On finite extensions of the field of meromorphic functions

Question

Let $\mathcal{M}$ be the field of meromorphic functions of one (complex) variable and $w = w(z)$ an analytic function satisfying a polynomial equation

$P(w; z) := w^n + a_{n-1}(z) w^{n-1} + \cdots + a_1(z) w + a_0(z) = 0$,

where $a_0(z), \ldots, a_{n-1}(z)$ are in $\mathcal{M}$ (actually, is suffices to consider the case where $a_j(z)$ is entire for $j = 0, \ldots, n-1$).

Suppose $w(z)$ has finitely many branch points.

In the (hyperelliptic) case $n=2$ it is clear that $\mathcal{M}(w) = \mathcal{M}(\sqrt{Q})$, where $Q(z)$ is a polynomial.

Is it always true that, under the above assumptions we have that $\mathcal{M}(w) = \mathcal{M}(\beta)$, where $\beta(z)$ is some algebraic function?

Answer 1

First, as you noticed, it is enough to consider the case that the equation has the form $$w^n+a_{n-1}(z)w^{n-1}+\ldots+a_0(z)=0,$$ where the coefficients are entire. Then $w$ is holomorphic on its Riemann surface, let us call this Riemann surface $S$. From your condition follows that $S$ is a compact Riemann surface with finitely many punctures. So it can be represented by an algebraic curve $K$ in $C^2$ given as a zero set of a polynomial $F(z,u)=0$. Suppose that this curve is non-singular. Let $m=\deg_u F$. We have an analytic function $w$ on $K$, so it can be extended to the whole $C^2$, So $w$ is a restriction of $K$ of an entire function $G(z,u)$. Now on $K$ we have $$G(z,u)=\sum_{k,j}a_{k,j}z^ku^j=\sum_{j=0}^{m-1} u^j\sum_{k,i=0}^\infty a_{k,j+im}z^k=\sum_{j=0}^{m-1}b_j(z)u^j,$$ where the rearrangement of the infinite sum is legitimate because of the absolute convergence. This proves your statement as $u$ is algebraic over $C(z)$.

It may happen that every realization of $S$ in $C^2$ is singular. In this case we realize $S$ as a non-singular curve $K$ in $C^n$ (I suppose one can take $n=3$ but this is irrelevant.) Let the coordinates in $C^n$ be $(z,u_1,\ldots,u_{n-1})$. Then $w$ can be represented by an entire function $G(z,u_1,\ldots,u_n)$ and the restrictions on $K$ of the coordinate functions $u_1,\ldots,u_{n-1}$ are algebraic functions of $z$, and by the theorem on the primitive element, they are all rational functions of $z$ and some $\beta$, where $\beta$ is an algebraic function of $z$. Then the same argument works.

Answer 2

This is not a complete answer but I think this might help you :

If $w(z)$ is meromorphic on a disk $\{|z-z_0| < r\}\subset \mathbb{C}$ then it is algebraic (over $\mathbb{C}(z)$) iff

$(1)$ there are finitely many points such that for $S = \mathbb{C} \setminus \{p_1,\ldots,p_n\}$ and every curve $\gamma : z_0 \to z_1\in S$ the analytic continuation of $w(z)$ along $\gamma$ is well-defined, call $w^\gamma(z)$ the resulting function analytic around $z_1$,

$(2)$ with $ \pi_1(S)$ the homothopy group then $\{ w^\gamma |\gamma \in \pi_1(S)\}$ is a finite set,

$(3)$ there is some $A$ such that each $w^\gamma(z) (z-p_k)^A$ is bounded around $p_k$ and $w^\gamma(z)z^{-A}$ is bounded around $\infty$.

In that case $G =\pi_1(S) / \{ \gamma \in S, w^\gamma = w\}$ is indeed the Galois group of $\mathbb{C}(z, \{ w^\gamma(z)\}_{\gamma\in G})/ \mathbb{C}(z)$.

Proof : look at the coefficients of $\prod_{\gamma \in G} (w^\gamma(z)-t)$, since they are fixed by $f(z) \mapsto f^\lambda(z), \lambda \in \pi_1(S)$ and they are locally analytic then they are globally analytic on $S$, and the condition $(3)$ shows they are meromorphic on the Riemann sphere, thus they are in $\mathbb{C}(z)$.

Without $(3)$ then $G$ is still the Galois group of $M_S( \{ w^\gamma(z)\}_{\gamma\in G})/ M_S$ where $M_S$ is the field of the meromorphic functions on $S$.