On the structure of order 4 class groups of $\mathbb{Q}(\sqrt{n^2+1})$

TL;DR

This paper investigates the structure of order 4 class groups in real quadratic fields of the form (+1) and provides conditions to determine their structure, along with computations of Dedekind zeta values and growth of class group size.

Contribution

It offers new sufficient conditions to determine the structure of class groups of order 4 in a specific family of real quadratic fields and analyzes their size growth.

Findings

Class groups of order 4 are either cyclic or product of two cyclic groups.

Conditions are provided to specify the structure of these class groups.

The size of the class group can be increased by adding more odd prime factors to n.

Abstract

Groups of order $4$ are isomorphic to either $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. We give certain sufficient conditions permitting to specify the structure of class groups of order $4$ in the family of real quadratic fields $\mathbb{Q}{(\sqrt{n^2+1})}$ as $n$ varies over positive integers. Further, we compute the values of Dedekind zeta function attached to these quadratic fields at the point $-1$. As a side result, we show that the size of the class group of this family could be made as large as possible by increasing the size of the number of distinct odd prime factors of $n$.

Equations

$$\log (h_{K,d}{R}_{K,d})\sim \log (\sqrt{{\Delta }_{K,d}}).$$

$${\zeta }_k(-1)=\frac{1}{60}\sum _{\begin{array}{c}|t|<\sqrt{D}\\ t^2\equiv D(\mathrm{mod}4)\end{array}}\sigma \left(\frac{D-t^2}{4}\right),$$

$$\delta (\mathfrak{a}):=r_1{r}_2^{\prime }-r_1^{\prime }{r}_2,$$

$$\epsilon \left[\begin{array}{c}{r}_1\\ r_2\end{array}\right]=M\left[\begin{array}{c}{r}_1\\ r_2\end{array}\right].$$

$${\zeta }_k(-1,\mathfrak{A})$$

$$\times S^3(a,c)+180a{c}^3(\text{textsl}sgnc)S^2(a,c)-240c^3(\text{textsl}sgnc)S^3(d,c)$$

$$\displaystyle+180dc^{3}(\textsl{sgn }c)S^{2}(d,c)\big{\}},$$

$$\left[\begin{array}{cc}Tr\left(\frac{r_1{r}_2^{\prime }\epsilon }{\delta (\mathfrak{a})}\right)& Tr\left(\frac{r_1{r}_1^{\prime }{\epsilon }^{\prime }}{\delta (\mathfrak{a})}\right)\text{vspace}\ast 2mm\\ Tr\left(\frac{r_2{r}_2^{\prime }\epsilon }{\delta (\mathfrak{a})}\right)& Tr\left(\frac{r_1{r}_2^{\prime }{\epsilon }^{\prime }}{\delta (\mathfrak{a})}\right)\end{array}\right]$$

$${\mathfrak{S}}_n:$$

$$\mathcal{N}:$$

$$h(d)\ge \{\begin{array}{ll}\mathcal{N}+2& \text{ if }d\equiv 1,2,6(\mathrm{mod}8),\\ \mathcal{N}+1& \text{ if }d\equiv 5(\mathrm{mod}8).\end{array}$$

$$(2)=\left(2,\frac{1+\sqrt{d}}{2}\right)\left(2,\frac{1-\sqrt{d}}{2}\right).$$

$$(p_i)=\left(p_i,\frac{1+\sqrt{d}}{2}\right)\left(p_i,\frac{1-\sqrt{d}}{2}\right).$$

$${\zeta }_{k_n}(-1,{\mathfrak{A}}_i)=\frac{n^3+n(4p_i^4+10p_i^2)}{360p_i^2},$$

$${\zeta }_{k_n}(-1,\mathfrak{B})=\frac{n^3+104n}{1440}.$$

$${\zeta }_{k_n}(-1,\mathfrak{P})=\frac{n^3+14n}{360}.$$

$${\zeta }_{k_n}(-1,\mathfrak{A})=\frac{n^3+n(4p^4+10p^2)}{360p^2},$$

$${\zeta }_{k_n}(-1,{\mathfrak{A}}^2)=\frac{n^3+n(4p^8+10p^4)}{360p^4}.$$

$${\zeta }_{k_n}(-1)={\zeta }_{k_n}(-1,\mathfrak{P})+{\zeta }_{k_n}(-1,\mathfrak{A})+{\zeta }_{k_n}(-1,{\mathfrak{A}}^2)+{\zeta }_{k_n}(-1,{\mathfrak{A}}^{-1}).$$

$${\zeta }_{k_n}(-1)=\frac{n^3+14n}{360}+\frac{n^3+n(4p^4+10p^2)}{180p^2}+\frac{n^3+n(4p^8+10p^4)}{360p^4}.$$

$${\zeta }_{k_n}(-1,\mathfrak{A})$$

$${\zeta }_{k_n}(-1,\mathfrak{B})$$

$${\zeta }_{k_n}(-1,\mathfrak{C})$$

$$n=4,2p,2q,2pq,4p,4q.$$

$${\zeta }_{k_n}(-1,\mathfrak{A})$$

$${\zeta }_{k_n}(-1,{\mathfrak{A}}^2)$$

$${\zeta }_{k_n}(-1)={\zeta }_{k_n}(-1,\mathfrak{P})+{\zeta }_{k_n}(-1,\mathfrak{A})+{\zeta }_{k_n}(-1,{\mathfrak{A}}^2)+{\zeta }_{k_n}(-1,{\mathfrak{A}}^{-1}).$$

$${\zeta }_{k_n}(-1)=\frac{n^3+14n}{360}+\frac{n^3+n(4p^4+10p^2)}{180p^2}+\frac{n^3+n(4p^8+10p^4)}{360p^4}.$$

$${\zeta }_{k_n}(-1,{\mathfrak{A}}^{-1})=\frac{n^3+n(4p^4+10p^2)}{360p^2},$$

$${\zeta }_{k_n}(-1,{\mathfrak{B}}^{-1})=\frac{n^3+n(4q^4+10q^2)}{360q^2}.$$

$$\frac{n^3+n(4p^8+10p^4)}{360p^4}=\frac{n^3+n(4q^4+10q^2)}{360q^2}.$$

$${\zeta }_{k_n}(-1,\mathfrak{A})=\frac{n^3+n(4p^4+10p^2)}{360p^2},$$

$${\zeta }_{k_n}(-1,\mathfrak{B})=\frac{n^3+n(4q^4+10q^2)}{360q^2},$$

$${\zeta }_{k_n}(-1,\mathfrak{C})=\frac{n^3+n(4{\ell }^4+10{\ell }^2)}{360{\ell }^2}.$$

$${\zeta }_{k_n}(-1,\mathfrak{P})=\frac{4n^3+11n}{180}.$$

Full text

Résumé. À isomorphisme près, il y a deux groupes possibles d’ordre $4$: $\mathbb{Z}/4\mathbb{Z}$ et $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. Nous donnons des conditions suffisantes permettant de spécifier la structure des groupes de classes d’ordre $4$ dans la famille des corps quadratiques réels $\mathbb{Q}{(\sqrt{n^{2}+1})}$ lorsque $n$ parcourt l’ensemble des entiers positifs. De plus, nous calculons la valeur de la fonction zêta de Dedekind attachée à ces corps au point $-1$. Comme résultat secondaire, nous montrons que la cardinalité du groupe de classes des corps de cette famille peut être aussi grande que possible en augmentant le nombre de facteurs premiers impairs distincts de $n$.

1. Introduction

Let $n$ be a positive integer. We are interested in the single parametric family of real quadratic fields, $k_{n}={\mathbb{Q}}(\sqrt{n^{2}+1}),$ which are popularly known as Richaud-Degert (R-D) type real quadratic fields. These fields have attracted the attention of many mathematicians over the years. Here we refer to few works which are connected to our investigation. Chowla and Friedlander [8] conjectured that if $n^{2}+1$ is a prime with $n>26$ then the class number of $k_{n}$ is greater than $1$. They also conjectured that if $n$ is even and $n^{2}+1$ is square-free then the class number of $k_{n}$ is greater than $1$ except for $n=2,4,6,10,14,26$. The first conjecture was proved by Mollin and Williams [16] under the generalized Riemann hypothesis, and the second one was settled by Biró in [2]. Chakraborty and Hoque [9] proved that the class number of $k_{n}$ is always greater than $1$ if $n=mp$, where $m$ is an odd integer and $p\equiv\pm 1\pmod{8}$ is a prime. They also proved an analogous result in [6] when $n=3m$ with $m$ an odd integer. On the other hand, Yokoi [18] showed that the class number of $k_{n}$ is $1$ if and only if ${\displaystyle\frac{n^{2}}{4}-t(t+1)}$ (with ${\displaystyle 1\leq t\leq\frac{n}{2}-1}$) is a prime. Furthermore, some interesting results on the class number one problem of R-D type (resp. non-R-D type) fields were obtained by Biró and Lapkova (resp. Hoque and Kotyada) (cf. [3, 13, 10]).

It is of considerable interest to explore the structure of class groups and in particular when its order is not prime. As a starting point, one can begin with the classification of class groups of order a prime power. One can conclude with the help of Brauer–Siegel theorem that the class number of $k_{n}$ can be made as large as possible. A simple version of Brauer–Siegel theorem is as follows:

Theorem A** ([12, Corollary]).**

Let $\Delta_{K,d}$, $R_{K,d}$ and $h_{K,d}$ be respectively, the discriminant, regulator and class number of the number field $K$ of degree $d$ over $\mathbb{Q}$. If $K$ runs over number fields of degree $d$, then

[TABLE]

Thus there are only finitely many $k_{n}$ of a given class number $h$. However, Theorem A is ineffective in finding out the exact values of $n$ such that class number of $k_{n}$ equals $h$. On the other hand, Byeon and Kim obtained certain necessary and sufficient conditions for the class number of $k_{n}$ to be $1$ in [4] (resp. $2$ in [5]). The authors obtained criteria for class number $3$ for $k_{n}$ in [7] following Byeon-Kim technique. In this paper, we give certain sufficient conditions (depending on the factorization of $n$) to specify the structure of the class groups of $k_{n}$ of order $4$. Note that $4$ is the smallest class number with the possibility of two choices of class groups up to isomorphism. We denote by $\mathfrak{C}(k_{n})$ and $\zeta_{k_{n}}(s)$ respectively, the class group of $k_{n}$ and the Dedekind zeta function attached to $k_{n}$. The results of this paper are presented below.

We discuss the results based on the class $n^{2}+1\equiv 1,2,5\pmod{8}$. We first consider the case when $n^{2}\equiv 0\pmod{8}$. Let $n=2^{m}n_{0}$ for some integers $m>1$ and $n_{0}\equiv 1\pmod{2}$. Then $n_{0}$ has at most two prime factors by Proposition 3.1.

Theorem 1.1**.**

Let $d=n^{2}+1\equiv 1\pmod{8}$ be a square-free integer, and let $h(d)=4$.

• (I)

If $n=2^{m}p^{t}$ with $m,t\geq 2$ and $p$ an odd prime, then

• (i)

$\mathfrak{C}(k_{n})\cong{\displaystyle\mathbb{Z}/4\mathbb{Z}}$,

• (ii)

${\displaystyle\zeta_{k_{n}}(-1)=\frac{n^{3}+14n}{360}+\frac{n^{3}+n(4p^{4}+10p^{2})}{180p^{2}}+\frac{n^{3}+n(4p^{8}+10p^{4})}{360p^{4}}}.$**

• (II)

If $n=2^{m}p^{s}q^{t}$ with $p,q$ distinct odd primes, and $m>1,s,t\geq 1$ three integers, then

• (i)

$\mathfrak{C}(k_{n})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$,

• (ii)

$\zeta_{k_{n}}(-1)={\displaystyle\frac{n^{3}+32n}{288}+\frac{n^{3}+n(4p^{4}+10p^{2})}{360p^{2}}+\frac{n^{3}+n(4q^{4}+10q^{2})}{360q^{2}}}.$**

We now consider $n^{2}\equiv 4\pmod{8}$. Let $n=2n_{0}$ for some odd integer $n_{0}\geq 1$. It follows from Proposition 3.1 that $n_{0}$ has at most three prime factors.

Theorem 1.2**.**

Let $d=n^{2}+1\equiv 5\pmod{8}$ be a square-free integer, and let $h(d)=4$.

• (I)

If $n=2p^{t}$ with $t\geq 3$ an integer, then

• (i)

$\mathfrak{C}(k_{n})\cong\mathbb{Z}/4\mathbb{Z}$,

• (ii)

$\zeta_{k_{n}}(-1)={\displaystyle\frac{n^{3}+14n}{360}+\frac{n^{3}+n(4p^{4}+10p^{2})}{180p^{2}}+\frac{n^{3}+n(4p^{8}+10p^{4})}{360p^{4}}}.$**

• (II)

If $n=2p^{s}q^{t}$ with $p,q$ distinct odd primes and $s,t$ two positive integers such that one of them is greater $2$ or both are greater than equal to $2$, then $\mathfrak{C}(k_{n})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

• (III)

If $n=2p^{r}q^{s}\ell^{t}$ with $p,q,\ell$ distinct odd primes and $r,s,t$ positive integers, then

• (i)

$\mathfrak{C}(k_{n})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$,

• (ii)

$\zeta_{k_{n}}(-1)={\displaystyle\frac{n^{3}+14n}{360}+\frac{n^{3}+n(4p^{4}+10p^{2})}{360p^{2}}+\frac{n^{3}+n(4q^{4}+10q^{2})}{360q^{2}}+\frac{n^{3}+n(4\ell^{4}+10\ell^{2})}{360\ell^{2}}}.$**

Finally the case $n^{2}\equiv 1\pmod{8}$ and in this case $n$ is an odd integer. Thus Proposition 3.1 implies that $n$ has at most two prime factors.

Theorem 1.3**.**

Let $d=n^{2}+1\equiv 2\pmod{4}$ be a square-free integer, and let $h(d)=4$.

• (I)

If $n$ is an odd positive integer with two distinct odd prime factors $p$ and $q$, then

• (i)

$\mathfrak{C}(k_{n})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$,

• (ii)

$\zeta_{k_{n}}(-1)={\displaystyle\frac{n^{3}+5n}{36}+\frac{4n^{3}+n(p^{4}+10p^{2})}{180p^{2}}+\frac{4n^{3}+n(q^{4}+10q^{2})}{180q^{2}}}$.

• (II)

If $n=p^{t}$ with $t\geq 2$ an integer and $h(d)=4$, then $\mathfrak{C}(k_{n})\cong\mathbb{Z}/4\mathbb{Z}$.

As expected, similar results for $d$ of the form $n^{2}+4$ are also obtained. We use partial zeta values attached to $k_{n}$ and generalized Dedekind sums which are discussed below. Technically speaking, the method adopted here should work for class groups of order $p^{2}$ for any prime number $p$, but it will be far more challenging to compute the zeta values, Dedekind sums, fundamental unit and appropriate ideal selection in such a situation.

2. Computation of partial Dedekind zeta values

Let $k$ be a real quadratic field, and $\zeta_{k}(s)$ be the Dedekind zeta function attached to $k$. By specializing Siegel’s formula [17] for $\zeta_{k}(1-2n)$ for general $k$, Zagier [19] described this formula by direct analytic methods when $k$ is a real quadratic field. This formula takes the following shape for $n=1$.

Theorem B** ([19, p. 69]).**

Let $k$ be a real quadratic field with discriminant $D$. Then

[TABLE]

where $\sigma(n)$ denotes the sum of divisors of $n$.

Lang [11] gave another method to calculate special values of $\zeta_{k}$. We briefly recall Lang’s formula. Recall that an integral basis for an integral ideal $\mathcal{I}$ of a number field $k$ is a set $\{\alpha_{1},\alpha_{2},\cdots,\alpha_{n}\}\subset\mathcal{I}$, where $n=[k:\mathbb{Q}]$, such that $\mathcal{I}=\mathbb{Z}\alpha_{1}\oplus\mathbb{Z}\alpha_{2}\oplus\cdots\oplus\mathbb{Z}\alpha_{n}$. For a given ideal class $\mathfrak{A}$ of $k$, let $\mathfrak{a}$ be an integral ideal in $\mathfrak{A}^{-1}$ with an integral basis $\{r_{1},r_{2}\}$, i.e. $\mathfrak{a}=\mathbb{Z}r_{1}\oplus\mathbb{Z}r_{2}$, where $r_{1},r_{2}\in\mathfrak{a}$. Set

[TABLE]

where $r_{1}^{\prime}$ and $r_{2}^{\prime}$ are the conjugates of $r_{1}$ and $r_{2}$ respectively.

Let $\varepsilon$ be the fundamental unit of $k$. Then $\{\varepsilon r_{1},\varepsilon r_{2}\}$ is also an integral basis of $\mathfrak{a}$, and thus we can find a matrix $M=\begin{bmatrix}a&b\\ c&d\end{bmatrix}$ with integer entries satisfying the following:

[TABLE]

The following result of Lang helps one to compute partial zeta value for $\mathfrak{A}$ at $-1$.

Theorem C** ([11, p. 159]).**

By keeping the above notations, we have

[TABLE]

where $N(\mathfrak{a})$ represents the norm of $\mathfrak{a}$ and $S^{i}(-,-)$ denotes the generalized Dedekind sum as defined in [1].

We need to determine the values of $a,b,c,d$ and generalized Dedekind sums in order to apply Theorem C. The following result (see, [11, p. 143, Eq. 2.15]) is helpful in determining the values of $a,b,c$ and $d$.

Lemma 2.1**.**

The matrix $M$ is given by

[TABLE]

Moreover, $\det(M)=N(\varepsilon)$ and $bc\neq 0$.

The following expressions (see, [11, pp. 155–157, Eq. 4.3–4.19]) involving special values of generalized Dedekind sums are also needed to compute partial zeta values for ideal classes of a real quadratic field.

Lemma 2.2**.**

For any positive integer $m$, we have

• (i)

$S^{3}(\pm 1,m)={\displaystyle\pm\frac{-m^{4}+5m^{2}-4}{120m^{3}}}.$**

• (ii)

$S^{2}(\pm 1,m)={\displaystyle\frac{m^{4}+10m^{2}-6}{180m^{3}}}.$**

Lemma 2.3**.**

For any positive even integer $m$, we have

• (i)

$S^{3}(m\pm 1,2m)=\pm S^{1}(m+1,2m)={\displaystyle\mp\frac{m^{4}-50m^{2}+4}{960m^{3}}}.$**

• (ii)

$S^{2}(m-1,2m)=S^{2}(m+1,2m)={\displaystyle\frac{m^{4}+100m^{2}-6}{1440m^{3}}}.$**

We determine Dedekind zeta values attached to $k_{n}$ in two ways using Theorem B and Theorem C and then compare these values and finally use elementary group theoretic arguments to establish our results.

3. Proof of the results

Throughout this section $d=n^{2}+1$ is square-free; $h(d)$ and $\mathfrak{C}(k_{n})$ denote the class number and the class group of $k_{n}=\mathbb{Q}(\sqrt{d})$ respectively. Note that $d\equiv 1,2,5,6\pmod{8}$ in this situation. For us $\mathfrak{P}$ will always refer to principal ideal class in the corresponding class group. Let

[TABLE]

The following proposition will be needed in the subsequent results.

Proposition 3.1**.**

If $\mathcal{N}\geq 3$ then

[TABLE]

Proof.

We will provide the complete proof for the case $d\equiv 1\pmod{8}$ and other cases can be handled along the same lines. In this case $2$ splits in $k_{n}$ as:

[TABLE]

Also $p_{i}\in\mathfrak{S}_{n}$ splits in $k_{n}$ as:

[TABLE]

Let $\mathfrak{A}_{i}$ and $\mathfrak{B}$ be two ideal classes in $k_{n}$ such that $\mathfrak{a}_{i}=\left(p_{i},\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{A}_{i}$ and $\mathfrak{b}\left(2,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{B}$. Then $\mathfrak{a}_{i}^{-1}=\left(p_{i},\frac{1-\sqrt{d}}{2}\right)\in\mathfrak{A}_{i}^{-1}$ and $\mathfrak{b}^{-1}=\left(2,\frac{1-\sqrt{d}}{2}\right)\in\mathfrak{B}^{-1}$. Consider a non-zero $\mathbb{Z}$-module M_{i}=\Big{[}p_{i},\frac{1-\sqrt{d}}{2}\Big{]} in $\mathcal{O}_{k_{n}}$. Then, by [14, Propositions 2.6 and 2.11], $M_{i}$ is an integral ideal and $N(M_{i})=p_{i}$. Also $N(\mathfrak{a}_{i}^{-1})=p_{i}$ and $M_{i}\subseteq\mathfrak{a}_{i}^{-1}$, therefore  $M_{i}=\mathfrak{a}_{i}^{-1}$. Hence $\{p_{i},\frac{1-\sqrt{d}}{2}\}$ is an integral basis for $\mathfrak{a}_{i}^{-1}$, i.e. $\mathfrak{a}_{i}^{-1}=\mathbb{Z}p_{i}\oplus\mathbb{Z}\frac{(1-\sqrt{d})}{2}$. Similarly, $\{2,\frac{(1-\sqrt{d})}{2}\}$ is an integral basis for $\mathfrak{b}^{-1}$, i.e. $\mathfrak{b}^{-1}=\mathbb{Z}2\oplus\mathbb{Z}\frac{(1-\sqrt{d})}{2}$. Now using Lemma 2.1, Lemma 2.2 and Theorem C, we obtain

[TABLE]

and

[TABLE]

Also by [4, Theorem 2.3],

[TABLE]

If $\zeta_{k_{n}}(-1,\mathfrak{A}_{i})=\zeta_{k_{n}}(-1,\mathfrak{P})$, then we must have $n=2p_{i}$. This contradicts the fact that $\mathcal{N}\geq 3$. Similarly, $\zeta_{k_{n}}(-1,\mathfrak{B})=\zeta_{k_{n}}(-1,\mathfrak{P})$ gives $n=4$, which is again a contradiction. Also, $\zeta_{k_{n}}(-1,\mathfrak{B})=\zeta_{k_{n}}(-1,\mathfrak{A}_{i})$ implies $n=4p_{i}$, which is not possible. Finally if $\zeta_{k_{n}}(-1,\mathfrak{A}_{i})=\zeta_{k_{n}}(-1,\mathfrak{A}_{j})$ for $i\neq j$, then $n=2p_{i}p_{j}$. This again contradicts the fact that $\mathcal{N}\geq 3$.

Therefore $\mathfrak{A}_{i}~{}(1\leq i\leq\mathcal{N})$ and $\mathcal{B}$ are distinct non-principal ideal classes in $k_{n}$, and thus $h(d)\geq\mathcal{N}+2$. ∎

As a consequence, we obtain the following interesting result.

Corollary 3.1**.**

$h(d)\to\infty$* as $\mathcal{N}\to\infty$.*

Proof of Theorem 1.1

We give the proof of the first part in details, and then we give the outline of the second part. In the case $n=2^{m}p^{t}$, both $2$ and $p$ split in $k_{n}$ as in (3.1) and (3.2) respectively.

Let $\mathfrak{A}$ be the ideal class in $k_{n}$ such that $\mathfrak{a}=\left(p,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{A}$. By multiplication formula for ideals in quadratic fields [15, p. 48], we will have $\left(p,\frac{1+\sqrt{d}}{2}\right)^{2}=\left(p^{2},\frac{1+\sqrt{d}}{2}\right)$ and $\left(p,\frac{1-\sqrt{d}}{2}\right)^{2}=\left(p^{2},\frac{1-\sqrt{d}}{2}\right)$. Then $\mathfrak{a}^{-1}=\left(p,\frac{1-\sqrt{d}}{2}\right)\in\mathfrak{A}^{-1}$, $\mathfrak{a}^{2}=\left(p^{2},\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{A}^{2}$ and $(\mathfrak{a}^{2})^{-1}=\left(p^{2},\frac{1-\sqrt{d}}{2}\right)\in(\mathfrak{A}^{2})^{-1}$. Consider nonzero $\mathbb{Z}$-modules M_{r}=\Big{[}p^{r},\frac{1-\sqrt{d}}{2}\Big{]} in $\mathcal{O}_{k_{n}}$, where $r\in\{1,2\}$. Then again by [14, Propositions 2.6 and 2.11], $M_{r}$ is an ideal and $N(M_{r})=p^{r}$, for all $r\in\{1,2\}$. Also $N((\mathfrak{a}^{r})^{-1})=p^{r}$ and $M_{r}\subseteq(\mathfrak{a}^{r})^{-1}$, for all $r\in\{1,2\}$. Therefore $M_{r}=(\mathfrak{a}^{r})^{-1}$, for all $r\in\{1,2\}$. Hence $\{p,\frac{1-\sqrt{d}}{2}\}$ and $\{p^{2},\frac{1-\sqrt{d}}{2}\}$ are an integral basis for $\mathfrak{a}^{-1}$ and $(\mathfrak{a}^{2})^{-1}$ respectively. Now using Lemma 2.1, Lemma 2.2 and Theorem C, we obtain

[TABLE]

and

[TABLE]

Also $\zeta_{k_{n}}(-1,\mathfrak{P})$ is given by (3.3). We claim that $\mathfrak{A}$ is a generator of $\mathfrak{C}(k_{n})$. To see this we pairwise equate $\zeta_{k_{n}}(-1,\mathfrak{P})$, $\zeta_{k_{n}}(-1,\mathfrak{A})$ and $\zeta_{k_{n}}(-1,\mathfrak{A}^{2})$ to find that $n=2p,2p^{2}$. These values are not possible since $m,t\geq 2.$ Thus order of $\mathfrak{A}$ is greater than $2$. This establishes our claim as $h(d)=4$. Therefore $\mathfrak{C}(k_{n})\cong\mathbb{Z}/4\mathbb{Z}$.

Now

[TABLE]

Thus,

[TABLE]

This completes the proof of (I) of Theorem 1.1.

In the case $n=2^{m}p^{s}q^{t}$, both $p$ and $q$ split as in (3.2), and $2$ splits as in (3.1). Let $\mathfrak{A},~{}\mathfrak{B}$ and $\mathfrak{C}$ be the three ideal classes in $k_{n}$ such that $\left(p,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{A}$, $\left(q,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{B}$ and $\left(2,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{C}$. Then analogous to the previous case, we obtain

[TABLE]

Equating pairwise these values as before, we get

[TABLE]

Since $h(d)=4$, $2^{2}|n$ and $s,t\geq 1$, these values of $n$ do not arise. Hence $\mathfrak{A},~{}\mathfrak{B}$ and $\mathfrak{C}$ are distinct non-principal ideal classes in $k_{n}$. Using these three non-principal ideal classes, we can prove (II) of Theorem 1.1 following similar arguments.

Proof of Theorem 1.2

We first consider the case $n=2p^{t}$ for some integer $t\geq 3$, and thus $p$ splits in $k_{n}$ similar to (3.2).

Let $\mathfrak{A}$ be an ideal class containing $\mathfrak{a}=\left(p,\frac{1+\sqrt{d}}{2}\right)$. Then again by multiplication formula for ideals in quadratic fields [15, p. 48], $\mathfrak{a}^{2}=\left(p,\frac{1+\sqrt{d}}{2}\right)^{2}=\left(p^{2},\frac{1+\sqrt{d}}{2}\right)$ and $(\mathfrak{a}^{2})^{-1}=\left(p,\frac{1-\sqrt{d}}{2}\right)^{2}=\left(p^{2},\frac{1-\sqrt{d}}{2}\right)$. Thus $\mathfrak{a}^{2}=\left(p^{2},\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{A}^{2}$, $\mathfrak{a}^{-1}=\left(p,\frac{1-\sqrt{d}}{2}\right)\in\mathfrak{A}^{-1}$ and $(\mathfrak{a}^{2})^{-1}=\left(p^{2},\frac{1-\sqrt{d}}{2}\right)\in(\mathfrak{A}^{2})^{-1}$. As proved in the proof of Theorem 1.1, $\{p,\frac{1-\sqrt{d}}{2}\}$ and $\{p^{2},\frac{1-\sqrt{d}}{2}\}$ are integral basis for $\mathfrak{a}^{-1}$ and $(\mathfrak{a}^{2})^{-1}$ respectively. Now by using Lemma 2.1, Lemma 2.2 and Theorem C we obtain:

[TABLE]

As in §2.1, we observe that $\mathfrak{A}$ and $\mathfrak{A}^{2}$ are distinct non-principal ideals classes in $k_{n}$ since $t\geq 3$. This implies that the order of $\mathfrak{A}$ in $\mathfrak{C}(k_{n})$ is greater than $2$. Therefore the order of $\mathfrak{A}$ in $\mathfrak{C}(k_{n})$ is $4$ and hence $\mathfrak{C}(k_{n})\cong\mathbb{Z}/4\mathbb{Z}$.

Since $\mathfrak{A}$ is a generator and $\mathfrak{P}$ is principal in $\mathfrak{C}(k_{n})$, we have

[TABLE]

Thus,

[TABLE]

This completes the proof of (I) of Theorem 1.2.

In the next case, both $p$ and $q$ split as in (3.2). Let $\mathfrak{A}$ and $\mathfrak{B}$ are two ideal classes in $k_{n}$ such that $\left(p,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{A}$ and $\left(q,\frac{1+\sqrt{d}}{2}\right)\in\mathfrak{B}$. As in §2.1, we see that both $\mathfrak{A}$ and $\mathfrak{B}$ are distinct non-principal in $\mathfrak{C}(k_{n})$ since either $s>2$ or $t>2$.

As $\left(p,\frac{1-\sqrt{d}}{2}\right)\in\mathfrak{A}^{-1}$ and $\left(q,\frac{1-\sqrt{d}}{2}\right)\in\mathfrak{B}^{-1}$, we have

[TABLE]

Also $\zeta_{k_{n}}(-1,\mathfrak{A}^{-1})=\zeta_{k_{n}}(-1,\mathfrak{A})$ and $\zeta_{k_{n}}(-1,\mathfrak{B}^{-1})=\zeta_{k_{n}}(-1\mathfrak{B})$. If $\mathfrak{C}(k_{n})\cong\mathbb{Z}/4\mathbb{Z}$, then either $\mathfrak{A}$ or $\mathfrak{B}$ is a generator. Let it be $\mathfrak{A}$. Then $\mathfrak{B}\neq\mathfrak{A}^{-1}$, as the corresponding partial Dedekind zeta values are not equal. This forces that $\mathfrak{A}^{2}=\mathfrak{B}$, and therefore $\zeta_{k_{n}}(-1,\mathfrak{A}^{2})=\zeta_{k_{n}}(-1,\mathfrak{B})$, i.e.

[TABLE]

On simplification, we get $n=2p^{2}q$, which contradicts the assumption on $n$, and hence contradiction to our assumption that $\mathfrak{C}(k_{n})\cong\mathbb{Z}/4\mathbb{Z}$. Therefore the only possibility is that, the order of each of $\mathfrak{A}$ and $\mathfrak{B}$ must be $2$. This completes the proof of (II) of Theorem 1.2.

In the last case, that is when $n=2p^{r}q^{s}\ell^{t}$, we see that $p,q$ and $\ell$ split in $k_{n}$ as in (3.2). If $\mathfrak{A},~{}\mathfrak{B}$ and $\mathfrak{C}$ are ideal classes in $k_{n}$ containing $\left(p,\frac{1+\sqrt{d}}{2}\right),~{}\left(q,\frac{1+\sqrt{d}}{2}\right)$ and $\left(\ell,\frac{1+\sqrt{d}}{2}\right)$ respectively, then similar to the previous cases,

[TABLE]

Proceeding as before one can show that these ideal classes are distinct and non-principal in $k_{n}$. Using these ideal classes, we can prove (III) of Theorem 1.2.

Proof of Theorem 1.3

Recall that $n$ has at most two prime factors. Also, the zeta value at $-1$ for $\mathfrak{P}$ is given by [4, Theorem 2.3], i.e.,

[TABLE]

We give the proof of (I) of Theorem 1.3, and the analogous arguments work for the remaining part.

Since $p,q\mid n$, so that both $p$ and $q$ split in $k_{n}$, that is,

[TABLE]

Also $(2)=(2,\sqrt{d})^{2}$. Let $\mathfrak{A},~{}\mathfrak{B}$ and $\mathfrak{C}$ be ideal classes in $k_{n}$ such that $(2,\sqrt{d})\in\mathfrak{A}$, $(p,1+\sqrt{d})\in\mathfrak{B}$ and $(q,1+\sqrt{d})\in\mathfrak{C}$. Now if we consider non-zero $\mathbb{Z}$-modules $N=[p,1-\sqrt{d}]$ and $N^{\prime}=[q,1-\sqrt{d}]$ in $\mathcal{O}_{k_{n}}$, then as before (using [14, Proposition 2.6 and 2.11]) one can prove that $N=(p,1-\sqrt{d})$ and $N^{\prime}=(q,1-\sqrt{d})$. Hence, $\{p,1-\sqrt{d}\}$ and $\{q,1-\sqrt{d}\}$ are an integral basis for $(p,1-\sqrt{d})$ and $(q,1-\sqrt{d})$ respectively. Now as before:

[TABLE]

Employing similar technique, we see that all these three ideal classes are distinct and non-principal in $k_{n}$.

Clearly $\mathfrak{C}$ is of order $2$. We claim that both $\mathfrak{A}$ and $\mathfrak{B}$ are of order $2$. We observe from (3.4) that $(p,1-\sqrt{d})\in\mathfrak{A}^{-1}$ and $(q,1-\sqrt{d})\in\mathfrak{B}^{-1}$. We see that $\zeta_{k_{n}}(-1,\mathfrak{A})=\zeta_{k_{n}}(-1,\mathfrak{A}^{-1})$ and $\zeta_{k_{n}}(-1,\mathfrak{B})=\zeta_{k_{n}}(-1,\mathfrak{B}^{-1})$. Also $\zeta_{k_{n}}(-1,\mathfrak{A}^{-1})\neq\zeta_{k_{n}}(-1,\mathfrak{B}^{-1})\neq\zeta_{k_{n}}(-1,\mathfrak{C})$ and hence our claim is proved. Therefore $\mathfrak{C}(k_{n})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

Also

[TABLE]

Therefore

[TABLE]

This completes the proof of (I) of Theorem 1.3.

In the case when $n$ is power of a single prime, i.e., $n=p^{t}$ for some integer $t\geq 3$, $p$ splits as in the last case. Following the previous arguments, one can prove (II) of Theorem 1.3.

4. Concluding remarks

Real quadratic fields of the form $k_{n,r}=\mathbb{Q}(\sqrt{n^{2}+r})$ (with $n^{2}+r$ square-free) are called Richaud-Degert(R-D) type if $r|4n$ for two integers $n$ and $r$. Furthermore, if $|r|=1,4$, then $k_{n,r}$ is known as narrow R-D type; otherwise $k_{n,r}$ is known as extended R-D type field. We have obtained sufficient conditions to specify the structure of order $4$ class groups in $k_{n,r}$ when $r=1$. Following the same method, we can obtain sufficient conditions to specify order $4$ class groups of $k_{n,r}$ when $r=4$. More precisely:

Theorem 4.1**.**

Let $d=n^{2}+4$ be a square-free integer, and let $h(d)=4$. The following statements hold:

• (i)

If $n=p^{t}$ with $t\geq 2$, then $\mathfrak{C}(k_{n,4})\cong\mathbb{Z}/4\mathbb{Z}$.

• (ii)

If $n=p^{r}q^{s}$ with $r\geq 2$ or $s\geq 2$ then $\mathfrak{C}(k_{n,4})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

• (iii)

If $n$ has more than two distinct odd prime factors, then $\mathfrak{C}(k_{n,4})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

Remark 4.1**.**

The method followed here may not work for $r=-1$ and $-4$. The ideals used here will not be helpful and it will not be easy to calculate generalized Dedekind sums for other ideals. However, it would be interesting to extend our method to extended R-D type real quadratic fields.

Acknowledgements

The authors would like to express their gratitude to Professor Claude Levesque for carefully reading this manuscript and for his useful comments. The second author is grateful to Professor Srinivas Kotyada for stimulating environment at The Institute of Mathematical Sciences, Chennai during his visiting period. The authors are thankful to the anonymous referees for their valuable comments and suggestions which have helped improving the presentation immensely. The second author acknowledges the grant SERB MATRICS Project (No. MTR/2017/00100). The third author is partially supported by ‘Infosys grant’.