  Let $S$ be a closed Riemann surface of genus $g \geq 2$ and $\varphi$ be a
conformal automorphism of $S$ of prime order $p$ such that $S/\langle \varphi
\rangle$ has genus zero. Let ${\mathbb K} \leq {\mathbb C}$ be a field of
definition of $S$. We provide an argument for the existence of a field
extension ${\mathbb F}$ of ${\mathbb K}$, of degree at most $2(p-1)$, for which
$S$ is definable by a curve of the form $y^{p}=F(x) \in {\mathbb F}[x]$, in
which case $\varphi$ corresponds to $(x,y) \mapsto (x,e^{2 \pi i/p} y)$. If,
moreover, $\varphi$ is also definable over ${\mathbb K}$, then ${\mathbb F}$
can be chosen to be at most a quadratic extension of ${\mathbb K}$. For $p=2$,
that is when $S$ is hyperelliptic and $\varphi$ is its hyperelliptic
involution, this fact is due to Mestre (for even genus) and Huggins and
Lercier-Ritzenthaler-Sijslingit in the case that ${\rm Aut}(S)/\varphi\rangle$
is non-trivial.
