\section{Faithful quotients and the lattice carried by a phase code}
\label{sec:quotient-extension}
The primitive hypothesis is mathematically useful, but it is not an intrinsic
restriction on which embedded tori can occur. It is a restriction on how the
torus has been parametrized. Removing it correctly requires retaining the
finite covering and the grid that the parametrization sends into the image.
Dividing an integer frequency by a common factor without changing the grid
does not perform that operation correctly.

In this section only, let $A\in\Z^{m\times d}$ have full column rank, without
assuming primitivity. Write
\[
 L_A=\sum_{i=1}^m\Z a_i^\top\subseteq\Z^d,
 \qquad h_A=[\Z^d:L_A].
\]
The row-lattice index $h_A$ is the product of the nonzero Smith invariants of
$A$, and is the gcd of its maximal-minor absolute values. For a modulus $Q$,
write
\[
 K_A(Q)=\{x\in(\Z/Q\Z)^d:Ax=0\pmod Q\},
 \qquad V_A(Q)=A(\Z/Q\Z)^d.
\]
The object $V_A(Q)$ is a set of distinct image words. Each of its elements has
exactly $|K_A(Q)|$ phase labels. This distinction is necessary even when the
weights are strictly positive.

\begin{proposition}[Exact primitive normalization]
\label{prop:primitive-normalization}
There exist an invertible integral $d\times d$ matrix $H$ and a primitive
$m\times d$ integer matrix $B$ such that
\[
 A=BH,\qquad |\det H|=h_A.
\]
The rows of $H$ may be any integer basis of $L_A$. The map
$\theta\mapsto H\theta$ on $\R^d/(2\pi\Z^d)$ is a covering of degree $h_A$,
and
\[
 F_{A,w}(\theta)=F_{B,w}(H\theta).
\]
Consequently the embedded image torus is unchanged by normalization, but the
image of the coordinate $Q$-grid becomes
\[
 \{F_{B,w}(2\pi y/Q):y\in H(\Z/Q\Z)^d\}.
\]
It is the full normalized $Q$-grid if and only if $\gcd(Q,h_A)=1$.
\end{proposition}
\begin{proof}
Each row of $A$ has unique integral coordinates in the row basis $H$; place
those coordinates in $B$. Since the rows of $A$ generate $L_A$, their
coordinate rows generate $\Z^d$, so $B$ is primitive. The determinant of a
full lattice basis gives its index. Substitution in the exponential formula
gives the identity between the two parametrizations.

The kernel of $H$ on the real torus is
$H^{-1}\Z^d/\Z^d$, which has $|\det H|$ elements. Surjectivity follows because
the real matrix $H$ is invertible. On the $Q$-torsion grid, $H$ is surjective
precisely when it is an invertible square matrix over $\Z/Q\Z$, equivalently
when its determinant is a unit. Finally, $B$ is faithful, so distinct
normalized grid parameters have distinct images. Thus no further finite
identification occurs after applying $B$.
\end{proof}

The full period lattice in the original real coordinates is
\[
 L_A^*=\{u\in\R^d:a_i\cdot u\in\Z\text{ for every }i\}
       =H^{-1}\Z^d.
\]
The actual embedded torus is therefore naturally parametrized by
$\R^d/(2\pi L_A^*)$. Its complete intrinsic $Q$-torsion is
$Q^{-1}L_A^*/L_A^*$ and always has $Q^d$ elements. The original coordinate
grid gives the subgroup
\[
 (Q^{-1}\Z^d+L_A^*)/L_A^*\subseteq Q^{-1}L_A^*/L_A^*.
\]
The inclusion is sometimes strict. The missing points arise from the finite
covering, rather than from a failure of the normalized character embedding.

\begin{proposition}[Finite covering exact sequence]
\label{prop:covering-sequence}
Let $K=\Z^d/H\Z^d$. There is an exact sequence
\begin{equation}\label{eq:covering-sequence}
 0\longrightarrow K[Q]\longrightarrow(\Z/Q\Z)^d
 \xrightarrow{\ H\ }(\Z/Q\Z)^d
 \longrightarrow K/QK\longrightarrow0.
\end{equation}
If $s_1\mid\cdots\mid s_d$ are the Smith invariants of $A$, then
\begin{equation}\label{eq:distinct-image-count}
 |K_A(Q)|=\prod_{j=1}^d\gcd(Q,s_j),\qquad
 |V_A(Q)|=\frac{Q^d}{\prod_{j=1}^d\gcd(Q,s_j)}.
\end{equation}
\end{proposition}
\begin{proof}
The cokernel of $H$ modulo $Q$ is
$\Z^d/(H\Z^d+Q\Z^d)=K/QK$. To identify the kernel, send an integral vector
$u$ with $Hu=Qz$ to $[z]\in K$. This class is killed by $Q$. Changing $u$ by
$Qv$ changes $z$ by $Hv$, so the map is well defined. If $[z]=0$, write
$z=Hv$; then $H(u-Qv)=0$ over the integers, giving $u=Qv$. Conversely, if
$Q[z]=0$ in $K$, write $Qz=Hu$ and obtain a kernel element modulo $Q$.
These constructions are inverse isomorphisms between the kernel and $K[Q]$.
They establish exactness.

Since $B$ has an integer left inverse, $BHx=0\pmod Q$ if and only if
$Hx=0\pmod Q$. The Smith invariants of $H$ equal those of the row lattice
generated by $A$. In Smith coordinates, multiplication by $s_j$ has
$\gcd(Q,s_j)$ zeroes modulo $Q$. Their product gives the kernel size; dividing
the domain size by that kernel size gives the image count.
\end{proof}

\subsection{A normalization example with a strict grid inclusion}
Take $A=(2,4)^\top$, so $H=(2)$, $B=(1,2)^\top$ and $h_A=2$. At $Q=6$ the
normalized coordinates attained by the original grid are $0,2,4$ modulo $6$.
There are three image points, each with two phase labels. The complete
normalized $6$-grid has six points. At $Q=5$, multiplication by $2$ is a
permutation and both codes have five points.

At $Q=2$, the original grid consists of a single image point. A minimum taken
over all nonzero input labels would be zero, although there are no two
distinct image points to compare. The correct distinct-image optimization
has an empty feasible set. Below its mathematical value is $+\infty$; the
software reports \texttt{minimum=None} and zero nearest neighbors. This
convention does not identify a singleton code with a code having two
coincident words.

Changing the row basis of $L_A$ replaces $H$ by $UH$ and $B$ by $BU^{-1}$ for
$U\in\mathrm{GL}_d(\Z)$. It transports the subgroup $H(\Z/Q\Z)^d$ by the same
unimodular coordinate change. Thus the complete pair consisting of the
normalized embedding and its attained grid is independent of the chosen
Hermite basis. Keeping $B$ while discarding this subgroup is not a
coordinate-invariant operation.

\section{Relative arithmetic rigidity and prime-power obstruction depths}
\label{sec:relative-rigidity}
The fixed-alphabet obstruction admits an exact form that applies to all
full-rank integer parametrizations. The correct comparison is between the
kernel of a row basis and the kernel already present in the complete
representation.

Let $\phi:\Z/Q\Z\to[0,\infty)$ satisfy
\[
 \phi(0)=0,\qquad \phi(r)>0\quad(r\ne0),
 \qquad \alpha=\min_{r\ne0}\phi(r),\quad
 \beta=\max_{r\ne0}\phi(r).
\]
Symmetry is unnecessary for the lower-bound theorem. Define
\[
 \delta_\phi(A,w;Q)=
 \min_{x\notin K_A(Q)}\sum_{i=1}^m w_i\phi((Ax)_i),
\]
with $\min\varnothing=+\infty$. The real cocircuit family and weighted
cogirth $\lambda_A(w)$ use exactly the real row configuration as before.

\begin{theorem}[Primitive-free arithmetic lower-bound classification]
\label{thm:relative-safe}
For a full-column-rank integral $A$ and $Q\ge2$, the following are equivalent.
\begin{enumerate}
\item For every $w\in(0,\infty)^m$,
      $\delta_\phi(A,w;Q)\ge\alpha\lambda_A(w)$.
\item For every real row basis $I$,
      $K_{A_I}(Q)=K_A(Q)$.
\item For every real row basis $I$, with row lattice $L_I$,
      $L_I+Q\Z^d=L_A+Q\Z^d$.
\item For every real row basis $I$,
\begin{equation}\label{eq:relative-smith-test}
 \prod_{j=1}^d\gcd(Q,s_j(A_I))=
 \prod_{j=1}^d\gcd(Q,s_j(A)).
\end{equation}
\end{enumerate}
The equivalence holds for every choice of strictly positive nonzero-residue
losses $\phi$. If it fails, a basis and a single nonzero image word certify
failure after an explicit choice of positive weights.
\end{theorem}
\begin{proof}
Assume (ii), and take $x\notin K_A(Q)$. The zero-row set
$R=\{i:(Ax)_i=0\pmod Q\}$ cannot have real rank $d$: otherwise it contains a
row basis $I$ and $x\in K_{A_I}(Q)=K_A(Q)$, a contradiction. Hence the nonzero
support of $Ax$ contains a real cocircuit $D$. Positivity gives
\[
 \sum_iw_i\phi((Ax)_i)
 \ge\alpha\sum_{i\in\operatorname{supp}(Ax)}w_i
 \ge\alpha\sum_{i\in D}w_i
 \ge\alpha\lambda_A(w).
\]
This proves (i), including the empty feasible-set case.

If (ii) fails, choose $x\in K_{A_I}(Q)\setminus K_A(Q)$. Place weight $W$ on
the $d$ basis rows and weight $1$ on the other rows, with
\[
 W=1+(m-d)\beta/\alpha.
\]
Every real cocircuit meets a real row basis, so $\lambda_A(w)\ge W$.
The image word $Ax$ is nonzero but contributes only off the basis. Thus
\[
 \delta_\phi(A,w;Q)\le(m-d)\beta<\alpha W
 \le\alpha\lambda_A(w).
\]
This disproves (i).

For (ii)$\Leftrightarrow$(iii), use the perfect pairing
$(u,x)\mapsto\exp(2\pi i\,u\cdot x/Q)$ on $(\Z/Q\Z)^d$. The annihilator of
the row submodule is precisely the associated modular kernel. A submodule
equals its double annihilator: Smith reduction proves this coordinatewise,
or finite character orthogonality proves that a subgroup and its
annihilator have cardinalities whose product is $Q^d$. Therefore two row
submodules have the same annihilator if and only if they coincide. Their
inverse images in $\Z^d$ are the two lattices in (iii).

Finally, $K_A(Q)\subseteq K_{A_I}(Q)$ always. Equality is therefore equivalent
to equality of their finite cardinalities, which the Smith kernel formula
expresses as (iv).
\end{proof}

We call a modulus satisfying these conditions \emph{relatively safe}. For a
primitive matrix the complete kernel is trivial, so relative safety reduces
exactly to the original basis-safe condition. For a nonprimitive matrix,
invertibility of every basis modulo $Q$ is sufficient but need not be
necessary.

\begin{corollary}[Different positive losses in different coordinates]
Suppose coordinate $i$ has a loss $\phi_i$ vanishing only at zero, and let
$\alpha_i=\min_{r\ne0}\phi_i(r)$. Then the universal lower bound is
\[
 \min_{x\notin K_A(Q)}\sum_iw_i\phi_i((Ax)_i)
 \ge \lambda_A(\alpha_1w_1,\ldots,\alpha_mw_m),
\]
for all positive weights if and only if $Q$ is relatively safe.
\end{corollary}
\begin{proof}
Replace the weights by $v_i=\alpha_iw_i$ and the coordinate losses by
$\phi_i/\alpha_i$. Each normalized loss is at least one off zero, so the
support argument is unchanged. For a bad basis choose $v_i=W$ on the basis
and $v_i=1$ elsewhere. The nonzero bad-basis image has cost at most
$\sum_{i\notin I}\max_r\phi_i(r)/\alpha_i$. Choosing $W$ larger than this
quantity gives the same strict counterexample.
\end{proof}

\subsection{The all-modulus relative class}
\begin{theorem}[Equimodularity of the quotient representation]
\label{thm:equimodular-relative}
For full-column-rank $A$, the following are equivalent:
every modulus is relatively safe; every real row basis generates $L_A$;
every nonzero maximal minor has absolute value $h_A$; and the primitive
normalization $B$ in $A=BH$ is full-minor unimodular.
\end{theorem}
\begin{proof}
If $L_I=L_A$, their sums with $Q\Z^d$ agree for all $Q$.
Conversely, fix $I$ and put $N=|\det A_I|$. The adjugate identity gives
$N\Z^d\subseteq L_I$. If $N\ge2$, relative safety at $Q=N$ yields
$L_A+N\Z^d=L_I+N\Z^d=L_I$, hence $L_A=L_I$. If $N=1$, the row basis already
generates $\Z^d$, so equality holds directly. Equality of the full-rank
lattices $L_I\subseteq L_A$ is equivalent to equality of their indices in
$\Z^d$, namely $|\det A_I|=h_A$. Finally,
$\det A_I=(\det B_I)(\det H)$, so these determinant equalities are equivalent
to $|\det B_I|=1$ for every nonsingular $B_I$.
\end{proof}

This theorem classifies a universal lower bound. It does not assert an exact
sharp formula for the original nonprimitive grid. For example, $A=(2)$ has
only one row basis and is relatively safe at every modulus. At $Q=4$, its
only nonzero image residue is $2$. For the symmetric loss
$\phi=(0,1,2,1)$ and weight one, its minimum is $2$, whereas
$\alpha\lambda_A=1$. The covering has skipped the residues where the loss
is smallest.

\subsection{An obstruction can first appear at a higher prime power}
For a basis $I$ and a prime $p$, define
\[
 \Delta_{I,p}(a)=
 \sum_{j=1}^d\min\{a,v_p(s_j(A_I))\}
 -\sum_{j=1}^d\min\{a,v_p(s_j(A))\},\qquad a\ge1.
\]
This is the base-$p$ logarithm of the kernel ratio
$|K_{A_I}(p^a)|/|K_A(p^a)|$, hence is a nonnegative integer. Put
\[
 \alpha_p(A)=\min\{a\ge1:\Delta_{I,p}(a)>0
                       \text{ for some row basis }I\},
\]
with value $+\infty$ if there is no such exponent.

\begin{theorem}[Exact obstruction-depth classification]
\label{thm:obstruction-depth}
Only finitely many primes have finite $\alpha_p(A)$. They are exactly the
prime divisors of at least one relative basis index
$|\det A_I|/h_A$. For every $Q\ge2$,
\begin{equation}\label{eq:relative-depth-law}
 Q\text{ is relatively safe}\quad\Longleftrightarrow\quad
 v_p(Q)<\alpha_p(A)\quad\text{for every prime with finite }\alpha_p(A).
\end{equation}
Once a basis has extra aliasing at $p^a$, it has extra aliasing at every
$p^{a+b}$, $b\ge0$. The depths are computable by exact Smith arithmetic.
\end{theorem}
\begin{proof}
The map $x\mapsto px$ from residues modulo $p^a$ to residues modulo
$p^{a+1}$ induces an injection
\[
 K_{A_I}(p^a)/K_A(p^a)\hookrightarrow
 K_{A_I}(p^{a+1})/K_A(p^{a+1}).
\]
It is well defined since multiplication by $p$ transports both kernel
conditions. If the image class belongs to $K_A(p^{a+1})$, then
$pAx=0\pmod {p^{a+1}}$, so $Ax=0\pmod {p^a}$ and the original quotient
class was zero. This proves injectivity and persistence of every defect.

For $a$ at least the largest $p$-valuation among the Smith invariants of both
matrices, the kernel ratio is
\[
 p^{v_p(|\det A_I|)-v_p(h_A)}
 =p^{v_p(|\det A_I|/h_A)}.
\]
Thus a defect occurs at some depth exactly when the relative index is
divisible by $p$. There are finitely many row bases and each relative index
is a positive integer, proving finiteness. The first positive value of
$\Delta_{I,p}(a)$ can be found within these finite valuation ranges.

The Chinese remainder decomposition expresses both complete and basis
kernels at $Q$ as products of their kernels at the prime-power factors of
$Q$. Every factor ratio is at least one. Their product is one precisely when
each factor ratio is one. Together with defect persistence, this gives
\eqref{eq:relative-depth-law}.
\end{proof}

For $A=(2,4)^\top$, the complete Smith invariant is $2$, and the two basis
invariants are $2$ and $4$. Therefore
\[
 \alpha_2(A)=2,\qquad
 Q\text{ relatively safe}\Longleftrightarrow 4\nmid Q.
\]
At $Q=2$, the complete image is a singleton. At $Q=6$, the image is
nontrivial and relative safety still holds. At $Q=4$, the basis character
$4$ vanishes everywhere, while character $2$ remains visible. The weights
$(1,100)$ and loss $(0,1,2,1)$ give minimum $2$, but real cogirth $101$.
This example explains why simply factoring maximal minors gives the wrong
obstruction set before the finite covering has been accounted for.

\section{Arithmetic rigidity is independent of the chordal loss profile}
\label{sec:loss-universality}
The sine function comes from Euclidean chords, but it is not responsible
for the exact combinatorial rigidity class. The same class governs every
symmetric positive loss on a cyclic alphabet, including profiles that are
not monotone, not convex and not minimized at a unit residue.

Let $A$ now be primitive and full-minor unimodular. Retain a loss $\phi$ as
above and additionally assume $\phi(-r)=\phi(r)$. Define
\[
 T_\phi=\{r\in\Z/Q\Z\setminus\{0\}:\phi(r)=\alpha\}.
\]
For every real cocircuit $D$, let $z_D\in\Z^d$ be its primitive unit
representative: $(Az_D)_i\in\{0,1,-1\}$, with support $D$.

\begin{theorem}[Exact law for every symmetric residue loss]
\label{thm:symmetric-loss}
For every positive weighting and every symmetric loss vanishing only at
zero,
\begin{equation}\label{eq:general-loss-law}
 \delta_\phi(A,w;Q)=\alpha\lambda_A(w).
\end{equation}
The minimizing phase differences are exactly
\begin{equation}\label{eq:general-loss-minimizers}
 \{t z_D\pmod Q:D\text{ a minimum-weight real cocircuit},\ t\in T_\phi\}.
\end{equation}
They are distinct as $D$ and $t$ vary, and their number is
$|T_\phi|N_*(A,w)$.
\end{theorem}
\begin{proof}
Unimodularity makes every modulus safe, so the lower bound follows from the
relative classification. For a minimum cocircuit $D$ and $t\in T_\phi$,
the residues of $Atz_D$ are zero off $D$ and $\pm t$ on $D$. Symmetry makes
their loss equal to $\alpha$. The phase residue $tz_D$ is nonzero because
some unit row-basis coordinate of $z_D$ equals one. This proves equality.

For the equality classification, a minimizer has support $S$ containing a
real cocircuit $D$. Equality in the lower-bound chain forces $S=D$, since
all weights are strictly positive. It also forces $D$ to have minimum weight
and every nonzero coordinate to attain the minimum loss.

Choose $d-1$ independent rows in the complement hyperplane of $D$ and one
row outside it. They form a unimodular square basis $U$. Since $x$ vanishes
on the first $d-1$ rows modulo $Q$, its basis coordinates have the form
$Ux=(0,\ldots,0,t)^\top$ modulo $Q$. Hence
$x=tU^{-1}e_d=tz_D$ up to the arbitrary sign chosen for $z_D$. Symmetry of
$T_\phi$ removes this sign ambiguity. The last coordinate condition gives
$t\ne0$, and minimality gives $t\in T_\phi$.

For fixed $D$, the map $t\mapsto tz_D$ is injective because applying $U$
recovers $t$. Different cocircuits have different nonzero coordinate
supports in the image word and therefore cannot coincide. This proves the
count and the complete classification.
\end{proof}

The choice $\phi(r)=4\sin^2(\pi r/Q)$ recovers the original chord law and its
one-or-two sign multiplicity. The Hamming profile
$\phi(r)=\mathbf1_{r\ne0}$ has $|T_\phi|=Q-1$, giving $(Q-1)N_*$ nearest
words. A Lee profile $\phi(r)=\min\{r,Q-r\}$ has the usual two minimizing
signs for $Q\ge3$. For a positive exponent $\gamma$, the chord-power profile
$\phi(r)=(4\sin^2(\pi r/Q))^\gamma$ gives
\[
 \delta_\phi=(4\sin^2(\pi/Q))^\gamma\lambda_A(w).
\]
This is a sum of coordinate chord powers; it is not being identified with
an arbitrary power of the total squared Euclidean distance.

A profile with $Q=6$ and values $(0,3,1,2,1,3)$ has minimizers $2,4$, neither
of which is a unit modulo $6$. The theorem still applies: a primitive
cocircuit direction multiplied by a nonzero residue remains nonzero. Thus
the loss theorem requires no invertibility assumption on its minimizing
residues.

\begin{corollary}[The universal rigidity class survives changing the loss]
Fix any family of symmetric positive losses $\phi_Q$, one for each prime
alphabet. For a primitive full-rank $A$, the universal lower bound
\[
 \delta_{\phi_Q}(A,w;Q)\ge
 \bigl(\min_{r\ne0}\phi_Q(r)\bigr)\lambda_A(w)
\]
for every positive weighting and every prime $Q$ holds if and only if $A$
is full-minor unimodular. In that case the sharp formula holds for every
modulus and every symmetric positive loss on that modulus.
\end{corollary}
\begin{proof}
If a nonzero maximal minor has absolute value greater than one, choose a
prime divisor. Its basis kernel is nontrivial, while primitivity makes the
complete kernel trivial. The relative classification supplies a violating
weighting for the prescribed loss at that prime. The converse and the
sharp formula follow from Theorem~\ref{thm:symmetric-loss}.
\end{proof}

\subsection{The Hamming specialization is sharper at a fixed safe alphabet}
\begin{proposition}[Fixed-alphabet Hamming equality]
\label{prop:hamming-fixed}
For primitive $A$, the weighted Hamming minimum equals $\lambda_A(w)$ for
every positive weighting if and only if $Q$ is basis-safe. Full-minor
unimodularity is not required when the modulus is fixed.
\end{proposition}
\begin{proof}
The forward implication follows from the lower-bound classification.
For the converse, take any real cocircuit $D$. Choose $d-1$ independent rows
in its complement hyperplane, append one outside row, and call the square
basis $U$. Since $Q$ is safe, $U$ is invertible modulo $Q$. Put
$x=U^{-1}e_d$ modulo $Q$.

For an arbitrary row $a_i^\top$, its value on $x$ is the determinant of $U$
with its last row replaced by $a_i^\top$, divided by $\det U$, up to the
fixed ordering sign. A row in the complement hyperplane gives determinant
zero. Every other row gives a nonsingular maximal minor, hence a unit
modulo $Q$. Thus $Ax$ has support exactly $D$. Choosing a minimum cocircuit
attains Hamming loss $\lambda_A(w)$, while safety supplies the reverse
inequality.
\end{proof}

This identifies the precise point where the Euclidean formula becomes
stronger than a support theorem: a safe alphabet realizes each cocircuit
support, but its nonzero residues may have different chord lengths.

Symmetry also has a precise role. With $A=(1,-1)^\top$, $Q=3$ and
$\phi=(0,1,4)$, both nonzero words have loss $5$ at unit weights. The minimum
cocircuit has weight $2$ and $\alpha=1$. The lower bound remains true, but
the sharp formula $\alpha\lambda_A=2$ fails. Opposite character signs can
have different costs for an asymmetric loss.

\section{An exact support polynomial from subsystem torsion}
\label{sec:support-polynomial}
The kernel information can be assembled into a complete polynomial that
counts modular support patterns. Its ingredients are classical Smith
normal form and inclusion--exclusion. Its quasi-polynomial behavior belongs
to the established theory of integral arrangements and arithmetic Tutte
invariants~\cite{kamiya,gtutte}. The purpose here is to connect those exact
counts to the weighted quantization obstruction and to implement the
connection without a full $Q^d$ grid search.

For $R\subseteq[m]$, set
\[
 Z_R(Q)=\#\{x:(Ax)_i=0\pmod Q\text{ for every }i\in R\}
       =Q^{d-r_R}\prod_{j=1}^{r_R}\gcd(Q,s_j(A_R)),
\]
where $r_R=\operatorname{rank}_{\R}(A_R)$, and $Z_\varnothing(Q)=Q^d$.
Let $n_S(Q)$ count input labels whose image has support exactly $S$. Let
$\kappa_Q=Z_{[m]}(Q)=|K_A(Q)|$. The corresponding number of distinct image
words is $\overline n_S(Q)=n_S(Q)/\kappa_Q$.

\begin{theorem}[Exact support distribution]
\label{thm:support-counts}
For every full-column-rank integer $A$ and $Q\ge2$,
\begin{equation}\label{eq:support-mobius}
 n_S(Q)=\sum_{T\subseteq S}(-1)^{|T|}Z_{S^c\cup T}(Q).
\end{equation}
Equivalently, the multiaffine polynomial of distinct image supports is
\begin{equation}\label{eq:support-polynomial}
 \mathcal S_{A,Q}(z_1,\ldots,z_m)
 =\frac1{\kappa_Q}\sum_{R\subseteq[m]} Z_R(Q)
       \prod_{i\in R}(1-z_i)\prod_{i\notin R}z_i.
\end{equation}
Its coefficients are nonnegative integers; its constant coefficient is
one; and $\mathcal S_{A,Q}(1,\ldots,1)=|V_A(Q)|$.
\end{theorem}
\begin{proof}
Requiring support $S$ means requiring every coordinate in $S^c$ to vanish
and excluding zero in every coordinate of $S$. Inclusion--exclusion over
those excluded events gives \eqref{eq:support-mobius}.

For a fixed input $x$, write $\epsilon_i=\mathbf1_{(Ax)_i=0}$. Its support
monomial is
\[
 \prod_{i=1}^m\bigl(z_i+(1-z_i)\epsilon_i\bigr).
\]
Expanding the product and summing over $x$ gives the numerator of
\eqref{eq:support-polynomial}. Every image word has precisely $\kappa_Q$
preimages, so dividing removes the uniform fiber multiplicity. This proves
the formula and integrality. Only the zero image word has empty support,
and evaluating all variables at one counts all distinct image words.
\end{proof}

\begin{corollary}[Support form of the obstruction]
A modulus is relatively safe if and only if every nonconstant monomial
with positive coefficient in $\mathcal S_{A,Q}$ has a support containing a
real cocircuit. The exact weighted Hamming minimum is
\[
 \min\left\{\sum_{i\in S}w_i:
             S\ne\varnothing,\ \overline n_S(Q)>0\right\},
\]
with value $+\infty$ when the set is empty.
\end{corollary}
\begin{proof}
Under relative safety the zero set of any nonzero image has rank below $d$,
so its complement contains a real cocircuit. Conversely, failure of
relative safety supplies a nonzero image vanishing on a row basis. Its
support cannot contain a real cocircuit, because every real cocircuit meets
that basis. The Hamming formula simply minimizes the support weight over
exactly the supports that occur.
\end{proof}

\subsection{Quasi-polynomial dependence on the alphabet}
Let
\[
 \rho=\operatorname{lcm}
 \{s_{r_R}(A_R):R\subseteq[m],\ r_R>0\},
\]
with $\rho=1$ if the collection is empty. All smaller Smith invariants divide
the last invariant of their subsystem. Thus every factor
$\gcd(Q,s_j(A_R))$ is periodic in $Q$ with period dividing $\rho$.

\begin{proposition}[A finite exact period for every support coefficient]
For each fixed support $S$, both $n_S(Q)$ and $\overline n_S(Q)$ are
quasi-polynomials of $Q$ of degree at most $d$, with a period dividing
$\rho$. In other words, for each residue class $c\pmod\rho$ there is a
polynomial with rational coefficients whose value at every integer
$Q\ge2$ in that class is the corresponding count.
\end{proposition}
\begin{proof}
On a fixed residue class, every gcd factor in $Z_R(Q)$ is constant, so
$Z_R(Q)$ is a constant multiple of $Q^{d-r_R}$. Formula
\eqref{eq:support-mobius} is a finite integer linear combination of these
polynomials, giving the result for $n_S$. The complete matrix has rank $d$,
so $\kappa_Q$ contains no free power of $Q$ and is itself a positive periodic
function with period dividing $\rho$. It is constant and nonzero on each
residue class. Division by that constant gives a rational-coefficient
polynomial for $\overline n_S$. The counting interpretation guarantees
integer values at admissible integers $Q$.
\end{proof}

The assertion is a periodicity statement about exact support multiplicities.
It does not assert polynomial or quasi-polynomial dependence of the actual
trigonometric minimum distance. The loss profile changes with $Q$ in the
Euclidean problem, and support data do not record the nonzero residues.

\subsection{Two complete support computations}
For $A=(2,3)^\top$ and $Q=6$, the four zero-subsystem counts are
\[
 Z_\varnothing=6,\quad Z_{\{1\}}=2,\quad
 Z_{\{2\}}=3,\quad Z_{\{1,2\}}=1.
\]
Inclusion--exclusion gives
\[
 \mathcal S_{A,6}=1+2z_1+z_2+2z_1z_2.
\]
The only real cocircuit is $\{1,2\}$, but modular singleton supports occur.
The weighted Hamming minimum is therefore $\min(w_1,w_2)$, rather than
$w_1+w_2$. Both singleton coefficients are explicit arithmetic obstruction
certificates.

For $A=(2,4)^\top$ and $Q=4$, the input-label polynomial is $2+2z_1$ and
$\kappa_Q=2$. The distinct-image polynomial is $1+z_1$. Dividing by the
complete kernel is necessary: otherwise a nearest-image count would be
incorrectly doubled.

Even the exact support polynomial at a fixed safe alphabet is insufficient
to recover Euclidean distances. At $Q=5$, compare $A=(1,1)^\top$ with
$A'=(1,2)^\top$, both at unit weights. Both are primitive and basis-safe, and
both have polynomial
\[
 \mathcal S_{A,5}=\mathcal S_{A',5}=1+4z_1z_2.
\]
Nevertheless,
\[
 D_5(A,(1,1))^2=8\sin^2(\pi/5)=5-\sqrt5,
 \qquad D_5(A',(1,1))^2=5.
\]
For the second identity, every nonzero phase gives the two squared sine
values $\sin^2(\pi/5)$ and $\sin^2(2\pi/5)$ in some order, whose sum is
$5/4$. The support polynomial correctly controls zero patterns and Hamming
loss; higher residue information is needed for a chordal spectrum.

\subsection{Exact implementation and its computational limits}
The release module \texttt{evefreq.quotient} implements the primitive
normalization, distinct image count, relative safety certificates,
prime-power depths, support coefficients and direct rational-loss minima.
The normalization uses Hermite normal form and checks both $A=BH$ and the
primitivity of $B$. The depth routine evaluates the exact Smith valuation
formula and returns a witness basis at the first unsafe exponent.

Support enumeration computes $2^m$ zero-subsystem counts and then applies a
subset M\"obius transform. Once those counts have been obtained, the transform
requires $O(m2^m)$ additions and subtractions. This avoids enumerating $Q^d$
phase labels, but it is still exponential in the number of rows. Moreover,
the inherited Smith routine enumerates determinantal divisors; it is a
small-instance certificate implementation, not a claim of an optimal Smith
algorithm. Explicit budgets stop oversized computations.

The direct loss verifier enumerates input labels, deduplicates their image
words, and performs exact rational comparisons. It accepts asymmetric
profiles as well, enabling direct checks of the symmetry boundary. Its
singleton output is explicit. Independent tests compare subset-derived
support counts against full grids, compare obstruction depths against
individual-modulus certificates, and check the full nearest-neighbor
multiplicity formula for several symmetric profiles. These are executable
audits of the stated arguments; they do not constitute a formal proof in a
proof assistant.