Exercise 2.6.9 (Complement to Hensel's lemma)

Proof. By hypothesis, $f(a)\equiv 0(\mathit{mod}{p}^j)$ for some $j\ge 1$. In particular $f(a)\equiv 0(\mathit{mod}p)$. Since $f^{\prime }(a)\not\equiv 0(\mathit{mod}p)$, the Hensel’s lamma shows that there exists for every positive integer $k$ a unique solution $a_k$ modulo $p^k$ of

which lifts $a$ modulo $p^k$.

Here $a\equiv a_j(\mathit{mod}{p}^j)$, by the unicity of the solution modulo $p^j$.

We define $c=a_{2j}$, so that $c$ is the solution modulo $p^{2j}$ of

A fortiori, $f(c)\equiv 0(\mathit{mod}{p}^j)$ (and $c\equiv a(\mathit{mod}p)$). The unicity of the solution modulo $p^j$ shows that

so that $c=a+t{p}^j$ for some integer $t$.

The Taylor’s expansion gives

where $n=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$, and $f^{(i)}(a)∕i!$ are integers for $0\le i\le n$ (see (2.3) page 86). Therefore

Since $f(c)=f(a+t{p}^j)\equiv 0(\mathit{mod}{p}^{2j})$,

By definition, $\overline{f^{\prime }(a)}$ is an integer chosen so that $f^{\prime }(a)\overline{f^{\prime }(a)}\equiv 1(\mathit{mod}{p}^{2j})$. Multiplying both members by $\overline{f^{\prime }(a)}$, we obtain

So

Since $c\equiv b(\mathit{mod}{p}^{2j})$, and $f(c)\equiv 0(\mathit{mod}{p}^{2j})$, we obtain

If $b=a-f(a)\overline{f^{\prime }(a)}$, then $f(b)\equiv 0(\mathit{mod}{p}^{2j})$. □