A five-dimensional W ⊇ span(1, u, u⁻¹) has three independent t with t, ut ∈ W iff it contains u², u⁻², 1/(u − a), or is H₀ + Q̄g + Q̄ug

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 $W \subseteq \mathbb{C}$ be a $\overline{\mathbb{Q}}$-vector subspace of dimension $5$ with $1, u, u^{-1} \in W$, that is, $H_0 \subseteq W$ for $H_0 = \mathrm{span}(1, u, u^{-1})$. Then the following are equivalent:

• there are linearly independent $t_0, t_1, t_2 \in \mathbb{C}$ with $t_j \in W$ and $ut_j \in W$ for every $j$ (condition (D1));
• $u^2 \in W$, or $u^{-2} \in W$, or $(u - a)^{-1} \in W$ for some $a \in \overline{\mathbb{Q}}$ with $a \neq 0$, or there is $g \in \mathbb{C}$ such that $1, u, u^{-1}, g, ug$ are linearly independent and $W = \mathrm{span}(1, u, u^{-1}, g, ug)$.

The alternatives before the last say that $W$ is visible: it contains $u^2$, $u^{-2}$ or $(u - a)^{-1}$, the elements of Theorem B (when $u\bar u$ is algebraic, $u^{-2}$ is an algebraic multiple of $\bar u^2$). The last is the degree-one space $H_0 + \overline{\mathbb{Q}} g + \overline{\mathbb{Q}} ug$.

Proof. $H_0$ has dimension $3$, $u$ being transcendental. ($\Rightarrow$) If every $t_j$ lies in $H_0$, their span is $H_0$, so $u$ is a combination of the $t_j$ and $u \cdot u = u^2 \in W$. Otherwise take $g = t_j \notin H_0$. If $(u - c)g \in H_0$ for some $c \in \overline{\mathbb{Q}}$, write $(u - c)g = \alpha + \beta u + \gamma u^{-1}$: for $c = 0$, $\gamma \neq 0$ (else $g \in H_0$) and $u^{-2} = \gamma^{-1}(g - \beta - \alpha u^{-1}) \in W$; for $c \neq 0$, with $r = \alpha + \beta c + \gamma/c$ one has $(u - c)(g - \beta + (\gamma/c)u^{-1}) = r$, so $r \neq 0$ (else $g \in H_0$) and $(u - c)^{-1} \in W$. If there is no such $c$, then $1, u, u^{-1}, g, ug$ are independent (a relation with a non-zero coefficient of $ug$ or of $g$ would put some $(u - c)g$ or $g$ in $H_0$), and they span $W$ by dimension. ($\Leftarrow$) Take $t = (1, u, u^{-1})$, $(1, u^{-1}, u^{-2})$, $(1, u^{-1}, (u - a)^{-1})$ (using $u(u - a)^{-1} = 1 + a(u - a)^{-1}$) or $(1, u^{-1}, g)$; independence comes from the transcendence of $u$ (a vanishing polynomial of degree at most $2$) or from that of $1, u, u^{-1}, g, ug$.

Novelty. Not found in the sources read; routine.