For the same u, a and z, W = Q̄ + Q̄u + Q̄ū + Q̄z + Q̄z̄ carries a 2×3 configuration, while no Q̄ + Q̄u + Q̄ū + Q̄w with w ∈ W outside Q̄ + Q̄u + Q̄ū does

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $\rho = u\bar u$ algebraic, let $a \in \overline{\mathbb{Q}}$, $a \neq 0$, and let $z$ be $u/(u^2 - a)$ if $a\bar a \neq \rho^2$, or $u/(u^2 - a)^2$ if $a\bar a = \rho^2$. Put $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ and $W = H_0 + \overline{\mathbb{Q}}z + \overline{\mathbb{Q}}\bar z$. Then $W$ carries a $2 \times 3$ configuration: $x_1, x_2$ independent over $\overline{\mathbb{Q}}$ and $y_1, y_2, y_3$ independent over $\overline{\mathbb{Q}}$ with all six products $x_iy_j$ in $W$. But for no $w \in W$ outside $H_0$ does $H_0 + \overline{\mathbb{Q}}w$ carry one.

So the configuration is invisible to Theorem B's four-dimensional spaces. It does not need all of $W$: its six products span the odd part $\overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u + \overline{\mathbb{Q}}z + \overline{\mathbb{Q}}\bar z$, which does not contain $1$. Since $\widetilde{\mathcal{L}}$ is closed under complex conjugation, a hypothesis $z \in \widetilde{\mathcal{L}}$ brings $\bar z$ with it, so $W$ is the space Roy's strong six exponentials theorem sees at a candidate; DiazModulus.circle_point_extension_two_by_three_configuration_iff (Theorem B) treats the four-dimensional spaces $H_0 + \overline{\mathbb{Q}}w$ only. The statement does not involve $e^u$, so it keeps its content if Diaz's conjecture holds.

Proof. The configuration is DiazModulus.circle_point_conjugate_pair_extension_carries_two_by_three_configuration. Let $w \in W \setminus H_0$ and suppose $H_0 + \overline{\mathbb{Q}}w$ carries one. By Theorem B, $w \in H_0 + \overline{\mathbb{Q}}t$ for $t$ one of $u^2$, $\bar u^2$, $1/(u - b)$ with $b$ algebraic and non-zero. Since $w \notin H_0$, exchange gives $t \in H_0 + \overline{\mathbb{Q}}w \subseteq W$, which DiazModulus.circle_point_conjugate_pair_extension_excludes_squares_and_reciprocals rules out.

Novelty. Not found in the sources read; it is short (parity plus Theorem B). With the strong six exponentials theorem, the configuration gives at a candidate exclusions that are already one substitution in Diaz (2004, Th. 2) and Diaz (2007, Th. 6(3)): DiazModulus.candidate_div_sq_sub_not_mem_logAlgTilde.