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\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

\[
r_{L/K}(\sigma^j)=0,\qquad 0\leq j<n.
\]

The reciprocity homomorphism and its defining prime norm give
\(jN\pi_\Sigma\in N_{L/K}A_L\). Set \(u=j(\pi_\Sigma-\pi_L)\in U_M\).
Then \(Nu\in N_{L/K}A_L\). Choose \(v\in A_L\) with \(Nv=Nu\). Its
valuation is zero because \(v_K(Nv)=v_L(v)\) in this totally ramified
extension; thus \(v\in U_L\).

Now \(N(v-u)=0\). Apply the norm-zero half of (1) to the cyclic
extension \(M/M_0\). There is \(a\in A_M\) with

\[
v-u=(\sigma-1)a,\qquad
j\pi_L+v=j\pi_\Sigma+(\sigma-1)a.
\tag{5}
\]

On the left of (5), \(s\) acts as \(\sigma\), since that side is in
\(A_L\). On the right, \(s\pi_\Sigma=\pi_\Sigma\). Applying \(s-1\), and
using commutation of \(s,\sigma\), shows that

\[
x=j\pi_L+v-(s-1)a
\]

is fixed by \(\sigma\), hence belongs to \(A_{M_0}\). The action
preserves \(w\), so \(w((s-1)a)=0\) and \(w(x)=j\). But \(M/M_0\) has
ramification index \(n\). The valuation-inclusion formula (9) of the
preceding lesson yields

\[
j=n v_{M_0}(x)\in nV\subset n\widehat{\mathbf Z}.
\]

An ordinary integer in \(n\widehat{\mathbf Z}\) is divisible by \(n\),
by reduction modulo \(n\). Since \(0\leq j<n\), we get \(j=0\). This
proves injectivity. The source and target both have order \(n\), the
latter by (1), so reciprocity is an isomorphism.

For a general cyclic \(L/K\), let \(M=L\cap\widetilde K\). Then \(M/K\)
is unramified and \(L/M\) is totally ramified cyclic. Their reciprocity
maps are isomorphisms, the first by Proposition 4.3 and the second by
the argument just given. In (3), the three groups have orders \([L:M]\),
\([L:K]\), and \([M:K]\), by (1). The kernel at the middle has order
\([L:M]\); hence its surjective norm map from the first group is
injective as well. The matching exact Galois sequence (4) now proves
that \(r_{L/K}\) is injective and surjective: lift a quotient element,
and use the isomorphism on the two kernels to correct the lift. This
proves cyclic reciprocity in full.

\subsection{4. From cyclic groups to every finite Galois
group}\label{from-cyclic-groups-to-every-finite-galois-group}

\subsubsection{Theorem 5.1. The general reciprocity
law}\label{theorem-5.1.-the-general-reciprocity-law}

For every finite Galois \(L/K\),

\[
r_{L/K}:\operatorname{Gal}(L/K)^{\mathrm{ab}}
\xrightarrow{\sim} A_K/N_{L/K}A_L.
\tag{6}
\]

\textbf{Proof for abelian groups.} Let \(T=\operatorname{Gal}(L/K)\) be
abelian. If an element of \(T\) has zero reciprocity class, its
restriction has zero reciprocity class in every cyclic quotient
extension. Cyclic reciprocity is injective, and the character argument
in section 2 says these restrictions detect every nonidentity element.
Thus \(r_{L/K}\) is injective.

Surjectivity follows by induction on \(|T|\). The cyclic case is proved.
If \(T\) is not cyclic, choose a nontrivial cyclic quotient and its
fixed field \(M\); then both \(M/K\) and \(L/M\) have smaller degree,
and the latter is abelian. Apply induction and the lifting argument in
(3)--(4). This proves the abelian case.

\textbf{Injectivity for arbitrary groups.} Put
\(T=\operatorname{Gal}(L/K)\), and let \(B\) be the fixed field of the
commutator subgroup \(T'\). Restriction identifies \(T^{\mathrm{ab}}\)
with \(\operatorname{Gal}(B/K)\). The composite

\[
T^{\mathrm{ab}}\xrightarrow{r_{L/K}}Q_{L/K}
\longrightarrow Q_{B/K}
\]

is the already proved abelian reciprocity isomorphism. Hence \(r_{L/K}\)
is injective. Equivalently, the kernel of the map on \(T\) is exactly
\(T'\): it contains \(T'\) because the target is abelian, and its
projection to \(T/T'\) is injective.

\textbf{Surjectivity for solvable groups.} Here a finite group is
solvable if its iterated commutator subgroups eventually become trivial.
Induct on its order. If \(T'=1\), the abelian case applies. Otherwise
\(1<T'<T\); with \(B\) as above, \(B/K\) is abelian and \(L/B\) has
smaller solvable Galois group. Their maps are surjective, so (3)--(4)
prove surjectivity for \(L/K\).

\textbf{The Sylow step.} First recall why finite \(p\)-groups are
solvable. Conjugacy classes outside the center have size divisible by
\(p\), so the class equation makes the center of a nontrivial
\(p\)-group nontrivial. Quotient by this center and induct on the order.
A group with an abelian normal subgroup and solvable quotient is
solvable: enough commutator iterations enter the normal subgroup and one
more makes them trivial. This proves the assertion about \(p\)-groups.

We need only the existence part of Sylow's theorem, which can also be
proved here. Write \(|T|=p^b u\), with \(p\nmid u\), and let \(T\) act
by left multiplication on the subsets of its underlying set having
\(p^b\) elements. Their number is \(\binom{p^b u}{p^b}\equiv u\pmod p\):
in \(\mathbf F_p[X]\), the identity \((1+X)^{p^b u}=(1+X^{p^b})^u\)
proves this by comparing coefficients. Thus some orbit has cardinality
prime to \(p\). Its stabilizer \(P\) has \(p\)-part of order \(p^b\), by
the orbit--stabilizer formula. On the selected subset, \(P\) acts
freely, since a nonidentity left multiplication fixes no group element.
Hence \(|P|\) divides the subset's cardinality \(p^b\). These two facts
imply \(|P|=p^b\), which is the required Sylow subgroup.

Now \(Q=Q_{L/K}\) is annihilated by \(m=|T|\). Indeed, inclusion sends
\(a\in A_K\) to \(A_L\), and \(N_{L/K}a=ma\). A group annihilated by
\(m=\prod_p p^{b_p}\) decomposes as \(Q=\bigoplus_p Q_p\), where
\(p^{b_p}Q_p=0\). To verify this even before knowing \(Q\) finite,
choose integers \(c_p\) which are 1 modulo \(p^{b_p}\) and 0 modulo the
other prime powers, by the Chinese remainder theorem. Multiplication by
\(c_p\) gives pairwise orthogonal projections whose sum is the identity
on \(Q\).

Fix \(p\mid m\), choose a Sylow \(p\)-subgroup \(P\subset T\), and let
\(F=L^P\). The extension \(F/K\) need not be Galois. The extension
\(L/F\) has Galois group \(P\), so its reciprocity isomorphism is proved
by the solvable case. Norm functoriality says that the image of

\[
N_{F/K}:A_F/N_{L/F}A_L\longrightarrow Q
\]

is contained in the image of \(r_{L/K}\). Inclusion also defines a map
\(i:Q\to A_F/N_{L/F}A_L\): every norm from \(L/K\) is a norm from
\(L/F\), by the right-coset decomposition in Proposition 4.4. On these
quotients,

\[
N_{F/K}i=[F:K].
\tag{7}
\]

The integer \([F:K]\) is prime to \(p\). Multiplication by it is an
automorphism of \(Q_p\), since an inverse modulo \(p^{b_p}\) acts as its
inverse there. Equation (7) therefore puts all of \(Q_p\) in the image
of \(N_{F/K}\), hence in the image of reciprocity. Do this for every
\(p\mid m\); the primary decomposition proves surjectivity onto all of
\(Q\). No solvability assumption on \(T\) was used. Together with
injectivity, this proves (6). \(\square\)

The argument also proves finiteness of \(Q\) in the general case.
Finiteness was not silently assumed when taking its primary parts.

\subsection{5. The norm residue symbol}\label{the-norm-residue-symbol}

Define the \textbf{norm residue symbol} by taking the inverse of (6):

\[
(\ ,L/K):A_K\longrightarrow\operatorname{Gal}(L/K)^{\mathrm{ab}},
\qquad \ker(\ ,L/K)=N_{L/K}A_L.
\tag{8}
\]

In additive notation the symbol is a homomorphism; in multiplicative
arithmetic notation it sends a product of elements to a product of
automorphisms.

\subsubsection{Proposition 5.2. Functoriality of the
symbol}\label{proposition-5.2.-functoriality-of-the-symbol}

For finite Galois \(L/K,L'/K'\) with \(K\subset K'\), \(L\subset L'\),
and \(a'\in A_{K'}\),

\[
(N_{K'/K}a',L/K)
=\operatorname{res}_{L'/L}(a',L'/K').
\tag{9}
\]

Conjugation by \(g\in G\) gives

\[
(ga,gL/gK)=g(a,L/K)g^{-1}.
\tag{10}
\]

For \(K'\subset L\) and \(a\in A_K\subset A_{K'}\),

\[
(a,L/K')=\operatorname{Ver}_{\operatorname{Gal}(L/K')}^{\operatorname{Gal}(L/K)}(a,L/K).
\tag{11}
\]

\textbf{Proof.} Each identity is the corresponding equality of
Proposition 4.4, composed with the inverse reciprocity isomorphisms. The
quotient norm, conjugation and inclusion were proved well defined there.
Thus these equalities hold for every representative in the stated
modules, not merely for a chosen prime element. \(\square\)

For an unramified extension of degree \(n\), Proposition 4.3 gives the
useful expression

\[
(a,L/K)=\varphi_{L/K}^{\,v_K(a)\bmod n}.
\tag{12}
\]

The exponent is the image in \(V/nV\simeq\mathbf Z/n\mathbf Z\). It need
not come from an integer-valued valuation before reduction.

\subsection{6. Norm limitation}\label{norm-limitation}

\subsubsection{Theorem 5.4. Only the maximal abelian subextension is
seen by
norms}\label{theorem-5.4.-only-the-maximal-abelian-subextension-is-seen-by-norms}

For any finite separable \(L/K\), let \(B=L\cap K^{\mathrm{ab}}\). Then

\[
N_{L/K}A_L=N_{B/K}A_B.
\tag{13}
\]

\textbf{Proof.} First suppose \(L/K\) is Galois. The left side is
contained in the right by norm transitivity. Theorem 5.1 gives the same
finite index \(|\operatorname{Gal}(L/K)^{\mathrm{ab}}|=[B:K]\) for both
subgroups, proving equality.

For a general finite separable extension, choose its finite Galois
closure \(E/K\). Put \(G=\operatorname{Gal}(E/K)\),
\(H=\operatorname{Gal}(E/L)\), and let \(G'\) be the commutator
subgroup. The group \(G'H\) is normal: it is the inverse image of the
subgroup \(\overline H\) of the abelian quotient \(G/G'\). Its fixed
field lies in \(L\) and is abelian over \(K\). Conversely, any abelian
intermediate field inside \(L\) has a fixing subgroup containing both
\(H\) and \(G'\), hence containing \(G'H\). Thus \(B=E^{G'H}\).

Use the already proved isomorphism \[
r_{E/K}:G/G'\xrightarrow{\sim}
A_K/N_{E/K}A_E.
\] The Galois extension \(E/L\) has group \(H\), even though \(L/K\)
need not be Galois. Surjectivity of \(r_{E/L}\), and norm functoriality
from Proposition 4.4, show that \[
\frac{N_{L/K}A_L}{N_{E/K}A_E}
=r_{E/K}(\overline H).
\] Indeed every element of \(A_L/N_{E/L}A_E\) is the reciprocity image
of an element of \(H\), and its norm to \(K\) is the reciprocity image
of that same element in \(G/G'\). This proves both inclusions in the
displayed equality.

Apply the identical argument to the Galois extension \(E/B\), whose
group is \(G'H\). Its image in \(G/G'\) is exactly \(\overline H\), so
\[
\frac{N_{B/K}A_B}{N_{E/K}A_E}
=r_{E/K}(\overline H)
=\frac{N_{L/K}A_L}{N_{E/K}A_E}.
\] Both norm subgroups contain \(N_{E/K}A_E\), by norm transitivity.
Equality of their images in this quotient is therefore equality of the
subgroups themselves. No reciprocity map for a nonnormal extension was
presumed. \(\square\)

For a nonabelian Galois extension, its full degree can therefore exceed
the norm index. More generally the norm index of a finite separable
extension is \([B:K]\), which need not equal \([L:K]\). For example, if
a nonnormal cubic extension has Galois closure with group \(S_3\), its
subgroup \(H\) has order two and \(G'H=S_3\); thus \(B=K\) and its norm
subgroup is all of \(A_K\). The quadratic subfield of the Galois closure
has a different norm subgroup. Passing to a Galois closure alone does
not preserve the norm subgroup.

\subsection{7. The class field
correspondence}\label{the-class-field-correspondence}

For finite abelian \(L/K\), abbreviate \(N_L=N_{L/K}A_L\). Equip \(A_K\)
with the \textbf{norm topology}, whose basic neighborhoods of zero are
the norm subgroups of finite Galois extensions. This is a group
topology: the norm subgroup from a compositum is contained in each of
the two given norm subgroups by norm transitivity, so the basic
neighborhoods are directed under intersection. Translates supply
neighborhoods of other elements. By norm limitation the same basis is
obtained using only finite abelian extensions. A subgroup is open
precisely when it contains one of these basic subgroups.

\subsubsection{Theorem 5.3. The
correspondence}\label{theorem-5.3.-the-correspondence}

The assignment \(L\mapsto N_L\) is a bijection between finite abelian
extensions of \(K\) in the fixed separable closure and open subgroups of
\(A_K\) in its norm topology. Moreover,

\[
\begin{aligned}
L_1\subset L_2&\quad\Longleftrightarrow\quad N_{L_1}\supset N_{L_2},\\
N_{L_1L_2}&=N_{L_1}\cap N_{L_2},\\
N_{L_1\cap L_2}&=N_{L_1}+N_{L_2}.
\end{aligned}
\tag{14}
\]

In multiplicative notation the last sum is the product
\(N_{L_1}N_{L_2}\). If \(N_L\subset H\subset A_K\), there is a unique
intermediate field \(M\subset L\) with \(H=N_M\).

\textbf{Proof.} Norm transitivity gives
\(N_{L_1L_2}\subset N_{L_1}\cap N_{L_2}\). Conversely, an element of
both subgroups has symbol in \(\operatorname{Gal}(L_1L_2/K)\)
restricting trivially to both fields, by (9). Those restrictions
determine the automorphism on the compositum, so the symbol is trivial
and the element belongs to \(N_{L_1L_2}\). This proves the intersection
formula.

Containment \(L_1\subset L_2\) implies the reverse norm containment by
transitivity. If \(N_{L_1}\supset N_{L_2}\), the intersection formula
says \(N_{L_1L_2}=N_{L_2}\). Theorem 5.1 gives \([L_1L_2:K]=[L_2:K]\),
hence \(L_1\subset L_2\). This proves order reversal and injectivity.

Let \(H\) contain \(N_L\) for an abelian \(L/K\). Under the isomorphism
\(A_K/N_L\simeq T=\operatorname{Gal}(L/K)\), its image is a subgroup
\(D\subset T\). Let \(M=L^D\). Restriction in (9) identifies the symbol
for \(M/K\) with the quotient map \(T\to T/D\). Since \(H\) contains its
original kernel \(N_L\), it is the full preimage of \(D\), hence exactly
the kernel \(N_M\). Uniqueness follows from order reversal. Every open
subgroup contains a norm subgroup, and norm limitation permits choosing
that subgroup from an abelian extension. Thus the correspondence is
surjective onto the open subgroups.

Finally let \(H=N_{L_1}+N_{L_2}\), which contains a norm subgroup and is
open. Write \(H=N_M\) by the result just proved. Since
\(H\supset N_{L_i}\), order reversal gives \(M\subset L_1\cap L_2\),
hence \(N_M\supset N_{L_1\cap L_2}\). The opposite containment follows
because \(L_1\cap L_2\subset L_i\) implies
\(N_{L_1\cap L_2}\supset N_{L_i}\). These two containments give the sum
formula in (14). \(\square\)

This abstract theorem does not say that every finite-index subgroup for
an unrelated preexisting topology is a norm subgroup. The norm topology
is the topology just defined. Later existence theorems identify its open
subgroups in the usual arithmetic topologies.

\subsection{8. The universal symbol and an
example}\label{the-universal-symbol-and-an-example}

As finite abelian \(L/K\) vary, the symbols are compatible under
restriction by (9). The profinite correspondence in the first lesson
identifies their inverse-limit Galois group with
\(\operatorname{Gal}(K^{\mathrm{ab}}/K)\). Therefore they define the
\textbf{universal norm residue symbol}

\[
(\ ,K^{\mathrm{ab}}/K):A_K\longrightarrow
\operatorname{Gal}(K^{\mathrm{ab}}/K).
\tag{15}
\]

Its kernel is \(\bigcap_L N_L\), the subgroup of universal norms. Its
image is dense: for any finite set of quotient conditions, take their
finite compositum and use the surjectivity of its finite symbol.
Surjectivity onto the whole inverse limit requires an additional
argument in an arithmetic application; it does not follow just from the
finite quotient surjections. Formula (12) gives on the unramified tower

\[
(a,\widetilde K/K)=\varphi_K^{v_K(a)},
\qquad d_K(a,\widetilde K/K)=v_K(a),
\tag{16}
\]

where the exponent is profinite. Equality follows by reducing it modulo
every positive integer.

For the finite-field degree formation, \(A_K=\mathbf Z\), all actions
are trivial, and a degree-\(n\) extension has norm \(n\mathbf Z\). The
cyclic Tate groups are \(\mathbf Z/n\mathbf Z\) and zero, so (1) holds.
The correspondence sends \(\mathbf F_{q^{mn}}/\mathbf F_{q^m}\) to
\(n\mathbf Z\). Compositum and intersection have relative degrees
\(\operatorname{lcm}(n_1,n_2)\) and \(\gcd(n_1,n_2)\), matching
intersection and sum of these subgroups. Its universal symbol is the
dense inclusion \(\mathbf Z\to\widehat{\mathbf Z}\); it is not
surjective. This is a concrete reason to keep density and surjectivity
distinct in (15).

\subsection{Exercises}\label{exercises}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \textbf{Easy.} Deduce \(N_{L_1L_2}=N_{L_1}\cap N_{L_2}\) for two
  finite abelian extensions directly from Theorem 5.1 and restriction of
  symbols.
\item
  \textbf{Medium.} Prove norm limitation for a finite Galois extension,
  and explain why its norm index is the order of the abelianization
  rather than its full degree.
\item
  \textbf{Medium.} If \(L/K\) is finite abelian and
  \(N_L\subset H\subset A_K\), construct the unique \(M\subset L\) with
  \(H=N_M\).
\item
  \textbf{Hard.} Carry out the Sylow step for
  \(T=\operatorname{Gal}(L/K)\simeq S_3\). Treat separately the
  subgroups of orders 2 and 3; identify the resulting norm quotient and
  the maximal abelian subfield.
\end{enumerate}

\subsection{Solutions}\label{solutions}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Transitivity gives the inclusion from the compositum's norm subgroup
  into each factor. If \(a\) is in both factors, its symbol in the
  compositum has identity restriction to both \(L_1,L_2\). An
  automorphism fixing both fixes the field they generate, so that symbol
  is identity. The kernel assertion of Theorem 5.1 places \(a\) in
  \(N_{L_1L_2}\), proving equality.
\item
  Let \(B\) be the fixed field of \(T'\). Norm transitivity gives
  \(N_L\subset N_B\). Theorem 5.1 computes
  \([A_K:N_L]=|T/T'|=[B:K]=[A_K:N_B]\). These finite indices force the
  contained subgroups to be equal. Thus when \(T\) is nonabelian the
  quotient records exactly its maximal abelian quotient, and its index
  can be strictly smaller than \(|T|\).
\item
  Map \(H\) into \(T\) by the surjective symbol for \(L/K\), and call
  its image \(D\). Set \(M=L^D\). Restriction identifies the symbol for
  \(M/K\) with the quotient by \(D\). Because \(H\supset N_L\), it
  equals the full preimage of \(D\), hence the kernel \(N_M\). If \(M'\)
  also has that norm subgroup, the order-reversing part of Theorem 5.3
  gives \(M\subset M'\) and \(M'\subset M\), so \(M=M'\).
\item
  The quotient \(Q=A_K/N_{L/K}A_L\) is killed by 6 and decomposes as
  \(Q_2\oplus Q_3\). For \(P_2=\langle(12)\rangle\), the fixed field
  \(F_2\) has degree 3 over \(K\) and need not be Galois. Cyclic
  reciprocity gives \(A_{F_2}/N_{L/F_2}A_L\simeq C_2\). Inclusion
  followed by norm acts on \(Q\) as multiplication by 3, which is an
  automorphism on \(Q_2\); thus \(Q_2\) lies in this norm's image and
  hence in the image of the transposition under \(r_{L/K}\). For
  \(P_3=A_3\), the fixed field \(F_3\) has degree 2 over \(K\), and
  \(L/F_3\) is cyclic of degree 3. Its norm map onto \(Q\), composed
  with inclusion, is multiplication by 2, an automorphism on \(Q_3\).
  However its image under reciprocity is zero: the 3-cycles lie in the
  commutator subgroup of \(S_3\). Indeed, sign gives an abelian quotient
  \(C_2\), while the commutator of two distinct transpositions is a
  3-cycle, proving \(S_3'=A_3\). Thus \(Q_3=0\). Restriction to the
  quadratic field \(F_3=L^{A_3}\) identifies \(T^{\mathrm{ab}}\) with
  \(C_2\), and cyclic reciprocity there makes the transposition's class
  nonzero in \(Q\). Consequently \(Q\simeq C_2\) and
  \(N_{L/K}A_L=N_{F_3/K}A_{F_3}\). This also shows explicitly why
  neither the degree-6 field nor its non-Galois cubic fixed field may be
  substituted for the maximal abelian subfield in norm limitation.
\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--4 prove the cyclic, abelian, solvable and Sylow steps
without assuming a finite norm quotient prematurely. Theorem 5.4 proves
norm limitation for every finite separable extension, using its maximal
abelian subextension. Sections 7--8 prove the norm-group correspondence,
its topology and the universal symbol.

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

\end{document}