\documentclass[11pt,a4paper]{article}
\usepackage[margin=25mm]{geometry}
\usepackage{fontspec}
\setmainfont{TeX Gyre Pagella}
\usepackage{amsmath,amssymb,mathtools}
\usepackage{graphicx}
\usepackage{xcolor}
\usepackage[unicode,colorlinks=true,linkcolor=blue,urlcolor=blue]{hyperref}
\usepackage{bookmark}
\usepackage{enumitem}
\setlist{itemsep=3pt,topsep=5pt}
\setlength{\emergencystretch}{3em}
\newcommand{\tightlist}{\setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}}
\newcommand{\passthrough}[1]{#1}
\newcommand{\pandocbounded}[1]{#1}
\makeatletter
\def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth\else\Gin@nat@width\fi}
\def\maxheight{\ifdim\Gin@nat@height>0.75\textheight0.75\textheight\else\Gin@nat@height\fi}
\makeatother
\setkeys{Gin}{width=\maxwidth,height=\maxheight,keepaspectratio}
\hypersetup{pdfauthor={OpenAI GPT-6.1 Sol, Codex, Ultra effort}}
\begin{document}
\section{Explicit local reciprocity and the existence
theorem}\label{explicit-local-reciprocity-and-the-existence-theorem}

\emph{Written by OpenAI GPT-6.1 Sol in Codex, Ultra effort, October
2026. Self-checked by the writing AI; no independent review is claimed.
Public domain (CC0).}

Lubin--Tate division fields turn local reciprocity into a formula. Write
an element of a local field as a unit times a power of a uniformizer.
Its valuation gives the arithmetic Frobenius on the unramified tower;
the inverse of its unit acts as a scalar on the division points. These
two actions account for every finite abelian extension.

We retain a nonarchimedean local field \(K\), valuation ring
\(\mathcal O\), uniformizer \(\pi\), and residue field \(\mathbf F_q\).
Write \(U=\mathcal O^\times\), \(U^{(n)}=1+\pi^n\mathcal O\) for
\(n\geq1\), and \(K_d/K\) for the unramified extension of degree \(d\).
The tower \(K^{\mathrm{ur}}\) has arithmetic Frobenius \(\varphi\),
inducing \(x\mapsto x^q\) on residues.

The proofs use the actual reciprocity construction from
\href{../frobenius-lifts-and-abstract-reciprocity.html}{Frobenius lifts
and abstract reciprocity}, the finite reciprocity isomorphism and
subgroup correspondence from
\href{../the-reciprocity-law-and-the-class-field-correspondence.html}{The
reciprocity law and the class field correspondence}, and its local
hypotheses and norm openness proved in
\href{../local-reciprocity-and-norm-groups.html}{Local reciprocity and
norm groups}. For \(L/K\) finite abelian, denote the arithmetic norm
residue symbol by \[
(a,L/K):K^\times\longrightarrow\operatorname{Gal}(L/K).
\] Its kernel is \(N_{L/K}L^\times\).

We also use Theorem 7.5 of
\href{../formal-groups-and-lubin-tate-modules.html}{Formal groups and
Lubin--Tate modules} and the division-field results of
\href{../lubin-tate-division-fields.html}{Lubin--Tate division fields}.
No explicit reciprocity formula is assumed.

\textbf{Prerequisite proof availability.} The named results below
identify specific programme lessons. The
\href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#lesson-9}{prerequisite
record} distinguishes published proofs, supplied owner texts awaiting
publication, and missing full proofs. A record or external reference is
not a supplied proof; arguments using an unavailable prerequisite retain
that dependency.

\subsection{1. Bringing a point back from a
completion}\label{bringing-a-point-back-from-a-completion}

The uniformizer-comparison series has coefficients in the completed
unramified field, so its value initially lies in a completion. The
Frobenius-lift definition, however, uses an algebraic fixed field. We
first establish the passage between them.

\subsubsection{Lemma. Separable algebraic elements in a completed
union}\label{lemma.-separable-algebraic-elements-in-a-completed-union}

Let \(M\) be a directed union of complete finite extensions of \(K\),
with their compatible valuations, and let \(\widehat M\) be its
completion. Any element of \(\widehat M\) that is separable algebraic
over \(M\) belongs to \(M\).

\textbf{Proof.} Let \(z\) be such an element. Multiplying it by an
element of \(K^\times\), we may suppose \(z\) is integral. A separable
polynomial over \(M\) vanishing at \(z\) can be multiplied by a nonzero
constant to obtain \(P\) with integral coefficients and \(P'(z)\ne0\).
Put \(s=v(P'(z))\geq0\).

Choose \(b\in M\) integral and sufficiently close to \(z\) that
\(v(b-z)>2s\). Taylor expansion with integral coefficients gives \[
v(P'(b))=s,\qquad v(P(b))>2s.
\tag{1}
\] Indeed \(P'(b)-P'(z)\) has valuation greater than \(s\); and
\(P(b)-P(z)\) is the sum of \(P'(z)(b-z)\) and terms containing
\((b-z)^2\).

Put all coefficients and \(b\) in one complete finite field
\(M_0\subset M\), and run Newton iteration there: \[
b_{j+1}=b_j-\frac{P(b_j)}{P'(b_j)},\qquad b_0=b.
\] If \(d_j=v(P(b_j))>2s\), its increment has valuation \(r_j=d_j-s>s\).
All iterates are integral, their derivatives retain valuation \(s\), and
cancellation of the linear Taylor term gives \[
d_{j+1}\geq2r_j,\qquad r_{j+1}\geq2r_j-s.
\] Thus \(r_j-s\) at least doubles; the sequence converges in \(M_0\) to
a root \(c\), with \(v(c-b)>s\). If an iterate is already a root, use it
as \(c\).

Both \(c\) and \(z\) are within valuation distance greater than \(s\) of
\(b\). Taylor expansion at \(z\) gives, if \(c\ne z\), \[
P(c)-P(z)=(c-z)\bigl(P'(z)+(c-z)Q\bigr),
\] where \(Q\) is integral because \(P,c,z\) are integral. The
parenthesis has valuation \(s\), so cannot be zero. Since both are
roots, this is impossible. Therefore \(z=c\in M_0\subset M\). Undo the
initial scaling. \(\square\)

\subsection{2. Computing the Frobenius
lift}\label{computing-the-frobenius-lift}

The reciprocity map \(r_{L/K}\) in lesson 4 runs in the opposite
direction to the norm residue symbol. If \(s\) is a positive Frobenius
lift of \(\sigma\in\operatorname{Gal}(L/K)\) to \(LK^{\mathrm{ur}}\),
and \(\Sigma\) is the fixed field of the closure of its powers, its
definition is \[
r_{L/K}(\sigma)
 =N_{\Sigma/K}(\pi_\Sigma)\pmod{N_{L/K}L^\times},
\tag{2}
\] with \(\pi_\Sigma\) any uniformizer of \(\Sigma\). Lesson 5 proves
that the symbol is the inverse of this isomorphism.

\subsubsection{Theorem 9.1. Explicit local
reciprocity}\label{theorem-9.1.-explicit-local-reciprocity}

For \(a=u\pi^r\), \(u\in U\), \(r\in\mathbf Z\), the compatible symbol
on \(K^{\mathrm{ur}}K_\pi\) acts as \[
(a,K^{\mathrm{ur}}K_\pi/K)|_{K^{\mathrm{ur}}}=\varphi^r,
\qquad
(a,K^{\mathrm{ur}}K_\pi/K)(x)=[u^{-1}]_f(x)
\tag{3}
\] for every division point \(x\) of a Lubin--Tate module for \(\pi\).

\textbf{Proof.} Fix \(n\geq1\), and put \(L=K_{\pi,n}\). Use the
polynomial series \[
f(X)=\pi X+X^q.
\] By Theorem 8.3 there is a unique \(\sigma\in\operatorname{Gal}(L/K)\)
acting on \(T_n(f)\) by \([u^{-1}]_f\).

Since \(L/K\) is totally ramified and \(K^{\mathrm{ur}}/K\) is
unramified, their intersection is \(K\); their finite levels are Galois
and linearly disjoint. Thus \[
\Gamma=\operatorname{Gal}(LK^{\mathrm{ur}}/K)
 \simeq\operatorname{Gal}(L/K)\times\widehat{\mathbf Z}.
\] Choose \(s=(\sigma,\varphi)\). It has degree 1. Its closed cyclic
subgroup is the graph \(t\mapsto(\sigma^t,t)\), where \(\sigma^t\) uses
the residue of \(t\) modulo the finite order of \(\sigma\). The quotient
has \(|\operatorname{Gal}(L/K)|=D_n\) elements. Its fixed field
\(\Sigma\) therefore has degree \(D_n\); the degree and ramification
calculation in lesson 4, equation (3), gives residue degree 1, so
\(\Sigma/K\) is totally ramified.

Set \(\pi'=u\pi\), and choose \(f'(X)=\pi'X+X^q\). Let
\(\widehat{\mathcal O}^{\,\mathrm{ur}}\) be the valuation ring of the
completion of \(K^{\mathrm{ur}}\). Theorem 7.5 supplies an
\(\mathcal O\)-linear isomorphism \[
\theta:F_f\longrightarrow F_{f'},\qquad
\theta\in\widehat{\mathcal O}^{\,\mathrm{ur}}[[X]],\qquad
\theta^\varphi=\theta\circ[u]_f.
\tag{4}
\] Its linear coefficient is a unit. The superscript applies \(\varphi\)
to the coefficients.

Let \(\lambda\) be a primitive point of \(T_n(f)\) and evaluate
\(y=\theta(\lambda)\) in the completion of \(M=LK^{\mathrm{ur}}\). The
isomorphism and its integral inverse converge there. Since \(\pi'=u\pi\)
differs by a unit, \(\mathcal O\)-linearity makes \(y\) a primitive
\(\pi'^n\)-division point for \(F_{f'}\). In particular it is a root of
the separable polynomial \(f'^{\circ n}\), by the root calculation in
Proposition 8.1. The preceding lemma gives \(y\in M\).

The valuation-preserving automorphism \(s\) extends continuously to the
completion. On the coefficients of \(\theta\) it acts as \(\varphi\),
and on \(\lambda\) as \([u^{-1}]_f\). Hence \[
s(y)=\theta^\varphi([u^{-1}]_f(\lambda))
     =\theta([u]_f([u^{-1}]_f(\lambda)))=y.
\tag{5}
\] Thus \(y\in\Sigma\). Theorem 8.3 applied to \(\pi'\) gives \[
[K(y):K]=D_n,\qquad K(y)=K_{\pi',n}.
\] This equals \([\Sigma:K]\), so \(\Sigma=K(y)\). Because \(f'\) is the
polynomial model, Proposition 8.4 gives the exact norm \[
N_{\Sigma/K}(-y)=\pi'=u\pi.
\tag{6}
\] The element \(-y\) is a uniformizer of \(\Sigma\). Substituting it in
(2), and using that \(\pi\) is a norm from \(L\), gives \[
r_{L/K}(\sigma)=u\pi\equiv u\pmod{N_{L/K}L^\times}.
\] Consequently \((u,L/K)=\sigma\), which is the inverse-unit action in
(3). The same norm fact gives \((\pi,L/K)=1\), so multiplying by any
integer power of \(\pi\) does not alter the action on this division
field.

The symbol on every finite unramified field is
\(\varphi^{v_K(a)}=\varphi^r\), by the unramified formula in lesson 6.
These formulas are compatible under restriction. Taking all \(n\) and
all finite unramified fields proves (3). Finally the integral comparison
between any two series for \(\pi\) is defined over \(K\) and commutes
with scalars; it transports the formula from the polynomial model to
every \(f\). \(\square\)

The inverse in \([u^{-1}]_f\) is forced by (4)--(5) and the direction of
the reciprocity isomorphism. Our normalization sends a uniformizer to
arithmetic Frobenius on the unramified tower.

\subsection{3. Which norm groups occur?}\label{which-norm-groups-occur}

For an integer \(d\geq1\), set \[
B_{d,n}=\langle\pi^d\rangle\,U^{(n)}
       =\{u\pi^r:u\in U^{(n)},\ d\mid r\}.
\tag{7}
\]

\subsubsection{Corollary 9.2. Norm groups of the explicit
fields}\label{corollary-9.2.-norm-groups-of-the-explicit-fields}

We have \[
N_{K_{\pi,n}/K}K_{\pi,n}^\times
 =\langle\pi\rangle U^{(n)},\qquad
N_{K_dK_{\pi,n}/K}(K_dK_{\pi,n})^\times=B_{d,n}.
\tag{8}
\]

\textbf{Proof.} The symbol on \(K_{\pi,n}\) is trivial exactly when
\([u^{-1}]_f\) acts identically on \(T_n(f)\). Proposition 8.1 says this
occurs exactly when \(u\equiv1\pmod{\pi^n}\). Finite local reciprocity
identifies this kernel with the norm group, giving the first equality.

On \(K_d\), the symbol is trivial exactly when \(d\mid r\). An
automorphism of the compositum is trivial exactly when both restrictions
are trivial. Its kernel is therefore \(B_{d,n}\), and reciprocity for
that finite abelian compositum gives the second equality. \(\square\)

\subsubsection{Theorem 9.3. The local existence
theorem}\label{theorem-9.3.-the-local-existence-theorem}

Every open subgroup \(H\subset K^\times\) of finite index is the norm
group of a unique finite abelian extension of \(K\).

\textbf{Proof.} Openness supplies \(n\geq1\) with \(U^{(n)}\subset H\).
Since \(K^\times/H\) is finite, the image of \(\pi\) has finite order
\(d\); then \(\pi^d\in H\). Thus \(B_{d,n}\subset H\).

Let \(A=K_dK_{\pi,n}\). Corollary 9.2 and finite reciprocity identify
\(K^\times/B_{d,n}\) with \(\operatorname{Gal}(A/K)\). Let \(J\) be the
image of \(H/B_{d,n}\), and put \(E=A^J\). Restriction of the
reciprocity symbol to \(E\) is the quotient by \(J\), by its
functoriality. Its kernel is exactly \(H\); finite reciprocity
identifies that kernel with \(N_{E/K}E^\times\). This constructs the
required abelian extension.

For uniqueness use the subgroup correspondence proved in Theorem 5.3:
finite abelian fields are determined by their norm groups, and norm
inclusion reverses field inclusion. Equivalently, if two such fields
have norm group \(H\), their compositum has the intersection of the two
norm groups, again \(H\). The norm-index formula makes the compositum
and each field have degree \([K^\times:H]\), so they coincide.
\(\square\)

The finite-index hypothesis matters. For example \(U^{(n)}\) alone is
open in \(K^\times\), but its quotient still has the infinite valuation
direction.

\subsection{4. The entire abelian
closure}\label{the-entire-abelian-closure}

\subsubsection{Theorem 9.4. The maximal abelian extension and
completion}\label{theorem-9.4.-the-maximal-abelian-extension-and-completion}

Inside a fixed separable closure, \[
K^{\mathrm{ab}}=K^{\mathrm{ur}}K_\pi.
\tag{9}
\] The reciprocity homomorphism \[
\operatorname{rec}_K:K^\times\longrightarrow
 \operatorname{Gal}(K^{\mathrm{ab}}/K)
\tag{10}
\] is continuous, injective and has dense image. It extends to an
isomorphism from the profinite completion taken with respect to
\textbf{open subgroups of finite index}. With the decomposition
\(K^\times=\pi^{\mathbf Z}\times U\), that completion is \[
\widehat{K^\times}_{\,\mathrm{open}}
 \simeq\widehat{\mathbf Z}\times U.
\tag{11}
\] This statement holds in both characteristic zero and characteristic
\(p\).

\textbf{Proof.} Every finite level of \(K^{\mathrm{ur}}K_\pi\) is
abelian, so that compositum lies in \(K^{\mathrm{ab}}\). Conversely let
\(E/K\) be finite abelian. Lesson 6 proves its norm group is open, and
finite reciprocity gives its finite index. As in Theorem 9.3 it contains
some \(B_{d,n}\). Since this is the norm group of \(A=K_dK_{\pi,n}\),
the reverse inclusion in the class field correspondence gives
\(E\subset A\). Taking the union over \(E\) proves (9).

The kernels on the cofinal fields \(A\) are \(B_{d,n}\). Each is open,
so the inverse-limit map (10) is continuous. If \(a=u\pi^r\) belongs to
its kernel, its unramified action is trivial for every degree \(d\).
Thus \(r\) is divisible by every positive integer, and \(r=0\). Its
action on every \(T_n\) is then trivial, so
\(u\in\bigcap_nU^{(n)}=\{1\}\). This proves injectivity.

Finite reciprocity is onto on each finite abelian field. Since finitely
many conditions in the Galois topology can be combined in their finite
compositum, every basic open subset of the inverse limit meets the image
of \(K^\times\). The image is dense.

The \(B_{d,n}\) are cofinal among open finite-index subgroups, as
already shown. Hence the required completion is \[
\varprojlim_{d,n}K^\times/B_{d,n}
 \simeq\varprojlim_{d,n}
       \bigl((\mathbf Z/d\mathbf Z)\times U/U^{(n)}\bigr)
 \simeq\widehat{\mathbf Z}\times U.
\] The last unit limit is \(U\) by the completeness argument in
Corollary 8.5. The finite reciprocity isomorphisms commute with
restriction, so their inverse limit is the asserted topological
isomorphism with the Galois group. \(\square\)

Under the action coordinates
\(\operatorname{Gal}(K^{\mathrm{ur}}/K)\simeq\widehat{\mathbf Z}\) and
\(\operatorname{Gal}(K_\pi/K)\simeq U\), the map is \[
u\pi^r\longmapsto(r,u^{-1}).
\tag{12}
\] The chosen uniformizer fixes these coordinates. Equation (9) shows
why the compositum is independent of that choice, even though \(K_\pi\)
itself need not be.

In positive characteristic the word ``open'' in the completion is
essential: one must use the local-field topology and its continuous
finite quotients, rather than silently adding every abstract finite
quotient. Formula (11) already uses the correct topology and requires no
characteristic-zero restriction.

\subsection{5. The local Kronecker--Weber
theorem}\label{the-local-kroneckerweber-theorem}

\subsubsection{\texorpdfstring{Corollary 9.5. Abelian extensions of
\(\mathbf Q_p\)}{Corollary 9.5. Abelian extensions of \textbackslash mathbf Q\_p}}\label{corollary-9.5.-abelian-extensions-of-mathbf-q_p}

Every finite abelian extension of \(\mathbf Q_p\) is contained in a
cyclotomic extension, and \[
\mathbf Q_p^{\mathrm{ab}}=\mathbf Q_p(\mu_\infty).
\tag{13}
\] For \(a=up^r\), \(u\in\mathbf Z_p^\times\), \[
(a,\mathbf Q_p(\zeta_{p^n})/\mathbf Q_p)(\zeta_{p^n})
 =\zeta_{p^n}^{\,u^{-1}\bmod p^n}.
\tag{14}
\] On prime-to-\(p\) roots of unity the symbol acts by arithmetic
Frobenius to the power \(r\).

\textbf{Proof.} The multiplicative Lubin--Tate module in lesson 8 gives
\[
(\mathbf Q_p)_p=\bigcup_n\mathbf Q_p(\zeta_{p^n}).
\] The unramified tower is generated by the prime-to-\(p\) roots of
unity: Corollary 3.1 of
\href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-07}{Unramified
and totally ramified extensions} constructs its degree-\(d\) field from
the roots of order \(p^d-1\), and shows that every prime-to-\(p\) root
lies in that tower. Theorem 9.4 now gives (13).

A finite extension contained in the union is generated by finitely many
elements. Each lies in some finite field
\(\mathbf Q_p(\zeta_{p^n},\zeta_m)\), \(p\nmid m\); taking a common
\(n,m\) puts them in \(\mathbf Q_p(\zeta_{p^nm})\). Thus the assertion
is containment in a finite cyclotomic field, not merely in an infinite
union.

The inverse-unit action from Theorem 9.1 and the exponent description in
lesson 8 give (14). On a root \(\zeta_m\) with \(p\nmid m\), arithmetic
Frobenius is \(\zeta_m\mapsto\zeta_m^p\). Thus the symbol is
\(\zeta_m\mapsto\zeta_m^{p^r\bmod m}\); for negative \(r\), the exponent
uses the inverse of \(p\) modulo \(m\). \(\square\)

For example, \(p\) acts trivially on every \(\mathbf Q_p(\zeta_{p^n})\),
while a unit \(u\) acts through its inverse residue modulo \(p^n\). The
class field of \(\langle p\rangle(1+p\mathbf Z_p)\) is
\(\mathbf Q_p(\zeta_p)\). When \(p=2\), that group is all of
\(\mathbf Q_2^\times\), and the first field is simply \(\mathbf Q_2\),
consistently with the degree formula.

\subsection{6. Exercises and complete
solutions}\label{exercises-and-complete-solutions}

\subsubsection{Exercise 1 --- easy}\label{exercise-1-easy}

Compute the symbol on \(\mathbf Q_3(\zeta_9)\) for \(a=3,2,4,7\).

\textbf{Solution.} The valuation part \(3\) is trivial on the division
field. The inverses of \(2,4,7\) modulo 9 are \(5,7,4\), respectively.
Consequently the four symbols send \(\zeta_9\), in the displayed order,
to \[
\zeta_9,\qquad\zeta_9^5,\qquad\zeta_9^7,\qquad\zeta_9^4.
\] These exponents all define automorphisms because they are units
modulo 9.

\subsubsection{Exercise 2 --- medium}\label{exercise-2-medium}

Derive both norm equalities in Corollary 9.2 directly from the explicit
symbol.

\textbf{Solution.} Write \(a=u\pi^r\). On \(K_{\pi,n}\), it is trivial
if and only if \(u^{-1}\) is the identity scalar modulo \(\pi^n\),
equivalently \(u\in U^{(n)}\). There is no restriction on \(r\), giving
the kernel \(\langle\pi\rangle U^{(n)}\). On \(K_d\) it is trivial if
and only if \(r\equiv0\pmod d\). Triviality on the compositum requires
both conditions, giving \(\langle\pi^d\rangle U^{(n)}\). Each field is
finite abelian, so its symbol kernel is its norm group by finite
reciprocity.

\subsubsection{Exercise 3 --- medium}\label{exercise-3-medium}

Construct the class field of an arbitrary open subgroup \(H\) of finite
index, and prove uniqueness among finite abelian extensions.

\textbf{Solution.} Choose \(U^{(n)}\subset H\), and choose \(d\) equal
to the order of \(\pi H\) in the finite group \(K^\times/H\). Then
\(B_{d,n}\subset H\). The explicit field \(A=K_dK_{\pi,n}\) has this
norm group, so its reciprocity isomorphism takes \(H/B_{d,n}\) to a
subgroup \(J\subset\operatorname{Gal}(A/K)\). For \(E=A^J\), the
restriction of the symbol has kernel exactly \(H\), hence
\(NE^\times=H\).

If \(E'\) is another finite abelian field with this kernel, the
compositum norm group is the intersection \(H\cap H=H\), by Theorem 5.3.
Therefore reciprocity gives \([EE':K]=[E:K]=[E':K]=[K^\times:H]\). The
inclusions force \(EE'=E=E'\). This also shows that the construction
does not depend on \(d,n\).

\subsubsection{Exercise 4 --- hard}\label{exercise-4-hard}

Compute Theorem 9.1 from the Frobenius-lift definition, explaining how
the auxiliary prime belongs to its algebraic fixed field.

\textbf{Solution.} Fix \(L=K_{\pi,n}\), use \(f=\pi X+X^q\), and let
\(\sigma=[u^{-1}]_f\) on its torsion. On \(M=LK^{\mathrm{ur}}\) choose
the lift \(s=(\sigma,\varphi)\); its fixed field \(\Sigma\) has degree
\(D_n\) and residue degree 1. Compare with the polynomial group for
\(\pi'=u\pi\), obtaining \(\theta^\varphi=\theta[u]_f\) as in (4). For a
primitive \(\lambda\), put \(y=\theta(\lambda)\).

The series converges in \(\widehat M\). Its value is primitive torsion
for \(\pi'\), hence satisfies the separable polynomial \(f'^{\circ n}\).
The complete Newton argument of the lemma descends it to \(M\). Equation
(5) makes it fixed by \(s\), so it lies in \(\Sigma\). The degree of
\(K(y)\) is \(D_n\), proving \(K(y)=\Sigma\); the polynomial norm
calculation gives \(N_{\Sigma/K}(-y)=u\pi\).

The Frobenius-lift definition thus gives \(r_{L/K}(\sigma)=u\pi\) modulo
norms. Since \(\pi=N_{L/K}(-\lambda)\), this class is \(u\). Inverting
reciprocity gives \((u,L/K)=\sigma\), and the norm of \(\pi\) makes
\((u\pi^r,L/K)=\sigma\) for every integer \(r\). On unramified fields
the already proved symbol is \(\varphi^r\). Passing through all levels
and transporting by the strict comparison for any other series gives the
full formula.

\subsection{Editable edition}\label{editable-edition}

The reading edition provides the complete LaTeX source of this lesson,
the cumulative course LaTeX and the editable source ZIP. The archive
contains all twenty-four Markdown lessons, complete LaTeX bodies,
original diagrams, metadata and reproduction instructions.

\subsection{References}\label{references}

The completed-ring Frobenius calculation and descent back to algebraic
fields give the explicit unit action. The remaining sections prove local
existence and the maximal abelian extension, with the sign conventions
kept explicit.

\begin{itemize}
\tightlist
\item
  \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne,
  Class Field Theory, version 4.03}.
\item
  \href{https://arxiv.org/abs/math/0606108v2}{Teruyoshi Yoshida, Local
  class field theory via Lubin--Tate theory, arXiv:math/0606108v2}.
\end{itemize}

The \href{../free-proofs.html}{proof guide} gives the lesson sequence
and the exact prerequisite record. External references accompany the
written arguments.

\end{document}