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)?

  • $\begingroup$ $v$ is compatible with "$\le$" if $0<b\le a$ implies $v(a)\le v(b)$. Which order are you using on $\mathbb{R}(x)$, because in the "canonical" order I can't see why is your $P$ the set of non-negative elements of $\mathbb{R}(x)$. $\endgroup$
    – user26857
    Jul 29, 2012 at 8:26
  • $\begingroup$ Thank you for the definition of compatible. I am using the positive cone $P$ as defined above, so the ordering is $f\!\leq\!g\Leftrightarrow g\!-\!f\!\in\!P$. $\endgroup$
    – Leo
    Jul 30, 2012 at 1:48

1 Answer 1


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):



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