Let $R$ be a Dedekind domain and $f \in R[X]$ irreducible over the field of fractions of $R$ such that $f$ and $f'$ generate the unit ideal.
I want to show that $S=R[X]/(f)$ is also a Dedekind domain, which is a Noetherian, integrally closed domain where each prime ideal is maximal.
My attempt was that I wanted show that any prime ideal of $S$ is already maximal. I was trying to do so by using the correspondence between ideals in $S$ and $R[X]$ and aiming at showing that adding any element that is not in the prime ideal already yields that the ideal contains $f'$. However, I didn't succeed. In particular I am not sure where the fact that $R$ itself is a Dedekind domain comes into play.
I would appreciate any hint.
The conditions you have imply that the module of differentials $\Omega_{S/R} = 0$ which is equivalent to that the morphism $\pi: R \to S$ is Etale. In particular, $\pi$ is unramified.
Now, let $P$ be a maximal ideal of $S$ and $p = P \cap R$. As $R$ is a Dedekind domain, $R_p$ is a regular local ring. In particular, $p_p$ is principal. As $\pi$ is unramified, $$ R_p \to S_P $$ is an unramified extension of local rings. Thus, $pS_P = PS_P$ is principal. This proves the statement.