Lemma 10.50.12. Let $A$ be a valuation ring with field of fractions $K$. Set $\Gamma = K^*/A^*$ (with group law written additively). For $\gamma , \gamma ' \in \Gamma $ define $\gamma \geq \gamma '$ if and only if $\gamma - \gamma '$ is in the image of $A - \{ 0\} \to \Gamma $. Then $(\Gamma , \geq )$ is a totally ordered abelian group.

**Proof.**
Omitted, but follows easily from Lemma 10.50.3. Note that in case $A = K$ we obtain the zero group $\Gamma = \{ 0\} $ endowed with its unique total ordering.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)