Let $\alpha \in \mathbb{C}$ be a zero of a monic irreducible polynomial $f \in \mathbb{Z}[X]$. Define $K = \mathbb{Q}[\alpha]$ and $A := \mathbb{Z}[\alpha]$, then $A \subseteq O_K$, where the latter denotes the ring of integers of the number field $K$.

Let $p$ be a prime number, and assume that $f$ factors modulo $p$ as
$$f \equiv g_1^{\alpha_1} \cdots g_k^{\alpha_k},$$
where the $g_i$ are monic and irreducible modulo $p$ (i.e., irreducible as elements in $(\mathbb{Z}/p\mathbb{Z})[X]$). Suppose further that there exist prime ideals $\mathfrak{q}_1,\ldots, \mathfrak{q}_k \subseteq O_K$ such that
$$p O_K = \mathfrak{q}_1^{\alpha_1} \cdots \mathfrak{q}_k^{\alpha_k}$$
and such that
$$\text{each } \mathfrak{q}_i \text{ has degree } \deg(g_i), i.e., [O_K/\mathfrak{q}_i:\mathbb{Z}/p\mathbb{Z}] = \deg(g_i).$$
(EDIT: If necessary, one may assume here that the above holds for *every* prime number $p$.)

**Is it then true$^\star$ that**

$$\mathfrak{q}_i = \langle p, g_i(\alpha) \rangle_{O_K} := p \cdot O_K + g_i(\alpha) \cdot O_K \text{ ?}$$

$\big($ $^\star$ This is meant up to ''reasonable'' permutation. E.g., if $g_i$ and $g_j$ have the same degree and $\alpha_i = \alpha_j$, then it could of course happen that $\mathfrak{q}_j = \langle p, g_i(\alpha) \rangle_{O_K}$ and $\mathfrak{q}_i = \langle p, g_j(\alpha) \rangle_{O_K}$.$\big)$

Of course, my claim holds for all but finitely many prime numbers $p$, which is precisely the **Kummer-Dedekind**-Theorem. To be concrete, it holds for all $p$ not dividing the index $[O_K : A]$. Hence I am asking here whether the generators of the prime ideals $\mathfrak{q}_i$ are nevertheless as predicted by Kummer-Dedekind, if we assume that $f$ and $pO_K$ have the same factorization pattern.

*A few observations:*

- The ideals $\mathfrak{p_i} := \langle p, g_i(\alpha) \rangle_A = p \cdot A + g_i(\alpha) \cdot A$ are precisely the prime ideals in $A$ containing $p$, which is due to the isomorphisms $$A/pA \cong \mathbb{Z}[X]/(f,p) \cong (\mathbb{Z}/p \mathbb{Z})[X]/(f).$$
- The ideals $\mathfrak{q}_i$ are precisely the prime ideals in $O_K$ containing $p$.
- Since $O_K \mid A$ is an integral ring extension, for every $\mathfrak{p}_i$ there exists a prime ideal in $O_K$ lying over it which (necessarily) contains $p$. In particular, by 1. and 2., there is a permutation $\sigma \in S_k$ such that $\mathfrak{p}_i = \mathfrak{q}_{\sigma(i)} \cap A$. However, (to my knowledge) this is not strong enough to conclude that $\mathfrak{q}_{\sigma(i)} = \mathfrak{p}_i \cdot O_K = \langle p, g_i(\alpha) \rangle_{O_K}$. At least, we may conclude that $\mathfrak{p}_i \cdot O_K$ is a power of $\mathfrak{q}_{\sigma(i)}$.
- By the isomorphisms from 1., we know that $|A/\mathfrak{p}_i| = |(\mathbb{Z}/p\mathbb{Z})[X]/(g_i)| = p^{\deg(g_i)}$. Further, our assumptions imply that $|O_K/\mathfrak{q}_i| = p^{\deg(g_i)}$. In the notation of 3., we have a canonical injection $A/\mathfrak{p}_i \to O_K/\mathfrak{q}_{\sigma(i)}$, hence $\deg(g_i) \leq \deg(g_{\sigma(i)})$ for every $1 \leq i \leq k$, hence these must be equalities, thus even $A/\mathfrak{p}_i \cong O_K/\mathfrak{q}_{\sigma(i)}$ for every $1 \leq i \leq k$.
- Let's now involve $p$-adic numbers. By Hensel's Lemma, the factorization $f \equiv g_1^{\alpha_1} \cdots g_k^{\alpha_k}$ modulo $p$ can be lifted to a decomposition of $f$ as a product $$f = G_1 \cdots G_k \text{ in } \mathbb{Z}_p[X],$$ where each $G_i$ is monic and reduces to $g_i^{\alpha_i}$ modulo $p$, in particular $\deg(G_i) = \alpha_i \cdot \deg(g_i)$.
- Using the isomorphism $$K \otimes_\mathbb{Q} \mathbb{Q}_p \cong \prod_{\substack{\mathfrak{p} \in \mathrm{Spec}(O_K) \\ \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}}} K_\mathfrak{p},$$ where $K_\mathfrak{p}$ denotes the completion of $K$ with respect to the $\mathfrak{p}$-adic topology, the factorization $$p O_K = \mathfrak{q}_1^{\alpha_1} \cdots \mathfrak{q}_k^{\alpha_k}$$ (togehter with the nice properties of étale algebras) implies that the factorization of $f$ over $\mathbb{Z}_p$ consists of precisely $k$ (monic) irreducible factors, whose degrees are given by $\deg(g_i) \cdot \alpha_i$, $1 \leq i \leq k$. This proves that $G_1, \ldots, G_k$ are irreducible.

Note that this is not a homework problem, hence my claim could possibly be wrong. Any help is appreciated!