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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 $\mathbb{C}$.

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maybe related: – Michael Bächtold Dec 13 '12 at 12:03
To the person voting to close: I suspect you are assuming that $\partial \Phi(w,z)/\partial w|_{(0,0)} \neq 0$. In this case, it's pretty straightforward: By the complex implicit function theorem, there is a small neighborhood of $0$ where there is an analytic function $f$ obeying $f(0)=0$ and $\Phi(f(z), z)=0$. Since the coefficients of $P$ are determined by a unique recursion in this case, they are the same as the coefficients of $f$ and $P$ is convergent on this small neighborhood. – David Speyer Dec 13 '12 at 14:10
But to do the case where $\Phi$ is singular at the origin (e.g. $\Phi(w,z) = w^2 + w^5 - z^2$) seems to me to require Weierstrass preparation, and I need to think about some of the details, although I am confident the answer is yes. To my mind, this makes the problem hard enough to definitely belong here. – David Speyer Dec 13 '12 at 14:13
What we need here is that the field of convergent Puiseux series is algebraically closed. This is probably a well-known result but I don't know a proof myself. – François Brunault Dec 13 '12 at 14:36
Ok, here is a reference – François Brunault Dec 13 '12 at 14:41
up vote 8 down vote accepted

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".

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``The equation Φ(w,z)=0 can be solved using Puiseux series.'' Is this true even if $\frac{\partial{\Phi}}{\partial{w}}\neq0$ ? And will your second statement be true, if you drop the uniqueness criteria? – Ritwik Dec 13 '12 at 14:24
Thanks Alexandre for correcting my first answer which was too vaguely (and thus erroneously) stated. I guess that the «finitely many» solutions correspond to different branches of ${\Phi=0}$, am I right? Since I was implicitly thinking about reduced equation I thought those solutions were unique. – Loïc Teyssier Dec 13 '12 at 14:58
In general, finitely many solutions correspond to different branches of the implicit function $w(z)$. The curve $\Phi(z,w)=0$ may have one or several branches near $(0,0)$. But in his situation, it is assumed that $w(z)$ is a series in integer powers, so such different series correspond to different branches of the curve $\Phi(z,w)=0$. – Alexandre Eremenko Dec 13 '12 at 15:38

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

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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.

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