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Say a field extension $E/F$ forces the order on an ordered $F$ if every positive $x$ in $F$ is a sum of squares in $E$. A real closure of $F$ does this. And $\mathbb{Q}$ forces its own sole ordering.

Does every ordered number field $F$ have some finite extension $E/F$ forcing the order? Is there ever such an extension, apart from the case of $\mathbb{Q}$?

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  • $\begingroup$ Every positive (or negative, for that matter) $x\in F$ is a sum of two squares in $E=F(\sqrt{-1})$. I suppose you should ask $E$ to be formally real to avoid this. Also, could you clarify the second question? $\endgroup$
    – Emil Jeřábek
    Commented Jan 27, 2014 at 16:10
  • $\begingroup$ In case you don't already know this, orderings of a number field $F$ are in bijection with embeddings $\sigma: F \to \mathbb{R}$. Proof: Clearly, every such an embedding gives an ordering. If an ordering comes from an embedding $\sigma$, then we can recover $\sigma(a)$ as $\mathrm{sup}(q \in \mathbb{Q} : q < a)$. $\endgroup$
    – David E Speyer
    Commented Jan 27, 2014 at 16:25
  • $\begingroup$ So it remains to show that every ordering of $F$ comes from an embedding. To this end, it is enough to show that every ordering of a number field is archimedean. Let $x \in F$ with minimal polynomial $x^d + \sum_{i<d} a_i x^i$. It is easy to write down an explicit $N$ in $\mathbb{Q}$ so that the minimal polynomial of $x-N$ has all positive coefficients, hence no positive roots, and thus $x < N$ in any ordering. $\endgroup$
    – David E Speyer
    Commented Jan 27, 2014 at 16:26

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I believe the answer to the first question is yes. Since ordered algebraic extensions of $\mathbb Q$ are archimedean, all orderings on a number field are induced by its real places, and in particular, there are only finitely many. Thus, there is a finite set $\{a_1,\dots,a_n\}\subseteq F$ such that every order is uniquely determined by the signs of $a_1,\dots,a_n$. For a given ordering $<$, let $E\subseteq\mathrm{rcl}(F,<)$ be an extension of $F$ containing $\sqrt{a_i}$ for $a_i>0$, and $\sqrt{-a_i}$ for $a_i<0$. Then every order on $E$ extends $<$, which implies that every $<$-positive element of $F$ is a sum of squares in $E$.

Also, if I am to interpret the second question as “do there exist ordered number fields other than $\mathbb Q$ that force their own order”, the correspondence of orderings with real embeddings shows that this happens if and only if the field has a unique real embedding. Examples abound, such as $\mathbb Q(\sqrt[3]2)$.

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  • $\begingroup$ Moreover, you only need four squares. Siegel proved the following (see mathoverflow.net/a/14473/297 ): Let $E$ be a number field and $x \in E$. Then we have $x = a^2+b^2+c^2+d^2$ with $a$, $b$, $c$, $d \in E$ if and only if $\sigma(x) \geq 0$ for every embedding $\sigma: E \to \mathbb{R}$. $\endgroup$
    – David E Speyer
    Commented Jan 27, 2014 at 16:30
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    $\begingroup$ The (more elementary, I believe) result I intended to use at the end is that for any field, elements positive under all its orderings are exactly those that are sums of squares. $\endgroup$
    – Emil Jeřábek
    Commented Jan 27, 2014 at 16:32
  • $\begingroup$ I realized that a little after writing the comment; hence the edit to my comment (which I made before seeing your reply). $\endgroup$
    – David E Speyer
    Commented Jan 27, 2014 at 16:33
  • $\begingroup$ All right. Thanks for the additional information. $\endgroup$
    – Emil Jeřábek
    Commented Jan 27, 2014 at 16:37

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