Totally ordered abelian group 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 $x=0$?
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.
It does follow that $\nu(-1)=0$.

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

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