1 Introduction
Let p be a prime number. The p-class field tower of a number field k is the sequence of fixed fields corresponding to the derived series of the Galois group of the maximal unramified pro-p-extension
$\widetilde {L}(k)$
of k. The structure of the metabelian quotient of
$\mathrm {Gal}(\widetilde {L}(k)/k)$
determines the principalization of ideals of k in unramified abelian p-extensions of k. In nonabelian Iwasawa theory of
$\mathbb Z_p$
-extensions [Reference Ozaki21], the Galois groups of p-class field towers have been studied along the cyclotomic
$\mathbb Z_p$
-extension
$k_{\infty }$
of k, where the
$\mathbb Z_p$
-extension
$k_{\infty }/k$
is regarded as a tower consisting of its subextensions
$k_n$
of degree
$p^n$
over k. A main object is the Galois group
$\widetilde {G}$
of the maximal unramified pro-p-extension
$\widetilde {L}(k_{\infty })$
of
$k_{\infty }$
, which is isomorphic to the projective limit
$\varprojlim \mathrm {Gal}(\widetilde {L}(k_n)/k_n)$
with respect to the restriction mappings. The extension
$\widetilde {L}(k_{\infty })$
is also studied as the maximal pro-p-extension of k which is unramified outside p and cyclotomically ramified at p (see e.g., [Reference Salle24]).
The abelianization of
$\widetilde {G}$
is the unramified Iwasawa module X, which is the Galois group of the maximal unramified abelian pro-p-extension
$L(k_{\infty })$
of
$k_{\infty }$
. When k is a totally real number field, it is conjectured by Greenberg [Reference Greenberg9] that X is finite. Then moreover, Greenberg’s conjecture implies that
$\widetilde {G}$
is a fab pro-p group, i.e., any open subgroups of
$\widetilde {G}$
has finite abelianization (see [Reference Mizusawa15, Proposition 1]). Toward this open problem, it seems a considerable problem to find various examples of fab
$\widetilde {G}$
even in the case of metabelian
$\widetilde {G}$
. In this paper, we focus on the case where
$p=2$
and k is a real quadratic field. Then there are a number of results and examples in the case of metacyclic
$\widetilde {G}$
(see [Reference Azizi, Rezzougui and Zekhnini2, Reference Mizusawa14–Reference Mizusawa16, Reference Mouhib18, Reference Mouhib and Movahhedi19] etc.). The following main theorem gives a family of real quadratic fields k with nonmetacyclic metabelian (possibly finite)
$\widetilde {G}$
.
Theorem 1.1 Let
$k_{\infty }$
be the cyclotomic
$\mathbb Z_2$
-extension of a real quadratic field
$k=\mathbb Q(\sqrt {\ell qr})$
with prime numbers
$\ell $
, q, r such that
$$ \begin{align} & \ell \equiv 9 \pmod{16},\ \ 2^{\frac{\ell -1}{4}} \equiv 1 \pmod{\ell},\ \ q \equiv 3 \pmod{8},\ \ q^{\frac{\ell -1}{2}} \equiv 1 \pmod{\ell}, \end{align} $$
$$ \begin{align} & r \equiv 5 \pmod{8}, \ \ r^{\frac{\ell -1}{2}} \equiv -1 \pmod{\ell}. \end{align} $$
Then the Galois group
$\widetilde {G}=\mathrm {Gal}(\widetilde {L}(k_{\infty })/k_{\infty })$
of the maximal unramified pro-
$2$
-extension
$\widetilde {L}(k_{\infty })$
of
$k_{\infty }$
is a metabelian pro-
$2$
group, and the following statements hold true:
-
(1)
$\widetilde {G}/[\widetilde {G},\widetilde {G}] \simeq [2,2,2]$
, i.e., the abelianization of
$\widetilde {G}$
is an elementary abelain 2-group of order
$8$
. -
(2) A maximal subgroup
$\mathrm {Gal}(\widetilde {L}(k_{\infty })/k_{\infty }(\sqrt {r}))$
of
$\widetilde {G}$
is an abelian pro-
$2$
group of 2-rank
$2$
. -
(3)
$\widetilde {G}$
is a finite 2-group if
$L(k^{\prime}_{\infty })/k^{\prime }_{\infty }$
is a finite extension, where
$L(k^{\prime }_{\infty })$
is the maximal unramified abelian pro-
$2$
-extension of the cyclotomic
$\mathbb Z_2$
-extension
$k^{\prime }_{\infty }$
of
$k'=\mathbb Q(\sqrt {\ell q})$
.
Remark 1.2 It is known that
$\widetilde {L}(k^{\prime }_{\infty })=L(k^{\prime }_{\infty })$
(see [Reference Mizusawa15, Lemma 9]). The finiteness of
$[L(k^{\prime }_{\infty }):k^{\prime }_{\infty }]$
is the assertion of Greenberg’s conjecture for
$k^{\prime }_{\infty }$
. Greenberg’s conjecture for
$\mathbb Z_2$
-extensions of real quadratic fields has been studied in various situations (see [Reference Fukuda and Komatsu6, Reference Kumakawa12, Reference Nishino20, Reference Ozaki and Taya22] etc.). One can find infinitely many examples of finite
$\widetilde {G}$
as in Example 3.2.
After proving Theorem 1.1, we also calculate the class-
$2$
quotients of
$\widetilde {G}$
and some examples. Moreover, we obtain the following explicit example of Theorem 1.1.
Example 1.3 (see Example 3.4 for the details)
For the triple
$(\ell ,q,r)=(73,3,5)$
, which satisfies the conditions (1.1) and (1.2),
$\widetilde {G}$
is a finite 2-group of order
$64$
and nilpotency class
$3$
having a presentation with three generators
$a,b,c$
and six relations
$$ \begin{align} [a,b]b^6c^2,\quad [a,c]c^2,\quad [b,c],\quad a^2b^4,\quad b^8,\quad c^4 , \end{align} $$
where
$[x,y]=x^{-1}y^{-1}xy$
.
Throughout the following sections, we assume
$p=2$
, and use the following notations. For a pro-
$2$
group G and its closed subgroup H,
$[H,G]$
denotes the minimal closed subgroup containing
$\{[h,g]\,|\,h \in H, g \in G\}$
. An abelian group
$\bigoplus _{i=1}^{d} \mathbb Z/2^{e_i}\mathbb Z$
is written as
$[2^{e_1},\ldots ,2^{e_d}]$
. For an abelian group A,
$|A|$
denotes the order of A, and
$A/2^{e}$
denotes the maximal quotient 2-group of exponent at most
$2^e$
. For a number field K and a finite set S of odd prime numbers,
$A_S(K)$
denotes the Sylow 2-subgroup of the ray class group of K modulo
$m=\prod _{l \in S}l$
, and
$\widetilde {L}_S(K)$
denotes the maximal pro-
$2$
-extension of K unramified outside S. Let
$L_S(K)$
be the maximal abelian subextension of
$\widetilde {L}_S(K)/K$
. Then
$A_S(K) \simeq \mathrm {Gal}(L_S(K)/K)$
via Artin map. Put
$A(K)=A_{\emptyset }(K)$
which is the Sylow 2-subgroup of the ideal class group of K, and put
$L(K)=L_{\emptyset }(K)$
which is the Hilbert 2-class field of K. Let
$\widetilde {L}'(K)$
be the maximal subextension of
$\widetilde {L}(K)/K$
in which any prime lying over 2 splits completely. Put
$L'(K)=L(K) \cap \widetilde {L}'(K)$
. Then
$A'(K)=A(K)/D(K) \simeq \mathrm {Gal}(L'(K)/K)$
via Artin map, where
$D(K)$
is the subgroup of
$A(K)$
generated by all ideal classes containing some products of prime ideals lying over
$2$
.
2 Proof of Theorem 1.1
Burnside’s basis theorem implies that a pro-
$2$
group G is pro-cyclic if
$G/[G,G]$
is pro-cyclic. Hence
$\widetilde {L}(K)=L(K)$
and
$|A(L(K))|=1$
if
$L(K)/K$
is a cyclic extension. Similarly,
$\widetilde {L}'(K)=L'(K)$
and
$|A'(L'(K))|=1$
if
$L'(K)/K$
is a cyclic extension. We often use these facts.
Suppose that
$k=\mathbb Q(\sqrt {\ell qr})$
with prime numbers
$\ell ,q,r$
satisfying (1.1) and (1.2). Then the quartic residue symbol
$\big (\frac {2}{\ell }\big )_4=1$
, and the quadratic residue symbols
$\big (\frac {q}{\ell }\big )=1$
,
$\big (\frac {r}{\ell }\big )=-1$
. For a number field K,
$\mathfrak {p}_K$
(resp.
$\mathfrak {l}_K$
,
$\mathfrak {q}_K$
,
$\mathfrak {r}_K$
) denotes a prime ideal of the integer ring
$\mathcal {O}_K$
of K lying over
$2$
(resp.
$\ell $
, q, r). Recall that
$K_n=K\mathbb Q_n$
if
$K \cap \mathbb Q_{\infty }=\mathbb Q$
, where
$\mathbb Q_n$
is the nth layer of the
$\mathbb Z_2$
-extension
$\mathbb Q_{\infty }$
of
$\mathbb Q$
. Note that
$\mathbb Q_1=\mathbb Q(\sqrt {2})$
.
Lemma 2.1
$A(\mathbb Q_n(\sqrt {\ell },\sqrt {r}))$
,
$A(\mathbb Q_n(\sqrt {q},\sqrt {r}))$
and
$A'(\mathbb Q_n(\sqrt {qr}))$
are trivial for all
$n \ge 0$
.
Proof. Since
$|A(\mathbb Q_1(\sqrt {\ell r}))|=2$
by [Reference Mizusawa17, Lemma 7.2],
$\widetilde {L}(\mathbb Q_1(\sqrt {\ell r}))=\mathbb Q_1(\sqrt {\ell },\sqrt {r})$
, and hence
$|A(\mathbb Q_1(\sqrt {\ell },\sqrt {r}))|=1$
. By Fukuda’s theorem [Reference Fukuda5, Theorem 1],
$|A(\mathbb Q_n(\sqrt {\ell },\sqrt {r}))|=1$
for all
$n \ge 0$
. Since the narrow genus field of
$\mathbb Q(\sqrt {qr})$
is
$\mathbb Q(\sqrt {-1}, \sqrt {-q}, \sqrt {r})$
,
$\mathbb Q(\sqrt {q},\sqrt {r})$
is the unique quadratic subextension of
$L(\mathbb Q(\sqrt {qr}))/\mathbb Q(\sqrt {qr})$
. Hence
$A(\mathbb Q(\sqrt {qr}))$
is cyclic. Since
$\mathfrak {p}_{\mathbb Q(\sqrt {qr})}$
is inert in
$\mathbb Q(\sqrt {q},\sqrt {r})$
, it is also inert in
$L(\mathbb Q(\sqrt {qr}))$
, i.e.,
$A(\mathbb Q(\sqrt {qr}))=D(\mathbb Q(\sqrt {qr}))$
. Since
$\mathfrak {p}_{\mathbb Q(\sqrt {qr})}^2$
is principal,
$|D(\mathbb Q(\sqrt {qr}))|=2$
. Hence
$\widetilde {L}(\mathbb Q(\sqrt {qr})) =\mathbb Q(\sqrt {q},\sqrt {r})$
, and
$|A(\mathbb Q(\sqrt {q},\sqrt {r}))|=1$
. Since
$\mathfrak {p}_{\mathbb Q(\sqrt {q},\sqrt {r})}$
is unique,
$|A(\mathbb Q_n(\sqrt {q},\sqrt {r}))|=1$
for all
$n \ge 0$
by Iwasawa’s theorem [Reference Iwasawa10]. Then
$\mathfrak {p}_{\mathbb Q_n(\sqrt {qr})}$
is inert in
$\widetilde {L}(\mathbb Q_n(\sqrt {qr})) =\mathbb Q_n(\sqrt {q},\sqrt {r})$
, and hence
$|A'(\mathbb Q_n(\sqrt {qr}))|=1$
for all
$n \ge 0$
.▪
Lemma 2.2
$A(k) \simeq [2,2]$
,
$|A'(k)|=2$
and
$A(k(\sqrt {\ell }))=D(k(\sqrt {\ell })) \simeq [2,2]$
.
Proof. Since the narrow genus field of k is
$\mathbb Q(\sqrt {-1}, \sqrt {\ell }, \sqrt {-q}, \sqrt {r})$
,
$K=k(\sqrt {\ell },\sqrt {q})$
is the maximal elementary abelian subextension of
$L(k)/k$
. Then
$A(k)/2 \simeq \mathrm {Gal}(K/k)$
via Artin map. Note that
$\mathfrak {p}_k^2$
and
$\mathfrak {r}_k^2$
are principal. Since
$\mathfrak {r}_k$
is inert in
$k(\sqrt {\ell })$
and
$\mathfrak {p}_{k(\sqrt {\ell })}$
is inert in K,
$A(k)$
is generated by the ideal classes of
$\mathfrak {p}_k$
and
$\mathfrak {r}_k$
, and hence
$A(k) \simeq [2,2]$
. Since
$|A'(k)|=2$
, we have
$\widetilde {L}'(k)=k(\sqrt {\ell })$
and
$|A'(k(\sqrt {\ell }))|=1$
. Put
$\widetilde {G}_0=\mathrm {Gal}(\widetilde {L}(k)/k)$
. Then
$H=\mathrm {Gal}(\widetilde {L}(k)/k(\sqrt {\ell }))$
is a maximal subgroup of
$\widetilde {G}_0$
such that
$A(k(\sqrt {\ell })) \simeq H/[H,H]$
. Since
$\widetilde {G}_0/[\widetilde {G}_0,\widetilde {G}_0] \simeq [2,2]$
,
$\widetilde {G}_0$
is either Klein four, dihedral, semidihedral, quaternion, or generalized quaternion (see e.g., [Reference Kisilevsky11]). Then
$H/[H,H] \simeq [2,2]$
or H is cyclic. Since
$|A(\mathbb Q(\sqrt {\ell }))|=1$
,
$A(k(\sqrt {\ell }))^{\tau +1} \simeq 0$
, where
$\tau $
is a generator of
$\mathrm {Gal}(k(\sqrt {\ell })/\mathbb Q(\sqrt {\ell }))$
. Since
$$ \begin{align*} |A(k(\sqrt{\ell}))/2| &=|\mathrm{Ker}(A(k(\sqrt{\ell})) \stackrel{\tau -1}{\longrightarrow} A(k(\sqrt{\ell})))| \\ &=\frac{2^5}{2 |\mathcal{O}_{\mathbb Q(\sqrt{\ell})}^{\times}/(\mathcal{O}_{\mathbb Q(\sqrt{\ell})}^{\times} \cap (k(\sqrt{\ell})^{\times})^{\tau +1})|} \ge 4 \end{align*} $$
by the genus formula for
$k(\sqrt {\ell })/\mathbb Q(\sqrt {\ell })$
(see e.g., [Reference Lemmermeyer13]), H is not cyclic, and hence
$A(k(\sqrt {\ell })) \simeq [2,2]$
.▪
Remark 2.3 It is well known that the 4-ranks of narrow class groups of quadratic fields can be calculated from Kronecker symbols of ramified prime numbers. The structure of
$\widetilde {G}_0$
can also be determined as in [Reference Benjamin and Snyder4]. Although we can use these results to prove Lemmas 2.1 and 2.2, we gave their proofs as above for the convenience.
Lemma 2.4
$A(k_1) \simeq [2,2,2]$
and
$A'(k_1) \simeq [2,2]$
.
Proof. By [Reference Azizi and Mouhib1, Théorème 1],
$A(k_1)/2 \simeq [2,2,2]$
. Since
$k_1(\sqrt {r}) \subset L(k_1)$
but
$k_1(\sqrt {r}) \not \subset L'(k_1)$
,
$D(k_1) \not \subset A(k_1)^2$
, and hence
$A'(k_1)/2 \simeq [2,2]$
. Let
$L/k_1$
be the unique
$[2,2]$
-subextension of
$L'(k_1)/k_1$
. Then
$A'(k_1)/2 \simeq \mathrm {Gal}(L/k_1)$
via Artin map. The octic Galois extension
$L/k$
is ramified only at
$\mathfrak {p}_k$
with the ramification index
$2$
. If
$L/k$
is abelian, the inertia field of
$\mathfrak {p}_k$
is a quartic subextension of
$L'(k)/k$
. However,
$L'(k)=k(\sqrt {\ell })$
by Lemma 2.2. This is a contradiction. Hence
$L/k$
is an octic dihedral extension. Since
$L/k_1(\sqrt {\ell })$
is unramified and
$k_1(\sqrt {\ell })/k(\sqrt {\ell })$
is ramified at
$\mathfrak {p}_{k(\sqrt {\ell })}$
,
$L/k(\sqrt {\ell })$
is not cyclic. Therefore
$\mathrm {Gal}(L/k(\sqrt {2\ell }))$
is the unique cyclic maximal subgroup of
$\mathrm {Gal}(L/k)$
. Since the class number of
$\mathbb Q_1$
is
$1$
,
$\mathfrak {p}_{k_1}^2$
,
$\mathfrak {l}_{k_1}^2$
and
$\mathfrak {r}_{k_1}^2$
are principal. Hence,
$|D(k_1)|=2$
. Note that
$\big (\frac {2\ell }{r}\big )=1$
. Since
$\mathfrak {r}_k$
is inert in
$k_1$
and splits in
$k(\sqrt {2\ell })$
,
$\mathfrak {r}_{k(\sqrt {2\ell })}$
is inert in
$k_1(\sqrt {\ell })$
, and hence it is also inert in L, i.e.,
$\mathfrak {r}_{k_1(\sqrt {\ell })}$
is inert in L. Since
$\big (\frac {qr}{\ell }\big )=-1$
,
$\mathfrak {l}_{k_1}$
is inert in
$k_1(\sqrt {\ell })$
. These imply that
$A'(k_1)$
is generated by the ideal classes of
$\mathfrak {l}_{k_1}$
and
$\mathfrak {r}_{k_1}$
. Hence
$L=L'(k_1)$
, i.e.,
$A'(k_1) \simeq [2,2]$
and
$A(k_1) \simeq [2,2,2]$
.▪
Remark 2.5 Recall that
$\mathfrak {r}_{k_1}$
does not splits completely in
$L'(k_1)$
, and
$\mathfrak {r}_k$
is inert in
$k_1$
and
$L'(k)=k(\sqrt {\ell })$
. Then lifting map
$A'(k) \rightarrow A'(k_1)$
is injective. This implies that we can not use [Reference Fukuda5, Theorem 2 (1)] to prove that
$|A(k_1)|=|A(k_2)|$
.
Lemma 2.6
$|A_{\{\ell \}}(\mathbb Q_n(\sqrt {r}))/2|=2$
for all
$n \ge 0$
.
Proof. Let L be the maximal elementary abelian subextension of
$L_{\{\ell \}}(\mathbb Q_n(\sqrt {r}))/\mathbb Q_n(\sqrt {r})$
. Then
$A_{\{\ell \}}(\mathbb Q_n(\sqrt {r}))/2 \simeq \mathrm {Gal}(L/\mathbb Q_n(\sqrt {r}))$
and
$\mathbb Q_n(\sqrt {\ell },\sqrt {r}) \subset L$
. The ramification indices of ramified primes in
$L/\mathbb Q_n(\sqrt {r})$
are
$2$
by the cyclicity of their inertia groups. Since
$\mathbb Q_n(\sqrt {\ell },\sqrt {r})/\mathbb Q_n(\sqrt {r})$
is ramified at any primes lying over
$\ell $
,
$L/\mathbb Q_n(\sqrt {\ell },\sqrt {r})$
is unramified, and hence
$L=\mathbb Q_n(\sqrt {\ell },\sqrt {r})$
by Lemma 2.1. Therefore,
$A_{\{\ell \}}(\mathbb Q_n(\sqrt {r}))$
is nontrivial cyclic.▪
Lemma 2.7
$|A_{\{q\}}(\mathbb Q_n(\sqrt {r}))|=1$
for all
$n \ge 0$
.
Proof. Note that
$\mathbb Q_n(\sqrt {r}) \subset L_{\{q\}}(\mathbb Q_n(\sqrt {r})) \subset \widetilde {L}_{\{q,r\}}(\mathbb Q_n)$
. By [Reference Mizusawa17, Proposition 6.2],
$\widetilde {L}_{\{q,r\}}(\mathbb Q_n)/\mathbb Q_n$
is a cyclic quartic extension, which is totally ramified at r. Hence
$L_{\{q\}}(\mathbb Q_n(\sqrt {r}))=\mathbb Q_n(\sqrt {r})$
.▪
For each prime ideal
$v \nmid 2$
of a number field K, let
$\varphi _{K,v} : \mathcal {O}_{K}^{\times } \rightarrow (\mathcal {O}_K/v)^{\times }/2 \simeq \mathbb Z/2\mathbb Z$
be the natural homomorphism induced from
$\mathcal {O}_{K}^{\times } \rightarrow (\mathcal {O}_K/v)^{\times } : \varepsilon \mapsto \varepsilon \bmod {v}$
. Then
$\varphi _{K,v}(\varepsilon )=0 \in \mathbb Z/2\mathbb Z$
if and only if
$\varepsilon $
is a quadratic residue modulo v. For a finite set S of odd prime numbers, a homomorphism
$$ \begin{align*} \varphi_{K,S} : \mathcal{O}_{K}^{\times} \rightarrow \textstyle{\bigoplus_{v|m}} (\mathbb Z/2\mathbb Z) : \varepsilon \mapsto (\varphi_{K,v}(\varepsilon))_{v|m} \end{align*} $$
is defined, where
$m=\prod _{l \in S}l$
. By the definition of ray class groups,
$$ \begin{align*} \mathrm{Coker}\,\varphi_{K,S} \simeq \mathrm{Ker}\,(A_S(K) \rightarrow A(K))/2 \end{align*} $$
as vector spaces over
$\mathbb F_2=\mathbb Z/2\mathbb Z$
.
Lemma 2.8
$|A_{\{\ell ,q\}}(\mathbb Q_n(\sqrt {r}))/2| \le 4$
for all
$n \ge 0$
.
Proof. Let F be the cyclic quartic subextension of
$\mathbb Q_2(\sqrt {r})/\mathbb Q$
different from
$\mathbb Q_2$
. Then
$\mathbb Q_2(\sqrt {r}) \subset L(F) \subset L_{\{q\}}(F) \subset L_{\{q\}}(\mathbb Q_2(\sqrt {r}))$
. Since
$q \equiv 3 \pmod {8}$
, q is inert in F, and
$\mathfrak {q}_F=q\mathcal {O}_F$
splits in
$\mathbb Q_2(\sqrt {r})$
. By Lemma 2.7,
$L_{\{q\}}(F)=L(F)=\mathbb Q_2(\sqrt {r})$
, and hence
$A_{\{q\}}(F) \simeq A(F)$
. Then
$\varphi _{F,\{q\}} : \mathcal {O}_{F}^{\times } \rightarrow \mathbb Z/2\mathbb Z$
is surjective, i.e.,
$\varphi _{F,\mathfrak {q}_F}(\eta _0)=1$
for some
$\eta _0 \in \mathcal {O}_{F}^{\times }$
. Since
$\ell \equiv 9 \pmod {16}$
and
$\big (\frac {r}{\ell }\big )=-1$
,
$\ell $
splits completely in F, and
$\mathfrak {l}_F$
is inert in
$\mathbb Q_2(\sqrt {r})$
, i.e.,
$\mathfrak {l}_{\mathbb Q_2(\sqrt {r})}=\mathfrak {l}_F \mathcal {O}_{\mathbb Q_2(\sqrt {r})}$
. Then the
$\mathbb F_2$
-linear map
$$ \begin{align*} \mathbb F_2 \simeq \mathbb F_{\ell}^{\times}/2 \simeq (\mathcal{O}_F/\mathfrak{l}_F)^{\times}/2 \rightarrow (\mathcal{O}_{\mathbb Q_2(\sqrt{r})}/\mathfrak{l}_{\mathbb Q_2(\sqrt{r})})^{\times}/2 \simeq \mathbb F_{\ell^2}^{\times}/2 \simeq \mathbb F_2 \end{align*} $$
is zero mapping
$1 \mapsto 0$
. Hence, the map
$$ \begin{align*} \varphi_{\mathbb Q_2(\sqrt{r}),\{\ell,q\}}=(\varphi_{\mathbb Q_2(\sqrt{r}),\{\ell\}},\varphi_{\mathbb Q_2(\sqrt{r}),\{q\}}) : \mathcal{O}_{\mathbb Q_2(\sqrt{r})}^{\times} \mapsto [2,2,2,2] \oplus [2,2] \end{align*} $$
sends
$\eta _0 \in \mathcal {O}_{F}^{\times }$
to
$$ \begin{align*} \varphi_{\mathbb Q_2(\sqrt{r}),\{\ell,q\}}(\eta_0)=(0,0,0,0,\varphi_{F,\mathfrak{q}_F}(\eta_0),\varphi_{F,\mathfrak{q}_F}(\eta_0)) = (0,0,0,0,1,1). \end{align*} $$
Since
$|\mathrm {Coker}\,\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell \}}|=2$
by Lemma 2.6, there are some
$\eta _1$
,
$\eta _2$
,
$\eta _3 \in \mathcal {O}_{\mathbb Q_2(\sqrt {r})}^{\times }$
such that
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell \}}(\eta _1)$
,
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell \}}(\eta _2)$
,
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell \}}(\eta _3)$
are linearly independent in the
$\mathbb F_2$
-vector space
$[2,2,2,2]$
. Then
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell ,q\}}(\eta _1)$
,
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell ,q\}}(\eta _2)$
,
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell ,q\}}(\eta _3)$
and
$\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell ,q\}}(\eta _0)$
are linearly independent in
$[2,2,2,2] \oplus [2,2]$
. Therefore
$|A_{\{\ell ,q\}}(\mathbb Q_2(\sqrt {r}))/2|=|\mathrm {Coker}\,\varphi _{\mathbb Q_2(\sqrt {r}),\{\ell ,q\}}| \le 4$
. By a version of Fukuda’s theorem [Reference Mizusawa17, Theorem 4.3],
$|A_{\{\ell ,q\}}(\mathbb Q_n(\sqrt {r}))/2| \le 4$
for all
$n \ge 0$
.▪
Lemma 2.9
$A(k_n(\sqrt {r}))/2 \simeq [2,2]$
for all
$n \ge 1$
.
Proof. Let L be the maximal elementary abelian subextension of
$L(k_n(\sqrt {r}))/k_n(\sqrt {r})$
for fixed
$n \ge 1$
. Then
$A(k_n(\sqrt {r}))/2 \simeq \mathrm {Gal}(L/k_n(\sqrt {r}))$
via Artin map. Let
$\sigma $
be a generator of
$\mathrm {Gal}(k_n(\sqrt {r})/\mathbb Q_n(\sqrt {r}))$
. Since
$A(\mathbb Q_n(\sqrt {r})) \simeq 0$
,
$A(k_n(\sqrt {r}))^{\sigma +1} \simeq 0$
. Then
$(A(k_n(\sqrt {r}))/2)^{\sigma -1} = (A(k_n(\sqrt {r}))/2)^{\sigma +1} \simeq 0$
. This implies that
$L/\mathbb Q_n(\sqrt {r})$
is abelian. Since
$L(\mathbb Q_n(\sqrt {r}))=\mathbb Q_n(\sqrt {r})$
,
$\mathrm {Gal}(L/\mathbb Q_n(\sqrt {r}))$
is generated by inertia groups. Since the order of each inertia group is at most
$2$
,
$L/\mathbb Q_n(\sqrt {r})$
is an elementary abelian 2-extension unramified outside
$\{2,\ell ,q\}$
. Let
$L'$
be the inertia field of the unique prime
$\mathfrak {p}_{\mathbb Q_n(\sqrt {r})}$
lying over
$2$
. Then
$[L:L']=2$
, and
$L'/\mathbb Q_n(\sqrt {r})$
is an elementary abelian
$2$
-extension unramified outside
$\{\ell ,q\}$
. By Lemma 2.8,
$|A(k_n(\sqrt {r}))/2|=[L:k_n(\sqrt {r})]=[L':\mathbb Q_n(\sqrt {r})] \le 4$
. Since
$L(k_1)k_n/k_n(\sqrt {r})$
is an unramified
$[2,2]$
-extension by Lemma 2.4, we have
$|A(k_n(\sqrt {r}))/2|=4$
.▪
Lemma 2.10
$\widetilde {L}(k_n)=L(k_n(\sqrt {r}))$
for all
$n \ge 0$
.
Proof. Put
$K=\mathbb Q(\sqrt {\ell },\sqrt {q},\sqrt {r})$
. Then
$k_n \subset k_n(\sqrt {r}) \subset K_n \subset L(k_n)$
. Let
$\sigma $
(resp.
$\tau $
) be a generator of
$\mathrm {Gal}(K_n/\mathbb Q_n(\sqrt {q},\sqrt {r}))$
(resp.
$\mathrm {Gal}(K_n/\mathbb Q_n(\sqrt {\ell },\sqrt {r}))$
). Then
$\sigma \tau $
is a generator of
$\mathrm {Gal}(K_n/k_n(\sqrt {r}))$
. By Lemma 2.1,
$A(K_n)^{\sigma +1} \simeq 0$
and
$A(K_n)^{\tau +1} \simeq 0$
, i.e.,
$\sigma $
and
$\tau $
act on
$A(K_n)$
as
$-1$
. Then
$\sigma \tau $
acts on
$A(K_n)$
trivially. This implies that
$\mathrm {Gal}(L(K_n)/k_n(\sqrt {r}))$
is abelian, i.e.,
$L(K_n)=L(k_n(\sqrt {r}))$
. By Lemma 2.9 and [Reference Benjamin, Lemmermeyer and Snyder3, Proposition 7],
$\widetilde {L}(k_n)=\widetilde {L}(k_n(\sqrt {r}))=L(k_n(\sqrt {r}))$
.▪
Lemma 2.11
$L'(k_n)=L'(k_n(\sqrt {\ell }))$
for all
$n \ge 0$
.
Proof. Note that
$k_n(\sqrt {\ell }) \subset L'(k_n) \subset L'(k_n(\sqrt {\ell }))$
. Let
$\sigma $
(resp.
$\tau $
) be a generator of
$\mathrm {Gal}(k_n(\sqrt {\ell })/\mathbb Q_n(\sqrt {qr}))$
(resp.
$\mathrm {Gal}(k_n(\sqrt {\ell })/\mathbb Q_n(\sqrt {\ell }))$
). By Lemma 2.1,
$A'(k_n(\sqrt {\ell }))^{\sigma +1} \simeq 0$
and
$A'(k_n(\sqrt {\ell }))^{\tau +1} \simeq 0$
. Then
$\sigma \tau $
acts on
$A'(k_n(\sqrt {\ell }))$
trivially, and hence
$L'(k_n)=L'(k_n(\sqrt {\ell }))$
.▪
Lemma 2.12
$A(k_n) \simeq [2,2,2]$
,
$A'(k_n) \simeq [2,2]$
,
$A(k_n(\sqrt {\ell })) \simeq [2,2,2]$
and
$|A'(k_n(\sqrt {\ell }))|=2$
for all
$n \ge 1$
.
Proof. By Lemma 2.4,
$L'(k_1)k_n/k_n$
is a
$[2,2]$
-subextension of
$L'(k_n)/k_n$
having three quadratic subextensions
$F^{(1)}$
,
$F^{(2)}$
and
$F^{(3)}=k_n(\sqrt {\ell })$
. Since
$L'(k_1)k_n \subset L'(k_n(\sqrt {\ell }))$
,
$A'(k_n(\sqrt {\ell }))$
is nontrivial. If
$F^{(1)}/k$
is a Galois extension,
$L'(k_1)k_n=F^{(1)}F^{(3)}$
is an abelian extension of k. Then the inertia field of
$\mathfrak {p}_k$
in
$L'(k_1)k_n/k$
is a quartic subextension of
$L'(k)/k$
. This implies
$|A'(k)| \ge 4$
, which contradicts Lemma 2.2. Hence,
$F^{(1)}/k$
is not Galois. Since
$L'(k_1)k_n$
and
$F^{(3)}$
are Galois over k,
$F^{(2)}/k$
is a conjugate of
$F^{(1)}/k$
, and hence
$A'(F^{(1)}) \simeq A'(F^{(2)})$
.
Put
$K=k(\sqrt {\ell },\sqrt {r})$
. Then
$k_n(\sqrt {l}) \subset K_n \subset L(k_n(\sqrt {l})) \subset L(K_n)$
for
$l \in \{\ell ,r\}$
. By Lemmas 2.9 and 2.10,
$|A(K_n)/2| \le 4$
, and hence
$|A(k_n(\sqrt {\ell }))/2| \le 8$
. Let M be the maximal elementary abelian subextension of
$L'(k_n(\sqrt {\ell }))/k_n(\sqrt {\ell })$
. Then
$[M:k_n(\sqrt {\ell })]=|A'(k_n(\sqrt {\ell }))/2| =2^t$
with some
$t \ge 1$
. By Lemma 2.2,
$L(k(\sqrt {\ell }))k_n/k_n(\sqrt {\ell })$
is a
$[2,2]$
-extension such that
$L(k(\sqrt {\ell }))k_n \cap L'(k_n(\sqrt {\ell })) =k_n(\sqrt {\ell })$
. Then
$L(k(\sqrt {\ell }))M/k_n(\sqrt {\ell })$
is an unramified elementary abelian
$2$
-extension of degree
$2^{t+2}$
. Since
$2^{t+2} \le |A(k_n(\sqrt {\ell }))/2| \le 8$
,
$t=1$
, i.e.,
$A'(F^{(3)}) \simeq \mathbb Z/2^m\mathbb Z$
with some
$m \ge 1$
. Since
$L'(k_n)=\widetilde {L}'(k_n(\sqrt {\ell }))$
by Lemma 2.11,
$\widetilde {L}'(k_n)=L'(k_n)$
and
$A'(k_n) \simeq [2,2^m]$
. Then
$\mathrm {Gal}(L'(k_n)/F^{(i)}) \simeq A'(F^{(i)})$
for each
$i \in \{1,2,3\}$
. If
$m \ge 2$
, one of either
$A'(F^{(1)})$
or
$A'(F^{(2)})$
is cyclic and the other is isomorphic to
$[2,2^{m-1}]$
. This is a contradiction. Therefore
$m=1$
, i.e.,
$|A'(k_n(\sqrt {\ell }))|=2$
and
$A'(k_n) \simeq [2,2]$
. Since
$\mathfrak {p}_{k_n}^2$
is principal and
$|A(k_n)/2| \ge 8$
by Lemma 2.4,
$A(k_n) \simeq [2,2,2]$
. Since
$\mathfrak {p}_{k_n(\sqrt {\ell })}$
has exactly two conjugates including itself,
$|D(k_n(\sqrt {\ell }))/2| \le 4$
. Since
$\mathfrak {p}_{k_n(\sqrt {\ell })}^2=\mathfrak {p}_{\mathbb Q_n(\sqrt {\ell })}\mathcal {O}_{k_n(\sqrt {\ell })}$
and
$|A(\mathbb Q_n(\sqrt {\ell }))|=1$
,
$|D(k_n(\sqrt {\ell }))| \le 4$
, and hence
$A(k_n(\sqrt {\ell })) \simeq [2,2,2]$
.▪
Lemma 2.13
$|A(L(k_n))| \le |A(k^{\prime }_n)|$
for all
$n \ge 0$
.
Proof. Let
$\sigma $
(resp.
$\delta $
) be a generator of
$\mathrm {Gal}(k_n(\sqrt {r})/\mathbb Q_n(\sqrt {r}))$
(resp.
$\mathrm {Gal}(k_n(\sqrt {r})/k^{\prime }_n)$
). Then
$\sigma \delta $
is a generator of
$\mathrm {Gal}(k_n(\sqrt {r})/k_n)$
, and
$\mathrm {Gal}(L(k_n)/k_n(\sqrt {r})) \simeq A(k_n(\sqrt {r}))/A(k_n(\sqrt {r}))^{\sigma \delta -1}$
. Since
$A(\mathbb Q_n(\sqrt {r}))) \simeq 0$
,
$A(k_n(\sqrt {r}))^{\sigma +1} \simeq 0$
, and hence
$A(k_n(\sqrt {r}))^{\sigma \delta -1}=A(k_n(\sqrt {r}))^{\delta +1}$
. Since the norm mapping
$A(k_n(\sqrt {r})) \twoheadrightarrow A(k^{\prime }_n)$
is surjective, the image of lifting map
$A(k^{\prime }_n) \rightarrow A(k_n(\sqrt {r}))$
is
$A(k_n(\sqrt {r}))^{\delta +1}$
. Then
$$ \begin{align*} |A(k^{\prime}_n)| &\ge |A(k_n(\sqrt{r}))^{\delta+1}|=|A(k_n(\sqrt{r}))|/[L(k_n):k_n(\sqrt{r})] \\ &=[L(k_n(\sqrt{r})):L(k_n)] =|A(L(k_n))|, \end{align*} $$
where we note that
$L(k_n(\sqrt {r}))=L(L(k_n))$
by Lemma 2.10.▪
For the cyclotomic
$\mathbb Z_2$
-extension
$K_{\infty }$
of a number field K, the norm mapping
$A(K_m) \rightarrow A(K_n)$
is surjective if
$m \ge n$
and n is sufficiently large. The unramified Iwasawa module of
$K_{\infty }$
is isomorphic to the projective limit
$\varprojlim A(K_n)$
with respect to the norm mappings. Then
$\varprojlim A(K_n)$
is finite if and only if
$|A(K_n)|$
is bounded as
$n \rightarrow \infty $
.
Now we complete the proof of Theorem 1.1. Since
$\widetilde {G}/[\widetilde {G},\widetilde {G}] =X \simeq \varprojlim A(k_n) \simeq [2,2,2]$
by Lemma 2.12, the statement (1) holds true. Since
$\mathrm {Gal}(\widetilde {L}(k_{\infty })/k_{\infty }(\sqrt {r})) \simeq \varprojlim \mathrm {Gal}(\widetilde {L}(k_n)/k_n(\sqrt {r})) \simeq \varprojlim A(k_n(\sqrt {r}))$
by Lemma 2.10, the statement (2) holds true by Lemma 2.9. Hence
$\widetilde {G}$
is metabelian. If
$\mathrm {Gal}(L(k^{\prime }_{\infty })/k^{\prime }_{\infty }) \simeq \varprojlim A(k^{\prime }_n)$
is finite, then
$[\widetilde {G},\widetilde {G}] \simeq \varprojlim A(L(k_n))$
is also finite by Lemma 2.13. Therefore the statement (3) holds true.
3 Class-2 quotients and examples
For a finite 2-group G, we put a condition (A) consisting of the following three properties:
-
•
$G/[G,G] \simeq [2,2,2]$
. -
• G has an abelian maximal subgroup
$H \simeq [4,4]$
. -
• G has a maximal subgroup N such that
$N/[N,N] \simeq [2,2,2]$
.
Let F be a free group on three letters
$a,b,c$
. For each pair
$(i,j)$
of
$i,j \in \{0,1\}$
, put
$G(i,j)=F/R$
with a normal subgroup
$R=R(i,j)$
of F which is normally generated by
$$ \begin{align*} [a,b]b^2c^{2i},\quad [a,c]c^2,\quad [b,c],\quad a^2c^{2j},\quad b^4,\quad c^4. \end{align*} $$
GAP [8] shows that the following statements hold true:
-
(i)
$G(i,j)$
is a finite 2-group satisfying (A) for any
$(i,j)$
. -
(ii) For each
$(i,j)$
, an abelian maximal subgroup H of
$G(i,j)$
is unique, and
$H= \langle b,c \rangle R/R \simeq [4,4]$
. -
(iii)
$G(1,0) \simeq G(1,1)$
, and any two of
$G(0,0)$
,
$G(0,1)$
,
$G(1,0)$
are not isomorphic. -
(iv) All of six nonabelian maximal subgroups N of
$G(0,0)$
satisfy
$N/[N,N] \simeq [2,2,2]$
. -
(v) If
$(i,j) \neq (0,0)$
, two of six nonabelian maximal subgroups N of
$G(i,j)$
satisfy
$N/[N,N] \simeq [2,2,2]$
, and other four satisfy
$N/[N,N] \simeq [2,4]$
.
Proposition 3.1 In the situation of Theorem
1.1
, assume that
$A(k_{n_0}(\sqrt {r}))/4 \simeq [4,4]$
for some
$n_0 \ge 1$
. Put
$\widetilde {G}_n=\mathrm {Gal}(\widetilde {L}(k_n)/k_n)$
for each n. Then the following statements hold true:
-
•
$\widetilde {G}/[[\widetilde {G},\widetilde {G}],\widetilde {G}] \simeq \widetilde {G}_n/[[\widetilde {G}_n,\widetilde {G}_n],\widetilde {G}_n]$
for all
$n \ge n_0$
. -
•
$\widetilde {G}_{n_0}/[[\widetilde {G}_{n_0},\widetilde {G}_{n_0}],\widetilde {G}_{n_0}] \simeq G(i_0,j_0)$
for some
$(i_0,j_0)$
. -
• If
$A'(k_{n_0}(\sqrt {r}))/4 \simeq [4,4]$
, then
$(i_0,j_0) \neq (0,1)$
.
Proof. Suppose
$n \ge n_0$
. By Lemmas 2.9 and 2.10,
$A=\mathrm {Gal}(\widetilde {L}(k_n)/k_n(\sqrt {r})) \simeq A(k_n(\sqrt {r}))$
is an abelian 2-group such that
$A/2 \simeq [2,2]$
. Let
$L^{(e)}$
be the fixed field of
$A^{2^e}$
. Then
$\mathrm {Gal}(L^{(2)}/k_n(\sqrt {r})) \simeq A/4$
. By Lemma 2.12,
$L^{(1)}=L(k_n) \subset L^{(2)}$
, and
$L(k_n(\sqrt {\ell }))/k_n(\sqrt {\ell })$
is a
$[2,2,2]$
-extension. Since
$L(k_n(\sqrt {\ell }))/k_n(\sqrt {\ell },\sqrt {r})$
is a
$[2,2]$
-extension, we have
$L(k_n(\sqrt {\ell })) \subset L^{(2)}$
. Since the restriction mapping
$A \twoheadrightarrow \mathrm {Gal}(\widetilde {L}(k_{n_0})/k_{n_0}(\sqrt {r})) \simeq A(k_{n_0}(\sqrt {r}))$
is surjective,
$L^{(2)}/k_n(\sqrt {r})$
is a
$[4,4]$
-extension. Then
$\mathrm {Gal}(L^{(2)}/k_n)$
satisfies (A). GAP [8] shows that there exist exactly three isomorphism classes of finite 2-groups G satisfying (A). By (i) and (iii),
$\mathrm {Gal}(L^{(2)}/k_n) \simeq G(i,j)$
with some
$(i,j)$
. Since
$G=G(i,j)$
has trivial
$[[G,G],G]$
, a surjective homomorphism
$$ \begin{align*} f : \widetilde{G}_n/[[\widetilde{G}_n,\widetilde{G}_n],\widetilde{G}_n] \twoheadrightarrow \mathrm{Gal}(L^{(2)}/k_n) \simeq G(i,j) \end{align*} $$
is induced from the restriction mapping
$\widetilde {G}_n \twoheadrightarrow \mathrm {Gal}(L^{(2)}/k_n)$
. Since the map
$\widetilde {G}_n \times \widetilde {G}_n \rightarrow [\widetilde {G}_n,\widetilde {G}_n] : (x,y) \mapsto [x,y]$
induces a bilinear surjective homomorphism
$$ \begin{align*} \widetilde{G}_n/[\widetilde{G}_n,\widetilde{G}_n] \times \widetilde{G}_n/[\widetilde{G}_n,\widetilde{G}_n] \twoheadrightarrow [\widetilde{G}_n,\widetilde{G}_n]/[[\widetilde{G}_n,\widetilde{G}_n],\widetilde{G}_n] , \end{align*} $$
the exponent of
$[\widetilde {G}_n,\widetilde {G}_n]/[[\widetilde {G}_n,\widetilde {G}_n],\widetilde {G}_n]$
is
$2$
. Since
$[\widetilde {G}_n,\widetilde {G}_n] \subset A$
, we have
$|[\widetilde {G}_n,\widetilde {G}_n]/[[\widetilde {G}_n,\widetilde {G}_n],\widetilde {G}_n]| \le 4$
, and hence f is an isomorphism. Under the isomorphism
$\mathrm {Gal}(L^{(2)}/k_n) \simeq G(i,j)=F/R$
, we have
$\mathrm {Gal}(L^{(2)}/k_n(\sqrt {r})) \simeq H=\langle b,c \rangle R/R$
by (ii), and
$\mathrm {Gal}(L^{(2)}/M^{(2)}) \simeq \langle b \rangle R/R$
with some cyclic quartic extension
$M^{(2)}/k_n(\sqrt {r})$
. Let
$M^{(1)}$
be the quadratic subextension of
$M^{(2)}/k_n(\sqrt {r})$
.
If
$i=0$
, we have
$a^{-1}ba \equiv b^{-1} \pmod {R}$
. Then
$\langle b \rangle R/R$
is a normal subgroup of
$G(0,j)$
, i.e.,
$M^{(2)}/k_n$
is a Galois extension. Moreover, if
$j=1$
, we have
$a^2 \equiv c^2 \pmod {\langle b \rangle R}$
. Then
$\mathrm {Gal}(M^{(2)}/k_n) \simeq F/\langle b \rangle R$
is isomorphic to a quaternion group. In particular, any maximal subgroup of
$\mathrm {Gal}(M^{(2)}/k_n)$
is cyclic. Since
$k_n(\sqrt {r}) \not \subset L'(k_n)$
,
$M^{(1)}/k_n$
is a
$[2,2]$
-extension such that
$M^{(1)} \not \subset L'(k_n)$
. Then
$M^{(0)}=M^{(1)} \cap L'(k_n)$
is a quadratic subextension of
$M^{(2)}/k_n$
. Since
$M^{(2)}/M^{(0)}$
is a cyclic quartic extension such that
$M^{(2)} \cap L'(M^{(0)}) = M^{(0)}$
, we have
$M^{(2)} \not \subset L'(M^{(1)})$
. This implies that
$L^{(2)} \not \subset L'(k_n(\sqrt {r}))$
, i.e.,
$A'(k_n(\sqrt {r}))/4 \not \simeq [4,4]$
. Therefore
$(i,j) \neq (0,1)$
if
$A'(k_n(\sqrt {r}))/4 \simeq [4,4]$
.
For all
$n \ge n_0$
, the norm mapping induces a surjective homomorphism
$A'(k_n(\sqrt {r}))/4 \twoheadrightarrow A'(k_{n_0}(\sqrt {r}))/4$
. Since
$\widetilde {G}_{n_0}/[[\widetilde {G}_{n_0},\widetilde {G}_{n_0}],\widetilde {G}_{n_0}] \simeq G(i_0,j_0)$
for some
$(i_0,j_0)$
, and the restriction mapping
$\widetilde {G}_n/[[\widetilde {G}_n,\widetilde {G}_n],\widetilde {G}_n] \twoheadrightarrow \widetilde {G}_{n_0}/[[\widetilde {G}_{n_0},\widetilde {G}_{n_0}],\widetilde {G}_{n_0}]$
is an isomorphism for each
$n \ge n_0$
, the statements hold true.▪
As in the following example, there exist infinitely many triples
$(\ell ,q,r)$
such that
$\widetilde {G}$
is a finite metabelian 2-group and
$\widetilde {G}/[[\widetilde {G},\widetilde {G}],\widetilde {G}] \simeq G(i,j)$
with some
$(i,j)$
.
Example 3.2 Put
$(\ell ,q)=(89,139)$
, which satisfies (1.1). Suppose that r satisfies (1.2). There exist infinitely many such r by Dirichlet’s prime number theorem. Since
$A(k^{\prime }_1) \simeq A(k^{\prime }_2) \simeq [8,4]$
(see [Reference Mizusawa15, Table 1]),
$A(k^{\prime }_n) \simeq [8,4]$
for all
$n \ge 1$
by Fukuda’s theorem [Reference Fukuda5]. Then
$L(k^{\prime }_{\infty })/k^{\prime }_{\infty }$
is a
$[8,4]$
-extension, in particular a finite extension. Hence
$\widetilde {G}$
is finite. Since the norm mapping
$A(k_n(\sqrt {r})) \twoheadrightarrow A(k^{\prime }_n)$
is surjective,
$A(k_n(\sqrt {r}))/4 \simeq [4,4]$
. Then
$\widetilde {G}/[[\widetilde {G},\widetilde {G}],\widetilde {G}] \simeq G(i,j)$
with some
$(i,j)$
.
Since the narrow class number of
$\mathbb Q_1$
is
$1$
, there exists a totally positive element
$\alpha \in \mathbb Z[\sqrt {2}]$
such that
$\mathfrak {l}_{\mathbb Q_1}=\alpha \mathbb Z[\sqrt {2}]$
. Put the conjugate
$\alpha '=\ell /\alpha $
.
Lemma 3.3
$L'(k_n)=k_n(\sqrt {\ell },\sqrt {\alpha q})$
and
$L(k_n)=k_n(\sqrt {\ell },\sqrt {q},\sqrt {\alpha })$
for all
$n \ge 1$
.
Proof. Since
$\ell \equiv 9 \pmod {16}$
,
$\alpha \equiv \pm 3$
,
$\pm (1+2\sqrt {2}) \pmod {4\sqrt {2}}$
. By [Reference Nishino20, Lemma 1 (2)],
$\mathbb Q_1(\sqrt {\alpha })/\mathbb Q_1$
is ramified at
$\mathfrak {p}_{\mathbb Q_1}$
, and hence
$\alpha \not \equiv 1$
,
$3+2\sqrt {2} \pmod {4}$
by [Reference Mizusawa15, Lemma 5]. Then
$\alpha \equiv 3$
,
$1+2\sqrt {2} \pmod {4\sqrt {2}}$
. Since
$\alpha q \equiv 1$
,
$3+2\sqrt {2} \pmod {4\sqrt {2}}$
,
$\mathfrak {p}_{\mathbb Q_1}$
splits in
$\mathbb Q_1(\sqrt {\alpha q})$
by [Reference Mizusawa15, Lemma 5]. Then
$k_n(\sqrt {\alpha q}) \subset L'(k_n)$
. Thus, we obtain the claim.▪
By using PARI/GP [23] and GAP [8], we obtain the following examples.
Example 3.4 Suppose
$(\ell ,q,r)=(73,3,5)$
which satisfies (1.1) and (1.2). Then
$\alpha =9+2\sqrt {2}$
. Put
$F=k^{\prime }_1(\sqrt {r\pi })$
with
$\pi =2+\sqrt {2}$
. Then
$\mathfrak {p}_F$
is inert in the unramified quadratic extension
$k_2(\sqrt {r})/F$
. Since
$A'(F) \simeq \texttt {[4,4]}$
, we have
$A'(k_2(\sqrt {r}))/4 \simeq [4,4]$
. Hence
$\widetilde {G}/[[\widetilde {G},\widetilde {G}],\widetilde {G}] \simeq \widetilde {G}_2/[[\widetilde {G}_2,\widetilde {G}_2],\widetilde {G}_2] \simeq G(i_0,j_0) \not \simeq G(0,1)$
. For
$n \ge 1$
,
$k_n(\sqrt {\alpha })/\mathbb Q_n(\sqrt {\alpha })$
is ramified at only prime divisors of
$\alpha 'qr$
. Since
$|A(\mathbb Q_2(\sqrt {\alpha }))|=\texttt {1}$
and the 2-rank of the ray class group of
$\mathbb Q_2(\sqrt {\alpha })$
modulo
$\alpha 'qr$
is
$\texttt {3}$
,
$A(k_2(\sqrt {\alpha }))/2 \simeq [2,2]$
by [Reference Mizusawa17, Theorem 3.1]. This implies that
$G(i_0,j_0)$
has a nonabelian maximal subgroup N such that the 2-rank of
$N/[N,N]$
is
$2$
. Hence,
$(i_0,j_0) \neq (0,0)$
by (iv). Therefore,
$\widetilde {G}/[[\widetilde {G},\widetilde {G}],\widetilde {G}] \simeq G(1,0) \simeq G(1,1)$
.
Note that
$\big (\frac {q}{r}\big )=-1$
. Recall that the ideal classes of
$\mathfrak {p}_k$
and
$\mathfrak {r}_k$
generate
$A(k) \simeq [2,2]$
. Since
$A(k(\sqrt {r})) \simeq \texttt {[8]}$
and
$\mathfrak {r}_{k(\sqrt {r})}$
is inert in
$L(k)$
,
$A(k(\sqrt {r}))$
is generated by an odd power of the ideal class of
$\mathfrak {r}_{k(\sqrt {r})}$
. Let
$N_{k(\sqrt {r})/k} : A(k(\sqrt {r})) \rightarrow A(k)$
be the norm map, and let
$j_{k(\sqrt {r})/k} : A(k) \rightarrow A(k(\sqrt {r}))$
be the lift map. Since
$\mathfrak {r}_{k}\mathcal {O}_{k(\sqrt {r})}=\mathfrak {r}_{\mathbb Q(\sqrt {r})}\mathcal {O}_{k(\sqrt {r})}$
is principal and
$\mathfrak {r}_{k}$
splits in
$k(\sqrt {r})$
, the ideal class of
$\mathfrak {r}_{k}$
is a nontrivial element of
$\mathrm {Ker}\,j_{k(\sqrt {r})/k} \cap N_{k(\sqrt {r})/k}\,A(k(\sqrt {r}))$
. Since PARI/GP shows that
$\mathfrak {p}_{k}\mathcal {O}_{k(\sqrt {r})}=\mathfrak {p}_{k(\sqrt {r})}$
is not principal,
$\widetilde {G}_0$
is isomorphic to a generalized quaternion group
$Q_{16}$
of order
$16$
(see [Reference Kisilevsky11], cf. [Reference Benjamin and Snyder4]). Since the nilpotency class of
$Q_{16}$
is
$3$
,
$[[\widetilde {G},\widetilde {G}],\widetilde {G}]$
is nontrivial. The p-group generation algorithm in the ANUPQ package [Reference Gamble, Nickel, O’Brien and Horn7] of GAP [8] shows that there exists uniquely an isomorphism class of a finite 2-group G such that
$G/[[G,G],G] \simeq G(1,0)$
and G has a quotient isomorphic to
$Q_{16}$
. Moreover, GAP [8] shows that such a unique isomorphism class contains
$F/\widetilde {R}$
, where
$\widetilde {R}$
is normally generated by the six relations (1.3). Therefore,
$\widetilde {G} \simeq F/\widetilde {R}$
. Since
$A(k^{\prime }_1) \simeq A(k^{\prime }_2) \simeq \texttt {[2,4]}$
, one can reconfirm that
$[\widetilde {G},\widetilde {G}] \simeq [2,4]$
by Fukuda’s theorem [Reference Fukuda5] and Lemma 2.13.
Remark 3.5 Also for each integer
$m \ge 5$
, it seems that there exists uniquely an isomorphism class of a finite 2-group G such that
$G/[[G,G],G] \simeq G(1,0)$
and G has a generalized quaternion quotient
$Q_{2^m}$
of order
$2^m$
. The author verified it for all
$m \le 12$
in the same way. A similar argument is used in the proofs of [Reference Mizusawa14, Theorem 2] and [Reference Mouhib and Movahhedi19, Theorem 3.1].
Example 3.6 Suppose
$(\ell ,q,r)=(73,19,5)$
which satisfies (1.1) and (1.2). By Lemma 2.12,
$A(k_1(\sqrt {\ell })) \simeq [2,2,2]$
. Since
$|A(\mathbb Q_1(\sqrt {\alpha }))|=\texttt {1}$
and the
$2$
-rank of the ray class group of
$\mathbb Q_1(\sqrt {\alpha })$
modulo
$\alpha 'qr$
is
$\texttt {4}$
,
$A(k_1(\sqrt {\alpha }))/2 \simeq [2,2,2]$
by [Reference Mizusawa17, Theorem 3.1]. Since
$k_1(\sqrt {\alpha '})$
is a conjugate of
$k_1(\sqrt {\alpha })$
, we have
$A(k_1(\sqrt {\alpha '}))/2 \simeq [2,2,2]$
. Assuming GRH,
$A'(k_1(\sqrt {r})) \simeq \texttt {[4,4]}$
. Then
$\widetilde {G}/[[\widetilde {G},\widetilde {G}],\widetilde {G}] \simeq \widetilde {G}_1/[[\widetilde {G}_1,\widetilde {G}_1],\widetilde {G}_1] \simeq G(0,0)$
by (v). Indeed,
$\widetilde {G}_1/[[\widetilde {G}_1,\widetilde {G}_1],\widetilde {G}_1]$
has at least three maximal subgroups N such that
$N/[N,N] \not \simeq [2,4]$
.
