An axis-parallel pair of unequal modulus has independent coordinates
ProvedDiaz.axis_triple_indepLet with and algebraic — a Diaz candidate. Let be real or purely imaginary (), put , and suppose is algebraic but , i.e. . Then , , are linearly independent over .
Where this sits. This is the linear-independence half of the last assertion of Theorem 2.3 (thm:axis-ratio, Axis-parallel rational-ratio rigidity) of the manuscript: "If these conditions fail, then are -linearly independent". The conditions that fail are the equivalent (i)–(iii), of which (ii) is ; that is the hypothesis hne.
Proof. Hermite–Lindemann puts off both coordinate axes, so and . Conjugating the rational relation and combining with it kills and forces in the real case and in the imaginary case, leaving
If this gives at once. If then with , and a direct expansion gives the manuscript's identity
If the two moduli agree, against hne. Otherwise is algebraic, hence so is ; it is non-zero, and its exponential is algebraic, so Hermite–Lindemann is contradicted.
What the hypotheses do and do not say. The manuscript takes with a second candidate, so is itself a non-zero logarithm. The proof never uses that: neither nor algebraicity of appears, and both have been dropped. What is used about is only that is algebraic and differs from .
What is deliberately not claimed. The equivalence (i) (ii) (iii) of Theorem 2.3, and the quadratic non-vanishing , both rest on Theorem 0.2 of Roy–Waldschmidt, which is not available in Mathlib. They are not published.
Possibly known; not checked against the literature.
Source. Carlo Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz (private manuscript, 15 August 2026). The mathematics is his; this node only records one step of it in Lean, and claims no novelty of its own.
import Mathlib import Definitions.Def_Diaz_Closure import Definitions.Def_Diaz_Instantiation open ComplexConjugate open Diaz
theorem Diaz.axis_triple_indep {u τ : ℂ}
(hu0 : u ≠ 0) (hexpu : IsAlgebraic ℚ (Complex.exp u))
(hρ : IsAlgebraic ℚ (u * conj u))
(hax : conj τ = τ ∨ conj τ = -τ)
(hq : IsAlgebraic ℚ ((u + τ) * conj (u + τ)))
(hne : (u + τ) * conj (u + τ) ≠ u * conj u)
{a b c : ℚ} (hrel : (a : ℂ) * u + (b : ℂ) * conj u + (c : ℂ) * τ = 0) :
a = 0 ∧ b = 0 ∧ c = 0 := by sorry