# proof of Bezout’s Theorem

Let $D$ be an integral domain with an Euclidean valuation. Let $a,b\in D$ not both 0. Let $(a,b)=\{ax+by|x,y\in D\}$. $(a,b)$ is an ideal in $D\ne \{0\}$. We choose $d\in (a,b)$ such that $\mu (d)$ is the smallest positive value. Then $(a,b)$ is generated by $d$ and has the property $d|a$ and $d|b$. Two elements $x$ and $y$ in $D$ are associate^{} if and only if $\mu (x)=\mu (y)$. So $d$ is unique up to a unit in $D$. Hence $d$ is the greatest common divisor^{} of $a$ and $b$.

Title | proof of Bezout’s Theorem |
---|---|

Canonical name | ProofOfBezoutsTheorem |

Date of creation | 2013-03-22 13:19:58 |

Last modified on | 2013-03-22 13:19:58 |

Owner | Thomas Heye (1234) |

Last modified by | Thomas Heye (1234) |

Numerical id | 7 |

Author | Thomas Heye (1234) |

Entry type | Proof |

Classification | msc 13F07 |