# Rational curve cover/Transcendental Galois field extension

Suppose the rational curve $C$ is a finite cover for the rational curve $D$ and the field of rational functions of $C$ is the purely transcendental extension $k(x)$ and that of $D$ is the subfield $k(t)$ where $t$ is a rational function in $x$. Assume the field extension is Galois so that there is a finite group $G$ acting on $k(x)$ with the subfield of invariants being $k(t)$.

I want to explicitly describe the group $G$. I also want to show that $G$ permutes the points on $C$ above a given point of $D$ and show that the ramification index of all the points above a given point of $D$ must be the same. Finally, I want to find all possible combinations of the ramification indices and find a group of the right order that realizes the ramification and acts on $\mathbb{P}^1$.

What I know and can do so far: it's a consequence of Luroth's Thoerem that the generators of $k(x)$ are the linear fractional transforms of $x$, ie. are all of the form $\frac{ax+b}{cx+d}$ where $ad-bc \neq 0$ and that $G$ will send $x$ to another generator of $k(x)$ (since the extension is Galois). So is the group in question simply $PGL_2(k)$ since this is the group acts as linear fractional transforms in projective space? I can also see how this group permutes $\mathbb{P}^1$...but after that, I am lost...

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$G$ is a finite subgroup of $\text{PGL}_2(k)$. What group it is depends on the covering map... I'm not sure exactly what you want to know. –  Qiaochu Yuan Apr 25 '12 at 1:31