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 10.50.4. Note that in case $A = K$ we obtain the zero group $\Gamma = \{ 0\} $ endowed with its unique total ordering.
$\square$

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