For the same u, a₁, a₂, W = Q̄ + Q̄u + Q̄ū + Q̄u/(u² − a₁) + Q̄u/(u² − a₂) 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 $a_1 \neq a_2$ be non-zero algebraic numbers, $z_i = u/(u^2 - a_i)$, $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ and $W = H_0 + \overline{\mathbb{Q}}z_1 + \overline{\mathbb{Q}}z_2$. Then $W$ carries a $2 \times 3$ configuration, but for no $w \in W$ outside $H_0$ does $H_0 + \overline{\mathbb{Q}}w$ carry one.

This removes the conjugation condition from DiazModulus.circle_point_conjugate_pair_configuration_invisible_to_four_dimensional_extensions (the case $a_2 = \rho^2/\bar a_1$), and covers the case $a_1\bar a_1 = a_2\bar a_2 = \rho^2$, where $W$ is stable under conjugation.

Proof. The configuration is DiazModulus.circle_point_two_pole_extension_carries_two_by_three_configuration. If $w \in W \setminus H_0$ and $H_0 + \overline{\mathbb{Q}}w$ carried one, Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff) would give $w \in H_0 + \overline{\mathbb{Q}}t$ with $t$ one of $u^2$, $\bar u^2$, $1/(u - b)$ ($b$ algebraic, non-zero); exchange then puts $t$ in $H_0 + \overline{\mathbb{Q}}w \subseteq W$, which DiazModulus.circle_point_two_pole_extension_excludes_squares_and_reciprocals rules out.

Novelty. Not found in the sources read; it is short (parity plus Theorem B).