\section{Rational signals on the exact unit support}\label{sec:signals}

We now add the Galois condition that is absent from the integral calculation.
If $3\nmid n$, Proposition~\ref{prop:unit-lattice} says
$\calL_{\le4}(\units n)=K(\units n)$.  The phase-ratio bridge then shows that
every surviving ratio is a translation, proving part (2) of
Theorem~\ref{thm:unit-main}.

Let $n=6m$.  Proposition~\ref{prop:unit-lattice} gives
$\calL_{\le5}(\units{6m})=\calL_{\le4}(\units{6m})$.  Hence the phase-ratio
bridge identifies the fourth- and fifth-order invisible ratios with the same
character group.  In the coordinates of (4.6), such a character has
\[
 r(v_t)=ab^t,\qquad r(-v_t)=a^{-1}b^{-t},\qquad b\in\mu_m.    \tag{5.1}
\]
Indeed, a character of $\Z\oplus C_m$ is freely specified by the image $a$ of
$(1,0)$ and by the image $b$ of $(0,1)$, with the sole condition $b^m=1$.
The negative formula follows from the inverse-pair row, not from an additional
reality assumption.  Thus (5.1) is the complete complex ambiguity before
translations and Galois descent.

A translation by $y$ multiplies $(a,b)$ by
$(\zeta_n^y,\zeta_m^y)$.  Choose $y$ modulo $m$ to make $b=1$.  Define
\[
 \eps(u)=\begin{cases}1,&u\equiv1\pmod6,\\-1,&u\equiv-1\pmod6.
 \end{cases}
\]
The normalized ratio is then
\[
 r_u=z^{\eps(u)}.                                            \tag{5.2}
\]

\begin{theorem}[Exact-unit rational descent]\label{thm:unit-descent}
Let $f,g:C_{6m}\to\Q$ have exact Fourier support $\units{6m}$ and equal
autocorrelations through order four.  After translating $g$, there are
$\alpha\in\Q(\zeta_{6m})^\times$ and $z\in\Q(\zeta_6)^\times$ such that
\[
 \widehat f(u)=\sigma_u(\alpha),\qquad
 \widehat g(u)=\sigma_u(z\alpha),\qquad z\overline z=1.     \tag{5.3}
\]
Conversely every such pair, extended by zero off the unit support, is
rational-valued and agrees through order five, hence also through order four.
Two normalized parameters are
translation-equivalent exactly modulo $\mu_6$, and order six agrees exactly
when $z^6=1$.
\end{theorem}

\begin{proof}
Rationality gives Galois covariance (3.3).  Applying it to the ratio at
$u=1$ yields
\[
 \sigma_a(z)=z^{\eps(a)}\qquad(a\in\units{6m}).              \tag{5.4}
\]
The kernel of $\eps:\units{6m}\to\{\pm1\}$ has index two and fixes
$\zeta_6$.  Its fixed field is therefore precisely $\Q(\zeta_6)$, so
$z\in\Q(\zeta_6)^\times$.  Complex conjugation corresponds to $a=-1$, and
(5.4) gives $\overline z=z^{-1}$.

To justify the fixed field statement without a degree shortcut, restriction
of automorphisms gives the surjection
\[
 \Gal(\Q(\zeta_{6m})/\Q)\longrightarrow
 \Gal(\Q(\zeta_6)/\Q)\cong\{\pm1\}.
\]
It is exactly $a\mapsto\eps(a)$.  Hence its kernel has fixed field
$\Q(\zeta_6)$.  Equation (5.4) says that every element of this kernel fixes
$z$, while the nontrivial coset sends it to its conjugate.

Put $\alpha=\widehat f(1)$.  Galois covariance gives the first equation in
(5.3), and (5.2)--(5.4) give the second.  Conversely, for arbitrary nonzero
$\alpha$ and norm-one $z$, the right side of (5.3) is Galois equivariant and
therefore defines rational signals with exact support $\units{6m}$.  Every
zero-sum row of length at most five has zero $\eps$-sum by the argument around
(4.6), so its ratio product is one.  Proposition~\ref{prop:ratio} proves
equality of the autocorrelations.

The exact-support assertion in the converse is immediate but worth recording:
every conjugate $\sigma_u(\alpha)$ and $\sigma_u(z\alpha)$ is nonzero, while
we set all nonunit coefficients to zero.  Galois equivariance preserves this
zero pattern.  Formula (3.4) then gives rational time-domain values, and the
automorphism $u=-1$ gives the conjugate symmetry needed for reality.

Translations preserving the normalization $b=1$ have $m\mid y$, and
$\zeta_{6m}^y$ then runs through $\mu_6$.  Every degree-six row has
$\eps$-sum $0$ or $\pm6$, while the closing row constructed after
Lemma~\ref{lem:sumset} has sum six.  Hence the sixth-order data read exactly
$z^6$.  They agree precisely for $z\in\mu_6$, which is exactly the residual
translation case.
\end{proof}

The norm-one parameter is a one-dimensional rational torus.  Since
$\Q(\zeta_6)=\Q(\sqrt{-3})$, every parameter other than $-1$ can be written
\[
 z=\frac{a+b\sqrt{-3}}{a-b\sqrt{-3}}qquad(a,b\in\Q),        \tag{5.5}
\]
with $-1$ obtained by $a=0$.  This parametrization is optional for the proof,
but it makes clear that the common fourth- and fifth-order ambiguity is
generally infinite even though the residual translation subgroup $\mu_6$ is
finite.  The sixth-order
test collapses the torus exactly to those six translation points.

This proves all signal-level assertions in Theorem~\ref{thm:unit-main}.  The
exact-support hypothesis is essential: it licenses every Fourier division in
Proposition~\ref{prop:ratio} and in (5.1).

\subsection{The \texorpdfstring{$C_{30}$}{C30} question}

At $n=30$, Theorem~\ref{thm:unit-descent} gives the order-four and order-five
classification stated in
Corollary~\ref{cor:u30-intro}.  The displayed construction of
Agulnick--Busick-Warner \cite{AgulnickBusickWarner2026} uses a common
amplitude in a quadratic subfield.  The complete classification permits
$\alpha$ to range over all of $\Q(\zeta_{30})^\times$, while the relative
phase remains in the same norm-one torus in $\Q(\zeta_6)$.  Thus the larger
field enlarges the common amplitude, not the ambiguity mechanism.  This is an
answer to that support-specific question, not a priority assertion.

For example, choose any primitive $\alpha\in\Q(\zeta_{30})^\times$ and any
rational point (5.5) with $z^6\ne1$.  The inverse transforms supplied by
(5.3) form a rational pair agreeing through order five, and therefore through
order four, but not order six.
Changing $\alpha$ changes both signals simultaneously and does not change
their ratio.  Changing $z$ moves along the actual ambiguity family.  This
separates the common-amplitude degree of freedom from the relative phase in a
way that is invisible if one works only with a single displayed example.

\begin{proofboundary}
Theorem~\ref{thm:unit-descent} classifies rational signal pairs only on the
exact unit support.  Later sections use the same lattice formalism on all 255
stable supports, but do not claim that every integral obstruction there has a
rational signal realization.
\end{proofboundary}