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 $f(x)$?


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