# On Constructing Extensions of Residually Isomorphic Characters

Samit Dasgupta111The author is supported by NSF grant DMS–2200787.

###### Abstract

This is an exposition of our joint work with Kakde, Silliman, and Wang, in
which we prove a version of Ribet’s Lemma for $\operatorname{GL}_{2}$ in the
residually indistinguishable case. We suppose we are given a Galois
representation taking values in the total ring of fractions of a complete
reduced Noetherian local ring $\mathbf{T}$, such that the characteristic
polynomial of the representation is reducible modulo some ideal
$I\subset\mathbf{T}$. We assume that the two characters that arise are
congruent modulo the maximal ideal of $\mathbf{T}$. We construct an associated
Galois cohomology class valued in a $\mathbf{T}$-module that is “large” in the
sense that its Fitting ideal is contained in $I$. We make some simplifying
assumptions that streamline the exposition—we assume the two characters are
actually equal, and we ignore the local conditions needed in arithmetic
applications.

###### Contents

  1. 1 Introduction
  2. 2 The DVR case
  3. 3 The Residually Distinguishable Case
  4. 4 The Residually Indistinguishable Case
    1. 4.1 Fitting Ideal
    2. 4.2 Construction of $M$
    3. 4.3 Explication of Fitting Ideal
    4. 4.4 Traces and Determinants
    5. 4.5 An Altered Matrix
    6. 4.6 An Example
  5. 5 Formal Variables
    1. 5.1 The ring $R$
    2. 5.2 Relation Ideal
    3. 5.3 Subring of traces and determinants
  6. 6 Matrix Invariant Theory and Rational Cohomology
    1. 6.1 Rational cohomology
    2. 6.2 Roadmap
    3. 6.3 Invariance
    4. 6.4 Kozsul complex
    5. 6.5 Embedding into an acyclic complex
    6. 6.6 A cascade of cohomology classes

##  1 Introduction

In 1970, Ribet proved the converse of Herbrand’s Theorem [ribet]. When I
learned about this in graduate school, Dick Gross recalled to me that the
methods introduced by Ribet “were like a thunderbolt” in the number theory
community at the time. Over 50 years later, Ribet’s method remains a central
force in algebraic number theory, particularly in Iwasawa Theory. Perhaps the
seminal work on the topic is the beautiful book of Joel Bellaïche and Gaetan
Chenevier [bc]. This book presents a very general form of Ribet’s approach and
also describes deep arithmetic applications. An introduction written for a
more general audience is given by Mazur [mazur].

The goal of Ribet’s method is to construct a nontrivial Galois cohomology
class from the knowledge that an $L$-function is appropriately divisible.
Typically, we will be given a special value of an $L$-function, which we
denote by $L$, lying in a ring $\mathbf{T}$ (e.g. $\mathbf{T}=\mathbf{Z}_{p}$
or $\mathbf{T}=\mathbf{Z}_{p}[[T]]$), and we assume assume that $L$ lies in
some ideal $I\subset\mathbf{T}$. The value $L$ will be associated to two (or
more) representations of the Galois group of a number field $F$ over $R$, say
$\rho_{1}$ and $\rho_{2}$. One then wants to construct a nontrivial class in

$H^{1}(G_{F},\overline{\rho}_{1}\otimes\overline{\rho}_{2}^{*}),$

where $\rho_{2}^{*}$ is the dual of $\rho_{2}$ and $\overline{\rho}_{i}$
denotes the reduction of ${\rho}_{i}$ modulo $I$. In Ribet’s original setting,
he had $\mathbf{T}=\mathbf{Z}_{p}$, $I=(p)$, and $\rho_{1},\rho_{2}$ one
dimensional characters of $G_{\mathbf{Q}}$. Specifically, $\rho_{1}$ was the
trivial character and $\rho_{2}$ a nontrivial character of ${\rm
Gal}(\mathbf{Q}(\mu_{p})/\mathbf{Q})$.

To produce an extension, Ribet constructed a cusp form congruent to the
Eisenstein series associated to $\rho_{1}$ and $\rho_{2}$. In his case, one
can show that the cusp form is an eigenform. One therefore obtains a
representation

$\rho\colon G_{\mathbf{Q}}\longrightarrow\operatorname{GL}_{2}(K),$

where $K$ is a finite extension of $\mathbf{Q}_{p}$, such that

$\operatorname{tr}(\rho)\equiv\rho_{1}+\rho_{2},\qquad\det(\rho)\equiv\rho_{1}\rho_{2}\pmod{I}.$
(1)

The representation $\rho$ can be conjugated to land in
$\operatorname{GL}_{2}(\mathcal{O}_{K})$, and the Brauer–Nesbitt Theorem
implies that one can choose a basis so that the reduction of $\rho$ modulo the
maximal ideal $\mathfrak{m}_{K}\subset\mathcal{O}_{K}$ has the form

$\overline{\rho}=\begin{pmatrix}{\overline{\rho}_{1}}&{*}\\\
{0}&{\overline{\rho}_{2}}\end{pmatrix}.$

Ribet then proves an important lemma which shows that the basis can be chosen
so that the image of $*$ in $\mathcal{O}_{K}/\mathfrak{m}_{K}$ defines a non-
trivial class in
$H^{1}(G_{F},\overline{\rho}_{1}\otimes\overline{\rho}_{2}^{-1})$.
Furthermore, Ribet proves certain local conditions satisfied by this non-
trivial class. It is elementary class field theory to prove that the existence
of this class implies the converse to Herbrand’s Theorem. We state Ribet’s
Lemma formally as follows.

######  Theorem 1.1 (Ribet’s Lemma, Version 1).

Let $\mathbf{T}$ be a complete DVR, and let $\mathfrak{m}$ denote its maximal
ideal. Let $\mathcal{G}$ be a compact group. Suppose we are given a continuous
irreducible representation

$\rho\colon\mathcal{G}\longrightarrow\operatorname{GL}_{2}(K),\qquad
K=\operatorname{Frac}(\mathbf{T}),$

such that

$\text{char}(\rho(g))\equiv(x-\chi_{1}(g))(x-\chi_{2}(g))\pmod{\mathfrak{m}}$
(2)

for two characters
$\chi_{1},\chi_{2}\colon\mathcal{G}\longrightarrow\mathbf{T}^{*}.$ Then there
exists a non-zero cohomology class

$\kappa\in
H^{1}(\mathcal{G},\mathbf{\mathbf{T}/\mathfrak{m}}(\chi_{2}^{-1}\chi_{1})).$

In more general settings, such as that employed by Mazur–Wiles [mw] and Wiles
[wiles] in their study of the Iwasawa Main Conjecture, the ring $\mathbf{T}$
cannot be assumed to be a DVR. It will usually be a complete local Noetherian
ring, perhaps reduced. An example of such a ring that is not a DVR is

$\mathbf{T}=\\{(a,b)\in\mathbf{Z}_{p}\times\mathbf{Z}_{p}\colon a\equiv
b\pmod{p}\\}.$

This example corresponds to two eigenforms with Hecke eigenvalues in
$\mathbf{Z}_{p}$ that are congruent to each other modulo $p$. The total ring
of fractions of $\mathbf{T}$ is
$K=\operatorname{Frac}(\mathbf{T})=\mathbf{Q}_{p}\times\mathbf{Q}_{p}$.

In addition, the ideal $I\subset\mathbf{T}$ modulo which the characteristic
polynomial of $\rho$ factors will in general not be the maximal ideal of
$\mathbf{T}$. One then wants to construct a cohomology class that generates a
module that (in a sense we will make precise in a moment) is “as large as”
$\mathbf{T}/I$. The first version of Ribet’s Lemma that applies in this case
was proven by Mazur–Wiles and Wiles. Their work was groundbreaking and had a
profound impact, leading to Wiles’ theory of pseudorepresentations. In the
statement below, a cohomology class $\kappa$ valued in a $\mathbf{T}$-module
$M$ is called surjective if the image of every representative cocycle
generates $M$ as a $\mathbf{T}$-module.

######  Theorem 1.2 (Ribet’s Lemma, Version 2).

Let $\mathbf{T}$ be a complete reduced local Noetherian ring, and let
$\mathfrak{m}$ denote its maximal ideal. Let $\mathcal{G}$ be a compact group.
Suppose we are given a continuous representation

$\rho\colon\mathcal{G}\longrightarrow\operatorname{GL}_{2}(K),\qquad
K=\operatorname{Frac}(\mathbf{T}),$

such that for any projection onto a field $K\rightarrow k$, the projection of
$\rho$ to a representation $\mathcal{G}\rightarrow\operatorname{GL}_{2}(k)$ is
irreducible. Let $I\subset\mathbf{T}$ be a proper ideal. Suppose that

$\text{char}(\rho(g))\equiv(x-\chi_{1}(g))(x-\chi_{2}(g))\pmod{I}$ (3)

for two characters
$\chi_{1},\chi_{2}\colon\mathcal{G}\longrightarrow\mathbf{T}^{*}$ satisfying
$\chi_{1}\not\equiv\chi_{2}\pmod{\mathfrak{m}}$. Then there exists a
fractional ideal of $\mathbf{T}$, $B\subset K$, and a surjective cohomology
class

$\kappa\in H^{1}(G,B/IB(\chi_{2}^{-1}\chi_{1})).$

The assumption $\chi_{1}\not\equiv\chi_{2}\pmod{\mathfrak{m}}$ is essential in
Wiles’ approach to Theorem 1.2, and it has important consequences. In the Main
Conjecture of Iwasawa Theory, one has $\chi_{1}$ equal to the trivial
character and $\chi_{2}$ equal to a totally odd character of a totally real
field. Let $c$ denote complex conjugation. When $p\neq 2$, we have
$\chi_{1}(c)=1\not\equiv-1=\chi_{2}(c)\pmod{p}$. However when $p=2$ we may
have $\chi_{1}\equiv\chi_{2}\pmod{p}$, and Theorem 1.2 cannot be applied. This
is the main reason that Wiles sets $p\neq 2$ in his proof of the Main
Conjecture.

The purpose of this exposition is to describe the main theorem of our paper
[jjm], joint with Mahesh Kakde, Jesse Silliman, and Jiuya Wang, in which we
establish a version of Ribet’s Lemma that holds even if
$\chi_{1}\equiv\chi_{2}\pmod{\mathfrak{m}}$. Here we describe the proof of the
following result, a simplified form of Theorem 2.1 of loc. cit.

######  Theorem 1.3 (Ribet’s Lemma, Version 3).

Let $\mathbf{T}$ be a complete reduced local Noetherian ring, and let
$\mathfrak{m}$ denote its maximal ideal. Let $\mathcal{G}$ be a compact group.
Suppose we are given a continuous representation

$\rho\colon\mathcal{G}\longrightarrow\operatorname{GL}_{2}(K),\qquad
K=\operatorname{Frac}(\mathbf{T}),$

such that for any projection onto a field $K\rightarrow k$, the projection of
$\rho$ to a representation $\mathcal{G}\rightarrow\operatorname{GL}_{2}(k)$ is
irreducible. Let $I\subset\mathbf{T}$ be a proper ideal. Suppose that

$\text{char}(\rho(g))\equiv(x-\chi_{1}(g))(x-\chi_{2}(g))\pmod{I}$ (4)

for two characters
$\chi_{1},\chi_{2}\colon\mathcal{G}\longrightarrow\mathbf{T}^{*}$. Then there
exists a finitely generated $\mathbf{T}$-module $M$ and a surjective
cohomology class

$\kappa\in H^{1}(G,M(\chi_{2}^{-1}\chi_{1}))$

such that

$\operatorname{Fitt}_{\mathbf{T}}(M)\subset I.$

The ideal $\operatorname{Fitt}_{\mathbf{T}}(M)$ is the 0th Fitting ideal of
the module $M$, which will be defined in §4.1. Intuitively, the inclusion
$\operatorname{Fitt}_{\mathbf{T}}(M)\subset I$ says that $M$ is “large.”

In [bsapet], we prove the Brumer–Stark conjecture at $p=2$ using a suitably
generalized version of Theorem 1.3. Previously, we proved the conjecture over
$\mathbf{Z}[1/2]$ in [bs], with the prime $p=2$ being avoided for reasons of
residual distinguishability. We hope that further strengthenings of Theorem
1.3 (for example, to groups other than $\operatorname{GL}_{2}$) could have
other arithmetic applications.

In this paper, we simplify notation by setting $\chi_{1}$ and $\chi_{2}$ to be
the trivial character; the case of general $\chi_{1},\chi_{2}$ requires no
extra ideas, but the notation is heavier. A more significant change in this
paper relative to [jjm] is that all the versions of Ribet’s method stated
above do not include local conditions on the cohomology classes constructed.
In arithmetic applications, ranging from Ribet’s original proof of the
converse to Herbrand’s theorem to our proof of the Brumer–Stark conjecture,
local conditions are always necessary. To prove the local properties we need
in the Brumer–Stark context, the argument presented here is generalized in
[jjm] using the Buchsbaum–Rim resolution of determinantal ideals; in this
paper, the simple Kozsul complex suffices. In loc. cit. we also give a mild
generalization to certain non-reduced Hecke algebras $\mathbf{T}$.

It is an honor to contribute this article to the memorial volume for Joël
Bellaïche. Joël was a wonderful collaborator and dear friend. We discussed the
residually indistinguishable case of Ribet’s Lemma in 2010, at which time both
of us felt the problem was intractable. It is a great sadness that I cannot
share this result with my colleague who perhaps would have appreciated it the
most.

##  2 The DVR case

In this section we prove Theorem 1.1, Ribet’s original setting. Let
$\mathbf{T}$ be a complete DVR and let $\mathfrak{m}$ denote its maximal
ideal. Let $\mathcal{G}$ be a compact group. We are given a continuous
irreducible representation

$\rho\colon\mathcal{G}\longrightarrow\operatorname{GL}_{2}(K),\qquad
K=\operatorname{Frac}(\mathbf{T}),$

such that

$\text{char}(\rho(g))\equiv(x-\chi_{1}(g))(x-\chi_{2}(g))\pmod{\mathfrak{m}}$
(5)

for two characters
$\chi_{1},\chi_{2}\colon\mathcal{G}\longrightarrow\mathbf{T}^{*}.$

######  Lemma 2.1.

The representation $\rho$ may be conjugated to have image contained in
$\operatorname{GL}_{2}(\mathbf{T})$, and such that the reduction
$\overline{\rho}$ has the shape

$\overline{\rho}=\begin{pmatrix}{\overline{\chi}_{1}}&{*}\\\
{0}&{\overline{\chi}_{2}}\end{pmatrix}.$

###### Proof.

The maximal compact subgroups of $\operatorname{GL}_{2}(K)$ are precisely the
conjugates of $\operatorname{GL}_{2}(\mathbf{T})$. Since $\mathcal{G}$ is
compact and $\rho$ is continuous, it follows $\rho$ that may be conjugated to
have image contained in $\operatorname{GL}_{2}(\mathbf{T})$. Now (5) states
that

$\text{char}(\overline{\rho}(g))=(x-\overline{\chi}_{1}(g))(x-\overline{\chi}_{2}(g))\text{
in }\mathbf{T}/\mathfrak{m}.$ (6)

The Brauer–Nesbitt Theorem [bn] states that if two representations over a
field have the same characteristic polynomial, then their semisimplifications
are isomorphic. Therefore, (6) implies that
$\overline{\rho}^{\operatorname{ss}}\cong\overline{\chi}_{1}\oplus\overline{\chi}_{2}$.
Hence we have either

$\overline{\rho}\cong\begin{pmatrix}{\overline{\chi}_{1}}&{*}\\\
{0}&{\overline{\chi}_{2}}\end{pmatrix}\qquad\text{ or
}\qquad\overline{\rho}\cong\begin{pmatrix}{\overline{\chi}_{2}}&{*}\\\
{0}&{\overline{\chi}_{1}}\end{pmatrix}.$ (7)

It is a pleasant exercise to prove (7) directly from (6) without reference to
the full strength of the Brauer–Nesbitt Theorem. Now, suppose we are in the
second case. Then we can conjugate $\rho$ by the matrix
$\begin{pmatrix}{\pi}&{0}\\\ {0}&{1}\end{pmatrix}$, where $\pi$ is a
uniformizer of $\mathbf{T}$, to obtain

$\overline{\rho}\cong\begin{pmatrix}{\overline{\chi}_{2}}&{0}\\\
{*}&{\overline{\chi}_{1}}\end{pmatrix}\cong\begin{pmatrix}{\overline{\chi}_{1}}&{*}\\\
{0}&{\overline{\chi}_{2}}\end{pmatrix}$

as desired. ∎

######  Lemma 2.2.

Suppose that the representation $\rho$ has been conjugated so that its image
lands in $\operatorname{GL}_{2}(\mathbf{T})$ and

$\overline{\rho}=\begin{pmatrix}{\overline{\chi}_{1}}&{\overline{b}}\\\
{0}&{\overline{\chi}_{2}}\end{pmatrix}.$

Then the function
$\kappa(\sigma)=\overline{\chi}_{2}^{-1}(\sigma)\overline{b}(\sigma)$ is a
1-cocycle in
$Z^{1}(\mathcal{G},\mathbf{T}/\mathfrak{m}(\chi_{2}^{-1}\chi_{1}))$.

###### Proof.

Since $\overline{\rho}$ is a matrix representation, we have

$\overline{b}(\sigma\tau)=\overline{\chi_{1}}(\sigma)\overline{b}(\tau)+\overline{b}(\sigma)\overline{\chi_{2}}(\tau).$

Multiplying by $\overline{\chi}_{2}^{-1}(\sigma\tau)$ gives the desired
1-cocycle formula for $\kappa=\overline{\chi}_{2}^{-1}\overline{b}$. ∎

What remains to prove Theorem 1.1 is to show that after conjugating $\rho$
further, we can arrange for the cohomology class represented by $\kappa$ to be
non-trivial.

###### Proof of Theorem 1.1.

In his 2008 lectures from the Clay Summer School in Hawaii, Bellaïche gives a
beautiful and conceptual proof of the fact that $\rho$ can be chosen so that
the cohomology class represented by $\kappa$ is non-trivial [clay]*Proposition
1.4. He attributes this proof to Serre. Here, we take the more computational
approach applied by Ribet.

Since we will be applying a recursive process, let $\rho_{1}=\rho$ and write

$\rho_{1}(\sigma)=\begin{pmatrix}{a_{1}(\sigma)}&{b_{1}(\sigma)}\\\
{c_{1}(\sigma)}&{d_{1}(\sigma}\end{pmatrix}\in\operatorname{GL}_{2}(\mathbf{T}).$

If $\pi$ denotes a uniformizer of $K=\operatorname{Frac}(\mathbf{T})$, then we
have

$\displaystyle a_{1}(\sigma)$ $\displaystyle\equiv\chi_{1}(\sigma)$
$\displaystyle\pmod{\pi},$ (8) $\displaystyle c_{1}(\sigma)$
$\displaystyle\equiv 0$ $\displaystyle\pmod{\pi},$ (9) $\displaystyle
d_{1}(\sigma)$ $\displaystyle\equiv\chi_{2}(\sigma)$ $\displaystyle\pmod{\pi}$
(10)

for all $\sigma\in\mathcal{G}$. Denote the cocycle constructed in Lemma 2.2 by
$\kappa_{1}$.

Suppose that $\kappa_{1}$ represents a trivial cohomology class; then there
exists $x_{1}\in\mathbf{T}$ such that

$\kappa_{1}(\sigma)\equiv(\chi_{2}^{-1}\chi_{1}(\sigma)-1)x_{1}\pmod{\pi}$

for all $\sigma\in\mathcal{G}$, or equivalently,

$b_{1}(\sigma)\equiv(\chi_{1}(\sigma)-\chi_{2}(\sigma))x_{1}\pmod{\pi}.$ (11)

Conjugating the representation $\rho_{1}$, we define

${\rho}_{2}(\sigma)=\begin{pmatrix}{a_{2}(\sigma)}&{b_{2}(\sigma)}\\\
{c_{2}(\sigma)}&{d_{2}(\sigma)}\end{pmatrix}=\begin{pmatrix}{1}&{x_{1}}\\\
{0}&{\pi}\end{pmatrix}\rho_{1}(\sigma)\begin{pmatrix}{1}&{x_{1}}\\\
{0}&{\pi}\end{pmatrix}^{-1}.$

Using the congruences (8)–(11), we find that
$\rho_{2}(\sigma)\in\operatorname{GL}_{2}(\mathbf{T})$ and furthermore that

$\displaystyle a_{2}(\sigma)$ $\displaystyle\equiv\chi_{1}(\sigma)$
$\displaystyle\pmod{\pi},$ $\displaystyle c_{2}(\sigma)$ $\displaystyle\equiv
0$ $\displaystyle\pmod{\pi^{2}},$ $\displaystyle d_{2}(\sigma)$
$\displaystyle\equiv\chi_{2}(\sigma)$ $\displaystyle\pmod{\pi}.$

We are therefore once again in the setting of Lemma 2.2 and obtain a cocycle

$\kappa_{2}(\sigma)=\overline{\chi}_{2}^{-1}(\sigma)\overline{b_{2}}(\sigma)\in
Z^{1}(\mathcal{G},\mathbf{T}/\mathfrak{m}(\chi_{2}^{-1}\chi_{1})).$

If $\kappa_{2}$ represents a nontrivial cohomology class, we are done. If not,
we may repeat this process and obtain another representation $\rho_{3}$, where
now $\pi^{3}\mid c_{3}(\sigma)$. This process continues.

If at any stage, we obtain a cocycle $\kappa_{i}$ that represents a non-
trivial class, then we are done. If the process continues forever, then one
checks that by defining $x=x_{1}+\pi x_{2}+\pi^{2}x_{3}+\cdots$, conjugating
the original representation $\rho_{1}$ by $\begin{pmatrix}{1}&{x}\\\
{0}&{1}\end{pmatrix}$ leaves a representation over $K$ with a 0 in the upper
right-hand corner. This contradicts the irreducibility of $\rho_{1}$. ∎

##  3 The Residually Distinguishable Case

In this section, we prove Theorem 1.2. Our complete local ring $\mathbf{T}$ is
no longer assumed to be a DVR, but we grant ourselves the assumption
$\chi_{1}\not\equiv\chi_{2}\pmod{\mathfrak{m}}$. Fix $\tau\in\mathcal{G}$ such
that $\chi_{1}(\tau)\not\equiv\chi_{2}(\tau)\pmod{\mathfrak{m}}$. By Hensel’s
Lemma, the congruence (3) implies that $\rho(\tau)$ has two distinct
eigenvalues $\lambda_{1},\lambda_{2}\in\mathbf{T}$ such that

$\lambda_{i}\equiv\chi_{i}(\tau)\pmod{I}$ (12)

for $i=1,2$. We choose a basis for $\rho$ over
$K=\operatorname{Frac}(\mathbf{T})$ such that

$\rho(\tau)=\begin{pmatrix}{\lambda_{1}}&{0}\\\
{0}&{\lambda_{2}}\end{pmatrix}.$

For $\sigma\in\mathcal{G}$, write

$\rho(\sigma)=\begin{pmatrix}{a(\sigma)}&{b(\sigma)}\\\
{c(\sigma)}&{d(\sigma)}\end{pmatrix}.$

Note that unlike the DVR case, we cannot ensure that $\rho$ takes values in
$\operatorname{GL}_{2}(\mathbf{T})$, only $\operatorname{GL}_{2}(K)$.
Nevertheless, we have the following.

######  Lemma 3.1.

We have $a(\sigma),d(\sigma)\in\mathbf{T}$ for all $\sigma\in\mathcal{G}$, and
furthermore

$\displaystyle a(\sigma)$ $\displaystyle\equiv\chi_{1}(\sigma)\pmod{I},$
$\displaystyle d(\sigma)$ $\displaystyle\equiv\chi_{2}(\sigma)\pmod{I}.$

###### Proof.

The congruence (3) implies that

$a(\sigma)+d(\sigma)\equiv\chi_{1}(\sigma)+\chi_{2}(\sigma)\pmod{I}.$ (13)

Applying this with $\sigma$ replaced by $\sigma\tau$, and noting that
$a(\sigma\tau)=a(\sigma)\lambda_{1}$ and $d(\sigma\tau)=d(\sigma)\lambda_{2}$
by our choice of basis, we obtain

$a(\sigma)\lambda_{1}+d(\sigma)\lambda_{2}\equiv\chi_{1}(\sigma)\chi_{1}(\tau)+\chi_{2}(\sigma)\chi_{2}(\tau).\pmod{I}$
(14)

Solving the congruences (13) and (14) using (12), we find that
$a(\sigma),d(\sigma)\in\mathbf{T}$ and furthermore
$a(\sigma)\equiv\chi_{1}(\sigma)\pmod{I}$ and
$d(\sigma)\equiv\chi_{2}(\sigma)\pmod{I}$ as desired. Note that this uses the
fact that $\chi_{1}(\tau)\not\equiv\chi_{2}(\tau)\pmod{\mathfrak{m}}$ in a
crucial way. ∎

We now let $B$ denote the $\mathbf{T}$-submodule of $K$ spanned by the
$b(\sigma)$ for $\sigma\in\mathcal{G}$. Note that the irreducibility
assumption on $\rho$ implies that $B$ is a fractional ideal of $\mathbf{T}$.
Indeed, if we write $K$ as a product of fields, then the projection of $\rho$
on to each field factor is irreducible, hence the projection of $B$ onto each
field factor is nonzero.

In view of the congruences of Lemma 3.1, the equation

$b(\sigma\sigma^{\prime})=a(\sigma)b(\sigma^{\prime})+b(\sigma)d(\sigma^{\prime})$

yields

$\overline{b}(\sigma\sigma^{\prime})\equiv\chi_{1}(\sigma)\overline{b}(\sigma^{\prime})+\chi_{2}(\sigma^{\prime})\overline{b}(\sigma)\quad\text{
in }B/IB.$

As in Lemma 2.2, multiplying by $\chi_{2}^{-1}(\sigma\sigma^{\prime})$ shows
that

$\kappa(\sigma)=\chi_{2}^{-1}(\sigma)\overline{b}(\sigma)\in B/IB$

defines an element of $Z^{1}(\mathcal{G},B/IB(\chi_{2}^{-1}\chi_{1})).$

It remains to prove that the cohomology class represented by $\kappa$ is
surjective. Therefore let

$\kappa^{\prime}(\sigma)=\kappa(\sigma)+(\chi_{2}^{-1}\chi_{1}(\sigma)-1)x$
(15)

be a cohomologous cocycle, where $x\in B/IB$. Let $B^{\prime}$ denote the
$\mathbf{T}$-submodule of $B/IB$ generated by the values of $\kappa^{\prime}$.
Applying (15) to $\tau$ and recalling that $\kappa(\tau)=0$, we see that

$\kappa^{\prime}(\tau)=(\chi_{2}^{-1}\chi_{1}(\tau)-1)x\in B^{\prime}.$

Since the item in parentheses in a unit in $\mathbf{T}$ by assumption, we find
that $x\in B^{\prime}$. Going back to (15), it follows that $\kappa(\sigma)\in
B^{\prime}$ for all $\sigma\in\mathcal{G}$. But it is clear from the
definitions of $B$ and $\kappa$ that $\kappa$ generates $B/IB$, whence
$B^{\prime}=B/IB$ as desired. This completes the proof of Theorem 1.2.

##  4 The Residually Indistinguishable Case

In the remainder of the paper we describe the proof of Theorem 1.3. Before
explaining the definition of the Fitting ideal that appears in the statement
of the theorem, we remark that the methods of the previous sections are not
directly applicable—in the DVR case, manipulations with the uniformizer $\pi$
were essential, and in the residually distinguishable case, the basis for
$\rho$ afforded by the special element $\tau$ was critical. Indeed, in the
residually indistinguishable case we do not know how to show that
$a(\sigma),d(\sigma)\in\mathbf{T}$ (let alone that the congruences of Lemma
3.1 hold) in any basis. A separate construction and proof is required.

As mentioned in the introduction, we will assume for the rest of the paper
that $\chi_{1}$ and $\chi_{2}$ are trivial; this offers notational
simplifications, but no significant changes to the argument.

###  4.1 Fitting Ideal

Let $R$ be a commutative ring. Let $M$ be a finitely presented $R$-module.
This means that there is a short exact sequence

${R^{n}}$${R^{m}}$${M}$${0.}$$\scriptstyle{f}$

######  Definition 4.1.

The 0th Fitting ideal of $M$ over $R$, which we denote
$\operatorname{Fitt}_{R}(M)$, is the ideal of $R$ generated by all $m\times m$
minors of the matrix representing the linear map $f$. By convention, if $n<m$,
then $\operatorname{Fitt}_{R}(M)=0$.

We leave the proof of the following basic facts about Fitting ideals as an
exercise for the reader (alternatively, she may consult [nickel]).

######  Proposition 4.2.

We have the following.

  * •

The Fitting ideal $\operatorname{Fitt}_{R}(M)$ depends only on $M$ up to
$R$-module isomorphism, and not on the particular presentation taken.

  * •

If $I\subset R$ is a finitely generated ideal, then
$\operatorname{Fitt}_{R}(R/I)=I$.

  * •

The Fitting ideal of $M$ is contained in the annihilator of $M$:
$\operatorname{Fitt}_{R}(M)\subset\operatorname{Ann}_{R}(M)$.

  * •

If $R=\mathbf{Z}$ and $M$ is a finitely generated abelian group, then
$\operatorname{Fitt}_{\mathbf{Z}}(M)=0$ if $M$ is infinite and
$\operatorname{Fitt}_{\mathbf{Z}}(M)=(\\#M)$, the ideal generated by the size
of $M$, if $M$ is finite.

  * •

If $M\twoheadrightarrow M^{\prime}$ is a surjection of finitely presented
$R$-modules, then
$\operatorname{Fitt}_{R}(M^{\prime})\supset\operatorname{Fitt}_{R}(M)$.

  * •

(Base Change) If $S$ is an $R$-algebra and $M$ is a finitely presented
$R$-module, then
$\operatorname{Fitt}_{S}(M\otimes_{R}S)=\operatorname{Fitt}_{R}(M)\cdot S.$

######  Corollary 4.3.

Let $\mathbf{T}$ and $B$ be as in Theorem 1.2. Then
$\operatorname{Fitt}_{\mathbf{T}}(B/IB)\subset I$.

###### Proof.

Since $B$ is a fractional ideal of $\mathbf{T}$, it is a faithful
$\mathbf{T}$-module, i.e. $\operatorname{Ann}_{\mathbf{T}}(B)=0$. It follows
that $\operatorname{Fitt}_{\mathbf{T}}(B)=0$. Therefore
$\operatorname{Fitt}_{\mathbf{T}/I}(B/IB)=0$, whence
$\operatorname{Fitt}_{\mathbf{T}}(B/IB)\subset I$. ∎

Motivated by this corollary, one of the insights of Theorem 1.3 is its
statement—in the residually indistinguishable case, instead of attempting to
necessarily construct a $\mathbf{T}$-module of the form $B/IB$ for a faithful
$\mathbf{T}$-module $B$, one should just attempt to construct some module $M$
such that $\operatorname{Fitt}_{\mathbf{T}}(M)\subset I$.

###  4.2 Construction of $M$

To prove Theorem 1.3, we must construct, under the assumptions of the theorem,
a finitely generated $\mathbf{T}$-module $M$ such that
$\operatorname{Fitt}_{\mathbf{T}}(M)\subset I$ and a surjective cohomology
class $\kappa\in H^{1}(\mathcal{G},M)$. Since we have specialized to
$\chi_{1}=\chi_{2}=1$, the $\mathcal{G}$-action on $M$ is trivial, whence
$H^{1}(\mathcal{G},M)$ is the group of continuous homomorphisms
$\mathcal{G}\rightarrow M$. Since $M$ is abelian, such a homomorphism
necessarily factors through the abelianization $\mathcal{G}^{\text{ab}}$ and
thereby induces a $\mathbf{T}$-module map

$\mathcal{G}^{\text{ab}}\otimes\mathbf{T}\longrightarrow M.$ (16)

The surjectivity of $\kappa$ is simply the statement that the
$\mathbf{T}$-module homomorphism (16) is surjective.

These considerations lead to a very natural construction of the module $M$.
Let $\Delta$ denote the augmentation ideal of $\mathcal{G}$ over $\mathbf{T}$,
i.e. the kernel of the $\mathbf{T}$-algebra homomorphism

$\mathbf{T}[\mathcal{G}]\longrightarrow\mathbf{T},\qquad\sum
a_{g}[g]\mapsto\sum a_{g}.$ (17)

It is then well-known that we have a $\mathbf{T}$-module isomorphism

$\Delta/\Delta^{2}\cong\mathcal{G}^{\text{ab}}\otimes\mathbf{T},\qquad\sum
a_{g}[g]\mapsto\sum g\otimes a_{g}.$

In order to apply the assumptions of Theorem 1.3, we must invoke the
continuous group representation
$\rho\colon\mathcal{G}\rightarrow\operatorname{GL}_{2}(K)$. Note that $\rho$
can be extended to a $\mathbf{T}$-algebra homomorphism (also denoted $\rho$)

$\mathbf{T}[\mathcal{G}]\longrightarrow M_{2}(K).$

We then define

$M=\rho(\Delta)/\rho(\Delta^{2}),$

so there is a canonical surjection

${\mathcal{G}^{\text{ab}}\otimes\mathbf{T}\cong\Delta/\Delta^{2}}$${M.}$$\scriptstyle{\rho}$

As explained above, this homomorphism represents a surjective cohomology class

$\kappa\in H^{1}(\mathcal{G},M).$

It remains to prove that $\operatorname{Fitt}_{\mathbf{T}}(M)\subset I$, and
this will take up the remainder of the paper.

###  4.3 Explication of Fitting Ideal

Since $\mathcal{G}$ is compact and $\rho$ is continuous, the image
$\rho(\mathcal{G})$ is a compact subset of $M_{2}(K)$, and hence
$\mathfrak{m}$-adically bounded. The same is therefore true of
$\rho(\mathbf{T}[\mathcal{G}])$ and $\rho(\Delta)$. It follows that
$\rho(\Delta)$ is a finitely generated $\mathbf{T}$-module, and hence that
$M=\rho(\Delta)/\rho(\Delta^{2})$ is finitely generated as well.

Let $\rho_{1},\dots,\rho_{r}$ denote a set of $\mathbf{T}$-module generators
for $\rho(\Delta)$, where $\rho_{i}=\rho(g_{i}-1)$ for elements
$g_{i}\in\mathcal{G}$. The images of the $\rho_{i}$ in $M$ are
$\mathbf{T}$-module generators, and there are two types of relations for these
generators in $M$:

  * •

We may have relations

$\sum_{i=1}^{r}\epsilon_{i}\rho_{i}=0\text{ in }M_{2}(K)$ (18)

for constants $\epsilon_{i}\in\mathbf{T}$. We call these relations of
$\epsilon$-type.

  * •

For each pair $1\leq i,j\leq r$, we may write

$\rho_{i}\rho_{j}=\sum_{k=1}^{r}\delta_{ijk}\rho_{k}$ (19)

for constants $\delta_{ijk}\in\mathbf{T}$. These expressions vanish in $M$; we
call these relations of $\delta$-type.

The Fitting ideal of $M$ is the ideal generated by all determinants $\det(D)$
where $D$ is an $r\times r$ matrix whose rows have the form
$(\epsilon_{1},\dotsc,\epsilon_{r})$ for relations of $\epsilon$-type or
$(\delta_{ij1},\delta_{ij2},\dots,\delta_{ijr})$ for relations of
$\delta$-type. We need to show that $\det(D)\in I$ for each such matrix $D$.

###  4.4 Traces and Determinants

In order to prove that $\det(D)\in I$, we need to generate expressions
involving the $\rho_{i}$ that are known to lie in $I$. This is given by the
following.

######  Lemma 4.4.

For each $A\in\rho(\Delta)$, we have $\operatorname{tr}(A)\in I$ and
$\det(A)\in I$.

###### Proof.

Let $\chi\colon\mathbf{T}[\mathcal{G}]\longrightarrow\mathbf{T}$ denote the
augmentation map defined in (17), i.e. the trivial character of $\mathcal{G}$
extended to a $\mathbf{T}$-algebra homomorphism on $\mathbf{T}[\mathcal{G}]$.
The congruence (4) implies that

$\operatorname{tr}(\rho(g))\equiv
2\chi(g)\pmod{I},\qquad\det(\rho(g))\equiv\chi(g)\pmod{I}$

for $g\in\mathcal{G}$. A simple argument shows that these congruences extend
to all $g\in\mathbf{T}[\mathcal{G}]$ (see [ow]*Lemma 3.1). By definition,
$\chi(g)=0$ for $g\in\Delta$. The result follows. ∎

###  4.5 An Altered Matrix

Write

$\rho_{i}=\begin{pmatrix}{a_{i}}&{b_{i}}\\\ {c_{i}}&{d_{i}}\end{pmatrix}.$

Given an $r\times r$ matrix $D$ of $\epsilon$-type and $\delta$-type relations
as in §4.3, we define an associated matrix $D^{\prime}$ obtained by altering
the rows of $D$ as follows.

  * •

No change for $\epsilon$-type rows.

  * •

For rows of $\delta$-type, replace $\delta_{ijj}$ by $\delta_{ijj}-a_{i}$,
replace $\delta_{iji}$ by $\delta_{iji}-d_{j}$, and leave the other
$\delta_{ijk}$ unchanged.

######  Lemma 4.5.

We have $\det(D^{\prime})=0.$

###### Proof.

Let

$w=(b_{1},b_{2},\dots,b_{r})^{T}\qquad\text{ (column vector). }$

We claim that $D^{\prime}w=0$. For a row of $\epsilon$-type, the corresponding
component of $D^{\prime}w$ is $\sum_{i=1}^{r}\epsilon_{i}b_{i}$. This is the
upper right coefficient of the matrix $\sum_{i=1}^{r}\epsilon_{i}\rho_{i}$,
and hence vanishes by the definition of the $\epsilon_{i}$ (see (18)).
Similarly for a row of $\delta$-type, the corresponding component of
$D^{\prime}w$ is

$\left(\sum_{k=1}^{r}\delta_{ijk}b_{k}\right)-(a_{i}b_{j}+b_{i}d_{j})=0.$ (20)

The quantity (20) vanishes because each expression in parentheses equals the
upper right coefficient of $\rho_{i}\rho_{j}$; this follows on the left by the
definition (19) of the $\delta_{ijk}$, and on the right by the definition of
matrix multiplication.

The ring $K=\operatorname{Frac}(\mathbf{T})$ is a product of fields. To prove
$\det(D^{\prime})=0$, it suffices to prove this on each field factor of $K$.
Now, in each field factor of $K$, the projection of some $b_{i}$ must be
nonzero; otherwise the projection of $\rho$ to that field factor would be
lower triangular and hence reducible, contrary to assumption. Therefore the
equation $D^{\prime}w=0$ implies $\det(D^{\prime})=0$ as desired. ∎

Our goal is to show that $\det(D)\in I$, and we have shown that
$\det(D^{\prime})=0$. It therefore suffices to show that
$\det(D^{\prime})-\det(D)\in I$. In other words, the alterations used to go
from $D$ to $D^{\prime}$ are small enough to leave the determinant unchanged
modulo $I$. Let us motivate our strategy to prove this with an example.

###  4.6 An Example

Suppose $r=2$. We consider a matrix with only rows of $\delta$-type, namely

$D=\begin{pmatrix}{\delta_{121}}&{\delta_{122}}\\\
{\delta_{211}}&{\delta_{212}}\end{pmatrix},\qquad\text{ whence }\qquad
D^{\prime}=\begin{pmatrix}{\delta_{121}-d_{2}}&{\delta_{122}-a_{1}}\\\
{\delta_{211}-a_{2}}&{\delta_{212}-d_{1}}\end{pmatrix}.$

As shorthand, write $t_{i}=a_{i}+d_{i}$ for the trace of $\rho_{i}$. We write
$d_{12}=c_{1}b_{2}+d_{1}d_{2}$ for the “$d$”-component of $\rho_{1}\rho_{2}$,
and $t_{12}=a_{12}+d_{12}$ for the trace of $\rho_{1}\rho_{2}$. By
multilinearity of the determinant, we have

$\det(D^{\prime})-\det(D)=-\det\begin{pmatrix}{d_{2}}&{a_{1}}\\\
{\delta_{211}}&{\delta_{212}}\end{pmatrix}-\det\begin{pmatrix}{\delta_{121}}&{\delta_{122}}\\\
{a_{2}}&{d_{1}}\end{pmatrix}+\det\begin{pmatrix}{d_{2}}&{a_{1}}\\\
{a_{2}}&{d_{1}}\end{pmatrix}.$ (21)

We evaluate the determinants on the right using the substitution
$t_{i}=a_{i}+d_{i}$ where convenient. Then

$\displaystyle\det\begin{pmatrix}{d_{2}}&{a_{1}}\\\
{\delta_{211}}&{\delta_{212}}\end{pmatrix}$
$\displaystyle=-t_{1}\delta_{211}+(d_{1}\delta_{211}+d_{2}\delta_{212})$
$\displaystyle=-t_{1}\delta_{211}+d_{21}$ (22)
$\displaystyle=-t_{1}\delta_{211}+(c_{2}b_{1}+d_{1}d_{2}).$ (23)

Note that equation (22) uses the definition of the $\delta_{ijk}$. Similarly

$\displaystyle\det\begin{pmatrix}{\delta_{121}}&{\delta_{122}}\\\
{a_{2}}&{d_{1}}\end{pmatrix}$
$\displaystyle=-t_{2}\delta_{122}+(d_{1}\delta_{121}+d_{2}\delta_{122})$
$\displaystyle=-t_{2}\delta_{122}+d_{12}$
$\displaystyle=-t_{2}\delta_{122}+(c_{1}b_{2}+d_{1}d_{2}).$ (24)

Combining (21), (23), and (24), we obtain

$\displaystyle\begin{split}\det(D^{\prime})-\det(D)&=t_{1}\delta_{211}-(c_{2}b_{1}+d_{2}d_{1})+\\\
&\ \ \ \ +t_{2}\delta_{122}-(c_{1}b_{2}+d_{1}d_{2})+\\\ &\ \ \ \
+(d_{1}d_{2}-a_{1}a_{2})\end{split}$
$\displaystyle=t_{1}\delta_{211}+t_{2}\delta_{122}-t_{12}.$ (25)

By Lemma 4.4, the expression (25) lies in $I$ as desired.

##  5 Formal Variables

We do not know how to establish explicit formulae such as (25) to prove that

$\det(D^{\prime})-\det(D)\in I$

in general. Instead, we will prove abstractly the existence of polynomial
identities such as (25) that express the difference $\det(D^{\prime})-\det(D)$
in terms of traces and determinants of elements of $\rho(\Delta)$.

For this, it is convenient to shift our perspective from working with the ring
$\mathbf{T}$ to working with formal polynomial rings. We will define a
polynomial algebra $R$ and a specialization homomorphism $\pi\colon
R\longrightarrow K=\operatorname{Frac}(\mathbf{T})$. We will show that the
identities we need hold in $R/\ker\pi$, so they apply in $\mathbf{T}$ as well
by applying $\pi$. The advantage of working in $R$ is that we can identify the
subring generated by traces and determinants of matrices mapping to
$\rho(\Delta)$ under $\pi$ as the subspace of invariants under a certain group
action, and use cohomological considerations to show that our element of
interest lies in this subspace. We now describe this in greater detail.

###  5.1 The ring $R$

Define

$R_{0}=\mathbf{Z}[\boldsymbol{\epsilon}_{i},\boldsymbol{\delta}_{ijk}],$

the commutative polynomial ring in $r^{2}$ free variables enumerated by the
entries of the matrix $D$. Define

$R=R_{0}[\boldsymbol{a}_{i},\boldsymbol{b}_{i},\boldsymbol{c}_{i},\boldsymbol{d}_{i}]_{i=1}^{r},$

a commutative polynomial ring in $r^{2}+4r$ free variables. Define a ring
homomorphism $R\longrightarrow K$ in the natural way indicated by our variable
names, i.e.

$\pi(\boldsymbol{\epsilon}_{i})=\epsilon_{i},\ \
\pi(\boldsymbol{\delta}_{ijk})=\delta_{ijk},\ \
\pi(\boldsymbol{a}_{i})=a_{i},\dotsc,\ \ \pi(\boldsymbol{d}_{i})=d_{i}.$

Note that $\pi(R_{0})\subset\mathbf{T}$.

Finally, let $\boldsymbol{D},\boldsymbol{D}^{\prime}\in M_{r}(R)$ denote the
natural matrices whose images under $\pi$ are equal $D,D^{\prime}$,
respectively, i.e. they are defined by making each entry bold. Our goal is to
prove that

$\pi(\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D}))=\det(D^{\prime})-\det(D)\in
I.$

###  5.2 Relation Ideal

We now define the polynomial relations that allow us to reduce
$\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})$ to an expression
involving traces and determinants, as in the example of §4.6. Define

$\boldsymbol{\rho_{i}}=\begin{pmatrix}{\boldsymbol{a}_{i}}&{\boldsymbol{b}_{i}}\\\
{\boldsymbol{c}_{i}}&{\boldsymbol{d}_{i}}\end{pmatrix}\in M_{2}(R).$

Let $J\subset R$ be the ideal generated by the following:

  * •

The 4 coefficients of

$\sum_{i=1}^{r}\boldsymbol{\epsilon}_{i}\boldsymbol{\rho}_{i}$ (26)

for each row of $\epsilon$-type in $D$.

  * •

The 4 coefficients of

$\boldsymbol{\rho}_{i}\boldsymbol{\rho}_{j}-\sum_{k=1}^{r}\boldsymbol{\delta}_{ijk}\boldsymbol{\rho}_{k}$
(27)

for each row of $\delta$-type in $D$.

It is clear that $J\subset\ker(\pi)$.

###  5.3 Subring of traces and determinants

Let $A\subset R$ denote the $R_{0}$-subalgebra generated by the traces and
determinants of all matrices in the noncommutative $\mathbf{Z}$-algebra
generated by the matrices $\boldsymbol{\rho}_{i}$. Denote by $\overline{A}$
the image of $A$ in $R/J$. We will show that in order to deduce our desired
result $\pi(\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D}))\in I$, it
suffices to prove that the image of
$\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})$ in $R/J$ lies in
$\overline{A}$. For this, it is important that
$\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})$ lies in the following
ideal of $R$:

$I_{R}=\langle\boldsymbol{a}_{i},\boldsymbol{b}_{i},\boldsymbol{c}_{i},\boldsymbol{d}_{i}\rangle.$

######  Lemma 5.1.

We have $\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})\in I_{R}.$

###### Proof.

This follows immediately from multilinearity of the determinant since every
entry of $\boldsymbol{D}^{\prime}-\boldsymbol{D}$ lies in $I_{R}$. ∎

######  Lemma 5.2.

We have $\pi(A\cap I_{R})\subset I$.

###### Proof.

Let $f\in\mathbf{Z}\langle X_{1},\dotsc,X_{r}\rangle$ be an element of the
free polynomial algebra in $r$ noncommuting variables. If $f$ has no constant
term, then

$\pi(f(\boldsymbol{\rho}_{1},\dotsc,\boldsymbol{\rho}_{r}))=f(\rho_{1},\dotsc,\rho_{r})\in\rho(\Delta),$

and hence the trace and determinant of this element lies in $I$ by Lemma 4.4.
The element $f(\boldsymbol{\rho}_{1},\dotsc,\boldsymbol{\rho}_{r})$ clearly
has coefficients lying in $I_{R}$. From these considerations, to prove the
lemma it suffices to prove that $\pi(R_{0}\cap I_{R})\subset I$. In fact, it
is clear that $R_{0}\cap I_{R}=0$, as

$R/I_{R}\cong R_{0},$

with
$\boldsymbol{a}_{i},\boldsymbol{b}_{i},\boldsymbol{c}_{i},\boldsymbol{d}_{i}\mapsto
0$. This concludes the proof. ∎

We summarize the result of this section: in order to prove the desired result

$\det(D^{\prime})-\det(D)\in I,$ (28)

it suffices to prove that the image of
$\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})$ in $R/J$ lies in the
subring $\overline{A}$, the image of $A$ in $R/J$. Indeed, if this condition
holds, then then

$\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})=a+j$ (29)

for some $a\in A,j\in J$. Since $J\subset I_{R}$, we have $a\in I_{R}$ by
Lemma 5.1 and hence $\pi(a)\in I$ by Lemma 5.2. Since $\pi(j)=0$, the desired
result (28) follows from applying $\pi$ to (29).

##  6 Matrix Invariant Theory and Rational Cohomology

The advantage of passing to the ring of formal variables $R$ (rather than
working directly with $\mathbf{T}$ and $K$) is that we may identify the
subring $A\subset R$ as the subspace invariant under a group action, rather
than needing to write down explicit formulae. To this end, we have the
following important classical result, called the fundamental theorem of matrix
invariant theory.

######  Theorem 6.1.

Endow the ring
$\mathbf{Z}[\boldsymbol{a}_{i},\boldsymbol{b}_{i},\boldsymbol{c}_{i},\boldsymbol{d}_{i}]_{i=1}^{r}$,
with an action of $\operatorname{GL}_{2}(\mathbf{Z})$ defined by simultaneous
conjugation on the matrices
$\boldsymbol{\rho}_{i}=\begin{pmatrix}{\boldsymbol{a}_{i}}&{\boldsymbol{b}_{i}}\\\
{\boldsymbol{c}_{i}}&{\boldsymbol{d}_{i}}\end{pmatrix}$. The subring of
invariant elements is generated over $\mathbf{Z}$ by the traces and
determinants of all matrices in the noncommutative algebra generated by the
$\boldsymbol{\rho}_{i}$.

With notation as in §5.1, we have the following corollary.

######  Corollary 6.2.

Endow the ring $R$ with an action of $\operatorname{GL}_{2}(\mathbf{Z})$
defined by simultaneous conjugation on the matrices
$\rho_{i}=\begin{pmatrix}{\boldsymbol{a}_{i}}&{\boldsymbol{b}_{i}}\\\
{\boldsymbol{c}_{i}}&{\boldsymbol{d}_{i}}\end{pmatrix}$. The subring of
invariant elements is equal to $A$.

###### Proof.

We may write
$R=\mathbf{Z}[\boldsymbol{a}_{i},\boldsymbol{b}_{i},\boldsymbol{c}_{i},\boldsymbol{d}_{i}]_{i=1}^{r}\otimes
R_{0}$, where $\operatorname{GL}_{2}(\mathbf{Z})$ acts trivially on $R_{0}$.
Since $R_{0}$ is $\mathbf{Z}$-flat, the result follows immediately from
Theorem 6.1. ∎

In our computations, it will be convenient to work with the Borel subgroup of
$\operatorname{GL}_{2}$ consisting of lower triangular matrices. We would like
restriction to the Borel to induce an isomorphism on cohomology. This is not
true in general for ordinary group cohomology, so for this reason, we must
work with the cohomology of algebraic groups called rational cohomology. Here
“rational” refers to actions defined by rational functions, rather than the
rational numbers; throughout, we work integrally over $\mathbf{Z}$.

###  6.1 Rational cohomology

For simplicity, we will describe algebraic groups and their cohomology through
their “functor of points” rather than via group schemes. We therefore view the
algebraic group $\mathbf{G}=\operatorname{GL}_{2}$ as the functor that
associates to any commutative ring $S$ the group $\operatorname{GL}_{2}(S)$.

######  Definition 6.3.

A rational $\mathbf{G}$-module is a $\mathbf{Z}$-module $V$ endowed with a
functorial action, for every ring $S$, of the group $\operatorname{GL}_{2}(S)$
on $V\otimes S$, such that the action of a matrix $\begin{pmatrix}{a}&{b}\\\
{c}&{d}\end{pmatrix}$ is given by rational functions in the variables
$a,b,c,d$. We say that $V$ is a rational $\mathbf{G}$-module over a
commutative ring $R_{0}$ if $V$ has a structure of $R_{0}$-module that
commutes with the $\mathbf{G}$-action.

######  Example 6.4.

We denote by
$\mathcal{A}=\mathbf{Z}A\oplus\mathbf{Z}B\oplus\mathbf{Z}C\oplus\mathbf{Z}D$
the $\mathbf{G}$-module given by the adjoint representation, i.e. if
$g=\begin{pmatrix}{a}&{b}\\\ {c}&{d}\end{pmatrix}$, then

$\begin{pmatrix}{g\cdot A}&{g\cdot B}\\\ {g\cdot C}&{g\cdot
D}\end{pmatrix}=\begin{pmatrix}{a}&{b}\\\
{c}&{d}\end{pmatrix}^{-1}\begin{pmatrix}{A}&{B}\\\
{C}&{D}\end{pmatrix}\begin{pmatrix}{a}&{b}\\\ {c}&{d}\end{pmatrix}.$

Let $\mathbf{B}\subset\mathbf{G}$ denote the algebraic subgroup of lower
triangular matrices. Note that for $g=\begin{pmatrix}{x}&{0}\\\
{y}&{z}\end{pmatrix}$ in $\mathbf{B}$, we have

$g\cdot A=A+\frac{y}{x}B,\qquad g\cdot B=\frac{z}{x}B,\qquad g\cdot
D=D-\frac{y}{x}B.$

In particular, $\mathbf{Z}B$ is a $\mathbf{B}$-submodule of $\mathcal{A}$.

######  Definition 6.5 ([jantzen] Ch. 4).

If $V$ is a $\mathbf{G}$-module, define the invariants

$\displaystyle H^{0}(\mathbf{G},V)=V^{\mathbf{G}}=\\{v\in V\colon\text{for all
commutative rings $A$ and $g\in\mathbf{G}(A)$, }g\cdot v=v\\}.$

The rational cohomology groups $H^{i}(\mathbf{G},-)$ are the right derived
functors of $H^{0}(\mathbf{G},-)$.

The same definitions hold with $\mathbf{G}$ replaced by $\mathbf{B}$. The
following is our motivation for using rational cohomology.

######  Theorem 6.6.

Let $V$ be a $\mathbf{G}$-module. The restriction map

$H^{i}(\mathbf{G},V)\rightarrow H^{i}(\mathbf{B},V)$

is an isomorphism for all $i\geq 0$.

###### Proof.

See [jjm]*Theorem 4.4 for the general case. In this paper we only need the
case that $i=0$ and $V$ is a finitely generated $\mathbf{Z}$-module, so we
describe the proof in this case. Injectivity is clear, so we need only prove
surjectivity. Let $\boldsymbol{V}$ denote the spectrum of the tensor algebra
of $V^{*}=\operatorname{Hom}(V,\mathbf{Z})$. This is the affine scheme whose
points over a ring $S$ are $\boldsymbol{V}(S)=V\otimes S$.

Let $x\in H^{0}(\mathbf{B},V)$. If $S$ is any ring, we can define a map
$\mathbf{G}(S)\longrightarrow\boldsymbol{V}(S)=V\otimes S$ by $g\mapsto g\cdot
x$. Since $x\in H^{0}(\mathbf{B},V)$, this map factors through
$G(S)/B(S)={\mathbf{P}}^{1}(S)$. This yields a morphism of schemes
${\mathbf{P}}^{1}\longrightarrow\boldsymbol{V}$. Since $\boldsymbol{V}$ is
affine, this morphism must be constant, i.e. $x\in H^{0}(\mathbf{G},V)$. ∎

######  Corollary 6.7.

We have $H^{0}(\mathbf{B},R)=A$.

###### Proof.

First note that $H^{0}(\mathbf{G},R)=A$. Indeed, the inclusion
$H^{0}(\mathbf{G},R)\supset A$ is clear. Meanwhile the inclusion
$H^{0}(\mathbf{G},R)\subset A$ follows from Corollary 6.2 and the observation
that invariance under $\mathbf{G}$ is stronger than invariance under its group
of points $\mathbf{G}(\mathbf{Z})$. The corollary then follows from Theorem
6.6. ∎

We conclude this section by stating a key input we will need from the
cohomology theory of algebraic groups. See [jjm]*§4.2 for a proof.

######  Theorem 6.8.

Let $\mathcal{A}$ be the adjoint representation defined in Example 6.4. For
any nonnegative integer $k$, the $\mathbf{B}$-module $\mathcal{A}^{\otimes
k}\otimes R$ is acyclic, i.e. $H^{i}(\mathbf{B},\mathcal{A}^{\otimes k}\otimes
R)=0$ for $i\geq 1$.

###  6.2 Roadmap

Let $\boldsymbol{e}=\det(\boldsymbol{D}^{\prime})-\det(\boldsymbol{D})$ and
denote by $\overline{\boldsymbol{e}}$ its image in $R/J$. Our strategy to
prove that $\overline{\boldsymbol{e}}$ lies in $\overline{A}$ is as follows.
We will first prove that the ideal $J$ is stable under the action of
$\mathbf{B}$. We will then show that

$\overline{\boldsymbol{e}}\in H^{0}(\mathbf{B},R/J).$

The long exact sequence in rational cohomology associated to

$0\longrightarrow J\longrightarrow R\longrightarrow R/J\longrightarrow 0$

yields an exact sequence

${H^{0}(\mathbf{B},R)=A}$${H^{0}(\mathbf{B},R/J)}$${H^{1}(\mathbf{B},J).}$$\scriptstyle{c_{J}}$
(30)

The equality on the left is Corollary 6.7. Let $\beta\in H^{1}(\mathbf{B},J)$
denote the image of $\overline{\boldsymbol{e}}$ under the connecting
homomorphism $c_{J}$. In view of (30), the desired result
$\overline{\boldsymbol{e}}\in\overline{A}$ will follow if we can show that
$\beta=0$.

For this, we will define a certain $\mathbf{B}$-stable subideal
$J^{\prime}\subset J$ and show that in fact

$\overline{\boldsymbol{e}}\in H^{0}(\mathbf{B},R/J^{\prime}).$

(This is a slight abuse of notation, as here $\overline{\boldsymbol{e}}$
denotes the reduction of $\boldsymbol{e}$ modulo $J^{\prime}$.)

We let $\alpha=c_{J^{\prime}}(\overline{\boldsymbol{e}})\in
H^{1}(\mathbf{B},J^{\prime})$ defined as above. Then $\beta=\iota_{*}(\alpha)$
where $\iota\colon J^{\prime}\rightarrow J$ is the inclusion and $\iota_{*}$
is the induced map on rational cohomology. To conclude, we will prove that the
map

$\iota_{*}\colon H^{1}(\mathbf{B},J^{\prime})\longrightarrow
H^{1}(\mathbf{B},J)$

vanishes. Therefore $\beta=0$, and our result follows.

###  6.3 Invariance

Let $J^{\prime}\subset J$ denote the subideal generated by the “$b$”
coefficients of the matrices in (26)–(27). To be precise, $J^{\prime}$ is
generated by:

  * •

The elements $\sum_{i=1}^{r}\boldsymbol{\epsilon}_{i}\boldsymbol{b}_{i}$ for
each row of $\epsilon$-type in $D$.

  * •

The elements
$\boldsymbol{a}_{i}\boldsymbol{b}_{j}+\boldsymbol{b}_{i}\boldsymbol{d}_{j}-\sum_{k=1}^{r}\boldsymbol{\delta}_{ijk}\boldsymbol{b}_{k}$
for each row of $\delta$-type in $D$.

######  Lemma 6.9.

The ideals $J^{\prime}$ and $J$ are stable under the action of $\mathbf{B}$.
More precisely, the $\mathbf{Z}$-module spanned by each set of 4 relations in
(26)–(27) is isomorphic as a $\mathbf{B}$-module to a copy of the adjoint
representation $\mathcal{A}$ (see Example 6.4).

###### Proof.

The relations defining $J$ are linear combinations of products of the
$\boldsymbol{\rho}_{i}$, with coefficients in $R_{0}$ (on which $\mathbf{B}$
acts trivially), and the definition of our action is by simultaneous
conjugation on $\boldsymbol{\rho}_{i}$. The second sentence of the Lemma
follows. The stability of $J^{\prime}$ under $\mathbf{B}$ follows since the
module of upper right entries $\mathbf{Z}B\subset\mathcal{A}$ is stable under
$\mathbf{B}$ (see Example 6.4). ∎

The goal of the rest of this subsection is to prove the following.

######  Lemma 6.10.

We have $\overline{\boldsymbol{e}}\in H^{0}(\mathbf{B},R/J^{\prime})$.

###### Proof.

The matrix $\boldsymbol{D}$ has coefficients in $R_{0}$, on which $\mathbf{B}$
acts trivially, so we must show that

$\overline{\det(\boldsymbol{D}^{\prime})}\in H^{0}(\mathbf{B},R/J^{\prime}).$

The group $\mathbf{B}$ is generated by elements of the form
$\sigma_{x,y}=\begin{pmatrix}{x}&{0}\\\ {0}&{y}\end{pmatrix}$ and
$\tau_{x}=\begin{pmatrix}{1}&{0}\\\ {x}&{1}\end{pmatrix}$. The matrix
$\boldsymbol{D}^{\prime}$ has coefficients in
$R_{0}[\boldsymbol{a}_{i},\boldsymbol{d}_{i}]$ on which $\sigma_{x,y}$ acts
trivially. Therefore we must only consider the action of $\tau_{x}$.

The difference
$\det(\tau_{x}(\boldsymbol{D}^{\prime}))-\det(\boldsymbol{D}^{\prime})$ is a
linear combination (with coefficients $\pm 1$) of determinants of all matrices
$M$ obtained by starting with $\boldsymbol{D}^{\prime}$ and replacing some
nonempty subset of the rows $w$ with $\tau_{x}(w)-w$. We will show $\det(M)\in
J^{\prime}$ for each such matrix $M$.

Rows of $\epsilon$-type in $\boldsymbol{D}^{\prime}$ contain only the elements
$\boldsymbol{\epsilon}_{i}\in R_{0}$ that are fixed by the action of
$\mathbf{B}$. So any matrix $M$ that contains a row $\tau_{x}(w)-w=0$ for a
row $w$ of $\epsilon$-type will have determinant 0. Therefore we need only
consider matrices $M$ that contain a row $\tau_{x}(w)-w$ for a row $w$ of
$\delta$-type. Conjugation by $\tau_{x}$ sends
$\boldsymbol{a}_{i}\mapsto\boldsymbol{a}_{i}+\boldsymbol{b}_{i}x$ and
$\boldsymbol{d}_{i}\mapsto\boldsymbol{d}_{i}-\boldsymbol{b}_{i}x$. Say $w$ is
associated to a pair $(i,j)$. Then

$\tau_{x}(w)-w=(0,0,\dots,\boldsymbol{b}_{j}x,0,\dots,-\boldsymbol{b}_{i}x,0,\dots,0),$

where there is $\boldsymbol{b}_{j}x$ in the $i$th slot and
$-\boldsymbol{b}_{i}x$ in the $j$th slot. Note that if $i=j$ then
$\tau_{x}(w)-w=0$ and hence $\det(M)=0$. So we assume $i\neq j$. We make the
following alterations to $M$ which do not change the determinant:

  * •

We replace $\tau_{x}(w)-w$ by

$(0,0,\dots,x,0,\dots,-x,0,\dots,0).$

At the same time, in every row other than $w$ we multiply the $j$th coordinate
by $\boldsymbol{b}_{j}$ and the $i$th coordinate by $\boldsymbol{b}_{i}$.

  * •

We then add the new $j$th column to the new $i$th column.

  * •

For each $1\leq k\leq r,k\neq i,j$, we add $\boldsymbol{b}_{k}$ times the
$k$th column to the new $i$th column.

In the matrix $M$ that results from these changes, the $i$th coordinate is
precisely the generator of $J^{\prime}$ associated to the row. Therefore $M$
has an entire column with elements lying in $J^{\prime}$, so its determinant
lies in $J^{\prime}$ as well. ∎

###  6.4 Kozsul complex

For simplicity, denote the $r$ elements of $J^{\prime}$ corresponding to the
rows of $\boldsymbol{D}^{\prime}$ by $B_{1},\dots,B_{r}$, i.e.

$\boldsymbol{D}^{\prime}\begin{pmatrix}\boldsymbol{b}_{1}\\\ \vdots\\\
\boldsymbol{b}_{r}\end{pmatrix}=\begin{pmatrix}B_{1}\\\ \vdots\\\
B_{r}\end{pmatrix}.$

The rank 1 modules $\mathbf{Z}B_{i}$ are rational $\mathbf{B}$-representations
isomorphic to $\mathbf{Z}B\subset\mathcal{A}$ as in Example 6.4. Let
$V=\oplus_{i=1}^{r}\mathbf{Z}B_{i}$. Consider the following complex of
rational $\mathbf{B}$-representations over $R$:

${0}$${\left(\bigwedge\nolimits^{r}V\right)\otimes
R}$${\left(\bigwedge\nolimits^{r-1}V\right)\otimes
R}$${\cdots\left(\bigwedge\nolimits^{2}V\right)\otimes R}$${V\otimes
R}$${R.}$$\scriptstyle{f_{r}}$$\scriptstyle{f_{r-1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{1}}$
(31)

Here all wedge products and tensor products are over $\mathbf{Z}$. The maps

$f_{i}\colon\left(\bigwedge\nolimits^{i}V\right)\otimes
R\rightarrow\left(\bigwedge\nolimits^{i-1}V\right)\otimes R$

are given by

$B_{k_{1}}\wedge B_{k_{2}}\wedge\cdots\wedge B_{k_{i}}\otimes
r\mapsto\sum_{j=1}^{i}(-1)^{j}B_{k_{1}}\wedge
B_{k_{2}}\wedge\cdots\wedge\hat{B}_{k_{j}}\wedge\cdots B_{k_{i}}\otimes
B_{k_{j}}r.$

Noting that each term $\left(\bigwedge\nolimits^{r}V\right)\otimes R$ in (31)
is isomorphic to $\bigwedge\nolimits^{r}_{R}\left(V\otimes R\right)$, the
sequence (31) is precisely the Kozsul complex for the free $R$-module
$V\otimes R$. It is therefore exact if we can prove that the elements
$B_{1},\dots,B_{r}$ form a regular sequence in $R$.

######  Lemma 6.11.

The elements $B_{1},\dots,B_{r}$ form a regular sequence in $R$.

After enacting the change of variables for rows of $\delta$-type:

$\delta_{ijk}^{\prime}=\begin{cases}\delta_{iji}-d_{j}&k=i\\\
\delta_{ijj}-a_{i}&k=j\\\ \delta_{ijk}&k\neq i,j,\end{cases}$

we see the lemma follows from the following.

######  Lemma 6.12.

Let $S$ be a commutative ring and let $T=S[x_{ij}]_{i,j=1}^{r}$. The elements
$L_{i}=\sum_{j=1}^{r}x_{ij}y_{j}$ form a regular sequence in
$T[y_{1},\dots,y_{r}]$.

We leave the proof of Lemma 6.12 as an exercise, referring the reader to
[jjm]*Proposition 5.13.

###  6.5 Embedding into an acyclic complex

######  Theorem 6.13.

There is a commutative diagram of complexes of rational $\mathbf{B}$-modules

${0}$${\left(\bigwedge\nolimits^{r}V\right)\otimes
R}$${\left(\bigwedge\nolimits^{r-1}V\right)\otimes
R}$${\cdots\left(\bigwedge\nolimits^{2}V\right)\otimes R}$${V\otimes
R}$${J^{\prime}}$${0}$${W_{r}\otimes R}$${W_{r-2}\otimes R}$${\cdots
W_{2}\otimes R}$${W_{1}\otimes
R}$${J,}$$\scriptstyle{f_{r}}$$\scriptstyle{\iota_{r}}$$\scriptstyle{f_{r-1}}$$\scriptstyle{\iota_{r-1}}$$\scriptstyle{f_{2}}$$\scriptstyle{\iota_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{\iota_{1}}$$\scriptstyle{\iota_{0}}$$\scriptstyle{g_{r}}$$\scriptstyle{g_{r-1}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{1}}$
(32)

where the $W_{i}\otimes R$ are acyclic $\mathbf{B}$-modules for $i\geq 1$.

###### Proof.

We have already constructed the complex in the top row (see (31)), noting that
the image of $f_{1}$ is the ideal generated by the $B_{i}$, namely
$J^{\prime}$. Next we define the bottom row. Fix an index $i=1,\dots,r$, and
suppose that the $i$th row of $D$ corresponds to a pair $(m,n)$, i.e. we have
$B_{i}=B_{mn}$ where

$\boldsymbol{\rho}_{m}\boldsymbol{\rho}_{n}-\sum_{k=1}^{r}\boldsymbol{\delta}_{mnk}\boldsymbol{\rho}_{k}=\begin{pmatrix}{A_{mn}}&{B_{mn}}\\\
{C_{mn}}&{D_{mn}}\end{pmatrix}.$

We then let

$\mathcal{A}_{i}=\mathbf{Z}A_{mn}\oplus\mathbf{Z}B_{mn}\oplus\mathbf{Z}C_{mn}\oplus\mathbf{Z}D_{mn}$

denote the corresponding copy of the adjoint. Define

$W_{i}=\bigoplus_{\\{k_{1},k_{2},\dots,k_{i}\\}\subset\\{1,\dotsc,r\\}}\mathcal{A}_{k_{1}}\otimes\cdots\otimes\mathcal{A}_{k_{i}}.$

The vertical maps $\iota_{i}$ are given by, for $k_{1}<k_{2}<\cdots<k_{i}$,

$B_{k_{1}}\wedge\cdots\wedge B_{k_{i}}\otimes r\mapsto
B_{k_{1}}\otimes\cdots\otimes B_{k_{i}}\otimes r.$

The maps $g_{i}$ are given by:

$X_{1}\otimes X_{2}\otimes\cdots\otimes X_{i}\otimes
r\mapsto\sum_{j=1}^{i}(-1)^{j}X_{1}\otimes
X_{2}\otimes\cdots\hat{X}_{j}\cdots\otimes X_{i}\otimes X_{j}r.$

The image of the map $g_{1}$ is precisely our ideal $J$. The fact that the
bottom row of (32) is a complex, as well as the commutativity of the diagram,
is clear. The $\mathbf{B}$-acyclicity of the $W_{i}\otimes R$ follows from
Theorem 6.8. ∎

###  6.6 A cascade of cohomology classes

######  Theorem 6.14.

Let $\iota\colon J^{\prime}\rightarrow J$ be the inclusion and let

$\iota_{*}\colon H^{1}(\mathbf{B},J^{\prime})\longrightarrow
H^{1}(\mathbf{B},J)$

be the induced map on first rational cohomology groups. Then $\iota_{*}=0$.

###### Proof.

Let the notation be as in Theorem 6.13. Recall
$\operatorname{im}(f_{1})=J^{\prime}$. Let

$\alpha_{1}\in
H^{1}(\mathbf{B},J^{\prime})=H^{1}(\mathbf{B},\operatorname{im}(f_{1})).$

We need to show that $\iota_{0,*}\alpha_{1}=0$. The long exact sequence in
cohomology associated to

$0\rightarrow\ker(f_{1})\rightarrow V\otimes
R\rightarrow\operatorname{im}(f_{1})\rightarrow 0$

yields a class $\alpha_{2}\in H^{2}(\mathbf{B},\ker(f_{1}))$ that represents
the obstruction to lifting $\alpha_{1}$ to a class in
$H^{1}(\mathbf{B},V\otimes R)$. Writing
$\ker(f_{1})=\operatorname{im}(f_{2})$, we can view $\alpha_{2}\in
H^{2}(\mathbf{B},\operatorname{im}(f_{2}))$ and repeat the process above,
using the coboundary in the long exact sequence associated to

$0\rightarrow\ker(f_{2})\rightarrow\bigwedge\nolimits^{2}V\otimes
R\rightarrow\operatorname{im}(f_{2})\rightarrow 0$

to obtain $\alpha_{3}\in H^{3}(\mathbf{B},\ker(f_{2}))$. Continuing in this
way we obtain

$\alpha_{i}\in
H^{i}(\mathbf{B},\ker(f_{i-1}))=H^{i}(\mathbf{B},\operatorname{im}(f_{i}))$

for $i=1,\dots,r+1$. Note that $\alpha_{r+1}=0$ since $\ker(f_{r})=0$.

For each $i=1,\dotsc,r+1$, define

$\beta_{i}=\iota_{i-1,*}\alpha_{i}\in
H^{i}(\mathbf{B},\operatorname{im}(g_{i})).$

In particular, we have $\beta_{1}=\iota_{0,*}\alpha_{1}$, which is the class
we are trying to show vanishes. The bottom row of (32) is a complex but we do
not claim it is exact. Nevertheless the obstruction to $\beta_{i}\in
H^{i}(\mathbf{B},\operatorname{im}(g_{i}))$ lifting to a class in
$H^{i}(\mathbf{B},W_{i}\otimes R)$ is precisely the image of $\beta_{i+1}\in
H^{i+1}(\mathbf{B},\operatorname{im}(g_{i+1}))$ in
$H^{i+1}(\mathbf{B},\ker(g_{i}))$. Now $\beta_{r+1}=0$ since $\alpha_{r+1}=0$,
and hence we conclude that $\beta_{r}$ lifts to a class in
$H^{r}(\mathbf{B},W_{r}\otimes R)$. However, by acyclicity we have
$H^{r}(\mathbf{B},W_{r}\otimes R)=0$ and hence $\beta_{r}=0$. Therefore,
$\beta_{r-1}$ lifts to a class in $H^{r-1}(\mathbf{B},W_{r-1}\otimes R)$;
again this cohomology group vanishes so $\beta_{r-1}=0$. This downward
cascading continues and we obtain $\beta_{i}=0$ for all $i$. In particular
$\beta_{1}=0$ as desired. ∎

As explained in §6.2, Theorem 6.14 implies that $\overline{\boldsymbol{e}}\in
H^{0}(\mathbf{B},R/J)$ lies in $\overline{A}\subset R/J$. From the discussion
of §5.3, it follows that $\det(D^{\prime})-\det(D)=-\det(D)\in I$. This
completes the proof of Theorem 1.3.

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