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 $f_\nu(x)$.