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Let $\Gamma$ be a totally ordered abelian group (written additively), and let $K$ be a field. A valuation of $K$ with values in $\Gamma$ is a mapping $v:K^* \to \Gamma$ such that

  • $1)$ $v(xy)=v(x)+v(y)$

  • $2)$ $v(x+y) \ge \min (v(x),v(y))$,
    for all $x,y \in K^*$.

Show that the set $A=\{x\in K^*:v(x) \ge 0\} \cup \{0\}$ is a valuation ring of $K$.

I want to show that $-1 \in A$. From $v(1)=v(1)+v(1)$, $v(1)=0$ and $v(1)=v(-1)+v(-1)$ so that $2v(-1)=0$. Is it possible to conclude that $v(-1)=0$?

In general for any totally ordered abelian group, does $2x=0$ imply $x=0$?

2 Answers 2

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Yes. If $x$ is positive then so is $nx$ for any $n \in \mathbb{Z}^+$ and if $x$ is negative then $-x$ is positive and $nx = 0$ iff $n(-x) = 0$.

In other words, a totally ordered abelian group is necessarily torsionfree. More interestingly, the converse also holds: any torsionfree abelian group can be totally ordered (in at least one way). See Section 17.2 of these notes for the proof.

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    Thanks! It's very nice. Actually I thought I found a counter-example $\mathbb{Z}_2$ with ordering $0 \ge 0, 1 \ge 0, 1 \ge 1$, but this ordering is not compatible with the addition, and thus not an ordered abelian group by definition.2011-05-20
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It does follow that $\nu(-1)=0$.

Here is an elementary argument. Suppose firstly that $\nu(-1)>0$. Then $2\nu(-1)=\nu(-1)+\nu(-1)>0$; now $2\nu(-1)=0$ implies $0>0$, contradiction.

The same argument, \textit{mutatis mutandis}, shows that you cannot have $\nu(-1)<0$.

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    @gobi: it is an easy consequence of the axioms of an ordered abelian group that the sum of two strictly positive elements is strictly positive. You should try to prove it, and perhaps ask another question if you get stuck.2011-05-20