# 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 $A$.

1. 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 $B$.

2. 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?

• If you write 1) as "$AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$" then that is exactly the general definition of a prime ideal of a ring (which is equivalent to the commutative version in commutative rings.) May 20 '13 at 17:51

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.

• I am not to sure about this. If $I=P^n$ for some prime $P$ why can $I$ not be irreducible. $I$ is clearly factorable which by the argument above should imply $I$ not irreducible. If the equivalence stated would be true every ideal in a Dedekind domain would have a factorization into distinct prime ideals, which I doubt to be true...? Jun 1 '13 at 16:58
• @harajm: Yes, you're right. I have corrected my answer. Jun 2 '13 at 15:44
• Yes, this I agree with. One last remark, it seems like you are saying that every primary ideal is an irreducible ideal in a Dedekind domain. (you state that Class (c)/the irreducible ideals consists precisely of the prime powers and these are precisly the primary ideals) Something that is not true in general Noetherian rings. I can not see that the argument you are presenting is proving this however, only the converse which was stated to be true already in the question. Jun 2 '13 at 19:43
• @harajm: That a prime power ideal in a Dedekind domain cannot be an intersection of two strictly larger ideals follows from the first boxed lemma. Otherwise put, the only ideals strictly containing $\mathfrak{p}^a$ are those of the form $\mathfrak{p}^b$ for $b < a$, and it is clear that $\mathfrak{p}^{b_1} \cap \mathfrak{p}^{b_2} = \mathfrak{p}^{\max b_1,b_2} \neq \mathfrak{p}^a$. Jun 3 '13 at 5:28