The Minimum Property of Krasner’s Constant^{∗}

Sudesh K. Khanduja

1. INTRODUCTION

Throughout v is a henselian valuation of field with value group G and v̄ is its unique prolongation to a fixed algebraic closure \overline{K} of K . For an element α of \overline{K}\K , which is separable over K , let ω_{K}(α) denote the well known Krasner’s constant given by

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

In 1946 Krasner proved that ω_{K}(α) satisfies the following principle known as Krasner’s Lemma (cf. [ 2, 16.8], [ 8, p.260], [10, p.141]). Krasner’s Lemma : Let (K, v) and (\overline{K}, v̄) be as above and α be an element of K^{sep}\K . If β ∈ \overline{K} is such that v̄(α − β) > ω_{K}(α), then K(α) ⊆ K(β).

The following question naturally arises: Is ω_{k}(α) the smallest among the elements λ of the divisible closure of G which have the property:

v̄(α − β) > λ, β ∈ \overline{K} implies K(α) ⊆ K(β) (†)

∗The research is partially supported by CSIR, New Delhi vide grant No.

25(95)/97/EMR-II

In this paper we prove that for each α ∈ K^{sep}\K , the answer to the above question is ”yes” when the rank of v is 1. In the general case, we prove that the answer is ”yes” for those α for which K(α)/K is a tame extension of valued fields. This leads to the following

\underline{Open Problem.} Suppose that (K, v) is a henselian valued field of rank > 1. Is the answer to the foregoing question ”yes” for all α ∈ K^{sep}\K? If not, can we characterize those α for which the answer is in the affirmative?

Recall that a finite extension (K′, v′)/(K, v) (or briefly K′/K when the underlying valuations are clear) is said to be tame if (a) it is defectless, i.e. [K′ : K] = ef, where e, f are the index of ramification and the residual degree of v′/v ; (b) the residue field of v′ is a separable extension of the residue field of v; (c) the index of ramification of v′/v is not divisible by the characteristic of the residue field of v. We shall prove THEOREM 1.1. Let v be a henselian real valuation of a field K and v̄ be its unique prolongation to an algebraic closure \overline{K} of K . Let α ∈ \overline{K}\K be separable over K . Suppose that λ is a real number with the property that whenever v̄(α − β) > λ for β ∈ \overline{K}, one has K(α) ⊆ K(β). Then λ ≥ ω_{K}(α). THEOREM 1.2. Let v be a henselian valuation of a field K with value group G. Let K(α) ⊆ \overline{K} be a tame extension of K of degree > 1. Suppose that λ is an element of a totally ordered abelian group containing G as an ordered subgroup, such that for β ∈ \overline{K} the inequality v̄(α − β) > λ implies K(α) ⊆ K(β). Then λ ≥ ω_{K}(α).

2. SOME PRELIMINARY RESULTS AND PROOF OF THEOREM 1.2

For α ∈ \overline{K}, we set

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

Note that if Δ_{K}(α) = ∞ is purely inseparable over K.

The following two lemmas are known, we omit their proofs (see [4,Theorem 1.2], [3, Theorem 2.3]). LEMMA 2.1. Let v be henselian valuation of a field K. Suppose that α is separable over K of degree a prime number, then Δ_{K}(α) = ω_{K}(α). LEMMA 2.2. Let (K′, v′) be a finite Galois tame extension of a henselian valued field (K, v). Then to each α ∈ K′\K, there corresponds a ∈ K such that v′(α − a) ≥ Δ_{K}(α).

Theorem 1.2 follows immediately from the following already known lemma [5, Theorem 1.3]. For reader’s convenience the proof of this lemma is included here. LEMMA 2.3. Let v be a henselian valuation of a field K and K(α) be a tame extension of K (with respect to the unique prolongation of v to K(α)) of degree n > 1. Then there exists δ ∈ K(α) such that [K(α) : K] < n and v̄(α − δ) = ω_{K}(α). PROOF. As K(α)/K is a tame extension, it must be separable. Let K′ be the smallest Galois extension of K containing α and v′ be the unique prolongation of v to K′. Since K(α)/K is a tame extension of K, the same is true of K′ (see [2, 21.2]). Let H denote the subgroup of Gal(K′/K) and L the subfield of K′ defined by

H = {σ ∈ Gal(K′/K)| v′(α − σα) ≥ ω_{K}(α)}.
L = the fixed field of H.  (2)

It is clear from the definition of ω_{K}(α) that Gal(K′/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 visualized that

Δ_{L}(α) = ω_{K}(α).  (3)

By virtue of Lemma 2.2, there exists δ ∈ L satisfying

v′(α − δ) ≥ Δ_{L}(α) = ω_{K}(α).

Since L is properly contained in K(α) it follows that [K(δ) : K] < [K(α) : K], and consequently the inequality is in fact an equality by virtue of Krasner’s Lemma. REMARK 2.4. It may be pointed out that the result of Theorem 1.2 holds when the hypothesis ”K(α)/K is a tame extension” is replaced by a weaker one, namely K(α)/M is a tame extension for some overfield M properly contained in K(α)) of L, where L is as given by (2). In this situation by Lemma 2.3 applied to K(α)/M, there exists δ ∈ K(α) such that

[M(δ) : M] < [M(α) : M], v̄(α − δ) = ω_{M}(α).  (4)

Using (4) and (3), we quickly conclude that

v̄(α − δ) = ω_{M}(α) ≥ Δ_{M}(α) ≥ Δ_{L}(α) = ω_{K}(α)  (5)

If ω_{K}(α) were strictly greater than λ , then (5) would imply that v̄(α−δ) > λ and consequently by property (†), K(α) would be contained in K(δ), which is impossible, because in view of (4), we have

[K(δ) : K] ≤ [M(δ) : K] = [M(δ) : M][M : K] < [M(α) : M] [M : K]
=  [K(α) : K]

This proves that λ ≥ ω_{K}(α) as desired.

We shall say that a finite extension (K′, v′)/(K, v) (or K′/K when the underlying valuations are clear) is unramified if the residue field of v′ is a separable extension of that of v and the value group of v′ equals the value group of v. LEMMA 2.5. Let v be a henselian valuation of a field K whose residue field has characteristic p > 0. Let n ≥ 2 be a natural number not divisible by p and ζ be a primitive nth root of unity. Then K(ζ) is an unramified defectless extension of K. PROOF. Let k′ denote the residue field of the prolongation v′ of v to K(ζ) and f(X) the minimal polynomial of ζ over K. Let f^{∗}(X) be the polynomial obtained by replacing the coefficients of f(X) by their corresponding v-residues in the residue field k of v. By a well known result [2, 16.6], f^{∗}(X) is a power of an irreducible polynomial of k[X]. Since char k does not divide n, it follow that X^{n} − 1 and hence its divisor f^{∗}(X) does not have repeated roots. It is now clear that f^{∗}(X) is irreducible over k and has no repeated roots. Consequently k′/k is a separable extension of degree [K(ζ) : K], which by virtue of the fundamental equality [2, 20.21] implies that (K(ζ), v′)/(K, v) is unramified and defectless. COROLLARY. Let v, K, n and ζ be as in the above lemma. Let c ∈ K and ξ be a root of the polynomial X^{n} − c = 0. Then K(ζ, ξ) is a tame extension of K. PROOF. The result follows from the above lemma and the fact that K(ζ, ξ)/ K(ζ), being of degree a divisor of n, is a tame extension. LEMMA 2.6. Let v be a henselian valuation of a field K and K(α) be a separable extension of K of degree a prime number p. Let Q(X) be a monic irreducible polynomial over K of degree p. If v̄(Q(α)) ≤ p ω_{K}(α) then for each root β of Q(X), v̄(α − β) = v̄(Q(α))/p. PROOF. If Q(X) = ∏_{i=1}^{n}(X − β_{i}), then

v̄(Q(α)) = ∑_{i} v̄(α − β_{i}) = pg (say)

so there exists β_{j} satisfying v̄(α − β_{j}) ≥ g. Then lemma is proved once we show that v̄(α − β_{i}) ≥ g for each i. For a fixed i, let σ be an automorphism of \overline{K}/K such that σ(β_{i}) = β_{j}. Since v̄(σα − α) ≥ ω_{K}(α) by virtue of Lemma 2.1 and g ≤ ω_{K}(α) by hypothesis, we have

v̄(α − β_{i}) = v̄(σα − β_{j}) ≥ min{v̄(σα − α), v̄(α − β_{j})} ≥ g.

NOTATION 2.7. For any ξ in the valuation ring of a valuation v′ extending the given valuation v of K, we shall denote by ξ^{∗} its v′-residue, i.e., the image of ξ under the canonical homomorphism from the valuation ring of v′ onto its residue field. For the sake of convenience, we shall henceforth write v̄(x) as v(x) for x ∈ \overline{K} . As before, G will denote the value group of v and k its residue field. When v is a real valuation, for α ∈ K^{sep}\K , we shall denote by δ_{K}(α) the real number defined by

δ_{K}(α) = sup\{v(α − β) | β ∈ \overline{K}, [K(β) : K] < [K(α) : K]\}  (6)

Observe that δ_{K}(α) ≤ ω_{K}(α) by virtue of Krasner’s Lemma. LEMMA 2.8. Let v be a henselian real valuation of field K and α be separable over K of degree n > 1 having v(α) = 0. Let g(X) = ∑ c_{i}X^{i} be a polynomial of degree m < n such that min_{i} v(c_{i}) = 0. Then v(g(α)) ≤ m δ_{K}(α). PROOF. If m = 0, the desired inequality is trivially true, because δ_{K}(α) ≥ v(α) ≥ 0 . For m ≥ 1, factorize g(X) as c_{m} ∏(X − β_{i}) . As [K(β) : K] ≤ m < n, we have v(α − β_{i}) ≤ δ_{K}(α). Hence the desired result is immediate if v(c_{m}) = 0. So assume v(c_{m}) > 0 and v(c_{j}) = 0 for some j ≤ m − 1. Since v(c_{j}/c_{m}) < 0, it is clear that at least one β_{i} has v-valuation negative, say v(βi) < 0 for 1 ≤ i ≤ l and v(β_{i}) ≥ 0 for l + 1 ≤ i ≤ m. Now using the strong triangle law, it can be easily seen that

v(∑ β_{i_{1}}...β_{i_{l}}) = v(β_{1}...β_{l})

where the summation on the left hand side is taken over all the tuples (i_{1}, ...i_{l}) such that 1 ≤ i_{1} < i_{2} < ... < i_{l} ≤ m. Therefore v(c_{m−l}/c_{m}) = v(β_{1}...β_{l}). Claim is that v(c_{m−l}) = 0 i.e.

v(β_{1}...β_{l}) = −v(c_{m}).  (7)

To prove the claim observe that for 1 ≤ i_{1} < i_{2} < ... < i_{r} ≤ m, v(β_{i_{1}}...β_{i_{r}}) ≥ v(β_{1}...β_{l}) and hence v(c_{m−r}/c_{m}) ≥ v(β_{1}...β_{l}). The inequality v(c_{m−r}) ≥ v(c_{m−l}) for 1 ≤ r ≤ m, together with hypothesis min{v(c_{i}} = 0 immediately implies that v(c_{m−l}) = 0 as claimed.

Keeping in view (7) and the fact that v(α) = 0 and v(β_{i}) < 0 for 1 ≤ i ≤ l, it can be easily seen that

v(g(α)) = v(c_{m}) + ∑_{i=1}^{m} v(α − β_{i}) ≤ v(c_{m}) + ∑_{i=1}^{l} v(α − β_{i}) + (m − 1)δ_{K}(α)

= (m − 1)δ_{K}(α) ≤ mδ_{K}(α))

DEFINITION. Let (K′, v′)/(K, v) be an extension of degree n. A basis x_{1}, ...x_{n} of K′/K is called valuation basis (w.r.t. v′/v) if for each x = ∑ a_{i}x_{i} in K′, a_{i} ∈ K, v′(x) = min_{i} v′(a_{i}x_{i}). It is clear if {x_{1}, ...x_{n}} is a valuation basis of (K′, v′)/(K, v) then so is {x_{1}/y, ...x_{n}/y} , where y is any non-zero element of K′.

Recall that an extension (K′, v′)/(K, v) is said to be immediate if v′, v have the same value group and the same residue field.

The first assertion of the following lemma is already known (cf.[11, Prop.2.5], [12, Lemma 1.1]); for the sake of completeness its proof is given here. LEMMA 2.9. Let v be a henselian valuation of a field K whose residue field k has characteristic p > 0. Let (K′, v′)/(K, v) be a separable extension of degree p.

(i) If (K′, v′)/(K, v) is defectless and if γ ∈ K is such that either v′(γ) does not belong to the value group G of v or v′(γ) = 0 and its v′ -residue γ^{∗} ∉ k, then

min{v(T r_{K′/K}(δ)) − v′(δ) | 0 ≠ δ ∈ K′} = (p − 1)[ω_{K}(γ) − v′(γ)].

(ii) If v is a real valuation and if γ ∈ K′\K with v′(γ) = 0, then for any δ ≠ 0 in K′

v(T r_{K′/K}(δ)) − v′(δ) ≥ (p − 1)[ω_{K}(γ) − (δ_{K}(γ)]  (8)

PROOF of (i). Let g(X) denote the minimal polynomial of γ over K and Tr stand for the trace map relative to the extension K′/K. As in [9, Lemma 3.9] or [6, Lemma 2.6] it can be easily verified that 1, γ, ...γ^{p−1} is a valuation basis of (K′, v′)/(K, v) and hence so is 1/g′(γ), γ/g′(γ), ...γ^{p−1}/g′(γ). For any δ = ∑ a_{i} \frac{γ^{i}}{g′(γ)} in K′, we have

v(T r(δ)) ≥ min_{i}{v(a_{i}) + v(T r(γ^{i}/g′(γ))}

and

v(δ) = min_{i}{v(a_{i}) + v(γ^{i}/g′(γ))}.

Therefore,

v(T r(δ)) − v(δ) ≥ min_{i}{v(T r(γ^{i}/g′(γ)) − v(γ^{i}/g′(γ))},

So the desired assertion is proved once it is shown that

min_{0≤i≤p−1}{v(T r(γ^{i}/g′(γ))} = (p − 1)[ω_{K}(γ) − v(γ)].  (9)

By an elementary result of field theory (See[7, Chap.VIII,Prop.5.5]), we have

T r(γ^{p−1}/g′(γ)) = 1, T r(γ^{i}/g′(γ)) = 0, for 0 ≤ i ≤ p − 2.  (10)

So the minimum on the left hand side of (9) is −v(γ^{p−1}/g′(γ)) which equals v(g′(γ))−(p−1)v(γ). Thus (9) is proved as soon as we verify that v(g′(γ)) = (p − 1)ω_{K}(γ). If γ = γ_{1}, ...γ_{p} are the K-conjugates of γ, then g′(γ) = ∏_{i=2}^{p}(γ − γ_{i}). By virtue of Lemma 2.1, v(γ − γ_{i}) = ω_{K}(γ) for each i ≥ 2. It follows that v(g′(γ)) = (p − 1)ω_{K}(γ) as desired. PROOF of (ii). Fix an element γ of K′|K with v′(γ) = 0. Let g(X) be the minimal polynomial of γ over K. For any non-zero δ in K′, write

δg′(γ) = ∑ a_{i}γ^{i}, a_{i} ∈ K.  (11)

Replacing δ by cδ for some c belonging to K, we may assume that

min_{i}(v(a_{i})) = 0.  (12)

By Lemma 2.8

v′(δg′(γ)) ≤ (p − 1)δ_{K}(γ).

Since v′(g′(γ)) = (p − 1)ω_{K}(γ), the above inequality can be rewritten as

v(δ) ≤ (p − 1)[δ_{K}(γ) − ω_{K}(γ)].  (13)

It follows quickly from (10) and (11) that

T r(δ) = v(a_{p−1}).

Keeping in view (12), we see that v(T r(δ)) ≥ 0; this inequality together with (13) immediately implies (8).

The following lemma is due to Ax [1, Lemma 6, Cor.1]. We shall omit its proof. LEMMA 2.10. Let v be a henselian valuation of a field K of characteristic p > 0 and \overline{K} be as before. If α ∈ \overline{K} is separable over K of degree p and v̄(α) ≥ 0, then to every natural number l > 1, there corresponds β ∈ \overline{K}, β purely inseparable over K, such that

v̄(α − β) ≥ (1 − 1/l)Δ_{K}(α),

where Δ_{K}(α) is as given by (1).

3. PROOF OF THEOREM 1.1

Observe that if λ is as in the theorem, then λ ≥ v(α), for otherwise v(α − 0) > λ would imply K(α) ⊆ K by virtue of property (†) of λ stated in the first section. So for β ∈ \overline{K} the inequality v(α − β) > λ holds if and only if v(α^{−1} −β^{−1}) > λ−2v(α). Keeping in view that ω_{K}(α^{−1}) = ω_{K}(α)−2v(α), we conclude that it is enough to prove the theorem when v(α) ≥ 0.

Let N be the smallest normal extension of K containing α and L ⊂ K(α) be the fixed field of the subgroup H of Gal(N /K ) given by

H = {σ ∈ Gal(N /K ) | v(α − σα) ≥ ω_{K} (α)}.

Observe that

Δ_{L}(α) = ω_{K}(α).  (14)

In view of Remark 2.4, it remains to prove the theorem when the characteristic p of the residue field of v is non-zero and K(α)/M is not a tame extension for any over field M of L properly contained in K(α).

Suppose to the contrary λ < ω_{K}(α). As said in the opening lines of this section 0 ≤ v(α) ≤ λ; consequently ω_{K}(α) > 0. With the above supposition K(α)/L is a normal extension, because for any L - conjugate α′ ≠ α of L, by virtue of (14) we have v(α − α′) = ω_{K}(α) which by property (†) of λ implies that K(α) ⊆ K(α′) and hence L(α) = L(α′). Let F_{0} be the fixed field of a p-Sylow subgroup of Gal(K (α)/L). So there exists a tower of fields F_{0} ⊂ F_{1} ⊂ ... ⊂ F_{s} = K(α) such that F_{i+1} is a normal extension of F_{i} of degree p for each i. We denote F_{s−1} by F and distinguish two cases.

Case I: char K = p > 0

We shall show that λ ≥ (1−1/m)ω_{K}(α) for every natural number m > 1. Suppose to the contrary, there exists l > 1 such that λ < (1 − 1/l)ω_{K}(α). Applying Lemma 2.10 to the extension K(α)/F, we see that there exists β ∈ \overline{K}, β purely inseperable over F such that

v(α − β) ≥ (1 − 1/l)Δ_{F} (α).  (15)

Keeping in view (14) and the fact that L ⊆ F , it can be easily seen that

ω_{K}(α) = Δ_{L}(α) = Δ_{F} (α).  (16)

It follows from (15), (16) and the above supposition that v(α − β) > λ; consequently K(α) ⊆ K(β) and hence F (α) ⊆ F (β) , which is impossible because F (α)/F is a separable extension of degree p, whereas F (β)/F is a purely inseparable extension.

Case II: char K = 0

Recall that we are dealing with the situation when K(α)/F is not a tame extension. Let T denote the compositum of all finite tame extensions of F contained in the algebraic closure \overline{K} of K. Then T (α)/T is not a tame extension and its degree being a divisor of p=[K(α) : F] is again p. It now follows by virtue of Lemma 2.1 that Δ_{F} (α) = ω_{F} (α) = ω_{T} (α). Arguing as for (16), we see that

ω_{K}(α) = Δ_{F} (α) = ω_{F} (α) = ω_{T} (α).  (17)

Observe that if β ∈ \overline{K} and [T (β) : T] < p, then T (β)/T is a tame extension and hence β ∈ T. It follows that if δ_{T} (α) is defined by the formula in (6), then

δ_{T} (α) = sup{v(α − t) | t ∈ T }.  (18)

Since we have assumed that λ < ω_{K}(α), there exists a natural number n > 1 not divisible by p such that

λ < (1 − 1/n)ω_{K}(α).  (19)

It will be shown that (19) leads us to a contradiction.

We shall sometimes denote T (α) by T. Let G_{T} ⊆ G′_{T} and k_{T} ⊆ k′_{T} denote respectively the value groups and the residue fields of the valuations obtained by restricting v̄ to T and T (α) respectively. \underline{In what follows T r will stand for} the trace relative to the extension T (α)/T. We divide this case into two subcases.

Subcase (i): T (α)/T is a defectless extension.

If G_{T} ≠ G′_{T}, choose an element x ∈ T (α) with v(x) ∉ G_{T} and if k_{T} ≠ k′_{T}, fix x ∈ T (α) such that v(x) = 0 and that the v̄-residue x^{∗} of x does not lie in k_{T}. In both the instances if t_{i} ∈ T, then

v(∑_{i=o}^{p−1} t_{i}x^{i}) = min(v(t_{i}) + iv(x)).

In particular if α is written as

α = a + ∑_{i=o}^{p−1} a_{i}x^{i}, a, a_{i} ∈ T,

then in view of the assumption v(α) ≥ 0, we have v(a) ≥ 0 and

v(α − a) = min_{i}(v(a_{i}) + iv(x)).

It is clear that when G_{T} ≠ G′_{T}, then v(α − a) ∉ G_{T}. If k_{T} ≠ k′_{T}, we choose b ∈ T such that

v(α − a) = min_{i}(v(a_{i}) + iv(x)) = min_{i}(v(a_{i})) = v(b),

so that

((α − a)/b)^{∗} = ∑_{i=1}^{p−1}(a_{i}/b)^{∗}(x^{∗})^{i} ∉ k_{T} .

Define γ ∈ T (α) = T′ by

γ = \{ \begin{matrix}α − a if G_{T} ≠ G′_{T} \\ (α − a)/b if k_{T} ≠ k′_{T} \end{matrix}

By Lemma 2.9(i) applied to T′/T, we have

min{v(T r_{T′/T} (δ)) − v(δ) | 0 ≠ δ ∈ T′} = (p − 1)[ω_{T} (γ)].  (20)

Assuming (19), we shall construct an element β ∈ T (α) such that

v(T r(α − a − β)) − v(α − a − β) < (p − 1)[ω_{T} (γ) − v(γ)]

which will contradict (20) and thus complete the proof of the theorem in this subcase.

Since K(α)/F and hence T (α)/T is a Galois extension of degree p, it follows by virtue of (17) that

pω_{K}(α) = v(c)

for some c ∈ F. Fix a root y of the polynomial X^{n} − c^{n−1}, where n is as in (19). By corollary to Lemma 2.5, y ∈ T. Let d be an element of T such that pv(α−a) = v(d) and let ξ be a root of the polynomial X^{1+pn} −d^{n(p−1)}. Again by corollary to Lemma 2.5, T (ξ)/T is a tame extension and hence ξ ∈ T. Recall that v(α − a) = v(b), b ∈ T in case k′_{T} ≠ k_{T}. Define z ∈ T by

z = \{ \begin{matrix}y/b^{p−1}, if k′_{T} ≠ k_{T} \\ y/ξ, if G′_{T} ≠ G_{T} \end{matrix}

so that

v(z) = {(1 − 1/n)pω_{K}(α) − (p − 1)v(α − a), k′_{T} ≠ k_{T}
{(1 − 1/n)pω_{K}(α) − [np(p − 1)/(1 + pn)]v(α − a), G′_{T} ≠ G_{T} .  (21)

Let P (X) = X^{p} +c_{1}X^{p−1} +...c_{p} denote the minimal polynomial of α −a over T. Define a polynomial Q(X) over T by

Q(X) = X^{p} + (c_{1} + z)X^{p−1} + c_{2}X^{p−2} + ... + c_{p}.

Then

v(Q(α − a)) = v(Q(α − a) − P (α − a)) = v(z(α − a)^{p−1}) = pg (say). (22)

Comparing (21) and (22), a simple calculation shows that

g = {(1 − 1/n)ω_{K}(α), if k′_{T} ≠ k_{T}
{(1 − 1/n)ω_{K}(α) + [(p − 1)/p(1 + pn)]v(α − a), otherwise.  (23)

Keeping in view that 0 ≤ v(α − a) ≤ ω_{T} (α) = ω_{K}(α) by virtue of (17) and ω_{K}(α) > 0, it follows immediately from (23) that (1 − 1/n)ω_{K}(α) ≤ g < ω_{K}(α); consequently by virtue of (19), we have

λ < g < ω_{K}(α).  (24)

If β_{1}, β_{2}, ..., β_{p} are the roots of Q(X), then (22) shows that ∑_{i} v(α−a−β_{i}) = pg. So there exists a root β (say) of Q(X) such that v(α − a − β) ≥ g. As shown in (24), g > λ, consequently by property (†) of λ, K(α) ⊆ K(α + β) and hence T (α) ⊆ T (α + β) = T (β). Since [T (α) : T] = p and β is a root of the polynomial Q(X) ∈ T [X], it follows that T (α) = T (β) and Q(X) is irreducible over T.Therefore by Lemma 2.6 we have

v(α − a − β) = g.  (25)

Recall that Tr stands for the trace from T (α) to T. Since T r(α − a − β) = T r(α − a) − T r(β) = z, keeping in view (22), we see that

v(T r(α − a − β)) = v(z) = pg − (p − 1)v(α − a).  (26)

If we take δ = α − a − β, then by (25) and (26)

v(T r(δ)) − v(δ) = (p − 1)[g − v(α − a)].

It now follows from (20) that

g − v(α − a) ≥ ω_{T} (γ) − v(γ).  (27)

Recall that when G′_{T} ≠ G_{T}, then γ = α − a and ω_{T} (α − a) = ω_{T} (α) = ω_{K}(α) by (17), so in this situation (27) asserts that g ≥ ω_{K}(α), which contradicts (24). This contradiction proves the theorem in the present situation. When k′_{T} ≠ k_{T}, then γ = (α − a)/b, v(γ) = 0 and

ω_{T} (γ) = ω_{T} (α − a) − v(b) = ω_{K}(α) − v(α − a).

Again on substitution in (27), we are led to g ≥ ω_{K}(α). This contradiction completes the proof of the theorem in subcase (i).

Subcase (ii): T (α)/T is an immediate extension.

Recall that n > 1 is an integer not divisible by p such that λ < (1 − 1/n)ω_{K}(α) and ω_{K}(α) > 0. Fix an integer m > n. As δ_{T} (α) is given by (18), so there exists a ∈ T such that

v(α − a) > δ_{T} (α) − ω_{K}(α)/m).  (28)

Since T (α)/T is an immediate extension, we can choose b ∈ T such that v(b) = v(α − a). As in subcase (i), choose c ∈ F such that pω_{K}(α) = v(c), and a root y of the polynomial X^{n} − c^{n−1}. Define z ∈ T by

z = y/b^{p−1}

Let P (X) = X^{p} + c_{1}X^{p−1} + ... + c_{p} be the minimal polynomial of α − a over T. Define Q(X) ∈ T [X] by

Q(X) = X^{p} + (c_{1} + z)X^{p−1} + c_{2}X^{p−2} + ... + c_{p}

Proceeding exactly as in the proof of subcase (i), we see that ∃β ∈ T (α) such that

v(α − a − β) = (1 − 1/n)ω_{K}(α)  (29)

and

v(tr(α − a − β)) = v(z) = p(1 − 1/n)ω_{K}(α) − (p − 1)v(α − a). (30)

Applying Lemma 2.9(ii) to the extension T (α)/T (choosing γ as (α − a)/b and δ as α − a − β), we conclude by virtue of (29), (30) that

(p − 1)[(1 − 1/n)ω_{K}(α) − v(α − a)] ≥ (p − 1)[ω_{T} ((α − a)/b) − δ_{T} ((α − a)/b)].(31)

It can be easily verified that

ω_{T} ((α − a)/b) = ω_{T} (α − a) − v(b) = ω_{T} (α) − v(b) = ω_{T} (α) − v(α − a),
δ_{T} ((α − a)/b) = δ_{T} (α) − v(b) = δ_{T} (α) − v(α − a).

Therefore in view of (17) and (28), we see that

ω_{T} ((α − a)/b) = ω_{K}(α) − v(α − a), δ_{T} ((α − a)/b) < \frac{ω_{K}(α)}{m}.  (32)

It follows immediately from (31) and (32) that

(1 − 1/n)ω_{K}(α) > (1 − 1/m)ω_{K}(α)

Since ω_{K}(α) > 0, the above inequality implies that m < n, which is not so. This contradiction completes the proof of the theorem.

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Title of the article^{1}: The Minimum Property of Krasner’s Constant

Name and address Sudesh K. Khanduja
of the author:  Professor

Department of Mathematics Panjab University Chandigarh-160014. INDIA.

e-mail address:  skhand@usa.net

Abstract: Let v be henselian valuation of a field K with value group G and v̄ be its unique prolongation to the algebraic closure \overline{K} of K. For an element α of \overline{K}\K, which is separable over K, let ω_{K}(α) denote the well known Krasner’s constant given by max{v̄(α − α′) | α′ ≠ α runs over K − conjugates of α}. In 1946 Krasner pointed out that ω_{K}(α) satisfies the following property : whenever β ∈ \overline{K} is such that v̄(α − β) > ω_{K}(α), then K(α) ⊆ K(β). In this paper we investigate whether ω_{K}(α) is the smallest element of the divisible closure of G which has the above property.

1996 Subject Classification No. : 12J10, 12J25, 13A18

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

1The research is partially supported by CSIR, New Delhi vide grant No.

25(95)/97/EMR-II