Homogeneous exhaustion of the forced conjugation plane of a candidate

Nothing homogeneous distinguishes a candidate inside its own forced plane.

Let $K \subseteq \mathbb{C}$ be a subfield, $u \neq 0$ transcendental over $K$ with $u \bar u \in K$ (the algebraic shape of a Diaz candidate: $\rho = u\bar u$ lies in the base field). If $c_0, \ldots, c_d \in K$ satisfy

then every $c_i$ vanishes.

Why. Because $u\bar u \in K$ and $u$ is transcendental over $K$, the ratio $u/\bar u = u^2/\rho$ is again transcendental over $K$: if it were algebraic then $u^2$ would be, and hence $u$. The binary-form lemma then applies.

Role. This is the coordinate form of the manuscript's Homogeneous exhaustion of the forced plane: a homogeneous polynomial over the base field vanishing at a tuple $x_j = a_j u + b_j \bar u$ of points of $\Lambda_u = \mathbb{Q}u + \mathbb{Q}\bar u$ already vanishes identically on the corresponding $\mathbb{Q}$-rational plane. Nothing is gained by adding homogeneous coordinates drawn from the plane that conjugation forces on a candidate; the plane is exhausted. Together with the coordinate-invariant homogenization barrier this is why the missing non-logarithmic homogenizer $1$ cannot be manufactured from inside the candidate's own data.

Source: Carlo Perassi's private manuscript on Diaz's modulus conjecture, section The precise open boundary, proposition Homogeneous exhaustion of the forced plane. No novelty is claimed; the argument is elementary given the transcendence hypothesis.