A valuation ring is recovered from its value group

The order relation of [L2] is well defined on cosets: if $x^{\mathrm{\prime }}=ux$ and $y^{\mathrm{\prime }}=vy$ with $u,v\in V^{\times }$, then $y^{\mathrm{\prime }}\mathrm{/}{x}^{\mathrm{\prime }}=(v\mathrm{/}u)(y\mathrm{/}x)$, and [A1] shows that $y^{\mathrm{\prime }}\mathrm{/}{x}^{\mathrm{\prime }}\in V$ exactly when $y\mathrm{/}x\in V$.

The order is reflexive because $x\mathrm{/}x=1\in V$. It is antisymmetric because if $\bar{x}\le \bar{y}$ and $\bar{y}\le \bar{x}$, then both $y\mathrm{/}x$ and $x\mathrm{/}y$ lie in $V$, so $y\mathrm{/}x$ is a unit of $V$. Hence $x$ and $y$ represent the same coset in $\mathrm{\Gamma }$.

The order is transitive because $y\mathrm{/}x\in V$ and $z\mathrm{/}y\in V$ imply $z\mathrm{/}x=(z\mathrm{/}y)(y\mathrm{/}x)\in V$. It is total because $V$ is a valuation ring: for any $x,y\in K^{\times }$, the quotient $y\mathrm{/}x$ either lies in $V$ or has inverse $x\mathrm{/}y$ in $V$. It is translation-invariant because $\bar{x}+\bar{z}\le \bar{y}+\bar{z}$ is equivalent to $(yz)\mathrm{/}(xz)=y\mathrm{/}x\in V$. Thus $\mathrm{\Gamma }$ is a totally ordered abelian group.

Define $v(0)=\mathrm{\infty }$ and $v(x)=\bar{x}$ for $x\mathrm{\ne }0$. Then $v(x)=\mathrm{\infty }$ exactly when $x=0$, and for $x,y\mathrm{\ne }0$ one has $v(xy)=\overline{xy}=\bar{x}+\bar{y}=v(x)+v(y)$. If $x+y=0$, then $v(x+y)=\mathrm{\infty }\ge \min \{v(x),v(y)\}$. Otherwise, after swapping $x$ and $y$ if needed, step 1.3 gives $v(x)\le v(y)$, so $y\mathrm{/}x\in V$ and $x+y=x(1+y\mathrm{/}x)$ with $1+y\mathrm{/}x\in V$; hence $v(x+y)\ge v(x)=\min \{v(x),v(y)\}$. Therefore $v$ satisfies the valuation axioms of [L1].

The nonnegative locus of $v$ is exactly $V$: for $x\in K^{\times }$, the condition $0\le v(x)$ means $\bar{1}\le \bar{x}$, which by [L2] is equivalent to $x\in V$. Since $0\in V$ as well, the valuation ring of $v$ is precisely $V$.