For the same u, a₁, a₂, the space Q̄ + Q̄u + Q̄ū + Q̄u/(u² − a₁) + Q̄u/(u² − a₂) contains none of u², ū² and 1/(u − b) with b ∈ Q̄ non-zero

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $\rho = u\bar u$ algebraic, let $a_1 \neq a_2$ be non-zero algebraic numbers, $z_i = u/(u^2 - a_i)$, $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$ and $W = H_0 + \overline{\mathbb{Q}}z_1 + \overline{\mathbb{Q}}z_2$. Then $W$ contains neither $u^2$ nor $\bar u^2$, nor $1/(u - b)$ for any non-zero $b \in \overline{\mathbb{Q}}$.

These are the elements $w$ for which, by Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff), a four-dimensional space $H_0 + \overline{\mathbb{Q}}w$ carries a configuration.

Proof. Both $z_i$ satisfy $z_i(u^2 - a_i) = u$. Clearing denominators (with $\bar u = \rho/u$) turns a membership $w \in W$ into $A(u^2) + uB(u^2) = 0$ with $A, B \in \overline{\mathbb{Q}}[X]$; the same identity at $-u$ gives $A = B = 0$, as $u^2$ is transcendental. For $u^2$ the odd part $B = (X - c_1)(X - a_1)(X - a_2)$ is non-zero ($c_1$ the coordinate on $u$); for $\bar u^2 = \rho^2/u^2$ the even part has $A(0) = \rho^2a_1a_2 \neq 0$; for $1/(u - b) = (u + b)/(u^2 - b^2)$ the odd part is $(X - a_1)(X - a_2)(b - c_1(X - b^2))$, non-zero because its last factor takes the value $b \neq 0$ at $X = b^2$. Here $c_1$ is the constant coordinate of the element of $W$.

Novelty. Not asserted separately; elementary (parity in $u$).