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Let $D$ be a Dedekind domain. Let $v:D \to \mathbb{R}$ a valuation. We know that for every prime ideal $\mathfrak p$ of $D$ the localization $D_{\mathfrak p}$ is DVR.

Does every valuation on $D$ directly arise from some valuation on some $D_{\mathfrak p}$ ?

How can we find all the valuations $v:D \to \mathbb{R}$ ?

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When you say valuations, do you mean discrete valuations? – user27126 Jan 13 '13 at 19:22
No. I mean function $v:D\to \mathbb{R}$ such that $v(xy)=v(x)+v(y)$ and $v(x+y) \geq min\{v(x),v(y)\}$. – colge Jan 13 '13 at 19:36
If $D\subseteq O_v$, then $v$ is discrete. – user26857 Jan 13 '13 at 19:40
The valuations you are considering are exactly the valuations of the field of fractions $K=\mathrm{Frac}(D)$, so you can't expect that it comes from some maximal ideal of $D$ (consider the $p$-adic valuation on $\mathbb Z[1/p]$) unless you impose $v$ to take non-negative values in $D$. – user18119 Jan 14 '13 at 16:10
And if $v$ only takes non-negative values in $D$ (i.e. $D\subseteq O_v$), then $v$ is a discrete valuation ring (see YACP's comment) and equals to some $D_p$. – user18119 Jan 14 '13 at 17:02
up vote 4 down vote accepted

If $D$ is the ring of integers of some number field, then the answer is yes by Ostrowski's theorem over $\mathbb Q$.

Otherwise, even for function fields, not all (non-Archimedean) valuations are discrete, so not all of them come from a localization $D_p$. For example, fix an irrational $\theta\in \mathbb R$ and a prime number $p$, and consider the Gauss valuation $$ \sum_{n\ge 0} a_n t^n \mapsto \min_n \{ v_p(a_n)+n\theta \}$$ on $\mathbb Q[t]$. Its values group $\mathbb Z+\theta \mathbb Z$ is dense in $\mathbb R$.

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