# A New Elementary Proof of Landau’s Prime Ideal Theorem, and Associated
Results

Alex Burgin  School of Mathematics, Georgia Institute of Technology, Atlanta,
GA<EMAIL_ADDRESS>
###### Abstract.

We give a new elementary proof of Landau’s Prime Ideal Theorem. The proof is
an extension of Richter’s proof of the Prime Number Theorem. The main result
contains other results related to the equidistribution of the prime ideal
counting function.

###### Contents

  1. 1 Introduction
    1. 1.1 Notation, Conventions, and Standard Definitions
    2. 1.2 Background
    3. 1.3 Density Constants
  2. 2 Building Towards Theorem 1
  3. 3 Proof of Proposition 3
    1. 3.1 Chebychev-Type Bounds for $\pi_{K}(x)$
    2. 3.2 Combinatorial Ingredients
  4. 4 Appendix A
  5. 5 Appendix B
  6. 6 Appendix C

##  1\. Introduction

We give a new elementary proof of Landau’s Prime Ideal Theorem [8], which is
in the spirit of Richter’s recent elementary proof of the Prime Number Theorem
[12]. That argument is phrased in the language of ‘orthogonality’ of
arithmetic functions.

Let $K$ be a number field. Unique factorization of elements in $K$ can fail,
but holds in the ring of Integral Ideals $I_{K}$. We can then speak of prime
ideals $\mathfrak{p}\in I_{K}$. For a general ideal $\mathfrak{m}$, let
$\Omega(\mathfrak{m})$ denote the number of prime ideals dividing
$\mathfrak{m}$, counting multiplicity. By unique factorization of $I_{K}$,
such a function is well-defined. Elements $\mathfrak{m}\in I_{K}$ also have a
norm $N(\mathfrak{m})$. Our Main Theorem is this.

######  Theorem 1.

Let $K$ be a number field. Then there exists a positive integer $N$ such that,
for any $g:\mathbb{N}_{0}\rightarrow\mathbb{C}$ satisfying $|g(n)|\leq 1$ for
all $n$, one has

(1) $\displaystyle\sum_{\mathfrak{m}\colon N(\mathfrak{m})\leq
x}g(\Omega(\mathfrak{m})+k_{1})=\sum_{\mathfrak{m}\colon N(\mathfrak{m})\leq
x}g(\Omega(\mathfrak{m})+k_{2})+o_{K}\Big{(}\sum_{\mathfrak{m}\colon
N(\mathfrak{m})\leq x}1\Big{)}$

for each $k_{1},k_{2}\geq N$.

This contains several results. For Landau’s Theorem, take $g(x)=(-1)^{x}$,
$k_{1}$ even, and $k_{2}$ odd. We see that asymptotically, the number of
ideals $\mathfrak{m}$ for which $\Omega(\mathfrak{m})$ is even is equal to
those for for which $\Omega(\mathfrak{m})$ is odd. That is an equivalent form
of Landau’s Theorem, see Lemma 1 and Appendix C.

Let $e(y)=e^{2\pi iy}$ for real $y$. For integer $q>2$, set $g(y)=e(ay/q)$,
for each $1\leq a<q$. It then follows that by choosing $k_{1}\geq N$ and
$k_{2}=k_{1}+1$ that

$\displaystyle(1-e(a/q))e(ak_{1}/q)\sum_{N(\mathfrak{m})\leq
x}e(a\Omega(\mathfrak{m})/q)$ $\displaystyle=o\Big{(}\sum_{N(\mathfrak{m})\leq
x}1\Big{)},$

and hence $\sum_{N(\mathfrak{m})\leq
x}e(a\Omega(\mathfrak{m})/q)=o\Big{(}\sum_{N(\mathfrak{m})\leq x}1\Big{)}$ for
each $1\leq a<q$. This extends a Theorem of Pillai [11] and Selberg [13],
establishing the equidistribution of
$(\Omega(\mathfrak{m}))_{N(\mathfrak{m})\rightarrow\infty}$ modulo $q$ for
each $q\geq 1$. See Appendix B for a concise explanation.

Take $g(x)=e(\alpha x)$ for $\alpha\in\mathbb{R}\setminus\mathbb{Q}$. With
choices of $k_{1}$ and $k_{2}$ depending upon $\alpha$, we can verify that

$\displaystyle\sum_{N(\mathfrak{m})\leq x}e(\alpha\Omega(\mathfrak{m}))$
$\displaystyle=o\Big{(}\sum_{N(\mathfrak{m})\leq x}1\Big{)}.$

This is a theorem of Erdős [5] and Delange [2], generalized to the number
field setting. Namely, that the fractional parts
$(\\{\Omega(\mathfrak{m})\alpha\\}\colon N(\mathfrak{m})\leq x)$ are
asymptotically uniformly distributed in $[0,1)$ for any irrational $\alpha$.

###  1.1. Notation, Conventions, and Standard Definitions

We use $\mathcal{O}_{K}$ to denote the ring of integers of the number field
$K$. Within $\mathcal{O}_{K}$, we let $I_{K}$ denote the set of integral
ideals, and denote ideals in $I_{K}$ with $\mathfrak{m},\mathfrak{n}$, and
letters in similar font. We use $\mathfrak{p}$ or $\mathfrak{q}$ to denote a
prime ideal, and define $N_{K/\mathbb{Q}}:I_{K}\rightarrow\mathbb{Z}$ to be
the ideal norm taking an ideal $\mathfrak{m}$ to its norm
$N_{K/\mathbb{Q}}(\mathfrak{m})$. Usually, we will simply drop the subscript
and just write $N(\mathfrak{m})$. $I_{K}$ is naturally endowed with a ring
structure via ideal addition and multiplication, and
$N:I_{K}\rightarrow\mathbb{Z}$ is a homomorphism from
$(I_{K},\cdot)\rightarrow(\mathbb{Z},\cdot)$.

Given some $m\in\mathbb{N}$, we say an integer sequence $(a_{n})_{n\geq 0}$ is
equidistributed modulo m if

$\displaystyle\lim_{x\rightarrow\infty}\frac{1}{x}\sum_{\begin{subarray}{c}0\leq
n<x\\\ a_{n}\equiv c\ (\text{mod
}m)\end{subarray}}1=\frac{1}{m}\quad\quad\text{for all
}c\in\mathbb{Z}/m\mathbb{Z}.$

Equidistribution of an integer sequence is closely related to the behavior of
associated exponential sums; for more explanation, see Appendix B.

When we write $\sum_{N(\mathfrak{m})\leq x}$, we mean that we are sum over all
ideals $\mathfrak{m}$ with norm no greater than $x$.

For a finite set of ideals $S\subset I_{K}$ and a function
$f:I_{K}\rightarrow\mathbb{C}$, we use
$\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{s}\in
S}f(\mathfrak{s})$ to denote the standard averaging operator

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{s}\in
S}f(\mathfrak{s}):=\frac{1}{|S|}\sum_{\mathfrak{s}\in S}f(\mathfrak{s})$

and $\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{s}\in
S}^{\log}f(\mathfrak{s})$ to denote the logarithmic (weighted) averaging
operator

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{s}\in
S}^{\log}f(\mathfrak{s}):=\frac{1}{\sum_{\mathfrak{s}\in
S}\frac{1}{N(\mathfrak{s})}}\sum_{\mathfrak{s}\in
S}\frac{f(\mathfrak{s})}{N(\mathfrak{s})}.$

These are analogues of the operators in [12].

We use the standard big-Oh notation $f(x)=O(g(x))$ if $|f(x)|\leq C|g(x)|$ for
a constant $C$; similarly, $f(x)=O_{\alpha,\beta,...}(g(x))$ reads that
$|f(x)|\leq C|g(x)|$ for some $C=C(\alpha,\beta,...)$. We use $f(x)=o(g(x))$
if the quotient $|f(x)/g(x)|\rightarrow 0$ as $x\rightarrow\infty$.

###  1.2. Background

The distribution of the prime-omega function over number fields $K$ is
intimately tied to the distribution of prime numbers, echoing the classical
case $K=\mathbb{Q}$. Recall the following Lemma:

######  Lemma 1.

Suppose that $\sum_{N(\mathfrak{m})\leq x}(-1)^{\Omega(\mathfrak{m})}=o(x)$,
i.e., that $\Omega$ is equidistributed modulo 2. Then
$\pi_{K}(x):=\sum_{N(\mathfrak{p})\leq x}1\sim\frac{x}{\log x}$.

An elementary proof of this Lemma (which is actually an equivalence, but the
proof of that will not be needed) follows from proving that the associated
function $M(x):=\sum_{N(\mathfrak{m})<x}\mu(\mathfrak{m})=o(x)$, where

$\mu(\mathfrak{m})=\begin{cases}0&\text{if there exists a nonunit ideal
}\mathfrak{p}\text{ such that }\mathfrak{p}^{2}|\mathfrak{m}\\\
(-1)^{\Omega(\mathfrak{m})}&\text{otherwise.}\end{cases}$

This function is analogous to the so-called Mertens function over the case
$K=\mathbb{Q}$, and showing sublinear growth for $M(x)$ implies that
$\pi_{K}(x)\sim\frac{x}{\log x}$. This is a well known implication for the
case of number fields, following from tight ideal density bounds (see §1.3 and
[3]). We will leave this implication unproven, but provide an elementary proof
in Appendix C that $L(x)=o(x)\implies M(x)=o(x)$. Then, with Theorem 1, this
recovers Landau’s prime ideal theorem.

###  1.3. Density Constants

A well-known and elementary111To see an explanation as to why this is true,
see [9, Ch. 6: The Distribution of Ideals in a Number Ring]. fact (in that it
requires no complex analysis) is that for any number field $K$ of degree $d$,
one has $\sum_{N(\mathfrak{m})\leq x}1=c_{K}x+O(x^{1-\frac{1}{d}})$ for a
constant $c_{K}>0$. This constant $c_{K}$ is often referred to as the ‘ideal
density’ of $\mathcal{O}_{K}$ (see, e.g., [10], §8.4). The power saving-nature
of the error term is important, but not essential–we note that these results
can be extended to derive PNTs and equidistribution results for many
multiplicative semigroups (or ‘Beurling number systems’) with suitably-quickly
decaying error term in density.

##  2\. Building Towards Theorem 1

The following result is fundamental to our paper:

######  Proposition 2.

Let $K$ be a number field of degree $d$, and let $c_{K}$ be the ideal density
of $\mathcal{O}_{K}$. Then for any $S\subset I_{K}$ finite and nonempty,

(2) $\displaystyle\frac{1}{x}\sum_{N(\mathfrak{m})\leq
x}\Big{|}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-\sum_{\mathfrak{n}\in
S}\frac{1}{N(\mathfrak{n})}\Big{|}^{2}=c_{K}\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+O(|S|^{1+\frac{1}{d}}\cdot
x^{-\frac{1}{d}})$

where
$\Phi(\mathfrak{n}_{1},\mathfrak{n_{2}})=N(\operatorname{gcd}(\mathfrak{n}_{1},\mathfrak{n}_{2}))-1$.

###### Proof.

Let $A$ denote the sum $\sum_{\mathfrak{n}\in
S}\frac{1}{N(\mathfrak{\mathfrak{n}})}$; then by expanding the square on the
left hand side of (2) one has

$\displaystyle\Big{|}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-A\Big{|}^{2}$
$\displaystyle=\Big{(}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}\Big{)}^{2}-2A\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}+A^{2},$

and summing this over all $\mathfrak{m}$ with $N(\mathfrak{m})\leq x$ and
dividing by $x$, one has

$\displaystyle\frac{1}{x}\sum_{N(\mathfrak{m})\leq
x}\Big{|}\sum_{\mathfrak{n}\in S}$
$\displaystyle\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-A\Big{|}^{2}\coloneqq
S_{1}-2AS_{2}+S_{3},$ $\displaystyle S_{1}$
$\displaystyle=\frac{1}{x}\sum_{N(\mathfrak{m})\leq
x}\Big{(}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}\Big{)}^{2},$ $\displaystyle S_{2}$
$\displaystyle=\frac{1}{x}\sum_{N(\mathfrak{m})\leq x}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}},$ $\displaystyle S_{3}$
$\displaystyle=\frac{A^{2}}{x}\sum_{N(\mathfrak{m})\leq x}1.$

_Calculation of $S_{3}$_. Using that $\sum_{N(\mathfrak{m})\leq
x}1=c_{K}x+O(x^{1-\frac{1}{d}})$ we see that
$S_{3}=c_{K}A^{2}+O(A^{2}x^{-\frac{1}{d}})$. Then, using that $A=O(\log|S|)$
(see Appendix A for a quick proof), we have
$S_{3}=c_{K}A^{2}+O(\log^{2}(|S|)x^{-\frac{1}{d}})$.

_Calculation of $S_{2}$_. For any fixed $\mathfrak{n}$, if
$\mathfrak{n}|\mathfrak{m}$ then one can write
$\mathfrak{m}=\mathfrak{n}\mathfrak{m}^{\prime}$ in a unique way, and hence

$\displaystyle\frac{1}{x}\sum_{\mathfrak{m}\colon N(\mathfrak{m})\leq
x}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}$
$\displaystyle=\frac{1}{x}\sum_{\mathfrak{m}\colon
N(\mathfrak{nm^{\prime}})\leq x}1$
$\displaystyle=\frac{1}{x}\sum_{\mathfrak{m}\colon
N(\mathfrak{m}^{\prime})\leq x/N(\mathfrak{n})}1$
$\displaystyle=\frac{c_{K}}{N(\mathfrak{n})}+O\Big{(}x^{-\frac{1}{d}}N(\mathbf{n})^{\frac{1}{d}-1}\Big{)}$

so that one has, after summing over $\mathfrak{n}\in S$,

$\displaystyle S_{2}=\sum_{\mathfrak{n}\in
S}\frac{c_{K}}{N(\mathfrak{\mathfrak{n}})}+O\Big{(}x^{-\frac{1}{d}}\sum_{\mathfrak{n}\in
S}\frac{1}{N(\mathfrak{n})^{1-\frac{1}{d}}}\Big{)}=c_{K}A+O\Big{(}x^{-\frac{1}{d}}\sum_{\mathfrak{n}\in
S}\frac{1}{N(\mathfrak{n})^{1-\frac{1}{d}}}\Big{)}.$

Then since

(3) $\displaystyle\sum_{\mathfrak{n}\in
S}\frac{1}{N(\mathfrak{n})^{1-\frac{1}{d}}}\ll|S|^{\frac{1}{d}}$

(for a proof of (3), see Appendix A), we then have
$S_{2}=c_{K}A+O(x^{-\frac{1}{d}}|S|^{\frac{1}{d}})$.

_Calculation of $S_{1}$_. We begin by noting that we can write

(4) $S_{1}=\frac{1}{x}\sum_{\mathfrak{m}\colon N(\mathfrak{m})\leq
x}\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\mathbf{1}_{\mathfrak{n}_{1}|\mathfrak{m}}\mathbf{1}_{\mathfrak{n}_{2}|\mathfrak{m}}$

For the moment, we view $\mathfrak{n}_{1},\mathfrak{n}_{2}$ as fixed. Then, we
note that since for any $\mathfrak{m}$ with $\mathfrak{n}_{1}|\mathfrak{m}$ we
can write $\mathfrak{m}=\mathfrak{m}^{\prime}\mathfrak{n}_{1}$ in a unique
way, so that

(5) $\displaystyle\frac{1}{x}\sum_{\mathfrak{m}\colon N(\mathfrak{m})\leq
x}\mathbf{1}_{\mathfrak{n}_{1}|\mathfrak{m}}\mathbf{1}_{\mathfrak{n}_{2}|\mathfrak{m}}$
$\displaystyle=\frac{1}{x}\sum_{\mathfrak{m}^{\prime}\colon
N(\mathfrak{m^{\prime}n}_{1})\leq
x}\mathbf{1}_{\mathfrak{n}_{2}|\mathfrak{m^{\prime}n}_{1}}$
$\displaystyle=\frac{1}{x}\sum_{\mathfrak{m}^{\prime}\colon
N(\mathfrak{m}^{\prime})\leq
x/N(\mathfrak{n}_{1})}\mathbf{1}_{\mathfrak{n}_{2}|\mathfrak{m}^{\prime}\mathfrak{n}_{1}}$

and then letting
$\mathfrak{d}=\operatorname{gcd}(\mathfrak{n}_{1},\mathfrak{n}_{2})$,
$\mathfrak{d}\mathfrak{d}_{i}=\mathfrak{n}_{i}$ ($i=1,2$) we have
$\mathfrak{n}_{2}|\mathfrak{m}^{\prime}\mathfrak{n}_{1}$ if and only if
$\mathfrak{d}_{2}|\mathfrak{m}^{\prime}\mathfrak{d}_{1}$, and hence
$\mathbf{1}_{\mathfrak{n}_{2}|\mathfrak{m}^{\prime}\mathfrak{n}_{1}}=\mathbf{1}_{\mathfrak{d}_{2}|\mathfrak{m^{\prime}}\mathfrak{d}_{1}}$.
Then since $\mathfrak{d}_{2},\mathfrak{d}_{1}$ are coprime, if
$\mathfrak{d}_{2}|\mathfrak{m}^{\prime}\mathfrak{d}_{1}$ we must have
$\mathfrak{d}_{2}|\mathfrak{m}^{\prime}$ so that, from (5),

$\displaystyle\frac{1}{x}\sum_{\mathfrak{m}\colon N(\mathfrak{m})\leq
x}\mathbf{1}_{\mathfrak{n}_{1}|m}1_{\mathfrak{n}_{2}|m}$
$\displaystyle=\frac{1}{x}\sum_{\mathfrak{m}^{\prime}\colon
N(\mathfrak{m}^{\prime})\leq
x/N(\mathfrak{n}_{1})}\mathbf{1}_{\mathfrak{d}_{2}|\mathfrak{m}^{\prime}}$
$\displaystyle=\frac{1}{x}\sum_{\mathfrak{m}^{\prime\prime}\colon
N(\mathfrak{m}^{\prime\prime})\leq x/N(\mathfrak{n}_{1}\mathfrak{d}_{2})}1$
$\displaystyle=\frac{c_{K}}{N(\mathfrak{n}_{1}\mathfrak{d}_{2})}+O(x^{-\frac{1}{d}}N(\mathfrak{n}_{1}\mathfrak{d}_{2})^{\frac{1}{d}-1})$
$\displaystyle=\frac{c_{K}\cdot
N(\operatorname{gcd}(\mathfrak{n}_{1},\mathfrak{n}_{2}))}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+O\Big{(}x^{-\frac{1}{d}}\frac{N(\operatorname{gcd}(\mathfrak{n}_{1},\mathfrak{n}_{2}))^{1-\frac{1}{d}}}{N(\mathfrak{n}_{1})^{1-\frac{1}{d}}N(\mathfrak{n}_{2})^{1-\frac{1}{d}}}\Big{)}.$

Defining
$\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})=N(\operatorname{gcd}(\mathfrak{n}_{1},\mathfrak{n}_{2}))-1$
and then summing over $\mathfrak{n}_{1},\mathfrak{n}_{2}\in S$ one then has

$\displaystyle S_{1}$
$\displaystyle=\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{c_{K}\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{c_{K}}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+E_{1}$
$\displaystyle=\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{c_{K}\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+c_{K}A^{2}+E_{1},$

where, using (3),

$\displaystyle E_{1}$ $\displaystyle\ll
x^{-\frac{1}{d}}\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\Big{(}\frac{N(\operatorname{gcd}(\mathfrak{n}_{1},\mathfrak{n}_{2}))}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}\Big{)}^{1-\frac{1}{d}}$
$\displaystyle\ll x^{-\frac{1}{d}}\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{1}{N(\mathfrak{n}_{2})^{1-\frac{1}{d}}}\ll
x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}}.$

We then have

$\displaystyle S_{1}=\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{c_{K}\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+c_{K}A^{2}+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}}).$

Now, let us denote $\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{c_{K}\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}$
with $S_{\Phi}$. One has

$\displaystyle S_{1}-2AS_{2}+S_{3}$
$\displaystyle=\Big{(}S_{\Phi}+c_{K}A^{2}+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}})\Big{)}$
$\displaystyle\quad-2A\Big{(}c_{K}A+O(x^{-\frac{1}{d}}|S|^{\frac{1}{d}})\Big{)}$
$\displaystyle\quad+\Big{(}c_{K}A^{2}+O(\log^{2}(|S|)x^{-\frac{1}{d}})\Big{)}$
$\displaystyle=S_{\Phi}+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}})$

since the $A^{2}$ term cancels. This is the desired estimate. ∎

We then want an equivalent formulation to Proposition 1 in terms of
expectation:

######  Corollary 2 (Equivalent Assertion to Prop. 2).

Let $K$ be a number field of degree $d$, and let $c_{K}$ be the ideal density
of $\mathcal{O}_{K}$. Suppose $S\subset K$ is finite and nonempty. Then

(6)
$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}\Big{|}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}\in
S}^{\log{}}(N(\mathfrak{n})\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-1)\Big{|}^{2}=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}\in
S}^{\log{}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}^{\prime}\in
S}^{\log{}}\Phi(\mathfrak{n},\mathfrak{n}^{\prime})+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}}),$

###### Proof.

We can begin by writing Proposition 2 as

$\sum_{N(\mathfrak{m})\leq x}\Big{|}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-A\Big{|}^{2}=c_{K}x\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+O(x^{1-\frac{1}{d}}|S|^{1+\frac{1}{d}}).$

Dividing both sides by $\sum_{N(\mathfrak{m})\leq
x}1=c_{K}x+O(x^{-\frac{1}{d}})$ gives then that

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}\Big{|}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-A\Big{|}^{2}$
$\displaystyle=\frac{1}{1+O(x^{-\frac{1}{d}})}\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}})$
$\displaystyle=\sum_{\mathfrak{n}_{1},\mathfrak{n}_{2}\in
S}\frac{\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})}{N(\mathfrak{n}_{1})N(\mathfrak{n}_{2})}+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}}).$

Dividing again by $A^{2}$ and noting that $\frac{1}{A}\sum_{\mathfrak{n}\in
S}\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}\in
S}^{\log{}}N(\mathfrak{n})\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}$, one has

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}\Big{|}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}\in
S}^{\log{}}N(\mathfrak{n})\mathbf{1}_{\mathfrak{n}|\mathfrak{m}}-1\Big{|}^{2}=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}_{1}\in
S}^{\log{}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{n}_{2}\in
S}^{\log}\Phi(\mathfrak{n}_{1},\mathfrak{n}_{2})+O(x^{-\frac{1}{d}}|S|^{1+\frac{1}{d}}).$

∎

The following Proposition is our main ingredient, analogous to Richter’s
Proposition 2.2:

######  Proposition 3.

For all $\eta>0$ there exists some $N=N(\eta)\in\mathbb{N}$ such that for all
$k\geq N$ there exist two nonempty sets of ideals
$S_{1}=S_{1}(\eta),S_{2}=S_{2}(\eta)\subset I_{K}$ for which the following are
true:

  1. (i)

All ideals in $S_{1}$ are prime and all ideals in $S_{2}$ are a product of
exactly $k$ prime ideals

  2. (ii)

The sets $S_{1}$ and $S_{2}$ have the same cardinality, and there exists an
enumeration
$S_{1}=\\{\mathfrak{p}_{1},\mathfrak{p}_{2},...,\mathfrak{p}_{k}\\}$ and
$S_{2}=\\{\mathfrak{m}_{1},\mathfrak{m}_{2},...,\mathfrak{m}_{k}\\}$ for which
$(1-\eta)N(\mathfrak{p}_{i})\leq
N(\mathfrak{m}_{i})\leq(1+\eta)N(\mathfrak{p}_{i})$ for all $i=1,...k$.

  3. (iii)

$\operatorname*{\scalerel*{\mathbb{E}}{\sum}}^{\log}_{\mathfrak{n}\in
S_{i}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}^{\log}_{\mathfrak{n}^{\prime}\in
S_{i}}\Phi(\mathfrak{n},\mathfrak{n}^{\prime})\leq\eta$ for $i=1,2$.

###### Proof of the Main Theorem 1 Assuming Proposition 3.

Fix $\eta>0$ and take $g:\mathbb{N}_{0}\rightarrow\mathbb{C}$ with $|g|\leq
1$. Choose $S_{1}=S_{1}(\eta)$ and $S_{2}=S_{2}(\eta)$ to the desired
specifications above. Note first that

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{i}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{m})}g(\Omega(\mathfrak{m}\mathfrak{n}))$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{i}}^{\log}\frac{1}{\sum_{N(\mathfrak{n})\leq
x/N(\mathfrak{m})}1}\sum_{N(\mathfrak{n})\leq
x/N(\mathfrak{m})}g(\Omega(\mathfrak{mn}))$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{i}}^{\log}\frac{1}{c_{K}\cdot\frac{x}{N(\mathfrak{m})}+O(x^{1-\frac{1}{d}})}\sum_{N(\mathfrak{n}^{\prime})\leq
x}1_{\mathfrak{m}|\mathfrak{n}^{\prime}}g(\Omega(\mathfrak{n}^{\prime}))$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{i}}^{\log}(1+O(x^{-\frac{1}{d}}))\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n^{\prime}})\leq
x}N(\mathfrak{m})1_{\mathfrak{m}|\mathfrak{n}^{\prime}}g(\Omega(\mathfrak{n}^{\prime}))$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{i}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n}^{\prime})\leq
x}N(\mathfrak{m})1_{\mathfrak{m}|\mathfrak{n}^{\prime}}g(\Omega(\mathfrak{n}^{\prime}))+O_{S_{i}}(x^{-\frac{1}{d}})$

(here we used 1-boundedness of $g$) so that

$\displaystyle\Big{|}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x}g(\Omega(\mathfrak{n}))-\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{m})}g(\Omega(\mathfrak{m}\mathfrak{n}))\Big{|}^{2}$
$\displaystyle\leq\Big{|}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x}g(\Omega(\mathfrak{n}))-\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n}^{\prime})\leq
x}N(\mathfrak{m})1_{\mathfrak{m}|\mathfrak{n}^{\prime}}g(\Omega(\mathfrak{n}^{\prime}))\Big{|}^{2}+O_{S_{2}}(x^{-\frac{1}{d}})$
$\displaystyle\leq\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x}\Big{|}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}(1-N(\mathfrak{m})1_{\mathfrak{m}|\mathfrak{n}})\Big{|}^{2}+O_{S_{2}}(x^{-\frac{1}{d}})$
$\displaystyle\leq\eta+O_{S_{2}}(x^{-\frac{1}{d}}).$

Here we used the bound $|g|\leq 1$ extensively alongside Proposition 1, the
Cauchy-Schwarz inequality, and the assumption on $S_{2}$. Since
$\Omega(\mathfrak{mn})=\Omega(\mathfrak{m})+\Omega(\mathfrak{n})$ we have

(7)
$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}g(\Omega(\mathfrak{m}))$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{m})}g(\Omega(\mathfrak{m})+\Omega(\mathfrak{n}))+O(\eta^{\frac{1}{2}})+O_{S_{2}}(x^{-\frac{1}{2d}})$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{m})}g(k+\Omega(\mathfrak{n}))+O(\eta^{\frac{1}{2}})+O_{S_{2}}(x^{-\frac{1}{2d}}).$

A similar argument with $S_{1}$ instead of $S_{2}$ and $g(n+k-1)$ in place of
$g(n)$ gives

(8)
$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}g(\Omega(\mathfrak{m})+k-1)$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}\in
S_{1}}^{\log{}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{p})}g(\Omega(\mathfrak{p})+\Omega(\mathfrak{n})+k-1)+O(\eta^{\frac{1}{2}})+O_{S_{1}}(x^{-\frac{1}{2d}})$
$\displaystyle=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}\in
S_{1}}^{\log{}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{p})}g(k+\Omega(\mathfrak{n}))+O(\eta^{\frac{1}{2}})+O_{S_{1}}(x^{-\frac{1}{2d}}).$

Finally, we know that there exists an enumeration of $S_{1},S_{2}$ such that
$S_{1}=\\{\mathfrak{p}_{1},\mathfrak{p}_{2},...,\mathfrak{p}_{j}\\}$ and
$S_{2}=\\{\mathfrak{m}_{1},\mathfrak{m}_{2},...,\mathfrak{m}_{j}\\}$ and
$(1-\eta)N(\mathfrak{p}_{i})\leq
N(\mathfrak{m}_{i})\leq(1+\eta)N(\mathfrak{p}_{i})$ for all $i=1,...k$. Then
it easily follows that
$\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{m}_{i})}g(k+\Omega(\mathfrak{n}))=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{p}_{i})}g(k+\Omega(\mathfrak{n}))+O(\eta)$. Taking logarithmic
averages over $S_{1}$ and $S_{2}$ yields

(9)
$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log{}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{\mathfrak{n}})\leq
x/N(\mathfrak{m})}g(k+\Omega(\mathfrak{n}))=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}\in
S_{1}}^{\log{}}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{n})\leq
x/N(\mathfrak{p})}g(k+\Omega(\mathfrak{n}))+O(\eta).$

From applying (9) to (7) and then applying (8) we then deduce that

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}g(\Omega(\mathfrak{m}))=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}g(\Omega(I)+k-1)+O(\eta^{\frac{1}{2}})+O_{S_{1},S_{2}}(x^{-\frac{1}{2d}}).$

This holds for all $k\geq N$, so that one then has

(10)
$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}g(\Omega(\mathfrak{m})+k_{1})=\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{N(\mathfrak{m})\leq
x}g(\Omega(I)+k_{2})+O(\eta^{\frac{1}{2}})+O_{S_{1},S_{2}}(x^{-\frac{1}{2d}})$

for all $k_{1},k_{2}\geq N$. The Theorem then follows upon taking
$x\rightarrow\infty$; then since $\eta>0$ was arbitrary, we are finished.

∎

##  3\. Proof of Proposition 3

###  3.1. Chebychev-Type Bounds for $\pi_{K}(x)$

We have this upper bound for prime ideals:

######  Lemma 3.

Let $\alpha\geq 1$. One then has

$\displaystyle\sum_{\mathfrak{p}\colon x<N(\mathfrak{p})\leq\alpha
x}1\leq(\alpha\log\alpha)\frac{x}{\log x}+O(x^{1-\frac{1}{d}}\log x).$

###### Proof.

The classical way to derive weak-type prime bounds is by studying the prime
factorizations of binomial coefficients; in particular, the central binomial
coefficient $\binom{2n}{n}=\frac{(2n)!}{(n!)^{2}}$ is at least as large as the
product of primes $n<p<2n$.

In the number field setting, we instead follow this line of reasoning. Let
$\mathfrak{p}$ be a prime ideal. We know that there are a total of
$\frac{c_{K}n}{N(\mathfrak{p})}+O(n^{1-\frac{1}{d}}N(\mathfrak{p})^{-1})$
ideals divisible by $\mathfrak{p}$ over the set of ideals with norm at most
$n$, and
$\frac{c_{K}n}{N(\mathfrak{p})^{2}}+O(n^{1-\frac{1}{d}}N(\mathfrak{p})^{-2})$
divisible by $\mathfrak{p}^{2}$, and so on. Then, the multiplicity
$x_{\mathfrak{p}}$ of a prime ideal $\mathfrak{p}$ in the ideal
$\prod_{N(\mathfrak{m})\leq n}\mathfrak{m}$ is given by

$\displaystyle x_{\mathfrak{p}}\Big{(}\prod_{N(\mathfrak{m})\leq
n}\mathfrak{m}\Big{)}$
$\displaystyle=c_{K}n\sum_{c=1}^{\log_{N(\mathfrak{p})}(n)}N(\mathfrak{p})^{-c}+O\Big{(}n^{1-\frac{1}{d}}\sum_{c=1}^{\log_{N(\mathfrak{p})}(n)}N(\mathfrak{p})^{-c}\Big{)}$
$\displaystyle=c_{K}n\sum_{c=1}^{\log_{N(\mathfrak{p})}(n)}N(\mathfrak{p})^{-c}+O\Big{(}\frac{n^{1-\frac{1}{d}}\log
n}{N(\mathfrak{p})\log N(\mathfrak{p})}\Big{)}.$

One can then estimate

$\displaystyle\log f(n)$ $\displaystyle\coloneqq\sum_{N(\mathfrak{p})\leq
n}x_{\mathfrak{p}}\Big{(}\prod_{N(\mathfrak{m})\leq
n}\mathfrak{m}\Big{)}\cdot\log N(\mathfrak{p})$
$\displaystyle=c_{K}n\sum_{N(\mathfrak{p})\leq n}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(n)}N(\mathfrak{p})^{-c}+O\Big{(}n^{1-\frac{1}{d}}\log
n\sum_{N(\mathfrak{p})\leq n}\frac{1}{N(\mathfrak{p})}\Big{)}$
$\displaystyle=c_{K}n\sum_{N(\mathfrak{p})\leq n}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(n)}N(\mathfrak{p})^{-c}+O\Big{(}n^{1-\frac{1}{d}}(\log
n)^{2}\Big{)}.$

We inspect the function $\theta_{\alpha}(x):=\frac{f(\alpha
x)}{f(x)^{\alpha}}$. Using the estimate derived for $\log f(n)$, we have
$\log\theta_{\alpha}(x)=\log f(\alpha x)-\alpha\log f(x)$ and one can compute

(11) $\displaystyle\log f(\alpha x)$ $\displaystyle=\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}+O(x^{1-\frac{1}{d}}(\log x)^{2}),$ (12)
$\displaystyle\alpha\log f(x)$ $\displaystyle=\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(x)}N(\mathfrak{p})^{-c}+O(x^{1-\frac{1}{d}}(\log
x)^{2})$

so that we have

$\displaystyle\log\theta_{\alpha}(x)$ $\displaystyle=\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(x)}N(\mathfrak{p})^{-c}$
$\displaystyle\quad+\alpha c_{K}x\sum_{N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1+\lfloor\log_{N(\mathfrak{p})}(x)\rfloor}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}$ $\displaystyle\quad-\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(x)}N(\mathfrak{p})^{-c}$
$\displaystyle\quad+O(x^{1-\frac{1}{d}}(\log x)^{2}).$

Then, subtracting third term from first above,

$\displaystyle\log\theta_{\alpha}(x)$ $\displaystyle=\alpha
c_{K}x\sum_{x<N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(x)}N(\mathfrak{p})^{-c}\quad$
$\displaystyle\quad+\alpha c_{K}x\sum_{N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1+\lfloor\log_{N(\mathfrak{p})}(x)\rfloor}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}$ $\displaystyle\quad+O(x^{1-\frac{1}{d}}(\log
x)^{2}).$

Notice that in the first term, since $x<N(\mathfrak{p})$ one has
$\log_{N(\mathfrak{p})}(x)<1$ and hence this term vanishes. We then obtain

$\displaystyle\log\theta_{\alpha}(x)=\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1+\lfloor\log_{N(\mathfrak{p})}(x)\rfloor}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}+O(x^{1-\frac{1}{d}}(\log x)^{2}).$

and truncating the sum to $x<N(\mathfrak{p})\leq\alpha x$ one surely has

$\displaystyle\log\theta_{\alpha}(x)$ $\displaystyle\geq\alpha
c_{K}x\sum_{x<N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}+O(x^{1-\frac{1}{d}}(\log x)^{2})$
$\displaystyle\geq\alpha c_{K}x\sum_{x<N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\cdot\frac{\lfloor\log_{N(\mathfrak{p})}(\alpha
x)\rfloor}{\alpha x}+O(x^{1-\frac{1}{d}}(\log x)^{2})$ $\displaystyle\geq
c_{K}\log x\sum_{x<N(\mathfrak{p})\leq\alpha x}1+O(x^{1-\frac{1}{d}}(\log
x)^{2}).$

Then using the fact that

$\displaystyle\log\theta_{\alpha}(x)$ $\displaystyle=c_{K}\alpha x\log(\alpha
x)-c_{K}\alpha x\log x+O(x^{1-\frac{1}{d}}(\log x)^{2})$ (13)
$\displaystyle=c_{K}(\alpha\log\alpha)x+O(x^{1-\frac{1}{d}}(\log x)^{2})$

one then has

$\displaystyle\sum_{x<N(\mathfrak{p})\leq\alpha
x}1\leq(\alpha\log\alpha)\frac{x}{\log x}+O(x^{1-\frac{1}{d}}\log x).$

∎

######  Lemma 4.

Let $\pi_{K}(x)$ be the number of prime ideals in $\mathcal{O}_{K}$ with norm
at most $x$. Then one has

$\displaystyle\liminf_{x\rightarrow\infty}\frac{\pi_{K}(x)}{x/\log x}\geq 1.$

######  Corollary 5.

Take $\epsilon>0$. Then for $x\gg_{\epsilon}1$ one has
$\pi_{K}(x)\geq(1-\epsilon)\frac{x}{\log x}$.

###### Proof.

As with the upper bound in the previous Lemma, this comes from estimating the
function $\theta_{\alpha}(x)$ for $\alpha>1$. Recall that
$\log\theta_{\alpha}(x)=\log f(\alpha x)-\alpha\log f(x)$ for
$f(x)=\prod_{N(\mathfrak{m})\leq n}N(\mathfrak{m})$. One can write , using
(11) and splitting into two sums,

$\displaystyle\log f(\alpha x)$ $\displaystyle=\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}$ $\displaystyle+\alpha
c_{K}x\sum_{x<N(\mathfrak{p})\leq\alpha x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}+O(x^{1-\frac{1}{d}}(\log x)^{2})$ (14)
$\displaystyle:=\alpha c_{K}n\sum_{N(\mathfrak{p})\leq n}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(\alpha
n)}N(\mathfrak{p})^{-c}+E_{\alpha,x}+O(x^{1-\frac{1}{d}}(\log x)^{2}),$

where $E_{\alpha,x}$ is the sum over $x<N(\mathfrak{p})\leq\alpha x$ and
satisfies

$\displaystyle E_{\alpha,x}\leq\alpha c_{K}\log(\alpha
x)\sum_{x<N(\mathfrak{p})\leq\alpha x}1.$

We also have, using (12)

(15) $\displaystyle\alpha\log f(n)=\alpha c_{K}x\sum_{N(\mathfrak{p})\leq
x}\log
N(\mathfrak{p})\sum_{c=1}^{\log_{N(\mathfrak{p})}(x)}N(\mathfrak{p})^{-c}+O(x^{1-\frac{1}{d}}(\log
x)^{2})$

so that, subtracting (15) from (14),

$\displaystyle\log\theta_{\alpha}(x)$ $\displaystyle=\alpha
c_{K}x\sum_{N(\mathfrak{p})\leq x}\log
N(\mathfrak{p})\sum_{\log_{N(\mathfrak{p})}(x)<c\leq\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}+E_{\alpha,n}+O(x^{1-\frac{1}{d}}(\log x)^{2}).$

Since

$\displaystyle\sum_{\log_{N(\mathfrak{p})}(x)<c\leq\log_{N(\mathfrak{p})}(\alpha
x)}N(\mathfrak{p})^{-c}\leq\frac{\log\alpha}{x}$

and $\log N(\mathfrak{p})\leq\log x$ one has

$\displaystyle\log\theta_{\alpha}(x)$
$\displaystyle\leq\alpha(\log\alpha)c_{K}(\log x)\sum_{N(\mathfrak{p})\leq
x}1+E_{\alpha,n}+O(x^{1-\frac{1}{d}}(\log x)^{2})$

so that we then deduce that

(16) $\displaystyle\sum_{N(\mathfrak{p})\leq
x}1\geq\frac{\log\theta_{\alpha}(x)-E_{\alpha,x}}{c_{K}\alpha(\log\alpha)(\log
x)}+O(x^{1-\frac{1}{d}}\log x).$

Recalling the expression (13) for $\log\theta_{\alpha}(x)$ given by

$\displaystyle\log\theta_{\alpha}(x)=(\alpha\log\alpha)c_{K}x+O(x^{1-\frac{1}{d}}\log
x)$

one has, by substituting this into (16) that

(17) $\displaystyle\sum_{N(\mathfrak{p})\leq x}1$
$\displaystyle\geq\frac{x}{\log x}-\frac{E_{\alpha,x}}{\log
x}+O(x^{1-\frac{1}{d}}\log x).$

Now, by definition we know that $0\leq E_{\alpha,x}\leq c_{K}\alpha\log(\alpha
x)\sum_{x<N(\mathfrak{p})\leq\alpha x}1$; hence under application of the
previous Lemma we can deduce that

$\displaystyle E_{\alpha,x}\leq c_{K}\alpha\log(\alpha
x)\cdot(\alpha\log\alpha)\frac{x}{\log x}+O(x^{1-\frac{1}{d}}\log x)$

so that, by (17),

$\displaystyle\sum_{N(\mathfrak{p})\leq x}1\geq\frac{x}{\log
x}-c_{K}\alpha^{2}\log\alpha\frac{x}{\log x}+O(x/(\log x)^{2})$

and hence

$\displaystyle\frac{\pi_{K}(x)}{x/\log x}\geq
1-c_{K}\alpha^{2}\log\alpha+O(1/\log x).$

Given $\epsilon>0$, choose $\alpha-1>0$ small so that
$c_{K}\alpha^{2}\log\alpha<\epsilon$; then the result follows as one takes
$x\rightarrow\infty$.

∎

Using these two Lemmas, we can now prove the following Proposition:

######  Proposition 4.

Let $\mathbb{P}$ be the set of prime ideals in $\mathcal{O}_{K}$, and let
$A(x,y)$ denote the set of ideals222Equivalently, $A(x,y)=N^{-1}((x,y])$
$\mathfrak{m}$ in $\mathcal{O}_{K}$ satisfying $x<N(\mathfrak{m})\leq y$. Then
there are $x_{0}\geq 1$ and $\epsilon_{0}>0$ such that

  1. (i)

$|\mathbb{P}\cap A(8^{x},8^{x+1})|\geq\frac{8^{x}}{x}$ for all $x\geq x_{0}$,
and

  2. (ii)

$|\mathbb{P}\cap A(8^{x},8^{x+\epsilon})|\leq\frac{\sqrt{\epsilon}8^{x}}{x}$
for all $x\geq x_{0}$ and $\epsilon\in(0,\epsilon_{0}]$.

###### Proof.

The upper bound (ii) is almost immediate by taking $\alpha=8^{\epsilon}$ in
Lemma 3, so that one has

$\displaystyle\sum_{8^{x}<N(\mathfrak{p})\leq 8^{x+\epsilon}}1$
$\displaystyle\leq\epsilon(\log 8)8^{\epsilon}\frac{8^{x}}{x\log
8}+O\Big{(}8^{x(1-\frac{1}{d})}x\Big{)}$ $\displaystyle\leq
2\epsilon\frac{8^{x}}{x}+O\Big{(}8^{x(1-\frac{1}{d})}x\Big{)}.$

Since $2\epsilon\leq\sqrt{\epsilon}$ for $\epsilon$ in a neighborhood of zero,
and since the order of growth of $8^{x}/x$ is much larger than that of
$8^{x(1-\frac{1}{d})}x$, the result follows.

The lower bound (i) follows from the fact that

$\displaystyle|\mathbb{P}\cap A(8^{x},8^{x+1})|$
$\displaystyle=|\mathbb{P}\cap A(0,8^{x+1})|-|\mathbb{P}\cap A(0,8^{x})|$
$\displaystyle=\pi_{K}(8^{x+1})-\pi_{K}(8^{x})$

and the dyadic decomposition
$(\frac{1}{2},8^{x}]=\bigcup_{k=0}^{3x}\Big{(}\frac{8^{x}}{2^{k+1}},\frac{8^{x}}{2^{k}}\Big{]}$
gives

$\displaystyle\pi_{K}(8^{x})$
$\displaystyle=\sum_{k=0}^{3x}\sum_{\frac{8^{x}}{2^{k+1}}<N(\mathfrak{p})\leq\frac{8^{x}}{2^{k}}}1$
$\displaystyle\leq O(1)+\sum_{k=0}^{3x-2}\Big{[}(2\log
2)\frac{8^{x}/2^{k+1}}{\log(8^{x}/2^{k+1})}+O\Big{(}(8^{x}/2^{k+1})^{1-\frac{1}{d}}x\Big{)}\Big{]}\quad(\text{from
(\ref{UpperBoundLemma}}))$
$\displaystyle=2\sum_{j=2}^{3x}\frac{2^{j}}{j}+O\Big{(}8^{x(1-\frac{1}{d})}x\Big{)}.$

We then note that

$\displaystyle\frac{x}{8^{x}}\sum_{j=2}^{3x}\frac{2^{j}}{j}$
$\displaystyle=\frac{x}{8^{x}}\sum_{2\leq j\leq
2x}\frac{2^{j}}{j}+\frac{x}{8^{x}}\sum_{2x<j\leq 3x}\frac{2^{j}}{j}$
$\displaystyle\leq x\sum_{2\leq j\leq 2x}2^{j-3x}+\frac{1}{2}\sum_{2x<j\leq
3x}2^{j-3x}$ $\displaystyle=1+o(1)$

and hence
$\pi_{K}(8^{x})\leq(2+o(1))\frac{8^{x}}{x}+O(8^{x(1-\frac{1}{d})}x)$. One then
has that, applying Lemma 4,

$\displaystyle\pi_{K}(8^{x+1})-\pi_{K}(8^{x})$
$\displaystyle\geq(1-\epsilon)\frac{8^{x+1}}{3(x+1)\log
2}-(2+o(1))\frac{8^{x}}{x}+O\Big{(}8^{x(1-\frac{1}{d})}x^{2}\Big{)}\quad(x\gg_{\epsilon}1)$
$\displaystyle=\Big{[}\frac{8(1-\epsilon)}{3\log
2}\frac{x}{x+1}-2+o(1)\Big{]}\cdot\frac{8^{x}}{x}+O(8^{x(1-\frac{1}{d})}x)$
$\displaystyle\geq\frac{8^{x}}{x}$

for $\epsilon$ small and sufficiently large $x$. This completes the proof of
Proposition 4.

∎

###  3.2. Combinatorial Ingredients

We include two combinatorial Lemmas, the first which can be easily extended
from Lemma 3.4 in [12] to the number field setting, and the second which is
precisely Lemma 3.5 in [12]:

######  Lemma 6.

Let $x_{0}$ be as in Proposition 4. There exists $\epsilon_{1}>0$ such that
for all $\epsilon\in(0,\epsilon_{1}]$ and all $\delta\in(0,1)$ there exists
$D=D(\epsilon,\delta)\in(0,1)$ with the following property: For all $n\geq
x_{0}$ there are $x,y\in[n,n+1)$ with $\epsilon^{4}<y-x<\epsilon$ such that

$\displaystyle|\mathbb{P}\cap
A(8^{x},8^{x+\delta})|\geq\frac{D8^{n}}{n},\quad\text{and}\quad|\mathbb{P}\cap
A(8^{y},8^{y+\delta})|\geq\frac{D8^{n}}{n}.$

######  Lemma 7 (Richter [12], Lemma 3.5 ).

Fix $x_{0}\geq 1$ and $0<\epsilon<\frac{1}{2}$. Suppose $\mathcal{X}$ is a
subset of $\mathbb{R}$ with the property that for every $n\geq x_{0}$, there
exist $x,y\in\mathcal{X}\cap[n,n+1)$ with $\epsilon^{4}<y-x<\epsilon$. Let
$j\geq\lceil 2/\epsilon^{4}\rceil$. Then for all
$n_{1},n_{2},...,n_{j}\in\mathbb{N}_{\geq x_{0}}$ there exist
$z,z_{1},...,z_{j}\in\mathcal{X}$ such that

  1. (I)

$z_{i}\in[n_{i},n_{i}+1)$ for all $1\leq i\leq j$

  2. (II)

$z_{1}+...+z_{j}\in[z,z+\epsilon)$.

We now have the tools necessary to prove Proposition 3.

###### Proof of Proposition 3.

Let $\eta\in(0,1)$ be given. Let $x_{0}$ and $\epsilon_{0}$ be as in
Proposition 4, and $\epsilon_{1}$ be as in Lemma 6. Pick any
$\epsilon<\min\\{\epsilon_{0},\epsilon_{1},\frac{\log(1+\eta)}{\log 64}\\}$.
Set $N=\lceil 2/\epsilon^{4}\rceil$. We will construct sets
$S_{1}=S_{1}(\eta)$ and $S_{2}=S_{2}(\eta)$ as specified in Proposition 3.

To do this, set $j\geq N$, and choose $\delta=\epsilon/j$, and let
$D=D(\epsilon,\delta)$ be as in Lemma 6. Define

$\displaystyle\mathcal{X}:=\Big{\\{}x\geq N:|\mathbb{P}\cap
A(8^{x},8^{x+\delta})|\geq D8^{\lfloor x\rfloor}/\lfloor x\rfloor\Big{\\}}.$

By Lemma 6, $\mathcal{X}$ is necessarily infinite and unbounded. For every
$x\in\mathcal{X}$ define $P_{x}$ to be a set satisfying the following
conditions:

(18) $\displaystyle P_{x}\subset\mathbb{P}\cap A(8^{x},8^{x+\delta})\text{ and
}|P_{x}|=\lfloor D8^{\lfloor x\rfloor}/\lfloor x\rfloor\rfloor$

We shall use the sets $P_{x}$ to construct the sets $S_{1}$ and $S_{2}$.

The set $\mathcal{X}$ satisfies the conditions of Lemma 7, which ensures, for
each tuple $\mathbf{m}:=(m_{1},...,m_{j})\in\mathbb{N}_{\geq x_{0}}^{j}$, the
existence of some
$\zeta_{\mathbf{m}},\zeta_{1,\mathbf{m}},...,\zeta_{j,\mathbf{m}}\in\mathcal{X}$
that satisfy the conditions

  1. (I.j)

$\zeta_{i,\mathbf{m}}\in[m_{i},m_{i}+1)$ for all $1\leq i\leq j$, and

  2. (II.j)

$\zeta_{1,\mathbf{m}}+....+\zeta_{j,\mathbf{m}}\in[\zeta_{\mathbf{m}},\zeta_{\mathbf{m}}+\epsilon)$.

By definition of the sets $P_{x}$, one has

(19) $\displaystyle|P_{\zeta_{i},\mathbf{m}}|=\lfloor
D8^{m_{i}}/m_{i}\rfloor.$

Now, let $M=M(D,\eta,j)$ be a constant that is to be determined later, and
define sets $A_{1},...,A_{j}\subset\mathbb{N}_{\geq x_{0}}$ inductively in the
following manner:

  1. (1)

Pick some $s_{1}\in\mathbb{N}$ with $s_{1}>\max\\{x_{0},2j\\}$ and let $A_{1}$
be a finite subset of $s_{1}\mathbb{N}=\\{s_{1}n:n\in\mathbb{N}\\}$ satisfying
$\sum_{n\in A_{1}}1/n\geq M$.

  2. (2)

Given $A_{i}$, form $A_{i+1}$ by taking any
$s_{i+1}>M+\max(A_{1})+...+\max(A_{i})$ and set $A_{i+1}$ to be a finite
subset of $s_{i+1}\mathbb{N}$ satisfying $\sum_{n\in A_{i+1}}1/n\geq M$.

We observe that these sets $A_{i}$ satisfy the following property:

  1. (A)

For any vector $\mathbf{q}=(q_{1},...,q_{j})\in A_{1}\times...\times A_{j}$
and $\mathbf{r}=(r_{1},...,r_{j})\in A_{1}\times...\times A_{j}$ with
$\mathbf{q}\neq\mathbf{r}$, one has the distance between the integers
$q_{1}+...+q_{j}$ and $r_{1}+...+r_{j}$ being at least $M$.

This follows because $\mathbf{q}\neq\mathbf{r}$ implies that there exists some
$1\leq i_{*}\leq j$ satisfying $q_{i_{*}}\neq r_{i_{*}}$. Choose such an
$i_{*}$ to be maximal, then one has

$\displaystyle\Big{|}\sum_{i=1}^{j}(q_{i}-r_{i})\Big{|}=\Big{|}\sum_{\begin{subarray}{c}1\leq
i\leq j\\\ q_{i}\neq r_{i}\end{subarray}}(q_{i}-r_{i})\Big{|}$
$\displaystyle=\Big{|}q_{i_{*}}-r_{i_{*}}+\sum_{\begin{subarray}{c}1\leq i\leq
j\\\ q_{i}\neq r_{i}\\\ i\neq i_{*}\end{subarray}}(q_{i}-r_{i})\Big{|}$
$\displaystyle\geq|q_{i_{*}}-r_{i_{*}}|-\Big{|}\sum_{\begin{subarray}{c}1\leq
i\leq j\\\ q_{i}\neq r_{i}\\\ i\neq i_{*}\end{subarray}}(q_{i}-r_{i})\Big{|}$
$\displaystyle=|q_{i_{*}}-r_{i_{*}}|-\sum_{\begin{subarray}{c}1\leq i<i_{*}\\\
q_{i}\neq r_{i}\end{subarray}}|q_{i}-r_{i}|,$

with the last step following from the maximality assumption. On the one hand,
since $q_{i_{*}}$ and $r_{i_{*}}$ are distinct elements of
$s_{i_{*}}\mathbb{N}$, they satisfy $|q_{i_{*}}-r_{i_{*}}|\geq s_{i_{*}}$. On
the other hand, if $1\leq i<i_{*}$, one has $|q_{i}-r_{i}|\leq j\leq\max
A_{i}$ since both $q_{i}$ and $r_{i}$ are positive integers, and hence we have

$\displaystyle\Big{|}\sum_{i=1}^{j}(q_{i}-r_{i})\Big{|}\geq
s_{i_{*}}-\sum_{1\leq i<i_{*}}\max A_{i}\geq M,$

which provides the proof for property (A).

The Construction

We can now construct the collections of ideals $S_{1}$ and $S_{2}$ in
Proposition 3. Set $j\geq N$. Recalling the definitions of the sets
$P_{\zeta_{i,\mathbf{m}}}$ as given in (18) and (19), we form $S_{2}$ as the
product set

$\displaystyle S_{2}:=\bigcup_{\mathbf{m}\in A_{1}\times\cdots\times
A_{j}}P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}=\\{\mathfrak{p}_{1}...\mathfrak{p}_{j}:\mathfrak{p}_{i}\in
P_{\zeta_{i,\mathbf{m}}}\text{ for all }1\leq i\leq j\\}.$

This then gives that

$\displaystyle|P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}|\leq\frac{D^{j}8^{m_{1}+\cdots+m_{j}}}{m_{1}\cdots
m_{j}}\leq\frac{D^{j}8^{m_{1}+\cdots+m_{j}}}{m_{1}+\cdots+m_{j}}<\frac{D8^{\lfloor\zeta_{\mathbf{m}}\rfloor}}{\lfloor\zeta_{\mathbf{m}}\rfloor}$

where the last step follows from the fact
$\zeta_{1,\mathbf{m}}+\cdots+\zeta_{j,\mathbf{m}}\geq\lfloor\zeta_{\mathbf{m}}\rfloor$
(as follows from (II.j)), and because $D^{j}<D$. Hence we have by definition
of $P_{\zeta_{\mathbf{m}}}$ that $|P_{\zeta_{1,\mathbf{m}}}\cdots
P_{\zeta_{j,\mathbf{m}}}|\leq|P_{\zeta_{\mathbf{m}}}|$. In particular, there
exists some collection of ideals $Q_{\mathbf{m}}\subseteq
P_{\zeta_{\mathbf{m}}}$ with $|Q_{\mathbf{m}}|=|P_{\zeta_{1,\mathbf{m}}}\cdots
P_{\zeta_{j,\mathbf{m}}}|$. We then form the collection of prime ideals

$\displaystyle S_{1}:=\bigcup_{\mathbf{m}\in A_{1}\times\cdots\times
A_{j}}Q_{\mathbf{m}}.$

###### Proof that $S_{1}$ and $S_{2}$ satisfy (i).

By construction, if $\mathfrak{m}\in S_{2}$ then one can write
$\mathfrak{m}=\mathfrak{p}_{\zeta_{1,\mathbf{m}}}\cdots\mathfrak{p}_{\zeta_{j,\mathbf{m}}}$
for some $\mathbf{m}\in A_{1}\times\cdots\times A_{j}$ and
$\mathfrak{p}_{\zeta_{1,\mathbf{m}}}\in P_{\zeta_{1,\mathbf{m}}}$. Then
$\Omega(\mathfrak{m})=j$, as desired. Whereas if $\mathfrak{p}\in S_{1}\subset
P_{\zeta_{\mathbf{m}}}$ then $\mathfrak{p}$ is by definition a prime ideal. ∎

Proof that $S_{1}$ and $S_{2}$ satisfy (ii). By definition, we have
$P_{\zeta_{i,\mathbf{m}}}\subset
A(8^{\zeta_{i,\mathbf{m}}},8^{\zeta_{i,\mathbf{m}}+\delta})$, so that

$\displaystyle P_{\zeta_{1,\mathbf{m}}}\cdots P_{\zeta_{j,\mathbf{m}}}$
$\displaystyle\subset
A(8^{\zeta_{1,\mathbf{m}}+\cdots+\zeta_{j,\mathbf{m}}},8^{\zeta_{1,\mathbf{m}}+...+\zeta_{j,\mathbf{m}}+j\delta})\subset
A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+2\epsilon}),$

with the last inclusion following from property (II.j) and the definition of
$\delta$ as $\epsilon/j$. Also, by definition we have $Q_{\mathbf{m}}\subset
P_{\zeta_{\mathbf{m}}}\subset
A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+\delta})$. Since
$\delta=\epsilon/j$ we clearly have $Q_{\mathbf{m}}\subset
A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+2\epsilon})$.

We first claim that each ideal in $S_{2}$ is distinct. It suffices to prove
that the sets
$\tilde{M}_{1}:=P_{\zeta_{1,\mathbf{m_{1}}}}...P_{\zeta_{j,\mathbf{m_{1}}}}$
and
$\tilde{M}_{2}:=P_{\zeta_{1,\mathbf{m_{2}}}}...P_{\zeta_{j,\mathbf{m_{2}}}}$
have empty intersection for $\mathbf{m_{1}}\neq\mathbf{m_{2}}$, both in
$A_{1}\times...\times A_{j}$ (by uniqueness of factorization of an ideal into
prime ideals, one has
$|P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}|=\prod_{k=1}^{j}|P_{\zeta_{k,\mathbf{m}}}|$).
Given some ideal $\mathfrak{u}\in
P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}$, we can write
$\mathfrak{u}=\mathfrak{p}_{\zeta_{1,\mathbf{m}}}...\mathfrak{p}_{\zeta_{j,\mathbf{m}}}$
for a collection of prime ideals $\mathfrak{p}_{\zeta_{i,\mathbf{m}}}$ with
each $\mathfrak{p}_{\zeta_{i,\mathbf{m}}}\in P_{\zeta_{1,\mathbf{m}}}$. Since
$\zeta_{i,\mathbf{m}}\in[m_{i},m_{i}+1)$ for all $1\leq i\leq j$ by point
(I.j), and
$\zeta_{1,\mathbf{m}}+...+\zeta_{j,\mathbf{m}}\in[\zeta_{\mathbf{m}},\zeta_{\mathbf{m}}+\epsilon)$,
we then discern that

$\displaystyle|m_{1}+...+m_{j}-\zeta_{\mathbf{m}}|$
$\displaystyle\leq\Big{|}-\zeta_{\mathbf{m}}+\sum_{k=1}^{j}\zeta_{k,\mathbf{m}}\Big{|}+\sum_{k=1}^{j}\Big{|}m_{i}-\zeta_{k,\mathbf{m}}\Big{|}<\epsilon+j,$

which alongside property (A) with the choice of $M\geq 2j+3$ gives that $u$
cannot possibly be in any other collection
$P_{\zeta_{1,\mathbf{m^{\prime}}}}...P_{\zeta_{j,\mathbf{m^{\prime}}}}$.

We next claim that each ideal in $S_{1}$ is distinct. We must prove that the
sets $Q_{\mathbf{m}}$ are disjoint for different $\mathbf{m}\in
A_{1}\times...\times A_{j}$; since $Q_{\mathbf{m}}\subset
P_{\zeta_{\mathbf{m}}}$, it suffices to prove that the
$P_{\zeta_{\mathbf{m}}}$ are disjoint for different $\mathbf{m}\in
A_{1}\times...\times A_{j}$. But this is easily apparent, since
$P_{\zeta_{\mathbf{m}}}\subset
A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+\delta})$ by definition, and
alongside property (A) with the above choice of $M$ and the bound above gives
that these annuli are disjoint.

Regarding cardinality, it is immediately clear that $S_{1}$ and $S_{2}$ have
the same cardinality since
$|Q_{\mathbf{m}}|=|P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}|$.
Also, since $Q_{\mathbf{m}}$ and
$P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}$ both lie in the annulus
$A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+2\epsilon})$, it is clear that
the ratio of the norm of any element of the latter with the norm of any
element of the former lies between $8^{-2\epsilon}$ and $8^{2\epsilon}$. Since
$8^{2\epsilon}\leq 1+\eta$, it then follows that by enumerating
$P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}=\\{P_{i}:N(P_{1})\leq...\leq
N(P_{r})\\}$ and $Q_{\mathbf{m}}=\\{Q_{i}:N(Q_{1})\leq...\leq N(Q_{r})\\}$ we
have that $(1-\eta)N(P_{j})\leq N(Q_{j})\leq(1+\eta)N(p_{j})$ for each
$j=1,...,r$. Using the fact that each $\mathbf{m}$ corresponds to a distinct
annulus $A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+\delta})$, we can then
extend this enumeration property to $S_{1}$ and $S_{2}$, so that for each
$Q_{i}\in S_{2}$ there exists a $P_{i}\in S_{1}$ for which
$(1-\eta)N(P_{i})\leq N(Q_{i})\leq(1+\eta)N(P_{i})$.
∎

###### Proof that $S_{1}$ and $S_{2}$ satisfy (iii)..

We first prove the coprimality condition for $S_{1}$, as it is simpler. We
must show that $\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}\in
S_{1}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}^{\prime}\in
S_{1}}^{\log}c_{K}\Phi(\mathfrak{p},\mathfrak{p}^{\prime})\leq\eta$, for
$\Phi(\mathfrak{p},\mathfrak{p}^{\prime}):=N(\operatorname{gcd}(\mathfrak{p},\mathfrak{p}^{\prime}))-1$.
The set $S_{1}$ only contains prime ideals, so it is clear that
$\Phi(\mathfrak{p},\mathfrak{p}^{\prime})=0$ for
$\mathfrak{p}\neq\mathfrak{p}^{\prime}$; hence we have

(20)
$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}\in
S_{1}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}^{\prime}\in
S_{1}}^{\log}\Phi(\mathfrak{p},\mathfrak{p}^{\prime})=\frac{\sum_{\mathfrak{p},\mathfrak{p}^{\prime}\in
S_{1}}\frac{\Phi(\mathfrak{p},\mathfrak{p}^{\prime})}{N(\mathfrak{p})N(\mathfrak{p}^{\prime})}}{\sum_{\mathfrak{p},\mathfrak{p}^{\prime}\in
S_{1}}\frac{1}{N(\mathfrak{p})N(\mathfrak{p}^{\prime})}}\leq\frac{\sum_{\mathfrak{p}\in
S_{1}}\frac{1}{N(\mathfrak{p})}}{\Big{(}\sum_{\mathfrak{p}\in
S_{1}}\frac{1}{N(\mathfrak{p})}\Big{)}^{2}}=\frac{1}{\sum_{\mathfrak{p}\in
S_{1}}\frac{1}{N(\mathfrak{p})}}.$

By definition of $S_{1}$ we evidently have $\sum_{\mathfrak{p}\in
S_{1}}\frac{1}{N(\mathfrak{p})}=\sum_{\mathbf{m}\in A_{1}\times...\times
A_{j}}\sum_{P\in Q_{\mathbf{m}}}\frac{1}{N(\mathfrak{p})}$, and since
$Q_{\mathbf{m}}\subset
A(8^{\zeta_{\mathbf{m}}},8^{\zeta_{\mathbf{m}}+2\epsilon})\subset
A(8^{\zeta_{\mathbf{m}}},8^{m_{1}+...+m_{j}+j+3\epsilon})$ one has
$N(\mathfrak{p})\leq 8^{m_{1}+...+m_{j}+j+1}$ for each $\mathfrak{p}\in
Q_{\mathbf{m}}$, so that

$\displaystyle\sum_{P\in
S_{1}}\frac{1}{N(\mathfrak{p})}\geq\sum_{\mathbf{m}\in A_{1}\times...\times
A_{j}}\frac{|Q_{\mathbf{m}}|}{8^{j+1}\cdot 8^{m_{1}+...+m_{j}}}.$

Then since
$|Q_{\mathbf{m}}|=|P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}|=\prod_{k=1}^{j}|P_{\zeta_{k,\mathbf{m}}}|$
one has $|Q_{\mathbf{m}}|\geq
D^{j}8^{-j}8^{m_{1}+...+m_{j}}/(m_{1}m_{2}...m_{j})$, so that

$\displaystyle\sum_{\mathfrak{p}\in
S_{1}}\frac{1}{N(\mathfrak{p})}\geq\sum_{\mathbf{m}\in A_{1}\times...\times
A_{j}}\frac{D^{j}}{8^{2j+1}m_{1}...m_{j}}\geq\frac{D^{j}M^{j}}{8^{2j+1}}.$

Then if $M$ is chosen sufficiently large, then
$\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}\in
S_{1}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{p}^{\prime}\in
S_{1}}^{\log}\Phi(\mathfrak{p},\mathfrak{p}^{\prime})\leq\eta$ immediately
follows.

Now, it suffices to show that
$\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}^{\prime}\in
S_{2}}^{\log}\Phi(\mathfrak{m},\mathfrak{m}^{\prime})\leq\eta$. By definition
of the set $S_{2}$, one has

$\displaystyle\sum_{\mathfrak{m},\mathfrak{m}^{\prime}\in
S_{2}}\frac{\Phi(\mathfrak{m},\mathfrak{m}^{\prime})}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}=\sum_{\mathbf{m},\mathbf{m^{\prime}}\in
A_{1}\times...\times A_{j}}\sum_{\mathfrak{m}\in
P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}}\sum_{\mathfrak{m}^{\prime}\in
P_{\zeta_{1,\mathbf{m_{2}}}}...P_{\zeta_{j,\mathbf{m^{\prime}}}}}\frac{\Phi(\mathfrak{m},\mathfrak{m}^{\prime})}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}.$

If some $\mathfrak{m}\in P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}$
and $\mathfrak{m}^{\prime}\in
P_{\zeta_{1,\mathbf{m^{\prime}}}}...P_{\zeta_{j,\mathbf{m^{\prime}}}}$ are
coprime then $\Phi(\mathfrak{m},\mathfrak{m}^{\prime})=0$, so do not
contribute towards the sum above. Whereas if $\mathfrak{m}$ and
$\mathfrak{m}^{\prime}$ are not coprime, set
$\operatorname{gcd}(\mathfrak{m},\mathfrak{m}^{\prime}):=\mathfrak{u}$ with
$\mathfrak{u}\in\prod_{i\in F}P_{\zeta_{i,\mathbf{m^{\prime}}}}$ for some
nonempty minimal subset $F\subset\\{1,...,j\\}$ and
$\mathfrak{u}^{\prime}\in\prod_{i\not\in F}P_{\zeta_{i,\mathbf{m^{\prime}}}}$
such that $\mathfrak{m}^{\prime}=\mathfrak{uu}^{\prime}$; this forces
$m_{i}=m_{i}^{\prime}$ for all $i\in F$ since $P_{\zeta_{i,\mathbf{m}}}$ and
$P_{\zeta_{i,\mathbf{m^{\prime}}}}$ are otherwise disjoint. One would then
have

$\displaystyle\frac{\Phi(\mathfrak{m},\mathfrak{m}^{\prime})}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}=\frac{N(\mathfrak{u})-1}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}\leq\frac{1}{N(\mathfrak{m})N(\mathfrak{u}^{\prime})}$

and as a result

$\displaystyle\sum_{\mathfrak{m},\mathfrak{m}^{\prime}\in
S_{2}}\frac{\Phi(\mathfrak{m},\mathfrak{m}^{\prime})}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}\leq\sum_{\begin{subarray}{c}F\subset\\{1,...,j\\}\\\
F\neq\varnothing\end{subarray}}\sum_{\begin{subarray}{c}\mathbf{m},\mathbf{m^{\prime}}\in
A_{1}\times...\times A_{j}\\\ m_{i}=m_{i}^{\prime},\ \forall i\in
F\end{subarray}}\sum_{\mathfrak{m}\in
P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}}\sum_{\mathfrak{u}^{\prime}\in\prod_{i\not\in
F}P_{\zeta_{i,\mathbf{m}^{\prime}}}}\frac{1}{N(\mathfrak{m})N(\mathfrak{u}^{\prime})}.$

Then, using the fact that $P_{\zeta_{i,\mathbf{m}}}\subset
A(8^{m_{i}},8^{m_{i}+1})$ and $P_{\zeta_{i,\mathbf{m^{\prime}}}}\subset
A(8^{m_{i}^{\prime}},8^{m_{i}^{\prime}+1})$ we then have

$\displaystyle\sum_{\mathfrak{m}\in
P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}}\sum_{\mathfrak{u}^{\prime}\in\prod_{i\not\in
F}P_{\zeta_{i,\mathbf{m}^{\prime}}}}\frac{1}{N(\mathfrak{m})N(\mathfrak{u}^{\prime})}$
$\displaystyle\leq\sum_{\mathfrak{m}\in
P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}}\sum_{\mathfrak{u}^{\prime}\in\prod_{i\not\in
F}P_{\zeta_{i,\mathbf{m}^{\prime}}}}\Big{(}\prod_{k=1}^{j}\frac{1}{8^{m_{i}}}\Big{)}\Big{(}\prod_{k\not\in
F}\frac{1}{8^{m_{i}^{\prime}}}\Big{)}$
$\displaystyle=\Big{(}\prod_{k=1}^{j}\frac{|P_{\zeta_{k,\mathbf{m}}}|}{8^{m_{i}}}\Big{)}\Big{(}\prod_{k\not\in
F}\frac{|P_{\zeta_{k,\mathbf{m^{\prime}}}}|}{8^{m_{k}^{\prime}}}\Big{)}.$

By definition, each set $P_{\zeta_{k,\mathbf{m}}}$ has cardinality $\lfloor
D8^{m_{i}}/m_{i}\rfloor$, so that

$\displaystyle\Big{(}\prod_{k=1}^{j}\frac{|P_{\zeta_{k,\mathbf{m}}}|}{8^{m_{i}}}\Big{)}\Big{(}\prod_{k\not\in
F}\frac{|P_{\zeta_{k,\mathbf{m^{\prime}}}}|}{8^{m_{k}^{\prime}}}\Big{)}\leq\Big{(}\prod_{k=1}^{j}\frac{D}{m_{i}}\Big{)}\Big{(}\prod_{k\not\in
F}\frac{D}{m_{k}^{\prime}}\Big{)},$

and hence we have

$\displaystyle\sum_{\mathfrak{m},\mathfrak{m}^{\prime}\in
S_{2}}\frac{\Phi(\mathfrak{m},\mathfrak{m}^{\prime})}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}$
$\displaystyle\leq\sum_{\begin{subarray}{c}F\subset\\{1,...,j\\}\\\
F\neq\varnothing\end{subarray}}\sum_{\begin{subarray}{c}\mathbf{m},\mathbf{m^{\prime}}\in
A_{1}\times...\times A_{j}\\\ m_{i}=m_{i}^{\prime},\ \forall i\in
F\end{subarray}}\Big{(}\prod_{k=1}^{j}\frac{D}{m_{k}}\Big{)}\Big{(}\prod_{k\not\in
F}\frac{D}{m_{k}^{\prime}}\Big{)}$
$\displaystyle=\sum_{\begin{subarray}{c}F\subset\\{1,...,j\\}\\\
F\neq\varnothing\end{subarray}}D^{2j-F}\Big{(}\prod_{k=1}^{j}\sum_{m\in
A_{k}}\frac{1}{m}\Big{)}\Big{(}\prod_{k\not\in F}\sum_{m\in
A_{k}}\frac{1}{m}\Big{)}.$

One also has that, since $\mathfrak{m}\in
P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}$ implies that
$N(\mathfrak{m})\leq 8^{m_{1}+...+m_{j}+j+1}$, we then have
$\sum_{\mathfrak{m}\in S_{2}}\frac{1}{N(\mathfrak{m})}\geq\sum_{\mathbf{m}\in
A_{1}\times...\times
A_{j}}\frac{|P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}|}{8^{m_{1}+...+m_{j}+j+1}}$
and since $|P_{\zeta_{1,\mathbf{m}}}...P_{\zeta_{j,\mathbf{m}}}|\geq
D^{j}8^{-j}8^{m_{1}+...+m_{j}}/(m_{1}m_{2}...m_{j})$ it follows that

$\displaystyle\sum_{\mathfrak{m}\in
S_{2}}\frac{1}{N(\mathfrak{m})}\geq\sum_{\mathbf{m}\in A_{1}\times...\times
A_{j}}\frac{D^{j}}{8^{2j+1}(m_{1}m_{2}...m_{j})}\geq\frac{D^{j}}{8^{3j}}\prod_{k=1}^{j}\sum_{m\in
A_{k}}\frac{1}{m}.$

We then deduce that

$\displaystyle\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}^{\prime}\in
S_{2}}^{\log}\Phi(\mathfrak{m},\mathfrak{m}^{\prime})$
$\displaystyle=\frac{\sum_{\mathfrak{m},\mathfrak{m}^{\prime}\in
S_{2}}\frac{\Phi(\mathfrak{m},\mathfrak{m}^{\prime})}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}}{\sum_{\mathfrak{m},\mathfrak{m}^{\prime}\in
S_{2}}\frac{1}{N(\mathfrak{m})N(\mathfrak{m}^{\prime})}}$
$\displaystyle\leq\frac{\sum_{\begin{subarray}{c}F\subset\\{1,...,j\\}\\\
F\neq\varnothing\end{subarray}}D^{2j-F}\Big{(}\prod_{k=1}^{j}\sum_{m\in
A_{k}}\frac{1}{m}\Big{)}\Big{(}\prod_{k\not\in F}\sum_{m\in
A_{k}}\frac{1}{m}\Big{)}}{\Big{(}\frac{D^{j}}{8^{3j}}\prod_{k=1}^{j}\sum_{m\in
A_{k}}\frac{1}{m}\Big{)}^{2}}$
$\displaystyle\leq\sum_{\begin{subarray}{c}F\subset\\{1,...,j\\}\\\
F\neq\varnothing\\}\end{subarray}}\frac{8^{6k}D^{-|F|}}{\Big{(}\prod_{k\in
F}\sum_{m\in A_{i}}\frac{1}{m}\Big{)}}.$

Finally, since $\sum_{m\in A_{i}}\frac{1}{m}\geq M$, we see that
$\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}\in
S_{2}}^{\log}\operatorname*{\scalerel*{\mathbb{E}}{\sum}}_{\mathfrak{m}^{\prime}\in
S_{2}}^{\log}\Phi(\mathfrak{m},\mathfrak{m}^{\prime})\leq\eta$ for
sufficiently large $M$. This completes the proof of Proposition 3. ∎

##  4\. Appendix A

We begin with an analogue of Stirling’s formula in the number field setting:

######  Lemma 8.

Let $K$ be a number field of degree $d$, and let
$f(x)=\prod_{N(\mathfrak{m})\leq x}N(\mathfrak{m})$. Then

$\displaystyle\log f(x)=c_{K}x\log x-c_{K}x+O(x^{1-\frac{1}{d}}\log x).$

###### Proof.

Let $\phi(n)$ denote the number of ideals with norm $n$ in $I_{K}$, then

$\displaystyle\log f(x)=\sum_{N(\mathfrak{m})\leq x}\log
N(\mathfrak{m})=\sum_{n=1}^{\lfloor x\rfloor}\phi(n)\log n.$

Via Abel summation, we can write this as

$\displaystyle\log f(x)$ $\displaystyle=\log
x\sum_{n=1}^{x}\phi(n)-\int_{1}^{x}\frac{\sum_{n\leq t}\phi(n)}{t}\ dt$
$\displaystyle=\log x\sum_{N(\mathfrak{m})\leq
x}1-\int_{1}^{x}\frac{\sum_{N(\mathfrak{m})\leq t}1}{t}\ dt$
$\displaystyle=\log
x\Big{(}c_{K}x+O(x^{1-\frac{1}{d}})\Big{)}-\int_{1}^{x}\frac{c_{K}t+O(t^{1-\frac{1}{d}})}{t}\
dt$ $\displaystyle=c_{K}x\log x-c_{K}x+O(x^{1-\frac{1}{d}}\log x).$

The result then immediately follows. ∎

The following Proposition generalizes this argument:

######  Proposition 5.

Let $K$ be a number field of degree $d$, and let
$F(x)=\sum_{N(\mathfrak{m})\leq x}f[N(\mathfrak{m})]$ for a function $f\in
C^{1}([1,\infty))$. Then one has

$\displaystyle F(x)=c_{K}f(1)\cdot xf(x)+c_{K}\int_{1}^{x}f(t)\
dt+c_{K}f(1)+E,$

where

$\displaystyle E\ll
f(x)x^{1-\frac{1}{d}}+\int_{1}^{x}|f^{\prime}(t)|t^{1-\frac{1}{d}}\ dt.$

###### Proof.

Let $\phi(n)$ denote the number of ideals with norm $n$ in $I_{K}$, then for
any $x\in\mathbb{N}_{0}$,

$\displaystyle F(x)=\sum_{N(\mathfrak{m})\leq
x}f(\mathfrak{m})=\sum_{n=1}^{x}\phi(n)f(n).$

Using partial summation, we can write this as

$\displaystyle F(x)$
$\displaystyle=f(x)\sum_{n=1}^{x}\phi(n)-\int_{1}^{x}f^{\prime}(t)\sum_{1\leq
n\leq t}\phi(n)\ dt$ $\displaystyle=f(x)\sum_{N(\mathfrak{m})\leq
x}1-\int_{1}^{x}f^{\prime}(t)\sum_{N(\mathfrak{m})\leq t}1\ dt$
$\displaystyle=f(x)\Big{(}c_{K}x+O(x^{1-\frac{1}{d}})\Big{)}-\int_{1}^{x}f^{\prime}(t)\Big{(}c_{K}t+O(t^{1-\frac{1}{d}})\Big{)}\
dt$ $\displaystyle=c_{K}xf(x)-c_{K}\int_{1}^{x}tf^{\prime}(t)\
dt+O\Big{(}f(x)x^{1-\frac{1}{d}}+\int_{1}^{x}|f^{\prime}(t)|t^{1-\frac{1}{d}}\
dt\Big{)}$ $\displaystyle:=c_{K}xf(x)-c_{K}\Big{(}xf(x)-f(1)-\int_{1}^{x}f(t)\
dt\Big{)}+E$ $\displaystyle=c_{K}f(1)\cdot xf(x)+c_{K}\int_{1}^{x}f(t)\
dt+c_{K}f(1)+E,$

where

$\displaystyle E\ll
f(x)x^{1-\frac{1}{d}}+\int_{1}^{x}|f^{\prime}(t)|t^{1-\frac{1}{d}}\ dt.$

∎

##  5\. Appendix B

In this section we recall several well-known facts about the connection
between equidistribution of a sequence and exponential sums. The most
fundamental is Weyl’s Criterion:

######  Lemma 9 (Weyl’s Criterion).

A sequence $(a_{n})\subset\mathbb{R}$ is uniformly distributed modulo 1 if and
only if for each $\ell\neq 0$, $\sum_{n\leq x}e(\ell a_{n})=o(x)$.

In the positive integer setting, one also has an associated equidistribution
criterion:

######  Lemma 10.

A sequence $(a_{n})\subset\mathbb{Z}_{\geq 0}$ is equidistributed modulo $m$
if and only if $\sum_{n\leq x}e(\ell a_{n}/m)=o(x)$ for each
$\ell\in\\{1,...,m-1\\}$.

###### Proof.

After expanding out the square and applying Parseval’s theorem over
$\mathbb{Z}_{m}$, one observes that

$\displaystyle\sum_{\ell\in\mathbb{Z}_{m}}\Big{|}-\frac{1}{m}+\frac{1}{x}\sum_{\begin{subarray}{c}n\leq
x\\\ a_{n}\equiv\ell\ (\text{mod
}m)\end{subarray}}1\Big{|}^{2}=\frac{1}{m}\sum_{\ell=1}^{m-1}\Big{|}\frac{1}{x}\sum_{n\leq
x}e(\ell a_{n}/m)\Big{|}^{2}.$

From this the Lemma is not hard to deduce. ∎

##  6\. Appendix C

In this section we show how $L(x)=o(x)$ implies that $M(x)=o(x)$, as defined
in the Introduction.

######  Lemma 11.

Let $K$ be an algebraic number field. Then $L(x)=o(x)$ implies that
$M(x)=o(x)$.

###### Proof.

We claim the following two statements are true (here $*$ denotes the Dirichlet
convolution operator):

  1. (i)

Suppose that $f$ is the indicator function for the square ideals. Then if
$f^{-1}$ denotes the Dirichlet inverse of $f$ (i.e. the function such that
$(f*f^{-1})(\mathfrak{m}):=\sum_{\mathfrak{d}_{1}\mathfrak{d}_{2}=\mathfrak{m}}f(\mathfrak{d}_{1})f^{-1}(\mathfrak{d}_{2})=\begin{cases}1&\mathfrak{m}=K\\\
0&\text{otherwise}\end{cases}$), then $f^{-1}$ is 1-bounded and $\text{supp
}f\supseteq\text{supp }f^{-1}$.

  2. (ii)

Let $g$ be the function such that $\mu=g*\lambda$. Then
$g=(\mathbf{1}_{square})^{-1}$, where $\mathbf{1}_{square}$ is the indicator
function for the square ideals.

Let us explain how these two statements together imply the result. Writing
$\mu=g*\lambda$, it is not hard to see then that

$\displaystyle M(x)=\sum_{N(\mathfrak{m})\leq
x}g(\mathfrak{m})L\Big{(}\frac{x}{N(\mathfrak{m})}\Big{)}.$

If $g=(\mathbf{1}_{square})^{-1}$ then by (i) $g$ is 1-bounded with support on
square ideals, so that

$\displaystyle\frac{M(x)}{x}=\frac{1}{x}\sum_{N(\mathfrak{m})^{2}\leq
x}g(\mathfrak{m}^{2})L\Big{(}\frac{x}{N(\mathfrak{m})^{2}}\Big{)}=\sum_{N(\mathfrak{m})^{2}\leq
x}\frac{g(\mathfrak{m}^{2})}{N(\mathfrak{m})^{2}}\cdot\frac{L(x/N(\mathfrak{m})^{2})}{x/N(\mathfrak{m})^{2}}.$

If $L(x)=o(x)$, then for each $\epsilon>0$ there exists some $C(\epsilon)$ for
which $x\geq C(\epsilon)\implies|L(x)/x|<\epsilon$, and hence

$\displaystyle\Big{|}\frac{M(x)}{x}\Big{|}$
$\displaystyle\leq\sum_{N(\mathfrak{m})^{2}\leq
x}\frac{1}{N(\mathfrak{m})^{2}}\Big{|}\frac{L(x/N(\mathfrak{m})^{2})}{x/N(\mathfrak{m})^{2}}\Big{|}$
$\displaystyle=\sum_{N(\mathfrak{m})^{2}\leq
x/C(\epsilon)}\frac{1}{N(\mathfrak{m})^{2}}\Big{|}\frac{L(x/N(\mathfrak{m})^{2})}{x/N(\mathfrak{m})^{2}}\Big{|}+\sum_{x/C(\epsilon)<N(\mathfrak{m})^{2}\leq
x}\frac{1}{N(\mathfrak{m})^{2}}\Big{|}\frac{L(x/N(\mathfrak{m})^{2})}{x/N(\mathfrak{m})^{2}}\Big{|}$
$\displaystyle<\epsilon\sum_{N(\mathfrak{m})^{2}\leq
x/C(\epsilon)}\frac{1}{N(\mathfrak{m})^{2}}+\sum_{x/C(\epsilon)<N(\mathfrak{m})^{2}\leq
x}\frac{1}{N(\mathfrak{m})^{2}}.$

Then since $\sum_{\mathfrak{m}}\frac{1}{N(\mathfrak{m})^{2}}<\infty$ (as can
be seen, say, from partial summation and §1.3), for sufficiently large $x$ we
can make this arbitrarily small to show that $M(x)=o(x)$.

Proof of (i). Since $f$ is multiplicative, one has, for
$\mathfrak{m}=\prod_{i}\mathfrak{p}_{i}^{\alpha_{i}}$ that
$f(\mathfrak{m})=\prod_{i}\mathbf{1}_{2|\alpha_{i}}$. By definition of
Dirichlet inverse, for each $\alpha_{i}\geq 1$ one has
$0=\sum_{\mathfrak{d}_{1}\mathfrak{d}_{2}=\mathfrak{p}_{i}^{\alpha_{i}}}f(\mathfrak{d}_{1})f^{-1}(\mathfrak{d}_{2})=\sum_{c=0}^{\alpha_{i}}f(\mathfrak{p}_{i}^{c})f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}-c})$.
Then

$\displaystyle
0=\sum_{c=0}^{\lfloor\alpha_{i}/2\rfloor}f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}-2c})$

At $\alpha_{i}=0$ one has $f^{-1}(1)=1$; at $\alpha_{i}=1$ one has
$f^{-1}(\mathfrak{p}_{i})=0$ by the above expression. Now let us induct on
$\alpha_{i}$ to show that $f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}})=0$ for
$\alpha_{i}\equiv 1\ (\text{mod }2)$. The base case is shown; suppose that
there exists some even $k$ for which $f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}})=0$
for each $\alpha_{i}\leq k$ odd. One has
$0=\sum_{c=0}^{\lfloor(k+1)/2\rfloor}f^{-1}(\mathfrak{p}_{i}^{k+1-2c})=f^{-1}(\mathfrak{p}_{i}^{k+1})$,
so that $f^{-1}(\mathfrak{p}_{i}^{k+1})=0$ as well. This provides that
$\text{supp }f^{-1}\subseteq\text{supp }f$, since $f^{-1}$ is multiplicative
(in general, the Dirichlet inverse of a multiplicative function is
multiplicative; see, e.g., Apostol’s Introduction to Analytic Number Theory,
Thm. 2.16).

To show boundedness, we again use the above expression for prime power ideals,
i.e. that
$0=\sum_{c=0}^{\lfloor\alpha_{i}/2\rfloor}f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}-2c})$.
It is clear that $f(1)=f^{-1}(1)=1$, hence evaluating at $\alpha_{i}=2$ gives
$f^{-1}(\mathfrak{p}_{i}^{2})=-1$. Then one has
$f^{-1}(\mathfrak{p}_{i}^{4})=-\sum_{1\leq c\leq
2}f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}-2c})=-(-1+1)=0$, and
$f^{-1}(\mathfrak{p}_{i}^{6})=-\sum_{1\leq c\leq
3}f^{-1}(\mathfrak{p}_{i}^{\alpha_{i}-2c})=-(0-1+1)=0$. It is not hard to see
that $f^{-1}(\mathfrak{p}_{i}^{2k})=\begin{cases}1&k=0\\\ -1&k=1\\\ 0&k\geq
2.\end{cases}$ by induction; then since $f^{-1}$ is multiplicative, this
provides a proof of boundedness since
$|f^{-1}(\mathfrak{m})|=\prod_{\mathfrak{p}|\mathfrak{m}}|f^{-1}(\mathfrak{p}^{x_{\mathfrak{p}}(\mathfrak{m})})|\leq
1$.

Proof of (ii). We first claim that $\lambda*1=\mathbf{1}_{square}$, where
$\mathbf{1}_{square}$ is the indicator function for the square ideals. Indeed,
it suffices to prove that $\mathbf{1}_{square}*\mu=\lambda$. One has

$\displaystyle(\mathbf{1}_{square}*\mu)(\mathfrak{m})=\sum_{\mathfrak{u\mathfrak{d}^{2}=\mathfrak{m}}}\mu(\mathfrak{u}).$

Write $\mathfrak{m}=\mathfrak{u}_{0}\mathfrak{d}_{max}^{2}$ for some
$\mathfrak{u}_{0}$ squarefree, then the above is the same as saying

$\displaystyle(\mathbf{1}_{square}*\mu)(\mathfrak{m})=\sum_{\mathfrak{d}_{1}\mathfrak{d}_{2}=\mathfrak{d}_{max}}\mu(\mathfrak{u}_{0}\mathfrak{d}_{2}^{2}).$

Each summand vanishes other than the case $\mathfrak{d}_{2}=1$, so that
$(\mathbf{1}_{square}*\mu)(N)=\mu(\mathfrak{u}_{0})=(-1)^{\Omega(\mathfrak{u}_{0})}$.
Then since $\Omega(\mathfrak{u}_{0})\equiv\Omega(\mathfrak{m})\ (\text{mod
}2)$, we have our desired equivalence.
Then convolving the equation $\lambda*1=\mathbf{1}_{square}$ with $g$ would
give $\lambda*g*1=g*\mathbf{1}_{square}$, and since $\lambda*g=\mu$ by
definition one would have $g*\mathbf{1}_{square}=\mu*1=\delta$. But then
$g=\mathbf{1}_{square}^{-1}$, by definition of our inverse. ∎

## Acknowledgements

The author extends gratitude for the idea from Michael Lacey, and for
discussions thereof. The author also thanks Alex Dunn and Vesselin Dimitrov
for information regarding number fields, and Florian Richter and Vitaly
Bergelson for their encouraging comments. Finally, the author thanks Rahul
Sethi and Junzhe Mao for close reading.

The author was partially supported by NSF grant award #2247254 under Michael
Lacey.

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