If  a formal power series over the complex numbers satisfies a polynomial identity, does it imply that the power series has a radius of convergence? Let $ P(z) $ be a  $\textit{formal}$  power series in $z$ that a priori  may not have a non zero radius of convergence. Assume that $P(0) =0$. 
Let $\Phi(w,z)$ be a polynomial in two variables, that is not identically zero. Assume that 
$\Phi(0,0) =0$. Suppose $\textbf{formally}$ we have the identity 
$$ \Phi(P(z),z) =0 $$ 
Can we conclude that $P(z)$ has a non zero radius of convergence?  
Everything is over the complex numbers $\mathbb{C}$. 
 A: The equation $\Phi(w,z)=0$  can be solved using Puiseux series. If $\frac{\partial{\Phi}}{\partial{w}}\not\equiv 0$  then there exist finitely many formal series  $f(z)=\sum_{n\geq0}a_nz^{n/p}$ such that formally $\Phi(w,z)=0$. All these series
are convergent. So the answer to your question is positive.
For the proof see any book titled "Algebraic functions".
A: The result holds allowing several variables $z_1,\dots,z_n$, by using Artin approximation. (The method of proof below applies verbatim over non-archimedean fields of any characteristic, where "analytification" below may be taken in the naive sense over such fields or in the sense of rigid-analytic geometry. A variant on the argument, again using Artin approximation -- or rather its generalization proved by Popescu -- shows that if $R$ is any excellent normal local noetherian domain then its henselization $R^{\rm{h}}$ is the subring of elements of  $\widehat{R}$ that satisfy a 1-variable polynomial equation over $R$ of positive degree;
recall that for any local noetherian ring $R$, $R^{\rm{h}}$ is local noetherian and the map $R \rightarrow R^{\rm{h}}$ induces an isomorphism between completions.) 
To make a precise statement about convergent power series, let $\Phi \in \mathbf{C}[w,z_1,\dots,z_n]$ involve $w$, and let $P \in \mathbf{C}[\![z_1,\dots,z_n]\!]$ be a formal power series such that $P(0,\dots,0) = 0$ and $\Phi(P,z_1,\dots,z_n) = 0$. We claim that $P$ converges on a ball around $(0,\dots,0)$ with positive radius.  Moreover, we claim that $P$ lies in the subring of $\mathbf{C}[\![z_1,\dots,z_n]\!]$ given by the henselization $R^{\rm{h}}$ of the algebraic local ring $R = \mathbf{C}[z_1,\dots,z_n]_{(z_1,\dots,z_n)}$. 
Since $\widehat{R}$ is a domain and $\Phi \in R[w]$ has positive $w$-degree, the equation $\Phi = 0$ has at most finitely many solutions in $\widehat{R}$. Thus, there is an exponent $e > 0$ such that distinct solutions in $\widehat{R}$ are distinct modulo the $e$th power of the maximal ideal $\mathfrak{m}$ of $\widehat{R}$.  By the Artin approximation theorem, for any $f \in \widehat{R}$ satisfying $\Phi(f,z_1,\dots,z_n)=0$ and any $m > 0$ there exists $f_m$ in the henselization $R^{\rm{h}}$ such that $\Phi(f_m,z_1,\dots,z_n)=0$ and $f_m \equiv f \bmod \mathfrak{m}^m$. Taking $m = e$, the solutions $f, f_e \in \widehat{R}$ to $\Phi=0$ must coincide!  In other words, all solutions to $\Phi=0$ in $\widehat{R}$ lie in $R^{\rm{h}}$.  
By construction, $R^{\rm{h}}$ is a direct limit of local-etale $R$-algebras, so there exists a local-etale map $R \rightarrow R'$ such that all solutions to $\Phi=0$ in $\widehat{R}$ lie in $R'$ (via the canonical isomorphism $\widehat{R} \rightarrow \widehat{R'}$ and the inclusion of $R'$ into its own completion).  By definition of "local-etale", there is an etale map $h:V \rightarrow \mathbf{A}^n_{\mathbf{C}}$ and a point $v \in h^{-1}(0)$ such that $O_{V,v} = R'$ as $R$-algebras.  (In particular, $V$ is smooth.) Since $h$ is etale, it follows from the Zariski local structure theorem for etale morphisms and the analytic inverse function theorem  in several complex variables that the analytification $h^{\rm{an}}$ is a local isomorphism.  In particular, $O_{V^{\rm{an}},v}$ is identified via $h^{\rm{an}}$-pullback with the local ring $O_{(\mathbf{A}^n_{\mathbf{C}})^{\rm{an}},0}$ of convergent power series in $z_1,\dots,z_n$ at the origin.  
Passing to completions on this identification of analytic local rings, we recover the identification of $O_{V,v}^{\wedge} = \widehat{R'}$ with $\widehat{R}$ induced by $h$, so it follows that under the inclusion 
$$R' = O_{V,v} \subset O_{V^{\rm{an}},v} = O_{(\mathbf{A}^n_{\mathbf{C}})^{\rm{an}},0}$$
the element of $R'$ that "is" $P$ (provided by Artin approximation) maps to a convergent power series near the origin that has Taylor expansion at the origin equal to $P$. Hence, $P$ has positive radius of convergence. QED 
A: The same holds if $\Phi(w,z)$ is a convergent power series $\neq 0$ in $1+n$ variables (i.e. $w$ is a single variable and $z = (z_{1},\dots, z_{n})$ a set of $n$ variables):
If $P$ is a formal power series satisfying $\Phi(P(z), z) =0$ and $P(0)=0$, then it is already convergent. This follows immediatly from the analytic version of Artin's Approxmiation theorem (which states that any formal implicit solution to the equation $F(w,z) = 0$ (where $F$ is a convergent power series) can be approximated in the $\mathfrak{m}$-adic topology by convergent solutions ) and the fact that the above equation has only finitely many solutions as a consequence of the Weierstrass division theorem:
Let $P(z)$ be a formal solution and set $Q(w,z)= (w - P(z))$, which is $w$-regular of order one so we can apply the Weierstrass division theorem to find $\Phi_{1}(w,z)$ and a formal series $R(z)$ so that $\Phi = Q\cdot \Phi_{1} + R$. Plugging $(P(z),z)$ into both sides yields that $0 = R$, so $\Phi = (w-P(z)) \Phi_{1}(w,z)$ and consequently $\mathbb{ord}(\Phi_{1}) = \mathbb{ord}(\Phi) -1$.
If $P_{2}$ is another formal implicit solution then it follows that $\Phi(P_{2},z) =0$, so we can repeat the factorization and receive $\Phi = (w-P)(w-P_{2}) \Phi_{2}$, where $\mathbb{ord}(P_{2}) = \mathbb{ord}(P_{1}) -2$. So we see that the number of implicit formal solutions of $\Phi(w,z) =0$ is bounded by $\mathbb{ord}(\Phi(w,z))$. Now given any formal solution $P$ there exists a sequence of convergent solution which converges to $P$ (in the m-adic topology), and since the sequence consists only of a finite number of elements it has to coincide ultimately with $P$, which is therefore convergent.
