In a Q̄-space of rational functions of u, the ratio x₁/x₀ of a 2×3 configuration is p(u)/q(u) with p, q coprime and not both constant

Let $u \in \mathbb{C}$ be transcendental, that is, $u \notin \overline{\mathbb{Q}}$ (the algebraic numbers); spans and linear independence are over $\overline{\mathbb{Q}}$. Let $W$ be a $\overline{\mathbb{Q}}$-subspace of $\mathbb{C}$ all of whose elements lie in $\overline{\mathbb{Q}}(u)$, the subfield of $\mathbb{C}$ generated by $\overline{\mathbb{Q}}$ and $u$, and let $x_0, x_1 \in \mathbb{C}$ be linearly independent and $y_0, y_1, y_2 \in \mathbb{C}$ linearly independent, with $x_iy_j \in W$ for all $i, j$ (a $2 \times 3$ configuration in $W$). Then there are coprime $p, q \in \overline{\mathbb{Q}}[X]$ with $\max(\deg p, \deg q) \geq 1$ and $x_1q(u) = x_0p(u)$.

That is, the ratio $x_1/x_0$ is $p(u)/q(u)$ in lowest terms, and it is not an algebraic number.

Proof. Both $x_0y_0$ and $x_1y_0$ lie in $W \subseteq \overline{\mathbb{Q}}(u)$, and $x_0, y_0 \neq 0$ because the families are independent, so $x_1/x_0 = (x_1y_0)/(x_0y_0)$ lies in the field $\overline{\mathbb{Q}}(u)$. So it is $r(u)/s(u)$ for polynomials $r, s$ over $\overline{\mathbb{Q}}$, with $s(u) \neq 0$ since $x_1/x_0 \neq 0$. Dividing $r$ and $s$ by $g = \gcd(r, s)$ gives coprime $p = r/g$ and $q = s/g$, and $x_1q(u) = x_0p(u)$ after cancelling $g(u)$, which is non-zero because $u$ is transcendental. If $p$ and $q$ were constants $\alpha$ and $\beta$, then $\alpha x_0 - \beta x_1 = 0$ would force $\alpha = \beta = 0$ by the independence of $x$, and $0, 0$ are not coprime; so $\max(\deg p, \deg q) \geq 1$.

Novelty. Not found in the sources read; routine.