TITLE: Factorizing polynomials in $\mathbf{Z}[[x]]$
QUESTION [12 upvotes]: Let $f(x)\in\mathbf{Z}[x]$ be a non-constant, irreducible polynomial, and let $\alpha \in\mathbf{C}$ be a root of $f(x)$. Denote by $\varphi_\alpha:\mathbf{Z}[x]\rightarrow\mathbf{C}$ the ring homomorphism sending $x$ to $\alpha$. Let now $\nu$ be a $p$-adic place of the number field $K=\mathbf{Q}(\alpha)$ such that $\alpha$ has strictly positive valuation with respect to $\nu$, i.e. such that $\alpha$ belongs to the maximal ideal of the integers of $K$ defined by $\nu$.
There are only a finite number of such places $\nu$, they are precisely those occurring with positive exponent in the prime factorization of the principal fractional ideal of $K$ generated by $\alpha$.
Then $\varphi_\alpha$ can be extended uniquely to a continuous map
$$\varphi_{\alpha,\nu}:\mathbf{Z}[[x]]\rightarrow K_\nu,$$
where $K_\nu$ denotes the completion of $K$ at the place $\nu$ (the topology considered here on $\mathbf{Z}[[x]]$ is the $x$--adic one). Let
$f_\nu(x)\in\mathbf{Z}[[x]]$ be a power series generating the kernel of $\varphi_{\alpha,\nu}$ (such ideal should indeed be principal, right?).
Is it true that the $f_\nu(x)$ can be chosen so that
$$f(x)=\prod_{\nu}f_\nu(x),$$
where $\nu$ ranges through the places of $K$ considered above?
EDIT: As suggested below, it would be more correct to ask that $f(x)$ be equal to the product $\prod_{\nu}f_\nu(x)$ only up to units, and for any choices of the $f_\nu(x)$.

REPLY [4 votes]: I think the answer to the question is yes. Here is the idea of a proof.
Let us assume that $\mathcal{O}_K=\mathbf{Z}[\alpha]$. I hope this is not too restrictive for you (hopefully, someone can extend the argument).
Put $(\alpha) = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_t^{a_t}$ where the $\mathfrak{p}_i$ are distinct maximal ideals of $\mathcal{O}_K$. Let $\nu_i$ be the place of $K$ associated to $\mathfrak{p}_i$. Let $K_i$ be the completion of $K$ with respect to $\mathfrak{p}_i$, and let $\mathcal{O}_i$ be the ring of integers of $K_i$. We have isomorphisms
\begin{equation*}
\mathbf{Z}[[X]]/(f) \cong \varprojlim_{n \geq 1} \mathbf{Z}[X]/(X^n,f) \cong  \varprojlim_{n \geq 1} \mathcal{O}_K/(\alpha^n)
\end{equation*}
\begin{equation*}
\cong \varprojlim_{n \geq 1} \prod_{i=1}^t \mathcal{O}_K/\mathfrak{p}_i^{a_i n} \cong \prod_{i=1}^t \mathcal{O}_{i}.
\end{equation*}
Now, convince yourself that the composition is just $(\varphi_{\alpha,\nu_1},\ldots,\varphi_{\alpha,\nu_t})$, using your notations (this is because $X$ is sent to $\alpha$).
Recall that $\mathbf{Z}[[X]]$ is a UFD. We already know (see my comment to the question) that $\ker(\varphi_{\alpha,\nu_i}) = (f_i)$ with $f_i \in \mathbf{Z}[[X]]$ irreducible. The ideals $(f_i)$ are pairwise distinct (they are the preimages of distinct ideals of $\prod_{i=1}^t \mathcal{O}_{\nu_i}$). So we get $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.