ARQMATH LAB
Math Stack Exchange

Definable orders

2
$\begingroup$

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

I know that $K \models 0 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 $K$?

abstract-algebra logic model-theory ordered-fields
asked 2012-10-19
user id:35458
16
22bronze badges
  • 1
    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. – 2012-10-19
  • 0
    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 – 2012-10-19
  • 1
    Do you not understand that the existence of a field with two different orders on it proves this is impossible? – 2012-10-19
  • 0
    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. – 2012-10-21

2 Answers 2

Related Posts

No Related Post Found