I think the answer to the question is yes. Here is an idea of a proof.

Let us assume that $\mathcal{O}_K=\mathbf{Z}[\alpha]$. I hope this is not too restrictive for you (hopefuly, 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 $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.

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 $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.

I think the answer to the question is yes. Here is an idea of a proof.

Let us assume that $\mathcal{O}_K=\mathbf{Z}[\alpha]$. I hope this is not too restrictive for you (hopefuly, someone can extend the argument).

Put $\alpha \mathcal{O}_K = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_t^{a_t}$ 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 $\mathcal{O}_{\nu_i}$ be the ring of integers of $K_{\nu_i}$. 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 \mathcal{O}_K \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}_{\nu_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 $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.

I think the answer to the question is yes. Here is an idea of a proof.

Let us assume that $\mathcal{O}_K=\mathbf{Z}[\alpha]$. I hope this is not too restrictive for you (hopefuly, 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 $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.