For u ∉ Q̄ with uū algebraic and distinct non-zero algebraic a₁, a₂, the space Q̄ + Q̄u + Q̄ū + Q̄u/(u² − a₁) + Q̄u/(u² − a₂) carries a 2×3 configuration

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 there are $x_1, x_2$, linearly independent over $\overline{\mathbb{Q}}$, and $y_1, y_2, y_3$, linearly independent over $\overline{\mathbb{Q}}$, with all six products $x_iy_j$ in $W$.

The configuration is $x = (1, u^2)$, $y = (b, bu^2, bu^4)$, $b = 1/(u(u^2 - a_1)(u^2 - a_2))$: $W$ contains the progression $b, bu^2, bu^4, bu^6$. For $a_2 = \rho^2/\bar a_1$ this is the first family of DiazModulus.circle_point_conjugate_pair_extension_carries_two_by_three_configuration; no conjugation condition is needed here.

Proof. Take $x = (1, u^2)$ and $y = (b, bu^2, bu^4)$ with $b = 1/(u(u^2 - a_1)(u^2 - a_2))$. The six products are $b, bu^2, bu^4, bu^6$, each in $W$ by partial fractions, using $u^{-1} = \bar u/\rho$: $b = (a_1a_2)^{-1}u^{-1} + (a_1(a_1 - a_2))^{-1}z_1 + (a_2(a_2 - a_1))^{-1}z_2$, $bu^2 = (z_1 - z_2)/(a_1 - a_2)$, $bu^4 = (a_1z_1 - a_2z_2)/(a_1 - a_2)$, $bu^6 = u + (a_1^2z_1 - a_2^2z_2)/(a_1 - a_2)$. Both families are free because $u$ is transcendental over $\overline{\mathbb{Q}}$.

Novelty. Not asserted for the shape: it is the progression of Fischler (2001, Lemma 6.1) and Diaz (2007, Th. 7(2)), with ratio $u^2$.