Hensel's lemma

For a complete discrete valuation ring $R$, a simple root of a polynomial modulo the maximal ideal lifts uniquely to a root in $R$.

A monic polynomial over a complete discrete valuation ring factors with prescribed coprime monic residue factors. Given monic lifts $g_r,h_r$ correct modulo ${\pi }^r$, the degree-bounded map $(u,v)\mapsto u{\bar{h}}_r+v{\bar{g}}_r$ is a bijection over the residue field, by coprimality and equality of dimensions. Solve it for the error $(f-g_r{h}_r)/{\pi }^r$, and replace the lifts by $g_r+{\pi }^{r}u,h_r+{\pi }^{r}v$. Their coefficients converge by completeness and the limiting product is $f$. The completeness assumption cannot be replaced just by localization.

For a complete discrete valuation ring $R$, let $f(a_1)\in (\pi )$ and let $f^{\prime }(a_1)$ be a unit. If $f(a_n)\in ({\pi }^n)$, choose $c$ modulo $\pi$ solving $f(a_n)/{\pi }^n+c{f}^{\prime }(a_n)\equiv 0$ and put $a_{n+1}=a_n+c{\pi }^n$. Taylor expansion gives $f(a_{n+1})\in ({\pi }^{n+1})$. The sequence is Cauchy and its limit is a root. If two roots lift the same residue, factoring $f(x)-f(y)$ leaves $x-y$ times a unit, proving uniqueness.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension of an absolute value to every finite extension. Factor a monic $g\in {\mathcal{O}}_K[X]$ into monic irreducible factors over $K$. All factors have integral coefficients because their roots have absolute value at most one. By pure-power reduction of a monic irreducible polynomial, each factor reduces to a power of a single irreducible polynomial. If $\overline{g}={\overline{g}}_1{\overline{g}}_2$ is a coprime monic factorization, assign each whole irreducible factor of $g$, with its multiplicity, to the unique side containing its residue factor. The resulting monic $g_1,g_2\in {\mathcal{O}}_K[X]$ satisfy $g=g_1{g}_2$ and reduce to the prescribed factors. Completeness of $K$ is not required for this proof.

For a non-Archimedean local field $K$, an element of $K^{\times }$ is an $m$th power for infinitely many positive integers $m$ exactly when it belongs to ${\mathcal{O}}_K^{\times }$. Valuations prove necessity; for sufficiency, choose infinitely many $m$ coprime to the residue characteristic and to $|k^{\times }|$, for which exponentiation by $m$ is an automorphism of the unit group.

The polynomial $X^4+9X^2-2$ has three irreducible factors over ${\mathbb{Q}}_2$, of degrees one, one, and two. Its quadratic in $X^2$ has one unit root congruent to one modulo eight, hence a square, and one root of valuation one, whose associated quadratic factor is Eisenstein.

Let $K$ be complete for a discrete valuation $v$. If $f\in {\mathcal{O}}_K[X]$ andthen Newton iteration over a valued field converges to a root $\alpha \in {\mathcal{O}}_K$ with $v(\alpha -a)>v(f^{\prime }(a))$.

Starting from $a_0$, Newton iteration sets $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Under the strong form of Hensel lemma inequality, the correction valuations grow at least geometrically, so $(a_n)$ is a Cauchy sequence converging to a root.