Suppose that $A$ is a Dedekind domain with fraction field $K$, $L$ is a finite separable extension of $K$, and $B$ is the integral closure of $A$ in $L$. Suppose that $t$ is a primitive element for $L/K$ with minimal polynomial $f(x)$. If $p$ is a maximal ideal of $A$, we can apply the Kummer-Dedekind theorem to see how $p$ factors in $B$, by factoring $f(x)$ modulo $p$; this works for all but finitely many $p$.

I would like to have an easily computable set $S$ of primes $p$ such that for $p\notin S$, the Kummer-Dedekind criterion can be applied. When $A=\mathbb Z$, one such set $S$ is the set of primes $p$ such that $p^2$ divides the discriminant of $f(x)$.

I am not sure whether this works when $A$ is an arbitrary Dedekind domain or even when it's a PID. Are there any references discussing the case of PID's?

The most general discussion I've found is in Neukirch's *Algebraic Number Theory*, Proposition 8.3 on page 47. Unfortunately, the condition given for the criterion to apply –- namely that $p$ does not divide a certain conductor-– is not, as far as I can see, easy to check in practice. Can this condition be replaced by something simpler, such as $p$ does not divide the discriminant of $f(x)$?