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 $\mathbb{Q}$?



Is any/every order on a number field forced by some finite extension?
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 $\mathbb Q(\sqrt[3]2)$.


