For f = P(u)/Q(u) in lowest terms, W(λ, f) = span(1, u, ū, f, f/(s + λu)) is visible to the four-dimensional extensions exactly when a degree bound holds

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$, $s \in \overline{\mathbb{Q}}$ with $s^2 = u\bar u$, and $\lambda \in \overline{\mathbb{Q}}$ with $|\lambda| = 1$. 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, $f = P(u)/Q(u)$, and suppose $1, u, \bar u, f, f/(s + \lambda u)$ are free over $\overline{\mathbb{Q}}$. Then $W(\lambda, f) = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, f, f/(s + \lambda u))$ is visible to the four-dimensional extensions if and only if

where $[\cdot]$ is $1$ when the condition holds and $0$ otherwise. For real $f$ these are the conjugation-stable spaces of DiazModulus.conj_stable_real_pencil_space_carries_two_by_three_configuration; the criterion does not need $f$ real.

Proof. Put $a_0 = -s\bar\lambda$ and $g = f/(s + \lambda u)$, so $s + \lambda u = \lambda(u - a_0)$ and $f = sg + \lambda ug$: the space is $\mathrm{span}(1, u, \bar u, g, ug)$, and these five are free. If $P(a_0) \neq 0$, $g = P(u)/\tilde Q(u)$ with $\tilde Q = \lambda Q(X - a_0)$ coprime to $P$; if $P(a_0) = 0$, $P = (X - a_0)P_1$ and $g = \lambda^{-1}P_1(u)/Q(u)$. DiazModulus.rational_degree_one_space_visible_to_four_dimensional_extensions_iff_natDegree_le applied to that fraction gives the bound in both cases.

Novelty. Not found in the sources read; routine. For Laurent $f$ it contains the invisibility in DiazModulus.laurent_real_pencil_configuration_invisible_to_four_dimensional_extensions.