The prime-grid reset and the three-color plateau

---

Standing

The maximal root-coverage reset occurs inside the actual distinct-prime family, already at its first proved jump. Thus restricting to prime side sizes does not supply the missing restart. There is also an exact three-color plateau and a sharper separation bound:

$$\alpha_3(S_3\otimes S_5\otimes S_7)=2/3, \qquad 3\leq\tau_3(S_3\otimes S_5\otimes S_7)\leq4.$$

Neither statement proves the unbounded coloring obligation. All algebra below is author-recorded; a separate reviewer checked it.

1. An exact root-closed partition at the prime jump

On $B=S_3\otimes S_5$, index rows by $r=0,1,2$, columns by $j=0,\ldots,4$, and use $f(r,j)=r+j\pmod3$. Its classes are

$$\begin{aligned} C_0&=\{(0,0),(0,3),(1,2),(2,1),(2,4)\},\\ C_1&=\{(0,1),(0,4),(1,0),(1,3),(2,2)\},\\ C_2&=\{(0,2),(1,1),(1,4),(2,0),(2,3)\}. \end{aligned}$$

Each class is rationally independent and rational-root-closed: $B\cap\operatorname{span}_{\mathbb Q}C_c=C_c$. Its complement graph in $K_{3,5}$ is connected and bridgeless: suppressing the five degree-two right vertices gives a triangle with edge multiplicities $(2,2,1)$. The proved cographic criterion applies.

In coordinates $(x_{11},x_{12},x_{13},x_{14},x_{21},x_{22},x_{23},x_{24})$, the three spans have equations

$$\begin{array}{ll} C_0:&x_{21}=x_{13},\quad x_{22}=x_{12},\quad x_{24}=x_{13},\\ C_1:&x_{14}=x_{11},\quad x_{22}=0,\quad x_{23}=x_{13}-x_{11},\\ C_2:&x_{12}=0,\quad x_{23}=x_{21}-x_{11},\quad x_{24}=x_{14}+x_{21}-x_{11}. \end{array}$$

Each pair of spans has exactly the same intersection

$$U=\{(a-b,0,a,a-b,a,0,b,a):a,b\in\mathbb Q\}.$$

Let $T$ exchange columns $j=3,4$. On each row of four standard coordinates it sends $(x_1,x_2,x_3,x_4)$ to $(x_1-x_4,x_2-x_4,x_3-x_4,-x_4)$. Thus

$$T(U)=\{(0,b-a,b,b-a,0,-a,b-a,-a):a,b\in\mathbb Q\}, \qquad U\cap T(U)=\{0\}.$$

Coordinates $12$ and $22$ prove the last assertion immediately.

Use six layer types in $B\otimes S_m$: the three cyclic relabelings of $f$, and the three cyclic relabelings of its $T$-image. For $m\geq6$, fill additional layers by repetitions. For any fixed color, deleting any single layer leaves at least two distinct original class types and two distinct transformed types. The remaining rational-span intersection is contained in $U\cap T(U)=0$.

The rational version of the layer criterion makes every resulting color class $D_c$ independent. Moreover a root $b\otimes s_j$ can belong to $\operatorname{span}D_c$ only if there is a common sum $u$ in all the other layer spans, with $u+b$ in layer $j$. This forces $u=0$, and base root closure forces that root to have color $c$. Therefore

$$(B\otimes S_m)\cap\operatorname{span}_{\mathbb Q}D_c=D_c.$$

At $m=7$, all three classes have 35 points and each root is omitted from exactly the other two colors' signed spans. Hence $\alpha_3=2/3$, the maximum allowed by its definition. The same witness works for arbitrary nonnegative weights on roots. The existing exact bound $\chi(B\otimes S_7)=3$ makes this a genuine postjump reset.

2. Rational thresholds and an ordinary plateau

For a finite transitive indexed vector configuration of $N$ columns and rank $d$, averaging a basis over the transitive group gives $\operatorname{rank}F\geq d|F|/N$ for every subfamily $F$. Rado--Horn therefore proves

$$\chi_{\mathbb Q\text{-ind}}(E)=\lceil N/d\rceil.$$

The precise indexed-list statement was checked in Lewko, Theorem 4.1, after its digest. Its necessity follows just by summing color-class ranks. For $q>N/d$, the rational separation threshold (independent layers, zero intersection of their rational spans) is consequently

$$\tau_q^{\mathbb Q}(E)= \left\lceil\frac q{q-N/d}\right\rceil.$$

Indeed $E\otimes S_p$ has density $Np/[d(p-1)]$; it has a rational $q$-coloring exactly when this is at most $q$, which rearranges to the displayed inequality. The rational layer criterion identifies this with the stated separation parameter. Indexed size-two simplexes are allowed in this rational statement.

For $Y=S_3\otimes S_5\otimes S_7$, $N/d=35/16$, so $\tau_3^{\mathbb Q}(Y)=\lceil48/13\rceil=4$. Rational separation is also ordinary signed separation. The already proved lower bound $\chi(Y)=3\Rightarrow\tau_3(Y)\geq3$ gives $3\leq\tau_3(Y)\leq4$.

Every full odd-prime prefix ending between 7 and 19 inclusive has ordinary dissociated chromatic number exactly three: it contains $Y$, while its rational density is below three.

At 23 only $3\leq\chi_{\rm diss}\leq4$ follows here. A rational chromatic jump does not prove an ordinary one.

3. Cyclic fiber colorings have a nonzero common signed sum

There is an algebraic reason that cyclic relabelings of the entire reset partition do not witness $\tau_3(Y)=3$.

Let $T\ne\varnothing$ be a finite vector set with $\sum_{t\in T}t=0$, partition it as $A_0\sqcup A_1\sqcup A_2$, and put $\sigma_a=\sum_{t\in A_a}t$. For $S_3=\{s_0,s_1,s_2\}$, define

$$D_c=\bigcup_{a=0}^2 s_{a+c}\otimes A_a\quad(c=0,1,2),$$

with indices modulo three. If these indexed classes are dissociated, then not all $\sigma_a$ vanish: a nonempty group with sum zero would already give a monochromatic signed relation. The nonzero vector $u=s_1\otimes\sigma_0-s_0\otimes\sigma_2$ satisfies

$$\begin{aligned} u&=-s_1\otimes\sigma_1+s_2\otimes\sigma_2\in\Sigma(D_0),\\ u&= s_1\otimes\sigma_0-s_0\otimes\sigma_2\in\Sigma(D_1),\\ u&=-s_2\otimes\sigma_0+s_0\otimes\sigma_1\in\Sigma(D_2). \end{aligned}$$

The identities use $s_0+s_1+s_2=0$ and $\sigma_0+\sigma_1+\sigma_2=0$. If $u=0$, independence of $s_0,s_1$ forces all three $\sigma_a=0$, a contradiction. All coefficients expand as $\pm1$ on disjoint groups.

The reset partition is of this form on its $S_3$-fibers, with $T=S_5\otimes S_7$. Three cyclic relabelings of any such partition cannot have zero common signed spans. This does not exclude three unrelated witness colorings, or arbitrary fiber permutations.

Finite discovery and the stopped test

The reset discovery checked at most the 120 permutations of the five base columns and stopped at permutation 2, using exact rational arithmetic. The displayed coordinate proof supersedes that search.

A separate fixed common-span checker constructs only the displayed seven-layer partition and its $96\times105$ equality matrix. It reported class sizes $(35,35,35)$, rank 83 and nullity 22. Its declared gate allowed ternary enumeration only at nullity at most 12, so no ternary enumeration ran. The algebraic cyclic-sum lemma above supersedes that unsuccessful route; no zero-intersection conclusion is drawn from the stopped computation.