unique factorization and ideals in ring of integers

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

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

We first state, that any prime number $\pi$ of $O$ generates aprime ideal $(\pi )$ of $O$.  For if  $(\pi )=𝔞𝔟$,  then we have the principal ideals  $𝔞=(\alpha )$  and $𝔟=(\beta )$.  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 $𝔞$ and $𝔟$ 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:  $\gamma ={\pi }_1\mathrm{\cdots }{\pi }_r={\varrho }_1\mathrm{\cdots }{\varrho }_s$.  Then also the principal ideal $(\gamma )$ splits to principal prime ideals in two ways:  $(\gamma )=({\pi }_1)\mathrm{\cdots }({\pi }_r)=({\varrho }_1)\mathrm{\cdots }({\varrho }_s)$.  Since the prime factorization of ideals is unique, the   $({\pi }_1),\mathrm{\dots },({\pi }_r)$  must be, up to the , identical with  $({\varrho }_1),\mathrm{\dots },({\varrho }_s)$  (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.

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

Consider any prime ideal $𝔭$ of $O$.  Let $\alpha$ be a non-zero element of $𝔭$ and let $\alpha$ have the prime factorization ${\pi }_1\mathrm{\cdots }{\pi }_n$.  Because $𝔭$ is a prime ideal and divides the ideal product $({\pi }_1)\mathrm{\cdots }({\pi }_n)$, $𝔭$ must divide one principal ideal  $({\pi }_i)=(\pi )$.  This means that  $\pi \in 𝔭$.  We write  $(\pi )=𝔭𝔞$, whence  $\pi \in 𝔭$  and  $\pi \in 𝔞$.  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

$$𝔭=(\pi ,\gamma ),𝔞=(\pi ,\delta ).$$

We multiply these, getting $𝔭𝔞=({\pi }^2,\pi \gamma ,\pi \delta ,\gamma \delta )$,  and so  $\gamma \delta \in 𝔭𝔞=(\pi )$.  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  $𝔞=(\pi ,{\delta }_1\pi )=(\pi )(1,{\delta }_1)=(\pi )(1)=(\pi )$;  thus we had  $𝔭𝔞=(\pi )=𝔭(\pi )$  which would imply the absurdity  $𝔭=(1)$.  But the former alternative means that  $\gamma ={\gamma }_1\pi$ (with  ${\gamma }_1\in O$),  which shows that

$$𝔭=(\pi ,{\gamma }_1\pi )=(\pi )(1,{\gamma }_1)=(\pi )(1)=(\pi ).$$

In other words, an arbitrary prime ideal $𝔭$ of $O$ is principal.  It follows that all ideals of $O$ are principal. Q.E.D.