# 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. ${1}^{\underset{\xaf}{o}}$. 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 $(\pi )=\U0001d51e\U0001d51f$, then we have the principal ideals^{} $\U0001d51e=(\alpha )$ and $\U0001d51f=(\beta )$. 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 $\U0001d51e$ and $\U0001d51f$ 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.

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

Consider any prime ideal $\U0001d52d$ of $O$. Let $\alpha $ be a non-zero element of $\U0001d52d$ and let $\alpha $ have the prime factorization ${\pi}_{1}\mathrm{\cdots}{\pi}_{n}$. Because $\U0001d52d$ is a prime ideal and divides the ideal product $({\pi}_{1})\mathrm{\cdots}({\pi}_{n})$, $\U0001d52d$ must divide one principal ideal $({\pi}_{i})=(\pi )$. This means that $\pi \in \U0001d52d$. We write $(\pi )=\U0001d52d\U0001d51e$, whence $\pi \in \U0001d52d$ and $\pi \in \U0001d51e$. 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

$$\U0001d52d=(\pi ,\gamma ),\U0001d51e=(\pi ,\delta ).$$ |

We multiply these, getting $\U0001d52d\U0001d51e=({\pi}^{2},\pi \gamma ,\pi \delta ,\gamma \delta )$, and so $\gamma \delta \in \U0001d52d\U0001d51e=(\pi )$. 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 $\U0001d51e=(\pi ,{\delta}_{1}\pi )=(\pi )(1,{\delta}_{1})=(\pi )(1)=(\pi )$; thus we had $\U0001d52d\U0001d51e=(\pi )=\U0001d52d(\pi )$ which would imply the absurdity $\U0001d52d=(1)$. But the former alternative means that $\gamma ={\gamma}_{1}\pi $ (with ${\gamma}_{1}\in O$), which shows that

$$\U0001d52d=(\pi ,{\gamma}_{1}\pi )=(\pi )(1,{\gamma}_{1})=(\pi )(1)=(\pi ).$$ |

In other words, an arbitrary prime ideal $\U0001d52d$ 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 |