Research proof index

Coefficient of an X^e-shift of a reduced element. For e < d, p < d and any x : R_q, the p-th coefficient of x·X^e is the (p-e)-th coefficient of x (no wrap, e ≤ p), or minus the (p+d-e)-th coefficient (one wrap past X^d = -1, p < e). This is the only place the proof needs the ring's multiplication; everything downstream is coef...

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.