Lemma: Using the setup and terminology of the statement of Hensel's Lemma, for $i\geq 0$ , \begin{eqnarray*} & \mbox{i) } & |f'(\alpha_i)| = |f'(\alpha_0)| \\ & \mbox{ii) } & \left|\frac{f(\alpha_i)}{f'(\alpha_i)^2}\right| \leq D^{2^i} \\ & \mbox{iii) } & | \alpha_i - \alpha_0 | \leq D \\ & \mbox{iv) } & \alpha_i \in \mathcal{O}_K \end{eqnarray*}where $ D=\left|\frac{f(\alpha_0)}{f'(\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 $\delta = \left|\frac{-f(\alpha_i)}{f'(\alpha_i)}\right|$ . Then $$f'(\alpha_{i+1}) = f'(\alpha_i+\delta) = f'(\alpha_i) + {\delta}u$$ $$f(\alpha_{i+1}) = f(\alpha_i+\delta) = f(\alpha_i) + f'(\alpha_i)\delta + {\delta^2}v$$ for $u, v \in \mathcal{O}_K$ . $|\delta| \leq D^{2^i}|f'(\alpha_i)|$ by induction, and since $D < 1$ , it follows that $|\delta| < |f'(\alpha_i)|$ . Since the norm is non-Archimedean, we see that $$f'(\alpha_{i+1}) = f'(\alpha_i)$$ proving i).

$f(\alpha_i)+f'(\alpha_i)\delta = 0$ by definition of $\delta$ , so $f(\alpha_{i+1}) = {\delta^2}v$ and hence $|f(\alpha_{i+1})| \leq |\delta^2|$ . Hence $$\left|\frac{f(\alpha_{i+1})}{f'(\alpha_{i+1})^2}\right| \leq \frac{|\delta|^2}{|f'(\alpha_{i+1})|^2} = \frac{|\delta|^2}{|f'(\alpha_i)|^2}=\left(\frac{|\delta|}{|f'(\alpha_i)|}\right)^2 = \left(\frac{|f(\alpha_i)|}{|f'(\alpha_i)|^2}\right)^2 \leq D^{2^{i+1}}$$ 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 $|\alpha_{i+1}-\alpha_i| \leq D^{2^i}|f'(\alpha_i)| = D^{2^i}|f'(\alpha_0) < D$ when $i>0$ since $D^2 < D = \left|\frac{f(\alpha_0)}{f'(\alpha_0)^2}\right|$ . So by induction, $|\alpha_{i+1}-\alpha_0| \leq D$ .

Finally, to prove iv) and complete the proof of the lemma, $\delta \in \mathcal{O}_K$ since $|\delta| < \left|\frac{f(\alpha_0)}{f'(\alpha_0)}\right| \leq 1$ 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 $|\delta| \leq D^{2^i}|f'(\alpha_0)|$ , so $\{\alpha_i\}$ converges to $\alpha \in \mathcal{O}_K$ since $K$ is complete. Thus $f(\alpha_i)\to f(\alpha)$ by continuity. But $|f(\alpha_i)| \leq |\delta^2| = D^{2^{i+1}}|f'(\alpha_0)|$ , so $|f(\alpha_i)| \to 0$ , so $f(\alpha)=0$ and the proof is complete.