Let $K$ be a finite field with $q$ elements then $Aut(K(x)/K)$ has $q^3 -q$ elements.

Question

Let $K$ be a finite field with $q$ elements.

$(a)$ Show that $Aut(K(x)/K)$ has $q^3 -q$ elements.

$(b)$ Show that $F(Aut(K(x)/K)) = K(\phi)$ where $$\phi(x) = \frac{(x^{q^2} -x)^{q+1}}{(x^q -x)^{q^2+1}} .$$

$(c)$ Show that if $U$ is the subgroup of $Aut(K(x)/K)$ which consists of all mappings $\sigma$ of the form $(\sigma \theta)(x) = \theta(ax+b)$ with $a \neq 0$ then $F(U) = K((x^q -x)^{q-1})$.

I am not getting any clue to solve the problem. Help Needed.

Here $F(U)$ is the fixed field of $U$.

Can someone give hints to the problem??

I think I have got the idea of the first part.

Answer 1

(a)It is well known that $G:=Aut(K(x)/K)=\{\frac{ax+b}{cx+d}| a,b,c,d \in K, \text{ and }ad-bc \neq 0 \} \cong PGL(2,K)$. Since $K=\mathbb{F}_q$, after counting $a,b,c,d \in \mathbb{F}_q$ s.t. $ad-bc \neq 0$, you arrive at $q^3-q$.

(b) I'll will use $K(x)^G$ to denote the fixed field $F(U)$ and use the rudiments of the Galois correspondence thm.

It can be checked that $\phi:=\frac{(x^{q^2}-x)^{q+1}}{(x^q-x)^{q^2+1}}$ is fixed by each of the the $q^3-q$ elements $\frac{ax+b}{cx+d}\in G:=Aut(K(x)/K)$. Thus $$K(\phi)\subseteq K(x)^G \subseteq K(x)$$ Set $\phi:=\frac{(x^{q^2}-x)^{q+1}}{(x^q-x)^{q^2+1}}=\frac{f(x)}{(x^q-x)^{q^2-q}},$ where $f(x)=\frac{(x^{q^2}-x)^{q+1}}{(x^q-x)^{q+1}}$. Thus $x$ is a root of the polynomial $$f(T)-\phi\cdot(T^q-T)^{q^2-q} \in K(\phi)[T]$$ of degree $=\deg f(T)=q^3-q$. Therefore, $[K(x):K(\phi)]\leq q^3-q$. On the other hand, we know that $[K(x):K(x)^G]=|G|=q^3-q$. Therefore, the inclusions in (*) give us $[K(x):K(\phi)]=q^3-q$, and then $K(x)^G=K(\phi)$, as desired.

(c) See an answer here (Show $F(U) = K((x^q -x)^{q-1})$.).