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For a ratio h of degree 2, pole fractions q(αq + βJ)/(u(p − λq)) at three distinct λ are linearly independent modulo span(1, u, u⁻¹)

Proved
DiazModulus.quadratic_ratio_pole_fractions_independent

by carlok · Oct 23, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theorytranscendence

Let u∈Cu \in \mathbb{C}u∈C be transcendental, that is, u∉Q‾u \notin \overline{\mathbb{Q}}u∈/Q​ (the algebraic numbers); spans and linear independence are over Q‾\overline{\mathbb{Q}}Q​. Let p,q∈Q‾[X]p, q \in \overline{\mathbb{Q}}[X]p,q∈Q​[X] be coprime with max⁡(deg⁡p,deg⁡q)=2\max(\deg p, \deg q) = 2max(degp,degq)=2, and J=p′q−pq′J = p'q - pq'J=p′q−pq′. Let λ0,λ1,λ2∈Q‾\lambda_0, \lambda_1, \lambda_2 \in \overline{\mathbb{Q}}λ0​,λ1​,λ2​∈Q​ be distinct, let αi,βi∈Q‾\alpha_i, \beta_i \in \overline{\mathbb{Q}}αi​,βi​∈Q​ with (αi,βi)≠(0,0)(\alpha_i, \beta_i) \neq (0, 0)(αi​,βi​)=(0,0) for each iii, and put gi=q(u)(αiq(u)+βiJ(u))/(u(p(u)−λiq(u)))g_i = q(u)(\alpha_iq(u) + \beta_iJ(u))/(u(p(u) - \lambda_iq(u)))gi​=q(u)(αi​q(u)+βi​J(u))/(u(p(u)−λi​q(u))). If a,b,c,d0,d1,d2∈Q‾a, b, c, d_0, d_1, d_2 \in \overline{\mathbb{Q}}a,b,c,d0​,d1​,d2​∈Q​ satisfy a+bu+cu−1+d0g0+d1g1+d2g2=0a + bu + cu^{-1} + d_0g_0 + d_1g_1 + d_2g_2 = 0a+bu+cu−1+d0​g0​+d1​g1​+d2​g2​=0, then a=b=c=0a = b = c = 0a=b=c=0 and d0=d1=d2=0d_0 = d_1 = d_2 = 0d0​=d1​=d2​=0.

So 1,u,u−1,g0,g1,g21, u, u^{-1}, g_0, g_1, g_21,u,u−1,g0​,g1​,g2​ are linearly independent: pole fractions at three distinct values of λ\lambdaλ are independent modulo H0=span(1,u,u−1)H_0 = \mathrm{span}(1, u, u^{-1})H0​=span(1,u,u−1).

Proof. The denominators are non-zero: u≠0u \neq 0u=0, and p−λq≠0p - \lambda q \neq 0p−λq=0 by coprimality and degree, so its value at the transcendental uuu is not 000. Write h=p(u)/q(u)h = p(u)/q(u)h=p(u)/q(u) and w=J(u)/q(u)w = J(u)/q(u)w=J(u)/q(u), so that gi=(q(u)/u)(αi+βiw)/(h−λi)g_i = (q(u)/u)(\alpha_i + \beta_iw)/(h - \lambda_i)gi​=(q(u)/u)(αi​+βi​w)/(h−λi​). The H0H_0H0​-part a+bu+cu−1a + bu + cu^{-1}a+bu+cu−1 is R(u)/uR(u)/uR(u)/u with R=c+aX+bX2R = c + aX + bX^2R=c+aX+bX2, and DiazModulus.quadratic_pencil_wronskian_span writes R=a′q+b′p+c′JR = a'q + b'p + c'JR=a′q+b′p+c′J. Dividing the relation by q(u)/uq(u)/uq(u)/u and multiplying by ∏j(h−λj)\prod_j(h - \lambda_j)∏j​(h−λj​) gives A(h)+B(h)w=0A(h) + B(h)w = 0A(h)+B(h)w=0 with A=(a′+b′X)Π+∑idiαiΠiA = (a' + b'X)\Pi + \sum_id_i\alpha_i\Pi_iA=(a′+b′X)Π+∑i​di​αi​Πi​ and B=c′Π+∑idiβiΠiB = c'\Pi + \sum_id_i\beta_i\Pi_iB=c′Π+∑i​di​βi​Πi​, where Π=∏j(X−λj)\Pi = \prod_j(X - \lambda_j)Π=∏j​(X−λj​) and Πi=∏j≠i(X−λj)\Pi_i = \prod_{j \neq i}(X - \lambda_j)Πi​=∏j=i​(X−λj​). Then DiazModulus.quadratic_ratio_wronskian_free gives A=B=0A = B = 0A=B=0. Evaluating at X=λiX = \lambda_iX=λi​, where Πi(λi)≠0\Pi_i(\lambda_i) \neq 0Πi​(λi​)=0 as the λj\lambda_jλj​ are distinct and every other term vanishes, leaves diαi=diβi=0d_i\alpha_i = d_i\beta_i = 0di​αi​=di​βi​=0, so di=0d_i = 0di​=0. Then a+bu+cu−1=0a + bu + cu^{-1} = 0a+bu+cu−1=0, and a=b=c=0a = b = c = 0a=b=c=0 because c+aX+bX2c + aX + bX^2c+aX+bX2 vanishes at uuu.

Novelty. Not found in the sources read; routine.

Preamble
import Definitions.Def_DiazModulus
Formal statement
open Complex ComplexConjugate

namespace DiazModulus

open Polynomial in
theorem quadratic_ratio_pole_fractions_independent (u : ℂ) (hu : u ∉ Qbar) (p q : (↥Qbar)[X])
    (hcop : IsCoprime p q) (hdeg : max p.natDegree q.natDegree = 2)
    (l α β : Fin 3 → ↥Qbar) (hl : Function.Injective l) (hαβ : ∀ i, α i ≠ 0 ∨ β i ≠ 0)
    (a b c : ↥Qbar) (d : Fin 3 → ↥Qbar)
    (hrel : (a : ℂ) + (b : ℂ) * u + (c : ℂ) * u⁻¹ + ∑ i, (d i : ℂ) *
        (aeval u q * ((α i : ℂ) * aeval u q + (β i : ℂ) * aeval u (derivative p * q - p * derivative q)) /
          (u * (aeval u p - (l i : ℂ) * aeval u q))) = 0) :
    a = 0 ∧ b = 0 ∧ c = 0 ∧ d = 0 := by
  sorry

end DiazModulus
Source
Not found in the sources read. Step 1 of the sheaf-free proof of Theorem NF. Uses DiazModulus.quadratic_pencil_wronskian_span and DiazModulus.quadratic_ratio_wronskian_free. R20 of the Diaz modulus mission. Formal proof: Diaz modulus mission, 23 October 2026 (C. Perassi).

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