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A complete discrete valuation ring is a discrete valuation ring complete for its maximal-ideal-adic topology. Equivalently, the canonical map to the inverse limit of the residue rings is an isomorphism: compatible residue classes determine a Cauchy sequence, and completeness gives its unique limit. Standard examples are ${\mathbb{Z}}_p$ and $k[[t]]$. This completeness is the hypothesis that makes Newton iteration over a valued field converge in the ring in the usual Hensel lemma.

A useful strong form of the Hensel lemma is the following. Let $F$ be a complete discretely valued field, with valuation ring $\mathcal{O}$, normalized valuation $v$, and $P\in \mathcal{O}[T]$. If $a_0\in \mathcal{O}$ satisfiesthen there is a unique root $a\in \mathcal{O}$ in the ball $v(a-a_0)>v(P^{\prime }(a_0))$. When $P(a_0)=0$ the assertion is immediate; otherwise the condition includes $P^{\prime }(a_0)\ne 0$.

Here is a proof by Newton iteration over a valued field. Write $c=v(P^{\prime }(a_0))$ and $t_0=v(P(a_0))>2c$, and setSuppose $v(P^{\prime }(a_j))=c$ and $t_j=v(P(a_j))>2c$. The correction has valuation $t_j-c>c\ge 0$, so $a_{j+1}\in \mathcal{O}$. Taylor expansion over $\mathcal{O}$ cancels the linear term and givesThus the derivative still has valuation $c$, and $t_{j+1}-2c\ge 2(t_j-2c)$. Unless an exact root is reached earlier, the corrections tend to zero with exponentially increasing valuations. Completeness gives $a_j\to a$, and continuity gives $P(a)=0$ and $v(a-a_0)=t_0-c>c$. For uniqueness, if $a,b$ are in this ball, polynomial expansion around $a_0$ givesThe second factor has valuation exactly $c$ and is nonzero, so two roots in the ball must be equal. In particular, a simple residue root lifts uniquely: $v(P(a_0))\ge 1$ and $v(P^{\prime }(a_0))=0$ satisfy the hypothesis.

For the square-class group of a p-adic field, write every element as $p^{r}u$ with $u\in {\mathbb{Z}}_p^{\times }$. Multiplication by a square changes $r$ by an even integer, so valuation parity gives one independent factor $C_2$.

For odd $p$, a unit $u$ is a square if and only if its residue is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency applies the simple-root Hensel lemma to $T^2-u$, whose derivative is a unit at a nonzero residue root. The cyclic group ${\mathbb{F}}_p^{\times }$ has square subgroup of index two. If $u_0$ is a unit with nonsquare residue, thenFor $p=2$, every odd square is one modulo eight. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $P(T)=T^2-u$ and $a_0=1$. Then $v_2(P(1))\ge 3>2v_2(P^{\prime }(1))=2$, so the strong Hensel lemma gives a square root. Thus unit square classes are exactly their residues $1,3,5,7$ modulo eight. The classes $-1$ and $5$ generate this group of order four independently, and valuation parity supplies the class $2$. Hence

A useful version is the strong form of Hensel lemma. Let $K$ be complete for a Non-Archimedean absolute value, let $f\in {\mathcal{O}}_K[X]$, and suppose $a\in {\mathcal{O}}_K$ satisfiesThen there is a unique root $\alpha$ in the ball $|\alpha -a|<|f^{\prime }(a)|$, and $|\alpha -a|\le |f(a)/f^{\prime }(a)|$.

Put $h=|f^{\prime }(a)|$ and $\eta =|f(a)|/h^2<1$. Starting with $x_0=a$, apply Newton iteration over a valued field $x_{n+1}=x_n-f(x_n)/f^{\prime }(x_n)$. Taylor expansion over the valuation ring givesfor integral $x,u$. Inductively $|f^{\prime }(x_n)|=h$ and $|f(x_n)|/h^2\le {\eta }^{2^n}$. Indeed the Newton increment has size at most $h{\eta }^{2^n}<h$, so the derivative size is unchanged, and the new function value is bounded by the square of the increment. The increments tend to zero, and the ultrametric inequality makes the sequence Cauchy. Completeness supplies a root with the stated distance bound.

If $b,c$ are distinct points of the ball, their divided difference is $f^{\prime }(a)$ plus a term of size at most $\max (|b-a|,|c-a|)<h$, so $(f(b)-f(c))/(b-c)$ is nonzero. Two roots in that ball are impossible. This proves the stated Hensel lemma. In particular a simple root modulo the maximal ideal lifts uniquely to a root in its residue class.

A strong lifting lemma. Let $R$ be a complete discrete valuation ring, with fraction field $F$ and normalized discrete valuation $v$. A useful strong form of Hensel lemma says that for $f\in R[T]$ and $a\in R$ satisfyingthere is a root $b\in R$ withIf $f(a)=0$, take $b=a$. Otherwise the hypothesis ensures that $f^{\prime }(a)\ne 0$. The root is unique among $b$ satisfying $v(b-a)>v(f^{\prime }(a))$. In particular, a simple root of the reduction of $f$ lifts uniquely with its specified residue.

Here is a proof by Newton iteration over a valued field. Put $a_0=a$ andWrite $A_j=v(f(a_j))$ and $B=v(f^{\prime }(a_0))$. Taylor expansion over $R$ shows, inductively, that the increment has valuation $A_j-B>B$, that $v(f^{\prime }(a_{j+1}))=B$, and thatIndeed, the constant and linear terms of $f(a_{j+1})$ cancel, leaving terms of valuation at least twice that of the increment. The change in $f^{\prime }$ has valuation greater than $B$, so its valuation stays $B$. Thus all iterates lie in $R$ and form a Cauchy sequence. Completeness and continuity give a root $b$. The later increments have strictly larger valuations than the first, proving the displayed distance formula. For uniqueness, if $b,c$ are in the stated ball, Taylor expansion factorsThe second factor is nonzero, so two roots must coincide. This proves Hensel's lemma in the form needed here.

Odd residue characteristic. For odd $p$, a p-adic unit $u$ is a square exactly when its reduction is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction. For sufficiency, a nonzero residue root of $T^2-u$ has derivative $2T$ invertible modulo $p$, so the Hensel lemma lifts it. The multiplicative group of a finite field is cyclic, and its subgroup of squares has index two. ThereforeRepresentatives are $1$ and any unit whose reduction is a quadratic nonresidue.

Residue characteristic two. An odd integer has square congruent to one modulo eight, so a square 2-adic unit must lie in $1+8{\mathbb{Z}}_2$. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $f(T)=T^2-u$ and $a=1$. Then $v_2(f(1))\ge 3>2v_2(f^{\prime }(1))=2$, and the strong form of Hensel lemma produces a square root. HenceThe four unit square classes of the p-adic integers have representatives $1,3,5,7$.

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.

One strong form of the Hensel lemma is this: if a complete discretely valued field $K$, a polynomial $f\in {\mathcal{O}}_K[X]$, and $a\in {\mathcal{O}}_K$ satisfythen there is a unique root $\alpha$ in the ball $v(\alpha -a)>v(f^{\prime }(a))$.

Set $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Taylor expansion shows that the valuation of the error at least doubles at each step, while $v(f^{\prime }(a_n))$ remains constant. Thus the corrections tend to zero geometrically, so completeness gives a limit $\alpha$. Continuity gives $f(\alpha )=0$. Applying the same Taylor estimate to two roots in the stated ball proves uniqueness. This is Newton iteration over a valued field.

One strong form of Hensel lemma is this: if $K$ is complete for a discrete valuation $v$, $f\in {\mathcal{O}}_K[X]$, and $a_0\in {\mathcal{O}}_K$ satisfiesthen there is a unique root $\alpha$ satisfying

Define the Newton iteration over a valued fieldTaylor expansion and the ultrametric inequality show inductively that $v(f^{\prime }(a_n))=v(f^{\prime }(a_0))$ andThus the valuations of the corrections $a_{n+1}-a_n$ tend to infinity, so $(a_n)$ is a Cauchy sequence. Completeness gives a limit $\alpha$, and continuity gives $f(\alpha )=0$. If $\beta$ is another root in the stated ball, Taylor expansion of $f(\beta )-f(\alpha )$ shows that the linear term has strictly smaller valuation than all higher terms unless $\beta =\alpha$, proving uniqueness.

The strong form of Hensel lemma says the following. Let $v$ be a discrete valuation on a complete field $K$. If $f\in {\mathcal{O}}_K[X]$ andthen $f$ has a root $x\in {\mathcal{O}}_K$ satisfying $v(x-x_0)>v(f^{\prime }(x_0))$.

To prove it, apply Newton iteration over a valued field:The initial inequality says that the first correction has valuation greater than $v(f^{\prime }(x_0))$. Taylor expansion then shows inductively that $v(f^{\prime }(x_n))=v(f^{\prime }(x_0))$, while the valuations of the corrections tend to infinity. Hence $(x_n)$ is a Cauchy sequence. Completeness gives a limit $x$, and continuity gives $f(x)=0$.

Now decompose the multiplicative group asThe P-adic valuation givesand cubing is the identity on $\{\pm 1\}$. Put $U_2=1+9{\mathbb{Z}}_3$. Expansion gives $U_1^3\subseteq U_2$. Conversely, for $u=1+9a\in U_2$, choose $b\equiv a(\mathrm{mod}3)$ and put $x_0=1+3b$. For $f(X)=X^3-u$,so the strong form of Hensel lemma produces a cube root in $U_1$. Thus $U_1^3=U_2$. Finally,is an isomorphism. Combining the valuation and principal-unit factors proves the cube-class group of the 3-adic numbers identity

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 $v(\alpha -a)>v(f^{\prime }(a))$.