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.

## Comments

## Thanks Pahio!

Thanks pahio for writing this, I will add a link to it on the algebraic number theory main entry.

A

## Re: Thanks Pahio!

I have learned this proof from my venerable teacher, the number-theorist Kustaa Inkeri (http://users.utu.fi/taumets/inkeri.htm).

Jussi