Mandatory root configurations with almost full height-two kernel rank

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Theorem and consequence for the live route

There are finite dissociated sets $C_t$ inside distinct odd-prime root grids with the following properties:

1. Their integer relation lattices have bases consisting of primitive height-two circuits, with connected support-intersection graph and supports covering every column. Their column images are saturated in the standard ambient integer tensor lattices.
2. Writing $k_t=\dim\ker_{\mathbb Q}C_t$,

$$\frac{k_t}{|C_t|}\longrightarrow1.$$

3. Every nonzero signed sum on $C_t$ has a mandatory coordinate: one column occurs in every representation with coefficients $0,\pm1$.
4. For a fresh odd prime $p$, a subset of $C_t\otimes S_p$ is dissociated if and only if it contains no full $p$-fiber. In particular $\chi_1(C_t\otimes S_p)=2$.

This disproves the proposed high-density short-relation descent criterion even for actual cyclotomic units and at coefficient height two. It strengthens the earlier density $1/20$ pivot construction and replaces the nonintegral diagnostic in the arithmetic-unit discussion.

It does not give a finite coloring of the full prime-root family, or exclude a mechanism coupling the actual classes of a coloring of that family. Saturation is included, so that additional hypothesis does not rescue the proposed rank-density descent. The main counterexample remains unproved.

1. A coefficient-two circuit of size linear in the largest prime

Let $p<q<r$ be distinct odd primes. Identify the labels of $S_p\otimes S_q$ with edges of $K_{p,q}$, oriented from row vertices $u_1,\ldots,u_p$ to column vertices $v_1,\ldots,v_q$. Its rational relation space is the space of graph gradients $t_u-t_v$: the row and column simplex relations span a space of dimension $p+q-1$, exactly the kernel dimension.

A label set is independent exactly when deleting its edges leaves a connected graph. Indeed a supported nonzero gradient exists precisely when a potential can take different constant values on the components of the remaining graph. Thus a bond (a minimal nonempty edge cut) is a circuit, with primitive cut coefficients $\pm1$.

Put $a=(u_1,v_1)$, $b=(u_2,v_2)$, and take the two bonds

$$O=\delta\{u_1,v_2\},\qquad S=\delta\bigl((\{u_1,\ldots,u_p\}\setminus\{u_3\}) \cup(\{v_1,\ldots,v_q\}\setminus\{v_1,v_2\})\bigr).$$

Both sides of each cut induce connected complete bipartite graphs (possibly stars). Their sizes are

$$|O|=p+q-2,\qquad |S|=2p+q-4.$$

Choose the cut potentials equal to one on the displayed sides. The coefficients at $a,b$ are $+1,-1$ for $O$, and $+1,+1$ for $S$.

Construct a subset of $(S_p\otimes S_q)\otimes S_r$ by layers:

• Layer zero: $O\setminus\{a\}$.
• Layer one: $S\setminus\{a,b\}$.
• Each of the remaining $r-2$ layers: $\{a,b\}$.

Every layer is independent. The cut relation on $O$ represents $a+b$ on its zero layer with coefficient $2$ at $b$ and coefficients $\pm1$ elsewhere. The cut relation on $S$ represents $a+b$ on layer one with all coefficients $\pm1$. The remaining layers use $a+b$ itself.

The common-sum intersection is exactly $\mathbb Q(a+b)$. To see this, the reference layers restrict it to $\operatorname{span}\{a,b\}$, whose intersection with $\operatorname{span}(S\setminus\{a,b\})$ is that line: any relation on $S$ is a multiple of its cut relation. The zero layer contains the line by its displayed representation. Since each layer is independent, coefficients are uniquely determined by the common sum. The simplex layer criterion now proves that the entire set is a circuit with one primitive coefficient of magnitude two and all other coefficients of magnitude one.

Its cardinality and rank are

$$n(p,q,r)=3p+2q+2r-13,\qquad d(p,q,r)=n(p,q,r)-1\le7r. \tag{1}$$

Realize the tensor columns faithfully by roots via the coprime cyclotomic field identification. Denote this circuit by $B\cup\{w\}$, with $|B|=d$, and reverse its signs if necessary to write

$$2w=\sum_{b\in B}\epsilon_b b,\qquad\epsilon_b\in\{\pm1\}. \tag{2}$$

Then $B$ is a rational basis. For any $b_0\in B$, the set

$$T=(B\setminus\{b_0\})\cup\{w\}$$

is an integer basis of $M=\mathbb Z(B\cup\{w\})$: equation (2) solves for $b_0$ integrally. The sublattice $L=\mathbb ZB$ has index two in $M$, since $w$ has exact order two modulo $L$.

The image $M$ is saturated in the standard ambient integer lattice. Indeed the graph-gradient, bond, and common-span arguments above work over every field, including characteristic two. The reference columns $a,b$ remain independent; the common intersection is still the nonzero line $\operatorname{span}(a+b)$. Thus the full circuit matrix has nullity one modulo every prime, with the displayed relation (whose coefficient two may reduce to zero). Removing a coefficient-$\pm1$ column leaves a rank-$d$ matrix over every such field. All its Smith invariant factors are therefore one, proving saturation of $\mathbb ZT=M$. The index-two sublattice $L$ itself is not saturated.

2. The staircase of column lattices

Choose such circuits $B_i\cup\{w_i\}$ on mutually disjoint prime triples, and choose corresponding bases $T_i$, lattices $L_i,M_i$, and ranks $d_i$. Fix $t$ and put $R=\prod_{i=1}^t d_i$.

Start with all $R$ columns in $B_1\cdots B_t$. For stage $j$, adjoin all columns

$$T_1\cdots T_{j-1}\,\{w_j\}\,B_{j+1}\cdots B_t. \tag{3}$$

Call the resulting set $C_t$. Products are injective because conductors are disjoint. The stages and the original base are disjoint: stage $j$ has rightmost distinguished coordinate $w_j$, whereas the original base has no distinguished coordinates. Thus

$$|C_t|=R+\sum_{j=1}^t\frac R{d_j},\qquad \operatorname{rank}_{\mathbb Q}C_t=R. \tag{4}$$

Before stage $j$, the generated lattice is

$$M_1\otimes\cdots\otimes M_{j-1}\otimes L_j\otimes\cdots\otimes L_t.$$

After that stage it is the same tensor with $L_j$ replaced by $M_j$. The quotient is

$$(\mathbb Z/2\mathbb Z)^{R/d_j}, \tag{5}$$

and the newly adjoined columns (3) are exactly a basis of this quotient. Consequently each new column has exact order two modulo all preceding columns, even after any earlier additions within its own stage.

Moreover, twice each new column is a signed sum of earlier actual columns: use (2) in its $j$-th factor. All required products $T_1\cdots T_{j-1}B_j\cdots B_t$ already occur. For a direct check, classify each such tuple by its rightmost $w_i$; if there is none it is in the initial base, and otherwise it appeared at stage $i<j$.

At the end, $\mathbb ZC_t=\bigotimes_iM_i$. Each $M_i$ is a direct summand of its ambient integer tensor lattice by the preceding saturation argument. Tensoring those direct summands shows that the image of $C_t$ is saturated too.

3. An integral height-two circuit basis and mandatory coordinates

Order the initial base first, then the new columns stage by stage. Equation (5) has this consequence:

In every nonzero integer relation, the coefficient at the greatest used column is even.

The greatest column cannot lie in the initial independent base. At a newly added column, its coefficient must annihilate its exact order-two coset modulo the preceding lattice, and therefore is even.

For every newly added column take its relation $2\,\mathrm{new}$ minus the signed old sum described above. These relations have height two and are an integer kernel basis. They are independent by their triangular last-column structure. Given any integer relation, its last coefficient is even, so subtract an integer multiple of the corresponding displayed relation to eliminate it; induction ends at the independent initial base.

Each basis relation is also a primitive rational circuit: its support is a fixed nonzero pure tensor times the factor circuit $B_j\cup\{w_j\}$. The supports cover every column. Their intersection graph is connected. After stage one, its components are indexed by $B_2\cdots B_t$. Inductively for $j\ge2$, after stage $j-1$, components are indexed by the untouched suffix in $B_j\cdots B_t$. A stage-$j$ circuit meets the components for every $b_j\in B_j$, joining them for each later suffix; its other circuits attach through old points. At the last stage only one component remains.

The even-last-coefficient property forbids any ternary relation, so $C_t$ is dissociated. If two signed representations of the same nonzero value had different greatest used columns, their difference would have last coefficient $\pm1$, impossible. Thus every representation has the same greatest used column, which is mandatory.

The pivot block of this integer kernel basis is triangular, not $2I$: earlier pivot entries can be odd. Hence the construction does not contradict the half-density ceiling for the older diagonal-pivot invariant, or any of the binary-code bounds.

4. Density tends to one

Use consecutive disjoint triples of odd primes. Their largest primes $r_i$ have divergent reciprocal sum. Indeed convergence of that every-third-prime subseries would imply convergence of the entire prime reciprocal series: the next block's three terms sum to at most $3/r_i$. The divergence of the prime reciprocal series also follows directly from Euler's finite products and divergence of the harmonic series.

By (1), $s_t=\sum_{i\le t}1/d_i\ge\frac17\sum_{i\le t}1/r_i\to\infty$. Equations (4) and the kernel basis give

$$k_t=Rs_t,\qquad \frac{k_t}{|C_t|}=\frac{s_t}{1+s_t}\longrightarrow1. \tag{6}$$

Thus the span of the height-two relations is the full kernel and occupies an arbitrarily large proportion of the column count, despite mandatory coordinates.

5. The next prime still has only fiber obstructions

A ternary relation on a subset of $C_t\otimes S_p$ has a common signed layer sum $u$. If $u=0$, every layer relation is trivial by base dissociation. If $u\ne0$, its mandatory base column occurs in every layer, giving a full $p$-fiber in the support. Conversely a full fiber is itself a ternary simplex relation.

Therefore precisely the fiber-proper subsets are dissociated. Splitting every fiber into two nonempty colors proves $\chi_1(C_t\otimes S_p)=2$. In particular, no height-two rank-density threshold strictly below one can guarantee a fiber-free relation in a one-prime extension, even for actual root configurations with a connected primitive circuit basis.

6. A staircase is finitely colorable at every fixed height

The selected staircase cannot supply infinite fixed-height coloring: for $t\ge1$ and every finite $h$,

$$\chi_h(C_t)\le\left\lceil\log_2(h+1)\right\rceil. \tag{7}$$

In particular $\chi_1(C_t)=1$ and $\chi_2(C_t)=\chi_3(C_t)=2$.

To prove this, change signs of the leaves in each rational basis so that $w_i=\frac12\sum_b e_{i,b}$. Individual column sign changes preserve every coefficient-height coloring number. Order product-basis rows lexicographically, with the distinguished deleted leaf $b_{0,i}$ largest in each factor. A column with $k$ pivot coordinates has leading row obtained by replacing those pivots by distinguished leaves, with leading coefficient $2^{-k}$.

For a fixed leading row $L$, let $Z$ be the positions where $L$ uses the distinguished leaf. The staircase columns with leading row $L$ have pivot sets exactly the initial segments of $Z$. Indeed before the last pivot, membership in the $T$-prefix forbids an un-replaced distinguished leaf; after it, the $B$-suffix forbids any pivot. Thus there is at most one column per pivot count $k$ at each leading row.

Choose $m=\lceil\log_2(h+1)\rceil$, so $2^m>h$, and color by pivot count modulo $m$. In a monochromatic height-$h$ relation, take the greatest participating leading row and, among its columns, the largest participating pivot count $K$. Its row equation, multiplied by $2^K$, makes the coefficient of that column divisible by $2^m$, because every other participating count differs by a positive multiple of $m$. That coefficient is nonzero and of magnitude at most $h<2^m$, a contradiction. This proves (7). A displayed height-two circuit supplies the lower bound two for $h=2,3$.

The union of staircases over all factor orderings is the full product of the factor circuits: put all pivot positions before all distinguished-leaf positions to include a given tuple. But the subfamily using only pivots and distinguished leaves is rationally independent, as a tensor of independent pairs. Incompatible orderings or same-leading-row collisions alone therefore do not force a relation. Any use of their union must also control the neutral leaves and their full expansions. No such actual-color argument was proved.

7. Parity cores and the limitation of separate lattice indices

Put $R_2=\mathbb Z_{(2)}$, the rationals with odd denominator. A finite configuration $C$ has an ordering in which every nonzero integer relation has even greatest coefficient if and only if every nonempty $S\subset C$ has an $x$ satisfying

$$x\notin\operatorname{span}_{R_2}(S\setminus\{x\}). \tag{8}$$

An odd coefficient at $x$ in an integer relation is equivalent to an $R_2$-expression of $x$ using the other columns, by clearing odd denominators. Thus an $R_2$-essential point can be removed and placed last, proving one direction inductively. In the other direction, the last point of any subset satisfies (8).

The repeated deletion of essential points has a unique terminal redundant core. Unions of subsets in which every point is generated by the other points over $R_2$ are again redundant. An essential point cannot belong to any such subset, so every deletion order leaves exactly their union. This is a criterion for the stronger parity-filtration property, not an equivalence with dissociation or with mandatory coordinates.

If every rank-dependent addition in a filtration has exact order two modulo the preceding lattice, its rank-increasing columns form a rational basis $B$ and

$$[\mathbb ZC:\mathbb ZB]=2^{|C|-\operatorname{rank}_{\mathbb Q}C}. \tag{9}$$

Every dependent addition multiplies the index by two; a rank-increasing addition extends both lattices by the same independent cyclic factor. For different color classes these indices need not combine into an index of the full-grid lattice.

One exact full-grid fact explains why. For $E=\bigotimes_{i=1}^kS_{p_i}$, deleting fewer than $2^k$ columns still leaves the whole standard integer lattice generated; the threshold is sharp. Over every field, the row code of a single simplex is the length-$p_i$ zero-sum code, of minimum support two. The tensor row code has minimum support $2^k$: in a nonzero tensor-code array a nonzero column forces at least the first code's minimum number of nonzero rows, each of at least the second code's minimum weight, and induction applies. Pure tensors of weight-two words attain the bound. Deleting fewer columns therefore preserves full rank over $\mathbb Q$ and over every prime field. Smith normal form implies integer generation. Deleting the support of a minimum-weight rational row word shows sharpness.

Thus a column essential within one class can be regenerated using other classes. No simultaneous-color contradiction was derived from (8), (9), or the deletion robustness.

The small circuits, quotient-basis argument, integer kernel elimination, mandatory property, saturation, and density limit are author-recorded; two separate reviewers checked them. The theorem is algebraic; no exploratory finite probe is an input. A separate reviewer also checked the pivot-depth coloring; the core and row-code claims have the direct proofs above and establish no new chromatic lower bound.