# On finite extensions of the field of meromorphic functions

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?

• Please clarify your statement "In the (hyperelliptic) case $n=2$ it is clear that $\mathcal{M}(w)=\mathcal{M}(\sqrt{\mathcal{Q}})$, where $\mathcal{Q}(z)$ is a polynomial". Let us consider an algebroid function (see en.wikipedia.org/wiki/Algebroid_function ) defined by $w^2-e^z=0$. This function has the only branch point at infinity and a series of branch cuts $\Im z=\pi +2\pi n, n\in \mathbb{Z}$. – user64494 Dec 22 '18 at 19:25
• Well, the equation you proposed has two entire solutions: $w_{1,2} = \pm e^{z/2}$. I cannot see any branch points. In fact we have $\mathcal{M}(w_j) = \mathcal{M}$. In general, what I mean by "hyperelliptic case" is the case $w^2 = g(z)^2 Q(z)$, where $g$ is entire and $Q$ is entire and square-free. Then, the branch points of $w$ are the zeros of $Q$. Thus the assumption of finitely many branch points implies that $Q$ is a polynomial. Consequently, $\mathcal{M}(w) = \mathcal{M(\sqrt{Q})}$. – vassilis papanicolaou Dec 22 '18 at 20:23
• The equation $w^2=exp(z)$ has algebroid solutions too. Their branch cuts are produced by the Maple command FunctionAdvisor(branch_cuts, sqrt(exp(z)),plot=2.); . – user64494 Dec 22 '18 at 20:32
• It is quite obvious that the quadratic equation $w^2 = e^z$ has two solutions for $w$, namely $\pm e^{z/2}$. Maple is probably confused. Probably it gives the correct answer in a very strange way (sqrt(exp(z)) is $\pm e^{z/2}$). – vassilis papanicolaou Dec 22 '18 at 20:40
• Well, here we are talking about analytic (therefore smooth) solutions $w$ defined on Riemann surfaces. The Dirichlet type functions are discontinuous. – vassilis papanicolaou Dec 22 '18 at 20:49

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.

• I would appreciate if you can send me a reference regarding the statement that $w$ on $K$ extends to an entire function on $C^2$. For example, is the book of Steven Krantz on several complex variables relevant? – vassilis papanicolaou Dec 24 '18 at 11:09
• @vassilis papanicolaou:This consequence of Cartan's "Theorem B" is mentioned in many textbooks. For example. H. Grauert and L. Fritzsche, Several Complex Variables, Springer 1976, Theorem 5.11 on p. 177. Sorry I am not familiar with Krantz's text. – Alexandre Eremenko Dec 24 '18 at 18:33

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 $$S$$.