Josnei Novacoski
email: josnei@ufscar.br

Mark Spivakovsky
email: mark.spivakovsky@math.univ-toulouse.fr

K ÄHLER DIFFERENTIALS, PURE EXTENSIONS AND MINIMAL KEY POLYNOMIALS

Keywords: 2010 Mathematics Subject Classification. Primary 13A18 Key polynomials, Kähler differentials, the defect

The main object of study in this paper is the module Ω of Kähler differentials of an extension of valuation rings. We show that in the case of pure extensions Ω has a very good description. Namely, it is isomorphic to the quotient of two fractional ideals of the value group of the valuation. This allows us to describe the annihilator of Ω and to give criteria for Ω to be finitely generated and to be finitely presented. We also discuss how to relate the module of Kähler differentials of pure extensions to minimal key polynomials.

Introduction

Our motivation for studying the module of Kähler differentials of extensions of valuation rings comes from [START_REF] Cutkosky | Kähler differentials of extensions of valuation rings and deeply ramified fields[END_REF] and [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF]. The main goal of that work is to present a simple characterization of deeply ramified fields. Their first step is to understand the module of Kähler differentials of an extension of valuation rings whose corresponding quotient field extension is Galois and of prime degree. Using this, the authors give a criterion for when the module of Kähler differentials of a valuation ring extension is zero, assuming that the corresponding extension of valued fields is Galois and finite (not necessarily of prime degree). Using this criterion, one can understand better when a given valued field is deeply ramified.

This work is the first step of the program whose goal is to understand the module Ω of Kähler differentials of an extension of valuation rings in the most general possible case. For instance, if Ω ̸ = (0), then it is interesting to have an explicit characterization for its structure. In order to understand the module of Kähler differentials of a ring extension B|A, the first step is to present a set of generators for it. If B = O L and A = O K are the valuation rings of an unramified simple valued field extension (L|K, v), then complete sets (defined in Section 2 below) naturally give rise to sets of generators of O L over O K . However, it is important to have a set of generators whose relations are as simple as possible. For this purpose, complete sets of key polynomials are very useful.

Let (L|K, v) be a simple algebraic extension of valued fields and Γ = vL -the value group of (L, v). For a subset S of Γ, we denote by I S the fractional ideal of L defined by S, that is, I S = {b ∈ L | there exists s ∈ S such that vb ≥ s}.

For any sequence of key polynomials Q for (L|K, v) (see Section 2 below), we define two final segments α and β of Γ. The main idea behind this work is that, under certain conditions, the O L -module Ω of Kähler differentials of the extension O L |O K is isomorphic to I α /I β . This allows us to present an explicit formula for the annihilator of Ω (Proposition 4.4). We also present conditions for Ω to be finitely generated and to be finitely presented (Proposition 4.8 and Corollary 4.9).

A natural situation when this idea can be applied is when (L|K, v) is a pure extension (see Definition 5.1). Such an extension admits a complete sequence of key polynomials of degree one. Hence, we can present simple formulas for α and β ((10) and (17), respectively) and therefore also for Ω. This allows us to present a simple characterization for the annihilator of Ω (see (18)). Also, it is easy to show that, in this case, Ω is not finitely generated and, in particular, not finitely presented.

By a plateau of key polynomials we mean the set of all the key polynomials of a constant degree. In Section 5, we extend the theory of pure extensions (as presented in [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF]). In [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF], only immediate extensions are considered. These can be pure defect or branched pure extensions. A pure defect extension can be understood as an extension consisting of only one plateau key polynomials and this plateau encodes the defect of the extension. Based on this, we consider the cases of purely inertial and purely ramified extensions. Analogously to the pure defect case, these extensions have only one plateau of key polynomials and this plateau encodes the inertia and ramification indices, respectively.

Finally, we apply the theory of minimal key polynomials to pure extensions (see Section 6). Minimal limit key polynomials are the main objects of study in [START_REF] Nart | Minimal limit key polynomials[END_REF]. For each plateau of key polynomials Ψ we define a finite subset B of N that characterizes the minimal key polynomials for Ψ. If the extension is pure, then it admits only the plateau consisting of linear key polynomials. We show that for any pure defect, branched pure or purely inertial extension (L|K, v), with p = char(Kv) > 0, we have

(1) Ω = (0) ⇐⇒ p ∤ B.
Let us give a more precise description of this result. If (L|K, v) is of pure defect, then, by the results of [START_REF] Nart | Minimal limit key polynomials[END_REF], p ∤ B if and only if 1 ∈ B. Then the above result follows from the fact (Proposition 6.6) that, under these assumptions, 1 ∈ B if and only if Ω = (0).

If (L|K, v) is branched pure, then each of the conditions in (1) is satisfied if and only if the extension is defectless. This is a consequence of Proposition 6.6 and of an explicit description of B presented in [START_REF] Nart | Minimal limit key polynomials[END_REF].

Finally, in the purely inertial case, we show that each of the conditions in (1) is satisfied if and only if the residue field extension is separable.

In the purely ramified case, we cannot obtain a set of generators for the O Kalgebra O L from sequences of key polynomials (unless we allow monomials in the key polynomials with integer, possibly negative exponents). However, we adapt previous results to provide a similar description of Ω (see Section 6.5).

To summarize, what we consider to be the main new results of the present paper compared to the earlier works on the subject are the following.

(i): We explicitly compute the O L -module Ω for all pure extensions of arbitrary degree as opposed to the defect Artin-Schreier and Kummer extensions of degree p as in [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF]. Our notion of pure extension is more general than that originally defined in [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF] as it also includes purely ramified and purely inertial branched extensions. The original pure extensions are called pure immediate extensions in our terminology. (ii): Our description of Ω allows us to explicitly describe the annihilator ann(Ω) for all pure extensions (not necessarily Galois) and, in particular, to characterize the situation when

(2) Ω = (0).

In [START_REF] Cutkosky | Kähler differentials of extensions of valuation rings and deeply ramified fields[END_REF], the equality ( 2) is characterized for arbitrary Galois extensions (not necessarily pure). (iii): We use the notion of minimal key polynomials as in [START_REF] Nart | Minimal limit key polynomials[END_REF] to characterize, in many cases, when the equality (2) is satisfied.

Acknowledgement. This paper was inspired by discussions with Franz-Viktor Kuhlmann and Hagen Knaf.

Preliminaries

Notation. In this paper we will use the letter v to denote a valuation of K or its extensions to algebraic extensions of K. For valuations of K[x], we will use the letters ν and µ. For a finite valued field extension (L|K, v) we will denote by e(L|K, v) and f (L|K, v) the ramification and inertial indices of (L|K, v), respectively:

e(L|K, v) = (vL : vK) and f (L|K, v) = [Lv : Kv].
We will denote by d(L|K, v) the defect of (L|K, v). This can be defined as follows. Let L h and K h be the henselizations of L and K, respectively, determined by a fixed extension of v to K. Then

d(L|K, v) := [L h : K h ] (vL : vK) • [Lv : Kv] .
Throughout this paper, when talking about a valued field extension (L|K, v), we will use the symbol Ω to denote the O L -module of Kähler differentials of the extension O L |O K .

2.1. Key polynomials. The problem of understanding all the possible extensions of a valuation on K to K[x] has been extensively studied since the beginning of the past century. Many objects have been introduced to "parametrize" these valuations.

Here, we will focus on one of them: key polynomials.

Even key polynomials appear in the literature in different ways. For instance, there are Mac Lane-Vaquié key polynomials and abstract key polynomials. The relation between these two objects is well-known and we will briefly discuss it here.

Roughly speaking, for a valuation ν on K[x], Mac Lane-Vaquié key polynomials tell the "future" of ν and abstract key polynomials tell its "past". In mathematical terms, a Mac Lane-Vaquié key polynomial for a valuation ν on K[x] allows us to construct another valuation µ such that

ν(f ) ≤ µ(f ) for every f ∈ K[x].
On the other hand, an abstract key polynomial for ν allows us to construct another valuation µ such that

ν(f ) ≥ µ(f ) for every f ∈ K[x].
This means that for treating a given problem, one of these objects may be more suitable than the other. For the goal of this paper, we believe that abstract key polynomials are more suitable. Hence, we will just refer to them as key polynomials.

For a valuation ν on K[x] we fix an extension of it to K[x] and denote it again by ν.

For f ∈ K[x] \ K we define δ(f ) = max{ν(x -a) | f (a) = 0}. A monic polynomial Q is a key polynomial for ν if for every f ∈ K[x] we have δ(f ) ≥ δ(Q) =⇒ deg(f ) ≥ deg(Q). For any polynomials f ∈ K[x] and q ∈ K[x] \ K, there exist uniquely determined f 0 , . . . , f r ∈ K[x] with deg(f ℓ ) < deg(q) for every ℓ, 0 ≤ ℓ ≤ r, such that (3) f = f 0 + f 1 q + . . . + f r q r .
This expression is called the q-expansion of f . For every monic polynomial

q ∈ K[x] \ K
we define the truncation of ν at q by

ν q (f ) = min 0≤ℓ≤r ν f ℓ q ℓ
where the notation is as in [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF]. We allow the possibility (4) ν(q) = ∞.

If q the has smallest degree among the polynomials satisfying (4), then ν q (f ) = ∞ if and only if f 0 = 0. If Q is a key polynomial for ν, then by [7, Proposition 2.6] ν Q is a valuation with

ν(f ) ≥ ν Q (f ) for every f ∈ K[x]. A subset Q of K[x] is called a complete set for a valuation ν on K[x], if for every f ∈ K[x] there exists q ∈ Q such that deg(q) ≤ deg(f ) and ν q (f ) = ν(f ).
We consider a valued field extension (L|K, v). We assume that this extension is simple and algebraic and write it as L = K(η) for some η ∈ L. Consider the valuation ν on K[x] with non-trivial support defined by v and η, that is,

ν(f ) := v(f (η)) for every f ∈ K[x].
The philosophy is that a complete set for ν should induce a set of generators for the extension O L |O K of valuation rings. This is true whenever (L|K, v) is unramified, that is, vL = vK (see Proposition 3.5 below).

In order to study the module of Kähler differentials, it is necessary to know the generators of the extension, but also the relations between these generators. Because of this, it is important to have a complete set with relations as simple as possible. For this purpose, complete sets formed by key polynomials are very handy.

Remark 2.1. We will use the term "complete sequence" instead of "complete set" when Q = {Q i } i∈I with I well-ordered and the map i → Q i is order preserving with respect to δ (i (1) It is not hard to prove from definitions that the set of all key polynomials is complete. A sequence as in Theorem 2.2 can be constructed as follows. For every m ∈ N consider the set Ψ m of all the key polynomials of degree m. For each m ∈ N such that Ψ m ̸ = ∅ we define Q m as follows. If ν(Ψ m ) has a last element ν(Q) with respect to δ, then put Q m = {Q}. If ν(Ψ m ) does not have a last element, take any sequence Q m whose images form an a cofinal subset of ν(Ψ m ). Then

< j =⇒ δ(Q i ) < δ(Q j )).
Q := m∈N Ψm̸ =∅ Q m
is the desired sequence of key polynomials.

(2) Assume that

supp(ν) := {f ∈ K[x] | ν(f ) = ∞} is a principal ideal (g) of K[x]
(this is the only situation considered in this paper). Then the complete sequence Q of key polynomials can be chosen so that g ∈ Q and ν(Q) < ∞ for all Q ∈ Q \ {g}. In particular, g is the last element of Q. From now on, we will assume that Q has this property without mentioning it explicitly.

Remark 2.4. Every linear monic polynomial is a key polynomial (in particular, Ψ 1 ̸ = ∅). If ν(Ψ 1 ) does not have a last element, then the cofinal families Q 1 in Ψ 1 are in bijection with the pseudo-convergent sequences on K induced by ν. This bijection can be presented explicitly:

Q 1 = {x -a i } i∈I -→ {a i } i∈I .

The generation of an extension of valuation rings

Let (K, v) be a valued field and take a valuation µ on K[x], extending v. For simplicity of notation, in this section we denote

Γ v = vK and Γ µ = µ(K[x]). Assume that e(µ/v) := (Γ µ : Γ v ) = 1. Definition 3.1. Two sets Q, Q ′ ⊂ K[x] are called K-proportional if there exists a bijection φ : Q -→ Q ′ such that for every q ∈ Q there exists a ∈ K \ {0} such that φ(q) = aq. Lemma 3.2. If Q ⊆ K[x]
is a complete set for µ, then any K-proportional set is also a complete set for µ.

Proof. For f, q ∈ K[x] and a ∈ K \ {0}, if f = f 0 + f 1 q + . . . + f n q n is the q-expansion of f , then f = f 0 + f 1 a • (aq) + . . . + f n a n • (aq) n is the aq-expansion of f . If µ q (f ) = µ(f ), then µ aq (f ) = µ(f ) for every a ∈ K. Moreover, deg(q) = deg(aq). The result follows immediately. □ Lemma 3.3. If e(µ/v) = 1
, then for every complete set Q of µ there exists a complete set Q for µ, K-proportional to Q, such that µ(q) = 0 for every q ∈ Q\{g}.

Proof. Since e(µ/v) = 1, for every q ∈ K[x] \ (g) there exists a ∈ K such that v(a -1 ) = µ(q).
Hence µ(aq) = 0 and we can apply the previous lemma. □

For a subset Q of K[x], we will denote by N Q 0 the set of mappings λ : Q -→ N 0 such that λ(q) = 0 for all but finitely many q ∈ Q (here N 0 denotes the set of the non-negative integers). For each λ ∈ N Q 0 we denote

Q λ = q∈Q q λ(q) . Theorem 3.4. Take a complete set Q ⊆ K[x] for µ such that µ(q) = 0 for every q ∈ Q \ {g}. For f ∈ K[x] \ (g), if µ(f ) ≥ 0, then there exist λ 1 , . . . , λ s ∈ N Q and a 1 , . . . , a s ∈ O K such that f = s ℓ=1 a ℓ Q λ ℓ and (5) µ(f ) = min 1≤ℓ≤s v(a ℓ ).
Proof. We proceed by induction on the degree of f . If deg(f ) = 1, then by our assumption there exists q ∈ Q of degree one (in this case

f = aq + b for a, b ∈ K), such that µ(f ) = min{µ(aq), µ(b)}.
Since 0 ≤ min{µ(aq), µ(b)} and µ(q) = 0, we have a, b ∈ O K and we are done. Now consider an integer n > 1 and assume that for every

f ∈ K[x] \ (g), if deg(f ) < n, then there exist λ 1 , . . . λ s ∈ N Q and a 1 , . . . , a s ∈ O K such that f = s ℓ=1 a ℓ Q λ ℓ and (5) holds. Take f ∈ K[x] \ (g) with deg(f ) = n and µ(f ) ≥ 0.
By our assumption on Q, there exists q ∈ Q such that deg(q) ≤ deg(f ) and µ(f ) = µ q (f ). This means that

f = f 0 + f 1 q + . . . + f r q r with deg(f l ) < deg(q) for every l, 0 ≤ l ≤ r, and 0 ≤ µ(f ) = min µ f l q l ≤ µ(f l ), for every l, 1 ≤ l ≤ r, because µ(q) = 0. Since deg(f l ) < deg(q) ≤ deg(f ) = n, there exist λ 0,1 , . . . , λ 0,s0 , . . . , λ r,1 , . . . , λ r,sr ∈ N Q and a 0,1 , . . . , a r,sr ∈ O K such that f l = s l ℓ=1 a l,ℓ Q λ l,ℓ and µ(f l ) = min 1≤ℓ≤s l v(a l,ℓ ) for every l, 0 ≤ l ≤ r. This implies that f = r l=0 s l ℓ=1 a l,ℓ Q λ l,ℓ q l = r l=0 s l ℓ=1 a l,ℓ Q λ ′ l,ℓ ,
where

λ ′ l,ℓ (q ′ ) = λ l,ℓ (q) + l if q = q ′ λ l,ℓ (q) if q ̸ = q ′ and µ(f ) = min 0≤l≤r 1≤ℓ≤s l v(a l,ℓ
). This concludes our proof. □

3.1. The algebraic case. Let (L|K, v) be a simple algebraic extensions of valued fields. Write L = K(η) and let ν be the valuation on K[x] with non-trivial support defined by v and η: ν(f

) := v(f (η)). For Q ⊆ K[x] we denote by Q(η) := {q(η) | q ∈ Q}. Proposition 3.5.
Assume that e(L|K, v) = 1. For any complete set of polynomials Q for ν there exists a K-proportional complete set Q such that

(6) O L = O K Q(η) .
Proof. By Lemma 3.3 there exists a set Q which is K-proportional such that ν(q) = 0 for every q ∈ Q \ (g).

By Lemma 3.2, Q is a also a complete set for ν. By definition, the right hand side of ( 6) is contained in the left hand side. For every b ∈ L there exists a polynomial f (

x) ∈ K[x] (with deg(f ) < deg(g)) such that b = f (η). If b ∈ O L , then 0 ≤ ν(f (x)). By Theorem 3.4, there exist a 1 , . . . , a r ∈ O K and λ 1 , . . . , λ r ∈ N Q such that b = f (η) = r ℓ=1 a ℓ Q(η) λ ℓ ∈ O K Q(η) .
This concludes the proof. □

A characterization of Ω

We consider a valued field extension (L|K, v) as before. Namely, L = K(η) for some algebraic η ∈ L, g is the minimal polynomial of η over K and ν is the valuation on K[x] with non-trivial support defined by v and η. In this case, g is a key polynomial for ν with

δ(g) = ∞ > δ(Q) for every key polynomial Q for ν such that Q ̸ = g.
Take a complete sequence of key polynomials Q ∪ {g} for ν, where Q = {Q i } i∈I . The initial element of I will be denoted by 0. Assume that (L|K, v) is unramified. For every i ∈ I choose a i ∈ K such that ν(Q i ) = v(a i ) and set Qi = Q i /a i . We set Q 0 = x. Replacing x by x a0 , we may assume that ν(x) = 0, in other words, Q 0 = Q0 = x. Also, we set I ∞ = I ∪ {∞} and Q ∞ = Q∞ = g. Definition 4.1. In the case described above, we will say that {Q i (η)} i∈I is a complete sequence of key polynomials for (L|K, v).

For an element f ∈ K[x] we denote by f ′ its formal derivative with respect to x. Let Γ denote the value group of v and fix a sequence of key polynomials {Q i (η)} i∈I for (L|K, v). For each i ∈ I consider the following values in Γ ∞ :

α i = ν(Q ′ i ) -ν(Q i ) and β i = ν(g ′ ) -ν i (g).
Here, we write ν i (g) = ν Qi (g) for simplicity.

Let α and β be the smallest final segments of Γ ∞ containing

{α i | i ∈ I} and {β i | i ∈ I},
respectively.

Let Ω denote the module of Kähler differentials for the extension O L |O K . It is not hard to show that the O L -module homomorphism [START_REF] Novacoski | Key polynomials and pseudo-convergent sequences[END_REF] Ω -→ I α /I β sending d Qi to Q ′ i /a i , i ∈ I, is well defined and surjective (below we will prove it in the special cases needed for this paper).

Question 4.2.

Under what conditions is the map (7) an isomorphism? Remark 4.3. In full generality, some conditions are definitely needed for (7) to be an isomorphism; it is not one unconditionally. The answer depends on the structure of the possible sequences of key polynomials for (L|K, v).

4.1.

Consequences of the isomorphism Ω ∼ = I α /I β . In this section we assume that we are in a situation where ( 7) is an isomorphism. We will show that this is the case for unramified pure extensions, for instance (see Section 5 below).

γ = {β ′ ∈ Γ | β ′ + α ⊆ β}.
Proof. Indeed, since two isomorphic O L -modules have the same annihilator, we deduce that ann(Ω) = ann (I α /I β ) .

Now an element

H(S) := {β ′ ∈ Γ | β ′ + S = S}. Remark 4.6. If α ̸ = β, then H(α) < γ.
Another important consequence of the isomorphism in ( 7) is about the finite generation of Ω.

Corollary 4.7. If Ω is finitely generated as O L -module, then it is generated by a single element.

Proof. Assume that I α /I β is finitely generated, say, by

{a 1 + I β , . . . , a n + I β } with a 1 , . . . , a n ∈ I α .
Suppose, without loss of generality, that v(a 1 ) ≤ v(a ℓ ) for every ℓ, 1 < ℓ ≤ n. Then for every such ℓ we have

a ℓ + I β = a ℓ a 1 (a 1 + I β ) .
Hence, I α /I β is generated by α 1 + I β . □

The above result can be interpreted as follows.

Proposition 4.8. Assume that Ω ̸ = (0). Then Ω is finitely generated as an O Lmodule if and only if α has a smallest element.

We can use the isomorphism [START_REF] Novacoski | Key polynomials and pseudo-convergent sequences[END_REF] to answer the question of when Ω is finitely presented.

Corollary 4.9. Assume that Ω ̸ = (0). Then Ω is finitely presented as an O Lmodule if and only if each of α and β has a smallest element.

Proof. If α does not have a smallest element (and Ω ̸ = (0)), then I α /I β is not finitely generated, so it is not finitely presented.

Suppose that α has a smallest element, say α 0 = va for some a ∈ L. In particular, β has a smallest element if and only if I α /I β is finitely presented. □

Pure extensions

The main goal of section is to show that ( 7) is an isomorphism for unramified pure extensions. Definition 5.1. Let (L|K, v) be a simple immediate extension of valued fields and take η ∈ L such that L = K(η). We say that the extension

(L|K, v) is pure immediate in η if for every f ∈ K[x], deg(f ) < [L : K], there exists c ∈ K such that for every b ∈ K we have (8) v(η -b) ≥ v(η -c) =⇒ v(f (b)) = v(f (c)).
Definition 5.2. When condition (8) is satisfied for some f , we say that v(f (c)) is the fixed value of f .

For b ∈ L we denote v(b -K) := {v(b -c) | c ∈ K}. Lemma 5.3. If the extension (L|K, v) is pure immediate in η, then v(η -K)
i } i∈I ⊂ K such that {v(η -c i )} i∈I is cofinal in v(η -K). For each i ∈ I take a i ∈ K such that v(η -c i ) = v(a i ). Then O L = O K η -c i a i i ∈ I .
Proof. Follows from Proposition 3.5.

□ Definition 5.6. The extension (L|K, v) is called unibranched if v is the unique extension of v| K to L.
The main interest in [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF] is to study pure extensions when (L|K, v) is algebraic and unibranched (hence d(L|K, v) = [L : K]). In view of Lemma 5.3, we extend the definition of pure extensions to the case of algebraic extensions that are not necessarily immediate. Definition 5.7. Assume that (L|K, v) is a simple algebraic extension of valued fields. Then we say that:

(i): (L|K, v) has pure defect if it is pure immediate and unibranched (hence d(L|K, v) = [L : K]). (ii): (L|K, v) is branched pure if it is pure immediate but not unibranched. (iii): (L|K, v) is purely ramified if vL/vK is cyclic and [L : K] = (vL : vK). (iv): (L|K, v) is purely inertial if Lv|Kv is simple and [L : K] = [Lv : Kv].
From now on, when referring to pure extensions we will mean any extension as one appearing in the definition above.

Proposition 5.8. For any extension of valued fields of the forms (i) to (iv) above there exists a generator η of L|K such that the set v(η -K) is a complete set of key polynomials for (L|K, v).

Proof. The cases (i) and (ii) follow from Lemma 5.3. In the case (iii) take η ∈ L such that vη generates vL over vK. In the case (iv) take η ∈ O L such that ηv generates Lv over Kv. In both cases, η generates L over K. Moreover, if ν is the valuation on K[x] induced by v and η, then it is easy to show that for every

f ∈ K[x], deg(f ) < [L : K], we have ν(f ) = ν x (f ).

□

In the case of purely inertial or purely ramified as above, we say that (L|K, v) is pure in η.

In what follows we present the proof of the isomorphism (9) Ω ∼ = I α /I β for unramified pure extensions. This proof is a simplified version of the general proof of this result which will be presented in future work.

5.1. Proof of (9) for pure defect, branched pure or purely inertial extensions. Suppose that (L|K, v) is a pure defect, branched pure or purely inertial extension in η.

Let n = [L : K]. Take a sequence Q = {Q i } i∈I where Q i = x -c i for some c i ∈ K, such that {v(η -c i )} i∈I is cofinal in v(η -K).
Remark 5.9. In the purely inertial case, we can choose Q to be a one-element set.

Consider the corresponding fractional ideals I α and I β .

Theorem 5.10. We have the isomorpihsm (9) of O L -modules.

Proof. It follows from Proposition 5.8 that {η -c i } i∈I is a complete sequence of key polynomials for (L|K, v). For each i ∈ I we have

α i = ν(Q ′ i ) -ν(Q i ) = v1 -v(η -c i ).
Consequently,

(10) α = -v(η -K).
For every i ∈ I take a i ,

d i ∈ K such that v(a i ) = v(η -c i ) and v(d i ) = ν i (g).
We may assume that vη = 0 and Q 0 (η) = η. Set Qi = x-ci ai . Take a set X = {X i | i ∈ I} of independent variables and consider the homomorphism For i, j ∈ I, the K-relations between Qj and Qi are generated by

O K [X i | i ∈ I] -→ O L
Qj = a i a j Qi + c i -c j a j .
Hence, all the O K -relations between them are generated by

(12) b ji Qj = Qi + c i -c j a i .
where b ji = a j /a i and j > i. All of these relations are generated (over K, but not over O K ) by ( 13)

a i Qi = Q0 + c i for i ∈ I.
Since v(a i ) = ν(x -c i ) and the (x -c i )-expansion of g is

g(x) = (x -c i ) n + . . . + g ′ (c i )(x -c i ) + g(c i )
we see that

d i = ν i (g) = min{v(a n i ), . . . , v(a i g ′ (c i )), v(g(c i ))}. This gives the K-relation 0 = g(η) = a n i Qi (η) n + . . . + a i g ′ (c i ) Qi (η) + g(c i ).
Hence, the ideal I is generated by the elements

(14) b ji X j -X i - c i -c j a i , j > i.
and the O K -relations of the form

a n i d i X n i + . . . + a i g ′ (c i ) d i X i + g(c i ) d i = g(a i X i + c i ) d i . Let g i := g(aiXi+ci) di ∈ O K [X i ] ⊂ O K [X].
: Ω -→ I α /I β s l=1 b l dX i l + J -→ s l=1 b l a i l + I β .
This can be seen as the map that sends dX 0 + J to 1 + I β and is extended to Ω by the relations a i dX i = dX 0 . The fact that the relations in ( 15) and ( 16) generate J guarantees that Ψ is well-defined and injective:

s l=1 b l dX i l ∈ J if and only if s l=1 b l a i l = a g ′ (η) d i for some a ∈ O L and i ∈ I, if and only if s l=1 b l a i l ∈ I β .
Since Ψ is defined on the generators of Ω and extended by O L -linearity, it is automatically a homomorphism of O L -modules. The surjectivity of Ψ follows by construction: if b ∈ I α , then there exists i ∈ I such that vb ≥ -v(a i ). Hence

Ψ (ba i dX i + J ) = ba i a i + I β = b + I β .
This concludes the proof of Theorem 5.10. □ 5.2. Computation of Ω for pure defect and branched pure extensions. Assume that (L|K, v) is a pure defect extension in η and consider a sequence Q as before. For i large enough, it follows from [1, Proposition 4.1] that ν i (g) = nν(Q i ).

Hence, for i large enough, we have

β i = v(g ′ (η)) -ν i (g) = v(g ′ (η)) -n • v(η -c i ).
Consequently,

(17) β = v(g ′ (η)) -n • v(η -K).
Combining this with (10), we can say precisely what the annihilator of Ω is:

(18) ann(Ω) = b ∈ O L v b g ′ (η) -v(η -K) ⊆ -n • v(η -K) .
For instance, if rk v = 1, then we may assume that vL ⊆ R and take

ρ := sup v(η -K) ∈ R.
Then, ann(Ω) can be written as

ann(Ω) = {b ∈ O L | vb ≥ (1 -n)ρ + v(g ′ (η))} .
In the case of a branched pure extension, we obtain a slightly different formula for β (for α we have the same formula (10)). Namely, by the defect formula ([5, Theorem 6.14]) we deduce that for large enough i ∈ I we have

ν i (g) = ν ∂ d g(a i )(x -c i ) d = ν(∂ d g(a i )) + dv(η -c i ), where d := d(L|K, v). If β d is the fixed value of ∂ d g, then (19) β = v(g ′ (η)) -β d -d • v(η -K).
Remark 5.12. In the case of a pure defect extension, we have n = d(L|K, v) = d, ∂ d g = 1 (as g is monic) and β d = 0. Thus formula (17) can be viewed as a special case of (19) when the number of distinct extensions of v to L is equal to one.

About [3, Proposition 4.1]

. In [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF] an alternative characterization of Ω is presented. We briefly compare this approach with ours.

Proposition 5.13. [3, Proposition 4.1] Let L|K be an algebraic field extension and suppose that A is a normal domain with quotient field K and B is a domain with quotient field L such that A ⊂ B is an integral extension. Suppose that there exist s α ∈ B, which are indexed by a totally ordered set S, such that

A[s α ] ⊂ A[s β ] if α < β and α∈S A[s α ] = B.
Further, suppose that there exist r α , r β ∈ A such that r β | r α if α ≤ β and for α ≤ β, there exist c α,β ∈ A and expressions

s α = r α r β s β + c α,β .
Let h α be the minimal polynomial of s α over K. Let U and V be the B-ideals

U = (r α | α ∈ S) and V = (h ′ α (s α ) | α ∈ S). Then we have a B-module isomorphism Ω B|A ≃ U/U V.
We now explain how to describe the elements r α , s α and c αβ of Proposition 5.13 using our terminology and notation.

Remark 5.14. It is well-known that for an algebraic extension (L|K, v) the valuation ring O K is normal and its integral closure in L is the intersection of all the valuation rings of L dominating O K . Hence, for the extension O L |O K to be integral, it is necessary for (L|K, v) to be unibranched.

Assume that (L|K, v) is a pure defect extension and consider the sequence of key polynomials {η -c i } and elements a i as before. Take r ∈ O L such that vr > v(η -K).

For each i ∈ I, write

r i = ra -1 i and s i = η -c i a i .
If i < j, then

s i = (η -c i ) a i = r i r j • s j + (c j -c i ) a i ∈ O K [s j ]. Since v(a i ) = v(η -c i ) < v(η -c j ) = v(a j
) we deduce that ra -1 i = r i is a multiple of r j = ra -1 j by an element of O L . In particular, the hypotheses of [START_REF] Cutkosky | Characterizations of Galois extensions with independent defect[END_REF]Proposition 4.1] are satisfied and we deduce that

Ω ≃ U/U V.
Let us compute U and V and compare with our results above. For each i ∈ I, we have

v(r i ) = v(ra -1 i ) = vr -v(a i ) and consequently (20) U = (r i | i ∈ I) = {b ∈ O L | vb > vr -v(a i ) for some i ∈ I} = I α+vr .
Let us calculate V . For each i, we have η = (η

-c i ) + c i . Hence (21) 0 = g(η) = g((η -c i ) + c i ) = g(c i ) + g ′ (c i )(η -c i ) + . . . + (η -c i ) n .
Remark 5.15. Since deg g ′ < deg g, for i sufficiently large the sequence

{v (g ′ (c i ))} i∈I
stabilizes and its stable value is equal to v(g ′ (η)). Of course, this includes the case g ′ = 0.

Since

s i = η-ci ai , (21) implies that 0 = g(c i ) a n i + g ′ (c i ) a n-1 i s i + . . . + s n i . Write (22) h i (x) = g(a i x + c i ) a n i = g(c i ) a n i + g ′ (c i ) a n-1 i x + . . . + x n ∈ K[x].
Since g(x) is irreducible over K, so is h i (x). Thus h i (x) is the minimal polynomial of s i .

In view of the expression (22) for h i in terms of g and the chain rule for differentiation, we have, for all i,

h ′ i (x) = g ′ (a i x + c i ) a n-1 i , so h ′ i (s i ) = g ′ (a i s i + c i ) a n-1 i = g ′ (η) a n-1 i .
Therefore,

V = v g ′ (η) a n-1 i | i ∈ I = {b ∈ O L | ∃i ∈ I such that v(b) ≥ v(g ′ (η)) + (1 -n)va i } = I v(g ′ (η))+(1-n)v(η-K) .

Minimal key polynomials

Let ν be a valuation on K[x] and for m ∈ N consider the set

Ψ m = {Q ∈ K[x] | deg(Q) = m and Q is a key polynomials for ν}.
If Ψ m ̸ = ∅ and F is a key polynomial of smallest degree strictly larger than m (we denote deg F by m + , so F ∈ Ψ m+ ), then we say that F is a key polynomial for Ψ m . If ν(Ψ m ) does not have a maximum, then every key polynomial for Ψ m will be called a limit key polynomial for Ψ m . For two polynomials f, q ∈ K[x], deg(q) > 0, denote by f = f r,q q r + . . . + f 0,q the q-expansion of f . For each m ∈ N such that Ψ m ̸ = ∅ we will choose a subset Q m as follows. If ν(Ψ m ) has a maximum, then set Q m = {Q m } where Q m ∈ Ψ m has this maximum value. If ν(Ψ m ) does not have a maximum, we choose Q m to be any well-ordered cofinal subset of Ψ m . Definition 6.1. A key polynomial F for Ψ m is said to be minimal if there exists

Q ∈ Q m and a subset B m of {1, . . . , r} such that for every R ∈ Q m , with ν(R) ≥ ν(Q), the polynomial F 0,R + ℓ∈Bm F ℓ,R R ℓ
is a key polynomial for Ψ m but for every s ∈ B m there exists a cofinal subset of Q m such for every R in it, the polynomial

F 0,R + ℓ∈Bm\{s} F ℓ,R R ℓ is not a key polynomial for Ψ m .
One can show that the set B m above does not depend on F or Q ∈ Ψ m . Remark 6.2. In the case when Ψ m contains a maximal element, the above definition becomes much simpler. Since the only possibility for Q is the unique element of Q m , the polynomial F is a minimal key polynomial for Ψ m if

F 0,Q + ℓ∈Bm F ℓ,Q Q ℓ is a key polynomial for Ψ m but for every s ∈ B m the polynomial F 0,Q + ℓ∈Bm\{s} F ℓ,Q Q ℓ is not a key polynomial for Ψ m .
For a pure extension (L|K, v) of degree n and a generator η for it, consider the valuation ν on K[x] defined by v and η. The only natural numbers m for which the corresponding set Ψ m is non-empty are m = 1 and m = n. Hence, the only case for which it makes sense to talk about minimal key polynomials is for the set Ψ 1 . Because of this, for the remaining of this section we will denote B 1 simply by B.

6.1. The purely inertial case. Assume that the extension (L|K, v) is purely inertial. Choosing η as in Proposition 5.8 we deduce that 0 = vη is the maximum of v(η -K) and ηv generates Lv|Kv.

The following lemma follows immediately from Proposition 3.5.

Lemma 6.3. We have O L = O K [η]. Proposition 6.4. If (L|K, v) is purely inertial, then Ω ≃ O L /(g ′ (η)).
Proof. In this case we can apply Theorem 5.10 for Q = {x}. Then we have

α = v1 -vη = 0 and β = v(g ′ (η)) -ν x (g) = v(g ′ (η)).
Hence, the result follows.

□ Let Q(y) = b 0 v + b 1 vy + . . . + y n ∈ Kv[y]
be the minimal polynomial of ηv over Kv. Then

Q = b 0 + b 1 x + . . . + x n = b 0 + k∈B b k x k ∈ K[x]
is a minimal key polynomial for Q 1 . Also, we deduce that

ν(g -Q) = ν x (g -Q) > ν x (g) = 0. Corollary 6.5. If (L|K, v) is purely inertial, then Ω = (0) ⇐⇒ Lv|Kv is separable ⇐⇒ p ∤ B.
Proof. By Proposition 6.4 we deduce that Ω = (0) if and only if v(g ′ (η)) = 0. This happens if and only if dQ dy ̸ = 0. This happens if and only if ηv is separable over Kv. Since

Q(y) = b 0 v + b 1 vy + . . . + y n = b 0 v + k∈B b k vy k we deduce that dQ dy ̸ = 0 ⇐⇒ p ∤ B. □ 6.2.
A description of Ω for pure defect and branched pure extensions. Proposition 6.6. Assume that (L/K, v) is a pure defect or a branched pure extensions.

Then

Ω = (0) ⇐⇒ 1 ∈ B.
Proof. Set n = deg(g). Take a cofinal well-ordered (with respect to ν) subset

Q 1 = {x -a i } i∈I of {x -a | a ∈ K} as before.
For each i ∈ I we have

α i = ν d dx (x -a i ) -ν(x -a i ) = ν(1) -ν(x -a i ) = -ν(x -a i ).
Since g is a polynomial of smallest degree which is Q 1 -unstable, there exists

i 0 ∈ I such that ν(g ′ ) = ν i (g ′ ) = ν(g ′ (a i )) for every i ∈ I with i ≥ i 0 .
Hence,

β i = ν(g ′ ) -ν i (g) for every i ∈ I with i ≥ i 0 .
In particular, {β i } i∈I is ultimately decreasing.

Clearly, {α i } i∈I is decreasing. Since {β i } i∈I is ultimately decreasing, there exists a final set I 0 ≥ i 0 of I such that {α i } i∈I0 and {β i } i∈I0 are decreasing.

For each i ∈ I, the (x -a i )-expansion of g is given by

g = g(a i ) + g ′ (a i )(x -a i ) + . . . + (x -a i ) n .
Then the condition α = β is equivalent to saying that for every i ∈ I 0 , there exists

j ∈ I 0 such that (23) ν(g ′ (a i )) -ν j (g) = ν(g ′ ) -ν j (g) = β j < α i = -ν(x -a i ).
This happens if and only if

ν (g ′ (a i )(x -a i )) < ν j (g).
This shows that Ω = (0) if and only if This proves the result for pure defect extensions.

If the extension is purely inertial, then the result was proved in Corollary 6.5. If (L|K, v) is a branched pure extension, then we can apply Proposition 6.6 to obtain that Ω = (0) if and only if 1 ∈ B. Since d ∈ B, if d = 1, then Ω = (0). On the other hand, it follows from [START_REF] Nart | Minimal limit key polynomials[END_REF]Theorem 6.26] that if 1 / ∈ B, then every element of B is a multiple of p. This concludes the proof of the theorem. □ 6.5. The purely ramified case. We discuss now one particular case where we cannot apply the above results. Assume that the extension (L|K, v) is purely ramified. Choosing η as in Proposition 5.8 we deduce that γ = vη is the maximum of v(η -K) and generates vL over vK.

Assume, without loss of generality, that γ > 0. Furthermore, let ∆ denote the greatest (in the sense of inclusion) isolated subgroup of vL such that ∆ < γ. Example 6.9. To illustrate the definition of ∆, we present an example where ∆ ̸ = (0). Let (L|K, v) be a valued field extension such that

vL = 1 2 Z × lex R ⊃ Z × lex R = vK and γ = 1 2 , 0 . Then ∆ = (0) × lex R.
Observe that in this case, vL >0 does not have a minimal element, but vL ∆ >0 does.

If vL ∆ >0 has a minimal element, then we will choose η in such a way that

γ + ∆ = min vL ∆ >0 . Set Q 1 = {x}. Let n = [K(η) : K].
Since the extension is purely ramified, there exists b 0 ∈ K such that

v (η n -b 0 ) > nvη = vb 0 .
In this case, Q = x n -b 0 is a minimal key polynomial for Q 1 . In particular, B = {n}. Lemma 6.10. Under the above assumptions we have We claim that there exists h ∈ K such that for all integers s, 1 ≤ s ≤ m, we have

(24) O L = O K η h h ∈ K and vh < γ .
(28) -v(b s ) ≤ svh < sγ.
Denote by ∆ + the smallest initial segment of vL containing ∆. If vL ∆ >0 contains a minimal element then, by the choice of γ, we have

-v(b s ) ∈ ∆ +
for all s, so we may take h to be any element of K such that

v(h) ∈ ∆ and v(h) ≥ max 1≤s≤m - v(b s ) s .
Assume that vL ∆ >0 does not contain a minimal element. Then every non-empty open interval in vL ∆ contains an element of vK + ∆. Hence, there exists h ∈ K such that

(29) max 1≤s≤m - v(b s ) s + ∆ < vh + ∆ < γ + ∆ so, in particular, ( 30
) max 1≤s≤m - v(b s ) s < vh < γ.
This completes the proof of the existence of h ∈ K satisfying (28). Then

b = b 0 + b 1 h η h + . . . + b m h m η h m ∈ O K η h .
This completes the proof of the lemma. □ Remark 6.11. Consider the final segment -∆ + of vL. Then the invariance subgroup of -∆ + is exactly ∆.

Write g = a 0 + a 1 x + . . . + x n . Remark 6.12. For any ℓ, 1 ≤ ℓ ≤ n -1, we have

(31) vℓ + va ℓ -(n -ℓ)γ > 0.
Indeed, since v(g(η)) = ∞ and γ, 2γ, (n -1)γ, nγ belong to distinct vK-cosets, we deduce that va 0 = nγ < min 1≤ℓ≤n-1 {va ℓ + ℓγ}.

The inequality (31) follows.

Proposition 6.13. Assume that the extension (L|K, v) is purely ramified.

(i):

If vL ∆ >0 contains a minimal element, then Ω ̸ = (0). (ii): If vL ∆ >0
does not contain a minimal element, then Ω = (0) if and only if there exists ℓ, 1 ≤ ℓ ≤ n, such that

(32) vℓ + va ℓ -(n -ℓ)γ ∈ ∆.
Proof. For h ∈ K with vh ≤ γ, let J h denote the fractional ideal of K defined by

J h = {b ∈ K | bg(hx) ∈ O K [x]}.
It follows from Lemma 6.10 that

(33) Ω = h∈K vh≤γ 1 h O L dx   bg ′ (η)dx b ∈ h∈K vh≤γ J h    . Remark 6.14. We have J h = 1 h n O K , so (33) becomes (34) Ω = h∈K vh≤γ 1 h O L dx   bg ′ (η)dx b ∈ h∈K vh≤γ 1 h n O K    .
The equality (34) says that Ω = (0) if and only if the following equality of fractional ideals in L holds:

(35) h∈K vh≤γ 1 h O L =   bg ′ (η) b ∈ h∈K vh≤γ 1 h n O K    O L .
Furthermore, we have

(36) v (g ′ (η)) = min 1≤ℓ≤n {vℓ + va ℓ + (ℓ -1)γ}.
In the rest of the proof of the proposition, we will consider cases (i) and (ii) separately.

Suppose that the hypothesis of (i) is satisfied. In this case the condition vh ≤ γ is equivalent to saying that vh ∈ ∆ + . Hence, the left hand side of (35) is equal to I -∆ + . On the other hand, for each h ∈ K with vh ≤ γ we have v(J h ) ⊂ -∆ + . In particular, the right hand side of (35) is contained in

I v(g ′ (η))-∆ + .
Since v(g ′ (η)) > ∆, by Remark 6.11, we deduce that

v(g ′ (η)) -∆ + ⫋ -∆ + .
Hence, the right hand side of ( 35) is strictly contained in its left hand side, so Ω ̸ = (0), as desired.

Suppose now that the hypothesis of (ii) is satisfied. Taking into account the equality (36), we see that the fractional ideal on the left hand side of (35) is I α , where

α = {-vh | h ∈ K, vh < γ} = vL >-γ
and the fractional ideal on the right hand side is I β , where

β = min 1≤ℓ≤n {vℓ + va ℓ + (ℓ -1)γ} -nvh h ∈ K, vh < γ . Rewrite β as β = min 1≤ℓ≤n {vℓ + va ℓ -(n -ℓ)γ} + n(γ -vh) -γ h ∈ K, vh < γ = min 1≤ℓ≤n {vℓ + va ℓ -(n -ℓ)γ} + vL >-γ ,
where the last equality holds because vL ∆ >0 does not contain a minimal element. A similar reasoning as Remark 6.11 shows that the invariance subgroup of vL >-γ is ∆. It follows that

α = β ⇐⇒ min 1≤ℓ≤n {vℓ + va ℓ -(n -ℓ)γ} ∈ ∆.
The latter condition is equivalent to saying that vℓ + va ℓ -(n -ℓ)γ ∈ ∆ for some ℓ, 1 ≤ ℓ ≤ n. This completes the proof of the proposition. □ Corollary 6.15. Assume that vL ∆ >0 does not have a minimum. If p ∤ B, then Ω = (0).

Proof. Since B = {n}, the condition p ∤ B is the same as p ∤ n. Hence, vn = 0. In particular, the condition (32) is satisfied for ℓ = n. Consequently, Ω = (0). □ Corollary 6.16. Assume that vL ∆ >0 does not have a minimum. If vp ∈ ∆, then Ω = (0).

Proof. If vp ∈ ∆, then vr ∈ ∆ for every r ∈ N. Hence, condition (32) is satisfied for ℓ = n and consequently, Ω = (0). □

Artin-Schreier and Kummer extensions

We apply the results of this paper to the particular cases of Kummer and Artin-Schreier extensions.

7.1. Artin-Schreier extensions. Assume that (L|K, v) is an Artin-Schreier extension of degree p = char(K) > 0. This means that L = K(η) where η is a root of an irreducible polynomial over K of the form g = x p -x -a. There are four cases to be considered.

If d(L|K, v) > 1, then (L|K, v) is a pure defect extension. Applying the results of Section 5.2, we obtain

(37) α = -v(η -K) and β = -p • v(η -K).
Hence, ann(Ω) = I γ ∩ O L , where γ is the final segment of Γ formed by all the elements β ′ in Γ for which there exists i ∈ I such that β ′ + pv(η -a i ) > v(η -a j ) for every j ∈ I.

In the rank one case, we set ρ = sup v(η -K). Then

ann(Ω) = {b ∈ O L | vb ≥ (1 -p)ρ}.
It follows from Proposition 4.8 that if α ̸ = β, then Ω is not finitely generated (and, consequently, not finitely presented).

If the extension is defectless and not unibranched, then it is branched pure and d(L|K, v) = 1. By Proposition 6.7 we deduce that Ω = (0).

For the rest of this section, assume that (L|K, v) is unibranched and defectless. Then, by the fundamental inequality, If vη < 0 (so vc > 0), then the residue field extension is purely inseparable and we have Ω ̸ = (0). If vη = 0, then the residue field extension is separable and Ω = (0) (in particular, ann(Ω) = O L ). It follows from Corollary 4.9 that Ω is finitely presented. If vK ̸ = vL, then by (38) [L : K] = (vL : vK) = p. In particular, (L|K, v) is purely ramified. Since p = (vL : vK) we deduce from Theorem 6.13 that Ω ̸ = (0). Indeed, if vL ∆ >0 has a minimum, then the result follows from Theorem 6.13 (i). Otherwise, observe that v(a 1 ) + (n -1)γ ≥ γ > ∆ and vp = ∞ > ∆ hence the result follows from Theorem 6.13 (ii). 7.2. Kummer extensions. Assume that (L|K, v) is a Kummer extension of degree q and let p = char(Kv) > 0. We can write L = K(η) where g = x q -a is the minimal polynomial of η over K.

If (L|K, v) is immediate and unibranched, then it is of pure defect. In particular, p = q = d(L|K, v). We can assume that vη = 0. By the results of Section 5.2 we obtain α = -v(η -K) and β = vp -p • v(η -K).

Analogously to the Artin-Schreier case, we deduce that ann(Ω) = I γ ∩ O L where γ is the final segment formed by all the elements β ′ in Γ for which there exists i ∈ I such that β ′ -vp + pv(η -a i ) > v(η -a j ) for every j ∈ I.

In rank one, this can be written as ann(Ω) = {b ∈ O L | vb ≥ vp + (1 -p)ρ} where ρ = sup v(η -K).

We can see that if α ̸ = β, then it follows from Proposition 4.8 that Ω is not finitely generated (and hence not finitely presented).

If the extension is branched pure, then by Proposition 6. For the remainder of this section, assume that (L|K, v) is unibranched and defectless. As in the Artin-Schreier case, if Lv ̸ = Kv, then (L|K, v) is purely inertial. By Theorem 6.8 we obtain that Ω = (0) ⇐⇒ p ∤ B ⇐⇒ Lv|Kv is separable.

More explicitly, taking c ∈ K such that vη = -vc, we have α = -vη = vc and β = v(qη q-1 ) -qv(η) = vq + (q -2)vc. Now α = β if and only if vη = -vc = vq q-1 . If Lv|Kv is not separable, then we have ann(Ω) = I β-α = {b ∈ O L | vb ≥ vq -(q -1)v(η)}.

It follows from Proposition 4.9 that Ω is finitely presented.

As in the Artin-Schreier case, if vK ̸ = vL, then (L|K, v) is purely ramified. As before, if vL ∆ >0 has a minimum, then Ω ̸ = (0). Assume that vL ∆ >0 does not have a minimum. If p ̸ = q, then vq = 0 and by Theorem 6.13 (ii) we deduce that Ω = (0). On the other hand, if p = q, then Theorem 6.13 (ii) implies that Ω = (0) ⇐⇒ vp ∈ ∆.

  For a valued field (K, v) we will denote by O K the valuation ring, by vK the value group and by Kv the residue field of v. Also, for b ∈ K we denote by vb or v(b) the value of b in vK. If b ∈ O K , then we denote by bv the residue of b in Kv.

Theorem 2 . 2 .

 22 [START_REF] Novacoski | Key polynomials and pseudo-convergent sequences[END_REF] Theorem 1.1] Every valuation ν on K[x] admits a complete sequence of key polynomials.Remark 2.3.

Proposition 4 . 4 .

 44 The annihilator of Ω is I γ ∩ O L where

  The O L -relations on {a + I β } are of the form b (a + I β ) = 0 for some b ∈ O L . This happens if and only if v(ab) ∈ β. Take finitely many b 1 , . . . , b r ∈ O L such that v(ab ℓ ) ∈ β for every ℓ, 1 ≤ ℓ ≤ r. Suppose, without loss of generality, that v(b 1 ) ≤ v(b ℓ ) for every ℓ, 1 ≤ ℓ ≤ r. Then all the relations b 1 (a + I β ) = 0, . . . , b r (a + I β ) = 0, are generated by b 1 (a + I β ) = 0.

  defined by X → Q(η). By Corollary 5.5, this homomorphism is surjective. Denote by I its kernel and take a set of generators {f k } k∈J of I. We can identifyΩ = i∈I O L dX i J where (11) J = (df k | k ∈ J)(see[START_REF] Matsumura | Commutative Algebra[END_REF], Chapter 10, Section 26, (26.E), Example 1 on pp. 183-184 and (26.I), Theorem 58 on pp. 187-188 -the second fundamental exact sequence -where we take A = O K [X], B = O L and m = I).

Proof.

  It is obvious that the right hand side of (24) is contained in the left hand side. To prove the opposite inclusion, fix an element b ∈ O L and write b = f (η) ∈ K[η] with deg(f ) = m < n: (25) b = b 0 + b 1 η + . . . + b m η m where b s = ∂ s f (0) ∈ K. The elements sγ, 0 ≤ s ≤ m, belong to different vK-cosets of vL. We deduce that (26) 0 ≤ v(b) = min 0≤s≤m {v(b s ) + sγ}. By (26), we have (27) -v(b s ) < sγ for every s, 1 ≤ s ≤ m.

  K] = (vL : vK)[Lv : Kv]. Since [L : K] is prime, if Lv ̸ = Kv, then (L|K, v) is purely inertial. We can apply Theorem 6.8 to obtain that Ω = (0) ⇐⇒ p ∤ B ⇐⇒ Lv|Kv is separable. Alternatively, we can present the computation directly from Theorem 5.10. Take c ∈ K with vη = -v(c). In this particular case, by Proposition 3.5 we have O L = O K [cη]. We have α = vc and β = pvc. In particular, we deduce that ann(Ω) = I β-α = {a ∈ O L | va ≥ (p -1)vc}.

Remark 7 . 1 .

 71 If the residue field extension is purely inseparable (vη < 0) we consider the "translation map"O L -→ I α /I β , a → ac + I β .Clearly this map is surjective and its kernel is cp-1 O L . Hence we deduce [2, Theorem 4.4].

  7 we have Ω = (0) if and only if d := d(L|K, v) = 1. Let β d be the fixed value of ∂ d g. By [6], if vp < r(β d + d • v(η -c)) for some r ∈ Z and c ∈ K, then Ω = (0).

Remark 7 . 2 .

 72 An alternative characterization for Ω = (0) for a purely ramified Kummer extension is presented in [2, Theorem 4.7].

  is a complete set of key polynomials for (L|K, v). The above Lemma provides a useful characterization of pure immediate extensions: an immediate extension is pure immediate if and only if all of the key polynomials for (L|K, v) are linear. Below we will partially extend this characterization to pure extensions that are not necessarily immediate.

	Proof. It follows from [7, Corollary 3.4].	□
	Remark 5.4.

Corollary 5.5. Assume that (L|K, v) is pure immediate in η and take {c

  Then the module J ⊂ , the element ∂gi ∂Xi dX i is congruent to g ′ (η) di dX 0 modulo the relations (15).

	generated by the elements of the form (15) b ji dX j -dX i , j > i for i, j ∈ I and (16) ∂g i Consider the map ∂X Remark 5.11. In Ψ	i∈I	O L dX i is

i dX i for i ∈ I. i∈I LdX i

  1 ∈ B. □ 6.3. The branched pure case. The next result is the equivalent of Corollary 6.5 for branched pure extensions.Proof. In this case we are in the situation of Proposition 6.6, so it is enough to show that 1 ∈ B if and only if d(L|K, v) = 1. This follows from [6, Thereom 6.22 and Theorem 6.26]. □ 6.4. A general result. The main goal of this section is to prove the following.

	Theorem 6.8. Assume that (L|K, v) is a pure defect, a branched pure or a purely
	inertial extension. Then Ω = (0) if and only if p ∤ B.
	Proof. If the extension is of pure defect, then it follows from Proposition 6.6 that
	Ω = (0) if and only if 1 ∈ B. On the other hand, it follows from [6, Theorem 5.10
	and Corollary 5.11] that
	B ⊆ {1, p, p 2 , . . .}.
	Proposition 6.7. Assume that (L|K, v) is a branched pure extension. Then
	Ω = (0) if and only if d(L|K, v) = 1.

During the realization of this project the first author was supported by a grant from Fundação de Amparo à Pesquisa do Estado de São Paulo (process number 2017/17835-9).