Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 28 (Local Fields)

Use the simple-root form of Hensel lemma. Let $R$ be a complete discrete valuation ring, let $\pi$ be a uniformizer, and let $F\in R[T]$. If $\bar{a}\in R/(\pi )$ is a root of $\bar{F}$ with ${\bar{F}}^{\prime }(\bar{a})\ne 0$, then there is exactly one root $a\in R$ reducing to $\bar{a}$.

Choose a lift $a_1$ of $\bar{a}$. Suppose $F(a_r)\in {\pi }^{r}R$. To construct a lift satisfying the next congruence, write $a_{r+1}=a_r+{\pi }^r{t}_r$. The polynomial expansion givesSince $F^{\prime }(a_r)$ is a unit, its residue is invertible in the residue field. There is a unique residue class of $t_r$ making this zero modulo ${\pi }^{r+1}$. Thus $a_{r+1}\equiv a_r(\mathrm{mod}{\pi }^r)$ and $F(a_{r+1})\equiv 0(\mathrm{mod}{\pi }^{r+1})$. This produces a Cauchy sequence in the complete discrete valuation ring. Its limit $a$ satisfies $F(a)=0$ by continuity of polynomial evaluation, and $a\equiv a_1(\mathrm{mod}\pi )$.

For uniqueness, suppose $a,b$ are two such roots. Polynomial division of their difference givesBecause $b-a\in \pi R$, the second factor is a unit. Hence $b-a=0$. The simple-root hypothesis is used both for constructing successive lifts and for this uniqueness argument.

First reduce the polynomial modulo seven. At residues $0,1,\dots ,6$, its values are respectively $4,2,6,1,0,2,6$. Its only residue root is therefore $4$, and $f^{\prime }(4)=45\equiv 3(\mathrm{mod}7)$ is a unit. By Hensel lemma, this residue root lifts uniquely to the 7-adic integers; any other root would have to reduce to the same residue root. Thus there is exactly one root in ${\mathbb{Z}}_7$, congruent to $4$ modulo seven.

Modulo five the values at $0,1,\dots ,4$ are $4,2,1,2,1$, so there is no residue root and hence no root in ${\mathbb{Z}}_5$. Modulo three the only possible residue root is $-1$. But writing $x=-1+3t$ givesThus that residue root cannot lift even modulo nine, and there is no root in ${\mathbb{Z}}_3$. A repeated residue root alone would not establish this obstruction; the second congruence is needed.

Modulo two both residues are roots. The even residue is simple because $f^{\prime }(0)=-3$ is a 2-adic unit, so Hensel lemma gives exactly one even root. For any odd 2-adic integer, $x^3\equiv x(\mathrm{mod}4)$, whence $f(x)\equiv -2x\equiv 2(\mathrm{mod}4)$. No odd root can exist. Consequently the full answer is

The two cosets of $2{\mathbb{Z}}_2$ are clopen sets in the 2-adic integers. The function is constant on each, so it is a locally constant function and therefore continuous. Its values on the nonnegative integers are $f(j)=(-1{)}^j$.

The Mahler coefficient at index $n$ is the nth forward difference operator applied at zero. HereThus the Mahler expansion of the parity function isTo verify the expansion directly, each binomial polynomial $\left(\genfrac{}{}{0ex}{}{x}{n}\right)$ takes values in ${\mathbb{Z}}_2$: it does so at nonnegative integers, which form a dense subset, and polynomial evaluation is continuous into ${\mathbb{Q}}_2$. Hence the nth summand has 2-adic absolute value at most $2^{-n}$ throughout the domain. The ultrametric inequality proves uniform convergence, and the sum is continuous. At each nonnegative integer $m$, the series becomes the finite binomial theorem identity $(1-2{)}^m=(-1{)}^m$. Equality on a dense subset and continuity prove equality everywhere. Finally, the coefficients in any such expansion are determined successively by its values at $0,1,2,\dots$, since $\left(\genfrac{}{}{0ex}{}{m}{n}\right)=0$ for $n>m$ and $\left(\genfrac{}{}{0ex}{}{m}{m}\right)=1$. This also verifies uniqueness without needing the general Mahler theorem.

All the local fields here carry normalized discrete valuations. Write ${\mathcal{O}}_F$ for the valuation ring, ${\mathfrak{m}}_F$ for its maximal ideal, and $k_F$ for its residue field. We use the basic degree identity $[B:A]=e(B/A)f(B/A)$ for finite extensions of non-Archimedean local fields, with $e$ the ramification index and $f$ the residue degree.

First construct the maximal unramified subextension of a local field extension inside an arbitrary finite extension $L/K$. Choose a primitive element of a field extension $\bar{z}$ for the finite residue field extension $k_L/k_K$. Let $\bar{h}$ be its monic irreducible minimal polynomial of an algebraic element, of degree $f$, and lift its coefficients to obtain a monic $h\in {\mathcal{O}}_K[T]$ of the same degree. Finite finite fields are perfect fields, so ${\bar{h}}^{\prime }(\bar{z})\ne 0$. By Hensel lemma applied in $L$, there is $z\in {\mathcal{O}}_L$ reducing to $\bar{z}$ with $h(z)=0$. For $U=K(z)$ we haveEvery inequality is therefore an equality. In particular $U/K$ is unramified, $k_U=k_L$, and $h$ is the minimal polynomial of an algebraic element $z$.

If $L/K$ is unramified, then $[L:K]=f$, so this construction gives $U=L$. Taking $x=z$ supplies the required generator, and $h^{\prime }(x)$ is a unit. Conversely suppose $L=K(x)$ with $x$ integral and its monic minimal polynomial of an algebraic element $g$ satisfying $g^{\prime }(x)\notin {\mathfrak{m}}_L$. Its coefficients lie in ${\mathcal{O}}_K$, since all conjugates of an integral element are integral and $K$ is complete. In the constructed unramified extension $U$ the element $\bar{x}\in k_U=k_L$ is a simple root of $\bar{g}$. Hensel lemma in $U$ produces a root $x^{\prime }\in U$ with residue $\bar{x}$. In $L$ both $x$ and $x^{\prime }$ lift this same simple residue root, so Hensel uniqueness gives $x=x^{\prime }$. Thus $L\subseteq U\subseteq L$, proving

For lifting residue-field embeddings of unramified extensions, use the generator $z$ and polynomial $h$ just constructed when $L/K$ is unramified. Given a $k_K$-field homomorphism $\iota :k_L\to k_M$, its value $\iota (\bar{z})$ is a simple root of $\bar{h}$. Hensel lemma in $M$ produces the unique root $z^{\prime }\in {\mathcal{O}}_M$ with that residue. The assignment $z\mapsto z^{\prime }$ defines a $K$-field homomorphism $K[T]/(h)\to M$; it induces $\iota$ because $\bar{z}$ generates $k_L$. Conversely any such field homomorphism preserves the unique extended valuation, hence induces a residue field homomorphism, and its image of $z$ is uniquely forced by Hensel uniqueness. This proves the asserted bijection. The same proof only requires that $M$ be a non-Archimedean local field containing $K$ with its compatible valuation; its finiteness over $K$ is not essential.

Now set $M=L$. The finite residue field extension has exactly $f=[L:K]$ automorphisms, cyclically generated by $a\mapsto a^{|k_K|}$. Their unique lifts give $[L:K]$ automorphisms of $L$. Consequently $L/K$ is a Galois extension, and reduction identifies its Galois group with that cyclic residue Galois group. Its generator is the arithmetic Frobenius:

For the compositum, choose a uniformizer ${\pi }_M$ of the totally ramified extension $M/K$. The Eisenstein generator of a totally ramified extension characterization, proved in the next solution, says that $M=K({\pi }_M)$ and its minimal polynomial of an algebraic element is Eisenstein over ${\mathcal{O}}_K$. A uniformizer of $K$ remains a uniformizer in the unramified extension $L$. All its Eisenstein coefficient conditions therefore remain true over ${\mathcal{O}}_L$. Thus the same Eisenstein polynomial is irreducible over $L$ and generates $LM=L({\pi }_M)$ as a totally ramified extension. This proves $LM/L$ is totally ramified, of degree $[M:K]$.

Here is an explicit example showing that the converse decomposition of an arbitrary extension can fail. Take $K={\mathbb{Q}}_3$ and $L=K(i)$ with $i^2=-1$. Since $T^2+1$ is irreducible and separable modulo three, the degree and residue-degree argument above shows that $L/K$ is unramified quadratic. PutThe element $1+i$ is a unit in ${\mathcal{O}}_L$, so $T^2-3(1+i)$ is Eisenstein over $L$. Hence $[N:L]=2$, $e(N/K)=2$, and $f(N/K)=2$. Moreover $i={\alpha }^2/3-1$ gives $N=K(\alpha )$, withWe verify its Galois group, since that is what rules out a totally ramified complement. By Hensel lemma there is $r\in K$ with $r^2=-2$: the residue root $1$ of $T^2+2$ is simple modulo three. Set $\beta =(1+i)/r$. ThenThe assignments $\sigma (i)=-i$ and $\sigma (\alpha )=\beta \alpha$ respect the defining relation, because $(\beta \alpha {)}^2=-3i(1+i)=3(1-i)$. They therefore define a $K$-automorphism of $N$. Its square satisfies ${\sigma }^2(i)=i$ and ${\sigma }^2(\alpha )=\bar{\beta }\beta \alpha =-\alpha$, so $\sigma$ has order four. There are thus four automorphisms of the degree-four extension, andThis is the cyclic quartic local extension without a totally ramified complement. Its unique quadratic subfield is $L$, fixed by ${\sigma }^2$, and is unramified. If $N=L^{\prime }{M}^{\prime }$ with $L^{\prime }/K$ unramified and $M^{\prime }/K$ totally ramified, then $[L^{\prime }:K]\le f(N/K)=2$ and $[M^{\prime }:K]\le e(N/K)=2$. To obtain the full degree four both would have degree two, forcing $M^{\prime }$ to be a totally ramified quadratic subfield. There is no such subfield. Thus $N/{\mathbb{Q}}_3$ is the required counterexample.

Let $d=[L:K]$, with normalized discrete valuations $v_K,v_L$. Suppose first that $L/K$ is totally ramified, so $e(L/K)=d$. Choose a uniformizer $x$ of $L$. The elements $1,x,\dots ,x^{d-1}$ are linearly independent over $K$: the nonzero terms $c_j{x}^j$ in any proposed relation have valuations $d{v}_K(c_j)+j$, distinct modulo $d$, and a sum with a unique least valuation cannot vanish by the ultrametric inequality. Hence $[K(x):K]\ge d$, proving $L=K(x)$.

Extend $v_K$ to an algebraic closure with values in $\mathbb{Q}$. Every conjugate of $x$ has valuation $1/d$, by uniqueness of the extended valuation on finite extensions of the complete field $K$. In its monic minimal polynomial of an algebraic element $g(T)=T^d+a_{d-1}{T}^{d-1}+\cdots +a_0$, the coefficient $a_{d-j}$ is a sum of products of $j$ conjugates, counted with multiplicity. Thus $v_K(a_{d-j})\ge j/d>0$ for $j\ge 1$, so every nonleading coefficient is in ${\mathfrak{m}}_K$. The constant coefficient is the product of all $d$ conjugates, so has valuation exactly one. This proves that $g$ is an Eisenstein polynomial. This argument also works for inseparable finite extensions, since the conjugates can be counted with their multiplicities.

Conversely let $g$ be an Eisenstein polynomial of degree $d$, with root $x$ and $L=K(x)$. The Eisenstein criterion proves irreducibility. For completeness, monic factors would have integral coefficients; their reductions would be positive-degree powers of $T$, so both constant coefficients would be divisible by the uniformizer of $K$, contradicting $v_K(g(0))=1$. To determine the root valuation, put $t=v_K(x)$ in the extended valuation. If $t<0$, the leading term is the unique term of least valuation; if $t=0$, the leading term again has uniquely least valuation zero. Both are impossible. If $t>0$, every nonconstant lower term has valuation at least $1+jt>1$. Cancellation in $g(x)=0$ therefore requires the leading term and constant term to have equal valuation:Since $t=v_L(x)/e(L/K)$ and $v_L(x)$ is a positive integer, $e(L/K)=d{v}_L(x)\ge d$. The opposite inequality follows from $[L:K]=e(L/K)f(L/K)=d$. Hence $e=d$, $f=1$, and $v_L(x)=1$. We have proved both the equivalence and the uniformizer assertion:

Now take $q=p^a$ with $a\ge 1$ and $\zeta ={\zeta }_{p^a}$. The translated cyclotomic polynomial isIn characteristic $p$, putting $U=T^{p^{a-1}}$ reduces this to ${\sum }_{j=0}^{p-1}(1+U{)}^j=U^{p-1}=T^{p^{a-1}(p-1)}$. Its constant coefficient over the integers is $p$. It is therefore Eisenstein at $p$, of degree $d=p^{a-1}(p-1)$. Consequently ${\mathbb{Q}}_p({\zeta }_{p^a})/{\mathbb{Q}}_p$ is totally ramified of degree $d$, with uniformizer ${\zeta }_{p^a}-1$. If $q=1$, the field is simply ${\mathbb{Q}}_p$, a trivial totally ramified extension.

For a general positive integer $n=p^{a}m$ with $p\nmid m$, the field generated by the prime-to-$p$ root of unity ${\zeta }_m$ is unramified. Indeed $T^m-1$ has distinct roots modulo $p$. In a finite unramified field whose residue field contains those roots, Hensel lemma lifts all of them uniquely. Reduction preserves their orders: a prime-to-$p$ root reducing to one must itself be one, by Hensel uniqueness at the root one of $T^m-1$. The arithmetic Frobenius therefore acts as ${\zeta }_m\mapsto {\zeta }_m^p$, and the extension degree is the multiplicative order of $p$ modulo $m$, with degree one when $m=1$.

The identityfollows by taking powers of a primitive nth root of unity and using the coprimality of $p^a,m$ in the opposite direction. The prime-to-$p$ field is unramified and the $p$-power field is totally ramified; unramified base change preserves total ramification proves that their compositum has the same residue degree as the prime-to-$p$ field. Thus the total ramification criterion for local cyclotomic extensions gives the full classification:This includes trivial extensions when $a=0$. For $p=2$ only $m=1$ is possible.

For a finite totally ramified Galois extension $L/K$ with Galois group $G$, define its lower ramification numbering byHere $G_0$ is the inertia group, which equals $G$, and $G_1$ is the wild inertia group. For a uniformizer $\pi$ the condition is equivalent to $v_L(\sigma (\pi )-\pi )\ge i+1$. To justify this uniformizer criterion for lower ramification groups, expand every $b\in {\mathcal{O}}_L$ as a convergent series ${\sum }_{j\ge 0}{c}_j{\pi }^j$ with the coefficients $c_j$ chosen from representatives of $k_K$ in ${\mathcal{O}}_K$. Such an expansion is obtained by successive subtraction of a residue representative and division by $\pi$, since total ramification gives $k_L=k_K$. The automorphism fixes each $c_j$, and $\sigma (\pi {)}^j-{\pi }^j$ is divisible by $\sigma (\pi )-\pi$. Thus the required valuation bound for $\pi$ implies it for every $b$, including the limit by completeness.

For $L={\mathbb{Q}}_p({\zeta }_{p^a})$, all primitive conjugates lie in $L$ and the Eisenstein degree is $\phi (p^a)$. HenceIf ${\sigma }_b\ne 1$, put $s=v_p(b-1)$, so $0\le s<a$. The element ${\zeta }^{b-1}$ is a primitive $p^{a-s}$th root of unity. Applying the already proved uniformizer result to its field, and multiplying its valuation by the ramification index of $L$ over that field, givesThe identity automorphism has infinite valuation. Therefore the lower ramification filtration of a prime-power cyclotomic extension is explicitlyThe order of the middle group is $p^{a-j}$. At odd $p$, $G_0/G_1$ has order $p-1$; at $p=2$, the condition $b\equiv 1(\mathrm{mod}2)$ is automatic, so $G_0=G_1$ when $a\ge 2$. The case $p=2,a=1$ has trivial Galois group, and the displayed formula remains valid. For $a=0$ all groups are trivial. The valuation calculation also provides the upper ramification numbering, if desired, by integrating the reciprocal indices with the Herbrand function; no change of numbering is implicit in the displayed answer.

Let $\pi \in K$ be a uniformizer. Because $L/K$ is unramified, the same $\pi$ is a uniformizer of $L$, and $v_L{|}_K=v_K$. The conjugates of $z\in L^{\times }$ all have the same valuation, soThis proves that every field norm has valuation divisible by $n$. Conversely $N_{L/K}({\pi }^r)={\pi }^{nr}$ for every integer $r$, so it remains to prove surjectivity of unit norms in unramified extensions.

Write $|k_K|=Q$, so $|k_L|=Q^n$. Reduction of the field norm on units is the finite-field normThe multiplicative group of a finite field is cyclic. The displayed power map on a group of order $Q^n-1$ has image of order $Q-1$, and is therefore onto $k_K^{\times }$. Given a unit $u\in {\mathcal{O}}_K^{\times }$, first choose $b_1\in {\mathcal{O}}_L^{\times }$ with $N(b_1)\equiv u(\mathrm{mod}\pi )$.

We can correct the norm at each subsequent power of $\pi$. For $r\ge 1$ and $a\in {\mathcal{O}}_L$, expanding the product over the cyclic Galois group givesAll products involving at least two correction terms are divisible by ${\pi }^{2r}$, hence by ${\pi }^{r+1}$. The residue field trace is onto even if the residue characteristic divides $n$. Indeed its polynomial $T(X)=X+X^Q+\cdots +X^{Q^{n-1}}$ is nonzero of degree $Q^{n-1}<Q^n$; it cannot vanish at every element of $k_L$. Its values lie in $k_K$, since raising $T(a)$ to $Q$ cyclically permutes the summands. Thus it is a nonzero $k_K$-linear map to the one-dimensional space $k_K$, and is surjective.

Suppose $N(b_r)\equiv u(\mathrm{mod}{\pi }^r)$. Choose ${\bar{a}}_r$ whose trace is the coefficient of ${\pi }^r$ in $u/N(b_r)-1$, and put $b_{r+1}=b_r(1+{\pi }^r{a}_r)$. Then $N(b_{r+1})\equiv u(\mathrm{mod}{\pi }^{r+1})$, and $b_{r+1}\equiv b_r(\mathrm{mod}{\pi }^r)$. By completeness the units $b_r$ converge to a unit $b$. The field norm is continuous, being a polynomial in coordinates of multiplication, so $N(b)=u$. Multiplying such a unit by a power of $\pi$ provesThe proof uses no local class field theory, and works in either characteristic.

The multiplicative Hilbert theorem 90 states that, for a finite Galois extension $L/K$ with Galois group $G$, every multiplicative cocycle $c:G\to L^{\times }$ satisfying $c_{\sigma \tau }=c_{\sigma }\sigma (c_{\tau })$ has the form $c_{\sigma }=\sigma (y)/y$ for some $y\in L^{\times }$. In the cyclic case $G=⟨\sigma ⟩$, it says

Here is the full existence argument. Distinct field homomorphisms $\sigma :L\to L$ are linearly independent as functions over $L$. To see this Artin independence theorem directly, assume a nonzero identically vanishing relation with the fewest nonzero coefficients. There are at least two terms. Choose two participating embeddings ${\sigma }_1,{\sigma }_j$ and an element $a$ with ${\sigma }_1(a)\ne {\sigma }_j(a)$. Evaluate the relation at $at$ and subtract ${\sigma }_j(a)$ times its value at $t$. The $j$th term disappears while the first remains nonzero, giving a shorter vanishing relation, a contradiction.

Apply this independence toThe coefficients are nonzero, so $A$ is not identically zero; choose $t$ with $y=A(t)\ne 0$. The cocycle identity gives $\tau (c_{\sigma }^{-1})=c_{\tau \sigma }^{-1}{c}_{\tau }$, and henceThis proves the general statement. In the cyclic case, given $N(x)=1$, set $c_1=1$ and $c_{{\sigma }^j}=x\sigma (x)\cdots {\sigma }^{j-1}(x)$ for $1\le j<n$. The norm-one identity makes this definition compatible with ${\sigma }^n=1$, and the cocycle identity follows by concatenating the products. The argument above gives $\sigma (y)/y=x$. Conversely that quotient has norm one because its conjugate product telescopes.

Finally let $L/K$ be unramified, and take its arithmetic Frobenius $\varphi$. For a unit $x$ of norm one, the cyclic theorem gives $b\in L^{\times }$ with $\varphi (b)/b=x$. Write $r=v_L(b)$ and choose the common uniformizer $\pi \in K$. Then $y={\pi }^{-r}b$ has valuation zero, and $\varphi (\pi )=\pi$ implies $\varphi (y)/y=\varphi (b)/b=x$. Thus the unit form of Hilbert theorem 90 isThe unramified hypothesis is what permits shifting an arbitrary integer valuation of $b$ by a power of a uniformizer belonging to $K$.

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.