# Can a given field have multiple value groups?

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?

• A field can have many different valuation rings. Take $p$-adic valuations on $\mathbb{Q}$. These do all have the same valuation group though. Feb 21 '19 at 19:41
• 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 $\Gamma = \lbrace 0 \rbrace$. Feb 22 '19 at 16:59

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 $$w=\phi\circ v$$.