Let $A$ be a Dedekind domain, $K$ its field of fractions. Let $L/K$ be a finite separable extension and $B$ its ring of integers. Further, let $\theta \in B$ be an integral primitive element of $L$ and denote by $f$ its minimal polynomial over $A$.
If $\mathfrak{p}$ is a prime ideal in $A$ such that $f$ is irreducible and separable modulo $\mathfrak{p}$, then $\mathfrak{p}$ is inert in $L$.
I started like this: Let $B' = A[\theta]$. We have the following surjective homomorphisms: $$A[X] \overset{\pi}{\to} A/\mathfrak{p}[X] \overset{\pi'}{\to}A/\mathfrak{p}[X]/(\bar{f})$$ and $$A[X] \to B'/\mathfrak{p}B'.$$ Since $\bar{f}$ is irreducible and $\mathfrak{p}$ is prime, hence maximal, $\ker(\pi'\circ \pi) = \pi^{-1}((\bar{f}))$ is maximal in $A[X]$. Because of this we have $\mathfrak{p}B' \subseteq \pi^{-1}((\bar{f}))$, hence we get a homomorphism $$B'/\mathfrak{p}B' \to A/\mathfrak{p}[X]/(\bar{f}).$$ Now I'd like to show that this is in fact an isomorphism, hence $\mathfrak{p}B'$ is a maximal ideal. Afterwards, I'd like to show that $B/\mathfrak{p}B \cong B'/\mathfrak{p}B'$ so $\mathfrak{p}B$ is a maximal ideal and so we're done.
Can someone help me?