# Elliptic curves and spin

Peter Koymans  ETH Zurich  Peter Vang Uttenthal  Aarhus University

###### Abstract

In the early 2000s, Ramakrishna asked the question: For the elliptic curve

$E:y^{2}=x^{3}-x,$

what is the density of primes $p$ for which the Fourier coefficient $a_{p}(E)$
is a cube modulo $p$? As a generalization of this question, Weston–Zaurova
[14] formulated conjectures concerning the distribution of power residues of
degree $m$ of the Fourier coefficients of elliptic curves $E/\mathbb{Q}$ with
complex multiplication. In this paper, we prove the conjecture of
Weston–Zaurova for cubic residues using the analytic theory of spin. Our proof
works for all elliptic curves $E$ with complex multiplication.

> It is interesting to recall that, in connexion with a lecture by Prof. R.
> Fueter at the 1932 Zurich Congress, Hilbert asserted that the theory of
> complex multiplication of elliptic modular functions which forms a powerful
> link between number theory and analysis, is not only the most beautiful part
> of mathematics but also of all science.

##  1 Introduction

For the elliptic curve

$E:y^{2}=x^{3}-x,$

with complex multiplication by $\mathbb{Q}(i)$ and error terms

$a_{p}(E)=p+1-|E(\mathbb{F}_{p})|,$

are there infinitely many primes $p\equiv 1\bmod 12$ such that $a_{p}(E)$ is a
cubic residue modulo $p$? This question was formulated by Ramakrishna in the
early 2000s. At that time, the question was perceived as interesting yet
inaccessible. Note that it is natural to restrict to primes $p\equiv 1\bmod
12$. Indeed, for $p\equiv 3\bmod 4$, we have $a_{p}(E)=p+1$ if $p\geq 5$, so
$a_{p}(E)$ is a cube modulo $p$. Furthermore, all elements of $\mathbb{F}_{p}$
are automatically cubes if $p\equiv 2\bmod 3$, so in particular $a_{p}(E)$ is
once more a cube modulo $p$.

In 2005, Weston [13] formulated a family of conjectures inspired by the
question, but only for elliptic curves without complex multiplication. In a
subsequent paper, Weston and Zaurova considered the analogous conjectures for
elliptic curves over $\mathbb{Q}$ with complex multiplication. Write $w_{K}$
for the number of roots of unity contained in a number field $K$.

######  Conjecture 1 (Weston–Zaurova [14, Conjecture 3.1]).

Let $E$ be an elliptic curve over $\mathbb{Q}$ with complex multiplication by
an imaginary quadratic field $K$. Let $m$ be an integer coprime with $w_{K}$.
Then the density of primes $p\equiv 1\bmod m$ with $a_{p}(E)\neq 0$ for which
$a_{p}(E)$ is an $m^{th}$ power residue modulo $p$ is $1/m$. Furthermore, the
density is independent of any Chebotarev class $\mathcal{P}$ contained in the
set of primes $p\equiv 1\bmod m$ with $a_{p}(E)\neq 0$ in the sense that if
the primes are restricted to $\mathcal{P}$, the relative density remains
$1/m$.

Note that, for all but finitely many primes, $p\equiv 1\bmod m$ and the
reduction of $E$ at $p$ being ordinary is equivalent to $p$ splitting
completely in $K(\zeta_{m})$.

In [13] and [14], Weston and Zaurova prove the special cases of the conjecture
to which classical techniques are applicable. In particular, they settle the
case when $K=\mathbb{Q}(\sqrt{-3})$ and $m=3$ by using that $K$ contains the
third roots of unity so that the classical cubic reciprocity law is available.
For this reason, we exclude the field $K=\mathbb{Q}(\sqrt{-3})$ in the
statements of this paper.

In contrast, the classical techniques do not apply to the case
$K=\mathbb{Q}(i)$ and $m=3$, and therefore the original question for the curve
$E:y^{2}=x^{3}-x$ remained open. In this paper, we answer Ramakrishna’s
question in the affirmative, and prove that among the primes $p\equiv 1\bmod
3$ for which $E:y^{2}=x^{3}-x$ has ordinary reduction (equivalently, $p\equiv
1\bmod 4$ for $p\geq 5$), the density of $a_{p}(E)$ being a cubic residue
modulo $p$ is $1/3$. In addition, we prove Conjecture 1 for all elliptic
curves with complex multiplication (CM) by an imaginary quadratic field
$K=\mathbb{Q}(\sqrt{-d})$ where $d\neq 3$ in the case of cubic power residues
(i.e. $m=3$). We do not need the curves to be defined over $\mathbb{Q}$, nor
do we need them to have CM by a maximal order; it suffices that they have CM
by some order $\mathcal{O}$ in $K$ and are defined over the ring class field
$L$ of $\mathcal{O}$.

To prove the main theorems of this paper, we extend the analytic theory of
spin to cubic symbols over totally imaginary biquadratic fields. Our results
are _unconditional_ , as we do not need to assume any standard conjectures on
short character sums. This is atypical for results on spin symbols, which are
most often conditional on such conjectures.

######  Theorem 1.1.

Let $K$ be an imaginary quadratic field satisfying $\gcd(3,w_{K})=1$, or
equivalently $K\neq\mathbb{Q}(\zeta_{3})$. For any prime $p$ that splits
completely in $K$ as $(\pi\overline{\pi})$, where $\overline{\pi}$ is the
conjugate of $\pi$, and where $\pi$ lies below a prime ideal $\mathfrak{p}$ in
$K(\zeta_{3})$, there is a well-defined cubic spin symbol of the form

$[\mathfrak{p}]=\left(\frac{\overline{\pi}}{\mathfrak{p}}\right)_{K(\zeta_{3}),3}.$

Furthermore, there is a constant $C>0$ such that for all $X\geqslant 100$

$\left|\sum_{N_{K(\zeta_{3})/\mathbb{Q}}(\mathfrak{p})\leqslant
X}[\mathfrak{p}]\right|\leqslant CX^{1-\frac{1}{3200}}.$

One readily checks that the symbol does not depend on the choice of generator
$\pi$, see Proposition 2.2. The last part of Theorem 1.1 lies much deeper and
its proof occupies the majority of this paper.

As a corollary, for any elliptic curve $E$ with complex multiplication by an
imaginary quadratic field $K$, we prove asymptotics with a power saving error
term for the primes $p$ that split completely in $K(\zeta_{3})$ such that
$a_{p}(E)$ is a cubic residue modulo $p$. Let us now explain what we mean by
$a_{p}(E)$ in this context.

Let $E$ be an elliptic curve with CM by an order $\mathcal{O}$ in an imaginary
quadratic field $K$ and let $L$ be the ring class field of $\mathcal{O}$.
Assume that $E$ is defined over $L$. If we take a prime $\mathfrak{p}$ above
$p$ where $E$ has good reduction, we may introduce the quantity

$\displaystyle a_{\mathfrak{p}}(E)=p+1-|E(O_{L}/\mathfrak{p})|.$ (1.1)

If $\mathcal{O}^{\times}=\\{\pm 1\\}$ and if $\mathfrak{q}$ is another prime
of $L$ above $p$, it is a classical result of Deuring that
$a_{\mathfrak{p}}(E)^{2}=a_{\mathfrak{q}}(E)^{2}$, i.e. $a_{\mathfrak{p}}(E)$
is determined up to sign (for a reference, see our Lemma 2.1). Therefore
$a_{\mathfrak{p}}(E)$ being a cube modulo $p$ does not depend on the choice of
prime $\mathfrak{p}$. A somewhat more involved argument gives the same result
if $\mathcal{O}^{\times}=\\{\pm 1,\pm i\\}$, see Proposition 2.2.

Since it is possible to impose arbitrary congruence conditions when handling
spin symbols, we are able to get density results even when imposing arbitrary
_abelian splitting conditions_.

######  Corollary 1.2.

Let $E$ be an elliptic curve with complex multiplication by an order
$\mathcal{O}$ in an imaginary quadratic field $K$ different from
$\mathbb{Q}(\zeta_{3})$. Suppose $E$ is defined over the ring class field $L$
of $\mathcal{O}$. Then

$\frac{\\#\\{p\leqslant X,p\equiv 1\bmod 3,p\textup{ splits in
}L:a_{\mathfrak{p}}(E)\textup{ is a cube modulo }p\\}}{\\#\\{p\leqslant
X,p\equiv 1\bmod 3,p\textup{ splits in
}L\\}}=\frac{1}{3}+O\left(\frac{1}{X^{1/3200}}\right).$

Here $a_{\mathfrak{p}}(E)$ is defined as in equation (1.1) with $\mathfrak{p}$
an arbitrary prime of $L$ above $p$.

If $\mathcal{P}$ is a Chebotarev class of rational primes defined by a
splitting condition in an abelian extension of $K(\zeta_{3})$ containing $L$
and projecting trivially in $\mathrm{Gal}(L/\mathbb{Q})$, then the result
remains valid when the set of primes is restricted to $p$ with
$p\in\mathcal{P}$.

For an imaginary quadratic field $K$ and a prime $p$, we let $K^{(m)}(p)$ be
the maximal abelian extension of $K$ killed by $m$ and unramified away from
$p$. We prove the following density theorem on the splitting of primes $p$ in
the number fields $K^{(m)}(p)$ that depend on $p$. In the statement of Theorem
1.3, the splitting condition on $p$ is imposed to control the size of the
maximal abelian extension of $K$ unramified away from $p$.

Let $m$ be a prime and let $\zeta_{m}$ be a primitive $m^{\operatorname{th}}$
root of unity. Let $F$ be a number field, and define

$V_{\emptyset}=\\{x\in F^{\times}:(x)=J^{m}\text{ for some fractional ideal
}J\\},$

where $(x)$ is the fractional ideal generated by $x\in F^{\times}$. We define
the $m$-governing field of $F$ to be the field
$F(\zeta_{m},\sqrt[m]{V_{\emptyset}}).$

######  Theorem 1.3 (Splitting in number fields that depend on the prime).

Let $K$ be an imaginary quadratic field and let $m\geq 3$ be a prime such that
$\gcd(m,w_{K})=1$. Let $\mathcal{O}$ be an order in $K$, and let $p$ be a
prime that splits completely in the ring class field of $\mathcal{O}$ and the
$m$-governing field of $K$. Let $K^{(m)}(p)$ be the maximal $m$-elementary
abelian extension of $K$ unramified away from $p$, and let
$f(p,K^{(m)}(p)/\mathbb{Q})$ be the inertial degree at $p$ in
$K^{(m)}(p)/\mathbb{Q}$. Then $[\mathfrak{p}]=1$ if and only if
$f(p,K^{(m)}(p)/\mathbb{Q})=1$.

In particular, for $m=3$, among the primes $p$ splitting completely in the
ring class field of $\mathcal{O}$ and the $3$-governing field, the set of
primes $p$ of residue field degree $1$ at any prime in $K^{(3)}(p)$ above $p$
has density $1/3$.

In this theorem, we are using implicitly that the density results on spin
symbols remain valid when the class of primes considered is restricted to
Chebotarev classes defined by Chebotarev conditions in abelian extensions of
$K(\zeta_{3})$. We expect that it is possible to prove the Weston–Zaurova
conjecture for all power residues $m$ by applying techniques from the joint
distribution of spins from Koymans–Milovic [4]. However, such results would be
conditional on short character sum conjectures.

Below, we state a corollary of Theorem 1.1.

######  Corollary 1.4.

Let $\ell$ be a prime, and let

$\rho:\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\longrightarrow\operatorname{GL}(2,\mathbb{Z}_{\ell})$

be a Galois representation arising from the Galois action on
$\ell^{\infty}$-torsion points of an elliptic curve over $\mathbb{Q}$ with
complex multiplication by an imaginary quadratic field $\mathbb{Q}(\sqrt{-d})$
where $d\neq-3$. Let $S$ be the smallest set of places containing $\ell$ and
the archimedean place of $\mathbb{Q}$ such that $\rho$ is unramified outside
$S$. For primes $p\not\in S$, let $\operatorname{Frob}_{p}$ be any lift of
Frobenius to $\mathrm{Gal}(\overline{\mathbb{Q}}_{p}/\mathbb{Q}_{p})$. Among
the primes $p\equiv 1\bmod 3$ splitting completely in the ring class field of
$\operatorname{End}_{\mathbb{C}}(E)$,

$\operatorname{trace}(\rho(\operatorname{Frob}_{p}))\in\mathbb{Z}$

is a cube modulo $p$ one third of the time.

Our method of proof of Theorem 1.1 relies on a version of Vinogradov’s sieve
extended to totally imaginary biquadratic fields. In [2],
Friedlander–Iwaniec–Mazur–Rubin introduced quadratic spin symbols of primes in
totally real, cyclic number fields (satisfying some additional technical
conditions) and proved equidistribution of such symbols by extending
Vinogradov’s sieve to totally real, cyclic fields. In general, their results
are conditional on standard conjectures on short character sums (denoted
$C_{n}$ in their paper where $n$ is the degree of the totally real cyclic
number field), and their results are only unconditional in the case $n=3$. In
this paper, we extend the methods in [2] to _cubic_ spin symbols over
biquadratic, totally imaginary fields of degree 4 over $\mathbb{Q}$. It is
natural to compare this to work of Merikoski [6], although we should warn the
reader that his spin symbol is not the same as ours.

Our main analytic accomplishment is to make our equidistribution results
unconditional in this setting. To make our results unconditional, we follow a
similar strategy strategy as employed previously by Koymans–Milovic [3] and
Piccolo [8]. However, unlike the works [3, 8], we do not assume that our
biquadratic field has class number $1$ and we deal with cubic symbols instead
of quartic symbols.

###  1.1 Acknowledgements

The authors are thankful to Ravi Ramakrishna and Fabien Pazuki for helpful
conversations. The first author gratefully acknowledges the support of Dr. Max
Rössler, the Walter Haefner Foundation and the ETH Zürich Foundation.

##  2 General spin symbols

In this section we introduce our cubic spin symbol of interest, and relate it
to $a_{p}(E)$. This symbol is a natural generalization of the spin symbol
introduced in [2]. In fact, it turns out that the algebraic part of our theory
readily generalizes to all $m$, not just $m=3$. We start by identifying a
family of spin symbols, indexed by $m$, capturing whether the error terms
$a_{p}(E)$ are power residues modulo $m$. This is achieved in Proposition 2.2.
We will then relate this to splitting of $p$ in a number field depending on
$p$, see Lemma 2.6.

###  2.1 Power residues of degree $m$ and spin

Let $K$ be an imaginary quadratic field with maximal order $\mathcal{O}_{K}$.
For any order $\mathcal{O}$ of conductor $f$ in $K$, there is a unique abelian
extension $L/K$ called the ring class field of $\mathcal{O}$. We refer to [1,
p. 179] for the definition of ring class fields.

We write $\sigma$ for the non-trivial element of $\mathrm{Gal}(K/\mathbb{Q})$.
To keep our notation brief, we shall frequently use conjugation to denote this
automorphism, so that we may speak of $\bar{\kappa}$ for elements $\kappa$ but
also of $\bar{\mathfrak{p}}$ for (prime) ideals $\mathfrak{p}$.

######  Lemma 2.1.

Let $K$ be an imaginary quadratic field and let $\mathcal{O}$ be an order in
$K$. Let $L$ be the ring class field of $\mathcal{O}$, and let $E$ be an
elliptic curve over $L$ with $\operatorname{End}_{\mathbb{C}}(E)=\mathcal{O}$.
Let $p$ be a prime that splits completely in $L$ and $\mathfrak{p}$ be a prime
in $L$ above $p$ (in particular, $O_{L}/\mathfrak{p}\simeq\mathbb{F}_{p}$).
Suppose $E$ has good reduction at $\mathfrak{p}$. Then there is
$\kappa\in\mathcal{O}$ such that $p=\kappa\overline{\kappa}$ and

$a_{\mathfrak{p}}(E)=p+1-|E(O_{L}/\mathfrak{p})|=\kappa+\overline{\kappa}.$

###### Proof.

See [1, Theorem 14.16]. ∎

######  Proposition 2.2.

Let $K=\mathbb{Q}(\sqrt{D})$ be an imaginary quadratic field and let $m\geq 3$
be any odd integer such that $\gcd(m,w_{K})=1$. Let
$M=\mathbb{Q}(\sqrt{D},\zeta_{m})$ and let

$\mathrm{Gal}(M/\mathbb{Q})\simeq\mathrm{Gal}(K/\mathbb{Q})\times\mathrm{Gal}(\mathbb{Q}(\zeta_{m})/\mathbb{Q})=\\{1,\sigma\\}\times(\mathbb{Z}/m\mathbb{Z})^{\times}.$

Let $\mathcal{O}$ be an order in $K$ and let $E$ be an elliptic curve defined
over the ring class field of $\mathcal{O}$ with
$\operatorname{End}_{\mathbb{C}}(E)=\mathcal{O}$. Let $p\equiv 1\bmod m$ be a
prime that splits completely in the ring class field of $\mathcal{O}$.
According to Lemma 2.1, there exists $\kappa\in\mathcal{O}$ such that

$p=\kappa\overline{\kappa}$

and

$a_{\mathfrak{P}}(E)=\kappa+\overline{\kappa},$

where $\mathfrak{P}$ is a prime of $L$ above $p$.

Let $\mathfrak{q}=(\kappa)$ and let $\mathfrak{p}$ be a prime in $M$ above
$\mathfrak{q}$.

${\underbrace{\prod_{\tau\in\mathrm{Gal}(M/K)}\tau(\mathfrak{p})}_{\mathfrak{q}\mathcal{O}_{M}}\underbrace{\prod_{\tau\in\mathrm{Gal}(M/K)}\tau(\sigma(\mathfrak{p}))}_{\overline{\mathfrak{q}}\mathcal{O}_{M}}}$${M}$${\underbrace{\mathfrak{q}\overline{\mathfrak{q}}}_{p\mathcal{O}}}$${\mathbb{Q}(\zeta_{m})}$${K}$${p}$${\mathbb{Q}}$$\scriptstyle{\langle\sigma\rangle}$$\scriptstyle{(\mathbb{Z}/m\mathbb{Z})^{\times}}$

Define the spin symbol

$[\mathfrak{p}]=\left(\frac{\overline{\kappa}}{\mathfrak{p}}\right)_{M,m}=\left(\frac{\sigma(\kappa)}{\mathfrak{p}}\right)_{M,m}.$

If $u\in\mathcal{O}^{\times},$ then

$\left(\frac{\overline{u\kappa}}{\mathfrak{p}}\right)_{M,m}=\left(\frac{\overline{\kappa}}{\mathfrak{p}}\right)_{M,m},$

so the spin symbol is well-defined, i.e. is independent on the choice of
generator $\kappa$ for $\mathfrak{q}$. Furthermore, $[\mathfrak{p}]=1$ if and
only if $a_{\mathfrak{P}}(E)$ is an $m^{\operatorname{th}}$ power residue
modulo $p$.

###### Proof.

Since $p\equiv 1\bmod m$, $p$ splits completely in $\mathbb{Q}(\zeta_{m})$.
Since we assume $p$ splits completely in $K,$ we conclude that $p$ splits
completely in $M$. Since $\mathfrak{p}$ lies above $\mathfrak{q},$
$\sigma(\mathfrak{p})$ lies above
$\sigma(\mathfrak{q})=\overline{\mathfrak{q}}$. In $M$, the primes
$\mathfrak{q}$ and $\overline{\mathfrak{q}}$ decompose as follows:

$\mathfrak{q}\mathcal{O}_{M}=\prod_{\tau\in\mathrm{Gal}(M/K)}\tau(\mathfrak{p}),\quad\overline{\mathfrak{q}}\mathcal{O}_{M}=\sigma(\mathfrak{q})\mathcal{O}_{M}=\prod_{\tau\in\mathrm{Gal}(M/K)}\tau(\sigma(\mathfrak{p})).$

Since $\gcd(m,w_{K})=1$ by assumption, we see that all elements of
$\mathcal{O}^{\times}$ are $m^{\operatorname{th}}$ powers in $K$. We conclude
that the spin symbol $[\mathfrak{p}]$ is independent of the choice of
$\kappa$. Using that $\kappa\equiv 0\bmod\mathfrak{p}$, we have by Lemma 2.1

$[\mathfrak{p}]=\left(\frac{\overline{\kappa}}{\mathfrak{p}}\right)_{M,m}=\left(\frac{\overline{\kappa}+\kappa}{\mathfrak{p}}\right)_{M,m}=\left(\frac{a_{\mathfrak{P}}(E)}{\mathfrak{p}}\right)_{M,m}.$

Hence $[\mathfrak{p}]=1$ if and only if $a_{\mathfrak{P}}(E)$ is an
$m^{\operatorname{th}}$ power residue modulo $\mathfrak{p}$. Now,
$\mathcal{O}_{M}/\mathfrak{p}\simeq\mathbb{F}_{p},$ so $a_{\mathfrak{P}}(E)$
is an $m^{\operatorname{th}}$ power modulo $\mathfrak{p}$ if and only if
$a_{\mathfrak{P}}(E)$ is an $m^{\operatorname{th}}$ power modulo $p$. ∎

###  2.2 Spin and inertial degrees of ramified primes

Let $m$ be a prime. For any set $S$ of tame places in a number field $F$,
define

$V_{S}=\\{x\in F^{\times}:(x)=J^{m}\text{ for some fractional ideal }J,\text{
and }x\in F_{v}^{\times m}\text{ }\forall v\in S\\}.$

Note that, by definition, $F^{\times m}\subseteq V_{S}$. Let
$\mathcal{O}_{F}^{\times}$ be the units in $F$ and $\operatorname{Cl}_{F}[m]$
the $m$-torsion in the class group of $F$. There is an exact sequence

$1\longrightarrow\mathcal{O}_{F}^{\times}\otimes\mathbb{F}_{m}\longrightarrow
V_{\emptyset}/F^{\times
m}\longrightarrow\operatorname{Cl}_{F}[m]\longrightarrow 1.$

The $m$-governing field of $F$ is the field
$F(\zeta_{m},\sqrt[m]{V_{\emptyset}})$, where $\zeta_{m}$ is a primitive
$m^{\operatorname{th}}$ root of unity.

Recall that $K^{(m)}(p)$ denotes the maximal $m$-elementary abelian extension
of $K$ unramified away from the primes above $p$. Write $H_{K}^{(m)}$ for the
maximal unramified $m$-elementary abelian extension of $K$.

######  Lemma 2.3.

Let $K$ be an imaginary quadratic field and let $m$ be a prime. Let $p$ be a
prime that splits completely in the $m$-governing field of $K$, and factor
$pO_{K}=\mathfrak{q}\overline{\mathfrak{q}}$. Then there exists a
$\mathbb{Z}/m\mathbb{Z}$-extension $K^{(m)}(\mathfrak{q})/K$ ramified exactly
at $\mathfrak{q}$ and a $\mathbb{Z}/m\mathbb{Z}$-extension
$K^{(m)}(\overline{\mathfrak{q}})/K$ ramified exactly at
$\overline{\mathfrak{q}}$. For any choice of $K^{(m)}(\mathfrak{q})/K$ and
$K^{(m)}(\overline{\mathfrak{q}})/K$ as above, we have

$K^{(m)}(p)=K^{(m)}(\mathfrak{q})K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)},$

and thus $[K^{(m)}(p):\mathbb{Q}]=2m^{2}\cdot|\mathrm{Cl}(K)[m]|$.
Diagramatically

${K^{(m)}(p)}$${K^{(m)}(\mathfrak{q})}$${H_{K}^{(m)}}$${K^{(m)}(\overline{\mathfrak{q}})}$${K}$${\mathbb{Q}.}$$\scriptstyle{\mathbb{Z}/m\mathbb{Z}}$$\scriptstyle{\mathbb{Z}/m\mathbb{Z}}$

###### Proof.

Since $p$ splits completely in the $m$-governing field and since the
$m$-governing field is ramified at $m$, it follows that $p\neq m$. Therefore,
all primes of $K$ above $p$ must be tamely ramified in $K^{(m)}(p)/K$.

By the tame Gras–Munnier theorem, there exists a
$\mathbb{Z}/m\mathbb{Z}$-extension of $K$ ramified exactly at $\mathfrak{q}$.
Likewise, there exists a $\mathbb{Z}/m\mathbb{Z}$-extension of $K$ ramified
exactly at $\overline{\mathfrak{q}}$. The compositum
$K^{(m)}(\mathfrak{q})K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}$ is an
$m$-elementary abelian extension of $K$ unramified away from $p$ with degree
$2m^{2}\cdot|\mathrm{Cl}(K)[m]|$. Evidently,
$K^{(m)}(\mathfrak{q})K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}$ is the
maximal such extension and hence
$K^{(m)}(p)=K^{(m)}(\mathfrak{q})K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}$.
∎

######  Lemma 2.4.

Let $K$ be an imaginary quadratic field and let $m$ be a prime. Let $p$ be a
prime that splits completely in the $m$-governing field of $K$. There are two
possibilities for the standard $(e,f,g)$-decomposition of $p$ in $K^{(m)}(p)$:

$e(p,K^{(m)}(p)/\mathbb{Q})$ | $f(p,K^{(m)}(p)/\mathbb{Q})$ | $g(p,K^{(m)}(p)/\mathbb{Q})$ | $[K^{(m)}(p):\mathbb{Q}]$
---|---|---|---
$m$ | $1$ | $2m\cdot|\mathrm{Cl}(K)[m]|$ | $2m^{2}\cdot|\mathrm{Cl}(K)[m]|$
$m$ | $m$ | $2\cdot|\mathrm{Cl}(K)[m]|$ | $2m^{2}\cdot|\mathrm{Cl}(K)[m]|$
Table 1: The $(e,f,g)$-decomposition of $p$ in $K^{(m)}(p)/\mathbb{Q}$

###### Proof.

Because $p$ is unramified in $K$ and tamely ramified in $K^{(m)}(p)/K$, and
because tame ramification is cyclic, we have $e(p,K^{(m)}(p)/\mathbb{Q})=m$.
Since $f(p,K/\mathbb{Q})=1,$ and since $\mathrm{Gal}(K^{(m)}(p)/K)$ is a
vector space over $\mathbb{F}_{m}$, it follows that
$f(p,K^{(m)}(p)/\mathbb{Q})$ is either 1 or $m$. By Lemma 2.3, we have

$e(p,K^{(m)}(p)/\mathbb{Q})f(p,K^{(m)}(p)/\mathbb{Q})g(p,K^{(m)}(p)/\mathbb{Q})=2m^{2}\cdot|\mathrm{Cl}(K)[m]|,$

and the result follows. ∎

######  Lemma 2.5.

Let $m$ be a prime. Suppose $p$ splits completely in the $m$-governing field
of a quadratic imaginary field $K$ and in $H_{K}^{(m)}$. Then we have for any
choice of $K^{(m)}(\overline{\mathfrak{q}})$

$f(p,K^{(m)}(p)/\mathbb{Q})=1\iff
f(\mathfrak{q},K^{(m)}(\overline{\mathfrak{q}})/K)=1.$

###### Proof.

Note that the compositum $K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}$ does
not depend on the choice of $K^{(m)}(\overline{\mathfrak{q}})/K$ as it is the
maximal $m$-elementary abelian extension of $K$ unramified away from
$\mathfrak{q}$. Since $p$ is assumed to split completely in $H_{K}^{(m)}$, we
conclude that $f(\mathfrak{q},K^{(m)}(\overline{\mathfrak{q}})/K)$ is
independent of this choice.

Since $f(p,K/\mathbb{Q})=1$, we have
$f(p,K^{(m)}(p)/\mathbb{Q})=f(\mathfrak{q},K^{(m)}(p)/K)$. Because
$K^{(m)}(p)=K^{(m)}(\mathfrak{q})K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}$
and because $p$ splits completely in $H_{K}^{(m)}$, we conclude that

$\displaystyle f(p,K^{(m)}(p)/\mathbb{Q})=1$ $\displaystyle\iff
f(\mathfrak{q},K^{(m)}(p)/K)=1$ $\displaystyle\iff
f(\mathfrak{q},K^{(m)}(\mathfrak{q})/K)=1\text{ and
}f(\mathfrak{q},K^{(m)}(\overline{\mathfrak{q}})/K)=1.$ (2.1)

Since $e(\mathfrak{q},K^{(m)}(\mathfrak{q})/K)=m$, the condition
$f(\mathfrak{q},K^{(m)}(\mathfrak{q})/K)=1$ is automatically satisfied.
Removing it from the equivalence (2.2) gives the lemma. ∎

######  Lemma 2.6.

Let $K$ be an imaginary quadratic field and let $m\geq 3$ be any odd integer
such that $\gcd(m,w_{K})=1$. Set $M:=K(\zeta_{m})$. Let $\mathcal{O}$ be an
order in $K$. Let $p$ be a prime that splits completely in the ring class
field of $\mathcal{O}$ and the $m$-governing field of $K$, and let
$\kappa\in\mathcal{O}$ with $p=\kappa\overline{\kappa}$. Let
$\mathfrak{q}=(\kappa)$, $\mathfrak{p}$ a prime in $M$ above $\mathfrak{q}$
and $[\mathfrak{p}]=(\overline{\kappa}/\mathfrak{p})_{M,m}$. Then we have for
any choice of $K^{(m)}(\overline{\mathfrak{q}})$

$f(\mathfrak{q},K^{(m)}(\overline{\mathfrak{q}})/K)=1\iff[\mathfrak{p}]=1.$

###### Proof.

Let $G(\overline{\mathfrak{q}})$ be the full ray class group of modulus
$\overline{\mathfrak{q}}$ above $K$ and let $K(\overline{\mathfrak{q}})$ be
the full ray class field of modulus $\overline{\mathfrak{q}}$ above $K$. If
$\psi_{K(\overline{\mathfrak{q}})/K}$ is the Artin reciprocity map, there is a
commutative diagram

${G(\overline{\mathfrak{q}})}$${\mathrm{Gal}(K(\overline{\mathfrak{q}})/K)}$${G(\overline{\mathfrak{q}})/G(\overline{\mathfrak{q}})^{m}}$${\mathrm{Gal}(K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}/K),}$$\scriptstyle{\psi_{K(\overline{\mathfrak{q}})/K}}$

where the horizontal arrows are isomorphisms. Let $\operatorname{Cl}_{K}$ be
the class group of $K$. By Theorem 1.7, Chapter V, in [7], there is a
commutative diagram

${1}$${(\mathcal{O}/\overline{\mathfrak{q}})^{\times}/\mathcal{O}^{\times}}$${G(\overline{\mathfrak{q}})}$${\operatorname{Cl}_{K}}$${1}$${1}$${(\mathcal{O}/\overline{\mathfrak{q}})^{\times}/(\mathcal{O}/\overline{\mathfrak{q}})^{\times
m}}$${G(\overline{\mathfrak{q}})/G(\overline{\mathfrak{q}})^{m}}$${\operatorname{Cl}_{K}/\operatorname{Cl}_{K}^{m}}$${1}$$\scriptstyle{\varphi}$

with exact horizontal rows. Here we used that
$\mathcal{O}^{\times}\subseteq(\mathcal{O}/\overline{\mathfrak{q}})^{\times
m}$, which follows from our assumption $\gcd(m,w_{K})=1$ as then every element
of $\mathcal{O}^{\times}$ must be an $m^{\operatorname{th}}$ power.

By definition, $\varphi(\kappa)=\mathfrak{q}$. Observe that $p$ certainly
splits in $H_{K}^{(m)}$, as $H_{K}^{(m)}$ is contained in the ring class field
of $\mathcal{O}$. Therefore we conclude that

$\displaystyle f(\kappa,K^{(m)}(\overline{\mathfrak{q}})/K)=1$
$\displaystyle\Longleftrightarrow
f(\kappa,K^{(m)}(\overline{\mathfrak{q}})H_{K}^{(m)}/K)=1$
$\displaystyle\Longleftrightarrow\kappa\text{ is an
}m^{\operatorname{th}}\text{ power modulo }\overline{\mathfrak{q}}.$

Now, $\kappa$ is an $m^{\operatorname{th}}$ power modulo
$\overline{\mathfrak{q}}$ if and only if $\kappa$ is an
$m^{\operatorname{th}}$ power modulo any prime in $\mathcal{O}_{M}$ above
$\overline{\mathfrak{q}}$ if and only if

$\left(\frac{\kappa}{\sigma(\mathfrak{p})}\right)_{M,m}=1.$

Since $\mathbb{Q}(\zeta_{m})$ is the fixed field of $\sigma$, we have the
equalities

$1=\left(\frac{\kappa}{\sigma(\mathfrak{p})}\right)_{M,m}\Longleftrightarrow
1=\sigma\left(\frac{\kappa}{\sigma(\mathfrak{p})}\right)_{M,m}=\left(\frac{\sigma(\kappa)}{\sigma^{2}(\mathfrak{p})}\right)_{M,m}=\left(\frac{\sigma(\kappa)}{\mathfrak{p}}\right)_{M,m}=\left(\frac{\overline{\kappa}}{\mathfrak{p}}\right)_{M,m}=[\mathfrak{p}],$

as desired. ∎

##  3 Analytic prerequisites

In this section we set up the analytic machinery that we will use in the next
two sections. Our approach is based on [3].

###  3.1 Cubic residue symbols

We shall briefly recall the required theory of cubic residue symbols that we
need. Let $K$ be a number field always containing a fixed primitive third root
of unity that we denote $\zeta_{3}$. For $\alpha\in O_{K}$ and $\mathfrak{p}$
a prime of $K$ not dividing $3$, we define
$\left(\frac{\alpha}{\mathfrak{p}}\right)_{K,3}$ as the unique element in
$\\{1,\zeta_{3},\zeta_{3}^{2},0\\}$ satisfying

$\left(\frac{\alpha}{\mathfrak{p}}\right)_{K,3}\equiv\alpha^{\frac{N_{K/\mathbb{Q}}(\mathfrak{p})-1}{3}}\bmod\mathfrak{p}.$

We multiplicatively extend this to all ideals $I$ coprime to $3$. We shall
need the following weak form of cubic reciprocity.

######  Proposition 3.1.

Let $M$ be a number field containing $\zeta_{3}$. Take elements
$\alpha,\beta\in O_{M}$ and assume that $\beta$ is coprime to $3$. Then the
cubic residue symbol $(\alpha/\beta)_{M,3}$ depends only on the congruence
class of $\beta$ modulo $27\alpha O_{M}$. Moreover, if $\alpha$ is also
coprime to $3$, we have

$\left(\frac{\alpha}{\beta}\right)_{M,3}=\mu\cdot\left(\frac{\beta}{\alpha}\right)_{M,3},$

where $\mu$ depends only on the congruence classes of $\alpha$ and $\beta$
modulo $27O_{M}$.

###  3.2 Definition of the spin symbol

Let $K=\mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic field with $d>0$
squarefree. We will assume that $d\neq 3$. The field $M:=K(\zeta_{3})$ will
play an important role throughout the paper. We fix the following data
associated to $M$

  * •

we let $\\{\eta_{1},\dots,\eta_{4}\\}$ be an integral basis for $O_{M}$ with
$\eta_{1}=1$,

  * •

the torsion subgroup of $O_{M}^{\ast}$ is generated by $\zeta_{12}$ if $d=1$,
and otherwise the torsion subgroup is generated by $\zeta_{6}$,

  * •

we pick two disjoint collections of prime ideals
$\mathfrak{p}_{1},\dots,\mathfrak{p}_{h}$ and
$\mathfrak{q}_{1},\dots,\mathfrak{q}_{h}$ coprime to $3$, where each
collection is a set of representatives for the class group $\mathrm{Cl}(M)$ of
$M$. We let

$\displaystyle\mathfrak{f}:=\prod_{i=1}^{h}\mathfrak{p}_{i}\prod_{j=1}^{h}\mathfrak{q}_{j},\quad\quad
f:=N_{M/\mathbb{Q}}(\mathfrak{f}).$ (3.1)

Then $\mathfrak{f}$ is principal. Furthermore, we may choose
$\mathfrak{p}_{1},\dots,\mathfrak{p}_{h},\mathfrak{q}_{1},\dots,\mathfrak{q}_{h}$
in such a way that the norm $f$ of $\mathfrak{f}$ is squarefree.

The field $M$ is a Galois extension of $\mathbb{Q}$ with Galois group
isomorphic to the Klein four group, say $\\{1,\sigma,\tau,\sigma\tau\\}$,
which we depict diagrammatically as

$\leavevmode\hbox to170.61pt{\vbox
to86.74pt{\pgfpicture\makeatletter\hbox{\hskip
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to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{}{}{}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-85.30515pt}{-43.36806pt}\pgfsys@invoke{
}\hbox{\vbox{\halign{\pgf@matrix@init@row\pgf@matrix@step@column{\pgf@matrix@startcell#\pgf@matrix@endcell}&#\pgf@matrix@padding&&\pgf@matrix@step@column{\pgf@matrix@startcell#\pgf@matrix@endcell}&#\pgf@matrix@padding\cr\hfil\hskip
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}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}}}&\hskip
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\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} {}{ {}{}{}}{}{ {}{}{}}
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}\pgfsys@setlinewidth{0.39998pt}\pgfsys@invoke{
}{}{}{}{{}}{}{}{{}}\pgfsys@moveto{-53.76344pt}{2.0122pt}\pgfsys@lineto{-18.59746pt}{20.24449pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}}}{{}{{}}{}{}{{}}{{{}}}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ }}{ } {{}{{}}}{{}{}}{}{{}{}} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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Let $\mathfrak{p}$ be a completely split prime ideal of $M$. If
$\mathfrak{p}\tau(\mathfrak{p})$ is a principal ideal of $K$, we define a
symbol

$[\mathfrak{p}]:=\left(\frac{\sigma(\pi)}{\mathfrak{p}}\right)_{M,3}\in\\{1,\zeta_{3},\zeta_{3}^{2}\\},\quad\quad(\pi)O_{K}=\mathfrak{p}\tau(\mathfrak{p})O_{K}.$

This does not depend on the choice of generator $\pi$ of the $O_{K}$-ideal
$\mathfrak{p}\tau(\mathfrak{p})O_{K}$, as changing $\pi$ by a unit changes the
total symbol by $\left(u/\mathfrak{p}\right)_{M,3}=1$ with $u\in\\{\pm 1,\pm
i\\}$ for $d=1$ and $u\in\\{\pm 1\\}$ for $d>1$. One directly checks that

$\sum_{\rho\in\mathrm{Gal}(M/\mathbb{Q})}[\rho(\mathfrak{p})]=\begin{cases}-2&\textup{if
}[\mathfrak{p}]\in\\{\zeta_{3},\zeta_{3}^{2}\\}\\\ 4&\textup{if
}[\mathfrak{p}]=1.\end{cases}$

More generally, if $\mathfrak{a}$ is any integral ideal of $O_{M}$ such that
$\mathfrak{a}\tau(\mathfrak{a})$ is a principal ideal of $K$, we define

$[\mathfrak{a}]:=\begin{cases}\left(\frac{\sigma(\alpha)}{\mathfrak{a}}\right)_{M,3}&\text{if
}\gcd(\mathfrak{a},(3))=1\\\ 0&\text{otherwise,}\end{cases}$

where $\alpha O_{K}$ is any generator of
$\mathfrak{a}\tau(\mathfrak{a})O_{K}$. This is once more independent of the
choice of generator $\alpha$ of the ideal $\mathfrak{a}$.

###  3.3 Field lowering

The coming results are variations of the results in [3, Subsection 3.2]. As
these results play a key role in making our results unconditional, we shall
provide full details for these variations. As the results in this subsection
hold in significant generality, we allow $K$ to be an arbitrary number field
in this subsection only, and we shall return to the setting
$K=\mathbb{Q}(\sqrt{-d})$ afterwards.

######  Lemma 3.2.

Let $K$ be a number field and let $\mathfrak{p}$ be a prime of $K$ coprime to
$3$. Assume that $L$ is a quadratic extension of $K$ such that $L$ contains
$\zeta_{3}$ and $\mathfrak{p}$ splits in $L$. Write $\sigma$ for the non-
trivial element of $\mathrm{Gal}(L/K)$. Then we have for all $\alpha\in O_{K}$

$\left(\frac{\alpha}{\mathfrak{p}O_{L}}\right)_{L,3}=\begin{cases}\left(\frac{\alpha}{\mathfrak{p}O_{K}}\right)_{K,3}^{2}&\textup{if
}\sigma\textup{ fixes }\zeta_{3}\\\
\mathbf{1}_{\mathfrak{p}\nmid\alpha}&\textup{if }\sigma\textup{ does not fix
}\zeta_{3}.\end{cases}$

###### Proof.

Since $\mathfrak{p}$ splits in $L$, we may write
$\mathfrak{p}O_{L}=\mathfrak{q}\sigma(\mathfrak{q})$ for some prime
$\mathfrak{q}$ of $L$. This gives

$\left(\frac{\alpha}{\mathfrak{p}O_{L}}\right)_{L,3}=\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}\left(\frac{\alpha}{\sigma(\mathfrak{q})}\right)_{L,3}=\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}\left(\frac{\sigma(\alpha)}{\sigma(\mathfrak{q})}\right)_{L,3}=\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}\sigma\left(\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}\right),$

where we used that $\sigma(\alpha)=\alpha$ thanks to the assumption $\alpha\in
O_{K}$. If $\sigma$ fixes $\zeta_{3}$, then the above becomes

$\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}\sigma\left(\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}\right)=\left(\frac{\alpha}{\mathfrak{q}}\right)_{L,3}^{2}=\left(\frac{\alpha}{\mathfrak{p}O_{K}}\right)_{K,3}^{2}.$

If $\sigma$ does not fix $\zeta_{3}$, then we get
$\mathbf{1}_{\mathfrak{p}\nmid\alpha}$ by checking all possibilities for the
cubic residue symbol
$(\alpha/\mathfrak{q})_{L,3}\in\\{1,\zeta_{3},\zeta_{3}^{2},0\\}$. ∎

######  Lemma 3.3.

Let $K$ be a number field and let $\mathfrak{p}$ be a prime of $K$ coprime to
$3$. Assume that $L$ is a quadratic extension of $K$ such that $L$ contains
$\zeta_{3}$ and assume that $\mathfrak{p}$ stays inert in $L$. Further assume
that $\mathfrak{p}$ has degree $1$ in $K$ and let $p$ be the prime of
$\mathbb{Q}$ lying below $\mathfrak{p}$. Then we have for all $\alpha\in
O_{K}$

$\left(\frac{\alpha}{\mathfrak{p}O_{L}}\right)_{L,3}=\begin{cases}\left(\frac{\alpha}{\mathfrak{p}O_{K}}\right)_{K,3}^{2}&\textup{if
}p\equiv 1\bmod 3\\\ \mathbf{1}_{\mathfrak{p}\nmid\alpha}&\textup{if }p\equiv
2\bmod 3.\end{cases}$

###### Proof.

We have the identities

$\left(\frac{\alpha}{\mathfrak{p}O_{L}}\right)_{L,3}\equiv\alpha^{\frac{N_{L/\mathbb{Q}}(\mathfrak{p})-1}{3}}\equiv\alpha^{\frac{p^{2}-1}{3}}.$

If $p\equiv 1\bmod 3$, we rewrite this as

$\alpha^{\frac{p^{2}-1}{3}}\equiv\left(\alpha^{\frac{p-1}{3}}\right)^{p+1}\equiv\left(\alpha^{\frac{N_{K/\mathbb{Q}}(\mathfrak{p})-1}{3}}\right)^{p+1}\equiv\left(\frac{\alpha}{\mathfrak{p}O_{K}}\right)_{K,3}^{p+1}\equiv\left(\frac{\alpha}{\mathfrak{p}O_{K}}\right)_{K,3}^{2}\bmod\mathfrak{p}.$

If $p\equiv 2\bmod 3$, we instead rewrite this as

$\alpha^{\frac{p^{2}-1}{3}}\equiv\left(\alpha^{p-1}\right)^{\frac{p+1}{3}}\equiv\left(\alpha^{N_{K/\mathbb{Q}}(\mathfrak{p})-1}\right)^{\frac{p+1}{3}}\equiv\mathbf{1}_{\mathfrak{p}\nmid\alpha}^{\frac{p+1}{3}}\equiv\mathbf{1}_{\mathfrak{p}\nmid\alpha}\bmod\mathfrak{p},$

as desired. ∎

######  Lemma 3.4.

Let $K$ be a number field and let $L$ be a quadratic extension of $K$. Write
$\sigma$ for the non-trivial element of $\mathrm{Gal}(L/K)$. Assume that
$\mathfrak{p}$ is a prime ideal of $K$ that stays unramified in $L$. Further
suppose that $\beta\in O_{L}$ satisfies
$\beta\equiv\sigma(\beta)\bmod\mathfrak{p}O_{L}$. Then there is
$\beta^{\prime}\in O_{K}$ such that
$\beta^{\prime}\equiv\beta\bmod\mathfrak{p}O_{L}$.

###### Proof.

This is proven in [3, Lemma 3.4]. ∎

###  3.4 A fundamental domain

Given an integral ideal $\mathfrak{a}$ of $M$, there are many possible
generators $\alpha$ of $\mathfrak{a}$. For our analytic purposes, it will be
useful to pick these generators in a systematic way. To this end, we decompose
$O_{M}^{\ast}$ as $T\times\langle\epsilon\rangle$, where $T$ is the torsion
subgroup and where $\epsilon$ is a fundamental unit. We write $V$ for the
group generated by $\epsilon$.

Using our fixed integral basis $\\{\eta_{1},\eta_{2},\eta_{3},\eta_{4}\\}$ of
$M$, we get an isomorphism, in the category of vector spaces,
$\iota:\mathbb{Q}^{4}\rightarrow M$ by sending $(a_{1},\dots,a_{4})$ to
$a_{1}\eta_{1}+a_{2}\eta_{2}+a_{3}\eta_{3}+a_{4}\eta_{4}$. We also view
$\mathbb{Q}^{4}$ as a subset of $\mathbb{R}^{4}$ in the natural way. For a
subset $S$ of $\mathbb{R}^{4}$ and an element
$\alpha=a_{1}\eta_{1}+a_{2}\eta_{2}+a_{3}\eta_{3}+a_{4}\eta_{4}\in M$, we will
abuse notation by writing $\alpha\in S$ to mean $(a_{1},\dots,a_{4})\in S$.
Given a subset $S$ of $\mathbb{R}^{4}$ and a real number $X>0$, we define
$S(X)$ to be the subset of $(s_{1},\dots,s_{4})\in S$ satisfying
$f(s_{1},\dots,s_{4})\leq X$, where $f$ is the norm polynomial
$f(X_{1},\dots,X_{4})=N_{M/\mathbb{Q}}(\eta_{1}X_{1}+\eta_{2}X_{2}+\eta_{3}X_{3}+\eta_{4}X_{4})\in\mathbb{Z}[X_{1},\dots,X_{4}]$.
We write $\mathcal{D}$ for the fundamental domain constructed in [3, Section
3.3]. The main properties of this fundamental domain can be found in [3, Lemma
3.5] that we summarize now.

######  Lemma 3.5.

There exists a subset $\mathcal{D}\subseteq\mathbb{R}^{4}$ with the following
properties

  1. (1)

for all non-zero $\alpha\in O_{M}$, there exists a unique unit $v\in V$ with
$v\alpha\in\mathcal{D}$. Moreover, we have $\\{u\in
O_{M}^{\ast}:u\alpha\in\mathcal{D}\\}=\\{vt:t\in T\\}$;

  2. (2)

the set $\mathcal{D}(1)$ has a $3$-Lipschitz parametrizable boundary.
Furthermore, there exists a constant $C>0$ such that the boundary of
$\mathcal{D}(1)$ intersects every line in at most $C$ points;

  3. (3)

there exists a constant $C>0$ such that for all
$\alpha=a_{1}\eta_{1}+a_{2}\eta_{2}+a_{3}\eta_{3}+a_{4}\eta_{4}\in\mathcal{D}$
(with $a_{i}\in\mathbb{Z}$), we have $|a_{i}|\leq
C|N_{M/\mathbb{Q}}(\alpha)|^{1/4}$.

###  3.5 Vinogradov’s sieve

We will now state the main sieve that we will use. This sieve was originally
developed by Vinogradov in the context of representing odd integers as sums of
three primes. Recall that $M:=K(\zeta_{3})$. From now on we shall abbreviate
$N_{M/\mathbb{Q}}$ simply as $N$ and our ideals will be ideals of $O_{M}$
unless otherwise indicated.

In order to prove our later results, we will handle a slightly more general
symbol than $[\mathfrak{p}]$. To this end, fix an ideal $\mathfrak{M}$ of $M$
and an invertible element $\mu\in O_{M}/\mathfrak{M}O_{M}$. We introduce the
function $r_{i}(\mathfrak{a},\mu,\mathfrak{M})$ for $i=1,\dots,h$, which is
the indicator function of $\mathfrak{a}\mathfrak{p}_{i}$ having a generator
$\alpha$ satisfying $\alpha\equiv\mu\bmod\mathfrak{M}$; recall that
$\mathfrak{p}_{i}$ is one of our fixed representatives for $\mathrm{Cl}(M)$.
We shall treat $\mu$ and $\mathfrak{M}$ as fixed throughout the paper, and we
shall abbreviate $r_{i}(\mathfrak{a}):=r_{i}(\mathfrak{a},\mu,\mathfrak{M})$.

######  Proposition 3.6.

For every $\epsilon>0$, there exists a constant $C>0$ such that

$\left|\sum_{\begin{subarray}{c}N(\mathfrak{a})\leq X\\\
\mathfrak{m}\mid\mathfrak{a}\end{subarray}}r_{i}(\mathfrak{a})[\mathfrak{a}]\right|\leq
CX^{1-\frac{1}{24}+\epsilon}$

uniformly for all non-zero integral ideals $\mathfrak{m}$ of $O_{M}$ and all
$X\geq 2$.

###### Proof.

The proof of this proposition is postponed until Section 4. ∎

######  Proposition 3.7.

For every $\epsilon>0$, there exists a constant $C>0$ such that

$\left|\sum_{N(\mathfrak{m})\leq M}\sum_{N(\mathfrak{n})\leq
N}\alpha_{\mathfrak{m}}\beta_{\mathfrak{n}}r_{i}(\mathfrak{m}\mathfrak{n})[\mathfrak{m}\mathfrak{n}]\right|\leq
C(M+N)^{\frac{1}{24}}(MN)^{1-\frac{1}{24}+\epsilon}$

uniformly for all $M,N\geq 2$ and all sequences of complex numbers
$|\alpha_{\mathfrak{m}}|\leq 1$ and $|\beta_{\mathfrak{n}}|\leq 1$.

###### Proof.

The proof of this proposition is postponed until Section 5. ∎

Once Propositions 3.6 and 3.7 are proven, Vinogradov’s sieve for number fields
gives the following result.

######  Theorem 3.8.

For every $\epsilon>0$, there exists a constant $C>0$ such that

$\left|\sum_{N(\mathfrak{p})\leq
X}r_{i}(\mathfrak{p})[\mathfrak{p}]\right|\leq CX^{1-\frac{1}{3136}+\epsilon}$

for all $X\geq 2$.

###### Proof.

Propositions 3.6 and 3.7 together with Vinogradov’s sieve for number fields as
presented in [2, Proposition 5.2] yield

$\left|\sum_{N(\mathfrak{a})\leq
X}r_{i}(\mathfrak{a})[\mathfrak{a}]\Lambda(\mathfrak{a})\right|\leq
CX^{1-\frac{1}{3136}+\epsilon},$

where $\Lambda$ is the natural generalization of the von Mangoldt function to
number fields. The theorem is then a direct consequence of partial summation.
∎

##  4 Sums of type I

This section is entirely devoted to the proof of Proposition 3.6. Our proof
follows among the same lines as [3, Section 4] and [2, Section 6]. From now on
we fix a non-zero integral ideal $\mathfrak{m}$ of $O_{M}$. Let $\mathfrak{M}$
be an ideal of $O_{M}$ and let $\mu$ be an element of $O_{M}$. Recall that we
introduced the function $r_{i}(\mathfrak{a})$ for $i=1,\dots,h$, which is the
indicator function of $\mathfrak{a}\mathfrak{p}_{i}$ having a generator
$\alpha$ satisfying $\alpha\equiv\mu\bmod\mathfrak{M}$ (recall that
$\mathfrak{p}_{i}$ is one of our fixed representatives for $\mathrm{Cl}(M)$).
We set

$F=2^{4}\cdot 3^{4}\cdot f\cdot N(\mathfrak{M})\cdot\Delta(M/\mathbb{Q}),$

where $f$ is defined in equation (3.1) and where $\Delta(M/\mathbb{Q})$ is the
absolute discriminant. Each non-zero ideal has exactly $6$ or $12$ generators
in $\mathcal{D}$, which is exactly the size of the torsion subgroup $T$.
Further splitting the sum into progressions modulo $F$, we unwrap the sum as

$\displaystyle\sum_{\begin{subarray}{c}N(\mathfrak{a})\leq X\\\
\mathfrak{m}\mid\mathfrak{a}\end{subarray}}r_{i}(\mathfrak{a})[\mathfrak{a}]$
$\displaystyle=\frac{1}{|T|}\sum_{\begin{subarray}{c}\alpha\in\mathcal{D}(XN(\mathfrak{p}_{i}))\\\
\alpha\equiv
0\bmod\mathfrak{m}\mathfrak{p}_{i}\end{subarray}}r_{i}((\alpha)/\mathfrak{p}_{i})[(\alpha)/\mathfrak{p}_{i}]$
$\displaystyle=\frac{1}{|T|}\sum_{\rho\bmod
F}\sum_{\begin{subarray}{c}\alpha\in\mathcal{D}(XN(\mathfrak{p}_{i}))\\\
\alpha\equiv 0\bmod\mathfrak{m}\mathfrak{p}_{i}\\\ \alpha\equiv\rho\bmod
F\end{subarray}}r_{i}((\alpha)/\mathfrak{p}_{i})[(\alpha)/\mathfrak{p}_{i}],$

where our first identity is justified due to the change of variables
$\mathfrak{a}\mathfrak{p}_{i}=(\alpha)$. At this point we insert the identity

$\displaystyle[(\alpha)/\mathfrak{p}_{i}]=[(\alpha)]\cdot\gamma_{\rho}$ (4.1)

for all $\alpha\equiv\rho\bmod F$, where $\gamma_{\rho}$ is some fixed number
in $\\{1,\zeta_{3},\zeta_{3}^{2},0\\}$ depending only on $\rho$ and $i$. We
also note that $r_{i}((\alpha)/\mathfrak{p}_{i})$ depends only on $\alpha\bmod
F$. Prompted by these calculations, we define for each $\rho\bmod F$ and each
ideal $\mathfrak{m}$

$A(X,\rho)=\sum_{\begin{subarray}{c}\alpha\in\mathcal{D}(X)\\\ \alpha\equiv
0\bmod\mathfrak{m}\\\ \alpha\equiv\rho\bmod F\end{subarray}}[(\alpha)].$

By the above manipulations it suffices to estimate each $A(X,\rho)$
individually. Furthermore, we may assume that $\gcd(3,\rho)=(1)$ as otherwise
the symbol $[(\alpha)]$ is constantly zero.

Recalling that $\eta_{1}=1$ by convention and writing
$\mathbb{M}=\mathbb{Z}\eta_{2}+\mathbb{Z}\eta_{3}+\mathbb{Z}\eta_{4}$ allows
us to decompose $O_{M}=\mathbb{Z}\oplus\mathbb{M}$. Therefore every $\alpha\in
O_{M}$ can uniquely be written as

$\alpha=a+\beta,\quad\quad a\in\mathbb{Z},\beta\in\mathbb{M}.$

This transforms the sum $A(X,\rho)$ as

$A(X,\rho)=\sum_{\begin{subarray}{c}a+\beta\in\mathcal{D}(X)\\\ a+\beta\equiv
0\bmod\mathfrak{m}\\\ a+\beta\equiv\rho\bmod
F\end{subarray}}\left(\frac{\sigma(a+\beta)}{a+\beta}\right)_{M,3}\left(\frac{\sigma\tau(a+\beta)}{a+\beta}\right)_{M,3}.$

Our aim is now to rewrite the cubic residue symbols. We have

$\left(\frac{\sigma(a+\beta)}{a+\beta}\right)_{M,3}=\left(\frac{a+\sigma(\beta)}{a+\beta}\right)_{M,3}=\left(\frac{\sigma(\beta)-\beta}{a+\beta}\right)_{M,3},$

and similarly for the other cubic residue symbol. If $\sigma(\beta)-\beta=0$
or if $\sigma\tau(\beta)-\beta=0$, then our cubic residue symbols are
constantly $0$ and we may safely remove such $\beta$ from consideration. In
all other cases, we are thus incentivized to factor $\sigma(\beta)-\beta$ and
$\sigma\tau(\beta)-\beta$. Define $\mathfrak{c}$ to be the largest divisor of
$\sigma(\beta)-\beta$ coprime to $F$ and define $\mathfrak{c}^{\prime}$ to be
the largest divisor of $\sigma\tau(\beta)-\beta$ coprime to $F$. After
applying cubic reciprocity as presented in Proposition 3.1 and following the
argument on [2, p. 726], we obtain

$\left(\frac{\sigma(\beta)-\beta}{a+\beta}\right)_{M,3}=\mu_{1}\cdot\left(\frac{a+\beta}{\mathfrak{c}}\right)_{M,3},\quad\quad\left(\frac{\sigma\tau(\beta)-\beta}{a+\beta}\right)_{M,3}=\mu_{2}\cdot\left(\frac{a+\beta}{\mathfrak{c}^{\prime}}\right)_{M,3}$

for some numbers $\mu_{1},\mu_{2}\in\\{1,\zeta_{3},\zeta_{3}^{2},0\\}$
depending on $\rho$ and $\beta$ but not on $a$. We record our progress by
applying the triangle inequality to deduce that

$|A(X,\rho)|\leq\sum_{\beta\in\mathbb{M}}|T(X,\rho,\beta)|,$

where

$T(X,\rho,\beta):=\sum_{\begin{subarray}{c}a\in\mathbb{Z}\\\
a+\beta\in\mathcal{D}(X)\\\ a+\beta\equiv 0\bmod\mathfrak{m}\\\
a+\beta\equiv\rho\bmod
F\end{subarray}}\left(\frac{a+\beta}{\mathfrak{c}}\right)_{M,3}\left(\frac{a+\beta}{\mathfrak{c}^{\prime}}\right)_{M,3}.$

We will treat $\beta$ as fixed and work towards estimating each
$T(X,\rho,\beta)$ individually. It is at this stage that we bring the material
from Subsection 3.3 into play. To this end, we remark that
$\mathfrak{c}^{\prime}$ is in fact the extension of an ideal of
$\mathbb{Q}(\sqrt{3d})$. Here we use that $\mathfrak{c}^{\prime}$ is coprime
to $F$ and thus the discriminant of $M$ and that $\mathfrak{c}^{\prime}$
divides an element of the shape $\sigma\tau(\beta)-\beta$. From now on we
shall view $\mathfrak{c}^{\prime}$ as an ideal of $O_{\mathbb{Q}(\sqrt{3d})}$.
We factor it as

$\mathfrak{c}^{\prime}O_{\mathbb{Q}(\sqrt{3d})}=\prod_{i=1}^{k}\mathfrak{p}_{i}^{e_{i}}$

for prime ideals $\mathfrak{p}_{i}$ of $O_{\mathbb{Q}(\sqrt{3d})}$, so that

$\left(\frac{a+\beta}{\mathfrak{c}^{\prime}O_{M}}\right)_{M,3}=\prod_{i=1}^{k}\left(\frac{a+\beta}{\mathfrak{p}_{i}O_{M}}\right)_{M,3}^{e_{i}}.$

Observe that the ideals $\mathfrak{p}_{i}$ are coprime to $F$ by construction
and hence unramified in the extension $M/\mathbb{Q}(\sqrt{3d})$. We now claim
that

$\displaystyle\left(\frac{a+\beta}{\mathfrak{p}_{i}O_{M}}\right)_{M,3}=\mathbf{1}_{\gcd(a+\beta,\mathfrak{p}_{i})=(1)}.$
(4.2)

Since $\mathfrak{p}_{i}$ divides $\mathfrak{c}^{\prime}$ and since
$\mathfrak{c}^{\prime}$ divides $\sigma\tau(\beta)-\beta$, it follows that
$\sigma\tau(a+\beta)\equiv a+\beta\bmod\mathfrak{p}_{i}O_{M}$. Therefore Lemma
3.4 allows us to replace $a+\beta$ by some $\beta^{\prime}\in
O_{\mathbb{Q}(\sqrt{3d})}$. Then Lemma 3.2 gives the claim in case
$\mathfrak{p}_{i}$ splits in $M$. Instead suppose that $\mathfrak{p}_{i}$
stays inert in $M$. If we write $p_{i}$ for the unique prime of $\mathbb{Z}$
below $\mathfrak{p}_{i}$, we certainly find that $p\equiv 2\bmod 3$. In this
case the claim is a consequence of Lemma 3.3.

Having proved our claim, we perform a similar operation on the other cubic
residue symbol. This residue symbol shall not disappear, but we will be able
to “lower it” to $\mathbb{Q}(\zeta_{3})$. This is the crucial step in making
our results unconditional compared to [2] despite working in a field of degree
$4$.

Arguing as before, we may view $\mathfrak{c}$ as the extension of an ideal of
$\mathbb{Z}[\zeta_{3}]$, and by abuse of notation we shall view $\mathfrak{c}$
as an ideal of $\mathbb{Z}[\zeta_{3}]$ from now on. In this case we factor
$\mathfrak{c}\mathbb{Z}[\zeta_{3}]=\mathfrak{g}\mathfrak{q}$, where
$\mathfrak{q}$ has squarefree norm, $\mathfrak{g}$ has squarefull norm and
$\gcd(N_{\mathbb{Q}(\zeta_{3})/\mathbb{Q}}(\mathfrak{g}),N_{\mathbb{Q}(\zeta_{3})/\mathbb{Q}}(\mathfrak{q}))=1$.
Furthermore, observe that $\mathfrak{q}$ and $\mathfrak{g}$ are both coprime
to $F$, so all prime divisors of $\mathfrak{q}$ and $\mathfrak{g}$ stay
unramified in $M/\mathbb{Q}(\zeta_{3})$. So far we have shown that

$\left(\frac{a+\beta}{\mathfrak{c}O_{M}}\right)_{M,3}=\left(\frac{a+\beta}{\mathfrak{g}O_{M}}\right)_{M,3}\left(\frac{a+\beta}{\mathfrak{q}O_{M}}\right)_{M,3}.$

Lemma 3.4 and the Chinese remainder theorem allow us to replace $\beta$ by
some $\beta^{\prime}\in\mathbb{Z}[\zeta_{3}]$. Combining Lemma 3.2 and Lemma
3.3 yields

$\displaystyle\left(\frac{a+\beta^{\prime}}{\mathfrak{q}O_{M}}\right)_{M,3}=\left(\frac{a+\beta^{\prime}}{\mathfrak{q}}\right)_{\mathbb{Z}[\zeta_{3}],3}^{2},$
(4.3)

as any prime $\mathfrak{p}$ in $\mathbb{Z}[\zeta_{3}]$ that stays inert in $M$
satisfies $p\equiv 1\bmod 3$ (writing $p:=\mathfrak{p}\cap\mathbb{Z}$).
Finally, as $q:=N_{\mathbb{Q}(\zeta_{3})/\mathbb{Q}}(\mathfrak{q})$ is
squarefree, the Chinese remainder theorem guarantees the existence of some
rational integer $b$ with $\beta^{\prime}\equiv b\bmod\mathfrak{q}$.

Thanks to equations (4.2) and (4.3), we have thus arrived at

$T(X,\rho,\beta)=\sum_{\begin{subarray}{c}a\in\mathbb{Z}\\\
a+\beta\in\mathcal{D}(X)\\\ a+\beta\equiv 0\bmod\mathfrak{m}\\\
a+\beta\equiv\rho\bmod
F\end{subarray}}\left(\frac{a+\beta}{\mathfrak{g}O_{M}}\right)_{M,3}\cdot\left(\frac{a+b}{\mathfrak{q}}\right)_{\mathbb{Z}[\zeta_{3}],3}^{2}\cdot\mathbf{1}_{\gcd(a+\beta,\mathfrak{c}^{\prime})=(1)}.$

Following the proof on [3], we fix $a$ modulo the radical $g_{0}$ of $g$ and
detect the condition $\mathbf{1}_{\gcd(a+\beta,\mathfrak{c}^{\prime})=(1)}$
using the Mobius function. Continuing in the footsteps of [3], we apply the
triangle inequality

$|T(X,\rho,\beta)|\leq\sum_{a_{0}\bmod
g_{0}}\sum_{\mathfrak{d}\mid\mathfrak{c}^{\prime}O_{M}}|T(X,\rho,\beta,a_{0},\mathfrak{d})|,$

where

$T(X,\rho,\beta,a_{0},\mathfrak{d}):=\sum_{\begin{subarray}{c}a\in\mathbb{Z}\\\
a+\beta\in\mathcal{D}(X)\\\ a+\beta\equiv 0\bmod\mathfrak{m}\\\
a+\beta\equiv\rho\bmod F\\\ a\equiv a_{0}\bmod g_{0}\\\ a+\beta\equiv
0\bmod\mathfrak{d}\end{subarray}}\left(\frac{a+b}{\mathfrak{q}}\right)_{\mathbb{Z}[\zeta_{3}],3}^{2}.$

Since squaring a cubic residue symbol is the same as conjugating it, we may
now remove the square factor without changing the absolute value of
$T(X,\rho,\beta,a_{0},\mathfrak{d})$. After doing so, the sum
$T(X,\rho,\beta,a_{0},\mathfrak{d})$ has the same shape as [3, Eq. (4.2)].
Having successfully applied our field lowering technique, the remainder of the
proof is identical to [3, p. 7422-7424].

##  5 Sums of type II

This section is entirely devoted to the proof of Proposition 3.7. This will be
relatively straightforward, as the required tools have already been developed
in the literature. This was started in [3, Proposition 3.6], which is
unfortunately not applicable here as the symbol is required to take values in
$\\{\pm 1,0\\}$. Smith [11] developed a very general large sieve that is
completely explicit in its dependence on the underlying number field. Another
interesting recent result in this direction can be found in recent work of
Santens [9]. Here we shall appeal to [5, Proposition 4.3].

In the notation of [5], we shall use [5, Proposition 4.3] with the integer $M$
on [5, p. 11] equal to $F$, with the number field $K$ equal to our
$M=K(\zeta_{3})$, with $n$ equal to $4$, with $\ell=3$, with $t_{1}=t_{2}=1$,
with the function $\gamma$ equal to

$\gamma(w,z):=\left(\frac{\sigma(w)}{z}\right)_{M,3}\left(\frac{\sigma(z)}{w}\right)_{M,3}\left(\frac{\sigma\tau(w)}{z}\right)_{M,3}\left(\frac{\sigma\tau(z)}{w}\right)_{M,3}$

and with $A_{\text{bad}}$ the set of squarefull integers. Then we can take
$C_{1}$ to be an absolute constant ($C_{1}=100$ suffices) and $C_{2}=1/2$. We
now check that this data satisfies the conditions (P1)-(P4) on [5, p. 11-12].

Condition (P1) is a consequence of Proposition 3.1. Condition (P2) follows
from the definition of the cubic residue symbol. Condition (P4) is readily
verified by counting squarefull integers in the usual way. Finally, (P3)
follows from the Chinese remainder theorem and orthogonality of characters.
Applying [5, Proposition 4.3], we exhibit, for any given $\epsilon>0$, the
existence of a constant $C(\epsilon)>0$ satisfying the inequality

$\displaystyle\sum_{\begin{subarray}{c}w\in\mathcal{D}(X)\\\
w\equiv\delta_{1}\bmod
F\end{subarray}}\sum_{\begin{subarray}{c}z\in\mathcal{D}(Y)\\\
z\equiv\delta_{2}\bmod F\end{subarray}}\alpha_{w}\beta_{z}\gamma(w,z)\leq
C(\epsilon)(X+Y)^{\frac{1}{24}}(XY)^{1-\frac{1}{24}+\epsilon}$ (5.1)

for all $\alpha_{w},\beta_{z}$ of magnitude bounded by $1$.

With this calculated, recall that our aim is to demonstrate the bound

$\left|\sum_{N(\mathfrak{m})\leq M}\sum_{N(\mathfrak{n})\leq
N}\alpha_{\mathfrak{m}}\beta_{\mathfrak{n}}r_{i}(\mathfrak{m}\mathfrak{n})[\mathfrak{m}\mathfrak{n}]\right|\leq
C(M+N)^{\frac{1}{24}}(MN)^{1-\frac{1}{24}+\epsilon}.$

Let us first deal with the case that $\mathfrak{m}$ and $\mathfrak{n}$ are
both principal. Observe that every principal ideal has exactly $|T|$
generators in $\mathcal{D}$ by Lemma 3.5, so our sum becomes

$\displaystyle\frac{1}{|T|^{2}}\sum_{w\in\mathcal{D}(M)}\sum_{z\in\mathcal{D}(N)}\alpha_{w}\beta_{z}r_{i}(wz)[wz].$
(5.2)

We note that $r_{i}(wz)$ depends only on $wz\bmod F$ and that

$[(wz)]=\gamma(w,z)\left(\frac{\sigma(w)}{w}\right)_{M,3}\left(\frac{\sigma(z)}{z}\right)_{M,3}\left(\frac{\sigma\tau(w)}{w}\right)_{M,3}\left(\frac{\sigma\tau(z)}{z}\right)_{M,3}.$

Absorbing the last four cubic residue symbols in the coefficients
$\alpha_{w},\beta_{z}$ and splitting the sum over congruence classes modulo
$F$ shows that we can bound the expression (5.2) by at most $F^{8}$ sums
appearing in the inequality (5.1).

This finishes the proof of Proposition 3.7 when the ideals $\mathfrak{m}$,
$\mathfrak{n}$ of that proposition are restricted to principal ideals. The
general case is handled by appealing to (4.1) and reducing to the case of
principal ideals.

##  6 Proof of main theorems

In this section, we gather our previous results to prove our main theorems.

###  6.1 Proof of Theorem 1.1

Define $\mathcal{I}$ to be the set of indices $1\leq i\leq h$ such that our
fixed generator $\mathfrak{p}_{i}$ of $\mathrm{Cl}(M)$ has principal norm in
$K$. Then Theorem 1.1 is an immediate consequence of Theorem 3.8 by taking
$\mathfrak{M}=O_{M}$, $\mu=1$ and summing over all $i\in\mathcal{I}$.

###  6.2 Proof of Corollary 1.2

We take $\mathcal{I}$ as above, and we write $f$ for the conductor of
$\mathcal{O}$. We also take $\mathfrak{M}=fO_{M}$. The first part of Corollary
1.2 follows from Proposition 2.2 and Theorem 3.8 by summing over all
$i\in\mathcal{I}$ and summing over all $\mu\in(O_{M}/fO_{M})^{\ast}$ whose
norm lies in the subgroup of $(O_{K}/fO_{K})^{\ast}$ generated by
$O_{K}^{\ast}$ and rational integers coprime to $f$.

The second part of Corollary 1.2 follows by enlarging $\mathfrak{M}$ and
choosing $\mu$ appropriately to account for the additional splitting
condition.

###  6.3 Proof of Theorem 1.3

The first part of Theorem 1.3 is an entirely algebraic statement which is
proven by combining Lemma 2.5 and Lemma 2.6. The last part of Theorem 1.3 is a
consequence of Theorem 3.8.

###  6.4 Proof of Corollary 1.4

We shall now give the necessary background in Galois representations to relate
Corollary 1.4 to Corollary 1.2.

###### Proof of Corollary 1.4.

For any elliptic curve $E$, the action of
$\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the
$\ell^{\infty}$-torsion points of $E$ gives rise to a Galois representation
$\rho$ ramified at finitely many places. By [10, Theorem V.2.3.1], at each
unramified place $p$, the error term $a_{\mathfrak{p}}(E)$ coincides with the
trace of the image under $\rho$ of any Frobenius lift at $p$. Therefore
Corollary 1.4 is a consequence of Corollary 1.2. ∎

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