a_{j})+jv(y) if and only if v(a_{i})+iv(x) ≤ v(a_{j})+jv(x). So, the map x ↦ y extends to a valuation preserving K-isomorphism of K(x) onto K(y). This is the desired embedding of (L, v) into (F, v).

Lemma 4.5: Let L/K be an extension of valued fields, finitely generated and of transcendence degree 1. Suppose that the factor group v(L^{×})/v(K^{×}) is torsion free. Then there exists x ∈ L, transcendental over K such that v(K(x)) = v(L). Moreover, if v(K) ≠ v(L), then v(x) has infinite order modulo v(K^{×}).

Proof: There is nothing to prove if v(K) = v(L). So suppose that v(K) ≠ v(L) and let x_{0} be an element of L such that v(x_{0}) ∉ v(K). By assumption v(x_{0}) has infinite index over v(K^{×}). In particular x_{0} is transcendental over K. Hence, by Lemma 4.3, v(K(x_{0})^{×}) = v(K^{×}) ⊕ ℤv(x_{0}). By assumption on L/K, the degree of L over K(x_{0}) is finite. Hence, v(K(x_{0})^{×}) has a finite index in v(L^{×}). Since, v(K(x_{0})^{×})/v(K^{×}) ≅ ℤ and v(L^{×})/v(K^{×}) is torsion free, this implies that v(L^{×}) ≅ ℤ. Choose x in L such that v(x) generates v(L^{×}) modulo v(K^{×}). Then x is transcendental over K, v(x) has infinite order modulo v(K^{×}), and v(K(x)) = v(L), as desired.

Proposition 4.6: Let (L, v)/(K, v) be an extension of p-adically valued fields such that K is Henselian, L is countable and v(L^{×})/v(K^{×}) is torsion free. Let (F, v) be an ℵ_{1}-saturated p-adically closed extension of (K, v). Then (L, v) can be K-embedded into (F, v).

Proof: We break to proof into several parts and reduce the proposition to the rational case which has already been handled above.

Part A: Suppose that Proposition 4.6 is true whenever trans.deg(L/K) = 1. Then Proposition 4.6 is true in general.

Since the value group does not change if we replace (L, v) by its Henselization (Proposition 2.6) we may assume that (L, v) is Henselian.

Consider the system of all triples (K_{1}, v, γ) where (K_{1}, v) is a valued field extension of (K, v) which is contained in (L, v), K_{1} is algebraically closed in L and γ: (K_{1}, v) → (F, v) is a K-embedding. By Zorn’s lemma, this system has a maximal element (K_{1}, v, γ). By Corollary 2.4, (K_{1}, v) is Henselian. By Lemma 3.3(a), v(L^{×})/v(K_{1}^{×}) is torsion free. If K_{1} ⊂ L, then any element x ∈ L − K_{1} is transcendental over K_{1}. Let L_{1} be the algebraic closure of K_{1}(x) in L. Then (L_{1}, v) is Henselian and v(L_{1}^{×})/v(K_{1}^{×}) is torsion free, and trans.deg(L_{1}/K_{1}) = 1. By assumption, applied to K_{1}, L_{1} instead of K, L, γ can be extended to a K_{1}-embedding of (L_{1}, v) into (F, v). This contradicts the maximality of (K_{1}, v, γ) and proves that K_{1} = L.

Part B: Suppose that every finitely generated subextension of L/K can be K-embed-ded into (F, v). Then (L, v) can be K-embedded into (F, v).

Indeed, the diagram of (L, v) (in a language which contains a constant symbol for each element of K) is countable. Each finite subset of the diagram has the form {φ_{1}(a), …, φ_{n}(a)}, where φ_{i}(X_{1}, …, X_{r}) is a quantifier free formula, i = 1, …, n, and a = (a_{1}, …, a_{r}) is an r-tuple of elements of L. All elements are contained in a finitely generated subextension of L/K, which can be K-embedded into (F, v). Hence there exist a′_{1}, …, a′_{r} ∈ F such that φ_{i}(a′) is true in (F, v), i = 1, …, n. Since (F, v) is ℵ_{1}- saturated, for each a ∈ L there exists a′ ∈ F such that if φ(X_{1}, …, X_{r}) is a quantifier free formula and a_{1}, …, a_{r} are elements in L such that (L, v) ⊧ φ(a_{1}, …, a_{r}), then (F, v) ⊧ φ(a′_{1}, …, a′_{r}). The map a ↦ a′ is then a K-embedding of (L, v) into (F, v).

Part C: Reduction to the case of rational function field.

By Parts A and B we may assume that L is of transcendence degree 1 and finitely generated. So, by Lemma 4.5, there exists x ∈ L transcendental over K such that v(K(x)) = v(L). Moreover, either v(K) = v(L) or v(x) has infinite order modulo v(K^{×}).

The Henselization of (K(x), v) is contained in the Henselization of (L, v). By Proposition 2.6, v(L^{h}) = v(L) = v(K(x)) = v(K(x)^{h}). Similarly, all residue fields are equal to 𝔽_{p}. Hence, by Proposition 2.8, K(x)^{h} = L^{h}, and therefore L ⊆ K(x)^{h}.

If v(K(x)) = v(K), then (K(x), v) can be K-embedded in (F, v) by Lemma 4.2. If x has infinite order modulo v(K^{×}), then this is so, by Lemma 4.4. As (F, v) is Henselian (Proposition 2.7), the K-embedding of (K(x), v) into (F, v) extends to an embedding of (K(x)^{h}, v) and therefore also of (L, v) into (F, v), as contended.

Proof of Theorem 4.1: By Lemma 3.3, v(L^{×})/v(K^{×}) is torsion free. By Proposition 2.7, (L, v) is Henselian. Hence, by Corollary 2.4, (K, v) is Henselian. So, Theorem 4.1 is a special case of Proposition 4.6.

5. Elimination of quantifiers for the theory of p-adically closed fields. This section collects the results of the preceding sections. We prove here the elementary equivalence theorem for p-adically closed fields and show that the theory of this family admits an elimination of quantifiers within a natural extension of the theory of valued fields.

Lemma 5.1: Let (L, v) be a p-adically closed field and let K be a subfield of L. Then K is algebraically closed in L if and only if (K, v) is p-adically closed.

Proof: Suppose first that K is algebraically closed in L. By Proposition 2.7, (L, v) is Henselian. So, by Corollary 2.4, so is (K, v). Obviously, \overline{K}_{v} = \overline{L}_{v} = 𝔽_{p}. Finally, by Proposition 2.7, v(L^{×}) is a ℤ-group. This means that the quotient group v(L^{×})/ℤ is divisible. As v(L^{×})/v(K^{×}) is torsion free, it follows that v(K^{×})/ℤ is divisible. Conclude from Proposition 2.7 that (K, v) is p-adically closed.

Conversely, suppose that K is p-adically closed. By the preceding paragraph, the algebraic closure \overline{K} of K in L is p-adically closed. Hence K = \overline{K} and therefore K is algebraically closed in L.

Lemma 5.1 enables us to combine the isomorphism theorem 3.1 with the embedding theorem for regular extensions 4.1 to a general embedding theorem:

Theorem 5.2: Let (K, v) be a p-adically valued field. Let (L, v) be a countable padically closed extension of (K, v). Let (F, v) be a p-adically closed extension of (K, v) which is ℵ_{1} saturated. Suppose that L^{n} ∩ K ⊆ F^{n} for each n ∈ ℕ. Then there is a K-embedding of (L, v) in (F, v).

Proof: Denote the algebraic closure of K in L by L_{0}. By Lemma 5.1 (L_{0}, v) is p-adically closed and therefore Henselian. Obviously L_{0}^{n} ∩ K = L^{n} ∩ K = F^{n}. Hence, by Proposition 3.5, there is a K-embedding of (L_{0}, v) into (F, v). By Lemma 3.3, v(L^{×})/v(L_{0}^{×}) is torsion free. Hence, by Proposition 4.6, we can extend φ_{0} to an embedding of (L, v) into (F, v).

To state the following theorem let us extend the language of valued fields by a sequence of unary predicate symbols P_{1}, P_{2}, P_{3}, . . .. We denote the extended language by L′(valued field). For a p-adically closed field (E, v) we always interpret P_{n} as the subset E^{n} = {x^{n}| x ∈ E} of E.

Theorem 5.3 (Elementary equivalence theorem): Let (E, v) and (F, v) be p-adically closed field extensions of a valued field (K, v). If E^{n} ∩ K = F^{n} ∩ K for each n ∈ ℕ, then (E, v) and (F, v) are elementarily equivalent in L′(valued field(K)).

Proof: Each sentence contains only finitely many elements of K. They are all contained in a finitely generated extension of ℚ. So, without loss, we may assume that K is countable.

Next replace (E, v) and (F, v) by nonprincipal ultraproducts, if necessary to assume that (E, v) and (F, v) are ℵ_{1}-saturated (Section 1).

Now use the Skolem–Löwenheim theorem [FJ, Prop. 6.4] to find a countable elementary valued subfield (L_{1}, v) of (E, v) which contains (K, v). In particular L_{1} is algebraically closed in E and therefore L_{1}^{n} ∩ K = E^{n} ∩ K. By Theorem 5.2, there is a K-isomorphism φ_{1} of (L_{1}, v) onto a subfield (M_{1}, v) of (F, v). Then (M_{1}, v) is p-adically closed. Hence, by Lemma 5.1, M_{1} is algebraically closed in F. Thus, for each n ∈ ℕ we have E^{n} ∩ L_{1} = L_{1}^{n} and F^{n} ∩ M_{1} = M_{1}^{n}. It follow that φ_{1} maps E^{n} ∩ L_{1} onto F^{n} ∩ M_{1}. Now choose a countable elementary valued subfield (M_{2}, v) of (F, v) which contains (M_{1}, v). Use Theorem 5.2 once more to extend φ_{1}^{−1} to an isomorphism φ_{2}^{−1} of (M_{2}, v) onto a valued subfield (L_{1}, v) of (E, v).

Continue by induction to construct a tower of compatible isomorphisms

φ_{i}: (L_{i}, v) → (M_{i}, v)

of valued subfields of (E, v) and (F, v), respectively such that (L_{2i+1}, v) is an elementary valued subfield of (E, v) and (M_{2i}, v) is an elementary valued subfield of (F, v), i = 1, 2, 3, …. Let L be the union of the L_{i}’s and let M be the union of the M_{i}’s. Then (L, v) is isomorphic over (K, v) to (M, v). Also (L, v) ≺ (E, v) and (M, v) ≺ (F, v) [FJ, Lemma 6.3(b)]. Hence (E, v) is elementary equivalent to (F, v) in L′(valued field).

We may combine Theorem 5.3 with Lemma 1.1 to get elimination of quantifiers:

Theorem 5.4: Let K be a p-adically valued field. Then the theory of p-adically closed fields which contain K admits an elimination of quantifiers in the language L′(valued field, K).

We rephrase Theorem 5.4 in term of “definable sets”: Each formula φ(X_{1}, …, X_{m}) of L(valued field, ℚ_{p}) defines a set

A = {a ∈ ℚ_{p}^{m}| ℚ_{p} ⊧ φ(a)}.

Theorem 5.5: Each definable set in L(valued field, ℚ_{p}) is a Boolean combination of sets of the form

(1)  {x ∈ ℚ_{p}^{m}| ∃y ∈ ℚ_{p}: f(x) = y^{n}},

with f ∈ ℤ_{p}[X_{1}, …, X_{m}] and n ∈ ℕ.

Proof: By Theorem 5.4, φ(X) is equivalent over ℚ_{p} to a quantifier free formula of the language L′(valued field(ℚ_{p})). The latter is a Boolean combination of formulas of one of the following forms: (2a) f(X) = 0, (2b) v(g(X)) ≥ v(h(X)), and (2c) P_{n}(q(X)), with f, g, h, q ∈ ℚ_{p}[X]. Note that (2c) is equivalent to ∃y: q(X) = y^{n}. So, multiplying the numerator and the denominators of the coefficients of f with a suitable power of p we may assume that q ∈ ℤ_{p}[X]. Obviously we can make the same assumption on f, g, h.

Thus, to end the proof it suffices to show that formulas (2a) and (2b) are equivalent to formulas of the form (2c).

Indeed, for a ∈ ℚ_{p} the equality a = 0 is equivalent to ∃y ∈ ℚ_{p}: pa^{2} = y^{2}. Also, if p ≠ 2 and a, b ∈ ℚ_{p}, then v(b) ≥ v(a) if and only if ∃y ∈ ℚ_{p}: a^{2} + pb^{2} = y^{2}.

Indeed, if v(b) ≥ v(a), then we may divide the latter equation by a^{2}, if necessary, to assume that a = 1. Then use Corollary 2.2 to deduce the existence of y ∈ ℚ_{p} such that a^{2} +pb^{2} = y^{2}. If, on the other hand, v(b) < v(a), then v(b) ≤ v(a)+1 and therefore v(pb^{2}) < v(a^{2}). It follows that 2v(y) = 1 + 2b(b), which is a contradiction.

Finally we claim that for p = 2 the inequality v(b) ≥ v(a) is equivalent to ∃y ∈ ℚ_{2}: a^{2} + 8b^{2} = y^{2}.

Indeed, if v(b) ≥ v(a), divide the equation by a^{2} and apply Corollary 2.2. If v(b) < v(a) and there exists y ∈ ℚ_{2} such that a^{2} + 8b^{2} = y^{2}, then v(2b) ≤ v(a). Divide by 4a^{2} to find c, z ∈ ℤ_{2} such that c^{2} + 2 = z^{2}. It follows that 2 ≡ 0 mod 8, a contradiction.

Lemma 5.6: For each positive integer k the quotient group ℚ_{p}^{×}/(ℚ_{p}^{×})^{k} is finite and can be represented by positive integers.

Proof: We may decompose ℚ_{p}^{×} as follows:

ℚ_{p}^{×} = ⟨p⟩ × U = ⟨p⟩ × W × U_{1} ≅ ℤ × 𝔽_{p}^{×} × U_{1},

where U is the group of p-adic units, W is the group of roots of unity of order prime to p and U_{1} = 1 + pℤ_{p} is the group of 1-units. The latter one has the subgroups U_{n} = 1 + p^{n}ℤ_{p}. They satisfy U_{n}/U_{n+1} ≅ ℤ_{p}. Hence (U_{1} : U_{n}) = p^{n}. Also, if n > 2v(k) then U_{n} ⊆ U_{1}^{k} (Corollary (2.2)). Hence U_{1}^{k} has a finite index in U_{1}. This concludes the proof of the first assertion of the lemma. The second one also follows from these arguments by taking positive integers which approximate representatives of U modulo kth powers to a higher degree of p than p^{2v(k)}.

Following Lemma 5.6, we choose for each k a finite set of integers C_{k} which represents the elements of ℚ_{p}^{×}/(ℚ_{p}^{×})^{k} and also includes 1 and p^{k}. We may choose C_{k} even canonically by choosing each element of C_{k} as the least possible positive integer. In particular, C_{k} has a natural ordering. Thus for each element z ∈ ℚ_{p} there exists a unique c ∈ C_{k} and there exists z_{1} ∈ ℚ_{p} such that z = cz_{1}^{k}. We call c the kth power residue of z.

Corollary 5.7: Each definable subset of ℚ_{p}^{m} is a finite disjunction of conjunctions of sets of the form

(3)  {x ∈ ℚ_{p}^{m}| ∃y ∈ ℚ_{p}: f(x) = cy^{n}},

with f ∈ ℤ_{p}[X_{1}, …, X_{m}], n ∈ ℕ, and c ∈ C_{n}.

Proof: It suffices to observe that the complement of a set of the form (1) is the disjoint union of set of the form (2), where c ranges over all elements of C_{n} different from p^{n} and 1.

6. Poincaré series and p-adic integrals. To each formula φ(X_{1}, …, X_{m}) in L(valued field)(ℤ_{p}) we associate two power series in t and two p-adic integrals in s with t = p^{−s}. Then we prove that the rationality of the series in t depends on the rationality of the corresponding integral in p^{−s}.

To define the series let

ν(φ, n) = #{x + p^{n}ℤ_{p}^{m}| ∃y ∈ ℤ_{p}^{m}: x ≡ y mod p^{n} and ℤ_{p} ⊧ φ(y)}

and

ν̃(φ, n) = #{x + p^{n}ℤ_{p}^{m}| ℤ/p^{n}ℤ ⊧ φ(x)}.

In particular

ν(φ, 0) = \{ \begin{matrix}1 & if there exists y ∈ ℤ_{p}^{m} such that ℤ_{p} ⊧ φ(y) \\ 0 & otherwise \end{matrix}

Then form the Poincaré series:

P (φ, t) = ∑_{n=0}^{∞} ν(φ, n)t^{n}  P̃(ν, t) = ∑_{n=0}^{∞} ν̃(φ, n)t^{n}.

Now consider the normalized absolute value on ℚ_{p}:

|a| = p^{−v(p)}

where v is the p-adic valuation of ℚ_{p}.

Consider also the Haar measure μ on ℚ_{p}^{m} which is normalized such that μ(ℤ_{p}^{m}) = 1. In particular μ is invariant under translations. That is, μ(a+A) = μ(A) for each a ∈ ℤ_{p}^{m} and each measurable set A. In particular, if A and B are measurable subgroups of ℚ_{p}^{m} and A is a subgroup of B of index n, then μ(A) = \frac{1}{n}μ(B). Also, if \overline{A} is a subset of (ℤ_{p}/p^{n}ℤ_{p})^{m} and A = \{x ∈ ℤ_{p}^{m}| x+p^{n}ℤ_{p} ∈ \overline{A}\} is the lifting of \overline{A}, then μ(\overline{A}) = p^{−mn}| \overline{A}|.

The uniqueness of the Haar measure up to a constant gives a rule for multiplicative translations of measurable sets: μ(cA) = |c|^{m}μ(A) for each c ∈ ℚ_{p}. Indeed, assume without loss that c ≠ 0. Then the function μ′ defined on the measurable sets by μ′(A) = μ(cA) is again a Haar measure of ℚ_{p}. Hence, there exists γ ∈ ℝ such that μ(cA) = γμ(A) for all measurable sets A. Let n = v(c). If n ≥ 0, then cℤ_{p}^{m} is an open subgroup of ℤ_{p}^{m} of index p^{mn}. If n < 0, then ℤ_{p}^{m} is an open subgroup of cℤ_{p}^{m} of index p^{−mn}. So, in both cases, μ(cℤ_{p}^{m}) = p^{−mn}μ(ℤ_{p}^{m}) = |c|^{m}μ(ℤ_{p}^{m}). Conclude that γ = |c|^{m}.

Each definable subset D of ℚ_{p}^{m} is measurable. Indeed, by Corollary 5.7 it suffices to consider only sets of the form

D = {x ∈ ℚ_{p}^{m}| ∃y ∈ ℚ_{p}: f(x) = cy^{n}},

If c = 0, then D is closed. If c ≠ 0, then D is open, by Lemma 2.2. So, D is measurable.

Note that μ is the mth power of the Haar measure on ℚ_{p} which we denote by the same letter μ. If f is an integrable complex valued function on a measurable subdomain D of ℚ_{p}^{m} and χ_{D} is the characteristic function of D, we write ∫_{D} f(x)|dx| for the integral ∫ fχ_{D}dμ. Here |dx| stands for |dx_{1}| ⋯ |dx_{n}|. In particular,

∫_{D} |dx| = μ(D).

The integral is a linear function in f:

∫_{D} ∑_{i=1}^{n} γ_{i}f_{i}(x)|dx| = ∑_{i=1}^{n} γ_{i} ∫_{D} f_{i}(x)|dx|.

Each real valued continuous function of ℚ_{p}^{m} with compact support is integrable. In particular, if D is a bounded definable subset of ℚ_{p}^{m} and g_{1}, g_{2} ∈ ℚ_{p}[X_{1}, …, X_{m}] such that g_{2}(x) ≠ 0 for each x ∈ D, and e is a positive integer, then

f(x) = |\frac{g_{1}(x)}{g_{2}(x)}|^{1/e}

is an integrable function on D.

The above two rules for μ imply the following rules for linear changes of variables in the integrals:

∫_{D} f(x)|dx| = ∫_{D−a} f(y + a)|dy|, ∫_{D} f(x)|dx| = |c|^{m} ∫_{c^{−1}D} f(cy)|dy|.

where a ∈ ℚ_{p}^{m} and c ∈ ℚ_{p}^{×}.

Fubini’s theorem allows us to change the order of integration in double integrals: Let C and D be measurable subdomains of, respectively, ℚ_{p}^{m} and ℚ_{p}^{n}. If f is a measurable function on C × D then

∫_{C×D} f(x, y)|d(x, y)| = ∫_{C} ∫_{D} f(x, y)|dx||dy| = ∫_{D} ∫_{C} f(x, y)|dy|dx|.

Lebesque’s theorem about monotonic convergence allows us to change the order of integration and infinite summation: Let f_{n}(x), n = 1, 2, 3, … be nonnegative real valued functions on a measurable subset D of ℚ_{p}^{m} such that f(x) = ∑_{n=1}^{∞} f_{n}(x) converges for each x ∈ D. If f(x) is integrable on D, then so is each f_{n}(x) and

∫_{D} f(x)|dx| = ∑_{n=1}^{∞} ∫_{D} f_{n}(x)|dx|.

Another useful rule is the σ-additivity of the integral: If f(x) is an integrable function on a subdomain D = ⋃ · _{n=1}^{∞} D_{n} of ℚ_{p}^{m} and each D_{n} is measurable, then f(x) is integrable on D_{n} and

∫_{D} f(x)|dx| = ∑_{n=1}^{∞} ∫_{D_{n}} f(x)|dx|.

Return now to the formula φ and consider the following two subdomains of ℤ_{p}^{m+1}:

D(φ) = {(x, z) ∈ ℤ_{p}^{m} × ℤ_{p}| ∃y ∈ ℤ_{p}^{m} : x ≡ y mod z and ℤ_{p} ⊧ φ(y)}

and

D̃(φ) = {(x, z) ∈ ℤ_{p}^{m} × ℤ_{p}| ℤ_{p}/zℤ_{p} ⊧ φ(x)}.

Associate with these domains the following integrals:

I(φ, s) = ∫_{D(φ)} |z|^{s}|dx||dz|  Ĩ(φ, s) = ∫_{_{D}(φ)} |z|^{s}|dx||dz|,  s > 0.

Lemma 6.1: With the above notation, we have

I(φ, s) = \frac{p − 1}{p}P (φ, p^{−m−1}p^{−s}) and Ĩ(φ, s) = \frac{p − 1}{p}P̃(φ, p^{−m−1}p^{−s})

Proof: We prove only the left identity. The proof of the right one is similar. The set D is a disjoint union of the sets

D_{n}(φ) = {(x, z) ∈ ℤ_{p}^{m} × ℤ_{p}| v(z) = n, ∃y ∈ ℤ_{p}^{m}: x ≡ y mod z and ℤ_{p} ⊧ φ(y)}
= A_{n} × S_{n}

where

A_{n} = {x ∈ ℤ_{p}^{m}| ∃y ∈ ℤ_{p}^{m}: x ≡ y mod p^{n} and ℤ_{p} ⊧ φ(y)}

and

S_{n} = {z ∈ ℤ_{p}| v(z) = n} = p^{n}ℤ_{p} − p^{n+1}ℤ_{p}.

Then, μ(A_{n}) = p^{−nm}ν(φ, n) and μ(S_{n}) = p^{−n} − p^{−n−1}. Thus,

I(φ, s) = ∑_{n=0}^{∞} ∫_{D_{n}(φ)} |z|^{s}|dx||dz|

= ∑_{n=0}^{∞} p^{−ns} ∫_{A_{n}} |dx| ∫_{S_{n}} |dz| = ∑_{n=0}^{∞} p^{−ns}p^{−mn}ν(φ, n)(p^{−n} − p^{−n−1})

= \frac{p − 1}{p} ∑_{n=0}^{∞} ν(φ, n)(p^{−m−1}p^{−s})^{n} = \frac{p − 1}{p}P (φ, p^{−m−1}p^{−s})

7. The absolute value of a definable function. A definable function from ℚ_{p}^{m} to ℚ_{p} is a function whose graph is a definable subset of ℚ_{p}^{m+1} (Section 5).

Let A be a definable subset of ℚ_{p}^{m}. A function θ: A → ℤ ∪ {∞} is simple if there exist a finite partition of A into definable sets A_{i}, polynomials f_{i}, g_{i} ∈ ℚ_{p}[X], and e_{i} ∈ ℕ such that for each x ∈ A_{i}

(1)  g_{i}(x) ≠ 0 and θ(x) = \frac{1}{e_{i}} v(\frac{f_{i}(x)}{g_{i}(x)}).

Note that if B_{1}, …, B_{n} are definable subsets of ℚ_{p}^{m} and θ is simple on each B_{i}, then θ is simple on B_{1} ∪ ⋯ ∪ B_{n}. Lemma 7.1: Let k be a positive integer.

(a) Let a, b be elements of ℚ_{p} such that v(a) = 0 and a ≡ b mod p^{2v(k)+1}. Then a and b have the same kth power residue.

(b) Let a = ∑_{i=1}^{n} a_{i} and suppose that v(a_{i}) − v(a_{1}) > 2v(k), i = 2, …, n. Then v(a) = v(a_{1}) and a and a_{1} have the same kth power residue.

Proof of (a): Apply Corollary 2.2 on the relation ab^{−1} ≡ 1 mod p^{2v(k)+1}. Proof of (b): Observe that a = a_{1} ∑_{i=1}^{n}(1 + a_{1}^{−1}a_{i}) and v(a_{1}^{−1}a_{i}) > 2v(k). Now apply Corollary 2.2. Proposition 7.2: Let S be a definable subset of ℚ_{p}^{m+1}. Suppose that for each x ∈ ℚ_{p}^{m}, the set {v(t)| (x, t) ∈ S} consists of exactly one element which we denote by θ(x). Then θ(x) is a simple function of x. Proof: It suffices to prove that θ is simple on the set

Z = {x ∈ ℚ_{p}^{m}| θ(x) ≠ 0, ∞}.

We break the proof that θ is simple on Z into several parts. Part A: Partition of Z. By Corollary 5.7, S is a disjunction of conjunction of sets of the form

S_{i} = {(x, t) ∈ ℚ_{p}^{m+1}| ∃y ∈ ℚ_{p}: f_{i}(x, t) = c_{i}y^{n_{i}}},

with f_{i} ∈ ℚ_{p}[X, T], c_{i}, n_{i} ∈ ℕ for i = 1, …, l.

Choose a common multiple n of n_{1}, …, n_{l} and write

f_{i}(X, T) = ∑_{j=0}^{d_{i}} a_{ij}(X)T^{j},

with a_{ij} ∈ ℚ_{p}[X]. We may partition Z in definable subsets A such that for each A and each (i, j) either

(2)  a_{ij}(x) = 0 for all x ∈ A

or

(3)  a_{ij}(x) ≠ 0 for all x ∈ A

It suffices to partition each such A.

Part B: Partition of A. For each α between 0 and 2v(n), each i between 1 and l and for all j_{1}, j_{2} such that j_{1} ≠ j_{2} and a_{ij_{1}} and a_{ij_{2}} satisfy (3) define

A_{α,i,j_{1},j_{2}} = {x ∈ A| ∃t ∈ ℚ_{p}^{×}: (x, t) ∈ S and v(a_{ij_{1}}(x)t^{j_{1}}) = v(a_{ij_{2}}(x)t^{j_{2}}) + α}.

Then, x ∈ A_{α,i,j_{1},j_{2}} satisfies

θ(x) = v(t) = \frac{1}{j_{1} − j_{2}} v(\frac{a_{ij_{2}}(x)p^{α}}{a_{ij_{1}}(x)} ).

Hence, θ is simple in A_{α,i,j_{1},j_{2}}.

It remains to prove that θ is simple on

B = A − ⋃ A_{α,i,j_{1},j_{2}},

where the union is over all α, i, j_{1}, j_{2} as above.

Part C: Partition of B. If x ∈ B, then there exists t ∈ ℚ_{p} and i between 1 and l such that (x, t) ∈ S_{i}. Then there exists j_{1} such that a_{i,j_{1}}(x) ≠ 0. Otherwise f_{i}(X, T) = 0 and therefore (x, t) ∈ S_{i} for each t ∈ ℚ_{p}, so that v(t) is not unique. Also, there exists j_{2} ≠ j_{1} such that a_{i,j_{2}}(x) ≠ 0. Otherwise f_{i}(X, T) = a_{i,j_{1}}(x)T^{j_{1}} and therefore f_{i}(x, t^{1+kn_{i}}) = a_{ij_{1}}(x)t^{(1+kn_{i})j_{1}} = c_{i}(yt^{kj_{1}})^{n_{i}}, so that (x, t^{1+kn_{i}}) ∈ S_{i} for each k ∈ ℤ. Since v(t) ≠ 0, v(t^{1+kn_{i}}) is not unique, a contradiction.

Suppose without loss that v(a_{i,j_{1}}(x)t^{j_{1}}) ≤ v(a_{i,j}(x)t^{j}) for all j. If j ≠ j_{1} and a_{i,j}(x) ≠ 0, then x ∉ A_{α,i,j_{1},j} and therefore v(a_{ij}(x)t^{j}) > v(a_{i,j_{1}}(x)t^{j_{1}}) + 2v(n). This inequality holds, of course, also in the case where a_{ij}(x) = 0. By Lemma 7.1(b), the nth power residue of f_{i}(x, t) is equal to that of a_{i,j_{1}}(x)t^{j_{1}}. Hence, there exists y_{i} ∈ ℚ_{p} such that

v(t^{j_{2}−j_{1}}) ≥ v(\frac{a_{i,j_{1}}(x)p^{2v(n)+1}}{a_{i,j_{2}}(x)} ) and c_{i}^{−1}a_{i,j_{1}}(x)t^{j_{1}} = y_{i}^{n_{i}}.

So we can make a disjunction over all possible combinations of (i, j_{1}, j_{2}) and the nth power residues of a_{i,j_{1}}(x)t^{j_{1}} to conclude that for x ∈ B the relation (x, t) ∈ S is equivalent to a disjunction of conjunctions of conditions of the form

(4)  v(t^{r}) ≥ v(\frac{b_{1}(x)}{b_{2}(x)}),

and

(5)  ∃y: b_{0}(x)t^{τ} = y^{k},

with b_{0}, b_{1}, b_{2}, ∈ ℚ_{p}[X], 0 ≠ r ∈ ℤ, τ ∈ ℕ, k|n, and b_{0}(x), b_{1}(x), b_{2}(x) ≠ 0 for all x ∈ B.

Making a disjunction over the different possibilities of the kth power residues of b_{0}(x) and t and on the residues of τ mod k, we obtain that for x ∈ B the relation (x, t) ∈ S is equivalent to a disjunction of conditions ψ_{i}, where each ψ_{i} is a conjunction of conditions of the form (4), definable conditions on b_{0}(x), and a single condition of the form

(6)  ∃y: t = cy^{k},

where c ∈ ℕ. Part D: Partition of B_{i}. Recall now that for each x ∈ ℚ_{p}^{m}, θ(x) is the unique value v(t) of the elements t ∈ ℚ_{p} for which (x, t) ∈ S. Hence it suffices to prove that θ is simple on the set B_{i} = {x ∈ B| ∃t ∈ ℚ_{p}: (x, t) satisfy ψ_{i}}.

If x ∈ B_{i} and θ_{0} = θ(x), then θ_{0} satisfies a conjunction of conditions of the form

(7)  rθ_{0} ≥ v(\frac{b_{1}(x)}{b_{2}(x)}),

and a single condition of the form

(8)  θ_{0} ≡ γ mod k,

where γ = v(c) and r, b_{1}, b_{2}, c and k are as in (5). If θ_{1} is another integer that satisfies the conditions (7) and (8), then θ_{1} ≡ θ_{0} mod k. Take t such that (x, t) satisfies ψ_{i}, and v(t) = θ(x) = θ_{0}. Suppose without loss that θ_{1} ≥ θ_{0}. Then (x, p^{θ_{1}−θ_{0}}t) also satisfies ψ_{i} and therefore belongs to S. Hence, by the uniqueness condition, θ_{1} = v(p^{θ_{1}−θ_{0}}t) = θ_{0}. So, for x ∈ B_{i}, θ(x) is the unique integer θ_{0} that satisfies conditions (7) and (8).

The conjunction of the conditions (7) can be rewritten as

(9)  \frac{1}{s} max_{j}\{v(g_{j}(x))\} ≤ θ(x) ≤ \frac{1}{s} min_{j}\{v(h_{j}(x))\}

with g_{j}, h_{j} ∈ ℚ_{p}(X) such that h_{j}(x) and g_{j}(x) are well defined, and s ∈ ℕ. Write θ(x) = γ + q(x)k. Then, for x ∈ B_{i} the unique value of q(x) must satisfy

q(x) = [\frac{min_{j}\{v(h_{j}(x))\} − γs}{sk}]

where [ ] denotes the greatest integer function. We have to prove that q(x) is simple on B_{i}. Part E: Conclusion of the proof. Cover B_{i} with definable subsets D_{i} on which a particular h_{j}(x) has minimal value and constant (sk)th power residue, say h_{j}(x) = c_{j}u_{j}p^{skλ(x)}, with c_{j} ∈ ℕ, v(u_{j}) = 0, and λ(x) ∈ ℤ. On such a subset

q(x) = [\frac{v(c_{j}) − γs + skλ(x)}{sk}] = [\frac{v(c_{j}) − γs}{sk}] + λ(x).

Hence

[\frac{v(c_{j}) − γs}{sk}] = ν_{j}

is an integer which is independent of x on D_{i}, and

q(x) = \frac{1}{sk} v(h_{j}(x)p^{skν_{j}−v(c_{j})}),

as desired. Corollary 7.3: Let f: ℚ_{p}^{m} → ℚ_{p} be a definable function. Then v(f(x)) is a simple function of x. Proof: In this case S = {(x, f(x))| x ∈ ℚ_{p}^{m}} and θ(x) = v(f(x)).

8. Partitions of ℚ_{p}^{m+1}. The integration of a function over ℚ_{p}^{m} becomes easier if we decompose the domain of integration to subdomains on which the function has a simpler form. We achieve such a decomposition in Proposition 8.4. Lemma 8.1: Let S be a definable subset of ℚ_{p}^{m+n} and let A be a definable subset of ℚ_{p}^{m}. For x ∈ ℚ_{p}^{m}, let

S_{x} = {y ∈ ℚ_{p}^{n}| (x, y) ∈ S}.

Suppose that there exists α ∈ ℕ such that for each x ∈ A

(1)  0 < #S_{x} ≤ α.

Then there exist definable functions f_{1}, …, f_{n} from A to ℚ_{p} such that (x, f_{1}(x), …, f_{n}(x)) ∈ S for all x ∈ A. Proof: We break the proof into four parts of which the three first are reduction steps. Part A: It suffices to prove the lemma for n = 1. Indeed, consider the projection of S on the first m + n − 1 coordinates: R = {(x, y) ∈ A × ℚ_{p}^{n−1}| ∃z ∈ ℚ_{p}: (x, y, z) ∈ S}. Then R_{x} = {y ∈ ℚ_{p}^{n−1}| (x, y) ∈ R} is the projection of S_{x} on the first n−1 coordinates. Hence, R_{x} is nonempty and of cardinality at most α. Use induction on n to find n − 1 definable functions f_{1}, …, f_{n−1} from A to ℚ_{p} such that (x, f_{1}(x), …, f_{n−1}(x)) ∈ R for each x ∈ A. Next consider the set

Q = {(x, f_{1}(x), …, f_{n−1}(x), z) ∈ A × ℚ_{p}^{n−1} × ℚ_{p}| (x, f_{1}(x), …, f_{n−1}(x), z) ∈ S}.

For each x ∈ A the set

Q_{(x,f_{1}(x),...,f_{n−1}(x))} = {z ∈ ℚ_{p}| (x, f_{1}(x), …, f_{n−1}(x), z) ∈ S}

is nonempty and is contained in the projection of S on the nth coordinate. Hence, its cardinality does not exceed α.

If the lemma is true for n = 1, then there is a definable function f_{n}: A → ℚ_{p} such that for each x ∈ A the point (x, f_{1}(x), …, f_{n}(x)) belongs to Q and therefore to S. So, from now on we assume that n = 1.

Part B: It suffices to find a positive integer k such that for each x ∈ A

(2) #S_{x} = 1, or

not all the elements of S_{x} have the same value, or

not all the elements of S_{x} have the same kth power residue.

Indeed, if α = 1, then define f(x) to be the unique y ∈ ℚ_{p} such that (x, y) ∈ S. Then f is definable and satisfies (x, f(x)) ∈ S.

Assume therefore that α > 1 and let B be the set of all (x, y_{0}) ∈ S such that v(y_{0}) is the smallest value among all values v(y) with y ∈ S_{x}, or if all elements of S_{x} have the same value, the kth power residue of y_{0} in C_{k} (Section 5) is the least one among the kth power residues of the elements of S_{x} in C_{k}. Then B is definable. Moreover, by (2), B_{x} is nonempty and has less that α elements. By induction on α there exists a definable function f such that for each x the point (x, f(x)) belongs to B and therefore to S.

Part C: We may suppose that the mean value of S_{x} is 0. Here the mean value of S_{x} is defined by

ȳ(x) = \frac{1}{#S_{x}} ∑_{y∈S_{x}} y.

By (1), ȳ is a definable function. Hence, S′ = {(x, y − ȳ(x)| (x, y) ∈ S} is a definable set and the mean value of S′_{x} is 0. If we find a definable function f such that (x, f(x)) belongs to S′ for each x ∈ A, then (x, f(x) + ȳ(x)) would belong to S.

Part D: Conclusion of the proof. Let κ = max_{1≤l≤α} v(l). We prove that k = φ(p^{κ+1}) = p^{k+1} − p^{k} satisfies (2).

Indeed, let x ∈ A. Then S_{x} = {y_{1}, y_{2}, …, y_{l}} for some l ≤ α. In particular

(3)  v(l) ≤ κ.

We have to prove that if l > 1 and v(y_{1}) = v(y_{2}) = ⋯ = v(y_{l}) = β, then not all y_{i} have the same kth power residue.

By assumption y_{i} = p^{β}y′_{i} with v(y′_{i}) = 0 and ∑_{i=1}^{l} y′_{i} = 0. It suffices to prove that all y′_{i} have the same kth power residue.

Otherwise we may find c, z_{i} ∈ ℚ_{p} with v(c) = v(z_{i}) = 0 such that y′_{i} = cz_{i}^{k}, i = 1, …, l. By Euler’s theorem, z_{i}^{k} ≡ 1 mod p^{κ+1} for all i. Hence

0 = ∑_{i=1}^{l} y′_{i} = c ∑_{i=1}^{l} z_{i}^{k} ≡ cl mod p^{κ+1}.

Which implies that p^{κ+1}|l. This contradiction to (3) concludes the proof. Lemma 8.2: (a) For each positive integer d, ℚ_{p} has only finitely many extensions of

degree at most d.

(b) Let K be a finite extension of ℚ_{p}. Then K/ℚ_{p} has a basis ξ_{1}, …, ξ_{k} with ξ_{1} = 1 such that for all z_{1}, …, z_{k} ∈ ℚ_{p}

(4)  v(∑_{i=1}^{k} z_{i}ξ_{i}) = min_{1≤i≤k} v(z_{i}ξ_{i}) and 0 ≤ v(ξ_{i}) < 1, 1, …, k.

Proof: Statement (a) can be proved by compactness arguments [L, p. 54]. To prove (b) let K_{0} be the maximal unramified extension of ℚ_{p} in K. Choose u_{1}, …, u_{m} in K_{0} whose residues modulo p form a basis for the residue field of K_{0} over 𝔽_{p} such that u_{1} = 1. Then v(K_{0}^{×}) = ℤ.

Let e = [K : K_{0}] and choose a prime element π for K. Then v(π) = 1/e and 1, π, …, π^{e−1} is a basis for K/K_{0}. The elements u_{i}π^{j}, i = 1, …, m and j = 0, …, e − 1 form a basis for K/ℚ_{p}. We prove that it has the desired property.

Let z_{ij} be elements of ℚ_{p}. Set y_{j} = ∑_{i=1}^{m} z_{ij}u_{u}. Then y_{j} ∈ K_{0} and therefore v(y_{j}π^{j}) ≠ v(y_{l}π^{l}) if j ≠ l and y_{j}, y_{l} ≠ 0. Hence

v( ∑_{j=0}^{e−1} ( ∑_{i=1}^{m} z_{ij}u_{i})π^{j}) = min_{0≤j≤e−1} v(y_{j}π^{j}).

It suffices therefore to prove that for a_{1}, …, a_{m} ∈ ℚ_{p} we have

v( ∑_{i=1}^{m} a_{i}u_{i}) = min_{1≤i≤m} v(a_{i}).

To this end, suppose without loss that a_{1} ≠ 0 and v(a_{1}) ≤ v(a_{i}), i = 1, 2, …, m. Then divide ∑_{i=1}^{m} a_{i}u_{i} by a_{1}, if necessary, to assume that a_{1} = 1 and v(a_{i}) ≥ 0 for i = 1, …, m. It follows that v(∑_{i=1}^{m} a_{i}u_{i}) = 0. Otherwise v(∑_{i=1}^{m} a_{i}u_{i}) > 0. Hence, reducing modulo p we would get a contradiction 1+ā_{2}ū_{2} +⋯+ā_{m}ū_{m} = 0 to the choice of u_{1}, …, u_{m}. Lemma 8.3: Let f(X, T) = a_{0}(X)T^{d} + a_{1}(X)T^{d−1} + ⋯ + a_{d}(X) ∈ ℚ_{p}[X_{1}, …, X_{m}, T] be a polynomial and let n ∈ ℕ. Then there exists a finite partition of ℚ_{p}^{m+1} into subsets A of the form

(5) A = {(x, t) ∈ ℚ_{p}^{m+1}| x ∈ C and v(t − c_{i}(x)) ⋄_{il} v(a_{il}(x)), for i ∈ I, l ∈ I_{i}}

where C is a definable subset of ℚ_{p}^{m}, ⋄_{il} denotes either ≤ or ≥, I and I_{i} are finite sets, and c_{i}, a_{il} are definable functions from ℚ_{p}^{m} to ℚ_{p}, such that for each (x, t) ∈ A we have

(6)  f(x, t) = u(x, t)^{n}h(x) ∏_{i∈I}(t − c_{i}(x))^{e_{i}},

with u(x, t) a unit in ℤ_{p}, h: ℚ_{p}^{m} → ℚ_{p} a definable function, and e_{i} ≥ 0. Proof: Consider the set Z = {(x, t) ∈ ℚ_{p}^{m+1}| a_{0}(x) = 0}. By induction on d it suffices to find a partition of Z′ = ℚ_{p}^{m+1} − Z. In several steps we partition Z′ into subsets which satisfy the conditions of the Lemma. Part A: Decomposition of f. Let K be the compositum of all extensions of ℚ_{p} of degree at most d. By Lemma 8.2(a), K is a finite extension of ℚ_{p}. Take a basis ξ_{1}, …, ξ_{k} for K/ℚ_{p} as in Lemma 8.2(b). Consider the following definable subset of ℚ_{p}^{m+kd}:

(7) R = {(x, y) ∈ ℚ_{p}^{m+kd}| a_{0}(x) ≠ 0 and f(x, T) = a_{0}(x) ∏_{i=1}^{d}(T − ∑_{j=1}^{k} y_{ij}ξ_{j})}.

For each (x, t) ∈ Z′ there exists y = (y_{1}, …, y_{d}) ∈ ℚ_{p}^{kd} with y_{i} = (y_{i1}, …, y_{ik}) such that (x, y) ∈ R, and y is uniquely determined up to a permutation of y_{1}, …, y_{d}. Hence, in the notation of Lemma 8.1, 0 < |R_{x}| < d!. It follows from Lemma 8.1 that there exist definable functions b_{ij}: Z′ → ℚ_{p} such that if a_{0}(x) ≠ 0, then

f(x, T) = a_{0}(x) ∏_{i=1}^{d}(T − b_{i1}(x) − ∑_{j=2}^{k} b_{ij}(x)ξ_{j}).

Part B: Partition of Z′. Let (I_{1}, I_{2}, I_{3}) range over all partitions {1, 2, …, d} = I_{1} ∪· I_{2} ∪· I_{3} of {1, 2, …, d}. By the choice of ξ_{1}, …, ξ_{k} each partition determines a subset B = B(I_{1}, I_{2}, I_{3}) of Z′ such that for (x, t) ∈ B we have:

(8)  v(t − b_{i1}(x) − ∑_{j=2}^{k} b_{ij}(x)ξ_{j}) = \{ \begin{matrix}v(t − b_{i1}(x)) ≠ ∞ & for i ∈ I_{1} \\ v(b_{i,j(i)}(x)) + v(ξ_{j}) ≠ ∞ & for i ∈ I_{2} \\ ∞ & for i ∈ I_{3} \end{matrix}

where

(9)

v(t − b_{i1}(x)) ≤ v(b_{ij}(x)) + v(ξ_{j}) for i ∈ I_{1}, j = 2, …, k
v(b_{i,j(i)}(x)) + v(ξ_{j(i)}) ≤ v(t − b_{i1}(x)), v(b_{ij}(x)) + v(ξ_{j}) for i ∈ I_{2}, j = 2, …, k.

Since 0 ≤ v(ξ_{j}) < 1 and the values of all other terms of (9) lie in ℤ we may improve (9) to

(10)

v(t − b_{i1}(x)) ≤ v(b_{ij}(x)) for i ∈ I_{1}, j = 2, …, k
v(b_{i,j(i)}(x)) ≤ v(t − b_{ij}(x)), b_{ij}(x) for i ∈ I_{2}, j = 2, …, k.

The collection of all B forms a partition of Z′. So, we have to partition each B separately to sets having the form (5).

Part C: Redecomposition of f. If I_{3} ≠ ∅, then f is zero on B and we may take h in (6) as the zero function. So, suppose that I_{3} = ∅.

Let (x, t) ∈ B. Then substitute t for T in (7) and (8) to rewrite f(x, t) as:

(11)  f(x, t) = a_{0}(x) ∏_{i∈I_{1}}(t − b_{i1}(x)) ∏_{i∈I_{2}} b_{i,j(i)}(x) ∏_{i=1}^{d} c_{i}(x, t),

where

(12)  c_{i}(x, t) = \{ \begin{matrix}1 + ∑_{j=2}^{k} \frac{b_{ij}(x)}{t−b_{i1}(x)}ξ_{j} & for i ∈ I_{1} \\ \frac{t−b_{i1}(x)}{b_{i,j(i)}(x)} − ∑_{j=2}^{k} \frac{b_{ij}(x)}{b_{i,j(i)}(x)}ξ_{j} & for i ∈ I_{2}, \end{matrix}

with (by (10))

(13)  \frac{b_{ij}(x)}{t − b_{i1}(x)} ∈ ℤ_{p} for i ∈ I_{1}, j = 2, …, k

and

(14)  \frac{t − b_{i1}(x)}{b_{i,j(i)}(x)} ∈ ℤ_{p},  \frac{b_{ij}(x)}{b_{i,j(i)}(x)} ∈ ℤ_{p} for i ∈ I_{2}, j = 2, …, k

Note that the coefficient of ξ_{j(i)} in the lower right hand side of (12) is 1. Hence by (4),

(15)  v(c_{i}(x, t)) = 0 for i ∈ I_{1} and 0 ≤ v(c_{i}(x, t)) < 1 for i ∈ I_{2}.
Let λ = 2v(n) + 1.

Part D: We can partition B into subsets A of the form (5) such that on each subset the residue classes of (13) and (14) modulo p^{λ+d} are fixed. Indeed, let i ∈ I_{1}. For each e between 0 and p^{λ+d} − 1 we have to express the condition

(16)  \frac{b_{ij}(x)}{t − b_{i1}(x)} ≡ e mod p^{λ+d},

in the form which is involved in (5). If e = 0, (16) is equivalent to v(t − b_{i1}(x)) ≤ v(p^{−λ−d}b_{ij}(x)). So, suppose that e ≠ 0. Then (16) implies

(17)  v(t − b_{i1}(x)) = v(b_{ij}(x)) − v(e).

A straightforward calculation (not using (17)) shows that (16) is equivalent to

(18)  v(t − b_{i1}(x) − e^{−1}b_{ij}(x)) ≥ λ + d − v(e) + v(t − b_{i1}(x)).

Substitute (17) in (18) to conclude that (16) is equivalent to the conjunction of (17) and

(19)  v(t − b_{i1}(x) − e^{−1}b_{ij}(x)) ≥ v(p^{λ+d}e^{−2}b_{ij}(x)).

Note that the value e_{i} = 0 is allowed in (6). Hence, we may multiply the right hand side of (11) by the factor (t − b_{i1}(x) − e^{−1}b_{ij}(x))^{0}.

If i ∈ I_{2}, then the condition

\frac{t − b_{i1}(x)}{b_{i,j(i)}(x)} ≡ e mod p^{λ+d}

is equivalent to v(t−b_{i1}(x)−b_{i,j(i)}(x)e) ≥ v(p^{λ+d}b_{i,j(i)}(x)). Again, we can multiply the right hand side of (11) by the appropriate factor to exponent 0. Finally, the condition

\frac{b_{ij}(x)}{b_{i,j(i)}(x)} ≡ e mod p^{λ+d}

is a definable condition on x and it is allowed in (5). Part E: Conclusion of the proof. It follows from (12), (4), and (15) that for (x, t) ∈ A and 1 ≤ i ≤ d we have 0 ≤ v(c_{i}(x, t)) < 1. Hence, on A we have c(x, t) = ∏_{i=1}^{d} c_{i}(x, t) ∈ ℤ_{p} and v(c(x, t)) < d. Similarly c(x, t) ≡ c_{0} mod p^{λ+d} for some fixed c_{0} ∈ ℤ with v(c_{0}) = v(c(x, t)) < d. Hence, v((c(x, t) − c_{0})c_{0}^{−1})) ≥ λ and, by Corollary 2.2, there exists a p-adic unit u(x, t) such that

(20)  c(x, t) = c_{0}(1 + \frac{c(x, t) − c_{0}}{c_{0}} ) = c_{0}u(x, t)^{n}.

Finally substitute (20) in (11) to present f(x, t) on A in the form (6).

The following result improves Lemma 8.3. Proposition 8.4: Let f_{i} ∈ ℚ_{p}[X_{1}, …, X_{m}, T], i = 1, …, r, be polynomials and fix n ∈ ℕ. Then there exists a finite partition of ℚ_{p}^{m+1} into subsets A of the form

(21)  A = {(x, t) ∈ ℚ_{p}^{m+1}| x ∈ C and v(a_{1}(x)) ⋄_{1} v(t − c(x)) ⋄_{2} v(a_{2}(x))},

where C is a definable subset of ℚ_{p}^{m}, each of the symbols ⋄_{1} and ⋄_{2} denotes either ≤, or <, or no condition, and a_{1}, a_{2}, c: ℚ_{p}^{m} → ℚ_{p} are definable functions, such that for each (x, t) ∈ A we have

f_{i}(x, t) = u_{i}(x, t)^{n}h_{i}(x)(t − c(x))^{ν_{i}}, i = 1, …, r,

with u_{i}(x, t) a unit in ℤ_{p}, h_{i}: ℚ_{p}^{m} → ℚ_{p} are definable functions, and ν_{i} ≥ 0. Proof: Use Lemma 8.3 to partition ℚ_{p}^{m+1} into sets of the form (5) separately with respect to each f_{i} and then to make a finer partition into sets A of the form (5) such that for each (x, t) ∈ A

f_{i}(x, t) = u_{i}(x, t)^{n}h_{i}(x) ∏_{j∈I}(t − c_{j}(x))^{e_{ij}}, i = 1, …, r,

with u_{i}(x, t) a unit in ℤ_{p}, h_{i}: ℚ_{p}^{m} → ℚ_{p}, and e_{ij} ≥ 0. Note that we may use the option to set e_{ij} = 0 in order to make I independent of i.

It suffices to prove that we can take I to contain only one element, say 1. In this case we may further partition each A to subsets B which have the form (21), where a_{1} and a_{2} are chosen among the functions a_{1l}, with l ∈ I.

Suppose that 1, 2 ∈ I. Thus t − c_{1}(x) and t − c_{2}(x) may appear in (5) and (6). We plan to eliminate from (5) and (6) either t − c_{1}(x) or t − c_{2}(x), or eliminate both and introduce a new t − c(x). Then we may use induction on |I| to conclude the proof.

After a partition of A into the set {(x, t) ∈ A| c_{1}(x) = c_{2}(x)} and its complement in A, we may suppose that c_{1}(x) ≠ c_{2}(x) for all (x, t) ∈ A. Set λ = 2v(n) + 1. By Corollary 2.2, if u ≡ 1 mod p^{λ}, then u in an nth power in ℚ_{p}.

Use the identity

(22)  \frac{p^{λ}(t − c_{1}(x))}{c_{2}(x) − c_{1}(x)} = \frac{p^{λ}(t − c_{2}(x))}{c_{2}(x) − c_{1}(x)} + p^{λ}

to distinguish between four cases for (x, t) ∈ A:

Case A: v(\frac{t−c_{1}(x)}{c_{2}(x)−c_{1}(x)}) ≥ λ. In this case

t − c_{2}(x) = t − c_{1}(x) − (c_{2}(x) − c_{1}(x))

= (t − c_{1}(x))(1 − \frac{c_{2}(x) − c_{1}(x)}{t − c_{1}(x)}) = (t − c_{1}(x))u(x, t)^{n},

with u(x, t) a unit. Thus we can eliminate t − c_{2}(x).

Case B: v(\frac{t−c_{1}(x)}{c_{2}(x)−c_{1}(x)}) < −λ. In this case

t − c_{2}(x) = t − c_{1}(x) − (c_{2}(x) − c_{1}(x))

= (t − c_{1}(x))(1 − \frac{c_{2}(x) − c_{1}(x)}{t − c_{1}(x)}) = (t − c_{1}(x))u(x, t)^{n},

with u(x, t) a unit. Thus we can eliminate t − c_{2}(x).

Case C: \frac{p^{λ}(t−c_{2}(x))}{c_{2}(x)−c_{1}(x)} ≡ 0 mod p^{2λ}. In this case

\frac{t − c_{1}(x)}{c_{2}(x) − c_{1}(x)} ≡ 1 mod p^{λ}.

Hence t − c_{1}(x) = (c_{2}(x) − c_{1}(x))u(x, t)^{n} with u(x, t) a unit. Thus we can eliminate t − c_{1}(x).

Suppose now that neither of the cases A, B, and C occur. Then the right hand side of (22) has a nonnegative value less than 2λ. Moreover, there exists a unique integer a modulo p^{3λ} such that a ≢ 0 mod p^{2λ} and a ≢ −p^{λ} mod p^{2λ} and

Case D_{a}: \frac{p^{λ}(t−c_{2}(x))}{c_{2}(x)−c_{1}(x)} ≡ a mod p^{3λ}. In this case v(p^{3λ}(a + p^{λ})^{−1}) ≥ 2λ and therefore there exists b ∈ ℤ_{p} and a unit u(x, t) of ℤ_{p} such that

\frac{p^{λ}(t − c_{2}(x))}{c_{2}(x) − c_{1}(x)} + p^{λ} = a + p^{λ} + bp^{3λ} = (a + p^{λ})(1 + \frac{bp^{3λ}}{a + p^{λ}} ) = (a + p^{λ})u(x, t)^{n}.

Hence, by (22)

(23)  t − c_{1}(x) = p^{−λ}(c_{2}(x) − c_{1}(x))(a + p^{λ})u(x, t)^{n}.

Similarly

\frac{p^{λ}(t − c_{2}(x))}{c_{2}(x) − c_{1}(x)} = a + bp^{3λ} = a(1 + \frac{bp^{3λ}}{a} ) = aw(x, t)^{n}

with w(x, t) a unit of ℤ_{p}. Hence,

(24)  t − c_{2}(x) = p^{−λ}(c_{2}(x) − c_{1}(x))aw(x, t)^{n}.

Now use (23) and (24) to eliminate both t − c_{1}(x) and t − c_{2}(x). However, to express the Condition D_{a} we need a new t − c(x). Specifically, Condition D_{a} is equivalent to

v(t − c_{2}(x) − ap^{−λ}(c_{2}(x) − c_{1}(x))) ≥ v(p^{2λ}(c_{2}(x) − c_{1}(x)).

9. Rationality of a power series. In Section 10 we show that the rationality of the Zeta function depends on the rationality of a power series of the form we treat in the following lemma.

Lemma 9.1 (Meuser – Denef): Let L be the set of all m-tuples (k_{1}, …, k_{m}) of integers that satisfy a finite system of linear inequalities with integral coefficients. Let A_{1}(s), …, A_{m}(s) be linear polynomials in s with integral coefficients. Let p ∈ ℝ, p > 1. Suppose that the series

(1)  J(s) = ∑_{k∈L} p^{−k_{1}A_{1}(s)−···−k_{m}A_{m}(s)}

converges in an open subset S of ℝ. Then J(s) is a rational function of p^{−s} on S.

Proof: We carry out the proof by induction on m. Its first four parts are reductions steps. Note that since the summands of the series in (1) are positive, each subseries converges on S. If all A_{i} are identically zero on S, there is nothing to prove. So, suppose that A_{m} is not identically zero.

Part A: We may suppose that k_{1}, …, k_{m} ≥ 0 if (k_{1}, …, k_{m}) ∈ L. Indeed, suppose that

L = {k ∈ ℤ^{m}| ∑_{j=1}^{m} a_{ij}k_{j} ≤ a_{i}, i = 1, …, r},

where a_{ij}, a_{i} ∈ ℤ. Separate L into a disjoint union of the sets

L_{ε} = {k ∈ L| sgn(k_{j}) = ε_{j}, j = 1, …, m},

where ε = (ε_{1}, …, ε_{m}) ranges over the set {−1, 0, 1}^{m}. For each such ε let k_{j}^{∗} = ε_{j}k_{j}, j = 1, …, m. Then k_{j}^{∗} ≥ 0, and ε_{k}k_{j}^{∗} = k_{j}, j = 1, …, m, and

L_{ε} = {k^{∗} ∈ ℤ^{m}| ∑_{j=1}^{m}(ε_{j}a_{ij})k_{j}^{∗} ≤ a_{i}, i = 1, …, r}.

Let also

J_{ε}(s) = ∑_{k^{∗}∈L_{ε}} p^{−k_{1}^{∗}(ε_{1}A_{1}(s))−···−k_{m}^{∗}(ε_{m}A_{m}(s))}.

Then J(s) = ∑_{ε∈{−1,0,1}^{m}} J_{ε}(s). So, if we prove that each J_{ε}(s) is a rational function of p^{−s}, then so will be J(s). Part B: Simplifying L. The system of inequalities which determines L consists of inequalities that do not involve k_{m} and inequalities of the form

(2a)  b_{i}k_{m} ≥ B_{i}(k_{1}, …, k_{m−1}), i = 1, …, l
(2b)  c_{j}k_{m} ≤ C_{j}(k_{1}, …, k_{m−1}), j = 1, …, l′

where b_{i}, c_{j} ∈ ℕ, and B_{i}, C_{j} linear polynomials with integral coefficients in k_{1}, …, k_{m−1}. Since the series in (1) converges for s ∈ S, condition (2) is nonempty, i.e., l ≥ 1 or l′ ≥ 1. If l = 0, we may, by Part A, take B_{1} = 0, and so assume that l ≥ 1. For j_{0} = 1, 2 let L_{j_{0}} be the set of all k ∈ ℤ^{m} which satisfy each inequality involved in L but (2b) for j = j_{0}.

Suppose that l′ ≥ 2. Then c_{1}, c_{2} > 0 and

L = \{k ∈ L_{2}| k_{m} ≤ \frac{1}{c_{1}} C_{1}(k_{1}, …, k_{m−1}), \frac{1}{c_{1}} C_{1}(k_{1}, …, k_{m−1}) ≤ \frac{1}{c_{2}} C_{2}(k_{1}, …, k_{m−1})\}

∪ · \{k ∈ L_{1}| k_{m} ≤ \frac{1}{c_{2}} C_{2}(k_{1}, …, k_{m−1}), \frac{1}{c_{2}} C_{2}(k_{1}, …, k_{m−1}) < \frac{1}{c_{1}} C_{1}(k_{1}, …, k_{m−1})\}

= {k ∈ L_{2}| c_{1}k_{m} ≤ C_{1}(k_{1}, …, k_{m−1}), c_{2}C_{1}(k_{1}, …, k_{m−1}) ≤ c_{1}C_{2}(k_{1}, …, k_{m−1})} ∪ · {k ∈ L_{1}| c_{2}k_{m} ≤ C_{2}(k_{1}, …, k_{m−1}), c_{1}C_{1}(k_{1}, …, k_{m−1}) + 1 ≤ c_{2}C_{1}(k_{1}, …, k_{m−1})}

So, we may separate the series in (1) into a sum of two series each with at most l′ − 1 inequalities of the form (2b) (but with an additional inequality in k_{1}, …, k_{m−1}). Likewise we may reduce the number of inequalities of the form (2a). After finitely many steps, we may suppose that either l′ = 0 or l′ = 1. Similarly we may suppose that l = 1. Write B for B_{1} and C for C_{1}.

If l′ = 1, multiply (2a) by c_{1} and (2b) by b_{1}, if necessary, to assume that b_{1} = c_{1} = c > 0. Then add the inequality B(k_{1}, …, k_{m−1}) ≤ C(k_{1}, …, k_{m−1}) to the inequalities in k_{1}, …, k_{m−1}, and note that J(s) is not changed because all other (k_{1}, …, k_{m−1}) do not contribute to the series. If l′ = 0, let c = b_{1} > 0.

After these changes inequalities (2) get the form

(3)  β_{0} + ∑_{i=1}^{m−1} β_{i}k_{i} ≤ ck_{m} ≤ γ_{0} + ∑_{i=1}^{m−1} γ_{i}k_{i},

with β_{i}, γ_{i} ∈ ℤ, i = 0, …, m − 1.

Part C: We may assume that c = 1. Write k_{i} = ck′_{i} + r_{i} with k′_{i} ∈ ℤ, k′_{i} ≥ 0, and 0 ≤ r_{i} < c, i = 1, …, m − 1. Then separate the series in (1) into partial series, each with a system of residues (r_{1}, …, r_{m−1}) fixed, if necessary, to assume that r_{1}, …, r_{m−1} are fixed for k ∈ L. Next let

β_{0} + ∑_{i=1}^{m−1} β_{i}r_{i} = βc + r and γ_{0} + ∑_{i=1}^{m−1} γ_{i}r_{i} = γc + s,

with 0 < r ≤ c and 0 ≤ s < c. Substitute these values in (3):

(4)  r + βc + ∑_{i=1}^{m−1} β_{i}ck′_{i} ≤ ck_{m} ≤ s + γc + ∑_{i=1}^{m−1} γ_{i}ck′_{i}.

Finally observe that (4) is equivalent to

(5)  β + ∑_{i=1}^{m−1} β_{i}k′_{i} ≤ k_{m} ≤ γ + ∑_{i=1}^{m−1} γ_{i}k′_{i}.

Note that if l′ = 0, then the right hand side of (5) does not exist.

Part D: Conclusion of the proof. Multiply J(s) by p^{r_{1}A_{1}(s)···+r_{m−1}A_{m−1}(s)} and A_{i}(s) by c if necessary to rewrite J(s) as

(6)  J(s) = ∑_{k′∈L′} p^{−k′_{1}A_{1}(s)−···−k′_{m−1}A_{m−1}(s)} ∑_{B(k′)≤k_{m}≤C(k′)} p^{−k_{m}A_{m}(s)},

where L′ is a set of (m − 1)-tuples k′ = (k_{1}, …, k_{m−1}) of integers that satisfy a finite system of linear inequalities with integral coefficients, B(k′) is the left hand side of (5), and C(k′) is its right hand side.

For s ∈ S_{0} = {s ∈ S| A_{m}(s) ≠ 0}, we have

(7) \begin{aligned}J(s) & = ( ∑_{k′∈L′} p^{−k′_{1}A_{1}(s)−···−k′_{m−1}A_{m−1}(s)−B(k′)A_{m}(s)})\frac{1}{1 − p^{−A_{m}(s)}} \\ & − ( ∑_{k′∈L′} p^{−k′_{1}A_{1}(s)−···−k′_{m−1}A_{m−1}(s)−C(k′)A_{m}(s)})p^{−A_{m}(s)} \frac{1}{1 − p^{−A_{m}(s)}} \end{aligned}

Note that if the right hand side of (5) does not exist, then, from the convergence of (6), it follows that A_{m}(s) > 0 on S and there is no second series in (7).

Both series in (7) converge because they are partial series of the converging series (6). Hence we can apply the induction hypothesis to conclude that J(s) is a continuous function of t = p^{−s} on S_{0} (which is S minus at most one point s_{0} and therefore also an open subset of ℝ). In addition we may rewrite J(s) from (1) as a Laurent series in t which converges in an annalus D with center in the origin except possibly in t_{0} = p^{−s_{0}}. From the convergence of J(s) on S, it follows that J(s), as a function of t, is analytic on D and in particular in a neighborhood of t_{0}. Hence, J(s) is a rational function of p^{−s} on all S.

10. Rationality of the Zeta function. We still need another lemma before we can prove the rationality of the zeta function from which the rationality of the Poincaré series of Section 6 follows.

Lemma 10.1: Let K be a field and d a positive integer. Then, any rational function of t^{1/d} with coefficients in K which can be expressed as a power series in t with finitely many negative terms is a rational function in t.

Proof: Consider the discrete valuation w of K(t) which is trivial on K and w(t) = 1. This valuation totally ramifies in K(t^{1/d}) but is unramified in K((t)). Hence, K(t^{1/d}) ∩ K((t)) = K, which implies the claim of the lemma.

Theorem 10.2: Let S be a definable subset of ℚ_{p}^{m} which is contained in a compact subset. Let h: ℚ_{p}^{m} → ℚ_{p} be a definable function such that |h(x)| is bounded on S. Let e ∈ ℕ. Suppose that v(h(x)) ∈ eℤ ∪ {∞} for all x ∈ S. Then

Z_{S}(s) = ∫_{S} |h(x|^{s/e}|dx|, s ∈ ℝ, s > 0

is a rational function of p^{−s}.

Proof: Let z = (x_{1}, …, x_{m−1}), t = x_{m} and x = (z, t). We first separate t in the integral Z_{S}(s).

By Corollary 7.3, v(h(x)) is a simple function on S. Hence, so is \frac{1}{e}v(h(x)). Since the integral on disjoint union of sets S_{i} is the sum of the integrals on S_{i}, we may suppose that there exist e′ ∈ ℕ and g_{1}, g_{2} ∈ ℚ_{p}[X] such that g_{2}(x) ≠ 0 and

(1)  |h(x)|^{1/e} = |\frac{g_{1}(x)}{g_{2}(x)}|^{1/e′} ,  for all x ∈ S.

By Corollary 5.7, S is a union of intersections of sets of the form (2) of Section 5. Since ∫_{A∪B} = ∫_{A} + ∫_{B} − ∫_{A∩B}, we may suppose that S is the set of all x ∈ ℚ_{p}^{m} satisfying

(2)  ∃y_{j}: f_{j}(x) = c_{j}y_{j}^{d_{j}},  for j = 1, …, l

where f_{j} ∈ ℚ_{p}[X], c_{j} ∈ ℤ, and d_{j} ∈ ℕ.

Apply Proposition 8.4 to the polynomials g_{1}, g_{2}, f_{1}, …, f_{l}, with m replaced by m − 1, and x replaced by z, and n = d_{1}d_{2} ⋯ d_{l}. This proposition gives a partition of S into subsets S ∩ A. Each A is the set of all x ∈ ℚ_{p}^{m} satisfying z ∈ C and

(3)  v(a_{1}(z)) ⋄_{1} v(t − c(z)) ⋄_{2} v(a_{2}(z)),

where C is a definable subset of ℚ_{p}^{m−1}, a_{1}, a_{2}, c: ℚ_{p}^{m−1} → ℚ_{p} are definable functions and each of the symbols ⋄_{i} denotes either ≤ or < or no condition. Moreover, for each x ∈ S ∩ A we have

g_{i}(x) = u′_{i}(z)^{n}h′_{i}(z)(t − c(z))^{ν′_{i}},  i = 1, 2
f_{j}(x) = u_{j}(z)^{n}h_{j}(z)(t − c(z))^{ν_{j}},  j = 1, …, l

where u′_{i}(z), u_{j}(z) are units of ℤ_{p}, h′_{i}, h_{j}: ℚ_{p}^{m−1} → ℚ_{p} are definable functions, and ν′_{i}, ν_{j} are nonnegative integers.

Hence

|h(x)|^{1/e} = |h_{0}(z)|^{1/e′}|t − c(z)|^{ν/e′}

with ν ∈ ℤ and h_{0}: ℚ_{p}^{m−1} → ℚ_{p} a definable function. Also, for all x ∈ S ∩ A condition (2) is equivalent to

(2′)  ∃y: h_{j}(z)(t − c(z))^{ν_{j}} = c_{j}y^{d_{j}},  j = 1, …, l.

We partition S ∩ A into subsets S′ on which h_{j}(z) and t − c(z) have constant nth power residue. Such an S′ is the set of all x ∈ ℚ_{p}^{m} satisfying (3), z ∈ D, and

(4)  ∃y: t − c(z) = by^{n},

where D is a definable subset of C and b is an element of ℚ_{p} such that (2′) holds. Then

\begin{aligned}Z_{S′}(s) & = ∫_{S′} |h(x)|^{s/e}|dx| \\ & = ∫_{D} |h_{0}(z)|^{s/e′} ∫_{(3),(4)} |t − c(z)|^{νs/e′}|dt||dz|. \end{aligned}

Change now the variable t in the inner integral to w = t − c(z). Rewrite conditions (3) and (4) as

(3′)  v(a_{1}(z)) ⋄_{1} v(w) ⋄_{2} v(a_{2}(z))
(4′)  ∃y: w = by^{n}

Then

Z′_{S}(s) = ∫_{D} |h_{0}(z)|^{s/e′} ∫_{(3′),(4′)} |w|^{νs/e′}|dw||dz|.

Condition (3′) is equivalent to the disjoint disjunction of conditions v(w) = k_{m} on all integers k_{m} satisfying

(5)  v(a_{1}(z)) ⋄_{1} k_{m} ⋄_{2} v(a_{2}(z)).

Hence,

(6)  Z_{S′}(s) = ∫_{D} |h_{0}(z)|^{s/e′} ( ∑_{{k_{m}∈ℤ \atop (5)}} p^{−k_{m}νs/e′} ∫_{{v(w)=k_{m} \atop (4′)}} |dw|) |dz|.

Make a multiplicative change in the variable w by w = p^{k_{m}}u. This changes (4′) to

(4′′)  ∃y: u = bp^{−k_{m}}y^{n}

and

(7)  ∫_{{v(w)=k_{m} \atop (4′)}} |dw| = p^{−k_{m}} ∫_{{v(u)=0 \atop (4′′)}} |du|.

The latter integral is zero unless k_{m} ≡ v(b) mod n. Thus b = b_{0}p^{k_{m}+qn} with v(b_{0}) = 0 and q ∈ ℤ. Then (4′′) is equivalent to the condition ∃y: u = b_{0}y^{n} which is independent of k_{m}. Hence the integral has value γ ∈ ℚ_{p} which is independent of k_{m}. So,

Z_{S′}(s) = γ ∫_{D} |h_{0}(z)|^{s/e′} (  ∑_{{k_{m}≡v(b) mod n \atop (5)}} p^{−k_{m} \frac{νs}{e′} −k_{m}} ) |dz|

= γ ∑_{k_{m}≡v(b) mod n} p^{−q_{m}k_{m}s−k_{m}} ∫_{D,(5)} |h_{0}(z)|^{s/e′}|dz|.

where q_{m} = ν/e′. Now apply the same procedure to the integral

∫_{D,(5)} |h_{0}(z)|^{s/e′}|dz|.

Note that D is independent of k_{m}. However when we separate the variable z_{m−1} = x_{m−1} we also apply Proposition 8.4 to a_{1}(z) and a_{2}(z). Condition (5) gives then a condition which depends on an integer k_{m−1} and has the form

v(a′_{1}(x_{1}, …, x_{m−2})) ⋄_{1} k_{m} + ν′k_{m−1} ⋄_{2} v(a′_{2}(x_{1}, …, x_{m−2})),

where a′_{1}, a′_{2}: ℚ_{p}^{m−2} → ℚ_{p} are definable functions and ν′ ∈ ℤ. We also get another condition of the form (5) on k_{m−1} and x_{1}, …, x_{m−2}.

Continue in this way to finally express Z_{S}(s) as a linear combination of convergent series of the form

(8)  ∑_{{k_{i}≡μ_{i} mod n_{i} \atop (k_{1},...,k_{m})∈L}} p^{(−q_{1}k_{1}−···−q_{m}k_{m})s−k_{1}−···−k_{m}},

where q_{1}, …, q_{m} ∈ ℚ, μ_{1}, …, μ_{m} are nonnegative integers, n_{1}, …, n_{m} ∈ ℕ, and L is the set of all (k_{1}, …, k_{m}) ∈ ℤ^{m} that satisfy a finite system of linear inequalities with integral coefficients.

Let d be a common denominator of the rational numbers q_{1}, …, q_{m}. Write s′ = s/d, q_{i} = q′_{i}/d with q′_{i} ∈ ℤ, and k_{i} = μ_{i} + n_{i}l_{i}, i = 1, …, m. Then L is equivalent to a finite system L′ of linear inequalities in l_{1}, …, l_{m} and (8) is equal to

(9)  (p^{−s′})^{∑_{i=1}^{m} q′_{i}μ_{i}}p^{∑_{i=1}^{m} μ_{i}} ∑_{(l_{1},...,l_{m})∈L′} p^{− ∑_{i=1}^{m} l_{i}(q′_{i}n_{i}s′+n_{i})}.

By Lemma 9.1, (9) is a rational function of p^{−s′}. Hence Z_{S}(s) is a rational function of t^{1/d} where t = p^{−s}.

On the other hand

(10)  Z_{S}(s) = ∑_{k∈ℤ} t^{k} ∫_{{v(h(x))=ek \atop x∈S}} |dx|.

Since |h(x)| is bounded on S, only finitely many negative k appear in (10). Conclude from Lemma 10.1 that Z_{S}(s) is a rational function of t.

References

[AK] J. Ax and S. Kochen, Diophantine problems over local fields II, American Journal of Mathematics 87 (1965), 631–648.

[D] J. Denef, The rationality of the Poincaré series associated to the p-adic points on a variety, Inventiones mathematicae 77 (1984), 1–23.

[FJ] M.D. Fried and M. Jarden, Field Arithmetic, Ergebnisse der Mathematik III 11, Springer, Heidelberg, 1986.

[J] M. Jarden, Intersection of local algebraic extensions of a Hilbertian field,

[L] S. Lang, Algebraic Number Theory, Addison-Wesley, Reading, 1970.

[PR] A. Prestel and P. Roquette, Formally p-adic Fields, Lecture Notes in Mathematics 1050, Springer, Berlin, 1984.

20 May, 2000