On the Existence of euclidean ideal class in quadratic, cubic, and quartic extensions

SriLakshmi Krishnamoorthy Srilakshmi Krishnamoorthy
INDIAN INSTITUTE OF SCIENCE EDUCATION AND RESEARCH, THIRUVANANTHAPURAM, INDIA. [email protected] and Sunil Kumar Pasupulati Sunil Kumar Pasupulati
INDIAN INSTITUTE OF SCIENCE EDUCATION AND RESEARCH, THIRUVANANTHAPURAM, INDIA. [email protected]

Abstract.

Lenstra introduced the notion of Euclidean ideal class. We prove the existence of the Euclidean ideal class in abelian quartic fields. As a corollary, we prove that any biquadratic field with class number two has a Euclidean ideal class. We also prove the existence of a Euclidean ideal class in certain cubic and quadratic extensions.

Key words and phrases:

Euclidean algorithm for number fields, Euclidean ideal class, Ideal class group, Hilbert class field.

2010 Mathematics Subject Classification:

Primary:11A05, Secondary 11R29.

1. Introduction and Results

Let $K$ be a number field. We denote the number ring and units of $K$ by $\mathcal{O}_{K}$ and $\mathcal{O}_{K}^{\times}$ respectively. The class group $Cl_{K}$ is defined as $J_{K}/P_{K}$ , where $J_{K}$ is the group of fractional ideals and $P_{K}$ is the group of principal fractional ideals of $K$ . The Hilbert class field of $K$ is denoted by H(K). Let $K/\mathbb{Q}$ be an abelian extension. The conductor of $K$ denoted as $f(K)$ is defined to be the smallest natural number such that $K\subseteq\mathbb{Q}(\zeta_{f(K)})$ . The conductor of $H(K)$ is also $f(K)$ whenever $H(K)/\mathbb{Q}$ is abelian. The compositum of the number fields $K_{1}$ and $K_{2}$ is denoted by $K_{1}K_{2}$ .

The connection between the Euclidean ideals and class group dates back to Weinberger. In 1961, assuming a generalized Riemann hypothesis (GRH), Weinberger [11] proved that for a number field $K$ with $\operatorname{rank}(\mathcal{O}_{K}^{\times})\geq 1$ , the number ring $\mathcal{O}_{K}$ is a Euclidean domain if and only if the class group $Cl_{K}$ is trivial. In 1979, Lenstra had generalized the the notion of Euclidean domain by introducing Euclidean ideal classes 1, for the purpose of capturing cyclic class groups. He established [9] that for a number field $K$ with $\operatorname{rank}(\mathcal{O}_{K}^{\times})\geq 1$ , the number ring $\mathcal{O}_{K}$ contains a Euclidean ideal if and only if the class group $Cl_{K}$ is cyclic, provided GRH holds.

Definition 1 (Lenstra [9]).

Let $R$ be a Dedekind domain and ${\mathbb{I}}$ be the set of non-zero integral ideals of $R$ . The ideal $C\in{\mathbb{I}}$ is called a Euclidean ideal if there exists a function $\Psi:{\mathbb{I}}\to W$ , where $W$ is a well-ordered set, such that for every $I\in{\mathbb{I}}$ and $x\in I^{-1}C\setminus C$ , there exists a $y\in C$ such that

\displaystyle\Psi\left((x-y)IC^{-1}\right)<\Psi(I).

We say $\Psi$ is a Euclidean algorithm for $C$ . If $C$ is a Euclidean ideal, then every ideal in the ideal class $[C]$ is also a Euclidean ideal, and the ideal class $[C]$ is called Euclidean ideal class.

The ideal $(1)$ is the Euclidean ideal class if and only if $R$ is a Euclidean domain. Lenstra [9] defined the idea of Euclidean ideal classes for integral domains and showed that if an integral domain has a Euclidean ideal class, it is automatically a Dedekind domain with a cyclic class group. In this paper, we will investigate the existence of ideal classes of the rings of integers of number fields.

Ram Murty and H.Graves [5] bypassed the assumption of GRH for a specific class of number fields, and they proved the following theorem,

Theorem 1 (Ram Murty and H.Graves [5]).

Let $K$ be a number field, Galois over $\mathbb{Q}$ . If its Hilbert class field $H(K)$ has an abelian Galois group over $\mathbb{Q}$ and if $\operatorname{rank}(\mathcal{O}_{K}^{\times})\geq 4$ , then

Cl_{K}=\langle[C]\rangle\ \text{ if and only if }\ [C]\ \text{is a Euclidean ideal class.}

Deshouillers, Gun, and Sivaraman [2] proved the existence of the Euclidean ideal class in number fields with $\operatorname{rank}(\mathcal{O}_{K}^{\times})\geq 3$ with an additional assumption the Galois group ${\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f(K)})/K\right)$ is cyclic. They proved the following theorem by using an error term of the linear sieve given by Iwaniec.

Theorem 2 (Deshouillers, Gun, and Sivaraman [2]).

Let $K$ be a number field with the rank $\left(\mathcal{O}_{K}^{\times}\right)\geq 3$ and the Hilbert class field $H(K)$ is abelian over $\mathbb{Q}$ . Also suppose that the conductor of $H(K)$ is $f$ and $\mathbb{Q}(\zeta_{f})$ over $K$ is cyclic. Then $K$ has a Euclidean ideal class.

We prove the existence of a Euclidean ideal class in abelian real quartic fields with prime class numbers, without the assumption $\mathbb{Q}(\zeta_{f})$ over $K$ is cyclic. We also prove the corollary 4 that settles the discussion on the existence of Euclidean ideal class in certain biquadratic fields with class number two.

Theorem 3.

Let $K$ be a real quartic extension with prime class number and the Hilbert class field $H(K)$ is abelian over $\mathbb{Q}$ . Also let $f=\operatorname{lcm}(16,f(K))$ . If $G$ be the Galois group of $\mathbb{Q}(\zeta_{f})$ over $K$ and $G_{l}$ be the Galois group of $\mathbb{Q}(\zeta_{f})$ over $\mathbb{Q}(\zeta_{l})$ and

\displaystyle G\not\subseteq\bigcup_{l}G_{l}\bigcup{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{H(K)}\right),

where $l$ is an odd prime dividing f or $l=4.$ Then $K$ has a Euclidean ideal class.

Several authors constructed new families of number fields having a non-principal Euclidean ideal class. To name a few, H.Graves has proved the existence of non principal Euclidean ideal in $\mathbb{Q}\left(\sqrt{2},\sqrt{35}\right)$ using the growth lemma. Certain families of biquadratic fields were constructed by Hsu [8] and Chattopadhyay and Muthukrishnan [1]. The second author [10] has proved existence of the Euclidean ideals in a generalized family of biquadratic fields. The following corollary generalizes the main results of [8],[1], and [10].

Corollary 4.

Let $K=\mathbb{Q}\left(\sqrt{q},\sqrt{rs}\right)$ be real biquadratic number field with class number two. If $p,q,r$ are the odd primes, then there exists a Euclidean ideal class in $Cl_{K}$ .

Gun and Sivaraman [6], also examined the existence of a Euclidean ideal in the case of abelian number fields $K$ with a prime class number, $\text{rank}(\mathcal{O}_{K}^{\times})\leq 2.$ They proved the following two theorems.

Theorem 5 (Gun and Sivaraman [6]).

Let $K_{1},K_{2}$ be distinct real cubic fields with prime class number having conductors $f_{1},f_{2}$ respectively. And also let $H(K_{i})$ be abelian over $\mathbb{Q}$ for $i=1,2$ and $f=lcm(f_{1},f_{2},16)$ . If

\displaystyle{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/K_{1}K_{2}\right)\not\subseteq\bigcup_{l\mid f}Gal\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\bigcup_{i=1}^{2}{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/H(K_{i})\right),

where $l$ is an odd prime or $l=4$ , then at least one of $K_{1},K_{2}$ has a Euclidean ideal class.

Theorem 6 (Gun and Sivaraman [6]).

Let $K_{1},K_{2},K_{3}$ be distinct real quadratic fields with prime class number having conductors $f_{1},f_{2},f_{3}$ respectively. And also let $H(K_{i})$ be abelian over $\mathbb{Q}$ for $i=1,2,3$ and $f=lcm(f_{1},f_{2},f_{3},16)$ . If

\displaystyle{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/K_{1}K_{2}K_{3}\right)\not\subseteq\bigcup_{l\mid f}Gal\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)\bigcup_{i=1}^{3}{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/H(K_{i})\right),

where $l$ is an odd prime or $l=4$ , then at least one of $K_{1},K_{2},K_{3}$ has a Euclidean ideal class.

We have examined the existence of a Euclidean ideal in quadratic and real cubic fields without assuming the prime class number. As a result, we have proved the following theorems.

Let $K_{1},K_{2}$ be abelian real cubic extensions with conductors $f_{1}$ and $f_{2}$ respectively. Set $H=H(K_{1})H(K_{2})$ , $K=K_{1}K_{2}$ , and $f=\operatorname{lcm}(16,f_{1},f_{2})$ .

Theorem 7.

Let $K_{1},K_{2}$ be distinct abelian real cubic extensions with $Cl_{K_{1}},Cl_{K_{2}}$ are cyclic and the Hilbert class field $H(K_{i})$ is abelian over $\mathbb{Q}$ for $i=1,2$ . Also let $f$ be as above and the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists a lift $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that

\displaystyle\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4.

and

\displaystyle H(K_{1})\cap K_{2}=K_{1}\cap K_{2}=H(K_{2})\cap K_{1}.

Then at least one of $K_{1},K_{2}$ has a Euclidean ideal class.

Let $K_{1},K_{2},K_{3}$ be distinct real quadratic extensions of $\mathbb{Q}$ . Denote the conductorof $K_{i}$ by $f_{i}$ . Set $H=H(K_{1})H(K_{2})H(K_{3})$ , $K=K_{1}K_{2}K_{3}$ , and $f=\operatorname{lcm}(16,f_{1},f_{2},f_{3})$ .

Theorem 8.

Let $K_{1},K_{2},K_{3}$ be distinct real quadratic extensions of $\mathbb{Q}$ and the Hilbert class field $H(K_{i})$ is abelian over $\mathbb{Q}$ for $i=1,2,3$ . Also let the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists a extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that

\displaystyle\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4.

and

\displaystyle H(K_{i})\cap K_{j}=K_{i}\cap K_{j}=H(K_{j})\cap K_{i}\ \text{for}\ i,j\in\{1,2,3\}.

Then at least one of $K_{1},K_{2},K_{3}$ has a Euclidean ideal class.

The article is arranged as follows. In Section 2, we write down some preliminaries which are useful for proving the main results. In Section 3, we prove Theorem 3 and 4. In Section 4, we prove Theorem 7 and in Section 5, we prove theorem 8. In the final section, we discuss a couple of consequences of the Elliott and Halberstam conjecture.

2. Preliminaries

Lemma 9.

The conductor of the quadratic field $K=\mathbb{Q}(\sqrt{d})$ is

\displaystyle f(K)=\begin{cases}d&d\equiv 1\mod{4}\\ 4d&d\equiv 2,3\mod{4}.\end{cases}

Lemma 10.

Let $L$ be compositum of $K_{1},K_{2}$ , then the conductor of the number field $L$ , $f(L)=\operatorname{lcm}(f(K_{1}),f(K_{2}))$ .

H.Graves [4] proved a useful growth result which gives a condition for the existence of a Euclidean ideal in the number fields without the assumption of GRH.

Theorem 11 (Graves [4]).

Suppose that K is a number field such that $\operatorname{rank}(\mathcal{O}_{K}^{\times})\geq 1$ and the ideal $C$ is a non-zero ideal of $\mathcal{O}_{K}$ . If $[C]$ generates the class group of $K$ and

\Biggr{|}\{\text{Prime\ ideal}\ \mathfrak{P}\subseteq\mathcal{O}_{K}:N(\mathfrak{P})\leq x,\ [\mathfrak{P}]=[C],\ \mathcal{O}_{K}^{\times}\twoheadrightarrow(\mathcal{O}_{K}/\mathfrak{p})^{\times}\}\Biggr{|}\gg\frac{x}{(\log x)^{2}},

then $[C]$ is a Euclidean ideal class.

Gun and Sivaraman proved the following generalized version of the above theorem.

Theorem 12 (Gun and Sivaraman [6]).

Suppose that $K$ is a number field with $\operatorname{rank}(\mathcal{O}_{K}^{\times})\geq 1$ and $Cl_{K}=\langle[\mathfrak{a}]\rangle$ . If there exists an unbounded inceasing sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

\displaystyle\Bigg{|}\Big{\{}\text{Prime ideal}\ \mathfrak{p}\subseteq\ \mathcal{O}_{K}:[\mathfrak{p}]=[\mathfrak{a}],\mathfrak{N}(\mathfrak{p})\leq x_{n},\mathcal{O}_{K}^{\times}\twoheadrightarrow(\mathcal{O}_{K}/\mathfrak{p})^{\times}\Big{\}}\Bigg{|}\gg\frac{x_{n}}{\log^{2}x_{n}},

then $[\mathfrak{a}]$ is a Euclidean ideal class.

We state a crucial lemma which will be used in the proofs of the main theorems.

Lemma 13 (Heath-Brown [7]).

Suppose that $u$ and $v$ are natural numbers with the following properties,

(u,v)=1,\quad v\equiv 0\mod{16},\quad\text{and}\quad\left(\frac{u-1}{2},v\right)=1.

Then there exist $b,c\in(1/4,1/2)$ with $b<c$ such that for any $\epsilon>0$ ,

	$\displaystyle\Bigg{\|}J_{\epsilon}(X):=$	$\displaystyle\Big{\{}p\equiv u\mod{v}:p\ \text{is a prime},p\in(X^{1-\epsilon},X)\text{such that}\ \frac{p-1}{2}\text{is either a}$
		$\displaystyle\text{prime or product of two primes}\ q_{1}q_{2}\ with\ X^{b}<q_{1}<X^{c}\Big{\}}\Bigg{\|}\gg\frac{X}{\log^{2}X}.$

Lemma 14.

Let $K$ be a totaly real number field and the non zero elements $e_{1},e_{2},e_{3}$ of $K$ be multiplicative independent. Then, for some $i\in\{1,2,3\}$ , either $e_{i},-e_{i}$ is a primitive root $\mod{\mathfrak{p}}$ for infinitely many primes ideals in the set $M_{\epsilon}$ . Let this set of prime ideals be called $P$ , and let $P(X)$ denotes the set of elements in $P$ of norm less than or equal to $X$ .Then there exists an increasing unbounded sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

P(x_{n})\gg\frac{x_{n}}{\log^{2}x_{n}}.

Proof.

Refer to [6, Lemma 16 ]. ∎

Now we end this section by stating a famous conjecture in Sieve Theory.

Conjecture 15 (Elliott and Halberstam conjecture [3]).

Let $a,q$ be natural numbers, $\phi$ be the Euler totient function, $\pi(Y,q,a):=\Big{\{}p\leq Y|\ p\ \text{is prime,}\,p\equiv a\mod{q}\Big{\}}$ , and $li(Y):=\int_{2}^{Y}\frac{1}{\log t}dt$ . For every real number $\theta<1$ and for every positive integer $k>0$ , one has

\displaystyle\sum_{q\leq X^{\theta}}\mathop{max}_{Y\leq X}\mathop{max}_{(a,q)=1}\left|\pi(Y,q,a)-\frac{\text{li}(Y)}{\phi(q)}\right|\ll\frac{X}{\log^{k}X},

for all real numbers $X>2$ .

3. Euclidean ideals in Real Quartic extensions

3.1. Proof of Theorem 3

Let $K$ be totally real quartic extension such that both $K,H(K)$ are abelian over $\mathbb{Q}$ . And let $f$ be the least common multiple of $16,f(K)$ . Suppose $G_{l}$ be Galois group of $\mathbb{Q}(\zeta_{f})$ over $\mathbb{Q}(\zeta_{l})$ .

Lemma 16.

Suppose that the Galios group $G$ of $\mathbb{Q}(\zeta_{f})$ over $K$ satisfying supposition of Theorem 3. Then there exists $a\in\mathbb{Z}$ such that any prime $p$ with $p\equiv a\mod{f}$ , splits completely in $K$ but does not split completely in $H(K)$ . Furthermore, there exist $b,c\in(1/4,1/2)$ such that for any $X,\epsilon>0$ we have

	$\displaystyle\Bigg{\|}J_{\epsilon}(X):=$	$\displaystyle\Big{\{}p\equiv a\mod{f}:p\ \text{is prime},p\in(X^{1-\epsilon},X)\text{such that}\ \frac{p-1}{2}\text{is either a }$
		$\displaystyle\text{prime or product of two primes}\ q_{1}q_{2}\ with\ X^{b}<q_{1}<X^{c}\Big{\}}\Bigg{\|}\gg\frac{X}{\log^{2}X}.$

Proof.

According to supposition of Theorem 3, we have

\displaystyle G\not\subseteq\bigcup_{l}G_{l}\bigcup{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{H(K)}\right).

Let $\sigma\in G\setminus\left(\bigcup_{l}G_{l}\bigcup{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{H(K)}\right)\right)$ . Choose $a$ to be the image of $\sigma$ under the isomorphism ${\mathrm{Gal}}(\mathbb{Q}(\zeta_{f}/\mathbb{Q})\to\left(\mathbb{Z}/f\mathbb{Z}\right)^{\times}$ . By the properties of the Artin symbol, whenever $p\equiv a\mod{f}$ then

\displaystyle\left(\frac{K/\mathbb{Q}}{p}\right)=\sigma|_{K}=id\ \text{ and}\ \left(\frac{H(K)/\mathbb{Q}}{p}\right)=\sigma|_{H(K)}\neq id.

Therefore $p$ splits completely in $K$ but does not split completely in $H(K)$ . The fact $\sigma\not\in\bigcup G_{l}$ for all prime $l$ dividing $f$ translates to $\left(\frac{a-1}{2},f\right)=1$ . By applying Lemma 13 for $u=a$ and $v=f$ , there exist $b,c\in\left(1/4,1/2\right)$ and for every $\epsilon\geq 0$ , we get

\displaystyle\Big{|}J_{\epsilon}(X)\Big{|}\gg\frac{X}{\log^{2}X}.

∎

For the number field $K$ in Theorem 3, we define

	$\displaystyle M_{\epsilon}:=$	$\displaystyle\Big{\{}\mathfrak{p}\subseteq\mathcal{O}_{K}\ \text{is a prime ideal}\ :\ \mathfrak{N}(\mathfrak{p})=p\ \text{ is a prime,}\ p\equiv a\mod{f},\ \frac{p-1}{2}\ \text{is}$
		$\displaystyle\text{ either a prime or product of two primes}\ q_{1}q_{2}\ with\ p^{b}<q_{1}<p^{{c}/{1-\epsilon}}\Big{\}}$

and

M_{\epsilon}(X)=\{\mathfrak{p}\in M_{\epsilon}:\mathfrak{N}(\mathfrak{p})\leq X\}.

Proof of Theorem 3

We have $K$ is real quartic field, $\mathcal{O}_{K}^{\times}$ contains three multiplicative independent elements, say $\epsilon_{1},\epsilon_{2},\epsilon_{3}$ . By Lemma 14, there exists $\eta\in\{\pm\epsilon_{1},\pm\epsilon_{2},\pm\epsilon_{3}\}$ and the set of primes $P$ such that $\eta$ is primitive root modulo $\mathfrak{p}$ for every $\mathfrak{p}\in P$ . We also get an increasing unbounded sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

\displaystyle\biggr{|}P(x_{n})\biggr{|}\geq\frac{cx_{n}}{\log^{2}x_{n}}.

There are only finitely many ideal classes in $Cl_{K}$ , we claim that there exists non-trivial ideal class $[\mathfrak{a}]$ such that

\displaystyle\Bigg{|}\Big{\{}\mathfrak{p}\in[\mathfrak{a}]:\mathfrak{p}\in P,\mathfrak{N}(\mathfrak{p})\leq y_{n}\Big{\}}\Bigg{|}\gg\frac{y_{n}}{h_{k}\log^{2}y_{n}}.

Otherwise we get that

\biggr{|}P(x_{n})\biggr{|}=o\left(\frac{y_{n}}{\log^{2}y_{n}}\right).

Since every prime $\mathfrak{p}\in M_{\epsilon}$ lie over some prime $p\equiv a\mod{f}$ , $\mathfrak{p}$ do not belong to the trivial ideal class. Therefore $[\mathfrak{a}]$ is a generator of $Cl_{K}$ . By Theorem 12, we have $[\mathfrak{a}]$ is the Euclidean ideal class. $\square$

Remark 1.

Consider the example $K=\mathbb{Q}(\sqrt{11},\sqrt{247})$ . Although we have $\text{rank}(\mathcal{O}_{K}^{\times})=3$ for this real quartic field $K$ , we can’t deduce that K has a Euclidean ideal by using Theorem 2. Because ${\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)\cong\mathbb{Z}/10\mathbb{Z}\times\mathbb{Z}/12\mathbb{Z}\times\mathbb{Z}/9\mathbb{Z}$ is not cyclic, where $f=10808$ is the conductor of the Hilbert class field of $K$ , one of the hypothesis of the Theorem 2 is not satisfied. However we can deduce that $K$ has a Euclidean ideal by using Theorem 3.

3.2. Proof of 4

Proposition 17.

Let $K=\mathbb{Q}\left(\sqrt{q},\sqrt{rs}\right)$ , where $p,q,r$ are odd primes, then

\displaystyle\mathbb{Q}\left(\sqrt{q},\sqrt{r},\sqrt{s}\right)\subseteq H(K).

Proof.

The number field $L=\mathbb{Q}\left(\sqrt{q},\sqrt{r},\sqrt{s}\right)$ is abelian over $K$ . It is enough to prove that all primes of $K$ are unramified over $L$ . The only primes of $K$ that can ramify in $L$ are the primes lie above $q,r,s,2.$ Now we observe that

\displaystyle L=K\mathbb{Q}(\sqrt{qr})=K\mathbb{Q}(\sqrt{sq})=K\mathbb{Q}(\sqrt{r})=K\mathbb{Q}(\sqrt{s}).

Therefore, by [10, Lemma 9], we have $L\subseteq H(K)$ . ∎

By Lemma 9 and Lemma 10, we have $f(K)=4qrs$ and set $f=\operatorname{lcm}(f(K),16)=16qrs$ .

Lemma 18.

Let $q,r,s$ be prime numbers and the number field $K=\mathbb{Q}(\sqrt{q},\sqrt{rs})$ has class number 2. If $Gal(H(K)/K)=\langle\sigma\rangle$ then there exists a lift $\hat{\sigma}\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/K\right)$ of $\sigma$ such that

\displaystyle\hat{\sigma}\not\in\bigcup_{l}{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)\bigcup{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{H(K)}\right).

for $l$ , an odd prime dividing f or $l=4$ .

Proof.

By 17, we have the Hilbert class field $H(K)=\mathbb{Q}\left(\sqrt{q},\sqrt{r},\sqrt{s}\right)$ . Clearly ${\mathrm{Gal}}\left(H(K)/K\right)=\langle\sigma\rangle$ , where $\sigma(\sqrt{r})=-\sqrt{r},\sigma(\sqrt{s})=-\sqrt{s}$ , and $\sigma(\sqrt{q})=\sqrt{q}$ .
Case (i) $r\equiv 1\mod{4},s\equiv 1\mod{4}$ :
Let $\tau$ be any lift of $\sigma$ to $Q(\zeta_{f})$ , then $\tau(\sqrt{r})=-\sqrt{r},\tau(\sqrt{s})=-\sqrt{s}$ . Therefore $\tau\not\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)$ , for $l=r,s$ .
Subcase(i) $q\equiv 1\mod{4}$ : Let $F=H(K)\mathbb{Q}(\zeta_{q})$ . Then we have $H(K)\cap\mathbb{Q}(\zeta_{q})=\mathbb{Q}(\sqrt{q})$ and ${\mathrm{Gal}}\left(F/\mathbb{Q}(\sqrt{q})\right)\cong{\mathrm{Gal}}\left(H(K)/\mathbb{Q}(\sqrt{q})\right)\times{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{q})/\mathbb{Q}(\sqrt{q})\right)$ . Therefore there exists $\sigma_{1}\in{\mathrm{Gal}}(F/K)$ a lift of $\sigma$ such that $\sigma_{1}(\zeta_{q})\neq\zeta_{q}$ . Observe that $F\cap\mathbb{Q}(i)=\mathbb{Q}$ , thus there exists $\sigma_{2}\in{\mathrm{Gal}}\left(F(i)/K\right)$ , which is a lift of $\sigma_{1}$ and $\sigma_{2}(i)=-i$ . Hence there exists a lift $\hat{\sigma}\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/K\right)$ of $\sigma$ and $\hat{\sigma}\not\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)$ , for $l=4,q,r,s$ .
Subcase(ii) $q\equiv 3\mod{4}$ : Let $F=H(K)\mathbb{Q}(i)$ . Clearly $H(K)\cap\mathbb{Q}(i)=\mathbb{Q}$ , therefore, we have $\sigma_{1}\in{\mathrm{Gal}}(F/\mathbb{Q})$ such that $\sigma_{1}(i)=-i$ and $\sigma_{1}|_{H(K)}=\sigma$ . Thus we have $\sigma_{1}(i\sqrt{q})=-i\sqrt{q}$ . Hence any lift of $\sigma_{1}$ does not belongs to ${\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f}/\zeta_{l})\right)$ for $l=4,q,r,s$ .
Case (ii) $r,s\equiv 3\mod{4}$ : Let $E=\mathbb{Q}(i)H(K)\mathbb{Q}(\zeta_{q})$ . Similar to case(i), there exists $\sigma_{1}\in{\mathrm{Gal}}\left(E/H(K)\right)$ , such that $\sigma_{1}(i)\neq i$ and $\sigma_{1}(\zeta_{q})\neq\zeta_{q}$ . Let $E_{1}=E\mathbb{Q}(\zeta_{r})$ , since $E\cap\mathbb{Q}(\zeta_{r})=Q(i\sqrt{r})$ we get ${\mathrm{Gal}}\left(E/\mathbb{Q}(i\sqrt{r})\right)\cong{\mathrm{Gal}}\left(H(K)/\mathbb{Q}(i\sqrt{r})\right)\times{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{r})/\mathbb{Q}(i\sqrt{r})\right)$ . Therefore there exists $\sigma_{2}\in{\mathrm{Gal}}(E_{1}/K)$ , which is a lift of $\sigma_{1}$ . Suppose $E_{2}=E_{1}\mathbb{Q}(\zeta_{s})$ then it is clear that $E_{2}\cap\mathbb{Q}(\zeta_{s})=\mathbb{Q}(i\sqrt{s})$ and ${\mathrm{Gal}}(E_{2}/\mathbb{Q}(i\sqrt{s}))\cong{\mathrm{Gal}}(E_{1}/\mathbb{Q}(i\sqrt{s}))\times{\mathrm{Gal}}(\mathbb{Q}(\zeta_{s})/\mathbb{Q}(i\sqrt{s}))$ . Thus there exists $\sigma_{3}\in{\mathrm{Gal}}(E_{2}/\mathbb{Q}(i\sqrt{s}))$ which is a lift of $\sigma_{2}$ and $\sigma_{3}(\zeta_{s})\neq\zeta_{s}$ . Hence, there exists $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ a lift of $\sigma$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l}))$ for $l=4,q,r,s$ .

Case(iii) Only one of $r$ or $s$ is congruent to $1$ modulo 4: Without loss of generality, we will consider $r\equiv 1\mod{4}$ and $s\equiv 3\mod{4}$ . We know that $\sigma(\sqrt{r})=-\sqrt{r}$ and $\sqrt{r}\subseteq\mathbb{Q}(\zeta_{r})$ . Thus $\tau(\zeta_{r})\neq\zeta_{r}$ for any lift $\tau$ of $\sigma$ . Suppose $L=H(K)\mathbb{Q}(i)\mathbb{Q}(\zeta_{q})$ then similar to case (i), there exists $\tau_{1}\in{\mathrm{Gal}}(L/K)$ such that $\tau_{1}|_{H(K)}=\sigma$ and $\tau_{1}\not\in{\mathrm{Gal}}(L/\mathbb{Q}(\zeta_{l}))$ , for $l=4,q$ . Let $L_{1}=L\mathbb{Q}(\zeta_{s})$ then we have ${\mathrm{Gal}}(L_{1}/\mathbb{Q}(i\sqrt{s}))\cong{\mathrm{Gal}}(L/\mathbb{Q}(i\sqrt{s}))\times{\mathrm{Gal}}(\mathbb{Q}(\zeta_{s})/\mathbb{Q}(i\sqrt{s}))$ . Thus there exists $\tau_{2}\in{\mathrm{Gal}}(L_{1}/K)$ a lift of $\tau_{1}$ and $\tau_{2}(\zeta_{s})\neq\zeta_{s}$ . Now choose $\hat{\sigma}$ to be a lift of $\tau_{2}$ to $\mathbb{Q}(\zeta_{f})$ , hence $\hat{\sigma}\not\in{\mathrm{Gal}}(\zeta_{f}/\zeta_{l})$ for any $l=4,q,r,s$ . ∎

Proof of 4

Let $K=\mathbb{Q}\left(\sqrt{q},\sqrt{rs}\right)$ with $q,r,s$ are primes. By 17, we have the Hilbert class field $H(K)=\mathbb{Q}\left(\sqrt{q},\sqrt{r},\sqrt{s}\right)$ . By Lemma 18, we have

\displaystyle Gal\left(\mathbb{Q}(\zeta_{f})/K\right)\not\subseteq\bigcup_{\begin{subarray}{c}l\mid f\\ l\ \text{is prime}\end{subarray}}{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)\bigcup{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/H(K)).

Then by Theorem 3, the biquadratic field $K$ has a Euclidean ideal class. $\square$

4. Euclidean ideals in Cubic fields

4.1. Proof of Theorem 7

Let $K_{1},K_{2}$ be abelian real cubic extensions and $H(K_{1})$ , $H(K_{2})$ are abelian over $\mathbb{Q}$ . Denote the conductors of $K_{i}$ by $f_{i}$ . Set $f=\operatorname{lcm}(16,f_{1},f_{2})$ and $K=K_{1}K_{2}$ . Clearly we have $H(K_{i})\subseteq\mathbb{Q}(\zeta_{f})$ for $i\in\{1,2\}$ .

Lemma 19.

Let $K_{1},K_{2}$ be abelian real cubic extensions, satisfying supposition of Theorem 7. Then there exists $a\in\mathbb{Z}$ with prime $p$ with $p\equiv a\mod{f}$ , If the prime $\mathfrak{P}$ of $K_{s}$ lies over $p$ then $[\mathfrak{P}]$ generates $Cl_{K_{s}}$ . Furthermore, there exists $b,c\in(1/4,1/2)$ such that for any $X,\epsilon>0$ we have

	$\displaystyle\Bigg{\|}J_{\epsilon}(X):=\Big{\{}p\equiv a\mod{f}:$	$\displaystyle\ p\ \text{is prime},p\in(X^{1-\epsilon},X)\ \text{such that}\ \frac{p-1}{2}\ \text{is}\text{ either a prime or}$
		$\displaystyle\text{product of two primes}\ q_{1}q_{2}\ \text{ with}\ X^{b}<q_{1}<X^{c}\Big{\}}\Bigg{\|}\gg\frac{X}{\log^{2}X}.$

Proof.

According to supposition of Theorem 7, we have $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that

\displaystyle\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4.

Let $a=\phi(\hat{\sigma})$ , where $\phi$ is the isomorphism ${\mathrm{Gal}}(\mathbb{Q}(\zeta_{f}/\mathbb{Q})\xrightarrow{\cong}\left(\mathbb{Z}/f\mathbb{Z}\right)^{\times}$ . Clearly $\sigma_{a}=\hat{\sigma}$ and $\sigma_{a}\mid_{H(K_{i})}\in Gal\left(H(K_{i})/K\right)$ for any $i$ . Observe that $Gal\left(H(K_{i})/K_{i}\right)=\langle\sigma_{a}\mid_{H(K_{1})}\rangle$ if and only if the fixed field of $\sigma_{a}\mid_{H(K_{1})}$ denoted by ${\mathcal{F}}(\sigma)$ is equal to $K_{i}$ .

Look at

\displaystyle{\mathcal{F}}(\sigma_{a}\mid_{H(K_{i})})={\mathcal{F}}(\sigma_{a})\cap H(K_{i})\subseteq K_{1}K_{2}\cap H(K_{i})\subseteq\left(K_{1}\cap H(K_{i})\right)\left(K_{2}\cap H(K_{i})\right)\subseteq K_{i}.

By the properties of the Artin symbol, whenever $p\equiv a\mod{f}$ , then $\left(\frac{K/\mathbb{Q}}{p}\right)=\sigma_{a}|_{K}=id$ . This implies that $p$ splits completely in $K$ , therefore for every prime $\mathfrak{p}$ of $K_{s}$ lying above $p\equiv a$ , we have $\mathfrak{N}(\mathfrak{p})=p$ . Hence

\left(\frac{H(K_{s})/Ks}{\mathfrak{p}}\right)=\sigma_{a}|_{H(K_{s})}\text{ is a generator of }{\mathrm{Gal}}\left(H(K_{s})/K_{s}\right).

Thus $[\mathfrak{p}]$ is a generator of $Cl_{K_{s}}$ .
By the assumption of the Theorem 7, $\sigma_{a}\not\in{\mathrm{Gal}}(\mathbb{Q}\left(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)$ . Therefore $(\frac{q-1}{2},l)=1$ for every odd prime $l$ dividing $f$ and $l=4$ , hence $(\frac{q-1}{2},f)=1$ . Now we can apply Lemma 13 for $u=a$ and $v=f$ there exist $b,c\in\left(1/4,1/2\right)$ and for every $\epsilon\geq 0$ . We get

\displaystyle\Big{|}J_{\epsilon}(X)\Big{|}\gg\frac{X}{\log^{2}X}.

∎

For the number field K in Theorem 7, we define

	$\displaystyle M_{\epsilon}:=$	$\displaystyle\Bigg{\{}\mathfrak{p}\subseteq\mathcal{O}_{K}\ \text{is prime}:\mathfrak{N}(\mathfrak{p})=p\ \text{ is prime,}\ p\equiv a\mod{f},\ \frac{p-1}{2}\ \text{is}$
		$\displaystyle\text{ either a prime or product of two primes}\ q_{1}q_{2}\ with\ p^{b}<q_{1}<p^{{c}/{1-\epsilon}}\Bigg{\}}$

and

\displaystyle M_{\epsilon}(x)=\{\mathfrak{p}\in M_{\epsilon}:\mathfrak{N}(\mathfrak{p})\leq X\}.

Proof of Theorem 7

Choose $\epsilon_{1},\epsilon_{2},\epsilon_{3}$ from $\mathcal{O}_{K_{1}}^{\times}\cup\mathcal{O}_{K_{2}}^{\times}$ such that the set $\epsilon_{1},\epsilon_{2},\epsilon_{3}$ is multiplicatively independent set. Let $P$ be as in Lemma 14. Then there exists $\eta\in\{\pm\epsilon_{1},\pm\epsilon_{2},\pm\epsilon_{3}\}$ such that $\eta$ is a primitive root modulo $\mathfrak{p}$ and we get an increasing unbounded sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

\biggr{|}P(x_{n})\biggr{|}\geq\frac{cx_{n}}{\log^{2}x_{n}}.

Observe that if $\mathfrak{P}\in P$ then $\mathfrak{p}=\mathfrak{P}\cap K_{s}$ generates class group of $K_{s}$ . Since there are only finitely many ideal classes, there exists an ideal class $[\mathfrak{a}]$ in $K_{s}$ such that

(1)

\displaystyle\biggr{|}\Big{\{}\mathfrak{P}\cap K_{s}\in[\mathfrak{a}]:\mathfrak{P}\in P,\mathfrak{N}(\mathfrak{P}\cap K_{s})\leq y_{n}\Big{\}}\Bigg{|}\gg\frac{y_{n}}{h_{K_{s}}\log^{2}y_{n}},

for some subsequence $\{y_{n}\}_{n\in\mathbb{N}}$ of $\{x_{n}\}_{n\in N}$ . Since $P\subseteq M_{\epsilon}$ , $[\mathfrak{a}]$ is a generator of $Cl_{K_{s}}$ . We have an increasing sequence $\{y_{n}\}_{n\in\mathbb{N}}$

	$\displaystyle\Biggr{\|}\Big{\{}\mathfrak{p}:\mathfrak{N}(\mathfrak{p})\leq y_{n},$	$\displaystyle[\mathfrak{p}]=[\mathfrak{a}],\mathcal{O}_{K_{s}}^{\times}/\mathfrak{p}\twoheadrightarrow\left(\mathcal{O}_{K_{s}}/\mathfrak{p}\right)^{\times}\Big{\}}\Biggr{\|}\geq$
		$\displaystyle\biggr{\|}\Big{\{}\mathfrak{P}\cap K_{s}\in[\mathfrak{a}]:\ \mathfrak{p}\in P,\mathfrak{N}(\mathfrak{p})\leq y_{n}\Big{\}}\Bigg{\|}\gg\frac{y_{n}}{h_{K_{s}}\log^{2}y_{n}}.$

Therefore, by Theorem 12, $K_{s}$ has a Euclidean ideal class. $\square$

Corollary 20.

Let $K_{1},K_{2}$ be abelian real cubic extensions with $Cl_{K_{1}},Cl_{K_{2}}$ are cyclic and the Hilbert class field $H(K_{i})$ is abelian over $\mathbb{Q}$ for $i=1,2$ . Also let the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists an extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that $\hat{\sigma}(\zeta_{l})\neq\zeta_{l}\ \text{for any odd prime }\ l\mid f\ or\ l=4$ . Also let $f_{1},f_{2}\ \text{are coprime}$ . Then at least one of $K_{1},K_{2}$ has a Euclidean ideal class.

Proof.

Suppose $f_{1},f_{2}$ are coprime then $\mathbb{Q}(\zeta_{f_{1}})\cap\mathbb{Q}(\zeta_{f_{2}})=\mathbb{Q}$ . We know that

\displaystyle\mathbb{Q}\subseteq K_{i}\subseteq H(K_{i})\subseteq\mathbb{Q}(\zeta_{f_{i}}).

Therefore we have

\displaystyle H(K_{1})\cap K_{2}=K_{1}\cap K_{2}=H(K_{2})\cap K_{1}.

Hence Theorem 7 implies that at least one of $K_{1},K_{2}$ has a Euclidean ideal. ∎

Corollary 21.

Let $K_{1},K_{2}$ be abelian real cubic extensions with class number one. The Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists a lift $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4$ . and

\displaystyle H(K_{1})\cap K_{2}=K_{1}\cap K_{2}=H(K_{2})\cap K_{1}.

Then at least one of $\mathcal{O}_{K_{1}},\mathcal{O}_{K_{2}}$ has a Euclidean domain.

5. Euclidean ideals in Real Quadratic extension

5.1. Proof of Theorem 8

Let $K_{1},K_{2},K_{3}$ be real quadratic extensions with the conductors $f_{1},f_{2},f_{3}$ respectively. Also assume that $H(K_{1}),H(K_{2}),H(K_{3})$ are abelian over $\mathbb{Q}$ . Set K= $K_{1}K_{2}K_{3}$ , and $f=\operatorname{lcm}(f_{1},f_{2},16)$ .

Lemma 22.

Let $K_{1},K_{2},K_{3}$ be totally real quadratic extensions, satisfying supposition of Theorem 8. Then there exists $a\in\mathbb{Z}$ with prime $p$ with $p\equiv a\mod{f}$ , If the prime $\mathfrak{P}$ of $K_{s}$ lies over $p$ then $[\mathfrak{P}]$ generates $Cl_{K_{s}}$ . Furthermore, there exist $b,c\in(1/4,1/2)$ such that for any $X,\epsilon>0$ , we have

	$\displaystyle\Bigg{\|}J_{\epsilon}(X):=\Big{\{}p\equiv a\mod{f}$	$\displaystyle\ :\ p\ \text{is prime},\ p\in(X^{1-\epsilon},X)\ \text{such that}\ \ \frac{p-1}{2}\text{is either a prime}$
		$\displaystyle\ \text{or}\ \text{product of two primes}\ q_{1}q_{2}\ \text{ with}\ X^{b}<q_{1}<X^{c}\Big{\}}\Bigg{\|}\gg\frac{X}{\log^{2}X}.$

Proof.

According to supposition of Theorem 8, the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and there exists $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)$ for any odd prime $l\mid f$ or $l=4$ . Let $a$ to be the image of $\hat{\sigma}$ under the isomorphism ${\mathrm{Gal}}(\mathbb{Q}(\zeta_{f}/\mathbb{Q})\to\left(\mathbb{Z}/f\mathbb{Z}\right)^{\times}$ . Clearly $\hat{\sigma}=\sigma_{a}$ and $\sigma_{a}\mid_{H(K_{i})}\in Gal\left(H(K_{i})/K_{i}\right)$ for any $i$ . Observe that $Gal\left(H(K_{i})/K_{i}\right)=\langle\sigma_{a}\mid_{H(K_{i})}\rangle$ if and only if fixed field of $\sigma_{a}\mid_{H(K_{i})}$ denoted by ${\mathcal{F}}(\sigma_{a})$ is equal to $K_{i}$ .

Look at

	$\displaystyle{\mathcal{F}}(\sigma_{a}\mid_{H(K_{i})})$	$\displaystyle={\mathcal{F}}(\sigma_{a})\cap H(K_{i})\subseteq K_{1}K_{2}K_{3}\cap H(K_{i})$
		$\displaystyle\subseteq\left(K_{1}\cap H(K_{i})\right)\left(K_{2}\cap H(K_{i})\right)\left(K_{3}\cap H(K_{i})\right)\subseteq K_{i}$

By the properties of the Artin symbol, if $p\equiv a\mod{f}$ , then $\left(\frac{K/\mathbb{Q}}{p}\right)=\sigma_{a}|_{K}=id,$ which implies that $p$ splits completely in $K$ . Therefore for every prime $\mathfrak{p}$ of $K_{s}$ lying above $p\equiv a$ , we have $\mathfrak{N}(\mathfrak{p})=p$ . Hence

\left(\frac{H(K_{s})/Ks}{\mathfrak{p}}\right)=\sigma_{a}|_{H(K_{s})}\text{ is a generator of }{\mathrm{Gal}}\left(H(K_{s})/K_{s}\right).

Thus $[\mathfrak{p}]$ is a generator of $Cl_{K_{s}}$ .
By the assumption of theorem 8, we have $\sigma\not\in{\mathrm{Gal}}(\mathbb{Q}\left(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)$ . Therefore $(\frac{q-1}{2},l)=1$ for every odd prime $l$ dividing $f$ and $l=4$ . Hence we get $(\frac{q-1}{2},f)=1$ . Now we can apply Lemma 13 for $u=a$ and $v=f$ there exist $b,c\in\left(1/4,1/2\right)$ and for every $\epsilon\geq 0$ . We get

\displaystyle\Big{|}J_{\epsilon}(X)\Big{|}\gg\frac{X}{\log^{2}X}.

∎

For the number field K, we define

	$\displaystyle M_{\epsilon}:=$	$\displaystyle\Bigg{\{}\mathfrak{p}\subseteq\mathcal{O}_{K}\ \text{is prime}\ :\ \mathfrak{N}(\mathfrak{p})=p\ \text{ is a prime,}\ p\equiv a\mod{f},\ \frac{p-1}{2}\ \text{is}$
		$\displaystyle\text{ either a prime or product of two primes}\ q_{1}q_{2}\ with\ p^{b}<q_{1}<p^{{c}/{1-\epsilon}}\Bigg{\}}$

and

M_{\epsilon}(x)=\{\mathfrak{p}\in M_{\epsilon}:\mathfrak{N}(\mathfrak{p})\leq X\}.

Proof of Theorem 8

Choose $e_{i}\in O_{K_{i}}^{\times}$ , such that $e_{1},e_{2},e_{3}$ multiplicatively independent set. Let $P$ be subset of $M_{\epsilon}$ such that there exists $\eta\in\{\pm e_{1},\pm e_{2},\pm e_{3}\}$ and $\eta$ is primitive root modulo $\mathfrak{p}$ , whenever $\mathfrak{p}\in P$ . By Lemma 14, we get an increasing unbounded sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

\biggr{|}P(x_{n})\biggr{|}\gg\frac{x_{n}}{\log^{2}x_{n}}.

Observe that if $\mathfrak{P}\in P$ then $\mathfrak{p}=\mathfrak{P}\cap K_{s}$ generates the class group of $K_{s}$ . Since there are only finitely many ideal classes, there exists an ideal class $[\mathfrak{a}]$ in $K_{s}$ such that

\displaystyle\biggr{|}\Big{\{}\mathfrak{P}\cap K_{s}\in[\mathfrak{a}]:\mathfrak{P}\in P,\mathfrak{N}(\mathfrak{P}\cap K_{s})\leq y_{n}\Big{\}}\Bigg{|}\gg\frac{y_{n}}{\log^{2}y_{n}},

for some subsequence $\{y_{n}\}_{n\in\mathbb{N}}$ of $\{x_{n}\}_{n\in N}$ . Since $P\subseteq M_{\epsilon}$ , the ideal class $[\mathfrak{a}]$ is a generator of $Cl_{K_{s}}$ . We have increasing sequence $\{y_{n}\}_{n\in\mathbb{N}}$

	$\displaystyle\Biggr{\|}\Big{\{}\mathfrak{p}:\mathfrak{N}(\mathfrak{p})\leq y_{n},[\mathfrak{p}]=[\mathfrak{a}],$	$\displaystyle\mathcal{O}_{K_{s}}^{\times}/\mathfrak{p}\twoheadrightarrow\left(\mathcal{O}_{K_{s}}/\mathfrak{p}\right)^{\times}\Big{\}}\Biggr{\|}\geq$
		$\displaystyle\biggr{\|}\Big{\{}\mathfrak{P}\cap K_{s}\in[\mathfrak{a}]:\ \mathfrak{p}\in P,\mathfrak{N}(\mathfrak{p})\leq y_{n}\Big{\}}\Bigg{\|}\gg\frac{y_{n}}{\log^{2}y_{n}}.$

Therefore, by Theorem 12, $K_{s}$ has a Euclidean ideal class. $\square$

Corollary 23.

Let $K_{1},K_{2},K_{3}$ be real quadratic extensions and the Hilbert class field $H(K_{i})$ is abelian over $\mathbb{Q}$ for $i=1,2,3$ . Also let the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and there exists an extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)$ for any odd prime $l\mid f$ or $l=4$ . Also let $f_{1},f_{2},f_{3}\ \text{are coprime}$ . Then at least one of $K_{1},K_{2},K_{3}$ has a Euclidean ideal class.

Corollary 24.

Let $K_{1},K_{2},K_{3}$ be real quadratic extensions with class number $1$ . The Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists an extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4$ . and

\displaystyle H(K_{i})\cap K_{j}=K_{i}\cap K_{j}=H(K_{j})\cap K_{i}\ \text{for}\ i,j\in\{1,2,3\}.

Then at least one of $K_{1},K_{2},K_{3}$ is a Euclidean domain.

6. Consequence of the Elliott and Halberstam Conjecture.

6.1. Real Quadratic fields

Let $K_{1},K_{2}$ be real quadratic extensions and also let $f_{1},f_{2}$ are the conductors $K_{1},K_{2}$ . Define $f$ to be least common multiple of $f_{1},f_{2},16$ . Clearly we have $H(K_{i})\subseteq\mathbb{Q}(\zeta_{f})$ for $i\in\{1,2\}$ . Set K= $K_{1}K_{2}$ and $H=H_{1}H_{2}$ .

Theorem 25.

Let $K_{1},K_{2}$ be real quadratic extensions with $Cl_{K_{1}},Cl_{K_{2}}$ are cyclic and the Hilbert class field $H(K_{i})$ is abelian over $\mathbb{Q}$ for $i=1,2$ . Also let the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists an extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}\left({\mathbb{Q}(\zeta_{f})}/{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4$ . Also let

\displaystyle H(K_{1})\cap K_{2}=K_{1}\cap K_{2}=H(K_{2})\cap K_{1}.

Then at least one of $K_{1},K_{2}$ has a Euclidean ideal class if Elliott and Halberstam Conjecture is true.

Proof.

We have the Galois group $G={\mathrm{Gal}}\left({H}/{K}\right)$ is cyclic, $G=\langle\sigma\rangle$ and also there exists an extension $\hat{\sigma}\in{\mathrm{Gal}}(\mathbb{Q}(\zeta_{f})/K)$ of $\sigma$ such that

\displaystyle\hat{\sigma}\not\in{\mathrm{Gal}}\left(\frac{\mathbb{Q}(\zeta_{f})}{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4.

Choose $a$ to be the image of $\hat{\sigma}$ under the isomorphism ${\mathrm{Gal}}(\mathbb{Q}(\zeta_{f}/\mathbb{Q})\xrightarrow{\cong}\left(\mathbb{Z}/f\mathbb{Z}\right)^{\times}$ . Clearly $\sigma_{a}=\hat{\sigma}$ , therefore ${\sigma_{a}}\not\in{\mathrm{Gal}}\left(\frac{\mathbb{Q}(\zeta_{f})}{\mathbb{Q}(\zeta_{l})}\right)\ \text{for any odd prime }\ l\mid f\ or\ l=4$ . Therefore we have $\left(\frac{a-1}{2},f\right)=1$ . Set $u=a$ and $v=f$ then we have $u$ and $v$ are natural numbers satisfying Lemma 13. Hence for every $\epsilon\geq 0$

	$\displaystyle\Bigg{\|}J_{\epsilon}(X):=\Big{\{}p\equiv a\mod{f}\ :\$	$\displaystyle p\ \text{is prime},\ p\in(X^{1-\epsilon},X)\ \text{such that}\ \ \frac{p-1}{2}\ \text{is either }$
	$\displaystyle\text{ a prime or}q_{1}q_{2}$	$\displaystyle\text{ with}\ X^{b}<q_{1}<X^{c}\Big{\}}\Bigg{\|}\gg\frac{X}{\log^{2}X}.$

Define

	$\displaystyle M_{\epsilon}:=$	$\displaystyle\Bigg{\{}\mathfrak{p}\subseteq\mathcal{O}_{K}\ \text{is prime}:\mathfrak{N}(\mathfrak{p})=p\ \text{ is a prime,}\ p\equiv a\mod{f},\ \frac{p-1}{2}\ \text{is}$
		$\displaystyle\text{ either a prime or product of two primes}\ q_{1}q_{2}\ with\ p^{b}<q_{1}<p^{\frac{c}{1-\epsilon}}\Bigg{\}}$

and

\displaystyle M_{\epsilon}(x)=\{\mathfrak{p}\in M_{\epsilon}:\mathfrak{N}(\mathfrak{p})\leq X\}.

Consider $e_{i}\in\mathcal{O}_{K_{i}}^{*}$ for every $i\in\{1,2\}$ such that $e_{1},e_{2}$ are multiplicative independent elements. Assuming Elliott and Halberstam Conjecture is true, by [6, Lemma 19 ], there exists a set $P\subseteq M_{\epsilon}$ such that for every $\mathfrak{p}\in P$ for some $\eta\in\{\pm e_{1},\pm e_{2}\}$ is primitive modulo $\mathfrak{p}$ and there exists an increasing sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

\displaystyle\left|P(x_{n})\right|\gg\frac{x_{n}}{\log x_{n}}.

If $\eta\in K_{s}$ . Set $R=\{\mathfrak{p}\cap K_{s}\ |\ \mathfrak{p}\in P\}$ , then

\displaystyle R(x_{n})\geq\frac{1}{[K:K_{s}]}P(x_{n})\gg\frac{1}{[K:K_{s}]}\frac{x_{n}}{\log x_{n}}.

Since $Cl_{K}$ is finite, we have a subsequence $\{y_{n}\}_{n\in\mathbb{N}}$ of $\{x_{n}\}_{n\in\mathbb{N}}$ and an ideal clas $[\mathfrak{a}]$ such that

\displaystyle\{p\in[\mathfrak{a}]\ |\ \mathfrak{p}\in R,\mathfrak{N}(p)\leq y_{n}\}\gg\frac{y_{n}}{\log y_{n}}.

Then $[\mathfrak{a}]$ is a Euclidean ideal class. Hence $K_{s}$ has a Euclidean ideal if Elliott and Halberstam Conjecture is true. ∎

Corollary 26.

Let $K_{1},K_{2}$ be real quadratic extensions with $Cl_{K_{1}},Cl_{K_{2}}$ are cyclic. Also let the Galois group $G={\mathrm{Gal}}\left(\frac{\mathbb{Q}(\zeta_{f})}{K}\right)$ is cyclic, where $K=K_{1}K_{2}$ . Also let $f_{1},f_{2}\ \text{are coprime}$ . Then at least one of $K_{1},K_{2}$ has a Euclidean ideal class if Elliott and Halberstam Conjecture is true.

6.2. Real Cubic fields

Let $K$ be abelian real cubic extension such that both $K,H(K)$ are abelian over $\mathbb{Q}$ . Set $f$ be the least common multiple of $16,f(K)$ . And let $G_{l}$ be Galois group of $\mathbb{Q}(\zeta_{f})$ over $\mathbb{Q}(\zeta_{l})$ for all odd prime $l$ dividing $f$ or $l=4$ .

Theorem 27.

Let $K$ be an abelian real cubic extension with $Cl_{K}$ is cyclic. Also let ${\mathrm{Gal}}(H(K)/K)=\langle\sigma\rangle$ and there exists an extension $\hat{\sigma}$ of $\sigma$ to $\mathbb{Q}(\zeta_{f})$ , such that $\hat{\sigma}\not\in G_{l}$ for any odd prime $l$ dividing $f$ and $l=4$ . Then $K$ has a Euclidean ideal class if Elliott and Halberstam Conjecture is true.

Proof.

It is clear that ${\mathrm{Gal}}(H(K)/K)=\langle\sigma\rangle$ , and the map $\hat{\sigma}\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}\right)$ , where $\hat{\sigma}$ is a lift of $\sigma$ . Therefore there exists $a<f$ and coprime to $f$ such that $\sigma_{a}=\hat{\sigma}.$ If $p\equiv a\mod{f}$ and $\mathfrak{p}$ lies over $p$ then the ideal class $[\mathfrak{p}]$ generates the class group $Cl_{K}$ . The assumption $\sigma\not\in G_{l}$ implies $(\frac{a-1}{2},l)=1$ . By the Lemma 13, with $u=a$ and $v=f$ , we get

	$\displaystyle\Bigg{\|}J_{\epsilon}(X):=$	$\displaystyle\Big{\{}p\equiv a\mod{f}\ :\ p\ \text{is prime},\ p\in(X^{1-\epsilon},X)\quad\text{such that}\ \frac{p-1}{2}\ \text{is either }$
		$\displaystyle\text{ prime or product of two primes}\ q_{1}q_{2}\ with\ X^{b}<q_{1}<X^{c}\Big{\}}\Bigg{\|}\gg\frac{X}{\log^{2}X}.$

∎

For the cubic number field $K$ in Theorem 27, we define

\displaystyle M_{\epsilon}:=

\displaystyle\Big{\{}\mathfrak{p}\subseteq\mathcal{O}_{K}\ \text{is prime}\ :\ \mathfrak{N}(\mathfrak{p})=p\ \text{ is a prime,}\ p\in J_{\epsilon}(X),\ \text{for some }X\Big{\}}

and

M_{\epsilon}(x)=\{\mathfrak{p}\in M_{\epsilon}:\mathfrak{N}(\mathfrak{p})\leq X\}.

Proof of Theorem 27

As K is real cubic field, $\mathcal{O}_{K}^{\times}$ contains two multiplicative independent elements, say $e_{1},e_{2}$ . Assuming Elliott and Halberstam Conjecture is true, by [6, Lemma 19], there exists a set $P\subseteq M_{\epsilon}$ such that for every $\mathfrak{p}\in P$ for some $\eta\in\{\pm e_{1},\pm e_{2}\}$ is primitive modulo $\mathfrak{p}$ and there exists an increasing sequence $\{x_{n}\}_{n\in\mathbb{N}}$ such that

\displaystyle\left|P(x_{n})\right|\gg\frac{x_{n}}{\log x_{n}}

There are only finitely many ideal classes in $Cl_{K}$ , we claim that there exists a non-trivial ideal class $[\mathfrak{a}]$ such that

\displaystyle\Bigg{|}\Big{\{}\mathfrak{p}\in[\mathfrak{a}]:\mathfrak{p}\in P,\mathfrak{N}(\mathfrak{p})\leq y_{n}\Big{\}}\Bigg{|}\gg\frac{y_{n}}{\log^{2}y_{n}},

where $(y_{n})_{n}$ is sub sequence of $(x_{n})_{n}$ Otherwise we get that

\biggr{|}P(x_{n})\biggr{|}=o\left(\frac{y_{n}}{\log^{2}y_{n}}\right).

Since every prime belongs to $M_{\epsilon}$ lie over some prime $p\equiv a\mod{f}$ , therefore for every $\mathfrak{p}\in M_{\epsilon}$ , $[\mathfrak{p}]$ is generator of ideal class group. Therefore $[\mathfrak{a}]$ is a generator of $Cl_{K}$ . By Theorem 12, we have $[\mathfrak{a}]$ is a Euclidean ideal class. $\square$

Corollary 28.

Let $K$ be a real cubic field with odd prime conductor $q$ . If the class number of $K$ is a prime such that $3\nmid p$ . Then $K$ has a Euclidean ideal class if Elliott and Halberstam’s conjecture is true.

Proof.

The order of the Galois group ${\mathrm{Gal}}\left(H(K)/\mathbb{Q}\right)$ is equal to $3p$ and $3\nmid p$ . By Sylow theorems, the group ${\mathrm{Gal}}\left(H(K)/\mathbb{Q}\right)$ is cyclic, therefore $H(K)$ is abelian. Observe that there exist a $\sigma_{1}\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{q})/K\right)$ which is lift of $\sigma$ , where ${\mathrm{Gal}}(H(K)/K)=\langle\sigma\rangle$ . Since $\mathbb{Q}(\zeta_{q})\cap\mathbb{Q}(\zeta_{4})=\mathbb{Q}$ , therefore there exists $\hat{\sigma}$ a lift of $\sigma_{1}$ such that $\hat{\sigma}\not\in{\mathrm{Gal}}\left(\mathbb{Q}(\zeta_{f})/\mathbb{Q}(\zeta_{l})\right)$ , for $l=q,4$ . Hence $K$ has a Euclidean ideal class if Elliott and Halberstam’s conjecture is true. ∎

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