Over K ≤ F, a 2×3 configuration in V₀ + Ku + Ku⁻¹, with V₀ ⊆ F and u transcendental over F, has its products in V₀

Let $K \subseteq F$ be subfields of $\mathbb{C}$, $V_0$ a $K$-subspace of $\mathbb{C}$ contained in $F$, and $u$ transcendental over $F$. If free $x_1, x_2$ and free $y_1, y_2, y_3$ over $K$ have all six products in $V_0 + Ku + Ku^{-1}$, then all six products lie in $V_0$.

Proof. Write $x_iy_j = c_{ij} + l_{ij}u + m_{ij}u^{-1}$ with $c_{ij} \in V_0$ and $l_{ij}, m_{ij} \in K$, and $P_{ij} = l_{ij}X^2 + c_{ij}X + m_{ij} \in F[X]$, so $P_{ij}(u) = ux_iy_j$. The $2 \times 2$ minors of $P$ vanish at $u$, hence vanish, and DiazModulus.rank_one_polynomial_matrix_factor gives $P_{ij} = a_ib_j$; the $a_i(u)$ are $K$-free, and so are the $b_j(u)$, being non-zero multiples of $x$ and $y$. If some $m_{ij} \neq 0$: when every $a_i$ is constant, the two $a_i(u)$ lie on one $K$-line; otherwise every $b_j$ has degree at most $1$ and, using that the constant and $X^2$ coefficients of $a_ib_j$ lie in $K$, the three $b_j(u)$ lie in a $K$-span of two numbers. Both are impossible, so every $m_{ij} = 0$; replacing $u$ by $u^{-1}$, which is also transcendental over $F$, gives every $l_{ij} = 0$. So $x_iy_j = c_{ij} \in V_0$.

Novelty. Not found in the sources read in this form. Waldschmidt (Diophantine Approximation on Linear Algebraic Groups, Prop. 12.2 and the Remark on p. 424) covers $V_0 = \overline{\mathbb{Q}}$ without $u^{-1}$; the factorisation and degree argument is that of DiazModulus.generic_circle_point_no_two_by_three_configuration. Routine.