Let $F$ be a number field with a prime ideal $\mathfrak{p}$. I saw a claim that the localization
$$\mathcal{O}_{F,(\mathfrak{p})} = \{\frac{a}{b} : a\in \mathcal{O}_F, b\in\mathcal{O}_F\setminus\frak{p}\}/\sim$$
when viewed as a subring of $F$, can be characterized as
$$\{x \in F : v_{\frak{p}}(x)\geq 0\}$$
It's easy to see that the localization is contained in this set, but I can't show the reverse inclusion. That is, suppose I have a fraction $\frac{a}{b}\in F$ with $a,b\in\mathcal{O}_F$, with $v_{\frak{p}}(\frac{a}{b})\geq 0$. There's certainly no guarantee that $b\notin \mathfrak{p}$, so generally we need to find an equivalent fraction $\frac{a'}{b'}$ with $b'\notin\frak{p}$. If $\mathcal{O}_F$ were a principle ideal domain, this would be straightforward because $\mathfrak{p}=(p)$ for some $p$ and I could just factor out powers of $p$. But $\mathcal{O}_F$ will only be a Dedekind domain, and then I can't factor out generators.
Ideally I would find some way to show this for, e.g., a Noetherian domain, but even if I assume it's a Dedekind domain and thus $\mathfrak{p}=\langle x,y\rangle$ for some $x,y$, I can't find any way to play with the fraction to help.
Is this statement true, and can anyone offer a proof?
That's because the ring of integers $\mathcal{O}_F$ of a number field $F$ is a Dedekind domain: among others, every (nonzero) fractional ideal can be written uniquely as a product of maximal ideals (or their inverses).
Then if $x\in F$, $v_{\mathfrak{p}}(x)$ is just the power to which $\mathfrak{p}$ appears in the decomposition of the fractional ideal $xA$. In particular, assume $v_{\mathfrak{p}}(x)\geq 0$: then $x=a/b$, where $a,b\in \mathcal{O}_F$, $a\mathcal{O}_F=\mathfrak{p}^{\alpha}I$, $b\mathcal{O}_F=\mathfrak{p}^\beta J$, where $I,J$ are ideals prime to $\mathfrak{p}$ (ie $I+\mathfrak{p}=J+\mathfrak{p}=\mathcal{O}_F$) and $\alpha-\beta=v(x) \geq 0$.
In particular, $Ja = J\mathfrak{p}^{\alpha}I = b\mathfrak{p}^{\alpha-\beta}I \subset \mathcal{O}_F$, so that $Jx \subset \mathcal{O}_F$, and thus $x \in \mathcal{O}_F[t^{-1}]$ for any $t \in J$. Since $J+\mathfrak{p}=\mathcal{O}_F$, we can choose $t \notin \mathfrak{p}$ and thus $x \in \mathcal{O}_F[t^{-1}] \subset \mathcal{O}_{F,(\mathfrak{p})}$.