Which fields have unique ordering?

Question

$\Bbb{R}$ has a unique ordering, as does $\Bbb{Q}$, and all real-closed fields. Furthermore, $\Bbb{Q}(\sqrt[3]{2})$ has a unique ordering, but this isn't trivial to see, and many other algebraic extensions of $\Bbb{Q}$ don't, such as $\Bbb{Q}(\sqrt{2})$ having two orderings. This seems like an eclectic bunch, so I would like to know if there is some order to which fields have this property. Is there a classification of fields that have a unique order? or some easy-to-test criterion for it?

Answer 1

By the Artin-Schreier theorem, every ordered field $F$ has a unique real closure $F_r$, which is the universal real closed field equipped with a morphism $F \to F_r$ of ordered fields (note that this is a genuine universal property, unlike the case with the algebraic closure!). Since the ordering on $F_r$ is unique, the ordering on $F$ is determined by this embedding. Note that our notation is a bit misleading here because $F_r$ depends on the choice of ordering.

The real closure of any ordered algebraic extension $F$ of $\mathbb{Q}$ must be the real algebraic numbers, which here I'll just write $\mathbb{Q}_r$. So we get a map from the set of orderings on $F$ to $\text{Hom}(F, \mathbb{Q}_r)$. It is surjective because every such homomorphism produces an order on $F$, and it is injective because if two orderings $<_1, <_2$ disagree on some pair of elements $a, b \in F$ then those two elements can't go to the same elements of $\mathbb{Q}_r$ (by uniqueness of the order), so the corresponding homomorphisms also disagree. So:

Corollary: The set of orderings on a number field $F$ (of possibly infinite degree) is in natural bijection with $\text{Hom}(F, \mathbb{Q}_r)$.

Since any map $F \to \mathbb{R}$ necessarily lands in $\mathbb{Q}_r$, we also get:

Corollary: The set of orderings on a number field $F$ is in natural bijection with $\text{Hom}(F, \mathbb{R})$.

This is well-understood. For example if $F$ is finite we can compute this homset by computing the tensor product

$$F \otimes \mathbb{R} \cong \mathbb{R}^{r_1} \times \mathbb{C}^{r_2}$$

where $r_1$ is the number of real embeddings $f \to \mathbb{R}$ and $r_2$ is the number of pairs of conjugate complex embeddings $f \to \mathbb{C}$, as famously occurs in the statement of Dirichlet's unit theorem. So $F$ has a unique ordering iff $r_1 = 1$.

As written this is just a restatement of the corollary, but if we're given more information about $F$ we can calculate $r_1$ and $r_2$ pretty explicitly. For example, as user8268 says in the comments, if

$$F = \mathbb{Q}(\alpha) \cong \mathbb{Q}[t]/m_{\alpha}(t)$$

has a primitive element $\alpha$ with minimal polynomial $m_{\alpha}$, then

$$F \otimes \mathbb{R} \cong \mathbb{R}[t]/m_{\alpha}(t)$$

from which it follows (by the Chinese remainder theorem) that $r_1$ is the number of real roots of $m_{\alpha}.$ Alternatively we can just argue directly that embeddings $F \to \mathbb{R}$ correspond to real roots of $m_{\alpha}.$ This neatly explains the difference between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{2})$, since $x^2 - 2$ has two real roots and $x^3 - 2$ has one, and in general it gives a classification, in some sense, of which number fields (finite this time) have the desired property.

This is essentially a question about the real spectrum, which is the set $\text{Spec}_r \, F$ of all orderings on $F$ equipped with a suitable topology. If $F$ is transcendental over $\mathbb{Q}$ I think it will generally have a large real spectrum (e.g. $\mathbb{Q}(t)$ has many orderings on it since there are many embeddings $\mathbb{Q}(t) \to \mathbb{R}$, and there are also non-Archimedean orderings that don't arise this way), unless it is itself real closed; I don't know if there's a nice precise criterion here.

Answer 2

I'll be taking from Real algebraic geometry by Bochnak, Coste, Roy.

If $P$ is the set of non-negative elements of some order on a field $F$, its called positive cone. The intersection of all positive cones on a field $F$ of characteristic $0$ is the set $\sum F^2 := \{y_1^2+...+y_n^2 : y_1, ..., y_n\in F\}$ of all sums of squares of $F$. It follows that if the order is unique, then $\sum F^2$ is a positive cone of that ordering.

And one can check that if $P_1, P_2$ are positive cones and $P_1\subseteq P_2$, then $P_1 = P_2$. It follows that if $\sum F^2$ is a positive cone, then $F$ has unique ordering.

In terminology of the book I'm citing, if $F$ has an ordering, then $\sum F^2$ is a proper cone. It's a positive cone iff for every $x\in F$, either $x\in \sum F^2$ or $-x\in \sum F^2$.

In order words $F$ has unique ordering iff $F$ has an ordering, and every element is either a sum of squares, or its negative is a sum of squares.

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For example, for $F = \mathbb{Q}(\sqrt[3]{2})$ one can check that $\sqrt[3]{2} = 2(\sqrt[3]{1/2})^2$.