6
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ The invariants of the order 2 automorphism are a number field, which needs to have a real place such that the completion of the original field with respect to this place is complex. $\endgroup$
    – eric
    Dec 14, 2013 at 16:22
  • $\begingroup$ Yes, but is there a criterion that does not require passing to the completion and invoking real and complex numbers? $\endgroup$ Dec 14, 2013 at 18:51
  • 2
    $\begingroup$ It's sort of equivalent, isn't it? for example forget order 2 -- when is the identity on a number field a complex conjugation? Precisely when the number field has a real place. $\endgroup$
    – eric
    Dec 14, 2013 at 23:01
  • $\begingroup$ @Eric I am asking for a criterion stated in terms of equations within $F$. For example it would be great if we could say: an involution on a number field $F$ extends to a complex conjugation iff it fixes every number in $F$ with negative norm. A counter example I gave for this in an earlier version of this comment actually fails but I am still working on that,. $\endgroup$ Dec 15, 2013 at 1:02

1 Answer 1

6
$\begingroup$

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 $F$.

$\endgroup$
1
  • 2
    $\begingroup$ This admits some simplification: $-1$ is not a sum of squares in the fixed field, but is in $F$. Am I wrong? $\endgroup$ Dec 15, 2013 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.