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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.

A suitable version of the Hensel lemma is this: if $R$ is a complete discrete valuation ring, $h\in R[X]$, and $\overline{a}$ is a root of $\overline{h}$ with ${\overline{h}}^{\prime }(\overline{a})\ne 0$, there is a unique $a\in R$ reducing to $\overline{a}$ with $h(a)=0$. Starting with any lift, Newton iteration over a valued field $a_{j+1}=a_j-h(a_j)/h^{\prime }(a_j)$ keeps the derivative a unit and doubles the error valuation, giving convergence. Uniqueness follows by factoring $h(u)-h(v)$ when two roots have the same residue.

The field in the rest of this question is not assumed complete, so that version cannot simply be applied to it. Instead use the stated uniqueness of the extension of an absolute value. Let ${\alpha }_i,{\alpha }_j$ be any two roots of a polynomial $f$, which is irreducible. The field embedding sending ${\alpha }_i$ to ${\alpha }_j$ identifies $K({\alpha }_i)$ with $K({\alpha }_j)$. Pulling the latter field's absolute value back gives an extension on $K({\alpha }_i)$, which must coincide with the given one. More generally, for every $h\in K[X]$,This proves equal absolute values of algebraic conjugates without assuming separability or a transitive action of a Galois group; repeated roots cause no difficulty.

Every root of the monic $f\in {\mathcal{O}}_K[X]$ has absolute value at most one. Indeed, if $|\alpha |>1$, then each lower term satisfies $|c_j{\alpha }^j|\le |\alpha {|}^j<|\alpha {|}^n$. The ultrametric inequality makes the leading term dominate their sum, contradicting $f(\alpha )=0$. Thus

All roots lie in the valuation ring ${\mathcal{O}}_M$, so their residues are defined in the residue field $k_M$. Let $\phi \in k_K[X]$ be the monic minimal polynomial of an algebraic element of ${\overline{\alpha }}_1$; this residue is algebraic because it satisfies the monic $\overline{f}$. Choose a coefficientwise lift $\Phi \in {\mathcal{O}}_K[X]$. Then $|\Phi ({\alpha }_1)|<1$, and the conjugacy argument gives $|\Phi ({\alpha }_i)|<1$ for every $i$. Hence every ${\overline{\alpha }}_i$ is a root of $\phi$.

Reduction of the splitting-field factorization, with multiplicities, givesEvery monic irreducible factor of $\overline{f}$ over $k_K$ has a root among these residues. That root also satisfies $\phi$, so its minimal polynomial is $\phi$. Unique factorization therefore proves the pure-power reduction of a monic irreducible polynomial:This does not assert that $\phi$ is separable or that the reduction is square-free.

Finally, factor the monic $g$ over $K$ as $g={\prod }_j{h}_j^{e_j}$ with distinct monic irreducible polynomials $h_j$. The same leading-term argument bounds every root of $g$ by one. Coefficients of any $h_j$ are elementary symmetric polynomials in some of these roots, so they lie in ${\mathcal{O}}_K$. By the preceding result each ${\overline{h}}_j$ is a power of a single monic irreducible residue polynomial ${\phi }_j$.

Since the prescribed ${\overline{g}}_1,{\overline{g}}_2$ are coprime, every ${\phi }_j$ occurs on exactly one side. Assign the entire factor $h_j^{e_j}$ to that side and form the two products. Comparing irreducible-factor multiplicities in the reduction proves the coprime factor lifting from valuation-extension uniqueness:Empty products are one, covering constant prescribed factors. The construction proves the required factor-lifting form of the Hensel lemma directly from extension uniqueness, without adding completeness as an unstated hypothesis.