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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.

An additive valuation on a field is a map $v:K^{\times }\to \Gamma$, where $\Gamma$ is a totally ordered abelian group, with $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ whenever $x+y\ne 0$. Set $v(0)=\infty$. We consider nontrivial valuations; if the trivial valuation is allowed, it must be listed separately from the asserted classification. Two valuations are equivalent when their valuation rings agree, or equivalently their ordered value-group images identify in a way compatible with the maps. Real-valued equivalent nontrivial valuations differ by positive scaling.

For a valuation on $\mathbb{Q}$, every integer has nonnegative value, because it is a sum of copies of $1$ or its negative. If all prime numbers had value zero, unique factorization would make the valuation trivial. Thus some prime $p$ has $v(p)>0$. There cannot be two such primes: for distinct $p,q$, Bézout's identity gives $ap+bq=1$ with integers $a,b$, and the valuation inequality would give $v(1)>0$. All other prime values therefore vanish. Factoring the numerator and denominator of a rational number givesThe image is the cyclic ordered group generated by $v(p)>0$, proving equivalence to the P-adic valuation. This is the classification of nontrivial valuations on the rational numbers; an Archimedean absolute value is not an additive valuation satisfying this ultrametric inequality.

The simple-root form of Hensel's lemma says that for a complete discrete valuation ring $\mathcal{O}$ with maximal ideal $\mathfrak{p}$, a polynomial $f\in \mathcal{O}[X]$ and $a\in \mathcal{O}$ with $f(a)\equiv 0(\mathrm{mod}\mathfrak{p})$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\mathfrak{p})$ have a unique root in $a+\mathfrak{p}$.

For $X^3+1$ over ${\mathbb{Q}}_7$, all roots are integral: a negative valuation would make the leading term the unique term of lowest valuation. Modulo $7$, the roots are $3,5,6$, and $3X^2$ is nonzero at each. Hensel's lemma gives three distinct lifts, and a cubic has no further roots. The number is $3$.

For $X^2+2X+4$ over ${\mathbb{Q}}_2$, a root is again integral and reduces to zero modulo $2$. Write it as $2y$ with $y\in {\mathbb{Z}}_2$. The equation becomes $y^2+y+1=0$, impossible modulo $2$. The number is $0$.

For $3X^3+X+3$ over ${\mathbb{Q}}_3$, a putative root of valuation $k<0$ gives term valuations $1+3k,k,1$. The first is uniquely minimal, a contradiction. Thus all roots are integral. Reduction modulo $3$ is $X$, whose root zero is simple since the derivative $9X^2+1$ is a unit everywhere. Hensel's lemma gives exactly one lift. The number is $1$.

The integral closure of $R$ in $T$ isIt is a subring: finitely many integral elements generate a finite $R$-module algebra, and the determinant trick shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.

A valuation ring is an integral domain $V$ such that, for every nonzero $z$ in its fraction field, either $z\in V$ or $z^{-1}\in V$. Equivalently, its ideals are totally ordered by inclusion. For principal ideals, comparability is precisely the condition on $a/b$; if two arbitrary ideals were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a ring is local. Its nonunits form an ideal: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique maximal ideal consists of those nonunits.

First, valuation rings are integrally closed. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its maximal ideal $\mathfrak{m}$. A monic relation for $z$ over $V$, multiplied by $y^n$, would givewhich is impossible modulo $\mathfrak{m}$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.

For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the valuation domination lemma: a local subring $(A,\mathfrak{n})$ of a field $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and ${\mathfrak{m}}_V\cap A=\mathfrak{n}$. Here is a proof, including the crucial maximality step.

Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the rings and of their maximal ideals. The union is a local ring: an element outside the union ideal is already a unit in a member of the chain, while an element in that ideal cannot become a unit in a later dominating member. The union still dominates $A$. Thus Zorn's lemma supplies a maximal pair $(V,\mathfrak{m})$.

For any $z\in K^{\times }$, at least one of $\mathfrak{m}V[z]$ and $\mathfrak{m}V[z^{-1}]$ is proper. Suppose otherwise. There would be relationswith $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\ge s$, the second relation givesRepeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak{m}$. This contradicts minimality of $n$ (or gives $1\in \mathfrak{m}$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.

Choose whichever extension has a proper extended ideal, then a maximal ideal containing it. Localizing that extension at the chosen maximal ideal produces a local ring dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its fraction field is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the valuation domination lemma.

Now let $x\in K$ be nonintegral over $R$, so $x\ne 0$, and set $y=x^{-1}$, $B=R[y]$. The ideal $yB$ is proper: otherwise $1=y(r_0+r_{1}y+\cdots +r_n{y}^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a maximal ideal $\mathfrak{n}$ of $B$ containing $y$, and apply the valuation domination lemma to $B_{\mathfrak{n}}\subset K$. Its dominating valuation ring $V$ contains $R$ and has $y\in {\mathfrak{m}}_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.

We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the integral closure as an intersection of valuation rings isNeither Noetherianity nor a discrete valuation is required for this separation argument.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfyingA Non-Archimedean absolute value satisfies the stronger ultrametric inequality $|x+y|\le \max (|x|,|y|)$. Two equivalent absolute values induce the same topology, or equivalently differ by a positive real power. The trivial absolute value on a field takes value one on every nonzero element. The rational classification and the compactness criterion below concern nontrivial absolute values on a field; the trivial exceptions are given explicitly.

Here the additive valuation has real values. For any $c>1$, the mutually inverse constructions areMultiplicativity becomes $v(xy)=v(x)+v(y)$, and the ultrametric inequality becomes $v(x+y)\ge \min (v(x),v(y))$. Equivalent real-valued valuations differ by positive scaling, so these constructions give the required bijection on equivalence classes. Changing $c$ merely rescales the valuation.

To justify the topology formulation, $|a|<1$ is equivalent to $a^n\to 0$. Thus two nontrivial absolute values on a field with the same topology give the same strict positivity relation on their additive valuations. Fix $t$ with $v_1(t)>0$. Comparing the signs of $m{v}_j(t)-n{v}_j(x)=v_j(t^m{x}^{-n})$, for integers $m$ and positive integers $n$, shows that $v_1(x)/v_1(t)$ and $v_2(x)/v_2(t)$ have identical rational cuts. They are equal, proving $v_2=c{v}_1$ for $c>0$. If one allows valuations in arbitrary ordered groups, the nontrivial classes arising this way are precisely rank-one valuations: higher-rank ordered value groups do not embed order-preservingly in $\mathbb{R}$.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, $|n|\le 1$ for every integer $n$, by repeatedly applying the ultrametric inequality to sums of ones. Some prime $p$ must have $|p|<1$, otherwise prime factorization and multiplicativity would make every nonzero rational have value one. There is at most one such prime: if both $|p|,|q|<1$, a Bezout identity $ap+bq=1$ contradicts the ultrametric inequality. If $p\nmid m$, another Bezout identity gives $|m|=1$. HenceThis proves the non-Archimedean part of the Ostrowski theorem. If the trivial absolute value on a field is admitted, it supplies one additional class and is not equivalent to any p-adic absolute value.

The valuation ring, its maximal ideal, and its residue field areSuppose the absolute value on a field is nontrivial and $R$ is compact. The ideal $\mathfrak{m}$ is an open additive subgroup of $R$, so $k$ is discrete; as a continuous image of a compact space it is finite. The subgroup $\mathfrak{m}$ is also closed, since all its cosets are open, and is therefore compact. The continuous function $|\cdot |$ attains a maximum $\rho$ on $\mathfrak{m}$, with $0<\rho <1$. Choose $\pi$ with $|\pi |=\rho$. Then the positive values of $v=-\log |\cdot |$ have least element $-\log \rho$. Division with remainder in this additive subgroup of $\mathbb{R}$ proves $v(K^{\times })=v(\pi )\mathbb{Z}$. Thus $K$ is a discretely valued field and $\pi$ is a uniformizer.

Conversely, normalize the discrete valuation by $v(\pi )=1$. If $k$ has $q$ elements, $R/{\pi }^{N}R$ has $q^N$ elements. Each quotient therefore supplies a finite cover by balls of radius tending to zero. The ring $R$ is a closed subset of the complete metric field $K$, hence complete and totally bounded, so it is compact. Equivalently,This is the local compactness criterion for a complete non-Archimedean field, with nontriviality understood. For the trivial absolute value on a field, $R=K$ has the discrete topology and is compact exactly when $K$ is a finite field; its value group is zero rather than a nonzero discrete cyclic group.

The valuation ring and local compactness. The opening, unlettered requests useThe ultrametric inequality makes ${\mathcal{O}}_K$ a valuation ring; its elements outside $\mathfrak{m}$ are exactly the units, so $\mathfrak{m}$ is its unique maximal ideal, and $k$ is the residue field.

Here the customary nontrivial-valuation hypothesis for a local field is necessary. With the trivial absolute value, any infinite field is discrete and locally compact, but its residue field is infinite and there is no nonzero uniformizer. We use the nontrivial valuation in the conclusions below.

Local compactness first gives a compact neighborhood of zero. Choose a sufficiently small nonzero $t$ so that the closed set $t{\mathcal{O}}_K$ is contained in that compact neighborhood. Then $t{\mathcal{O}}_K$, and hence ${\mathcal{O}}_K$, is compact. Since $\mathfrak{m}$ is open, its cosets give a discrete compact quotient $k$, which must be finite. Write $|k|=q$. The subgroup $\mathfrak{m}$ now has finite index and is closed in ${\mathcal{O}}_K$, hence compact. The continuous function $|\cdot |$ attains on $\mathfrak{m}$ a maximum $c$ with $0<c<1$. Choose $\pi$ with $|\pi |=c$. Every $x\in \mathfrak{m}$ has $|x/\pi |\le 1$, soSuccessive division by $\pi$ shows that every nonzero element of ${\mathcal{O}}_K$ is a unit times a power of $\pi$: division must terminate because $|\pi {|}^j\to 0$. Thus the value group is discrete. A Cauchy sequence eventually lies in a translate of the compact ring ${\mathcal{O}}_K$, has a convergent subsequence, and consequently converges itself. These arguments prove the discrete valuation from nontrivial local compactness and the completeness of locally compact nontrivially valued fields.

The iterated-power limit. Normalize the discrete valuation by $v(\pi )=1$, and write $q={\ell }^f$, where $\ell$ is the residue characteristic. For $u\in 1+{\pi }^s{\mathcal{O}}_K$ with $s\ge 1$, the binomial theorem givesThe linear term gains a factor $q\in \mathfrak{m}$ (or is zero in positive characteristic), while all terms of degree at least two have valuation at least $2s\ge s+1$.

For $a\in {\mathcal{O}}_K^{\times }$, reduction gives $a^{q-1}\in 1+\mathfrak{m}$, because $k^{\times }$ has order $q-1$. Iterating the inequality yieldsThe sequence is Cauchy and converges to a unit $\omega (a)$. Since the reduction of $a^{q^j}$ is always $\bar{a}$, its limit has the same residue. Taking limits after shifting the sequence gives $\omega (a{)}^q=\omega (a)$, whenceThis is the Teichmuller representative. The Hensel lemma applied to $T^{q-1}-1$, whose derivative is a unit at every nonzero residue, shows uniqueness for each residue. Consequently the prime-to-$\ell$ roots give a canonical splitting

Normalize the discrete valuation on $L$ by $v_L(L^{\times })=\mathbb{Z}$, set $v_L(0)=+\infty$, and write ${\mathcal{O}}_L$ for its valuation ring. For a finite Galois extension of local fields, define $G_{-1}=G=\mathrm{Gal}(L/K)$ and, for integers $i\ge 0$,Equivalently, $G_i$ is the kernel of the action on ${\mathcal{O}}_L/{\mathfrak{m}}_L^{i+1}$. These are the lower ramification numbering of the ramification groups. In particular, $G_0$ is the inertia group, the kernel of the action on the residue field, and $G_1$ is the wild inertia group. They form a decreasing sequence of normal subgroups, eventually trivial.

For a real index $t\ge -1$, extend by $G_t=G_{\lceil t\rceil }$. Define the Herbrand function byIt is continuous, strictly increasing and piecewise linear; denote its inverse by ${\psi }_{L/K}$. The upper ramification numbering isThe definition includes the placement of the groups at break endpoints: the group at a break is the group before the drop. Lower numbering is compatible with subgroups; upper numbering is the one compatible with quotients. In a totally ramified extension, the uniformizer criterion for lower ramification groups permits testing the defining valuation inequality on one uniformizer instead of on every integral element.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

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.

Fix a non-Archimedean local field $K$, its valuation ring $R={\mathcal{O}}_K$, a uniformizer $\pi$, and a residue field with cardinality $q$. The construction below applies both in mixed and equal characteristic. A one-dimensional commutative formal group law is $F(X,Y)\in R[[X,Y]]$ with identity, associativity and commutativity, and $F(X,Y)=X+Y+$ higher-degree terms. It has a unique formal inverse. Its morphisms are series $h(T)\in TR[[T]]$ with $h(F(X,Y))=G(h(X),h(Y))$; a unit linear coefficient makes such a morphism invertible by recursive coefficient comparison. Evaluating its integral series on the maximal ideal of a finite extension of $K$ is legitimate because the series converge there.

Choose a Lubin–Tate series $f(T)$ satisfyingA useful choice is $f(T)=\pi T+T^q$. The Lubin–Tate functional equation lemma constructs a unique series with any prescribed linear form $\ell$ and satisfyingHere is its coefficient argument. If terms below degree $d$ have been found, the unknown homogeneous degree-$d$ term appears with multiplier $\pi -{\pi }^d$. The already known error has every coefficient divisible by $\pi$: after reduction, the equation becomes $H(X{)}^q=H(X_1^q,\dots ,X_r^q)$, valid because the residue coefficients lie in ${\mathbb{F}}_q$. Since $\pi -{\pi }^d$ is $\pi$ times a unit for $d\ge 2$, the next term is uniquely determined and integral. Recursion proves existence and uniqueness.

Use $\ell =X+Y$ to construct $F_f$, and $\ell =aT$ to construct $[a{]}_f$ for every $a\in R$. Uniqueness gives $F_f(X,0)=X$ and $F_f(0,Y)=Y$. Comparing $[a{]}_f(F_f(X,Y))$ with $F_f([a{]}_f(X),[a{]}_f(Y))$, which satisfy the same functional equation and have the same linear form, proves that $[a{]}_f$ is an endomorphism of the law. The same uniqueness applied to three variables proves associativity: $F_f(F_f(X,Y),Z)$ and $F_f(X,F_f(Y,Z))$ have the same linear form and functional equation. It also proves commutativity and the identitiesThus this Lubin–Tate formal group is equipped with an action of the whole ring $R$, not only its integers. For a unit $a$, the inverse of $[a{]}_f$ is $[a^{-1}{]}_f$. The same recursion, with two series for the fixed $\pi$, gives a unique strict Lubin–Tate change of series. Its integral series and inverse converge on torsion points, so it identifies their torsion fields. We may compute with $f(T)=\pi T+T^q$ without losing the general construction for that uniformizer.

Write $f_0(T)=T$ and $f_n=f^{\circ n}$. The Lubin–Tate torsion of level $n$ isFor $n\ge 1$, its primitive points are the roots ofThis is a monic polynomial of degree $D_n=(q-1)q^{n-1}$. Modulo $\pi$ it is $T^{D_n}$, and its constant coefficient is exactly $\pi$. Thus the Eisenstein layers of Lubin–Tate torsion are Eisenstein polynomials. A root ${\lambda }_n$ generates a totally ramified extension of degree $D_n$, and has $v_K({\lambda }_n)=1/D_n$ in the uniquely extended valuation. Also $f_{n-1}({\lambda }_n{)}^{q-1}=-\pi \ne 0$, so it has exact level $n$.

The mapis well-defined and injective. If $a={\pi }^{r}u$ with $r<n$ and $u$ a unit, then $[a]{\lambda }_n=[u]f_r({\lambda }_n)\ne 0$, because $[u]$ is invertible and ${\lambda }_n$ is primitive. Conversely every multiple of ${\pi }^n$ kills it. The image therefore contains $q^n$ distinct roots of the degree-$q^n$ polynomial $f_n$, so it is all of ${\Lambda }_n$. This proves, without merely assuming the torsion cardinality,Its primitive generators are exactly $[a]{\lambda }_n$ with $a\in (R/{\pi }^{n}R{)}^{\times }$.

Define the Lubin–Tate extension $K_{\pi ,n}=K({\Lambda }_n)$. Every integral endomorphism series evaluated at ${\lambda }_n$ converges in the complete field $K({\lambda }_n)$, so all torsion points already lie there. HenceIt is the splitting field of $f_n$, which has $q^n$ distinct roots by the module calculation. Thus it is Galois. Every K-automorphism preserves the uniquely extended absolute value on a field, hence is continuous and commutes with limits of the integral endomorphism series. Its automorphisms therefore commute with the endomorphism action and send a primitive generator to a primitive generator. There is a unique unit class $a_{\sigma }$ such that $\sigma ({\lambda }_n)=[a_{\sigma }]{\lambda }_n$. The Lubin–Tate Galois action on primitive torsion gives an injective homomorphismBoth sides have order $D_n=(q-1)q^{n-1}$, so it is an isomorphism. In particular every finite layer is abelian and totally ramified.

For an example, take $K={\mathbb{Q}}_p$, $\pi =p$, and the equally admissible series $f(T)=(1+T{)}^p-1$. Its formal group law is the formal multiplicative group $F(X,Y)=X+Y+XY$, with $[a](T)=(1+T{)}^a-1$ for $a\in {\mathbb{Z}}_p$. The binomial polynomials make its coefficients integral. Its level-$n$ torsion points are $\zeta -1$ with ${\zeta }^{p^n}=1$. Thus the Lubin–Tate cyclotomic example gives $K_{p,n}={\mathbb{Q}}_p({\mu }_{p^n})$ and Galois action $\zeta \mapsto {\zeta }^a$. The order formula includes $p=2,n=1$, when the first field and its Galois group are trivial over ${\mathbb{Q}}_2$.

Finally choose primitive points compatibly, $[\pi ]{\lambda }_{n+1}={\lambda }_n$. They exist by taking a root of $f(T)-{\lambda }_n$ in the algebraic closure; its roots have positive valuation and exact level $n+1$. Then $K_{\pi ,n}\subseteq K_{\pi ,n+1}$, and restriction of Galois automorphisms corresponds to reduction of unit classes. The Lubin–Tate tower $K_{\pi }={\bigcup }_n{K}_{\pi ,n}$ consequently has the topological Galois groupThe final isomorphism uses completeness of the discrete valuation ring $R$. This computes the finite-layer and infinite-tower Galois groups directly from the formal action and the primitive Eisenstein polynomials.

Use the simple-root form of the Hensel lemma: for $F\in {\mathcal{O}}_K[T]$ and $\bar{a}\in k_K$ satisfying $\overline{F}(\bar{a})=0$ and $\overline{F^{\prime }}(\bar{a})\ne 0$, there is a unique $a\in {\mathcal{O}}_K$ with $\bar{a}$ as its reduction and $F(a)=0$.

Let $d=[\ell :k_K]$. The residue field $k_K$ is finite, so $\ell /k_K$ is separable, and the primitive element theorem gives $\ell =k_K(\bar{\alpha })$. Let $\bar{F}$ be the monic irreducible polynomial of $\bar{\alpha }$, of degree $d$, and choose a monic coefficient lift $F\in {\mathcal{O}}_K[T]$. Irreducible reduction implies that $F$ is irreducible over $K$: any monic factorization over $K$ has integral coefficients, since all its roots are integral over the valuation ring, and would reduce to a factorization of $\bar{F}$.

Adjoin a root $\alpha$ and put $L=K(\alpha )$. Then $[L:K]=d$ and $\alpha \in {\mathcal{O}}_L$. Its reduction satisfies $\bar{F}$, so $k_L$ contains a copy of $\ell$ over $k_K$. For finite extensions of non-Archimedean local fields, the fundamental degree relation isThe residue degree is at least $d$, while the left side equals $d$, henceThis constructs the required unramified extension.

For uniqueness, let $L^{\prime }/K$ be another unramified extension with a chosen identification $k_{L^{\prime }}\cong \ell$. The element corresponding to $\bar{\alpha }$ is a simple residue root of $F$, because finite fields are perfect. The Hensel lemma in $L^{\prime }$ gives a root ${\alpha }^{\prime }\in {\mathcal{O}}_{L^{\prime }}$ with this reduction. Since $F$ is irreducible of degree $d$, $[K({\alpha }^{\prime }):K]=d=[L^{\prime }:K]$, so $L^{\prime }=K({\alpha }^{\prime })$. Sending $\alpha$ to ${\alpha }^{\prime }$ yields a $K$-isomorphism $L\to L^{\prime }$, compatible with the prescribed residue identification. ThusThis proves the existence and uniqueness of unramified local extensions using the stated version of Hensel's lemma, without assuming the classification in advance.

The completion of a number field at a prime ideal does not change either invariant. More precisely, ${\hat{B}}_q$ is the valuation ring of $L_q$, its maximal ideal is $q{\hat{B}}_q$, andThus the completed extension has the same ramification index and the same residue-field extension, hence the same residue-field degree $f_{q/p}$. This is completion preserves ramification and residue data.

For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

First take $K=\mathbb{Q}$. The neighborhoodof zero meets the diagonal copy of $\mathbb{Q}$ only in zero: a rational number lying in every ${\mathbb{Z}}_p$ is an integer, and the only integer in the indicated real interval is zero. Thus $\mathbb{Q}$ is discrete in ${\mathbb{A}}_{\mathbb{Q}}$.

Every rational adele is congruent modulo $\mathbb{Q}$ to an element ofIndeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the Chinese remainder theorem; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every ${\mathbb{Z}}_p$, and the Tychonoff theorem. Its image covers the quotient, so ${\mathbb{A}}_{\mathbb{Q}}/\mathbb{Q}$ is compact.

Now choose a $\mathbb{Q}$-basis of the number field $K$. The given topological isomorphismidentifies the additive pair $({\mathbb{A}}_K,K)$ with $({\mathbb{A}}_{\mathbb{Q}}^n,{\mathbb{Q}}^n)$. A finite product of discrete subgroups is discrete, andis compact.

The idele group iswhere the distinguished subgroup at a finite place is ${\mathcal{O}}_v^{\times }$. It carries the corresponding restricted product topology on the idele group. The inclusion $j:J_K\to {\mathbb{A}}_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^{\times }$, and outside finitely many places every idele component already belongs to ${\mathcal{O}}_v^{\times }\subseteq {\mathcal{O}}_v$.

It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to 1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in {\mathbb{Z}}_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhoodcontains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.

For the discrete valuation $v$ corresponding to $|\cdot |$, the valuation ring and its maximal ideal areThe first is a subring of the field $K$, hence an integral domain. An element of ${\mathcal{O}}_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak{m}$ and $\mathfrak{m}$ is the unique maximal ideal.

Choose a uniformizer $\pi$ with $v(\pi )=1$. If $I\ne 0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u{\pi }^n$ for a unit $u$, so $(x)=({\pi }^n)$. Every $y\in I$ has $v(y)\ge n$, hence $y/x\in {\mathcal{O}}_K$, and therefore $I=(x)$. Thus ${\mathcal{O}}_K$ is a discrete valuation ring, in particular a principal ideal domain.

The valuation ring and its maximal ideal areIf ${\mathcal{O}}_K$ is Noetherian, its maximal ideal is finitely generated. Ideals in a valuation ring are totally ordered, so every finitely generated ideal is generated by one of its generators; write $\mathfrak{m}=(\pi )$. Then $v(\pi )$ is the smallest positive element of the value group. Subtracting integral multiples of this value shows that every value is an integral multiple of $v(\pi )$, so $v$ is discrete.

Now assume $K$ is complete and discretely valued. Parts i and ii, applied to $v$ and to the other discrete valuation $v^{\prime }$, giveIf $v(a)=0$ and $\pi$ is a uniformizer for $v$, then $\pi a^r\in {\mathfrak{m}}_v$ for every $r\in \mathbb{Z}$. Hencefor every integer $r$, forcing $v^{\prime }(a)=0$ and in particular $v^{\prime }(a)\ge 0$. Every nonzero $x$ has the form $x={\pi }^{m}u$ with $v(u)=0$, soThe proportionality constant is positive because $v^{\prime }$ is nontrivial. Thus the valuations, and their associated absolute values, are equivalent.

Let $R$ be a valuation ring with fraction field $K$. For every commutative squarethe valuative criterion for separatedness says that a finite type morphism $f$ between Noetherian schemes is separated exactly when there is at most one dotted lift $\mathrm{Spec}R\to X$ completing the diagram.

Under the same finiteness hypotheses, the valuative criterion for properness says that $f$ is proper exactly when every such square has a unique lift. Thus separatedness supplies uniqueness, while properness supplies existence as well.

This can be false. In a valued field with value group $\mathbb{Q}$, such as the Hahn series field $k((t^{\mathbb{Q}}))$, the maximal idealof the valuation ring is not principal: if $v(a)=q>0$, an element of valuation $q/2$ belongs to $\mathfrak{m}$ but not to $(a)$. Thus the valuation ring need not be a principal ideal domain.

A valuation ring for a nontrivial real-valued rank-one valuation is a principal ideal domain exactly when its value group is discrete. A least positive value makes the value group cyclic and lets every nonzero ideal be generated by an element of smallest valuation. Conversely, if the maximal ideal is principal, its generator has the least positive value. Nontriviality matters: a field with the trivial valuation is a principal ideal domain, but not a discrete valuation ring in the usual nontrivial convention.

For a degree-$n$ totally ramified Galois extension and a uniformizer $\pi$, let $f$ be its minimal polynomial. Dividing the ramification polynomial $f(\pi (1+X))/{\pi }^n$ by its known factor $X$ gives the reduced ramification polynomial. It is monic of degree $n-1$, integral over the extension's valuation ring, and has roots $(\sigma (\pi )-\pi )/\pi$ for nonidentity automorphisms. Its constant coefficient is nonzero in a separable extension. Removing the zero root allows the usual Newton polygon root valuation theorem to apply directly.