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I'm currently learning the basic machinery of valuation theory from "ordered exponential fields" by Kuhlmann, and I have a question regarding an assertion made on page 1.

She gives the standard axioms for a valuation $\nu:M\rightarrow\Gamma\cup\{+\infty\}$ with $M$ a left $R$-module and $\Gamma$ a totally ordered set order extended to $\Gamma\cup\{+\infty\}$. Namely, these axioms are

  1. $\nu$ is surjective,
  2. $\nu(x)=+\infty\iff x=0,$
  3. $\nu(rx)=\nu(x)$ for all $x\in M$ and $0\neq r\in R$,
  4. $\nu(x-y)\geq\min\{\nu(x),\nu(y)\}$ for all $x,y\in M$.

She then goes on to assert that these axioms imply the following three predicates for all $x,y\in M$:

  1. $\nu(-x)=\nu(x)$,
  2. $\nu(x)\neq\nu(y)\implies\nu(x+y)=\min\{\nu(x),\nu(y)\}$,
  3. $\nu(x+y)>\nu(x)\implies\nu(x)\neq\nu(y).$

It is not clear to me how these predicates follow, and I believe we can construct a specific example where they do not.

Let $\mathbb{Q}(t)$ be the field of of rational functions over $\mathbb{Q}$ viewed as a $\mathbb{Q}$-module, and for all $\frac{p}{q}\in\mathbb{Q}(t)$ let $\uparrow^p$ and $\uparrow^q$ be the highest exponents with nonzero coefficients in $p$ and $q$ respectively. We then define a mapping $\nu:\mathbb{Q}(t)\rightarrow\mathbb{Z}\cup\{+\infty\}$ given by $$\nu(\frac{p}{q})=\uparrow^p-\uparrow^q\ \text{if}\ \frac{p}{q}\neq0,$$ $$\nu(0)=+\infty.$$ We then have that

  1. $n=\nu(\frac{t^n}{1})$ and $-n=\nu(\frac{1}{t^n})$ for all $n\in\mathbb{N}$, so $\nu$ is surjective.
  2. $\nu(x)=+\infty\iff x=0$ is immediate.
  3. $\nu(x\cdot\frac{p}{q})=\nu(\frac{p}{q})$ for all $0\neq x\in\mathbb{Q}$ is trivial.
  4. Let $\frac{a}{b},\frac{c}{d}\in\mathbb{Q}(t)$, and w.l.o.g. suppose that $\uparrow^a-\uparrow^b\leq\uparrow^c-\uparrow^d$, so $\uparrow^a+\uparrow^d\leq\uparrow^c+\uparrow^b$. We then have that $$\nu(\frac{a}{b}-\frac{c}{d})=\nu(\frac{ad-bc}{bd})=\uparrow^{ad-bc}-\uparrow^{bd}=\max\{\uparrow^{ad},\uparrow^{bc}\}-\uparrow^{bd}$$ $$=\max\{\uparrow^a+\uparrow^d,\uparrow^b+\uparrow^c\}-(\uparrow^b+\uparrow^d)=\uparrow^c+\uparrow^b-\uparrow^b-\uparrow^d=\uparrow^c-\uparrow^d\geq\uparrow^a-\uparrow^b$$ $$=\min\{\uparrow^a-\uparrow^b,\uparrow^c-\uparrow^d\}=\min\{\nu(\frac{a}{b}),\nu(\frac{c}{d})\},$$ $$\therefore \nu(\frac{a}{b}-\frac{c}{d})\geq\min\{\nu(\frac{a}{b}),\nu(\frac{c}{d})\}.$$

So this mapping satisfies axioms $1-4$ above and is consequently a valuation on $\mathbb{Q}(t)$, but it does not satisfy conditions $2$ or $3$ on the following list.

Am I wrong that this is a valuation, and if not what additional criteria are needed on a valuation to make the 'following predicates' list actually follow?

Alec Rhea
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1 Answers1

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Note that, with your definition, one would have $$\begin{align}\nu((1+t)-t)&=0\\ \nu(1+t)&=1\\ \nu(t)&=1\\ \therefore \nu((1+t)-t) &<\nu(1+t)\wedge \nu(t)\text{.} \end{align}$$

I think your definition is off by a sign. Let $k$ be a field, and let $k((t^{-1}))$ be the ring of formal Laurent series (with only finitely nonvanishing terms of positive degree, but possibly infinitely many nonvanishing terms of negative degree). Then $k(t)$ embeds in $k((t^{-1}))$ (by taking the Laurent series at infinity), and $$\mathrm{ord}_{\infty}f \stackrel{\text{def}}{=}\text{negative of the largest }m\text{ such that }t^m\text{ appears with nonzero coefficient in }f$$ is a valuation. In fact $k[[t^{-1}]]$ is a discrete valuation ring with field of fractions $k((t^{-1}))$.

K B Dave
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  • This is exactly why I was wrong, thank you. Nlab suggests a valuation on rational functions that I believe is equivalent but phrased differently than the one you suggest here, I just couldn't see why it worked but mine did not. – Alec Rhea Feb 27 '18 at 04:08
  • @AlecRhea : Maybe it'd be helpful to explain my thought process. For me, a "valuation" is something that takes "a function" and gives you "the order of zero it has at a point", with negative values corresponding to poles—instead of remembering the formal definition, I just work out what's supposed to happen in the case of functions. In this particular case, we want the "order of the zero as $t$ approaches infinity", which is exactly the negative of the degree. – K B Dave Feb 27 '18 at 04:16
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    Very cool! I'm using valuation theory to explore non-Archimidean ordered fields built from large ordinals, so there are elements like $\frac{\omega_1+\omega^\omega-3}{5\omega^2-4}$ and so on which do not canonically admit representations as functions, but I wonder if there is some interpretation for functions with zeroes or poles of transfinite degree, maybe over the surreals. – Alec Rhea Feb 27 '18 at 04:23