TITLE: Meromorphic function on $\mathbb{C}$ algebraic over $\mathbb{C}(z)$
QUESTION [16 upvotes]: Let $f$ be a meromorphic function on $\mathbb{C}$ which is algebraic over the field of rational functions $\mathbb{C}(z)$ (i.e. satisfies a nontrivial equation $\sum a_i(z)f(z)^{i}=0$, with $a_i(z)\in \mathbb{C}(z)$). Is $f$ actually rational?

REPLY [2 votes]: Using only basic tools of complex analysis

$f(1/z)$ is meromorphic at $0$:
$\sum_{n=0}^N a_n(z)f(z)^n=0$ with $a_N\ne 0$, take $k$ such that $m=v_0(a_N(1/z)z^{-kN}) \le v_0(a_n(1/z)z^{-kn})$ (order of zero at $0$, negative if a pole) then $\sum_{n=0}^N b_n(z) (z^k f(1/z))^n = 0$ with the $b_n(z)=a_n(1/z)z^{-m-kn}$ holomorphic at $0$ and $b_N(0)\ne 0$. This implies that $z^k f(1/z)$ is bounded (and holomorphic) for $0<|z|< r$ small enough. You'll get that $z^{2+k} f(1/z)$ is holomorphic on $|z|<r$.

So $f$ is meromorphic on the Riemann sphere. It has finitely many poles. Substracting some rational functions $g_j \in \{\frac{c}{(z-b)^d}, c z^d,b,c\in \Bbb{C},d\in \Bbb{Z}_{\ge 1}\}$ you'll get that $f-\sum_j g_j$ is holomorphic on the Riemann sphere, it attains its maximum modulus somwehere, and the maximum modulus principle implies that it is constant.