Ordering of algebraic extensions over $\mathbb{Q}$

Question

This question is closely related to the question asked and answered here. We can define two different orders on $\mathbb{Q}(\sqrt{2})$.

Can we prove that in $\mathbb{Q}(\sqrt{2})$ there are only two orders.
Can we prove that any finite algebraic extension of $\mathbb{Q}$ has more than one ordering? ( UPDATE: I have missed the case for $\mathbb{Q}(\imath)$. What I have in my mind is the cases which does not contain $\imath$ as an element of the extended field. For instance what happens for $\mathbb{Q}(\sqrt{2}, \sqrt{3})$.)

Advanced thanks for any help.

$\mathbb{Q}(\sqrt{-d})$ for $d > 0$ is totally imaginary that's why it has no ordering — reuns, Aug 31 '17 at 06:02

score 2 · Accepted Answer · answered Aug 31 '17 at 05:54

2

There are only two orderings on $\Bbb Q(\sqrt 2)$. Any countable ordered field can be embedded in $\Bbb R$ in an order-preserving way, so these correspond to embeddings taking $\sqrt2$ to $\sqrt2$ and $\sqrt2$ to $-\sqrt2$.

$\Bbb Q(i)$ cannot be made into an ordered field.

In general, if $K=\Bbb Q(\alpha)$ is a number field, with $\alpha$ having minimal polynomial $f$, the orderings of $K$ correspond to the real zeros of $f$.

answered Aug 31 '17 at 05:54

Angina Seng

161,540

And those real embeddings and closure of number fields have non-trivial properties – reuns Aug 31 '17 at 06:12

score 1 · Answer 2 · answered Aug 31 '17 at 05:54

Yes. Any ordering of $\Bbb Q[\sqrt 2]$ gives rise to an embedding into $\Bbb R$ and there are only two if these. Concretely, whether $a+b\sqrt 2$ is positive can be determined from whether or not $\sqrt 2$ is positive.
No. $\Bbb Q[i]$ has no ordering at all.

Ordering of algebraic extensions over $\mathbb{Q}$

2 Answers2