Are there non-Archimedean fields without associated valuation or being a non-archimedan field implies it is a valuation field? I understand that a non-Archimedean field is a field which does not satisfy the Archimedean property.
Non-Archimedean Fields
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field-theory
valuation-theory
ordered-fields
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5Can you include your definition of "non-archimedean field"? – 2012-08-24
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1Usually, "non-archimedean" refers to the valuation. Hence, I don't see how a (non-)archimedean field could be so without it being also a valuation field. – 2012-08-24
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0@MTurgeon: Perhaps iago is thinking of non-Archimedean ordered fields, like the hyperreals and surreals. – 2012-08-24
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0@BrianM.Scott Perhaps. That's why we need his definition of NA field in order to properly answer the question :) – 2012-08-24
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0@MTurgeon I added my definition of NA field after the question. Should I specify anything more? – 2012-08-24
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0@iago With this definition, then you should look at Brian's comment. – 2012-08-24
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2A non-archimedian ordered field has a natural valuation on it. (Of course the values of that valuation need not be real numbers, but belong to some ordered abelian group.) – 2012-08-24
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1For example, [N. Alling has shown](http://en.wikipedia.org/wiki/Surreal_number#Hahn_series) that the surreal numbers are isomorphic (as an ordered and valued field) to the field of [Hahn series](http://en.wikipedia.org/wiki/Hahn_series) with real coefficients and value group the surreal numbers themselves. – 2012-08-24
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1@iago: To make the problem more clear, $\mathbb{Q}(x)$ with the ordering defined by making $x$ larger than every integer is non-Archimedean. However, $\mathbb{Q}(x)$ with the ordering defined by making $x = \pi$ (and using the ordering on the real numbers) is Archimedean. So the question "Is $\mathbb{Q}(x)$ Archimedean?" clearly cannot make sense. – 2013-07-31