Embedding ordered number fields into $\Bbb R$

Question

Consider a number field $K$ equipped with a total order $\le$. Is there always a field homomorphism $\phi:K \rightarrow \Bbb R$ which respects the total order, i. e. with $\phi(x)>\phi(y)$ for $x>y$?

Answer 1

Here is Theorem 11.6 from N. Jacobson's Basic Algebra II:

Let $F$ be an algebraic number field, and let $\mathbb{R}_0$ be the field of real algebraic numbers [i.e., those real numbers which are algebraic over $\mathbb{Q}$]. Then we have a $1-1$ correspondence between the set of orderings of $F$ and the set of monomorphisms of $F$ into $\mathbb{R}_0$. The ordering determined by the monomorphism $\sigma$ is that in which $a > 0$ for $a \in F$ if $\sigma a > 0$ in $\mathbb{R}_0$.

It follows from this that the orderings on a number field $F$ are in bijection with the number of roots of any minimal polynomial $P$ for $F$ in $\mathbb{Q}_0$. One knows -- e.g. by model completeness of the theory of real-closed fields -- that this is the same as the number of roots of $P$ in $\mathbb{R}$ and thus the same as the number of embeddings of $F$ into $\mathbb{R}$. Thus the number of (total, compatible with the field structure) orderings of a number field $F$ is equal to the number of embeddings of $F$ into $\mathbb{R}$.

Added: If you don't like the model completeness bit, you can replace it with the fact that any homomorphism of real-closed fields must be order-preserving. Indeed, if not then by restriction one would get a different ordering on the smaller real-closed field, but the ordering on a real-closed field is unique: the positive elements are precisely the nonzero squares.

If you are interested in the proof of this theorem (and don't have ready access to Jacobson's book), please let me know.

Answer 2

Take $L$ the Dedekind completion of $K$. Because the rational nmbrrs are dense in $K$ we have that $L$ must be isomorphic to the reals (as an ordered field). Now restrict the isomorphism to $K$ and we are done.

Of course one has to make a severe distinction between orderable and ordered, one only requires that an order exists, and the other specifies such order. Pete L. Clark has answered regarding to how many orders are compatible with a number field.