Arithmetic units, not equal norms, are needed for height descent

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Standing

The saturated prime staircase supplies actual root units with mandatory coordinates and full height-two kernel rank of density tending to one. Consequently integrality, even with saturation, does not rescue the high-density descent proposed here. The nonintegral example and binary restrictions below retain their stated scopes.

The prime-pivot example leaves open a high-density short-relation descent specific to roots of unity. This page proves that equal norms, even at every field embedding, cannot supply it. It then proves arithmetic restrictions that distinguish odd roots from the example. Those restrictions do not yet establish descent or unbounded coloring.

The subsequent binary-circuit lifting theorem sharpens the distance-five bound to the exact distance nineteen on distinct-prime grids, using singleton-layer descent and a graph-potential argument. The simpler residue proof remains useful for isolating the role of integrality.

1. An equal-modulus obstruction of density tending to one

Put

$$\alpha=(1+i\sqrt{15})/4,\qquad C_n=\{1,\alpha,\ldots,\alpha^{n-1}\},\quad n\geq3.$$

The primitive irreducible polynomial of $\alpha$ is $f(t)=2t^2-t+2$; its two complex roots have modulus one. The monic minimal polynomial $t^2-t/2+1$ is not integral, so $\alpha$ is not an algebraic integer and in particular not a root of unity. Its powers are distinct. Both field embeddings send every column to a point of modulus one; the normalized trace inner product also gives every column norm one.

The integer relation lattice of the indexed family $C_n$ has basis

$$z_j=2e_j-e_{j+1}+2e_{j+2},\qquad 0\leq j\leq n-3. \tag{1}$$

Indeed an integer polynomial of degree at most $n-1$ vanishes at $\alpha$ exactly when it is divisible by $f$. Primitivity and Gauss's lemma make the quotient integral. Thus (1) is an integral basis, and the rational rank is two. Each basis vector is a primitive height-two circuit: a dependence on two powers would make some nonzero power of $\alpha$ rational, necessarily $\pm1$ by modulus one, which would make $\alpha$ a root of unity. The basis supports cover every column and have connected intersection graph.

Every nonzero integer relation polynomial has even lowest nonzero coefficient, because $f(0)=2$. Consequently no nonzero ternary relation exists: $C_n$ is dissociated. Moreover, two ternary representations of a nonzero sum cannot have different least used indices. Their difference would be a relation with lowest coefficient $\pm1$. The shared least index is therefore present in every representation: it is a mandatory coordinate.

The layer argument now gives

$$D\subseteq C_n\otimes S_p\text{ dissociated} \quad\Longleftrightarrow\quad D\text{ contains no full }p\text{-fiber}. \tag{2}$$

For a ternary relation the layer sums agree. A common zero gives a trivial relation by base dissociation. A common nonzero sum forces its mandatory coordinate into every layer. Thus every fiber-proper two-coloring is proper. Nevertheless the span of the height-two relations has density $(n-2)/n\to1$. This example is not a root configuration.

2. Binary relation codes of odd roots

Let $C$ be a finite set of $n$ distinct odd-order roots, let

$$L=\{a\in\mathbb Z^C:\sum_{x\in C}a_xx=0\},\qquad k=\operatorname{rank}_{\mathbb Z}L,$$

and let $\overline L\subseteq\mathbb F_2^C$ be its reduction. The kernel $L$ is saturated: if $2a\in L$, then $a\in L$. Hence reduction induces an injection $L/2L\to\mathbb F_2^C$, and $\dim\overline L=k$. Explicitly, an integral basis cannot become dependent modulo two, since dividing such a dependence by two and reexpanding in the basis would force all its coefficients to have been even.

Work in the number field generated by $C$, and reduce its algebraic integers modulo a prime over two. All roots are units, and reduction is injective on the finite group of odd-order roots in that field. For a direct check of injectivity, a nontrivial odd-order root $z$ reducing to one would give $0=1+z+\cdots+z^{m-1}\equiv m\equiv1\pmod2$.

For the support $S$ of any word in $\overline L$, reduction of its integer relation and of the complex-conjugate relation gives

$$\sum_{x\in S}\bar x=0, \qquad \sum_{x\in S}\bar x^{-1}=0. \tag{3}$$

There are no words of weight one or two: the reduced roots are nonzero and distinct. For weight three, normalize the reduced roots to $1,a,b$. Equation (3) gives $a+b=1$ and $ab=1$. They are exactly the three cube roots of unity. Injectivity of odd-root reduction lifts their ratios, so the original three roots form a regular three-gon and sum to zero.

Weight four is impossible for any $C$. If the four reduced roots are $r_i$, their elementary symmetric functions satisfy $e_1=e_3=0$ by (3), since their product is nonzero. The polynomial with these four roots would be $t^4+e_2t^2+e_4$, whose derivative in characteristic two is zero. It cannot have four distinct roots.

If $C$ is dissociated, weight three is excluded as well. Thus $\overline L$ has minimum distance at least five. The radius-two Hamming balls about its $2^k$ codewords are disjoint, proving

$$2^k\left(1+n+\binom n2\right)\leq2^n. \tag{4}$$

In particular $\operatorname{rank}_{\mathbb Q}C\geq \log_2(1+n+\binom n2)$. For the even-$H$, $H I$-pivot invariant, the code is supported on the $n-k$ nonpivot coordinates; the sharper bound is

$$2^k\left(1+(n-k)+\binom{n-k}{2}\right)\leq2^{n-k}. \tag{5}$$

3. The remaining arithmetic obligation

The basis (1) reduces to individual middle-coordinate vectors of weight one, violating even the simplest root-unit restriction. Equivalently, $2x-y+2z=0$ among roots would make the unit $y$ divisible by two in the ring of algebraic integers. Thus the obstruction in Section 1 cannot be embedded relation-preservingly into roots.

But (4) allows nullity density tending to one, and (5) only sharpens the already known pivot half-density bound by a lower-order term. Neither proves a root-specific high-density descent, let alone its required actual-color matching. The reduced integral kernel must also not be confused with the entire modular kernel of a nonsaturated column image. Equality holds when the column image is saturated at two, meaning that its saturation quotient has no two-torsion; full saturation is stronger than needed. No such hypothesis is imposed by the binary-circuit lifting theorem.

These arguments were independently internally checked. No numerical search, theorem-prover run, or external review is claimed.