Let $R=\mathbb C\{x\_1,...,x\_n\}\subset S=\mathbb C [[x\_1,...,x\_n]]$ denote the ring of convergent, respectively formal, power series over $\mathbb C$.  

Suppose $f\in R$ is irreducible in $R$. Does it remain irreducible in $S$?



Factorization in formal power series versus in convergent power series over the complexes
As Arno Fehm points out, this follows from results in Nagata's [Some Remarks on Local Rings II](https://projecteuclid.org/euclid.kjm/1250777424). Both $R$ and $S$ are UFD's, so $f$ is irreducible, in $R$ or $S$ respectively, if and only if the ideal it generates, in $R$ or $S$ respectively, is prime. At the bottom of page 1 of Nagata's paper, he states that, if $\mathfrak{p}$ is a prime ideal of $R$, then $\mathfrak{p}S$ is prime as well.


I found it hard to absorb all of Nagata's vocabulary; here is a route to get the desired claim from his results while skipping some of the sophisticated language.


Let $R'$ and $S'$ denote the versions of $R$ and $S$ with $n-1$ variables. Suppose that $f=gh$ for $f \in R$ and $g$, $h \in S$ nonunits. Use the [Weierstrass preparation theorem](https://ncatlab.org/nlab/show/Weierstrass+preparation+theorem) to factor $f = pu$, $g=qv$ and $h = rw$ where $p \in R'[x\_n]$, $q \in S'[x\_n]$ and $r \in S'[x\_n]$ are [Weierstrass polynomials](https://ncatlab.org/nlab/show/Weierstrass+preparation+theorem#weierstrass_polynomial) and $u \in R^{\times}$, $v \in S^{\times}$ and $w \in S^{\times}$ are units. Then $qr$ is a Weierstrass polynomial of $S$, and $vw \in S^{\times}$, so $f = (qr) (vw)$ and $f = pu$ are both Weierstrass factorizations in $S$. Since such factorizations are unique, we have $p = qr$ and $u = vw$.


Write $p(x\_n) = x\_n^a + \sum\_{i=0}^{a-1} p\_i x\_n^i$, $q(x\_n) = x\_n^b + \sum\_{i=0}^{a-1} q\_i x\_n^i$, $r(x\_n) = x\_n^c + \sum\_{i=0}^{c-1} r\_i x\_n^i$. Then the $q\_i$ and $r\_i$ are polynomial combinations of the roots of $p$, so the $q\_i$ and $r\_i$ are integral over $R'$. But Nagata, in his proof of Theorem 5, shows that $R'$ is integrally closed in $S'$, and this proof is extremely concrete. So that shows that the $q\_i$ and $r\_i$ land in $R'$. Thus $q$ and $r \in R$, and we deduce that $f$ factors in $R$ as well.


