For u ∉ Q̄ and real algebraic s with s² = uū, the cyclic space with double poles at ±s is conjugation-stable, spanned by a 2×3 configuration and invisible to Theorem B

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 $v = u/s$, put $G_1 = 1/(v - 1) + 2/(v + 1)$ and $G_2 = 1/(v - 1)^2 + 1/(v - 1) - 2/(v + 1)^2 + 2/(v + 1)$, and let $W$ be the $\overline{\mathbb{Q}}$-span of $1, u, \bar u, G_1, G_2$ (poles at $u = \pm s$, each a double pole). Then $W$ is stable under conjugation, the six products of $(1, u^2)$ and $(\bar u, G_1, G_2)$ span it, and for no $w \in W$ outside $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ does $H_0 + \overline{\mathbb{Q}}w$ carry a configuration.

$W$ is the cyclic member of the classification of conjugation-stable invisible spaces with a configuration factor of degree two (research notes R12, Q1.7): the factor $(1, u^2)$ acts on $G_1, G_2$ modulo $H_0$ as a Jordan block, where R14's two-orbit spaces have two eigenvalues.

Proof. Write $\bar u = s^2/u$. Then $\bar G_1 = 1 - G_1$ and $\bar G_2 = G_2$ (clear denominators and reduce modulo $\bar u u = s^2$), so conjugation maps the five generators into $W$, and induction on the span finishes. The six products of $x = (1, u^2)$ and $y = (\bar u, G_1, G_2)$ are $\bar u$, $G_1$, $G_2$, $u^2\bar u = s^2u$, $u^2G_1 = 3su - s^2 + s^2G_1$ and $u^2G_2 = s^2G_2 + 3su - 2s^2 + 2s^2G_1$: all in $W$, and they give back $u = u^2\bar u/s^2$ and $1 = 3u/s + G_1 - u^2G_1/s^2$, so they span $W$. The pair $x$ is free as $u^2 \notin W$, and $y$ is free as the five generators are, both by DiazModulus.cyclic_pair_space_excludes_squares_and_reciprocals. Invisibility: if $H_0 + \overline{\mathbb{Q}}w$ carries a configuration for some $w \in W$ outside $H_0$, Theorem B puts $w$ in $H_0 + \overline{\mathbb{Q}}t$ with $t = u^2$, $\bar u^2$ or $(u - e)^{-1}$; by exchange $t \in H_0 + \overline{\mathbb{Q}}w \subseteq W$, which the same node excludes.

Novelty. The configuration is printed: rescaled by $u^{\mp 1}$, it is the configuration $(v, \bar v) \otimes (1, w, \bar w)$ of Diaz (2007, Th. 7(1)) at $v = u$, $w = uG_2$ (credited here on 16 October 2026). That no four-dimensional extension of $H_0$ inside $W$ carries a configuration was not found in the sources read; routine.