# proof that a Euclidean domain is a PID

Let $D$ be a Euclidean domain^{}, and let $\U0001d51e\subseteq D$ be a nonzero ideal. We show that $\U0001d51e$ is principal. Let

$$A=\{\nu (x):x\in \U0001d51e,x\ne 0\}$$ |

be the set of Euclidean valuations of the non-zero elements of $\U0001d51e$. Since $A$ is a non-empty set of non-negative integers, it has a minimum $m$. Choose $d\in \U0001d51e$ such that $\nu (d)=m$. Claim that $\U0001d51e=(d)$. Clearly $(d)\subseteq \U0001d51e$. To see the reverse inclusion, choose $x\in \U0001d51e$. Since $D$ is a Euclidean domain, there exist elements $y,r\in D$ such that

$$x=yd+r$$ |

with $$ or $r=0$. Since $r\in \U0001d51e$ and $\nu (d)$ is minimal in $A$, we must have $r=0$. Thus $d|x$ and $x\in (d)$.

Title | proof that a Euclidean domain is a PID |
---|---|

Canonical name | ProofThatAEuclideanDomainIsAPID |

Date of creation | 2013-03-22 12:43:11 |

Last modified on | 2013-03-22 12:43:11 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 7 |

Author | rm50 (10146) |

Entry type | Result |

Classification | msc 13F07 |

Related topic | PID |

Related topic | UFD |

Related topic | IntegralDomain |

Related topic | EuclideanValuation |