No holomorphic stabilizer: a Möbius map over $K$ fixing a transcendental point is the identity

Source. This is Carlo Perassi's mathematics, from his unpublished note on Diaz's modulus conjecture, section The conjugation-degree framework, statement Lemma (No holomorphic stabilizer). Published on his mission with his permission. No novelty is claimed for it here; the argument is elementary, and it is possibly known — it has not been checked against the literature.

Statement. Let $K \subset \mathbb{C}$ be a subfield and $z$ transcendental over $K$. If $g(t) = (at+b)/(ct+d)$ has coefficients in $K$ and fixes $z$, then $c = b = 0$ and $a = d$ — that is, $g$ is the identity of $\mathrm{PGL}(2,K)$.

What it is for. The conjugation-degree framework studies a candidate through the action of $\Gamma = \mathrm{PGL}(2,\overline{\mathbb{Q}})$ and its extension by complex conjugation on $\mathbb{C}\setminus\overline{\mathbb{Q}}$. This lemma says the holomorphic part of that action is free on transcendental points, which is what makes the anti-holomorphic stabilizer of a point of conjugation degree one exactly $\{\mathrm{id}, R\}$ with $R$ the reflection in its canonical circle — and hence what makes the reflection $R_\rho(z) = \rho/\bar z$ the unique non-identity element of the stabilizer of a Diaz candidate. Note that non-degeneracy of $g$ is not needed: the conclusion already forces $ad - bc = a^2$.

Proof. Clearing the denominator, $az + b = z(cz+d)$, i.e. $cz^2 + (d-a)z + (-b) = 0$, a quadratic relation for $z$ with coefficients in $K$. Since $z$ is transcendental over $K$, the polynomial $cX^2 + (d-a)X - b \in K[X]$ is zero, so $c = 0$, $d = a$ and $b = 0$.