1 Introduction
Let $X_f$ be the multiplicative group of the $\phi (f)$ Dirichlet characters modulo $f>2$ . Let $X_f^- =\{\chi \in X_f;\ \chi (-1)=-1\}$ be the set of the $\phi (f)/2$ odd Dirichlet characters modulo f. Let $L(s,\chi )$ be the Dirichlet L-function associated with $\chi \in X_f$ . Let H denote a subgroup of index m in the multiplicative group $G:=({\mathbb Z}/f{\mathbb Z})^*$ . We assume that $-1\not \in H$ . Hence, m is even. We set $X_f(H) =\{\chi \in X_f;\ \chi _{/H}=1\}$ , a subgroup of order m of $X_f$ isomorphic to the group of Dirichlet characters of the abelian quotient group $G/H$ of order m. Define $X_f^-(H)=\{\chi \in X_f^-;\ \chi _{/H}=1\}$ , a set of cardinal $m/2$ . Let K be an abelian number field of degree m and prime conductor $p\geq 3$ , i.e., let K be a subfield of the cyclotomic number field ${\mathbb Q}(\zeta _p)$ (Kronecker–Weber’s theorem). The Galois group $\mathrm {Gal}({\mathbb Q}(\zeta _p)/{\mathbb Q})$ is canonically isomorphic to the multiplicative cyclic group $({\mathbb Z}/p{\mathbb Z})^*$ and $H :=\mathrm {Gal}({\mathbb Q}(\zeta _p)/K)$ is a subgroup of $({\mathbb Z}/p{\mathbb Z})^*$ of index m and order
Now, assume that K is imaginary. Then d is odd, m is even, $-1\not \in H$ and the set
is of cardinal $(p-1)/(2d) =m/2$ . Let $K^+$ be the maximal real subfield of K of degree $m/2$ fixed by the complex conjugation. The class number $h_{K^+}$ of $K^+$ divides the class number $h_K$ of K. The relative class number of K is defined by $h_K^- =h_K/h_{K^+}$ . We refer the reader to [Reference SerreSer, Was] for such basic knowledge. The mean square value of $L(1,\chi )$ as $\chi $ ranges in $X_f^-(H)$ is defined by
The analytic class number formula and the arithmetic–geometric mean inequality give
where $w_K$ is the number of complex roots of unity in K. Hence, $w_K=2p$ for $K ={\mathbb Q}(\zeta _p)$ and $w_K=2$ otherwise. In [Reference Louboutin and MunschLM21, Theorem 1.1], we proved that
as p tends to infinity uniformly over subgroups H of $({\mathbb Z}/p{\mathbb Z})^*$ of odd order $d\leq \frac {\log p}{3(\log \log p)}$ Footnote 1 . Hence, by (1.2), we have
In some situations, it is even possible to give an explicit formula for $M(p,H)$ implying a completely explicit bound for $h_K^-$ . Indeed, by [Reference WalumMet, Wal] (see also (4.2)), we have
Hence,
We refer the reader to [Reference GranvilleGra] for more information about the expected size of $h_{{\mathbb Q}(\zeta _p)}^-$ . The only other situation where a similar explicit result is known is the following one (see Theorem 6.6 for a new proof).
Theorem (SeeFootnote 2 [Reference LouboutinLou16, Theorem 1])
Let $p\equiv 1\ \ \pmod 6$ be a prime integer. Let K be the imaginary subfield of degree $(p-1)/3$ of the cyclotomic number field ${\mathbb Q}(\zeta _p)$ . Let H be the subgroup of order $3$ of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . We have (compare with ( 1.5 ) and ( 1.6 ))
In [Reference LouboutinLou94] (see also [Reference LouboutinLou11]), the following simple argument allowed to improve on (1.6). Let $d_0>1$ be a given integer. Assume that $\gcd (d_0,f)=1$ . For $\chi $ modulo f, let $\chi '$ be the character modulo $d_0f$ induced by $\chi $ . Then
(throughout the paper, this notation means that q runs over the distinct prime divisors of $d_0$ ). Let H be a subgroup of order d of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ , with $-1\not \in H$ . We define
andFootnote 3
Clearly, there is no restriction in assuming from now on that $d_0$ is square- free. Let now H be of odd order d in the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . Using (1.8), we obtain (compare with (1.2))
Let $d=o(\log p)$ as $p\rightarrow \infty $ . Then, by Corollary 2.4, we have
and we expect that
Hence, (1.11) should indeed improve on (1.2). The aim of this paper is twofold. First, in Theorem 1.1, we give an asymptotic formula for $M_{d_0}(p,H)$ when d satisfies the same restriction as in (1.3) allowing us to improve on the bound (1.4). Second, we treat the case of groups of orders $1$ and $3$ for small $d_0$ ’s as well as the case of Mersenne primes and groups of size $\approx \log p$ . In both cases, an explicit description of these subgroups allows us to obtain explicit formulas for $M_{d_0}(p,H)$ . Our main result is the following.
Theorem 1.1 Let $d_0\geq 1$ be a given square-free integer. As $p\rightarrow +\infty $ , we have the following asymptotic formula:
uniformly over subgroups H of $({\mathbb Z}/p{\mathbb Z})^*$ of odd order $d\leq \frac {\log p}{3(\log \log p)}.$ Moreover, let K be an imaginary abelian number field of prime conductor p and of degree $m=(p-1)/d$ . Let $C<4\pi ^2=39.478..$ be any positive constant. If p is sufficiently large and $m\geq 3\frac {(p-1) \log \log p} {\log p}$ , then we have
Remark 1.2 The second result in Theorem 1.1 improves on (1.4), (1.6), and (1.7). It follows from the first result in Theorem 1.1, and by using (1.11) and (2.2), where we take $d_0$ as the product of sufficiently many consecutive first primes.
The special case $d_0=1$ was proved in [Reference Louboutin and MunschLM21, Theorem $1.1$ ]. Note that the restriction on d cannot be extended further to the range $d=O(\log p)$ as shown by Theorem 5.2. Moreover, the constant C in (1.13) cannot be taken larger than $4\pi ^2$ (see the discussion about Kummer’s conjecture in [Reference Murty and PetridisMP01].
In the first part of the paper, the presentation goes as follows:
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• In Section 2, we explain the condition about the prime divisors of $d_0$ and prove that $D_{d_0}(p,H) =1+o(1)$ .
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• In Section 3, we review some results on Dedekind sums and prove a new bound of independent interest for Dedekind sums $s(h,f)$ with h being of small order modulo f (see Theorem 3.1). To do so, we use techniques from uniform distribution and discrepancy theory. Then we relate $M_{d_0}(p,H)$ to twisted moments of L- functions which we further express in terms of Dedekind sums. For the sake of clarity, we first treat separately the case $H=\{1\}$ . Note that we found that this case is related to elementary sums of maxima that we could not estimate directly (see Section 3.4). Using our estimates on Dedekind sums, we deduce the asymptotic formula of Theorem 1.1 and the related class number bounds.
In the second part of the paper, we focus on the explicit aspects. Let us describe briefly our presentation:
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• In Section 4.1, we establish a formula for $M_{d_0}(f,\{1\})$ , $d_0>2$ , provided that all the prime factors q of f satisfy $q\equiv \pm 1\ \ \pmod {d_0}$ . In particular, we get formulas for $M_{d_0}(f,\{1\})$ for $d_0\in \{1,2,3,6\}$ and $\gcd (d_0,f)=1$ (such formulae become harder to come by as $d_0$ gets larger). For example, for $p\geq 5$ and $d_0=6$ , using Theorem 4.1, we obtain the following formula for $M_6(p,\{1\})$ :
$$ \begin{align*}M_6(p,\{1\}) ={\pi^2\over 9}\left (1-\frac{c_p}{p}\right ) \leq\frac{\pi^2}{9}, \mbox{ where } c_p =\begin{cases} 1,&\mbox{if } p\equiv 1\ \ \pmod 3,\\ 0,&\mbox{if } p\equiv 2\ \ \pmod 3,\\ \end{cases}\end{align*} $$which by (1.11) and Corollary 2.4 give improvements on (1.6) (see also [Reference FengFeng])$$ \begin{align*}h_{{\mathbb Q}(\zeta_p)}^- \leq 3p\left ({p\over 36}\right )^{(p-1)/4}.\end{align*} $$See also [Reference LouboutinLou23, Theorem 5.2] for even better bounds. In Section 4.3, we obtain an explicit formula of the form(1.14) $$ \begin{align} M_{d_0}(p,H) =\frac{\pi^2}{6} \left\{\prod_{q\mid d_0}\left (1-\frac{1}{q^2}\right )\right\} \left (1+\frac{N_{d_0}(p,H)}{p}\right ){\kern-1.5pt}, \end{align} $$where $N_{d_0}(p,H)$ defined in (4.5) is an explicit average of Dedekind sums. In Proposition 4.6, we prove that $N_{d_0}(p,\{1\})\in {\mathbb Q}$ depends only on p modulo $d_0$ and is easily computable. -
• For $H\neq \{1\}$ explicit formulae for $M_{d_0}(p,H)$ seem difficult to come by. In Section 5, we focus on Mersenne primes $p=2^d-1$ , with d odd. We take $H=\{2^k;\ 0\leq k\leq d-1\}$ , a subgroup of odd order d of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . For $d_0\in \{1,3,15\}$ , we prove in Theorem 5.4 that
$$ \begin{align*}M_{d_0}(p,H) =\frac{\pi^2}{2}\left\{\prod_{q\mid d_0}\left (1-\frac{1}{q^2}\right )\right\} \left (1+\frac{N_{d_0}'(p,H)}{p}\right ){\kern-1.5pt},\end{align*} $$where $N_{d_0}'(p,H)=a_1(p)d+a_0(p)$ with $a_1(p),a_0(p)\in {\mathbb Q}$ depending only on $p=2^d-1$ modulo $d_0$ and easily computable. In the range $d \gg \log p$ , we see that $M_{d_0}(p,H)$ has a different asymptotic behavior than the one in Theorem 1.1. -
• In Section 6, we turn to the specific case of subgroups of order $3$ . Writing $f=a^2+ab+b^2$ not necessarily prime, and taking $H=\{1,a/b,b/a\}$ , the subgroup of order $3$ of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ , we prove in Proposition 6.4 that $N_{d_0}(f,H) =O(\sqrt f)$ in (1.14) for $d_0\in \{1,2,3,6\}$ . To do so, we obtain bounds for the Dedekind sums stronger than the one in Theorem 3.1. Note that this cannot be expected in general for subgroups of order $3$ modulo composite f (see Remarks 3.4 and 6.2). Furthermore, we show that these bounds are sharp in the case of primes $p=a^2+a+1$ , in accordance with Conjecture 7.1.
2 Preliminaries
2.1 Algebraic considerations
Take $a\in {\mathbb Z}$ with $\gcd (a,f)=1$ . There are infinitely many prime integers in the arithmetic progressions $a+f{\mathbb Z}$ . Taking a prime $p\in a+f{\mathbb Z}$ with $p>d_0f$ , we have $s_{d_0}(p)=a$ , where $s_{d_0}:({\mathbb Z}/d_0f{\mathbb Z})^*\longrightarrow ({\mathbb Z}/f{\mathbb Z})^*$ is the canonical morphism. Therefore, $s_{d_0}$ surjective and its kernel is of order $\phi (d_0)$ . Let H be a subgroup of $({\mathbb Z}/f{\mathbb Z})^*$ of order d. Then $H_{d_0} =s_{d_0}^{-1}(H)$ is a subgroup of order $\phi (d_0)d$ of $({\mathbb Z}/d_0f{\mathbb Z})^*$ and as $\chi $ runs over $X_f^-(H)$ the $\chi '$ ’s run over $X_{d_0f}^-(H_{d_0}),$ and by (1.1) and (1.9), we have
The following Lemma is probably well known but we found no reference in the literature.
Lemma 2.1 Let $f>2$ . Let H be a subgroup of index $m =(G:H)$ in the multiplicative group $G:=({\mathbb Z}/f{\mathbb Z})^*$ . Then $\# X_f(H) =m$ and $H=\cap _{\chi \in X_f(H)}\ker \chi $ . Moreover, if $-1\not \in H$ , then m is even, $\# X_f^-(H)=m/2$ and $H=\cap _{\chi \in X_f^-(H)}\ker \chi $ .
Proof Since $X_f(H)$ is isomorphic to the group of Dirichlet characters of the abelian quotient group $G/H$ , it is of order m, by [Reference SerreSer, Chapter VI, Proposition 2]. Clearly, $H\subseteq \cap _{\chi \in X_f(H)}\ker \chi $ . Conversely, take $g\not \in H$ , of order $n\geq 2$ in the abelian quotient group $G/H$ . Define a character $\chi $ of the subgroup $\langle g,H\rangle $ of G generated by g and H by $\chi (g^kh) =\exp (2\pi ik/n)$ , $(k,h)\in {\mathbb Z}\times H$ . It extends to a character of G still denoted $\chi $ , by [Reference SerreSer, Chapter VI, Proposition 1]. Since $g\not \in \ker \chi $ and $\chi \in X_f(H)$ , we have $g\not \in \cap _{\chi \in X_f(H)}\ker \chi $ , i.e., $ \cap _{\chi \in X_f(H)}\ker \chi \subseteq H$ .
Now, assume that $-1\not \in H$ . Set $H'=\langle -1,H\rangle $ , of index $m/2$ in G. Then $X_f^-(H) =X_f(H)\setminus X_f(H')$ is indeed of order $m-m/2 =m/2$ , by the first assertion. Clearly, $H\subseteq \cap _{\chi \in X_f^-(H)}\ker \chi $ . Conversely, take $g\not \in H$ . Set $H" :=\langle g,H\rangle =\{g^kh;\ k\in {\mathbb Z},\ h\in H\}$ , of index $m"$ in G, with $m>m"$ . If $-1=g^kh\in H"$ , then clearly $\chi (g)\neq 1$ for $\chi \in X_f^-(H)$ ; hence, $g\not \in \cap _{\chi \in X_f^-(H)}\ker \chi $ . If $-1\not \in H"$ and $\chi \in X_f^-(H)\setminus X_f^-(H")$ , a nonempty set or cardinal $m/2-m"/2 =(H":H)/2\geq 1$ , then clearly $\chi (g)\neq 1$ ; hence, $g\not \in \cap _{\chi \in X_f^-(H)}\ker \chi $ . Therefore, $\cap _{\chi \in X_f^-(H)}\ker \chi \subseteq H$ .
2.2 On the size of $\Pi _{d_0}(f,H)$ and $D_{d_0}(f,H)$ defined in (1.10)
Lemma 2.3 Let H be a subgroup of order $d\geq 1$ of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ , where $f>2$ . Assume that $-1\not \in H$ . Let g be the order of a given prime integer q in the multiplicative quotient group $({\mathbb Z}/f{\mathbb Z})^*/H$ . Let $X_f(H)$ be the multiplicative group of the $\phi (f)/d$ Dirichlet characters modulo f for which $\chi _{/H} =1$ . Define $X_f^-(H) = \{\chi \in X_f(H);\ \chi (-1)=-1\}$ , a set of cardinal $\phi (f)/(2d)$ . Then
Proof Let $\alpha $ be of order g in an abelian group A of order n. Let $B=\langle \alpha \rangle $ be the cyclic group generated by $\alpha $ . Let $\hat B$ be the group of the g characters of B. Then $P_{B}(X) :=\prod _{\chi \in \hat B} (X-\chi (\alpha )) =X^g-1$ . Now, the restriction map $\chi \in \hat A\rightarrow \chi _{/B}\in \hat B$ is surjective, by [Reference SerreSer, Proposition 1], and of kernel isomorphic to $\widehat {A/B}$ of order $n/g$ , by [Reference SerreSer, Proposition 2]. Therefore, $P_{A}(X) :=\prod _{\chi \in \hat A} (X-\chi (\alpha )) =P_B(X)^{n/g} =(X^g-1)^{n/g}$ . With $A=({\mathbb Z}/f{\mathbb Z})^*/H$ of order $n=\phi (f)/d$ , we have $\hat A =X_f(H)$ and
Let $H'$ be the subgroup of order $2d$ generated by $-1$ and H. With $A'=({\mathbb Z}/f{\mathbb Z})^*/H'$ of order $n'=\phi (f)/(2d)$ , we have $\hat A' =X_f(H') =X_f^+(H):=\{\chi \in X_f(H);\ \chi (-1)=+1\}$ and
where q is of order $g'$ in $A'$ .
Since $X_f^-(H) =X_f(H)\setminus X_f^+(H)$ , it follows that
Since $q^g\in H$ , we have $q^g\in H'$ and $g'$ divides g. Since $q^{g'}\in H'=\{\pm h;\ h\in H\}$ , we have $q^{2g'}\in H$ and g divides $2g'$ . Hence, $g=g'$ or $g =2g'$ and $g=2g'$ if and only if g is even and $q^{g/2}=q^{g'}\in H'\setminus H=\{-h;\ h\in H\}$ . The assertion follows.
Corollary 2.4 Fix $d_0>1$ square-free. Let $p\geq 3$ run over the prime integers that do not divide $d_0$ . Let H a subgroup of odd order d of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . Then
where $\omega (d_0)$ stands for the number of prime divisors of $d_0$ . In particular when $d =o(\log p)$ , we have
Moreover,
In particular, $\Pi _{6}(p,\{1\}) \geq 2/3 \,\, \mbox {for } p\geq 5.$
Proof Let q be a prime divisor of $d_0$ . Let g be the order of q in the multiplicative quotient group $({\mathbb Z}/p{\mathbb Z})^*/H$ . Then
by (1.10) and Lemma 2.3, with $f=p$ , $\phi (f)=p-1$ and $m =(p-1)/d$ . Either $q^g\equiv 1\ \ \pmod p$ , in which case $q^g\geq p+1$ , or $q^g\equiv h\ \ \pmod p$ for some $h\in \{2,\ldots ,p-1\}\cap H$ , in which case p divides $S:=1+h+\cdots +h^{d-1}$ which satisfies $p\leq S\leq 2h^{d-1}$ . Therefore, in both cases, we have $q^{g}\geq (p/2)^{\frac {1}{d-1}}$ . Hence,
where we used for $x=q^{-g}$ the fact that $\log (1-x) \geq -2(\log 2)x$ in $\left [0,1/2\right ]$ ,
where we used the fact that $e^{-x} \geq 1-x$ . Therefore, we have
where we used the inequality $(1-x)^n \geq 1-nx$ for $x \leq 1$ and $n\in \mathbb {N}$ . A similar reasoning gives an explicit upper bound $D_{d_0}(p,H) \leq 1+ c\omega (d_0)p^{-1/2(d-1)}$ for some constant $c>0$ . Therefore, we do get (2.2). Finally, $p^{1/(d-1)}$ tends to infinity in the range $d =o(\log p)$ and (2.3) follows.
Notice that if $p=2^d-1$ runs over the Mersenne primes and $H=\langle 2\rangle $ , we have $d=O(\log p)$ but $D_2(p,H) =\left (1-\frac {1}{2}\right )^2$ does not satisfy (2.3).
Now, assume that $H=\{1\}$ . Then $K={\mathbb Q}(\zeta _p)$ and $q^g\geq p+1$ . Hence,
The desired lower bound easily follows.
3 Dedekind sums and mean square values of L-functions
3.1 Dedekind sums and Dedekind–Rademacher sums
The Dedekind sums is the rational number defined by
with the convention $s(c,-1)=s(c,1)=0$ for $c\in {\mathbb Z}$ (see [Reference ApostolApo] or [Reference Rademacher and GrosswaldRG] where it is, however, assumed that $d>1$ ). It depends only on c mod $\vert d\vert $ and $c\mapsto s(c,d)$ can therefore be seen as a mapping from $({\mathbb Z}/\vert d\vert {\mathbb Z})^*$ to ${\mathbb Q}$ . Notice that
(make the change of variables $n\mapsto nc$ in $s(c^*,d)$ ). Recall the reciprocity law for Dedekind sums
In particular,
For $b,c\in {\mathbb Z},\ d\in {\mathbb Z}\setminus \{-1,0,1\}$ such that $\gcd (b,d) =\gcd (c,d)=1$ , the Dedekind–Rademacher sum is the rational number defined by
with the convention $s(b,c,-1)=s(b,c,1)=0$ for $b,c\in {\mathbb Z}$ . Hence, $s(c,d) =s(1,c,d)$ , if $\alpha \in ({\mathbb Z}/\vert d\vert {\mathbb Z})^*$ is represented as $\alpha =b/c$ with $\gcd (b,d)=\gcd (c,d) =1$ , then $s(\alpha ,d) =s(b,c,d)$ , and
For $\gcd ( b,c) =\gcd (c,d) =\gcd (d,b) =1$ , we have a reciprocity law for Dedekind–Rademacher sums (see [Reference RademacherRad] or [Reference Bayad and RaoujBR])
The Cauchy–Schwarz inequality and (3.4) yield
3.2 Nontrivial bounds on Dedekind sums
In this section, we will use the alternative definition of the Dedekind sums given by
where $ \left (\left (\right )\right ):\mathbb {R} \rightarrow \mathbb {R}$ stands for the sawtooth function defined by
In order to prove Theorem 1.1, we need general bounds on Dedekind sums depending on the multiplicative order of the argument. This is a new type of bounds for Dedekind sums and the following result that improves upon (3.7) when the order is $o\left (\frac {\log p}{\log \log p}\right )$ might be of independent interest (see also Conjecture 7.1 for further discussions).
Theorem 3.1 Let $p>1$ be a prime integer and assume that h has order $k\geq 3$ in the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ We have
Remark 3.2 Let us notice that by a result of Vardi [Reference VardiVar], for any function f such that $\lim _{n\to +\infty } f(n)=+\infty $ we have $s(c,d) \ll f(d)\log d$ for almost all $(c,d)$ with $\gcd (c,d)=1$ . However, Dedekind sums take also very large values (see, for instance, [Reference Conrey, Fransen and KleinCEK, Reference GirstmairGir03] for more information).
Our proof builds from ideas of the proof of [Reference Louboutin and MunschLM21, Theorem $4.1$ ] where some tools from equidistribution theory and the theory of pseudo-random generators were used. We refer for more information to [Reference KorobovKor, Nied77], or the book of Konyagin and Shparlinski [Reference Konyagin and ShparlinskiKS, Chapter $12$ ] (see [Reference Louboutin and MunschLM21, Section $4$ ] for more details and references). Let us recall some notations. For any fixed integer s, we consider the s-dimensional cube $I_s=\left [0,1\right ]^s$ equipped with its s-dimensional Lebesgue measure $\lambda _s$ . We denote by $\mathcal {B}$ the set of rectangular boxes of the form
where $0\leq \alpha _i<\beta _i\leq 1.$ If S is a finite subset of $I^s$ , we define the discrepancy $D(S)$ by
Let us introduce the following set of points:
For good choice of h, the points are equidistributed and we expect for “nice” functions f
Lemma 3.3 For any h of order $k\geq 3$ , we have the following discrepancy bound:
Proof It follows from the proof of [Reference Louboutin and MunschLM21, Theorem 4.1] where the bound was obtained as a consequence of Erdős–Turan inequality and tools from pseudo-random generators theory.
3.2.1 Proof of Theorem 3.1
Observe that
where $f(x,y)= ((x))((y))$ . By Koksma–Hlawka inequality [Reference Drmota and TichyDT, Theorem $1.14$ ], we have
where $V(f)$ is the Hardy–Krause variation of f. Moreover, we have
The readers can easily convince themselves that $V(f) \ll 1$ . Hence, the result follows from Lemma 3.3.
Remark 3.4 The same method used to bound the discrepancy leads to a similar bound for composite f. Indeed, for $h \in ({\mathbb Z}/f{\mathbb Z})^*$ of order $k\geq 3$ , we have $s(h,f)= O\left ((\log f)^2 f/E(f)\right )$ with $E(f)=\max \{P^{+}(f)^{1/\phi (k^*)},\textrm {rad}(f)^{1/k}\}$ where $P^{+}(f)$ is the largest prime factor of f, $k^*$ is the order of h modulo $P^+(f)$ and $\displaystyle {\mathrm{rad}(f)=\prod _{\ell \mid f \atop \ell \textrm {prime}}\ell }$ is the radical of f. If $f=h^3-1$ is square-free, then we have $E(f) = f^{1/3}$ and $s(h,f)=O\left ((\log f)^2 f^{2/3}\right )$ which is close to the truth by a logarithmic factor (see Remark 6.2).
For $\gcd (b,p)=\gcd (c,p)=1$ , we recall the other definition of Dedekind–Rademacher sums
A similar argument as in the proof of Theorem 3.1 leads to a bound on these generalized sums.
Theorem 3.5 Let $q_1$ , $q_2$ and $k\geq 3$ be given natural integers. Let p run over the primes and h over the elements of order k in the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . Then, we have
Proof The proof follows exactly the same lines as the proof of Theorem 3.1 except for the fact that the function f is replaced by the function $g(x,y)= ((q_1 x))((q_2 y))$ . Hence, we have
and by symmetry we remark that
Again, $V(g) \ll 1$ and the result follows from Lemma 3.3 and Koksma–Hlawka inequality.
3.3 Twisted second moment of L- functions and Dedekind sums
We illustrate the link between Dedekind sums and twisted moments of L- functions by first proving Theorem 1.1 in the case $H=\{1\}$ with a stronger error term. For any integers $q_1,q_2 \geq 1$ and any prime $p\geq 3$ , we define the twisted moment
The following formula (see [Reference LouboutinLou94, Proposition 1]) will help us to relate L- functions to Dedekind sums:
Theorem 3.6 Let $q_1$ and $q_2$ be given coprime integers. Then, when p goes to infinity
Remark 3.7 It is worth to notice that in the case $q_2=1$ , explicit formulas are known by [Reference LouboutinLou15, Theorem $4$ ] (see also [Reference Lee and LeeLee17]). This also gives a new and simpler proof of [Reference Lee and LeeLee19, Theorem $1.1$ ] in a special case.
Proof Let us define
For p large enough, we have $\gcd (q_1,p)=\gcd (q_2,p)=1$ . Hence, using orthogonality relations and (3.9), we arrive at
When $q_1$ and $q_2$ are fixed coprime integers and p goes to infinity, we infer from (3.6) and (3.7) that
The result follows immediately.
Corollary 3.8 Let $q_1$ and $q_2$ be given natural integers. Then, when p goes to infinity
Proof Let $\delta = \gcd (q_1,q_2)$ . We clearly have $M_{q_1,q_2}(p)=M_{q_1/\delta ,q_2/\delta }(p)$ and the result follows from Theorem 3.6.
The proof of Theorem 1.1 in the case of the trivial subgroup follows easily.
Corollary 3.9 Let $d_0$ be a given square-free integer. When p goes to infinity, we have the following asymptotic formula:
Proof For $\chi $ modulo p, let $\chi '$ be the character modulo $d_0p$ induced by $\chi $ . By (1.8) and Corollary 3.8, we have
3.4 An interesting link with sums of maxima
Before turning to the general case of Theorem 1.1, we explain how to use Theorem 3.6 to estimate the seemingly innocuous sumFootnote 4 defined for any integers $q_1,q_2 \geq 1$ by
where, here and below, $q_1 x, q_2 x$ denote the representatives modulo p taken in $\left [1,p\right ]$ .
Theorem 3.10 Let $q_1$ and $q_2$ be natural integers such that $q_1 \neq q_2$ . Then, we have the following asymptotic formula:
Remark 3.11 In the special case $q_1=1$ , we are able to evaluate the sum directly without the need of Dedekind sums and L- functions. However, we could not prove Theorem 3.10 in the general case using elementary counting methods.
Remark 3.12 Let us notice that $\int _{0}^{1}\int _{0}^{1} \max (x,y) dx dy = 2/3$ . Hence, using the same method as in Section 3.2, we can show that if the points $\left (\left \{\frac {x}{p}\right \},\left \{\frac {qx}{p}\right \}\right )$ are equidistributed in the square $[0,1]^2,$ then
For q fixed and $p \rightarrow +\infty $ , the points are not equidistributed in the square and we see that the correcting factor $\frac {\gcd (q_1,q_2)^2}{12q_1q_2}$ from equidistribution is related to the Dedekind sum $s(q_1,q_2,p)$ .
We need the following result of [Reference Louboutin and MunschLM21, Theorem $2.1$ ].
Proposition 3.13 Let $\chi $ be a primitive Dirichlet character modulo $f>2$ , its conductor. Set $\displaystyle {S(k,\chi ) =\sum _{l=0}^k\chi (l)}$ . Then
3.4.1 Proof of Theorem 3.10
We follow a strategy similar to the proof of [Reference Louboutin and MunschLM21, Corollary $2.2$ ]. We denote by $\chi _0$ the trivial character. Using Proposition 3.13 and recalling the definition (3.8), we arrive at
Adding the contribution of the trivial character
we obtain
For sufficiently large p, using the fact that $q_1 \neq q_2 \bmod p$ and the orthogonality relations, we have
We now follow the method used in the proof of [Reference Louboutin and MunschLM21, Theorem $4.1$ ] (see also [Reference ElmaElma]) with some needed changes to treat the left-hand side of (3.10). Again by orthogonality, we obtain
where
Changing the order of summation and making the change of variables $n_1=q_2m_1$ , we arrive at
By symmetry, injecting this into (3.10), we arrive at
Hence, comparing the terms of order $p^3$ in the above formula (3.11) and using Corollary 3.8, we have
where
This concludes the proof.
We know turn to the general case of Theorem 1.1. Let $d_0$ be a given square-free integer such that $\gcd (d_0,p)=1$ . For $\chi $ modulo p, let $\chi '$ be the character modulo $d_0p$ induced by $\chi $ . Recall that we want to show for H a subgroup of $\left (\mathbb {Z}/p\mathbb {Z}\right )^*$ of odd order $d \ll \frac {\log p}{\log \log p}$ that
3.5 Twisted average of L- functions over subgroups
For any integers $q_1,q_2 \geq 1$ and any prime $p\geq 3$ , we define
Our main result is the following.
Theorem 3.14 Let $q_1$ and $q_2$ be given coprime integers. When H runs over the subgroups of $\left (\mathbb {Z}/p\mathbb {Z}\right )^*$ of odd order d, we have the following asymptotic formula:
Proof The proof follows the same lines as the proof of Theorem 3.6. Let us define
Hence, we obtain similarly
where we used Theorem 3.5 in the last line and noticed that $\phi (k)$ divides $\phi (d)$ whenever k divides d.
Remark 3.15 The error term is negligible as soon as $d\leq \frac {\log p}{3(\log \log p)}$ .
Corollary 3.16 Let $q_1$ and $ q_2$ be given integers. When H runs over the subgroups of $\left (\mathbb {Z}/p\mathbb {Z}\right )^*$ of odd order d, we have the following asymptotic formula:
3.6 Proof of Theorem 1.1
As in the proof of Corollary 3.9 and using Corollary 3.16,
using the condition on d.
4 Explicit formulas for $M_{d_0}(f,H)$
Recall that by (3.9)
Hence, using the definition of Dedekind sums, we obtain (see [Reference LouboutinLou16, Proof of Theorem 2])
4.1 A formula for $M_{d_0}(f,\{1\})$ for $d_0=1,2,3,6$
The first consequence of (4.1) is a short proof of [Reference LouboutinLou94, Théorèmes 2 and 3] by taking $H=\{1\}$ , the trivial subgroup of the multiplicative group $({\mathbb Z}/f{\mathbb Z}^*)$ . Indeed, (4.1) and (3.4) give
The arithmetic functions $f\mapsto \sum _{\delta \mid f}\mu (\delta )\delta ^k$ being multiplicative, we obtain (see also [Reference QiQi])
Now, it is clear by (2.1) that for $d_0$ odd and square-free and f odd, we have
Hence, on applying (4.2) to $2f$ instead of f, we therefore obtain
For $d_0\in \{3,6\}$ , the following explicit formula holds true for any f coprime with $d_0$ . It generalizes [Reference LouboutinLou94, Théorème 4] to composite moduli.
Theorem 4.1 Let $d_0>2$ be a given square-free integer. Set
For $n\in {\mathbb Z}$ , set $\varepsilon (n) =+1$ if $n\equiv +1\ \ \pmod {d_0}$ and $\varepsilon (n) =-1$ if $n\equiv -1\ \ \pmod {d_0}$ . Then, for $f>2$ such that all its prime divisors q satisfy $q\equiv \pm 1\ \ \pmod {d_0}$ , we have
In particular, for $f>2$ such that all its prime divisors q satisfy $q\equiv 1\ \ \pmod {d_0}$ , we have
Proof With the notation of [Reference LouboutinLou11, Lemma 2], we have $M_{d_0}(f,\{1\})) =4\pi ^2S(d_0,f)$ . Hence, by [Reference LouboutinLou11, Lemmas 3 and 6], we have
where the $A(d_0,f/d)$ ’s are rational numbers such that $A(d_0,f/d) =\varepsilon A(d_0,1)$ if $f/d\equiv \varepsilon \ \ \pmod {d_0}$ with $\varepsilon \in \{\pm 1\}$ (see (4.13)). If all the prime divisors q of f satisfy $q\equiv \pm 1\ \ \pmod {d_0},$ then $f/d\equiv \varepsilon (f/d)\ \ \pmod {d_0}$ and $A(d_0,f/d) =\varepsilon (f/d)A(d_0,1) =\varepsilon (f)A(d_0,1)\varepsilon (d)$ and
Hence, we finally get
The desired formula for $M_{d_0}(f,\{1\})$ follows by using the explicit formula
given in [Reference LouboutinLou11, Lemma 6].
4.2 A formula for $M(p,H)$
The second immediate consequence of (4.1) and (3.4) is:
Proposition 4.2 For $f>2$ and H a subgroup of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ , set
Then, for $p\geq 3$ a prime and H a subgroup of odd order of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ , we have
Remark 4.3 In particular, $N(f,\{1\}) =-3+2/f$ and (4.4) implies (1.5). Notice also that $N(p,H)\in {\mathbb Z}$ for $H\neq \{1\}$ , by [Reference LouboutinLou19, Theorem 6]. Moreover, by [Reference Louboutin and MunschLM21, Theorem $1.1$ ], the asymptotic formula $M(p,H) =\frac {\pi ^2}{6}+o(1)$ holds as p tends to infinity and H runs over the subgroup of $({\mathbb Z}/p{\mathbb Z})^*$ of odd order $d\leq \frac {\log p}{\log \log p}$ . Hence, we have $N(p,H)=o(p)$ under this restriction.
4.3 A formula for $M_{d_0}(p,H)$
We will now derive a third consequence of (4.1): a formula for the mean square value $M_{d_0}(f,H)$ defined in (1.9) when f is prime.
Theorem 4.4 Let $d_0>1$ be a square-free integer. Let $f>2$ be coprime with $d_0$ . Let H be a subgroup of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Whenever $\delta $ divides $d_0$ , let $s_\delta :({\mathbb Z}/\delta f{\mathbb Z})^*\longrightarrow ({\mathbb Z}/f{\mathbb Z})^*$ be the canonical surjective morphism and set $H_{\delta }=s_\delta ^{-1}(H)$ and $H_{\delta }' =s_\delta ^{-1}(H\setminus \{1\})$ . Define the rational number
Then, for $p\geq 3$ a prime which does not divide $d_0$ and H a subgroup of odd order of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ , we have
where
and
Moreover,
where
Proof Using (2.1) and by making the change of variables $\delta \mapsto d_0f/\delta $ in (4.1), we obtain
Since $\{\delta ;\ \delta \mid d_0p\}$ is the disjoint union of $\{\delta ;\ \delta \mid d_0\}$ and $\{\delta p;\ \delta \mid d_0\}$ , by (4.10), we obtain
Now, $S:=\sum _{h\in H_{d_0}}s(h,\delta )=0$ whenever $\delta \mid d_0$ , which gives
and implies (4.7). Indeed, let $\sigma :({\mathbb Z}/d_0f{\mathbb Z})^*\longrightarrow ({\mathbb Z}/\delta {\mathbb Z})^*$ be the canonical surjective morphism. Its restriction $\tau $ to the subgroup $H_{d_0}$ is surjective, by the Chinese reminder theorem. Hence, $S =(H_{d_0}:\ker \tau ) \times S'$ , where $S':=\sum _{c\in ({\mathbb Z}/\delta {\mathbb Z})^*}s(c,\delta ) =\sum _{c\in ({\mathbb Z}/\delta {\mathbb Z})^*}s(-c,\delta ) =-S'$ yields $S'=0$ . In the same way, whenever $\delta \mid d_0$ , the kernel of the canonical surjective morphism $s:({\mathbb Z}/d_0f{\mathbb Z})^*\longrightarrow ({\mathbb Z}/\delta f{\mathbb Z})^*$ being a subgroup of order $\phi (d_0f)/\phi (\delta f) =\phi (d_0)/\phi (\delta )$ , we have
and (4.6) follows from (4.11) and (4.12).
Then, (4.7) is a direct consequence of (4.6) and (4.5). Finally, (4.9) is an immediate consequence of (4.5) and (4.12).
4.3.1 A new proof of Theorem 1.1
We split the sum in (4.11) into two cases depending whether $h=1$ or not. By (3.4), we have $s(1,\delta p)=\frac {p\delta }{12}+O(1)$ giving a contribution to the sum of order
When $h\neq 1$ and $h\in H_{d_0}$ , it is clear that the order of h modulo p is between $3$ and d. Hence, it follows from Theorem 3.1 (see the Remark after) that $s(h,\delta p)=O((\log p)^2 p^{1-\frac {1}{\phi (d)}})$ . The integer $d_0$ being fixed, we can sum up these error terms and the proof is finished.
4.4 An explicit way to compute $N_{d_0}(f,\{1\})$
Lemma 4.5 Let $d_0>1$ be a square-free integer. Let $f>2$ be coprime with $d_0$ . Recall that $H_{d_0}(f) =\{h\in ({\mathbb Z}/d_0f{\mathbb Z})^*,\ h\equiv 1\ \ \pmod f\}$ and set
and
a rational number depending only on f modulo $d_0$ . Then $U(d_0,f) =fA(d_0,f).$
Proof As in [Reference LouboutinLou11, Lemma 3], set
where $F(x,y)=1+\cot (\pi x)\cot (\pi y)$ . On the one hand, since $\gcd (d_0f,n)=1$ if and only if $\gcd (d_0,n)=\gcd (f,n)=1$ and $\displaystyle {\sum _{d\mid f\atop d\mid n}\mu (d)}$ is equal to $1$ if $\gcd (f,n)=1$ and is equal to $0$ otherwise, we have
On the other hand, the canonical morphism $\sigma :H_{d_0}(f) \rightarrow H_{d_0}(f/d)$ is surjective and both groups have order $\phi (d_0f)/\phi (f) =\phi (d_0(f/d))/\phi (f/d) =\phi (d_0)$ . Hence, $\sigma $ is bijective and
Using [Reference LouboutinLou11, Lemma 6] and Möbius’ inversion formula, we finally do obtain
where we set $\delta '=d/\delta $ .
Proposition 4.6 Let $d_0>1$ be a square-free integer. Set $B=\prod _{q\mid d_0}(q^2-1)$ . For $f>2$ and $\gcd (d_0,f)=1,$ we have
Consequently, $N_{d_0}(f,\{1\})$ is a rational number depending only on f modulo $d_0$ .
Proof Set $H=H_{d_0}(f):=\{h\in ({\mathbb Z}/d_0f{\mathbb Z})^*,\ h\equiv 1\ \ \pmod f\}$ . By (4.5), we have
Using (3.4) to evaluate the contribution of $h=1$ in this expression and $\sum _{\delta \mid d_0}\mu (\delta ) =0$ , we get
and
by (3.1) and by noticing that $\# H =\phi (d_0)$ . Therefore,
(make the change of variable $\delta \mapsto d_0/\delta $ ). Lemma 4.5 gives the desired result.
Remark 4.7 As a consequence, we obtain $M_{d_0}(p,\{1\}) =\frac {\pi ^2}{6}\prod _{q\mid d_0}\left (1-\frac {1}{q^2}\right ) +O(p^{-1})$ , using (4.7) and the fact that $N_{d_0}(p,\{1\})$ depends only on p modulo $d_0$ . This gives in this extreme situation another proof of Theorem 1.1 with a better error term. Moreover, in that situation, we have $K ={\mathbb Q}(\zeta _p)$ and in (1.11) the term $\Pi _{d_0}(p,\{1\})$ is bounded from below by a constant independent of p, by Corollary 2.4.
5 The case where $f=a^{d-1}+\cdots +a^2+a+1$
In this specific case, we are able to obtain explicit formulas for $M_{d_0}(f,H)$ when the subgroup H is defined in terms of the parameter a defining the modulus. For a general subgroup H, it seems unrealistic to be more explicit than the formula involving Dedekind sums given in Theorem 4.4. It might be interesting to explore formulas involving continued fraction expansions in view of their link to Dedekind sums [Reference HickersonHic].
5.1 Explicit formulas for $d_0=1,2$
Lemma 5.1 Let $f>1$ be a rational integer of the form $f=(a^d-1)/(a-1)$ for some $a\neq -1,0,1$ and some odd integer $d\geq 3$ . Hence, f is odd. Set $H=\{a^k;\ 0\leq k\leq d-1\}$ , a subgroup of order d of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Then
and
Proof We have $S(H,f) =\sum _{k=0}^{d-1}s(a^k,f)$ . Moreover, $S(H_2,2f) =\sum _{k=0}^{d-1}s(a^k,2f)$ if a is odd and $S(H_2,2f) =s(1,2f)+\sum _{k=1}^{d-1}s(a^k+f,2f)$ if a is even. Now, we claim that for $0\leq k\leq d-1$ , we have
and that for $1\leq k\leq d-1$ , we have
Noticing that $\sum _{k=1}^{d-1}a^k =f-1$ and $\sum _{k=1}^{d-1}a^{-k} =\frac {f-1}{(a-1)f+1}$ , we then get the assertions on $S(H,f)$ and $S(H_2,2f)$ . Now, let us, for example, prove the third claim. Hence, assume that a is even and that $1\leq k\leq d-1$ . Then $f_k :=(a^k-1)/(a-1)$ is odd, $\mathrm {sign}(f_k)=\mathrm {sign}(a)^k$ and $a^k+f>0$ . First, since $2f\equiv -2a^{k}\ \ \pmod {a^k+f}$ , using (3.3), we have
Second, noticing that $a^k+f\equiv f_k\ \ \pmod {2a^{k}}$ and using (3.3), we have
Finally, noticing that $2a^{k}\equiv 2\ \ \pmod {f_k}$ and using (3.3) and (3.4), we have
After some simplifications, we obtain the desired formula for $s(a^k+f,2f)$ .
Notice that for $d=3$ , we obtain $S(H,f) =\frac {f-1}{12}$ , in accordance with (6.1).
Using (4.6) and Lemma 5.1, we readily obtain:
Theorem 5.2 Let $d\geq 3$ be a prime integer. Let $p\equiv 1\ \ \pmod {2d}$ be a prime integer of the form $p=(a^d-1)/(a-1)$ for some $a\neq -1,0,1$ . Let K be the imaginary subfield of degree $(p-1)/d$ of the cyclotomic field ${\mathbb Q}(\zeta _p)$ . Set $H=\{a^k;\ 0\leq k\leq d-1\}$ , a subgroup of order d of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . We have the mean square value formulas
and
Consequently, for a given d, as $p\rightarrow \infty $ , we have
On the other hand, for a given a, as $p\rightarrow \infty $ , we have
Remark 5.3 Assertion (5.1) was initially provedFootnote 5 in [Reference LouboutinLou16, Theorem 5] for $d=5$ and then generalized in [Reference Louboutin and MunschLM21, Proposition 3.1] to any $d\geq 3$ . However, (5.1) is much simpler than [Reference Louboutin and MunschLM21, equation (22)]. Notice that if p runs over the prime of the form $p=(a^d-1)/(a-1)$ with $a\neq 0,2$ even then $M_2(p,H) =\frac {6}{8}\times \frac {a-1}{a+1}\times M(p,H)$ and the asymptotic (1.12) is not satisfied.
5.2 The case where p is a Mersenne prime and $d_0=1,3,15$
In the setting of Theorem 5.4, we have $2\in H$ . Hence, by Remark 2.2, we assume that $d_0$ is odd.
Theorem 5.4 Let $p=2^d-1>3$ be a Mersenne prime. Hence, d is odd and $H=\{2^k;\ 0\leq k\leq d-1\}$ is a subgroup of odd order d of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . Let K be the imaginary subfield of degree $m =(p-1)/d$ of ${\mathbb Q}(\zeta _p)$ . Then
and
In particular, for $d\equiv 3\ \ \pmod 4$ , we have $h_K^-\leq 2\left (\frac {8p}{75}\right )^{m/4}.$
Proof By (4.6), we have
where for H a subgroup of odd order of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ , we set
The formulas for $M(p,H),M_3(p,H)$ , and $M_{15}(p,H)$ follow from (5.3)) and Lemma 5.5. The upper bounds on $h_K^-$ follow from (1.11) and Lemma 2.3 according to which $\Pi _q(p,H)\geq 1$ if q is of even order in the quotient group $G/H$ , where $G=({\mathbb Z}/p{\mathbb Z})^*$ , hence, if q is of even order in the group G. Now, since $p\equiv 3\ \ \pmod 4$ the group G is of order $p-1=2N$ with N odd and q is of even order in G if and only $q^N=-1$ in G, i.e., if and only if the Legendre symbol $\left (\frac {q}{p}\right )$ is equal to $-1$ . Now, since $p=2^d-1\equiv -1\equiv 3 \ \ \pmod 4$ for $d\geq 3$ , the law of quadratic reciprocity gives $\left (\frac {3}{p}\right ) =-\left (\frac {p}{3}\right ) =-\left (\frac {1}{3}\right ) =-1$ , as $p\equiv (-1)^d-1\equiv -2\equiv 1\ \ \pmod 3$ . Hence, $\Pi _3(p,H)\geq 1$ . In the same way, if $d\equiv 3\ \ \pmod 4$ then $p=2^d-1 =2\cdot 4^{\frac {d-1}{2}}-1\equiv 2\cdot (-1)^{\frac {d-1}{2}}-1\equiv -3\equiv 2\ \ \pmod 5$ and $\left (\frac {5}{p}\right ) =\left (\frac {p}{5}\right ) =\left (\frac {2}{5}\right ) =-1$ and $\Pi _5(p,H)\geq 1$ .
Lemma 5.5 Set $f=2^d-1$ and $\varepsilon _d =(-1)^{(d-1)/2}$ with $d\geq 2$ odd. Hence, $\gcd (f,15)=1$ . Set $H =\{2^k;\ 0\leq k\leq d-1\}$ , a subgroup of order d of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Then,
Proof The first assertion is the special case $a=2$ of Lemma 5.1. Let us now deal with the second assertion. Here, $H_3 =\{2^k;\ 0\leq k\leq d-1\} \cup \{2^k+(-1)^kf;\ 0\leq k\leq d-1\}$ . We assume that $0\leq k\leq d-1$ . Hence, $\mathrm {sign}(2^k+(-1)^kf)=(-1)^k$ .
1. Noticing that $3f\equiv -3\ \ \pmod {2^k}$ , by (3.3), we obtain
Noticing that $2^k\equiv (-1)^k\ \ \pmod 3$ , by (3.3) and (3.4), we obtain
Hence,
2. Noticing that $3f\equiv -3\cdot (-1)^k2^k\ \ \pmod {2^k+(-1)^kf}$ , by (3.3), we obtain
and noticing that $2^k+(-1)^kf\equiv (-1)^{k-1}\ \ \pmod {3\cdot 2^k}$ , by (3.3), we obtain
Using (3.4), we finally obtain
3. Using $\sum _{k=0}^{d-1}2^k =f$ , $\sum _{k=0}^{d-1}2^{-k}=\frac {2f}{f+1}$ and $\sum _{k=0}^{d-1}(-1)^k=1$ , we obtain
Hence, we do obtain
and $N_3'(f,H) =-d$ , by (5.4).
Let us finally deal with the third and fourth assertions. The proof involves tedious and repetitive computations. For this reason, we will restrict ourselves to a specific case. Let us, for example, give some details for the proof of (5.8) in the case that $d\equiv 1\ \ \pmod 4$ . We have $f=2^d-1\equiv 1\ \ \pmod {30}$ and $H_{15}=\cup _{l=0}^{14}E_l$ , where $E_l := \{2^k+lf;\ 0\leq k\leq d-1,\ \gcd (2^k+l,15)=1\}$ for $0\leq l\leq 14$ . We have to compute the sums $s_l:=\sum _{n\in E_l}s(n,15f)$ . Let us, for example, give some details in the case that $l=1$ . We have $\gcd (2^k+1,15)=1$ if and only if $k\equiv 0\ \ \pmod 4$ . Hence, $s_1=\sum _{k=0}^{(d-1)/4}s(16^{k}+f,15f)$ . Using (3.3) and (3.4), we obtain
Finally, using $\sum _{k=0}^{(d-1)/4}16^k =\frac {8f+7}{15}$ and $\sum _{k=0}^{(d-1)/4}16^{-k} =\frac {2(8f+7)}{15(f+1)}$ , we obtain
We conclude this section with the following result for $d_0 =3\times 5\times 7 =105$ , whose long proof we omit:Footnote 6
Lemma 5.6 Set $f =2^d-1$ with $d>1$ odd. Assume $\gcd (f,105)=1$ , i.e., that $d\equiv 1, 5, 7, 11\ \ \pmod {12}$ . Set $H =\{2^k;\ 0\leq k\leq d-1\}$ , a subgroup of order d of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Then
Lemmas 5.5 and 5.6 show that the following conjecture holds true for $d_0\in \{1,3,5,15,105\}$ :
Conjecture 5.7 Let $d_0\geq 1$ be odd and square-free. Let N be the order of $2$ in the multiplicative group $({\mathbb Z}/d_0{\mathbb Z})^*$ . Set $f =2^d-1$ with $d>1$ odd and $H =\{2^k;\ 0\leq k\leq d-1\}$ , a subgroup of order d of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Assume $\gcd (f,d_0)=1$ . Then $N_{d_0}'(f,H) =A_1(d)d+A_0(d)$ , where $A_1(d)$ and $A_0(d)$ are rational numbers which depend only on d modulo N, i.e., only on f modulo $d_0$ . Hence, for a prime $p\geq 3$ , we expect
confirming again that the restriction on d in Theorem 1.1 should be sharp.
There is apparently no theoretical obstruction preventing us to prove Conjecture 5.7. Indeed, for a fixed $d_0$ , the formulas for $A_0(d)$ and $A_1(d)$ could be guessed using numerous examples on a computer algebra system. However, for large $d_0$ ’s, the set of cases to consider grows linearly and a more unified approach seems to be required to give a complete proof.
6 The case of subgroups of order $d=3$
6.1 Formulas for $d_0=1,2,6$
Let $p\equiv 1\ \ \pmod 6$ be a prime integer. Let K be the imaginary subfield of degree $m=(p-1)/3$ of the cyclotomic field ${\mathbb Q}(\zeta _p)$ . Since p splits completely in the quadratic field ${\mathbb Q}(\sqrt {-3})$ of class number one, there exists an algebraic integer $\alpha =a+b\frac {1+\sqrt {-3}}{2}$ with $a,b\in {\mathbb Z}$ such that $p =N_{{\mathbb Q}(\sqrt {-3})/{\mathbb Q}}(\alpha ) =a^2+ab+b^2$ . Then, $H=\{1,a/b,b/a\}$ is the unique subgroup of order $3$ of the cyclic multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . So we consider the integers $f>3$ of the form $f=a^2+ab+b^2$ , with $a,b\in {\mathbb Z}\setminus \{0\}$ and $\gcd (a,b)=1$ , which implies $\gcd (a,f) =\gcd (b,f)=1$ and the oddness of f. We have the following explicit formula.
Lemma 6.1 Let $f>3$ be of the form $f=a^2+ab+b^2$ , with $a,b\in {\mathbb Z}$ and $\gcd (a,b)=1$ . Set $H=\left \{1,a/b,b/a\right \}$ , a subgroup of order $3$ of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Then
Proof Noticing that $s(b,f,a) =s(b,b^2,a) =s(1,b,a) =s(b,a)$ , by (3.5), and $s(f,a,b) =s(a^2,a,b) =s(a,1,b) =s(a,b)$ , and using (3.3), we obtain
Finally, $S(H,f)=s(1,f)+s(a,b,f)+s(b,a,f) =s(1,f)+2s(a,b,f)$ and use (3.4) and (4.9).
Remark 6.2 Take $f_1=A^2+AB+B^2>0$ , where $3\nmid f_1$ and $\gcd (A,B)=1$ . Set $f=(f_1+1)^3-1$ . Then $f=a^2+ab+b^2$ , where $a=Af_1+A-B$ , $b=Bf_1+A+2B$ and $\gcd (a,b)=1$ . By Lemmas 6.1, we have an infinite family of moduli f for which the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ contains at the same time an element $h=a/b$ of order $d=3$ for which $s(h,f)$ is asymptotic to $1/12$ and an element $h'=f_1+1$ of order $d=3$ for which $s(h',f)$ is asymptotic to $f^{2/3}/12$ . Indeed, by (3.3) and (3.4), for $f=(f_1+1)^3-1$ , we have $s(h',f) =\frac {h^{\prime 5}+h^{\prime 4} - 6h^{\prime 3}+6}{12f}$ .
To deal with the case $d_0>1$ , we notice that by (4.9), we have the following proposition.
Proposition 6.3 Let $d_0\geq 1$ be a given square-free integer. Take $f>3$ odd of the form $f =a^2+ab+b^2$ , where $\gcd (a,b)=1$ and $\gcd (d_0,f)=1$ . Set $H=\{1,a/b,b/a\}$ , a subgroup of order $3$ of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Let $N_{d_0}(f,H)$ be the rational number defined in ( 4.5 ). Then
where $N_{d_0}(f,\{1\})$ is a rational number which depends only on f modulo $d_0$ , by Proposition 4.6, and where
It seems that there are no explicit formulas for $S(a,b,\delta f)$ , $S(H_\delta ,\delta f)$ , or $N_\delta (f,H)$ for $\delta>1$ (however, assuming that $b=1$ , we will obtain such formulas in Section 6.2 for $\delta \in \{2,3,6\}$ ). Instead, our aim is to prove in Proposition 6.4 that $N_\delta (f,H) =O(\sqrt f)$ for $\delta \in \{2,3,6\}$ .
Let $f>3$ be of the form $f=a^2+ab+b^2$ , with $a,b\in {\mathbb Z}$ and $\gcd (a,b)=1$ . Hence, a or b is odd. Since $a^2+ab+b^2 =a^{\prime 2} + a'b' + b^{\prime 2} = a^{\prime \prime 2} + a" b" + b^{\prime \prime 2}$ and $a'/b' =a/b$ and $a"/b" =a/b$ in $({\mathbb Z}/f{\mathbb Z})^*$ , where $(a',b') =(-b,a+b)$ and $(a",b") =(-a-b,a)$ , we may assume that both a and b are odd. Moreover, assume that $\gcd (3,f)=1$ . If $3\nmid ab$ , by swapping a and b as needed, which does not change neither H nor $S(a,b,H)$ , we may assume that $a\equiv -1\ \ \pmod 6$ and $b\equiv 1\ \ \pmod 6$ . If $3\mid ab$ , by swapping a and b and then changing both a and b to their opposites as needed, which does not change neither H nor $S(a,b,H)$ , we may assume that $a\equiv 3\ \ \pmod 6$ and $b\equiv 1\ \ \pmod 6$ . So in Proposition 6.3, we may restrict ourselves to the integers of the form
Proposition 6.4 Let $\delta \in \{2,3,6\}$ be given. Let f be as in ( 6.2 ), with $\gcd (f,\delta )=1$ . Then, $s(h,\delta f) =O(\sqrt f)$ for any $h\in ({\mathbb Z}/\delta f{\mathbb Z})^*$ such that $h\equiv a/b\ \ \pmod f$ . Consequently, for a given $d_0\in \{1,2,3,6\}$ , in Proposition 6.3, we have $N_{d_0}(f,H) =O(\sqrt f)$ , and we cannot expect great improvements on these bounds, by ( 6.11 ), ( 6.13 ), and ( 6.15 ).
Proof First, by (6.1), we have
Second, f being odd, recalling (4.13), we have $A(2,f) =A(2,1) =0$ , $N_2(f,\{1\}) =-1$ ,
and
Third, assume that $d_0\in \{3,6\}$ . Then $\gcd (f,3)=1$ . Hence, $f\equiv 1\ \ \pmod 6$ . Therefore, $A(3,f) =A(3,1) =4/3$ , $A(6,f) =A(6,1) =-4$ , $N_3(f,\{1\}) =N_6(f,\{1\}) =-1$ ,
and
If $a\equiv -1\ \ \pmod 6$ , $b\equiv 1\ \ \pmod 6,$ and $\delta \in \{1,2\}$ , then $\{h\in ({\mathbb Z}/3\delta f{\mathbb Z})^*;\ h\equiv a/b\ \ \pmod f\} =\{a/b,(a+2f)/b\}$ and
If $a\equiv 3\ \ \pmod 6$ , $b\equiv 1\ \ \pmod 6,$ and $\delta \in \{1,2\}$ , then $\{h\in ({\mathbb Z}/3\delta f{\mathbb Z})^*;\ h\equiv a/b\ \ \pmod f\} =\{(a-\delta f)/b,(a+\delta f)/b\}$ and
Let us now bound the Dedekind–Rademacher sums in (6.3)–(6.5). We will need the bounds,
Indeed, $4f-(\vert a\vert +\vert b\vert )^2 \geq 3(\vert a\vert -\vert b\vert )^2\geq 0$ and $f\leq a^2+a^2b^2+b^2 =3a^2b^2$ .
First, we deal with the Dedekind–Rademacher sums $s(a,b,\delta f)$ in (6.3) and (6.4), where $\delta \in \{2,3,6\}$ . Here, $\gcd (a,b) =\gcd (a,\delta f)=\gcd (b,\delta f) =1$ . Then (3.7) and (6.6) enable us to write (3.6) as follows:
Hence, in (6.3) and (6.4), we have $s(a,b,2f),\ s(a,b,3f),\ s(a,b,6f)=O(\sqrt f)$ .
Second, the remaining and more complicated Dedekind–Rademacher sums in (6.4) and (6.5) are of the form $s(a+\varepsilon \delta f,b,3\delta f)$ , where $\varepsilon \in \{\pm 1\}$ , $\delta \in \{1,2\}$ and $\gcd (a+\varepsilon \delta f,3\delta f) =\gcd (b,3\delta f)=1$ . Set $\delta ' =\gcd (a+\varepsilon \delta f,b)$ . Then $\gcd (\delta ',3\delta f)=1$ . Thus, $s(a+\varepsilon \delta f,b,3\delta f)$ $=s((a+\varepsilon \delta f)/\delta ',b/\delta ',3\delta f)$ , where now the three terms in this latter Dedekind–Rademacher are pairwise coprime. Then (3.7) and (6.6) enable us to write (3.6) as follows:
Now, $3\delta f\equiv -3\varepsilon a\ \ \pmod {a+\varepsilon \delta f}$ gives $s(b/\delta ',3\delta f,(a+\varepsilon \delta f)/\delta ') =-\varepsilon s(b/\delta ',3a,(a+\varepsilon \delta f)/\delta ')$ . Since the three rational integers in this latter Dedekind–Rademacher are pairwise coprime, the bounds (6.6) and (3.7) enable us to write (3.6) as follows:
It follows that $s(a+\varepsilon \delta f,b,3\delta f) =s((a+\varepsilon \delta f)/\delta ',b/\delta ',3\delta f) =O(\sqrt f)$ , i.e., in (6.4) and (6.5), we have $s(a+2f,b,6f),\ s(a-2f,b,6f),\ s(a+2f,b,3f), s(a-f,b,3f), s(a+f,b,3f) =O(\sqrt f)$ .
Conjecture 6.5 Let $\delta $ be a given square-free integer. Let $f>3$ run over the odd integers of the form $f=a^2+ab+b^2$ with $\gcd (a,b)=1$ and $\gcd (\delta ,f)=1$ . Then $s(h,\delta f) =O(\sqrt f)$ for any $h\in ({\mathbb Z}/\delta f{\mathbb Z})^*$ such that $h\equiv a/b\ \ \pmod f$ . Consequently, for a given square-free integer $d_0$ , in Proposition 6.3, we would have $N_{d_0}(f,H) =O(\sqrt f)$ for $\gcd (d_0,f)=1$ .
Putting everything together, we obtain:
Theorem 6.6 Let $p\equiv 1\ \ \pmod 6$ be a prime integer. Let K be the imaginary subfield of degree $(p-1)/3$ of the cyclotomic field ${\mathbb Q}(\zeta _p)$ . Let H be the subgroup of order $3$ of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . We have
and the following effective asymptotics and upper bounds
6.2 The special case $p=a^2+a+1$ and $d_0=1,2,6$
Let $f>3$ be of the form $f=a^2+a+1$ , $a\in {\mathbb Z}$ . Then $\gcd (f,6)=1$ if and only if $a\equiv 0,2,3,5\ \ \pmod 6$ . We define $c_a^{\prime }$ , $c_a^{\prime \prime }$ , $c_a^{\prime \prime \prime }$ and $c_a =(-1-2c_a^{\prime } -c_a^{\prime \prime }+2c_a^{\prime \prime \prime })/12$ , as follows:
Theorem 6.7 Let $p\equiv 1\ \ \pmod 6$ be a prime integer of the form $p=a^2+a+1$ with $a\in {\mathbb Z}$ . Let K be the imaginary subfield of degree $(p-1)/3$ of the cyclotomic field ${\mathbb Q}(\zeta _p)$ . Let H be the subgroup of order $3$ of the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ . We have
and
showing that the error term in ( 6.7 ) is optimal.
Proof The formula (6.8) is a special case of (5.2) for $d=3$ . By (4.7), we have
Lemma 6.8 Let $f>3$ be of the form $f=a^2+a+1$ , $a\in {\mathbb Z}$ . Set $H=\{1,a,a^2\}$ , a subgroup of order $3$ of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . We have
and
Proof Apply Lemma 5.1 with $d=3$ and $f=a^2+a+1$ to get (6.10). Then, using (4.8), we get $N_2(f,H) =-f-4S(H,f)+8S(H_2,2f) =\frac {2c_a^{\prime }+1}{3} =(-1)^{a-1}(2a+1)$ .
Lemma 6.9 Let $f>3$ be of the form $f=a^2+a+1$ , $a\in {\mathbb Z}$ . Assume that $\gcd (f,6)=1$ , i.e., that $a\equiv 0, 2, 3, 5\ \ \pmod 6$ . Set $H=\{1,a,a^2\}$ , a subgroup of order $3$ of the multiplicative group $({\mathbb Z}/f{\mathbb Z})^*$ . Then
and
Proof Let us, for example, detail the computation of $S(H_6,6f)$ in the case that $a\equiv 0\ \ \pmod 6$ . We have $f\equiv 1\ \ \pmod 6$ and $H_6=\{1,1+4f,a+f,a+5f,a^2+f,a^2+5f\}$ . Since $a^2+f =(a+f)^{-1}$ and $a^2+5f=(a+5f)^{-1}$ in $({\mathbb Z}/f{\mathbb Z})^*$ , we have $S(H_6,6f) =s(1,6f)+s(1+4f,6f)+2s(a+f,6f)+2s(a+5f,6f)$ , by (3.2). Using (3.3) and (3.4), we obtain $s(1,6f) =\frac {18f^2-9f+1}{36f}$ , $s(1+4f,6f) =\frac {2f^2-13f+1}{36f}$ , $s(a+f,6f) =-\frac {(3a-21)f+1}{72f}$ , and $s(a+5f,6f) =-\frac {(35a+19)f+1}{72f}$ . Formula (6.14) follows.
By (4.8), we have
and
Using (6.1), (6.10), and (6.12), we obtain $N_3(f,H) =\frac {c_a^{\prime \prime }+1}{4}$ and (6.13). Using (6.1), (6.10), (6.12), and (6.14), we obtain $N_6(f,H) =\frac {-1-2c_a^{\prime }-c_a^{\prime \prime }+2c_a^{\prime \prime \prime }}{12} =c_a$ and (6.15).
7 Conclusion and a conjecture
The proof of Lemma 5.1 gives for $d\geq 3$ odd and $a\neq 0,\pm 1$
Our numerical computations suggest the following stronger version of Theorem 3.1:
Conjecture 7.1 There exists $C>0$ such that for any $d>2$ dividing $p-1$ and any h of order d in the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ , we have
Indeed, for $p\leq 2\cdot 10^5$ , we checked on a desk computer that any $d>2$ dividing $p-1$ and any h of order d in the multiplicative group $({\mathbb Z}/p{\mathbb Z})^*$ , we have
The estimate (7.2) would allow to slightly extend the range of validity of Theorem 1.1 to $d \leq (1-\varepsilon )\frac {\log p}{\log \log p}$ . Moreover, the choice $a=2$ in (7.1) for which $s(2,f)$ is asymptotic to $\frac {1}{24}f$ with $f=2^d-1$ shows that $s(h,p) =o(p)$ cannot hold true in the range $d \asymp \log p$ . Notice that we cannot expect a better bound than (7.2), by (7.1). Finally, the restriction that p be prime in (7.2) is paramount by Remark 6.2 where $s(a,f) \sim f^{2/3}/12$ for a of order $3$ in $({\mathbb Z}/(a^3-1){\mathbb Z})^*$ .