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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

Proved
DiazModulus.rank_one_config_card_le_finrank_add_one

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

diaz-modulus-leanlinear-algebratranscendence

A p×qp \times qp×q configuration over a field KKK in a KKK-subspace V⊆CV \subseteq \mathbb{C}V⊆C is a pair x1,…,xpx_1, \dots, x_px1​,…,xp​ and y1,…,yqy_1, \dots, y_qy1​,…,yq​ of complex numbers, each family linearly independent over KKK, with every product xiyjx_iy_jxi​yj​ in VVV; it is the input of Roy's strong six exponentials theorem when p=2p = 2p=2, q=3q = 3q=3, K=Q‾K = \overline{\mathbb{Q}}K=Q​ and V=L~V = \widetilde{\mathcal{L}}V=L.

Let KKK be a subfield of C\mathbb{C}C, uuu transcendental over KKK, and V0⊆K(u)V_0 \subseteq K(u)V0​⊆K(u) a finite-dimensional KKK-subspace. Every p×qp \times qp×q configuration over KKK in V0V_0V0​ with p,q≥1p, q \ge 1p,q≥1 has p+q≤dim⁡KV0+1p + q \le \dim_K V_0 + 1p+q≤dimK​V0​+1.

For V0=K+Ku+KuˉV_0 = K + Ku + K\bar uV0​=K+Ku+Kuˉ with uuˉ∈Ku\bar u \in Kuuˉ∈K (dimension 333) it gives p+q≤4p + q \le 4p+q≤4, so no 2×32 \times 32×3 configuration.

Proof. Clear denominators: there is ggg with g(u)≠0g(u) \neq 0g(u)=0 and polynomials AijA_{ij}Aij​ with Aij(u)=g(u)xiyjA_{ij}(u) = g(u)x_iy_jAij​(u)=g(u)xi​yj​. Evaluation at uuu is injective, so Ai1A1j=A11AijA_{i1}A_{1j} = A_{11}A_{ij}Ai1​A1j​=A11​Aij​ in K[X]K[X]K[X]. Then P=span⁡{Ai1}P = \operatorname{span}\{A_{i1}\}P=span{Ai1​} and Q=span⁡{A1j}Q = \operatorname{span}\{A_{1j}\}Q=span{A1j​} have dimensions ppp and qqq, PQ⊆A11⋅span⁡{Aij}PQ \subseteq A_{11}\cdot\operatorname{span}\{A_{ij}\}PQ⊆A11​⋅span{Aij​}, and v↦v(u)/g(u)v \mapsto v(u)/g(u)v↦v(u)/g(u) embeds span⁡{Aij}\operatorname{span}\{A_{ij}\}span{Aij​} into V0V_0V0​. Linear Cauchy–Davenport (DiazModulus.polynomial_submodule_mul_finrank_ge) gives p+q≤dim⁡PQ+1≤dim⁡V0+1p + q \le \dim PQ + 1 \le \dim V_0 + 1p+q≤dimPQ+1≤dimV0​+1.

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.

Preamble
import Definitions.Def_DiazModulus
Formal statement
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
Source
Linear Cauchy–Davenport (S. Eliahou, C. Lecouvey, Th. 6.2, as stated in C. Bachoc, O. Serra, G. Zémor, An analogue of Vosper's theorem for extension fields, Math. Proc. Cambridge Philos. Soc. (2017), arXiv:1501.00602, Th. 2) applied to configurations in K(u). R6 of the Diaz modulus mission (the polar-degree note). Formal proof: Diaz modulus mission, 8 October 2026 (C. Perassi).

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