TITLE: Automorphism of the Field of rational functions
QUESTION [35 upvotes]: Let $K$ be a field and let $K(x)$ be the field of rational functions in $x$ whose coefficients are in $K$. Let $\theta(x)$ $\in \operatorname{Aut}(K(x))$ such that $\theta|_K = \operatorname{id}_K$.  Show that $\theta(x) =\frac{ax+b}{cx+d}$, with $ad\neq bc$.  
Here is my attempt.
Let $\theta(x) = \frac{f}{g}$, $f,g \in K[x]$, with $\gcd(f, g)=1$. let $h \in K(x)$. Then  $h(\frac{f}{g}) = x$ and $\frac{f(x)}{g(x)}\neq\frac{f(y)}{g(y)}$ if $x\neq y$ . Suppose $\deg f \gt 1$ or $\deg g \gt 1$, then the equation $g(x)b=f(x)$, $b\in K$,  will have more than one solution for $x$. Hence $f$ and $g$ have degrees at most $1$ and thus $\theta(x)=\frac{ax+b}{cx+d}$ with inverse function $\theta(x) =\frac{dx-b}{a-cx}$.
What I'd like to know is if Ive approached the problem in the right manner. Please any input will be very much appreciated. Thanks.

REPLY [14 votes]: This and related results are frequently called "Lüroth's Theorem". Searching on that term will turn up much of interest, e.g. this stimulating exercise set of George Bergman, which I append below for those who may not have easy access to a postscript viewer. See also the MO thread on elementary proofs.