Capitulation kernels of class groups over Z_p^d-extensions

TL;DR

This paper extends Iwasawa's theorem on class groups from $ ext{Z}_p$-extensions to more complex $ ext{Z}_p^d$-extensions, broadening the understanding of class group behavior in number theory.

Contribution

It introduces a generalization of Iwasawa's theorem applicable to all $ ext{Z}_p^d$-extensions, expanding the scope of class group analysis.

Findings

Generalization of Iwasawa's theorem to $ ext{Z}_p^d$-extensions

New insights into capitulation kernels in higher-dimensional extensions

Enhanced understanding of class group structure in complex extensions

Abstract

We generalize Iwasawa's theorem on class group over $\Z_p$-extensions to all $\Z_p^d$-extensions.

Equations

$${\dot{A}}_E=\bigcup _{E\subset E^{\prime }\subset K}\ker (c_{E^{\prime }/E}),{\dot{A}}_E^{\prime }=\bigcup _{E\subset E^{\prime }\subset K}\ker (c_{E^{\prime }/E}^{\prime }),$$

$$c_{E^{\prime }/E}^{\prime }:A_E^{\prime }⟶A_{E^{\prime }}^{\prime }$$

$${\dot{X}}_K=\underset{k\subset E\subset K}{\underleftarrow{\lim }}{\dot{A}}_E,{\dot{X}}_K^{\prime }=\underset{k\subset E\subset K}{\underleftarrow{\lim }}{\dot{A}}_E^{\prime },$$

$$W_K:=\underset{k\subset E\subset K}{\underleftarrow{\lim }}\mathrm{Hom}(A_E,{\mathbb{Q}}_p/{\mathbb{Z}}_p),W_K^{\prime }:=\underset{k\subset E\subset K}{\underleftarrow{\lim }}\mathrm{Hom}(A_E^{\prime },{\mathbb{Q}}_p/{\mathbb{Z}}_p),$$

$$X_K:=\underset{k\subset E\subset K}{\underleftarrow{\lim }}{A}_E,X_K^{\prime }:=\underset{k\subset E\subset K}{\underleftarrow{\lim }}{A}_E^{\prime }.$$

$${}^{♯}:{\Lambda }_{\Gamma }⟶{\Lambda }_{\Gamma }$$

$$W_K\sim X_K^{♯},W_K^{\prime }\sim X_K^{\prime ♯}.$$

$$\mathfrak{D}=\underset{\chi }{⨁}{\mathfrak{D}}_{\chi },$$

$${\mathfrak{D}}_{\chi }=\{x\in \mathfrak{D}\mid \text{tensor}{[}^{\sigma }]x=\chi (\sigma )\cdot x,\text{for all}\sigma \in \Theta \}.$$

$$W_{\chi }\sim (X_{{\chi }^{-1}}{)}^{♯},W_{\chi }^{\prime }\sim X_{{\chi }^{-1}}^{\prime ♯}.$$

$${\mathrm{CH}}_{\Gamma }(W_{\chi })={\mathrm{CH}}_{\Gamma }(X_{{\chi }^{-1}}^{♯}),{\mathrm{CH}}_{\Gamma }(W_{\chi }^{\prime })={\mathrm{CH}}_{\Gamma }(X_{{\chi }^{-1}}^{\prime ♯}).$$

$${\mathrm{CH}}_{\Gamma }(W_K^{\prime })={\mathrm{CH}}_{\Gamma }(X_K^{\prime ♯}).$$

$${\Delta }_{\theta }:=\{\chi \in \hat{\Gamma }\mid \chi (\theta )=0\}.$$

$$\chi ({\xi }_j)={\zeta }_j,j=1,...,m,$$

$${\Delta }_f=Z_1\cup \cdots \cup Z_l.$$

$${\omega }_{\sigma ,n}:={\sigma }^{p^n}-1,$$

$${\nu }_{\sigma ,n,m}:={\omega }_{\sigma ,m}/{\omega }_{\sigma ,n}$$

$${\mathcal{J}}_n:=({\nu }_{{\tau }_1,0,n},...,{\nu }_{{\tau }_c,0,n})\subset {\Lambda }_{\Gamma },$$

$${\mathcal{I}}_{\underline{r},\underline{n}}:=({\nu }_{{\sigma }_1,r_1,n_1},{\nu }_{{\sigma }_2,r_2,n_2},...,{\nu }_{{\sigma }_d,r_d,n_d})\subset {\Lambda }_{\Gamma },$$

$${\Delta }_I:=\bigcap _{\theta \in I}{\Delta }_{\theta }=\bigcap _{i=1}^l{\Delta }_{{\theta }_i}.$$

$$Z\subset {\Delta }_{({\nu }_{{\xi }_1,r_1-1,r_1},\cdots ,{\nu }_{{\xi }_m,r_m-1,r_m})}={\Delta }_{{\nu }_{{\xi }_1,r_1-1,r_1}}\cap \cdots \cap {\Delta }_{{\nu }_{{\xi }_m,r_m-1,r_m}}.$$

$$x=\sum _{i=0}^{\delta -1}{y}_i\cdot T^i,y_0,...,y_{\delta -1}\in {\Lambda }_{{\Gamma }^{\prime }}.$$

$$0=\chi (x)=\sum _{i=0}^{\delta -1}\bar{\chi }(y_i)\cdot \chi (T{)}^i.$$

$$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 6.75pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{\hbox{\kern-6.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{{\mathfrak{Y}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 13.02777pt\raise 4.28406pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-1.28406pt\hbox{$\scriptstyle{\sim}$}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 30.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 30.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{\bar{\mathfrak{Y}}:=\varprojlim_{n}{\mathfrak{Y}}/\mathscr{J}_{n}{\mathfrak{Y}}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$$

$${\nu }_{n,m}:=\prod _{i=1}^d{\nu }_{{\sigma }_i,n,m}\in {\Lambda }_{\Gamma }.$$

$$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 12.73622pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{\hbox{\kern-12.73622pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{{\mathfrak{Y}}/\mathfrak{I}_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 16.49524pt\raise 5.32639pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.6875pt\hbox{$\scriptstyle{\nu_{n,m}}$}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 36.73622pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 36.73622pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{{\mathfrak{Y}}/\mathfrak{I}_{m}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$$

$$\ddot{\mathfrak{Y}}_{(n,m)}:=\ker(\,\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 17.25012pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{\hbox{\kern-17.25012pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{{\mathfrak{Y}}/\mathscr{J}_{n}{\mathfrak{Y}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 21.00914pt\raise 5.32639pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.6875pt\hbox{$\scriptstyle{\nu_{n,m}}$}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 41.25012pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 41.25012pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{$\textstyle{{\mathfrak{Y}}/\mathscr{J}_{m}{\mathfrak{Y}}}$}}}}}}}\ignorespaces}}}}\ignorespaces\,),$$

$${\ddot{\mathfrak{Y}}}_{(n)}:=\bigcup _{m\ge n}{\ddot{\mathfrak{Y}}}_{(n,m)},$$

$$\ddot{\mathfrak{Y}}:=\underset{n}{\underleftarrow{\lim }}{\ddot{\mathfrak{Y}}}_{(n)}.$$

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Full text

† The author was supported in part by the Ministry of Science and Technology of Taiwan, MOST 103-2115-M-002 -008 -MY2, MOST 105-2115-M-002 -009 -MY2.

It is our pleasure to thank NCTS/TPE for supporting a number of meetings of the authors in National Taiwan University.

1. Introduction

Let $k$ be a number field and let $K/k$ be an abelian extension. If $K/k$ is a finite extension, we define a map from the group of ideals of $k$ to that of $K$ by taking $\mathfrak{a}$ to the fractional ideal $\mathfrak{a}{\mathcal{O}}_{K}$. This map induces the homomorphism $c_{K/k}:C_{k}\longrightarrow C_{K}$ of class groups which is called the capitulation map. Previous studies indicate that there is certain connection between the order of $\ker(c_{K/k})$ and the ramification of $K/k$. A famous theorem of Suzuki [Suz91] that generalizes both Hilbert’s theorem 94 and the principal ideal theorem of class field theory says if $K/k$ is unramified, then the order of $\ker(c_{K/k})$ is divisible by the degree $[K:k]$. More results in this aspect can be found in [Gon07]. In contrast, Iwasawa [Iwa73] proves that if $K/k$ is a ${\mathbb{Z}}_{p}$-extension, so it is almost totally ramified at some place dividing $p$, then the the capitulation kernel $\dot{X}_{K}$ (see below) is pseudo-null. In this note, we generalize this result of Iwasawa to every ${\mathbb{Z}}_{p}^{d}$-extension of $k$, and then use it to establish a pseudo-isomorphism of Iwasawa modules that generalizes [Iwa73, Theorem 11].

From now on let $K/k$ be a ${\mathbb{Z}}_{p}^{d}$-extension. For a number field $E\subset K$, let $A_{E}$ denote the Sylow $p$-subgroup of the class group ${C}_{E}$ of $E$ and let $A^{\prime}_{E}$ be the quotient of $A_{E}$ modulo the subgroup generated by ideals above $p$. Define

[TABLE]

where $E^{\prime}/E$ runs through finite sub-extensions in $K/E$ and

[TABLE]

is induced from $c_{E/E^{\prime}}$. Note that $\dot{A}_{E}$ is a subgroup of $A_{E}$. Put

[TABLE]

with the limit taken over the norm maps. Let $\sim$ denote pseudo-isomorphism.

Theorem 1**.**

We have $\dot{X}_{K}\sim 0$ and $\dot{X}^{\prime}_{K}\sim 0$.

Theorem 1 is proved in §2.3 by following the path of Iwasawa [Iwa73, Theorem 10]. For $d\geq 2$, the Iwasawa algebra is much more complicated than that of a ${\mathbb{Z}}_{p}$-extension, this might be the reason why the content of the theorem has seldom discussed since the publication of Iwasawa’s paper. The breakthrough in our proof is the duality in Lemma 2.1.3(c). Having it, we use Monsky’s theorem [Mon81] (see §2.1) to show Proposition 2.2.3, which then leads to the proof of the theorem.

Write $\Gamma$ for the Galois group of $K/k$ and $\Lambda_{\Gamma}$ for the Iwasawa algebra ${\mathbb{Z}}_{p}[[\Gamma]]$. Define the $\Lambda_{\Gamma}$-modules

[TABLE]

with the limits taken over the homomorphisms dual to the capitulation maps, and

[TABLE]

The following theorem might be known to experts, for the convenience of the readers, we include a proof of it in §3.3.

Theorem 2**.**

The $\Lambda_{\Gamma}$-modules $W_{K}$ and $W^{\prime}_{K}$ are finitely generated.

Let

[TABLE]

be the involution of ${\mathbb{Z}}_{p}$-algebra induced from $\Gamma\longrightarrow\Gamma$, $\gamma\mapsto\gamma^{-1}$. For a $\Lambda_{\Gamma}$-module $\mathfrak{D}$, let $\mathfrak{D}^{\sharp}$ denote the module with $\mathfrak{D}$ as the underlying ${\mathbb{Z}}_{p}$-module while $\Gamma$ acts on $\mathfrak{D}$ via ${}^{\sharp}:\Gamma\longrightarrow\Gamma$.

Theorem 3**.**

We have

[TABLE]

Suppose $k_{0}$ is a subfield of $k$ such that $K/k_{0}$ is an abelian extension having Galois group $\operatorname{Gal}(K/k_{0})=\Gamma\times\Theta$ with $\Theta$ finite of order prime to $p$. Then every pro-$p$ $\operatorname{Gal}(K/k_{0})$-module $\mathfrak{D}$ can be written as

[TABLE]

where $\chi$ runs through all ${\bar{{\mathbb{Q}}}_{p}}^{*}$-valued characters of $\Theta$ and

[TABLE]

It is clear that $(\mathfrak{D}^{\sharp})_{\chi}=(\mathfrak{D}_{\chi^{-1}})^{\sharp}$, and we have (see the last paragraph of §3.3)

[TABLE]

These pseudo-isomorphisms imply the equalities of characteristic ideals

[TABLE]

Theorem 3, proved in §3.3, is not entirely new, when $d=1$, it is by Iwasawa [Iwa73, Theorem 11]. Suppose $K/k$ has ramification locus $S$. Nekovář defines in [Nek06, §9.5] a morphism from $X_{K}$ to the Iwasawa adjoint $E^{1}(W_{K}^{\sharp})$ of $W_{K}^{\sharp}$ (they are pseudo-isomorphic) and shows the cokernel is pseudo-null under the condition: (Dec 2): For every $v\in S$, the decomposition group $\Gamma_{v}\simeq{\mathbb{Z}}_{p}^{r(v)}$, with $r(v)\geq 2$. A similar result for $X_{K}^{\prime}$ and $W_{K}^{\prime\sharp}$ is given in [Nek06, §9.4], while Vauclair, in collaboration with Nekovář, proves in [Vau09, Theorem 7.6] that there is a natural morphism $\alpha:W_{K}^{\prime}\longrightarrow E^{1}(X_{K}^{\prime\sharp})$ with pseudo-isomorphic kernel and cokernel, and if (Dec 2) holds then $\alpha$ is a pseudo-isomorphism. This also implies unconditionally the equality of characteristic ideals as in (2) (see [Vau09, Corollary 6.8])

[TABLE]

Our proof of Theorem 3 uses the theory of $\Gamma$-system established in [LLTT18]. It turns out Theorem 1 is the key ingredient for having the theory work in our situation. Indeed, a sensible homomorphism $X_{K}\longrightarrow W_{K}^{\sharp}$ should be compatible with the duality between $A_{E}$ and $\operatorname{Hom}(A_{E},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})$ at all finite layers, and hence must factors through the quotient of $X_{K}$ modulo $\dot{X}_{K}$. Therefore, the pseudo isomorphism between $X_{K}$ and $W_{K}^{\sharp}$ cannot established without having $\dot{X}_{K}\sim 0$, the same with $X^{\prime}_{K}$ and $W_{K}^{\prime\sharp}$.

It is worthwhile to mention that over global function fields of characteristic $p$ our theorems hold trivially, because the global unit group $U_{K}$ has trivial $p$-primary part (cf. §3.1 and §3.3, especially the exact sequence (17), Lemma 3.1.1 and the proof of Theorem 3).

We thank I. Longhi for helping us with the proof of Lemma 2.1.4, we also thank D. Vauclair for a communication on our and his work.

1.1. Notation and preliminary remarks

Let $E$ stand for a finite intermediate extension of $K/k$ and denote $\Gamma_{E}=\operatorname{Gal}(K/E)$. Put $\Gamma^{(n)}:=\Gamma^{p^{n}}$, $k_{n}:=K^{\Gamma^{(n)}}$, the $n$th layer of $K/k$. Let $\mathscr{I}_{\Gamma}$ denote the augmentation ideal of $\Lambda_{\Gamma}$ and put $\mathscr{I}_{n}:=\ker(\Lambda_{\Gamma}\longrightarrow{\mathbb{Z}}_{p}[\operatorname{Gal}(k_{n}/k)])$ so that $\mathscr{I}_{0}=\mathscr{I}_{\Gamma}$.

For a set $T$ of places of $k$, let $T_{E}$ denote the set of places of $E$ sitting over $T$. Let ${\mathcal{P}}$ be the set of places of $k$ above $p$. For a Galois group $G$ of number field extension, let $G_{w}$ and $G_{w}^{0}$ denote respectively the decomposition subgroup and the inertia subgroup at a place $w$.

For a topological group $B$, let $B^{\vee}$ denote $\operatorname{Hom}(B,{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})$, where ${\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}$ is endowed with the discrete topology. In most cases, $B$ is pro-$p$ or $p$-primary and discrete so that $B^{\vee}$ is its Pontryagin dual. We also identify $B^{\vee}$ with $\operatorname{H}^{1}(B,{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})$ when $B$ is a compact group. For a functor $\lozenge$ on the category of finite intermediate extension of $K/k$, we abbreviate $\lozenge_{k_{n}}=\lozenge_{n}$. If $\mathfrak{Q}\longrightarrow\mathfrak{D}$ is a morphism of two such functotors with $\mathfrak{Q}_{E}$, $\mathfrak{D}_{E}$ compact and $\mathfrak{Q}_{E}\longrightarrow\mathfrak{D}_{E}$ surjective for every $E$, then $\mathfrak{Q}_{K}\longrightarrow\mathfrak{D}_{K}$ is also surjective. To see this, we apply the dual

\textstyle{\mathfrak{D}_{E}^{\vee}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{Q}_{E}^{\vee}} , which implies the injectivity of

\textstyle{\mathfrak{D}_{K}^{\vee}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{Q}_{K}^{\vee}} , and then obtain the desired surjectivity via the duality [Kap50].

Since $X_{K}$ is torsion, it is annihilated by some non-zero $f\in\Lambda_{\Gamma}$. Then $W_{K}$ is annihilated by $f^{\sharp}$, whence torsion. To see this, we may assume that for every intermediate extension $E$ of $K/k$, the extension $K/E$ contains no nontrivial unramified intermediate extension. Then the restriction map $\operatorname{Hom}(A_{E},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})\longrightarrow\operatorname{Hom}(A_{E^{\prime}},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})$ is injective for $E\subset E^{\prime}\subset K$, and hence $X_{K}\longrightarrow A_{E}$ is surjective. This implies $f\cdot A_{E}=0$, so $f^{\sharp}\cdot\operatorname{Hom}(A_{E},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})=0$ and $f^{\sharp}\cdot W_{K}=0$.

2. Towers of $\Gamma$-modules

2.1. Monsky’s Theorem

Endow \mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{\infty}} with the discrete topology. Let $\hat{\Gamma}$ denote all continuous characters from $\Gamma$ to \mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{\infty}}. Every $\chi\in\hat{\Gamma}$ is of finite order, it factors through $\Gamma\longrightarrow\Gamma/\Gamma^{(n)}$, for some $n$, and hence extends uniquely to a continuous ${\mathbb{Z}}_{p}$-algebra homomorphism $\chi:\Lambda_{\Gamma}\longrightarrow{\mathcal{O}}$, where ${\mathcal{O}}$ is the ring of integers of $\bar{\mathbb{Q}}_{p}$.

Definition 2.1.1**.**

1. (1)

Define the zero set of an element $\theta\in\Lambda_{\Gamma}$ to be

[TABLE] 2. (2)

A ${\mathbb{Z}}_{p}$-flat $Z$ of codimension $m$ is a subset of $\hat{\Gamma}$ consisting of solutions $\chi$ to the system of equations

[TABLE]

*where $\xi_{1},...,\xi_{m}$ are elements of $\Gamma$, extendable to a ${\mathbb{Z}}_{p}$-basis, \zeta_{1},...,\zeta_{m}\in\mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{\infty}}. * 3. (3)

We say a set $\tau_{1},....\tau_{c}$ of ${\mathbb{Z}}_{p}$-generators of $\Gamma$ is tight, if each $\tau_{i}\notin\Gamma^{(1)}$ and the topological closure of $<\tau_{i}>$ and $<\tau_{j}>$ are distinct for $i\not=j$.

Theorem 2.1.2** (Monsky, [Mon81, Theorem 2.2,2.6] ).**

The zero set $\Delta_{f}$ of a non-zero $f\in\Lambda_{\Gamma}$ is a proper subset of $\hat{\Gamma}$. There are ${\mathbb{Z}}_{p}$-flats $Z_{1},...,Z_{l}$ such that

[TABLE]

Let us introduce more notation. For $n\geq 0$, $\sigma\in\Gamma$, $\sigma\not=id$, denote

[TABLE]

which is regarded as an element of $\Lambda_{\Gamma}$. Put $\omega_{\sigma,-1}=1$. For $m\geq n\geq-1$, write

[TABLE]

which is also an element of $\Lambda_{\Gamma}$. Because $\nu_{\sigma,r,r+1}$ is the $p^{r+1}$th cyclotomic polynomial in $\sigma$, it is irreducible in ${\mathbb{Z}}_{p}[\sigma]$. Hence in $\Lambda_{\Gamma}$, if $r\not=r^{\prime}$, then $\nu_{\sigma,r^{\prime},r^{\prime}+1}$ is relatively prime to $\nu_{\sigma,r,r+1}$. Also, if $\sigma$ and $\sigma^{\prime}$ are linearly independent over ${\mathbb{Z}}_{p}$, then $\nu_{\sigma,n,m}$ and $\nu_{\sigma^{\prime},n^{\prime},m^{\prime}}$ are relatively prime in $\Lambda_{\Gamma}$.

Due to technical reason, we will need to deal with ideals other than $\mathscr{I}_{n}$, especially when $d\geq 2$. We introduce the ideals

[TABLE]

where $\tau_{1},....\tau_{c}$ is a chosen set of tight ${\mathbb{Z}}_{p}$-generators of $\Gamma$. It is clear that for $m\geq n$, the inclusion $\mathscr{J}_{m}\subset\mathscr{J}_{n}$ holds. Since $\omega_{\tau,0}\cdot\nu_{\tau,0,n}=\omega_{\tau,n}$, the ideal $\mathscr{I}_{n}=(\omega_{\tau_{1},n},...,\omega_{\tau_{c},n})$ is inside $\mathscr{J}_{n}$. For a chosen ${\mathbb{Z}}_{p}$-basis $\sigma_{1},...,\sigma_{d}$ of $\Gamma$, we also need to consider ideals,

[TABLE]

with $\underline{n}=(n_{1},...,n_{d})$, $\underline{r}=(r_{1},...,r_{d})\in{\mathbb{Z}}^{d}$ such that $r_{i}\geq-1$ and $n_{i}>r_{i}$, for every $i$, the latter condition will be abbreviated as $\underline{n}>\underline{r}$.

For an ideal $I=(\theta_{1},...,\theta_{l})\subset\Lambda_{\Gamma}$, denote

[TABLE]

Write $\Delta_{\underline{r},\underline{n}}$ for $\Delta_{\mathscr{I}_{\underline{r},\underline{n}}}$. If in (3) each $\zeta_{j}$ is of order $p^{r_{j}}$, then

[TABLE]

Lemma 2.1.3**.**

The following statements hold true:

1. (a)

If the order of $\chi\in\hat{\Gamma}$ does not exceed $p^{n}$ or $p^{n_{i}}$, for every $i=1,...,d$, then for all $x\in\mathscr{I}_{\underline{r},\underline{n}}+\mathscr{J}_{n}$ the value $\chi(x)$ is divisible by the minimum of $p^{n}$ and $p^{n_{1}-r_{1}},...,p^{n_{d}-r_{d}}$. 2. (b)

For a fixed $\underline{r}$, the intersection $\displaystyle{\bigcap_{\underline{n}>\underline{r},n>0}(\mathscr{I}_{\underline{r},\underline{n}}+\mathscr{J}_{n}})=0$. 3. (c)

Let $x\in\Lambda_{\Gamma}$. Then $x\in\mathscr{I}_{\underline{r},\underline{n}}$ if and only if $\chi(x)=0$ for all $\chi\in\Delta_{\underline{r},\underline{n}}$.

Proof.

Let $p^{\alpha}$ denote the order of $\chi(\sigma)$. For $m\geq\alpha$, we have $\chi(\nu_{\sigma,r,m})=0$, if $\alpha>r$, while $\chi(\nu_{\sigma,r,m})=p^{m-r}$, if $\alpha\leq r$. The assertion (a) follows.

To prove (b), let $x$ be an element of the intersection in question. In view of Theorem 2.1.2, we have to show $\chi(x)=0$ for all $\chi\in\hat{\Gamma}$. Let $\alpha$ denote the maximum of $r_{1},...,r_{d}$ . For an integer $\beta>\alpha$, put $\underline{n}=(\beta,...,\beta)$, $n=\beta$. Then by (a), the value $\chi(x)$ is divisible by $p^{\beta-\alpha}$. Since this holds for all $\beta$, we must have $\chi(x)=0$.

We prove (c) by induction on $d$. The statement trivially holds for $d=0$. Suppose $d>0$, let $\Gamma^{\prime}\subset\Gamma$ be the subgroup topologically generated by elements of the basis $\sigma_{1},...,\sigma_{d}$ other than $\sigma_{1}$. If $d=1$, then both $\Gamma^{\prime}$ and $\hat{\Gamma^{\prime}}$ are the trivial group and $\Lambda_{\Gamma^{\prime}}={\mathbb{Z}}_{p}$. In this case, put $\mathscr{I}_{\underline{r^{\prime}},\underline{n^{\prime}}}=(0)$ and $\Delta_{\underline{r^{\prime}},\underline{n^{\prime}}}=\hat{\Gamma^{\prime}}$. If $d\geq 2$, let $\underline{r^{\prime}}=(r_{2},...,r_{d})$ and $\underline{n^{\prime}}=(n_{2},...,n_{d})$. Put $T:=\sigma_{1}-1\in\Lambda_{\Gamma}$. Then $\nu_{\sigma_{1},r_{1},n_{1}}$ is a distinguished polynomial in $T$ of degree $\delta:=p^{n_{1}}-[p^{r_{1}}]$, where $[\cdot]$ is the Gauss symbol. The natural map $\hat{\Gamma}\longrightarrow\hat{\Gamma^{\prime}}$, $\chi\mapsto\bar{\chi}$, induces a surjection $\rho:\Delta_{\underline{r},\underline{n}}\longrightarrow\Delta_{\underline{r^{\prime}},\underline{n^{\prime}}}$ such that the fibre $\rho^{-1}(\bar{\chi})$ for each $\bar{\chi}\in\Delta_{\underline{r^{\prime}},\underline{n^{\prime}}}$ is of cardinality $\delta$ and a character $\chi\in\rho^{-1}(\bar{\chi})$ is determined by the value $\chi(\sigma_{1})$, or equivalently, by $\chi(T)$. By the Weierstrass division theorem [Bou72, VII, §3, Proposition 5], we may assume that

[TABLE]

Thus, for $\chi\in\rho^{-1}(\bar{\chi})$,

[TABLE]

This means the polynomial $\sum_{i=0}^{\delta-1}\bar{\chi}(y_{i})\cdot X^{i}$ in $X$ has $\delta$ distinct roots, hence must be trivial. Therefore, each $y_{i}$ is annihilated by characters in $\Delta_{\underline{r^{\prime}},\underline{n^{\prime}}}$. The induction hypothesis implies $y_{i}\in\mathscr{I}_{\underline{r^{\prime}},\underline{n^{\prime}}}$ for every $i$, so $x\in\mathscr{I}_{\underline{r},\underline{n}}$. ∎

Lemma 2.1.4**.**

Suppose ${\mathfrak{Y}}$ is a finitely generated $\Lambda_{\Gamma}$-module. Then $\bigcap_{n}\mathscr{J}_{n}{\mathfrak{Y}}=0$ so that the natural maps ${\mathfrak{Y}}\longrightarrow{\mathfrak{Y}}/\mathscr{J}_{n}{\mathfrak{Y}}$ induce the isomorphism

[TABLE]

Proof.

The homomorphism ${\mathfrak{Y}}\longrightarrow\bar{\mathfrak{Y}}$ has dense image, and it is surjective because both ${\mathfrak{Y}}$ and $\bar{\mathfrak{Y}}$ are compact. It is sufficient to show $\bigcap_{n}\mathscr{J}_{n}{\mathfrak{Y}}=0$. Lemma 2.1.3 (b) says the assertion holds for ${\mathfrak{Y}}=\Lambda_{\Gamma}$. In general, we have a surjective homomorphism $\phi:\bigoplus_{i=1}^{l}\Lambda_{\Gamma}\longrightarrow{\mathfrak{Y}}$, with $\mathscr{J}_{n}{\mathfrak{Y}}=\phi(\bigoplus_{i=1}^{l}\mathscr{J}_{n})$. Suppose $x\in\bigcap_{n}\mathscr{J}_{n}{\mathfrak{Y}}$. Denote $X=\phi^{-1}(x)$ which is a compact subset of $\bigoplus_{i=1}^{l}\Lambda_{\Gamma}$. Since $X\cap\bigoplus_{i=1}^{l}\mathscr{J}_{n}\not=\emptyset$, for every $n$, and $\bigcap_{n}\bigoplus_{i=1}^{l}\mathscr{J}_{n}=0$, we must have $0\in X$. Therefore, $x=\phi(0)=0$. ∎

2.2. The norm maps

Fix a basis $\sigma_{1},...,\sigma_{d}$ of $\Gamma$ and set

[TABLE]

Since $\mathscr{I}_{n}$ is gernerated by $\omega_{\sigma_{i},n}$ and $\nu_{\sigma_{i},n,m}\cdot\omega_{\sigma_{i},n}=\omega_{\sigma_{i},m}$, we have $\nu_{n,m}\cdot\mathscr{I}_{n}\subset\mathscr{I}_{m}$. For a $\Lambda_{\Gamma}$-module ${\mathfrak{Y}}$, the map sending $y\in{\mathfrak{Y}}$ to $\nu_{n,m}\cdot y$ induces an endomorphism \textstyle{{\mathfrak{Y}}/\mathscr{I}_{m}{\mathfrak{Y}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu_{n,m}}$$\textstyle{{\mathfrak{Y}}/\mathscr{I}_{m}{\mathfrak{Y}}} , which turns out to be the norm map of $\Gamma^{(n)}/\Gamma^{(m)}$ acting on ${\mathfrak{Y}}/\mathscr{I}_{m}{\mathfrak{Y}}$, hence independent of the choice of the basis.

In general, if for each $n$, there associates a $\Lambda_{\Gamma}$ submodule $\mathfrak{I}_{n}\subset{\mathfrak{Y}}$ containing $\mathscr{I}_{n}{\mathfrak{Y}}$ such that $\nu_{n,m}\cdot\mathfrak{I}_{n}\subset\mathfrak{I}_{m}$ for $m\geq n$, we shall also denote the $\Lambda_{\Gamma}$-map induced from $\nu_{n,m}$ as

[TABLE]

Let $\mathscr{J}_{n}$ be defined by choosing a tight set $\tau_{1},...,\tau_{c}$ of generators of $\Gamma$.

Lemma 2.2.1**.**

We have $\nu_{n,m}\cdot\mathscr{J}_{n}\subset\mathscr{J}_{m}$.

Proof.

We have to show that $\nu_{n,m}\cdot\nu_{\tau_{i},0,n}\in\mathscr{J}_{m}$. Since modulo $\mathscr{I}_{m}$, the image $\nu_{n,m}\cdot\nu_{\tau_{i},0,n}$ is independent of the choice of the basis, we may assume that $\tau_{i}=\sigma_{1}$. Then $\nu_{n,m}(\nu_{\tau_{i},0,n})=\nu_{\sigma_{1},0,m}\cdot\nu_{\sigma_{2},n,m}\cdot\cdots\cdot\nu_{\sigma_{d},n,m}\in\mathscr{J}_{m}$. ∎

Definition 2.2.2**.**

For a $\Lambda_{\Gamma}$-module ${\mathfrak{Y}}$, define

[TABLE]

[TABLE]

and

[TABLE]

Proposition 2.2.3**.**

Suppose ${\mathfrak{Y}}$ is a finitely generated $\Lambda_{\Gamma}$-module. Then $\ddot{\mathfrak{Y}}\sim 0$.

Proof.

Consider an exact sequence

[TABLE]

with $\mathfrak{N}$, $\mathfrak{N}^{\prime}$ pseudo-null and $\mathfrak{Z}$ a direct sum $\oplus_{i=1}^{r}\Lambda_{\Gamma}/(f_{i})$ where $f_{i}$ is either [math] or a power of an irreducible element of $\Lambda_{\Gamma}$. Since $\mathfrak{N}\oplus\mathfrak{N}^{\prime}$ is pseudo-null, there are relatively prime $g_{1},g_{2}\in\Lambda_{\Gamma}$ annihilating both $\mathfrak{N}$ and $\mathfrak{N}^{\prime}$. These would lead to an exact seuqence

[TABLE]

with both $\mathfrak{a}$ and $\mathfrak{b}$ annihilated by $g_{1}^{2}$ and $g_{2}^{2}$, hence pseudo-null. Thus, it remains to show that $\ddot{\mathfrak{Y}}=0$ for ${\mathfrak{Y}}=\Lambda_{\Gamma}/(f)$.

Let $\bar{x}\in\ddot{\mathfrak{Y}}\subset{\mathfrak{Y}}$ (Lemma 2.1.4) and let $x\in\Lambda_{\Gamma}$ be a lifting of $\bar{x}$. For each $n$, there is $m>n$ such that

[TABLE]

for some $a\in\Lambda_{\Gamma}$. Put $\omega:=\omega_{\tau_{1},0}\cdot\cdots\cdot\omega_{\tau_{c},0}$, $y:=\omega x$. Since $\omega\cdot\mathscr{J}_{m}\subset\mathscr{I}_{m}$,

[TABLE]

Suppose $f=0$. For every $\chi\in\Delta_{\mathscr{I}_{n}}$, we have $\chi(\nu_{n,m})\not=0$, while $\chi(\mathscr{I}_{m})=0$, so $\chi(y)=0$. Then it follows that $y\in\mathscr{I}_{n}$ (Lemma 2.1.3(c)). Since this holds for all $n$, Lemma 2.1.4 says $y=0$, whence $x=0$ as desired.

Suppose $f\not=0$ and let $Z_{1}$,…,$Z_{l}$ be the ${\mathbb{Z}}_{p}$-flats in Theorem 2.1.2. For each $i$, let $\chi(\xi_{i})=\zeta_{i}$ be one of the defining equations (3) of $Z_{i}$, so that (4) says $Z_{i}\subset\Delta_{\epsilon_{i}}$, if $\zeta_{i}$ is of order $p^{r_{i}}$ and $\epsilon_{i}:=\nu_{\xi_{i},r_{i}-1,r_{i}}$. Then we put $\epsilon:=\epsilon_{1}\cdot\cdots\cdot\epsilon_{l}$ to have

[TABLE]

Denote $\nu_{n,m}^{(i)}:=\nu_{\sigma_{i+1},n,m}\cdot\cdots\cdot\nu_{\sigma_{d},n,m}$, $\underline{-1}=(-1,...,-1)$, and $\underline{m}^{(i)}=(m_{1}^{(i)},...,m_{d}^{(d)})$, with $m_{j}^{(i)}=n$ for $j\leq i$; $m_{j}^{(i)}=m$, for $j>i$. We claim that if

[TABLE]

holds for $t_{i},b_{i}\in\Lambda_{\Gamma}$, then there is some $b_{i+1}\in\Lambda_{\Gamma}$ such that

[TABLE]

Then, beginning with (6), for which $i=0$, by repeatedly applying the above implication, we can deduce

[TABLE]

which means $\omega\epsilon^{d-1}x\in\mathscr{I}_{n}+(f)$. As this happens for every $n$, Lemma 2.1.4 says $\omega\epsilon^{d-1}x\in(f)$. The proposition is proved, if $f$ is relatively prime to $\omega\epsilon$, and it remains to treat the case where $f$ is a power of $\nu_{\sigma,\alpha-1,\alpha}$, $\alpha\geq 0$, $\sigma$ extendable to a ${\mathbb{Z}}_{p}$-basis of $\Gamma$, because every irreducible factor of $\omega\epsilon$ is of such type. We may also assume that $\sigma=\sigma_{1}$. Because $\tau_{1},...,\tau_{c}$ form a tight generating set, we can write

[TABLE]

with $\delta=0,1$, and $\omega^{\prime}$ relatively prime to $f$. Note that if $\delta=1$, then $\alpha=0$ and $\sigma_{1}=\tau_{j}$, for some $j$, we may assume that the basis $\sigma_{1},...,\sigma_{d}$ are taken from $\{\tau_{1},...,\tau_{c}\}$. Put $z=\omega^{\prime}x$, $\underline{\mathfrak{a}}=(\mathfrak{a}_{1},...,\mathfrak{a}_{d})$ with $\mathfrak{a}_{1}=\alpha$, $\mathfrak{a}_{j}=-1$ for $j\not=1$. By (5),

[TABLE]

Again, we claim that for $n>\alpha$, if

[TABLE]

holds for $s_{i},c_{i}\in\Lambda_{\Gamma}$, then there is some $c_{i+1}\in\Lambda_{\Gamma}$ such that

[TABLE]

Then we can deduce that $\omega^{\prime}x\in\mathscr{I}_{\underline{\mathfrak{a}},\underline{n}^{(0)}}+(f)$, for all $n>\alpha$, and hence $\omega^{\prime}x\in(f)$. Therefore, $\bar{x}=0$, since $\omega^{\prime}$ is relatively prime to $f$.

The proofs of the claims rely on Lemma 2.1.3(c). Denote $\underline{\mathfrak{b}}^{(i)}:=(u_{1},...,u_{d})$, with $u_{i+1}=n$ and $u_{j}=-1$ for $j\not=i+1$. Let $\chi\in\Delta_{\underline{\mathfrak{b}}^{(i)},\underline{m}^{(i)}}$. The inclusion

[TABLE]

leads to $\chi(\nu_{\sigma_{i+1},n,m})=0=\chi(\mathscr{I}_{\underline{-1},\underline{m}^{(i)}})$, so (8) yields $\chi(b_{i}\cdot f)=0$. Now, if $\chi(f)=0$, then (7) implies $\chi(\epsilon)=0$; otherwise, we have $\chi(b_{i})=0$. Therefore, $\chi(\epsilon b_{i})=0$ always holds. Lemma 2.1.3(c) says $\epsilon b_{i}\in\mathscr{I}_{\underline{\mathfrak{b}}^{(i)},\underline{m}^{(i)}}$. Then we can write

[TABLE]

with $b_{i+1}^{\prime}\in(\nu_{\sigma_{1},-1,n},...,\nu_{\sigma_{i,-1,n}},\nu_{\sigma_{i+2},-1,m},...,\nu_{\sigma_{d},-1,m})\subset\mathscr{I}_{\underline{-1},\underline{m}^{(i+1)}}$. By (8) again,

[TABLE]

In view of Lemma 2.1.3, this proves the first claim, because $\Delta_{\nu_{\sigma_{i+1},n,m}}\cap\Delta_{\underline{-1},\underline{m}^{(i+1)}}=\emptyset$, if $\chi\in\Delta_{\underline{-1},\underline{m}^{(i+1)}}$, then $\chi(\nu_{\sigma_{i+1},n,m})\not=0$ and consequently $\chi(\nu_{n,m}^{(i+1)}\cdot t_{i}-b_{i+1}\cdot f)=0$.

The proof of the second claims is similar to the previous one, the basic difference lies in that all characters applied will be in $\Delta_{\nu_{\sigma,\alpha,\beta}}$, for some $\beta>\alpha$, so that $\chi(f)$ is never zero. Let $\underline{\mathfrak{a}}^{(i)}=(\mathfrak{a}_{1}^{(i)},...,\mathfrak{a}_{d}^{(i)})$ be such that $\mathfrak{a}_{j}^{(i)}=\mathfrak{a}_{j}$, for $j\not=i+1$, $\mathfrak{a}_{i+1}^{(i)}=n$. Let $\chi\in\Delta_{\underline{\mathfrak{a}}^{(i)},\underline{m}^{(i)}}$. Since $\chi(\nu_{\sigma_{i+1},n,m})=0=\chi(\mathscr{I}_{\underline{\mathfrak{a}},\underline{m}^{(i)}})$, the congruence (10) yields $\chi(c_{i}\cdot f)=0$, hence $\chi(c_{i})=0$. Lemma 2.1.3(c) says $c_{i}\in\mathscr{I}_{\underline{\mathfrak{a}}^{(i)},\underline{m}^{(i)}}$, and we can write

[TABLE]

with $c_{i+1}^{\prime}\in(\nu_{\sigma_{1},\mathfrak{a}_{1},n},...,\nu_{\sigma_{i,\mathfrak{a}_{i},n}},\nu_{\sigma_{i+2},\mathfrak{a}_{d+2},m},...,\nu_{\sigma_{d},\mathfrak{a}_{d},m})\subset\mathscr{I}_{\underline{\mathfrak{a}},\underline{m}^{(i+1)}}$. Thus, by (10),

[TABLE]

But $\chi(\nu_{\sigma_{i+1},n,m})\not=0$, for all $\chi\in\Delta_{\underline{\mathfrak{a}},\underline{m}^{(i+1)}}$, it follows that $\nu_{n,m}^{(i+1)}\cdot s_{i}-c_{i+1}\cdot f$ is contained in $\mathscr{I}_{\underline{\mathfrak{a}},\underline{m}^{(i+1)}}$. ∎

Remark 2.2.4**.**

If we replace $\mathscr{J}_{n}{\mathfrak{Y}}$, $\mathscr{J}_{m}{\mathfrak{Y}}$ in Definition 2.2.2 by smaller $\mathfrak{I}_{n}$, $\mathfrak{I}_{m}$ satisfying $\nu_{n,m}(\mathfrak{I}_{n})\subset\mathfrak{I}_{m}$, for $m>n$, and let $\dot{\dot{{\mathfrak{Y}}}}_{(n)}$, $\dot{\dot{{\mathfrak{Y}}}}$ be the resulting counterpart of $\ddot{{\mathfrak{Y}}}_{(n)}$, $\ddot{{\mathfrak{Y}}}$, then $\dot{\dot{{\mathfrak{Y}}}}\subset\ddot{{\mathfrak{Y}}}\subset{\mathfrak{Y}}$. Therefore, $\dot{\dot{{\mathfrak{Y}}}}\sim 0$ as well. See the proof of Theorem 1.

2.3. The module $\dot{X}_{K}$

Now we prove Theorem 1. By replacing $k$ by $k_{n}$ if necessary, we may assume that for each $v$ in the ramification locus $S$, there is some integer $e$ such that

[TABLE]

and

[TABLE]

Let $K_{n}$ be the maximal unramified abelian $p$-extension over $k_{n}$ so that by Class Field Theory, $A_{n}:=A_{k_{n}}=\operatorname{Gal}(K_{n}/k_{n})$. For $m>n$, the norm map $A_{m}\longrightarrow A_{n}$ is compatible with the restriction of Galois action $\operatorname{Gal}(K_{m}/k_{m})\longrightarrow\operatorname{Gal}(K_{n}/k_{n})$, we have $X_{K}=\operatorname{Gal}(L/K)$, for $L=\bigcup_{n}K_{n}$. Denote $G:=\operatorname{Gal}(L/k)$. Then $G/X_{K}=\Gamma$ and since $L/K$ is unramified, at every place $v$ of $k$, we have $G_{v}^{0}\simeq\Gamma_{v}^{0}$, a commutative group.

[TABLE]

Suppose $S=\{v_{1},...,v_{s}\}$. For each $j$, choose a place $u_{j}$ of $L$ sitting over $v_{j}$ and then choose a ${\mathbb{Z}}_{p}$-basis $\tilde{\xi}_{1}^{(j)},...,\tilde{\xi}_{d_{j}}^{(j)}$ of $G_{u_{j}}^{0}$. By (11) and (12), we can choose these bases to have the union of their images under $G\longrightarrow\Gamma$ form a tight set $\mathfrak{g}:=\{\tau_{1},...,\tau_{c}\}$ of generators of $\Gamma$. Then among $\mathfrak{g}$, we choose a ${\mathbb{Z}}_{p}$-basis $\{\sigma_{1},...,\sigma_{d}\}$ of $\Gamma$. We lift each $\sigma_{i}$ to some $\tilde{\xi}_{l}^{(j)}$ and denote it by $\tilde{\sigma}_{i}$. Every $g\in G$ can be uniquely written as

[TABLE]

Let $J_{ram}$ denote the ${\mathbb{Z}}_{p}$-submodule of $X_{K}$ generated by

[TABLE]

The commutator $\tilde{\sigma}_{i}\cdot\tilde{\sigma}_{j}\cdot\tilde{\sigma}_{i}^{-1}\cdot\tilde{\sigma}_{j}^{-1}\in J_{ram}$, because $g=\tilde{\sigma}_{i}\cdot\tilde{\sigma}_{j}\cdot\tilde{\sigma}_{i}^{-1}$ in (13) yields

[TABLE]

The canonical action of $\Gamma$ on $X_{K}$ coincides with the conjugation in $G$, namely, if $\gamma\in\Gamma$, $x\in X_{K}$, and $\tilde{\gamma}\in G$ is a lifting of $\gamma$, then

[TABLE]

Hence $\tensor[^{\gamma-1}]x{}$ is the same as the commutator $\tilde{\gamma}\cdot x\cdot\tilde{\gamma}^{-1}\cdot x^{-1}$ in $G$. Set

[TABLE]

Note that $J$ is a $\Gamma$-module, since $\Gamma$ acts trivially on $X_{K}/\mathscr{I}_{0}X_{K}$. Let $\tilde{J}\subset G$ denote the closed subgroup generated by $J$ and $\{\tilde{\sigma}_{1},...,\tilde{\sigma}_{d}\}$. Then for every place $u$ of $L$, the inertia subgroup $G_{u}^{0}\subset\tilde{J}$, because $u=\tensor[^{\gamma}]u{{}_{j}}$ for some $j$, and hence $\tilde{J}$ contains the ${\mathbb{Z}}_{p}$-basis $\tilde{\gamma}\cdot\tilde{\xi}_{i}^{(j)}\cdot\tilde{\gamma}^{-1}$, $i=1,...,d_{j}$, of $G_{u}^{0}$. Thus, $\tilde{J}$ is the closed subgroup of $G$ topologically generated by all commutators and all inertia subgroups of $G$. Therefore, the fixed field $L^{\tilde{J}}$ is the maximal unramified abelian $p$-extension $K_{0}$ of $k$, with $\operatorname{Gal}(K_{0}/k)=A_{0}$. By the uniqueness of the expression (13), we have

[TABLE]

Also, (12) implies $K\cap K_{0}=k$, hence $\operatorname{Gal}(KK_{0}/K)\simeq\operatorname{Gal}(K_{0}/k)$. By (14),

[TABLE]

To apply the above argument to $k_{n}$, let $G^{(n)}$ denote the pre-image of $\Gamma^{(n)}$ under $G\longrightarrow\Gamma$ and set $\tilde{\sigma}_{i,n}:=\tilde{\sigma}_{i}^{p^{n}}$. Every element $g\in G^{(n)}$ can be uniquely expressed as

[TABLE]

Write $\tilde{\xi}_{1,n}^{(j)},...,\tilde{\xi}_{d_{j},n}^{(j)}$ for $(\tilde{\xi}_{1}^{(j)})^{p^{n}},...,(\tilde{\xi}_{d_{j}}^{(j)})^{p^{n}}$. They form a ${\mathbb{Z}}_{p}$-basis of the inertia subgroup of $G_{u_{j}}^{(n)}$. Let $J_{ram}^{(n)}$ denote the ${\mathbb{Z}}_{p}$-submodule of $X_{K}$ generated by

[TABLE]

Set

[TABLE]

Let $\tilde{J}_{n}\subset G$ denote the closed subgroup generated by $J_{n}$ and $\{\tilde{\sigma}_{1,n},...,\tilde{\sigma}_{d,n}\}$.

Lemma 2.3.1**.**

The fixed field of $\tilde{J}_{n}\subset G^{(n)}$ is $K_{n}$. We have

[TABLE]

Furthermore, $J_{n}$ is a $\Gamma$-module, hence a $\Lambda_{\Gamma}$-module.

Proof.

It remains to show that $J_{n}$ is invariant under the action of $\Gamma$, or equivalently $K_{n}/k$ is a Galois extension. But this follows from the fact that $k_{n}/k$ is Galois and $K_{n}$ is the maximal unramified abelian $p$-extension over $k_{n}$. ∎

The next lemma follows [Iwa73, Theorem 7].

Lemma 2.3.2**.**

If $m\geq n$, then $\nu_{n,m}(J_{n})\subset J_{m}$ and that induces the commutative diagram

[TABLE]

Proof.

Since (12) says $K/k$ contains no non-trivial unramified subextension, the restriction of Galois action $X_{K}\longrightarrow A_{n}$ is surjective for every $n$. Let $x_{n}\in A_{n}$ and let $x\in X_{K}$ such that $x\mid_{k_{n}}=x_{n}$. Denote $x\mid_{k_{m}}:=x_{m}$. Let $[\;]_{m}$ denote the Artin map and let $\mathfrak{L}$ be an ideal in ${\mathcal{O}}_{k_{m}}$ with $[\mathfrak{L}]_{m}=x_{m}$. If $\mathfrak{l}=\operatorname{N}_{k_{m}/k_{n}}(\mathfrak{L})$, then $x_{n}=[\mathfrak{l}]_{n}$, hence

[TABLE]

In particular, if $x\in J_{n}$, then the left-hand side is trivial, hence $\nu_{n,m}(x)\in J_{m}$. ∎

Proof of Theorem 1.

By Lemma 2.3.2, we can write for the capitulation kernel

[TABLE]

This makes it possible to apply the technique developed in the previous sections. Because $J\subset X_{K}$ is of finite index, for convenience, we replace $X_{K}$ by $J$. To be more precise, since $\mathscr{I}_{n}\cdot X_{K}\subset J_{n}$, we have $\mathscr{I}_{n}\cdot J\subset J_{n}$, put \dot{J}_{(n,m)}:=\ker(\,\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern 15.63199pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern 0.0pt\offinterlineskip\halign{\entry@#!@&&\entry@@#!@\cr&\crcr}}}\ignorespaces{\hbox{\kern-15.63199pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{\textstyle{J/J_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 19.391pt\raise 5.32639pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.6875pt\hbox{\scriptstyle{\nu_{n,m}}}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 39.63199pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 39.63199pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise 0.0pt\hbox{\textstyle{J/J_{m}}}}}}}}}\ignorespaces}}}}\ignorespaces\,), $\dot{J}_{(n)}:=\bigcup_{m\geq n}\dot{J}_{(n,m)}$, and $\dot{J}:=\varprojlim_{n}\dot{J}_{(n)}$. Then it is sufficient to show $\dot{J}\sim 0$.

Take ${\mathfrak{Y}}:=J$, $\mathfrak{I}_{n}:=J_{n}\cap\mathscr{J}_{n}\cdot{\mathfrak{Y}}$ and let $\dot{\dot{\mathfrak{Y}}}_{(n)}$, $\dot{\dot{\mathfrak{Y}}}$ be as in Remark 2.2.4. The homomorphisms ${\mathfrak{Y}}/\mathfrak{I}_{n}\longrightarrow{\mathfrak{Y}}/\mathscr{J}_{n}\cdot{\mathfrak{Y}}$, for all $n$, yield a homomorphism $\dot{\dot{\mathfrak{Y}}}\longrightarrow\ddot{\mathfrak{Y}}$ fitting into the commutative diagram

[TABLE]

Since the down-arrows are injective, Proposition 2.2.3 implies $\dot{\dot{\mathfrak{Y}}}\sim 0$.

To see the difference between $\mathfrak{I}_{n}$ and $J_{n}$, we first check that for $x\in X_{K}$, the element $\omega_{\sigma_{i},n}\cdot x=\nu_{\sigma_{i},0,n}\cdot\omega_{\sigma_{i},0}\cdot x\in\mathscr{J}_{n}\cdot J$, because $\sigma_{i}\in\mathfrak{g}$ and $\omega_{\sigma_{i},0}\cdot x\in\mathscr{I}_{0}\cdot X_{K}\subset J$. This shows $\mathscr{I}_{n}\cdot X_{K}\subset\mathscr{J}_{n}\cdot J\cap J_{n}=\mathfrak{I}_{n}$. Denote $\tilde{\xi}_{i}^{(j)}=:g$, $\tilde{\xi}_{i,n}^{(j)}=:g_{n}$ and let $\tau\in\mathfrak{g}$ be the image of $g$ under $G\longrightarrow\Gamma$. Let $x=x_{g}\in J_{ram}$ so that $g=\bar{g}\cdot x$ with $\bar{g}=\tilde{\sigma}_{1}^{a_{1}}\cdot\cdots\cdot\tilde{\sigma}_{d}^{a_{d}}$. Then

[TABLE]

Similarly, if $g^{\prime}=\tilde{\gamma}\cdot g\cdot\tilde{\gamma}^{-1}=\bar{g}\cdot x^{\prime}$, with $x^{\prime}=x_{g^{\prime}}\in J_{ram}$, and $g^{\prime}_{n}=\tilde{\gamma}\cdot g_{n}\cdot\tilde{\gamma}^{-1}$, then

[TABLE]

Let $\tilde{\rho}:=(\tilde{\sigma}_{1,n}^{a_{1}}\cdot\cdots\cdot\tilde{\sigma}_{d,n}^{a_{d}})^{-1}\bar{g}^{p^{n}}$ and write $X_{K}$ additively. Then, since $x_{g_{n}},x_{g_{n}^{\prime}}\in J_{ram}^{(n)}$,

[TABLE]

Therefore, we have shown that the ${\mathbb{Z}}_{p}$-module $J_{n}/\mathfrak{I}_{n}$ is generated by the classes of $x_{g_{n}}$, with $g=\tilde{\xi}_{i}^{(j)}$, $j=1,...,s$, $i=1,...d_{j}$. Then we observe that $x_{g_{n}}$ is trivial if the above $\tau=\sigma_{i}$. This shows the $p$-rank of $J_{n}/\mathfrak{I}_{n}$ is at most $c-d$, which equals [math] if $d=1$. The exact sequence

[TABLE]

gives rise to the exact sequence

[TABLE]

with $p$-ranks of both $\mathfrak{a}_{n}$ and $\mathfrak{b}_{n}$ bounded by $c-d$. Consequently, the cokernel of the induced map $\dot{\dot{\mathfrak{Y}}}\longrightarrow\dot{J}$ is of $p$-ranks bounded by $c-d$. Hence $\dot{J}$ is pseudo-null and the theorem is proved for $\dot{X}_{K}$. The corresponding assertion for $\dot{X}_{K}^{\prime}$ can be proved by similar approach, of which we only give a sketch. Write

[TABLE]

where ${\mathcal{P}}_{1}$ consists of place $v$ such that $\Gamma_{w}^{0}\not=\Gamma_{w}$. For each $v\in{\mathcal{P}}$, choose a place $u$ of $L$ above $v$ and denote $w=u\mid_{{}_{K}}$, if $v\in S$, let $u$ be as before. If $v\in{\mathcal{P}}_{1}$, then by (11), we have non-canonically $\Gamma_{w}=Z_{p}\times\Gamma_{w}^{0}$, whence $\Gamma_{w}\simeq G_{u}$ and $w$ splits completely over $L$. In this case, choose an $\eta_{u}\in G_{u}$ representing a topological generator of $G_{u}/G_{u}^{0}$ such that the map $G\longrightarrow\Gamma$ sends $\eta_{u}$ into $\mathfrak{g}$. Write $\eta_{u,n}$ for $\eta_{u}^{p^{n}}$ and let $J_{un}^{(n)}$ denote the $Z_{p}$-module generated by

[TABLE]

If $v\in{\mathcal{P}}_{2}$, then we have canonically $G_{u}\simeq\operatorname{Gal}(L_{u}/K_{w})\times\Gamma_{w}$ with $\operatorname{Gal}(L_{u}/K_{w})\subset X_{K}$, $G_{u}^{0}\simeq\Gamma_{w}$. Let $[X_{K}]$ be the $\Gamma$-submodule of $X_{K}$ generated by $\operatorname{Gal}(L_{u}/K_{w})$, for all $v\in{\mathcal{P}}_{2}$. Set

[TABLE]

let $\widetilde{\{J_{n}\}}$ denote the closed subgroup of $G$ generated by $\{J_{n}\}$ and $\{\tilde{\sigma}_{1,n},...,\tilde{\sigma}_{d,n}\}$, and let $K_{n}^{\prime}$ be the fixed field of $\widetilde{\{J_{n}\}}$. Then $K_{n}^{\prime}$ is the maximal unramified $p$-extension of $k_{n}$ with all places above $p$ splitting completely, so $\operatorname{Gal}(K_{n}^{\prime}/k_{n})=A_{n}^{\prime}$. Let $[A]_{n}$ be the image of $\{J_{n}\}$ under $X_{K}\longrightarrow A_{n}=X_{K}/J_{n}$. Then the exact sequences

[TABLE]

and the fact that

[TABLE]

yield $X_{K}^{\prime}=X_{K}/[X_{K}]$. Let $J_{n}^{\prime}$ be the image of $\{J_{n}\}$ under $X_{K}\longrightarrow X_{K}^{\prime}$. Then $(X^{\prime}_{K},J_{n}^{\prime},K_{n}^{\prime})$-version of Lemma 2.3.1 holds, namely,

[TABLE]

Again by the calss field theory, the $(X^{\prime}_{K},J^{\prime}_{n},A^{\prime}_{n})$-version of Lemma 2.3.2 holds. Then take ${\mathfrak{Y}}^{\prime}:=J_{0}^{\prime}$, $\mathfrak{I}_{n}^{\prime}:=J_{n}^{\prime}\cap\mathscr{J}_{n}\cdot{\mathfrak{Y}}^{\prime}$ and proceed as above. ∎

3. Cohomology groups of global units

The proof of Theorem 2 involves cohomology of unit groups. Denote the group of global units of $E$ by $U_{E}:={\mathcal{O}}_{E}^{*}$ and put $U_{K}:=\bigcup_{E}U_{E}$. The cohomology groups $\operatorname{H}^{i}(\Gamma,U_{K})$, $i=1,2$, has been studied in [Iwa83, Yam84] for $d=1$ case. We are going to show that for general $d$, they are co-finitely generated ${\mathbb{Z}}_{p}$-modules. Put

[TABLE]

and

[TABLE]

where $\bar{E}^{ab,p}$ is the maximal pro-$p$ abelian extension of $E$, unramified outside $p$.

3.1. An exact sequence

Let $\operatorname{Div}_{E}$ and $P_{E}$ denote the groups of divisors and principal divisors of $E$. In [Iwa83, Proposition 1], Iwasawa deduces the seven-term exact sequence:

[TABLE]

with $G:=\operatorname{Gal}(E/k)$ and $b_{E}:=\ker(\operatorname{H}^{2}(G,E^{*})\longrightarrow\operatorname{H}^{2}(G,\operatorname{Div}_{E}))$.

The exact sequence identifies $\operatorname{H}^{1}(G,U_{E})$ with the kernel of $\alpha_{E/k}$. The restriction of $\alpha_{E/k}$ to $C_{k}\subset\operatorname{Div}_{E}^{G}/P_{k}$ is the capitulation homomorphism $c_{E/k}$. Therefore, we have the exact sequence

[TABLE]

If $v\in S$ and $w$ is a place of $E$ sitting over $v$, then the sum

[TABLE]

is fixed by $G$. Set $g_{E/k,v}:=|G_{v}^{0}|$. In $\operatorname{Div}_{E}$, we have $g_{E/k,v}\cdot\epsilon_{v}=v$. The sequence

[TABLE]

with $l_{E/k}$ the projection onto the $S_{E}$-component, is exact. Combining (17) and (18), we obtain the exact sequence

[TABLE]

By taking the injective limit of the above sequence, one can show that $\operatorname{H}^{1}(\Gamma,U_{K})$ is co-finitely generated over ${\mathbb{Z}}_{p}$. As for $\operatorname{H}^{2}(\Gamma,U_{K})$, we have the following. Denote

[TABLE]

Lemma 3.1.1**.**

The following statements are equivalent:

1. (a)

The $\Lambda_{\Gamma}$-module $W_{K}$ is finitely generated. 2. (b)

The abelian group $A_{K}^{\Gamma}$ has finite $p$-rank. 3. (c)

The abelian group $\operatorname{H}^{2}(\Gamma,U_{K})$ has finite $p$-rank.

Proof.

Since $A_{K}^{\Gamma}[p]$ is the Pontryagin dual of $W_{K}/(p\Lambda_{\Gamma}+\mathscr{I}_{\Gamma})W_{K}$, the equivalence between (a) and (b) follows from Nakayama’s Lemma.

Denote $Q_{E}:=\operatorname{Im}(\alpha_{E/k})\cap A_{E}^{\Gamma}$. Then (16) induces the exact sequence

[TABLE]

Now, the $p$-rank of $Q_{K}$ is $\leq$ the $p$-rank of $A_{k}$ plus the cardinality of $S$. Also, if $v\not\in S$, then the composition of \textstyle{\operatorname{H}^{2}(\Gamma,U_{K})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\textstyle{\operatorname{H}^{2}(k,K^{*})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{H}^{2}(k_{v},K_{v}^{*})} is the trivial homomorphism. Hence $\operatorname{Im}(\beta)$ is embedded into $\prod_{v\in S}{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}$ by the local invariant maps and is of $p$-rank bounded by the cardinality of $S$. Therefore, (b) and (c) are equivalent. ∎

3.2. The structure of $\mathcal{U}_{K}^{\vee}$

Recall that $U_{K}$ denotes the group of global units of $K$ and $\mathcal{U}_{K}^{\vee}=\operatorname{Hom}({\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}\otimes U_{K},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})$.

Let $\mathcal{V}_{E}$ denote the group of ${\mathcal{P}}_{E}$-units of $E$. Kummer’s theory yields the commutative diagram

[TABLE]

Since $E^{*}/{\mathcal{V}}_{E}$ and ${\mathcal{V}}_{E}/U_{E}$ are torsion free, arrows in the upper row of the diagram are injective. Let $n$ goes to $\infty$. We obtain the commutative diagram

[TABLE]

Lemma 3.2.1**.**

The restriction map \textstyle{\operatorname{H}^{1}(\mathfrak{M}_{E},\mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{\infty}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{E}}$$\textstyle{\operatorname{H}^{1}(\mathfrak{M}_{K},\mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{\infty}})^{\Gamma_{E}}} has finite kernel and cokernel.

Proof.

[Tan10, Lemma 3.2.1]. ∎

Lemma 3.2.2**.**

*The cokernel of \textstyle{{\mathcal{V}}_{E}\otimes{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{H}^{1}(\mathfrak{M}_{E},\mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{\infty}})}

is finite.*

Proof.

We show that as $n\rightarrow\infty$, the corkernel of

[TABLE]

remains bounded. By Kummer’s theory, an element of \operatorname{H}^{1}(\mathfrak{M}_{E},\mbox{\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{\color[rgb]{1,1,1}.}}\raisebox{3.7889pt}{\color[rgb]{1,1,1}.}\hskip 4.60007pt}{}_{p^{n}}) is represented by an $f\in E^{*}$ modulo $(E^{*})^{p^{n}}$ such that $p^{n}\mid\operatorname{ord}_{v}(f)$ for every $v\notin{\mathcal{P}}_{E}$. Let $w_{1},...,w_{r}$ be a set of generators of $A_{E}$ and let $p^{m}$ be the exponent of $A_{E}$ so that each $w_{i}^{p^{m}}=(f_{i})$ for some $f_{i}\in E^{*}$. If $n>m$, then modulo $(E^{*})^{p^{n}}$, $f$ can be expressed as a product of powers of $f_{1}^{p^{n-m}},...,f_{r}^{p^{n-m}}$ together with elements of ${\mathcal{V}}_{E}$. Hence the cokernel in question has order bounded by $p^{mr}$. ∎

Corollary 3.2.3**.**

The natural map $\mathcal{U}_{E}\longrightarrow\mathcal{U}_{K}^{\Gamma_{E}}$ has finite kernel, its cokernel is cofinitely generated over ${\mathbb{Z}}_{p}$ of corank bounded by $|S_{E}|$.

Proof.

The exact sequence

[TABLE]

where $\tilde{\mho}_{E}$ is defined by taking valuations at all $v\in{\mathcal{P}}_{E}$ induces the commutative diagram of complexes

[TABLE]

Write $Q_{E}:=\ker(\mho_{E})$, $R_{E}:=\ker(\mho_{K}\circ\beta_{E})$. Then $Q_{E}/\mathcal{U}_{E}$ is finite. In view of the diagram (20) and Lemma 3.2.1, 3.2.2, we need to show that the cokernel of

\textstyle{Q_{E}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R_{E}} has corank bounded by $|S_{E}|$. If $w$ is a place of $K$ sitting over $v$ of $E$ with $v\not\in S_{E}$, then $w$ is unramified under $K/k$, and hence the map

[TABLE]

is injective. Therefore, we have the exact sequence

[TABLE]

∎

Lemma 3.2.4**.**

Let $\mathfrak{Q}_{K}$ be a finitely generated $\Lambda_{\Gamma}$-module. If $\mathfrak{Q}_{K}$ is of rank $r$ over $\Lambda_{\Gamma}$, then $\mathfrak{Q}_{K}/\mathscr{I}_{n}\mathfrak{Q}_{K}$ has ${\mathbb{Z}}_{p}$-rank

[TABLE]

Proof.

There exists an exact sequence

[TABLE]

Since $T$ is torsion, [Tan14, Lemma 1.13] says $T/\mathscr{I}_{n}T$ has ${\mathbb{Z}}_{p}$-rank $=O(p^{(d-1)n})$. Hence

[TABLE]

To obtain an inequality in the other direction, we use an exact sequence

[TABLE]

with $T^{\prime}$ torsion. ∎

Let $r_{1}$ and $r_{2}$ denote the number of real and complex places of $K$.

Proposition 3.2.5**.**

The $\Lambda_{\Gamma}$-module $\mathcal{U}_{K}^{\vee}$ is finitely generated, of rank $r_{1}+r_{2}$.

Proof.

Since $|S_{k_{n}}|=O(p^{(d-1)n})$, Corollary 3.2.3 says $\mathcal{U}_{K}^{\vee}/\mathscr{I}_{n}\mathcal{U}_{K}^{\vee}$, which is the Pontryagin dual of $\mathcal{U}_{K}^{\Gamma^{(n)}}$, has ${\mathbb{Z}}_{p}$-rank equals $(r_{1}+r_{2})\cdot p^{dn}+O(p^{(d-1)n})$. In particular, $\mathcal{U}_{K}^{\vee}/(p\Lambda_{\Gamma}+\mathscr{I}_{\Gamma})\mathcal{U}_{K}^{\vee}$ is finite. Hence $\mathcal{U}_{K}^{\vee}$ is finitely generated. By Lemma 3.2.4, the rank of $\mathcal{U}_{K}^{\vee}$ equals $r_{1}+r_{2}$. ∎

3.3. The module $W_{K}$

Let $K^{\prime}/k$ be an intermediate ${\mathbb{Z}}_{p}^{d-1}$-extension of $K/k$ with Galois group $\Gamma^{\prime}$. We shall fix a ${\mathbb{Z}}_{p}$-extension $k^{(\infty)}/k$ linearly disjoint from $K^{\prime}/k$ so that $K=K^{\prime}k^{(\infty)}$ and $K^{\prime}\cap k^{(\infty)}=k$. Let $k^{(n)}$ denote the $n$th layer of $k^{(\infty)}/k$. From now on, the symbol $F$ will denote a finite intermediate extension of $K^{\prime}/k$, while $E$ denotes a finite intermediate extension of $K/k$ such that $E\cap K^{\prime}=F$ and $E=Fk^{(n)}$ for some $n$. Put $F^{(\infty)}:=Fk^{(\infty)}$. Denote $\Psi=\operatorname{Gal}(K/K^{\prime})$ and let $\psi$ be a topological generator of $\Psi$.

[TABLE]

For each $F$ and for $m\geq n$, we have the commutative diagram

[TABLE]

where the left down-arrow is induced by the natural map ${\mathbb{Z}}[\frac{1}{p^{n}}]\longrightarrow{\mathbb{Z}}[\frac{1}{p^{m}}]$ and the right down-arrow is the inflation map. This yields

[TABLE]

Proof of Theorem 2.

Since $W_{K}^{\prime}\subset W_{K}$, it is sufficient to treat $W_{K}$. Lemma 3.1.1 says the theorem is equivalent to the assertion that $\operatorname{H}^{2}(\Gamma,U_{K})$ has finite $p$-rank. For $d=1$, the value of this $p$-rank is shown in [Iwa83, Yam84], the equality (21) also implies it is bounded by the rank of $U_{k}$. We prove the theorem by induction on $d$.

By (16), we have the exact sequence

[TABLE]

where $\dot{C}_{E/F}$ is generated by $\operatorname{Div}_{E}^{\Psi}$. Denote $\mathcal{C}_{K}:=\varinjlim C_{E}$. Let $E$ goes to $K$ and take the direct limit to obtain the exact sequence

[TABLE]

Since the $p$-part of $\mathcal{C}_{K}^{\Gamma}$ is Pontryagin dual to $W_{K}/\mathscr{I}_{\Gamma}W_{K}$, it remains to show both $\dot{\mathcal{C}}_{K/K^{\prime}}^{\Gamma^{\prime}}$ and $\operatorname{H}^{2}(\Psi,U_{K})^{\Gamma^{\prime}}$ are of finite $p$-ranks.

The surjection (21) says $\operatorname{H}^{2}(\Psi,U_{K})$ is a quotient of $\mathcal{U}_{K^{\prime}}$. Hence its Pontryagin dual, denoted $\mathfrak{J}_{K^{\prime}}$, is a $\Lambda_{\Gamma^{\prime}}$ submodule of $\mathcal{U}_{K^{\prime}}^{\vee}$. Proposition 3.2.5 implies $\mathfrak{J}_{K^{\prime}}$ is $\Lambda_{\Gamma^{\prime}}$ finitely generated. Then $\operatorname{H}^{2}(\Psi,U_{K})^{\Gamma^{\prime}}$, being the Pontryagin dual of $\mathfrak{J}_{K^{\prime}}/\mathscr{I}_{\Gamma^{\prime}}\mathfrak{J}_{K^{\prime}}$, must have finite $p$-rank.

Let $T\subset S$ be the subset consisting of places $v$ with $\Gamma_{v}^{0}$ of rank $d$. We choose $K^{\prime}$ such that $K/K^{\prime}$ is unramified outside $T_{K^{\prime}}$, which is a finite set. Apply (18) to the case where $k=F$ and then let $E$ goes to $K$ to obtain the exact sequence

[TABLE]

where $\bar{\mathcal{C}}_{K^{\prime}}$ is the image of the capitulation homomorphism $\mathcal{C}_{K^{\prime}}\longrightarrow\mathcal{C}_{K}$ and $V$ is a quotient of $\prod_{w\in T_{K^{\prime}}}{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}\cdot w$. Obviously, the $p$-rank of $V$ is finite. Now $\bar{\mathcal{C}}_{K^{\prime}}^{\vee}$ is a $\Lambda_{\Gamma^{\prime}}$ submodule of $W_{K^{\prime}}$, whence finitely generated by the induction hypothesis. Consequently, $\bar{\mathcal{C}}_{K^{\prime}}^{\Gamma^{\prime}}$, the Pontryagin dual of $\bar{\mathcal{C}}_{K^{\prime}}^{\vee}/\mathscr{I}_{\Gamma^{\prime}}\bar{\mathcal{C}}_{K^{\prime}}^{\vee}$, must have finite $p$-rank. The above exact sequence implies the $p$-rank of $\dot{\mathcal{C}}_{K/K^{\prime}}^{\Gamma^{\prime}}$ is finite. ∎

Write $\mathfrak{a}_{E}:=A_{E}$, $\mathfrak{b}_{E}:=\operatorname{Hom}(A_{E},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p})$. For $E\subset E^{\prime}\subset K$, let $\mathfrak{r}^{E^{\prime}}_{E}:\mathfrak{b}_{E}\longrightarrow\mathfrak{b}_{E^{\prime}}$ and $\mathfrak{k}^{E^{\prime}}_{E}:\mathfrak{b}_{E^{\prime}}\longrightarrow\mathfrak{b}_{E}$ denote respectively the restriction and the corestriction. Extend them to homomorphisms

[TABLE]

and

[TABLE]

such that the restrictions to the first factors are respectively the capitulation and the norm. Then

[TABLE]

the norm map of $\operatorname{Gal}(E^{\prime}/E)$-action. Let $\langle\;,\;\rangle_{E}:\mathfrak{a}_{E}\times\mathfrak{b}_{E}\longrightarrow{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}$ be the natural pairing. For $(a,b)\in\mathfrak{a}_{E}\times\mathfrak{b}_{E}$, $(a^{\prime},b^{\prime})\in\mathfrak{a}_{E^{\prime}}\times\mathfrak{b}_{E^{\prime}}$, we have

[TABLE]

Thus, in the terminology of [LLTT18, §3.1.1], the collection

[TABLE]

form a complete $\Gamma$-system. Indeed, it is a T-system [op. cit.], because for every intermediate ${\mathbb{Z}}_{p}^{e}$-extension $M/k$ of $K/k$, $d\geq e\geq 1$, if

[TABLE]

then $\mathfrak{a}_{M}\times\mathfrak{b}_{M}$ is finitely generated torsion over the Iwasawa algebra of $M/k$.

Proof of Theorem 3.

In view of [LLTT18, Theorem 1], for proving $W_{K}\sim X_{K}^{\sharp}$, we need to show the system $\mathfrak{A}$ is pseudo-control, in the sense that

[TABLE]

is pseudo-null. That $\dot{\mathfrak{a}}_{K}\sim 0$ is by Theorem 1. To show $\dot{\mathfrak{b}}_{K}\sim 0$, we may assume that $K/k$ has no nontrivial unramified extension. In this case \textstyle{\mathfrak{b}_{E}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathfrak{r}^{E^{\prime}}_{E}}$$\textstyle{\mathfrak{b}_{E^{\prime}}} is injective, whence $\dot{\mathfrak{b}}_{K}=0$.

The assertion $W^{\prime}_{K}\sim X_{K}^{\prime\sharp}$ is proved similarly. We first form the $\Gamma$-system $\mathfrak{A}^{\prime}$ with $\mathfrak{a}_{E}^{\prime}=A_{E}^{\prime}$, $\mathfrak{b}_{E}^{\prime}=\operatorname{Hom}(\mathfrak{a}_{E}^{\prime},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{P})$. Since $\mathfrak{a}_{M}^{\prime}$ is quotient of $\mathfrak{a}_{M}$, $\mathfrak{b}_{M}^{\prime}\subset\mathfrak{b}_{M}$, $\mathfrak{A}^{\prime}$ is a $T$-system. Theorem 1 says $\dot{\mathfrak{a}}_{K}^{\prime}\sim 0$, and since $\dot{\mathfrak{b}}_{K}^{\prime}\subset\dot{\mathfrak{b}}_{K}\sim 0$, $\mathfrak{A}^{\prime}$ is pseudo-controlled. Then apply [LLTT18, Theorem 1]. ∎

As for (1), form the $\Gamma$-system $\mathfrak{A}_{\chi}=\{\mathfrak{a}_{\chi,E},\mathfrak{b}_{\chi,E},\mathfrak{r}_{E}^{E^{\prime}},\mathfrak{k}_{E}^{E^{\prime}}\}$ with $\mathfrak{a}_{\chi,E}=(A_{E})_{\chi^{-1}}$, $\mathfrak{b}_{E,\chi}=\operatorname{Hom}(\mathfrak{a}_{\chi,E},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{P})$, it is a pseudo-controlled $T$-system, and hence the first pseudo-isomorphism follows. The second is proved similarly.