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The independence upgrade at the Diaz-locus quadric

Proved
Diaz.quadric_trdeg_two

by carlok · Sep 8, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theory

Source. The hypothesis hIU is the Lemma Independence upgrade, statement 37 of C. Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz, unpublished manuscript, 15 August 2026, the predecessor manuscript of Carlo Perassi's note on Diaz's modulus conjecture. That manuscript states explicitly that the lemma is not its own: "it is stated explicitly in [Waldschmidt, Diophantine Approximation on Linear Algebraic Groups, p. 593], immediately after Theorem 15.30", in the form that Q\mathbb QQ-linearly independent elements of L\mathcal LL generating a field of transcendence degree one satisfy Q(λ1,…,λn)≠0Q(\lambda_1,\dots,\lambda_n)\ne0Q(λ1​,…,λn​)=0 for every non-zero homogeneous rational QQQ of degree two. In the manuscript it is derived from Roy and Waldschmidt's Théorème 0.2.

The cited lemma, as carried. Let λ1,…,λn∈L\lambda_1,\dots,\lambda_n\in\mathcal Lλ1​,…,λn​∈L be linearly independent over Q\mathbb QQ and let P∈Q[X1,…,Xn]P\in\mathbb Q[X_1,\dots,X_n]P∈Q[X1​,…,Xn​] be a non-zero homogeneous polynomial of degree at most two with P(λ1,…,λn)=0P(\lambda_1,\dots,\lambda_n)=0P(λ1​,…,λn​)=0. Then trdeg⁡QQ(λ1,…,λn)≥2\operatorname{trdeg}_{\mathbb Q}\mathbb Q(\lambda_1,\dots,\lambda_n)\ge2trdegQ​Q(λ1​,…,λn​)≥2.

This statement was transcribed from C. Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz, unpublished manuscript, 15 August 2026 and has not been verified against its original (Waldschmidt's book, or the Roy–Waldschmidt paper); neither source is held on this mission. It is carried as the explicit hypothesis hIU rather than asserted. The node asserts only the consequence below.

The consequence. Let K⊂CK\subset\mathbb CK⊂C be a subfield and let u,vu,vu,v be such that u,u‾,v,v‾u,\overline u,v,\overline vu,u,v,v are all logarithms of elements of KKK, with squared moduli rationally commensurable,

vv‾  =  m (uu‾),m∈Q×,v\overline v \;=\; m\,(u\overline u),\qquad m\in\mathbb Q^\times,vv=m(uu),m∈Q×,

and with u,u‾,v,v‾u,\overline u,v,\overline vu,u,v,v linearly independent over Q\mathbb QQ. Then

trdeg⁡QQ(u,u‾,v,v‾) ≥ 2.\operatorname{trdeg}_{\mathbb Q}\mathbb Q\bigl(u,\overline u,v,\overline v\bigr)\ \ge\ 2 .trdegQ​Q(u,u,v,v) ≥ 2.

This is the manuscript's own reading of the lemma on the Diaz locus: two Diaz candidates with rational squared-modulus ratio have a Q\mathbb QQ-linearly independent conjugate quadruple (that independence is the mission's Diaz.indep_quadruple) lying on the rational quadric mX1X2−X3X4=0mX_1X_2-X_3X_4=0mX1​X2​−X3​X4​=0, so the lemma applies at full strength.

What is verified. The instantiation itself: that P=mX1X2−X3X4P=mX_1X_2-X_3X_4P=mX1​X2​−X3​X4​ is a non-zero homogeneous polynomial of degree two over Q\mathbb QQ (non-zero because its value at (1,1,0,0)(1,1,0,0)(1,1,0,0) is m≠0m\ne0m=0), that it vanishes at the quadruple exactly when vv‾=m uu‾v\overline v=m\,u\overline uvv=muu, and that the Q\mathbb QQ-linear independence hypothesis of the lemma is the one supplied. Everything deeper sits inside hIU.

Formalization note. The lemma is carried over Fin n with MvPolynomial (Fin n) ℚ, MvPolynomial.IsHomogeneous, and evaluation by MvPolynomial.aeval; transcendence degree is Algebra.trdeg ℚ of Algebra.adjoin ℚ of the relevant set. #print axioms on the submitted proof: [propext, Classical.choice, Quot.sound].

Preamble
import Mathlib

open ComplexConjugate
Formal statement
theorem Diaz.quadric_trdeg_two {K : Subfield ℂ}
    (hIU : ∀ (n : ℕ) (l : Fin n → ℂ) (P : MvPolynomial (Fin n) ℚ),
      (∀ i, Complex.exp (l i) ∈ K) →
      (∀ c : Fin n → ℚ, (∑ i, (c i : ℂ) * l i = 0) → c = 0) →
      P ≠ 0 → (∃ d, d ≤ 2 ∧ P.IsHomogeneous d) →
      MvPolynomial.aeval l P = 0 →
      2 ≤ Algebra.trdeg ℚ ↥(Algebra.adjoin ℚ (Set.range l)))
    {u v : ℂ} {m : ℚ}
    (hu : Complex.exp u ∈ K) (huc : Complex.exp (conj u) ∈ K)
    (hv : Complex.exp v ∈ K) (hvc : Complex.exp (conj v) ∈ K)
    (hm : m ≠ 0) (hquad : v * conj v = (m : ℂ) * (u * conj u))
    (hindep : ∀ a b c d : ℚ,
      (a : ℂ) * u + (b : ℂ) * conj u + (c : ℂ) * v + (d : ℂ) * conj v = 0 →
      a = 0 ∧ b = 0 ∧ c = 0 ∧ d = 0) :
    2 ≤ Algebra.trdeg ℚ ↥(Algebra.adjoin ℚ ({u, conj u, v, conj v} : Set ℂ)) := by sorry

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