For u transcendental over a subfield K of ℂ and a finite-dimensional K-space V₀ ⊆ K(u), every p×q configuration over K in V₀ has p + q ≤ dim V₀ + 1
ProvedDiazModulus.rank_one_config_card_le_finrank_add_oneA configuration over a field in a -subspace is a pair and of complex numbers, each family linearly independent over , with every product in ; it is the input of Roy's strong six exponentials theorem when , , and .
Let be a subfield of , transcendental over , and a finite-dimensional -subspace. Every configuration over in with has .
For with (dimension ) it gives , so no configuration.
Proof. Clear denominators: there is with and polynomials with . Evaluation at is injective, so in . Then and have dimensions and , , and embeds into . Linear Cauchy–Davenport (DiazModulus.polynomial_submodule_mul_finrank_ge) gives .
Novelty. Not found in the sources read in this form; it is short: linear Cauchy–Davenport (DiazModulus.polynomial_submodule_mul_finrank_ge) after clearing denominators.
import Definitions.Def_DiazModulus
open Complex ComplexConjugate
namespace DiazModulus
theorem rank_one_config_card_le_finrank_add_one (K : Subfield ℂ) (u : ℂ)
(hT : Transcendental K u) (V₀ : Submodule K ℂ) [FiniteDimensional K V₀]
(hV₀ : ∀ v ∈ V₀, v ∈ IntermediateField.adjoin K ({u} : Set ℂ))
(p q : ℕ) (hp : 1 ≤ p) (hq : 1 ≤ q) (x : Fin p → ℂ) (y : Fin q → ℂ)
(hx : LinearIndependent K x) (hy : LinearIndependent K y)
(hxy : ∀ i j, x i * y j ∈ V₀) :
p + q ≤ Module.finrank K V₀ + 1 := by
sorry
end DiazModulus