# A representative of $R\Gamma(N,T)$
for higher dimensional twists of $\mathbb{Z}_{p}^{r}(1)$

Alessandro Cobbe

###### Abstract

Let $N/K$ be a Galois extension of $p$-adic number fields and let $V$ be a
$p$-adic representation of the absolute Galois group $G_{K}$ of $K$. The
equivariant local $\varepsilon$-constant conjecture $C_{EP}^{na}(N/K,V)$ is
related to the compatibility of the equivariant Tamagawa number conjecture
with the functional equation of Artin $L$-functions and it can be formulated
as the vanishing of a certain element $R_{N/K}$ in
$K_{0}(\mathbb{Z}_{p}[G],\mathbb{Q}_{p}^{c}[G])$. One of the main technical
difficulties in the computation of $R_{N/K}$ arises from the so-called
cohomological term $C_{N/K}$, which requires the construction of a bounded
complex $C_{N,T}^{\bullet}$ of cohomologically trivial modules which
represents $R\Gamma(N,T)$ for a full $G_{K}$-stable
$\mathbb{Z}_{p}$-sublattice $T$ of $V$. In this paper we generalize the
construction of $C_{N,T}^{\bullet}$ in [BC17, Thm. 2] to the case of a higher
dimensional $T$.

## Introduction

Let $K$ be a $p$-adic number field and let
$\rho^{\mathrm{nr}}:G_{K}\to\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ be an unramified
representation of the Galois group $G_{K}=\mathrm{Gal}(K^{c}/K)$. To such a
representation we can associate the image of the Frobenius element $F_{K}$,
i.e. the matrix $\rho^{\mathrm{nr}}(F_{K})\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$.
We will see that this is actually a one to one correspondence between
unramified $r$-dimensional representations of $G_{K}$ and matrices in
$\mathrm{Gl}_{r}(\mathbb{Z}_{p})$. We will be mainly interested in the case
where $\rho^{\mathrm{nr}}$ is the restriction to $G_{K}$ of some unramified
representation
$\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}:G_{\mathbb{Q}_{p}}\to\mathrm{Gl}_{r}(\mathbb{Z}_{p})$,
which, as above, can be described by a matrix
$u\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$. More concretely, our object of study
will be the $G_{K}$-module $T=\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}})$, i.e.
the twist of $\mathbb{Z}_{p}^{r}$, considered with the trivial action, by the
cyclotomic character $\chi_{\mathrm{cyc}}$ and by $\rho^{\mathrm{nr}}$.

The module $T$ arises in a very natural way as the Tate module of a particular
formal group, as follows. By [Haz78, Sec. 13.3] there is a unique
$r$-dimensional Lubin-Tate formal group $\mathcal{F}=\mathcal{F}_{pu^{-1}}$
attached to the parameter $pu^{-1}$. We will show that the $p$-adic Tate
module $T_{p}\mathcal{F}$ of $\mathcal{F}$ is isomorphic to $T$ over the
completion $\overline{{\mathbb{Q}_{p}^{\mathrm{nr}}}}$ of the maximal
unramified extension of $\mathbb{Q}_{p}$.

A module $T$ as above arises also in a more geometrical setting. If
$A/\mathbb{Q}_{p}$ is an abelian variety of dimension $r$ with good ordinary
reduction, then we will show that the Tate module of the associated formal
group $\hat{A}$ is isomorphic to
$\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}})$ for an
appropriate choice of $\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}$. Here it is worth
to remark that the converse is not true, i.e. not every module
$\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}})$ comes from an
abelian variety.

Let $N$ be a Galois extension of $K$ with Galois group $G$ and let
$R\Gamma(N,T)$ be the complex of the $G_{N}$-invariants of the standard
resolution of $T$. In this paper we construct a bounded complex of
cohomologically trivial modules which represents $R\Gamma(N,T)$, generalizing
[BC17, Thm. 2] to the higher dimensional case $r>1$. More concretely we will
define $\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})$ similarly to [BC17], endow it
with a natural action of $\mathrm{Gal}(K_{\mathrm{nr}}/K)\times G$ and prove
the following theorem.

######  Theorem 1.

The complex

$C^{\bullet}_{N,T}:=\left[\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\right]$

with non-trivial modules only in degree 0 and 1 represents $R\Gamma(N,T)$.

Specializing our results to $r=1$, we will of course recover the corresponding
results of [BC17], and as a by-product we will eliminate the distinction
between the cases $\chi^{\mathrm{nr}}|_{G_{N}}=1$ and
$\chi^{\mathrm{nr}}|_{G_{N}}\neq 1$, which had been studied separately in
[BC17].

The main motivation for studying $R\Gamma(N,T)$ comes from the equivariant
epsilon constant conjecture, which we denote by $C_{EP}^{na}(N/K,V)$, where
$V=\mathbb{Q}_{p}\otimes_{\mathbb{Z}_{p}}T$. For a formulation see for example
Conjecture 3.1.1 in [BC17]; for more details and some remarks on the history
of the conjecture we refer the interested reader to the introduction and
Section 3.1 of [BC17] or to the introduction of [BC16]. The main idea in
[BC17] was to translate the conjecture to the language of $K$-theory in the
spirit of [Bre04] and prove it to be equivalent to the vanishing of an element

$R_{N/K}=C_{N/K}+U_{\mathrm{cris}}+\partial^{1}_{\mathbb{Z}_{p}[G],B_{\mathrm{dR}}[G]}(t)-rU_{N/K}+\partial^{1}_{\mathbb{Z}_{p}[G],B_{\mathrm{dR}}[G]}(\varepsilon_{D}(N/K,V))$

in the relative algebraic $K$-group
$K_{0}(\mathbb{Z}_{p}[G],B_{\mathrm{dR}}[G])$.

We recall that for $r=1$ and representations $\rho^{\mathrm{nr}}$ which are
restrictions of unramified extensions
$\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}:G_{\mathbb{Q}_{p}}\longrightarrow\mathbb{Z}_{p}^{\times}$
Izychev and Venjakob in [IV16] have proven the validity of
$C_{EP}^{na}(N/K,V)$ for tame extensions $N/K$. The main result of [BC17]
shows that $C_{EP}^{na}(N/K,V)$ holds for certain weakly and wildly ramified
finite abelian extensions $N/K$. We will generalize both results in an
upcoming work with Werner Bley:

######  Theorem 2.

Let $N/K$ be a tame extension of $p$-adic number fields and let

$\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}\colon
G_{\mathbb{Q}_{p}}\longrightarrow\mathrm{Gl}_{r}(\mathbb{Z}_{p})$

be an unramified representation of $G_{\mathbb{Q}_{p}}$. Let
$\rho^{\mathrm{nr}}$ denote the restriction of
$\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}$ to $G_{K}$. Then $C_{EP}^{na}(N/K,V)$ is
true for $N/K$ and $V=\mathbb{Q}_{p}^{r}(1)(\rho^{\mathrm{nr}})$, if
$\det(\rho^{\mathrm{nr}}(F_{N})-1)\neq 0$.

In particular, this includes the case in which $T$ is the $p$-adic Tate module
of the formal group of an abelian variety defined over $\mathbb{Q}_{p}$. We
will also prove $C_{EP}^{na}(N/K,V)$ in some special cases of wild
ramification, including for example the following statement.

######  Theorem 3.

Let $p$ be an odd prime, let $A/\mathbb{Q}_{p}$ be an $r$-dimensional abelian
variety with good ordinary reduction and let $K/\mathbb{Q}_{p}$ be the
unramified extension of degree $m$. Let $N/K$ be a weakly and wildly ramified
finite abelian extension with cyclic ramification group. Let $d$ denote the
inertia degree of $N/K$ and assume that $m$ and $dp^{r-1}t(r)$ are relatively
prime, where $t=t(r):=\prod_{i=1}^{r}\left(p^{i}-1\right)$ or $t=1$ according
to whether $r>1$ or $r=1$. Then $C_{EP}^{na}(N/K,V)$ is true for $N/K$ and
$V=\mathbb{Q}_{p}\otimes_{\mathbb{Z}_{p}}T_{p}\hat{A}$, where $T_{p}\hat{A}$
is the Tate module of the formal group of $A$.

The results of the present paper will play a central role in the proof of both
theorems. Actually, in the weakly ramified setting we will prove a slightly
more general statement, but we renounce to formulate it here in order to avoid
some technicalities.

Notation. We will mostly rely on the notation of [BC17]. In particular $L/K$
will always denote a finite Galois extension, $L_{0}$ will be the completion
of the maximal unramified extension $L^{\mathrm{nr}}$ of $L$ and
$\widehat{L_{0}^{\times}}$ the $p$-completion of $L_{0}^{\times}$. We will
denote by $e_{L/K}$ and $d_{L/K}$ the ramification index and the inertia
degree of $L/K$, $\mathcal{O}_{L}$ will be the ring of integers of $L$ and
$U_{L}$ will be its group of units. We also set
$\Lambda_{L}=\prod_{r}\widehat{L_{0}^{\times}}(\rho^{\mathrm{nr}})$,
$\Upsilon_{L}=\prod_{r}\widehat{U_{L_{0}}}(\rho^{\mathrm{nr}})$ and
$\mathcal{Z}=\mathbb{Z}_{p}^{r}(\rho^{\mathrm{nr}})$ and we will use an
additive notation for the (twisted) action of the absolute Galois group
$G_{L}$. The elements fixed by the action of $G_{L}$ will be denoted by
$\Lambda_{L}^{G_{L}}$, $\Upsilon_{L}^{G_{L}}$ and $\mathcal{Z}^{G_{L}}$
respectively.

If $A$ is a ring (and so in particular also if it is a field), then we denote
$A^{r}=\\{(a_{1},\dots,a_{r})\ |\ a_{i}\in A\\}$; if $A$ is a multiplicative
group, then we use the same notation for $A^{r}=\\{a^{r}\ |\ a\in A\\}$.
Context will be strong enough to prevent any confusion. All vectors are
intended to be columns, even if we write them horizontally for typographical
reasons.

Let $\varphi$ be the absolute Frobenius automorphism, let $F_{L}$ be the
Frobenius automorphism of $L$ and let $F=F_{K}$ be the Frobenius of $K$.

##  1 Unramified $p$-adic representations

The aim of this section is to give a characterization of the unramified
representations of $G_{K}$.

######  Lemma 1.1.

For $r,n\geq 1$ one has

$\\#\mathrm{Gl}_{r}(\mathbb{Z}_{p}/p^{n}\mathbb{Z}_{p})=p^{(n-1)r^{2}}\cdot
p^{s}\cdot t$

with

$s=s(r):=(r-1)r/2,\quad t=t(r):=\prod_{i=1}^{r}\left(p^{i}-1\right).$

###### Proof.

For $n=1$

$\\#\mathrm{Gl}_{r}(\mathbb{Z}_{p}/p\mathbb{Z}_{p})=\prod_{j=0}^{r-1}\left(p^{r}-p^{j}\right)=p^{\sum_{j=0}^{r-1}j}\prod_{i=1}^{r}\left(p^{i}-1\right)=p^{s}\cdot
t.$

Each element of $\mathrm{Gl}_{r}(\mathbb{Z}_{p}/p\mathbb{Z}_{p})$ has
$p^{(n-1)r^{2}}$ different lifts to
$\mathrm{Gl}_{r}(\mathbb{Z}_{p}/p^{n}\mathbb{Z}_{p})$. ∎

######  Lemma 1.2.

Each matrix $U\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ can be uniquely written in
the form $U=NV$ with matrices $N,V\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ such
that

$N^{t}=1,\quad V^{p^{s}}\equiv 1\pmod{p}\text{ and }NV=VN.$

Moreover, we can further decompose $N$ into a product of matrices $N_{q}$ of
$q$-power order, one for each $q$ dividing $t$, which commute pairwise and
with $V$.

###### Proof.

For all $n\geq 1$ there exist integers $a_{n},b_{n}$ such that
$a_{n}p^{s+(n-1)r^{2}}+b_{n}t=1$. Note that $a_{n}p^{s+(n-1)r^{2}}\equiv
1\equiv a_{n+1}p^{s+nr^{2}}\pmod{t}$ and hence also $a_{n}\equiv
a_{n+1}p^{r^{2}}\pmod{t}$. Since $U^{p^{s+(n-1)r^{2}}}$ has order dividing $t$
modulo $p^{n}$ we deduce that

$U^{a_{n}p^{s+(n-1)r^{2}}}=(U^{p^{s+(n-1)r^{2}}})^{a_{n}}\equiv(U^{p^{s+(n-1)r^{2}}})^{a_{n+1}p^{r^{2}}}=U^{a_{n+1}p^{s+nr^{2}}}\pmod{p^{n}}.$

Similarly $b_{n}t\equiv b_{n+1}t\pmod{p^{s+(n-1)r^{2}}}$, i.e. $b_{n}\equiv
b_{n+1}\pmod{p^{s+(n-1)r^{2}}}$, and we obtain

$U^{b_{n}t}\equiv U^{b_{n+1}t}\pmod{p^{n}}.$

Therefore we can define

$N:=\lim_{n\to\infty}U^{a_{n}p^{s+(n-1)r^{2}}}\qquad\text{and}\qquad
V:=\lim_{n\to\infty}U^{b_{n}t}.$

It is now clear that $V^{p^{s}}\equiv 1\pmod{p}$ and that all the other
requirements hold modulo $p^{n}$ for any $n\geq 1$, hence they must also hold
in $\mathbb{Z}_{p}$.

The last assertion can be proved similarly, but actually the proof is easier
since the order of $N$ is finite. The pairwise commutativity follows from the
fact that all the $N_{q}$ are some powers of $N$.

It remains to prove uniqueness. Let $N_{1},V_{1}$ and $N_{2},V_{2}$ be as
above and let $a_{n}\in\mathbb{N}$ be such that $ta_{n}\equiv
1\pmod{p^{s+(n-1)r^{2}}}$. Then

$V_{1}\equiv
V_{1}^{ta_{n}}=(N_{1}V_{1})^{ta_{n}}=(N_{2}V_{2})^{ta_{n}}=V_{2}^{ta_{n}}\equiv
V_{2}\pmod{p^{n}}.$

Since this is true for all $n$, we deduce that $V_{1}=V_{2}$ and hence also
that $N_{1}=N_{2}$. ∎

######  Lemma 1.3.

For $q\neq p$ there is a one to one correspondence of
$\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{q},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
with the set of all elements in $\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ of $q$-power
order. This correspondence associates
$\varphi\in\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{q},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
with $\varphi(1)\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$. In particular if $q\nmid
p\cdot t(r)$, then
$\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{q},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))=0$.

###### Proof.

Let
$\varphi\in\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{q},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$.
By continuity, for any $n\geq 1$, the preimage of
$\\{M\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})\mid M\equiv I\pmod{p^{n}}\\}$ must be
a neighborhood of $0$ in $\mathbb{Z}_{q}$. In other words there must exist
$m\geq 1$ such that $\varphi(q^{m})=\varphi(1)^{q^{m}}\equiv I\pmod{p^{n}}$
and we can clearly assume that $q^{m}$ is always the maximal power of $q$
dividing $t$. Hence there exists a fixed $m$ such that for every $n\geq 1$ we
have $\varphi(1)^{q^{m}}\equiv I\pmod{p^{n}}$ and it follows that
$\varphi(1)^{q^{m}}=I$.

Conversely, let $U\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ be of order $q^{m}$,
then we can set $\varphi(x)=U^{x}$ for all $x\in\mathbb{Z}_{q}$. The two maps
are clearly inverse to one another. ∎

######  Lemma 1.4.

There is a one to one correspondence of
$\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{p},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
with the set of all elements in $\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ which
project to elements of $p$-power order in
$\mathrm{Gl}_{r}(\mathbb{Z}_{p}/p\mathbb{Z}_{p})$. This correspondence
associates
$\varphi\in\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{p},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
with $\varphi(1)\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$.

###### Proof.

Let
$\varphi\in\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{p},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$.
As in the proof of the previous lemma, we deduce that $\varphi(1)$ is of
$p$-power order modulo $p$.

Conversely let $U\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ be such that its order
modulo $p$ is some power of $p$. By Lemma 1.1 the order of $U$ modulo $p$ must
divide $p^{s}$ and, moreover, its order modulo $p^{n}$ must divide
$p^{s+(n-1)r^{2}}$. Therefore for $x\in\mathbb{Z}_{p}$, $U^{x}$ is well-
defined modulo $p^{n}$ for all $n$ and hence it is also well-defined as an
element of $\mathrm{Gl}_{r}(\mathbb{Z}_{p})$. This implies that we can set
$\varphi(x)=U^{x}$. ∎

Let $\hat{\mathbb{Z}}$ denote the Prüfer ring. We write

$\hat{\mathbb{Z}}=\prod_{q\nmid pt}\mathbb{Z}_{q}\times\prod_{q\mid
t}\mathbb{Z}_{q}\times\mathbb{Z}_{p}$

and write $1\in\hat{\mathbb{Z}}$ in the form
$1=e^{\prime}+e_{t}+e_{p}=e^{\prime}+\sum_{q|t}e_{q}+e_{p}$ according to the
above decomposition. In other words, the elements $e_{p}$ and $e_{q}$ in
$\prod_{q\nmid pt}\mathbb{Z}_{q}\times\prod_{q\mid
t}\mathbb{Z}_{q}\times\mathbb{Z}_{p}$ correspond to $1\in\mathbb{Z}_{p}$ and
$1\in\mathbb{Z}_{q}$ respectively.

######  Lemma 1.5.

There is a one to one correspondence of
$\mathrm{Hom}_{\mathrm{cont}}(\hat{\mathbb{Z}},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
with the set of all elements in $\mathrm{Gl}_{r}(\mathbb{Z}_{p})$, which
associates
$\varphi\in\mathrm{Hom}_{\mathrm{cont}}(\hat{\mathbb{Z}},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
with $\varphi(1)\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$.

###### Proof.

By Lemma 1.2, each matrix $U\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ can be
decomposed as a product of the form $U=\left(\prod_{q|t}N_{q}\right)\cdot V$
with matrices $N_{q},V\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$, where $N_{q}$ has a
$q$-power-order and the order of $V$ modulo $p$ is a power of $p$. Moreover
all these matrices commute with one another. By Lemma 1.3 and Lemma 1.4 we can
define
$\varphi_{q}\in\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{q},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
by $\varphi_{q}(e_{q})=N_{q}$ and
$\varphi_{p}\in\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{p},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))$
by $\varphi_{p}(e_{p})=V$. Note that

$\mathrm{Hom}_{\mathrm{cont}}(\hat{\mathbb{Z}},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))=\prod_{q|t}\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{q},\mathrm{Gl}_{r}(\mathbb{Z}_{p}))\times\mathrm{Hom}_{\mathrm{cont}}(\mathbb{Z}_{p},\mathrm{Gl}_{r}(\mathbb{Z}_{p})).$

So we can set $\varphi=\prod_{q|t}\varphi_{q}\cdot\varphi_{p}$ and note that
$\varphi(1)=\prod_{q|t}\varphi_{q}(e_{q})\cdot\varphi_{p}(e_{p})=U$, which
proves that the considered correspondence is one to one. ∎

We can reformulate this result as follows.

######  Proposition 1.6.

Let $K/\mathbb{Q}_{p}$ be a $p$-adic number field and let
$U\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$.

a) Then the assignment $\rho(F_{K}):=U$ defines an unramified representation

$\rho=\rho_{U,K}\colon G_{K}\longrightarrow\mathrm{Gl}_{r}(\mathbb{Z}_{p}).$

b) The map $U\mapsto\rho_{U,K}$ is a one-to-one correspondence between
$\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ and $r$-dimensional unramified $p$-adic
representations of $G_{K}$. The inverse is given by $\rho\mapsto\rho(F_{K})$.

######  Corollary 1.7.

Let $K$ be a $p$-adic number field and let
$U\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$. Then there exists a finite unramified
extension $E/K$ of degree dividing $t$ such that

$\rho_{U}(F_{E})^{p^{s(r)}}\equiv 1\pmod{p}.$

###### Proof.

Write $U=NV$ as above and let $m$ denote the order of $N$. Then $m$ divides
$t(r)$ and the unramified extension $E$ of degree $m$ satisfies the assertions
of the corollary since $F_{E}=F_{K}^{m}$ and so
$\rho_{U}(F_{E})^{p^{s(r)}}=(NV)^{mp^{s(r)}}=V^{mp^{s(r)}}\equiv 1\pmod{p}$. ∎

Next we will show that unramified representations appear naturally in the
context of (higher dimensional) Lubin-Tate formal groups.

Let $u\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ and let
$\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}=\rho_{u}\colon
G_{\mathbb{Q}_{p}}\longrightarrow\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ denote the
unramified representation attached to $u$ by Proposition 1.6. By [Haz78, Sec.
13.3] there is a unique $r$-dimensional Lubin-Tate formal group
$\mathcal{F}=\mathcal{F}_{pu^{-1}}$ attached to the parameter $pu^{-1}$. We
want to construct a formal isomorphism with the $r$-dimensional multiplicative
group $\mathbb{G}_{m}^{r}$ over the completion
$\overline{\mathbb{Z}_{p}^{\mathrm{nr}}}$ of the ring of integers in the
maximal unramified extension of $\mathbb{Q}_{p}$.

We first need the following lemma.

######  Lemma 1.8.

Let $K_{m}/\mathbb{Q}_{p}$ be the unramified extension of degree $m$ of
$\mathbb{Q}_{p}$ and let $n\in\mathbb{N}$. Then

$H^{1}(K_{m}/\mathbb{Q}_{p},\mathrm{Gl}_{r}(\mathcal{O}_{K_{m}}/p^{n}\mathcal{O}_{K_{m}}))=1.$

###### Proof.

We prove this by induction on $n$. For $n=1$ the result is proved in [Ser79,
Ch. X, Prop. 3]. For the inductive step we consider the exact sequence

$0\to
M_{r}(p^{n}\mathcal{O}_{K_{m}}/p^{n+1}\mathcal{O}_{K_{m}})\to\mathrm{Gl}_{r}(\mathcal{O}_{K_{m}}/p^{n+1}\mathcal{O}_{K_{m}})\to\mathrm{Gl}_{r}(\mathcal{O}_{K_{m}}/p^{n}\mathcal{O}_{K_{m}})\to
1.$

Note also that $p^{n}\mathcal{O}_{K_{m}}/p^{n+1}\mathcal{O}_{K_{m}}$ is
isomorphic to the additive group of the finite field
$\mathbb{F}_{m}=\mathcal{O}_{K_{m}}/p\mathcal{O}_{K_{m}}$. By the existence of
a normal basis, the $\mathrm{Gal}(K_{m}/\mathbb{Q}_{p})$-module
$M_{r}(\mathbb{F}_{m})$ is induced in the sense of [Ser79, Ch. VII, Sect. 1]
and hence has trivial cohomology. Therefore, using the long exact cohomology
sequence, from the inductive assumption
$H^{1}(K_{m}/\mathbb{Q}_{p},\mathrm{Gl}_{r}(\mathcal{O}_{K_{m}}/p^{n}\mathcal{O}_{K_{m}}))=1$
we deduce that

$H^{1}(K_{m}/\mathbb{Q}_{p},\mathrm{Gl}_{r}(\mathcal{O}_{K_{m}}/p^{n+1}\mathcal{O}_{K_{m}}))=1.$

∎

######  Lemma 1.9.

There exists
$\varepsilon\in\mathrm{Gl}_{r}(\overline{\mathbb{Z}_{p}^{\mathrm{nr}}})$ such
that $\varphi(\varepsilon^{-1})\varepsilon=u^{-1}$. Moreover if $\varepsilon$
and $E$ are two solutions modulo $p^{n}$, then

$\varphi(E^{-1}\varepsilon)\equiv E^{-1}\varepsilon\pmod{p^{n}}.$

###### Proof.

For all $n\in\mathbb{N}$, let $m(n)\in\mathbb{N}$ be the order of
$\overline{u}\in\mathrm{Gl}_{r}(\mathcal{\mathbb{Z}}/p^{n}\mathbb{Z})$ and let
$K_{m(n)}/\mathbb{Q}_{p}$ be the unramified extension of degree $m(n)$ of
$\mathbb{Q}_{p}$. Then

$\prod_{i=0}^{m(n)-1}\varphi^{i}(u^{-1})=u^{-m(n)}\equiv I\pmod{p^{n}}.$

It follows that the map
$a:\mathrm{Gal}(K_{m(n)}/\mathbb{Q}_{p})\to\mathrm{Gl}_{r}(\mathcal{\mathbb{Z}}/p^{n}\mathbb{Z})$
defined by
$a(\varphi^{j})=\prod_{i=0}^{j-1}\varphi^{i}(\bar{u}^{-1})=\bar{u}^{-j}$ is a
cocycle. By Lemma 1.8 it must therefore also be a coboundary, i.e. there
exists a matrix
$\bar{\varepsilon}\in\mathrm{Gl}_{r}(\mathcal{O}_{K_{m(n)}}/p^{n}\mathcal{O}_{K_{m(n)}})$
satisfying
$\overline{u}^{-1}=a(\varphi)=\varphi(\bar{\varepsilon}^{-1})(\bar{\varepsilon}^{-1})^{-1}$.
Therefore for every $n>0$ the set

$V_{n}=\\{\bar{\varepsilon}\in\mathrm{Gl}_{r}(\mathcal{O}_{K_{m(n)}}/p^{n}\mathcal{O}_{K_{m(n)}}):\
\varphi(\bar{\varepsilon}^{-1})\cdot\bar{\varepsilon}=\overline{u}^{-1}\\}$

is non-empty. We want to define a map $\pi_{n+1,n}:V_{n+1}\to V_{n}$. Let us
consider the projection
$\tilde{\pi}_{n+1,n}:\mathrm{Gl}_{r}(\mathcal{O}_{K_{m(n+1)}}/p^{n+1}\mathcal{O}_{K_{m(n+1)}})\to\mathrm{Gl}_{r}(\mathcal{O}_{K_{m(n+1)}}/p^{n}\mathcal{O}_{K_{m(n+1)}})$
and let $\bar{\varepsilon}\in V_{n+1}$. We want to show that
$\tilde{\pi}_{n+1,n}(\bar{\varepsilon})\in V_{n}$, in order to define
$\pi_{n+1,n}(\bar{\varepsilon})=\tilde{\pi}_{n+1,n}(\bar{\varepsilon})$. Since
$V_{n}\neq\emptyset$, we can choose $\bar{E}\in V_{n}$. Then
$\varphi(\bar{\varepsilon}^{-1})\cdot\bar{\varepsilon}\equiv
u^{-1}\equiv\varphi(\bar{E}^{-1})\cdot\bar{E}\pmod{p^{n}}$ and we deduce that

$\varphi(\bar{E}\bar{\varepsilon}^{-1})\equiv\bar{E}\bar{\varepsilon}^{-1}\pmod{p^{n}}.$

It follows that $\bar{E}\bar{\varepsilon}^{-1}\equiv c\pmod{p^{n}}$ for some
$c\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$. Therefore

$\bar{\varepsilon}^{-1}=\bar{E}^{-1}\bar{E}\bar{\varepsilon}^{-1}\equiv\bar{E}^{-1}c\pmod{p^{n}},$

i.e. $c^{-1}\bar{E}\in\mathcal{O}_{K_{m(n)}}$ represents the same element in
$\mathrm{Gl}_{r}(\mathcal{O}_{K_{m(n+1)}}/p^{n}\mathcal{O}_{K_{m(n+1)}})$ as
$\bar{\varepsilon}$, which proves that
$\tilde{\pi}_{n+1,n}(\bar{\varepsilon})\in V_{n}$.

We have shown that the $V_{n}$ form a projective system of non-empty finite
sets and by standard arguments there must exist an element in their projective
limit. This element $\varepsilon$ satisfies the conditions in the statement of
the lemma. ∎

######  Proposition 1.10.

There is an isomorphism $\theta:\mathcal{F}\to\mathbb{G}_{m}^{r}$ defined over
$\overline{\mathbb{Z}}_{p}^{\mathrm{nr}}$ such that

$\theta(X)=\varepsilon^{-1}X+\ldots\quad\text{and}\quad\theta^{\varphi}\circ\theta^{-1}=u^{-1}.$

###### Proof.

Using the matrix
$\varepsilon\in\mathrm{Gl}_{r}(\overline{\mathbb{Z}_{p}^{\mathrm{nr}}})$
defined in Lemma 1.9, this is a higher dimensional version of [Neu92, Korollar
V.2.3]. ∎

######  Proposition 1.11.

Let $T:=\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}})$ and, as usual, we write
$T_{p}\mathcal{F}$ for the $p$-adic Tate module of $\mathcal{F}$. Then the
homomorphism $\theta$ induces an isomorphism of $p$-adic representations,

$T\cong T_{p}\mathcal{F}.$

###### Proof.

Clearly $T\cong T_{p}\mathbb{G}_{m}^{r}(\rho^{\mathrm{nr}})$. Let

$\displaystyle\psi\colon T_{p}\mathcal{F}$ $\displaystyle\longrightarrow$
$\displaystyle T_{p}\mathbb{G}_{m}^{r}(\rho^{\mathrm{nr}}),$ $\displaystyle
z=\\{z_{n}\\}_{n=1}^{\infty}$ $\displaystyle\mapsto$
$\displaystyle\psi(z):=\\{\theta(z_{n})\\}_{n=1}^{\infty}.$

Then, for $\tau\in I_{\mathbb{Q}_{p}}$, the fact that $\theta$ is defined over
$\overline{{\mathbb{Q}_{p}^{\mathrm{nr}}}}$ implies that

$\tau\cdot\psi(z)=\\{\theta(\tau(z_{n}))\\}_{n=1}^{\infty}=\psi(\tau(z)),$

and for $\varphi=F_{\mathbb{Q}_{p}}$ we have

$\varphi\cdot\psi(z)=\\{u\theta^{\varphi}(\varphi(z_{n}))\\}_{n=1}^{\infty}=\\{(\theta\circ\theta^{-\varphi}\circ\theta^{\varphi})(\varphi(z_{n}))\\}_{n=1}^{\infty}=\\{\theta(\varphi(z_{n}))\\}_{n=1}^{\infty}=\psi(\varphi(z)).$

∎

We conclude this section by choosing a more geometric point of view in order
to show that a module $T=\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}})$ can also
arise in a natural way from an abelian variety. More specifically, let
$A/\mathbb{Q}_{p}$ be an abelian variety of dimension $r$ with good ordinary
reduction, then the associated formal group $\mathcal{F}=\hat{A}$ is toroidal
(see [LR78]) in the sense that there exists a unique isomorphism of formal
groups $f\colon\mathcal{F}\longrightarrow\mathbb{G}_{m}^{r}$ defined over the
maximal unramified extension ${\mathbb{Q}_{p}^{\mathrm{nr}}}/\mathbb{Q}_{p}$.
We let $u\in\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ denote the twist matrix defined
by

$\textstyle{\mathbb{G}_{m}^{r}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{-1}}$$\scriptstyle{u}$$\textstyle{\hat{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{F_{\mathbb{Q}_{p}}}}$$\textstyle{\mathbb{G}_{m}^{r},}$

######  Proposition 1.12.

Let $\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}\colon
G_{\mathbb{Q}_{p}}\longrightarrow\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ denote the
unramified $p$-adic representation defined by
$\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}(F_{\mathbb{Q}_{p}}):=u^{-1}$. Then we
have:

  1. (a)

The $p$-adic representations $T_{p}\hat{A}$ and
$\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}})$ are isomorphic
(as representations of $G_{\mathbb{Q}_{p}}$).

  2. (b)

Set $U:=u^{d_{K}}$ and
$\rho^{\mathrm{nr}}:=\rho^{\mathrm{nr}}_{\mathbb{Q}_{p}}|_{G_{K}}$. Then
$\rho^{\mathrm{nr}}=\rho_{U^{-1},K}$ and the $p$-adic representations
$T_{p}\hat{A}$ and $\mathbb{Z}_{p}^{r}(1)(\rho^{\mathrm{nr}})$ are isomorphic
(as representations of $G_{K}$).

###### Proof.

As in the proof of Proposition 1.11. ∎

##  2 Some preliminary results

In this section we formulate and prove some technicals lemmas, which will be
needed in the proof of Theorem 1. We start with a generalized version of
[BC17, Lemma 4.1.1].

######  Lemma 2.1.

Let $L/K$ denote a finite Galois extension. Then there is an isomorphism

$f_{\mathcal{F},L}\colon\mathcal{F}(\mathfrak{p}_{L})\times\mathcal{Z}^{G_{L}}\longrightarrow\Lambda_{L}^{G_{L}}.$

Explicitly $f_{\mathcal{F},L}$ is given by a power series of the form
$1+\varepsilon^{-1}X+(\deg\geq 2)$ on $\mathcal{F}(\mathfrak{p}_{L})$, where
$\varepsilon\in\mathrm{Gl}_{r}(\overline{\mathbb{Z}_{p}^{\mathrm{nr}}})$ is as
above (here $1$ is a vector consisting only of ones).

###### Proof.

We have $L_{0}^{\times}\cong\pi_{L}^{\mathbb{Z}}\times\kappa^{\times}\times
U_{L_{0}}^{(1)}$, where $\kappa$ denotes the residue class field of $L_{0}$.
Since any element of $\kappa^{\times}$ has coprime to $p$ order and
$U_{L_{0}}^{(1)}$ is $p$-complete we obtain
$\widehat{L_{0}^{\times}}\cong\mathbb{Z}_{p}\times U_{L_{0}}^{(1)}$. By [LR78,
Lemma page 237] and recalling the obvious isomorphism
$U_{L_{0}}^{(1)}\cong\mathbb{G}_{m}(\mathfrak{p}_{L_{0}})$, the formal
isomorphism $\theta:\mathcal{F}\to\mathbb{G}_{m}^{r}$ induces an isomorphism

$\tilde{f}\colon\mathcal{F}(\mathfrak{p}_{L})\longrightarrow\left(\prod_{r}U_{L_{0}}^{(1)}(\rho^{\mathrm{nr}})\right)^{G_{L}}.$

By the construction of $\theta$, this isomorphism is given by a power series
$1+\varepsilon^{-1}X+(\deg\geq 2)$. To conclude, it is enough to twist and
take $G_{L}$-fixed points in
$\prod_{r}\widehat{L_{0}^{\times}}\cong\prod_{r}\mathbb{Z}_{p}\times\prod_{r}U_{L_{0}}^{(1)}$.
∎

We set $\mathcal{G}:=\mathrm{Gal}(K_{0}/K)\times\mathrm{Gal}(L/K)$. In analogy
to [Chi85, (6.2)] and [BC17] we set

$\mathcal{I}_{L/K}(\rho^{\mathrm{nr}}):=\mathrm{Ind}_{\mathrm{Gal}(L_{0}/K)}^{\mathcal{G}}\Lambda_{L}$

and we usually identify $\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$ with
$\prod_{d_{L/K}}\prod_{r}\widehat{L_{0}^{\times}}$. Note that the
$\mathcal{G}$-module structure on $\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$ is
characterized by

$(F\times
1)\cdot[x_{1},x_{2},\dots,x_{d_{L/K}}]=[F_{L}x_{d_{L/K}},x_{1},x_{2},\dots,x_{d_{L/K}-1}],$
(1)

and

$(F^{-n}\times\sigma)\cdot[x_{1},x_{2},\dots,x_{d_{L/K}}]=[\tilde{\sigma}x_{1},\tilde{\sigma}x_{2},\dots,\tilde{\sigma}x_{d_{L/K}}],$
(2)

where $x_{i}\in\Lambda_{L}$, the elements $F^{-n}$ and
$\sigma\in\mathrm{Gal}(L/K)$ have the same restriction to $L\cap K_{0}$ and
$\tilde{\sigma}\in\mathrm{Gal}(L_{0}/K)$ is uniquely defined by
$\tilde{\sigma}|_{K_{0}}=F^{-n}$ and $\tilde{\sigma}|_{L}=\sigma$.

We also have a $G_{K}$\- and a $\mathrm{Gal}(L/K)$-action on
$\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$, via the natural maps
$G_{K}\to\mathrm{Gal}(L_{0}/K)\hookrightarrow\mathcal{G}$ and
$\mathrm{Gal}(L/K)\hookrightarrow\mathcal{G}$. The following lemma clarifies
the structure of $\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$ as a
$\mathrm{Gal}(L/K)$-module.

######  Lemma 2.2.

If we identify the inertia group $I_{L/K}$ with $\mathrm{Gal}(L_{0}/K_{0})$,
then there is a $\mathrm{Gal}(L/K)$-isomorphism

$\tilde{f}_{\mathcal{F},L}\colon\mathrm{Ind}_{I_{L/K}}^{\mathrm{Gal}(L/K)}\left(\prod_{r}\widehat{L_{0}^{\times}}\right)\longrightarrow\mathcal{I}_{L/K}(\rho^{\mathrm{nr}}).$

In particular, $\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$ is
$\mathrm{Gal}(L/K)$-cohomologically trivial.

###### Proof.

The proof is the same as in [BC17, Lemma 4.1.2], up to some obvious
modifications. ∎

The following lemma will be used to prove Lemma 2.4, which will be one of the
key ingredients in the proof of our main theorem.

######  Lemma 2.3.

Let $K$ be an algebraically closed field of characteristic $p$, let $q$ be a
power of $p$, let $A$, $B\in K^{n,n}$ for some $n\in\mathbb{N}$ and let
$f:K\to K^{n}$ be a function whose components are polynomials involving only
terms of degree larger than $q$ and constants. Assuming that $B$ is
invertible, the system of equations

$f(x_{1})+Ax^{q}+Bx=0$ (3)

has a solution $x=(x_{1},\dots,x_{n})\in K^{n}$, where the $q$-th power
$x^{q}$ is taken componentwise.

###### Proof.

We prove this lemma by induction on $n$. For $n=1$, (3) consists of a single
non-constant polynomial equation, since there must be a non-trivial term of
degree $1$ ($B$ is assumed to be invertible). Since $K$ is algebraically
closed, it follows that (3) has a solution.

So let us assume the result holds for $n-1\geq 1$ and let us prove it for $n$.
Up to some elementary row operations we can assume that the first row of $A$
has at most one nonzero entry $a_{1,1}$ (note that we allow also $a_{1,1}=0$).
Then the first equation of our system takes the form

$f_{1}(x_{1})+a_{1,1}x_{1}^{q}+\sum_{j=1}^{n}b_{1,j}x_{j}=0,$

where some of the $b_{1,j}$ are non-zero. Here we need to distinguish two
cases.

Case 1: $b_{1,j}=0$ for all $j>1$ and $b_{1,1}\neq 0$. This means that $B$
takes the form

$B=\begin{pmatrix}b_{1,1}&0\\\ \tilde{b}&\tilde{B}\end{pmatrix},$

where $\tilde{B}\in K^{n-1,n-1}$ is invertible and $\tilde{b}\in K^{n-1,1}$.
In particular, the first equation is a non-trivial polynomial in the only
variable $x_{1}$, which has a solution since $K$ is algebraically closed.
Substituting this value for $x_{1}$ in all the other equations we obtain a new
system of the shape

$c+\tilde{A}\tilde{x}^{q}+\tilde{B}\tilde{x}=0,$

where $\tilde{x}=(x_{2},\dots,x_{n})$ and $c$ is a constant vector in
$K^{n-1}$, which we can interpret as a constant function $K\to K^{n-1}$. Note
also that $\tilde{B}$ is invertible as observed above. Therefore we can apply
the inductive hypothesis to find a solution $\tilde{x}$; together with the
already computed first component $x_{1}$, we obtain a solution of (3).

Case 2: $b_{1,j}\neq 0$ for some $j>1$. Up to a permutation of the variables
and normalizing the last coefficient, we can assume $b_{1,n}=1$. Up to some
further elementary row operations (which do not modify the first row) we may
assume that $b_{i,n}=0$ for $i>1$. So the matrix $B$ has the shape

$B=\begin{pmatrix}\tilde{b}&1\\\ \tilde{B}&0\end{pmatrix},$

where $\tilde{B}\in K^{n-1,n-1}$ is clearly invertible and $\tilde{b}\in
K^{1,n-1}$. Hence we can rewrite the first equation as

$x_{n}=-f_{1}(x_{1})-a_{1,1}x_{1}^{q}-\sum_{j=1}^{n-1}b_{1,j}x_{j}.$ (4)

Substituting this into the other equations of (3) we obtain a new system of
the type

$\tilde{f}(x_{1})+\tilde{A}\tilde{x}^{q}+\tilde{B}\tilde{x}=0,$ (5)

where $\tilde{x}=(x_{1},\dots,x_{n-1})$ and $\tilde{B}$ is as above. Note that
(5) has the same shape as (3), since, as already mentioned, $\tilde{B}$ is
invertible. Note also that here we used that $q$ is a multiple of the
characteristic of the field, so that, when computing
$x_{n}^{q}=(-f_{1}(x_{1})-a_{1,1}x_{1}^{q}-\sum_{j=1}^{n-1}b_{1,j}x_{j})^{q}$,
all the mixed terms vanish. Therefore we can use the inductive hypothesis to
find $x_{1},\dots,x_{n-1}$ and then take $x_{n}$ as in (4). This will be a
solution of (3). ∎

Now we are in a position to prove the following result which generalizes a
well-known lemma, see e.g. [Neu92, Lemma V.2.1].

######  Lemma 2.4.

Given an element $c\in\Upsilon_{L}$, there exists $x\in\Upsilon_{L}$ such that
$(F_{L}-1)\cdot x=c$. If $\tilde{x}$ is such that
$(F_{L}-1)\cdot\tilde{x}\equiv c\pmod{\mathfrak{p}_{L_{0}}^{i}}$ for some $i$,
then we can assume $x\equiv\tilde{x}\pmod{\mathfrak{p}_{L_{0}}^{i}}$. In
particular, if $c\in\prod_{r}U_{L_{0}}^{(1)}(\rho^{\mathrm{nr}})$, then we can
assume $x\in\prod_{r}U_{L_{0}}^{(1)}(\rho^{\mathrm{nr}})$.

###### Proof.

Let $I$ be the $r\times r$ identity matrix and let $q\in\mathbb{N}$ be such
that $x^{F_{L}}\equiv x^{q}\pmod{\mathfrak{p}_{L_{0}}}$ for all
$x\in\mathcal{O}_{L_{0}}$. The determinant of the matrix
$q\rho^{\mathrm{nr}}(F_{L})-I$ is obviously a unit in $\mathbb{Z}_{p}$.
Therefore there exists an inverse matrix $\mathcal{M}$ of
$q\rho^{\mathrm{nr}}(F_{L})-I$ with coefficients in $\mathbb{Z}_{p}$. Then we
take $x_{1}=\mathcal{M}\cdot c$ and we have $(F_{L}-1)\cdot x_{1}\equiv
c\pmod{\mathfrak{p}_{L_{0}}}$.

Let us assume we have $x_{i}$ such that $(F_{L}-1)\cdot x_{i}\equiv
c\pmod{\mathfrak{p}_{L_{0}}^{i}}$, i.e. such that there exists
$a_{i}\in\prod_{r}U_{L_{0}}^{(i)}(\rho^{\mathrm{nr}})$ with $((F_{L}-1)\cdot
x_{i})\cdot a_{i}=c$, where the product of $((F_{L}-1)\cdot x_{i})$ and
$a_{i}$ is done componentwise. We take $b_{i}\in\prod_{r}\mathcal{O}_{L_{0}}$
so that $a_{i}=1+\pi_{L}^{i}b_{i}$, where $\pi_{L}$ is an uniformizing element
of $L$ (and hence also of $L_{0}$) and $1$ is the vector consisting only of
ones. Since the residue field of $L_{0}$ is algebraically closed, we can apply
Lemma 2.3 and obtain a $y_{i}\in\prod_{r}\mathcal{O}_{L_{0}}$ such that

$\rho^{\mathrm{nr}}(F_{L})y_{i}^{q}-y_{i}-b_{i}\equiv
0\pmod{\mathfrak{p}_{L_{0}}}.$

We set $x_{i+1}=x_{i}(1+y_{i}\pi_{L}^{i})$ and we calculate

$\begin{split}(F_{L}-1)\cdot x_{i+1}&=((F_{L}-1)\cdot
x_{i})\cdot((F_{L}-1)\cdot(1+y_{i}\pi_{L}^{i}))\\\ &\equiv
ca_{i}^{-1}(1+\rho^{\mathrm{nr}}(F_{L})y_{i}^{q}\pi_{L}^{i})(1-y_{i}\pi_{L}^{i})\pmod{\mathfrak{p}_{L_{0}}^{i+1}}\\\
&\equiv
ca_{i}^{-1}(1+\rho^{\mathrm{nr}}(F_{L})y_{i}^{q}\pi_{L}^{i}-y_{i}\pi_{L}^{i})\pmod{\mathfrak{p}_{L_{0}}^{i+1}}\\\
&\equiv ca_{i}^{-1}(1+b_{i}\pi_{L}^{i})\pmod{\mathfrak{p}_{L_{0}}^{i+1}}\\\
&\equiv c\pmod{\mathfrak{p}_{L_{0}}^{i+1}}.\end{split}$

Therefore we have an inductive construction for the $x_{i}$, which can be
started with $x_{1}=1$ in the case that
$c\in\prod_{r}U_{L_{0}}^{(1)}(\rho^{\mathrm{nr}})$. Then the conclusions of
the lemma hold for $x=\lim_{n\to\infty}x_{n}$. ∎

To conclude this section we prove some technical results which will be needed
to prove our main theorem.

######  Lemma 2.5.

Let $R$ be a discrete valuation ring with uniformizing element $\pi$, let $K$
be its field of quotients and let $f\in\mathrm{End}_{R}(R^{r})$. Then there
exists $\omega\in\mathbb{N}$ such that for all $n\geq\omega$,

$K\cdot\mathrm{im}(f)\cap\pi^{n}R^{r}\subseteq\pi^{n-\omega}\mathrm{im}(f).$

###### Proof.

We set $\mathcal{U}=\mathrm{im}(f)$ and $\tilde{\mathcal{U}}=R^{r}\cap
K\cdot\mathcal{U}$. Since $R^{r}$ is a noetherian ring, the ascending chain of
ideals $R^{r}\cap\frac{1}{\pi^{n}}\mathcal{U}$ for $n>0$ stabilizes. Therefore
there exists $s\geq 0$ such that
$\tilde{\mathcal{U}}=R^{r}\cap\frac{1}{\pi^{s}}\mathcal{U}$ and hence we have

$\pi^{s}\tilde{\mathcal{U}}\subseteq\mathcal{U}.$

By the Theorem of Artin-Rees there exists $k\in\mathbb{N}$ such that for all
$n\geq k$,

$\tilde{\mathcal{U}}\cap\pi^{n}R^{r}\subseteq\pi^{n-k}(\tilde{\mathcal{U}}\cap(\pi^{k}R^{r})).$

Therefore setting $\omega=s+k$ we deduce that for all $n\geq\omega$,

$K\cdot\mathcal{U}\cap\pi^{n}R^{r}\subseteq\tilde{\mathcal{U}}\cap\pi^{n}R^{r}\subseteq\pi^{n-k}(\tilde{\mathcal{U}}\cap(\pi^{k}R^{r}))\subseteq\pi^{n-\omega}\pi^{s}\tilde{\mathcal{U}}\subseteq\pi^{n-\omega}\mathcal{U}.$

∎

######  Lemma 2.6.

Keeping all the notations of the previous lemma, for all $m,n\in\mathbb{N}$
such that $m>n+\omega$ we have

$((\mathrm{im}(f)\cap\pi^{n}R^{r})+\pi^{m}R^{r})/(\pi^{n}\mathrm{im}(f)+\pi^{m}R^{r})\cong(\mathrm{im}(f)\cap\pi^{n}R^{r})/\pi^{n}\mathrm{im}(f).$

###### Proof.

The natural map

$\mathrm{im}(f)\cap\pi^{n}R^{r}\to((\mathrm{im}(f)\cap\pi^{n}R^{r})+\pi^{m}R^{r})/(\pi^{n}\mathrm{im}(f)+\pi^{m}R^{r})$
(6)

is obviously surjective. Its kernel is

$\mathrm{im}(f)\cap\pi^{n}R^{r}\cap(\pi^{n}\mathrm{im}(f)+\pi^{m}R^{r}).$

We want to show that this is equal to $\pi^{n}\mathrm{im}(f)$. Note that

$\pi^{n}\mathrm{im}(f)+\pi^{m}R^{r}\subseteq\pi^{n}R^{r}$

and, recalling Lemma 2.5, we get

$\begin{split}\mathrm{im}(f)\cap\pi^{n}R^{r}\cap(\pi^{n}\mathrm{im}(f)+\pi^{m}R^{r})&=\mathrm{im}(f)\cap(\pi^{n}\mathrm{im}(f)+\pi^{m}R^{r})\\\
&=\pi^{n}\mathrm{im}(f)+(\mathrm{im}(f)\cap\pi^{m}R^{r})\\\
&=\pi^{n}\left(\mathrm{im}(f)+\left(\frac{1}{\pi^{n}}\mathrm{im}(f)\cap\pi^{m-n}R^{r}\right)\right)\\\
&=\pi^{n}\mathrm{im}(f).\end{split}$

∎

From now on we set $\omega_{L}$ to be the natural number obtained by setting
$R=\mathbb{Z}_{p}$ and considering the endomorphism $f$ of
$\mathbb{Z}_{p}^{r}$ induced by the matrix
$\mathcal{M}_{L}=\rho^{\mathrm{nr}}(F_{L})-I$ in Lemma 2.5.

######  Lemma 2.7.

If $m,n\in\mathbb{N}$ are such that $m>n+\omega_{L}$, then

$\begin{split}&\left(\left((F_{L}-1)\cdot\Lambda_{L}\cap
p^{n}\Lambda_{L}\right)+p^{m}\Lambda_{L}\right)/\left(p^{n}(F_{L}-1)\cdot\Lambda_{L}+p^{m}\Lambda_{L}\right)\\\
&\qquad\cong\left((F_{L}-1)\cdot\Lambda_{L}\cap
p^{n}\Lambda_{L}\right)/\left(p^{n}(F_{L}-1)\cdot\Lambda_{L}\right).\end{split}$

###### Proof.

Let $e:\mathcal{Z}\to\Lambda_{L}$ be defined by
$e(a_{1},\dots,a_{r})=(\pi_{L}^{a_{1}},\dots,\pi_{L}^{a_{r}})$. Then

$\Lambda_{L}=\Upsilon_{L}\times e(\mathcal{Z})$

and all the computations may be split into those two components. Let us
consider $\Upsilon_{L}$ first. By Lemma 2.4,
$(F_{L}-1)\cdot\Upsilon_{L}=\Upsilon_{L}$ and it is straightforward to check
that

$\begin{split}&\left(\left((F_{L}-1)\cdot\Upsilon_{L}\cap
p^{n}\Upsilon_{L}\right)+p^{m}\Upsilon_{L}\right)/\left(p^{n}(F_{L}-1)\cdot\Upsilon_{L}+p^{m}\Upsilon_{L}\right)\\\
&\qquad\cong\left((F_{L}-1)\cdot\Upsilon_{L}\cap
p^{n}\Upsilon_{L}\right)/\left(p^{n}(F_{L}-1)\cdot\Upsilon_{L}\right),\end{split}$

since both sides are trivial. So it remains the $e(\mathcal{Z})$-component and
what has to be proved is equivalent to the statement of Lemma 2.6 for
$\mathcal{M}_{L}=\rho^{\mathrm{nr}}(F_{L})-I$. ∎

##  3 The proof of Theorem 1

The goal of this section is to prove our main theorem, generalizing the
approach of [BC17, Sect. 4.2 and 4.3] to the case in which $T$ is an
unramified twist of $\mathbb{Z}_{p}^{r}(1)$, for any $r\geq 1$.

Let $W_{n}$ and $V_{n}$ be respectively the kernel and the image of the map

$F_{L}-1:\Lambda_{L}/p^{n}\Lambda_{L}\to\Lambda_{L}/p^{n}\Lambda_{L}.$

######  Lemma 3.1.

For $m>n$, we have a commutative diagram of exact sequences:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{W_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau_{m,n}}$$\textstyle{\Lambda_{L}/p^{m}\Lambda_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F_{L}-1}$$\scriptstyle{\pi_{m,n}}$$\textstyle{V_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{m,n}}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{W_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{L}/p^{n}\Lambda_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F_{L}-1}$$\textstyle{V_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

Here $\pi_{m,n}$ denotes the canonical projection and $\tau_{m,n}$ and
$\sigma_{m,n}$ are induced by $\pi_{m,n}$. Moreover, $\pi_{m,n}$ and
$\sigma_{m,n}$ are surjective and the projective system
$\left(W_{n}\right)_{n}$ satisfies the Mittag-Leffler condition.

###### Proof.

The exactness of the rows and the commutativity of the diagram as well as the
surjectivity of $\pi_{m,n}$ and $\sigma_{m,n}$ are evident. It remains to show
that the projective system $(W_{n})_{n}$ satisfies the Mittag-Leffler
condition. To that end we prove that $\mathrm{coker}(\tau_{m,n})$ stabilizes
for any $n\in\mathbb{N}$, as $m$ tends to infinity.

By the Snake Lemma we obtain

$p^{n}\Lambda_{L}/p^{m}\Lambda_{L}\xrightarrow{F_{L}-1}\ker(\sigma_{m,n})\longrightarrow\mathrm{coker}(\tau_{m,n})\longrightarrow
0$

i.e.

$\mathrm{coker}(\tau_{m,n})\cong\mathrm{coker}\left(p^{n}\Lambda_{L}/p^{m}\Lambda_{L}\xrightarrow{F_{L}-1}\ker(\sigma_{m,n})\right).$

Note that

$V_{n}=\left((F_{L}-1)\cdot\Lambda_{L}+p^{n}\Lambda_{L}\right)/p^{n}\Lambda_{L}$

and

$\begin{split}\ker\sigma_{m,n}&=\left(\left((F_{L}-1)\cdot\Lambda_{L}+p^{m}\Lambda_{L}\right)\cap
p^{n}\Lambda_{L}\right)/p^{m}\Lambda_{L}\\\
&=\left(\left((F_{L}-1)\cdot\Lambda_{L}\cap
p^{n}\Lambda_{L}\right)+p^{m}\Lambda_{L}\right)/p^{m}\Lambda_{L}.\end{split}$

Therefore

$\begin{split}&\mathrm{coker}(\tau_{m,n})\cong\left(\left((F_{L}-1)\cdot\Lambda_{L}\cap
p^{n}\Lambda_{L}\right)+p^{m}\Lambda_{L}\right)/\left(p^{n}(F_{L}-1)\cdot\Lambda_{L}+p^{m}\Lambda_{L}\right).\end{split}$

It is an immediate consequence of Lemma 2.7 that $\mathrm{coker}(\tau_{m,n})$
is independent of $m$, provided $m>n+\omega_{L}$. ∎

######  Lemma 3.2.

We have a commutative diagram of short exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{m,n}}$$\textstyle{\Lambda_{L}/p^{m}\Lambda_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu_{L,m}}$$\scriptstyle{\pi_{m,n}}$$\textstyle{\mathcal{Z}/((F_{L}-1)\cdot\mathcal{Z}+p^{m}\mathcal{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{L}/p^{n}\Lambda_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu_{L,n}}$$\textstyle{\mathcal{Z}/((F_{L}-1)\cdot\mathcal{Z}+p^{n}\mathcal{Z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

The vertical maps are all surjective and so, in particular, they satisfy the
Mittag-Leffler condition.

###### Proof.

The commutativity of the diagram, the injectivity of the left inclusions and
the surjectivity of the maps on the right and of the vertical ones are
obvious.

Let us consider the homomorphism $e:\mathcal{Z}\to\Lambda_{L}$ defined by
$e(a_{1},\dots,a_{r})=(\pi_{L}^{a_{1}},\dots,\pi_{L}^{a_{r}})$. Then,
recalling Lemma 2.4 and using an additive notation as usual,

$\begin{split}\ker\nu_{L,n}&=\Upsilon_{L}/p^{n}\Upsilon_{L}\times
e((F_{L}-1)\cdot\mathcal{Z}+p^{n}\mathcal{Z})/e(p^{n}\mathcal{Z})\\\
&=\left((F_{L}-1)\cdot\Upsilon_{L}+p^{n}\Upsilon_{L}\right)/p^{n}\Upsilon_{L}\times((F_{L}-1)\cdot
e(\mathcal{Z})+p^{n}e(\mathcal{Z}))/p^{n}e(\mathbb{Z}_{p}^{r})\\\
&=\left((F_{L}-1)\cdot\Lambda_{L}+p^{n}\Lambda_{L}\right)/p^{n}\Lambda_{L}=V_{n}.\end{split}$

Therefore there is exactness also at $\Lambda_{L}/p^{m}\Lambda_{L}$. ∎

We define

$W:=\varprojlim_{n}W_{n},\quad V:=\varprojlim_{n}V_{n},$

and thus, using the above results, we obtain by [Wei94, Prop. 3.5.7] two short
exact sequences

$0\longrightarrow W\longrightarrow\Lambda_{L}\longrightarrow V\longrightarrow
0,$ $0\longrightarrow
V\longrightarrow\Lambda_{L}\longrightarrow\mathcal{Z}/(F_{L}-1)\cdot\mathcal{Z}\longrightarrow
0.$

Splicing together these two sequences we obtain

$0\longrightarrow
W\longrightarrow\Lambda_{L}\xrightarrow{F_{L}-1}\Lambda_{L}\stackrel{{\scriptstyle\nu_{L}}}{{\longrightarrow}}\mathcal{Z}/(F_{L}-1)\cdot\mathcal{Z}\longrightarrow
0.$ (7)

We clearly have $W\cong\Lambda_{L}^{G_{L}}$ which by Lemma 2.1 is isomorphic
to $\mathcal{F}(\mathfrak{p}_{L})\times\mathcal{Z}^{G_{L}}$.

Note that we also have a natural $\mathrm{Gal}(L/K)$-action on
$\Lambda_{L}^{G_{L}}$ and on $\mathcal{Z}/(F_{L}-1)\cdot\mathcal{Z}$, defined
by extending an element $\sigma\in\mathrm{Gal}(L/K)$ to an element
$\tilde{\sigma}\in\mathrm{Gal}(L_{0}/K)$.

So we can formulate the following theorem, which generalizes a result by
Serre.

######  Theorem 3.3.

We have an exact sequence

$0\to\Lambda_{L}^{G_{L}}\longrightarrow\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\stackrel{{\scriptstyle
w_{L}}}{{\longrightarrow}}\mathcal{Z}/(F_{L}-1)\cdot\mathcal{Z}\to 0$

of $\mathbb{Z}_{p}[\mathrm{Gal}(L/K)]$-modules. Here $w_{L}$ denotes the sum
of the componentwise valuations $\nu_{L}$.

###### Proof.

The calculations are very similar to those in the proof of [BC17, Thm. 4.2.3].
The main ingredients are the definitions and equation (7). ∎

We notice that the above theorem can also be considered as an improvement of
[BC17, Thm. 4.2.3] in the sense that it does not require any assumptions on
$\rho^{\mathrm{nr}}(F_{L})$. Therefore the distinction made in [BC17] between
the cases $\chi^{\mathrm{nr}}(F_{L})=1$ and $\chi^{\mathrm{nr}}(F_{L})\neq 1$
turns out to be unnecessary. This allows us also to formulate the next lemma
without considering separately the different possibilities.

We let $M/L/K$ be finite extensions such that both $M/K$ and $L/K$ are Galois.
We have a canonical inclusion

$\displaystyle\iota=\iota_{M/L}\colon\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$
$\displaystyle\longrightarrow$
$\displaystyle\mathcal{I}_{M/K}(\rho^{\mathrm{nr}}),$ $\displaystyle x$
$\displaystyle\mapsto$
$\displaystyle[F_{L}^{d_{M/L}-1}x,F_{L}^{d_{M/L}-2}x,\dots,x],$

where $x=[x_{1},\dots,x_{d_{L/K}}]\in\prod_{d_{L/K}}\Lambda_{L}$.

######  Lemma 3.4.

For extensions $M/L/K$ there is a commutatiove diagram

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{L}^{G_{L}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\textstyle{\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(F-1)\times
1}$$\scriptstyle{{\iota_{M/L}}}$$\textstyle{\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{w_{L}}$$\scriptstyle{{\iota_{M/L}}}$$\textstyle{\mathcal{Z}/(F_{L}-1)\cdot\mathcal{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{M/L}\sum_{i=0}^{d_{M/L-1}}F_{L}^{i}}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{M}^{G_{M}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{I}_{M/K}(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(F-1)\times
1}$$\textstyle{\mathcal{I}_{M/K}(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{w_{M}}$$\textstyle{\mathcal{Z}/(F_{M}-1)\mathcal{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

with exact rows, where $e_{M/L}$ and $d_{M/L}$ are the ramification index and
the inertia degree of $M/L$.

###### Proof.

The proof is a long but straightforward computation. ∎

Following [BC17], the next step is to take a direct limit in the previous
diagram. In order to get explicit results, we need a couple of lemmas.

######  Lemma 3.5.

There exists an extension $\tilde{K}$ of $K$ such that
$\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}=\bigcap_{L/K}\mathcal{M}_{L}\mathbb{Q}_{p}^{r}$.
We can assume $N\subseteq\tilde{K}$.

###### Proof.

Let us define a sequence of fields $L_{i}$ by induction, starting with
$L_{0}=N$. We assume that we have already defined $L_{i}$ for some
$i\in\mathbb{N}$ and that there exists an extension $L/K$ such that
$\mathcal{M}_{L_{i}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{L}\mathbb{Q}_{p}^{r}\subsetneq\mathcal{M}_{L_{i}}\mathbb{Q}_{p}^{r}$.
Then we define $L_{i+1}=L_{i}L$ and we note that
$\mathcal{M}_{L_{i+1}}\mathbb{Q}_{p}^{r}\subseteq\mathcal{M}_{L_{i}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{L}\mathbb{Q}_{p}^{r}\subsetneq\mathcal{M}_{L_{i}}\mathbb{Q}_{p}^{r}$.
In particular this process must terminate since the dimension of
$\mathcal{M}_{L_{i}}\mathbb{Q}_{p}^{r}$ is finite and decreasing at each step.
We call $\tilde{K}$ the last of the $L_{i}$. By construction it must satisfy
the desired properties. ∎

From now on we will fix a field $\tilde{K}$ as in the previous lemma.

######  Lemma 3.6.

Let $M/L$ be an extension of local fields containing $\tilde{K}$, let
$(a_{1},\dots,a_{r})\in\mathcal{Z}$. Then there exists an extension
$\tilde{M}$ of $M$ such that the classes of the elements

$e_{M/L}\sum_{i=0}^{d_{M/L}-1}F_{L}^{i}\cdot(a_{1},\dots,a_{r})$

and

$[M:L](a_{1},\dots,a_{r})$

have the same image under the map

$e_{\tilde{M}/M}\sum_{i=0}^{d_{\tilde{M}/M}-1}F_{M}^{i}:\mathcal{Z}/(F_{M}-1)\mathcal{Z}\to\mathcal{Z}/(F_{\tilde{M}}-1)\mathcal{Z}$

of Lemma 3.4.

###### Proof.

We calculate

$\begin{split}&e_{M/L}\sum_{i=0}^{d_{M/L}-1}F_{L}^{i}\cdot(a_{1},\dots,a_{r})-[M:L](a_{1},\dots,a_{r})\\\
&\qquad\qquad=e_{M/L}\sum_{i=0}^{d_{M/L}-1}(F_{L}^{i}-1)\cdot(a_{1},\dots,a_{r})\\\
&\qquad\qquad=e_{M/L}(F_{L}-1)\sum_{i=0}^{d_{M/L}-1}\sum_{j=0}^{i-1}F_{L}^{j}\cdot(a_{1},\dots,a_{r})\\\
&\qquad\qquad=e_{M/L}\mathcal{M}_{L}\sum_{i=0}^{d_{M/L}-1}\sum_{j=0}^{i-1}\rho^{\mathrm{nr}}(F_{L})^{j}(a_{1},\dots,a_{r}).\end{split}$

Since $L$ and $M$ both contain $\tilde{K}$, it follows that
$\mathcal{M}_{L}\mathbb{Q}_{p}^{r}=\mathcal{M}_{M}\mathbb{Q}_{p}^{r}$. Hence
there exists a vector $v\in\mathbb{Q}_{p}^{r}$ such that

$e_{M/L}\sum_{i=0}^{d_{M/L}-1}F_{L}^{i}\cdot(a_{1},\dots,a_{r})-[M:L](a_{1},\dots,a_{r})=\mathcal{M}_{M}v.$

Clearly, for some $e$, we have $p^{e}v\in\mathbb{Z}_{p}^{r}$. Let $\tilde{M}$
be a totally ramified extension of $M$ of degree $p^{e}$. Then we obtain

$\begin{split}&e_{\tilde{M}/M}\sum_{i=0}^{d_{\tilde{M}/M}-1}F_{M}^{i}\cdot\left(e_{M/L}\sum_{i=0}^{d_{M/L}-1}F_{L}^{i}\cdot(a_{1},\dots,a_{r})-[M:L](a_{1},\dots,a_{r})\right)\\\
&\qquad=p^{e}\mathcal{M}_{M}v=(F_{M}-1)\cdot(p^{e}v)=(F_{\tilde{M}}-1)\cdot(p^{e}v)\in(F_{\tilde{M}}-1)\cdot\mathcal{Z}.\end{split}$

∎

######  Lemma 3.7.

There is an isomorphism

$\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}})\cong\varinjlim_{L}\mathcal{Z}/(F_{L}-1)\cdot\mathcal{Z},$

where the morphisms of the direct system are those of Lemma 3.4.

###### Proof.

We consider a tower of totally ramified extensions
$\tilde{K}\subseteq\tilde{K}_{1}\subseteq\tilde{K}_{2}\subseteq\dots$, each
one of degree $p$. Then we define
$\Phi:\mathbb{Q}_{p}^{r}\to\varinjlim_{L}\mathbb{Z}_{p}^{r}/\mathcal{M}_{L}\mathbb{Z}_{p}^{r}(\rho^{\mathrm{nr}})$
by

$\Phi\left(\frac{a_{1}}{p^{e}},\dots,\frac{a_{r}}{p^{e}}\right)=((a_{1},\dots,a_{r})\pmod{\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}})_{\tilde{K}_{e}},$

where $a_{1},\dots,a_{r}\in\mathbb{Z}_{p}$ and $e\in\mathbb{N}$. This is
clearly well-defined. To prove surjectivity, let
$((a_{1},\dots,a_{r})\pmod{\mathcal{M}_{L}\mathbb{Z}_{p}^{r}})_{L}$, where we
may assume that $L\supseteq\tilde{K}$. Let $e\in\mathbb{N}$ and
$\alpha\in\mathbb{Z}$ be coprime to $p$ and such that
$[L:\tilde{K}]=p^{e}\alpha$. By construction the element
$((\alpha^{-1}a_{1},\dots,\alpha^{-1}a_{r})\pmod{\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}})_{\tilde{K}_{e}}$
is in the image of $\Phi$ and, by Lemma 3.6, it represents the same element as

$((\alpha^{-1}[L\tilde{K}_{e}:\tilde{K}_{e}]a_{1},\dots,\alpha^{-1}[L\tilde{K}_{e}:\tilde{K}_{e}]a_{r})\pmod{\mathcal{M}_{L\tilde{K}_{e}}\mathbb{Z}_{p}^{r}})_{L\tilde{K}_{e}}$

in
$\varinjlim_{L}\mathbb{Z}_{p}^{r}/\mathcal{M}_{L}\mathbb{Z}_{p}^{r}(\rho^{\mathrm{nr}})$.
Note that

$\alpha^{-1}[L\tilde{K}_{e}:\tilde{K}_{e}]=\frac{[L\tilde{K}_{e}:\tilde{K}]}{\alpha[\tilde{K}_{e}:\tilde{K}]}=\frac{[L\tilde{K}_{e}:\tilde{K}]}{\alpha
p^{e}}=\frac{[L\tilde{K}_{e}:\tilde{K}]}{[L:\tilde{K}]}=[L\tilde{K}_{e}:L]$

and therefore

$(([L\tilde{K}_{e}:L]a_{1},\dots,[L\tilde{K}_{e}:L]a_{r})\pmod{\mathcal{M}_{L\tilde{K}_{e}}\mathbb{Z}_{p}^{r}})_{L\tilde{K}_{e}}$

is in the image of $\Phi$. Again by Lemma 3.6, this coincides with the element

$((a_{1},\dots,a_{r})\pmod{\mathcal{M}_{L}\mathbb{Z}_{p}^{r}})_{L},$

and this concludes the proof of surjectivity.

It remains to be proved that
$\ker\Phi=\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$. Let us assume that
$\left(\frac{a_{1}}{p^{e}},\dots,\frac{a_{r}}{p^{e}}\right)\in\ker\Phi$. Then
$((a_{1},\dots,a_{r})\pmod{\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}})_{\tilde{K}_{e}}=0$
in
$\varinjlim_{L}\mathbb{Z}_{p}^{r}/\mathcal{M}_{L}\mathbb{Z}_{p}^{r}(\rho^{\mathrm{nr}})$.
This means that there exists an extension $L/\tilde{K}_{e}$ such that

$e_{L/\tilde{K}_{e}}\sum_{i=1}^{d_{L/\tilde{K}_{e}}-1}F_{\tilde{K}_{e}}^{i}\cdot(a_{1},\dots,a_{r})\equiv
0\pmod{\mathcal{M}_{L}\mathbb{Z}_{p}^{r}}.$

By the definition of $\mathcal{M}_{\tilde{K}_{e}}$ we also know that

$\begin{split}e_{L/\tilde{K}_{e}}\sum_{i=1}^{d_{L/\tilde{K}_{e}}-1}F_{\tilde{K}_{e}}^{i}\cdot(a_{1},\dots,a_{r})&\equiv
e_{L/\tilde{K}_{e}}d_{L/\tilde{K}_{e}}(a_{1},\dots,a_{r})\pmod{\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}}\\\
&\equiv[L:\tilde{K}_{e}](a_{1},\dots,a_{r})\pmod{\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}}.\end{split}$

Since
$\mathcal{M}_{L}\mathbb{Z}_{p}^{r}\subseteq\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}$,
we deduce that
$[L:\tilde{K}_{e}](a_{1},\dots,a_{r})\in\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}$
and therefore
$(a_{1},\dots,a_{r})\in\mathcal{M}_{\tilde{K}_{e}}\mathbb{Q}_{p}^{r}=\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$.
This proves the inclusion
$\ker\Phi\subseteq\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$.

Conversely, let
$\mathcal{M}_{\tilde{K}}\left(\frac{a_{1}}{p^{e}},\dots,\frac{a_{r}}{p^{e}}\right)\in\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$,
then

$\Phi\left(\mathcal{M}_{\tilde{K}}\left(\frac{a_{1}}{p^{e}},\dots,\frac{a_{r}}{p^{e}}\right)\right)=(\mathcal{M}_{\tilde{K}}(a_{1},\dots,a_{r})\pmod{\mathcal{M}_{\tilde{K}_{e}}\mathbb{Z}_{p}^{r}})_{\tilde{K}_{e}},$

By the properties of $\tilde{K}$, we know that there exists a vector
$v\in\mathbb{Q}_{p}^{r}$ such that

$\mathcal{M}_{\tilde{K}}(a_{1},\dots,a_{r})=\mathcal{M}_{\tilde{K}_{e}}v.$

It is also clear that there exists an $\tilde{e}\in\mathbb{N}$ such that
$p^{\tilde{e}}v\in\mathbb{Z}_{p}^{r}$. Then we consider a totally ramified
extension $\tilde{L}$ of $\tilde{K}_{e}$ of degree $p^{\tilde{e}}$. Because of
total ramification, we have
$\mathcal{M}_{\tilde{L}}=\mathcal{M}_{\tilde{K}_{e}}$, and hence we obtain

$\begin{split}\Phi\left(\mathcal{M}_{\tilde{K}}\left(\frac{a_{1}}{p^{e}},\dots,\frac{a_{r}}{p^{e}}\right)\right)&=(\mathcal{M}_{\tilde{K}_{e}}(p^{\tilde{e}}v)\pmod{\mathcal{M}_{\tilde{L}}\mathbb{Z}_{p}^{r}})_{\tilde{L}}\\\
&=(\mathcal{M}_{\tilde{L}}(p^{\tilde{e}}v)\pmod{\mathcal{M}_{\tilde{L}}\mathbb{Z}_{p}^{r}})_{\tilde{L}}=0.\end{split}$

Therefore we also have the inclusion
$\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\subseteq\ker\Phi$. ∎

######  Lemma 3.8.

We have an exact sequence

$0\rightarrow\varinjlim_{L}\Lambda_{L}^{G_{L}}\rightarrow\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\rightarrow\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}})\rightarrow
0.$

###### Proof.

We use Lemma 3.4, the fact that the direct limit functor is exact and Lemma
3.7. ∎

Our next step will be to apply the $G_{N}$-fixed point functor to a complex
obtained from the above exact sequence in order to find an explicit
representative of $R\Gamma\left(N,\Lambda_{N}^{G_{N}}\right)$.

######  Lemma 3.9.

We have

$\left(\mathcal{I}_{M/K}(\rho^{\mathrm{nr}})\right)^{\mathrm{Gal}(M/L)}=\iota_{M/L}\left(\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\right).$

In particular,

$\left(\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\right)^{G_{N}}=\mathcal{I}_{N/K}(\rho^{\mathrm{nr}}),$
(8)

where the direct limit is taken over all finite Galois extensions $L/K$.

###### Proof.

Up to some obvious modifications, the proof is the same as in [BC17, Lemma
4.2.6]. ∎

######  Lemma 3.10.

We have

$(\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}}))^{G_{N}}=\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}})\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}),$
(9)

where by $\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$ we
mean all the elements $x\in\mathbb{Q}_{p}^{r}$ such that
$\mathcal{M}_{N}x\in\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$.

###### Proof.

An element
$[x]\in\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}})$
is fixed by $G_{N}$ if and only if
$\mathcal{M}_{N}x\in\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}$, which implies
the result of the lemma. ∎

######  Lemma 3.11.

The $G_{K}$-modules $\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})$ and
$\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}})$
are acyclic with respect to the fixed point functor $A\mapsto A^{G_{N}}$.

###### Proof.

By [NSW08, Prop. 1.5.1] we have for $i>0$

$H^{i}(N,\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}}))=\varinjlim_{L}H^{i}(\mathrm{Gal}(L/N),\mathcal{I}_{L/K}(\rho^{\mathrm{nr}}))$

which is trivial by Lemma 2.2. The module
$\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}})$
is cohomologically trivial since it is divisible (see [NSW08, Prop. 1.6.2]). ∎

We can now state and prove the following theorem. The proof works as in [BC17,
Thm. 4.2.8], but we prefer to rewrite it in detail for clarity.

######  Theorem 3.12.

The complex

$C^{\bullet}_{N,\rho^{\mathrm{nr}}}\\!=\\!\\!\left[\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\\!\\!\xrightarrow{(F-1)\times
1}\\!\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\\!\\!\to\\!\\!\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r})(\rho^{\mathrm{nr}})\right]$

with non-trivial modules in degree $0$, $1$ and $2$ represents
$R\Gamma\left(N,\Lambda_{N}^{G_{N}}\right)$.

###### Proof.

We write $X^{\bullet}$ for the standard resolution (as defined in [NSW08, Sec.
1.2]) and $I^{\bullet}$ for an injective resolution of
$\varinjlim_{L}\Lambda_{L}^{G_{L}}$. Let $C^{\bullet}$ be the complex

$\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\varinjlim_{L}\mathcal{I}_{L/K}(\rho^{\mathrm{nr}})\rightarrow\mathbb{Q}_{p}^{r}/\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}(\rho^{\mathrm{nr}}).$

The $G$-modules $\mathrm{Map}(G^{n},\varinjlim_{L}\Lambda_{L}^{G_{L}})$
appearing in $X^{\bullet}$ are induced in the sense of [NSW08, Sec. 1.3]. By
Proposition 1.3.7 of loc.cit. the standard resolution consists therefore of
acyclic modules with respect to the fixed point functor. By [Lan02, Th.
XX.6.2] there exists a morphism of complexes $X^{\bullet}\longrightarrow
I^{\bullet}$ inducing the identity on $\varinjlim_{L}\Lambda_{L}^{G_{L}}$ and
isomorphisms on cohomology. By Lemma 3.8 and 3.11 we may apply [Lan02, Th.
XX.6.2] once again and obtain a morphism of complexes
$C^{\bullet}\longrightarrow I^{\bullet}$, also inducing the identity on
$\varinjlim_{L}\Lambda_{L}^{G_{L}}$ and isomorphisms on cohomology. Therefore
applying the $G_{N}$-fixed point functor together with (8) and (9) shows that
$C_{N,\rho^{\mathrm{nr}}}^{\bullet}=\left(C^{\bullet}\right)^{G_{N}}$ is
quasi-isomorphic to $R\Gamma\left(N,\Lambda_{N}^{G_{N}}\right)$. ∎

The following corollary is inspired by the analogous statement appearing in
the first part of the proof of [Bre04, Thm. 4.20] and by [BC17, Coroll.
4.2.10].

######  Corollary 3.13.

There exists a complex

$P^{\bullet}:=[P^{-1}\longrightarrow P^{0}\longrightarrow
P^{1}\longrightarrow\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r})(\rho^{\mathrm{nr}})]$

which is quasi-isomorphic to

$R\Gamma\left(N,\Lambda_{N}^{G_{N}}\right)$

and such that $P^{-1},P^{0},P^{1}$ are finitely generated
$\mathbb{Z}_{p}[G]$-projective.

###### Proof.

By [Lan02, Prop. XXI.1.1] there exists a quasi-isomorphism

$f\colon
K^{\bullet}\longrightarrow\left[\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\right],$

where $K^{\bullet}:=[A\longrightarrow B]$ is centered in degree $0$ and $1$
with $B$ a $\mathbb{Z}_{p}[G]$-projective module. Since by Theorem 3.3 and
Lemma 2.1, the complex

$\left[\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\right]$

has finitely generated cohomology groups, the proof of [Lan02, Prop. XXI.1.1]
actually shows that we can assume that $A$ and $B$ are both finitely
generated. By our Lemma 2.2 the module $\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})$
is cohomologically trivial. We hence apply [Lan02, Prop. XXI.1.2] with
$\mathfrak{F}$ being the family of cohomologically trivial modules. We obtain
that $A$ is cohomologically trivial and hence has a two term resolution
$0\longrightarrow P^{-1}\longrightarrow P^{0}\longrightarrow A\longrightarrow
0$ with finitely generated projective $\mathbb{Z}_{p}[G]$-modules $P^{-1}$ and
$P^{0}$. It follows that
$\left[\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\right]$ is quasi-isomorphic to
$[P^{-1}\longrightarrow P^{0}\longrightarrow P^{1}]$ with $P^{1}=B$. The map

$\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\to\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r})(\rho^{\mathrm{nr}})$

induces a map
$P^{1}\to\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r})(\rho^{\mathrm{nr}})$,
so that the following diagram commutes:

$\textstyle{\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(F-1)\times
1}$$\textstyle{\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r})(\rho^{\mathrm{nr}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{=}$$\textstyle{P^{-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}/(\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r}\cap\mathcal{M}_{N}^{-1}\mathcal{M}_{\tilde{K}}\mathbb{Q}_{p}^{r})(\rho^{\mathrm{nr}})}$

and the second row becomes a complex. It is also straightforward to check that
the vertical arrows give a quasi-isomorphism of complexes. To conclude we need
only to recall Theorem 3.12. ∎

The following lemma is analogous to [BC17, Lemma. 4.3.4], but we need to pay
some attention in the proof, since in the present situation $P^{\bullet}$ does
not comprise only projectives. We will focus on the peculiarity of the present
setting without reproducing all the details of the proof which can be checked
in [BC17].

######  Lemma 3.14.

There exist quasi-isomorphisms of complexes

$\varphi_{n}\colon P^{\bullet}/p^{n}\longrightarrow
R\Gamma(N,\mathcal{F}[p^{n}])[1]$

which are compatible with the inverse systems.

###### Proof.

As in the first part of the proof of [BC17, Lemma. 4.3.4], we can construct
quasi-isomorphisms

$\varphi_{n}:P^{\bullet}/p^{n}\to R\Gamma(N,\mathcal{F}[p^{n}])[1]$

such that

$(\varphi_{n-1}\pi_{n}^{P}-\pi_{n}^{Q}\varphi_{n})\pi_{n}=0$

in $D(\mathbb{Z}_{p}[G])$, where

$\pi_{n}:P^{\bullet}\to P^{\bullet}/p^{n},$ $\pi_{n}^{P}:P^{\bullet}/p^{n}\to
P^{\bullet}/p^{n-1}$

and

$\pi_{n}^{Q}:R\Gamma(N,\mathcal{F}[p^{n}])[1]\to
R\Gamma(N,\mathcal{F}[p^{n-1}])[1]$

are the canonical projections. Let us consider the complex

$\tilde{P}^{\bullet}=[P^{-1}\to P^{0}\to P^{1}].$

Note that there is a natural inclusion

$\iota:\tilde{P}^{\bullet}\to P^{\bullet}$

and, since $P^{\bullet}/p^{n}=\tilde{P}^{\bullet}/p^{n}$, also a natural
projection

$\tilde{\pi}_{n}:\tilde{P}^{\bullet}\to P^{\bullet}/p^{n}$

such that the diagram

$\textstyle{\tilde{P}^{\bullet}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\scriptstyle{\tilde{\pi}_{n}}$$\textstyle{P^{\bullet}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{n}}$$\textstyle{P^{\bullet}/p^{n}}$

commutes. We deduce that

$(\varphi_{n-1}\pi_{n}^{P}-\pi_{n}^{Q}\varphi_{n})\pi_{n}\iota=(\varphi_{n-1}\pi_{n}^{P}-\pi_{n}^{Q}\varphi_{n})\tilde{\pi}_{n}=0$

in $D(\mathbb{Z}_{p}[G])$. Since $\tilde{P}^{\bullet}$ is bounded from above
and comprising only projectives, we can apply [BC17, Lemma 4.3.3] and deduce
that $(\varphi_{n-1}\pi_{n}^{P}-\pi_{n}^{Q}\varphi_{n})\tilde{\pi}_{n}$ is
homotopic to zero. As in the proof of [BC17, Lemma. 4.3.4], we can now use the
surjectivity of $\tilde{\pi}_{n}$ to deduce that also
$\varphi_{n-1}\pi_{n}^{P}-\pi_{n}^{Q}\varphi_{n}$ is homotopic to zero. At
this point we are again in the same situation of [BF98, page 1367], as it was
the case in the proof of [BC17, Lemma. 4.3.4], and we can easily conclude. ∎

We are now ready to prove Theorem 1, of which we recall the formulation.

######  Theorem 3.15.

The complex

$C^{\bullet}_{N,\mathcal{F}}:=\left[\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\xrightarrow{(F-1)\times
1}\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})\right]$

with non-trivial modules in degree 0 and 1 represents $R\Gamma(N,T)$.

###### Proof.

The proof is the same as that of [BC17, Thm. 4.3.1], using the corresponding
results we have obtained in this section. ∎

We can easily deduce the following corollary.

######  Corollary 3.16.

With the above notation we have

  1. (i)

$H^{1}(N,T)\cong\Lambda_{N}^{G_{N}}\cong\mathcal{F}(\mathfrak{p}_{N})\times\mathcal{Z}^{G_{N}}$,

  2. (ii)

$H^{2}(N,T)\cong\mathcal{Z}/(F_{N}-1)\mathcal{Z}$,

  3. (iii)

$H^{i}(N,T)=0\text{ for }i\neq 1,2$.

###### Proof.

This is straightforward by Theorem 3.3, Theorem 3.15 and Lemma 2.1. ∎

######  Example 3.17.

The special case in which the representation $\rho^{\mathrm{nr}}$ is totally
decomposable into a product of characters $\chi^{\mathrm{nr}}_{i}$ is very
easy, already using the results of [BC17]. To that aim it is enough to notice
that also $\mathcal{I}_{N/K}(\rho^{\mathrm{nr}})$ decomposes into a direct
product of the corresponding $\mathcal{I}_{N/K}(\chi^{\mathrm{nr}})$ defined
in [BC17] and the same holds for the modules appearing in $R\Gamma(N,T)$.
Therefore Theorem 3.15 is in this case an easy consequence of [BC17, Thm. 2].

######  Example 3.18.

Let us consider a different example, which can not be handled with the results
of [BC17]. Let $\rho^{\mathrm{nr}}=\rho_{u}\colon
G_{\mathbb{Q}_{p}}\longrightarrow\mathrm{Gl}_{r}(\mathbb{Z}_{p})$ be the
unramified representation attached to

$u=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$

by Proposition 1.6. Then

$\mathcal{M}_{N}=u^{d_{N}}-I=\begin{pmatrix}0&d_{N}\\\ 0&0\end{pmatrix},$

where $d_{N}$ is the inertia degree of $N/\mathbb{Q}_{p}$. By Corollary 3.16
and recalling Lemma 2.1, we obtain

  1. (i)

$H^{1}(N,T)\cong\mathcal{F}(\mathfrak{p}_{N})\times\mathbb{Z}_{p}$,

  2. (ii)

$H^{2}(N,T)\cong\mathbb{Z}_{p}/d_{N}\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.

## Acknowledgements

I would like to thank Werner Bley for encouraging me to generalize some of the
results from our common paper [BC17], for sharing some of his insights and for
suggesting some improvements to the final version of the manuscript. I am also
very grateful to Cornelius Greither for his comments and for helping me to fix
some problems.

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