Let $\mathcal{O}$ be an order in a number field $K$.
Question: Is specifying an invertible ideal $I$ of $\mathcal{O}$ equivalent to specifying invertible ideals $I_p$ of $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathcal{O}$ for each prime $p \in \mathbb{Z}$, where $I_p$ is the unit ideal for all but finitely many primes $p$?
Partial Answer: If $\mathcal{O} = \mathcal{O}_K$ is the ring of integers, then the answer is yes by the unique factorization of ideals in Dedekind domains. Otherwise, it is known (see Neukirch's Algebraic Number Theory, Proposition 12.6) that specifying an invertible ideal $I$ of $\mathcal{O}$ is equivalent to specifying invertible ideals $I_{\mathfrak{p}}$ of the completion $\mathcal{O}_{\mathfrak{p}}$ for each nonzero prime ideal $\mathfrak{p}$ of $\mathcal{O}$. So it is equivalent to answer the following:
Question$'$: Let $p \in \mathbb{Z}$ be a prime, and let $S$ be the set of nonzero prime ideals of $\mathcal{O}$ lying above $p$. Is specifying an invertible ideal $I_p$ of $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathcal{O}$ equivalent to specifying invertible ideals $I_{\mathfrak{p}}$ of the completion $\mathcal{O}_{\mathfrak{p}}$ for each prime ideal $\mathfrak{p} \in S$?
I guess answering Question$'$ boils down to understanding the relationship between the rings $\mathcal{O}_{\mathfrak{p}}$ for $\mathfrak{p} \in S$ and the ring $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathcal{O}$, but it's not clear to me whether this relationship admits a simple description that works for any order in any number field.