Let $C$ be a connected curve of arithmetic genus $g$
over algebraically closed field $k$ of characteristic zero having only nodes
as singularities together with finite morphism
$f: C \to \mathbb{P}^1$.


In [3264 and All That](https://www.cambridge.org/core/books/3264-and-all-that/DC062983CC5F8B7CDD37CFEBCCA5FEA4) by Eisenbud and Harris is claimed on page 313 that the the
automorphism group of the map $f$ — that is, automorphisms $\phi$ of $C$ such that
$f \circ \phi =f$ — is finite.


**Q1**: Why and how to prove it? If $C$ is smooth, then it follows immediately
from Hurwitz's automorphisms theorem. Clearly it suffice to check
that for every $q \in f^{-1}(p)$ the stabilizer of $q$ is finite. Why is
it true?


**Q2**: Is the automorphism group of $f$ also finite if we weaken the assumptions
on base field $k$? For example if we allow positive cahracteristic?



Finiteness of automorphism group of finite map $f: C \to \mathbb{P}^1$
Q1: since maps of curves are determined by their action on the function field, we get an injection $\mathrm{Aut}(C/\mathbb{P}^1) \to \mathrm{Aut}(k(C)/k(t))$ but the latter is just a Galois group (or automorphism group of a field extension if you don't like this terminology for a non-Galois extension) of a finite extension of fields and hence is finite.


In particular, this shows that $|\mathrm{Aut}(C/\mathbb{P}^1)| \le \deg{f}$ where equality is achieved for so-called Galois covers (since for $C$ smooth it is equivalent to $k(C) / k(\mathbb{P}^1)$ Galois). This argument does not depend on the base curve being $\mathbb{P}^1$, just that you have a nonconstant map of (possibly singular) curves.


Q2: the above proof works over any field.


