proof of Hensel’s lemma

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

i)$|f^{\prime }({\alpha }_i)|=|f^{\prime }({\alpha }_0)|$

ii)$\left|{\displaystyle \frac{f({\alpha }_i)}{f^{\prime }{({\alpha }_i)}^2}}\right|\le D^{2^i}$

iii)$|{\alpha }_i-{\alpha }_0|\le D$

iv)${\alpha }_i\in {𝒪}_K$

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 $u,v\in {𝒪}_K$. $|\delta |\le D^{2^i}|f^{\prime }({\alpha }_i)|$ by induction, and since , it follows that . 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 $|f({\alpha }_{i+1})|\le |{\delta }^2|$. Hence

$$\left|\frac{f({\alpha }_{i+1})}{f^{\prime }{({\alpha }_{i+1})}^2}\right|\le \frac{{|\delta |}^2}{{|f^{\prime }({\alpha }_{i+1})|}^2}=\frac{{|\delta |}^2}{{|f^{\prime }({\alpha }_i)|}^2}={\left(\frac{|\delta |}{|f^{\prime }({\alpha }_i)|}\right)}^2={\left(\frac{|f({\alpha }_i)|}{{|f^{\prime }({\alpha }_i)|}^2}\right)}^2\le D^{2^{i+1}}$$

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 when $i>0$ since . So by induction, $|{\alpha }_{i+1}-{\alpha }_0|\le D$.

Finally, to prove iv) and the proof of the lemma, $\delta \in {𝒪}_K$ since and hence is in the valuation ring of $K$. So by induction, ${\alpha }_{i+1}={\alpha }_i+\delta \in {𝒪}_K$.

Proof of Hensel’s Lemma:

To prove Hensel’s lemma from the above lemma, note that $\delta ={\delta }_i\to 0$ since $|\delta |\le D^{2^i}|f^{\prime }({\alpha }_0)|$, so $\{{\alpha }_i\}$ converges to $\alpha \in {𝒪}_K$ since $K$ is complete. Thus $f({\alpha }_i)\to f(\alpha )$ by continuity. But $|f({\alpha }_i)|\le |{\delta }^2|=D^{2^{i+1}}|f^{\prime }({\alpha }_0)|$, so $|f({\alpha }_i)|\to 0$, so $f(\alpha )=0$ and the proof is complete.