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Suppose $v : K \rightarrow Γ$ is a surjective valuation, where $K$ is the field and $Γ$ is the value group. The set $A = \{x : v(x) \geq 0\}$ is a valuation ring and $U=\{x :v(x)=0\}$ are the units of $A$. I can see how $Γ$ is isomorphic to $K/U$, so $f: K\rightarrow K/U$ being the natural homomorphism is the same as $v$. So $(K,Γ,v)$ is isomorphic to $(K,K/U,f)$. But the problem I'm having is that apparently given any valuation ring $R$, the units of $R$, $V$, will make the same valued field. I am not sure why this is. If this is true, does this mean a field can only have one valuation ring?

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    $\begingroup$ A field can have many different valuation rings. Take $p$-adic valuations on $\mathbb{Q}$. These do all have the same valuation group though. $\endgroup$ – 0x539 Feb 21 at 19:41
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    $\begingroup$ Maybe you mean $f:K^\times \rightarrow K^\times /U$. -- As for your question, remember that every field, on top of interesting valuations it might have, has the trivial valuation $v(x) := \begin{cases} 0 \text{ if } x \neq 0 \\\infty \text{ if } x=0\end{cases}$, for which $U=K^\times$ and $\Gamma = \lbrace 0 \rbrace$. $\endgroup$ – Torsten Schoeneberg Feb 22 at 16:59
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Let $R$ be a valuation ring and let $K$ be its field of fractions. Then the natural map

$ v:K^\times\rightarrow K^\times/R^\times $

is a valuation of $K$ and $R$ is its valuation ring. The abelian group on the right hand side is ordered by

$ xR^\times\leq yR^\times :\Leftrightarrow yx^{-1}\in R. $

This fact together with what you already formulated yields that there is a bijection between the set of all valuation rings of $K$ and the set of equivalence classes of surjective valuations of $K$. Here two valuations $v:K^\times\rightarrow\Gamma_v$ and $w:K^\times\rightarrow\Gamma_w$ are equivalent iff there exists an order-preserving isomorphism $\phi:\Gamma_v\rightarrow\Gamma_w$ such that $w=\phi\circ v$.

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