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:* 


1. 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 
$$A/pA \cong \mathbb{Z}[X]/(f,p) \cong (\mathbb{Z}/p \mathbb{Z})[X]/(f).$$
2. The ideals $\mathfrak{q}\_i$ are precisely the prime ideals in $O\_K$ containing $p$.
3. 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 $\mathfrak{q}\_{\sigma(i)}$.
4. 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 $1 \leq i \leq k$.
5. 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 $\deg(G\_i) = \alpha\_i \cdot \deg(g\_i)$.
6. 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!



Generators of prime ideals and factorization of polynomials
$\newcommand{\Z}{\mathbb{Z}}$No.


It is easier to construct a counter-example by starting from the local case.


Let $f\in\Z\_p[X]$ be monic irreducible of degree $e>1$ such that $F = \mathbb{Q}\_p[X]/f$ is totally ramified, let $\alpha = X \bmod f$, and let $v$ be the normalised valuation on $F$.


Assume that $f = X^e \bmod p$, i.e. $\alpha$ is nilpotent mod $p$, i.e. $v(\alpha)\ge 1$. The local question corresponding to yours is: is $(p,\alpha)$ the maximal ideal of $\Z\_F$? Equivalently, is $\alpha$ a uniformiser, i.e. $v(\alpha)=1$?


From this local formulation it is easy to find a counter-example, for instance $f(X) = X^3-p^2$, which then also works globally.


