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|cdotright| a non-archimedean absolute value on K for which K is complete, mathcalO the associated valuation ring, mathcalM the maximal ideal, pi a uniformizer. Let PhiiinmathcalO[X1,,cdots,Xn] for 1leqileqn and consider the map Phi=(Phi1,,cdots,Phin):mathcalOntomathcalOn. Let J be the Jacobian det(partialPhii/partialXj)inmathcalO[X1,,cdots,Xn].
Theorem. If x0inmathcalOn, y0=Phi(x0) and J(x0)neq0, then for any Rin(0,left|J(x0)right|), Phi induces a bijection overlineB(x0,R)toy0+(DPhi)(x0)overlineB(0,R)
(I use the standard notations overlineB and Bcirc for closed and open balls respectively.)
The proof uses in an essential way the Picard fixed point theorem.
Corollary 1. Take n=1, Phi1=P, x0=alpha, varepsilonin(0,1). Suppose that left|P(alpha)right|leqvarepsilonleft|P′(alpha)right|2. Then there exists a unique betainmathcalO such that P(beta)=0 and left|beta−alpharight|leqvarepsilonleft|P′(alpha)right|. (We take R=varepsilonleft|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−alpharight|<left|P′(alpha)right|.
As an even more special case, if P′(alpha)inmathcalOtimes and left|P′(alpha)right|<1, there exists betainmathcalO such that P(beta)=0 and left|beta−alpharight|<1. Restating this in terms of the residue field: a simple zero in the residue field can be lifted to a real zero in mathcalO. 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,hinmathcalO[X] such that degg=n, degh=m and degf=degg+degh=n+m. Assume that there exists varepsilonin(0,1) such that left|f−ghright|leqvarepsilonleft|textRes(g,h)right|2 and deg(f−gh)leqm+n−1. Then there exist G,HinmathcalO[X] such that f=GH, deg(G−g)leqn−1, deg(H−h)leqm−1, and also left|G−gright|leqvarepsilonleft|textRes(g,h)right| and left|H−hright|leqvarepsilonleft|textRes(g,h)right|. (Obviously textRes denotes the resultant here, and left|cdotright| the Gauss norm.)
To prove this: write G=g+xi and H=h+eta where xi and eta are polynomials with coefficients in mathcalO and have degrees leqn−1 and leqm−1 respectively. Then f=GH if and only if f=(g+xi)(h+eta). It can be seen as a map from mathcalOntimesmathcalOmtomathcalOn+m given by polynomials. So consider the map Phi:(xi,eta)mapsto(g+xi)(h+eta)−f. We have also textRes(g,h)=det((xi,eta)mapstogxi+heta)). It is easy to see that the theorem above then gives the result.
As a corollary: if f, g and h satisfy overlinef=overlinegoverlineh - where overlinef is f reduced modulo mathcalM et cetera - and if overlineg and overlineh are coprime (this is a condition on the resultant!) then there exist G,HinO[X] satisfying the following conditions: f=GH, deg(G−g)leqn−1, deg(H−h)leqm−1, overlineG=g and overlineH=h. Hence "a factorization over the residue field lifts to a factorization over mathcalO" (under the right conditions).
Corollary 3. Finally, let us come to the motivation for the question: the more general result is that if PinK[X] is irreducible, then left|Pright| (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)=sumni=0aiXn−iinK[X]. Suppose WLOG that left|Pright|=1. Let mathbbF be the residue field and let overlineP be the image of P modulo mathcalM. Set r=minn:overlinean−rneq0. Then we have in the residue field the factorization overlineP(X)=Xrleft(overlinean−r+overlinean−r−1X+cdots+overlinea0Xn−rright) 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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