# PID and UFD are equivalent in a Dedekind domain

This article shows that if $A$ is a Dedekind domain^{}, then $A$ is a UFD if and only if it is a PID. Note that this result implies the more specific result given in the article unique factorization and ideals in ring of integers.

Since any PID is a UFD, we need only prove the other direction. So assume $A$ is a UFD, let $\U0001d52d$ be a nonzero (proper) prime ideal^{}, and choose $0\ne x\in \U0001d52d$. Note that $x$ is a nonunit since $\U0001d52d$ is a proper ideal^{}. Since $A$ is a UFD, we may write $x$ uniquely (up to units) as $x={p}_{1}^{{a}_{1}}\mathrm{\cdots}{p}_{k}^{{a}_{k}}$ where the ${p}_{i}$ are distinct irreducibles^{} in $A$, the ${a}_{i}$ are positive integers, and $k>0$ since $x$ is not a unit. Since $\U0001d52d$ is prime and $x\in \U0001d52d$, it follows that some ${p}_{i}$, say ${p}_{1}$, is in $\U0001d52d$. Then $({p}_{1})\subset \U0001d52d$. But $({p}_{1})$ is prime since clearly in a UFD any ideal generated by an irreducible is prime. Since $A$ is Dedekind and thus has Krull dimension 1, it must be that $({p}_{1})=\U0001d52d$ and thus $\U0001d52d$ is principal.

Title | PID and UFD are equivalent in a Dedekind domain |
---|---|

Canonical name | PIDAndUFDAreEquivalentInADedekindDomain |

Date of creation | 2013-03-22 17:53:45 |

Last modified on | 2013-03-22 17:53:45 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 7 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 16D25 |

Classification | msc 13G05 |

Classification | msc 13A15 |

Classification | msc 11N80 |