# 亨泽尔引理

## 定理内容

${\displaystyle f(x)}$係數多項式${\displaystyle k}$為不少於2的整數，${\displaystyle p}$質數。若整數${\displaystyle r}$是下面同餘式的根：

${\displaystyle f(r)\equiv 0{\pmod {p^{k-1}}}.}$

${\displaystyle f(r+tp^{k-1})\equiv 0{\pmod {p^{k}}}}$ （I）

，則有：

• ${\displaystyle f'(r)\not \equiv 0{\pmod {p}}}$，則存在唯一的整數${\displaystyle 0\leq t\leq p-1}$使得（I）成立。
${\displaystyle tf'(r)\equiv -(f(r)/p^{k-1}){\pmod {p}}.\,}$
• ${\displaystyle f'(r)\equiv 0{\pmod {p}}}$${\displaystyle f(r)\equiv 0{\pmod {p^{k}}}}$ ，則（I）對任意整數t成立。
• ${\displaystyle f'(r)\equiv 0{\pmod {p}}}$${\displaystyle f(r)\not \equiv 0{\pmod {p^{k}}}}$，則（I）無整數解。

## 證明

Hensel引理可用泰勒公式證明。

${\displaystyle f(r+tp^{k-1})=f(r)+tp^{k-1}f'(r)+{\frac {1}{2}}t^{2}p^{2(k-1)}f''(r)+{\frac {1}{6}}t^{3}p^{3(k-1)}f'''(r)+...}$

${\displaystyle f(r+tp^{k-1})\equiv f(r)+tp^{k-1}f'(r){\pmod {p^{k}}}}$

## 推廣

${\displaystyle K}$為完備局域。設 ${\displaystyle {\mathcal {O}}_{K}}$${\displaystyle K}$的整數環，設${\displaystyle f(x)}$為係數在 ${\displaystyle {\mathcal {O}}_{K}}$的多項式，若存在 ${\displaystyle \alpha _{0}\in {\mathcal {O}}_{K}}$使得

${\displaystyle |f(\alpha _{0})|<|f'(\alpha _{0})|^{2}}$

${\displaystyle f(x)}$有根${\displaystyle \alpha \in K}$

1. ${\displaystyle \alpha _{i+1}=\alpha _{i}-{\frac {f(\alpha _{i})}{f'(\alpha _{i})}}}$ 趨近${\displaystyle \alpha }$
2. ${\displaystyle |\alpha -\alpha _{0}|\leq |{\frac {f(\alpha _{i})}{f'(\alpha _{i})}}|<1}$