For subfields K ≤ F of ℂ, a K-space V₀ ⊆ F and w transcendental over F: a 2×2 configuration over K with all products in V₀ + Kw has all products in V₀

A $p \times q$ configuration over a field $K$ in a $K$-subspace $V \subseteq \mathbb{C}$ is a pair $x_1, \dots, x_p$ and $y_1, \dots, y_q$ of complex numbers, each family linearly independent over $K$, with every product $x_iy_j$ in $V$; it is the input of Roy's strong six exponentials theorem when $p = 2$, $q = 3$, $K = \overline{\mathbb{Q}}$ and $V = \widetilde{\mathcal{L}}$.

Let $K \subseteq F$ be subfields of $\mathbb{C}$, $V_0 \subseteq F$ a $K$-subspace, and $w$ transcendental over $F$. If $x_1, x_2$ and $y_1, y_2$ form a $2 \times 2$ configuration over $K$ in $V_0 + Kw$, then every $x_iy_j$ lies in $V_0$: the transcendental number never enters. The coefficients of $w$ must lie in $K$, not in $F$.

Proof. Write $x_iy_j = c_{ij} + \ell_{ij}w$ with $c_{ij} \in V_0 \subseteq F$ and $\ell_{ij} \in K$. Since $(x_1y_1)(x_2y_2) = (x_1y_2)(x_2y_1)$, the polynomial $\det(C + XL)$ over $F$ vanishes at $w$, so its leading coefficient $\det L$ is $0$. If a column $j$ of $L$ were non-zero, $z = \ell_{2j}x_1 - \ell_{1j}x_2$ would be non-zero ($x$ is $K$-independent) with $zy_1, zy_2 \in F$ (because $\det L = 0$). For each row $i$, $(zy_1)(x_iy_2) = (zy_2)(x_iy_1)$ is linear in $w$ over $F$, so $z(\ell_{i2}y_1 - \ell_{i1}y_2) = 0$, and $K$-independence of $y$ makes the whole row of $L$ vanish, a contradiction. So $L = 0$ and $x_iy_j = c_{ij} \in V_0$.

Novelty. Not found in the sources read; it is short. The step that DiazModulus.rank_one_config_separation iterates.