Research proof index

Conjugation negates a square-root-of--1 monomial: if X^{2j} = -1 then σ_{-1}(X^j) = X^{j·conjExp} = -X^j. The exponent j·conjExp ≡ -j, so X^{j·conjExp} is the inverse of X^j; and X^j·(-X^j) = -(X^j)² = -X^{2j} = 1, so the inverse is -X^j.

The key relation X^{2^α} = -1 (X^d = -1), proven via the quotient: X^d + 1 is the modulus, so it vanishes in the quotient.

X^c - 1 is a unit whenever some X^{c·2^t} = -1. Then (∑_{j<2^t} (X^c)^j)·(X^c - 1) = (X^c)^{2^t} - 1 = -2, a unit (isUnit_two), so X^c - 1 is a unit. This is the crux of the trace-of-monomial vanishing (Hachi [NOZ26, §3, Claim 2]): it makes the geometric sum collapse without a domain.

The split. Evaluating a multilinear polynomial equals the vector–matrix–vector product of its reshaped coefficient matrix with the monomial bases of the low and high evaluation points (Hachi [NOZ26, §4]).

The split (Lagrange representation). Evaluating a multilinear polynomial given by its hypercube values equals the vector–matrix–vector product of its reshaped value matrix with the Lagrange bases of the low and high evaluation points.

Coordinate isolation (Hachi Lemma 8, case (C), the c5 subtract-and-divide crux): if c ≡ⱼ c', the challenge-difference sum collapses to the j-th block, tensorG (c − c') x = (cⱼ − c'ⱼ) •ᵥ (G_k *ᵥ xⱼ).