How to understand whether a given ideal is a prime ideal in a Dedekind domain

Question

$\DeclareMathOperator{\Norm}{Norm}$I have started Algebraic Number Theory, and I have a basic doubt about how to understand whether a given ideal is prime or not in a Dedekind domain. What I thought is, if ideal $I\subset J$, then if $\Norm(I)\mid \Norm(J)$, then by showing $\Norm(P)$ to be prime, we can prove ideal $P$ as prime. But I am confused whether, $I\subset J\rightarrow \Norm(I)\mid \Norm(J)$ is true or not.

Answer 1

I assume you are working with a Dedekind domain that is the integral closure of some other Dedekind domain in a finite field extension, and the norm is with respect to that field extension.

By definition $N(I)$ is the ideal generated by norms of elements of $I$. Thus $I \subset J$ implies $N(I) \subset N(J)$, in which case $I = J$ iff $N(I) = N(J)$. (This last fact requires properties of Dedekind domains.)

Hence $J \mid I$ implies $N(J) \mid N(I)$, and $N(I)$ prime implies $I$ prime.

(But the converse, $I$ prime implies $N(I)$ prime, is false in general.)