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A conjugation hull in the polynomial, involution or real pencil class is five-dimensional

Proved
DiazModulus.laurent_conj_hull_class_finrank_eq_five

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​ and s∈Q‾s \in \overline{\mathbb{Q}}s∈Q​ with s2=uuˉs^2 = u\bar us2=uuˉ (so s=±∣u∣s = \pm|u|s=±∣u∣ is real), and put v=u/sv = u/sv=u/s, so that vˉ=1/v\bar v = 1/vvˉ=1/v. Let k≥0k \geq 0k≥0, let cj∈Cc_j \in \mathbb{C}cj​∈C for all j∈Zj \in \mathbb{Z}j∈Z (they need not be algebraic), and put z=∑j=−kkcjvjz = \sum_{j=-k}^{k} c_jv^jz=∑j=−kk​cj​vj and W=spanQ‾(1,u,uˉ,z,zˉ)W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, z, \bar z)W=spanQ​​(1,u,uˉ,z,zˉ). If WWW is in one of the classes

  • (P) αu2+βu3∈W\alpha u^2 + \beta u^3 \in Wαu2+βu3∈W for some α,β∈Q‾\alpha, \beta \in \overline{\mathbb{Q}}α,β∈Q​, not both zero;
  • (Q) W=spanQ‾(1,u,uˉ,vˉ2+ev2,βvˉ3+ζv2+ev3)W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, \bar v^2 + ev^2, \beta\bar v^3 + \zeta v^2 + ev^3)W=spanQ​​(1,u,uˉ,vˉ2+ev2,βvˉ3+ζv2+ev3) for some e,β,ζ∈Q‾e, \beta, \zeta \in \overline{\mathbb{Q}}e,β,ζ∈Q​ with eeˉ=1e\bar e = 1eeˉ=1, ββˉ=1\beta\bar\beta = 1ββˉ​=1 and ζ=−βe2ζˉ\zeta = -\beta e^2\bar\zetaζ=−βe2ζˉ​;
  • (R) W=spanQ‾(1,u,uˉ,g,gˉ)W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, g, \bar g)W=spanQ​​(1,u,uˉ,g,gˉ​) with g=G(v)+κ vG(v)‾g = G(v) + \kappa\,\overline{vG(v)}g=G(v)+κvG(v)​, for some κ∈Q‾\kappa \in \overline{\mathbb{Q}}κ∈Q​ with κκˉ=1\kappa\bar\kappa = 1κκˉ=1 and some G∈Q‾[X]G \in \overline{\mathbb{Q}}[X]G∈Q​[X] with G(0)=0G(0) = 0G(0)=0 and deg⁡G≥2\deg G \geq 2degG≥2,

then dim⁡Q‾W=5\dim_{\overline{\mathbb{Q}}} W = 5dimQ​​W=5.

Proof. Work in v=u/sv = u/sv=u/s, which is transcendental with vˉ=1/v\bar v = 1/vvˉ=1/v since sss is real. In class (P), with m=αu2+βu3∈Wm = \alpha u^2 + \beta u^3 \in Wm=αu2+βu3∈W, also mˉ∈W\bar m \in Wmˉ∈W, as the hull is conjugation-stable; 1,u,uˉ,m,mˉ1, u, \bar u, m, \bar m1,u,uˉ,m,mˉ are free, because a relation cleared by v3v^3v3 is a polynomial of degree 666 in vvv whose seven coefficients vanish, and WWW has five generators. In class (Q) the normal-form generators are free in the same way, using β≠0\beta \neq 0β=0 from ββˉ=1\beta\bar\beta = 1ββˉ​=1. In class (R), with n=deg⁡G≥2n = \deg G \geq 2n=degG≥2 and H=∑GˉiXn−iH = \sum\bar G_iX^{n-i}H=∑Gˉi​Xn−i, so that H(v)=vnG(v)‾H(v) = v^n\overline{G(v)}H(v)=vnG(v)​ and H(0)=Gˉn≠0H(0) = \bar G_n \neq 0H(0)=Gˉn​=0, a relation cleared by vn+1v^{n+1}vn+1 and evaluated at 000 loses its ggg coordinate, its conjugate loses the gˉ\bar ggˉ​ one, and 1,u,uˉ1, u, \bar u1,u,uˉ are free.

Novelty. Not found in the sources read; routine.

Preamble
import Definitions.Def_DiazModulus
Formal statement
open Complex ComplexConjugate

namespace DiazModulus

theorem laurent_conj_hull_class_finrank_eq_five
    (u s : ℂ) (hu : u ∉ Qbar) (hs : s ∈ Qbar) (hss : s * s = u * conj u)
    (k : ℕ) (c : ℤ → ℂ)
    (hcls : (∃ α β : ℂ, α ∈ Qbar ∧ β ∈ Qbar ∧ (α ≠ 0 ∨ β ≠ 0) ∧ α * u ^ 2 + β * u ^ 3 ∈ Submodule.span Qbar ({1, u, conj u, (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j), conj (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j)} : Set ℂ)) ∨
      (∃ e β ζ : ℂ, e ∈ Qbar ∧ β ∈ Qbar ∧ ζ ∈ Qbar ∧ e * conj e = 1 ∧ β * conj β = 1 ∧
        ζ = -(β * e ^ 2 * conj ζ) ∧
        Submodule.span Qbar ({1, u, conj u, (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j), conj (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j)} : Set ℂ) =
          Submodule.span Qbar ({1, u, conj u, conj (u / s) ^ 2 + e * (u / s) ^ 2, β * conj (u / s) ^ 3 + ζ * (u / s) ^ 2 + e * (u / s) ^ 3} : Set ℂ)) ∨
      (∃ κ : ℂ, κ ∈ Qbar ∧ κ * conj κ = 1 ∧ ∃ G : Polynomial ↥Qbar, G.coeff 0 = 0 ∧ 2 ≤ G.natDegree ∧
        Submodule.span Qbar ({1, u, conj u, (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j), conj (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j)} : Set ℂ) =
          Submodule.span Qbar ({1, u, conj u, Polynomial.aeval (u / s) G + κ * conj ((u / s) * Polynomial.aeval (u / s) G), conj (Polynomial.aeval (u / s) G + κ * conj ((u / s) * Polynomial.aeval (u / s) G))} : Set ℂ))) :
    Module.finrank (↥Qbar) (Submodule.span Qbar ({1, u, conj u, (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j), conj (∑ j ∈ Finset.Icc (-(k : ℤ)) k, c j * (u / s) ^ j)} : Set ℂ)) = 5 := by
  sorry

end DiazModulus
Source
Not found in the sources read. R18 of the Diaz modulus mission. Formal proof: Diaz modulus mission, 18 October 2026 (C. Perassi).

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