Exact varying-height escape inside actual odd roots

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Statement and purpose

There are finite sets $X_i=\{1\}\cup Y_i$ of odd squarefree-order roots, in pairwise coprime conductor groups, such that:

• Each $X_i$ is a rational circuit with a unique primitive relation whose magnitudes are $\{1,H_i\}$, where $H_i=2^{2^{i-1}}$.
• Every finite product $X_1\cdots X_n$ is dissociated.
• The nested tail-product sets

$$D_j=\bigcup_{n\ge j}X_jX_{j+1}\cdots X_n$$

have least nonzero integer-relation height exactly $H_j$.

• For each $x\in D_j$, sufficiently late $k$ satisfies $xD_k\subset D_j$.

Thus height-free flattening fails even for actual root units with injective products and fresh conductors, not just for unrelated scalar configurations. The construction realizes the precise escape in the new coefficient-growth bound. It is not a coloring of the whole root group, and does not obstruct color-adaptive selection under stronger quantitative hypotheses.

1. Moving the coefficient-two seed to arbitrary prime triples

The eight-point seed in $S_3^{\otimes3}$ has primitive kernel generator

$$(-1,1,-1,1,-1,1,-1,-2)$$

on labels $000,012,021,102,110,211,222,001$. Its only coefficient of magnitude two is at $001$.

For any distinct odd primes $p,q,r$, enlarge the three simplex factors by duplicating the following layers:

1. First-coordinate layer $2$ into $p-2$ copies.
2. Second-coordinate layer $2$ into $q-2$ copies.
3. Third-coordinate layer $0$ into $r-2$ copies.

The seed's layer-duplication lemma preserves the entire one-dimensional integer kernel: all layer supports are independent, and their coefficients are uniquely determined by their common vector sum. None of these layers contains the distinguished point $001$. Consequently the enlarged set is a circuit in $S_p\otimes S_q\otimes S_r$, with one coefficient $-2$ and all other coefficients $\pm1$.

The usual coprime cyclotomic tensor identification realizes these columns faithfully by roots. Divide all roots by the distinguished root. The resulting circuit has the form $\{1\}\cup L$, with $L$ rationally independent and

$$2=\sum_{u\in L}\epsilon_u u,\qquad \epsilon_u\in\{\pm1\}. \tag{1}$$

All roots are distinct and $1\notin L$.

This construction works on arbitrarily fresh disjoint prime triples. It uses the already proved algebraic seed, not a new search or certificate.

2. Arbitrary powers of two at a unique pivot

Take $s$ versions of (1) on disjoint conductor triples, and let $Y$ be the full product of their leaf sets $L$. Multiplication identifies the tensor of their cyclotomic fields with the compositum. Therefore $Y$ is rationally independent, and multiplication of (1) gives

$$2^s=\sum_{y\in Y}\epsilon_y y,\qquad \epsilon_y\in\{\pm1\}. \tag{2}$$

The product map is injective. In particular $1\notin Y$: equality to one would force every factor to equal one, although none belongs to its leaf set. Thus $\{1\}\cup Y$ is a circuit. Its displayed relation is primitive and unique up to integer multiples, because the leaves form a basis of their span and their coefficients in (2) are all nonzero.

For factor $i$, take $s_i=2^{i-1}$ such seeds. Reserve distinct prime triples for all seeds in all factors. This gives

$$X_i=\{1\}\cup Y_i,\qquad H_i=2^{s_i}=\sum_{y\in Y_i}\epsilon_i(y)y. \tag{3}$$

3. The full products really are independent at the claimed height

Fix a finite index set $I\subset\{j,j+1,\ldots\}$, and a relation with coefficients $a_x\in[-H_j+1,H_j-1]\cap\mathbb Z$ on $\prod_{i\in I}X_i$. The tensor leaf products are a rational basis of the span. For a leaf tuple $y=(y_i)_{i\in I}$ and $S\subset I$, let $x^S(y)$ be the root tuple using $1$ at positions in $S$ and $y_i$ at the other positions. Expand $1=H_i^{-1}\sum\epsilon_i(y)y$ and multiply the coefficient of the basis tuple $y$ by $\prod_iH_i$. It is

$$0=\sum_{S\subset I}a_{x^S(y)} \left(\prod_{i\in S}\epsilon_i(y_i)\right) \prod_{i\in I\setminus S}H_i. \tag{4}$$

Put $B=H_j$. The last products in (4) are distinct nonnegative integer powers of $B$, since their exponents are subset sums of the distinct powers $2^{i-j}$. An integer combination of distinct powers of $B$, with coefficient magnitudes at most $B-1$, cannot vanish unless every coefficient vanishes: divide by its least participating power and reduce modulo $B$.

Hence $a_{x^S(y)}=0$ for every $S,y$. Every product column occurs in this form (fill the unused leaf positions arbitrarily), so the original relation is trivial. Each such product is $(H_j-1)$-independent.

Because every $X_i$ contains $1$, the finite products defining $D_j$ are nested. Every relation has finite support, so $D_j$ is $(H_j-1)$-independent. The relation on $X_j\subset D_j$ has height $H_j$, proving the exact minimum. In particular $D_1$ is dissociated.

For $x\in D_j$, take $k$ beyond the largest factor used in $x$. Then $xD_k\subset D_j$ simply uses disjoint sets of factors. Also the $D_j$ eventually avoid every prescribed finite conductor-prime set.

4. Rescaling does not manufacture Hall tightness

Every nonzero relation on $X_i$ is a multiple of (3). Its magnitude pair is therefore $\{t_i,t_iH_i\}$ for some positive integer $t_i$. View these pairs as edges joining positive real labels. A cycle among distinct edges would imply

$$\prod_iH_i^{\sigma_i}=1, \qquad \sigma_i\in\{-1,0,1\},\quad \sigma\ne0,$$

and thus $\sum_i\sigma_i2^{i-1}=0$, impossible. This also excludes parallel edges. The graph is a forest, so every nonempty subfamily has strictly more incident vertices than edges. No rescaling produces the Hall-tight condition needed for determinant flattening.

This is compatible with the stronger subset-counting theorem: the associated scalar model attains equality at its exact threshold, as computed on that page.

5. Scope

The supports above have product recurrence and a rational dependence in every tail, but they do not cover the whole root group. Indeed $D_1$ is not piecewise syndetic. Given finite multipliers $F$, take a factor whose conductor is disjoint from all conductors in $F$. Its projection of $FD_1$ is the proper subset $X_i$ of its full conductor group. It cannot contain a translate of that finite subgroup, so $FD_1$ is not thick. The properness follows, for example, from dissociation of $X_i$ and a full prime polygon in the conductor group.

The Hare-based recurrent construction strengthens this diagnostic by making the rational coloring number infinite in every tail. That additional property still does not imply piecewise syndeticity or the actual-color growth inequality.

Fresh-prime duplication, product injectivity, the basis expansion, and tail independence received independent internal checks. No computation was run.