For u ∉ Q̄ and the pencil of any Möbius involution, the space Q̄ + Q̄u + Q̄u⁻¹ + Σ Q̄ q₁(u)ⁱq₂(u)³⁻ⁱ/(u l₁(u)l₂(u)) carries a 2×3 configuration

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$. Let $q_1, q_2 \in \overline{\mathbb{Q}}[X]$ be coprime of degree at most $2$, one of them of degree exactly $2$, so that $\Lambda = \overline{\mathbb{Q}}q_1 + \overline{\mathbb{Q}}q_2$ is the pencil of quadratics of a Möbius involution $\tau$ of the $u$-line (its members vanish on the $\tau$-orbits). Let $l_1, l_2 \in \Lambda$ be non-zero, and let $W$ be the $\overline{\mathbb{Q}}$-span of $1, u, u^{-1}$ and of $q_1(u)^iq_2(u)^{3-i}/(u\,l_1(u)l_2(u))$, $i = 0, \dots, 3$. Then there are $x_1, x_2$, free over $\overline{\mathbb{Q}}$, and $y_1, y_2, y_3$, free over $\overline{\mathbb{Q}}$, with all six products $x_iy_j$ in $W$: $W$ contains the progression $c, cr, cr^2, cr^3$ with ratio $r = q_1(u)/q_2(u)$.

Proof. Put $r = q_1(u)/q_2(u)$ and $c = q_2(u)^3/(u\,l_1(u)l_2(u))$; then $c\,r^k$, $k = 0, \dots, 3$, are the four extra spanning elements, so $x = (1, r)$, $y = (c, cr, cr^2)$ has all six products in $W$. Since $u$ is transcendental, $q_2(u)$, $l_i(u)$ and $u$ are non-zero, and $r \notin \overline{\mathbb{Q}}$: otherwise $q_1 = r q_2$ as polynomials, against coprimality and the degree condition. So $r$ is transcendental, and $1, r, r^2$ are free.

Novelty. Not asserted for the shape: it is the progression of Fischler (2001, Lemma 6.1) and Diaz (2007, Th. 7(2)); that a configuration whose products span four dimensions is a progression is Bachoc–Serra–Zémor (2017, Lemma 4).