\section{Cubic Frobenius conditions}
\label{sec:frobenius}

We deduce Theorem~\ref{thm:reservoir} from the arithmetic progression
law and the cubic trace estimate. Let
$K=F(b_1^{1/3},\ldots,b_s^{1/3})$ and $H=\Gal(K/F)$ as in that theorem.
Kummer theory identifies $H$ with an elementary abelian $3$-group.
Complex conjugation fixes the real cube roots and inverts $\zeta_3$;
therefore
\[
 \Gal(K/\Q)=H\rtimes\langle\sigma\rangle,\qquad
 \sigma h\sigma^{-1}=h^{-1}.
\]
The commutators with $\sigma$ generate $H$, since $h^{-2}=h$ in a
group of exponent three. If $H\ne1$, the maximal abelian subextension
of $K/\Q$ is consequently $F$. In particular,
\begin{equation}\label{eq:intersection}
 K\cap\Q(\zeta_M)=F\qquad(3\mid M).
\end{equation}
For $H=1$ this is immediate as well.

Every nontrivial character $\eta\in\widehat H$ is, by class field
theory, a cubic Hecke character of $F$ with $\eta^\sigma=\eta^{-1}$.
Let $\mathcal R$ contain one character from each pair
$\{\eta,\eta^{-1}\}$ of nontrivial characters. Finite Fourier inversion
on $H$ gives, for an inversion-stable $\mathcal C$,
\begin{equation}\label{eq:finite-fourier}
 \ind_{\mathcal C}(h)=\frac{|\mathcal C|}{|H|}
 +\sum_{\eta\in\mathcal R}c_\eta
       \bigl(\eta(h)+\eta(h)^{-1}\bigr),\qquad
 c_\eta=\frac1{|H|}\sum_{g\in\mathcal C}\overline{\eta(g)}.
\end{equation}
The coefficients $c_\eta$ are real because $\mathcal C$ is
inversion-stable. At an unramified split rational prime the paired
character value is exactly $t_\eta(p)$.

Expand the fixed residue-class indicator as
\begin{equation}\label{eq:dirichlet-expand}
 \ind_{p\bmod M\in U}
 =\sum_{\chi\bmod M}b_\chi\chi(p),\qquad
 b_\chi=\frac1{\varphi(M)}\sum_{u\in U}\overline{\chi(u)}.
\end{equation}
This identity retains the zero extension on nonunits. Inserting
\eqref{eq:finite-fourier} and \eqref{eq:dirichlet-expand},
Theorem~\ref{thm:trace} makes every nonconstant term $o(x/L)$.
Proposition~\ref{prop:progressions}, the prime number theorem in fixed
progressions, and \eqref{eq:c-eps} evaluate the constant term as
\[
 \left(\frac{|U|}{\varphi(M)}\frac{|\mathcal C|}{|H|}
 c_\eps\int\Phi+o(1)\right)\frac{x}{L}.
\]
This proves Theorem~\ref{thm:reservoir}. Relation~\eqref{eq:intersection}
also shows directly why no further compatibility factor occurs between
the rational congruence and the cubic condition.

\section{The finite obstructions and the final sieve}
\label{sec:final}

\subsection{Choosing the quadratic progression}
We give the elementary choice of progression explicitly; it is the
quadratic reciprocity argument of~\cite[Lemma~1]{AnwarPappalardi}.
Let $V$ be the $\F_2$-subspace of $\Q^\times/\Q^{\times2}$ generated
by the classes of the $a_i$. Failure of \textup{(A)} is equivalent to
the existence of a linear functional $\ell:V\to\F_2$ such that
\[
 \ell([a_i])=1\qquad(1\leq i\leq r).
\]
Indeed, the only obstruction to specifying these values is a relation
among an odd number of the $[a_i]$.

Enlarge $V$ by adjoining $[-1],[2],[3]$ and the classes of the primes
dividing $a_1\cdots a_r$, and extend $\ell$ to this space. If possible,
choose such an extension with $\ell([-3])=1$. If this is impossible,
$[-3]$ belongs to $V$ and every admissible functional takes the value
zero on it. Expressing $[-3]$ as a sum of classes $[a_i]$ shows that
it is the product of an even number of the $a_i$ modulo squares.

Let $\mathcal T$ consist of $3$ and the odd prime divisors of the bases,
and put
\[
 M_0=8\prod_{q\in\mathcal T}q.
\]
Choose a unit class modulo $8$ with the prescribed Legendre values for
$-1$ and $2$. For each $q\in\mathcal T$, quadratic reciprocity prescribes
one of the two possible Legendre values of $u\bmod q$ in order to have
$({q}/{p})=(-1)^{\ell([q])}$. When $q>3$, either choice has
$(q-1)/2\geq2$ representatives, so we may also require $u\not\equiv1$
modulo $q$. At $q=3$ there is a unique choice for the prescribed sign.
The Chinese remainder theorem gives $u_0\in(\Z/M_0\Z)^\times$ such that
every prime $p\equiv u_0\pmod{M_0}$ satisfies
\begin{equation}\label{eq:quadratic-choice}
 \left(\frac{a_i}{p}\right)=-1\quad(1\leq i\leq r),
 \qquad p\not\equiv1\pmod q\quad(q\in\mathcal T,\ q>3).
\end{equation}
Moreover $p\equiv2\pmod3$ if $\ell([-3])=1$, and $p\equiv1\pmod3$
otherwise, because $({-3}/{p})$ is the quadratic character modulo $3$.

In the first case no cubic condition is needed. In the second case,
failure of \textup{(B)} supplies a prime $\ell_0\equiv1\pmod3$ for
which none of the $a_i$ is a cube. This prime divides none of the bases,
since zero is a cube. Its Frobenius in
\[
 K=F(a_1^{1/3},\ldots,a_r^{1/3})
\]
shows that the set
\begin{equation}\label{eq:good-cubic-set}
 \mathcal C=
 \{h\in\Gal(K/F):h(a_i^{1/3})\ne a_i^{1/3}
       \text{ for all }i\}
\end{equation}
is nonempty. It is inversion-stable. At a split unramified prime its
Frobenius condition says that every $a_i$ is a noncube. In this case put
$d_{\mathcal C}=|\mathcal C|/|H|>0$; in the case with no cubic condition
put $d_{\mathcal C}=1$.

\subsection{The uniform splitting estimate}
For $p\nmid a$ write
\[
 i_p(a)=\frac{p-1}{\ord_p(a)}.
\]
For a prime $q$ and $p\nmid aq$, the condition $q\mid i_p(a)$ is
equivalent to complete splitting in $\Q(\zeta_q,a^{1/q})$.
The field-degree and smooth prime-ideal estimates in
\cite[Lemmas~9.1--9.2]{AR}, as combined before the restricted-range
specialization in the proof of its Proposition~2.2, give the following
form. There is a common $\delta_0=10^{-6}$ such that, for every fixed
$a\in\Z\setminus\{0,1,-1\}$ and every sufficiently large prime $q$
(depending only on $a$),
\begin{equation}\label{eq:splitting}
 \#\{x<p\leq2x:q\mid i_p(a)\}
 \ll_a\frac{x}{q(q-1)\log x}
       +\frac{x^{1-\delta_0}\log(2q)}{\log x}.
\end{equation}
The implied constant is independent of $q$. A fixed smooth majorant
for the interval gives the same upper bound for sharp endpoints.

Choose $0<\eps<\min(\delta_0/4,1/4)$ in \eqref{eq:giant-def}.
Let $Y$ exceed the fixed exceptional primes for each of the finitely
many bases and every prime dividing $M_0$. Enlarge the progression by
putting
\[
 M_Y=M_0\prod_{\substack{3<q\leq Y\\q\nmid M_0}}q,
\]
where here and below $q$ in such products is prime. Let $U_Y$ be all
classes modulo $M_Y$ that equal $u_0$ modulo $M_0$ and avoid $0$ and
$1$ modulo every new $q$. Their relative density among primes is
\begin{equation}\label{eq:density-Y}
 d_Y=\frac{|U_Y|}{\varphi(M_Y)}
 =\frac1{\varphi(M_0)}
  \prod_{\substack{3<q\leq Y\\q\nmid M_0}}\frac{q-2}{q-1}
 \asymp_S\frac1{\log Y}.
\end{equation}
For the last assertion use Mertens' theorem and
\[
 \frac{q-2}{q-1}
 =\left(1-\frac1q\right)
  \left(1-\frac1{(q-1)^2}\right).
\]
The second product converges to a positive limit.

Summing \eqref{eq:splitting} over the bases and over
$Y<q\leq2x^\eps$ yields
\begin{equation}\label{eq:intermediate}
 \#\{x<p\leq2x:\text{some }q\in(Y,2x^\eps]
                \text{ divides some }i_p(a_i)\}
 \leq C_S\frac{x}{Y\log x}+O_S(x^{1-\delta_0+\eps}).
\end{equation}
Here $C_S$ is independent of $Y$. Fix $Y$ large enough that
\[
 C_S/Y<\tfrac14 d_Yd_{\mathcal C}c_\eps\int\Phi.
\]
Such a choice is possible by \eqref{eq:density-Y}. From now on $Y$,
$M_Y$, and $U_Y$ remain fixed.

\subsection{Proof of Theorem~\ref{thm:main}}
Theorem~\ref{thm:reservoir}, with the cubic condition
\eqref{eq:good-cubic-set} when necessary, gives
\begin{equation}\label{eq:positive-reservoir}
 \left(d_Yd_{\mathcal C}c_\eps\int\Phi+o(1)\right)\frac{x}{\log x}
\end{equation}
as the weighted count of suitable primes with
$p-1=kQ$, $Q\in I_x$ prime, and $k<2x^\eps$. Compact support inside
$(1,2)$ ensures that these primes lie in $(x,2x]$ for all sufficiently
large $x$. The weights are at most one.

No prime $q\leq Y$ divides any $i_p(a_i)$: the prime $2$ is excluded
by \eqref{eq:quadratic-choice}; the prime $3$ either does not divide
$p-1$ or is excluded by the noncube condition; every larger such $q$
is excluded by $p\not\equiv1\pmod q$. Equation~\eqref{eq:intermediate}
removes all remaining index primes at most $2x^\eps$, at a cost smaller
than one quarter of the main term in \eqref{eq:positive-reservoir},
up to $o(x/\log x)$.

If an index prime remains, it must be $Q$, since it divides $kQ$ and
is larger than $k$. The condition $Q\mid i_p(a_i)$ implies
$a_i^k\equiv1\pmod p$. Every such $p$ is therefore a divisor of the
nonzero integer
\[
 \mathcal N_x=
 \prod_{i=1}^r\prod_{1\leq j\leq\lfloor2x^\eps\rfloor}|a_i^j-1|.
\]
Since $|a_i|\geq2$,
\[
 \log\mathcal N_x\ll_S\sum_{j\leq2x^\eps}j\ll_S x^{2\eps}.
\]
It follows that at most $O_S(x^{2\eps}/\log x)=o(x/\log x)$ primes
larger than $x$ divide $\mathcal N_x$. Subtracting these losses from
\eqref{eq:positive-reservoir} proves \eqref{eq:main}, for example with
$c_S$ equal to one quarter of its positive leading constant. Applying
the dyadic bound to $(x/2,x]$ and using the prime number theorem gives
the positive lower relative density.

Finally suppose \textup{(A)} holds. At an odd prime where all the bases
are primitive roots, all their Legendre symbols equal $-1$. Their
product over the indicated odd subset is then $-1$, contradicting the
square relation. If \textup{(B)} holds, the even product instead forces
$({-3}/{p})=1$, hence $p\equiv1\pmod3$. The second part of
\textup{(B)} makes at least one base a cube at $p$, so that its order
divides $(p-1)/3$. The finitely many primes dividing the relations or
the bases can be discarded. Thus $\mathcal P_S$ is finite in either
case, completing the proof.\hfill$\square$

\begin{example}
For $S=\{2,-6\}$ the relation $2(-6)=-3\cdot2^2$ forces the split
case, but both bases are primitive roots modulo $13$. The classes of
$2$ and $-6$ are independent in $F^\times/F^{\times3}$: their
valuations at the primes above $2$ and $3$ show this successively.
Hence $H\simeq(\Z/3\Z)^2$, and $\mathcal C$ has four elements.
If $\eta_1,\eta_2$ are the two coordinate characters, then
\[
 \ind_{\mathcal C}
 =\frac{4-2t_{\eta_1}-2t_{\eta_2}
                  +t_{\eta_1\eta_2}+t_{\eta_1\eta_2^{-1}}}{9}
\]
at split primes. This displays explicitly the four nonconstant trace
terms handled by Theorem~\ref{thm:trace}.
\end{example}