Theorem. Let $O$ be the maximal order, i.e. the ring of integers of an algebraic number field. Then $O$ is a unique factorization domain if and only if $O$ is a principal ideal domain.

Proof. $1^{\underline{o}}$ . Suppose that $O$ is a PID.

We first state, that any prime number $\pi$ of $O$ generates a prime ideal $(\pi)$ of $O$ . For if $(\pi) = \mathfrak{ab}$ , then we have the principal ideals $\mathfrak{a} = (\alpha)$ and $\mathfrak{b} = (\beta)$ . It follows that $(\pi) = (\alpha\beta)$ , i.e. $\pi = \lambda\alpha\beta$ with some $\lambda\in O$ , and since $\pi$ is prime, one of $\alpha$ and $\beta$ must be a unit of $O$ . Thus one of $\mathfrak{a}$ and $\mathfrak{b}$ is the unit ideal $O$ , and accordingly $(\pi)$ is a maximal ideal of $O$ , so also a prime ideal.

Let a non-zero element $\gamma$ of $O$ be split to prime number factors $\pi_i$ , $\varrho_j$ in two ways: $\gamma = \pi_1\cdots\pi_r = \varrho_1\cdots\varrho_s$ . Then also the principal ideal $(\gamma)$ splits to principal prime ideals in two ways: $(\gamma) = (\pi_1)\cdots(\pi_r) = (\varrho_1)\cdots(\varrho_s)$ . Since the prime factorization of ideals is unique, the sequence $(\pi_1),\,\ldots,\,(\pi_r)$ must be, up to the order, identical with $(\varrho_1),\,\ldots,\,(\varrho_s)$ (and $r = s$ ). Let $(\pi_1) = (\varrho_{j_1})$ . Then $\pi_1$ and $\varrho_{j_1}$ are associates of each other; the same may be said of all pairs $(\pi_i,\,\varrho_{j_i})$ . So we have seen that the factorization in $O$ is unique.

$2^{\underline{o}}$ . Suppose then that $O$ is a UFD.

Consider any prime ideal $\mathfrak{p}$ of $O$ . Let $\alpha$ be a non-zero element of $\mathfrak{p}$ and let $\alpha$ have the prime factorization $\pi_1\cdots\pi_n$ . Because $\mathfrak{p}$ is a prime ideal and divides the ideal product $(\pi_1)\cdots(\pi_n)$ , $\mathfrak{p}$ must divide one principal ideal $(\pi_i) = (\pi)$ . This means that $\pi \in \mathfrak{p}$ . We write $(\pi) = \mathfrak{pa}$ , whence $\pi\in \mathfrak{p}$ and $\pi\in \mathfrak{a}$ . Since $O$ is a Dedekind domain, every its ideal can be generated by two elements, one of which may be chosen freely (see the two-generator property). Therefore we can write $$\mathfrak{p} = (\pi,\,\gamma),\,\,\, \mathfrak{a} = (\pi,\,\delta).$$ We multiply these, getting $\mathfrak{pa} = (\pi^2,\,\pi\gamma,\,\pi\delta,\,\gamma\delta)$ , and so $\gamma\delta\in \mathfrak{pa} = (\pi)$ . Thus $\gamma\delta = \lambda\pi$ with some $\lambda\in O$ . According to the unique factorization, we have $\pi\,|\,\gamma$ or $\pi\,|\,\delta$ .

The latter alternative means that $\delta = \delta_1\pi$ (with $\delta_1\in O$ ), whence $\mathfrak{a} = (\pi,\,\delta_1\pi) = (\pi)(1,\,\delta_1) = (\pi)(1) = (\pi)$ ; thus we had $\mathfrak{pa} = (\pi) = \mathfrak{p}(\pi)$ which would imply the absurdity $\mathfrak{p} = (1)$ . But the former alternative means that $\gamma = \gamma_1\pi$ (with $\gamma_1\in O$ ), which shows that $$\mathfrak{p} = (\pi,\,\gamma_1\pi) = (\pi)(1,\,\gamma_1) = (\pi)(1) = (\pi).$$ In other words, an arbitrary prime ideal $\mathfrak{p}$ of $O$ is principal. It follows that all ideals of $O$ are principal. Q.E.D.