Corollary E(a): if a polynomial in u and ū has an exponent of absolute value at least four, its hull carries a 2×3 configuration exactly in the real pencil class

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 $k \geq 0$, let $c_j \in \overline{\mathbb{Q}}$ for all $j \in \mathbb{Z}$, and put $z = \sum_{j=-k}^{k} c_jv^j$ and $W = \mathrm{span}_{\overline{\mathbb{Q}}}(1, u, \bar u, z, \bar z)$. Suppose $c_j \neq 0$ for some $j$ with $4 \leq |j| \leq k$. Then $W$ 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) if and only if $W$ is in the class

• (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. If: class (R) gives $\dim W = 5$ (DiazModulus.laurent_conj_hull_class_finrank_eq_five), and Theorem E (DiazModulus.laurent_conj_hull_two_by_three_configuration_iff) gives the configuration. Only if: Theorem E gives $\dim W = 5$ and one of the classes (P), (Q), (R). In (Q), $z \in W$ lies in the window $\mathrm{span}(v^{-3}, \dots, v^3)$. In (P), $m = \alpha u^2 + \beta u^3 \in W$ is itself a Laurent polynomial in $v$ of class (P), so its hull, contained in $W$, is five-dimensional by DiazModulus.laurent_conj_hull_class_finrank_eq_five, hence equal to $W$, and again $z$ lies in the window. Clearing by $v^k$ and comparing the coefficients of $\sum c_jX^{j+k}$ then gives $c_j = 0$ for $|j| \geq 4$, a contradiction.

Novelty. Not found in the sources read; routine, given Theorem E (DiazModulus.laurent_conj_hull_two_by_three_configuration_iff).