For z ≡ a v̄³ + b v̄² + c v² + d v³ mod span(1, v, v̄) with a five-dimensional hull, the hull contains a non-zero α v² + β v³ exactly when a c̄ = b d̄

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $a, b, c, d \in \overline{\mathbb{Q}}$ and let $z \in \mathbb{C}$ with $z - (a\bar v^3 + b\bar v^2 + cv^2 + dv^3) \in \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v)$, and let $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, z, \bar z)$ with $\dim_{\overline{\mathbb{Q}}} W = 5$. Then $\alpha v^2 + \beta v^3 \in W$ for some $\alpha, \beta \in \overline{\mathbb{Q}}$, not both zero, if and only if $a\bar c - b\bar d = 0$.

Proof. Let $m = a\bar v^3 + b\bar v^2 + cv^2 + dv^3$. As $\mathrm{span}(1, v, \bar v)$ is conjugation-stable, $\bar z \equiv \bar m = \bar d\bar v^3 + \bar c\bar v^2 + \bar bv^2 + \bar av^3$, so $W = \mathrm{span}(1, v, \bar v, m, \bar m)$ and $\dim W = 5$ makes these five free; also $v^{-3}, \dots, v^3$ are free over $\overline{\mathbb{Q}}$, as $v$ is transcendental. So an element $x_0 + x_1v + x_2\bar v + xm + y\bar m$ of $W$ equals some $\alpha v^2 + \beta v^3$ only if its $\bar v^3$ and $\bar v^2$ coefficients $xa + y\bar d$ and $xb + y\bar c$ vanish. If such an element with $(\alpha, \beta) \neq (0, 0)$ exists, then $(x, y) \neq (0, 0)$, and Cramer's rule gives $(a\bar c - b\bar d)x = (a\bar c - b\bar d)y = 0$, so $a\bar c - b\bar d = 0$. Conversely, if $a\bar c = b\bar d$, take $(x, y) = (\bar d, -a)$, or $(\bar c, -b)$ when $a = d = 0$; then $xm + y\bar m = \alpha v^2 + \beta v^3$ with $(\alpha, \beta) \neq (0, 0)$, because $m, \bar m$ are free.

Novelty. Not found in the sources read; routine.