# Krull valuation domain

###### Theorem.

Any Krull valuation $|\cdot |$ of a field $K$ determines a unique valuation domain $R=\{a\in K:|x|\leqq 1\}$, whose field of fraction^{} is $K$.

Proof. We first see that $1\in R$ since $|1|=1$. Let then $a,b$ be any two elements of $R$. The non-archimedean triangle inequality shows that $|a-b|\leqq \mathrm{max}\{|a|,|b|\}\leqq 1$, i.e. that the difference $a-b$ belongs to $R$. Using the multiplication rule (http://planetmath.org/OrderedGroup) 4 of inequalities we obtain

$$|ab|=|a|\cdot |b|\leqq 1\cdot 1=1$$ |

which shows that also the product $ab$ is element of $R$. Thus, $R$ is a subring of the field $K$, and so an integral domain^{}. Let now $c$ be an arbitrary element of $K$ not belonging to $R$. This implies that $$, whence $$ (see the inverse rule (http://planetmath.org/OrderedGroup) 5). Consequently, the inverse ${c}^{-1}$ belongs to $R$, and we conclude that $R$ is a valuation domain. The $a=\frac{a}{1}$ and $c=\frac{1}{{c}^{-1}}$ make evident that $K$ is the field of fractions of $R$.

Title | Krull valuation domain |
---|---|

Canonical name | KrullValuationDomain |

Date of creation | 2013-03-22 14:55:01 |

Last modified on | 2013-03-22 14:55:01 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13F30 |

Classification | msc 13A18 |

Classification | msc 12J20 |

Classification | msc 11R99 |

Related topic | ValuationDeterminedByValuationDomain |