Let $R$ be a dedekind domain, $K$ the field of fractions of $R$ and $f \in R[X]$ irreducible as polynomial in $K[X]$ s.t. $(f,f') = (1) = R[X]$. I want to check that $S = R[X]/(f)$ is also a dedekind domain.
Since $R$ is dedekind and $f$ irreducible, it follows that $R[X]/(f)$ is a noetherian integral domain, right?
I wanted to show that all prime ideals are maximal but failed. Is this the right path or is it possible to show that $S_q$ is a DVR for all $q \in Spec(S)$ wich would also imply that $S$ is dedekind?
Edit: This question has already been asked here. However, the given answer contains the use of things like the module of differentials, which I'm not familiar with. Also, I don't have the assumption that $f$ is monic (Would this be necessary?).
There may be different approach.
For $P$ a maximal ideal of $R$ then $R/P$ is a field so $f= \prod_j f_j^{e_j}\in R/P[X]$ with $f_j$ irreducible. $e_j=1$ by $(f,f')=(1)\in R/P[X]$.
$(f_j)$ is a maximal ideal of $R/P[X]$ so $(P,f_j)$ is a maximal ideal of $S$.
The comaximality of those ideals gives that $P S=\prod_j (P,f_j)$.
$\qquad$ (as $(P,f_1)(P,f_2)=(P(P,f_1,f_2),f_1f_2) = (P(1),f_1f_2)=(P,f_1f_2)$ and so on)
$R$ is Dedekind means that $P $ is invertible so $PS$ is invertible and $(P,f_j)$ is invertible.
The invertibility of those maximal ideals $(P,f_j)$ (and the noetherian-ness) implies the unique factorization in product of maximal ideals and that $(S-(P,f_j))^{-1}S$ is a DVR.