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In following post, I describe the "classical" example of $\mathbb{Q}(\sqrt{2})$ that can be ordered in two distinct ways.

More generally, if $(k,P)$ is an ordered field, $R$ a real closure of $(k,P)$ and $K/k$ an algebraic extension, how can we describe the orders of $K$ that extend $P$?

Can one produce a case where the number of orders of $K$ that extend $P$ is infinite?

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Orders on $K$ extending $P$ are in a 1–1 correspondence with $k$-embeddings of $K$ in $R$.

In particular, there can be infinitely many only if $K/k$ has infinite degree; a simple example is $k=\mathbb Q$, $K=\mathbb Q(\{\sqrt p:p\text{ prime}\})$.

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  • $\begingroup$ Thanks Emil for your response. While I have an idea to prove that to a $k$-embedding of $K$ in $R$ is associated an order of $K$, I don't see how to prove that there is not more orders. Do you have a hint for that? $\endgroup$
    – mathcounterexamples.net
    Commented Sep 30, 2014 at 4:48
  • $\begingroup$ This follows from the existence and uniqueness of real closures. If $(K,Q)$ is an order extending $P$, let $S$ be its real closure. Since $K/k$ is algebraic, $S$ is also a real closure of $(k,P)$, hence by uniqueness, there exists a $k$-isomorphism $S\to R$. It restricts to an order-preserving $k$-embedding $K\to R$. By a similar argument, the fact that two different embeddings cannot induce the same order follows from uniqueness of the $K$-isomorphism of two real closures of $K$. $\endgroup$
    – Emil Jeřábek
    Commented Sep 30, 2014 at 9:20

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