A 2×3 configuration whose products times u^N are values at u of polynomials in H factors through coprime a₀, a₁ and free b₀, b₁, b₂ with aᵢbⱼ ∈ H

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$, let $H \subseteq \overline{\mathbb{Q}}[X]$ be a $\overline{\mathbb{Q}}$-subspace and $N \geq 0$. Let $x_0, x_1 \in \mathbb{C}$ be linearly independent over $\overline{\mathbb{Q}}$, and $y_0, y_1, y_2 \in \mathbb{C}$ too, such that for all $i, j$ there is $Q_{ij} \in H$ with $Q_{ij}(u) = u^Nx_iy_j$. Then there are coprime $a_0, a_1 \in \overline{\mathbb{Q}}[X]$ and linearly independent $b_0, b_1, b_2 \in \overline{\mathbb{Q}}[X]$ with $a_ib_j \in H$ for all $i, j$ and $x_1a_0(u) = x_0a_1(u)$.

Proof. As $u$ is transcendental over $\overline{\mathbb{Q}}$, evaluation at $u$ is injective on $\overline{\mathbb{Q}}[X]$. Each $Q_{ij}$ is non-zero, and the $2 \times 2$ minors $Q_{ij}Q_{i'j'} - Q_{ij'}Q_{i'j}$ vanish at $u$, hence vanish; DiazModulus.rank_one_polynomial_matrix_factor gives $Q_{ij} = a_ib_j$ with $a_0, a_1$ coprime, so $a_ib_j \in H$. The $b_j(u) = (u^Nx_0/a_0(u))\,y_j$ are a non-zero multiple of the free family $y$, so $b$ is free; and $x_1a_0(u)b_0(u) = u^Nx_0x_1y_0 = x_0a_1(u)b_0(u)$, where $b_0(u) \neq 0$ cancels.

Novelty. Not found in the sources read in this form; routine. It is the normalisation of DiazModulus.config_shifted_polynomial_spans, sharpened by the rank-one factorisation DiazModulus.rank_one_polynomial_matrix_factor.