For g = P(u)/Q(u) in lowest terms, span(1, u, ū, g, ug) is visible to the four-dimensional extensions exactly when max(deg P, deg Q) ≤ 2 + [Q(0) = 0]

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 call $W \supseteq H_0$ 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 $P, Q \in \overline{\mathbb{Q}}[X]$ be coprime, $g = P(u)/Q(u)$, and suppose $1, u, \bar u, g, ug$ are 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 $\max(\deg P, \deg Q) \leq 2 + [Q(0) = 0]$, where $[Q(0) = 0]$ is $1$ if $Q(0) = 0$ and $0$ otherwise.

Proof. $Q \neq 0$, since otherwise $g = 0$. By DiazModulus.degree_one_space_visible_to_four_dimensional_extensions_iff_cubic_over_quadratic the space is visible exactly when $u\,R(u)\,P(u)/Q(u) = N(u)$ for such $R, N$; as evaluation at the transcendental $u$ is injective, that is $XRP = NQ$ in $\overline{\mathbb{Q}}[X]$. DiazModulus.coprime_fraction_cubic_over_quadratic_iff_natDegree_le turns this into the degree bound.

Novelty. Not found in the sources read; routine.