Hensel's Lemma/First Form
Theorem
Let $p$ be a prime number.
Let $k > 0$ be a positive integer.
Let $f \left({X}\right) \in \Z \left[{X}\right]$ be a polynomial.
Let $x_k \in \Z$ such that:
- $f \left({x_k}\right) \equiv 0 \pmod {p^k}$
- $f' \left({x_k}\right) \not \equiv 0 \pmod p$
Then for every integer $l \ge 0$, there exists an integer $x_{k + l}$ such that:
- $f \left({x_{k + l} }\right) \equiv 0 \pmod {p^{k + l} }$
- $x_{k + l}\equiv x_k \pmod {p^k}$
and any two integers satisfying these congruences are congruent modulo $p^{k + l}$.
Moreover, for all $l\geq0$ and any solutions $x_{k + l}$ and $x_{k + l + 1}$:
- $x_{k + l + 1} \equiv x_{k + l} - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod {p^{k + l + 1} }$
- $x_{k + l + 1} \equiv x_{k + l} \pmod {p^{k + l} }$
Proof
We use induction on $l$.
The base case $l = 0$ is trivial.
Let $l \ge 0$ be such that a solution $x_{k + l}$ exists and is unique up to a multiple of $p^{k + l}$.
Choose a solution $x_{k + l}$ satisfying:
- $f \left({x_{k + l} }\right) \equiv 0 \pmod {p^{k + l} }$
- $x_{k + l} \equiv x_k \pmod {p^k}$
By Congruence by Divisor of Modulus, each solution $x_{k + l + 1}$ is also a solution of the previous congruence.
By uniqueness, it has to satisfy $x_{k + l + 1}\equiv x_{k + l} \pmod {p^{k + l} }$, hence is of the form $x_{k+ l} + t p^{k + l}$ with $t \in \Z$.
Let $d = \deg f$.
We have, for all $t \in \Z$:
| \(\ds f \left({x_{k + l} + t p^{k + l} }\right)\) | \(=\) | \(\ds f \left({x_{k + l} }\right) + t p^{k + l} f' \left({x_{k + l} }\right) + \left({t p^{k + l} }\right)^2 m\) | for some $m \in \Z$, by Taylor Expansion for Polynomials: Order 1 | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds f \left({x_{k+l} }\right) + t p^{k + l} f' \left({x_{k + l} }\right) \pmod {p^{k + l + 1} }\) |
Because $f' \left({x_{k + l} }\right) \equiv f' \left({x_k}\right) \not \equiv 0 \pmod p$, $f' \left({x_{k + l} }\right)$ is invertible modulo $p$.
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Thus $x_{k + l} + t p^{k + l}$ is a solution modulo $p^{k + l + 1}$ if and only if:
- $t \equiv - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod p$
Thus, necessarily:
- $x_{k + l + 1} \equiv x_{k + l} - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod {p^{k + l + 1} }$
which proves the existence and uniqueness.
By induction, we have shown uniqueness and existence for all $l \ge 0$, as well as the relations:
- $x_{k + l + 1} \equiv x_{k + l} - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod {p^{k + l + 1} }$
- $x_{k + l + 1} \equiv x_{k + l} \pmod {p^{k + l} }$
$\blacksquare$