Irreducible ideal implies prime ideal in Dedekind Domains? 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 $A$.


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*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? 
 A: You ask about three different classes of nonzero ideals in a Dedekind domain $R$:
(a) Prime ideals.
(b) Unfactorable ideals: if $\mathfrak{p} = IJ$ then $I = R$ or $J = R$.
(c) Irreducible ideals: If $\mathfrak{p} = I \cap J$ then $I = \mathfrak{p}$ or $J = \mathfrak{p}$.
Classes (a) and (b) are the same: their equivalence follows easily from the fact that every nonzero ideal in a Dedekind domain factors uniquely as a product of primes.
Class(c) consists precisely of the prime powers $\mathfrak{p}^a$ for $a \in \mathbb{N}$.  This follows from the following more general result about intersections of ideals in a Dedekind domain:

Lemma. Let $I = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_r^{a_r}$, $J = \mathfrak{p}_1^{b_1} \cdots \mathfrak{p}_r^{b_r}$ be nonzero ideals in a Dedekind domain: here $a_i, b_i$ are non-negative integers.  Then $I \cap J = \mathfrak{p}_1^{\max (a_1, b_1)} \cdots \mathfrak{p}_r^{\max (a_r, b_r)}$.

Let me know if you need help in proving this. If you know about localizations, this allows for a very easy proof: you can reduce to the case in which $R$ has exactly one nonzero prime ideal, and in this case the result is trivial.
Finally, one has the following simple result.

Lemma. If $\mathfrak{p}$ is an ideal in a commutative ring $R$, the following are equivalent:
(i) For any ideals $I$, $J$ in $R$, if $\mathfrak{p} \supset IJ$, then $\mathfrak{p} \supset I$ or $\mathfrak{p} \supset J$.
(ii) $\mathfrak{p}$ is prime.

Proof: $\neg$ (i) $\implies$ $\neg$ (ii): The hypotheses give the existence of $x \in I \setminus \mathfrak{p}$ and $y \in J \setminus \mathfrak{p}$ such that $xy \in \mathfrak{p}$.  Thus $\mathfrak{p}$ is not prime.
$\neg$ (ii) $\implies$ $\neg$ (i): If $\mathfrak{p}$ is not prime, then there are $x,y \in R \setminus \mathfrak{p}$ with $xy \in \mathfrak{p}$.  Take $I = (x)$, $J = (y)$.
I believe this answers all of your questions, but let me know if I missed something.
