Theorem: Let \(I\) be an ideal of a Dedekind domain \(R\). Then there exists an ideal \(J\) with \(I J\) principal.

Proof: Take any nonzero \(\alpha \in I\), and let \(J = \{\beta\in R: \beta I\subset (\alpha)\}\). Then \(J\) is a nonzero ideal, and we have \(I J \subset (\alpha)\). We wish to show that this is in fact an equality. We require two lemmas:

Lemma: In a Dedekind domain, every ideal contains the product of prime ideals.

Proof: Suppose not. Then consider the set of ideals that do not contain products of prime ideals. Because a Dedekind domain is Noetherian, this set has a maximal member \(M\). \(M\) cannot be prime (it does not contain the product of primes) so there exist \(r,s\in R\setminus M\) with \(r s \in M\). Now \(M+(r),M+(s)\) are strictly bigger than \(M\) so they contain products of primes, but this implies \((M+(r))(M+(s))\) does too. This is a contradiction since this ideal is contained in \(M\).

Lemma: Let \(A\) be a proper ideal in a Dedekind domain \(R\) with field of fractions \(K\). Then there exists \(\gamma \in K\setminus R\) with \(\gamma A \subset R\).

Proof: Fix any nonzero \(a \in A\). By the previous lemma the principal ideal \((a)\) contains a product of primes, so let \(P_1 P_2 ... P_r \subset (a)\) where \(r\) is as small as possible. Every proper ideal is contained in a maximal ideal, and since a maximal ideal is prime, this means \(A \subset P\) for some prime ideal \(P\). Then \(P\) also contains this product of primes. This implies \(P\) contains some \(P_i\): if not, then if we take an \(a_i \in P_i \setminus P\) for all \(i\) and consider the product \(a_1...a_r\), then since \(P\) is prime, it must contain one of the \(a_i\), which is a contradiction. Relabel so that \(P_1 \subset P\). But since every prime ideal is maximal, we have \(P = P_1\). By the minimality of \(r\), there exists \(b\in(P_2 P_3...P_r) \setminus(a)\). Then take \(\gamma = b/a \in K\setminus R\) and we have \(\gamma A \subset R\).

Now we can complete the proof of the theorem. Consider the set \(A = \frac{1}{\alpha} I J\). Since \(I J \subset (\alpha)\) we have \(A \subset R\), and in fact \(A\) is an ideal. If \(A = R\), then \(I J = (\alpha)\), otherwise, \(A\) is a proper ideal of \(R\). Applying the latter of the above lemmas, we have \(\gamma A \subset R\) for some \(\gamma \in K \setminus R\).

Since \(\alpha \in I\) we have \(J \subset A\), thus \(\gamma J subset \gamma A \subset R\). This implies \(\gamma J \subset J\).

Now pick generators \(\alpha_1,...,\alpha_m\) for \(J\). We have

for some \(m\times m\) matrix \(M\) over \(R\). We can then argue that \(det(\gamma I - M) = 0\) and obtain a monic polynomial over \(R\) having \(\gamma\) as a root. Thus \(\gamma \in R\) by integral closure, which is a contradiction.

Corollary: The ideal classes in a Dedekind domain form a group.

Corollary (Cancellation Law): If \(A,B,C\) are ideals in a Dedekind domain with \(A B = A C\), then \(B=C\).

Proof: There exists an ideal \(J\) with \(A J\) principal. Let \(A J = (\alpha)\). Then \(\alpha B = \alpha C\) whence \(B = C\).

Corollary: If \(A,B\) are ideals in a Dedekind domain \(R\), then \(A|B\) if and only if \(A \supset B\).

Proof: The forward direction is trivial. Assuming \(A \supset B\), find a \(J\) with \(A J\) principal. Set \(A J = (\alpha)\). Then since \(J B \subset A J\) we have that \(C = (1/\alpha)J B\) is an ideal in \(R\). Lastly, note we have \(A C = B\).

Theorem: Every ideal in a Dedekind domain \(R\) is uniquely representable as a product of prime ideals.

Proof: First we shall prove that every ideal can be represented as a product of prime ideals. Suppose not. Then consider the set of ideals that cannot be represented as a product of primes. Take any maximal member \(M\) (which is guaranteed to exist in a Dedekind domain). By convention, \(R\) is the empty product so \(M \ne R\) (or instead we may restrict the set to proper ideals of \(R\)).

As in the proof of one of the above lemmas, this implies \(M\) is contained in some prime ideal \(P\). Thus by the above corollary, \(M = P I\) for some ideal \(I\). So \(I\) contains \(M\). Furthermore, this contaiment is strict, for \(I = M\) implies \(R M = P M\) which by the cancellation law implies \(R = P\), a contradiction. Since \(M\) was chosen to be maximal, we must have \(I\) a product of prime ideals, which then implies \(M\) is also a product of primes, a contradiction.

Now we handle uniqueness. Suppose \(P_1 P_2 ... P_r = Q_1 Q_2 ... Q_s\). Then \(P_1 \supset Q_1 ... Q_s\). As in the proof of one of the above lemmas, this means \(P_1 \supset Q_i\) for some \(i\). Relabel so that \(P_1 \supset Q_1\). Then we must have \(P_1 = Q_1\) because every prime ideal is maximal. By the cancellation law we obtain \(P_2 ... P_r = Q_2 ... Q_s\). Repeating this argument shows that \(r = s\) and \(P_i = Q_i\) for all \(i\).

Since every number ring is a Dedekind domain, we immediate obtain the following:

Corollary: The ideals in a number ring factor uniquely into prime ideals.

Example: Consider \((2), (3)\) in the number ring \(R = \mathbb{Z}[\sqrt{-5}]\). Then it turns out that

All the ideals on the right-hand side are primes: observe that \(|R/(2)| = 4\), so \(|R/(2,1+\sqrt{-5})|\) divides 4. It must be 2, because \((2,1+\sqrt{-5})\) properly contains \((2)\) and cannot be all of \(R\). Thus \((2,1+\sqrt{-5})\) is maximal as an additive subgroup. Thus it is maximal and hence prime. Similarly the factors of \((3)\) are primes.

As elements, we have the non-unique factorization into primes \(6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\), but as ideals, we find that this becomes a unique factorization once it is verified that \((1+\sqrt{-5})=(2,1+\sqrt{-5})(3,1+\sqrt{-5})\) and \((1-\sqrt{-5})=(2,1+\sqrt{=5})(3,1-\sqrt{-5})\).