For u ∉ Q̄ with uū algebraic, each space Q̄ + Q̄u + Q̄ū + Q̄w with w = u², ū² or 1/(u − a), a ∈ Q̄ non-zero, carries a 2×3 configuration

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $u\bar u$ algebraic, and let $w$ be $u^2$, $\bar u^2$, or $1/(u - a)$ with $a \in \overline{\mathbb{Q}}$, $a \neq 0$. Then there are $x_1, x_2$, linearly independent over $\overline{\mathbb{Q}}$, and $y_1, y_2, y_3$, linearly independent over $\overline{\mathbb{Q}}$, with all six products $x_i y_j$ in $\overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u + \overline{\mathbb{Q}}w$.

The configurations are $x = (1, u)$, $y = (u^{-1}, 1, u)$ for $w = u^2$; $x = (1, u^{-1})$, $y = (u, 1, u^{-1})$ for $w = \bar u^2$; and $x = (u^{-1}, (u - a)^{-1})$, $y = (1, u, u^2 - au)$ for $w = 1/(u - a)$. At a candidate $u$, Roy's strong six exponentials theorem turns them into Diaz's exclusions (2007, Cor. 5(1) and 5(4)): none of $u^2$, $\bar u^2$, $1/(u - a)$ lies in $\widetilde{\mathcal{L}}$. The statement itself does not involve $e^u$, so it keeps its content if Diaz's conjecture holds.

Proof. Since $u^{-1} = \bar u/(u\bar u)$ with $u\bar u$ algebraic and non-zero, every product listed lies in the space; for instance $u/(u - a) = 1 + a/(u - a)$ and $(u^2 - au)/(u - a) = u$. Independence comes from the transcendence of $u$ over $\overline{\mathbb{Q}}$: after multiplying by $u$, or by $u(u - a)$, a vanishing combination becomes a polynomial of degree at most $2$ in $u$ with algebraic coefficients, which must be zero; for the last family $a \neq 0$ is used.

Novelty. Not asserted. These are the configurations behind Diaz (2007), Cor. 5(1) and 5(4); the first also appears in DiazModulus.power_hull_strong_six_exp_configuration_iff.