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With the arithmetic Frobenius normalization, a chosen uniformizer $\pi$ and Lubin–Tate formal group give these actions on the Lubin–Tate tower and the maximal unramified extension. The two fields are linearly disjoint and their compositum is the maximal abelian extension. The inverse on units is essential: using geometric Frobenius would invert the entire Local Artin map. The finite kernels on a compositum of residue degree $d$ and torsion level $r$ are ${\pi }^{d\mathbb{Z}}(1+{\pi }^r{\mathcal{O}}_K)$. They are a basis of finite-index open subgroups, yielding the profinite completion of $K^{\times }$ as the abelian Galois group.

Write $\mathcal{O}$ for the valuation ring of $K$, $\pi$ for a uniformizer, $v(\pi )=1$, and $k={\mathbb{F}}_q$ for the residue field. A useful strong form of Hensel lemma is the following: if $f\in \mathcal{O}[X]$ and $a_0\in \mathcal{O}$ satisfythen there is a root $a\in \mathcal{O}$ with $v(a-a_0)>v(f^{\prime }(a_0))$, unique among roots in that ball. If $f(a_0)\ne 0$, its distance satisfies $v(a-a_0)=v(f(a_0))-v(f^{\prime }(a_0))$.

Here is a proof by successive correction. Put $s=v(f^{\prime }(a_0))$, $d_j=v(f(a_j))$, andStop if a root has already been reached. Otherwise the correction has valuation $d_j-s>s$. Integral Taylor expansions show that $v(f^{\prime }(a_{j+1}))=s$ andThus $d_j-2s$ grows at least geometrically, the corrections tend to zero, and completeness gives a limit $a$ with $f(a)=0$. The first correction has strictly smaller valuation than every later nonzero correction, proving the asserted distance. If $x,y$ are in the ball $v(x-a_0),v(y-a_0)>s$, thenThe second factor cannot vanish. Two roots in the ball must therefore coincide. This proves Hensel lemma; in particular a simple root modulo $\pi$ has a unique lift with that residue.

We use this to establish existence and uniqueness of unramified local extensions. For each integer $d\ge 1$, choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $d$ and lift it to a monic $f\in \mathcal{O}[X]$. Reduction and Gauss lemma show that $f$ remains irreducible over $K$. LetThe ring $A$ is finite free and complete over $\mathcal{O}$, and $A/\pi A\cong {\mathbb{F}}_{q^d}$. An element outside $\pi A$ is invertible: lift an inverse from the residue field and invert the remaining $1+\pi z$ by a convergent geometric series. Every nonzero element of $A$ is ${\pi }^t$ times such a unit, since ${\bigcap }_t{\pi }^{t}A=0$ in a finite free module. Thus $A$ is a discrete valuation ring, its fraction field is $L$, andThis constructs an unramified extension of each degree.

Conversely, let $M/K$ be an unramified extension of degree $d$. Its residue field is the unique degree-$d$ finite field extension ${\mathbb{F}}_{q^d}/{\mathbb{F}}_q$. Choose a residue root of $\bar{f}$ in it. The root is simple, so Hensel lemma lifts it to a root of $f$ in $M$. This embeds $L$ into $M$, and equality of degrees makes it an isomorphism. All $d$ residue conjugates lift in $L$ itself; hence $L/K$ is Galois, and reduction identifies its Galois group with the cyclic Galois group of the residue field extension. Its distinguished Frobenius automorphism reduces to $x\mapsto x^q$.

There is exactly one unramified extension $K_d$ of each degree inside a fixed algebraic closure, andFor the middle equality, every nonzero element of ${\mathbb{F}}_{q^d}$ lifts uniquely to a $(q^d-1)$st root of unity, because that exponent is prime to $p$. A primitive such residue generates the finite field extension, so its lift generates $K_d$. Taking the union gives the maximal unramified extension, with Galois group $\hat{\mathbb{Z}}$.

Let $K$ be a field, $K_s$ a separable closure and $G_K$ its absolute Galois group. A discrete Galois module $A$ is an abelian group with a continuous $G_K$-action, so each element has an open stabilizer. Galois cohomology uses continuous cochains. Its zeroth group is $H^0(K,A)=A^{G_K}$, and its first group iswhereThus cocycles measure a failure of descent, and coboundaries account for changing a chosen lift. A short exact sequence of Galois modules gives a long exact sequence in group cohomology. Its connecting map sends an invariant element of a quotient to the cocycle $\sigma b-b$ of a chosen lift $b$; changing the lift changes this by a coboundary. These constructions provide the bridge between arithmetic divisibility and cohomology.

For Kummer theory, assume the characteristic does not divide $n$. The multiplicative exact sequence isThe final arrow is surjective because a root of $T^n-a$ is separable for $a\ne 0$. Hilbert's theorem 90 says $H^1(K,K_s^{\times })=0$. Here is its usual finite-extension proof. For a multiplicative cocycle $c_{\sigma }$ on a finite Galois extension $L/K$, choose $a\in L$ so thatSuch an $a$ exists by the Artin independence theorem. The cocycle identity gives $\tau A=c_{\tau }^{-1}A$. With $b=A^{-1}$ this becomes $c_{\tau }=\tau b/b$, a coboundary. Every continuous cocycle with values in $K_s^{\times }$ is defined over a sufficiently large finite Galois extension, so the same result holds for the absolute group.

The long exact sequence in group cohomology now givesThe class of $a$ maps to $c_{\sigma }=\sigma (\alpha )/\alpha$, where ${\alpha }^n=a$. This cohomological statement does not require ${\mu }_n\subset K$; it retains the actual Galois action on ${\mu }_n$. If all $n$th roots of unity do lie in $K$, the action is trivial, and after choosing a primitive root of unity the cocycles are characters into $\mathbb{Z}/n\mathbb{Z}$. Thus adjoining an $n$th root of $a$ describes a cyclic extension of degree dividing $n$, and Kummer classes classify the corresponding characters. For $n=2$, this gives the familiar quadratic extensions and the square-class group of a field.

Now let $K$ be a number field and $E/K$ an elliptic curve. The multiplication-by-n morphism is a surjective isogeny on $E(\overline{K})$, with kernel $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$. The Kummer exact sequence of an elliptic curveyieldsExplicitly, choose $Q$ with $nQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. This lies in $E[n]$, satisfies the cocycle identity and is unchanged as a class if $Q$ is replaced by another division point. If it is a coboundary $\sigma T-T$ for $T\in E[n]$, then $Q-T$ is rational over $K$ and $P=n(Q-T)$. Conversely such a rational division point makes the cocycle zero. This proves the injectivity of the Kummer map of an elliptic curve on $E(K)/nE(K)$.

To prove the Weak Mordell-Weil theorem, it remains to prove that this particular cohomological image is finite. The full group $H^1(K,E[n])$ need not be finite, so the injection alone is not enough. Choose a finite set $S$ of finite places containing the bad-reduction places and those dividing $n$. At any other place, the construction in Question 3 works over the local field: reduce $P$, find an $n$-division point over a finite residue extension, lift it by the Hensel lemma, and correct its error by the inverse formal multiplication. Thus $P$ has an $n$-division point over the maximal unramified extension. Choosing that lift locally makes the cocycle zero on inertia. For an arbitrary lift its restriction is a coboundary, so the cohomology class is unramified. Therefore the image lies in first Galois cohomology unramified outside a finite set:We prove the required finiteness of this group. Take a finite Galois extension $L/K$ containing all of $E[n]$ and ${\mu }_n$, and enlarge $S$ to include any remaining ramification of $L/K$. Let $S_L$ be the places above it. The inflation-restriction exact sequence bounds the kernel of restriction to $L$ bya finite group, since both its group of arguments and its group of values are finite. Over $L$ the torsion module is constant and can be identified with ${\mu }_n^2$, so Kummer theory givesRestrictions of our unramified classes remain unramified outside $S_L$. If a Kummer class $[a]$ is unramified at such a place $v$, its $n$th root can be taken in an unramified local extension. The extended valuation still has integer values, and $nv(\sqrt[n]{a})=v(a)$. Hence $v(a)$ is divisible by $n$. Each coordinate therefore lies in the S-unramified power class groupThis group is finite. Indeed, the ideal of $a$ away from $S_L$ is ${\mathfrak{a}}^n$. Sending $[a]$ to $[\mathfrak{a}]$ yields the exact sequenceFor its kernel, a principal $\mathfrak{a}$ lets us divide $a$ by an $n$th power to obtain an $S_L$-unit. For surjectivity, an ideal class killed by $n$ has an $n$th power which is principal away from $S_L$, giving such an $a$. The S-unit group is finitely generated and the ideal class group is finite; hence both end groups are finite, proving finiteness of S-unramified Kummer classes. Restriction has finite image in $L(S_L,n{)}^2$ and finite kernel, so $H_S^1(K,E[n])$ is finite. We concludewhich is the Weak Mordell-Weil theorem and its Kummer-theoretic proof of the weak Mordell-Weil theorem.

This framework also defines the Selmer group of an elliptic curve by imposing that a class in $H^1(K,E[n])$ lie in the local Kummer image at every completion. Its classes are unramified outside a fixed finite set, so the same finiteness proof applies. Its relation to rational points isThe Tate–Shafarevich group here measures the difference between locally soluble torsors and globally soluble ones. Finally, the weak theorem gives finite divisibility quotients; adding the height descent lemma and bounded-height finiteness proves the stronger Mordell-Weil theorem. Those height inputs are additional to the cohomological finiteness just established.

Fix a non-Archimedean local field $K$, write $Q=|k_K|$, and use the arithmetic Frobenius normalization throughout: a uniformizer maps to the automorphism inducing $a\mapsto a^Q$ on the maximal unramified extension. The Local Artin map, or local reciprocity map, is the canonical continuous group homomorphismwhere $K^{\mathrm{ab}}$ is the maximal abelian extension inside a fixed algebraic closure with its separable part understood. Its finite projection to every finite abelian extension $L/K$ is onto and has kernel exactly the norm subgroup of a local field extension:In particular, the quotient has order $[L:K]$. This is how local class field theory turns abelian extension problems into computations inside a multiplicative group.

One precise definition of the finite reciprocity map uses the cup-product construction of local reciprocity. For a finite Galois extension with group $G$, the local fundamental class $u_{L/K}\in H^2(G,L^{\times })$ is specified by local Brauer invariant $1/[L:K]$. Cup product gives an isomorphismThe finite Local Artin map is its inverse, composed with the quotient map from $K^{\times }$. For abelian $G$ the target is $G$ itself. Taking the compatible inverse limit over finite abelian extensions defines ${\mathrm{Art}}_K$.

The essential inputs to this cohomological construction have concrete arithmetic content. The local Brauer invariant identifies $\mathrm{Br}(K)$ with $\mathbb{Q}/\mathbb{Z}$; restriction to $L$ multiplies the invariant by $[L:K]$, so its kernel is cyclic of that order and has the distinguished generator $u_{L/K}$. For every subgroup $H\subseteq G$, Hilbert theorem 90 gives $H^1(H,L^{\times })=0$, while the same invariant calculation over the fixed field gives $H^2(H,L^{\times })$ cyclic of order $|H|$. These are the hypotheses of the algebraic cup-product isomorphism in Tate cohomology of a finite group. It is obtained by dimension shifting in group cohomology with a projective resolution of the trivial module and the extension representing the fundamental class. Thus the norm-kernel statement comes from the degree-zero definition of Tate cohomology of a finite group, rather than from an arbitrary assignment of an automorphism to each element. In an unramified extension of degree $d$, the fundamental class is the cyclic algebra with arithmetic Frobenius generator and parameter a uniformizer; the isomorphism sends that Frobenius to the class of the uniformizer. This explains the chosen sign of reciprocity.

Several properties follow immediately from this construction. If $L\supseteq E\supseteq K$ are finite abelian extensions, restricting ${\mathrm{Art}}_{L/K}(a)$ to $E$ gives ${\mathrm{Art}}_{E/K}(a)$. The Local Artin map is therefore compatible with the finite projections defining its infinite target. For any finite separable extension $E/K$ and $a\in E^{\times }$, its norm compatibility isHere $K^{\mathrm{ab}}E/E$ is abelian, so the displayed restriction makes sense. An outline of the compatibility proof is to use the corestriction map in group cohomology: on multiplicative coefficients its degree-zero map is the norm. The cup-product projection formula, together with restriction and corestriction of the local fundamental class, gives exactly the commuting reciprocity diagram. The compatibility is an arithmetic instance of the functoriality of the cohomological definition.

The Local Artin map has dense image, because every finite abelian quotient receives a surjective map. Its kernel is trivial, and it extends to an isomorphismwhere this profinite completion uses the open finite-index subgroups of the topological multiplicative group. The original map is generally not onto: on the unramified quotient it sends the integer valuation into $\hat{\mathbb{Z}}$, rather than producing every profinite integer. Its units map onto the abelian inertia group. For a finite abelian extension, the corresponding unit image is its inertia group, while the valuation accounts for its residue extension. Indeed $v_K(Nz)=f(L/K)v_L(z)$, so the valuation quotient has order $f$; the full norm quotient has order $ef$, leaving unit image of order $e$. Since unit images act trivially on the residue field, they give precisely the inertia group. More finely, the image of the higher principal-unit group $1+{\mathfrak{m}}_K^r$ in a finite abelian extension is its rth upper ramification group, for each integer $r\ge 1$. The use of upper ramification numbering is essential for compatibility with quotients.

The existence theorem of local class field theory states that every open finite-index subgroup $H\subseteq K^{\times }$ is $N_{L/K}(L^{\times })$ for a unique finite abelian extension $L/K$. Inclusions are reversed: larger extensions have smaller norm subgroups. In the established reciprocity isomorphism, the proof is simply to take the fixed field of the open subgroup corresponding to $H$. Conversely a finite extension gives an open kernel of a finite continuous projection. Uniqueness follows because that kernel specifies its fixed field. The substantive existence problem is therefore to prove that the reciprocity isomorphism realizes all open finite-index subgroups; it is not just this final fixed-field argument.

The Lubin–Tate description of local reciprocity provides an explicit route to that existence problem. Choose a uniformizer $\pi$ and a Lubin–Tate series $f(T)=\pi T+T^Q$, with its associated Lubin–Tate formal group $F$. Solving $f(F(X,Y))=F(f(X),f(Y))$ successively by total degree constructs the law with linear part $X+Y$; the analogous equations construct the endomorphisms $[a]$ with linear coefficient $a$. The congruence $f(T)\equiv T^Q(\mathrm{mod}\pi )$ supplies the divisibility needed at each step. These endomorphisms satisfy $[a][b]=[ab]$ and make the torsion an ${\mathcal{O}}_K$-module.

If $f_r$ denotes the rfold iterate, the primitive level-r Lubin–Tate torsion points are roots ofReduction modulo $\pi$ is $T^{Q^{r-1}(Q-1)}$, and the constant term is $\pi$, so this is an Eisenstein polynomial. Its roots therefore generate a totally ramified extension $K_{\pi ,r}$ of degree $(Q-1)Q^{r-1}$. A primitive point generates the entire torsion module, and all points are its images under the integral convergent endomorphisms $[a]$. The torsion field is therefore a Galois extension of $K$. The action map into $({\mathcal{O}}_K/{\pi }^r{)}^{\times }$ is injective, and the computed degree equals the order of this unit group, proving it is an isomorphism. Passing to the inverse limit gives $\mathrm{Gal}(K_{\pi }/K)\simeq {\mathcal{O}}_K^{\times }$ for the Lubin–Tate tower $K_{\pi }$.

The unramified tower and $K_{\pi }$ are linearly disjoint field extensions: a finite common subextension would be both unramified and totally ramified, hence trivial. The Lubin–Tate description of local reciprocity on their compositum isThe inverse on the unit action goes with the arithmetic Frobenius convention; changing to geometric Frobenius inverts the whole reciprocity map. The reciprocity theorem identifies this explicit action with the finite cup-product maps, independently of the chosen uniformizer. Its proof compares the fundamental class with the norm relations between consecutive torsion fields. The construction makes the possible finite kernels concrete: on the compositum of the degree-d unramified field with $K_{\pi ,r}$ the kernel isThese subgroups are cofinal among open finite-index subgroups. Indeed any such $H$ contains $1+{\pi }^r{\mathcal{O}}_K$ for sufficiently large $r$, and contains ${\pi }^d$ for some positive $d$, since the class of $\pi$ in $K^{\times }/H$ has finite order. Taking the appropriate fixed subgroup in this explicit finite compositum realizes $H$. This proves the open-subgroup part of existence once the norm compatibility of the explicit and cohomological constructions is established. Together with the finite reciprocity maps it also shows that $K^{\mathrm{ab}}=K^{\mathrm{nr}}{K}_{\pi }$.

As a direct check, on a finite unramified extension the construction reduces toThe earlier elementary unit-norm proof shows its kernel is exactly ${\pi }^{d\mathbb{Z}}{\mathcal{O}}_K^{\times }$. At $K={\mathbb{Q}}_p$, the Lubin–Tate cyclotomic example takes $[p](T)=(1+T{)}^p-1$ and $F(X,Y)=X+Y+XY$. Its torsion is $\zeta -1$, so a unit $u$ acts by $\zeta \mapsto {\zeta }^{u^{-1}}$. These examples exhibit both the valuation and unit parts of local reciprocity with consistent signs.

Let $K^{\mathrm{ur}}$ be the maximal unramified extension of $K$, with residue field $\overline{k}$. First show that every geometric solution of $nQ=P$ lies in $E(K^{\mathrm{ur}})$.

The multiplication-by-n morphism on the reduced elliptic curve is surjective over $\overline{k}$. Choose $\overline{Q}$ with $n\overline{Q}=\overline{P}$, defined over some finite finite field extension $k^{\prime }/k$. Let $L/K$ be the corresponding finite unramified extension. By the surjectivity of good reduction over a local field, lift $\overline{Q}$ to $Q_0\in E(L)$. Then $P-n{Q}_0\in E_1(L)$. The multiplication isomorphism of a formal group law gives a unique $R\in E_1(L)$ with $nR=P-n{Q}_0$, so $Q_0+R$ is an $n$-division point of $P$ defined over $L$.

Taking $P=O$ gives every element of $E[n]$ over $K^{\mathrm{ur}}$: reduction is injective on this torsion point of an elliptic curve group because $[n]$ is injective on the formal kernel, and each reduced $n$-torsion point has the unique corrected lift just constructed. Every other division point differs from one lift by an element of $E[n]$. Hence the entire field $K([n{]}^{-1}P)$ is unramified, not merely a field containing one choice of $Q$.

To bound the composite uniformly in $P$, put $f=[K(E[n]):K]$. This division field of an elliptic curve is unramified. The Galois group of $K^{\mathrm{ur}}/K$ is procyclic, generated topologically by its Frobenius element $\sigma$, and ${\sigma }^f$ fixes every element of $E[n]$. If $nQ=P\in E(K)$, thenIteration gives ${\sigma }^{jf}Q=Q+jT$. In particular ${\sigma }^{nf}Q=Q$. Thus every such $Q$, for every $P$ together, lies in the unique unramified extension of degree $nf$. If $M$ is the composite of all these fields, the uniform unramified division field over a local field givesThe prime-to-$p$ hypothesis is essential to the inverse-series argument; no assertion about arbitrary $p$-division is being made.

Let $I(L/K)$ be the inertia group and let $\phi$ denote Frobenius on the residue-field extension. The Relative Weil group isIts Weil-group topology makes $I(L/K)$ an open profinite group with its usual topology and gives $W(L/K)/I(L/K)$ the discrete topology. Thus every inertia coset is an open copy of $I(L/K)$.

Take $L=K^{\mathrm{nr}}$, the maximal unramified extension of $K$. ThenThe subgroup $\{0\}\subset \mathbb{Z}$ is open in the discrete Weil-group topology, but it is not open in the profinite topology inherited from $\hat{\mathbb{Z}}$.

Let $E/K$ have good reduction over a local field of residue characteristic $p$, with $p\nmid n$. All solutions of $nQ=P$, for all $P\in E(K)$ together, lie in a single finite unramified extension of degree at most $nf$, where $f=[K(E[n]):K]$. First lift a solution over the algebraic closure of the residue field to a finite unramified extension; the multiplication isomorphism of a formal group law uniquely corrects the error in $E_1$. This also puts $E[n]$ in the maximal unramified extension. If $\sigma$ is its Frobenius generator, ${\sigma }^f$ fixes $E[n]$. For $nQ=P$, set $T={\sigma }^{f}Q-Q\in E[n]$. Then ${\sigma }^{nf}Q=Q+nT=Q$, uniformly in $P$.