Timeline for Is any/every order on a number field forced by some finite extension?

6 events

when toggle format	what		by	license	comment
Jan 27, 2014 at 17:33	vote	accept	Colin McLarty
Jan 27, 2014 at 16:27	answer	added	Emil Jeřábek		timeline score: 3
Jan 27, 2014 at 16:26	comment	added	David E Speyer		So it remains to show that every ordering of $F$ comes from an embedding. To this end, it is enough to show that every ordering of a number field is archimedean. Let $x \in F$ with minimal polynomial $x^d + \sum_{i<d} a_i x^i$. It is easy to write down an explicit $N$ in $\mathbb{Q}$ so that the minimal polynomial of $x-N$ has all positive coefficients, hence no positive roots, and thus $x < N$ in any ordering.
Jan 27, 2014 at 16:25	comment	added	David E Speyer		In case you don't already know this, orderings of a number field $F$ are in bijection with embeddings $\sigma: F \to \mathbb{R}$. Proof: Clearly, every such an embedding gives an ordering. If an ordering comes from an embedding $\sigma$, then we can recover $\sigma(a)$ as $\mathrm{sup}(q \in \mathbb{Q} : q < a)$.
Jan 27, 2014 at 16:10	comment	added	Emil Jeřábek		Every positive (or negative, for that matter) $x\in F$ is a sum of two squares in $E=F(\sqrt{-1})$. I suppose you should ask $E$ to be formally real to avoid this. Also, could you clarify the second question?
Jan 27, 2014 at 16:02	history	asked	Colin McLarty	CC BY-SA 3.0