# every PID is a UFD - alternative proof

Proposition. If $R$ is a principal ideal domain^{}, then $R$ is a unique factorization domain^{}.

Proof. Recall, that due to Kaplansky Theorem (see this article (http://planetmath.org/EquivalentDefinitionsForUFD) for details) it is enough to show that every nonzero prime ideal^{} in $R$ contains a prime element^{}.

On the other hand, recall that an element $p\in R$ is prime if and only if an ideal $(p)$ generated by $p$ is nonzero and prime.

Thus, if $P$ is a nonzero prime ideal in $R$, then (since $R$ is a PID) there exists $p\in R$ such that $P=(p)$. This completes the proof. $\mathrm{\square}$

Title | every PID is a UFD - alternative proof |
---|---|

Canonical name | EveryPIDIsAUFDAlternativeProof |

Date of creation | 2013-03-22 19:04:26 |

Last modified on | 2013-03-22 19:04:26 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 13F07 |

Classification | msc 16D25 |

Classification | msc 13G05 |

Classification | msc 11N80 |

Classification | msc 13A15 |