The hull span(1, v, v̄, z, z̄) of a Laurent polynomial z in v, if of dimension at most four, carries no 2×3 configuration

Let $v \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ with $v\bar v = 1$. Let $k \geq 0$, let $c_j \in \overline{\mathbb{Q}}$ for all $j \in \mathbb{Z}$, and put $z = \sum_{j=-k}^{k} c_jv^j$ and $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, v, \bar v, z, \bar z)$. If $\dim_{\overline{\mathbb{Q}}} W \leq 4$, then $W$ carries no $2 \times 3$ 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).

Proof. Let $H_0 = \mathrm{span}(1, v, \bar v)$. If $z \in H_0$, then $\bar z \in H_0$ and $W \subseteq H_0$, the polynomial hull of $P = 0$, where Theorem C(iii) (DiazModulus.polynomial_hull_no_two_by_three_configuration) excludes a configuration. Otherwise $1, v, \bar v, z$ are free, so $W = H_0 + \overline{\mathbb{Q}}z$ has dimension $4$, and a configuration would, by Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff), put $z$ in $H_0 + \overline{\mathbb{Q}}w$ with a non-zero coefficient $\delta$ on some $w \in \{v^2, \bar v^2, (v - a)^{-1}\}$, $a \neq 0$. For $w = v^2$ or $\bar v^2$, the relation $\bar z \in \mathrm{span}(1, v, \bar v, z)$ times $v^2$ is a quartic identity at the transcendental $v$ whose constant, resp. top, coefficient is $\bar\delta$, so $\delta = 0$. For $w = (v - a)^{-1}$, $v^kz = F(v)$ gives $X(X - a)F = (X - a)B + \delta X^{k+1}$ in $\overline{\mathbb{Q}}[X]$, which at $X = a$ reads $0 = \delta a^{k+1}$.

Novelty. Not found in the sources read; routine, as a consequence of Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff) and Theorem C(iii) (DiazModulus.polynomial_hull_no_two_by_three_configuration).