I'm having some hard times trying to prove this point of the exercise. We have $X$ a curve of genus $2$ over a field with characteristic $0$. $G=\text{Aut}(X)$ (we know it is a finite group, let's call its order $n$) acts on K(X) and let $L$ be the fixed field. Then the field extension $L \subseteq K(X)$ correponds to a finite morphism of curves $f:X \rightarrow Y$ of degree $n$.

I need to prove the following: if $P \in X$ is a ramification point, and $e_P=r$ is the ramification index, then $f^{-1}f(P)$ consists of exactly $\frac{n}{r}$ points, each having ramification index $r$.

I want to use the fact that $\sum_{Q \in \{f^{-1}f(P)\}} e_Q(f) = \text{deg}(f)=n$, so that I just need to prove that each poin in the pre-image has the same ramification index $r$. I think I need to use the action on $G$ on $X$ but I'm not sure how to proceed. Is the action transitive on the points of $\{f^{-1}f(P)\}$, what information can I get about ramification index?

  • 1
    $\begingroup$ Yes. Points in $f^{-1}f(P)$ corresponds to primes lying over $f(P)$ and the automorphism group acts transitively on them, hence same ramification index. $\endgroup$ Oct 1, 2023 at 18:06

1 Answer 1


First of all, it suffices to show that $G = \operatorname{Aut}(X)$ acts transitively on the fibers $f^{-1}(Q)$. Indeed, we will then have for any $P, P' \in f^{-1}(Q)$, an isomorphism of local rings $\mathcal{O}_{X, P} \cong \mathcal{O}_{X, P'}$ as $\mathcal{O}_{Y, Q}$ algebras. This implies $e_P = e_{P'}$.

Now let's show the orbit of any $P \in f^{-1}(Q)$ is all of $f^{-1}(Q)$.

First of all, since any automorphism in $G$ is an automorphism over $Y$, it follows that the orbit is contained in $f^{-1}(Q)$. Conversely, let $t_P \in K(X)$ be a uniformizing parameter for $P$. Then, for each $g \in G$, $(t_P)^{g^{-1}}$ is a uniformizing parameter for $g\cdot P$.

Consider $F = \prod_{g \in G} (t_P)^{g^{-1}}$. Note that this only vanishes along the $G$-orbit of $P$.

But now, $F$ is invariant under the action of $G$, so it is in the subfield $K(Y) = K(X)^G \subset K(X)$. If we view this as a function on $Y$, we see that $F(Q) = 0$. Otherwise, $F$ is invertible in $\mathcal{O}_{Y, Q}$ so its image in$\mathcal{O}_{X, P}$ is invertible.

Since $F(Q) = 0$ on $Y$, we see that when viewed as a function on $X$, $F$ vanishes along the entire preimage of $Q$. Hence, the $G$ orbit of $P$ contains the entire preimage of $Q$, as required.


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