Collision-free private lifts and their color limitation

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Standing

Pairwise coprime conductors are sufficient but not necessary in determinant flattening. This page removes that formal restriction under an exact injectivity hypothesis. It does not remove the requirement of one actual support color. All arguments are algebraic and use no computation.

Private projections suffice

Let $M,n_1,\ldots,n_r$ be pairwise coprime, and let $X_i\subset\mu_{Mn_i}$. If projection of $X_i$ to its private $\mu_{n_i}$-coordinate is injective, multiplication on $X_1\times\cdots\times X_r$ is injective: project an equality of products onto $\mu_{n_i}$ to recover its $i$-th factor.

The existing determinant proof therefore works for factors sharing the entire conductor $M$, provided their private projections are injective and the surviving determinant support is monochromatic. The disjoint magnitude groups still make different matching blocks disjoint.

A universal exact lift

Given finite $X\subset\mu_M$, assign to each $x$ a distinct odd prime $p_x\nmid M$, and put

$$L_x=\{x\zeta_{p_x}^{\,j}:1\leq j<p_x\},\qquad L(X)=\bigcup_{x\in X}L_x.$$

Its integral relations are exactly

$$b_{x,j}=a_x\text{ for all }j, \qquad \sum_{x\in X}a_xx=0. \tag{1}$$

To prove this, fix $x$ and let $F$ be the compositum of the old conductor field and all private-prime fields other than the $p_x$-field. The $p_x-1$ nonidentity $p_x$-th roots are an $F$-basis, by the coprime cyclotomic degree formula. Write all other terms of a relation as one $u\in F$, and use $1=-\sum_{j=1}^{p_x-1}\zeta_{p_x}^j$. The coefficient equations $xb_{x,j}-u=0$ make all $b_{x,j}$ equal. Do this for each block; their sums are $-a_xx$, proving (1). The converse follows from the same root-sum identity.

Thus the entire integer relation lattice, primitivity, and coefficient heights are preserved by coefficient duplication. The private projection is injective: the order of a private component identifies its prime $p_x$, hence its block, and then its exponent identifies the point. Independently lifted factors can share all old conductor primes and still have injective products.

Color limitation. The lift is always the union of two rationally independent sets. Color one point of every $L_x$ red and its nonempty remainder blue. Equation (1), also for rational relations after clearing denominators, forces every nonzero relation to use all points of some block. Neither color does so. Exact preservation of the integer kernel therefore does not preserve chromatic forcing.

Moreover a proper dissociated coloring cannot preserve the color of $x$ on every member of $L_x$, since $x+\sum_{y\in L_x}y=0$ would be monochromatic.

Why prime powers and translations do not supply automatic copies

Put $R=\operatorname{rad}(N)$. The $N/R$ representatives of $\mu_R$-cosets in $\mu_N$ span $\mathbb Q(\zeta_N)$ over $\mathbb Q(\zeta_R)$. That extension has degree $\varphi(N)/\varphi(R)=N/R$, so the representatives form a basis. Every rational relation therefore separates by these cosets. For every coefficient height $h$,

$$\chi_h(\mu_N)=\chi_h(\mu_R).$$

Restriction proves one inequality; using the same proper palette in all the additively independent rotated cosets proves the other.

An irredundant relation lies in one such coset. If $r$ such supports in this fixed prime-power tower have injective products, normalizing each by its coset representative puts every product in $\mu_R$, so $\prod_i|X_i|\leq R$. Arbitrarily high prime powers do not give new multiplicatively independent copies of the old relation.

Scalar translations do not repair collisions either: $\prod_i(t_ix_i)=\prod_i(t_iy_i)$ if and only if $\prod_i x_i=\prod_i y_i$. For translates of one fixed grouped relation, the formal group-ring rows are scalar monomial multiples of the same row; their determinant is identically zero even before evaluation. There need not be a nonzero formal support left to color.

Fresh private primes force growing supports

Normalize an irredundant relation into its squarefree relative conductor. For an active prime $p$, grouping by its $p$-coordinate makes all $p$ layer sums equal over the other-prime field. If one layer is empty, all sums vanish; more than one occupied layer would split the relation. Thus an active prime has all $p$ layers occupied and the support has at least $p$ points.

In particular, an irredundant relation whose private projection is injective and whose private primes all exceed $B$ has more than $B$ points. Some private coordinate must vary, since a one-point relation is impossible. Fixed-size relation copies cannot be moved into arbitrarily fresh private primes. This explains another loss in whole-support profile focusing, but does not prove that all adaptive focusing must fail.