For a ratio h = p(u)/q(u) of degree 2, 1 and w = J(u)/q(u), with J the Wronskian, are linearly independent over Q̄(h)

Let $u \in \mathbb{C}$ be transcendental, that is, $u \notin \overline{\mathbb{Q}}$ (the algebraic numbers); spans and linear independence are over $\overline{\mathbb{Q}}$. Let $p, q \in \overline{\mathbb{Q}}[X]$ be coprime with $\max(\deg p, \deg q) = 2$, let $J = p'q - pq'$ be their Wronskian, and put $h = p(u)/q(u)$ and $w = J(u)/q(u)$. If $A, B \in \overline{\mathbb{Q}}[X]$ satisfy $A(h) + B(h)\,w = 0$, then $A = 0$ and $B = 0$.

Clearing denominators, this says that $1$ and $w$ are linearly independent over the field $\overline{\mathbb{Q}}(h)$.

Proof. Coprimality and degree give $q \neq 0$, so $q(u) \neq 0$. Also $h \notin \overline{\mathbb{Q}}$: otherwise $p - hq$ is a polynomial over $\overline{\mathbb{Q}}$ vanishing at the transcendental $u$, so $p = hq$, coprimality makes $q$ a unit, and both have degree $0$; hence $h$ is transcendental. Evaluating the second identity of DiazModulus.quadratic_pencil_wronskian_identities at $u$ and dividing by $q(u)^2$ gives $w^2 = \Delta(h)$, with $\Delta = d_2X^2 + d_1X + d_0$ built from the coefficients of that identity. If $A(h) + B(h)w = 0$, then $A(h)^2 = B(h)^2\Delta(h)$, so $A^2 = B^2\Delta$ in $\overline{\mathbb{Q}}[X]$, $h$ being transcendental. If $B = 0$ this gives $A = 0$. Otherwise $B^2$ divides $A^2$, so $B$ divides $A$, and $\Delta = R^2$ with $R = r_1X + r_0$ of degree at most $1$. Comparing coefficients gives $d_1^2 - 4d_2d_0 = 0$; but $d_1^2 - 4d_2d_0 = 16\,\mathrm{Res}$, with $\mathrm{Res} = [20]^2 - [21][10]$, which DiazModulus.quadratic_pencil_resultant_ne_zero says is non-zero.

Novelty. Not found in the sources read; routine.