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A candidate is exactly a point of the Diaz locus

Proved
DiazModulus.diaz_locus_dictionary

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

diaz-modulus-leannumber-theory

The two vocabularies used on this mission name the same set.

The conjecture tree is written in terms of IsCandidate u: u≠0u \neq 0u=0, ∣u∣|u|∣u∣ algebraic, eue^{u}eu algebraic. The results transcribed from the author's manuscripts are written in terms of the Diaz locus

D  =  { u∈L∖{0}  :  uuˉ∈Qˉ },\mathcal{D} \;=\; \{\, u \in \mathcal{L} \setminus \{0\} \;:\; u\bar{u} \in \bar{\mathbb{Q}} \,\},D={u∈L∖{0}:uuˉ∈Qˉ​},

whose defining data appear inlined in those statements as the three hypotheses u≠0u \neq 0u=0, IsAlgebraic ℚ (Complex.exp u), IsAlgebraic ℚ (u * conj u). Nothing on the board records that these are the same condition, so no result stated in one vocabulary can presently be cited by a proof written in the other.

They are the same condition, and unconditionally so: the equivalence needs no transcendence input at all — not Hermite–Lindemann, not six exponentials — and holds for every complex uuu, whether or not a candidate exists.

The two clauses. The second conjunct is the content. Its whole mathematical substance is the chain ∣u∣∈Qˉ  ⟺  ∣u∣2∈Qˉ  ⟺  uuˉ∈Qˉ|u| \in \bar{\mathbb{Q}} \iff |u|^{2} \in \bar{\mathbb{Q}} \iff u\bar{u} \in \bar{\mathbb{Q}}∣u∣∈Qˉ​⟺∣u∣2∈Qˉ​⟺uuˉ∈Qˉ​, which is already the first conjunct of the published Diaz.normal_form; what is added here is only that algebraicity over Q\mathbb{Q}Q does not notice the embedding R↪C\mathbb{R} \hookrightarrow \mathbb{C}R↪C, so that the conjecture tree's IsAlgebraic ℚ ((‖u‖ : ℝ) : ℂ) and the locus's IsAlgebraic ℚ ‖u‖ say the same thing. The proof imports Diaz.normal_form and uses exactly that.

The first conjunct is the other half of the translation, and it is not decoration. Each family carries its own copy of the algebraic numbers — DiazModulus.Qbar and Diaz.Qbar — declared independently, each as (algebraicClosure ℚ ℂ).toSubfield. They are literally the same subfield, the equation holding by rfl; but until it is stated, a DiazModulus hypothesis cannot be fed to a Diaz result quantified over Diaz.Qbar, and the instance Diaz.QbarIsAlgebraic — the one the manuscripts' own audit found missing — does not fire on the conjecture tree's base field.

What this is, and what it is not. It is a dictionary, not a decomposition. It makes no part of Diaz's conjecture easier, proves nothing new about candidates, and asserts no existence. Its value is entirely structural: with it, the roughly thirty results of the Diaz.* family that are stated over an arbitrary conjugation-stable subfield KKK with hypotheses Transcendental K u and u * conj u ∈ K become directly applicable at the mission's own base field, and the conjugation, independence and rigidity statements proved for the locus become statements about candidates.

The same equivalence in words: Diaz's conjecture asserts D=∅\mathcal{D} = \emptysetD=∅; IsCandidate is the predicate whose non-emptiness is the negation of the conjecture. Naming both without saying they agree is the gap this closes.

Preamble
import Definitions.Def_DiazModulus
import Definitions.Def_Diaz_Instantiation

open Complex ComplexConjugate
Formal statement
namespace DiazModulus
theorem diaz_locus_dictionary :
    Qbar = Diaz.Qbar ∧
      ∀ u : ℂ, IsCandidate u ↔
        (u ≠ 0 ∧ IsAlgebraic ℚ (Complex.exp u) ∧ IsAlgebraic ℚ (u * conj u)) := by sorry
end DiazModulus

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