On the polynomial Ramanujan sums over finite fields

Abstract

The polynomial Ramanujan sum was first introduced by Carlitz (Duke Math J 14:1105–1120, 1947), and a generalized version by Cohen (Duke Math J 16:85–90, 1949). In this paper, we study the arithmetical and analytic properties of these sums, deriving various fundamental identities, such as Hölder formula, reciprocity formula, orthogonality relation, and Davenport–Hasse type formula. In particular, we show that the special Dirichlet series involving the polynomial Ramanujan sums are, indeed, the entire functions on the whole complex plane, and we also give a square mean values estimation. The main results of this paper are new appearance to us, which indicate the particularity of the polynomial Ramanujan sums.

1 Introduction and results statement

One hundred years ago, Ramanujan was the first to appreciate the importance of the following exponential sum:

$$\begin{aligned} c(m, n) = \sum _{\begin{array}{c} d=1,\\ (d,n)=1 \end{array}}^{n}e\left( \frac{md}{n}\right) , \end{aligned}$$

(1.1)

where $e(x) = e^{2\pi i x}$; n and m are positive integers. His interest in this sum originated in his desire to obtain expression for a variety of well-known arithmetical functions in the form of a series $\sum _n a_nc(m,n)$; in particular, he obtained some very famous identities, for example (see [29] and [41])

$$\begin{aligned} \sum _{n=1}^{+\infty }\frac{c(m,n)}{n}=0, \quad \sum _{n=1}^{+\infty }\frac{c(m,n)}{m}=-\Lambda (n) \quad \text{ and } \quad c(m,n)=\sum _{d|n, d|m}d \mu \left( \frac{n}{d}\right) , \end{aligned}$$

(1.2)

where $\mu (n)$ is the Möbius function; $\Lambda (n)$ is the Mangoldt function, and the last equality is usually said to be the Kluyver’s equality. Following him, many other authors were also interested in this fascinating sum [6, 7, 36], especially it makes surprising appearances in singular series of the Hardy–Littlewood asymptotic formula for Waring problems and in the asymptotic formula of Vingradov on sums of three primes. For details, the reader is referred to [18].

Many mathematicians later tried to generalize this sum to find more and more applications. One of the most popular generalization was given by Cohen [10,11,12, 14] (or see [46]) that

$$\begin{aligned} c_k(m, n) = \sum _{\begin{array}{c} d=1,\\ (d,n^k)_k=1 \end{array}}^{n}e\left( \frac{md}{n^k}\right) , \end{aligned}$$

(1.3)

where the g.c.d function $(a,b)_k$ is the greatest common kth power divisors of the integers a and b. In [10] and [12], Cohen presented an analogue of the Kluyver and Hölder formula for the above generalized Ramanujan sums

$$\begin{aligned} c_k(m, n) = \sum _{d|n, d^k|m}d^k \mu \left( \frac{n}{d}\right) = \phi _k(n)\mu (N)\phi _k^{-1}(N), \end{aligned}$$

(1.4)

where $N^k = \frac{n^k}{(m, n^k)_k}$, and $\phi _k(n)$ is the Jordan totient function given by the following product expression (z is a complex number):

$$\begin{aligned} \phi _z(n) = n^z \prod _{\begin{array}{c} p|n\\ p\,\mathrm{prime} \end{array}}\left( 1-\frac{1}{p^z}\right) . \end{aligned}$$

(1.5)

Various other generalizations were discussed in many papers including that of [3, 9, 16, 32, 35, 42, 50]. Here, we mention another interesting result, namely reciprocity formula. In [26, 27], Johenson showed that

$$\begin{aligned} \frac{\mu (\bar{m})c(nm^*, m)}{m^*} = \frac{\mu (\bar{n})c(mn^*, n)}{n^*}, \end{aligned}$$

(1.6)

where $\bar{n}$ denotes the largest square-free divisions of n, and $n^* = \frac{n}{\bar{n}}$.

In more recent years, people have more and more interests in this sum; it appeared in various other seemingly unrelated problems. In Algebra, Ramanujan sums as the super characters exhibit a new application of the theory of super characters [20], which recently developed by Diaconis–Issacs and Andre’ [15], and Ramanujan sums in arithmetical semigroup [22]. In number theory, Ramanujan sums appeared in the study of Waring type formula [30], the distribution of rational numbers in short interval [28], equirepartition modulo odd integers [5], the distribution of average of Ramanujan sums [1, 2, 8, 21, 34, 37, 44, 49], graph theory [17], symmetry classes of tensors [47], combinatorics [43], cyclotomic polynomials [19, 33, 48], and Mahler matrices [31]. In physics, Ramanujan sums have applications in the processing of low-frequency noise [39], and of long-period sequences [40], and in the study of quantum phase locking [38].

The main purpose of this paper is to generalize the Ramanujan sums to the polynomial case, and discuss various analogue properties of the classical case. The polynomial Ramanujan sum was first introduced by Carlitz [7], and a generalized version by Cohen [10]. To state their definition, let $\mathbb {F}_q$ be a finite field with $q=p^l$ elements, where p is a prime number; $\mathbb {F}_q[x]$ be the polynomial ring. Suppose H is a fixed polynomial in $\mathbb {F}_q[x]$, and $H \ne 0$, we first choose a complex-valued character of the additive group of the residue class ring $\mathbb {F}_q[x]/\langle H \rangle$; these characters are said to be the additive characters modulo H on $\mathbb {F}_q[x]$. If A is in $\mathbb {F}_q[x]$, let $A \equiv a_{m-1}x^{m-1} + \cdots + a_1x + a_0 ~({{\mathrm{mod}}}H)$, where $m=\deg (H)$, we set an additive function modulo H on $\mathbb {F}_q[x]$ by $t(A) = a_{m-1}$. Then, for any A and B in $\mathbb {F}_q[x]$, we have $t(A+B) = t(A) + t(B)$, and $t(A) = t(B)$ if $A \equiv B~({{\mathrm{mod}}}H)$, in particular, $t(A) = 0$ whenever H|A. To generalize this t-function modulo H, for any given polynomial G in $\mathbb {F}_q[x]$, we let $t_G(A) = t(GA)$. Clearly, $t_G$ is also an additive function modulo H, that is

$$\begin{aligned} t_G(A+B) = t_G(A) + t_G(B)\quad \text{ and } \quad t_G(A) = t_G(B) \text{ if } \quad A \equiv B~({{\mathrm{mod}}}H). \end{aligned}$$

(1.7)

Next, let $\lambda$ be a fixed non-principal character on $\mathbb {F}_q$, for example, one may choose $\lambda (a) = e\big (\frac{{{\mathrm{tr}}}(a)}{p}\big )$ for $a\in \mathbb {F}_q$, where ${{\mathrm{tr}}}(a)$ is the trace map from $\mathbb {F}_q$ to $\mathbb {F}_p$. We define a complex-valued function E(G, H) on $F_q[x]$ by

$$\begin{aligned} E(G, H)(A) = \lambda (t_G(A)). \end{aligned}$$

(1.8)

It is easy to see that E(G, H) is an additive character modulo H on $\mathbb {F}_q[x]$, and

$$\begin{aligned} E(G, H)(A) = E(A, H)(G). \end{aligned}$$

(1.9)

The polynomial Ramanujan sum modulo H on $\mathbb {F}_q[x]$ is given by (see Carlitz [7, 4.1])

$$\begin{aligned} \eta (G, H) = \sum _{\begin{array}{c} D {{\mathrm{mod}}}H,\\ (D,H)=1 \end{array}} E(G, H)(D), \end{aligned}$$

(1.10)

where the summation extends over a complete residue system modulo H in $\mathbb {F}_q[x]$. If $k\ge 1$ is a fixed integer, the generalized version of Cohen (see [10, 3.3]) is the following:

$$\begin{aligned} \eta _k(G, H) = \sum _{\begin{array}{c} D {{\mathrm{mod}}}H^k,\\ (D,H^k)_k=1 \end{array}} E(G, H^k)(D), \end{aligned}$$

(1.11)

where the summation ranges over a complete residue system modulo $H^k$ in $\mathbb {F}_q[x]$, and the g.c.d. function $(A, B)_k$ denotes the largest kth power common divisor (monic) of the polynomials A and B in $\mathbb {F}_q[x]$. We set $(A, B)_1 = (A, B)$ the usual g.c.d. function. If $D {{\mathrm{mod}}}H^k$ with $(D,H^k)_k=1$, which is said to be a k-reduced residue system modulo H according to Cohen [11, 12]. Clearly, $\eta _1(G, H) = \eta (G, H)$.

By the above notations, it is easy to verify (see [10, (3.4) and (3.5)]) that

$$\begin{aligned} \eta _k(G_1, H) = \eta _k(G_2, H), \quad \text{ if } G_1 \equiv G_2~({{\mathrm{mod}}}H^k), \end{aligned}$$

(1.12)

$$\begin{aligned} \eta _k(G, H_1H_2) = \eta _k(G, H_1)\eta _k(G, H_2),\quad \text{ if } (H_1, H_2)=1, \end{aligned}$$

(1.13)

and

$$\begin{aligned} \eta _k(G, H) = \sum _{D|H, D^k|G}|D|^k \mu \left( \frac{H}{D}\right) , \end{aligned}$$

(1.14)

where and later, $\mu (H)$ is the Möbius function on $\mathbb {F}_q[x]$, $|D| = q^{\deg (D)}$ is the absolute value function on $\mathbb {F}_q[x]$, and D|H means D is a monic divisor of H, and $\sum _{D|H}$ means D extending over all of monic divisors of H.

We note that the additive characters E(G, H) given by (1.8) are, indeed, all of the additive characters $\psi$ modulo H; in other words, for any additive character $\psi$ modulo H, there exists a unique polynomial G in $\mathbb {F}_q[x]$, such that $\psi = E(G, H)$, and $\deg (G) < \deg (H)$ (See Lemma 2.1). Therefore, the polynomial Ramanujan sums $\eta (G, H)$ and $\eta _k(G,H)$ coincide with the classical sums c(m, n) and $c_k(m,n)$ respectively.

The first result of this paper is to derive an analogue of Hölder formula for the polynomial sums. There is no essential difficulty to do this, but we make use of a simpler method to show (see Theorem 2.1) that

$$\begin{aligned} \eta (G, H) =&\,\phi (H)\mu \left( \frac{H}{(G, H)}\right) \phi ^{-1}\left( \frac{H}{(G, H)}\right) ,\nonumber \\&\text{ and } \quad \eta _k(G, H)=\phi _k(H)\mu (N)\phi _k^{-1}(N), \end{aligned}$$

(1.15)

where $\phi (H)$ is the Euler totient function, and $\phi _k(H)$ is the Jordan totient function on $\mathbb {F}_q[x]$, and $N^k = \frac{H^k}{(G, H)_k}$. As we know, in the classical case, the proof of (1.4) by Cohen is more complicated (see Theorem 1 of [12]).

The second result is to present an analogue of the reciprocity formula for the polynomial Ramanujan sums (see Theorem 3.1). Let $\bar{H}$ be the largest square-free divisor of H, and $H^* = \frac{H}{\bar{H}}$, then we have

$$\begin{aligned} \frac{\mu (\bar{H})\eta (GH^*, H)}{|H^*|} = \frac{\mu (\bar{G})\eta (HG^*, G)}{|G^*|}. \end{aligned}$$

(1.16)

The generalized sums $\eta _k(G, H)$ seemingly cannot share this kind of formula when $k>1$; however, we derive the second reciprocity formula for $\eta _k(G, H)$ (see Theorem 3.3): if $D_1^{k}|H$ and $D_2^k | H$, then

$$\begin{aligned} \phi _k(D_2)\eta _k\left( \frac{H}{D_2^k}, D_1\right) = \phi _k(D_1)\eta _k\left( \frac{H}{D_1^k}, D_2\right) . \end{aligned}$$

(1.17)

The importance of (1.17) lies in the fact that it is equivalent to the Hölder formula (1.15), and plays a role in the proof of the orthogonal relation formula (4.13).

The main results of this paper are the following theorems on the special Dirichlet series involving in the polynomial Ramanujan sums. Let $\mathbb {A}$ be the set of monic polynomials of $\mathbb {F}_q[x]$. We define

$$\begin{aligned} \delta _k(s, G) = \sum _{H \in \mathbb {A}}\frac{\eta _k(G, H)}{|H|^s} = \sum _{n=0}^{+\infty }A(n)q^{-ns} \end{aligned}$$

(1.18)

and

$$\begin{aligned} \tau _k(s, H) = \sum _{G \in \mathbb {A}}\frac{\eta _k(G, H)}{|G|^s} = \sum _{n=0}^{+\infty }B(n)q^{-ns}, \end{aligned}$$

(1.18')

where $s = c+it$ is a complex number, and

$$\begin{aligned} A(n) = \sum _{\begin{array}{c} H\in \mathbb {A} \\ \deg (H)=n \end{array}}\eta _k(G, H), \quad B(n) = \sum _{\begin{array}{c} G\in \mathbb {A} \\ \deg (G)=n \end{array}}\eta _k(G, H). \end{aligned}$$

The last two infinite series in (1.18) and (1.18’) are the definitions of $\delta _k(s, G)$ and $\tau _k(s, H)$, which tell us how to understand the special Dirichlet series in the middle of (1.18) and (1.18’). We show that

Theorem 1.1

If $G \ne 0$, then $\delta _k(s, G)$ is an entire function on the whole complex plane, and we have

$$\begin{aligned} \delta _k(s, G) = (1-q^{1-s})\sum _{D^k|G}|D|^{k-s}. \end{aligned}$$

(1.19)

In particular, we have

$$\begin{aligned} \delta _k(1, G) = \sum _{H \in \mathbb {A}}\frac{\eta _k(G,H)}{|H|}=0. \end{aligned}$$

(1.20)

Furthermore, for any real numbers c and $T>0$, we also have the following square mean value estimation:

$$\begin{aligned} \frac{1}{2T}\int _{-T}^{T}|\delta _k(c+it, G)|^2 \mathrm{d}t =&\,\Big (1+q^{2(1-c)}\Big )\sigma _0(2(k-c), G)\nonumber \\&- 2q^{1-k}\sigma _1(2(k-c), G)+O\left( \frac{1}{T}\right) , \end{aligned}$$

(1.21)

where (x is a real number)

$$\begin{aligned} \sigma _0(x, G) = \sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ \deg (D_1) = \deg (D_2) \end{array}}|D_1|^x,\quad \text{ and } \quad \sigma _1(x, G)=\sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ \deg (D_1) = \deg (D_2)+1 \end{array}}|D_1|^x, \end{aligned}$$

(1.22)

and the constant implied by “O” depends on q, G, and c only.

If $G=0$, then $E(G, H^k)$ is the principal additive character modulo $H^k$, and $\eta _k(G, H) = \phi _k(H)$ by (1.11). It is easy to see that

$$\begin{aligned} \delta _k(s, G) = \zeta _{\mathbb {A}}^{-1}(s) \zeta _{\mathbb {A}}(s-k), \quad \text{ if } {{\mathrm{Re}}}(s)>k+1, \end{aligned}$$

(1.23)

where $\zeta _{\mathbb {A}}$ is the zeta function on $\mathbb {F}_q[x]$ given by

$$\begin{aligned} \zeta _{\mathbb {A}}(s) = \sum _{H\in \mathbb {A}}\frac{1}{|H|^s}. \end{aligned}$$

(1.24)

It is well known (see [45, Chap. 2]) that

$$\begin{aligned} \zeta _{\mathbb {A}}(s) = \Big (1-q^{1-s}\Big )^{-1}, \quad \text{ if } {{\mathrm{Re}}}(s)>1. \end{aligned}$$

(1.25)

Therefore, $\eta _k(s, G)$ has a simple pole $s=k+1$ with residue $\frac{1}{\zeta _{\mathbb {A}}(k+1)\log q}$ at $s=k+1$, when $G=0$.

Theorem 1.2

If H is a positive degree polynomial in $\mathbb {F}_q[x]$, then $\tau _k(s, H)$ is an entire function on the whole complex plane, and we have

$$\begin{aligned} \tau _k(s, H) = (1-q^{1-s})^{-1}\phi _{k(1-s)}(H), \end{aligned}$$

(1.26)

where $\phi _{k(1-s)}(H)$ is the generalized Jordan totient function given by

$$\begin{aligned} \phi _z(H) = |H|^z \prod _{\begin{array}{c} P|H\\ P ~\mathrm{irreducible} \end{array}}\left( 1-\frac{1}{|P|^z}\right) . \end{aligned}$$

(1.27)

Especially, if $s=1$, we have

$$\begin{aligned} \tau _k(1, H) = \sum _{G\in \mathbb {A}}\frac{\eta _k(G, H)}{|G|} = -\frac{k\Lambda (H)}{\log q}, \end{aligned}$$

(1.28)

where $\Lambda (H)$ is the Mangoldt function on $\mathbb {F}_q[x]$. Moreover, for any real numbers c and T with $c\ne 1$ and $T>0$, we have

$$\begin{aligned} \frac{1}{2T}\int _{-T}^{T}|\tau _k(c+it, H)|^2 \mathrm{d}t =&\big |1-q^{2c(1-c)}\big |^{-1}\nonumber \\&\times \sum _{\begin{array}{c} D_1|H, D_2|H\\ \deg (D_1)=\deg (D_2) \end{array}}\mu \left( \frac{H}{D_1}\right) \mu \left( \frac{H}{D_2}\right) |D_1|^{2(1-c)} \nonumber \\&+2|1-q^{2(1-c)}|^{-1}\nonumber \\&\times \sum _{\begin{array}{c} D_1|H, D_2|H\\ \deg (D_1)>\deg (D_2) \end{array}} \mu \left( \frac{H}{D_1}\right) \mu \left( \frac{H}{D_2}\right) \nonumber \\&\times |D_1|^{(k+2)(1-c)}|D_2|^{(2-k)(1-c)} + O\left( \frac{1}{T}\right) , \end{aligned}$$

(1.29)

where the constant implied by “O” depends on q, H, and c only.

If $\deg (H)=0$, then $\eta _k(G, H) = 1$ for any G in $\mathbb {F}_q[x]$, it follows that $\tau _k(s, H) = \zeta _{\mathbb {A}}(s)$, which has a simple pole $s=1$ with residue $\frac{1}{\log q}$ at $s=1$. If $c=1$, the square mean value estimation is more complicated, and we will present it in another place.

The polynomial Ramanujan sum $\eta (G, H)$ in fact is a special type of Gauss sum on $\mathbb {F}_q[x]$. In [51], we presented an analogue of Davenport–Hasse’s theorem for the polynomial Gauss sums (see [51, Theorem 1.3]). In the last section of this paper, we show that the generalized polynomial Ramanujan sums $\eta _k(G, H)$ also share this kind of Davenport–Hasse’s formula (see Theorem 7.1).

Throughout this paper, P denotes an irreducible polynomial in $\mathbb {F}_q[x]$, D|H means that D is a monic divisor of H; $\sum _{D|H}$ means D extending over all of monic divisors of H, and $|H| = q^{\deg (H)}$ is the absolute value function on $\mathbb {F}_q[x]$.

2 Preliminaries

We start this section by determining the construction of the additive character group modulo H on $\mathbb {F}_q[x]$ via E(G, H).

Lemma 2.1

For any $\psi$, an additive character modulo H on $\mathbb {F}_q[x]$, there exists a unique polynomial G in $\mathbb {F}_q[x]$ such that $\psi =E(G, H)$, and $\deg (G)< \deg (H)$.

Proof

For the sake of convenience, we write $\psi _G = E(G, H)$. By (1.8), we have $\psi _{G_1} = \psi _{G_2}$, if $G_1 \equiv G_2~({{\mathrm{mod}}}H)$, and hence we may set G in a complete residue system modulo H in $\mathbb {F}_q[x]$, so that $\deg (G)<\deg (H)$. Moreover, we have

$$\begin{aligned} \psi _{G_1 + G_2} = \psi _{G_1} \cdot \psi _{G_2} \quad \text{ and } \quad \bar{\psi }_G = \psi _{-G}, \end{aligned}$$

(2.1)

where $\bar{\psi }_G$ is the usual conjugation of a complex number $\psi _G$. Since $\psi _G = \psi _0$, the principal additive character modulo H, if $G=0$, or H|G. Conversely, we have $\psi _G = \psi _0$ if and only if H|G. To show that statement, we see that $\lambda$ is a non-principal additive character on $\mathbb {F}_q$ by assumption, then there is an element a in $\mathbb {F}_q$, so that $\lambda (a)\ne 1$. If $H\not \mid G$, we let

$$\begin{aligned} R = (G, H) = x^k + a_{k-1}x^k + \cdots + a_1x + a_0\in \mathbb {F}_q[x], \end{aligned}$$

(2.2)

where $0\le k \le m-1$, and $m=\deg (H)$. It follows that

$$\begin{aligned} ax^{m-1-k}R = ax^{m-1}+ \cdots . \end{aligned}$$

We note the following congruent equation in variable T that

$$\begin{aligned} GT \equiv ax^{m-1-k}R ~({{\mathrm{mod}}}H) \end{aligned}$$

(2.3)

is solvable in $\mathbb {F}_q[x]$; therefore, there exists a polynomial A in $\mathbb {F}_q[x]$ such that

$$\begin{aligned} GA \equiv ax^{m-1}+\cdots . ~({{\mathrm{mod}}}H), \end{aligned}$$

(2.4)

and $t_G(A) = t(GA) = a$, so $\psi _G(A) = \lambda (t_G(A)) = \lambda (a)\ne 1$, and $\psi _G \ne \psi _0$. By (2.1), we have immediately

$$\begin{aligned} \psi _{G_1}\ne \psi _{G_2}, \quad \text{ if } G_1 \not \equiv G_2~({{\mathrm{mod}}}H) \end{aligned}$$

(2.5)

because if $\psi _{G_1} = \psi _{G_2}$ then $\psi _{G_1-G_2} = \psi _0$, and $G_1 \equiv G_2~({{\mathrm{mod}}}H)$. This shows that $\psi _G$ are different from each other when G running through a complete residue system of modulo H. Hence, there are exactly $|H|=q^m$ different characters $\psi _G$, but the number of additive characters modulo H on $\mathbb {F}_q[x]$ is exactly $q^m$, thus every character $\psi$ is just of the form $\psi _G$. We complete the proof of Lemma 2.1. $\square$

Next two lemmas are not new; one may find them in Carlitz [7] (see [7, (2.4), (2.5), and (2.6)]), but we give a more explicit expression here.

Lemma 2.2

If A is a monic polynomial in $\mathbb {F}_q[x]$, then we have

$$\begin{aligned} E(GA, HA) = E(G, H). \end{aligned}$$

(2.6)

Proof

For any $B \in \mathbb {F}_q[x]$, let

$$\begin{aligned} GB \equiv a_{m-1}x^{m-1} + \cdots + a_1x+a_0~({{\mathrm{mod}}}H), \end{aligned}$$

then

$$\begin{aligned} AGB \equiv A(a_{m-1}x^{m-1} + \cdots +a_0)~({{\mathrm{mod}}}AH). \end{aligned}$$

Because A is a monic polynomial, we see that the function $t_{GA}$ modulo HA just is the function $t_G$ modulo H. It follows that

$$\begin{aligned} E(GA, HA)(B) = \lambda (t_G(B)) = E(G, H)(B), \end{aligned}$$

(2.7)

and the lemma follows immediately. $\square$

Lemma 2.3

Suppose $A\in \mathbb {F}_q[x]$, then we have

$$\begin{aligned} \sum _{G{{\mathrm{mod}}}H}E(G, H)(A) = \left\{ \begin{array}{lll} |{ H}|, \quad \;\; \text{ if } {} { H}|{ A},\\ 0, \quad \qquad \text{ otherwise }, \end{array}\right. \end{aligned}$$

(2.8)

where the summation extends over a complete residue system modulo H.

Proof

By Lemma 2.1, it just is the orthogonal relation formula; we have the lemma at once. $\square$

The following lemma is a more complicated orthogonal relation, which will be used in the next section.

Lemma 2.4

If $D_1 | H$, $D_2 |H$, $(X, D_1^k)_k = 1$, and $(Y, D_2^k)_k=1$, where X and Y are two polynomials in $\mathbb {F}_q[x]$ with $\deg (X)<k\deg (D_1)$ and $\deg (Y)<k\deg (D_2)$, then we have

$$\begin{aligned}&\sum _{A+B\equiv G~({{\mathrm{mod}}}H^k)}E(X, D_1^k)(A) E(Y, D_2^k)(B)\nonumber \\&\quad =\left\{ \begin{array}{lll} |H|^kE(X, D^k), \text {if } X=Y, D_1=D_2=D,\\ 0, \quad \quad \quad \qquad \qquad \text{ otherwise }, \end{array} \right. \end{aligned}$$

(2.9)

where the summation ranges over a complete residue system modulo $H^k$.

Proof

We first note that $E(G, H) = E(G, a^{-1}H)$, where a is the leading coefficient of H. Without loss of generality, we may suppose that H is a monic polynomial, and write $D_1R_1 = H$, $D_2R_2 = H$, and $B=G-A~({{\mathrm{mod}}}H^k)$, then the left side of (2.9) is (see (1.9)) that

$$\begin{aligned}&\sum _{A{{\mathrm{mod}}}H^k}E\big (X, D_1^k\big )(A)E\big (Y, D_2^k\big )(G-A)\nonumber \\&\quad =E(Y, D_2^k)(G) \sum _{A{{\mathrm{mod}}}H^k} E\big (XR_1^k, H^k\big )(A)E\big (YR_2^k, H^k\big )(-A)\nonumber \\&\quad =E(Y, D_2^k)(G) \sum _{A{{\mathrm{mod}}}H^k} E\big (A, H^k\big )\big (XR_1^k-YR_2^k\big ). \end{aligned}$$

(2.10)

By Lemma 2.3, the inner sum in the above equality is zero, if $XR_1^k-YR_2^k \not \equiv 0 ~({{\mathrm{mod}}}H^k)$, and $|H|^k$, if $XR_1^k\equiv YR_2^k ~({{\mathrm{mod}}}H^k)$. Since $\deg (X)< k \deg (D_1)$ and $\deg (Y)<k \deg (D_2)$, we have $XR_1^k = YR_2^k$, if $XR_1^k\equiv YR_2^k ~({{\mathrm{mod}}}H^k)$. It follows that $XD_2^k = YD_1^k$, and $D_1 = D_2$, $X=Y$, since $(X, D_1^k)_k = (Y, D_2^k)_k = 1$. We complete the proof of this lemma. $\square$

Remark 2.1

According to Definition 3.3 of [24], we may give a modern way to approach the additive characters modulo H on $\mathbb {F}_q[x]$ (see (1.8)). Let $K = \mathbb {F}_q(x)$ be the field of rational functions over $\mathbb {F}_q$, and $K_{\infty }$ be the completion of K with respect to the infinite place. Then every $\alpha \in K_{\infty }$ can be expressed in a unique way in Laurent series of the form

$$\begin{aligned} \alpha = \sum _{i=n}^{+\infty } a_i \left( \frac{1}{x}\right) ^i,\quad n \in \mathbb {Z},\quad a_i \in \mathbb {F}_q\quad \text{ and } \quad a_n \ne 0. \end{aligned}$$

Let ${{\mathrm{res}}}_{\infty }(\alpha ) = a_1$, which is said to be the residue of $\alpha$ at the infinite place. Let H be a fixed polynomial in $\mathbb {F}_q[x]$, $H \ne 0$, and G be in $\mathbb {F}_q[x]$ with $\deg (G)< \deg (H)$. We may define an additive character E(G, H) on $K_{\infty }^+$ by (see also Sect. 3 of [25])

$$\begin{aligned} E(G, H)(\alpha ) = \lambda \left( {{\mathrm{res}}}_{\infty }\left( \frac{G\alpha }{H} \right) \right) , \end{aligned}$$

where $K_{\infty }^+$ is the additive group of the local field $K_{\infty }$, and $\lambda$ is a fixed non-principal character on $\mathbb {F}_q$ as before. It is easy to see that the restriction of E(G, H) on $\mathbb {F}_q[x]$ is an additive character modulo H on $\mathbb {F}_q[x]$. By this definition, we see that Lemma 2.2 is trivial.

Next, we give a simple proof of (1.14) and (1.15). In order to prove the Kluyver and Hölder’s formula in the polynomial case, we first make some slight modification of the Möbius inversion formula on $\mathbb {F}_q[x]$.

Definition 2.1

A non-zero mapping $\delta$ from $\mathbb {F}_q[x]$ to the complex plane is said to be an arithmetical function on $\mathbb {F}_q[x]$, if $\delta (aA) = \delta (A)$ for any $a\in \mathbb {F}_q^*$ and $A\in \mathbb {F}_q[x]$. It is said to be a multiplicative function, if $\delta (AB) = \delta (A)\cdot \delta (B)$, whenever $(A, B)=1$, and a complex multiplicative function, if $\delta (AB) = \delta (A)\cdot \delta (B)$ for any A, B in $\mathbb {F}_q[x]$.

We start with the following a few important examples of the arithmetical functions on $\mathbb {F}_q[x]$.

Möbius function $\mu (H)$: Let $\mu (0) = 0$, and

$$\begin{aligned} \mu (H) = \left\{ \begin{array}{lll} 1, \qquad \qquad \;\text {if } H\in \mathbb {F}_q^*,\\ 0, \qquad \qquad \;\text {if there exists a} \;P \;\text{ such } \text{ that }\; P^2|H,\\ (-1)^t, \qquad \text {if } H=P_1P_2 \cdots P_t, \text {where} \; P_j \;\text {are different}. \end{array}\right. \end{aligned}$$

(2.11)

It is easy to see that $\mu (H)$ is a multiplicative function on $\mathbb {F}_q[x]$. Moreover, we have the following identities that

$$\begin{aligned} \sum _{D|H}\mu (D)=\left\{ \begin{array}{lll} 1, \quad \text {if } \deg (H)=0,\\ 0, \quad \text {if } \deg (H)\ge 1, \end{array}\right. \end{aligned}$$

(2.12)

and

$$\begin{aligned} \sum _{D^k|H}\mu (D)=\left\{ \begin{array}{llll} 1, \quad \text{ if } D \;\text{ is } \; k \text{ th } \text{ power } \text{ divisors-free },\\ 0,\quad \text{ otherwise }. \end{array}\right. \end{aligned}$$

(2.13)

Euler totient function $\phi (H)$: If $H\ne 0$, we define $\phi (H)$ to be the number of polynomials of degree less than $\deg (H)$ that are coprime to H, and $\phi (0)=0$.

Jordan totient function $\phi _k(H)$: If $H\ne 0$, $\phi _k(H)$ is the number of polynomials of degree less than $k\cdot \deg (H)$ that have no common kth power divisors other than one with $H^k$.

Clearly, $\phi _1(H) = \phi (H)$, and $\phi _k(H)$ is a multiplicative function on $\mathbb {F}_q[x]$. The following equalities are easy to verify:

$$\begin{aligned} \sum _{D|H}\phi (D) = |H|\quad \text{ and } \quad \sum _{D|H}\phi _k(D) = |H|^k. \end{aligned}$$

(2.14)

Mangoldt function $\Lambda (H)$: We define the Mangoldt function $\Lambda (H)$ on $\mathbb {F}_q[x]$ by

$$\begin{aligned} \Lambda (H)=\left\{ \begin{array}{llll} \log |P|, \text {if}\; H=P^t, t\ge 1,\\ 0, \quad \qquad \text {otherwise}. \end{array} \right. \end{aligned}$$

(2.15)

It is easy to see that

$$\begin{aligned} \sum _{D|H}\Lambda (D) = \log |D|. \end{aligned}$$

(2.16)

Lemma 2.5

(Möbius Inversion Formula) If $\Delta (H)$ and $\delta (H)$ are two arithmetical functions on $\mathbb {F}_q[x]$, then

$$\begin{aligned} \Delta (H) = \sum _{D|H}\delta (D), \quad \text{ for } \text{ all } H\ne 0, \end{aligned}$$

(2.17)

if and only if

$$\begin{aligned} \delta (H) = \sum _{D|H}\mu (D)\Delta \left( \frac{H}{D}\right) , \quad \text{ for } \text{ all } H \ne 0. \end{aligned}$$

(2.18)

Proof

The proof is similar to the classical case; from (2.12), we have the lemma immediately. $\square$

As a direct consequence of Lemma 2.5, from (2.14) and (2.16), we have

$$\begin{aligned} \phi (H) = \sum _{D|H}\mu (D)\left| \frac{H}{D}\right| = |H|\prod _{P|H}\left( 1-\frac{1}{|P|}\right) , \end{aligned}$$

(2.19)

$$\begin{aligned} \phi _k(H) = \sum _{D|H}\mu (D)\left| \frac{H}{D}\right| ^k = |H|^k\prod _{P|H}\left( 1-\frac{1}{|P|^k}\right) , \end{aligned}$$

(2.20)

and

$$\begin{aligned} \Lambda (H) = \sum _{D|H}\log |D| \cdot \mu \left( \frac{H}{D}\right) . \end{aligned}$$

(2.21)

We give a simple proof of (1.14) and (1.15) now. The method here is an analogue of Anderson and Apostol [3], or see Apostol [4].

Lemma 2.6

(Kluyver’s formula) Let $\eta _k(G, H)$ be the polynomial Ramanujan sums given by (1.11), and $H \ne 0$, then we have

$$\begin{aligned} \eta _k(G, H) = \sum _{D|H, D^k|G}|D|^k \mu \left( \frac{H}{D}\right) . \end{aligned}$$

(2.22)

Proof

By (2.13), we have

$$\begin{aligned} \eta _k(G, H) =&\sum _{R {{\mathrm{mod}}}H^k} E(G, H^k)(R) \sum _{D^k | (R, H^k)_k}\mu (D)\nonumber \\ =&\sum _{D|H}\mu (D) \sum _{\begin{array}{c} R {{\mathrm{mod}}}H^k\\ D^k|R \end{array}}E(G, H^k)(R)\nonumber \\ =&\sum _{D|H}\mu (D) \sum _{\begin{array}{c} R {{\mathrm{mod}}}H^k\\ D^k|R \end{array}}E(R, H^k)(G). \end{aligned}$$

(2.23)

We write $R=A\cdot D^k$, and note that $E(R, H^k) = E(A, \left( \frac{H}{D}\right) ^k)$ by Lemma 2.2. It follows that

$$\begin{aligned} \eta _k(G, H) =&\sum _{D|H} \mu (D) \sum _{A {{\mathrm{mod}}}\left( \frac{H}{D}\right) ^k}E\left( A, \left( \frac{H}{D}\right) ^k\right) (G)\nonumber \\ =&\sum _{D|H}\mu \left( \frac{H}{D}\right) \sum _{A {{\mathrm{mod}}}D^k}E(A, D^k)(G)\nonumber \\ =&\sum _{\begin{array}{c} D|H\\ D^k|G \end{array}}\mu \left( \frac{H}{D}\right) |D|^k. \end{aligned}$$

(2.24)

We complete the proof of Lemma 2.6. $\square$

Theorem 2.1

(Hölder formula) For any G and H in $\mathbb {F}_q[x]$ and $H\ne 0$, we have

$$\begin{aligned} \eta _k(G, H) = \phi _k(H)\mu (N)\phi _k^{-1}(N), \quad \text{ where } N^k = \frac{H^k}{(G, H^k)_k}. \end{aligned}$$

(2.25)

Proof

Let $(G, H^k)_k = A^k$, then D|H and $D^k|G$ if and only if D|A. Let $N=\frac{H}{A}$, then by Lemma 2.6, we have

$$\begin{aligned} \eta _k(G, H) =&\sum _{D|A}|D|^k \mu \left( \frac{NA}{D}\right) = \sum _{D|A}\mu (ND)\left| \frac{A}{D}\right| ^k\nonumber \\ =&\mu (N)\sum _{\begin{array}{c} D|A\\ (D, N)=1 \end{array}}\mu (D)\left| \frac{A}{D}\right| ^k \nonumber \\ =&\mu (N)|A|^k \prod _{\begin{array}{c} P|A\\ P\not \mid N \end{array}}(1-|P|^{-k})\nonumber \\ =&\frac{\mu (N)|A|^k \prod _{P|A\cdot N}(1-|P|^{-k})}{\prod _{P|N}(1-|P|^{-k})}\nonumber \\ =&\frac{\mu (N)|H|^k \prod _{P|H}(1-|P|^{-k})}{|N|^k\prod _{P|N}(1-|P|^{-k})}\nonumber \\ =&\phi _k(H)\mu (N)\phi _k^{-1}(N), \end{aligned}$$

(2.26)

where $N^k = \frac{H^k}{A^k} = \frac{H^k}{(G, H^k)_k}$. We complete the proof of Theorem 2.1. $\square$

By (2.25), (1.15) and (1.14) follow immediately; in particular, we have

$$\begin{aligned} \eta _k(G, H) = \phi _k(H) \text{ if } H^k | G, \end{aligned}$$

(2.27)

and

$$\begin{aligned} \eta _k(G, H) = \mu (H) \text{ if } (G, H^k)_k = 1. \end{aligned}$$

(2.28)

3 Reciprocity formula

In this section, we give two reciprocity formulas for the polynomial Ramanujan sums. We start with the following lemmas.

Lemma 3.1

Let $\eta (G, H)$ be the polynomial Ramanujan sum given by (1.10), and H be a square-free polynomial, then $\mu (H)\eta (G, H)$ is a multiplicative function in variable G.

Proof

Let $f(G) = \mu (H) \eta (G, H)$, $H=P_1P_2\cdots P_n$, and $G = G_1\cdot G_2$, where $(G_1, G_2) =1$. We set

$$\begin{aligned} (G, H) = \prod _{i\in \gamma } P_i, \quad (G_1, H) = \prod _{i\in \gamma _1}P_i,\quad (G_2, H) = \prod _{i\in \gamma _2}P_i, \end{aligned}$$

(3.1)

where $\gamma _1 \cup \gamma _2 = \gamma \subset \{1, 2, \ldots , n\}$, and $\gamma _1 \cap \gamma _2$ is an empty set. Then by the Hölder formula (2.25) ($k=1$), we have

$$\begin{aligned} f(G) = (-1)^{-|\gamma |}\mu ^2(H)\prod _{i\in \gamma }\phi (P_i), \end{aligned}$$

(3.2)

$$\begin{aligned} f(G_1) = (-1)^{-|\gamma _1|}\mu ^2(H)\prod _{i\in \gamma _1}\phi (P_i), \end{aligned}$$

(3.3)

and

$$\begin{aligned} f(G_2) = (-1)^{-|\gamma _2|}\mu ^2(H)\prod _{i\in \gamma _2}\phi (P_i). \end{aligned}$$

(3.4)

Since $|\gamma _1| + |\gamma _2| = |\gamma |$, and $\mu ^2(H) = \mu ^4(H)$, then $f(G) = f(G_1)\cdot f(G_2)$, and the lemma follows. $\square$

Lemma 3.2

If H is square-free, then we have

$$\begin{aligned} \mu (H)\eta (G, H) = \sum _{D|(G, H)}|D|\mu (D). \end{aligned}$$

(3.5)

Proof

Since the both sides of (3.5) are multiplicative in G, it suffices to prove that when $G=P^s$. Let $H = P_1P_2\cdots P_n$. If $P \ne P_j$, $1\le j \le n$, then the both sides of (3.5) are one, so we may let $P=P_j$. It follows that

$$\begin{aligned} \mu (H)\eta (G, H) = -\mu ^2(H)\phi (P_j) = 1-|P_j|, \end{aligned}$$

(3.6)

and

$$\begin{aligned} \sum _{D|(G, H)} |D| \mu (D) = 1-|P_j|. \end{aligned}$$

(3.7)

The lemma follows at once. $\square$

We note that the sum in (3.5) is symmetric in G and H, and hence as a direct consequence of Lemma 3.2, we have

Corollary 3.1

Suppose that both G and H are square-free, then

$$\begin{aligned} \mu (H)\eta (G, H) = \mu (G)\eta (H, G). \end{aligned}$$

(3.8)

Theorem 3.1

(The first reciprocity formula) Let $\bar{H}$ be the largest square-free divisor of H, and $H^* = \frac{H}{\bar{H}}$, then

$$\begin{aligned} \frac{\mu (\bar{H})\eta (GH^*, H)}{|H^*|} = \frac{\mu (\bar{G})\eta (HG^*, G)}{|G^*|}. \end{aligned}$$

(3.9)

Proof

It is easy to verify the observation of Hardy that

$$\begin{aligned} \eta (G, H)=0 \text{ if } H^*\not \mid G \quad \text{ and } \quad \phi (H)= |H^*|\phi (\bar{H}). \end{aligned}$$

(3.10)

In order to prove (3.9), we first show that

$$\begin{aligned} \eta (GH^*, H) = |H^*| \eta (G, \bar{H}). \end{aligned}$$

(3.11)

By (2.25)($k=1$), one has

$$\begin{aligned} \eta (GH^*, H)&= \phi (H)\mu \left( \frac{H}{(GH^*, H)}\right) \phi ^{-1}\left( \frac{H}{(GH^*, H)}\right) \\&= |H^*|\phi (\bar{H})\mu \left( \frac{\bar{H}}{(G, \bar{H})}\right) \phi ^{-1}\left( \frac{\bar{H}}{(G, \bar{H})}\right) \\&= |H^*|\eta (G,\bar{H}), \end{aligned}$$

and (3.11) follows. We note that $\eta (G, \bar{H}) = \eta (\bar{G}, \bar{H})$; it follows by Corollary 3.1 that

$$\begin{aligned} \frac{\mu (\bar{H})\eta (GH^*, H)}{|H^*|}&= \mu (\bar{H})\eta (G, \bar{H}) = \mu (\bar{H})\eta (\bar{G}, \bar{H})\\&=\mu (\bar{G})\eta (\bar{H}, \bar{G}) = \mu (\bar{G})\eta (H, \bar{G}) \\&=\frac{\mu (\bar{G})\eta (HG^*, G)}{|G^*|}. \end{aligned}$$

We complete the proof of Theorem 3.1. $\square$

The next reciprocity property first appeared in [20, Theorem 3.8] in the rational case, and here, we present an analogue in the polynomial case.

Theorem 3.2

(The second reciprocity formula) If $D_1|H$, and $D_2|H$, then we have

$$\begin{aligned} \phi (D_2)\eta \left( \frac{H}{D_2}, D_1\right) = \phi (D_1)\eta \left( \frac{H}{D_1}, D_2\right) . \end{aligned}$$

(3.12)

Proof

By (2.25), then

$$\begin{aligned} \eta \left( \frac{H}{D_2}, D_1\right) = \phi (D_1)\mu \left( \frac{D_1}{\left( \frac{H}{D_2}, D_1\right) }\right) \phi ^{-1}\left( \frac{D_1}{\left( \frac{H}{D_2},D_1\right) }\right) \end{aligned}$$

(3.13)

and

$$\begin{aligned} \eta \left( \frac{H}{D_1}, D_2\right) = \phi (D_2)\mu \left( \frac{D_2}{\left( \frac{H}{D_1}, D_2\right) }\right) \phi ^{-1}\left( \frac{D_2}{\left( \frac{H}{D_1},D_2\right) }\right) . \end{aligned}$$

(3.14)

We write $D=D_1D_2$, then

$$\begin{aligned} \frac{D_1}{\left( \frac{H}{D_2},D_1\right) } = \frac{D}{(H,D)} = \frac{D_2}{\left( \frac{H}{D_1},D_2\right) }, \end{aligned}$$

(3.15)

and the theorem follows at once. $\square$

The importance of (3.12) lies in the fact that it is equivalent to the Hölder formula (2.25) ($k=1$). If we take $D_1 = H$, and $D_2 = \frac{H}{(G, H)}$ in (3.12), and note that

$$\begin{aligned} \eta (G, H) = \eta ((G, H), H), \end{aligned}$$

(3.16)

and $\eta (1, H) = \mu (H)$, then

$$\begin{aligned} \eta (G, H) = \phi (H)\mu \left( \frac{H}{(G, H)}\right) \phi ^{-1}\left( \frac{H}{(G, H)}\right) , \end{aligned}$$

(3.17)

which is the Hölder formula of $\eta (G, H)$.

The generalized polynomial Ramanujan sum $\eta _k(G, H)$ seemingly cannot share the first reciprocity formula when $k>1$, since G and H are not symmetric in a generalized version of (3.5). But, indeed, $\eta _k(G, H)$ shares the second reciprocity formula.

Theorem 3.3

If $D_1^k|H$ and $D_2^k|H$, then

$$\begin{aligned} \phi _k(D_2)\eta _k\left( \frac{H}{D_2^k}, D_1\right) = \phi _k(D_1)\eta _k\left( \frac{H}{D_1^k}, D_2\right) . \end{aligned}$$

(3.18)

Proof

By (2.25), we have

$$\begin{aligned} \eta _k\left( \frac{H}{D_2^k}, D_1\right) = \phi _k(D_1)\mu (N_1)\phi _k^{-1}(N_1) \end{aligned}$$

(3.19)

and

$$\begin{aligned} \eta _k\left( \frac{H}{D_1^k}, D_2\right) = \phi _k(D_2)\mu (N_2)\phi _k^{-1}(N_2), \end{aligned}$$

(3.20)

where $N_1^k = \frac{D_1^k}{(\frac{H}{D_2^k}, D_1^k)_k}$ and $N_2^k = \frac{D_2^k}{(\frac{H}{D_1^k}, D_2^k)_k}$.

We write $D=D_1D_2$, then

$$\begin{aligned} N_1^k = \frac{D^k}{(H,D^k)_k} = N_2^k. \end{aligned}$$

(3.21)

It follows that $N_1 = N_2$, and we have Theorem 3.3. $\square$

As the case of $k=1$, (3.18) is equivalent to (2.25). If we replace H by $H^k$ in (3.18), then if $D_1|H$ and $D_2|H$, we have

$$\begin{aligned} \phi _k(D_2)\eta _k\left( \frac{H^k}{D_2^k}, D_1\right) = \phi _k(D_1)\eta _k\left( \frac{H^k}{D_1^k}, D_2\right) . \end{aligned}$$

(3.22)

As an analogue of the above equality in the rational case, one may see Lemma 1 of [12]. Taking $D_1 = H$, and $D_2=\frac{H}{A}$, where $A^k = (G, H^k)_k$ in (3.22), we note that

$$\begin{aligned} \eta _k(G, H) = \eta _k(A^k, H) \quad \text{ and } \quad \eta _k(1, H) = \mu (H). \end{aligned}$$

(3.23)

It follows that

$$\begin{aligned} \eta _k(G, H) = \phi _k(H)\mu (N)\phi _k^{-1}(N), \end{aligned}$$

where $N=\frac{H}{A}$, and $N^k=\frac{H^k}{A^k} = \frac{H^k}{(G, H^k)_k}$, we have (2.25) at once.

To make applications of the first reciprocity formula, one may follow Johnson [26] to consider the C-series representations and the $C'$-series representation for the arithmetical function on $\mathbb {F}_q[x]$, and show that the two classes of representation are equivalent under certain conditions. Here we consider a special Dirichlet series and derive its values by using Theorem 3.1. We set

$$\begin{aligned} \sigma (s, H) = \sum _{\begin{array}{c} G\in \mathbb {A}\\ G~{\text {square-free}} \end{array}}\frac{\eta (G, H)}{|G|^s}, \end{aligned}$$

(3.24)

where $\mathbb {A}$ is the set of monic polynomials of $\mathbb {F}_q[x]$, and s is a complex variable. If H is square-free, and ${{\mathrm{Re}}}(s)>1$, we have the following formula:

$$\begin{aligned} \mu (H)\sigma (s, H) = \zeta _{\mathbb {A}}(s)\zeta ^{-1}_{\mathbb {A}}(2s)\prod _{P|H}\frac{1-|P|+|P|^s}{1+|P|^s}, \end{aligned}$$

(3.25)

where $\zeta _{\mathbb {A}}(s)$ is the zeta function on $\mathbb {F}_q[x]$ given by (1.24). If ${{\mathrm{Re}}}(s)>1$, then $\zeta _{\mathbb {A}}(s)$ has the following Euler product formula:

$$\begin{aligned} \zeta _{\mathbb {A}}(s) = \prod _{P\in \mathbb {A}}(1-|P|^{-s})^{-1}. \end{aligned}$$

(3.26)

To prove (3.25), by (3.8), we have

$$\begin{aligned} \mu (H)\sigma (s, H) = \sum _{\begin{array}{c} G\in \mathbb {A},\\ G~{\text {square-free}} \end{array}}\frac{\mu (G)\eta (H, G)}{|G|^s}, \end{aligned}$$

(3.27)

where $f(G) = \mu (G)\eta (H, G)$ is a multiplicative function in G, so we may make use of Euler product and obtain that

$$\begin{aligned} \mu (H)\sigma (s, H) = \prod _{P\in \mathbb {A}}\left( 1-\frac{\eta (H, P)}{|P|^s}\right) . \end{aligned}$$

It is easy to see that $\eta (H, P)=\mu (P)$, if $P \not \mid H$, and $\eta (H, P) = \phi (P)$, if P|H. Then, we have

$$\begin{aligned} \mu (H)\sigma (s, H)&= \prod _{\begin{array}{c} P\in \mathbb {A},\\ P\not \mid H \end{array}}\left( 1+\frac{1}{|P|^s}\right) \prod _{P|H}\left( 1-\frac{\phi (P)}{|P|^s}\right) \\&= \zeta _{\mathbb {A}}(s)\zeta _{\mathbb {A}}^{-1}(2s)\prod _{P|H}\frac{1-|P|+|P|^s}{1+|P|^s}, \end{aligned}$$

which is (3.25).

An application of the second reciprocity formula appears in the next section (See Lemma 4.4).

4 Orthogonality relation

In this section, we derive some more complicated orthogonal formula and a number of corollaries for the polynomial Ramanujan sums. In the rational case, one may find the analogues in Cohen [11, 12]. We begin by proving

Lemma 4.1

If $D_1|H$ and $D_2|H$, then for any G in $\mathbb {F}_q[x]$, we have

$$\begin{aligned} \sum _{A+B\equiv G~({{\mathrm{mod}}}H^k)}\eta _k(A, D_1)\eta _k(B, D_2)=\left\{ \begin{array}{lll} |H|^k\eta _k(G, D), \;\;\text {if}\; D_1=D_2=D,\\ 0, \qquad \qquad \qquad \qquad \text {otherwise}, \end{array} \right. \end{aligned}$$

(4.1)

where the summation extends over a complete residue system modulo $H^k$.

Proof

By Lemma 2.4, the left-hand side of (4.1) is

$$\begin{aligned}&\sum _{(X, D_1^k)_k=1}\sum _{(Y, D_2^k)_k=1}\sum _{A+B\equiv G~({{\mathrm{mod}}}H^k)}E(A, D_1^k)(X)E(B, D_2^k)(Y) \nonumber \\&\quad =|H|^k \sum _{(X, D^k)=1}E(X, D^k)(G) = |H|^k \eta _k(G, D), \end{aligned}$$

(4.2)

if $D_1=D_2=D$. Otherwise it is zero. We have the lemma immediately. $\square$

Definition 4.1

A k-reduced residue system modulo H is that D ranges over modulo $H^k$ such that $(D, H^k)_k = 1$.

Lemma 4.2

A complete residue system modulo $H^k$ is given by $A=R\big (\frac{H}{D}\big )^k$, where D ranges over the divisors of H, and for each D, R ranges over a k-reduced residue system modulo D.

Proof

See Cohen [13, Lemma 4]. $\square$

The next lemma is a more generalized form of the second reciprocity formula of $\eta _k(G, H)$.

Lemma 4.3

If $D_1^k|H$, and $D_2^k|H$, then for any R in $\mathbb {F}_q[x]$, we have

$$\begin{aligned} \phi _k(D_2)\eta _k\left( \frac{RH}{D_2^k}, D_1\right) = \phi _k(D_1)\eta _k\left( \frac{RH}{D_1^k}, D_2\right) . \end{aligned}$$

(4.3)

Proof

It follows directly from Theorem 3.3. $\square$

If we replace H by $H^k$ in the above equality, then we have the following corollary.

Corollary 4.1

If $D_1|H$, and $D_2|H$, then for any polynomial R in $\mathbb {F}_q[x]$ we have (comparing with [12, Lemma 1]) that

$$\begin{aligned} \phi _k(D_2)\eta _k\left( R\left( \frac{H}{D_2}\right) ^k, D_1\right) = \phi _k(D_1)\eta _k\left( R\left( \frac{H}{D_1}\right) ^k, D_2\right) . \end{aligned}$$

(4.4)

Lemma 4.4

If D|H, $D_1|H$, and $R\in \mathbb {F}_q[x]$ such that $(R, D^k)_k=1$, then we have

$$\begin{aligned} \eta _k\left( R\left( \frac{H}{D}\right) ^k, D_1 \right) = \eta _k\left( \left( \frac{H}{D}\right) ^k, D_1 \right) . \end{aligned}$$

(4.5)

Proof

By (4.3), then

$$\begin{aligned} \eta _k\left( R\left( \frac{H}{D}\right) ^k, D_1 \right) = \frac{\phi _k(D_1)}{\phi _k(D)}\eta _k\left( R\left( \frac{H}{D_1}\right) ^k, D \right) . \end{aligned}$$

(4.6)

Because of $(R, D^k)_k=1$, it is easy to see by (2.25) that

$$\begin{aligned} \eta _k\left( R\left( \frac{H}{D_1}\right) ^k, D \right) = \eta _k\left( \left( \frac{H}{D_1}\right) ^k, D\right) . \end{aligned}$$

(4.7)

By (4.3) once again, we have

$$\begin{aligned} \eta _k\left( \left( \frac{H}{D_1}\right) ^k, D \right) = \frac{\phi _k(D)}{\phi _k(D_1)}\eta _k\left( \left( \frac{H}{D}\right) ^k, D_1 \right) , \end{aligned}$$

(4.8)

and the lemma follows at once. $\square$

Now we state and prove the main result of this section.

Theorem 4.1

If $D_1|H$, and $D_2|H$, then

$$\begin{aligned} \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, D_1\right) \eta _k\left( \left( \frac{H}{D_2}\right) ^k, D\right) =\left\{ \begin{array}{llll} |H|^k, \text {if} \; D_1=D_2,\\ 0, \quad \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.9)

Proof

By Lemma 4.2, $A{{\mathrm{mod}}}H^k$ is given by $A = R\big (\frac{H}{D}\big )^k$, where D ranges over the divisors of H, and for each D, R ranges over a k-reduced residue system modulo D. Therefore, by (4.5), the left side of (4.1) may be written as

$$\begin{aligned} \sum _{A{{\mathrm{mod}}}H^k}&\eta _k(A, D_1)\eta _k(G-A, D_2) \nonumber \\&= \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, D_1 \right) \sum _{\begin{array}{c} R{{\mathrm{mod}}}D^k\\ (R, D^k)_k=1 \end{array}}\eta _k\left( G-R\left( \frac{H}{D}\right) ^k, D_2 \right) . \end{aligned}$$

(4.10)

We consider the inner sum of the right side of (4.10) separately, and denote this sum by S, then

$$\begin{aligned} S&= \sum _{\begin{array}{c} R{{\mathrm{mod}}}D^k\\ (R, D^k)_k=1 \end{array}}\sum _{\begin{array}{c} A{{\mathrm{mod}}}D_2^k\\ (A, D_2^k)_k=1 \end{array}}E\left( G-R\left( \frac{H}{D}\right) ^k, D_2^k\right) (A)\nonumber \\&=\sum _{\begin{array}{c} A{{\mathrm{mod}}}D_2^k\\ (A, D_2^k)_k=1 \end{array}}E(G, D_2^k)(A)\sum _{\begin{array}{c} R{{\mathrm{mod}}}D^k\\ (R, D^k)_k=1 \end{array}}E\left( -A\left( \frac{H}{D}\right) ^k, D_2^k\right) (R). \end{aligned}$$

(4.11)

We write $D_2 \cdot B = H$, by (2.6) of Lemma 2.2, then

$$\begin{aligned} E(-A\left( \frac{H}{D}\right) ^k, D_2^k)&= E\left( -A\left( \frac{H}{D}\right) ^k B^k, H^k\right) \nonumber \\&= E\left( -A\left( \frac{H}{D_2}\right) ^k \left( \frac{H}{D}\right) ^k, H^k\right) \nonumber \\&= E\left( -A\left( \frac{H}{D_2}\right) ^k, D^k\right) . \end{aligned}$$

(4.12)

It follows that

$$\begin{aligned} S&=\sum _{\begin{array}{c} A{{\mathrm{mod}}}D_2^k\\ (A, D_2^k)_k=1 \end{array}}E(G, D_2^k)(A)\sum _{\begin{array}{c} R{{\mathrm{mod}}}D^k\\ (R, D^k)_k=1 \end{array}}E\left( -A\left( \frac{H}{D_2}\right) ^k, D^k\right) (R)\nonumber \\&=\sum _{\begin{array}{c} A{{\mathrm{mod}}}D_2^k\\ (A, D_2^k)_k=1 \end{array}}E(G, D_2^k)(A) \eta _k\left( -A\left( \frac{H}{D_2}\right) ^k, D\right) \nonumber \\&=\sum _{\begin{array}{c} A{{\mathrm{mod}}}D_2^k\\ (A, D_2^k)_k=1 \end{array}}E(G, D_2^k)(A) \eta _k\left( \left( \frac{H}{D_2}\right) ^k, D\right) \nonumber \\&=\eta _k(G, D_2)\eta _k\left( \left( \frac{H}{D_2}\right) ^k, D\right) . \end{aligned}$$

(4.13)

By Lemma 4.1 and (4.10), we have

$$\begin{aligned}&\eta _k(G, D_2) \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, D_1\right) \eta _k\left( \left( \frac{H}{D_2}\right) ^k, D\right) \nonumber \\&\quad =\left\{ \begin{array}{lll} |H|^k\eta _k(G, D), \;\text {if}\, D_1=D_2=D,\\ 0, \qquad \qquad \qquad \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.14)

If we take $G=0$ in the above equality, and note that $\eta _k(0, D_2) = \phi _k(D_2)$ (see (2.27)), then $\eta _k(0, D_2) \ne 0$, and we have

$$\begin{aligned} \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, D_1\right) \eta _k\left( \left( \frac{H}{D_2}\right) ^k, D\right) =\left\{ \begin{array}{llll} |H|^k,\qquad \; \text {if} \; D_1=D_2=D,\\ 0, \qquad \qquad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.15)

We complete the proof of Theorem 4.1. $\square$

Next, we deduce a number of the arithmetical relations, most of which are the straightforward consequences of (4.1) and (4.9).

Corollary 4.2

$$\begin{aligned} \sum _{G {{\mathrm{mod}}}H^k}\eta _k(G, H)=\left\{ \begin{array}{lll} 1, \quad \text {if} \;\deg (H)=0,\\ 0, \quad \text {if}\; \deg (H)\ge 1. \end{array} \right. \end{aligned}$$

(4.16)

Proof

Let $D_1 = H$, and $D_2=1$ in (4.1), and we have this corollary at once. $\square$

Corollary 4.3

$$\begin{aligned} \sum _{D|H}\eta _k(G, D)=\left\{ \begin{array}{llll} |H|^k, \text {if }\; G\equiv 0~({{\mathrm{mod}}}H^k),\\ 0, \quad \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.17)

Proof

Taking $D_1=1$ in Theorem 4.1, we have (4.17) immediately. $\square$

Corollary 4.4

$$\begin{aligned} \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, H\right) \eta _k(G, D)=\left\{ \begin{array}{lll} |H|^k, \text {if} (G, H^k)_k=1,\\ 0, \quad \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.18)

Proof

Let $D_1=H$ in (4.9), and by (4.5) we have this corollary. $\square$

Corollary 4.5

If $D_1|H$, then we have

$$\begin{aligned} \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, D_1\right) \phi _k(D)=\left\{ \begin{array}{llll} |H|^k, \text {if}\; D_1=1,\\ 0, \quad \quad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.19)

Proof

If we take $D_2=1$ in (4.9), and note that $\eta _k(H^k, D)=\phi _k(D)$, then (4.19) follows immediately. $\square$

Using (4.3) with $R=1$, we may reformulate Theorem 4.1 as follows.

Corollary 4.6

If $D_1|H$ and $D_2|H$, then

$$\begin{aligned} \sum _{D|H}\frac{1}{\phi _k(D)}\eta _k\left( \left( \frac{H}{D_1}\right) ^k, D\right) \eta _k\left( \left( \frac{H}{D_2}\right) ^k, D\right) =\left\{ \begin{array}{lll} \frac{|H|^k}{\phi _k(D')}, \text {if} \; D_1=D_2=D',\\ 0, \quad \qquad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.20)

In particular, for any $D_1|H$, we have

$$\begin{aligned} \sum _{D|H}\frac{1}{\phi _k(D)}\eta _k\left( \left( \frac{H}{D_1}\right) ^k, D\right) ^2 = \frac{|H|^k}{\phi _k(D_1)}. \end{aligned}$$

(4.21)

If we make use of the Hölder formula in the above equality, it yields

Corollary 4.7

If $D_1D_1'=H$, then

$$\begin{aligned} \sum _{\begin{array}{c} D|H\\ N(D_1', D)=D \end{array}}\phi _k(D)\left( \frac{\mu (N)}{\phi _k(N)}\right) ^2 = \frac{|H|^k}{\phi _k(D_1)}. \end{aligned}$$

(4.22)

The special cases of $D_1=1$ and $D_1=H$ lead to the following corollaries, respectively

Corollary 4.8

(see (2.14))

$$\begin{aligned} \sum _{D|H}\phi _k(D) = |H|^k \end{aligned}$$

(4.23)

and

Corollary 4.9

$$\begin{aligned} \sum _{D|H}\frac{|\mu (D)|}{\phi _k(D)} = \frac{|H|^k}{\phi _k(H)}. \end{aligned}$$

(4.24)

In fact, we have the following more generalized conclusions.

Lemma 4.5

If R|H, then we have

$$\begin{aligned} \alpha _k(R, H)=\sum _{\begin{array}{c} D|\frac{H}{R}\\ (D, R)=1 \end{array}}\frac{|\mu (D)|}{\phi _k(D)} = \frac{\phi _k(R)}{\phi _k(H)}\left| \frac{H}{R}\right| ^k. \end{aligned}$$

(4.25)

Proof

The left side of (4.25) has the following product expression:

$$\begin{aligned} \alpha _k(R, H)&= \prod _{\begin{array}{c} P|\frac{H}{R}\\ P\not \mid R \end{array}}\left( 1+\frac{1}{\phi _k(P)}\right) \\&= \prod _{P|H}\left( 1+\frac{1}{\phi _k(P)}\right) \prod _{P|R}\left( 1+\frac{1}{\phi _k(P)}\right) ^{-1}\\&= \frac{\phi _k(R)}{\phi _k(H)}\left| \frac{H}{R}\right| ^k, \end{aligned}$$

and the lemma follows. $\square$

Next, we generalize the Jordan totient function $\phi _k(H)$ to $\phi _s(H)$, where s is a complex number variable. We define (see (1.5) in the rational case)

$$\begin{aligned} \phi _s(H) = \sum _{D|H} |D|^s \mu \left( \frac{H}{D}\right) = |H|^s\prod _{P|H}\left( 1-\frac{1}{|P|^s}\right) . \end{aligned}$$

(4.26)

Lemma 4.6

If s is an arbitrary complex number, then

$$\begin{aligned} \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, H\right) \phi _{ks}(D) = |H|^{ks}\phi _{k(1-s)}(H). \end{aligned}$$

(4.27)

Proof

By (2.25) and (4.26), we have

$$\begin{aligned} \sum _{D|H}\eta _k\left( \left( \frac{H}{D}\right) ^k, H\right) \phi _{ks}(D)&= \phi _k(H)\sum _{D|H}\frac{\mu (D)}{\phi _k(D)}\phi _{ks}(D)\nonumber \\&=\phi _k(H)\sum _{D|H}\frac{\mu (D)}{\phi _k(D)}\sum _{R|D}|R|^{ks}\mu \left( \frac{D}{R}\right) \nonumber \\&=\phi _k(H)\sum _{D|H}\sum _{\begin{array}{c} R\cdot E = D\\ (R, E)=1 \end{array}}\frac{\mu ^2(E)|R|^{ks}\mu (R)}{\phi _k(R)\phi _k(E)}\nonumber \\&=\phi _k(H)\sum _{R|H}\frac{|R|^{ks}\mu (R)}{\phi _k(R)}\sum _{\begin{array}{c} E|\frac{H}{R}\\ (E, R)=1 \end{array}}\frac{|\mu (E)|}{\phi _k(E)}\nonumber \\&=|H|^{ks}\phi _{k(1-s)}(H), \end{aligned}$$

(4.28)

and the lemma follows. $\square$

We may obtain more orthogonal relation formulas from (4.1) and (4.9). For example, if we take $G=0$ in (4.1), and note that $\eta _k(-B, D_2) = \eta _k(B, D_2)$, if follows that

Corollary 4.10

If $D_1|H$ and $D_2|H$, then

$$\begin{aligned} \sum _{A{{\mathrm{mod}}}H^k}\eta _k(A, D_1)\eta _k(A, D_2)=\left\{ \begin{array}{lll} |H|^k\phi _k(D), \;\;\text {if}\; D_1=D_2=D,\\ 0, \,\qquad \qquad \qquad \text {otherwise}. \end{array} \right. \end{aligned}$$

(4.29)

In particular, if D|H, then

$$\begin{aligned} \sum _{A{{\mathrm{mod}}}H^k}\eta _k(A, D)^2 = |H|^k \phi _k(D). \end{aligned}$$

(4.30)

5 The series $\delta _k(s, G)$

In this section, we prove Theorem 1.1. Recall that the Dirichlet series $\delta _k(s, G)$ is given by (see (1.18))

$$\begin{aligned} \delta _k(s, G) = \sum _{H\in \mathbb {A}}\frac{\eta _k(G, H)}{|H|^s}. \end{aligned}$$

(5.1)

Lemma 5.1

If ${{\mathrm{Re}}}(s)>k+1$, $G \ne 0$, then we have

$$\begin{aligned} \delta _k(s, G) = (1-q^{1-s})\sum _{D^k|G}|D|^{k-s}. \end{aligned}$$

(5.2)

Proof

Since ${{\mathrm{Re}}}(s) > k+1$, (5.1) is absolutely convergent, and thus by (2.22) we have

$$\begin{aligned} \delta _k(s, G)&= \sum _{H\in \mathbb {A}}\frac{1}{|H|^s}\sum _{D|H, D^k|G}|D|^k \mu \left( \frac{H}{D}\right) \nonumber \\&= \sum _{D^k|G}|D|^k\sum _{\begin{array}{c} H\in \mathbb {A}\\ D|H \end{array}}\frac{\mu \left( \frac{H}{D}\right) }{|H|^s}\nonumber \\&= \sum _{D^k|G}|D|^{k-s}\sum _{H \in \mathbb {A}}\frac{\mu (H)}{|H|^s}. \end{aligned}$$

(5.3)

If ${{\mathrm{Re}}}(s)>1$, we have (see [45, Proposition 2.6])

$$\begin{aligned} \sum _{H\in \mathbb {A}} \frac{\mu (H)}{|H|^s} = \zeta _{\mathbb {A}}^{-1}(s) = (1-q^{1-s}), \end{aligned}$$

(5.4)

and the lemma follows at once. $\square$

Proof of Theorem 1.1

We first show that the series $\sum _{n=0}^{+\infty } A(n) q^{-ns}$ in (1.18) converges for all s to an entire function, in fact a polynomial in $q^{-s}$. We let

$$\begin{aligned} A(n) = \sum _{\begin{array}{c} H\in \mathbb {A}\\ \deg (H)=n \end{array}}\eta _k(G, H). \end{aligned}$$

(5.5)

By definition (1.18), we have

$$\begin{aligned} \delta _k(s, G)=\sum _{n=0}^{+\infty }A(n)q^{-ns}=\sum _{n=0}^{+\infty }A(n)u^n, \end{aligned}$$

(5.6)

where $u=q^{-s}$. On the other hand, if $G\ne 0$, and ${{\mathrm{Re}}}(s)>k+1$, by Lemma 5.1,

$$\begin{aligned} \delta _k(s, G)&= (1-q^{1-s})\sum _{D^k|G}q^{dk}q^{-ds} \nonumber \\&= (1-qu)\sum _{D^k|G}q^{dk}u^d\nonumber \\&= \sum _{D^k|G}q^{dk}u^d - \sum _{D^k|G}q^{dk+1}u^{d+1}, \end{aligned}$$

(5.7)

where $d=\deg (D)$. We set

$$\begin{aligned} \gamma (G, n) = \sum _{\begin{array}{c} D^k|G\\ \deg (D)=n \end{array}}1. \end{aligned}$$

(5.8)

By (5.7), we have

$$\begin{aligned} \delta _k(s, G)&= \sum _{n=0}^{+\infty }q^{nk}\gamma (G, n)u^n - \sum _{n=0}^{+\infty }q^{nk+1}\gamma (G, n)u^{n+1}\nonumber \\&=\sum _{n=0}^{+\infty }q^{nk}\gamma (G, n)u^n - \sum _{n=1}^{+\infty }q^{(n-1)k+1}\gamma (G, n-1)u^{n}\nonumber \\&=\sum _{n=0}^{+\infty }q^{nk}\left( \gamma (G, n) -q^{1-k}\gamma (G, n-1)\right) u^n. \end{aligned}$$

(5.9)

Comparing the coefficients of $u^n$ of (5.6) and (5.9), we have

$$\begin{aligned} A(n) = q^{nk}\left( \gamma (G, n)-q^{1-k}\gamma (G, n-1) \right) . \end{aligned}$$

(5.10)

By the definition of $\gamma (G, n)$, if $n-1 > \frac{\deg (G)}{k}$, it is easy to see that $\gamma (G, n) = \gamma (G, n-1) = 0$, and it follows that

$$\begin{aligned} A(n)=0, \text{ whenever } n-1 > \frac{\deg (G)}{k}. \end{aligned}$$

(5.11)

We note that the definition of A(n) is independent on the choice of s, and thus for any s, we have

$$\begin{aligned} \delta _k(s, G) = \sum _{0\le n \le \frac{\deg (G)}{2}+1}A(n)u^n = \sum _{\begin{array}{c} H\in \mathbb {A}\\ \deg (H)\le \frac{\deg (G)}{2}+1 \end{array}}\frac{\eta _k(G, H)}{|H|^s}, \end{aligned}$$

(5.12)

which indicates that $\delta _k(s, G)$ is, indeed, a finite summand; therefore, $\delta _k(s, G)$ is an entire function, and on the whole complex plane, we have

$$\begin{aligned} \delta _k(s, G) = (1-q^{1-s})\sum _{D^k|G}|D|^{k-s}. \end{aligned}$$

(5.13)

In particular, if $s=1$, then we have

$$\begin{aligned} \delta _k(1, G) =\sum _{H\in \mathbb {A}}\frac{\eta _k(G, H)}{|H|} =0. \end{aligned}$$

To complete the proof of Theorem 1.1, next we show the square mean values estimate (1.22). If c and T are two given real numbers and $T>0$, by (5.13), we have

$$\begin{aligned} \int _{-T}^{T}|\delta _k(c+it, G)|^2 \mathrm{d}t =&\sum _{D_1^k|G, D_2^k|G}q^{(d_1+d_2)(k-c)}\nonumber \\&\times \int _{-T}^{T}(1-q^{1-c-it})(1-q^{1-c+it})q^{(d_2-d_1)it} \mathrm{d}t, \end{aligned}$$

(5.14)

where $d_1 = \deg (D_1)$ and $d_2 = \deg (D_2)$. We denote the inner integral of (5.14) by $S(d_1, d_2)$,

$$\begin{aligned} S(d_1, d_2) = \int _{-T}^T (1-q^{1-c-it})(1-q^{1-c+it})q^{(d_2-d_1)it} \mathrm{d}t. \end{aligned}$$

(5.15)

Making substitution $u=q^{it}$, it follows that

$$\begin{aligned} S(d_1, d_2) = \frac{1}{i\log q}\int _{q^{-iT}}^{q^{iT}} \left( 1+q^{2(1-c)}-q^{1-c}u^{-1} - q^{1-c}u\right) u^{d_2-d_1-1}\mathrm{d}u. \end{aligned}$$

(5.16)

Let $n=d_2-d_1-1$, if $n\ne -2, -1, 0$, then

$$\begin{aligned} S(d_1, d_2) =&\frac{2}{\log q}\left[ \frac{1+q^{2(1-c)}}{n+1} \left( q^{i(n+1)T}-q^{-i(n+1)T}\right) \right. \\&- \left. \frac{q^{1-c}}{n+2}\left( q^{i(n+2)T} - q^{-i(n+2)T}\right) - \frac{q^{1-c}}{n}\left( q^{inT} - q^{-inT}\right) \right] , \end{aligned}$$

which yields the following estimate:

$$\begin{aligned} |S(d_1, d_2)| \le \frac{4}{\log q}\left( \frac{1+q^{2(1-c)}}{n+1} + \frac{q^{1-c}}{n+2} + \frac{q^{1-c}}{n}\right) . \end{aligned}$$

(5.17)

If $n=d_2-d_1-1=0$, then $d_2 = 1+d_1$, by (5.16), we have

$$\begin{aligned} S(d_1, d_2) =&\frac{1}{i\log q}\left[ \big (1+q^{2(1-c)}\big )\big (q^{iT}-q^{-iT}\big )-2iTq^{1-c}\log q \right. \nonumber \\&\left. - \frac{q^{1-c}}{2}\big (q^{2iT}-q^{-2iT}\big ) \right] \nonumber \\ =&-2Tq^{1-c}+O(1). \end{aligned}$$

(5.18)

If $n=d_2-d_1-1=-1$, then $d_2=d_1$, and

$$\begin{aligned} S(d_1, d_2)&= \frac{1}{i\log q}\left[ 2iT\big (1+q^{2(1-c)}\big )\log q - 2q^{1-c}\big (q^{iT} -q^{-iT}\big )\right] \nonumber \\&= 2T\big (1+q^{2(1-c)}\big )+O(1). \end{aligned}$$

(5.19)

If $n=d_2-d_1-1=-2$, we also have

$$\begin{aligned} S(d_1, d_2) = -2Tq^{1-c}+O(1). \end{aligned}$$

(5.20)

Putting the above equalities together, by (5.14), we have

$$\begin{aligned} \frac{1}{2T}\int _{-T}^{T}|\delta _k(c+it, G)|^2 \mathrm{d}t =&\,\big (1+q^{2(1-c)}\big )\sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ d_1=d_2=d \end{array}}q^{2d(k-c)} - 2q^{1-c} \\&\times \sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ d_1=d_2+1 \end{array}}q^{(d_1+d_2)(k-c)}+O\left( \frac{1}{T}\right) \\ =&(1+q^{2(1-c)})\sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ d_1=d_2=d \end{array}}|D_1|^{2(k-c)} - 2q^{1-k}\\&\times \sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ d_1=d_2+1 \end{array}}|D_1|^{2(k-c)}+O\left( \frac{1}{T}\right) . \end{aligned}$$

Let

$$\begin{aligned} \sigma _0(x, G) = \sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ d_1=d_2 \end{array}}|D_1|^x \quad \text{ and } \quad \sigma _1(x, G) = \sum _{\begin{array}{c} D_1^k|G, D_2^k|G\\ d_1=d_2+1 \end{array}}|D_1|^x, \end{aligned}$$

(5.21)

then we finally obtain

$$\begin{aligned} \frac{1}{2T}\int _{-T}^{T}|\delta _k(c+it, G)|^2 \mathrm{d}t = (1+q^{2(1-c)})\sigma _0(2(k-c)) - 2q^{1-k}\sigma _1(2(k-c))+O\left( \frac{1}{T}\right) . \end{aligned}$$

We complete the proof of Theorem 1.1.

6 The series $\tau _k(s, H)$

In this section, we prove Theorem 1.2. Recalling the Dirichlet series $\tau _k(s, H)$ is given by (see (1.18’)) that

$$\begin{aligned} \tau _k(s, H) = \sum _{G \in \mathbb {A}}\frac{\eta _k(G, H)}{|G|^s}. \end{aligned}$$

(6.1)

This series is more complicated than $\delta _k(s, G)$, because that $\eta _k(G, H)$ is not multiplicative in G, and thus we cannot make use of Euler product directly. Our treatment begins with the following auxiliary series:

$$\begin{aligned} \tau _k'(s, H) = \sum _{G\in \mathbb {A}}\frac{\chi (G)}{|G|^s}, \end{aligned}$$

(6.2)

where $\chi (G) = 1$, if $(G, H^k)_k = 1$, and $\chi (G)=0$, if $(G, H^k)_k>1$. Since $\chi (G)$ is a multiplicative function in G, we have the following equality by using Euler product (if ${{\mathrm{Re}}}(s)>1$):

$$\begin{aligned} \tau _k'(s, H)&= \prod _{P\in \mathbb {A}}(1-|P|^{-s})^{-1} \prod _{P|H}(1-|P|^{-ks})\nonumber \\&= \zeta _{\mathbb {A}}(s)\phi _{ks}(H)|H|^{-ks}. \end{aligned}$$

(6.3)

First, we show a precise analogue of Theorem 13 of [12] with the following lemma.

Lemma 6.1

If H is positive polynomial and ${{\mathrm{Re}}}(s)>k+1$, then

$$\begin{aligned} \tau _k(s, H) = (1-q^{1-s})^{-1}\phi _{k(1-s)}(H). \end{aligned}$$

(6.4)

Proof

Let $A^k = (G, H^k)_k$. By (3.23), we have $\eta _k(G, H) = \eta _k(A^k, H)$, and it follows that

$$\begin{aligned} \tau _k(s, H)&= \sum _{A|H}\eta _k(A^k, H) \sum _{\begin{array}{c} G\in \mathbb {A}\\ A^k = (G, H^k)_k \end{array}}\frac{1}{|G|^s} \\&= \sum _{A|H}\frac{\eta _k(A^k, H)}{|A|^{sk}} \sum _{\begin{array}{c} G_1\in \mathbb {A}\\ (G_1, (\frac{H}{A})^k)_k=1 \end{array}}\frac{1}{|G_1|^s} \\&= \zeta _{\mathbb {A}}(s)|H|^{-ks}\sum _{A|H}\eta _k(A^k, H)\phi _{ks}\left( \frac{H}{A}\right) . \end{aligned}$$

By Lemma 4.6, we have

$$\begin{aligned} \tau _k(s, H) = \zeta _{\mathbb {A}}(s) \phi _{k(1-s)}(H) = (1-q^{1-s})^{-1}\phi _{k(1-s)}(H). \end{aligned}$$

We complete the proof of Lemma 6.1. $\square$

Proof of Theorem 1.2

We now return to the proof of Theorem 1.2. For any integer $n\ge 0$, we let

$$\begin{aligned} B(n) = \sum _{\begin{array}{c} G\in \mathbb {A}\\ \deg (G)=n \end{array}} \eta _k(G, H). \end{aligned}$$

(6.5)

By definition (1.18’), we have

$$\begin{aligned} \tau _k(s, H) = \sum _{n=0}^{+\infty }B(n)q^{-ns} = \sum _{n=0}^{+\infty }B(n)u^n, \end{aligned}$$

(6.6)

where $u=q^{-s}$. If ${{\mathrm{Re}}}(s)>k+1$, by Lemma 6.1

$$\begin{aligned} \tau _k(s, H)&= (1-q^{1-s})^{-1}\phi _{k(1-s)}(H) = (1-q^{1-s})^{-1}\sum _{D|H}|D|^{k(1-s)}\mu \left( \frac{H}{D}\right) \nonumber \\&= (1-qu)^{-1}\sum _{D|H}\mu \left( \frac{H}{D}\right) q^{dk}u^{dk}, \end{aligned}$$

(6.7)

where $u=q^{-s}$ and $d=\deg (D)$. Because of $|qu| < 1$, then $(1-qu)^{-1}$ has a geometric series expression, we have

$$\begin{aligned} \tau _k(s, H)&= \sum _{D|H}\mu \left( \frac{H}{D}\right) \sum _{n=0}^{+\infty }q^{n+dk}u^{n+dk}\nonumber \\&= \sum _{D|H}\mu \left( \frac{H}{D}\right) \sum _{n=dk}^{+\infty }q^{n}u^{n}. \end{aligned}$$

(6.8)

We set

$$\begin{aligned} J_k(D,n)=\left\{ \begin{array}{lll} 1, \quad \text {if} \;n \ge k \deg (D),\\ 0, \quad \text {if} \; n < k \deg (D). \end{array} \right. \end{aligned}$$

(6.9)

It follows from (6.8) that

$$\begin{aligned} \tau _k(s, H) = \sum _{n=0}^{+\infty }\left( \sum _{D|H}\mu \left( \frac{H}{D}\right) J_k(D, n) \right) q^n u^n. \end{aligned}$$

(6.10)

Comparing coefficients of $u^n$ of (6.6) and (6.10), we have

$$\begin{aligned} B(n) = \sum _{D|H}\mu \left( \frac{H}{D}\right) J_k(D, n)\cdot q^n. \end{aligned}$$

(6.11)

If $n\ge k \deg (H)$, then $J_k(D, n)=1$ for all of D that D|H, and hence

$$\begin{aligned} B(n) = q^n \sum _{D|H}\mu \left( \frac{H}{D}\right) = 0. \end{aligned}$$

(6.12)

The last equality of (6.12) follows from (2.12) and $\deg (H)\ge 1$. We note that B(n) is independent on the choice of complex number s by the definition of B(n), and therefore, for any complex number s, we have $B(n)=0$, whenever $n\ge k\cdot \deg (H)$. It follows from (6.6) that

$$\begin{aligned} \tau _k(s, H) = \sum _{n \le k\deg (H)}B(n)u^n = \sum _{\begin{array}{c} G\in \mathbb {A}\\ \deg (G)\le k \cdot \deg (H) \end{array}}\frac{\eta _k(G, H)}{|G|^s}, \end{aligned}$$

(6.13)

which indicates that $\tau _k(s, H)$ is, indeed, a finite summand, and thus, $\tau _k(s, H)$ is an entire function on the whole complex plane. Moreover, by the principle of analytic continuation, on the whole complex plane, we have

$$\begin{aligned} \tau _k(s, H) = (1-q^{1-s})^{-1}\phi _{k(1-s)}(H). \end{aligned}$$

(6.14)

In particular, if $s=1$, by L’Hôpital’s rule we have

$$\begin{aligned} \tau _k(1, H)&= \lim _{s\rightarrow 1}(1-q^{1-s})^{-1}\phi _{k(1-s)}(H)\nonumber \\&= -\frac{k\sum _{D|H}\log |D| \mu \left( \frac{H}{D}\right) }{\log q} = \frac{-k\Lambda (H)}{\log q}, \end{aligned}$$

(6.15)

which is the equality (1.28), and is an analogue of Ramanujan’s identity (1.2).

In order to complete the proof of Theorem 1.2, it remains to prove the square mean value estimate. If T and c are any real numbers that $T>0$, $c\ne 1$, by (6.14) we have

$$\begin{aligned} \int _{-T}^T |\tau _k(c+it, H)|^2 \mathrm{d}t =&\, \sum _{D_1|H, D_2|H} \mu \left( \frac{H}{D_1}\right) \mu \left( \frac{H}{D_2}\right) |D_1D_2|^{k(1-c)} \nonumber \\&\times \int _{-T}^T \frac{q^{k(d_2-d_1)it} \mathrm{d}t}{(1-q^{1-c-it})(1-q^{1-c+it})}. \end{aligned}$$

(6.16)

We denote

$$\begin{aligned} S_1(d_1, d_2) = \int _{-T}^T \frac{q^{k(d_2-d_1)it} \mathrm{d}t}{(1-q^{1-c-it})(1-q^{1-c+it})}, \end{aligned}$$

(6.17)

and make the substitution of $u=-t$, it follows that

$$\begin{aligned} S_1(d_1, d_2) = \int _{-T}^T \frac{q^{k(d_1-d_2)it} \mathrm{d}t}{(1-q^{1-c-it})(1-q^{1-c+it})}, \end{aligned}$$

(6.18)

which shows that $d_1$ and $d_2$ are symmetric in (6.16). Therefore, we may suppose that $d_2 \ge d_1$, and make the substitution $u=q^{it}$ in (6.17), then

$$\begin{aligned} S_1(d_1, d_2) =\frac{i}{q^{1-c}\log q} \int _{q^{-iT}}^{q^{iT}} \frac{u^{k(d_2-d_1)}\mathrm{d}u}{(u-q^{1-c})(u-q^{c-1})}. \end{aligned}$$

(6.19)

If $d_1 = d_2$, then we have

$$\begin{aligned} S_1(d_1, d_2) =&\,\frac{i}{(q^{2(1-c)}-1)\log q} \big [\log \big (q^{iT}-q^{1-c}\big ) - \log \big (q^{-iT}-q^{1-c}\big ) \nonumber \\&- \log \big (q^{iT}-q^{c-1}\big ) + \log \big (q^{-iT}-q^{c-1}\big )\big ]. \end{aligned}$$

(6.20)

We note that $1-c$ and $c-1$ are symmetric in the above: first we let $c>1$, and then

$$\begin{aligned}&\log (q^{iT}-q^{1-c}) - \log (q^{-iT}-q^{1-c}) \nonumber \\&\quad = 2iT\log q + \log (1-q^{-iT}q^{1-c}) - \log (1-q^{iT}q^{1-c}). \end{aligned}$$

(6.21)

Let $z=q^{1-c}q^{\pm iT}$, then $|z| =q^{1-c}<1$ for $c>1$, and we have the following power series expansion that

$$\begin{aligned} -\log (1-z) = z+\frac{1}{2}z^2 + \frac{1}{3}z^3 + \cdots . \end{aligned}$$

It follows that

$$\begin{aligned} |\log (1-z)| \le |z|+ \frac{1}{2}|z|^2 + \cdots \le \log \frac{1}{1-|z|} = \log \frac{1}{1-q^{1-c}}, \end{aligned}$$

(6.22)

and by (6.21), we have

$$\begin{aligned} \log (q^{iT}-q^{1-c}) - \log (q^{-iT}-q^{1-c}) = 2iT\log q+O(1). \end{aligned}$$

(6.23)

The remaining part of (6.20) is

$$\begin{aligned} \log (q^{iT}-q^{c-1}) - \log (q^{iT}-q^{c-1})&= \log \frac{1-q^{1-c}q^{iT}}{1-q^{1-c}q^{-iT}}\\&= \log (1-q^{1-c}q^{iT}) - \log (1-q^{1-c}q^{-iT}). \end{aligned}$$

By (6.22), we have

$$\begin{aligned} \big |\log \big (1-q^{1-c}q^{iT}\big ) - \log \big (1-q^{1-c}q^{-iT}\big )\big | \le 2\log \frac{1}{1-q^{1-c}}. \end{aligned}$$

(6.24)

Hence, if $d_2=d_1$, and $c>1$, we obtain

$$\begin{aligned} S_1(d_1, d_2) = \frac{2T}{1-q^{2(1-c)}} + O(1). \end{aligned}$$

(6.25)

If $d_1=d_2$, and $c<1$, the same method yields the following estimate

$$\begin{aligned} S_1(d_1, d_2) = \frac{2T}{q^{2(1-c)}-1}+O(1). \end{aligned}$$

(6.26)

Therefore, if $d_1=d_2$, we have

$$\begin{aligned} S_1(d_1, d_2) = \frac{2T}{|1-q^{2(1-c)}|}+O(1). \end{aligned}$$

(6.27)

Next, we consider $d_2>d_1$, and let $n=(d_2-d_1)k$. By (6.19), then

$$\begin{aligned} S_1(d_1, d_2) = \frac{i}{(q^{2(1-c)}-1)\log q}&\left[ \int _{q^{-iT}-q^{1-c}}^{q^{iT}-q^{1-c}}\frac{(u+q^{1-c})^n}{u}\mathrm{d}u \right. \nonumber \\&\left. - \int _{q^{-iT}-q^{c-1}}^{q^{iT}-q^{c-1}}\frac{(u+q^{c-1})^n}{u}\mathrm{d}u \right] . \end{aligned}$$

(6.28)

We write

$$\begin{aligned} (u+q^{\pm (1-c)})^n = \sum _{j=0}^n\left( {\begin{array}{c}n\\ j\end{array}}\right) u^j q^{\pm (1-c)(n-j)}. \end{aligned}$$

(6.29)

If $j\ne 0$, then it is easy to verify that the inner integral in (6.28) is O(1). if $j=0$, there is a similar argument like the case of $d_1=d_2$, which yields

$$\begin{aligned} S_1(d_1, d_2) = \frac{q^{k(d_2-d_1)(1-c)}}{|1-q^{2(1-c)}|}2T + O(1). \end{aligned}$$

(6.30)

By (6.16), we finally obtain

$$\begin{aligned}&\frac{1}{2T}\int _{-T}^T |\tau _k(c+it, H)|^2 \mathrm{d}t \nonumber \\&\quad = \frac{1}{|1-q^{2(1-c)}|}\sum _{\begin{array}{c} D_1|H, D_2|H\\ \deg (D_1)= \deg (D_2) \end{array}}\mu \left( \frac{H}{D_1}\right) \mu \left( \frac{H}{D_2}\right) |D_1|^{2(1-c)}\nonumber \\&\qquad + \frac{2}{|1-q^{2(1-c)}|}\sum _{\begin{array}{c} D_1|H, D_2|H\\ \deg (D_1)> \deg (D_2) \end{array}} \left( \mu \left( \frac{H}{D_1}\right) \mu \left( \frac{H}{D_2}\right) \right. \nonumber \\&\qquad \times \left. |D_1|^{(k+2)(1-c)}|D_2|^{(2-k)(1-c)}\right) +O\left( \frac{1}{T}\right) . \end{aligned}$$

(6.31)

We complete the proof of Theorem 1.2.

7 Davenport–Hasse type formula

The polynomial Ramanujan sum $\eta (G, H)$ essentially is a special type of Gauss sum on $\mathbb {F}_q[x]$. Let $\chi$ be a multiplicative character modulo H on $\mathbb {F}_q[x]$, and $\psi _G= E(G, H)$ be the additive character modulo H given by (1.8); the Gauss sum $G(\chi , \psi _G)$ modulo H on $\mathbb {F}_q[x]$ is defined by

$$\begin{aligned} G(\chi , \psi _G) = \sum _{D{{\mathrm{mod}}}H}\chi (D)\psi _G(D), \end{aligned}$$

(7.1)

where D extends over a complete residue system modulo H in $\mathbb {F}_q[x]$. Let $\chi _0$ be the principal multiplicative character: it is $\chi _0(D)=1$ if $(D, H)=1$, and $\chi _0(D)=0$ if $(D, H)>1$, and then we see that $\eta (G, H)=G(\chi _0, \psi _G)$.

In an upcoming paper [51], we presented an analogue of Davenport–Hasse’s theorem for the polynomial Gauss sums (see [51, Theorem 1.3]). To state this result, let $\mathbb {F}_{q^n}/\mathbb {F}_q$ be a finite extension over $\mathbb {F}_q$ of degree n, and ${{\mathrm{tr}}}(a)$ and N(a) be the trace map and norm from $\mathbb {F}_{q^n}$ to $\mathbb {F}_q$, respectively,

$$\begin{aligned} {{\mathrm{tr}}}(a) = \sum _{i=1}^n \sigma ^i(a) \quad \text{ and } \quad {{\mathrm{N}}}(a) = \prod _{i=1}^n \sigma ^i(a), \end{aligned}$$

(7.2)

where $\sigma (a) = a^q$ for a in $\mathbb {F}_{q^n}$ is the Frobenius of $\mathbb {F}_{q^n}$. If A is a polynomial in $\mathbb {F}_{q^n}[x]$, $A=a_kx^k + a_{k-1}x^{k-1} + \cdots + a_1x + a_0$, the trace map and norm can be extended to $\mathbb {F}_{q^n}[x]$ by

$$\begin{aligned} {{\mathrm{tr}}}(A) = \sum _{i=1}^n \sigma ^i(A) \quad \text{ and } \quad {{\mathrm{N}}}(A) = \prod _{i=1}^n \sigma ^i(A), \end{aligned}$$

(7.3)

where $\sigma (A) = \sum _{i=0}^k \sigma (a_i)x^i$.

Let H be a polynomial in $\mathbb {F}_q[x]$ and, therefore, also a polynomial in $\mathbb {F}_{q^n}[x]$. To define a Gauss sum modulo H on $\mathbb {F}_{q^n}[x]$, for any A in $\mathbb {F}_{q^n}[x]$, we set

$$\begin{aligned} \psi _G^{(n)}(A) = \psi _G({{\mathrm{tr}}}(A))\quad \text{ and } \quad \chi ^{(n)}(A) = \chi (N(A)), \end{aligned}$$

(7.4)

and thus, the Gauss sums $G(\chi ^{(n)}, \psi _G^{(n)})$ modulo H on $\mathbb {F}_{q^n}[x]$ is given by

$$\begin{aligned} \psi (\chi ^{(n)}, \psi _{G}^{(n)}) = \sum _{\begin{array}{c} D\in \mathbb {F}_{q^n}[x]\\ D {{\mathrm{mod}}}H \end{array}}\chi ^{(n)}(D)\psi _G^{(n)}(D), \end{aligned}$$

(7.5)

where the summation extends over a complete residue system modulo H in $\mathbb {F}_{q^n}[x]$.

By the above notations, we may define a polynomial Ramanujan sum $\eta ^{(n)}(G, H)$ modulo H on $\mathbb {F}_{q^n}[x]$ by

$$\begin{aligned} \eta ^{(n)}(G, H) = \sum _{\begin{array}{c} D{{\mathrm{mod}}}H\\ D\in \mathbb {F}_{q^n}[x] \end{array}}\chi _0^{(n)}(D)\psi _G^{(n)}(D), \end{aligned}$$

(7.6)

and a generalized version $\eta ^{(n)}_k(G, H)$ by

$$\begin{aligned} \eta _k^{(n)}(G, H) = \sum _{\begin{array}{c} D{{\mathrm{mod}}}H^k\\ (D, H^k)_k=1 \end{array}}\psi _G^{(n)}(D), \end{aligned}$$

(7.7)

where the summation ranges over a complete residue system modulo $H^k$ in $\mathbb {F}_{q^n}[x]$.

If $\chi$ and $\psi _G$ are not both principal, in [51] we showed the following Davenport–Hasse type formula:

$$\begin{aligned} (-1)^{m-m_1}\frac{\phi ^{(n)}(N)}{\phi ^{(n)}(H)}G(\chi ^{(n)}, \psi _G^{(n)}) = \left( (-1)^{m-m_1}\frac{\phi (N)}{\phi (H)}G(\chi , \psi _G) \right) ^n, \end{aligned}$$

(7.8)

where $\phi (H)$ is the Euler totient function on $\mathbb {F}_{q}[x]$, $\phi ^{(n)}(H)$ is the function on $\mathbb {F}_{q^n}[x]$, $N=\frac{H}{(G, H)}$, $m=\deg (H)$, and $m_1=\deg (G, H)$.

As a direct consequence of (7.8), if $H \not \mid G$, then $\psi _G$ is not principal, and we have

$$\begin{aligned} (-1)^{m-m_1}\frac{\phi ^{(n)}(N)}{\phi ^{(n)}(H)}\eta ^{(n)}(G, H) = \left( (-1)^{m-m_1}\frac{\phi (N)}{\phi (H)}\eta (G, H) \right) ^n. \end{aligned}$$

(7.9)

The main purpose of this section is to show that the generalized version $\eta _k(H, G)$ also shares this kind of Davenport–Hasse type formula. We have

Theorem 7.1

If H and G are any polynomials in $\mathbb {F}_q[x]$ such that $H^k \not \mid G$ and $H \ne 0$, then

$$\begin{aligned} (-1)^{m-m_1}\frac{\phi _k^{(n)}(N)}{\phi _k^{(n)}(H)}\eta _k^{(n)}(G, H) = \left( (-1)^{m-m_1}\frac{\phi _k(N)}{\phi _k(H)}\eta _k(G, H) \right) ^n, \end{aligned}$$

(7.10)

where $\phi _k(H)$ is the Jordan totient function on $\mathbb {F}_q[x]$, and $\phi _k^{(n)}(H)$ is the function on $\mathbb {F}_{q^n}[x]$, $N=\frac{H}{A}$, $A^k = (G, H^k)_k$, $m=\deg (H)$, and $m_1 = \deg (A)$.

Proof

If $A^k = (G, H^k)_k$ in $\mathbb {F}_q[x]$, it is easy to verify that $A^k = (G, H^k)_k$ holds in $\mathbb {F}_{q^n}[x]$. By (2.25), we have

$$\begin{aligned} \eta _k^{(n)}(G, H) = \phi _k^{(n)}(H)\mu ^{(n)}(N)\left( \phi _k^{(n)}(N) \right) ^{-1}, \end{aligned}$$

(7.11)

and

$$\begin{aligned} \eta _k(G, H) = \phi _k(H)\mu (N)\phi _k^{-1}(N), \end{aligned}$$

(7.12)

where $\mu ^{(n)}(H)$ is the Möbius function on $\mathbb {F}_{q^n}[x]$. To prove (7.10), it suffices to show that

$$\begin{aligned} (-1)^{m+m_1}\mu ^{(n)}(N) = \left( (-1)^{m+m_1}\mu (N) \right) ^n, \end{aligned}$$

(7.13)

where $N=\frac{H}{A}$, and $A^k = (G, H^k)_k$.

We note that both sides of (7.13) are multiplicative in H, so it suffices to prove (7.13) when $H=P^t$, where P is an irreducible in $\mathbb {F}_q[x]$ and $t \ge 1$. If $t \ge 1$ and $A=P^{t_1}$ with $t_1< t-1$, then both sides of (7.13) are zero. Therefore, we may suppose $A=P^{t-1}$, and $N=P$. It is well known (see Hayes [23], for example) that P is product of exactly (h, n) irreducibles in $\mathbb {F}_{q^n}[x]$, where $h = \deg (P)$, so (7.13) becomes that

$$\begin{aligned} (-1)^{th+(t-1)h+(h, n)} = (-1)^{n(th+(t-1)h+1)}, \end{aligned}$$

(7.14)

which is equivalent to

$$\begin{aligned} h + (h, n) \equiv n(h+1) ~({{\mathrm{mod}}}2). \end{aligned}$$

(7.15)

It is easy to verify that (7.15) is true for any positive integers n and h, and we complete the proof of Theorem 7.1. $\square$