# What is a valuation associated to an ordering on a field?

If $(K,\leq)$ is a totally ordered field with $P\!=\!\{\alpha\!\in\!K;\, 0\!\leq\!\alpha\}$, how is the valuation associated to $P$ defined?

I was searching through Prestel & Delzell's Positive Polynomials and Engler & Prestel's Valued Fields, but didn't find anything. Perhaps I didn't search thoroughly enough. Google also didn't provide much. Any reference is welcome.

I must calculate the valuation $v$ on $\mathbb{R}(x)$ associated to $P\!=\!\{x^kf(x);\, k\!\in\!\mathbb{Z}, f\!\in\!\mathbb{R}(x), 0\!<\!f(0)\!<\!\infty\}$.

Furthermore, given an ordering $\leq$ and valuation $v$ on $K$, when is $v$ compatible with $\leq$ (definition)?

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 $v$ is compatible with "$\le$" if $0 ## 1 Answer This doesn't really deserve to be an answer but I don't have enough points to make a comment, so here it is: As navigetor23 says, a valuation is order compatible if$0<a\leq b$implies$v(a)\geq v(b)$. But, for me at least, a nicer equivalent formulation is that the valuation ring$\mathcal{O}_v$corresponding to$v\$ is convex.

There are three more equivalent conditions on page 3 of the following link (along with other useful results):