Hensel's Lemma/P-adic Integers/Lemma 1

Theorem
Let $\Z_p$ be the $p$-adic integers for some prime $p$.

Let $\map F X \in \Z_p \sqbrk X$ be a polynomial.

Let $\map {F'} X$ be the (formal) derivative of $F$.

Let $\alpha_0 \in \Z_p$ be a $p$-adic integer:
 * $\map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
 * $\map {F'} {\alpha_0} \not\equiv 0 \pmod {p\Z_p}$

Then:
 * There exists a unique $p$-adic expansion $\ds \sum_{n = 0}^\infty d_n p^n$:
 * $\forall k : a_k = \ds \sum_{n = 0}^k d_n p^n$ satisfies:
 * $(1) \quad \map F {a_k} \equiv 0 \pmod {p^{k+1}\Z_p}$
 * $(2) \quad a_k \equiv \alpha_0 \pmod {p\Z_p}$

Proof
The Second Principle of Recursive Definition is used to construct the sequence $\sequence {b_n}$.

Let $T$ be the set of $p$-adic digits.

For each $k \in \N_{>0}$, let:
 * $S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {\ds \sum_{n = 0}^{k-1} b_n p^n } \equiv 0 \pmod{p^k\Z_p} \quad \text{and} \quad \ds \sum_{n = 0}^{k-1} b_n p^n \equiv \alpha_0 \pmod{p\Z_p}}$

Let $d_0$ be the first $p$-adic digit of the canonical expansion of $\alpha_0$.

Lemma 5
For $k \in \N_{>0}$, let $G_k : T^k \to T$ be the mapping defined by:
 * $\map {G_k} {b_0, b_1, \ldots, b_{k-1}} = \begin{cases}

\map b {b_0, b_1, \ldots, b_{k-1}} & : \tuple {b_0, b_1, \ldots, b_{k-1}} \in S_k\\ 0 & : \tuple {b_0, b_1, \ldots, b_{k-1}} \notin S_k \end{cases}$

From Lemma 5:
 * $G_k$ is well-defined for all $k \in \N$

and
 * $\forall \tuple {b_0, b_1, \ldots, b_{k-1}} \in S_k$:
 * $\tuple {b_0, b_1, \ldots, b_{k-1}, \map {G_k} {b_0, b_1, \ldots, b_k} } \in S_{k+1}$

From Second Principle of Recursive Definition, there exists exactly one mapping $d: \N \to T$ such that:
 * $\forall x \in \N: \map d x = \begin{cases}

d_0 & : x = 0 \\ \map {G_n} {\map d 0, \ldots, \map d n} & : x = n + 1 \end{cases}$

We have:
 * $(a)\quad \tuple{d_0} \in S_0$
 * $(b)\quad \tuple{d_0, d_1, \ldots, d_{k-1}} \in S_k \implies \tuple{d_0, d_1, \ldots, d_k} \in S_{k+1}$

From Principle of Mathematical Induction:
 * $\forall k \in N_{>0}: \tuple{d_0, d_1, \ldots, d_{k-1}} \in S_k$

That is, for all $k \in N$:
 * $a_k = \ds \sum_{n = 0}^k d_n p_n$ satisfies:
 * $(a) \quad \map F {a_k} \equiv 0 \pmod {p^{k+1}\Z_p}$
 * $(b) \quad a_k \equiv \alpha_0 \pmod {p\Z_p}$