The distribution of ideals whose norm divides $n$ in the Gaussian ring

1.   Introduction

For each positive integer $n$, denote by $\tau(n)$ the number of divisors of $n$ and let $\Omega_{n} = \{d_{1}, d_{2}, \cdots, d_{\tau(n)}\}$ be the set of divisors of $n$. Let $\mathfrak{S}_{n}$ be the set of all subsets of $\Omega_{n}$ and let $\mu_{n}$ be the uniform probability measure on $\Omega_{n}$:

It is easily verified that $(\Omega_{n}, \mathfrak{S}_{n}, \mu_{n})$ is a probability space. Consider the random variable $D_{n}$:

The distribution function $F_{n}$ of $D_{n}$ is given by

It is clear that the sequence $\{F_{n}\}_{n = 1}^{\infty}$ does not converge pointwise on $[0, 1]$ since

However, Deshouillers, Dress, and Tenenbaum ^[3] proved that its $\mathrm{Ces\grave{a}ro}$ means is uniformly convergent on $[0, 1]$. No less remarkable, this limit is the distribution function of a probability law well known to specialists: the arcsine law, with density $1/\left(\pi\sqrt{u(1-u)}\right)$. More precisely,

holds uniformly for $x\geq 2$ and $0\leq t\leq1$, and the error term in (1.1) is optimal.

Subsequently, Cui and Wu^[1], Feng^[6], and Feng and Wu^[4] studied the related issues of the Deshouillers-Dress-Tenenbaum (DDT) theorem. Recently, Leung^[9] proved that factorization of integers into $k$ parts follows the Dirichlet distribution Dir$(\frac{1}{k}, \cdots, \frac{1}{k})$ by multidimensional contour integration, thereby generalizing the DDT arcsine law on divisors where $k = 2$. Their results were obtained in $\mathbb{Z}$.

In this paper, we consider a similar problem in the Gaussian ring, unless otherwise stated, and throughout this paper $K$, $O_{K}$, $s$, and $\sigma_{0}(\tau)$ will be the Gaussian field, the Gaussian ring(of the form $a+bi$, where $a, b\in\mathbb{Z}$ and $i^{2} = -1$), $\sigma+i\tau$, and $c_{0}/\log(q(\lvert \tau\rvert+1))$. For each positive integer $n$, let $\Xi_{n} = \left\{\mathfrak{a}\in O_{K}: N(\mathfrak{a})\; \text{divides}\; n\right\}$. Denoting by $\xi_{K}(n)$ the number of ideals in $\Xi_{n}$, then

where $a_{K}(n)$ is the number of integral ideals with norm $n$ in $O_{K}$. Since $a_{K}(n)$ is multiplicative, so is $\xi_{K}(n)$.

Let $\mathfrak{S}_{n}$ be the set of all subsets of $\Xi_{n}$ and let $\mu_{n}$ be the uniform probability:

It is easily verified that $(\Xi_{n}, \mathfrak{S}_{n}, \mu_{n})$ is a probability space. Consider the random variable $\mathfrak{D}_{n}$:

The distribution function of $\mathfrak{D}_{n}$ is given by

It is clear that the sequence $\left\{F_{K, n}\right\}_{n = 1}^{\infty}$ does not converge pointwise on $[0, +\infty)$, since

$(1)$ If $p\equiv1(\bmod4)$, then $\Xi_{p} = \{\mathfrak{a}_{0}, \mathfrak{a}_{1}, \mathfrak{a}_{2}\in\mathbb{Z}[i]: N(\mathfrak{a}_{0}) = 1, N(\mathfrak{a}_{1}) = N(\mathfrak{a}_{2}) = p\}$, and so we have

$(2)$ If $p\equiv3(\bmod4)$, then $\Xi_{p} = \left\{\mathfrak{a}_{0}\in\mathbb{Z}[i]: N(\mathfrak{a}_{0}) = 1\right\}$, and so we have $F_{K, p}(t) = 1$.

$(3)$If $p = 2$, $\Xi_{p} = \{\mathfrak{a}_{0}, (1+i): N(\mathfrak{a}_{0}) = 1, N(1+i) = 2\}$, then

However, we shall see that

is uniformly convergent on $[0, 1]$, and we get the following result.

Theorem 1.1. Uniformly for $x\geq2$ and $0\leq t\leq1$,

holds, where

is beta function. So, as $x\rightarrow +\infty$, $x^{-1}\sum_{n\leq x}F_{K, n}(t)$ is subject to beta distribution with density $\sqrt{3}/\left(2\pi\sqrt[3]{u^{2}(1-u)}\right)$, since $B(1/3, 2/3) = \Gamma(1/3)\Gamma(2/3)/\Gamma(1) = 2\pi/\sqrt{3}$.

This distribution is not arcsince law. Feng and Wu^[5] also gave a special case that satisfies the beta distribution.

2.   Preliminaries

In order to study $x^{-1}\sum_{n\leq x}F_{K, n}(t)$, we need to consider $\sum_{n\leq x}1/\xi_{K}(nN(\mathfrak{a}))$. Let's start with the properties of $\xi_{K}(n)$. Dedekind defined the Dedekind zeta function of $K$ as follows:

where $\mathfrak{a}$ runs over all nonzero integral ideals in $O_{K}$. According to [7, Theorem 2.8], we have $\zeta_{K}(s) = \zeta(s)L(s, \chi)$, where $\chi$ is the primitive character modulo $4$. Easily, we get $a_{K}(n) = \sum_{d\mid n}\chi(d)$, so formula (1.2) can be converted to

When $n = p^{m}$, $p$ is prime. We have

Second, we will get the mean value of $1/\xi_{K}(nN(\mathfrak{a}))$ based on the Selberg-Delange method. The method was developed by Selberg^[10] and Delange^[2,3]. For more details, the reader is referred to the book of Tenenbaum^[11].

It's necessary to use Hankel contour when applying the method. For each value of the positive parameter $r$, we designate the Hankel contour as the path consisting of the circle $\lvert s\rvert = r$ excluding the point $s = -r$ and of the half-line $(-\infty, -r]$ covered twice, with respective arguments $+\pi$ and $-\pi$. The brief introduction of Hankel's formula follows.

Lemma 2.1 (Hankel's formula). For each $X > 1$, let $H(X)$ denote the part of the Hankel contour situated in the half-plane $\sigma > -X$, then we have, uniformly for $z\in\mathbb{C}$,

Proof. For a detailed description of this lemma, see [11, p.179, Theorem 0.17, Corollary 0.18].□

The proof of Theorem 1.1 depends on the following two lemmas.

Lemma 2.2. For any integral ideal $\mathfrak{a}\in O_{K}$,

holds uniformly for $x\geq2$, where

Proof. In order to get the mean value of $1/\xi_{K}(nN(\mathfrak{a}))$, we first consider its Dirichlet series $\sum_{n = 1}^{+\infty}\xi_{K}(nN(\mathfrak{a}))^{-1}n^{-s}$. Let $\upsilon_{p}(n)$ denote the $p$-adic valuation of $n$. By using the formula

we write for $\Re s > 1$:

where $\chi_{0}$ is the principal character mod $4$, $\chi$ is the primitive character mod $4$, and

converges absolutely for $\Re s > 1/2$.

Let $\mathcal{G}_{\mathfrak{a}}\left(s; \frac{2}{3}, -\frac{1}{3}\right) = \mathcal{G}_{1}\left(s; \frac{2}{3}, -\frac{1}{3}\right)\mathcal{G}_{2}\left(s; \frac{2}{3}, -\frac{1}{3}\right) \mathcal{G}_{3}\left(s; \frac{2}{3}, -\frac{1}{3}\right)\mathcal{G}_{4}\left(s; \frac{2}{3}, -\frac{1}{3}\right) \mathcal{G}_{5}\left(s; \frac{2}{3}, -\frac{1}{3}\right)$, where

where $\tau(N(\mathfrak{a})) = (t+1)\prod_{p^{\upsilon}\parallel N(\mathfrak{a}), p\equiv1(\bmod4)}(\upsilon+1)\prod_{p^{2\nu}\parallel N(\mathfrak{a}), p\equiv3(\bmod4)}(2\nu+1)$.

When $\Re s = \sigma > 1/2+\varepsilon$,

To deal with the estimation of $\mathcal{F}_{\mathfrak{a}}(s)$ near $1$, we introduce the function $Z(s; z_{1})$. The function

is holomorphic in the $|s-1| < 1$, and admits the Taylor series expansion

where $\gamma_{j}(z_{1})$ is an entire function, for all $\varepsilon > 0$,

Now, let $z_{1} = 2/3$. The function $Z(s; 2/3)\mathcal{G}_{\mathfrak{a}}(s; 2/3, -1/3)L(s, \chi)^{-1/3}$ is holomorphic in the disc $\lvert s-1\rvert < (1-\hat{\beta})/2$, where $\hat{\beta} = \beta_{0}$ when $L(s, \chi)$ has a real zero $\beta_{0}$, $\hat{\beta} = 1-\sigma_{0}(\tau)$ when $L(s, \chi)$ has no real zero $\beta_{0}$, and

for $\lvert s-1\rvert < (1-\hat{\beta})/2$. Thus, for $\lvert s-1\rvert < (1-\hat{\beta})/2$, we can write

where

We can apply Perron's formula with the choice of parameters $\sigma_{a} = 1, A(n) = n^{\varepsilon}, \alpha = 0$ to write

where $b = 1+2/\log x$ and $100\leq T\leq x$, such that $L(\sigma+iT, \chi)\neq0$ for $0 < \sigma < 1$.

Let $\mathcal{L}_{T}$ be the boundary of the modified rectangle with vertices $1/2+\varepsilon\pm iT$ and $b\pm iT$, where

● $\varepsilon > 0$ is a small constant chosen such that $L(1/2+\varepsilon+i\gamma, \chi)\neq0$ for $|\gamma| < T$. Let $l_{1}$ be the horizontal line segment with the imaginary part $T$ and the real part $1/2+\varepsilon$ to $b$, and let $l_{2}$ be the horizontal line segment with the imaginary part $-T$ and the real part $b$ to $1/2+\varepsilon$. Let $l_{3}$ be the vertical line segment with the real part $1/2+\varepsilon$ and the imaginary part $0^{+}$ to $T$, and let $l_{4}$ be the vertical line segment with the real part $1/2+\varepsilon$ and the imaginary part $-T$ to $0^{-}$.

● The zeros of $L(s, \chi)$ of the form $\rho = \beta+i\gamma$ with $\beta > 1/2+\varepsilon$ and $\lvert \gamma\rvert < T$ are avoided by $\Gamma_{\rho}$ that horizontal cut drawn from the critical line inside this rectangle to $\rho = \beta+i\gamma$.

● $L(s, \chi)$ has a possible Siegel zero. The possible Siegel zero $\beta_{0}$ of $L(s, \chi)$ is avoided by contour $\Gamma_{0}$ (its upper part is made up of an arc surrounding the point $s = \beta_{0}$ with radius $r = 1/\log x$ and a line segment joining $\beta_{0}-r$ to $1/2+\varepsilon$).

● The pole of $L(s, \chi_{0})$ at the points $s = 1$ is avoided by the truncated Hankel contour $\Gamma$ (its upper part is made up of an arc surrounding the point $s = 1$ with radius $r = 1/\log x$ and a line segment joining $1-r$ to $\tilde{\beta}$), where

Clearly the function $\mathcal{F}_{\mathfrak{a}}(s)$ is analytic inside $\mathcal{L}_{T}$. By the Cauchy residue theorem, we can write

where

and

A. Evaluation of $I$.

Let $0 < c < (1-\hat{\beta})/10$ be a small constant. Since $Z(s; 2/3)\mathcal{G}_{\mathfrak{a}}(s; 2/3, -1/3)L(s, \chi)^{-\frac{1}{3}}$ is holomorphic and $O(M)$ in the disc $\lvert s-1\rvert\leq c$, the Cauchy formula implies that

where $g_{l}(2/3)$ is defined as in (2.3). From this and (2.3), it is easy to deduce that for $\lvert s-1\rvert\leq c/2$,

where

So, we have

Let $s-1 = \omega/\log x$. According to Lemma 2.1, we have

On the other hand,

According to (2.5) and (2.6), we have

B. Evaluation of $I_{1}$ and $I_{2}$.

It is well known that

From (2.8) and [12, Lemma 2.1], we deduce that

for $1/2\leq\sigma\leq1+\log^{-1}\lvert \tau\rvert$ and $\lvert \tau\rvert\geq3$, and

for $\sigma > 1/2$. In view of (2.9) and (2.10), we have

C. Evaluation of $I_{3}$ and $I_{4}$.

Let $\sigma_{0} = 1/2+\varepsilon$, $\tau_{0} = \lvert \tau\rvert+3$, for $s = \sigma_{0}+i\tau$ with $0\leq\lvert \tau\rvert\leq T$, In view of (2.9) and (2.10), we have

D. Evaluation of $I_{\rho}$.

For $s = \sigma+i\gamma$ with $1/2+\varepsilon\leq\sigma\leq\beta\leq1-\sigma_{0}(\gamma)$, we have

then we deduce that

Denote by $N(\sigma, T)$ the number of $L(s, \chi_{0})$ in the region $\Re s\geq\sigma$ and $\lvert\Im s\rvert\leq T$. We have

According to Huxley^[8],

for $1/2+\varepsilon\leq\sigma\leq1$, and $T\geq2$. Thus,

E. Evaluation of $I_{0}$.

If $L(s, \chi)$ has no Siegel zero, then $I_{0} = 0$. If it has Siegel zero $\beta_{0}$, then $L(s, \chi) = (s-\beta_{0})V(s)$, $V(\beta_{0})\neq0$. For $\lvert s-\beta_{0}\rvert\leq1/\log x$, we can write

where $C(\beta_{0})$ is a constant depending on $\beta_{0}$, then

Taking $T = e^{\sqrt{\log x}}$ and inserting (2.11)–(2.14) and (2.7) into (2.4), we have

Lemma 2.3. For any $n\in\mathbb{Z}^{+}$, we have that

holds uniformly for $x\geq2$, where $g(n)$ and $h$ are defined in Lemma 2.2.

□

Proof. In order to get the mean value of $g(n)a_{K}(n)$, we first consider its Dirichlet series $\sum_{n = 1}^{+\infty}g(n)a_{K}(n)n^{-s}$. Since $g(n)$, $a_{K}(n)$ is multiplicative, the Dirichlet series has Euler expansion. When $\Re s > 1$,

where

converges absolutely and is $O(M)$ for $\Re s > 1/2$. Since

then

Applying the Selberg-Delange theorem [11, p.281, Theorem 5.2], we have the formula

□

3.   Proof of Theorem 1.1

We only need to consider $0\leq t\leq1$. Now, we have

When $0\leq t\leq1/2$, we will first calculate $S$. According to Lemma 2.2,

Since $\log(x/N(\mathfrak{a})) = \log x-\log N(\mathfrak{a})\geq\log x-\log x^{t} = (1-t)\log x\geq1/2\log x$, we have

Next, we calculate $R$. According to Lemma 2.2,

When $0\leq t\leq1/2$, $(1-t)/t\geq1$, and since $N(\mathfrak{a})\geq2$, $u = \log\left(N(\mathfrak{a})^{\frac{1-t}{t}}\right)\geq\log N(\mathfrak{a})\geq\log2\approx0.693$. Let $y = (\pi-1)u-1$. Since $(\pi-1)\log\left(N(\mathfrak{a})^{\frac{1-t}{t}}\right)-1\geq(\pi-1)\log2-1 > 0$, $\pi\log\left(N(\mathfrak{a})^{\frac{1-t}{t}}\right)\geq\log\left(N(\mathfrak{a})^{\frac{1-t}{t}}\right)+1$, $R$ has the following estimates,

Therefore,

Since

we have

Let $G(x) = \sum_{n\leq x}g(n)a_{K}(n)$. According to Lemma 2.3 and using the Abelian Summation formula, we have

so we can obtain

Let

Clearly, $\Delta(t)$ is symmetric with respect to $t = 1/2$, and

then when $1/2\leq t\leq1$, $S(x, t)$ is the same as (3.1). This completes the proof.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The authors declare no competing interest.