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I am self-studying Miranda's book Algebraic Curves and Riemann Surfaces and I am looking for a hint for a problem. For those with the book, it is on page 178, and is Problem VI.$1$.L. The problem is the following:

Let $G$ be a finite group acting effectively on an algebraic curve $X$.

(i) Show that $G$ acts on the function field $\mathcal{M}(X)$.

(ii) Show that the function field of the quotient Riemann surface $X/G$ is the field of invariants $\mathcal{M}(X)^G$.

(iii) Show that $X/G$ is an algebraic curve.

Parts (i) and (ii) I found rather easy, so that for part (iii) I have $\mathcal{M}(X)^G\cong \mathcal{M}(X/G)$. According to Miranda, an algebraic curve is a compact Riemann surface $X$ so that $\mathcal{M}(X)$ separates points and separates tangents. So, I am trying to demonstrate that $\mathcal{M}(X)^G$ separates points on $X/G$.

Now, by assumption $\mathcal{M}(X)$ separates points on $X$, the issue however is finding a function $f\in \mathcal{M}(X)^G$ separating points $x$ and $y$ lying in distinct $G$-orbits. I tried fixing $f\in \mathcal{M}(X)$ such that $f(x)\ne f(y)$. Then I produced a $G$-invariant function $\overline{f}$ by $$ \overline{f}=\frac{1}{\lvert G\rvert}\sum_{g\in G} f\circ g$$ where I identify $G$ with a subgroup of $\operatorname{Aut}(X)$. I suspect I might be able to show that this function has the desired property, but I have not succeeded. I am having similar difficulties with separation of tangents.

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I think $\overline f$ has no reason to separate $x$ from $y$ since there might be some cancellations.

However here is an easy way to separe them : by Lemma 1.13 from the same chapter, for any finite set of points $q_i$ and another point $p$ there is a function $f$ with a pole at $p$ and a zero at each $q_i$. We pick a function with a pole at $y$ and a zero at $G \cdot x \cup G\cdot y \backslash\{y\}.$ This way, $\overline f$ certainly has a pole at $G \cdot y$ and a zero at $G \cdot x$.

The argument is similar for separating tangents : let $x \in X$. There is a function with a pole of order $1$ at $x$ since $X$ is an algebraic curve. By adding suitable functions we can also assume that $f$ is zero on $G \cdot x \backslash \{x\}$. Now it is clear that $\overline f$ separates tangents at $G \cdot x$.

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