Hensel's Lemma/P-adic Integers/Lemma 5

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:
 * $\map F {\alpha_0} \equiv 0 \pmod {p\Z_p}$
 * $\map {F'} {\alpha_0} \not\equiv 0 \pmod {p\Z_p}$

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:
 * $\tuple{b_0, b_1, \ldots, b_{k-1}} \in S_k$.

Then there exists a unique $p$-adic digit $\map b {b_0, b_1, \ldots, b_{k-1}}$:
 * $\tuple{b_0, b_1, \ldots, b_{k-1}, \map b {b_0, b_1, \ldots, b_{k-1}}} \in S_{k+1}$

Proof
Let:
 * $a_{k-1} = \ds \sum_{n = 0}^{k-1} b_n p_n$

By hypothesis:
 * $\map F {a_{k-1}} \equiv 0 \pmod{p^k\Z_p}$

By definition of congruence modulo an ideal:
 * $\map F {a_{k-1}} \in p^k\Z_p$

By definition of principal ideal:
 * $\exists \beta_k \in \Z_p : \map F {a_{k-1}} = \beta_k p^k$

By hypothesis:
 * $a_{k-1} \equiv \alpha_0 \pmod{p\Z_p}$

From User:Leigh.Samphier/Sandbox/Polynomials of Congruent Ring Elements are Congruent:
 * $\map {F'} {a_{k-1}} \equiv \map {F'} {\alpha_0} \pmod{p\Z_p}$

By hypothesis:
 * $\map {F'} {\alpha_0} \not\equiv 0 \pmod{p\Z_p}$

Hence:
 * $\map {F'} {a_{k-1}} \not\equiv 0 \pmod{p\Z_p}$

It follows that:
 * $\map {F'} {a_{k-1}} \ne 0$

Lemma 6

 * $\exists b_k \in T : b_k p^k \equiv \dfrac {-\beta_k p^k} {\map {F'} {a_{k-1}}} \pmod{p^{k+1}\Z_p}$

Let $a_k = a_{k-1} + b_kp^k$.

We have:

Let $\map F X = \ds \sum_{j \mathop = 0}^r c_j X^j$ where $X$ is the indeterminate and $c_0, c_1, \ldots, c_r \in \Z_p$.

Then:
 * $\map {F'} X = \ds \sum_{j \mathop = 1}^r j c_j X^j$

We have:

Let $\map b {b_0, b_1, \ldots, b_{k-1}} = b_k$.

It follows that:
 * $\tuple{b_0, b_1, \ldots, b_{k-1}, \map b {b_0, b_1, \ldots, b_{k-1}}} \in S_{k+1}$