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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 $K$, 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. 9 added 67 characters in body; [made Community Wiki] 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$G,H \in O[X]$such that 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. 8 deleted 10 characters in body A far more general theorem 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 all texts on$p$-adic analysis - and I learned it from my number theory professor last semester (Jean-Benoît Bost). This theorem is very 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$f = GH$,$\deg(G - g) \leq n - 1$,$\deg(H - h) \leq m - 1$,$\left\|G - g\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$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$G,H \in O[X]$such that$f = GH$,$\deg(G - g) \leq n - 1$,$\deg(H - h)\leq m - 1$,$\overline{G} = g$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$\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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