A sufficient and necessary condition for $\mathbb{C}(f(x),g(x))=\mathbb{C}(x)$?

Solution 1:

Here goes. Let $\mathbb{C}[f(x),g(x)]\subset A\subset \mathbb{C}[x]$, where $A$ is the integral closure of $\mathbb{C}[f,g]$ in its fraction field. Then it is well known that $A=\mathbb{C}[u(x)]$ for some polynomial $u(x)$. Now, let $c\in\mathbb{C}$ such that the gcd of $p=f(x)-f(c), q= g(x)-g(c)$ is precisely $x-c$. This implies $p,q\in A$ are not relatively prime and so some non-constant polynomial $h(u)$ divides both. Then $h(u(x))$ divides both $p,q$ in $\mathbb{C}[x]$, and so by assumption, $h(u(x))$ must have degree one in $x$. Then, $u$ must also be of degree one in $x$ and thus $A=\mathbb{C}[x]$. Since the fraction fileds of $\mathbb{C}[f,g]$ is the same as that of $A$, we are done.