Fixed-height chromatic growth suffices for full dissociation

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Standing

The piecewise-syndetic reformulation is equivalent to the fixed-height premise below. That page also constructs an ordinary idempotent containing a dissociated root set, with infinite rational coloring number in every member. Thus rational dependence in all members, even with idempotence, does not replace one common finite height.

This is a positive, elementary focusing reduction for the exact gap in the main argument. It does not establish its hypothesis. Unlike a bare monochromatic-rectangle assertion, it explicitly pays for the whole-profile palette and for every excluded conductor prime.

Let

$$G=\bigoplus_{p\text{ odd prime}}\mu_p$$

be the multiplicative group of odd squarefree-order complex roots. Additive relations always refer to their values in $\mathbb C$. For $h\geq1$, write $\chi_h(X)$ for the least number of classes with no nonzero integer relation having every coefficient in $[-h,h]$. Thus $\chi_1$ is the dissociated coloring number. This is a bound on coefficient magnitude, not the fixed-order notion of $h$-freeness of Nešetřil–Rödl–Sales.

Theorem

If $\chi_h(G)=\infty$ for one fixed finite $h$, then $\chi_1(G)=\infty$. Consequently

$$\chi_1(G)<\infty \quad\Longleftrightarrow\quad \chi_h(G)<\infty\text{ for every finite }h. \tag{1}$$

The forward implication is special to this multiplicatively closed root family and the focusing argument below; it is not asserted for arbitrary integer sets. In fact, for each fixed finite $h$, finiteness of $\chi_1(G)$ and of $\chi_h(G)$ are equivalent.

Finite witnesses and conductor tails

For a finite set $S$ of odd primes, put $T_S=\bigoplus_{p\notin S,\ p\text{ odd}}\mu_p$. If $T_S$ had a finite $h$-independent coloring, using separate palettes on its finitely many multiplicative cosets would color $G$: multiplication by a nonzero root preserves every integer relation. Therefore

$$\chi_h(G)=\infty\quad\Longrightarrow\quad \chi_h(T_S)=\infty\quad\text{for every finite }S. \tag{2}$$

An infinite chromatic number has finite witnesses. Indeed, if every finite subset of a countable set had a proper $q$-coloring, take colorings of nested finite exhaustions and successively stabilize each vertex color along infinite subsequences. The resulting coloring avoids every forbidden relation, because its support is finite. Apply the contrapositive for each $q$.

Focusing induction with a fixed height

Assume $\chi_h(G)=\infty$. We prove the following statement simultaneously for every finite $S$, every palette size $q$, and every $r\geq1$:

Every $q$-coloring of $T_S$ contains finite height-$h$ dependent sets $X_1,\ldots,X_r$ in pairwise coprime conductor groups, disjoint from $S$, whose full product $X_1\cdots X_r$ is monochromatic.

For $r=1$, use (2). Suppose the statement holds for $r-1$, with all finite palettes. Choose a finite $F\subset T_S$ with $\chi_h(F)>q$, and let $J$ be its conductor-prime set. Given $f:T_S\to[q]$, color $T_{S\cup J}$ by the complete finite profiles

$$y\longmapsto (f(xy))_{x\in F}\in[q]^F.$$

There are at most $q^{|F|}$ profile colors. The induction hypothesis at that palette supplies $r-1$ conductor-disjoint height-$h$ dependent sets whose product $Y$ has one profile $g:F\to[q]$. Since $\chi_h(F)>q$, one color class of $g$ contains a nonzero height-$h$ relation. Take its support as $X_1$. Then $f(xy)$ is one fixed color for every $x\in X_1$, $y\in Y$. The conductors of $X_1$ and all the other factors are coprime by construction. This completes the induction without any bound on $|F|$ or the profile palette: the hypothesis permits every finite palette at the same height.

For $h\geq2$, take $r=h$ and apply the proved Hall/determinant flattening lemma. It gives a nonzero monochromatic ternary relation. This holds for every $q$, so $\chi_1(G)=\infty$. The case $h=1$ is immediate. Finally $\chi_h\geq\chi_1$, proving the other direction of (1).

Exact unresolved quantifier

The previously proved bounded-height result has the form

$$\forall q\ \exists h=h(q)\quad \chi_h(G)>q, \qquad h(q)=\lceil e^{3q}\rceil\text{ is sufficient}. \tag{3}$$

The focusing theorem instead needs

$$\exists h<\infty\ \forall q\quad\chi_h(G)>q. \tag{4}$$

There is no exchange of these quantifiers. Finite numbers $\chi_h(G)$ growing with $h$ satisfy (3) without satisfying (4). In particular, applying (3) at $q^{|F|}$ changes the height and invalidates the fixed-height induction.

A proof of (4), even for $h=2$, would be a genuine route to completing the required counterexample: the theorem gives unbounded ternary coloring on finite odd-prime grids, and the already proved uniform extraction, exact integer encoding, and separated-block assembly then apply. No such fixed height is currently known in this package.

Finite quantitative formulation

If finite conductor-disjoint factors $F_i$, $n_i=|F_i|$, satisfy

$$\chi_h(F_i)>q^{\,n_1\cdots n_{i-1}}, \tag{5}$$

with empty product one, every $q$-coloring of their product has a monochromatic product of height-$h$ dependent subsets. Start with the last factor, color it by all prefix profiles, and select a dependent profile class. The common profile is a coloring of the shorter prefix; repeat. For $h$ factors, determinant flattening applies.

Conditionally on $\chi_h(G)=\infty$, put $\kappa_h(t)=\min\{k:\chi_h(E_k)>t\}$, and let $N_k=|E_k|$. Define $k_0=0$, $N_0=1$, and

$$k_i=\kappa_h\bigl(N_{k_{i-1}}q^{N_{k_{i-1}}}\bigr).$$

These indices increase: $\chi_h(E_k)\leq N_k$ excludes earlier ones. The consecutive prime blocks $F_i$ satisfy (5), since coloring their $N_{k_{i-1}}$ prefix slices separately gives $\chi_h(E_{k_i})\leq N_{k_{i-1}}\chi_h(F_i)$. Thus $\chi_1(E_{k_h})>q$. This is a bound in terms of an unproved finite witness function, not a new numerical chromatic estimate.

A conditional idempotent formulation

This formulation records a second rigorous focusing mechanism; no existence or minimality theorem for ultrafilters is used. Let $\mathcal U$ be an ultrafilter on $G$, with $R_B=\{x:x^{-1}B\in\mathcal U\}$, and suppose it is idempotent, meaning $B\in\mathcal U$ if and only if $R_B\in\mathcal U$.

Proposition. If one finite $h$ works so that every member of $\mathcal U$ contains a height-$h$ dependence, every member contains a ternary dependence. Consequently, if $\mathcal U$ contains a dissociated set, it contains nested $h$-independent sets for all finite $h$.

For $A\in\mathcal U$, set $A^*=A\cap R_A$. It belongs to $\mathcal U$; for $x\in A^*$, both $x^{-1}A$ and $x^{-1}R_A=R_{x^{-1}A}$ belong to $\mathcal U$, so $x^{-1}A^*\in\mathcal U$. Every finite-prime tail $T_S$ also belongs to $\mathcal U$: the ultrafilter selects a coset $aT_S$ in the finite quotient, and $R_{aT_S}=T_S$; apply idempotence.

Construct $h$ factors inductively. If $P$ is the finite set of all nonempty products taking at most one point from the preceding factors, and $S$ contains their conductor primes, choose the next dependent support in

$$T_S\cap A^*\cap\bigcap_{s\in P}s^{-1}A^*\in\mathcal U.$$

The invariant $P\subset A^*$ persists. All conductors are disjoint, and the full product lies in $A^*$. Determinant flattening proves the proposition. Its contrapositive supplies an $h$-independent member for each $h$; intersect these successively with the dissociated member to make them nested.

For $h=1$, the proposition is immediate from its hypothesis; the determinant step is needed only for $h\geq2$.

Covering information does not supply uniform height

Call $FB$ thick if, for every finite $E\subset G$, some translate $tE$ is contained in $FB$. If this holds with finite $F$, $k=|F|$, and $B$ is $h$-independent, assigning each $x\in E$ an $f\in F$ with $f^{-1}tx\in B$ properly $h$-colors $E$ with $k$ colors. The existing small-column inequality, on all odd-prime prefixes and with $P\to\infty$, gives

$$\log(h+1)\leq2k.$$

The explicit earlier height $\lceil e^{3k}\rceil$ therefore forces a dependence in a set with such a $k$-element thick-cover witness.

But these cover sizes cannot be uniformly bounded inside an idempotent ultrafilter, even after fixing one member $A$. For finite $S$, $A\cap T_S\in\mathcal U$, and any finite $F$ with $F(A\cap T_S)$ thick has

$$|F|\geq[G:T_S]=\prod_{p\in S}p.$$

A thick set projects onto the entire finite quotient, while the displayed product projects into the image of $F$. This only disproves a uniform bound on these covering witnesses; it does not independently prove that minimal relation heights grow. The missing exchange is still from member-dependent heights to one height for every member.

The elementary finite focusing theorem and its finite quantitative form received an independent internal audit. All new proofs here are algebraic or finite-color compactness arguments, with no numerical experiment.