For u ∉ Q̄, c ≠ 0 and h ∉ Q̄: if 1, u, u⁻¹ lie in the Q̄-span of c, ch, ch², ck, chk, then u is a root of a non-zero quadratic over Q̄[h]

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and $c, h, k \in \mathbb{C}$ with $c \neq 0$, and suppose $1, u, u^{-1}$ lie in the $\overline{\mathbb{Q}}$-span of $c, ch, ch^2, ck, chk$. If $h \notin \overline{\mathbb{Q}}$, there are $A, B, C \in \overline{\mathbb{Q}}[X]$, not all zero, with $A(h)u^2 + B(h)u + C(h) = 0$.

A configuration $(1, h) \otimes Y$ whose products span a five-dimensional $W \supseteq \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}u^{-1}$ gives, after normalisation, $W = c\cdot\mathrm{span}(1, h, h^2, k, hk)$ (research notes R12, Q1; not formalised). So no such configuration has a ratio of degree $3$ or more.

Proof. Unpacking the three memberships gives $1 = c(a_0(h) + k\,b_0(h))$, $u = c(a_1(h) + k\,b_1(h))$ and $u^{-1} = c(a_2(h) + k\,b_2(h))$ with $a_i, b_i \in \overline{\mathbb{Q}}[X]$: three vectors in the two-dimensional $\overline{\mathbb{Q}}(h)$-span of $c$ and $ck$. Write $D_{ij} = a_ib_j - a_jb_i$. If $D_{01} \neq 0$, the identity $D_{12}\,w_0 - D_{02}\,w_1 + D_{01}\,w_2 = 0$ for $w_i = a_i(h) + k\,b_i(h)$, together with $uu^{-1} = 1$, gives $(A, B, C) = (-D_{02}, D_{12}, D_{01})$. If $D_{01} = 0$, then $a_0(h)u = a_1(h)$ and $b_0(h)u = b_1(h)$, and $a_0$ or $b_0$ is non-zero (otherwise $1 = 0$), which gives a relation of degree one. Only $u \neq 0$ is used; $h \notin \overline{\mathbb{Q}}$ makes the relation non-trivial as an equation in $u$.

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