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Diaz's conjecture (Qr2) in transcendence degree one: for logarithms ℓ₀, ℓ₁ off both axes, a real or purely imaginary ℓ₁/ℓ₀ is rational

Proved
DiazModulus.diaz_2007_qr2_of_trdeg_one

by carlok · Oct 2, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theorytranscendence

Let ℓ0,ℓ1\ell_0, \ell_1ℓ0​,ℓ1​ be logarithms of algebraic numbers, neither of them real or purely imaginary, such that ℓ0,ℓˉ0,ℓ1,ℓˉ1\ell_0, \bar\ell_0, \ell_1, \bar\ell_1ℓ0​,ℓˉ0​,ℓ1​,ℓˉ1​ generate a Q\mathbb{Q}Q-algebra of transcendence degree at most 111. If ℓ1/ℓ0\ell_1/\ell_0ℓ1​/ℓ0​ is real or purely imaginary, then ℓ1/ℓ0∈Q\ell_1/\ell_0 \in \mathbb{Q}ℓ1​/ℓ0​∈Q; in particular it is never purely imaginary.

This is Diaz's conjecture (Qr2) (2007, p. 376) under the hypothesis on the transcendence degree. Diaz states (Qr2) as a consequence of the four exponentials conjecture; here that conjecture is replaced by the theorem that holds in transcendence degree one.

Proof. Put λ=ℓˉ0\lambda = \bar\ell_0λ=ℓˉ0​ and μ=ℓ1\mu = \ell_1μ=ℓ1​, both non-zero logarithms of algebraic numbers. Then λμ=∣ℓ0∣2 ℓ1/ℓ0\lambda\mu = |\ell_0|^2\,\ell_1/\ell_0λμ=∣ℓ0​∣2ℓ1​/ℓ0​, so λμ\lambda\muλμ is real, or purely imaginary, exactly when ℓ1/ℓ0\ell_1/\ell_0ℓ1​/ℓ0​ is. In the real case, DiazModulus.log_mul_real_trichotomy_of_trdeg_one leaves only μ∈Qλˉ=Qℓ0\mu \in \mathbb{Q}\bar\lambda = \mathbb{Q}\ell_0μ∈Qλˉ=Qℓ0​, since its other two cases put ℓ0\ell_0ℓ0​ on an axis. In the imaginary case, DiazModulus.log_mul_imaginary_mixed_of_trdeg_one would put ℓˉ0\bar\ell_0ℓˉ0​, hence ℓ0\ell_0ℓ0​, on an axis.

Novelty. Not asserted. The statement is the transcendence-degree-one case of Diaz's conjecture (Qr2) (2007, p. 376), and the proof is Diaz's own argument for (Qr2) from the four exponentials conjecture (p. 377), with that conjecture replaced by the theorem of Waldschmidt (1973) and Brownawell (1974) in transcendence degree one, which Diaz quotes on p. 386. That combination was not found in the sources read.

Preamble
import Mathlib
Formal statement
open Complex ComplexConjugate

namespace DiazModulus

theorem diaz_2007_qr2_of_trdeg_one (l₀ l₁ : ℂ)
    (he₀ : IsAlgebraic ℚ (Complex.exp l₀)) (he₁ : IsAlgebraic ℚ (Complex.exp l₁))
    (h₀ : l₀.re ≠ 0 ∧ l₀.im ≠ 0) (h₁ : l₁.re ≠ 0 ∧ l₁.im ≠ 0)
    (htr : Algebra.trdeg ℚ ↥(Algebra.adjoin ℚ ({l₀, l₁, conj l₀, conj l₁} : Set ℂ)) ≤ 1)
    (hax : (l₁ / l₀).im = 0 ∨ (l₁ / l₀).re = 0) :
    ∃ q : ℚ, l₁ / l₀ = q := by
  sorry

end DiazModulus
Source
The transcendence-degree-one case of Diaz's conjecture (Qr2) (G. Diaz, Produits et quotients de combinaisons linéaires de logarithmes de nombres algébriques : conjectures et résultats partiels, J. Théor. Nombres Bordeaux 19 (2007), 373–391, p. 376: for logarithms ℓ₀, ℓ₁ off both axes, ℓ₁/ℓ₀ ∈ ℝ ∪ iℝ implies ℓ₁/ℓ₀ ∈ ℚ). Diaz derives (Qr2) from the four exponentials conjecture with the same 2×2 matrix (p. 377); here the conjecture is replaced by the four exponentials theorem in transcendence degree one (M. Waldschmidt, Solution du huitième problème de Schneider, J. Number Theory 5 (1973), 191–202, Cor. 4; W. D. Brownawell, The algebraic independence of certain numbers related by the exponential function, J. Number Theory 6 (1974), 22–31, Cor. 7; M. Waldschmidt, Diophantine Approximation on Linear Algebraic Groups, Grundlehren 326, Springer, 2000, Cor. 15.28(c)), which Diaz quotes on p. 386 without combining it with (Qr2). His Cor. 6(2) (p. 387) proves (Qr2) under ℓ₁/ℓ₀ ∈ ℒ̃ instead. The transcendence-degree-one form was not found in the sources read. Formal proof: Diaz modulus mission, 2 October 2026 (C. Perassi).

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