Fields Institute Communications Volume 00, 0000

A Note on Tame Fields

Kamal Aghigh

Department of Mathematics Khaje Nassir-Al-Deen Toosi University Tehran Iran

                          kaghigh@yahoo.com

Sudesh K. Khanduja

Department of Mathematics Panjab University Chandigarh-160014 India

                           skhand@pu.ac.in

Abstract. Let v be a henselian valuation of any rank of a field K and v̄ be the extension of v to a fixed algebraic closure \overline{K} of K. For any α belonging to \overline{K} \ K, let Δ_{K}(α) (resp. ω_{K}(α)) denote the invariants defined to be the minimum (resp. maximum) of the set {v̄(α − α′) | α′ ≠ α runs over K-conjugates of α}. In 1998, it was proved that (K, v) is a tame field, i.e., every finite extension of (K, v) is tamely ramified and defectless, if and only if to each α ∈ \overline{K} \K, there corresponds a ∈ K such that v̄(α − a) ≥ Δ_{K}(α) (see [6]). In this paper, we give similar characterizations of tame fields using the invariants ω_{K}(α) and δ_{K}(α), where δ_{K}(α) = sup\{v̄(α − β) | β ∈ \overline{K}, [K(β) : K] < [K(α) : K]\}.

1 Introduction

Throughout v is a henselian valuation of any rank of a field K with residue field k_{v} and value group G_{v}, and v̄ is a unique prolongation of v to a fixed algebraic closure \overline{K} of K. A finite extension (K′, v′)/(K, v) (or briefly K′/K) will be called tame if

(a) it is defectless, i.e., [K′ : K] = ef, where e, f are respectively the index of ramification and the residual degree of v′/v;

(b) the residue field of v′ is a separable extension of k_{v};

(c) the ramification index of v′/v is not divisible by the characreristic of k_{v}.

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society

2000 Mathematics Subject Classification. Primary 12J10, 12J25, 13A18.

Key words and phrases. Valued fields, Valuations and their generalizations.

All correspondence may be addressed to the second author.

The research is partially supported by CSIR, New Delhi vide grant No. 25(95)/97/EMR-II.

A field (K, v) will be called tame (resp. defectless) if each finite extension of (K, v) is tame (resp. defectless).

In 1970, James Ax [1, §2, Proposition 2′] pointed out that if K is a perfect field of nonzero characteristic and v is of rank one, then to each α ∈ \overline{K} \ K there corresponds a ∈ K satisfying v̄(α − a) ≥ Δ_{K}(α), where

Δ_{K}(α) = min{v̄(α − α′) | α′ runs over K-conjugates of α}.  (1)

In 1991, a counterexample was given to show that this result is false (see [5]). In 1998, while giving a characterization of those fields for which Ax’s result holds, Khanduja [6, Theorem 1.1] proved the following result.

Theorem A Let (K, v) be a henselian valued field of any rank and (\overline{K}, v̄) be as above. Then to each α ∈ \overline{K} \ K, there corresponds a ∈ K satisfying v̄(α − a) ≥ Δ_{K}(α) if and only if (K, v) is a tame field.

As regards the Krasner’s constant ω_{K}(α) given by

ω_{K}(α) = max{v̄(α − α′) | α′ ≠ α runs over K-conjugates of α},

the following analogous result is known (see [6, Theorem 1.3]).

Theorem B Let v be a henselian valuation of a field K with residue field k_{v} and let v′ be the prolongation of v to a finite Galois extension K′ of K of degree n with residue field k_{v′}. Then the following statements are equivalent.

I) k_{v′} ≠ k_{v} and to each α ∈ K′ \ K, there corresponds a ∈ K such that

v′(α − a) = ω_{K}(α). II) n is a prime number and k_{v′}/k_{v} a Galois extension of degree n.

In this note, we wish to point out that it is possible to give a similar characterization of tame fields by means of another invariant defined below.

For an element α of \overline{K}, deg α will stand for the degree of the extension K(α)/K. If α ∈ \overline{K} \ K, we set

δ_{K}(α) = sup\{v̄(α − β) | β ∈ \overline{K}, deg β < deg α\}  (2)

where for the sake of supremum, the value group of v̄ may be viewed as a subset of its Dedekind order completion as defined in [2, Chap. III, Sec. 1, Ex. 15] and [8, Chap. 2]). It follows quickly from the well-known Krasner’s Lemma [3, 16.8] that δ_{K}(α) ≤ ω_{K}(α) when α is separable over K. In 1999, it was proved that δ_{K}(α) = ω_{K}(α) when K(α)/K is a tame extension (cf. [7, Theorem 1.3]. So the following question naturally arises:

If (K, v) is a henselian valued field such that δ_{K}(α) = ω_{K}(α) for each α ∈ \overline{K}\K, then is it true that (K, v) is a tame field?

In this paper, we give an example to show that the answer to the above question is “no” in general and it is “yes” if and only if we have the additional hypothesis that (K, v) is a defectless field. In fact we prove

Theorem 1.1 Let (K, v) be a Henselian valued field of any rank, and (\overline{K}, v̄) be as above. Then the following statements are equivalent.

i) (K, v) is a tame field.

ii) To every α ∈ \overline{K} \ K, there corresponds a ∈ K satisfying v̄(α − a) ≥ Δ_{K}(α). iii) To every α ∈ \overline{K} \ K, there corresponds β ∈ \overline{K} with deg β < deg α such that

v̄(α − β) ≥ Δ_{K}(α). iv) To every α ∈ \overline{K} \ K, there corresponds β ∈ \overline{K} with deg β < deg α such that

v̄(α − β) ≥ ω_{K}(α).

v) For each α ∈ \overline{K} \ K, K(α)/K is a defectless extension and δ_{K}(α) = ω_{K}(α). vi) (K, v) is a defectless valued field and δ_{K}(α) = ω_{K}(α) for all α ∈ \overline{K} \ K.

2 Some Preliminary Results

Lemma 2.1 Let K, v, \overline{K} and v̄ be as in the foregoing text. If α, β are elements of \overline{K} such that v̄(α − β) ≥ Δ_{K}(α), then Δ_{K}(β) ≥ Δ_{K}(α).

Proof Let σ be any automorphism of \overline{K}/K. Then

v̄(σβ − β) = v̄(σβ − σα + σα − α + α − β) ≥ min{v̄(σ(β − α)), v̄(σα − α), v̄(α − β)}.

Since v̄(σ(β − α)) = v̄(α − β) ≥ Δ_{K}(α), and v̄(σα − α) ≥ Δ_{K}(α), we see that

v̄(σβ − β) ≥ Δ_{K}(α).

As the above inequality holds for all automorphisms σ of \overline{K}/K, the lemma follows.

Notation. Let (K, v) and (\overline{K}, v̄) be as above. For an overfield L of K contained in \overline{K}, R(L) and G(L) will respectively stand for the residue field and the value group of the valuation of L obtained by restricting v̄. For α ∈ \overline{K} \K, M(α, K) will denote the subset of the value group of v̄ defined by

M(α, K) = \{v̄(α − β) | β ∈ \overline{K}, [K(β) : K] < [K(α) : K]\}.

Observe that δ_{K}(α) defined by (2) is the supremum of M(α, K).

The result of the following lemma is proved in [7, Theorem 1.1]. Its proof will be omitted.

Lemma 2.2 Let (K, v) and (\overline{K}, v̄) be as above. Let β, β_{1} ∈ \overline{K} be such that v̄(β − β_{1}) > v̄(β − γ) for every γ ∈ \overline{K} satisfying [K(γ) : K] < [K(β) : K], then

G(K(β)) ⊆ G(K(β_{1})), R(K(β)) ⊆ R(K(β_{1})).

Lemma 2.3 Let (K, v) be a henselian valued field and let α be algebraic over K of degree n > 1. If K(α)/K is defectless and every simple extension of K of degree less than n is also defectless, then δ_{K}(α) ∈ M(α, K).

Proof Suppose to the contrary that δ_{K}(α) ∉ M(α, K), i.e., M(α, K) does not have maximum element. We shall obtain the desired contradiction by showing that K(α)/K is not defectless. Since M(α, K) is totally ordered without last element, it contains a well-ordered cofinal subset (see [4, p. 68]). So we can choose a net {δ_{i}}_{i∈I} in M(α, K) satisfying

1. {δ_{i}}_{i∈I} is cofinal in M(α, K) and δ_{i} < δ_{j}, for i < j, i, j ∈ I.

2. δ_{i} = v̄(α − β_{i}), β_{i} ∈ \overline{K} is such that deg β_{i} < n and whenever γ ∈ \overline{K} has

degree less than deg β_{i}, then v̄(α − γ) < δ_{i}.

If necessary on replacing {δ_{i}}_{i∈I} by a subnet, we may assume that all β_{i} are of the same degree (say s) over K. Keeping in view that δ_{i} < δ_{j} for i < j, it follows that

v̄(β_{i} − β_{j}) = min{v̄(β_{i} − α), v̄(α − β_{j})} = δ_{i}.  (3)

For any γ ∈ \overline{K} with [K(γ) : K] < s, by virtue of condition 2 we have

v̄(α − γ) < δ_{i} = v̄(α − β_{i});

therefore using (3) and keeping in view that i < j, we have

v̄(β_{i} − γ) = v̄(α − γ) < δ_{i} = v̄(β_{i} − β_{j}).

Keeping in view the above inequality, it follows immediately from Lemma 2.2 that

G(K(β_{i})) ⊆ G(K(β_{j})), i < j,  (4)
R(K(β_{i})) ⊆ R(K(β_{j})), i < j.  (5)

As all the extensions K(β_{i})/K are defectless by hypothesis and of the same degree s < n, it is clear that equality holds in (4) and (5). Thus for any j ∈ I,

⋃_{i∈I} G(K(β_{i})) = G(K(β_{j})), ⋃_{i∈I} R(K(β_{i})) = R(K(β_{j})).  (6)

We are going to prove that

G(K(α)) = ⋃_{i∈I} G(K(β_{i})), R(K(α)) = ⋃_{i∈I} R(K(β_{i})).  (7)

As the extension K(α)/K is of degree n > s, (6) and (7) immediately imply that K(α)/K is not defectless, leading to the desired contradiction.

To prove (7), let F (x) ∈ K[x] be any polynomial of degree less than n. It is enough to prove that there exists k ∈ I such that v̄(F (α) − F (β_{k})) > v̄(F (β_{k})). Let γ be a root of F (x). Since v̄(α − γ) belongs to M(α, K) and the net {δ_{i}}_{i∈I} is cofinal in M(α, K), there exists k ∈ I such that v̄(α − γ) < δ_{k}. Choosing k sufficiently large, we may assume that

v̄(α − γ_{t}) < δ_{k}  (8)

for each root γ_{t} of F (x). Write F (x) = c ∏(x − γ_{t}). Then

\frac{F (α)}{F (β_{k})} = ∏_{t} \frac{(α − γ_{t})}{(β_{k} − γ_{t})} = ∏_{t}(1 + \frac{(α − β_{k})}{(β_{k} − γ_{t})}).  (9)

Since v̄(α − γ_{t}) < δ_{k} by (8) and v̄(α − β_{k}) = δ_{k} by choice of δ_{k}, we have on using the strong triangle law

v̄(β_{k} − γ_{t}) = min{v̄(β_{k} − α), v̄(α − γ_{t})} = v̄(α − γ_{t});

consequently (9) shows that v̄(\frac{F (α)}{F (β_{k})} − 1) > 0, which proves (7) and completes the

proof of the lemma.

Lemma 2.4 Let K be a perfect field of characteristic p > 0 with a real henselian valuation v and (\overline{K}, v̄) be as above. Then δ_{K}(α) = ω_{K}(α) for each α ∈ \overline{K} \ K.

Proof Let N be the smallest Galois extension of K containing α. Let H denote the subgroup of Gal(N/K) and L the subfield of N defined by

H = {σ ∈ Gal(N/K) | v̄(α − σ(α)) ≥ ω_{K}(α)}  (10)
L = the fixed field of H.  (11)

It is clear from the definition of ω_{K}(α) that Gal(N/K(α)) is properly contained in H. Therefore L is properly contained in K(α). If Δ_{L}(α) is defined on replacing K by L in (1), then it can be quickly visualized that

Δ_{L}(α) = ω_{K}(α).  (12)

Two cases are distinguished.

Case I v̄(α) ≥ 0.

Using [1, Proposition 2], we see that to any integer m ≥ 1 there corresponds β ∈ Ł such that v̄(α−β) ≥ (1−1/m)Δ_{L}(α). So by (12), v̄(α−β) ≥ (1−1/m)ω_{K}(α). Since β ∈ Ł and [L : K] < [K(α) : K], it follows that δ_{K}(α) ≥ (1 − 1/m)ω_{K}(α). As it is true for all positive integers m, we see that δ_{K}(α) ≥ ω_{K}(α). By virtue of Krasner’s Lemma, δ_{K}(α) ≤ ω_{K}(α) and hence equality holds.

Case II v̄(α) < 0.

It is easily verified that for α ≠ 0 in \overline{K} \ K, we have

δ_{K}(α^{−1}) = δ_{K}(α) − 2v̄(α)
ω_{K}(α^{−1}) = ω_{K}(α) − 2v̄(α).

Therefore the desired equality holds by virtue of Case I.

3 Proof of Theorem 1.1

The equivalence of i) and ii) is established by Theorem A. We first prove the equivalence of ii) and iii). For this it only needs to be shown that iii) implies ii), as the reverse implication is obvious. We apply induction on the degree of the extension K(α)/K. If deg α = 2, then nothing to prove. Let n ≥ 3 be an integer. Suppose that to each β ∈ \overline{K} \ K, which is algebraic over K of degree less than n, there corresponds a ∈ K such that v̄(β − a) ≥ Δ_{K}(β). Let α ∈ \overline{K} be such that [K(α) : K] = n. By virtue of iii), there exists β ∈ \overline{K} with deg β < deg α such that

v̄(α − β) ≥ Δ_{K}(α).  (13)

By induction there exists a ∈ K such that

v̄(β − a) ≥ Δ_{K}(β).  (14)

Now by virtue of (13) and Lemma 2.1, we have

Δ_{K}(β) ≥ Δ_{K}(α).  (15)

It follows from (13), (14) and (15) that

v̄(α − a) ≥ min{v̄(α − β), v̄(β − a)} ≥ Δ_{K}(α)

as desired.

Next we show that i)–iii) are equivalent to iv). For this we only need to check that iii) implies iv). Assume that iii) holds. Since i)–iii) are equivalent, it follows that (K, v) is a tame field. In particular K is a perfect field. Let α ∈ \overline{K} \ K, and let L be the subfield of K(α) defined by (10) and (11). Then as pointed out in (12) we have Δ_{L}(α) = ω_{K}(α). By virtue of i), we see that K and hence L is a tame field. So by ii) there exists β ∈ L satisfying

v̄(α − β) ≥ Δ_{L}(α) = ω_{K}(α).

Since [K(β) : K] ≤ [L : K] < [K(α) : K], assertion iv) follows. Thus we have shown that i)–iv) are equivalent.

To complete the proof of the theorem, clearly it is enough to prove the equivalence of v) with any of i)–iv).

Assume first that i) holds. In particular (K, v) is a perfect defectless field. So for any α ∈ \overline{K} \ K, δ_{K}(α) ≤ ω_{K}(α) by virtue of Krasner’s Lemma. Also iv) implies that δ_{K}(α) ≥ ω_{K}(α). Hence δ_{K}(α) = ω_{K}(α) and v) holds. Conversely, let us assume that v) holds. So by Lemma 2.3, δ_{K}(α) ∈ M(α, K), i.e., there exists β ∈ \overline{K}, deg β < deg α such that v̄(α − β) = δ_{K}(α). By virtue of v), we have δ_{K}(α) = ω_{K}(α). Hence v̄(α − β) = ω_{K}(α), which proves iv). This completes the proof of the theorem.

We give an example to show that there are perfect henselian valued fields (K, v) such that to every α ∈ \overline{K} \ K, δ_{K}(α) = ω_{K}(α) but (K, v) is not a defectless field.

Let k_{0} be the algebraic closure of the finite field F_{2} of 2 elements and K_{0} = k_{0}((T)) be the field of Laurent series in T with valuation v_{0} given by v_{0}(T) = 1. Let v̄_{0} be the extension of v_{0} to an algebraic closure \overline{K_{0}} of K_{0}. Let K be the inseparable closure of K_{0} in \overline{K_{0}} with valuation v which is the restriction of v̄_{0}. Then (K, v) being an algebraic extension of a complete rank 1 valued field is henselian. Let β be a root of the polynomial x^{2} − x − T^{−1} = 0. As shown in [5], [K(β) : K] = 2 and there does not exist any a ∈ K such that v̄(β − a) ≥ 0, but δ_{K}(β) = ω_{K}(β) = 0 by virtue of Lemma 2.4. So δ_{K}(β) ∉ M(β, K). It now follows from Lemma 2.3 that K(β)/K is not a defectless extension. Thus it has been shown that (K, v) is not a defectless field. But Lemma 2.4 implies that δ_{K}(α) = ω_{K}(α) for every α ∈ \overline{K} \ K.

References

[1] J. Ax, Zeros of polynomials over local fields—The Galois action, J. Algebra 15 (1970), 417–428.

[2] N. Bourbaki, Element of Mathematics, Theory of Sets, Herman, 1968.

[3] O. Endler, Valuation Theory, Springer-Verlag, 1972.

[4] P. R. Halmos, Naive Set Theory, Springer-Verlag, 1974.

[5] S. K. Khanduja, A note on a result of James Ax, J. Algebra 140 (1991), 360–361.

[6] S. K. Khanduja, Tame Fields and Tame Extensions, J. Algebra 201 (1998), 647–655.

[7] S. K. Khanduja and J. Saha, A Generalized Fundamental principle, Mathematika 46 (1999), 83–92.

[8] F. V. Kuhlmann, Valuation theory of fields, abelian groups and modules, to appear in the algebra, logic and application series, Gordon and Breach.