For a candidate u, the numbers 1, u, ū are linearly independent over the algebraic numbers

Let $u$ be a candidate: $u \neq 0$ with $|u|$ and $e^{u}$ algebraic. If $p, q$ are algebraic numbers and $pu + q\bar u$ is algebraic, then $p = q = 0$. Equivalently, $1, u, \bar u$ are linearly independent over $\overline{\mathbb{Q}}$.

The conclusion is the hypothesis hbaker of DiazModulus.no_first_order_arithmetic_operator, here proved.

Proof. A candidate lies on neither axis: on an axis $u^2 = \pm|u|^2$ would be algebraic, against Hermite–Lindemann. Hence $u$ and $\bar u$, both logarithms of algebraic numbers, are linearly independent over $\mathbb{Q}$, since $su + t\bar u = 0$ gives $(s + t)\,\mathrm{Re}\,u = 0$ and $(s - t)\,\mathrm{Im}\,u = 0$. Baker's theorem with a constant term (Schanuel.baker_linear_forms_in_logarithms, with $n = 2$) applied to $-(pu + q\bar u) + p\,u + q\,\bar u = 0$ then forces every coefficient to vanish.

Novelty. None: an instance of Baker's theorem.