Every automorphism of an algebraic number field $F$ extends to an automorphism of $\mathbb{\overline{Q}}$, but an order 2 automorphism of $F$ need not extend to one of order 2 on $\mathbb{\overline{Q}}$, 


Is there an algebraic criterion, just looking at the action on $F$, to tell whether a given involution on $F$ will extend to a complex conjugation? 


Trivially, when $F$ is quadratic, an automorphism extends to a complex conjugation iff it fixes every number with negative norm. That just says either the automorphism is the identity or the field is imaginary. 


Does something like this generalize to non-trivial cases?



Can you identify complex conjugations in a number field?
An involution $\sigma$ is a complex conjugation if not every element of $F^\sigma$ is a sum of squares of elements of $F^\sigma$ and, if $\sigma$ is nontrivial, some element of $F^\sigma$ which is not a sum of squares of elements of $F^\sigma$ is a sum of squares of elements of $F$.


Proof that every complex conjugation satisfies this: $F^\sigma$ has an embedding into the reals. Since every negative real number is not a sum of squares, some element in $F^\sigma$ is not a sum of squares. If $\sigma$ is nontrivial, there is a complex number in $F$ of the form $a+bi$ with $|b|>|a|$. Then $(a+bi)^2 + (a-bi)^2 = 2a^2-2b^2 <0$, so some negative number is a sum of squares in $F$. Of course, negative numbers are not a sum of squares in $F^\sigma$.


Proof that every $\sigma$ that satisfies this is a complex conjugation: $F^\sigma$ is a formally real field, hence it can be given an ordering, and hence an embedding into $\mathbb R$. Moreover, if $x\in F^\sigma$ is an element that is not a sum of squares in $F^\sigma$, then it can be given an ordering such that $x$ is negative. This is because the preordering consisting of the sums of squares can be extended to a preordering including $x$, and thus by Zorn's lemma to an ordering of the field. For $x\in F^\sigma$ an element which is not a sum of squares in $F^\sigma$ but is a sum of squares in $F$, applying this construction gives a real embedding of $F^\sigma$ that extends to a complex embedding of $F$.


