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? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:03:05Z http://mathoverflow.net/feeds/question/116269 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116269/if-a-formal-power-series-over-the-complex-numbers-satisfies-a-polynomial-identit 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? Ritwik 2012-12-13T11:38:43Z 2012-12-14T12:59:17Z <p>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 $P(0) =0$. </p> <p>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 </p> <p>$$\Phi(P(z),z) =0$$ </p> <p>Can we conclude that $P(z)$ has a non zero radius of convergence? </p> <p>Everything is over the complex numbers $\mathbb{C}$. </p> http://mathoverflow.net/questions/116269/if-a-formal-power-series-over-the-complex-numbers-satisfies-a-polynomial-identit/116278#116278 Answer by Loïc Teyssier for 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? Loïc Teyssier 2012-12-13T13:45:36Z 2012-12-13T14:43:36Z <p>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.</p> <p>For the proof see any book titled "Algebraic functions".</p> http://mathoverflow.net/questions/116269/if-a-formal-power-series-over-the-complex-numbers-satisfies-a-polynomial-identit/116368#116368 Answer by kreck for 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? kreck 2012-12-14T12:59:17Z 2012-12-14T12:59:17Z <p>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.) </p> <p>To make a precise statement about convergent power series, let $\Phi \in \mathbf{C}[w,z_1,\dots,z_n]$ involve $w$, and let <code>$P \in \mathbf{C}[\![z_1,\dots,z_n]\!]$</code> 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 <code>$\mathbf{C}[\![z_1,\dots,z_n]\!]$</code> given by the henselization $R^{\rm{h}}$ of the algebraic local ring $R = \mathbf{C}[z_1,\dots,z_n]_{(z_1,\dots,z_n)}$. </p> <p>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 $R^{\rm{h}}$. </p> <p>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. </p> <p>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 </p>