If span(1, v, v̄, z, z̄) contains v² P(v) with P ≠ 0 and carries a 2×3 configuration, it contains a non-zero α v² + β v³

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $z \in \mathbb{C}$ and $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, z, \bar z)$. Suppose that $v^2P(v) \in W$ for some non-zero $P \in \overline{\mathbb{Q}}[X]$, and that $W$ carries a $2 \times 3$ configuration (free $x_0, x_1$ and free $y_0, y_1, y_2$ over $\overline{\mathbb{Q}}$ with all six products $x_iy_j$ in the space). Then $\alpha v^2 + \beta v^3 \in W$ for some $\alpha, \beta \in \overline{\mathbb{Q}}$, not both zero.

Proof. Put $A = v^2P(v)$; as $W$ is conjugation-stable, $W' = \mathrm{span}(1, v, \bar v, A, \bar A) \subseteq W$. A relation among these five, multiplied by $v^{d+2}$ with $d = \deg P$, is a polynomial identity at the transcendental $v$ (with $\bar v = 1/v$ and the reflected polynomial for $\overline{P(v)}$); its coefficient at $X^{2d+4}$ kills the coordinate of $A$, the conjugate relation kills that of $\bar A$, and the rest is trivial, so $\dim W' = 5$. As $\dim W \leq 5$, $W' = W$, and the configuration lies in the polynomial hull of $Q = X^2P$ at $u := v$, where $u\bar u = 1$ is algebraic. Theorem C (DiazModulus.polynomial_hull_two_by_three_configuration_iff) gives $\deg Q \in \{2, 3\}$, so $P = \alpha + \beta X$ with $(\alpha, \beta) \neq (0, 0)$, and $\alpha v^2 + \beta v^3 = v^2P(v) \in W$.

Novelty. Not found in the sources read; routine, as a consequence of Theorem C (DiazModulus.polynomial_hull_two_by_three_configuration_iff).