proof of Hensel’s lemma

Lemma:  Using the setup and terminology of the statement of Hensel’s Lemma, for i0,

i) |f(αi)|=|f(α0)|
ii) |f(αi)f(αi)2|D2i
iii) |αi-α0|D
iv) αi𝒪K

where D=|f(α0)f(α0)2|.

Proof: All four statements clearly hold when i=0. Suppose they are true for i. The proof for i+1 essentially uses Taylor’s formulaMathworldPlanetmathPlanetmath. Let δ=|-f(αi)f(αi)|. Then


for u,v𝒪K. |δ|D2i|f(αi)| by inductionMathworldPlanetmath, and since D<1, it follows that |δ|<|f(αi)|. Since the norm is non-Archimedean, we see that


proving i).

f(αi)+f(αi)δ=0 by definition of δ, so f(αi+1)=δ2v and hence |f(αi+1)||δ2|. Hence


where the last equality follows by induction. This proves ii).

To prove iii), note that |αi+1-αi|=|δ| by the definitions of δ and αi+1, so |αi+1-αi|D2i|f(αi)|=D2i|f(α0)<D when i>0 since D2<D=|f(α0)f(α0)2|. So by induction, |αi+1-α0|D.

Finally, to prove iv) and the proof of the lemma, δ𝒪K since |δ|<|f(α0)f(α0)|1 and hence is in the valuation ringMathworldPlanetmathPlanetmath of K. So by induction, αi+1=αi+δ𝒪K.

Proof of Hensel’s Lemma:

To prove Hensel’s lemma from the above lemma, note that δ=δi0 since |δ|D2i|f(α0)|, so {αi} converges to α𝒪K since K is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Thus f(αi)f(α) by continuity. But |f(αi)||δ2|=D2i+1|f(α0)|, so |f(αi)|0, so f(α)=0 and the proof is complete.

Title proof of Hensel’s lemma
Canonical name ProofOfHenselsLemma
Date of creation 2013-03-22 15:32:16
Last modified on 2013-03-22 15:32:16
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 5
Author rm50 (10146)
Entry type Proof
Classification msc 13H99
Classification msc 12J99
Classification msc 11S99