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If span(1, u, ū, g, ug) contains u², ū² or some 1/(u − a), then u·R(u)·g = N(u) with R ≠ 0, deg R ≤ 2, deg N ≤ 3

Proved
DiazModulus.degree_one_space_cubic_over_quadratic_of_mem_square_or_reciprocal

by carlok · Oct 18, 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​ with ρ=uuˉ\rho = u\bar uρ=uuˉ algebraic (so uˉ=ρ/u\bar u = \rho/uuˉ=ρ/u). Let g∈Cg \in \mathbb{C}g∈C and Wg=spanQ‾(1,u,uˉ,g,ug)W_g = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, g, ug)Wg​=spanQ​​(1,u,uˉ,g,ug). If WgW_gWg​ contains u2u^2u2, uˉ2\bar u^2uˉ2 or some 1/(u−a)1/(u - a)1/(u−a) with a∈Q‾×a \in \overline{\mathbb{Q}}^\timesa∈Q​×, then u R(u) g=N(u)u\,R(u)\,g = N(u)uR(u)g=N(u) for some R,N∈Q‾[X]R, N \in \overline{\mathbb{Q}}[X]R,N∈Q​[X] with R≠0R \neq 0R=0, deg⁡R≤2\deg R \leq 2degR≤2 and deg⁡N≤3\deg N \leq 3degN≤3.

Proof. Write the element as t=c0+c1u+c2uˉ+αg+βugt = c_0 + c_1u + c_2\bar u + \alpha g + \beta ugt=c0​+c1​u+c2​uˉ+αg+βug, replace uˉ\bar uuˉ by ρ/u\rho/uρ/u and clear denominators. For t=u2t = u^2t=u2 take R=α+βXR = \alpha + \beta XR=α+βX, N=X3−c1X2−c0X−c2ρN = X^3 - c_1X^2 - c_0X - c_2\rhoN=X3−c1​X2−c0​X−c2​ρ; for t=uˉ2t = \bar u^2t=uˉ2, R=X(α+βX)R = X(\alpha + \beta X)R=X(α+βX), N=ρ2−X(c0X+c1X2+c2ρ)N = \rho^2 - X(c_0X + c_1X^2 + c_2\rho)N=ρ2−X(c0​X+c1​X2+c2​ρ); for t=1/(u−a)t = 1/(u - a)t=1/(u−a), R=(X−a)(α+βX)R = (X - a)(\alpha + \beta X)R=(X−a)(α+βX), N=X−(X−a)(c0X+c1X2+c2ρ)N = X - (X - a)(c_0X + c_1X^2 + c_2\rho)N=X−(X−a)(c0​X+c1​X2+c2​ρ). In each case N≠0N \neq 0N=0 (monic, N(0)=ρ2N(0) = \rho^2N(0)=ρ2, N(a)=aN(a) = aN(a)=a), so R≠0R \neq 0R=0: otherwise N(u)=0N(u) = 0N(u)=0, while uuu is transcendental over Q‾\overline{\mathbb{Q}}Q​.

Novelty. Not found in the sources read; routine.

Preamble
import Definitions.Def_DiazModulus
Formal statement
open Complex ComplexConjugate

namespace DiazModulus

open Polynomial in
theorem degree_one_space_cubic_over_quadratic_of_mem_square_or_reciprocal
    (u g : ℂ) (hu : u ∉ Qbar) (hρ : IsAlgebraic ℚ (u * conj u))
    (ht : u ^ 2 ∈ Submodule.span Qbar ({1, u, conj u, g, u * g} : Set ℂ) ∨
        conj u ^ 2 ∈ Submodule.span Qbar ({1, u, conj u, g, u * g} : Set ℂ) ∨
        ∃ a ∈ Qbar, a ≠ 0 ∧ (u - a)⁻¹ ∈ Submodule.span Qbar ({1, u, conj u, g, u * g} : Set ℂ)) :
    ∃ R N : (↥Qbar)[X], R ≠ 0 ∧ R.natDegree ≤ 2 ∧ N.natDegree ≤ 3 ∧ u * aeval u R * g = aeval u N := by
  sorry

end DiazModulus
Source
Not found in the sources read. R17 of the Diaz modulus mission. Formal proof: Diaz modulus mission, 18 October 2026 (C. Perassi).

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