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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?

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    $\begingroup$ $f$ is algebraic, so at worst we have a branch point at $z=\infty$. In particular, poles don't accumulate there, so the singularity is isolated and also non-essential. So $f$ is meromorphic on the sphere and thus rational. $\endgroup$ Nov 18 at 17:14
  • $\begingroup$ @Christian Remling: I am confused by your argument: to say that $f$ is meromorphic at $\infty$, you don't seem to use that $f$ is meromorphic on the whole of $\mathbb{C}$. Would you say that the function $\sqrt{z}$ (say, for $\operatorname{Re}(z)>0 $, and $=1$ for $z=1$) is meromorphic at $\infty$? $\endgroup$
    – abx
    Nov 18 at 17:55
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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. Returing to $f(z)$, we conclude that $f(z)$ is a rational function.

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    $\begingroup$ This is convincing, thanks to you and to @Christian Remling. $\endgroup$
    – abx
    Nov 18 at 20:12
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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.

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    $\begingroup$ Very simple indeed, I should have thought of that! In geometric terms, the extension of fields $\mathbb{C}(z)\rightarrow \mathbb{C}(z,f(z))$ corresponds to a finite map of compact Riemann surfaces $X\rightarrow \mathbb{P}^1$; this map has a section, hence is an isomorphism. Thanks again! $\endgroup$
    – abx
    Nov 19 at 6:40
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    $\begingroup$ @abx: This is interesting, a bit too algebraic already to be in my comfort zone. One could actually also state the key fact in elementary terms: if $p$ is irreducible, then one can get from any solution $w_1$ of $p(z,w)=0$ to any other solution $w_2$ by holomorphic continuation, but of course a meromorphic $f$ has no continuations other than itself. I don't know though how one would establish this fact without referring to the Riemann surface. $\endgroup$ Nov 19 at 17:03
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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.

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    $\begingroup$ Is there a way to justify your last "therefore $f$ is rational" without appealing to something like great Picard theorem? $\endgroup$
    – Wojowu
    Nov 19 at 14:50
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    $\begingroup$ Sure. With a few extra words you can appeal to Casoratti-Weierstrass theorem. $\endgroup$ Nov 19 at 18:32
  • $\begingroup$ Lovely, thank you for elaborating! $\endgroup$
    – Wojowu
    Nov 19 at 19:14
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    $\begingroup$ Nice. Your proof is similar to the argument in my post, but of course subtly different. I should add that for Casorati-Sochotski-Weierstrass you probably need that $f(1/z)$ has an isolated singularity at $z=0$. This is clear of course, since $f$ has at most $d$ poles (applying your argument for $a=0$ and for $1/f$ in place of $f$). $\endgroup$
    – GH from MO
    Nov 19 at 19:14

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