An unfamiliar (to me) form of Hensel's Lemma

Question

In his very nice article

Peter Roquette, History of valuation theory. I. (English summary) Valuation theory and its applications, Vol. I (Saskatoon, SK, 1999), 291--355, Fields Inst. Commun., 32, Amer. Math. Soc., Providence, RI, 2002

Roquette states the following result, which he attributes to Kurschak:

Hensel-Kurschak Lemma: Let $(K,|\ |)$ be a complete, non-Archimedean normed field. Let $f(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \in K[x]$ be a polynomial. Assume (i) $f(x)$ is irreducible and (ii) $|a_0| \leq 1$. Then $|a_i| \leq 1$ for all $0 < i < n$.

He says that this result is today called Hensel's Lemma and that Hensel's standard proof applies.

This is an interesting result: Roquette explains how it can be used to give a very simple proof of the fact that, with $K$ as above, if $L/K$ is an algebraic field extension, there exists a unique norm on $L$ extending $| \ |$ on $K$. This is in fact the argument I gave in a course on local fields that I am currently teaching.

It was my initial thought that the Hensel-Kurschak Lemma would follow easily from one of the more standard forms of Hensel's Lemma. Indeed, in class last week I claimed that it would follow from

Hensel's Lemma, version 1: Let $(K,| \ |)$ be a complete non-Archimedean normed field with valuation ring $R$, and let $f(x) \in R[x]$ be a polynomial. If there exists $\alpha \in R$ such that $|f(\alpha)| < 1$ and $|f'(\alpha)| = 0$, then there exists $\beta \in R$ with $f(\beta) = 0$ and $|\alpha - \beta| < 1$.

Then in yesterday's class I went back and tried to prove this...without success. (I was not at my sharpest that day, and I don't at all mean to claim that it is not possible to deduce Hensel-Kurschak from HLv1; only that I tried the obvious thing -- rescale $f$ to make it a primitive polynomial -- and that after 5-10 minutes, neither I nor any of the students saw how to proceed.) I am now wondering if maybe I should be trying to deduce it from a different version of Hensel's Lemma (e.g. one of the versions which speaks explicitly about factorizations modulo the maximal ideal).

This brings me to a second question. There are of course many results which go by the name Hensel's Lemma. Nowadays we have the notion of a Henselian normed field, i.e., a non-Archimedean normed field in which the exended norm in any finite dimensional extension is unique. (There are many other equivalent conditions; that's rather the point.) Therefore, whenever I state a result -- let us restrict attention to results about univariate polynomials, to fix ideas -- as "Hensel's Lemma", I feel honorbound to inquire as to whether this result holds in a non-Archimedean normed field if and only if the field is Henselian, i.e., that it is equivalent to all the standard Hensel's Lemmata.

Is it true that the conclusion of the Hensel-Kurschak Lemma holds in a non-Archimedean valued field iff the field is Henselian?

More generally, what is a good, reasonably comprehensive reference for the various Hensel's Lemmata and their equivalence in the above sense?

Answer 1

What you say at the beginning of your post is right: Hensel-Kurschak's lemma may be deduced from some refined version of Hensel's lemma. Actually, it's what Neukirch does in Algebraic Number Theory (see chapter II, corollary 4.7). His proof relies on the following (see 4.6)

Hensel's lemma: Let $(K,|.|)$ be a complete valued field with valuation ring $R$, maximal ideal $\mathfrak{m}$. Let $f(x) \in R[x]$ be a primitive polynomial (ie $f\ne 0$ mod $\mathfrak{m}$). Suppose $f=\bar{g}\bar{h}$ mod $\mathfrak{m}$, with $\bar{g}$ and $\bar{h}$ relatively prime. Then you can lift $\bar{g}$ and $\bar{h}$ to polynomials $g$ and $h$ in $R[x]$ such that $\textrm{deg}(g)=\textrm{deg}(\bar{g})$ and $f=gh$.

Neukirch goes on with proving that the valuation on $K$ extends uniquely to any algebraic extension (see corollary 4.7), as you say Roquette does.

As regards your last question, you may want to have a look at chapter II, paragraph 6 (appropriately called Henselian Fields) in the book of Neukirch. His definition of Henselian field is that it should satisfy Hensel-Kurschak's lemma. In theorem 6.6, he shows that this property is equivalent to the unique extension of the valuation to algebraic extensions.

Answer 2

A far more general result is the "non-archimedean inverse function theorem". I haven't looked at Roquette's reference, so maybe he is mentioning it. But it is something which I didn't really find in the standard number theory textbooks - probably you can find it in texts on $p$-adic analysis - and I learned it from my number theory professor last semester (Jean-Benoît Bost). This theorem is powerful - and I find it fascinating and surprising - and all versions of Hensel's lemma which one usually encounters while learning number theory are immediate consequences.

Let $K$ be a field, $\left| \cdot \right|$ a non-archimedean absolute value on $K$ for which $K$ is complete, $\mathcal{O}$ the associated valuation ring, $\mathcal{M}$ the maximal ideal, $\pi$ a uniformizer. Let $\Phi_i \in \mathcal{O}[X_1,\,\cdots,X_n]$ for $1 \leq i \leq n$ and consider the map $\Phi = (\Phi_1,\,\cdots,\Phi_n) : \mathcal{O}^n \to \mathcal{O}^n$. Let $J$ be the Jacobian $\det(\partial \Phi_i / \partial X_j) \in \mathcal{O}[X_1,\,\cdots,X_n]$.

Theorem. If $x_0 \in \mathcal{O}^n$, $y_0 = \Phi(x_0)$ and $J(x_0) \neq 0$, then for any $R \in (0, \left|J(x_0)\right|)$, $\Phi$ induces a bijection $$\overline{B}(x_0,R) \to y_0 + (D\Phi)(x_0) \overline{B}(0,R)$$ (where $D\Phi$ is the derivative we all know!) and furthermore we have a bijection $$B^\circ(x_0,\left|J(x_0)\right| \to y_0 + (D\Phi)(x_0) B^\circ(0,\left|J(x_0)\right|).$$

(I use the standard notations $\overline{B}$ and $B^\circ$ for closed and open balls respectively.)

The proof uses in an essential way the Picard fixed point theorem.

Corollary 1. Take $n = 1$, $\Phi_1 = P$, $x_0 = \alpha$, $\varepsilon \in (0,1)$. Suppose that $\left|P(\alpha)\right| \leq \varepsilon \left|P'(\alpha)\right|^2$. Then there exists a unique $\beta \in \mathcal{O}$ such that $P(\beta) = 0$ and $\left|\beta - \alpha\right| \leq \varepsilon \left|P'(\alpha)\right|$. (We take $R = \varepsilon \left|P'(\alpha)\right|$ in the first bijection.)

Hence, as a special case, if $\left|P(\alpha)\right| < \left|P'(\alpha)\right|^2$, we find $\left|\beta - \alpha\right| < \left|P'(\alpha)\right|$.

As an even more special case, if $P'(\alpha) \in \mathcal{O}^\times$ and $\left|P'(\alpha)\right| <1$, there exists $\beta \in \mathcal{O}$ such that $P(\beta) = 0$ and $\left|\beta - \alpha\right| < 1$. Restating this in terms of the residue field: a simple zero in the residue field can be lifted to a real zero in $\mathcal{O}$. This is the really known version of Hensel's lemma, I guess.

[Definition: the Gauss norm of a polynomial with coefficients in $K$ is defined as the maximum of the absolute values of its coefficients. It is very easy to check that the Gauss norm is multiplicative.]

Corollary 2. Take $f,g,h \in \mathcal{O}[X]$ such that $\deg g = n$, $\deg h = m$ and $\deg f = \deg g + \deg h = n + m$. Assume that there exists $\varepsilon \in (0,1)$ such that $\left\|f - gh\right\| \leq \varepsilon\left|\text{Res}(g,h)\right|^2$ and $\deg(f - gh) \leq m + n - 1$. Then there exist $G, H \in \mathcal{O}[X]$ such that $f = GH$, $\deg(G - g) \leq n - 1$, $\deg(H - h) \leq m - 1$, and also $\left\|G - g\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$ and $\left\|H - h\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$. (Obviously $\text{Res}$ denotes the resultant here, and $\left\|\cdot\right\|$ the Gauss norm.)

To prove this: write $G = g + \xi$ and $H = h + \eta$ where $\xi$ and $\eta$ are polynomials with coefficients in $\mathcal{O}$ and have degrees $\leq n - 1$ and $\leq m - 1$ respectively. Then $f = GH$ if and only if $f = (g + \xi)(h + \eta)$. It can be seen as a map from $\mathcal{O}^n \times \mathcal{O}^m \to \mathcal{O}^{n + m}$ given by polynomials. So consider the map $\Phi: (\xi, \eta) \mapsto (g + \xi)(h + \eta) - f$. We have also $\text{Res}(g,h) = \det((\xi, \eta) \mapsto g \xi + h \eta))$. It is easy to see that the theorem above then gives the result.

As a corollary: if $f$, $g$ and $h$ satisfy $\overline{f} = \overline{g} \overline{h}$ - where $\overline{f}$ is $f$ reduced modulo $\mathcal{M}$ et cetera - and if $\overline{g}$ and $\overline{h}$ are coprime (this is a condition on the resultant!) then there exist $G,H \in O[X]$ satisfying the following conditions: $f = GH$, $\deg(G - g) \leq n - 1$, $\deg(H - h)\leq m - 1$, $\overline{G} = g$ and $\overline{H} = h$. Hence "a factorization over the residue field lifts to a factorization over $\mathcal{O}$" (under the right conditions).

Corollary 3. Finally, let us come to the motivation for the question: the more general result is that if $P \in K[X]$ is irreducible, then $\left\|P\right\|$ (Gauss) is the maximum of the absolute values of the leading coefficient and the constant coefficient. (As a special case, we find the result which Pete L. Clark cites as the Hensel-Kurschak lemma.)

Indeed, let $P(X) = \sum_{i = 0}^n a_i X^{n - i} \in K[X]$. Suppose WLOG that $\left\|P\right\| = 1$. Let $\mathbb{F}$ be the residue field and let $\overline{P}$ be the image of $P$ modulo $\mathcal{M}$. Set $r = \min \{n : \overline{a_{n - r}} \neq 0\}$. Then we have in the residue field the factorization $\overline{P}(X) = X^r \left(\overline{a_{n - r}} + \overline{a_{n - r - 1}}X + \cdots + \overline{a_0} X^{n - r}\right)$ and we can lift the factorization by Corollary 2, contradicting irreducibility.

I know this is quite some digression; but I find the whole discussion about the various forms of Hensel's lemma very interesting, and I thought this could add something to the discussion.

Answer 3

An exposition of various versions of Hensel's Lemma can be found in

• P. Ribenboim, Equivalent forms of Hensel's lemma, Expos. Math. 3 (1985), 3-24