Proof. We have to show that for all $b \in B$

\[ {\rm inv}_v(\alpha(b)(x)) = \partial_v \alpha(b)(\textrm{Frob}_{x \bmod{ \mathfrak{m}_v}}). \]

This follows from functoriality of the residue map [Reference Colliot-Thélène and SkorobogatovCTS21, Theorem 3.7.4], in particular the commutativity of the following diagram.

The composition of the bottom row is ${\rm inv}_v$ by combining [Reference Colliot-Thélène and SkorobogatovCTS21, Theorem 1.4.10, Lemma 2.3.3] and the definition of ${\rm inv}_v$.

Proof. Let $P := x \bmod \mathfrak {m}_v$. We have $P \in \pi _{b^\vee }(Y_{b^\vee }(\mathbb {F}_v))$ if and only if $\pi _{b^\vee }^{ -1}(P)$ has a $\mathbb {F}_v$-point. This fiber is the $B^\vee$-torsor corresponding to the composition

\[ \pi_1(\mathbb{F}_v) = \textrm{Gal}(\overline{\mathbb{F}}_v/\mathbb{F}_v) \xrightarrow{P_*} \pi_1(\mathcal{X}_{\mathbb{F}_v}) \xrightarrow{\partial^v \alpha - \chi_{b^\vee} } B^\vee. \]

A torsor over a field has a rational point if and only if it splits. Thus, this torsor has a $\mathbb {F}_v$-point if and only if $(\partial ^v \alpha - \chi _{b^\vee })(\textrm {Frob}_P) = 0$. In other words $\partial ^v \alpha (\textrm {Frob}_P) = b^\vee$ and we are done by Lemma 2.4.

Proof. The first statement follows from standard homological algebra. To prove the second statement we note that the set of connected components of $Y_{b^\vee }$ has a transitive $B^\vee$-action with stabilizer $\textrm {im}(\partial ^v \alpha - \chi _{b^\vee })$, analogously the set of connected components of $\overline {Y}$ has a transitive $B^{\vee }$ action with stabilizer $\textrm {im}(\overline {\partial ^v \alpha })$. Every connected component is thus geometrically connected if and only if the natural map $\textrm {im}(\overline {\partial ^v \alpha }) \to \textrm {im}(\partial ^v \alpha - \chi _{b^\vee })$ is an isomorphism. The snake lemma implies that this happens if and only if $\delta ^{b^\vee } = 0$.

Proof. Let $P \in X$ be the cone point. Consider the blow-up $X'$ at this point with exceptional divisor $E$. It is a $\mathbb {P}^1$-bundle over $Y$ and $X^{\textrm {sm}} = X \setminus P = X' \setminus E$. Thus, $X^{\textrm {sm}}$ is the complement of a relative normal crossing divisor $E$ in the smooth proper $Y$-variety $X'$. The desired statement follows from [Reference GrothendieckGro71, Examples XIII.4.4].

Proof. By Lemma 2.6 and Remark 2.7(i) it suffices to show that if all of the connected components of $Y_{b^\vee }$ are geometrically connected then it has a $\mathbb {F}_v$-point. By Lemma 2.8 there exists an étale cover $D \to C$ such that $Y_{b^\vee } = D \times _C \mathcal {X}_{\mathbb {F}_v}$. We show that $D(\mathbb {F}_v) \neq \emptyset$. Every connected component of $D$ is a connected étale $\textrm {im}(\overline {\partial ^v \alpha })$-torsor over $C$. By the Hurwitz formula this is a curve of genus $g'$. The Weil conjectures [Reference DeligneDel80] together with condition (2.1) give us what we want.

Proof. Dually, the map $\overline {\partial ^v \alpha }$ is surjective so $\textrm {coker}(\overline {\partial ^v \alpha }) = 0$ and the condition in the theorem is automatically satisfied.

Proof. This is a combination of the second bullet point of [Reference Colliot-Thélène and SkorobogatovCTS21, p. 37] and applying [Reference Colliot-Thélène and SkorobogatovCTS21, Lemma 2.33] to compare the different notions of residue.

Proof. By the Grothendieck purity theorem [Reference GrothendieckGro68, III: Theorem 6.1] it suffices to prove that each of these Brauer elements is unramified at all the divisors of $X_{\mathbf {a}}$. Note that $\mathcal {A}$ is the pullback along $\alpha$ of the following Brauer element described in [Reference Colliot-Thélène and XuCTX09, § 5.8]:

(3.2)\begin{equation} \bigg(\frac{f}{Y_3}, a_0 a_1 a_2 a_3\bigg) \in \operatorname{Br}(Y_{\mathbf{a}} \setminus \{Y_3 = 0 \}). \end{equation}

This is independent of the choice of $f$ as an element of $\operatorname {Br}(Y_{\mathbf {a}} \setminus \{Y_3 = 0 \})/\operatorname {Br}(\mathbb {Q})$ which shows that $\mathcal {A}$ is well-defined as an element of $\operatorname {Br}(\mathbb {Q}(X_{\mathbf {a}}))/\operatorname {Br}(\mathbb {Q})$. It follows that $\mathcal {A}$ is unramified except possibly at the divisor $\{X_3 = 0\}$. But $f$ is ramified at $\{X_3 = 0\}$ of order 2 so by [Reference Garibaldi, Merkurjev and SerreGMS03, Proposition 8.2] $\mathcal {A}$ is also unramified at this divisor.

We now show that $\gamma _{\sqrt {\theta _{\mathbf {a}}}}$ is indeed a morphism. To prove this we may work over the algebraic closure $\overline {\mathbb {Q}}$ by Galois descent. The given equations define $\gamma$ except when $X_m^2 = \sqrt {-({a_n}/{a_m})} X_n^2$ and $X_k^2 = -({\sqrt {\theta _{\mathbf {a}}}}/{a_k a_n}) \sqrt {-({a_n}/{a_m})} X_\ell ^2$ for any choice of the square root $\sqrt {-({a_n}/{a_m})}$. Note that this includes the cases $X_k = X_\ell = 0$, $X_n = X_m = 0$. Multiplying all the coordinates by $(X_k^2 - ({\sqrt {\theta _{\mathbf {a}}}}/{a_k a_n}) \sqrt {-({a_n}/{a_m})} X_\ell ^2)/(X_m^2 - \sqrt {-({a_n}/{a_m})} X_n^2)$ gives coordinates which are defined in this case. Indeed,

(3.3) \begin{align} \frac{X_k^2 X_m^2 - ({\sqrt{\theta_{\mathbf{a}}}}/{a_k a_m}) X_\ell^2 X_n^2}{X_m^2 - \sqrt{-({a_n}/{a_m})} X_n^2} & = X_k^2 + X_n^2\frac{\sqrt{-({a_n}/{a_m})}X_k^2 - ({\sqrt{\theta_{\mathbf{a}}}}/{a_k a_m}) X_\ell^2}{X_m^2 - \sqrt{-({a_n}/{a_m})}X_n^2} \nonumber\\ & = X_k^2 + \frac{a_m}{a_k} \sqrt{-\frac{a_n}{a_m}}X_n^2 \frac{X_m^2 + \sqrt{-({a_n}/{a_m})}X_n^2}{X_k^2 - ({\sqrt{\theta_{\mathbf{a}}}}/{a_k a_n}) \sqrt{-({a_n}/{a_m})} X_\ell^2}, \nonumber\\ \frac{({\sqrt{\theta_{\mathbf{a}}}}/{a_\ell a_m}) X_k^2 X_n^2 + X_\ell^2 X_m^2}{X_m^2 - \sqrt{-({a_n}/{a_m})} X_n^2} & = X_\ell^2 + X_n^2\frac{\sqrt{-({a_n}/{a_m})} X_\ell^2 + ({\sqrt{\theta_{\mathbf{a}}}}/{a_\ell a_m}) X_k^2 }{X_m^2 - \sqrt{-({a_n}/{a_m})}X_n^2}\nonumber\\ & = X_\ell^2 + \frac{\sqrt{\theta_{\mathbf{a}}}}{a_k a_\ell } X_n^2 \frac{X_m^2 + \sqrt{-({a_n}/{a_m})}X_n^2}{X_k^2 - ({\sqrt{\theta_{\mathbf{a}}}}/{a_k a_n}) \sqrt{-({a_n}/{a_m})}X_\ell^2}, \nonumber\\ \frac{X_m^4 + ({a_n}/{a_m})X_n^4}{X_m^2 - \sqrt{-({a_n}/{a_m})} X_n^2} & = X_m^2 + \sqrt{-\frac{a_n}{a_m}} X_n^2. \end{align}

The divisor of $h$ is a square so $\mathcal {B}_{k \ell m; \sqrt {\theta _{\mathbf {a}}}}$ can only be ramified at the divisors contained in $\gamma ^*\{Z_m = 0\}$. By the definition of $\gamma$ and (3.3) this is the union of

\[ \bigg\{X_m^2 = \sqrt{-\frac{a_n}{a_m}} X_n^2,\ X_k^2 = \frac{\sqrt{\theta_{\mathbf{a}}}}{a_k a_n} \sqrt{-\frac{a_n}{a_m}} X_\ell^2\bigg\} \]

for the two different choices of $\sqrt {-({a_m}/{a_n})}$. So $-a_k a_m \sqrt {\theta _{\mathbf {a}}}$ is a square in the residue field.

If we have two choices of tangent lines defined by $g_1, g_2$, then their quotient $g_1/g_2$ is a rational function with square divisor and is thus a square up to multiplication by a constant. The associated quaternion algebras thus differ by a constant algebra.

Proof. If $X_{\mathbf {a}}(\mathbb {Q}_p) \neq \emptyset$, then by fixing a solution and then taking an equivalent surface while appropriately changing the solution we may assume that $a_0 \in \mathbb {Z}_p^{\times }$, $a_1, a_2, a_3 \in \mathbb {Z}_p$ and that there exist $x_0, x_1, x_2, x_3 \in \mathbb {Z}_p^{\times }$ such that $a_0 x_0^4 + a_1 x_1^4 + a_2 x_2^4 + a_3 x_3^4 = 0$. By reducing modulo $p$ we see that at least one of the other coefficients is a $p$-adic unit. If there is exactly one other coefficient $a_k$ which is a $p$-adic unit, then $-({a_k}/{a_0}) \equiv ({x_k}/{x_0})^4 \not \equiv 0 \pmod {p}$, so by Hensel's lemma $-({a_k}/{a_0}) \in \mathbb {Q}_p^{\times 4}$ and the first condition holds. Otherwise the second condition holds.

If the first condition holds, then without loss of generality $k = 0, \ell = 1$ and $[1: \sqrt [4]{-a_1/a_0}: 0: 0] \in X_{\mathbf {a}}(\mathbb {Q}_p)$. For the second condition take an equivalent surface such that $a_k, a_\ell, a_m \in \mathbb {Z}_p^{\times }$. In this case the equation $a_k X_k^4 + a_\ell X_\ell ^4 + a_m X_m^4 = 0$ defines a smooth proper curve of genus $3$ which has a $\mathbb {F}_p$-point if $p > 33$ because of the Weil conjectures [Reference DeligneDel80]. It thus has a $\mathbb {Z}_p$-point by Hensel's lemma.

Proof. This is the same as [Reference BrightBri15, Corolllary 7.10] except that it holds for more primes. The proof is the same except that we can bound $|\operatorname {Br}(X_{\mathbf {a}})/\operatorname {Br}(\mathbb {Q})|$ by $32$ because in this case $\operatorname {Br}_1(X_{\mathbf {a}}) = \operatorname {Br}(X_{\mathbf {a}})$ due to [Reference Ieronymou and SkorobogatovIS15, Theorem 1.1] and [Reference Gvirtz and SkorobogatovGvSk22, Main theorem]. In addition $|\operatorname {Br}_1(X_{\mathbf {a}})/\operatorname {Br}(\mathbb {Q})| \leq 32$ by [Reference BrightBri02, Appendix A].

Proof. In this situation $\operatorname {Br}_1(X_{\mathbf {a}}) = \operatorname {Br}(X_{\mathbf {a}})$ because of [Reference Ieronymou and SkorobogatovIS15, Theorem 1.1] and [Reference Gvirtz and SkorobogatovGvSk22, Main theorem]. The only possible case of [Reference BrightBri02, Appendix A] is A131, hence $\operatorname {Br}(X_{\mathbf {a}})/\operatorname {Br}(\mathbb {Q}) \cong \mathbb {Z}/ 2\mathbb {Z}$. We prove that $\mathcal {B}_{k \ell m; \sqrt {\theta _{\mathbf {a}}}}$ is prolific. This suffices because in that case $\mathcal {B}_{k \ell m; \sqrt {\theta _{\mathbf {a}}}}$ generates $\operatorname {Br}(X_{\mathbf {a}})/\operatorname {Br}(\mathbb {Q})$.

We apply Corollary 2.10. Let $\mathcal {X}$ be the smooth locus of the integral model given by the same equation. The special fiber $\mathcal {X}_{\mathbb {F}_p}$ is the smooth locus of the cone over the curve $C: a_k X_k^4 + a_\ell X_\ell ^4 + a_m X_m^4 = 0$. To compute $\partial _p \mathcal {B}_{k \ell m ; \sqrt {\theta _{\mathbf {a}}}}$ we may work over the maximal unramified extension of $\mathbb {Q}_p$ by Remark 3.2. We may thus assume that $a_k = a_\ell = a_m = 1$ and take $h = Z_k + \sqrt {-1} Z_m = 0$ as the equation defining a tangent line of $Z_{\mathbf {a}}^{k \ell m}$. Then

\[ \gamma^*_{\sqrt{\theta_{\mathbf{a}}}} \frac{h}{Z_m} \equiv \frac{X_k^2 X_m^2 + \sqrt{-1}X_m^4}{X_m^4} \equiv \frac{X_k^2}{X_m^2} + \sqrt{-1} \pmod{p}. \]

Lemma 2.11 implies that $\partial _p \mathcal {B}_{k \ell m ; \sqrt {\theta _{\mathbf {a}}}} = {X_k^2}/{X_m^2} + \sqrt {-1}$ since $v_p(\theta _{\mathbf {a}})$ is odd.

We check that the étale cover of $C$ defined by $Y^2 = X_k^2 + \sqrt {-1} X_m^2$ is geometrically irreducible using the Magma function IsAbsolutelyIrreducible(). The full script can be found on the authors webpage. Hence, $\mathcal {B}_{k \ell m ; \sqrt {\theta _{\mathbf {a}}}}$ is prolific by Corollary 2.10.

Proof. Let $\mathcal {X}$ be the integral model given by the same equation. It follows from looking at primitive solutions that $\mathcal {X}^{\textrm {sm}}(\mathbb {Z}_p) = X_{\mathbf {a}}(\mathbb {Q}_p)$. The special fiber $\mathcal {X}^{\textrm {sm}}_{\mathbb {F}_v}$ is irreducible, so $v_p$ can be extended to $\mathbb {Q}_p(X)$. By Lemma 2.4 it suffices to show that $\partial _p \mathcal {A} = 0$. Since $\mathcal {A}$ is normalized $\alpha ^* f$ will be non-zero when reduced modulo $p$. The reduction $\alpha ^* \tilde {f}$ cannot be zero on the special fiber because it has smaller degree than the defining equation so $v_p({\alpha ^* f}/{X_3^2}) = 0$. We are done by applying Lemma 2.11.

Proof. Because $-({a_k}/{a_\ell }) \in \mathbb {Q}_p^{\times 4}$ and $\theta _{\mathbf {a}}\not \in \mathbb {Q}_p^2$ we must have $-({a_m}/{a_n}) \not \in \mathbb {Q}_p^{\times 2}$. Let $\mathcal {X}$ be the integral model defined by the same equation. By looking at primitive solutions modulo $p^4$ we see that $\mathcal {X}^{\textrm {sm}}(\mathbb {Z}_p) = X_{\mathbf {a}}(\mathbb {Q}_p)$. Let $\mathcal {Y}$ be the integral model of $Y_{\mathbf {a}}$ given by the same equation. For the same reason $\mathcal {Y}^{\textrm {sm}}(\mathbb {Z}_p) = Y_{\mathbf {a}}(\mathbb {Q}_p)$.

Take a $b \in \mathbb {Z}_p^\times$ such that $-({a_k}/{a_\ell }) = b^4$. The special fiber $\mathcal {X}^{\textrm {sm}}_{\mathbb {F}_p}$ is equal to $\{X_k^4 = b^4 X_\ell ^4\} \setminus \{X_k = 0 = X_\ell \}$. This has four geometrically connected components, $\{X_k = \sqrt {-1}^t b X_\ell \}$ for $t = 0,1,2,3$. It thus has four or three connected components for $p \equiv 1 \pmod {4}$ or $p \equiv 3 \pmod {4}$, respectively.

The plane defined by $f$ is tangent to a $\mathbb {Q}$-point of $Y_{\mathbf {a}}$, so it is tangent to a $\mathbb {Z}_p$-point of $\mathcal {Y}^{\textrm {sm}}$. The reduction of this plane modulo $p$ is defined by $\tilde {f}$ because $\mathcal {A}$ is normalized. However, $\mathcal {Y}^{\textrm {sm}}_{\mathbb {F}_p} = \{Y_k^2 = b^4Y_\ell ^2\} \setminus \{Y_k = 0 = Y_\ell \}$ is a disjoint union of $2$ planes, so $\{ \tilde {f} = 0\}$ must be one of these two planes. Over each of these two planes lie two geometric components of $\mathcal {X}^{\textrm {sm}}_{\mathbb {F}_p}$.

Let $D$ be one of the connected components of $\mathcal {X}^{\textrm {sm}}_{\mathbb {F}_p}$ and $D^c$ the complement in $\mathcal {X}^{\textrm {sm}}_{\mathbb {F}_p}$. Let $\mathcal {X}^D := \mathcal {X}^{\textrm {sm}} \setminus D^c$. The special fiber $\mathcal {X}^D_{\mathbb {F}_p} = D$ is connected. By Lemma 2.4 it suffices to compute the residue $\partial _D \mathcal {A}$ for $\mathcal {X}^D$. We can extend $v_p$ to a valuation $v_D$ of $\mathbb {Q}_p(X)$. Since $v_p(\theta _{\mathbf {a}})$ is even and $\theta _{\mathbf {a}}\not \in \mathbb {Q}_p^{\times 2}$ we can apply Lemma 2.11 to find that

\[ \partial_D \mathcal{A} = \frac{1}{2} v_D\bigg(\frac{\alpha^* f}{X_3^2}\bigg) \in \operatorname{H}^1(\mathbb{F}_p, \mathbb{Z}/2\mathbb{Z}) \subset \operatorname{H}^1(D, \mathbb{Q}/\mathbb{Z}). \]

If $D \not \subset \{\alpha ^*\tilde {f} = 0\}$, then $v_D({\alpha ^*f}/{X_3^2}) = 0$ so $\partial _p \mathcal {A} = 0$.

If we change $\mathcal {A}$ by a constant algebra, then $\partial _D \mathcal {A}$ changes by a constant independent of $D$. It thus suffices to show that for a single choice of $f$ we have $\partial _D \mathcal {A} = \frac {1}{2}$ if $D \subset \{\alpha ^*\tilde {f} = 0\}$. By Remark 3.2 we may take $f = Y_k - b^2Y_\ell$ and, thus, $\alpha ^* f = X_k^2 - b^2X_\ell ^2$. Then

\[ \frac{X_k^2 - b^2X_\ell^2}{X_3^2} = -\frac{a_m X_n^4 + a_n X_n^4}{a_k(X_k^2 + b^2 X_\ell^2)X_3^2} = -p^i \frac{({a_m}/{p^i}) X_m^4 + ({a_n}/{p^i}) X_n^4}{a_k(X_k^2 +b^2 X_\ell^2)X_3^2}, \]

where $i = \min (v_p(a_m), v_p(a_n))$, so $i$ is odd, ${a_m}/{p^i}, {a_n}/{p^i} \in \mathbb {Z}_p$ and at least one of them is a unit. Because $D \not \subset \{X_k^2 = - b^2X_\ell ^2\}, \{({a_m}/{p^i}) X_m^4 + ({a_n}/{p^i})X_n^4 = 0\}$ we get $\partial _D(\mathcal {A}) = {i}/{2} = \frac {1}{2}$.

Note that $\partial _D( \mathcal {A})$ changes if we change $D$, so $\mathcal {A} \not \in \operatorname {Br}_0(X)$. The last statement follows by Hensel's lemma because $\alpha ^*\{ Y_k = -b^2 Y_\ell \} = \{X_k^2 = - b^2 X_\ell ^2\}$ has $\mathbb {F}_p$-points if and only if $p \equiv 3 \pmod {4}$.

Proof. If $\theta _{\mathbf {a}}\in \mathbb {Q}_p^2$, then $\mathcal {A} = 1$ as an element of $\operatorname {Br}(X_{\mathbb {Q}_p})$ so ${\rm inv}_p(\mathcal {A}(\cdot ))$ will be constant.

By taking an equivalent surface we may assume that $p \nmid a_k a_\ell$, $v_p(a_m), v_p(a_n) \in \{1, 3\}$ and $-({a_k}/{a_\ell }) \in \mathbb {Q}_p^{\times 4}$. We may then also assume that $\mathcal {A}$ is $p$-normalized. The only thing to show after applying Lemma 3.7 is the formula for ${\rm inv}(\mathcal {A_{\mathbf {u}}}(\cdot ))$. As in the lemma there exists a $t$ such that $\alpha ^* \tilde {f} = X_k^2 - (-1)^{t} X_\ell ^2$. Let $\mathcal {X}_{\mathbf {a} \mathbf {u}^2}$ be the obvious integral model. By definition, the reduction of $\alpha ^* \tilde {f}$ is equal to $\alpha ^* \tilde {f}_\mathbf {u} = u_k X_k^2 - (-1)^t u_\ell X_\ell ^2$. The desired statement follows after applying Lemma 3.7 to $X_{\mathbf {a} \mathbf {u}^2}.$

Proof. Assume that $u_0$ is the non-unit. Let $[y_0: y_1: y_2 : y_3] \in Y_{\mathbf {a}}(\mathbb {Q}_p)$ be such that $f = \sum _{i=0}^3 y_i a_i Y_i$ and, thus, $f_{\mathbf {u}} = \sum _{i=0}^3 y_i u_i a_i Y_i$. If $\mathcal {A}$ is $p$-normalized and $\mathcal {A}_{\mathbf {u}}$ is not, then $v_p(y_0) = 0$ but $v_p(y_i) \geq 1$ for $i \neq 1$. On the other hand, if $\mathcal {A}_{\mathbf {u}}$ is $p$-normalized but $\mathcal {A}$ is not, then $v_p(y_0) = - v_p(u_0) \leq -1$ but $v_p(y_i) \geq 1$ for $i \neq 1$. Both are impossible because $\sum _{i = 0}^3 a_i y_i^2 = 0$.

Proof. It has local solutions because it has a rational point $[2:1:0:0] \in X(\mathbb {Q})$.

The quadric $Y$ has a rational point $[4:1:0:0]$ so we have a normalized $\mathcal {A} = (1 + 4({X_1^2}/{X_0^2}), -1)$. The left-hand side is always positive so ${\rm inv}_{\infty }(\mathcal {A}(\cdot )) = 0$. By Lemma 3.6 ${\rm inv}_{p}(\mathcal {A}(\cdot )) = 0$ for $p \neq 2,7$. Because $7 \equiv 3 \pmod {4}$ Lemma 3.7 implies that ${\rm inv}_{7}(\mathcal {A}(\cdot ))$ is constant. Lastly, by looking at primitive solutions $\mathbf {x} = [x_0: x_1: x_2: x_3] \in X(\mathbb {Q}_2)$ modulo $4$ we see that $x_0 \in \mathbb {Z}_2^{\times }$ so we have $1 + 4({x_1^2}/{x_0^2}) \equiv 1 \pmod {4}$. The invariant ${\rm inv}_{2}(\mathcal {A}(\mathbf {x}))$ is thus always equal to $(1 + 4({x_1^2}/{x_0^2}),-1)_2 = (1, -1)_2 = 0$.

Proof. We first show the required bounds, let $H_1, C_{1, \varepsilon }$ and $H_2, C_{2, \varepsilon }$ be the required constants for $\rho _1$ and $\rho _2$, respectively. Then by the divisor bound

\begin{align*} |\rho_1 \rho_2 (\mathfrak{p}^k)| &= |\rho_1(\mathfrak{p}^{k})| |\rho_2(\mathfrak{p}^{k})| \leq (H_1 H_2)^k, \\ |\rho_1 \rho_2 (\mathfrak{n})| &= |\rho_1(\mathfrak{n})| |\rho_2(\mathfrak{n})| \leq C_{1, \varepsilon} C_{2, \varepsilon} \mathbf{N}(\mathfrak{n})^{2 \varepsilon} , \\ |\rho_1 * \rho_2(\mathfrak{p}^k)| &\leq \sum_{i = 0}^k |\rho_1(\mathfrak{p}^{i})||\rho_2(\mathfrak{p}^{k-i})| \leq \sum_{i = 0}^k H_1^{i} H_2^{k - i} \leq (H_1 + H_2)^k , \\ |\rho_1 * \rho_2(\mathfrak{n})| &\leq \sum_{\mathfrak{d}| \mathfrak{n}} |\rho_1(\mathfrak{d})| \bigg|\rho_2\bigg(\frac{\mathfrak{n}}{\mathfrak{d}}\bigg)\bigg| \leq \tau(\mathfrak{n}) C_{1, \varepsilon} C_{2, \varepsilon} \mathbf{N}(\mathfrak{d})^{\varepsilon} \mathbf{N}\bigg(\frac{\mathfrak{n}}{\mathfrak{d}}\bigg)^{\varepsilon} \ll_{\varepsilon} C_{1, \varepsilon} C_{2, \varepsilon} \mathbf{N}(\mathfrak{n})^{2 \varepsilon}. \end{align*}

If $\mathfrak {p}$ is totally split, then $\rho _1 \rho _2(\mathfrak {p}) = \rho _1(\mathfrak {p}) \rho _2(\mathfrak {p}), \rho _1 * \rho _2(\mathfrak {p}) = \rho _1(\mathfrak {p}) + \rho _2(\mathfrak {p})$. The class function of $\rho _1 \rho _2$ and $\rho _1 * \rho _2$ is thus the product and the sum of the class functions, respectively, of $\rho _1$ and $\rho _2$.

Proof. For $K = \mathbb {Q}$ this is the first part of the proof of [Reference Loughran and MatthiesenLM19, Lemma 2.4]. The argument generalizes to general number fields. To spell out the details, let $L'/L$ be an extension such that $L'/K$ is Galois and $\chi$ corresponds via class field theory to a character of $\Gamma := \textrm {Gal}(L'/K)$, which we also denote by $\chi$. Let $\varphi : \Gamma \to \mathbb {C}$ be the class function defining $\rho$. The subgroup $N := \textrm {Gal}(L'/ L) \subset \Gamma$ is normal and $\varphi$ is invariant under translation by $N$. Then

\[ |\Gamma|m(\chi \rho) = \sum_{g \in \Gamma} \chi(g) \varphi(g) = \sum_{gN \in \Gamma/N} \varphi(gN) \sum_{h \in gN} \chi(h) = \sum_{gN \in \Gamma/N} \varphi(gN) \sum_{h' \in N} \chi(gh'). \]

The sum $\sum _{h' \in N} \chi (gh')$ is $0$ unless $\chi$ is trivial on $N$. In that case $\chi$ is a character of $\Gamma /N = \textrm {Gal}(L/K)$ as desired.

Proof. As $\varphi$ is a class function we can write it as a sum over the irreducible characters of $\Gamma$:

\[ \varphi = \sum_{\chi} \lambda_\chi \chi. \]

We have the trivial inequality $|\lambda _\chi | \leq \chi (1)H = O_{\Gamma, H}(1)$.

This leads to the equality

\[ L(\rho, s) = G(s)\prod_{\chi} L(\chi, s)^{\lambda_\chi}, \]

where $L(\chi, s)$ is the Artin $L$-function of $\chi$ and

\[ G(s) = \prod_{\mathfrak{p}} \bigg(1 + \frac{\rho(\mathfrak{p})}{\mathbf{N}(\mathfrak{p})^s} + \frac{\rho(\mathfrak{p}^2)}{\mathbf{N}(\mathfrak{p})^{2s}} + \cdots\bigg) \prod_{\chi}\textrm{det}\bigg(1- \frac{\pi_{\chi}(\textrm{Frob}_{\mathfrak{p}})}{\mathbf{N}(\mathfrak{p})^s}\bigg)^{\lambda_\chi}. \]

Here $\pi _\chi$ is the representation corresponding to $\chi$.

Let us consider the local factors of $G(s)$, there are three cases. We use the standard notation $s = \sigma + it$.

1. (i) If $\mathfrak {p} \in S$, then the contribution of the factors $\textrm {det}(1- {\pi _{\chi }(\textrm {Frob}_{\mathfrak {p}})}/{\mathbf {N}(\mathfrak {p})^s})$ is $e^{O_{n, \Gamma, H}(|S|)}$. For the other factors we have a bound for any $\varepsilon > 0$ of

   \[ 1 + \frac{|\rho(\mathfrak{p})|}{\mathbf{N}(\mathfrak{p})^{\sigma}} + \cdots \leq 1 + C_{\varepsilon} \frac{\mathbf{N}(\mathfrak{p})^{\varepsilon}}{\mathbf{N}(\mathfrak{p})^{\sigma}} + \cdots = 1 + C_{\varepsilon} \mathbf{N}(\mathfrak{p})^{\varepsilon - \sigma}(1 - \mathbf{N}(\mathfrak{p})^{\varepsilon - \sigma})^{-1}. \]

   The total contribution of these parts in the vertical strip $\sigma > \frac {3}{4}$ is thus $e^{ O_{n,\Gamma, H, C_{\varepsilon }}(|S|)}$.

2. (ii) The contribution of the primes $\mathfrak {p}$ which are not totally split is absolutely convergent for $\sigma > \frac {1}{2}$ and in the vertical strip $\sigma > \frac {3}{4}$ is $O_{n, \Gamma, H, C_{\varepsilon }}(1)$.

3. (iii) For the other primes $\mathfrak {p}$ we have, by definition, $\rho (\mathfrak {p}) = \sum _\chi \lambda _\chi \textrm {Tr}(\pi _{\chi }(\textrm {Frob}_{\mathfrak {p}}))$ so the associated factor is

   \[ 1 + O_{\Gamma, H}\bigg(\frac{|\rho(\mathfrak{p})| + |\rho(\mathfrak{p}^2)|}{\mathbf{N}(\mathfrak{p})^{2 \sigma}} + \frac{|\rho(\mathfrak{p})| + |\rho(\mathfrak{p}^2)| + |\rho(\mathfrak{p}^3)|}{\mathbf{N}(\mathfrak{p}^{3\sigma})} + \cdots \bigg). \]
   For all $\sigma > \varepsilon > 0$, this is bounded by
   \[ 1 + C_{\varepsilon} O_{\Gamma, H}\bigg(\frac{\mathbf{N}(\mathfrak{p})^{2 \varepsilon}}{\mathbf{N}(\mathfrak{p})^{2 \sigma}} + \cdots \bigg) = 1 + C_{\varepsilon}\mathbf{N}(\mathfrak{p})^{2\varepsilon - 2\sigma} O_{\Gamma}\big((1 - \mathbf{N}(\mathfrak{p})^{ \epsilon - \sigma})^{-1}\big). \]
   The total contribution of all of these primes is thus absolutely convergent for $\sigma > \frac {1}{2}$ and is $O_{n, \Gamma, H,C_{\varepsilon }}( 1)$ in the vertical strip $\sigma > \frac {3}{4}$.

Combining everything we see that $G(s) = e^{O_{n, \Gamma, H, C_{\varepsilon }}(|S|)}$ in the vertical strip $\sigma > \frac {3}{4}$.

By the Brauer-induction theorem [Reference SerreSer77, Theorem 18] we may assume that the $\chi$ are Hecke characters such that the $L$-series $L(\chi, s)$ has conductor at most $q$ at the cost of changing $\lambda _\chi$. However, $\lambda _\chi$ will still be $O_{\Gamma, H}(1)$.

Let $T = e^{\sqrt {\log x}}$. Choose $1 > 1 - \delta \geq \frac {3}{4}$ such that $s$ has at least a distance $\delta$ from every zero of $L(\chi, s)$ for all $s$ in the region

(4.3)\begin{equation} \sigma \geq 1 -\delta,\quad |t| \leq T. \end{equation}

We can choose $\delta$ such that $\delta \gg _A q^{-{1}/{2A}} \geq 1/\log T = 1/\sqrt {\log x}$ by the standard zero-free region of Hadamard and de la Vallée Poussin and Siegel's theorem for Hecke characters [Reference FogelsFog62]. We have the following inequality in the region (4.3):

(4.4)\begin{equation} | \log L(\chi, s)| \ll_A \log \log q t \ll_A \log \log T. \end{equation}

The case of Dirichlet characters is [Reference Montgomery and VaughanMV07, Theorem 11.4]. The proof for general Hecke characters is the same.

Exponentiating we obtain

\[ |L(\chi, s)^{\lambda_\chi}| = (\log T )^{O_{A, \Gamma, H}(1)} = (\log x )^{{O_{A, \Gamma, H}(1)}} \]

and, thus,

(4.5)\begin{equation} |F(s)| = e^{O_{n, \Gamma, H, C_{\epsilon}}(|S|)} (\log x)^{O_{A, \Gamma, H}(1)}. \end{equation}

We now apply Perron's formula [Reference TenenbaumTen95, Theorem II.3.3] for $\kappa = 1 + {1}/{\log x}$:

\[ \sum_{\mathbf{N}(\mathfrak{n}) \leq x} \rho(\mathfrak{n}) \log \bigg(\frac{x}{\mathbf{N}(\mathfrak{n})}\bigg) = \frac{1}{2 \pi i} \int_{\kappa - i \infty}^{\kappa + i\infty} F(s) x^s \frac{ds}{s^2}. \]

The function $F(s)$ has an analytic continuation to the region (4.3). We can, thus, change the contour to be the straight lines between $\kappa - i \infty$, $\kappa - iT$, $1 - {\delta }/{\log (1 + T)} - iT$, $1 - {\delta }/{\log (1 + T)} + iT$, $\kappa + iT$, $\kappa + i\infty$. The integrals over everything except the line between $1 - {\delta }/{\log (1 + T)} - iT$, $1 - {\delta }/{\log (1 + T)} + iT$ are

\[ \ll e^{O_{n, \Gamma, C_{\epsilon}}(|S|)} (\log x)^{O_{A, \Gamma, H}(1)} \frac{x^{\kappa}}{T} \ll_{A, \Gamma, H} e^{O_{n,\Gamma, C_{\epsilon}}(|S|)} x e^{- ({1}/{2}) \sqrt{ \log x}}. \]

The contribution of this last line segment is

\[ \ll e^{O_{n,\Gamma, H, C_{\epsilon}}(|S|)} (\log x)^{O_{A, H}(1)} \frac{x^{1 - \delta}}{\delta} \ll_{A, \Gamma, H} e^{O_{n,\Gamma, H, C_{\epsilon}}(|S|)} x e^{- ({1}/{2}) \sqrt{ \log x}}. \]

Equation (4.9) follows from partial summation.

Proof. Consider the induced frobenian multiplicative function $\textrm {Ind}^{\mathbb {Q}}_{K}(\rho )$. We may take its $H$ constant to be $H [K :\mathbb {Q}]$ by (4.1) and its $C_{{1}/{3}}$ constant may be taken to be $\ll C_{{1}/{6}}$. The proof thus reduces to proving that if $\rho$ is a frobenian multiplicative function over $\mathbb {Q}$, then

\[ \sum_{n \leq x} |\rho(n)| \ll C_{1/3} x (\log x)^{H}. \]

Write $n = m k^2$ with $m$ square-free. We claim that $|\rho (n)| \leq C_{1/3} \tau _H(m) k$. To prove this we reduce to the case when $n$ is a prime power. If $n = p$, then $|\rho (p)| \leq H = \tau _H(p)$. If $n = p^i$ for $i > 1$, then $|\rho (p^i)| \leq C_{1/3} p^{i/3} \leq p^{\lfloor i \rfloor /2} = \sqrt {k}$.

We thus get an upper bound

\[ \leq \sum_{m k^2 \leq x} |\rho(m k^2)| \leq C_{1/3} \sum_{m k^2 \leq x} \tau_H(m) k \leq C_{1/3 } x \sum_{m \leq x} \frac{\tau_H(m)}{m} \leq C_{1/3} x (\log x)^{H}. \]

Proof. If $q(\rho ) \leq (\log x)^{([K :\mathbb {Q}] H + C)\varepsilon ^{-1}}$, then this follows from Lemma 4.9. Otherwise it follows from Lemma 4.10.

Proof. By considering the induced frobenian multiplicative function we may assume that $K = \mathbb {Q}$, note that $|S|$ may increase at most by the amount of prime divisors $\omega (q)$ of the conductor $q$, but by the divisor bound $e^{O(\omega (q))} \leq \tau _{O(1)}(q) \ll _{\varepsilon } q^{\varepsilon }$.

The decomposition of $L$-series $L(\rho,s) = L_n(\rho, s) \zeta (s)^{m(\rho )}$ corresponds to writing $\rho$ as a convolution $\rho _n * \tau _{m(\rho )}$. Here $\rho _n = \rho * \tau _{-m(\rho )}$. The function $\tau _{-m(\rho )}$ is frobenian multiplicative of mean $-m(\rho )$ by the divisor bound, so $\rho _n$ is frobenian multiplicative of mean $0$ by Lemma 4.3.

By the hyperbola method we have

(4.8)\begin{align} \sum_{n \leq x} \rho(n) &= \sum_{n \leq \sqrt{x}} \rho_n(n) \sum_{m \leq {x}/{n}} \tau_{m(\rho)}(m) + \sum_{m \leq \sqrt{x}} \tau_{m(\rho)}(m) \sum_{n \leq {x}/{m}} \rho_n(n) \nonumber\\ &\quad -\sum_{n \leq \sqrt{x}} \rho_n(n) \sum_{m \leq \sqrt{x}} \tau_{m(\rho)}(m). \end{align}

It follows from the Selberg–Delange method [Reference TenenbaumTen95] that for any $M \in \mathbb {N}, y \geq 2$,

(4.9)\begin{align} \sum_{m \leq y} \tau_{m(\rho)}(m) &= y (\log y)^{m(\rho) - 1} \bigg(\sum_{j = 0}^M \frac{\gamma_j(m(\rho))}{\Gamma(m(\rho) - j) j! (\log y)^j}\bigg) \nonumber\\ &\quad + O_M((\log y)^{m(\rho) -M - 2}). \end{align}

From this, Corollary 4.11, Lemma 4.10 and partial summation, we deduce that the contribution of the latter two sums is

\[ \ll_{ B, \varepsilon, H, n} q^{\varepsilon} x (\log x)^{m(\rho) - 1 - B }. \]

This is part of the error term of (4.7) if we choose $B= N + 1$.

Filling (4.9) into the first sum of (4.8) gives an error term

\[ \ll x (\log x)^{m(\rho) - M - 2} \sum_{n \leq \sqrt{x}} \frac{|\rho_n(n)|}{n} \ll x (\log x)^{m(\rho) - M - 2} (\log x)^{H + |m(\rho)|}. \]

With the choice $M = H + 2\lceil |m(\rho )| \rceil + N$ this is part of the error term of (4.7).

The remaining part to deal with is the sum over the following terms:

\[ \frac{\gamma_j(m(\rho)) x}{\Gamma(m(\rho) - j) j!} \sum_{n \leq \sqrt{x}} \frac{\rho_n(n)}{n} \bigg(\log \frac{x}{n}\bigg)^{m(\rho) - 1 - j}. \]

We expand $(\log ({x}/{n}))^{m(\rho ) - 1 - j} = (\log x- \log n)^{m(\rho ) - 1 - j}$ to obtain

\[ \bigg(\log \frac{x}{n}\bigg)^{m(\rho) - 1 - j} = \sum_{h = 0}^{\infty} (-1)^h \binom{m(\rho) - 1 - j}{h}(\log x)^{m(\rho) - 1 - j - h} (\log n)^h. \]

Using the identity $\binom {m(\rho ) - 1 - j}{h} = {\Gamma (m(\rho ) - j)}/{\Gamma (m(\rho ) - j - h) h!}$, we have the sum over $j,h$ of

\[ \frac{\gamma_j(m(\rho)) x (\log x)^{m(\rho) - 1 - j - h}}{\Gamma(m(\rho) - j - h) j! h!} \sum_{n \leq \sqrt{x}} \frac{\rho_n(n)}{n} (\log n)^{h}. \]

The sum over $n$ converges by Corollary 4.11 and partial summation. The total series is thus equal to $L_n^{(h)}(\rho,1)$ by Abel's theorem. The resulting error bounds are

\begin{align*} \sum_{n > \sqrt{x}} \frac{\rho_n(n)}{n} (\log n)^{h} & \ll_{C, \varepsilon} (h + 1) e^{|S|} q^{\varepsilon} (\log x)^{-C + h}, \\ \sum_{n = 1}^{\infty} \frac{\rho_n(n)}{n} (\log n)^{h} & \ll_{C, \varepsilon} (h + 1)e^{|S|} q^{\varepsilon}. \end{align*}

We use the first bound if $j + h \leq N$ and the second bound if $j + h > N$. Taking $C \geq 2 N + 1$, using the bounds $\Gamma (m(\rho ) - j - h) \gg 1$ and [Reference TenenbaumTen95, II.5.(7)] proves the theorem.

Proof. The first statement is clear if $U$ is infinite.

If it is finite, choose $\mathfrak {a_i}$ such that $L_U = \prod _{i \in I} \mathbf {N}(\mathfrak {a_i})$. Recall that for any number field $K$ there exists a constant $C_K$ such that $\#\{ \mathfrak {n} \in I_{K}: \mathbf {N}(\mathfrak {n}) \leq T\} \sim C_K T$ as $T \to \infty$. Now note that $U$ contains the set $\{(\mathfrak {u}_i)_{i \in I} : \mathbf {N}(\mathfrak {u_i}) \leq \mathbf {N}(\mathfrak {a_i}) \}$ which has size $\gg \prod _{i \in I} \mathbf {N}(\mathfrak {a_i}) = L_U$. It is also contained in the set $\{(\mathfrak {u}_i)_{i \in I} : \prod _{i \in I}\mathbf {N}(\mathfrak {u_i}) \leq L_U \}$ which is of size $\ll L_U (\log L_U)^{|I| -1}$.

The latter statements follow directly from the definitions.

Proof. We prove this via induction on $n$: if $n = 1$, then the result is equal to the assumption; otherwise, we can write

\[ \sum_{\mathbf{u} \in U} \prod_{i \in I} f_i(\mathfrak{u}_i) = \sum_{\mathfrak{u}_1} f_1(\mathfrak{u}_1) \sum_{\mathbf{u} \in U[\mathfrak{u}_1]} \prod_{i \in I} f_i(\mathfrak{u}_i). \]

By the induction hypothesis we can approximate the sum over the slice $U[\mathfrak {u}_1]$, because it is also downward-closed. The error term is

\begin{align*} & \sum_{\mathfrak{u}_1} f_1(\mathfrak{u}_1) \sum_{J \subsetneq \{2, \ldots, n\}} \int_{U_{\mathbb{R}}[\mathbf{N}(\mathfrak{u}_1)]} \prod_{j \in J} g_j(x_j) \,dx_j \prod_{i \in \{2, \ldots, n\} \setminus J} h_i(x_i) \,d x_i \nonumber\\ &\quad = \sum_{J \subsetneq \{2, \ldots, n\}} \int \sum_{\mathbf{N}(\mathfrak{u}_1) \in U_{\mathbb{R}}[x_i : i \in \{2, \ldots, n\}]} f_1(\mathfrak{u}_1) \prod_{j \in J} g_j(x_j) \,dx_j \prod_{i \in \{2, \ldots, n\} \setminus J} h_i(x_i)\,d x_i. \end{align*}

As $U_{\mathbb {R}}[x_i : i \in \{2, \ldots, n\}]$ is downward-closed, there exists an $x > 0$ such that it is equal to $[0, x]$ or $[0,x)$. Applying the inequality $\sum _{\mathbf {N}(\mathfrak {u}_1) \leq x} f_1(\mathfrak {u}_1) \leq \int _0^x g_1(x_1)\,d x_1 +\int _0^x h_1(x_1)\,dx_1$ we obtain

(4.10)\begin{equation} \sum_{ \{2, \ldots, n\} \neq J \subsetneq \{1, \ldots, n\}} \int_{U_{\mathbb{R}}} \prod_{j \in J} g_j(x_j)\,dx_j \prod_{i \in \{1, \ldots, n\} \setminus J} h_i(x_i)\,d x_i. \end{equation}

The main term is given by

\[ \sum_{\mathfrak{u}_1} f_1(\mathfrak{u}_1) \int_{U_{\mathbb{R}}[\mathbf{N}(\mathfrak{u}_1)]} \prod_{i = 2}^n g_i(x_i)\,dx_j = \int \bigg(\sum_{\mathbf{N}(\mathfrak{u}_1) \in U_{\mathbb{R}}[x_i : i \in \{2, \ldots, n\}]} f_1(\mathfrak{u}_1)\bigg) \prod_{i = 2}^s g_i(x_i)\, dx_i. \]

Using again that $U_{\mathbb {R}}[x_i : i \in \{2, \ldots, n\}]$ is downward-closed, we can approximate the sum over $\mathfrak {u}_1$ using the assumption. The resulting error term is

\[ \int_{U_{\mathbb{R}}} h_1(x_1)\,d x_1 \prod_{i = 2}^s g_i(x_i)\,dx_i, \]

which together with (4.10) gives the required error term. The main term is

\[ \int_{\mathbf{x} \in U_{\mathbb{R}}} \prod_{i = 1}^n g_i(x_i)\,dx_i \]

as desired.

Proof. That $\alpha$ is bilinear is clear. In what follows, we write $\mathfrak {u}$ for one of $n, m, \mathfrak {n}, \mathfrak {m}$ and $\mathfrak {v}$ for the other variable of $\alpha$ to keep the arguments uniform. If we fix the variable $\mathfrak {v}$, then $\alpha (\mathfrak {v}, \cdot )$, respectively $\alpha (\cdot, \mathfrak {v})$ is a product of a Hecke characters modulo $\mathfrak {v}$ or a Hecke character of the field extension $\mathbb {Q}(\sqrt {-1}, \sqrt [4]{\mathbf {N}(\mathfrak {v})})$ which has discriminant dividing $2^{24}\mathbf {N}(\mathfrak {v})^6$. The norm of the modulus of this Hecke character must divide this discriminant. It only remains to show the non-principality.

We first handle the case that $\mathfrak {u}$ is an odd totally split prime $\mathfrak {p}$. It then suffices to show that there exists a single $\mathfrak {v} \equiv 1 \pmod {2^{24}}$ such that the value of $\alpha (\mathfrak {p}, \mathfrak {v})$, respectively $\alpha (\mathfrak {v},\mathfrak {p})$ is not $1$. Except in the cases 5, 6 and 8 this can be done by choosing $\mathfrak {v}$ as a totally split prime and applying Chebotarev density.

Consider now the cases of 5, 6 and 8 where we have to choose an ideal $\mathfrak {v} \in I_{\mathbb {Z}[i]}$. Start by finding a prime $\mathfrak {q} \equiv 1 \pmod {2^{24}}$ such that $\bigl (\frac {\mathbf {N}(\mathfrak {p})}{\mathfrak {q}}\bigr )_{4}$ is equal to $i$ or $-i$, i.e. $\mathbf {N}(\mathfrak {p})$ is not a square modulo $\mathbf {N}(\mathfrak {q})$. This is possible because of Dirichlet's theorem. Then $\alpha (\mathfrak {p}, \mathfrak {v})$, respectively $\alpha (\mathfrak {v},\mathfrak {p})$ is not equal to $1$ for one of two choices $\mathfrak {v} = \mathfrak {q}, \overline {\mathfrak {q}}$.

The only remaining case is to find some $n \in \mathbb {N}$ such that $\bigl (\frac {n}{\mathfrak {p}}\bigr )_{4} \bigl (\frac {\mathbf {N}(\mathfrak {p})}{n}\bigr ) \neq 1$. For this choose $n \equiv 1 \pmod {2^{24}}$ which is a not a square modulo $\mathbf {N}(\mathfrak {p})$. By quadratic reciprocity

\[ \bigg(\frac{n}{\mathfrak{p}}\bigg)_{4} \bigg(\frac{\mathbf{N}(\mathfrak{p})}{n}\bigg) = \bigg(\frac{n}{\mathfrak{p}}\bigg)_{4} \bigg(\frac{n}{\mathbf{N}(\mathfrak{p})}\bigg) = \bigg(\frac{n^3}{\mathfrak{p}}\bigg)_{4} \neq 1. \]

To deduce the general case assume that $p$ is an odd prime such that $v_p(\mathbf {N}(\mathfrak {m})) = 1$. In this case we can write $\mathfrak {m} = \mathfrak {p} \mathfrak {m}'$ such that $\mathbf {N}(\mathfrak {p}) = p$ and $p, \mathbf {N}(\mathfrak {m}')$ are coprime. In this case $\alpha (\mathfrak {m}, \cdot ) = \alpha (\mathfrak {p}, \cdot ) \alpha (\mathfrak {m}', \cdot )$ is a product of a non-principal Hecke character such that $p$ divides the conductor and a Hecke character whose conductor is coprime to $p$. The conductor of such a product has to be divisible by $p$. The case $v_p(\mathbf {N}(\mathfrak {n})) = 1$ is completely analogous.

Proof. Write $B(M, N)$ for the left-hand side of the expression in Proposition 4.21. We may assume by symmetry that $M \leq N$. Let $k$ be a natural number to be chosen later. By applying Hölder's inequality to the sum over $\mathfrak {m}$ and switching the order of summation we obtain

\begin{align*} |B(M,N)|^{k} &\ll M^{k-1} \sum_{\mathbf{N}(\mathfrak{m}) \leq M} \bigg\lvert\sum_{\substack{(\mathfrak{m}, \mathfrak{n}) \in U \\ \mathbf{N}(\mathfrak{n}) \leq N}} b_{\mathfrak{n}} \alpha(\mathfrak{m}, \mathfrak{n})\bigg\rvert^{k} \\ &\leq M^{k-1} \sum_{\mathbf{N}(\mathfrak{n}_1), \ldots, \mathbf{N}(\mathfrak{n}_k) \leq N} \bigg\lvert \sum_{\substack{(\mathfrak{m}, \mathfrak{n}_i) \in U \\ \mathbf{N}(\mathfrak{m}) \leq M}} c_{\mathfrak{m}} \alpha(\mathfrak{m}, \mathfrak{n}_1 \cdots \mathfrak{n}_k) \bigg\rvert, \end{align*}

where $|c_\mathfrak {m}| = 1$ is the complex number such that $c_\mathfrak {m} (\sum b_{\mathfrak {n}} \alpha (\mathfrak {m}, \mathfrak {n}))^{k}$ is a positive real.

At the cost of a factor of $k$ we may sum only over those tuples such that $\mathbf {N}(\mathfrak {n}_1), \ldots, \mathbf {N}(\mathfrak {n}_{k -1}) \leq \mathbf {N}(\mathfrak {n}_k) \leq N$. Note that the condition $(\mathfrak {m}, \mathfrak {n}_i) \in U$ is then implied by $(\mathfrak {m}, \mathfrak {n}_k) \in U$ because $U$ is downward-closed. Write $\mathfrak {l} = \mathfrak {n}_1 \cdots \mathfrak {n}_{k-1}$. The amount of tuples giving the same $\mathfrak {l}$ is

\[ \sum_{\substack{\mathbf{N}(\mathfrak{n}_1), \ldots, \mathbf{N}(\mathfrak{n}_{k-1}) \leq \mathbf{N}(\mathfrak{n}_k) \\ \mathfrak{n}_1 \cdots \mathfrak{n}_{k-1} = \mathfrak{l}}} 1 \leq \tau_{k-1}(\mathfrak{l}). \]

Applying Cauchy–Schwarz gives that $|B(M,N)|^{2k}$ is bounded by

\begin{align*} & k^2 M^{2k-2} \bigg(\sum_{\mathbf{N}(\mathfrak{n}_k) \leq N} \sum_{\mathbf{N}(\mathfrak{l}) \leq N^{k-1}} \tau_{k-1}(\mathfrak{l})^2\bigg) \bigg(\sum_{\mathbf{N}(\mathfrak{n}_k) \leq N} \sum_{\mathbf{N}(\mathfrak{m}_1), \mathbf{N}(\mathfrak{m}_2) \leq M} \bigg\lvert \sum_{\mathfrak{l}} \alpha(\mathfrak{m}_1 , \mathfrak{l}) \overline{\alpha(\mathfrak{m}_2 , \mathfrak{l})}\bigg\rvert\bigg) \\ &\quad \ll k^2 M^{2k - 2} N^{k + 1} (\log N)^{k^2 -2 k} \sum_{\mathbf{N}(\mathfrak{m}_1), \mathbf{N}(\mathfrak{m}_2) \leq M} \bigg\lvert \sum_{\mathbf{N}(\mathfrak{l}) \leq N^{k-1}} \alpha(\mathfrak{m}_1 , \mathfrak{l}) \overline{\alpha(\mathfrak{m}_2 , \mathfrak{l})}\bigg\rvert. \end{align*}

Every pair $\mathfrak {m}_1, \mathfrak {m}_2$ such that $v_p(\mathbf {N}_K(\mathfrak {m}_1 \mathfrak {m}_2)) \neq 1$ for all $p \nmid q$ has the property that their product can be written (non-uniquely) as $\mathfrak {m}_1 \mathfrak {m}_2 = \mathfrak {u} \mathfrak {v}^2 \mathfrak {w}^3$ with $\mathbf {N}(\mathfrak {u})| q$. The number of such pairs is by the divisor bound

\[ \ll \sum_{\substack{\mathbf{N}(\mathfrak{u} \mathfrak{v}^2 \mathfrak{w}^3) \leq M^2\\ \mathbf{N}(\mathfrak{u}) \mid q}} \tau(\mathfrak{u} \mathfrak{v}^2 \mathfrak{w}^3) \ll \sum_{\mathbf{N}(\mathfrak{u}) \mid q} \mathbf{N}(\mathfrak{u})^{{1}/{2}} \sum_{\mathbf{N}(\mathfrak{v}) \mathbf{N}(\mathfrak{w})^{{3}/{2}} \leq M} \mathbf{N}(\mathfrak{v} \mathfrak{w})^{{1}/{2}}\ll q M^{{3}/{2}}. \]

The total contribution of this case is, thus, $\ll k^2 q M^{2k-{1}/{2}} N^{2k} (\log N)^{k^2-2k}$.

We claim that in the other cases $\alpha (\mathfrak {m}_1 , \cdot ) \overline {\alpha (\mathfrak {m}_2, \cdot })$ is a non-principal character. As $\alpha$ is an oscillating bilinear character the norm of its modulus is at most $q \mathbf {N}_K(\mathfrak {m}_1 \mathfrak {m_2})^{A}$. The sum over $\mathfrak {l}$ is, in this case,

\[ \ll q^{{1}/{(d_L + 1})} \mathbf{N}_K(\mathfrak{m}_1 \mathfrak{m_2})^{{A}/{(d_L +1)}} (\log M)^{d_L}N^{(k-1)({(d_L-1)}/{(d_L + 1)})} \]

by Polya–Vinogradov for number fields [Reference LandauLan18].

The total contribution of this part is, thus,

\[ \ll k^2 q M^{2k+ {2 A}/{(d_L + 1)}} (\log M)^{d_L} N^{{( 2k d_L + 2)}/{(d_L + 1})} (\log N)^{k^2 -2k}. \]

Choosing $k$ sufficiently large such that $ {(2k d_L + 2 + 2A)}/{(d_L + 1)} \leq 2k - \frac {1}{2}$ and $d_L \leq 2k$, taking $\delta = {1}/{4k}$ and using that $M \leq N$ gives the desired bound.

It remains to prove the claim. We may assume that there exists a prime $p \nmid q$ such that $v_{p}(\mathbf {N}(\mathfrak {m}_1)) = 1$ and $p \nmid \mathbf {N}(\mathfrak {m}_2)$. There exists a natural number $d$ such that $\overline {\alpha (\mathfrak {m}_2 , \cdot )} = \alpha (\mathfrak {m}_2 , \cdot )^d$ because it is a Hecke character. We thus have to show that $\alpha (\mathfrak {m}_1 \mathfrak {m}_2^d, \cdot )$ is non-principal. However, $v_p(\mathbf {N}(\mathfrak {m}_1 \mathfrak {m}_2^d)) = 1$ so this is true by definition.