\section{Introduction}\label{In}
Let $p$ be a prime number and $k$ a number field.
In Iwasawa theory, one studies the arithmetic of the multiple $\mathbb{Z}_p$-extensions over $k$,
which are galois groups being topologically
isomorphic to direct products of copies of the additive group $\mathbb{Z}_p$ of $p$-adic integers.
In this paper, we consider the maximal multiple $\mathbb{Z}_p$-extension field $\widetilde{k}$ of $k$.
By class field theory, there exists a positive integer $r$ with $r\geq r_2(k)+1$ such that
$\mathrm{Gal}(\widetilde{k}/k)$ is isomorphic to $\mathbb{Z}_p^{\oplus r}$, where
$r_2(k)$ is the number of complex places of $k$.
Let $L_{\widetilde{k}}$
be the maximal unramified pro-$p$ abelian extension field of $\widetilde{k}$.
We denote the galois group $\mathrm{Gal}(L_{\widetilde{k}}/\widetilde{k})$
by $X_{\widetilde{k}}$ and
call it the Iwasawa module of $\widetilde{k}/k$.
Iwasawa \cite{Iw} and Greenberg \cite{Gre} proved that Iwasawa modules are finitely generated torsion
$\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-modules.
This module has a very important role in Iwasawa theory.
The usual form of Iwasawa's main conjecture states that
Iwasawa modules, regarded as algebraic objects,
have a relation with analytic objects;
roughly speaking, the characteristic ideal of a Iwasawa module
should coincide with the ideal generated by a certain $p$-adic $L$-function.
Under several assumptions,
Mazur-Wiles \cite{M-W} and Wiles \cite{Wiles} proved this conjecture
for the cyclotomic $\mathbb{Z}_p$-extensions of real abelian fields and
of totally real number fields, respectively.
\par
Concerning the structure of Iwasawa modules,
Greenberg formulated a conjecture which is called Greenberg's generalized conjecture (or GGC, for short):
\begin{GGC}[\cite{Gre1}, Greenberg's generalized conjecture (GGC)]
\begin{rm}
For each prime number $p$ and number field $k$,
the $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-module $X_{\widetilde{k}}$ is pseudo-null,
in other words, the height of the annihilator ideal
$\mathrm{Ann}_{\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]}(X_{\widetilde{k}})$ is greater than one.
\end{rm}
\end{GGC}
Here a $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-module is said to be pseudo-null
if there exist two relatively prime annihilators.
If $k$ is a totally real number field and if Leopoldt's conjecture holds for $p$ and $k$,
then this conjecture is equivalent to the usual form of Greenberg's conjecture,
which states that the Iwasawa module is finite (\cite{Gre2}).
Furthermore, by the structure theorem (see Section \ref{Preli}),
the usual form of Greenberg's conjecture holds
if and only if
the Iwasawa $\lambda$- and $\mu$-invariants of the
cyclotomic $\mathbb{Z}_p$-extension over $k$ vanish.
\par
In this paper, we consider a weaker form of this conjecture.
We call the following conjecture weak Greenberg's generalized conjecture (or weak GGC, for short):
\begin{WGGC}[Weak Greenberg's generalized conjecture (weak GGC)]
\begin{rm}
Let $p$ be a prime number and $k$ a number field.
Assume that $X_{\widetilde{k}}$ is not trivial.
Then $X_{\widetilde{k}}$ has a non-trivial pseudo-null $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-submodule.
\end{rm}
\end{WGGC}
We note that weak GGC is proposed by Nguyen Quang Do \cite{Quang1, Quang2} for totally real number fields
and by Wingberg  \cite{Wing} for arbitrary number fields.
In a very recent paper,
Kataoka \cite{Kata20} proved that
$X_{\widetilde{k}}$ has no non-trivial finite $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-submodule
for imaginary quadratic fields in which $p$ splits.
\par
Throughout this paper, we suppose that $p$ is an odd prime number and that $k$ is an imaginary quadratic field.
We assume that $p$ splits in $k$ into $\mathfrak{p}$ and $\mathfrak{p}^{\ast}$.
Under this assumption, there exists a uniquely defined $\mathbb{Z}_p$-extension $N_{\infty}/k$ (respectively, $N_{\infty}^{\ast}/k$) 
such that the prime ideal $\mathfrak{p}^{\ast}$ (respectively, $\mathfrak{p}$) does not ramify.
We also denote by $k_{\infty}^c$ the cyclotomic $\mathbb{Z}_p$-extension field of $k$.\par
To state our main theorem,
we introduce the notion of $p$-split $p$-rational fields.
\begin{Defn}\label{rational}
For an imaginary quadratic field $k$ in which $p$ splits into $\mathfrak{p}$ and $\mathfrak{p}^{\ast}$,
$k$ is said to be $p$-split $p$-rational
if $k$ satisfies that $L_k \subset \widetilde{k}$, $\widetilde{k}^{\mathfrak{D_p}} \neq k$,
and that $\widetilde{k}^{\mathfrak{D_p}} \subset L_k$,
where $\mathfrak{D_p}$ is the decomposition group of the prime $\mathfrak{p}$
in $\mathrm{Gal}(\widetilde{k}/k)$,
$\widetilde{k}^{\mathfrak{D_p}}$ is the fixed field of $\widetilde{k}$ by $\mathfrak{D_p}$,
and $L_k$ is the $p$-Hilbert class field of $k$.
\end{Defn}
If we suppose that
$\widetilde{k}^{\mathfrak{D_p}} \subset L_k$, then
$\mathfrak{D}_{\mathfrak{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$  (Lemma \ref{cond normal}).
Hence we have ${\mathfrak{D_p}}={\mathfrak{D_{p^{\ast}}}}$.
Therefore, the definition of a 
$p$-split $p$-rational field does not depend on the choice of $\mathfrak{p}$.
If $k$ is $p$-split $p$-rational, the $p$-Sylow subgroup of the ideal class group of $k$ is cyclic
from the assumption that $L_k \subset \widetilde{k}$.
In Section \ref{B}, we will give a necessary and sufficient condition for $k$ to be $p$-split $p$-rational (Proposition \ref{eqrational}).\par
In this paper, we prove the following
\begin{thm}\label{main thm}
Let $p$ be an odd prime number and $k$ an imaginary quadratic field
which is not $p$-split $p$-rational.
Assume the following conditions:\par
$(\mathrm{i})$ The Iwasawa $\lambda$-invariant of $N_{\infty}/k$ is zero.\par
$(\mathrm{ii})$ The characteristic ideal of $X_{k_{\infty}^c}$
has a generator which is square-free.\\
Then weak {\rm{GGC}} holds for $p$ and $k$.
\end{thm}
On the $\mathbb{Z}_p$-extension $N_{\infty}/k$, it is known that the Iwasawa $\mu$-invariant of $N_{\infty}/k$ is zero.
This was proved by Gillard \cite{Gi} and Schneps \cite{sch} for $p \geq 5$ and
Oukhaba-Vigui\'{e} \cite{OV} for $p=2, 3$.
By \cite[Proposition 1.C]{Mi},
the Iwasawa $\lambda$-invariant of $N_{\infty}/k$ is zero if and only if
every ideal class of the $p$-part of the ideal class group of $k$
becomes principal in $N_{\infty}$. 
We note that vanishing of the Iwasawa $\lambda$-invariant of $N_{\infty}/k$ here is similar to that in the usual form of Greenberg's conjecture
in the case where $\mathfrak{p}$ is totally ramified in $N_{\infty}/k$.
No counter examples of the assumption (i) have been found yet.
Fukuda-Komatsu \cite{F-K1, F-K2} checked that the Iwasawa $\lambda$-invariants of $N_{\infty}/k$ vanish
for some imaginary quadratic fields with $p=3$ 
under the assumption that $\mathfrak{p}$ is totally ramified in $N_{\infty}/k$.
On the assumption (ii) in Theorem \ref{main thm},
we note that we do not need this condition in the case where the Iwasawa module
$X_{\widetilde{k}}$ is cyclic as a $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-module (Remark \ref{frem}).
\par
There are two key new ideas of the proof of this theorem.
One is to find an annihilator of $X_{\widetilde{k}}$,
which is introduced in \cite[Lemma $3.3$]{Mu}.
We fix this annihilator and denote it by $f(S,T)$
(Lemma \ref{f(S,T)}).
Using this power series, we can 
prove that the Iwasawa module $X_{\widetilde{k}}$ is pseudo-isomorphic to
a certain $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-cyclic module,
provided that weak GGC does not hold for $p$ and $k$
(Corollary \ref{psisom}).
The other is  to consider a certain subfield of $L_{\widetilde{k}}$, which is denoted by $M_{\mathfrak{p}^{\ast}}(N_{\infty})$.
From the $\mathbb{Z}_p$-rank of $\mathrm{Gal}(M_{\mathfrak{p}^{\ast}}(N_{\infty})/\widetilde{k})$,
we consider two inequalities (A) and (B), which are stated in the end of Section \ref{decomp}.
These inequalities contradict each other, which imply Theorem \ref{main thm}.
We will prove them,
assuming that weak GGC does not hold for $p$ and $k$.
\par
In the case where $k$ is a $p$-split $p$-rational field, we know that the Iwasawa $\lambda$-invariant of $N_{\infty}/k$ is zero
by genus formula (see for example \cite[Remark $3.2$]{Mu}).
Furthermore, we see that $X_{\widetilde{k}}$ is cyclic as a $\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/k)]]$-module
(\cite[Proposition 3.8]{Mu}).
To treat this case, we consider the $\mathbb{Z}_p$-extension $N_{\infty} N_{s+1}^{*}/ N_{s+1}^{*}$,
where $s$ is the positive integer satisfying $p^s = [L_k :k]$ and $N_{s+1}^{*}$ is the $(s+1)$-st layer of $N_{\infty}^{*}/k$.
We put $H= N_{s+1}^{*}$ and $H_{\infty}=N_{\infty} N_{s+1}^{*}$.
Then $H_{\infty}/H$ is a $\mathbb{Z}_p$-extension unramified outside all prime ideals of $H$ lying above $\mathfrak{p}$.
We prove the following theorem
under the assumption that the Iwasawa $\lambda$-invariant for $H_{\infty}/ H$ vanishes, which is similar to that
in the usual form of Greenberg's conjecture mentioned above.
\begin{thm}\label{main thm 2}
Let $p$ be an odd prime number and $k$ an imaginary quadratic field
which is $p$-split $p$-rational.
Suppose that the Iwasawa $\lambda$-invariant of $H_{\infty}/ H$ is zero.
Then weak {\rm{GGC}} holds for $p$ and $k$.
\end{thm}
There is a relation between weak GGC and GGC.
Fujii verified that weak {\rm{GGC}} for $p$ and $k$ implies {\rm{GGC}} for $p$ and $k$
if the characteristic ideal of
the $\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)$-coinvariant module of $X_{\widetilde{k}}$
as a $\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}^c/k)]]$-module
is a prime ideal (\cite[Proposition $3.1$]{Fuj10}).
Hence we obtain the following corollary from Theorems \ref{main thm} and \ref{main thm 2}.
\begin{cor} \label{cor GGC}
Assume either the assumption of Theorem \ref{main thm} or that of Theorem \ref{main thm 2} holds.
Assume also that the characteristic ideal of
the $\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)$-coinvariant module of $X_{\widetilde{k}}$
as a $\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}^c/k)]]$-module
is a prime ideal.
Then {\rm{GGC}} holds for $p$ and $k$.
\end{cor}
Furthermore, we apply Theorem \ref{main thm} and Corollary \ref{cor GGC}
to imaginary quadratic fields
$k=\mathbb{Q}(\sqrt{-d})$.
Fujii and Shirakawa gave criteria of  GGC
for imaginary quadratic fields (\cite[Theorem 5.1]{Fuj10}, \cite[Theorem 5.2]{Shi}).
If $1< d < 10^3$
and $d \equiv 2 ~\mathrm{mod}~3$,
they verified the conjecture
using the first and the second layers of a certain $\mathbb{Z}_p$-extension
except for $d=971$.
Combining Corollary \ref{cor GGC} with Fujii's criterion,
we can verify the conjecture including the case of $d=971$.
We will give more precise calculation in Section \ref{example}.
Therefore we obtain the following
\begin{cor} \label{thm GGC}
Let $d$ be a positive integer and
$k=\mathbb{Q}(\sqrt{-d})$ an imaginary quadratic field in which $p$ splits.
If $1< d < 10^3$ and $d \equiv 2 ~\mathrm{mod}~3$, then
{\rm{GGC}} holds for $p=3$ and $k$.
\end{cor}
Concerning GGC, the following is known.
Let $p$ be an odd prime number.
Minardi \cite[Proposition $3.\mathrm{B}$]{Mi} verified GGC for an imaginary quadratic field $k$ in which $p$ splits
under the assumption that
the class number of $k$ is not divisible by $p$.
This result is generalized by Itoh \cite{Ito} and Fujii \cite{Fuj17}.
They verified the conjecture for imaginary abelian quartic fields (\cite{Ito})
and for CM-fields of degree greater than or equal to $4$ (\cite{Fuj17})
under
the assumption that $p$ splits completely in the base field and that
the class number of the base field is not divisible by $p$ , and
that all Iwasawa invariants of the cyclotomic $\mathbb{Z}_p$-extension over the maximal totally real subfield of the base field
are trivial.
Using a method of Minardi in \cite{Mi},
Ozaki \cite[Theorem B]{Oz}
proved GGC under the assumption that
at least one prime ideal lying above $p$ does not split in $N_{\infty}$
and that the Iwasawa $\lambda$-invariant of $N_{\infty}/k$ is zero
for an imaginary quadratic field $k$ in which $p$ splits.
GGC has some important consequences.
For example, if this conjecture holds for a totally real number field $k$, then
this conjecture gives a simple proof of
the usual form of Iwasawa's main conjecture for $p$ and $k$ (\cite{Gre2}).
Ozaki \cite{Oz} and Kataoka \cite{Kata17} studied the behavior of Iwasawa $\lambda$- and $\mu$-invariants
of infinitely many $\mathbb{Z}_p$-extensions over a fixed number field.
They determined many Iwasawa $\lambda$- and
$\mu$-invariants of $\mathbb{Z}_p$-extensions
under the assumption that GGC holds for the base field.
\par
The outline of this paper is as follows.
In Section \ref{Preli}, we provide some basic definitions and notation.
In Section \ref{decomp}, we study the decomposition of the prime number $p$
in $\widetilde{k}/\mathbb{Q}$.
Furthermore, in the end of this section,
we give two inequalities $(\mathrm{A})$ and $(\mathrm{B})$,
which imply Theorem \ref{main thm}.
These inequalities give an upper bound and a lower bound of the $\mathbb{Z}_p$-rank
of a certain Galois group.
In Sections \ref{A} and \ref{B},
we give the proof of the inequalities (A) and (B), respectively,
assuming that weak GGC does not hold for $p$ and $k$.
In Section \ref{rational case}, we consider the case where $k$ is a $p$-split $p$-rational and prove
Theorem \ref{main thm 2}.
In Section \ref{example}, we introduce some numerical examples including $p=3$ and $k=\mathbb{Q}(\sqrt{-971})$. 
\section{Preliminary}\label{Preli}
\subsection~
Let $p$ be an odd prime number and $k$ an imaginary quadratic field in which $p$ splits
into $\mathfrak{p}$ and $\mathfrak{p}^{\ast}$.
Let $K$ be a $\mathbb{Z}_p$-extension or the $\mathbb{Z}_p^{\oplus 2}$-extension of $k$.
We denote by $L_{K}/K$ the maximal unramified pro-$p$ abelian extension
and put $X_K=\mathrm{Gal}(L_{K}/K) $.
Since the Galois group $\mathrm{Gal}(K/k)$ acts naturally on $X_{K}$,
it becomes a $\mathbb{Z}_p[[\mathrm{Gal}(K/k)]]$-module.
It is known that $X_{K}$ is a finitely generated torsion $\mathbb{Z}_p[[\mathrm{Gal}(K/k) ]]$-module
(\cite{Gre}, \cite{Iw}).\par
Since we have $\mathrm{Gal}(\widetilde{k}/k) \cong \mathbb{Z}_p^{\oplus 2}$,
$k$ has two independent $\mathbb{Z}_p$-extensions.
For example, the cyclotomic $\mathbb{Z}_p$-extension $k_{\infty}^c$ and the anti-cyclotomic $\mathbb{Z}_p$-extension $k_{\infty}^a$
are disjoint over $k$ and satisfy $\widetilde{k}=k_{\infty}^ck_{\infty}^a$.
Furthermore, since we suppose that $p$ splits in $k$,
there exist two $\mathbb{Z}_p$-extensions in which one of the prime ideals of $k$ lying above $p$ does not ramify.
Let $N_{\infty}/k$ (respectively, $N_{\infty}^{\ast}/k$) be the $\mathbb{Z}_p$-extension of $k$
in which $\mathfrak{p}^{\ast}$ (respectively, $\mathfrak{p}$) does not ramify.
We note that $N_{\infty}$ (respectively, $N_{\infty}^{\ast}$) coincides with the fixed field of $\widetilde{k}$ by
the inertia subgroup of $\mathrm{Gal}(\widetilde{k}/k)$ for the prime ideal $\mathfrak{p}^{\ast}$ (respectively, $\mathfrak{p}$) of $k$
(\cite[Lemma 3.2]{Mi}).\par
Let $\sigma$ and $\tau $ be topological generators of $\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)$ and $\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)$, respectively.
By the isomorphism
$$
\mathrm{Gal}(\widetilde{k}/k) \cong \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \times \mathrm{Gal}(\widetilde{k}/k_{\infty}^a),
$$
we fix an isomorphism 
\begin{eqnarray}
\mathbb{Z}_p[[ \mathrm{Gal}(\widetilde{k}/k)]]  \cong \mathbb{Z}_p[[S,T]] \qquad  (\sigma \leftrightarrow 1+S, ~\tau \leftrightarrow 1+T). \label{isom ring}
\end{eqnarray}
We put $\Lambda=\mathbb{Z}_p[[S,T]]$. By this isomorphism, we regard $X_{\widetilde{k}}$ as a $\Lambda$-module.
We note that $\Lambda$ is a unique factorization domain and 
a noetherian local integral domain with the maximal ideal $(S,T,p)$.\par
\subsection~
For a ring $R$, we denote by $R^{\times}$ the unit group of $R$.
We suppose that $R$ is the formal power series ring in one variable
over a discrete valuation ring or suppose that
$R$ is the formal power series ring $\Lambda$.
For a finitely generated torsion $R$-module $M$,
we define the characteristic ideal of $M$ as follows.
By the structure theorem of $R$-modules (\cite[Proposition 5.1.7]{NSW}), there exists an $R$-homomorphism 
\[
M \rightarrow \bigoplus_{i=1}^{l} R/{\mathfrak{p}_i}^{m_i}
\]
with pseudo-null kernel and pseudo-null cokernel,
where $\mathfrak{p}_i$'s are prime ideals of height one and
$l$ is a non-negative integer, and $m_i$'s are positive integers.
Here, for an $R$-module $P$,
$P$ is said to be pseudo-null if
there are two relatively prime annihilators of $P$.
We note that a pseudo-null $R$-module is an $R$-module of finite length
if $R$ is the formal power series ring in one variable
over a discrete valuation ring.
Then, we define the characteristic ideal of $M$ by
\[
\mathrm{char}_{R}(M)=\left( \prod_{i=1}^{l} {\mathfrak{p}_i}^{m_i} \right),
\]
which is an ideal in $R$.\par
Let $G$ be a profinite group. For any topological $G$-module $M$, we denote by $M^{G}$
the subset of elements of $M$ invariant under the action of $G$.
We also denote by $M_{G}$ the largest quotient module of $M$ on which $G$ acts trivially, namely,
$$
M_{G}= M/M', \quad M'= \overline{\langle (g-1)m~|~g \in G, m \in M \rangle },
$$
where $\overline{\langle (g-1)m~|~g \in G, m \in M \rangle }$ is the topological closure of $\langle (g-1)m~|~g \in G, m \in M \rangle $ in $M$.
\section{The decomposition of the prime $p$}\label{decomp}
In this section, we roughly describe the decomposition of the prime $p$ in $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
We first have the following
\begin{lem}[{\rm{See for example} \cite[Lemma 1]{Oz}}] \label{Ozaki}
Let $k$ be an imaginary quadratic field in which $p$ splits and
$k_{\infty}$ a $\mathbb{Z}_p$-extension different from $N_{\infty}$ and $N_{\infty}^{\ast}$.
Then there exists an exact sequence
of $\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}/k)]]$-modules:
$$
0 \rightarrow (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty})}
\rightarrow X_{k_{\infty}} \rightarrow
\mathrm{Gal}(\widetilde{k}/k_{\infty}) \rightarrow 0.
$$
\end{lem}
For a $\mathbb{Z}_p$-extension $k_{\infty}/k$, we denote by $\lambda(k_{\infty}/k)$, $\mu(k_{\infty}/k)$
the Iwasawa $\lambda$-invariant, $\mu$-invariant of $k_{\infty}/k$, respectively.
Since we suppose that $p$ splits in $k$,
we have $\lambda(k_{\infty}^c/k) \geq 1$.
\begin{lem}\label{trivial}
The Iwasawa module $X_{\widetilde{k}}$ is trivial if and only if $\lambda(k_{\infty}^c/k)=1$.
\end{lem}
\begin{proof}
In the case of $\lambda(k_{\infty}^c/k)=1$,
we have $(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)}=0$ using Lemma \ref{Ozaki} and using the fact that
$X_{k_{\infty}^c}$ has no non-trivial finite $\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}^c/k)]]$-submodule (\cite[Corollary 13.29]{Wa}).
By Nakayama's lemma, we obtain $X_{\widetilde{k}}=0$.
If we suppose that $X_{\widetilde{k}}=0$, then we have $X_{k_{\infty}^c} \cong \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \cong \mathbb{Z}_p$.
This implies that $\lambda(k_{\infty}^c/k)=1$. Thus we get the conclusion.
\end{proof}
We put $\lambda^{*}=\lambda(k_{\infty}^c/k) -1$.
We obtain an annihilator of $X_{\widetilde{k}}$ from the following 
\begin{lem}[{\cite[Lemma 3.3]{Mu}}]\label{f(S,T)}
Suppose that $\lambda^{\ast}\geq 1$, where $\lambda^{\ast}$ is the integer defined above.
Then there exist power series $f(S,T) \in \mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})$ and
$g_i(S) \in \mathbb{Z}_p[[S]]$ $(i=0, \dots , \lambda^{\ast}-1)$
such that
$$
f(S,T) = T^{\lambda^{\ast}} + g_{\lambda^{\ast}-1}(S) T^{\lambda^{\ast}-1} + \dots + g_{1}(S)T +g_{0}(S).
$$
\end{lem}
\begin{proof}
We note that
$X_{k_{\infty}^c}$ has no non-trivial finite $\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}^c/k)]]$-submodule.
Applying Lemma  \ref{Ozaki} to the case of $k_{\infty}=k_{\infty}^c$,
we have
$$
  X_{\widetilde{k}}/SX_{\widetilde{k}} \cong (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} \cong \mathbb{Z}_p^{\oplus \lambda^{\ast} }
$$
by the isomorphism (\ref{isom ring}).
Using Nakayama's lemma, there exist $x_i \in X_{\widetilde{k}} ~(i=1, \dots, \lambda^{\ast})$ such that $X_{\widetilde{k}}= \langle x_1, \dots, x_{\lambda^{\ast}}
\rangle_{\mathbb{Z}_p[[S]]}$.
Then there exist $f_{ij}(S) \in \mathbb{Z}_p[[S]]~(i,j=1,\dots , \lambda^{\ast})$ such that
\begin{eqnarray*}
Tx_{i}= \sum_{j=1}^{\lambda^{\ast}} f_{ij}(S) x_j , \quad (i=1,\dots , \lambda^{\ast}).
\end{eqnarray*}
By these relations, we have the following matrix
\begin{eqnarray*}
A=
~\left(
\begin{array}{ccccc}
T-f_{11}(S) &  -f_{12}(S) & \dots & -f_{1 \lambda^{\ast}}(S)\\
-f_{21}(S) &T-f_{22}(S) &  \dots & -f_{2 \lambda^{\ast}}(S) \\
  \dots &  \dots  &   \dots      &  \dots \\
-f_{\lambda^{\ast} 1}(S)  &  -f_{\lambda^{\ast} 2}(S) & \dots & T-f_{\lambda^{\ast} \lambda^{\ast}}(S)
\end{array}
\right).
\end{eqnarray*}
We denote by $\mathrm{det}(A)$ the determinant of the matrix $A$.
We put $f(S,T)=\mathrm{det}(A)$. Then we obtain
$$
f(S,T) = T^{\lambda^{\ast}} + g_{\lambda^{\ast}-1}(S) T^{\lambda^{\ast}-1} + \dots + g_{1}(S)T +g_{0}(S)
$$
for some $g_i(S) \in \mathbb{Z}_p[[S]]$ $(i=0, \dots , \lambda^{\ast}-1)$.
It is easy to see that $f(S,T)X_{\widetilde{k}}=0$.
Thus we get the conclusion.
\end{proof}
\begin{Rem}\label{uni of f}
\begin{rm}
The uniqueness of the power series $f(S,T)$ is not known.
In this paper, we fix this power series in the case where $\lambda(k_{\infty}^c/k) \geq 2$.
\end{rm}
\end{Rem}
We fix an isomorphism
\begin{eqnarray}\label{isom cyc}
\mathbb{Z}_p[[ \mathrm{Gal}(k_{\infty}^c/k)]]  \cong \mathbb{Z}_p[[T]] \qquad
(\tau \mathrm{Gal}(\widetilde{k}/k_{\infty}^c)  \leftrightarrow 1+T).
\end{eqnarray}
By this isomorphism, we identify these rings.\par
There exists a relation between the power series $f(S,T)$
and the characteristic ideal of $X_{k_{\infty}^c}$ as follows.
\begin{prop}\label{char cyc}
Assume that the characteristic ideal 
of $X_{k_{\infty}^c}$
has a generator which is square-free
or $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module.
Suppose that $\lambda(k_{\infty}^c/k) \geq 2$.
Then we have
\[
\mathrm{char}_{\mathbb{Z}_p[[T]]} (X_{k_{\infty}^c})
= (Tf(0,T)),
\]
where $f(S,T)$ is the same power series defined in Lemma \ref{f(S,T)}.
\end{prop}
\begin{proof}
Since $X_{\widetilde{k}}$ is a finitely generated $\Lambda$-module,  
there exists a positive integer $r$ such that
$\left( \Lambda /f(S,T) \Lambda \right)^{\oplus r} \rightarrow  X_{\widetilde{k}}$
is surjective.
We can take $r=1$ if $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module.
Hence we get a surjective homomorphism
$\left( \mathbb{Z}_p[[T]] /f(0,T) \mathbb{Z}_p[[T]] \right)^{\oplus r}
\rightarrow  X_{\widetilde{k}}/S X_{\widetilde{k}}$.
This implies that
\[
(Tf(0,T)^r) \subset  \mathrm{char}_{{\mathbb{Z}_p}[[T]]}(X_{k_{\infty}^c}).
\]
By Lemma \ref{f(S,T)}, we note that
$f(0,T)$ is a polynomial with $\mathrm{deg}(f(0,T))=\lambda^{*}$,
where  $\mathrm{deg}(f(0,T))$ is the degree
of the polynomial $f(0,T)$.
Since  $\mathrm{char}_{{\mathbb{Z}_p}[[T]]}(X_{k_{\infty}^c})$
has a generator which is square-free in the case of $r\geq 2$,
we obtain
\[
(Tf(0,T)) \subset  \mathrm{char}_{{\mathbb{Z}_p}[[T]]}(X_{k_{\infty}^c}).
\]
By the definition of $\lambda^{*}$, we have
$(Tf(0,T)) =  \mathrm{char}_{{\mathbb{Z}_p}[[T]]}(X_{k_{\infty}^c})$.
\end{proof}
Using the structure theorem, we can calculate the order of $\left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)}$.
\begin{lem}\label{Momo}
Suppose that $\lambda(k_{\infty}^c/k) \geq 2$.
With the same notation as above, we have the following:
\[
\# \left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)} = \# (\mathbb{Z}_p/g_0(0)\mathbb{Z}_p),
\]
where $g_0(S)$ is the same power series defined in Lemma \ref{f(S,T)}.
\end{lem}
\begin{proof}
Using the structure theorem, we have an injective pseudo-isomorphism
\[
\Phi : (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)}
\rightarrow 
\bigoplus_{j=1}^{l} \mathbb{Z}_p[[T]]/(f_{j}(T)^{n_j}),
\]
where $l$ and $n_j$'s are positive integers and $f_j(T)$'s are irreducible elements of $\mathbb{Z}_p[[T]]$.
By Lemma \ref{Ozaki} and Proposition \ref{char cyc},
the characteristic ideal of $(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)}$
is generated by $f(0,T)$.
Then we have a prime factorization $f(0,T)=\displaystyle{\prod_{j=1}^{l}} f_j(T)^{n_j}$.
By \cite[Proposition $6$]{J-S}, $f(0,T)$ is not divisible by $T$.
Hence we have a commutative diagram:
\begin{eqnarray*}\label{diag}
\begin{CD}
0 @>>> (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} @>\Phi>> \displaystyle{\bigoplus_{j=1}^{l} \mathbb{Z}_p[[T]]/(f_{j}(T)^{n_j})}
 @>>> C @>>> 0
\\
& & @V  ~T\times VV  @V ~T\times VV   @V  ~T\times VV & 
\\
0 @>>> (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} @>\Phi>>\displaystyle{\bigoplus_{j=1}^{l} \mathbb{Z}_p[[T]]/(f_{j}(T)^{n_j})}
@>>> C @>>> 0,
\end{CD}
\end{eqnarray*}
where the vertical maps are multiplication by $T$ and $C$ is the cokernel of $\Phi$.
Since $C$ is finite, we obtain
\[
\# (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k)} = \# (\mathbb{Z}_p/f(0,0) \mathbb{Z}_p) = \# (\mathbb{Z}_p/g_0(0) \mathbb{Z}_p).
\]
\end{proof}
In the following lemmas and propositions,
we consider the decomposition of the prime $p$ in $\widetilde{k}/\mathbb{Q}$.
We first note that the prime number $p$ is only finitely decomposed in $\widetilde{k}/\mathbb{Q}$.
This fact was verified by Minardi in his thesis using class field theory:
\begin{lem}\cite[Lemma 3.1]{Mi}\label{Minardi}
Let $\mathfrak{D_{p}}$ be the decomposition group of $\mathfrak{p}$ in $\mathrm{Gal}(\widetilde{k}/k)$.
Then $\mathfrak{D_p}$ has finite index in $\mathrm{Gal}(\widetilde{k}/k)$.
\end{lem}
We give a necessary and sufficient condition
for the decomposition group $\mathfrak{D_{p}}$ to be normal in $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
\begin{lem}\label{cond normal}
The decomposition group $\mathfrak{D_p}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$
if and only if
$[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] \leq  [L_{k}\cap \widetilde{k}:k].$
\end{lem}
\begin{proof}
First, we note that $k_{\infty}^a /\mathbb{Q}$ is a galois extension. Furthermore, the complex conjugation, namely,
the generator of $\mathrm{Gal}(k/\mathbb{Q})$ acts on $\mathrm{Gal}(k_{\infty}^a /k)$ as inverse. 
Then we have $L_{k} \cap \widetilde{k} \subset k_{\infty}^a$.
By the definition of $N_{\infty}$,
$k_{\infty}^a \cap N_{\infty}$ coincides with $ L_{k} \cap \widetilde{k}$.
Since $\mathfrak{p}^{\ast}$ splits completely in $\widetilde{k}^{\mathfrak{D}_{\mathfrak{p}^{\ast}}}$,
we have $\widetilde{k}^{\mathfrak{D}_{\mathfrak{p}^{\ast}}} \subset N_{\infty}$.\par
We suppose that
$ \mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Then we have $\widetilde{k}^{\mathfrak{D_{p}}}= \widetilde{k}^{\mathfrak{D}_{\mathfrak{p}^{\ast}}}$.
Hence $p$ splits completely in $\widetilde{k}^{\mathfrak{D_{p}}}$
and $\widetilde{k}^{\mathfrak{D_{p}}}$ is a subfield of $k_{\infty}^a$.
This implies that $[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] \leq  [L_{k}\cap \widetilde{k}:k]$.
Conversely, we suppose that $[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] \leq  [L_{k}\cap \widetilde{k}:k].$
Since $L_{k}\cap \widetilde{k}$ and $\widetilde{k}^{\mathfrak{D}_{\mathfrak{p}}}$ are subfields of the $\mathbb{Z}_p$-extension
field $N_{\infty}^{\ast}$,
we have $\widetilde{k}^{\mathfrak{D}_{\mathfrak{p}}} \subset L_{k}\cap \widetilde{k}$.
Then $\widetilde{k}^{\mathfrak{D}_{\mathfrak{p}}}/\mathbb{Q}$ is a galois extension.
This implies that $\mathfrak{D_p}$ is normal.
\end{proof}
From the following proposition,
we see that the $p$-adic valuation of the constant term of $f(S,T)$ gives an upper bound of the number of prime ideals of $\widetilde{k}$
lying above $p$. 
\begin{prop}\label{index}
Suppose that $\lambda(k_{\infty}^c/k) \geq 2$.
We have the following:
\begin{equation*}
[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] 
\leq 
\mathrm{min}
\{
[L_{k}\cap \widetilde{k}:k], \# \left( \mathbb{Z}_p/g_0(0)\mathbb{Z}_p \right)
\}
\end{equation*}
\hspace{3.5cm} if $\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$,
\begin{equation*}
 [L_{k}\cap \widetilde{k}:k] < [\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] \leq \# \left( \mathbb{Z}_p/g_0(0)\mathbb{Z}_p \right)
\end{equation*}
\hspace{3.5cm} if $ \mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q}).$
\end{prop}
\begin{proof}
We will prove that $[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] \leq \# \left( \mathbb{Z}_p/g_0(0)\mathbb{Z}_p \right)$.
By Lemma \ref{Ozaki}, we have an exact sequence
\begin{eqnarray*}
0 \rightarrow (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)}
\rightarrow X_{k_{\infty}^c} \rightarrow \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \rightarrow 0 \label{ex cyc}
\end{eqnarray*}
as $\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}^c/k)]]$-modules. 
The snake lemma gives the exact sequence
\begin{eqnarray*}
0 &\rightarrow & \left( (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} \right)^{\mathrm{Gal}(k_{\infty}^c/k)}
\rightarrow \left( X_{k_{\infty}^c} \right)^{\mathrm{Gal}(k_{\infty}^c/k)}
\rightarrow \left( \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \right)^{\mathrm{Gal}(k_{\infty}^c/k)} \\ 
& \rightarrow &
\left( (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} \right)_{\mathrm{Gal}(k_{\infty}^c/k)}
\rightarrow  \left( X_{k_{\infty}^c} \right)_{\mathrm{Gal}(k_{\infty}^c/k)} \rightarrow
\left( \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \right)_{\mathrm{Gal}(k_{\infty}^c/k)} \rightarrow 0. \nonumber
\end{eqnarray*}
Then we have
$\left( (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} \right)^{\mathrm{Gal}(k_{\infty}^c/k)}=0$.
Indeed,
$T$ does not divide a generator of
$\mathrm{char}_{\mathbb{Z}_p[[T]]}\left((X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)} \right)$.
Furthermore, we have $\left( X_{k_{\infty}^c} \right)^{\mathrm{Gal}(k_{\infty}^c/k)}=D_{k_{\infty}^c}$ by \cite[Lemma 4.1]{Okano},
where $D_{k_{\infty}^c} $ is the decomposition group of a prime ideal of $k_{\infty}^c$ lying above $p$ in
$X_{k_{\infty}^c} =\mathrm{Gal}(L_{k_{\infty}^c}/k_{\infty}^c)$.
Therefore we obtain an exact sequence
\begin{eqnarray}\label{ex index}
0 &\rightarrow &  D_{k_{\infty}^c}
\rightarrow  \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \\  \label{snake ex} 
& \rightarrow &  (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k)} 
\rightarrow  \left( X_{k_{\infty}^c} \right)_{\mathrm{Gal}(k_{\infty}^c/k)}  \rightarrow
\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)  \rightarrow 0. \nonumber
\end{eqnarray}
This implies that
\begin{eqnarray*}
\mathrm{Coker} (D_{k_{\infty}^c} \rightarrow \mathrm{Gal}(\widetilde{k}/k_{\infty}^c))
&\cong &
\mathrm{Gal}(\widetilde{k}/k_{\infty}^c) / \mathrm{ker}(\mathrm{Gal}({\widetilde{k}/k_{\infty}^c})
\rightarrow \left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)})\\
&\cong &
\mathrm{Image}(\mathrm{Gal}({\widetilde{k}/k_{\infty}^c})
\rightarrow \left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)}).
\end{eqnarray*}
We note that
\[
\mathrm{Image} (D_{k_{\infty}^c}  \rightarrow \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) )
=
\mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \cap \mathfrak{D_{p}}.
\]
From the exact sequence (\ref{ex index}), we obtain
\begin{eqnarray*}
 [\mathrm{Gal}(\widetilde{k}/k) : \mathfrak{D_{p}}] 
&=&
[\mathrm{Gal}(\widetilde{k}/k_{\infty}^c) \mathfrak{D_{p}} : \mathfrak{D_{p}}] \\
&= &
[\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)  : \mathrm{Gal}(\widetilde{k}/k_{\infty}^c)  \cap \mathfrak{D_{p}}] \\
&=&
\# \mathrm{Coker} (D_{k_{\infty}^c} \rightarrow \mathrm{Gal}(\widetilde{k}/k_{\infty}^c))\\
&\leq &
\#  (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k)}.
\end{eqnarray*}
By Lemma \ref{Momo}, we have completed the proof.
\end{proof}
Let $D_{\mathfrak{p}}$ (respectively, $D_{\mathfrak{p}^\ast}$)
be the decomposition group of $\mathfrak{p}$
(respectively, $\mathfrak{p}^\ast$) in $\mathrm{Gal}(N_{\infty}/k)$ (respectively, $\mathrm{Gal}(N_{\infty}^{\ast}/k))$.
\begin{lem}\label{indexD}
With the same notation as above, we have the following:
\begin{eqnarray*}\label{D}
[\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p}}] &=& [\mathrm{Gal}(N_{\infty}^{\ast}/k):D_{\mathfrak{p}}],\\
~[\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p^{\ast}}}] &=& [\mathrm{Gal}(N_{\infty}/k):D_{\mathfrak{p}^{\ast}}].
\end{eqnarray*}
\end{lem}
\begin{proof}
From the natural exact sequence
\begin{eqnarray*}
0 &\rightarrow &
\mathrm{Gal}(\widetilde{k}/N_{\infty})
\rightarrow \mathrm{Gal}(\widetilde{k}/k)
\rightarrow \mathrm{Gal}(N_{\infty}/k)
\rightarrow 0,
\end{eqnarray*}
we have
\begin{eqnarray*}
0 &\rightarrow & \mathrm{Gal}(\widetilde{k}/N_{\infty}) \cap \mathfrak{D}_{\mathfrak{p}^{\ast}}
\rightarrow  \mathfrak{D}_{\mathfrak{p}^{\ast}}
\rightarrow D_{\mathfrak{p}^{\ast}}
\rightarrow 0.
\end{eqnarray*}
Since all prime ideals of $N_{\infty}$ lying above $\mathfrak{p}^{\ast}$ ramify in
$\widetilde{k}/N_{\infty}$, we have $\mathrm{Gal}(\widetilde{k}/N_{\infty}) \cap \mathfrak{D}_{\mathfrak{p}^{\ast}}=\mathrm{Gal}(\widetilde{k}/N_{\infty})$.
Thus we obtain
$\mathrm{Gal}(\widetilde{k}/k)/\mathfrak{D}_{\mathfrak{p}^{\ast}} \cong \mathrm{Gal}(N_{\infty}/k)/D_{\mathfrak{p}^{\ast}}$.
Hence we have proved the former part.
We can prove that
$[\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p}}] = [\mathrm{Gal}(N_{\infty}^{\ast}/k):D_{\mathfrak{p}}]$
in the same way as above.
Thus we get the conclusion.
\end{proof}
\begin{prop}\label{split}
Suppose that
$ \mathfrak{D_{p^{\ast}}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q}).$
Then,
$p$ splits completely in $L_k \cap \widetilde{k}$.
\end{prop}
\begin{proof}
We have
$[\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p^{\ast}}}] = [\mathrm{Gal}(N_{\infty}/k):D_{\mathfrak{p}^{\ast}}]$
by Lemma \ref{indexD}.
Thus we have $N_{\infty}^{D_{\mathfrak{p^\ast}}} = \widetilde{k}^{\mathfrak{D_{p^\ast}}}$.
Since ${D_{\mathfrak{p^\ast}}}$ is not a normal subgroup of $\mathrm{Gal}(N_{\infty}/\mathbb{Q})$, we have
\[
N_{\infty}^{D_{\mathfrak{p^\ast}}} \supset L_{k} \cap \widetilde{k},
\quad N_{\infty}^{D_{\mathfrak{p^\ast}}} \neq L_{k} \cap \widetilde{k}.
\]
Indeed, if we suppose that $N_{\infty}^{D_{\mathfrak{p^\ast}}} \subset L_{k} \cap \widetilde{k}$,
then $N_{\infty}^{D_{\mathfrak{p^\ast}}}$ is a subfield of $k_{\infty}^a$.
Hence $N_{\infty}^{D_{\mathfrak{p^\ast}}}$ is galois over $\mathbb{Q}$.
Therefore ${{\mathfrak{D}}_{\mathfrak{p^\ast}}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
This is a contradiction.
By the same reason, the inclusion
$N_{\infty}^{D_{\mathfrak{p}}}  \supset L_{k} \cap \widetilde{k}$ 
is proper.
These imply that $\mathfrak{p}$ and $\mathfrak{p^{\ast}}$ split completely in $L_k \cap \widetilde{k}$.
Thus we get the conclusion.
\end{proof}
For an algebraic extension $K/k$,
we denote by $M_{\mathfrak{p}}(K)$ (respectively, $M_{\mathfrak{p}^{\ast}}(K)$) the maximal pro-$p$ abelian extension of $K$
unramified outside all prime ideals of $K$ lying above
$\mathfrak{p}$ (respectively, $\mathfrak{p}^{\ast}$).
To prove Theorem \ref{main thm}, we will show two inequalities below
under the assumption that weak GGC does not hold for $p$ and $k$:
\begin{eqnarray*}\label{AB}
\begin{cases}
(\mathrm{A}) ~\mathrm{rank}_{\mathbb{Z}_p} (\mathrm{Gal} (M_{\mathfrak{p^{\ast}}}(N_{\infty})/ \widetilde{k}))
\leq \mathrm{min}\{ [L_k \cap \widetilde{k}:k], [\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p^\ast}}] \} -1,\\
(\mathrm{B})~\mathrm{rank}_{\mathbb{Z}_p} (\mathrm{Gal} (M_{\mathfrak{p^{\ast}}}(N_{\infty})/ \widetilde{k}))
\geq 
\mathrm{min}\{ [L_k \cap \widetilde{k}:k], [\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p^\ast}}] \},
\end{cases}
\end{eqnarray*}
where we denote by $\mathrm{rank}_{\mathbb{Z}_p}(\ast)$ the $\mathbb{Z}_p$-rank of $\ast$. 
If both (A) and (B) hold, then this is a contradiction.
Thus weak GGC holds for $p$ and $k$.
In Sections \ref{A} and \ref{B}, we will prove these inequalities.\par
Finally, we introduce the following proposition needed later. For convenience we include a proof.
\begin{prop}{\rm{(\cite[Theorem 2]{Fuj})}}\label{Fuj}
The following two conditions are equivalent:\\
$\mathrm{(i)}$
The Iwasawa module $X_{\widetilde{k}}$ has a non-trivial pseudo-null $\Lambda$-submodule.\\
$\mathrm{(ii)}$ We have $M_{\mathfrak{p^\ast}}(\widetilde{k}) \neq L_{\widetilde{k}}$.
\end{prop}
\begin{proof}
We first suppose that $M_{\mathfrak{p^\ast}}(\widetilde{k}) = L_{\widetilde{k}}$.
Then we have $X_{\frak{p}^{\ast}}(\widetilde{k})=X_{\widetilde{k}}$,
where $X_{\mathfrak{p}^{\ast}}(\widetilde{k})$ is the galois group of the extension
$M_{\mathfrak{p^\ast}}(\widetilde{k}) / \widetilde{k}$.
By (\cite{PR}), $X_{\mathfrak{p}^{\ast}}(\widetilde{k})$ has no non-trivial pseudo-null $\Lambda$-submodule.
Hence $\mathrm{(i)}$ implies that $\mathrm{(ii)}$ holds.\par
Next we suppose that $\mathrm{(ii)}$ holds.
Using Lemma $2$ in \cite{Fuj},
we obtain
$M_{\mathfrak{p^{\ast}}}(k_{\infty}^c) =L_{k_{\infty}^c}$.
Hence we have a commutative diagram of $\mathbb{Z}_p[[T]]$-modules:
\begin{eqnarray*}\label{diag}
\begin{CD}
0 @>>> X_{\mathfrak{p^{\ast}}}(\widetilde{k})/S  X_{\mathfrak{p^{\ast}}}(\widetilde{k})  @>>> X_{k_{\infty}^c} @>>> \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) @>>> 0
\\
& & @V   VV @| @\vert    & 
\\
0 @>>>X_{\widetilde{k}}/S X_{\widetilde{k}}@>>> X_{k_{\infty}^c} @>>> \mathrm{Gal}(\widetilde{k}/k_{\infty}^c) @>>> 0.
\end{CD}
\end{eqnarray*}
This implies that
\begin{eqnarray}\label{LCT}
X_{\mathfrak{p^{\ast}}}(\widetilde{k})/S  X_{\mathfrak{p^{\ast}}}(\widetilde{k}) 
\cong X_{\widetilde{k}}/S X_{\widetilde{k}} \cong \mathbb{Z}_p^{\oplus \lambda^{\ast}}.
\end{eqnarray}
In the case of $\lambda(k_{\infty}^c /k) =1$, we have proved that
$X_{\widetilde{k}}=0$ by Lemma \ref{trivial}.
Then we have $X_{\mathfrak{p^{\ast}}}(\widetilde{k}) =0$ by (\ref{LCT}).
This is a contradiction.
Hence we have $\lambda(k_{\infty}^c /k) \geq 2$.
In this case, we see that $X_{\widetilde{k}}$ is not trivial.
Hence, without loss of generality, we may assume that $f(S,T)$ annihilates $X_{\mathfrak{p^{\ast}}}(\widetilde{k})$,
where $f(S,T)$ is the same power series defined in Lemma \ref{f(S,T)}.
We put $Y=\mathrm{Gal}(M_{\mathfrak{p^\ast}}(\widetilde{k}) / L_{\widetilde{k}})$.
Then we have a commutative diagram of $\Lambda$-modules:
\begin{eqnarray*}\label{diag}
\begin{CD}
0 @>>> Y @>>>X_{\mathfrak{p^{\ast}}}(\widetilde{k}) @>>>X_{\widetilde{k}} @>>> 0
\\
& & @V ~S\times VV  @V ~S\times VV   @V  ~S\times VV & 
\\
0 @>>> Y @>>> X_{\mathfrak{p^{\ast}}}(\widetilde{k}) @>>> X_{\widetilde{k}} @>>> 0,
\end{CD}
\end{eqnarray*}
where the vertical maps are multiplication by $S$.
Using (\ref{LCT}), we have a surjective homomorphism
\[
X_{\widetilde{k}} [S] \rightarrow Y/SY,
\]
where $X_{\widetilde{k}} [S]$ is the $S$-torsion subgroup of $X_{\widetilde{k}}$.
By Nakayama's lemma, $X_{\widetilde{k}} [S] $ is not trivial because of the assumption that $Y \neq 0$.
Furthermore, $X_{\widetilde{k}} [S]$ is pseudo-null since $S$ is coprime to $f(S,T)$.
Therefore $X_{\widetilde{k}}$ has a non-trivial pseudo-null submodule.
Thus we get the conclusion.
\end{proof}
\section{Proof of the inequality (A)} \label{A}
In this section, we will prove the inequality (A):
\begin{eqnarray*}
\mathrm{rank}_{\mathbb{Z}_p} (\mathrm{Gal} (M_{\mathfrak{p^{\ast}}}(N_{\infty})/ \widetilde{k}))
 \leq \mathrm{min}\{ [L_k \cap \widetilde{k}:k], [\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p^\ast}}] \} -1
\end{eqnarray*}
under the assumption
of Theorem \ref{main thm} and assuming that
weak GGC does not hold for $p$ and $k$.\par
By Lemma \ref{Minardi}, the prime ideal $\frak{p}^{\ast}$ finitely decomposes in $\widetilde{k}/k$.
We put $p^{n_0} = [\mathrm{Gal}(\widetilde{k}/k) : \mathfrak{D_{p^{\ast}}}] $.
Using Lemma \ref{indexD}, we have
$$
p^{n_0}
= [\mathrm{Gal}(\widetilde{k}/k) : \mathfrak{D_{p^{\ast}}}] 
= [\mathrm{Gal}(N_{\infty}/k) : D_{\mathfrak{p^{\ast}}}].
$$
Let $k_{\infty}/k$ be a $\mathbb{Z}_p$-extension.
For each $n\geq 0 $, we denote by $k_n$ the intermediate field of the $\mathbb{Z}_p$-extension $k_\infty$
such that $k_n$ is the unique cyclic extension over $k$ of degree $p^n$.
Let $K/k$ be an algebraic extension.
For a prime ideal $\mathfrak{q}$ of $K$,
let $K_{\mathfrak{q}}$ be the completion of $K$ at $\mathfrak{q}$.
We denote by $U_{\mathfrak{q}}^{(1)}$ the principal local unit group with respect to $\mathfrak{q}$
in $K_{\mathfrak{q}}$.\\
\par
\subsection{Normal case}~
In this subsection, we will prove the inequality (A)
in the case where $\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.\par
\begin{prop}\label{rank uper}
Assume the same condition $\mathrm{(i)}$ as in Theorem \ref{main thm}. 
Assume also that weak {\rm{GGC}} does not hold for $p$ and $k$
and that
$\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Then the inequality $\mathrm{(A)}$ holds.
\end{prop}
\begin{proof}
We assume that weak GGC does not hold for $p$ and $k$.
By Proposition \ref{Fuj}, we have $M_{\mathfrak{p^\ast}}(\widetilde{k}) = L_{\widetilde{k}}$.
By Lemma \ref{cond normal},
$p$ splits completely in $\widetilde{k}^{\mathfrak{D}_{{\mathfrak{p}}}}$.
Since $\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$,
we have 
$\widetilde{k}^{\mathfrak{D}_{{\mathfrak{p}}}}=\widetilde{k}^{\mathfrak{D}_{{\mathfrak{p}}^{\ast}}}$.
Let
\[
\{ \mathfrak{p}_{\infty,i}^{\ast} ~|~ 1 \leq  i \leq p^{n_0}\}
\]
be the set of prime ideals of $N_{\infty}$ lying above $\mathfrak{p}^\ast$.
We denote by $ I_{\mathfrak{p}_{\infty,i}^{\ast}}$ the inertia subgroup of $\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/N_{\infty})$
for the prime ideal $\mathfrak{p}_{\infty,i}^{\ast}$.
Then we have an exact sequence
\[
0 \rightarrow  \displaystyle{\sum_{i=1}^{p^{n_0}} I_{\mathfrak{p}_{\infty,i}^{\ast}}} \rightarrow \mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/ N_{\infty})
\rightarrow \mathrm{Gal}(L_{N_{\infty}}/ N_{\infty}) \rightarrow 0.
\]
Since $M_{\mathfrak{p^{\ast}}}(N_{\infty}) / \widetilde{k}$ is an unramified extension,
we have $\mathrm{Gal}(M_{\mathfrak{p^\ast}}(N_{\infty})/\widetilde{k}) \cap I_{\mathfrak{p}_{\infty,i}^{\ast}} =1$.
Hence $I_{\mathfrak{p}_{\infty,i}^{\ast}}$ is isomorphic to $\mathbb{Z}_p$ for each $i$.
From \cite{Gi, sch, OV},
we obtain $\mu(N_{\infty}/k)=0.$
Furthermore,
since we suppose that $\lambda({N_{\infty}}/k)=0$, $\mathrm{Gal}(L_{N_{\infty}}/N_{\infty})$ is finite.
This implies that
$\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/ N_{\infty})$ is a finitely generated $\mathbb{Z}_p$-module.
Therefore we obtain
$\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/ \widetilde{k})) \leq p^{n_0}-1$.
By Lemma \ref{cond normal},
we have $p^{n_0} \leq [L_k \cap \widetilde{k} :k]$.
Thus we get the conclusion.
\end{proof}~\\
\subsection{Non-normal case}~
In this subsection, we will prove the inequality (A) in the case where $\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
\begin{prop}\label{rank n0}
With the same notation as above, we have
\[
\mathrm{rank}_{\mathbb{Z}_p} (\mathrm{Gal} (M_{{\mathfrak{p^{\ast}}}}(N_{n_0})/ N_{n_0})) =1.
\]
\end{prop}
\begin{proof}
Let $\mathcal{O}_{N_{n_0}}$ be the ring of integers in $N_{n_0}$ and
\[
\{ \mathfrak{p}_{n_0,i}^{\ast} ~|~ 1 \leq  i \leq p^{n_0}\}
\]
be the set of prime ideals of $N_{n_0}$ lying above $\mathfrak{p}^\ast$.
We put $E_{N_{n_0}}=\{ u \in \mathcal{O}_{N_{n_0}}^{\times}~|~u \equiv 1 ~\mathrm{mod}~\frak{p}_{n_0,i}^{\ast}~(1 \leq i \leq p^{n_0}) \}$.
Since $N_{n_0}/k$ is an abelian extension,
Leopoldt's conjecture holds for $p$ and $N_{n_0}$ (\cite{Bu}).
Using class field theory, we obtain
\begin{eqnarray*}
\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{n_0})/N_{n_0}))
&=&\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{n_0})/L_{N_{n_0}}))\\
&=&\mathrm{rank}_{\mathbb{Z}_p}\left(
\prod_{i=1}^{p^{n_0}} U_{\mathfrak{p}_{n_0,i}^{\ast}}^{(1)}/\overline{\varphi (E_{N_{n_0}})} \right)\\
&=&\mathrm{rank}_{\mathbb{Z}_p} \left(
\sum_{i=1}^{p^{n_0}}[(N_{n_0})_{\mathfrak{p}_{n_0,i}^{\ast}} : \mathbb{Q}_p] \right)
- \mathrm{rank}_{\mathbb{Z}_p}(\overline{\varphi (E_{N_{n_0}})})\\
&=&1,
\end{eqnarray*}
where
$\varphi : E_{N_{n_0}} \rightarrow \displaystyle{\prod_{i=1}^{p^{n_0}}U_{\mathfrak{p}_{n_0,i}^{\ast}}^{(1)}}$
is the diagonal map,
and $\overline{\varphi (E_{N_{n_0}})}$ is the topological closure of $\varphi (E_{N_{n_0}})$
in $\displaystyle{\prod_{i=1}^{p^{n_0}} U_{\mathfrak{p}_{n_0,i}^{\ast}}^{(1)}}$.
Thus we get the conclusion.
\end{proof}
We obtain the inequality (A) by the following
\begin{prop}\label{(A)}
Assume the same condition $\mathrm{(i)}$ as in Theorem \ref{main thm}.
Assume also that weak {\rm{GGC}} does not hold for $p$ and $k$
and that
$\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Then we have the inequality $\mathrm{(A)}$.
\end{prop}
\begin{proof}
We note that $\mathfrak{p^{\ast}}$ splits completely in $N_{\infty}^{ D_{\mathfrak{p^{\ast}}}} = N_{n_0}$
and that all prime ideals of $ N_{n_0}$
lying above $\mathfrak{p}^{\ast}$ do not split in $N_{\infty}$.
Let
\[
\{ \mathfrak{p}_{\infty,i}^{\ast} ~|~ 1 \leq  i \leq p^{n_0}\}
\]
be the set of prime ideals of $N_{\infty}$ lying above $\mathfrak{p}^\ast$
and
$I_{\mathfrak{p}_{\infty,i}^{\ast}}$ the inertia subgroup of
$\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/ N_{\infty})$
for the prime ideal
$\mathfrak{p}_{\infty,i}^{\ast}$.
By the same reason as in the proof of Proposition \ref{rank uper},
we have
$I_{\mathfrak{p}_{\infty,i}^{\ast}} \cong \mathbb{Z}_p$ and $\mu(N_{\infty}/k)=0$.
Furthermore, the fixed field of
$M_{\mathfrak{p^{\ast}}}(N_{\infty})$ by $\displaystyle{\sum_{i=1}^{p^{n_0}} I_{\mathfrak{p}_{\infty,i}^{\ast}}}$
coincides with $L_{N_{\infty}}$.
Thus $\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/ N_{\infty})$ is a finitely generated $\mathbb{Z}_p$-module,
since we suppose that $\lambda(N_{\infty}/k)=0$.
Let $M'$ be the submodule of $\mathbb{Z}_p$-torsion in $\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/N_{\infty})$.
Hence we have an exact sequence
\[
0\rightarrow M' \rightarrow \mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/N_{\infty}) \rightarrow
\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/N_{\infty})/M' \rightarrow 0 
\]
as $\mathbb{Z}_p[[\mathrm{Gal}(N_{\infty}/k)]]$-modules.
Since $M'$ is a finite $\mathbb{Z}_p$-module, we have
\begin{eqnarray}\label{char}
&~&
\mathrm{char}_{\mathbb{Z}_p[[\mathrm{Gal}(N_{\infty}/k)]]}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(N_{\infty})/N_{\infty})) \\
&=&\mathrm{char}_{\mathbb{Z}_p[[\mathrm{Gal}(N_{\infty}/k)]]}
\left(\displaystyle{\sum_{i=1}^{p^{n_0}} I_{\mathfrak{p}_{\infty,i}^{\ast}}}\right).\nonumber
\end{eqnarray}
For each $i$ with $1 \leq i \leq p^{n_0}$, the decomposition group $D_{\mathfrak{p}^{\ast}}$ acts on
$I_{\mathfrak{p}_{\infty,i}^{\ast}} $
because  all prime ideals of $ N_{n_0}$
lying above $\mathfrak{p}^{\ast}$ do not split in $N_{\infty}$.
Since we have $\mathrm{Gal}(M_{\mathfrak{p^\ast}}(N_{\infty})/\widetilde{k}) \cap I_{\mathfrak{p}_{\infty,i}^{\ast}} =1$,
there exists an injective homomorphism
\begin{eqnarray}\label{inj}
 I_{\mathfrak{p}_{\infty,i}^{\ast}}  \rightarrow \mathrm{Gal}(\widetilde{k}/N_{\infty})
\end{eqnarray}
as $\mathbb{Z}_p[[\mathrm{Gal}(N_{\infty}/N_{n_0})]]$-modules.
Then $D_{\mathfrak{p}^{\ast}}$ acts on $I_{\mathfrak{p}_{\infty,i}^{\ast}}$ trivially
because  $\widetilde{k}/N_{n_0}$ is an abelian extension.
Hence $D_{\mathfrak{p}^{\ast}}$ acts on
$\displaystyle{\sum_{i=1}^{p^{n_0}} I_{\mathfrak{p}_{\infty,i}^{\ast}}}$ trivially.
Let $F$ be the fixed field of $M_{\mathfrak{p^{\ast}}}(N_{\infty})$
by $M'$.
By (\ref{char}) and (\ref{inj}),
$F/N_{n_0}$ is an abelian extension.\par
Let $s$ be a non-negative integer such that $p^s= [L_{k} \cap \widetilde{k}:k]$.
By Proposition \ref{split}, $p$ splits completely in $L_{k} \cap \widetilde{k}$.
We note that all prime ideals of $L_k \cap \widetilde{k}$ lying above $\mathfrak{p}$
do not split in $N_{\infty}$.
Let
\[
\{ \mathfrak{p}_{n_0,i}~|~1 \leq  i \leq p^{s}\}
\]
be the set of prime ideals of $N_{n_0}$ lying above $\mathfrak{p}$
and
$I_{\mathfrak{p}_{n_0,i}}$ the inertia subgroup of $\mathrm{Gal}(F/ N_{n_0})$
for the prime ideal $\mathfrak{p}_{n_0,i}$.
Furthermore, by the same method as in the proof of Proposition \ref{rank uper},
we have $I_{\mathfrak{p}_{n_{0},i}} \cong \mathbb{Z}_p$ for each $i$.
Since the fixed field in $F$
by $\displaystyle{\sum_{i=1}^{p^{s}} I_{\mathfrak{p}_{n_0,i}}}$
is contained in $M_{\mathfrak{p}^{\ast}}(N_{n_0})$,
we obtain
\[
\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p}^{\ast}}(N_{\infty})/N_{n_0})) =
\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(F/N_{n_0})) \leq p^s +1
\]
by Proposition \ref{rank n0}.
Using Proposition \ref{index}, we have $p^s < p^{n_0}$.
Thus we get the conclusion.
\end{proof}
\section{Proof of the inequality (B)} \label{B}
In this section, we will prove the inequality (B):
\begin{eqnarray*}
\mathrm{rank}_{\mathbb{Z}_p} (\mathrm{Gal} (M_{\mathfrak{p^{\ast}}}(N_{\infty})/ \widetilde{k})) \geq
\mathrm{min}\{ [L_k \cap \widetilde{k}:k], [\mathrm{Gal}(\widetilde{k}/k):\mathfrak{D_{p^\ast}}] \}
\end{eqnarray*}
under the assumption
of Theorem \ref{main thm} and assuming that
weak GGC does not hold for $p$ and $k$.~\\
\subsection{Module theory}~
In this subsection, we will prove some module theoretical properties needed later
provided that
weak GGC does not hold for $p$ and $k$.
From Lemma \ref{trivial},
we have $\lambda(k_{\infty}^c/k) \geq 2$ if weak GGC does not hold for $p$ and $k$.
Let $\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})$ be the set of associated prime ideals of $X_{\widetilde{k}}$.
In other words, we put
\[
\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})
= \{\mathfrak{p} : \mathrm{~prime~ideal~}|~\mathfrak{p}=\mathrm{Ann}_{\Lambda}(x) ~\mathrm{for~some~element~} x \mathrm{~of}~X_{\widetilde{k}} \},
\]
where we write $\mathrm{Ann}_{\Lambda}(x)=\{ a \in \Lambda ~|~ ax=0\}$.
\begin{lem}\label{principal}
Assume that weak {\rm{GGC}} does not hold for $p$ and $k$.
Let $Y$ be a $\Lambda$-submodule of $X_{\widetilde{k}}$.
If $Y$ is cyclic as a $\Lambda$-module, in other words,
$Y$ is isomorphic to $\Lambda/\mathrm{Ann}_{\Lambda}(Y)$, then $\mathrm{Ann}_{\Lambda}(Y)$ is a principal ideal.
In particular, an associated prime ideal of $X_{\widetilde{k}}$ is principal.
\end{lem}
\begin{proof}
We suppose that $Y$ is isomorphic to $\Lambda/\mathrm{Ann}_{\Lambda}(Y)$.
Let $g,h$ be elements of $\mathrm{Ann}_{\Lambda}(Y)$.
We denote by $G$ the greatest common divisor for $g$ and $h$.
Then there exist $g',h' \in \Lambda$ such that $g=g'G, h=h'G$, and $g'$ is coprime to $h'$.
Then $GY$ is a pseudo-null $\Lambda$-submodule of $X_{\widetilde{k}}$.
Since $X_{\widetilde{k}}$ has no non-trivial pseudo-null $\Lambda$-submodule, we get $GY=0$.
Hence we have $G \in \mathrm{Ann}_{\Lambda}(Y)$.
This implies that $(g,h) \subset (G) \subset \mathrm{Ann}_{\Lambda}(Y)$ as an ideal of $\Lambda$.
Since $\mathrm{Ann}_{\Lambda}(Y)$ is a finitely generated $\Lambda$-module,
we can prove that
$\mathrm{Ann}_{\Lambda}(Y)$ is a principal ideal, inductively.
\end{proof}
If $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module, we can determine the isomorphism class
of $X_{\widetilde{k}}$.
\begin{prop}\label{isom str}
Assume that $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module.
Assume also that weak {\rm{GGC}} does not hold for $p$ and $k$. Then we have
\[
X_{\widetilde{k}} \cong \Lambda/f(S,T) \Lambda
\]
as $\Lambda$-modules, where $f(S,T)$ is the same annihilator of $X_{\widetilde{k}}$ defined in Lemma \ref{f(S,T)}.
\end{prop}
\begin{proof}
Since $X_{\widetilde{k}}$ is cyclic,
we have $X_{\widetilde{k}} \cong \Lambda/ \mathrm{Ann}_{\Lambda}(X_{\widetilde{k}}).$
By Lemma \ref{principal}, $\mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})$ is a principal ideal.
Let $G(S,T) \in \Lambda$ be a generator of $\mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})$.
Using Lemma \ref{Ozaki}, we have $\mathrm{char}_{\mathbb{Z}_p[[T]]}(X_{k_{\infty}^c}) = (TG(0,T))$.
By Lemma \ref{f(S,T)}, there exists a power series $H(S,T) \in \Lambda$ such that
$f(S,T) =G(S,T)H(S,T).$
Using Proposition \ref{char cyc}, we have $\mathrm{char}_{\mathbb{Z}_p[[T]]}(X_{k_{\infty}^c}) = (Tf(0,T))$.
This implies that $H(S,T) \in \Lambda^{\times}$.
Therefore we have $\mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})=(f(S,T))$.
Thus we get the conclusion.
\end{proof}
To consider the case where $X_{\widetilde{k}}$ is not $\Lambda$-cyclic, we prove the following
\begin{lem}\label{an}
Suppose that $\lambda(k_{\infty}^c/k) \geq 2$
and that $X_{\widetilde{k}}$ is not cyclic as a $\Lambda$-module.
Let
\begin{eqnarray*}\label{factor}
\displaystyle{f(S,T)} = \prod_{i=1}^{l} f_{i}(S,T)^{n_i}
\end{eqnarray*}
be a prime factorization, in other words,
$l$ and $n_i$'s are positive integers and
$f_{i}(S,T)$'s are irreducible elements of $\Lambda$.
Assume that
the characteristic ideal of $X_{k_{\infty}^c}$
has a generator which is square-free.
Then $f_{i}(S,T)$ is a zero-divisor of $X_{\widetilde{k}}$ for each $i$.
\end{lem}
\begin{proof}
We note that
$\mathrm{char}_{\Lambda}(X_{\widetilde{k}})$ has a square-free generator
because
the characteristic ideal
of
$X_{k_{\infty}^c}$
also has a square-free generator.
Hence we have $n_i=1$ for each $i$.
In the case of $l =1$, in other words, $f(S,T)$ is an irreducible element,
$f(S,T)$ is a zero-divisor of $X_{\widetilde{k}}$ by Lemma \ref{f(S,T)}.
Thus we get the conclusion.
In the following, we suppose that $l \geq 2$.
We put $F_i(S,T)=\displaystyle{\prod_{j=1, j \neq i}^{l} f_{j}(S,T)}$.
We prove that
$F_{i}(S,T)  X_{\widetilde{k}} \neq 0$.
Suppose that
$F_{i}(S,T)  X_{\widetilde{k}} =0$.
Since $X_{\widetilde{k}}$ is a finitely generated $\Lambda$-module, there exists a positive integer $r$ and
a surjective homomorphism
\[
\left( \Lambda/ F_{i}(S,T) \Lambda \right)^{\oplus r}
\rightarrow 
X_{\widetilde{k}}.
\]
This homomorphism induces
\[
\left( \mathbb{Z}_p[[T]]/ F_{i}(0,T) \mathbb{Z}_p[[T]] \right)^{\oplus r}
\rightarrow 
X_{\widetilde{k}}/S X_{\widetilde{k}}.
\]
Since the characteristic ideal of $X_{k_{\infty}^c}$
has a square-free generator, we have
\[
(T F_{i}(0,T)) \subset
\mathrm{char}_{{\mathbb{Z}_p}[[\mathrm{Gal}(k_{\infty}^c/k)]]}(X_{k_{\infty}^c}) = (Tf(0,T))
\]
by Proposition \ref{char cyc}. 
Hence we have
$\displaystyle{ \mathrm{deg}(F_{i}(0,T)) \geq  \mathrm{deg}(f(0,T))}$.
This implies that $f_i(S,T)$ is a unit in $\Lambda$
because we have $f(0,T)=F_i(0,T)f_i(0,T)$.
This is a contradiction.
Thus we get the conclusion.
\end{proof}
\begin{prop}\label{cycsub}
Assume that weak {\rm{GGC}} does not hold for $p$ and $k$.
Assume also that
the characteristic ideal 
of $X_{k_{\infty}^c}$
has a generator which is square-free
or $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module.
Then there exists a $\Lambda$-submodule $Y$ of $X_{\widetilde{k}}$ such that $Y$ is isomorphic to $\Lambda/(f(S,T))$.
\end{prop}
\begin{proof}
If $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module,
then $X_{\widetilde{k}}$ is isomorphic to $\Lambda/(f(S,T))$ by Proposition \ref{isom str}.
Hence we can take $Y=X_{\widetilde{k}}$.
Thus we get the conclusion.
We suppose that $X_{\widetilde{k}}$ is not cyclic as a $\Lambda$-module.
Since $\Lambda$ is a noetherian ring, $\displaystyle{\bigcup_{\mathfrak{p}\in \mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})} \mathfrak{p}}$
is the set of zero-divisors of $X_{\widetilde{k}}$ by \cite[Theorem $6.1$]{Mat}.
Let $\mathfrak{p}$ be an associated prime ideal of $X_{\widetilde{k}}$.
Let $\displaystyle{f(S,T)} = \prod_{i=1}^{l} f_{i}(S,T)$
be the same prime factorization as in Lemma \ref{an}.
Since $f_i(S,T)$ is an irreducible element of $\Lambda$, $(f_i(S,T))$ is a prime ideal.
We claim that $(f_{i}(S,T))$ is an associated prime ideal of $X_{\widetilde{k}}$ for each $i$.
Indeed, $f_{i}(S,T)$ is a zero-divisor of $X_{\widetilde{k}}$
by Lemma \ref{an}.
Hence we have $\displaystyle{ (f_{i}(S,T)) \subset \bigcup_{\mathfrak{p}\in \mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})} \mathfrak{p}}$.
By the prime avoidance theorem,
$(f_{i}(S,T)) \subset \mathfrak{p}$ for some associated prime ideal $\mathfrak{p}$.
By Lemma \ref{principal}, $\mathfrak{p}$ is a principal ideal.
Therefore we obtain $\mathfrak{p}=(f_{i}(S,T))$.
Furthermore, we can prove that
\[
\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})=
\{
(f_{i}(S,T)) ~|~1\leq i\leq l
\}.
\]
Indeed, let $\mathfrak{p}$ be
an associated prime ideal of $X_{\widetilde{k}}$.
There exists an element $x$ of $X_{\widetilde{k}}$ such that
$\mathfrak{p} = \mathrm{Ann}_{\Lambda}(x)$.
By Lemma \ref{principal}, $\mathfrak{p}$ is a principal ideal of $\Lambda$.
We put $\mathfrak{p}=(g)$ for some irreducible element $g$ in $\Lambda$.
If $g$ is coprime to $f(S,T)$, then
$\Lambda x$ is a non-trivial pseudo-null $\Lambda$-submodule of $X_{\widetilde{k}}$.
This is a contradiction.
Therefore there exists an integer $i$ such that
$(g)=(f_i(S,T))$.\par
By the structure theorem,
there exists a pseudo-isomorphism
\begin{eqnarray}\label{pshom}
\varphi: \bigoplus_{i=1}^{l'} \Lambda/\mathfrak{q}_i \rightarrow X_{\widetilde{k}},
\end{eqnarray}
where $l'$ is a positive integer and
$\mathfrak{q}_i$'s are prime ideals of height one.
We note that
$\mathrm{char}_{\Lambda}(X_{\widetilde{k}})$ has a square-free generator
by the same reason as in the proof of Lemma \ref{an}.
Hence we obtain $\mathfrak{q}_i \neq \mathfrak{q}_j$ for $i \neq j$.
Since $\displaystyle{\bigoplus_{i=1}^{l'} \Lambda/\mathfrak{q}_i}$ has no non-trivial pseudo-null
$\Lambda$-submodule, this homomorphism is injective. 
In the following, we will prove that $\mathfrak{q}_i$ is an associated prime ideal for each $i$ and that
$\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})$ coincides with
the set $\{\mathfrak{q}_i ~|~1\leq i \leq l' \}$.
Indeed, for each $j$ with $1\leq j\leq l'$,
we have a homomorphism
\[
\pi_j:\Lambda/\mathfrak{q}_j \rightarrow \bigoplus_{i=1}^{l'} \Lambda/\mathfrak{q}_i \xrightarrow{\varphi } X_{\widetilde{k}},
\]
where $\displaystyle{\Lambda/\mathfrak{q}_j \rightarrow \bigoplus_{i=1}^{l'} \Lambda/\mathfrak{q}_i}$ is the
natural injective homomorphism
such that, for an element $a$ of $\Lambda/\mathfrak{q}_j$, we define $a \mapsto (a_i)_i$ satisfying that $a_i =0$ for $i \neq j$ and $a_j=a$.
Hence
$\pi_j$ is an injective homomorphism.
Thus $\mathfrak{q}_j$ is an associated prime ideal.
This implies that
$\{\mathfrak{q}_i| 1\leq i \leq l' \}$ is contained in $\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})$.
Next we prove that $\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})$ is contained in $\{\mathfrak{q}_i| 1\leq i \leq l' \}$.
We assume that $\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}}) \not\subset \{\mathfrak{q}_i| 1\leq i \leq l' \}$.
Then there exists an associated prime ideal $\mathfrak{p}$
such that
$\mathfrak{p} \not\in \{\mathfrak{q}_i| 1\leq i \leq l' \}$.
We note that $\mathfrak{p}$ is a prime ideal of height one by Lemma \ref{principal}.
Using localization with respect to
$\mathfrak{p}$, we obtain
\[
(X_{\widetilde{k}})_{\mathfrak{p}} \cong \left(  \bigoplus_{i=1}^{l'} \Lambda/\mathfrak{q}_i \right)_{\mathfrak{p}} = 0
\]
from (\ref{pshom}).
Hence $\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}})$
is not contained in
the support of $X_{\widetilde{k}}$.
This is a contradiction. Thus we get
\[
\mathrm{Ass}_{\Lambda}(X_{\widetilde{k}}) =  \{\mathfrak{q}_i~|~ 1\leq i \leq l' \} = \{ (f_i(S,T))~|~ 1\leq i \leq l \}.
\]
This implies that $l = l'$.
Therefore we have a natural injective homomorphism
\[
\Lambda/(f(S,T)) \rightarrow \bigoplus_{i=1}^{l} \Lambda/(f_{i}(S,T)) = \bigoplus_{i=1}^{l} \Lambda/\mathfrak{q}_i
\rightarrow X_{\widetilde{k}}
\]
since we have a prime factorization $\displaystyle{f(S,T)} = \prod_{i=1}^{l} f_{i}(S,T)$.
Thus we obtain the conclusion.
\end{proof}
By the proposition above, we obtain the following
\begin{cor}\label{psisom}
Let $M$ be a finitely generated torsion $\Lambda$-module.
Assume that $M/SM$ is a free $\mathbb{Z}_p$-module of finite rank
and that
$\mathrm{char}_{\mathbb{Z}_p[[T]]}(M/SM)$
has a generator which is square-free.
Assume also that
$M$ has no non-trivial pseudo-null $\Lambda$-submodule.
Then we have a pseudo-isomorphism
\[
M \rightarrow \Lambda / \mathrm{char}_{\Lambda}(M).
\]
\end{cor}
Let $k_{\infty}/k$ be a $\mathbb{Z}_p$-extension. Then there exists a pair $(\alpha, \beta) \in \mathbb{Z}_p^{ \oplus 2} - p\mathbb{Z}_p^{\oplus 2}$
such that $k_{\infty}=\widetilde{k}^{\overline{\langle \sigma^{\alpha} \tau^{\beta} \rangle}}$,
where $\sigma$ and $\tau$ are the fixed topological generators defined in Section \ref{Preli}.
It is easy to check the following
\begin{lem}\label{Zp}
Let $k_{\infty}/k$ be a $\mathbb{Z}_p$-extension with $k_{\infty} \cap k_{\infty}^a = k_{m}^a$,
where $ k_{m}^a$ is the $m$-th layer of $ k_{\infty}^a$
and $m$ is a non-negative integer.
Then there exists a unit $u \in \mathbb{Z}_p^{\times}$ such that
$k_{\infty}=\widetilde{k}^{\overline{\langle \sigma^{up^{m}} \tau \rangle}}$.
\end{lem}
We put $\Gamma=\mathrm{Gal}(\widetilde{k}/N_{\infty})$.
Recall that $p^s = [L_{k}\cap \widetilde{k}:k]$.
Hence we have $N_{\infty} \cap k_{\infty}^a = L_k \cap \widetilde{k} = k_s^a$.
Then there exists a unit $u$ such that
\[
\mathrm{Gal}(\widetilde{k}/N_{\infty}) = \overline{ \langle \sigma^{up^{s}} \tau \rangle}
\]
by the lemma above.
We put $T_{s}=(1+S)^{up^s}(1+T)-1.$\par
The following proposition plays an important role
to obtain a lower bound of the $\mathbb{Z}_p$-rank of
$\mathrm{Gal} (M_{\mathfrak{p^{\ast}}}(N_{\infty})/ \widetilde{k}) $.
\begin{prop}\label{rank}
Under the same assumption as in Proposition \ref{cycsub},
we have $\mathrm{rank}_{\mathbb{Z}_p}((X_{\widetilde{k}})_{\Gamma}) \geq \mathrm{rank}_{\mathbb{Z}_p}(Y_{\Gamma})$,
where $Y$ is the same cyclic $\Lambda$-submodule of $X_{\widetilde{k}}$ as in Proposition \ref{cycsub}.
\end{prop}
\begin{proof}
If $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module, then we have $Y=X_{\widetilde{k}}$.
Thus we get the conclusion.
We suppose that $X_{\widetilde{k}}$ is not cyclic as a $\Lambda$-module.
By Proposition \ref{cycsub}, we have an exact sequence
\[
0 \rightarrow Y \rightarrow X_{\widetilde{k}} \rightarrow \mathrm{Coker} \rightarrow 0,
\]
where we put $\mathrm{Coker} = X_{\widetilde{k}}/Y$.
Using the snake lemma, we get
\begin{eqnarray*}
0 \rightarrow   Y^{\Gamma} \rightarrow (X_{\widetilde{k}})^{\Gamma}
\rightarrow   \mathrm{Coker}^{\Gamma} 
  \rightarrow   Y_{\Gamma} \rightarrow (X_{\widetilde{k}})_{\Gamma} \rightarrow 
\mathrm{Coker}_{\Gamma} \rightarrow 0.
\end{eqnarray*}
Since $X_{\widetilde{k}}$ has no non-trivial pseudo-null $\Lambda$-submodule,
we have $(X_{\widetilde{k}})^{\Gamma}=0$.
Indeed, we have $T_s (X_{\widetilde{k}})^{\Gamma}=f(S,T)(X_{\widetilde{k}})^{\Gamma}=0$.
Since the characteristic ideal of $X_{k_{\infty}^c}$
is generated by a power series which is square-free,
$T_s$ is coprime to $f(S,T)$.
This implies that $(X_{\widetilde{k}})^{\Gamma}$ is pseudo-null.
Hence we have $(X_{\widetilde{k}})^{\Gamma}=0$.
Then we obtain an exact sequence
\begin{eqnarray}\label{museq}
0 \rightarrow   \mathrm{Coker}^{\Gamma} 
  \rightarrow   Y_{\Gamma} \rightarrow (X_{\widetilde{k}})_{\Gamma} \rightarrow 
\mathrm{Coker}_{\Gamma} \rightarrow 0.
\end{eqnarray}
We identify the following rings below,
using isomorphisms
\begin{eqnarray}\label{is}
\Lambda/T_s \Lambda \cong \mathbb{Z}_p[[S]]
\cong \mathbb{Z}_p[[\mathrm{Gal}(N_{\infty}/k)]],
\end{eqnarray}
where we define
\begin{eqnarray*}
G(S,T) \mapsto G(S,(1+S)^{-up^s}-1) \mapsto G(\sigma \mathrm{Gal}(\widetilde{k}/N_{\infty})-1,\sigma^{-up^s} \mathrm{Gal}(\widetilde{k}/N_{\infty})-1).
\end{eqnarray*}
Here we note that $\mathbb{Z}_p[[S,T_s]]=\mathbb{Z}_p[[S,T]]$
because we have $T=(1+T_s)(1+S)^{-up^s}-1$.
We will prove that
$Y_{\Gamma}$ is a finitely generated $\mathbb{Z}_p$-module.
Indeed,
$\mathrm{Coker}_{\Gamma}$ is a finitely generated $\mathbb{Z}_p$-module
since $(X_{\widetilde{k}})_{\Gamma}$, which is isomorphic to $\mathrm{Gal}(M_{\mathfrak{p}^{\ast}}(N_{\infty})/\widetilde{k})$,
is a finitely generated $\mathbb{Z}_p$-module.
Furthermore, we can show that
$\mathrm{Coker}^{\Gamma}$ is a pseudo-null $\Lambda$-module
by the same method as above.
From (\ref{is}), we have
\begin{eqnarray*}
&~&f(S,T) \equiv f(S,(1+S)^{-up^s}-1)~~\mathrm{~mod~}T_s, \\
&~&f(S,(1+S)^{-up^s}-1) \mathrm{Coker}^{\Gamma}=0.
\end{eqnarray*}
We note that $f(S,(1+S)^{-up^s}-1) \neq 0$ since $f(0,0) \neq 0$.
Hence $\mathrm{Coker}^{\Gamma}$ is a torsion $\mathbb{Z}_p[[S]]$-module.
By \cite[I,1.3, Lemma 4]{PR}, we have
\[
\mathrm{char}_{\mathbb{Z}_p[[S]]}(\mathrm{Coker}_{\Gamma}) =
\pi(\mathrm{char}_{\Lambda}(\mathrm{Coker})) \mathrm{char}_{\mathbb{Z}_p[[S]]}(\mathrm{Coker}^{\Gamma}),
\]
where $\pi$ is the natural projection from $\Lambda$ to $\Lambda/T_s \Lambda$.
This implies that
$Y_{\Gamma}$ is a finitely generated $\mathbb{Z}_p$-module.
Hence all terms in (\ref{museq}) have finite $\mathbb{Z}_p$-rank.
Therefore we obtain
\begin{eqnarray*}
\mathrm{rank}_{\mathbb{Z}_p}( (X_{\widetilde{k}})_{\Gamma})
&=& \mathrm{rank}_{\mathbb{Z}_p}( Y_{\Gamma})
+ \mathrm{rank}_{\mathbb{Z}_p}( \mathrm{Coker}_{\Gamma}) -
\mathrm{rank}_{\mathbb{Z}_p}(  \mathrm{Coker}^{\Gamma})\\ \nonumber
&\geq & \mathrm{rank}_{\mathbb{Z}_p}( Y_{\Gamma}).
\end{eqnarray*}
Thus we get the conclusion.
\end{proof}~\\
\subsection{Normal case}~
In this subsection, we prove the inequality (B)
in the case where $k$ is not $p$-split $p$-rational
assuming that weak GGC does not hold for $p$ and $k$.
We first give a necessary and sufficient condition for $k$ to be $p$-split $p$-rational.
\begin{prop}\label{eqrational}
The following two conditions are equivalent:\\
$\mathrm{(i)}$ The decomposition group $\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$,
$\displaystyle{[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}]}
= \# \left( \mathbb{Z}_p/g_0(0)\mathbb{Z}_p \right)$,
and $\lambda(k_{\infty}^c/k)\geq 2$, where $g_0(S)$ is the same power series defined in Lemma \ref{f(S,T)}.\\
$\mathrm{(ii)}$ The imaginary quadratic field $k$ is $p$-split $p$-rational.
\end{prop}
\begin{proof}
We suppose that (i) holds.
Let $L$ be the fixed field of $L_{k_{\infty}^c}$ by $TX_{k_{\infty}^c}$.
We note that
$\displaystyle{\mathrm{Gal}(L/k_{\infty}^c)=\left( X_{k_{\infty}^c} \right)_{\mathrm{Gal}(k_{\infty}^c/k)}}$.
Using the exact sequence (\ref{ex index}) in the proof of Proposition \ref{index},
we have
\begin{eqnarray*}
\#\mathrm{Image}( \mathrm{Gal}(\widetilde{k}/k_{\infty}^c)\rightarrow \left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)})
=\#  (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k)}

 
\end{eqnarray*}
from $\displaystyle{[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}]}
= \# \left( \mathbb{Z}_p/g_0(0)\mathbb{Z}_p \right)$
and from Lemma \ref{Momo}.
Hence the homomorphism
$\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)\rightarrow \left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)}$
is surjective.
This implies that $L=\widetilde{k}$ from (\ref{ex index}).
Since $L_k k_{\infty}^c/k$ is an abelian extension, $L_k k_{\infty}^c$ is contained in $L$.
Therefore, we obtain $L_k \subset \widetilde{k}$.
Then it is easy to check that $M_p(k)=\widetilde{k}$,
where $M_p(k)$ is the maximal pro-$p$ abelian extension field of $k$ unramified outside all prime ideals lying above $p$.
By Lemma \ref{cond normal}, we have $\widetilde{k}^{\mathfrak{D_p}} \subset L_k$
since $\mathfrak{D_p}$ is a normal subgroup.
Furthermore, we have $\widetilde{k}^{\mathfrak{D_p}} \neq k$
from $\lambda(k_{\infty}^c/k)\geq 2$.
Hence $k$ is $p$-split $p$-rational.
\par
Conversely we suppose that $k$ is $p$-split $p$-rational.
By Definition \ref{rational}, we have
\[
L_k \subset \widetilde{k},\quad \widetilde{k}^{\mathfrak{D_{p}}} \neq k ,\quad \widetilde{k}^{\mathfrak{D_{p}}} \subset L_k.
\]
This implies that
$M_p(k) = \widetilde{k}$
and that
$\mathfrak{p}$ splits in $\widetilde{k}$.
Hence the homomorphism
$\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)\rightarrow \left( X_{\widetilde{k}}\right)_{\mathrm{Gal}(\widetilde{k}/k)}$ above
is surjective.
We note that $\lambda(k_{\infty}^c/k)\geq 2$.
By the exact sequence (\ref{ex index}) in the proof of Proposition \ref{index}, we obtain
\[
[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}]
=\#  (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k)}
=\# \left( \mathbb{Z}_p/g_0(0) \mathbb{Z}_p \right).
\]
Since $\widetilde{k}^{\mathfrak{D_{p}}}$ is contained in $L_k$, we have
$\displaystyle{[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] \leq [L_k :k]}$.
By Lemma \ref{cond normal}, $\mathfrak{D_p}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Thus we get the conclusion.
\end{proof}
For a power series $V(S) = \displaystyle{\sum_{i=0}^{\infty} b_i S^i }\in \mathbb{Z}_p[[S]]$,
we put
\begin{eqnarray*}
\mu(V(S)) &=& \mathrm{sup} \{ ~i~|~V(S) \equiv 0 \mathrm{~mod~}p^i~ ~\},\\
\lambda(V(S)) &=& \mathrm{inf} \{ ~i~|~b_i \not\equiv 0  \mathrm{~mod~}p~\}.
\end{eqnarray*}
Then we call $\mu(V(S)), \lambda(V(S))$ the $\mu$- and $\lambda$-invariant of $V(S)$, respectively.
\par
Next, we get a lower bound of the $\lambda$-invariant of $g_0(S)$ in the case where
the $\mu$-invariant of $g_0(S)$ is zero. 
Recall that $p^{n_0} = \displaystyle{[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p^{\ast}}}]}$.
\begin{prop}\label{normal g}
Assume that
$\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Assume also that $k$ is not $p$-split $p$-rational
and that $\lambda(k_{\infty}^c/k)\geq 2$.
If $\mu(g_0(S))$ is zero, then we have $\lambda(g_0(S)) \geq  p^{n_0}.$ 
\end{prop}
\begin{proof}
Since $k_{n_0}^a/k$ is an abelian extension, Leopoldt's conjecture holds for $p$ and $k_{n_0}^a$ (\cite{Bu}).
Hence we have $\mathrm{Gal}(\widetilde{k_{n_0}^a}/k_{n_0}^a) \cong \mathbb{Z}_p^{\oplus{p^{n_0}+1}}$,
where $\widetilde{k_{n_0}^a}$ is the composite of all $\mathbb{Z}_p$-extension fields over $k_{n_0}^a$.
Since $p$ splits in $k_{n_0}^a$, $\widetilde{k_{n_0}^a}/k_{\infty}^a$ is an unramified abelian extension.
Thus we have a surjective homomorphism
\[
X_{k_{\infty}^a} \rightarrow \mathrm{Gal}(\widetilde{k_{n_0}^a}/k_{\infty}^a).
\]
Furthermore, we see that
\[
 \mathrm{Gal}(\widetilde{k_{n_0}^a}/k_{\infty}^a) \cong  \mathbb{Z}_p[[S]]/((1+S)^{p^{n_0}}-1).
\]
Hence we have $\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) \subset ((1+S)^{p^{n_0}}-1)$.
On the other hand, there exists a surjective homomorphism $(\Lambda/f(S,T)\Lambda)^{\oplus r} \rightarrow X_{\widetilde{k}}$,
where $r$ is a positive integer and
$f(S,T)$ is the same power series defined in Lemma \ref{f(S,T)}.
Then we have a surjective homomorphism
$(\mathbb{Z}_p[[S]]/g_0(S) \mathbb{Z}_p[[S]])^{\oplus r} \rightarrow X_{\widetilde{k}}/TX_{\widetilde{k}}$.
This implies that
\[
(Sg_0(S)^r) \subset \mathrm{char}_{ \mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) \subset ((1+S)^{p^{n_0}}-1).
\]
Since $(1+S)^{p^{n_0}}-1$ is square-free, we obtain $(Sg_0(S)) \subset ((1+S)^{p^{n_0}}-1).$
Then there exists a power series $\widetilde{g_0}(S) \in \mathbb{Z}_p[[S]]$ such that
$Sg_0(S) = \{(1+S)^{p^{n_0}}-1\}\widetilde{g_0}(S)$.
We have
$p^{n_0}=[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D_{p}}] < \# \left( \mathbb{Z}_p/g_0(0)\mathbb{Z}_p \right)$
by Propositions \ref{index} and \ref{eqrational}.
Hence, we get
$\mathrm{ord}_p(\widetilde{g_0}(0)) > 0$,
where $\mathrm{ord}_p$ is the normalized additive valuation on $\mathbb{Q}_p$
of $p$-adic numbers such that $\mathrm{ord}_p(p)=1$.
Therefore, we obtain $\lambda(g_0(S)) \geq p^{n_0}$ because $\mu(g_0(S))$ is zero.
\end{proof}
By the proposition above, we can prove the inequality (B).
\begin{prop}\label{normal rankXM}
Assume the same condition as in Proposition \ref{cycsub}.
Assume also that $\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$
and that $k$ is not $p$-split $p$-rational.
Then
the inequality $\mathrm{(B)}$ holds.
\end{prop}
\begin{proof}
We have $\lambda(k_{\infty}^c /k)\geq 2$ since weak GGC does not hold.
By Proposition \ref{cycsub}, there exists a $\Lambda$-submodule $Y$ of $X_{\widetilde{k}}$
such that $Y$ is isomorphic to $\Lambda/(f(S,T))$.
We note that $Y_{\Gamma}$ is a finitely generated $\mathbb{Z}_p$-module
by the proof of Proposition \ref{rank}.
We claim that $\mathrm{rank}_{\mathbb{Z}_p}(Y_{\Gamma}) \geq p^{n_0}$.
By Lemma \ref{Zp}, there exists a unit $u$ such that
\[
\Gamma=\mathrm{Gal}(\widetilde{k}/N_{\infty}) = \overline{ \langle \sigma^{up^{s}} \tau \rangle.}
\]
Then we have
\begin{eqnarray*}
Y_{\Gamma} &\cong & \Lambda/(f(S,T), (1+S)^{up^s}(1+T)-1)\\
&\cong & \mathbb{Z}_p[[S]] /(f(S, (1+S)^{-up^s}-1)).
\end{eqnarray*}
By the definition of $f(S,T)$,
we obtain
\begin{eqnarray*}
&~&f(S,  (1+S)^{-up^s}-1)\\
&=&
\{(1+S)^{-up^s}-1\}^{\lambda^{\ast}} + \sum_{i=0}^{\lambda^{\ast}-1} g_{i}(S) \{ (1+S)^{-up^s}-1\}^{i}\\
&\equiv &
\left\{ \sum_{j=1}^{\infty} \dbinom{-u}{j}S^{jp^s} \right\}^{\lambda^{\ast}}+
\sum_{i=0}^{\lambda^{\ast}-1} g_{i}(S) \left\{ \sum_{j=1}^{\infty} \dbinom{-u}{j}S^{jp^s} \right\}^{i}
\quad \mathrm{~mod~}p\\
&\equiv &
\begin{cases}
g_0(S)  \quad \mathrm{~mod~}(p, S^{p^s}) \quad &\mathrm{~if~} \mu(g_0(S))=0,\\
0      \hspace{6.3mm} \quad \mathrm{~mod~}(p, S^{p^s}) \quad &\mathrm{~if~} \mu(g_0(S))>0.
\end{cases}
\end{eqnarray*}
By Proposition \ref{normal g}, we obtain $\lambda(f(S,  (1+S)^{-up^s}-1)) \geq p^{n_0}$.
This implies that
$\mathrm{rank}_{\mathbb{Z}_p}(Y_{\Gamma}) \geq p^{n_0}$.
Thus we obtain 
$\mathrm{rank}_{\mathbb{Z}_p}((X_{\widetilde{k}})_{\Gamma}) \geq p^{n_0}$
by Proposition \ref{rank}.
Using Proposition \ref{index}, we get the conclusion.
\end{proof}~\\
\subsection{Non-normal case}~
In this subsection, we prove  the inequality (B) in
the case where $\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$
under the assumption that weak GGC does not hold for $p$ and $k$.
We first show the following two propositions.
\begin{prop}\label{rankXM}
Assume the same condition as in Proposition \ref{cycsub}.
Assume also that $\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Suppose that $\mu(k_{\infty}^a/k)=0$,
then we have $\lambda(k_{\infty}^a/k) \geq [L_{k} \cap \widetilde{k}:k] + 1$.
\end{prop}
\begin{proof}
We suppose that $\mu(k_{\infty}^a/k)=0$.
By Proposition \ref{split},
$p$ splits completely in $L_k \cap \widetilde{k}$.
Hence we have $\lambda(k_{\infty}^a/k) \geq [L_{k} \cap \widetilde{k}:k]=p^s$.
We assume
that $\lambda(k_{\infty}^a/k) = [L_{k} \cap \widetilde{k}:k]$.
By the same reason as in the proof of Proposition \ref{normal g},
we have a surjective homomorphism
$
X_{k_{\infty}^a} \rightarrow \mathrm{Gal}(\widetilde{k_s^a}/k_{\infty}^a).
$
Then we have $\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a})$ $=\mathrm{char}_{\mathbb{Z}_p[[S]]}(\mathrm{Gal}(\widetilde{k_s^a}/k_{\infty}^a))$
because we suppose that $\lambda(k_{\infty}^a/k)$ $= [L_{k} \cap \widetilde{k}:k]$.
Using
$\mathrm{char}_{\mathbb{Z}_p[[S]]}(\mathrm{Gal}(\widetilde{k_s^a}/k_{\infty}^a)) = ((1+S)^{p^s}-1)$,
we get  $\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) =((1+S)^{p^s}-1).$
By Proposition \ref{cycsub}, we have a $\Lambda$-submodule $Y$ of $X_{\widetilde{k}}$
such that
$Y$
is isomorphic to $\Lambda/(f(S,T))$
and have a exact sequence
\[
0 \rightarrow Y \rightarrow X_{\widetilde{k}} \rightarrow \mathrm{Coker} \rightarrow 0.
\]
The $\Lambda$-modules
$(X_{\widetilde{k}})^{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)}$ and
$\mathrm{Coker}^{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)}$
are pseudo-null as $\Lambda$-modules
because
$T$ is coprime to $f(S,T)$.
Hence we get the exact sequence
\begin{eqnarray*}
0
\rightarrow   \mathrm{Coker}^{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)}
\rightarrow    Y_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)} \rightarrow (X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)} \rightarrow 
\mathrm{Coker}_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)} \rightarrow 0.
\end{eqnarray*}
Since we suppose that $\mu(k_{\infty}^a/k)=0$, $(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)}$ is a finitely
generated $\mathbb{Z}_p$-module.
By the same method as in the proof of Proposition \ref{rank},
we see that $Y_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^a)} $ is finitely generated as a $\mathbb{Z}_p$-module.
This implies that $\mu(g_0(S))=0$ and that
\[
\mathrm{rank}_{\mathbb{Z}_p}(X_{\widetilde{k}}/T X_{\widetilde{k}})
\geq \mathrm{rank}_{\mathbb{Z}_p}(\mathbb{Z}_p[[S]]/ g_0(S) \mathbb{Z}_p[[S]]).
\]
By Lemma \ref{f(S,T)}, there exists a surjective homomorphism
$\left( \Lambda /f(S,T) \Lambda \right)^{\oplus r} \rightarrow  X_{\widetilde{k}}$,
where $r$ is a positive integer.
Hence
this homomorphism induces a surjective homomorphism
$\left( \mathbb{Z}_p[[S]] /g_0(S) \mathbb{Z}_p[[S]] \right)^{\oplus r}
\rightarrow   X_{\widetilde{k}}/T X_{\widetilde{k}}$.
Then we have $(g_0(S)^r) \subset \mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{\widetilde{k}}/T X_{\widetilde{k}})$.
Since $(1+S)^{p^s}-1$ is square-free, we obtain
$\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{\widetilde{k}}/T X_{\widetilde{k}})$ 
$\subset (g_0(S)).$
This implies that
\[
\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) \subset (Sg_0(S)).
\]
Therefore there exists a power series $v(S)$ such that
$(1+S)^{p^s}-1=Sg_0(S)v(S)$.
Hence we have $p^s=g_0(0)v(0)$ and $s \geq \mathrm{ord}_p(g_0(0))$.
This contradicts Proposition \ref{index}.
Thus we get the conclusion.
\end{proof}
\begin{prop}\label{g0}
Assume the same condition as in Proposition \ref{cycsub}.
Assume also that $\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Suppose that $\mu(g_{0}(S)) =0$, then we have $\lambda(g_{0}(S)) \geq [L_{k} \cap \widetilde{k}:k]$.
\end{prop}
\begin{proof}
By the same argument in Proposition \ref{rankXM}, we have
\[
\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a})  \subset
\mathrm{char}_{\mathbb{Z}_p[[S]]}(\mathrm{Gal}(\widetilde{{k_{s}^a}}/k_{\infty}^a)) = ((1+S)^{p^s}-1)
\]
and have a surjective homomorphism
\[
\left( \mathbb{Z}_p[[S]] /g_0(S) \mathbb{Z}_p[[S]] \right)^{\oplus r}
\rightarrow  X_{\widetilde{k}}/T X_{\widetilde{k}}
\]
for some positive integer $r$.
Thus we get
\[
(Sg_0(S)^r) \subset \mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) \subset ((1+S)^{p^s}-1).
\]
Since $(1+S)^{p^s}-1$ is square-free, we have
$(Sg_0(S)) \subset  ((1+S)^{p^s}-1)$.
By the $p$-adic Weierstrass preparation theorem (\cite[Theorem 7.3]{Wa}),
there exist a distinguished polynomial $\widetilde{g_0}(S) \in \mathbb{Z}_p[[S]]$
and a unit $u(S) \in \mathbb{Z}_p[[S]]^{\times}$ such that 
\[
g_0(S) = \frac{(1+S)^{p^s}-1}{S}\widetilde{g_0}(S)u(S).
\]
Thus we obtain $\mathrm{ord}_p(\widetilde{g_0}(0)) >0$ by Proposition \ref{index}.
This implies that $\lambda(g_0(S)) \geq p^s$ if $\mu(g_{0}(S)) =0$.
\end{proof}
By the following proposition, we can prove the inequality $\mathrm{(B)}$.
\begin{prop}\label{fprop}
Assume the same condition as in Proposition \ref{cycsub}.
Assume also that $\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$.
Then the inequality $\mathrm{(B)}$ holds.
\end{prop}
\begin{proof}
We use the same method as in the proof of Proposition \ref{normal rankXM}.
By Proposition \ref{cycsub}, there exists a $\Lambda$-submodule $Y$ of $X_{\widetilde{k}}$
such that $Y$
is isomorphic to $\Lambda/(f(S,T))$.
We will prove that $\mathrm{rank}_{\mathbb{Z}_p}(Y_{\Gamma}) \geq p^s$.
There exists a unit $u$ such that
\[
\mathrm{Gal}(\widetilde{k}/N_{\infty}) = \overline{ \langle \sigma^{up^{s}} \tau \rangle.}
\]
Then we have
\begin{eqnarray*}
Y_{\Gamma}
\cong  \mathbb{Z}_p[[S]] /(f(S, (1+S)^{-up^s}-1)).
\end{eqnarray*}
By the proof of Proposition \ref{cycsub},
$Y_{\Gamma}$ is a finitely generated $\mathbb{Z}_p$-module.
Hence we have $\mu(f(S,(1+S)^{-up^s}-1))=0$.
Furthermore, we have
\begin{eqnarray*}
&~&f(S,  (1+S)^{-up^s}-1)
\equiv 
\begin{cases}
g_0(S)  \quad \mathrm{~mod~}(p, S^{p^s}) \quad &\mathrm{~if~} \mu(g_0(S))=0,\\
0       \hspace{6.3mm} \quad \mathrm{~mod~}(p, S^{p^s}) \quad &\mathrm{~if~} \mu(g_0(S))>0.
\end{cases} 
\end{eqnarray*}
Using Proposition \ref{g0}, we have $\mathrm{rank}_{\mathbb{Z}_p}(Y_{\Gamma}) \geq p^s$.
By Proposition \ref{rank}, we obtain  $\mathrm{rank}_{\mathbb{Z}_p}(X_{\widetilde{k}}) \geq p^s$.
This implies the inequality (B) from Proposition \ref{index}.
Thus we get the conclusion.
\end{proof}
Combining the results of Sections \ref{A} and \ref{B},
we have completed the proof of Theorem \ref{main thm}. 
\begin{Rem}\label{frem}
\begin{rm}
Theorem \ref{main thm} holds without the assumption (ii) if $X_{\widetilde{k}}$ is not cyclic as a $\Lambda$-module.
Indeed, we have proved the inequality (A) under the assumption (i) in Theorem \ref{main thm}.
Furthermore, we have verified the inequality (B) in Propositions \ref{normal rankXM} and \ref{fprop}.
In these propositions, we assumed (ii) in the case where $X_{\widetilde{k}}$ is not cyclic as a $\Lambda$-module.
\end{rm}
\end{Rem}
\section{Proof of Theorem \ref{main thm 2}}\label{rational case}
In this section, we prove Theorem \ref{main thm 2}.
In the following, we prove some lemmas and propositions needed later.
Let $k$ be a $p$-split $p$-rational field.
By Definition \ref{rational}, we have
$\widetilde{k}^{\mathfrak{D}_{{\mathfrak{p}}}}=\widetilde{k}^{\mathfrak{D}_{{\mathfrak{p}}^{\ast}}}$
and $[\mathrm{Gal}(\widetilde{k}/k): \mathfrak{D}_{{\mathfrak{p}}}] \geq p$.
We put
$\nu_{m}(S)=\displaystyle{((1+S)^{p^m}-1)/S}$ for a non-negative integer $m$.
As we stated in Section \ref{In}, we see that $X_{\widetilde{k}}$ is $\Lambda$-cyclic
by the following 
\begin{prop}[{\cite[Propositions 3.5 and 3.8]{Mu}}]\label{g00}
Assume that $L_k \subset \widetilde{k}$
and that $\lambda(k_{\infty}^c/k) \geq 2$.
Then we have a surjective homomorphism
$$
\Lambda/f(S,T) \Lambda \rightarrow X_{\widetilde{k}}
$$
as a $\Lambda$-module, where $f(S,T)$ is the same power series defined in Lemma \ref{f(S,T)}.
Furthermore,
if $\mathfrak{D_{p}}$ is a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$,
then there exists a power series $u(S) \in \mathbb{Z}_p[[S]]^{\times}$ such that
\[
f(S,0) =
\nu_{n_0}(S)u(S),
\]
where $n_0$ is the non-negative integer satisfying $[\mathrm{Gal}(\widetilde{k}/k) :\mathfrak{D}_{{\mathfrak{p}}}]=p^{n_0}$.
\end{prop}
We have $\lambda(N_{\infty}/k)=0$ by genus formula
(see for example \cite[Remark $3.2$]{Mu}).
By Proposition \ref{g00},
we can determine the isomorphism class of $X_{\widetilde{k}}$ as a $\Lambda$-module
provided that weak GGC does not hold for $p$ and $k$.
In fact, by Proposition \ref{isom str}, we have an isomorphism
\[
X_{\widetilde{k}} \cong \Lambda/ f(S,T) \Lambda.
\]
Using the isomorphism above,
we can determine the characteristic ideal of $X_{k_{\infty}^a}$.
\begin{prop}\label{ant-rational}
Let $k$ be a $p$-split $p$-rational field.
Assume that weak {\rm{GGC}} does not hold for $p$ and $k$. Then we have
\[
\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) = ((1+S)^{p^{n_0}}-1).
\]
\end{prop}
\begin{proof}
By Proposition \ref{isom str}, we have
\[
X_{\widetilde{k}}/TX_{\widetilde{k}} \cong \mathbb{Z}_p[[S]]/g_0(S) \mathbb{Z}_p[[S]].
\]
We note that $f(S,0)=g_0(S)$.
Hence we obtain $\mathrm{char}_{\mathbb{Z}_p[[S]]}(X_{k_{\infty}^a}) = ((1+S)^{p^{n_0}}-1)$
by Lemma \ref{Ozaki} and Proposition \ref{g00}.
\end{proof}
As in Section \ref{B}, we put $p^s = [L_{k} :k]$.
Then there exists a unit $u \in \mathbb{Z}_p^{\times}$ such that
\[
\mathrm{Gal}(\widetilde{k}/N_{\infty}) = \overline{ \langle \sigma^{up^{s}} \tau \rangle}
\]
by Lemma \ref{Zp}.\par
By the same methods as Sections \ref{A} and \ref{B},
we have the following
\begin{prop}\label{Zprank}
Under the same assumption as in Proposition \ref{ant-rational},
we have
\[
(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/N_{\infty})}
 \cong
\mathbb{Z}_p[[S]]/\nu_{n_0}(S) \mathbb{Z}_p[[S]].
\]
\end{prop}
\begin{proof}
By the same reason as in the proof of Proposition \ref{fprop},
we have
\[
(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/N_{\infty})} \cong \mathbb{Z}_p[[S]] /f(S,(1+S)^{-up^s}-1) \mathbb{Z}_p[[S]].
\]
We note that $\mu(f(S,(1+S)^{-up^s}-1))=0$.
Using Proposition \ref{rank uper}, we have
$\mathrm{rank}_{\mathbb{Z}_p}((X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/N_{\infty})}) \leq p^{n_0}-1$.
By the same method as in the proof of Propositions \ref{normal rankXM}, \ref{fprop}, and \ref{g00}, we obtain
$\lambda(f(S,(1+S)^{-up^s}-1)) \geq p^{n_0}-1$.
Thus we obtain $\lambda(f(S,(1+S)^{-up^s}-1)) = p^{n_0}-1$.
In the same way as Proposition \ref{(A)}, we can verify that
$M_{\mathfrak{p}^{\ast}}(N_{\infty})/N_{n_0}$ is an abelian extension.
This implies that $M_{\mathfrak{p}^{\ast}}(N_{\infty})=\widetilde{k_{n_0}^a}$.
Furthermore, we have $\widetilde{k_{n_0}^a} = L_{k_{\infty}^a}$
since $X_{k_{\infty}^a}$ is a free $\mathbb{Z}_p$-module
from the proof of Proposition \ref{ant-rational}.
Therefore we have
\[
(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/N_{\infty})} \cong \mathrm{Gal}( L_{k_{\infty}^a}/\widetilde{k})
\cong
\mathbb{Z}_p[[S]] / g_0(S) \mathbb{Z}_p[[S]].
\]
Thus we get the conclusion.
\end{proof}
We put $\rho = \sigma^{up^s}\tau $.
We fix an isomorphism
\begin{eqnarray}
\mathbb{Z}_p[[\mathrm{Gal}(\widetilde{k}/N_{\infty})]] \cong \mathbb{Z}_p[[T_s]] \quad (\rho \leftrightarrow 1+T_s). \label{rho}
\end{eqnarray}
Then we have $T_s =(1+S)^{up^s}(1+T)-1.$
We note that $\Lambda=\mathbb{Z}_p[[S,T_s]]$ since we have
$T=(1+T_s)(1+S)^{-up^s}-1$.
By the same method as Lemma \ref{f(S,T)}, we can find an annihilator of $X_{\widetilde{k}}$
as follows.
\begin{lem}\label{q(S,V)}
Assume the same conditions as in Proposition \ref{ant-rational}.
Then there exist power series $q(S,T_s) \in \mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})$ and
$r_i(T_s) \in \mathbb{Z}_p[[T_s]]$ $(i=0, \dots , p^{n_0}-2)$
such that
$$
q(S,T_s) = S^{p^{n_0}-1} + r_{p^{n_0}-2}(T_s) S^{p^{n_0}-2} + \dots + r_{1}(T_s)S +r_{0}(T_s).
$$
\end{lem}
\begin{proof}
Using the isomorphism (\ref{rho}) and Proposition \ref{Zprank}, we have
\[
X_{\widetilde{k}}/T_s X_{\widetilde{k}} \cong \mathbb{Z}_p^{\oplus p^{n_0}-1}.
\]
By Nakayama's lemma, there exist $y_i \in X_{\widetilde{k}} ~~(i=1,2, \dots, p^{n_0}-1)$
such that
$X_{\widetilde{k}}= \langle y_1, \dots, y_{p^{n_0}-1}
\rangle_{\mathbb{Z}_p[[T_s]]}$.
By the same method as Lemma \ref{f(S,T)}, we have relations
\[
Sy_i = \sum_{j=1}^{p^{n_0}-1}h_{ij}(T_s)y_j \quad (i=1,2, \dots, p^{n_0}-1)
\]
for some $h_{ij}(T_s) \in \mathbb{Z}_p[[T_s]]$.
We put $c=p^{n_0}-1$.
From theses relations, we have a matrix
\begin{eqnarray*}
A=
~\left(
\begin{array}{ccccc}
S-h_{11}(T_s) &  -h_{12}(T_s) & \dots & -h_{1 c}(T_s)\\
-h_{21}(T_s) &S-h_{22}(T_s) &  \dots & -h_{2 c}(T_s) \\
  \dots &  \dots  &   \dots      &  \dots \\
-h_{c 1}(T_s)  &  -h_{c 2}(T_s) & \dots & S-h_{c c}(T_s)
\end{array}
\right).
\end{eqnarray*}
We put $q(S,T_s) = \mathrm{det}(A)$.
As in Lemma \ref{f(S,T)}, $q(S,T_s)$ satisfies the desired result.
Thus we get the conclusion.
\end{proof}
We note that the uniqueness of the power series $q(S,T_s)$ is not known.
In this paper, we fix this power series.
\begin{lem}\label{q=f}
Assume the same conditions as in Proposition \ref{ant-rational}.
Then we have
\[
\mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})=(q(S,T_s)),
\]
where $q(S,T_s)$ is the same power series defined in Lemma \ref{q(S,V)}.
\end{lem}
\begin{proof}
By Proposition \ref{isom str} and Lemma \ref{q(S,V)},
we have $q(S,T_s) \in \mathrm{Ann}_{\Lambda}(X_{\widetilde{k}})=(f(S,T))$.
Then, there exists a power series $Q(S,T_s) \in \Lambda$ such that
\[
q(S,T_s) =f(S,(1+T_s)(1+S)^{-up^s}-1) Q(S,T_s).
\]
Hence we have $q(S,0)=f(S,(1+S)^{-up^s}-1)Q(S,0)$.
By the definition of $q(S,T_s)$, we have  $\mu(q(S,0))=0$ and $\lambda(q(S,0)) \leq p^{n_0}-1$.
By Lemma \ref{q(S,V)} and Proposition \ref{g00}, we have a surjective homomorphism
$
\Lambda/q(S,T_s) \Lambda \rightarrow X_{\widetilde{k}}.
$
This homomorphism induces a surjective homomorphism
$
\mathbb{Z}_p[[S]]/q(S,0) \mathbb{Z}_p[[S]] \rightarrow
\mathbb{Z}_p[[S]]/f(S,(1+S)^{-up^s}-1) \mathbb{Z}_p[[S]].
$
Using Lemma \ref{Zprank},
we obtain
\[
\mathrm{rank}_{\mathbb{Z}_p}(q(S,0)) \geq p^{n_0}-1.
\]
Thus the surjective homomorphism above is an isomorphism.
Hence we obtain $Q(S,0) \in \mathbb{Z}_p[[S]]^{\times}$.
This implies that $Q(S,T)$ is a unit in $\Lambda$.
Thus we get the conclusion.
\end{proof}
We put $H=N_{s+1}^{\ast}$ and $H_{\infty}=N_{\infty} N_{s+1}^{\ast}$.
Then $H_{\infty}/H$ is a $\mathbb{Z}_p$-extension of $H$
unramified outside all prime ideals of $H$ lying above $\mathfrak{p}$.
We prove the following
\begin{prop}\label{Zprank2}
Under the same assumption as in Proposition \ref{ant-rational},
we have
\[
\mathrm{rank}_{\mathbb{Z}_p}((X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/H_{\infty})}) = p(p^{n_0}-1).
\]
\end{prop}
\begin{proof}
We note that
$\mathrm{Gal}(\widetilde{k}/ H_{\infty}) = \overline{ \langle \rho^p \rangle}$.
By Lemma \ref{q=f}, we have
\begin{eqnarray*}
(X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/H_{\infty})}
& \cong & \mathbb{Z}_p[[S,T_s]] /(q(S,T_s), (1+T_s)^p -1)\\
& \cong & (\mathbb{Z}_p[[T_s]]/((1+T_s)^p-1)\mathbb{Z}_p[[T_s]])^{\oplus p^{n_0}-1}\\
& \cong & \mathbb{Z}_p^{\oplus p(p^{n_0}-1)}.
\end{eqnarray*}
Thus we get the conclusion.
\end{proof}
We will prove Theorem \ref{main thm 2}.
\begin{proof}[Proof of Theorem \ref{main thm 2}]
We assume that weak {\rm{GGC}} does not hold for $p$ and $k$.
From Proposition \ref{Fuj}, $M_{\mathfrak{p^\ast}}(\widetilde{k})$ coincides with $L_{\widetilde{k}}$.
We note that $\mathfrak{p^{\ast}}$ splits completely in $N_{\infty}^{ D_{\mathfrak{p^{\ast}}}} = N_{n_0}$.
Hence all prime ideals of $H$
lying above $\mathfrak{p}^{\ast}$ do not split in $H_{\infty}$.
Let
\[
\{ \mathfrak{P}_{\infty,i}^{\ast}~|~1 \leq  i \leq p^{n_0} \}
\]
be the set of prime ideals of
$H_{\infty}$ lying above $\mathfrak{p}^\ast$.
We denote by $I_{\mathfrak{P}_{\infty,i}^{\ast}}$ the inertia subgroup of
$\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(H_{\infty}) /H_{\infty})$
for the prime $\mathfrak{P}_{\infty,i}^{\ast}$,
where $M_{\mathfrak{p^{\ast}}}(H_{\infty})$ is the maximal abelian pro-$p$ extension of $H_{\infty}$
unramified outside all prime ideals of $H_{\infty}$ lying above $\mathfrak{p}^{\ast}$.
We have $\mathrm{Gal}(M_{\mathfrak{p^\ast}}(H_{\infty})/\widetilde{k}) \cap I_{\mathfrak{P}_{\infty,i}^{\ast}} =1$ for each $i$
because $M_{\mathfrak{p^{\ast}}}(H_{\infty} ) / \widetilde{k}$ is unramified.
Hence we have $I_{\mathfrak{P}_{\infty,i}^{\ast}} \cong \mathbb{Z}_p$.
Using \cite{Gi, sch, OV},
we obtain $\mu(H_{\infty} / H)=0$
since $H/k$ is an abelian extension.
This implies that
$\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(H_{\infty} )/ H_{\infty})$ is a finitely generated as a $\mathbb{Z}_p$-module.
Furthermore, since we suppose that
$\lambda(H_{\infty}/ H)=0$, $L_{H_{\infty} }/H_{\infty}$ is a finite extension.
Thus we have
$\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(H_{\infty} )/ H_{\infty})) \leq  p^{n_0}$.
Therefore we obtain
\[
\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(H_{\infty})/ \widetilde{k})) \leq p^{n_0}-1.
\]
On the other hand, we have
\begin{eqnarray*}
\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M_{\mathfrak{p^{\ast}}}(H_{\infty} )/ \widetilde{k}))
&=&
\mathrm{rank}_{\mathbb{Z}_p}((X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/H_{\infty})})\\
&=& 
p(p^{n_0}-1)\\
&>&
p^{n_0}-1
\end{eqnarray*}
by Proposition \ref{Zprank2}.
This is a contradiction.
Thus we have completed the proof.
\end{proof} 
\section{Numerical examples}\label{example}
In this section, we introduce some numerical examples which were computed using PARI/GP.\par
Let $A_{k_{n}^c}$ be the $p$-Sylow subgroup of the ideal class group of $k_{n}^{c}$.
Then, by class field theory, we have
$\displaystyle{X_{k_{\infty}^{c}} \cong
\plim[n]A_{k_n^{c}}
}$, where the inverse limit is
taken with respect to the relative norms. 
As in Sections \ref{Preli} and \ref{decomp}, 
$X_{k_{\infty}^{c}}$ is a finitely generated torsion $\mathbb{Z}_p[[T]]$-module
via an fixed isomorphism (\ref{isom cyc}) in Section \ref{decomp}.
Let $h(T)$ be the distinguished polynomial which generates
the characteristic ideal of $X_{k_{\infty}^{c}}$.
We can calculate the polynomial $h(T)$ mod $p^n$ for small $n$ numerically
using PARI/GP.
By this method,
we can check whether the characteristic ideal of $X_{k_{\infty}^c}$ has a square-free generator.
\par
We first give a criterion whether weak GGC holds,
using the characteristic ideal of $X_{k_{\infty}^{c}}$:
\begin{prop}\label{ex1}
Assume that $L_k \subset \widetilde{k}$, $\lambda(k_{\infty}^c/k)\geq 2$, and that
$\mathrm{ord}_p(g_0(0)) > [L_k :k]$,
where $g_0(S)$ is the same power series defined in Lemma \ref{f(S,T)}.
Assume also that
the characteristic ideal of $X_{k_{\infty}^c}$
has a square-free generator.
Then weak {\rm{GGC}} holds for $p$ and $k$.
\end{prop}
\begin{proof}
We have $\lambda(N_{\infty}/k)=0$ by $L_k \subset \widetilde{k}$ from genus formula
(see for example \cite[Remark $3.2$]{Mu}).
By Proposition \ref{g00},
we have $\mathrm{ord}_p(g_0(0)) = [\mathrm{Gal}(\widetilde{k}/k) :\mathfrak{D}_{{\mathfrak{p}}}]$.
From $\mathrm{ord}_p(g_0(0)) > [L_k :k]$,
$\mathfrak{D_{p}}$ is not a normal subgroup of $\mathrm{Gal}(\widetilde{k}/\mathbb{Q})$
by Lemma \ref{cond normal}.
Using Proposition \ref{eqrational}, we see that $k$ is not a $p$-split $p$-rational
field.
Furthermore, $X_{\widetilde{k}}$ is cyclic as a $\Lambda$-module by Proposition \ref{g00}.
Therefore weak GGC holds for $p$ and $k$
by Theorem \ref{main thm} and Remark \ref{frem}.
\end{proof}~\par
We use the following criteria to check whether $L_{k}$ is contained in $\widetilde{k}$.
\begin{lem}[{\cite[Corollary of Proposition 6.B]{Mi} and \cite[Theorem 1]{Br0}}]\label{brink}
Let $k=\mathbb{Q}(\sqrt{-d})$ with a square-free positive integer $d$.\\
$\mathrm{(i)}$ 
If $p=3$ and $d \not\equiv 3 \mathrm{~mod~}9$, then
$L_k \subset \widetilde{k}$
if and only if 
the class number of $\mathbb{Q}(\sqrt{3d})$ is not divisible by $3$.\\
$\mathrm{(ii)}$ Assume that $p\geq 5$ and
that $k$ has the same $p$-class number as $k({\zeta_p})$,
then
$L_k \subset \widetilde{k}$,
where $\zeta_p$ is a primitive $p$-th root of unity.
\end{lem}
As we stated in Section \ref{In}, we can prove GGC from weak GGC if the following condition is satisfied.
\begin{prop}[{\cite[Proposition $3.1$]{Fuj10}}]\label{Fu}
Suppose that weak {\rm{GGC}} holds for $p$ and $k$
and that
$\mathrm{char}_{\mathbb{Z}_p[[\mathrm{Gal}(k_{\infty}^c/k)]]}((X_{\widetilde{k}})_{\mathrm{Gal}(\widetilde{k}/k_{\infty}^c)})$
is a prime ideal.
Then {\rm{GGC}} holds for $p$ and $k$.
\end{prop}
Using Propositions \ref{ex1}, \ref{Fu}, and Lemma \ref{brink} above,
we can prove GGC for $(p,k)=(3,\mathbb{Q}(\sqrt{-971}))$, $(3,\mathbb{Q}(\sqrt{-17291}))$,
and $(5,\mathbb{Q}(\sqrt{-2239}))$.~\\
\begin{Ex}~\label{971}
\begin{rm}
Put $p=3$ and put $k=\mathbb{Q}(\sqrt{-971})$.
Then $3$ splits completely in $k$.
By PARI/GP, we have
$
A_k
\cong \mathbb{Z}/3\mathbb{Z}$.
We can check that $L_k \subset \widetilde{k}$ by Lemma \ref{brink}.
Indeed, the class number of $\mathbb{Q}(\sqrt{2913})$ is $7$. 
Hence, we have $\lambda(N_{\infty}/k)=0$.
Using PARI/GP,
we have 
\[
h(T)\equiv T^2+64638T~~\mathrm{mod}~3^{11}.
\]
By Hensel's lemma, there exists an integer $\alpha \in \mathbb{Z}_3$ such that
\[
h(T)=T(T+\alpha ),
\]
where $\alpha$ satisfies that $\alpha \equiv 486$ mod $3^6$.
Hence the characteristic ideal of $X_{k_{\infty}^c}$
has a square-free generator.
Let $f(S,T)$ be the same power series defined in Lemma \ref{f(S,T)}.
By Proposition \ref{char cyc},
we have $\mathrm{ord}_3(f(0,0))=\mathrm{ord}_3(\alpha)=5$.
Hence
$k$ is not $p$-split $p$-rational
and
weak GGC holds by Proposition \ref{ex1}.
Furthermore, by Proposition \ref{Fu}, GGC holds since we have $\lambda(k_{\infty}^c/k)=2$.
\end{rm}
\end{Ex}~\par
\begin{Ex}~\label{17291}
\begin{rm}
Put $p=3$ and put $k=\mathbb{Q}(\sqrt{-17291})$.
Then $3$ splits completely in $k$.
By PARI/GP, we have $A_k \cong \mathbb{Z}/3\mathbb{Z}$.
We can check that $L_k \subset \widetilde{k}$ by Lemma \ref{brink}.
Indeed, the class number of $\mathbb{Q}(\sqrt{51873})$ is $1$. 
Hence we have $\lambda(N_{\infty}/k)=0$.
Furthermore we have 
\[
h(T) \equiv T^4 + 405T^3 + 72T^2 + 522T ~~\mathrm{mod}~3^7.
\]
This implies that there exists an irreducible polynomial $g(T) \in \mathbb{Z}_p[[T]]$ such that
$h(T)=Tg(T).$
Hence the characteristic ideal of $X_{k_{\infty}^c}$
has a square-free generator.
Let $f(S,T)$ be the same power series defined in Lemma \ref{f(S,T)}.
By Proposition \ref{char cyc},
we have $\mathrm{ord}_3(f(0,0))=\mathrm{ord}_3(522)=2$.
Hence
$k$ is not $p$-split $p$-rational
and
weak GGC holds by Proposition \ref{ex1}.
Furthermore, by Proposition \ref{Fu}, GGC holds 
since $g(T)$ is irreducible.
\end{rm}
\end{Ex}~\par
\begin{Ex}~\label{2239}
\begin{rm}
Put $p=5$ and put $k=\mathbb{Q}(\sqrt{-2239})$.
Then $5$ splits completely in $k$.
By PARI/GP, we have $A_k \cong \mathbb{Z}/5\mathbb{Z}$.
We can check that $L_k \subset \widetilde{k}$ by Lemma \ref{brink}.
Indeed, the class number of $k(\zeta_5)$ is $560$. 
Hence we have $\lambda(N_{\infty}/k)=0$.
Furthermore we have 
\[
h(T)\equiv T^2+3100T~~\mathrm{mod}~5^5.
\]
By Hensel's lemma, there exists an integer $\alpha \in \mathbb{Z}_5$ such that
\[
h(T)=T(T+\alpha ),
\]
where $\alpha$ satisfies that $\alpha \equiv 100$ mod $5^3$.
Hence the characteristic ideal of $X_{k_{\infty}^c}$
has a square-free generator.
Let $f(S,T)$ be the same power series defined in Lemma \ref{f(S,T)}.
By Proposition \ref{char cyc},
we have $\mathrm{ord}_5(f(0,0))=\mathrm{ord}_5(\alpha)=2$.
Hence
$k$ is not $p$-split $p$-rational.
From Proposition \ref{ex1},
weak GGC for $p$ and $k$ holds.
Furthermore, by Proposition \ref{Fu}, GGC holds for $k$ 
since $\lambda(k_{\infty}^c/k)=2$.
\end{rm}
\end{Ex}~\par
Let $k_{\infty}/k$ be any $\mathbb{Z}_p$-extension.
For a non-negative integer $n$, 
let $k_n$ be the $n$-th layer of $k_{\infty}/k$ and $\mathcal{O}_{k_n}$ the ring of integers of $k_n$.
We denote by $A_{k_n}$ the $p$-Sylow subgroup of the ideal class group of $k_{n}$.
We define a homomorphism $i_{k_n}:A_{k} \rightarrow A_{k_n}$
by sending the ideal class of $\mathfrak{a}$ to the class of $\mathfrak{a}\mathcal{O}_{k_n}$
for every ideal $\mathfrak{a}$ of $k$.\par
We use the following criteria to check whether $\lambda(N_{\infty}/k)=0$.
\begin{prop}[{\cite[Proposition 1.C]{Mi}}]\label{Mi-cri}
The Iwasawa $\lambda$-invariant of $k_{\infty}/k$ is zero
if and only if
$A_{k} = \displaystyle{\bigcup_{n\geq 0} \mathrm{Ker}(i_n)}$, in other words,
every ideal class of $A_{k}$ capitulates in $k_{\infty}$.
\end{prop}
\begin{prop}[{\cite[Theorem 1]{Fuku}}]\label{Fukuda}
Let $k_{\infty}/k$ be any $\mathbb{Z}_p$-extension.
Let $c$ be a non-negative integer such that any prime ideal of $k$ which is ramified
in $k_{\infty}/k$ totally ramified in $k_{\infty}/k_{c}$.
If there exists an integer $n \geq c$ such that $\# A_{k_{n}} = \# A_{k_{n+1}}$,
then $\# A_{k_{m}} = \# A_{k_{n}}$ for all $m \geq n$.
In particular, we see that $\mu(k_{\infty}/k)=\lambda(k_{\infty}/k)=0$.
\end{prop}~\\
We can prove GGC for $(p,k)=(3,\mathbb{Q}(\sqrt{-5069}))$ by Theorem \ref{main thm} and propositions above.~\\
\begin{Ex}~\label{5069}
\begin{rm}
Put $p=3$ and put $k=\mathbb{Q}(\sqrt{-5069})$.
Then $3$ splits completely in $k$.
Using PARI/GP, we can check that
$A_k \cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$
and that
$\mathrm{Cl}_{k}$ is generated by the classes of the prime ideals lying above $5$ and $19$.
Using Fujii's criteria (\cite[Lemma 4.3]{Fuj10}), we have
$\mathrm{Gal}(\widetilde{k}/L_k \cap \widetilde{k}) \cong \mathbb{Z}/3\mathbb{Z}$.
Then we have 
\[
h(T)\equiv T(T +1989) ~~\mathrm{mod}~3^7.
\]
There exists an integer $\alpha \in \mathbb{Z}_3$ such that
$f(T)=T(T+\alpha)$
with $\alpha \equiv 45$ mod $3^5$.
Hence the characteristic ideal of $X_{k_{\infty}^c}$ 
is generated by a square-free generator.\par
Next we prove that $\lambda(N_{\infty}/k)=0$.
Using criteria in \cite[Theorem $2$]{Br}, we obtain,
\begin{eqnarray*}
x^6 - 2x^5 - 33x^4 - 70x^3 + 5462x^2 - 38784x + 83808
\end{eqnarray*}
as a defining polynomial of $k_{1}^{a}$ over $\mathbb{Q}$.
Then we see that $3$ splits completely in $k_1^{a}$.
Furthermore, we obtain
\begin{eqnarray*}
&~&x^{18} - 30x^{16} - 604x^{15} - 7705x^{14} - 45030x^{13} + 13536x^{12}+ 2239858x^{11}\\
&+& 34485072x^{10}+305384350x^9 + 2021144762x^8 + 11280115140x^7\\
&+& 53497712456x^6 + 198613209492x^5+ 669494384271x^4 + 1792552517970x^3\\
&+& 3479920548408x^2 + 6719313226248x + 8090872119549
\end{eqnarray*}
as a defining polynomial of $N_{2}$ over $\mathbb{Q}$.
Then all prime ideal classes above $5$ and $19$ become principal in $N_2$.
Hence $\lambda(N_{\infty}/k)=0$ by Proposition \ref{Mi-cri}.
We can also check this from Proposition \ref{Fukuda}.
Indeed, using PARI/GP, we see that
any prime ideal of $k$ which is ramified
in $N_{\infty}/k$ totally ramified in $N_{\infty}/k_{1}^a$.
Furthermore, $k_1^a$ and $N_2$ has the same $p$-class number
since $\mathrm{ord}_3(\# A_{k_1^a})=\mathrm{ord}_3(\# A_{N_2})=3$.
Hence we get $\lambda(N_{\infty}/k)=0$ by Proposition \ref{Fukuda}.
Therefore weak GGC holds by Theorem \ref{main thm}.
Furthermore, by Proposition \ref{Fu}, GGC holds
since $\lambda(k_{\infty}^c/k)=2$.
\end{rm}
\end{Ex}
\section*{Acknowledgements}
The author is sincerely grateful to
Professor Masato Kurihara and Professor Yoshitaka Hachimori
for their kind encouragement and valuable advice.
The author expresses his thanks to Professor
Satoshi Fujii and Doctor John Minardi for their beautiful studies in
\cite{Fuj10, Fuj} and \cite{Mi}.
The author would like to express his deepest appreciation
to Toshihiro Shirakawa
for his valuable advice about calculation
in Section \ref{example}.
