My question is related to this one, but it has some important differences.
Consider a Noetherian, local domain $A$ and fix a prime ideal $\mathfrak p \subset A$.
Notation: The symbol $\widehat R$ indicates the completion of $R$ with respect to its maximal ideal if $R$ is a local domain.
Consider the embedding $A\to \hat A$ and let $\mathfrak P\subset\hat A $ a prime ideal lying over $\mathfrak p$ (I mean that $\mathfrak P\cap A=\mathfrak p$).
What is the relationship between $\widehat{A_\mathfrak p}$ and $\hat A_\mathfrak P$?
If $$\text{Br}(\mathfrak p):=\{\mathfrak P_1,\ldots, \mathfrak P_n\}$$ is the set of ideals lying over $\mathfrak p$, my bet is that:
$$\widehat{A_\mathfrak p}=\prod_{i}\hat A_{\mathfrak P_i}$$
Is the above statement true? If yes, can you give a reference for the proof?
I talked with a professor about this topic and he was skeptical that there would be anything you can say in general or even in nice cases.
First, in my initial comment, I lied a little -- $\widehat{A_\mathfrak{p}} \not \simeq A_\mathfrak{p}\otimes_A \hat{A}$ in general.
Furthermore, I think you run into issues when you want to compare $\widehat{A_\mathfrak{p}}$ and $\hat{A}_\mathfrak{P}$. Notably, $\widehat{A_\mathfrak{p}}$ is a complete ring whereas $\hat{A}_\mathfrak{P}$ is not complete.
Another issue; say $(A,\mathfrak{p})$ is an normal local domain. Then $\widehat{A_\mathfrak{p}}$ is also a domain, but $\prod_i \hat{A}_{\mathfrak{P}_i}$ cannot be a domain (as products of rings are never domains).
Another thing; you seem to assume that any prime in $A$ should only have finitely many primes lying over it in $\hat{A}$, but there's no reason to expect that.