\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} \title{Class field theory} \author{OpenAI GPT-6.1 Sol in Codex, Ultra effort} \date{October 2026} \maketitle \noindent Twenty-four lesson drafts and ninety-six solved exercises. Original programme text and figures are dedicated to CC0 1.0. The exact prerequisite record identifies incomplete companion proofs; no independent review or formal verification is claimed. \tableofcontents \clearpage \section{Profinite groups and infinite Galois theory}\label{profinite-groups-and-infinite-galois-theory} \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).} An automorphism of an infinite algebraic extension is determined by its actions on finite subextensions. These actions cannot be chosen independently: restricting an action from a larger field must recover the action already chosen on a smaller one. This compatibility turns an infinite Galois group into a compact topological group. Class field theory will describe its abelian quotients by arithmetic data. We use field operations, polynomial division, finite-dimensional linear algebra and the basic definitions of compact Hausdorff spaces and product topology. Section 0 proves the finite Galois correspondence and the Chinese remainder theorem used below. We also prove the required compactness and multiquadratic facts. The irreducibility and finite Galois groups of cyclotomic polynomials are proved in \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/cyclotomic-polynomials-and-their-automorphisms.html}{Cyclotomic fields}, Theorem 12.1. Here we use that theorem to pass to infinite extensions. Freely accessible comparisons are Milne's \emph{Fields and Galois Theory} and Kedlaya's \emph{Notes on class field theory}. \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-1}{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{0. The finite algebra behind the limits}\label{the-finite-algebra-behind-the-limits} We first establish the algebraic results needed to recover an infinite field from its finite pieces. \textbf{Finite automorphism lemma.} If a finite group \(H\) of distinct automorphisms acts on a field \(E\), and \(F=E^H\), then \([E:F]=|H|\). Moreover \(E/F\) is separable and normal, and \(H\) is its full automorphism group. \textbf{Proof.} Put \(n=|H|\). Given \(n+1\) elements \(x_1,\ldots,x_{n+1}\) of \(E\), form the matrix with entries \(h(x_j)\), indexed by \(h\in H\). Its homogeneous system has a nonzero solution over \(E\). Choose a solution \(c=(c_j)\) with least possible nonzero support and scale one nonzero coordinate to 1. Applying any \(h_0\in H\) to every equation permutes the rows, so \(h_0(c)\) is another solution. The difference \(h_0(c)-c\) has smaller support, because the normalized coordinate vanishes. Minimality forces this difference to be zero. Thus every \(c_j\) belongs to \(F\), and the identity row gives a nontrivial \(F\)-linear dependence among the \(x_j\). Consequently \(d=[E:F]\leq n\). Distinct field homomorphisms are linearly independent over the target field. Indeed, a nontrivial relation with least number of terms cannot have one term. If \(h_1\) and \(h_r\) occur, select \(y\) with \(h_1(y)\ne h_r(y)\). Subtract \(h_r(y)\) times the relation evaluated at \(x\) from the relation evaluated at \(yx\). This removes the last term and leaves the first nonzero, contradicting minimality. Each member of \(H\) is \(F\)-linear, while \(\operatorname{Hom}_F(E,E)\) has dimension \(d\) over \(E\): evaluation on an \(F\)-basis identifies it with \(E^d\). Hence \(n\leq d\), proving equality. For \(x\in E\), multiply \(X-z\) over the distinct members \(z\) of its \(H\)-orbit. The resulting polynomial is fixed coefficientwise by \(H\), hence belongs to \(F[X]\), and has distinct roots in \(E\). The minimal polynomial of \(x\) over \(F\) divides it. Thus that minimal polynomial is separable and splits in \(E\). This proves both separability and normality. Finally a finite extension of degree \(d\) has at most \(d\) embeddings into an algebraic closure. To prove the bound, choose finitely many generators and adjoin them successively; at each step an embedding has at most the minimal-polynomial degree many extensions, since the new generator must go to one of its roots. Multiplying these bounds gives the extension degree by the tower law. Since the \(n=d\) members of \(H\) already attain the bound, there are no other automorphisms. \(\square\) For a finite separable extension the same generator argument gives exactly the degree many embeddings: at each stage the transformed minimal polynomial has its full number of distinct roots in an algebraic closure, and sending the generator to any of them defines an extension of the embedding. If the extension is normal, all these embeddings preserve the field and are automorphisms. Preservation implies surjectivity because the image has the same finite dimension over the ground field. \textbf{Finite Galois correspondence.} Let \(E/K\) be finite, separable and normal, with group \(G\). Subgroups \(H\subset G\) correspond bijectively and in reverse inclusion order to intermediate fields, by \[ H\longmapsto E^H,\qquad M\longmapsto\operatorname{Gal}(E/M). \] The degree is \([E^H:K]=[G:H]\). Normal subgroups correspond to normal extensions of \(K\), with quotient group \(G/H\). \textbf{Proof.} The automorphism lemma gives \(\operatorname{Gal}(E/E^H)=H\) and \([E:E^H]=|H|\). For an intermediate field \(M\), the extension \(E/M\) is still separable and normal: the minimal polynomial over \(M\) of any element of \(E\) divides its split separable polynomial over \(K\). The embedding count gives \(|\operatorname{Gal}(E/M)|=[E:M]\). Its fixed field contains \(M\), and the automorphism lemma gives it the same relative degree as \(M\), so the fields are equal. The tower law proves the displayed degree formula. A \(K\)-embedding of \(M\) into an algebraic closure extends to an embedding of \(E\) by the successive root construction. Normality of \(E/K\) makes the extension an element of \(G\). It follows that \(M/K\) is normal exactly when \(M\) is preserved by every element of \(G\). The equality \(g(E^H)=E^{gHg^{-1}}\), and the bijection just proved, make this equivalent to normality of \(H\). In that case restriction maps \(G\) onto \(\operatorname{Gal}(M/K)\), with kernel \(H\), proving the quotient assertion. \(\square\) \textbf{Primitive element lemma.} A finite separable extension \(E/K\) is generated by one element. For infinite \(K\), write \(E=K(a_1,\ldots,a_r)\), and list its \(d=[E:K]\) embeddings into an algebraic closure, as counted above. For each pair of different embeddings, the linear polynomial \(\sum_j c_j(\sigma(a_j)-\tau(a_j))\) is nonzero. Their product is a nonzero polynomial in the \(c_j\). A nonzero polynomial over an extension of an infinite field cannot vanish on every tuple from that field: induction on the number of variables chooses values at which a nonzero coefficient stays nonzero, and then avoids finitely many roots in the last variable. Choose \(c_j\in K\) making this product nonzero. The element \(a=\sum_j c_ja_j\) then has \(d\) distinct images. Its minimal polynomial has degree at least \(d\), and at most \([E:K]=d\); thus \(K(a)=E\). For finite \(K\), the field \(E\) is finite. Its multiplicative group is cyclic. To prove this, let \(m\) be its exponent. For each prime power dividing \(m\), choose an element attaining that prime-power order by taking a suitable power of an element whose order contains it. The product of these elements has order \(m\), because they commute and their orders are pairwise coprime. Every nonzero element is a root of \(X^m-1\), so the polynomial root bound gives \(|E^\times|\leq m\). Since \(m\) divides the group order, equality follows. A generator of \(E^\times\) consequently generates \(E\) as a field. This proves the lemma in both cases. \textbf{Chinese remainder lemma.} For pairwise coprime positive integers \(n_1,\ldots,n_r\), reduction gives \[ \mathbf Z/(n_1\cdots n_r)\mathbf Z \simeq\prod_{j=1}^r\mathbf Z/n_j\mathbf Z. \] Put \(N=n_1\cdots n_r\). For each \(j\), Bezout's identity supplies integers \(b_j,c_j\) with \(b_j(N/n_j)+c_jn_j=1\). Then \(e_j=b_jN/n_j\) has residue 1 at \(n_j\) and 0 at the other moduli. Thus \(\sum_j a_je_j\) lifts any desired tuple of residues. An integer in the kernel is divisible by every \(n_j\), hence by their product, proving injectivity. The divisibility assertion follows successively from Bezout: if \(a,b\) are coprime and both divide \(x\), then \(b\) divides \(x/a\). Bezout itself follows by choosing the least positive integer in the ideal \((a,b)\) and applying integer division to show that it divides both generators. This also proves the construction without assuming the remainder lemma in advance. The same construction proves the \textbf{polynomial remainder lemma} used later for completions: if \(f_1,\ldots,f_r\in k[X]\) are pairwise coprime, then \[ k[X]/(f_1\cdots f_r)\simeq\prod_j k[X]/(f_j). \] Polynomial division shows that an ideal generated by two polynomials is generated by a polynomial of least nonzero degree: division by that polynomial leaves a smaller-degree member of the ideal, hence zero. Coprimality therefore gives a polynomial Bezout identity. Apply it to \(f_j\) and the product of the other factors to construct the same idempotent lifts as above. An element in the kernel is divisible by each factor, and successive Bezout identities make it divisible by the product. This proves surjectivity and injectivity over every field \(k\). \subsection{1. What compatibility means}\label{what-compatibility-means} For every positive integer \(n\), put \[ \widehat{\mathbf Z}=\varprojlim_n\mathbf Z/n\mathbf Z. \] The indexing order is divisibility. An element is a family \((a_n)_n\), with \(a_n\in\mathbf Z/n\mathbf Z\), satisfying \(a_m\bmod n=a_n\) whenever \(n\mid m\). Addition and multiplication are coordinatewise. Give each finite ring its discrete topology and the limit the subspace topology of their product. The map \(\mathbf Z\to\widehat{\mathbf Z}\) is injective: an integer divisible by every positive integer is zero. Its image is dense. Indeed, finitely many compatible congruence conditions are all satisfied by an integer representing the coordinate at the least common multiple of their moduli. Over \(\mathbf F_q\), restriction from \(\mathbf F_{q^m}\) to \(\mathbf F_{q^n}\), for \(n\mid m\), takes the exponent of the Frobenius automorphism modulo \(m\) to that exponent modulo \(n\). Thus compatible Frobenius exponents give exactly this inverse limit. Notice that the algebraic cyclic subgroup generated by Frobenius contains only integral exponents. Taking its closure adds all compatible profinite exponents. The same construction works for an inverse system of finite groups \(G_i\) with homomorphisms \(G_j\to G_i\). Its inverse limit consists of the compatible tuples in \(\prod_iG_i\). A \textbf{profinite group} is a topological group isomorphic to such an inverse limit. \begin{figure} \centering \pandocbounded{\includegraphics[keepaspectratio,alt={Three finite stages of the 2-adic inverse limit. Each residue modulo 8 maps to its residue modulo 4, then modulo 2. The highlighted compatible residues are 3, 3 and 1.}]{assets/compatible-residues.png}} \caption{Three finite stages of the 2-adic inverse limit. Each residue modulo 8 maps to its residue modulo 4, then modulo 2. The highlighted compatible residues are 3, 3 and 1.} \end{figure} \emph{Figure 1. Finite approximations to \(\mathbf Z_2\). Each edge is reduction of a residue. The highlighted branch is the image of the integer 3. Continuing the tree indefinitely gives \(\mathbf Z_2\); this finite drawing is a schematic of the compatibility condition.} \subsection{2. Finite quotients recognize compact groups}\label{finite-quotients-recognize-compact-groups} \subsubsection{Proposition 1.1. Characterization of profinite groups}\label{proposition-1.1.-characterization-of-profinite-groups} A topological group is profinite if and only if it is compact, Hausdorff and totally disconnected. In a profinite group, open normal subgroups form a neighborhood basis at the identity. Every open subgroup is closed and has finite index. Conversely, every closed subgroup of finite index is open. \textbf{Proof.} First, a product of finite discrete nonempty spaces is compact. To see this directly, take a family of closed subsets with the finite intersection property. The filter it generates extends to a maximal proper filter, or ultrafilter, by Zorn's lemma. In a finite partition, exactly one part belongs to an ultrafilter: two disjoint parts cannot both belong, and if no part belonged then all their complements would belong, giving an empty intersection. Apply this to each coordinate partition and choose its selected coordinate value. The tuple so obtained has every basic neighborhood in the ultrafilter. It belongs to each original closed set, because otherwise its open complement would contain such a neighborhood and contradict membership of that closed set in the filter. Thus every closed family with the finite intersection property has nonempty intersection, the equivalent closed-set formulation of compactness. A product containing an empty factor is empty and compact as well. Products of Hausdorff spaces are Hausdorff, since a differing coordinate separates two points. An inverse limit of finite discrete groups is a closed subset of their product: each compatibility condition is an equality of two continuous maps to a finite Hausdorff space. It is consequently compact and Hausdorff. Any two distinct tuples differ at a coordinate, and that coordinate separates them by a clopen set. A connected subset can therefore have only one point. Finite intersections of coordinate kernels form a basis of open normal subgroups. For the converse, we first justify the clopen neighborhoods that total disconnectedness supplies. In a compact Hausdorff space \(X\), let \(Q_x\) be the intersection of all clopen sets containing \(x\). This intersection is connected. Otherwise write it as two disjoint nonempty compact sets, with \(x\) in the first, and choose disjoint open neighborhoods \(U,V\) of those sets. Each point outside \(U\cup V\) is excluded by a clopen neighborhood of \(x\). Compactness gives finitely many such neighborhoods whose intersection \(C\) lies in \(U\cup V\). Both \(C\cap U\) and \(C\cap V\) are clopen in \(X\). The first contains \(x\) and excludes the second part of \(Q_x\), contradicting the definition of \(Q_x\). In a totally disconnected space it follows that \(Q_x=\{x\}\). Applying compactness once more to the complement of any neighborhood of \(x\) produces a clopen neighborhood inside it. Now let \(G\) be compact, Hausdorff and totally disconnected, and choose a clopen neighborhood \(C\) of 1 inside a prescribed neighborhood. Compactness of \(C\) and continuity of multiplication give an identity neighborhood \(V\) such that \(VC\subset C\). Replace \(V\) by a symmetric smaller neighborhood. For \(g\in V\), both \(gC\subset C\) and \(g^{-1}C\subset C\), so \(gC=C\). The stabilizer \[ H=\{g\in G:gC=C\} \] is a subgroup containing \(V\), hence open; also \(H\subset C\), since \(1\in C\). Its cosets form an open cover of the compact space \(G\), so there are finitely many. The core \(N=\bigcap_{g\in G}gHg^{-1}\) is an intersection of finitely many conjugates, and is an open normal subgroup inside \(C\). The natural map \[ G\longrightarrow\varprojlim_{N\text{ open normal}}G/N \] is injective because these \(N\) form a basis and \(G\) is Hausdorff. It is surjective: the cosets prescribed by a compatible tuple have the finite intersection property, and compactness supplies a point in their intersection. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Finally, an open subgroup has open cosets; its complement is a union of them, so it is closed. Compactness makes its index finite. If a subgroup is closed and has finite index, its complement is a finite union of closed cosets, so it is open. \(\square\) The word \emph{closed} in the last assertion matters. We will construct a subgroup of index 2 which is dense and proper. \subsection{3. The correspondence for an infinite extension}\label{the-correspondence-for-an-infinite-extension} Let \(L/K\) be algebraic, normal and separable. Every finite collection of elements of \(L\) lies in a finite Galois extension \(E/K\) contained in \(L\): take the splitting fields of their minimal polynomials and their compositum. Give \(G=\operatorname{Gal}(L/K)\) the \textbf{Krull topology}, with the groups \(\operatorname{Gal}(L/E)\), for such \(E\), as an identity neighborhood basis. Restriction induces a topological isomorphism \[ G\simeq\varprojlim_{E/K\text{ finite Galois in }L}\operatorname{Gal}(E/K). \] It is injective because the fields \(E\) exhaust \(L\). Conversely, a compatible family defines an automorphism on their union; its inverse is the family of inverse automorphisms. This also proves that the topology just defined is the inverse-limit topology. In particular, \(G\) is profinite. Here is the embedding extension argument in the infinite case. In an algebraic closure \(\Omega\) of \(K\), partially order embeddings of subfields containing an intermediate field \(M\), extending a specified \(K\)-embedding of \(M\), by extension of both domain and map. The union along a chain is an embedding of its union field. Zorn's lemma supplies a maximal embedding. If its domain omits \(x\in\Omega\), transform the minimal polynomial of \(x\) by that embedding and choose a root in \(\Omega\). Polynomial division identifies the adjoining field with the quotient by the minimal polynomial, so this choice extends the embedding to include \(x\), a contradiction. Its domain is therefore all of \(\Omega\). The image is algebraically closed: pull a polynomial's coefficients back to \(\Omega\), split it there, and apply the embedding to its roots. It contains \(K\), and every element of \(\Omega\) is algebraic over \(K\). Its polynomial over \(K\) already splits in this image, so that element belongs to the image. Hence the extended embedding is an automorphism of \(\Omega\). Normality makes it and its inverse preserve \(L\), so restriction gives an automorphism of \(L\). In particular, \(G\to\operatorname{Gal}(E/K)\) is surjective for every finite Galois \(E/K\) inside \(L\). \subsubsection{Theorem 1.2. Infinite Galois correspondence}\label{theorem-1.2.-infinite-galois-correspondence} The assignments \[ H\longmapsto L^H, \qquad M\longmapsto\operatorname{Gal}(L/M) \] are inverse, inclusion-reversing bijections between closed subgroups of \(G\) and fields \(K\subset M\subset L\). An open subgroup corresponds to a finite extension of \(K\), with \([M:K]=[G:H]\). A closed normal subgroup corresponds to a Galois extension of \(K\), and \[ \operatorname{Gal}(M/K)\simeq G/H \] as topological groups. For any subgroup \(H\), closed or otherwise, \[ \operatorname{Gal}(L/L^H)=\overline H. \] \textbf{Proof.} The group fixing \(M\) is closed: it is the intersection of the stabilizers of its elements, and each stabilizer is open and closed. If \(x\in L\setminus M\), the separable minimal polynomial of \(x\) over \(M\) has another root in \(L\), since \(L/M\) is normal. The embedding sending \(x\) to that root extends to an automorphism of \(L/M\). Thus the fixed field of \(\operatorname{Gal}(L/M)\) is exactly \(M\). The action on each element of \(L\) is locally constant, so \(H\) and \(\overline H\) have the same fixed field. If \(\sigma\notin\overline H\), choose finite Galois \(E/K\) such that the coset \(\sigma\operatorname{Gal}(L/E)\) misses \(H\). The restriction \(\sigma|_E\) is outside the image \(H_E\) of \(H\) in \(\operatorname{Gal}(E/K)\). Finite Galois theory supplies an element of \(E^{H_E}\) moved by \(\sigma\). That element lies in \(L^H\). This proves the last displayed equality and the bijection for closed subgroups. If \(H\) is open, it contains \(\operatorname{Gal}(L/E)\) for some finite Galois \(E/K\). Hence \(L^H\subset E\), and finite Galois theory computes its degree as \([G:H]\). Conversely, if \(M/K\) is finite, finitely many generators have an open common stabilizer, namely \(\operatorname{Gal}(L/M)\). Conjugation gives \(L^{\sigma H\sigma^{-1}}=\sigma(L^H)\). Thus \(H\) is normal exactly when \(M=L^H\) is stable under all \(K\)-automorphisms of \(L\). By the extension fact, stability is equivalent to normality of \(M/K\); separability is inherited. Restriction to \(M\) is then a continuous surjection with kernel \(H\). Its quotient isomorphism is a homeomorphism by compactness. \(\square\) For two intermediate fields, taking their compositum corresponds to intersecting their closed subgroups. Taking their intersection corresponds to the closure of the subgroup generated by the two subgroups. These follow directly from the inclusion-reversing bijection. \subsection{4. Frobenius and cyclotomic characters}\label{frobenius-and-cyclotomic-characters} Inside an algebraic closure of \(\mathbf F_q\), the roots of \(X^{q^n}-X\) form a field: raising to \(q^n\) preserves sums and products in characteristic \(p\), and also preserves inverses of nonzero roots. The derivative is \(-1\), so there are exactly \(q^n\) roots. This gives a field of degree \(n\) over \(\mathbf F_q\). Conversely, a degree-\(n\) extension has \(q^n\) elements and every nonzero element satisfies \(x^{q^n-1}=1\) by the finite-group order theorem. It is therefore exactly this root field. Its Frobenius has order \(n\): its \(n\)-th power is the identity, while a smaller positive power \(d0\), and put \(\wp(x)=x^p-x\). This is an additive map whose kernel is \(\mathbf F_p\). Every polynomial \(X^p-X-a\) is separable, since its derivative is \(-1\). For a root \(\alpha\), the function \(g\mapsto g(\alpha)-\alpha\) takes values in \(\mathbf F_p\) and is a continuous homomorphism. Changing the root adds a constant in \(\mathbf F_p\), and changing \(a\) by \(\wp(b)\), for \(b\in K\), replaces a root by \(\alpha+b\). Thus this construction depends only on \(a\bmod\wp(K)\). \subsubsection{Proposition 3.4. Artin--Schreier correspondence}\label{proposition-3.4.-artinschreier-correspondence} There is an isomorphism of \(\mathbf F_p\)-vector spaces \[ K/\wp(K)\simeq\operatorname{Hom}_{\mathrm{cont}}(G_K,\mathbf F_p). \] Subspaces \(D\subset K/\wp(K)\) correspond, preserving inclusion, to abelian Galois extensions of exponent dividing \(p\), by adjoining roots of \(X^p-X-a\) for the classes \(a\in D\). Their Galois groups are \[ \operatorname{Hom}_{\mathbf F_p}(D,\mathbf F_p), \] with the product topology. A finite-dimensional \(D\) gives degree \(p^{\dim D}\). \textbf{Proof.} A class maps to zero exactly when its root is fixed by \(G_K\), hence lies in \(K\); that is exactly \(a\in\wp(K)\). A continuous homomorphism to \(\mathbf F_p\) factors through some finite Galois \(E/K\). Additive Hilbert 90 realizes it as \(g(\alpha)-\alpha\), with \(\alpha\in E\). Then \(\wp(\alpha)\) is fixed and lies in \(K\), proving surjectivity. For a finite-dimensional subspace, its root field is finite Galois, and its action injects into the vector-space dual of \(D\). The translation pairing injects \(D\) into the dual of the Galois group, since a root fixed by that group lies in \(K\). Equal dimensions, or the two resulting cardinality inequalities, make both maps isomorphisms. Conversely, additive Hilbert 90 realizes all characters of a finite elementary abelian \(p\)-group by roots; they separate the group and therefore generate its field. This proves the finite correspondence and its inverse. Taking unions of finite-dimensional subspaces and inverse limits of their Galois groups proves the general case, exactly as in Theorem 3.3. \(\square\) For \(K=\mathbf F_p(t)\), the class of \(t^{-1}\) is nonzero in \(K/\wp(K)\). A pole of order \(m>0\) in \(b\) becomes a pole of order \(pm\) in \(b^p-b\); if \(b\) has no pole at \(t=0\), neither does \(b^p-b\). Neither possibility gives the pole of order 1 of \(t^{-1}\). Hence \(X^p-X=t^{-1}\) defines a cyclic extension of degree \(p\). \subsection{Exercises}\label{exercises} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item \textbf{Easy.} In \(\mathbf Q(i)\), find every \(b\ne0\) satisfying \(\sigma(b)/b=(3+4i)/5\), where \(\sigma(i)=-i\). \item \textbf{Medium.} Describe all quadratic and biquadratic extensions of \(\mathbf Q\) in terms of square classes. Apply the description to the two classes \(-2\) and 7. \item \textbf{Medium.} For a finite subgroup \(D\subset K^\times/K^{\times n}\), prove that the Kummer pairing of Theorem 3.3 is perfect without assuming a degree formula for radical extensions. \item \textbf{Hard.} Prove additive Hilbert 90 in arbitrary characteristic using a trace-one element. Use it to derive the Artin--Schreier character isomorphism and the correspondence for infinite elementary abelian extensions. \end{enumerate} \subsection{Solutions}\label{solutions} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item The element \(b_0=2-i\) satisfies the equation by direct multiplication. If \(b\) is another solution, then \(\sigma(b/b_0)=b/b_0\), so \(b/b_0\in\mathbf Q^\times\). All solutions are therefore \(b=r(2-i)\), with \(r\in\mathbf Q^\times\). \item A quadratic extension is determined by a one-dimensional subspace of \(\mathbf Q^\times/\mathbf Q^{\times2}\), and has the form \(\mathbf Q(\sqrt d)\), where \(d\) is a nontrivial square class. It has a unique squarefree integer representative. A biquadratic extension is determined by a two-dimensional subspace, generated by two independent classes \(a,b\); its field is \(\mathbf Q(\sqrt a,\sqrt b)\), and its three quadratic subfields correspond to \(a,b,ab\). Different bases of the same subspace give the same extension. The classes \(-2\) and 7 are independent by their valuations at 2 and 7. The resulting field has degree 4 and quadratic subfields \(\mathbf Q(\sqrt{-2})\), \(\mathbf Q(\sqrt7)\) and \(\mathbf Q(\sqrt{-14})\). \item Let \(L\) be the root field and \(G=\operatorname{Gal}(L/K)\). The map \(G\to\operatorname{Hom}(D,\mu_n)\) is injective because the roots generate \(L\). This also proves that \(G\) is abelian of exponent dividing \(n\). The map in the other direction is injective because a root fixed by all of \(G\) lies in \(K\). The finite character lemma above proves that each of these dual groups has the same order as its group. Thus \(|G|\leq|D|\leq|G|\), and both injections are surjective. This proves perfectness and simultaneously establishes the degree formula. \item Independence of the distinct embeddings makes the trace a nonzero \(K\)-linear map to \(K\), so choose \(t\) of trace 1. For an additive cocycle, \(S=\sum_h c(h)h(t)\) satisfies \(gS=S-c(g)\), and \(b=-S\) gives \(c(g)=gb-b\). No division by \(|G|\) occurs. Given a continuous character \(G_K\to\mathbf F_p\), factor it through finite Galois \(E/K\) and use this construction to obtain \(\alpha\) with \(g\alpha-\alpha=\chi(g)\). Then \(\alpha^p-\alpha\in K\). Its class maps to the character; the kernel consists exactly of \(\wp(K)\). The finite-dimensional translation pairing is perfect by the two injections and equality of dual dimensions. Any subspace is a union of finite-dimensional ones; their root fields have the corresponding union, and their Galois groups have the inverse limit. Conversely, every element of an infinite elementary abelian extension lies in a finite Galois subextension. Applying the finite inverse assignments there proves the infinite correspondence with no finiteness assumption on \(K/\wp(K)\). \end{enumerate} \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} Sections 1--3 prove independence and both forms of Hilbert 90 in every characteristic. The finite character lemma proves dual size and separation before the full finite and infinite Kummer and Artin--Schreier correspondences. \begin{itemize} \tightlist \item \href{https://www.jmilne.org/math/CourseNotes/FT.pdf}{J. S. Milne, Fields and Galois Theory, version 5.10}. \item \href{https://stacks.math.columbia.edu/}{The Stacks project, independence, Kummer and Artin--Schreier sections}. \end{itemize} The Stacks comparisons are Tags \href{https://stacks.math.columbia.edu/tag/0CKK}{0CKK}, \href{https://stacks.math.columbia.edu/tag/09I6}{09I6} and \href{https://stacks.math.columbia.edu/tag/09I7}{09I7}. Related independently written programme material belongs to the AI Integrated Stacks Project; it has not been reviewed by the official Stacks maintainers. The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Frobenius lifts and abstract reciprocity}\label{frobenius-lifts-and-abstract-reciprocity} \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).} The quotient by norms is the arithmetic object we want to understand. For an unramified extension of degree \(n\), valuation already labels its classes by \(\mathbf Z/n\mathbf Z\), and Frobenius supplies the matching cyclic generator. A ramified extension asks for something more: how can an automorphism still produce a class when it is not itself a residue-field Frobenius? We answer that question by making the automorphism into Frobenius over a different finite field inside an enlarged unramified tower. The proposed class is then the norm of a prime element. The proof has three distinct jobs: calibrate that prescription on unramified extensions, show that choices and products change it only by norms, and establish the two different functorial operations. These jobs determine the order below. Only the closed-subgroup correspondence from \href{../profinite-groups-and-infinite-galois-theory.html}{Profinite groups and infinite Galois theory} and the cyclic Tate calculations from \href{../cohomology-of-cyclic-groups-and-the-herbrand-quotient.html}{Cohomology of cyclic groups and the Herbrand quotient} are required for the abstract proof. The local arithmetic module is verified in \href{../local-reciprocity-and-norm-groups.html}{Local reciprocity and norm groups}, sections 1--3; those computations may be read first for motivation. Sections 4--5 of that lesson apply the abstract theorem after it has been proved. This separates the example from the proof it is meant to illustrate. \subsection{1. What a reciprocity class must do}\label{what-a-reciprocity-class-must-do} Consider a cyclic unramified local extension \(L/K\) of degree \(n\). Unit norms are surjective and \(v_K(Nb)=n v_L(b)\). Therefore the valuation map gives \[ K^\times/NL^\times\simeq\mathbf Z/n\mathbf Z, \qquad [\pi_K]\longmapsto1. \] For the kernel assertion, if \(v_K(a)=nm\), divide \(a\) by the norm of \(\pi_K^m\), viewed in \(L\). The remainder is a unit, hence a unit norm. This calculation uses only the unramified norm theorem, not reciprocity. It suggests the unique map taking arithmetic Frobenius to \([\pi_K]\). There are two different changes of field to track. Restricting an automorphism should correspond to taking a norm of the arithmetic element. Including an arithmetic element in a larger ground field should correspond to transfer of the automorphism. For example, in an unramified cyclic extension of degree six, the degree-three intermediate field has Frobenius equal to the third power of ground-field Frobenius. Transfer from the order-six group to its order-two subgroup sends a generator to that third power. On multiplicative groups a ground-field uniformizer stays a uniformizer, whereas its norm down a degree-three unramified extension has valuation three. Thus the two operations cannot be interchanged. We construct the map from automorphisms to norm classes. The next lesson proves that it is an isomorphism once the class field axiom holds; its inverse is the arithmetic Artin map. Throughout this lesson, neither existence of that inverse nor a reciprocity theorem is an assumption. \subsection{2. A degree map supplies the unramified direction}\label{a-degree-map-supplies-the-unramified-direction} Let \(G\) be profinite and let \[ d:G\longrightarrow\widehat{\mathbf Z} \] be a continuous surjection. We use field notation for its subgroup lattice: the ground field \(k\) means \(G_k=G\), a finite extension \(K/k\) means an open subgroup \(G_K\), and \(K\subset L\) means \(G_L\subset G_K\). Its degree is \([L:K]=[G_K:G_L]\). Composita mean intersections of closed subgroups. In applications these will be actual fields and Galois groups; all the arguments here use only the indicated subgroup operations. Put \(I=\ker d\) and let \(\widetilde k\) denote its fixed object. For finite \(K/k\), define \[ f_K=[\widehat{\mathbf Z}:d(G_K)],\qquad I_K=G_K\cap I,\qquad d_K=f_K^{-1}d|_{G_K}. \] The image \(d(G_K)\) is the open subgroup \(f_K\widehat{\mathbf Z}\). Division by \(f_K\) therefore defines a continuous surjection onto \(\widehat{\mathbf Z}\), with kernel \(I_K\). The fixed object of \(I_K\) is \(\widetilde K=K\widetilde k\), and \[ \operatorname{Gal}(\widetilde K/K)\simeq\widehat{\mathbf Z}. \] Its element of degree 1 is the \textbf{arithmetic Frobenius} \(\varphi_K\). For finite \(L/K\), set \[ f_{L/K}=[d(G_K):d(G_L)],\qquad e_{L/K}=[I_K:I_L]. \] Both are multiplicative in towers, and \[ [L:K]=e_{L/K}f_{L/K},\qquad f_L=f_Kf_{L/K},\qquad d_K|_{G_L}=f_{L/K}d_L. \tag{1} \] Indeed, \(I_KG_L\) is a subgroup of \(G_K\), and its two successive indices are \(f_{L/K}\) and \(e_{L/K}\). This proves the first equality without assuming that \(L/K\) is Galois. The other two follow directly from the images of \(d\). Call \(L/K\) \textbf{unramified} when \(e_{L/K}=1\), equivalently \(L\subset\widetilde K\). It is then cyclic and \([L:K]=f_{L/K}\). Call it \textbf{totally ramified} when \(f_{L/K}=1\). The largest unramified subextension of \(L/K\) is \(L\cap\widetilde K\), of degree \(f_{L/K}\). Fix for now a finite Galois \(L/K\). Write \[ \Gamma=\operatorname{Gal}(\widetilde L/K),\qquad J=\operatorname{Gal}(\widetilde L/\widetilde K). \] Here \(J\) is finite of order \(e_{L/K}\), and \[ 1\longrightarrow J\longrightarrow\Gamma \xrightarrow{d_K}\widehat{\mathbf Z}\longrightarrow1. \tag{2} \] Choose \(\phi\in\Gamma\) of degree 1. The map from \(\widehat{\mathbf Z}\) to the closure of its powers, sending 1 to \(\phi\), is an isomorphism: composing it with \(d_K\) gives the identity. Consequently every element of \(\Gamma\) has a unique expression \(j\phi^z\), with \(j\in J\), \(z\in\widehat{\mathbf Z}\). This splitting is a tool for the proof, not an extra datum in the final map. A \textbf{positive Frobenius lift} is \(s\in\Gamma\) with \(d_K(s)=n\) a positive ordinary integer. Its closed cyclic subgroup \(S=\overline{\langle s\rangle}\) is a copy of \(\widehat{\mathbf Z}\): the degree map on it is multiplication by \(n\), which is injective. In particular \(S\cap J=1\). If \(\Sigma\) is its fixed object, then \[ [\Sigma:K]=n|J|,\qquad f_{\Sigma/K}=n, \qquad\widetilde\Sigma=\widetilde L, \qquad s=\varphi_\Sigma. \tag{3} \] For the degree, the cosets of \(S\) in \(\Gamma\) are represented by \(\phi^i j\), \(0\leq id_K(s)\), and set \(h=s^{-1}s'\). This is a positive lift restricting to the identity on \(L\). Its fixed field contains \(L\), so its defining norm belongs to \(N_{L/K}A_L\), and \(R(h)=0\). Equation (13) now gives \(R(s')=R(s)+R(h)=R(s)\). \(\square\) \subsubsection{Proposition 4.2. The reciprocity homomorphism}\label{proposition-4.2.-the-reciprocity-homomorphism} The prescription \[ r_{L/K}(g)=R(s),\qquad s|_L=g, \] defines a homomorphism \[ r_{L/K}:\operatorname{Gal}(L/K)^{\mathrm{ab}} \longrightarrow A_K/N_{L/K}A_L. \tag{15} \] For a multiplicatively written \(A\), equation (13) says exactly that \(r\) is multiplicative. \textbf{Proof.} Positive lifts exist by section 2, and Proposition 4.1 makes the prescription independent of them. The product of lifts of \(g_1,g_2\) is a positive lift of \(g_1g_2\), so (13) proves the homomorphism property. The target is abelian, so the homomorphism kills commutators and factors through the finite group's abelianization. \(\square\) \subsection{7. Norms, conjugation and transfer}\label{norms-conjugation-and-transfer} \subsubsection{Proposition 4.4. Functoriality}\label{proposition-4.4.-functoriality} The maps (15) have the following three compatibilities. \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item If \(K\subset K'\), \(L\subset L'\), and both \(L/K\), \(L'/K'\) are finite Galois, then \[ N_{K'/K}\,r_{L'/K'}(g')=r_{L/K}(g'|_L). \tag{16} \] The norm on the left is taken on the corresponding quotient modules. \item For \(g\in G\), \[ r_{gL/gK}(g\sigma g^{-1})=g\,r_{L/K}(\sigma). \tag{17} \] \item If \(K'\) is intermediate in a finite Galois \(L/K\), the inclusion of modules satisfies \[ r_{L/K}(\sigma)\text{ viewed in }A_{K'}/N_{L/K'}A_L =r_{L/K'}(\operatorname{Ver}\sigma). \tag{18} \] Here \(\operatorname{Ver}:\operatorname{Gal}(L/K)^{\mathrm{ab}}\to \operatorname{Gal}(L/K')^{\mathrm{ab}}\) is transfer. \end{enumerate} \textbf{Proof of (16).} Lift \(g'\) to a positive \(s'\in\operatorname{Gal}(\widetilde L'/K')\), of degree \(n'\). Its restriction \(s\) to \(\widetilde L\) is positive, of degree \(f_{K'/K}n'\). If \(\Sigma'\) and \(\Sigma\) are their fixed fields, then \(\Sigma=\Sigma'\cap\widetilde L\). Their absolute residue degrees are both \(f_{K'}n'\), so \(f_{\Sigma'/\Sigma}=1\). For a prime \(\pi'\in A_{\Sigma'}\), equation (8) makes \(\pi=N_{\Sigma'/\Sigma}\pi'\) a prime of \(\Sigma\). Norm transitivity gives the exact equality \[ N_{\Sigma/K}\pi =N_{\Sigma'/K}\pi' =N_{K'/K}N_{\Sigma'/K'}\pi'. \] This gives (16). The quotient norm is well defined because \(N_{K'/K}N_{L'/K'}A_{L'}=N_{L'/K}A_{L'}\subset N_{L/K}A_L\). \textbf{Proof of (17).} A lift \(s\) with fixed field \(\Sigma\) becomes \(gsg^{-1}\) with fixed field \(g\Sigma\). Its degree remains the same, since the degree map has abelian target. The element \(g\pi_\Sigma\) is prime by the conjugation identity for valuations. Equation (5) gives (17) directly. For (18), we first establish the needed group-theoretic formula. Let \(T\) be finite and \(H\subset T\). Choose representatives \(R\) for left cosets \(T/H\). For \(\sigma\in T\), write \[ \sigma r=r' h_r(\sigma),\qquad r'\in R,\ h_r(\sigma)\in H. \] Then \(\operatorname{Ver}(\sigma)=\prod_{r\in R}h_r(\sigma)\) in \(H^{\mathrm{ab}}\). Replacing \(r\) by \(rk_r\) replaces its factor by \(k_{r'}^{-1}h_r(\sigma)k_r\); the \(k_r\) cancel in the abelian product. Thus transfer does not depend on representatives. The factors for \(\sigma\tau\) are those for \(\sigma\) at the cosets permuted by \(\tau\), followed by those for \(\tau\). Their products prove that transfer is a homomorphism, hence factors through \(T^{\mathrm{ab}}\). On an orbit of \(\sigma\) in \(T/H\), choose representatives \(r,\sigma r,\ldots,\sigma^{a-1}r\). The contribution is \(r^{-1}\sigma^a r\), where \(a\) is that orbit's length. Therefore \[ \operatorname{Ver}(\sigma)=\prod_r r^{-1}\sigma^{a_r}r \quad\text{in }H^{\mathrm{ab}}, \tag{19} \] with one representative per orbit. Now take \(T=\operatorname{Gal}(L/K)\), \(H=\operatorname{Gal}(L/K')\). The inclusion in (18) is well defined: decomposing \(T\) into right cosets \(H t\) shows that \(N_{L/K}a=N_{L/K'}(\sum_t t a)\), so every norm from \(L/K\) is a norm from \(L/K'\). Choose a positive lift \(s\in\Gamma\) of \(\sigma\), with fixed field \(\Sigma\), and write \(S=\overline{\langle s\rangle}\), \(\Gamma'=\operatorname{Gal}(\widetilde L/K')\). The finite coset spaces \(\Gamma/\Gamma'\) and \(T/H\) agree. Representatives for the \(S\)-orbits may therefore be chosen to lift those in (19). For each, put \[ s_r=r^{-1}s^{a_r}r,\qquad S_r=\Gamma'\cap r^{-1}Sr=\overline{\langle s_r\rangle}. \] The equality follows because the open subgroup of the procyclic group \(S\) stabilizing that coset has index \(a_r\). The element \(s_r\) is a positive lift over \(K'\): its degree is \(a_r d_K(s)/f_{K'/K}\), a positive integer since its restriction belongs to \(\Gamma'\). Let \(\Sigma_r\) be its fixed field. Then \[ \Sigma_r=(r^{-1}\Sigma)K', \qquad \Sigma_r/(r^{-1}\Sigma)\text{ is unramified of degree }a_r. \] Thus \(\pi_r=r^{-1}\pi_\Sigma\) is still prime in \(\Sigma_r\). Finally, the double cosets \(\Gamma'\backslash\Gamma/S\) partition the cosets used in the norm from \(\Sigma\). The part represented by \(r^{-1}\) has stabilizer \(S_r\) in \(\Gamma'\), so summing it gives \[ N_{\Sigma/K}\pi_\Sigma =\sum_r N_{\Sigma_r/K'}\pi_r \quad\text{in }A_{K'}. \tag{20} \] Each summand represents \(r_{L/K'}(r^{-1}\sigma^{a_r}r)\). Since reciprocity is a homomorphism, (19) and (20) prove (18). \(\square\) The arrows have different arithmetic meanings. Restricting a Galois automorphism corresponds to taking a norm. Transfer corresponds to including an arithmetic element in a larger field. Confusing those two operations reverses a reciprocity diagram. \subsection{8. Two complete degree models}\label{two-complete-degree-models} The simplest complete model isolates the degree contribution. After checking it, compare it with the local-field application: the degree is still present, but units now carry additional information and the descent lemma controls that information. Take \(G=\operatorname{Gal}(\overline{\mathbf F}_q/\mathbf F_q)=\widehat{\mathbf Z}\), with \(d\) the identity. Let \(A=\mathbf Z\) have trivial action, and let \(v\) be the inclusion in \(\widehat{\mathbf Z}\). Every \(A_K\) is \(\mathbf Z\). For \(K=\mathbf F_{q^m}\), the norm to the ground field is multiplication by \(m\), so (7) holds with \(V=\mathbf Z\), and \(v_K\) is the identity. Every extension is unramified and every unit module is zero, so (10) holds. For \(L/K\) of degree \(n\), the quotient module is \(\mathbf Z/n\mathbf Z\), and reciprocity sends arithmetic Frobenius to 1. A positive lift of degree \(b\) has fixed field of degree \(b\) over \(K\); the norm of its prime element 1 is \(b\), as required. This is the degree formation associated with finite fields. It uses \(\mathbf Z\), not the multiplicative group of the finite field, whose norm is surjective and could not give this quotient. For local fields the arithmetic module will instead be the multiplicative group of a separable closure, and the degree direction will be the unramified tower. The local reciprocity lesson will verify the valuation and unit hypotheses. The construction above already identifies the map once those hypotheses are verified. The same model can be built with \(A=\widehat{\mathbf Z}\), still treated as a discrete module with trivial action, and \(v\) the identity. A degree-\(m\) norm is multiplication by \(m\), so its image is \(m\widehat{\mathbf Z}\). All normalized valuations are the identity, the unit modules are zero, and the quotient for degree \(n\) is again \(\mathbf Z/n\mathbf Z\). An element of \(\widehat{\mathbf Z}\) need not be an ordinary integral multiple of 1, but its class in this finite quotient is. This explains why the finite-quotient condition on \(V\), rather than an integer-valued hypothesis, is the correct input to the proof. \subsection{Exercises}\label{exercises} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item \textbf{Easy.} Work out the degree formation of section 8, including its normalized valuations, norm groups and reciprocity for every \(\mathbf F_{q^{mn}}/\mathbf F_{q^m}\). \item \textbf{Medium.} Prove that changing the prime of a positive lift's fixed field does not change its reciprocity class. Identify the unramified extension on which the unit axiom is applied. \item \textbf{Medium.} Prove conjugation compatibility (17) directly from the norm definition, including preservation of prime elements. \item \textbf{Hard.} Starting with the unit descent lemma, prove the composition formula (13). Check explicitly the identity \(t_3=t_4t_1\) and why every term of (14) is a difference of units. \end{enumerate} \subsection{Solutions}\label{solutions} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item The subgroup for \(\mathbf F_{q^m}\) is \(m\widehat{\mathbf Z}\), so \(f_K=m\), \(d_K=d/m\), and \(I_K=0\). Trivial action gives \(A_K=\mathbf Z\). A norm along a degree-\(n\) extension is multiplication by \(n\), hence its group is \(n\mathbf Z\). Since \(N_{K/k}\) is multiplication by \(m\), \(v_K=m^{-1}vN_{K/k}\) is the identity. The only prime element is 1, the unit module is zero, and both Tate groups in the unit axiom vanish. The cyclic Galois group of order \(n\) maps isomorphically to \(\mathbf Z/n\mathbf Z\), with Frobenius going to 1 and its \(b\)-th power to \(b\bmod n\). \item Two primes differ by \(u\in U_\Sigma\). Equation (3) puts \(L\Sigma\) inside \(\widetilde\Sigma\), so \(L\Sigma/\Sigma\) is finite unramified cyclic. The norm-surjectivity half of (10) gives \(u=N_{L\Sigma/\Sigma}b\), with \(b\) a unit. Taking its norm to \(K\) gives \(N_{\Sigma/K}u=N_{L/K}N_{L\Sigma/L}b\), which is zero in the quotient by norms from \(L\). The extension required is \(L\Sigma/\Sigma\), not a possibly ramified extension \(L/K\). \item Representatives \(t\) for \(G_K/G_\Sigma\) become \(gtg^{-1}\) for \(G_{gK}/G_{g\Sigma}\), so the norm of \(ga\) is \(gN_{\Sigma/K}a\). The same identity for the norm to \(k\), together with unchanged \(f_\Sigma\), yields \(v_{g\Sigma}(ga)=v_\Sigma(a)\). Thus \(g\pi_\Sigma\) is prime. A positive lift becomes \(gsg^{-1}\), of the same degree and with fixed field \(g\Sigma\). Applying its defining norm proves (17), and conjugation also carries the denominator norm subgroup to the corresponding denominator. \item Choose \(s_4=\phi^{n_1}s_2\phi^{-n_1}\), \(\pi_4=\phi^{n_1}\pi_2\). Then \[ t_4t_1 =\phi^{n_2}\phi^{n_1}s_2^{-1}\phi^{-n_1} \phi^{n_1}s_1^{-1} =\phi^{n_1+n_2}(s_1s_2)^{-1}=t_3. \] Form \(u\) as in section 6. Each prime has valuation 1, so \(w(u)=0\). Applying \(\phi-1\) replaces \(P_{n_i}(\phi)\pi_i\) by \((t_i-1)\pi_i\). Substitute \(\pi_3=a+\pi_4\), \(\pi_1=b+\pi_4\). The remaining coefficient of \(\pi_4\) is \(t_3-t_1-t_4+1=(t_4-1)(t_1-1)\), giving (14). Here \(a,b\), and \((t_1-1)\pi_4\) all have valuation zero. Thus \((\phi-1)u\in I_JU_{\widetilde L}\). Unit descent makes \(N_Ju\) a norm from \(L\); (11) identifies its class with \(R(s_1s_2)-R(s_1)-R(s_4)\), and (5) identifies \(R(s_4)=R(s_2)\). This proves composition with the chosen left-action convention. \end{enumerate} \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} Sections 2--3 give the full abstract degree and valuation hypotheses, including the general permitted value group. Sections 5--7 prove unit descent, independence, composition and all three functorialities. The central finite unramified tower used in descent is constructed explicitly. \begin{itemize} \tightlist \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The abstract hypotheses allow the full value group specified in section 3. The proof does not assume a splitting of that valuation sequence. The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{The reciprocity law and the class field correspondence}\label{the-reciprocity-law-and-the-class-field-correspondence} \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).} The preceding lesson constructed a homomorphism from Galois automorphisms to classes modulo norms. The class field axiom will make it an isomorphism. Its inverse, the norm residue symbol, expresses an automorphism in terms of an arithmetic element of the ground field. The resulting correspondence classifies finite abelian extensions by their norm subgroups. The cyclic axiom does not restrict the conclusion to solvable extensions. We will prove the cyclic case by a valuation argument, handle abelian extensions by cyclic quotients, and then use Sylow subgroups to obtain the theorem for every finite Galois group. This order keeps the group-theoretic reduction explicit. We retain the additive module notation, the degree map, the henselian valuation with image \(V\subset\widehat{\mathbf Z}\), and normalized valuations from \href{../frobenius-lifts-and-abstract-reciprocity.html}{Frobenius lifts and abstract reciprocity}. Thus \(V\) contains \(\mathbf Z\), and \(V/nV\simeq\mathbf Z/n\mathbf Z\). For multiplicative modules, sums below mean products. Cyclic Tate groups and their exact hexagon were proved in \href{../cohomology-of-cyclic-groups-and-the-herbrand-quotient.html}{Cohomology of cyclic groups and the Herbrand quotient}. \subsection{1. The class field axiom implies the unit axiom}\label{the-class-field-axiom-implies-the-unit-axiom} The \textbf{class field axiom} is the following condition for every finite cyclic extension \(L/K\), with \(T=\operatorname{Gal}(L/K)\): \[ |\widehat H^0(T,A_L)|=[L:K],\qquad \widehat H^{-1}(T,A_L)=0. \tag{1} \] The first assertion says that \(A_K/N_{L/K}A_L\) has exactly the extension degree many elements. The second says that every norm-zero element is a difference \((\sigma-1)a\), for a generator \(\sigma\) of \(T\). Knowing only the quotient of these two orders would not give either assertion separately. For an unramified cyclic extension of degree \(n\), (1) implies both parts of the unit axiom used in the preceding lesson. Indeed, valuation induces a surjection \[ A_K/N_{L/K}A_L\longrightarrow V/nV. \] Both groups have order \(n\), so it is an isomorphism. If \(u\in U_K\), it follows that \(u=N_{L/K}a\) for some \(a\in A_L\). The equation \(0=v_K(u)=n v_L(a)\) implies \(a\in U_L\), since \(\widehat{\mathbf Z}\) is torsion free. Thus unit norms are surjective. If \(u\in U_L\) has norm zero, (1) gives \(u=(\sigma-1)a\), \(a\in A_L\). Choose \(b\in A_K\) with \(v_K(b)=v_L(a)\). This is possible because \(v_K\) is onto \(V\), and unramified inclusion preserves normalized valuation. Then \(a-b\in U_L\) and \((\sigma-1)(a-b)=u\). Hence the norm-zero units are precisely the differences of units. This proof works for the full allowed value group \(V\); it does not require expressing \(a\) as an integer multiple of a prime plus a unit. Consequently the complete construction and functoriality of the preceding lesson apply. Write \[ Q_{L/K}=A_K/N_{L/K}A_L,\qquad r_{L/K}:\operatorname{Gal}(L/K)^{\mathrm{ab}}\longrightarrow Q_{L/K}. \tag{2} \] \subsection{2. Two elementary reduction tools}\label{two-elementary-reduction-tools} Suppose \(M/K\) is a Galois intermediate extension in a finite Galois \(L/K\). Norm transitivity gives an exact sequence \[ A_M/N_{L/M}A_L \xrightarrow{N_{M/K}} Q_{L/K} \longrightarrow Q_{M/K}\longrightarrow0. \tag{3} \] The last map is the quotient map. Its kernel is \(N_{M/K}A_M/N_{L/K}A_L\), exactly the image of the first map; the first map need not be injective. The corresponding Galois sequence is \[ \operatorname{Gal}(L/M)\longrightarrow \operatorname{Gal}(L/K)\longrightarrow \operatorname{Gal}(M/K)\longrightarrow1. \tag{4} \] Norm functoriality from Proposition 4.4 makes reciprocity commute with these two sequences, viewing each Galois group through its map to the abelianization when needed. In particular, if \(r_{M/K}\) and \(r_{L/M}\) are surjective, so is \(r_{L/K}\): lift a quotient class using the former, subtract its reciprocity value, and lift the remainder in the kernel of (3) using the latter. We also need that cyclic quotients detect elements of a finite abelian group \(B\). Here is a direct proof. Given \(b\ne0\), define a character \(\langle b\rangle\to\mathbf Q/\mathbf Z\) taking \(b\) to an element of its exact order. Extend a character \(\chi\) from a subgroup \(D\) to \(D+\langle x\rangle\) as follows. Let \(m\) be the least positive integer with \(mx\in D\), and choose \(c\in\mathbf Q/\mathbf Z\) with \(mc=\chi(mx)\). Define the extension by \(\chi(d+jx)=\chi(d)+jc\). Any relation between two such expressions has \(j-j'\) divisible by \(m\), which proves well-definedness. Repeating this finite process extends the character to \(B\). Its image is a finite subgroup of \(\mathbf Q/\mathbf Z\), hence cyclic: a common denominator embeds it in a cyclic group. Thus some cyclic quotient of \(B\) detects \(b\). \subsection{3. The cyclic reciprocity theorem}\label{the-cyclic-reciprocity-theorem} First suppose \(L/K\) is cyclic of degree \(n\), and totally ramified. We prove injectivity of \(r_{L/K}\) directly. Put \(\Gamma=\operatorname{Gal}(\widetilde L/K)\). Since \(L\cap\widetilde K=K\), restriction gives \[ \Gamma\simeq\operatorname{Gal}(L/K)\times \operatorname{Gal}(\widetilde K/K). \] Choose a generator \(\sigma\) of the first factor, acting trivially on \(\widetilde K\), and let \(\phi\) be the arithmetic Frobenius of the second, acting trivially on \(L\). They commute. The positive lift \(s=\sigma\phi\) has degree 1. Its fixed field \(\Sigma\) is a totally ramified cyclic extension of \(K\) of degree \(n\), by the degree and index calculation in section 2 of the preceding lesson. Take prime elements \(\pi_\Sigma,\pi_L\). Let \(M=L\Sigma\subset\widetilde L\). It is finite Galois over \(K\), because \(\Gamma\) is abelian, and is unramified over both \(L\) and \(\Sigma\). If \(M_0=M\cap\widetilde K\), then \(M/M_0\) is cyclic of degree \(n\), generated by the restriction of \(\sigma\). This follows because the inertia subgroup \(\langle\sigma\rangle\) has trivial intersection with the stabilizer of \(\Sigma\). Its restrictions identify it with both \(\operatorname{Gal}(L/K)\) and \(\operatorname{Gal}(\Sigma/K)\). Therefore its norm operator \(N=1+\sigma+\cdots+\sigma^{n-1}\) restricts to \(N_{L/K}\) on \(A_L\) and to \(N_{\Sigma/K}\) on \(A_\Sigma\). Use the normalized valuation \(w=v_M\). Both chosen primes have \(w=1\). Suppose \[ r_{L/K}(\sigma^j)=0,\qquad 0\leq jj\), so they are Cauchy. Completeness gives a limit \(B\in M\). The residuals \(u_{j+1}\) tend to 1 by the neighborhood-basis assumption. From \(u=(PB_j)u_{j+1}\) and continuity of \(P\) we obtain \(u=PB\). Thus both required kernel-image equalities hold. \(\square\) Apply the lemma to \(M_j=W_{r+j}\). The preceding lattice estimates provide closedness, completeness and the neighborhood basis, while (8) provides the quotient vanishings. Consequently \[ \widehat H^0(T,W_r)=\widehat H^{-1}(T,W_r)=0. \tag{9} \] No claim that \(W_r\) itself is an induced multiplicative module was needed. It is its successive additive quotients that have the regular permutation basis. \subsubsection{Proposition 6.2. The local class field axiom}\label{proposition-6.2.-the-local-class-field-axiom} For cyclic \(L/K\) of degree \(n\), \[ h(T,U_L)=1,\qquad |K^\times/N_{L/K}L^\times|=n,\qquad \widehat H^{-1}(T,L^\times)=0. \tag{10} \] \textbf{Proof.} The exact sequence \(1\to W_r\to U_L\to U_L/W_r\to1\) has finite final module. Its Herbrand quotient is 1, and (9) gives \(h(T,W_r)=1\). Multiplicativity from Proposition 2.2 implies \(h(T,U_L)=1\), including finiteness of both Tate groups. The valuation sequence \[ 1\longrightarrow U_L\longrightarrow L^\times \xrightarrow{v_L}\mathbf Z\longrightarrow0 \] has trivial \(T\)-action on \(\mathbf Z\), whose quotient is \(h(T,\mathbf Z)=n\). Hence \(h(T,L^\times)=n\). Hilbert 90 gives the last vanishing in (10), so the numerator Tate group has order \(n\). This proves both parts of the class field axiom. \(\square\) The proof treats all local characteristics and all ramification. In positive characteristic, replacing the filtration by a logarithm would be invalid. Even in characteristic zero, writing \(1+\pi_K^r\Lambda\) does not on its own give an additive module isomorphism; (8) and the convergence argument are what justify the cohomology calculation. \subsection{4. Local reciprocity}\label{local-reciprocity} \subsubsection{Theorem 6.3. The local reciprocity law}\label{theorem-6.3.-the-local-reciprocity-law} For every finite Galois extension \(L/K\), the norm residue symbol induces a canonical isomorphism \[ (\ ,L/K):K^\times/N_{L/K}L^\times \xrightarrow{\sim}\operatorname{Gal}(L/K)^{\mathrm{ab}}. \tag{11} \] It respects norms, conjugation and inclusion with transfer as in Proposition 5.2. \textbf{Proof.} The pair in section 1 satisfies the degree and henselian-valuation conditions. Proposition 6.2 verifies the class field axiom at every finite cyclic extension of every finite base extension of \(K\), since each such base is again complete discretely valued with finite residue field. Theorem 5.1 applies, and inversion of its reciprocity isomorphism gives (11). Proposition 5.2 supplies its compatibilities. \(\square\) The inverse map is particularly concrete. For \(\tau\in\operatorname{Gal}(L/K)\), choose a positive Frobenius lift \(s\) to \(\operatorname{Gal}(L^{\mathrm{ur}}/K)\), let \(\Sigma\) be its fixed field, and take a uniformizer \(\pi_\Sigma\). Then \[ r_{L/K}(\tau)=N_{\Sigma/K}(\pi_\Sigma) \pmod{N_{L/K}L^\times}. \] The first abstract lesson proved all choice independence, including independence of that lift. Thus the construction determines the finite symbols canonically. \subsubsection{Proposition 6.4. The unramified formula}\label{proposition-6.4.-the-unramified-formula} If \(L/K\) is unramified, then \[ (a,L/K)=\varphi_{L/K}^{\,v_K(a)}. \tag{12} \] \textbf{Proof.} Every unit is a norm by Proposition 6.1, and Proposition 4.3 identifies the class of a uniformizer with arithmetic Frobenius. Write \(a=\pi_K^{v_K(a)}u\) and use multiplicativity of the symbol. Equivalently, this is formula (12) of the preceding lesson with the ordinary integer valuation. \(\square\) For the unramified quadratic extension of any \(\mathbf Q_p\), its symbol is nonidentity exactly when the valuation is odd. In particular, a unit cannot detect that unramified quadratic automorphism. \subsection{5. A direct proof that norms contain deep units}\label{a-direct-proof-that-norms-contain-deep-units} Finite index alone is not an openness argument in positive characteristic. We prove openness by a polynomial calculation valid in every characteristic. Let \(L/K\) be finite Galois. The trace map is not zero: it is the sum of its distinct automorphisms, and independence of field maps prevents that sum from being the zero map. This holds even if the residue characteristic divides \([L:K]\). The trace is \(K\)-linear with values in \(K\), so choose \(b\in L\) with \(\operatorname{Tr}_{L/K}b=1\). Expand its norm polynomial: \[ N_{L/K}(1+tb)=1+t+\sum_{j=2}^n a_jt^j, \qquad a_j\in K. \tag{13} \] Indeed, it is the product of \(1+t g(b)\), \(g\in T\), whose coefficients are invariant; its linear coefficient is the trace. If \(n=1\), the sum is empty and the conclusion below is immediate. Use an absolute value on \(L\) that extends the one on \(K\) literally. Choose \(\rho=|\pi_K|^s\), \(s\geq1\), so small that \[ \rho |b|_L<1,\qquad \lambda=\max_{2\leq j\leq n}|a_j|\rho^{j-1}<1. \tag{14} \] For any \(z\in K\) with \(|z|\leq\rho\), consider on the closed ball \(|t|\leq\rho\) the map \[ T_z(t)=z-\sum_{j=2}^n a_jt^j. \] It maps the ball into itself. Factoring \(t^j-u^j\) and applying the nonarchimedean inequality gives \[ |T_z(t)-T_z(u)|\leq\lambda|t-u|. \] Start with \(t_0=0\) and iterate \(t_{i+1}=T_z(t_i)\). The successive distances are at most \(\lambda^i\rho\), so the sequence is Cauchy; in an ultrametric space this also bounds all subsequent distances. Completeness gives a limit \(t\) in the closed ball, and continuity makes it a fixed point. Substitution in (13) gives \(N(1+tb)=1+z\). The first inequality of (14) makes \(1+tb\) a unit in \(L\). We have proved \[ 1+\mathfrak m_K^s\subset N_{L/K}U_L. \tag{15} \] \subsubsection{Proposition 6.5. Openness of norm groups}\label{proposition-6.5.-openness-of-norm-groups} Every norm subgroup of a finite separable extension of local fields is open and of finite index in \(K^\times\). \textbf{Proof.} For Galois extensions, (15) proves openness, while Theorem 6.3 gives finite index. For an arbitrary finite separable \(L/K\), take a finite Galois closure \(E/K\). Norm transitivity gives \(N_{E/K}E^\times\subset N_{L/K}L^\times\). The larger subgroup therefore also contains a deep unit neighborhood and has finite index. \(\square\) In particular, a norm group is open in the usual valuation topology. The converse existence assertion for every open subgroup of finite index is still ahead: it will be proved with the explicit Lubin--Tate extensions. The abstract class field correspondence already classifies extensions by the norm subgroups that occur. \subsection{\texorpdfstring{6. Four symbols over \(\mathbf Q_2\)}{6. Four symbols over \textbackslash mathbf Q\_2}}\label{four-symbols-over-mathbf-q_2} The polynomial \((X+1)^2+1=X^2+2X+2\) is Eisenstein, so Theorem 5.1 of the unramified lesson makes \(L=\mathbf Q_2(i)\) a ramified quadratic field. Its norm is \[ N(x+iy)=x^2+y^2. \] In particular \(N(1+i)=2\), \(N(1+2i)=5\). To exclude \(-1\) and 3, we check all possible denominators, not just integral sums of squares. For \(x,y\in\mathbf Q_2\), not both zero, let \(m\) be the smaller of their valuations, disregarding a zero entry. Write \(x=2^m a\), \(y=2^m b\), with \(a,b\in\mathbf Z_2\) and at least one odd. If exactly one is odd, \(a^2+b^2\) is odd and congruent to 1 modulo 4. If both are odd, their squares are 1 modulo 8, so \(a^2+b^2\) has valuation 1 and its quotient by 2 is again 1 modulo 4. Thus every nonzero norm has the form \[ 2^j u,\qquad u\equiv1\pmod4. \tag{16} \] The two units \(-1,3\) are 3 modulo 4 and cannot be norms. The group in (16) has index 2 in \(\mathbf Q_2^\times\), since the two unit classes modulo 4 are represented by 1 and 3. The norm group also has index 2 by Theorem 6.3 and is contained in this group, so they are equal. Let \(c\) denote the nontrivial automorphism \(i\mapsto-i\). We obtain \[ (2,L/\mathbf Q_2)=(5,L/\mathbf Q_2)=1,\qquad (-1,L/\mathbf Q_2)=(3,L/\mathbf Q_2)=c. \tag{17} \] The calculation uses a ramified extension, so its nontrivial symbol need not come from an odd valuation: here the uniformizer 2 is itself a norm. \begin{figure} \centering \pandocbounded{\includegraphics[keepaspectratio,alt={Norm membership for the unramified quadratic extension of Q₂ and for Q₂(i), by valuation and odd unit class modulo 4.}]{assets/quadratic-norms.png}} \caption{Norm membership for the unramified quadratic extension of Q₂ and for Q₂(i), by valuation and odd unit class modulo 4.} \end{figure} \emph{The two norm subgroups, projected onto the valuation \(k\in\mathbf Z\) and odd unit class modulo 4. Each cell represents every \(2^k u\) with that residue class, not one selected number. The displayed window is \(-2\leq k\leq2\); the conditions continue for every integer \(k\). Blue cells are exactly norms, by Proposition 6.4 and equation (16). The unramified condition concerns valuation, while the ramified condition concerns the unit. Original diagram by the writing AI.} \subsection{Exercises}\label{exercises} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item \textbf{Easy.} For an unramified degree-\(n\) extension, compute \(N_{L/K}L^\times\) and its index directly, without using the general reciprocity theorem. \item \textbf{Medium.} Compute the symbols of \(-1,2,3,5\) for \(\mathbf Q_2(i)/\mathbf Q_2\), proving both positive and negative norm assertions. \item \textbf{Medium.} Prove \(h(T,U_L)=1\) for cyclic \(L/K\) by constructing a cohomologically trivial open unit subgroup. Explain exactly where a normal basis and completeness enter. \item \textbf{Hard.} Prove that a norm subgroup is open in positive characteristic by solving the norm polynomial near 1. Account for extensions whose degree is divisible by the characteristic. \end{enumerate} \subsection{Solutions}\label{solutions} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item Unramified unit norms are onto by the independently proved Proposition 4.1 of the unramified lesson. A ground-field uniformizer is an upper uniformizer, and its norm is \(\pi_K^n\). Every upper element is a power of it times a unit, so \(N L^\times=\pi_K^{n\mathbf Z}U_K\). The valuation map identifies the quotient with \(\mathbf Z/n\mathbf Z\); its index is \(n\). \item The norms of \(1+i\) and \(1+2i\) are 2 and 5, respectively. For an arbitrary potential norm \(x^2+y^2\), remove the common factor \(2^{2m}\) as in section 6. One odd entry gives an odd unit 1 modulo 4; two odd entries give exactly one factor of 2 and an odd unit 1 modulo 4. This excludes both \(-1\) and 3, including rational 2-adic denominators. Since the Galois group has order 2, the kernel is the norm group and each nonnorm has symbol \(c\). The four values are those of (17). \item Choose the integral normal basis from section 2 and its lattice \(\Lambda\). Pick \(c\) with \(\pi_K^c\mathcal O_L\subset\Lambda\), and choose \(r\geq c+1\). The product estimate (7) makes \(W_r=1+\pi_K^r\Lambda\) a subgroup and identifies each \(W_j/W_{j+1}\), \(j\geq r\), with the induced additive residue module. Both its Tate groups vanish. For invariant \(u\), choose norm preimages successively modulo \(W_{j+1}\); for norm-one \(u\), choose \(\sigma-1\) preimages in the same way. The residuals stay invariant or norm one, respectively. Completeness makes the products of corrections converge, and continuity of norm and \(\sigma\) proves both vanishing assertions for \(W_r\). It is open of finite index because it contains \(1+\pi_K^{r+c}\mathcal O_L\). The quotient \(U_L/W_r\) is finite, of Herbrand quotient 1; multiplicativity gives \(h(T,U_L)=1\). The basis is used on the additive successive quotients, not as an asserted multiplicative linearization. \item First treat a Galois extension. Independence of its distinct automorphisms makes \(\operatorname{Tr}:L\to K\) nonzero, hence onto. Choose trace-one \(b\); tracing 1 would fail when the characteristic divides the degree, which is why that shortcut is not used. Expand \(N(1+tb)=1+t+\sum_{j\geq2}a_jt^j\). Choose \(\rho=|\pi_K|^s\) with \(\rho|b|_L<1\) and every \(|a_j|\rho^{j-1}<1\). For \(|z|\leq\rho\), iterating \(t\mapsto z-\sum_{j\geq2}a_jt^j\) stays in the ball and contracts distances by a fixed factor below 1. The complete field supplies a fixed point with \(N(1+tb)=1+z\), and \(1+tb\) is an upper unit. Thus all of \(1+\mathfrak m_K^s\) consists of norms. For a separable extension that is not Galois, apply this to its Galois closure and use the norm tower. This proves openness by an actual neighborhood in the norm group, without assuming continuity of an abstract finite quotient. \end{enumerate} \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} Sections 1--3 verify both cyclic axioms by a proved normal basis and a complete filtered lifting lemma. This argument includes equal characteristic; it does not use an exponential or logarithm there. Sections 4--5 give reciprocity, norm limitation and openness. \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://www.mathi.uni-heidelberg.de/~schmidt/Neukirch-en/Neukirch_cft_02_may15.pdf}{Jürgen Neukirch, Class Field Theory --- The Bonn Lectures, Online Edition 2.0 (May 2015), edited by Alexander Schmidt}. \end{itemize} The filtered lifting argument above supplies the equal-characteristic case as well as mixed characteristic. The characteristic-zero exponential argument is not needed. The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Formal groups and Lubin--Tate modules}\label{formal-groups-and-lubintate-modules} \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).} The multiplicative expression \(X+Y+XY\) becomes a group law near zero because \(1+(X+Y+XY)=(1+X)(1+Y)\). A formal group retains such a law as a power series, before choosing the elements on which it will be evaluated. Lubin--Tate theory builds one whose multiplication by a uniformizer reduces to Frobenius. Its division points will furnish explicit abelian extensions in the next lesson. Two congruences drive the construction. The linear coefficient is a uniformizer \(\pi\), while reduction of the entire series is \(X^q\). One controls the coefficient to be solved; the other makes the obstruction divisible by \(\pi\). We prove that coefficient calculation, derive the group and its scalar action, and then compare different uniformizers using Frobenius on the completed unramified tower. We use \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-02}{Completions, the p-adic numbers and complete discretely valued fields}, Theorem 2.1 and Proposition 2.2, for completion, extension of isometries and integer approximation in \(\mathbf Z_p\); \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-04}{Extensions of complete valued fields}, Theorem 1.2, for completeness of finite extensions and their uniquely extended valuation; and \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-07}{Unramified and totally ramified extensions}, Theorem 2.1 and Corollary 3.1, for the unramified tower and its arithmetic Frobenius. The construction is algebraic in its power-series variables; convergence will enter only when we evaluate them or complete a coefficient ring. \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-7}{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. Formal laws and their inverses}\label{formal-laws-and-their-inverses} Let \(R\) be a commutative ring with 1. A \textbf{one-dimensional commutative formal group law} is \(F(X,Y)\in R[[X,Y]]\) with \[ \begin{gathered} F(X,Y)=X+Y+\text{terms of total degree at least 2},\\ F(X,0)=X,\quad F(0,Y)=Y,\quad F(X,Y)=F(Y,X),\\ F(F(X,Y),Z)=F(X,F(Y,Z)). \end{gathered} \tag{1} \] Substitution of series with zero constant term is well defined: a coefficient of a given total degree involves only finitely many terms. No analytic convergence is needed for these identities. A series \(h(X)=cX+\cdots\) with \(c\in R^\times\) has a unique compositional inverse. Solve \(h(k(X))=X\) degree by degree: \(k_1=c^{-1}\), and at degree \(m\) the unknown coefficient of \(k\) has coefficient \(c\), hence has a unique value. The same recursion constructs a left inverse; associativity of substitution identifies left and right inverses. If one starts with the linear, commutative and associative conditions alone, the identity conditions in (1) follow: \(h(X)=F(X,0)\) satisfies \(h\circ h=h\), and its invertible linear coefficient is 1, so composing with its inverse gives \(h=X\). The other identity follows by commutativity. \subsubsection{Proposition 7.1. The formal group inverse}\label{proposition-7.1.-the-formal-group-inverse} There is a unique \(\iota_F(X)=-X+\cdots\in R[[X]]\) satisfying \[ F(X,\iota_F(X))=0. \tag{2} \] It also satisfies \(F(\iota_F(X),X)=0\). \textbf{Proof.} The degree-one term forces \(-X\). Once coefficients below degree \(m\) are chosen, adding a new term \(bX^m\) to that truncation adds exactly \(bX^m\) to \(F(X,\iota_F(X))\) in degree \(m\). The coefficient comes from the linear \(Y\) term; every nonlinear term involving the change has higher degree. Choose \(b\) to cancel the known coefficient. This recursion works over any \(R\), proves uniqueness, and commutativity proves the other identity. \(\square\) If \(S\) is a separated complete \(R\)-algebra for the powers of an ideal \(I\), evaluation on \(x,y\in I\) converges in the \(I\)-adic topology. Its degree-\(m\) terms lie in \(I^m\). The identities (1)--(2) make \(I\) an abelian group, denoted \(F(I)\), with operation \(x+_F y=F(x,y)\). In a finite extension of a complete valued field, integral coefficients likewise converge on its maximal ideal, since terms of increasing degree have valuation tending to infinity. The additive law is \(G_a(X,Y)=X+Y\). The multiplicative law is \(G_m(X,Y)=X+Y+XY\); associativity and commutativity follow from multiplication of \(1+X,1+Y,1+Z\), and its inverse is \(-X/(1+X)\). A homomorphism \(h:F\to H\) is a zero-constant series satisfying \[ h(F(X,Y))=H(h(X),h(Y)). \] It is an isomorphism if its linear coefficient is a unit, since its compositional inverse then satisfies the reverse homomorphism identity by substitution. The endomorphisms of \(F\) form a ring, with sum \(F(h(X),k(X))\), product \(h\circ k\), zero series 0 and identity \(X\). Associativity of \(F\) supplies additive associativity; \(\iota_F\circ h\) supplies the negative; the homomorphism identity and associativity of composition give distributivity. This ring need not be commutative for an arbitrary formal group. \subsection{2. The coefficient lemma}\label{the-coefficient-lemma} Let \(K\) be a nonarchimedean local field, \(\mathcal O=\mathcal O_K\), \(\pi\) a uniformizer, and \(\mathcal O/\pi\mathcal O=\mathbf F_q\). A \textbf{Lubin--Tate series for \(\pi\)} is \(f\in\mathcal O[[X]]\) with \[ f(X)\equiv\pi X\pmod{\text{degree }2}, \qquad f(X)\equiv X^q\pmod\pi. \tag{3} \] Write \(\mathcal F_\pi\) for their set. The series \(\pi X+X^q\) is an example. \subsubsection{Lemma 7.2. The Lubin--Tate coefficient lemma}\label{lemma-7.2.-the-lubintate-coefficient-lemma} For \(f,g\in\mathcal F_\pi\) and \(L(X_1,\ldots,X_d)=\sum_i a_iX_i\), \(a_i\in\mathcal O\), there is a unique zero-constant series \(H\in\mathcal O[[X_1,\ldots,X_d]]\) with linear part \(L\) and \[ f(H(X))=H(g(X_1),\ldots,g(X_d)). \tag{4} \] \textbf{Proof.} Suppose a polynomial \(H_{0\), the unique solution of \(a-\alpha\varphi(a)=b\) is \[ a=\sum_{j\geq0}\alpha^j\varphi^j(b). \tag{11} \] The coefficients \(\alpha\) are fixed by \(\varphi\), so substitution telescopes and proves the equation. The terms tend to zero and the ring is complete. A difference of two solutions would satisfy \(a=\alpha\varphi(a)\), impossible for a nonzero element because \(\varphi\) preserves valuation. This proves existence and uniqueness in every homogeneous degree. Condition (9) is exactly the degree-one equation. We have proved (10) without appealing to a formal logarithm. \subsubsection{Theorem 7.5. Comparison over the completed unramified extension}\label{theorem-7.5.-comparison-over-the-completed-unramified-extension} Suppose \(\pi'=u\pi\), \(u\in\mathcal O_K^\times\). For any \(f\in\mathcal F_\pi\) and \(f'\in\mathcal F_{\pi'}\), there is an isomorphism of formal \(\mathcal O_K\)-modules \[ \theta:F_f\longrightarrow F_{f'} \quad\text{over }\widehat{\mathcal O} \] with linear term \(\varepsilon X\), where \(\varepsilon\) is a unit satisfying \[ \varphi(\varepsilon)=u\varepsilon. \] It satisfies \[ \theta^\varphi=\theta\circ[u]_f. \tag{12} \] \textbf{Proof.} Lemma 7.4 supplies \(\varepsilon\). Equation (9) holds for \(L=\varepsilon X\), so the twisted lemma constructs \(\theta\) satisfying \(f'\theta=\theta^\varphi f\). Compare \(\theta(F_f(X,Y))\) and \(F_{f'}(\theta(X),\theta(Y))\). The coefficients of \(F_f,F_{f'}\) are fixed by \(\varphi\). Their commuting identities and \(f'\theta=\theta^\varphi f\) therefore show that both compared series satisfy (10), with linear part \(\varepsilon(X+Y)\). Uniqueness proves the formal group homomorphism identity. Likewise, \(\theta[a]_f\) and \([a]_{f'}\theta\) satisfy (10) with linear term \(\varepsilon aX\), so they agree for every \(a\in\mathcal O_K\). Since \(\varepsilon\) is a unit, the formal compositional inverse exists and makes \(\theta\) an isomorphism. Finally, \(\theta^\varphi\) satisfies (10) by applying \(\varphi\) to its defining equality. The series \(\theta[u]_f\) satisfies the same equation because \([u]_f\) commutes with \(f\) and has fixed coefficients. Their linear terms agree, \(\varphi(\varepsilon)X=\varepsilon uX\), so uniqueness gives (12). \(\square\) The direction matters: with \(\pi'=u\pi\), the source of \(\theta\) is the group for \(\pi\), and \([u]_f\) in (12) acts on that source. Inverting the direction changes the corresponding unit equation. \subsection{\texorpdfstring{6. The multiplicative example over \(\mathbf Z_p\)}{6. The multiplicative example over \textbackslash mathbf Z\_p}}\label{the-multiplicative-example-over-mathbf-z_p} For \(a\in\mathbf Z_p\), the binomial series \[ [a](X)=(1+X)^a-1=\sum_{j\geq1}\binom aj X^j \tag{13} \] has integral coefficients. For fixed \(j\), \(\binom aj\) is a continuous polynomial function of \(a\) over \(\mathbf Q_p\). Approximate \(a\) by nonnegative integers, whose binomial coefficients are integral; closedness of \(\mathbf Z_p\) proves the assertion. For nonnegative integers, (13) respects \(G_m\) by the ordinary power identity. Taking coefficientwise limits proves the homomorphism identity for \(a\in\mathbf Z_p\), since each coefficient of a substituted identity depends on finitely many coefficients. The same reasoning applied to integer pairs gives \[ [a+b]=G_m([a],[b]),\qquad [ab]=[a]\circ[b]. \] The series \(f=(1+X)^p-1\) has linear term \(pX\) and reduction \(X^p\), so belongs to \(\mathcal F_p\). The group \(G_m\) and its series (13) satisfy the defining equations of Theorem 7.3; uniqueness identifies them with \(F_f\) and \([a]_f\). The other choice \(pX+X^p\) therefore yields a formal group isomorphic to \(G_m\) over \(\mathbf Z_p\). Every endomorphism \(h\) of \(G_m\) over \(\mathbf Z_p\) commutes with its integer multiplication \([p]\), because a homomorphism preserves a repeated group sum. If its linear coefficient is \(a\), Lemma 7.2 makes it the unique series commuting with \(f=[p]\) and starting with \(aX\). Thus it is (13), proving \[ \operatorname{End}_{\mathbf Z_p}(G_m)=\mathbf Z_p \] with the ring operations already described. This argument does not assert an analogous derivative classification for arbitrary formal groups in positive characteristic. \subsection{Exercises}\label{exercises} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item \textbf{Easy.} Verify the formal group identities for \(G_m\), its inverse, and the integral scalar series \((1+X)^a-1\), \(a\in\mathbf Z_p\). \item \textbf{Medium.} Construct the formal inverse of an arbitrary \(F\) over an arbitrary commutative ring, proving uniqueness without division by integers. \item \textbf{Medium.} Prove that every endomorphism of \(G_m\) over \(\mathbf Z_p\) is \((1+X)^a-1\) for a unique \(a\in\mathbf Z_p\). \item \textbf{Hard.} Prove both surjectivities in Lemma 7.4, identify both kernels, and explain why completing the unramified union is part of the argument. \end{enumerate} \subsection{Solutions}\label{solutions} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item Multiplying \(1+X,1+Y\) yields \(1+G_m(X,Y)\), so commutativity, associativity and the identity follow from multiplication in \(R[[X,Y,Z]]\). The inverse is \(-X/(1+X)=-X+X^2-\cdots\). For fixed \(j\), the continuous polynomial \(\binom aj\) takes integral values at nonnegative integers dense in \(\mathbf Z_p\); its value at \(a\) is therefore integral. The integer power identity gives the scalar homomorphism identity, and coefficientwise continuity extends it to every \(a\). The same extension for addition and composition gives the scalar ring action. \item Write the desired inverse as \(-X+\sum_{m\geq2}b_mX^m\). After coefficients below \(m\) are set, the degree-\(m\) coefficient of \(F(X,\iota(X))\) is a known element plus \(b_m\). A new \(b_mX^m\) entering a nonlinear term has degree at least \(m+1\), so its only contribution at degree \(m\) comes from the linear \(Y\). Set \(b_m\) to the negative of the known coefficient. This recursion uses no division and gives existence and uniqueness in every degree. Commutativity gives the left inverse identity as well. \item An endomorphism preserves \(p\) repeated sums, hence commutes with \(f=(1+X)^p-1\). Its linear coefficient \(a\) lies in \(\mathbf Z_p\). Lemma 7.2 with one variable says that exactly one series with linear part \(aX\) commutes with \(f\). The integral series (13) is such a series, so it is the endomorphism. Conversely all those series are endomorphisms by solution 1. Linear coefficients distinguish \(a\), and the addition and composition identities identify the endomorphism ring with \(\mathbf Z_p\). \item Reduce an additive target modulo \(\pi\) and solve \(y^q-y=c\) in \(\overline{\mathbf F}_q\); successive corrections \(\pi^r y_r\) solve the next residue error, producing a Cauchy sequence. For a unit target, first solve \(y^{q-1}=c\) with \(y\ne0\), then correct with factors \(1+\pi^r y_r\). Their Frobenius quotients have residue correction \(y_r^q-y_r\), so this also improves the error one depth at a time. The product is Cauchy and remains a unit; continuity gives the exact solution. A fixed integral element has residue in \(\mathbf F_q\); subtract its ground-field representative, divide by \(\pi\), and repeat. Its resulting ground-field digit series converges in \(K\), proving the additive kernel is \(\mathcal O_K\), and the unit kernel is \(\mathcal O_K^\times\). The correction residues need not stay in one finite residue extension, so their limit is guaranteed in the completion, not in the uncompleted algebraic union. \end{enumerate} \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} Sections 1--3 prove inversion, coefficient correction, every formal-group identity and scalar compatibility. The first nonlinear coefficients are computed explicitly. Sections 4--6 prove the Frobenius equations and the twisted construction over the completed unramified ring. \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. \clearpage \section{Lubin--Tate division fields}\label{lubintate-division-fields} \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).} A Lubin--Tate formal module gives a way to divide by a uniformizer. The points killed by \(\pi^n\) form one cyclic module over \(\mathcal O_K/\pi^n\). Its generators generate a field, and the units of that quotient act as every automorphism of the field. We prove these assertions by counting polynomial roots and applying Eisenstein's criterion. Fix a nonarchimedean local field \(K\), its valuation ring \(\mathcal O\), a uniformizer \(\pi\), and residue field \(\mathbf F_q\). Normalize \(v_K(\pi)=1\) and extend this valuation to a separable closure \(K^s\). For a Lubin--Tate series \(f\in\mathcal F_\pi\), write \(F_f\) and \([a]_f\) for the formal group and scalar action constructed in \href{../formal-groups-and-lubin-tate-modules.html}{Formal groups and Lubin--Tate modules}, Theorem 7.3. We also use \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-07}{Unramified and totally ramified extensions}, Theorem 5.1: a root of a monic Eisenstein polynomial of degree \(d\) over a complete discretely valued field generates a totally ramified extension of degree \(d\), and that root is a uniformizer. Finite extensions are complete with the uniquely extended valuation, by \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-04}{Extensions of complete valued fields}, Theorem 1.2. \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-8}{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. Evaluating the formal module}\label{evaluating-the-formal-module} Put \[ \mathfrak m_s=\{x\in K^s:v_K(x)>0\}. \] Here \(0\) has valuation \(+\infty\). For finitely many elements of \(\mathfrak m_s\), choose a finite extension of \(K\) containing them. Integral power series converge there, since their terms of degree tending to infinity have valuation tending to infinity. Thus \(F_f\), its inverse and every \([a]_f\) can be evaluated on \(\mathfrak m_s\). This does not require the separable closure itself to be complete. The identities proved in lesson 7 give \(\mathfrak m_s\) an \(\mathcal O\)-module structure, with addition \(F_f\). Define \[ T_n(f)=F_f[\pi^n] =\{x\in\mathfrak m_s:f^{\circ n}(x)=0\},\qquad T_0(f)=\{0\}. \tag{1} \] The superscript \(\circ n\) denotes composition, not the \(n\)th power of a series. A point is \textbf{primitive at level \(n\)} if it belongs to \(T_n(f)\setminus T_{n-1}(f)\). The particularly useful series \[ f_0(X)=\pi X+X^q \tag{2} \] is a polynomial. It permits a finite root calculation. For an arbitrary \(f\), Theorem 7.3 supplies an integral isomorphism \[ h=[1]_{f,f_0}:F_{f_0}\longrightarrow F_f,\qquad h(X)=X+\text{higher terms},\qquad f\circ h=h\circ f_0. \tag{3} \] Its inverse is integral, and both series can be evaluated in every finite extension on its maximal ideal. Consequently \(h\) identifies \(T_n(f_0)\) with \(T_n(f)\), including their scalar actions and primitive points. \subsubsection{Proposition 8.1. The division module}\label{proposition-8.1.-the-division-module} For each \(n\geq1\), \(T_n(f)\) has \(q^n\) elements and is a free module of rank one over \(\mathcal O/\pi^n\mathcal O\). If \(\lambda\) is any primitive point, the isomorphism is \[ \mathcal O/\pi^n\mathcal O\longrightarrow T_n(f),\qquad a\longmapsto[a]_f(\lambda). \tag{4} \] \textbf{Proof.} First use \(f_0\). The polynomial \(f_0^{\circ n}\) is monic of degree \(q^n\). All its roots have positive valuation. Indeed, for \(v_K(x)<0\), \[ v_K(f_0(x))=qv_K(x)<0, \] because the two summands have distinct valuations \(1+v_K(x)\) and \(qv_K(x)\). For \(v_K(x)=0\), its value likewise has valuation zero. Neither type of point can reach zero under iteration. At every point of positive valuation, \[ f_0'(x)=\pi+q x^{q-1}\ne0. \tag{5} \] In characteristic \(p\), the second summand is zero. In characteristic zero, \(v_K(q)\geq1\), so that summand has valuation strictly greater than 1, whereas \(\pi\) has valuation 1. This also gives \(f_0'(0)=\pi\ne0\). The derivative of the iterate is the product of these nonzero derivatives along the orbit of a root. Thus all \(q^n\) roots are simple and lie in \(K^s\). We have \(|T_n(f_0)|=q^n\). The nesting \(T_{n-1}\subset T_n\) is strict, so primitive points exist. Fix one, \(\lambda\). The kernel of \(a\mapsto[a]_{f_0}(\lambda)\) contains \(\pi^n\mathcal O\). Every nonzero \(a\in\mathcal O\) has the form \(\pi^r u\) with \(u\) a unit. The series \([u]_{f_0}\) is an invertible module map. If \(r0\). In \(P_n(\lambda)\), the leading term has valuation \(dt\), the constant term has valuation 1, and every intervening term has valuation at least \(1+it>1\). A sum equal to zero cannot have a unique term of least valuation: after dividing by that term, reduction in the residue field would give \(1=0\). Therefore \(dt=1\). If \(e\) is the ramification index of \(K(\lambda)/K\), its value group in this normalization is \(e^{-1}\mathbf Z\); thus \(d\) divides \(e\). The degree formula for complete discretely valued fields gives \(d\leq e\leq[K(\lambda):K]\), whereas the polynomial \(P_n\) gives the reverse bound \([K(\lambda):K]\leq d\). All these quantities are consequently \(d\), the residue degree is 1, and \(\lambda\) has normalized valuation 1 in its field. In particular \[ [K(\lambda):K]=D_n, \] total ramification, and the uniformizer assertion. By (8), this is the entire division field. Its Galois group and the right side of (10) both have \(D_n\) elements; the injection is consequently onto. For general \(f\), the strict isomorphism (3) preserves the generated field and the valuation of each nonzero point: if \(v_K(x)>0\), every higher term of \(h(x)-x\) has valuation strictly greater than \(v_K(x)\). A primitive \(h(\lambda)\) therefore remains a uniformizer. This proves all the assertions. \(\square\) The polynomial argument uses the specific choice (2). An arbitrary Lubin--Tate series can have infinitely many nonzero coefficients; it need not supply a polynomial quotient as in (11). The integral isomorphism transfers the field and module conclusions without making such an assumption. \subsubsection{Proposition 8.4. A uniformizer is a norm}\label{proposition-8.4.-a-uniformizer-is-a-norm} For every \(n\geq1\), \[ \pi\in N_{K_{\pi,n}/K}(K_{\pi,n}^{\times}). \tag{12} \] \textbf{Proof.} Choose a primitive \(\lambda\) for \(f_0\). Its minimal polynomial is \(P_n\), with constant coefficient \(\pi\), so \[ N_{K_{\pi,n}/K}(\lambda)=(-1)^{D_n}\pi. \] Since \(N(-1)=(-1)^{D_n}\), we obtain \[ N_{K_{\pi,n}/K}(-\lambda)=\pi. \tag{13} \] The field is independent of the chosen series, so (12) holds for it in every presentation. \(\square\) When \(q\) is even and \(n=1\), \(D_1=q-1\) is odd, and \(N(\lambda)=-\pi\). Formula (13) includes this case. A primitive point for a different series is still a uniformizer, but its norm need not have the particular value of the polynomial coordinate just used. \subsection{3. The infinite tower}\label{the-infinite-tower} The nesting of the division modules gives fields \[ K_{\pi,1}\subset K_{\pi,2}\subset\cdots,\qquad K_\pi=\bigcup_{n\geq1}K_{\pi,n}. \tag{14} \] \subsubsection{Corollary 8.5. The Galois group of the tower}\label{corollary-8.5.-the-galois-group-of-the-tower} There is a canonical isomorphism of topological groups, characterized by its action on division points, \[ \operatorname{Gal}(K_\pi/K)\simeq\mathcal O^\times. \tag{15} \] \textbf{Proof.} Multiplication by \(\pi\) maps \(T_{n+1}(f)\) onto \(T_n(f)\): in the cyclic module \(\mathcal O/\pi^{n+1}\), its image has \(q^n\) elements, lies in \(T_n\), and hence equals it. Choose primitive points compatibly with \([\pi]_f(\lambda_{n+1})=\lambda_n\). A preimage of a primitive level-\(n\) point must be primitive at level \(n+1\). If \(\sigma(\lambda_{n+1})=[u]_{f}(\lambda_{n+1})\), applying \([\pi]_f\) shows \(\sigma(\lambda_n)=[u]_f(\lambda_n)\). Thus the restriction maps in Theorem 8.3 correspond to reduction of unit classes. Infinite Galois theory from lesson 1 now yields \[ \operatorname{Gal}(K_\pi/K) \simeq\varprojlim_n(\mathcal O/\pi^n\mathcal O)^\times. \] For completeness, this last inverse limit is \(\mathcal O^\times\). Given compatible classes, choose representatives \(u_n\in\mathcal O^\times\). They satisfy \(u_{n+1}-u_n\in\pi^n\mathcal O\), so form a Cauchy sequence and have a limit \(u\in\mathcal O\). Its nonzero residue makes it a unit, and it represents all the classes. Uniqueness follows from \(\bigcap_n\pi^n\mathcal O=\{0\}\). Reduction is continuous, and its kernels \(1+\pi^n\mathcal O\) form a neighborhood basis, so the bijection and its inverse are continuous. \(\square\) The isomorphism uses scalar action and is independent of the chosen compatible generators. All finite levels are abelian, as is the tower. \subsection{4. Two concrete descriptions}\label{two-concrete-descriptions} For \(K=\mathbf Q_p\), \(\pi=p\), choose \[ f(X)=(1+X)^p-1,\qquad F_f(X,Y)=X+Y+XY. \] Lesson 7 proves \([a]_f(X)=(1+X)^a-1\) for \(a\in\mathbf Z_p\). Iterating gives \(f^{\circ n}(X)=(1+X)^{p^n}-1\). Therefore \[ T_n(f)=\{\zeta-1:\zeta^{p^n}=1\},\qquad K_{p,n}=\mathbf Q_p(\zeta_{p^n}). \tag{16} \] The root count and positive-valuation assertion can also be obtained by transferring from \(f_0\), as above. For a primitive \(p^n\)th root \(\zeta\), (10) says precisely \[ \sigma(\zeta)=\zeta^u,\qquad u\in(\mathbf Z/p^n\mathbf Z)^\times. \tag{17} \] For a \(p\)-adic scalar, choose an ordinary integer congruent to it modulo \(p^n\); Proposition 8.1 makes their actions identical on \(T_n\). Thus the exponent in (17) has an unambiguous meaning. For the polynomial \(f_0\) over general \(K\), its nonzero first-level points satisfy \[ \lambda^{q-1}=-\pi,\qquad K_{\pi,1}=K\bigl((-\pi)^{1/(q-1)}\bigr). \tag{18} \] This is a totally ramified extension of degree \(q-1\). If \(q=2\), its degree is one and \(\lambda=-\pi\) already belongs to \(K\). Changing the series for a fixed uniformizer preserves the tower. Changing the uniformizer can change even its first field. For example, over \(\mathbf Q_3\), the choices \(3\) and \(-3\) yield \(\mathbf Q_3(\sqrt{-3})\) and \(\mathbf Q_3(\sqrt3)\); Exercise 4 proves they are different. The next lesson will explain why adjoining the entire unramified tower removes this dependence. \subsection{5. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- easy}\label{exercise-1-easy} For \(K=\mathbf Q_p\), verify the torsion and Galois descriptions (16)--(17) using the multiplicative formal group. \textbf{Solution.} Composition of \(f(X)=(1+X)^p-1\) multiplies the exponent by \(p\), giving \(f^{\circ n}(X)=(1+X)^{p^n}-1\) by induction. Its zeros are exactly \(\zeta-1\). They lie in the positive-valuation domain because this torsion set is the image under (3) of the polynomial torsion set. A primitive point corresponds to a primitive root: otherwise its exponent order would divide \(p^{n-1}\). All other roots are powers of it, so the generated field is \(\mathbf Q_p(\zeta_{p^n})\). For integer \(a\), \([a]_f(\zeta-1)=\zeta^a-1\) follows from the multiplicative law. For any \(a\in\mathbf Z_p\), choose that integer modulo \(p^n\); their difference kills the division module. Theorem 8.3 makes every unit class a Galois automorphism, giving (17) and degree \((p-1)p^{n-1}\), including \(p=2\). \subsubsection{Exercise 2 --- medium}\label{exercise-2-medium} Prove that the Galois action on \(T_n(f)\) commutes with every scalar \([a]_f\). \textbf{Solution.} Put a division point and its conjugate in a finite Galois extension \(L/K\). The coefficients of \([a]_f\) belong to \(\mathcal O\subset K\). For each finite truncation, applying \(\sigma\) to its value equals evaluating it at \(\sigma x\). The unique extended valuation on \(L\) is \(\sigma\)-invariant, so \(\sigma\) is an isometry and preserves limits. Passing to the convergent series gives (9). Applying the same argument to \(F_f\) proves additivity. In particular \(\sigma\) takes points killed by \(\pi^n\) to points killed by \(\pi^n\), and is an \(\mathcal O\)-module automorphism there. \subsubsection{Exercise 3 --- medium}\label{exercise-3-medium} For the polynomial choice \(f_0=\pi X+X^q\), prove that \(f_0^{\circ n}/f_0^{\circ(n-1)}\) is Eisenstein. Explain the role of this choice for a general series. \textbf{Solution.} With \(A=f_0^{\circ(n-1)}\), composition is \(f_0(A)=\pi A+A^q\). The quotient is the polynomial \(\pi+A^{q-1}\). The polynomial \(A\) is monic of degree \(q^{n-1}\), has constant coefficient zero, and reduces to \(X^{q^{n-1}}\) modulo \(\pi\). Hence the quotient is monic of degree \((q-1)q^{n-1}\), has constant coefficient exactly \(\pi\), and reduces to the pure power \(X^{(q-1)q^{n-1}}\). These are all the Eisenstein conditions. For a general power series in \(\mathcal F_\pi\), its iterate is not necessarily a polynomial. The assertions about torsion and fields follow instead from the integral strict isomorphism with \(f_0\), which preserves primitive points, their fields and their valuations. This supplies the same degree and ramification conclusion without assuming finite polynomial degree for an infinite series. \subsubsection{Exercise 4 --- hard}\label{exercise-4-hard} Let \(K=\mathbf Q_p\) with \(p\) odd. For uniformizers \(\pi,\pi'\), prove \[ K_{\pi,1}=K_{\pi',1} \quad\Longleftrightarrow\quad \pi'/\pi\equiv1\pmod p. \tag{19} \] \textbf{Solution.} Put \(m=p-1\), \(u=\pi'/\pi\), and choose \(\alpha^m=-\pi\), \(\beta^m=-\pi'\). Equation (18) gives the two fields. If they are equal to \(L\), both \(\alpha,\beta\) are uniformizers of the same totally ramified extension. Their ratio is a unit of \(L\), and \[ (\beta/\alpha)^m=u. \] The residue field of \(L\) is \(\mathbf F_p\). Every nonzero residue has \(m\)th power 1, so reducing this equality gives \(u\equiv1\pmod p\). Conversely suppose \(u\in1+p\mathbf Z_p\). We construct \(a\in1+p\mathbf Z_p\) with \(a^m=u\). Start \(a_0=1\), and put \[ a_{j+1}=a_j+\frac{u-a_j^m}{m a_j^{m-1}}. \] The denominator is a \(p\)-adic unit. If \(a_j\in1+p\mathbf Z_p\) and the error \(u-a_j^m\) has valuation at least \(r\geq1\), the increment has valuation at least \(r\). Expanding the \(m\)th power cancels its linear error; all remaining terms contain at least the square of the increment, and have integral coefficients. The next error has valuation at least \(2r\). Thus the increments tend to zero with valuations at least \(2^j\), the sequence converges in \(1+p\mathbf Z_p\), and its limit \(a\) satisfies \(a^m=u\). Now \(a\alpha\) is a root of \(X^m+\pi'\). It generates \(K(\alpha)\), since \(a\in K^\times\); hence the two fields are equal. This proves (19). \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} Sections 1--2 prove root counts, the division module, independence of the defining series and the full Galois group. The valuation calculation in Theorem 8.3 proves the degree and total ramification directly. The norm sign and the whole tower are retained. \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. \clearpage \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. \clearpage \section{Abelian ramification, conductors and Hasse--Arf}\label{abelian-ramification-conductors-and-hassearf} \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).} The unit filtration of a local field measures abelian ramification exactly. Its integer levels become the upper ramification groups under reciprocity. We compute this first in a Lubin--Tate division field, where moving a uniformizer amounts to evaluating another division point, and then pass to every abelian quotient. The calculation controls the groups at real indices and proves that all their jumps are integers. Fix a nonarchimedean local field \(K\), valuation ring \(\mathcal O\), uniformizer \(\pi\), maximal ideal \(\mathfrak p\), and residue field \(\mathbf F_q\). Put \(U^{(0)}=U=\mathcal O^\times\) and \(U^{(j)}=1+\mathfrak p^j\) for \(j\geq1\). The explicit symbol and local existence theorem are proved in \href{../explicit-local-reciprocity-and-existence.html}{Explicit local reciprocity and the existence theorem}. We use the actual written ramification prerequisites in \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-09}{Ramification groups and the different of a local extension}, Proposition 1.1, equation (2.2) and Theorem 3.1, and \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-10}{Herbrand's function and the upper numbering}, Proposition 1.1 and Theorem 3.1. In particular, the latter's quotient theorem was proved without Hasse--Arf; its reference to a planned proof of Hasse--Arf is fulfilled here and is not a hypothesis of our argument. \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-10}{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. Conventions at a jump}\label{conventions-at-a-jump} For a finite Galois extension \(L/K\), normalize \(v_L(L^\times)=\mathbf Z\) and put \(G=\operatorname{Gal}(L/K)\). Its motion number and lower groups are \[ i_G(\sigma)=\min_{x\in\mathcal O_L}v_L(\sigma x-x),\qquad G_j=\{\sigma:i_G(\sigma)\geq j+1\}. \] Identity has infinite motion number. Use the real-index convention \[ G_t=G_{\lceil t\rceil}\quad(t\geq-1),\qquad \phi(t)=\int_0^t\frac{|G_s|}{|G_0|}\,ds\quad(t\geq0). \tag{1} \] On \([-1,0]\), put \(\phi(t)=t\). Let \(\psi=\phi^{-1}\), and \(G^v=G_{\psi(v)}\). Thus a group retains its value at a jump and decreases immediately afterward. The group \(G^0=G_0\) is inertia. Herbrand's quotient theorem says that for \(H\triangleleft G\), \[ (G/H)^v=G^vH/H\qquad(v\geq-1). \tag{2} \] Hilbert's different formula is \[ \delta_{L/K}:=v_L(\mathfrak D_{L/K}) =\sum_{j\geq0}(|G_j|-1). \tag{3} \] All sums of nonzero terms here are finite. \subsection{2. Motion in a Lubin--Tate field}\label{motion-in-a-lubintate-field} Let \(L=K_{\pi,n}\), \(n\geq1\), and choose a primitive division point \(\lambda\) for \(f(X)=\pi X+X^q\). Lesson 8 proves that \(\lambda\) is a uniformizer and \[ G\simeq(\mathcal O/\pi^n\mathcal O)^\times,\qquad |G|=D_n=(q-1)q^{n-1}. \] Write \(\sigma_u(\lambda)=[u]_f(\lambda)\). This is the scalar-action coordinate; the reciprocity symbol of a unit \(u\) is \(\sigma_{u^{-1}}\). For a nonidentity class, write \(u-1=\pi^k a\), where \(a\in U\) and \(0\leq k0\), \[ G^v=\{\sigma_u:u\in U^{(\lceil v\rceil)}\}, \tag{6} \] where unit classes are taken modulo \(\pi^n\); a level at least \(n\) gives identity. For \(-1\leq v\leq0\), the group is \(G\). \textbf{Proof.} Put \(b_k=q^k-1\), so \(b_0=0\). Formula (5) and the definition of \(G_j\) give \[ G_t=\operatorname{image}(U^{(k)}) \quad\text{for }b_{k-1}2\), since the residue-unit quotient has order \(q-1\). When \(q=2\), \(U=U^{(1)}\), so zero is not a jump. In particular \(q=2,n=1\) gives a trivial extension. \subsection{3. Every abelian extension}\label{every-abelian-extension} \subsubsection{Theorem 10.2. Units and upper ramification}\label{theorem-10.2.-units-and-upper-ramification} Let \(L/K\) be finite abelian, with Galois group \(G\). Then \[ (U,L/K)=G^0,\qquad (U^{(j)},L/K)=G^j\quad(j\geq1). \tag{9} \] More precisely, at every real \(v>0\), \[ G^v=(U^{(\lceil v\rceil)},L/K). \tag{10} \] On the maximal abelian extension the corresponding upper group is the closure of the reciprocity image of the same unit group. \textbf{Proof.} By Theorem 9.4, \(L\) lies in \(A=K_dK_{\pi,n}\) for some \(d,n\). Its Galois group is the product of the unramified group and the division-field group. The inertia group is the latter factor. The extension \(A/K_{\pi,n}\) is unramified, so the same primitive point \(\lambda\) is a uniformizer of \(A\), with the same normalized values on the division field. The maximal unramified subfield of \(A/K\) is \(K_d\). Over it, the uniformizer formula for inertial motion applies to \(\lambda\), and (4)--(5) remain unchanged. Thus all nonnegative lower groups and the denominator \(|G_0|=D_n\) in Herbrand's integral are those of the division field. Proposition 10.1 gives their upper filtration, with trivial unramified coordinate. The explicit reciprocity law sends units to precisely these groups: inversion preserves each \(U^{(j)}\). This proves (9)--(10) for \(A/K\). Restriction to \(L\) carries the symbol to its symbol on \(L\), and Herbrand's theorem (2) carries each real upper group onto that of \(L\). This proves the result for \(L/K\). For the infinite assertion, the upper groups are the inverse limits of the finite upper groups, by Proposition 4.1 of the Herbrand lesson. On every finite collection of levels, their prescribed elements can be tested in the finite compositum, where (9)--(10) give surjectivity from the relevant unit group. Its image is therefore dense in the inverse-limit upper group. It is contained in that closed group, proving the closure statement. \(\square\) \subsubsection{Corollary 10.3. Hasse--Arf}\label{corollary-10.3.-hassearf} Every upper jump of a finite abelian extension of local fields is an integer. \textbf{Proof.} Formula (10) makes the filtration constant on each \(k-1c_\chi-1\). Its last upper jump is therefore \(c_\chi-1\). If \(a(\chi)=0\), it is trivial on inertia and has no nonnegative ramification jump. \(\square\) For the full cyclotomic extension \(\mathbf Q_p(\zeta_{p^n})/\mathbf Q_p\), its norm group is \(p^{\mathbf Z}(1+p^n\mathbf Z_p)\). The conductor exponent is \(n\), except for \(p=2,n=1\), when the extension is trivial and its exponent is zero. \subsection{5. Discriminants as sums of conductors}\label{discriminants-as-sums-of-conductors} Let \(\mathfrak d_{L/K}\) denote the discriminant ideal, defined by the determinant of the trace matrix in an integral \(\mathcal O_K\)-basis. We first record explicitly its relation to the different: \[ v_K(\mathfrak d_{L/K})=f(L/K)\,\delta_{L/K}. \tag{14} \] Indeed, if \(T\) is that invertible trace matrix, the trace dual is \(T^{-1}\mathcal O_K^{[L:K]}\), while the integral lattice is \(\mathcal O_K^{[L:K]}\). The quotient of the former by the latter has length \(v_K(\det T)\). To see the length assertion over a DVR, choose an entry of \(T\) of smallest valuation, move it to the first position, and use integral row and column operations to eliminate its row and column: it divides every entry. Repeat on the smaller matrix. This gives a diagonal matrix up to invertible integral operations; the quotient length and determinant valuation are both the sum of its diagonal valuations. By the definition of the different, the trace dual is \(\mathfrak P_L^{-\delta}\). Its quotient by \(\mathcal O_L\) has \(\delta\) successive quotients isomorphic to \(\kappa_L\), each of length \(f(L/K)\) over \(\mathcal O_K\). This proves (14) directly, including its residue-degree factor. \subsubsection{Theorem 10.5. The local conductor--discriminant formula}\label{theorem-10.5.-the-local-conductordiscriminant-formula} For finite abelian \(L/K\), \[ v_K(\mathfrak d_{L/K})=\sum_{\chi\in\widehat G}a(\chi). \tag{15} \] \textbf{Proof.} Put \(\epsilon_\chi(H)=1\) if \(\chi\) is nontrivial on \(H\), and 0 otherwise. Proposition 10.4 and Hasse--Arf give \[ a(\chi)=\epsilon_\chi(G_0) +\int_0^\infty\epsilon_\chi(G^v)\,dv. \] For a ramified character the integral is the length of the interval from zero to its last upper jump, namely \(a(\chi)-1\); for an unramified character both terms vanish. Substitute \(v=\phi(t)\) and use (1), whose ceiling convention makes \(G_t=G_j\) on \(j-12\), a unit with residue different from 1 shows that \(U\) is not contained, while \(U^{(1)}\) is; the exponent is 1. If \(n=1,p=2\), all units already lie in \(U^{(1)}\), the norm group is the entire multiplicative group, and the exponent is 0. Thus the conductor ideal is \(p^n\mathbf Z_p\) except in this trivial case, when it is \(\mathbf Z_2\). \subsubsection{Exercise 2 --- medium}\label{exercise-2-medium} Compute the norm groups and conductors of all quadratic extensions of \(\mathbf Q_2\). \textbf{Solution.} First, an odd \(2\)-adic unit is a square exactly when it is \(1\) modulo 8. Necessity follows by squaring odd residues. For sufficiency, start a root modulo 8 and inductively lift a root \(z\) modulo \(2^r\), \(r\geq3\): replacing \(z\) by \(z+2^{r-1}\) changes its square by \(2^rz\) modulo \(2^{r+1}\), toggling the required next bit. These compatible corrections converge and square to the given unit. Valuation parity and the four odd residues modulo 8 therefore give eight square classes, with the seven nonsquare representatives in the following table. For \(d\) in its first column put \(L_d=\mathbf Q_2(\sqrt d)\). {\def\LTcaptype{none} % do not increment counter \begin{longtable}[]{@{} >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3000}} >{\raggedright\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.3000}} >{\raggedleft\arraybackslash}p{(\linewidth - 4\tabcolsep) * \real{0.4000}}@{}} \toprule\noalign{} \begin{minipage}[b]{\linewidth}\raggedright \(d\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright Norm group \(N_{L_d/\mathbf Q_2}L_d^\times\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft Conductor exponent \end{minipage} \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot \(5\) & \(2^{2\mathbf Z}\mathbf Z_2^\times\) & 0 \\ \(-1\) & \(2^{\mathbf Z}(1+4\mathbf Z_2)\) & 2 \\ \(-5\) & \(6^{\mathbf Z}(1+4\mathbf Z_2)\) & 2 \\ \(2\) & \((-2)^{\mathbf Z}\{u\mid u\bmod8\in\{1,7\}\}\) & 3 \\ \(-2\) & \(2^{\mathbf Z}\{u\mid u\bmod8\in\{1,3\}\}\) & 3 \\ \(10\) & \((-10)^{\mathbf Z}\{u\mid u\bmod8\in\{1,7\}\}\) & 3 \\ \(-10\) & \(10^{\mathbf Z}\{u\mid u\bmod8\in\{1,3\}\}\) & 3 \\ \end{longtable} } In the last four rows the displayed \(u\) ranges over \(\mathbf Z_2^\times\). We justify every row. The field \(L_5\) is the unique unramified quadratic extension, as proved in the unramified-extension lesson, so its norm group is the first row. The others are ramified: they are distinct quadratic fields by the Kummer correspondence in lesson 3 and the distinct square classes, and cannot be that unique unramified field. For any ramified quadratic field, norm valuations are all integers, since \(v_2(Nx)=f\,v_L(x)=v_L(x)\). Finite reciprocity gives norm index 2; adjusting valuations by a norm shows that its unit norm subgroup has index 2 in \(\mathbf Z_2^\times\). For \(d=-1,-5\), a norm is \(x^2+cy^2\) with \(c=1,5\). If the minimum of \(v_2(x),v_2(y)\) is a negative integer \(r\), factor out \(2^{2r}\). The remaining norm has valuation 0 when just one scaled coordinate is odd, and valuation 1 when both are odd, since \(1+c\equiv2\pmod4\). It cannot give a unit. Thus a unit norm has integral coordinates, exactly one odd, and is 1 modulo 4. This residue condition has index 2, so the preceding index argument proves that all \(1+4\mathbf Z_2\) are unit norms. The elements \(1+i\) and \(1+\sqrt{-5}\) have norms 2 and 6, giving the two full norm groups. For \(d=2\varepsilon\), \(\varepsilon\) odd, the terms in \(x^2-2\varepsilon y^2\) have valuations \(2v_2(x)\) and \(1+2v_2(y)\), which are distinct. A unit norm requires \(x\) odd and \(y\) integral. Modulo 8 its residue is 1 if \(y\) is even, and \(1-2\varepsilon\) if \(y\) is odd. For \(\varepsilon=1,5\) these are \(\{1,7\}\), and for \(\varepsilon=-1,-5\) they are \(\{1,3\}\). Each inverse image is an index-two subgroup of the unit group, so is the exact unit norm subgroup. The norm of \(\sqrt d\) is \(-d\), with valuation 1, supplying each displayed generator. Finally \(U^{(1)}=U\) over \(\mathbf Q_2\). The odd ramified rows contain \(U^{(2)}\), but not all units, so have exponent 2. The even rows contain \(U^{(3)}\) and exclude the unit 5 in \(U^{(2)}\), so have exponent 3. The unramified row has exponent 0. These also equal the discriminant exponents. For the odd ramified rows, \(1+\sqrt d\) is a uniformizer, and integral uniformizer generation gives \(\mathcal O_{L_d}=\mathbf Z_2[\sqrt d]\). For the even rows, \(X^2-d\) is Eisenstein and gives the same integral basis \(1,\sqrt d\). Its trace matrix is diagonal with entries \(2,2d\), so its discriminant is \(4d\), of valuation 2 for the odd rows and 3 for the even rows. The unramified different, hence discriminant, is trivial by (3) and (14). Theorem 10.5 gives the same values from the sole nontrivial character. \subsubsection{Exercise 3 --- medium}\label{exercise-3-medium} Verify the conductor--discriminant formula for \(\mathbf Q_3(\zeta_9)\). \textbf{Solution.} Its group has order 6, inertia is the whole group, and the first upper group has order 3; the second is identity. Exactly two characters are trivial on that order-three subgroup. They have exponents 0 and 1. The other four have exponent 2, so their total is 9. Independently (5) gives lower groups of orders \(6,3,3\) at indices \(0,1,2\), then identity. Hilbert's formula yields \(\delta=(6-1)+(3-1)+(3-1)=9\). The extension is totally ramified, so \(f=1\), and (14) gives discriminant exponent 9. This agrees with the character sum. \subsubsection{Exercise 4 --- hard}\label{exercise-4-hard} Deduce Hasse--Arf for every finite abelian extension using the unit-filtration theorem, including real indices. \textbf{Solution.} Theorem 10.2 proves that for every real \(v>0\) the group is the image of \(U^{(\lceil v\rceil)}\). If \(v\) lies strictly between consecutive integers, choose \(\epsilon>0\) small enough that \(v+\epsilon\) remains in the same interval. Their ceilings, and hence their groups, agree, so \(v\) cannot be a jump. At a positive integer the group may decrease immediately afterward; these are precisely the permitted locations. The filtration on \((-1,0]\) is constantly inertia, leaving only the integer locations \(-1\) and 0 there. This proves the assertion for all jumps. The real-index formula comes from the explicit Lubin--Tate rescaling and Herbrand's quotient theorem, both proved before invoking Hasse--Arf. Thus there is no circular use of integral jumps in obtaining it. \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 division-point motion calculation determines all real-index ramification groups. The quotient filtration gives Hasse--Arf for every finite abelian local extension. The conductor, different and character-sum arguments retain their exact local-field prerequisites. \begin{itemize} \tightlist \item \href{https://arxiv.org/abs/math/0606108v2}{Teruyoshi Yoshida, Local class field theory via Lubin--Tate theory, arXiv:math/0606108v2}. \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Hilbert symbols and local conics}\label{hilbert-symbols-and-local-conics} \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).} Kummer theory labels abelian extensions by radicals, while reciprocity labels their automorphisms by elements of the ground field. The Hilbert symbol records the action of the latter on the former. This produces a perfect pairing, a norm criterion, and explicit tests for whether a quadratic conic has a local point. Initially let \(K\) be a nonarchimedean local field containing \(\mu_n\), where \(n\geq1\) is prime to \(\operatorname{char}K\). In characteristic zero, this allows \(n\) divisible by the residue characteristic. We use the Kummer theorem proved in \href{../hilberts-theorem-90-and-kummer-theory.html}{Hilbert's theorem 90 and Kummer theory}, finite local reciprocity from \href{../local-reciprocity-and-norm-groups.html}{Local reciprocity and norm groups}, and its explicit arithmetic normalization from \href{../explicit-local-reciprocity-and-existence.html}{Explicit local reciprocity and the existence theorem}. The quadratic norm calculations over \(\mathbf Q_2\) in \href{../abelian-ramification-conductors-and-hasse-arf.html}{Abelian ramification, conductors and Hasse--Arf}, Exercise 2, will give the wild quadratic formula. Real quadratic symbols are treated in section 4. \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-11}{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. Definition and norm identities}\label{definition-and-norm-identities} For \(a,b\in K^\times\), choose \(\lambda\) with \(\lambda^n=b\). The field \(L_b=K(\lambda)\) is finite cyclic: all roots are \(\zeta\lambda\), \(\zeta\in\mu_n\subset K\), and its Galois group embeds in \(\mu_n\). Define \[ (a,b)_n=\frac{(a,L_b/K)(\lambda)}{\lambda}\in\mu_n. \tag{1} \] The \textbf{first} argument enters reciprocity; the \textbf{second} is put under the root. Replacing \(\lambda\) by \(\zeta\lambda\) has no effect because \(\zeta\in K\). This fixes the convention even when \(n>2\). \subsubsection{Proposition 11.1. Algebraic properties}\label{proposition-11.1.-algebraic-properties} The symbol is multiplicative in each argument, factors through \(K^\times/K^{\times n}\) in each, and satisfies \[ \begin{gathered} (a,b)_n=1\ \Longleftrightarrow\ a\in N_{L_b/K}L_b^\times,\\ (a,-a)_n=1,\qquad (a,1-a)_n=1\quad(a\ne1),\\ (a,b)_n(b,a)_n=1. \end{gathered} \tag{2} \] The elements in each displayed symbol are nonzero. \textbf{Proof.} Multiplicativity in \(a\) is the homomorphism property of the norm residue symbol. For the second argument, put roots of \(b,c\) in their finite abelian compositum. Restriction compatibility says its reciprocity automorphism acts on their product as the product of its actions. That product is an \(n\)th root of \(bc\), proving \((a,bc)_n=(a,b)_n(a,c)_n\). An \(n\)th power in the first argument gives an \(n\)th power in \(\mu_n\), hence 1; multiplying the second argument by an \(n\)th power only multiplies its root by an element of \(K\). Thus the pairing factors through both quotients. An automorphism of \(L_b\) fixes \(\lambda\) exactly when it is identity. Finite reciprocity identifies the identity kernel with the norm group, proving the first equivalence. We prove the two special norm identities without assuming \([L_b:K]=n\). The separable polynomial \(T^n-b\) defines a finite algebra \[ A_b=K[T]/(T^n-b). \] Factoring it into relatively prime irreducible polynomials and applying the polynomial Chinese remainder theorem expresses \(A_b\) as a product of fields. Every factor is \(K\)-isomorphic to \(L_b\), because its root is some \(\zeta\lambda\) and generates the same field. The determinant norm of an element of this product is the product of its field norms, and therefore lies in \(N_{L_b/K}L_b^\times\) when the element is invertible. Multiplication by \(T\) has eigenvalues the \(n\) roots of \(T^n-b\), so \[ N_{A_b/K}(-T)=-b,\qquad N_{A_b/K}(1-T)=1-b. \tag{3} \] The first follows by multiplying their negatives, and the second by evaluating the monic polynomial at 1. These elements are invertible when their displayed norms are nonzero. Taking \(b=-a\) proves that \(a\) is a norm from \(L_{-a}\); taking \(b=1-a\) proves that \(a\) is a norm from \(L_{1-a}\). These are the middle identities in (2). Finally expand \(1=(ab,-ab)_n\). In its first factor use \(-ab=(-a)b\), and in its second use \(-ab=(-b)a\). Multiplicativity yields \[ 1=(a,-a)_n(a,b)_n(b,-b)_n(b,a)_n =(a,b)_n(b,a)_n. \] This proves the last identity. \(\square\) For \(n=2\), inversion of a sign changes nothing, so the pairing is symmetric. It need not be alternating: \((a,a)_n=(a,-1)_n^{-1}\), and \((-1,-1)_2\) over \(\mathbf Q_2\) will be \(-1\). \subsection{2. Nondegeneracy}\label{nondegeneracy} \subsubsection{Theorem 11.2. The perfect Hilbert pairing}\label{theorem-11.2.-the-perfect-hilbert-pairing} The group \(B=K^\times/K^{\times n}\) is finite, and the pairing induces an isomorphism \[ B\longrightarrow\operatorname{Hom}(B,\mu_n),\qquad b\longmapsto\bigl(a\mapsto(a,b)_n\bigr). \tag{4} \] In particular it is nondegenerate in both arguments. \textbf{Proof.} First verify finiteness, including the wild case. Put \(s=v_K(n)\), which is finite since \(n\ne0\) in \(K\). If \(v_K(t)>2s\), the polynomial \[ P(X)=X^n-(1+t) \] has \(v_K(P(1))>2s\) and \(v_K(P'(1))=s\). The Newton estimates proved in the descent lemma of the preceding reciprocity lesson apply starting at 1: an increment of valuation \(r>s\) gives a next increment of valuation at least \(2r-s\), while the derivative keeps valuation \(s\). The iterates converge in \(K\) to a unit whose \(n\)th power is \(1+t\). Hence sufficiently deep units are \(n\)th powers. The unit group modulo a deep unit subgroup is finite, and valuation modulo \(n\) has only \(n\) classes. Thus \(B\) is finite. It has exponent dividing \(n\). If \(b\) is nontrivial in \(B\), the field \(L_b\) is nontrivial. Finite reciprocity is onto its Galois group, so some \(a\) gives an automorphism that moves its generating root. Therefore \((a,b)_n\ne1\). This proves injectivity in (4). A finite abelian group of exponent dividing \(n\) has exactly as many \(\mu_n\)-valued characters as elements. The finite character lemma in lesson 3, section 4, proves this directly, together with character separation, for every finite abelian group killed by \(n\). Consequently the injection (4) is onto. If \(a\) pairs trivially with every \(b\), it is annihilated by every character of \(B\), and character separation gives \(a=1\) in \(B\). This proves the other nondegeneracy. \(\square\) Thus being a norm from every radical extension of exponent dividing \(n\) is equivalent to being an \(n\)th power in \(K\). \subsection{3. The tame formula and its direction}\label{the-tame-formula-and-its-direction} Now assume the residue characteristic \(p\) does not divide \(n\). Write \(q=|\kappa_K|\), and \[ a=\pi^\alpha u,\qquad b=\pi^\beta v,\qquad u,v\in\mathcal O_K^\times. \] Reduction maps \(\mu_n\) injectively to \(\kappa_K^\times\). Indeed, if \(\zeta=1+x\) is an \(n\)th root with \(v_K(x)>0\), the term \(nx\) in \((1+x)^n-1\) has strictly smaller valuation than every higher term, so cannot cancel unless \(x=0\). Hence \(n\mid q-1\). \subsubsection{Proposition 11.3. The tame Hilbert symbol}\label{proposition-11.3.-the-tame-hilbert-symbol} The symbol is the unique root in \(\mu_n\) whose reduction is \[ \overline{(a,b)_n} =\left((-1)^{\alpha\beta} \frac{\bar v^{\,\alpha}}{\bar u^{\,\beta}}\right)^{(q-1)/n} =\left(\overline{(-1)^{\alpha\beta}b^\alpha/a^\beta}\right)^{(q-1)/n}. \tag{5} \] \textbf{Proof.} An \(n\)th root of a unit \(v\) lies in a finite unramified extension. Choose such an extension whose residue field contains a root of \(X^n-\bar v\); its derivative is a unit there. Lifting that residue root and applying the Newton iteration with derivative valuation zero constructs a root of \(X^n-v\) in the complete unramified field. Existence of the field with that residue extension is Theorem 2.1 of \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-LOC-07}{Unramified and totally ramified extensions}. The symbol of a unit acts trivially on an unramified extension, so \((u,v)_n=1\). The symbol of \(\pi\) there is arithmetic Frobenius. For \(z^n=v\), its action has ratio with reduction \[ \overline{\varphi(z)/z}=\bar z^{\,q-1} =\bar v^{(q-1)/n}. \] Thus \((\pi,v)_n\) has that reduction. Antisymmetry gives \(\overline{(u,\pi)_n}=\bar u^{-(q-1)/n}\). Finally \((\pi,-\pi)_n=1\) implies \[ \overline{(\pi,\pi)_n}=(-1)^{(q-1)/n}; \] the inverse of this sign equals itself. Expanding the pairing on \(\pi^\alpha u,\pi^\beta v\) proves (5), also for negative integer exponents. Injectivity of reduction on \(\mu_n\) identifies the symbol itself. \(\square\) In particular, under arithmetic normalization the numerator is \(b^\alpha\). Swapping the two arguments inverts the symbol; this distinction is visible for orders greater than two. \subsection{4. Quadratic formulas}\label{quadratic-formulas} Use the notation \((a,b)_p\) for the quadratic symbol over \(\mathbf Q_p\). For odd \(p\), let \((u/p)\) be the sign \(\bar u^{(p-1)/2}\) in the residue field, identified with \(\{\pm1\}\); this is the Legendre symbol of a unit. For odd \(2\)-adic units define, modulo 2, \[ \epsilon(u)=\frac{u-1}{2},\qquad \omega(u)=\frac{u^2-1}{8}. \tag{6} \] These integral expressions depend only on the odd residue modulo 8. \subsubsection{Theorem 11.4. Explicit quadratic symbols}\label{theorem-11.4.-explicit-quadratic-symbols} For \(a=p^\alpha u,\ b=p^\beta v\), with units \(u,v\), one has \[ (a,b)_p=(-1)^{\alpha\beta(p-1)/2} (u/p)^\beta(v/p)^\alpha\quad(p\text{ odd}), \tag{7} \] and \[ (a,b)_2=(-1)^{\epsilon(u)\epsilon(v) +\alpha\omega(v)+\beta\omega(u)}. \tag{8} \] Over \(\mathbf R\), \[ (a,b)_\infty=-1 \quad\Longleftrightarrow\quad a<0\text{ and }b<0. \tag{9} \] \textbf{Proof.} Formula (7) is (5) for \(n=2,q=p\); the inverse of each Legendre sign equals the sign. For (8), the square-class calculation in Exercise 2 of the ramification lesson proves that \(-1,2,5\) form a basis of the eight square classes of \(\mathbf Q_2\). Its independently computed norm groups give all the following pairings: \[ \begin{array}{c|ccc} (\ ,\ )_2&-1&2&5\\ \hline -1&-1&1&1\\ 2&1&1&-1\\ 5&1&-1&1 \end{array}. \tag{10} \] Here \((-1,-1)_2=-1\) because \(-1\) is not in \(1+4\mathbf Z_2\), the unit norm subgroup from \(\mathbf Q_2(i)\). The units \(-1\) and 5 are norms as appropriate for the other entries: \(-1\) is a unit norm from \(\mathbf Q_2(\sqrt2)\), and every unit is a norm from the unramified field \(\mathbf Q_2(\sqrt5)\). Also \(2=N(2+\sqrt2)\), giving \((2,2)_2=1\). The element 2 has odd valuation and is not a norm from that unramified quadratic field, giving \((2,5)_2=-1\). Symmetry supplies the reflected entries, and \((5,5)_2=1\) is again an unramified unit norm. For every odd unit, \[ u\equiv(-1)^{\epsilon(u)}5^{\omega(u)} \pmod{\mathbf Q_2^{\times2}}. \tag{11} \] Check its four residues: \(1,3,5,7\) have exponent pairs \((0,0),(1,1),(0,1),(1,0)\). Their proposed representatives have exactly these residues modulo 8, whose equality is the square criterion. Express \(a,b\) with (11), reduce \(\alpha,\beta\) modulo 2, and expand by (10). The only negative basis pairings are the \((-1,-1)\) entry and the two \(2,5\) entries. Their exponents are exactly those in (8). For the real case, the quadratic reciprocity map on \(\mathbf R^\times\) takes positive elements to identity and negative elements to complex conjugation. Its kernel is the norm image from \(\mathbf C^\times\): \(N(x+iy)=x^2+y^2\) gives precisely the positive real numbers. If \(b>0\), its square root is real, so the symbol is 1. If \(b<0\), its root generates \(\mathbf C\), and a negative \(a\) acts by conjugation and changes the root's sign. This proves (9). It also shows the real pairing is nondegenerate on the two square classes. Over \(\mathbf C\) all elements are squares and the quadratic symbol is 1. \(\square\) \subsection{5. The conic criterion}\label{the-conic-criterion} \subsubsection{Proposition 11.5. Local points on a diagonal conic}\label{proposition-11.5.-local-points-on-a-diagonal-conic} For \(K=\mathbf Q_p\) or \(\mathbf R\), and \(a,b\in K^\times\), \[ (a,b)_K=1 \quad\Longleftrightarrow\quad z^2=ax^2+by^2 \text{ has a solution }(x,y,z)\ne(0,0,0)\text{ in }K^3. \tag{12} \] \textbf{Proof.} If \(b\) is a square, the symbol is 1 and \((0,1,\sqrt b)\) is a nonzero solution. Otherwise \(L=K(\sqrt b)\) is quadratic. The norm criterion, including the real norm calculation, says the symbol is 1 exactly when \(a=s^2-bt^2\) for some \(s,t\in K\). Such an equality supplies the conic point \((1,t,s)\). Conversely, if a nonzero solution has \(x=0\), then \(z^2=by^2\); nonzeroness forces \(y\ne0\), contradicting that \(b\) is nonsquare. Thus \(x\ne0\), and division by \(x^2\) gives \[ a=(z/x)^2-b(y/x)^2, \] which is the norm of \(z/x+(y/x)\sqrt b\). Since \(a\ne0\), this is a nonzero norm, proving the symbol is 1. \(\square\) This is a test over one completion. A global assertion that a rational conic has a rational point requires an additional local-to-global theorem. \subsection{6. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- easy}\label{exercise-1-easy} Compute \((2,3)_3\), \((-1,-1)_2\), and \((2,5)_5\). \textbf{Solution.} For \((2,3)_3\), the first argument is a unit, the second has valuation 1, so (7) gives the Legendre sign \((2/3)=-1\). For \((-1,-1)_2\), both valuations are zero and both \(\epsilon(-1)\) equal 1 modulo 2; (8) gives \(-1\). For \((2,5)_5\), formula (7) gives \((2/5)=-1\), because the nonzero squares modulo 5 are 1 and 4. \subsubsection{Exercise 2 --- medium}\label{exercise-2-medium} Prove the conic criterion, including solutions with a zero coordinate. \textbf{Solution.} If \(b\) is square, the point \((0,1,\sqrt b)\) proves solubility and the radical extension is trivial. If \(b\) is nonsquare, a nonzero point cannot have \(x=0\): that would give either a square root of \(b\), or all coordinates zero. Divide its equation by \(x^2\) to express \(a\) as a quadratic norm, so the symbol is 1. Conversely write a nonzero norm \(a=s^2-bt^2\) and take \((x,y,z)=(1,t,s)\). This reasoning allows \(t=0\) or \(s=0\) and applies over both the nonarchimedean fields and \(\mathbf R\). \subsubsection{Exercise 3 --- medium}\label{exercise-3-medium} Prove the Steinberg relation \((a,1-a)_n=1\) for \(a\ne0,1\). \textbf{Solution.} Put \(b=1-a\) and form \(A_b=K[T]/(T^n-b)\). Separability and \(\mu_n\subset K\) make each field factor isomorphic to \(L_b=K(b^{1/n})\). The determinant norm of \(1-T\) is \[ \prod_{\lambda^n=b}(1-\lambda)=1-b=a. \] It is nonzero, so \(1-T\) is invertible in the product. Its component norms multiply to \(a\), and their product is a single norm from \(L_b\), by norm multiplicativity. The symbol's norm criterion proves the relation. Using the product algebra accounts for the case where the radical field has degree smaller than \(n\). \subsubsection{Exercise 4 --- hard}\label{exercise-4-hard} Derive the complete quadratic formula (8) from square classes and norm groups. \textbf{Solution.} The odd-unit square criterion identifies the square classes with \(2^\alpha(-1)^e5^w\), for \(\alpha,e,w\in\{0,1\}\). For an odd unit \(u\), its four residues modulo 8 give \(e=\epsilon(u)\), \(w=\omega(u)\), as in (11). The exact quadratic norm groups from the ramification lesson give the symmetric table (10), with negative entries only at \((-1,-1),(2,5),(5,2)\). If \(a\) has exponent vector \((\epsilon(u),\alpha,\omega(u))\) in the ordered basis \((-1,2,5)\), and \(b\) has vector \((\epsilon(v),\beta,\omega(v))\), multiplicativity gives their sign exponent \[ \epsilon(u)\epsilon(v)+\alpha\omega(v)+\omega(u)\beta. \] This proves (8) for all classes. Squares in either argument do not affect the symbol, so the calculation applies to arbitrary integer valuations and all \(2\)-adic units. \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 norm characterization and finite perfect pairing lead to the tame, real and dyadic quadratic formulas. The product-algebra proof handles local conics, including the split case. All displayed symbol normalizations are proved in the lesson. \begin{itemize} \tightlist \item \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne, Class Field Theory, version 4.03}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Weil groups and one-dimensional representations}\label{weil-groups-and-one-dimensional-representations} \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).} Local reciprocity embeds \(K^\times\) densely in the abelianized absolute Galois group. Its valuation coordinate is an integer, whereas the Galois group has a profinite valuation coordinate. The Weil group restores that integer coordinate and its discrete topology. Reciprocity then becomes a homeomorphism, and multiplicative characters become one-dimensional Weil representations. We use \href{../explicit-local-reciprocity-and-existence.html}{Explicit local reciprocity and the existence theorem}, especially Theorem 9.4, and the unit-filtration theorem in \href{../abelian-ramification-conductors-and-hasse-arf.html}{Abelian ramification, conductors and Hasse--Arf}. The analytic character factors come from \emph{Tate's local theory at the finite places} and \emph{Tate's local theory at the infinite places} in the course \emph{Adèles and L-functions}, lessons 7 and 8. We supply the additional Fourier argument needed in positive characteristic. Sections 1--4 concern a nonarchimedean local field of either characteristic. Section 5 treats the real and complex fields; section 6 translates Frobenius conventions, and section 7 connects the local construction to global reciprocity. \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-12}{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. Inertia open, Frobenius discrete}\label{inertia-open-frobenius-discrete} Let \(G_K=\operatorname{Gal}(K^{\mathrm{sep}}/K)\), and let \(I_K\) be inertia. The unramified-extension theorem gives \[ 1\longrightarrow I_K\longrightarrow G_K \xrightarrow{d}\widehat{\mathbf Z}\longrightarrow1. \tag{1} \] We choose \(d\) so that an arithmetic Frobenius, acting by \(x\mapsto x^q\) on the residue field, has degree 1. Define \[ W_K=d^{-1}(\mathbf Z). \tag{2} \] Give \(I_K\) its profinite topology and make it open in \(W_K\). Every coset of \(I_K\) has the topology transported from \(I_K\) by translation. \subsubsection{Proposition 12.1. The local Weil group}\label{proposition-12.1.-the-local-weil-group} The topology just specified makes \(W_K\) a locally compact, totally disconnected group. Its quotient by \(I_K\) is the discrete group \(\mathbf Z\). The inclusion \(W_K\hookrightarrow G_K\) is continuous and has dense image. The Weil topology is strictly finer than the topology induced by \(G_K\). \textbf{Proof.} Choose \(\phi\in G_K\) with \(d(\phi)=1\). Integer powers of any element of a profinite group extend continuously to profinite powers: in each finite quotient reduce the exponent modulo that element's finite order, and take the compatible inverse limit. Thus \(t\mapsto\phi^t\), \(t\in\widehat{\mathbf Z}\), is a continuous homomorphism and \(d(\phi^t)=t\). Every element of \(G_K\) has a unique expression \(i\phi^t\). The corresponding bijection \[ I_K\rtimes\widehat{\mathbf Z}\longrightarrow G_K \] is a homeomorphism, since its domain is compact and its target Hausdorff. On \(W_K\) this expression has \(t\in\mathbf Z\). The prescribed topology is precisely the product topology of \(I_K\rtimes\mathbf Z\), with \(\mathbf Z\) discrete. The action of each \(\phi^m\) on \(I_K\) is continuous, so multiplication and inversion are continuous. The open subgroup \(I_K\) is compact and totally disconnected, proving the asserted local properties. The inclusion is continuous on every open coset. Since \(\mathbf Z\) is dense in \(\widehat{\mathbf Z}\), the product description proves density. In the subspace topology the set \(\{\phi^m:m\in\mathbf Z\}\) has the profinite, rather than discrete, topology. Indeed, every open neighborhood of 1 in \(G_K\) contains a nonidentity \(\phi^m\): pass to an open normal subgroup and take a positive multiple of the order of \(\phi\) in its finite quotient. Such a power has degree \(m\ne0\), so it is outside inertia. Consequently \(I_K\) is not open in the subspace topology. ∎ The choice of \(\phi\) splits the group but is not part of its definition. A different lift changes the splitting, not \(W_K\) or its topology. \subsection{2. A topological reciprocity isomorphism}\label{a-topological-reciprocity-isomorphism} For any topological group \(H\), write \[ H^{\mathrm{ab}}=H/\overline{[H,H]}. \] The closure is essential: this is the maximal \textbf{Hausdorff} abelian quotient. \subsubsection{Theorem 12.2. Reciprocity for the Weil group}\label{theorem-12.2.-reciprocity-for-the-weil-group} Arithmetic local reciprocity induces a topological isomorphism \[ \operatorname{rec}_K:K^\times\xrightarrow{\ \sim\ }W_K^{\mathrm{ab}}. \tag{3} \] It carries \(\mathcal O_K^\times\) onto the image of \(I_K\), and a uniformizer onto a class of arithmetic degree 1. \textbf{Proof.} Put \(D=\overline{[G_K,G_K]}\). Since \(d\) has abelian target, \(D\subset I_K\). Density of \(W_K\) implies \[ \overline{[W_K,W_K]}^{\,G_K}=D: \tag{4} \] approximate each of the finitely many entries in a product of commutators by elements of \(W_K\), using continuity of the commutator map. Both commutator subgroups lie in \(I_K\), where the two topologies agree. Thus their closure in \(W_K\) is also \(D\). We obtain an injective map \(W_K/D\to G_K/D\), whose image is exactly the inverse image of \(\mathbf Z\) under the induced degree map. Theorem 9.4 identifies \(G_K^{\mathrm{ab}}\) with the completion of \(K^\times=\pi^{\mathbf Z}\times\mathcal O_K^\times\) for its open subgroups of finite index: \[ G_K^{\mathrm{ab}}\simeq\widehat{\mathbf Z}\times\mathcal O_K^\times. \tag{5} \] The degree is the first coordinate, and reciprocity embeds \(K^\times\) as \(\mathbf Z\times\mathcal O_K^\times\). By (4), this is the underlying group \(W_K^{\mathrm{ab}}\). It remains to check the topology. The subgroup \(I_K/D\) is compact and open in \(W_K/D\). Its map onto the degree-zero subgroup in (5) is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Reciprocity is therefore a homeomorphism on the unit group. Each integer-degree coset in both groups is open and is a translate of that unit group. Translating the homeomorphism on each coset proves (3). ∎ This argument does not put the subspace topology from \(G_K^{\mathrm{ab}}\) on \(W_K^{\mathrm{ab}}\). With that topology its integer coordinate would again fail to be discrete. \subsection{3. Characters, conductors and Euler factors}\label{characters-conductors-and-euler-factors} A character below means a continuous homomorphism to \(\mathbf C^\times\); it need not be unitary or have finite image. \subsubsection{Proposition 12.3. Local reciprocity in dimension one}\label{proposition-12.3.-local-reciprocity-in-dimension-one} Composition with (3) gives a bijection \[ \{\text{characters of }W_K\} \longleftrightarrow \{\text{characters of }K^\times\},\qquad \chi(a)=\rho(\operatorname{rec}_K(a)). \tag{6} \] The restriction of \(\rho\) to inertia has finite image. Unramified characters correspond to characters trivial on \(\mathcal O_K^\times\). Their conductors and their arithmetic Euler factors agree. More precisely, put \(U^0=\mathcal O_K^\times\) and \(U^j=1+\mathfrak p_K^j\) for \(j\geq1\). The conductor is \[ a(\chi)=\min\{j\geq0:\chi(U^j)=1\} =a(\rho). \tag{7} \] For \(q^{-s}=\exp(-s\log q)\), define \[ L_{\mathrm{ar}}(s,\rho) =\det(1-q^{-s}\rho(\phi)\mid V^{I_K})^{-1}. \] In dimension one this equals \[ L(s,\chi)= \begin{cases} (1-\chi(\pi)q^{-s})^{-1},&a(\chi)=0,\\ 1,&a(\chi)>0. \end{cases} \tag{8} \] \textbf{Proof.} A character kills commutators and their closure, because its target is Hausdorff. It therefore factors through \(W_K^{\mathrm{ab}}\), and Theorem 12.2 proves the bijection. The group \(\mathbf C^\times\) has a neighborhood of 1 containing no nontrivial subgroup. One may use an annulus with argument restricted to a small arc: powers of an element of nonunit modulus leave the annulus; powers of a nontrivial unit complex number leave the arc. Continuity gives an open subgroup \(J\) of the profinite group \(I_K\) whose image is inside that neighborhood. Its image is a subgroup, hence trivial. Thus \(\rho(I_K)\) is finite. To compare conductors without assuming that \(\chi\) has finite image, write \[ \chi(\pi^m u)=\chi(\pi)^m\chi_0(u),\qquad \chi_0(\pi)=1. \] The unit character has finite image, so \(\chi_0\) is finite order. By the existence theorem it factors through a finite abelian extension. The unit-filtration theorem of lesson 10 identifies the images of \(U^j\) with its upper ramification groups at integer \(j\). This proves (7), with conductor zero exactly in the unramified case. The unramified factor \(\chi(\pi)^m\) does not affect inertia or the conductor. If inertia acts nontrivially in dimension one, \(V^{I_K}=0\) and the determinant is 1. Otherwise \(V^{I_K}=V\), and all degree-one lifts give the scalar \(\chi(\pi)\): the units are killed. This proves (8). ∎ For every \(\lambda\in\mathbf C^\times\), \[ \chi_\lambda(\pi^m u)=\lambda^m \] is an unramified character. Its Weil representation factors through the discrete quotient \(W_K/I_K=\mathbf Z\). When \(\lambda=q^{-1}\), it cannot extend continuously to a character of the compact group \(G_K\): its image contains the unbounded sequence \(q^m\), \(m\geq0\). \subsection{4. Rank-one local constants in either characteristic}\label{rank-one-local-constants-in-either-characteristic} We record the analytic data as well as the representation. Choose a nontrivial continuous additive character \(\psi:K\to\mathbf T\) and a positive additive Haar measure \(dx\). Write \[ N=\max\{r\in\mathbf Z:\psi(\pi^{-r}\mathcal O_K)=1\}, \qquad m=\operatorname{vol}_{dx}(\mathcal O_K). \tag{9} \] Continuity and nontriviality make \(N\) a finite integer. Normalize \(d^\times x\) to give the units volume 1. Here is the Fourier foundation for all nonarchimedean local fields, including positive characteristic. Such a \(\psi\) exists: take a nontrivial additive character of the finite residue field on \(\mathcal O_K\), and extend successively to \(\pi^{-r}\mathcal O_K\). At each step the quotient is finite abelian. A circle-valued character extends across a finite cyclic enlargement by choosing a root of the prescribed value of a multiple of its generator; adjoining finitely many generators gives the required extension. The union is \(K\), and the kernel still contains \(\pi\mathcal O_K\), so the resulting character is continuous. Additive Haar measure can be constructed by assigning mass \(m q^{-r}\) to each coset of \(\pi^r\mathcal O_K\). These assignments are consistent under subdivision into \(q\) smaller cosets, define the measure on compact open sets, and then on their Borel completion. Translation invariance and the scaling identity \(d(ax)=|a|\,dx\) follow on these cosets and hence on integrable functions. Let \(\mathcal S(K)\) be the locally constant compactly supported functions, and use the positive kernel \(\widehat f(y)=\int f(x)\psi(xy)\,dx\). The annihilator of \(\pi^r\mathcal O_K\) is \(\pi^{-r-N}\mathcal O_K\): multiplying this ideal by \(y\) is an ideal on which \(\psi\) is trivial exactly when \(v(y)+r\geq-N\). Translation and character orthogonality give \[ \widehat{\mathbf1_{a+\pi^r\mathcal O_K}}(y) =m q^{-r}\psi(ay)\mathbf1_{\pi^{-r-N}\mathcal O_K}(y). \tag{10} \] Applying this formula twice gives \[ \widehat{\widehat f}(x)=m^2q^N f(-x). \tag{11} \] Every Schwartz function is a finite linear combination of these indicators, so (10) proves preservation of \(\mathcal S(K)\) and (11) for every \(f\). The self-dual measure is characterized by \(m=q^{-N/2}\). For a multiplicative character \(c\), put \(\check c=c^{-1}|\cdot|\), and set \[ Z(f,c)=\int_{K^\times}f(x)c(x)\,d^\times x. \] The integral converges when \(|c(\pi)|<1\). Splitting into shells \(\pi^j\mathcal O_K^\times\) gives a finite Laurent polynomial plus a tail. Local constancy makes the tail \(f(0)c(\pi)^J/(1-c(\pi))\) in the unramified case; unit-character orthogonality makes it zero in the ramified case. This gives rational continuation in \(c(\pi)\). The functional equation follows by the same two-variable calculation in every characteristic. In the strip \(q^{-1}<|c(\pi)|<1\), both \(c\) and \(\check c\) have absolutely convergent zeta integrals. With \(d^\times x=\kappa\,dx/|x|\), the substitution \(y=xz\) transforms \[ Z(f,c)Z(\widehat g,\check c) =\kappa^2\int_{K^\times}c(z)^{-1} \left(\int_K f(x)\widehat g(zx)\,dx\right)dz. \tag{12} \] The absolute-value double integral is finite by reversing that substitution. For each fixed \(z\), Fubini on the compact supports gives \[ \int f(x)\widehat g(zx)\,dx=\int\widehat f(zt)g(t)\,dt. \] The resulting double integral is also absolutely convergent, as the reverse substitution now gives the product for \(g,\widehat f\). Hence (12) equals \(Z(g,c)Z(\widehat f,\check c)\). A test \(g=c|_{\mathcal O_K^\times}^{-1}\mathbf1_{\mathcal O_K^\times}\) has zeta integral 1, so this proves \[ Z(\widehat f,\check c) =\epsilon(c,\psi,dx)\frac{L(\check c)}{L(c)}Z(f,c). \tag{13} \] Here \(L(c)=(1-c(\pi))^{-1}\) when \(c\) is unramified and is 1 otherwise. Rational continuation extends (13) beyond the strip. Formula (11), applied twice, shows that its factor is not identically zero. For completeness, its finite calculation is \[ \epsilon(c,\psi,dx)= \begin{cases} m q^N c(\pi)^N,&a(c)=0,\\ m q^N c(b)\displaystyle\sum_{u\in(\mathcal O_K/\mathfrak p_K^a)^\times} c(u)^{-1}\psi(u/b),&a(c)=a\geq1, \end{cases} \quad v(b)=a+N. \tag{14} \] In the first case use \(f=\mathbf1_{\mathcal O_K}\) and (10). In the second use the preceding unit-character test. Translation by \(\mathfrak p_K^a\) forces its transform to be supported on \(\pi^{-a-N}\mathcal O_K\). It vanishes at every strictly higher valuation: for \(a\geq2\) multiply the unit variable by \(v\in U^{a-1}\) with \(c(v)\ne1\), which leaves the additive factor unchanged; for \(a=1\), ordinary orthogonality on the units suffices. At \(y=w/b\) the transform is \[ m q^{-a}c(w)\sum_u c(u)^{-1}\psi(u/b). \] Its dual zeta integral on that one shell multiplies this by \(c(b)|b|^{-1}c(w)^{-1}\), proving (14). Replacing \(b\) by \(vb\), \(v\) a unit, multiplies its two factors by \(c(v)\) and \(c(v)^{-1}\), respectively. The Gauss sum is nonzero. Indeed, applying (11) again to this test gives the product of the sums for the unit character and its inverse equal to \(c(-1)q^a\); complex conjugation gives absolute value \(q^{a/2}\) for either sum. This proves nonvanishing in (14) for every \(c(\pi)\ne0\), not just meromorphically. All steps used finite quotients, compact open cosets and (10); none required characteristic zero. They also explain exactly how the finite-place Tate formulas extend to function-field completions. For the corresponding Weil character define \[ \epsilon_{\mathrm{ar}}(s,\rho,\psi,dx) =\epsilon(\chi|\cdot|^s,\psi,dx). \tag{15} \] This transports the full functional equation, rather than just an unspecified scalar. In particular, with the measure fixed, \[ \epsilon(c,\psi,t\,dx)=t\epsilon(c,\psi,dx),\qquad \epsilon(c,\psi_r,dx)=c(r)|r|^{-1}\epsilon(c,\psi,dx). \tag{16} \] These follow by scaling the Fourier transform and by substituting \(y\mapsto ry\) in its zeta integral. For self-dual measure, (11) and cancellation of the two \(L\)-ratios give \[ \epsilon(c,\psi,dx_\psi)\epsilon(c^{-1}|\cdot|,\psi,dx_\psi)=c(-1). \tag{17} \] Thus duality involves an inverse character \textbf{and} a norm twist; it is not a rule that simply inverts epsilon. The higher-dimensional extension is Deligne's Theorem 4.1: there is a unique family of nonzero constants for finite-dimensional smooth complex Weil representations, multiplicative in exact sequences, scaled by \(t^{\dim V}\) under \(dx\mapsto t\,dx\), compatible with induction of virtual representations of dimension zero through finite separable extensions using \(\psi\circ\operatorname{Tr}\), and equal to Tate's constants in dimension one under geometric reciprocity. Its representation-theoretic proof belongs to \href{https://kokunoyumeto.github.io/open-math-courses/courses/LG-GAL/local-l-factors-and-epsilon-factors.html\#3-existence-and-uniqueness-of-local-constants}{Local L-factors and epsilon-factors of Weil group representations}, section 3. Theorem 3.0 and sections 3A–3D supply the existence argument, including the Brauer-relation check and arbitrary smooth representations; Theorem 3.2 supplies uniqueness. Our proofs of (3), (6)--(8) and (13)--(17) do not use that higher-dimensional existence assertion. \subsection{5. The real and complex Weil groups}\label{the-real-and-complex-weil-groups} \subsubsection{Proposition 12.4. Archimedean abelianizations}\label{proposition-12.4.-archimedean-abelianizations} Set \(W_{\mathbf C}=\mathbf C^\times\) with its usual topology. The real Weil group has two components \[ W_{\mathbf R}=\mathbf C^\times\sqcup j\mathbf C^\times,\qquad j^2=-1,\qquad jzj^{-1}=\bar z. \tag{20} \] It is a locally compact group, and \[ W_{\mathbf C}^{\mathrm{ab}}=\mathbf C^\times,\qquad W_{\mathbf R}^{\mathrm{ab}}\simeq\mathbf R^\times, \quad z\longmapsto|z|^2,\quad j\longmapsto-1 \tag{21} \] are topological isomorphisms. \textbf{Proof.} Realize \(z\) as the complex matrix \(\operatorname{diag}(z,\bar z)\) and \(j\) as \(\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\). Their two sets of matrices are closed under multiplication and satisfy (20), so associativity follows from matrix multiplication. Each component has the usual topology of \(\mathbf C^\times\), and the relations show continuity of the group operations. The indicated map \(p:W_{\mathbf R}\to\mathbf R^\times\) is a surjective homomorphism: the relation \(j^2=-1\) maps to \((-1)^2=|-1|^2=1\), and conjugation preserves \(|z|^2\). Its kernel is the circle \(S^1\) in the first component. The commutators \([j,z]=\bar z/z\) fill that circle, while every commutator lies in the kernel because the quotient is abelian. Thus the commutator subgroup equals \(S^1\), already closed. The inverse map on quotient classes is represented by \(\sqrt{x}\) for \(x>0\) and \(j\sqrt{-x}\) for \(x<0\); these representatives vary continuously on the two open components. This proves the real topological assertion. The complex assertion is immediate because \(\mathbf C^\times\) is abelian. ∎ The real sign character is 1 on \(\mathbf C^\times\) and is \(-1\) on \(j\mathbf C^\times\). The map \(z\mapsto|z|^2\) in (21) also displays compatibility with the norm \(\mathbf C^\times\to\mathbf R^\times\). All real multiplicative characters have the form \(\operatorname{sgn}(x)^e|x|^t\), \(e\in\{0,1\}\), \(t\in\mathbf C\). All complex ones have the form \((z/|z|)^\ell|z|_{\mathbf C}^t\), \(\ell\in\mathbf Z\), with \(|z|_{\mathbf C}=|z|^2\). To see this, a continuous character of \(\mathbf R_{>0}\) becomes a character of \((\mathbf R,+)\) under logarithm; lifting near 1 and subdividing intervals gives \(x\mapsto e^{ax}\). A character of \(S^1\) lifted along \(e^{i\theta}\) has form \(e^{ib\theta}\), and periodicity forces \(b\in\mathbf Z\). A character of \(\{\pm1\}\) has values \(1,(-1)^e\). Transporting the proved archimedean Tate factors through (21), for the negative additive characters \(\psi_{\mathbf R}(x)=e^{-2\pi ix}\), \(\psi_{\mathbf C}(z)=e^{-4\pi i\operatorname{Re}z}\), and self-dual measures \(dx\), \(2\,d(\operatorname{Re}z)\,d(\operatorname{Im}z)\), respectively, gives \[ \begin{array}{c|c|c} \chi&L(s,\chi)&\epsilon(s,\chi)\\ \hline \operatorname{sgn}^e|\cdot|^t& \Gamma_{\mathbf R}(s+t+e)&(-i)^e\\ (z/|z|)^\ell|\cdot|_{\mathbf C}^t& \Gamma_{\mathbf C}(s+t+|\ell|/2)&(-i)^{|\ell|} \end{array} \tag{22} \] where \(\Gamma_{\mathbf R}(u)=\pi^{-u/2}\Gamma(u/2)\) and \(\Gamma_{\mathbf C}(u)=2(2\pi)^{-u}\Gamma(u)\). The Gaussian and polynomial Fourier proofs are \emph{Tate's local theory at the infinite places}, Propositions 8.2--8.3 and Theorem 8.1. Replacing these negative additive characters by their positives multiplies epsilon by \(\chi(-1)\), giving \(i^e\) and \(i^{|\ell|}\). This phase change is independent of a Frobenius convention. \subsection{6. Translating Frobenius conventions}\label{translating-frobenius-conventions} \subsubsection{Proposition 12.5. Arithmetic and geometric reciprocity}\label{proposition-12.5.-arithmetic-and-geometric-reciprocity} For nonarchimedean \(K\), geometric reciprocity is \[ \operatorname{rec}^{\mathrm{geo}}_K(a)=\operatorname{rec}_K(a)^{-1}. \tag{18} \] It sends a uniformizer to a geometric Frobenius class. For a fixed Weil character \(\rho\), its multiplicative character under (18) is \(\chi^{-1}\). For a fixed multiplicative character \(\chi\), the corresponding Weil characters in the two conventions are dual. Conductors are unchanged. For any complex Weil representation with finite inertia image, \[ L_{\mathrm{geo}}(s,V)=L_{\mathrm{ar}}(s,V^\vee), \tag{19} \] where each side uses the indicated Frobenius on the same underlying representation or its dual. \textbf{Proof.} Inversion is a continuous automorphism of the abelian group \(W_K^{\mathrm{ab}}\), and it reverses degree. Evaluating \(\rho\) on (18) proves the character statement. Inverse characters have the same kernel on each unit group, proving the conductor statement. For (19), averaging over the finite inertia image identifies \((V^\vee)^{I_K}\) with the dual of \(V^{I_K}\): the averaging idempotent has image \(V^{I_K}\), and its dual has image the dual of that summand. Frobenius preserves both summands. Its operator on the dual invariants is the transpose inverse of its operator on the original invariants, which has the same determinant polynomial as the inverse operator. A geometric lift is the inverse of an arithmetic lift up to inertia, which is trivial on these invariants. This proves (19). ∎ For example an unramified fixed representation with arithmetic eigenvalue \(\lambda\) has arithmetic factor \((1-\lambda q^{-s})^{-1}\) and geometric factor \((1-\lambda^{-1}q^{-s})^{-1}\). If one changes the reciprocity map and also dualizes the representation so that the multiplicative character is held fixed, its Tate factor stays the same. Deligne fixes geometric reciprocity in §2.3 and explains its cohomological Euler-factor convention in §3.6. When using his factors, evaluate the multiplicative character through his reciprocity map; do not keep an arithmetic character identification while inserting a geometric Frobenius. Additive characters and measures are separate choices, governed by (16). At archimedean places there is no residue-field Frobenius and hence no arithmetic/geometric choice of this kind. \subsection{7. Global Weil groups and the idèle class group}\label{global-weil-groups-and-the-iduxe8le-class-group} Let \(C_F=\mathbb A_F^\times/F^\times\) for a global field \(F\). The local theorem has a global analogue \[ C_F\simeq W_F^{\mathrm{ab}}, \tag{23} \] compatible with the maps \(F_v^\times\to C_F\) and local Weil groups. For a function field over its full finite constant field, the group is again a subgroup of the absolute Galois group: take the inverse image of the integral powers of constant-field Frobenius and make its geometric kernel open. The proof of (23) uses global reciprocity, the compact degree-zero subgroup of \(C_F\), and its integral degree quotient. The complete proof is \href{../global-existence-and-the-idelic-class-field-correspondence.html}{Global existence and the idèlic class field correspondence}, sections 5--8, equations (14)--(34). It proves injectivity even for characteristic-power extensions, identifies the compact degree-zero kernel, and verifies that the Hausdorff commutator quotient has exactly the required topology and local compatibility. A number field has no finite constant field whose Frobenius supplies such a subgroup. Its global Weil group is instead constructed from relative extensions \[ 1\longrightarrow C_L\longrightarrow W_{L/F} \longrightarrow\operatorname{Gal}(L/F)\longrightarrow1 \tag{24} \] defined by the global class-formation fundamental classes, with compatible transition maps as \(L/F\) runs through finite Galois extensions. \href{../brauer-groups-of-local-and-global-fields.html}{Brauer groups of local and global fields}, Theorem 24.6 and section 7, proves the fundamental-class theorem, constructs these extensions and their coherent quotient maps, and proves local compactness, the Hausdorff abelianization and local compatibility in (23). Its equations (29)--(39) give the complete number-field construction. The function-field construction is the preceding proof in lesson 17. These global arguments use reciprocity established later in the course; the local proofs above are independent of them. The classical number-field construction is due to Weil; Deligne §2.5 identifies this source rather than constructing it there. \subsection{8. Exercises with solutions}\label{exercises-with-solutions} \subsubsection{Exercise 1. Density and the topology --- easy}\label{exercise-1.-density-and-the-topology-easy} Prove density of \(W_K\) and openness of inertia. Show why density does not give the Weil group the subspace topology. \textbf{Solution.} In the splitting \(G_K=I_K\rtimes\widehat{\mathbf Z}\), the subset \(W_K\) is \(I_K\rtimes\mathbf Z\). Every product neighborhood meets it because an open subset of \(\widehat{\mathbf Z}\) contains an integer. In the Weil topology \(I_K\) is open by construction. In the Galois subspace topology every identity neighborhood contains a nonzero integral power of \(\phi\), by taking a multiple of its order in a finite quotient. Therefore inertia is not open in that topology. \subsubsection{Exercise 2. The real abelianization --- medium}\label{exercise-2.-the-real-abelianization-medium} Compute the closed commutator subgroup of \(W_{\mathbf R}\) and recover the sign representation. \textbf{Solution.} For \(z=re^{i\theta}\), \([j,z]=e^{-2i\theta}\), so these commutators fill \(S^1\). The homomorphism \(z\mapsto r^2\), \(j\mapsto-1\) has precisely this kernel; hence all commutators are in \(S^1\) and the closed commutator subgroup equals \(S^1\). The quotient is \(\mathbf R^\times\), with the continuous inverse classes described in Proposition 12.4. Pulling back sign gives 1 on the first component and \(-1\) on the second; the relation \(j^2=-1\) is respected since both sides have character value 1. \subsubsection{Exercise 3. Matching conductors --- medium}\label{exercise-3.-matching-conductors-medium} Let \(\rho\) correspond to \(\chi\). Prove equality of their conductors even when \(\chi(\pi)\) has infinite order. \textbf{Solution.} The unit restriction has finite image by the no-small-subgroup argument. Remove the unramified character taking \(\pi\) to \(\chi(\pi)\), leaving a finite-order character \(\chi_0\) with the same restriction to every \(U^j\). It factors through a finite abelian extension by local existence. Reciprocity sends \(U^j\) to its integer upper ramification group, so the least \(j\) on which \(\chi_0\) is trivial equals the Weil conductor. Restoring the unramified factor changes neither side. \subsubsection{Exercise 4. Proving the homeomorphism --- hard}\label{exercise-4.-proving-the-homeomorphism-hard} A continuous bijection need not be a homeomorphism. Supply the topological step in Theorem 12.2 and contrast the embedding in \(G_K^{\mathrm{ab}}\). \textbf{Solution.} Write \(D=\overline{[G_K,G_K]}\subset I_K\). In \(W_K/D\), the subgroup \(I_K/D\) is compact and open. Reciprocity restricts to a continuous bijection \(\mathcal O_K^\times\to I_K/D\); compactness and the Hausdorff property make it a homeomorphism. Both groups are disjoint unions of open translates indexed by \(\mathbf Z\). On every translate, multiplication transports that homeomorphism, so the inverse map is continuous on an open cover and hence everywhere. In \(G_K^{\mathrm{ab}}\) the degree coordinate instead lies in \(\widehat{\mathbf Z}\). Its integer subgroup is dense and has a nondiscrete induced topology, so the original dense embedding is not the same topological assertion. \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 local Weil construction and one-dimensional formulas use the written rank-one Fourier prerequisites. Global Weil constructions are supplied in lessons 17 and 24. General higher-dimensional epsilon existence is supplied by LG-GAL-06, Theorem 3.0 and sections 3A–3D; its uniqueness theorem is Theorem 3.2. These are programme proofs, distinct from the external reading sources. \begin{itemize} \tightlist \item \href{https://math.mit.edu/~poonen/786/notes.pdf}{Bjorn Poonen, Tate's Thesis, MIT 18.786 lecture notes (2015)}. \item \href{https://ncatlab.org/nlab/files/TateNumberTheory.pdf}{John Tate, Number theoretic background (1979), freely available paper}. \item \href{https://publications.ias.edu/sites/default/files/Number20.pdf}{Pierre Deligne, Les constantes des équations fonctionnelles des fonctions L (1973), IAS archive}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Idèles in extensions and their cohomology}\label{iduxe8les-in-extensions-and-their-cohomology} \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).} A field extension replaces a place by the places above it. Its idèle norm multiplies the corresponding local norms. This lets us compute cyclic cohomology one place at a time. The resulting quotient of the idèle group is usually infinite; the principal idèles impose relations between those local obstructions. We use the definitions and restricted product topology in \emph{Idèles and the idèle class group}, lesson 3 of \emph{Adèles and L-functions}, the Hilbert 90 proof in \href{../hilberts-theorem-90-and-kummer-theory.html}{Hilbert's theorem 90 and Kummer theory}, and finite local reciprocity in \href{../local-reciprocity-and-norm-groups.html}{Local reciprocity and norm groups}. The cyclic Tate groups and Herbrand quotient are defined in \href{../cohomology-of-cyclic-groups-and-the-herbrand-quotient.html}{Cohomology of cyclic groups and the Herbrand quotient}. Let \(K\) be a global field and \(L/K\) a finite separable extension. A global field means a number field or a finite extension of a finite-field rational function field. For a place \(v\) of \(K\), write \(w\mid v\) for its extensions to \(L\). The arguments below apply to either kind of global field. There are no archimedean places in the function-field case. \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-13}{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. Embeddings and norms}\label{embeddings-and-norms} Write \[ J_K=\prod_v'K_v^\times,\qquad C_K=J_K/K^\times, \] where the restricted product uses \(U_v=\mathcal O_v^\times\) at finite places. If \(S\) is a finite set containing the archimedean places, let \(S_L\) be its inverse image and put \[ J_L^S=\prod_{w\in S_L}L_w^\times\times \prod_{w\notin S_L}U_w. \tag{1} \] These are open subgroups, and their union is \(J_L\). We recall the local decomposition in a form that also fixes the norm: \[ L\otimes_K K_v\simeq\prod_{w\mid v}L_w,\qquad \sum_{w\mid v}[L_w:K_v]=[L:K]. \tag{2} \] Indeed, write \(L=K[T]/(f)\) by a primitive element, factor its separable polynomial over \(K_v\), and apply the polynomial Chinese remainder theorem. Its irreducible factors give the finite extensions of \(K_v\). The image of \(K[T]/(f)\) is dense in each such field, since \(K\) is dense in \(K_v\) and powers of its generator span that field. Thus those factors are exactly the completions at the extensions of \(v\), proving (2). In the Galois case, \(L_w\) contains the images of all conjugates of the primitive element, since they already belong to \(L\). Thus it splits \(f\), and \(L_w/K_v\) is Galois. Any local automorphism sends that primitive element to a global conjugate, so it preserves the dense subfield \(L\) and restricts to an element of \(G\) preserving \(w\). Conversely every such global element extends continuously. This identifies the local group with \(G_w\). Finally every other local factor has a root of \(f\), which is some global conjugate in \(L_w\); the resulting embedding is the embedding of \(L\) twisted by that global automorphism. Its induced place is a conjugate of \(w\). Hence \(G\) acts transitively on the places above \(v\). Only finitely many places ramify. One way to see the finiteness needed here is to choose a primitive element integral after excluding finitely many denominators. Its nonzero discriminant is a unit outside a finite further set. At such a finite place the reduction of its polynomial is separable; lifting its relatively prime factors gives unramified local factors. For a number field use integer rings; for a function field use the integral closure of a finite-field polynomial ring and include the places over infinity. This argument proves the same finite exceptional-set assertion in both cases. \subsubsection{Proposition 13.1. Extension and norm maps}\label{proposition-13.1.-extension-and-norm-maps} The diagonal embeddings at the places above \(v\) define a continuous injection \(J_K\hookrightarrow J_L\). The continuous norm is \[ (N_{L/K}x)_v=\prod_{w\mid v}N_{L_w/K_v}(x_w). \tag{3} \] It sends principal idèles to their ordinary field norms, is transitive in towers, and sends an embedded \(a\in J_K\) to \(a^{[L:K]}\). It induces maps \[ C_K\hookrightarrow C_L,\qquad N_{L/K}:C_L\longrightarrow C_K. \tag{4} \] \textbf{Proof.} A local unit stays a unit on extension, and the norm of a local unit is a unit. Thus both formulas give idèles and are continuous on every restricted product chart. The diagonal map is injective because each component embedding is injective. In (2), multiplication by \((x_w)\) is the product of the multiplication operators on the factors, so its determinant is (3). For a principal element \(x\in L^\times\), this operator is the scalar extension of multiplication by \(x\) on the \(K\)-vector space \(L\). Its determinant is therefore \(N_{L/K}(x)\), independently of \(v\). Transitivity follows from field-norm transitivity at the completions and regrouping the places. For \(a_v\in K_v^\times\), the determinant is \(a_v^{[L:K]}\) by (2). To prove the first injection in (4), we need \(J_K\cap L^\times=K^\times\). Embed \(L\) in a finite Galois extension \(M/K\). A diagonal idèle from \(J_K\) is fixed by \(\operatorname{Gal}(M/K)\). If it is the principal idèle of \(x\in L^\times\), each automorphism therefore fixes \(x\) in every completion and hence fixes \(x\) as an element of \(M\). Thus \(x\in K^\times\). The norm map descends because of the principal-element calculation. ∎ For Galois \(L/K\), the norm (3) is also the group-module norm \(\prod_{g\in G}g(x)\). This follows either from the determinant or by grouping the embeddings according to the place they induce. \subsection{2. Galois descent for classes}\label{galois-descent-for-classes} Assume \(L/K\) Galois, with \(G=\operatorname{Gal}(L/K)\). Its action on idèles is \[ (g x)_w=g(x_{g^{-1}w}). \tag{5} \] Here the last \(g\) denotes the induced isomorphism of completions. \subsubsection{Proposition 13.2. Fixed idèles and fixed classes}\label{proposition-13.2.-fixed-iduxe8les-and-fixed-classes} With the embeddings just defined, \[ J_L^G=J_K,\qquad C_L^G=C_K. \tag{6} \] \textbf{Proof.} Fix \(w\mid v\) and its decomposition group \(G_w\). The local field \(L_w/K_v\) is Galois with group \(G_w\); this follows also from (2), since the global automorphisms permute its factors transitively and the stabilizer acts on one factor. A fixed idèle has \(x_w\in L_w^{G_w}=K_v\). Transitivity on the places above \(v\) then makes all their components the image of the same element of \(K_v^\times\). Almost all of these elements are units, proving the first equality. A fixed class need not initially have a fixed representative. If \([x]\in C_L^G\), define \[ a_g=\frac{g(x)}x\in L^\times. \] These principal elements satisfy \(a_{gh}=a_g\,g(a_h)\). The full, noncyclic Hilbert 90 theorem proved in lesson 3 gives \(a_g=g(b)/b\) for some \(b\in L^\times\). Consequently \(x/b\) is fixed by every \(g\), hence belongs to \(J_K\), and represents \([x]\). Proposition 13.1 gives injectivity, proving the second equality. ∎ This is the invariant part of the exact sequence \(1\to L^\times\to J_L\to C_L\to1\). Hilbert 90 supplies the surjectivity that invariants alone would not preserve. \subsection{3. A cyclic induction calculation}\label{a-cyclic-induction-calculation} Suppose \(G=\langle\sigma\rangle\) has order \(n\). Let \(H=\langle\tau\rangle\), \(\tau=\sigma^r\), be a subgroup, with \(r=[G:H]\). We use additive notation for an abelian \(H\)-module \(M\). Its induced module is \(B=M^r\), with \[ \sigma(x_0,\ldots,x_{r-1})=(\tau x_{r-1},x_0,\ldots,x_{r-2}). \tag{7} \] \textbf{Lemma.} For \(i=0,-1\), \[ \widehat H^i(G,B)\simeq\widehat H^i(H,M). \tag{8} \] \textbf{Proof.} Invariants in (7) are diagonal vectors \((a,\ldots,a)\) with \(a\in M^H\). If \(N_H=1+\tau+\cdots+\tau^{n/r-1}\), then \[ N_G(x_0,\ldots,x_{r-1}) =(N_H\textstyle\sum x_i,\ldots,N_H\sum x_i). \tag{9} \] Thus invariants modulo norms give (8) in degree zero. Summing coordinates modulo \((\tau-1)M\) identifies \[ B/(\sigma-1)B\simeq M/(\tau-1)M. \tag{10} \] To check the kernel as well as surjectivity, in the left quotient each coordinate vector is equivalent to a vector in the first coordinate, by successive applications of \(\sigma\). Returning after \(r\) steps identifies its value with its image under \(\tau\). These are precisely the relations in the right quotient; placing a value in the first coordinate gives the inverse. Under (10), the norm to invariants is exactly \(N_H\), by (9). Its kernel is \(\widehat H^{-1}\), proving the other case. ∎ For a chosen \(w\mid v\), the identifications \(L_w\simeq L_{\sigma^i w}\) identify \[ B_v=\prod_{w'\mid v}L_{w'}^\times \simeq\operatorname{Ind}_{G_w}^G L_w^\times. \tag{11} \] The same holds for the unit groups at finite places. This is (7) written multiplicatively, not a trivial permutation action on the factors. \subsection{4. The restricted product contributes a direct sum}\label{the-restricted-product-contributes-a-direct-sum} For a cyclic unramified local extension, both Tate groups of its units vanish. The unit norm is surjective by the unramified norm theorem used in lesson 6. For the other group, Hilbert 90 writes a norm-one unit as \(\tau(b)/b\). Since an unramified extension has the same valuation group and a ground-field uniformizer, divide \(b\) by the ground-field uniformizer to the power \(v(b)\). This makes \(b\) a unit without changing its coboundary. Thus \[ \widehat H^0(G_w,U_w)=\widehat H^{-1}(G_w,U_w)=0 \quad\text{at an unramified finite place.} \tag{12} \] \subsubsection{Proposition 13.3. Local decomposition of idèle cohomology}\label{proposition-13.3.-local-decomposition-of-iduxe8le-cohomology} For cyclic \(L/K\) and \(i=0,-1\), restriction to the local factors gives \[ \widehat H^i(G,J_L) \simeq\bigoplus_v\widehat H^i(G_w,L_w^\times). \tag{13} \] In particular, \[ J_K/N_{L/K}J_L \simeq\bigoplus_v K_v^\times/N_{L_w/K_v}L_w^\times, \qquad \widehat H^{-1}(G,J_L)=0. \tag{14} \] One place \(w\) above each \(v\) is chosen; changing it transports the same local quotient by a \(K_v\)-isomorphism. \textbf{Proof.} Take finite \(S\) containing the archimedean and ramified places. Invariants, the norm kernel and the image of \(\sigma-1\) commute with arbitrary products of modules: the assertions are componentwise, and for the image choose one preimage in each component. Hence (8), (11) and (12) give \[ \widehat H^i(G,J_L^S) =\prod_{v\in S}\widehat H^i(G_w,L_w^\times). \tag{15} \] There is no tail contribution. Passing to the union over finite \(S\) changes this finite product into a direct sum. More explicitly, a norm-zero or fixed representative belongs to some \(J_L^S\), and a relation exhibiting a norm or coboundary also belongs to some such chart; only finitely many group operations occur. Thus kernels and the relevant images pass to the filtered union, proving (13). Degree zero is invariants modulo norms, and (6) identifies invariants with \(J_K\). Local degree zero is \(K_v^\times/N L_w^\times\), proving the first formula in (14). The local degree-minus-one group is zero by Hilbert 90, including at archimedean fields. This proves the last assertion. ∎ There is a concrete interpretation of the first formula. A fixed idèle is a unit at almost all unramified places, and those units are local norms. Thus it has only finitely many nonzero obstruction classes. Conversely any finite list of local classes can be represented by an idèle. If all its local classes vanish, choose norm preimages that are units outside a finite set, using (12); these preimages form an idèle whose norm is the given one. \subsection{5. The quotient for a finite set of places}\label{the-quotient-for-a-finite-set-of-places} \subsubsection{\texorpdfstring{Proposition 13.4. The Herbrand quotient of \(S\)-idèles}{Proposition 13.4. The Herbrand quotient of S-idèles}}\label{proposition-13.4.-the-herbrand-quotient-of-s-iduxe8les} For cyclic \(L/K\), with \(S\) finite and containing all archimedean and ramified places, \[ h(G,J_L^S)=\prod_{v\in S}[L_w:K_v]. \tag{16} \] \textbf{Proof.} Formula (15) makes both Tate groups finite. For a finite cyclic extension of nonarchimedean local fields, finite reciprocity gives norm index \([L_w:K_v]\), and Hilbert 90 gives degree-minus-one group zero. Their Herbrand quotient is therefore \([L_w:K_v]\). At a real place that becomes complex, the norm image is \(\mathbf R_{>0}\), of index 2, and a norm-one complex number is \(z/\bar z\); hence the same quotient is 2. Every other archimedean local degree is 1 and contributes 1. Multiply these local orders in (15), with no tail contribution by (12). ∎ The quotient \(h(G,J_L)\) need not be defined: its degree-zero Tate group may be infinite. Formula (16) concerns the specified open subgroup, not the whole idèle group. \subsection{6. An infinite local quotient and a finite class quotient}\label{an-infinite-local-quotient-and-a-finite-class-quotient} Consider \(L=\mathbf Q(i)\), \(K=\mathbf Q\). At an odd prime \(p\), the polynomial \(T^2+1\) splits if \(p\equiv1\pmod4\), and otherwise gives the unramified quadratic extension. This follows from cyclicity of the residue-field units and the simple-root lifting theorem. At 2 its completion is the ramified quadratic field \(\mathbf Q_2(i)\). The real completion becomes complex. The norm groups proved in lessons 6 and 10 therefore give \[ J_{\mathbf Q}/N J_{\mathbf Q(i)} \simeq \mathbf Z/2\ \oplus\ \mathbf Z/2\ \oplus\!\!\bigoplus_{p\equiv3\ (4)}\mathbf Z/2. \tag{17} \] The first coordinate is the real sign. The second sends \(x_2=2^\alpha u\) to the class of its odd unit \(u\) modulo \(1+4\mathbf Z_2\). The \(p\)-coordinate is \(v_p(x_p)\) modulo 2. Split primes contribute zero. There are infinitely many primes \(3\pmod4\): from a putative finite list take \(4\prod p-1\). It is \(3\pmod4\), none of the listed primes divides it, and some prime factor is \(3\pmod4\), since a product of primes all \(1\pmod4\) cannot be \(3\pmod4\). Thus (17) is infinite. We can also compute the class quotient here without assuming global reciprocity. Let \(e_\infty,e_2,e_p\) be the coordinate generators in (17). The image of the principal idèle \(-1\) is \(e_\infty+e_2\); that of a prime \(p\equiv3\pmod4\) is \(e_2+e_p\). Principal primes \(1\pmod4\) and the prime 2 have zero image. These elements generate the full principal image, since \(\mathbf Q^\times\) is generated by \(-1\) and its primes. The quotient by these relations identifies every coordinate generator with \(e_2\), and the sum of all coordinates remains a nonzero invariant. Consequently \[ C_{\mathbf Q}/N C_{\mathbf Q(i)}\simeq\mathbf Z/2. \tag{18} \] Every sum here has finite support. The passage from (17) to (18) displays the relations imposed by principal idèles. \subsection{7. Exercises with solutions}\label{exercises-with-solutions} \subsubsection{Exercise 1. The norm image is open --- easy}\label{exercise-1.-the-norm-image-is-open-easy} For finite separable \(L/K\), prove that \(N J_L\) is open in \(J_K\). \textbf{Solution.} A norm subgroup for any finite extension of nonarchimedean local fields is open: embed it in a finite Galois extension, whose norm subgroup is open by local reciprocity, and use norm transitivity to put that open subgroup inside the original norm image. At infinity a norm image is the whole local multiplicative group or the positive real subgroup; both are open. Choose finite \(S\) containing ramification and infinity. At each \(v\in S\), the product of the local norm images is an open subgroup \(V_v\subset K_v^\times\). Outside \(S\), every unit is a norm of a unit from any one unramified factor. Thus \[ \prod_{v\in S}V_v\times\prod_{v\notin S}U_v \subset N J_L. \] Choose the corresponding preimages componentwise; their unit tails make them idèles. The displayed group is an open identity neighborhood, so the subgroup \(N J_L\) is open. Infinite index is compatible with openness. \subsubsection{Exercise 2. Fixed classes --- medium}\label{exercise-2.-fixed-classes-medium} Prove that every \(G\)-fixed idèle class has a fixed representative. Identify where Hilbert 90 enters. \textbf{Solution.} Choose representative \(x\). Fixedness means \(g(x)/x=a_g\in L^\times\), and the action gives \(a_{gh}=a_g g(a_h)\). Hilbert 90 for the full finite Galois group gives \(a_g=g(b)/b\). Then \(g(x/b)=x/b\), so the representative is in \(J_L^G=J_K\). Its class is unchanged. Injection follows from \(J_K\cap L^\times=K^\times\). The cocycle could not be removed merely by applying an exactness claim to invariants; Hilbert 90 is precisely the missing surjectivity step. \subsubsection{Exercise 3. The quadratic rational example --- medium}\label{exercise-3.-the-quadratic-rational-example-medium} Describe the quotient in (17), provide a representative for each coordinate generator, and derive (18). \textbf{Solution.} For \(e_\infty\), use real component \(-1\) and all finite components 1. For \(e_2\), use component 3 at 2 and all other components 1. For \(e_p\), use a uniformizer \(p\) at that inert prime and 1 elsewhere. Their local norms fail exactly at the indicated place. Local unramified norms have even valuation and arbitrary unit; the norm group at 2 is \(2^{\mathbf Z}(1+4\mathbf Z_2)\); the real norm is positive; at split primes it is surjective. This proves the coordinate description, its independence and its exhaustiveness by (14). On principal elements, \(-1\) gives \(e_\infty+e_2\), each inert prime gives \(e_2+e_p\), and the remaining prime generators give zero. Quotienting by their span identifies all generators, leaving a single group of order 2 detected by the sum of coordinates. This is the class quotient because \(N C_L\) is the image of \(N J_L\) after quotienting by \(\mathbf Q^\times\). \subsubsection{Exercise 4. Proving the decomposition directly --- hard}\label{exercise-4.-proving-the-decomposition-directly-hard} Prove both degrees of (13), keeping track of restricted-product support and the twisted action on a block. \textbf{Solution.} For a place block choose the cosets \(1,\sigma,\ldots,\sigma^{r-1}\) of \(G_w=\langle\sigma^r\rangle\), and identify the components using these automorphisms. The action is the twisted cycle (7). Its invariants are diagonal \(G_w\)-invariants; its norm is (9). Thus degree zero reduces to the local degree-zero quotient. The sum map (10), with inverse given by the first coordinate, reduces the kernel of the norm on coinvariants to \(\ker N_{G_w}/(\sigma^r-1)L_w^\times\); this proves degree minus one, too. An idèle belongs to a chart with only finitely many unrestricted places. At every remaining place the unit Tate groups vanish by unramified unit norms and the unit-adjusted Hilbert 90 argument. For the product of those tails, choose the local preimages all in unit blocks, so they form a restricted-product preimage. Hence the chart's cohomology is the finite product in (15). A representative or a finite algebraic relation involves only one sufficiently large chart. Taking their union therefore produces the direct sum over all places, not an unrestricted product of local obstruction groups. \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} Section 1 gives the local tensor decomposition from the primitive element and polynomial remainder lemmas of lesson 1. Subsequent sections prove norms, invariant class descent and the induced local calculation in the restricted product. \begin{itemize} \tightlist \item \href{https://www.mathi.uni-heidelberg.de/~schmidt/Neukirch-en/Neukirch_cft_02_may15.pdf}{Jürgen Neukirch, Class Field Theory --- The Bonn Lectures, Online Edition 2.0 (May 2015), edited by Alexander Schmidt}. \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{The Herbrand quotient of the idèle class group}\label{the-herbrand-quotient-of-the-iduxe8le-class-group} \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).} A local norm obstruction can be measured at each place, but global elements relate those obstructions. Which part survives after principal idèles are removed? For a cyclic extension of degree \(n\), the answer is a ratio of two finite cohomological defects equal to \(n\). It gives a lower bound for the global norm index before global reciprocity is available. We begin with a calculation in which all the places and units can be seen. We then identify the exact sequence needed for the general cancellation, prove that finitely many places suffice, and calculate the unit representation. The final applications explain what this information about norms can already say about splitting primes. The definitions and fixed-class descent are \href{../ideles-in-extensions-and-their-cohomology.html}{Idèles in extensions and their cohomology}. Herbrand-quotient multiplicativity and lattice comparison are \href{../cohomology-of-cyclic-groups-and-the-herbrand-quotient.html}{Cohomology of cyclic groups and the Herbrand quotient}. Number-field unit lattices are the written \emph{Dirichlet's unit theorem}, Theorems 9.2 and 9.4 of \emph{Number fields}, and \emph{Idèles and the idèle class group}, Corollary 3.4 of \emph{Adèles and L-functions}. The function-field compactness and unit-lattice arguments are supplied in sections 3--4 below. Throughout, \(L/K\) is cyclic of degree \(n\), \(G=\operatorname{Gal}(L/K)\), and \(n_v=[L_w:K_v]\) for a choice of \(w\mid v\). \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-14}{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. A calculation with six constant-field automorphisms}\label{a-calculation-with-six-constant-field-automorphisms} Take \(K=\mathbf F_2(t)\), \(L=\mathbf F_{64}(t)\), and let \(S\) consist of the places defined by \[ t,\qquad t^2+t+1,\qquad t^3+t+1. \] The second polynomial has no root in \(\mathbf F_2\); neither does the cubic, so they are irreducible. Their residue degrees are 1, 2 and 3. In a constant extension of degree six, a place of degree \(d\) has \(\gcd(6,d)\) places above it, each with local degree \(6/\gcd(6,d)\). This follows by factoring the finite-field tensor product, or by the orbits of the sixth iterate of binary Frobenius on its roots. The three decomposition-group orders are therefore 6, 3 and 2. There are six places of \(L\) above \(S\), all rational over \(\mathbf F_{64}\). Write them as \(t-a_1,\ldots,t-a_6\). An \(S\)-unit has divisor supported there, and the product formula says that its six exponents sum to zero. Conversely every such exponent vector is realized by a product of the ratios \((t-a_j)/(t-a_6)\). Thus, modulo the finite constant units, the unit lattice is exactly the augmentation kernel of the permutation lattice on these six places. The group permutes them in orbits of sizes 1, 2 and 3. For a cyclic group, a permutation orbit with stabilizer of order \(m\) has Herbrand quotient \(m\). Consequently this permutation lattice has quotient \(6\cdot3\cdot2=36\). Its trivial augmentation target has quotient 6, so its kernel has quotient 6. Finite constant units have quotient 1 even when their order is not prime to six. The \(S\)-idèle quotient is 36 by the local computation of lesson 13. To check that these places suffice, let \(P\) be any monic irreducible polynomial over \(\mathbf F_{64}\). Its function has divisor \([P]-\deg(P)[\infty]\). Subtracting such principal divisors from an arbitrary divisor leaves its degree times \([\infty]\). The divisor of \(t-a_1\) is \([t-a_1]-[\infty]\), so the remaining class can be represented at this place in \(S\). In particular every degree-zero divisor is principal. Applied to the finite valuation support of an idèle, this gives \(J_L=L^\times J_L^S\). Dividing the two quotients gives \[ h(G,C_L)=36/6=6. \] The infinite product of local factors has disappeared from the calculation. The principal elements removed a sum-zero lattice; the one trivial direction they did not remove is responsible for the extension degree. \subsection{2. The three modules that have to be compared}\label{the-three-modules-that-have-to-be-compared} For finite \(S\), let \(J_L^S\) allow arbitrary components above \(S\) and units elsewhere, and let \(E_{L,S}=L^\times\cap J_L^S\). Once \(S\) is large enough that \(J_L=L^\times J_L^S\), these groups fit into \[ 1\longrightarrow E_{L,S}\longrightarrow J_L^S\longrightarrow C_L\longrightarrow1. \] The calculation to be proved is \[ \begin{array}{c|c|c} \text{module}&\text{Herbrand quotient}&\text{reason}\\\hline J_L^S&\prod_{v\in S}n_v&\text{local norms and induced modules}\\ E_{L,S}&n^{-1}\prod_{v\in S}n_v&\text{the sum-zero logarithmic lattice}\\ C_L&n&\text{the exact cyclic hexagon}. \end{array} \] The first line is Proposition 13.4. The second is Proposition 14.2 below. Multiplicativity gives the third only after both groups have finite Tate defects. We therefore need to prove the choice of \(S\), the full-lattice assertion and that finiteness; the table itself is not a substitute for those arguments. Adding a finite unramified place of local degree \(m\) to a permissible set \(S\) multiplies both numerator and denominator by \(m\). For the idèles this is the newly unrestricted local factor; for the units it is the additional orbit in the permutation representation. The global answer is unchanged. This is a useful check that the result describes \(C_L\), rather than the auxiliary set of places. \subsection{3. Compactness and a finite set of places}\label{compactness-and-a-finite-set-of-places} For a global field \(F\), normalize the finite absolute values by \(|x|_w=q_w^{-v_w(x)}\), and the real and complex ones by ordinary modulus and its square. The product formula makes \[ |x|=\prod_w|x_w|_w,\qquad C_F^1=\ker\bigl(|\cdot|:C_F\to\mathbf R_{>0}\bigr) \tag{1} \] well-defined. For number fields \(C_F^1\) is compact by the full box argument in \emph{Adèles and L-functions}, Theorem 3.3. The norm has a continuous section at any archimedean place, so \(C_F=C_F^1\times\mathbf R_{>0}\) after choosing that section. \subsubsection{The function-field compactness argument}\label{the-function-field-compactness-argument} We give the additional support for \(F\) a function field over its full constant field \(k=\mathbf F_q\). First \(F\) has a separating rational parameter \(t\), so \(F/k(t)\) is finite separable. Here is the elementary characteristic-\(p\) justification. For any rational parameter \(u\), Frobenius and the degree formula give \([F:F^p]=p\): both \([F:k(u)]\) and \([F^p:k(u)^p]\) are the same, while \([k(u):k(u)^p]=p\). Choose \(t\notin F^p\). Then \(F=F^p(t)\). A finite extension of a one-variable field over a perfect field is inseparable only if it contains the \(p\)th root of that field's rational parameter. To verify this criterion, take its maximal separable subfield \(E\). It too satisfies \([E:E^p]=p\), and \(t\notin E^p\), since a separable extension cannot contain the nontrivial purely inseparable element \(t^{1/p}\). Hence \(E=E^p(t)\) and \(E^{1/p}=E(t^{1/p})\). A nontrivial finite purely inseparable extension of \(E\) contains \(E^{1/p}\): if an element has minimal purely inseparable exponent \(r\geq1\), its \(p^{r-1}\) power generates a degree-\(p\) extension, which is the entire degree-\(p\) field \(E^{1/p}\). This proves the criterion. Since \(t\) is not a \(p\)th power in \(F\), \(F/k(t)\) is separable. For \(E=k(t)\), partial fractions and the polynomial Chinese remainder theorem give a compact set of additive representatives for \(\mathbb A_E/E\): \[ D_E=\prod_{P\text{ finite}}\mathcal O_P \times t^{-1}k[[t^{-1}]]. \tag{2} \] Indeed, finitely many finite principal parts can be removed by a rational function with poles only at those irreducible polynomials \(P\). Subtract a polynomial to remove the nonnegative powers of \(t\) at infinity. The remainder lies in (2). Also \(E\cap D_E=0\): a rational function integral at every finite place is a polynomial, and one vanishing at infinity is zero. Since \(D_E\) is open, \(E\) is discrete; since it is compact and supplies representatives, the quotient is compact. An \(E\)-basis of \(F\) identifies \(\mathbb A_F\) topologically with \(\mathbb A_E^{[F:E]}\), and \(F\) with \(E^{[F:E]}\). To check the tail topology, take an integral primitive basis and exclude its denominators and discriminant. Outside those finitely many primes it is an integral basis of the local product: the trace-pairing matrix has unit determinant, so the coordinates of any integral element are integral. At the remaining finitely many completions it is a vector-space isomorphism. Thus \(F\) is discrete in \(\mathbb A_F\), and \(\mathbb A_F/F\) is compact. The function-field product formula also follows from this rational parameter: local determinant norms give \[ \prod_{w\mid v}|a|_w=|N_{F/E}(a)|_v. \] The rational formula is just equality of the degree of a polynomial with the sum of the degrees of its prime factors, including the opposite valuation at infinity. Multiplying the displayed norm formula proves the product formula for \(F\). Give \(\mathbb A_F\) additive Haar measure and let \(c\) be the finite positive volume of its compact quotient by the discrete subgroup \(F\). For an idèle \(\beta\), the compact open box \(B_\beta=\prod_w\beta_w\mathcal O_w\) has volume \(b|\beta|\), where \(b=\operatorname{vol}(B_1)>0\). If this volume exceeds \(c\), there are two distinct box elements differing by a nonzero element of \(F\). For clarity, integration of \(\sum_{a\in F}\mathbf1_{B_\beta}(x+a)\) over the quotient gives \(\operatorname{vol}(B_\beta)\). If every coset met the box at most once, that integral would be at most \(c\). The integration identity follows by partitioning into local slices on which the discrete quotient is injective. Since the box is an additive subgroup, their difference belongs to the box. Thus \[ |\beta|>c/b\quad\Longrightarrow\quad 0\ne a\in F,\quad |a|_w\leq|\beta_w|_w\ \text{for every }w. \tag{3} \] The norm-one idèles are closed in \(\mathbb A_F\), and their additive and idèle subspace topologies agree. We include the topology check needed to use (3). For a norm-one idèle, constrain finitely many coordinates to have product absolute value less than 2 and leave integral tails. A new tail nonunit costs a factor at most \(1/2\), so a norm-one element in that additive neighborhood must still have unit tails. This proves equality of the two topologies. To prove closedness, a zero component, or infinitely many tail nonunits, allows a finite coordinate product less than 1, excluding norm one. For an idèle of norm \(d<1\), the same argument works. For norm \(d>1\), include all places with \(q_w\leq2d\) and constrain the finite product to lie between 1 and \(2d\); a tail nonunit then forces the total below 1, and a unit tail leaves it above 1. There are only finitely many such places: each lies over a prime of \(k(t)\) of bounded residue degree, and that rational field has only finitely many irreducible polynomials of bounded degree. Choose \(\eta\) with \(|\eta|>c/b\). Apply (3) to \(\eta x^{-1}\) for \(x\in J_F^1\). The resulting \(a x\) is still norm one and belongs to the fixed compact box \(B_\eta\). Its intersection with \(J_F^1\) is compact by the topology check. Every norm-one class has a representative there, proving compactness of \(C_F^1\). \subsubsection{\texorpdfstring{Choosing \(S\)}{Choosing S}}\label{choosing-s} For either kind of global field, \(C_F/\operatorname{image}(J_F^S)\) is finite when \(S\) is finite, contains infinity, and is nonempty. The image is open. In a number field it has the full norm image because of an archimedean component, so the quotient is an image of compact \(C_F^1\), and is discrete, hence finite. In a function field the norm image is a subgroup of \(q^{\mathbf Z}\); the norm image of \(J_F^S\) contains a nonzero power of \(q\), so it has finite index in the full norm image. Each of finitely many norm cosets is then accounted for by a compact norm-one quotient. Again the discrete quotient is finite. Take representatives of this finite quotient and enlarge \(S\) to include their nonunit components. The enlarged set satisfies \(J_F=F^\times J_F^S\). Applying this to \(F=K,L\), and taking the finite images of the needed places of \(L\) in \(K\), we can choose a single finite \(S\) containing ramification and infinity with \[ J_K=K^\times J_K^S,\qquad J_L=L^\times J_L^S. \tag{4} \] For function fields keep \(S\) nonempty. Define \[ E_{L,S}=L^\times\cap J_L^S =\{a\in L^\times:v_w(a)=0\text{ for }w\notin S_L\}. \] Equation (4) gives the exact sequence of \(G\)-modules \[ 1\longrightarrow E_{L,S}\longrightarrow J_L^S \longrightarrow C_L\longrightarrow1. \tag{5} \] \subsection{\texorpdfstring{4. The \(S\)-unit representation}{4. The S-unit representation}}\label{the-s-unit-representation} The logarithm map \[ \ell:E_{L,S}\longrightarrow H=\left\{(x_w)_{w\in S_L}\in\mathbf R^{S_L}: \sum_w x_w=0\right\}, \quad a\longmapsto(\log|a|_w)_w \tag{6} \] has finite kernel \(\mu_L\), and its image is a full lattice in \(H\). For number fields this is the written \(S\)-logarithmic theorem cited above. In function fields, here is the proof from section 3. The group \(J_L^{S,1}/E_{L,S}\) identifies with the image of \(J_L^{S,1}\) in compact \(C_L^1\). It is an open subgroup, hence closed of finite index and compact. The local valuation map sends \(J_L^{S,1}\) onto the discrete lattice \[ \Delta=\left\{(-m_w\log q_w)_w:m_w\in\mathbf Z,\ \sum_w m_w\log q_w=0\right\}. \] Since all \(q_w\) are powers of the full constant-field cardinality, \(\Delta\) spans \(H\). Its kernel is the compact product of all local unit groups. Compactness of the preceding quotient makes \(\Delta/\ell(E_{L,S})\) finite, so the logarithmic image is a full lattice. Its kernel on principal elements is finite, being the intersection of a discrete closed subgroup with that compact product. It consists exactly of the constants \(k_L^\times\), or equivalently the roots of unity: a finite-order element is algebraic over the finite constant field, and every nonzero constant is a unit everywhere. This proves (6) with all its assertions in this case, too. The action on logarithms is a permutation: \[ \ell(g a)_w=\ell(a)_{g^{-1}w}. \tag{7} \] There is no extra residue-degree factor: the normalized absolute value is transported by the isomorphism \(L_{g^{-1}w}\to L_w\). \subsubsection{\texorpdfstring{Proposition 14.2. The \(S\)-unit quotient}{Proposition 14.2. The S-unit quotient}}\label{proposition-14.2.-the-s-unit-quotient} For cyclic \(L/K\), \[ h(G,E_{L,S})=\frac1n\prod_{v\in S}n_v. \tag{8} \] \textbf{Proof.} Finite torsion has Herbrand quotient 1, so we may replace \(E_{L,S}\) by its logarithmic lattice \(\Lambda\) in \(H\). Adjoin the trivial lattice \(\mathbf Z e\), \(e=(1,\ldots,1)\), to form \(\Lambda\oplus\mathbf Z e\). Its real representation is the full permutation space \(\mathbf R^{S_L}\). The lattice comparison theorem in lesson 2 therefore gives the same Herbrand quotient as the permutation lattice \[ P=\mathbf Z^{S_L} =\bigoplus_{v\in S}\mathbf Z[G/G_w]. \] That comparison does not require the two lattices to have rational coordinates in a common real embedding. Their integral representations are real-isomorphic; the rational commuting-map equations consequently have a real solution with nonzero determinant. The determinant is a nonzero polynomial on that rational solution space, so it has a rational nonzero value. Clearing denominators gives an equivariant injection with finite cokernel, exactly the argument proved in lesson 2. The explicit induced cyclic calculation in lesson 13 gives \(h(G,\mathbf Z[G/G_w])=h(G_w,\mathbf Z)=|G_w|=n_v\). The trivial lattice \(\mathbf Z e\) has quotient \(n\). Multiplicativity now gives \[ n\,h(G,E_{L,S})=h(G,\Lambda\oplus\mathbf Z e) =h(G,P)=\prod_{v\in S}n_v, \] proving (8). ∎ \subsection{5. The quotient of the idèle class group}\label{the-quotient-of-the-iduxe8le-class-group} \subsubsection{Theorem 14.1. The degree survives}\label{theorem-14.1.-the-degree-survives} For every cyclic extension \(L/K\) of global fields, \[ h(G,C_L)=n. \tag{9} \] Both Tate groups defining this quotient are finite. \textbf{Proof.} Choose \(S\) as in (4). Proposition 13.4 gives \(h(G,J_L^S)=\prod_{v\in S}n_v\); Proposition 14.2 gives (8), with finite Tate groups. Apply the full exact cyclic hexagon and its quotient multiplicativity to (5). This gives finiteness for the third module, and \[ h(G,C_L)=\frac{h(G,J_L^S)}{h(G,E_{L,S})} =\frac{\prod_{v\in S}n_v}{(\prod_{v\in S}n_v)/n}=n. \] No Herbrand quotient of the unrestricted idèle group is used. ∎ \subsubsection{Corollary 14.3. The norm index is at least the degree}\label{corollary-14.3.-the-norm-index-is-at-least-the-degree} For cyclic \(L/K\), \[ [C_K:N_{L/K}C_L] =n\,|\widehat H^{-1}(G,C_L)|\geq n. \tag{10} \] \textbf{Proof.} Proposition 13.2 identifies \(C_L^G=C_K\), and the module norm is the idèle-class norm. Thus degree zero is \(C_K/N C_L\). Equation (9) says that the size of this finite group is \(n\) times the size of degree minus one. ∎ The later norm inequality will show equality and kill the remaining Tate group. Equation (10) alone makes no such assertion. \subsection{6. Primes cannot all split}\label{primes-cannot-all-split} We use the fully proved weak approximation theorem of \emph{Local fields}, \emph{Absolute values, valuations and Ostrowski's theorem}, Theorem 5.2. Its hypothesis is a finite list of pairwise inequivalent nontrivial absolute values on any field; it therefore applies to both global-field cases here. Density of the field in its individual completions turns its original field-valued targets into arbitrary completion-valued targets. \textbf{Lemma.} A nontrivial cyclic extension \(L/K\) has infinitely many finite places that do not split completely. \textbf{Proof.} Suppose only finitely many finite places fail to split, and include these and infinity in a finite set \(T\). Given \(x\in J_K\), each local norm image at a place of \(T\) is open. Weak approximation chooses \(a\in K^\times\) close enough to \(x_v\) that \(x_v/a\) is a local norm for every \(v\in T\). Choose the neighborhoods to exclude zero. Outside \(T\), complete splitting makes the local norm surjective. Proposition 13.3 then says \(x/a\in N J_L\). Thus every idèle class is a norm, giving \([C_K:N C_L]=1\), contrary to (10) and \(n>1\). ∎ There is a stronger conclusion for cyclic extensions of prime-power degree: infinitely many finite places are \textbf{inert}, meaning they are unramified and have just one place above them. To prove it, let \(G\) have order \(\ell^r>1\) and let \(H\) be its unique subgroup of index \(\ell\). Every proper subgroup is contained in \(H\). If only finitely many places were inert, every other unramified place would have proper decomposition group and would therefore split completely in the degree-\(\ell\) field \(L^H/K\). This contradicts the preceding lemma for that field. Inertness and failure of complete splitting are different conditions. For a noncyclic Galois extension, an unramified decomposition group is cyclic, so it cannot equal the whole group. There are consequently no inert unramified places, although Corollary 14.4 below gives infinitely many places that fail to split completely. Only the finitely many ramified places can remain a single place for another reason. \subsubsection{Corollary 14.4. Complete splitting detects the field}\label{corollary-14.4.-complete-splitting-detects-the-field} If almost all finite places of \(K\) split completely in a finite separable extension \(M/K\), then \(M=K\). \textbf{Proof.} Let \(L\) be its Galois closure. At a completely split place of \(M/K\), its primitive-element polynomial splits into linear factors over \(K_v\), by the local decomposition (2) of lesson 13. Its splitting field therefore embeds in \(K_v\). Since \(L/K\) is Galois, all the completions of \(L\) above \(v\) have degree one. Thus almost all places of \(K\) split completely in \(L\). If \(L\ne K\), its finite group contains an element of prime order: take any nonidentity element of order \(m\), a prime \(\ell\mid m\), and its \(m/\ell\) power. Let \(K'\) be the fixed field of that order-\(\ell\) subgroup. Then \(L/K'\) is a nontrivial cyclic extension. Every place of \(K'\) above a completely split place of \(L/K\) is completely split in \(L/K'\); only finitely many places of \(K'\) lie above the exceptional finite set. This contradicts the lemma. Hence \(L=K\) and \(M=K\). ∎ Taking the contrapositive gives infinitely many nonsplit primes in every nontrivial finite separable extension. The proof uses a cyclic extension over an intermediate base field, so it also works when the original Galois group has no nontrivial cyclic quotient. \textbf{Separating cyclic extensions.} Let \(L_1,\ldots,L_r\) be cyclic extensions of the same prime degree \(\ell\), and put \(M=L_2\cdots L_r\). If \(L_1\not\subset M\), there are infinitely many finite primes of \(K\) that split completely in \(M\) and are inert in \(L_1\). \textbf{Proof.} Put \(E=L_1M\). The hypothesis makes \(E/M\) cyclic of degree \(\ell\). It has infinitely many inert primes by the prime-power conclusion above. Exclude those over the finitely many primes ramified in \(E/K\). At a remaining prime \(v\) of \(K\), the decomposition group \(D\) in \(\operatorname{Gal}(E/K)\) is cyclic. This Galois group embeds in a product of groups of order \(\ell\), so every nonidentity element has order \(\ell\); hence \(|D|\leq\ell\). If its image in \(\operatorname{Gal}(M/K)\) were nontrivial, \(D\cap\operatorname{Gal}(E/M)\) would be trivial, contradicting the selected prime's inertness in \(E/M\). Thus \(D\) maps trivially to the group of \(M/K\), which makes \(v\) completely split in \(M\), and \(D=\operatorname{Gal}(E/M)\) maps nontrivially to \(\operatorname{Gal}(L_1/K)\), making \(v\) inert in \(L_1\). Infinitely many primes of \(M\) lie over infinitely many primes of \(K\), since every fiber is finite. ∎ Pairwise intersections \(L_i\cap L_j=K\) do not imply the hypothesis just used. For example, \(\mathbf Q(\sqrt6)\), \(\mathbf Q(\sqrt2)\), and \(\mathbf Q(\sqrt3)\) have pairwise intersection \(\mathbf Q\), but the first is contained in the compositum of the latter two. Away from ramification, splitting in both latter fields forces splitting in the first. The required condition is independence of the first field from the \textbf{whole} compositum. \subsection{7. Returning to a number-field example}\label{returning-to-a-number-field-example} For \(L=\mathbf Q(i)\), \(K=\mathbf Q\), take \(S=\{2,\infty\}\). The Gaussian integers are Euclidean, as proved in \emph{Idèles and the idèle class group}, so all their ideal classes are trivial and this \(S\) satisfies (4). Both local degrees are 2. Thus \[ h(G,J_L^S)=4,\qquad h(G,E_{L,S})=2,\qquad h(G,C_L)=2. \tag{11} \] Concretely \(E_{L,S}=\mu_4(1+i)^{\mathbf Z}\). To verify this description, an \(S\)-unit's Gaussian fractional ideal is supported only on \(1+i\); dividing by the corresponding power leaves a Gaussian unit, which is in \(\mu_4\). Conjugation sends \(1+i\) to \(1-i=-i(1+i)\), so it acts trivially on the infinite cyclic quotient by torsion. That quotient has Herbrand quotient 2. The explicit computation in lesson 13 gives \([C_{\mathbf Q}:N C_{\mathbf Q(i)}]=2\), consistent with (10), and in this example \(\widehat H^{-1}(G,C_L)=0\). \subsection{8. Exercises with solutions}\label{exercises-with-solutions} \subsubsection{Exercise 1. Permutation and augmentation lattices --- easy}\label{exercise-1.-permutation-and-augmentation-lattices-easy} Let \(G=C_6\) act on three orbits with stabilizers of orders 6, 3 and 2. Compute the Herbrand quotient of their permutation lattice and its augmentation kernel. Then give the formula for one orbit \(G/H\) in a cyclic group of order \(n\), and for the augmentation kernel of the regular lattice. \textbf{Solution.} The three-orbit permutation quotient is \(6\cdot3\cdot2=36\). Its augmentation is onto a trivial integer lattice of quotient 6, so the kernel has quotient 6, exactly the unit calculation in section 1. In general an orbit lattice is induced from the trivial \(H\)-module \(\mathbf Z\). Its invariant norm quotient has order \(|H|\), and its degree-minus-one group is zero, by the twisted-cycle calculation, so the quotient is \(|H|\). For the regular lattice \(h=1\). The augmentation is surjective, and its target is a trivial lattice of quotient \(n\). Multiplicativity gives \(h\) of the augmentation kernel equal to \(1/n\), including \(n=1\), when the kernel is zero. \subsubsection{Exercise 2. Quadratic fields directly --- medium}\label{exercise-2.-quadratic-fields-directly-medium} For quadratic \(L/\mathbf Q\), prove (8) directly from the plus and minus eigenspaces of conjugation. \textbf{Solution.} Let \(s\) places in \(S\) split and \(t\) not split. A split place contributes a swapped pair of coordinates to \(\mathbf R^{S_L}\); a nonsplit place contributes one fixed coordinate. The plus eigenspace has dimension \(s+t\), and the minus eigenspace dimension \(s\). Removing the trivial sum line gives dimensions \(a=s+t-1\) and \(b=s\) for the unit representation. By integral lattice comparison, its quotient equals that of \(a\) trivial lattices and \(b\) sign lattices. A trivial rank-one lattice for a group of order 2 has quotient 2; a sign lattice has invariants zero, norm zero and \((\sigma-1)\mathbf Z=2\mathbf Z\), hence quotient \(1/2\). Finite torsion contributes 1. Consequently \(h(E_{L,S})=2^{a-b}=2^{t-1}\). Exactly the \(t\) nonsplit local degrees are 2, so this is \((\prod n_v)/2\). This includes both real and imaginary quadratic fields. \subsubsection{Exercise 3. Arbitrary extensions --- medium}\label{exercise-3.-arbitrary-extensions-medium} Deduce Corollary 14.4 from the norm inequality without assuming the Galois group has a cyclic quotient. \textbf{Solution.} Complete splitting in \(M\) makes its defining separable polynomial split over the completion, hence makes its Galois closure \(L\) split there too. If \(L\ne K\), choose a prime-order subgroup of \(\operatorname{Gal}(L/K)\), with fixed field \(K'\). Almost all places of \(K'\) split in the nontrivial cyclic extension \(L/K'\). Weak approximation at the finitely many exceptional places then makes every \(K'\)-idèle class a norm: local norm groups are open there, and all other local norms are surjective. This gives norm index 1, violating (10) for that cyclic extension. Therefore \(L=K\). \subsubsection{Exercise 4. The general unit calculation --- hard}\label{exercise-4.-the-general-unit-calculation-hard} Show why enlarging a permissible set of places cannot change the quotient in Theorem 14.1. Your argument should prove (8) and explain why a real logarithmic lattice suffices for integral cohomology. \textbf{Solution.} Remove the finite torsion of \(E_{L,S}\); its quotient is the full logarithmic lattice \(\Lambda\) in the sum-zero hyperplane. Adjoin a trivial rank-one lattice. The resulting real representation is the permutation representation on \(S_L\). To compare integral lattices, solve the intertwining matrix equations over \(\mathbf Q\). A real isomorphism gives a point where the determinant polynomial is nonzero; a nonzero polynomial over \(\mathbf Q\) cannot vanish at every rational point, as follows by induction on its number of variables. A rational invertible intertwiner therefore exists. Clearing denominators embeds one lattice in the other with finite cokernel, which has quotient 1. Thus the Herbrand quotients agree. Decompose the permutation lattice into \(\mathbf Z[G/G_w]\), one for each \(v\in S\). Their quotients are \(|G_w|=n_v\) by Exercise 1. The adjoined trivial line has quotient \(n\); divide by it to obtain (8). Adding a place of local degree \(m\) inserts its permutation orbit, so (8) multiplies the unit quotient by \(m\). Proposition 13.4 multiplies the idèle quotient by the same factor. Their ratio, and hence \(h(G,C_L)\), remains \(n\). \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 local induced-module quotients and the proved S-unit representation give the idèle-class Herbrand quotient. The function-field compactness and lattice argument and the constant-field example are written separately with their full hypotheses. \begin{itemize} \tightlist \item \href{https://www.mathi.uni-heidelberg.de/~schmidt/Neukirch-en/Neukirch_cft_02_may15.pdf}{Jürgen Neukirch, Class Field Theory --- The Bonn Lectures, Online Edition 2.0 (May 2015), edited by Alexander Schmidt}. \item \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne, Class Field Theory, version 4.03}. \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{The norm index bound and Hasse's norm theorem}\label{the-norm-index-bound-and-hasses-norm-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).} The preceding lesson proved that a cyclic extension of degree \(n\) has idèle-class norm index at least \(n\). We now prove the reverse bound. Kummer theory supplies a subgroup of precisely the required index; restriction and norm reduce arbitrary Galois groups to prime cyclic extensions. In function-field characteristic \(p\), an adelic argument supplies the case that roots of unity cannot reach. The equality of the two bounds gives Hasse's norm theorem and the local-global principle for conics. After the ray-prime theorem is proved in lesson 22, its section 6A develops the full Hasse--Minkowski theorem from these conic results. We use \href{../hilberts-theorem-90-and-kummer-theory.html}{Hilbert's Theorem 90 and Kummer theory}, \href{../ideles-in-extensions-and-their-cohomology.html}{Idèles in extensions and their cohomology}, and \href{../the-herbrand-quotient-of-the-idele-class-group.html}{The Herbrand quotient of the idèle class group}. Write \(J_F\) for the idèles, \(C_F=J_F/F^\times\), \(J_F^S\) for unrestricted components in \(S\) and unit components elsewhere, and \(E_S=F^\times\cap J_F^S\). \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-15}{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. The bound and the class field axiom}\label{the-bound-and-the-class-field-axiom} \textbf{Theorem 15.1.} If \(L/K\) is a finite Galois extension of global fields, then \[ [C_K:N_{L/K}C_L]\leq [L:K]. \tag{1} \] If the extension is cyclic of degree \(n\), then \[ \bigl|\widehat H^0(\operatorname{Gal}(L/K),C_L)\bigr|=n, \qquad \widehat H^{-1}(\operatorname{Gal}(L/K),C_L)=0. \tag{2} \] Thus idèle classes satisfy the cyclic class field axiom. We first note that the quotient in (1) is finite, without using global reciprocity. The idèle norm image is open, by the local norm theorem and surjectivity on units outside finitely many places. Its image in \(C_K\) is open. Also \[ |N_{L/K}x|_K=|x|_L. \tag{3} \] For number fields the right side takes every positive real value. For function fields it takes a nonzero subgroup of the discrete group of possible contents, so has finite index there. The compactness of \(C_K^1\), proved in the preceding lesson, now makes the remaining discrete quotient finite. Moreover this quotient is killed by \([L:K]\), because the norm of an embedded \(K\)-idèle is its \([L:K]\)th power. After proving (1), (2) follows immediately from the preceding lesson: \[ [C_K:NC_L] =n\,\bigl|\widehat H^{-1}(G,C_L)\bigr|. \tag{4} \] The argument below proves (1) algebraically whenever the prime cyclic degrees differ from the characteristic. Sections 5--7 prove the remaining function-field case. \subsection{2. Local power indices}\label{local-power-indices} Let \(n=\ell^m\), with \(\ell\ne\operatorname{char}K\), and suppose \(\mu_n\subset K\). For every place \(v\), \[ |K_v^\times/K_v^{\times n}|=\frac{n^2}{|n|_v}. \tag{5} \] At a finite place where \(n\) is a unit, deep principal units are uniquely \(n\)-divisible: solve \((1+z)^n=1+t\) by successive residue corrections, since the linear coefficient is a unit. Reduction identifies the unit quotient with the residue-field power quotient, of order \(n\); the \(n\) distinct roots of unity remain distinct in the residue field. Valuations supply the second factor \(n\). At a finite place dividing \(n\), put \(s_v=v(n)\). For \(k>s_v\), the map \[ U^k\xrightarrow{(\cdot)^n}U^{k+s_v} \tag{6} \] is bijective. Indeed, on \(z\in\pi^k\mathcal O_v\), the equation \[ z=\frac{t-\sum_{j\geq2}\binom nj z^j}{n} \] is a contraction when \(t\in\pi^{k+s_v}\mathcal O_v\): the nonlinear part has Lipschitz valuation at least \(k-s_v>0\). The same estimate proves injectivity. No nontrivial \(n\)th root of unity lies in \(U^k\). The map \(U/U^k\to U^n/U^{k+s_v}\) therefore has kernel \(\mu_n\), of order \(n\). Counting its image gives \[ [U:U^n]=nq_v^{s_v}. \] Together with valuations this proves (5). In \(\mathbf C\) the quotient is trivial and \(|n|_v=n^2\). A real place permits \(n>1\) only when \(n=2\); its quotient has order \(2=4/|2|_v\). Choose a finite, nonempty \(S\) containing the archimedean places, the places dividing \(n\), and any specified ramified places, enlarged so that \(J_K=K^\times J_K^S\). Such an \(S\) exists by the preceding lesson. Put \(s=|S|\). The product formula applied to \(n\) gives \[ \left|\prod_{v\in S}K_v^\times/K_v^{\times n}\right|=n^{2s}. \tag{7} \] The \(S\)-unit theorem gives \[ E_S/E_S^n\simeq(\mathbf Z/n)^s. \tag{8} \] Here the rank is \(s-1\), and the finite cyclic torsion contains \(\mu_n\). Also \(E_S/E_S^n\to K^\times/K^{\times n}\) is injective: an \(n\)th root of an \(S\)-unit has zero valuations outside \(S\). \subsection{3. The Kummer subgroup}\label{the-kummer-subgroup} \textbf{Proposition 15.2.} Suppose \(\mu_n\subset K\), \(n=\ell^m\) is prime to the characteristic, and \[ \operatorname{Gal}(L/K)\simeq(\mathbf Z/n)^r. \] With \(S\) as above, including every ramified place, there is a set \(T\) of \(s-r\) further finite places, all splitting completely in \(L\), such that \[ \Delta=\{u\in E_S:u^{1/n}\in L\} =\ker\left(E_S\longrightarrow \prod_{v\in T}K_v^\times/K_v^{\times n}\right). \tag{9} \] For \[ I(S,T)= \prod_{v\in S}K_v^{\times n} \times\prod_{v\in T}K_v^\times \times\prod_{v\notin S\cup T}U_v, \qquad C(S,T)=\operatorname{im}(I(S,T)\to C_K), \tag{10} \] one has \[ [C_K:C(S,T)]=n^r=[L:K], \qquad C(S,T)\subseteq N_{L/K}C_L. \tag{11} \] \textbf{Proof.} Every Kummer radical class for \(L/K\) has an \(S\)-unit representative. If \(a^{1/n}\in L\), then at an unramified place outside \(S\), valuation in the unramified local extension shows \(n\mid v(a)\). Form an idèle having valuation \(v(a)/n\) outside \(S\), with components \(1\) in \(S\). Write it as \(b j\), with \(b\in K^\times\) and \(j\in J_K^S\). Then \(a/b^n\in E_S\). Let \(N=K(E_S^{1/n})\). By (8) and the perfect Kummer pairing proved in lesson 3, \[ \operatorname{Gal}(N/K)\simeq(\mathbf Z/n)^s,\qquad L\subseteq N. \] The restriction onto \(\operatorname{Gal}(L/K)\), a free \(\mathbf Z/n\)-module, splits by lifting a basis. Its kernel \(H\) is free of rank \(s-r\). For completeness, over \(\mathbf Z/\ell^m\), a surjective matrix onto a free module has a unit pivot; elementary row and column operations remove one pivot at a time, leaving a free kernel. Choose a basis \(\sigma_1,\ldots,\sigma_{s-r}\) of \(H\). Set \(F_i=N^{\langle\sigma_i\rangle}\). The cyclic prime-power extension \(N/F_i\) has infinitely many unramified inert primes, by the stronger splitting consequence proved in lesson 14. Choose one above a place \(v_i\) of \(K\), avoiding \(S\) and previously chosen places. Its decomposition group in the abelian extension \(N/K\) is cyclic of order at most \(n\), and contains \(\langle\sigma_i\rangle\), which already has order \(n\). Hence it equals \(\langle\sigma_i\rangle\). In particular \(v_i\) splits in \(L\). For \(u\in E_S\), its \(n\)th root is in \(K_{v_i}\) precisely when its Kummer character kills \(\sigma_i\). Thus \(T=\{v_i\}\) gives (9). Each \(v_i\) is outside the divisors of \(n\), and \(u\) is a unit there. The kernel index is \(n^{s-r}\), which equals the order of \(\prod_{v\in T}U_v/U_v^n\). Therefore \[ E_S\longrightarrow\prod_{v\in T}U_v/U_v^n \quad\text{is surjective}. \tag{12} \] The case \(s=r\) simply has \(T=\varnothing\). The crucial principal intersection is \[ K^\times\cap I(S,T)=E_{S\cup T}^{\,n}. \tag{13} \] One inclusion is clear. For the other, take \(y\) in the left side and put \(M=K(y^{1/n})\), a cyclic extension of degree dividing \(n\). Given any idèle class, represent it by \(x\in J_K^S\). Its components in \(T\) are units. By (12), choose \(a\in E_S\) so that \(x/a\) is an \(n\)th power at every place in \(T\). At places in \(S\), the extension \(M\) splits, because \(y\) is already an \(n\)th power. Outside \(S\cup T\), \(y\) is a unit and \(n\) is invertible, so \(M\) is unramified and the unit \(x/a\) is a local norm. At \(T\), its being an \(n\)th power makes it a local norm as well: every local degree divides \(n\). Choose unit norm preimages outside finitely many places to obtain an idèle norm. Thus \(NC_M=C_K\). The cyclic lower bound from lesson 14 forces \([M:K]=1\). Consequently \(y=b^n\), and its valuations show \(b\in E_{S\cup T}\), proving (13). Now \(J_K=K^\times J_K^{S\cup T}\), and \[ [J_K^{S\cup T}:I(S,T)]=n^{2s},\qquad [E_{S\cup T}:E_{S\cup T}^{\,n}]=n^{s+|T|}=n^{2s-r}. \] Dividing these indices using (13) proves the first assertion of (11). Finally, every component in (10) is a local norm from \(L\). At \(S\), the local abelian Galois group has exponent dividing \(n\); local reciprocity identifies it with the local norm quotient, so that quotient kills \(n\)th powers. This uses the exponent, not the potentially larger local degree. At \(T\), the extension splits. Elsewhere it is unramified and unit norms are surjective. Unit preimages at almost all places give an idèle norm, proving the second assertion. \(\square\) \subsection{4. Descent and arbitrary Galois groups}\label{descent-and-arbitrary-galois-groups} First let \(L/K\) be cyclic of prime degree \(\ell\ne\operatorname{char}K\). Put \(K'=K(\mu_\ell)\), \(L'=LK'\), and \(d=[K':K]\). Then \(d\mid\ell-1\), \(L\cap K'=K\), and \(L'/K'\) is cyclic of degree \(\ell\). Proposition 15.2 gives index at most \(\ell\) over \(K'\). Restriction of idèles and norm back induce maps \[ C_K/NC_L\longrightarrow C_{K'}/NC_{L'} \longrightarrow C_K/NC_L \tag{14} \] whose composite is \(x\mapsto x^d\). These maps are well-defined by base change for field norms and transitivity; the same identities hold componentwise for idèles and principal elements. The first quotient is killed by \(\ell\), so the first map is injective. Its order is therefore at most \(\ell\). In any tower \(K\subset M\subset L\), \[ [C_K:N_{L/K}C_L]\leq [C_K:N_{M/K}C_M]\,[C_M:N_{L/M}C_L]. \tag{15} \] Indeed the kernel of the map between the first two quotients is the image, under \(N_{M/K}\), of \(C_M/N_{L/M}C_L\). Prime towers prove (1) for cyclic extensions and for Galois groups of prime-power order, provided the relevant prime cyclic case is known. Here is the reduction for every finite Galois group \(G\); a composition series with cyclic factors would not suffice. Write \(|G|=p^a m\), \(p\nmid m\). There is a subgroup \(P\) of order \(p^a\). An elementary proof is to let \(G\) act on its subsets of size \(p^a\). The number \(\binom{p^a m}{p^a}\) is congruent to \(m\) modulo \(p\), by taking the coefficient of \(X^{p^a}\) in \((1+X)^{p^a m}\). Some orbit has size prime to \(p\). Its stabilizer has order divisible by \(p^a\), and acts freely on that subset, so its order also divides \(p^a\). This is the required \(P\). A nontrivial \(p\)-group has nontrivial center by its conjugacy-class equation. Taking a central element of order \(p\) and inducting on the quotient supplies a tower of prime cyclic extensions. Thus (15) bounds the norm quotient for \(L/M\), where \(M=L^P\), by \(|P|\). Restriction to \(M\) and norm back have composite \(x\mapsto x^{[M:K]}\). Restriction is well-defined on the norm quotients: on \(J_L\), group the product \(\prod_{g\in G}g(y)\) into the cosets \(Pr\) to obtain a \(P\)-norm. Because \([M:K]\) is prime to \(p\), restriction injects the \(p\)-primary part of \(C_K/NC_L\) into \(C_M/N_{L/M}C_L\). Hence that primary part has order at most \(p^a\). Multiplying over the primes dividing \(|G|\) proves (1). This completes the algebraic argument for number fields and for prime cyclic degrees different from the function-field characteristic. We next supply that missing characteristic case without assuming global class field theory. \subsection{5. Additive duality for function fields}\label{additive-duality-for-function-fields} Let \(F\) have full constant field \(k=\mathbf F_q\) of characteristic \(p\). The preceding lesson gave a separating parameter, a finite separable extension \(F/E\) with \(E=k(t)\), and discreteness and compact quotient for \(F\subset\mathbb A_F\). On \(\mathbb A_E\), define \[ \Psi_E(x)= \exp\left(\frac{2\pi i}{p} \operatorname{Tr}_{k/\mathbf F_p} \sum_v\operatorname{res}_v(x_v\,dt)\right). \tag{16} \] At a finite place the residue includes the residue-field trace to \(k\). Only finitely many terms are nonzero. Partial fractions prove the residue theorem for rational differentials on \(k(t)\): each finite principal part contributes the negative of its contribution at infinity, and a polynomial contributes zero. Thus \(\Psi_E\) is trivial on \(E\). We need the stronger assertion \[ E^\perp=E \quad\text{under }(x,y)\longmapsto\Psi_E(xy). \tag{17} \] Subtract from \(y\in E^\perp\) a rational function to put its remainder in the compact representative set \[ \prod_{P\text{ finite}}\mathcal O_P\times t^{-1}k[[t^{-1}]] \] constructed in lesson 14. Write its infinity component as \(\sum_{j\geq1}c_jt^{-j}\). Testing against \(a t^{j-1}\), \(a\in k\), gives \(\operatorname{Tr}_{k/\mathbf F_p}(a c_j)=0\), so all \(c_j\) vanish. Now test against arbitrary proper fractions \(a(t)/P^r\). Other places give zero, and these fractions represent every principal part at \(P\). The local residue pairing between principal parts and \(\mathcal O_P\) is nondegenerate: expand in the parameter \(P\), and use nondegeneracy of the finite-field trace for the first nonzero coefficient. The derivative \(P'(t)\) is a unit at \(P\), so replacing \(dP\) by \(dt\) does not change this assertion. Each finite component therefore vanishes. This proves (17). The pairing also gives every continuous additive character of \(\mathbb A_E\). Locally, a character is a linear map to the \(p\)th roots of unity, killed by some parameter-power ideal. Coefficient expansion and the finite-field trace identify it uniquely with a Laurent series through the residue pairing. Globally, continuity kills all unit components outside a finite set, so the representing Laurent series form an adèle. Conversely an adèle gives a continuous character. This establishes self-duality, not just nondegeneracy. Put \(\Psi_F(x)=\Psi_E(\operatorname{Tr}_{F/E}x)\). Choose trace-dual \(E\)-bases of \(F\). Under the topological isomorphism \(\mathbb A_F\simeq\mathbb A_E^{[F:E]}\) proved in lesson 14, the pairing \(\Psi_F(xy)\) is the coordinatewise pairing between those dual bases. Thus \(\mathbb A_F\) is self-dual and \[ F^\perp=F. \tag{18} \] This argument avoids assuming a geometric residue or duality theorem for \(F\). For each \(w\), let \(N_w\) be the largest integer such that the local character is trivial on \(\pi_w^{-N_w}\mathcal O_w\). It is nontrivial at each place by the nondegenerate local trace. Outside finitely many places the extension of \(E\) is unramified, its integral trace pairing is perfect, and the base character has conductor \(N_w=0\). Hence \[ \kappa=\sum_w N_w w \tag{19} \] is a finite divisor. The local annihilator calculation from the rank-one Fourier theory in lesson 12 gives, for every divisor \(D=\sum d_w w\), \[ B(D)=\prod_w\pi_w^{-d_w}\mathcal O_w,\qquad B(D)^\perp=B(\kappa-D). \tag{20} \] \subsection{6. Riemann--Roch and finite-order Hecke functions}\label{riemannroch-and-finite-order-hecke-functions} Define \(H^0(D)=F\cap B(D)\), and \(l(D)=\dim_k H^0(D)\). It is finite: a discrete subgroup meets a compact set in finitely many points. The quotient \(\mathbb A_F/(F+B(D))\) is finite because \(\mathbb A_F/F\) is compact and the image of \(B(D)\) is open. By (18)--(20), its characters are exactly \(H^0(\kappa-D)\). A finite abelian group has as many characters as elements, so that quotient has order \(q^{l(\kappa-D)}\). Let \(c\) be the Haar volume of \(\mathbb A_F/F\). The image of \(B(D)\) has volume \(\operatorname{vol}(B(D))/q^{l(D)}\). Counting its cosets gives \[ \frac{\operatorname{vol}(B(D))}{c} =q^{l(D)-l(\kappa-D)}. \tag{21} \] The ratio of the left sides for \(D\) and \(0\) is \(q^{\deg D}\). Functions integral everywhere are the constants: they form a finite \(k\)-algebra inside \(F\), hence a finite field contained in the full constant field \(k\). Thus \(l(0)=1\). Put \(g=l(\kappa)\). Equation (21) yields \[ l(D)-l(\kappa-D)=\deg D+1-g. \tag{22} \] Applying it to \(\kappa\) gives \(\deg\kappa=2g-2\). A negative-degree divisor has no nonzero section, by the product formula. Consequently \[ \deg D>2g-2\quad\Longrightarrow\quad l(D)=\deg D+1-g. \tag{23} \] This is the needed Riemann--Roch theorem, proved from the additive quotient and its exact duality. Now let \(\chi\) be a finite-order character of \(C_F\), with conductor divisor \(\mathfrak f\). Let \(U(\mathfrak f)\) be deep principal units at its support and full units elsewhere. The ray group \[ \Gamma_{\mathfrak f}=C_F/\operatorname{im}U(\mathfrak f) \] is discrete, and its degree-zero subgroup is finite by compactness of \(C_F^1\). Denote its order by \(h_{\mathfrak f}\), and write the degree image as \(\delta\mathbf Z\). No assertion \(\delta=1\) is needed here. Weak approximation makes any idèle principal times an idèle congruent to \(1\) at \(\mathfrak f\); taking its other valuations shows that every ray class is represented by a divisor prime to \(\mathfrak f\). For \(\mathfrak f>0\), fix such a representative \(A\) of large degree \(d\). The functions producing effective divisors in its ray class are \[ a\in H^0(A),\qquad a\equiv1\pmod{\mathfrak f}. \] Evaluation onto \(\mathcal O/\mathfrak f\) is surjective: its kernel is \(H^0(A-\mathfrak f)\), and (23) makes the dimension difference \(\deg\mathfrak f\), the dimension of the target. The indicated fiber therefore has \[ q^{d-\deg\mathfrak f+1-g} \tag{24} \] elements. Different functions in this fiber give different divisors, since the only constant congruent to \(1\) is \(1\). If \(\mathfrak f=0\), divide the nonzero sections by \(k^\times\); the count instead is \[ \frac{q^{d+1-g}-1}{q-1}. \tag{25} \] These counts depend only on the degree, not on the ray class. The Euler product \[ L(\chi,T)=\prod_{w\notin\operatorname{supp}\mathfrak f} (1-\chi(w)T^{\deg w})^{-1} \tag{26} \] is the generating function for these effective divisors with their character weights. If \(\chi\) is nontrivial on the finite degree-zero subgroup of \(\Gamma_{\mathfrak f}\), character summation on each fixed-degree coset makes all sufficiently large coefficients zero. Hence \(L(\chi,T)\) is a polynomial. Otherwise \(\chi\) is a degree character, say \(\chi(D)=\lambda^{\deg D}\), with \(|\lambda|=1\). It is unramified, so its minimal conductor is zero. Formula (25) shows that \[ \zeta_F(T) =\text{a polynomial}+ \sum_{\substack{d\geq d_0\\d\in\delta\mathbf Z}} h_0\frac{q^{d+1-g}-1}{q-1}T^d. \tag{27} \] Thus its only possible denominator factors are \(1-T^\delta\) and \(1-q^\delta T^\delta\). At \(T=q^{-1}\) it has a simple pole with positive nonzero leading coefficient. A nontrivial degree character has \(\lambda^\delta\ne1\), so \(L(\chi,T)=\zeta_F(\lambda T)\) is holomorphic there. These divisor counts also justify absolute convergence of the Euler products for \(|T|1\), \[ \log\prod_{\chi\in A^\vee}L_S(s,\chi) =|A|\sum_{\substack{v\notin S\\a_v=1}}(Nv)^{-s}+O(1). \tag{28} \] Indeed the terms of first degree have character sum \(|A|\) or \(0\); all higher powers have bounded total near \(1\), since \(\sum_v(Nv)^{-2s}\) converges there. The logarithm defined by the Euler products is real, and their product is positive. The principal factor has a simple pole. Each nonprincipal factor is holomorphic at \(1\), by section 6 for function fields or the cited internal analytic theorem for number fields. If its zero order there is \(m_\chi\geq0\), taking real logarithms of absolute values in (28) gives \[ |A|\sum_{a_v=1}(Nv)^{-s} =\left(1-\sum_{\chi\ne1}m_\chi\right) \log\frac1{s-1}+O(1). \tag{29} \] We have not assumed nonvanishing. For the primes splitting completely in \(L\), the Euler logarithm of \(\zeta_L\) gives \[ n\sum_{v\text{ split completely}}(Nv)^{-s} =\log\frac1{s-1}+O(1). \tag{30} \] At a nonsplit unramified prime every residue degree is at least \(2\), so its contribution is bounded by a constant times \((Nv)^{-2s}\); ramified primes are finite. Every completely split prime has \(a_v=1\), since its uniformizer is an idèle norm. Comparing (29) and (30) proves \[ \frac{|A|}{n}\leq1-\sum_{\chi\ne1}m_\chi\leq1. \] This proves (1) for all global fields, including characteristic-\(p\) extensions of degree divisible by \(p\), and completes Theorem 15.1. \subsection{8. Hasse's norm theorem and conics}\label{hasses-norm-theorem-and-conics} \textbf{Theorem 15.3 (Hasse's norm theorem).} For a finite cyclic extension \(L/K\) of global fields, an element \(a\in K^\times\) is a field norm if and only if it is a norm from \(L\otimes_K K_v\) at every place \(v\). \textbf{Proof.} Apply the cyclic Tate hexagon to \[ 1\longrightarrow L^\times\longrightarrow J_L \longrightarrow C_L\longrightarrow1. \] Its relevant exact segment is \[ \widehat H^{-1}(G,C_L)\longrightarrow K^\times/NL^\times\longrightarrow J_K/NJ_L. \tag{31} \] The first group is zero by (2). Thus the last map is injective. An element in all local norm images gives an idèle norm: outside finitely many ramified places it is a unit and has a unit norm preimage. Its image in the last group is zero, so its field-norm class is zero. The reverse implication follows by completion. \(\square\) Cyclicity is essential; the theorem need not hold for general noncyclic abelian extensions. This cyclic result is also the input for the planned lesson \emph{The principal genus theorem} in the Noether course, where the more general cohomological comparison is developed. \textbf{Corollary 15.4.} A smooth conic over a global field has a rational point if and only if it has a point over every completion. In characteristic different from \(2\), this applies to \[ aX^2+bY^2=Z^2,\qquad a,b\in K^\times, \tag{32} \] has a \(K\)-point if and only if it has a point over every \(K_v\). \textbf{Proof.} If \(b\) is a square, \(X=0\) gives a point. Otherwise put \(L=K(\sqrt b)\). A point in (32) must have \(X\ne0\), and the equation becomes \[ a=N_{L/K}\left(Z/X+(Y/X)\sqrt b\right). \] Over a completion where \(b\) becomes a square the split norm is surjective; the conic has a point automatically. At other completions the same equivalence holds. Theorem 15.3 proves the assertion. \(\square\) This treats every smooth conic in characteristic different from \(2\): diagonalize its nondegenerate ternary form and scale the last coefficient to obtain (32). Diagonal ternary equations in characteristic \(2\) are not smooth conics. The corresponding smooth equation \[ X^2+XY+bY^2=aZ^2 \tag{33} \] also has the local-global principle in characteristic \(2\): if \(T^2+T+b\) splits there is an immediate point; otherwise the left side is the norm of \(X+Y\theta\) from the cyclic Artin--Schreier quadratic field. A point then has \(Z\ne0\), and Theorem 15.3 applies. Every smooth characteristic-\(2\) conic either already has a rational point or has this form. Its polar form has rank \(2\); choose a paired basis and a radical vector to write its equation as \(A X^2+XY+B Y^2+C Z^2=0\). Smoothness forces \(C\ne0\), since a zero of the quadratic form on its polar radical would be a singular point. If \(A=0\), there is a rational point. Otherwise divide by \(A\) and replace \(Y\) by \(A Y'\), obtaining (33) with \(b=AB\) and \(a=C/A\). This completes the characteristic-\(2\) case as well. For example, in \(\mathbf Q(i)\) every odd prime \(p\equiv3\pmod4\) must have even valuation in a norm. At \(2\), the norm subgroup is \[ 2^{\mathbf Z}(1+4\mathbf Z_2), \] as computed in the preceding idèle lesson. Thus \(3\) fails at both \(3\) and \(2\), and \(7\) fails at both \(7\) and \(2\). The number \(5=1^2+2^2\) is a norm globally. \subsection{9. From conics to quadratic forms}\label{from-conics-to-quadratic-forms} The norm theorem proves the local-to-global principle for conics here. The full Hasse--Minkowski theorem for nondegenerate quadratic forms over number fields is proved in \href{the-chebotarev-density-theorem.md\#6a-general-quadratic-forms}{The Chebotarev density theorem, section 6A}, after reciprocity, existence and the ray-prime theorem. That argument retains all dimensions and uses the conic theorem above for its ternary base case. \subsection{10. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- Gaussian norms (easy)}\label{exercise-1-gaussian-norms-easy} Decide whether \(3,5,7\) are norms from \(\mathbf Q(i)\), identifying local obstructions. \textbf{Solution.} At an inert odd prime, norm valuations are even. The valuations of \(3\) at \(3\), and \(7\) at \(7\), are odd, so both fail. Their odd \(2\)-adic units are \(3\pmod4\), so they also fail at \(2\). The element \(1+2i\) has norm \(5\); equivalently all its local tests pass and Hasse's norm theorem applies. Positive sign at the real place is satisfied by all three. \subsubsection{Exercise 2 --- The cohomological injection (medium)}\label{exercise-2-the-cohomological-injection-medium} Derive Hasse's norm theorem directly from \(\widehat H^{-1}(G,C_L)=0\). \textbf{Solution.} Invariants in \(L^\times\) and \(J_L\) are \(K^\times\) and \(J_K\). The Tate hexagon for their exact sequence with \(C_L\) gives (31). Vanishing makes the field-norm quotient inject into the idèle-norm quotient. Local norm preimages can be chosen to be units almost everywhere, so a locally norm element dies in the latter quotient; injectivity kills its field-norm class. This explains both the local-to-idèle step and the exact cohomological step. \subsubsection{Exercise 3 --- The Kummer index (medium)}\label{exercise-3-the-kummer-index-medium} In Proposition 15.2, prove the exact index of \(C(S,T)\), including the principal intersection. \textbf{Solution.} The kernel of \(E_S\to\prod_T U_v/U_v^n\) is the radical \(\Delta\), so its index is \(n^{s-r}\); this equals the target order and proves surjectivity. For \(y\in K^\times\cap I(S,T)\), the cyclic field \(K(y^{1/n})\) splits at \(S\), is unramified outside \(S\cup T\), and has every idèle class as a norm after multiplying a representative by the \(S\)-unit supplied by that surjectivity at \(T\). The lower norm bound forces this field to be \(K\), and the root is an \(S\cup T\)-unit. Hence the principal intersection is exactly \(E_{S\cup T}^n\). The unrestricted \(S\)-local power quotients have product order \(n^{2s}\) by (5) and the product formula; principal \(S\cup T\)-units remove a subgroup of order \(n^{s+|T|}=n^{2s-r}\). The resulting index is \(n^r\). Merely counting local factors without proving the principal intersection would not establish it. \subsubsection{Exercise 4 --- The analytic norm bound (hard)}\label{exercise-4-the-analytic-norm-bound-hard} Assume the simple pole of the global zeta functions at \(1\) and holomorphy there of every nonprincipal finite-order Hecke function. Prove (1) without assuming their nonvanishing. \textbf{Solution.} The finite norm quotient \(A\) has \(|A|\) characters. Orthogonality in its Euler products gives (28). If the nonprincipal functions have total zero order \(m\), their product with the principal one has logarithm \((1-m)\log(1/(s-1))+O(1)\). Thus the primes with trivial uniformizer class have logarithmic density \((1-m)/|A|\). Independently the Euler product of \(\zeta_L\) gives density \(1/n\) for completely split primes: exactly those have \(n\) terms of norm \(Nv\), and all other unramified terms have degree at least \(2\). Complete splitting implies trivial norm-quotient class. Therefore \[ \frac1n\leq\frac{1-m}{|A|}\leq\frac1{|A|}, \] proving \(|A|\leq n\). This also forces \(m=0\) for these characters. Finite omitted Euler factors are nonzero at \(1\), so they change neither zero orders nor the argument. \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 Kummer norm-index bound includes Sylow descent. The function-field proof treats prime-to-characteristic radicals and characteristic-p additive duality separately. The cyclic Hasse norm theorem and the quadratic-form deduction retain their exact analytic and approximation prerequisites. \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://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{The global reciprocity law}\label{the-global-reciprocity-law} \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).} A local symbol tells how one completion acts on an abelian extension. The global question is whether the symbols at all completions can be assembled into an invariant of an idèle class. Multiplying them defines a map on idèles immediately. It does not immediately define a map on classes: a principal element appears at every place, and its local contributions must cancel. We first inspect that cancellation in two explicit examples. The proof then has two tasks. Construct a map that is already defined on classes, using the abstract reciprocity theorem and the global cyclic norm axiom. Identify it with the local product by comparing one local norm quotient at a time. The identification proves both the principal product law and the exact global norm kernel. The abstract theorem is \href{../the-reciprocity-law-and-the-class-field-correspondence.html}{The reciprocity law and the class field correspondence}, Theorem 5.1. Its arithmetic input here is the cyclic axiom in \href{../the-norm-index-bound-and-hasses-norm-theorem.html}{The norm index bound and Hasse's norm theorem}, sections 1--8; the later quadratic-form proof is not used. Local symbols and their explicit cyclotomic convention were proved in lessons 6 and 9. Fixed idèle classes and compact norm-one classes were proved in lessons 13 and 14. The data will use integer degree for function fields and a profinite degree for number fields, so the valuation hypothesis must allow both. \subsection{1. The cancellation that the theorem must explain}\label{the-cancellation-that-the-theorem-must-explain} Take the principal element \(-6\) in \(\mathbf Q(\zeta_5)/\mathbf Q\). At 2 and 3 its local symbols act on \(\zeta_5\) with exponents 2 and 3. At 5 it is a unit, whose exponent is \((-6)^{-1}\equiv4\pmod5\). At infinity it is negative, whose exponent is \(-1\equiv4\pmod5\). All other finite components are unramified units and act trivially. The total exponent is \[ 2\cdot3\cdot4\cdot4\equiv1\pmod5. \] Each ramified or infinite correction matters. Looking only at the prime divisors of 6 would give exponent 1 modulo 5 in this particular case, but it would hide two nontrivial compensating actions. The individual contributions, rather than their coincidental partial product, are what the theorem must organize. For a function-field example, take \(K=\mathbf F_7(t)\), a constant extension of degree six, and \(a=(t^2+1)/t^2\). Since \(-1\) is not a square in \(\mathbf F_7\), the numerator defines a degree-two place. Its valuation is 1, while the degree-one place \(t\) has valuation \(-2\), and infinity has valuation zero. Their constant-field Frobenius exponents add to \(2\cdot1+1\cdot(-2)=0\). The product is again the identity. Here cancellation is a degree equation rather than an archimedean sign correction. These examples establish the rule only for cyclotomic or constant extensions, using the explicit local results. For arbitrary abelian extensions the principal product law is still to be proved. In particular, we cannot assume a map on \(C_K\) merely because we have written a finite product on \(J_K\). \subsection{2. The product on idèles}\label{the-product-on-iduxe8les} For a finite abelian extension \(L/K\), choose \(w\mid v\). The local Galois group is the decomposition group \(D_w\subseteq\operatorname{Gal}(L/K)\). Since the latter is abelian, changing \(w\) conjugates this subgroup and its map trivially. Define \[ [x,L/K]=\prod_v\operatorname{rec}_{L_w/K_v}(x_v), \qquad x\in J_K. \tag{1} \] At almost all finite places the extension is unramified and \(x_v\) is a unit, so the factor is \(1\). The arithmetic convention sends a uniformizer in an unramified extension to residue-field Frobenius \(z\mapsto z^{Nv}\). The map is continuous: finitely many local open norm kernels, together with units outside a finite set, give an open subgroup of its kernel. An idèle norm lies in the kernel, because each local factor is a norm from the product of the completions over that place. \textbf{Proposition 16.1.} Suppose \(K\subseteq K'\), \(L\subseteq L'\), and both \(L/K\) and \(L'/K'\) are finite abelian. Then \[ [N_{K'/K}x,L/K] =[x,L'/K']|_L. \tag{2} \] The symbols also commute with conjugation and restriction in towers. Each map (1) is surjective. \textbf{Proof.} The \(v\)-component of \(N_{K'/K}x\) is \(\prod_{v'\mid v}N_{K'_{v'}/K_v}x_{v'}\). Local norm functoriality, proved in \href{../local-reciprocity-and-norm-groups.html}{Local reciprocity and norm groups}, identifies its symbol with the product of the restrictions of the \(v'\)-symbols. Multiplication over \(v\) proves (2). Local conjugation and tower compatibility give the other two identities. The image of (1) contains each decomposition group, since local reciprocity is onto and we may vary just that idèle component. The fixed field of the image therefore splits completely at every place of \(K\). Corollary 14.4 says that a nontrivial finite separable extension has infinitely many places failing complete splitting. The fixed field must be \(K\), so the image is the whole Galois group. \(\square\) For an infinite abelian extension, define the symbol by these compatible finite quotients. This is equivalently a convergent product of local symbols in the profinite group: in any fixed finite quotient only finitely many factors are nontrivial. At an infinite local extension we use the union of its finite localizations, not an assertion that an algebraic union is complete. \subsection{3. Principal cyclotomic and constant-field symbols}\label{principal-cyclotomic-and-constant-field-symbols} \textbf{Proposition 16.2.} If \(K\) is a number field, \(a\in K^\times\), and \(\zeta\) is a root of unity, then \[ [a,K(\zeta)/K]=1. \tag{3} \] For a function field, every principal idèle likewise has trivial symbol in every finite constant extension. \textbf{Proof for number fields.} By (2), restriction of the left side to \(\mathbf Q(\zeta)\) equals \([N_{K/\mathbf Q}a,\mathbf Q(\zeta)/\mathbf Q]\). Restriction is injective on \(\operatorname{Gal}(K(\zeta)/K)\), so it suffices to work over \(\mathbf Q\). Split the order of \(\zeta\) into prime powers; it suffices to treat order \(\ell^m\). Write \(a=u_\ell\ell^{v_\ell(a)}\). The explicit local theorem in \href{../explicit-local-reciprocity-and-existence.html}{Explicit local reciprocity and existence}, Corollary 9.5, gives the exponents acting on \(\zeta_{\ell^m}\): \[ \begin{array}{c|c} \text{place}&\text{exponent}\\ \hline p\ne\ell,\ \ p\text{ finite}&p^{v_p(a)}\\ \ell&u_\ell^{-1}\pmod{\ell^m}\\ \infty&\operatorname{sgn}(a). \end{array} \tag{4} \] Negative valuations use inverse powers modulo \(\ell^m\). Since \[ u_\ell=\operatorname{sgn}(a)\prod_{p\ne\ell}p^{v_p(a)}, \] the product of the exponents in (4) is \(1\) in \((\mathbf Z/\ell^m)^\times\). This proves (3), including the real sign. \textbf{Proof for function fields.} Let \(k=\mathbf F_q\) be the full constant field of \(K\). In \(K_m=K\mathbf F_{q^m}\), the local extension at \(v\) is unramified, with residue degree \(m/\gcd(m,\deg v)\). Its uniformizer symbol acts on constants as the \(\deg v\)th power of \(q\)-Frobenius. Thus the product symbol of a principal element is Frobenius to the power \(\sum_v\deg(v)v(a)=0\), by the product formula. \(\square\) Here \([K_m:K]=m\). Indeed an irreducible finite-field polynomial remains irreducible over \(K\): a monic factor would have coefficients in the algebraic constant field as well as in \(K\), hence in \(k\), contradicting irreducibility. Its roots generate the constant extension. This also shows that the full constant field of \(K_m\) is \(\mathbf F_{q^m}\). For example, take \(a=-1\), \(L=\mathbf Q(i)\). At \(2\) the inverse unit \(-1\) gives complex conjugation, and at infinity the negative sign gives the same automorphism. Every odd place contributes \(1\). The two conjugations multiply to \(1\). \subsection{4. Constructing a map on idèle classes}\label{constructing-a-map-on-iduxe8le-classes} The special cases provide exactly the degree direction needed for a map on classes. The remaining input is the cyclic norm axiom, already proved without global reciprocity. It is useful to view these inputs as a checklist: fixed classes, a procyclic Galois degree, a compatible valuation of norm images, and the two cyclic Tate conditions. \subsubsection{The module and its fixed classes}\label{the-module-and-its-fixed-classes} Fix a separable closure of a ground global field \(k_0\), and put \[ A=\varinjlim_{K/k_0\text{ finite separable}}C_K. \tag{5} \] The transition maps are the diagonal embeddings. They are injective by lesson 13. Give \(A\) the discrete topology for its Galois-module structure; every element is defined over a finite extension, so its stabilizer is open. This does not change the ordinary locally compact topology used on each \(C_K\). One has \(A^{G_K}=C_K\). To check this, represent a fixed element in \(C_M\) with \(M/K\) finite Galois. Invariance in the direct limit is invariance in \(C_M\), by injectivity. The invariant descent \(C_M^{\operatorname{Gal}(M/K)}=C_K\) was proved using Hilbert 90 in lesson 13. The group-theoretic module norm is exactly the idèle-class norm, since both are the product of Galois conjugates. Thus the cyclic class field axiom of Theorem 15.1 is the precise axiom required by the abstract theorem. What remains is its degree and valuation data. \subsubsection{Integer degree for function fields}\label{integer-degree-for-function-fields} For a function field \(k_0\) with full constants \(\mathbf F_q\), let \(\widetilde{k_0}=k_0\overline{\mathbf F}_q\). Arithmetic constant-field Frobenius supplies \(d:G_{k_0}\to\widehat{\mathbf Z}\). For a finite separable \(K/k_0\), \(f_K\) is the degree of its full constant field over \(\mathbf F_q\), and \(d_K\) is Frobenius normalized for those full constants. The degree map \[ v_K(c)=\deg_K c=\sum_v[\kappa(v):k_K]v(x_v), \qquad c=[x]\in C_K, \tag{8} \] is well-defined by the product formula. It is onto \(\mathbf Z\). To prove the last assertion without a point-count estimate, let \(\delta\) be the greatest common divisor of all place degrees. If a prime \(\ell\mid\delta\), every residue field contains the degree-\(\ell\) constant field. The nontrivial constant extension \(K\mathbf F_{|k_K|^\ell}/K\) then splits completely at every place, contrary to Corollary 14.4. Hence \(\delta=1\). Bézout applied to finitely many of the degrees produces an idèle of degree \(1\). For every finite separable \(L/K\), the local valuation formula for norms gives \[ \deg_K(N_{L/K}c)=[k_L:k_K]\deg_L(c). \tag{9} \] Indeed the coefficient at \(v\) is the sum of \([\kappa(v):k_K]f(w/v)v_w(x_w) =[k_L:k_K][\kappa(w):k_L]v_w(x_w)\). Here \([k_L:k_K]=f_{L/K}\), the index of the constant-field degree images. Thus the henselian conditions hold with \(V=\mathbf Z\subset\widehat{\mathbf Z}\); surjectivity means onto \(\mathbf Z\), not onto its profinite completion. By Proposition 16.2 the constant-extension symbol is precisely Frobenius to this integer degree. \subsubsection{Profinite degree for number fields}\label{profinite-degree-for-number-fields} Take \(k_0=\mathbf Q\). In lesson 1 we proved \[ \operatorname{Gal}(\mathbf Q(\mu_\infty)/\mathbf Q) \simeq\widehat{\mathbf Z}^{\,\times} \simeq\widehat{\mathbf Z}\times T, \] where \(T\) is a product of finite torsion factors. The closure of the finite-order elements is \(T\). It matters that we take the closure: the product \(T\) can contain elements of infinite order. Its fixed field \(\widetilde{\mathbf Q}\) has Galois group \(\widehat{\mathbf Z}\). Choose an isomorphism once, obtaining \(d:G_{\mathbf Q}\to\widehat{\mathbf Z}\). For a number field \(K\), set \[ f_K=[K\cap\widetilde{\mathbf Q}:\mathbf Q], \qquad d_K=f_K^{-1}d|_{G_K}, \qquad\widetilde K=K\widetilde{\mathbf Q}. \] Proposition 16.2 lets the product symbol in \(\widetilde K/K\) descend to \(C_K\). Define \[ v_K(c)=d_K[c,\widetilde K/K]\in\widehat{\mathbf Z}. \tag{6} \] Here \(d_K\) denotes its induced isomorphism on that procyclic Galois group. This map is onto. Proposition 16.1 makes its image onto every finite quotient, hence dense. Split \(C_K=C_K^1\times\mathbf R_{>0}\) using a chosen archimedean component. The positive-real factor is divisible and every finite quotient kills it: take an \([L:K]\)th root before applying a finite symbol. The image consequently comes from compact \(C_K^1\), is closed, and is all of \(\widehat{\mathbf Z}\). For every finite extension \(L/K\), norm functoriality gives \[ v_K(N_{L/K}c)=f_{L/K}v_L(c), \qquad f_{L/K}=[d(G_K):d(G_L)]. \tag{7} \] Since \(v_L\) is onto, the norm image under \(v_K\) is exactly \(f_{L/K}\widehat{\mathbf Z}\). The normalization also gives \(v_K=f_K^{-1}v_{\mathbf Q}N_{K/\mathbf Q}\). These are exactly the henselian valuation conditions of lesson 4, with value group \(V=\widehat{\mathbf Z}\); an integer-only condition would be inappropriate here. \textbf{Proposition 16.3.} The Galois module (5), the cyclotomic degree and valuation (6) for number fields, or the constant-field degree and valuation (8) for function fields, form a class field theory in the sense of lessons 4--5. \textbf{Proof.} The invariant-module and norm identifications are those just proved for (5). The degree maps are continuous surjections, with normalized images for finite subfields as in lesson 4. The two valuation constructions above prove the henselian norm-image conditions. Both value groups satisfy \(V/nV\simeq\mathbf Z/n\). The cyclic axiom is Theorem 15.1. These verify every input to Theorem 5.1. \(\square\) \subsection{5. Comparing the maps on local norm quotients}\label{comparing-the-maps-on-local-norm-quotients} The map on classes supplied by the abstract theorem has the correct norm kernel, but that does not yet identify the action of a single completion. The local product is the map whose completion-by-completion behavior we want. For both maps a local norm dies, so the comparison reduces to a finite quotient. The following proof makes every class of that quotient accessible from an extension on which the comparison is already known. Apply the abstract theorem. For every finite Galois \(L/K\) it gives \[ r_{L/K}:\operatorname{Gal}(L/K)^{\mathrm{ab}} \xrightarrow{\sim}C_K/N_{L/K}C_L. \tag{10} \] Write \(\rho_{L/K}\) for its inverse on \(C_K\). It is continuous because its kernel is the open norm subgroup. Norm, conjugation and tower compatibility are Proposition 5.2. For finite subextensions of \(\widetilde K/K\), its formula is \[ \rho_{L/K}(c)=\varphi_{L/K}^{\,v_K(c)\bmod[L:K]}. \tag{11} \] By the definition of \(v_K\), this equals \([c,L/K]\). We must extend the comparison to every abelian extension. \subsubsection{Making a local quotient accessible from a degree extension}\label{making-a-local-quotient-accessible-from-a-degree-extension} At a finite place of a number field, every cyclotomic \(\mathbf Z_p\)-extension has infinite local degree. If the residue characteristic is \(p\), the \(p\)-power cyclotomic local degrees grow without bound by the totally ramified division fields of lesson 8; adjoining the fixed finite completion cannot bound them. If the residue characteristic is \(\ell\ne p\), the extension is unramified and its Frobenius projects to the \(p\)-adic unit \(\ell^{f_v}\) modulo torsion. This is not torsion, since no positive power of the positive rational integer \(\ell^{f_v}>1\) equals \(1\). Its closed cyclic image in \(\mathbf Z_p\) is therefore a nonzero open subgroup. These observations also apply when \(p=2\), using the finite sign factor. For function fields the constant \(\mathbf Z_p\)-extension has local degree image \(\deg(v)\mathbf Z_p\), also nonzero and open. Thus in either case there is a \(\mathbf Z_p\)-subextension \(T/K\) of \(\widetilde K/K\) with infinite local degree at the specified finite place. Suppose now that \(L/K\) is cyclic of \(p\)-power degree, generated by \(\sigma\), and its decomposition group at the specified place is the whole group. Put \(M=LT\), and let \(D\subseteq\operatorname{Gal}(M/K)\) be the decomposition subgroup there. It maps onto \(\operatorname{Gal}(L/K)\), and its image in \(\operatorname{Gal}(T/K)=\mathbf Z_p\) is open. The image of the kernel of \(D\to\operatorname{Gal}(L/K)\) is still open, since that kernel has finite index in \(D\). Choose \(g\in D\) restricting to \(\sigma\), and adjust it by this kernel so its \(\mathbf Z_p\)-coordinate is a nonzero positive ordinary integer. This is possible because the allowed coordinates form a coset of an open subgroup \(p^b\mathbf Z_p\), and every such coset contains a positive integer. Let \(H=\overline{\langle g\rangle}\), and \(K'=M^H\). The coordinate map on \(H\) is injective, with open image, so \(H\simeq\mathbf Z_p\) and \([K':K]<\infty\). Also \(H\cap\operatorname{Gal}(M/T)=1\), whence \[ M=K'T,\qquad L'=K'L\subseteq\widetilde{K'}. \tag{12} \] Because \(H\subseteq D\), the chosen place has full decomposition group in \(M/K'\). The finite local group for \(L'/K'\) maps onto \(\operatorname{Gal}(L/K)\), sending the restriction of \(g\) to \(\sigma\). This proves the required auxiliary construction, including its local property. \subsubsection{Comparing at finite places}\label{comparing-at-finite-places} For abelian \(L/K\), compose \(K_v^\times\to C_K\) with \(\rho_{L/K}\), and compare it with local reciprocity followed by \(D_v\hookrightarrow\operatorname{Gal}(L/K)\). Both maps kill the local norm group: a local norm gives a global idèle norm supported at that place. They therefore factor through the finite local norm quotient \(D_v\). It suffices to compare on its prime-power elements. Take such an element \(\sigma\), and replace the base by \(K_0=L^{\langle\sigma\rangle}\). At the chosen place \(L/K_0\) has full cyclic decomposition group \(\langle\sigma\rangle\). Apply the auxiliary construction just proved to get \(L'/K'\) as in (12). Comparison for \(L'/K'\) is already known by (11), because it is cyclotomic or constant. Local reciprocity supplies an element \(a'\) whose local symbol is the lift of \(\sigma\). Norm it down to \(K_{0,v_0}\), then to \(K_v\). Global norm functoriality for \(\rho\) and local norm functoriality give \(\sigma\) on both sides. Since both maps already factor through \(D_v\), this proves their equality. Every element of a finite abelian group is a product of prime-power elements, so the comparison holds on all \(K_v^\times\). \subsubsection{Comparing at infinite places}\label{comparing-at-infinite-places} Only a real place becoming complex has nontrivial local group. First take the special extension \(F(i)/F\) at a real place of a number field \(F\). Proposition 16.2 makes its product symbol a map on \(C_F\); it kills norms and is onto, since a negative component at that place maps to conjugation. Its norm quotient has order \(2\) by Theorem 15.1. There is just one isomorphism between groups of order \(2\), so this product map is \(\rho_{F(i)/F}\). For a general complexification, let \(\sigma\) be its conjugation element and pass to \(K_0=L^{\langle\sigma\rangle}\). In \(M=L(i)\), extend \(\sigma\) by the chosen complex conjugation, and let \(K'\) be its fixed field. The chosen completion of \(K'\) is real, and \[ L'=K'L=M=K'(i). \] The equality follows since both \(L\) and \(i\) are moved by that conjugation and \(M/K'\) has degree \(2\). Norm functoriality now reduces the comparison to the special case, exactly as above. At real places staying real and at complex places both maps are trivial because the local norm group is all of \(K_v^\times\). \subsection{6. Artin reciprocity}\label{artin-reciprocity} \textbf{Theorem 16.4 (Artin reciprocity).} For every finite abelian extension \(L/K\) of global fields, the product (1) induces an isomorphism \[ C_K/N_{L/K}C_L\xrightarrow{\sim}\operatorname{Gal}(L/K). \tag{13} \] In particular \[ \prod_v\operatorname{rec}_{L_w/K_v}(a)=1 \quad(a\in K^\times), \tag{14} \] and \[ J_K/(K^\times N_{L/K}J_L)\simeq\operatorname{Gal}(L/K). \tag{15} \] \textbf{Proof.} Section 5 identifies the product and \(\rho\) on every idèle supported at one place. Finite products of such idèles are dense in the restricted product: a basic neighborhood restricts only finitely many components, with unit conditions on the tail. Both maps are continuous to a finite group, so they agree everywhere on \(J_K\). The abstract map factors through \(C_K\), proving (14), and has kernel \(NC_L\), proving (13)--(15). \(\square\) For an unramified finite place, the global symbol of its idèle uniformizer is its arithmetic Frobenius. Thus this theorem includes the usual Frobenius prescription at primes, as well as every ramified and archimedean component. For nonabelian finite Galois extensions (10) remains valid, and the norm group equals that of the maximal abelian subextension by norm limitation in lesson 5. For an infinite abelian extension the map on \(C_K\) has dense image, because it is onto every finite quotient. Over number fields it is onto: the positive-real factor is killed and the image of compact \(C_K^1\) is closed. Over function fields its constant-field projection is the integer degree subgroup of \(\widehat{\mathbf Z}\), so it is not onto the infinite constant extension. The next lesson describes the precise topology and the existence correspondence. The construction of the abstract number-field degree involved a choice of an isomorphism with \(\widehat{\mathbf Z}\). The final map is independent of that choice, since the comparison identifies it with the product of the fixed arithmetic local maps. \subsection{7. Quadratic reciprocity as a product formula}\label{quadratic-reciprocity-as-a-product-formula} Let \(p,q\) be distinct odd rational primes, and put \(p^*=(-1)^{(p-1)/2}p\). Take \(L=\mathbf Q(\sqrt{p^*})\) and the principal element \(q\). By the explicit Hilbert symbol formulas of \href{../hilbert-symbols-and-local-conics.html}{Hilbert symbols and local conics}, its local actions on \(\sqrt{p^*}\) are \[ \begin{array}{c|c} v&\text{multiplying sign}\\ \hline q&(p^*/q)\\ p&(q/p)\\ 2&1\\ \infty&1. \end{array} \tag{16} \] At \(q\) the extension is unramified and Frobenius gives the residue square character. At \(p\), the odd-place Hilbert formula has valuations \(0\) and \(1\), giving \((q/p)\). At \(2\), both entries are odd and \(p^*\equiv1\pmod4\), so the odd-unit \(2\)-adic Hilbert formula gives \(1\). At infinity \(q>0\). All other places have an unramified unit symbol. Theorem 16.4 now gives \((p^*/q)(q/p)=1\). Since \((-1/q)=(-1)^{(q-1)/2}\), this is \[ (p/q)(q/p)=(-1)^{(p-1)(q-1)/4}. \tag{17} \] No prior global quadratic reciprocity theorem was used: only the already proved local Hilbert formulas and the global product theorem. \subsection{8. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- The real sign (easy)}\label{exercise-1-the-real-sign-easy} Verify the product formula for \(-1\) in \(\mathbf Q(i)/\mathbf Q\). \textbf{Solution.} At \(2\), Corollary 9.5 acts on \(i\) by the inverse of the unit \(-1\) modulo \(4\), hence sends \(i\) to \(-i\). At infinity, \(-1\) gives complex conjugation. At every odd prime the extension is unramified or split and \(-1\) is a unit, so the symbol is trivial. The product is conjugation squared, hence \(1\). Omitting infinity would give the wrong product. \subsubsection{Exercise 2 --- A fifth root of unity (medium)}\label{exercise-2-a-fifth-root-of-unity-medium} Verify the product formula for \(3\) in \(\mathbf Q(\zeta_5)/\mathbf Q\). \textbf{Solution.} At \(3\), arithmetic Frobenius sends \(\zeta_5\) to \(\zeta_5^3\). At \(5\), the unit \(3\) acts through its inverse modulo \(5\), which is \(2\). The positive real component is trivial, and every other finite component is an unramified unit. Composing the two nontrivial actions gives exponent \(3\cdot2=6\equiv1\pmod5\). \subsubsection{Exercise 3 --- The quadratic law (medium)}\label{exercise-3-the-quadratic-law-medium} Derive quadratic reciprocity for distinct odd primes from Theorem 16.4. \textbf{Solution.} Use \(L=\mathbf Q(\sqrt{p^*})\), \(a=q\), and the four local calculations in (16). The product is \(1\), so \((p^*/q)=(q/p)\). Expanding \(p^*=(-1)^{(p-1)/2}p\) and evaluating the finite-field character of \(-1\) gives (17). The special choice \(p^*\equiv1\pmod4\) makes the \(2\)-adic contribution trivial; using \(p\) without that sign adjustment would require retaining it. \subsubsection{Exercise 4 --- Verify the class field theory data (hard)}\label{exercise-4-verify-the-class-field-theory-data-hard} Prove Proposition 16.3, including its value-group distinction. \textbf{Solution.} Injectivity and invariant descent from lesson 13 identify the discrete union module's \(G_K\)-invariants with \(C_K\), and the coset norm with the idèle norm. Theorem 15.1 supplies both cyclic Tate groups required by the abstract axiom. For number fields, take the torsion-closure quotient of the cyclotomic Galois group as the degree map. Proposition 16.2 kills principal elements in its product symbol, so (6) is defined on \(C_K\). Proposition 16.1 makes its image dense in \(\widehat{\mathbf Z}\). The divisible positive-real factor maps trivially to every finite quotient, and compact \(C_K^1\) has closed image; the map is onto. Norm compatibility gives precisely \(f_{L/K}\widehat{\mathbf Z}\) as the valuation of the norm image. For function fields, take constant Frobenius as the degree map, and use integer idèle degree as valuation. If its image had greatest common divisor \(\delta>1\), a prime dividing \(\delta\) would give a nontrivial constant extension split at every place, contradicting lesson 14. Thus the image is \(\mathbf Z\). Local norm valuations yield (9), with \(f_{L/K}\) the full-constant-field degree. Both \(\mathbf Z\) and \(\widehat{\mathbf Z}\) have quotient \(\mathbf Z/n\) by their \(n\)-multiples. These verify all hypotheses; falsely requiring the function-field degree to be onto \(\widehat{\mathbf Z}\), or the number-field valuation to be integer-valued, would not. \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 principal cancellation, local comparison and auxiliary procyclic construction are proved here. Number-field cyclotomic degree and function-field constant degree are constructed separately; the permitted profinite valuation image is retained. \begin{itemize} \tightlist \item \href{https://www.mathi.uni-heidelberg.de/~schmidt/Neukirch-en/Neukirch_cft_02_may15.pdf}{Jürgen Neukirch, Class Field Theory --- The Bonn Lectures, Online Edition 2.0 (May 2015), edited by Alexander Schmidt}. \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Global existence and the idèlic class field correspondence}\label{global-existence-and-the-iduxe8lic-class-field-correspondence} \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).} \textbf{Lesson 17.} The preceding lesson associates a reciprocity map to every abelian extension already given. Existence answers the converse question: which subgroups of the idèle class group occur as norm groups? Every open subgroup of finite index does. Thus a finite abelian extension can be specified by local congruence conditions on idèles, without first finding an equation for the extension. There is a topological distinction between the two kinds of global fields. For a number field, reciprocity maps onto the abelianized absolute Galois group and kills exactly the connected component of the idèle class group. For a function field it is injective, and its image has integral constant-field degree. Giving those integer cosets the discrete topology identifies the class group with an abelianized Weil group. We prove the function-field assertion, including the wild characteristic case. The prerequisites are \href{../the-norm-index-bound-and-hasses-norm-theorem.html}{The norm index bound and Hasse's norm theorem}, especially its full S-unit radical and additive duality, and \href{../the-global-reciprocity-law.html}{The global reciprocity law}. We use the compactness of norm-one classes from lesson 14 and the local class field correspondence from lesson 6. The higher-dimensional Hasse--Minkowski theorem is not an input. Write \(J_K\) for idèles, \(C_K=J_K/K^\times\), and \(B_K=G_K^{\mathrm{ab}}=\operatorname{Gal}(K^{\mathrm{ab}}/K)\). All reciprocity maps use arithmetic Frobenius. For a finite extension put \[ N_{L/K}C_L=\operatorname{im}(N_{L/K}:C_L\longrightarrow C_K). \] The symbol \(D_K\) will denote the connected component of \(1\) in \(C_K\). When temporarily discussing universal norms, we use a different symbol \(\mathcal N_K\). \subsection{1. Norm groups and the existence theorem}\label{norm-groups-and-the-existence-theorem} \textbf{Theorem 17.1 (global existence).} If \(K\) is a global field and \(H\subseteq C_K\) is open of finite index, there is a unique finite abelian extension \(L/K\), inside the fixed separable closure, with \[ H=N_{L/K}C_L. \tag{1} \] An open subgroup is closed, since its other cosets are open. Conversely a closed subgroup of finite index is open: its complement is a finite union of closed cosets. These observations explain why either formulation is often used. The finite-index hypothesis remains essential for a \emph{finite} class field; for example the degree-zero subgroup of a function-field class group has infinite index. We first prove existence for number fields. The function-field proof, including characteristic-power indices, is given in sections 5--8. \subsubsection{Cofinal Kummer norm groups}\label{cofinal-kummer-norm-groups} Let \(n=\ell^a\) and suppose \(\mu_n\subset K\). Choose a finite set \(S\) containing the infinite places and the places dividing \(n\), large enough that \(J_K=K^\times J_K^S\). With \(E_S\) the S-unit group, put \[ M=K(E_S^{1/n}),\qquad I(S)=\prod_{v\in S}K_v^{\times n}\times\prod_{v\notin S}\mathcal O_v^\times, \qquad C(S)=\operatorname{im}(I(S)\to C_K). \tag{2} \] If \(s=|S|\), the S-unit theorem and Kummer pairing give \(\operatorname{Gal}(M/K)\simeq(\mathbf Z/n)^s\). This full radical is unramified outside \(S\): adjoining roots of units of order prime to the residue characteristic is unramified there. Proposition 15.2 applies with \(r=s\), so its auxiliary set \(T\) is empty. It proves \[ C(S)\subseteq N_{M/K}C_M,\qquad [C_K:C(S)]=n^s=[M:K]. \] Reciprocity, Theorem 16.4, gives the same index for the norm group. Consequently \[ C(S)=N_{M/K}C_M. \tag{3} \] Suppose \(C_K/H\) is an \(\ell\)-group of exponent dividing \(n\). Then \(C_K^n\subseteq H\). Openness of the inverse image of \(H\) in \(J_K\) also puts \[ U^S=\{x\in J_K:x_v=1\ (v\in S),\ x_v\in\mathcal O_v^\times\ (v\notin S)\} \tag{4} \] inside that inverse image after enlarging \(S\). Every element of \(I(S)\) is a product of an idèle \(n\)th power and an element of \(U^S\). Therefore (3) is a norm group contained in \(H\). If \(\mu_n\) is not in \(K\), set \(K'=K(\mu_n)\). Enlarge \(S\) until the full set \(S'\) above it satisfies \(J_{K'}=K'^\times J_{K'}^{S'}\). This is possible by adding the projections of finitely many class-group generating primes of \(K'\). Apply (3) over \(K'\) to \(M=K'(E_{S'}^{1/n})\). The local norm of an \(n\)th power is an \(n\)th power, and the local norm of a unit is a unit. Hence \[ N_{K'/K}C(S')\subseteq C(S),\qquad N_{M/K}C_M\subseteq C(S)\subseteq H. \tag{5} \] Replace \(M\) by a Galois closure \(\widetilde M/K\); tower norms only shrink the image, so it still lies in \(H\). For a general finite quotient \(A=C_K/H\), let \(H_\ell\) be the kernel of its projection to the \(\ell\)-primary factor. Construct a finite Galois norm group inside each \(H_\ell\). The compositum of these finitely many extensions has its norm group inside \(\bigcap_\ell H_\ell=H\). \subsubsection{From a contained norm group to the desired field}\label{from-a-contained-norm-group-to-the-desired-field} We have obtained a finite Galois \(M/K\) with \(N_{M/K}C_M\subseteq H\). Norm limitation, proved in lesson 5 and applied in lesson 16, replaces \(M\) by its maximal abelian subextension \(A/K\), without changing that norm group. Reciprocity identifies \[ C_K/N_{A/K}C_A\simeq\operatorname{Gal}(A/K). \] Let \(L\) be the fixed field of the subgroup corresponding to \(H/N_{A/K}C_A\). Restriction compatibility and Theorem 16.4 identify the kernel for \(L/K\) with exactly \(H\). This proves existence over number fields. Uniqueness follows from the same reciprocity kernels: the fixed field of the closure of the image of \(H\) in \(B_K\) is determined by \(H\). This argument also treats function-field quotients of order prime to the characteristic. It does not treat characteristic-power quotients by pretending that the missing roots of unity exist. Sections 5--8 supply that case. \subsection{2. The class field correspondence and local information}\label{the-class-field-correspondence-and-local-information} \textbf{Theorem 17.2.} Finite abelian extensions of \(K\) correspond, with reversed inclusions, to open subgroups of finite index of \(C_K\). The group for \(L/K\) is its norm group, and \[ \operatorname{Gal}(L/K)\simeq C_K/N_{L/K}C_L. \tag{6} \] If \(i_v:K_v^\times\to C_K\) inserts a component at \(v\), then \[ \begin{aligned} v\text{ is unramified in }L&\iff i_v(\mathcal O_v^\times)\subseteq N_{L/K}C_L &&\text{for finite }v,\\ v\text{ splits completely in }L&\iff i_v(K_v^\times)\subseteq N_{L/K}C_L. \end{aligned} \tag{7} \] At a real place, ramification means becoming complex; this happens exactly when the class of a negative local element is not in the norm group. \textbf{Proof.} Existence and uniqueness are Theorem 17.1. If \(L_1\subseteq L_2\), tower norms put \(N_{L_2/K}C_{L_2}\subseteq N_{L_1/K}C_{L_1}\). Conversely the inclusion of these kernels reverses the inclusion of their fixed fields in \(K^{\mathrm{ab}}\). Equation (6) is reciprocity. Local compatibility in Theorem 16.4 identifies the image of \(K_v^\times\) with the decomposition group and the image of \(\mathcal O_v^\times\) with its inertia subgroup. Their triviality is exactly complete splitting and absence of ramification. At a real place the local quotient for \(\mathbf C/\mathbf R\) is the sign group. These arguments concern the \emph{images} under \(i_v\); they require no injectivity assertion about that map. \(\square\) For example, requiring all finite unit groups to lie in \(H\) produces an extension unramified at every finite prime. Requiring positive real components rather than all real components allows real places to become complex. Deep principal-unit conditions will give the ray class fields in the next lesson. \subsection{3. Connected components and number-field reciprocity}\label{connected-components-and-number-field-reciprocity} We record two elementary topological facts, so that the kernel description does not hide a topology theorem. \textbf{Compact-group lemma.} In a compact Hausdorff abelian group \(A\), the identity component \(A^0\) is the intersection of the open subgroups of finite index. The quotient \(A/A^0\) is profinite. If a locally compact abelian group has a neighborhood basis of compact open profinite subgroups, every quotient by a closed subgroup is totally disconnected and has such a basis. \textbf{Proof.} First, in a compact Hausdorff space, the intersection \(Q\) of all clopen neighborhoods of a point is its connected component. To see the nontrivial inclusion, suppose \(Q\) is separated into two nonempty closed parts, and choose disjoint open neighborhoods of those compact parts. The compact complement of their union is excluded by finitely many clopen neighborhoods of the original point. Their intersection is a clopen set containing \(Q\) and lying in that union. Its portion in the open neighborhood of the part containing the point is clopen and misses the other part, contradicting the definition of \(Q\). Thus \(Q\) is connected. For \(a\notin A^0\), choose a clopen set \(V\) containing \(1\) and missing \(a\). Its translation stabilizer is an open subgroup: compactness of \(V\) and its complement gives a neighborhood of \(1\) whose translations preserve both sets. More explicitly, cover each of the two compact sets by finitely many neighborhoods on which all translations from a common sufficiently small identity neighborhood stay in the same set. This common neighborhood lies in the stabilizer. The stabilizer has finite index by compactness, and misses \(a\), since an element stabilizing \(V\) carries \(1\in V\) into \(V\). Connected sets cannot cross its clopen cosets. This proves the asserted intersection. The map of \(A/A^0\) into the product of these finite quotients is continuous and injective; compactness makes it a homeomorphism onto a closed subgroup, proving profiniteness. For the last assertion let \(q:G\to G/H\) be the quotient by a closed subgroup. If \(P\subseteq G\) is compact open profinite, then \(q(P)\) is compact and open, and equals \(P/(P\cap H)\), a profinite group. A quotient of a profinite abelian group by a closed subgroup is profinite: its closed subgroup is separated from any outside point by a finite quotient, using compactness and a basis of open finite-index subgroups. Given a neighborhood in \(G/H\), choose \(P\) inside its inverse image. The images \(q(P)\) therefore form the required basis. In particular the quotient is totally disconnected. \(\square\) \textbf{Proposition 17.3.} For a number field, the map \[ \operatorname{rec}_K:C_K\longrightarrow B_K \] is onto and has kernel \[ D_K=\bigcap_{H\text{ open, finite index}}H =\overline{\operatorname{im}\left(\prod_{v\mid\infty}(K_v^\times)^0\longrightarrow C_K\right)}. \tag{8} \] \textbf{Proof.} Surjectivity was proved in lesson 16: finite-quotient surjectivity gives density, the divisible positive-real content factor is killed, and the remaining norm-one group \(C_K^1\) is compact. Existence and finite reciprocity identify the kernel with the intersection in (8). The decomposition \(C_K=C_K^1\times\mathbf R_{>0}\) reduces this intersection to the compact-group lemma. Indeed an open finite-index subgroup contains the connected positive-real factor, and open finite-index subgroups of \(C_K^1\) detect its identity component. Thus the intersection is precisely \(D_K\). Let \(A_\infty^0=\prod_{v\mid\infty}(K_v^\times)^0\), embedded as archimedean idèles, and let \(D_\infty\) be the closure of its image. Its image is connected, so \(D_\infty\subseteq D_K\). The quotient of \(J_K\) by \(A_\infty^0\) is \[ \{\pm1\}^{r_1}\times J_{K,f}. \tag{9} \] This group has a basis of compact open profinite subgroups: at finitely many finite places use deep units, elsewhere full units, and use the discrete sign factors. Now \(C_K/D_\infty\) is (9) modulo the closure of the projected principal subgroup. The last part of the lemma makes it totally disconnected. The image of the connected group \(D_K\) in that quotient is trivial. Therefore \(D_K\subseteq D_\infty\), proving equality. It is a \emph{closure}: this argument does not assert that the archimedean image is closed. \(\square\) \subsection{4. Rational idèles and a quadratic class field}\label{rational-iduxe8les-and-a-quadratic-class-field} There is a useful explicit decomposition \[ C_{\mathbf Q}\simeq\mathbf R_{>0}\times\widehat{\mathbf Z}^{\,\times}. \tag{10} \] For an idèle \(x\), divide it by the positive rational number \(r=\prod_p p^{v_p(x_p)}\). All finite components become units. Multiply the representative by \(-1\) if necessary to make its real component positive. The resulting positive real component and unit tuple are unique: a rational number which is a unit at every finite prime is \(\pm1\), and positivity fixes the sign. This normalization and its inverse are continuous on restricted-product neighborhoods, proving (10). The unit group in (10) is profinite. Hence \[ D_{\mathbf Q}=\mathbf R_{>0}. \tag{11} \] For the extension \(\mathbf Q(\zeta_m)/\mathbf Q\), the normalized unit tuple \(u\) acts through \(\zeta_m\mapsto\zeta_m^{u^{-1}\bmod m}\). This is the local cyclotomic formula of Corollary 9.5, multiplied over the prime divisors of \(m\); the positive real component is trivial. Consider \[ H=\mathbf R_{>0}\times \{u\in\widehat{\mathbf Z}^{\,\times}:u_5\equiv\pm1\pmod5\}. \tag{12} \] It has index \(2\). Its image in \(\operatorname{Gal}(\mathbf Q(\zeta_5)/\mathbf Q)\) is the subgroup \(\{\pm1\}\); taking inverses does not change that subgroup. Its fixed field is \[ \mathbf Q(\zeta_5+\zeta_5^{-1})=\mathbf Q(\sqrt5). \tag{13} \] Indeed \(z=\zeta_5+\zeta_5^{-1}\) satisfies \(z^2+z-1=0\), so \((2z+1)^2=5\). Thus (12) is the norm group of \(\mathbf Q(\sqrt5)\). This example uses one cyclotomic extension, not the later assertion that every abelian extension of \(\mathbf Q\) is cyclotomic. \subsection{\texorpdfstring{5. Logarithmic differentials in characteristic \(p\)}{5. Logarithmic differentials in characteristic p}}\label{logarithmic-differentials-in-characteristic-p} For the rest of the proof let \(F\) be a global function field with full constant field \(k=\mathbf F_q\), of characteristic \(p\). Put \(C_F^1=\ker(\deg:C_F\to\mathbf Z)\). It is compact by lesson 14, and \(\deg\) is onto by lesson 16. We need to show that the universal norm subgroup \[ \mathcal N_F=\ker(\operatorname{rec}_F:C_F\to B_F) =\bigcap_{L/F\text{ finite separable}}N_{L/F}C_L \tag{14} \] is trivial. Equality follows from finite reciprocity and norm limitation: Galois closures are cofinal, and their norms equal those of their maximal abelian subextensions. All constant-extension norms occur in this intersection. The norm degree formula forces the degree of an element of \(\mathcal N_F\) to be divisible by every positive integer. Thus \(\mathcal N_F\) is a closed subgroup of compact \(C_F^1\). Here is the needed algebraic logarithmic-differential criterion, with its proof. If a characteristic-\(p\) field \(E\) satisfies \([E:E^p]=p\), choose a p-basis \(t\), so \(E=\bigoplus_{i=0}^{p-1}E^p t^i\). Let \(\partial t=1\), \(\partial E^p=0\). For \[ y=\sum_{i=0}^{p-1}y_i^p t^i, \qquad \omega=y\,dt, \] define the Cartier operator by \[ \mathcal C(\omega)=y_{p-1}\,dt. \tag{15} \] \textbf{Logarithmic-differential lemma.} This definition is independent of the p-basis, and \[ \mathcal C(\omega)=\omega\quad\Longleftrightarrow\quad \omega=db/b\text{ for some }b\in E^\times. \tag{16} \] \textbf{Proof.} We have \(\partial^p=0\). Let \(T=\partial-y\), an \(E^p\)-linear operator on the \(p\)-dimensional space \(E\). For multiplication by \(f\), its commutator with \(T\) is multiplication by \(\partial f\). In characteristic \(p\), \((\operatorname{ad}T)^p=\operatorname{ad}(T^p)\), by expanding the repeated commutator and cancelling the intermediate binomial coefficients. Hence \(T^p\) commutes with every multiplication operator and is itself multiplication by \(T^p(1)\). We compute that scalar explicitly. For a derivation \(\partial\) and a multiplication term \(h\), repeated application of \(\partial+h\) to \(1\) gives \[ \sum_{\sum j m_j=n} \frac{n!}{\prod_j m_j!\,(j!)^{m_j}} \prod_j(\partial^{j-1}h)^{m_j}. \tag{17} \] One proof counts partitions of \(n\) labelled elements: applying the multiplication term creates a new singleton block, and differentiating an existing factor adds the new element to that block. This gives the recurrence \(P_{n+1}=\partial P_n+hP_n\), with \(P_0=1\), and the displayed partition counts. At \(n=p\), all coefficients are divisible by \(p\), except the partition into \(p\) singletons and the single block of size \(p\); every other denominator is prime to \(p\). Taking \(h=-y\) gives \[ T^p=-\bigl(y^p+\partial^{p-1}y\bigr) =-\bigl(y^p-y_{p-1}^p\bigr). \tag{18} \] The last equality uses \((p-1)!=-1\) in \(\mathbf F_p\); pairing each nonzero element with its inverse proves this, with only \(1,-1\) left unpaired (and gives the same identity for \(p=2\)). Thus \(\mathcal C\omega=\omega\) exactly when \(T^p=0\). A nilpotent endomorphism of a nonzero finite-dimensional space has a nonzero kernel vector \(b\). For that vector \(\partial b=yb\), giving \(db/b=\omega\). Conversely, if \(y=\partial b/b\), then \(T=b\partial b^{-1}\), so \(T^p=0\), proving (16). The definition gives \(\mathcal C(df)=0\) and \(\mathcal C(f^p\omega)=f\mathcal C(\omega)\). For any other p-basis \(u\), (16) gives \(\mathcal C(d\log u)=d\log u\), and hence \(\mathcal C(u^{p-1}du)=du\). These properties determine the operator in the \(u\)-basis: the terms \(u^i du\), \(i0,\ p\nmid d}a_d u^{-d}, \qquad a_d\in\kappa, \tag{20} \] with finitely many summands. Eliminate a highest pole whose order is divisible by \(p\) by subtracting \(\wp(cu^{-d/p})\); perfection of \(\kappa\) supplies its coefficient root and the remaining pole has smaller order. The positive-power part \(a_+\) is \(\wp(-\sum_{r\geq0}a_+^{p^r})\), a convergent series. This proves the reduction. The representative (20) is rational over \(\kappa\). Form its explicit Artin--Schreier extension of \(\kappa(u)\), as in lesson 3. At every place \(P\ne(u)\), including infinity, \(a\) is integral. The extension is unramified there, and arithmetic Frobenius translates a root by \(\operatorname{Tr}_{\kappa(P)/\mathbf F_p}(a(P))\): iterating \(\theta^p=\theta+a(P)\) through \([\kappa(P):\mathbf F_p]\) steps gives this identity. Thus the symbol of a rational \(b\) at \(P\) is its valuation times this trace. On the other hand, \[ \operatorname{res}_P(a\,db/b)=v_P(b)a(P) \quad(P\ne(u)), \] because the logarithmic derivative of a local unit is an integral differential. The partial-fraction residue theorem on \(\kappa(u)\), proved in lesson 15, says that the sum of the residue traces, including infinity, is zero. The global principal product of Theorem 16.4 says that the sum of the local translation symbols is zero. Comparing these two sums proves (19) at \(u=0\) for rational \(b\). Rational elements are dense in the completion; both the symbol and the residue functional are continuous in \(b\), so the formula holds for all local \(b\). It remains to justify invariance when replacing \(a\) by its representative. In Laurent coordinates, \[ \mathcal C\left(\sum_j c_j u^jdu\right) =\sum_m c_{pm+p-1}^{1/p}u^mdu. \tag{21} \] For \(h\in E\), coefficient multiplication gives \[ \operatorname{res}(h^p\omega) =\bigl(\operatorname{res}(h\mathcal C\omega)\bigr)^p. \] Indeed if \(h=\sum b_j u^j\), the two sides before taking the \(p\)th power have coefficients \(\sum b_j c_{-pj-1}^{1/p}\). The residue sum is finite. Taking the finite-field trace removes the \(p\)th power. Since \(\mathcal C(d\log b)=d\log b\), the residue trace of \((h^p-h)d\log b\) is zero. Changing \(a\) by \(\wp h\) also changes the chosen root by \(h\) and leaves its translation character unchanged. This completes the proof of (19). We now obtain the global residue pairing without assuming a differential trace-residue theorem. Choose a separating \(t\in F\), and set \(\omega_0=d\log t\). It is nonzero in every completion. Define an additive character of \(\mathbb A_F\) by \[ \Phi(x)=\exp\left(\frac{2\pi i}{p} \sum_v\operatorname{Tr}_{\kappa(v)/\mathbf F_p} \operatorname{res}_v(x_v\omega_0)\right). \tag{22} \] Only finitely many terms are nonzero. Outside finitely many places, \(t\) is a unit and \(dt\) is a unit differential: this follows at the unramified places over \(k(t)\), where an irreducible prime polynomial has unit derivative. Thus (22) is continuous, and its local component is nontrivial at every place by the Laurent coefficient pairing and nondegeneracy of the finite-field trace. For \(a\in F\), apply (19) to the explicit global extension \(\theta^p-\theta=a\) and to the principal element \(t\). Global reciprocity makes the sum in (22) zero. Therefore \(\Phi\) is trivial on \(F\). Section 5 of lesson 15 proved a trace-based self-duality \((x,z)\mapsto\Psi_F(xz)\) of \(\mathbb A_F\), with exact annihilator \(F\). By that self-duality \(\Phi(x)=\Psi_F(cx)\) for a unique adèle \(c\). Triviality on \(F\) forces \(c\in F\); nontriviality forces \(c\ne0\). Multiplication by this global nonzero scalar preserves \(F\). Consequently \[ \mathbb A_F\text{ is self-dual under }(x,z)\mapsto\Phi(xz), \qquad F^\perp=F. \tag{23} \] This deduction explains why no unproved residue-trace compatibility is hidden in the argument. \subsection{7. Universal norms are trivial}\label{universal-norms-are-trivial} Take \([x]\in\mathcal N_F\), represented by an idèle \(x\). For every \(a\in F\), its reciprocity symbol in \(F(\wp^{-1}a)/F\) is trivial. Formula (19) gives \[ \sum_v\operatorname{Tr}_{\kappa(v)/\mathbf F_p} \operatorname{res}_v(a\,d\log x_v)=0. \tag{24} \] Write \(d\log x_v=y_v\omega_0\). The \(y_v\) form an adèle: at almost all places \(x_v\) is a unit and \(\omega_0\) is a unit differential, so \(y_v\) is integral. Equations (23)--(24) force \(y\in F\). We have therefore obtained a single global differential \(\omega=y\omega_0\) whose local image is \(d\log x_v\) everywhere. The local logarithmic differentials are Cartier-fixed, and global-local compatibility of (15) shows \(\mathcal C\omega=\omega\). Lemma (16) supplies \(b\in F^\times\) with \(\omega=d\log b\). Hence \[ d\log(x_v/b)=0\quad\text{at every }v. \] The kernel of the local derivative is \(F_v^p\), as the Laurent expansion shows. Thus each \(x_v/b\) has a unique \(p\)th root. These roots are units almost everywhere, and form an idèle. We have proved \[ \mathcal N_F\subseteq C_F^p. \tag{25} \] This alone does not yet prove that \(\mathcal N_F\) is \(p\)-divisible \emph{inside itself}. The following compactness step supplies that distinction. \textbf{Universal-norm lifting lemma.} For every finite separable \(E/F\), norm maps \(\mathcal N_E\) onto \(\mathcal N_F\). \textbf{Proof.} The norm of a universal norm is a universal norm: to test against a finite extension of \(F\), form its compositum with \(E\) and use tower norms. For surjectivity fix \(x\in\mathcal N_F\). For each finite separable \(L/E\), consider \[ Y_L=\{y\in N_{L/E}C_L^1:N_{E/F}y=x\}. \tag{26} \] This is a compact set, since \(C_L^1\) is compact. It is nonempty: \(x\) is a norm from \(L/F\), and its norm preimage has degree zero by the formula \(\deg_F N_{L/F}z=[k_L:k_F]\deg_L z\). Its norm to \(E\) belongs to (26). A compositum contains any finite collection of the \(L\)'s and gives a set contained in their intersection. Compactness in \(C_E^1\) gives a point of all the \(Y_L\), which is a universal norm in \(E\) mapping to \(x\). \(\square\) Apply (25) over every finite \(E/F\). Lift \(x\in\mathcal N_F\) to \(y\in\mathcal N_E\), write \(y=z^p\), and norm \(z\) down. Degrees show \(z\in C_E^1\). Thus the compact sets \[ R_E=\{r\in N_{E/F}C_E^1:r^p=x\} \tag{27} \] are nonempty. They have the finite-intersection property by composita. Their intersection contains a \(p\)th root of \(x\) belonging to every finite norm image, hence to \(\mathcal N_F\). This proves internal \(p\)-divisibility. For a prime \(\ell\ne p\), first work over a finite extension \(E\) containing \(\mu_\ell\). For every sufficiently large finite \(S\), (3), valid also for these function fields, gives \[ \mathcal N_E\subseteq C_E^\ell\operatorname{im}(U^S). \tag{28} \] The tail groups \(U^S\) approach \(1\): every idèle neighborhood restricts only finitely many places, and choosing \(S\) to include them makes their components exactly \(1\). All their classes have degree zero. If an element of \(\mathcal N_E\) is a product in (28), its power factor also has degree zero, so its root has degree zero. The image \((C_E^1)^\ell\) is compact and closed. Intersecting (28) over these \(S\) therefore proves \(\mathcal N_E\subseteq(C_E^1)^\ell\): a point outside that closed power image has a neighborhood disjoint from it, which a sufficiently small tail group cannot bridge. Universal-norm lifting and the compact root-set argument (27), with \(\ell\) in place of \(p\), now prove internal \(\ell\)-divisibility of \(\mathcal N_F\). Extensions containing \(\mu_\ell\) are cofinal among finite separable extensions, so the intersection still tests all norm images. Finally \(C_F\) is totally disconnected. The finite idèle group has compact open profinite neighborhood subgroups, the principal subgroup is closed, and the quotient assertion of the compact-group lemma applies. Its compact subgroup \(C_F^1\), and hence the closed subgroup \(\mathcal N_F\), are profinite. A profinite group divisible by every prime has no nontrivial finite quotient: if a finite quotient has order \(m\), divisibility by the prime factors of \(m\) makes the \(m\)-power map onto, whereas that map is identically \(1\). Finite quotients separate points of a profinite group. Therefore \[ \mathcal N_F=1. \tag{29} \] Reciprocity for function fields is injective, including all characteristic-power extensions. \subsection{8. Function-field existence and Weil topology}\label{function-field-existence-and-weil-topology} Choose a class \(c\in C_F\) of degree \(1\), whose existence was proved in lesson 16, and put \(\gamma=\operatorname{rec}_F(c)\). Let \[ \delta:B_F\longrightarrow\widehat{\mathbf Z} \] be restriction to the full constant-field extension. The continuous power map \(z\mapsto\gamma^z\) on \(\widehat{\mathbf Z}\) is a section of \(\delta\), since \(\delta(\gamma)=1\). In particular it is injective. The compact subgroup \(\operatorname{rec}_F(C_F^1)\), multiplied by this section, is closed and contains the dense reciprocity image from lesson 16. It is therefore all of \(B_F\). Taking constant-degree kernels gives \[ C_F^1\xrightarrow{\sim}\ker\delta, \qquad \operatorname{rec}_F(C_F)=\delta^{-1}(\mathbf Z). \tag{30} \] The first map is a homeomorphism, by (29) and compactness. Give the group on the right of the second equality the topology with \(\ker\delta\) compact open and its integer cosets discrete. Then (30) is a homeomorphism from \(C_F=C_F^1\times c^{\mathbf Z}\). This topology is finer than its subspace topology in the profinite group \(B_F\). Its image is dense and proper in \(B_F\), because \(\mathbf Z\) is dense and proper in \(\widehat{\mathbf Z}\). Now let \(H\subseteq C_F\) be open of finite index. Put \(H_0=H\cap C_F^1\); it is compact open in \(C_F^1\), and \(\deg H=d\mathbf Z\) for a positive integer \(d\). Choose \(h\in H\) of degree \(d\). Every element of \(H\) is uniquely a product of an element of \(H_0\) and a power of \(h\). In \(B_F\), \[ \overline{\operatorname{rec}_F(H)} =\operatorname{rec}_F(H_0)\cdot \{\operatorname{rec}_F(h)^z:z\in\widehat{\mathbf Z}\}. \tag{31} \] The power group is compact and its projection is \(d\widehat{\mathbf Z}\); multiplication by \(d\) there is injective, so its constant-degree kernel is trivial. Thus (31) has kernel \(\operatorname{rec}_F(H_0)\) under constant degree and is an open finite-index subgroup of \(B_F\). To verify openness explicitly, (30) splits \(B_F\) as \(\ker\delta\times\widehat{\mathbf Z}\); on the clopen part with second coordinate in \(d\widehat{\mathbf Z}\), division by \(d\) is continuous and (31) imposes membership in the open subgroup \(\operatorname{rec}_F(H_0)\) after a continuous translation. Its inverse image in \(C_F\) is exactly \(H\). Indeed an integer in \(d\widehat{\mathbf Z}\) belongs to \(d\mathbf Z\), and its uniquely determined exponent in the power group of (31) is that ordinary integer divided by \(d\). The remaining factor must lie in \(H_0\). Galois correspondence gives a finite abelian extension \(L/F\) with Galois kernel (31). Finite reciprocity then gives \(N_{L/F}C_L=H\). This finishes the function-field proof of Theorem 17.1 and the correspondence of Theorem 17.2. \subsubsection{The absolute Weil group}\label{the-absolute-weil-group} Let \(G=G_F\), let \(I\) be the kernel of its constant-field map to \(\widehat{\mathbf Z}\), and choose \(\varphi\in G\) of degree \(1\). Define \[ W_F=\{g\in G:\deg g\in\mathbf Z\}=I\rtimes\varphi^{\mathbf Z}, \tag{32} \] with \(I\) compact open and the integer quotient discrete. It is dense in \(G\). Let \(P=\overline{[G,G]}\). All commutators have degree zero, so \(P\subseteq I\). Density and continuity of commutators show \[ \overline{[W_F,W_F]}^{\,W_F}=P. \tag{33} \] For clarity, approximate each of the finitely many factors in a product of commutators in \(G\) by elements of \(W_F\). Their commutators converge inside \(I\), whose topology is the same in both groups. This proves that the two closures are equal; the reverse containment is immediate. The quotient \(W_F/P\) is consequently the Hausdorff abelianization of \(W_F\). Its map to \(G/P=B_F\) is injective with image \(\delta^{-1}(\mathbf Z)\). Its degree-zero subgroup is \(I/P\simeq\ker\delta\), a compact open group, and its degree cosets are discrete. Comparing with (30) proves the full topological identification \[ C_F\xrightarrow{\sim}W_F^{\mathrm{ab}}. \tag{34} \] It is the \emph{abelianization}, not the generally nonabelian group \(W_F\), that is identified with \(C_F\). This proves the global function-field Weil assertion used in \href{../weil-groups-and-one-dimensional-representations.html}{Weil groups and one-dimensional representations}. The local compatibility also follows. A chosen embedding of a local decomposition group takes a local Weil element of integer Frobenius degree \(m\) to global constant degree \(m\deg(v)\), so it maps into (32). On Hausdorff abelianizations its map agrees with insertion \(F_v^\times\to C_F\): Theorem 16.4 proves agreement in every finite abelian quotient, and those quotients separate the image in \(B_F\). Thus (34) respects local reciprocity, local inertia and the arithmetic Frobenius normalization. \subsection{9. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- Congruence neighborhoods over the rationals (easy)}\label{exercise-1-congruence-neighborhoods-over-the-rationals-easy} Show that every open finite-index subgroup of \(C_{\mathbf Q}\) contains the image of \[ U_m=\mathbf R_{>0}\times \prod_{p\mid m}(1+p^{v_p(m)}\mathbf Z_p) \times\prod_{p\nmid m}\mathbf Z_p^\times \subseteq J_{\mathbf Q} \] for some positive integer \(m\). \textbf{Solution.} The connected positive-real factor is contained in the subgroup, since its map to the discrete finite quotient is trivial. Under (10), openness in the unit product supplies a neighborhood restricting finitely many primes. At each restricted prime choose \(1+p^{a_p}\mathbf Z_p\) inside that neighborhood, increasing \(a_p\geq1\) if necessary. With \(m=\prod p^{a_p}\), its product with the unrestricted unit factors and the positive-real factor is the stated image of \(U_m\). One may take \(m=1\) when no prime restrictions are needed. \subsubsection{Exercise 2 --- The kernel as an intersection (medium)}\label{exercise-2-the-kernel-as-an-intersection-medium} For either kind of global field prove \[ \ker\operatorname{rec}_K=\bigcap_{H\text{ open, finite index}}H. \] Explain why it is not the same nontrivial group in both cases. \textbf{Solution.} The profinite group \(B_K\) is the inverse limit of its finite abelian Galois quotients. Their kernels pulled back to \(C_K\) are exactly the finite abelian norm groups, by Theorem 16.4. Theorem 17.1 says that these are precisely all open finite-index subgroups. Intersecting their kernels gives the assertion. Over number fields Proposition 17.3 identifies the intersection with the connected component, which contains the positive-real content factor. Over function fields (29) makes it trivial; the idèle class group is totally disconnected and its reciprocity map is injective, although not onto \(B_K\). \subsubsection{Exercise 3 --- The real quadratic field of conductor five (medium)}\label{exercise-3-the-real-quadratic-field-of-conductor-five-medium} Determine the class field of the subgroup (12). \textbf{Solution.} Reciprocity on \(\mathbf Q(\zeta_5)\) sends the normalized unit tuple to its inverse modulo \(5\). The inverse image of the subgroup \(\{1,-1\}\subseteq(\mathbf Z/5)^\times\) is exactly (12). Its fixed field is generated by \(z=\zeta_5+\zeta_5^{-1}\). Dividing \(1+\zeta_5+\cdots+\zeta_5^4=0\) by \(\zeta_5^2\) gives \(z^2+z-1=0\), whence \((2z+1)^2=5\). This quadratic field is \(\mathbf Q(\sqrt5)\), and finite reciprocity identifies (12) with its norm group. The index is \(4/2=2\), as required. \subsubsection{Exercise 4 --- Existence with roots of unity (hard)}\label{exercise-4-existence-with-roots-of-unity-hard} Let \(K\) be a global field, \(n=\ell^a\) prime to its characteristic, and \(\mu_n\subset K\). Suppose \(H\subseteq C_K\) is open of finite index and \(C_K^n\subseteq H\). Prove existence of its class field using the full S-unit radical. \textbf{Solution.} Choose \(S\) containing the infinite places, the divisors of \(n\), class-group generators, and every component restricted by an identity neighborhood in the inverse image of \(H\). Then \(U^S\) lies in that inverse image and \(J_K=K^\times J_K^S\). Every element of \(I(S)\) in (2) is an idèle \(n\)th power times an element of \(U^S\), so \(C(S)\subseteq H\). S-unit Kummer theory gives \(M=K(E_S^{1/n})\) with group \((\mathbf Z/n)^{|S|}\). Proposition 15.2 with \(T=\varnothing\) proves that \(C(S)\) is contained in its norm group and has index \([M:K]\); finite reciprocity makes these groups equal. In \(\operatorname{Gal}(M/K)\), take the subgroup corresponding to \(H/C(S)\), and let \(L\) be its fixed field. Restriction of reciprocity has kernel exactly \(H\), and its norm-kernel theorem gives \(H=N_{L/K}C_L\). This proof uses the exact principal-intersection and index calculation of Proposition 15.2; just saying that Kummer radicals exist would not identify the norm group. \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 number-field S-unit existence argument and norm descent are retained. The characteristic-p radical and residue separation proof is written here, together with the complete function-field Weil extension and its topology. \begin{itemize} \tightlist \item \href{https://www.mathi.uni-heidelberg.de/~schmidt/Neukirch-en/Neukirch_cft_02_may15.pdf}{Jürgen Neukirch, Class Field Theory --- The Bonn Lectures, Online Edition 2.0 (May 2015), edited by Alexander Schmidt}. \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \item \href{https://ncatlab.org/nlab/files/TateNumberTheory.pdf}{John Tate, Number theoretic background (1979), freely available paper}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Ray class fields, conductors and ideal reciprocity}\label{ray-class-fields-conductors-and-ideal-reciprocity} \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).} \textbf{Lesson 18.} A local uniformizer at an unramified prime acts by Frobenius. Multiplying these uniformizer idèles turns global reciprocity into a map on fractional ideals. The remaining idèle components record the congruences and signs that a principal generator must satisfy. Their quotient is the ray class group, and the global existence theorem constructs its class field. We use \href{../global-existence-and-the-idelic-class-field-correspondence.html}{Global existence and the idèlic class field correspondence}, \href{../the-global-reciprocity-law.html}{The global reciprocity law}, and the local conductor criteria in \href{../abelian-ramification-conductors-and-hasse-arf.html}{Abelian ramification, conductors and Hasse--Arf}, Proposition 10.4. The explicit quadratic Hilbert symbols of lesson 11 will determine the examples. Throughout sections 1--7, \(K\) is a number field. Section 8 explains the function-field degree distinction. \subsection{1. Moduli and ray classes}\label{moduli-and-ray-classes} A \textbf{modulus} is a pair \(\mathfrak m=\mathfrak m_f\mathfrak m_\infty\), where \(\mathfrak m_f\) is a nonzero integral ideal and \(\mathfrak m_\infty\) is a set of real places, each occurring once. Complex places impose no additional condition. Divisibility means ideal divisibility together with inclusion of the selected real places. Write \(m_v=v(\mathfrak m_f)\) at finite places, and define \[ U_v(\mathfrak m)= \begin{cases} 1+\mathfrak p_v^{m_v}\mathcal O_v,&v\text{ finite},\ m_v>0,\\ \mathcal O_v^\times,&v\text{ finite},\ m_v=0,\\ \mathbf R_{>0},&v\in\mathfrak m_\infty,\\ K_v^\times,&v\mid\infty,\ v\notin\mathfrak m_\infty. \end{cases} \tag{1} \] Put \(U(\mathfrak m)=\prod_v U_v(\mathfrak m)\subset J_K\), and let \(C(\mathfrak m)\) be its image in \(C_K\). Let \(I^{\mathfrak m}\) be the group of fractional ideals supported on finite primes not dividing \(\mathfrak m_f\). Define \(P_{\mathfrak m,1}\) to consist of the principal ideals \((a)\) for which \[ a\in1+\mathfrak p_v^{m_v}\mathcal O_v\quad(v\mid\mathfrak m_f), \qquad a_v>0\quad(v\in\mathfrak m_\infty). \tag{2} \] The finite condition is a multiplicative local congruence, denoted \(a\equiv1\pmod{\mathfrak m_f}\); it forces \(a\) to be a unit at the primes in question, but does not require it to be integral everywhere. The \textbf{ray class group} is \[ \operatorname{Cl}_{\mathfrak m}=I^{\mathfrak m}/P_{\mathfrak m,1}. \tag{3} \] \subsubsection{From ideal classes to idèle classes}\label{from-ideal-classes-to-iduxe8le-classes} Choose a uniformizer \(\pi_{\mathfrak p}\) at each prime outside \(\mathfrak m_f\). An ideal \(\mathfrak a=\prod\mathfrak p^{e_{\mathfrak p}}\) gives the idèle with components \(\pi_{\mathfrak p}^{e_{\mathfrak p}}\) there and \(1\) elsewhere. Changing a uniformizer multiplies by units already in (1). This defines a canonical map \[ I^{\mathfrak m}\longrightarrow C_K/C(\mathfrak m). \] It is onto. Given an idèle \(x\), weak approximation chooses \(a\in K^\times\) such that \(ax_v\) belongs to the deep unit group in (1) at each finite divisor of \(\mathfrak m_f\), and is positive at every selected real place. Outside these places, form the ideal with valuations \(v(ax_v)\). Removing its uniformizer idèle leaves an element of \(U(\mathfrak m)\). Its kernel is \(P_{\mathfrak m,1}\). Indeed if a uniformizer idèle \(j\) equals \(a u\), with \(u\in U(\mathfrak m)\), its component \(1\) at each modulus prime forces \(a\) to satisfy (2), and its outside valuations give exactly the ideal \((a)\). The same argument at selected real places forces positivity. Conversely a generator satisfying (2) differs from its ideal's uniformizer idèle by an element of \(U(\mathfrak m)\). Hence \[ \operatorname{Cl}_{\mathfrak m}\xrightarrow{\sim}C_K/C(\mathfrak m). \tag{4} \] The group is finite. More explicitly, with \(E_K=\mathcal O_K^\times\), there is an exact sequence \[ E_K\longrightarrow (\mathcal O_K/\mathfrak m_f)^\times\times\{\pm1\}^{|\mathfrak m_\infty|} \longrightarrow\operatorname{Cl}_{\mathfrak m} \longrightarrow\operatorname{Cl}_K\longrightarrow1. \tag{5} \] When \(\mathfrak m_f=(1)\), the residue factor means the trivial group. The first map is reduction and the selected real signs. For the second, use weak approximation to choose an element with the specified residue units and signs and take its principal ideal class. Changing this element changes the ideal by a generator satisfying (2). Its ray class is trivial exactly when its residues and signs come from a global unit: divide by a ray-principal generator and use equality of the principal ideals. Finally every ordinary ideal class has a representative prime to \(\mathfrak m_f\), by weak approximation at those finitely many primes. The kernel of forgetting the ray condition is precisely the principal ideals just described. This proves (5), and finiteness follows from the finite residue and sign groups and the finite ordinary class group. Under our convention \(\operatorname{Cl}_{(1)}=\operatorname{Cl}_K\). Selecting all real places, with finite modulus \((1)\), gives the narrow class group. Some accounts build total positivity into every modulus; converting their convention requires adding all the real places to ours. Keeping this choice explicit prevents exchanging the ordinary and narrow class fields. \subsection{2. Frobenius and the ideal Artin map}\label{frobenius-and-the-ideal-artin-map} Let \(L/K\) be finite abelian and unramified at every finite prime outside \(\mathfrak m_f\). Define \[ \psi_{L/K}:I^{\mathfrak m}\longrightarrow\operatorname{Gal}(L/K), \qquad \mathfrak p\longmapsto\operatorname{Frob}_{\mathfrak p}, \tag{6} \] extended multiplicatively, allowing negative ideal exponents. Arithmetic Frobenius acts on the residue field by \(z\mapsto z^{N\mathfrak p}\). Since the extension is abelian, the automorphism is independent of the prime chosen above \(\mathfrak p\). \textbf{Proposition 18.1.} The ideal symbol of \(\mathfrak a\) in (6) is the global reciprocity symbol of its uniformizer idèle. \textbf{Proof.} At an unramified prime, local reciprocity takes any uniformizer to arithmetic Frobenius, and takes units to \(1\). Theorem 16.4 identifies insertion of this local symbol with the global symbol. Multiplying over the support of \(\mathfrak a\) gives (6). Negative exponents give inverse Frobenius powers, as required for fractional ideals. \(\square\) If \(C(\mathfrak m)\) is contained in the norm group of \(L\), (4) and global reciprocity therefore make (6) onto and kill \(P_{\mathfrak m,1}\). We next determine exactly when that containment holds. \subsection{3. The conductor, including real places}\label{the-conductor-including-real-places} For a finite place \(v\), choose \(w\mid v\) and let \(a_v\) be the least integer \(a\geq0\) such that \[ U_v^{(a)}\subseteq N_{L_w/K_v}L_w^\times, \qquad U_v^{(0)}=\mathcal O_v^\times. \tag{7} \] This is the local conductor exponent of Proposition 10.4. It is finite by continuity, independent of \(w\), and zero at unramified places. At a real place put \(a_v=1\) if it becomes complex, and \(a_v=0\) otherwise; complex places have exponent zero. \textbf{Proposition 18.3 (local description of the global conductor).} The smallest modulus \(\mathfrak f(L/K)\) satisfying \(C(\mathfrak m)\subseteq N_{L/K}C_L\) is \[ \mathfrak f(L/K)= \prod_{v\text{ finite}}\mathfrak p_v^{a_v} \prod_{v\text{ real},\ L_w=\mathbf C}v. \tag{8} \] More precisely, \[ C(\mathfrak m)\subseteq N_{L/K}C_L \quad\Longleftrightarrow\quad\mathfrak f(L/K)\mid\mathfrak m. \tag{9} \] A place ramifies in \(L\) exactly when it occurs in (8), where real ramification means complexification. \textbf{Proof.} A local element inserted at a single place has trivial global symbol exactly when its local symbol is trivial: the local decomposition group embeds in the global Galois group. Thus containment on the left of (9) forces every group \(U_v(\mathfrak m)\) into its local norm group. Conversely, if this holds at every place, all the local symbols of an idèle in \(U(\mathfrak m)\) are trivial, and its global product symbol is trivial. The norm-kernel theorem gives the containment. The least finite level is precisely (7). At a complexified real place the norm group is \(\mathbf R_{>0}\), so that place must be selected; at every other infinite place all local elements are norms. This proves (8)--(9). Proposition 10.4 identifies positive finite conductor with nontrivial inertia, proving the final assertion. \(\square\) The argument tests \emph{each local group} inside the subgroup. One particular global norm class need not have each of its components a local norm, since its local symbols can cancel. For example the principal idèle \(-1\) in \(\mathbf Q(i)/\mathbf Q\) has nontrivial symbols at \(2\) and infinity, whose product is \(1\). Testing single-place insertions is what makes (9) valid. Equivalently, the finite exponent \(a_v\) is the maximum of the conductor exponents of the restrictions of all characters of \(\operatorname{Gal}(L/K)\) to its local decomposition group. Characters separate that finite abelian group, so killing all these restrictions is exactly killing the group-valued local symbol. Proposition 10.4 proves the character formula and its upper-ramification interpretation, including wild ramification. \subsection{4. Artin reciprocity for ideals}\label{artin-reciprocity-for-ideals} Let \(I_L^{\mathfrak m}\) denote fractional ideals of \(L\) prime to every prime above \(\mathfrak m_f\). Its norm is the usual ideal norm; in particular \(N_{L/K}\mathfrak q=\mathfrak p^{f(\mathfrak q/\mathfrak p)}\). \textbf{Theorem 18.2 (ideal reciprocity).} If \(\mathfrak f(L/K)\mid\mathfrak m\), then \[ \ker\psi_{L/K}=P_{\mathfrak m,1}\,N_{L/K}I_L^{\mathfrak m}, \qquad I^{\mathfrak m}/\bigl(P_{\mathfrak m,1}N_{L/K}I_L^{\mathfrak m}\bigr) \xrightarrow{\sim}\operatorname{Gal}(L/K). \tag{10} \] \textbf{Proof.} By (9), the global symbol factors through (4); Proposition 18.1 identifies that factor with the ideal symbol, which is therefore onto. It remains to identify the image of the norm subgroup in the ray quotient. Take \(y\in J_L\). Multiplying by a principal element of \(L\), weak approximation makes its components above \(\mathfrak m_f\) sufficiently close to \(1\) that their local norms belong to (1). It also makes its real components above the selected real places positive; norms from complex components are positive automatically. This normalization is possible simultaneously at these finitely many places, and does not change its norm class. The ideal \(\mathfrak b=\prod_{\mathfrak q\nmid\mathfrak m_f}\mathfrak q^{v_{\mathfrak q}(y_{\mathfrak q})}\) is now prime to \(\mathfrak m_f\). The local valuation formula \[ v_{\mathfrak p}\bigl((Ny)_{\mathfrak p}\bigr) =\sum_{\mathfrak q\mid\mathfrak p}f(\mathfrak q/\mathfrak p) v_{\mathfrak q}(y_{\mathfrak q}) \tag{11} \] shows that the image of \(Ny\) under (4) is the ray class of \(N\mathfrak b\). Conversely a uniformizer idèle for any \(\mathfrak b\in I_L^{\mathfrak m}\) has norm with exactly that ray class, by (11), and with component \(1\) at the modulus primes and at infinity. Thus the image of \(NC_L\) is \(P_{\mathfrak m,1}NI_L^{\mathfrak m}/P_{\mathfrak m,1}\). Applying the norm-kernel theorem proves (10). \(\square\) In particular, if \(\mathfrak p\nmid\mathfrak m_f\), its residue degree in \(L\) is the order of its class in the quotient in (10); it has \([L:K]/f\) primes above it. Its Frobenius is trivial exactly when it splits completely. This converts a splitting problem into a test in an explicit ideal group. \subsection{5. Ray class fields and Takagi's formulation}\label{ray-class-fields-and-takagis-formulation} \textbf{Theorem 18.4.} For every modulus \(\mathfrak m\) there is a unique finite abelian \textbf{ray class field} \(K_{\mathfrak m}/K\) with norm group \(C(\mathfrak m)\), and \[ \operatorname{Gal}(K_{\mathfrak m}/K)\simeq\operatorname{Cl}_{\mathfrak m}. \tag{12} \] A finite abelian \(L/K\) lies in \(K_{\mathfrak m}\) exactly when its conductor divides \(\mathfrak m\). If \(\mathfrak m\mid\mathfrak n\), then \(K_{\mathfrak m}\subseteq K_{\mathfrak n}\). \textbf{Proof.} The group \(U(\mathfrak m)\) is open in the restricted product, so its image \(C(\mathfrak m)\) is open. Equation (4) and finiteness from (5) give finite index. Theorem 17.1 constructs its unique class field, and Theorem 16.4 gives (12). The reversed-inclusion correspondence and (9) prove the conductor criterion. Finally (1) gives \(C(\mathfrak n)\subseteq C(\mathfrak m)\), so field inclusion reverses that subgroup inclusion. \(\square\) The conductor of \(K_{\mathfrak m}\) divides \(\mathfrak m\); equality need not hold. Different moduli can give the same ray class field. For instance \(\operatorname{Cl}_{2\infty}(\mathbf Q)\) is trivial, as (5) or the unit decomposition of lesson 17 shows, so this modulus gives \(\mathbf Q\) itself. \textbf{Proposition 18.5 (Takagi's congruence subgroups).} For fixed \(\mathfrak m\), the subgroups \[ P_{\mathfrak m,1}\subseteq H\subseteq I^{\mathfrak m} \tag{13} \] correspond with reversed inclusions to the subextensions of \(K_{\mathfrak m}/K\). Their class field \(L\) has ideal Artin kernel \(H\) and \[ \operatorname{Gal}(L/K)\simeq I^{\mathfrak m}/H. \tag{14} \] Every finite abelian extension appears for some modulus. \textbf{Proof.} Use (4) to regard \(H/P_{\mathfrak m,1}\) as a subgroup of the finite quotient \(C_K/C(\mathfrak m)\). Its inverse image \(\widetilde H\subset C_K\) is open of finite index and contains \(C(\mathfrak m)\). Theorem 17.1 gives a unique class field \(L\), and reversed inclusion places it in \(K_{\mathfrak m}\). Proposition 18.1 and finite reciprocity show that its ideal kernel is exactly \(H\), giving (14). Conversely each subextension yields such a subgroup. A finite abelian extension has the finite conductor (8), hence lies in its conductor ray field. This proves the final assertion and equivalence with the idèlic formulation. \(\square\) For a larger modulus \(\mathfrak n\) divisible by \(\mathfrak m\), the same field is represented by \(H\cap I^{\mathfrak n}\). Indeed the ideal Artin map for \(\mathfrak n\) is the restriction of (6), and its kernel is this intersection. The surjection \(\operatorname{Cl}_{\mathfrak n}\to\operatorname{Cl}_{\mathfrak m}\) is equally the quotient map from \(C(\mathfrak n)\subseteq C(\mathfrak m)\). Thus changing the modulus preserves the field when the congruence subgroup is changed compatibly. \subsection{6. Quadratic conductors over the rationals}\label{quadratic-conductors-over-the-rationals} Let \(d\ne1\) be a nonzero squarefree integer and \(L=\mathbf Q(\sqrt d)\). Its fundamental discriminant is \[ D=\begin{cases}d,&d\equiv1\pmod4,\\4d,&d\not\equiv1\pmod4.\end{cases} \tag{15} \] We claim \[ \mathfrak f(L/\mathbf Q)= \begin{cases} |D|,&d>0,\\ |D|\infty,&d<0. \end{cases} \tag{16} \] Here \(\infty\) is the real place, a factor of the modulus, not a numeric multiplier. We verify every local factor, independently of a later global conductor--discriminant formula. The quadratic local norm character on \(\mathbf Q_p^\times\) is \(b\mapsto(b,d)_p\). At an odd prime dividing \(d\), the odd-place formula of Theorem 11.4 restricts on units to \(u\mapsto(u/p)\). It is nontrivial on all units and trivial on \(1+p\mathbf Z_p\), so its conductor exponent is \(1\). At an odd prime not dividing \(d\), the formula is trivial on units and the exponent is \(0\). For an odd \(2\)-adic unit write \[ \epsilon(u)=\frac{u-1}{2}\pmod2, \qquad\omega(u)=\frac{u^2-1}{8}\pmod2. \] If \(d\) is odd, the unit character is \[ (u,d)_2=(-1)^{\epsilon(u)\epsilon(d)}. \tag{17} \] For \(d\equiv1\pmod4\) this is trivial, giving exponent \(0\). For \(d\equiv3\pmod4\) it is nontrivial on units, including \(u=3\), but trivial on \(1+4\mathbf Z_2\). Its exponent is \(2\), since \(U_2^{(1)}=\mathbf Z_2^\times\). If \(d=2e\) with \(e\) odd, the character is \[ (u,2e)_2=(-1)^{\epsilon(u)\epsilon(e)+\omega(u)}. \tag{18} \] It is trivial on \(1+8\mathbf Z_2\), while \(u=5\in1+4\mathbf Z_2\) gives \(-1\). Its exponent is \(3\). These are exactly the \(2\)-adic exponents of \(|D|\) in (15). Finally the real extension is trivial for \(d>0\) and complex for \(d<0\). Proposition 18.3 now proves (16), including its sign condition. For instance \(\mathbf Q(\sqrt5)\) has conductor \(5\), \(\mathbf Q(i)\) has conductor \(4\infty\), and \(\mathbf Q(\sqrt{-3})\) has conductor \(3\infty\). Keeping the real factor distinguishes the latter two fields from the corresponding real ray fields. \subsection{7. A ray field over the Gaussian rationals}\label{a-ray-field-over-the-gaussian-rationals} Take \(K=\mathbf Q(i)\) and modulus \((3)\). The ring \(\mathbf Z[i]\) is Euclidean: approximate the real and imaginary parts of a complex quotient by integers, leaving remainder norm at most one half the divisor norm. Thus its class group is trivial. Since \(X^2+1\) is irreducible over \(\mathbf F_3\), \[ \mathbf Z[i]/(3)\simeq\mathbf F_9, \qquad (\mathbf Z[i]/(3))^\times\simeq\mathbf Z/8. \] The four units \(\{\pm1,\pm i\}\) have distinct reductions. Sequence (5) gives \[ \operatorname{Cl}_{(3)}\simeq \mathbf F_9^\times/\{\pm1,\pm i\}\simeq\mathbf Z/2. \tag{19} \] Its ray field is \[ K_{(3)}=\mathbf Q(i,\sqrt3)=\mathbf Q(\zeta_{12}). \tag{20} \] To identify it, not merely predict its degree, let \(L=K(\sqrt3)\). The element \(3\) is not a square in \(K\): squaring \(a+bi\), with \(a,b\in\mathbf Q\), forces \(ab=0\), and neither remaining rational square equation is possible. Thus \([L:K]=2\). At the prime \((3)\), \(X^2-3\) is Eisenstein over the unramified quadratic completion of \(\mathbf Q_3\), and the quadratic extension is tame. Its conductor exponent is \(1\), by Proposition 10.4 and the tame unit symbol of Proposition 11.3. At the prime above \(2\), adjoining \(\sqrt3\) is the same as adjoining \(\sqrt{-3}\), since \(i\in K\). The field \(\mathbf Q_2(\sqrt{-3})=\mathbf Q_2(\sqrt5)\) is the unramified quadratic extension: their defining units differ by a square because \(-3/5\equiv1\pmod8\). The odd-unit square criterion and unramified quadratic field were proved in lessons 10--11. Base change of an unramified local extension remains unramified or split, by lifting its separable residue-field polynomial. Hence the prime over \(2\) contributes no conductor. At every other finite prime, the unit radical of order \(2\) gives an unramified or split extension; the archimedean places of \(K\) are complex. Thus \(\mathfrak f(L/K)=(3)\). Theorem 18.4 places \(L\) in \(K_{(3)}\). Both have degree \(2\), by (19), so they coincide. Finally \(\zeta_{12}=(\sqrt3+i)/2\), while \(i=\zeta_{12}^3\) and \(\sqrt3=\zeta_{12}+\zeta_{12}^{-1}\); this proves the second equality in (20). \subsection{8. What changes over a function field}\label{what-changes-over-a-function-field} For a function field over full constants \(\mathbf F_q\), use divisors instead of fractional ideals, and finite place powers instead of an archimedean sign set. The uniformizer map and weak-approximation proof of (4) give the divisor ray group as \(C_K/C(\mathfrak m)\). Here \(C(\mathfrak m)\) has degree zero. Consequently there is an exact sequence \[ 0\longrightarrow C_K^1/C(\mathfrak m) \longrightarrow C_K/C(\mathfrak m) \xrightarrow{\deg}\mathbf Z\longrightarrow0. \tag{21} \] The left group is finite because \(C_K^1\) is compact and \(C(\mathfrak m)\) is open. The right group is infinite. Thus this entire divisor ray group does not correspond to a finite ray extension: constant extensions of every degree satisfy the same unit conditions. To specify a finite class field, impose an additional finite-index degree condition. For example, choose a degree-one class \(c\) and take the subgroup generated by \(C(\mathfrak m)\) and \(c^d\), with \(d>0\); (21) shows that its index is \(d\,|C_K^1/C(\mathfrak m)|\), so Theorem 17.1 applies. Other subgroups of finite index give the corresponding general extensions. The local conductor and ideal-symbol arguments remain valid in this setting, with no real factors and with divisors prime to the finite modulus. This degree condition is why the number-field finiteness proof of a ray class group should not simply be repeated for the full function-field divisor group. \subsection{9. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- Quadratic conductors (easy)}\label{exercise-1-quadratic-conductors-easy} Determine the conductor of \(\mathbf Q(\sqrt d)/\mathbf Q\) for a nontrivial squarefree \(d\). \textbf{Solution.} Each odd divisor of \(d\) has unit character \((u/p)\), giving exponent \(1\); every other odd prime gives exponent \(0\). At \(2\), formula (17) gives exponent \(0\) when \(d\equiv1\pmod4\) and exponent \(2\) when \(d\equiv3\pmod4\). For even \(d\), formula (18) kills \(U_2^{(3)}\) and is nontrivial on \(5\in U_2^{(2)}\), giving exponent \(3\). Their product is \(|D|\), with \(D\) as in (15). Add the real place exactly when \(d<0\). This is (16); omitting that place for an imaginary quadratic extension would violate the local norm condition at infinity. \subsubsection{Exercise 2 --- The Gaussian ray group (medium)}\label{exercise-2-the-gaussian-ray-group-medium} Compute \(\operatorname{Cl}_{(3)}(\mathbf Q(i))\) and identify its class field. \textbf{Solution.} Euclideanity gives ordinary class number \(1\). The residue unit group at the inert prime \((3)\) has order \(8\), and the global units have four distinct residues. Sequence (5) gives a cyclic quotient of order \(2\). The quadratic extension \(K(\sqrt3)/K\) has tame conductor exponent \(1\) at \((3)\), is unramified at \(2\) because it is the base change of the unramified extension \(\mathbf Q_2(\sqrt{-3})\), and is unramified at every other finite prime. There are no real places. Its conductor is \((3)\), so it lies in the degree-two ray field and equals it. The explicit identities following (20) identify it with \(\mathbf Q(\zeta_{12})\). \subsubsection{Exercise 3 --- Increasing the modulus (medium)}\label{exercise-3-increasing-the-modulus-medium} Prove \(K_{\mathfrak m}\subseteq K_{\mathfrak n}\) for \(\mathfrak m\mid\mathfrak n\), and describe the resulting map on ray groups. \textbf{Solution.} Increasing finite exponents shrinks the principal-unit groups, and adding a real place replaces \(\mathbf R^\times\) by \(\mathbf R_{>0}\). Thus \(U(\mathfrak n)\subseteq U(\mathfrak m)\) and \(C(\mathfrak n)\subseteq C(\mathfrak m)\). The class field correspondence reverses that inclusion. Restriction of Galois automorphisms is the surjective quotient map \(C_K/C(\mathfrak n)\to C_K/C(\mathfrak m)\). Under (4) this is the ideal map induced by \(I^{\mathfrak n}\subseteq I^{\mathfrak m}\). Its surjectivity is already proved by the idèle quotient description; no unsupported assertion about moving ideals past new modulus primes is needed. \subsubsection{Exercise 4 --- Congruence subgroups (hard)}\label{exercise-4-congruence-subgroups-hard} Prove Proposition 18.5, including its converse and its compatibility with the idèlic correspondence. \textbf{Solution.} Given \(P_{\mathfrak m,1}\subseteq H\subseteq I^{\mathfrak m}\), use the isomorphism (4) to lift \(H/P_{\mathfrak m,1}\) to a subgroup \(\widetilde H\) of \(C_K\) containing \(C(\mathfrak m)\). It has finite index and is open, since the quotient ray group is finite and \(C(\mathfrak m)\) is open. Existence gives its unique abelian class field \(L\). Containment of norm groups puts \(L\) inside \(K_{\mathfrak m}\). On a prime away from the modulus the quotient map is the Frobenius map, by Proposition 18.1; therefore its ideal kernel is exactly \(H\), and its quotient is (14). Conversely for a subextension \(L\), take its norm group, pass to the quotient by \(C(\mathfrak m)\), and pull back through (4); these operations reverse the construction. They also reverse inclusions and agree with restriction and norm. Every finite abelian \(L\) has conductor (8), so the construction includes it after choosing that modulus. \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 local conductor criterion, optional real modulus, idèle-to-ideal ray identification, norm kernel, Takagi correspondence and decomposition law are proved with the stated fractional-ideal conventions. \begin{itemize} \tightlist \item \href{https://www.mathi.uni-heidelberg.de/~schmidt/Neukirch-en/Neukirch_cft_02_may15.pdf}{Jürgen Neukirch, Class Field Theory --- The Bonn Lectures, Online Edition 2.0 (May 2015), edited by Alexander Schmidt}. \item \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne, Class Field Theory, version 4.03}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Hilbert and ring class fields, and quadratic prime forms}\label{hilbert-and-ring-class-fields-and-quadratic-prime-forms} \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).} \textbf{Lesson 19.} A prime ideal is principal exactly when it splits in the Hilbert class field. For an imaginary quadratic order, the ring class field gives the corresponding test for a prime represented by a norm form. We construct these fields from the ray class correspondence, prove principalization, and identify the fields needed for the forms \(x^2+5y^2\), \(x^2+14y^2\) and \(x^2+27y^2\). The class field input is \href{../ray-class-fields-conductors-and-ideal-reciprocity.html}{Ray class fields, conductors and ideal reciprocity}. The arithmetic of quadratic ideal classes and orders is supplied by \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ANT-10}{Quadratic fields: ideal classes and binary quadratic forms}, Theorems 10.1--10.2 and Proposition 10.3, and \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ANT-11}{Orders in number fields and their Picard groups}, Theorems 11.3--11.4 and Corollary 11.5. Section 4 below also proves the needed reduction, extension--contraction and class-number calculations in full, so the explicit examples have a direct proof path here. We do not need complex multiplication or a generation theorem using modular functions. \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-19}{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. The ordinary and narrow Hilbert class fields}\label{the-ordinary-and-narrow-hilbert-class-fields} \textbf{Theorem 19.1.} A number field \(K\) has a unique maximal finite abelian extension \(H/K\) unramified at every place, including the real places, and \[ H=K_{(1)},\qquad \operatorname{Gal}(H/K)\simeq\operatorname{Cl}_K. \tag{1} \] A finite prime splits completely in \(H\) exactly when its ideal is principal. The \textbf{narrow Hilbert class field} \(H^+\) is maximal among abelian extensions unramified at the finite places, and \[ \operatorname{Gal}(H^+/K)\simeq\operatorname{Cl}_K^+. \tag{2} \] It is the ray field whose finite modulus is \((1)\) and whose real modulus includes all real places. Its real places may become complex. \textbf{Proof.} Theorem 18.4 constructs both fields. Proposition 18.3 says that absence of ramification at all places is equivalent to conductor \((1)\), so every such abelian extension lies in \(K_{(1)}\). Equation (4) of lesson 18 identifies this ray group with the ordinary ideal class group. In its ideal Artin map, the class of a prime is its Frobenius; triviality of that class is precisely complete splitting. Selecting every real place permits the sign quotient, while imposing full unit groups at every finite place forbids finite ramification. The same conductor criterion gives maximality of this field, and the positivity condition on principal generators identifies its ray group with the narrow class group. \(\square\) These fields are equal for imaginary number fields. In a real quadratic field, Proposition 10.3 of the quadratic-fields lesson gives \([H^+:H]=1\) when there is a unit of norm \(-1\), and \([H^+:H]=2\) otherwise. This statement about the two full class groups does not, by itself, give their quotients by squares; genus theory will require a separate argument. \subsection{2. Transfer and the principal ideal theorem}\label{transfer-and-the-principal-ideal-theorem} For a finite-index subgroup \(V\subset G\), choose representatives \(r\) of left cosets \(G/V\). If \(gr=r'h_r\), with \(h_r\in V\), the \textbf{transfer} is \[ \operatorname{Ver}:G^{\mathrm{ab}}\longrightarrow V^{\mathrm{ab}}, \qquad g\longmapsto\prod_r h_r\pmod{[V,V]}. \tag{3} \] The product is independent of the representatives and is multiplicative: in a product \(g_1g_2\), the cosets for \(g_2\) are merely permuted before applying \(g_1\), and changing representatives inserts cancelling factors. This is the transfer used in Proposition 5.2, equation (11), of \href{../the-reciprocity-law-and-the-class-field-correspondence.html}{The reciprocity law and the class field correspondence}. \subsubsection{The transfer to the commutator subgroup}\label{the-transfer-to-the-commutator-subgroup} We prove the group-theoretic theorem completely. For a group \(G\), let \(I_G\) be the augmentation ideal in \(\mathbf Z[G]\), and write \(\delta g=g-1\). \textbf{Transfer lemma.} If \(G\) is finitely generated and \(A=G/[G,G]\) is finite, then \[ \operatorname{Ver}:A\longrightarrow [G,G]/[[G,G],[G,G]] \quad\text{is trivial}. \tag{4} \] \textbf{Proof.} Put \(V=[G,G]\), \(R=\mathbf Z[A]\), and \(M=I_G/(I_GI_V)\), viewed as a right \(R\)-module. The action factors through \(A\): multiplying an element of \(I_G\) by a difference of representatives modulo \(V\) gives an element of \(I_GI_V\). The ring \(R\) is commutative. First identify transfer in this module. The abelian group \(T=\mathbf Z[G]I_V\) has basis \(r(h-1)\), for coset representatives \(r\) and \(h\in V\setminus\{1\}\). Send this basis element to the class of \(h\) in \(V^{\mathrm{ab}}\), using additive notation in the target. This map kills \(I_GI_V\). Indeed if \(gr=r'h'\), then \[ gr(h-1)=r'\bigl(\delta(h'h)-\delta h'\bigr), \] whose image is \([h'h]-[h']=[h]\), the same as the image of \(r(h-1)\). Their difference generates the required relations. Conversely \(h\mapsto\delta h\) modulo \(I_GI_V\) is a homomorphism, since \(\delta(hk)=\delta h+\delta k+\delta h\delta k\) and \(I_V^2\subseteq I_GI_V\). The two maps are inverse, because \(r\delta h\equiv\delta h\). Thus \[ T/(I_GI_V)\simeq V^{\mathrm{ab}}. \tag{5} \] Summing \(\delta(gr)=\delta g\,r+\delta r=\delta r'+r'\delta h_r\) over the representatives cancels the \(\delta r\) terms by permutation. Modulo \(I_GI_V\), it gives \[ \delta g\sum_r r\equiv\sum_r\delta h_r. \tag{6} \] Under (5), the right side is exactly (3). Let \(g_1,\ldots,g_n\) generate \(G\). The kernel of \(\mathbf Z^n\to A\), sending the standard basis to these generators, is a full lattice. Choose its basis matrix \((m_{ik})\) with positive determinant \(|A|\); integer lattice index gives this determinant. The words \(w_k=\prod_i g_i^{m_{ik}}\) belong to \(V\). The identities \[ \delta(xy)=\delta x\,y+\delta y, \qquad\delta(x^{-1})=-\delta x\,x^{-1} \tag{7} \] expand every word as a right-linear combination of the \(\delta g_i\). The augmentation of the coefficient of \(\delta g_i\) is its total exponent in the word. Each \(w_k\) also has an expression as a product of commutators of generator words, with total exponent zero for each generator. Subtract these two expansions of the same \(\delta w_k\) and pass to \(M\). We obtain a matrix \((\mu_{ik})\) over \(R\) with \[ \sum_i\delta g_i\,\mu_{ik}=0, \qquad \operatorname{aug}(\mu_{ik})=m_{ik}. \tag{8} \] This explains both the existence of the matrix and its augmentation, including negative exponents. Let \(\Delta=\det(\mu_{ik})\in R\). Multiplication by its adjugate gives \(\delta g_i\Delta=0\) in \(M\) for every \(i\). These \(\delta g_i\) generate \(I_G\) as a right module, by (7), so \(\delta g\Delta=0\) for all \(g\). Projecting to \(R\) gives \((a-1)\Delta=0\) for all \(a\in A\). Thus the coefficients of \(\Delta\) are all equal: translation by \(A\) permutes its basis regularly. Hence \[ \Delta=c\sum_{a\in A}a, \qquad c|A|=\operatorname{aug}(\Delta)=\det(m_{ik})=|A|, \] so \(c=1\). Therefore \(\delta g\sum_r r=0\) in \(M\). Equations (5)--(6) make every transfer value trivial, proving (4). \(\square\) \textbf{Theorem 19.2 (principal ideal theorem).} Every fractional ideal of \(K\) becomes principal in its ordinary Hilbert class field \(H\). \textbf{Proof.} Let \(H_2\) be the Hilbert class field of \(H\). Uniqueness makes \(H_2/K\) Galois: every conjugation of \(H\) sends its maximal abelian everywhere-unramified extension to itself. Unramified extensions compose in towers, and real places still split, so \(H_2/K\) is everywhere unramified. Its maximal abelian subextension is \(H\), by Theorem 19.1. Consequently for \(G=\operatorname{Gal}(H_2/K)\), \[ V=\operatorname{Gal}(H_2/H)=[G,G], \] and \(V\) is abelian. Reciprocity identifies \(\operatorname{Cl}_K\) with \(G/V\) and \(\operatorname{Cl}_H\) with \(V\). Inclusion of idèle classes corresponds to transfer by Proposition 5.2; passing to their ordinary ideal quotients makes the same map extension of ideals \(\mathfrak a\mapsto\mathfrak a\mathcal O_H\). Lemma (4) makes this map zero. Thus the extended ideal has trivial class in \(\operatorname{Cl}_H\), which is exactly principality. \(\square\) The ideal map used here is compatible with extension because a finite local component inserted into all completions above it has valuations multiplied by the ramification indices. The associated ideal is therefore the extended ideal. This verifies the arithmetic interpretation of the transfer, not just the formal Galois reduction. \subsection{3. Quadratic genus fields and the real-field distinction}\label{quadratic-genus-fields-and-the-real-field-distinction} Let \(K=\mathbf Q(\sqrt D)\), with nontrivial fundamental discriminant \(D\). Factor it into prime discriminants \[ D=d_1\cdots d_t, \qquad p^*=(-1)^{(p-1)/2}p\text{ for odd }p, \tag{9} \] with the single dyadic factor, if present, equal to \(-4\), \(8\), or \(-8\). The integer \(t\) is the number of rational primes dividing \(|D|\). These signed discriminants include each prime exactly once. \textbf{Proposition 19.3 (genus field).} The maximal subfield of \(H^+\) abelian over \(\mathbf Q\) is \[ M^+=\mathbf Q(\sqrt{d_1},\ldots,\sqrt{d_t}), \qquad \operatorname{Gal}(M^+/K)\simeq \operatorname{Cl}_K^+/(\operatorname{Cl}_K^+)^2 \simeq(\mathbf Z/2)^{t-1}. \tag{10} \] For imaginary \(K\), this is also the ordinary genus field. For real \(K\), the ordinary genus field is the maximal totally real subfield of \(M^+\). Consequently \[ |\operatorname{Cl}_K/\operatorname{Cl}_K^2|= \begin{cases} 2^{t-1},&D<0,\\ 2^{t-1},&D>0\text{ and every }d_i>0,\\ 2^{t-2},&D>0\text{ and some }d_i<0. \end{cases} \tag{11} \] The last case has at least two negative factors, so its exponent is nonnegative. \textbf{Proof.} The square classes of the \(d_i\) are independent: each odd prime valuation detects its own factor, and the remaining dyadic square class is nontrivial. Kummer theory gives \([M^+:\mathbf Q]=2^t\), and (9) places \(K\) in \(M^+\). At a finite prime, only one of the quadratic fields \(\mathbf Q(\sqrt{d_i})\) ramifies, and its local ramified extension has degree \(2\); all the other factors are unramified. Thus the absolute inertia in their compositum has order \(2\) and maps nontrivially to the inertia in \(K/\mathbf Q\). Its kernel over \(K\) is trivial. This holds at \(2\) as well: odd fundamental discriminants give unramified or split \(2\)-adic extensions, and the unique dyadic factor supplies the ramified one. Therefore \(M^+/K\) is unramified at every finite place, and is abelian, so it lies in \(H^+\). Put \(A=\operatorname{Cl}_K^+\). Conjugation of ideals induces inversion on \(A\), since \(\mathfrak a\bar{\mathfrak a}=(N\mathfrak a)\) has a positive rational generator. The narrow class field is stable under the nontrivial automorphism of \(K/\mathbf Q\), by uniqueness. Choose a finite prime ramified in \(K\). Since \(H^+/K\) is unramified there, its absolute inertia has order \(2\) and maps onto \(\operatorname{Gal}(K/\mathbf Q)\). This supplies an order-two lift of that automorphism. Thus \[ \operatorname{Gal}(H^+/\mathbf Q)=A\rtimes\mathbf Z/2, \] where the lift acts by inversion. Its commutator subgroup is \(A^2\), as the commutators with the lift are precisely the squares. The maximal abelian subfield is therefore multiquadratic, with its relative Galois group \(A/A^2\). We show that every quadratic subfield of this abelian subfield is already in \(M^+\). At a ramified finite prime its character on local units is either the character of \(K\) or trivial, because absolute inertia has order \(2\) and maps isomorphically to that of \(K\). At an unramified prime it is trivial on units. Select exactly those prime discriminants \(d_i\) where the first choice occurs. The quadratic character of their product has exactly the same restrictions on every finite unit group: all other prime discriminants are locally unramified there. The quotient of these characters defines an extension of \(\mathbf Q\) unramified at every finite prime. Such an abelian extension is trivial, since the narrow Hilbert class group of \(\mathbf Q\) is trivial---every rational ideal has a positive generator. Hence the two characters coincide. This proves maximality of \(M^+\), and its already computed degree gives (10). For real \(K\), the ordinary Hilbert field requires all real places to remain real. Thus its maximal abelian-over-\(\mathbf Q\) subfield is exactly the real part of \(M^+\): that real part is unramified at every place over \(K\), and every ordinary genus field lies in it. If all \(d_i\) are positive, nothing is removed. Otherwise complex conjugation is a nontrivial order-two element on \(M^+\), so its real part has half its degree. Applying the same inversion-and-commutator calculation to the ordinary class group proves (11). The imaginary case has no real place condition and \(H=H^+\). \(\square\) For a real quadratic field, the middle case of (11) is also equivalent to \(-1\in N_{K/\mathbf Q}K^\times\). The Hilbert symbol at each odd ramified prime tests \((-1/p)\); it is \(1\) exactly when \(p\equiv1\pmod4\). If all these tests pass, the \(2\)-adic formula of lesson 11 also gives \(1\), and infinity gives \(1\). Hasse's cyclic norm theorem then supplies the global norm. These conditions are precisely positivity of all the factors in (9). The assertion concerns a norm of a field element, which need not be a unit. For example \(D=12=(-4)(-3)\), so \(M^+=\mathbf Q(i,\sqrt{-3})\). The field \(K=\mathbf Q(\sqrt3)\) has narrow genus quotient of order \(2\), but ordinary genus quotient of order \(1\). This agrees with its ordinary class number \(1\) and narrow class number \(2\). It shows why a formula \(2^{t-1}\) for every ordinary quadratic class group would be false. \subsection{4. Ring class fields of imaginary quadratic orders}\label{ring-class-fields-of-imaginary-quadratic-orders} Let \(K\) be imaginary quadratic and \(\mathcal O=\mathbf Z+f\mathcal O_K\) an order of conductor integer \(f\geq1\). Its Picard group consists of invertible fractional \(\mathcal O\)-ideals modulo principal ideals. Theorem 11.3 of the orders lesson proves extension and contraction of ideals prime to \(f\), with equal norms, and that every invertible class has such a representative. This identifies \[ \operatorname{Pic}(\mathcal O)\simeq I_K^{(f)}/P_{K,\mathbf Z}(f), \tag{12} \] where \(P_{K,\mathbf Z}(f)\) is generated by principal ideals \((\alpha)\) with \(\alpha\equiv a\pmod{f\mathcal O_K}\) locally at \(f\), for an integer \(a\) coprime to \(f\). Here is the kernel verification in (12). A prime-to-conductor ideal has trivial Picard class exactly when its contraction is principal. At every conductor prime, a generator of such a principal ideal must be a unit in the localized order. Its residue is therefore in \((\mathcal O/f\mathcal O_K)^\times=(\mathbf Z/f)^\times\). Conversely a generator with such a residue is an order unit at those primes, so contraction of its full-ring principal ideal agrees locally everywhere with its order principal ideal. The equality follows from local detection of ideal membership, proved in the orders lesson. Fractional generators may have denominators outside \(f\); their residue is still represented by an integer coprime to \(f\). Passing to ratios gives exactly the generated subgroup in (12). When \(f=1\), these conditions impose no restriction. \textbf{Theorem 19.4.} There is a unique finite abelian \textbf{ring class field} \(L_{\mathcal O}/K\) whose ideal Artin kernel in \(I_K^{(f)}\) is \(P_{K,\mathbf Z}(f)\), and \[ \operatorname{Gal}(L_{\mathcal O}/K)\simeq\operatorname{Pic}(\mathcal O). \tag{13} \] It is Galois over \(\mathbf Q\), with \[ \operatorname{Gal}(L_{\mathcal O}/\mathbf Q) \simeq\operatorname{Pic}(\mathcal O)\rtimes\mathbf Z/2, \tag{14} \] where the nontrivial element acts by inversion. Its conductor over \(K\) divides \(f\mathcal O_K\). The maximal-order case is the Hilbert class field. \textbf{Proof.} The subgroup in (12) contains the ray principal subgroup modulo \(f\), and has finite index by the Picard finiteness theorem, Theorem 11.4 of the orders lesson. Takagi's correspondence, Proposition 18.5, therefore constructs the field and gives (13), with the stated conductor bound. Complex conjugation preserves both the modulus and the subgroup in (12). Uniqueness and conjugation compatibility of reciprocity imply \(\overline{L_{\mathcal O}}=L_{\mathcal O}\), so the field is Galois over \(\mathbf Q\). Complex conjugation is an order-two automorphism restricting nontrivially to \(K\), and therefore splits its Galois exact sequence. On ideal classes it acts by inversion: \(\mathfrak a\bar{\mathfrak a}=(N\mathfrak a)\), with rational generator prime to \(f\), is trivial in (12). This proves (14). For \(f=1\), (12) is the ordinary ideal class group and Theorem 19.1 identifies the field. \(\square\) \subsubsection{Order arithmetic used in the examples}\label{order-arithmetic-used-in-the-examples} Here are the ideal and form arguments needed in sections 4--8. They also give a direct reading route through those examples. For a freely readable comparison of proper ideals and their inverses, see A. Sutherland, \href{https://math.mit.edu/classes/18.783/2023/LectureNotes17.pdf}{\emph{The CM torsor}, §§17.1--17.3, Theorem 17.10}. Only the order arithmetic is used here. \textbf{Ideals away from the conductor.} Put \(B=\mathcal O_K\) and \(O=\mathbf Z+fB\). Then \(fB\subset O\). At a rational prime \(\ell\nmid f\), localization therefore gives \(O\otimes\mathbf Z_\ell=B\otimes\mathbf Z_\ell\). An integral ideal prime to \(f\) has localization equal to the full ring at every \(\ell\mid f\). Thus extension and contraction preserve its localizations everywhere and are mutually inverse. Equality of these lattices can be detected locally: after clearing a common denominator, a nonzero quotient is a finitely generated abelian group and has a nonzero localization at some prime. The same local argument proves preservation of ideal norms, since the quotients at conductor primes are zero and the other quotients agree. Every invertible fractional \(O\)-ideal has a representative prime to \(f\). Indeed an invertible ideal is locally principal: if \(II^{-1}=O\), write \(1=\sum a_jb_j\), and at a maximal ideal some \(a_jb_j\) is a unit; this \(a_j\) generates the localized ideal. Over \(O\otimes\mathbf Z_\ell\), choose one generator by the Chinese remainder theorem on its finitely many maximal ideals. For the finitely many \(\ell\mid f\), let these generators be \(\lambda_\ell\). Choose \(a\in K^\times\) sufficiently close to \(\lambda_\ell^{-1}\) at all those primes. Such approximation follows by clearing denominators and applying the Chinese remainder theorem to the rank-two integer lattice \(B\); the required congruences are open and nonempty. Then \(aI\) equals the local order at every conductor prime. Multiplying by an integer prime to \(f\) clears its remaining denominators. This gives the desired integral representative. The argument applies as well to fractional \(B\)-ideals. It proves the representative and extension--contraction assertions used in (12). There is also an exact sequence \[ B^\times\longrightarrow \frac{(B/fB)^\times}{(\mathbf Z/f\mathbf Z)^\times} \longrightarrow\operatorname{Pic}(O) \longrightarrow\operatorname{Cl}(B)\longrightarrow1. \] For \(f=1\), the middle residue group is trivial. For \(f>1\), lift a residue unit to \(\alpha\in B\) prime to \(f\) and contract \(\alpha B\); its invertible order-ideal class defines the middle map. Two lifts have quotient congruent to 1 locally modulo \(fB\), hence their contracted ideals differ by an order principal ideal. Multiplication by an integer residue unit has the same property. Surjectivity on the right follows from the prime-to-conductor representatives. A class in its kernel extends to \(\alpha B\); after clearing denominators prime to \(f\), \(\alpha\) gives a residue unit, so the class comes from the middle map. Finally this class is order principal exactly when a generator can be changed by a \(B\)-unit to be an order unit at the conductor. Its residue is then an integer unit modulo \(f\). This is exactly the image of the first map, proving exactness. These arguments also verify the principal subgroup in (12), including fractional generators. Norms of invertible ideals are multiplicative. Locally such an ideal is generated by one element; multiplication by that element has determinant its field norm on the rank-two lattice. The local index of a product therefore multiplies by the same determinant. Comparing indices at every rational prime proves the global identity, and clearing denominators extends it to fractional ideals. \textbf{Forms and ideal classes.} Let \(D<0\) be the discriminant of \(O\), choose \(\sqrt D\) with positive imaginary part, and orient ideal bases by that embedding. A primitive positive form \[ Q(x,y)=ax^2+bxy+cy^2,\qquad b^2-4ac=D, \] gives the ideal \(I=[a,(b+\sqrt D)/2]\). Its norm is \(a\), because its second generator differs from a fixed generator of \(O\) by an integer. If \(\tau=(b+\sqrt D)/(2a)\), its primitive integral polynomial is \(aX^2-bX+c\). An element preserving \([1,\tau]\) is \(m+n\tau\), and preserving its product with \(\tau\) requires \(a\mid nb\) and \(a\mid nc\). As \(\gcd(a,b,c)=1\), these two requirements imply \(a\mid n\). Thus the multiplier ring is \([1,a\tau]=O\). Further, \[ I\bar I=a[a,(b+\sqrt D)/2,(b-\sqrt D)/2,c]=aO: \] the integer generators \(a,b,c\) span 1. Hence \(I\) is invertible and \(I^{-1}=\bar I/a\). Its normalized norm form is exactly \(Q\). Conversely, for an invertible ideal with oriented basis \((\alpha,\beta)\), \[ Q_I(x,y)=\frac{N_{K/\mathbf Q}(x\alpha+y\beta)}{N(I)} \] is integral, primitive and positive, with discriminant \(D\). Integrality follows because \((z)I^{-1}\) is an integral ideal for \(z\in I\), and its norm is \(N(z)/N(I)\). For primitivity, at each prime \(\ell\) choose an element of \(I\) congruent to a local generator modulo \(\ell I\). Its normalized norm is an \(\ell\)-adic unit. No prime can therefore divide every value, or all three coefficients, of \(Q_I\). The determinant of the ideal basis has absolute value \(N(I)\) times that of the order basis, so the discriminant of its norm form is \(D\). Write its first two coefficients as \(a,b\). The orientation gives \(\beta/\alpha=(b+\sqrt D)/(2a)\), so scaling the ideal by \(a/\alpha\) recovers \([a,(b+\sqrt D)/2]\). Changing an oriented basis is an \(\mathrm{SL}_2(\mathbf Z)\) change of variables. Multiplying the ideal by \(\lambda\in K^\times\) preserves orientation, scales its norm by \(N(\lambda)>0\), and leaves the normalized form unchanged. The two constructions just given are inverse. Thus proper form classes are precisely invertible ideal classes. Conjugation replaces \(b\) by \(-b\) and takes the inverse class, by \(I\bar I=aO\). \textbf{Reduction and the three maximal-order class numbers.} A shear first puts \(|b|\leq a\). If \(c0\), \(K=\mathbf Q(\sqrt{-n})\), and \(\mathcal O=\mathbf Z[\sqrt{-n}]\). This is an order even when \(n\) has square factors. Its discriminant is \(-4n=f^2D_K\), and let \(L\) be its ring class field. \textbf{Theorem 19.5.} For an odd prime \(p\nmid n\), \[ p=x^2+ny^2\text{ with }x,y\in\mathbf Z \quad\Longleftrightarrow\quad p\text{ splits completely in }L. \tag{15} \] There is a real algebraic integer \(\alpha\) with \(L=K(\alpha)\). Its minimal polynomial \(f_n\in\mathbf Z[X]\) has degree \(|\operatorname{Pic}(\mathcal O)|\), and for odd \(p\nmid n\operatorname{disc}(f_n)\), \[ p=x^2+ny^2 \quad\Longleftrightarrow\quad (-n/p)=1\text{ and }f_n\text{ has a root in }\mathbf F_p. \tag{16} \] \textbf{Proof.} Since \(p\nmid-4n\), the prime is unramified in \(K\) and prime to the conductor of \(\mathcal O\). A representation gives \(\beta=x+y\sqrt{-n}\in\mathcal O\) of norm \(p\). Its principal ideal has norm \(p\), by the ideal norm and extension--contraction theorem, so it is a prime ideal above a split rational prime. Its invertible order ideal class is trivial. Conversely a split prime whose contracted order ideal is principal has a generator \(\beta\in\mathcal O\) with norm \(p\): the ideal is contained in \(\mathcal O\), hence its generator belongs to it, and the imaginary norm is positive. Writing \(\beta=x+y\sqrt{-n}\) gives the representation. By (13) triviality of this class means trivial Frobenius in \(L/K\). The conjugate prime has inverse class and is trivial as well. As \(L/\mathbf Q\) is Galois, this is complete rational splitting, proving (15). Let \(c\) be complex conjugation in (14), and \(R=L^{\langle c\rangle}\). It is a real field of degree \(|\operatorname{Pic}(\mathcal O)|\) over \(\mathbf Q\). A primitive element can be chosen integral: for two separable generators, all but finitely many rational linear combinations distinguish their embeddings; repeat and clear a rational denominator. Choose such an \(\alpha\in R\). Since \(K\) is imaginary quadratic and \(R\) is real, \(K\cap R=\mathbf Q\), so \(KR=L\). Thus \(L=K(\alpha)\) and its monic minimal polynomial over \(\mathbf Q\) has the asserted degree and integer coefficients. At a prime avoiding the polynomial discriminant, its roots have distinct reductions in the splitting field. Frobenius fixes a reduced root exactly when that root lies in \(\mathbf F_p\); distinctness makes this equivalent to fixing its algebraic conjugate. If \((-n/p)=1\), Frobenius lies in \(A=\operatorname{Gal}(L/K)\). Its action on the \(A\)-orbit of \(\alpha\) is regular, because \(\alpha\) generates \(L/K\). Therefore fixing any one of those conjugates forces Frobenius to be the identity. The full \(\mathbf Q\)-orbit is that same orbit, by its degree. Combining this with (15) proves (16). The discriminant exclusion is necessary for this distinct-root argument. \(\square\) \subsection{6. The forms with coefficients five and fourteen}\label{the-forms-with-coefficients-five-and-fourteen} The quadratic-fields lesson enumerates the two reduced forms of discriminant \(-20\), so \(K=\mathbf Q(\sqrt{-5})\) has class number \(2\). Its prime discriminants are \(-4\) and \(5\). Proposition 19.3 gives a genus field of relative degree \(2\), already the whole Hilbert class field: \[ H=\mathbf Q(\sqrt{-5},i)=\mathbf Q(i,\sqrt5). \tag{17} \] For an odd prime \(p\ne5\), complete splitting in this biquadratic field means \((-1/p)=(5/p)=1\). The quadratic reciprocity law proved in lesson 16 gives \((5/p)=(p/5)\). Thus (15) gives \[ p=x^2+5y^2\quad\Longleftrightarrow\quad p\equiv1,9\pmod{20}. \tag{18} \] The conditions are \(p\equiv1\pmod4\) and \(p\equiv\pm1\pmod5\); the Chinese remainder theorem gives the two displayed residues. The excluded prime \(5\) is represented by \(0^2+5\cdot1^2\), whereas \(2\) is not represented. Thus for every rational prime the condition is \(p=5\) or \(p\equiv1,9\pmod{20}\). The restriction in (18) must not be omitted. For \(K=\mathbf Q(\sqrt{-14})\), the reduced forms of discriminant \(-56\) give class group \(\mathbf Z/4\), as proved in the quadratic-fields lesson. We claim \[ H=K(\alpha),\qquad \alpha=\sqrt{2\sqrt2-1}>0, \qquad f_{14}(X)=X^4+2X^2-7. \tag{19} \] We verify degree, cyclicity and unramifiedness, including the dyadic place. The norm of \(-1+2\sqrt2\) from \(\mathbf Q(\sqrt2)\) is \(-7\), so it is not a square there. Hence \([\mathbf Q(\alpha):\mathbf Q]=4\). This field is real and intersects the imaginary quadratic \(K\) in \(\mathbf Q\); therefore \([K(\alpha):K]=4\). The field \(L=K(\alpha)\) contains \[ \sqrt2=(\alpha^2+1)/2,\qquad \sqrt{-7}=\sqrt{-14}/\sqrt2,\qquad \beta=\sqrt{-7}/\alpha, \] with \(\beta^2=-1-2\sqrt2\). Thus it contains all four roots \(\pm\alpha,\pm\beta\) of \(f_{14}\), so it is a splitting field. An automorphism over \(K\) taking \(\alpha\) to \(\beta\) takes \(\sqrt2\) and \(\sqrt{-7}\) to their negatives; it consequently takes \(\beta\) to \(-\alpha\). It has order \(4\), proving that \(L/K\) is cyclic. The resultant with \(f_{14}'=4X(X^2+1)\) gives \[ \operatorname{disc}(f_{14})=-2^{14}\cdot7. \tag{20} \] Indeed the product of its roots is \(-7\), while the product of their \((X^2+1)\) values is \(f_{14}(i)f_{14}(-i)=64\), and the derivative contributes \(4^4\). Therefore no finite prime except \(2\) or \(7\) can ramify in the splitting field. This follows also directly from distinct residue roots: inertia fixes each integral root's residue, and when the discriminant is a unit the roots have distinct residues, forcing inertia to fix all roots. At \(7\), choose the \(7\)-adic square root \(\sqrt2\equiv3\pmod7\), supplied by Hensel's lemma. Then \(\alpha^2\equiv5\pmod7\) is a nonsquare unit, so adjoining \(\alpha\) is unramified quadratic. The completion of \(K\) is \(\mathbf Q_7(\sqrt{-7})\), and that of \(L\) is obtained by adjoining this unramified radical. Base change remains unramified. At \(2\), \(-7\equiv1\pmod8\) is a square in \(\mathbf Q_2\), so \(K_v=E=\mathbf Q_2(\sqrt2)\). Put \(u=-1+2\sqrt2\). The polynomial \[ 25r^2+5r+2 \] has value \(32\) and odd derivative \(55\) at \(r=1\). Hensel's lemma gives a root \(r\equiv1\pmod8\), which is a square \(a^2\) in \(\mathbf Q_2\) by the odd-unit square criterion. Set \(b=1/(5a)\). Its defining equation gives \[ a^2+2b^2=-1/5,\qquad ab=1/5, \qquad (a+b\sqrt2)^2=u/5. \tag{21} \] Consequently \(E(\alpha)=E(\sqrt5)\), the base change of the unramified quadratic extension of \(\mathbf Q_2\). It is unramified over \(E\). All infinite places of \(K\) are complex. We have proved that \(L/K\) is everywhere unramified and cyclic of degree \(4\). The Hilbert class field has degree \(4\), so \(L=H\), proving (19). Equations (16), (19) and (20) now give the explicit test, for odd \(p\ne7\), \[ p=x^2+14y^2\quad\Longleftrightarrow\quad (-14/p)=1\text{ and }X^4+2X^2-7\text{ has a root modulo }p. \tag{22} \] \subsection{\texorpdfstring{7. The cubic criterion for \(x^2+27y^2\)}{7. The cubic criterion for x\^{}2+27y\^{}2}}\label{the-cubic-criterion-for-x227y2} The order \(\mathcal O=\mathbf Z[\sqrt{-27}]\) has conductor \(6\) in \(K=\mathbf Q(\sqrt{-3})\) and Picard group of order \(3\), computed in Corollary 11.5 and Exercise 3 of the orders lesson. Its ring class field \(L\) is therefore cyclic cubic over \(K\), and (14) makes \(L/\mathbf Q\) an \(S_3\)-extension. Its finite ramification is supported above \(2\) and \(3\). We identify \(L\) without a table of class fields. Let \(\sigma\) generate \(\operatorname{Gal}(L/K)\), let \(\zeta\) be a primitive cube root of unity, and choose a real integral primitive element \(\alpha\), as in Theorem 19.5. Define \[ u_j=\alpha+\zeta^j\sigma^{-1}\alpha+\zeta^{2j}\sigma^{-2}\alpha, \qquad j=0,1,2. \tag{23} \] Then \(\sigma u_j=\zeta^j u_j\). Complex conjugation fixes \(\alpha\), inverts \(\zeta\), and changes \(\sigma\) into \(\sigma^{-1}\); it interchanges the last two terms and fixes each \(u_j\). Thus \(u_j^3\) is a real element of \(K\), hence a rational integer. At least one of \(u_1,u_2\) is nonzero: otherwise the inverse three-term Fourier transform would give \(3\alpha=u_0\), making \(\alpha\) rational. A nonzero such \(u_j\) is not in \(K\), by its nontrivial eigenvalue, and generates \(L/K\). Removing integer cube factors and changing its sign gives \[ L=K(\sqrt[3]m),\qquad m>0\text{ cube-free}. \tag{24} \] Every rational prime dividing \(m\) ramifies in this radical extension. Its valuation in \(K\) is its exponent \(1\) or \(2\), multiplied by the base ramification index, which is \(2\) only at \(3\). This is never divisible by \(3\). A cube root therefore has fractional valuation and forces relative ramification. Since only \(2,3\) can ramify, the possible \(m\)'s are \(2,3,4,6,9,12,18,36\). Replacing a radical by its square pairs these into four fields: \[ K(\sqrt[3]2),\quad K(\sqrt[3]3),\quad K(\sqrt[3]6),\quad K(\sqrt[3]{12}). \tag{25} \] But \(31=2^2+27\) splits completely in \(L\), by (15). It splits in \(K\), since \(11^2\equiv-3\pmod{31}\). Its nonzero cubic residues are precisely \[ 1,2,4,8,15,16,23,27,29,30. \] At a split prime away from \(6\), the extension in (24) splits locally exactly when \(m\) is a cube in \(\mathbf F_{31}\): the simple residue root lifts, and \(\mu_3\) is already present. This excludes \(m=3,6,12\) in (25). Therefore \[ L=K(\sqrt[3]2),\qquad f_{27}(X)=X^3-2. \tag{26} \] Its polynomial discriminant is \(-108\). For \(p>3\), the quadratic reciprocity law of lesson 16 gives \((-3/p)=1\) exactly when \(p\equiv1\pmod3\). The root test (16) consequently proves Gauss's criterion \[ p=x^2+27y^2\quad\Longleftrightarrow\quad p\equiv1\pmod3\text{ and }2\text{ is a cube modulo }p. \tag{27} \] The exclusion of \(2,3\) agrees with the explicit discriminant calculation. \subsection{\texorpdfstring{8. A cubic generator for the Hilbert field of discriminant \(-23\)}{8. A cubic generator for the Hilbert field of discriminant -23}}\label{a-cubic-generator-for-the-hilbert-field-of-discriminant--23} Reduction gives the three forms \((1,1,6),(2,1,3),(2,-1,3)\) of discriminant \(-23\), so \(K=\mathbf Q(\sqrt{-23})\) has class number \(3\). Let \(\alpha\) be the real root of \[ f(X)=X^3-X-1. \tag{28} \] This cubic has no root modulo \(2\), hence is irreducible over \(\mathbf Q\), and its discriminant is \(-23\). Its splitting field \(S\) has Galois group \(S_3\): a transitive cubic group is \(C_3\) or \(S_3\), and the square root of the discriminant changes by the sign of a root permutation. Its nonsquare discriminant excludes \(C_3\), and its quadratic fixed field is exactly \(K\). Hence \(S=K(\alpha)\) and \([S:K]=3\). Outside \(23\), distinct root reductions show that \(S/\mathbf Q\) is unramified. At \(23\), \[ f(X)\equiv(X-10)^2(X-3)\pmod{23}. \] The simple root \(3\) lifts to \(\mathbf Q_{23}\), so the local decomposition group fixes one cubic root and is contained in its order-two stabilizer. The quadratic field \(K\) is ramified there, since its discriminant is \(-23\). Thus absolute inertia has order \(2\) and maps nontrivially to \(K\); its relative kernel over \(K\) is trivial. Therefore \(S/K\) is unramified at \(23\) as well. There are no real places of \(K\). The degree-three abelian extension \(S/K\) is everywhere unramified and has the Hilbert field's degree, proving \[ H(\mathbf Q(\sqrt{-23}))=K(\alpha). \tag{29} \] \subsection{9. Iterating the Hilbert field}\label{iterating-the-hilbert-field} Starting with \(K_0=K\), set \(K_{j+1}=H(K_j)\). The principal ideal theorem proved in section 2 makes the ideal classes of \(K_j\) principal in \(K_{j+1}\). This allows further class groups and further Hilbert fields. \href{brauer-groups-of-local-and-global-fields.md\#theorem-24-9-an-infinite-hilbert-class-field-tower}{Brauer groups of local and global fields, Theorem 24.9} proves that the tower of \[ K=\mathbf Q\bigl(\sqrt{-5\cdot13\cdot17\cdot29\cdot37\cdot41}\bigr) \] is infinite. Its proof combines the genus construction of section 3 with the complete filtered Golod--Shafarevich proof and the unit-cohomology relation bound developed in that lesson. Follow that theorem after the cohomology has been established. \subsection{10. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- A quadratic Hilbert field (easy)}\label{exercise-1-a-quadratic-hilbert-field-easy} Show that \(H(\mathbf Q(\sqrt{-5}))=\mathbf Q(\sqrt{-5},i)\). \textbf{Solution.} The two reduced positive forms of discriminant \(-20\) are \((1,0,5)\) and \((2,2,3)\), so the class number is \(2\), by the quadratic-fields lesson. The prime discriminants are \(-4\) and \(5\). Proposition 19.3 proves that \(M=\mathbf Q(i,\sqrt5)\) is unramified over \(K=\mathbf Q(\sqrt{-5})\) and has relative degree \(2\). It lies in the Hilbert class field by Theorem 19.1. The latter also has degree \(2\), so they are equal. Since \(i\sqrt5=\sqrt{-5}\), the two descriptions of \(M\) agree. \subsubsection{Exercise 2 --- The congruence test and its exceptional prime (medium)}\label{exercise-2-the-congruence-test-and-its-exceptional-prime-medium} For an odd prime \(p\ne5\), prove \(p=x^2+5y^2\Longleftrightarrow p\equiv1,9\pmod{20}\). Then include the primes excluded by this statement. \textbf{Solution.} The preceding exercise identifies the Hilbert field with the biquadratic field \(\mathbf Q(i,\sqrt5)\). For \(p\nmid10\), Theorem 19.5 says that representation is equivalent to complete splitting there. Its two independent quadratic characters give \((-1/p)=(5/p)=1\). The first condition is \(p\equiv1\pmod4\). Quadratic reciprocity gives \((5/p)=(p/5)\), so the second is \(p\equiv1\) or \(4\pmod5\). Solving these two simultaneous congruences gives \(p\equiv1\) or \(9\pmod{20}\). Finally \(5=0^2+5\cdot1^2\), while a representation of \(2\) would have \(y=0\) and then require an integer square equal to \(2\). Thus the condition for all primes is \(p=5\) or those two residue classes. \subsubsection{Exercise 3 --- A cubic Hilbert-field generator (medium)}\label{exercise-3-a-cubic-hilbert-field-generator-medium} For \(K=\mathbf Q(\sqrt{-23})\), show that adjoining the real root of \(X^3-X-1\) gives its Hilbert class field. \textbf{Solution.} The reduced-form bound \(a\leq\sqrt{23/3}<3\) leaves \((1,1,6),(2,1,3),(2,-1,3)\), so \(h_K=3\). The cubic has no root modulo \(2\), hence is irreducible, and discriminant \(-23\). Its splitting field has group \(S_3\), with quadratic fixed field \(K\); adjoining one cubic root to \(K\) is the entire splitting field, cyclic cubic over \(K\). It is unramified away from \(23\), since the root differences have unit discriminant there. At \(23\) the factorization \((X-10)^2(X-3)\) has the simple root \(3\), which Hensel lifts. The local decomposition group therefore lies in a stabilizer of order \(2\). Its inertia maps nontrivially to the ramified quadratic field \(K\), so relative inertia over \(K\) is trivial. The field is everywhere unramified over \(K\), which has no real places. Its degree is the Hilbert degree \(3\), proving the assertion. \subsubsection{Exercise 4 --- The transfer theorem (hard)}\label{exercise-4-the-transfer-theorem-hard} Prove that transfer from a finitely generated group with finite abelianization to the abelianization of its commutator subgroup is trivial. Explain its use in Theorem 19.2. \textbf{Solution.} Set \(V=[G,G]\), \(A=G/V\), \(R=\mathbf Z[A]\) and \(M=I_G/(I_GI_V)\). The map \(r(h-1)\mapsto[h]\) identifies \(\mathbf Z[G]I_V/(I_GI_V)\) with \(V^{\mathrm{ab}}\); left multiplication merely permutes cosets and does not change \([h]\). Under this identification, \(\delta g\sum_r r\) is transfer, as the coset-permutation terms cancel in (6). Choose generators \(g_i\) and a basis \((m_{ik})\) of the relation lattice \(\ker(\mathbf Z^n\to A)\), with determinant \(|A|\). Expand \(\prod_i g_i^{m_{ik}}\) both as that word and as a product of commutators. Their difference, using (7), gives \(\sum_i\delta g_i\mu_{ik}=0\) in \(M\), with \(\operatorname{aug}(\mu_{ik})=m_{ik}\). The adjugate of this matrix annihilates each \(\delta g_i\) by \(\Delta=\det(\mu_{ik})\), and hence annihilates every \(\delta g\). Projection to \(R\) gives \((a-1)\Delta=0\) for each \(a\in A\), so \(\Delta=c\sum_A a\). Taking augmentation gives \(c|A|=|A|\), hence \(c=1\). Transfer is therefore zero. Each step applies to negative exponents by the inverse identity in (7). For \(G=\operatorname{Gal}(H_2/K)\), the commutator subgroup is \(\operatorname{Gal}(H_2/H)\), which is abelian. The ideal-extension map \(\operatorname{Cl}_K\to\operatorname{Cl}_H\) is this transfer by reciprocity. Its vanishing means that each extended ideal is principal. The argument does not assert that \(\operatorname{Cl}_H\) itself vanishes. \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} Sections 1--3 prove Hilbert fields, the augmentation-ideal transfer theorem, principalization and the ordinary/narrow genus distinction. Section 4 proves order arithmetic and reduction. Sections 6--8 verify the explicit quadratic prime-form fields by degrees, ramification and residue calculations; lesson 24 supplies the tower arguments. \begin{itemize} \tightlist \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \item \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne, Class Field Theory, version 4.03}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Kronecker--Weber and the maximal abelian extension of the rationals}\label{kroneckerweber-and-the-maximal-abelian-extension-of-the-rationals} \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).} \textbf{Lesson 20.} Every finite abelian extension of the rationals is contained in a cyclotomic field. The ray fields of \(\mathbf Q\) make this theorem a direct consequence of global class field theory. We identify those fields, compute the Artin map on arbitrary rational idèles, and derive the conductor--discriminant formula. These identifications also specify the Galois action on the roots of unity appearing in the Bost--Connes system. We use \href{../ray-class-fields-conductors-and-ideal-reciprocity.html}{Ray class fields, conductors and ideal reciprocity}, the explicit local cyclotomic symbol in Corollary 9.5 of \href{../explicit-local-reciprocity-and-existence.html}{Explicit local reciprocity and existence}, and the local conductor--discriminant theorem, Theorem 10.5 of \href{../abelian-ramification-conductors-and-hasse-arf.html}{Abelian ramification, conductors and Hasse--Arf}. Theorem 12.1 and Theorem 12.3 of \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ANT-12}{Cyclotomic fields} prove cyclotomic irreducibility, degrees, inertia and arithmetic Frobenius. \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ADL-06}{Quasi-characters and Hecke characters}, especially Theorem 6.3, supplies the character terminology and its ideal/idèle convention. All these providers are written lessons. Fix \(\zeta_m=e^{2\pi i/m}\), put \(U_m=(\mathbf Z/m\mathbf Z)^\times\), and write \[ \sigma_a(\zeta_m)=\zeta_m^a\quad(a\in U_m). \tag{1} \] The groups \(U_1,U_2\) are trivial. Arithmetic reciprocity sends an unramified uniformizer to arithmetic Frobenius. Geometric reciprocity is its inverse. The cyclotomic character always records the exponent in (1); its definition does not change when reciprocity is inverted. \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-20}{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. The rational ray class fields}\label{the-rational-ray-class-fields} For an integer \(m\geq1\), the moduli \(m\infty\) and \(m\) differ by the real place condition. In the first, principal generators must be positive; in the second, either sign is allowed. \textbf{Proposition 20.1.} Their ray groups and fields are \[ \begin{aligned} \operatorname{Cl}_{m\infty}(\mathbf Q)&\simeq U_m,& \mathbf Q_{m\infty}&=\mathbf Q(\zeta_m),\\ \operatorname{Cl}_{m}(\mathbf Q)&\simeq U_m/\langle-1\rangle,& \mathbf Q_m&=\mathbf Q(\zeta_m)^+. \end{aligned} \tag{2} \] Here the superscript \(+\) means the subfield fixed by complex conjugation. The formulas include \(m=1,2\). \textbf{Proof.} Every fractional ideal of \(\mathbf Q\) prime to \(m\) has a unique positive rational generator \(a\), whose numerator and denominator are coprime to \(m\). Send that ideal to \(a\bmod m\). Every unit residue occurs, by taking a positive integral representative. Its kernel consists exactly of the ideals with positive generator congruent to \(1\) modulo \(m\), giving the first group in (2). If the generator may have either sign, an ideal is ray principal exactly when its positive generator is congruent to \(1\) or \(-1\); this gives the second group. Let \(E=\mathbf Q(\zeta_m)\). The local symbol proves its conductor divides \(m\infty\). Indeed, write \(m=p^a b\), \(p\nmid b\). At \(p\), the field generated by the \(b\)-th roots is unramified, so local units act trivially on them. Units congruent to \(1\pmod{p^a}\) act trivially on the \(p^a\)-th roots by Corollary 9.5. When \(a=0\), all units suffice. At the real place, positive numbers act trivially. Thus the entire modulus subgroup is in the global reciprocity kernel, by local compatibility. Theorem 18.4 places \(E\) in \(\mathbf Q_{m\infty}\). Both degrees are \(\varphi(m)\), the first by cyclotomic irreducibility and the second by the ray-group computation. They are equal. Dropping the real place adds the class of a negative real component to the norm subgroup. Its symbol on \(E\) is complex conjugation: local reciprocity at \(\mathbf R\) identifies the sign quotient with \(\operatorname{Gal}(\mathbf C/\mathbf R)\). Therefore the corresponding subfield is \(E^+\). Its group is the quotient by \(-1\) in (1). This proves the remaining field assertion, also when conjugation is already trivial. \(\square\) The modulus in (2) need not be the field's least conductor. For example \(\mathbf Q(\zeta_2)=\mathbf Q\), and for odd \(m\) the fields of orders \(m\) and \(2m\) are equal. A conductor is a minimal condition, whereas a ray modulus is a specified condition. \subsection{2. The Kronecker--Weber theorem}\label{the-kroneckerweber-theorem} \textbf{Theorem 20.2 (Kronecker--Weber).} Every finite abelian \(L/\mathbf Q\) lies in \(\mathbf Q(\zeta_m)\) for some positive integer \(m\). Consequently \[ \mathbf Q^{\mathrm{ab}}=\mathbf Q^{\mathrm{cycl}} =\bigcup_{m\geq1}\mathbf Q(\zeta_m)=\mathbf Q(\mu_\infty). \tag{3} \] \textbf{Proof.} The extension has a finite global conductor, by Proposition 18.3. Enlarge it, if necessary, to a modulus \(m\infty\). The ray-field containment theorem and (2) give \(L\subseteq\mathbf Q_{m\infty}=\mathbf Q(\zeta_m)\). Every cyclotomic field is abelian by (1), so their union is contained in the maximal abelian extension. The reverse inclusion follows from the finite assertion, since every algebraic element in \(\mathbf Q^{\mathrm{ab}}\) belongs to a finite abelian extension. \(\square\) This proof identifies the containing field through its ray conditions. Section 7 gives a second proof using local Kronecker--Weber and finite inertia groups. \subsection{3. The explicit rational Artin map}\label{the-explicit-rational-artin-map} The normalization of idèles in lesson 17 gives a topological isomorphism \[ C_{\mathbf Q}\simeq\mathbf R_{>0}\times\widehat{\mathbf Z}^{\times}. \tag{4} \] Explicitly, for an idèle \(x=(x_\infty,(x_p))\), set \[ q=\prod_p p^{-v_p(x_p)}>0,\quad \epsilon=\operatorname{sgn}(x_\infty),\quad t=\epsilon qx_\infty>0,\quad u_p=\epsilon qx_p\in\mathbf Z_p^\times. \tag{5} \] Only finitely many factors define \(q\). Multiplication by the principal idèle \(\epsilon q\) yields the normalized representative \((t,u)\). A rational number that is a unit at every prime is \(\pm1\), and positivity fixes its sign; hence the representative is unique. Continuity holds on the restricted-product neighborhoods where the finitely many valuations and the real sign are fixed. \textbf{Proposition 20.3.} For \(u\in\widehat{\mathbf Z}^{\times}\), compatible exponent notation gives \[ \operatorname{rec}_{\mathrm{arith}}(t,u)=\sigma_{u^{-1}}, \qquad \operatorname{rec}_{\mathrm{geom}}(t,u)=\sigma_u. \tag{6} \] For a prime \(p\), the idèle \(\varpi_p\) equal to \(p\) at \(p\) and \(1\) elsewhere acts on \(\mathbf Q(\zeta_m)\), \(p\nmid m\), as \(\sigma_p\) in the arithmetic convention. The idèle equal to \(-1\) at the real place and \(1\) at every finite place acts as complex conjugation. \textbf{Proof.} A positive real component acts trivially by real local reciprocity. On \(p^a\)-th roots, the local component \(u_p\) acts with exponent \(u_p^{-1}\), by Corollary 9.5; every other finite unit component is unramified on those roots and acts trivially. Combining the prime-power roots gives the first formula on \(\mathbf Q(\zeta_m)\), for every \(m\). Local compatibility and (3) make it the full global formula. Inverting gives geometric reciprocity. For \(\varpi_p\), (5) gives \(t=1/p\), \(u_p=1\), and \(u_\ell=1/p\) for \(\ell\ne p\). Thus \(u^{-1}\equiv p\pmod m\) when \(p\nmid m\), proving the Frobenius assertion. No such residue \(p\pmod m\) defines a unit when \(p\mid m\); at the \(p\)-primary roots this particular idèle acts trivially, and at the prime-to-\(p\) roots it acts with exponent \(p\). For the negative real idèle, (5) gives \(t=1\), \(u_p=-1\) at every finite prime. Its inverse is itself, so (6) is \(\sigma_{-1}\), complex conjugation. \(\square\) \textbf{Corollary 20.4.} The cyclotomic character is a canonical topological isomorphism \[ \chi_{\mathrm{cyc}}:\operatorname{Gal}(\mathbf Q^{\mathrm{ab}}/\mathbf Q) \xrightarrow{\sim}\widehat{\mathbf Z}^{\times}, \qquad \sigma_a\longmapsto a. \tag{7} \] Arithmetic reciprocity induces an isomorphism from \(C_{\mathbf Q}/\mathbf R_{>0}\), and its composite with (7) is \(u\mapsto u^{-1}\). \textbf{Proof.} The finite isomorphisms (1) are compatible with reduction of exponents. Taking inverse limits, as proved in Proposition 1.3, gives (7) for the cyclotomic union, which is (3). Formula (6) is onto and has kernel exactly the positive real factor: a unit tuple acting trivially on every root has residue \(1\) modulo every integer. Both (7) and its inverse are continuous by their finite quotient coordinates. \(\square\) \subsection{4. Dirichlet, Galois and Hecke characters}\label{dirichlet-galois-and-hecke-characters} A Dirichlet character means a character of some \(U_m\), with characters identified when they are inflations of the same primitive character. Its \textbf{finite conductor} \(f(\chi)\) is the least modulus of that primitive character. When values on integers are used, the primitive character is extended by zero precisely at integers not coprime to \(f(\chi)\). \textbf{Proposition 20.5.} Dirichlet characters correspond bijectively to continuous complex characters of \(\operatorname{Gal}(\mathbf Q^{\mathrm{ab}}/\mathbf Q)\), and to finite-order Hecke characters of \(\mathbf Q\), by \[ \rho_\chi(\sigma_a)=\chi(a),\qquad \omega_\chi(t,u)=\chi(u^{-1}). \tag{8} \] The ideal character at an unramified prime is \(\omega_\chi(\varpi_p)=\chi(p)\). Their finite conductor exponents agree. If \(L/\mathbf Q\) is finite abelian and \(X(L)\) is its character group inflated to (7), then \[ f_L=\operatorname{lcm}_{\chi\in X(L)}f(\chi) \tag{9} \] is its finite conductor and its least cyclotomic modulus. Its full global conductor includes \(\infty\) exactly when \(L\) is not totally real. \textbf{Proof.} A continuous character from a profinite group to \(\mathbf C^\times\) has open kernel. To check this, choose a small neighborhood of \(1\) containing no nontrivial subgroup, for example \(1/2<|z|<2\), \(|\arg z|<\pi/2\). All powers force a subgroup element's modulus to be \(1\), and a nonidentity circle element has a power outside that arc. A sufficiently small open subgroup of the profinite domain maps into this neighborhood and hence trivially. The resulting quotient is finite. Applying this to (7) shows that every character factors through some \(U_m\). A finite-order Hecke character kills \(\mathbf R_{>0}\), since a connected group's continuous image in a finite group is trivial. Conversely a finite-image character on the unit factor in (4) defines such a Hecke character. Formula (6) gives exactly the inverse in (8), and gives \(\chi(p)\) at \(p\nmid f(\chi)\). Its real sign is \(\chi(-1)\). For the conductor, the Chinese remainder decomposition of \(U_m\) gives a least exponent in each prime-power factor through which a character factors. These are the least exponents killing the local principal-unit groups; inversion changes neither the kernel nor the exponent. This proves agreement with local and Hecke conductors. The finite character lemma proved before section 4 of \href{../hilberts-theorem-90-and-kummer-theory.html}{Hilbert's theorem 90 and Kummer theory} says that the characters of \(\operatorname{Gal}(L/\mathbf Q)\) separate its elements. Thus a local unit subgroup kills the field exactly when it kills every character. Its least exponent is their maximum, proving (9). Equivalently \(L\subseteq\mathbf Q(\zeta_m)\) precisely when every character factors modulo \(m\), by the annihilator description of a fixed field. The least local exponents show that this is equivalent to \(f_L\mid m\), proving minimality without assuming a chosen ambient modulus. Complex conjugation is \(-1\) under (7). It is trivial on the field exactly when every \(\chi\in X(L)\) is even, which is precisely that \(L\) is totally real. The real conductor statement follows. \(\square\) Multiplication in \(X(L)\) means multiplication after inflation to a common modulus, followed by taking the primitive representative. Multiplying zero-extended arithmetic functions would impose an incorrect conductor on the result. \subsection{5. The global conductor--discriminant formula}\label{the-global-conductordiscriminant-formula} \textbf{Theorem 20.6.} For a finite abelian number field \(L/\mathbf Q\), \[ |d_L|=\prod_{\chi\in X(L)}f(\chi). \tag{10} \] The trivial character contributes \(1\). These are finite conductors; an infinity factor is not an integer factor of the discriminant. \textbf{Proof.} Fix a rational prime \(p\), and let \(D\) be its decomposition group in \(G=\operatorname{Gal}(L/\mathbf Q)\). Its completion \(E/\mathbf Q_p\) has group \(D\), and there are \(g=[G:D]\) primes above \(p\). Restriction \(\widehat G\to\widehat D\) is onto and each fiber has size \(g\): a character of a subgroup of a finite abelian group extends by successively choosing a root for the value of a newly adjoined generator. Its kernel is \(\widehat{G/D}\). The local conductor is computed by restriction to \(D\). Theorem 10.5 therefore gives \[ \sum_{\chi\in X(L)}v_p(f(\chi)) =g\sum_{\eta\in\widehat D}a(\eta) =g\,v_p(\mathfrak d_{E/\mathbf Q_p}). \tag{11} \] This is also \(v_p(d_L)\). To verify the local/global identification directly, tensor an integral basis and its trace matrix with \(\mathbf Z_p\). The Chinese remainder theorem on the quotients by \(p^n\), followed by completion, gives \[ \mathcal O_L\otimes\mathbf Z_p\simeq \prod_{\mathfrak P\mid p}\mathcal O_{L_{\mathfrak P}}. \] Trace on this product is the sum of the local traces. Choosing a basis in each factor makes the trace pairing block diagonal, so its determinant valuation is the sum of the local discriminant valuations. Integral changes of \(\mathbf Z_p\)-basis have unit determinant and do not alter that valuation. In the Galois case all \(g\) local factors are isomorphic, giving the last term of (11). Equality of prime exponents proves (10); absolute value removes the archimedean signature sign. \(\square\) For \(L=\mathbf Q(\zeta_5)\), its four characters have conductors \(1,5,5,5\), so \(|d_L|=125\). For \(L=\mathbf Q(\zeta_{12})\), the four characters are \(1,\chi_{-3},\chi_{-4},\chi_{12}\), with conductors \(1,3,4,12\); their product is \(144\). Both agree with the discriminants calculated independently in the cyclotomic-fields lesson. \subsection{6. Cyclotomic values and the adelic component space}\label{cyclotomic-values-and-the-adelic-component-space} Put \(e(r)=\exp(2\pi ir)\) when it denotes a complex number. In the Bost--Connes algebra the same notation labels a diagonal arithmetic generator; its value in a cooled extremal phase indexed by \(v\in\widehat{\mathbf Z}^{\times}\) is \(e(vr)\). The construction and classification of those states are proved in \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#QSM-L04}{Phase transition in the Bost--Connes system}, sections 3, 4, 6 and 18. We now specify the number theory in that formula. \textbf{Proposition 20.7.} The values \(e(a/N)=\zeta_N^a\), for rational residues \(a/N\), generate \(\mathbf Q^{\mathrm{ab}}\). For every compatible unit \(b\), \[ \sigma_b\bigl(e(a/N)\bigr)=e(ba/N),\qquad \sigma_b\bigl(e(va/N)\bigr)=e(bva/N). \tag{12} \] The adelic component space used for the cyclotomic tower has a canonical topological identification \[ \mathbf Q^\times\backslash \bigl(\mathbf A_f^\times\times\{\pm1\}\bigr) \simeq \mathbf A_f^\times/\mathbf Q_{>0}^\times \simeq\widehat{\mathbf Z}^{\times}. \tag{13} \] Here a rational \(q\) acts by \((a,\epsilon)\mapsto(qa,\operatorname{sgn}(q)\epsilon)\). It is the zero-dimensional space denoted \(\operatorname{Sh}(\mathrm{GL}_1,\{\pm1\})\) in the Bost--Connes discussion. With the geometric reciprocity convention, a normalized unit idèle \(b\) gives \(\sigma_b\); its action on cyclotomic embedding labels in (13) is multiplication by \(b\). \textbf{Proof.} The values contain \(\zeta_N=e(1/N)\) for every \(N\), and every displayed value is a root of unity. Thus their generated field is the cyclotomic union, equal to (3). Equations (12) are exactly exponent multiplication in (1), including after replacing \(a/N\) by a residue indexed by \(v\). For (13), every orbit has a representative with sign \(+1\), obtained by multiplication by \(-1\) when necessary. Two such representatives differ by a positive rational. For \(a\in\mathbf A_f^\times\), multiplication by \(q=\prod_p p^{-v_p(a_p)}\) produces a unit tuple. A positive rational unit everywhere is \(1\), so that tuple is unique. The normalization is continuous on neighborhoods with fixed valuations, and its inverse is the inclusion of the compact unit tuples. This proves the topological identifications, not just a set bijection. The unit group is totally disconnected, so every connected component of this space is a point. At level \(N\), reduction of its unit coordinate is the label \(v\in U_N\) of the embedding \(\zeta_N\mapsto\zeta_N^v\). This identifies its inverse system with the complex points of the cyclotomic tower \(\operatorname{Spec}\mathbf Q(\zeta_N)\). Composition with \(\sigma_b\) sends that embedding to the one labeled \(bv\), proving the Galois action. Deligne's chosen reciprocity sends a uniformizer to geometric Frobenius, so it is \(\operatorname{rec}_{\mathrm{geom}}\). Formula (6) identifies its unit coordinate with precisely \(\sigma_b\), as asserted. \(\square\) For the arithmetic normalization the same \(\sigma_b\) is the image of the unit idèle \(b^{-1}\). Thus (12) uses the cyclotomic exponent \(b\); substituting an arithmetic idèle coordinate for that exponent requires the inverse. The arithmetic symmetry formulas in \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#QSM-L03}{The Bost--Connes Hecke algebra}, section 6, use the exponent coordinate. Their operator algebra and state proofs remain in those lessons. Hilbert's twelfth problem asks for explicit generators of abelian extensions through special values of functions. Roots of unity solve it over \(\mathbf Q\) by (3). Imaginary quadratic fields motivate elliptic functions and complex multiplication; their general generation theory is beyond this lesson. The preceding lesson obtains its particular ring class fields by independently checked radical equations. \subsection{7. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- The least cyclotomic modulus (easy)}\label{exercise-1-the-least-cyclotomic-modulus-easy} Find the smallest \(m\) containing \(\mathbf Q(\sqrt{-7})\) and \(\mathbf Q(\sqrt3)\) in \(\mathbf Q(\zeta_m)\). \textbf{Solution.} Their fundamental discriminants are \(-7\) and \(12\). The quadratic conductor calculation in lesson 18 gives finite conductors \(7\) and \(12\), respectively. Proposition 20.5 proves that each is its least cyclotomic modulus. Inclusion at \(7\) is also exhibited by the quadratic Gauss sum \(\sum_{a\bmod7}(a/7)\zeta_7^a\), whose square is \(-7\), proved in Proposition 12.6 of \emph{Cyclotomic fields}. At \(12\), \(\zeta_{12}+\zeta_{12}^{-1}=\sqrt3\). The real factor of the first field's conductor is \(\infty\); it does not change the finite least modulus. Similarly \(\sqrt2=\zeta_8+\zeta_8^{-1}\), and \(\sqrt5\) belongs to \(\mathbf Q(\zeta_5)\) by its quadratic Gauss sum. Their least moduli are \(8\) and \(5\), as their conductors predict. \subsubsection{Exercise 2 --- The negative real idèle (medium)}\label{exercise-2-the-negative-real-iduxe8le-medium} Compute the Artin image of the idèle equal to \(-1\) at infinity and \(1\) at every finite prime, in both conventions. \textbf{Solution.} Its normalized representative under (5) is \((1,(-1)_p)\). Formula (6) sends it to \(\sigma_{-1}\) in either convention, because this unit tuple is its own inverse. It sends every root of unity to its inverse, hence is complex conjugation. This idèle is not the diagonal principal element \(-1\): the latter has \(-1\) at the finite primes too and has trivial global symbol. Their product displays explicitly the cancellation of the real and finite signs on a principal idèle. \subsubsection{Exercise 3 --- The real ray field (medium)}\label{exercise-3-the-real-ray-field-medium} Prove that \(\mathbf Q(\zeta_m)^+\) is the ray field for modulus \(m\), including \(m=1,2\). \textbf{Solution.} In the ideal ray group, a positive generator represents the same ideal as its negative. Removing the positivity condition therefore replaces \(U_m\) by \(U_m/\langle-1\rangle\). The full ray field for \(m\infty\) is \(\mathbf Q(\zeta_m)\). Passing to modulus \(m\) kills the image of the real sign, whose Artin symbol is \(\sigma_{-1}\). The fixed field of that subgroup is exactly the real subfield. When \(m>2\), this subgroup has order \(2\), so the degree is \(\varphi(m)/2\). When \(m=1,2\), both groups and fields are trivial, and taking a half of \(\varphi(m)\) would be incorrect. \subsubsection{Exercise 4 --- A proof from local Kronecker--Weber (hard)}\label{exercise-4-a-proof-from-local-kroneckerweber-hard} Prove the finite assertion in Theorem 20.2 using local Kronecker--Weber and the absence of a nontrivial abelian extension of \(\mathbf Q\) unramified at every finite prime. \textbf{Solution.} Let \(S\) contain the finite primes ramified in \(L/\mathbf Q\). At \(p\in S\), local Kronecker--Weber puts each completion of \(L\) in \(\mathbf Q_p(\zeta_{p^{n_p}},\zeta_{b_p})\), with \(p\nmid b_p\), for a sufficiently large \(n_p\). The second factor is unramified. Therefore, after adjoining the \(p^{n_p}\)-th roots, the compositum of the completion with that cyclotomic local field is unramified over it: it is a subextension of the base change of the unramified factor. Put \(m=\prod_{p\in S}p^{n_p}\), \(M=\mathbf Q(\zeta_m)\), and \(T=LM\). At \(p\in S\), the completion of \(M\) is the product of the totally ramified \(p\)-primary cyclotomic extension and unramified prime-to-\(p\) factors. The preceding observation makes \(T/M\) unramified there. Outside \(S\), both \(L\) and \(M\) are unramified, so their compositum is too. All these local assertions follow from separable residue-field extensions and their preservation under base change. The group \(G=\operatorname{Gal}(T/\mathbf Q)\) is abelian. Its inertia subgroup \(I_p\) maps isomorphically onto the inertia of \(M/\mathbf Q\): restriction is onto, and its kernel is relative inertia for \(T/M\), already trivial. The cyclotomic inertia theorem gives \(|I_p|=\varphi(p^{n_p})\). The subgroup \(J\) generated by these finite inertia groups has fixed field unramified at every finite prime over \(\mathbf Q\). Its fixed field is abelian, and is therefore \(\mathbf Q\), since the narrow Hilbert class group of \(\mathbf Q\) is trivial. Hence \(J=G\). Because \(G\) is abelian, multiplication of the finitely many inertia subgroups is onto, and \[ |G|\leq\prod_{p\in S}|I_p|=\prod_{p\in S}\varphi(p^{n_p}) =\varphi(m)=[M:\mathbf Q]. \] But \(M\subseteq T\) gives the reverse degree inequality. Thus \(T=M\), so \(L\subseteq M\), as required. If \(S\) is empty, the narrow Hilbert argument already gives \(L=\mathbf Q\). This proof uses no assertion about the class number of \(M\); relative unramifiedness by itself would not imply \(T=M\). \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 lesson proves both Kronecker--Weber routes, the rational Artin map, conductors and discriminants. Section 6 proves the cyclotomic values and the canonical topological component identification. The Bost--Connes state construction retains its separately identified programme prerequisite. \begin{itemize} \tightlist \item \href{https://www.jmilne.org/math/CourseNotes/ANT.pdf}{J. S. Milne, Algebraic Number Theory}. \item \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne, Class Field Theory, version 4.03}. \item \href{https://alainconnes.org/wp-content/uploads/bookwebfinal-2.pdf}{Alain Connes and Matilde Marcolli, Noncommutative Geometry, Quantum Fields and Motives, author-posted draft}. \end{itemize} The arithmetic and geometric Frobenius conventions are compared in equations (6) and (13). The Connes--Marcolli reference is the author-posted draft, with its own section numbering and pagination. The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Artin L-functions, conductors and discriminants}\label{artin-l-functions-conductors-and-discriminants} \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).} \textbf{Lesson 21.} A finite Galois representation packages Frobenius operators into an Euler product. Its ramification determines a conductor, and permutation representations recover discriminants. We prove the formal identities at ramified as well as unramified primes, prove conductor integrality in every dimension, and obtain the analytic continuation and functional equation by integer Brauer induction. The splitting field of \(X^3-2\) provides a concrete nonabelian example. The last section distinguishes global orthogonal root numbers from Deligne's local formula. Our representation-theoretic inputs are written in \href{https://kokunoyumeto.github.io/open-math-courses/courses/RT-FIN/characters-and-the-orthogonality-relations.html\#theorem-3-2}{Characters and the orthogonality relations}, Theorem 3.2; \href{https://kokunoyumeto.github.io/open-math-courses/courses/RT-FIN/induced-representations-and-frobenius-reciprocity.html\#1-one-copy-for-each-coset}{Induced representations and Frobenius reciprocity}, §§1--4; and \href{https://kokunoyumeto.github.io/open-math-courses/courses/RT-FIN/brauers-induction-theorem.html\#theorem-5-1}{Brauer's induction theorem}, Theorem 5.1. In particular, Brauer's theorem expresses every character as an \textbf{integer} linear combination of characters induced from one-dimensional characters of subgroups. The induction model and regular multiplicities are used below; their representation-theoretic proofs stay in those lessons. For ramification we use Hilbert's different formula and Herbrand's quotient theorem, with the conventions of \href{../abelian-ramification-conductors-and-hasse-arf.html}{Abelian ramification, conductors and Hasse--Arf}. \href{../ray-class-fields-conductors-and-ideal-reciprocity.html}{Ray class fields, conductors and ideal reciprocity} identifies rank-one Galois and idèle characters. The written number-field analytic input is Theorem 10.1 of \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ADL-10}{Hecke L-functions and the Dedekind zeta function}. Section 6 also supplies the function-field analytic argument from the adelic duality proved in lessons 15 and 17 and the local Fourier calculations of lesson 12. \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-21}{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. Euler factors and induction}\label{euler-factors-and-induction} Let \(L/K\) be a finite Galois extension of global fields, \(G=\operatorname{Gal}(L/K)\), and let \(V\) be a finite-dimensional complex representation of \(G\). At a finite place \(v\) of \(K\), choose a place \(w\) of \(L\), decomposition group \(D_w\), and inertia group \(I_w\). Put \(q_v=|\kappa(v)|\). An arithmetic Frobenius \(\phi_w\in D_w/I_w\) acts on the residue field by \(x\mapsto x^{q_v}\). It acts unambiguously on \(V^{I_w}\), even though its lift to \(D_w\) is not unique. Define \[ L_v(s,V)=\det(1-q_v^{-s}\phi_w\mid V^{I_w})^{-1}, \qquad L_K(s,V)=\prod_{v\text{ finite}}L_v(s,V). \tag{1} \] Changing \(w\) conjugates the relevant groups and operator, so does not change the determinant. Eigenvalues have absolute value one, since \(G\) is finite. The logarithmic expansion is \[ \log L_K(s,V)=\sum_v\sum_{m\geq1} \frac{\operatorname{tr}(\phi_w^m\mid V^{I_w})}{m q_v^{ms}}. \tag{2} \] Its absolute value is bounded termwise by \(\dim V\) times the logarithmic series for \(\zeta_K(\operatorname{Re}s)\). The latter converges for \(\operatorname{Re}s>1\). Thus (1) converges absolutely there, locally uniformly, and has no zeros there. For a virtual representation, take quotients of these products. \textbf{Proposition 21.1.} Artin L-functions satisfy \[ \begin{aligned} L_K(s,V\oplus W)&=L_K(s,V)L_K(s,W),\\ L_K(s,\operatorname{Ind}_H^G W)&=L_E(s,W),\quad E=L^H,\\ L_K(s,\operatorname{Inf}_{G/N}^G U)&=L_K(s,U),\quad N\triangleleft G. \end{aligned} \tag{3} \] The last function on the right is computed in \(L^N/K\). Also \[ \zeta_E(s)=L_K(s,\mathbf C[G/H]),\qquad \zeta_L(s)=\prod_{\rho\in\operatorname{Irr}(G)}L_K(s,\rho)^{\dim\rho}. \tag{4} \] \textbf{Proof.} Direct sums give block-diagonal determinants. Under passage to a quotient, decomposition and inertia map onto their counterparts, and Frobenius maps to Frobenius. The invariant space and its operator are consequently unchanged, proving inflation invariance. We prove induction place by place. Restrict \(\mathbf C[G]\otimes_{\mathbf C[H]}W\) to \(D=D_w\). Splitting its coset basis into \(D\)-orbits gives \[ \operatorname{Res}_D\operatorname{Ind}_H^G W \simeq\bigoplus_{g\in D\backslash G/H} \operatorname{Ind}_{J_g}^{D}W_g, \quad J_g=D\cap gHg^{-1}. \tag{5} \] Here \(W_g\) is the conjugate representation on the same vector space. The double cosets correspond to the places \(u\mid v\) of \(E\), and \(J_g\) is the local Galois group over \(E_u\). Let \(I=I_w\). For one summand of (5), its \(I\)-invariants have one copy of \(W_g^{I\cap J_g}\) for each \(I\)-orbit in \(D/J_g\). This follows directly in the coset model: an invariant vector on an orbit is determined by its value at a representative, and that value must be fixed by its stabilizer. Because \(D/I\) is cyclic, Frobenius permutes these orbits cyclically, in \(f=[D:IJ_g]=[\kappa(u):\kappa(v)]\) slots. After one circuit its action on the initial copy is arithmetic Frobenius of \(E_u\): \(\phi_w^f\) can be adjusted by an element of \(I\) to lie in \(J_g\), and that adjustment does not affect the invariant vectors. For a cyclic block operator with wrap-around operator \(A\), \[ \det(1-T\Phi)=\det(1-T^f A). \tag{6} \] To see this, successively change bases in the slots so that all transition maps except the last are identities. Eliminating the first \(f-1\) variables from \((1-T\Phi)x\) leaves the last block \(1-T^f A\); the other diagonal blocks are identities. This is a polynomial identity, so the computation over \(\mathbf C(T)\) proves it for every \(T\). Since \(q_u=q_v^f\), (6) proves that this summand contributes exactly \(L_u(s,W)\). Taking all summands proves the middle identity of (3), including ramified places. The trivial representation has Euler factor \((1-q_v^{-s})^{-1}\), hence L-function \(\zeta_K\). Apply induction to \(W=\mathbf1\) to obtain the first identity of (4). For \(H=1\), the permutation representation is the regular representation. The written regular-multiplicity theorem decomposes it as \(\bigoplus_\rho\rho^{\oplus\dim\rho}\). Additivity gives the second identity. \(\square\) \subsection{2. Rank one and primitive Hecke characters}\label{rank-one-and-primitive-hecke-characters} Write \(\operatorname{rec}_{\mathrm{arith}}:C_K\to G^{\mathrm{ab}}\) for arithmetic global reciprocity. For a one-dimensional character \(\chi:G\to\mathbf C^\times\), define \[ \omega=\chi\circ\operatorname{rec}_{\mathrm{arith}}. \tag{7} \] It is a finite-order Hecke character. Its finite conductor is the least ideal whose local principal unit groups it kills; real sign conditions are recorded separately in its infinite type. \textbf{Theorem 21.2.} With the primitive Euler product, \(L_K(s,\chi)=L_K(s,\omega)\). \textbf{Proof.} Local compatibility sends units onto inertia. If \(\chi\) is nontrivial on inertia at \(v\), its invariant space is zero, so the Artin factor is 1. The local Hecke character is ramified and its primitive factor is also 1. If \(\chi\) is trivial on inertia, then \(\omega_v(\pi_v)=\chi(\phi_w)\), and both factors are \((1-\omega_v(\pi_v)q_v^{-s})^{-1}\). This proves equality in the convergence region, hence wherever continuation is defined. One may equivalently replace \(L\) by the abelian field fixed by \(\ker\chi\); inflation invariance proves that this replacement changes no factor. \(\square\) A prime ramified in \(L\) can be unramified for \(\chi\). Its factor must then be retained. Omitting every prime dividing the conductor of the ambient extension produces an imprimitive Hecke function, generally different from (1). The relevant modulus is the character's conductor. Deligne's Euler factors use geometric Frobenius. On a finite-image representation they give our arithmetic factors for \(V^\vee\). In rank one the corresponding reciprocity map is geometric as well. These two changes must be made together. For self-dual representations the L-functions themselves coincide in either Frobenius convention. \subsection{3. The local conductor and its integrality}\label{the-local-conductor-and-its-integrality} Let \(T/F\) be a finite Galois extension of nonarchimedean local fields, with group \(G\), residue degree \(f\), and lower ramification groups \(G_j\). For \(\sigma\ne1\), let \(i_G(\sigma)\) be its motion number as in lesson 10; it is zero off inertia. Define the class function \[ a_G(\sigma)=-f i_G(\sigma)\quad(\sigma\ne1),\qquad a_G(1)=f\sum_{\sigma\ne1}i_G(\sigma). \tag{8} \] For a representation \(V\), its conductor exponent is \[ a_F(V)=\langle a_G,\chi_V\rangle_G =\sum_{j\geq0}\frac{|G_j|}{|G_0|} \operatorname{codim}V^{G_j}. \tag{9} \] Indeed \(i_G(\sigma)\) counts the indices \(j\geq0\) for which \(\sigma\in G_j\). Insert this count into the character inner product and use \(|G_j|^{-1}\sum_{\sigma\in G_j}\overline{\chi_V(\sigma)}=\dim V^{G_j}\) and \(|G|=f|G_0|\). The expression in (9) is nonnegative rational; its integrality requires proof. With the ceiling convention of lesson 10, change variables by Herbrand's function to obtain \[ a_F(V)=\operatorname{codim}V^{G^0} +\int_0^\infty\operatorname{codim}V^{G^u}\,du. \tag{10} \] Herbrand's quotient theorem shows that (10), hence (9), is unchanged if the representation is realized in a larger finite Galois extension. It therefore defines the exponent intrinsically for any finite-image local representation. It is additive on direct sums, and extends additively to virtual representations. \textbf{Proposition 21.4.} For a one-dimensional \(\chi\), \(a_F(\chi)\) is the least integer \(a\geq0\) for which its multiplicative character under reciprocity kills \(U_F^{(a)}\), where \(U_F^{(0)}=\mathcal O_F^\times\). In particular it is an integer. \textbf{Proof.} Pass to the abelian quotient through which \(\chi\) factors, using (10). Lesson 10 identifies its upper groups with the images of \(U_F^{(\lceil u\rceil)}\) for \(u>0\) and with the units at \(u=0\). If the character is unramified, every codimension in (10) is zero. Otherwise its least unit level is some integer \(a\geq1\), and its invariant space is zero at \(u=0\) and for \(0a-1\). Endpoints have no effect on the integral. Formula (10) gives \(1+(a-1)=a\), also for tame characters with \(a=1\). The integer filtration theorem used here is the proved Hasse--Arf theorem of lesson 10. \(\square\) To pass to higher dimension, we need the different in a tower. Here and below \(\mathfrak D\) is the different and \(\mathfrak d\) the discriminant ideal. \textbf{Different in a tower.} For finite separable local extensions \(T/E/F\), \[ \mathfrak D_{T/F}=\mathfrak D_{T/E}\mathfrak D_{E/F}\mathcal O_T, \quad \delta_{T/F}=\delta_{T/E}+e(T/E)\delta_{E/F}. \tag{11} \] For completeness, identify the inverse different with the trace-dual lattice: \(\mathfrak D_{E/F}^{-1}=\{x:\operatorname{Tr}_{E/F}(x\mathcal O_E)\subseteq\mathcal O_F\}\). There is an adjunction \[ \operatorname{Hom}_{\mathcal O_F}(\mathcal O_T,\mathcal O_F) \simeq \operatorname{Hom}_{\mathcal O_E} \bigl(\mathcal O_T,\operatorname{Hom}_{\mathcal O_F}(\mathcal O_E,\mathcal O_F)\bigr), \tag{12} \] sending \(\ell\) to the map \(z\mapsto(b\mapsto\ell(bz))\); its inverse evaluates at \(b=1\). Under trace identifications the middle lattice is \(\mathfrak D_{E/F}^{-1}\). It is a principal fractional \(\mathcal O_E\)-ideal, so it factors out of the final Hom in (12). Trace transitivity then identifies the left lattice with \(\mathfrak D_{T/E}^{-1}\mathfrak D_{E/F}^{-1}\mathcal O_T\). Inverting and taking valuations proves (11). All trace identifications are isomorphisms because the extensions are separable. \textbf{Induction formula.} If \(E=T^H\), with \(H\leq G\), and \(W\) is a representation of \(H\), then \[ a_F(\operatorname{Ind}_H^G W) =(\dim W)v_F(\mathfrak d_{E/F})+f(E/F)a_E(W). \tag{13} \] This does not require \(H\) to be normal. \textbf{Proof.} The lower groups of \(T/E\) are \(H_j=H\cap G_j\), since their definition uses \(v_T\) and \(\mathcal O_T\) in both cases. Since \(G_j\) is normal in \(G\), its orbits on \(G/H\) all have size \(|G_j|/|H_j|\). An invariant vector on each orbit is determined by a vector fixed by \(H_j\). Thus \[ \dim(\operatorname{Ind}W)^{G_j}=[G:G_jH]\dim W^{H_j}. \tag{14} \] Separating the missing orbit dimensions from the missing dimensions within each orbit in (9) gives \[ a_F(\operatorname{Ind}W) =(\dim W)a_F(\operatorname{Ind}\mathbf1) +f(E/F)a_E(W), \tag{15} \] because \(\frac{|G_j|}{|G_0|}[G:G_jH] =f(E/F)\frac{|H_j|}{|H_0|}\). Here \(f(E/F)=|G||H_0|/(|G_0||H|)\). For the permutation term, direct substitution in (9) yields \[ a_F(\operatorname{Ind}\mathbf1) =\frac{f(E/F)}{|H_0|}\sum_{j\geq0}(|G_j|-|H_j|). \tag{16} \] Hilbert's different formula identifies the sum with \(\delta_{T/F}-\delta_{T/E}\). By (11) this is \(|H_0|\delta_{E/F}\). Finally the local trace-dual discriminant formula, proved in lesson 10, is \(v_F(\mathfrak d_{E/F})=f(E/F)\delta_{E/F}\). This proves (13). \(\square\) \textbf{Artin integrality in every dimension.} Apply the written integer Brauer induction theorem to \(\chi_V\): \[ \chi_V=\sum_i n_i\operatorname{Ind}_{H_i}^G\chi_i, \qquad n_i\in\mathbf Z,\quad\dim\chi_i=1. \tag{17} \] The conductor is a linear functional on characters. Each induced term has integral conductor by (13) and Proposition 21.4. Hence \(a_F(V)\in\mathbf Z\); (9) makes it nonnegative for an actual representation. This proves integrality without assuming it in the definition. Moreover (8) is itself a character: its multiplicity on each irreducible is the nonnegative integer just proved, and irreducible characters form an orthonormal basis of class functions. \subsection{4. Global conductors and the discriminant formula}\label{global-conductors-and-the-discriminant-formula} At each finite place use the restricted local representation, and set \[ \mathfrak f_K(V)=\prod_v\mathfrak p_v^{a_{K_v}(V)}. \tag{18} \] Only finitely many exponents are nonzero. This is an integral ideal for an actual representation; for virtual representations it is a fractional ideal. Infinite signs contribute to the gamma factors below rather than to (18). For a finite separable global \(E/K\), Mackey's decomposition (5) and (13) give \[ \mathfrak f_K(\operatorname{Ind}_{E}^{K}W) =\mathfrak d_{E/K}^{\dim W}\,N_{E/K}\mathfrak f_E(W). \tag{19} \] Indeed, at \(v\) sum (13) over \(u\mid v\). The second term is precisely the valuation of the ideal norm. The first is the valuation of the global discriminant: completion decomposes the integral lattice into the local integer rings, its trace pairing is block diagonal, and determinant valuations add. This localization argument was also given explicitly in lesson 20. No normality assumption on \(E/K\) is needed in (19). \textbf{Theorem 21.5 (conductor--discriminant).} For finite Galois \(L/K\), \[ \mathfrak d_{L/K} =\prod_{\rho\in\operatorname{Irr}(G)}\mathfrak f_K(\rho)^{\dim\rho}. \tag{20} \] More generally \(\mathfrak f_K(\mathbf C[G/H])=\mathfrak d_{L^H/K}\). \textbf{Proof.} Set \(W=\mathbf1\) in (19). Its conductor over \(E\) is the unit ideal, proving the permutation assertion. For \(H=1\), use the regular decomposition in Proposition 21.1 and additivity of every local exponent. This proves (20) as an equality of ideals. Alternatively (16) with \(H=1\) gives the exponent \(f(L_w/K_v)\delta_{L_w/K_v}\) for one local regular representation, and the global restriction has one such copy for every place above \(v\). This is exactly the completed global trace discriminant. \(\square\) \subsection{5. Meromorphy and the functional equation over number fields}\label{meromorphy-and-the-functional-equation-over-number-fields} Suppose first that \(K\) is a number field, of absolute discriminant \(d_K\). Put \[ Q_K(V)=|d_K|^{\dim V}N_{K/\mathbf Q}\mathfrak f_K(V),\quad \Gamma_{\mathbf R}(s)=\pi^{-s/2}\Gamma(s/2),\quad \Gamma_{\mathbf C}(s)=2(2\pi)^{-s}\Gamma(s). \tag{21} \] At a real place, let \(d_v^+,d_v^-\) be the multiplicities of \(+1,-1\) for complex conjugation. At a complex place there is one gamma factor of complex type per dimension. Define \[ \Gamma_K(s,V)= \prod_{v\text{ real}}\Gamma_{\mathbf R}(s)^{d_v^+} \Gamma_{\mathbf R}(s+1)^{d_v^-} \prod_{v\text{ complex}}\Gamma_{\mathbf C}(s)^{\dim V}, \quad \Lambda_K(s,V)=Q_K(V)^{s/2}\Gamma_K(s,V)L_K(s,V). \tag{22} \] For virtual representations, dimensions and multiplicities in these formulas are signed integers and \(Q_K(V)\) is a positive real number. Both completion factors respect induction. For \(Q\), use (19) and \[ |d_E|=|d_K|^{[E:K]}N_{K/\mathbf Q}\mathfrak d_{E/K}, \quad Q_K(\operatorname{Ind}_E^K W)=Q_E(W). \tag{23} \] The discriminant tower identity follows by taking norms of (11) at every finite place, or by trace adjunction for the corresponding global lattices. For the gamma factor, a real place of \(E\) over a real place of \(K\) retains its sign type, whereas a complex place over a real place contributes one \(+1\) and one \(-1\) per dimension. The gamma duplication identity gives \(\Gamma_{\mathbf R}(s)\Gamma_{\mathbf R}(s+1)=\Gamma_{\mathbf C}(s)\). At a complex place induction simply adds the dimensions for the places above it. Therefore \[ \Lambda_K(s,\operatorname{Ind}_E^K W)=\Lambda_E(s,W). \tag{24} \] \textbf{Theorem 21.3.} Every finite-image Artin L-function is meromorphic on \(\mathbf C\). With the above completion it satisfies \[ \Lambda_K(s,V)=W_K(V)\Lambda_K(1-s,V^\vee), \qquad |W_K(V)|=1. \tag{25} \] The function-field version is proved in section 6. \textbf{Proof over number fields.} Use (17) for the global Galois group, and put \(E_i=L^{H_i}\). Propositions 21.1 and Theorem 21.2 imply, initially in \(\operatorname{Re}s>1\), \[ L_K(s,V)=\prod_i L_{E_i}(s,\omega_i)^{n_i}, \quad \Lambda_K(s,V)=\prod_i\Lambda_{E_i}(s,\omega_i)^{n_i}, \tag{26} \] where \(\omega_i=\chi_i\circ\operatorname{rec}_{\mathrm{arith}}\). The second equality uses (24) and additivity of conductors, dimensions and gamma multiplicities. Theorem 10.1 of the written Hecke lesson proves continuation and the completed functional equation for each finite-order \(\omega_i\), with unit absolute-value root number. Integer powers in (26) give meromorphic functions, including when some \(n_i\) are negative. Taking their functional equations gives (25), with \(W_K(V)=\prod_i W_{E_i}(\omega_i)^{n_i}\). The dual of an induced representation is induced from the dual, as follows by dualizing its finite coset model, so the function on the right is the required \(V^\vee\). The resulting function agrees with (1) on its convergence half-plane, hence is independent of the Brauer expression by analytic uniqueness. The constant is independent as well: on a nonempty open set where both completed functions are finite and nonzero it is their ratio in (25). Thus two expressions give the same constant. \(\square\) Negative powers explain why this argument proves meromorphy rather than holomorphy. \textbf{Artin's conjecture} for number fields asserts that the L-function of every nontrivial irreducible finite-image representation is entire. It is proved in dimension one by the Hecke theorem; (26) does not prove it in general. The trivial representation has the zeta pole and is deliberately excluded. Nothing here asserts the conjecture for all solvable Galois groups. \subsection{6. The analytic argument over function fields}\label{the-analytic-argument-over-function-fields} Let \(K\) now be a global function field. We give the analytic input needed for the same Brauer argument, rather than applying a number-field theorem to it. Lessons 15 and 17 proved that \(\mathbb A_K/K\) is compact and that there is a continuous additive character \(\psi\), trivial on \(K\), for which the pairing \(\psi(xy)\) identifies the adèles with their dual and has exact annihilator \(K\). Use the product of the local self-dual measures of lesson 12 and the positive Fourier kernel. Schwartz functions here are locally constant with compact support. First the additive covolume of \(K\) is 1. Here is a finite counting proof, also fixing the normalization in Poisson summation. Choose a compact open additive subgroup \(U\) with \(U\cap K=0\), and put \(V=U^\perp\). The finite quotient \(\mathbb A_K/(K+U)\) has dual \(K\cap V\), by the exact annihilator assertion; write their common order as \(N\). The map from \(U\) into the compact quotient \(\mathbb A_K/K\) is injective and its image has index \(N\), so the covolume is \(N\operatorname{vol}(U)\). Also \(K+V=\mathbb A_K\): this open subgroup is closed and its quotient has annihilator \(K\cap U=0\), so that quotient is trivial, since characters separate a compact abelian quotient. Consequently the covolume is \(\operatorname{vol}(V)/N\). Local Fourier inversion gives \(\operatorname{vol}(U)\operatorname{vol}(V)=1\). The two covolume formulas therefore give covolume squared 1, and positivity gives 1. The finite duality used here is the character duality already proved for the relevant adelic compact quotients in lesson 15. Periodize a Schwartz function on \(\mathbb A_K/K\). The periodization is locally constant, so factors through a finite quotient of this compact group. Its Fourier coefficients, indexed by the annihilator \(K\), are \(\widehat f(\alpha)\), since the covolume is 1; unfolding the integral proves this assertion directly. Finite Fourier inversion at zero gives \[ \sum_{\alpha\in K}f(\alpha x) =|x|^{-1}\sum_{\alpha\in K}\widehat f(\alpha/x),\qquad x\in\mathbb A_K^\times. \tag{27} \] The scaling factor follows by the substitution \(z\mapsto zx\) in additive measure. Each sum is finite on compact sets of idèles. This follows from discreteness of \(K\) and compactness of the supports. The content image is a nontrivial discrete cyclic subgroup of \(\mathbf R_{>0}\), say \(R^{\mathbf Z}\) with \(R>1\). Choose an idèle class \(c\) with \(|c|=R\). No assertion that a degree-one place exists is required. The norm-one group \(C_K^1\) is compact, as proved in lesson 14. Fix its quotient Haar measure, of volume \(\kappa\), so that each norm level is a translate with that same measure. Let \(\omega\) be a finite-order Hecke character and put \[ \Theta_f(x)=\sum_{\alpha\in K^\times}f(\alpha x),\qquad Z(f,\omega,s)=\int_{C_K}\Theta_f(x)\omega(x)|x|^s\,d^\times x. \tag{28} \] For \(\operatorname{Re}s>1\), this converges and unfolds to the usual idèle zeta integral. To check convergence directly, (27) bounds \(\Theta_f\) on negative norm levels by \(C(|x|^{-1}+1)\): the inverted classes have positive norm, where the Fourier theta sum vanishes beyond a fixed bound, and the remaining finitely many levels are compact. Summing the resulting geometric bound gives convergence for \(\operatorname{Re}s>1\). Write \(P_{\geq}(f,\omega,s)\) and \(P_>(f,\omega,s)\) for the portions of (28) with norm at least 1 and strictly greater than 1. Both are finite Laurent polynomials in \(R^{-s}\). Indeed a compact additive support bounds the module of every idèle lying in it: outside a finite set its components are integral, and inside that set their absolute values are bounded. Since \(|\alpha|=1\) for \(\alpha\in K^\times\), the condition \(\alpha x\in\operatorname{supp}f\) bounds \(|x|\). Thus only finitely many positive levels occur; their integrals are finite by compactness of \(C_K^1\). Apply (27) to the negative levels in (28), removing the zero summands and substituting \(y=x^{-1}\). It gives \[ Z(f,\omega,s)=P_{\geq}(f,\omega,s) +P_>(\widehat f,\omega^{-1},1-s)+C(f,\omega,s). \tag{29} \] If \(\omega\) is nontrivial on \(C_K^1\), the correction \(C\) is zero by character orthogonality. Otherwise set \(\lambda=\omega(c)\). Summing the negative-level zero terms gives exactly \[ C(f,\omega,s)=\kappa\left( \widehat f(0)\frac{\lambda^{-1}R^{1-s}}{1-\lambda^{-1}R^{1-s}} -f(0)\frac{\lambda^{-1}R^{-s}}{1-\lambda^{-1}R^{-s}} \right). \tag{30} \] This was derived in the convergence region by geometric series and is now a rational continuation in \(R^{-s}\). The continued integral satisfies \[ Z(f,\omega,s)=Z(\widehat f,\omega^{-1},1-s). \tag{31} \] Here is the check at norm zero, which avoids any overlap assumption on convergence regions. The two strictly positive polynomial parts in (29) exchange. The norm-one terms satisfy, by (27), \[ P_0(f,\omega,s)-P_0(\widehat f,\omega^{-1},1-s) =\begin{cases}\kappa(\widehat f(0)-f(0)),&\omega|_{C_K^1}=1,\\0,&\text{otherwise}. \end{cases} \tag{32} \] When applying Poisson to the inverted norm-one class, use that inversion preserves Haar measure and that \(\Theta_{\widehat{\widehat f}}=\Theta_f\), since \(\widehat{\widehat f}(z)=f(-z)\) and \(-1\in K^\times\). If \(\omega\) is trivial on \(C_K^1\), write \(A=\lambda^{-1}R^{1-s}\) and \(B=\lambda^{-1}R^{-s}\). The corresponding variables in the dual correction are \(1/B,1/A\). The identity \(z/(1-z)+(1/z)/(1-1/z)=-1\) shows that the difference of the two corrections is \(-\kappa(\widehat f(0)-f(0))\), canceling (32). If \(\omega\) is nontrivial there, both corrections and the difference in (32) vanish. This proves (31) as a rational identity. To extract the L-function, take at every unramified place \(f_v=\mathbf1_{\mathcal O_v}\), and at each ramified place \(f_v=\omega_v|_{\mathcal O_v^\times}^{-1}\mathbf1_{\mathcal O_v^\times}\). These form a restricted tensor product Schwartz function. With multiplicative unit volume 1 its local zeta integrals are respectively the primitive Euler factor and 1. Absolute factorization in \(\operatorname{Re}s>1\) therefore gives \(Z(f,\omega,s)=L_K(s,\omega)\). Let \(n_v\) be the local additive conductor in the convention that \(\psi_v\) is trivial on \(\mathfrak p_v^{-n_v}\), and set \[ D_K(\psi)=\prod_vq_v^{n_v},\qquad Q_K(V)=D_K(\psi)^{\dim V}N\mathfrak f_K(V),\qquad \Lambda_K(s,V)=Q_K(V)^{s/2}L_K(s,V). \tag{33} \] Only finitely many \(n_v\) are nonzero. The actual finite Fourier calculation of lesson 12, equations (13)--(17), gives \[ \epsilon_v(s,\omega_v)=W_v(\omega_v)q_v^{(a_v+n_v)(1/2-s)}, \qquad |W_v(\omega_v)|=1. \tag{34} \] This holds in positive characteristic too. Factor the dual integral in its own convergence region, use the local equations, and then use the rational continuations (29). Equations (31) and (34) imply \(L_K(s,\omega)=W_K(\omega)Q_K(\omega)^{1/2-s}L_K(1-s,\omega^{-1})\), where \(W_K(\omega)=\prod_v W_v(\omega_v)\). Hence (25) holds in rank one, with no infinite gamma factors. In particular the trivial character has \(W_K(\mathbf1)=1\), since its local central constants in (34) are all 1. For \(E/K\), use \(\psi_E=\psi\circ\operatorname{Tr}_{E/K}\). The trace-dual formula (12) gives \(n_u=e(E_u/K_v)n_v+\delta_{E_u/K_v}\). Multiplying the corresponding powers of residue cardinalities gives \[ D_E(\psi_E)=D_K(\psi)^{[E:K]}N\mathfrak d_{E/K}. \tag{35} \] Thus (19) makes \(Q\) respect induction here as well. The Brauer proof (26) now proves meromorphy and (25) for every finite-image representation over a function field. It also proves rationality in \(q^{-s}\), where \(q\) is the cardinality of the full constant field: every \(R\) and every residue norm is an integral power of \(q\), and the same is true for the constant fields of the finite extensions used in (26). The completed function may include the elementary exponential factor \(Q^{s/2}\); the uncompleted L-function is rational in \(q^{-s}\). The scale \(D_K(\psi)\) is independent of multiplying the global character by a principal nonzero scalar: the local conductor changes multiply it by the inverse global module of that scalar, which is 1. Formula (33) suffices here without invoking a separate theorem about degrees of canonical divisors. \subsection{7. The cubic field and its two-dimensional representation}\label{the-cubic-field-and-its-two-dimensional-representation} Put \(\alpha=\sqrt[3]2\), \(E=\mathbf Q(\alpha)\), and \(L=\mathbf Q(\alpha,\zeta_3)\). The polynomial is irreducible by Eisenstein at 2. Its discriminant is \(-108\), computed as \((-1)^3N_{E/\mathbf Q}(3\alpha^2)\). Since this is not a square, the transitive cubic Galois group is \(S_3\). Its quadratic subfield is \(\mathbf Q(\sqrt{-3})\). The polynomial order is already maximal. At 2 the polynomial is Eisenstein, so \(\alpha\) is a local uniformizer and \(\mathcal O_{E_2}=\mathbf Z_2[\alpha]\). At 3, \(\beta=\alpha+1\) satisfies \[ \beta^3-3\beta^2+3\beta-3=0, \tag{36} \] an Eisenstein equation, so \(\mathcal O_{E_3}=\mathbf Z_3[\beta]=\mathbf Z_3[\alpha]\). The local integer-ring assertion for an Eisenstein generator follows by expanding an integer in powers of the uniformizer with residue representatives from the base field, then reducing by its monic equation; the resulting degree-less-than-three coefficients converge in \(\mathbf Z_p\). At other primes the polynomial discriminant is a unit, so the order's trace dual is itself and no index can occur. Thus \(\mathcal O_E=\mathbf Z[\alpha]\) and \(d_E=-108\). Let \(\rho_2\) be the standard two-dimensional representation of \(S_3\), the complement of the constants in its action on three letters. The transposition stabilizer has index three, hence \[ \mathbf C[S_3/S_2]=\mathbf1\oplus\rho_2, \qquad \mathbf C[S_3]=\mathbf1\oplus\operatorname{sgn}\oplus\rho_2^{\oplus2}. \tag{37} \] The sign character is the quadratic character \(\chi_{-3}\). Consequently \[ \zeta_E(s)=\zeta_{\mathbf Q}(s)L_{\mathbf Q}(s,\rho_2),\qquad \zeta_L(s)=\zeta_{\mathbf Q}(s)L_{\mathbf Q}(s,\chi_{-3})L_{\mathbf Q}(s,\rho_2)^2. \tag{38} \] The non-Galois permutation assertion in Theorem 21.5 immediately gives \[ \mathfrak f_{\mathbf Q}(\rho_2)=(108)=(2^2 3^3). \tag{39} \] Here the trivial summand has conductor 1. We check both local exponents directly. At 2, adjoining \(\zeta_3\) is unramified of degree two, since \(X^2+X+1\) is irreducible modulo 2. Adjoining \(\alpha\) is totally ramified of degree three and tame. The local splitting field has degree six, decomposition group \(S_3\), inertia \(C_3\), and no positive ramification group. A three-cycle has eigenvalues \(\zeta_3,\zeta_3^2\) on \(\rho_2\), so no invariant vectors. Formula (9) gives \(a_2(\rho_2)=2\). At 3, the cubic extension is totally ramified and \(\alpha\) is a unit. Its different is generated by \(3\alpha^2\), so \(\delta_{E_3/\mathbf Q_3}=3\). The quadratic extension \(\mathbf Q_3(\zeta_3)\) is ramified of degree two. In the compositum the ramification index is divisible by both 3 and 2; its degree is six, so it is totally ramified of degree six. Its extension over \(E_3\) is tame quadratic, with different exponent 1. Formula (11) gives \(\delta_{L_3/\mathbf Q_3}=1+2\cdot3=7\). Now \(G_0=S_3\), while its wild inertia is its unique Sylow 3-subgroup \(C_3\). Hilbert's formula reads \(7=5+\sum_{j\geq1}(|G_j|-1)\). The positive groups are either \(C_3\) or trivial, so exactly one is nontrivial. Thus \(G_1=C_3,G_2=1\). Neither \(S_3\) nor \(C_3\) has invariants on \(\rho_2\), and \[ a_3(\rho_2)=2+\frac36\,2=3. \tag{40} \] This verifies (39) with the lower-group weights, independently of the global numerical prediction. \subsection{8. Orthogonal root numbers}\label{orthogonal-root-numbers} An orthogonal representation here means the complexification of a finite-image real representation with an invariant positive definite inner product. Its global root number is the constant in (25). A local root number is the central epsilon constant for a nontrivial local additive character and self-dual measure. These local constants need not all be 1 even when the global constant is 1. The higher-dimensional local epsilon family is supplied by \href{https://kokunoyumeto.github.io/open-math-courses/courses/LG-GAL/local-l-factors-and-epsilon-factors.html\#3-existence-and-uniqueness-of-local-constants}{Local L-factors and epsilon-factors of Weil group representations, Theorem 3.0}, with its existence argument in sections 3A–3D and uniqueness in Theorem 3.2. The \href{https://kokunoyumeto.github.io/open-math-courses/courses/LG-GAL/local-l-factors-and-epsilon-factors.html\#7-sharp-twisting-and-orthogonal-root-numbers}{orthogonal formula, Theorem 7.4}, has its real-induction, Clifford-class and dihedral Fourier arguments in Lemmas 7.5–7.8. The analytic proof of Theorem 21.3 above uses the rank-one theory and Brauer induction and does not depend on these higher-dimensional local arguments. \textbf{Deligne's local orthogonal theorem.} If \(A\) is a virtual finite-image real representation of the absolute Galois group of a local field \(F\), with virtual dimension zero and determinant one, then \[ W_F(A)=\exp\bigl(2\pi i\operatorname{inv}_F(\operatorname{cl}(w_2(A)))\bigr). \tag{41} \] The total Stiefel--Whitney class extends to virtual representations by \(w(V-W)=w(V)w(W)^{-1}\); its degree-two part is \(w_2(A)\in H^2(G_F,\mathbf Z/2)\). Inflate from a finite quotient and map the coefficient \(1\) to \(-1\in\bar F^\times\). This gives \(\operatorname{cl}(w_2(A))\in\operatorname{Br}(F)\). The invariant is in \(\mathbf Q/\mathbf Z\); at a real place its nonzero value is \(1/2\), and at a complex place it is zero. In characteristic two the coefficient map is trivial and the right side is 1. Dimension zero removes measure dependence and determinant one removes additive-character dependence. The specified theorem and class construction are the ones owned by §7 of the provider, corresponding to Deligne (1976), (1.3)--(1.5). They are not proved by Brauer meromorphy alone. Here is the global deduction with its precise inputs. It uses the local existence and orthogonal theorems just identified and \href{../brauer-groups-of-local-and-global-fields.html}{Brauer groups of local and global fields}, section 4 and Theorem 24.4. That lesson proves that for every global Brauer class \(b\), all but finitely many local invariants vanish and \(\sum_v\operatorname{inv}_v(b_v)=0\), including arbitrary splitting groups and both characteristics. This is the required part of the Albert--Brauer--Hasse--Noether theorem. \textbf{Fröhlich--Queyrut global theorem.} For a finite-image orthogonal global representation \(V\), \[ W_K(V)=1. \tag{42} \] \textbf{Deduction from those local and Brauer inputs.} First verify that the global constant in (25) is the product of local central constants. Interpret the provider's geometric local constants on \(V^\vee\) to match our arithmetic convention. For orthogonal representations this dualization changes nothing. Write \(P_K(V)=\prod_v W_{K_v}(V)\); only finitely many factors differ from 1. Rank-one Tate theory, the written Hecke theorem, and section 6 show \(P_K(\chi)=W_K(\chi)\). For a permutation representation \(P=\operatorname{Ind}_E^K\mathbf1\), let \(n=\dim P\), \(\eta=\det P\), and \(A=P-\eta-(n-1)\mathbf1\). This is a global real virtual representation of dimension zero and determinant one. Its Stiefel--Whitney class gives a global Brauer class. Formula (41) and the invariant sum law give \(P_K(A)=1\). The trivial character has global root number 1. Every quadratic character \(\eta\) also has global root number 1: for its quadratic field \(M\), (24) and Proposition 21.1 identify its completed L-function with \(\Lambda_M(s,\mathbf1)/\Lambda_K(s,\mathbf1)\), and both zeta completions have root number 1. The latter is the number-field zeta theorem in the Hecke provider and, for function fields, the trivial-character case of (34). Consequently \(P_K(P)=1\). Local induction for virtual dimension zero implies, on subtracting \((\dim W)\mathbf1\), \[ W_F(\operatorname{Ind}_{E_u}^F W) =\lambda(E_u/F)^{\dim W}W_{E_u}(W),\quad \lambda(E_u/F)=\frac{W_F(\operatorname{Ind}\mathbf1)}{W_{E_u}(\mathbf1)}. \tag{43} \] Using all completions above each place, the product of these \(\lambda\)'s is \(P_K(P)/P_E(\mathbf1)=1\). Thus \(P_K(\operatorname{Ind}_E^K\chi)=P_E(\chi)=W_E(\chi)\). Integer Brauer induction and (26) give \(P_K(V)=W_K(V)\) for every finite-image representation. This proves the needed factorization from the stated local inputs, rather than assuming a separate global factorization theorem. Finally for an arbitrary orthogonal \(V\) of dimension \(n\), put \(A=V-\det V-(n-1)\mathbf1\). Apply (41) at every place and the global invariant sum law to its global class. They give \(W_K(A)=P_K(A)=1\). The determinant is trivial or quadratic and has root number 1, as above, and so does the trivial character. Multiplicativity gives (42). \(\square\) The assertion is global. For example the real sign character, with the negative exponential additive character used in lesson 12, has local central constant \(-i\). Its global quadratic counterpart has product 1, so the other factors cancel that phase. Nor does (42) apply to every self-dual complex representation: a symplectic representation need not be the complexification of a real orthogonal one. \subsection{9. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- The cubic zeta factor (easy)}\label{exercise-1-the-cubic-zeta-factor-easy} Prove \(\zeta_{\mathbf Q(\sqrt[3]2)}(s)=\zeta_{\mathbf Q}(s)L_{\mathbf Q}(s,\rho_2)\), with every ramified Euler factor included. \textbf{Solution.} Irreducibility at 2 and the nonsquare cubic discriminant show that the splitting field has group \(S_3\). The cubic field is fixed by the subgroup stabilizing one root, of order two. Its coset permutation representation is the three-letter representation. Its constant line is trivial, and the coordinate-sum-zero plane is \(\rho_2\), giving \(\operatorname{Ind}_{S_2}^{S_3}\mathbf1=\mathbf1\oplus\rho_2\). The ramified induction calculation (5)--(6) identifies its Euler factor with the product of the cubic field's factors above each prime. Additivity and \(L(s,\mathbf1)=\zeta_{\mathbf Q}(s)\) prove the identity. No factors at 2 or 3 have been deleted. \subsubsection{Exercise 2 --- The two local conductor exponents (medium)}\label{exercise-2-the-two-local-conductor-exponents-medium} Compute \(a_2(\rho_2)\) and \(a_3(\rho_2)\) directly from the lower ramification groups, and recover conductor 108. \textbf{Solution.} At 2 the local cubic field is tame totally ramified of degree three, and the quadratic roots of unity form the unramified quadratic extension. Thus the splitting field has \(G_0=C_3\), \(G_j=1\) for \(j\geq1\). A three-cycle has eigenvalues \(\zeta_3,\zeta_3^2\) on the coordinate-sum-zero plane, so \(a_2=2\). At 3 use the Eisenstein generator \(\beta\) of (36). Since \(\alpha\) is a unit, the derivative \(3\alpha^2\) has valuation 3 in the cubic field, so its different exponent is 3. The splitting field has total ramification degree six; its relative quadratic extension over the cubic field is tame, of different exponent 1. Different transitivity gives exponent \(1+2\cdot3=7\). In \(S_3\), the wild inertia is \(C_3\). Hilbert's formula forces \(G_0=S_3,G_1=C_3,G_2=1\), because the zero group contributes 5 and each nontrivial positive group contributes 2. Both nontrivial groups have zero invariants on \(\rho_2\), so (9) gives \(a_3=2+(3/6)2=3\). All other places are unramified, since the polynomial discriminant is a unit there. Hence the conductor is \(2^2 3^3=108\). \subsubsection{Exercise 3 --- Induction at a ramified prime (medium)}\label{exercise-3-induction-at-a-ramified-prime-medium} Prove the induction identity in Proposition 21.1 without assuming the base prime is unramified. \textbf{Solution.} Fix its local decomposition and inertia groups \(D,I\). Split the restricted induced representation according to \(D\backslash G/H\); these double cosets are precisely the places of \(E=L^H\) over the base place. On a summand \(\operatorname{Ind}_J^D W_g\), taking \(I\)-invariants leaves copies of \(W_g^{I\cap J}\) indexed by \(I\backslash D/J\). A vector on an inertia orbit is determined by a stabilizer-invariant initial value, which proves this statement even when \(I\) is nontrivial. The arithmetic Frobenius cycles the \(f=[D:IJ]\) copies; after \(f\) steps it is the arithmetic Frobenius at the corresponding place of \(E\). The cyclic block determinant (6) changes \(q_v^{-s}\) into \(q_v^{-fs}=q_u^{-s}\). Multiplying over the double cosets gives the full local Euler identity, and absolute convergence then gives the global identity. At no point was \(I=1\) required. \subsubsection{Exercise 4 --- Discriminants from ramification (hard)}\label{exercise-4-discriminants-from-ramification-hard} Derive the conductor--discriminant formula directly from the different formula, including the multiplicities of global irreducible characters. \textbf{Solution.} At a fixed completion \(T/F\), its regular representation has dimension \(|G|\), and its \(G_j\)-invariants have dimension \(|G|/|G_j|\). Thus \[ a_F(\mathbf C[G]) =\frac{|G|}{|G_0|}\sum_{j\geq0}(|G_j|-1) =f(T/F)\delta_{T/F} =v_F(\mathfrak d_{T/F}). \tag{44} \] For a global Galois \(L/K\), the restriction of \(\mathbf C[\operatorname{Gal}(L/K)]\) to a decomposition group is the direct sum of \([G:D]\) copies of its regular representation: partition its basis into left \(D\)-orbits. Summing (44) therefore gives \([G:D]f(T/F)\delta_{T/F}\). The completed global integer lattice is the product of the \([G:D]\) local integer rings, and its trace determinant is the product of their trace determinants; hence that sum is exactly the valuation of \(\mathfrak d_{L/K}\). Globally the regular character is \(\sum_{\rho\in\operatorname{Irr}(G)}(\dim\rho)\chi_\rho\). Since (9) is linear in the character, the same discriminant exponent is \(\sum_\rho(\dim\rho)a_{K_v}(\rho)\). Taking the ideal product over all finite places proves (20), with precisely the displayed multiplicities. General conductor integrality was proved in section 3, so the factors are actual integral ideals. For a non-Galois subfield, the analogous permutation computation (16) uses different transitivity to give its discriminant as well. \subsection{What this lesson does not prove}\label{what-this-lesson-does-not-prove} Finite-group character theory, induction and integer Brauer induction are supplied by the three exact written representation-theory lessons identified at the beginning. The number-field rank-one analytic theorem is supplied by the written Hecke lesson; the function-field argument is proved in section 6. Hilbert's different formula and Herbrand's quotient theorem have the written local-field providers identified in lesson 10. The central Artin conductor integrality and discriminant arguments themselves are proved here. The general local epsilon existence theorem and Deligne’s orthogonal theorem have the exact programme providers \href{https://kokunoyumeto.github.io/open-math-courses/courses/LG-GAL/local-l-factors-and-epsilon-factors.html\#3-existence-and-uniqueness-of-local-constants}{LG-GAL-06, Theorem 3.0 and sections 3A–3D} and \href{https://kokunoyumeto.github.io/open-math-courses/courses/LG-GAL/local-l-factors-and-epsilon-factors.html\#7-sharp-twisting-and-orthogonal-root-numbers}{Theorem 7.4 with Lemmas 7.5–7.8}. The global invariant sum law used in section 8 is proved in this course’s lesson 24, Theorem 24.4. Equation (42) uses those local results and that sum law through the deduction above; the local theorem (41) is proved in its provider rather than repeated here. Artin’s number-field conjecture remains a conjecture. \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 ramified induction identities, conductor integrality, conductor--discriminant identity and meromorphic functional equation are proved using the named internal character and Fourier lessons. The function-field analytic argument is written in section 6. The orthogonal deduction uses LG-GAL-06, Theorem 7.4, with its real-induction, Clifford-class and dihedral Fourier lemmas 7.5–7.8. \begin{itemize} \tightlist \item \href{https://math.mit.edu/~poonen/786/notes.pdf}{Bjorn Poonen, Tate's Thesis, MIT 18.786 lecture notes (2015)}. \item \href{https://www.wwli.asia/downloads/YAlg1.pdf}{Wen-Wei Li, Yanqi Lake Lectures on Algebra: Part 1}. \item \href{https://publications.ias.edu/sites/default/files/Number20.pdf}{Pierre Deligne, Les constantes des équations fonctionnelles des fonctions L (1973), IAS archive}. \item \href{https://publications.ias.edu/sites/default/files/Number26.pdf}{Pierre Deligne, Les constantes locales de l'équation fonctionnelle de la fonction L d'Artin d'une représentation orthogonale (1976), IAS archive}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{The Chebotarev density theorem}\label{the-chebotarev-density-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).} \textbf{Lesson 22.} Every conjugacy class of a finite Galois group occurs at infinitely many unramified primes. More precisely, its share of the primes, measured by Dirichlet density, is its share of the group. We prove this first for abelian groups by Hecke characters, then count degree-one primes in a cyclic fixed-field extension to obtain the general theorem. Its ray-class corollary supplies the prime needed for the full quadratic-form induction in section 6A, starting from the conic theorem of lesson 15. The arithmetic Frobenius and Artin L-function conventions are those of \href{../artin-l-functions-conductors-and-discriminants.html}{Artin L-functions, conductors and discriminants}. The analytic number-field inputs are the written Theorem 10.1 of \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ADL-10}{Hecke L-functions and the Dedekind zeta function}, which makes a nontrivial finite-order Hecke L-function holomorphic at 1, and Theorem 16.2 of \href{https://kokunoyumeto.github.io/open-math-courses/courses/NT-CFT/proof-dependencies.html\#NT-ANT-16}{The Dedekind zeta function and the analytic class number formula}, which gives the positive simple zeta pole there. Neither input uses Chebotarev. The function-field versions needed here follow directly from section 6 of the preceding lesson, as checked below. We use the written existence and ray correspondence of lessons 17--18. \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-22}{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. Which density is being computed?}\label{which-density-is-being-computed} For a global field \(K\), let \(\mathcal P_K\) denote its finite places and write \(N\mathfrak p\) for the residue cardinality. For \(S\subseteq\mathcal P_K\), define its \textbf{Dirichlet density}, when the limit exists, by \[ \delta_K(S)=\lim_{s\downarrow1} \frac{\sum_{\mathfrak p\in S}(N\mathfrak p)^{-s}} {\log(1/(s-1))}. \tag{1} \] Here \(s\) is real and tends to 1 from above. Its \textbf{natural density}, when it exists, is \[ \lim_{X\to\infty} \frac{\#\{\mathfrak p\in S:N\mathfrak p\leq X\}} {\#\{\mathfrak p\in\mathcal P_K:N\mathfrak p\leq X\}}. \tag{2} \] The two are different definitions. Our Chebotarev theorem asserts (1). \textbf{Proposition 22.1.} The full set of primes has Dirichlet density 1. Densities lie in \([0,1]\), are monotone, are unchanged by finite modifications, and are finitely additive on disjoint sets whose densities exist. Complements have density \(1-\delta\). More generally a set whose prime sum remains bounded as \(s\downarrow1\) has density zero. Whenever the natural density (2) exists, its Dirichlet density exists and equals it. \textbf{Proof.} The zeta Euler product gives \[ \log\zeta_K(s)=\sum_{\mathfrak p}(N\mathfrak p)^{-s} +\sum_{\mathfrak p}\sum_{m\geq2}\frac{(N\mathfrak p)^{-ms}}m. \tag{3} \] For \(s\geq1\), the last sum is at most \(2\sum_{\mathfrak p}(N\mathfrak p)^{-2}\), a convergent subseries of \(\zeta_K(2)\). The positive simple pole at 1 gives \[ \sum_{\mathfrak p}(N\mathfrak p)^{-s} =\log\frac1{s-1}+O_K(1). \tag{4} \] For number fields the zeta input is the exact written theorem identified above. For function fields, use the test \(f=\prod_v\mathbf1_{\mathcal O_v}\) and the trivial character in (29)--(30) of the preceding lesson. Its zeta integral is \(\zeta_K(s)\). The positive-level parts are finite Laurent polynomials. At \(s=1\) the first correction has a simple pole with residue \(\kappa\widehat f(0)/\log R>0\); the second correction is regular. Indeed \(\widehat f(0)=\operatorname{vol}(\prod_v\mathcal O_v)>0\), with only finitely many local volumes different from 1. Thus the pole is simple and positive, proving (4) here too. Replacing the denominator of (1) by the full prime sum consequently gives the same limit. All assertions about bounds, inclusion, disjoint sums, complements and bounded contributions now follow from nonnegative summands and elementary limit laws. Finite contributions are bounded. Countable additivity is not asserted: a countable union of individual zero-density primes is the full prime set. For the last assertion, write \(A_S(t),A_K(t)\) for the two prime counts and suppose \(A_S(t)/A_K(t)\to d\). Partial summation gives \[ \sum_{\mathfrak p\in S}(N\mathfrak p)^{-s} =s\int_1^\infty A_S(t)t^{-s-1}\,dt \qquad(s>1). \tag{5} \] The endpoint at infinity vanishes: choose \(1<\sigma0\), choose \(T\) beyond which \(|A_S(t)-dA_K(t)|\leq\epsilon A_K(t)\). The integral of the difference on \([1,T]\) stays bounded as \(s\downarrow1\); the tail is at most \(\epsilon\) times the full prime sum. Divide by that diverging sum in (4) and let \(\epsilon\to0\). This proves the equality of densities. \(\square\) A useful relative version of a zero-density observation will be used in the proof. If \(E/K\) is finite and \(u\) is a prime of \(E\) of residue degree \(f(u/v)>1\) over \(K\), then \(Nu=(Nv)^{f(u/v)}\). There are at most \([E:K]\) primes over \(v\), and therefore \[ \sum_{\substack{u\in\mathcal P_E\\f(u/v)>1}}(Nu)^{-s} \leq[E:K]\sum_{v\in\mathcal P_K}(Nv)^{-2s}=O_{E/K}(1) \quad(s\downarrow1). \tag{6} \] Thus these relative degree-greater-than-one primes contribute no Dirichlet density. In particular this argument does not discard degree-one primes merely because their degree over the prime field is larger than one. \subsection{2. Finite-order Hecke characters do not vanish at 1}\label{finite-order-hecke-characters-do-not-vanish-at-1} \textbf{Proposition 22.2.} If \(\omega:C_K\to\mathbf C^\times\) is a nontrivial finite-order continuous Hecke character, then its primitive L-function is holomorphic and nonzero at 1. \textbf{Proof.} The kernel is open of finite index. Global existence gives its finite abelian class field \(L/K\), and reciprocity identifies \(\omega\) with a nontrivial character of \(G=\operatorname{Gal}(L/K)\). For every nontrivial character \(\chi\) of \(G\), Theorem 21.2 identifies its Artin L-function with its primitive finite-order Hecke L-function. For number fields the Hecke theorem makes it holomorphic at 1: a nontrivial finite-order character cannot be a pure imaginary norm twist, since the norm direction is connected and its finite image must be trivial. For function fields the same holomorphy follows from the explicit corrections in the preceding lesson. If \(\chi\) is nontrivial on \(C_K^1\), its correction is zero and its zeta integral is a finite Laurent polynomial. If it is trivial there, it has \(\chi(c)=\lambda\ne1\), because \(c\) generates the norm quotient and the global character is nontrivial. At \(s=1\), both denominators \(1-\lambda^{-1}\) and \(1-\lambda^{-1}R^{-1}\) are nonzero. Thus its integral and primitive L-function are holomorphic there. This checks explicitly the possible constant-field characters. Because \(G\) is abelian, all irreducible characters have dimension one. Proposition 21.1 gives \[ \zeta_L(s)=\zeta_K(s)\prod_{\substack{\chi\in\widehat G\\\chi\ne1}}L_K(s,\chi). \tag{7} \] Both zeta functions have simple poles with positive residues at 1. Their quotient is holomorphic and has nonzero value equal to the quotient of those residues. Every factor on the right of the quotient identity is holomorphic at 1, so none can vanish there. In particular \(L_K(1,\omega)\ne0\). This uses no density theorem or assumption about primes in a ray class. \(\square\) If extra primes are omitted to describe an imprimitive character modulo a larger ray modulus, its L-function is multiplied by finitely many factors \(1-\chi(\operatorname{Frob}_v)(Nv)^{-s}\). At 1 these are nonzero, because \(|\chi(\operatorname{Frob}_v)|=1{\raggedright\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2143}} >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2857}} >{\raggedright\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2143}} >{\raggedleft\arraybackslash}p{(\linewidth - 6\tabcolsep) * \real{0.2857}}@{}} \toprule\noalign{} \begin{minipage}[b]{\linewidth}\raggedright Frobenius class \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft Number of elements \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright Factor degrees of \(X^3-2\bmod p\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft Dirichlet density \end{minipage} \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot Identity & 1 & \(1+1+1\) & \(1/6\) \\ Transpositions & 3 & \(1+2\) & \(1/2\) \\ Three-cycles & 2 & \(3\) & \(1/3\) \\ \end{longtable} } In the first row the prime splits completely in the splitting field. In the third row the cubic is irreducible modulo \(p\). The finite exceptional primes affect none of these densities. \subsection{6A. General quadratic forms}\label{a.-general-quadratic-forms} \textbf{Theorem 22.7 (Hasse--Minkowski).} A nondegenerate quadratic form over a number field is isotropic over that field if and only if it is isotropic over every completion. \textbf{Proof.} The inputs are now available: Hasse's norm theorem and the conic principle in lesson 15; global reciprocity with local compatibility in lesson 16; existence in lesson 17; ray fields in lesson 18; and Corollary 22.6 above. The latter lets us prescribe a Frobenius element outside any finite set. Their proofs do not use the quadratic-form theorem. Dimension \(1\) is immediate. In dimension \(2\), local isotropy says a fixed element is a square everywhere. Its quadratic extension would split completely at every prime, contradicting Corollary 14.4 unless the extension is trivial. Dimension \(3\) is Corollary 15.4 of \href{../the-norm-index-bound-and-hasses-norm-theorem.html}{The norm index bound and Hasse's norm theorem}. Suppose \(n\geq4\), diagonalize, and write \(q=q_1+q_2\), with \(q_1\) binary and \(q_2\) of dimension \(n-2\geq2\). At every place, choose a nonzero \(t_v\) represented by both \(q_1\) and \(-q_2\). Local isotropy supplies such a value unless its two part-values are zero. In that case at least one part is isotropic and contains a hyperbolic plane, hence represents every scalar; choose a nonzero value of the other part. The simultaneous value set contains a neighborhood of \(t_v\). At a representation of a nonzero value, some coordinate \(z\ne0\). Changing just that coordinate changes the value by \(2czh+ch^2\), which maps a sufficiently small ball onto a neighborhood by contraction at finite places, and the usual one-variable inverse function argument at archimedean places. At a real place all multiples by positive scalars are represented, by scaling coordinates; at a complex place every nonzero scalar is represented. Choose \(S\) containing the archimedean and dyadic places, all nonunit diagonal coefficients, and the ramified places of the binary norm fields. If \(q_1=aX^2+bY^2\), its field is \(M_1=K(\sqrt{-b/a})\), omitting it if already split. If \(q_2\) is binary, also take the field for \(-q_2\), with leading coefficient \(a_2\). Outside \(S\), a binary unit form represents every unit: over the odd residue field it is either split or a scalar times the quadratic norm, which is surjective; a nonsingular residue representation lifts. A unit form of dimension at least \(3\) is isotropic locally. To see this, in three variables the two sets of values \(c_1x^2\) and \(-c_3-c_2y^2\), each of size \((q_v+1)/2\), intersect; the solution with third coordinate \(1\) lifts. An isotropic nondegenerate form contains a hyperbolic plane and represents all scalars. Let \(x\) be the idèle with components \(t_v\) in \(S\) and \(1\) elsewhere. For each binary norm field \(M_i\), representation says \(t_v/a_i\) is a local norm at \(S\). Global reciprocity for the principal element \(a_i\), whose other local symbols are trivial since it is a unit at unramified places, gives \[ \operatorname{rec}_{M_i/K}(x) =\prod_{v\in S}\operatorname{rec}_{M_i/K,v}(t_v) =\prod_{v\in S}\operatorname{rec}_{M_i/K,v}(a_i)=1. \tag{34} \] Choose a product subgroup \(U\subset J_K\), consisting of sufficiently deep principal units at the finite places in \(S\), positive reals at real places, full complex groups, and units outside \(S\). Require that \(t_vU_v\) lies in both local value sets and \(U_v\) lies in each binary local norm group. Its image \(H\subset C_K\) is open of finite index: it has full content image and its remaining quotient comes from compact \(C_K^1\). The ray class correspondence gives its abelian class field \(A/K\), containing the \(M_i\). The ray-prime theorem chooses a prime \(\mathfrak p\notin S\) with \[ \operatorname{Frob}_{\mathfrak p}(A/K) =\operatorname{rec}_{A/K}(x)^{-1}. \] By (34) it splits in every \(M_i\). If \(\pi_{\mathfrak p}\) is its idèle uniformizer, then \(x\pi_{\mathfrak p}=t u\) for some \(t\in K^\times\) and \(u\in U\). Thus \(t\) is in the chosen simultaneous value neighborhood at \(S\), is a unit outside \(S\cup\{\mathfrak p\}\), and has valuation \(1\) at \(\mathfrak p\). At this last prime every binary part is split, and every larger part is isotropic. Consequently \(t\) is represented by \(q_1\) and by \(-q_2\) at every place. Apply induction to \(q_1-tZ^2\) and \(q_2+tZ^2\), of dimensions \(3\) and \(n-1\). Each is globally isotropic. If its \(Z\)-coordinate is nonzero, division produces the required global representation. If that coordinate is zero, its original part is isotropic and represents every scalar via a hyperbolic plane, so again it represents the required value. Combining the representations gives a nonzero global isotropic vector of \(q\). This completes the induction. The hyperbolic-plane assertion used here is elementary: given an isotropic \(e\ne0\), choose \(f\) with the symmetric bilinear form \(B(x,y)=(q(x+y)-q(x)-q(y))/2\) and \(B(e,f)=1\), and replace \(f\) by \(f-q(f)e/2\). Their plane has form \(2XY\). \(\square\) \subsection{7. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- Complete splitting in the cubic splitting field (easy)}\label{exercise-1-complete-splitting-in-the-cubic-splitting-field-easy} Compute the density of primes splitting completely in \(\mathbf Q(\sqrt[3]2,\zeta_3)\). \textbf{Solution.} Eisenstein irreducibility and discriminant \(-108\), a nonsquare, give splitting group \(S_3\) of order six. At an unramified prime complete splitting means trivial decomposition group, equivalently identity Frobenius. The identity class has one element, so Theorem 22.4 gives density \(1/6\). Removing the ramified primes 2 and 3 makes no difference to the limit. \subsubsection{Exercise 2 --- Irreducible cubic reductions (medium)}\label{exercise-2-irreducible-cubic-reductions-medium} Compute the density of rational primes at which \(X^3-2\) is irreducible modulo \(p\). \textbf{Solution.} At \(p\ne2,3\), the polynomial is separable modulo \(p\). Its factor degrees are the lengths of Frobenius orbits on its three roots, by the finite-field orbit argument in section 6. It is irreducible exactly when Frobenius is a three-cycle. The two three-cycles form one conjugacy class in \(S_3\); hence its density is \(2/6=1/3\). At 2 the reduction is \(X^3\), and at 3 it is \((X+1)^3\), so neither is irreducible, and their exclusion in any event contributes zero density. \subsubsection{Exercise 3 --- Recovering a Galois field from split primes (medium)}\label{exercise-3-recovering-a-galois-field-from-split-primes-medium} Prove the equality assertion in Bauer's theorem directly from Chebotarev. \textbf{Solution.} Put the two Galois fields \(E,F\) in their Galois compositum \(N\). Let \(H_E,H_F\) be the kernels of restriction from \(G=\operatorname{Gal}(N/K)\). If the fields are distinct, the kernels are distinct, so some element belongs to one and not the other. Both kernels are normal, so its entire conjugacy class belongs to that same difference. Chebotarev gives that class positive density and therefore infinitely many unramified primes. Every one splits completely in the first field and fails to split completely in the second. This contradicts agreement of the splitting sets outside a finite set. Thus the kernels and their fixed fields are equal. This argument also explains why a mere finite list of matching split primes cannot determine the field. \subsubsection{Exercise 4 --- The pole argument without a hidden zero (hard)}\label{exercise-4-the-pole-argument-without-a-hidden-zero-hard} Prove Proposition 22.2 in detail, explaining why the zeta factorization forbids a zero of any nontrivial character and why extra omitted Euler factors do not cause an exception. \textbf{Solution.} Let \(L/K\) be the class field of \(\ker\omega\). Its group is finite abelian, and the primitive Artin--Hecke comparison identifies every group character \(\chi\) with its finite-order Hecke L-function. The analytic input gives holomorphy at 1 for every nontrivial character. Over number fields this is the finite-order case of the written Hecke theorem. Over function fields it follows from the finite-polynomial and correction formulas: a character nontrivial on norm-one classes has no correction, and a nontrivial norm character has \(\lambda\ne1\), so its denominators are nonzero at 1. The trivial character is precisely \(\zeta_K\), which has a simple pole with positive residue \(c_K\). The same zeta argument for \(L\) gives a positive simple pole with residue \(c_L\). In a punctured neighborhood of 1, divide (7) by \(\zeta_K\). The quotient extends holomorphically with value \(c_L/c_K\ne0\). Thus the finite product of the nontrivial holomorphic character L-functions has a nonzero value. If any one factor vanished, the product would vanish; other factors cannot cancel it by poles because they have no poles at 1. Equivalently, the total order at 1 in (7) is \(-1=-1+\sum_{\chi\ne1}\operatorname{ord}_1L(s,\chi)\), with every summand nonnegative, so all are zero. This proves nonvanishing for each character, including \(\omega\). A larger ray modulus deletes finitely many primitive factors, multiplying by \(1-\chi(\operatorname{Frob}_v)(Nv)^{-s}\) at character-unramified omitted primes. Each has absolute-value input \((Nv)^{-1}<1\) at 1 and cannot be zero there. Character-ramified primes already have primitive factor 1. Consequently neither holomorphy nor nonvanishing at 1 is altered by these extra omissions. No prime-density or ray-prime existence assertion was used in this argument. \subsection{What this lesson does not prove}\label{what-this-lesson-does-not-prove} The number-field Hecke continuation and positive zeta-pole theorems have the exact written internal providers identified at the beginning; the necessary function-field consequences were checked explicitly from the preceding lesson's proof. The global class-field and ray-field constructions were proved in lessons 17--18. The present lesson proves nonvanishing by their zeta factorization and supplies the character sums, cyclic fixed-field count, densities and applications in full. The theorem proved here is Dirichlet-density Chebotarev. We make no assertion of natural density or an effective error term. Those require additional prime-counting analysis. Proposition 22.1 proves only that an existing natural density must agree with the Dirichlet density; it supplies no converse. \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 lesson proves nonvanishing at 1, abelian character density and the cyclic fixed-field coset count, then Bauer and primes in ray classes. It proves Dirichlet density over both sorts of global field; no natural-density or effective assertion is substituted. \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://math.mit.edu/~poonen/786/notes.pdf}{Bjorn Poonen, Tate's Thesis, MIT 18.786 lecture notes (2015)}. \item \href{https://kskedlaya.org/cft/sec_abstractcft1.html}{Kiran S. Kedlaya, Notes on class field theory, author-hosted HTML edition}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Power residue symbols and reciprocity laws}\label{power-residue-symbols-and-reciprocity-laws} \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).} \textbf{Lesson 23.} A power residue symbol measures the Frobenius action on a radical. The global product of local Hilbert symbols then compares two power residue symbols with their arguments reversed. We recover quadratic reciprocity and both supplements, prove a number-field primary form of quadratic reciprocity, and compute the local factors needed for cubic and quartic reciprocity. The primary normalizations are specified separately for each law. Use the Hilbert symbol of \href{../hilbert-symbols-and-local-conics.html}{Hilbert symbols and local conics}: the first argument enters arithmetic reciprocity and the second is placed under the radical. Thus \[ (a,b)_{v,n}=\frac{\operatorname{rec}_{K_v,\mathrm{arith}}(a)(\sqrt[n]b)}{\sqrt[n]b}. \tag{1} \] Proposition 11.1 proves multiplicativity, the norm criterion and \((a,b)_{v,n}=(b,a)_{v,n}^{-1}\); Proposition 11.3 proves the tame formula. \href{../the-global-reciprocity-law.html}{The global reciprocity law}, especially its principal product and local compatibility, supplies the global input. No power reciprocity law is assumed in those proofs. \subsection{1. Residue symbols and their Frobenius interpretation}\label{residue-symbols-and-their-frobenius-interpretation} Let \(K\) be a global field containing \(\mu_n\), with \(n\geq2\) prime to its characteristic. At a finite prime \(\mathfrak p\nmid n\), put \(q=N\mathfrak p\), and suppose \(a\in K^\times\) is a local unit there. Reduction embeds \(\mu_n\) into \(\kappa(\mathfrak p)^\times\): a nontrivial \(n\)-th root cannot be congruent to 1, by the derivative-unit argument of lesson 11. In particular \(n\mid q-1\). Define the \textbf{power residue symbol} by the unique root \[ \left(\frac a{\mathfrak p}\right)_n\in\mu_n, \qquad \left(\frac a{\mathfrak p}\right)_n\equiv a^{(q-1)/n}\pmod{\mathfrak p}. \tag{2} \] The displayed residue is an \(n\)-th root, by the finite-field multiplicative group order. \textbf{Proposition 23.1.} The symbol depends only on the nonzero residue of \(a\), is multiplicative in \(a\), and equals 1 exactly when that residue is an \(n\)-th power. For a fractional ideal \(\mathfrak b=\prod\mathfrak p^{m_{\mathfrak p}}\), supported away from \(n\) and at places where \(a\) is a unit, the definition \[ \left(\frac a{\mathfrak b}\right)_n =\prod_{\mathfrak p}\left(\frac a{\mathfrak p}\right)_n^{m_{\mathfrak p}} \tag{3} \] is multiplicative in both its numerator and ideal denominator. \textbf{Proof.} Uniqueness in (2) gives residue dependence and multiplicativity. The finite-field group is cyclic. One elementary proof is to let \(m\) be the exponent of this finite abelian group: for every prime power dividing \(m\), an element attaining that prime-power order exists; their product has order \(m\). Every group element is a root of \(X^m-1\), so the root bound gives group order at most \(m\), and hence equal to \(m\). Thus it has a generator. If \(g\) is a generator of order \(q-1\), then \((g^r)^{(q-1)/n}=1\) exactly when \(n\mid r\), exactly when \(g^r\) is an \(n\)-th power. Formula (3) and its multiplicativity follow from unique prime ideal factorization, with negative exponents permitted for fractional ideals. \(\square\) For a principal denominator write \((a/b)_n=(a/(b))_n\). A composite denominator loses information: even for \(n=2\), a Jacobi symbol of 1 need not mean a square modulo that denominator. For example \((2/15)_2=(2/3)_2(2/5)_2=1\), whereas 2 is a square modulo neither 3 nor 5. \textbf{Proposition 23.2.} Let \(\beta^n=a\). The extension \(K(\beta)/K\) is unramified at \(\mathfrak p\) as above, and its arithmetic Frobenius satisfies \[ \frac{\operatorname{Frob}_{\mathfrak p}(\beta)}\beta =\left(\frac a{\mathfrak p}\right)_n =(\pi_{\mathfrak p},a)_{\mathfrak p,n} =(a,\pi_{\mathfrak p})_{\mathfrak p,n}^{-1}. \tag{4} \] The uniformizer can be any local prime element. \textbf{Proof.} In a finite residue extension containing an \(n\)-th root of \(\bar a\), the derivative of \(X^n-a\) at that unit root is a unit. The unramified lifting theorem and Hensel's iteration give a root in the corresponding unramified local extension. Hence the completion of \(K(\beta)\) is unramified. Its Frobenius ratio reduces to \(\bar\beta^{q-1}=\bar a^{(q-1)/n}\). That ratio is in \(\mu_n\), so reduction injectivity gives the first equality of (4). Arithmetic local reciprocity sends a uniformizer to precisely this Frobenius, giving the second equality through (1). Skew symmetry gives the inverse in the last equality. A change of uniformizer is a unit, which acts trivially on the unramified radical extension. The argument does not assume its degree equals \(n\). \(\square\) \subsection{2. The global product and general reciprocity}\label{the-global-product-and-general-reciprocity} \textbf{Theorem 23.3 (Hilbert's product formula).} For \(a,b\in K^\times\), \[ \prod_v(a,b)_{v,n}=1. \tag{5} \] Only finitely many factors differ from 1. \textbf{Proof.} The extension \(L=K(\sqrt[n]b)\) is finite cyclic, since \(\mu_n\subset K\); its degree may divide \(n\). At almost every finite place, \(a\) and \(b\) are units and \(n\) is a unit, so the radical extension is unramified and reciprocity of \(a\) is trivial. For the finitely many other places, global local compatibility says that the product of the local reciprocity automorphisms is the global symbol of the principal idèle \(a\). This is identity by global reciprocity. Evaluate it on a fixed global root of \(b\), using compatible embeddings in each completion. Its ratio is the product in (5), so that product is 1. \(\square\) At a tame finite place write \(a=\pi^r u\), \(b=\pi^t z\), with units \(u,z\). The precise tame formula of lesson 11 is \[ \overline{(a,b)_{v,n}} =\left((-1)^{rt}\frac{\bar z^{\,r}}{\bar u^{\,t}}\right)^{(Nv-1)/n}. \tag{6} \] In particular two units pair trivially. \textbf{Theorem 23.4 (general power reciprocity).} Suppose \(a,b\in\mathcal O_K\) are nonzero, their principal ideals are coprime, and both are prime to \(n\). Then \[ \left(\frac a b\right)_n \left(\frac b a\right)_n^{-1} =\prod_{v\mid n\infty}(a,b)_{v,n}. \tag{7} \] For function fields, the same statement uses the chosen valuation supports in place of \(\mathcal O_K\); outside the indicated exceptional places they must be disjoint. It also holds for fractional principal ideals with disjoint supports and unit components at the exceptional finite places. In positive characteristic prime to \(n\), there are no places dividing \(n\) and no infinite places, so the right side is empty and equals 1. \textbf{Proof.} At a prime of \((b)\), the numerator \(a\) is a unit and (4) and multiplicativity give \((a,b)_{v,n}=(a/v)_n^{-v(b)}\). At a prime of \((a)\), similarly \((a,b)_{v,n}=(b/v)_n^{v(a)}\). At all other tame places both are units, and (6) gives 1. Separate these factors in (5) to obtain \[ 1=\left(\frac b a\right)_n \left(\frac a b\right)_n^{-1} \prod_{v\mid n\infty}(a,b)_{v,n}, \] which is (7). The same argument works with negative valuations for fractional ideals. \(\square\) Thus a Frobenius residue symbol is \((\pi,a)\), whereas reversing the entries to \((a,\pi)\) inserts an inverse. For quadratic signs this inverse is invisible; for higher powers it is essential. \subsection{3. Quadratic reciprocity over the rationals}\label{quadratic-reciprocity-over-the-rationals} For distinct odd positive primes \(p,q\), (7) with \(n=2\) has only the 2-adic and real exceptional factors. The real factor is 1 because both entries are positive. The explicit 2-adic unit formula proved in lesson 11 gives \[ (p,q)_{2,2}=(-1)^{((p-1)/2)((q-1)/2)}. \tag{8} \] \textbf{Corollary 23.5.} The Legendre symbols satisfy \[ \left(\frac p q\right)\left(\frac q p\right) =(-1)^{((p-1)/2)((q-1)/2)},\quad \left(\frac{-1}p\right)=(-1)^{(p-1)/2},\quad \left(\frac2p\right)=(-1)^{(p^2-1)/8}. \tag{9} \] \textbf{Proof.} For the first equation, inversion equals identity on signs, so (7) and (8) give the assertion. For \((-1/p)\), use (5) on \((-1,p)\). At \(p\), (4) gives the inverse of \((-1/p)\), hence the same sign. The real factor is 1 because \(p>0\). Other odd places have two unit entries and factor 1. The 2-adic factor, by the unit formula, is \((-1)^{(p-1)/2}\). Their product is 1, giving the first supplement. For \((2/p)\), use (5) on \((2,p)\). The only possible nontrivial factors are at 2 and \(p\); the real factor is again 1. At \(p\) the factor is \((2/p)^{-1}=(2/p)\). The full 2-adic formula, with valuation of the first entry 1 and the second entry an odd unit, gives \((2,p)_{2,2}=(-1)^{(p^2-1)/8}\). The product proves the second supplement. \(\square\) For example 2 is a square modulo an odd prime exactly when the prime is 1 or 7 modulo 8. This conclusion comes from the local obstruction at 2, rather than from an unexplained global sign. \subsection{4. A primary quadratic law over any number field}\label{a-primary-quadratic-law-over-any-number-field} In this section a number \(\alpha\in\mathcal O_K\) is \textbf{primary for the quadratic law} if it is odd and \(\alpha\equiv\xi^2\pmod{4\mathcal O_K}\) for some \(\xi\in\mathcal O_K\). This definition belongs to this quadratic statement; the cubic and quartic normalizations below are different. For odd, coprime \(\alpha,\beta\), with at least one primary in this sense, the general quadratic law takes the particularly simple form \[ \left(\frac\alpha\beta\right)_2 \left(\frac\beta\alpha\right)_2 =(-1)^{\#\{v\text{ real}:\alpha_v<0,\ \beta_v<0\}}. \tag{10} \] In particular it is symmetric if at least one of the entries is totally positive. This is the primary quadratic reciprocity law in the notation of our symbols. \textbf{Proof.} Suppose \(\alpha\) is primary. At each dyadic completion \(F\), \(\xi\) is a unit and \(\alpha/\xi^2=1+4c\), \(c\in\mathcal O_F\). An element \(\theta\) satisfying \(\theta^2+\theta=c\) gives \((1+2\theta)^2=1+4c\). Its equation reduces to a separable quadratic over the finite residue field, since its derivative \(2\theta+1\) is a unit. If the reduced equation has a root it lifts in \(F\); otherwise it lifts in the unramified quadratic extension. Hence \(F(\sqrt\alpha)/F\) is trivial or unramified. The unit \(\beta\) has trivial reciprocity on it, so \((\beta,\alpha)_{F,2}=1\), and skew symmetry gives \((\alpha,\beta)_{F,2}=1\). Thus every dyadic factor in (7) is 1. Complex factors are 1. At a real place the quadratic Hilbert symbol is \(-1\) exactly when both entries are negative, as proved in lesson 11. Their product is (10). The argument with \(\alpha,\beta\) exchanged is identical. \(\square\) The proof above uses the unramified local quadratic equation and the already proved global product, with no hypothesis that the field's different is principal. The more general law without a primary entry remains (7), retaining its dyadic factors. \subsection{5. Cubic reciprocity and its local calculation}\label{cubic-reciprocity-and-its-local-calculation} Put \(\omega=e^{2\pi i/3}\), \(K=\mathbf Q(\omega)\), and \(\lambda=1-\omega\). Its integer ring is \(\mathbf Z[\omega]\); \(\lambda^2=-3\omega\), and the unique prime over 3 has residue field \(\mathbf F_3\). We call an element prime to 3 \textbf{primary for the cubic law} if it is congruent to \(-1\pmod3\). Each prime away from 3 has a unique associate with this normalization. Indeed the six units \(\{\pm1,\pm\omega,\pm\omega^2\}\) have distinct residues and exhaust the six units modulo \(\lambda^2\); multiplication by exactly one gives residue \(-1\). Their distinctness follows because differences between distinct cube roots have valuation 1, and changing sign gives a unit difference. The norm \(a^2-ab+b^2\) also shows directly that these are all the global units. Allowing either sign modulo 3 gives the alternate two-associate convention; \(-1\) is a cube, so it changes no cubic symbol. The necessary wild local assertion will be proved explicitly. Let \(F=\mathbf Q_3(\omega)\), with \(U^j=1+\lambda^j\mathcal O_F\). \textbf{Cubic local lemma.} If \(a,b\in U^2\), then \((a,b)_{F,3}=1\). \textbf{Proof.} Write \(b=1+\lambda^2 c\). If \(b\) is a cube, the assertion is immediate. Otherwise \(L=F(\gamma)\), \(\gamma^3=b\), is cyclic of degree three. For \(t\in\mathcal O_F\), \[ N_{L/F}(1+t(\gamma-1))=(1-t)^3+bt^3 =1+\lambda^2 P(t), \quad P(t)=\omega^2t-\omega^2t^2+ct^3. \tag{11} \] The first identity is the product over the three conjugates of \(\gamma\); the second uses \(3=-\omega^2\lambda^2\). Since \(P'(0)=\omega^2\) is a unit, for each \(d\in\lambda\mathcal O_F\) Hensel's iteration at zero solves \(P(t)=d\) with \(t\in\lambda\mathcal O_F\). Thus every element of \(U^3\) is a norm by (11). Modulo \(\lambda\), \(P(1)=c\) and \(P(-1)=1-c\). At least one of these residues is nonzero in \(\mathbf F_3\). The corresponding norm in (11) lies in \(U^2\) and has nonzero image in \(U^2/U^3\simeq(\mathbf F_3,+)\), so its image generates that quotient. The norm group already contains all of \(U^3\), hence contains all of \(U^2\): adjust any target's residue by a power of that norm and its remaining part is in \(U^3\). Every element used in (11) is nonzero, since its norm is a unit. The norm criterion of Proposition 11.1 now gives \((a,b)_{F,3}=1\). \(\square\) \textbf{Cubic reciprocity.} For nonassociate primary primes \(\pi,\rho\) away from 3, \[ \left(\frac\pi\rho\right)_3=\left(\frac\rho\pi\right)_3. \tag{12} \] The same holds for coprime primary elements. \textbf{Proof.} Their negatives lie in \(U^2\) at \(\lambda\), and their signs are cubes, so the lemma makes the local exceptional factor 1. All infinite places are complex. Therefore (7) has right side 1 and gives (12). Multiplicativity gives the version for coprime primary elements directly from the same local argument. \(\square\) The use of the full residue cardinality matters. For a rational prime remaining prime in \(\mathbf Z[\omega]\), that cardinality is \(p^2\), not \(p\). Formula (2) works without distinguishing split and inert primes. No unspecified local cubic symbol is left in the proof. \subsection{6. Quartic reciprocity and a two-generator local pairing}\label{quartic-reciprocity-and-a-two-generator-local-pairing} Now put \(K=\mathbf Q(i)\), \(\lambda=1+i\). An odd Gaussian element is \textbf{primary for the quartic law} if it is 1 modulo \(\lambda^3\), equivalently modulo \(2+2i\), since these generators are associates. The four units \(\mu_4\) give all four distinct unit residues modulo \(\lambda^3\), so every odd Gaussian prime has a unique primary associate. For \(a+bi\) this congruence says that \(a\) is odd, \(b\) is even, and \(a+b\equiv1\pmod4\). Work locally in \(F=\mathbf Q_2(i)\), with \(U^j=1+\lambda^j\mathcal O_F\). The integer ring is \(\mathbf Z_2[i]\), as follows from the Eisenstein equation for \(\lambda\). For \(u\in U^3\), its norm is 1 modulo 4, by the coordinate description just given. Define \[ e(u)=\frac{N_{F/\mathbf Q_2}(u)-1}{4}\pmod2. \tag{13} \] It is a homomorphism on \(U^3\): multiply two norms congruent to 1 modulo 4 and reduce the quotient by 4 modulo 2. \textbf{Quartic local lemma.} For \(u,v\in U^3\), \[ (u,v)_{F,4}=(-1)^{e(u)e(v)}. \tag{14} \] \textbf{Proof.} First \((U^3)^2=U^5\). If \(x\in\lambda^3\mathcal O_F\), then \((1+x)^2-1=2x+x^2\in\lambda^5\mathcal O_F\). Conversely for \(y\in\lambda^5\mathcal O_F\), solve \(x=y/2-x^2/2\) by contraction in \(\lambda^3\mathcal O_F\). The constant term has valuation at least 3, the quadratic term at least 4, and the difference at two arguments is multiplied by \((x+z)/2\), of valuation at least 1. The iterates converge to the required square root in \(U^3\). The quotient \(U^3/U^5\) has order four, since both successive residue quotients have order two. It has basis \[ g=-1+2i,\qquad h=5. \tag{15} \] Indeed \(v_F(g-1)=3\), \(v_F(h-1)=4\), and \(g^2-1=-4-4i\), \(h^2-1=24\) have valuations 5 and 6. Thus both have order two in the quotient and are independent by their distinct leading levels. We compute three quartic pairings using the already proved global product formula over \(\mathbf Q(i)\) and the tame formula. This uses no quartic reciprocity theorem. The only odd places involved lie over 5, generated by \(g\) and \(\bar g=-1-2i\); both have norm 5, and \(5=g\bar g\). For \((g,g)_4\), the tame factor at \(g\) is \(-1\), and every other odd factor is 1. For \((h,h)_4\), each of the two places over 5 has tame factor \(-1\), with product 1. All infinite factors are trivial. Hence (5) gives \[ (g,g)_{F,4}=-1,\qquad(h,h)_{F,4}=1. \tag{16} \] For \((g,h)_4\), at \(g\) formula (6) gives the residue \(-\bar g=2\pmod g\). Here \(i=3\pmod g\), so that factor is \(-i\). At \(\bar g\), it gives \(g^{-1}=2\pmod{\bar g}\), and \(i=2\pmod{\bar g}\), so that factor is \(i\). Their product is 1, yielding \[ (g,h)_{F,4}=1. \tag{17} \] Compatibility of radicals gives \((a,b)_4^2=(a,b)_2\): the square of a fourth root of \(b\) is a square root. The squares of all pairings in (16)--(17) are 1, and the quadratic symbol is skew symmetric. Since \(U^5=(U^3)^2\), its square classes are already accounted for by the two generators (15). Bilinearity therefore shows that the quadratic pairing is trivial on all of \(U^3\times U^3\). Consequently all quartic values there are signs. Also \(U^5\) is in the radical of that restricted quartic pairing, since for \(z,v\in U^3\), \((z^2,v)_4=(z,v)_4^2=(z,v)_2=1\), and likewise in the other argument. Thus (16)--(17) describe the entire restricted pairing on the two-dimensional binary quotient \(U^3/U^5\): its only negative basis pairing is the \(g,g\) entry. Finally \(e(g)=1\), \(e(h)=0\), and \(e\) kills \(U^5\), since the square of a norm congruent to 1 modulo 4 is 1 modulo 8. Its value is precisely the coefficient of \(g\) in this quotient. This proves (14). \(\square\) \textbf{Quartic reciprocity.} For distinct primary Gaussian primes \(\pi,\rho\), \[ \left(\frac\pi\rho\right)_4 =\left(\frac\rho\pi\right)_4 (-1)^{((N\pi-1)/4)((N\rho-1)/4)}. \tag{18} \] \textbf{Proof.} Both local entries at the unique dyadic prime lie in \(U^3\). Their local norms are their ordinary Gaussian norms, so (13)--(14) give exactly the sign in (18). This is the only exceptional finite factor in (7), and the infinite places are complex. Substitute it into (7) to obtain (18). \(\square\) The term biquadratic reciprocity also denotes this fourth-power law. Its sign is absent from cubic reciprocity because the corresponding primary cubic local pairing vanishes. The local unit quotient computation above explains the quartic sign rather than imposing it as a convention. \subsection{7. Exercises and complete solutions}\label{exercises-and-complete-solutions} \subsubsection{Exercise 1 --- The second quadratic supplement (easy)}\label{exercise-1-the-second-quadratic-supplement-easy} Derive \((2/p)=(-1)^{(p^2-1)/8}\) for an odd positive prime from the product formula. \textbf{Solution.} In \(\prod_v(2,p)_{v,2}=1\), every odd place other than \(p\) has two unit entries and tame factor 1. The real factor is 1. The factor at \(p\) is the inverse of \((2/p)\) by (4), which is the same sign. The explicit dyadic formula gives \((2,p)_{2,2}=(-1)^{(p^2-1)/8}\). Thus those two remaining signs multiply to 1 and are equal. The exponent is even for \(p\equiv1,7\pmod8\) and odd for \(p\equiv3,5\pmod8\), giving the familiar residue criterion. \subsubsection{Exercise 2 --- Recovering the symbol from Frobenius (medium)}\label{exercise-2-recovering-the-symbol-from-frobenius-medium} Show that \((a/\mathfrak p)_n\) is determined by the Frobenius in \(K(\sqrt[n]a)/K\), including when the radical extension has degree smaller than \(n\). \textbf{Solution.} Suppose \(a\) is a unit and \(\mathfrak p\nmid n\). A residue root lifts in an unramified local extension because the derivative is a unit. Choose a global root \(\beta\) and a place above \(\mathfrak p\). Arithmetic Frobenius sends its residue to \(\bar\beta^q\), so its ratio with \(\beta\) is a root in \(\mu_n\) reducing to \(\bar a^{(q-1)/n}\). Reduction injectivity on \(\mu_n\) makes this ratio exactly (2). Replacing \(\beta\) by another root multiplies it by an element of \(\mu_n\subset K\), fixed by Frobenius, so changes no ratio. Conjugating the chosen place changes no action in this abelian radical extension. Nothing in these steps asserts degree \(n\); the Frobenius may have smaller order and its ratio then lies in the corresponding subgroup of \(\mu_n\). Finally it is \((\pi,a)_n\), and equals the inverse of \((a,\pi)_n\) with the convention (1). \subsubsection{\texorpdfstring{Exercise 3 --- The quartic symbol of \(i\) (medium)}{Exercise 3 --- The quartic symbol of i (medium)}}\label{exercise-3-the-quartic-symbol-of-i-medium} For an odd Gaussian prime \(\pi\), compute \((i/\pi)_4\). Give examples with values \(i\) and \(-1\). \textbf{Solution.} The residue cardinality satisfies \(4\mid N\pi-1\), since \(\mu_4\) injects into its multiplicative group. Formula (2) says the desired root reduces to \(i^{(N\pi-1)/4}\). This is already a global fourth root of unity, so injectivity gives the exact equality \[ \left(\frac i\pi\right)_4=i^{(N\pi-1)/4}. \tag{19} \] For \(\pi=-1+2i\), norm 5, the exponent is 1 and the value is \(i\). For the inert Gaussian prime \(\pi=-3\), norm 9, it is 2 and the value is \(-1\). Both chosen associates are primary, although (19) requires no primary normalization. Using norm 3 instead of 9 for the second prime would not even give an integer exponent and would be incorrect. \subsubsection{Exercise 4 --- The primary cubic local norm calculation (hard)}\label{exercise-4-the-primary-cubic-local-norm-calculation-hard} Prove cubic reciprocity for nonassociate primary primes by computing the exceptional Hilbert symbol at \(1-\omega\). \textbf{Solution.} Primary primes are \(-1\pmod3\). Their negatives are in \(U^2\subset\mathbf Q_3(\omega)^\times\), and \(-1\) is a cube. It is enough to prove every element of \(U^2\) is a norm from the radical field of any \(b\in U^2\). If \(b\) is a cube this is automatic. Otherwise write \(b=1+\lambda^2c\) and \(\gamma^3=b\). Compute the norm of \(1+t(\gamma-1)\) as in (11). The derivative \(P'(0)=\omega^2\) is a unit, so for every \(d\in\lambda\mathcal O_F\) the equation \(P(t)=d\) has a root near zero. Consequently all of \(U^3\) is in the norm group. Among \(P(1)\) and \(P(-1)\), the residues are \(c\) and \(1-c\); at least one is nonzero and its norm generates the three-element quotient \(U^2/U^3\). The norm group therefore contains all of \(U^2\). The norm criterion proves that the Hilbert symbol of the two primary entries is 1 at \(\lambda\). This is the only exceptional finite place, and complex symbols are trivial. The general reciprocity equation (7) gives (12). This supplies the local calculation itself, rather than citing a wild-symbol formula without its proof. \subsection{What this lesson does not prove}\label{what-this-lesson-does-not-prove} The Kummer, local Hilbert-symbol and global principal reciprocity results are the exact earlier written course inputs specified at the beginning. All five numbered results, the arbitrary-number-field primary quadratic theorem, and the cubic and quartic primary laws are proved here. The necessary wild primary local calculations are given explicitly in sections 5--6. Further unit and ramified-prime supplementary laws for higher powers are not asserted here. \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 global Hilbert product and the tame formula give the general power-residue identity. The primary quadratic law keeps its real sign. The explicit cubic norm polynomial and quartic local two-generator pairing prove the primary cubic and quartic laws in the lesson. \begin{itemize} \tightlist \item \href{https://www.jmilne.org/math/CourseNotes/CFT.pdf}{J. S. Milne, Class Field Theory, version 4.03}. \end{itemize} The \href{../free-proofs.html}{proof guide} gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments. \clearpage \section{Brauer groups of local and global fields}\label{brauer-groups-of-local-and-global-fields} \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).} \textbf{Lesson 24.} A Brauer class measures the obstruction to making a central simple algebra into a matrix algebra. Over a nonarchimedean local field, one rational number modulo integers measures the entire obstruction. Over a global field, the local numbers determine the algebra, and their sum must vanish. We prove both assertions, then use their cohomology to construct fundamental classes, Weil extensions and the class-field-tower bound. The algebraic prerequisites are the written Noether-course lessons \href{https://kokunoyumeto.github.io/open-math-courses/courses/AG-LTF/support/NOE-HYP--NOE-HYP-04.html\#4-maximal-subfields-and-splitting-fields}{Central simple algebras and the Brauer group}, Theorems 4.2--4.3 and 5.1, and \href{https://kokunoyumeto.github.io/open-math-courses/courses/AG-LTF/support/NOE-HYP--NOE-HYP-05.html\#3-multiplication-in-the-relative-brauer-group}{Crossed products and factor systems}, Theorems 3.2 and 4.1 and Proposition 6.2. In particular, every class has a finite Galois splitting field, and, for finite Galois \(L/K\) with group \(G\), \[ \operatorname{Br}(L/K)=H^2(G,L^\times),\qquad \operatorname{Br}(K)=H^2_{\mathrm{cont}}(G_K,(K^s)^\times). \tag{1} \] These are actual preceding algebra proofs, including the convention on crossed products. We use ordinary cochains and Shapiro from \href{https://kokunoyumeto.github.io/open-math-courses/courses/ag-etale-cohomology/galois-cohomology-and-the-etale-cohomology-of-a-field.html}{Galois cohomology and the étale cohomology of a field}, §§2--3 and 5.1--5.2. The finite-group Tate constructions needed beyond those sections are developed below. Our arithmetic prerequisites are \href{../local-reciprocity-and-norm-groups.html}{Local reciprocity and norm groups}, \href{../the-norm-index-bound-and-hasses-norm-theorem.html}{The norm-index bound and Hasse's norm theorem}, and \href{../the-global-reciprocity-law.html}{The global reciprocity law}. Their proofs precede the present Brauer reformulation. The invariant is normalized by \textbf{arithmetic} Frobenius: an unramified cyclic algebra with Frobenius generator and uniformizer parameter has invariant \(1/n\). \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-24}{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. Cohomological tools for a finite extension}\label{cohomological-tools-for-a-finite-extension} We first isolate the arguments that turn cyclic computations into statements about arbitrary finite Galois groups. Coefficients in this section are written additively. \subsubsection{Restriction, norms and inflation}\label{restriction-norms-and-inflation} For \(H\subseteq G\), restriction and corestriction satisfy \[ \operatorname{Cor}_H^G\operatorname{Res}_H^G=[G:H]. \tag{2} \] Here is a construction, rather than an appeal to the identity. With \(\operatorname{Coind}_H^G A=\{f:G\to A:f(hg)=hf(g)\}\), the maps \[ A\longrightarrow\operatorname{Coind}_H^G A,\quad a\longmapsto(g\mapsto ga), \qquad f\longmapsto\sum_{g\in H\backslash G}g^{-1}f(g) \tag{3} \] are \(G\)-linear, and their composite is \([G:H]\). The summand is independent of the coset representative. Replacing \(g\) by \(gx\) proves equivariance of the sum. Under Shapiro the first map induces restriction, so the second defines corestriction and proves (2). Consequently restriction to a Sylow \(p\)-subgroup injects the \(p\)-primary part of positive cohomology. Every positive cohomology class is killed by \(|G|\): in homogeneous cochains the sum of the contractions inserting each element of \(G\) is an equivariant homotopy for multiplication by \(|G|\). For \(H\triangleleft G\), \(Q=G/H\), we shall also use \[ 0\to H^1(Q,A^H)\to H^1(G,A)\to H^1(H,A), \tag{4} \] and, when \(H^1(H,A)=0\), \[ 0\to H^2(Q,A^H)\xrightarrow{\mathrm{Inf}}H^2(G,A) \xrightarrow{\mathrm{Res}}H^2(H,A). \tag{5} \] To verify these low-degree assertions, take an injective \(G\)-resolution \(I^\bullet\) of \(A\). Restriction makes its terms \(H\)-injective, because induction is an exact left adjoint. The terms \((I^j)^H\) are \(Q\)-injective, because inflation is an exact left adjoint to taking \(H\)-invariants. Apply the free bar resolution of \(Q\) to the complex \((I^\bullet)^H\). The two filtrations of the resulting first-quadrant double complex have respectively cohomology \(H^*(G,A)\) and second page \[ H^i(Q,H^j(H,A)). \tag{6} \] Indeed the first filtration kills positive horizontal degrees termwise by injectivity; the second first takes the vertical cohomology \(H^j(H,A)\). Each total degree has finitely many terms, so the filtrations converge without a limit issue. The entries of total degree one give (4). If the row \(j=1\) vanishes, the edge in total degree two gives exactly (5). If \(H^j(H,A)=0\) for \(1\leq j\leq m\), inflation is an isomorphism in degrees \(1,\ldots,m\). The edge maps are the displayed cochain inflation and restriction: they agree in degree zero and on connecting maps. This supplies the particular spectral-sequence calculation used here. \subsubsection{A cyclic-to-general lemma}\label{a-cyclic-to-general-lemma} Suppose a system of modules \(A_L\) indexed by finite Galois field extensions has \(A_L^H=A_{L^H}\), and that for every cyclic extension of prime degree \(p\), \[ H^1(C_p,A_L)=0,\qquad |H^2(C_p,A_L)|=p. \tag{7} \] Then for arbitrary finite Galois \(L/K\), \[ H^1(G,A_L)=0,\qquad |H^2(G,A_L)|\text{ divides }|G|. \tag{8} \] \textbf{Proof.} First take a \(p\)-group and induct on its order. It has a normal subgroup \(H\) of index \(p\): a nontrivial finite \(p\)-group has a nontrivial center, and induction on the quotient by a central order-\(p\) subgroup gives an order-\(p\) quotient. Equations (4) and (7) kill \(H^1(G,A_L)\). Equation (5) embeds a group of order \(p\) as the kernel of restriction in degree two, whose image has order dividing \(|H|\) by induction. Thus the order divides \(p|H|\). For a general group, restrict to its Sylow groups and use (2): \(H^1\) vanishes primary part by primary part, and the \(p\)-primary part of \(H^2\) embeds in a group of order dividing the order of the Sylow group. These injections also prove finiteness. Multiplying the bounds proves (8). \(\square\) For \(A_L=L^\times\), Hilbert 90 and cyclic local reciprocity prove (7). For \(A_L=C_L\), fixed classes descend by lesson 13. The cyclic Herbrand quotient of lesson 14 and the exact norm index of lesson 16 give \(H^1(C_p,C_L)=0\); cyclic periodicity gives \(H^2(C_p,C_L)=C_K/NC_L\) of order \(p\). Thus (8) applies to both arithmetic systems, in both characteristics. \subsection{2. Unramified algebras and the invariant}\label{unramified-algebras-and-the-invariant} Let \(K\) be a nonarchimedean local field with finite residue field, \(K_n/K\) its unramified extension of degree \(n\), and \(F\) arithmetic Frobenius. Valuations are normalized to have image \(\mathbf Z\). \subsubsection{Proposition 24.1. The unramified invariant}\label{proposition-24.1.-the-unramified-invariant} Valuation induces an isomorphism \[ H^2(\operatorname{Gal}(K_n/K),K_n^\times) \xrightarrow{\sim}H^2(C_n,\mathbf Z) \xrightarrow{\sim}\tfrac1n\mathbf Z/\mathbf Z. \tag{9} \] The cyclic algebra \((K_n/K,F,a)\), defined by \(ux=F(x)u\), \(u^n=a\), maps to \(v_K(a)/n\). Inflation between unramified extensions preserves this rational invariant. Their union therefore gives \[ \operatorname{Br}(K^{\mathrm{ur}}/K)\simeq\mathbf Q/\mathbf Z. \tag{10} \] \textbf{Proof.} The unit module \(U_{K_n}\) has zero Tate cohomology in every degree. In degree zero this is surjectivity of unramified unit norms. In degree minus one, Hilbert 90 writes a norm-one unit as \(F(b)/b\); multiplication of \(b\) by a power of a uniformizer in \(K\) makes \(b\) a unit. The same statements hold for every subgroup. Cyclic periodicity proves the assertion in all degrees. The exact sequence \[ 1\to U_{K_n}\to K_n^\times\xrightarrow{v_{K_n}}\mathbf Z\to0 \] therefore makes valuation an isomorphism in degree two. The cyclic resolution identifies \(H^2(C_n,\mathbf Z)\) with \(\mathbf Z/n\mathbf Z\). Equivalently the boundary from \(0\to\mathbf Z\to\mathbf Q\to\mathbf Q/\mathbf Z\to0\) identifies it with characters of \(C_n\); evaluation at \(F\) gives (9). The parameter cocycle is 1 except at a carry in adding Frobenius exponents, where it is \(a\). Its valuation is the carry cocycle with parameter \(v_K(a)\). This proves the stated formula. Inflation pulls back the character on Frobenius, keeping its value unchanged. All rational numbers modulo integers occur at some finite level. \(\square\) There is an important base-change calculation before passing to the whole Brauer group. If \(E/K\) is any finite extension, with ramification index \(e\) and residue degree \(f\), an unramified class of invariant \(r\) restricts to an unramified class over \(E\) of invariant \[ \operatorname{inv}_E(\operatorname{Res}_{E/K}b)=efr=[E:K]r. \tag{11} \] To see this on the preceding construction, restrict the character to the unramified base change. Its Frobenius restricts to \(F^f\), multiplying the character value by \(f\). Changing normalized valuations multiplies the valuation cocycle by \(e\). Boundary maps commute with both operations, yielding (11). This argument also permits \(E/K\) to be non-Galois. \subsubsection{Theorem 24.2. Local Brauer groups}\label{theorem-24.2.-local-brauer-groups} For every nonarchimedean local field with finite residue field, \[ \operatorname{inv}_K:\operatorname{Br}(K)\xrightarrow{\sim}\mathbf Q/\mathbf Z. \tag{12} \] For \(L/K\) finite Galois of degree \(n\), its relative group is exactly the subgroup \(\tfrac1n\mathbf Z/\mathbf Z\). Restriction multiplies invariants by the extension degree; corestriction preserves them. Also \[ \operatorname{Br}(\mathbf R)=\tfrac12\mathbf Z/\mathbf Z, \qquad \operatorname{Br}(\mathbf C)=0. \tag{13} \] \textbf{Proof.} Put \(M=LK_n\). Every class split by \(K_n\) restricts to zero over \(L\): its invariant over \(L\), calculated in the unramified extension \(M/L\), is \(n\) times its original invariant by (11). Equation (5), applied to \(M/L/K\) and Hilbert 90, then places it in \(H^2(G,L^\times)\). This gives a subgroup of order \(n\) there. The bound (8) says that the whole group has order dividing \(n\), so it is precisely that subgroup. Every Brauer class has a finite Galois splitting field by (1), proving (12) and the relative assertion. Formula (11) now applies to every class. For corestriction, restriction \(\operatorname{Br}(K)\to\operatorname{Br}(E)\) is surjective because multiplication by \([E:K]\) on \(\mathbf Q/\mathbf Z\) is surjective. Write a class \(c\) over \(E\) as \(\operatorname{Res}b\). Then (2) gives \(\operatorname{inv}_K(\operatorname{Cor}c)=[E:K]\operatorname{inv}_K(b)=\operatorname{inv}_E(c)\). Over \(\mathbf C\), all central simple algebras split. Over \(\mathbf R\), every class is split by \(\mathbf C\), so (1) and the cyclic resolution give \(H^2(C_2,\mathbf C^\times)=\mathbf R^\times/N\mathbf C^\times\). Complex norms are exactly the positive reals; this group has two elements, with negative parameter giving Hamilton's algebra. Assign that class invariant \(1/2\). This proves (13) without importing a classification of real division algebras. \(\square\) \subsubsection{Cyclic algebras and the Artin character}\label{cyclic-algebras-and-the-artin-character} If \(L/K\) is cyclic, \(\sigma\) a generator, and \(\chi(\sigma)=1/n\), then \[ \operatorname{inv}_K(L/K,\sigma,a) =\chi(\operatorname{rec}_{K,\mathrm{arith}}(a)). \tag{14} \] One must check this normalization for ramified extensions as well. Regard the class as \(\delta\chi\cup a\); rational lifts of \(\chi\) give exactly the carry cocycle used in the cyclic algebra. The cochain cup product and transfer satisfy the projection formula: distribute the coset sum (3) over a cup product whose other factor is restricted from \(G\); the restricted factor is pulled outside each summand. In degree zero the coefficient transfer is the field norm. Now choose the positive Frobenius lift of \(\sigma\) and its fixed field \(\Sigma\) in the enlarged unramified extension, as in \href{../frobenius-lifts-and-abstract-reciprocity.html}{Frobenius lifts and abstract reciprocity}, sections 2 and 5--6. Over \(\Sigma\), the character becomes unramified and its arithmetic Frobenius has value \(\chi(\sigma)\). For a uniformizer \(\pi_\Sigma\), Proposition 24.1 gives \(\operatorname{inv}_\Sigma(\delta\chi|_\Sigma\cup\pi_\Sigma)=\chi(\sigma)\). Corestriction preserves the invariant by Theorem 24.2, so the projection formula gives that same invariant for parameter \(N_{\Sigma/K}\pi_\Sigma\). The Frobenius-lift definition in lesson 4 says that this parameter represents the inverse reciprocity image of \(\sigma\). Its class generates \(K^\times/NL^\times\). Both sides of (14) are homomorphisms on that cyclic quotient, so agreement on this generator proves (14). The real version is the same sign calculation, and the complex version is zero. \subsection{3. Tate cohomology and the fundamental class}\label{tate-cohomology-and-the-fundamental-class} For a finite group, extend ordinary cohomology to all integer degrees by a complete resolution. Concretely, splice the free bar resolution of \(\mathbf Z\) to its integral dual, using the norm between the two degree-zero terms. The bar resolution is exact by insertion of the identity on underlying abelian groups; its dual is exact because those underlying sequences split over \(\mathbf Z\). The splice is exact at the middle since the kernel of augmentation is the augmentation ideal and the invariant elements of \(\mathbf Z[G]\) are multiples of the norm. Applying \(\operatorname{Hom}_G(-,A)\) defines \(\widehat H^q(G,A)\). It agrees with ordinary \(H^q\) for \(q>0\), and \[ \widehat H^0=A^G/N_GA,\quad \widehat H^{-1}=\ker N_G/I_GA,\quad \widehat H^{-q-1}(G,\mathbf Z)=H_q(G,\mathbf Z)\quad(q\geq1). \tag{15} \] In particular \(\widehat H^{-2}(G,\mathbf Z)=G^{\mathrm{ab}}\). The degree-one bar boundary imposes precisely \([gh]=[g]+[h]\), giving this identification. Short exact sequences give long exact Tate sequences: all the resolution terms are free, so taking cochains is termwise exact. Induced modules \(\mathbf Z[G]\otimes B\) have zero Tate cohomology, including after restriction to any subgroup. Identify their cochains with the underlying split bar complex by evaluation on a coset coordinate; the insertion contraction and its dual prove this assertion also in the two middle degrees. The maps (3) and their bar comparisons extend restriction, corestriction and (2) to Tate degrees. Cup products extend by the bar diagonal and dimension shifting; on nonnegative cochains they are \[ (c\cup d)(g_1,\ldots,g_{r+s}) =c(g_1,\ldots,g_r)\otimes(g_1\cdots g_r)d(g_{r+1},\ldots,g_{r+s}). \tag{16} \] Splitting the alternating boundary at the join proves its Leibniz rule. The long exact sequences extend this product to negative degrees and make connecting maps commute with it, with the usual degree sign. We place the degree-two class on the right; its even degree introduces no sign in the isomorphisms below. \subsubsection{Tate's two-degree criterion}\label{tates-two-degree-criterion} If \(\widehat H^j(H,B)=\widehat H^{j+1}(H,B)=0\) for every subgroup \(H\), then all its Tate groups vanish. \textbf{Proof.} There are shifts in both directions. The surjection \(\mathbf Z[G]\otimes B\to B\), \(g\otimes b\mapsto gb\), and the injection \(B\to\operatorname{Maps}(G,B)\), \(b\mapsto(g\mapsto gb)\), have Tate-acyclic middle modules; their kernels and cokernels shift cohomology by one, simultaneously for all subgroups. Shift to \(j=1\). It suffices to prove vanishing in degrees 0 and 3; iteration then proves all degrees. Induct on \(|G|\). For a group that is not a \(p\)-group, all Sylow subgroups are proper, and (2) kills the two groups primary part by primary part. For a \(p\)-group take \(H\triangleleft G\) of index \(p\). Induction gives vanishing for \(H\) in degrees 0 through 3. The inflation consequence of (6) gives isomorphisms in degrees 1, 2 and 3 with the corresponding groups of \(C_p\) on \(B^H\). Cyclic periodicity gives degree-three vanishing from degree one and \((B^H)^{C_p}=N_{C_p}B^H\) from degree-two vanishing. Degree-zero vanishing for \(H\) says \(B^H=N_HB\), so \(B^G=N_GB\). This proves degree-zero vanishing for \(G\) and completes the induction. \(\square\) \subsubsection{Tate's cup-product theorem}\label{tates-cup-product-theorem} Suppose for every \(H\subseteq G\), \(H^1(H,A)=0\) and \(H^2(H,A)\) is cyclic of order \(|H|\). If \(u\) generates \(H^2(G,A)\), then \[ \widehat H^q(G,\mathbf Z)\xrightarrow{x\mapsto x\cup u} \widehat H^{q+2}(G,A) \tag{17} \] is an isomorphism for every integer \(q\). \textbf{Proof.} First restriction of \(u\) generates each \(H^2(H,A)\). Its order is at least \(|H|\), because corestriction of it is \([G:H]u\), which has exactly that order by (2); the target itself has order \(|H|\). Twice take the cokernel of the injection into the induced module just used. The resulting module \(B\) has natural shifts \(\widehat H^q(H,B)=\widehat H^{q+2}(H,A)\). Thus \(\widehat H^{-1}(H,B)=0\) and \(\widehat H^0(H,B)\) is cyclic of order \(|H|\). Lift the shifted class of \(u\) to \(b\in B^G\). Adjoin \(\mathbf Z[G]\) to \(B\), which does not change any Tate group, and inject \[ \mathbf Z\longrightarrow B\oplus\mathbf Z[G], \qquad 1\longmapsto(b,N_G). \] The second coordinate proves injectivity. On degree-zero cohomology of every subgroup this is an isomorphism: the image of 1 is the restricted generator just proved. In the long exact sequence for its cokernel \(D\), the facts \(\widehat H^{-1}(H,B)=0\) and \(H^1(H,\mathbf Z)=0\) consequently give \(\widehat H^{-1}(H,D)=\widehat H^0(H,D)=0\). The two-degree criterion makes \(D\) Tate-acyclic. The injection of \(\mathbf Z\) therefore induces isomorphisms in every degree. Its first coordinate is multiplication by the shifted class of \(u\); commuting the two connecting maps with (16) gives precisely (17). \(\square\) \subsubsection{Proposition 24.3. Local fundamental classes and reciprocity}\label{proposition-24.3.-local-fundamental-classes-and-reciprocity} For \(L/K\) finite Galois of degree \(n\), let \(u_{L/K}\) be the unique class with local invariant \(1/n\). Cup product gives \[ \widehat H^q(G,\mathbf Z)\simeq\widehat H^{q+2}(G,L^\times), \qquad G^{\mathrm{ab}}\simeq K^\times/N_{L/K}L^\times. \tag{18} \] The last map is the inverse of \textbf{arithmetic} reciprocity. \textbf{Proof.} Hilbert 90 and Theorem 24.2 verify every hypothesis of (17), including for subgroups. For \(H\subset G\) with fixed field \(E\), restriction multiplies the invariant by \([E:K]\), giving \(1/[L:E]\); it is exactly \(u_{L/E}\). In a cyclic group with generator \(\sigma\), the bar and periodic resolutions give \([\sigma]\cup[a]=a\) in the norm quotient: summing the carry cocycle once round the cycle yields its parameter. Formula (14) therefore identifies this map with inverse arithmetic reciprocity. For arbitrary \(G\), take \(H=\langle g\rangle\). Corestriction on \(H_1\) in the negative Tate notation is induced by inclusion, and on \(\widehat H^0\) it is \(N_{E/K}\). The projection formula and the restriction identity for \(u\) reduce the image of \([g]\) to the cyclic result over \(E\). Norm functoriality of lesson 5 gives precisely \(g\) in \(G^{\mathrm{ab}}\). Every abelianized element is represented by such a \(g\), proving the identification without an undetermined inverse. \(\square\) \subsection{4. The local--global theorem}\label{the-localglobal-theorem} Let \(K\) be a global field. At a real place use the invariant in (13); at a complex place use zero. Only finitely many localizations of a Brauer class are nonzero. Here is a direct justification of this support assertion using (1). Represent the class by a cocycle on a finite Galois splitting field. Outside finitely many places that extension is unramified and all its finitely many cocycle values are units. Localization lies in the cohomology of the unramified unit module, which vanishes by Proposition 24.1. No global sum law is being assumed in this argument. \subsubsection{General injectivity from the cyclic norm theorem}\label{general-injectivity-from-the-cyclic-norm-theorem} If a global Brauer class is zero at every place, it is zero globally. \textbf{Proof.} Induct on the order of a finite Galois splitting group \(G\). For a cyclic extension, the parameter is a local norm everywhere; Hasse's norm theorem in lesson 15 makes it a global norm, and the cyclic algebra splits. If \(G\) is a \(p\)-group of larger order, choose \(H\triangleleft G\) of index \(p\). Over \(E=L^H\) the class is locally zero, so induction kills its restriction. Equation (5) and Hilbert 90 make it a cyclic relative class for \(E/K\), to which the first argument applies. If \(G\) is not a \(p\)-group, every Sylow subgroup is proper; restriction to its fixed field vanishes by induction. Equation (2) kills every primary part. Thus the original class vanishes. \(\square\) \subsubsection{A supply of cyclic splitting extensions}\label{a-supply-of-cyclic-splitting-extensions} Given finitely many finite places \(S\) and a prescribed power \(p^{a_v}\) at each, there is a cyclic extension whose local degree at every \(v\in S\) is divisible by \(p^{a_v}\). For \(p=2\) it can additionally make any specified real places complex. \textbf{Proof for number fields.} Start with the cyclotomic \(\mathbf Z_p\)-extension of \(\mathbf Q\) supplied by lesson 20 and restrict its characters to \(G_K\). Its image is an open subgroup of \(\mathbf Z_p\), because a number field intersects that tower in a finite field. At a place over \(p\), the local image is open by total cyclotomic ramification and the finite degree of \(K_v\). At \(v\nmid p\), Frobenius has image the principal-unit component of the integer \(Nv>1\). This component is nonzero: if it were zero, \(Nv\) would be a \(p\)-adic root of unity, hence \((Nv)^{p-1}=1\) for odd \(p\), or \((Nv)^2=1\) for \(p=2\), impossible for that positive integer. A nonzero subgroup generated in \(\mathbf Z_p\) is open. Thus finite cyclic quotients at sufficiently high levels have arbitrarily large local \(p\)-power degrees at every place of the finite set. For \(p=2\), this tower is real. Take a finite quotient character of order \(2^N\) and multiply it by the quadratic character of \(\mathbf Q(i)\), also restricted to \(G_K\). At a real place complex conjugation now maps to \(-1\). Choose the initial level to give finite local image order at least \(\max(4,2^{a_v})\). An element of order \(2^b\), \(b\geq2\), in the cyclic group of order \(2^N\) has exponent of valuation \(N-b\); adding 0 or \(2^{N-1}\) does not change that valuation. Hence twisting does not reduce the required local orders. The image of the twisted character is still cyclic, and its fixed field has all the asserted properties. Odd-primary local classes have no real obstruction. For a function field with full constant field \(\mathbf F_q\), use a constant extension of degree \(n\). At a place of residue degree \(d_v\), the local degree is \(n/\gcd(n,d_v)\): the tensor product of the two finite residue fields has that degree in each factor. Choose \(n\) divisible by \(p^{a_v}d_v\) for every \(v\in S\). Then its local degree is divisible by \(p^{a_v}\). There are no archimedean places. \(\square\) \subsubsection{Theorem 24.4. Albert--Brauer--Hasse--Noether}\label{theorem-24.4.-albertbrauerhassenoether} For every global field the following sequence is exact: \[ 0\longrightarrow\operatorname{Br}(K) \longrightarrow\bigoplus_v\operatorname{Br}(K_v) \xrightarrow{\ \sum_v\operatorname{inv}_v\ }\mathbf Q/\mathbf Z \longrightarrow0. \tag{19} \] \textbf{Proof.} Injectivity has just been proved for arbitrary splitting groups. To prove the sum law, first take a \(p\)-primary global class \(b\). Choose a cyclic extension whose local degrees at its finite support are divisible by all local orders, and which complexifies any real support when \(p=2\). Restriction multiplies local invariants by these degrees, so \(b\) becomes locally zero over the extension. General injectivity over that global field makes it globally zero. Thus \(b\) is represented by a cyclic algebra \((L/K,\sigma,a)\), by (1). Formula (14), including its archimedean cases and restricted characters at each completion, gives \[ \sum_v\operatorname{inv}_v(b) =\sum_v\chi(\operatorname{rec}_{K_v}(a)) =\chi(\operatorname{rec}_K(a))=0. \tag{20} \] The last equality is the principal product law already proved in lesson 16. Every Brauer class is torsion by the finite splitting field and (2), so summing its finitely many primary components proves (20) generally. In particular, cyclic splitting was proved here, not presumed as an external classification theorem. Conversely take a finite tuple \((r_v)\) with sum zero, and split it into its primary tuples. For a single primary tuple choose a cyclic \(L/K\), of degree \(n\), with local decomposition orders divisible by the denominators at its support. Fix a faithful character \(\chi:G\to\tfrac1n\mathbf Z/\mathbf Z\). At each support place local reciprocity is onto the decomposition group, whose character image contains \(r_v\); choose \(a_v\in K_v^\times\) giving that value. Put \(a_v=1\) elsewhere. This is an idèle. Its global Artin image is trivial, since its faithful character value is \(\sum r_v=0\). Lesson 16's exact norm kernel gives \[ a=xN_{L/K}y,\qquad x\in K^\times,\quad y\in J_L. \] The cyclic algebra with parameter \(x\) has local invariant \(r_v\) at every place by (14); local norms contribute zero, also outside the original support. Add the primary classes to realize the tuple. Finally the sum map is onto: choose any nonarchimedean place and the local invariant desired in \(\mathbf Q/\mathbf Z\). This proves every part of (19). \(\square\) \subsection{5. Quaternion algebras and ramification}\label{quaternion-algebras-and-ramification} Assume characteristic different from 2 for the symbol presentation \((a,b)_K\). At a nonarchimedean or real place, the Hilbert symbol of lesson 11 gives \[ \operatorname{inv}_v(a,b)= \begin{cases}0,&(a,b)_v=1,\\[2pt]1/2,&(a,b)_v=-1.\end{cases} \tag{21} \] Indeed a quaternion algebra is the quadratic cyclic algebra with parameter \(b\); it splits exactly when \(b\) is a norm from \(K_v(\sqrt a)\). The first-entry reciprocity convention of lesson 11 gives the same split criterion, irrespective of the interchangeable entries of a quadratic symbol. \subsubsection{\texorpdfstring{Corollary 24.5. Quaternion algebras over \(\mathbf Q\)}{Corollary 24.5. Quaternion algebras over \textbackslash mathbf Q}}\label{corollary-24.5.-quaternion-algebras-over-mathbf-q} The ramification set of a quaternion algebra over \(\mathbf Q\) is a finite set of places of even cardinality. Every such set occurs, and it determines the algebra up to isomorphism. \textbf{Proof.} Equation (21) and the sum law imply finiteness and even cardinality. For an even finite set, (19) realizes the tuple having invariant \(1/2\) there and zero elsewhere. Choose a rational \(d\) that is a nonsquare in every specified finite completion and negative if the real place is specified. This is possible by weak approximation: each finite nonsquare class contains an open neighborhood, and negativity is open at infinity. If the set is nonempty this makes \(\mathbf Q(\sqrt d)\) quadratic and locally of degree two throughout the set. The realized class restricts to zero everywhere over that field, hence globally by injectivity. The algebraic splitting criterion in the Noether lesson makes its index divide two. A nonzero class is therefore a division algebra of degree two containing that quadratic field, hence its cyclic crossed product is a quaternion algebra. The empty set is represented by \(M_2(\mathbf Q)\). If two quaternion algebras have the same set, (19) makes their classes equal, and Wedderburn uniqueness together with their equal degree two makes the algebras isomorphic. \(\square\) For Hamilton's algebra \((-1,-1)_{\mathbf Q}\), every odd finite place has invariant zero by the odd-prime unit formula of lesson 11. At 2 the odd-unit formula gives \((-1,-1)_2=-1\), and at the real place both entries are negative, giving \(-1\). Its ramification set is exactly \(\{2,\infty\}\), in agreement with the even-set theorem. \subsection{6. Global fundamental classes}\label{global-fundamental-classes} The global formation uses \(C_L\), not \(L^\times\). The two coefficient modules have quite different degree-two cohomology. We now construct the invariant for the former. \subsubsection{The cyclic class-group invariant}\label{the-cyclic-class-group-invariant} For a cyclic extension \(L/K\), of degree \(n\), the idèle sequence and Hilbert 90 give \[ 0\to H^2(G,L^\times)\to H^2(G,J_L) \to H^2(G,C_L)\to0. \tag{22} \] The term on the right after this sequence is \(H^3(G,L^\times)=H^1(G,L^\times)=0\), by cyclic periodicity. The fixed-class and \(H^1\)-vanishing arguments in §§1 and lesson 13 justify the injection on the left. Shapiro and the unit-tail argument of lesson 13, now applied to degree two, identify the middle term as \[ \bigoplus_v H^2(D_w,L_w^\times). \tag{23} \] The proof for the tail works in this degree because every unramified local unit Tate group vanishes; a finite cocycle and a cochain relation belong to one finite chart, and all remaining primitives can be chosen as units. By Theorem 24.4, the image on the left of (22) consists exactly of the tuples in (23) whose invariant sum is zero. For the converse part of this claim, realize such a tuple as a global Brauer class. Its restriction to \(L\) has every local invariant zero because each denominator divides its local extension degree. Global injectivity over \(L\) makes it a relative class. Thus the invariant sum identifies \(H^2(G,C_L)\) with \(\tfrac1n\mathbf Z/\mathbf Z\). Its image is that entire group: Chebotarev in lesson 22 gives an unramified place whose Frobenius generates the cyclic group, hence local degree \(n\). With generator \(\sigma\) and \(\chi(\sigma)=1/n\), the periodic-resolution identification gives \[ H^2(G,C_L)=C_K/NC_L, \qquad \operatorname{inv}(c)=\chi(\operatorname{rec}_K(c)). \tag{24} \] Indeed lift a representative class to an idèle; (14) computes the sum in (22), and the principal product law makes it independent of the lift. \subsubsection{Theorem 24.6. The global class formation}\label{theorem-24.6.-the-global-class-formation} For finite Galois \(L/K\) of degree \(n\), there is a canonical invariant \[ H^2(G,C_L)\simeq\tfrac1n\mathbf Z/\mathbf Z, \qquad H^1(G,C_L)=0. \tag{25} \] Inflation preserves the invariant and restriction to a finite base extension multiplies it by the extension degree. The fundamental class \(u_{L/K}\), of invariant \(1/n\), gives \[ \widehat H^q(G,\mathbf Z)\simeq\widehat H^{q+2}(G,C_L). \tag{26} \] In degree \(-2\), this is inverse arithmetic global reciprocity. \textbf{Proof.} The \(H^1\)-vanishing and order bound \(|H^2(G,C_L)|\mid n\) are already proved in §1. Choose a cyclic extension \(T/K\) of degree \(n\). For function fields take constants. For number fields choose a rational prime \(\ell\equiv1\pmod n\) unramified in \(K\), using the rational ray-class prime result in lesson 22. The degree-\(n\) subfield of \(\mathbf Q(\zeta_\ell)\) intersects \(K\) trivially: it is totally ramified at \(\ell\), whereas every subfield of \(K\) is unramified there. Its compositum with \(K\) supplies \(T\). Put \(N=LT\). Inflation of both groups \(H^2(T/K,C_T)\) and \(H^2(L/K,C_L)\) into \(H^2(N/K,C_N)\) is injective by (5) and the \(H^1\)-vanishing. Every class from the cyclic field \(T\) lifts to \(H^2(T/K,J_T)\), by (22). After restricting to \(L\), its idelic invariant sum is multiplied by \([L:K]=n\): at each base place, sum the local degrees \([L_{v'}:K_v]\), which sum to \(n\). Its original invariant lies in \(\tfrac1n\mathbf Z/\mathbf Z\), so the new sum is zero. The extension \(N/L\) is cyclic. The cyclic case (22) therefore kills the restricted class in \(H^2(N/L,C_N)\). By (5) the class descends to \(H^2(L/K,C_L)\). The cyclic subgroup from \(T\) has order \(n\); the order bound makes the two inflated subgroups equal. To make the invariant independent of this auxiliary field, put \(\mathcal J=\bigcup_L J_L\) and \(\mathcal C=\bigcup_L C_L\), with the extension embeddings of lesson 13, regarded as discrete \(G_K\)-modules. Their fixed modules are \(J_L\) and \(C_L\). Continuous cochains have finite image, so their cohomology is the direct limit of finite-extension cohomology. The same finite-chart argument as (23) gives \[ H^2(G_K,\mathcal J)=\bigoplus_v\operatorname{Br}(K_v). \tag{27} \] There is no missing local class in this limit: for a finite local tuple, the cyclic-extension construction of §4 supplies one global extension whose local degrees split every entry. Thus the finite relative groups in (23) exhaust the displayed direct sum. The auxiliary-field comparison just proved shows that every class in \(H^2(G_K,\mathcal C)\) comes from an idelic class: at a common finite level it equals a class from a cyclic extension, and those lift by (22). The exact sequence \(1\to(K^s)^\times\to\mathcal J\to\mathcal C\to1\) therefore identifies \[ H^2(G_K,\mathcal C) =\left(\bigoplus_v\operatorname{Br}(K_v)\right)/\operatorname{Br}(K) \xrightarrow{\sum\operatorname{inv}_v}\mathbf Q/\mathbf Z. \tag{28} \] Exactness uses \(H^1(G_K,\mathcal C)=0\), which follows from the finite vanishing. Formula (19) proves the last isomorphism. Finite \(H^2(G,C_L)\) injects into (28) by (5), has order \(n\), and hence is the unique subgroup \(\tfrac1n\mathbf Z/\mathbf Z\). This defines the canonical invariant. Inflation compatibility is built into this injection. Restriction compatibility follows from (27) and local degree multiplication, summed over places. The same reasoning gives preservation of the invariant by corestriction, using local corestriction and its sum. For \(H\subset G\), restriction of \(u_{L/K}\) has invariant \([L^H:K]/n=1/|H|\), so it is \(u_{L/L^H}\). Tate's cup-product theorem now proves (26). In a cyclic group the parameter computation of §3 and (24) give inverse reciprocity with the asserted sign. For an arbitrary \(g\in G\), restrict to \(\langle g\rangle\), apply the cyclic result and corestrict back. The projection formula and the norm functoriality of lesson 16 give \(g\) in \(G^{\mathrm{ab}}\), exactly as in the local argument for (18). This proves the full assertion. \(\square\) \subsection{7. Relative Weil groups and their transition maps}\label{relative-weil-groups-and-their-transition-maps} For this section \(K\) is a number field. Write a multiplicative normalized cocycle \(f\) for \(u_{L/K}\). Define \[ W_{L/K}=C_L\times G, \qquad (a,g)(b,h)=(a\,g(b)\,f(g,h),gh). \tag{29} \] The cocycle identity proves associativity. Give this group the topology consisting of finitely many copies of the locally compact group \(C_L\); the action and multiplication are continuous. Changing the cocycle by a coboundary gives the continuous isomorphism that changes the chosen lifts by that cochain. Thus \[ 1\to C_L\to W_{L/K}\to G\to1 \tag{30} \] is the extension represented by the fundamental class. \subsubsection{Transfer and abelianization of a fundamental extension}\label{transfer-and-abelianization-of-a-fundamental-extension} For an extension \(1\to A\to E\to G\to1\), with \(G\) finite and \(A\) abelian, transfer to \(A\) lands in \(A^G\). Choose lifts \(s(t)\), and for \(x\) mapping to \(g\), write \(xs(t)=s(gt)a_t\). The transfer is \(\prod_t a_t\). A change of lifts cancels by the permutation of \(G\); multiplying two elements gives the product of their transfers. Conjugation consequently leaves the transfer unchanged, while its effect on the target is the given \(G\)-action. This proves that the value is invariant. On \(A\) it is \(N_G\). For a fundamental class the resulting map \[ E^{\mathrm{ab}}\longrightarrow A^G \tag{31} \] is an isomorphism. To verify its kernel, the degree-one bar calculation for this extension gives the exact sequence \[ H_2(G,\mathbf Z)\longrightarrow A_G \longrightarrow E_{\mathrm{ab}}\longrightarrow G_{\mathrm{ab}}\to0. \tag{32} \] The first arrow evaluates the extension cocycle on a bar two-cycle. Indeed lift the successive edges of that cycle to \(E\); their boundary products cancel in the quotient, and the remaining factors are precisely its cocycle values, modulo conjugate differences. Thus its image in \(A_G\) is the image of \(\widehat H^{-3}(G,\mathbf Z)\xrightarrow{\cup u}\widehat H^{-1}(G,A)\). Tate's theorem makes that image all of \(\ker(N_G:A_G\to A^G)\). On the final quotient of (32), the transfer is \(G_{\mathrm{ab}}\xrightarrow{\cup u}A^G/N_GA\): substitute the lift products in the transfer formula, the same bar evaluation as (18). This is again an isomorphism. Hence transfer is injective and surjective in (31). For \(A=C_L\), the field norm \(C_L\to NC_L\) is open and has compact kernel. Its modulus map identifies the noncompact positive-real direction, while its norm-one kernel lies in the compact norm-one idèle class group of lesson 14. Openness follows either from the local norm charts of lesson 13 and quotient topology, or from their induced open map on class groups. Transfer, restricted to each component of \(E\), is a translate of this norm map. It is therefore open onto \(C_K\), whose norm subgroup has finite index, and has closed kernel. The algebraic commutator kernel found in (32) is consequently already closed. Thus (31) is a topological isomorphism \[ W_{L/K}^{\mathrm{ab}}\simeq C_K. \tag{33} \] It is inverse arithmetic reciprocity after projection to \(G^{\mathrm{ab}}\). \subsubsection{The quotient in a tower}\label{the-quotient-in-a-tower} Let \(M/L/K\) be finite Galois over \(K\), \(H=\operatorname{Gal}(M/L)\), and \(E=W_{M/K}\). The inverse image \(E_H\) of \(H\) has fundamental class \(u_{M/L}\). Its transfer identifies \(E_H^{\mathrm{ab}}\) with \(C_L\). Consequently \(F=E/[E_H,E_H]\) is an extension of \(\operatorname{Gal}(L/K)\) by \(C_L\). Its class is exactly \(u_{L/K}\), as follows. If \(v\) denotes the class of \(F\), inflate it to \(\operatorname{Gal}(M/K)\) and include its coefficients \(C_L\subset C_M\). Pushing (29) through the transfer on its kernel says that this class is \(N_Hu_{M/K}\). Since \(H\) is normal, \(N_H\) is central in \(\mathbf Z[G]\). A central group-ring element acts on \(\operatorname{Ext}_{\mathbf Z[G]}^*(\mathbf Z,A)\) by its augmentation: multiplication on a free resolution lifts its augmentation on \(\mathbf Z\), and the comparison homotopy between those two lifts gives the equality on cohomology. Here this gives \(N_Hu_{M/K}=|H|u_{M/K}\), of invariant \(1/[L:K]\). Inflation of \(u_{L/K}\) has that same invariant by (25). Inflation with the coefficient inclusion is injective by \(H^1(H,C_M)=0\) and (5); hence \(v=u_{L/K}\). Choose an isomorphism of this quotient with \(W_{L/K}\). This provides a surjective transition homomorphism \[ W_{M/K}\longrightarrow W_{L/K}, \tag{34} \] whose restriction to \(C_M\) is \(N_{M/L}\). Its kernel is the closed commutator subgroup of \(E_H\), and is compact: the transfer kernel meets each of its finitely many components in a translate of the compact norm kernel. This also proves that (34) is proper. Nested quotients give transitive transitions. Isomorphism choices differ by a cocycle with coefficients \(C_L\); \(H^1(G,C_L)=0\) makes those differences inner coefficient conjugations. \subsubsection{The absolute group and its topology}\label{the-absolute-group-and-its-topology} A number field and its separable closure are countable. Enumerate the finite Galois extensions and take successive composita to obtain a cofinal tower \(K=L_0\subset L_1\subset\cdots\). Choose the adjacent quotient isomorphisms (34), and use their composites for all tower transitions. This is a coherent system, rather than a claim that cocycles have canonical rigid representatives. Define \[ W_K=\varprojlim_i W_{L_i/K}. \tag{35} \] Every projection is surjective: over a fixed finite-level point the higher fibres form a system of nonempty compact sets, and their finite compatibility conditions have a common solution. The same compactness argument says that the inverse image of a compact set at any level is compact. The inverse image of a compact identity neighborhood is therefore a compact neighborhood in the limit, proving local compactness. Projection to the finite Galois quotients gives a continuous map \(W_K\to G_K\) with dense image, since it is onto every finite quotient. The projection to \(W_{K/K}=C_K\) is the compatible transfer. Its kernel is exactly the closure of the commutator subgroup of \(W_K\). One inclusion follows because \(C_K\) is abelian. For the other, take an element in that kernel and a basic neighborhood determined at level \(i\). Its component lies in the commutator kernel of (33). Approximate that component by a finite product of commutators, and lift each factor to \(W_K\), using projection surjectivity. Their commutator product belongs to the neighborhood. This proves the required density in the kernel. The projection is open: a basic open set from level \(i\) has image under the open transfer at that level. Thus \[ W_K^{\mathrm{ab}}\xrightarrow{\sim}C_K \tag{36} \] is a homeomorphism. Choice independence can be proved without assuming surjectivity of coefficient norms. For two systems at a common field \(L\), isomorphisms fixing \(C_L\) and the Galois quotient form a nonempty torsor for \(C_L/C_K\): their differences are cocycles, killed by \(H^1(G,C_L)=0\), and an inner coefficient conjugation is trivial exactly on \(C_L^G=C_K\). This quotient is compact, since the embedded positive modulus direction of \(C_K\) covers that of \(C_L\). Quotienting commutator subgroups gives continuous transition maps on these compact isomorphism spaces. Every finite list of compatibility conditions is met at one higher common field. Compactness supplies an inverse-limit isomorphism, continuous with continuous inverse. Aligning two cofinal towers by composita applies this argument to (35). This proves the asserted topological choice independence with the coefficient and Galois maps preserved. \subsubsection{Local-to-global compatibility}\label{local-to-global-compatibility} Fix a place \(v\) and an embedding of separable closures realizing its decomposition group. There is a continuous homomorphism \[ \theta_v:W_{K_v}\longrightarrow W_K \tag{37} \] compatible with that embedding on Galois groups and with the idèle inclusion \(K_v^\times\to C_K\) on Hausdorff abelianizations. We give the compatibility argument, including the inverse-limit choice. First the local group of lesson 12 has relative quotient \(W_{K_v}/\overline{[W_{L_w},W_{L_w}]}\) represented by the local fundamental class. Its kernel is \(L_w^\times\) by Theorem 12.2. Transfer is the inclusion on multiplicative abelianizations, since the finite transfer identities of lesson 5 commute with the continuous local reciprocity homeomorphisms of lesson 12. For an unramified cyclic extension, a Frobenius lift raised to its degree is the Frobenius of \(L_w\), whose abelianized image is a uniformizer; the carry cocycle has invariant \(1/[L_w:K_v]\). For a general local extension compare with the unramified extension of the same degree in a common finite Galois extension. The quotient-by-commutators identity and transfer give the inflation identity \(\mathrm{Inf}\,\alpha_{L_w/K_v}=[M_w:L_w]\alpha_{M_w/K_v}\) for its extension classes: the coefficient map followed by inclusion is the central norm operator, exactly as in the proof of (34). The equal-degree comparison therefore identifies the two inflated classes. Hilbert 90 makes inflation injective, proving that the general class is the normalized \(u_{L_w/K_v}\). For real and complex places, (20)--(21) of lesson 12 give respectively the parameter \(-1\) class and the trivial class, with the same invariants. For global \(L/K\), let \(D=D_w\) and \(E=L^D\). The map \(L_w^\times\to C_L\), inserting a component at \(w\), takes the local fundamental class to the restriction of \(u_{L/K}\). Indeed the resulting class is an idelic class over \(E\) supported at its distinguished place, with sum of invariants \(1/|D|\); (25) identifies it with \(u_{L/E}\). Equality of these cocycle classes gives a continuous homomorphism from the local relative extension to the inverse image of \(D\) in \(W_{L/K}\), with the specified coefficient and quotient maps. An explicit construction changes one lifted cocycle by the cochain whose boundary is their quotient, then uses (29). Let \(I_L\) be the set of those homomorphisms. It is nonempty. Two members differ by a \(D\)-cocycle in \(C_L\); \(H^1(D,C_L)=0\) makes it an inner conjugation. Thus \(I_L\) is a torsor for \[ C_L/C_L^D=C_L/C_E. \tag{38} \] This quotient is compact: the positive modulus direction of \(C_E\) covers that of \(C_L\), leaving a quotient of the compact norm-one group. It supplies \(I_L\) with a compact topology. If \(M\supset L\), a member of \(I_M\) descends to a member of \(I_L\): local commutators over \(L_w\) map into the global commutator kernel in (34). The induced map is continuous, its change under inner conjugation being the norm \(C_M\to C_L\). Finite compatibility conditions can all be met by choosing a member at one higher common level and descending it. Compactness therefore supplies an element of \(\varprojlim I_L\), without requiring any individual norm map to be onto. The resulting compatible homomorphisms define (37); continuity follows componentwise. Finally transfer commutes with these local maps. Within the subgroup over \(D\), choose its coset lifts as the images of local lifts. Formula (31) then shows that transfer of the local map is exactly the coefficient inclusion on the local transfer: on abelianizations it is \(K_v^\times\to C_E\), inserted at the distinguished place of \(E=L^D\), whose completion is \(K_v\). Inclusion of that subgroup into the full relative Weil group induces the field norm \(C_E\to C_K\). To check this last identity, the transfer formula, decomposed into cosets of \(D\), gives \[ V_G(x)=\prod_{gD\in G/D}g\bigl(V_D(x)\bigr) \qquad(x\text{ in the subgroup over }D). \] This is the transfer double-coset identity when the target normal subgroup is the common coefficient group: multiply the coefficients within each block, and the changes of lifts telescope between permuted blocks. The displayed product is the field norm on the \(D\)-fixed coefficient group. The norm of the distinguished single-place idèle is the same single component in \(C_K\). Passing to the limit therefore proves the commuting diagram \[ \begin{array}{ccc} W_{K_v}^{\mathrm{ab}}&\xrightarrow{\theta_v^{\mathrm{ab}}}&W_K^{\mathrm{ab}}\\ \downarrow\wr&&\downarrow\wr\\ K_v^\times&\longrightarrow&C_K. \end{array} \tag{39} \] Together with (35)--(36), this is the number-field construction and compatibility asserted in lesson 12. Choosing geometric reciprocity inverts both vertical identifications. Weil's construction and Tate's formulation of its local relationship are credited in the references; the cocycle, compactness and normalization arguments here supply the proof. \subsection{8. Unit cohomology and infinite class field towers}\label{unit-cohomology-and-infinite-class-field-towers} We finish the tower argument stated in lesson 19. Here \(K\) is a number field, \(p\) a prime, and an everywhere-unramified extension includes splitting at real places. For a finite \(p\)-group \(G\), put \[ d=\dim_{\mathbf F_p}H^1(G,\mathbf F_p),\qquad r=\dim_{\mathbf F_p}H^2(G,\mathbf F_p). \tag{40} \] Both coefficients have trivial action. \subsubsection{Theorem 24.7. The finite-group Golod--Shafarevich bound}\label{theorem-24.7.-the-finite-group-golodshafarevich-bound} If \(G\) is a nontrivial finite \(p\)-group, then \[ r>\frac{d^2}{4}. \tag{41} \] \textbf{Proof.} We supply the presentation and filtered-algebra argument. The Frattini quotient \(G/(G^p[G,G])\) is an elementary abelian group of rank \(d\). Lifts of a basis generate \(G\): a proper generated subgroup is contained in a maximal subgroup, which in a finite \(p\)-group is normal of index \(p\), and that subgroup cannot contain a basis of the quotient. To check the index assertion, induct using the nontrivial center; if the center is not contained in the maximal subgroup its product with that subgroup is the whole group, forcing index \(p\); otherwise pass to the central quotient. Take the free pro-\(p\) group \(F\) on these \(d\) generators, with kernel \(R\) onto \(G\). Its completed group algebra is the noncommutative power-series algebra \[ T=\mathbf F_p\langle\!\langle X_1,\ldots,X_d\rangle\!\rangle, \qquad x_i\longmapsto1+X_i. \tag{42} \] Here is the universal construction behind this identification. In every finite nilpotent augmentation quotient of \(T\), the group \(1+(X_1,\ldots,X_d)\) is a finite \(p\)-group, so freeness gives the map from \(F\). Conversely substitute \(x_i-1\) in power series in the completed group algebra. Both continuous algebra maps are inverse on their generators and on the dense polynomial algebra. The augmentation topology agrees with the finite-group-quotient topology: in a finite \(p\)-group algebra the augmentation ideal is nilpotent. Indeed choose a central \(z\) of order \(p\); \((z-1)^p=0\), and induction on \(G/\langle z\rangle\) puts a sufficiently high augmentation power in \((z-1)\), whose \(p\)th power is zero. Conversely the augmentation quotients on \(d\) generators have finite dimension, bounded by the number of words of bounded length. This proves the topology assertion used in (42). The augmentation ideal in \(T\), as a right module, is freely \(\bigoplus_i X_iT\), by the unique first letter of each word. Thus \(0\to T^d\to T\to\mathbf F_p\to0\) is a free resolution, proving \(H^2(F,\mathbf F_p)=0\). The low-degree sequence (6) for \(1\to R\to F\to G\to1\), valid also for continuous cochains by finite-quotient passage, identifies \[ H^2(G,\mathbf F_p) =\operatorname{Hom}_{\mathrm{cont}} (R/\overline{R^p[R,F]},\mathbf F_p). \tag{43} \] The preceding \(H^1(G)\to H^1(F)\) is an isomorphism because the chosen generators are minimal. Hence \(R/\overline{R^p[R,F]}\) has rank \(r\). Choose \(r\) relators \(\rho_j\) lifting a basis. They normally generate \(R\). To prove this last claim, quotient by their closed normal closure. If the remaining image of \(R\) were nontrivial, it would be detected in a finite \(p\)-group quotient. Its nonzero Frattini quotient has nonzero coinvariants under the ambient \(p\)-group: the augmentation ideal of that finite group algebra is nilpotent, so a nonzero finite module cannot equal its augmentation multiple. Those coinvariants would be a nonzero quotient of (43)'s relation module with all the chosen basis classes zero, a contradiction. Put \(f_j=\rho_j-1\in T\). Minimality puts each \(f_j\) in augmentation degree at least two, since \(R\subseteq\overline{F^p[F,F]}\). The algebra \(A=\mathbf F_p[G]\) is \(T\) modulo the closed two-sided ideal of these \(r\) series. Let \(I\) be its augmentation ideal, and \[ b_n=\dim_{\mathbf F_p}(A/I^{n+1}),\qquad b_{-1}=b_{-2}=0. \] Write \(f_j=\sum_i X_i f_{ij}\); every \(f_{ij}\) has degree at least one. There is an exact sequence of right \(A\)-modules \[ A^r\xrightarrow{(f_{ij})}A^d \xrightarrow{(v_i)\mapsto\sum X_iv_i}I\to0. \tag{44} \] For exactness in the middle, lift a kernel vector to \(T^d\); the series \(\sum X_iv_i\) lies in the relator ideal. The first-letter derivative obeys \(\partial_i(uv)=(\partial_i u)v+\epsilon(u)\partial_i v\). Differentiating a sum of \(u f_j v\) and reducing modulo the relator ideal leaves exactly combinations of the columns \((f_{ij})\). Passage to limits is valid because the target \(A^d\) is finite-dimensional. This proves (44), including the absence of an additional hidden relation module. The induced surjection \(A^d/I^nA^d\to I/I^{n+1}\) has kernel of dimension \(d b_{n-1}-(b_n-1)\). Any vector in that kernel lifts to an actual kernel vector: subtract a vector in \(I^nA^d\) expressing its error, using \(I^{n+1}=\sum_i X_iI^n\). Equation (44) generates these lifts, and its column entries lie in \(I\), so its image modulo \(I^nA^d\) has dimension at most \(r b_{n-2}\). Therefore, for every \(n\geq0\), \[ b_n-d b_{n-1}+r b_{n-2}\geq1. \tag{45} \] For \(n=0\) this is \(b_0=1\). Let \(H_A(t)=\sum_{n\geq0}\dim(I^n/I^{n+1})t^n\). Since \(I\) is nilpotent it is a polynomial with positive constant term, and \(\sum b_nt^n=H_A(t)/(1-t)\). Sum (45) for \(01/4\). This proves (41) in every case. \(\square\) \subsubsection{Theorem 24.8. The terminal unramified relation bound}\label{theorem-24.8.-the-terminal-unramified-relation-bound} If the maximal everywhere-unramified pro-\(p\) extension \(M/K\) is finite, with group \(G\), then \[ r-d\leq r_1(K)+r_2(K)\leq[K:\mathbf Q]. \tag{47} \] \textbf{Proof.} The class group of \(M\) has order prime to \(p\). Otherwise its Hilbert class field supplies an everywhere-unramified cyclic degree-\(p\) extension of \(M\). The compositum of its finitely many conjugates over \(K\) is still everywhere unramified, and its group over \(K\) is a \(p\)-group, contradicting maximality. Let \(U_M\) be the product of all finite local unit groups and all archimedean multiplicative groups, and \(E_M=\mathcal O_M^\times\). Every finite decomposition group is unramified cyclic, and its local unit Tate groups vanish as in Proposition 24.1. Every archimedean decomposition group is trivial because real places split. Shapiro makes \(U_M\) Tate-acyclic. Infinite products introduce no obstruction: the finite-group complete resolution has finitely generated free terms in each degree, so its cochains commute with these products; choose the local primitives componentwise. The finite class group \(\operatorname{Cl}(M)\) is also Tate-acyclic for the \(p\)-group, since its order is prime to \(p\) and every Tate group is killed by \(|G|\). Break the idelic ideal sequence into \[ 1\to E_M\to U_M\to C_M^{\mathrm{unit}}\to1, \qquad 1\to C_M^{\mathrm{unit}}\to C_M\to\operatorname{Cl}(M)\to1. \] Its Tate sequences give \(\widehat H^q(G,C_M)=\widehat H^{q+1}(G,E_M)\). In degree minus one, (26) consequently gives \[ H_2(G,\mathbf Z) =\widehat H^{-3}(G,\mathbf Z) \simeq\widehat H^{-1}(G,C_M) \simeq E_K/N_{M/K}E_M. \tag{48} \] The unit group \(E_K\) has free rank \(r_1+r_2-1\) and cyclic finite torsion, by the unit theorem already used in lesson 14. Thus the finite \(p\)-group on the right of (48) needs at most \(r_1+r_2\) generators. For completeness, the coefficient sequence \(0\to\mathbf Z\xrightarrow{p}\mathbf Z\to\mathbf F_p\to0\) gives \[ 0\to H^2(G,\mathbf Z)/p\to H^2(G,\mathbf F_p) \to H^3(G,\mathbf Z)[p]\to0. \tag{49} \] The rational coefficient sequence identifies \(H^2(G,\mathbf Z)=\operatorname{Hom}(G^{\mathrm{ab}},\mathbf Q/\mathbf Z)\) and \(H^3(G,\mathbf Z)=\operatorname{Hom}(H_2(G,\mathbf Z),\mathbf Q/\mathbf Z)\). Indeed positive rational cohomology vanishes by averaging, and applying \(\operatorname{Hom}(-,\mathbf Q/\mathbf Z)\) to the free bar chain complex takes its homology to its cohomology, since a character extends from a subgroup by divisibility. Finite \(p\)-groups and their character duals have the same number of cyclic factors. The first term of (49) therefore has dimension \(d\), and the last has dimension at most \(r_1+r_2\) by (48). This proves (47). \(\square\) \subsubsection{Theorem 24.9. An infinite Hilbert class field tower}\label{theorem-24.9.-an-infinite-hilbert-class-field-tower} The ordinary Hilbert class field tower is infinite for \[ K=\mathbf Q\!\left(\sqrt{-5\cdot13\cdot17\cdot29\cdot37\cdot41}\right), \] \textbf{Proof.} Proposition 19.3 gives genus rank six from the seven prime discriminants, \(-4\) and the six positive primes displayed above. Hence the maximal unramified pro-2 group has \(d\geq6\). If it were finite, (47) would give \(r\leq d+2\), whereas (41) gives \(r>d^2/4\); for every \(d\geq6\), \(d^2/4>d+2\). This is impossible. Its pro-2 extension is infinite. The ordinary Hilbert class field tower is therefore infinite as well: a stabilized finite top field has class number one, but base change of the infinite unramified pro-2 extension gives an infinite pro-2 extension of that top field. Such a group has a nontrivial finite \(p\)-group quotient and therefore a cyclic order-\(p\) quotient, contradicting class number one. This closes the tower argument using the genus calculation of lesson 19 and the proofs (41), (47) here. \(\square\) \subsection{9. Exercises with solutions}\label{exercises-with-solutions} \subsubsection{Exercise 1. Local quaternion invariants --- easy}\label{exercise-1.-local-quaternion-invariants-easy} Compute the invariant of \((a,b)_{\mathbf Q_p}\) from the Hilbert symbol, including a formula that can be used at every prime. \textbf{Solution.} It is zero for symbol 1 and \(1/2\) for symbol \(-1\), by (21). For odd \(p\), write \(a=p^\alpha u\), \(b=p^\beta v\), with units \(u,v\). Lesson 11 proves \[ (a,b)_p=(-1)^{\alpha\beta(p-1)/2} \left(\frac{\bar u}{p}\right)^\beta \left(\frac{\bar v}{p}\right)^\alpha. \] At 2, with odd units \(u,v\), its exponent of \(-1\), modulo two, is \[ \frac{u-1}{2}\frac{v-1}{2} +\alpha\frac{v^2-1}{8}+\beta\frac{u^2-1}{8}. \] Each quotient is integral and depends only on the appropriate odd residue class. Choose invariant \(1/2\) exactly when the indicated parity is odd. Negative valuations cause no change in these parity computations. This includes every \(a,b\ne0\). \subsubsection{Exercise 2. The infinite places --- medium}\label{exercise-2.-the-infinite-places-medium} Prove \(\operatorname{Br}(\mathbf R)=\mathbf Z/2\) and \(\operatorname{Br}(\mathbf C)=0\), and identify the invariant of Hamilton's algebra. \textbf{Solution.} Over an algebraically closed field a division representative is the field: factor the minimal polynomial of an element into linear factors and use absence of zero divisors. Thus every complex central simple algebra is a matrix algebra. Every real class is consequently split by \(\mathbf C\); the relative crossed-product theorem and cyclic resolution give \(\operatorname{Br}(\mathbf R)=\mathbf R^\times/N\mathbf C^\times\). The norm \(z\mapsto |z|^2\) has image precisely \(\mathbf R_{>0}\), so the quotient has order two. Its negative parameter gives relations \(i^2=j^2=-1\), \(ji=-ij\), hence Hamilton's algebra. It is the nonzero class, assigned invariant \(1/2\). No classification of arbitrary real division algebras was needed. \subsubsection{Exercise 3. Hamilton's ramification --- medium}\label{exercise-3.-hamiltons-ramification-medium} Determine every ramified place of \((-1,-1)_{\mathbf Q}\) directly from local formulas. \textbf{Solution.} For odd \(p\), both valuations are zero, and the odd-prime formula of Exercise 1 is 1. At 2, the unit cross term is \((-2/2)(-2/2)=1\) modulo two; the valuation terms are zero, so the symbol is \(-1\). At infinity two negative entries give symbol \(-1\). Thus exactly 2 and infinity have invariant \(1/2\); their sum is 1, zero in \(\mathbf Q/\mathbf Z\). All other places have invariant zero. This both determines the set and checks the global sum law in this example. \subsubsection{Exercise 4. The local degree-two bound --- hard}\label{exercise-4.-the-local-degree-two-bound-hard} For every finite Galois extension of nonarchimedean local fields, prove that \(|H^2(G,L^\times)|\) divides \([L:K]\), including positive characteristic and noncyclic groups. Explain why that bound alone is not the local invariant theorem. \textbf{Solution.} In a prime cyclic extension, Hilbert 90 kills \(H^1\), and cyclic periodicity identifies \(H^2\) with the norm quotient. Local reciprocity gives its order \(p\). Induct in a \(p\)-group using a normal index-\(p\) subgroup. Equation (5) makes the restriction kernel a group of order \(p\), while its image has order dividing the smaller group order. Hence the total order divides \(|G|\). For a general group, restriction injects each primary part into its Sylow-group cohomology by (2), giving the corresponding Sylow bound. Multiplying gives the stated divisibility and finiteness. This argument uses neither characteristic-zero logarithms nor solvability. To obtain an invariant isomorphism, one still has to exhibit a subgroup of the full degree: Theorem 24.2 does this by comparing with the unramified extension of equal degree and using (11). The bound then forces equality. \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 local invariant, global Brauer exact sequence and Tate fundamental-class theorem are proved using the exact internal algebra and cochain prerequisites. The number-field Weil construction, complete filtered group-algebra inequality and arithmetic unit-cohomology relation bound are supplied here; a survey statement is not counted as their proof. \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://ncatlab.org/nlab/files/TateNumberTheory.pdf}{John Tate, Number theoretic background (1979), freely available paper}. \item \href{https://m-ershov.github.io/Research/gssurvey_revised.pdf}{Mikhail Ershov, Golod--Shafarevich groups: a survey, author preprint (2012)}. \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}