When are finite maps quotients by finite groups?

Question

Let $f: X \to Y$ be a finite map of projective varieties.

I'm trying to understand when and how often should i expect $f$ to be a quotient map by a finite group acting on $X$. Even more strictly let $G=Aut(X/Y)$.

1. When is $f$ isomorphic to the quotient map by an action $X \to X^G$?

If $X$ and $Y$ are smooth projective curves then I think this always holds but I haven't been able to formalize this. Are there any useful conditions which ensure that (1) holds?

EDIT: I deleted the irrelevant part of the question. I state it as a different question: When is $Aut(X/Y)=Aut(k(X)/k(Y))$?

Answer 1

If $X$ is normal and $G$ acts on $k(X)/k(Y)$ then the $G$ acts also on $X/Y$ (in a way consistent with its action on $k(X)$).

Because $X$ is integral, its ring of functions on each affine open embeds into $k(X)$, so the action of $G$ on an affine open is determined by the action on the field of fractions. So the $G$-action is unique. Because of this, the existence is a local question, so we may assume $X$ and $Y$ affine.

In this case $k[X]$ consists of all elements of $k(X)$ that are integral over $k[X]$. Because each element of $k[X]$ is integral over $k[Y]$ (by the finiteness assumption), $k[X]$ consists of all elements of $k(X)$ that are integral over $k[Y]$.

From this definition it is clear that every automorphism of $k(X)$ that fixes $k(Y)$, and so fixes $k[Y]$, necessarily fixes $k[X]$, and thus acts on $X$.

If $Y$ is also normal, and $k(X)^G= k(Y)$ then $Y= X/G$.

There is certainly a map $X/G \to Y$. Whether or not this map is an isomorphism is again a question local on $Y$

As we saw before, the elements of $k[X]$ are exactly the elements of $k(X)$ that are integral over $k[Y]$. Among these, the $G$-invariant elements are those that lie in $k(Y)$. But by assumption on $Y$, all the elements of $k(Y)$ that are integral over $k[Y]$ lie in $k[Y]$.

So $k[X/G] = k[X]^G=k[Y]$.