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 $f\_\nu(x)$.



Factorizing polynomials in $\mathbf{Z}[[x]]$
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 $(f)=\cap\_{i=1}^t (f\_i)=(f\_1 \cdots f\_t)$.


