# Group action of $\text{Aut}(X)$ on ramification points (Hartshorne, exercise IV.2.5)

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?

• 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. Oct 1, 2023 at 18:06

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.