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For u ∉ Q̄ and coprime p, q ∈ Q̄[X] of maximal degree 2: a five-dimensional Q̄ + Q̄u + Q̄u⁻¹ + Σ Q̄ c (p(u)/q(u))ᵏ is R9's involution-pencil space for (p, q)

Proved
DiazModulus.quadratic_ratio_progression_space_eq_involution_pencil_space

by carlok · Oct 12, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theorytranscendence

Let u∈C∖Q‾u \in \mathbb{C} \setminus \overline{\mathbb{Q}}u∈C∖Q​, let p,q∈Q‾[X]p, q \in \overline{\mathbb{Q}}[X]p,q∈Q​[X] be coprime, h=p(u)/q(u)h = p(u)/q(u)h=p(u)/q(u), let c∈Cc \in \mathbb{C}c∈C be non-zero, and let WWW be the Q‾\overline{\mathbb{Q}}Q​-span of 1,u,u−11, u, u^{-1}1,u,u−1 and c,ch,ch2,ch3c, ch, ch^2, ch^3c,ch,ch2,ch3. Suppose dim⁡W=5\dim W = 5dimW=5. If max⁡(deg⁡p,deg⁡q)=2\max(\deg p, \deg q) = 2max(degp,degq)=2, then WWW is the Q‾\overline{\mathbb{Q}}Q​-span of 1,u,u−11, u, u^{-1}1,u,u−1 and p(u)iq(u)3−i/(u l1(u)l2(u))p(u)^iq(u)^{3-i}/(u\,l_1(u)l_2(u))p(u)iq(u)3−i/(ul1​(u)l2​(u)), i=0,…,3i = 0, \dots, 3i=0,…,3, for some non-zero l1,l2∈Q‾p+Q‾ql_1, l_2 \in \overline{\mathbb{Q}}p + \overline{\mathbb{Q}}ql1​,l2​∈Q​p+Q​q.

ppp and qqq span the pencil of a Möbius involution of the uuu-line, and these are exactly the spaces of R9, which carry a configuration invisible to Theorem B.

Proof. hhh is transcendental, so c,ch,ch2,ch3c, ch, ch^2, ch^3c,ch,ch2,ch3 are free, and dim⁡W=5\dim W = 5dimW=5 gives two free elements c φi(h)c\,\varphi_i(h)cφi​(h) of Q‾+Q‾u+Q‾u−1\overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}u^{-1}Q​+Q​u+Q​u−1 with deg⁡φi≤3\deg\varphi_i \le 3degφi​≤3: cu φi(h)=Ri(u)cu\,\varphi_i(h) = R_i(u)cuφi​(h)=Ri​(u), deg⁡Ri≤2\deg R_i \le 2degRi​≤2. Write φi=gψi\varphi_i = g\psi_iφi​=gψi​ with ψ1,ψ2\psi_1, \psi_2ψ1​,ψ2​ coprime of degree at most m≥1m \geq 1m≥1, and Ψi=∑jψijpjqm−j\Psi_i = \sum_j \psi_{ij}p^jq^{m-j}Ψi​=∑j​ψij​pjqm−j. Then Ψ1R2=Ψ2R1\Psi_1R_2 = \Psi_2R_1Ψ1​R2​=Ψ2​R1​, and Ψ1,Ψ2\Psi_1, \Psi_2Ψ1​,Ψ2​ have no common zero, so Ri=ρΨiR_i = \rho\Psi_iRi​=ρΨi​. One Ψi\Psi_iΨi​ has degree 2m2m2m, so m=1m = 1m=1 and ρ=κ\rho = \kappaρ=κ is a non-zero constant: cu g(h)=κq(u)cu\,g(h) = \kappa q(u)cug(h)=κq(u) with deg⁡g≤2\deg g \le 2degg≤2. Splitting ggg over \overline{\mathbb{Q}} gives the pencil members l1,l2l_1, l_2l1​,l2​ with chk=κp(u)kq(u)3−k/(u l1(u)l2(u))ch^k = \kappa p(u)^kq(u)^{3-k}/(u\,l_1(u)l_2(u))chk=κp(u)kq(u)3−k/(ul1​(u)l2​(u)).

Novelty. Not found in the sources read; it is short.

Preamble
import Definitions.Def_DiazModulus
Formal statement
open Complex ComplexConjugate

namespace DiazModulus

theorem quadratic_ratio_progression_space_eq_involution_pencil_space
    (u : ℂ) (hu : u ∉ Qbar) (p q : Polynomial ↥Qbar) (hcop : IsCoprime p q)
    (hdeg : max p.natDegree q.natDegree = 2) (c : ℂ) (hc : c ≠ 0)
    (hdim : Module.finrank (↥Qbar) (Submodule.span Qbar (({1, u, u⁻¹} : Set ℂ) ∪
      Set.range (fun k : Fin 4 => c * (Polynomial.aeval u p / Polynomial.aeval u q) ^ (k : ℕ)))) = 5) :
    ∃ l₁ l₂ : Polynomial ↥Qbar,
      l₁ ∈ Submodule.span ↥Qbar ({p, q} : Set (Polynomial ↥Qbar)) ∧ l₁ ≠ 0 ∧
      l₂ ∈ Submodule.span ↥Qbar ({p, q} : Set (Polynomial ↥Qbar)) ∧ l₂ ≠ 0 ∧
      Submodule.span Qbar (({1, u, u⁻¹} : Set ℂ) ∪
      Set.range (fun k : Fin 4 => c * (Polynomial.aeval u p / Polynomial.aeval u q) ^ (k : ℕ))) =
        Submodule.span Qbar (({1, u, u⁻¹} : Set ℂ) ∪ Set.range (fun i : Fin 4 =>
          Polynomial.aeval u p ^ (i : ℕ) * Polynomial.aeval u q ^ (3 - (i : ℕ)) /
            (u * Polynomial.aeval u l₁ * Polynomial.aeval u l₂))) := by
  sorry

end DiazModulus
Source
Not found in the sources read. R13 of the Diaz modulus mission. Formal proof: Diaz modulus mission, 12 October 2026 (C. Perassi).

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