# unique factorization and ideals in ring of integers

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   $(\pi_{1}),\,\ldots,\,(\pi_{r})$  must be, up to the , 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.

 Title unique factorization and ideals in ring of integers Canonical name UniqueFactorizationAndIdealsInRingOfIntegers Date of creation 2015-05-06 15:32:53 Last modified on 2015-05-06 15:32:53 Owner pahio (2872) Last modified by pahio (2872) Numerical id 17 Author pahio (2872) Entry type Theorem Classification msc 13B22 Classification msc 11R27 Synonym equivalence of UFD and PID Related topic ProductOfFinitelyGeneratedIdeals Related topic PIDsAreUFDs Related topic NumberFieldThatIsNotNormEuclidean Related topic DivisorTheory Related topic FundamentalTheoremOfIdealTheory Related topic EquivalentDefinitionsForUFD