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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 $Aut(X/Y)=Aut(k(X)/k(Y))$?

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  • $\begingroup$ How did you come to conclude 1? This situation can not be better than Galois theory for filed extensions. $\endgroup$
    – Mohan
    Commented Dec 5, 2016 at 20:38
  • $\begingroup$ @Mohan What do you mean? $\endgroup$
    – Saal Hardali
    Commented Dec 5, 2016 at 20:40
  • $\begingroup$ Mohan means that your statements "Generically such an action always exists" and "If $X$ and $Y$ are curves then I think this is always the case" both seem wrong unless you are using a different definition of one of the terms than usual. $\endgroup$
    – Will Sawin
    Commented Dec 5, 2016 at 20:46
  • $\begingroup$ For either of these to be true, there must be an action of $G$ on $k(X)$. But $G$ acts on $\tilde{k(X)}$, which is usually not equal to $k(X)$. $\endgroup$
    – Will Sawin
    Commented Dec 5, 2016 at 20:48

1 Answer 1

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

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