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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 $x>y$?

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If the number field is contained in $\Bbb{R}$, shouldn't such a $\phi$ always exist? – user38268 Feb 8 '13 at 11:14
@BenjaLim: The question is, whether the only possible orderings of $K$ are those inherited from an embedding in $\Bbb R$. I see no appareant reason why their couldn't exist other "exotic" orderings of $K$. – Dominik Feb 8 '13 at 11:18
Ok, then use the fact that a field can be totally ordered iff $a_1^2+a_2^2+\dots+a_n^2=0 \implies a_1=a_2=\dots=0$. – Berci Feb 8 '13 at 11:47
@Berci: I don't see how this fact relates to the question how to extend the total order of $K$. – Dominik Feb 8 '13 at 13:13
Your question is a little bit vague, but I assume you mean that $K$ is a number field that is also an ordered field, whose order is total. That is the order structure and the field structure play nice. At any rate the key term here is formally real field the answer may be somewhere in these notes. – JSchlather Feb 8 '13 at 19:39
up vote 4 down vote accepted

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 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.

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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.

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How do you know the rationals are dense in $K$? (It's true, but I'm wondering whether it's clear.) – Pete L. Clark Feb 10 '13 at 3:18
@Pete Well, number fields are Archemedian. I don't see an immediate argument better than "obviously", but I am already in bed. I'll give it some thought during the first two dreams, and the last one. I hope it turns up nicely in the morning. :-) – Asaf Karagila Feb 10 '13 at 3:22
$@$Asaf: the natural orderings on number fields are Archimedean, which, because the answer to this question is **yes**, are all the orderings. But if you contemplate an arbitrary ordering of a number field? I feel confident that if $L/K$ is a finite degree field extension and $\sigma$ is an ordering of $L$, then $(L,\sigma)$ is Archimedean iff $(K,\sigma|K)$ is Archimedean, but this too seems to require proof. – Pete L. Clark Feb 10 '13 at 3:25
Hmm, I think I see how to prove that: the real closure of an ordered field $K$ is unique up to $K$-isomorphism, so if $L/K$ is a finite degree ordered extension it must be realizable as an ordered subfield of the real closure. Since $\mathbb{R}_0$ is a subfield of the Archimedean ordered field $\mathbb{R}$, it must itself be Archimedean. – Pete L. Clark Feb 10 '13 at 3:30
For a simpler proof that $(L,\sigma)$ is Archimedean iff $(K,\sigma|K)$ is see David Speyer's comments at… (he does it for $K=\mathbb{Q}$ but the proof goes through for this as well). And $L/K$ only needs to be algebraic, not necessarily finite. – Jarek Kuben Oct 18 '15 at 0:44

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