By the fundamental theorem of algebra, the algebraic closure $\mathbb{K}$ of $\mathbb{Q}$ decomposes as $\mathbb{K} = F \oplus i F$ where $F = \mathbb{R} \cap \mathbb{K}$ (the intersection is in $\mathbb{C}$). I want to know if there is a purely algebraic way to characterize $F$, i.e. without invoking any analysis, topology, or transcendental number theory. I am asking this because I noticed that it is often convenient when working with examples in characteristic 0 algebraic number theory to give preference to the real roots of a polynomial, and I am wondering if there is a canonical algebraic way to formulate this preference. It doesn't seem like an object built out of lots and lots of transcendental extensions should be so fundamental to purely algebraic examples.

Here are some specific questions that I have been playing with.

Is there a purely algebraic way to distinguish between the splitting fields of $x^2 + 2$ and $x^2 - 2$?

Is there a purely algebraic way to distinguish the real root among the three roots of $x^3 - 2$ in a splitting field?

Of course, the relevant algebraic structures can't be invariant under $\mathbb{Q}$ automorphisms. But I don't see why one can't just be a little bit imaginative.

(Examples edited, changing 1 to 2)

notalways possible to extend a number field to a CM field. Any Galois extension of Q is totally real or totally complex, but the totally complex Galois extensions of Q aren't all CM fields since in a CM field the intrinsic complex conjugation is an order 2 aut. that commutes with all other automorphisms in the Galois group. The splitting field of x^3 - 2 over Q has Galois group S_3, whose auts. of order 2 aren't in the center, so this S_3-Galois extension of Q is totally complex but is not CM. $\endgroup$ – KConrad May 24 '10 at 2:22allreal numbers. In particular, if a number in a CM field has one Q-conjugate in R then all of its Q-conjugates are in R. This is why a cube root of 2 is not in any CM field. $\endgroup$ – KConrad May 24 '10 at 2:28