No vanishing statement over $\bar{\mathbb{Q}}$ separates a Diaz candidate from an ordinary point of its circle

No vanishing statement with algebraic coefficients distinguishes a candidate from an ordinary point of the same circle.

Call $u \in \mathbb{C}$ a candidate for Diaz's conjecture when $u \neq 0$, $e^{u}$ is algebraic over $\mathbb{Q}$, and $u\bar u = r^{2}$ for some non-zero real algebraic $r$; then $r = |u|$. Diaz's conjecture is that no candidate exists, and nothing here bears on that.

The claim is that there exist a point $t$ of the same circle — $t \neq 0$, $t$ transcendental over $\bar{\mathbb{Q}}$, and $t\bar t = r^{2}$ — and a ring homomorphism $\Phi : \mathbb{C} \to \mathbb{C}$ with

such that for every matrix $M$ with entries in $\bar{\mathbb{Q}}(u)$ and all vectors $w$, $v$ with entries in $\bar{\mathbb{Q}}$, the transported matrix $\Phi(M)$ again has entries in $\bar{\mathbb{Q}}(t)$ and

Here $\bar{\mathbb{Q}}(u)$ is the smallest subfield of $\mathbb{C}$ containing the algebraic numbers and $u$, written hull Qbar u.

Reading. The four clauses pinning $\Phi$ down are what makes the last one say something. A ring homomorphism of $\mathbb{C}$ fixing $\bar{\mathbb{Q}}$ and sending $u$ to $t$ is determined on $\bar{\mathbb{Q}}(u)$ (Diaz.eqOn_hull), so an entry of $\Phi(M)$ is the expression defining the corresponding entry of $M$, read with $t$ in place of $u$ — conjugations included, since $\Phi$ intertwines conjugation there. So the two sides of the equivalence are literally the same vanishing statement, evaluated at the candidate and at the ordinary point. No such statement can separate them.

Proof. From Diaz.candidate_indistinguishable one gets $t$ and $\Phi$ with the first three clauses. That $\Phi$ maps $\bar{\mathbb{Q}}(u)$ into $\bar{\mathbb{Q}}(t)$ is a closure argument: $\Phi^{-1}\bigl(\bar{\mathbb{Q}}(t)\bigr)$ is a subfield of $\mathbb{C}$ containing $\bar{\mathbb{Q}}$ and $u$, and $\bar{\mathbb{Q}}(u)$ is the smallest such. That $t\bar t = r^{2}$ comes from applying $\Phi$ to $u\bar u = r^{2}$: the right side is algebraic and so fixed, and the left side goes to $\Phi(u)\Phi(\bar u) = t\,\bar t$ by the intertwining. The equivalence is Corollary 7 of the note, applied to this $\Phi$ with $K = \bar{\mathbb{Q}}$.

Relation to what is already on the platform. Diaz.candidate_no_vanishing_coeff_Qbar is the specialised case: one specific $2 \times 2$ matrix, $H = \begin{pmatrix} u & r \\ r & \bar u\end{pmatrix}$, with $\mathrm{Fin}\,2$ coefficient vectors, and a non-vanishing conclusion. Diaz.coeff_transfer is the forward identity for coefficients. This node is the statement those two were specialisations and fragments of, and is the sentence the source note asserts in prose and does not formalise. Note also that $t\bar t = r^{2}$ is stated in the note's Theorem 6 but absent from the platform node Diaz.candidate_indistinguishable; it is recovered here, and without it the statement could not say "the same circle".

What is deliberately not claimed. Nothing about $e^{t}$. If $e^{t}$ were algebraic then $t$ would itself be a candidate, so asserting its transcendence would be asserting an instance of the conjecture. And nothing here advances or retreats from Diaz's conjecture; it constrains the methods that could settle it.

Attribution. The mathematics is Carlo Perassi's, from a note predating this mission (Corollary 7, cor:transfer, and the paragraph immediately following it). No novelty is claimed for it; the contribution is the formalisation.