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 generalisation of the spin symbol introduced in [ Reference Friedlander, Iwaniec, Mazur and Rubin2 ]. In fact, it turns out that the algebraic part of our theory readily generalises 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 2·3. We will then relate this to splitting of p in a number field depending on p, see Lemma 2·10.

2·1. Power residue symbols

We shall briefly recall the required theory of power residue symbols that we need. Let M be a number field always containing a fixed primitive $m{\operatorname{th}}$ root of unity that we denote $\zeta_m$ . For $\alpha \in O_M$ and $\mathfrak{p}$ a prime of M not dividing m, we define $(\alpha/\mathfrak{p})_{M, m}$ as the unique element in $\langle \zeta_m \rangle \cup \{0\}$ satisfying

\[\left(\frac{\alpha}{\mathfrak{p}}\right)_{M, m} \equiv \alpha^{\frac{N_{M/\mathbb{Q}}(\mathfrak{p}) - 1}{m}} \bmod \mathfrak{p}.\]

We multiplicatively extend this to all ideals I coprime to m.

For our application, the case $m = 3$ is particularly important. We will use the following weak form of cubic reciprocity.

Proposition 2·1. 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 $27O_M$ .

2·2. 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 [ Reference Cox1 , p. 179] for the definition of ring class fields.

We write $\sigma$ for the non-trivial element of $\textrm{Gal}(K/\mathbb{Q})$ . 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 $\bar{\mathfrak{q}}$ for (prime) ideals $\mathfrak{q}$ .

Lemma 2·2. 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 [ Reference Cox1 , theorem 14·16].

Proposition 2·3. 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

\[\operatorname{Gal}\!(M/\mathbb{Q}) \simeq \operatorname{Gal}\!(K/\mathbb{Q}) \times \operatorname{Gal}\!(\mathbb{Q}(\zeta_m)/\mathbb{Q})= \{1, \sigma \} \times (\mathbb{Z}/m\mathbb{Z})^\times.\]

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 2·2, 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 $\mathfrak{q}$ .

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:

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

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 2·2

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

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.

2·3. Definition of the spin symbol

Let $K = \mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic field with $d \gt 0$ 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:

1. (1) we let $\{\eta_1, \dots, \eta_4\}$ be an integral basis for $O_M$ with $\eta_1 = 1$ ;

2. (2) the torsion subgroup of $O_M^\times$ is generated by $\zeta_{12}$ if $d = 1$ , and otherwise the torsion subgroup is generated by $\zeta_6$ ;

3. (3) 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 6d, where each collection is a set of representatives for the class group $\textrm{Cl}(M)$ of M. We let

   (2·1) \begin{align}\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}).\end{align}
   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

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 $d \gt 1$ . One directly checks that

\[\sum_{\rho \in \textrm{Gal}(M/\mathbb{Q})} [\rho(\mathfrak{p})]=\begin{cases}-2 &\text{if } [\mathfrak{p}] \in \{\zeta_3, \zeta_3^2\} \\ 4 &\text{if } [\mathfrak{p}] = 1.\end{cases}\]

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

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

where $g O_K$ 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 $\mathfrak{a}$ . In particular, we have the useful formula

(2·2) \begin{align}[(\alpha)] = \left(\frac{\sigma(\alpha) \sigma \tau(\alpha)}{\mathfrak{a}}\right)_{M, 3}\end{align}

for all $\alpha \in O_M$ coprime to 3.

2·4. Field lowering

The coming results are variations of the results in [ Reference Koymans and Milovic5 , 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 2·4. Let K be a number field and let $\mathfrak{q}$ be a prime of K coprime to 3. Assume that M is a quadratic extension of K such that M contains $\zeta_3$ and $\mathfrak{q}$ splits in M. Write $\sigma$ for the non-trivial element of $\operatorname{Gal}\!(M/K)$ . Then we have for all $\alpha \in O_K$

\[\left(\frac{\alpha}{\mathfrak{q}O_M}\right)_{M, 3} =\begin{cases}\left(\frac{\alpha}{\mathfrak{q}O_K}\right)_{K, 3}^2 &\text{if } \sigma \text{ fixes } \zeta_3 \\ \textbf{1}_{\mathfrak{q} \nmid \alpha} &\text{if } \sigma \text{ does not fix } \zeta_3.\end{cases}\]

Proof. Since $\mathfrak{q}$ splits in M, we may write $\mathfrak{q} O_M = \mathfrak{Q} \sigma(\mathfrak{Q})$ for some prime $\mathfrak{Q}$ of M. This gives

\[\left(\frac{\alpha}{\mathfrak{q}O_M}\right)_{M, 3} = \left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3} \left(\frac{\alpha}{\sigma(\mathfrak{Q})}\right)_{M, 3} = \left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3} \left(\frac{\sigma(\alpha)}{\sigma(\mathfrak{Q})}\right)_{M, 3} = \left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3} \sigma\left(\left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3}\right),\]

where we used that $\sigma(\alpha) = \alpha$ thanks to the assumption $\alpha \in O_K$ . If $\sigma$ fixes $\zeta_3$ , then the above becomes

\[\left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3} \sigma\left(\left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3}\right) = \left(\frac{\alpha}{\mathfrak{Q}}\right)_{M, 3}^2= \left(\frac{\alpha}{\mathfrak{q}O_K}\right)_{K, 3}^2.\]

If $\sigma$ does not fix $\zeta_3$ , then we get $\textbf{1}_{\mathfrak{q} \nmid \alpha}$ by checking all possibilities for the cubic residue symbol $(\alpha/\mathfrak{Q})_{M, 3} \in \{1, \zeta_3, \zeta_3^2, 0\}$ .

Lemma 2·5. Let K be a number field and let $\mathfrak{q}$ be a prime of K coprime to 3. Assume that M is a quadratic extension of K such that M contains $\zeta_3$ and assume that $\mathfrak{q}$ stays inert in M. Further assume that $\mathfrak{q}$ has degree 1 in K and let q be the prime of $\mathbb{Q}$ lying below $\mathfrak{q}$ . Then we have for all $\alpha \in O_K$

\[\left(\frac{\alpha}{\mathfrak{q}O_M}\right)_{M, 3} =\begin{cases}\left(\frac{\alpha}{\mathfrak{q}O_K}\right)_{K, 3}^2 &\text{if } q \equiv 1 \bmod 3 \\ \textbf{1}_{\mathfrak{q} \nmid \alpha} &\text{if } q \equiv 2 \bmod 3.\end{cases}\]

Proof. We have the identities

\[\left(\frac{\alpha}{\mathfrak{q}O_M}\right)_{M, 3} \equiv \alpha^{\frac{N_{M/\mathbb{Q}}(\mathfrak{q}) - 1}{3}} \equiv \alpha^{\frac{q^2 - 1}{3}} \bmod \mathfrak{q}O_M.\]

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

\[\alpha^{\frac{q^2 - 1}{3}} \equiv \left(\alpha^{\frac{q - 1}{3}}\right)^{q + 1} \equiv \left(\alpha^{\frac{N_{K/\mathbb{Q}}(\mathfrak{q}) - 1}{3}}\right)^{q + 1} \equiv \left(\frac{\alpha}{\mathfrak{q}O_K}\right)_{K, 3}^{q + 1} \equiv \left(\frac{\alpha}{\mathfrak{q}O_K}\right)_{K, 3}^2 \bmod \mathfrak{q}.\]

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

\[\alpha^{\frac{q^2 - 1}{3}} \equiv \left(\alpha^{q - 1}\right)^{\frac{q + 1}{3}} \equiv \left(\alpha^{N_{K/\mathbb{Q}}(\mathfrak{q}) - 1}\right)^{\frac{q + 1}{3}} \equiv \textbf{1}_{\mathfrak{q} \nmid \alpha}^{\frac{q + 1}{3}} \equiv \textbf{1}_{\mathfrak{q} \nmid \alpha} \bmod \mathfrak{q}O_M,\]

as desired.

Lemma 2·6. Let K be a number field and let M be a quadratic extension of K. Write $\sigma$ for the non-trivial element of $\textrm{Gal}(M/K)$ . Assume that $\mathfrak{q}$ is a prime ideal of K that stays unramified in M. Further suppose that $\beta \in O_M$ satisfies $\beta \equiv \sigma(\beta) \bmod \mathfrak{q} O_M$ . Then there is $\beta' \in O_K$ such that $\beta' \equiv \beta \bmod \mathfrak{q} O_M$ .

Proof. This is proven in [ Reference Koymans and Milovic5 , lemma 3·4].

2·5. 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 2·7. 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 $[K^{(m)}(p) \;:\; \mathbb{Q}] = 2m^2 \cdot |\textrm{Cl}(K)[m]|$ . Diagramatically

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 $m^2 \cdot |\textrm{Cl}(K)[m]|$ over K, and hence has degree $2m^2 \cdot |\textrm{Cl}(K)[m]|$ over $\mathbb{Q}$ . We will now show that this implies $K^{(m)}(p) = K^{(m)}(\mathfrak{q}) K^{(m)}(\overline{\mathfrak{q}}) H_K^{(m)}$ .

The inclusion $K^{(m)}(\mathfrak{q}) K^{(m)}(\overline{\mathfrak{q}}) H_K^{(m)} \subseteq K^{(m)}(p)$ is by definition, so it suffices to show that the degree of $K^{(m)}(p)$ over K is at most $m^2 \cdot |\textrm{Cl}(K)[m]|$ . By class field theory, we have a canonical isomorphism

\[\frac{\textrm{Cl}(K, p)}{m \textrm{Cl}(K, p)} = \textrm{Gal}(K^{(m)}(p)/K),\]

where $\textrm{Cl}(K, p)$ denotes the ray class group of conductor p. By the ray class group sequence, we have

\[0 \rightarrow \frac{(O_K/p)^\times}{\text{im } O_K^\times} \rightarrow \textrm{Cl}(K, p) \rightarrow \textrm{Cl}(K) \rightarrow 0.\]

The functor sending an abelian group A to $A/mA$ is right exact, so we get a new exact sequence

\[\frac{(O_K/p)^\times}{(O_K/p)^{\times m} \cdot \text{im } O_K^\times} \rightarrow \frac{\textrm{Cl}(K, p)}{m \textrm{Cl}(K, p)} \rightarrow \frac{\textrm{Cl}(K)}{m \textrm{Cl}(K)} \rightarrow 0.\]

This gives the bound

(2·3) \begin{align}[K^{(m)}(p) \;:\; K] &= \left|\frac{\textrm{Cl}(K, p)}{m \textrm{Cl}(K, p)}\right| \leq \left|\frac{(O_K/p)^\times}{(O_K/p)^{\times m} \cdot \text{im } O_K^\times}\right| \cdot \left|\frac{\textrm{Cl}(K)}{m \textrm{Cl}(K)}\right| \nonumber \\ &= \left|\frac{(O_K/p)^\times}{(O_K/p)^{\times m} \cdot \text{im } O_K^\times}\right| \cdot |\textrm{Cl}(K)[m]|.\end{align}

Finally, since $pO_K = \mathfrak{q} \overline{\mathfrak{q}}$ , we get an isomorphism

\[(O_K/p)^\times \cong (O_K/\mathfrak{q})^\times \times (O_K/\overline{\mathfrak{q}})^\times \cong \mathbb{F}_p^\times \times \mathbb{F}_p^\times.\]

Since $\mathbb{F}_p^\times$ is cyclic, this gives the upper bounds

\[\left|\frac{(O_K/p)^\times}{(O_K/p)^{\times m} \cdot \text{im } O_K^\times}\right| \leq \left|\frac{(O_K/p)^\times}{(O_K/p)^{\times m}}\right| \leq m^2.\]

Combining this with (2·3) shows that the degree of $K^{(m)}(p)$ over K is at most $m^2 \cdot |\textrm{Cl}(K)[m]|$ , as desired.

Table 1. The (e, f, g)-decomposition of p in $K^{(m)}(p)/\mathbb{Q}$

Lemma 2·8. 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 $K^{(m)}(p)$ (Table 1):

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 $\textrm{Gal}(K^{(m)}(p)/K)$ 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 2·7, we have

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

and the result follows.

Lemma 2·9. 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

(2·4) \begin{align}f(p, K^{(m)}(p)/\mathbb{Q}) = 1 &\iff f(\mathfrak{q}, K^{(m)}(p)/K) = 1 \nonumber \\ &\iff f(\mathfrak{q}, K^{(m)}(\mathfrak{q})/K) = 1 \text{ and }f(\mathfrak{q}, K^{(m)}(\overline{\mathfrak{q}})/K) = 1.\end{align}

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 (2·4) gives the lemma.

Lemma 2·10. 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

where the horizontal arrows are isomorphisms. Let $\operatorname{Cl}_K$ be the class group of K. By [ Reference Milne9 , theorem 1.7, chapter V], there is a commutative diagram

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

\begin{align*}f(\kappa, K^{(m)}(\overline{\mathfrak{q}})/K) = 1 &\Longleftrightarrow f(\kappa, K^{(m)}(\overline{\mathfrak{q}}) H_K^{(m)}/K) = 1 \\ &\Longleftrightarrow \kappa \text{ is an } m{\operatorname{th}} \text{ power modulo } \overline{\mathfrak{q}}.\end{align*}

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

\[1 = \left(\frac{\kappa}{\sigma(\mathfrak{p})}\right)_{M, m}\Longleftrightarrow1 =\sigma \left(\frac{\kappa}{\sigma(\mathfrak{p})}\right)_{M, m}= \left(\frac{\sigma(\kappa)}{\sigma^2(\mathfrak{p})}\right)_{M, m}= \left(\frac{\sigma(\kappa)}{\mathfrak{p}}\right)_{M, m}= \left(\frac{\overline{\kappa}}{\mathfrak{p}}\right)_{M, m}= [\mathfrak{p}],\]

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 [ Reference Koymans and Milovic5 ].

3·2. 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 $\textrm{Cl}(M)$ . We shall treat $\mu$ and $\mathfrak{M}$ as fixed throughout the paper, and we shall abbreviate $r_i(\mathfrak{a}) \;:\!=\; r_i(\mathfrak{a}, \mu, \mathfrak{M})$ . We will assume throughout the paper that $\mathfrak{M}$ is divisible by f, see equation (2·1).

Proposition 3·2. For every $\epsilon \gt 0$ , there exists a constant $C \gt 0$ such that

\[\left|\sum_{\substack{N(\mathfrak{a}) \leq X \\ \mathfrak{m} \mid \mathfrak{a}}} r_i(\mathfrak{a}) [\mathfrak{a}]\right| \leq C X^{1 - \frac{1}{24} + \epsilon}\]

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 3·3. For every $\epsilon \gt 0$ , there exists a constant $C \gt 0$ 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 3·2 and 3·3 are proven, Vinogradov’s sieve for number fields gives the following result.

Theorem 3·4. For every $\epsilon \gt 0$ , there exists a constant $C \gt 0$ such that

\[\left|\sum_{N(\mathfrak{p}) \leq X} r_i(\mathfrak{p}) [\mathfrak{p}]\right| \leq C X^{1 - \frac{1}{3136} + \epsilon}\]

for all $X \geq 2$ .

Proof. Propositions 3·2 and 3·3 together with Vinogradov’s sieve for number fields as presented in [ Reference Friedlander, Iwaniec, Mazur and Rubin2 , proposition 5·2] yield

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

where $\Lambda$ is the natural generalisation 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 3·2. Our proof follows among the same lines as [ Reference Koymans and Milovic5 , section 4] and [ Reference Friedlander, Iwaniec, Mazur and Rubin2 , section 6]. From now on we fix a non-zero integral ideal $\mathfrak{m}$ of $O_M$ . Let $\mathfrak{M}$ be an ideal of $O_M$ divisible by f and let $\mu$ be an element of $O_M$ coprime to $\mathfrak{M}$ . 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 $\textrm{Cl}(M)$ ). We set

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

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

\begin{align*}\sum_{\substack{N(\mathfrak{a}) \leq X \\ \mathfrak{m} \mid \mathfrak{a}}} r_i(\mathfrak{a}) [\mathfrak{a}] &= \frac{1}{|T|} \sum_{\substack{\alpha \in \mathcal{D}(X N(\mathfrak{p}_i)) \\ \alpha \equiv 0 \bmod \mathfrak{m} \mathfrak{p}_i}} r_i((\alpha)/\mathfrak{p}_i) [(\alpha)/\mathfrak{p}_i] \\ &= \frac{1}{|T|} \sum_{\rho \bmod F} \sum_{\substack{\alpha \in \mathcal{D}(X N(\mathfrak{p}_i)) \\ \alpha \equiv 0 \bmod \mathfrak{m} \mathfrak{p}_i \\ \alpha \equiv \rho \bmod F}} r_i((\alpha)/\mathfrak{p}_i) [(\alpha)/\mathfrak{p}_i],\end{align*}

where our first identity is justified due to the change of variables $\mathfrak{a} \mathfrak{p}_i = (\alpha)$ . We claim that

(4·1) \begin{align}[(\alpha)/\mathfrak{p}_i] = [(\alpha)] \cdot \gamma_\rho\end{align}

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. To prove the claim, first note that the claim holds if $\sigma(\mathfrak{p}_i) \sigma \tau(\mathfrak{p}_i)$ is not a principal ideal of K as we may simply take $\gamma_\rho = 0$ . Henceforth we assume that $\sigma(\mathfrak{p}_i) \sigma \tau(\mathfrak{p}_i)$ is principal and generated by some element $g \in O_K$ , and we may similarly assume that $\alpha$ is coprime to 3. Recalling the identity (2·2), we see that

(4·2) \begin{align}[(\alpha)/\mathfrak{p}_i] &= \left(\frac{\sigma(\alpha) \sigma \tau(\alpha)/g}{(\alpha)/\mathfrak{p}_i}\right)_{M, 3} = \left(\frac{\sigma(\alpha) \sigma \tau(\alpha) g^2}{(\alpha) \mathfrak{p}_i^2}\right)_{M, 3} \nonumber \\ &= [(\alpha)] \left(\frac{g}{\alpha}\right)_{M, 3}^2 \left(\frac{\sigma(\alpha) \sigma \tau(\alpha)}{\mathfrak{p}_i}\right)_{M, 3}^2 \left(\frac{g^2}{\mathfrak{p}_i^2}\right)_{M, 3}.\end{align}

In the first line of the above display, we have used that $\alpha$ is coprime to f (as $\mathfrak{M}$ is divisible by f by assumption), and hence to $N_{M/\mathbb{Q}}(\mathfrak{p}_i)$ , and that $\mathfrak{p}_i$ is an unramified prime of degree 1 so that $\mathfrak{p}_i$ is coprime to g. Applying Proposition 2·1 to $(g^2/\alpha)_{M, 3}$ , it follows that each of the individual symbols in (4·2) depends only on $\rho$ and i. Hence the claim follows.

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 $\mathfrak{m}$

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

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

\[A(X, \rho) = \sum_{\substack{a + \beta \in \mathcal{D}(X) \\ a + \beta \equiv 0 \bmod \mathfrak{m} \\ a + \beta \equiv \rho \bmod F}} \left(\frac{\sigma(a + \beta)}{a + \beta}\right)_{M, 3} \left(\frac{\sigma \tau(a + \beta)}{a + \beta}\right)_{M, 3}\]

due to equation (2·2). 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 incentivised 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 $\mathfrak{c}'$ to be the largest divisor of $\sigma \tau(\beta) - \beta$ coprime to F. We claim that

(4·3) \begin{align}\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}'}\right)_{M ,3}\end{align}

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 will prove the claim for $(\sigma(\beta) - \beta)/a + \beta)_{M, 3}$ , the other residue symbol can be treated similarly. We choose a factorisation

\[(\sigma(\beta) - \beta) = (t_3^3) \mathfrak{c} \mathfrak{t}_1 \mathfrak{t}_2\]

with the properties:

1. (1) $\mathfrak{t}_1 \mid F^\infty$ and is coprime to 3;

2. (2) $\mathfrak{t}_2 \mid 27^h$ , where we recall that h denotes the class number of K, and $t_3 \mid 3^\infty$ .

We emphasise that such a factorization always exists although may not be unique. We denote by $\mathfrak{C}$ , $\mathfrak{T}_1$ and $\mathfrak{T}_2$ the fixed inverses in the class group of respectively $\mathfrak{c}$ , $\mathfrak{t}_1$ and $\mathfrak{t}_2$ that we have chosen in Subsection 2.3. Then we have $\mathfrak{c} \mathfrak{C} = (g_0)$ , $\mathfrak{t}_1 \mathfrak{T}_1 = (g_1)$ and $\mathfrak{t}_2 \mathfrak{T}_2 = (g_2)$ . Because $\mathfrak{c} \mathfrak{t}_1 \mathfrak{t}_2$ is principal, it follows that $\mathfrak{C} \mathfrak{T}_1 \mathfrak{T}_2 = (g_3)$ must also be principal. For a good choice of generators $g_i$ , it follows that

\[\sigma(\beta) - \beta = t_3^3 g_0 g_1 g_2 g_3^{-1}.\]

We conclude that

\[\left(\frac{\sigma(\beta) - \beta}{a + \beta}\right)_{M, 3} = \left(\frac{g_0 g_1 g_2 g_3^{-1}}{a + \beta}\right)_{M, 3} = \left(\frac{g_0 g_1 g_2 g_3^2}{a + \beta}\right)_{M, 3} = \left(\frac{g_0 g_1 g_3^2}{a + \beta}\right)_{M, 3} \left(\frac{g_2}{a + \beta}\right)_{M, 3},\]

where we used that $a + \beta$ is coprime to 3f. Since $(g_2) = \mathfrak{t}_2 \mathfrak{T}_2$ divides $F/27$ , Proposition 2·1 shows that the last symbol depends only on $\rho$ and $\beta$ . For the other symbol, we also apply Proposition 2·1 to obtain

\[\left(\frac{g_0 g_1 g_3^2}{a + \beta}\right)_{M, 3} = \mu_3 \cdot \left(\frac{a + \beta}{g_0 g_1 g_3^2}\right)_{M, 3}\]

for some number $\mu_3 \in \{1, \zeta_3, \zeta_3^2\}$ depending on $\rho$ and $\beta$ but not on a. We finally observe that

\[\left(\frac{a + \beta}{g_0 g_1 g_3^2}\right)_{M, 3} = \left(\frac{a + \beta}{\mathfrak{c}}\right)_{M, 3} \left(\frac{a + \beta}{\mathfrak{t}_1 \mathfrak{T}_2^2}\right)_{M, 3}.\]

Since the last factor only depends on $\rho$ and $\beta$ , this proves our claim (4·3). 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

\[T(X, \rho, \beta) \;:\!=\; \sum_{\substack{a \in \mathbb{Z} \\ a + \beta \in \mathcal{D}(X) \\ a + \beta \equiv 0 \bmod \mathfrak{m} \\ a + \beta \equiv \rho \bmod F}} \left(\frac{a + \beta}{\mathfrak{c}}\right)_{M, 3} \left(\frac{a + \beta}{\mathfrak{c}'}\right)_{M ,3}.\]

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 2.4 into play. To this end, we remark that $\mathfrak{c}'$ is in fact the extension of an ideal of $\mathbb{Q}(\sqrt{3d})$ . Here we use that $\mathfrak{c}'$ is coprime to F and thus the discriminant of M and that $\mathfrak{c}'$ is the largest such divisor of $\sigma \tau(\beta) - \beta$ . From now on we shall view $\mathfrak{c}'$ as an ideal of $O_{\mathbb{Q}(\sqrt{3d})}$ . We factor it as

\[\mathfrak{c}' O_{\mathbb{Q}(\sqrt{3d})} = \prod_{i = 1}^k \mathfrak{p}_i^{e_i}\]

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

\[\left(\frac{a + \beta}{\mathfrak{c}' O_M}\right)_{M ,3} = \prod_{i = 1}^k \left(\frac{a + \beta}{\mathfrak{p}_iO_M}\right)_{M ,3}^{e_i}.\]

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

(4·4) \begin{align}\left(\frac{a + \beta}{\mathfrak{p}_iO_M}\right)_{M ,3} = \textbf{1}_{\gcd(a + \beta, \mathfrak{p}_i) = (1)}.\end{align}

Since $\mathfrak{p}_i$ divides $\mathfrak{c}'$ and since $\mathfrak{c}'$ divides $\sigma \tau(\beta) - \beta$ , it follows that $\sigma \tau(a + \beta) \equiv a + \beta \bmod \mathfrak{p}_i O_M$ . Therefore Lemma 2·6 allows us to replace $a + \beta$ by some $\beta' \in O_{\mathbb{Q}(\sqrt{3d})}$ . Then Lemma 2·4 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 2·5.

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 [ Reference Friedlander, Iwaniec, Mazur and Rubin2 ] 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 2·6 and the Chinese remainder theorem allow us to replace $\beta$ by some $\beta' \in \mathbb{Z}[\zeta_3]$ . Combining Lemma 2·4 and Lemma 2·5 yields

(4·5) \begin{align}\left(\frac{a + \beta'}{\mathfrak{q} O_M}\right)_{M ,3} = \left(\frac{a + \beta'}{\mathfrak{q}}\right)_{\mathbb{Z}[\zeta_3] ,3}^2,\end{align}

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 $\beta' \equiv b \bmod \mathfrak{q}$ .

Thanks to equations (4·4) and (4·5), we have thus arrived at

\[T(X, \rho, \beta) = \sum_{\substack{a \in \mathbb{Z} \\ a + \beta \in \mathcal{D}(X) \\ a + \beta \equiv 0 \bmod \mathfrak{m} \\ a + \beta \equiv \rho \bmod F}} \left(\frac{a + \beta}{\mathfrak{g} O_M}\right)_{M ,3} \cdot \left(\frac{a + b}{\mathfrak{q}}\right)_{\mathbb{Z}[\zeta_3] ,3}^2 \cdot \textbf{1}_{\gcd(a + \beta, \mathfrak{c}') = (1)}.\]

Following the proof on [ Reference Koymans and Milovic5 ], we fix a modulo the radical $g_0$ of g and detect the condition $\textbf{1}_{\gcd(a + \beta, \mathfrak{c}') = (1)}$ using the Mobius function. Continuing in the footsteps of [ Reference Koymans and Milovic5 ], we apply the triangle inequality

\[|T(X, \rho, \beta)| \leq \sum_{a_0 \bmod g_0} \sum_{\mathfrak{d} \mid \mathfrak{c}'O_M} |T(X, \rho, \beta, a_0, \mathfrak{d})|,\]

where

\[T(X, \rho, \beta, a_0, \mathfrak{d}) \;:\!=\; \sum_{\substack{a \in \mathbb{Z} \\ a + \beta \in \mathcal{D}(X) \\ a + \beta \equiv 0 \bmod \mathfrak{m} \\ a + \beta \equiv \rho \bmod F \\ a \equiv a_0 \bmod g_0 \\ a + \beta \equiv 0 \bmod \mathfrak{d}}} \left(\frac{a + b}{\mathfrak{q}}\right)_{\mathbb{Z}[\zeta_3] ,3}^2.\]

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 [ Reference Koymans and Milovic5 , equation (4·2)]. Having successfully applied our field lowering technique, the remainder of the proof is identical to [ Reference Koymans and Milovic5 , p. 7422-7424].

5. Sums of type II

This section is entirely devoted to the proof of Proposition 3·3. This will be relatively straightforward, as the required tools have already been developed in the literature. This was started in [ Reference Koymans and Milovic5 , proposition 3·6], which is unfortunately not applicable here as the symbol is required to take values in $\{\pm 1, 0\}$ . Smith [ Reference Smith14 ] 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 [ Reference Santens12 ]. Here we shall appeal to [ Reference Koymans and Rome7 , proposition 4·3].

In the notation of [ Reference Koymans and Rome7 ], we shall use [ Reference Koymans and Rome7 , proposition 4·3] with the integer M on [ Reference Koymans and Rome7 , 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) form [ Reference Koymans and Rome7 , p. 11-12].

Condition (P1) is a consequence of Proposition 2·1. 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 [ Reference Koymans and Rome7 , proposition 4·3], we exhibit, for any given $\epsilon \gt 0$ , the existence of a constant $C(\epsilon) \gt 0$ satisfying the inequality

(5·1) \begin{align}\sum_{\substack{w \in \mathcal{D}(X) \\ w \equiv \delta_1 \bmod F}} \sum_{\substack{z \in \mathcal{D}(Y) \\ z \equiv \delta_2 \bmod F}} \alpha_w \beta_z \gamma(w, z) \leq C(\epsilon) (X + Y)^{\frac{1}{24}} (XY)^{1 - \frac{1}{24} + \epsilon}\end{align}

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 3·1, so our sum becomes

(5·2) \begin{align}\frac{1}{|T|^2} \sum_{w \in \mathcal{D}(M)} \sum_{z \in \mathcal{D}(N)} \alpha_w \beta_z r_i(wz) [wz].\end{align}

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 3·3 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.