Minimal Polynomials of Some Eta-Quotients Evaluated at CM Points

Abstract

We study certain eta-quotients of weight zero evaluated at CM points of imaginary quadratic orders. Using the theory of extended form class groups, we show that these special values generate the corresponding ring class fields and we provide explicit descriptions of their minimal polynomials. Finally, we apply these results to certain Diophantine problems.

1. Introduction

Let $\mathbb{H}$ denote the complex upper half-plane, namely,

$$\mathbb{H}=\{\tau \in \mathbb{C}|\mathrm{Im}\left(\tau \right)>0\}.$$

The Dedekind eta-function on $\mathbb{H}$ is defined by the infinite product

$$\eta \left(\tau \right)=q^{1/24}\prod _{m=1}^{\infty }(1-q^m)(\tau \in \mathbb{H},q=e^{2\pi \mathrm{i}\tau }).$$

(1)

For a positive integer N, a function of the form

$$\prod _{d|N}\eta {\left(d\tau \right)}^{m_d}(m_d\in \mathbb{Z})$$

is called an eta-quotient, whose modularity was investigated by Newman [1,2] and Gordon-Sinor [3]. Such eta-quotients can be used to construct bases for certain spaces of modular forms for the congruence subgroup

$${\Gamma }_0\left(N\right)=\left\{\gamma =\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in {\mathrm{SL}}_2\left(\mathbb{Z}\right)|c\equiv 0\left(\mathrm{mod}N\right)\right\}$$

(see [4,5,6]).

The aim of this paper is to enrich the explicit class field theory for imaginary quadratic fields in terms of eta-quotients. More precisely, we consider eta-quotients of weight 0 evaluated at CM points and study their minimal polynomials.

Let K be an imaginary quadratic field with the ring of integers ${\mathcal{O}}_K$, and let $\mathcal{O}$ be an order in K with conductor ${\ell }_{\mathcal{O}}$ and discriminant $D_{\mathcal{O}}$. Define the element ${\tau }_{\mathcal{O}}$ in $\mathbb{H}$ by

$${\tau }_{\mathcal{O}}=\left\{\begin{array}{cc}{\displaystyle \frac{-1+\sqrt{D_{\mathcal{O}}}}{2}}& \mathrm{if}{D}_{\mathcal{O}}\equiv 1\left(\mathrm{mod}4\right),\hfill \\ {\displaystyle \frac{\sqrt{D_{\mathcal{O}}}}{2}}& \mathrm{if}{D}_{\mathcal{O}}\equiv 0\left(\mathrm{mod}4\right)\hfill \end{array}\right.$$

(2)

so that $\mathcal{O}=\mathbb{Z}{\tau }_{\mathcal{O}}+\mathbb{Z}$. Eum et al. proved that if $N\equiv 0\left(\mathrm{mod}4\right)$ and ${\eta }_N$ is the eta-quotient given by

$${\eta }_N\left(\tau \right)=4^4{\displaystyle \frac{\eta {\left(N\tau \right)}^8}{\eta {\left((N/4)\tau \right)}^8}}(\tau \in \mathbb{H}),$$

then the special value ${\eta }_N\left({\tau }_{{\mathcal{O}}_K}\right)$ is a real algebraic integer and generates the ring class field of the order of conductor N over K ([7], Theorem 4.5) (see also [8]). In this paper, we extend this result to CM points associated with arbitrary quadratic orders and we further describe the Galois actions concisely by means of the extended form class group. Our main theorem is as follows.

Theorem 1

(Theorem 3). Assume that $N\equiv 0\left(\mathrm{mod}4\right)$.

(i)

The special value ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ is a real algebraic integer.

(ii)

It generates the ring class field of the order of conductor $N{\ell }_{\mathcal{O}}$ over K.

(iii)

Let $Q_1,Q_2,\dots ,Q_s$ be all the reduced binary quadratic forms of discriminant $D_{\mathcal{O}}$. Furthermore, let ${\gamma }_1,{\gamma }_2,\dots ,{\gamma }_t$ be a complete set of representatives for the left cosets of ${\Gamma }_0\left(N\right)$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$. If $D_{\mathcal{O}}\ne -3,-4$, then the minimal polynomial of ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ over K is given by

$$\prod _{(i,k)\in S_{\mathcal{O},N}}(x-{\eta }_N\left(\widehat{{\gamma }_k}(-\overline{{\omega }_{Q_i}})\right))(\in \mathbb{Z}\left[x\right]),$$

where

$$S_{\mathcal{O},N}=\left\{(i,k)\in \mathbb{Z}\times \mathbb{Z}|1\le i\le s,1\le k\le t,Q_i\left({\gamma }_k\left[\begin{array}{c}x\\ y\end{array}\right]\right)\in \mathcal{Q}(D_{\mathcal{O}},N)\right\}.$$

For the extended form class group $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_0\left(N\right)}$, developed in [9,10], see Section 3. Furthermore, for $\gamma =\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in {\mathrm{SL}}_2\left(\mathbb{Z}\right)$, we denote by $\widehat{\gamma }=\left[\begin{array}{cc}d& b\\ c& a\end{array}\right]$. We note that Theorem 1 (iii) can be viewed as a refinement of ([11], Section 6). We then present a couple of examples illustrating Theorem 1 (iii) and conclude with an application of our results to the primes of the form $x^2+n{y}^2$.

2. Class Fields for Orders

Throughout this paper, we use the following notation:

• K: An imaginary quadratic field.
• $\mathcal{O}$: An order in K.
• ${\ell }_{\mathcal{O}}$: The conductor of $\mathcal{O}$.
• $D_{\mathcal{O}}$: The discriminant of $\mathcal{O}$.
• N: A positive integer.

In this section, we shall introduce class fields for $\mathcal{O}$ and their construction.

Let $I\left(\mathcal{O}\right)$ be the group of proper fractional $\mathcal{O}$-ideals, and let $P\left(\mathcal{O}\right)$ be its subgroup consisting of principal fractional $\mathcal{O}$-ideals (cf. [12], Section 7.A). For each subgroup G of ${(\mathbb{Z}/N\mathbb{Z})}^{*}$, we define

$${\mathcal{C}}_G(\mathcal{O},N)=I(\mathcal{O},N)/P_G(\mathcal{O},N)$$

where $I(\mathcal{O},N)$ and $P_G(\mathcal{O},N)$ are subgroups of $I\left(\mathcal{O}\right)$ and $P\left(\mathcal{O}\right)$, respectively, given by

$$\begin{array}{cc}\hfill I(\mathcal{O},N)& =\langle \mathfrak{a}|\mathfrak{a},\mathrm{a}\mathrm{nontrivial}\mathrm{proper}\mathcal{O}-\mathrm{ideal}\mathrm{prime}\mathrm{to}N\rangle ;\hfill \\ \hfill P_G(\mathcal{O},N)& =\langle \nu \mathcal{O}|\nu \in \mathcal{O}\setminus \{0\}\mathrm{and}\nu \equiv t(\mathrm{mod}N\mathcal{O})\mathrm{for}\mathrm{some}t\in \mathbb{Z}\mathrm{satisfying}t+N\mathbb{Z}\in G\rangle .\hfill \end{array}$$

As shown in ([9], Corollary 2.8), the group ${\mathcal{C}}_G(\mathcal{O},N)$ is isomorphic to a generalized ideal class group of K modulo $N{\ell }_{\mathcal{O}}{\mathcal{O}}_K$. Thus, by the existence theorem of class field theory (cf. [13], V.8), there exists a unique abelian extension $K_{\mathcal{O},G}$ of K such that

(i)

Every prime of K ramified in $K_{\mathcal{O},G}$ divides ${\ell }_{\mathcal{O}}N{\mathcal{O}}_K$;

(ii)

$\mathrm{Gal}(K_{\mathcal{O},G}/K)$ is isomorphic to the generalized ideal class group via the Artin map for the modulus ${\ell }_{\mathcal{O}}N{\mathcal{O}}_K$.

In particular, if $N=1$ (so $G={(\mathbb{Z}/\mathbb{Z})}^{*}$), then ${\mathcal{C}}_G(\mathcal{O},N)$ is the usual $\mathcal{O}$-ideal class group $I\left(\mathcal{O}\right)/P\left(\mathcal{O}\right)$. Note further that

$$I\left(\mathcal{O}\right)/P\left(\mathcal{O}\right)\simeq I({\mathcal{O}}_K,{\ell }_{\mathcal{O}})/P_{{(\mathbb{Z}/{\ell }_{\mathcal{O}}\mathbb{Z})}^{*}}({\mathcal{O}}_K,{\ell }_{\mathcal{O}})$$

(3)

(cf. [12], Proposition 7.22). In this case, $K_{\mathcal{O},G}$ is called the ring class field of the order $\mathcal{O}$, simply denoted by $K_{\mathcal{O}}$. Let j be the elliptic modular function with Fourier expansion

$$j\left(\tau \right)={\displaystyle \frac{1}{q}}+744+196884q+21493760q^2+\cdots (\tau \in \mathbb{H}).$$

As a consequence of the first main theorem of the theory of complex multiplication, we obtain the following.

Proposition 1.

With ${\tau }_{\mathcal{O}}$ as in (2), the singular modulus $j\left(N{\tau }_{\mathcal{O}}\right)=j\left(N{\ell }_{\mathcal{O}}{\tau }_{{\mathcal{O}}_K}\right)$ generates $K_{{\mathcal{O}}^{\prime }}$ over K, where ${\mathcal{O}}^{\prime }$ is the order of conductor $N{\ell }_{\mathcal{O}}$ in K.

Proof. 

Observe that

$$\mathbb{Z}\left(N{\tau }_{\mathcal{O}}\right)+\mathbb{Z}=\mathbb{Z}\left(N{\ell }_{\mathcal{O}}{\tau }_{{\mathcal{O}}_K}\right)+\mathbb{Z}={\mathcal{O}}^{\prime }.$$

The proposition follows from ([14], Theorem 5 in Chapter 10). □

Remark 1.

Note that $j\left(N\tau \right)\in {\mathcal{F}}_{{\Gamma }_0\left(N\right),\mathbb{Q}}$ (cf. [12], Theorem 11.9).

For a congruence subgroup $\Gamma$ of ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$, let ${\mathcal{F}}_{\Gamma ,\mathbb{Q}}$ denote the field of meromorphic modular functions for $\Gamma$ with rational Fourier coefficients (cf. [15], Section 2.1). Define

$${\Gamma }_G=\left\{\gamma \in {\mathrm{SL}}_2\left(\mathbb{Z}\right)|\gamma \equiv \left[\begin{array}{cc}t& *\\ 0& *\end{array}\right]\left(\mathrm{mod}N{M}_2\left(\mathbb{Z}\right)\right)\mathrm{for}\mathrm{some}\mathrm{t}\in \mathbb{Z}\mathrm{such}\mathrm{that}t+N\mathbb{Z}\in G\right\}.$$

Here, for matrices $\gamma$, $\delta \in M_2\left(\mathbb{Z}\right)$, the notation $\gamma \equiv \delta (modN{M}_2\left(\mathbb{Z}\right))$ means that $\gamma -\delta \in N{M}_2\left(\mathbb{Z}\right)$.

Using the theory of Shimura’s canonical models ([15], Chapter 6), we obtain the following.

Proposition 2.

We have

$$K_{\mathcal{O},G}=K\left(f\left({\tau }_{\mathcal{O}}\right)\right|f\in {\mathcal{F}}_{{\Gamma }_G,\mathbb{Q}}is finite at{\tau }_{\mathcal{O}}).$$

Proof. 

See ([9], Theorem 3.5). □

3. Extended Form Class Groups

Following [9,10], we shall review explicit class field theory via extended form class groups.

Let D be a negative integer such that $D\equiv 0$ or $1\left(\mathrm{mod}4\right)$. Define

$$\mathcal{Q}(D,N)=\left\{Q(x,y)=Q\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)=a{x}^2+bxy+c{y}^2\in \mathbb{Z}[x,y]|\begin{array}{c}gcd(a,b,c)=1,\hfill \\ b^2-4ac=D,a>0,\hfill \\ gcd(a,N)=1\hfill \end{array}\right\}.$$

The congruence subgroup ${\Gamma }_G$ naturally acts on the set $\mathcal{Q}(D,N)$, inducing the equivalence relation ${\sim }_{{\Gamma }_G}$ defined by the following:

• For $Q,Q^{\prime }\in \mathcal{Q}(D,N)$,

  $$Q{\sim }_{{\Gamma }_G}{Q}^{\prime }⟺Q^{\prime }\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)=Q{\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)}^{\gamma }=Q\left(\gamma \left[\begin{array}{c}x\\ y\end{array}\right]\right)\mathrm{for}\mathrm{some}\gamma \in {\Gamma }_G.$$

For $Q(x,y)=a{x}^2+bxy+c{y}^2\in \mathcal{Q}(D,N)$, let ${\omega }_Q$ be the root of the quadratic polynomial $Q(x,1)=a{x}^2+bx+c$ lying in $\mathbb{H}$, i.e.,

$${\omega }_Q={\displaystyle \frac{-b+\sqrt{D}}{2a}}.$$

Proposition 3.

The set of equivalence classes $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$ can be endowed with a group structure such that the map

$$\begin{array}{cc}\hfill \mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}& \to \mathrm{Gal}(K_{\mathcal{O},G}/K)\hfill \\ \hfill \left[Q\right]& \mapsto \left(f\left({\tau }_{\mathcal{O}}\right)\mapsto f(-\overline{{\omega }_Q})|f\in {\mathcal{F}}_{{\Gamma }_G,\mathbb{Q}}is finite at{\tau }_{\mathcal{O}}\right)\hfill \end{array}$$

is a well-defined isomorphism. Here, $\overline{·}$ denotes the complex conjugation.

Proof. 

See Proposition 2 and ([9], Corollary 5.5). □

We call the group $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$ in Proposition 3 an extended form class group of discriminant $D_{\mathcal{O}}$ and level N, which extends the classical form class group $\mathcal{Q}(D_{\mathcal{O}},1)/{\sim }_{{\mathrm{SL}}_2\left(\mathbb{Z}\right)}$ due to Gauss (cf. [12], Section 7.B).

Remark 2.

(i)    Moreover, the map

$$\begin{array}{cc}\hfill \mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}& \to {\mathcal{C}}_G(\mathcal{O},N)\hfill \\ \hfill \left[Q\right]& \mapsto [\mathbb{Z}{\omega }_Q+\mathbb{Z}]\hfill \end{array}$$

is a well-defined isomorphism ([9], Theorem 5.4).

(ii) 

Let $f\in {\mathcal{F}}_{{\Gamma }_G,\mathbb{Q}}$ be finite at ${\tau }_{\mathcal{O}}$. By Proposition 3, we have

$$\mathrm{irr}(f\left({\tau }_{\mathcal{O}}\right),K)=\prod _{i=1}^h(x-f(-\overline{{\omega }_{Q_i}})),$$

where $Q_1,Q_2,\dots ,Q_h\in \mathcal{Q}(D_{\mathcal{O}},N)$ form a complete set of representatives for the elements of the extended form class group $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$. Since $e^{2\pi \mathrm{i}{\tau }_{\mathcal{O}}}$ is real, $f\left({\tau }_{\mathcal{O}}\right)$ is real as well. Hence $\mathrm{irr}(f\left({\tau }_{\mathcal{O}}\right),K)$ has coefficients in $\mathbb{Q}$.

4. Representatives for the Elements of an Extended Form Class Group

Using the argument of ([16], Section 6), we shall present a complete set of representatives for the elements of the extended form class group $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$ under the assumption that $D_{\mathcal{O}}\ne -3,-4$.

Lemma 1.

If $D_{\mathcal{O}}\ne -3,-4$, then the isotropy group of $Q\in \mathcal{Q}(D_{\mathcal{O}},1)$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$ is $\{\pm I_2\}$.

Proof. 

See ([17], Proposition 1.5 (c)). □

Theorem 2.

Let $Q_1,Q_2,\dots ,Q_s$ be all the reduced binary quadratic forms of discriminant $D_{\mathcal{O}}$. Furthermore, let ${\gamma }_1,{\gamma }_2,\dots ,{\gamma }_t$ be a complete set of representatives for the left cosets of the subgroup $\pm {\Gamma }_G$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$, i.e.,

$${\mathrm{SL}}_2\left(\mathbb{Z}\right)={\gamma }_1\Gamma {\displaystyle ⊍}{\gamma }_2\Gamma {\displaystyle ⊍}\cdots {\displaystyle ⊍}{\gamma }_t\Gamma with\Gamma =\pm {\Gamma }_G.$$

(4)

If $D_{\mathcal{O}}\ne -3,-4$, then the forms

$$Q_i^{{\gamma }_k}(1\le i\le s,1\le k\le tsuch that{Q}_i^{{\gamma }_k}=Q_i\left({\gamma }_k\left[\begin{array}{c}x\\ y\end{array}\right]\right)belongs to\mathcal{Q}(D_{\mathcal{O}},N))$$

(5)

represent all distinct elements of the extended form class group $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$.

Proof. 

Let $C\in \mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$. Then

$$C=\left[Q\right]\mathrm{for}\mathrm{some}Q\in \mathcal{Q}(D_{\mathcal{O}},N).$$

(6)

Since

$$\mathcal{Q}(D_{\mathcal{O}},1)/{\sim }_{{\mathrm{SL}}_2\left(\mathbb{Z}\right)}=\{\left[Q_1\right],\left[Q_2\right],\dots ,\left[Q_s\right]\}\mathrm{of}\mathrm{order}s$$

(7)

(cf. [12], Theorem 2.8), we have

$$Q=Q_i^{\alpha }\mathrm{for}\mathrm{some}\alpha \in {\mathrm{SL}}_2\left(\mathbb{Z}\right).$$

(8)

From the left coset decomposition (4) of ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$, it follows that

$$\alpha ={\gamma }_k\beta \mathrm{for}\mathrm{some}1\le k\le t\mathrm{and}\beta \in \Gamma =\pm {\Gamma }_G.$$

(9)

By (6), (8), and (9), we obtain

$$C=\left[Q_i^{{\gamma }_k\beta }\right]=\left[{\left(Q_i^{{\gamma }_k\beta }\right)}^{{\beta }^{-1}}\right]=\left[Q_i^{{\gamma }_k}\right].$$

Hence every element of $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$ can be represented by a form of the type given in (5).

On the other hand, suppose that in the form class group $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$

$$\left[Q_i^{{\gamma }_k}\right]=\left[Q_{i^{\prime }}^{{\gamma }_{k^{\prime }}}\right]\mathrm{for}\mathrm{some}1\le \mathrm{i},{\mathrm{i}}^{\prime }\le \mathrm{s}\mathrm{and}1\le \mathrm{k},{\mathrm{k}}^{\prime }\le \mathrm{t}\mathrm{such}\mathrm{that}{\mathrm{Q}}_{\mathrm{i}}^{{\gamma }_{\mathrm{k}}},{\mathrm{Q}}_{{\mathrm{i}}^{\prime }}^{{\gamma }_{{\mathrm{k}}^{\prime }}}\in \mathcal{Q}({\mathrm{D}}_{\mathcal{O}},\mathrm{N}).$$

Then

$$Q_{i^{\prime }}^{{\gamma }_{k^{\prime }}}={\left(Q_i^{{\gamma }_k}\right)}^{\gamma }\mathrm{for}\mathrm{some}\gamma \in {\Gamma }_G.$$

This forces—by (7)—that $i^{\prime }=i$, and hence,

$$Q_i^{{\gamma }_{k^{\prime }}}=Q_i^{{\gamma }_k\gamma }.$$

Since the isotropy group of $Q_i$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$ is $\{\pm I_2\}$ by Lemma 1, we obtain

$${\gamma }_{k^{\prime }}={\gamma }_k\gamma \mathrm{or}{\gamma }_{k^{\prime }}=-{\gamma }_k\gamma .$$

Therefore, ${\gamma }_{k^{\prime }}\Gamma ={\gamma }_k\Gamma$ with $\Gamma =\pm {\Gamma }_G$. By (4), it follows that $k^{\prime }=k$, which shows that no two forms in (5) represent the same element of $\mathcal{Q}(D_{\mathcal{O}},N)/{\sim }_{{\Gamma }_G}$. □

Remark 3.

(i)    Recall that a form $Q(x,y)=a{x}^2+bxy+c{y}^2\in \mathcal{Q}(D_{\mathcal{O}},1)$ is said to be reduced if

$$\left|b\right|\le a\le candb\ge 0if either\left|b\right|=a\mathrm{or}a=c.$$

One sees that if $Q(x,y)$ is reduced, then $a\le \sqrt{{\displaystyle \frac{|D_{\mathcal{O}}|}{3}}}$.

(ii) 

Consider the case where $G={(\mathbb{Z}/N\mathbb{Z})}^{*}$. Then we have $\pm {\Gamma }_G={\Gamma }_0\left(N\right)$. Let ${\mathrm{SL}}_2\left(\mathbb{Z}\right)/{\Gamma }_0\left(N\right)$ denote the set of left cosets of ${\Gamma }_0\left(N\right)$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$. There is a well-known bijection ([15], Proposition 1.43):

$$\begin{array}{cc}\hfill \{(c,d)\in {\mathbb{Z}}^2|gcd(N,c,d)=1\}/\sim & \to {\mathrm{SL}}_2\left(\mathbb{Z}\right)/{\Gamma }_0\left(N\right)\hfill \\ \hfill \left[\right(c,d\left)\right]& \mapsto [any\gamma \in {\mathrm{SL}}_2\left(\mathbb{Z}\right)satisfying\gamma \equiv \left[\begin{array}{cc}*& *\\ c& d\end{array}\right]\left(\mathrm{mod}N{M}_2\left(\mathbb{Z}\right)\right)],\hfill \end{array}$$

where ∼ is the equivalence relation defined by

$$(c,d)\sim (c^{\prime },d^{\prime })⟺(c,d)\equiv u(c^{\prime },d^{\prime })\left(\mathrm{mod}N{\mathbb{Z}}^2\right)for someu\in {(\mathbb{Z}/N\mathbb{Z})}^{*}.$$

Thus, we have $[{\mathrm{SL}}_2\left(\mathbb{Z}\right):{\Gamma }_0\left(N\right)]=N{\prod }_{p|N}\left(1+{\displaystyle \frac{1}{p}}\right)$.

5. Special Values of Some Eta-Quotients

In this section, we shall prove our main theorem concerning special values of eta-quotients.

Lemma 2.

If ${\left\{m_d\right\}}_{d|N}$ is a family of integers such that

(i)

${\sum }_{d|N}{m}_d=0$,

(ii)

${\sum }_{d|N}d{m}_d\equiv {\sum }_{d|N}(N/d)m_d\equiv 0\left(\mathrm{mod}24\right)$,

(iii)

${\prod }_{d|N}{d}^{m_d}$ is a rational square,

then the associated eta-quotient ${\prod }_{d|N}\eta {\left(d\tau \right)}^{m_d}$ is a weakly holomorphic modular function for ${\Gamma }_0\left(N\right)$ with rational Fourier coefficients.

Proof. 

See [2] or [18]. □

We focus on the specific eta-quotient

$${\eta }_N\left(\tau \right)=4^4{\displaystyle \frac{\eta {\left(N\tau \right)}^8}{\eta {\left((N/4)\tau \right)}^8}}(\tau \in \mathbb{H})$$

(10)

when $N\equiv 0\left(\mathrm{mod}4\right)$. It is a weakly holomorphic modular function for ${\Gamma }_0\left(N\right)$ with rational Fourier coefficients by the definition (1) and Lemma 2.

Lemma 3.

If $N\equiv 0\left(\mathrm{mod}4\right)$, then there is a pair of polynomials $A\left(x\right),B\left(x\right)\in \mathbb{Q}\left[x\right]$ satisfying

(i)

$j\left(N\tau \right)=A\left({\eta }_N\left(\tau \right)\right)/B\left({\eta }_N\left(\tau \right)\right)$,

(ii)

$B\left({\eta }_N\left({\tau }_{\mathcal{O}}\right)\right)\ne 0$.

Proof. 

(i)    Note that if we set $g\left(\tau \right)=4^4\eta {\left(4\tau \right)}^8/\eta {\left(\tau \right)}^8$, then the field ${\mathcal{F}}_{{\Gamma }_0\left(4\right),\mathbb{Q}}=\mathbb{Q}\left(g\left(\tau \right)\right)$ (see [7], Lemma 3.5). Since $j\left(4\tau \right)\in {\mathcal{F}}_{{\Gamma }_0\left(4\right),\mathbb{Q}}$, there exist relatively prime polynomials $A\left(x\right),B\left(x\right)\in \mathbb{Q}\left[x\right]$ such that

$$j\left(4\tau \right)={\displaystyle \frac{A\left(g\right(\tau \left)\right)}{B\left(g\right(\tau \left)\right)}}.$$

By substituting $(N/4)\tau$ for $\tau$, we obtain the assertion.

(ii)

From Proposition 2, we know that ${\eta }_N\left({\tau }_{\mathcal{O}}\right)\in K_{\mathcal{O},{(\mathbb{Z}/N\mathbb{Z})}^{*}}$. If $B\left({\eta }_N\left({\tau }_{\mathcal{O}}\right)\right)=0$, then $A\left({\eta }_N\left({\tau }_{\mathcal{O}}\right)\right)=0$, since $j\left(N\tau \right)$ is weakly holomorphic. However, this would imply that the minimal polynomial of ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ over $\mathbb{Q}$ divides both $A\left(x\right)$ and $B\left(x\right)$, which is impossible.

See also ([7], Lemmas 3.5 and 4.3). □

Lemma 4.

If M is a positive integer and ${\tau }_0\in \mathbb{H}$ is an imaginary quadratic argument, then the special value $M\eta {\left(M{\tau }_0\right)}^2/\eta {\left({\tau }_0\right)}^2$ is an algebraic integer dividing M.

Proof. 

See ([14], Theorem 4 in Chapter 12). □

Theorem 3.

Assume that $N\equiv 0\left(\mathrm{mod}4\right)$:

(i)

The special value ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ is a real algebraic integer.

(ii)

It generates the ring class field of the order of conductor $N{\ell }_{\mathcal{O}}$ over K.

(iii)

Let $Q_1,Q_2,\dots ,Q_s$ be all the reduced binary quadratic forms of discriminant $D_{\mathcal{O}}$. Furthermore, let ${\gamma }_1,{\gamma }_2,\dots ,{\gamma }_t$ be a complete set of representatives for the left cosets of ${\Gamma }_0\left(N\right)$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$. If $D_{\mathcal{O}}\ne -3,-4$, then the minimal polynomial of ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ over K is given by

$$\prod _{(i,k)\in S_{\mathcal{O},N}}(x-{\eta }_N\left(\widehat{{\gamma }_k}(-\overline{{\omega }_{Q_i}})\right))(\in \mathbb{Z}\left[x\right]),$$

where

$$S_{\mathcal{O},N}=\left\{(i,k)\in \mathbb{Z}\times \mathbb{Z}|1\le i\le s,1\le k\le t,Q_i\left({\gamma }_k\left[\begin{array}{c}x\\ y\end{array}\right]\right)\in \mathcal{Q}(D_{\mathcal{O}},N)\right\}.$$

Proof. 

(i)    From Remark 2 (ii), we observe that the special value ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ is a real number. Moreover, since $(N/4){\tau }_{\mathcal{O}}\in \mathbb{H}$, the special value

$$4\eta {(4·(N/4){\tau }_{\mathcal{O}})}^2/\eta {\left((N/4){\tau }_{\mathcal{O}}\right)}^2$$

is an algebraic integer by Lemma 4. Consequently, by the definition (10), ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ is a real algebraic integer.

(ii)

We deduce that

$$\begin{array}{cc}\hfill K_{\mathcal{O},{(\mathbb{Z}/N\mathbb{Z})}^{*}}& =K\left(f\left({\tau }_{\mathcal{O}}\right)\right|f\in {\mathcal{F}}_{{\Gamma }_0\left(N\right),\mathbb{Q}}\mathrm{is}\mathrm{finite}\mathrm{at}{\tau }_{\mathcal{O}})\mathrm{by}\mathrm{Proposition}2\hfill \\ \hfill & =K\left(f\left(\left[\begin{array}{cc}1& u\\ 0& 1\end{array}\right]\left({\ell }_{\mathcal{O}}{\tau }_{{\mathcal{O}}_K}\right)\right)\right|f\in {\mathcal{F}}_{{\Gamma }_0\left(N\right),\mathbb{Q}}\mathrm{is}\mathrm{finite}\mathrm{at}{\tau }_{\mathcal{O}})\hfill \\ \hfill & \phantom{a}\mathrm{where}\mathrm{u}\mathrm{is}\mathrm{an}\mathrm{integer}\mathrm{satisfying}{\tau }_{\mathcal{O}}={\ell }_{\mathcal{O}}{\tau }_{O_K}+u\hfill \\ \hfill & =K\left(f\left({\ell }_{\mathcal{O}}{\tau }_{{\mathcal{O}}_K}\right)\right|f\in {\mathcal{F}}_{{\Gamma }_0\left(N\right),\mathbb{Q}}\mathrm{is}\mathrm{finite}\mathrm{at}{\tau }_{\mathcal{O}})\mathrm{because}\left[\begin{array}{cc}1& u\\ 0& 1\end{array}\right]\in {\Gamma }_0\left(N\right)\hfill \\ \hfill & \subseteq K\left(g\left({\tau }_{{\mathcal{O}}_K}\right)\right|g\in {\mathcal{F}}_{{\Gamma }_0\left(N{\ell }_{\mathcal{O}}\right),\mathbb{Q}}\mathrm{is}\mathrm{finite}\mathrm{at}{\tau }_{\mathcal{O}})\hfill \\ \hfill & \phantom{a}\mathrm{since}f\left({\ell }_{\mathcal{O}}\tau \right)=f\left(\left[\begin{array}{cc}{\ell }_{\mathcal{O}}& 0\\ 0& 1\end{array}\right]\left(\tau \right)\right)\in {\mathcal{F}}_{{\Gamma }_0\left(N{\ell }_{\mathcal{O}}\right),\mathbb{Q}}\mathrm{for}\mathrm{all}f\in {\mathcal{F}}_{{\Gamma }_0\left(N\right),\mathbb{Q}}\hfill \\ \hfill & =K_{{\mathcal{O}}_K,{(\mathbb{Z}/N{\ell }_{\mathcal{O}}\mathbb{Z})}^{*}}\mathrm{by}\mathrm{Proposition}2\hfill \\ \hfill & =K_{{\mathcal{O}}^{\prime }}\mathrm{where}{\mathcal{O}}^{\prime }\mathrm{is}\mathrm{the}\mathrm{order}\mathrm{of}\mathrm{conductor}N{\ell }_{\mathcal{O}}\mathrm{in}\mathrm{K}\hfill \\ \hfill & =K\left(j\left(N{\tau }_{\mathcal{O}}\right)\right)\mathrm{by}\mathrm{Proposition}1\hfill \\ \hfill & \subseteq K\left({\eta }_N\left({\tau }_{\mathcal{O}}\right)\right)\mathrm{by}\mathrm{Lemma}3\hfill \\ \hfill & \subseteq K_{\mathcal{O},{(\mathbb{Z}/N\mathbb{Z})}^{*}}\mathrm{by}\mathrm{Lemma}2\mathrm{and}\mathrm{Proposition}2.\hfill \end{array}$$

This argument yields the following chain of inclusions:

$$K_{\mathcal{O},{(\mathbb{Z}/N\mathbb{Z})}^{*}}\subseteq K_{{\mathcal{O}}^{\prime }}\subseteq K\left({\eta }_N\left({\tau }_{\mathcal{O}}\right)\right)\subseteq K_{\mathcal{O},{(\mathbb{Z}/N\mathbb{Z})}^{*}}.$$

Since the two ends of this chain coincide, we obtain that

$$K\left({\eta }_N\left({\tau }_{\mathcal{O}}\right)\right)=K_{\mathcal{O},{(\mathbb{Z}/N\mathbb{Z})}^{*}}=K_{{\mathcal{O}}^{\prime }}.$$

(11)

(iii)

The result follows by (i), (ii), (11), Remark 2 (ii), Theorem 2 and the observation $-\overline{{\omega }_{Q_i^{{\gamma }_k}}}=\widehat{{\gamma }_k}(-\overline{{\omega }_{Q_i}})((i,k)\in S_{\mathcal{O},N}).$

□

In the next section, we present several examples. The computations were carried out using the MAPLE software (Version 2025.1). The procedures for finding reduced quadratic forms and coset representatives were based on Remark 3. From a computational perspective, eta-quotients are advantageous because they are defined by relatively simple q-products and are classical modular functions in number theory with rational Fourier coefficients. In our computations with MAPLE, we truncate the product expansion of the eta function up to $m=500$ and determine the minimal polynomial of ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ over K by combining the approximate values with the theoretical fact that the coefficients are integers.

6. Examples

In this section, we shall present a couple of examples of Theorem 3 (iii).

Example 1.

Let $K=\mathbb{Q}\left(\sqrt{-2}\right)$, $\mathcal{O}$ be the order of conductor 3 in K, $N=4$. There are two reduced forms of discriminant $D_{\mathcal{O}}=-72$:

$$\begin{array}{cc}{Q}_1=x^2+18y^2,\hfill & Q_2=2x^2+9y^2.\hfill \end{array}$$

By Remark 3 (iii), we determine a complete set of representatives for the left cosets of ${\Gamma }_0\left(4\right)$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$ as follows:

$$\begin{array}{cc}\hfill & {\gamma }_1=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],{\gamma }_2=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right],{\gamma }_3=\left[\begin{array}{cc}1& -1\\ 1& 0\end{array}\right],\hfill \\ \hfill & {\gamma }_4=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right],{\gamma }_5=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right],{\gamma }_6=\left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right].\hfill \end{array}$$

By using these representatives, we compute

$$\begin{array}{c}\hfill \begin{array}{ccc}{Q}_1^{{\gamma }_1}=x^2+18y^2,\hfill & Q_1^{{\gamma }_2}=18x^2+y^2,\hfill & Q_1^{{\gamma }_3}=19x^2-2xy+y^2,\hfill \\ Q_1^{{\gamma }_4}=73x^2+72xy+18y^2,\hfill & Q_1^{{\gamma }_5}=163x^2+108xy+18y^2,\hfill & Q_1^{{\gamma }_6}=22x^2-4xy+y^2,\hfill \\ Q_2^{{\gamma }_1}=2x^2+9y^2,\hfill & Q_2^{{\gamma }_2}=9x^2+2y^2,\hfill & Q_2^{{\gamma }_3}=11x^2-4xy+2y^2,\hfill \\ Q_2^{{\gamma }_4}=38x^2+36xy+9y^2,\hfill & Q_2^{{\gamma }_5}=83x^2+54xy+9y^2,\hfill & Q_1^{{\gamma }_6}=17x^2-8xy+2y^2.\hfill \end{array}\end{array}$$

From this computation, we obtain

$$\begin{array}{c}\hfill S_{\mathcal{O},4}=\{(1,1),(1,3),(1,4),(1,5),(2,2),(2,3),(2,5),(2,6)\}.\end{array}$$

Applying Theorem 3 (iii), we deduce that the minimal polynomial of ${\eta }_4\left({\tau }_{\mathcal{O}}\right)={\eta }_4\left(\sqrt{-18}\right)$ over K is

$$\begin{array}{cc}\hfill & \prod _{(i,k)\in S_{\mathcal{O},4}}\left(x-{\eta }_4\left(\widehat{{\gamma }_k}(-\overline{{\omega }_{Q_i}})\right)\right)\hfill \\ \hfill & =\left(x-{\eta }_4\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left(\sqrt{-18}\right)\right)\right)\left(x-{\eta }_4\left(\left[\begin{array}{cc}0& -1\\ 1& 1\end{array}\right]\left(\sqrt{-18}\right)\right)\right)\left(x-{\eta }_4\left(\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left(\sqrt{-18}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_4\left(\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]\left(\sqrt{-18}\right)\right)\right)\left(x-{\eta }_4\left(\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left(\frac{\sqrt{-18}}{2}\right)\right)\right)\left(x-{\eta }_4\left(\left[\begin{array}{cc}0& -1\\ 1& 1\end{array}\right]\left(\frac{\sqrt{-18}}{2}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_4\left(\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]\left(\frac{\sqrt{-18}}{2}\right)\right)\right)\left(x-{\eta }_4\left(\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]\left(\frac{\sqrt{-18}}{2}\right)\right)\right)\hfill \\ \hfill & =x^8+64x^7+616160x^6+29518336x^5+500985984x^4+3441954816x^3-370278512640x^2\hfill \\ \hfill & -6042786430976x+4096.\hfill \end{array}$$

Example 2.

Let $K=\mathbb{Q}\left(\sqrt{-15}\right)$, $\mathcal{O}$ be the order of conductor 2 in K, $N=12$. There are two reduced forms of discriminant $D_{\mathcal{O}}=-60$:

$$\begin{array}{cc}{Q}_1=x^2+15y^2,\hfill & Q_2=3x^2+5y^2.\hfill \end{array}$$

A complete set of the left coset representatives of ${\Gamma }_0\left(12\right)$ in ${\mathrm{SL}}_2\left(\mathbb{Z}\right)$ is given by

$$\begin{array}{cc}\hfill & {\gamma }_1=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],{\gamma }_2=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right],{\gamma }_3=\left[\begin{array}{cc}1& -1\\ 1& 0\end{array}\right],{\gamma }_4=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right],{\gamma }_5=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right],{\gamma }_6=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right],\hfill \\ \hfill & {\gamma }_7=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right],{\gamma }_8=\left[\begin{array}{cc}1& 0\\ 6& 1\end{array}\right],{\gamma }_9=\left[\begin{array}{cc}1& 0\\ 7& 1\end{array}\right],{\gamma }_{10}=\left[\begin{array}{cc}1& 0\\ 8& 1\end{array}\right],{\gamma }_{11}=\left[\begin{array}{cc}1& 0\\ 9& 1\end{array}\right],{\gamma }_{12}=\left[\begin{array}{cc}1& 0\\ 10& 1\end{array}\right],\hfill \\ \hfill & {\gamma }_{13}=\left[\begin{array}{cc}1& 0\\ 11& 1\end{array}\right],{\gamma }_{14}=\left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right],{\gamma }_{15}=\left[\begin{array}{cc}2& -1\\ 3& -1\end{array}\right],{\gamma }_{16}=\left[\begin{array}{cc}2& -1\\ 5& -2\end{array}\right],{\gamma }_{17}=\left[\begin{array}{cc}3& -1\\ 1& 0\end{array}\right],{\gamma }_{18}=\left[\begin{array}{cc}3& 1\\ 2& 1\end{array}\right],\hfill \\ \hfill & {\gamma }_{19}=\left[\begin{array}{cc}3& -1\\ 4& -1\end{array}\right],{\gamma }_{20}=\left[\begin{array}{cc}3& -1\\ 7& -2\end{array}\right],{\gamma }_{21}=\left[\begin{array}{cc}4& -1\\ 1& 0\end{array}\right],{\gamma }_{22}=\left[\begin{array}{cc}4& 1\\ 3& 1\end{array}\right],{\gamma }_{23}=\left[\begin{array}{cc}4& -1\\ 5& -1\end{array}\right],{\gamma }_{24}=\left[\begin{array}{cc}6& -1\\ 1& 0\end{array}\right].\hfill \end{array}$$

From these, one can compute that

$$\begin{array}{c}\hfill \begin{array}{ccc}{Q}_1^{{\gamma }_1}=x^2+15y^2,\hfill & Q_1^{{\gamma }_2}=15x^2+y^2,\hfill & Q_1^{{\gamma }_3}=16x^2-2xy+y^2,\hfill \\ Q_1^{{\gamma }_4}=61x^2+60xy+15y^2,\hfill & Q_1^{{\gamma }_5}=136x^2+90xy+15y^2,\hfill & Q_1^{{\gamma }_6}=241x^2+120xy+15y^2,\hfill \\ Q_1^{{\gamma }_7}=376x^2+150xy+15y^2,\hfill & Q_1^{{\gamma }_8}=541x^2+180xy+15y^2,\hfill & Q_1^{{\gamma }_9}=736x^2+210xy+15y^2,\hfill \\ Q_1^{{\gamma }_{10}}=961x^2+240xy+15y^2,\hfill & Q_1^{{\gamma }_{11}}=1216x^2+270xy+15y^2,\hfill & Q_1^{{\gamma }_{12}}=1501x^2+300xy+15y^2,\hfill \\ Q_1^{{\gamma }_{13}}=1816x^2+330xy+15y^2,\hfill & Q_1^{{\gamma }_{14}}=19x^2-4xy+y^2,\hfill & Q_1^{{\gamma }_{15}}=139x^2-94xy+16y^2,\hfill \\ Q_1^{{\gamma }_{16}}=379x^2-304xy+61y^2,\hfill & Q_1^{{\gamma }_{17}}=24x^2-6xy+y^2,\hfill & Q_1^{{\gamma }_{18}}=69x^2+66xy+16y^2,\hfill \\ Q_1^{{\gamma }_{19}}=249x^2-126xy+16y^2,\hfill & Q_1^{{\gamma }_{20}}=744x^2-426xy+61y^2,\hfill & Q_1^{{\gamma }_{21}}=31x^2-8xy+y^2,\hfill \\ Q_1^{{\gamma }_{22}}=151x^2+98xy+16y^2,\hfill & Q_1^{{\gamma }_{23}}=391x^2-158xy+16y^2,\hfill & Q_1^{{\gamma }_{24}}=51x^2-12xy+y^2,\hfill \\ Q_2^{{\gamma }_1}=3x^2+5y^2,\hfill & Q_2^{{\gamma }_2}=5x^2+3y^2,\hfill & Q_2^{{\gamma }_3}=8x^2-6xy+3y^2,\hfill \\ Q_2^{{\gamma }_4}=23x^2+20xy+5y^2,\hfill & Q_2^{{\gamma }_5}=48x^2+30xy+5y^2,\hfill & Q_2^{{\gamma }_6}=83x^2+40xy+5y^2,\hfill \\ Q_2^{{\gamma }_7}=128x^2+50xy+5y^2,\hfill & Q_2^{{\gamma }_8}=183x^2+60xy+5y^2,\hfill & Q_2^{{\gamma }_9}=248x^2+70xy+5y^2,\hfill \\ Q_2^{{\gamma }_{10}}=323x^2+80xy+5y^2,\hfill & Q_2^{{\gamma }_{11}}=408x^2+90xy+5y^2,\hfill & Q_1^{{\gamma }_{12}}=503x^2+100xy+5y^2,\hfill \\ Q_2^{{\gamma }_{13}}=608x^2+110xy+5y^2,\hfill & Q_2^{{\gamma }_{14}}=17x^2-12xy+3y^2,\hfill & Q_2^{{\gamma }_{15}}=57x^2-42xy+8y^2,\hfill \\ Q_2^{{\gamma }_{16}}=137x^2-112xy+23y^2,\hfill & Q_2^{{\gamma }_{17}}=32x^2-18xy+3y^2,\hfill & Q_2^{{\gamma }_{18}}=47x^2+38xy+8y^2,\hfill \\ Q_2^{{\gamma }_{19}}=107x^2-58xy+8y^2,\hfill & Q_2^{{\gamma }_{20}}=272x^2-158xy+23y^2,\hfill & Q_2^{{\gamma }_{21}}=53x^2-24xy+3y^2,\hfill \\ Q_2^{{\gamma }_{22}}=93x^2+54xy+8y^2,\hfill & Q_2^{{\gamma }_{23}}=173x^2-74xy+8y^2,\hfill & Q_1^{{\gamma }_{24}}=113x^2-36xy+3y^2.\hfill \end{array}\end{array}$$

In this case, the set $S_{\mathcal{O},12}$ is given precisely by

$$\begin{array}{cc}\hfill S_{\mathcal{O},12}=& \left\{\right(1,1),(1,4),(1,6),(1,8),(1,10),(1,12),(1,14),(1,15),\hfill \\ \hfill & (1,16),(1,21),(1,22),(1,23),(2,2),(2,4),(2,6),(2,10),\hfill \\ \hfill & (2,12),(2,14),(2,16),(2,18),(2,19),(2,21),(2,23),(2,24)\}.\hfill \end{array}$$

By Theorem 3 (iii), the minimal polynomial of ${\eta }_{12}\left({\tau }_{\mathcal{O}}\right)={\eta }_{12}\left(\sqrt{-15}\right)$ over K is

$$\begin{array}{cc}\hfill & \prod _{(i,k)\in S_{\mathcal{O},12}}\left(x-{\eta }_{12}\left(\widehat{{\gamma }_k}(-\overline{{\omega }_{Q_i}})\right)\right)\hfill \\ \hfill & =\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 6& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 8& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 10& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}-1& -1\\ 3& 2\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}-2& -1\\ 5& 2\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}0& -1\\ 4& 1\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 1\\ 3& 4\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}-1& -1\\ 5& 4\end{array}\right]\left(\sqrt{-15}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 8& 1\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 0\\ 10& 1\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}-2& -1\\ 5& 2\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}1& 1\\ 2& 3\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}-1& -1\\ 4& 3\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\hfill \\ \hfill & \times \left(x-{\eta }_{12}\left(\left[\begin{array}{cc}0& -1\\ 1& 4\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}-1& -1\\ 5& 4\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left(x-{\eta }_{12}\left(\left[\begin{array}{cc}0& -1\\ 1& 6\end{array}\right]\left(\frac{\sqrt{-15}}{3}\right)\right)\right)\left[3\right]\hfill \\ \hfill & =x^{24}+192x^{23}-7122306993261792x^{22}-1253526030816147968x^{21}-91383204195139439529x^{20}\left[3\right]\hfill \\ \hfill & -3389719435105910475168x^{19}-49401241850013269672816x^{18}\left[3\right]\hfill \\ \hfill & +1228473321987438668964096x^{17}+86840132355875889487145349x^{16}\left[3\right]\hfill \\ \hfill & +2457144449583692060476097152x^{15}+44234812305221946123591380928x^{14}\left[3\right]\hfill \\ \hfill & +552735876182379232822999458810x^{13}+4841799881653100583964377978842x^{12}\left[3\right]\hfill \\ \hfill & +28308281222344201581487804712803x^{11}+90141810550325076912679521259347x^{10}\left[3\right]\hfill \\ \hfill & -37037392168787459577963652303623x^9-1550236147193962873251193610591518x^8\left[3\right]\hfill \\ \hfill & -3639162775938143122457492966909150x^7+18821613358222160529419402240943524x^6\left[3\right]\hfill \\ \hfill & +136332738915817934864732450527911904x^5+298872342513781468168883205086012077x^4\left[3\right]\hfill \\ \hfill & +22964197003252385546057560757637927x^3+1694808824718030833349954509125564x^2\left[3\right]\hfill \\ \hfill & -198153347290972848225794251004x+1\hfill \end{array}$$

Additionally, we observe that ${\eta }_{12}\left({\tau }_{\mathcal{O}}\right)={\eta }_{12}\left(\sqrt{-15}\right)$ is a unit.

Remark 4.

The computations in the above examples were carried out with the MAPLE software, using 100 significant digits for numerical approximations. For instance, in Example 1, the approximate value of the coefficient of $x^7$ is

$$\begin{array}{c}\hfill 63.999\dots 9760481790273634068827\end{array}$$

from which the corresponding integer coefficient is determined to be 64. For all other coefficients, the error between the approximate value and the corresponding integer is less than ${10}^{-60}$.

7. Application to Primes of the Form ${\mathit{x}}^{\mathbf{2}}+{\mathit{n}\mathit{y}}^{\mathbf{2}}$

Let n be a positive integer. In this final section, we present our recent joint work with Koo, Shin and Yoon [9], which extends Cho’s study ([12], Theorem 15.19) on primes of the form $x^2+n{y}^2$ under the additional conditions $x\equiv 1(modN)$ and $y\equiv 0(modN)$. We further apply these results to the minimal polynomial of ${\eta }_N\left({\tau }_{\mathcal{O}}\right)$ over K.

It is worth noting that the study of primes of this type has a long history. In his two-squared theorem, Gauss showed that class field theory is useful:

$$p=x^2+y^2⟺p=2\mathrm{or}p\mathrm{splits}\mathrm{completely}\mathrm{in}\mathbb{Q}\left(\sqrt{-1}\right).$$

Here, $\mathbb{Q}\left(\sqrt{-1}\right)$ is a class field (an abelian extension) of $\mathbb{Q}$. Later, Weber, Hilbert, and Artin extended this idea to primes of the form $x^2+n{y}^2$ for positive integers n (see [7]).

For a prime p, we let $\left({\displaystyle \frac{·}{p}}\right)$ be the Kronecker symbol.

Proposition 4.

Let n be a positive integer, $K=\mathbb{Q}\left(\sqrt{-n}\right)$ and $\mathcal{O}=\mathbb{Z}\left[\sqrt{-n}\right]$. Let $\mathfrak{v}$ be a real algebraic integer which generates $K_{\mathcal{O},G}$ over K and $f\left(X\right)\in \mathbb{Z}\left[X\right]$ be its minimal polynomial over K. If p is a prime dividing neither $2nN$ nor the discriminant of $f\left(X\right)$, then

$$\begin{array}{cc}\hfill & psplitscompletelyin{K}_{\mathcal{O},G}\hfill \\ \hfill ⟺& p=x^2+n{y}^{2}forsomex,y\in \mathbb{Z}suchthatx+N\mathbb{Z}\in Gandy\equiv 0\left(\mathrm{mod}\right)N\hfill \\ \hfill ⟺& \left({\displaystyle \frac{-n}{p}}\right)=1\mathit{and}f\left(X\right)\equiv 0\left(\mathrm{mod}p\right)\mathit{has}\mathit{an}\mathit{integer}\mathit{solution}.\hfill \end{array}$$

Example 3.

Let $n=18$ and $N=4$. In this case, $K=\mathbb{Q}\left(\sqrt{-18}\right)=\mathbb{Q}\left(\sqrt{-2}\right)$ and $\mathcal{O}=\mathbb{Z}\left[\sqrt{-18}\right]$ is the order of conductor 3 in K as in Example 1. By Theorem 3 (ii), the special value ${\eta }_4\left({\tau }_{\mathcal{O}}\right)$ generates the field $K_{\mathcal{O},{(\mathbb{Z}/4\mathbb{Z})}^{*}}$ over K.

Now, let $F_1\left(X\right)$ denote the minimal polynomial of ${\eta }_4\left({\tau }_{\mathcal{O}}\right)$ computed in Example 1. The discriminant of $F_1\left(X\right)$ is

$$\begin{array}{c}\hfill 2^{140}\times 3^4\times 5^{24}\times 7^{24}\times {13}^4\times {23}^2\times {29}^4\times {31}^4\times {47}^2\times {53}^4\times {61}^4\times {71}^2\end{array}$$

By Proposition 4, we obtain that if p is a prime dividing neither $2\times 18\times 4=2^4\times 3^2$ nor the discriminant of $F_1\left(X\right)$, then

$$\begin{array}{cc}\hfill & p=x^2+18y^2\mathit{for}\mathit{some}x,y\in \mathbb{Z}\mathit{such}\mathit{that}x+4\mathbb{Z}\in {(\mathbb{Z}/4\mathbb{Z})}^{*}\mathit{and}y\equiv 0\left(\mathrm{mod}\right)4\hfill \\ \hfill ⟺& \left({\displaystyle \frac{-18}{p}}\right)=\left({\displaystyle \frac{-2}{p}}\right)=1\mathit{and}{F}_1\left(X\right)\equiv 0\left(\mathrm{mod}p\right)\mathit{has}\mathit{an}\mathit{integer}\mathit{solution}.\hfill \end{array}$$

Example 4.

Now suppose that $n=15$ and $N=12$. In this setting, $K=\mathbb{Q}\left(\sqrt{-15}\right)$ and $\mathcal{O}=\mathbb{Z}\left[\sqrt{-15}\right]$ is the order of conductor 2 in K as in Example 2. According to Theorem 3 (ii), the special value ${\eta }_{12}\left({\tau }_{\mathcal{O}}\right)$ generates the field $K_{\mathcal{O},{(\mathbb{Z}/12\mathbb{Z})}^{*}}$ over K.

Let $F_2\left(X\right)$ be the minimal polynomial of ${\eta }_{12}\left({\tau }_{\mathcal{O}}\right)$ obtained in Example 2. Its discriminant is

$$\begin{array}{cc}\hfill & 1632307284832228718158746061289157482137253227993306249138258190141743\hfill \\ \hfill & 4443886904187699419233968544129371609537666746058020201645256459023316\hfill \\ \hfill & 1450976209285388060263902555546664209814035196870467272938837811559698\hfill \\ \hfill & 4646777031790006030223836505343256035536977578802290009208146418232314\hfill \\ \hfill & 7207023344755523235848212080602604501510776930718153742770446681908313\hfill \\ \hfill & 1121066322695831455501223560962300752968883605012639239629618085337152\hfill \\ \hfill & 4135622077689093458736503200226802006277447862764466920648473690147947\hfill \\ \hfill & 8838276826852283173788448512760066665196740033344822814786500932531274\hfill \\ \hfill & 3066197568526166959058190073392084856601495255137071381658875036689443\hfill \\ \hfill & 7316416266743663550022019921472794570589619274430792993430690949100309\hfill \\ \hfill & 3301211813098054734935429130659739199585880351314003697305749253438887\hfill \\ \hfill & 8125460909721029258210543309243456065960236418896059032355518538069598\hfill \\ \hfill & 8769704753247909041266437556413094470271294756321425909670008432149825\hfill \\ \hfill & 6632191525267779790953881718905128776564373037893414184701110326854965\hfill \\ \hfill & 3620150694730593478351814158637819770323066261877876743430503705793406\hfill \\ \hfill & 9288576995423061523788622204848624863398597089386590680850993070553066\hfill \\ \hfill & 136720240398698093565827667947828132747941752323889.\hfill \end{array}$$

Proposition 4 implies that if p is a prime dividing neither $2\times 15\times 12=2^3\times 3^2\times 5$ nor the discriminant of $F_2\left(X\right)$, then

$$\begin{array}{cc}\hfill & p=x^2+15y^2\mathit{for}\mathit{some}x,y\in \mathbb{Z}\mathit{such}\mathit{that}x+12\mathbb{Z}\in {(\mathbb{Z}/12\mathbb{Z})}^{*}\mathit{and}y\equiv 0\left(\mathrm{mod}12\right)\hfill \\ \hfill ⟺& \left({\displaystyle \frac{-15}{p}}\right)=1\mathit{and}{F}_2\left(X\right)\equiv 0\left(\mathrm{mod}p\right)\mathit{has}\mathit{an}\mathit{integer}\mathit{solution}.\hfill \end{array}$$