Let $K/\mathbb{Q}$ be a number field and $v$ a finite valuation of $K$. We can consider the completion $K_v$, which is a finite extension of $\mathbb{Q}_v$.
We can define a "$v$-adic ring of integers" by $$\mathcal{O}_v=\{x\in K_v : P_{\text{min},K_v/\mathbb{Q}_v, x}\in \mathbb{Z}_p[X]\}.$$ The "regular" ring of integers $\mathcal{O}$ of $K/\mathbb{Q}$ is of course contained in $$\mathcal{O}':=K\cap\left(\bigcap \mathcal{O}_v\right).$$ Does the other inclusion hold (i.e. $\mathcal{O}=\mathcal{O}'$) ?
Being given that an element of $\mathbb{Q}[X]$ that belongs to $\mathbb{Z}_p[X]$ for all $p$ belongs to $\mathbb{Z}[X]$, I think that this is true, but I did not succeed in formalizing this.
Indeed, if $x\in \mathcal{O}'$, we only have the relationship $\mathbb{Z}_p[X]\ni P_{\text{min},K_v/\mathbb{Q}_v,x}\mid P_{\text{min},K/\mathbb{Q},x}$ in $\mathbb{Q}_p[X]$, so we could not say that for example $P_{\text{min},K/\mathbb{Q},x}\in\mathbb{Z}_p[X]$.
As explained in the comments, the answer is yes. The key is that by the generalization of Ostrowski's theorem for number fields, we have $$v=v_\mathfrak{P}$$ for some prime ideal $\mathfrak{P}$ of $\mathcal{O}_K$. But it is obvious that an element $x\in K$ belongs to $\mathcal{O}_K$ if and only if $$v_\mathfrak{P}(x)\ge 0$$ for all $\mathfrak{P}\in\text{Spec}(\mathcal{O}_K)$.