Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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

share|cite|improve this answer
Oops, I assumed implicitly that $a>0$ above. But do you have any idea how to actually impose such an order? – Dani Hobbes Apr 4 '11 at 7:56
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. – shamovic Apr 4 '11 at 9:04
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. – Dani Hobbes Apr 4 '11 at 9:21
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 – shamovic Apr 4 '11 at 10:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.