Value groups, residue fields and bad places of rational function fields

Franz-Viktor Kuhlmann

22. 6. 2003

Contents

1 Introduction 1

2 Notation and valuation theoretical preliminaries 6 2.1 Pseudo Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Valuation independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.3 Construction of valued field extensions with prescribed value groups and residue fields . . 10 2.4 Orderings and valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.5 A version of Krasner’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Valuations on K(x) 16 3.1 A basic classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.2 Pure and weakly pure extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.3 Construction of nasty examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.4 All valuations on K(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.5 Prescribed implicit constant fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Rational function fields of higher transcendence degree 29

5 Homogeneous sequences 33 5.1 Homogeneous approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 5.2 Homogeneous sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5.3 Conditions for the existence of homogeneous sequences . . . . . . . . . . . . . . . . . . . . 37

6 Applications 39

1 Introduction

In this paper, we denote a valued field by (K, v), its value group by vK, and its residue field by Kv. When we write (L|K, v) we mean a field extension L|K endowed with a valuation v on L and its restriction on K.

In many recent applications of valuation theory, valuations on algebraic function fields play a main role. To mention only a short and incomplete list of applications and references: local uniformization and resolution of singularities ([C], [CP], [S], [KU3], [KKU1,2]), model theory of valued fields ([KU1,2,4]), study of curves via constant reduction ([GMP1,2], [PL]), classification of all extensions of an ordering from a base field to a rational function field ([KUKMZ]), Gröbner bases ([SW], [MOSW1,2]).

In many cases, a basic tool is the classification of all extensions of a valuation from a base field to a function field. As the classification of all extensions of a valuation from a field to an algebraic extension is taken care of by general ramification theory (cf. [E], [KU2]), a crucial step in the classification is the case of rational function fields. Among the first papers describing valuations on rational function fields systematically were [M] and [MS]. Since then, an impressive number of papers have been written about the construction of such valuations and about their properties; the following list is by no means exhaustive: [AP], [APZ1-3], [KH1-10], [KHG1-6], [KHPR], [MO1,2], [MOSW1], [O1-3], [PP], [V]. From the paper [APZ3] the reader may get a good idea of how MacLane’s original approach has been developed further. Since then, the notion of “minimal pairs” has been adopted and studied by several authors (see, e.g., [KHPR]). In the present paper, we will develop a new approach to this subject. It serves to determine in full generality which value groups and which residue fields can possibly occur. This question has recently played a role in two other papers:

1) In [KU3], we prove the existence of “bad places” on rational function fields of transcendence degree ≥ 2. These are places whose value group is not finitely generated, or whose residue field is not finitely generated over the base field. The existence of such places has been shown by MacLane and Schilling ([MS]) and by Zariski and Samuel ([ZS], ch. VI, §15, Examples 3 and 4). However, our approach in [KU3] using Hensel’s Lemma seems to be new, and the present paper contains a further refinement of it. The following theorem of [MS] and [ZS] is a special case of a result which we will prove by this refinement:

Theorem 1.1 Let K be any field. Take Γ to be any non-trivial ordered abelian group of finite rational rank ρ, and k to be any countably generated extension of K of finite transcendence degree τ. Choose any integer n > ρ + τ. Then the rational function field in n variables over K admits a valuation whose restriction to K is trivial, whose value group is Γ and whose residue field is k.

In particular, every additive subgroup of ℚ and every countably generated algebraic extension of K can be realized as value group and residue field of a place of the rational function field K(x, y)|K whose restriction to K is the identity.

The rational rank of an abelian group Γ is the dimension of the ℚ-vector space ℚ⊗_{ℤ}Γ. We denote it by rr Γ. It is equal to the cardinality of any maximal set of rationally independent elements in Γ.

Bad places on function fields are indeed bad: the value group or residue field not being finitely generated constitutes a major hurdle for the attempt to prove local uniformization or other results which are related to resolution of singularities (cf. [CP]). Another hurdle is the phenomenon of defect which can appear when the residue characteristic of a valued field is positive, even if the field itself seems to be quite simple. Indeed, we will prove in Section 3.3, and by a different method in Section 3.5:

Theorem 1.2 Let K be any algebraically closed field of positive characteristic. Then there exists a valuation v on the rational function field K(x, y)|K whose restriction to K is trivial, such that (K(x, y), v) admits an infinite chain of immediate Galois extensions of degree p and defect p.

An extension (L′|L, v) of valued fields is called immediate if the canonical embeddings of vL in vL′ and of Lv in L′v are surjective (which we will express by writing vL′ = vL and L′v = Lv). For a finite immediate extension (L′|L, v), its defect is equal to its degree if and only if the extension of v from L to L′ is unique (or equivalently, L′|L is linearly disjoint from some (or every) henselization of (L, v)).

One of the examples we shall construct for the proof of the above theorem is essentially the same as in Section 7 of [CP], but we use a different and more direct approach (while the construction in [CP] is more intricate since it serves an additional purpose).

2) In [KUKMZ], the classification of all extensions of an ordering to a rational function field is considered in the context of power series fields, and the above question is partially answered in this setting. In the present paper, we will consider the question without referring to power series fields (see Theorem 1.8 below).

During the preparation of [KUKMZ], we found that the construction of an extension of the valuation v from K to the rational function field K(x) with prescribed value group vK(x) and residue field K(x)v is tightly connected with the determination of the relative algebraic closure of K in a henselization K(x)^{h} of K(x) with respect to v. In earlier papers, we have introduced the name “henselian function field” for the henselizations of valued function fields (although these are not function fields, unless the valuation is trivial). In the same vein, one can view the relative algebraic closure as being the (exact) constant field of the henselian function field (K(x)^{h}|K, v). We will call it the implicit constant field of (K(x)|K, v) and denote it by IC (K(x)|K, v). Clearly, the henselization K(x)^{h} depends on the valuation which has been fixed on the algebraic closure K̃(x). So whenever we will talk about the implicit constant field, we will do it in a setting where the valuation on K̃(x) has been fixed. However, since the henselization L^{h} of any valued field (L, v) is unique up to valuation preserving isomorphism over L, the implicit constant field is unique up to valuation preserving isomorphism over K. If L_{0} is a subfield of L, then L^{h} contains a (unique) henselization of L_{0} . Hence, IC (K(x)|K, v) contains a henselization of K and is itself henselian. Further, L^{h}|L is a separable-algebraic extension; thus, K(x)^{h}|K is separable. Therefore, IC (K(x)|K, v) is a separable-algebraic extension of K.

In the present paper, we answer the above question on value groups and residue fields by determining which prescribed separable-algebraic extensions of K can be realized as implicit constant fields. The following result shows in particular that every countably generated separable-algebraic extension of a henselian base field can be realized:

Theorem 1.3 Let (K_{1}|K, v) be a countable separable-algebraic extension of non-trivially valued fields. Then there is an extension of v from K_{1} to the algebraic closure K̃_{1}(x) = K̃(x) of the rational function field K(x) such that, upon taking henselizations in (K̃(x), v),

K_{1}^{h} = IC (K(x)|K, v) .  (1)

In Section 3.1 we introduce a basic classification (“value-transcendental” – “residuetranscendental” – “valuation-algebraic”) of all possible extensions of v from K to K(x). In Section 3.2 we introduce a class of extensions (K(x)|K, v) for which IC (K(x)|K, v) = K^{h} holds. Building on this, we prove Theorem 1.3 in Section 3.5. In fact, we prove a more detailed version: we show under which additional conditions the extension can be chosen in a prescribed class of the basic classification. This yields the following

Theorem 1.4 Take any valued field (K, v), an ordered abelian group extension Γ_{0} of vK such that Γ_{0}/vK is a torsion group, and an algebraic extension k_{0} of Kv. Further, take Γ to be the abelian group Γ_{0} ⊕ ℤ endowed with any extension of the ordering of Γ_{0} .

Assume first that Γ_{0}/vK and k_{0}|Kv are finite. If v is trivial on K, then assume in addition that k_{0}|Kv is simple. Then there is an extension of v from K to the rational function field K(x) which has value group Γ and residue field k_{0}. If v is non-trivial on K, then there is also an extension which has value group Γ_{0} and as residue field a rational function field in one variable over k_{0} .

Now assume that v is non-trivial on K and that Γ_{0}/vK and k_{0}|Kv are countably generated. Suppose that at least one of them is infinite or that (K, v) admits an immediate transcendental extension. Then there is an extension of v from K to K(x) which has value group Γ_{0} and residue field k_{0}.

Here is the converse:

Theorem 1.5 Let (K(x)|K, v) be a valued rational function field. Then one and only one of the following three cases holds:

1) vK(x) ≃ Γ_{0} ⊕ ℤ, where Γ_{0}|vK is a finite extension of ordered abelian groups, and K(x)v|Kv is finite;

2) vK(x)/vK is finite, and K(x)v is a rational function field in one variable over a finite extension of Kv;

3) vK(x)/vK is a torsion group and K(x)v|Kv is algebraic. In all cases, vK(x)/vK is countable and K(x)v|Kv is countably generated.

In 2), we use a fact which was proved by J. Ohm [O2] and is known as the “Ruled Residue Theorem”: If K(x)v|Kv is transcendental, then K(x)v is a rational function field in one variable over a finite extension of Kv. For the countability assertion, see Theorem 2.9.

In Section 3.4 we give an explicit description of all possible extensions of v from K to K(x) (Theorem 3.9).

Theorem 1.4 is used in the proof of our next theorem:

Theorem 1.6 Let (K, v) be any valued field, n, ρ, τ non-negative integers, n ≥ 1, Γ ≠ {0} an ordered abelian group extension of vK such that Γ/vK is of rational rank ρ, and k|Kv a field extension of transcendence degree τ. Part A. Suppose that n > ρ + τ and that A1) Γ/vK and k|Kv are countably generated, A2) Γ/vK or k|Kv is infinite. Then there is an extension of v to the rational function field K(x_{1}, …, x_{n}) in n variables such that

vK(x_{1}, …, x_{n}) = Γ and K(x_{1}, …, x_{n})v = k .  (2)

Part B. Suppose that n ≥ ρ + τ and that B1) Γ/vK and k|Kv are finitely generated, B2) if v is trivial on K, n = ρ + τ and ρ = 1, then k is a simple algebraic extension of a rational function field in τ variables over Kv (or of Kv itself if τ = 0), or a rational function field in one variable over a finitely generated field extension of Kv of transcendence degree τ − 1, B3) if n = τ, then k is a rational function field in one variable over a finitely generated field extension of Kv of transcendence degree τ − 1, B4) if ρ = 0 = τ, then there is an immediate extension of (K, v) which is either infinite separable-algebraic or of transcendence degree at least n. Then again there is an extension of v to K(x_{1}, …, x_{n}) such that (2) holds.

Theorem 1.1 is the special case of Part A for v trivial on K. The following converse holds:

Theorem 1.7 Let v be a valuation on the rational function field F = K(x_{1}, …, x_{n}), n ≥ 1. Set ρ = rr vF/vK and τ = trdeg F v|Kv. Then n ≥ ρ + τ, vF/vK is countable, and F v|Kv is countably generated.

If n = ρ + τ, then vF/vK is finitely generated and F v|Kv is a finitely generated field extension. Assertions B2) and B3) of Theorem 1.6 hold for k = F v, and if ρ = 0 = τ, then there is an immediate extension of (K̃, v) of transcendence degree n (for any extension of v from K to K̃).

There is a gap between Theorem 1.6 and this converse for the case of ρ = 0 = τ, as the former talks about K and the latter talks about the algebraic closure K̃ of K. This gap can be closed if (K, v) has residue characteristic 0 or is a Kaplansky field; because the maximal immediate extension of such fields is unique up to isomorphism, one can show that K̃ can be replaced by K. But in the case where (K, v) is not such a field, we do not know enough about the behaviour of maximal immediate extensions under algebraic field extensions. This question should be considered in future research.

A valuation on an ordered field is called convex if the associated valuation ring is convex. For the case of ordered fields with convex valuations, we can derive from Theorem 1.6 the existence of convex extensions of the valuation with prescribed value groups and residue fields in the frame given by Theorem 1.7, provided that a natural additional condition for the residue fields is satisfied:

Theorem 1.8 In the setting of Theorem 1.6, assume in addition that K is ordered and that v is convex w.r.t. the ordering. Assume further that k is equipped with an extension of the ordering induced by < on Kv. Then this extension can be lifted through v to K(x_{1}, …, x_{n}) in such a way that the lifted ordering extends <. It follows that v is convex w.r.t. this lifted ordering on K(x_{1}, …, x_{n}).

In Section 5 we shall introduce “homogeneous sequences”. In the “tame case”, they can be used to determine the implicit constant field of a valued rational function field, and also to characterize this “tame case”. In Section 6 we shall show how to apply our results to power series, in the spirit of [MS] and [ZS] (Theorem 6.1). We will also use our approach to give proofs of two well known facts in p-adic algebra: that the algebraic closure of ℚ_{p} is not complete and that its completion is not maximal.

Finally, let us mention that we use our criteria for IC (K(x)|K, v) = K^{h} in Section 3.2 to give an example for the following fact: Suppose that K is relatively algebraically closed in a henselian valued field (L, v) such that vL/vK is a torsion group. Then it is not necessarily true that vL = vK, even if v has residue characteristic 0.

I would like to thank Murray Marshall and Salma Kuhlmann for the very inspiring joint seminar; without this seminar, this paper would not have been written. I also feel very much endebted to Sudesh Kaur Khanduja for many ideas she has shared with me. Finally, I would like to thank Roland Auer for finding some mistakes and asking critical questions.

2 Notation and valuation theoretical preliminaries

For an arbitrary field K, will will denote by K^{sep} the separable-algebraic closure of K, and by K̃ the algebraic closure of K. By Gal K we mean the absolute Galois group Gal (K̃|K) = Gal (K^{sep}|K). For a valuation v on K, we let O_{K} denote the valuation ring of v on K.

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

n ≥ ∑_{i=1}^{g} e_{i}f_{i}  (3)

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

In analogy to field theory, an extension Γ ⊂ Δ of abelian groups will also be written as Δ|Γ, and it will be called algebraic if Δ/Γ is a torsion group. The fundamental inequality implies the following well known fact:

Lemma 2.1 If (L|K, v) is finite, then so are vL/vK and Lv|Kv. If (L|K, v) is algebraic, then so are vL/vK and Lv|Kv.

Given two subextensions M|K and L|K within a fixed extension N|K, the field compositum M.L is defined to be the smallest subfield of N which contains both M and L. If L|K is algebraic, the compositum is uniquely determined by taking N = M̃ and specifying a K-embedding of L in M̃.

Lemma 2.2 Let (M|K, v) be an immediate extension of valued fields, and (L|K, v) a finite extension such that [L : K] = (vL : vK)[Lv : Kv]. Then for every K-embedding of L in M̃ and every extension of v from M to M̃, the extension (M.L|L, v) is immediate.

Proof: Via the embedding, we identify L with a subfield of M̃. Pick any extension of v from M to M̃. This will also be an extension of v from L to M̃ because by the fundamental inequality, the extension of v from K to L is unique. We consider the extension (M.L|M, v).

It is clear that vL ⊆ v(M.L) and Lv ⊆ (M.L)v; therefore, (v(M.L) : vK) ≥ (vL : vK) and [(M.L)v : Kv] ≥ [Lv : Kv]. Since (M|K, v) is immediate, we have

[M.L : M] ≥ (v(M.L) : vM)[(M.L)v : Mv] = (v(M.L) : vK)[(M.L)v : Kv]
≥ (vL : vK)[Lv : Kv] = [L : K] ≥ [M.L : M] .

This shows that (v(M.L) : vK) = (vL : vK) and [(M.L)v : Kv] = [Lv : Kv], that is, v(M.L) = vL and (M.L)v = Lv. □

2.1 Pseudo Cauchy sequences

We assume the reader to be familiar with the theory of pseudo Cauchy sequences as presented in [KA]. Recall that a pseudo Cauchy sequence A = (a_{ν})_{ν<λ} in (K, v) (where λ is some limit ordinal) is of transcendental type if for every g(x) ∈ K(x), the value vg(a_{ν}) is eventually constant, that is, there is some ν_{0} < λ such that

vg(a_{ν}) = vg(a_{ν_{0}}) for all ν ≥ ν_{0} , ν < λ .  (4)

Otherwise, A if of algebraic type.

Take a pseudo Cauchy sequence A in (K, v) of transcendental type. We define an extension v_{A} of v from K to the rational function field K(x) as follows. For each g(x) ∈ K[x], we choose ν_{0} < λ such that (4) holds. Then we set

v_{A} g(x) := vg(a_{ν_{0}}) .

We extend v_{A} to K(x) by setting v_{A}(g/h) := v_{A}g − v_{A}h. The following is Theorem 2 of [KA]:

Theorem 2.3 Let A be a pseudo Cauchy sequence in (K, v) of transcendental type. Then v_{A} is a valuation on the rational function field K(x). The extension (K(x)|K, v_{A}) is immediate, and x is a pseudo limit of A in (K(x), v_{A}). If (K(y), w) is any other valued extension of (K, v) such that y is a pseudo limit of A in (K(y), w), then x ↦ y induces a valuation preserving K-isomorphism from (K(x), v_{A}) onto (K(y), w).

From this theorem we deduce:

Lemma 2.4 Suppose that in some valued field extension of (K, v), x is the pseudo limit of a pseudo Cauchy sequence in (K, v) of transcendental type. Then (K(x)|K, v) is immediate and x is transcendental over K.

Proof: Assume that (a_{ν})_{ν<λ} is a pseudo Cauchy sequence in (K, v) of transcendental type. Then by Theorem 2.3 there is an immediate extension w of v to the rational function field K(y) such that y becomes a pseudo limit of (a_{ν})_{ν<λ}; moreover, if also x is a pseudo limit of (a_{ν})_{ν<λ} in (K(x), v), then x ↦ y induces a valuation preserving isomorphism from K(x) onto K(y) over K. Hence, (K(x)|K, v) is immediate and x is transcendental over K. □

Lemma 2.5 A pseudo Cauchy sequence of transcendental type in a valued field remains a pseudo Cauchy sequence of transcendental type in every algebraic valued field extension of that field.

Proof: Assume that (a_{ν})_{ν<λ} is a pseudo Cauchy sequence in (K, v) of transcendental type and that (L|K, v) is an algebraic extension. If (a_{ν})_{ν<λ} were of algebraic type over (L, v), then by Theorem 3 of [KA] there would be an algebraic extension L(y)|L and an immediate extension of v to L(y) such that y is a pseudo limit of (a_{ν})_{ν<λ} in (L(y), v). But then, y is also a pseudo limit of (a_{ν})_{ν<λ} in (K(y), v). Hence by the foregoing lemma, y must be transcendental over K. This is a contradiction to the fact that L(y)|L and L|K are algebraic. □

2.2 Valuation independence

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

Lemma 2.6 Let (L|K, v) be an extension of valued fields. Take elements x_{i}, y_{j} ∈ L, i ∈ I, j ∈ J, such that the values vx_{i}, i ∈ I, are rationally independent over vK, and the residues y_{j}v, j ∈ J, are algebraically independent over Kv. 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

vf = min_{k} v c_{k} ∏_{i∈I} x_{i}^{μ_{k,i}} ∏_{j∈J} y_{j}^{ν_{k,j}} = min_{k} vc_{k} + ∑_{i∈I} μ_{k,i}vx_{i} .  (5)

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

vK(x_{i}, y_{j} | i ∈ I, j ∈ J) = vK ⊕ ⨁_{i∈I} ℤvx_{i}
K(x_{i}, y_{j} | i ∈ I, j ∈ J)v = Kv (y_{j}v | j ∈ J) .

Moreover, the valuation v on K(x_{i}, y_{j} | i ∈ I, j ∈ J) is uniquely determined by its restriction to K, the values vx_{i} and the residues y_{j}v.

Conversely, if (K, v) is any valued field and we assign to the vx_{i} any values in an ordered group extension of vK which are rationally independent, then (5) defines a valuation on L, and the residues y_{j}v, j ∈ J, are algebraically independent over Kv.

Corollary 2.7 Let (L|K, v) be an extension of finite transcendence degree of valued fields. Then

trdeg L|K ≥ trdeg Lv|Kv + rr (vL/vK) .  (6)

If in addition L|K is a function field and if equality holds in (6), then the extensions vL|vK and Lv|Kv are finitely generated.

Proof: Choose elements x_{1}, …, x_{ρ}, y_{1}, …, y_{τ} ∈ L such that the values vx_{1}, …, vx_{ρ} are rationally independent over vK and the residues y_{1}v, …, y_{τ}v are algebraically independent over Kv. Then by the foregoing lemma, ρ + τ ≤ trdeg L|K. This proves that trdeg Lv|Kv and the rational rank of vL/vK are finite. Therefore, we may choose the elements x_{i}, y_{j} such that τ = trdeg Lv|Kv and ρ = rr (vL/vK) to obtain inequality (6).

Assume that equality holds in (6). This means that L is an algebraic extension of L_{0} := K(x_{1}, …, x_{ρ}, y_{1}, …, y_{τ}). Since L|K is finitely generated, it follows that L|L_{0} is finite; hence by Lemma 2.1, also vL/vL_{0} and Lv|L_{0}v are finite. Since already vL_{0}|vK and L_{0}v|Kv are finitely generated by the foregoing lemma, it follows that also vL|vK and Lv|Kv are finitely generated. □

The algebraic analogue of the transcendental case discussed in Lemma 2.6 is the following lemma (see [R] or [E]):

Lemma 2.8 Let (L|K, v) be an extension of valued fields. Take η_{i} ∈ L such that vη_{i}, i ∈ I, belong to distinct cosets modulo vK. Further, take ϑ_{j} ∈ O_{L}, j ∈ J, such that ϑ_{j}v are Kv-linearly independent. Then the elements η_{i}ϑ_{j}, i ∈ I, j ∈ J, are K-linearly independent, and for every choice of elements c_{ij} ∈ K, only finitely many of them nonzero, we have that

v ∑_{i,j} c_{ij}η_{i}ϑ_{j} = min_{i,j} vc_{ij}η_{i}ϑ_{j} = min_{i,j} (vc_{ij} + vη_{i}) .

If the elements η_{i}ϑ_{j} form a K-basis of L, then vη_{i}, i ∈ I, is a system of representatives of the cosets of vL modulo vK, and ϑ_{j}v, j ∈ J, is a basis of Lv|Kv.

The following is an application which is important for our description of all possible value groups and residue fields of valuations on K(x). The result has been proved with a different method in [APZ3] (Corollary 5.2); cf. our remark in Section 3.1.

Theorem 2.9 Let K be any field and v any valuation of the rational function field K(x). Then vK(x)/vK is countable, and K(x)v|Kv is countably generated.

Proof: Since K(x) is the quotient field of K[x], we have that vK(x) = vK[x] − vK[x]. Hence, to show that vK(x)/vK is countable, it suffices to show that {α+vK | α ∈ vK[x]} is countable. If this were not true, then by Lemma 2.8 (applied with J = {1} and ϑ_{1} = 1), we would have that K[x] contains uncountably many K-linearly independent elements. But this is not true, as K[x] admits the countable K-basis {x^{i} | i ≥ 0}.

Now assume that K(x)v|Kv is not countably generated. Then by Corollary 2.7, K(x)v|Kv must be algebraic. It also follows that K(x)v has uncountable Kv-dimension. Pick an uncountable set κ and elements f_{i}(x)/g_{i}(x), i ∈ κ, with f_{i}(x), g_{i}(x) ∈ K[x] and vf_{i}(x) = vg_{i}(x) for all i, such that their residues are Kv-linearly independent. As vK(x)/vK is countable, there must be some uncountable subset λ ⊂ κ such that for all i ∈ λ, the values vf_{i}(x) = vg_{i}(x) lie in the same coset modulo vK, say vh(x) + vK with h(x) ∈ K[x]. The residues (f_{i}(x)/g_{i}(x))v, i ∈ λ, generate an algebraic extension of uncountable dimension. Choosing suitable elements c_{i} ∈ K such that

vc_{i}f_{i}(x) = vh(x) = vc_{i}g_{i}(x) ,

we can write

\frac{f_{i}(x)}{g_{i}(x)} = \frac{c_{i}f_{i}(x)}{h(x)} · \frac{h(x)}{c_{i}g_{i}(x)} = \frac{c_{i}f_{i}(x)}{h(x)} · (\frac{c_{i}g_{i}(x)}{h(x)})^{−1}

for all i ∈ λ. Therefore,

\frac{f_{i}(x)}{g_{i}(x)} v = (\frac{c_{i}f_{i}(x)}{h(x)} v) · (\frac{c_{i}g_{i}(x)}{h(x)} v)^{−1}

for all i ∈ λ. In order that these elements generate an algebraic extension of Kv of uncountable dimension, the same must already be true for the elements (c_{i}f_{i}(x)/h(x))v, i ∈ λ, or for the elements (c_{i}g_{i}(x)/h(x))v, i ∈ λ. It follows that at least one of these two sets contains uncountably many Kv-linearly independent elements. But then by Lemma 2.8 (applied with I = {1} and η_{1} = 1), there are uncountably many K-linearly independent elements in the set

\frac{1}{h(x)} K[x]

and hence also in K[x], a contradiction. □

Finally, let us mention the following lemma which combines the algebraic and the transcendental case. We leave its easy proof to the reader.

Lemma 2.10 Let (L|K, v) be an extension of valued fields. Take x ∈ L. Suppose that for some e ∈ ℕ there exists an element d ∈ K such that vdx^{e} = 0 and dx^{e}v is transcendental over Kv. Let e be minimal with this property. Then for every f = c_{n}x^{n} + … + c_{0} ∈ K[x],

vf = min_{1≤i≤n} vc_{i}x^{i} .

Moreover, K(x)v = Kv(dx^{e}v) is a rational function field over Kv, and we have

vK(x) = vK + ℤvx with (vK(x) : vK) = e .

2.3 Construction of valued field extensions with prescribed value groups and residue fields

In this section, we will deal with the following problem. Suppose that (K, v) is a valued field, Γ|vK is an extension of ordered abelian groups and k|Kv is a field extension. Does there exist an extension (L|K, v) of valued fields such that vL = Γ and Lv = k? We include the case of (K, v) being trivially valued; this amounts to the construction of a valued field with given value group and residue field. Throughout, we use the well known fact that if (K, v) is any valued field and L is any extension field of K, then there is at least one extension of v to L (cf. [E], [R]).

Let us adjust the following notion to our purposes. Usually, when one speaks of an Artin-Schreier extension then one means an extension of a field K generated by a root of an irreducible polynomial of the form X^{p} − X − c, provided that p = char K. We will replace this by the weaker condition “p = char Kv”. In fact, such extensions also play an important role in the mixed characteristic case, where char K = 0.

Every Artin-Schreier polynomial X^{p} − X − c is separable since its derivative does not vanish. The following is a simple but very useful observation:

Lemma 2.11 Let (K, v) be a valued field and c ∈ K such that vc < 0. If a ∈ K̃ such that a^{p} − a = c, then for every extension of v from K to K(a),

0 > v(a^{p} − c) = va > pva = vc .

Proof: Take any extension of v from K to K(a). Necessarily, va < 0 since otherwise, ∞ = v(a^{p}−a−c) = min{pva, va, vc} = vc, a contradiction. It follows that va^{p} = pva < va and thus,

vc = min{v(a^{p} − a − c), v(a^{p} − a)} = v(a^{p} − a) = min{pva, va} = pva .
□

Lemma 2.12 Let (K, v) be a non-trivially valued field, p a prime and α an element of the divisible hull of vK such that pα ∈ vK, α ∉ vK. Choose an element a ∈ K̃ such that a^{p} ∈ K and va^{p} = pα. Then v extends in a unique way from K to K(a). It satisfies

va = α , [K(a) : K] = (vK(a) : vK) , vK(a) = vK + ℤα and K(a)v = Kv . (7)

If char K = char Kv = p, then this extension K(a)|K is purely inseparable. On the other hand, if char Kv = p, then there is always an Artin-Schreier extension K(a)|K with properties (7); if α < 0, then a itself can be chosen to be the root of an Artin-Schreier polynomial over K.

Proof: Take c ∈ K such that vc = pα and a ∈ K̃ such that a^{p} = c. Choose any extension of v from K to K(a). Then pva = vc = pα. Consequently, va = α and (vK(a) : vK) ≥ (vK + ℤα : vK) = p. On the other hand, the fundamental inequality (3) shows that

p = [K(a) : K] ≥ (vK(a) : vK) · [K(a)v : Kv] ≥ (vK(a) : vK) ≥ p .

Hence, equality holds everywhere, and we find that (vK(a) : vK) = p and [K(a)v : Kv] = 1. That is, vK(a) = vK + ℤα and K(a)v = Kv. Further, the fundamental inequality implies that the extension of v from K to K(a) is unique.

Now suppose that char Kv = p. Choose c ∈ K such that vc = −|pα| < 0. By the foregoing lemma, every root b of the Artin-Schreier polynomial X^{p} − X − c must satisfy pvb = vc. Now we set a = b if α < 0, and a = 1/b if α > 0 (but note that then 1/a is in general not the root of an Artin-Schreier polynomial). Then as before one shows that (7) holds. □

For f ∈ O_{K}[X], we define the reduction fv ∈ Kv[X] to be the polynomial obtained from f through replacing every coefficient by its residue.

Lemma 2.13 Let (K, v) be a valued field and ζ an element of the algebraic closure of Kv. Choose a monic polynomial f ∈ O_{K}[X] whose reduction fv is the minimal polynomial of ζ over Kv. Further, choose a root b ∈ K̃ of f. Then there is a unique extension of v from K to K(b) and a corresponding extension of the residue map such that

bv = ζ , [K(b) : K] = [Kv(ζ) : Kv] , vK(b) = vK and K(b)v = Kv(ζ) . (8)

In all cases, f can be chosen separable, provided that the valuation v is non-trivial. On the other hand, if char K = char Kv = p > 0 and ζ is purely inseparable over Kv, then b can be chosen purely inseparable over K.

If v is non-trivial, char Kv = p > 0 and ζ^{p} ∈ Kv, ζ ∉ Kv, then there is also an Artin-Schreier extension K(b)|K such that (8) holds and db is the root of an Artin-Schreier polynomial over K, for a suitable d ∈ K.

Proof: We choose an extension of v from K to K(b). Since f is monic with integral coefficients, b must also be integral for this extension, and bv must be a root of fv. We may compose the residue map with an isomorphism in Gal Kv which sends this root to ζ. Doing so, we obtain a residue map (still associated with v) that satisfies bv = ζ. Now ζ ∈ K(b)v and consequently, [K(b)v : Kv] ≥ [Kv(ζ) : Kv] = deg fv = deg f. On the other hand, the fundamental inequality shows that

deg f = [K(b) : K] ≥ (vK(b) : vK) · [K(b)v : Kv] ≥ [K(b)v : Kv] ≥ deg f .

Hence, equality holds everywhere, and we find that [K(b)v : Kv] = [Kv(ζ) : Kv] = [K(b) : K] and (vK(b) : vK) = 1. That is, vK(b) = vK and K(b)v = Kv(ζ). Further, the uniqueness of v on K(a) follows from the fundamental inequality.

If fv is separable, then so is f. Even if fv is not separable but v is nontrivial on K, then f can still be chosen separable since we can add a summand cX with c ≠ 0, vc > 0 (we use that v is non-trivial) without changing the reduction of f. On the other hand, if fv is purely inseparable and hence of the form X^{p^{ν}} − cv, then we can choose f = X^{p^{ν}} − c which also is purely inseparable if char K = p.

Now suppose that char Kv = p > 0 and ζ^{p} ∈ Kv, ζ ∉ Kv. Choose c ∈ K such that cv = ζ^{p}. To construct an Artin-Schreier extension, choose any d ∈ K with vd < 0, and let b_{0} be a root of the Artin-Schreier polynomial X^{p} − X − d^{p}c. Since vd^{p}c = pvd < 0, Lemma 2.11 shows that v(b_{0}^{p}−d^{p}c) = vb_{0} > vb_{0}^{p}. Consequently, v((b_{0}/d)^{p}−c) > v(b_{0}/d)^{p} = vc = 0, whence (b_{0}/d)^{p}v = cv and (b_{0}/d)v = (cv)^{1/p} = ζ. We set b = b_{0}/d; so K(b) = K(b_{0}). As before, it follows that vK(b) = vK and K(b)v = Kv(ζ). □

Theorem 2.14 Let (K, v) be an arbitrary valued field. For every extension Γ|vK of ordered abelian groups and every field extension k|Kv, there is an extension (L, v) of (K, v) such that vL = Γ and Lv = k. If Γ|vK and k|Kv are algebraic, then L|K can be chosen to be algebraic, with a unique extension of v from K to L. If ρ = rr Γ/vK and τ = trdeg k|Kv are finite, then L|K can be chosen of transcendence degree ρ + τ. If Γ ≠ {0}, then L can be chosen to be a separable extension of K.

If both Γ|vK and k|Kv are finite, then L|K can be taken to be a finite extension such that [L : K] = (Γ : vK)[k : Kv]. If in addition v is non-trivial on K, then L|K can be chosen to be a simple separable-algebraic extension.

If Γ/vK is countable and k|Kv is countably generated, then L|K can be taken to be a countably generated extension.

Proof: For the proof, we assume that Γ ≠ {0} (the other case is trivial). Let α_{i}, i ∈ I, be a maximal set of elements in Γ rationally independent over vK. Then by Lemma 2.6 there is an extension (K_{1}, v) := (K(x_{i} | i ∈ I), v) of (K, v) such that vK_{1} = vK ⊕ ⨁_{i∈I} ℤα_{i} and K_{1}v = Kv. Next, choose a transcendence basis ζ_{j}, j ∈ J, of k|Kv. Then by Lemma 2.6 there is an extension (K_{2}, v) := (K_{1}(y_{j} | j ∈ J), v) of (K_{1}, v) such that vK_{2} = vK_{1} and K_{2}v = Kv(ζ_{j} | j ∈ J). If Γ|vK and k|Kv are algebraic, then I = J = ∅ and K_{2} = K.

If we have an ascending chain of valued fields whose value groups are subgroups of Γ and whose residue fields are subfields of k, then the union over this chain is again a valued field whose value group is a subgroup of Γ and whose residue field is a subfield of k. So a standard argument using Zorn’s Lemma together with the transitivity of separable extensions shows that there are maximal separable-algebraic extension fields of (K_{2}, v) with these properties. Choose one of them and call it (L, v). We have to show that vL = Γ and Lv = k. Since already Γ|vK_{2} and k|K_{2}v are algebraic, the same holds for Γ|vL and k|Lv. If vL is a proper subgroup of Γ, then there is some prime p and some element α ∈ Γ \ vL such that pα ∈ vL. But then, Lemma 2.12 shows that there exists a proper separable-algebraic extension (L′, v) of (L, v) such that vL′ = vL + ℤα ⊂ Γ and L′v = Lv ⊂ k, which contradicts the maximality of L. If Lv is a proper subfield of k, then there is some element ζ ∈ k \ Lv algebraic over Lv. But then, Lemma 2.13 shows that there exists a proper separable-algebraic extension (L′, v) of (L, v) such that vL′ = vL ⊂ Γ and L′v = Lv(ζ) ⊂ k, which again contradicts the maximality of L (here we have used Γ ≠ {0}, which implies that v is not trivial on L). This proves that vL = Γ and Lv = k, and (L, v) is the required extension of (K, v). Since K_{2} is generated over K by a set of elements which are algebraically independent over K, we know that K_{2}|K is separable. Since also L|K_{2} is separable, we find that L|K is separable. Since L|K_{2} is algebraic, {x_{i} , y_{j} | i ∈ I , j ∈ J} is a transcendence basis of (K_{2}|K, v). If Γ|vK and k|Kv are algebraic, then I = J = ∅ and L is an algebraic extension of K = K_{2} .

If Γ|vK and k|Kv are finite, then L can be constructed by a finite number of applications of Lemma 2.12 and Lemma 2.13. Since extension degree, ramification index and inertia degree are multiplicative, we obtain that [L : K] = (vL : vK)[Lv : Kv] = (Γ : vK)[k : Kv]. If in addition v is non-trivial, then L|K can be chosen to be a separable extension. Since it is finite, it is simple.

If Γ|vK and k|Kv are countably generated algebraic, then they are unions over a countable chain of algebraic extensions. Hence also L can be constructed as a union over a countable chain of algebraic extensions and will thus be a countably generated extension of K.

If Γ|vK and k|Kv are countably generated, then the sets I and J in our above construction are both countable, that is, K_{2}|K is countably generated. Moreover, the extensions Γ|vK_{2} and k|K_{2}v are countably generated algebraic. Hence by what we have just shown, L can be taken to be a countably generated extension of K_{2} , and thus also of K. □

Every ordered abelian group is an extension of the trivial group {0} as well as of the ordered abelian group ℤ. Every field of characteristic 0 is an extension of ℚ, and every field of characteristic p > 0 is an extension of 𝔽_{p}. Let Γ ≠ 0 be an ordered abelian group and k a field. If char k = 0, then ℚ endowed with the trivial valuation v will satisfy vℚ = {0} ⊂ Γ and ℚv = ℚ ⊂ k. If char k = p > 0, then we can choose v to be the p-adic valuation on ℚ to obtain that vℚ = ℤ ⊂ Γ and ℚv = 𝔽_{p} ⊂ k. But also 𝔽_{p} endowed with the trivial valuation v will satisfy v𝔽_{p} = {0} ⊂ Γ and 𝔽_{p}v = 𝔽_{p} ⊂ k. An application of the foregoing theorem now proves:

Corollary 2.15 Let Γ ≠ 0 be an ordered abelian group and k a field. Then there is a valued field (L, v) with vL = Γ and Lv = k. If char k = p > 0, then L can be chosen to be of characteristic 0 (mixed characteristic case) or of characteristic p (equal characteristic case).

For the sake of completeness, we add the following information. From the fundamental inequality it follows that vK̃|vK, vK^{sep}|vK, K̃v|Kv and K^{sep}v|Kv are algebraic extensions. On the other hand, Lemma 2.12 shows that the value group of a separablealgebraically closed field must be divisible. Similarly, it follows from Lemma 2.13 that the residue field of a separable-algebraically closed non-trivially valued field must be algebraically closed. This proves:

Lemma 2.16 Let (K, v) be a non-trivially valued field and extend v to K̃. Then the value groups vK̃ and vK^{sep} are equal to the divisible hull of vK, and the residue fields K̃v and K^{sep}v are equal to the algebraic closure of Kv.

A valued field (K, v) of residue characteristic p > 0 is called Artin-Schreier closed if every Artin-Schreier polynomial with coefficients in K admits a root in K. Recall that if char K = p, then this means that every Artin-Schreier polynomial with coefficients in K splits into linear factors over K. If K is Artin-Schreier closed, then so is Kv. As a corollary to Lemmas 2.12 and 2.13, we obtain:

Corollary 2.17 Every Artin-Schreier closed non-trivially valued field of residue characteristic p > 0 has p-divisible value group and perfect Artin-Schreier closed residue field.

2.4 Orderings and valuations

We will assume the reader to be familiar with the basic theory of convex valuations, which can be found in [L] and [PR].

Proposition 2.18 Suppose that (K, <) is an ordered field with convex valuation v, and denote by <_{r} the ordering induced by < through v on Kv. Let (L|K, v) be an extension of valued fields. If 2vL ∩ vK = 2vK, then each extension <′_{r} of <_{r} to an ordering of Lv can be lifted through v to an ordering of L which extends the ordering < of K.

Proof: We fix a section from vK/2vK to K^{×}/K^{×}^{2}. Since 2vL ∩ vK = 2vK, this section can be extended to a section from vL/2vL to L^{×}/L^{×}^{2}. Now there is a bijection between the set of all liftings of <_{r} through v to orderings of K, and the set of all group characters of vK/2vK; see [PR], (7.5) to (7.9). The same construction yields a bijection between the set of all liftings of <′_{r} through v to orderings of L, and the set of all group characters of vL/2vL. Since we use an extension of the section vK/2vK → K^{×}/K^{×}^{2} to a section vL/2vL → L^{×}/L^{×}^{2}, the bijection maps commute with the restriction from L to K of any lifting. That is, if a lifting <′ of <′_{r} to L corresponds to a character χ of vL/2vL, then the restriction of <′ to K is the unique lifting of <_{r} to K which corresponds to the restriction of χ to vK/2vK.

As the given ordering < of K is a lifting of <_{r}, it corresponds to a unique group character of vK/2vK. Since 2vL ∩ vK = 2vK, we can extend it to a group character of vL/2vL. Take the lifting of <′_{r} through v to an ordering <′ of L which corresponds to this group character of vL/2vL. Then its restriction to K is <. □

The following was proved by Knebusch and Wright [KW] and by Prestel (cf. [PR]); see Theorem 5.6 of [L].

Theorem 2.19 Let v be a convex valuation on the ordered field (K, <), <_{r} the ordering induced by < on Kv, and R a real closure of (K, <). Then there exists a unique extension of v to a convex valuation of R. Denoting this extension again by v, we have that (R, v) is henselian, vR is divisible, and Rv with the ordering induced by < is a real closure of (Kv, <_{r}).

Corollary 2.20 Let v be a convex valuation on the ordered field (K, <) and R a real closure of (K, <), endowed with the unique convex extension of v. Further, let Γ|vK be an algebraic extension of ordered abelian groups, and k|Kv a subextension of some real closure of Kv. Then there is a (separable-algebraic) subextension (L|K, v) of (R|K, v) such that vL = Γ and Lv = k. If both Γ|vK and k|Kv are finite, then L|K can be taken to be a finite simple extension of the form K(a)|K such that [K(a) : K] = (Γ : vK)[k : Kv].

We leave the proof of the corollary as an exercise to the reader. It is a straightforward application of Hensel’s Lemma, using the fact that (R, v) is henselian. One also uses the fact that all real closures of Kv are isomorphic over Kv, so by passing to an equivalent residue map (place), one passes from Rv to the real closure given in the hypothesis.

2.5 A version of Krasner’s Lemma

Let (K, v) be any valued field. If a ∈ K̃ \ K is not purely inseparable over K, we choose some extension of v from K to K̃ and define

kras(a, K) := max{v(τa − σa) | σ, τ ∈ Gal K and τa ≠ σa} ∈ vK̃

and call it the Krasner constant of a over K. Since all extensions of v from K to K̃ are conjugate, this does not depend on the choice of the particular extension of v. For the same reason, over a henselian field (K, v) our Krasner constant kras(a, K) coincides with the Krasner constant

max{v(a − σa) | σ ∈ Gal K and a ≠ σa}

as defined by S. K. Khanduja in [KH11,12]. The following is a variant of the well-known Krasner’s Lemma (cf. [R]).

Lemma 2.21 Take K(a)|K to be a separable-algebraic extension, and (K(a, b), v) to be any valued field extension of (K, v) such that

v(b − a) > kras(a, K) .  (9)

Then for every extension of v from K(a, b) to its algebraic closure K̃(a, b) = K̃(b), the element a lies in the henselization of (K(b), v) in (K̃(b), v).

Proof: Take any extension of v from K(a, b) to K̃(b) and denote by K(b)^{h} the henselization of (K(b), v) in (K̃(b), v). Since a is separable-algebraic over K, it is also separablealgebraic over K(b)^{h}. Since for every ρ ∈ Gal K(b)^{h} we have that ρa = ρ|K̃a and ρ|K̃ ∈ Gal K, we find that

{v(a − ρa) | ρ ∈ Gal K(b)^{h} and a ≠ ρa} ⊆ {v(a − σa) | σ ∈ Gal K and a ≠ σa}
⊆ {v(τa − σa) | σ, τ ∈ Gal K and τa ≠ σa} .

This implies that

kras(a, K(b)^{h}) ≤ kras(a, K) ,

and consequently, v(b − a) > kras(a, K(b)^{h}). Now a ∈ K(b)^{h} follows from the usual Krasner’s Lemma. □

3 Valuations on K(x)

3.1 A basic classification

In this section, we wish to classify all extensions of the valuation v of K to a valuation of the rational function field K(x). As

1 = trdeg K(x)|K ≥ rr vK(x)/vK + trdeg K(x)v|Kv  (10)

holds by Lemma 2.6, there are the following mutually exclusive cases:

• (K(x)|K, v) is valuation-algebraic:

vK(x)/vK is a torsion group and K(x)v|Kv is algebraic,

• (K(x)|K, v) is value-transcendental:

vK(x)/vK has rational rank 1, but K(x)v|Kv is algebraic,

• (K(x)|K, v) is residue-transcendental:

K(x)v|Kv has transcendence degree 1, but vK(x)/vK is a torsion group.

We will combine the value-transcendental case and the residue-transcendental case by saying that

• (K(x)|K, v) is valuation-transcendental:

vK(x)/vK has rational rank 1, or K(x)v|Kv has transcendence degree 1.

A special case of the valuation-algebraic case is the following:

• (K(x)|K, v) is immediate:

vK(x) = vK and K(x)v = Kv.

Remark 3.1 It was observed by several authors that a valuation-algebraic extension of v from K to K(x) can be represented as a limit of an infinite sequence of residuetranscendental extensions. See, e.g., [APZ3], where the authors also derive the assertion of our Theorem 2.9 from this fact. A “higher form” of this approach is found in [S]. The approach is particularly important because residue-transcendental extensions behave better than valuation-algebraic extensions: the corresponding extensions of value group and residue field are finitely generated (Corollary 2.7), and they do not generate a defect: see the Generalized Stability Theorem (Theorem 3.1) and its application in [KKU1].

If K is algebraically closed, then the residue field Kv is algebraically closed, and the value group vK is divisible. So we see that for an extension (K(x)|K, v) with algebraically closed K, there are only the following mutually exclusive cases: (K(x)|K, v) is immediate: vK(x) = vK and K(x)v = Kv, (K(x)|K, v) is value-transcendental: rr vK(x)/vK = 1, but K(x)v = Kv, (K(x)|K, v) is residue-transcendental: trdeg K(x)v|Kv = 1, but vK(x) = vK.

Let us fix an arbitrary extension of v to K̃. Every valuation w on K(x) can be extended to a valuation on K̃(x). If v and w agree on K, then this extension can be chosen in such a way that its restriction to K̃ coincides with v. Indeed, if w′ is any extension of w to K̃(x) and v′ is its restriction to K̃, then there is an automorphism τ of K̃|K such that v′τ = v on K̃. We choose σ to be the (unique) automorphism of K̃(x)|K(x) whose restriction to K̃ is τ and which satisfies σx = x. Then w′σ is an extension of w from K(x) to K̃(x) whose restriction to K̃ is v. We conclude:

Lemma 3.2 Take any extension of v from K to its algebraic closure K̃. Then every extension of v from K to K(x) is the restriction of some extension of v from K̃ to K̃(x).

Now extend v to K̃(x). We know that vK̃(x)/vK(x) and vK̃/vK are torsion groups, and also vK̃(x)/vK(x) ⊂ vK̃(x)/vK(x) is a torsion group. Hence,

rr vK̃(x)/vK̃ = rr vK(x)/vK .

Since vK̃ is divisible, vK(x)/vK is a torsion group if and only if vK̃(x) = vK̃.

Further, the extensions K̃(x)v|K(x)v and K̃v|Kv are algebraic, and also the subextension K̃(x)v|K(x)v of K̃(x)v|K(x)v is algebraic. Hence,

trdeg K̃(x)v|K̃v = trdeg K(x)v|Kv .

Since K̃v is algebraically closed, K(x)v|Kv is algebraic if and only if K̃(x)v = K̃v. We have proved:

Lemma 3.3 (K(x)|K, v) is valuation-algebraic if and only if (K̃(x)|K̃, v) is immediate. (K(x)|K, v) is valuation-transcendental if and only if (K̃(x)|K̃, v) is not immediate. (K(x)|K, v) is value-transcendental if and only if (K̃(x)|K̃, v) is value-transcendental. (K(x)|K, v) is residue-transcendental if and only if (K̃(x)|K̃, v) is residue-transcendental.

The proof can easily be generalized to show:

Lemma 3.4 Let (F |K, v) be any valued field extension. Then vF |vK and F v|Kv are algebraic if and only if (F.K̃|K̃, v) is immediate, for some (or any) extension of v from F to F.K̃.

3.2 Pure and weakly pure extensions

Take t ∈ K(x). If vt is not a torsion element modulo vK, then t will be called a valuetranscendental element. If vt = 0 and tv is transcendental over Kv, then t will be called a residue-transcendental element. An element will be called a valuationtranscendental element if it is value-transcendental or residue-transcendental. We will call the extension (K(x)|K, v) pure (in x) if one of the following cases holds:

• for some c, d ∈ K, d · (x − c) is valuation-transcendental,

• x is the pseudo limit of some pseudo Cauchy sequence in (K, v) of transcendental type. We leave it as an exercise to the reader to prove that (K(x)|K, v) is pure in x if and only if it is pure in any other generator of K(x) over K; we will not need this fact in the present paper.

If (K(x)|K, v) is pure in x then it follows from Lemma 2.6 and Lemma 2.4 that x is transcendental over K. If d · (x − c) is value-transcendental, then vK(x) = vK ⊕ ℤv(x − c) and K(x)v = Kv by Lemma 2.6 (in this case, we may in fact choose d = 1). If d · (x − c) is residue-transcendental, then again by Lemma 2.6, we have vK(x) = vK and that K(x)v = Kv(d(x − c)v) is a rational function field over Kv. If x is the pseudo limit of some pseudo Cauchy sequence in (K, v) of transcendental type, then (K(x)|K, v) is immediate by Lemma 2.4. This proves:

Lemma 3.5 If (K(x)|K, v) is pure, then vK is pure in vK(x) (i.e., vK(x)/vK is torsion free), and Kv is relatively algebraically closed in K(x)v.

Here is the “prototype” of pure extensions:

Lemma 3.6 If K is algebraically closed and x ∉ K, then (K(x)|K, v) is pure.

Proof: Suppose that the set

v(x − K) := {v(x − b) | b ∈ K}  (11)

has no maximum. Then there is a pseudo Cauchy sequence in (K, v) with pseudo limit x, but without a pseudo limit in K. Since K is algebraically closed, Theorem 3 of [KA] shows that this pseudo Cauchy sequence must be of transcendental type. The extension therefore satisfies the third condition for being pure.

Now assume that the set v(x−K) has a maximum, say, v(x−c) with c ∈ K. If v(x−c) is a torsion element over vK, then v(x−c) ∈ vK because vK is divisible as K is algebraically closed. It then follows that there is some d ∈ K such that vd(x − c) = 0. If d(x − c)v were algebraic over Kv, then it were in Kv since K and thus also Kv is algebraically closed. But then, there were some b_{0} ∈ K such that v(d(x − c) − b_{0}) > 0. Putting b := c + d^{−1}b_{0} , we would then obtain that v(x − b) = v((x − c) − d^{−1}b_{0}) > −vd = v(x − c), a contradiction to the maximality of v(x − c). So we see that either v(x − c) is non-torsion over vK, or there is some d ∈ K such vd(x − c) = 0 and d(x − c)v is transcendental over Kv. In both cases, this shows that (K(x)|K, v) is pure. □

We will call the extension (K(x)|K, v) weakly pure (in x) if it is pure in x or if there are c, d ∈ K and e ∈ ℕ such that vd(x − c)^{e} = 0 and d(x − c)^{e}v is transcendental over Kv.

Lemma 3.7 Assume that the extension (K(x)|K, v) is weakly pure. If we take any extension of v to K̃(x) and take K^{h} to be the henselization of K in (K̃(x), v), then K^{h} is the implicit constant field of this extension:

K^{h} = IC (K(x)|K, v) .

Proof: As we have noted already in the introduction, K^{h} is contained in IC (K(x)|K, v). Since K(x)^{h} is the fixed field of the decomposition group G_{x}^{d} := G^{d}(K(x)^{sep}|K(x), v) in the separable-algebraic closure K(x)^{sep} of K(x), we know that IC (K(x)|K, v) is the fixed field in K^{sep} of the subgroup

G_{res} := {σ|_{K^{sep}} | σ ∈ G_{x}^{d}}

of Gal K. In order to show our assertion, it suffices to show that IC (K(x)|K, v) ⊆ K^{h}, that is, that the decomposition group G^{d} := G^{d}(K^{sep}|K, v) is contained in G_{res} . So we have to show: if τ is an automorphism of K^{sep}|K such that vτ = v on K^{sep}, then τ can be lifted to an automorphism σ of K(x)^{sep}|K(x) such that vσ = v on K(x)^{sep}. In fact, it suffices to show that τ can be lifted to an automorphism σ of K^{sep}(x)|K(x) such that vσ = v on K^{sep}(x). Indeed, then we take any extension σ′ of σ from K^{sep}(x) to K(x)^{sep}. Since the extensions vσ′ and v of v from K^{sep}(x) to K(x)^{sep} are conjugate, there is some ρ ∈ Gal (K(x)^{sep}|K^{sep}(x)) such that vσ′ρ = v on K(x)^{sep}. Thus, σ := σ′ρ ∈ G^{d} is the desired lifting of τ to K(x)^{sep}.

We take σ on K^{sep}(x) to be the unique automorphism which satisfies σx = x and σ|_{K^{sep}} = τ. Using that (K(x)|K, v) is weakly pure, we have to show that vσ = v on K^{sep}(x). Assume first that for some c, d ∈ K and e ∈ ℕ, d(x−c)^{e} is valuation-transcendental. Since K(x − c) = K(x), we may assume w.l.o.g. that c = 0. Every element of K^{sep}(x) can be written as a quotient of polynomials in x with coefficients from K^{sep}. For every polynomial f(x) = a_{n}x^{n} + … + a_{1}x + a_{0} ∈ K^{sep}[x],

vσf(x) = v (σ(a_{n})x^{n} + … + σ(a_{1})x + σ(a_{0}))
= min_{i}(vσ(a_{i}) + ivx) = min_{i}(vτ(a_{i}) + ivx)
= min_{i}(va_{i} + ivx) = vf(x) ,

where the second equality holds by Lemma 2.6 and Lemma 2.10. This shows that vσ = v on K^{sep}(x).

Now assume that x is the pseudo limit of a pseudo Cauchy sequence in (K, v) of transcendental type. By Lemma 2.5, this pseudo Cauchy sequence is also of transcendental type over (K^{sep}, v). Observe that x is still a pseudo limit of this pseudo Cauchy sequence in (K^{sep}(x), vσ), because vσ(x − a) = v(σx − σa) = v(x − a) for all a ∈ K. But vσ = vτ = v on K^{sep}, and the extension of v from K^{sep} to K^{sep}(x) is uniquely determined by the pseudo Cauchy sequence (cf. Theorem 2.3). Consequently, vσ = v on K^{sep}(x). □

3.3 Construction of nasty examples

We are now able to give the Proof of Theorem 1.2: Let K be any algebraically closed field of characteristic p > 0. On K(x), we take v to be the x-adic valuation. We work in the power series field K((\frac{1}{p^{∞}} ℤ)) of all power series in x with exponents in \frac{1}{p^{∞}} ℤ, the p-divisible hull of ℤ. We choose y to be a power series

y = ∑_{i=1}^{∞} x^{−p^{−e_{i}}}  (12)

where e_{i} is any increasing sequence of natural numbers such that e_{i+1} ≥ e_{i} + i for all i. We then restrict the canonical (x-adic) valuation of K((\frac{1}{p^{∞}} ℤ)) to K(x, y) and call it again v. We show first that vK(x, y) = \frac{1}{p^{∞}} ℤ. Indeed, taking p^{e_{j}}-th powers and using that the characteristic of K is p, we find

y^{p^{e_{j}}} − ∑_{i=1}^{j} x^{−p^{e_{j} −e_{i}}} = ∑_{i=j+1}^{∞} x^{−p^{e_{j} −e_{i}}} .

Since e_{j} − e_{i} ≥ 0 for i ≤ j, the left hand side is an element in K(x, y). The right hand side has value

−p^{e_{j}−e_{j+1}}vx ;

since e_{j} − e_{j+1} ≤ −j, we see that \frac{1}{p^{j}} vx lies in vK(x, y). Hence, \frac{1}{p^{∞}} ℤ ⊆ vK(x, y). On the other hand, vK(x, y) ⊆ vK((\frac{1}{p^{∞}} ℤ)) = \frac{1}{p^{∞}} ℤ and therefore, vK(x, y) = \frac{1}{p^{∞}} ℤ.

By definition, y is a pseudo limit of the pseudo Cauchy sequence

(∑_{i=1}^{ℓ} x^{−p^{−e_{i}}} )_{ℓ∈ℕ}

in the field L = K(x^{1/p^{i}} | i ∈ ℕ) ⊂ K((\frac{1}{p^{∞}} ℤ)). Suppose it were of algebraic type. Then by [KA], Theorem 3, there would exist some algebraic extension (L(b)|L, v) with b a pseudo limit of the sequence. But then b would also be algebraic over K(x) and hence the extension K(x, b)|K(x) would be finite. On the other hand, since b is a pseudo limit of the above pseudo Cauchy sequence, it can be shown as before that vK(x, b) = \frac{1}{p^{∞}} ℤ and thus, (vK(x, b) : vK(x)) = ∞. This contradiction to the fundamental inequality shows that the sequence must be of transcendental type. Hence by Lemma 3.7, L^{h} is relatively algebraically closed in L(y)^{h}. Since L^{h} = L.K(x)^{h} is a purely inseparable extension of K(x)^{h} and K(x, y)^{h}|K(x)^{h} is separable, this shows that K(x)^{h} is relatively algebraically closed in K(x, y)^{h}.

Now we set η_{0} := \frac{1}{x}, and by induction on i we choose η_{i} ∈ K̃(x) such that η_{i}^{p}−η_{i} = η_{i−1} . Then we have

vη_{i} = −\frac{1}{p^{i}} vx

for every i. Since vK(x)^{h} = vK(x) = ℤvx, this shows that the extension K(x)^{h}(η_{i})|K(x)^{h} has ramification index at least p^{i}. On the other hand, it has degree at most p^{i} and therefore, it must have degree and ramification index equal to p^{i}. Note that for all i ≥ 0, K(x, η_{i}) = K(x, η_{1}, …, η_{i}) and every extension K(x, η_{i+1})|K(x, η_{i}) is a Galois extension of degree and ramification index p. By what we have shown, the chain of these extensions is linearly disjoint from K(x)^{h}|K(x). Since K(x)^{h} is relatively algebraically closed in K(x, y)^{h} and the extensions are separable, it follows that the chain is also linearly disjoint from K(x, y)^{h}|K(x).

We will now show that all extensions K(x, y, η_{i})|K(x, y) are immediate. First, we note that K(x, y)v = K since K ⊂ K(x, y) ⊂ K((\frac{1}{p^{∞}} ℤ)) and K((\frac{1}{p^{∞}} ℤ))v = K. Since K is algebraically closed, the inertia degree of the extensions must be 1. Further, as the ramification index of a Galois extension is always a divisor of the extension degree, the ramification index of these extensions must be a power of p. But the value group of K(x, y) is p-divisible, which yields that the ramification index of the extensions is 1. By what we have proved above, they are linearly disjoint from K(x, y)^{h}|K(x, y), that is, the extension of the valuation is unique. This shows that the defect of each extension K(x, y, η_{i})|K(x, y) is equal to its degree p^{i}. □

Remark Instead of defining y as in (12), we could also use any power series

y = ∑_{i=1}^{∞} x^{n_{i}p^{−e_{i}}}  (13)

where n_{i} ∈ ℤ are prime to p and the sequence n_{i}p^{−e_{i}} is strictly increasing. The example in [CP] is of this form. But in this example, the field K(x, y) is an extension of degree p^{2} of a field K(u, v) such that the extension of the valuation from K(u, v) to K(x, y) is unique. Since the value group of K(x, y) is \frac{1}{p^{∞}} ℤ, it must be equal to that of K(u, v). Since K is algebraically closed, both have the same residue field. Therefore, the extension has defect p^{2}. This shows that we can also use subfields instead of field extensions to produce defect extensions, in quite the same way.

A special case of (13) is the power series

y = ∑_{i=1}^{∞} x^{i−p^{−e_{i}}} = ∑_{i=1}^{∞} x^{i}x^{−p^{−e_{i}}}  (14)

which now has a support that is cofinal in \frac{1}{p^{∞}} ℤ.

To conclude this section, we use Lemma 3.7 to construct an example about relatively closed subfields in henselian fields. The following fact is well known: Suppose that K is relatively closed in a henselian valued field (L, v) of residue characteristic 0 and that Lv|Kv is algebraic. Then vL/vK is torsion free. We show that the assumption “Lv|Kv is algebraic” is necessary. Example On the rational function field ℚ(x), we take v to be the x-adic valuation. We extend v to the rational function field ℚ(x, y) in such a way that vy = 0 and yv is transcendental over ℚ(x)v = ℚ. So by Lemma 2.6 we have vℚ(x, y) = vℚ(x) = ℤvx and ℚ(x, y)v = ℚ(yv). We pick any integer n > 1. Then vℚ(x^{n}) = ℤnvx and ℚ(x^{n})v = ℚ. Further, vℚ(x^{n}, xy) = ℤvx since vx = vxy ∈ vℚ(x^{n}, xy) ⊆ ℤvx. Also, ℚ(x^{n}, xy)v = ℚ(y^{n}v) by Lemma 2.10. From Lemma 3.7 we infer that ℚ(x^{n})^{h} is relatively algebraically closed in ℚ(x^{n}, xy)^{h}. But

vℚ(x^{n}, xy)^{h}/vℚ(x^{n})^{h} = vℚ(x^{n}, xy)/vℚ(x^{n}) = ℤvx/ℤnvx ≃ ℤ/nℤ

is a non-trivial torsion group. □

3.4 All valuations on K(x)

In this section, we will explicitly define extensions of a given valuation on K to a valuation on K(x). First, we define valuation-transcendental extensions, using the idea of valuation independence. Let (K, v) be an arbitrary valued field, and x transcendental over K. Take a ∈ K and an element γ in some ordered abelian group extension of vK. We define a map v_{a,γ} : K(x)^{×} → vK + ℤγ as follows. Given any g(x) ∈ K[x] of degree n, we can write

g(x) = ∑_{i=0}^{n} c_{i}(x − a)^{i} .  (15)

Then we set

v_{a,γ} g(x) := min_{0≤i≤n} vc_{i} + iγ .  (16)

We extend v_{a,γ} to K(x) by setting v_{a,γ}(g/h) := v_{a,γ}g − v_{a,γ}h.

For example, the valuation v_{0,0} is called Gauß valuation or functional valuation and is given by

v_{0,0} (c_{n}x^{n} + … + c_{1}x + c_{0}) = min_{0≤i≤n} vc_{i} .

Lemma 3.8 v_{a,γ} is a valuation which extends v from K to K(x). It satisfies:

1) If γ is non-torsion over vK, then v_{a,γ}K(x) = vK ⊕ ℤγ and K(x)v_{a,γ} = Kv.

2) If γ is torsion over vK, e is the smallest positive integer such that eγ ∈ vK and d ∈ K is some element such that vd = −eγ, then d(x − a)^{e}v_{a,γ} is transcendental over Kv, K(x)v_{a,γ} = Kv(d(x − a)^{e}v_{a,γ}) and v_{a,γ}K(x) = vK + ℤγ. In particular, if γ = 0 then (x − a)v_{a,γ} is transcendental over Kv, K(x)v_{a,γ} = Kv ((x − a)v_{a,γ}) and v_{a,γ}K(x) = vK.

Proof: It is a straightforward exercise to prove that v_{a,γ} is a valuation and that 1) and 2) hold. However, one can also deduce this from Lemma 2.6. It says that if we assign a non-torsion value γ to x − a then we obtain a unique valuation which satisfies (16). Since this defines a unique map v_{a,γ} on K[x], we see that v_{a,γ} must coincide with the valuation given by Lemma 2.6, which in turn satisfies assertion 1). Similarly, if γ ∈ vK, d ∈ K with vd = −γ and we assign a transcendental residue to d(x − a), then Lemma 2.6 gives us a valuation on K(x) which satisfies (16) and hence must coincide with v_{a,γ}. This shows that v_{a,γ} is a valuation and satisfies 2).

If e > 1, then we can first use Lemma 2.6 to see that v_{a,γ} is a valuation on the subfield K(d(x − a)^{e}) of K(x) and that v_{a,γ}K(d(x − a)^{e}) = vK and K(d(x − a)^{e})v_{a,γ} = Kv(d(x − a)^{e}v_{a,γ}) with d(x − a)^{e}v_{a,γ} transcendental over Kv. We know that there is an extension w of v_{a,γ} to K(x). It must satisfy w(x − a) = −vd/e = γ. So 0, w(x − a), w(x − a)^{2}, …, w(x − a)^{e−1} lie in distinct cosets modulo vK. From Lemma 2.8 it follows that w satisfies (16) on K(x), hence it must coincide with the valuation v_{a,γ} on K(x). Assertion 2) for this case follows from Lemma 2.8. □

Now we are able to prove:

Theorem 3.9 Take any valued field (K, v). Then all extensions of v to the rational function field K̃(x) are of the form • ṽ_{a,γ} where a ∈ K̃ and γ is an element of some ordered group extension of vK, or

• ṽ_{A} where A is a pseudo Cauchy sequence in (K̃, ṽ) of transcendental type, where ṽ runs through all extensions of v to K̃. The extension is of the form ṽ_{a,γ} with γ ∉ ṽK̃ if and only if it is value-transcendental, and with γ ∈ ṽK̃ if and only if it is residue-transcendental. The extension is of the form ṽ_{A} if and only if it is valuationalgebraic.

All extensions of v to K(x) are obtained by restricting the above extensions, already from just one fixed extension ṽ of v to K̃.

Proof: By Lemma 3.8 and Theorem 2.3, ṽ_{a,γ} and ṽ_{A} are extensions of ṽ to K̃(x). For the converse, let w be any extension of v to K̃(x) and set ṽ = w|K̃ . From Lemma 3.6 we know that (K̃(x)|K̃, w) is always pure. Hence, either d(x − c) is valuation-transcendental for some c, d ∈ K̃, or x is the pseudo limit of some pseudo Cauchy sequence A in (K̃, ṽ) of transcendental type. In the first case, Lemma 2.6 shows that

w ∑_{i=0}^{n} d_{i}(d(x − c))^{i} = min_{0≤i≤n} vd_{i} + iwd(x − c) = min_{0≤i≤n} vd_{i}d^{i} + iw(x − c)

for all d_{i} ∈ K̃. This shows that w = ṽ_{c,γ} for γ = w(x − c). If this value is not in ṽK̃, then it is non-torsion over ṽK̃ and thus, the extension of ṽ to K̃(x), and hence also the extension of v to K(x), is value-transcendental. If it is in ṽK̃, then the residue of d(x − c) is not in K̃ṽ, and the extension of ṽ to K̃(x), and hence also the extension of v to K(x), is residue-transcendental.

In the second case, we know from Theorem 2.3 that A induces an extension ṽ_{A} of ṽ to K̃(x) such that x is a pseudo limit of A in (K̃(x), ṽ_{A}). Since x is also a pseudo limit of A in (K̃(x), w), we can infer from Lemma 2.4 that w = ṽ_{A} . It also follows from Theorem 2.3 that (K̃(x)|K̃, ṽ_{A}) is immediate and consequently, (K̃(x)|K, ṽ_{A}) is valuation-algebraic.

For the last assertion, we invoke Lemma 3.2. Now it just remains to show that it suffices to take the restrictions of the valuations ṽ_{a,γ} and ṽ_{A} for one fixed ṽ. Suppose that w̃ is another extension of v to K̃. Since all such extensions are conjugate, there is σ ∈ Gal K such that w̃ = ṽ ∘ σ. Let g(x) ∈ K[x] be given as in (15). Extend σ to an automorphism of K̃(x) which satisfies σx = x. Since g has coefficients in K, we then have

g(x) = σg(x) = ∑_{i} σc_{i}(x − σa)^{i}

and therefore,

w̃_{a,γ} g(x) = min_{i}(w̃c_{i} + iγ) = min_{i}(ṽσc_{i} + iγ) = ṽ_{σa,γ} g(x) .

This shows that w̃_{a,γ} = ṽ_{σa,γ} on K(x).

Given a pseudo Cauchy sequence A in (K̃, w̃), we set A_{σ} = (σa_{ν})_{ν<λ}. This is a pseudo Cauchy sequence in (K̃, ṽ) since ṽ(σa_{μ} − σa_{ν}) = ṽσ(a_{μ} − a_{ν}) = w̃(a_{μ} − a_{ν}). For every polynomial f(x) ∈ K̃[x], we have ṽf(σa_{ν}) = w̃σ^{−1}(f(σa_{ν})) = w̃(σ^{−1}(f))(a_{ν}), where σ^{−1}(f) denotes the polynomial obtained from f(x) by applying σ^{−1} to the coefficients. So we see that A_{σ} is of transcendental type if and only if A is. If g(x) ∈ K[x], then σ^{−1}(g) = g and the above computation shows that ṽg(σa_{ν}) = w̃g(a_{ν}). This implies that w̃_{A} = ṽ_{A_{σ}} on K(x). □

Remark 3.10 If v is trivial on K, hence Kv = K (modulo an isomorphism), and if we choose γ > 0, then the restriction w of ṽ_{a,γ} to K(x) will satisfy xw = aw = a. It follows that K(x)w = K(a). Further, wK(x) ⊆ ℤγ and thus, wK(x) ≃ ℤ.

3.5 Prescribed implicit constant fields

If not stated otherwise, we will always assume that (K, v) is any valued field. The following is an immediate consequence of our version of Krasner’s Lemma:

Lemma 3.11 Assume that K(a)|K is a separable-algebraic extension. Assume further that K(x)|K is a rational function field and v is a valuation on K̃(x) such that

v(x − a) > kras(a, K) .  (17)

Then K(a) ⊆ (K(x)|K, v), and consequently,

vK(a) ⊆ vK(x) and K(a)v ⊆ K(x)v .

Proof: By Lemma 2.21, condition (17) implies that K(a) ⊆ K(x)^{h}, the henselization being chosen in (K̃(x), v). Consequently, K(a) ⊆ IC (K(x)|K, v), vK(a) ⊆ vK(x)^{h} = vK(x) and K(a)v ⊆ K(x)^{h}v = K(x)v. □

Proposition 3.12 Assume that (K(a)|K, v) is a separable-algebraic extension of valued fields. Further, take Γ to be the abelian group vK(a) ⊕ ℤ endowed with any extension of the ordering of vK(a), and take k to be the rational function field in one variable over K(a)v. Then there exists an extension v_{1} of v from K(a) to K̃(x) such that

v_{1}K(x) = Γ and K(x)v_{1} = K(a)v and K(a)^{h} = IC (K(x)|K, v_{1}) .  (18)

If v is non-trivial on K, then there exists an extension v_{2} of v from K(a) to K̃(x) such that

v_{2}K(x) = vK(a) and K(x)v_{2} = k and K(a)^{h} = IC (K(x)|K, v_{2}) .  (19)

If (K(a), v) admits a transcendental immediate extension, then there is also an extension v_{3} such that

v_{3}K(x) = vK(a) and K(x)v_{3} = K(a)v .  (20)

If in addition (K(a), v) admits a pseudo Cauchy sequence of transcendental type, then v_{3} can be chosen such that K(a)^{h} = IC (K(x)|K, v_{3}).

Suppose that (K(a), <) is an ordered field and that v is convex. Denote by <_{r} the ordering induced by < on the residue field K(a)v. Then <_{r} can be lifted through v_{1} and through v_{3} to K(x) in such a way that the lifted orderings extend < (from K). If <′_{r} is an extension of <_{r} to k, then <′_{r} can be lifted through v_{2} to K(x) in such a way that the lifted ordering extends < (from K).

Proof: If K(a) = K, then we do the following. We take a generator γ of Γ over vK and let v_{1} be any extension of v_{0,γ} to K̃(x). Further, we take v_{2} to be any extension of v_{0,0} to K̃(x). Then the desired properties of v_{1} and v_{2} follow from Lemma 3.8 and Lemma 3.7. To construct v_{3}, we send x to some transcendental element in the given immediate extension of (K, v). The so obtained embedding induces an extension v_{3} of v to K(x) such that (K(x)|K, v_{3}) is immediate. We extend v_{3} further to K̃(x). If (K, v) admits a pseudo Cauchy sequence of transcendental type, then we can use Theorem 2.3 to construct an immediate extension v_{3} of v to K(x) such that x is a pseudo limit of this pseudo Cauchy sequence. Then by Lemma 3.7, K^{h} = IC (K(x)|K, v_{3}).

Now assume that a ∉ K. If v′ is any extension of v to K̃(x) such that v′(x − a) > kras(a, K), then Lemma 3.11 shows that a ∈ K(x)^{h}, where the henselization is taken in (K̃(x), v′). Thus,

K(x)^{h} = K(a, x)^{h} ,  (21)

and consequently,

\begin{matrix}v′K(x) & = & v′K(x)^{h} = v′K(a, x)^{h} = v′K(a, x) \\ K(x)v′ & = & K(x)^{h}v′ = K(a, x)^{h}v′ = K(a, x)v′ . \end{matrix} \}  (22)

If in addition we know that

K(a)^{h} = IC (K(a, x)|K(a), v′) ,  (23)

then

K(a)^{h} ⊆ IC (K(x)|K, v′) ⊆ IC (K(a, x)|K(a), v′) = K(a)^{h} ,

which yields that K(a)^{h} = IC (K(x)|K, v′).

As vK is cofinal in its divisible hull vK̃, we can choose some α ∈ vK such that α ≥ kras(a, K).

To construct v_{1} , we choose any positive generator β of Γ over vK(a). Then also γ := α + β is a generator of Γ over vK(a). Now we take v_{1} to be any extension of v_{a,γ} to K̃(x). From Lemma 3.8 we know that v_{a,γ}K(a, x) = vK(a) ⊕ ℤγ = Γ and K(a, x)v_{a,γ} = K(a)v. Since

v_{a,γ}(x − a) = γ > kras(a, K) ,  (24)

equations (22) hold for v′ = v_{a,γ}, and we obtain:

v_{1}K(x) = v_{a,γ}K(a, x) = vK(a) + ℤγ = Γ
K(x)v_{1} = K(a, x)v_{a,γ} = K(a)v .

From Lemma 3.7 we get that (23) holds for v′ = v_{a,γ}. Hence, K(a)^{h} = IC (K(x)|K, v_{1}).

To construct v_{2} , we make use of our assumption that v is non-trivial on K and choose some positive β ∈ vK. We set γ := α + β and take v_{2} to be any extension of v_{a,γ} to K̃(x). From Lemma 3.8 we know that v_{a,γ}K(a, x) = vK(a) = Γ and K(a, x)v_{a,γ} is a rational function field in one variable over K(a)v; in this sense, K(a, x)v_{a,γ} = k. Again we have (24), and equations (22) hold for v′ = v_{a,γ}; so we obtain

v_{2}K(x) = v_{a,γ}K(a, x) = vK(a)
K(x)v_{2} = K(a, x)v_{a,γ} = k .

From Lemma 3.7 we get that (23) holds for v′ = v_{a,γ}. Hence, K(a)^{h} = IC (K(x)|K, v_{2}).

To construct v_{3} , we take (L|K(a), v) to be the transcendental immediate extension which exists by hypothesis. We choose some y ∈ L, transcendental over K(a) and such that γ := vy > kras(a, K). Then (K(a, y)|K(a), v) is immediate. Now the isomorphism K(a, y) ≃ K(a, x) induced by y ↦ x − a induces on K(a, x) a valuation v′ such that (K(a, x)|K(a), v′) is immediate and v′(x − a) = γ. We take v_{3} to be any extension of v′ to K̃(x). Again, we have (22), and we obtain

v_{3}K(x) = v′K(a, x) = vK(a)
K(x)v_{3} = K(a, x)v′ = K(a)v .

If (K(a), v) admits a pseudo Cauchy sequence of transcendental type, then we can use Theorem 2.3 to construct an immediate extension (K(a, z)|K(a), v). Multiplying z with a suitable element from K will give us y such that vy > kras(a, K). By our above construction, also x will be a pseudo limit of a pseudo Cauchy sequence of transcendental type. Then by Lemma 3.7, K(a)^{h} = IC (K(x)|K, v_{3}).

Finally, suppose that (K(a), <) is an ordered field and that <_{r} is the ordering induced by < on K(a)v. We have that vK(a) = v_{2}K(a, x) = v_{3}K(a, x), so it is trivially true that 2v_{2}K(a, x) ∩ vK(a) = 2vK(a) = 2v_{3}K(a, x) ∩ vK(a). Further, v_{1}K(a, x) = vK(a) ⊕ ℤγ, hence also in this case, 2v_{1}K(a, x) ∩ vK(a) = 2vK(a). Therefore, Proposition 2.18 shows that the given orderings on K(a)v and k can be lifted through v_{1} , v_{2} and v_{3} respectively, to orderings on K(a, x) which extend the ordering < of K(a). □

This proposition proves Theorem 1.3 in the case where K_{1}^{h}|K^{h} is finite since then, there is some a ∈ K_{1} such that K_{1}^{h} = K(a)^{h}. Further, we obtain:

Proposition 3.13 Take any finite ordered abelian group extension Γ_{0} of vK and any finite field extension k_{0} of Kv. Further, take Γ to be the abelian group Γ_{0} ⊕ ℤ endowed with any extension of the ordering of Γ_{0} , and take k to be the rational function field in one variable over k_{0} . If v is trivial on K, then assume in addition that k_{0}|Kv is separable. Then there exists an extension v_{1} of v from K to K(x) such that

v_{1}K(x) = Γ and K(x)v_{1} = k_{0} .  (25)

If v is non-trivial on K, then there exists an extension v_{2} of v from K to K(x) such that

v_{2}K(x) = Γ_{0} and K(x)v_{2} = k .  (26)

If (K, v) admits a transcendental immediate extension, then there is also an extension v_{3} such that

v_{3}K(x) = Γ_{0} and K(x)v_{3} = k_{0} .  (27)

Suppose that (K, <) is an ordered field and that v is convex. Suppose further that k_{0} and k are equipped with extensions of the ordering induced by < on Kv. Then these extensions can be lifted through v_{1} , v_{2} , v_{3} to K(x) in such a way that the lifted orderings extend <.

Proof: We choose any finite separable extension (K(a)|K, v) such that vK(a) = Γ_{0} , K(a)v = k_{0} and [K(a) : K] = (vK(a) : vK)[K(a)v : Kv]. If v is non-trivial on K, then such an extension exists by Theorem 2.14; otherwise, K(a) is just equal to k_{0} , up to the isomorphism induced by the residue map of the trivial valuation v. If (K, v) admits a transcendental immediate extension, then by Lemma 2.2, also (K(a), v) admits a transcendental immediate extension. Now the first part of our theorem follows from Proposition 3.12.

Suppose that (K, <) is an ordered field with v convex. Then by Corollary 2.20 we can choose the field K(a) to be a subfield of a real closure (R, <) of K, equipped with a convex extension of v, in such a way that the given ordering on k_{0} is induced by < through this extension of v. Now again, our assertion for the ordered case follows from Proposition 3.12. □

We turn to the realization of countably infinite separable-algebraic extensions as implicit constant fields.

Proposition 3.14 Let (K_{1}|K, v) be a countably infinite separable-algebraic extension of non-trivially valued henselian fields. Then (K_{1}, v) admits a pseudo Cauchy sequence of transcendental type. In particular, there is an extension v_{3} of v to K̃(x) such that (K_{1}(x)|K_{1}, v) is immediate, with x being the pseudo limit of this pseudo Cauchy sequence. Moreover, K_{1} = IC (K(x)|K, v_{3}).

Proof: K_{1}|K is a countably infinite union of finite subextensions. Thus, we can choose a sequence a_{i}, i ∈ ℕ such that K(a_{i})|K is separable-algebraic and K(a_{i}) ⊂_{≠} K(a_{i+1}) for all i, and such that K_{1} = ⋃_{i∈ℕ} K(a_{i}). Through multiplication with elements from K it is possible to choose each a_{i} in such a way that

va_{i+1} > max{va_{i}, kras(a_{i}, K(a_{1}, …, a_{i−1}))} .  (28)

We set

b_{i} := ∑_{j=1}^{i} a_{j} .

By construction, (b_{i})_{i∈ℕ} is a pseudo Cauchy sequence. Suppose that x is a pseudo limit of it, for some extension of v from K_{1} to K_{1}(x). Then by induction on i, we show that a_{i} ∈ K(x)^{h}, where K(x)^{h} is chosen in some henselization of (K_{1}(x), v). First, by Lemma 2.21, v(x − a_{1}) = v(x − b_{1}) = va_{2} > kras(a_{1}, K) implies that a_{1} ∈ K(x)^{h}. If we have already shown that a_{1}, …, a_{i} ∈ K(x)^{h}, then b_{i} ∈ K(x)^{h} and v(x − b_{i} − a_{i+1}) = v(x−b_{i+1}) = va_{i+2} > kras(a_{i+1}, K(a_{1}, …, a_{i})) implies that a_{i+1} ∈ K(a_{1}, …, a_{i})(x−b_{i})^{h} = K(a_{1}, …, a_{i})(x)^{h} = K(x)^{h}.

This proves that K_{1} ⊆ K(x)^{h}. Since K_{1}|K is infinite, this also proves that x must be transcendental. As a pseudo Cauchy sequence of algebraic type would admit an algebraic pseudo limit ([KA], Theorem 3), this yields that (b_{i})_{i∈ℕ} is a pseudo Cauchy sequence of transcendental type in (K_{1}, v).

By Theorem 2.3 we can now extend v to K_{1}(x) such that (K_{1}(x)|K_{1}, v) is immediate and x is a pseudo limit of (b_{i})_{i∈ℕ}. We choose any extension v_{3} of v from K_{1}(x) to K̃(x).

By Lemma 3.7 we know that K_{1} = IC (K_{1}(x)|K_{1}, v_{3}). Hence,

K_{1} ⊆ K(x)^{h} ⊆ IC (K(x)|K, v_{3}) ⊆ IC (K_{1}(x)|K_{1}, v_{3}) = K_{1} ,

which shows that K_{1} = IC (K(x)|K, v_{3}). □

Since IC (K(x)|K, v) = IC (K^{h}(x)|K^{h}, v), this proposition proves Theorem 1.3 in the case where K_{1}^{h}|K^{h} is countably infinite.

Proposition 3.15 Take any non-trivially valued field (K, v), any countably generated ordered abelian group extension Γ_{0} of vK such that Γ_{0}/vK is a torsion group, and any countably generated algebraic field extension k_{0} of Kv. Assume that Γ_{0}/vK or k_{0}|Kv is infinite. Then there exists an extension v_{3} of v from K to K(x) such that equations (27) hold.

Suppose that (K, <) is an ordered field and that v is convex. If k_{0} is equipped with an extension of the ordering induced by < on Kv, then this can be lifted through v_{3} to an ordering of K(x) which extends <.

Proof: We fix an extension of v to K̃. By Theorem 2.14 there is a countably generated separable-algebraic extension K_{1}|K^{h} such that vK_{1} = Γ_{0} and K_{1}v = k_{0} . Since at least one of the extensions Γ_{0}|vK and k_{0}|Kv is infinite, K_{1}^{h}|K^{h} is infinite too, taking the henselizations in (K̃, v). Hence by Proposition 3.14 there is an extension v_{3} of v to K_{1}(x) such that K_{1} ⊂ K(x)^{h} and that (K_{1}(x)|K_{1}, v_{3}) is immediate. Consequently, K_{1}(x)^{h} = K(x)^{h}, which gives us v_{3}K(x) = v_{3}K(x)^{h} = v_{3}K_{1}(x) = vK_{1} = Γ_{0} and K(x)v_{3} = K(x)^{h}v_{3} = K_{1}(x)v_{3} = K_{1}v = k_{0} .

Suppose that (K, <) is ordered, v is convex and k_{0} is equipped with an extension of the ordering induced by < on Kv. Take any real closure (R, <) of (K, <) with an extension of v to a convex valuation of R. By Corollary 2.20 we can choose the extension K_{1}|K as a subextension of R|K, with the ordering on k_{0} induced by < through v. Since (K_{1}(x)|K_{1}, v_{3}) is immediate, Proposition 2.18 shows that there is a lifting of the ordering of k_{0} through v to an ordering of K_{1}(x) which extends the ordering < of K_{1} . Its restriction to K(x) is an extension of the ordering < of K. □

To conclude this section, we now give an alternative Proof of Theorem 1.2: Let K be an algebraically closed field of characteristic p > 0. On K(x), we again take v to be the x-adic valuation. We assume that K contains an element t which is transcendental over the prime field of K. Then it can be proved that K(x)^{h} admits two infinite linearly disjoint towers of Galois extensions of degree p and ramification index p. They can be defined as follows. For the first tower, we set η_{0} = x^{−1}, take η_{i+1} to be a root of X^{p}−X−η_{i}, and set L_{i} := K(η_{i}). For the second tower, we set ϑ_{0} = tx^{−1}, take ϑ_{i+1} to be a root of X^{p} − X − tϑ_{i}, and set N_{i} := K(ϑ_{i}). Note that L_{0} = N_{0} = K(x), η_{i} ∈ L_{i+1} and ϑ_{i} ∈ N_{i+1} . For each i ≥ 0, we have that vη_{i} = vϑ_{i} = −\frac{1}{p^{i}} vx, and the extensions L_{i+1}|L_{i} and N_{i+1}|N_{i} are Galois of degree p with ramification index p. Consequently, the same is true for the extensions L_{i+1}^{h}|L_{i}^{h} and N_{i+1}^{h}|N_{i}^{h} (note that L_{i}^{h} = L_{i}.K(x)^{h} and N_{i}^{h} = N_{i}.K(x)^{h}).

We set L := ⋃_{i∈ℕ} L_{i} and N := ⋃_{i∈ℕ} N_{i}. By the above, L and N are linearly disjoint from K(x)^{h} over K(x). Thus, L^{h} = L.K(x)^{h} and N^{h} = N.K(x)^{h} are countably infinite separable-algebraic extensions of K(x)^{h}, and it can be proved that they are linearly disjoint over K(x)^{h}. We use Proposition 3.14 (with K replaced by K(x)^{h} and x replaced by y) to obtain an extension of v from K(x)^{h} to K(x)^{h}(y) such that vK(x)^{h}(y) = vL^{h} = \frac{1}{p^{∞}} ℤ and K(x)^{h}(y)v = L^{h}v = K(x)^{h}v = K, and such that the extension (K(x)^{h}(y)|K(x)^{h}, v) has implicit constant field L^{h} (i.e., L^{h} is relatively algebraically closed in K(x, y)^{h} = (K(x)^{h}(y))^{h}). Since K(x) is relatively algebraically closed in K(x, y), we see that N is linearly disjoint from K(x, y) over K(x) and therefore, N.K(x, y)|K(x, y) is again an infinite tower of Galois extensions of degree p. Since N is linearly disjoint from K(x)^{h} over K(x) and N^{h} = N.K(x)^{h} is linearly disjoint from L^{h} over K(x)^{h}, we see that N is linearly disjoint from L^{h} over K(x). Since L^{h} is relatively algebraically closed in K(x, y)^{h}, this implies that N is linearly disjoint from K(x, y)^{h} over K(x) and hence, N.K(x, y) is linearly disjoint from K(x, y)^{h} over K(x, y). Therefore, the extension of v from K(x, y) to N.K(x, y) is unique. Since (K(x)^{h}(y))^{h} = K(x, y)^{h}, we see that (K(x)^{h}(y)|K(x, y), v) is immediate. So we have vK(x, y) = \frac{1}{p^{∞}} ℤ, which is p-divisible, and K(x, y)v = K, which is algebraically closed. Hence, the extension (N.K(x, y)|K(x, y), v) is immediate, and so it is an infinite tower of Galois extensions of degree p and defect p. □

Remark For the above defined η_{i}, we have that the roots of X^{p} − X − η_{i−1} are η_{i}, η_{i} + 1, …, η_{i} + p − 1. Therefore,

kras(η_{i}, K(η_{i−1})) = 0

for all i. Setting a_{i} := x^{i}η_{i}, we obtain that

kras(a_{i}, K(x)^{h}(a_{1}, …, a_{i−1})) = kras(a_{i}, K(x)^{h}(η_{i−1})) = ivx + kras(η_{i}, K(x)^{h}(η_{i−1})) = ivx

and va_{i+1} = (i + 1)vx − \frac{1}{p^{i+1}} vx > ivx. This shows that va_{i+1} > va_{i} and that (28) is satisfied. So we can take y to be a pseudo limit of the Cauchy sequence (∑_{j=1}^{i} a_{j})_{i∈ℕ}. That is,

y = ∑_{i=1}^{∞} x^{i}η_{i} .  (29)

Comparing this with (14), we see that we have replaced the term x^{−p^{−e_{i}}} by η_{i} which has an infinite expansion in powers of x, starting with x^{−p^{−i}}.

4 Rational function fields of higher transcendence degree

This section is devoted to the proof of Theorems 1.6 and 1.8. We will make use of the following theorem which we prove in [KU5]:

Theorem 4.1 Let (L|K, v) be a valued field extension of finite transcendence degree ≥ 0, with v non-trivial on L. Assume that one of the following two cases holds: \underline{transcendental case}: vL/vK has rational rank at least 1 or Lv|Kv is transcendental; \underline{separable-algebraic case}: L|K contains a separable-algebraic subextension L_{0}|K such that within some henselization of L, the corresponding extension L_{0}^{h}|K^{h} is infinite. Then each maximal immediate extension of (L, v) has infinite transcendence degree over L.

Note: The assertion need not be true for an infinite purely inseparable extension L|K. Now we can give the Proof of Theorem 1.6: Assume that (K, v) is a valued field, n, ρ, τ, ℓ are non-

negative integers, Γ is an ordered abelian group extension of vK such that Γ/vK is of rational rank ρ, and k|Kv is a field extension of transcendence degree τ. We pick a maximal set of elements α_{1}, …, α_{ρ} in Γ rationally independent over vK, and a transcendence basis ζ_{1}, …, ζ_{τ} of k|Kv.

We prove Part A of Theorem 1.6 first. So assume further that n > ρ + τ, Γ|vK and k|Kv are countably generated, and at least one of them is infinite. By Lemma 2.6 there is a unique extension of v from K to K(x_{1}, …, x_{ρ+τ}) such that vx_{i} = α_{i} for 1 ≤ i ≤ ρ, and x_{ρ+i}v = ζ_{i} for 1 ≤ i ≤ τ. Since Γ|vK and k|Kv are countably generated, vK(x_{1}, …, x_{ρ+τ}) contains α_{1}, …, α_{ρ} and K(x_{1}, …, x_{ρ+τ})v contains ζ_{1}, …, ζ_{τ}, the extensions Γ|vK(x_{1}, …, x_{ρ+τ}) and k|K(x_{1}, …, x_{ρ+τ})v are countably generated and algebraic.

Suppose first that at least one of these extensions is infinite. From our assumption that Γ ≠ {0} it follows that v is non-trivial on K(x_{1}, …, x_{ρ+τ}). Hence by Proposition 3.15 we can find an extension of v from K(x_{1}, …, x_{ρ+τ}) to K(x_{1}, …, x_{ρ+τ+1}) such that vK(x_{1}, …, x_{ρ+τ+1}) = Γ and K(x_{1}, …, x_{ρ+τ+1})v = k, and such that the implicit constant field of (K(x_{1}, …, x_{ρ+τ+1})|K(x_{1}, …, x_{ρ+τ}), v) is an infinite separable-algebraic extension of K(x_{1}, …, x_{ρ+τ})^{h}. That is, K(x_{1}, …, x_{ρ+τ+1})^{h}|K(x_{1}, …, x_{ρ+τ})^{h} contains an infinite separable-algebraic subextension. Hence by the separable-algebraic case of Theorem 4.1, each maximal immediate extension of K(x_{1}, …, x_{ρ+τ+1})^{h} is of infinite transcendence degree. This shows that we can find an immediate extension of v from K(x_{1}, …, x_{ρ+τ+1})^{h} to K(x_{1}, …, x_{ρ+τ+1})^{h}(x_{ρ+τ+2}, …, x_{n}). Its restriction to K(x_{1}, …, x_{n}) is an immediate extension of v from K(x_{1}, …, x_{ρ+τ+1}).

Now suppose that both extensions Γ|vK(x_{1}, …, x_{ρ+τ}) and k|K(x_{1}, …, x_{ρ+τ})v are finite. Then from our hypothesis that at least one of the extensions Γ|vK and k|Kv is infinite, it follows that ρ > 0 or τ > 0. Hence by the transcendental case of Theorem 4.1, any maximal immediate extension of (K(x_{1}, …, x_{ρ+τ}), v) is of infinite transcendence degree. In combination with Proposition 3.13, this shows that there is an extension of v from K(x_{1}, …, x_{ρ+τ}) to K(x_{1}, …, x_{ρ+τ+1}) such that vK(x_{1}, …, x_{ρ+τ+1}) = Γ and K(x_{1}, …, x_{ρ+τ+1})v = k. Again by the transcendental case of Theorem 4.1, every maximal immediate extension of (K(x_{1}, …, x_{ρ+τ+1}), v) is of infinite transcendence degree. So we can find an extension of v to K(x_{1}, …, x_{n}) as in the previous case.

In both cases we have that vK(x_{1}, …, x_{n}) = Γ and K(x_{1}, …, x_{n})v = k, as required.

Now we prove Part B. So assume that n ≥ ρ+τ, and that Γ|vK and k|Kv are finitely generated.

I) First, we consider the case of ρ > 0. By Lemma 2.6 there is an extension of v from K to K(x_{1}, …, x_{ρ−1+τ}) such that vx_{i} = α_{i} for 1 ≤ i ≤ ρ − 1, and x_{ρ−1+i}v = ζ_{i} for 1 ≤ i ≤ τ. Since Γ|vK and k|Kv are finitely generated, vK(x_{1}, …, x_{ρ−1+τ}) contains α_{1}, …, α_{ρ−1} and K(x_{1}, …, x_{ρ−1+τ})v contains ζ_{1}, …, ζ_{τ}, the extension Γ|vK(x_{1}, …, x_{ρ−1+τ}) is finitely generated of rational rank 1 (and thus, Γ is of the form Γ_{0} ⊕ℤ with Γ_{0}/vK(x_{1}, …, x_{ρ−1+τ}) finite), and the extension k|K(x_{1}, …, x_{ρ−1+τ})v is finite.

If v is not trivial on K(x_{1}, …, x_{ρ−1+τ}), then we can apply Proposition 3.13 to obtain an extension of v from K(x_{1}, …, x_{ρ−1+τ}) to K(x_{1}, …, x_{ρ+τ}) such that vK(x_{1}, …, x_{ρ+τ}) = Γ and K(x_{1}, …, x_{ρ+τ})v = k. From the transcendental case of Theorem 4.1 we see that any maximal immediate extension of K(x_{1}, …, x_{ρ+τ}), v) is of infinite transcendence degree. This shows that we can find an immediate extension of v from K(x_{1}, …, x_{ρ+τ}) to K(x_{1}, …, x_{n}).

Now suppose that v is trivial on K(x_{1}, …, x_{ρ−1+τ}). Then ρ = 1 and v is trivial on K. It follows that Γ ≃ ℤ, and we pick a generator γ of Γ.

If n > 1 + τ, we modify the above procedure in such a way that we use Lemma 2.6 to choose an extension of v from K(x_{1}, …, x_{τ}) to K(x_{1}, …, x_{1+τ}) so that the latter has value group ℤγ = Γ and residue field Kv(ζ_{1}, …, ζ_{τ}). Then we use Proposition 3.13 in combination with the transcendental case of Theorem 4.1 to find an extension of v to K(x_{1}, …, x_{n}) with value group Γ and residue field k.

If n = 1+τ, then by our assumption B2), k is a simple algebraic extension of a rational function field k′ in τ variables over Kv (or of Kv itself if τ = 0), or a rational function field in one variable over a finitely generated field extension k_{0} of Kv of transcendence degree τ − 1.

Assume the first case. The extension of v to the rational function field K(x_{1}, …, x_{τ}) can be chosen such that k′ = Kv(x_{1}v, …, x_{τ}v) = K(x_{1}, …, x_{τ})v. Then the extension k|K(x_{1}, …, x_{τ})v is simple algebraic, and according to the remark following Theorem 3.9 there is an extension of v to the rational function field K(x_{1}, …, x_{τ})(x_{1+τ}) which satisfies vK(x_{1}, …, x_{1+τ}) = ℤγ = Γ and K(x_{1}, …, x_{1+τ})v = k.

In the second case, we know that τ ≥ 1. The elements ζ_{i} ∈ k can be chosen in such a way that ζ_{1}, …, ζ_{τ−1} form a transcendence basis of k_{0}|Kv. We pick γ ∈ Γ, γ ≠ 0. By Lemma 2.6 there is an extension of v from K to K(x_{1}, …, x_{τ}) such that x_{i}v = ζ_{i} for 1 ≤ i ≤ τ − 1, and vx_{τ} = γ. Since K(x_{1}, …, x_{τ})v = Kv(ζ_{1}, …, ζ_{τ−1}), we see that k is a rational function field in one variable over a finite extension of K(x_{1}, …, x_{τ})v. Further, vK(x_{1}, …, x_{τ}) = Γ. Now we use Proposition 3.13 to find an extension of v to K(x_{1}, …, x_{1+τ}) with value group Γ and residue field k.

II) Second, suppose that ρ = 0. Then because of Γ ≠ {0}, we know that v is nontrivial on K. If τ > 0, then we proceed as follows. The case of n > τ is covered by Part A. So we assume that n = τ, and that k is a rational function field in one variable over a finitely generated field extension k_{0} of Kv of transcendence degree τ − 1. Again we choose ζ_{1}, …, ζ_{τ−1} to form a transcendence basis of k_{0}|Kv, and use Lemma 2.6 to find an extension of v from K to K(x_{1}, …, x_{τ−1}) such that x_{i}v = ζ_{i} for 1 ≤ i ≤ τ − 1. Again we obtain that k is a rational function field in one variable over a finite extension of K(x_{1}, …, x_{τ−1})v. By assumption, Γ/vK(x_{1}, …, x_{τ−1}) = Γ/vK is finite. Hence by Proposition 3.13 there is an extension of v from K(x_{1}, …, x_{τ−1}) to K(x_{1}, …, x_{τ}) such that vK(x_{1}, …, x_{τ}) = Γ and K(x_{1}, …, x_{τ})v = k.

Finally, suppose that ρ = 0 = τ and that there is an immediate extension (K′, v) of (K, v) which is either infinite separable-algebraic or of transcendence degree at least n. If the former holds, then we obtain from the separable-algebraic case of Theorem 4.1 that every maximal immediate extension of (K′, v) has infinite transcendence degree. But a maximal immediate extension of (K′, v) is also a maximal immediate extension of (K, v). Thus in all cases, we have the existence of immediate extensions of (K, v) of transcendence degree at least n. Therefore, we can choose an immediate extension of v from K to K(x_{1}, …, x_{n−1}) such that (K(x_{1}, …, x_{n−1}), v) still admits a transcendental immediate extension. The extensions Γ|vK = Γ|vK(x_{1}, …, x_{n−1}) and k|Kv = k|K(x_{1}, …, x_{n−1})v are both finite since they are finitely generated and algebraic. Hence by Proposition 3.13 there is an extension of v from K(x_{1}, …, x_{n−1}) to K(x_{1}, …, x_{n}) such that vK(x_{1}, …, x_{n}) = Γ and K(x_{1}, …, x_{n})v = k. □

Now we give the

Proof of Theorem 1.7: Corollary 2.7 shows that n ≥ ρ + τ. If n = ρ + τ, then Corollary 2.7 tells us that vF |vK and F v|Kv are finitely generated extensions. The fact that vF |vK and F v|Kv are always countably generated follows from Theorem 2.9 by induction on the transcendence degree n.

Suppose that F admits a transcendence basis x_{1}, …, x_{n} such that F v|K(x_{1}, …, x_{n−1})v is of transcendence degree 1. (This is in particular the case when n = τ.) Then by Ohm’s Ruled Residue Theorem, F v is a rational function field in one variable over a finite extension of K(x_{1}, …, x_{n−1})v. Whenever equality holds in (6) for an extension L|K of finite transcendence degree, it will hold in the respective inequality for any subextension of L|K. Hence if n = ρ+τ then by Corollary 2.7, K(x_{1}, …, x_{n−1})v|Kv is a finitely generated field extension, and if n = τ, this extension is of transcendence degree τ − 1. This yields that B3) holds for k = F v.

Now suppose that there does not exist such a transcendence basis, and that n = ρ + τ with ρ = 1 and v is trivial on K. We pick any transcendence basis x_{1}, …, x_{n}. Since F v|K(x_{1}, …, x_{n−1})v is algebraic and n = ρ + τ, we must have that the rational rank of vF/vK(x_{1}, …, x_{n−1}) is 1. Since ρ = 1 and v is trivial on K, this means that v is also trivial on K(x_{1}, …, x_{n−1}). Hence, Kv = K and K(x_{1}, …, x_{n−1})v = K(x_{1}, …, x_{n−1}) (modulo an isomorphism). The remark following Theorem 3.9 now shows that F v is a simple algebraic extension of K(x_{1}, …, x_{n−1}), which in turn is a rational function field in τ variables over K = Kv, or equal to Kv if τ = 0. Together with what we have shown before, this proves that B2) holds for k = F v.

To conclude this proof, assume that ρ = 0 = τ. Then by Lemma 3.4, (F.K̃|K̃, v) is immediate, for any extension of v from F to F.K̃. This shows that (K̃, v) admits an immediate extension of transcendence degree n. □

Finally, we give the

Proof of Theorem 1.8: We have to show that whenever we construct an extension of the form (L_{2}|L_{1}, v) with Kv ⊆ L_{1}v ⊆ L_{2}v ⊆ k in the previous proof, then the ordering of L_{2}v induced by the given ordering of k can be lifted to an extension of the lifting that we have already obtained on L_{1} . Whenever we apply Propositions 3.12, 3.13 and 3.15, we obtain this already from the assertions of these propositions. Whenever we apply Lemma 2.6, we obtain this from Proposition 2.18 because in this case, we always have that vL_{2} is generated over vL_{1} by rationally independent values and therefore, 2vL_{2} ∩ vL_{1} = 2vL_{1} . Finally, whenever (L_{2}|L_{1}, v) is an immediate extension, we can also apply Proposition 2.18 because 2vL_{2} ∩ vL_{1} = 2vL_{1} ∩ vL_{1} = 2vL_{1} . □

5 Homogeneous sequences

In this section, we will develop special sequences which under certain tameness conditions can be used to determine implicit constant fields.

5.1 Homogeneous approximations

Let (K, v) be any valued field and a, b elements in some valued field extension (L, v) of (K, v). We will say that a is strongly homogeneous over (K, v) if a ∈ K^{sep} \ K, the extension of v from K to K(a) is unique (or equivalently, K(a)|K is linearly disjoint from all henselizations of (K, v)), and

va = kras(a, K) .  (30)

Note that in this case, va = v(σa − a) for all automorphisms such that σa ≠ a; indeed, we have that vσa = va and therefore, va ≥ v(σa − a) ≥ va.

We will say that a is homogeneous over (K, v) if there is some d ∈ K such that a−d is strongly homogeneous over (K, v), i.e.,

v(a − d) = kras(a − d, K) = kras(a, K) .

We call a ∈ L a homogeneous approximation of b over K if a is homogeneous over K and v(b − a) > vb (and consequently, va = vb). From Corollary 2.21 we obtain:

Lemma 5.1 If a ∈ L is a homogeneous approximation of b then a lies in the henselization of K(b) w.r.t. every extension of the valuation v from K(a, b) to K̃(b).

We will also exploit the following easy observation:

Lemma 5.2 Let (K′, v) be any henselian extension field of (K, v) such that a ∉ K′. If a is homogeneous over (K, v), then it is also homogeneous over (K′, v), and kras(a, K) = kras(a, K′). If a is strongly homogeneous over (K, v), then it is also strongly homogeneous over (K′, v).

Proof: Suppose that a − d is separable-algebraic over K and v(a − d) = kras(a − d, K) for some d ∈ K. Then a−d is also separable-algebraic over K′. Further, kras(a−d, K′) ≤ kras(a − d, K) since restriction to K̃ is a map sending {σ ∈ Gal K′ | σa ≠ a} into {σ ∈ Gal K | σa ≠ a}. Hence, v(a−d) ≤ kras(a−d, K′) ≤ kras(a−d, K) = v(a−d), which shows that equality holds everywhere. Thus, kras(a, K′) = kras(a − d, K′) = kras(a − d, K) = kras(a, K). Since (K′, v) is henselian by assumption, the extension of v from K′ to K′(a) is unique. This shows that a − d is strongly homogeneous over (K′, v), and concludes the proof of our assertions. □

The following gives the crucial criterion for an element to be (strongly) homogeneous over (K, v):

Lemma 5.3 Suppose that a ∈ K̃ and that there is some extension of v from K to K(a) such that if e is the least positive integer for which eva ∈ vK, then

a) e is not divisible by char Kv,

b) there exists some c ∈ K such that vca^{e} = 0, ca^{e}v is separable-algebraic over Kv, and the degree of ca^{e} over K is equal to the degree f of ca^{e}v over Kv. Then [K(a) : K] = ef and if a ∉ K, then a is strongly homogeneous over (K, v).

Proof: We have

ef ≥ [K(a) : K(a^{e})] · [K(a^{e}) : K] = [K(a) : K] ≥ (vK(a) : vK) · [K(a)v : Kv] ≥ ef .

So equality holds everywhere, and we obtain [K(a) : K] = ef, (vK(a) : vK) = e and [K(a)v : Kv] = f. By the fundamental inequality, the latter implies that the extension of v from K to K(a) is unique.

Now assume that a ∉ K. Take two distinct conjugates σa ≠ τa of a and set η := σa/τa ≠ 1. If σa^{e} ≠ τa^{e}, then cσa^{e} = σca^{e} and cτa^{e} = τca^{e} are distinct conjugates of ca^{e}. By hypothesis, their residues are also distinct and therefore, the residue of σa^{e}/τa^{e} = η^{e} is not 1. It follows that the residue of η is not 1. If σa^{e} = τa^{e}, then η is an e-th root of unity. Since e is not divisible by the residue characteristic, it again follows that the residue of η is not equal to 1. Hence in both cases, we obtain that v(η − 1) = 0, which shows that v(σa − τa) = vτa = va. We have now proved (30). □

Lemma 5.4 Assume that b is an element in some algebraically closed valued field extension (L, v) of (K, v). Suppose that there is some e ∈ ℕ not divisible by char Kv, and some c ∈ K such that vcb^{e} = 0 and cb^{e}v is separable-algebraic over Kv. If e > 0 or cb^{e}v ∉ Kv, then we can find in L a strongly homogeneous approximation of b over K.

Proof: Take a monic polynomial g over K with v-integral coefficients whose reduction modulo v is the minimal polynomial of cb^{e}v over Kv. Then let d ∈ K̃ be the root of g whose residue is cb^{e}v. The degree of d over K is the same as that of cb^{e}v over Kv. We have that v(\frac{d}{cb^{e}} − 1) > 0. So there exists a_{0} ∈ K̃ with residue 1 and such that a_{0}^{e} = \frac{d}{cb^{e}} . Then for a := a_{0}b, we find that v(a − b) = vb + v(a_{0} − 1) > vb and ca^{e} = d. By the foregoing lemma, this shows that a is a strongly homogeneous approximation of b. □

5.2 Homogeneous sequences

Let (K(x)|K, v) be any extension of valued fields. We fix an extension of v to K̃(x).

Let S be an initial segment of ℕ, that is, S = ℕ or S = {1, …, n} for some n ∈ ℕ or S = ∅. A sequence

𝔖 := (a_{i})_{i∈S}

of elements in K̃ will be called a homogeneous sequence for x if the following conditions are satisfied for all i ∈ S (where we set a_{0} := 0):

(HS) a_{i} − a_{i−1} is a homogeneous approximation of x − a_{i−1} over K(a_{0}, …, a_{i−1}). Recall that then by definition of “homogeneous”, a_{i} ∉ K(a_{0}, …, a_{i−1})^{h}. We call S the support of the sequence 𝔖. We set

K_{𝔖} := K(a_{i} | i ∈ S) .

If 𝔖 is the empty sequence, then K_{𝔖} = K.

From this definition, we obtain:

Lemma 5.5 If 1 ≤ i < j ∈ S, then

v(x − a_{j}) > v(x − a_{i}) = v(a_{i+1} − a_{i}) .  (31)

If S = ℕ then (a_{i})_{i∈S} is a pseudo Cauchy sequence with pseudo limit x.

Proof: If 1 ≤ i ∈ S, then a_{i} − a_{i−1} is a homogeneous approximation of x − a_{i−1} . Hence by definition,

v(x − a_{i}) = v(x − a_{i−1} − (a_{i} − a_{i−1})) > v(x − a_{i−1}) ,

whence v(a_{i} − a_{i−1}) = min{v(x − a_{i}), v(x − a_{i−1})} = v(x − a_{i−1}). If i < j ∈ S, then by induction, v(x − a_{j}) > v(x − a_{i}).

Suppose that S = ℕ. Then it follows by induction that for all k > j > i ≥ 1,

v(x − a_{k}) > v(x − a_{j}) > v(x − a_{i})

and therefore,

v(a_{k} − a_{j}) = min{v(x − a_{k}), v(x − a_{j})} = v(x − a_{j}) > v(x − a_{i})
= min{v(x − a_{j}), v(x − a_{i})} = v(a_{j} − a_{i}) .

This shows that (a_{i})_{i∈S} is a pseudo Cauchy sequence. The equality in (31) shows that x is a pseudo limit of this sequence. □

Let us also observe the following:

Lemma 5.6 Take x, x′ ∈ L. If a ∈ L is a homogeneous approximation of x over K and if v(x − x′) > vx, then a is also a homogeneous approximation of x′ over K. If (a_{i})_{i∈S} is a homogeneous sequence for x over K and if v(x − x′) > v(x − a_{k}) for all k ∈ S, then (a_{i})_{i∈S} is also a homogeneous sequence for x′ over K.

In particular, for each k ∈ S such that k > 1, (a_{i})_{i<k} is a homogeneous sequence for a_{k} over K.

Proof: Suppose that a is a homogeneous approximation of x over K. Then v(x−a) > vx. If also v(x − x′) > vx, then vx′ = vx and v(x′ − a) ≥ max{v(x − x′), v(x − a)} > vx = vx′. This yields the first assertion.

Now assume that (a_{i})_{i∈S} is a homogeneous sequence for x over K and that v(x − x′) > v(x−a_{k}) for all k ∈ S. Then for all k ∈ S, v(x′−a_{k}) = min{v(x′−x), v(x−a_{k})} = v(x−a_{k}). Hence,

v(x′ − a_{k−1} − (a_{k} − a_{k−1})) = v(x′ − a_{k}) = v(x − a_{k}) > v(x − a_{k−1}) = v(x′ − a_{k−1})

showing that a_{k} − a_{k−1} is also a homogeneous approximation of x′ − a_{k−1} . Hence, (a_{i})_{i∈S} is a homogeneous sequence for x′ over K. □

What is special about homogeneous sequences is described by the following lemma:

Lemma 5.7 Assume that (a_{i})_{i∈S} is a homogeneous sequence for x over K. Then

K_{𝔖} ⊂ K(x)^{h} .  (32)

For every k ∈ S, a_{1}, …, a_{k} ∈ K(a_{k})^{h}. If S = {1, …, n}, then

K_{𝔖}^{h} = K(a_{n})^{h} .  (33)

Proof: We use induction on k ∈ S. Suppose that we have already shown that a_{k−1} ∈ K(x)^{h}. As a_{k}−a_{k−1} is a homogeneous approximation of x−a_{k−1} , we know from Lemma 5.1 that

a_{k} − a_{k−1} ∈ K(x − a_{k−1})^{h} ⊆ K(x)^{h} .

This proves (32). Now all other assertions follow when we replace x by a_{k} in the above argument, using the fact that by the previous lemma, (a_{i})_{i<n} is a homogeneous sequence for a_{n} over K. □

Proposition 5.8 Assume that 𝔖 = (a_{i})_{i∈S} is a homogeneous sequence for x over K with support S = ℕ. Then (a_{i})_{i∈ℕ} is a pseudo Cauchy sequence of transcendental type in (K_{𝔖}, v) with pseudo limit x, and (K_{𝔖}(x)|K_{𝔖}, v) is immediate and pure.

Proof: By Lemma 5.5, (a_{i})_{i∈ℕ} is a pseudo Cauchy sequence with pseudo limit x. Suppose it were of algebraic type. Then by [KA], Theorem 3, there would exist some algebraic extension (K_{𝔖}(b)|K_{𝔖}, v) with b a pseudo limit of the sequence. But then v(x − b) > v(x − a_{k}) for all k ∈ S and by Lemma 5.6, (a_{i})_{i∈S} is also a homogeneous sequence for b over K. Hence by Lemma 5.7, K_{𝔖}^{h} ⊂ K(b)^{h} = K^{h}(b). Since b is algebraic over K, the extension K^{h}(b)|K^{h} is finite. On the other hand, K_{𝔖}^{h}|K^{h} is infinite since by the definition of homogeneous elements, a_{k} ∉ K(a_{i} | 1 ≤ i < k)^{h} for every k ∈ ℕ and therefore, each extension K(a_{i} | 1 ≤ i ≤ k + 1)^{h}|K(a_{i} | 1 ≤ i ≤ k)^{h} is non-trivial. This contradiction shows that the sequence is of transcendental type. Hence by definition, (K_{𝔖}(x)|K_{𝔖}, v) is pure. Further, it follows from Lemma 2.4 that (K_{𝔖}(x)|K_{𝔖}, v) is immediate. □

This proposition leads to the following definition. A homogeneous sequence 𝔖 for x over K will be called (weakly) pure homogeneous sequence if (K_{𝔖}(x)|K_{𝔖}, v) is (weakly) pure in x. Hence if S = ℕ, then 𝔖 is always a pure homogeneous sequence. The empty sequence is a (weakly) pure homogeneous sequence for x over K if and only if already (K(x)|K, v) is (weakly) pure in x.

Theorem 5.9 Suppose that 𝔖 is a (weakly) pure homogeneous sequence for x over K. Then

K_{𝔖}^{h} = IC (K(x)|K, v) .

Further, K_{𝔖}v is the relative algebraic closure of Kv in K(x)v, and the torsion subgroup of vK(x)/vK_{𝔖} is finite. If 𝔖 is pure, then vK_{𝔖} is the relative divisible closure of vK in vK(x).

Proof: The assertions follow from Lemma 3.5, Lemma 2.10 and Lemma 3.7, together with the fact that because K_{𝔖}|K is algebraic, the same holds for vK_{𝔖}|vK and K_{𝔖}v|Kv by Lemma 2.1. □

5.3 Conditions for the existence of homogeneous sequences

Now we have to discuss for which extensions (K(x)|K, v) there exist homogeneous sequences.

An algebraic extension (L|K, v) of henselian fields is called tame if the following conditions hold: (TE1) Lv|Kv is separable, (TE2) if char Kv = p > 0, then the order of every element in vL/vK is prime to p, (TE3) for every finite subextension K′|K of L|K, [K′ : K] = (vK′ : vK)[K′v : Kv]. Condition (TE3) means that equality holds in the fundamental inequality (3). If L′|K is any subextension of L|K, then (L|K, v) is a tame extension if and only if (L|L′, v) and (L′|K, v) are (this is easy to prove if L|K is finite). Further, it is well known that for (K, v) henselian, the ramification field of the extension (K^{sep}|K, v) is the unique maximal tame extension of (K, v) (cf. [E]). A henselian valued field (K, v) is called a tame field if all its algebraic extensions are tame, or equivalently, the following conditions hold: (T1) Kv is perfect, (T2) if char Kv = p > 0, then vK is p-divisible, (T3) for every finite extension K′|K, [K′ : K] = (vK′ : vK)[K′v : Kv]. Note that every valued field with a residue field of characteristic zero is tame; this is a consequence of the Lemma of Ostrowski (cf. [R]). It follows directly from the definition together with the multiplicativity of ramification index and inertia degree that every finite extension of a tame field is again a tame field. If (K, v) is a tame field, then condition (T3) shows that (K, v) does not admit any proper immediate algebraic extensions; hence by Theorem 3 of [KA], every pseudo Cauchy sequence in (K, v) without a pseudo limit in K must be of transcendental type.

If an element a ∈ K̃ satisfies the conditions of Lemma 5.3, then (K(a)|K, v) is a tame extension. The following implication is also true, as was noticed by Sudesh K. Khanduja (cf. [KH11], Theorem 1.2):

Proposition 5.10 Suppose that (K, v) is henselian. If a is homogeneous over (K, v), then (K(a)|K, v) is a tame extension. If 𝔖 is a homogeneous sequence over (K, v), then K_{𝔖} is a tame extension of K.

Proof: Since K(a − d) = K(a) for d ∈ K, we may assume w.l.o.g. that a is strongly homogeneous over (K, v). If (K(a)|K, v) were not a tame extension, then a would not lie in the ramification field of the extension (K^{sep}|K, v). So there would exist an automorphism σ in the ramification group such that σa ≠ a. But by the definition of the ramification group,

v(σa − a) > va = kras(a, K) ,

a contradiction.

The second assertion is proved using the first assertion and the fact that a (possibly infinite) tower of tame extensions is itself a tame extension. □

In fact, it can also be shown that va = kras(a, K) implies that a satisfies the conditions of Lemma 5.3.

We can give the following characterization of elements in tame extensions:

Proposition 5.11 An element b ∈ K̃ belongs to a tame extension of the henselian field (K, v) if and only if there is a finite homogeneous sequence a_{1}, …, a_{k} for a over (K, v) such that b ∈ K(a_{k}).

Proof: Suppose that such a sequence exists. By the foregoing proposition, K_{𝔖} is a tame extension of K. Since K(a_{1}, …, a_{k}) = K(a_{k}) by Lemma 5.7, it contains b.

For the converse, let b be an element in some tame extension of (K, v). Then b satisfies the assumptions of Lemma 5.4 and hence there is a homogeneous approximation a_{1} ∈ K̃ of b over K. By the foregoing proposition, K(a_{1}) is a tame extension of K and therefore, by the general facts we have noted following the definition of tame extensions, K(a_{1}, b − a_{1}) is a tame extension of K(a_{1}). We repeat this step, replacing b by b − a_{1} . By induction, we build a homogeneous sequence for b over K. It cannot be infinite since b is algebraic over K (cf. Proposition 5.8). Hence it stops with some element a_{k}. Our construction shows that this can only happen if b ∈ K(a_{1}, …, a_{k}) = K(a_{k}). □

Proposition 5.12 Assume that (K, v) is a henselian field. Then (K, v) is a tame field if and only if for every element x in any extension (L, v) of (K, v) there exists a weakly pure homogeneous sequence for x over K, provided that x is transcendental over K.

Proof: First, let us assume that (K, v) is a tame field and that x is an element in some extension (L, v) of (K, v), transcendental over K. We set a_{0} = 0. We assume that k ≥ 0 and that a_{i} for i ≤ k are already constructed. Like K, also the finite extension K_{k} := K(a_{0}, …, a_{k}) is a tame field. Therefore, if x is the pseudo limit of a pseudo Cauchy sequence in K_{k}, then this pseudo Cauchy sequence must be of transcendental type, and K_{k}(x)|K_{k} is pure and hence weakly pure in x.

If K_{k}(x)|K_{k} is weakly pure in x, then we take a_{k} to be the last element of 𝔖 if k > 0, and 𝔖 to be empty if k = 0.

Assume that this is not the case. Then x cannot be the pseudo limit of a pseudo Cauchy sequence without pseudo limit in K_{k}. So the set v(x − a_{k} − K_{k}) must have a maximum, say x − a_{k} − d with d ∈ K_{k}. Since we assume that K_{k}(x)|K_{k} is not weakly pure in x, there exist e ∈ ℕ and c ∈ K_{k} such that vc(x − a_{k} − d)^{e} = 0 and c(x − a_{k} − d)^{e} is algebraic over K_{k}v. Conditions (T1) and (T2) that e can be chosen to be prime to char Kv and that c(x − a_{k} − d)^{e}v is separable-algebraic over K_{k}v. Since v(x − a_{k} − d) is maximal in v(x − a_{k} − K_{k}), we must have that e > 1 or c(x − a_{k} − d)^{e}v ∉ K_{k}v.

Now Lemma 5.4 shows that there exists a homogeneous approximation a ∈ K̃ of x−a_{k} −d over K_{k}; so a+d is a homogeneous approximation of x−a_{k} over K_{k}, and we set a_{k+1} := a_{k} + a + d. This completes our induction step. If our construction stops at some k, then K_{k}(x)|K_{k} is weakly pure in x and we have obtained a weakly pure homogeneous sequence. If the construction does not stop, then S = ℕ and the obtained sequence is pure homogeneous.

For the converse, assume that (K, v) is not a tame field. We choose an element b ∈ K̃ such that K(b)|K is not a tame extension. On K(b, x) we take the valuation v_{b,γ} with γ an element in some ordered abelian group extension such that γ > vK. Choose any extension of v to K̃(x). Since vK is cofinal in in vK̃, we have that γ > vK̃. Since b ∈ K̃, we find γ ∈ vK̃(x). Hence, (K̃(x)|K̃, v) is value-transcendental.

Now suppose that there exists a weakly pure homogeneous sequence 𝔖 for x over K. By Lemma 3.3, also (K_{𝔖}(x)|K_{𝔖}, v) is value-transcendental. Since (K_{𝔖}(x)|K_{𝔖}, v) is also weakly pure, it follows that there must be some c ∈ K_{𝔖} such that x − c is a value-transcendental element (all other cases in the definition of “weakly pure” lead to immediate or residue-transcendental extensions). But if c ≠ b then v(b − c) ∈ vK̃ and thus, v(c − b) < γ. This implies v(x − c) = min{v(x − b), v(b − c)} = v(b − c) ∈ vK̃, a contradiction. This shows that b = c ∈ K_{𝔖}. On the other hand, K_{𝔖} is a tame extension of K by Proposition 5.10 and cannot contain b. This contradiction shows that there cannot exist a weakly pure homogeneous sequence for x over K. □

6 Applications

Let us show how to apply our results to power series fields. We denote by k((G)) the field of power series with coefficients in the field k and exponents in the ordered abelian group G.

Theorem 6.1 Let (K, v) be a henselian subfield of a power series field k((G)) such that v is the restriction of the canonical valuation of k((G)). Suppose that K contains all monomials of the form ct^{γ} for c ∈ Kv ⊆ k and γ ∈ vK ⊆ Γ. Consider a power series

z = ∑_{i∈ℕ} c_{i}t^{γ_{i}}

where (γ_{i})_{i∈ℕ} is a strictly increasing sequence in G, all c_{i} ∈ k are separable-algebraic over Kv, and for each i there is an integer e_{i} > 0 prime to the characteristic of Kv and such that e_{i}γ_{i} ∈ vK. Then, upon taking the henselization of K(z) in k((G)), we obtain that K(c_{i}t^{γ_{i}} | i ∈ ℕ) ⊆ K(z)^{h}. Consequently, vK(z) contains all γ_{i}, and if γ_{i} ∈ vK for all i ∈ ℕ, then K(z)v contains all c_{i}.

If vK + ∑_{i=1}^{∞} ℤγ_{i}/vK or Kv(c_{i} | i ∈ ℕ)|Kv is infinite, then z is transcendental over K, we have IC (K(z)|K, v) = K(c_{i}t^{γ_{i}} | i ∈ ℕ), and a) vK(z) is the group generated over vK by the elements γ_{i},

b) if γ_{i} ∈ vK for all i ∈ ℕ, then K(z)v = Kv(c_{i} | i ∈ ℕ).

Proof: We derive a homogeneous sequence 𝔖 from z as follows. We set a_{0} = 0. If all c_{i} are in Kv and all γ_{i} are in vK, then we take 𝔖 to be the empty sequence. Otherwise, having chosen i_{j} ∈ ℕ and defined

a_{j} := ∑_{1≤i≤i_{j}} c_{i}t^{γ_{i}}

for all j < m, we proceed as follows. We let ℓ be the first index in the power series z −a_{m−1} for which c_{ℓ}t^{γ_{ℓ}} ∉ K(a_{1}, …, a_{m−1}); if such an index does not exist, we let a_{m−1} be the last element of 𝔖. Otherwise, we set i_{m} := ℓ and a_{m} := ∑_{1≤i≤i_{m}} c_{i}t^{γ_{i}}. We have that

c_{ℓ}t^{γ_{ℓ}} = a_{m} − a_{m−1} − d with d = ∑_{i_{m−1}<i<ℓ} c_{i}t^{γ_{i}} ∈ K(a_{1}, …, a_{m−1}) .

By assumption, e_{ℓ}vc_{ℓ}t^{γ_{ℓ}} = e_{ℓ}γ_{ℓ} ∈ vK and hence, c := t^{−e_{ℓ}γ_{ℓ}} ∈ K. We have that c(c_{ℓ}t^{γ_{ℓ}})^{e_{ℓ}} = c_{ℓ}^{e_{ℓ}}. Since c_{ℓ} is separable-algebraic over Kv, the same holds for c_{ℓ}^{e_{ℓ}}. Since v is trivial on k, the degree of c_{ℓ}^{e_{ℓ}}v over Kv is equal to that of c_{ℓ}^{e_{ℓ}} over K. It now follows by Lemma 5.3 that c_{ℓ}t^{γ_{ℓ}} is strongly homogeneous over K. By Lemma 5.2 it follows that it is also strongly homogeneous over the henselian field K(a_{1}, …, a_{m−1}). Therefore, a_{m} − a_{m−1} is homogeneous over K(a_{1}, …, a_{m−1}). Further, v(z − a_{m−1} − (a_{m} − a_{m−1})) = v(z − a_{m}) = γ_{ℓ+1} > γ_{ℓ} ≥ v(z − a_{m−1}). This proves that a_{m} − a_{m−1} is a homogeneous approximation of z − a_{m−1} over K(a_{0}, …, a_{m−1}). By induction, we obtain a homogeneous sequence 𝔖 for z in k((G)). It now follows from Lemma 5.7 that K(c_{i}t^{γ_{i}} | i ∈ ℕ) = K(a_{j} | j ∈ S) ⊆ K(z)^{h}; thus, γ_{i} = vc_{i}t^{γ_{i}} ∈ vK(z) for all i ∈ ℕ. If γ_{i} ∈ vK and hence t^{γ_{i}} ∈ K, then c_{i} ∈ K(z)^{h}; since the residue map is the identity on elements of k, this implies that c_{i} ∈ K(z)v.

Now assume that vK + ∑_{i=1}^{∞} ℤγ_{i}/vK or Kv(c_{i} | i ∈ ℕ)|Kv is infinite. Then 𝔖 must be infinite, and it follows from Proposition 5.8 that z is a pseudo limit of the pseudo Cauchy sequence (a_{i})_{i∈ℕ} of transcendental type. Thus, z is transcendental over K by Lemma 2.4. Theorem 5.9 now shows that IC (K(z)|K, v) = K(c_{i}t^{γ_{i}} | i ∈ ℕ), vK(z) = vK(c_{i}t^{γ_{i}} | i ∈ ℕ) and K(z)v = K(c_{i}t^{γ_{i}} | i ∈ ℕ)v. Since K(c_{i}t^{γ_{i}} | i ∈ ℕ) ⊆ K(c_{i}, t^{γ_{i}} | i ∈ ℕ), for the proof of assertion a) it now suffices to show that the value group of the latter field is generated over vK by the γ_{i}, and that its residue field is generated over Kv by the c_{i}. As t^{γ} ∈ K for every γ ∈ vK, we have that

(vK + ∑_{i=1}^{ℓ} ℤγ_{i} : vK) = [K(t^{γ_{i}} | 1 ≤ i ≤ ℓ) : K] .

On the other hand,

[K(t^{γ_{i}} | 1 ≤ i ≤ ℓ) : K] ≥ (vK(t^{γ_{i}} | 1 ≤ i ≤ ℓ) : vK) ≥ (vK + ∑_{i=1}^{ℓ} ℤγ_{i} : vK) ,

where the last inequality holds since γ_{i} ∈ vK(t^{γ_{i}} | 1 ≤ i ≤ ℓ) for i = 1, …, ℓ. We obtain that vK(t^{γ_{i}} | 1 ≤ i ≤ ℓ) = vK + ∑_{i=1}^{ℓ} ℤγ_{i} and that K(t^{γ_{i}} | 1 ≤ i ≤ ℓ)v = Kv for all ℓ ∈ ℕ. This implies that vK(t^{γ_{i}} | i ∈ ℕ) = vK + ∑_{i=1}^{∞} ℤγ_{i} and K(t^{γ_{i}} | i ∈ ℕ)v = Kv.

Set K′ := K(t^{γ_{i}} | i ∈ ℕ). As K′v = Kv ⊂ K ⊂ K′, we have that

[K′v(c_{i} | 1 ≤ i ≤ ℓ) : K′v] ≥ [K′(c_{i} | 1 ≤ i ≤ ℓ) : K′] .

On the other hand,

[K′(c_{i} | 1 ≤ i ≤ ℓ) : K′] ≥ [K′(c_{i} | 1 ≤ i ≤ ℓ)v : K′v] ≥ [K′v(c_{i} | 1 ≤ i ≤ ℓ) : K′v] ,

where the last inequality holds since c_{i} ∈ K′(c_{i} | 1 ≤ i ≤ ℓ)v for i = 1, …, ℓ. We obtain that K′(c_{i} | 1 ≤ i ≤ ℓ)v = K′v(c_{i} | 1 ≤ i ≤ ℓ) and that vK′(c_{i} | 1 ≤ i ≤ ℓ) = vK′. This implies that vK(c_{i}, t^{γ_{i}} | i ∈ ℕ) = vK′(c_{i} | i ∈ ℕ) = vK′ = vK + ∑_{i=1}^{∞} ℤγ_{i} and K(c_{i}, t^{γ_{i}} | i ∈ ℕ)v = K′(c_{i} | i ∈ ℕ)v = K′v(c_{i} | i ∈ ℕ) = Kv(c_{i} | i ∈ ℕ). This proves assertions a) and b). □

Remark 6.2 Assertions a) and b) of the previous theorem will also hold if K = Kv((vK)). Indeed, if G_{0} denotes the subgroup of G generated by the γ_{i} over Kv, and k_{0} = Kv(c_{i} | i ∈ ℕ), then K(z) ⊂ k_{0}((G_{0})) and therefore, vK(z) ⊆ G_{0} and K(z)v ⊆ k_{0} .

Our methods also yield an alternative proof of the following well known fact:

Theorem 6.3 The algebraic closure of the field ℚ_{p} of p-adic numbers is not complete.

Proof: Choose any compatible system of n-th roots p^{1/n} of p, i.e., such that (p^{1/mn})^{m} = p^{1/n} for all m, n ∈ ℕ. For i ∈ ℕ, choose any n_{i} ∈ ℕ not divisible by p, and set γ_{i} := i + \frac{1}{n_{i}} if n_{i} > 1, and γ_{i} = i otherwise. Further, choose c_{i} in some fixed set of representatives in ℚ̃_{p} of its residue field 𝔽̃_{p} such that the degree of c_{i}^{n_{i}} over ℚ_{p} is equal to the degree of c_{i}^{n_{i}}v_{p} over ℚ_{p}v_{p} = 𝔽_{p}. Then set

b_{i} := ∑_{1≤j≤i} c_{j}p^{γ_{j}} .  (34)

Since γ_{i} < γ_{i+1} for all i and the sequence (γ_{i})_{i∈ℕ} is cofinal in the value group ℚ of ℚ̃_{p}, the sequence (b_{i})_{i∈ℕ} is a Cauchy sequence in (ℚ̃_{p}, v_{p}). If this field were complete, it would contain a pseudo limit z to every such Cauchy sequence.

On the other hand, as in the proof of Theorem 6.1 one shows that ℚ_{p}(c_{i}p^{γ_{i}} | i ∈ ℕ) ⊆ ℚ_{p}(z)^{h}, and that

a) vℚ_{p}(z) is the group generated over ℤ by the elements γ_{i},

b) if γ_{i} ∈ ℤ for all i ∈ ℕ, then ℚ_{p}(z)v = 𝔽_{p}(c_{i}v_{p} | i ∈ ℕ),

c) if vℚ_{p}(z)/ℤ or 𝔽_{p}(c_{i}v_{p} | i ∈ ℕ)|𝔽_{p} is infinite, then z is transcendental over ℚ_{p}. Hence, if we choose (n_{i})_{i∈ℕ} to be a strictly increasing sequence and c_{i} = 1 for all i, or if we choose n_{i} = 1 for all i and the elements c_{i} of increasing degree over ℚ_{p}, then z will be transcendental over ℚ_{p}. Since z lies in the completion of ℚ̃_{p}, we have now proved that this completion is transcendental over ℚ̃_{p}. □

With the same method, we can also prove another well known result:

Theorem 6.4 The completion ℂ_{p} of ℚ̃_{p} admits a pseudo Cauchy sequence without a pseudo limit in ℂ_{p}. Hence, ℂ_{p} is not maximal and not spherically complete.

Proof: In the same setting as in the foregoing proof, we now choose c_{i} = 1 for all i. Further, we choose (n_{i})_{i∈ℕ} to be a strictly increasing sequence and set γ_{i} := 1 − \frac{1}{n_{i}}. Then (b_{i})_{i∈ℕ} is a pseudo Cauchy sequence. Suppose it would admit a pseudo limit y in ℂ_{p}. Then, using that ℚ̃_{p} is dense in ℂ_{p}, we could choose z ∈ ℚ̃_{p} such that v_{p}(y − z) ≥ 1. Since 1 > γ_{i} for all i, it would follow that also z is a pseudo limit of (b_{i})_{i∈ℕ}. But as in the foregoing proof one shows that z must be transcendental over ℚ_{p}. This contradiction shows that (b_{i})_{i∈ℕ} cannot have a pseudo limit in ℂ_{p}. By the results of [KA], this implies that ℂ_{p} admits a proper immediate extension, which shows that ℂ_{p} is not maximal. With the elements b_{i} defined as in (34), we also find that the intersection of the nest ({a ∈ ℂ_{p} | v(a − b_{i}) ≥ γ_{i+1}})_{i∈ℕ} of balls is empty. Hence, ℂ_{p} is not spherically complete. □

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Mathematical Sciences Group, University of Saskatchewan, 106 Wiggins Road, Saskatoon, Saskatchewan, Canada S7N 5E6 email: fvk@math.usask.ca