\section{Galois realization on every stable support}\label{sec:galois}

The preceding sections determine the lattice closing degree on every
$\Aut(C_{2pq})$-stable support.  We now prove that none of those lower bounds
disappears when the time-domain signals are required to be rational or
integer-valued.  The division of ownership is important: the Cubic and
Quartic Skeletons and the depth-five and depth-six presentations supply the
already established lattice classifier; on every support of depth greater
than one, the argument below supplies a Galois-equivariant, nontranslation
phase ratio immediately below its closing degree.  The sole depth-one support
is direct.  Proposition~\ref{prop:direct-realization} then turns each ratio
into rational and integer signal pairs.

We continue to write $n=2pq$ with $p<q$ odd primes and
$\sigma_u(\zeta_n)=\zeta_n^u$.  The box notation is that of
Section~\ref{sec:boxes}.  If the zero orbit is present, every ratio below is
extended by $r_0=1$.

\subsection{Non-torsion orbit phases and the low-depth layers}

The phase families used at depths three and four come from an elementary
norm-one construction.

\begin{lemma}[Non-torsion cyclotomic phase]\label{lem:nontorsion-phase}
For every divisor $m\mid n$ with $m>2$, there is
$z_m\in\Q(\zeta_m)^\times$ such that
$z_m\overline{z_m}=1$ and $z_m$ is not a root of unity.  If $s_m=n/m$, then
\[
 \psi_m(us_m)=\sigma_u(z_m)\qquad(us_m\in\Orb m)             \tag{10.1}
\]
is well defined, Galois equivariant, and satisfies
$\psi_m(-s)=\psi_m(s)^{-1}$.  It is not the restriction of a translation
ratio.
\end{lemma}

\begin{proof}
For $t\in\Q$ put
\[
 z_m(t)=\frac{t-\zeta_m}{t-\zeta_m^{-1}}.
\]
Conjugation inverts this value.  The map $t\mapsto z_m(t)$ is injective on
$\Q$, since equality at $t,u$ gives
$(t-u)(\zeta_m-\zeta_m^{-1})=0$.  A number field contains only finitely many
roots of unity, so some rational $t$ gives the required non-torsion $z_m$.

Two units carrying $s_m$ to the same point have the same restriction to
$\Q(\zeta_m)$, so (10.1) is well defined.  Equivariance is built into the
definition, and the unit $-1$ gives the inverse identity.  Every value of a
translation ratio is a root of unity, whereas every conjugate of $z_m$ is
non-torsion.  Hence $\psi_m$ is not a translation.
\end{proof}

The following observation isolates the logical step used repeatedly below.  It
also makes clear which part comes from the existing lattice classifier and
which part is new in this section.

\begin{lemma}[Closing-degree witness]\label{lem:closing-witness}
Let $S$ be stable and suppose
\(
 \calL_{\le d-1}(S)\subsetneq\calL_{\le d}(S)=K(S).
\)
If a Galois-equivariant ratio $r=(r_s)_{s\in S}$ satisfies the inverse
condition, kills every row of $\calL_{\le d-1}(S)$, and is not a translation,
then it is detected by a row of degree $d$.  Moreover, it is realized by
exact-support rational and integer signal pairs that agree through order
$d-1$ and fail to agree at order $d$.
\end{lemma}

\begin{proof}
If $r$ killed $\calL_{\le d}(S)=K(S)$, the phase-ratio criterion of
Proposition~\ref{prop:ratio} would make it a translation, contrary to the
hypothesis.  Hence some generator introduced at degree $d$ has nontrivial
product.  Proposition~\ref{prop:direct-realization} realizes $r$ by taking
$\widehat f=1$ and $\widehat g=r$ on $S$; its common-denominator rescaling
preserves support, equality of lower autocorrelations, and failure at degree
$d$.
\end{proof}

The zero orbit separates at every degree.  Indeed, $e_0$ is a length-one
zero-sum row and, as in (9.4a),
\[
 K(S\cup\{0\})=K(S)\oplus\Z e_0,\qquad
 \calL_{\le d}(S\cup\{0\})=\calL_{\le d}(S)\oplus\Z e_0. \tag{10.2}
\]
Thus adjoining $\Orb1$ changes neither a witness nor its depth.  The support
$\Orb1$ alone has depth one.

For the depth-two base $E=\Orb2$, take $r_E=2$.  It is rational and invariant,
there is no degree-one relation on $E$, and a translation takes only the
values $\pm1$ there.  The pair row $E+E=0$ closes at degree two, so
Proposition~\ref{prop:direct-realization} proves exact rational and integer
depth two, with or without $\Orb1$.

Each of the 77 depth-three bases other than $E$ contains an orbit $\Orb m$
with $m>2$.  Put $r=\psi_m$ on one such orbit and $r=1$ on all other boxes.
The only relations through degree two are the zero row and inverse pairs, so
Lemma~\ref{lem:nontorsion-phase} makes $r$ invisible through degree two and
nontranslational.  The Cubic Skeleton closes the base at degree three.  This
realizes all 154 raw depth-three rows after the optional zero orbit is restored.
Notice that the construction does not need a choice compatible across
different bases: the theorem asks for a witness on each fixed support.  Once
the orbit $\Orb m$ is chosen, however, all of its Fourier values are forced by
one cyclotomic number and Galois equivariance, so the resulting inverse DFT is
rational rather than merely real.

\subsection{The nine depth-four families}

For a point in a box with $p$-coordinate $a$ or $q$-coordinate $b$, write
$\psi_p(a)$ or $\psi_q(b)$ for the corresponding value in (10.1).  On boxes
not named in the phase column, the ratio is one.  Table~\ref{tab:galois-nine}
is a canonical first-match cover of every base that fails the cubic criterion
and closes at degree four.  The last column records the generic and
prime-three first-match counts; those counts are checks on the routing, not
premises of the cancellation proof.

\begin{table}[ht]
\centering
\scriptsize
\begin{tabularx}{\textwidth}{@{}l p{2.25cm} p{3.35cm} Y c@{}}
\toprule
Family & Mask condition & Nontrivial ratio & Cubic types seeing it &
$d4$ gen./$p=3$\\
\midrule
$N1$ & $G,\neg F,\neg P$ & $\psi_p(a)$ on $P1,G$ & $QGG,QP1G$ & $16/12$\\
$N2$ & $G,\neg F,\neg Q$ & $\psi_q(b)$ on $Q1,G$ & $PGG,PQ1G$ & $8/8$\\
$N3$ & $P1,\neg F,\neg P$ & $\psi_p(a)$ on $P1,G$ & $QGG,QP1G$ & $6/4$\\
$N4$ & $Q1,\neg F,\neg Q$ & $\psi_q(b)$ on $Q1,G$ & $PGG,PQ1G$ & $5/5$\\
$N5$ & $P1,\neg G,\neg P,\neg Q1$ & $\psi_p(a)$ on $P1$ & none & $4/4$\\
$N6$ & $Q1,\neg G,\neg P1,\neg Q$ & $\psi_q(b)$ on $Q1$ & none & $4/4$\\
$N8$ & $\begin{array}{c}E,G\\ \neg F,\neg P1,\neg Q1\end{array}$ & $-1$ on $E$ & none & $1/1$\\
$N9$ & $\begin{array}{c}E,Q1\\ \neg G,\neg P,\neg Q\end{array}$ & $-1$ on $E$ & none & $1/1$\\
$N10$ & $\begin{array}{c}P1,Q1\\ \neg E,\neg F,\neg G\end{array}$ & $-1$ on $P1$ & $PP1P1$; two signs & $1/1$\\
\midrule
Total & & & & $46/40$\\
\bottomrule
\end{tabularx}
\caption{The nine canonical depth-four Galois lifts.}
\label{tab:galois-nine}
\end{table}

We first justify that the table is complete without invoking a finite search.
The necessity half of Theorem~\ref{thm:cubic} covers the complement of its ten
positive clauses by $N1,\ldots,N10$.  The omitted valid family is
\[
 N7:\quad \neg E,\neg F,\neg P,\neg Q.
\]
A nonempty $N7$ mask uses only $P1,Q1,G$.  If it contains $G$, it is in $N1$;
if it does not contain $G$ but contains $P1$, it is in $N3$; otherwise it is
the $Q1$ case of $N4$.  Hence
\[
 N7\subset N1\cup N3\cup N4,                               \tag{10.3}
\]
and the nine displayed families cover every cubic-negative mask.

Next we verify invisibility through degree three.  Inverse pairs cancel by
Lemma~\ref{lem:nontorsion-phase}.  For $N1,N3$, the exhaustive triple list in
Theorem~\ref{thm:cubic} shows that only $QGG$ and $QP1G$ can see the marked
$p$-phase.  Their zero $p$-coordinate forces the two marked arguments to be
$a,-a$.  The $N2,N4$ argument is the same on the $q$-coordinate, using
$PGG,PQ1G$.  The exclusions in $N5,N6$ remove every triple with an unbalanced
marked factor.  In $N8,N9$, every triple containing $E$ would require an
excluded box.  Finally, $N10$ sees only $PP1P1$, where the sign occurs twice.
Thus every row of Table~\ref{tab:galois-nine} kills
$\calL_{\le3}$ and is Galois equivariant.

The six cyclotomic rows are nontranslations by
Lemma~\ref{lem:nontorsion-phase}.  The three sign rows require separate
arguments.  In $N8$, the ratio is one on the generator $1\in G$, forcing a
putative translation parameter to be zero, but it is $-1$ on $E$.  In $N9$,
the point $p\in Q1$ has ratio one, forcing the translation parameter to be
even, whereas $-1$ on $E=pq$ requires it to be odd.  In $N10$, the point
$q\in P1$ with ratio $-1$ forces an odd parameter, while the point $p\in Q1$
with ratio one forces an even parameter.

The Quartic Skeleton says that every base counted in the last column has
$\calL_{\le4}=K$.  Lemma~\ref{lem:closing-witness} therefore applies to each
displayed ratio: a degree-four row detects it, and the rational and integer
realizations have exact depth four.  There are $46$ generic or $40$
prime-three bases, and (10.2) doubles these to $92$ or $80$ supports.  Thus the
last column of Table~\ref{tab:galois-nine} is used only after the symbolic
cancellation and nontranslation arguments have been completed; it is never a
substitute for either argument.

\subsection{Depth five and six}

The two depth-five bases are
\[
 S_p=P\cup P1\cup Q1,\qquad S_q=Q\cup P1\cup Q1.
\]
On either base put $r=-1$ on $P1$ and $r=1$ elsewhere.  This is the $N10$
sign and is not a translation.  If a zero-sum row of length at most four had
product $-1$, it would contain odd numbers of both parity-one boxes.  On
$S_p$, if $Q1$ occurs once, its nonzero $q$-coordinate cannot be cancelled by
the $P$ and $P1$ boxes.  If $Q1$ occurs three times, the length bound leaves
exactly one $P1$ point and no $P$ point, so its nonzero $p$-coordinate cannot
be cancelled.  The argument on $S_q$ is symmetric, with $P1,Q1$ and $p,q$
exchanged.  Hence the sign is invisible through degree four.  It is detected
by the five-term rows
\[
\begin{aligned}
 S_p:&\quad P[-a]+P1[a]+Q1[1]+Q1[1]+Q1[-2],\\
 S_q:&\quad P1[1]+P1[1]+P1[-2]+Q[-b]+Q1[b].
\end{aligned}                                               \tag{10.4}
\]
All shown coordinates are nonzero.  When $p=3$, the first-axis triple
$1,1,-2$ is simply $1,1,1$.  Thus both bridge bases, and their zero-orbit
copies, have exact rational and integer depth five.

Suppose now that $p,q\ge5$.  The only depth-six base is $P1\cup Q1$.  Use the
same sign on $P1$.  A negative product through degree five would require odd
cardinalities in both boxes, so the total length would be even and at most
four.  The preceding $1+1$ and $3+1$ obstruction again rules this out.  The
six-term row
\[
 P1[1]+P1[1]+P1[-2]+Q1[1]+Q1[1]+Q1[-2]                   \tag{10.5}
\]
is zero-sum and contains three negative factors.  It detects the ratio and
proves exact depth six for the base and its zero-orbit copy.

It remains to realize the seven prime-three critical bases.  Put
\[
 A=P1=\Orb6,\quad B=Q1=\Orb{2q},\quad C=Q=\Orb q,\quad
 D=G=\Orb{6q}.                                             \tag{10.6}
\]
They are
\[
 A, A\cup C, A\cup B, D, C\cup D, A\cup D,
 A\cup C\cup D.                                           \tag{10.7}
\]
Choose a non-torsion norm-one $z\in\Q(\zeta_6)^\times$.  On $A,D$, let
$\eps(s)\in\{1,-1\}$ be the nonzero $\F_3$ coordinate and set
$r_s=z^{\eps(s)}$; set $r=1$ on $B,C$.  Every unit modulo $6q$ is
$1$ or $-1$ modulo six.  It acts on $z$ as the identity or conjugation and on
$\eps$ by the same sign, so the ratio is Galois equivariant.

For a zero-sum multiset, let $\eta$ be the sum of the marked exponents.  Its
$\F_3$ coordinate gives $3\mid\eta$.  Except on $A\cup B$, parity also makes
$\eta$ even.  Thus a row of length at most five has $6\mid\eta$ and
$|\eta|\le5$, hence $\eta=0$.  On $A\cup B$, a value $\eta=\pm3$ would need
three $A$ points and an odd number of $B$ points.  One $B$ point cannot cancel
the $q$-coordinate, while three already make total length six.  The ratio is
therefore invisible through degree five on all seven bases.

Degree six detects it explicitly.  Six positive $A$ points work whenever
$A$ is present outside $A\cup B$.  If only $D$ is marked, use three pairs
$D[1,b],D[1,-b]$.  On $A\cup B$, use three positive $A$ points together with
$B[1],B[1],B[-2]$.  The resulting products are $z^6$ or $z^3$, neither one
because $z$ is non-torsion.  The same non-torsion value also excludes a
translation.  Hence all seven bases and their zero-orbit copies have exact
rational and integer depth six.

For reference, the detector assignment is displayed without suppressing any
of the seven cases:
\begin{center}
\small
\begin{tabularx}{0.92\textwidth}{@{}l Y c@{}}
\toprule
Critical base & Degree-six zero-sum detector & Product\\
\midrule
$A$ & six positive $A$ points & $z^6$\\
$A\cup C$ & six positive $A$ points & $z^6$\\
$A\cup B$ & three positive $A$ points and $B[1]+B[1]+B[-2]$ & $z^3$\\
$D$ & $D[1,b]+D[1,-b]$, repeated three times & $z^6$\\
$C\cup D$ & $D[1,b]+D[1,-b]$, repeated three times & $z^6$\\
$A\cup D$ & six positive $A$ points & $z^6$\\
$A\cup C\cup D$ & six positive $A$ points & $z^6$\\
\bottomrule
\end{tabularx}
\end{center}
The coordinates in the middle column sum to zero in both CRT components.
Because $z$ is non-torsion, the last column is nontrivial in every row; because
each listed base already has $\calL_{\le6}=K$, these are genuine closing rows,
not merely upper-bound certificates.

\subsection{The realized pairs and the proof audit}

For every ratio chosen above, the rational pair is completely explicit in
Fourier coordinates.  Namely,
\[
 f_r(x)=\frac1n\sum_{s\in S}\zeta_n^{sx},\qquad
 g_r(x)=\frac1n\sum_{s\in S}r_s\zeta_n^{sx}.
\]
Galois equivariance fixes both sums under every element of
$\Gal(\Q(\zeta_n)/\Q)$; hence $f_r,g_r\in\Q^{C_n}$.  Their Fourier transforms
are respectively $1$ and
$r_s$ at every point of $S$ and zero off $S$, so their support is exactly $S$.
If $D$ clears all time-domain denominators, then $Df_r,Dg_r$ are integer
signals with the same exact Fourier support.  Multiplying both signals by $D$
multiplies each order-$k$ autocorrelation by $D^k$, so it neither creates nor
removes the first separating order.  It also leaves the phase ratio unchanged,
and therefore cannot turn a nontranslation pair into a translation pair.

The proof obligations are summarized in the following table.  Its middle
columns cite the human argument that makes the ratio invisible below the
claimed depth and the independent lattice statement that closes at that
depth.  This separation prevents the finite profile enumeration from being
used as a surrogate for the new realization argument.

\begin{center}
\small
\begin{tabularx}{\textwidth}{@{}c >{\raggedright\arraybackslash}p{2.55cm}
>{\raggedright\arraybackslash}p{3.4cm} Y@{}}
\toprule
Depth & Nonzero bases & Lower-degree cancellation & Closing input or detector\\
\midrule
$1$ & none; zero orbit only & vacuous & the singleton row $e_0$\\
$2$ & $E$ & no degree-one row & the inverse pair $E+E$\\
$3$ & 77 Cubic-Skeleton bases & inverse-pair cancellation for one $\psi_m$ &
the Cubic Skeleton, $\calL_{\le3}=K$\\
$4$ & 46 generic or 40 prime-three bases & the nine rows of
Table~\ref{tab:galois-nine} against the exhaustive cubic types &
the Quartic Skeleton, $\calL_{\le4}=K$\\
$5$ & $S_p,S_q$ & parity plus the $1+1$ and $3+1$ CRT obstructions &
the two rows in (10.4)\\
$6$, generic & $P1\cup Q1$ & parity reduces length at most five to the
same CRT obstruction & the row in (10.5)\\
$6$, $p=3$ & seven bases in (10.7) & $3\mid\eta$, parity, and the separate
$A\cup B$ argument & the seven detectors displayed above\\
\bottomrule
\end{tabularx}
\end{center}

The depth-one row is direct, and the depth-two row uses
Proposition~\ref{prop:direct-realization}, whose hypotheses here are Galois
equivariance, lower-degree cancellation, and nontranslation.  Every row from
depth three onward has supplied all four ingredients required by
Lemma~\ref{lem:closing-witness}: Galois equivariance, inverse symmetry,
lower-degree cancellation, and nontranslation.  The zero-orbit splitting in
(10.2) then transfers each nonzero-base construction to its optional-zero
copy without altering the applicable ingredients.

\subsection{Assembly}

We can now prove the signal-level theorem without changing the lattice
classifier.

\begin{proof}[Proof of Theorem~\ref{thm:uniform-descent}]
The inclusion of signal classes and Proposition~\ref{prop:ratio} give
\[
 \depth_{\Z}(S)\le\depth_{\Q}(S)\le\depth(S).              \tag{10.8}
\]
The depth-one support was handled directly.  For every support with
$\depth(S)>1$, the preceding construction gives a Galois-equivariant ratio
that is not a translation and kills
$\calL_{\le\depth(S)-1}(S)$.  Proposition~\ref{prop:direct-realization}
therefore gives both reverse inequalities.  The three depths are equal.

The underlying nonzero bases are partitioned as
\[
\begin{array}{c|ccccc}
 &d=2&d=3&d=4&d=5&d=6\\
\midrule
p=3&1&77&40&2&7\\
p,q\ge5&1&77&46&2&1.
\end{array}                                                  \tag{10.9}
\]
Restoring the optional zero orbit doubles every base and adds the sole
depth-one support.  This gives exactly the two profiles in
Theorem~\ref{thm:profiles} for both rational and integer signals.
\end{proof}

The theorem establishes thresholds, not a complete moduli description.  The
exact unit-support torus is classified in Section~\ref{sec:signals}; on a
stable support of depth greater than one, the construction above deliberately
records only the Galois-equivariant lower-bound ratio needed for exactness.