If a non-zero α u² + β u³ (α, β algebraic) lies in the conjugation hull of z = Σ c_j (u/s)^j, the hull is Theorem F's two-orbit space D2a for the pencil (X², 1)

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 $s \in \overline{\mathbb{Q}}$ with $s^2 = u\bar u$, and put $v = u/s$. Let $k \geq 0$, let $c_j \in \mathbb{C}$ for $j \in \mathbb{Z}$ be arbitrary, put $z = \sum_{j=-k}^{k} c_jv^j$ and $W = \mathrm{span}(1, u, \bar u, z, \bar z)$. Let $\alpha, \beta \in \overline{\mathbb{Q}}$, not both zero, with $\alpha u^2 + \beta u^3 \in W$ (Theorem E's class (P)). Then there are $P', Q', S_1, S_2 \in \overline{\mathbb{Q}}[X]$ with $P'(u) = u^2$, $Q'(u) = 1$, $S_1(u) = \bar\beta s^6 + \bar\alpha s^4u$ and $S_2(u) = \alpha u + \beta u^2$, such that $P', Q'$ are coprime, $\max(\deg P', \deg Q') = 2$, $\deg S_1, \deg S_2 \leq 2$, and $W = \mathrm{span}(1, u, u^{-1}, Q'(u)S_1(u)/(uP'(u)), P'(u)S_2(u)/(uQ'(u)))$.

That is, $W$ has Theorem F's shape (D2a) with $P' = X^2$ and $Q' = 1$ ($u$ being transcendental, the values at $u$ determine the polynomials), and $W = \mathrm{span}(1, u, u^{-1}, \bar w, w)$ for $w = \alpha u^2 + \beta u^3$. No dimension hypothesis is made, and the $c_j$ are not assumed algebraic, as the proof does not need it.

Proof. As $u \neq 0$ and $s^2 = u\bar u$, $s \neq 0$ and $\bar u = s^2/u$. Write $w = \alpha u^2 + \beta u^3$; then $\bar w = (\bar\beta s^6 + \bar\alpha s^4u)/u^3$. Take $P' = X^2$ and $Q' = 1$ (coprime, of maximal degree $2$), $S_1 = \bar\beta s^6 + \bar\alpha s^4X$ and $S_2 = \alpha X + \beta X^2$, with coefficients in $\overline{\mathbb{Q}}$ because complex conjugation maps $\overline{\mathbb{Q}}$ to itself; then $Q'(u)S_1(u)/(uP'(u)) = \bar w$ and $P'(u)S_2(u)/(uQ'(u)) = w$. (1) The five vectors $1, u, u^{-1}, \bar w, w$ lie in $W$: $u^{-1} = s^{-2}\bar u$; $w \in W$ by hypothesis; and $W$ is stable under complex conjugation, since conjugation permutes its generators $1, u, \bar u, z, \bar z$ and maps $\overline{\mathbb{Q}}$ to itself, so $\bar w \in W$. (2) They are linearly independent: multiplying a relation $x_0 + x_1u + x_2u^{-1} + x_3\bar w + x_4w = 0$ by $u^3$ gives a polynomial in $u$ of degree at most $6$ with coefficients $x_3\bar\beta s^6, x_3\bar\alpha s^4, x_2, x_0, x_1, x_4\alpha, x_4\beta$, all algebraic; as $u$ is transcendental they vanish, and $(\alpha, \beta) \neq (0, 0)$, $s \neq 0$ give $x_3 = x_4 = 0$. (3) $W$ is spanned by five vectors, so $\dim W \leq 5$, and the inclusion of the five-dimensional span of (1) in $W$ is an equality. Nothing about $z$ is used beyond the hull, so the $c_j$ need not be algebraic.

Novelty. Not found in the sources read; routine. It writes Theorem E's class (P) in the coordinates of Theorem F: every hull of class (P) has shape (D2a), with the pencil $(X^2, 1)$. The configuration of class (P) is in print: Diaz (2007, Théorème 7(1)), already in Diaz (2004, Théorème 2).