Chain circuits with unbounded box complexity and one finite coloring

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This finite algebraic observation complements the bounded-box theorem, which proves that every fixed number of disjoint constant-sign product boxes is finitely colorable. It shows that relations whose box complexity grows without bound can still all be split by one finite palette. It is not another candidate counterexample.

The chain family

Fix distinct odd primes $p_1,\ldots,p_k$, $k\geq2$, and choose $\alpha_j\in\mu_{p_j}\setminus\{1\}$. In the product grid, put

$$\begin{aligned} B_1&=(\mu_{p_1}\setminus\{\alpha_1\})\times\{1\}^{k-1},\\ B_j&=\{\alpha_1\}\times\cdots\times\{\alpha_{j-1}\} \times(\mu_{p_j}\setminus\{1,\alpha_j\})\times\{1\}^{k-j} && (1<j<k),\\ B_k&=\{\alpha_1\}\times\cdots\times\{\alpha_{k-1}\} \times(\mu_{p_k}\setminus\{1\}). \end{aligned}$$

Give $B_j$ coefficient sign $(-1)^{j-1}$. Their disjoint union supports a ternary rational circuit of size

$$|B_1\cup\cdots\cup B_k|=2+\sum_{j=1}^k(p_j-2). \tag{1}$$

For a direct proof, start with the full $p_1$-cycle. Replace its selected term $v$ by $-v\sum_{a\in\mu_p\setminus\{1\}}a$ at a fresh prime, and repeat, selecting the term with the next prescribed $\alpha_j$. This produces exactly the displayed boxes and signs. It preserves being a rational circuit: in a relation on the new support, each nonidentity new-prime layer contains one scalar multiple of $v$. Equal layer sums force all their coefficients to agree. The identity layer and their common value then give precisely a relation on the old circuit. The new kernel has dimension one, and every displayed coefficient is nonzero. Formula (1) follows by counting the boxes.

Exactly k boxes are needed

The exact minimum number of product boxes whose union is precisely this support is $k$, even without requiring constant signs or disjointness. A box contained in the support cannot meet both $B_j$ and $B_l$, $j<l$: take the first-departure coordinate from a point of $B_j$ and the nonidentity $l$-coordinate from a point of $B_l$. The resulting point would also be in the product box, but its first departure at $j$ is followed by a nonidentity coordinate and it belongs to none of the $B_i$. Thus at least one separate box is needed for each $B_j$; the displayed partition attains that number.

One finite coloring splits every chain

Nevertheless all these chains admit one finite coloring. Record the complete symbols in prime coordinates 3 and 5 (when present), and color the remaining coordinates by the parity of membership in fixed almost-equal binary parts. Choose each counted part to exclude the default symbol $1$, so this parity sum is finite on every finite-support tuple. This uses at most $3\cdot5\cdot2=30$ colors on the finite-support grid. If a chain has an active small-prime coordinate, its full projection onto that coordinate already prevents monochromaticity. Otherwise all active primes are at least 7. Its first box contains $p_1-1$ points in one prime fiber, more than either binary part, so parity splits that box.

Thus (1) coexists with unbounded box complexity and a bounded palette. No claim that all bounded-excess relations have this form, or have a bounded coloring, follows. A general bounded-conductor-excess argument, which would need a singleton-layer or recursive alternative, is not established and is not imported as a theorem; the elementary chain construction above is proved directly, without the Conway–Jones reduced-conductor theorem. This page is author-recorded.