The space span(1, u, ū, g, ug) is visible to the four-dimensional extensions exactly when u·R(u)·g = N(u) with R ≠ 0, deg R ≤ 2, deg N ≤ 3

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $\rho = u\bar u$ algebraic (so $\bar u = \rho/u$). Let $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$, and say that a space $W \supseteq H_0$ is visible to the four-dimensional extensions if some $w \in W \setminus H_0$ makes $H_0 + \overline{\mathbb{Q}}w$ carry a $2 \times 3$ 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 the space). Let $g \in \mathbb{C}$ with $1, u, \bar u, g, ug$ free over $\overline{\mathbb{Q}}$. Then $W_g = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, g, ug)$ is visible to the four-dimensional extensions if and only if $u\,R(u)\,g = N(u)$ for some $R, N \in \overline{\mathbb{Q}}[X]$ with $R \neq 0$, $\deg R \leq 2$ and $\deg N \leq 3$.

Every $W_g$ carries the configuration $(1, u) \otimes (\bar u, 1, g)$; this says when Theorem B already sees $W_g$ through one of its four-dimensional subspaces.

Proof. By Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff), $H_0 + \overline{\mathbb{Q}}w$ with $w \notin H_0$ carries a configuration exactly when it contains $u^2$, $\bar u^2$ or some $1/(u - a)$ with $a \in \overline{\mathbb{Q}}^\times$. If $W_g$ is visible, exchange puts such an element in $H_0 + \overline{\mathbb{Q}}w \subseteq W_g$, and DiazModulus.degree_one_space_cubic_over_quadratic_of_mem_square_or_reciprocal gives $R$ and $N$. Conversely DiazModulus.degree_one_space_mem_square_or_reciprocal_of_cubic_over_quadratic puts such an element $t$ in $W_g$; it is not in $H_0$, since $u$ is transcendental, and $w = t$ works.

Novelty. Not found in the sources read; routine.