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\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.

\end{document}