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Let $(K, <)$ be an order field, can I define the order "<" in $K$ ?

I know that $K \models 0<a \;$ if and only if there is $b$ in the real closure of $K$ such that $b*b = a$. Can I "interpret" the real closure of $K$ in $K$?

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The precise result here is that if $K$ is an orderable field, and $a$ is an element of $K$ that is not a sum of squares, then there's an ordering of $K$ that makes $a$ negative. Thus the only ordered fields in which the order is determined by the field structure are those where every positive element is a sum of squares, like for example $\mathbb{Q}$ and $\mathbb{R}$ but unlike the examples given in the answers. –  Chris Eagle Oct 19 '12 at 15:23
    
Maybe I wasn't clear, the problem is not finding differents orders in a field. The question is if we have a fixed field K with an order < fixed, Is there a formula in the langague of rings who defines the order <? If this is not always the case, when is it? –  Samy Oct 19 '12 at 16:09
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Do you not understand that the existence of a field with two different orders on it proves this is impossible? –  Chris Eagle Oct 19 '12 at 16:13
    
I don't see how it's impossible. For example, you have two different roots for a square, but in order fields, you can define a particular one. Likewise, having two orders on a field doesn't necesarily means you can't define a particular one. –  Rafa Oct 21 '12 at 20:33

2 Answers 2

A slightly more convoluted solution than Hagen's:

Consider the field $K=\mathbb Q(t)$, any interpretation of $t$ as a transcendental real number will define an embedding of $\Bbb Q(t)$ into $\Bbb R$, and by that an ordering and real-closure. If we map $t=\pi$ and $t=-e$ we get that $t>0$ and $t<0$ respectively.

Hence, definability of the real-closure or the order is impossible.

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Consider $K=\mathbb Q[X]/(X^2-2)$. There are two orders on $K$, depending on if we map $X\mapsto\sqrt 2 $ or $X\mapsto-\sqrt 2$. Hence $<$ cannot be recovered from the field $K$ alone.

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