Paul: you ask if there is a way to algebraically characterize the field of real algebraic numbers. As a specific field in C$\mathbf C$, 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 Q$\mathbf Q$. Any two real closures of Q $\mathbf Q$ 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. :)
