\section{Local reductions and uniform connector schemata}\label{app:connectors}

In CRT coordinates write
\[
\begin{gathered}
P[a]=(0,a,0),\quad Q[b]=(0,0,b),\\
P1[a]=(1,a,0),\quad Q1[b]=(1,0,b),\\
F[a,b]=(0,a,b),\quad G[a,b]=(1,a,b),\quad E=(1,0,0).
\end{gathered}
\]

\subsection{Expanded local reductions}

We record the endpoint details suppressed from
Proposition~\ref{prop:local-boxes}.  Let a target-valued function kill the
appropriate rows.

On a parity-zero punctured line, pairs give $f(-x)=-f(x)$ and triples give
\[
 f(x+y)=f(x)+f(y)\qquad(x,y,x+y\ne0).                        \tag{A.1}
\]
Fixing a generator of the prime cycle and walking through its nonzero
multiples proves $f(x)=xu$, $ru=0$.  On a parity-one line there are no odd
rows.  For $r\ge5$, the quadruple $(x,1,1,-x-2)$ and its pair reductions give
the recurrence
\[
 f(x+2)-f(x)=2f(1).                                         \tag{A.2}
\]
The translation-by-two cycle on $C_r$ becomes a path after deleting zero, so
(A.2) determines $f(x)=c+xu$.  Opposite pairs give $2c=0$.
For $r\ge7$, the row $(2,2,2,r-6)$ gives $4c+ru=0$ and therefore
$ru=0$.  For $r=5$, the separate row $(1,1,1,2)$ gives $4c+5u=0$ and
therefore $5u=0$.  Conversely $c+xu$ kills every even zero-sum row.  For
$r=3$, each quartic row has two
points from each sign class, so the quotient remains $\Z$.

For the parity-zero rectangle $F$, suppose first $p,q\ge5$.  Pairs and
triples give partial additivity.  For $s\in C_p$ define
\[
 \Delta_s(x,y)=f(x+s,y)-f(x,y)
\]
when both endpoints are in the punctured rectangle.  Given two base points,
choose $c$ outside
\[
 \{0,-s,-x-s,-x'-s\}
\]
and $d$ outside $\{0,-y,-y'\}$.  The choices exist for $p\ge5,q\ge7$;
at $p=5$, at most four classes are forbidden and the fifth remains.  Applying
partial additivity to
\[
 (x+s,y)+(c,d)=(x,y)+(c+s,d)                                \tag{A.3}
\]
and to the primed version makes $\Delta_s$ base-point independent.  A second
auxiliary point proves $\Delta_{s+t}=\Delta_s+\Delta_t$.  Repeating vertically
gives $f(a,b)=c+au+bv$, $pu=qv=0$.  An opposite pair gives $2c=0$ and the
triple $(1,1),(1,1),(-2,-2)$ gives $3c=0$, hence $c=0$.

When $p=3$, set $h(b)=f(1,b)$ and recover the other sheet from pairs.  A
triple has zero $C_3$ coordinate precisely when its three signs agree.  Thus
pair sums having the same nonzero total have the same $h$-sum.  To cross a
zero total, choose a bridge point outside at most four $q$-classes; this works
at $q=5$.  The resulting difference equation yields
$h(b)=u+bv$, $3u=qv=0$, and hence $f(a,b)=au+bv$.

For the parity-one rectangle $G$ in the generic branch, the four points
\[
 (1,x+s,y),\ (1,-x,-y),\ (1,x',y'),\ (1,-x'-s,-y')          \tag{A.4}
\]
sum to zero whenever their coordinates are nonzero.  Pair reduction of (A.4)
makes horizontal differences base-point independent; the vertical argument
is identical.  This gives $c+au+bv$ with $2c=pu=qv=0$.  If $p=3$, every
quartic row contains two signs of each kind and becomes a Freiman relation on
$C_q^\times$.  Therefore $h(b)=u+bv$ with $u$ free and $qv=0$.  These
arguments prove every row of Table~\ref{tab:local}, including the smallest
generic prime.

\subsection{The fourteen structural types}

Every displayed parameter is nonzero, and every derived parameter is required
to be nonzero.  The following table gives the fourteen structural types; an
exchanged orientation means $p\leftrightarrow q$ and $P\leftrightarrow Q$,
$P1\leftrightarrow Q1$.

\begin{longtable}{@{}p{1.25cm}p{2.85cm}p{9.0cm}@{}}
\toprule
Type & Box count & Uniform zero-sum schema\\
\midrule
\endfirsthead
\toprule
Type & Box count & Uniform zero-sum schema\\
\midrule
\endhead
X1 & $0000012$ & $F[-2a,-2b]+G[a,b]+G[a,b]$\\
X2 & $0102000$ & $P[-2a]+2P1[a]$; exchange\\
X3 & $0100002$ & $P[-2a]+G[a,b]+G[a,-b]$; exchange\\
X4 & $0001011$ & $P1[a]+F[c,b]+G[-a-c,-b]$; choose
$a,c,a+c\ne0$; exchange\\
X5 & $1000011$ & $E+F[a,b]+G[-a,-b]$\\
X6 & $0100020$ & $P[-2a]+F[a,b]+F[a,-b]$; exchange\\
X7 & $0001110$ & $P1[a]+Q1[b]+F[-a,-b]$\\
X8 & $1101000$ & $E+P[a]+P1[-a]$; exchange\\
X9 & $0011001$ & $Q[b]+P1[a]+G[-a,-b]$; exchange\\
X10A & $0001003$ &
$P1[-c]+G[a,b]+G[-a,b]+G[c,-2b]$; exchange\\
X10B & $0001003$ &
$P1[a]+G[-a,b]+G[c,d]+G[-c,-b-d]$; exchange\\
X11 & $1000003$ & $E+G[a,b]+G[a,b]+G[-2a,-2b]$\\
X12 & $1003000$ & $E+P1[a]+P1[a]+P1[-2a]$; exchange\\
X13 & $0002020$ & $2P1[a]+F[-a,b]+F[-a,-b]$; exchange\\
X14 & $0111100$ & $P[a]+Q[b]+P1[-a]+Q1[-b]$\\
\bottomrule
\end{longtable}

A count string uses the order $(E,P,Q,P1,Q1,F,G)$.  Substitution proves each
row coordinate by coordinate.  Multiplication by $-1$ or $-2$ preserves
nonzero coordinates over odd primes.  For X4 one may choose $a=c=1$; then
$a+c=2\ne0$.  For X10B choose $a=c=b=d=1$; the derived entries are $-1$ and
$-2$.  Thus every schema is available at the endpoints $p=3$ and $q=5$.

The A and B choices of X10 share a box-count type but have different images
in the local presentation.  X10B is needed for the $P1-G$ slope edge in (8.3)
and for the integral relation $u=g$ in (8.4).  This parameter distinction is
why a count-pattern catalogue alone is insufficient for the Quartic
Skeleton.

For completeness, the connector functions are partitioned as follows.
X2, X3, and X6 attach prime-axis slopes to odd lines or rectangles.  X1, X4,
X7, X9, X10B, X13, and X14 compare rectangle slopes and cross-box parity.
X5, X8, X10A/B, X11, and X12 anchor parity constants.  Since two is invertible
in both odd-primary layers, a doubled slope difference imposes the same
equality as an undoubled one.  In the $p=3$ integral sign layer the integer
coefficients are retained, producing precisely the rows (8.4).

\subsection{Connector images in the prime-three layer}

For $p=3$, discard the already separated $C_q$ component and use the
generators $(e,p,u,c,f,g)$ of Section~\ref{sec:quartic}.  The required images
are
\[
\begin{array}{c|c|c}
\text{type}&\text{support}&\text{integral image}\\
\midrule
X1&F,G&f+2g\\
X2&P,P1&p+2u\\
X3&P,G&p+2g\\
X6&P,F&p+2f\\
X13&P1,F&2u-2f\\
X10B&P1,G&u-g\\
X11&E,G&e+3g\\
X12&E,P1&e+3u\\
X12^*&E,Q1&e+c\\
X10B^*&Q1,G&c+3g\\
X7&P1,Q1,F&u+c-f\\
X14&P,Q,P1,Q1&p-u+c.
\end{array}                                                   \tag{A.5}
\]
Substitution of the local forms into each parameterized zero-sum row gives
this table.  Every image vanishes under
$(e,p,u,c,f,g)\mapsto(3,4,1,3,4,1)$ in $C_6$.  If a chosen schema also has a
$C_q$-slope component, two parameter values isolate it by
Lemma~\ref{lem:isolation}.  Thus (A.5) is an integral presentation, not the
result of evaluating at one value of $q$.

In the generic branch, primary projection turns the same substitutions into
graph edges.  The $p$-slope graph on $P,P1,F,G$ and the $q$-slope graph on
$Q,Q1,F,G$ are complete.  On parity coordinates, X12 and X11 connect $E$ to
$P1,Q1,G$; X10B and its exchange connect the odd lines to $G$; X7 or X14
provides the remaining $P1-Q1$ comparison under the support conditions stated
in Section~\ref{sec:quartic}.  These images account for every use of the
connector table in the proof.

\subsection{Two exact closing rows}

The generic critical pair $P1\cup Q1$ closes by combining the two triples
$(1,1,-2)$, one in each odd coordinate line.  For the prime-three family, six
positive $P1$ or $G$ points give the $6u$ generator.  On $P1\cup Q1$, three
positive $P1$ points and three $Q1$ points with zero $q$-sum give $3u+c$.
These are existence constructions over every admissible prime, not residue
samples.