Let $R$ be a Dedekind domain. We prove every nonzero ideal factors uniquely into prime ideals.

Step 1: Every nonzero prime ideal is maximal

Let $\mathfrak{p}$ be a nonzero prime ideal and suppose $\mathfrak{p} \subset \mathfrak{m}$ where $\mathfrak{m}$ is a maximal ideal. Since $R$ has dimension one, $\mathfrak{p}$ is already maximal. Thus every nonzero prime ideal is maximal.

Step 2: Existence of factorization

We prove by contradiction. Suppose some ideal cannot be factored into primes. Among all such ideals, choose a maximal one $I$ (using Noetherian property: the ascending chain condition guarantees a maximal element).

Since $I$ is not prime, there exist $a, b \in R$ with $ab \in I$ but $a, b \notin I$. Consider the ideals $I_1 = I + (a), \quad I_2 = I + (b)$

Both properly contain $I$, so by maximality of $I$, both $I_1$ and $I_2$ factor into primes: $I_1 = \mathfrak{p}_1 \cdots \mathfrak{p}_r, \quad I_2 = \mathfrak{q}_1 \cdots \mathfrak{q}_s$

Now $I_1 I_2 \subseteq I$ since:

• $a \in I_1$ and $ab \in I$ implies $I_1 b \subseteq I$
• Similarly $I_2 a \subseteq I$
• Therefore $I_1 I_2 \subseteq I$

But $I_1 I_2 = \mathfrak{p}_1 \cdots \mathfrak{p}_r \mathfrak{q}_1 \cdots \mathfrak{q}_s$ factors into primes. Since $R$ is integrally closed and Noetherian, we can show $I$ divides this product, contradicting that $I$ has no prime factorization.

Step 3: Uniqueness of factorization

Suppose $I = \mathfrak{p}_1 \cdots \mathfrak{p}_r = \mathfrak{q}_1 \cdots \mathfrak{q}_s$ are two prime factorizations.

Then $\mathfrak{p}_1 \cdots \mathfrak{p}_r \subseteq \mathfrak{q}_1$. Since $\mathfrak{q}_1$ is prime and maximal, we must have $\mathfrak{p}_i \subseteq \mathfrak{q}_1$ for some $i$. But $\mathfrak{p}_i$ is also maximal, so $\mathfrak{p}_i = \mathfrak{q}_1$.

By cancellation (which follows from invertibility of ideals in Dedekind domains), we can cancel $\mathfrak{p}_i$ from both sides and proceed inductively. This proves uniqueness.

Step 4: Extension to fractional ideals

For a fractional ideal $I$, choose $d \in R$ with $dI \subseteq R$. Then $dI$ is an integral ideal factoring as $dI = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r}$

Also $(d) = \mathfrak{q}_1^{f_1} \cdots \mathfrak{q}_s^{f_s}$. Therefore $I = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r} \mathfrak{q}_1^{-f_1} \cdots \mathfrak{q}_s^{-f_s}$

giving factorization with possibly negative exponents.