# proof of Hensel’s lemma

Lemma:  Using the setup and terminology of the statement of Hensel’s Lemma, for $i\geq 0$,

 i) $\displaystyle|f^{\prime}(\alpha_{i})|=|f^{\prime}(\alpha_{0})|$ ii) $\displaystyle\left|\frac{f(\alpha_{i})}{f^{\prime}(\alpha_{i})^{2}}\right|\leq D% ^{2^{i}}$ iii) $\displaystyle|\alpha_{i}-\alpha_{0}|\leq D$ iv) $\displaystyle\alpha_{i}\in\mathcal{O}_{K}$

where $D=\left|\frac{f(\alpha_{0})}{f^{\prime}(\alpha_{0})^{2}}\right|$.

Proof: All four statements clearly hold when $i=0$. Suppose they are true for $i$. The proof for $i+1$ essentially uses Taylor’s formula. Let $\delta=\left|\frac{-f(\alpha_{i})}{f^{\prime}(\alpha_{i})}\right|$. Then

 $f^{\prime}(\alpha_{i+1})=f^{\prime}(\alpha_{i}+\delta)=f^{\prime}(\alpha_{i})+% {\delta}u$
 $f(\alpha_{i+1})=f(\alpha_{i}+\delta)=f(\alpha_{i})+f^{\prime}(\alpha_{i})% \delta+{\delta^{2}}v$

for $u,v\in\mathcal{O}_{K}$. $|\delta|\leq D^{2^{i}}|f^{\prime}(\alpha_{i})|$ by induction, and since $D<1$, it follows that $|\delta|<|f^{\prime}(\alpha_{i})|$. Since the norm is non-Archimedean, we see that

 $f^{\prime}(\alpha_{i+1})=f^{\prime}(\alpha_{i})$

proving i).

$f(\alpha_{i})+f^{\prime}(\alpha_{i})\delta=0$ by definition of $\delta$, so $f(\alpha_{i+1})={\delta^{2}}v$ and hence $|f(\alpha_{i+1})|\leq|\delta^{2}|$. Hence

 $\left|\frac{f(\alpha_{i+1})}{f^{\prime}(\alpha_{i+1})^{2}}\right|\leq\frac{|% \delta|^{2}}{|f^{\prime}(\alpha_{i+1})|^{2}}=\frac{|\delta|^{2}}{|f^{\prime}(% \alpha_{i})|^{2}}=\left(\frac{|\delta|}{|f^{\prime}(\alpha_{i})|}\right)^{2}=% \left(\frac{|f(\alpha_{i})|}{|f^{\prime}(\alpha_{i})|^{2}}\right)^{2}\leq D^{2% ^{i+1}}$

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

To prove iii), note that $|\alpha_{i+1}-\alpha_{i}|=|\delta|$ by the definitions of $\delta$ and $\alpha_{i+1}$, so $|\alpha_{i+1}-\alpha_{i}|\leq D^{2^{i}}|f^{\prime}(\alpha_{i})|=D^{2^{i}}|f^{% \prime}(\alpha_{0}) when $i>0$ since $D^{2}. So by induction, $|\alpha_{i+1}-\alpha_{0}|\leq D$.

Finally, to prove iv) and the proof of the lemma, $\delta\in\mathcal{O}_{K}$ since $|\delta|<\left|\frac{f(\alpha_{0})}{f^{\prime}(\alpha_{0})}\right|\leq 1$ and hence is in the valuation ring of $K$. So by induction, $\alpha_{i+1}=\alpha_{i}+\delta\in\mathcal{O}_{K}$.

Proof of Hensel’s Lemma:

To prove Hensel’s lemma from the above lemma, note that $\delta=\delta_{i}\to 0$ since $|\delta|\leq D^{2^{i}}|f^{\prime}(\alpha_{0})|$, so $\{\alpha_{i}\}$ converges to $\alpha\in\mathcal{O}_{K}$ since $K$ is complete. Thus $f(\alpha_{i})\to f(\alpha)$ by continuity. But $|f(\alpha_{i})|\leq|\delta^{2}|=D^{2^{i+1}}|f^{\prime}(\alpha_{0})|$, so $|f(\alpha_{i})|\to 0$, so $f(\alpha)=0$ and the proof is complete.

Title proof of Hensel’s lemma ProofOfHenselsLemma 2013-03-22 15:32:16 2013-03-22 15:32:16 rm50 (10146) rm50 (10146) 5 rm50 (10146) Proof msc 13H99 msc 12J99 msc 11S99