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It is quite easy to show that if $A$ is a Dedekind domain and $\mathfrak{p}\in \operatorname{Spec} A$, then if $A_{\mathfrak{p}}$ is the completion of $A$ at $\mathfrak{p}$ and $A_{(\mathfrak{p})}=(A\setminus \mathfrak{p})^{-1}A$ is the localization of $A$ at $\mathfrak{p}$, then $$A_{(\mathfrak{p})}=A_{\mathfrak{p}}\cap K$$ where $K$ is the field of fractions of $A$ (essentially, by using the notion of order at $\mathfrak{p}$ of the elements of $K$). I would like to know if the same equality remains true in more general cases, assuming $A$ to be for example an order in some number field. Without prime factorization available I find the equality not easy to prove, and I wonder if it remains true or if there are contrexamples?

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  • $\begingroup$ TeX note: use $\operatorname{Spec} A$ $\operatorname{Spec} A$, not $Spec$ $A$ $Spec$ $A$. I have edited accordingly. $\endgroup$
    – LSpice
    Commented May 6, 2022 at 10:06
  • $\begingroup$ There is an ambiguity: what is the completion $A_{\mathfrak p}$? There are two possibilities: either take the $\mathfrak p$-adic completion, or first localize at $\mathfrak p$ then complete the local ring. It seems that A.G's answer is taking the former, but the later seems more reasonable. If you take the later, you could start with a local domain, and ask whether it is some sort of intersection of the completion and its fraction field? $\endgroup$
    – Z. M
    Commented May 6, 2022 at 19:22

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No. If $A$ is a complete noetherian local ring, then $A$ is complete for the adic topology defined by any other prime ideal.

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  • $\begingroup$ Thank you for the answer, actually I was interested in something more specific, like an order in a number field. I should probably have specified better in my question, which I have now modified accordingly $\endgroup$
    – Hair80
    Commented May 6, 2022 at 14:25
  • $\begingroup$ What I said was that, more generally, the completion at a prime does not determine the localization at that prime. But I do not understand the meaning of the equality that you suggest. $\endgroup$
    – A.G
    Commented May 6, 2022 at 14:48
  • $\begingroup$ @Hair80 It does not make sense to modify the question so that an answer is "invalidated". $\endgroup$
    – Z. M
    Commented May 6, 2022 at 16:00
  • $\begingroup$ That intersection could be interpreted if $A_{\mathfrak p}$ is a domain, or maybe weakened, so that the injective ring map $A\to A_{\mathfrak p}$ (injectivity by Krull's intersection theorem) induces an injective map $K\to Q(A_{\mathfrak p})$ where $Q(\cdots)$ is the total ring of fractions, where the intersection is taken. $\endgroup$
    – Z. M
    Commented May 6, 2022 at 16:05
  • $\begingroup$ @Z.M It is not ideal, but it sometimes necessary. $\endgroup$
    – Will Sawin
    Commented May 6, 2022 at 16:36

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