I've been trying to prove this claim that I think is true, but I'm getting more and more sure it's not true. Here's the set up: Let $\mathcal{O}$ be a Dedekind domain, and $K$ its field of fractions. Given a discrete valuation $v$ of $K$, I'm trying to show that the valuation ring $\mathcal{O}_v := [x \in K | v(x) \geq 0]$ contains the original Dedekind domain $\mathcal{O}$.
My current attempts tried exploiting the fact that they both contain an image of $\mathbb{Z}$, and seeing where I can go from there. Spoiler alert: nowhere, really. Is my claim even true? If it is, am I at least on the right track? If it isn't, is there a modification to make it true?
It is false: the Dedekind domain (DVR in fact) $\mathbb{Z}_{(2)}=\{a/b|a,b\in \mathbb{Z},b\equiv 1 \bmod 2\}$ has field of fractions $\mathbb{Q}$. The valuation $v_p$ of $\mathbb{Q}$ for $p\ne 2$ has valuation ring $\mathbb{Z}_{(p)}=\{a/b|a,b\in \mathbb{Z},b\not\equiv 0 \bmod p\}$, and $\mathbb{Z}_{(2)}\not\subset \mathbb{Z}_{(p)}$.