A 2×3 configuration whose products, times u^N, are values at u of polynomials in H gives polynomial spans with f·g ∈ P₀·H

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$, $H \subseteq \overline{\mathbb{Q}}[X]$ a subspace, $N \geq 0$, and let free $x_1, x_2$ and free $y_1, y_2, y_3$ over $\overline{\mathbb{Q}}$ satisfy $u^Nx_iy_j = Q_{ij}(u)$ with $Q_{ij} \in H$. Then there are free $a_1, a_2$, free $b_1, b_2, b_3$ and $P_0 \neq 0$ in $\overline{\mathbb{Q}}[X]$ such that for every $f$ in the span of the $a_i$ and every $g$ in the span of the $b_j$, $fg = P_0h$ for some $h \in H$.

Proof. Take $a_i = Q_{i0}$, $b_j = Q_{0j}$ and $P_0 = Q_{00}$. As $u$ is transcendental, evaluation at $u$ is injective on $\overline{\mathbb{Q}}[X]$. $P_0 \neq 0$ because $P_0(u) = u^Nx_0y_0 \neq 0$; $a_i(u) = (u^Ny_0)x_i$ and $b_j(u) = (u^Nx_0)y_j$, so $a$ and $b$ are free. For $f = \sum c_ia_i$ and $g = \sum d_jb_j$ put $h = \sum c_id_jQ_{ij}$: at $u$ both $fg$ and $P_0h$ equal $u^{2N}x_0y_0(\sum c_ix_i)(\sum d_jy_j)$, so $fg = P_0h$.

Novelty. Not found in the sources read in this form; it is the normalisation in the proof of DiazModulus.laurent_hull_config_iff, for any polynomial hull. Routine.