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A rational combination au+buˉau+b\bar uau+buˉ lies off both rational rays

Proved
Diaz.conj_combination_off_rays

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

diaz-modulus-leannumber-theory

Let u∈Cu \in \mathbb{C}u∈C lie off the two axes, that is uˉ≠u\bar u \neq uuˉ=u and uˉ≠−u\bar u \neq -uuˉ=−u (equivalently Im⁡u≠0\operatorname{Im} u \neq 0Imu=0 and Re⁡u≠0\operatorname{Re} u \neq 0Reu=0), and let a,b∈Q×a,b \in \mathbb{Q}^\timesa,b∈Q×. Then

μ  =  au+buˉ  ∉  Qu ∪ Quˉ.\mu \;=\; au + b\bar u \;\notin\; \mathbb{Q}u \,\cup\, \mathbb{Q}\bar u .μ=au+buˉ∈/Qu∪Quˉ.

In particular μ≠0\mu \neq 0μ=0, which is the instance c=0c = 0c=0 of either clause.

Where this sits. This is the closing clause of Theorem 3.9 (thm:mixed-rigidity, Mixed-coordinate rigidity at a Diaz point) of the manuscript. Theorem 3.9 takes uuu on the Diaz locus, puts ρ=uuˉ\rho = u\bar uρ=uuˉ, and for μ∈L∖(Qu∪Quˉ)\mu \in \mathcal{L} \setminus (\mathbb{Q}u \cup \mathbb{Q}\bar u)μ∈L∖(Qu∪Quˉ) concludes

trdeg⁡QQ(u,μ,eρ/μ)  ≥  2.\operatorname{trdeg}_{\mathbb{Q}}\mathbb{Q}\bigl(u,\mu,e^{\rho/\mu}\bigr) \;\ge\; 2 .trdegQ​Q(u,μ,eρ/μ)≥2.

Its final paragraph verifies that the headline instance μ=au+buˉ\mu = au + b\bar uμ=au+buˉ with a,b∈Q×a,b \in \mathbb{Q}^\timesa,b∈Q× really does satisfy that hypothesis: a relation au+buˉ∈Qu∪Quˉau + b\bar u \in \mathbb{Q}u \cup \mathbb{Q}\bar uau+buˉ∈Qu∪Quˉ would force uˉ/u∈Q\bar u/u \in \mathbb{Q}uˉ/u∈Q and put uuu on the axes. This node is exactly that verification, and it is unconditional — the deep input of Theorem 3.9 is Théorème 7.1 of Roy–Waldschmidt (1997), which is not available in this Mathlib revision and is not on the mission graph.

Proof. Off the axes, uuu and uˉ\bar uuˉ are linearly independent over Q\mathbb{Q}Q (Diaz.indep_of_not_axis: from Re⁡u≠0\operatorname{Re} u \neq 0Reu=0 and Im⁡u≠0\operatorname{Im} u \neq 0Imu=0, a relation xu+yuˉ=0xu + y\bar u = 0xu+yuˉ=0 with x,y∈Qx,y \in \mathbb{Q}x,y∈Q gives x+y=0x + y = 0x+y=0 on real parts and x−y=0x - y = 0x−y=0 on imaginary parts, hence x=y=0x = y = 0x=y=0).

  • If au+buˉ=cuau + b\bar u = cuau+buˉ=cu with c∈Qc \in \mathbb{Q}c∈Q, then (a−c)u+buˉ=0(a-c)u + b\bar u = 0(a−c)u+buˉ=0, so b=0b = 0b=0 — contradicting b≠0b \neq 0b=0.
  • If au+buˉ=cuˉau + b\bar u = c\bar uau+buˉ=cuˉ with c∈Qc \in \mathbb{Q}c∈Q, then au+(b−c)uˉ=0au + (b-c)\bar u = 0au+(b−c)uˉ=0, so a=0a = 0a=0 — contradicting a≠0a \neq 0a=0.

Formalisation notes. The hypothesis is stated in the off-axes form u‾≠u∧u‾≠−u\overline u \neq u \wedge \overline u \neq -uu=u∧u=−u, which is the weakest form and is exactly what Diaz.indep_of_not_axis consumes; a consumer working on the Diaz locus obtains it from Diaz.not_on_axes. The two ray exclusions quantify over all c∈Qc \in \mathbb{Q}c∈Q (not merely Q×\mathbb{Q}^\timesQ×), so the non-vanishing μ≠0\mu \neq 0μ=0 that the manuscript notes separately is the case c=0c = 0c=0 and is not stated as a third conjunct.

Source. Carlo Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz (private manuscript, 15 August 2026), Theorem 3.9, closing paragraph of the proof. The mathematics is his; this node records one step of it in Lean and claims no novelty of its own. Elementary; possibly known, not checked against the literature.

Preamble
import Mathlib
import Definitions.Def_Diaz_Closure
import Definitions.Def_Diaz_Instantiation

open ComplexConjugate
open Diaz
Formal statement
theorem Diaz.conj_combination_off_rays {u : ℂ}
    (h1 : conj u ≠ u) (h2 : conj u ≠ -u)
    {a b : ℚ} (ha : a ≠ 0) (hb : b ≠ 0) :
    (∀ c : ℚ, (a : ℂ) * u + (b : ℂ) * conj u ≠ (c : ℂ) * u) ∧
      (∀ c : ℚ, (a : ℂ) * u + (b : ℂ) * conj u ≠ (c : ℂ) * conj u) := by sorry
Source
Carlo Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz (private manuscript, 15 August 2026), Theorem 3.9 (thm:mixed-rigidity), closing paragraph of the proof.

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