Definition of a quotient curve? If $C \subseteq \mathbb{P}^m(\mathbb{\overline{F_q}})$ is an irreducible projective curve and $H \leq Aut(C)$ is a finite subgroup of the group of the automorphisms of $C$, then I know that the set $C/H = \{ Orb_H(x) | x \in C\}$ is again a projective curve. However I don't know what exactly are its projective coordinates or even its function field. Can some one tell me, please? I've searched through the internet about this, but with no success...
 A: First, it may be important to note that a projective variety does not have well-defined "projective coordinates". The projective coordinate ring may change depending on the embedding.
However, for now let $R$ be the projective coordinate ring of your curve $C\subseteq\Bbb P^m$ (with a fixed embedding) and let $K$ be its function field (which does not depend on the embedding). Then, $H$ acts on $K$ by precomposition, i.e. for $h\in H$ and $f\in K$ you have $(h.f)(x)=f(h^{-1}x)$. 
Let us consider an easy case first: If the elements of $H$ are linear functions, i.e. if all automorphisms in $H$ extend to $\Bbb P^m$, then $H$ also acts on $R$ in the same manner (essentially because $f$ and $f\circ h^{-1}$ have the same degree for all homogeneous $f\in R$ and all $h\in H$) and the quotient curve $C/H$ is realized as $\operatorname{Proj}(R^H)$, where
$$
R^H = \{ f \in R \mid \forall h\in H\colon\, h.f = f \}
$$
is the ring of invariants. The finite morphism $C\to C/H$ corresponds to the inclusion of $R^H$ in $R$.
In general, we can consider any orbit $X=H.x$ of $H$ in $C$. Since $X$ is finite, there is some homogeneous form $f$ which does not vanish at any point of $X$. Then, $f_H := \sum_{h\in H} f\circ h$ is $H$-invariant and also does not vanish anywhere on $X$. The open subset $D(f_H)$ is affine, $H$-invariant, and contains $X$. Hence, we can cover $C$ by affine, $H$-invariant, principal open subsets $D=\operatorname{Spec}(A)$ for which the quotient is always given by $D/H=\operatorname{Spec}(A^H)$. These affine quotiens glue to $C/H$. Unfortunately, I am not quite sure how $C/H$ canonically embeds into a projective space, though.
What will be the function field $L$ of $C/H$? Since $L=\operatorname{Frac}(A^H)$ for any affine open $D=\operatorname{Spec}(A)$ as above and since $K=\operatorname{Frac}(A)$, it is clear that $L$ is contained in $K^H$. However, equality is not clear to me in general, but I might be missing something blatantly obvious because the group is finite (not used to those, sorry).
What little I have to share is the following: If $H$ is unipotent, there is a neat trick to show  $K^H\subseteq L$. I will work on an affine open $D=\operatorname{Spec}(A)$ as above. Pick any $H$-invariant element $f\in K=\operatorname{Frac}(A)$ and consider $I:=\{q\in A \mid fq\in A \}$. Since $I$ is an $H$-module and $H$ is unipotent, there is an $H$-invariant element $q\in I$ (Humphreys, Linear Algebraic Groups, 17.5) such that $p:=fq\in A$. Hence, $f=\frac pq$ is a quotient of two $H$-invariant elements.
