No subspace of ℂ is in two of the polynomial, involution and real pencil classes of conjugation hulls

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and $s \in \overline{\mathbb{Q}}$ with $s^2 = u\bar u$ (so $s = \pm|u|$ is real), and put $v = u/s$, so that $\bar v = 1/v$. Let $W \subseteq \mathbb{C}$ be any $\overline{\mathbb{Q}}$-subspace. Then no two of the following hold:

• (P) $\alpha u^2 + \beta u^3 \in W$ for some $\alpha, \beta \in \overline{\mathbb{Q}}$, not both zero;
• (Q) $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, \bar v^2 + ev^2, \beta\bar v^3 + \zeta v^2 + ev^3)$ for some $e, \beta, \zeta \in \overline{\mathbb{Q}}$ with $e\bar e = 1$, $\beta\bar\beta = 1$ and $\zeta = -\beta e^2\bar\zeta$;
• (R) $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, g, \bar g)$ with $g = G(v) + \kappa\,\overline{vG(v)}$, for some $\kappa \in \overline{\mathbb{Q}}$ with $\kappa\bar\kappa = 1$ and some $G \in \overline{\mathbb{Q}}[X]$ with $G(0) = 0$ and $\deg G \geq 2$.

Proof. As $s$ is real, $v = u/s$ is transcendental with $\bar v = 1/v$, $u = sv$ and $\bar u = s/v$, so a relation in $v$ and $1/v$ with algebraic coefficients clears to a polynomial identity and can be compared coefficientwise. (P) against (Q): writing $\alpha u^2 + \beta u^3$ in the (Q) basis and multiplying by $v^3$ gives a degree-$6$ identity whose coefficients at $1, v, v^5, v^6$ force the (Q) coordinates, and then $\alpha, \beta$, to vanish. For (R), with $n = \deg G \geq 2$ and $H = \sum_{i \leq n}\bar G_iX^{n-i}$, so that $H(v) = v^n\overline{G(v)}$ and $H(0) = \bar G_n \neq 0$, evaluating a cleared identity at $0$ kills the coordinate of $g$, and then clearing by $v^n$ kills that of $\bar g$. So (P) and (R) exclude each other, since both coordinates of $\alpha u^2 + \beta u^3$ vanish and then $\alpha = \beta = 0$; and so do (Q) and (R), since applied to $\bar v^2 + ev^2$ and its conjugate both coordinates vanish, while $\bar v^2 + ev^2 \notin \mathrm{span}(1, v, \bar v)$.

Novelty. Not found in the sources read; routine.