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This article is cited in 1 scientific paper (total in 1 paper) Arithmetic of certain L. V. Kuz'min National Research Centre "Kurchatov Institute", Moscow
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     § 1. IntroductionLet $\ell$ be a regular odd prime number, $k=\mathbb Q(\zeta_0)$, where $\zeta_0$ is a primitive root of unity of degree $\ell$, and $K=k(\sqrt[\ell]{a}\,)$, where $a$ is a natural number of the form $a=p_1^{r_1}p_2^{r_2}p_3^{r_3}$ such that its prime divisors $p_1$, $p_2$, $p_3$ remain prime in the cyclotomic $\mathbb Z_\ell$-extension $k_\infty$ of $k$ the field of type 2.1, according to the language of [1], or $a=p^rq^s$, where $p$ and $q$ are primes such that $p$ splits in $k_\infty$ into a product of two primes and $q$ remains prime (the field of type 2.2, according to the language of [1]). The arithmetic of $K$, which has many interesting features, was the subject of our investigations in [1]–[3]. The present paper continues these studies. In § 2, we give the necessary definitions and explain our notation. In § 3, we show how one can use the analogue of the Riemann–Hurwitz formula to get a certain information about the Tate module (the Iwasawa module) of $K_\infty$, where $K_\infty=k_\infty\cdot K$, and the definition of the Tate module $T_\ell(K_\infty)$ is given in § 2. The most interesting results are obtained in the case, where there are exactly three places ramified in the extension $K_\infty/k_\infty$ (in particular, this is the case of the fields of type 2.1 and 2.2). In this case, the module $T_\ell(K_\infty)$ is either infinite (and in this case, the exact action of the Galois group $\Gamma=G(K_\infty/K)$ on $T_\ell(K_\infty)$ is known; see Theorem 5.1 in [1]), or $T_\ell(K_\infty)$ is finite. Our main goal (which was not achieved in the present paper) was to prove the existence of fields $K$ of the above form with infinite module $T_\ell(K_\infty)$. We will give in § 2 the precise definition of the module $T_\ell(K_\infty)$ of and the modules $\overline T_\ell(K_\infty)$ and $R_\ell(K_\infty)$ to be used below. In this way, the most powerful results are obtained for $\ell= 3$, and in this case, the field $K$ is analogous, in a sense, to the field of rational functions on an elliptic curve. The meaning of this analogy will be explained in more detail in our forthcoming paper. Namely, we will explain the structure of the maximal unramified $\ell$-extension of $K$ for $\ell=3$. In § 4, we consider the structure of the $\ell$-component of the class group $\mathrm{Cl}_\ell(K)$ in the case $\ell=3$ and if $K$ is of type 2.2. We prove that either the group $\mathrm{Cl}_\ell(K)$ is generated by the prime divisors of $\ell$, or any such divisor is zero in this group. The group $\overline T_\ell(K_\infty)$, whose definition is also given in § 2, always has two generators, even though the group $\mathrm{Cl}_\ell(K)$ may be cyclic. If the Tate module is finite, we have $R_\ell(K_\infty)\cong (\mathbb Z/3\mathbb Z)^2$, where $R_\ell(K_\infty)$ is the subgroup of $\overline T_\ell(K_\infty)$ generated by the decomposition subgroups of all places over $\ell$. In this case, $|T_\ell(K_\infty)|=\ell^r$ for some even $r$. Note that these results are similar to those obtained in [2] for Case 2.1. In § 5, we consider the case of the field $K$ of type 2.1 and with finite $T_\ell(K_\infty)$. In this case, taking into account the action of the group $\Delta=G(k/\mathbb Q)$ on the group of cohomologies $H^0(H,U(K))$, where $H=G(K/k)$ and $U(K)$ is the group of units, we obtain some restrictions on the possible order of the group $\mathrm{Cl}_\ell(K)$. Namely, we show in Theorem 5.1 that $|\mathrm{Cl}_\ell(K)|=\ell^j$, where $j\leqslant \ell-1$. In addition, $j\neq \ell-2$, and $j$ is odd if $j<\ell-1$. In combination with the main result of Theorem 6.1 in [3], which asserts that $j\geqslant 2$, this result yields $j\geqslant 3$ (see Theorem 5.2 below). In the case $\ell=5$, we always have $j=4$. § 2. Notation and definitionsWe will mainly follow the notation of [1]–[3]. Let $\ell$ be a regular odd prime and $\zeta_n$ be a primitive root of unity of degree $\ell^{n+1}$. We put $k=\mathbb Q(\zeta_0)$ and define $k_\infty=\bigcup_{n=1}^\infty k_n$, where $k_n=k(\zeta_n)$. Let $K=k(\sqrt[\ell]{a}\,)$, where $a$ is a natural number such that the place $v$ of the field $k$ lying over $\ell$ splits completely in the extension $K/k$. This means that $a^{\ell-1}\equiv 1\pmod{\ell^2}$. In addition, we assume that there are exactly three places that ramify in $K_\infty/k_\infty$. Correspondingly, we assume that either $a=p_1^{r_1}p_2^{r_2}p_3^{r_3}$ or $a=p^rq^s$. In the first case, $p_1$, $p_2$, $p_3$ are primes that remain prime in the extension $k_\infty/\mathbb Q$, and in the second case, the prime $p$ splits in the only quadratic subfield $F$ of the field $k$ into the product $(p)=\mathfrak{p_1p_2}$, and each divisor $\mathfrak p_i$ remains prime in the extension $k_\infty/F$. The divisor $\mathfrak q=(q)$ remains prime in the extension $k_\infty/\mathbb Q$. So, we will speak about extensions of type 2.1 and 2.2. We denote the Galois group $G(K/\mathbb Q)$ by $G$, denote the group $G(K/k)$ by $H$, and denote the group $G(k/\mathbb Q)$ by $\Delta$. So, $G$ is a semi-direct product of $H$ and $\Delta$, and the group $\Delta$ acts on $H$ via the Teichmuller character $\omega\colon \Delta\to(\mathbb Z/\ell\mathbb Z)^\times$. If $A$ is some $G$-module, then $\Delta$ acts on the Tate group of cohomologies $H^i(H,A)$ for any $i$. We denote by $h'$ and $\delta'$ some fixed generators of the groups $H$ and $\Delta$, respectively. We denote by $\mathbb F_\ell(i)$ the group $\mathbb Z/\ell\mathbb Z$, such that $\Delta$ acts on it as $\omega^i$. The index $i$ is well-defined modulo $\ell-1$. Let $A$ be a finite $G$-module which is cyclic as an $H$-module and such that $N_H(A)=0$, where $N_H=\sum_{h\in H}h$ is the norm operator. Let $A=A_0\supseteq A_1\supseteq\dots \supseteq A_n=0$ be the low central series of an $H$-module $A$. If $A_0/A_1\cong \mathbb F_\ell(i)$ and $A_{n-1}\cong \mathbb F_\ell(j)$, then we say that $A$ starts with $\mathbb F_\ell(i)$ and ends with $\mathbb F_\ell(j)$. In this case, using Lemma 3.2 in [1], we have $A_k/A_{k+1}\cong \mathbb F_\ell(i+k)$ for any $k<n$. If $|A|=\ell^r$, then $r\equiv i-j\pmod{\ell-1}$. For an Abelian group $A$, we let $A[\ell]$ denote the pro-$\ell$-completion of $A$. If $K$ is a field of type 2.1, then, in parallel with $K$, we consider the field $L$ of the form $k(\sqrt[\ell]{b}\,)$, where $b=p_1^{s_1}p_2^{s_2}$ and $b^{\ell-1}\equiv 1\pmod{\ell^2}$. The prime numbers $p_1$, $p_2$ determine the field $L$ uniquely. It is clear that $K L/K$ is an Abelian unramified extension of degree $\ell$ such that all places in $S$ split completely in it, where $S$ is the set of all places over $\ell$. For an arbitrary algebraic number field $K$ (or its completion $K_v$ relative to some place $v$), we denote by $U(K)$ (or by $U(K_v)$) the group of units of this field. By $\mu(K)$ we denote the group of all roots of unity in $K$, by $\mu_\ell(K)$ we denote the $\ell$-component of $\mu(K)$, and by $\overline U(K)$ we denote the group $U(K)/\mu(K)$. We use analogous notation also for the field $K_v$. By $U^{(1)}(K_v)$ (respectively, $\overline U^{(1)}(K_v)$),we denote the group of principal units of $K_v$ (respectively, the group $U^{(1)}(K_v)/\mu_\ell(K_v)$). We also define
$$
\begin{equation*}
\mathcal{A}(K)=\prod_{v\in S}U^{(1)}(K_v),\qquad \overline{\mathcal{A}}(K)=\prod_{v\in S}\overline U^{(1)}(K_v)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathcal{A}(K_\infty)=\varprojlim \mathcal{A}(K_n),\qquad \overline{\mathcal{A}}(K_\infty)=\varprojlim\overline{\mathcal{A}}(K_n),
\end{equation*}
\notag
$$
where $K_n=k_n\cdot K$ and the limits are taken relative to the norm maps. We denote by $\mathscr D(K)$ the group of divisors of $K$ and by $\mathscr D^0(K)$ the subgroup of principal divisors. In the present paper, we will use an analogue of the Riemann–Hurwitz formula which was proved by the author in [4] (the improved proof is given in [5]). This formula, which is given in the next section, connects some linear combinations of the Iwasawa $\lambda$-invariants of certain Galois modules connected with some finite $\ell$-extension of the fields $L'/L$. Let us now give the definition of these Galois modules. Note that all the fields we consider here are $\ell$-extensions of the field $k$. Hence all the Galois modules under consideration have zero Iwasawa $\mu$-invariants. Let $L$ be an algebraic number field and $L_\infty$ be the cyclotomic $\mathbb Z_\ell$-extension of the field $L$. Next, let $\overline N$ be the maximal unramified Abelian $\ell$-extension of $L_\infty$ and $N$ be the maximal subfield of $\overline N$ such that all places of $S$ split completely in $N/L_\infty$. We denote the Galois groups of the extensions $N/L_\infty$ and $\overline N/L_\infty$ by $T_\ell(L_\infty)$ and $\overline T_\ell(L_\infty)$, respectively. These groups are torsion compact Noetherian modules relative to the action of the group $\Gamma=G(L_\infty/L)$, where $\Gamma\cong \mathbb Z_\ell$. We assume that some topological generator $\gamma_0$ of $\Gamma$ is fixed. So, these modules are modules relative to the action of the Iwasawa algebra $\Lambda=\mathbb Z_\ell[[\Gamma]]=\varprojlim\mathbb Z_\ell[\Gamma/\Gamma_n]$, where $\Gamma_n$ is the unique subgroup of $\Gamma$ of index $\ell^n$. By $R_\ell(L_\infty)$ we denote the kernel of the natural map $\overline T_\ell(L_\infty)\to T_\ell(L_\infty)$, that is, $R_\ell(L_\infty)$ is the subgroup of $\overline T_\ell(L_\infty)$ generated by the decomposition subgroups of all places in $S$. Let $M$ be the maximal Abelian $\ell$-extension of $L_\infty$ unramified outside $S$ and $X(L_\infty)=G(M/L_\infty)$. Then $X(L_\infty)$ is a $\Lambda$-module whose submodule of $\Lambda$-torsion is denoted by $\operatorname{Tors}X(L_\infty)$. The natural maps $X(L_\infty)\to \overline T_\ell(L_\infty)$ and $X(L_\infty)\to T_\ell(L_\infty)$ induce the maps $\operatorname{Tors}X(L_\infty)\to \overline T_\ell(L_\infty)$ and $\operatorname{Tors}X(L_\infty)\to T_\ell(L_\infty)$. We denote the images of these maps by $\overline T'_\ell(L_\infty) $ and $T'_\ell(L_\infty)$, respectively. We put $T''_\ell(L_\infty)=T_\ell(L_\infty)/T'_\ell(L_\infty)$, and denote the $\lambda$-invariants of $T'_\ell(L_\infty)$ and $T''_\ell(L_\infty)$ by $\lambda'(L_\infty)$ and $\lambda''(L_\infty)$, respectively. We put $R'(L_\infty)=R(L_\infty)\cap\overline T_\ell'(L_\infty)$ and define $R''(L_\infty)=R(L_\infty)/R'_\ell(L_\infty)$. By $r''(L_\infty)$ we denote the $\lambda$-invariant of $R''(L_\infty)$. In addition, we need another invariant $d(L_\infty)$, which is a $\lambda$-invariant of the $\Lambda$-module $D(L_\infty)$. This module is defined if $L_\infty$ is Abelian over $\ell$ and if $k\subset L$. For the definition of $D(L_\infty)$, see [1], § 6, where one can also found the definitions of the modules $V(L_\infty), V^+(L_\infty)$ and $V^-(L_\infty)$ that we use for the definition of $D(L_\infty)$. For a detailed description of this module and its properties, see [5]. Now we note only that $D(L_\infty)$ is defined as $V(L_\infty)/(V^+(L_\infty)\oplus V^-(L_\infty))$. In addition, note that $V(L_\infty)$, $V^+(L_\infty)$ and $V^-(L_\infty)$ are free $\Lambda$-modules. Note also that we often, without special mention, use additive notation for multiplication, since we never use the addition operation in this paper. § 3. Application of the analogue of the Riemann–Hurwitz formulaSo, let $\ell$ be a regular odd prime number, $k=\mathbb Q(\zeta_0)$ be the $\ell$th cyclotomic field, and $K/k$ be a cyclic extension of degree $\ell$. We also assume that $K$ is Abelian over $\ell$. This means that, for any place $v$ over $\ell$, the completion $K_v$ of the field $K$ is an Abelian extension of $\mathbb Q_\ell$. Let $k_\infty=\bigcup_n\mathbb Q(\zeta_n)$ be the cyclotomic $\mathbb Z_\ell$-extension of $k$ and $K_\infty=K\cdot k_\infty$ be the cyclotomic $\mathbb Z_\ell$-extension of the field $K$. Propositions 2.1–2.4 of [1] characterize the fields $K$ such that $K$ is Galois over $\mathbb Q$, $K$ is Abelian over $\ell$, and in the extension $K_\infty/k_\infty$, only three places $\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak p_3$ are ramified and do not lie over $\ell$. Such fields will be referred to as fields of type 2.$i$ or as Cases 2.$i$ for $i=1,2,3,4$. So, in $K_\infty/k_\infty$ there ramify only three places not lying over $\ell$ if and only if we have one of the following cases. Case 2.1. The places $\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak p_3$ lie over distinct prime numbers $p_1$, $p_2$, $p_3$, respectively. In this case, $K=k(\sqrt[\ell]{a}\,)$, where $a=p_1^{r_1}p_2^{r_2}p_3^{r_3}$, $r_i\not\equiv 0\pmod\ell$ for $i=1,2,3$ and $a^{\ell-1}\equiv 1\pmod{\ell^2}$. Note that $p_i$ remains prime in $k_\infty$, that is, $(p_i)=\mathfrak p_i$ if and only if $p_i$ is a primitive root modulo $\ell^2$. For any such triple $p_1$, $p_2$, $p_3$, there are exactly $\ell-2$ fields $K$ of such form. Case 2.2. The places $\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak p_3=q$ are such that $\mathfrak p_1$ and $\mathfrak p_2$ lie over the same prime $p\neq \ell$, whereas $\mathfrak q$ lies over prime $q\neq p,\ell$. In this case, $K=k(\sqrt[\ell]{a}\,)$, where $a$ is a natural number of the form $a=p^{r_1}q^{r_2}$, $r_1r_2\not\equiv 0\pmod\ell$. The prime $p$ splits in the unique quadratic subfield $F$ of the field $k$ into the product of two distinct prime divisors $\mathfrak p_1$ and $\mathfrak p_2$, each of which remains prime in the extension $k_\infty/F$. The divisor $\mathfrak q=(q)$ remains prime in the extension $k_\infty/\mathbb Q$. The number $a$ satisfies the congruence $a^{\ell-1}\equiv 1\pmod{\ell^2}$. Any such pair of primes $p$, $q$ defines the unique field $K$. Case 2.3. The ramified places $\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak p_3$ of the extension $K_\infty/k_\infty$ lie over the same prime $p$ and $\ell>3$. In this case, $\ell\equiv 1\pmod3$, and $p$ satisfies the congruence $p\equiv d^3\pmod{\ell^2}$, where $d$ is some primitive root modulo $\ell^2$. For each $p$ of this form, there are two such fields $K'/k$ and $K''/k$. The exact form of these fields is given in [1], formula (2.12). Case 2.4. Let $\ell=3$ and $p$ be a prime such that $(p)$ has exactly three prime divisors in $k_\infty$. This means that $p$ splits completely in the extension $\mathbb Q_1/\mathbb Q$, where $\mathbb Q_1$ is the first layer of the cyclotomic $\mathbb Z_\ell$-extension $\mathbb Q_\infty/\mathbb Q$, into the product of three distinct prime divisors $\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak p_3$, each of which remains prime in the extension $\mathbb Q_\infty/\mathbb Q_1$. The primes $p$ with this type of decomposition are characterized by the condition $p\equiv 8,17\pmod{27}$. The field $K_\infty$ that corresponds to a given $p$ is defined uniquely and is of the form $K_\infty=K\cdot k_\infty$, where $K=k(\sqrt[3]{p}\,)$. Thus, for $\ell=3$, only Cases 2.1, 2.2, and 2.4 are possible, and for $\ell>3$, only Cases 2.1, 2.2, and 2.3 are possible. In what follows, we need an analogue of the Riemann–Hurwitz formula that was obtained by the author in [4]. This formula gives a relation between Iwasawa $\lambda$-invariants of the Galois modules defined in § 2 for the fields $L_\infty'$ and $L_\infty$, where $L'/L$ is a finite $\ell$-extension of algebraic number fields and the fields $L'$ and $L$ satisfy some additional conditions. These conditions are as follows: the fields $L$ and $L'$ are Abelian over $\ell$, the field $L$ contains $\zeta_0$, that is, $L\supset k$ and all Galois modules entering the Riemann–Hurwitz formula have zero Iwasawa $\mu$-invariants. The field $k_\infty$ satisfies all these conditions since the corresponding Galois modules vanish. So, for any finite $\ell$-extension $L_\infty/k_\infty$, the corresponding Galois modules of the field $L_\infty$ have zero $\mu$-invariants. For the field $L_\infty$, we put
$$
\begin{equation}
g(L_\infty)=\frac{1}{2} d(L_\infty)+\lambda'(L_\infty) +2\lambda''(L_\infty)+2r''(L_\infty).
\end{equation}
\tag{3.1}
$$
The analogue of the Riemann–Hurwitz formula connects the invariant $2g(L_\infty)$ and the invariant $2g(L'_\infty)$ (defined in the same way) for the finite $\ell$-extension $L'/L$. Namely, under the above assumptions on the fields $L_\infty$ and $L'_\infty$,
$$
\begin{equation}
2g(L'_\infty)-2=[L'_\infty:L_\infty][2g(L_\infty)-2] +\sum_{v\nmid\ell}(e_v-1).
\end{equation}
\tag{3.2}
$$
This formula is an analogue of the Riemann–Hurwitz formula. Here $v$ runs through all places of $L_\infty'$ that are not over $\ell$, and $e_v$ means the ramification index of $v$ in the extension $L'_\infty/L_\infty$. Remark 3.1. If we consider an analogy between formula (3.2) and usual Riemann–Hurwitz formula, then the invariant $g(L_\infty)$ (see (3.1)) is an analogue of the genus of a curve. If this analogy is correct, we can hope that $g(L_\infty)$ assumes only integer non-negative values. Indeed, according to [4], on $D(L_\infty)$, there is a non-degenerate skew-symmetric product under the assumption that $L$ is Abelian over $\ell$. Hence the number $d(L_\infty)$ is always even. Remark 3.2. It is clear that $g(k_\infty)=0$. In the case $\ell=3$, applying (3.2) to the extension $K_\infty/k_\infty$, where $K_\infty/k_\infty$ is a cyclic extension of degree $\ell$ with exactly three ramified places not lying above $\ell$, we obtain $g(K_\infty)=1$. The following result is a direct consequence of (3.2) and Remark 3.2 Proposition 3.1. Let $\ell=3$ and $L_\infty/K_\infty$ be a finite unramified $\ell$-extension. Then $g(L_\infty)=1$. We also have the following result. Proposition 3.2. Let $\ell=3$, $K_\infty$ have the same meaning as in Remark 3.2, and $L_\infty/K_\infty$ be a finite unramified $\ell$-extension. Then at least one of the following two alternatives holds: If $L_\infty/K_\infty$ is a finite unramified extension, then both fields are of type (A) or are of type (B). Proof. The fact that the field $K_\infty$ satisfies one of the two alternatives (A), (B), was verified in Proposition 6.5 of [4]. This is also true for any $\ell$.
Assume that $\ell=3$. Consider, for example, the invariant $\lambda''(L_\infty)$. There are natural maps $i\colon T_\ell''(K_\infty)\to T_\ell''(L_\infty)$ and $N\colon T_\ell''(L_\infty)\to T_\ell''(K_\infty)$, of which the first one is induced by the inclusion $K_\infty\hookrightarrow L_\infty$, and the second one is induced by the norm map from $L_\infty$ into $K_\infty$. Since $N\circ i=[L_\infty:K_\infty]$, we have $\lambda''(L_\infty)\geqslant \lambda''(K_\infty)$. Analogous inequalities also hold for other $\lambda$-invariants from (3.1). In particular, if $K_\infty$ is of type (A), then $d(L_\infty)\geqslant 2(\ell-1)$. Hence by Proposition 3.1, the exact equality $d(L_\infty)=2(\ell-1)$ holds, and the remaining $\lambda$-invariants of $L_\infty$ vanish, that is, $L_\infty$ is of the type (A). Similarly, if $K_\infty$ is of type (B), then $\lambda'(L_\infty)\geqslant \ell-1$. Hence by Proposition 3.1, the exact equality $\lambda'(L_\infty)=\ell-1$ holds. The remaining $\lambda$-invariants of the field $L_\infty$ vanish, that is, $L_\infty$ is of type (B). This proves the proposition. Proposition 3.3. Let $L$ be an Abelian field over $\ell$, $k\subseteq L$, and the field $L_\infty$ satisfies $\lambda''(L_\infty)=r''(L_\infty)=d(L_\infty)=0$. Then the modulus $\overline T_\ell(L_\infty)$ contains no any non-trivial finite submodule. Proof. The module $D(L_\infty)$ is a factormodule of two free $\Lambda$-modules $V(L_\infty)$ and $V^+(L_\infty)\oplus V^-(L_\infty)$. Hence $D(L_\infty)$ contains no non-trivial finite submodule. Therefore, the condition $d(L_\infty)=0$ implies that $D(L_\infty)=0$, that is, For the field $L_\infty$, there is the exact sequence of $\Lambda$-modules where $\mathscr W(L_\infty)$ is the submodule of $X(L_\infty)$ generated by the inertia subgroups of all places over $\ell$.
For any intermediate field $L_n$ of the extension $L_\infty/L$, there is the exact sequence of Galois modules
$$
\begin{equation}
0\to\mathscr W(L_n)\to X(L_n)\to \mathrm{Cl}_\ell(L_n)\to 0,
\end{equation}
\tag{3.5}
$$
which is similar to (3.4). Here, $X(L_n)$ is the Galois group of the maximal Abelian $\ell$-extension of $L_n$ unramified outside $\ell$, $\mathrm{Cl}_\ell(L_n)$ is the Galois group of the maximal Abelian unramified $\ell$-extension of $L_n$ canonically isomorphic to the $\ell$-component of the class group of the field $L_n$, and $\mathscr W(L_n)$ is the Galois submodule of $X(L_n)$ generated by the inertia subgroups of all places over $\ell$.
The exact sequence (3.4) can be obtained from (3.5) by passing to the inverse limit relative to the norm maps $N_{L_m/L_n}$ for all pairs $m>n$. The resulting sequence remains exact since all modules entering (3.5) are compact. According to the global class field theory, $\mathscr W(L_n)\cong \mathcal{A}(L_n)/U(L_n)[\ell]$. Passing to inverse limit relative to the norm maps again, we obtain the exact sequence
$$
\begin{equation}
0\to U(L_\infty)\to\mathcal{A}(L_\infty)\to \mathscr W(L_\infty)\to 0.
\end{equation}
\tag{3.6}
$$
We put ${\bf D}(L_\infty)=\bigl(\varprojlim \prod_{v|\ell}\mu_\ell(L_{n,v})\bigr)/ (\varprojlim \mu_\ell(L_n))$, where the limits are taken relative to the norm maps. Setting $\overline{\mathscr W}(L_\infty)=\mathscr W(L_\infty)/{\bf D}(L_\infty)$, we verify that $\overline {\mathscr W}(L_\infty)$ is contained in the exact sequence of $\Lambda$-modules
$$
\begin{equation}
0\to \overline U(L_\infty)\to\overline {\mathcal{A}}(L_\infty) \to \overline{\mathscr W}(L_\infty)\to 0.
\end{equation}
\tag{3.7}
$$
By [5], formula (7.5), we have $r''(L_\infty)=\lambda (V^+(L_\infty)/\overline{\mathcal{A}}^+(L_\infty))$, where $\overline{\mathcal{A}}^+(L_\infty) =\overline{\mathcal{A}}(L_\infty)\cap V^+(L_\infty)$. The condition $r''(L_\infty)=0$ means that $V^+(L_\infty)/\mathcal{A}^+(L_\infty)$ is finite, but both modules $V^+(L_\infty)$ and $\mathcal{A}^+(L_\infty)$ are free as $\Lambda$-modules. Hence On the other hand, $\lambda(\mathcal{A}^+(L_\infty)/\overline U(L_\infty))=0$ and both modules $\mathcal{A}^+(L_\infty)$ and $\overline U(L_\infty)$ are free as $\Lambda$-modules. Hence $\overline U(L_\infty)=\mathcal{A}^+(L_\infty)=V^+(L_\infty)$. Now (3.7) implies the exact sequence
$$
\begin{equation*}
0\to\mathcal{A}(L_\infty)/V^+(L_\infty)\to V(L_\infty)/V^+(L_\infty) \to V(L_\infty)/\mathcal{A}(L_\infty)\to 0.
\end{equation*}
\notag
$$
The module $V(L_\infty)/V^+(L_\infty)$ is $\Lambda$-free by (3.3), and $V(L_\infty)/\mathcal{A}(L_\infty)\cong \mathbb Z_\ell^s$ as a $\mathbb Z_\ell$-module, where $s$ is the number of places over $\ell$ in the field $L_\infty$. Hence $\overline{\mathscr W}(L_\infty)$ is a free $\Lambda$-module. Thus we have the exact sequence
$$
\begin{equation*}
0\to \overline{\mathscr W}(L_\infty) \to \overline X(L_\infty) \to \overline T_\ell(L_\infty)\to 0,
\end{equation*}
\notag
$$
which contains the free module $\overline{\mathscr W}(L_\infty)$. Therefore, $\overline T_\ell(L_\infty)$ contains a non-trivial finite submodule if and only if $\overline X(L_\infty)$ contains a non-trivial finite submodule. But this is impossible because of Theorem 7.1 of [6]. This proves Proposition 3.3. § 4. The Tate module for the fields of type 2.2In what follows, we need to generalize for $\ell=3$ and fields of type 2.2 some results that were obtained for the fields of type 2.1 in [1] and [2]. The next assertion refines Theorem 4.2 of [1] to the case of $\ell=3$ and extensions of type 2.2. Proposition 4.1. Let $\ell=3$ and let $K/k$ be an extension of type 2.2. Then $\overline T_\ell(K_\infty)\neq 0$, and, either any prime divisor of $\ell$ is principal in the group $\mathrm{Cl}_\ell(K)$, or the group $\mathrm{Cl}_\ell(K)$ is generated by the images of all prime divisors of $\ell$. Proof. Under the assumptions of the proposition, $k=F=\mathbb Q(\sqrt{-3})$. Let $K=k(\sqrt[3]{a}\,)$, where $a=p^{r_1}q^{r_2}$, and $p$ splits in $k$ into a product of two primes: $p=\pi_1\pi_2$, where $q$ remains prime. We can assume that $\pi_1$ and $\pi_2$ are conjugate by the unique automorphism $\delta$ of $k$. We can also consider $\delta$ as an automorphism of the local field $\mathbb Q_3(\sqrt{-3})$. Let $U$ be the group of units of $\mathbb Q_3(\sqrt{-3})$. The action of $\delta$ implies the decomposition $U=U^+\oplus U^-$, where $\delta$ acts on $U^+$ and $U^-$ as the multiplication by $+1$ and $-1$, respectively. (Recall that we use additive notation for multiplication.)
The element $b\colon=\pi_1\pi_2^{-1}$ belongs to the group $U^-$, which (as a $\mathbb Z_3$-module) is isomorphic to $\mathbb Z_3\oplus (\mathbb Z/3\mathbb Z)$. Here, the direct summand $\mathbb Z_3$ is generated by a primary element, and the direct summand $(\mathbb Z/3\mathbb Z)$ is generated by a cubic root of unity. The group $U^+$ is isomorphic to $\mathbb Z_3^\times\cong \{\pm 1\}\oplus (1+3\mathbb Z_3)$. For any $u\in U$, we have the decomposition $u=u^+\oplus u^-$. Let $u^+$ satisfy $u^+\equiv 1\pmod 3$ and $u^+\not\equiv 1\pmod 9$. By the Chebotarev density theorem, for any $u\in U$, there are infinitely many primes $\pi\in k$ such that $(\pi)=\mathfrak p$ is a prime divisor prime with $3$ and the image of $\pi$ in $U$ coincides with the given element $u=u(\pi)$ modulo $U^3$. We put $\pi=\pi_1$, $\pi_2=\delta(\pi)$ and $p=\pi_1\pi_2$. With our choice of $u^+$, $\pi_1$ and $\pi_2$ remain prime in the extension $k_\infty/k$. Let us consider the component $u^-$. In the case where $u^-$ is a cube or a cube multiplied by a root of unity in the field $\mathbb Q_3(\sqrt{-3})$, the extension $K_\infty(\sqrt[3]{b}\,)/K_\infty$ is unramified, and all places over $\ell$ split in it. In the case where $u^-$ is a primary element or a primary element up to multiplication by a root of unity, the extension $K(\sqrt[3]{b}\,)/K$ is unramified, and any place $v$ of the field $K$ over $\ell$ remains prime in this extension. In other words, the Galois group of the extension $K_\infty(\sqrt[3]{b}\,)/K_\infty$ coincides with the decomposition subgroup of any place of $K_\infty$ over $\ell$. In either of this two cases, $0\neq \overline T_\ell(K_\infty)=R_\ell(K_\infty)$. This proves the proposition. In what follows, we assume that the element $b$ is chosen so that the extension $K(\sqrt[3]{b}\,)/K$ is unramified. As explained in the above proof, this can be always ensured by multiplying $b$ by a suitable root of unity. The next assertion refines Proposition 4.4 of [1]. Proposition 4.2. Let $\ell=3$ and $K/k$ be an extension of type 2.2. Let, as above, $K=k(\sqrt[3]{a}\,)$ and an element $b=\pi_1\pi_2^{-1}$ be defined so that $K(\sqrt[3]{b}\,)$ is an unramified extension of $K$. Next, let $L=k(\sqrt[3]{b}\,)$ and $L=L_0\cdot k$, where $L_0$ is a cyclic cubic extension of $\mathbb Q$. If $q$ splits completely in $L_0/\mathbb Q$, then $\mathrm{Cl}_\ell(K)\cong(\mathbb Z/3\mathbb Z)^2$. If $b$ is a cube in the local field $\mathbb Q_3(\sqrt{-3}\,)$ and $P$ is the Hilbert $3$-class field of $K$, then all prime places over $\ell$ split completely in $P/K$. If $b$ is a primary element in $\mathbb Q_3(\sqrt{-3}\,)$, then the group $\mathrm{Cl}_\ell(K)\cong G(P/K)$ is generated by the images of prime divisors of $\ell$. If $q$ remains prime in the extension $L_0/\mathbb Q$ then $\mathrm{Cl}_\ell(K)\cong\mathbb Z/3\mathbb Z$. Again, if $b$ is a cube in $\mathbb Q_3(\sqrt{-3}\,)$, then all places over $\ell$ split completely in $P/K$. If $b$ is a primary element in $\mathbb Q_3(\sqrt{-3}\,)$, then the group $\mathrm{Cl}_\ell(K)$ is generated by the prime divisors of $\ell$. Proof. The case where $b$ is a cube in $\mathbb Q_3(\sqrt{-3}\,)$ and $q$ splits completely in $L_0/\mathbb Q$ was considered in Proposition 4.4 of [1], where it was proved that the group $\mathrm{Cl}_\ell(K)$ has at least two generators in this case. On the other hand, by Corollary 3.1 in [1], the group $\mathrm{Cl}_\ell(K)$ is of period at most $\ell$. In this case, $\mathrm{Cl}_\ell(K)\cong (\mathbb Z/3\mathbb Z)^2$, and all divisors that divide $\ell$ are principal in $\mathrm{Cl}_\ell(K)$. The last fact was verified in the proof of Proposition 4.4 of [1].
If $q$ splits completely and $b$ is a primary element in $\mathbb Q_3(\sqrt{-3}\,)$, then the argument repeats that of the proof of Proposition 4.4 in [1]. Now suppose that $\mathfrak q=(q)$ remains prime in the extension $L_\infty/\mathbb Q$. This means that the Frobenius automorphism corresponding to the divisor $\mathfrak Q=\mathfrak q^{1/\ell}$ generates the Galois group of the unramified extension $L/K$, that is, $\mathfrak Q$ is not principal in the class group $\mathrm{Cl}_\ell(K)$. Moreover, there is an epimorphism $\overline T_\ell(K_\infty)\to\mathrm{Cl}_\ell(K)$, and, by virtue Theorem 4.1 of [1], $\overline T_\ell(K_\infty)$ is a cyclic $H$-module. Hence $\mathrm{Cl}_\ell(K)$ is also a cyclic $H$-module, that is, the class $\operatorname{cls}\mathfrak Q$ generates $\mathrm{Cl}_\ell(K)$ as an $H$-module. But the divisor $\mathfrak Q$ is fixed under the action of any automorphism of $K$. Therefore, the Galois group $G=G(K/\mathbb Q)$ acts trivially on $\mathrm{Cl}_\ell(K)$. Hence $\mathrm{Cl}_\ell(K)\cong\mathbb Z/3\mathbb Z$. This proves the proposition. Proposition 4.3. In Case 2.2 for $\ell=3$, the group $\overline T_\ell(K_\infty)$ has two generators as a $\mathbb Z_3$-module. Proof. By Theorem 3.1 of [2], the module $\overline T_\ell(K_\infty)$ is contained in the cyclic $H$- module $E'(K_\infty)$, which has at most $\ell-1$ generators as a $\mathbb Z_\ell$-module. So it suffices to verify that the group $\overline T_\ell(K_\infty)$ has at least two generators. This claim is clear if the group $\mathrm{Cl}_\ell(K)$ has two generators. So, we only need to examine the case where $\mathrm{Cl}_\ell(K)\cong \mathbb Z/3\mathbb Z$. In this case, according to the proof of Proposition 4.2, the divisor $\mathfrak Q$ is not principal. We denote by $\mathfrak P_1$ and $\mathfrak P_2$ prime divisors in $K$ of the divisors $\mathfrak p_1$ and $\mathfrak p_2$, respectively. By Proposition 3.1 of [3], $|H^0(H,U(K))|= 3$, whence $|H^1(H,U(K))|=3^2$. Therefore, the divisors $\mathfrak Q$, $\mathfrak P_1$, $\mathfrak P_2$, which are ramified in $K/k$, generate the cyclic subgroup in $\mathrm{Cl}_\ell(K)$. Note that $\delta'$ acts trivially on $\mathfrak Q$ and acts by inversion on $\mathfrak P_1\mathfrak P_2^{-1}$. In addition, the divisor $\mathfrak Q$ is not principal. Hence the divisor $(b^{1/\ell})=\mathfrak P_1\mathfrak P_2^{-1}$ is principal in $K$. Hence $b$ is of the form where $(c)=\mathfrak P_1 \mathfrak P_2^{-1}$ and $u$ is some unit of the field $K$ such that $u$ is not a cube in $K$. But $u$ is either a cube in the completion $K_v$ of the field $K$ for any place $v$ over $\ell$, or $u$ is a primary element in $K_v$ for any $v \,|\, \ell$ depending on whether either all places over $\ell$ split in $L/K$ or remain prime.
In any case, the analogue of Proposition 4.2 of [2] does not hold for $K$, that is, the natural map $\varphi_0\colon U(K)[\ell]\to P(K)$ is not an epimorphism. Recall that the group $P(K)$ was defined as the kernel of the natural map $\prod_{v\,|\,\ell}\Gamma_v\to\Gamma$, where $\Gamma=G(K_\infty/K)$ and $\Gamma_v=G(K_{\infty,v}/K_v)$. Now we follow the proof of Proposition 4.2 of [2] for the cases A.1 and A.2, that is, for the cases where $\mathrm{Cl}_\ell(K)\cong (\mathbb Z/\ell\mathbb Z)^{\ell-1}$. Similarly to formula (4.2) of [2], we have the exact sequence
$$
\begin{equation}
0\to V_3\to G(F_2K_\infty/K_\infty)\to\mathrm{Cl}_\ell(K)\to0,
\end{equation}
\tag{4.2}
$$
where
In the present case, the order of $V_3$ is at least $3$, the order of $\mathrm{Cl}_\ell(K)$ is at least $3$, the group $G(F_2K_\infty/K_\infty)$ is of period $3$ since it is a factorgroup of $\overline T_\ell(K_\infty)$. The automorphism $\gamma_0$ acts by multiplication by $-2$ on this last group, whereas the group $\Gamma$ acts trivially on all groups in (4.2). Therefore, $G(F_2K_\infty/K_\infty)$ has at least two generators. Since $\overline T_\ell(K_\infty)$ has at most two generators, $\overline T_\ell(K_\infty)$ has exactly two generators and the order of $V_3$ is exactly $3$. This proves the proposition. The next assertion is a generalization of Theorem 4.1 of [2] to Case 2.2. Proposition 4.4. Let $\ell=3$, $K/k$ be an extension of type 2.2, and the module $\overline T_\ell(K_\infty)$ be finite. Then $R_\ell(K_\infty)\cong (\mathbb Z/3\mathbb Z)^2$. Proof. If $T_\ell(K_\infty)=0$, then $\overline T_\ell(K_\infty)=R_\ell(K_\infty)$. Hence by Proposition 4.3 the group $R_\ell(K_\infty)$ has two generators. Since the group $R_\ell(K_\infty)$ is generated by the Frobenius automorphisms of the prime divisors of $\ell$, the group $\Gamma$ acts trivially on $R_\ell(K_\infty)$. On the other hand, by Corollary 3.1 of [2], the group $\Gamma$ acts on $\overline T_\ell(K_\infty)$ via multiplication by $\sqrt{\varkappa}=-2$. Hence $R_\ell(K_\infty)$ is of period $3$, which proves the proposition in the case $T_\ell(K_\infty)=0$.
Now consider the case $T_\ell(K_\infty)\neq 0$. In this case, by Proposition 4.3, $T_\ell(K_\infty)$ has exactly two generators as a $\mathbb Z_3$-module. Applying again Corollary 3.1 of [2], we get that $T_\ell(K_\infty)^\Gamma$ is the subgroup of all elements in $T_\ell(K_\infty)$ annihilated by multiplication by $\ell$. So, $T_\ell(K_\infty)^\Gamma\cong (\mathbb Z/3\mathbb Z)^2$. As in the proof of Theorem 4.1 of [2], we use Proposition 7.5 of [6], which asserts that the group $T_\ell(K_\infty)^\Gamma$ is naturally isomorphic to the factorgroup of the subgroup of local universal norms in the group $\overline U_S(K)[\ell]$ by the subgroup of global universal norms. To be more precise: an element $x\in \overline U_S(K)[\ell]$ is called a local universal norm in $K_\infty/K$ if, for any place $v|\ell$ of the field $K$ and any $n$, there is an element $x_{n,v}\in \overline K_{n,v}^\times[\ell]$ in the local field $K_{n,v}$ such that the norm of $x_{n,v}$ in the extension of local fields $K_{n,v}/K_v$ is $x$. Similarly, $x$ is called a global universal norm if, for any $n$, there is an element $x_n\in \overline U_S(K_n)[\ell]$ such that $N_{K_n/K}(x_n)=x$. We denote the group of the local universal norms by $\mathcal{U}_2(K)$ and denote the group of the global universal norms by $\mathcal{U}_1(K)$. Remark 4.1. The groups $\overline K^\times_{n,v}[\ell]$ and $\overline U_S(K_n)[\ell]$ are compact, and hence our definition of local and global universal norms is equivalent to the existence of coherent chains $\{x_{n,v}\}$, $n\geqslant 0$, $\{x_n\}$, $n\geqslant 0$, where $v$ is any place over $\ell$, where $x_{0,v}=x_0=x$, and all elements are compatible with respect to the norm maps. According to the local class field theory, an element $x\in\overline U_S(K)[\ell]$ is a local universal norm if and only if $\varphi_0(x)=0$, where $\varphi_0\colon \overline U_S(K)[\ell]\to P(K)$ is a natural map. As shown in the proof of Proposition 4.3, if $\mathrm{Cl}_\ell(K)\cong\mathbb Z/3\mathbb Z$, then the group $V_3$ is of order $3$. Hence the map $\varphi_0$ injects the group $\overline U(K)[\ell]$ into $P(K)$, and the image of this injection is a subgroup of index $3$ in $P(K)$. This means that $\overline U(K)[\ell]\cap \mathcal{U}_2(K)=1$. In other words, any element $x\in \mathcal{U}_2(K)$ is defined uniquely by its divisor. Now let us show that $\varphi_0(\overline U_S(K)[\ell])=\varphi_0(\overline U(K)[\ell])$. In other words, this means that by multiplying any $S$-unit $x$ by a suitable unit $u_1$, we can transform the element $xu_1$ into a local universal norm. Assume on the contrary that there is $x\in \overline U_S(K)[\ell]$ such that $\varphi_0(x)^3=u$, where $u$ is the element of (4.1). In this case, $x$ is not a local universal norm, but $x^3u^{-1}$ is a local universal norm. Let $\mathscr D_S(K_n)$ be the group of divisors of $K_n$ with supports in $S$. Then $\mathscr D_S(K_n)$ is a free Abelian group generated by all (three) places of $K_n$ over $\ell$. Since all such places are purely ramified in $K_\infty/K$, the norm map induces, for any $m>n\geqslant 0$, an isomorphism $\mathscr D_S(K_m)\cong \mathscr D_S(K_n)$. For any $n$, the period of $R_\ell(K_n)$ is at most $\ell$. Hence there is a sequence of elements $\{\alpha_n\}$, $n\geqslant 0$, $\alpha_n\in\overline K^\times_n[\ell]$, such that the principal divisors $(\alpha_n)$ satisfy $N_{K_n/K}((\alpha_n))=(x^\ell)$. Consequently, the sequence $\{N_{K_n/K}(\alpha_n)\}$ has a limit point $\alpha$ in $\overline U_S(K)[\ell]$, that is, $\alpha\in \mathcal{U}_1(K)$. But this contradicts the fact that $\mathcal{U}_2(K)/\mathcal{U}_1(K)\cong(\mathbb Z/3\mathbb Z)^2$. This proves Proposition 4.4. § 5. Restrictions on the order of an $H$-module $\mathrm{Cl}_\ell(K)$In this section we assume that $K$ is of type 2.1 or 2.2 and $\ell>3$. Hence either the divisors $\mathfrak p_1=(p_1)$, $\mathfrak p_2=(p_2)$, $\mathfrak p_3=(p_3)$ ramify in $K/k$, or the divisors $\mathfrak p_1$, $\mathfrak p_2$ ramify in $K/k$, where $\mathfrak p_1\mathfrak p_2=(p)$ and $\mathfrak q=(q)$, as explained in the beginning of § 3. Proposition 5.1. Assume that all three divisors $\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak p_3$ ($\mathfrak p_1$, $\mathfrak p_2$, $\mathfrak q$ in Case 2.2) present zero element of the group $\mathrm{Cl}_\ell(K)$. Then $|H^1(H,U(K))|=\ell^3$. Proof. In Case 2.1, this result follows from consideration of the exact sequence (3.3) of [1]. In Case 2.2, the proof goes along the same lines. This completes the proof. Proof. In Case 2.1, the claim is Proposition 3.3 of [3]. In Case 2.2, the proof is the same. This completes the proof. By the Dirichlet theorem, we have $|H^1(H,U(K))|=\ell|H^0(H,U(K))|$. Therefore, $|H^0(H,\overline U(K))|=1,\ell$, and this order is 1 if and only if, among the divisors that ramify in $K/k$, there is a divisor that presents a non-zero element of $\mathrm{Cl}_\ell(K)$. Suppose that $|H^0(H,\overline U(K))|=\ell$. Since the group $H^0(H,\overline U(K))$ is a $\Delta$-module, we have for some index $i$, which will be referred to as the degree of the corresponding class of cohomologies. Note that (5.1) defines the index $i$ uniquely modulo $\ell-1$.Proposition 5.3. Let the index $i$ be defined by (5.1). Then $i$ is even and $i\not\equiv 0\pmod{\ell-1}$. Proof. The group $\overline U(K)^H$ coincides with $\overline U(k)$. Therefore, there is an epimorphism $\overline U(k)/\overline U(k)^\ell\to H^0(H,\overline U(K))$, but $\overline U(k)/\overline U(k)^\ell \cong \bigoplus_{i=1}^{(\ell-3)/2}\mathbb F_\ell(2i)$. This proves the proposition. We have the diagonal injection $\overline U(K)\hookrightarrow \overline{\mathcal{A}}(K)$. Since $\overline{\mathcal{A}}(K)$ is a cohomologically trivial $H$-module, this injection induces the isomorphism
$$
\begin{equation}
\delta_0\colon H^{-1}(H,\overline{\mathcal{A}}(K)/\overline U(K)) \cong H^0(H,\overline U(K)).
\end{equation}
\tag{5.2}
$$
Proposition 5.4. Isomorphism (5.2) induces the isomorphism of $\Delta$-modules Proof. We have only to check that the connecting homomorphism $\delta_0$ in (5.2) commutes with the action of $\Delta$.
Note that this is not so in general for the exact sequence of $G$-modules $0\to A\to B\to C\to 0$ and the connecting homomorphism $\delta_1\colon H^0(H,C)\to H^1(H,A)$. The map $\delta_0$ is given explicitly as follows. Let $x\in \overline{\mathcal{A}}(K)/\overline U(K)$ be an element that presents a non-zero class in $H^{-1}(H,\overline{\mathcal{A}}(K)/\overline U(K))$. This means that $N_H(x)=0$, but $x\notin (h'-1)(\overline{\mathcal{A}}(K)/\overline U(K))$. Let $\overline x$ be a representative of $x$ in the group $\overline{\mathcal{A}}(K)$. Then $N_H(\overline x)\in \overline U(K)^H$ and $\delta_0 (x)=N_H(\overline x)$. To verify that $\delta_0$ commutes with the action of $\Delta$, we note that, for any $\delta\in\Delta$, where $h^\delta=\delta h\delta^{-1}$. If $h$ runs through $H$, then the elements $h^\delta$ vary over $H$, and so $\delta(N_H(\overline x))=N_H(\delta(\overline x))$, that is, the connecting homomorphism $\delta_0$ preserves the degree of the class of cohomologies. This proves the proposition.Theorem 5.1. In Case 2.1, assume that $|\mathrm{Cl}_\ell(K)|<\ell^{\ell-1}$. Then $|\mathrm{Cl}_\ell(K)|=\ell^{i-1}$, where $i$ has the same meaning as in (5.1). In other words $|\mathrm{Cl}_\ell(K)|=\ell^j$, where $j$ is odd, and $j\not\equiv \ell-2\pmod{\ell-1}$. Proof. Let $\widetilde K$ be the maximal Abelian $\ell$-extension of $K$ unramified outside $\ell$. Let $\widetilde A$ be the Galois group of the extension $\widetilde K/K$. Thus, $\widetilde A$ has a natural structure of $H$-module. According to the class field theory, $\widetilde A$ is contained in the exact sequence of $H$-modules Let $\widetilde A_1$ be the subgroup of $\widetilde A$ generated by the inertia subgroups of all places over $\ell$. Then $\widetilde A_1=\mathcal{A}(K)/U(K)[\ell]$, that is, $\widetilde A_1$ coincides with the kernel of $\psi$. In turn, $\widetilde A_1$ is contained in the exact sequence of $H$-modules where $\widetilde A_2=\bigl(\prod_{v|\ell}\mu_\ell(K_v)\bigr)/\mu_\ell(K)$ and $\widetilde A_3=\overline{\mathcal{A}}(K)/(\overline U(K)[\ell])$.
Let $\widetilde k$ be the maximal Abelian $\ell$-extension of the field $k$ unramified outside $\ell$, $\widetilde K_1$ be the maximal subfield of $\widetilde K$ that is Abelian over $k$, and $\widetilde B$ be the subgroup of $\widetilde A$ that fixes each element of $\widetilde K_1$. Then $\widetilde B=I_H\widetilde A$ and $N_H$ maps isomorphically $\widetilde A$ onto $G(\widetilde K_1/K)$. Since $H^{-1}(H,A)=\operatorname{Ker}_{N_H}A/I_H A$ for any $H$-module $A$ and $G(\widetilde k/k)= G(K\cdot\widetilde k/K)$, we get the exact sequence of $\Delta$-modules
$$
\begin{equation}
0\to H^{-1}(H, \widetilde A)\to \widetilde A/\widetilde B \to G(\widetilde k/k)\to 0,
\end{equation}
\tag{5.5}
$$
where $G(\widetilde k/k)$ is a free $\mathbb Z_\ell$-module of rank $(\ell+1)/2$. Hence the exact sequence (5.5) splits, and the group $H^{-1}(H,\widetilde A)$ coincides with the torsion subgroup of $\widetilde A/\widetilde B$. In particular, this subgroup has period $\ell$. For its precise description, we will employ the Kummer theory.
Let $k'/k$ be a cyclic extension of degree $\ell$. Then $k'=k(\sqrt[\ell]{x}\,)$ for some $x\in k^\times$. The inclusion $k'\subset \widetilde k$ is equivalent to the fact that $x$ is of the form $x=x_1y^\ell$, where $x_1\in U_S(k)$ and $y\in k^\times$, and the inclusion $k'\subset \widetilde K_1$ is equivalent to the fact that $x=x_1x_2y^\ell$, where $x_1$, $y$ are of the same meaning as above and $x_2$ is a product of primes $p_1$, $p_2$, $p_3$ with some exponents. In particular, if $x_1=1$ and $x_2=a$, then $k'=K$. Hence $|H^{-1}(H,\widetilde A)|=\ell^2$, and there is a natural isomorphism of the subgroup $H^{-1}(H,\widetilde A)$ in (5.5) onto the Galois group of the extension $k(\sqrt[\ell]{b},\sqrt[\ell]{p_3}\,)/k$. Since $p_3$ remains prime in $k_\infty$, we have $(p_3,\zeta_0)_\ell(k_v)=\zeta_0^z$ for some $z\not\equiv 0\pmod \ell$, where $(x,y)_\ell(k_v)$ is the norm residue symbol in the local field $k_v$ ($v$ is the place over $\ell$) with values in the group $\mu_\ell(k_v)$. Let $\mathfrak P_3$ be a prime divisor of $\mathfrak p_3=(p_3)$ in the field $K$, that is, $\mathfrak P_3^\ell=\mathfrak p_3$. Putting $\mathbf{k}=k(\sqrt[\ell]{p_3}\,)$, we get
$$
\begin{equation*}
(\sqrt[\ell]{p_3},\zeta_0)_\ell(\mathbf{k}) =(N_{\mathbf{k}/k}(\sqrt[\ell]{p_3}\,),\zeta_0)_\ell(\mathbf{k}) =(p_3,\zeta_0)_\ell(k_v)=\zeta_0^z
\end{equation*}
\notag
$$
for some $z\not\equiv 0 \pmod\ell$. This means that the group $\widetilde A_2$ in (5.4) acts non-trivially on $\mathbf{k}$, or, in other words, that the exact sequence (5.5) induces the injection $\mathbb F_\ell=H^{-1}(H,\widetilde A_2)\hookrightarrow H^{-1}(H,\widetilde A)$, whose cokernel is isomorphic $\mathbb F_\ell$.
Let $\widetilde K_2=\widetilde K^{\widetilde A_2}$ and let $\widetilde A'=\widetilde A/\widetilde A_2$ be the Galois group of the extension $\widetilde K_2/K$. Since $\widetilde K_2\supset \widetilde k$, we have the exact sequence of $G$-modules
$$
\begin{equation*}
0\to H^{-1}(H,\widetilde A')\to \widetilde A'/I_H\widetilde A' \to G(\widetilde k/k)\to 0,
\end{equation*}
\notag
$$
which is analogous to (5.5). This means that there is the exact sequence of $G$-modules
$$
\begin{equation}
0\to\widetilde A_3\to\widetilde A'\stackrel{\alpha}{\to}\mathrm{Cl}_\ell(K)\to 0,
\end{equation}
\tag{5.6}
$$
and the natural map $H^{-1}(H,\widetilde A_3)\to H^{-1}(H,\widetilde A')$ induced by this sequence is zero.
Let $x\in \operatorname{Ker}N_H(\widetilde A_3)$ presents a non-zero element of $H^{-1}(H,\widetilde A_3)$. Then $x=(h'-1)y$ for some $y\in \widetilde A'$. If $\overline y$ is the image of $y$ in $\mathrm{Cl}_\ell(K)$ under the map $\alpha$ of (5.6), then $\overline y\in \mathrm{Cl}_\ell(K)^H$ and the multiplication by $(h-1)$ defines an isomorphism $H^0(H,\mathrm{Cl}_\ell(K))\to H^{-1}(H,\widetilde A_3)$ of degree $+1$ relative to the action of $\Delta$. In other words, if $H^{-1}(H,\widetilde A_3)\cong \mathbb F_\ell(i)$ as a $\Delta$-module, where $i$ has the same meaning as in Proposition 5.1, then, as by Lemma 3.2 of [1], we have $H^0(H,\mathrm{Cl}_\ell(K))\cong \mathbb F_\ell(i-1)$. Thus the $H$-module $\mathrm{Cl}_\ell(K)$ begins with $\mathbb F_\ell(1)$ and ends with $\mathbb F_\ell(i-1)$. Therefore, $|\mathrm{Cl}_\ell(K)| =\ell^{i-1}$, and the assertion of the theorem is secured by Proposition 5.3. This proves Theorem 5.1. As an immediate consequence of this theorem and Theorem 6.2 of [3], we get the following result. Theorem 5.2. Let $K$ be a field of type 2.1 and $\ell>3$. We put $|\mathrm{Cl}_\ell(K)|=\ell^j$. Then $j\geqslant 3$. If $j<\ell-1$, then $j$ is odd and $j\not\equiv \ell-2\pmod{\ell-1}$. In particular, $j=4$ for $\ell=5$. 
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