Fields Institute Communications Volume 00, 0000

A generalization of Hensel’s Lemma

Hervé Perdry

Équipe de Mathématiques de Besançon, UMR CNRS 6623, UFR Sciences et techniques, Université de Franche-Comté, 25030 Besançon cedex

                              perdry@math.univ-fcomte.fr

Abstract. Sudesh Kaur Khanduja and Jayanti Saha gave in [5] an elegant generalization of Hensel’s Lemma for residually transcendental extensions of a valuation v of K to K(x), allowing to deduce a factorization of a polynomial from an “approximate” factorization. However, their proof was restricted to the case of a complete valued field of rank 1. We give here a proof of this lemma for any Henselian field, and for value transcendental and residue transcendental extensions of v. We give a geometric interpretation of this lemma, and another version allowing to derive a root of a polynomial from an “approximate” root, together with a multidimensional version of this last result.

1 Definitions, notations & statements

Let K be a Henselian field, and K^{ac} be its algebraic closure. The value group of K is denoted by G, and the one of K^{ac} by \overline{G}, the divisible closure of G. The valuation on K^{ac} is always denoted by v; the valuation ring of K is denoted by V, the one of K^{ac} by V^{ac}. The residue fields of K and K^{ac} are denoted by k and k^{ac}. The galois group of K is denoted by Gal(K). For all α ∈ K^{ac} and all σ ∈ Gal(K), we have v(σα) = v(α). For general definitions and statements, cf [4].

Let H be an ordered abelian group which contains \overline{G}. Take (a, δ) ∈ K^{ac} × H. We use it to define an extension of v to a simple transcendental extension K^{ac}(x) of K^{ac}:

w_{a,δ} (∑_{i} c_{i}(x − a)^{i}) = min_{i} {vc_{i} + iδ}.

If δ ∈ \overline{G}, the valuation w_{a,δ} defines a residually transcendental extension of v (cf. [7] or [3], §10). The residue fields of K(x) and K^{ac}(x) for such a valuation are k′(t) and k^{ac}(t), where k′ is a finite extension of k and t is transcendental over k; this property is the definition of a residually transcendental extension. For every residually transcendental extension w on K(x), there exists (a, δ) such that w is

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

2000 Mathematics Subject Classification. Primary 12J25; secondary 12D05, 13J15.

.

defined by the above formula (cf. [2]). This pair (a, δ) is not unique; we shall later see exactly when w_{a,δ} = w_{a′,δ′}.

The pair (0, 0) defines the so-called Gaussian extension v^{g} of v; we have v^{g}(∑_{i} c_{i}x^{i}) = min_{i} vc_{i}.

A value transcendental valuation w is defined by a pair (a, δ) where a ∈ K^{ac} and δ is defined by a cut in \overline{G}: we have two subsets of \overline{G}, \overline{G}^{<δ} and \overline{G}^{>δ} such that \overline{G} = \overline{G}^{<δ} ∪\overline{G}^{>δ} and for all a ∈ \overline{G}^{<δ} and b ∈ \overline{G}^{>δ}, we have a < b. We consider more over that in that case we have a < δ < b, and this is sufficient to give an order on \overline{G} + ℤ · δ. The residue field of K(x) for this valuation is k′, a finite extension of k, and the value group is contained in \overline{G} + ℤ · δ.

One of the classical versions of Hensel’s Lemma is as follows (cf. [8]). We denote by lc(f) the leading coefficient of the polynomial f.

Theorem 1.1 (Classical Hensel’s Lemma) Let f, g_{0}, h_{0} be polynomials in K[X] such that the conditions H_{1}, H_{2}, H_{3} are satisfied:

• H_{1} v^{g}(f − g_{0} · h_{0}) > v^{g}(f)

• H_{2} v(lc(g_{0})) = v^{g}(g_{0})
• H_{3} ∃a, b ∈ K[x] such that v^{g}(a · g_{0}) ≥ 0, v^{g}(b · h_{0}) ≥ 0
and v^{g}(a · g_{0} + b · h_{0} − 1) > 0.

Then there exist g, h ∈ K[x] such that the following conditions H_{a}, H_{b}, H_{c} are satisfied; moreover, these polynomials are unique.

• H_{a} f = g · h

• H_{b} deg g = deg g_{0} and lc(g) = lc(g_{0})
• H_{c} v^{g}(g − g_{0}) > v^{g}(g_{0}).

As soon as the conditions H_{a}, H_{b}, H_{c} are satisfied, we have v^{g}(h − h_{0}) > v^{g}(h_{0}).

Remark 1.2 We denote by a ↦ \overline{a} the canonical map from V to k. If v^{g}(f) = v^{g}(g_{0}) = v^{g}(h_{0}) = 0, the hypotheses can be reformulated as follows:

• H_{1} \overline{f} = \overline{g_{0}} · \overline{h_{0}}
• H_{2} deg \overline{g_{0}} = deg g_{0}
• H_{3} \overline{g_{0}} and \overline{h_{0}} are relatively prime.

And the conclusion becomes:

• H_{a} f = g · h

• H_{b} deg g = deg g_{0} and lc(g) = lc(g_{0})
• H_{c} \overline{g} = \overline{g_{0}}.

If the conditions of the conclusion are verified, we have clearly \overline{h} = \overline{h_{0}}.

The following theorem, in the case of a residually transcendental extension w_{a,δ} of v (with the additional assumption that the valuation v has rank 1), is exactly the generalization given by Sudesh K. Khanduja and Jayanti Saha in [5].

Theorem 1.3 (Generalized Hensel’s Lemma) Let w_{a,δ} be either a residually transcendental valuation or a value transcendental extension of v. Let f, g_{0}, h_{0} be polynomials in K[X] such that the conditions GH_{1}, GH_{2}, GH_{3} are satisfied:

• GH_{1} w_{a,δ}(f − g_{0} · h_{0}) > w_{a,δ}(f)

• GH_{2} v(lc(g_{0})) + deg(g_{0}) · δ = w_{a,δ}(g_{0})
• GH_{3} ∃a, b ∈ K[x] such that w_{a,δ}(a · g_{0}) ≥ 0, w_{a,δ}(b · h_{0}) ≥ 0
and w_{a,δ}(a · g_{0} + b · h_{0} − 1) > 0,

Then there exists g, h ∈ K[x] such that the following conditions GH_{a}, GH_{b}, GH_{c} are satisfied; moreover, these polynomials are unique.

• GH_{a} f = g · h

• GH_{b} deg g = deg g_{0} and lc(g) = lc(g_{0})
• GH_{c} w_{a,δ}(g − g_{0}) > w_{a,δ}(g_{0}),

As soon as the conditions GH_{a}, GH_{b}, GH_{c} are satisfied, we have w_{a,δ}(h − h_{0}) > w_{a,δ}(h_{0}).

This generalization is very natural; the only point where there might be a problem is the second hypothesis, denoted by H_{2} in the Classical Lemma, and by GH_{2} in the Generalized Lemma. Where does this − deg(g_{0}) · δ come from? We shall see that this is in fact as natural as the other modifications.

We are going to prove that every Henselian field satisfies this Generalized Hensel’s Lemma.

2 The residually transcendental case

A residually transcendental valuation w_{0} = w_{a,δ}, defined by (a, δ) ∈ K^{ac} × \overline{G} is fixed throughout this section.

Let L be an algebraic extension of K[a, d] where d ∈ K^{sep} is such that v(d) = δ (if δ = v(a)/n with a in K, take e.g. d to be one of the roots of X^{n} + a · X + a). Being an algebraic extension of a Henselian field, L is Henselian.

We denote by f ↦ f̂ and f ↦ f̌ the following automorphisms of L[x]:

f(x) ↦ f̂(x) = f(d · x + a) f(x) ↦ f̌(x) = f(\frac{1}{d}(x − a)).

These automorphisms are obviously inverse of each other.

• Fact 1: For every f(x) ∈ L[x] we have w_{a,δ} (f(x)) = v^{g} (f̂(x)).

• Fact 2: Let f, g_{0}, h_{0}, g, h ∈ L[x]. We have the following equivalences:

f, g_{0}, h_{0} satisfy GH_{1}, GH_{2}, GH_{3} ⟺ f̂, ĝ_{0}, ĥ_{0} satisfy H_{1}, H_{2}, H_{3}.
f, g, h satisfy H_{a}, H_{b}, H_{c} ⟺ f̌, ǧ, ȟ satisfy GH_{a}, GH_{b}, GH_{c}.

Of course, we have other such equivalences using the fact the two transformations “hat” (f ↦ f̂) and “check” (f ↦ f̌) are inverse of each other. The proofs are straightforward and left to the reader. Note that fact 1 allows to prove in a short way that w_{0} is a valuation on K^{ac}[x]; from fact 2, one understands better why one must have v(lc(g_{0})) = − deg(g_{0}) · δ + w_{a,δ}(g_{0}) in the hypotheses GH_{2} of the generalized lemma.

Before starting the proof, there is still a bit of preparation: we said that two different pairs (a, δ) and (a′, δ′) in K^{ac} × \overline{G} may define the same valuation w_{a,δ} = w_{a′,δ′}. How can we characterize such pairs? The answer of this question can be found in [1]; this is the first point of the next lemma.

Lemma 2.1 We write B_{a,δ} = {α ∈ K^{ac} : v(a − α) ≥ δ}.

• Let (a, δ) and (a′, δ′) be in K^{ac} × \overline{G}. We have w_{a,δ} = w_{a′,δ′} iff B_{a,δ} = B_{a′,δ′}.

• If g_{0} ∈ K[x] is a polynomial such that the hypothesis GH_{2} holds, then all roots of g_{0} are in B_{a,δ}.

• In this case, we can choose b ∈ K^{sep} such that we have w_{a,δ} = w_{b,δ} = w_{σb,δ}, for all σ ∈ Gal(K).

Proof Note that given (a, δ), we have B_{a,δ} = B_{a′,δ′} iff δ′ = δ and a′ ∈ B_{a,δ}. Now note that w_{a,δ} and w_{a′,δ′} coincide on the field K ⊆ K[x]; thus as soon

as they have the same valuation ring, they are equal. Using fact 1 we write the following equivalences, where d, d′ are elements of K^{sep} such that v(d) = δ and v(d′) = δ′.

∀f ∈ L[x] w_{a,δ}(f) = w_{a′,δ′}(f) ⟺

[∀f ∈ L[x] f(dx + a) ∈ V[x] ⟺ f(d′x + a′) ∈ V[x]]  ⟺
[ ∀f ∈ L[x] f(dx + a) ∈ V[x] ⟺ f (d · (\frac{d′}{d} x + \frac{a′ − a}{d}) + a) ∈ V[x]]  ⟺
[ ∀g ∈ L[x] g(x) ∈ V[x] ⟺ g (\frac{d′}{d} x + \frac{a′ − a}{d}) ∈ V[x]]  ⟺

v (\frac{d′}{d} ) = 0 and v (\frac{a′ − a}{d}) ≥ 0

The second assertion of the lemma is very simple to prove. Note that ĝ_{0} ∈ K^{ac}[x] has leading coefficient with valuation v^{g}(ĝ_{0}); hence its roots are in V^{ac}.

Now if ζ is a root of g_{0}, \frac{ζ − a}{d} is a root of ĝ_{0}. From v (\frac{ζ − a}{d}) ≥ 0 follows

ζ ∈ B_{a,δ}.

Before proving the third assertion, note that if we replace g_{0}(x) by g_{0}(x) + c · x, with v(c) large enough, the hypotheses of the lemma are still satisfied; hence one can suppose that g_{0} is separable, and so the roots of g_{0} are in K^{sep}.

Now for a root ζ of g_{0} as above, we have B_{a,δ} = B_{ζ,δ}; we have σζ ∈ B_{ζ,δ} for all σ ∈ Gal(K). Thus we can take b equal to a root of g_{0}, and by first asserstion we have w_{a,δ} = w_{b,δ} = w_{σb,δ}, for all σ ∈ Gal(K).

We are now going to prove the generalized lemma. We assume that f, g_{0}, h_{0} are polynomials satisfying the hypotheses GH_{1}, GH_{2}, GH_{3}. In this situation, the previous lemma allows us to assume that w_{σa,δ} = w_{a,δ}, for all σ ∈ Gal(K).

We first prove a uniqueness statement, which will be useful in the proof of existence.

Lemma 2.2 (Uniqueness statement) In a Henselian field L such that a, d ∈ L, let f, g_{0}, h_{0} be polynomials satisfying the hypotheses GH_{1}, GH_{2}, GH_{3} of the generalized Hensel’s lemma. If there are polynomials g, h satisfying its conclusion GH_{a}, GH_{b}, GH_{c} then they are unique.

Proof We use the unicity of Classical Hensel’s lemma: f̂, ĝ_{0}, ĥ_{0} satisfy the hypotheses H_{1}, H_{2}, H_{3} of the classical lemma, and ĝ, ĥ satisfy its conclusion H_{a}, H_{b}, H_{c}. But there are no other polynomials than ĝ, ĥ satisfying this conclusion, hence there are no other polynomials than g, h satisfying GH_{a}, GH_{b}, GH_{c}.

Existence proof Of course the idea is to put L = K[a, d] ⊆ K^{sep}, and to say that if f, g_{0}, h_{0} satisfy GH_{1}, GH_{2}, GH_{3}, then f̂, ĝ_{0}, ĥ_{0} satisfy H_{1}, H_{2}, H_{3}, to use the Classical Hensel’s lemma in L to find (unique) polynomials g_{1}, h_{1} satisfying H_{a}, H_{b}, H_{c}, and then just do the reverse transformation to obtain g = ǧ_{1}, and h = ȟ_{1} satisfying GH_{a}, GH_{b}, GH_{c}. But now these polynomials are in L[x]!

It suffices to show that g, h are fixed by every element σ of Gal(K), so that they are in K[x]. Using the uniqueness statement (in L), it suffices to show that for all σ ∈ Aut(L|K), σg and σh satisfy the conclusion GH_{a}, GH_{b}, GH_{c}.

This is clear for GH_{a} and GH_{b}. Now let g_{0} = ∑_{i} r_{i}(a)(x − a)^{i} and g = ∑_{i} ρ_{i}(a, d)(x − a)^{i}, with r_{i} ∈ K[x] and ρ_{i} ∈ K[x, y]. We have

g_{0}(x) − g(x) = ∑_{i}(r_{i}(a) − ρ_{i}(a, d))(x − a)^{i}

and

g_{0}(x) − σg(x) = σg_{0}(x) − σg(x) = ∑_{i}(r_{i}(σa) − ρ_{i}(σa, σd))(x − σa)^{i}.

K being Henselian, we have v(r_{i}(σa) − ρ_{i}(σa, σd)) = v(r_{i}(a) − ρ_{i}(a, d)), hence w_{σa,δ}(σg_{0} −σg) = w_{a,δ}(g_{0} −g). But we said that one can assume that w_{σa,δ} = w_{a,δ}. Then we have w_{a,δ}(σg_{0} − σg) = w_{a,δ}(g_{0} − g) > w_{a,δ}(g_{0}). This proves that g and h are in K[x].

From v^{g}(ĥ_{0} − h_{1}) > v^{g}(ĥ_{0}) we conclude that w_{a,δ}(h_{0} − h) > w_{a,δ}(h_{0}). This concludes the proof.

3 Geometric interpretation

We give here a geometric interpretation of the previous result.

For any ζ ∈ K^{ac} we denote by B_{ζ,0}^{∘} = {α ∈ K^{ac} : v(α − ζ) > 0} the open ball of center ζ; the closed ball V^{ac} = B_{0,0} is a disjoint union of such open balls. Note that there is bijection between the set of open balls B_{α,0}^{∘} ⊆ B_{0,0} and the residue field of K^{ac} (given by B_{α,0}^{∘} ↦ \overline{α}).

Let f, f_{0}, g_{0} be polynomials satisfying the hypotheses of the Classical Hensel’s Lemma. As soon as there are roots of f in a ball B_{α,0}^{∘} ⊆ B_{0,0}, then there are roots of g_{0} or h_{0}. If there are some roots of g_{0} (resp. h_{0}) in a ball B_{α,0}^{∘}, then there are exactly the same number of roots of f. All the roots of g_{0} are in B_{0,0}. Moreover, in any ball B_{α,0}^{∘} ⊆ B_{0,0} there can not be roots of g_{0} together with roots of h_{0}.

These geometric conditions are equivalent to H_{1}, H_{2}, H_{3}.

If g, h are the factors of f provided by the conclusion of the lemma, the roots of g are exactly the roots of f which are in the open balls B_{α,0}^{∘} where the roots of g_{0} can be found; the same holds for integral roots of h, and the non integral roots of f (the roots not in B_{0,0}) are exactly the non integral roots of h.

Now for the Generalized Lemma, in the case of a residually transcendantal valuation w_{a,δ}, almost the same holds: just replace B_{0,0} by B_{a,δ}, and the open balls B_{α,0}^{∘} by open balls B_{α,δ}^{∘} ⊆ B_{a,δ}. We think that these geometric interpretations are of some interest for a better understanding of the proof given in the previous section; they are certainly useful for the proof given in the next section.

The following lemma gathers all these results in a very formal manner.

Lemma 3.1 Let w_{a,δ} be a residually transcendental valuation.

• For p(x) = ∑_{i} c_{i} · (x − a)^{i} ∈ K[x], write M_{a,δ}(p) = max{i : v(c_{i}) + i · δ = w_{a,δ}(p)}, and m_{a,δ}(p) = min{i : v(c_{i}) + i · δ = w_{a,δ}(p)}. Then, in K^{ac}, there are exactly (counted with multiplicities) M_{a,δ}(p) roots of p(x) in B_{a,δ} and m_{a,δ}(p) in B_{a,δ}^{∘}.

• For p, q ∈ K[x], we have M_{a,δ}(p · q) = M_{a,δ}(p) + M_{a,δ}(q), and m_{a,δ}(p · q) = m_{a,δ}(p)+m_{a,δ}(q). If w_{a,δ}(f −g_{0} ·h_{0}) > w_{a,δ}(f), then M_{a,δ}(f) = M_{a,δ}(g_{0} ·h_{0}) and m_{a,δ}(f) = m_{a,δ}(g_{0} · h_{0}).

• If there are a, b ∈ K[x] such that w_{a,δ}(a · g_{0}) ≥ 0, w_{a,δ}(b · h_{0}) ≥ 0 and w_{a,δ}(a · g_{0} + b · h_{0} − 1) > 0, then min(m_{a,δ}(g_{0}), m_{a,δ}(h_{0})) = 0.

Proof It’s easy to verify all this in the case where w_{a,δ} = w_{0,0} = v^{g} — it comes from the fact that if v(c) = v^{g}(p), then we can write \overline{p(x)/c} = \overline{c′_{M}}·x^{M}+⋯+\overline{c′_{m}}·x^{m} ∈ k[x]. For the general case use the case of the Gaussian extension and the “hat” and “check” transformations (f ↦ f̂ and f ↦ f̌).

Remark 3.2 If B_{a,δ} = B_{b,δ}, we don’t necessarily have B_{a,δ}^{∘} = B_{b,δ}^{∘}; hence the value of m_{a,δ}(p) for a polynomial p depends not only on the choice of w_{a,δ}, but on the choice of the pair (a, δ), whereas the value of M_{a,δ}(p) is the same for each pair defining the same valuation.

Example 3.3 In K = ℚ_{7}, this figure describes the situation for the polynomials f = 7x^{6} + x^{5} + 5x^{4} + 3x^{3} + 9x^{2} − 4x + 7, g_{0} = x^{3} + x^{2} − x + 6, h_{0} = x^{2} + 4x + 7. The gray disks are the roots of f in B_{0,0}, the white ones the roots of g_{0} and the crosses the roots of f_{0}. We have B_{0,0} = B_{0,0}^{∘} ∪ B_{1,0}^{∘} ∪ ⋯ ∪ B_{6,0}^{∘}.

4 The value transcendental case

We prove Theorem 1.3 in the case where w_{a,δ} is a value transcendental valuation. In that case δ defines a cut in \overline{G} as stated in the first part of the paper.

Proof of theorem 1.3 Consider a value transcendental valuation w_{a,δ}.

Suppose that f, g_{0}, h_{0} in K[x] verify the hypotheses GH_{1}, GH_{2}, GH_{3}. We write

g_{0} · h_{0} = ∑_{i=0}^{n} c_{i} · (x − a)^{i} and f = ∑_{i=0}^{k} a_{i} · (x − a)^{i},

g_{0} = ∑_{i=0}^{m} r_{i} · (x − a)^{i} and h_{0} = ∑_{i} s_{i} · (x − a)^{i}.

Note that GH_{2} implies that v(r_{m}) + m · δ = w_{a,δ}(g_{0}).

Since we have δ ∉ \overline{G}, there exists a unique ℓ ≤ k such that w(f) = v(a_{ℓ}) + ℓ · δ. With the notations of lemma 3.1, we have ℓ = M_{a,δ}(f) = m_{a,δ}(f). Now GH_{1} is equivalent to

\begin{cases}v(a_{ℓ} − c_{ℓ}) & > & v(a_{ℓ}) \\ δ & > & \frac{v(a_{ℓ}) − v(a_{i} − c_{i})}{i − ℓ} & for i > ℓ \\ δ & < & \frac{v(a_{ℓ}) − v(a_{i} − c_{i})}{i − ℓ} & for i < ℓ. \end{cases}

An equivalence with other systems of strict inequalities can be written for GH_{2}, GH_{3}. We conclude that there exist ε_{1}, ε_{2} in \overline{G} such that for all δ′ ∈ (ε_{1}, ε_{2}) := \{δ′ ∈ \overline{G} : ε_{1} < δ′ < ε_{2}\}, the hypotheses GH_{1}, GH_{2}, GH_{3} hold for the residually transcendental valuation w_{a,δ′}. In the case where \overline{G}^{<δ} = ∅ we can even take ε_{1} = −∞, and in the case where \overline{G}^{>δ} = ∅ we can take ε_{2} = +∞.

Take δ_{1} < δ_{2} ∈ (ε_{1}, ε_{2}), denote w_{a,δ_{1}} and w_{a,δ_{2}} the associated residually transcendental valuations.

Following the geometric interpretation given in the previous section, we see that all roots of g_{0} are in B_{a,δ_{1}} ⊆ B_{a,δ_{2}}^{∘}. Then we see that all roots of g_{0} are in B_{a,δ′}^{∘} for all δ′ ∈ (ε_{1}, ε_{2}).

In the case \overline{G}^{>δ} = ∅, that proves that g has no roots in B_{a,δ′}^{∘}, for all δ′ ∈ (ε_{1}, +∞), that is m_{a,δ′} = 0. Hence, by definition of m_{a,δ′}, for all δ′ ∈ (ε_{1}, +∞), we have w_{a,δ′}(g_{0}) = v(r_{0}), and that is true for δ as well. Since we have also v(lc(g_{0})) + deg(g_{0}) · δ = w_{a,δ}(g_{0}), we have deg g_{0} = 0 and the theorem is trivial. We can then assume \overline{G}^{>δ} ≠ ∅

Still using the geometric interpretation, we see also that h_{0} has not roots in B_{a,δ_{2}}^{∘} (for roots of g_{0} are in it) which contains B_{a,δ_{1}}.

Then for all δ′ ∈ (ε_{1}, ε_{2}), h_{0} has no roots in B_{a,δ′}. That is M_{a,δ′}(h_{0}) = 0, for all δ′ ∈ (ε_{1}, ε_{2}). We conclude that v(s_{0}) = w_{a,δ′}(h_{0}) for all δ′ ∈ (ε_{1}, ε_{2}), and now it’s easy to show that v(s_{0}) = w_{a,δ}(h_{0}).

In the case \overline{G}^{<δ} = ∅, that proves that deg h_{0} = 0. Moreover, from GH_{1} you can now conclude that deg g_{0} = deg g and the theorem is trivial. We can then assume that \overline{G}^{<δ} ≠ ∅.

Now take δ_{1} ∈ (ε_{1}, δ) and δ_{2} ∈ (δ, ε_{2}).

Applying the residually transcendental case, we get unique polynomials g_{1}, h_{1} and g_{2}, h_{2} satisfying the conclusion of the theorem, for w_{a,δ_{1}} and w_{a,δ_{2}} respectively.

We write g_{2} = ∑_{i=0}^{m} r′_{i} · (x − a)^{i}. Now w_{a,δ_{2}}(g_{2} − g_{0}) > w_{a,δ_{2}}(g_{0}) is equivalent to

\begin{cases}v(r_{m} − r′_{m}) & > & v(r_{m}) \\ δ_{2} & < & \frac{v(r′_{i} − r_{i}) − v(r_{m})}{m − i}, & ∀ i < m. \end{cases}

(Remember that v(r_{m}) + m · δ′ = w_{a,δ′}(g_{0}) for all δ′ ∈ (ε_{1}, ε_{2}).) For we have δ_{1} < δ < δ_{2}, the same system of inequalities remains true if we replace δ_{2} by δ or δ_{1}, and we conclude that w_{a,δ}(g_{2} − g_{0}) > w_{a,δ}(g_{0}) and w_{a,δ_{1}}(g_{2} − g_{0}) > w_{a,δ_{1}}(g_{0}).

Now g_{2}, h_{2} satisfy the conditions GH_{a}, GH_{b}, GH_{c} for the valuation w_{a,δ_{1}}, and by unicity we have g_{1} = g_{2} and h_{1} = h_{2}.

We take g = g_{1} = g_{2} and h = h_{1} = h_{2} = ∑_{i} s′_{i}·(x−a)^{i}. Of course GH_{a} and GH_{b} hold, and we just proved that GH_{c} holds. It remains to show that w_{a,δ}(h − h_{0}) > w_{a,δ}(h_{0}).

The inequality w_{a,δ_{1}}(h − h_{0}) > w_{a,δ_{1}}(h_{0}) holds, and, since we have w_{a,δ_{1}}(h_{0}) = v(s_{0}), is equivalent to

\begin{cases}v(s′_{0} − s_{0}) & > & v(s_{0}) \\ δ_{1} & > & \frac{v(s_{0}) − v(s′_{i} − s_{i})}{i}, & ∀ i > 0. \end{cases}

From δ > δ_{1} we conclude that w_{a,δ}(h−h_{0}) > w_{a,δ}(h_{0}), and this ends the proof.

Remark 4.1 During the proof, we have proved that if we keep the definitions of lemma 3.1 for M_{a,δ} and m_{a,δ}, for all p(x) ∈ K[x], we have M_{a,δ}(p) = m_{a,δ}(p); we proved that the three statements of lemma 3.1 were still true. Thus if f, g_{0}, h_{0} satisfy GH_{1}, GH_{2}, GH_{3}, then deg h_{0} = 0 and deg g_{0} = M_{a,δ}(f).

5 A multidimensional lemma

We thank Sudesh Kaur Khanduja for suggesting us to try to state a multivariate version of this result; we will do this now. Here we consider only residually transcendental extensions of v.

It was an abuse to speak of the Classical Hensel’s Lemma: there is another very classical statement of Hensel’s Lemma about zeros of a polynomial (cf. [8]). There is a well-known equivalent form, the multi-dimensional Hensel Lemma dealing with zeros of systems of polynomials (cf. [6]). We restate these two results below.

For a system (f) = (f_{1}, …, f_{n}) of polynomials in V[x_{1}, …, x_{n}], we denote by

J_{f} the Jacobian matrix (\frac{∂f_{i}}{∂x_{j}} )_{ij}.

Theorem 5.1 (Another Classical Hensel’s Lemma) In a Henselian field, the following statements hold:

1. Let f(x) ∈ V[x] and ζ_{0} ∈ V, such that v(f(ζ_{0})) > 0 and v(f′(ζ_{0})) = 0. Then there exists a unique ζ ∈ V such that f(ζ) = 0 and v(ζ − ζ_{0}) > 0.

2. Let (f) = (f_{1}, …, f_{n}) be polynomials in V[x_{1}, …, x_{n}], and ζ^{0} ∈ V^{n} be a n-tuple (ζ_{1}^{0}, …, ζ_{n}^{0}) of integers such that v(f_{i}(ζ^{0})) > 0 for all i. We assume moreover that v(det J_{f}(ζ^{0})) = 0. Then there exists a unique ζ = (ζ_{1}, …, ζ_{n}) ∈ V^{n} such that f_{i}(ζ) = 0 for all i, and v(ζ_{j} − ζ_{j}^{0}) > 0 for all j.

Remark 5.2 The geometric interpretation of the multi-dimensional lemma is that there is one and only one root of (f) in B_{ζ_{1}^{0},0}^{∘} × ⋯ × B_{ζ_{n}^{0},0}^{∘}. The conclusion asserts that this root is in K^{n}.

We fix a n-tuple \overline{a} = (a_{1}, …, a_{n}) of (K^{ac})^{n} and \overline{δ} = (δ_{1}, …, δ_{n}) ∈ \overline{G}^{n}. We extend the valuation v of K^{ac} by a valuation w_{\overline{a},\overline{δ}} of K^{ac}[x_{1}, …, x_{n}]: first we extend v to K[x_{1}] using the pair (a_{1}, δ_{1}), and then to K[x_{1}, x_{2}] using (a_{2}, δ_{2}), and so on.

For each δ_{i}, we fix a d_{i} ∈ K^{sep} such that v(d_{i}) = δ_{i}. We denote by D the closed polydisk B_{a_{1},δ_{1}} × ⋯ × B_{a_{n},δ_{n}}.

Theorem 5.3 (Another Generalized Hensel’s Lemma) In a Henselian field, the following statements hold:

1. Let f(x) be of non-negative w_{a,δ}-value and ξ_{0} ∈ B_{a,δ} ∩ K, such that v(f(ξ_{0})) > 0 and v(f′(ξ_{0})) = −δ. Then there exists a unique ξ ∈ B_{a,δ} ∩ K

such that f(ξ) = 0, and v(ξ − ξ_{0}) > δ.

2. Let (f) = (f_{1}, …, f_{n}) be polynomials in K[x_{1}, …, x_{n}] with non-negative w_{\overline{a},\overline{δ}}-value. Let ξ^{0} = (ξ_{1}^{0}, …, ξ_{n}^{0}) ∈ D ∩ K^{n} be such that v(f_{i}(ξ^{0})) > 0 for all i. We assume moreover that v(det J_{f}(ξ^{0})) = − ∑_{i} δ_{i}. Then there exists a unique ξ = (ξ_{1}, …, ξ_{n}) ∈ D ∩ K^{n} such that f_{i}(ξ) = 0 for all i, and v(ξ_{j} − ξ_{j}^{0}) > δ_{j} for all j.

Remark 5.4 Now, the geometric interpretation for this modified hypotheses is that there is one and only one root of (f) in B_{ξ_{1}^{0},δ}^{∘} × ⋯ × B_{ξ_{n}^{0},δ}^{∘}.

Proof We prove the second statement (the first is a particular case, or could be deduced from Theorem 1.3).

We note that D ∩ (K^{sep})^{n} being non-empty, we can assume w.l.o.g. that a_{i} ∈ K^{sep} for all i. We set g_{i} = f_{i}(d_{1}x_{1} + a_{1}, …, d_{n}x_{n} + a_{n}) ∈ K^{sep}[x_{1}, …, x_{n}] for all i

and ζ_{j}^{0} = \frac{ξ_{j}^{0} − a_{j}}{d_{j}}. The hypothesis w_{\overline{a},\overline{δ}}(f_{i}) ≥ 0 leads to g_{i} ∈ V^{sep}[x_{1}, …, x_{n}]. The

tuple ζ^{0} = (ζ_{1}^{0}, …, ζ_{n}^{0}) is in V^{n}, and we have that v(g_{i}(ζ^{0})) > 0 for all i.

The Jacobian matrix J_{g} is obtained from J_{f} simply by multiplying the j-th column by d_{j}, hence det J_{g}(ζ^{0}) = d_{1} ⋯ d_{n} · det J_{f}(ξ^{0}) has value 0.

We apply the Classical Lemma to get ζ ∈ (V^{sep})^{n}, and we set ξ_{j} = d_{j}ζ_{j} + a_{j}. We have f_{i}(ξ) = 0 for all i and ξ ∈ D. The inequality v(ζ_{j} − ζ_{j}^{0}) > 0 leads to v(ξ_{j} − ξ_{j}^{0}) > δ_{j}. As in the previous section, the unicity in the statement of the Classical Lemma leads to the unicity of such a tuple ξ in K^{sep}. Then we show that for all σ ∈ Gal(K), we have σξ_{j} = ξ_{j}, just by remarking that the tuple σξ is another solution; we conclude that ξ ∈ K^{n}.

One could call these generalizations De-centralized Hensel’s Lemmas, because they replace the classical ball B_{0,0} = V by any other ball.

Acknowledgments

I wish to thank gratefully Franz-Viktor Kuhlmann for his invitation to Saskatoon for the Conference on Valuation Theory, for suggesting this problem, and for many readings and corrections on earlier versions of this paper; many thanks go as well to Sudesh Kaur Khanduja for her suggestions and encouragements. I don’t forget Salma Kuhlmann and Murray Marshall for their hospitality and kindness in Saskatoon.

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