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 $K$, $\mathcal{O}$ 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.