Valuation Theory and Hensel's Lemma

Definition of $p$ -adic numbers. For a prime number $p$ , define

Z_{p} = {\sum_{i = 0}^{+ \infty} d_{i} p^{i} : d_{i} \in {0, 1, \dots, p - 1}}, Q_{p} = {\sum_{i = m}^{+ \infty} d_{i} p^{i} : d_{i} \in {0, 1, \dots, p - 1}, m \in Z} .

$Z_{p}$ is a ring and $Q_{p}$ is a field, under the normal addition and multiplication of formal series. There is a canonical embedding $Q ↪ Q_{p}$ .

Analytic Approach

Define a valuation ${| \cdot |}_{p} : Q \to R_{\geq 0}$ (i.e. non-negative, multiplicative, triangle inequality) for a prime $p$ as follows: $| x |_{p} = p^{- e}$ where $x = p^{e} \cdot \frac{a}{b}$ for integers $a$ and $b$ coprime to $p$ . We define $Q_{p}$ to be the completion (i.e. set of Cauchy sequences modulo the equivalence relation $x \sim y$ defined by $lim | x_{i} - y_{i} |_{p} = 0$ ) of $Q$ under ${| \cdot |}_{p}$ . In this sense $Z_{p} = {x \in Q_{p} : | x |_{p} \leq 1}$ .

Lemma. $Z_{p}$ is the integral closure of $Z$ in $Q_{p}$ .

There's a canonical embedding $Z ↪ Z_{p}$ .

Lemma. The canonical embedding induces a ring isomorphism $Z / p^{n} Z ≃ Z_{p} / p^{n} Z_{p}$ . Hence

Z_{p} = \underset{n}{\lim_{\leftarrow}} {Z / p^{n} Z}

Proposition (special case of Hensel's lemma). Suppose $f \in Z_{p} [X]$ . Let $\bar{f} = f mod p \in F_{p} [X]$ . If $α \in F_{p}$ with $\bar{f} (α) = 0$ and $\bar{f} (α) \neq 0$ , there is a unique $x \in Z_{p}$ with $x \equiv α (\mod p)$ and $f (x) \equiv 0$ .

Proof. Existence. Fix $x_{0} \in Z_{p}$ such that $x_{0} \equiv α (\mod p)$ . Then $f (x_{0}) \in p Z_{p}$ and $f^{'} (x_{0}) \in Z_{p}^{\times}$ . Let

x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}

Then

\begin{aligned} f (x_{1}) = & f (x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}) \\ = & f (x_{0}) - f^{'} (x_{0}) \cdot \frac{f (x_{0})}{f^{'} (x_{0})} + O ({(\frac{f (x_{0})}{f^{'} (x_{0})})}^{2}) \end{aligned}

hence $f (x_{1}) \in p^{2} Z$ and $f^{'} (x_{1}) \in Z_{p}^{\times}$ . Repeating this process we get a convergent sequence $x_{0}, x_{1}, \dots$ . Suppose its limit is $x$ , then we have $f (x) = 0$ .

Uniqueness is trivial (notice that $α$ is a simple root).

Corollary. There is a unique group homomorphism $ω : F_{p}^{\times} \to Z_{p}^{\times}$ such that $ω (α) \equiv α (\mod p)$ for every $α \in F_{p}^{\times}$ .

Proof. Apply the proposition for $f (x) = x^{p - 1} - 1$ .