A generalization of Hensel’s Lemma, following a previous result of S.K Khanduja and J. Saha

Hervé Perdry Équipe de Mathématiques de Besançon, UMR CNRS 6623 UFR Sciences et techniques, Université de Franche-Comté 25030 Besançon cedex e-mail : perdry@math.univ-fcomte.fr

Introduction

Sudesh Kaur Khanduja and Jayanti Saha gave in [KS] 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 in the case of a complete valued field of rank 1. We give here a proof of this lemma for any Henselian field.

We give a geometric interpretation of this lemma, and an other version allowing to deduce a root of a polynomial from an “approximate” root together with a multidimensional version of this last result.

I. Definitions, notations & statements

Let K be a Henselian field, and K^{ac} be its algebraic closure. The valuation group of K is G, and the one of K^{ac} is \overline{G}, the divisible closure of G. The valuation on K^{ac} is always denoted by v ; the valuation ring of K is V, the one of K^{ac} is V^{ac}. The residue fields of K and K^{ac} are k and \overline{k}. The galois group of K is Gal(K). For general defintions and statements, cf [E].

Let (a, δ) ∈ K^{ac} × \overline{G}. We use it to define a residually transcendental extension of v to K^{ac}(x) (cf [PP] or [B], §10):

w_{a,δ} (∑_{i} f_{i}(x − a)^{i}) = min_{i} {vf_{i} + iδ}

The residue fields of K[X] and K^{ac}[X] for such a valuation are k[X] and \overline{k}[X] ; 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 defined by the above formula (cf [APZ]). 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} f_{i}x^{i}) = min_{i} vf_{i}.

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

A generalization of Hensel’s lemma

1. Classical Hensel’s Lemma Given f, g_{0}, h_{0} with v-valuation zero such that the conditions H_{1}, H_{2}, H_{3} are satisfied ; then there exists f, g ∈ V[x] such that H_{a}, H_{b}, H_{c} are satisfied ; moreover, these polynomials are unique.

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

v^{g}(a.g_{0} + b.h_{0} − 1) > 0.

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}) > 0, v^{g}(h − h_{0}) > 0

The generalization given by Sudesh K. Khanduja and Jayanti Saha in [KS] for a residually transcendental valuation w_{a,δ} is the following:

2. Generalized Hensel’s Lemma Given f, g_{0}, h_{0} ∈ K[x] with w_{a,δ}-valuation zero, such that the conditions GH_{1}, GH_{2}, GH_{3} are satisfied ; then there exists f, g ∈ K[x], with w_{a,δ}-valuation zero, such that GH_{a}, GH_{b}, GH_{c} are satisfied ; moreover, these polynomials are unique.

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

w_{a,δ}(a.g_{0} + b.h_{0} − 1) > 0.

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}) > 0, w_{a,δ}(h − h_{0}) > 0

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

We are going to give a proof that every Henselian field satisfies this generalized Hensel lemma.

II. The proof

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

Let L be an extension of K[a, d] where d ∈ K^{ac} is such that v(d) = δ. 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 one of the other. The following two facts are of some interest, almost obvious, and the reader shall guess the main idea of our proof just by reading them:

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

Of course, we have other such equivalences using the fact the two transformation “hat” (f ↦ f̂) and “check” (f ↦ f̌) are inverse one of the other. The proofs are straightforward and left to the reader. Remark that the fact 1 allows to prove really shortly that w is a valuation on K^{ac}[X] ; from the fact 2, one understand better why one must have v(lc(g_{0})) = − deg(g_{0}).δ in the hypothesis 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 find in [AP]; this is the first point of the next lemma.

Lemma: We denote B_{a,δ} = {x ∈ K^{ac} : v(a − x) ≥ 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 there exists a polynomial g_{0} ∈ K[x] satisfying w_{a,δ}(g_{0}) ≥ 0 and the hypothesis GH_{2}:v(lc(g_{0})) = − deg(g_{0}).δ then all the roots of g_{0} are in B_{a,δ}.

• In this case, one can suppose w.l.o.g. that for all σ ∈ Gal(K), we have w_{σa,δ} = w_{a,δ}.

Proof: Remark that given (a, δ), we have B_{a,δ} = B_{a′,δ′} iff δ′ = δ and a′ ∈ B_{a,δ}. The proof of the first affirmation of the lemma is easy, using the fact 1: we have the following equivalences, letting d, d′ ∈ K^{ac} such that v(d) = δ and v(d′) = δ′.

v^{g}(f(dx + a)) = v^{g}(f(d′x + a′)) ∀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] ∀f ∈ L[x]]

⇔ [g(x) ∈ V[x] ⇔ g (\frac{d′}{d} x + \frac{a′ − a}{d}) ∈ V[X] ∀g ∈ L[x]]

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

The next affirmation of the lemma may seem surprising despite of its simplicity:

It suffices to remark that ĝ_{0} is in V^{ac}[x], with a unit as leading coefficient; 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}.

For the third affirmation, remark that for any root ζ of g_{0} as above, B_{a,δ} = B_{ζ,δ}; we have σζ ∈ B_{ζ,δ} for all σ ∈ Gal(K). Then we can replace a by ζ.

We are now going to prove the generalized lemma. We assume that f, g_{0}, h_{0} are polynomials with w_{a,δ}- valuation zero satisfying the hypothesis GH_{1}, GH_{2}, GH_{3}. In this situation, the previous lemma allows to assume that a, d ∈ K^{sep} and w_{σa,δ} = w_{a,δ}, for all σ ∈ Gal(K).

We first prove an unicity statement, which will be useful in the proof of existence.

UnicityIn a Henselian field L such that a, d ∈ L, let f, g_{0}, h_{0} be polynomials satisfying the hypothesis 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 hypothesis 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^{ac}, 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 classical Hensel’s lemma in L to find (unique) polynomials g_{1}, h_{1} verifying 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, K being Henselian, they are in K. We use the unicity statement (in L), and it suffices to show that for all σ ∈ Aut(L|K), σg and σh satisfy the conclusion GH_{a}, GH_{b}, GH_{c}.

It’s clear for GH_{a} and GH_{b}. Now let g_{0} = ∑_{i} c_{i}(a)(x − a)^{i} and g = ∑_{i} γ_{i}(a, d)(x − a)^{i}, with c_{i} ∈ K[x] and γ_{i} ∈ K[x, y]. We have

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

and

g_{0}(x) − σg(x) = σg_{0}(x) − σg(x) = ∑_{i}(c_{i}(σa) − γ_{i}(σa, σd))(x − σa)^{i}

K being Henselian, we have v(c_{i}(σa)−γ_{i}(σa, σd)) = v(c_{i}(a)−γ_{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) > 0. In the same way, we have w_{a,δ}(σh_{0} − σh) > 0.

This allows us to conclude the proof.

III. Geometric interpretation & multidimensionnal lemma

We thank Sudesh Kaur Khanduja for suggesting us to try to state a multivariate version of this result; that’s what we are doing now.

We first give a geometric interpretation of the previous result. What is the geometric interpretation of the Classical Lemma? It’s well known but a bit long to describe:

For any ζ ∈ K^{ac} we denote by b_{ζ,0} = {x ∈ K^{ac} : v(x − 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_{x,0} ⊆ B_{0,0} and the residue field of K^{ac} (given by b_{x,0} ↦ residue of x).

Let f, f_{0}, g_{0} be polynomials satisfying the hypothesis of the Classical Hensel’s Lemma. All the roots of g_{0} are in B_{0,0}; in any ball b_{x,0} ⊆ B_{0,0} there can not be roots of g_{0} together with roots of h_{0}. If there are some roots of g_{0} (resp. h_{0}) in a ball b_{x,0}, then there are exactly the same number of roots of f. As soon as there are roots of f in a ball b_{x,0} ⊆ B_{0,0}, then there are roots of g_{0} or h_{0}. These geometric conditions are equivalents 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_{x,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, almost the same holds: just replace B_{0,0} by B_{a,δ}, and the open balls b_{x,0} by open balls b_{x,δ} ⊆ B_{a,δ}. We think that these geometric interpretations are of some interest for a better understanding of the proofs given in the previous section.

Example : In K = ℚ_{7}, this figure describe 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 ares 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}.

All this leads us to a question: the geometric interpretation of Hensel’s Lemma is more usually quoted when dealing with roots of polynomial. Can we say something about that? In fact, we have been cheating by talking of “classical” Hensel’s Lemma: there is an other very classical statement of Hensel’s Lemma about zeros of polynomials (cf [R]). There is a well-known equivalent form, the multi-dimensionnal Hensel Lemma dealing with zeros of systems of polynomials (cf [K]). We restate these two results below.

For a system (f) = (f_{1}, …, f_{n}) of polynomials in V[x_{1}, …, x_{n}], we denote J_{f} the Jacobian matrix (\frac{∂f_{i}}{∂x_{j}} )_{ij}.

1.An other Classical Hensel’s Lemma

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

• Let (f) = (f_{1}, …, f_{n}) be polynomials in V[x_{1}, …, x_{n}], and ζ^{0} = (ζ_{1}^{0}, …, ζ_{n}^{0}) ∈ V^{n} be a n-tuple 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_{i} − ζ_{i}^{0}) > 0 for all i.

Remark: The geometric interpretation of the multi-dimensionnal lemma is that there is one an 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_{0}, …, 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}}.

2.An other Generalized Hensel’s Lemma

• Let f(x) of w_{a,δ}-valuation nonnegative and ξ_{0} ∈ B_{a,δ} ∩ K, such that v(f(ξ_{0})) > 0 and v(f′(ξ_{0})) = −δ. Then there exist a unique ξ ∈ B_{a,δ} ∩ K such that f(ξ) = 0, and v(ξ − ξ_{0}) > δ.

• Let (f) = (f_{1}, …, f_{n}) be polynomials in K[x_{1}, …, x_{n}] with w_{\overline{a},\overline{δ}}-valuation nonnegative. 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_{i} − ξ_{i}^{0}) > δ_{i} for all i.

Remark: Now, the geometric interpretation for this modified hypothesis is that there is one an 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 our first result).

We remark that D ∩ K^{n} being no-empty, we can assume w.l.o.g. a_{i} ∈ K 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_{i}^{0} = \frac{ξ_{i}^{0}−a_{i}}{d_{i}}. 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 clearly v(g_{i}(ζ^{0})) > 0 for all i.

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

We apply the classical lemma to get ζ ∈ (V^{sep})^{n}, and we set ξ_{i} = d_{i}ζ_{i} + a_{i}. We have f_{i}(ξ) = 0 for all i and ξ ∈ D. The inequality v(ζ_{i} − ζ_{i}^{0}) > 0 leads to v(ξ_{i} − ξ_{i}^{0}) > δ_{i}. 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_{i} = ξ_{i}, just by remarking that the tuple σξ is an other solution; we conclude that ξ ∈ K^{n}.

It could be nice to call these generalizations De-localized Hensel’s Lemmas, because they replace the classical ball B_{0,0} = V by any other ball.

IV. Conclusion

2nd point Following an idea of Franz-Viktor Kuhlmann, we announce that we can deduce from this generalized lemma the same statement where w is a value transcendental extension of v. In that case w = w_{a,δ} where δ is not in \overline{G}. The value group is \overline{G}(δ) ; and δ defines a cut in \overline{G}.

Sketch of proof: we replace δ by an element ϵ of \overline{G} verifying ϵ < δ but close to δ enough to have all roots of g_{0} in B_{a,ϵ}, and to assure that the hypothesis are still verified for w_{a,ϵ}. The previous lemma applies. The polynomials f, g it provides are our solution.

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