Why does Hensel's lemma imply that $\sqrt{2}\in\mathbb{Q_7}$?
I understand Hensel's lemma, namely:
Let $f(x)$ be a polynomial with integer coefficients, and let $m$, $k$ be positive integers such that $m \leq k$. If $r$ is an integer such that $f(r) \equiv 0 \pmod{p^k}$ and $f'(r) \not\equiv 0 \pmod{p}$ then there exists an integer $s$ such that $f(s) \equiv 0 \pmod{p^{k+m}}$ and $r \equiv s \pmod{p^{k}}$.
But I don't see how this has anything to do with $\sqrt{2}\in\mathbb{Q_7}$?
I know a $7$-adic number $\alpha$ is a $7$-adically Cauchy sequence $a_n$ of rational numbers. We write $\mathbb{Q}_7$ for the set of $7$-adic numbers.
A sequence $a_n$ of rational numbers is $p$-adically Cauchy if $|a_{m}-a_n|_p \to 0$ as $n \to \infty$.
How do we show $\sqrt{2}\in\mathbb{Q_7}$?
Best Answer
Here's an easier way to understand Hensel's lemma: Let $f(X) \in \mathbb{Z}_p[X]$ be a primitive polynomial such that $\overline{f} = \overline{g} \cdot \overline{h}$ in $\mathbb{Z}/p\mathbb{Z}[X] \cong \mathbb{Z}_p / p\mathbb{Z}_p[X]$, and $\overline{g}$ and $\overline{h}$ are coprime.
Then $f$ splits as $f = g \cdot h$ in $\mathbb{Z}_p[X]$, where $\deg g = \deg \overline{g}$ or $\deg h = \deg \overline{h}$, and $g \equiv \overline{g}, \; h \equiv \overline{h}$ mod $p$.
Use this on $f(X) = X^2 - 2$. Since this splits in $\mathbb{Z}/7\mathbb{Z}$ as $(X-3)(X+3)$, it splits into linear factors in $\mathbb{Z}_7[X]$.