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?



Meromorphic function on $\mathbb{C}$ algebraic over $\mathbb{C}(z)$
The following argument is based on Christian Remling's proof (given in a [comment](https://mathoverflow.net/questions/408861/meromorphic-function-on-mathbbc-algebraic-over-mathbbcz#comment1048802_408861)), 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](https://en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_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.


