Let $p$ be a prime number, and let $F_{p}$ be the finite field with $p$ elements. Let $F=F_{p}(t)$ be the field of rational functions over $F_{p}$ . Consider all subfields of $F$ such that $F/C$ is a finite Galois extension.
- Show that among such subfields, there is a smallest one $C_{0}$ , i.e. $C_{0}$ is contained in any other $C$.
- What is the degree of $F/C_{0}$?
$$Aut(k(t)/k) = PGL_2(k)$$ For $k=\Bbb{F}_p$ this is a finite group and $Aut(k(t)/k)=Aut(k(t))$, so $C_0$ is the subfield fixed by $PGL_2(\Bbb{F}_p)$ and $$[\Bbb{F}_p(t):C_0]= |PGL_2(\Bbb{F}_p)| =\frac{|GL_2(\Bbb{F}_p)|}{|\Bbb{F}_p^{\times }|}= \frac{(p^2-1)(p^2-p)}{p-1}=N_p$$
Write $$\prod_{g\in PGL_2(\Bbb{F}_p)} (X-g(t)) = \sum_{n=0}^{N_p} a_n X^n$$ then $C_0=\Bbb{F}_p(a_n)$ for any non-constant coefficient $a_n$ (this is because $a_n$ has at most $N_p$ poles so $[\Bbb{F}_p(t):\Bbb{F}_p(a_n)]\le N_p$)
$n=0$ never works and $n=N_p-1$ won't work for $p$ odd, it doesn't have to be obvious which $n=n(p)$ works.