I am reviewing materials in polynomial ring and came across this problem from Lang's Algebra.
Let $F$ be a field and $F(X)$ be the quotient field of $F[X]$. Show that every automorphism of $F(X)$ which induces the identity on $F$ is of type $X \mapsto \frac{aX+b}{cX+d}$ where $a,b,c,d \in F$ and the norm of the image is nonzero, i.e $ad-bc \neq 0$.
I don't quite have the intuition for this since I haven't touched this for a long time. Any suggestion or guide is tremendously appreciated.
I'll use that poles of rational functions behave nicely (of course) under multiplication and inverse, but also under addition.
Note $\text{div}(\frac1x) = [\infty]-[0] \qquad $ ($f(x) = \frac1x$ has a simple pole at $0$ and a simple zero at $\infty$) and for any $f \in F(x)$, $\#$ poles $= \#$ zeros.
If $\phi \in F(x),f \in F(X)$ and $\phi$ has two poles at $p,q$ then write $$\phi(x) = (x-p)^{-k} (x-q)^{-l} \Phi(x),\qquad f(\phi(x)) = \frac{\sum_{i=0}^d c_i \phi(x)^i}{\sum_{j=0}^e b_j \phi(x)^j}= (x-p)^{k(e-d)} (x-q)^{l(e-d)} h(x)$$ where $h$ and $\Phi$ have no pole or zero at $p,q$. In particular $f(\phi(x)) \ne \frac1x$ and $x,\frac1x \not \in F(\phi(x))$.
Therefore $x \mapsto \phi(x)$ is an automorphism of $F(x)$ implies $\phi$ has only one simple pole ie. $\text{div}(\phi(x)) = [q]-[p]$ with $p\ne q$ and hence $\phi(x) = C \frac{x-q}{x-p}$ is a Möbius transformation.