On generalized Mersenne Primes and class-numbers of equivalent quadratic fields and cyclotomic fields

Abstract

In this paper we define equivalent quadratic fields and prove that generalized Mersenne primes generate a family of infinitely many equivalent quadratic fields with equivalent index $2$ and whose class numbers are divisible by 3. We also prove that the class-number of the cyclotomic field $\mathbb {Q}\big ( \zeta _m \big )$, where $m\in \mathbb {N}$ and $\zeta _m$ is a primitive $m$-th root of unity, is divisible by a certain integer $g$.

1 Introduction

The class number problem of quadratic field and cyclotomic field is one of the most intriguing unsolved problems in algebraic number theory and it has been the object of attention of many years among researchers. It was proved by Nagel [6] that there are infinitely many quadratic number fields each with class number divisible by a given positive integer. Weinberger [10] showed that for all positive integers $n$, there are infinitely many real quadratic fields with class numbers divisible by $n$. Ankeny and Chowla [1] proved that there exist infinitely many imaginary quadratic fields each with class number divisible by $g$ where $g$ is any given rational integer. Hartung [3] constructed a family of infinitely many imaginary quadratic fields whose class number is divisible by 3. Ankeny et al.[2], Lang [5], Takeuchi [8] and Osada [7] independently proved that the class-number of the maximal real subfield of a cyclotomic field is greater than $1$. Osada [7] also proved that this class-number is divisible by a certain integer $n$. Watabe [9] deduced some results on the divisibility of the class-numbers of certain cyclotomic fields. Recently, we generalized the results of Takeuchi [8] in one of our papers.

In this paper, we introduce the notion of equivalent quadratic fields. We denote by $\Delta _F$ and $\mathrm{rad(F)}$ respectively the discriminant and radicand of the quadratic field $F$. Two quadartic fields $F_1$ and $F_2$ are said to be equivalent if there exists an integer $n$ satisfying $1<n< \min \lbrace |\mathrm{rad}(\mathrm{F}_1)|, |\mathrm{rad}(\mathrm{F}_2)| \rbrace$ such that $\Delta _{F_1}\equiv \Delta _{F_2}({\mathrm {mod}} \ n)$. Such a smallest positive integer $n$ is called the equivalent index of $F_1$ and $F_2$. We write $F_1 \triangleq F_2$ to represent $F_1$ is equivalent to $F_2$ or simply $F_1$ and $F_2$ are equivalent. Also by $(F_1:F_2)$ we mean the equivalent index of $F_1\triangleq F_2$. We show that generalized Mersenne primes generate a family of infinitely many equivalent quadratic fields with equivalent index 2 and whose class numbers are divisible by 3. We also establish that the class-number of the cyclotomic field $\mathbb {Q}\big ( \zeta _m \big )$, where $m\in \mathbb {N}$ and $\zeta _m$ is a primitive $m$-th root of unity, is divisible by a certain integer $g$.

Example Consider the quadratic fields $F_1=\mathbb {Q}(\sqrt{13})$, $F_2=\mathbb {Q}(\sqrt{15})$ and $F_3=\mathbb {Q}(\sqrt{21})$. Then $\mathrm{rad}(\mathrm{F}_1)=13$, $\mathrm{rad}(\mathrm{F}_2)=15$, $\mathrm{rad}(\mathrm{F}_3)=21$, $\Delta _{F_1}=13$, $\Delta _{F_2}=60$ and $\Delta _{F_3}=21$. Thus $F_1\triangleq F_3$ with $(F_1:F_3)=2$ and $F_2\triangleq F_3$ with $(F_2:F_3)=3$. But $F_1$ and $F_2$ are not equivalent.

2 Main results

The concept of generalized Mersenne Prime(GMP) was defined by Hoque and Saikia [4] as a prime of the form:

$$\begin{aligned} M_{p,q}=p^q-p+1 \end{aligned}$$

where $p$ and $q$ are positive integers.

Throughout this paper we consider the GMP, $M_{p,q}$ with the restrictions: $p$ is odd prime and $q$ is odd integer.

We see that $M_{p,q}\equiv 1({\mathrm {mod}}\ 3)$ and $M_{p,q}\equiv 1({\mathrm {mod}}\ 4)$.

We consider the following trinomials:

$$\begin{aligned} f_1(x)&=x^3-M_{p,q}x + p\\ f_2(x)&=x^3-px + M_{p, q} \end{aligned}$$

The discriminants of $f_1(x)$ and $f_2(x)$ are respectively $D(f_1)=4M_{p,q}^3-27p^2$ and $D(f_2)=4p^3-27M_{p,q}^2$. For all odd primes $p$ and odd integers $q>1$, we have $D(f_1)>0$ and $D(f_2)<0$. Now

$$\begin{aligned} f_1(x)&\equiv x^3 -x \pm 1 \ ({\mathrm {mod}}\ 3)\\ f_2(x)&\equiv x^3 \pm x + 1 \ ({\mathrm {mod}} \ 3) \end{aligned}$$

We see both the trinomials in the right sides are irreducible ${\mathrm {mod}} \ 3$. Thus both $f_1(x)$ and $f_2(x)$ are irreducible over $\mathbb {Q}$.

Theorem 2.1

The class number of the real quadratic field $\mathbb {Q}(\sqrt{D(f_1)})$ and the imaginary quadratic field $\mathbb {Q}(\sqrt{D(f_2)})$ are divisible by 3.

Proof

Let $F_1(M_{p,q})=\mathbb {Q}(\sqrt{D(f_1)})=\mathbb {Q}(\sqrt{4M_{p,q}^3-27p^2})$ and $F_2(-M_{p,q})=\mathbb {Q}(\sqrt{D(f_2)})=\mathbb {Q}(\sqrt{4p^3-27M_{p,q}^2})$. Let $K_i$ be the splitting field of $f_i(x)$ and $G_i$ be the Galois group of $K_i \ (i=1, 2)$ over $\mathbb {Q}$. Since $f_1(x)$ and $f_2(x)$ are irreducible over $\mathbb {Q}$, $G_1$ and $G_2$ are isomorphic to $S_3$. Also $(2M_{p, q}, 3p)=1$ and $(2p, 3M_{p,q})=1$. Thus by Proposition 1 in [12], $K_i$ over $F_i\ (i=1,2)$ is unramified. Also the Galois group of $K_i$ over $F_i\ (i=1,2)$ is cubic cyclic group. Thus $K_i$ over $F_i\ (i=1,2)$ is cubic cyclic unramified extension. By class field theory, the Hilbert class field of $F_i$ contains $K_i \ (i=1,2)$ and thus the class number of $F_i\ (i=1, 2)$ is divisible by 3. $\square$

Theorem 2.2

There are infinitely many equivalent quadratic fields with index $2$ whose class numbers are divisible by $3$.

Proof

Since $D(f_1)=4M_{p,q}^3-27p^2\equiv 1 ({\mathrm {mod}}\ 4)$, the discriminant of the real quadratic field is $\Delta _{F_1(M_{p,q})}=D(f_1)$.

Also, $D(f_2)=4p^3-27M_{p,q}^2\equiv 1 ({\mathrm {mod}}\ 4)$ gives the discriminant of the imaginary quadratic field is $\Delta _{F_2(-M_{p,q})}=D(f_2)$.

Now

$$\begin{aligned} \Delta _{F_1(M_{p,q})}-\Delta _{F_2(-M_{p,q})}=4\left( M_{p,q}^3-p^3\right) +27\left( M_{p,q}^2-p^2 \right) \equiv 0({\mathrm {mod}}\ 2) \end{aligned}$$

Thus the quadratic fields $F_1(M_{p,q})\triangleq F_2(-M_{p,q})$ with $(F_1(M_{p,q}): F_2(-M_{p,q}))= 2$.

Now we count the number of real quadratic fields of the form $F_1(M_{p,q})$.

Under the truth of the Conjecture 2.1 [4], we can find infinitely many GMPs such that $D(f_1)$’s are not perfect square and hence infinitely many quadratic fields of the form $F_1(M_{p, q})=\mathbb {Q}(\sqrt{D( f _1)})$. Let $S$ be the set of all $D(f_1)$’s which give rise to same fields more than once. We fix $p$. Let $q_0$ be give one member in $S$. Suppose $4M_{p,q_0}^3-27p^2=a^2d$ for some fixed square-free integer $d$ and an integer $a$. Then if $F_1(M_{p,q_0})=F_1(M_{p,q})$, there exists an integer $c$ such that $4M_{p,q}^3-27p^2=c^2a^2d$. Thus $(M_{p,q},\ ca)$ is an integral solution of

$$\begin{aligned} 4X^3=dY^2+27p^2 \end{aligned}$$

(1)

By Siegel’s theorem, the algebraic curve given by the Diophantine equation (1) has only finite numbers of integral solutions. Thus $\#S$ is finite and hence there are infinitely many real quadratic fields of the form $F_1(M_{p,q})$. Also corresponding to each value of $M_{p,q}$ that contributes a real quadratic field of the form $F_1(M_{p,q})$, we get one imaginary quadratic field of the form $F_2(-M_{p,q})$. Thus there are infinitely many imaginary quadratic fields of the form $F_2(-M_{p,q})$. By the Theorem 2.1, we complete the proof. $\square$

Yamaguchi [11] showed the following result.

Lemma 2.3

[11] If $\phi (m)>4$, then $h(\mathbb {Q}(\sqrt{m}))|H(\mathbb {Q}(\zeta _{4m}+\zeta ^{-1} _{4m}))$, where $\phi$ stands for the Euler’s function and $m>0$ is an integer.

Now from Theorem 2.1 and this lemma, we obtain the following result.

Theorem 2.4

Let $m= D(f_1)$ such that $\phi (m)>4$. Then $3|H(\mathbb {Q}(\zeta _{4m}+\zeta ^{-1} _{4m}))$.

Lemma 2.5

[9] If $u\ge 5$ is a prime and $v$ is a prime with $v\equiv 1({\mathrm {mod}}\ u)$, then

• (i) $u^{\frac{u-3}{2}}\mid h(\mathbb {Q}(\zeta _{uv}))$.

• (ii) $2^{\frac{u-3}{2}}\mid h\left( \mathbb {Q}\left( \sqrt{(-1)^{\frac{v-1}{2}}v}, \zeta _{u}\right) \right)$.

• (iii) $2^{\frac{u-3}{2}}\mid h(\mathbb {Q}(\zeta _{uv}))$.

Applying this lemma we obtain the following result.

Theorem 2.6

For any prime $u\ge 5$, the following statements hold:

• (i) $u^{\frac{u-3}{2}}\mid h(\mathbb {Q}(\zeta _{u M_{p, q} }))$.

• (ii) $2^{\frac{u-3}{2}}\mid h\left( \mathbb {Q}\left( \sqrt{M_{p,q}}, \zeta _{u}\right) \right)$.

• (iii) $2^{\frac{u-3}{2}}\mid h(\mathbb {Q}(\zeta _{uM_{p,q}}))$.

We now characterise units and irreducible elements in the real quadratic field $F_1(M_{p,q})$.

Theorem 2.7

Let $D>1$ be an square-free integer such that $D\equiv 1({\mathrm {mod}}\ 4)$. Let $\alpha =1\pm u$, where $u$ is a unit in the real quadratic field $\mathbb {Q}(\sqrt{D})$. Then the following hold:

• (i) If $N(u)=-1$, then $\alpha$ is irreducible if and only if $u$ is expressible as $u=\frac{a+b\sqrt{D}}{2}$ such that $a$ is an odd prime and $b$ is an odd integer.

• (ii) If $N(u)=1$, then $\alpha$ is irreducible if and only if $u$ is expressible as $u=\frac{a+b\sqrt{D}}{2}$ such that $a\pm 2$ is an odd prime and $b$ is an odd integer.

Proof

(i) Since $D \equiv 1({\mathrm {mod}}\ 4)$ and $u$ is a unit in $\mathbb {Q}(\sqrt{\mathrm {D}})$, $u$ is of the form

$$\begin{aligned} \dfrac{a+b\sqrt{D}}{2} \end{aligned}$$

where $a$ and $b$ are integers.

Now $N(\alpha )=\frac{(2\pm a)^2-Db^2}{4}=N(u)\pm a+1=\pm a$.

For $\alpha$ to be irreducible in $\mathbb {Q}(\sqrt{\mathrm {D}})$, $a$ should be a rational prime.

If $a=2$, then $N(u)=-1$ gives a contradiction to D$\equiv 1({\mathrm {mod}} \ 4)$.

Thus $\alpha$ to be irreducible in $\mathbb {Q}(\sqrt{\mathrm {D}})$, $a$ should be a rational odd prime.

Again, $N(u)=-1\Rightarrow a^2-db^2=-4\Rightarrow b\equiv 1({\mathrm {mod}} \ 2)$.

Thus $b$ is a rational odd integer.

The converse part is obvious as $N(\alpha )=\pm a$, a rational odd prime.

Similarly we can prove (ii).

As a consequence we have the following result. $\square$

Corollary 2.8

Let $u$ be a unit in $F_1(M_{p,n})$ and $\alpha =1\pm u$. Then the following hold:

• (i) If $N(u)=1$, then $\alpha$ is irreducible if and only if $u$ is expressible as $u=\frac{a+b\sqrt{D(f_1)}}{2}$ such that $a$ is a rational odd prime and $b$ is a rational odd integer.

• (ii) If $N(u)=1$, then $\alpha$ is irreducible if and only if $u$ is expressible as $u=\frac{a+b\sqrt{D(f_1)}}{2}$ such that $a\pm 2$ is a rational odd prime and $b$ is a rational odd integer.