For u ∉ Q̄, Q̄u⁻¹ + Q̄ + Q̄u + Q̄u³ + Q̄u⁴ is spanned by the products of (1, u) and (u⁻¹, 1, u³), and contains none of u², u⁻², 1/(u − b)

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$ and $V = \overline{\mathbb{Q}}u^{-1} + \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}u^3 + \overline{\mathbb{Q}}u^4$. Then $(1, u)$ and $(u^{-1}, 1, u^3)$ are free over $\overline{\mathbb{Q}}$, their six products span $V$, and $V$ contains none of $u^2$, $u^{-2}$ and $1/(u - b)$ ($b \in \overline{\mathbb{Q}}$ non-zero).

So $V$ carries a $2 \times 3$ configuration whose products fill the whole five-dimensional space (the second case of the classification, not a geometric progression), and no four-dimensional $H_0 + \overline{\mathbb{Q}}w \subseteq V$ is one of Theorem B's. $V$ is not stable under conjugation, so it says nothing at a candidate; it shows that the conjugation condition is needed in the conjecture that every such space is either visible to Theorem B or contains a progression.

Proof. Distinct powers of the transcendental $u$ are free, so $(1, u)$ and $(u^{-1}, 1, u^3)$ are free; the six products are $u^{-1}, 1, u^3, 1, u, u^4$. A membership of $u^2$, $u^{-2}$ or $1/(u - b)$ in the span of $u^{-1}, 1, u, u^3, u^4$ clears to a polynomial identity: for $u^2$ the coefficient of $X^3$ is $-1$, for $u^{-2}$ the constant term is $1$, and for $1/(u - b)$ the identity $u = (u - b)P(u)$ fails at $b \neq 0$.

Novelty. Not found in the sources read; it is short.