## An unfamiliar (to me) form of Hensel’s Lemma

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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?

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@Harry: if $(R,\mathfrak{m})$ is a valuation ring, it is $\mathfrak{m}$-adically separated iff it is Noetherian iff it is a DVR. So your first claim is not correct. It is true that Henselianity can be characterized in terms of etale morphisms (Milne, Etale Cohomology, Section 1.4), but what does this have to do with the Hensel-Kurschak Lemma? – Pete L. Clark Feb 18 2010 at 6:33

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@Harry Gindi: Sorry, I'm not going to play the disappearing comments game with you tonight. Please give an answer if you have one. – Pete L. Clark Feb 18 2010 at 6:45

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@Everyone: in case it was not apparent, this was a rather embarrassing question to ask: I should have worked this out myself, but I wanted a quick answer for my students' sake. In fact I got, in addition to exactly the information I wanted/needed, lots of other interesting stuff. The moral of the story is....what is it, exactly? You can use MO to help teach your graduate courses? :) – Pete L. Clark Feb 18 2010 at 13:53

## 3 Answers

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.

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Thanks, Jerome!  I see now that if I guessed the right "standard" version of Hensel's Lemma, the proof would have come out easily.  (Why did I choose the one that I did?  Ridiculous but true answer: I hadn't talked about residue fields yet.)  Thanks also for the reference for H-K is equivalent to Henselian.   – Pete L. Clark Feb 18 2010 at 8:21

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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.

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@AS: Yes, it's interesting! [When you say Ribenboim, do you really mean Ribenboim -- as in the paper that Franz cited -- or Roquette, as in the paper that I cited?] I will say though that I do not believe that what you have written is the most general possible Hensel's Lemma, a quick corollary of my belief that the most general possible HL does not exist. For instance, this version of HL applies to a complete valuation ring; in (e.g.) Eisenbud's book there is a form of HL which applies to a complete, m-adically separated local ring. Neither class of rings contains the other... – Pete L. Clark Feb 18 2010 at 13:02

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...(See my first comment above). As a working arithmetic geometer, the version of HL that I use most often is the following: if K is a complete local field with valuation ring R and residue field k and X/R is a finite-type proper regular flat R-scheme, then the image of the reduction map X(K) = X(R) -> X(k) is precisely the set of smooth points in X(k). (This version is a consequence of Bost's nice formulation.) – Pete L. Clark Feb 18 2010 at 13:11
Hm, I like your last statement (I am not yet an arithmetic geometer :)). Could you give me some reference for it? In fact I expected to find this somewhere in Liu's book, but my first attempt to find it (using the index of the book) has been unsuccessful. – Wanderer Feb 18 2010 at 13:24
@AS: for instance, it is Lemma 9 in math.uga.edu/~pete/plclarkarxiv7.pdf. If you look there, you will be referred back to a 1985 paper of Jordan and Livne on Shimura curves, which is where I first learned about it (since my thesis work was on Shimura curves). I think the result is known to most people who seriously study curves over DVRs, so it should go back to the 1960s if not earlier. – Pete L. Clark Feb 18 2010 at 13:39

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@AS: That's a great answer! Let me just point out that with virtually the same method, you can prove the same results for a larger class of fields (thus proving they are Henselian), namely fraction fields of local rings of analytic spaces. They are not Banach rings but inductive limits of them (for any r>0, take the completion of the ring of functions that converge on the disc of radius r with center at your point). If your space in Archimedean though, you won't have such a precise form of the theorem (you can say exactly what the radii are only thanks to the ultrametric triangle inequality). – Jérôme Poineau Feb 19 2010 at 16:49
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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
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+1: So I learned that this journal exists earlier this week and now for a completely unrelated question I get referred to a paper in this journal. It's amazing to me how often this sort of thing happens. – Pete L. Clark Feb 18 2010 at 18:56
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