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Definition Proof Convergent Sequence Induction Valuation Equality Norm Orders In A Number Field Henselian Field Hensels Lemma

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Proof Of Hensels Lemma

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

	i)	$\displaystyle \vert f'(\alpha_i)\vert = \vert f'(\alpha_0)\vert$
	ii)	$\displaystyle \left\vert\frac{f(\alpha_i)}{f'(\alpha_i)^2}\right\vert \leq D^{2^i}$
	iii)	$\displaystyle \vert \alpha_i - \alpha_0 \vert \leq D$
	iv)	$\displaystyle \alpha_i \in \mathcal{O}_K$

where $D=\left\vert\frac{f(\alpha_0)}{f'(\alpha_0)^2}\right\vert$ .

Proof: All four statements clearly hold when . Suppose they are true for . The proof for essentially uses Taylor's formula. Let $\delta = \left\vert\frac{-f(\alpha_i)}{f'(\alpha_i)}\right\vert$ . Then

$\displaystyle f'(\alpha_{i+1}) = f'(\alpha_i+\delta) = f'(\alpha_i) + {\delta}u$

$\displaystyle f(\alpha_{i+1}) = f(\alpha_i+\delta) = f(\alpha_i) + f'(\alpha_i)\delta + {\delta^2}v$

for $u, v \in \mathcal{O}_K$ . $\vert\delta\vert \leq D^{2^i}\vert f'(\alpha_i)\vert$ by induction, and since

, it follows that $\vert\delta\vert < \vert f'(\alpha_i)\vert$ . Since the norm is non-Archimedean, we see that

$\displaystyle f'(\alpha_{i+1}) = f'(\alpha_i)$

proving i).

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

$\displaystyle \left\vert\frac{f(\alpha_{i+1})}{f'(\alpha_{i+1})^2}\right\vert \... ...ac{\vert f(\alpha_i)\vert}{\vert f'(\alpha_i)\vert^2}\right)^2 \leq D^{2^{i+1}}$

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

To prove iii), note that $\vert\alpha_{i+1}-\alpha_i\vert = \vert\delta\vert$ by the definitions of $\delta$ and $\alpha_{i+1}$ , so $\vert\alpha_{i+1}-\alpha_i\vert \leq D^{2^i}\vert f'(\alpha_i)\vert = D^{2^i}\vert f'(\alpha_0) < D$ when since $D^2 < D = \left\vert\frac{f(\alpha_0)}{f'(\alpha_0)^2}\right\vert$ . So by induction, $\vert\alpha_{i+1}-\alpha_0\vert \leq D$ .

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

1	Equality
1	Definition
1	Proof
1	Convergent Sequence
1	Induction
1	Valuation
3	Orders In A Number Field
3	Norm
7	Henselian Field
13	Hensels Lemma