Hensel's Lemma/P-adic Integers/Lemma 9

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:
 * $\paren{x, y \in T : \tuple{b_0, b_1, \ldots, b_{k-1}, x}, \tuple{b_0, b_1, \ldots, b_{k-1}, y} \in S_{k+1}} \implies x = y$

Lemma 8
Let:
 * $a = \ds \sum_{n = 0}^{k-1} b_n p^n$

We have:
 * $\map F a \equiv 0 \pmod{p^k\Z_p}$

and
 * $a \equiv \alpha_0 \pmod{p\Z_p}$

Let:
 * $x' = a + xp^k$

and
 * $y' = a + yp^k$

We have:
 * $\map F {x'} \equiv \map F {y'} \equiv 0 \pmod{p^{k+1}\Z_p}$

and
 * $x' \equiv y' \equiv \alpha_0 \pmod{p\Z_p}$

From Lemma 8:
 * $\map F {x'} = \map F {a + xp^k} \equiv \map F a + xp^k \map {F'} a \pmod{p^{k+1}\Z_p}$
 * $\map F {y'} = \map F {a + yp^k} \equiv \map F a + yp^k \map {F'} a \pmod{p^{k+1}\Z_p}$


 * $0 \equiv \map F {x'} - \map F {y'} \equiv xp^k \map {F'} a - yp^k \map {F'} a \pmod{p^{k+1}\Z_p}$


 * $xp^k \map {F'} a - yp^k \map {F'} a = p^k\paren{x\map {F'} a - y\map {F'} a} \in p^{k+1}\Z_p$


 * $x\map {F'} a - y\map {F'} a \in p\Z_p$


 * $\map {F'} a \paren{x - y} \in p\Z_p$

From === Lemma 7:
 * $\exists b \in T : \map {F'} a \equiv b \pmod{p\Z_p}$


 * $a \equiv \alpha_0 \pmod{p\Z_p} \implies \map {F'} a \equiv \map {F'} {\alpha_0} \pmod{p\Z_p}$


 * $\map {F'} a \equiv b \pmod{p\Z_p} \implies b \equiv \map {F'} {\alpha_0} \pmod{p\Z_p}$


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


 * $\map {F'} a \equiv b \pmod{p\Z_p} \implies \map {F'} a \paren{x - y} \equiv b \paren{x - y} \pmod {p\Z_p} \implies b \paren{x - y} \equiv 0 \pmod {p\Z_p} \implies b \paren{x - y} \in p\Z_p \implies p \divides b \paren{x - y}$


 * $b \not\equiv 0 \pmod {p\Z_p} \implies p \nmid b$


 * $\leadsto p \divides x - y$


 * $\leadsto x = y$