For u ∉ Q̄ and real algebraic s with s² = uū, the span of 1, u, ū, u³/s³ + u²/s², ū³/s³ + ū²/s² carries no 2×3 configuration whose ratio x₂/x₁ is a Möbius function of u

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and let $s \in \overline{\mathbb{Q}}$ be real with $s^2 = u\bar u$ (so $\bar u = s^2/u$; with $v = u/s$, conjugation acts as $v \mapsto 1/v$ on coefficients). Put $F = u^3/s^3 + u^2/s^2$ and let $W_1$ be the $\overline{\mathbb{Q}}$-span of $1, u, \bar u, F, \bar F$ (in $v$: $v^{-1}, 1, v, v^3 + v^2, v^{-3} + v^{-2}$). Then there are no free $x_1, x_2$ and free $y_1, y_2, y_3$ with all six products in $W_1$ and $x_2/x_1 = (\alpha u + \beta)/(\gamma u + \delta)$, $\alpha, \beta, \gamma, \delta \in \overline{\mathbb{Q}}$, $\alpha\delta - \beta\gamma \neq 0$.

$W_1$ answers a question left open in R10: a conjugation-stable five-dimensional space containing $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ can carry a configuration that Theorem B does not see without having a configuration factor of degree one, so it is not one of R11's real pencil spaces. Every real pencil space has a configuration of degree one (DiazModulus.conj_stable_real_pencil_space_carries_two_by_three_configuration), so $W_1$ is none of them.

Proof. Suppose $x_1(\gamma u + \delta) = x_0(\alpha u + \beta)$ and all $x_iy_j \in W_1$. Put $c = \alpha x_0 - \gamma x_1$, non-zero because $x$ is free and $\alpha\delta - \beta\gamma \neq 0$. Then $t_j = cy_j = \alpha x_0y_j - \gamma x_1y_j \in W_1$ and $ut_j = \delta x_1y_j - \beta x_0y_j \in W_1$. But $t, ut \in W_1$ forces $t \in \overline{\mathbb{Q}} + \overline{\mathbb{Q}}\bar u$ (comparing the coefficients of $u\cdot s^3u^3t = s^3u^3\cdot ut$, a polynomial identity of degree $7$), and the three free $t_j$ cannot lie in a two-dimensional space.

Novelty. Not found in the sources read; it is short.