#  Hilbert genus fields of some number fields with high degrees

Mohamed Mahmoud Chems-Eddin  Mohamed Mahmoud Chems-Eddin: Sidi Mohamed Ben
Abdellah University, Faculty of Sciences Dhar El Mahraz, Departement of
Mathematics, Fez, Morocco<EMAIL_ADDRESS>,  Moulay Ahmed Hajjami
Moulay Ahmed Hajjami: Moulay Ismail University of Meknes, Faculty of Sciences
and Technology, Department of Mathematics, Errachidia, Morocco.
<EMAIL_ADDRESS>and  Mohammed Taous  Mohammed Taous: Moulay Ismail
University of Meknes, Faculty of Sciences, Departement of Mathematics, Meknes,
Morocco<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to give some properties of Hilbert genus fields and
construct the Hilbert genus fields of the fields
$L_{m,d}:=\mathbb{Q}(\zeta_{2^{m}},\sqrt{d})$, where $m\geq 3$ is a positive
integer and $d$ is a square-free integer whose prime divisors are congruent to
$\pm 3\pmod{8}$ or $9\pmod{16}$.

###### Key words and phrases:

Unramified extensions, Hilbert genus fields.

###### 2010 Mathematics Subject Classification:

11R16, 11R29, 11R27, 11R04, 11R37

##  1\. Introduction

Let $d$ be a square-free integer and $m\geq 3$ a positive integer. The fields
$L_{m,d}:=\mathbb{Q}(\zeta_{2^{m}},\sqrt{d})$ represent the layers of the
cyclotomic $\mathbb{Z}_{2}$-extension of the special Dirichlet fields
$\mathbb{Q}(\sqrt{-1},\sqrt{d})$ and they were the subject of some recent
studies, which gave interest to their $2$-class groups (cf. [2, 3]). In the
present, work we shall construct the Hilbert genus fields of these fields.

This question have been investigated by many mathematicians for fields of
small degrees (eg. $2$ and $4$). For example, Bae and Yue studied the Hilbert
genus field of the fields $\mathbb{Q}(\sqrt{p},\sqrt{d})$, for a prime number
$p$ such that, $p=2$ or $p\equiv 1\pmod{4}$, and a positive square-free
integer $d$ (cf. [1]).

Thereafter, Ouyang and Zhang have determined the Hilbert genus field of the
imaginary biquadratic fields $\mathbb{Q}(\sqrt{\delta},\sqrt{d})$, where
$\delta=-1,-2$ or $-p$ with $p\equiv 3\pmod{4}$ a prime number and $d$ any
square-free integer. Thereafter, they constructed the Hilbert genus field of
real biquadratic fields $\mathbb{Q}(\sqrt{\delta},\sqrt{d})$, for any positive
square-free integer $d$, and $\delta=p,2p$ or $p_{1}p_{2}$ where $p$, $p_{1}$
and $p_{2}$ are prime numbers congruent to $3\pmod{4}$, such that the class
number of $\mathbb{Q}(\sqrt{\delta})$ is odd (cf. [10, 11]). For more works on
this subject, we refer the reader to the papers [11, 12, 5, 14].

In this paper, we shall construct the Hilbert genus field of the fields
$L_{m,d}$, $m\geq 3$, whenever all the prime divisors of $d$ are congruent to
$\pm 3\pmod{8}$ or $9\pmod{16}$. Our results generalize some results of Ouyang
and Zhang on the fields $\mathbb{Q}(\sqrt{-1},\sqrt{d})$ (cf. [10]).

The plan of this paper is as follows; in Section 2, we introduce some
properties of the Hilbert genus field. In the last section, we give the list
of Hilbert genus fields of the fields $L_{m,d}$.

Let us stick the following notations: Let $k$ be a number field.
$\mathcal{O}_{k}$ the ring of integers $k$. $E(k)$ the Hilbert genus field of
$k$. $N_{k/k^{\prime}}$ denotes the norm map of an extension $k/k^{\prime}$.
Let $r_{2}(\mathrm{G})$ denote the $2$-rank of an abelian finite group
$\mathrm{G}$. Denote by $\zeta_{n}$ an $n$th primitive root of unity. For more
notations see the beginning of each section below.

##  2\. Some results on Hilbert genus fields

Let us start this section by recalling some definitions and results. Let $k$
be a number field and denote by $\mathbf{C}l(k)$ its class group. It is know
by class field theory that we have
$\mathrm{G}:=\mathrm{Gal}(H(k)/k)\simeq\mathbf{C}l(k)$, where $H(k)$ is the
Hilbert class field. The Hilbert genus field of $k$ is defined as the
invariant field $E(k)$ of $\mathrm{G}^{2}$. Then, we have:

$\mathbf{C}l(k)/\mathbf{C}l(k)^{2}\simeq\mathrm{G}/\mathrm{G}^{2}\simeq\mathrm{Gal}(E(k)/k),$

and thus, $r_{2}(\mathbf{C}l(k))$ = $r_{2}(\mathrm{Gal}(E(k)/k))$. On the
other hand, $E(k)/k$ is the maximal unramified Kummer extension of exponent
$2$. Thus, by Kummer theory (cf. [9, p. 14]), there exists a unique
multiplicative group $\Delta(k)$ such that

$E(k)=H(k)\cap k(\sqrt{k^{*}})=k(\sqrt{\Delta(k)})\text{ and
}{k^{*}}^{2}\subset{\Delta(k)}\subset k^{*}.$

Therefore, the construction of the Hilbert genus field of $k$ is equivalent to
the determination of a set of generators for the finite group
$\Delta(k)/k{{}^{*}}^{2}$.

We need the following proposition which is a particular case of the result in
[8, p. 239].

######  Proposition 2.1 ([8], Theorem 5.20).

Let $k/k^{\prime}$ be a quadratic extension of number fields and $\alpha$ a
element of $k^{\prime}$, coprime with $2$, such that
$k=k^{\prime}(\sqrt{\alpha})$. The extension $k/k^{\prime}$ is unramified at
all finite primes of $k^{\prime}$ if and only if the two following items hold:

  1. 1.

the ideal generated by $\alpha$ is the square of the fractional ideal of
$k^{\prime}$, and

  2. 2.

there exists a nonzero number $\xi$ of $k^{\prime}$ verifying
$\alpha\equiv\xi^{2}\pmod{4}$.

######  Corollary 2.2.

Let $k/k^{\prime}$ be an extension of number fields and let $\alpha\in
k^{\prime}$ be a square-free in $k$. If $k^{\prime}(\sqrt{\alpha})/k^{\prime}$
is unramified, then $k(\sqrt{\alpha})/k$ is also unramified.

###### Proof.

As $k^{\prime}(\sqrt{\alpha})/k^{\prime}$ is unramified, then there exist an
ideal $\mathfrak{b}$ of $\mathcal{O}_{k^{\prime}}$ such that
$\alpha\mathcal{O}_{k^{\prime}}=\mathfrak{b}^{2}$ and $\xi\in k^{\prime}$
verifying $\alpha\equiv\xi^{2}\pmod{4}$. Therefore,
$\alpha\mathcal{O}_{k}=(\mathfrak{b}\mathcal{O}_{k})^{2}$ and $\xi\in
k^{\prime}$ verifying $\alpha\equiv\xi^{2}\pmod{4}$. So the result by
Proposition 2.1. ∎

Let us put the following definition.

######  Definition 2.3.

A number field $k$ is said QO-field if it is a quadratic extension of certain
number field $k^{\prime}$ whose class number is odd. We shall call
$k^{\prime}$ a base field of the QO-field $k$ and the $k/k^{\prime}$ is a QO-
extension.

######  Lemma 2.4.

Let $k/k^{\prime}$ be a QO-extension. Let $\Delta(k)$ denote the
multiplicative group such that ${k^{*}}^{2}\subset{\Delta(k)}\subset k^{*}$
and $k(\sqrt{\Delta(k)})$ the Hilbert genus field of $k$ $($see the beginning
of this section$)$. Then

$r_{2}(\Delta(k)/{k^{*}}^{2})=t-e-1,$

with $t$ is the number of ramified primes (finite or infinite) in the
extension $k/k^{\prime}$ and $e$ is defined by
$2^{e}=[E_{k^{\prime}}:E_{k^{\prime}}\cap N_{k/k^{\prime}}(k^{*})]$, where
$E_{k^{\prime}}$ is the unit group of $k^{\prime}$.

###### Proof.

Put $\mathrm{G}=\mathrm{Gal}(H(k)/k)$. By the definition of $E(k)$ $($see the
beginning of this section$)$ we have:

$\mathbf{C}l(k)/\mathbf{C}l(k)^{2}\simeq\mathrm{G}/\mathrm{G}^{2}\simeq\mathrm{Gal}(E(k)/k),$

Since $E(k)=k(\sqrt{\Delta(k)})$, then by class field theory:

$r_{2}(\Delta(k)/{k^{*}}^{2})=\log_{2}[k(\sqrt{\Delta(k)}):k]=\log_{2}[E(k):k]=r_{2}(\mathbf{C}l(k)).$

So the result by the well known ambiguous class number formula (cf. [13, Lemma
2.4]). ∎

######  Theorem 2.5.

Let $k=k^{\prime}(\sqrt{\mu})$ be a ramified quadratic extension of QO-fields
that have the same base field. Assume furthermore, that
$r_{2}(\mathbf{C}l(k))=r_{2}(\mathbf{C}l(k^{\prime}))$. Then we have

$E(k)=k(E(k^{\prime})).$

###### Proof.

Let $\gamma_{1}$, …, $\gamma_{r}$ be $r=r_{2}(\mathbf{C}l(k^{\prime}))$
elements of $k^{\prime}$ such that,

$E(k^{\prime})=k^{\prime}(\sqrt{\gamma_{1}},...,\sqrt{\gamma_{r}}).$

By Corollary 2.2, $k(\sqrt{\gamma_{i}})/k$ is unramified for all $i$.
Therefore, $\\{{\gamma_{1}},...,{\gamma_{r}}\\}$ is a representative set of
$\Delta(k)/k^{2}$. In fact, it suffices to prove that it is linearly
independent modulo ${k^{*}}^{2}$. Put
$\beta={\gamma_{1}}{\gamma_{2}}^{a_{2}}...{\gamma_{r}}^{a_{r}}$, for some
$a_{i}\in\\{0,1\\}$, and assume that $\beta\in k^{2}$. One can easily check
that we have $\beta\in k^{\prime 2}$ or $\mu\beta\in k^{\prime 2}$.

  1. $\bullet$

Since $\\{{\gamma_{1}},...,{\gamma_{r}}\\}$ is a representative set of
$\Delta(k^{\prime})/k^{\prime 2}$, the case $\beta\in k^{2}$, is clearly
impossible.

  2. $\bullet$

Assume that $\mu\beta\in k^{\prime 2}$. Thus,
$\mu{\gamma_{1}}{\gamma_{2}}^{a_{2}}...{\gamma_{r}}^{a_{r}}=\alpha^{2}$, for
some $\alpha\in k^{\prime}$. Since $k/k^{\prime}$ is ramified, this equality
implies that there is a square-free element $\mu^{\prime}$ of $k^{\prime}$
such that $\mu^{\prime}=\alpha^{\prime 2}$, for some $\alpha^{\prime}\in
k^{\prime}$, which is impossible.

Since $r_{2}(\mathbf{C}l(k))=r$, we have the result. ∎

Thus, we have the following corollary.

######  Corollary 2.6.

Let $k$ be a QO-number field, $k^{\prime}$ its base field and $k_{n}$ be the
$n$th layer of its cyclotomic $\mathbb{Z}_{2}$-extension. Let $n_{0}$ be an
integer such that every prime which ramify in $k_{\infty}/k_{n_{0}}$ is
totally ramified and
$r_{2}(\mathbf{C}l(k_{n_{0}}))=r_{2}(\mathbf{C}l(k_{n_{0}+1}))=r$. Assume
furthermore that $k_{n_{0}+1}$ is a QO-field whose base field is the
$(n_{0}+1)$th layer of the cyclotomic $\mathbb{Z}_{2}$-extension of
$k^{\prime}$. If $E(k)=k_{n_{0}}(\sqrt{\gamma_{1}},...,\sqrt{\gamma_{r}})$ and
$k_{n_{0}}^{\prime}$ has odd class number, then

$E(k_{n})=k_{n}(\sqrt{\gamma_{1}},...,\sqrt{\gamma_{r}}),$

for all $n\geq n_{0}$. Furthermore,
$E(k_{\infty})=k_{\infty}(\sqrt{\gamma_{1}},...,\sqrt{\gamma_{r}}).$

###### Proof.

Under the hypothesis of the corollary, one can use the well known result of
Fukuda (cf. [4, Theorem 1]) to check what follows:

  1. $\bullet$

The class number of $k^{\prime}_{n}$, the $n$th layer of cyclotomic
$\mathbb{Z}_{2}$-extension of $k^{\prime}$, is odd for all $n$.

  2. $\bullet$

$r_{2}(\mathbf{C}l(k_{n}))=r$, for all $n\geq n_{0}$.

Therefore, a recurrent application of Theorem 2.5 gives the corollary. ∎

##  3\. The Hilbert genus field of
$L_{m,d}=\mathbb{Q}(\zeta_{2^{m}},\sqrt{d})$

Keep notations fixed in the previous sections. Now we state the main result of
the paper. For the existance of solutions of the diophantine equations
mentioned on the theorem below, one can see [10] and [7, p. 324].

######  Theorem 3.1.

Let $d$ be a square-free integer whose prime divisors are congruent to $\pm
3\pmod{8}$ or $9\pmod{16}$. Let $n$ be the number of prime divisors of $d$,
$s$ the number of prime divisors of $d$ which are congruent $5\pmod{8}$, $t$
the number of prime divisors of $d$ which are congruent $3\pmod{8}$ and $r$
the number of those which are congruent $9\pmod{16}$. We have $n=r+t+s$. Let
$d=\ell_{1}...\ell_{r}p_{1}...p_{s}q_{1}...q_{t}$ be the factorization of $d$
such that $\ell_{i}\equiv 9\pmod{16}$ and $p_{i}\equiv 5\equiv-q_{i}\pmod{8}$.
Put $L_{m,d}=\mathbb{Q}(\zeta_{2^{m}},\sqrt{d}),$ $m\geq 3$.

  1. 1.

Assume that $r=t=0$ and $s\geq 2$. Let $(x_{i},y_{i})$ be a primitive solution
of $p_{1}p_{i}=x^{2}+y^{2}$ $(i\geq 2)$ such that $x_{i}\equiv 1\pmod{2}$ and
$y_{i}\equiv 0\pmod{4}$. Then for all $m\geq 3$,

$E(L_{m,d})=L_{m,d}(\sqrt{p_{1}},...,\sqrt{p_{n-1}},\sqrt{\gamma_{2}},...,\sqrt{\gamma_{n}}),$

where $\gamma_{i}=x_{i}+y_{j}\sqrt{-1}$.

  2. 2.

Assume that $r=s=0$ and $t\geq 2$. Let $(x_{i}^{\prime},y_{i}^{\prime})$ be a
primitive solution of $q_{1}q_{i}=x^{2}+2y^{2}$ $(i\geq 2)$ such that
$x_{i}^{\prime}\equiv 1\pmod{2}$ and $y_{i}^{\prime}\equiv 0\pmod{2}$ . Then
for all $m\geq 1$,

$E(L_{m,d})=L_{m,d}(\sqrt{q_{1}},...,\sqrt{q_{n-1}},\sqrt{\alpha_{2}},...,\sqrt{\alpha_{n}}),$

where $\alpha_{i}=x_{i}^{\prime}+y_{i}^{\prime}\sqrt{-2}$.

  3. 3.

Assume that $r=0$ and $t,s\geq 2$. Then for all $m\geq 3$,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{s}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=1}^{t-1}\mathbb{Q}(\sqrt{q_{i}})\right)\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right)\left(\prod_{i=2}^{t}\mathbb{Q}(\sqrt{\alpha_{i}})\right),$

where $\alpha_{i}$ and $\gamma_{i}$ are defined in the first two items.

  4. 4.

Assume that $r\geq 1$ and $t,s\geq 2$. Let $(a_{i},b_{i})$, $(e_{i},f_{i})$
and $(u_{i},v_{i})$ such
$\ell_{i}=a_{i}^{2}+16b_{i}^{2}=e_{i}^{2}-32f_{i}^{2}=u_{i}^{2}+8v_{i}^{2}$
and $a_{i}\equiv e_{i}\equiv u_{i}\equiv\pm 1\pmod{4}$. Then for all $m\geq
3$,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{s}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=1}^{t-1}\mathbb{Q}(\sqrt{q_{i}})\right)$
$\displaystyle\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right)\left(\prod_{i=2}^{t}\mathbb{Q}(\sqrt{\alpha_{i}})\right),$

where $\alpha_{i}$ and $\gamma_{i}$ are defined in the first two items, and
$\pi_{1,i}=a_{i}+4ib_{i}$, $\pi_{2,i}=e_{i}+4f_{i}\sqrt{2}$ and
$\pi_{3,i}=u_{i}+2v_{i}\sqrt{-2}$.

  5. 5.

Assume that $t=0$, $r\geq 1$ and $s\geq 2$. For all $m\geq 3$, we have

    1. $\bullet$

If $\exists i\in\\{1,\dots,r\\}$ such that
$\left(\frac{2}{l_{i}}\right)_{4}=1$, then

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{l_{i}})\right)\left(\prod_{i=1}^{s-1}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right),$

    2. $\bullet$

else,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{s-1}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right),$

where $\gamma_{i}$ , $\pi_{1,i}$, $\pi_{2,i}$ and $\pi_{3,i}$ are defined as
above.

  6. 6.

Assume that $s=0$, $r\geq 1$ and $t\geq 2$. For all $m\geq 3$, we have:

    1. $\bullet$

If $\exists i\in\\{1,\dots,r\\}$ such that
$\left(\frac{2}{l_{i}}\right)_{4}=-1$, then

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{l_{i}})\right)\left(\prod_{i=1}^{t-1}\mathbb{Q}(\sqrt{q_{i}})\right)\left(\prod_{i=2}^{t}\mathbb{Q}(\sqrt{\alpha_{i}})\right),$

    2. $\bullet$

else,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{t-1}\mathbb{Q}(\sqrt{q_{i}})\right)\left(\prod_{i=2}^{t}\mathbb{Q}(\sqrt{\alpha_{i}})\right),$

where $\alpha_{i}$ , $\pi_{1,i}$, $\pi_{2,i}$ and $\pi_{3,i}$ are defined as
above.

  7. 7.

Assume that $t=s=0$ and $r\geq 2$. For all $m\geq 3$, we have:

    1. $\bullet$

If there exist $i,j\in\\{1,\dots,r\\}$ such that
$\left(\frac{2}{l_{i}}\right)_{4}\neq\left(\frac{2}{l_{j}}\right)_{4}$, then

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right),$

    2. $\bullet$

else,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right),$

where $\pi_{1,i}$, $\pi_{2,i}$ and $\pi_{3,i}$ are defined as above.

  8. 8.

Assume that $s=1$, $r\geq 1$ and $t\geq 2$. For all $m\geq 3$, we have

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{t}\mathbb{Q}(\sqrt{q_{i}})\right)\left(\prod_{i=2}^{t}\mathbb{Q}(\sqrt{\alpha_{i}})\right).$

  9. 9.

Assume that $t=1$, $r\geq 1$ and $s\geq 2$. For all $m\geq 3$, we have

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{s}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right).$

  10. 10.

Assume that $s=t=1$ and $r\geq 1$. For all $m\geq 3$, we have

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\mathbb{Q}(\sqrt{p_{1}}).$

  11. 11.

Assume that $t=0$, $s=1$ and $r\geq 1$. For all $m\geq 3$, we have

    1. $\bullet$

If $\exists i\in\\{1,\dots,r\\}$ such that
$\left(\frac{2}{l_{i}}\right)_{4}=1$, then

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{l_{i}})\right).$

    2. $\bullet$

else,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right).$

  12. 12.

Assume that $s=0$, $t=1$ and $r\geq 1$. For all $m\geq 3$, we have

    1. $\bullet$

If $\exists i\in\\{1,\dots,r\\}$ such that
$\left(\frac{2}{l_{i}}\right)_{4}=-1$, then

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{l_{i}})\right).$

    2. $\bullet$

else,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{l_{i}},\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right).$

  13. 13.

Assume that $r=0$ and $s=t=1$. For all $m\geq 3$, we have

$E(L_{m,d})=L_{m,d}(\sqrt{p_{1}})=L_{m,d}(\sqrt{q_{1}}).$

  14. 14.

Assume that $r=s=0$ and $t=1$ or $r=t=0$ and $s=1$. For all $m\geq 3$, we have

$E(L_{m,d})=L_{m,d}.$

  15. 15.

Assume that $t=s=0$ and $r=1$. For all $m\geq 3$, we have

$E(L_{m,d})=L_{m,d}(\sqrt{\pi_{1,1}},\sqrt{\pi_{2,1}}),$

where $\pi_{1,1}=a_{1}+4ib_{1}$ and $\pi_{2,1}=e_{1}+4f_{1}\sqrt{2}$ such that
$l_{1}=a_{1}^{2}+16b_{1}^{2}=e_{1}^{2}-32f_{1}^{2}.$

###### Proof.

To simplify notations let us put
$L=L_{3,d}=\mathbb{Q}(\zeta_{8},\sqrt{d})=K(\sqrt{d})$, with
$K=\mathbb{Q}(\zeta_{8})=\mathbb{Q}(\sqrt{-1},\sqrt{2})$.

  1. 1.

As the couple $(x_{i},y_{i})$ is a primitive solution of
$p_{1}p_{i}=x^{2}+y^{2}$ $(i\geq 2)$ such that $x_{i}\equiv 1\pmod{2}$ and
$y_{i}\equiv 0\pmod{4}$, then $\gamma_{i}=x_{i}+y_{i}\sqrt{-1}\equiv
x_{i}\pmod{4}\equiv\pm 1\pmod{4}$. Since $i=\sqrt{-1}\in L$, the equation
$\gamma_{i}\equiv\xi^{2}\pmod{4}$ has solutions in $L$, furthermore the prime
ideals of $K$ above $p_{1}$ and $p_{i}$ are ramified in $L/K$, so
$(\gamma_{i})$ is the square of a fractional ideal of $L$. It follows by
Proposition 2.1, that $\forall i\in\\{2,...,n\\}$, $L(\sqrt{\gamma_{i}})/L$ is
an unramified extension and it is very clear that the extension
$L(\sqrt{p_{i}})/L$, $1\leqslant i\leqslant n-1$ is also unramified.
On the other hand by [2, Theorem 1], we have $r_{2}(\mathbf{C}l(L))=2n-2$.
Consider the following set :

$\eta=\\{p_{1},...,p_{n-1},\gamma_{2},...,\gamma_{n}\\}.$

Let us show that the elements of $\eta$ are linearly independents modulo
${L^{*}}^{2}$. Put
$\beta=\left(\displaystyle\prod_{i=1}^{n-1}{p_{i}}^{a_{i}}\right)\left(\displaystyle\prod_{j=2}^{n}{\gamma_{j}}^{b_{j}}\right)$,
where $a_{i},b_{j}\in\\{0,1\\}$ are not all zero. Assume that
$\beta\in{L^{*}}^{2}.$ Thus, $\beta\in{K^{*}}^{2}$ or $d\beta\in{K^{*}}^{2}.$

Note that $\exists\,j\in\\{2,3,\dots,n\\}$ with $\,b_{j}\not=0$ (else, we get
$\beta=\displaystyle\prod_{i=1}^{n-1}{p_{i}}^{a_{i}}\in{K^{*}}^{2}$, then
$\sqrt{\beta}\in{K^{*}}$. So $\mathbb{Q}(\sqrt{\beta})$ is a quadratic
subfield of $K=\mathbb{Q}(\zeta_{8})=\mathbb{Q}(\sqrt{-1},\sqrt{2})$. But this
is impossible, since the only quadratic subfields of $K=\mathbb{Q}(\zeta_{8})$
are $k=\mathbb{Q}(\sqrt{-1})$, $k^{\prime}=\mathbb{Q}(\sqrt{2})$ and
$k^{\prime\prime}=\mathbb{Q}(\sqrt{-2})$). We therefore distinguish two cases
:

    1. $\bullet$

If $\beta\in{K^{*}}^{2}.$ Then
$N_{K/k}(\gamma_{j})=x_{j}^{2}+y_{j}^{2}=p_{1}p_{j}$. Thus,

$N_{K/k}(\beta)=\left(\displaystyle\prod_{i=1}^{n-1}{p_{i}}^{a_{i}}\right)^{2}\left(\displaystyle\prod_{j=2}^{n}{p_{1}p_{j}}^{b_{j}}\right)=p_{1}^{\ell}\left(\displaystyle\prod_{j=2}^{n}{p_{j}}^{b_{j}}\right)Z^{2},$

such that $\ell\in\\{0,1\\}$ and $Z\in\mathbb{Q}$. This clearly implies that
$p_{1}^{\ell}\left(\displaystyle\prod_{j=2}^{n}{p_{j}}^{b_{j}}\right)$ is a
square in $k$, which is impossible. Thus, $\beta\notin{K^{*}}^{2}.$

    2. $\bullet$

If $d\beta\in{K^{*}}^{2},$ then $N_{K/k}(d\beta)=d^{2}N_{K/k}(\beta)$. As
above we show that this implies that
$p_{1}^{\ell}\left(\displaystyle\prod_{j=2}^{n}{p_{j}}^{b_{j}}\right)\in
k^{2}$, which is also impossible. So $\beta$ can not be a square in $L$.
Therefore, the elements of $\eta$ are linearly independents modulo $L^{2}$. If
follows that, $\eta$ is a representative set of $\Delta(L)/L$. Hence,

$E(L_{3,d})=\mathbb{Q}(\zeta_{8},\sqrt{d},\sqrt{p_{1}},...,\sqrt{p_{n-1}},\sqrt{\gamma_{2}}...\sqrt{\gamma_{n}}).$

Note that by [2, Theorem 1], we have $r_{2}(\mathbf{C}l(L_{m,d}))=2n-2$, for
all $m\geq 3$. Therefore, by Corollary 2.6, we get :

$E(L_{m,d})=L_{m,d}(\sqrt{p_{1}},...,\sqrt{p_{n-1}},\sqrt{\gamma_{2}},...,\sqrt{\gamma_{n}}).$

  2. 2.

By [2, Theorem 1], we have $r_{2}(\mathbf{C}l(L))=2n-2$. Consider the set:

$\eta=\\{q_{1},...,q_{n-1},\alpha_{2},...,\alpha_{n}\\},$

Since $\forall i\in\\{2,\dots,n\\}$:
$\alpha_{i}=x_{i}^{\prime}+y_{i}^{\prime}\sqrt{-2}$, where the couple
$(x_{i}^{\prime},y_{i}^{\prime})$ is a primitive solution of
$q_{1}q_{i}=x^{2}+2y^{2}$ $(i\geq 2)$ such that $x_{i}^{\prime}\equiv
1\pmod{2}$ and $y_{i}^{\prime}\equiv 0\pmod{2}$, then $\alpha_{i}\equiv
x_{i}^{\prime}+y_{i}^{\prime}+\frac{-1+\sqrt{-2}}{2}2y_{i}^{\prime}\pmod{4}\equiv
x_{i}^{\prime}+y_{i}^{\prime}\pmod{4}\equiv\pm 1\pmod{4}.$ Proceeding as above
we show that the extensions $L(\sqrt{\alpha_{i}})/L$ and $L(\sqrt{q_{i}})/L$
are unramified and the elements of $\eta$ are linearly independents modulo
$L^{2}$. Thus, $\eta$ is the set of generators of $\Delta(L^{*})/{L^{*}}^{2}$.
Therefore,

$E(L)=L_{3,d}(\sqrt{q_{1}},...,\sqrt{q_{n-1}},\sqrt{\alpha_{2}},...,\sqrt{\alpha_{n}}),$

and so by [2, Theorem 1] and Corollary 2.6, we get :

$E(L_{m,d})=L_{m,d}(\sqrt{q_{1}},...,\sqrt{q_{n-1}},\sqrt{\alpha_{2}},...,\sqrt{\alpha_{n}}).$

  3. 3.

By [2, Theorem 1], we have $r_{2}(\mathbf{C}l(L))=2n-3$. Put

$\eta=\\{{p_{1}},...,{p_{s}},{q_{1}},...,{q_{t-1}},{\gamma_{2}},...,{\gamma_{s}},{\alpha_{2}},...,{\alpha_{t}}\\},$

Let us show that the elements of $\eta$ are linearly independents modulo
${L^{*}}^{2}$. Set
$\beta=\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{a_{i}}\right)\left(\displaystyle\prod_{i=1}^{t-1}{q_{i}}^{a_{i}^{\prime}}\right)\left(\displaystyle\prod_{j=2}^{s}{\gamma_{j}}^{b_{j}}\right)\left(\displaystyle\prod_{j=2}^{t}{\alpha_{j}}^{b_{j}^{\prime}}\right)$,
where $a_{i},b_{j},a_{i}^{\prime},b_{j}^{\prime}\in\\{0,1\\}$ are not all
zero. Assume that $\beta\in{L^{*}}^{2}.$ Then, $\beta\in{K^{*}}^{2}$ or
$d\beta\in{K^{*}}^{2}.$ Note that $\exists j\in\\{2,\dots,s\\},\;b_{j}\neq 0$
or $\exists j\in\\{2,\dots,t\\},\;b_{j}^{\prime}\neq 0$.

    1. $\bullet$

Assume that $\beta\in{K^{*}}^{2}$. Since
$N_{K/k^{\prime\prime}}(\alpha_{j})={x_{j}^{\prime}}^{2}+2{y_{j}^{\prime}}^{2}=q_{1}q_{j}$
and $N_{K/k}(\gamma_{j})=x_{j}^{2}+y_{j}^{2}=p_{1}p_{j}$, we have

$N_{K/k^{\prime\prime}}(\beta)=\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{a_{i}}\prod_{i=1}^{t-1}{q_{i}}^{a_{i}^{\prime}}\prod_{j=2}^{s}{\gamma_{j}}^{b_{j}}\right)^{2}\left(\displaystyle\prod_{j=2}^{t}{(q_{1}q_{j})}^{b_{j}^{\prime}}\right)=q_{1}^{\ell}\left(\displaystyle\prod_{j=2}^{t}{q_{j}}^{b_{j}^{\prime}}\right)Z_{1}^{2},$
$N_{K/k}(\beta)=\left(\displaystyle\prod_{i=1}^{s-1}{p_{i}}^{a_{i}}\prod_{i=1}^{t-1}{q_{i}}^{a_{i}^{\prime}}\prod_{j=2}^{s}{\alpha_{j}}^{b_{j}^{\prime}}\right)^{2}\left(\displaystyle\prod_{j=2}^{t}{(p_{1}p_{j})}^{b_{j}}\right)=p_{1}^{\ell^{\prime}}\left(\displaystyle\prod_{j=2}^{t}{p_{j}}^{b_{j}}\right)Z_{2}^{2},$

for some $\ell,\ell^{\prime}\in\\{0,1\\}$ and $Z_{i}\in\mathbb{Q}$. This
implies that
$q_{1}^{\ell}\left(\displaystyle\prod_{j=2}^{t}{q_{j}}^{b_{j}^{\prime}}\right)$
is a square in $k^{\prime\prime}$ and
$p_{1}^{\ell^{\prime}}\left(\displaystyle\prod_{j=2}^{t}{p_{j}}^{b_{j}}\right)$
is a square in $k$, which is impossible. Thus, $\beta\notin{K^{*}}^{2}.$

    2. $\bullet$

Analogously, we check that $d\beta\in{K^{*}}^{2}$ is impossible.

Since the extensions $L(\sqrt{p_{i}})/L$, $L(\sqrt{q_{i}})/L$,
$L(\sqrt{\gamma_{j}})/L$ and $L(\sqrt{\alpha_{j}})/L$ are unramified, then the
elements of $\eta$ are generators of $\Delta(L^{*})/{L^{*}}^{2}$. Hence,

$\displaystyle E(L_{m,d})$ $\displaystyle=$ $\displaystyle
L_{m,d}(\sqrt{p_{1}},...,\sqrt{p_{s}},\sqrt{q_{1}},...,\sqrt{q_{t-1}},\sqrt{\gamma_{2}},...,\sqrt{\gamma_{s}},\sqrt{\alpha_{2}},...,\sqrt{\alpha_{t}})$
$\displaystyle=$ $\displaystyle
L_{m,d}\left(\prod_{i=1}^{s}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=1}^{t-1}\mathbb{Q}(\sqrt{q_{i}})\right)\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right)\left(\prod_{i=2}^{t}\mathbb{Q}(\sqrt{\alpha_{i}})\right).$

  4. 4.

In this case, we have $r_{2}(\mathbf{C}l(L))=n-3=4r+2s+2t-3$ . Consider the
set:

$\eta=\\{l_{1},...,l_{r},{p_{1}},...,{p_{s}},{q_{1}},...,{q_{t-1}},{\pi_{1,1}},...,{\pi_{1,r}},\pi_{2,1},...,{\pi_{2,r}},\pi_{3,1},...,{\pi_{3,r}},,{\gamma_{2}},...,{\gamma_{s}},{\alpha_{2}},...,{\alpha_{t}}\\},$

Put

$\beta=\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{1}}{\pi_{1,i}}^{\theta_{2}}{\pi_{2,i}}^{\theta_{3}}{\pi_{3,i}}^{\theta_{4}}\right)\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{\theta_{1}^{\prime}}\right)\left(\displaystyle\prod_{i=1}^{t-1}{q_{i}}^{\theta_{2}^{\prime}}\right)\left(\displaystyle\prod_{j=2}^{s}{\gamma_{j}}^{\theta_{3}^{\prime}}\right)\left(\displaystyle\prod_{j=2}^{t}{\alpha_{j}}^{\theta_{4}^{\prime}}\right),$

where
${\theta_{1,i}},{\theta_{2,i}},{\theta_{3,i}},{\theta_{4,i}},{\theta_{1,i}^{\prime}},{\theta_{2,i}^{\prime}},{\theta_{3,i}^{\prime}},{\theta_{4,i}^{\prime}}\in\\{0,1\\}$
are not all zero.
Assume that $\beta\in{L^{*}}^{2}$, then $\beta\in{K^{*}}^{2}$ or
$d\beta\in{K^{*}}^{2}$ (note that, as in the first item, the exponents
${\theta_{2,i}},{\theta_{3,i}},{\theta_{4,i}},{\theta_{3,i}^{\prime}},{\theta_{4,i}^{\prime}}$
can not all vanish, since elsewhere we get
$\beta=\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{1,i}}\right)\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{\theta_{1,i}^{\prime}}\right)\left(\displaystyle\prod_{i=1}^{t-1}{q_{i}}^{\theta_{2,i}^{\prime}}\right)$
is a square in $K^{*}$, which implies that $\mathbb{Q}(\sqrt{\beta})$ is a
quadratic subfield of $K$ which is impossible).

    1. $\bullet$

Let $\beta\in{K^{*}}^{2}.$ Since,
$N_{K/k}(\pi_{1,i})=a_{i}^{2}+16b_{i}^{2}=l_{i}$,
$N_{K/k}(\gamma_{j})=x_{j}^{2}+y_{j}^{2}=p_{1}p_{j}$,
$N_{K/k^{{}^{\prime}}}(\pi_{2,i})=e_{i}^{2}-32b_{i}^{2}=l_{i}$,
$N_{K/k^{{}^{\prime\prime}}}(\pi_{3,i})=u_{i}^{2}+8v_{i}^{2}=l_{i}$ and
$N_{K/k^{{}^{\prime\prime}}}(\alpha_{j})={x_{j}^{\prime}}^{2}+2{y_{j}^{\prime}}^{2}=q_{1}q_{j}$,
then

$\displaystyle N_{K/k}(\beta)$ $\displaystyle=$
$\displaystyle\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{1,i}}{\pi_{2,i}}^{\theta_{3,i}}{\pi_{3,i}}^{\theta_{4,i}}\right)^{2}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{2,i}}\right)\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{\theta_{1,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{i=1}^{t-1}{q_{i}}^{\theta_{2,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{j=2}^{s}({p_{1}p_{j}})^{\theta_{3,i}^{\prime}}\right)\left(\displaystyle\prod_{j=2}^{t}{\alpha_{j}}^{\theta_{4,i}^{\prime}}\right)^{2}$
$\displaystyle=$ $\displaystyle
p_{1}^{\ell}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{2,i}}\right)\left(\displaystyle\prod_{j=2}^{t}{p_{j}}^{\theta_{3,i}^{\prime}}\right)Z_{1}^{2}.$

$\displaystyle N_{K/k^{{}^{\prime}}}(\beta)$ $\displaystyle=$
$\displaystyle\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{1,i}}{\pi_{1,i}}^{\theta_{2,i}}{\pi_{3,i}}^{\theta_{4,i}}\right)^{2}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{3,i}}\right)\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{\theta_{1,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{i=1}^{t-1}{q_{i}}^{\theta_{2,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{j=2}^{s}{\gamma_{j}}^{\theta_{3,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{j=2}^{t}{\alpha_{j}}^{\theta_{4,i}^{\prime}}\right)^{2}$
$\displaystyle=$
$\displaystyle\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{3,i}}\right)Z_{2}^{2}.$

$\displaystyle N_{K/k^{{}^{\prime\prime}}}(\beta)$ $\displaystyle=$
$\displaystyle\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{1,i}}{\pi_{1,i}}^{\theta_{2,i}}{\pi_{2,i}}^{\theta_{3,i}}\right)^{2}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{4,i}}\right)\left(\displaystyle\prod_{i=1}^{s}{p_{i}}^{\theta_{1,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{i=1}^{t-1}{q_{i}}^{\theta_{2,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{j=2}^{s}{\gamma_{j}}^{\theta_{3,i}^{\prime}}\right)^{2}\left(\displaystyle\prod_{j=2}^{t}({q_{1}q_{j}})^{\theta_{4,i}^{\prime}}\right)$
$\displaystyle=$ $\displaystyle
q_{1}^{\ell^{{}^{\prime}}}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{4,i}}\right)\left(\displaystyle\prod_{j=2}^{t}{q_{j}}^{\theta_{4,i}^{\prime}}\right)Z_{3}^{2}.$

such that $\ell,\ell^{{}^{\prime}},\in\\{0,1\\}$ and $Z_{i}\in\mathbb{Q}$.
Thus,
$p_{1}^{\ell}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{2,i}}\right)\left(\displaystyle\prod_{j=2}^{t}{p_{j}}^{\theta_{3,i}^{\prime}}\right)$
is a square in $k$,
$\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{3,i}}\right)$ is a square
in $k^{\prime}$ and
$q_{1}^{\ell^{{}^{\prime}}}\left(\displaystyle\prod_{i=1}^{r}{l_{i}}^{\theta_{4,i}}\right)\left(\displaystyle\prod_{j=2}^{t}{q_{j}}^{\theta_{4,i}^{\prime}}\right)$
is a square in $k^{{}^{\prime\prime}}$, then the above three equations are
impossible.

    2. $\bullet$

We similarly show that $d\beta\in{K^{*}}^{2}$ is impossible. Thus, the
elements of $\eta$ are linearly independents modulo ${L^{*}}^{2}$.

On the other hand as above, we check that the primes of $L$ generated
respectively by $\pi_{1,i}$, ${\pi_{2,i}}$ and $\pi_{3,i}$ are the squares of
certain fractional ideals of $L$ and as $a_{i}\equiv e_{i}\equiv\pm
1\pmod{4}$, since they are odd, the equations $\pi_{1,i}\equiv\xi^{2}\pmod{4}$
and $\pi_{2,i}\equiv\xi^{2}\pmod{4}$ have solutions in $L$ (since $i\in L$).
Furthermore, we have
$\pi_{3,i}=u_{i}+2v_{i}\sqrt{-2}=u_{i}+2v_{i}+4v_{i}\frac{-1+\sqrt{-2}}{2}\equiv
u_{i}+2v_{i}\pmod{4}\equiv\pm 1\pmod{4}$ ($u_{i}$ is odd), then the equation
$\pi_{3,i}\equiv\xi^{2}\pmod{4}$ also has solutions $L$. Thus, by Proposition
2.1, the extensions $L(\sqrt{\pi_{1,i}})/L$, $L(\sqrt{\pi_{2,i}})/L$ and
$L(\sqrt{{\pi_{3,i}}})/L$ are unramified for $i\in\\{1,\dots,k\\}$. It follows
that $\eta$ is a set of generators of $\Delta(L^{*})/{L^{*}}^{2}$. Hence, we
have the fourth item.

  5. 5.

By [2, Theorem 1], if $\exists i\in\\{1,\dots,r\\}$ such that
$\left(\frac{2}{l_{i}}\right)_{4}=-1$, then we have
$r_{2}(\mathbf{C}l(L))=4r+2s-3$. Put:

$\eta=\\{l_{1},...,l_{r-1},{p_{1}},...,{p_{s}},{\pi_{1,1}},...,{\pi_{1,r}},\pi_{2,1},...,{\pi_{2,r}},\pi_{3,1},...,{\pi_{3,r}},,{\gamma_{2}},...,{\gamma_{s}}\\},$

As above we show that $\eta$ is a set of generators of
$\Delta(L)/{L^{*}}^{2}$. Therefore,

$\displaystyle
E(L_{m,d})=L_{m,d}\left(\prod_{i=1}^{r}\mathbb{Q}(\sqrt{\pi_{1,i}},\sqrt{\pi_{2,i}},\sqrt{\pi_{3,i}})\right)\left(\prod_{i=1}^{r-1}\mathbb{Q}(\sqrt{l_{i}})\right)\left(\prod_{i=1}^{s-1}\mathbb{Q}(\sqrt{p_{i}})\right)\left(\prod_{i=2}^{s}\mathbb{Q}(\sqrt{\gamma_{i}})\right).$

With analogous reasoning, we prove the rest of our theorem. ∎

## Acknowledgment

The authors are so grateful to the unknown referee for his/her several helpful
suggestions that helped us to improve our paper, and for calling our attention
to the missing details.

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