Under the strong six exponentials theorem, ℒ̃ + Q̄u + Q̄ū carries no 2×3 configuration when uū is algebraic and u is transcendental over the field of ℒ̃

Assume Roy's strong six exponentials theorem: no $2 \times 3$ configuration (free $x_1, x_2$ and free $y_1, y_2, y_3$ over $\overline{\mathbb{Q}}$ with all six products $x_iy_j$ in $\widetilde{\mathcal L}$). Let $u \in \mathbb{C}$ with $u\bar u$ algebraic and $u$ transcendental over the field generated by $\widetilde{\mathcal L}$. Then $\widetilde{\mathcal L} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ carries no such configuration either.

So Roy's theorem, applied to $\widetilde{\mathcal L} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$, cannot exclude a point of a circle of algebraic radius that is transcendental over the field of $\widetilde{\mathcal L}$.

Proof. Take $K = \overline{\mathbb{Q}}$, $F$ the field generated by $\widetilde{\mathcal L}$ and $V_0 = \widetilde{\mathcal L}$; $F \supseteq \overline{\mathbb{Q}}$ since $\widetilde{\mathcal L}$ is a $\overline{\mathbb{Q}}$-space containing $1$. As $\bar u = \rho u^{-1}$, a configuration in $\widetilde{\mathcal L} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ lies in $\widetilde{\mathcal L} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}u^{-1}$, and DiazModulus.rank_one_config_inverse_adjoin puts its products in $\widetilde{\mathcal L}$, which the strong six exponentials hypothesis forbids.

Novelty. Not found in the sources read; routine, from DiazModulus.rank_one_config_inverse_adjoin.