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

\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).}

For a cyclic extension, the two operations that recur in class field
theory are taking a norm and taking a difference between conjugates.
Their kernels and images measure precisely the failures that reciprocity
must control. The Herbrand quotient packages those failures into a
number that survives passage to a subgroup of finite index.

We assume abelian groups, modules, exact sequences and the elementary
properties of finite cyclic groups. The topology in
\href{../profinite-groups-and-infinite-galois-theory.html}{Profinite
groups and infinite Galois theory} explains the Galois groups to which
these calculations will apply, but this lesson concerns finite groups.
Freely accessible comparisons are Milne's \emph{Class Field Theory} and
the free electronic Bonn lectures, credited below.

\subsection{1. Two operators, two
obstructions}\label{two-operators-two-obstructions}

Let a finite group \(G\) act on an abelian group \(A\), written
additively. Put

\[
N_G=\sum_{g\in G}g,
\qquad I_GA=\langle ga-a:g\in G,\ a\in A\rangle.
\]

Here \(N_G\) is an endomorphism of \(A\). Its image lies in \(A^G\), and
it vanishes on \(I_GA\). The two Tate groups needed in this course are

\[
\widehat H^0(G,A)=A^G/N_GA,
\qquad
\widehat H^{-1}(G,A)=\ker N_G/I_GA.
\]

The first asks whether every invariant is a norm. The second asks
whether every norm-zero element is a sum of conjugate differences. For a
multiplicative module, replace sums by products, zero by 1 and \(ga-a\)
by \(g(a)/a\).

Now let \(G=\langle\sigma\rangle\) have order \(n\), and write

\[
D=\sigma-1,
\qquad N=1+\sigma+\cdots+\sigma^{n-1}.
\]

We have \(DN=ND=0\), \(A^G=\ker D\) and \(I_GA=DA\). The last equality
follows from

\[
\sigma^j-1=(\sigma-1)(1+\sigma+\cdots+\sigma^{j-1}).
\]

Thus the two groups are the two alternating homology groups of the
periodic sequence

\[
\cdots\xrightarrow{N}A\xrightarrow{D}A\xrightarrow{N}A\xrightarrow{D}A\xrightarrow{N}\cdots.
\]

A \textbf{crossed homomorphism}, or 1-cocycle, is a function
\(c:G\to A\) satisfying \(c(gh)=c(g)+g c(h)\). A coboundary has the form
\(c(g)=ga-a\) for some \(a\in A\). Their quotient is \(H^1(G,A)\). A
cocycle for the cyclic group is determined by \(b=c(\sigma)\), since

\[
c(\sigma^j)=\sum_{i=0}^{j-1}\sigma^i b.
\]

The relation \(\sigma^n=1\) is exactly \(Nb=0\), and this condition also
suffices to define a cocycle. Coboundaries correspond to \(b\in DA\).
Consequently

\[
H^1(G,A)\simeq\widehat H^{-1}(G,A).
\]

This identification uses the selected generator \(\sigma\). Changing
that generator changes the map evaluating a cocycle.

The alternating operators also come from an exact sequence of free
modules, rather than merely from the relation \(DN=0\). Let
\(R=\mathbf Z[G]\), with augmentation
\(\varepsilon(\sum a_j\sigma^j)=\sum a_j\). Then \[
\cdots\xrightarrow{D}R\xrightarrow{N}R
\xrightarrow{D}R\xrightarrow{\varepsilon}\mathbf Z\longrightarrow0
\] is exact, where the rightmost \(R\) has degree zero. Here is a
coefficient proof. For \(x=\sum_{j=0}^{n-1}a_j\sigma^j\), the
coefficient of \(\sigma^j\) in \(Dx\) is \(a_{j-1}-a_j\), with indices
modulo \(n\). Thus \(Dx=0\) exactly when all coefficients agree, or
\(x=cN\); this is precisely the image of multiplication by \(N\). On the
other hand \(Nx=(\sum_j a_j)N\), so \(\ker N=\ker\varepsilon\). If
\(\sum_j a_j=0\), choose \(b_0=0\) and successively solve
\(b_{j-1}-b_j=a_j\) for \(1\leq j<n\). The missing equation at \(j=0\)
follows from the zero sum. Hence \(x=D\sum_j b_j\sigma^j\). This proves
\(\ker N=DR=\ker\varepsilon\) at every required position.

Applying \(\operatorname{Hom}_R(-,A)\) identifies each term with \(A\)
by evaluation at 1 and gives the alternating cochain operators \(D,N\).
In particular its first two positive-degree cohomology groups are
\(\ker N/DA\) and \(\ker D/NA\). This is the explicit cyclic free
resolution used for the later cyclic factor-system computation; its
exactness has been proved here. The norm and difference descriptions of
the two Tate degrees require no appeal to an external resolution
theorem.

\subsection{2. How an exact sequence passes around the
cycle}\label{how-an-exact-sequence-passes-around-the-cycle}

\subsubsection{Proposition 2.1. The exact
hexagon}\label{proposition-2.1.-the-exact-hexagon}

A short exact sequence of \(G\)-modules

\[
0\longrightarrow A\xrightarrow{i}B\xrightarrow{p}C\longrightarrow0
\]

for cyclic \(G\) gives the following exact cycle. In the diagram, write
\(\widehat H^i(A)\) for \(\widehat H^i(G,A)\), and likewise for \(B,C\).

\[
\begin{array}{ccccc}
\widehat H^0(A)&\longrightarrow&\widehat H^0(B)&\longrightarrow&\widehat H^0(C)\\
{\scriptstyle\delta_{-1}}\uparrow&&&&\downarrow{\scriptstyle\delta_0}\\
\widehat H^{-1}(C)&\longleftarrow&\widehat H^{-1}(B)&\longleftarrow&\widehat H^{-1}(A)
\end{array}
\]

Traversing the diagram clockwise gives the repeated exact sequence. The
upward connecting arrow closes it; no zero terms are asserted.

\textbf{Proof.} Identify \(A\) with its image in \(B\). If \(c\in C^G\),
lift it to \(b\in B\). Then \(Db\in A\) and \(NDb=0\). Define

\[
\delta_0(c\bmod NC)=Db\bmod DA.
\]

Changing \(b\) by an element of \(A\) adds an element of \(DA\).
Changing \(c\) by \(Nc'\), and its lift by \(Nb'\), leaves \(Db\)
unchanged. Thus this map is well defined.

If \(Nc=0\), lift \(c\) to \(b\in B\). Then \(Nb\in A^G\), and set

\[
\delta_{-1}(c\bmod DC)=Nb\bmod NA.
\]

Changing the lift adds a norm from \(A\); changing \(c\) by \(Dc'\)
changes the lift by \(Db'\), whose norm is zero. This map is also well
defined. The four remaining arrows are induced by \(i\) and \(p\).

Here are the six exactness checks, so that the closing arrow is
included. An invariant \(a\in A\) becomes a norm in \(B\) precisely when
\(a=Nb\); then \(p(b)\) has norm zero and maps to its class under
\(\delta_{-1}\). An invariant \(b\in B\) whose class maps to zero in
\(\widehat H^0(G,C)\) has \(p(b)=Nc'\). Subtracting \(Nb'\), for a lift
\(b'\) of \(c'\), leaves an invariant in \(A\). An invariant \(c\in C\)
has \(\delta_0(c)=0\) precisely when a lift \(b\) has \(Db=Da\) for some
\(a\in A\); then \(b-a\) is an invariant lift.

Next, a norm-zero \(a\in A\) becomes a difference in \(B\) precisely
when \(a=Db\). In that case \(p(b)\) is invariant and maps to its class
under \(\delta_0\). A norm-zero \(b\in B\) whose image is \(Dc'\) can be
adjusted by \(Db'\) to a norm-zero element of \(A\). Finally, a
norm-zero \(c\in C\) has \(\delta_{-1}(c)=0\) precisely when a lift
satisfies \(Nb=Na\), with \(a\in A\); then \(b-a\) is a norm-zero lift.
Each image is visibly in the next kernel. These checks establish
equality at every position. \(\square\)

The connecting arrows express a useful principle. An invariant
downstairs may fail to have an invariant lift upstairs; its obstruction
is a norm-zero class in the kernel module. A norm-zero element
downstairs has the complementary obstruction, an invariant norm class in
that module.

\subsection{3. Counting the
obstructions}\label{counting-the-obstructions}

When both Tate groups are finite, define the \textbf{Herbrand quotient}

\[
h_G(A)=\frac{|\widehat H^0(G,A)|}{|\widehat H^{-1}(G,A)|}.
\]

A trivial group has order 1 in this formula. The quotient is a positive
rational number; it need not be an integer.

\subsubsection{Proposition 2.2.
Multiplicativity}\label{proposition-2.2.-multiplicativity}

For the short exact sequence of Proposition 2.1, if the Herbrand
quotients are defined for any two of \(A,B,C\), they are defined for the
third, and

\[
h_G(B)=h_G(A)h_G(C).
\]

\textbf{Proof.} In the hexagon, a group whose two adjacent groups are
finite is an extension of a finite image by a finite image. When the
Tate groups of any two modules are finite, this applies to both groups
of the remaining module. Now name the six consecutive images
\(J_1,\ldots,J_6\), beginning with the arrow out of
\(\widehat H^0(G,A)\). Exactness gives

\[
\begin{array}{lll}
|\widehat H^0(G,A)|=|J_6||J_1|,&
|\widehat H^0(G,B)|=|J_1||J_2|,&
|\widehat H^0(G,C)|=|J_2||J_3|,\\
|\widehat H^{-1}(G,A)|=|J_3||J_4|,&
|\widehat H^{-1}(G,B)|=|J_4||J_5|,&
|\widehat H^{-1}(G,C)|=|J_5||J_6|.
\end{array}
\]

Substitution and cancellation prove the formula. \(\square\)

\subsubsection{Proposition 2.3. Finite modules and induced
modules}\label{proposition-2.3.-finite-modules-and-induced-modules}

For cyclic \(G\) of order \(n\):

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  A finite \(G\)-module \(A\) has \(h_G(A)=1\).
\item
  With trivial action, \(h_G(\mathbf Z)=n\).
\item
  For \(H\subset G\) and an \(H\)-module \(B\), induction gives
  \(\widehat H^i(G,\operatorname{Ind}_H^G B)\simeq\widehat H^i(H,B)\)
  for \(i=0,-1\). Hence \(h_G(\operatorname{Ind}_H^G B)=h_H(B)\)
  whenever these quotients are defined.
\item
  In particular, \(h_G(\mathbf Z[G/H])=|H|\).
\end{enumerate}

\textbf{Proof.} For finite \(A\), the two identities

\[
|A|=|\ker D|\,|DA|=|\ker N|\,|NA|
\]

give the first assertion. For \(\mathbf Z\), \(D=0\) and \(N=n\); the
two Tate groups are \(\mathbf Z/n\mathbf Z\) and zero. This proves the
second.

For induction, put \(r=[G:H]\), so \(H=\langle\tau\rangle\) with
\(\tau=\sigma^r\), and write

\[
P=\operatorname{Ind}_H^G B=
\bigoplus_{j=0}^{r-1}\sigma^j B.
\]

The action of \(\sigma\) shifts each summand to the next and takes
\(\sigma^{r-1}b\) to \(\tau b\) in the first summand. An invariant is
therefore a constant tuple \((b,\ldots,b)\), with \(b\in B^H\). The norm
of an element supported in the first summand is the constant tuple with
value \(N_Hb\). These elements generate \(P\), so
\(P^G/N_GP\simeq B^H/N_HB\).

For degree \(-1\), pass first to coinvariants. The relations
\(\sigma x-x\) identify all summands and impose exactly the relation
\(\tau b-b\) on the first. Thus

\[
P_G=P/(\sigma-1)P\simeq B/(\tau-1)B=B_H.
\]

Under this identification and the invariant identification just proved,
the norm map \(P_G\to P^G\) becomes \(N_H:B_H\to B^H\). Its kernel is
\(\widehat H^{-1}\), proving the second induction isomorphism. Taking
\(B=\mathbf Z\) proves the last assertion. \(\square\)

In particular, \(\mathbf Z[G]\) has both Tate groups zero. Its
invariants are the multiples of \(\sum_g g\), all of which are norms.
Its norm-zero elements are precisely the vectors whose coefficients sum
to zero, and these are conjugate differences.

If a submodule \(A\subset B\) has finite index, Proposition 2.2 and the
finite-module calculation show that \(h_G(A)=h_G(B)\), whenever one is
defined. Finite torsion likewise has no effect on the quotient.

\subsection{4. A calculation from an orbit
decomposition}\label{a-calculation-from-an-orbit-decomposition}

Let \(G\) be cyclic of order 6, and let \(P\) be the permutation lattice
on the disjoint union of an orbit of size 2 and an orbit of size 3. The
stabilizers have orders 3 and 2, respectively. Therefore

\[
h_G(P)=3\cdot2=6.
\]

Let \(P_0\) be the kernel of the sum of all five coordinates. The
augmentation is surjective and gives

\[
0\longrightarrow P_0\longrightarrow P\longrightarrow\mathbf Z\longrightarrow0.
\]

Consequently \(h_G(P_0)=6/6=1\). If instead \(P=\mathbf Z[G]\), its
augmentation kernel has Herbrand quotient \(1/6\). This example shows
why the dimension of a lattice alone cannot determine the quotient: the
representation carried by it matters.

\subsection{5. Changing a lattice without changing its
quotient}\label{changing-a-lattice-without-changing-its-quotient}

A \textbf{\(G\)-lattice} is a finitely generated free abelian group with
a \(G\)-action. Its Tate groups are finite. To see this, they are
finitely generated quotients of subgroups of a lattice and are killed by
\(n=|G|\). For degree 0, \(Na=na\) for invariant \(a\). For degree
\(-1\), if \(Na=0\), then

\[
na=\sum_{j=0}^{n-1}(a-\sigma^j a)\in DA.
\]

A finitely generated abelian group killed by \(n\) is finite.

\subsubsection{Theorem 2.4. The real representation determines the
quotient}\label{theorem-2.4.-the-real-representation-determines-the-quotient}

If \(M,M'\) are \(G\)-lattices for cyclic \(G\), and

\[
M\otimes_{\mathbf Z}\mathbf R\simeq M'\otimes_{\mathbf Z}\mathbf R
\]

as real \(G\)-representations, then \(h_G(M)=h_G(M')\).

\textbf{Proof.} Choose integral bases. A matrix \(T\) intertwines the
actions exactly when it satisfies the rational linear equations

\[
T\rho(g)=\rho'(g)T\quad(g\in G).
\]

Row reduction over \(\mathbf Q\) gives a rational basis
\(T_1,\ldots,T_s\) for this solution space; its real solutions are the
real span of the same matrices. An invertible real intertwiner exists by
hypothesis. Therefore the polynomial

\[
\det(t_1T_1+\cdots+t_sT_s)\in\mathbf Q[t_1,\ldots,t_s]
\]

is not the zero polynomial. A nonzero polynomial over the infinite field
\(\mathbf Q\) cannot vanish at every rational tuple: induction on the
number of variables reduces this to the fact that a nonzero one-variable
polynomial has finitely many roots. Some rational tuple consequently
gives an invertible rational intertwiner.

Multiply that matrix by a positive integer clearing denominators. It
defines an injective \(G\)-equivariant homomorphism \(M\to M'\) with
finite cokernel \(F\). By Propositions 2.2 and 2.3,

\[
h_G(M')=h_G(M)h_G(F)=h_G(M).
\]

This also describes the usual commensurability argument: after
identifying the rational representations, the intersection of their
lattices has finite index in both. \(\square\)

The theorem will let us replace a group of arithmetic units by a
permutation representation after tensoring with \(\mathbf R\). It
computes a ratio, not either Tate group separately. Vanishing of
\(H^1\), such as Hilbert's Theorem 90, is a separate input.

\subsection{Exercises}\label{exercises}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \textbf{Easy.} For \(G\) cyclic of order \(n\), compute both Tate
  groups of \(\mathbf Z\) with trivial action and of \(\mathbf Z[G]\)
  with its regular action.
\item
  \textbf{Medium.} Prove \(h_G(A)=1\) for finite \(A\) by counting the
  fibers of \(D\) and \(N\). Explain why this does not imply that either
  Tate group is zero.
\item
  \textbf{Medium.} Let
  \(G=\operatorname{Gal}(\mathbf F_{q^n}/\mathbf F_q)\) act on
  \(A=\mathbf F_{q^n}^{\times}\). Compute both Tate groups and
  \(h_G(A)\).
\item
  \textbf{Hard.} Prove that an isomorphism of the real representations
  of two \(G\)-lattices yields an injective equivariant map between the
  lattices with finite cokernel, and deduce equality of their Herbrand
  quotients. Identify the step where infinitude of \(\mathbf Q\) is
  used.
\end{enumerate}

\subsection{Solutions}\label{solutions}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  For \(\mathbf Z\), \(D=0\) and \(N=n\), so
  \(\widehat H^0=\mathbf Z/n\mathbf Z\) and \(\widehat H^{-1}=0\). In
  the regular module, an invariant has every coefficient equal, hence is
  a multiple of \(S=\sum_g g=N(1)\). The norm of \(\sum a_g g\) is
  \((\sum a_g)S\). Its kernel is generated by the differences
  \(\sigma^j-1\), all in \(D\mathbf Z[G]\). Both Tate groups are
  therefore zero.
\item
  The fiber counts give \(|\ker D||DA|=|A|=|\ker N||NA|\). Dividing
  gives \((|\ker D|/|NA|)/(|\ker N|/|DA|)=1\). For a counterexample to
  vanishing, let \(G\) have prime order \(p\) and act trivially on
  \(A=\mathbf Z/p\mathbf Z\). Then \(D=N=0\), and both Tate groups equal
  \(A\). Their equal orders still give quotient 1.
\item
  The multiplicative group of a finite field is cyclic. Frobenius is
  \(x\mapsto x^q\), and the norm is \(x\mapsto x^{(q^n-1)/(q-1)}\). Its
  image has order \(q-1\), hence is all of \(\mathbf F_q^\times=A^G\).
  Its kernel has order \((q^n-1)/(q-1)\). The difference map is now
  \(x\mapsto\sigma(x)/x=x^{q-1}\); its image has that same order and
  lies in the norm kernel. The two are equal. Both Tate groups are
  trivial and \(h_G(A)=1\).
\item
  The intertwining matrices solve a rational linear system. Express its
  solution space in a rational basis \(T_j\). Existence of an invertible
  real solution says that \(\det(\sum t_jT_j)\) is a nonzero rational
  polynomial. Infinitude of \(\mathbf Q\) ensures a rational tuple at
  which it is nonzero, so an invertible rational intertwiner exists.
  Clearing its denominators gives an injective map of the lattices. An
  injective integral matrix of full rank has finite cokernel, of order
  the absolute value of its determinant. That cokernel is a finite
  \(G\)-module with quotient 1; multiplicativity proves the required
  equality.
\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}

Section 1 proves the exact cyclic integral resolution. Sections 2--4
prove the connecting maps, the complete six-term sequence,
multiplicativity, induced-module calculations and lattice invariance.

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

\end{document}