Here is something I've been wondering about recently.

Suppose you have an arbitrary ordered field $F$, and let $F(\sqrt{a})$ be a field extension with $a>0$ in $F$. Is there then some way to order $F(\sqrt{a})$ such that $F(\sqrt{a})$ is also an ordered field? Here $F(\sqrt{a})$ is the set of all $p+q\sqrt{a}$ for $p,q,a\in F$. I'm hoping to satisfy existence of a positive cone in $F(\sqrt{a})$.

I was interested because for field extensions like $\mathbb{Q}(\sqrt{2})$, we know that $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{R}$, which itself is ordered, so we can take the positive cone of $\mathbb{Q}(\sqrt{2})$ to be those elements which are positive in $\mathbb{R}$. But if we don't know a larger ordered field containing the extension, what do we do instead? Thanks.


1 Answer 1


The answer is yes if and only if $a$ is positive in $F$. For the a non-example consider $\mathbb{C}/\mathbb{R}$. Here $\mathbb{C} = \mathbb{R}(\sqrt{-1})$ and $\mathbb{C}$ is not an ordered field

  • $\begingroup$ Oops, I assumed implicitly that $a>0$ above. But do you have any idea how to actually impose such an order? $\endgroup$ Apr 4, 2011 at 7:56
  • $\begingroup$ The order is in general non-unique. In your example $\mathbb{Q}(\sqrt{2})$ can be embedded into $\mathbb{R}$ in two different ways obtaining two orders, which are not equivalent. However in general one considers the set $\sum P (F(\sqrt{a})^{\times})^2$, It doesn't contain $-1$ hence extends to an ordering on $F(\sqrt{a})$. The proof I know is via Zorn's lemma so it is not very constructive. $\endgroup$
    – shamovic
    Apr 4, 2011 at 9:04
  • $\begingroup$ Do you have a reference? I'm just interested that some order exists. I'm familiar with Zorn's Lemma, so maybe I could understand it. $\endgroup$ Apr 4, 2011 at 9:21
  • $\begingroup$ Sure, a few books. I've studied some of field theory from "Valuation, Orderings and Milnor $K$-Theory" by I. Efrat. I think you can also find basics in Lang's Algebra or any other book that treats Artin-Schreier theory $\endgroup$
    – shamovic
    Apr 4, 2011 at 10:13

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