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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 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 $\mathbb{R}(x)$.2012-07-29
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    Thank you for the definition of *compatible*. I am using the positive cone $P$ as defined above, so the ordering is $f\!\leq\!g\Leftrightarrow g\!-\!f\!\in\!P$.2012-07-30

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