Meromorphic function on $\mathbb{C}$ algebraic over $\mathbb{C}(z)$ 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?
 A: Using only basic tools of complex analysis

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*$f(1/z)$ 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.
A: A more abstract argument is also possible: $f$ satisfies $p(z,f(z))=0$, and let's for convenience assume that $p$ is irreducible (but the argument works in general). We have two meromorphic maps on the associated Riemann surface $R=\{ (z,w): p(z,w)=0\}$: the standard map $(z,w)\mapsto w$ and also $(z,w)\mapsto f(z)$, this being the composition of $(z,w)\mapsto z$ with $f$.
These maps agree on an open subset of $R$, so are identical. It follows that $p$ is of degree one in $w$, so $f$ is rational.
A: The following argument is based on Christian Remling's proof (given in a comment), but is more elementary. Let us examine the behavior of $f(1/z)$ as $z\to 0$. The function $f(1/z)$ is algebraic over $\mathbb{C}(z)$, hence there are complex polynomials $p_n(z)$ such that
$$\sum_{n=0}^N p_n(z)f(1/z)^n=0.$$
Here $N$ is a positive integer. Without loss of generality, $p_N(z)$ is not identically zero, and the
the constant terms $p_n(0)$ are also not all zero. Rewriting the
equation as
$$p_N(z)=-\sum_{n=0}^{N-1} p_n(z)f(1/z)^{n-N},\qquad f(1/z)\neq 0,$$
we see that the set of poles of $f(1/z)$ is contained in the set of zeros of $p_N(z)$. In particular $f(1/z)$ is holomorphic in some punctured disk $\dot D(0,r)$ around the origin. Let $(z_k)\subset\dot D(0,r)$ be any sequence tending to zero such that $f(1/z_k)$ tends to a finite limit $w\in\mathbb{C}$. Then we have
$$\sum_{n=0}^N p_n(0)w^n=0.$$
That is, there are at most $N$ possible values for the finite limit $w\in\mathbb{C}$. By the Casorati-Weierstrass theorem, we conclude that $f(1/z)$ does not have an essential singularity at $z=0$. That is, both $f(z)$ and $f(1/z)$ are meromorphic on $\mathbb{C}$, which implies that $f(z)$ is a rational function.
Remark. The last sentence is also elementary and can be explained as follows. The poles of $f(z)$ are contained in the disk $D(0,1/r)$, so there are finitely many poles, and we can subtract from $f(z)$ the principal parts of its Laurent series at the various poles. The resulting function $g(z)$ is entire, and $g(1/z)$ does not have an essential singularity at $z=0$. Expanding $g(z)$ into a power series around $z=0$, and then replacing $z$ by $1/z$, we see that $g(z)$ is a polynomial. Returning to $f(z)$, we conclude that $f(z)$ is a rational function.
A: Here is a more elementary proof. Suppose $F(z,f(z))=0$ where $F$ is a polynomial in two variables. How many solutions can the equation $f(z)=a$ for generic $a$ have? Pugging $f(z)=a$ we obtain $F(z,a)=0$ which has at most $d=\deg F$ solutions. So all equations $f(z)=a$ have at most $d$ solutions, therefore $f$ is rational.
I was asked in the comment to stay with completely elementary means, and to avoid Picard's theorems. Let $a$ be a point with maximal number $d$ of solutions of $f(z)=a$. Let these solutions be $z_1,\ldots,z_d$. Since $f$ is open, the union of small disks around $z_j$ contains all $d$ solutions of $f(z)=a'$ for all $a'$ sufficiently close to $a$. Now apply Casorati-Sochotski-Weierstrass theorem which is completely elementary.
