1 Introduction
For a number field K, we denote its class group, class number, discriminant, and Hilbert class field by
$C\ell (K)$
,
$h_K$
,
$D_K$
, and
$H(K)$
, respectively. If
$K/F$
is a Galois extension of number fields, we write
$G_{K/F}$
for its Galois group. For any integer
$n> 1$
, let
$\zeta _n$
denote a primitive nth root of unity, and define the maximal real subfield of
$\mathbb {Q}(\zeta _n)$
as
$\mathbb {Q}(\zeta _n)^+ = \mathbb {Q}(\zeta _n + \zeta _n^{-1})$
, called a real cyclotomic field. For any group G, we write
$|G|$
to denote its cardinality.
Given an integer
$t>1$
, we say that the t-rank of the class group of a number field K is r if
$C\ell (K)$
contains exactly r-copies of
$\mathbb {Z}/{t\mathbb {Z}}$
. The
$2$
-rank of class groups of quadratic fields is completely known by Gauss’s genus theory, and using this one can easily construct quadratic fields whose class group has a large
$2$
-rank. Motivated by this, many researchers have explored families of number fields whose class groups have large t-rank for integers
$t> 1$
. Several works [Reference Ankeny and Chowla1, Reference Ichimura11, Reference Louboutin15, Reference Mestre21, Reference Murty23, Reference Soundararajan28, Reference Yamamoto33] provide a family of quadratic fields whose class groups contain
$\mathbb {Z}/{t\mathbb {Z}}$
as a subgroup.
The main objective of this article is to use the known results regarding the t-rank of class group of a field K to obtain a lower bound on the t-rank of class groups of any cyclotomic fields (or real cyclotomic fields) that contain this K. In this context, we prove the following theorem.
Theorem 1.1 Let
$n = p_1^{e_1} p_2^{e_2} \ldots p_s^{e_s}$
be a product of powers of distinct odd primes
$p_i$
with
$e_i \geq 1$
, and set
$d = p_1 p_2 \ldots p_s$
. Define
$$\begin{align*}K = \begin{cases} \mathbb{Q}(\sqrt{d}), & \text{if } d \equiv 1\quad \pmod{4}, \\ \mathbb{Q}(\sqrt{-d}), & \text{if } d \equiv 3\quad \pmod{4}. \end{cases} \end{align*}$$
Let
$m> 1$
be the largest odd divisor of
$h_K$
, and suppose
$t> 1$
divides m such that the t-rank of
$C\ell (K)$
is r. Then:
-
(1) If
$d \equiv 1\ \pmod {4}$
, the t-rank of the class groups of
$\mathbb {Q}(\zeta _n)^+$
and
$\mathbb {Q}(\zeta _n)$
is at least r, and m divides the class number of both fields. -
(2) If
$d \equiv 3\ \pmod {4}$
, the t-rank of the class group of
$\mathbb {Q}(\zeta _n)$
is at least r, and m divides its class number.
Theorem 1.1 highlights the important role of class groups of quadratic fields in deducing arithmetic information about the class groups of cyclotomic fields. For instance, by combining Theorem 1.1 of [Reference Byeon4] with Theorem 1.1, one can conclude that there exist infinitely many primes
$p \equiv q \equiv 3\ \pmod {4}$
such that the class number of
$\mathbb {Q}(\zeta _{pq})^+$
is divisible by
$3$
.
Furthermore, for any odd integer
$t> 1$
, Theorem 1.1 provides lower bounds on the t-rank of class groups of various fields of the form
$\mathbb {Q}(\zeta _n)^+.$
Even though for any odd integer
$t> 5$
, no infinite family of quadratic fields is currently known whose class groups have t-rank at least
$3$
. However, Theorem 1.1 can be used to construct cyclotomic (or real cyclotomic) fields whose class groups have large t-rank. For example, one can find square-free positive integers
$d_1, d_2, \ldots , d_s$
such that
$(d_i, d_j) = 1$
for
$i \neq j$
and
$d_i \equiv 3\ \pmod {4}$
. For each i, the class group of
$K_i = \mathbb {Q}(\sqrt {-d_i})$
contains at least one copy of
$\mathbb {Z}/t\mathbb {Z}$
. Theorem 1.1, together with Proposition 2.9, then implies that the class group of
$\mathbb {Q}(\zeta _{d_1 d_2 \ldots d_s})$
has t-rank at least s. We prove one such application of Theorem 1.1, which is the following result.
Corollary 1.2 Let
$t> 1$
and
$a> 1$
be two odd integers, and let
$u \geq 2$
be an integer. Assume that
$-d_{u,t}$
and
$d^{\prime }_{a,t}$
are the square-free parts of
$1 - 4u^t$
and
$a^{2t} + 4$
, respectively. Suppose
$a = 17b$
and
$u = a^2$
, where
$b \geq 1$
is an integer. Then, the t-rank of the class group of
$\mathbb {Q}(\zeta _{d_{u,a,t}})$
is at least
$2$
, where
$d_{u,a,t} = d_{u,t} \cdot d^{\prime }_{a,t}$
, and the class group of
$\mathbb {Q}(\zeta _{d^{\prime }_{a,t}})^+$
has a subgroup isomorphic to
$\mathbb {Z}/{t\mathbb {Z}}$
.
There are many results providing families of fields
$\mathbb {Q}(\zeta _n)^+$
whose class numbers are greater than one. For example, Ankeny, Chowla, and Hasse [Reference Ankeny, Chowla and Hasse2] proved that if
$n = (2vq)^2 + 1$
is prime, where q is prime and
$v> 1$
is an integer, then the class number of
$\mathbb {Q}(\zeta _n)^+$
is greater than one. For more results in this direction, see [Reference Chakraborty and Hoque6, Reference Cornell and Washington8, Reference Hoque and Saikia10, Reference Lang13, Reference Osada24, Reference Ram25, Reference Takeuchi29, Reference Yamaguchi32]. In Corollary 1.2, we provide a family of maximal real subfields
$\mathbb {Q}(\zeta _{d^{\prime }_{a,t}})^+$
whose class groups contain a subgroup isomorphic to
$\mathbb {Z}/{t\mathbb {Z}}$
.
The next theorem shows that there exist infinitely many real cyclotomic fields whose class numbers are even. In particular, we establish the following result.
Corollary 1.3 Let s be a positive integer relatively prime to
$3$
. Then, there exist infinitely many real cyclotomic fields whose class numbers are divisible by s.
Combining Corollaries 1.2 and 1.3, we obtain the following remarkable conclusion:
$$ \begin{align*} &\textit{There are infinitely many real cyclotomic fields and cyclotomic fields whose}\\ &\textit{class numbers are divisible by any given positive integer m not divisible by 6.} \end{align*} $$
We remark that this conclusion also follows from [Reference Mishra, Schoof and Washington22], although our methods are different.
Studying class numbers of cyclotomic fields and the class numbers of maximal real subfields of cyclotomic fields is still a difficult and mysterious problem after a long research era. However, there is a nice result on the class numbers of cyclotomic fields, which is the following.
Let
$n \not \equiv 2\ \pmod {4}$
be a positive integer. Then, the class number of
$\mathbb {Q}(\zeta _n)$
is one if and only if
$$ \begin{align} n \in \left\{ \begin{array}{@{}c@{}} 1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, \\ 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84 \end{array} \right\}. \end{align} $$
In [Reference Masley17], Masley proved that the class number of
$\mathbb {Q}(\zeta _n)$
equals
$2$
if and only if
$n = 39$
or
$n = 56$
, provided
$n \not \equiv 2\ \pmod {4}$
. For more results on class numbers of cyclotomic fields, we refer the reader to [Reference Masley16, Reference Masley18–Reference Masley and Montgomery20, Reference Ram27].
In this context, we prove the following theorem, which provides a restriction on possible values of the class number of certain real cyclotomic fields.
Theorem 1.4 Let
$K = \mathbb {Q}(\zeta _{p^e})^+$
, where p is a prime and
$e \geq 1$
is an integer. Suppose q is a prime such that
$q - 1$
is relatively prime to
$\frac {p^{e-1}(p - 1)}{2}$
. Then,
$h_K \neq q$
.
As an immediate consequence, we obtain the following corollary.
Corollary 1.5 Let p be a prime and
$e \geq 1$
be an integer. Then, the class number of
$\mathbb {Q}(\zeta _{p^e})^+$
is not equal to
$2$
.
Finally, we provide an upper bound for the class number of real cyclotomic fields in terms of the relative Pólya group, as defined by Chabert in [Reference Chabert5].
Theorem 1.6 Let
$n \not \equiv 2\ \pmod {4}$
be a positive integer, and let
$L = \mathbb {Q}(\zeta _n)$
and
$K = \mathbb {Q}(\zeta _n)^+$
. Then,
$$\begin{align*}h_K \leq \begin{cases} |\mathcal{P}o(L/K)| & \text{if } n \text{ is a prime power}, \\ 2 |\mathcal{P}o(L/K)| & \text{otherwise}, \end{cases} \end{align*}$$
where
$\mathcal {P}o(L/K)$
denotes the relative Pólya group of the extension
$L/K$
. Moreover, if
$h_K = 1$
, then
$\mathcal {P}o(L/K)$
is trivial. The converse of the last assertion holds provided
$h_L$
is odd.
Theorem 1.6 ensures the following asymptotic behavior:
The structure of the article is as follows: In Section 2, we present some preliminary results. Section 3 is devoted to the proof of Theorem 1.1. In Section 4, we establish Corollaries 1.2 and 1.3. Finally, Sections 5 and 6 contain the proofs of Theorems 1.4 and 1.6, respectively.
2 Preliminaries
The purpose of this section is to present the necessary background results that will be used in the proofs of the results. We begin with the following theorem due to Louboutin [Reference Louboutin15].
Theorem 2.1 Let
$t> 1$
be an odd integer. Then, for any integer
$u \geq 2$
, the ideal class group of the imaginary quadratic field
$\mathbb {Q}\left (\sqrt {1 - 4u^t}\right )$
contains an element of order t.
We now recall a result due to Ichimura [Reference Ichimura11].
Theorem 2.2 Let
$t \geq 2$
be an integer, and let
$a \geq 3$
be an odd integer. Then, the class group of the real quadratic field
$\mathbb {Q}(\sqrt {a^{2t} + 4})$
contains an element of order t.
We next present a well-known result regarding Galois extensions.
Lemma 2.3 If
$K/\mathbb {Q}$
is a Galois extension, then the Hilbert class field
$H(K)$
over
$\mathbb {Q}$
is also a Galois extension.
For a Galois extension
$K/\mathbb {Q}$
, there exists an exact sequence:
$$ \begin{align} 1 \longrightarrow \mathrm{C\ell}(K) \longrightarrow \mathrm{G}_{H(K)/\mathbb{Q}} \longrightarrow \mathrm{G}_{K/\mathbb{Q}} \longrightarrow 1. \end{align} $$
This sequence is said to split if there exists a group homomorphism
$\psi : \mathrm {G}_{K/\mathbb {Q}} \rightarrow \mathrm {Aut}(\mathrm {C\ell }(K))$
such that
$$\begin{align*}\mathrm{G}_{H(K)/\mathbb{Q}} \simeq \mathrm{C\ell}(K) \rtimes_\psi \mathrm{G}_{K/\mathbb{Q}}. \end{align*}$$
Regarding the splitting of the sequence (2.1), the following result was proved by Gold [Reference Gold9].
Theorem 2.4 Let
$K/\mathbb {Q}$
be a Galois extension of number fields. Suppose s is the least common multiple of the ramification indices of all finite primes in the extension
$K/\mathbb {Q}$
. If
$s = [K : \mathbb {Q}]$
, then the exact sequence
$$\begin{align*}1 \longrightarrow \mathrm{C\ell}(K) \longrightarrow \mathrm{G}_{H(K)/\mathbb{Q}} \longrightarrow \mathrm{G}_{K/\mathbb{Q}} \longrightarrow 1 \end{align*}$$
splits.
We now state the following result from Ankeny, Chowla, and Hasse [Reference Ankeny, Chowla and Hasse2].
Theorem 2.5 Let
$L/K$
be an extension of number fields. Suppose there exists no intermediate field
$F \neq K$
such that
$F/K$
is abelian and unramified at all primes of K. Then, the class number
$h_K$
divides
$h_L$
.
We next recall a result from [Reference Washington31, Proposition 2.15], useful for identifying ramified primes in real cyclotomic subfields.
Proposition 2.6 Let
$n \not \equiv 2\ \pmod {4}$
be a positive integer. If
$n = p^a$
for some prime p and integer
$a \geq 1$
, then the extension
$\mathbb {Q}(\zeta _n)/\mathbb {Q}(\zeta _n)^+$
is ramified only at the prime of
$\mathbb {Q}(\zeta _n)^+$
lying above p and at the infinite primes of
$\mathbb {Q}(\zeta _n)^+$
. If n is not a prime power, then only the infinite primes of
$\mathbb {Q}(\zeta _n)^+$
are ramified in
$\mathbb {Q}(\zeta _n)$
.
We also record the following result from [Reference Washington31, Lemma 6.15].
Theorem 2.7 Let
$m\not \equiv 2\ \pmod {4}$
and
$n \not \equiv 2\ \pmod {4}$
be two positive integers greater than
$1$
such that
$\mathbb {Q}(\zeta _m) \subset \mathbb {Q}(\zeta _n)$
. Suppose K is a number field with
$\mathbb {Q}(\zeta _m) \subseteq K \subseteq \mathbb {Q}(\zeta _n)$
. If
$K/\mathbb {Q}(\zeta _m)$
is unramified at all primes of
$\mathbb {Q}(\zeta _m)$
, then
$K = \mathbb {Q}(\zeta _m)$
.
Following the proof of Theorem 1.1 in [Reference Ram26] (or see also [Reference Cornell7, p. 2]), one can obtain the next result.
Proposition 2.8 Let
$n = p_1^{e_1}p_2^{e_2} \ldots p_s^{e_s} \not \equiv 2\ \pmod {4}$
, where
$p_1, p_2, \ldots , p_s$
are distinct primes and
$e_i \in \mathbb {N}$
. Then, the class group of
$\mathbb {Q}(\zeta _n)$
contains a subgroup isomorphic to
$\mathrm {C\ell }(K_1) \oplus \mathrm {C\ell }(K_2) \oplus \cdots \oplus \mathrm {C\ell }(K_s)$
, where
$K_i = \mathbb {Q}(\zeta _{p_i^{e_i}})$
for
$1 \leq i \leq s$
.
Moreover, the following result holds.
Proposition 2.9 Let
$d_1, d_2$
be two coprime positive integers with
$d_1\not \equiv 2\ \pmod {4}$
and
$d_2 \not \equiv 2\ \pmod {4}$
. Then, the class group of
$\mathbb {Q}(\zeta _{d_1d_2})$
contains a subgroup isomorphic to
$\mathrm {C\ell }(\mathbb {Q}(\zeta _{d_1})) \oplus \mathrm {C\ell }(\mathbb {Q}(\zeta _{d_2}))$
.
Finally, we recall a classical result concerning the genus field of an abelian extension from [Reference Leopoldt14] (or see [Reference Ishida12, p. 52]). The Genus field of a number field K, denoted by
$G(K)$
, is defined as the maximal abelian extension of K that is unramified at all finite primes of K and is the compositum of K with an abelian extension of
$\mathbb {Q}$
.
Theorem 2.10 Let
$K/\mathbb {Q}$
be an abelian extension, and let
$G(K)$
be the genus field of K. Then,
$$\begin{align*}[G(K) : K] = \frac{\prod_p e(p)}{[K : \mathbb{Q}]}, \end{align*}$$
where the product runs over all rational primes p, and
$e(p)$
denotes the ramification index of p in K.
3 Proof of Theorem 1.1
We begin this section with the following result, which will be used in the next result. It is of independent interest as well.
Theorem 3.1 Let
$K/\mathbb {Q}$
be an abelian extension of number fields. Suppose
$m> 1$
is the largest positive divisor of
$h_K$
such that
$\gcd (m, [K : \mathbb {Q}]) = 1$
. Consider the subfield L of the Hilbert class field
$H(K)$
containing K with
$[L : K] = m$
. Then,
$L/\mathbb {Q}$
is not abelian. Moreover, for any intermediate field
$L'$
with
$K \subsetneq L' \subseteq L$
, the extension
$L'/\mathbb {Q}$
is also not abelian.
Proof (Theorem 3.1) Assume, by contradiction, that
$L/\mathbb {Q}$
is abelian. Let
$G(K)$
be the genus field of K. Since
$L \subset H(K)$
and L contains K, it follows that
$L \subset G(K)$
. Hence,
$[L : K]$
divides
$[G(K) : K]$
. By Theorem 2.10, we know that
$$\begin{align*}[G(K) : K] = \frac{\prod_p e(p)}{[K : \mathbb{Q}]}, \end{align*}$$
where the product is over all primes p, and
$e(p)$
denotes the ramification index of p in K. Therefore, m divides
$\frac {\prod _p e(p)}{[K : \mathbb {Q}]}$
, contradicting the assumption that
$m> 1$
and
$\gcd (m, [K : \mathbb {Q}]) = 1$
. Thus,
$L/\mathbb {Q}$
cannot be abelian.
A similar argument shows that the same conclusion holds for any proper intermediate field
$L'$
with
$K \subsetneq L' \subset L$
, thereby completing the proof.
Note that in Theorem 3.1, such a subfield L of
$H(K)$
is unique. If
$L'/\mathbb {Q}$
is not a Galois extension, then we do not need to show that
$L'/\mathbb {Q}$
is non-abelian.
Using the above theorem, we establish the following result, which will be used in the sequel. Although it can be obtained from [Reference Yokoyama34], we present a self-contained proof here using a different approach for the sake of completeness.
Theorem 3.2 Let
$L/\mathbb {Q}$
be an abelian extension of number fields containing a proper subfield
$K \ne \mathbb {Q}$
. Let
$m> 1$
be the largest positive divisor of
$h_K$
such that
$\gcd (m, [K:\mathbb {Q}]) = 1$
. If
$t> 1$
divides m and the t-rank of the class group of K is r, then the t-rank of
$C\ell (L)$
is at least r. Moreover, m divides
$h_L$
.
Proof (Theorem 3.2) It is known that
$C\ell (K)$
is a finite abelian group, and from class field theory, we have
$C\ell (K) \simeq G_{H(K)/K}$
. Hence, the Galois group
$G_{H(K)/K}$
has a unique subgroup H of order
$h_K/m$
. Let
$L'$
denote the fixed field of H in
$H(K)$
. Then,
$L'$
is the unique subfield of
$H(K)$
containing K such that
$[L' : K] = m$
.
Since
$L'/K$
is abelian and unramified at all primes of K, it follows that the extension
$LL'/L$
is also abelian and unramified at all primes of L. Therefore, the class group
$C\ell (L)$
contains a subgroup isomorphic to
$G_{LL'/L}$
.
As
$L/K$
and
$L'/K$
are both finite Galois extensions, we have
$$ \begin{align} [LL' : K] = \frac{[L : K] \cdot [L' : K]}{[L \cap L' : K]}. \end{align} $$
By hypothesis,
$L/\mathbb {Q}$
is an abelian extension and
$L \cap L'$
is a subfield of both L and
$L'$
containing K, Theorem 3.1 implies that
$L \cap L' = K$
. Consequently, (3.1) becomes
$$ \begin{align} [LL' : K] = [L : K] \cdot [L' : K]. \end{align} $$
On the other hand, we also have
$$ \begin{align} [LL' : K] = [LL' : L] \cdot [L : K]. \end{align} $$
Comparing (3.2) and (3.3), we deduce
$$ \begin{align} [LL' : L] = [L' : K] = m. \end{align} $$
Thus,
$G_{LL'/L}$
is a group of order m, and as noted earlier, is isomorphic to a subgroup of
$C\ell (L)$
. Hence, m divides
$h_L$
.
Define the map
$$\begin{align*}\psi : G_{LL'/L} \longrightarrow G_{L'/K} ~~\text{ by } \quad \sigma \mapsto \sigma|_{L'}. \end{align*}$$
This map is clearly an injective group homomorphism. Therefore, by (3.4), it is a group isomorphism:
$$\begin{align*}G_{LL'/L} \simeq G_{L'/K}. \end{align*}$$
Now, since
$t \mid m$
and
$C\ell (K)$
contains a unique subgroup of order m isomorphic to
$G_{L'/K}$
, it follows that the t-rank of
$G_{L'/K}$
is r; because
$C\ell (K)$
has t-rank r by assumption. Therefore,
$G_{LL'/L}$
contains at least r copies of
$\mathbb {Z}/t\mathbb {Z}$
, implying that
$C\ell (L)$
also contains at least r copies of
$\mathbb {Z}/t\mathbb {Z}$
. This completes the proof.
Proof (Theorem 1.1) Let
$L = \mathbb {Q}(\zeta _n)$
and let
$L^+ = \mathbb {Q}(\zeta _n)^+$
denote its maximal real subfield.
We first consider the case where
$d \equiv 1\ \pmod {4}$
, so that
$K = \mathbb {Q}(\sqrt {d})$
. In this case, K is contained in
$L^+$
, and since
$[K : \mathbb {Q}] = 2$
, it follows that
$K/\mathbb {Q}$
is a nontrivial abelian extension.
Given that m is odd, it is relatively prime to
$[K : \mathbb {Q}] = 2$
. Therefore, by applying Theorem 3.2 with
$K \subset L^+$
, we conclude that the t-rank of the class group of
$L^+$
is at least r, and that m divides the class number of
$L^+$
. Since
$C\ell (L^+)$
is isomorphic to a subgroup of
$C\ell (L)$
, these conclusions also hold for L. This completes the proof for the case
$d \equiv 1\ \pmod {4}$
.
Next, consider the case where
$d \equiv 3\ \pmod {4}$
, so that
$K = \mathbb {Q}(\sqrt {-d})$
. Then, K is contained in L, and as before,
$[K : \mathbb {Q}] = 2$
, implying that
$K \neq \mathbb {Q}$
and
$K/\mathbb {Q}$
is abelian. Since m is odd, m is again coprime with
$[K : \mathbb {Q}]$
. Applying Theorem 3.2 with
$K \subset L$
, we obtain that the t-rank of the class group of L is at least r, and that m divides the class number of L. This completes the proof of the theorem.
4 Proof of Corollaries 1.2 and 1.3
Combining Theorems 1.1, 2.5, and 2.7, we obtain the following result.
Theorem 4.1 Let
$d \not \equiv 2\ \pmod {4}$
be a square-free positive integer. Define
$$\begin{align*}K = \begin{cases} \mathbb{Q}(\sqrt{d}) & \text{if } d \equiv 1\quad \pmod{4}, \\ \mathbb{Q}(\sqrt{-d}) & \text{if } d \equiv 3\quad \pmod{4}. \end{cases} \end{align*}$$
Suppose
$m> 1$
is the largest odd divisor of
$h_K$
, and let
$t> 1$
be a divisor of m such that the t-rank of
$C\ell (K)$
is r. Let
$n> 1$
be any integer such that
$d \mid n$
. Then, the class group of
$\mathbb {Q}(\zeta _n)$
contains at least r copies of
$\mathbb {Z} / t\mathbb {Z}$
.
Proof (Corollary 1.2) From Theorem 2.2, we know that the class group of
$\mathbb {Q}(\sqrt {d^{\prime }_{a,t}})$
contains a subgroup isomorphic to
$\mathbb {Z} / t\mathbb {Z}$
. Hence, the t-rank of the class group of
$\mathbb {Q}(\sqrt {d^{\prime }_{a,t}})$
is at least
$1$
.
Since a is odd and
$d^{\prime }_{a,t}$
is the square-free part of
$a^{2t} + 4$
, it follows that
$d^{\prime }_{a,t} \equiv 1\ \pmod {4}$
. Moreover, as t is assumed to be odd, t is relatively prime to
$[\mathbb {Q}(\sqrt {d^{\prime }_{a,t}}) : \mathbb {Q}] = 2$
. Therefore, by Theorem 1.1, we conclude that both the fields
$\mathbb {Q}(\zeta _{d^{\prime }_{a,t}})$
and
$\mathbb {Q}(\zeta _{d^{\prime }_{a,t}})^+$
have class groups containing a subgroup isomorphic to
$\mathbb {Z} / t\mathbb {Z}$
.
Similarly, Theorem 2.1 implies that the class group of
$\mathbb {Q}(\sqrt {-d_{u,t}})$
also contains a subgroup isomorphic to
$\mathbb {Z} / t\mathbb {Z}$
, so the t-rank of its class group is at least
$1$
. Since
$-d_{u,t}$
is the square-free part of
$1 - 4u^t$
, it follows that
$d_{u,t} \equiv 3\ \pmod {4}$
. Given that t is odd, t is again relatively prime to
$[\mathbb {Q}(\sqrt {-d_{u,t}}) : \mathbb {Q}] = 2$
. Therefore, Theorem 1.1 applies, and we conclude that the class group of
$\mathbb {Q}(\zeta _{d_{u,t}})$
contains a subgroup isomorphic to
$\mathbb {Z} / t\mathbb {Z}$
.
Now, since
$a = 17b$
and
$u = a^2$
, it follows that
$d^{\prime }_{a,t}$
and
$d_{u,t}$
are relatively prime. Applying Proposition 2.9, we deduce that the class group of
$\mathbb {Q}(\zeta _{d_{u,a,t}})$
contains a subgroup isomorphic to
$ C\ell \big (\mathbb {Q}(\zeta _{d^{\prime }_{a,t}})\big ) \oplus C\ell \big (\mathbb {Q}(\zeta _{d_{u,t}})\big ). $
We have already shown that each of the above groups contains a subgroup isomorphic to
$\mathbb {Z} / t\mathbb {Z}$
. Therefore, the class group of
$\mathbb {Q}(\zeta _{d_{u,a,t}})$
contains a subgroup isomorphic to
$ \mathbb {Z} / t\mathbb {Z} \oplus \mathbb {Z} / t\mathbb {Z}. $
This completes the proof.
Remark 4.2 Theorem 4.1 further implies that if a cyclotomic field contains
$\mathbb {Q}(\zeta _{d_{u,a,t}})$
as a subfield, then the t-rank of its class group is also at least
$2$
.
The following result, cited from [Reference Uchida30], will be used to prove Corollary 1.3.
Theorem 4.3 Let m and a be positive integers such that a is coprime to
$6$
, and a has a prime factor q for which either
$2$
or
$3$
is not a quadratic residue modulo q. Furthermore, for every odd prime divisor
$\ell $
of m, assume that a has two prime factors
$q_1$
and
$q_2$
such that:
-
•
$2$
is not an
$\ell $
-th power residue modulo
$q_1$
and modulo
$q_2$
, and -
•
$3$
is an
$\ell $
-th power residue modulo
$q_1$
but not modulo
$q_2$
.
Let
$K = \mathbb {Q}(\theta )$
, where
$\theta $
is a root of the polynomial
$$\begin{align*}f(x) = x^3 + ux^2 + 2ux + u, \end{align*}$$
with integer
$u = \frac {a^{2m} + 27}{4}$
. Then,
$K/\mathbb {Q}$
is a cyclic cubic extension, and its class number is divisible by m. Moreover, there are infinitely many such fields K.
Proof (Corollary 1.3) Let m, a, and K be as in Theorem 4.3. From Theorem 4.3, we know that the class number
$h_K$
is divisible by m. Let s be a positive divisor of m such that
$\gcd (s,3) = 1$
. Since
$K/\mathbb {Q}$
is an abelian extension, the Kronecker–Weber theorem implies that
$K \subset \mathbb {Q}(\zeta _f)$
for some positive integer f. Let f be the minimal such positive integer. Therefore, keeping in mind the fact that
$[K : \mathbb {Q}] = 3$
, we see that
$K \subset \mathbb {Q}(\zeta _f)^+$
.
Define
$L = \mathbb {Q}(\zeta _f)^+$
. Applying Theorem 3.2 for the fields
$L, K$
, and the integer s, we conclude that s divides
$h_L$
. Consequently, s divides the class number of
$\mathbb {Q}(\zeta _f)$
.
By Theorem 4.3, there are infinitely many such fields K, and thus infinitely many corresponding fields
$L = \mathbb {Q}(\zeta _f)^+$
whose class numbers are divisible by s. This completes the proof.
5 Proof of Theorem 1.4
We first prove an auxiliary result, analogous to Theorem 2.7, which will be used in the proof of Theorem 1.4.
Proposition 5.1 Let
$m \not \equiv 2\ \pmod {4}$
and
$n \not \equiv 2\ \pmod {4}$
be two positive integers greater than
$1$
such that
$\mathbb {Q}(\zeta _m)^+ \subset \mathbb {Q}(\zeta _n)$
. Suppose K is a number field with
$$\begin{align*}\mathbb{Q}(\zeta_m)^+ \subseteq K \subseteq \mathbb{Q}(\zeta_n), \end{align*}$$
and the extension
$K/\mathbb {Q}(\zeta _m)^+$
is unramified at all primes of
$\mathbb {Q}(\zeta _m)^+$
. Then,
$K = \mathbb {Q}(\zeta _m)^+$
.
Proof (Proposition 5.1) Assume, by contradiction, that
$K/\mathbb {Q}(\zeta _m)^+$
is a nontrivial extension unramified at all primes of
$\mathbb {Q}(\zeta _m)^+$
and
$$\begin{align*}\mathbb{Q}(\zeta_m)^+ \subset K \subseteq \mathbb{Q}(\zeta_n). \end{align*}$$
Then, the compositum
$K\mathbb {Q}(\zeta _m)$
is contained in
$\mathbb {Q}(\zeta _n)$
, since both K and
$\mathbb {Q}(\zeta _m)$
are subfields of
$\mathbb {Q}(\zeta _n)$
. Moreover, since
$K/\mathbb {Q}(\zeta _m)^+$
is unramified at all primes of
$\mathbb {Q}(\zeta _m)^+$
, so is
$K\mathbb {Q}(\zeta _m)/\mathbb {Q}(\zeta _m)$
at all primes of
$\mathbb {Q}(\zeta _m)$
. By Theorem 2.7, we conclude that
$K\mathbb {Q}(\zeta _m) = \mathbb {Q}(\zeta _m)$
. Hence,
$K \subseteq \mathbb {Q}(\zeta _m)$
, and since
$\mathbb {Q}(\zeta _m)^+ \subset K$
, we must have
$$\begin{align*}\mathbb{Q}(\zeta_m)^+ \subset K \subseteq \mathbb{Q}(\zeta_m). \end{align*}$$
But
$[\mathbb {Q}(\zeta _m) : \mathbb {Q}(\zeta _m)^+] = 2$
, so the only possibility is
$K = \mathbb {Q}(\zeta _m)$
. This is not possible, otherwise the assumption that
$K/{\mathbb {Q}(\zeta _m)^+}$
is unramified at all primes of
$\mathbb {Q}(\zeta _m)^+$
contradicts Proposition 2.6. This contradiction proves that
$K=\mathbb {Q}(\zeta _m)^+$
, thereby completing the proof.
One can prove the following result using a similar approach as in the proof of Proposition 5.1.
Proposition 5.2 Let
$m \not \equiv 2\ \pmod {4}$
and
$n \not \equiv 2\ \pmod {4}$
be positive integers greater than
$1$
such that
$\mathbb {Q}(\zeta _m)^+ \subset \mathbb {Q}(\zeta _n)^+$
. Suppose K is a number field satisfying
$$\begin{align*}\mathbb{Q}(\zeta_m)^+ \subseteq K \subseteq \mathbb{Q}(\zeta_n)^+. \end{align*}$$
If the extension
$K/\mathbb {Q}(\zeta _m)^+$
is unramified at all primes of
$\mathbb {Q}(\zeta _m)^+$
, then
$K = \mathbb {Q}(\zeta _m)^+$
.
We can now deduce the following result as a direct consequence of Theorem 2.5 and Proposition 5.2.
Proposition 5.3 Let
$n \not \equiv 2\ \pmod {4}$
be a positive integer, and let p be a prime divisor of n. Then, the class number of
$\mathbb {Q}(\zeta _p)^+$
divides the class number of
$\mathbb {Q}(\zeta _n)^+$
. Moreover, for any positive divisor d of n, the class group of
$\mathbb {Q}(\zeta _n)^+$
contains a subgroup isomorphic to the class group of
$\mathbb {Q}(\zeta _d)^+$
.
In particular, from Corollary 1.2 and Proposition 5.3, we observe that if
$d^{\prime }_{a,t}$
is a divisor of a positive integer n, then the class group of
$\mathbb {Q}(\zeta _n)^+$
has a subgroup isomorphic to
$\mathbb {Z}/t\mathbb {Z}$
.
Proof (Theorem 1.4) Assume, by contradiction, that the class number
$h_K = q$
. Since
$K/\mathbb {Q}$
is a Galois extension and p is the only ramified prime in
$K/\mathbb {Q}$
with ramification index equal to
$[K : \mathbb {Q}]$
, we apply Theorem 2.4, which gives a group homomorphism
$$\begin{align*}\psi \colon G_{K/\mathbb{Q}} \to \operatorname{Aut}(C\ell(K)) \end{align*}$$
such that
$$ \begin{align} G_{H(K)/\mathbb{Q}} \simeq C\ell(K) \rtimes_\psi G_{K/\mathbb{Q}}. \end{align} $$
Since
$C\ell (K)$
is of order q, a prime, we have that the order of
$\operatorname {Aut}(C\ell (K))$
is
$q-1$
. The assumption that
$\gcd (q-1, \frac {p^{e-1}(p-1)}{2}) = 1$
implies that
$\psi $
is a trivial homomorphism. Consequently, the semidirect product becomes a direct product:
$$ \begin{align} C\ell(K) \rtimes_\psi G_{K/\mathbb{Q}} \simeq C\ell(K) \times G_{K/\mathbb{Q}}. \end{align} $$
Since both
$G_{K/\mathbb {Q}}$
and
$C\ell (K)$
are abelian groups, it follows that
$C\ell (K) \times G_{K/\mathbb {Q}}$
is also abelian. So, by (5.1) and (5.2), we see that
$G_{H(K)/\mathbb {Q}}$
is abelian, and hence the Hilbert class field
$H(K)/\mathbb {Q}$
is an abelian extension.
Therefore, there exists a positive integer v such that
$H(K) \subseteq \mathbb {Q}(\zeta _v)$
. Since the extension
$H(K)/\mathbb {Q}(\zeta _{p^e})^+$
is unramified at all primes of
$\mathbb {Q}(\zeta _{p^e})^+$
and satisfies
$$\begin{align*}\mathbb{Q}(\zeta_{p^e})^+ \subset H(K) \subseteq \mathbb{Q}(\zeta_v), \end{align*}$$
by Proposition 5.1, it follows that
$H(K) = \mathbb {Q}(\zeta _{p^e})^+$
. Hence,
$h_K = 1$
, leading to a contradiction. This completes the proof.
Remark 5.4 Using Corollary 1.5 and Proposition 5.3, we observe the following: if there exist infinitely many positive integers
$n \not \equiv 2\ \pmod {4}$
such that the class number of
$\mathbb {Q}(\zeta _n)^+$
is
$2$
, then we obtain infinitely many primes p for which the class number of
$\mathbb {Q}(\zeta _p)^+$
is
$1$
. Determining the density of such primes p for which
$h_{\mathbb {Q}(\zeta _p)^+} = 1$
remains an open problem. Therefore, exploring whether there exist infinitely many positive integers n with
$h_{\mathbb {Q}(\zeta _n)^+} = 2$
might be a promising direction for further research.
Along similar lines to the proof of Theorem 1.4, one can establish the following result.
Theorem 5.5 Let p be a prime and
$e \geq 1$
be an integer. Then, the class number of
$\mathbb {Q}(\zeta _{p^e})$
cannot be equal to
$2$
.
Remark 5.6 Using the list in (1.1), Proposition 2.8, and Theorem 5.5, we observe the following: for any positive integer
$n \not \equiv 2\ \pmod {4}$
, if
$h_{\mathbb {Q}(\zeta _n)} = 2$
, then every prime divisor p of n must lie in the set
$\{2, 3, 5, 7, 11, 13, 17, 19\}$
. By applying Theorem 1.1, Theorems 2.5–2.7, Proposition 2.9, and Theorem 5.5, along with the known data regarding the class numbers of cyclotomic fields
$\mathbb {Q}(\zeta _v)$
for
$1 \leq v \leq 160$
and performing some computations, one can conclude that
$$\begin{align*}h_{\mathbb{Q}(\zeta_n)} = 2 \quad \text{if and only if} \quad n = 39 \text{ or } n = 56. \end{align*}$$
This conclusion was also obtained by Masley in [Reference Masley17].
6 Proof of Theorem 1.6
We first state two results which will be used in the proof of Theorem 1.6. The following proposition is taken from [Reference Brumer and Rosen3, Proposition 3.2].
Theorem 6.1 Let
$L/K$
be a Galois extension of number fields of prime degree p. Suppose w is the number of infinite primes of K that are unramified in L. Then, the order of the first cohomology group
$H^1(G_{L/K}, \mathcal {O}_L^\times )$
divides
$p^{w+1}$
.
The next result describes the size of the first cohomology group (cf. [Reference Chabert5, Proposition 4.4]).
Theorem 6.2 Let
$L/K$
be a Galois extension of number fields. Then
$$\begin{align*}|\mathcal{P}o(L/K)| = \frac{h_K \times \prod_{\mathfrak{p}} e(\mathfrak{p})}{|H^1(G_{L/K}, \mathcal{O}_L^\times)|}, \end{align*}$$
where the product runs over all the finite prime ideals
$\mathfrak {p}$
of K that ramify in L, and
$e(\mathfrak {p})$
denotes the ramification index of
$\mathfrak {p}$
in the extension
$L/K$
.
Proof (Theorem 1.6) Applying Proposition 2.6, we obtain
$$\begin{align*}\prod_{\mathfrak{p}} e({\mathfrak{p}}) = \begin{cases} 2 & \text{if } n \text{ is a prime power}, \\ 1 & \text{otherwise}, \end{cases} \end{align*}$$
where the product runs over all finite prime ideals
$\mathfrak {p}$
of K that ramify in L, and
$e(\mathfrak {p})$
denotes the ramification index of
$\mathfrak {p}$
in the extension
$L/K$
.
Since
$L/K$
is a Galois extension of degree
$2$
and all infinite primes of K ramify in L, Theorem 6.1 implies that
$|H^1(G_{L/K}, \mathcal {O}_L^\times )|$
divides
$2.$
Thus, by Theorem 6.2, we conclude
$$ \begin{align} h_K \leq \begin{cases} |\mathcal{P}o(L/K)| & \text{if } n \text{ is a prime power}, \\ 2 \cdot |\mathcal{P}o(L/K)| & \text{otherwise}. \end{cases} \end{align} $$
Now assume that
$h_K = 1$
. We aim to show that the relative Pólya group
$\mathcal {P}o(L/K)$
is trivial.
Let
$\mathfrak {p}$
be a prime ideal of
$\mathcal {O}_K$
with residue degree f in the extension
$L/K$
. Define
$$\begin{align*}\prod_{\mathfrak{p}^f}(L/K) = \prod_{\substack{\mathfrak{P} \in \mathrm{Max}(\mathcal{O}_L) \\ \mathfrak{P} \cap \mathcal{O}_K = \mathfrak{p}}} \mathfrak{P}, \end{align*}$$
where
$\mathrm {Max}(\mathcal {O}_L)$
denotes the set of all maximal ideals (i.e., prime ideals) of
$\mathcal {O}_L$
. Since
$L/K$
is a Galois extension of degree
$2$
, it follows that the residue degree f is either
$1$
or
$2$
.
Case I: Suppose
$n = p^a$
is a prime power. Then, by Proposition 2.6, we have
$$\begin{align*}\prod_{\mathfrak{p}^f}(L/K) = \begin{cases} \mathfrak{P} & \text{if } p \in \mathfrak{p} \text{ and } f = 1, \\ \mathfrak{p} \mathcal{O}_L = \mathfrak{P}_1 \mathfrak{P}_2 & \text{if } p \notin \mathfrak{p} \text{ and } f = 1, \\ \mathfrak{p} \mathcal{O}_L & \text{if } p \notin \mathfrak{p} \text{ and } f = 2. \end{cases} \end{align*}$$
Case II: Suppose n is not a prime power. Again, by Proposition 2.6, we have
$$\begin{align*}\prod_{\mathfrak{p}^f}(L/K) = \begin{cases} \mathfrak{p} \mathcal{O}_L = \mathfrak{P}_1 \mathfrak{P}_2 & \text{if } f = 1, \\ \mathfrak{p} \mathcal{O}_L & \text{if } f = 2. \end{cases} \end{align*}$$
Since
$h_K = 1$
, the ideal
$\mathfrak {p} \mathcal {O}_L$
is principal in
$\mathcal {O}_L$
for each prime ideal
$\mathfrak {p}$
of
$\mathcal {O}_K$
. Also,
$\mathfrak {P}$
is principal due to L being a Pólya field. Using these facts in both cases, it follows that
$\prod _{\mathfrak {p}^f}(L/K)$
is a principal ideal of
$\mathcal {O}_L$
for every prime ideal
$\mathfrak {p}$
of
$\mathcal {O}_K$
, but the relative Pólya group
$\mathcal {P}o(L/K)$
is generated by the ideal classes containing the ideals of the form
$\prod _{\mathfrak {p}^f}(L/K)$
. Therefore, we see that the group
$\mathcal {P}o(L/K)$
is trivial.
Now assume that
$\mathcal {P}o(L/K)$
is trivial and the class number
$h_L$
is odd. Using inequality (6.1), we deduce that
$h_K \in \{1, 2\}$
. However, since
$h_K$
divides
$h_L$
and
$h_L$
is odd,
$h_K$
must be odd. Hence,
$h_K \neq 2$
, so
$h_K = 1$
. This completes the proof.
