Short-relation rank and a genuine height-descent template

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Standing

These elementary lemmas refine the remaining bounded-height focusing problem. They distinguish algebraic descent from the missing color matching. The six-column template below is not asserted to embed in the prime grids or occur monochromatically in an arbitrary coloring.

The actual prime-grid obstruction shows that arbitrarily large connected full short-relation rank, with saturated image, can still leave only full fibers as minimal ternary supports in the next simplex extension. Its explicit nullity density is $1/20$; the common-pivot mechanism has an absolute half-density ceiling. A quantitative high-density descent hypothesis was left open there. The saturated staircase excludes that repair as well: its full integral height-two circuit rank has density tending to one, with mandatory coordinates still preventing fiber-free descent. The exact algebraic template below remains valid, but requires the stated arrangement and actual-color matching.

1. Counting the span of short relations

Let $C\subset\mathbb Z^d$ have $s$ columns, each of norm at most $P$. Let $W_h(C)$ be the rational span of its integer relation vectors with coefficient magnitudes at most $h$, and let $r=\dim W_h(C)$. Then

$$r\geq s-\frac{d\log(1+hPs/d)}{\log(h+1)}. \tag{1}$$

Choose $r$ coordinate labels whose coordinate projection is injective on $W_h(C)$, and delete those labels. The remaining set $C'$ is $h$-degree independent: any short relation supported there belongs to $W_h(C)$ and has zero chosen coordinates. Its $(h+1)^{s-r}$ coefficient sums in $\{0,\ldots,h\}$ are distinct. Coordinate-box counting gives $(h+1)^{s-r}\leq(1+hP(s-r)/d)^d\leq(1+hPs/d)^d$, proving (1). This concerns the span of short integer relations, not all rational dependencies.

2. Large connected short-relation rank does not itself descend

For $H=h+1$ and $n\geq2$, the scalar set $C_n=\{1,H,\ldots,H^{n-1}\}$ is $h$-degree independent, by successive reduction modulo $H$. Its height-$H$ relations $He_i-e_{i+1}$ form a connected system and span its full $(n-1)$-dimensional rational kernel.

Nevertheless, any nonzero $u\in\Sigma_h(C_n)$ has a mandatory coordinate in every such bounded representation. If $H^j$ is the largest power of $H$ dividing $u$, reduction modulo $H$ forces all coefficients below $j$ to vanish and the coefficient at $j$ to be nonzero. The argument uses divisibility by powers of $H$, whether or not $H$ is prime.

In a height-at-most-$h$ relation on $C_n\otimes S_p$, all layer sums equal one $u$. If $u=0$, base independence makes the relation trivial; otherwise every layer uses that mandatory coordinate. Thus every such relation contains a full $p$-fiber. This rules out a height-descent argument from connected short-relation rank alone. These scalar examples are not prime-grid subconfigurations.

3. Two arranged directions do descend one coefficient-height step

Fix $h\geq1$, $H=h+1$. In $\mathbb Z^4$, take

$$\begin{aligned} c_1=e_1,\quad c_2=e_2,\quad c_3=e_3,\quad c_6=e_4,\\ c_4=-c_1-Hc_2-hc_3+c_6,\qquad c_5=-Hc_1-c_2+hc_3+c_6. \end{aligned}$$

Their integral kernel is generated by

$$d_1=(H,1,-h,0,1,-1),\qquad d_2=(1,H,h,1,0,-1).$$

Indeed $c_1,c_2,c_3,c_6$ are a basis; the coefficients on $c_5,c_4$ read off the integer coefficients $\lambda,\mu$ of any kernel vector $\lambda d_1+\mu d_2$. This proves saturation as well. There is no nonzero kernel vector of height at most $h$. If just one parameter is nonzero, a first or second coordinate has magnitude at least $H$; the same is true if both have the same sign. If they have opposite signs, the third coordinate has magnitude $h(|\lambda|+|\mu|)\geq2h\geq H$.

On the other hand, the three coefficient vectors

$$\begin{aligned} v_1&=(1,1,0,1,1,0),\\ v_2&=(-h,0,h,1,0,1),\\ v_3&=(0,-h,-h,0,1,1) \end{aligned}$$

have height at most $h$, have empty total support intersection, and all represent $u=-Hc_1-Hc_2+2c_6\ne0$. Use each type at least once as the layers of $C_h\otimes S_p$, repeating types for $p>3$. The layer sums all equal $u$, so they form a nonzero height-at-most-$h$ relation, but no full $p$-fiber lies in its support. At $h=1$ this is an actual ternary relation over a dissociated base.

This is a positive algebraic template for $H\to H-1$ descent. Its required arrangement is stronger than having two independent short relations. The one-line obstruction explains why a single short-relation direction cannot supply such a fiber-free step.

4. Color matching cannot be omitted

For any $h$-degree-independent base $C$, color one whole layer of $C\otimes S_p$ blue and all other layers red. Each color omits a simplex direction, and each constituent base layer is $h$-degree independent, so both colors are $h$-degree independent. A selected old layer can retain its original color.

Thus an arbitrary-color descent cannot be forced by working solely inside one previously monochromatic independent support and leaving its new layer colors unrestricted. It must couple actual old color classes, force the specific matching of representation supports, or supply a monochromatic determinant support as in the main bounded-height route. None of those forcing statements is proved here.