An ideal is irreducible if it can not be written as the finite intersection of strictly larger ideals. In a Noetherian ring every irreducible ideal is primary, but the converse doesn't hold. I wonder about the situation in a Dedekind Domain. In a Dedekind Domain every ideal can be factored uniquely into a product of prime ideals. And every primary ideal is a power of a prime ideal. Hence in a Dedekind domain every irreducible ideal is a power of a prime ideal. Furthermore, a prime ideal is always irreducible.
Let $R$ be a Dedekind Domain.
If $A$ and $B$ are ideals in a Dedekind domain then $B$ is said to divide $A$ if there exists an ideal $C$ such that $A=BC$. We have that $A\subseteq B$ iff $B$ divides $A$.
Define $M$ as the set of ideals of $R$ that is such that if $P$ divides $AB$ then this implies that $P$ divides $A$ or $B$.
Let $K$ be those ideals in $R$ that do not have any non-trivial factorization.
In the notation of the book I am using (Mollin) the set $M$ is defined as the set of prime ideals of a number field and the set $K$ is defined as the set of irreducible ideals in a Dedekind domain. Furthermore it is stated that an ideal $A$ belongs to $K$ if and only if $A$ belongs to $M$.
So my natural question are: Is the set of prime ideals in a Dedekind Domain equal to $M$? Is the set of irreducible ideals in a Dedekind Domain equal to $K$? I.e, by proving that $A$ belongs to $K$ if and only if $A$ belongs to $M$ do we prove that an ideal in a Dedekind domain is irreducible if and only if it is prime?