Proving that $[K(x):L]$ is finite where x is an indeterminate over the field K and L is an intermediate field of $K(x)/K$ such that $L\neq K$.

Question

Let $G=Aut(E/K)$ and let $H\leq J\leq G$. Then, I know that if $[H:J]$ is finite, so is $[L:M]$ where $L\subset M$ are the subfields of $E$ that are fixed by $J$ and $H$, respectively. I also know that $Aut(K(x)/K)$ is the set of automorphisms that map $x$ to $\frac{ax+b}{cx+d}$ with $ad\neq bc$.

Trivially, the subgroup of $Aut(K(x)/K)$ that fixes $K(x)$ is $\{1\}$. So, the question boils down to showing that the subgroup that fixes any intermediate field $L$, is finite but I don't have any idea about the subfields of $K(x)$, to begin with.

Answer 1

Let $L$ an intermediate field in the extension $K(x)/K$. Suppose $L\neq K$, so that, there is $p(x)/q(x)$, with $p(x), q(x)$ both relatively primes and one of them of degree at least one. Define $$F(z)=p(z)-\frac{p(x)}{q(x)}q(z)\in L[z].$$ Since $x$ is a root of $F(z)$ which have degree at least one, we have that $x$ is algebraic over $L$, then $K(x)/L$ is finitely generated and algebraic, so that finite.