I am studying the following exercise:
Show that the meromorphic functions on the Riemann sphere have the form $p(z)/q(z)$, where $p$, $q$ are coprime polynomials.
Is an exercise in Donaldson Riemann Surface.
I thought the following way to solve:
Let $f$ be a meromorphic function on the Riemann sphere. Let $\lambda_1,\ldots,\lambda_n$ be a points in $f^{-1}(0)$ such that $\lambda_i \neq \infty$. Let $\mu_1,\ldots,\mu_m$ be the points in $f^{-1}(\infty)$ in the same way. Then $$\frac{f(z)(z- \mu_1)\cdots(z-\mu_m)}{(z-\lambda_1)\cdots(z-\lambda_n)} =:g(z)$$ defines a meromorphic function with no zeros (except possibly at $\infty$) and no poles (except possibly $\infty$). Either way $g(z)$ either lacks zeros it lacks poles. So $\deg(g)=0$ and hence $g$ is constant. Then $f=p/q$ for some polynomials $p$ and $q$.
I wonder if you're reasonable. Any adjustment ??
$\endgroup$ 51 Answer
$\begingroup$To show that a function that is meromorphic in $\mathbb C\cup\{\infty\}$ and has no poles or zeros is constant, you can argue as follows.
First say that it's holomorphic in all of $\mathbb C$ because of the lack of poles. Then, since it's holomorphic at $\infty$, it has neither a pole nor an essential singuarity at $\infty$; hence it has some finite value there. Now you have a function that is continuous on the compact space $\mathbb C\cup\{\infty\}$ and that never takes the value $\infty$, so there is some open neighborhood of $\infty$ in which the function has no values. Thus it is bounded and holomorphic on $\mathbb C.$ Then use Liouville's theorem.
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