1 Introduction

Let p be an odd prime, and let S be a subset of \(\mathbb {Z}\) such that \(\{ 0, S, -S\}\) is a complete set of representatives modulo p. Let \(s\in S\) and let \(a\in \mathbb {Z}\) be coprime to p. Then, there exist \(\epsilon (a, s)\in \{\pm 1\}\) and \(s_a\in S\) such that \(as=\epsilon (a, s)s_a\). Gauss’ lemma states that the Legendre symbol \(\left( \frac{a}{p}\right) \) can be written as follows:

$$\begin{aligned} \left( \frac{a}{p}\right) =\prod _{s\in S}\epsilon (a, s). \end{aligned}$$

For distinct odd primes p and q, it holds that

$$\begin{aligned} \left( \frac{p}{q}\right) \left( \frac{q}{p}\right) =(-1)^{\frac{p-1}{2}\frac{q-1}{2}}, \end{aligned}$$

which is called the quadratic reciprocity law in the field of rational numbers.

A quadratic reciprocity law for a quadratic number field was first established by Gauss, who provided the quadratic reciprocity law for \(\mathbb {Q}(\sqrt{-1})\). The quadratic reciprocity law in imaginary quadratic number fields using elliptic functions was put forth by Herglotz, Niemeyer, Bayad [2], and Hajir and Villegas [8]. In particular, Bayad [2] constructed certain elliptic functions and proved the product formulas for them. As an application, he used the product formulas to establish the quadratic reciprocity law in imaginary quadratic number fields. Hayashi [11] corrected and refined Bayad’s reciprocity law.

The analogies between number fields and function fields have several interesting aspects. Artin [1] established a rational function field analog of the quadratic reciprocity law, and Schmidt [16] proved a more general reciprocity law over rational function fields. Carlitz [3,4,5] provided another proof of the general reciprocity law using an analog of Gauss’ lemma. In [10], we generalized the analog of Gauss’ lemma over the rational function fields, and provided another proof of the general reciprocity law for power residue symbols. For details of the general reciprocity law over rational function fields, we refer to [15, 17]. The purpose of this paper is to provide an analog of Gauss’ lemma over general function fields, and establish a reciprocity law for power residue symbols. As an application, a reciprocity law for power residue symbols in totally imaginary function fields is established.

In Section 2, Gauss’ lemma is generalized over function fields. In Section 3, a reciprocity law in function fields is established using Gauss’ lemma. The last section is devoted to the proof of the theorems in the previous section.

2 Gauss’ lemma

Let F be a function field in one variable over the field of constants \(\mathbb {F}_q\), a finite field of q elements. Let \(\infty \) be a place of F. We express R as the ring of elements of F that are regular outside \(\infty \). Let \(F_{\infty }\) be the completion of F with respect to \(\infty \), and let \(C_{\infty }\) be the completion of an algebraic closure of \(F_{\infty }\) with respect to \(\infty \). In this section, we introduce certain symbols to establish Gauss’ lemma. The ideas of Reichardt are used (see [12, 13]).

2.1 Generalized Gauss’ lemma

We assume that F contains a primitive n-th root of unity \(\zeta _n\). This implies that n divides \(q-1\). The group \(\mathbb {F}_q^{*}\) contains the n-th roots of unity \(\mu _n:=\{ 1, \zeta _n, \ldots , \zeta _n^{n-1}\}\). Let \(\mathfrak {p}\) be a prime ideal of R, and let \(\varphi (\mathfrak {p})\) be the order of the unit group \((R/\mathfrak {p})^{*}\). When \(\alpha \in R\) is coprime to \(\mathfrak {p}\), there exists a unique element \(\left( \frac{\alpha }{\mathfrak {p}}\right) _n\in \mu _n\) such that

$$\begin{aligned} \alpha ^{\varphi (\mathfrak {p})/n}\equiv \left( \frac{\alpha }{\mathfrak {p}}\right) _n\ (\text {mod}\ \mathfrak {p}). \end{aligned}$$

When \(\alpha \in R\) is contained in \(\mathfrak {p}\), let \(\left( \frac{\alpha }{\mathfrak {p}}\right) _n=0\). For any ideal \(\mathfrak {a}\) in R having the prime ideal decomposition \(\mathfrak {a}=\mathfrak {p}_1^{e_1}\cdots \mathfrak {p}_r^{e_r}\), we extend the above symbol multiplicatively by setting \(\left( \frac{\alpha }{\mathfrak {a}}\right) _n =\left( \frac{\alpha }{\mathfrak {p}_1}\right) _n^{e_1}\cdots \left( \frac{\alpha }{\mathfrak {p}_r}\right) _n^{e_r}\). This symbol is called the n-th power residue symbol.

The function field F contains the n-th roots of unity \(\mu _n=\{ 1, \zeta _n, \ldots , \zeta _n^{n-1}\}\). Let \(\mathfrak {a}\) be a non-zero ideal of R. A subset \(S=\{ s_1, \ldots , s_m\}\) of R such that \(\{ 0, S, \zeta _nS, \ldots , \zeta _n^{n-1}S\}\) is a complete set of representatives modulo \(\mathfrak {a}\) is called a 1/n-system modulo \(\mathfrak {a}\). Let \(\alpha \in R\) be coprime to \(\mathfrak {a}\) and let S be a 1/n-system modulo \(\mathfrak {a}\). There exists a permutation \(\pi \) of \(\{ 1, \ldots , m\}\) such that for any \(s_j\in S\), there exists an element \(\zeta _n^{a(j)}\in \mu _n\) such that

$$\begin{aligned} \alpha s_j\equiv \zeta _n^{a(j)} s_{\pi (j)}\quad (\text {mod}\ \mathfrak {a}). \end{aligned}$$
(2.1)

We write \(\epsilon (\alpha , s_j)\) for \(\zeta _n^{a(j)}\).

The following theorem is an analog of Gauss’ lemma:

Theorem 2.1

(Generalized Gauss’ lemma). For any element \(\alpha \in R\) coprime to \(\mathfrak {a}\),

$$\begin{aligned} \left( \frac{\alpha }{\mathfrak {a}}\right) _n =\prod _{s\in S}\epsilon (\alpha , s), \end{aligned}$$

where S is a 1/n-system modulo \(\mathfrak {a}\).

2.2 Proof of Theorem 2.1

Set

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n =\prod _{s\in S}\epsilon (\alpha , s), \end{aligned}$$

which is independent of the choice of S. Indeed, let \(S'=\{ s_1', \ldots , s_m'\}\) be another 1/n-system modulo \(\mathfrak {a}\). There exist permutations \(\pi \) and \(\pi '\) of \(\{ 1, \ldots , m\}\) such that for any j,

$$\begin{aligned} \alpha s_j\equiv \epsilon (\alpha , s_j)s_{\pi (j)}\quad (\text {mod}\ \mathfrak {a}),\quad \alpha s_j'\equiv \epsilon (\alpha , s_j')s_{\pi ' (j)}\quad (\text {mod}\ \mathfrak {a}). \end{aligned}$$

There exists a permutation \(\sigma \) of \(\{ 1, \ldots , m\}\) such that for any j,

$$\begin{aligned} s_j\equiv \zeta _n^{b(j)}s_{\sigma (j)}'\quad (\text {mod}\ \mathfrak {a}),\quad \alpha s_j\equiv \zeta _n^{b(j)}\alpha s_{\sigma (j)}' \equiv \zeta _n^{b(j)}\epsilon (\alpha , s'_{\sigma (j)})s'_{\pi '\sigma (j)}\quad (\text {mod}\ \mathfrak {a}). \end{aligned}$$

Applying these congruences to the exponential function \(e_{\mathfrak {a}}\) for \(\mathfrak {a}\), we obtain

$$\begin{aligned}{} & {} e_{\mathfrak {a}}(\alpha s_j)=\epsilon (\alpha , s_j)e_{\mathfrak {a}}(s_{\pi (j)}),\quad e_{\mathfrak {a}}(\alpha s_j')=\epsilon (\alpha , s_j')e_{\mathfrak {a}}(s_{\pi ' (j)}),\\{} & {} e_{\mathfrak {a}}(\alpha s_j)=\zeta _n^{b(j)}\epsilon (\alpha , s_{\sigma (j)}')e_{\mathfrak {a}}(s_{\pi 7\sigma (j)}). \end{aligned}$$

Hence, we have

$$\begin{aligned} \prod _{j=1}^m\epsilon (\alpha , s_j) = \prod _{j=1}^m\frac{e_{\mathfrak {a}}(\alpha s_j)}{e_{\mathfrak {a}}(s_j)} = \prod _{j=1}^m\frac{\zeta _n^{b(j)}\epsilon (\alpha , s_{\sigma (j)}')e_{\mathfrak {a}}(s_{\pi '\sigma (j)}')}{\zeta _n^{b(j)}e_{\mathfrak {a}}(s_{\sigma (j)}')} = \prod _{j=1}^m\epsilon (\alpha , s_j'). \end{aligned}$$

Let \(S^{*}=\{s_j\in S\ |\ s_j\ \text {coprime to}\ \mathfrak {a}\}\), and set

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n^{*} =\prod _{s\in S^{*}}\epsilon (\alpha , s), \end{aligned}$$

which is independent of the choice of S. This can be proven in the similar way as the case \(\left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n\). For any ideal \(\mathfrak {b}\) of R containing \(\mathfrak {a}\), set

$$\begin{aligned} S_{\mathfrak {b}}=\{s_j\in S\ |\ s_jR+\mathfrak {a}=\mathfrak {b}\}. \end{aligned}$$

We observe that \(S_{\mathfrak {a}}=\phi \) and \(S_{R}=S^{*}\). The set S is a disjoint union of \(S_{\mathfrak {b}}\ (\mathfrak {a}\subset \mathfrak {b})\). A subset T of R such that \(\{ T, \zeta _nT, \ldots , \zeta _n^{n-1}T\}\) is a complete set of the representatives of \((R/\mathfrak {a})^{*}\) is referred to as a prime 1/n-system modulo \(\mathfrak {a}\). The following lemma is required to prove Gauss’ lemma.

Lemma 2.2

Let \(\mathfrak {a}\) be a non-zero ideal of R. For any element \(\alpha \in R\) coprime to \(\mathfrak {a}\),

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n =\prod _{\mathfrak {b}|\mathfrak {a}} \left\{ \frac{\alpha }{\mathfrak {b}}\right\} _n^{*}. \end{aligned}$$

Proof

Let \(S_{\mathfrak {b}}=\{t_1, \ldots , t_k\}\). There exists an element \(u_1\in \mathfrak {b}^{-1}\) such that \(t_1u_1\equiv 1\ (\text {mod}\ \mathfrak {a}\mathfrak {b}^{-1})\).

It holds that \(\{t_1u_1, \ldots , t_ku_1\}\) is a prime 1/n-system modulo \( \mathfrak {a}\mathfrak {b}^{-1}\). Indeed, we take any \(\alpha \in R\) coprime to \(\mathfrak {a}\mathfrak {b}^{-1}\). For each i, there exists \(\zeta _n^{a(i)}\in \mu _n\) and \(s_j\in S\) such that

$$\begin{aligned} \alpha t_i\equiv \zeta _n^{a(i)}s_j\quad (\text {mod}\ \mathfrak {a}). \end{aligned}$$
(2.2)

We see easily that \(s_j\in S_{\mathfrak {b}}\) and that

$$\begin{aligned} \alpha t_iu_1\equiv \zeta _n^{a(i)}s_ju_1\quad (\text {mod}\ \mathfrak {a}\mathfrak {b}^{-1}). \end{aligned}$$
(2.3)

Next, we assume that there exist \(\zeta , \zeta '\in \mu _n\) and \(t_i, t_j\in S_{\mathfrak {b}}\) such that \(\zeta t_iu_1\equiv \zeta 't_ju_1\ (\text {mod}\ \mathfrak {a}\mathfrak {b}^{-1})\). Multiplying this congruence by \(t_1\), we obtain \(\zeta t_i\equiv \zeta 't_j\ (\text {mod}\ \mathfrak {a})\), which implies \(\zeta =\zeta '\) and \(t_i=t_j\).

Observing that (2.2) and (2.3) are equivalent and that \(S=\bigcup _{\mathfrak {b}\subset \mathfrak {a}}S_{\mathfrak {b}}\),

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n =\prod _{\mathfrak {b}|\mathfrak {a}} \left\{ \frac{\alpha }{\mathfrak {a}\mathfrak {b}^{-1}}\right\} _n^{*} =\prod _{\mathfrak {b}|\mathfrak {a}} \left\{ \frac{\alpha }{\mathfrak {b}}\right\} _n^{*}. \end{aligned}$$

\(\square \)

Lemma 2.3

Let \(\mathfrak {a}\) be a non-zero ideal of R. For any element \(\alpha \in R\) coprime to \(\mathfrak {a}\),

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n^{*} \equiv \alpha ^{\varphi (\mathfrak {a})/n}\quad (\textrm{mod}\ \mathfrak {a}). \end{aligned}$$
(2.4)

Proof

Since \(\{S^{*}, \zeta _nS^{*}, \ldots , \zeta _n^{n-1}S^{*}\}\) is a complete set of representatives for \((R/ \mathfrak {a})^{*}\), \(|S^{*}|=\varphi (\mathfrak {a})/n\). Using (2.1), we have

$$\begin{aligned} \left( \prod _{s_j\in S^{*}}s_j\right) \alpha ^{\varphi (\mathfrak {a})/n}\equiv & {} \prod _{s_j\in S^{*}}\alpha s_j \equiv \prod _{s_j\in S^{*}}\epsilon (\alpha , s_j)s_{\pi (j)} \quad (\text {mod}\ \mathfrak {a})\\\equiv & {} \left( \prod _{s_j\in S^{*}}s_{\pi (j)}\right) \left( \prod _{s_j\in S^{*}}\epsilon (\alpha , s_j)\right) \quad (\text {mod}\ \mathfrak {a})\\\equiv & {} \left( \prod _{s_j\in S^{*}}s_j\right) \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n^{*} \quad (\text {mod}\ \mathfrak {a}), \end{aligned}$$

where \(\pi \) is a permutation of \(\{ 1, \ldots , \varphi (\mathfrak {a})/n\}\). This yields (2.4). \(\square \)

For any non-zero ideal \(\mathfrak {b}\) of R, it holds that

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {b}}\right\} _n^{*} =\left\{ \begin{array}{ll} \left( \frac{\alpha }{\mathfrak {p}}\right) _n &{} \text {if}\ \mathfrak {b}=\mathfrak {p}^a,\\ 1 &{} \text {otherwise}.\end{array} \right. \end{aligned}$$
(2.5)

Indeed, when there exist coprime ideals \(\mathfrak {c}, \mathfrak {d}\) of R such that \(\mathfrak {b}=\mathfrak {cd}\), \(\alpha ^{\varphi (\mathfrak {c})}\equiv 1\ (\text {mod}\ \mathfrak {c})\) implies \(\alpha ^{\varphi (\mathfrak {b})/n}\equiv \left( \alpha ^{\varphi (\mathfrak {c})}\right) ^{\varphi (\mathfrak {d})/n} \equiv 1\ (\text {mod}\ \mathfrak {c})\). Similarly, we have \(\alpha ^{\varphi (\mathfrak {b})/n}\equiv 1\ (\text {mod}\ \mathfrak {d})\), which yields \(\alpha ^{\varphi (\mathfrak {b})/n}\equiv 1\ (\text {mod}\ \mathfrak {b})\). Next, we consider a case in which there exists a prime ideal \(\mathfrak {p}\) and a positive number a such that \(\mathfrak {b}=\mathfrak {p}^a\). If the cardinality of \(R/\mathfrak {p}\) is \(p^f\), then \(\varphi (\mathfrak {b})=p^{f(a-1)}\left( p^f-1\right) \). As \(p^f\equiv 1\ (\text {mod}\ n)\), using Lemma 2.3,

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {b}}\right\} _n^{*}\equiv \alpha ^{\varphi (\mathfrak {b})/n}\equiv \left( \frac{\alpha }{\mathfrak {p}}\right) _n^{p^{f(a-1)}} \equiv \left( \frac{\alpha }{\mathfrak {p}}\right) _n \quad (\text {mod}\ \mathfrak {p}), \end{aligned}$$

which yields (2.5).

When \(\mathfrak {a}=\mathfrak {p}_1^{e_1}\cdots \mathfrak {p}_r^{e_r}\), using Lemma 2.2 and (2.5), we obtain

$$\begin{aligned} \left\{ \frac{\alpha }{\mathfrak {a}}\right\} _n =\prod _{\mathfrak {b}|\mathfrak {a}}\left\{ \frac{\alpha }{\mathfrak {b}}\right\} _n^{*} =\prod _{i=1}^r\prod _{j=1}^{e_i}\left\{ \frac{\alpha }{\mathfrak {p}_i^j}\right\} _n^{*} =\prod _{i=1}^r\left( \frac{\alpha }{\mathfrak {p}_i}\right) _n^{e_i} =\left( \frac{\alpha }{\mathfrak {a}}\right) _n. \end{aligned}$$

This completes the proof of Theorem 2.1. \(\Box \)

3 Reciprocity laws

Let \(\phi \) be a rank one Drinfeld R-module corresponding to the R-lattice \(L=\xi R\). For coprime \(\alpha , \beta \in R\), let \(\left( \frac{\alpha }{\beta }\right) _n=\left( \frac{\alpha }{\beta R}\right) _n\). The following is a reciprocity law for the n-th power residue symbol.

Theorem 3.1

For coprime \(\alpha , \beta \in R\),

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) _n \left( \frac{\beta }{\alpha }\right) ^{-1}_n =(-1)^{\frac{N(\alpha )-1}{n}\frac{N(\beta )-1}{n}} \nu (\alpha )^{\frac{N(\beta )-1}{n}}\nu (\beta )^{-\frac{N(\alpha )-1}{n}}, \end{aligned}$$
(3.1)

where \(N(\alpha )=q^{\deg (\alpha )}\) is the norm of \(\alpha \), and \(\nu (\alpha )\) is the leading coefficient of \(\phi _{\alpha }\).

Remark 3.2

The value \(\nu (\alpha )^{\frac{N(\beta )-1}{n}}\nu (\beta )^{-\frac{N(\alpha )-1}{n}}\) in the above theorem belongs to \(\mu _n\). Indeed, for \(\alpha , \beta \in R{\setminus }\{ 0\}\), the value \(\phi _{\alpha \beta }\) of the Drinfeld module \(\phi \) provides \(\phi _{\alpha \beta }=\phi _{\alpha }\phi _{\beta }=\phi _{\beta }\phi _{\alpha }\), which yields \(\nu (\alpha \beta )=\nu (\alpha )\nu (\beta )^{N(\alpha )}=\nu (\alpha )^{N(\beta )}\nu (\beta )\). Therefore, the claim follows from this.

From Theorem 3.1, we obtain another type of reciprocity law:

Theorem 3.3

There exists a multiplicative function \(\kappa : R{\setminus }\{ 0\}\rightarrow C_{\infty }^{*}\) such that for coprime \(\alpha , \beta \in R\),

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) _n \left( \frac{\beta }{\alpha }\right) ^{-1}_n =(-1)^{\frac{N(\alpha )-1}{n}\frac{N(\beta )-1}{n}} \kappa (\alpha )^{\frac{N(\beta )-1}{n}}\kappa (\beta )^{-\frac{N(\alpha )-1}{n}}. \end{aligned}$$
(3.2)

Remark 3.4

Let \(A=\mathbb {F}_q[T]\) be the polynomial ring over \(\mathbb {F}_q\), and let \(K=\mathbb {F}_q(T)\) be its quotient field. Let \(K_{\infty }=\mathbb {F}_q((1/T))\) be the completion of K with respect to \(\infty =(1/T)\), and let \(C_{\infty }\) be the completion of an algebraic closure of \(K_{\infty }\) with respect to \(\infty \). A separable extension F/K is called totally imaginary if \(\infty \) has only one prime over F. For details of such extensions, we refer the reader to the papers Gekeler [6], Rosen [14], and Hamahata [9]. Let F be a totally imaginary extension of K, and let \(O_F\) be the integral closure of A in F. Using Theorem 3.1 for \(R=O_F\), we have the analog of Bayad [2, Théorème 2.9].

4 Proof of Theorems 3.1 and 3.3

4.1 Proof of Theorem 3.1

Let \(T_{\beta }=\{y_1, \ldots , y_m\}\) be a 1/n-system modulo \(\beta R\), and set \(S_{\beta }=\xi \beta ^{-1}T_{\beta }\). For each i, it holds that \(\alpha y_i\equiv \epsilon (\alpha , y_i)y_{\pi (i)}\ (\text {mod}\ \beta R)\) if and only if \(\alpha \xi \beta ^{-1} y_i\equiv \epsilon (\alpha , y_i)\xi \beta ^{-1}y_{\pi (i)}\ (\text {mod}\ L)\). Using \(e_L(\alpha \xi \beta ^{-1}y_i)=\epsilon (\alpha , y_i)e_L(\xi \beta ^{-1}y_{\pi (i)})\) and Theorem 2.1,

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) _n =\prod _{z\in S_{\beta }}\frac{e_{L}(\alpha z)}{e_{L}(z)} =\prod _{z\in S_{\beta }}\frac{\phi _{\alpha }(e_L(z))}{e_L(z)}. \end{aligned}$$

We can identify \(C_{\infty }\{\tau \}\) with the non-commutative ring of additive polynomials of X with coefficients in \(C_{\infty }\), where the product is the composition of maps. For \(\phi _{\alpha }\in C_{\infty }\{\tau \}\), we write \(\phi _{\alpha }(X)=\alpha X+\cdots +\nu (\alpha )X^{N(\alpha )}\). As

$$\begin{aligned} \phi _{\alpha }(X)= & {} \nu (\alpha )\prod _{y\in L/\alpha L}\left( X-e_{L}(y/\alpha )\right) \\= & {} \nu (\alpha )X\prod _{\zeta \in \mu _n} \prod _{y\in T_{\alpha }}\left( X-\zeta e_{L}(y/\alpha )\right) \\= & {} \nu (\alpha )X\prod _{z\in S_{\alpha }}\left( X^n-e_{L}(z)^n\right) , \\ \left( \frac{\alpha }{\beta }\right) _n= & {} \prod _{x\in S_{\beta }}\nu (\alpha )\prod _{z\in S_{\alpha }} \left( e_{L}(x)^n-e_{L}(z)^n\right) \\= & {} \nu (\alpha )^{\frac{N(\beta )-1}{n}} (-1)^{\frac{N(\alpha )-1}{n}\frac{N(\beta )-1}{n}} \prod _{x\in S_{\beta }}\prod _{z\in S_{\alpha }} \left( e_{L}(z)^n-e_{L}(x)^n\right) . \end{aligned}$$

Similarly, we have

$$\begin{aligned} \left( \frac{\beta }{\alpha }\right) _n =\nu (\beta )^{\frac{N(\alpha )-1}{n}} \prod _{x\in S_{\beta }}\prod _{z\in S_{\alpha }} \left( e_{L}(z)^n-e_{L}(x)^n\right) , \end{aligned}$$

which yields (3.1), as desired.\(\Box \)

4.2 Proof of Theorem 3.3

We retain the notations used in Theorem 3.1. Fix a sign function \(\text {sgn}:F_{\infty }^{*}\rightarrow \mathbb {F}_{\infty }^{*}\), where \(\mathbb {F}_{\infty }\) is the field of constants of \(F_{\infty }\). Let \(\text {sgn}(0)=0\). There exists an element \(c\in C_{\infty }\) such that \(\psi :=c\phi c^{-1}\) is a sgn-normalized Drinfeld R-module. When \(\kappa (\alpha )\) is the leading coefficient of \(\psi _{\alpha }(X)\), \(\kappa :R\rightarrow \mathbb {F}_{\infty }\) is a twisting of \(\text {sgn}\). From \(\kappa (\alpha )=c^{1-N(\alpha )}\nu (\alpha )\ (\alpha \in R)\), we have

$$\begin{aligned} \nu (\alpha )^{\frac{N(\beta )-1}{n}}\nu (\beta )^{-\frac{N(\alpha )-1}{n}} = \kappa (\alpha )^{\frac{N(\beta )-1}{n}}\kappa (\beta )^{-\frac{N(\alpha )-1}{n}}, \end{aligned}$$

which yields (3.2), as desired. \(\square \)