Abhyankar places admit local uniformization in any characteristic

— preliminary version —

Franz–Viktor Kuhlmann and Peter Roquette

17. 5. 2000

Abstract We prove that every place of an algebraic function field F |K of arbitrary characteristic admits local uniformization, provided that the sum of the rational rank of its value group and the transcendence degree of its residue field over K is equal to the transcendence degree of F |K. Further, we show that finitely many such places admit simultaneous local uniformization if they have isomorphic value groups. We also prove the arithmetic analogues for extensions of p-adic places, and a generalization.

1 Introduction

In [Z], Zariski proved the Local Uniformization Theorem for places of algebraic function fields over base fields of characteristic 0. In [Z3], he uses this theorem to prove resolution of singularities for surfaces in characteristic 0. As the resolution of singularities for arbitrary dimensions in positive characteristic is still an open problem, one is interested in generalizations of the Local Uniformization Theorem to positive characteristic.

Throughout this paper, “function field” will always mean “algebraic function field”. By a “place of F |K” we mean a place whose restriction to K is the identity. The valuation associated with P will be denoted by v_{P} , its value group on any field F by v_{P} F and its residue field on F by F v_{P} or F P. The v_{P} -value of an element a will be denoted by v_{P} a and its v_{P} -residue by av_{P} or aP.

For every place P of a function field F |K, we have the following inequality:

trdeg F |K ≥ trdeg F P |K + dim_{ℚ} ℚ ⊗ v_{P} F .  (1)

This is a special case of the Abhyankar inequality. Note that dim_{ℚ} ℚ⊗v_{P} F is the rational rank of the value group v_{P} F, i.e., the maximal number of rationally independent elements in v_{P} F. We call P an Abhyankar place if equality holds in (1).

In this paper, we will prove the Local Uniformization Theorem for Abhyankar places in arbitrary characteristic:

Theorem 1.1 Let F |K be a function field of arbitrary characteristic, and P an Abhyankar place of F |K such that F P |K is separable. Further, take any elements ζ_{1}, …, ζ_{m} in the valuation ring O_{F} of P on F. Then there exists a model of F |K on which P is centered at a smooth point whose local ring contains ζ_{1}, …, ζ_{m}.

Theorem 1.1 follows from the following structure theorem for function fields with Abhyankar places.

Theorem 1.2 Under the assumptions of Theorem 1.1, there exist a transcendence basis {t_{1}, …, t_{s}} ⊂ O_{F} of F |K, elements z ∈ K[t_{1}, …, t_{s}] and η_{1}, …, η_{n} ∈ O_{F}, and polynomials h_{i}(X) ∈ K[t_{1}, …, t_{s}, z^{−1}][X], 1 ≤ i ≤ n, such that

a) v_{P} z = 0, hence K[t_{1}, …, t_{s}, z^{−1}, η_{1}, …, η_{n}] ⊂ O_{F} ,

b) h_{i}(η_{i}) = 0 and h′_{i}(η_{i})P ≠ 0 for 1 ≤ i ≤ n,

c) F = K(t_{1}, …, t_{s}, η_{1}),

d) ζ_{1}, …, ζ_{m} ∈ K[t_{1}, …, t_{s}, z^{−1}, η_{1}, …, η_{n}].

From condition b) one can deduce

h_{i}P (η_{i}P) = h_{i}(η_{i})P = 0 and (h_{i}P)′(η_{i}P) = h′_{i}(η_{i})P ≠ 0 ,

where h_{i}P denotes the P -reduction of h_{i}, i.e., the polynomial obtained from h_{i} through replacing every coefficient by its P -residue. This shows that η_{i}P is a simple root of h_{i}P. Since this is the condition of Hensel’s Lemma, such an element η_{i} is called v_{P} -henselian. We will discuss henselian elements in Section 5.

Suppose that the assertion of Theorem 1.2 holds for the place P of F |K. Choose polynomials f_{i}(X_{1}, …, X_{n+s+1}) ∈ K[X_{1}, …, X_{n+s+1}], 1 ≤ i ≤ n + 1, such that

1) f_{1}(t_{1}, …, t_{s}, X_{s+1}, …, X_{n+s+1}) = zX_{s+1} − 1,

2) f_{i}(t_{1}, …, t_{s}, z^{−1}, X_{s+2}, …, X_{n+s+1}) = h_{i−1}(X_{s+i}) for 2 ≤ i ≤ n + 1. Then

J := (\frac{∂f_{i}}{∂X_{j}} )_{s+1≤j≤s+n+1^{1≤i≤n+1}} = ( \begin{matrix}z & & & 0 \\ ⋆ & \frac{∂f_{2}}{∂X_{s+2}} \\ ⋮ & & ... \\ ⋆ & 0 & & \frac{∂f_{n+1}}{∂X_{s+n+1}} \end{matrix} ) .

Therefore,

det J(t_{1}P, …, t_{s}P, z^{−1}P, η_{1}P, …, η_{n}P) =

det ( \begin{matrix}zP & & & 0 \\ ⋆ & h′_{1}(η_{1})P \\ ⋮ & & ... \\ ⋆ & 0 & & h′_{n}(η_{n})P \end{matrix} ) = zP · h′_{1}(η_{1})P · . . . · h′_{n}(η_{n})P ≠ 0 .

This means that the place P is centered on K[t_{1}, …, t_{s}, z^{−1}, η_{1}, …, η_{n}] at the smooth point (t_{1}P, …, t_{s}P, z^{−1}P, η_{1}P, …, η_{n}P). Therefore, Theorem 1.1 follows from Theorem 1.2.

In [K3], we show that the Abhyankar places lie dense in the Zariski space of all places of F |K, with respect to a “Zariski patch topology”. This topology is finer than the Zariski topology (but still compact); its basic open sets are the sets of the form

{P | P a place of F |K such that a_{1}P ≠ 0, …, a_{k}P ≠ 0 ; b_{1}P = 0, …, b_{ℓ}P = 0}

with a_{1}, …, a_{k}, b_{1}, …, b_{ℓ} ∈ F \ {0}. (The density actually holds for even stronger topologies; cf. [K3].)

Corollary 1.3 The uniformizable places of F |K lie dense in the Zariski space of F |K, with respect to the Zariski patch topology.

In order to formulate an arithmetic analogue of the foregoing results, we introduce the following notion. Assume that F is a field extension of ℚ of finite transcendence degree. A place P of F will be called an arithmetic Abhyankar place of F if it extends the p-adic place of ℚ and satisfies

trdeg F |ℚ + 1 = trdeg F P |𝔽_{p} + dim_{ℚ} ℚ ⊗ v_{P} F ,  (2)

for some prime p. We will prove the following arithmetic analogue of Theorem 1.2:

Theorem 1.4 Let F be a function field over ℚ and P an arithmetic Abhyankar place of F |ℚ. Further, take any elements ζ_{1}, …, ζ_{m} in the valuation ring O_{F} of P on F. Set s = trdeg F |ℚ. Then there exist a transcendence basis {t_{1}, …, t_{s}} ⊂ O_{F} of F |K, elements t_{s+1} ∈ O_{F} , z ∈ ℤ[t_{1}, …, t_{s+1}] and η_{1}, …, η_{n} ∈ O_{F}, and polynomials h_{i}(X) ∈ ℤ[t_{1}, …, t_{s+1}, z^{−1}][X], 1 ≤ i ≤ n, such that

a) v_{P} z = 0, hence ℤ[t_{1}, …, t_{s+1}, z^{−1}, η_{1}, …, η_{n}] ⊂ O_{F} ,

b) h_{i}(η_{i}) = 0 and h′_{i}(η_{i})P ≠ 0 for 1 ≤ i ≤ n,

c) F = ℚ(t_{1}, …, t_{s+1}, η_{1}),

d) ζ_{1}, …, ζ_{m} ∈ ℤ[t_{1}, …, t_{s+1}, z^{−1}, η_{1}, …, η_{n}].

Both Theorem 1.2 and Theorem 1.4 follow from a more general structure theorem which we will state now. See Section 3 for the definition of “defectless”.

Theorem 1.5 Assume that R is an integral domain with quotient field K, and that F is a function field over K. Let P be a place of F such that F P |KP is separable, (K, P) is a defectless field with R ⊂ O_{K} , and

trdeg F |K = trdeg F P |KP + dim_{ℚ} ℚ ⊗ (v_{P} F/v_{P} K) .  (3)

Further, take any elements ζ_{1}, …, ζ_{m} in the valuation ring O_{F} of P on F. Then there exist a transcendence basis {t_{1}, …, t_{s}} ⊂ O_{F} of F |K, elements t_{s+1}, …, t_{s+k} ∈ O_{F} , z ∈ R[t_{1}, …, t_{s+k}] and η_{1}, …, η_{n} ∈ O_{F}, and polynomials h_{i}(X) ∈ R[t_{1}, …, t_{s+k}, z^{−1}][X], 1 ≤ i ≤ n, such that

a) v_{P} z = 0, hence R[t_{1}, …, t_{s+k}, z^{−1}, η_{1}, …, η_{n}] ⊂ O_{F} ,

b) h_{i}(η_{i}) = 0 and h′_{i}(η_{i})P ≠ 0 for 1 ≤ i ≤ n,

c) F = K(t_{1}, …, t_{s+k}, η_{1}),

d) ζ_{1}, …, ζ_{m} ∈ R[t_{1}, …, t_{s+k}, z^{−1}, η_{1}, …, η_{n}]. The number k is bounded from above by the maximal number of rationally independent elements in v_{P} K which are bigger than some positive element in v_{P} F \v_{P} K, if this number is finite. If K is of finite transcendence degree k_{0} over its prime field, then k ≤ k_{0} if P is trivial on the prime field, and k ≤ k_{0} + 1 otherwise.

Theorem 1.2 follows from this theorem with R = K and k = 0; note that in this case (K, P) is defectless since P is trivial on K.

Theorem 1.4 follows with R = ℤ and k = 1; in fact, one can take k = 0 and set t_{s+1} = 1, say, if each positive element of v_{P} F \ v_{P} ℚ is bigger than every positive multiple of v_{P} p (P extending the p-adic valuation of ℚ). The latter is equivalent to saying that ℤv_{P} p is a convex subgroup of v_{P} F. Note that F P is automatically separable over ℚP = 𝔽_{p}, and that ℚ together with any p-adic valuation is a defectless field.

Remark 1.6 Here are two criteria for (K, P) to be a defectless field (they follow from Theorem 3.1 below):

a) Suppose that K is a finitely generated field extension of ℚ. If P extends the p-adic place of ℚ for some prime p and satisfies

trdeg K|ℚ + 1 = trdeg KP |𝔽_{p} + dim_{ℚ} ℚ ⊗ v_{P} K ,  (4)

then (K, P) is a defectless field.

b) Suppose that K|K_{0} is a finitely generated field extension. If P is trivial on K_{0} and satisfies

trdeg K|K_{0} = trdeg KP |K_{0} + dim_{ℚ} ℚ ⊗ v_{P} K ,  (5)

then (K, P) is a defectless field.

Remark 1.7 Assume that P is a place of F and Q is a coarsening of P, i.e., a place of F whose valuation ring contains that of P. Then for every element a ∈ F, aP ≠ 0 implies aQ ≠ 0. Thus, assertions a), b), c), d) carry over to every coarsening of P. This shows that if P admits local uniformization, then so does Q.

We conclude with a generalization of Theorem 1.5. It gives simultaneous uniformization for finitely many places which satisfy the assumptions of Theorem 1.5 and have isomorphic value groups.

Theorem 1.8 Assume that R is an integral domain with quotient field K, and that F is a function field over K. Let P be a place of K such that (K, P) is a defectless field with R ⊂ O_{K} . Further, let P_{1}, …, P_{ℓ} be extensions of P to F such that F P_{j}|KP is separable and

trdeg F |K = trdeg F P_{j}|KP + dim_{ℚ} ℚ ⊗ (v_{P_{j}}F/v_{P} K)  (6)

for every j. Assume also that all value groups v_{P_{j}}F are isomorphic over v_{P} K (as ordered groups). Then there exist a transcendence basis {t_{1}, …, t_{s}} ⊂ O of F |K, elements t_{s+1}, …, t_{s+k} ∈ O := ⋂_{j=1}^{ℓ} O_{P_{j}} , z ∈ R[t_{1}, …, t_{s+k}] and η ∈ O, and a polynomial h(X) ∈ R[t_{1}, …, t_{s+k}, z^{−1}][X], such that

a) z is a unit in O, hence R[t_{1}, …, t_{s+k}, z^{−1}, η] ⊂ O,

b) h(η) = 0 and h′(η)P_{j} ≠ 0 for 1 ≤ j ≤ ℓ,

c) F = K(t_{1}, …, t_{s+k}, η). The number k is bounded from above as in Theorem 1.5.

Using this theorem and results from [K3], we deduce that every algebraic variety admits “simultaneous local resolution of singularities at finitely many points”:

Theorem 1.9 Let K be a perfect field and V an affine variety defined over K, with function field F and coordinate ring A. Let 𝔓_{1}, …, 𝔓_{ℓ} be prime ideals of A of equal dimension d. Suppose that trdeg F |K > d′ ≥ d. Then there are places P_{1}, …, P_{ℓ} of F of dimension d′ centered on A at 𝔓_{1}, …, 𝔓_{ℓ} respectively, and a model of F on which P_{1}, …, P_{ℓ} all are centered at smooth points.

Here, the dimension of P is the transcendence degree of F P over K. The places P_{1}, …, P_{ℓ} can be chosen of “maximal rank” trdeg F |K − d′.

2 Valuation independence

For the easy proof of the following theorem, see [B], Chapter VI, §10.3, Theorem 1.

Theorem 2.1 Let (F |K, P) be an extension of valued fields. Take elements x_{i}, y_{j} ∈ F, i ∈ I, j ∈ J, such that the values v_{P} x_{i}, i ∈ I, are rationally independent over v_{P} K, and the residues y_{j}P, j ∈ J, are algebraically independent over KP. Then the elements x_{i}, y_{j}, i ∈ I, j ∈ J, are algebraically independent over K.

Moreover, if we write

f = ∑_{k} c_{k} ∏_{i∈I} x_{i}^{μ_{k,i}} ∏_{j∈J} y_{j}^{ν_{k,j}} ∈ K[x_{i}, y_{j} | i ∈ I, j ∈ J]

in such a way that for every k ≠ ℓ there is some i s.t. μ_{k,i} ≠ μ_{ℓ,i} or some j s.t. ν_{k,j} ≠ ν_{ℓ,j}, then

v_{P} f = min_{k} v_{P} c_{k} ∏_{i∈I} x_{i}^{μ_{k,i}} ∏_{j∈J} y_{j}^{ν_{k,j}} = min_{k} (v_{P} c_{k} + ∑_{i∈I} μ_{k,i}v_{P} x_{i}) .

That is, the value of the polynomial f is equal to the least of the values of its monomials. In particular, this implies:

v_{P} K(x_{i}, y_{j} | i ∈ I, j ∈ J) = v_{P} K ⊕ ⨁_{i∈I} ℤv_{P} x_{i}
K(x_{i}, y_{j} | i ∈ I, j ∈ J)P = KP (y_{j}P | j ∈ J) .

It also implies that the valuation v_{P} and the place P on K(x_{i}, y_{j} | i ∈ I, j ∈ J) are uniquely determined by their restrictions to K, the values v_{P} x_{i} and the residues y_{j}P.

Every finite extension (L|K, P) of valued fields satisfies the fundamental inequality (cf. [EN]):

n ≥ ∑_{i=1}^{g} e_{i}f_{i}  (7)

where n = [L : K] is the degree of the extension, P_{1}, …, P_{g} are the distinct extensions of P from K to L, e_{i} = (v_{P_{i}}L : v_{P} K) are the respective ramification indices and f_{i} = [LP_{i} : KP] are the respective inertia degrees. Note that g = 1 if (K, P) is henselian.

Corollary 2.2 Let (F |K, P) be an extension of valued fields of finite transcendence degree. Then the following inequality holds:

trdeg F |K ≥ trdeg F P |KP + dim_{ℚ} ℚ ⊗ (v_{P} F/v_{P} K) .  (8)

If in addition F |K is a function field, and if equality holds in (8), then the extensions v_{P} F |v_{P} K and F P |KP are finitely generated. In particular, if P is trivial on K, then v_{P} F is a product of finitely many copies of ℤ, and F P is again a function field over K.

Proof: Choose elements x_{1}, …, x_{ρ}, y_{1}, …, y_{τ} ∈ F such that the values v_{P} x_{1}, …, v_{P} x_{ρ} are rationally independent over v_{P} K and the residues y_{1}P, …, y_{τ}P are algebraically independent over KP. Then by the foregoing lemma, ρ + τ ≤ trdeg F |K. This proves that trdeg F P |KP and the rational rank of v_{P} F/v_{P} K are finite. Therefore, we may choose the elements x_{i}, y_{j} such that τ = trdeg F P |KP and ρ = dim_{ℚ} ℚ ⊗ (v_{P} F/v_{P} K) to obtain inequality (8).

Assume that this is an equality. This means that for F_{0} := K(x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}), the extension F |F_{0} is algebraic. Since F |K is finitely generated, it follows that F |F_{0} is finite. By the fundamental inequality, this yields that v_{P} F |v_{P} F_{0} and F P |F_{0}P are finite extensions. Since already v_{P} F_{0}|v_{P} K and F_{0}P |KP are finitely generated by the foregoing theorem, it follows that also v_{P} F |v_{P} K and F P |KP are finitely generated. □

Note that (1) is the special version of (8) for the case of a valued function field with trivially valued base field. We will say that (F |K, P) is without transcendence defect if equality holds in (8).

3 Inertially generated function fields

We will make essential use of a theorem which we proved in [K1] (cf. also [K2]). It is a generalization of the “Grauert–Remmert Stability Theorem”. To state it, we introduce a fundamental notion. A valued field (K, P) is called defectless (or stable) if equality holds in the fundamental inequality (7) for every finite extension L|K. If char KP = 0, then (K, P) is defectless (this is a consequence of the “Lemma of Ostrowski”, cf. [EN], [R]).

Now we are able to state the Stability Theorem, which deals with an arbitrary valued function field (F |K, P), where P need not be the identity on K.

Theorem 3.1 Let (F |K, P) be a valued function field without transcendence defect. If (K, P) is a defectless field, then also (F, P) is a defectless field.

Whenever we talk of a valued field K, we will implicitly assume the valuation and place to be extended to the algebraic closure. Therefore, we will talk of the henselization K^{h}, and of the absolute inertia field of K, which we define to be the inertia field of the normal separable extension K^{sep}|K with respect to the given valuation; here, K^{sep} denotes the separable-algebraic closure of K.

The following lemma is proved in [K2] (and partially also in [EN]):

Lemma 3.2 A valued field (K, P) is defectless if and only if its henselization (K^{h}, P) is.

An extension (L|K, P) of valued fields is called immediate if the canonical embeddings KP → LP and v_{P} K → v_{P} L are onto. Corollary 3.3 If (K, P) is defectless, then (K^{h}, P) does not admit proper immediate algebraic extensions. Proof: If (K, P) is defectless, then so is (K^{h}, P), by the foregoing lemma. Suppose that (L|K^{h}, P) is a finite immediate algebraic extension. Hence, (v_{P} L : v_{P} K^{h}) = 1 = [LP : K^{h}P]. Since (K^{h}, P) is a henselian field, there is a unique extension of v_{P} from K^{h} to L. Since (K^{h}, P) is defectless, we have that [L : K^{h}] = (v_{P} L : v_{P} K^{h})[LP : K^{h}P] = 1, showing that L = K^{h}. As every proper immediate extension would contain a proper finite immediate extension, it follows that (K^{h}, P) does not admit any proper immediate algebraic extension. □

From these facts, we deduce: Theorem 3.4 Assume that F |K is a function field and P is a place of F such that F P |KP is a separable extension, (K, P) is a defectless field and equality holds in (8). Then (F |K, P) is inertially generated, by which we mean that there is a transcendence basis T = {x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}} such that

a) v_{P} F = v_{P} K ⊕ ℤv_{P} x_{1} ⊕ … ⊕ ℤv_{P} x_{ρ},

b) y_{1}P, …, y_{τ}P is a separating transcendence basis of F P |KP,

c) (F, P) lies in the absolute inertia field of (K(T), P). Assertion c) holds for each transcendence basis T which satisfies assertions a) and b). Proof: By Corollary 2.2, value group and residue field of (F, P) are finitely generated. We choose x_{1}, …, x_{ρ} ∈ F such that v_{P} F = ℤv_{P} x_{1} ⊕ … ⊕ ℤv_{P} x_{ρ} (where ρ = dim_{ℚ} ℚ ⊗ (v_{P} F/v_{P} K). Since F P |KP is a finitely generated separable extension, it is separably generated. Therefore, we can choose y_{1}, …, y_{τ} ∈ F such that F P |KP (y_{1}P, …, y_{τ}P) is separable-algebraic (where τ = trdeg F P |KP). We set T := {x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}} and F_{0} := K(T).

Now we can choose some a ∈ F P such that F P = KP (y_{1}P, …, y_{τ}P, a). Since a is separable-algebraic over KP (y_{1}P, …, y_{τ}P), by Hensel’s Lemma there exists an element η in the henselization of (F, P) such that ηP = a and that the reduction of the minimal polynomial of η over F_{0} is the minimal polynomial of a over KP (y_{1}P, …, y_{τ}P). Then η lies in the absolute inertia field of F_{0} . Now the field F_{0}(η) has the same value group and residue field as F, and it is contained in the henselization F^{h} of F . As henselizations are immediate extensions and the henselization F_{0}(η)^{h} of F_{0}(η) can be chosen inside of F^{h}, we obtain an immediate algebraic extension (F^{h}|F_{0}(η)^{h}, P). On the other hand, (K, P) is assumed to be a defectless field. By construction, (F_{0}|K, P) is without transcendence defect, and the same is true for (F_{0}(η)|K, P) since this property is preserved by algebraic extensions. Hence we know from Theorem 3.1 that (F_{0}(η), P) is a defectless field. Now Corollary 3.3 shows that the extension F^{h}|F_{0}(η)^{h} must be trivial. Therefore, F is contained in F_{0}(η)^{h}, which in turn is a subfield of the absolute inertia field of F_{0}. This proves our theorem. □

4 Abhyankar places on rational function fields

The following lemma was proved (but not explicitly stated) by Zariski in [Z] for subgroups of ℝ, using the algorithm of Perron. We leave it as an easy exercise to the reader to prove the general case by induction on the rank of the ordered abelian group. However, an instant proof of the lemma can also be found in [EL] (Theorem 2.2).

Lemma 4.1 Let Γ be a finitely generated ordered abelian group. Take any non-negative elements α_{1}, …, α_{ℓ} ∈ Γ. Then there exist positive elements γ_{1}, …, γ_{ρ} ∈ Γ such that Γ = ℤγ_{1} ⊕ … ⊕ ℤγ_{ρ} and every α_{i} can be written as a sum ∑_{j} n_{ij}γ_{j} with non-negative integers n_{ij}.

The foregoing lemma and Theorem 2.1 are the main ingredients in the proof of the next proposition. We consider the following situation:

\begin{matrix}F |K a function field, \\ P a place of F, \\ x_{1}, …, x_{ρ}, y_{1}, …, y_{τ} ∈ O_{F} such that \\ 1) v_{P} x_{1}, …, v_{P} x_{ρ} form a maximal set of elements in v_{P} F \\ rationally independent over v_{P} K, \\ 2) y_{1}P, …, y_{τ}P form a transcendence basis of F P over KP, \\ E := K(x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}) , \\ ζ_{1}, …, ζ_{m} ∈ O_{E} . \end{matrix} \}  (9)

Proposition 4.2 In the situation described in (9), assume in addition that v_{P} K is a convex subgroup of v_{P} E (this always holds if P is trivial on K.) Then there are x′_{1}, …, x′_{ρ} ∈ O_{E} and z ∈ O_{K}[x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}] of value vz = 0 such that

a) O_{K}[x_{1}, …, x_{ρ}] ⊆ O_{K}[x′_{1}, …, x′_{ρ}] and K(x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}) = E,

b) v_{P} E = v_{P} K ⊕ ⨁_{i=1}^{ρ} ℤv_{P} x′_{i},

c) ζ_{1}, …, ζ_{m} ∈ O_{K}[x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}, z^{−1}]. The number k is bounded from above as in Theorem 1.5.

Proof: By Theorem 2.1, we know that

v_{P} E = v_{P} K ⊕ ℤv_{P} x_{1} ⊕ … ⊕ ℤv_{P} x_{ρ} .

We may assume that v_{P} x_{i} > 0 for 1 ≤ i ≤ ρ. The map

v_{P} E/v_{P} K ∋ μ_{1}(v_{P} x_{1} +v_{P} K)+…+μ_{ρ}(v_{P} x_{ρ} +v_{P} K) ↦ x_{1}^{μ_{1}} ·. . .·x_{ρ}^{μ_{ρ}} (μ_{1}, …, μ_{ρ} ∈ ℤ)

is an isomorphism from v_{P} E/v_{P} K onto the multiplicative subgroup of E^{×} generated by x_{1}, …, x_{ρ}. We denote this group by X.

Take ζ to be any of the elements ζ_{1}, …, ζ_{m} ∈ O_{E} and write ζ = f/g with polynomials f, g ∈ K[x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}]. Write

f = ∑_{i=1}^{d} c_{i} x_{1}^{μ_{1,i}} · . . . · x_{ρ}^{μ_{ρ,i}} · y_{1}^{ν_{1,i}} · . . . · y_{τ}^{ν_{τ,i}} and g = ∑_{i=1}^{d′} c′_{i} x_{1}^{μ′_{1,i}} · . . . · x_{ρ}^{μ′_{ρ,i}} · y_{1}^{ν′_{1,i}} · . . . · y_{τ}^{ν′_{τ,i}}

as sums of monomials (in such a way that in either polynomial, two different monomials differ in at least one exponent). Then by Theorem 2.1, the value of f is equal to the least of the values of its monomials, say, to the value of the first. Similarly for g. So we can write

ζ = \frac{∑_{i} \frac{c_{i}}{c′_{1}} x_{1}^{μ_{1,i}−μ′_{1,1}} · . . . · x_{ρ}^{μ_{ρ,i}−μ′_{ρ,1}} · y_{1}^{ν_{1,i}} · . . . · y_{τ}^{ν_{τ,i}}}{∑_{i} \frac{c′_{i}}{c′_{1}} x_{1}^{μ′_{1,i}−μ′_{1,1}} · . . . · x_{ρ}^{μ′_{ρ,i}−μ′_{ρ,1}} · y_{1}^{ν′_{1,i}} · . . . · y_{τ}^{ν′_{τ,i}}}  (10)

where the denominator has value 0 and the summands appearing in it all have value ≥ 0. Since ζ ∈ O_{E}, the numerator has value ≥ 0, and the same must thus be true for all its summands. The only obstruction is that some of the x_{i}’s may appear with negative exponents in some summands, and that, if P is non-trivial on K, some of the c_{i}/c′_{1} or c′_{i}/c′_{1} may have negative value.

We collect all summands of the form h = cx_{1}^{μ_{1}} ·. . .·x_{ρ}^{μ_{ρ}} ·y_{1}^{ν_{1}} ·. . .·y_{τ}^{ν_{τ}} which appear in the numerator or denominator of (10), and all products of the form ξ = x_{1}^{μ_{1}} ·. . .·x_{ρ}^{μ_{ρ}} ∈ X which appear in these summands. We do the same for all elements ζ_{1}, …, ζ_{m}. We also add the elements x_{1}, …, x_{ρ} to the list of elements so obtained. In this way, we get finitely many elements h_{1}, …, h_{l}, all of them having non-negative value, and corresponding elements ξ_{1}, …, ξ_{l} ∈ X. Since v_{P} x_{1}, …, v_{P} x_{ρ} are among ξ_{1}, …, ξ_{l}, these elements generate X. We note that v_{P} ξ_{j} ∈ v_{P} h_{j} + v_{P} K, and as v_{P} K is a convex subgroup of v_{P} E and v_{P} h_{j} is non-negative, it follows that v_{P} ξ_{j} + v_{P} K is non-negative in v_{P} E/v_{P} K, for the induced order. At this point, we apply Lemma 4.1 to v_{P} ξ_{1} + v_{P} K, …, v_{P} ξ_{ℓ} + v_{P} K. Pulling the result back through the above isomorphism, we find generators x′′_{1}, …, x′′_{ρ} of X for which the values v_{P} x′′_{j} + v_{P} K in v_{P} E/v_{P} K are positive, and such that each of the ξ’s can be written as a (unique) product of the x′′_{j} with non-negative exponents.

Let α ∈ v_{P} K be the minimum of the values of the coefficients c ∈ K appearing in the monomials h_{1}, …, h_{k}. We take c̃ ∈ K such that v_{P} c̃ = α if α < 0, and c̃ = 1 if α ≥ 0. We set x′_{j} := c̃x′′_{j}. It follows that x_{1}, …, x_{ρ} ∈ O_{K}[x′_{1}, …, x′_{ρ}] and therefore, assertion a) holds. It also follows that h_{1}, …, h_{k} ∈ O_{K}[x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}]. Since the values v_{P} x′′_{j} + v_{P} K in v_{P} E/v_{P} K were positive, all values v_{P} x′_{j} are positive.

As we can read off from (10), every ζ_{j} can be written as ζ′_{j}/ζ′′_{j}, where ζ′_{j}, ζ′′_{j} lie in O_{K}[x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}] , with v_{P} ζ′′_{j} = 0. We set z = ∏_{j=1}^{m} ζ′′_{j}, and we are done. □

Proposition 4.3 In the situation described in (9), assume in addition that K is the quotient field of the ring R and that R ⊂ O_{E}. Then there are x′_{1}, …, x′_{ρ+k} ∈ O_{E} and z ∈ R[x′_{1}, …, x′_{ρ+k}, y_{1}, …, y_{τ}] of value vz = 0 such that

a) K(x′_{1}, …, x′_{ρ+k}, y_{1}, …, y_{τ}) = E,

b) v_{P} E = v_{P} K + ⨁_{i=1}^{ρ+k} ℤv_{P} x′_{i},

c) ζ_{1}, …, ζ_{m} ∈ R[x′_{1}, …, x′_{ρ+k}, y_{1}, …, y_{τ}, z^{−1}]. The number k is bounded from above as described in Theorem 1.5.

Proof: We start as in the foregoing proof and write ζ = f/g. As K is the quotient field of R, we can choose c_{i}, c′_{i} ∈ R. We write down such a representation for each ζ_{i}, 1 ≤ i ≤ m. We collect all coefficients c_{i}, c′_{i} from these representations. The values of these finitely many elements generate a finitely generated subgroup Γ of v_{P} K. This subgroup can be written in the form

ℤγ_{1} ⊕ … ⊕ ℤγ_{k} ⊕ ℤγ_{k+1} ⊕ … ⊕ ℤγ_{k′}

where

i) γ_{j} > 0 for all j, ii) there is some N ∈ ℕ and a positive element δ in

Δ := ℤv_{P} x_{1} ⊕ … ⊕ ℤv_{P} x_{ρ}

such that Nγ_{j} > δ for 1 ≤ j ≤ k, and iii) Nγ_{j} < δ for all N ∈ ℕ and all positive δ ∈ Δ. In other words, γ_{1}, …, γ_{k} are archimedean comparable to some elements in Δ, and γ_{k+1}, …, γ_{k′} are infinitely smaller than all positive elements in Δ. As δ ∈ v_{P} F \ v_{P} K and Nγ_{1}, …, Nγ_{k} are rationally independent elements in v_{P} K, we find that k is bounded from above by the maximal number of rationally independent elements in v_{P} K which are bigger than some positive element in v_{P} F \ v_{P} K.

Now we can rewrite the representations using polynomials of the form

f = ∑_{i=1}^{d} r_{i} x_{1}^{μ_{1,i}} · . . . · x_{ρ+k}^{μ_{ρ+k,i}} · y_{1}^{ν_{1,i}} · . . . · y_{τ}^{ν_{τ,i}} and g = ∑_{i=1}^{d′} r′_{i} x_{1}^{μ′_{1,i}} · . . . · x_{ρ+k}^{μ′_{ρ+k,i}} · y_{1}^{ν′_{1,i}} · . . . · y_{τ}^{ν′_{τ,i}}

where

r_{i} , r′_{i} ∈ R with v_{P} r_{i} , v_{P} r′_{i} ∈ Γ_{0} := ℤγ_{k+1} ⊕ … ⊕ ℤγ_{k′} .

We obtain

ζ = \frac{∑_{i} \frac{r_{i}}{r′_{1}} x_{1}^{μ_{1,i}−μ′_{1,1}} · . . . · x_{ρ+k}^{μ_{ρ+k,i}−μ′_{ρ,1}} · y_{1}^{ν_{1,i}} · . . . · y_{τ}^{ν_{τ,i}}}{∑_{i} \frac{r′_{i}}{r′_{1}} x_{1}^{μ′_{1,i}−μ′_{1,1}} · . . . · x_{ρ+k}^{μ′_{ρ,i}−μ′_{ρ+k,1}} · y_{1}^{ν′_{1,i}} · . . . · y_{τ}^{ν′_{τ,i}}}  (11)

From now on, we proceed as in the foregoing proof, with the elements x_{1}, …, x_{ρ+k} in the place of x_{1}, …, x_{ρ}. Also, we replace v_{P} E by Δ + Γ, and v_{P} K by Γ_{0} and note that by construction, Γ_{0} is a convex subgroup of Δ + Γ. In this way, we obtain the elements x′′_{j}, 1 ≤ i ≤ ρ + k.

As before, let α be the minimum of the values of the coefficients c ∈ R appearing in the monomials h_{1}, …, h_{k}. Then α ∈ Γ_{0} . We take c̃ ∈ R such that v_{P} c̃ = α if α < 0, and c̃ = 1 if α ≥ 0. We set x′_{j} := c̃x′′_{j}, 1 ≤ i ≤ ρ + k. It follows that x_{1}, …, x_{ρ+k} ∈ R[x′_{1}, …, x′_{ρ+k}] and therefore, assertion a) holds. It also follows that h_{1}, …, h_{k} ∈ R[x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}]. Since the values v_{P} x′′_{j} + Γ_{0} in Δ + Γ were positive, all values v_{P} x′_{j} are positive. Every ζ_{j} can be written as ζ′_{j}/ζ′′_{j} with ζ′_{j}, ζ′′_{j} ∈ R[x′_{1}, …, x′_{ρ}, y_{1}, …, y_{τ}] and v_{P} ζ′′_{j} = 0. We set z = ∏_{j=1}^{m} ζ′′_{j}, and we are done. □

5 Henselian elements

Throughout this section, let L|K be an algebraic extension and v a valuation on the algebraic closure L̃ of L. We will always assume that (L, v) is contained in the absolute inertia field of (K, v). By O_{L} we denote the valuation ring of v on L, and by O_{K} the valuation ring of v on K. An element a ∈ O_{L} is called v-henselian over K if there is a (not necessarily monic) polynomial h(X) ∈ O_{K}[X] such that

h(a) = 0 and vh′(a) = 0 ,

or in other words, the residue av is a simple root of the reduction hv of h modulo v. We will say that h is an associated polynomial for a. It may not be monic, and it also may not be irreducible over K.

5.1 The basic results

We show:

Lemma 5.1 Assume that L|K is an algebraic extension and v a valuation on L̃ such that (L, v) is contained in the absolute inertia field of (K, v). Then O_{L} is generated as an O_{K}-algebra by v-henselian elements.

Clearly, we only have to prove this assertion for the case where L|K is finite. The proof for this case uses two results on the construction of henselian elements.

Lemma 5.2 Let a ∈ O_{L} be v-henselian and v(a) = 0. Then a^{−1} is v-henselian too.

Proof: Let h(X) ∈ O_{K}[X] and h(a) = 0. Put g(X) = X^{n}h(X^{−1}) where n is the degree of h(X). Then g(a^{−1}) = 0. Moreover

g′(X) = nX^{n−1}h(X^{−1}) − X^{n−2}h′(X^{−1})

and hence

g′(a^{−1}) = −a^{−(n−2)}h′(a) .
□

In what follows, we denote by W the finite set of all valuations w ≠ v of L which coincide with v on K. The letter w will always denote a valuation from W. Since L|K is algebraic, we know that the valuations in the set W ∪ {v} are pairwise incomparable.

Lemma 5.3 Assume that (L|K, v) is a finite extension of valued fields such that (L, v) is contained in the absolute inertia field of (K, v). Let ϑ be a non-zero generator of the residue field extension, i.e., Lv = Kv(ϑ). Assume that z ∈ L is such that zv = ϑ and w(z) ≠ 0 for every w ∈ W. Then z is v-henselian over K. Moreover, L = K(z).

Proof: Let σ_{1}, …, σ_{n} denote the isomorphic mappings of L|K into the algebraic closure K̃ of K, where n = [L : K]. We regard the valuation v of L to be extended to K̃. The mappings x ↦ v_{i}(x) := v(σ_{i}x) are all the valuations of L which coincide with v on K, but they are not necessarily distinct. After changing the enumeration we may assume that v_{1} = … = v_{k} = v but v_{i} ≠ v for i > k. Then k = [Lv : Kv] = [LK^{h} : K^{h}]. Each of these σ_{i} for i ≤ k induces an isomorphism of Lv|Kv into its algebraic closure (which we also denote by σ_{i}), and we have

(σ_{i}(z))v = σ_{i}(zv) for 1 ≤ i ≤ k .

These are the conjugates of ϑ over Kv, and they are all distinct since ϑ generates the separable extension Lv|Kv.

We enumerate the other isomorphisms σ_{k+1}, …, σ_{n} in such a way that

v_{ν}z > 0  hence (σ_{ν}z)v = 0  for k < ν ≤ r
v_{μ}z < 0  hence (σ_{μ}z)v = ∞  for r < μ ≤ n .

The monic minimal polynomial of z over K is

h_{0}(X) = ∏_{1≤i≤k}(X − σ_{i}z) ∏_{k<ν≤r}(X − σ_{ν}z) ∏_{r<μ≤n}(X − σ_{μ}z) .

Dividing h_{0}(X) by a coefficient c of minimal v-value, we obtain a polynomial h(X) ∈ K[X] which is v-primitive, i.e., every coeffcient has v-value ≥ 0 and there is at least one coefficient of v-value = 0. Now consider the polynomial

h_{1}(X) = ∏_{1≤i≤k}(X − σ_{i}z) ∏_{k<ν≤r}(X − σ_{ν}z) ∏_{r<μ≤n}((σ_{μ}z)^{−1}X − 1)

which is obtained by dividing h_{0}(X) by the factor ∏_{μ}(σ_{μ}z)^{−1} ∈ K̃. The two polynomials h(X), h_{1}(X) differ by a constant factor d ∈ K̃, and since both are v-primitive we have vd = 0. (Observe that h_{1}(X) is v-primitive because by construction, each linear factor is v-primitive; use Gauss’ lemma.) Thus we have

h(X) = d · h_{1}(X) with d ∈ K̃, dv ≠ 0, ∞

and thus

hv(X) = ±dv · g(X) · X^{ℓ} .

This shows that ϑ is a simple root of hv(X) and thus h′(z)v ≠ 0 .

Finally, we see that σ_{i}z ≠ z for every σ_{i} ≠ id, which shows that h_{0} is irreducible and therefore, L = K(z). □

By Theorem 11.14 of [EN], an element z as in the foregoing lemma will always exist. Hence, we have proved:

Corollary 5.4 Assume that (L|K, v) is a finite extension of valued fields such that (L, v) is contained in the absolute inertia field of (K, v). Then there is z ∈ L such that z is v-henselian over K, L = K(z) and Lv = Kv(zv).

Now we give a Proof of Lemma 5.1: Let a ∈ O_{L} be arbitrary. We are going to show that a is contained in a ring which is generated over O_{K} by finitely many henselian elements.

We start with a henselian element z ∈ O_{L} as in Corollary 5.4. The residue av ∈ Lv can be written as a polynomial in ϑ with coefficients in Kv. Consider an inverse image in O_{K}[X] of that polynomial, say f(X). Then av = f(z)v. We put

b = a − f(z) + z so that bv = zv .

We have

a ∈ O_{K}[z, b] .

Hence it suffices to show that b is contained in a ring generated over O_{K} by finitely many henselian elements.

Employing again Theorem 11.14 of [EN], we can find an element c ∈ L with the following properties:

v(c − 1) > 0 so that cv = 1
wc > 0 if wb ≥ 0
wc = 0 if wb < 0

for all w ∈ W. Now, if we put d = bc then we have:

dv = bv = zv
wd > 0 if wb ≥ 0
wd < 0 if wb < 0

and hence wd ≠ 0 for all w ∈ W. According to Lemma 5.3 we conclude that d is henselian. On the other hand, the element e = cd also satisfies these relations, hence e is henselian too. Therefore, e^{−1} is henselian by Lemma 5.2. Since

b = d · c^{−1} = d^{2} · e^{−1} ∈ O_{K}[d, e^{−1}]

we see that b and hence a is contained in a ring generated over O_{K} by henselian elements.

□

5.2 A simultaneous version for finitely many places

Lemma 5.5 Suppose that L|K is a finite extension, v a valuation of K and v_{1}, …, v_{ℓ} distinct extensions of v to L. Assume that Kv is an infinite field. Further, assume that for each i, (L, v_{i}) is contained in the absolute inertia field of (K, v). Then there exists an element η ∈ O := ⋂_{i=1}^{ℓ} O_{(L,v_{i})} which is v_{i}-henselian over (K, v) for all i and such that L = K(η).

Proof: We adapt the proof of Lemma 5.3. We take an extension of v to K̃ and σ_{1}, …, σ_{n} as in that proof, this time choosing the enumeration in such a way that for 1 ≤ i ≤ ℓ, v_{i} = v ∘σ_{j} for k_{i−1} < j ≤ k_{i} (where k_{0} = 0). Then we denote by v_{ℓ+1}, …, v_{k} the remaining distinct extensions of v to L which are of the form v ∘ σ_{j} for ℓ < j ≤ n. For 1 ≤ i ≤ ℓ, we choose non-zero generators ϑ_{i} of the finite separable extensions Lv_{i}|Kv. After multiplying them with suitable elements from Kv if necessary, we may assume that ϑ_{i} is not conjugate to ϑ_{j} over Kv if i ≠ j; this is possible since Kv is infinite by assumption. Being distinct extensions of v to the algebraic extension L of K, the valuations v_{1}, …, v_{k} are pairwise incomparable. By Theorem 11.14 of [EN], there is an element z ∈ O such that zv_{i} = ϑ_{i} for 1 ≤ i ≤ ℓ and v_{i}z > 0 for ℓ < i ≤ k. The monic minimal polynomial of z over K is

h(X) = ( ∏_{1≤i≤ℓ} ∏_{k_{i−1}<j≤k_{i}}(X − σ_{j}z)) ∏_{ℓ<ν≤k}(X − σ_{ν}z) .

The reduction of this polynomial modulo v is

hv(X) = ( ∏_{1≤i≤ℓ} ∏_{k_{i−1}<j≤k_{i}}(X − σ_{j}ϑ_{i})) · X^{k−ℓ} .

By our choice of the elements ϑ_{i}, the elements σ_{j}ϑ_{i} are distinct. Since the coefficients of h are fixed under all σ_{j}, we have that hv_{i} = hv for all i. Hence for 1 ≤ i ≤ ℓ, the residue zv_{i} = ϑ_{i} is a simple root of hv_{i}. As before, σ_{i}z ≠ z for every σ_{i} ≠ id, which shows that L = K(z). □

For the sake of completeness, we include the following result.

Lemma 5.6 Suppose that L|K is a finite extension and that v_{1}, …, v_{ℓ} are valuations of L whose restrictions to K are pairwise incomparable. Further, assume that for each i, v_{i} is extended to L such that (L, v_{i}) is contained in the absolute inertia field of (K, v_{i}). Take any elements ζ_{1}, …, ζ_{m} ∈ O := ⋂_{i=1}^{ℓ} O_{(L,v_{i})} . Then there are η_{1}, …, η_{2m+1} ∈ O which are henselian over all (K, v_{i}) and such that

a) L = K(η_{1}),

b) ζ_{1}, …, ζ_{m} ∈ O_{(K,v_{i})}[η_{1}, …, η_{2m+1}] for all i.

Proof: We denote by W_{i} the finite set of valuations of L which differ from v_{i} on L but coincide with v_{i} on K. Since v_{1}, …, v_{ℓ} are pairwise incomparable on K, the valuations in {v_{1}, …, v_{ℓ}} ∪ ⋃_{i=1}^{m} W_{i} are pairwise incomparable on L. We will now copy the proof of Lemma 5.1, simultaneously for v_{1}, …, v_{ℓ}.

Applying Theorem 11.14 of [EN], for every i we choose some z_{i} ∈ L such that Lv_{i} = Kv_{i}(z_{i}v_{i}) and wz_{i} > 0 for all w ∈ W_{i}. By Lemma 5.3, z_{i} is v_{i}-henselian over K. Again by Theorem 11.14 of [EN], we find an element η_{1} ∈ L such that for all i, v_{i}(η_{1} − z_{i}) > 0 and wη_{1} > 0 for w ∈ W_{i}. It follows from Lemma 5.3 that η_{1} is v_{i}-henselian over K for all i, and that L = K(η_{1}).

Now take j ∈ {1, …, m} and set a = ζ_{j}. For 1 ≤ i ≤ ℓ, choose f_{i}(X) ∈ O_{(K,v_{i})}[X] such that av_{i} = f_{i}(η_{1})v_{i}. After adding coefficients of value > 0 if necessary we may assume that all f_{i} have the same degree. Applying Theorem 11.14 of [EN] to the coefficients of the f_{i} (using that v_{1}, …, v_{ℓ} are pairwise incomparable on the field K), we find a polynomial f(X) ∈ O[X] such that av_{i} = f(η_{1})v_{i} for all i. We set b = a − f(z) + z.

Using Theorem 11.14 of [EN] once more, we find an element c ∈ L with the following properties:

v_{i}(c − 1) > 0  for 1 ≤ i ≤ ℓ
wc > 0 if wb ≥ 0
wc = 0 if wb < 0

for all w ∈ ⋃_{i=1}^{ℓ} W_{i}. We put d = bc. According to Lemma 5.3 we obtain that d and e = cd are v_{i}-henselian over K for all i. Therefore, e^{−1} is also v_{i}-henselian by Lemma 5.2, for all i. By construction, we have that ζ_{j} ∈ O_{K}[η_{1}, d, e^{−1}]. We set η_{2j} = d and η_{2j+1} = e^{−1}. Repeating this for all j ∈ {1, …, m}, we obtain elements η_{1}, …, η_{2m+1} which are v_{i}- henselian over K for all i and such that ζ_{1}, …, ζ_{m} ∈ O_{K}[η_{1} …, η_{2m+1}]. □

6 Proof of Theorem 1.5

Assume that F |K, R, P and ζ_{1}, …, ζ_{m} satisfy the assumptions of Theorem 1.5. By Theorem 3.4, there is a transcendence basis x_{1}, …, x_{ρ}, y_{1}, …, y_{τ} of F |K such that assertions a), b), c) of the theorem are satisfied. We set E := K(x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}). Applying Lemma 5.1 to the finite extension F |E, we find v_{P} -henselian elements η_{1}, …, η_{n} ∈ O_{F} with associated polynomials h_{i} ∈ O_{E}[X], such that ζ_{1}, …, ζ_{m} ∈ O_{E}[η_{1}, …, η_{n}]. By Lemma 5.3, we can choose η_{1} such that F = E(η_{1}). We collect all coefficients of the polynomials h_{1}, …, h_{n} and all elements from O_{E} which are necessary to generate ζ_{1}, …, ζ_{m} from η_{1}, …, η_{n}. The finitely many elements so-obtained will be called ζ_{1}^{∗}, …, ζ_{m^{∗}}^{∗}, say.

Now we set s = ρ + τ and apply Proposition 4.3 with E = K(T) to these elements. We obtain a set {t_{1}, …, t_{s+k}} := {x′_{1}, …, x′_{ρ+k}, y_{1}, …, y_{τ}} ⊂ O_{F} and an element z ∈ R[t_{1}, …, t_{s+k}] of value v_{P} z = 0 such that E = K(t_{1}, …, t_{s+k}) and ζ_{1}^{∗}, …, ζ_{m^{∗}}^{∗} ∈ R[t_{1}, …, t_{s}, z^{−1}]. It follows that h_{i}(X) ∈ R[t_{1}, …, t_{s+k}, z^{−1}][X], 1 ≤ i ≤ n, and that ζ_{1}, …, ζ_{m} ∈ R[t_{1}, …, t_{s}, z^{−1}, η_{1}, …, η_{n}]. Since {t_{1}, …, t_{s+k}} contains a transcendence basis of F |K, we can choose the enumeration such that t_{1}, …, t_{s} form a transcendence basis. This completes the proof of Theorem 1.5. □

7 Proof of Theorems 1.8 and 1.9

Throughout this section, we assume that the hypotheses of Theorem 1.8 hold. We set v_{j} = v_{P_{j}} . In view of Remark 1.7, we may assume that the valuations v_{1}, …, v_{ℓ} are pairwise incomparable. Since their value groups are assumed to be isomorphic over v_{P} K, ρ and consequently also τ are the same for all v_{j}. Moreover, passing to equivalent valuations if necessary, we may assume that v_{1}F = … = v_{ℓ}F. Since this group is finitely generated over vK by Corollary 2.2 we can write v_{1}F = v_{P} K ⊕ ⨁_{i=1}^{ρ} ℤα_{i} with positive α_{i}.

If τ > 0, we proceed as follows. We choose a separating transcendence basis ξ_{1}, …, ξ_{τ} of the separable extension F P_{1}|KP. By passing to equivalent places if necessary, we may assume that ξ_{1}, …, ξ_{τ} is also the separating transcendence basis ξ_{1}, …, ξ_{τ} of the separable extensions F P_{j}|KP, 0 < j ≤ ℓ.

For this situation, we prove:

Proposition 7.1 There are elements x_{1}, …, x_{ρ}, y_{1}, …, y_{τ} ∈ O such that for 1 ≤ j ≤ ℓ,

i) v_{j}x_{i} = α_{i} for 1 ≤ i ≤ ρ, ii) v_{j}y_{i} = 0 and y_{i}P_{j} = ξ_{i} for 1 ≤ i ≤ τ. Moreover, the valuations v_{1}, …, v_{ℓ} and the places P_{1}, …, P_{ℓ} coincide on the rational function field K(x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}).

Proof: For each i ∈ {1, …, ρ}, we employ the Approximation Theorem for pairwise incomparable valuations ([R], Théorème 1, p. 135) to find elements x_{i} ∈ O such that v_{j}x_{i} = α_{i} for 1 ≤ j ≤ ℓ. Since α_{i} does not depend on the valuation v_{i}, the compatibility condition of the theorem is satisfied.

For each i ∈ {1, …, τ}, we employ Theorem 11.14 of [EN] to find an element y_{i} ∈ O such that y_{i}P_{j} = ξ_{i} for 1 ≤ j ≤ ℓ.

The last assertion is a direct consequence of Theorem 2.1. □

Remark 7.2 The following more general assertion can be proved for pairwise incomparable valuations with not necessarily equal value groups (using the [R], Théorème 3, p. 136): There are elements z_{1}, …, z_{s} ∈ O which simultaneously form a transcendence basis of (F |K, P_{i}) as described in Theorem 3.4, for 1 ≤ i ≤ ℓ.

• Proof of Theorem 1.8: Choose T = {x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}} according to Proposition 7.1. Let P denote the common restriction of P_{1}, …, P_{ℓ} to K(T). By Theorem 3.4, for each j, (F, P_{j}) lies in the absolute inertia field of (K(T), P).

We choose an element η for the extension F |K(T) according to Lemma 5.5. We collect all coefficients of its associated polynomial h and call them ζ_{1}^{∗}, …, ζ_{k}^{∗}, say. Now the rest of the proof can be taken over from the proof of Theorem 1.5. □

• Proof of Theorem 1.9: We show in [K3] that there are places P_{1}, …, P_{ℓ} of F of dimension d′ centered on A at 𝔓_{1}, …, 𝔓_{ℓ} respectively, and having value group ℤ ⊕ … ⊕ ℤ, ordered lexicograpically and of rational rank ρ = trdeg F |K − d′. Since d′ = τ, these places are Abhyankar places, and they have equal value group. Now our assertion follows from Theorem 1.8. □

References

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[A2] Abhyankar, S. S. : An algorithm on polynomials in one indeterminate with coefficients in a two dimensional regular local domain, Ann. Mat. Pura Appl.

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[K4] Kuhlmann, F.–V. : Valuation theoretic and model theoretic aspects of local uniformization, in: Resolution of Singularities - A Research Textbook in Tribute to Oscar Zariski. Herwig Hauser, Joseph Lipman, Frans Oort, Adolfo Quiros (eds.), Progress in Mathematics Vol. 181, Birkhäuser Verlag Basel (2000), 381–456

[K5] Kuhlmann, F.–V. : On local uniformization in arbitrary characteristic, The Fields Institute Preprint Series, Toronto (1997)

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Department of Mathematics and Statistics, University of Saskatchewan, 106 Wiggins Road, Saskatoon, Saskatchewan, Canada S7N 5E6 email: fvk@math.usask.ca — home page: http://math.usask.ca/˜ fvk/index.html

Mathematisches Institut, Universität Heidelberg, Im Neuenheimer Feld 288, D–69120 Heidelberg, Germany email: roquette@uni-hd.de — home page: http://www.rzuser.uni-heidelberg.de/˜ ci3