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Let $L/K$ be an extension of number fields with $L=K(\alpha)$, where $\alpha\in \mathcal{O}_L$. Let $f\in \mathcal{O}_K[x]$ be the minimal polynomial of $\alpha$ over $K$. Let $\mathfrak{p}\triangleleft\mathcal{O}_K$ be a prime of $K$.

What is the most general condition under which $\mathfrak{p}\mathcal{O}_L$’s factorization into primes of $L$ is given by $f$’s factorization mod $\mathfrak{p}$?

(More precise formulation given below.)

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**Motivation:** I ask because I know that this is guaranteed to happen if the rational prime under $\mathfrak{p}$ does not divide the index of $\mathcal{O}_K[\alpha]$ in $\mathcal{O}_L$ (cf. e.g. Marcus, *Number Fields*, theorem 27 on p. 79), but I also know that this is guaranteed to happen if $L/K$ is Galois and $f$ has distinct roots mod $\mathfrak{p}$, regardless of any condition on $\mathcal{O}_K[\alpha]$’s relationship to $\mathcal{O}_L$ (cf. David Cox, *Primes of the form $x^2+ny^2$*, proposition 5.11(ii) on p. 102). On the other hand, I know that it does not always happen, e.g. take $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{-3})$, $\mathfrak{p}=(2)$, $\alpha = \sqrt{-3}$, $f=x^2+3$; then $\mathfrak{p}=(2)$ is unramified, in fact inert, in $L$ even though $x^2+3=x^2+1$ is a square mod $2$. Is there some more unified understanding we have of exactly when this happens and when it doesn’t?

**More precise formulation of the question:** With all notation as above, let $\bar f$ be $f$’s residue in $k(\mathfrak{p})[x]$ (where $k(\mathfrak{p})=\mathcal{O}_K/\mathfrak{p}$ is $\mathfrak{p}$’s residue field). Let $f_1,\dots,f_r\in \mathcal{O}_K[x]$ and $e_1,\dots,e_r\in \mathbb{Z}^{\geq 0}$ be such that

$$\bar f = \bar f_1^{e_1}\dots \bar f_r^{e_r}$$

is the factorization of $\bar f$ into irreducible factors over $k(\mathfrak{p})$. What is the most general condition in which

$$\mathfrak{p}\mathcal{O}_L = \mathfrak{P}_1^{e_1}\dots \mathfrak{P}_r^{e_r}$$

where $\mathfrak{P}_i = \mathfrak{p}\mathcal{O}_L + f_i(\alpha)\mathcal{O}_L$ and each $\mathfrak{P}_i$ is a prime of $\mathcal{O}_L$?

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