extension of valuation from complete base field
Here the valuations^{} are of rank one, and it may be supposed that the values are real numbers.

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Assume a finite field extension $K/k$ and a valuation of $K$. If the base field^{} is complete^{} (http://planetmath.org/Complete) with regard to this valuation, so is also the extension field.

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If $K/k$ is an algebraic field extension and if the base field $k$ is complete (http://planetmath.org/Complete) with regard to its valuation $\cdot $, then this valuation has one and only one extension^{} to the field $K$. This extension is determined by
$$\alpha =\sqrt[n]{N(\alpha )}\mathit{\hspace{1em}}(\alpha \in K),$$ where $N(\alpha )$ is the norm of the element $\alpha $ in the simple field extension $k(\alpha )/k$ and $n$ is the degree of this field extension.
These theorems concern also Archimedean valuations.
Title  extension of valuation from complete base field 
Canonical name  ExtensionOfValuationFromCompleteBaseField 
Date of creation  20130322 15:01:01 
Last modified on  20130322 15:01:01 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  9 
Author  pahio (2872) 
Entry type  Theorem 
Classification  msc 13F30 
Classification  msc 13A18 
Classification  msc 12J20 
Classification  msc 11R99 
Related topic  CompleteUltrametricField 
Related topic  ValueGroupOfCompletion 
Related topic  NthRoot 