For u ∉ Q̄ and non-zero algebraic a, b, α, β, γ, δ with a² ≠ b², the span of 1, u, u⁻¹, α/(u − a) + β/(u + a), γ/(u − b) + δ/(u + b) is five-dimensional and excludes u², u⁻², 1/(u − c)

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$, let $a, b \in \overline{\mathbb{Q}}^\times$ with $a^2 \neq b^2$ and $\alpha, \beta, \gamma, \delta \in \overline{\mathbb{Q}}^\times$, put $g_1 = \alpha/(u - a) + \beta/(u + a)$, $g_2 = \gamma/(u - b) + \delta/(u + b)$, and let $W$ be the $\overline{\mathbb{Q}}$-span of $1, u, u^{-1}, g_1, g_2$. Then the five numbers are free over $\overline{\mathbb{Q}}$, and $W$ contains none of $u^2$, $u^{-2}$ and $1/(u - c)$ for non-zero $c \in \overline{\mathbb{Q}}$.

Proof. One residue lemma: if $c_0 + c_1u + c_2u^{-1} + d/(u - p) + \sum_i e_i/(u - q_i) = 0$ with the $q_i \neq p$ and all coefficients in \overline{\mathbb{Q}}, clearing denominators gives a polynomial vanishing at the transcendental $u$, hence zero, and its value at $p$ gives $d\prod_i(p - q_i) = 0$. Applied at the five distinct poles $0, \pm a, \pm b$: in a vanishing combination the residues $c_2$, $\lambda\alpha$, $\mu\gamma$ vanish, so the five numbers are free. A membership of $u^2$ or $u^{-2}$ would leave $u$ algebraic or a residue $-1$ at $0$; one of $1/(u - e)$ forces, at $e = \pm a$ (or $\pm b$), both residues of that pair to match $(1, 0)$, impossible since $\alpha, \beta \neq 0$, and for any other $e$ a sixth pole with residue $-1$.

Novelty. Not found in the sources read; it is short.