Is there a purely algebraic criterion which characterizes the real algebraic numbers? 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)
 A: One of your questions does have  a positive answer. 


*

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


The field $Q(\sqrt 2)$ can be ordered while the field $Q(i\sqrt 2)$ can not since
-1 is a sum of squares.
A: The real root of $x^3-1$ satisfies $x=1$, which the non-real roots do not, but I cannot imagine any purely algebraic way to distinguish the real root of $x^3-2$ from the others. This may say more about my imagination than it does about the cube roots of two, but it seems to me that any purely algebraic equation satisfied by any one of the roots is satisfied by all three, making it impossible to distinguish them in any purely algebraic way. 
A: As other correspondents have pointed out there is no algebraic way
to distinguish the elements of $\mathbb{Q}^{alg}\cap\mathbb{R}$ but there
is an algebraic way of distinguishing totally real algebraic numbers
(those whose conjugates are all real). An algebraic number $\alpha$
is totally real iff $n-r=1$ where $n$ is the degree of $K=\mathbb{Q}(\alpha)$
and $r$ is the rank of the unit group of the integral closure of
$\mathbb{Z}$ in $K$. This follows from Dirichlet's units theorem.
A: Paul:  you ask if there is a way to algebraically characterize the field of real algebraic numbers. As a specific field in $\mathbf C$, no there's not a good algebraic characterization, but as an abstract field yes there is a characterization.  This field is one particular example (and the only concrete one at that) of a real closure of $\mathbf Q$.  Any two real closures of $\mathbf Q$ are isomorphic to each other.  
If you pick a number field $K$ other than the rationals, you can contemplate its real closures: maximal algebraic extensions of $K$ which admit an ordering.  Assuming this is possible at least once (e.g., ${\mathbf Q}(i)$ has no real closure), then you can ask if the real closures of $K$ are all isomorphic to each other respecting the embedding of $K$ into them. Nope.
Consider $K = {\mathbf Q}(a)$ where $a^2 = 2$.  If we stuff $K$ into the real algebraic numbers by sending $a$ to $\sqrt{2}$ then $a$ is a square in the real algebraic numbers (it's the square of $\sqrt[4]{2}$, which is a real algebraic number: we're talking about concrete real numbers that are algebraic). But if we stuff $K$ into the real algebraic numbers by sending $a$ to $-\sqrt{2}$ then $a$ is not a square in the real algebraic numbers (all squares in the real numbers are positive).  Therefore these two embeddings of $K$ into the real alg. numbers are not compatible with each other as extensions of $K$.  That is, there is no automorphism of the real algebraic numbers which commutes with these two embeddings of $K$ into it.  In other words, real closures of $K$ are not all isomorphic as extensions of $K$.
Theorem: Let $K$ be a number field.  Every real closure of $K$, up to isomorphism as an  extension of $K$, looks like the real algebraic numbers using some real embedding of $K$, and different real embeddings lead to non-isomorphic real closures as extensions of $K$.
I think I got that right.  If I screwed up I'm sure BCnrd will let me know. :)
A: The answer to your title question is "no", assuming "purely algebraic criterion" means something like a first order statement using only ring operations.  This is because there are ring-theoretic automorphisms of the algebraic numbers that take the real algebraic numbers to some other real closed subfield (e.g., choose one that permutes the cube roots of 2 transitively, i.e., any element of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ whose image in $Gal(\mathbb{Q}[\sqrt[3]{2}, \frac{1+\sqrt{-3}}{2}]/\mathbb{Q})$ has order 3).
