Uniform coefficient height and conditional flattening

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Standing

The coefficient-growth theorem extends the algebraic flattening criterion beyond Hall-tight magnitude sets. Actual monochromatic factors satisfying its inequality have not been forced. Root circuit products show that height-free flattening fails even for actual roots.

This isolates a quantitative coefficient gap in the unrestricted coloring problem. For each fixed palette size, a monochromatic relation with coefficients bounded by one uniform integer is unavoidable. A separate lemma converts sufficiently many such dependent factors into a relation with coefficients in $\{-1,0,1\}$, provided their full Cartesian product is monochromatic. That hypothesis has not been forced for arbitrary colorings. Thus this page is not a proof of unbounded dissociated coloring.

All proofs are elementary and require no computation or external theorem.

Fixed-height focusing proves a valid color-adaptive use of the determinant: unbounded chromatic number at one fixed coefficient height would suffice. The available height bound depends on the palette and does not establish that hypothesis. Prescribed-support splitting and private conductor lifts delimit two shortcuts without ruling out adaptive focusing.

1. Radical counting gives many small columns

Fix a prime $\ell$, $P\geq2$, and

$$E=\bigotimes_{\substack{p\leq P\\p\ne\ell}}S_p, \qquad d=\prod_{\substack{p\leq P\\p\ne\ell}}(p-1), \qquad H=\sum_{\substack{n\leq P\\\ell\nmid n}}\frac1n.$$

The case $\ell=2$ is the main odd-prime prefix. Use the standard integer coordinates for each simplex. For a column, let $a$ be the product of the primes at which its last simplex symbol was chosen. Its $\ell_1$-norm is $\varphi(a)$, and exactly $d/\varphi(a)$ columns have this index $a$. Consequently

$$\#\{v\in E:\|v\|_1\leq P\} \geq d\sum_{\substack{a\leq P\text{ squarefree}\\\ell\nmid a}} \frac1{\varphi(a)}\geq dH. \tag{1}$$

For the second inequality, group positive integers by their radical and use

$$\sum_{\operatorname{rad}(n)=a}\frac1n =\prod_{p\mid a}\sum_{j\geq1}\frac1{p^j} =\frac1{\varphi(a)}.$$

This includes $a=1$, with its single integer $n=1$. Every integer $n\leq P$ has $\operatorname{rad}(n)\leq P$, so the grouped infinite sums dominate the required finite harmonic sum. Finally,

$$H=H_{\lfloor P\rfloor}-\ell^{-1}H_{\lfloor P/\ell\rfloor} \geq(1-1/\ell)\log P\geq\tfrac12\log P. \tag{2}$$

2. Every fixed palette forces bounded-height dependence

Say a set is $h$-degree independent when it has no nonzero integer relation with all coefficient magnitudes at most $h$. For an integer $h\geq1$, if $E$ has a partition into $q$ such sets, then

$$\log(h+1)\leq\frac qH\log\left(1+\frac{hPH}q\right) \leq\frac{2q}{\log P} \log\left(1+\frac{hP\log P}{2q}\right). \tag{3}$$

Indeed some color contains $s\geq dH/q$ of the small columns from (1). Its $(h+1)^s$ sums with coefficients in $\{0,\ldots,h\}$ are distinct. Coordinate widths total at most $hPs$, hence

$$(h+1)^s\leq(1+hPs/d)^d.$$

The function $t\mapsto\log(1+hPt)/t$ is decreasing for $t>0$, as follows from $\log(1+x)>x/(1+x)$. Insert $t=s/d\geq H/q\geq\log P/(2q)$ to get (3).

In particular, put

$$h_q=\lceil e^{3q}\rceil,\qquad P\geq e^{8(q+1)}. \tag{4}$$

Every $q$-coloring of $E$ has a nonzero monochromatic integer relation with coefficient magnitudes at most $h_q$.

To check the constants, set $u=\log P\geq8(q+1)$. Since $he^u u\geq2$, the right side of (3) is at most

$$2q+\frac{2q}{u}(\log h+\log u).$$

Here $\log h<3q+1$. The function $(3q+1+\log u)/u$ decreases on this range, and at $u_0=8(q+1)$ it is less than $1/2$, since $\log(8(q+1))<q+3$. The displayed upper bound is therefore less than $3q$, whereas $\log(h+1)>3q$, contradicting (3).

For the prime-selector formulation, take $q=\ell$ on the conductor coprime to $\ell$. A hypothetical dissociated selector already has, at the finite stage in (4), an integer relation whose primitive coefficient height lies between 2 and $\lceil e^{3\ell}\rceil$. Dividing a relation's coefficient gcd normalizes it, but does not make it a rational circuit or a ternary relation.

3. Coefficient-safe finite templates

Suppose a dissociated color class supports a nonzero relation $\sum_x a_xx=0$ of height at most $h$. Divide out its coefficient gcd, and group by coefficient magnitude:

$$w_m=\sum_{|a_x|=m}\operatorname{sign}(a_x)x, \qquad I=\{m:w_m\text{ has a nonempty defining support}\}.$$

Each $w_m$ is nonzero, the indexed family $(w_m)_{m\in I}$ is dissociated, and

$$\sum_{m\in I}m w_m=0. \tag{5}$$

A signed relation between these sums expands over disjoint original supports, keeping all coefficient magnitudes at most one; dissociation rules it out. Thus $2\leq|I|\leq h$. The finite possibilities for $I\subset[h]$ give a finite set of relation templates, but not a finite set of actual root supports, whole color profiles, or rational isomorphism types. Additional rational dependencies can occur among the $w_m$.

4. Hall/determinant flattening

Let $h\geq2$. Let $X_1,\ldots,X_h$ be finite sets of roots in pairwise coprime conductor groups, each supporting a nonzero integer relation of height at most $h$. If their entire multiplicative Cartesian product has one color, that color is not dissociated.

More generally, group one relation in factor $i$ by its nonzero coefficient magnitudes $I_i$, as in (5). It suffices that some nonempty subfamily $R$ satisfy

$$\left|\bigcup_{i\in R}I_i\right|\leq |R|. \tag{6}$$

The uniform-height assertion follows because $h$ subsets of $[h]$ necessarily have such a subfamily.

Proof

If any factor already has a ternary relation, hold all other coordinates fixed. Otherwise every magnitude group is a nonzero signed sum, and each $I_i$ has at least two elements. Choose $R$ minimal nonempty satisfying (6), and write $U=\bigcup_{i\in R}I_i$, $r=|R|$. Every nonempty proper $J\subset R$ has $|\bigcup_{i\in J}I_i|>|J|$. Deleting one index from $R$ therefore gives $|U|\geq r$; (6) gives equality. These inequalities are Hall's matching condition, so there is a bijection from $R$ to $U$ assigning every row $i$ one of its magnitudes in $I_i$.

For completeness, the needed finite matching implication is elementary. Take a maximum matching. If a row is unmatched, explore alternating paths from it. Every reached column is matched, or the matching could be enlarged. Let $J$ be the reached rows and $K$ all their neighboring columns. Matched columns correspond bijectively to the reached rows other than the initial one, so $|K|=|J|-1$, contrary to the stated inequalities. Hence the matching is perfect.

Form the square complex matrix

$$W=(w_{i,m})_{i\in R,m\in U},\qquad w_{i,m}=\sum_{|a_{i,x}|=m}\operatorname{sign}(a_{i,x})x,$$

putting zero in absent groups. Each row annihilates the same nonzero column vector $(m)_{m\in U}$, since $\sum_m m w_{i,m}=0$. Thus

$$0=\det W=\sum_{\pi:R\to U\ {\rm bijective}} \operatorname{sign}(\pi)\prod_{i\in R}w_{i,\pi(i)}. \tag{7}$$

Expand every signed block sum. Distinct permutations choose disjoint Cartesian product blocks, because in some factor their magnitude groups differ. Within each block the expansion coefficients are also $\pm1$. The matching guarantees at least one nonempty block. Multiplication is injective on root tuples from pairwise coprime conductor groups: their multiplicative subgroups have direct product. Therefore (7) is a genuinely nonzero formal ternary relation on distinct root products. Fix arbitrary roots in factors outside $R$ to keep its support within the original monochromatic product.

This proof makes no rational-circuit assumption about the grouped sums, and it does not require the maps from universal coefficient templates to be injective. Disjoint original supports and coprime conductors are what ensure that no nonzero formal coefficient is lost. The result and the uniform-height theorem are author-recorded; separate reviewers checked each.

The proof in fact needs only the union of the nonempty permutation-product blocks in (7) to have one color (after fixing any other factors). A full monochromatic product is a stronger convenient sufficient condition. Forcing this determinant support is therefore a narrower possible research target.

This remains a conditional use: the algebra supplies a nonzero formal relation on the specified support, not its monochromaticity. The minimal subfamily is Hall-tight (equal union and factor sizes), and Hall's matching theorem is satisfied, not contradicted.

5. Exact outstanding focusing step

The uniform height in (4) does not give the monochromatic rectangle required in Section 4. Coloring a multiplier $y$ by its entire pattern $(f(xy))_{x\in X}$ can use $q^{|X|}$ colors. Reapplying (4) to that palette makes the height bound depend on $q^{|X|}$, not just on the original $q$. Repeating the argument has not provided $h$ factors with one common controlled height $h$, or more generally the matching obstruction (6).

There is a second quantitative loss when conductors must stay coprime. If a finite set $S$ of already used primes is excluded, the same radical count uses $H_S(P)=\sum_{n\leq P,\ (n,\prod_{p\in S}p)=1}1/n$, with $H_S(P)\geq\alpha_S\log P$, $\alpha_S=\prod_{p\in S}(1-1/p)$. To prove this bound inductively, use $H_{S\cup\{p\}}(P)=H_S(P)-p^{-1}H_S(P/p)\geq(1-1/p)H_S(P)$. The necessary inequality (3) then has $1/2$ replaced by $\alpha_S$. The available eventual height condition is $\log(h+1)>q/\alpha_S$, so the single-excluded-prime bound cannot be reapplied unchanged to arbitrarily many disjoint conductors.

Nor can an arbitrary dependence simply be flattened without height control: the abstract scalar factors $\{1,2^{2^{i-1}}\}$ are each rationally dependent, but their indexed tensor product is $\{2^a:0\leq a<2^k\}$, which is dissociated. This is an obstruction to an abstract height-free lemma, not a coloring of the root family.

Likewise, disjoint primitive-root expansions of a point cannot automatically preserve its color: if a point of a dissociated class were a signed sum of other points of that class, it would already give a forbidden relation. The next proof obligation is a quantitatively controlled focusing or height-descent mechanism, not an assumption that rational dependence suffices. It must address actual colors on the surviving determinant blocks, as well as their magnitude-set condition. No such forcing statement is established.

A single short-relation direction cannot supply one-prime descent

Let $C$ be $h$-degree independent. Suppose all its integer relation vectors of height at most $2h$ lie in one rational line (or there are none). Then every height-at-most-$h$ relation in $C\otimes S_p$ has a full $p$-fiber in its support.

Indeed let $v_j\in[-h,h]^C\cap\mathbb Z^C$ be the layer coefficient vectors. Their vector sums in the base are all equal, so every difference $v_j-v_0$ is a height-at-most-$2h$ base relation. If they all vanish, any nonzero coordinate of $v_0$ gives a full fiber. Otherwise let $a$ be the primitive integer generator of the relation line and write $v_j=v_0+t_j a$, with $t_j\in\mathbb Z$. Its height $H$ is strictly greater than $h$. The layer bounds give $(\max t_j-\min t_j)H\leq2h$, so the integer parameters have at most two consecutive values. At a coordinate with $|a_b|=H$, the two layer coefficients differ by $H>h$. Neither can be zero, since both lie in $[-h,h]$. This is the required full fiber.

Consequently a coloring making each $p$-fiber nonmonochromatic prevents all such descent. To avoid this elementary polygon obstruction, a one-prime step confined to the old support needs at least two independent short relation directions, not merely one monochromatic rational circuit.