# 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?

• $f$ 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. Nov 18 at 17:14
• @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 $\infty$?
– abx
Nov 18 at 17:55

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.

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

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.

• 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!
– abx
Nov 19 at 6:40
• @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. Nov 19 at 17:03

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.

• Is there a way to justify your last "therefore $f$ is rational" without appealing to something like great Picard theorem? Nov 19 at 14:50
• Sure. With a few extra words you can appeal to Casoratti-Weierstrass theorem. Nov 19 at 18:32
• Lovely, thank you for elaborating! Nov 19 at 19:14
• 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 $f$). Nov 19 at 19:14