In following [post](http://www.mathcounterexamples.net/a-field-that-can-be-ordered-in-two-distinct-ways/), 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?



Ways to order an algebraic extension
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 $k=\mathbb Q$, $K=\mathbb Q(\{\sqrt p:p\text{ prime}\})$.


