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π² is transcendental

Proved
DiazModulus.pi_sq_transcendental

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

diaz-modulus-leannumber-theory

Statement. π2\pi^{2}π2 is transcendental over Q\mathbb{Q}Q.

Source and attribution. All the mathematics of this section is Carlo Perassi's, in his manuscript C. Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz, unpublished manuscript, 15 August 2026, §Polar coordinates and the discreteness of the period. No novelty is claimed. There this fact is used unnamed inside two proofs --- Theorem Rational-translate rigidity (thm:q-translate), "two distinct non-zero elements would give π2(r−r′)∈Qˉ\pi^{2}(r-r')\in\bar{\mathbb{Q}}π2(r−r′)∈Qˉ​, contradicting the transcendence of π\piπ", and Theorem Period-plane classification (thm:period-plane), "Lindemann makes π2∉Qˉ\pi^{2}\notin\bar{\mathbb{Q}}π2∈/Qˉ​". The statement itself is Lindemann's theorem and is entirely classical; nothing here is new.

Why it is on the board. Several nodes of this mission carry (π:C)2∉K(\pi:\mathbb{C})^{2}\notin K(π:C)2∈/K as an explicit hypothesis over an abstract subfield K≤CK\le\mathbb{C}K≤C --- Diaz.q_translate_unique is the clearest case, and Diaz.nonreal_two_point_fibre_pi_sq is another. At K=QˉK=\bar{\mathbb{Q}}K=Qˉ​ that hypothesis is exactly this statement, and nothing on the board discharged it, so those nodes could not be instantiated at the algebraic numbers. This node supplies the missing instance, and it is used by Diaz.plane_normSq_algebraic_iff.

Proof. One line from the platform's own DiazModulus.pi_transcendental: if π2\pi^{2}π2 were algebraic then π\piπ, a square root of it, would be algebraic --- IsAlgebraic.of_pow at n=2n = 2n=2.

Preamble
import Definitions.Def_DiazModulus

open Complex ComplexConjugate
Formal statement
namespace DiazModulus
theorem pi_sq_transcendental : Transcendental ℚ ((Real.pi ^ 2 : ℝ) : ℂ) := by sorry
end DiazModulus

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