# Roots of polynomials those coefficients are analytic functions

We say a power series in $\mathbb{C}[[z]]$ is analytic, if it has a positive radius of convergence.

Let $P(T) = \sum_{i=0}^n u_iT^i$ be a polynomial those coefficients $u_i \in \mathbb{C}[[z]]$ are analytic power series and let $f \in \mathbb{C}[[z]]$ be a root of $P$.

Question: Is $f$ analytic ?

Background: If the answer is "yes", any non-analytic power series would be transcendental over the ring of analytic power series. This would produce a counter-example to the second question in https://mathoverflow.net/questions/266198/how-to-pass-from-algebraic-power-series-to-the-analytic-ones/266216#266216.

• It is enough to show the root is holomorphic somewhere. Apr 3 '17 at 0:53
• If you replace $T(z)$ by $T(z_0)+T'(z_0)(z-z_0)+o(z-z_0)$ the same for $u_i(z) = u_i(z_0)+u_i'(z_0)(z-z_0)+o(z-z_0)$ and solve the obtained algebraic equation in $T(z_0),T'(z_0)$, is it enough for showing $T(z)$ is complex differentiable, and then holomorphic ? Apr 3 '17 at 1:04
• "Over what field": Over the field of functions that are meromorphic in an open disc around the origin. Apr 3 '17 at 1:09
• Yes, it is enough to show that $f$ is holomorphic around the origin. Apr 3 '17 at 1:14
• It won't have to be : $z^{1/2}$ the solution of $T^2-z = 0$ has a branch point at $z=0$. But it is "locally holomorphic" (whatever it means) Apr 3 '17 at 1:20

Proof: Let $F$ be the field of Puiseux series over $\mathbb{C}$. Each $g \in F$ can be written as $g(z)=h(z^{1/r})$ where $h(z)=z^k\sum_{n=0}^\infty a_nz^n$ and $k$ an integer. $g$ is called an analytic Puiseux series, if $\sum_{n=0}^\infty a_nz^n$ is analytic. Denote the set of analytic Puiseux series by $F_a$. We will use the Newton-Puiseux Theorem from "Nowak: Some elementary proofs of Puiseux's Theorem" (in the paper "analytic" is called "convergent"):
$F$ and $F_a$ are algebraically closed fields.
$P$ (from the question) is a polynomial over $F_a\subseteq F$, so its roots $f_1,...,f_n$ are in $F_a$. Since $f \in \mathbb{C}[[z]] \subseteq F$ is also a root of $P$, there is an $i$ such that $f=f_i$. Hence $f \in F_a$ is an analytic Puiseux series and because $f$ is a power series, it's an analytic power series. Q.E.D.