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A 2×2 block of products whose determinant vanishes as a form has an algebraic row or column relation

Proved
DiazModulus.qbar_dependent_of_det_linear_forms

by carlok · Sep 25, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leanlinear-algebratranscendence

Let e1,…,e4∈Ce_1, \dots, e_4 \in \mathbb{C}e1​,…,e4​∈C, and let x0,x1,yj0,yj1∈Cx_0, x_1, y_{j_0}, y_{j_1} \in \mathbb{C}x0​,x1​,yj0​​,yj1​​∈C with x0≠0x_0 \neq 0x0​=0 and yj0≠0y_{j_0} \neq 0yj0​​=0. Suppose that each product is an algebraic linear form in the eke_kek​,

xi yj=Lij(e)=∑kcijk ek,cijk∈Q‾,x_i\, y_j = L_{ij}(e) = \sum_{k} c_{ijk}\, e_k, \qquad c_{ijk} \in \overline{\mathbb{Q}},xi​yj​=Lij​(e)=k∑​cijk​ek​,cijk​∈Q​,

and that the determinant of the block vanishes identically as a quadratic form:

L0j0(z) L1j1(z)−L0j1(z) L1j0(z)=0for all z∈C4.L_{0 j_0}(z)\, L_{1 j_1}(z) - L_{0 j_1}(z)\, L_{1 j_0}(z) = 0 \quad \text{for all } z \in \mathbb{C}^{4}.L0j0​​(z)L1j1​​(z)−L0j1​​(z)L1j0​​(z)=0for all z∈C4.

Then x0,x1x_0, x_1x0​,x1​ or yj0,yj1y_{j_0}, y_{j_1}yj0​​,yj1​​ are linearly dependent over Q‾\overline{\mathbb{Q}}Q​.

The numerical determinant x0yj0⋅x1yj1−x0yj1⋅x1yj0x_0 y_{j_0} \cdot x_1 y_{j_1} - x_0 y_{j_1} \cdot x_1 y_{j_0}x0​yj0​​⋅x1​yj1​​−x0​yj1​​⋅x1​yj0​​ is always zero; the hypothesis is the identity of forms. It arises when a quadratic relation among the eke_kek​ is known to be a multiple of a fixed form, as in DiazModulus.generic_no_strong_six_exp_configuration.

Preamble
import Definitions.Def_DiazModulus
Formal statement
namespace DiazModulus

theorem qbar_dependent_of_det_linear_forms {m : ℕ} (e : Fin 4 → ℂ) (C : Fin 2 → Fin m → Fin 4 → ℂ)
    (hC : ∀ i j k, IsAlgebraic ℚ (C i j k)) (x : Fin 2 → ℂ) (y : Fin m → ℂ) (j0 j1 : Fin m)
    (hx0 : x 0 ≠ 0) (hy0 : y j0 ≠ 0)
    (hxy : ∀ i j, x i * y j = ∑ k, C i j k * e k)
    (hdet : ∀ z : Fin 4 → ℂ, (∑ k, C 0 j0 k * z k) * (∑ k, C 1 j1 k * z k)
      - (∑ k, C 0 j1 k * z k) * (∑ k, C 1 j0 k * z k) = 0) :
    (∃ p q : ↥Qbar, ¬(p = 0 ∧ q = 0) ∧ (p : ℂ) * x 0 + (q : ℂ) * x 1 = 0) ∨
      (∃ p q : ↥Qbar, ¬(p = 0 ∧ q = 0) ∧ (p : ℂ) * y j0 + (q : ℂ) * y j1 = 0) := by
  sorry

end DiazModulus
Source
Standard linear algebra. Formal proof: Diaz modulus mission, 25 September 2026 (C. Perassi); extracted from the proofs of DiazModulus.generic_no_strong_six_exp_configuration and DiazModulus.generic_qbar_homogeneous_four_exp_barrier.

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