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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?

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notsure
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  • Since $\iota: A \rightarrow \hat{A}$ is such that $\iota(\mathfrak{p}) \subset \mathfrak{P}$, certainly we have a map $\widehat{A_{\mathfrak{p}}} \simeq A_\mathfrak{p} \otimes_A \hat{A} \rightarrow \hat{A}_\mathfrak{P}$. It seems that this map is not likely to be nice (i.e., not integral so that lying over theorems will not apply). – walkar Feb 07 '17 at 17:44
  • Maybe http://stacks.math.columbia.edu/tag/07N9, can be useful – notsure Feb 07 '17 at 17:48
  • That requires an integral map (I assume finite ring map means $S$ is module finite over $R$) to begin with, and I don't see a natural one floating around there. – walkar Feb 07 '17 at 17:51
  • I think we just need that the induced (and reversing arrows) map between affine schemes ha finite fibres. They quote lemma 10.35.21 (in the above link) . – notsure Feb 07 '17 at 17:54
  • Do you already know that if $\mathfrak{p}$ is maximal, then $\hat{A}{\hat{\mathfrak{p}}}=\widehat{A\mathfrak{p}}$? – Andres Mejia Feb 07 '17 at 18:12
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    Perhaps http://math.stackexchange.com/questions/38152/do-localization-and-completion-commute will help – Andres Mejia Feb 07 '17 at 18:14
  • Yes this is true for maximal ideals. I wanted a similar results for prime ideals in general. The point is that $\hat {\mathfrak p}$ is not prime in general, so we have to deal with the primes lying over $\mathfrak p$ – notsure Feb 07 '17 at 18:30

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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.

walkar
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