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Natural valuation has a convex valuation ring, in fact the valuation ring is the convex hull of $\mathbb{Z}$.

How to prove that every ordered field $K$ has a natural valuation $v$, whose residue field is an archimedean ordered field.

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  • $\begingroup$ Well you can first try to prove that the convex hull of $\mathbb{Z}$ in $K$ is a valuation ring of $K$. $\endgroup$ – nombre Dec 2 '18 at 13:40
  • $\begingroup$ There is a theorem that gives that result right away. Convex hull of any subring in ordred field K is valuation ring. Therefore this one too. $\endgroup$ – XYZ Dec 2 '18 at 14:03
  • $\begingroup$ Okay, then do you know how the order on this residue field is defined? If so, given $x \in \operatorname{Hull}(\mathbb{Z})$, can you find $n \in \mathbb{N}$ with $\overline{x} \leq \overline{n}$? (where $\overline{a}= a+\mathfrak{m}$ and $\mathfrak{m}$ is the maximal ideal of the valuation ring) $\endgroup$ – nombre Dec 2 '18 at 15:28
  • $\begingroup$ I know that $P=\{\bar a : a\in \mathbb{Z} , a\geq 0 \}$ is an ordering in residue field. How to use it to find $n$? $\endgroup$ – XYZ Dec 2 '18 at 16:18
  • $\begingroup$ Correction: $a$ in ordering P is in convex hull of $\mathbb{Z}$ $\endgroup$ – XYZ Dec 2 '18 at 16:29

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