Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x=0$?

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
The converse uses the axiom of choice (every $\mathbb{Q}$-module has a basis) and implies that every set can be totally ordered. – Ricky Demer May 20 '11 at 10:45
@Ricky: sure, okay. (You won't make it very far in commutative algebra without using AC...) – Pete L. Clark May 20 '11 at 11:41
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. – Gobi May 20 '11 at 14:43

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 $\nu(-1)<0$.

share|cite|improve this answer
Does it hold that $v(-1)>0$ and $v(-1)>0$ implies $v(-1)+v(-1)>0$? By definition of an ordered abelian group, $x_1 \ge x_2$ and $y_1 \ge y_2$ implies $x_1+y_1 \ge y_1+y_2$, so I think we can only get $v(-1)+v(-1) \ge 0$. – Gobi May 20 '11 at 15:01
@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. – Pete L. Clark May 20 '11 at 16:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.