For u ∉ Q̄ and a Möbius function h of u: if Q̄ + Q̄u + Q̄u⁻¹ + Σ Q̄ c hᵏ (k ≤ 3) is five-dimensional, it contains u², u⁻² or some 1/(u − b), b ≠ 0

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$, let $\alpha, \beta, \gamma, \delta \in \overline{\mathbb{Q}}$ with $\alpha\delta - \beta\gamma \neq 0$, put $h = (\alpha u + \beta)/(\gamma u + \delta)$, and let $c \in \mathbb{C}$ be non-zero. If the $\overline{\mathbb{Q}}$-span $W$ of $1, u, u^{-1}$ and $c, ch, ch^2, ch^3$ has dimension $5$, then $W$ contains $u^2$, $u^{-2}$ or $1/(u - b)$ for some non-zero $b \in \overline{\mathbb{Q}}$.

This is the index-one half of the classification of the progression spaces (Case I): when the ratio $h$ of the progression generates $\overline{\mathbb{Q}}(u)$, the space contains one of the elements of Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff), so when $u\bar u$ is algebraic Theorem B sees it. With R9's involution spaces (the index-two half) it accounts for every five-dimensional space containing a $4$-term progression; that the index is $1$ or $2$ is proved in the research notes and is not formalised.

Proof. Put $A = \alpha u + \beta$, $G = \gamma u + \delta \neq 0$ and $c' = c/G^3$, so $ch^k = c'A^kG^{3-k}$. From $\delta A - \beta G = Du$ and $\alpha G - \gamma A = D$ ($D = \alpha\delta - \beta\gamma$) every $c'u^j$, $j \leq 3$, lies in $W$. The spaces $H = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}u^{-1}$ and $P = \sum_{j \le 3} \overline{\mathbb{Q}}c'u^j$ have dimensions $3$ and $4$, and $H + P \subseteq W$ has dimension at most $5$, so $H \cap P \neq 0$: there are non-zero polynomials $p$, $r$ of degrees at most $3$ and $2$ with $F\,p(u) = r(u)$, $F = c'u$. Induction on $\deg p$: if $p$ is constant, $F$ is a polynomial in $u$ and $u^3 \in F\cdot\overline{\mathbb{Q}}[u]_{\le 3} + \overline{\mathbb{Q}}[u]_{\le 2}$; otherwise $p$ has a root $b \in \overline{\mathbb{Q}}$, and either $r(b) = 0$ and $u - b$ cancels, or $(u - b)^{-1} \in F\cdot\overline{\mathbb{Q}}[u]_{\le 3} + \overline{\mathbb{Q}}[u]_{\le 2}$. Dividing by $u$ gives $u^2$, $u^{-2}$ ($b = 0$) or $1/(u(u - b))$, hence $1/(u - b)$, in $W$. Only the dimension bound is used, not the equality.

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